Introduction To Tensor Calculus

Introduction to Tensor Calculus and Continuum Mechanics

by J.H. Heinbockel Department of Mathematics and Statistics Old Dominion University

PREFACE This is an introductory text which presents fundamental concepts from the subject areas of tensor calculus, differential geometry and continuum mechanics. The material presented is suitable for a two semester course in applied mathematics and is flexible enough to be presented to either upper level undergraduate or beginning graduate students majoring in applied mathematics, engineering or physics. The presentation assumes the students have some knowledge from the areas of matrix theory, linear algebra and advanced calculus. Each section includes many illustrative worked examples. At the end of each section there is a large collection of exercises which range in difficulty. Many new ideas are presented in the exercises and so the students should be encouraged to read all the exercises. The purpose of preparing these notes is to condense into an introductory text the basic definitions and techniques arising in tensor calculus, differential geometry and continuum mechanics. In particular, the material is presented to (i) develop a physical understanding of the mathematical concepts associated with tensor calculus and (ii) develop the basic equations of tensor calculus, differential geometry and continuum mechanics which arise in engineering applications. From these basic equations one can go on to develop more sophisticated models of applied mathematics. The material is presented in an informal manner and uses mathematics which minimizes excessive formalism. The material has been divided into two parts. The first part deals with an introduction to tensor calculus and differential geometry which covers such things as the indicial notation, tensor algebra, covariant differentiation, dual tensors, bilinear and multilinear forms, special tensors, the Riemann Christoffel tensor, space curves, surface curves, curvature and fundamental quadratic forms. The second part emphasizes the application of tensor algebra and calculus to a wide variety of applied areas from engineering and physics. The selected applications are from the areas of dynamics, elasticity, fluids and electromagnetic theory. The continuum mechanics portion focuses on an introduction of the basic concepts from linear elasticity and fluids. The Appendix A contains units of measurements from the Syst`eme International d’Unit`es along with some selected physical constants. The Appendix B contains a listing of Christoffel symbols of the second kind associated with various coordinate systems. The Appendix C is a summary of useful vector identities.

J.H. Heinbockel, 1996

c Copyright
1996 by J.H. Heinbockel. All rights reserved. Reproduction and distribution of these notes is allowable provided it is for non-profit purposes only.

INTRODUCTION TO TENSOR CALCULUS AND CONTINUUM MECHANICS PART 1: INTRODUCTION TO TENSOR CALCULUS §1.1 INDEX NOTATION . . . . . . . . . . . . . . Exercise 1.1 . . . . . . . . . . . . . . . . . . . . . §1.2 TENSOR CONCEPTS AND TRANSFORMATIONS Exercise 1.2 . . . . . . . . . . . . . . . . . . . . . . §1.3 SPECIAL TENSORS . . . . . . . . . . . . . . Exercise 1.3 . . . . . . . . . . . . . . . . . . . . . . §1.4 DERIVATIVE OF A TENSOR . . . . . . . . . . Exercise 1.4 . . . . . . . . . . . . . . . . . . . . . . §1.5 DIFFERENTIAL GEOMETRY AND RELATIVITY Exercise 1.5 . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . .

1 28 35 54 65 101 108 123 129 162

PART 2: INTRODUCTION TO CONTINUUM MECHANICS §2.1 TENSOR NOTATION FOR VECTOR QUANTITIES . . . . Exercise 2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . §2.2 DYNAMICS . . . . . . . . . . . . . . . . . . . . . . Exercise 2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . §2.3 BASIC EQUATIONS OF CONTINUUM MECHANICS . . . Exercise 2.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . §2.4 CONTINUUM MECHANICS (SOLIDS) . . . . . . . . . Exercise 2.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . §2.5 CONTINUUM MECHANICS (FLUIDS) . . . . . . . . . Exercise 2.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . §2.6 ELECTRIC AND MAGNETIC FIELDS . . . . . . . . . . Exercise 2.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . BIBLIOGRAPHY . . . . . . . . . . . . . . . . . . . . . APPENDIX A UNITS OF MEASUREMENT . . . . . . . APPENDIX B CHRISTOFFEL SYMBOLS OF SECOND KIND APPENDIX C VECTOR IDENTITIES . . . . . . . . . . INDEX . . . . . . . . . . . . . . . . . . . . . . . . . .

171 182 187 206 211 238 243 272 282 317 325 347 352 353 355 362 363

1 PART 1: INTRODUCTION TO TENSOR CALCULUS

A scalar field describes a one-to-one correspondence between a single scalar number and a point. An ndimensional vector field is described by a one-to-one correspondence between n-numbers and a point. Let us generalize these concepts by assigning n-squared numbers to a single point or n-cubed numbers to a single point. When these numbers obey certain transformation laws they become examples of tensor fields. In general, scalar fields are referred to as tensor fields of rank or order zero whereas vector fields are called tensor fields of rank or order one. Closely associated with tensor calculus is the indicial or index notation. In section 1 the indicial notation is defined and illustrated. We also define and investigate scalar, vector and tensor fields when they are subjected to various coordinate transformations. It turns out that tensors have certain properties which are independent of the coordinate system used to describe the tensor. Because of these useful properties, we can use tensors to represent various fundamental laws occurring in physics, engineering, science and mathematics. These representations are extremely useful as they are independent of the coordinate systems considered. §1.1 INDEX NOTATION
and B
can be expressed in the component form Two vectors A
= A1
A e1 + A2
e2 + A3
e3

and


= B1
e1 + B2
e2 + B3
e3 , B


and e2 and
e3 are orthogonal unit basis vectors. Often when no confusion arises, the vectors A where
e1 ,

are expressed for brevity sake as number triples. For example, we can write B
= (A1 , A2 , A3 ) A

and


= (B1 , B2 , B3 ) B


and B
are given. The unit vectors would where it is understood that only the components of the vectors A be represented
e1 = (1, 0, 0),


e2 = (0, 1, 0),


e3 = (0, 0, 1).


and B
is the index or indicial notation. In the index notation, A still shorter notation, depicting the vectors A the quantities Ai ,

i = 1, 2, 3

and

Bp ,

p = 1, 2, 3


and B.
This notation focuses attention only on the components of represent the components of the vectors A the vectors and employs a dummy subscript whose range over the integers is specified. The symbol Ai refers
simultaneously. The dummy subscript i can have any of the integer to all of the components of the vector A
Setting i = 2 focuses values 1, 2 or 3. For i = 1 we focus attention on the A1 component of the vector A.
and similarly when i = 3 we can focus attention on attention on the second component A2 of the vector A
The subscript i is a dummy subscript and may be replaced by another letter, say the third component of A. p, so long as one specifies the integer values that this dummy subscript can have.

2 It is also convenient at this time to mention that higher dimensional vectors may be defined as ordered n−tuples. For example, the vector
= (X1 , X2 , . . . , XN ) X with components Xi , i = 1, 2, . . . , N is called a N −dimensional vector. Another notation used to represent this vector is
= X1
X e1 + X2
e2 + · · · + XN
eN where
e1 ,
e2 , . . . ,
eN are linearly independent unit base vectors. Note that many of the operations that occur in the use of the index notation apply not only for three dimensional vectors, but also for N −dimensional vectors. In future sections it is necessary to define quantities which can be represented by a letter with subscripts or superscripts attached. Such quantities are referred to as systems. When these quantities obey certain transformation laws they are referred to as tensor systems. For example, quantities like Akij

eijk

δij

δij

Ai

Bj

aij .

The subscripts or superscripts are referred to as indices or suffixes. When such quantities arise, the indices must conform to the following rules: 1. They are lower case Latin or Greek letters. 2. The letters at the end of the alphabet (u, v, w, x, y, z) are never employed as indices. The number of subscripts and superscripts determines the order of the system. A system with one index is a first order system. A system with two indices is called a second order system. In general, a system with N indices is called a N th order system. A system with no indices is called a scalar or zeroth order system. The type of system depends upon the number of subscripts or superscripts occurring in an expression. m For example, Aijk and Bst , (all indices range 1 to N), are of the same type because they have the same

number of subscripts and superscripts. In contrast, the systems Aijk and Cpmn are not of the same type because one system has two superscripts and the other system has only one superscript. For certain systems the number of subscripts and superscripts is important. In other systems it is not of importance. The meaning and importance attached to sub- and superscripts will be addressed later in this section. In the use of superscripts one must not confuse “powers ”of a quantity with the superscripts. For example, if we replace the independent variables (x, y, z) by the symbols (x1 , x2 , x3 ), then we are letting y = x2 where x2 is a variable and not x raised to a power. Similarly, the substitution z = x3 is the replacement of z by the variable x3 and this should not be confused with x raised to a power. In order to write a superscript quantity to a power, use parentheses. For example, (x2 )3 is the variable x2 cubed. One of the reasons for introducing the superscript variables is that many equations of mathematics and physics can be made to take on a concise and compact form. There is a range convention associated with the indices. This convention states that whenever there is an expression where the indices occur unrepeated it is to be understood that each of the subscripts or superscripts can take on any of the integer values 1, 2, . . . , N where N is a specified integer. For example,

3 the Kronecker delta symbol δij , defined by δij = 1 if i = j and δij = 0 for i = j, with i, j ranging over the values 1,2,3, represents the 9 quantities δ11 = 1

δ12 = 0

δ13 = 0

δ21 = 0

δ22 = 1

δ23 = 0

δ31 = 0

δ32 = 0

δ33 = 1.

The symbol δij refers to all of the components of the system simultaneously. As another example, consider the equation
em ·
en = δmn

m, n = 1, 2, 3

(1.1.1)

the subscripts m, n occur unrepeated on the left side of the equation and hence must also occur on the right hand side of the equation. These indices are called “free ”indices and can take on any of the values 1, 2 or 3 as specified by the range. Since there are three choices for the value for m and three choices for a value of n we find that equation (1.1.1) represents nine equations simultaneously. These nine equations are
e1 = 1 e1 ·



e1 ·
e2 = 0


e1 ·
e3 = 0


e2 ·
e1 = 0


e2 ·
e2 = 1


e2 ·
e3 = 0


e3 ·
e1 = 0


e3 ·
e2 = 0


e3 ·
e3 = 1.

Symmetric and Skew-Symmetric Systems A system defined by subscripts and superscripts ranging over a set of values is said to be symmetric in two of its indices if the components are unchanged when the indices are interchanged. For example, the third order system Tijk is symmetric in the indices i and k if Tijk = Tkji

for all values of i, j and k.

A system defined by subscripts and superscripts is said to be skew-symmetric in two of its indices if the components change sign when the indices are interchanged. For example, the fourth order system Tijkl is skew-symmetric in the indices i and l if Tijkl = −Tljki

for all values of ijk and l.

As another example, consider the third order system aprs , p, r, s = 1, 2, 3 which is completely skewsymmetric in all of its indices. We would then have aprs = −apsr = aspr = −asrp = arsp = −arps . It is left as an exercise to show this completely skew- symmetric systems has 27 elements, 21 of which are zero. The 6 nonzero elements are all related to one another thru the above equations when (p, r, s) = (1, 2, 3). This is expressed as saying that the above system has only one independent component.

4 Summation Convention The summation convention states that whenever there arises an expression where there is an index which occurs twice on the same side of any equation, or term within an equation, it is understood to represent a summation on these repeated indices. The summation being over the integer values specified by the range. A repeated index is called a summation index, while an unrepeated index is called a free index. The summation convention requires that one must never allow a summation index to appear more than twice in any given expression. Because of this rule it is sometimes necessary to replace one dummy summation symbol by some other dummy symbol in order to avoid having three or more indices occurring on the same side of the equation. The index notation is a very powerful notation and can be used to concisely represent many complex equations. For the remainder of this section there is presented additional definitions and examples to illustrated the power of the indicial notation. This notation is then employed to define tensor components and associated operations with tensors. EXAMPLE 1.1-1 The two equations y1 = a11 x1 + a12 x2 y2 = a21 x1 + a22 x2 can be represented as one equation by introducing a dummy index, say k, and expressing the above equations as yk = ak1 x1 + ak2 x2 ,

k = 1, 2.

The range convention states that k is free to have any one of the values 1 or 2, (k is a free index). This equation can now be written in the form yk =

2 

aki xi = ak1 x1 + ak2 x2

i=1

where i is the dummy summation index. When the summation sign is removed and the summation convention is adopted we have yk = aki xi

i, k = 1, 2.

Since the subscript i repeats itself, the summation convention requires that a summation be performed by letting the summation subscript take on the values specified by the range and then summing the results. The index k which appears only once on the left and only once on the right hand side of the equation is called a free index. It should be noted that both k and i are dummy subscripts and can be replaced by other letters. For example, we can write yn = anm xm

n, m = 1, 2

where m is the summation index and n is the free index. Summing on m produces yn = an1 x1 + an2 x2 and letting the free index n take on the values of 1 and 2 we produce the original two equations.

5 EXAMPLE 1.1-2. For yi = aij xj , i, j = 1, 2, 3 and xi = bij zj , i, j = 1, 2, 3 solve for the y variables in terms of the z variables. Solution: In matrix form the given equations can be expressed: 

  y1 a11  y2  =  a21 y3 a31

a12 a22 a32

  x1 a13 a23   x2  a33 x3

 and

  x1 b11  x2  =  b21 x3 b31

b12 b22 b32

  z1 b13 b23   z2  . b33 z3

Now solve for the y variables in terms of the z variables and obtain 

  y1 a11  y2  =  a21 y3 a31

a12 a22 a32

 b11 a13 a23   b21 a33 b31

b12 b22 b32

  z1 b13 b23   z2  . b33 z3

The index notation employs indices that are dummy indices and so we can write yn = anm xm ,

n, m = 1, 2, 3 and xm = bmj zj ,

m, j = 1, 2, 3.

Here we have purposely changed the indices so that when we substitute for xm , from one equation into the other, a summation index does not repeat itself more than twice. Substituting we find the indicial form of the above matrix equation as yn = anm bmj zj ,

m, n, j = 1, 2, 3

where n is the free index and m, j are the dummy summation indices. It is left as an exercise to expand both the matrix equation and the indicial equation and verify that they are different ways of representing the same thing.

EXAMPLE 1.1-3.

The dot product of two vectors Aq , q = 1, 2, 3 and Bj , j = 1, 2, 3 can be represented

Since the with the index notation by the product Ai Bi = AB cos θ i = 1, 2, 3, A = |A|, B = |B|.

subscript i is repeated it is understood to represent a summation index. Summing on i over the range specified, there results A1 B1 + A2 B2 + A3 B3 = AB cos θ. Observe that the index notation employs dummy indices. At times these indices are altered in order to conform to the above summation rules, without attention being brought to the change. As in this example, the indices q and j are dummy indices and can be changed to other letters if one desires. Also, in the future, if the range of the indices is not stated it is assumed that the range is over the integer values 1, 2 and 3.

To systems containing subscripts and superscripts one can apply certain algebraic operations. We present in an informal way the operations of addition, multiplication and contraction.

6 Addition, Multiplication and Contraction The algebraic operation of addition or subtraction applies to systems of the same type and order. That i is again a is we can add or subtract like components in systems. For example, the sum of Aijk and Bjk i i = Aijk + Bjk , where like components are added. system of the same type and is denoted by Cjk

The product of two systems is obtained by multiplying each component of the first system with each component of the second system. Such a product is called an outer product. The order of the resulting product system is the sum of the orders of the two systems involved in forming the product. For example, if Aij is a second order system and B mnl is a third order system, with all indices having the range 1 to N, then the product system is fifth order and is denoted Cjimnl = Aij B mnl . The product system represents N 5 terms constructed from all possible products of the components from Aij with the components from B mnl . The operation of contraction occurs when a lower index is set equal to an upper index and the summation convention is invoked. For example, if we have a fifth order system Cjimnl and we set i = j and sum, then we form the system N mnl . C mnl = Cjjmnl = C11mnl + C22mnl + · · · + CN

Here the symbol C mnl is used to represent the third order system that results when the contraction is performed. Whenever a contraction is performed, the resulting system is always of order 2 less than the original system. Under certain special conditions it is permissible to perform a contraction on two lower case indices. These special conditions will be considered later in the section. The above operations will be more formally defined after we have explained what tensors are. The e-permutation symbol and Kronecker delta Two symbols that are used quite frequently with the indicial notation are the e-permutation symbol and the Kronecker delta. The e-permutation symbol is sometimes referred to as the alternating tensor. The e-permutation symbol, as the name suggests, deals with permutations. A permutation is an arrangement of things. When the order of the arrangement is changed, a new permutation results. A transposition is an interchange of two consecutive terms in an arrangement. As an example, let us change the digits 1 2 3 to 3 2 1 by making a sequence of transpositions. Starting with the digits in the order 1 2 3 we interchange 2 and 3 (first transposition) to obtain 1 3 2. Next, interchange the digits 1 and 3 ( second transposition) to obtain 3 1 2. Finally, interchange the digits 1 and 2 (third transposition) to achieve 3 2 1. Here the total number of transpositions of 1 2 3 to 3 2 1 is three, an odd number. Other transpositions of 1 2 3 to 3 2 1 can also be written. However, these are also an odd number of transpositions.

7 EXAMPLE 1.1-4.

The total number of possible ways of arranging the digits 1 2 3 is six. We have

three choices for the first digit. Having chosen the first digit, there are only two choices left for the second digit. Hence the remaining number is for the last digit. The product (3)(2)(1) = 3! = 6 is the number of permutations of the digits 1, 2 and 3. These six permutations are 1 2 3 even permutation 1 3 2 odd permutation 3 1 2 even permutation 3 2 1 odd permutation 2 3 1 even permutation 2 1 3 odd permutation. Here a permutation of 1 2 3 is called even or odd depending upon whether there is an even or odd number of transpositions of the digits. A mnemonic device to remember the even and odd permutations of 123 is illustrated in the figure 1.1-1. Note that even permutations of 123 are obtained by selecting any three consecutive numbers from the sequence 123123 and the odd permutations result by selecting any three consecutive numbers from the sequence 321321.

Figure 1.1-1. Permutations of 123.

In general, the number of permutations of n things taken m at a time is given by the relation P (n, m) = n(n − 1)(n − 2) · · · (n − m + 1). By selecting a subset of m objects from a collection of n objects, m ≤ n, without regard to the ordering is called a combination of n objects taken m at a time. For example, combinations of 3 numbers taken from the set {1, 2, 3, 4} are (123), (124), (134), (234). Note that ordering of a combination is not considered. That is, the permutations (123), (132), (231), (213), (312), (321) are considered equal. In general, the number of n n

n! where m combinations of n objects taken m at a time is given by C(n, m) = = are the m m!(n − m)! binomial coefficients which occur in the expansion (a + b)n =

n   n  n−m m a b . m m=0

8 The definition of permutations can be used to define the e-permutation symbol.

Definition: (e-Permutation symbol or alternating tensor) The e-permutation symbol is defined  if ijk . . . l is an even permutation of the integers 123 . . . n  1 ijk...l if ijk . . . l is an odd permutation of the integers 123 . . . n = eijk...l = −1 e   0 in all other cases

EXAMPLE 1.1-5.

Find e612453 .

Solution: To determine whether 612453 is an even or odd permutation of 123456 we write down the given numbers and below them we write the integers 1 through 6. Like numbers are then connected by a line and we obtain figure 1.1-2.

Figure 1.1-2. Permutations of 123456. In figure 1.1-2, there are seven intersections of the lines connecting like numbers. The number of intersections is an odd number and shows that an odd number of transpositions must be performed. These results imply e612453 = −1.

Another definition used quite frequently in the representation of mathematical and engineering quantities is the Kronecker delta which we now define in terms of both subscripts and superscripts. Definition: (Kronecker delta)  δij = δij =

1 0

The Kronecker delta is defined:

if i equals j if i is different from j

9 EXAMPLE 1.1-6. Some examples of the e−permutation symbol and Kronecker delta are: e123 = e123 = +1

δ11 = 1

δ12 = 0

e213 = e213 = −1

δ21 = 0

δ22 = 1

δ31

δ32 = 0.

e112 = e

EXAMPLE 1.1-7.

112

=0

=0

When an index of the Kronecker delta δij is involved in the summation convention,

the effect is that of replacing one index with a different index. For example, let aij denote the elements of an N × N matrix. Here i and j are allowed to range over the integer values 1, 2, . . . , N. Consider the product aij δik where the range of i, j, k is 1, 2, . . . , N. The index i is repeated and therefore it is understood to represent a summation over the range. The index i is called a summation index. The other indices j and k are free indices. They are free to be assigned any values from the range of the indices. They are not involved in any summations and their values, whatever you choose to assign them, are fixed. Let us assign a value of j and k to the values of j and k. The underscore is to remind you that these values for j and k are fixed and not to be summed. When we perform the summation over the summation index i we assign values to i from the range and then sum over these values. Performing the indicated summation we obtain aij δik = a1j δ1k + a2j δ2k + · · · + akj δkk + · · · + aN j δN k . In this summation the Kronecker delta is zero everywhere the subscripts are different and equals one where the subscripts are the same. There is only one term in this summation which is nonzero. It is that term where the summation index i was equal to the fixed value k This gives the result akj δkk = akj where the underscore is to remind you that the quantities have fixed values and are not to be summed. Dropping the underscores we write aij δik = akj Here we have substituted the index i by k and so when the Kronecker delta is used in a summation process it is known as a substitution operator. This substitution property of the Kronecker delta can be used to simplify a variety of expressions involving the index notation. Some examples are: Bij δjs = Bis δjk δkm = δjm eijk δim δjn δkp = emnp . Some texts adopt the notation that if indices are capital letters, then no summation is to be performed. For example, aKJ δKK = aKJ

10 as δKK represents a single term because of the capital letters. Another notation which is used to denote no summation of the indices is to put parenthesis about the indices which are not to be summed. For example, a(k)j δ(k)(k) = akj , since δ(k)(k) represents a single term and the parentheses indicate that no summation is to be performed. At any time we may employ either the underscore notation, the capital letter notation or the parenthesis notation to denote that no summation of the indices is to be performed. To avoid confusion altogether, one can write out parenthetical expressions such as “(no summation on k)”.

EXAMPLE 1.1-8. In the Kronecker delta symbol δji we set j equal to i and perform a summation. This operation is called a contraction. There results δii , which is to be summed over the range of the index i. Utilizing the range 1, 2, . . . , N we have N δii = δ11 + δ22 + · · · + δN

δii = 1 + 1 + · · · + 1 δii = N. In three dimension we have δji , i, j = 1, 2, 3 and δkk = δ11 + δ22 + δ33 = 3. In certain circumstances the Kronecker delta can be written with only subscripts. δij ,

For example,

i, j = 1, 2, 3. We shall find that these circumstances allow us to perform a contraction on the lower

indices so that δii = 3.

EXAMPLE 1.1-9.

The determinant of a matrix A = (aij ) can be represented in the indicial notation.

Employing the e-permutation symbol the determinant of an N × N matrix is expressed |A| = eij...k a1i a2j · · · aN k where eij...k is an N th order system. In the special case of a 2 × 2 matrix we write |A| = eij a1i a2j where the summation is over the range 1,2 and the e-permutation symbol is of order 2. In the special case of a 3 × 3 matrix we have   a11  |A| =  a21  a31

a12 a22 a32

 a13  a23  = eijk ai1 aj2 ak3 = eijk a1i a2j a3k a33 

where i, j, k are the summation indices and the summation is over the range 1,2,3. Here eijk denotes the e-permutation symbol of order 3. Note that by interchanging the rows of the 3 × 3 matrix we can obtain

11 more general results. Consider (p, q, r) as some permutation of the integers (1, 2, 3), and observe that the determinant can be expressed

  ap1  ∆ =  aq1  ar1

ap2 aq2 ar2

 ap3  aq3  = eijk api aqj ark . ar3 

If (p, q, r)

is an even permutation of (1, 2, 3) then

∆ = |A|

If (p, q, r)

is an odd permutation of (1, 2, 3) then

∆ = −|A|

If (p, q, r)

is not a permutation of (1, 2, 3) then

∆ = 0.

We can then write eijk api aqj ark = epqr |A|. Each of the above results can be verified by performing the indicated summations. A more formal proof of the above result is given in EXAMPLE 1.1-25, later in this section.

EXAMPLE 1.1-10.

The expression eijk Bij Ci is meaningless since the index i repeats itself more than

twice and the summation convention does not allow this. If you really did want to sum over an index which occurs more than twice, then one must use a summation sign. For example the above expression would be n  written eijk Bij Ci . i=1

EXAMPLE 1.1-11. The cross product of the unit vectors  ek  

ei ×
ej = −
ek   0


e1 ,
e2 ,
e3 can be represented in the index notation by if (i, j, k) is an even permutation of (1, 2, 3) if (i, j, k) is an odd permutation of (1, 2, 3) in all other cases

This result can be written in the form
ei ×
ek . This later result can be verified by summing on the ej = ekij
index k and writing out all 9 possible combinations for i and j.

EXAMPLE 1.1-12.

Given the vectors Ap , p = 1, 2, 3 and Bp , p = 1, 2, 3 the cross product of these two

vectors is a vector Cp , p = 1, 2, 3 with components Ci = eijk Aj Bk ,

i, j, k = 1, 2, 3.

(1.1.2)

The quantities Ci represent the components of the cross product vector
=A
×B
= C1
e1 + C2
e2 + C3
e3 . C
is to be summed over each of the indices which The equation (1.1.2), which defines the components of C, repeats itself. We have summing on the index k Ci = eij1 Aj B1 + eij2 Aj B2 + eij3 Aj B3 .

(1.1.3)

12 We next sum on the index j which repeats itself in each term of equation (1.1.3). This gives Ci = ei11 A1 B1 + ei21 A2 B1 + ei31 A3 B1 + ei12 A1 B2 + ei22 A2 B2 + ei32 A3 B2

(1.1.4)

+ ei13 A1 B3 + ei23 A2 B3 + ei33 A3 B3 . Now we are left with i being a free index which can have any of the values of 1, 2 or 3. Letting i = 1, then letting i = 2, and finally letting i = 3 produces the cross product components C1 = A2 B3 − A3 B2 C2 = A3 B1 − A1 B3 C3 = A1 B2 − A2 B1 .
×B
= eijk Aj Bk
ei . This result can be verified by The cross product can also be expressed in the form A summing over the indices i,j and k.

EXAMPLE 1.1-13.

Show eijk = −eikj = ejki

for

i, j, k = 1, 2, 3

Solution: The array i k j represents an odd number of transpositions of the indices i j k and to each transposition there is a sign change of the e-permutation symbol. Similarly, j k i is an even transposition of i j k and so there is no sign change of the e-permutation symbol. The above holds regardless of the numerical values assigned to the indices i, j, k.

The e-δ Identity An identity relating the e-permutation symbol and the Kronecker delta, which is useful in the simplification of tensor expressions, is the e-δ identity. This identity can be expressed in different forms. The subscript form for this identity is eijk eimn = δjm δkn − δjn δkm ,

i, j, k, m, n = 1, 2, 3

where i is the summation index and j, k, m, n are free indices. A device used to remember the positions of the subscripts is given in the figure 1.1-3. The subscripts on the four Kronecker delta’s on the right-hand side of the e-δ identity then are read (first)(second)-(outer)(inner). This refers to the positions following the summation index. Thus, j, m are the first indices after the summation index and k, n are the second indices after the summation index. The indices j, n are outer indices when compared to the inner indices k, m as the indices are viewed as written on the left-hand side of the identity.

13

Figure 1.1-3. Mnemonic device for position of subscripts. Another form of this identity employs both subscripts and superscripts and has the form j k k eijk eimn = δm δn − δnj δm .

(1.1.5)

One way of proving this identity is to observe the equation (1.1.5) has the free indices j, k, m, n. Each of these indices can have any of the values of 1, 2 or 3. There are 3 choices we can assign to each of j, k, m or n and this gives a total of 34 = 81 possible equations represented by the identity from equation (1.1.5). By writing out all 81 of these equations we can verify that the identity is true for all possible combinations that can be assigned to the free indices. An alternate proof of the e − δ identity is  1  δ1 δ21  2  δ δ2 2  13  δ1 δ23

to consider   δ31   1 δ32  =  0 δ33   0

the determinant  0 0  1 0  = 1. 0 1

By performing a permutation of the rows of this matrix we can use the permutation symbol and write   i  δ1 δ2i δ3i    j j j  ijk δ  k1 δk2 δk3  = e .  δ1 δ2 δ3  By performing a permutation of the columns,  i  δr  j δ  kr  δr

we can write  δsi δti  δsj δtj  = eijk erst . δsk δtk 

Now perform a contraction on the indices i and r to obtain  i   δi δsi δti   j  j  ijk j δ  i δs δt  = e eist .  δk δk δk  s t i Summing on i we have δii = δ11 + δ22 + δ33 = 3 and expand the determinant to obtain the desired result δsj δtk − δtj δsk = eijk eist .

14 Generalized Kronecker delta The generalized Kronecker delta is defined by the (n × n) determinant

ij...k δmn...p

 i  δm  j  δm  = .  ..   δk m

δni δnj .. . δnk

 · · · δpi  · · · δpj  . . .. . ..  · · · δpk 

For example, in three dimensions we can write ijk δmnp

 i  δm  j =  δm  δk m

δni δnj δnk

 δpi  δpj  = eijk emnp . δpk 

Performing a contraction on the indices k and p we obtain the fourth order system rs rsp r s s δmn = δmnp = ersp emnp = eprs epmn = δm δn − δnr δm .

As an exercise one can verify that the definition of the e-permutation symbol can also be defined in terms of the generalized Kronecker delta as ··· N ej1 j2 j3 ···jN = δj11 j22 j33 ···j . N

Additional definitions and results employing the generalized Kronecker delta are found in the exercises. In section 1.3 we shall show that the Kronecker delta and epsilon permutation symbol are numerical tensors which have fixed components in every coordinate system. Additional Applications of the Indicial Notation The indicial notation, together with the e − δ identity, can be used to prove various vector identities. EXAMPLE 1.1-14. Solution: Let


×B
= −B
×A
Show, using the index notation, that A
=A
×B
= C1
C e1 + C2
e2 + C3
e3 = Ci
ei

and let


=B
×A
= D1
e1 + D2
e2 + D3
e3 = Di
ei . D We have shown that the components of the cross products can be represented in the index notation by Ci = eijk Aj Bk

and Di = eijk Bj Ak .

We desire to show that Di = −Ci for all values of i. Consider the following manipulations: Let Bj = Bs δsj and Ak = Am δmk and write Di = eijk Bj Ak = eijk Bs δsj Am δmk

(1.1.6)

where all indices have the range 1, 2, 3. In the expression (1.1.6) note that no summation index appears more than twice because if an index appeared more than twice the summation convention would become meaningless. By rearranging terms in equation (1.1.6) we have Di = eijk δsj δmk Bs Am = eism Bs Am .

15 In this expression the indices s and m are dummy summation indices and can be replaced by any other letters. We replace s by k and m by j to obtain Di = eikj Aj Bk = −eijk Aj Bk = −Ci .
= −C
or B
×A
= −A
× B.
That is, D
= Di

ei = −Ci
ei = −C. Consequently, we find that D Note 1. The expressions Ci = eijk Aj Bk

and

Cm = emnp An Bp

with all indices having the range 1, 2, 3, appear to be different because different letters are used as subscripts. It must be remembered that certain indices are summed according to the summation convention and the other indices are free indices and can take on any values from the assigned range. Thus, after summation, when numerical values are substituted for the indices involved, none of the dummy letters used to represent the components appear in the answer. Note 2. A second important point is that when one is working with expressions involving the index notation, the indices can be changed directly. For example, in the above expression for Di we could have replaced j by k and k by j simultaneously (so that no index repeats itself more than twice) to obtain Di = eijk Bj Ak = eikj Bk Aj = −eijk Aj Bk = −Ci . Note 3. Be careful in switching back and forth between the vector notation and index notation. Observe that a
can be represented vector A
= Ai
A ei or its components can be represented
·
A ei = Ai ,

i = 1, 2, 3.


= Ai as this is a Do not set a vector equal to a scalar. That is, do not make the mistake of writing A misuse of the equal sign. It is not possible for a vector to equal a scalar because they are two entirely different quantities. A vector has both magnitude and direction while a scalar has only magnitude.

EXAMPLE 1.1-15.

Verify the vector identity
· (B
× C)
=B
· (C
× A)
A

Solution: Let


×C
=D
= Di
B ei

where

Di = eijk Bj Ck


×A
= F
= Fi
ei C

where

Fi = eijk Cj Ak

where all indices have the range 1, 2, 3. To prove the above identity, we have
· (B
× C)
=A
·D
= Ai Di = Ai eijk Bj Ck A = Bj (eijk Ai Ck ) = Bj (ejki Ck Ai )

and let

16 since eijk = ejki . We also observe from the expression Fi = eijk Cj Ak that we may obtain, by permuting the symbols, the equivalent expression Fj = ejki Ck Ai . This allows us to write
· (B
× C)
= Bj Fj = B
· F
= B
· (C
× A)
A which was to be shown.
· (B
× C)
is called a triple scalar product. The above index representation of the triple The quantity A scalar product implies that it can be represented as a determinant (See example 1.1-9). We can write   A1 


A · (B × C) =  B1  C1

A2 B2 C2

 A3  B3  = eijk Ai Bj Ck C3 

A physical interpretation that can be assigned to this triple scalar product is that its absolute value represents
B,
C.
The absolute value is the volume of the parallelepiped formed by the three noncoplaner vectors A, needed because sometimes the triple scalar product is negative. This physical interpretation can be obtained from an analysis of the figure 1.1-4.

Figure 1.1-4. Triple scalar product and volume

17
× C|
is the area of the parallelogram P QRS. (ii) the unit vector In figure 1.1-4 observe that: (i) |B
en =


×C
B
× C|
|B


and C.
(iii) The dot product is normal to the plane containing the vectors B  

   
·

· B×C =h A en  = A
× C|
 |B
on
equals the projection of A en which represents the height of the parallelepiped. These results demonstrate that

EXAMPLE 1.1-16.

  


× C|
h = (area of base)(height) = volume.
 = |B A · (B × C)

Verify the vector identity
× B)
× (C
× D)
= C(
D
·A
× B)
− D(
C
·A
× B)
(A


×B
= Fi

=C
×D
= Ei
ei and E ei . These vectors have the components Solution: Let F
= A Fi = eijk Aj Bk

and

Em = emnp Cn Dp


= F
× E
= Gi
ei has the components where all indices have the range 1, 2, 3. The vector G Gq = eqim Fi Em = eqim eijk emnp Aj Bk Cn Dp . From the identity eqim = emqi this can be expressed Gq = (emqi emnp )eijk Aj Bk Cn Dp which is now in a form where we can use the e − δ identity applied to the term in parentheses to produce Gq = (δqn δip − δqp δin )eijk Aj Bk Cn Dp . Simplifying this expression we have: Gq = eijk [(Dp δip )(Cn δqn )Aj Bk − (Dp δqp )(Cn δin )Aj Bk ] = eijk [Di Cq Aj Bk − Dq Ci Aj Bk ] = Cq [Di eijk Aj Bk ] − Dq [Ci eijk Aj Bk ] which are the vector components of the vector
D
·A
× B)
− D(
C
·A
× B).
C(

18 Transformation Equations Consider two sets of N independent variables which are denoted by the barred and unbarred symbols i

x and xi with i = 1, . . . , N. The independent variables xi , i = 1, . . . , N can be thought of as defining the coordinates of a point in a N −dimensional space. Similarly, the independent barred variables define a point in some other N −dimensional space. These coordinates are assumed to be real quantities and are not complex quantities. Further, we assume that these variables are related by a set of transformation equations. xi = xi (x1 , x2 , . . . , xN )

i = 1, . . . , N.

(1.1.7)

It is assumed that these transformation equations are independent. A necessary and sufficient condition that these transformation equations be independent is that the Jacobian determinant be different from zero, that  1  ∂x1  2  i   ∂x  ∂x   ∂x x ∂x1    J( ) =  j  =  . x ∂x ¯  ..  N  ∂x 1

is

∂x

∂x1 ∂x2 ∂x2 ∂x2

··· ··· .. . ···

.. .

∂xN ∂x2

      ..  = 0. .  ∂xN  N ∂x1 ∂xN ∂x2 ∂xN

∂x

This assumption allows us to obtain a set of inverse relations xi = xi (x1 , x2 , . . . , xN )

i = 1, . . . , N,

(1.1.8)

where the x s are determined in terms of the x s. Throughout our discussions it is to be understood that the given transformation equations are real and continuous. Further all derivatives that appear in our discussions are assumed to exist and be continuous in the domain of the variables considered. EXAMPLE 1.1-17.

The following is an example of a set of transformation equations of the form

defined by equations (1.1.7) and (1.1.8) in the case N = 3. Consider the transformation from cylindrical coordinates (r, α, z) to spherical coordinates (ρ, β, α). From the geometry of the figure 1.1-5 we can find the transformation equations r = ρ sin β α=α

0 < α < 2π

z = ρ cos β with inverse transformation ρ=

0<β<π

 r2 + z 2

α=α r β = arctan( ) z Now make the substitutions (x1 , x2 , x3 ) = (r, α, z)

and

(x1 , x2 , x3 ) = (ρ, β, α).

19

Figure 1.1-5. Cylindrical and Spherical Coordinates The resulting transformations then have the forms of the equations (1.1.7) and (1.1.8).

Calculation of Derivatives We now consider the chain rule applied to the differentiation of a function of the bar variables. We represent this differentiation in the indicial notation. Let Φ = Φ(x1 , x2 , . . . , xn ) be a scalar function of the variables xi ,

i = 1, . . . , N and let these variables be related to the set of variables xi , with i = 1, . . . , N by

the transformation equations (1.1.7) and (1.1.8). The partial derivatives of Φ with respect to the variables xi can be expressed in the indicial notation as ∂Φ ∂Φ ∂xj ∂Φ ∂x1 ∂Φ ∂x2 ∂Φ ∂xN = = + + · · · + ∂xi ∂x1 ∂xi ∂x2 ∂xi ∂xj ∂xi ∂xN ∂xi for any fixed value of i satisfying 1 ≤ i ≤ N.

(1.1.9)

The second partial derivatives of Φ can also be expressed in the index notation. Differentiation of equation (1.1.9) partially with respect to xm produces

  ∂Φ ∂xj ∂Φ ∂ 2 xj ∂ ∂2Φ = + . ∂xi ∂xm ∂xm ∂xj ∂xi ∂xj ∂xi ∂xm

(1.1.10)

This result is nothing more than an application of the general rule for differentiating a product of two quantities. To evaluate the derivative of the bracketed term in equation (1.1.10) it must be remembered that the quantity inside the brackets is a function of the bar variables. Let ∂Φ = G(x1 , x2 , . . . , xN ) G= ∂xj to emphasize this dependence upon the bar variables, then the derivative of G is ∂G ∂G ∂xk ∂ 2 Φ ∂xk = = . (1.1.11) ∂xm ∂xk ∂xm ∂xj ∂xk ∂xm This is just an application of the basic rule from equation (1.1.9) with Φ replaced by G. Hence the derivative from equation (1.1.10) can be expressed ∂Φ ∂ 2 xj ∂ 2 Φ ∂xj ∂xk ∂2Φ = + j i m i m ∂x ∂x ∂x ∂x ∂x ∂xj ∂xk ∂xi ∂xm where i, m are free indices and j, k are dummy summation indices.

(1.1.12)

20 EXAMPLE 1.1-18.

Let Φ = Φ(r, θ) where r, θ are polar coordinates related to the Cartesian coordinates ∂2Φ ∂Φ and (x, y) by the transformation equations x = r cos θ y = r sin θ. Find the partial derivatives ∂x ∂x2 Solution: The partial derivative of Φ with respect to x is found from the relation (1.1.9) and can be written ∂Φ ∂r ∂Φ ∂θ ∂Φ = + . ∂x ∂r ∂x ∂θ ∂x

(1.1.13)

The second partial derivative is obtained by differentiating the first partial derivative. From the product rule for differentiation we can write     ∂2Φ ∂Φ ∂ 2 r ∂r ∂ ∂Φ ∂θ ∂ ∂Φ ∂Φ ∂ 2 θ = + + + . ∂x2 ∂r ∂x2 ∂x ∂x ∂r ∂θ ∂x2 ∂x ∂x ∂θ

(1.1.14)

To further simplify (1.1.14) it must be remembered that the terms inside the brackets are to be treated as functions of the variables r and θ and that the derivative of these terms can be evaluated by reapplying the basic rule from equation (1.1.13) with Φ replaced by

∂Φ ∂r

and then Φ replaced by

∂Φ ∂θ .

This gives

  ∂ 2 Φ ∂θ ∂ 2Φ ∂Φ ∂ 2 r ∂r ∂ 2 Φ ∂r + = + ∂x2 ∂r ∂x2 ∂x ∂r2 ∂x ∂r∂θ ∂x   ∂ 2 Φ ∂θ ∂θ ∂ 2 Φ ∂r ∂Φ ∂ 2 θ + + + . ∂θ ∂x2 ∂x ∂θ∂r ∂x ∂θ2 ∂x

(1.1.15)

y and from x these relations we can calculate all the necessary derivatives needed for the simplification of the equations From the transformation equations we obtain the relations r2 = x2 + y 2

and

tan θ =

(1.1.13) and (1.1.15). These derivatives are: ∂r = 2x or ∂x y ∂θ = − 2 or sec2 θ ∂x x sin2 θ ∂2r ∂θ = = − sin θ ∂x2 ∂x r 2r

∂r x = = cos θ ∂x r y ∂θ sin θ =− 2 =− ∂x r r ∂θ ∂r + sin θ ∂x −r cos θ ∂x ∂2θ 2 sin θ cos θ = = . ∂x2 r2 r2

Therefore, the derivatives from equations (1.1.13) and (1.1.15) can be expressed in the form ∂Φ ∂Φ ∂Φ sin θ = cos θ − ∂x ∂r ∂θ r 2 2 ∂Φ sin θ cos θ ∂ 2 Φ ∂Φ sin θ ∂ 2 Φ cos θ sin θ ∂ 2 Φ sin2 θ ∂ Φ +2 + = + cos2 θ − 2 . 2 2 2 ∂x ∂r r ∂θ r ∂r ∂r∂θ r ∂θ2 r2 By letting x1 = r, x2 = θ, x1 = x, x2 = y and performing the indicated summations in the equations (1.1.9) and (1.1.12) there is produced the same results as above.

Vector Identities in Cartesian Coordinates Employing the substitutions x1 = x, x2 = y, x3 = z, where superscript variables are employed and e2 ,
e3 , we illustrated how various vector operations denoting the unit vectors in Cartesian coordinates by
e1 ,
are written by using the index notation.

21 Gradient.

In Cartesian coordinates the gradient of a scalar field is grad φ =

∂φ ∂φ ∂φ


e1 + e2 + e3 . ∂x ∂y ∂z

The index notation focuses attention only on the components of the gradient. In Cartesian coordinates these components are represented using a comma subscript to denote the derivative
ej · grad φ = φ,j =

∂φ , ∂xj

j = 1, 2, 3.

The comma notation will be discussed in section 4. For now we use it to denote derivatives. For example ∂φ ∂2φ , φ ,jk = , etc. φ ,j = j ∂x ∂xj ∂xk Divergence.


is a scalar field and can be In Cartesian coordinates the divergence of a vector field A

represented
= div A
= ∂A1 + ∂A2 + ∂A3 . ∇·A ∂x ∂y ∂z Employing the summation convention and index notation, the divergence in Cartesian coordinates can be represented
= div A
= Ai,i = ∇·A

∂Ai ∂A1 ∂A2 ∂A3 = + + ∂xi ∂x1 ∂x2 ∂x3

where i is the dummy summation index.
= curl A
= ∇×A
in Cartesian coordinates, we note that the index Curl. To represent the vector B
can notation focuses attention only on the components of this vector. The components Bi , i = 1, 2, 3 of B be represented
= eijk Ak,j , Bi =
ei · curl A

for

i, j, k = 1, 2, 3

k where eijk is the permutation symbol introduced earlier and Ak,j = ∂A ∂xj . To verify this representation of the
we need only perform the summations indicated by the repeated indices. We have summing on j that curl A

Bi = ei1k Ak,1 + ei2k Ak,2 + ei3k Ak,3 . Now summing each term on the repeated index k gives us Bi = ei12 A2,1 + ei13 A3,1 + ei21 A1,2 + ei23 A3,2 + ei31 A1,3 + ei32 A2,3 Here i is a free index which can take on any of the values 1, 2 or 3. Consequently, we have ∂A3 ∂A2 − ∂x2 ∂x3 ∂A1 ∂A3 = − ∂x3 ∂x1 ∂A2 ∂A1 = − ∂x1 ∂x2

For

i = 1,

B1 = A3,2 − A2,3 =

For

i = 2,

B2 = A1,3 − A3,1

For

i = 3,

B3 = A2,1 − A1,2


in Cartesian coordinates. which verifies the index notation representation of curl A

22 Other Operations. The following examples illustrate how the index notation can be used to represent additional vector operators in Cartesian coordinates.
· ∇)A
are 1. In index notation the components of the vector (B
· ∇)A}
·
{(B ep = Ap,q Bq

p, q = 1, 2, 3

This can be verified by performing the indicated summations. We have by summing on the repeated index q Ap,q Bq = Ap,1 B1 + Ap,2 B2 + Ap,3 B3 . The index p is now a free index which can have any of the values 1, 2 or 3. We have: for

p = 1,

for

p = 2,

for

p = 3,

A1,q Bq = A1,1 B1 + A1,2 B2 + A1,3 B3 ∂A1 ∂A1 ∂A1 = B1 + B2 + B3 ∂x1 ∂x2 ∂x3 A2,q Bq = A2,1 B1 + A2,2 B2 + A2,3 B3 ∂A2 ∂A2 ∂A2 = B1 + B2 + B3 1 2 ∂x ∂x ∂x3 A3,q Bq = A3,1 B1 + A3,2 B2 + A3,3 B3 ∂A3 ∂A3 ∂A3 = B1 + B2 + B3 1 2 ∂x ∂x ∂x3


· ∇)φ has the following form when expressed in the index notation: 2. The scalar (B
· ∇)φ = Bi φ,i = B1 φ,1 + B2 φ,2 + B3 φ,3 (B ∂φ ∂φ ∂φ = B1 1 + B2 2 + B3 3 . ∂x ∂x ∂x
× ∇)φ is expressed in the index notation by 3. The components of the vector (B  
× ∇)φ = eijk Bj φ,k .
ei · (B This can be verified by performing the indicated summations and is left as an exercise.
× ∇) · A
may be expressed in the index notation. It has the form 4. The scalar (B
× ∇) · A
= eijk Bj Ai,k . (B This can also be verified by performing the indicated summations and is left as an exercise.
in the index notation are represented 5. The vector components of ∇2 A
= Ap,qq .
ep · ∇2 A The proof of this is left as an exercise.

23 EXAMPLE 1.1-19. In Cartesian coordinates prove the vector identity
= ∇ × (f A)
= (∇f ) × A
+ f (∇ × A).
curl (f A)
= curl (f A)
and write the components as Solution: Let B Bi = eijk (f Ak ),j = eijk [f Ak,j + f,j Ak ] = f eijk Ak,j + eijk f,j Ak . This index form can now be expressed in the vector form
= curl (f A)
= f (∇ × A)
+ (∇f ) × A
B


+ B)
= ∇·A
+∇·B
EXAMPLE 1.1-20. Prove the vector identity ∇ · (A
+B
=C
and write this vector equation in the index notation as Ai + Bi = Ci . We then Solution: Let A have
+ ∇ · B.

= Ci,i = (Ai + Bi ),i = Ai,i + Bi,i = ∇ · A ∇·C


· ∇)f = A
· ∇f EXAMPLE 1.1-21. In Cartesian coordinates prove the vector identity (A Solution: In the index notation we write
· ∇)f = Ai f,i = A1 f,1 + A2 f,2 + A3 f,3 (A ∂f ∂f ∂f
· ∇f. = A1 1 + A2 2 + A3 3 = A ∂x ∂x ∂x

EXAMPLE 1.1-22. In Cartesian coordinates prove the vector identity
× B)
= A(∇

− B(∇

+ (B
· ∇)A
− (A
· ∇)B
∇ × (A · B) · A)
× B)
is Solution: The pth component of the vector ∇ × (A
× B)]
= epqk [ekji Aj Bi ],q
ep · [∇ × (A = epqk ekji Aj Bi,q + epqk ekji Aj,q Bi By applying the e − δ identity, the above expression simplifies to the desired result. That is,
× B)]
= (δpj δqi − δpi δqj )Aj Bi,q + (δpj δqi − δpi δqj )Aj,q Bi
ep · [∇ × (A = Ap Bi,i − Aq Bp,q + Ap,q Bq − Aq,q Bp In vector form this is expressed
× B)
= A(∇

− (A
· ∇)B
+ (B
· ∇)A
− B(∇

∇ × (A · B) · A)

24

= ∇(∇ · A)
− ∇2 A EXAMPLE 1.1-23. In Cartesian coordinates prove the vector identity ∇ × (∇ × A)
is given by

= eijk Ak,j and consequently the Solution: We have for the ith component of ∇ × A ei · [∇ × A]
is pth component of ∇ × (∇ × A)
= epqr [erjk Ak,j ],q
ep · [∇ × (∇ × A)] = epqr erjk Ak,jq . The e − δ identity produces
= (δpj δqk − δpk δqj )Ak,jq
ep · [∇ × (∇ × A)] = Ak,pk − Ap,qq .

= ∇(∇ · A)
− ∇2 A. Expressing this result in vector form we have ∇ × (∇ × A)

Indicial Form of Integral Theorems The divergence theorem, in both vector and indicial notation, can be written    
dσ F
· n div · F
dτ = Fi,i dτ = Fi ni dσ i = 1, 2, 3 V

S

V

(1.1.16)

S

where ni are the direction cosines of the unit exterior normal to the surface, dτ is a volume element and dσ is an element of surface area. Note that in using the indicial notation the volume and surface integrals are to be extended over the range specified by the indices. This suggests that the divergence theorem can be applied to vectors in n−dimensional spaces. The vector form and indicial notation for the Stokes theorem are    
dσ = F
· d
r (∇ × F
) · n eijk Fk,j ni dσ = Fi dxi S

C

S

i, j, k = 1, 2, 3

(1.1.17)

C

and the Green’s theorem in the plane, which is a special case of the Stoke’s theorem, can be expressed  

∂F2 ∂F1 − ∂x ∂y









F1 dx + F2 dy

dxdy = C

Fi dxi

e3jk Fk,j dS = S

i, j, k = 1, 2

(1.1.18)

C

Other forms of the above integral theorems are  
dσ ∇φ dτ = φn V

S


where C
is a constant vector. By replacing F
by obtained from the divergence theorem by letting F
= φC
in the divergence theorem one can derive F
× C    
F
×
n dσ. ∇ × F dτ = − V

S

In the divergence theorem make the substitution F
= φ∇ψ to obtain     2
dσ. ψ + (∇φ) · (∇ψ) dτ = (φ∇ψ) · n (φ∇ V

S

25 The Green’s identity



2

φ∇ ψ − ψ∇2 φ dτ = V


dσ (φ∇ψ − ψ∇φ) · n S

is obtained by first letting F
= φ∇ψ in the divergence theorem and then letting F
= ψ∇φ in the divergence theorem and then subtracting the results. Determinants, Cofactors For A = (aij ), i, j = 1, . . . , n an n × n matrix, the determinant of A can be written as det A = |A| = ei1 i2 i3 ...in a1i1 a2i2 a3i3 . . . anin . This gives a summation of the n! permutations of products formed from the elements of the matrix A. The result is a single number called the determinant of A. EXAMPLE 1.1-24.

In the case n = 2 we have    a11 a12    = enm a1n a2m |A| =  a21 a22  = e1m a11 a2m + e2m a12 a2m = e12 a11 a22 + e21 a12 a21 = a11 a22 − a12 a21

EXAMPLE 1.1-25.

In the case  a11 A =  a21 a31

n = 3 we can use either of the notations    1 a12 a13 a1 a12 a13 a22 a23  or A =  a21 a22 a23  a32 a33 a31 a32 a33

and represent the determinant of A in any of the forms det A = eijk a1i a2j a3k det A = eijk ai1 aj2 ak3 det A = eijk ai1 aj2 ak3 det A = eijk a1i a2j a3k . These represent row and column expansions of the determinant. An important identity results if we examine the quantity Brst = eijk air ajs akt . It is an easy exercise to change the dummy summation indices and rearrange terms in this expression. For example, Brst = eijk air ajs akt = ekji akr ajs ait = ekji ait ajs akr = −eijk ait ajs akr = −Btsr , and by considering other permutations of the indices, one can establish that Brst is completely skewsymmetric. In the exercises it is shown that any third order completely skew-symmetric system satisfies Brst = B123 erst . But B123 = det A and so we arrive at the identity Brst = eijk air ajs akt = |A|erst .

26 Other forms of this identity are eijk ari asj atk = |A|erst

and eijk air ajs akt = |A|erst .

Consider the representation of the determinant  1 a  1 |A| =  a21  a31

a12 a22 a32

(1.1.19)

 a13  a23  a33 

by use of the indicial notation. By column expansions, this determinant can be represented |A| = erst ar1 as2 at3

(1.1.20)

and if one uses row expansions the determinant can be expressed as |A| = eijk a1i a2j a3k .

(1.1.21)

Define Aim as the cofactor of the element am i in the determinant |A|. From the equation (1.1.20) the cofactor of ar1 is obtained by deleting this element and we find A1r = erst as2 at3 .

(1.1.22)

The result (1.1.20) can then be expressed in the form |A| = ar1 A1r = a11 A11 + a21 A12 + a31 A13 .

(1.1.23)

That is, the determinant |A| is obtained by multiplying each element in the first column by its corresponding cofactor and summing the result. Observe also that from the equation (1.1.20) we find the additional cofactors A2s = erst ar1 at3

and

A3t = erst ar1 as2 .

(1.1.24)

Hence, the equation (1.1.20) can also be expressed in one of the forms |A| = as2 A2s = a12 A21 + a22 A22 + a32 A23 |A| = at3 A3t = a13 A31 + a23 A32 + a33 A33 The results from equations (1.1.22) and (1.1.24) can be written in a slightly different form with the indicial notation. From the notation for a generalized Kronecker delta defined by ijk eijk elmn = δlmn ,

the above cofactors can be written in the form 1 1jk 1 1jk s t e erst asj atk = δrst aj ak 2! 2! 1 1 2jk s t A2r = e123 esrt as1 at3 = e2jk erst asj atk = δrst aj ak 2! 2! 1 1 3jk s t A3r = e123 etsr at1 as2 = e3jk erst asj atk = δrst aj ak . 2! 2! A1r = e123 erst as2 at3 =

27 These cofactors are then combined into the single equation Air =

1 ijk s t δ a a 2! rst j k

(1.1.25)

which represents the cofactor of ari . When the elements from any row (or column) are multiplied by their corresponding cofactors, and the results summed, we obtain the value of the determinant. Whenever the elements from any row (or column) are multiplied by the cofactor elements from a different row (or column), and the results summed, we get zero. This can be illustrated by considering the summation 1 ijk s t m 1 s t δmst aj ak ar = eijk emst am r aj ak 2! 2! 1 1 ijk = eijk erjk |A| = δrjk |A| = δri |A| 2! 2!

i am r Am =

Here we have used the e − δ identity to obtain ijk δrjk = eijk erjk = ejik ejrk = δri δkk − δki δrk = 3δri − δri = 2δri

which was used to simplify the above result. As an exercise one can show that an alternate form of the above summation of elements by its cofactors is r arm Am i = |A|δi .

EXAMPLE 1.1-26.

N In N-dimensions the quantity δkj11jk22...j ...kN is called a generalized Kronecker delta. It

can be defined in terms of permutation symbols as N ej1 j2 ...jN ek1 k2 ...kN = δkj11jk22...j ...kN

(1.1.26)

Observe that k1 k2 ...kN N δkj11jk22...j = (N !) ej1 j2 ...jN ...kN e

This follows because ek1 k2 ...kN is skew-symmetric in all pairs of its superscripts. The left-hand side denotes a summation of N ! terms. The first term in the summation has superscripts j1 j2 . . . jN and all other terms have superscripts which are some permutation of this ordering with minus signs associated with those terms having an odd permutation. Because ej1 j2 ...jN is completely skew-symmetric we find that all terms in the summation have the value +ej1 j2 ...jN . We thus obtain N ! of these terms.

28 EXERCISE 1.1
1.

Simplify each of the following by employing the summation property of the Kronecker delta. Perform

sums on the summation indices only if your are unsure of the result.


2.

(a) eijk δkn

(c) eijk δis δjm δkn

(e) δij δjn

(b) eijk δis δjm

(d)

(f ) δij δjn δni

aij δin

Simplify and perform the indicated summations over the range 1, 2, 3 (a)

δii

(b) δij δij

(c) eijk Ai Aj Ak

(e) eijk δjk

(d)

(f ) Ai Bj δji − Bm An δmn

eijk eijk


3.


= Ai Express each of the following in index notation. Be careful of the notation you use. Note that A
·
is an incorrect notation because a vector can not equal a scalar. The notation A ei = Ai should be used to express the ith component of a vector.
· (B
× C)
(a) A


A
· C)
(c) B(


× (B
× C)
(b) A

(d)


A
· C)
− C(
A
· B)
B( (b) eijk = −ejik = −eikj = −ekji


4.

Show the e permutation symbol satisfies: (a)


5.


× (B
× C)
= B(
A
· C)
− C(
A
· B)
Use index notation to verify the vector identity A


6.

Let yi = aij xj and xm = aim zi where the range of the indices is 1, 2

eijk = ejki = ekij

(a) Solve for yi in terms of zi using the indicial notation and check your result to be sure that no index repeats itself more than twice. (b) Perform the indicated summations and write out expressions for y1 , y2 in terms of z1 , z2 (c)

Express the above equations in matrix form. Expand the matrix equations and check the solution obtained in part (b).


7.

Use the e − δ identity to simplify (a)


8.

Prove the following vector identities: (a)

eijk ejik

(b)

eijk ejki


· (B
× C)
=B
· (C
× A)
=C
· (A
× B)
A triple scalar product


× B)
×C
= B(
A
· C)
− A(
B
· C)
(b) (A
9.

Prove the following vector identities:
× B)
· (C
× D)
= (A
· C)(
B
· D)
− (A
· D)(
B
· C)
(a) (A
× (B
× C)
+B
× (C
× A)
+C
× (A
× B)
=
0 (b) A (c)


× B)
× (C
× D)
= B(
A
·C
× D)
− A(
B
·C
× D)
(A

29
10.


= (1, −1, 0) and B
= (4, −3, 2) find using the index notation, For A (a)

Ci = eijk Aj Bk ,

i = 1, 2, 3

(b) Ai Bi (c)
11.

What do the results in (a) and (b) represent?

Represent the differential equations

dy1 = a11 y1 + a12 y2 dt

dy2 = a21 y1 + a22 y2 dt

and

using the index notation.
12. Let Φ = Φ(r, θ) where r, θ are polar coordinates related to Cartesian coordinates (x, y) by the transformation equations

x = r cos θ

and y = r sin θ. ∂2Φ ∂Φ , and (a) Find the partial derivatives ∂y ∂y 2 (b) Combine the result in part (a) with the result from EXAMPLE 1.1-18 to calculate the Laplacian ∇2 Φ =

∂2Φ ∂2Φ + ∂x2 ∂y 2

in polar coordinates.
13.

(Index notation) Let a11 = 3,

a12 = 4,

a21 = 5,

a22 = 6.

Calculate the quantity C = aij aij , i, j = 1, 2.
14.

Show the moments of inertia Iij defined by  (y 2 + z 2 )ρ(x, y, z) dτ I11 =

 I23 = I32 = −

R 

R  2

I22 =

2

(x + z )ρ(x, y, z) dτ

I12 = I21 = −

R

xyρ(x, y, z) dτ R



 (x2 + y 2 )ρ(x, y, z) dτ

I33 =

yzρ(x, y, z) dτ

I13 = I31 = −

R

xzρ(x, y, z) dτ, R

 can be represented in the index notation as Iij =

m m

x x δij − xi xj ρ dτ, where ρ is the density,

R

x1 = x, x2 = y, x3 = z and dτ = dxdydz is an element of volume.
15.

Determine if the following relation is true or false. Justify your answer.
ei · (
ej ×
ek ) = (
ei ×
ej ) ·
ek = eijk ,

i, j, k = 1, 2, 3.

Hint: Let
em = (δ1m , δ2m , δ3m ).
16.

Without substituting values for i, l = 1, 2, 3 calculate all nine terms of the given quantities (a)


17.

B il = (δji Ak + δki Aj )ejkl

(b)

Ail = (δim B k + δik B m )emlk

Let Amn xm y n = 0 for arbitrary xi and y i , i = 1, 2, 3, and show that Aij = 0 for all values of i, j.

30
18. (a) For amn , m, n = 1, 2, 3 skew-symmetric, show that amn xm xn = 0. (b) Let amn xm xn = 0,

m, n = 1, 2, 3 for all values of xi , i = 1, 2, 3 and show that amn must be skew-

symmetric.
19.

Let A and B denote 3 × 3 matrices with elements aij and bij respectively. Show that if C = AB is a

matrix product, then det(C) = det(A) · det(B). Hint: Use the result from example 1.1-9.
20. (a) Let u1 , u2 , u3 be functions of the variables s1 , s2 , s3 . Further, assume that s1 , s2 , s3 are in turn each  ∂um  ∂(u1 , u2 , u3 ) denote the Jacobian of the u s with functions of the variables x1 , x2 , x3 . Let  n  = ∂x ∂(x1 , x2 , x3 ) respect to the x s. Show that       i  i  ∂u   ∂u ∂sj   ∂ui   ∂sj  = = ·    ∂xm   ∂sj ∂xm   ∂sj   ∂xm  . ∂xi ∂ x ¯j ∂xi i = = δm and show that J( xx¯ )·J( xx¯ ) = 1, where J( xx¯ ) is the Jacobian determinant j m ∂x ¯ ∂x ∂xm of the transformation (1.1.7).

(b) Note that


21.

A third order system amn with 7, m, n = 1, 2, 3 is said to be symmetric in two of its subscripts if the

components are unaltered when these subscripts are interchanged. When amn is completely symmetric then amn = amn = anm = amn = anm = anm . Whenever this third order system is completely symmetric, then: (i) How many components are there? (ii) How many of these components are distinct? Hint: Consider the three cases (i) 7 = m = n
22.

(ii) 7 = m = n

(iii) 7 = m = n.

A third order system bmn with 7, m, n = 1, 2, 3 is said to be skew-symmetric in two of its subscripts

if the components change sign when the subscripts are interchanged. A completely skew-symmetric third order system satisfies bmn = −bmn = bmn = −bnm = bnm = −bnm . (i) How many components does a completely skew-symmetric system have? (ii) How many of these components are zero? (iii) How many components can be different from zero? (iv) Show that there is one distinct component b123 and that bmn = emn b123 . Hint: Consider the three cases (i) 7 = m = n
23.

(ii) 7 = m = n

(iii) 7 = m = n.

Let i, j, k = 1, 2, 3 and assume that eijk σjk = 0 for all values of i. What does this equation tell you

about the values σij , i, j = 1, 2, 3?
24.

Assume that Amn and Bmn are symmetric for m, n = 1, 2, 3. Let Amn xm xn = Bmn xm xn for arbitrary

values of xi , i = 1, 2, 3, and show that Aij = Bij for all values of i and j.
25.

Assume Bmn is symmetric and Bmn xm xn = 0 for arbitrary values of xi , i = 1, 2, 3, show that Bij = 0.

31
26.

(Generalized Kronecker delta)

ij...k δmn...p

 i  δm  j  δm  = .  ..   δk m

(a)

δni δnj .. . δnk

Define the generalized Kronecker delta as the n × n determinant  · · · δpi  · · · δpj  .  .. . ..  · · · δpk 

where δsr is the Kronecker delta.

123 eijk = δijk

Show

(b) Show

ijk eijk = δ123

(c)

Show

ij δmn = eij emn

(d)

Define

rs rsp δmn = δmnp

and show

(summation on p)

rs r s s δmn = δm δn − δnr δm

Note that by combining the above result with the result from part (c) we obtain the two dimensional form of the e − δ identity r rn = 12 δmn (e) Define δm

 1  a1  i r Let Ar denote the cofactor of ai in the determinant  a21  a31 (a) Show erst Air = eijk asj atk


28.

rst δpst = 2δpr

(summation on n) and show

rst δrst = 3!

(f ) Show


27.

r s s ers emn = δm δn − δnr δm .

a12 a22 a32

 a13  a23  as given by equation (1.1.25). a33 

(b) Show erst Ari = eijk ajs akt

(a) Show that if Aijk = Ajik , i, j, k = 1, 2, 3 there is a total of 27 elements, but only 18 are distinct.

(b) Show that for i, j, k = 1, 2, . . . , N there are N 3 elements, but only N 2 (N + 1)/2 are distinct.
29.
30.


31.

Let aij = Bi Bj for i, j = 1, 2, 3 where B1 , B2 , B3 are arbitrary constants. Calculate det(aij ) = |A|.

(a)

For

A = (aij ), i, j = 1, 2, 3,

show

|A| = eijk ai1 aj2 ak3 .

(b)

For

A = (aij ), i, j = 1, 2, 3,

show

|A| = eijk ai1 aj2 ak3 .

(c)

For

A = (aij ), i, j = 1, 2, 3,

show

|A| = eijk a1i a2j a3k .

(d)

For

I = (δji ), i, j = 1, 2, 3,

show

|I| = 1.

Let |A| = eijk ai1 aj2 ak3 and define Aim as the cofactor of aim . Show the determinant can be

expressed in any of the forms: (a) |A| = Ai1 ai1

where Ai1 = eijk aj2 ak3

(b) |A| = Aj2 aj2

where Ai2 = ejik aj1 ak3

(c) |A| = Ak3 ak3

where

Ai3 = ejki aj1 ak2

32
32.

Show the results in problem 31 can be written in the forms:

Ai1 =


33.

1 e1st eijk ajs akt , 2!

Ai2 =

1 e2st eijk ajs akt , 2!

Ai3 =

1 e3st eijk ajs akt , 2!

or Aim =

1 emst eijk ajs akt 2!

Use the results in problems 31 and 32 to prove that apm Aim = |A|δip . 


34.

1 Let (aij ) =  1 2


35.

Let

 2 1 0 3  and calculate C = aij aij , i, j = 1, 2, 3. 3 2 a111 = −1, a211 = 1,

a112 = 3, a212 = 5,

a121 = 4,

a122 = 2

a221 = 2,

a222 = −2

a1121 = 3,

a1122 = 1

and calculate the quantity C = aijk aijk , i, j, k = 1, 2.
36.

Let a1111 = 2,

a1112 = 1,

a1211 = 5,

a1212 = −2,

a2111 = 1,

a2112 = 0,

a2211 = −2,

a2212 = 1,

a1221 = 4,

a1222 = −2

a2121 = −2,

a2122 = −1

a2221 = 2,

a2222 = 2

and calculate the quantity C = aijkl aijkl , i, j, k, l = 1, 2.
37.


38.

Simplify the expressions: (a) (Aijkl + Ajkli + Aklij + Alijk )xi xj xk xl

(c)

∂xi ∂xj

(b) (Pijk + Pjki + Pkij )xi xj xk

(d)

aij

Let g denote the determinant of   g1r  (a) g erst =  g2r  g3r

∂ 2 xi ∂xj ∂ 2 xm ∂xi r − ami t s ∂x ∂x ∂x ∂xs ∂xt ∂xr

the matrix having the components gij , i, j    gir gis g1s g1t   (b) g erst eijk =  gjr gjs g2s g2t   gkr gks g3s g3t 

 i  δm  j =  δm  δk m

δni δnj δnk

= 1, 2, 3. Show that  git  gjt  gkt 

 δpi  δpj  δpk 


39.

ijk Show that eijk emnp = δmnp


40.

Show that eijk emnp Amnp = Aijk − Aikj + Akij − Ajik + Ajki − Akji Hint: Use the results from problem 39.


41.

Show that (a)

eij eij = 2!

(c)

eijkl eijkl = 4!

(b)

eijk eijk = 3!

(d)

Guess at the result

ei1 i2 ...in ei1 i2 ...in

33
42.

Determine if the following statement is true or false. Justify your answer. eijk Ai Bj Ck = eijk Aj Bk Ci .


43.

Let aij , i, j = 1, 2 denote the components of a 2× 2 matrix A, which are functions of time t.   a11 a12   to verify that these representations are the same. (a) Expand both |A| = eij ai1 aj2 and |A| =  a21 a22  (b) Verify the equivalence of the derivative relations d|A| dai1 daj2 = eij aj2 + eij ai1 dt dt dt

and

 d|A|  dadt11 = a21 dt

da12 dt a22

    a11  +  da   21 dt

 a12  da22  dt

(c) Let aij , i, j = 1, 2, 3 denote the components of a 3 × 3 matrix A, which are functions of time t. Develop appropriate relations, expand them and verify, similar to parts (a) and (b) above, the representation of a determinant and its derivative.
44.

For f = f (x1 , x2 , x3 ) and φ = φ(f ) differentiable scalar functions, use the indicial notation to find a

formula to calculate grad φ .
45.

Use the indicial notation to prove (a) ∇ × ∇φ =
0


46.

If Aij is symmetric and Bij is skew-symmetric, i, j = 1, 2, 3, then calculate C = Aij Bij .


47. Amn
48.


=0 (b) ∇ · ∇ × A

Assume Aij = Aij (x1 , x2 , x3 ) and Aij = Aij (x1 , x2 , x3 ) for i, j = 1, 2, 3 are related by the expression ∂xi ∂xj ∂Amn = Aij m n . Calculate the derivative . ∂x ∂x ∂xk Prove that if any two rows (or two columns) of a matrix are interchanged, then the value of the

determinant of the matrix is multiplied by minus one. Construct your proof using 3 × 3 matrices.
49.

Prove that if two rows (or columns) of a matrix are proportional, then the value of the determinant

of the matrix is zero. Construct your proof using 3 × 3 matrices.
50.

Prove that if a row (or column) of a matrix is altered by adding some constant multiple of some other

row (or column), then the value of the determinant of the matrix remains unchanged. Construct your proof using 3 × 3 matrices.
51.

Simplify the expression φ = eijk emn Ai Ajm Akn .


52.

Let Aijk denote a third order system where i, j, k = 1, 2. (a) How many components does this system

have? (b) Let Aijk be skew-symmetric in the last pair of indices, how many independent components does the system have?
53.

Let Aijk denote a third order system where i, j, k = 1, 2, 3. (a) How many components does this

system have? (b) In addition let Aijk = Ajik and Aikj = −Aijk and determine the number of distinct nonzero components for Aijk .

34
54.

Show that every second order system Tij can be expressed as the sum of a symmetric system Aij and

skew-symmetric system Bij . Find Aij and Bij in terms of the components of Tij .
55.

Consider the system Aijk ,

i, j, k = 1, 2, 3, 4.

(a) How many components does this system have? (b) Assume Aijk is skew-symmetric in the last pair of indices, how many independent components does this system have? (c) Assume that in addition to being skew-symmetric in the last pair of indices, Aijk + Ajki + Akij = 0 is satisfied for all values of i, j, and k, then how many independent components does the system have?
56.


in indicial form. (b) Write the equation of the plane (a) Write the equation of a line
r =
r0 + t A


n · (
r −
r0 ) = 0 in indicial form. (c) Write the equation of a general line in scalar form. (d) Write the equation of a plane in scalar form. (e) Find the equation of the line defined by the intersection of the planes 2x + 3y + 6z = 12 and 6x + 3y + z = 6. (f) Find the equation of the plane through the points (5, 3, 2), (3, 1, 5), (1, 3, 3). Find also the normal to this plane.
57.

The angle 0 ≤ θ ≤ π between two skew lines in space is defined as the angle between their direction

vectors when these vectors are placed at the origin. Show that for two lines with direction numbers ai and bi i = 1, 2, 3, the cosine of the angle between these lines satisfies ai b i √ cos θ = √ ai ai b i b i
58.
59.
60.

Let aij = −aji for i, j = 1, 2, . . . , N and prove that for N odd det(aij ) = 0. ∂λ ∂2λ (b) Let λ = Aij xi xj where Aij = Aji and calculate (a) ∂xm ∂xm ∂xk Given an arbitrary nonzero vector Uk , k = 1, 2, 3, define the matrix elements aij = eijk Uk , where eijk

is the e-permutation symbol. Determine if aij is symmetric or skew-symmetric. Suppose Uk is defined by the above equation for arbitrary nonzero aij , then solve for Uk in terms of the aij .
61.

If Aij = Ai Bj = 0 for all i, j values and Aij = Aji for i, j = 1, 2, . . . , N , show that Aij = λBi Bj

where λ is a constant. State what λ is.
62.

Assume that Aijkm , with i, j, k, m = 1, 2, 3, is completely skew-symmetric. How many independent

components does this quantity have?
63.

Consider Rijkm , i, j, k, m = 1, 2, 3, 4. (a) How many components does this quantity have? (b) If

Rijkm = −Rijmk = −Rjikm then how many independent components does Rijkm have? (c) If in addition Rijkm = Rkmij determine the number of independent components.
64.

¯j , i, j = 1, 2, 3 denote a change of variables from a barred system of coordinates to an Let xi = aij x unbarred system of coordinates and assume that A¯i = aij Aj where aij are constants, A¯i is a function of the ∂ A¯i . x ¯j variables and Aj is a function of the xj variables. Calculate ∂x ¯m

35 §1.2 TENSOR CONCEPTS AND TRANSFORMATIONS
as e2 ,
e3 independent orthogonal unit vectors (base vectors), we may write any vector A For
e1 ,

= A1
A e1 + A2
e2 + A3
e3
relative to the base vectors chosen. These components are the where (A1 , A2 , A3 ) are the coordinates of A
onto the base vectors and projection of A
= (A
·

·

·
A e1 )
e1 + (A e2 )
e2 + (A e3 )
e3 .
1, E
2, E
3 ), not necessarily of unit length, we can then Select any three independent orthogonal vectors, (E write
e1 =


1 E ,
1| |E


e2 =


2 E ,
2| |E


e3 =


3 E ,
3| |E


can be expressed as and consequently, the vector A      
·E
2
·E
3
·E
1 A A A
1 +
2 +
3.
= E E E A
1 · E
2 · E
3 · E
1
2
3 E E E Here we say that
·E
(i) A ,
(i) · E
(i) E

i = 1, 2, 3


2, E
3 . Recall that the parenthesis about
relative to the chosen base vectors E
1, E are the components of A the subscript i denotes that there is no summation on this subscript. It is then treated as a free subscript which can have any of the values 1, 2 or 3. Reciprocal Basis
1, E
2, E
3 ) which are not necessarily orthogonal, nor of Consider a set of any three independent vectors (E
in terms of these vectors we must find components (A1 , A2 , A3 ) unit length. In order to represent the vector A such that
= A1 E
1 + A2 E
2 + A3 E
3. A This can be done by taking appropriate projections and obtaining three equations and three unknowns from which the components are determined. A much easier way to find the components (A1 , A2 , A3 ) is to construct
2, E
3 ). Recall that two bases (E
1, E
2, E
3 ) and (E
1, E
2, E
3 ) are said to be reciprocal
1, E a reciprocal basis (E if they satisfy the condition


i · E
j = δj = E i

1 0

if i = j . if i =  j


1 = δ21 = 0 and E
3 · E
1 = δ31 = 0 so that the vector E
1 is perpendicular to both the
2 · E Note that E
2 and E
3 . (i.e. A vector from one basis is orthogonal to two of the vectors from the other basis.) vectors E
2 × E
3 where V is a constant to be determined. By taking the dot
1 = V −1 E We can therefore write E
1 · (E
2 × E
3 ) is the volume
1 we find that V = E product of both sides of this equation with the vector E
2, E
3 when their origins are made to coincide. In a
1, E of the parallelepiped formed by the three vectors E

36
1, E
2, E
3 ) a given set of basis vectors, then the reciprocal similar manner it can be demonstrated that for (E basis vectors are determined from the relations
1 = 1 E
2 × E
3 × E
1 × E
3,
2 = 1 E
1,
3 = 1 E
2, E E E V V V
2 × E
3 ) = 0 is a triple scalar product and represents the volume of the parallelepiped
1 · (E where V = E having the basis vectors for its sides.
1, E
2, E
3 ) and (E
1, E
2, E
3 ) denote a system of reciprocal bases. We can represent any vector A
Let (E
2, E
3 ) and represent A
in the form
1, E with respect to either of these bases. If we select the basis (E
1 + A2 E
2 + A3 E
3,
= A1 E A

(1.2.1)


relative to the basis vectors (E
1, E
2, E
3 ) are called the contravariant then the components (A1 , A2 , A3 ) of A
These components can be determined from the equations components of A.
·E
1 = A1 , A


·E
2 = A2 , A


·E
3 = A3 . A


1, E
2, E
3 ) and represent A
in the form Similarly, if we choose the reciprocal basis (E
= A1 E
1 + A2 E
2 + A3 E
3, A

(1.2.2)


1, E
2, E
3 ) are called the covariant components of then the components (A1 , A2 , A3 ) relative to the basis (E
These components can be determined from the relations A.
·E
1 = A1 , A


·E
2 = A2 , A


·E
3 = A3 . A

The contravariant and covariant components are different ways of representing the same vector with respect to a set of reciprocal basis vectors. There is a simple relationship between these components which we now develop. We introduce the notation
i · E
j = gij = gji , E


i · E
j = g ij = g ji E

and

(1.2.3)

where gij are called the metric components of the space and g ij are called the conjugate metric components of the space. We can then write
·E
1 = A1 (E
1 · E
1 ) + A2 (E
2 · E
1 ) + A3 (E
3 · E
1 ) = A1 A
·E
1 = A1 (E
1 · E
1 ) + A2 (E
2 · E
1 ) + A3 (E
3 · E
1 ) = A1 A or A1 = A1 g11 + A2 g12 + A3 g13 .

(1.2.4)


·E
2 and A
·E
3 one can establish the results In a similar manner, by considering the dot products A A2 = A1 g21 + A2 g22 + A3 g23

A3 = A1 g31 + A2 g32 + A3 g33 .

These results can be expressed with the index notation as Ai = gik Ak .
·E
1, Forming the dot products A


·E
2, A

(1.2.6)


·E
3 it can be verified that A Ai = g ik Ak .

(1.2.7)

The equations (1.2.6) and (1.2.7) are relations which exist between the contravariant and covariant compo
Similarly, if for some value j we have E
j = αE
1 + β E
2 + γ E
3 , then one can show nents of the vector A. j ij
=g E
i . This is left as an exercise. that E

37 Coordinate Transformations Consider a coordinate transformation from a set of coordinates (x, y, z) to (u, v, w) defined by a set of transformation equations x = x(u, v, w) (1.2.8)

y = y(u, v, w) z = z(u, v, w)

It is assumed that these transformations are single valued, continuous and possess the inverse transformation u = u(x, y, z) v = v(x, y, z)

(1.2.9)

w = w(x, y, z). These transformation equations define a set of coordinate surfaces and coordinate curves. The coordinate surfaces are defined by the equations u(x, y, z) = c1 v(x, y, z) = c2

(1.2.10)

w(x, y, z) = c3 where c1 , c2 , c3 are constants. These surfaces intersect in the coordinate curves
r(u, c2 , c3 ),


r(c1 , v, c3 ),


r(c1 , c2 , w),

(1.2.11)

where
r(u, v, w) = x(u, v, w)
e1 + y(u, v, w)
e2 + z(u, v, w)
e3 . The general situation is illustrated in the figure 1.2-1. Consider the vectors
1 = grad u = ∇u, E


2 = grad v = ∇v, E


3 = grad w = ∇w E

(1.2.12)

evaluated at the common point of intersection (c1 , c2 , c3 ) of the coordinate surfaces. The system of vectors
2, E
3 ) can be selected as a system of basis vectors which are normal to the coordinate surfaces.
1, E (E Similarly, the vectors
1 = ∂
r , E ∂u


2 = ∂
r , E ∂v


3 = ∂
r E ∂w

(1.2.13)


1, E
2, E
3 ) which when evaluated at the common point of intersection (c1 , c2 , c3 ) forms a system of vectors (E we can select as a basis. This basis is a set of tangent vectors to the coordinate curves. It is now demonstrated
2, E
3 ) and the tangential basis (E
1, E
2, E
3 ) are a set of reciprocal bases.
1, E that the normal basis (E e2 + z
e3 denotes the position vector of a variable point. By substitution for Recall that
r = x
e1 + y
x, y, z from (1.2.8) there results
r =
r(u, v, w) = x(u, v, w)
e1 + y(u, v, w)
e2 + z(u, v, w)
e3 .

(1.2.14)

38

Figure 1.2-1. Coordinate curves and coordinate surfaces. A small change in
r is denoted d
r = dx
e1 + dy
e2 + dz
e3 =

∂
r ∂
r ∂
r du + dv + dw ∂u ∂v ∂w

(1.2.15)

where

∂
r ∂x ∂y ∂z


= e1 + e2 + e3 ∂u ∂u ∂u ∂u ∂x ∂
r ∂y ∂z (1.2.16)


= e1 + e2 + e3 ∂v ∂v ∂v ∂v ∂x ∂y ∂z ∂
r


= e1 + e2 + e3 . ∂w ∂w ∂w ∂w In terms of the u, v, w coordinates, this change can be thought of as moving along the diagonal of a paral∂
r ∂
r ∂
r du, dv, and dw. lelepiped having the vector sides ∂u ∂v ∂w Assume u = u(x, y, z) is defined by equation (1.2.9) and differentiate this relation to obtain du =

∂u ∂u ∂u dx + dy + dz. ∂x ∂y ∂z

(1.2.17)

The equation (1.2.15) enables us to represent this differential in the form: du = grad u · d
r   ∂
r ∂
r ∂
r du = grad u · du + dv + dw ∂u ∂v ∂w       ∂
r ∂
r ∂
r du = grad u · du + grad u · dv + grad u · dw. ∂u ∂v ∂w

(1.2.18)

By comparing like terms in this last equation we find that
1 · E
1 = 1, E


1 · E
2 = 0, E


1 · E
3 = 0. E

Similarly, from the other equations in equation (1.2.9) which define v = v(x, y, z), can be demonstrated that dv =

(1.2.19) and w = w(x, y, z) it

      ∂
r ∂
r ∂
r grad v · du + grad v · dv + grad v · dw ∂u ∂v ∂w

(1.2.20)

39 and dw =

      ∂
r ∂
r ∂
r grad w · du + grad w · dv + grad w · dw. ∂u ∂v ∂w

(1.2.21)

By comparing like terms in equations (1.2.20) and (1.2.21) we find
1 = 0,
2 · E E


2 · E
2 = 1, E


2 · E
3 = 0 E


3 · E
1 = 0, E


3 · E
2 = 0, E


3 · E
3 = 1. E

(1.2.22)

The equations (1.2.22) and (1.2.19) show us that the basis vectors defined by equations (1.2.12) and (1.2.13) are reciprocal. Introducing the notation (x1 , x2 , x3 ) = (u, v, w)

(y 1 , y 2 , y 3 ) = (x, y, z)

(1.2.23)

where the x s denote the generalized coordinates and the y  s denote the rectangular Cartesian coordinates, the above equations can be expressed in a more concise form with the index notation. For example, if xi = xi (x, y, z) = xi (y 1 , y 2 , y 3 ),

and y i = y i (u, v, w) = y i (x1 , x2 , x3 ),

i = 1, 2, 3

(1.2.24)

then the reciprocal basis vectors can be represented
i = grad xi , E

i = 1, 2, 3

(1.2.25)

and
i = ∂
r , E ∂xi

i = 1, 2, 3.

(1.2.26)

We now show that these basis vectors are reciprocal. Observe that
r =
r(x1 , x2 , x3 ) with d
r =

∂
r dxm ∂xm

(1.2.27)

and consequently dxi = grad xi · d
r = grad xi ·

  ∂
r m i
i · E
m dxm = δm E dx = dxm , ∂xm

i = 1, 2, 3

(1.2.28)

Comparing like terms in this last equation establishes the result that
i · E
m = δi , E m

i, m = 1, 2, 3

which demonstrates that the basis vectors are reciprocal.

(1.2.29)

40 Scalars, Vectors and Tensors Tensors are quantities which obey certain transformation laws. That is, scalars, vectors, matrices and higher order arrays can be thought of as components of a tensor quantity. We shall be interested in finding how these components are represented in various coordinate systems. We desire knowledge of these transformation laws in order that we can represent various physical laws in a form which is independent of the coordinate system chosen. Before defining different types of tensors let us examine what we mean by a coordinate transformation. Coordinate transformations of the type found in equations (1.2.8) and (1.2.9) can be generalized to higher dimensions. Let xi , i = 1, 2, . . . , N denote N variables. These quantities can be thought of as representing a variable point (x1 , x2 , . . . , xN ) in an N dimensional space VN . Another set of N quantities, call them barred quantities, xi , i = 1, 2, . . . , N, can be used to represent a variable point (x1 , x2 , . . . , xN ) in an N dimensional space V N . When the x s are related to the x s by equations of the form xi = xi (x1 , x2 , . . . , xN ),

i = 1, 2, . . . , N

(1.2.30)

then a transformation is said to exist between the coordinates xi and xi , i = 1, 2, . . . , N. Whenever the relations (1.2.30) are functionally independent, single valued and possess partial derivatives such that the Jacobian of the transformation

J

x x

 =J

1

2

N

x ,x ,...,x x1 , x2 , . . . , xN



 ∂x1  1  ∂x  =  ...  N  ∂x

∂x1 ∂x2

.. .

∂xN ∂x2

∂x1

... ... ...

   ..  .  ∂xN  ∂x1 ∂xN

(1.2.31)

∂xN

is different from zero, then there exists an inverse transformation xi = xi (x1 , x2 , . . . , xN ),

i = 1, 2, . . . , N.

(1.2.32)

For brevity the transformation equations (1.2.30) and (1.2.32) are sometimes expressed by the notation xi = xi (x), i = 1, . . . , N

and

xi = xi (x), i = 1, . . . , N.

(1.2.33)

¯ coordinates. For simplicity Consider a sequence of transformations from x to x ¯ and then from x¯ to x ¯ = z. If we denote by T1 , T2 and T3 the transformations let x ¯ = y and x T1 : T2 :

y i = y i (x1 , . . . , xN ) i = 1, . . . , N i

i

1

N

z = z (y , . . . , y ) i = 1, . . . , N

or

T1 x = y

or T2 y = z

Then the transformation T3 obtained by substituting T1 into T2 is called the product of two successive transformations and is written T3 :

z i = z i (y 1 (x1 , . . . , xN ), . . . , y N (x1 , . . . , xN ))

i = 1, . . . , N

or T3 x = T2 T1 x = z.

This product transformation is denoted symbolically by T3 = T2 T1 . The Jacobian of the product transformation is equal to the product of Jacobians associated with the product transformation and J3 = J2 J1 .

41 Transformations Form a Group A group G is a nonempty set of elements together with a law, for combining the elements. The combined elements are denoted by a product. Thus, if a and b are elements in G then no matter how you define the law for combining elements, the product combination is denoted ab. The set G and combining law forms a group if the following properties are satisfied: (i) For all a, b ∈ G, then ab ∈ G. This is called the closure property. (ii) There exists an identity element I such that for all a ∈ G we have Ia = aI = a. (iii) There exists an inverse element. That is, for all a ∈ G there exists an inverse element a−1 such that a a−1 = a−1 a = I. (iv) The associative law holds under the combining law and a(bc) = (ab)c for all a, b, c ∈ G. For example, the set of elements G = {1, −1, i, −i}, where i2 = −1 together with the combining law of ordinary multiplication, forms a group. This can be seen from the multiplication table.

× 1 -1 -i i

1 1 -1 -i i

-1 -1 1 i -i

i i -i 1 -1

-i -i i -1 1

The set of all coordinate transformations of the form found in equation (1.2.30), with Jacobian different from zero, forms a group because: (i) The product transformation, which consists of two successive transformations, belongs to the set of transformations. (closure) (ii) The identity transformation exists in the special case that x and x are the same coordinates. (iii) The inverse transformation exists because the Jacobian of each individual transformation is different from zero. (iv) The associative law is satisfied in that the transformations satisfy the property T3 (T2 T1 ) = (T3 T2 )T1 . When the given transformation equations contain a parameter the combining law is often times represented as a product of symbolic operators. For example, we denote by Tα a transformation of coordinates having a parameter α. The inverse transformation can be denoted by Tα−1 and one can write Tα x = x or x = Tα−1 x. We let Tβ denote the same transformation, but with a parameter β, then the transitive property is expressed symbolically by Tα Tβ = Tγ where the product Tα Tβ represents the result of performing two successive transformations. The first coordinate transformation uses the given transformation equations and uses the parameter α in these equations. This transformation is then followed by another coordinate transformation using the same set of transformation equations, but this time the parameter value is β. The above symbolic product is used to demonstrate that the result of applying two successive transformations produces a result which is equivalent to performing a single transformation of coordinates having the parameter value γ. Usually some relationship can then be established between the parameter values α, β and γ.

42

Figure 1.2-2. Cylindrical coordinates. In this symbolic notation, we let Tθ denote the identity transformation. That is, using the parameter value of θ in the given set of transformation equations produces the identity transformation. The inverse transformation can then be expressed in the form of finding the parameter value β such that Tα Tβ = Tθ . Cartesian Coordinates At times it is convenient to introduce an orthogonal Cartesian coordinate system having coordinates i

y,

i = 1, 2, . . . , N. This space is denoted EN and represents an N-dimensional Euclidean space. Whenever

the generalized independent coordinates xi , i = 1, . . . , N are functions of the y  s, and these equations are functionally independent, then there exists independent transformation equations y i = y i (x1 , x2 , . . . , xN ),

i = 1, 2, . . . , N,

(1.2.34)

with Jacobian different from zero. Similarly, if there is some other set of generalized coordinates, say a barred system xi , i = 1, . . . , N where the x s are independent functions of the y  s, then there will exist another set of independent transformation equations y i = y i (x1 , x2 , . . . , xN ),

i = 1, 2, . . . , N,

(1.2.35)

with Jacobian different from zero. The transformations found in the equations (1.2.34) and (1.2.35) imply that there exists relations between the x s and x s of the form (1.2.30) with inverse transformations of the form (1.2.32). It should be remembered that the concepts and ideas developed in this section can be applied to a space VN of any finite dimension. Two dimensional surfaces (N = 2) and three dimensional spaces (N = 3) will occupy most of our applications. In relativity, one must consider spaces where N = 4. EXAMPLE 1.2-1. (cylindrical coordinates (r, θ, z)) Consider the transformation x = x(r, θ, z) = r cos θ

y = y(r, θ, z) = r sin θ

z = z(r, θ, z) = z

from rectangular coordinates (x, y, z) to cylindrical coordinates (r, θ, z), illustrated in the figure 1.2-2. By letting y 1 = x,

y 2 = y,

y3 = z

x1 = r,

x2 = θ,

x3 = z

the above set of equations are examples of the transformation equations (1.2.8) with u = r, v = θ, w = z as the generalized coordinates.

43

EXAMPLE 1.2.2. (Spherical Coordinates) (ρ, θ, φ) Consider the transformation x = x(ρ, θ, φ) = ρ sin θ cos φ

y = y(ρ, θ, φ) = ρ sin θ sin φ

z = z(ρ, θ, φ) = ρ cos θ

from rectangular coordinates (x, y, z) to spherical coordinates (ρ, θ, φ). By letting y 1 = x, y 2 = y, y 3 = z

x1 = ρ, x2 = θ , x3 = φ

the above set of equations has the form found in equation (1.2.8) with u = ρ, v = θ, w = φ the generalized coordinates. One could place bars over the x s in this example in order to distinguish these coordinates from the x s of the previous example. The spherical coordinates (ρ, θ, φ) are illustrated in the figure 1.2-3.

Figure 1.2-3. Spherical coordinates.

Scalar Functions and Invariance We are now at a point where we can begin to define what tensor quantities are. The first definition is for a scalar invariant or tensor of order zero.

44 Definition: ( Absolute scalar field) Assume there exists a coordinate transformation of the type (1.2.30) with Jacobian J different from zero. Let the scalar function f = f (x1 , x2 , . . . , xN )

(1.2.36)

be a function of the coordinates xi , i = 1, . . . , N in a space VN . Whenever there exists a function f = f (x1 , x2 , . . . , xN )

(1.2.37)

which is a function of the coordinates xi , i = 1, . . . , N such that f = J W f, then f is called a tensor of rank or order zero of weight W in the space VN . Whenever W = 0, the scalar f is called the component of an absolute scalar field and is referred to as an absolute tensor of rank or order zero.

That is, an absolute scalar field is an invariant object in the space VN with respect to the group of coordinate transformations. It has a single component in each coordinate system. For any scalar function of the type defined by equation (1.2.36), we can substitute the transformation equations (1.2.30) and obtain f = f (x1 , . . . , xN ) = f (x1 (x), . . . , xN (x)) = f (x1 , . . . , xN ).

(1.2.38)

Vector Transformation, Contravariant Components In VN consider a curve C defined by the set of parametric equations C:

xi = xi (t),

i = 1, . . . , N

where t is a parameter. The tangent vector to the curve C is the vector  T
=

dx1 dx2 dxN , ,..., dt dt dt

 .

In index notation, which focuses attention on the components, this tangent vector is denoted Ti =

dxi , dt

i = 1, . . . , N.

For a coordinate transformation of the type defined by equation (1.2.30) with its inverse transformation defined by equation (1.2.32), the curve C is represented in the barred space by xi = xi (x1 (t), x2 (t), . . . , xN (t)) = xi (t),

i = 1, . . . , N,

with t unchanged. The tangent to the curve in the barred system of coordinates is represented by dxi ∂xi dxj = , dt ∂xj dt

i = 1, . . . , N.

(1.2.39)

45 i

Letting T , i = 1, . . . , N denote the components of this tangent vector in the barred system of coordinates, the equation (1.2.39) can then be expressed in the form i

T =

∂xi j T , ∂xj

i, j = 1, . . . , N.

(1.2.40)

This equation is said to define the transformation law associated with an absolute contravariant tensor of rank or order one. In the case N = 3 the matrix form of this transformation is represented  1 ∂x 1 T ∂x1  2  T  =  ∂x21 ∂x 3 ∂x3 T 1 

∂x

∂x1 ∂x2 ∂x2 ∂x2 ∂x3 ∂x2

∂x1 ∂x3 ∂x2 ∂x3 ∂x3 ∂x3



 T1  2  T T3

(1.2.41)

A more general definition is

Definition: (Contravariant tensor) Whenever N quantities Ai in i

a coordinate system (x1 , . . . , xN ) are related to N quantities A in a coordinate system (x1 , . . . , xN ) such that the Jacobian J is different from zero, then if the transformation law i

A = JW

∂xi j A ∂xj

is satisfied, these quantities are called the components of a relative tensor of rank or order one with weight W . Whenever W = 0 these quantities are called the components of an absolute tensor of rank or order one. We see that the above transformation law satisfies the group properties. EXAMPLE 1.2-3. (Transitive Property of Contravariant Transformation) Show that successive contravariant transformations is also a contravariant transformation. Solution: Consider the transformation of a vector from an unbarred to a barred system of coordinates. A vector or absolute tensor of rank one Ai = Ai (x), i = 1, . . . , N will transform like the equation (1.2.40) and i

A (x) =

∂xi j A (x). ∂xj

(1.2.42)

Another transformation from x → x coordinates will produce the components i

i

A (x) =

∂x j A (x) ∂xj

(1.2.43)

Here we have used the notation Aj (x) to emphasize the dependence of the components Aj upon the x coordinates. Changing indices and substituting equation (1.2.42) into (1.2.43) we find i

A (x) =

i

∂x ∂xj m A (x). ∂xj ∂xm

(1.2.44)

46 From the fact that

i

i

∂x ∂x ∂xj = , ∂xm ∂xj ∂xm the equation (1.2.44) simplifies to i

i

A (x) =

∂x m A (x) ∂xm

(1.2.45)

and hence this transformation is also contravariant. We express this by saying that the above are transitive with respect to the group of coordinate transformations. Note that from the chain rule one can write ∂xm ∂x1 ∂xm ∂x2 ∂xm ∂x3 ∂xm ∂xm ∂xj m 1 ∂xn + 2 ∂xn + 3 ∂xn = ∂xn = δn . j ∂xn = ∂x ∂x ∂x ∂x Do not make the mistake of writing ∂xm ∂x2 ∂xm 2 ∂xn = ∂xn ∂x

∂xm ∂x3 ∂xm 3 ∂xn = ∂xn ∂x

or

as these expressions are incorrect. Note that there are no summations in these terms, whereas there is a summation index in the representation of the chain rule.

Vector Transformation, Covariant Components Consider a scalar invariant A(x) = A(x) which is a shorthand notation for the equation A(x1 , x2 , . . . , xn ) = A(x1 , x2 , . . . , xn ) involving the coordinate transformation of equation (1.2.30). By the chain rule we differentiate this invariant and find that the components of the gradient must satisfy ∂A ∂A ∂xj . i = ∂xj ∂x ∂xi

(1.2.46)

Let Aj =

∂A ∂xj

and

Ai =

∂A , ∂xi

then equation (1.2.46) can be expressed as the transformation law Ai = Aj

∂xj . ∂xi

(1.2.47)

This is the transformation law for an absolute covariant tensor of rank or order one. A more general definition is

47 Definition: (Covariant tensor) 1

Whenever N quantities Ai in a

N

coordinate system (x , . . . , x ) are related to N quantities Ai in a coordinate system (x1 , . . . , xN ), with Jacobian J different from zero, such that the transformation law Ai = J W

∂xj Aj ∂xi

(1.2.48)

is satisfied, then these quantities are called the components of a relative covariant tensor of rank or order one having a weight of W . Whenever W = 0, these quantities are called the components of an absolute covariant tensor of rank or order one.

Again we note that the above transformation satisfies the group properties. Absolute tensors of rank or order one are referred to as vectors while absolute tensors of rank or order zero are referred to as scalars. EXAMPLE 1.2-4. (Transitive Property of Covariant Transformation) Consider a sequence of transformation laws of the type defined by the equation (1.2.47) x→x x→x

∂xj ∂xi ∂xm Ak (x) = Am (x) k ∂x Ai (x) = Aj (x)

We can therefore express the transformation of the components associated with the coordinate transformation x → x and

  ∂xj ∂xm ∂xj Ak (x) = Aj (x) m = A (x) , j k k ∂x ∂x ∂x

which demonstrates the transitive property of a covariant transformation.

Higher Order Tensors We have shown that first order tensors are quantities which obey certain transformation laws. Higher order tensors are defined in a similar manner and also satisfy the group properties. We assume that we are given transformations of the type illustrated in equations (1.2.30) and (1.2.32) which are single valued and continuous with Jacobian J different from zero. Further, the quantities xi and xi , i = 1, . . . , n represent the coordinates in any two coordinate systems. The following transformation laws define second order and third order tensors.

48

Definition: (Second order contravariant tensor) Whenever N-squared quantities Aij mn

in a coordinate system (x1 , . . . , xN ) are related to N-squared quantities A 1

in a coordinate

N

system (x , . . . , x ) such that the transformation law mn

A

(x) = Aij (x)J W

∂xm ∂xn ∂xi ∂xj

(1.2.49)

is satisfied, then these quantities are called components of a relative contravariant tensor of rank or order two with weight W . Whenever W = 0 these quantities are called the components of an absolute contravariant tensor of rank or order two.

Definition: (Second order covariant tensor) 1

Whenever N-squared quantities

N

Aij in a coordinate system (x , . . . , x ) are related to N-squared quantities Amn in a coordinate system (x1 , . . . , xN ) such that the transformation law Amn (x) = Aij (x)J W

∂xi ∂xj ∂xm ∂xn

(1.2.50)

is satisfied, then these quantities are called components of a relative covariant tensor of rank or order two with weight W . Whenever W = 0 these quantities are called the components of an absolute covariant tensor of rank or order two.

Definition: (Second order mixed tensor) Aij

1

Whenever N-squared quantities m

N

in a coordinate system (x , . . . , x ) are related to N-squared quantities An in

a coordinate system (x1 , . . . , xN ) such that the transformation law m

An (x) = Aij (x)J W

∂xm ∂xj ∂xi ∂xn

(1.2.51)

is satisfied, then these quantities are called components of a relative mixed tensor of rank or order two with weight W . Whenever W = 0 these quantities are called the components of an absolute mixed tensor of rank or order two. It is contravariant of order one and covariant of order one.

Higher order tensors are defined in a similar manner. For example, if we can find N-cubed quantities Am np

such that

∂xi ∂xα ∂xβ (1.2.52) ∂xγ ∂xj ∂xk then this is a relative mixed tensor of order three with weight W . It is contravariant of order one and i

Ajk (x) = Aγαβ (x)J W

covariant of order two.

49 General Definition In general a mixed tensor of rank or order (m + n) ...im Tji11ji22...j n

(1.2.53)

is contravariant of order m and covariant of order n if it obeys the transformation law   x W i1 ∂xi2 ∂xim ∂xb1 ∂xb2 ∂xbn i1 i2 ...im ...am ∂x T j1 j2 ...jn = J Tba11ba22...b · · · · · · · n x ∂xa1 ∂xa2 ∂xam ∂xj1 ∂xj2 ∂xjn

(1.2.54)

   ∂x  ∂(x1 , x2 , . . . , xN ) =   = J x ∂x ∂(x1 , x2 , . . . , xN )

where

x

is the Jacobian of the transformation. When W = 0 the tensor is called an absolute tensor, otherwise it is called a relative tensor of weight W. Here superscripts are used to denote contravariant components and subscripts are used to denote covariant components. Thus, if we are given the tensor components in one coordinate system, then the components in any other coordinate system are determined by the transformation law of equation (1.2.54). Throughout the remainder of this text one should treat all tensors as absolute tensors unless specified otherwise. Dyads and Polyads Note that vectors can be represented in bold face type with the notation A = Ai Ei This notation can also be generalized to tensor quantities. Higher order tensors can also be denoted by bold face type. For example the tensor components Tij and Bijk can be represented in terms of the basis vectors Ei , i = 1, . . . , N by using a notation which is similar to that for the representation of vectors. For example, T = Tij Ei Ej B = Bijk Ei Ej Ek . Here T denotes a tensor with components Tij and B denotes a tensor with components Bijk . The quantities Ei Ej are called unit dyads and Ei Ej Ek are called unit triads. There is no multiplication sign between the basis vectors. This notation is called a polyad notation. A further generalization of this notation is the representation of an arbitrary tensor using the basis and reciprocal basis vectors in bold type. For example, a mixed tensor would have the polyadic representation ij...k T = Tlm...n Ei Ej . . . Ek El Em . . . En .

A dyadic is formed by the outer or direct product of two vectors. For example, the outer product of the vectors a = a1 E 1 + a2 E 2 + a3 E 3

and b = b1 E1 + b2 E2 + b3 E3

50 gives the dyad ab =a1 b1 E1 E1 + a1 b2 E1 E2 + a1 b3 E1 E3 a2 b 1 E 2 E 1 + a2 b 2 E 2 E 2 + a2 b 3 E 2 E 3 a3 b 1 E 3 E 1 + a3 b 2 E 3 E 2 + a3 b 3 E 3 E 3 . In general, a dyad can be represented A = Aij Ei Ej

i, j = 1, . . . , N

where the summation convention is in effect for the repeated indices. The coefficients Aij are called the coefficients of the dyad. When the coefficients are written as an N × N array it is called a matrix. Every second order tensor can be written as a linear combination of dyads. The dyads form a basis for the second order tensors. As the example above illustrates, the nine dyads {E1 E1 , E1 E2 , . . . , E3 E3 }, associated with the outer products of three dimensional base vectors, constitute a basis for the second order tensor A = ab having the components Aij = ai bj with i, j = 1, 2, 3. Similarly, a triad has the form T = Tijk Ei Ej Ek

Sum on repeated indices

where i, j, k have the range 1, 2, . . . , N. The set of outer or direct products { Ei Ej Ek }, with i, j, k = 1, . . . , N i constitutes a basis for all third order tensors. Tensor components with mixed suffixes like Cjk are associated

with triad basis of the form i C = Cjk Ei Ej Ek

where i, j, k have the range 1, 2, . . . N. Dyads are associated with the outer product of two vectors, while triads, tetrads,... are associated with higher-order outer products. These higher-order outer or direct products are referred to as polyads. The polyad notation is a generalization of the vector notation. The subject of how polyad components transform between coordinate systems is the subject of tensor calculus. In Cartesian coordinates we have Ei = Ei =
ei and a dyadic with components called dyads is written ei
ej or A = Aij


A =A11
e1
e1 + A12
e1
e2 + A13
e1
e3 e2
e1 + A22
e2
e2 + A23
e2
e3 A21
e3
e1 + A32
e3
e2 + A33
e3
e3 A31


ej are called unit dyads. Note that a dyadic has nine components as compared with a where the terms
ei
vector which has only three components. The conjugate dyadic Ac is defined by a transposition of the unit vectors in A, to obtain

e1
e1 + A12
e2
e1 + A13
e3
e1 Ac =A11
e1
e2 + A22
e2
e2 + A23
e3
e2 A21
e1
e3 + A32
e2
e3 + A33
e3
e3 A31


51 If a dyadic equals its conjugate A = Ac , then Aij = Aji and the dyadic is called symmetric. If a dyadic equals the negative of its conjugate A = −Ac , then Aij = −Aji and the dyadic is called skew-symmetric. A special dyadic called the identical dyadic or idemfactor is defined by e1 +
e2 +
e3 . e2
e3
J=
e1

produces the This dyadic has the property that pre or post dot product multiplication of J with a vector V same vector V
. For example,
· J = (V1
V e1 + V2
e2 + V3
e3 ) · J
e1 ·
e1 + V2
e2 ·
e2 + V3
e3 ·
e3 = V = V1
e1
e2
e3

= J · (V1
e1 + V2
e2 + V3
e3 ) and J · V
e1
e1 ·
e2
e2 ·
e3
e3 ·
e1 + V2
e2 + V3
e3 = V = V1
A dyadic operation often used in physics and chemistry is the double dot product A : B where A and B are both dyadics. Here both dyadics are expanded using the distributive law of multiplication, and then ej :
en are combined according to the rule each unit dyad pair
ei
em

ei
ej :
en = (
em
ei ·
em )(
ej ·
en ). ei
ej and B = Bij
ei
ej , then the double dot product A : B is calculated as follows. For example, if A = Aij
ei
ej ) : (Bmn
em
en ) = Aij Bmn (
ej :
en ) = Aij Bmn (
A : B = (Aij
ei
em
ei ·
em )(
ej ·
en ) = Aij Bmn δim δjn = Amj Bmj = A11 B11 + A12 B12 + A13 B13 + A21 B21 + A22 B22 + A23 B23 + A31 B31 + A32 B32 + A33 B33 When operating with dyads, triads and polyads, there is a definite order to the way vectors and polyad
= Ai

= Bi
components are represented. For example, for A ei and B ei vectors with outer product
B
= Am Bn
A em
en = φ there is produced the dyadic φ with components Am Bn . In comparison, the outer product
A
= Bm An
B em
en = ψ produces the dyadic ψ with components Bm An . That is
B
=A1 B1
e1
e1 + A1 B2
e1
e2 + A1 B3
e1
e3 φ=A e2
e1 + A2 B2
e2
e2 + A2 B3
e2
e3 A2 B1
e3
e1 + A3 B2
e3
e2 + A3 B3
e3
e3 A3 B1

A
=B1 A1
e1
e1 + B1 A2
e1
e2 + B1 A3
e1
e3 and ψ = B e2
e1 + B2 A2
e2
e2 + B2 A3
e2
e3 B2 A1
e3
e1 + B3 A2
e3
e2 + B3 A3
e3
e3 B3 A1
are different dyadics.
is defined for both pre and post multiplication as The scalar dot product of a dyad with a vector C
=A
B
·C
=A(
B
· C)
φ·C
·φ=C
·A
B
=(C
· A)
B
C These products are, in general, not equal.

52 Operations Using Tensors The following are some important tensor operations which are used to derive special equations and to prove various identities. Addition and Subtraction Tensors of the same type and weight can be added or subtracted. For example, two third order mixed i denote two third order tensors, when added, produce another third order mixed tensor. Let Aijk and Bjk

mixed tensors. Their sum is denoted i i = Aijk + Bjk . Cjk

That is, like components are added. The sum is also a mixed tensor as we now verify. By hypothesis Aijk i and Bjk are third order mixed tensors and hence must obey the transformation laws i

∂xi ∂xn ∂xp ∂xm ∂xj ∂xk i n p m ∂x ∂x ∂x = Bnp . ∂xm ∂xj ∂xk

Ajk = Am np i

B jk i

i

i

We let C jk = Ajk + B jk denote the sum in the transformed coordinates. Then the addition of the above transformation equations produces   i i n p

∂xi ∂xn ∂xp i i m m ∂x ∂x ∂x C jk = Ajk + B jk = Am + B = C . np np np ∂xm ∂xj ∂xk ∂xm ∂xj ∂xk Consequently, the sum transforms as a mixed third order tensor. Multiplication (Outer Product) The product of two tensors is also a tensor. The rank or order of the resulting tensor is the sum of the ranks of the tensors occurring in the multiplication. As an example, let Aijk denote a mixed third order l tensor and let Bm denote a mixed second order tensor. The outer product of these two tensors is the fifth

order tensor il l = Aijk Bm , i, j, k, l, m = 1, 2, . . . , N. Cjkm i

l

Here all indices are free indices as i, j, k, l, m take on any of the integer values 1, 2, . . . , N. Let Ajk and B m il

denote the components of the given tensors in the barred system of coordinates. We define C jkm as the il l is a tensor for by hypothesis Aijk and Bm are tensors outer product of these components. Observe that Cjkm

and hence obey the transformation laws ∂xα ∂xj ∂xk ∂xi ∂xβ ∂xγ δ m δ l ∂x ∂x B # = Bm . ∂xl ∂x# The outer product of these components produces α

Aβγ = Aijk

∂xα ∂xj ∂xk ∂xδ ∂xm ∂xi ∂xβ ∂xγ ∂xl ∂x# (1.2.56) ∂xα ∂xj ∂xk ∂xδ ∂xm il = Cjkm i ∂x ∂xβ ∂xγ ∂xl ∂x# transforms as a mixed fifth order absolute tensor. Other outer products are αδ

α

δ

l C βγ# = Aβγ B # = Aijk Bm

il which demonstrates that Cjkm

analyzed in a similar way.

(1.2.55)

53 Contraction The operation of contraction on any mixed tensor of rank m is performed when an upper index is set equal to a lower index and the summation convention is invoked. When the summation is performed over the repeated indices the resulting quantity is also a tensor of rank or order (m − 2). For example, let Aijk , i, j, k = 1, 2, . . . , N denote a mixed tensor and perform a contraction by setting j equal to i. We obtain Aiik = A11k + A22k + · · · + AN N k = Ak where k is a free index. To show that Ak is a tensor, we let transformed components of

Aijk .

By hypothesis

Aijk

i Aik

(1.2.57)

= Ak denote the contraction on the

is a mixed tensor and hence the components must

satisfy the transformation law ∂xi ∂xn ∂xp . ∂xm ∂xj ∂xk Now execute a contraction by setting j equal to i and perform a summation over the repeated index. We i

Ajk = Am np

find

∂xi ∂xn ∂xp ∂xn ∂xp = Am np i k m ∂x ∂x ∂x ∂xm ∂xk (1.2.58) p p ∂xp m n ∂x n ∂x = Anp δm k = Anp k = Ap k . ∂x ∂x ∂x Hence, the contraction produces a tensor of rank two less than the original tensor. Contractions on other i

Aik = Ak = Am np

mixed tensors can be analyzed in a similar manner. New tensors can be constructed from old tensors by performing a contraction on an upper and lower index. This process can be repeated as long as there is an upper and lower index upon which to perform the contraction. Each time a contraction is performed the rank of the resulting tensor is two less than the rank of the original tensor. Multiplication (Inner Product) The inner product of two tensors is obtained by: (i) first taking the outer product of the given tensors and (ii) performing a contraction on two of the indices. EXAMPLE 1.2-5. (Inner product) Let Ai and Bj denote the components of two first order tensors (vectors). The outer product of these tensors is Cji = Ai Bj , i, j = 1, 2, . . . , N. The inner product of these tensors is the scalar C = Ai Bi = A1 B1 + A2 B2 + · · · + AN BN . Note that in some situations the inner product is performed by employing only subscript indices. For example, the above inner product is sometimes expressed as C = Ai Bi = A1 B1 + A2 B2 + · · · AN BN . This notation is discussed later when Cartesian tensors are considered.

54 Quotient Law Assume Brqs and Cps are arbitrary absolute tensors. Further assume we have a quantity A(ijk) which we think might be a third order mixed tensor Aijk . By showing that the equation Arqp Brqs = Cps is satisfied, then it follows that Arqp must be a tensor. This is an example of the quotient law. Obviously, this result can be generalized to apply to tensors of any order or rank. To prove the above assertion we shall show from the above equation that Aijk is a tensor. Let xi and xi denote a barred and unbarred system of coordinates which are related by transformations of the form defined by equation (1.2.30). In the barred system, we assume that r

qs

s

Aqp B r = C p

(1.2.59)

l are arbitrary absolute tensors and therefore must satisfy the transformation where by hypothesis Bkij and Cm

equations ∂xq ∂xs ∂xk ∂xi ∂xj ∂xr s m s l ∂x ∂x C p = Cm . l ∂x ∂xp qs

B r = Bkij

qs

s

We substitute for B r and C p in the equation (1.2.59) and obtain the equation     q s s k m r ij ∂x ∂x ∂x l ∂x ∂x Aqp Bk = Cm l ∂xi ∂xj ∂xr ∂x ∂xp ∂xs ∂xm = Arqm Brql l . ∂x ∂xp Since the summation indices are dummy indices they can be replaced by other symbols. We change l to j, q to i and r to k and write the above equation as   q k m ∂xs r ∂x ∂x k ∂x Aqp i − Aim p Bkij = 0. ∂xj ∂x ∂xr ∂x Use inner multiplication by

∂xn ∂xs

and simplify this equation to the form   q k m r ∂x ∂x n k ∂x δj Aqp i − Aim p Bkij = 0 ∂x ∂xr ∂x   q k m ∂x r ∂x ∂x k Aqp i − Aim p Bkin = 0. ∂x ∂xr ∂x

or

Because Bkin is an arbitrary tensor, the quantity inside the brackets is zero and therefore r

Aqp

m ∂xq ∂xk k ∂x = 0. r − Aim i ∂x ∂x ∂xp

This equation is simplified by inner multiplication by r

∂xi ∂xl ∂xj ∂xk

to obtain

∂xm ∂xi ∂xl =0 ∂xp ∂xj ∂xk ∂xm ∂xi ∂xl = Akim p ∂x ∂xj ∂xk

δjq δrl Aqp − Akim l

Ajp

which is the transformation law for a third order mixed tensor.

or

55 EXERCISE 1.2
1.

Consider the transformation equations representing a rotation of axes through an angle α.  Tα :

x1

= x1 cos α − x2 sin α

x2

= x1 sin α + x2 cos α

Treat α as a parameter and show this set of transformations constitutes a group by finding the value of α which: (i) gives the identity transformation. (ii) gives the inverse transformation. (iii) show the transformation is transitive in that a transformation with α = θ1 followed by a transformation with α = θ2 is equivalent to the transformation using α = θ1 + θ2 .
2.



Show the transformation Tα :

x1 x2

= αx1 = α1 x2

forms a group with α as a parameter. Find the value of α such that: (i) the identity transformation exists. (ii) the inverse transformation exists. (iii) the transitive property is satisfied.
3.

Show the given transformation forms a group with parameter α.  Tα :


4.

x1

=

x1 1−αx1

x2

=

x2 1−αx1

Consider the Lorentz transformation from relativity theory having the velocity parameter V, c is the

speed of light and x4 = t is time.

 1 x        x2  x3       x4

TV :

1

4

x −V x =  V2 1−

c2

= x2 = x3 1

x4 − Vcx2

= 

2

1− V2 c

Show this set of transformations constitutes a group, by establishing: (i) V = 0 gives the identity transformation T0 . (ii) TV2 · TV1 = T0 requires that V2 = −V1 . (iii) TV2 · TV1 = TV3 requires that V3 =
5.

V1 + V2 . 1 + V1c2V2


1, E
2, E
3 ) an arbitrary independent basis, (a) Verify that For (E
1 = 1 E
2 × E
3, E V


3 × E
2 = 1 E
1, E V


1 · (E
2 × E
3) is a reciprocal basis, where V = E


1 × E
3 = 1 E
2 E V


i.
j = g ij E (b) Show that E

56

Figure 1.2-4. Cylindrical coordinates (r, β, z).
6.

For the cylindrical coordinates (r, β, z) illustrated in the figure 1.2-4.

(a) Write out the transformation equations from rectangular (x, y, z) coordinates to cylindrical (r, β, z) coordinates. Also write out the inverse transformation. (b) Determine the following basis vectors in cylindrical coordinates and represent your results in terms of cylindrical coordinates.
2, E
3 . (ii)The normal basis E
1, E
2, E
3 . (iii) e
1, E ˆr , e ˆβ , e ˆz (i) The tangential basis E ˆr , e ˆβ , e ˆz are normalized vectors in the directions of the tangential basis. where e
= Ax
e1 + Ay
e2 + Az
e3 can be represented in any of the forms: (c) A vector A
1 + A2 E
2 + A3 E
3
= A1 E A
= A1 E
1 + A2 E
2 + A3 E
3 A
= Ar e ˆr + Aβ e ˆβ + Az e ˆz A depending upon the basis vectors selected . In terms of the components Ax , Ay , Az (i) Solve for the contravariant components A1 , A2 , A3 . (ii) Solve for the covariant components A1 , A2 , A3 . (iii) Solve for the components Ar , Aβ , Az . Express all results in cylindrical coordinates. (Note the components Ar , Aβ , Az are referred to as physical components. Physical components are considered in more detail in a later section.)

57

Figure 1.2-5. Spherical coordinates (ρ, α, β).
7.

For the spherical coordinates (ρ, α, β) illustrated in the figure 1.2-5.

(a) Write out the transformation equations from rectangular (x, y, z) coordinates to spherical (ρ, α, β) coordinates. Also write out the equations which describe the inverse transformation. (b) Determine the following basis vectors in spherical coordinates
1, E
2, E
3. (i) The tangential basis E
2, E
3.
1, E (ii) The normal basis E ˆρ , e ˆα , e ˆβ which are normalized vectors in the directions of the tangential basis. Express all results (iii) e in terms of spherical coordinates.
= Ax
e1 + Ay
e2 + Az
e3 can be represented in any of the forms: (c) A vector A
= A1 E
1 + A2 E
2 + A3 E
3 A
= A1 E
1 + A2 E
2 + A3 E
3 A
= Aρ e ˆρ + Aα e ˆα + Aβ e ˆβ A depending upon the basis vectors selected . Calculate, in terms of the coordinates (ρ, α, β) and the components Ax , Ay , Az (i) The contravariant components A1 , A2 , A3 . (ii) The covariant components A1 , A2 , A3 . (iii) The components Aρ , Aα , Aβ which are called physical components.
8.

Work the problems 6,7 and then let (x1 , x2 , x3 ) = (r, β, z) denote the coordinates in the cylindrical

system and let (x1 , x2 , x3 ) = (ρ, α, β) denote the coordinates in the spherical system. (a) Write the transformation equations x → x from cylindrical to spherical coordinates. Also find the inverse transformations.

( Hint: See the figures 1.2-4 and 1.2-5.)

(b) Use the results from part (a) and the results from problems 6,7 to verify that Ai = Aj

∂xj ∂xi

for

i = 1, 2, 3.

(i.e. Substitute Aj from problem 6 to get A¯i given in problem 7.)

58 (c) Use the results from part (a) and the results from problems 6,7 to verify that i

A = Aj

∂xi ∂xj

for

i = 1, 2, 3.

(i.e. Substitute Aj from problem 6 to get A¯i given by problem 7.)
9.

Pick two arbitrary noncolinear vectors in the x, y plane, say
1 = 5
V e1 +
e2


2 =
and V e1 + 5
e2


3 =

2 . The vectors V
1 and V
2 can be thought of and let V e3 be a unit vector perpendicular to both V
1 and V as defining an oblique coordinate system, as illustrated in the figure 1.2-6.
1 , V
2 , V
3 ). (a) Find the reciprocal basis (V (b) Let
3 e2 + z
e3 = αV
1 + β V
2 + γ V
r = x
e1 + y
and show that

y 5x − 24 24 5y x β=− + 24 24 γ=z

α=

(c) Show x = 5α + β y = α + 5β z=γ (d) For γ = γ0 constant, show the coordinate lines are described by α = constant

and

and sketch some of these coordinate lines. (See figure 1.2-6.) (e) Find the metrics gij and conjugate metrices g ij associated with the (α, β, γ) space.

Figure 1.2-6. Oblique coordinates.

β = constant,

59
10.

Consider the transformation equations x = x(u, v, w) y = y(u, v, w) z = z(u, v, w)

substituted into the position vector e2 + z
e3 .
r = x
e1 + y


Define the basis vectors
1, E
2, E
3) = (E

∂
r ∂
r ∂
r , , ∂u ∂v ∂w



with the reciprocal basis
1 = 1 E
2 × E
3, E V


3 × E
2 = 1 E
1, E V


1 × E
3 = 1 E
2. E V

where
1 · (E
2 × E
3 ). V =E
2 × E
3 ) and show that v · V = 1.
1 · (E Let v = E
11.

Given the coordinate transformation x = −u − 2v

y = −u − v

z=z

(a) Find and illustrate graphically some of the coordinate curves. (b) For
r =
r(u, v, z) a position vector, define the basis vectors
1 = ∂
r , E ∂u


2 = ∂
r , E ∂v


3 = ∂
r . E ∂z


1, E
2, E
3. Calculate these vectors and then calculate the reciprocal basis E (c) With respect to the basis vectors in (b) find the contravariant components Ai associated with the vector
= α1
A e1 + α2
e2 + α3
e3 where (α1 , α2 , α3 ) are constants.
given in part (c). (d) Find the covariant components Ai associated with the vector A (e) Calculate the metric tensor gij and conjugate metric tensor g ij . (f) From the results (e), verify that gij g jk = δik (g) Use the results from (c)(d) and (e) to verify that Ai = gik Ak (h) Use the results from (c)(d) and (e) to verify that Ai = g ik Ak
on unit vectors in the directions E
1, E
2, E
3. (i) Find the projection of the vector A
2, E
3.
on unit vectors the directions E
1, E (j) Find the projection of the vector A

60 ei where y i = y i (x1 , x2 , x3 ), i = 1, 2, 3 we have by definition For
r = y i



12.

i
j = ∂
r = ∂y
E ei . From this relation show that ∂xj ∂xj

m
m = ∂x
ej E ∂y j

and consequently m m
i · E
j = ∂y ∂y , gij = E ∂xi ∂xj


13.

i j
i · E
j = ∂x ∂x , and g ij = E ∂y m ∂y m

i, j, m = 1, . . . , 3

Consider the set of all coordinate transformations of the form y i = aij xj + bi

where aij and bi are constants and the determinant of aij is different from zero. Show this set of transformations forms a group.
14.

For αi , βi constants and t a parameter, xi = αi + t βi ,i = 1, 2, 3 is the parametric representation of

a straight line. Find the parametric equation of the line which passes through the two points (1, 2, 3) and (14, 7, −3). What does the vector
15.

d' r dt

represent?

A surface can be represented using two parameters u, v by introducing the parametric equations xi = xi (u, v),

i = 1, 2, 3,

a < u < b and c < v < d.

The parameters u, v are called the curvilinear coordinates of a point on the surface. A point on the surface can be represented by the position vector
r =
r(u, v) = x1 (u, v)
e1 + x2 (u, v)
e2 + x3 (u, v)
e3 . The vectors and

∂' r ∂v

∂' r ∂u

are tangent vectors to the coordinate surface curves
r(u, c2 ) and
r(c1 , v) respectively. An element of

surface area dS on the surface is defined as the area of the elemental parallelogram having the vector sides ∂' r ∂u du

and

∂' r ∂v dv.

Show that dS = |

 ∂
r ∂
r × | dudv = g11 g22 − (g12 )2 dudv ∂u ∂v

where g11 =

∂
r ∂
r · ∂u ∂u

g12 =

∂
r ∂
r · ∂u ∂v

g22 =

∂
r ∂
r · . ∂v ∂v


× B)
· (A
× B)
= |A
× B|
2 See Exercise 1.1, problem 9(c). Hint: (A
16. (a) Use the results from problem 15 and find the element of surface area of the circular cone x = u sin α cos v α a constant (b) Find the surface area of the above cone.

y = u sin α sin v 0≤u≤b

z = u cos α

0 ≤ v ≤ 2π

61
17.

The equation of a plane is defined in terms of two parameters u and v and has the form xi = αi u + βi v + γi

i = 1, 2, 3,

where αi βi and γi are constants. Find the equation of the plane which passes through the points (1, 2, 3), (14, 7, −3) and (5, 5, 5). What does this problem have to do with the position vector
r(u, v), the vectors ∂' r ∂' r ∂u , ∂v


18.

and
r(0, 0)? Hint: See problem 15. Determine the points of intersection of the curve x1 = t, x2 = (t)2 , x3 = (t)3 with the plane 8 x1 − 5 x2 + x3 − 4 = 0.


19.


k = E
i × E
j where v = E
1 · (E
2 × E
3 ) and and v −1 eijk E


20.

¯i = cij xj , where cij are constants Let x ¯i and xi , i = 1, 2, 3 be related by the linear transformation x


k = E
i × E
j Verify the relations V eijk E
1 · (E
2 × E
3 ).. V =E

n denote the cofactor of cm such that the determinant c = det(cij ) is different from zero. Let γm n divided by

the determinant c. (a) Show that cij γkj = γji cjk = δki . ¯j . (b) Show the inverse transformation can be expressed xi = γji x (c) Show that if Ai is a contravariant vector, then its transformed components are A¯p = cpq Aq . (d) Show that if Ai is a covariant vector, then its transformed components are A¯i = γ p Ap . i


21.

Show that the outer product of two contravariant vectors Ai and B i , i = 1, 2, 3 results in a second

order contravariant tensor. Show that for the position vector
r = y i (x1 , x2 , x3 )
ei the element of arc length squared is m m ∂y ∂y 2 i j
i · E
j = ds = d
r · d
r = gij dx dx where gij = E . ∂xi ∂xj


22.


23.

i

k

i

p i absolute tensors, show that if Aijk Bnk = Cjn then Ajk B n = C jn . For Aijk , Bnm and Ctq


24.

Let Aij denote an absolute covariant tensor of order 2. Show that the determinant A = det(Aij ) is  an invariant of weight 2 and (A) is an invariant of weight 1.


25.

Let B ij denote an absolute contravariant tensor of order 2. Show that the determinant B = det(B ij ) √ is an invariant of weight −2 and B is an invariant of weight −1.


26. (a) Write out the contravariant components of the following vectors
1 (i) E


2 (ii) E


3 (iii) E

where


i = ∂
r E ∂xi

for i = 1, 2, 3.

(b) Write out the covariant components of the following vectors
1 (i) E


2 (ii) E


3 (ii) E


i = grad xi , where E

for i = 1, 2, 3.

62
27.

Let Aij and Aij denote absolute second order tensors. Show that λ = Aij Aij is a scalar invariant.


28.

Assume that aij , i, j = 1, 2, 3, 4 is a skew-symmetric second order absolute tensor. (a) Show that bijk =

∂ajk ∂aki ∂aij + + i j ∂x ∂x ∂xk

is a third order tensor. (b) Show bijk is skew-symmetric in all pairs of indices and (c) determine the number of independent components this tensor has.
29.

Show the linear forms A1 x + B1 y + C1 and A2 x + B2 y + C2 , with respect to the group of rotations

and translations x = x cos θ − y sin θ + h and y = x sin θ + y cos θ + k, have the forms A1 x + B 1 y + C 1 and A2 x + B 2 y + C 2 . Also show that the quantities A1 B2 − A2 B1 and A1 A2 + B1 B2 are invariants.
30.

Show that the curvature of a curve y = f (x) is κ = ± y  (1 + y 2 )−3/2 and that this curvature remains

invariant under the group of rotations given in the problem 1. Hint: Calculate

dy dx

=

dy dx dx dx .


31.

Show that when the equation of a curve is given in the parametric form x = x(t), y = y(t), then x¨ ˙ y − y˙ x¨ the curvature is κ = ± 2 and remains invariant under the change of parameter t = t(t), where (x˙ + y˙ 2 )3/2 x˙ = dx dt , etc.


32.

ij Let Aij k denote a third order mixed tensor. (a) Show that the contraction Ai is a first order

contravariant tensor. (b) Show that contraction of i and j produces Aii k which is not a tensor. This shows that in general, the process of contraction does not always apply to indices at the same level.
33.

Let φ = φ(x1 , x2 , . . . , xN ) denote an absolute scalar invariant. (a) Is the quantity

Is the quantity
34.

2

∂ φ ∂xi ∂xj

∂φ ∂xi

a tensor? (b)

a tensor?

Consider the second order absolute tensor aij , i, j = 1, 2 where a11 = 1, a12 = 2, a21 = 3 and a22 = 4.

Find the components of aij under the transformation of coordinates x1 = x1 + x2 and x2 = x1 − x2 .
35.

Let Ai , Bi denote the components of two covariant absolute tensors of order one. Show that

Cij = Ai Bj is an absolute second order covariant tensor.
36.

Let Ai denote the components of an absolute contravariant tensor of order one and let Bi denote the

components of an absolute covariant tensor of order one, show that Cji = Ai Bj transforms as an absolute mixed tensor of order two.
37.

(a) Show the sum and difference of two tensors of the same kind is also a tensor of this kind. (b) Show

that the outer product of two tensors is a tensor. Do parts (a) (b) in the special case where one tensor Ai is a relative tensor of weight 4 and the other tensor Bkj is a relative tensor of weight 3. What is the weight of the outer product tensor Tkij = Ai Bkj in this special case?
38.

ij j Let Aij km denote the components of a mixed tensor of weight M . Form the contraction Bm = Aim

j and determine how Bm transforms. What is its weight?


39.

Let Aij denote the components of an absolute mixed tensor of order two. Show that the scalar

contraction S = Aii is an invariant.

63
40.

Let Ai = Ai (x1 , x2 , . . . , xN ) denote the components of an absolute contravariant tensor. Form the

quantity Bji =

∂Ai ∂xj

and determine if Bji transforms like a tensor.

∂Ai ∂Aj Let Ai denote the components of a covariant vector. (a) Show that aij = − are the j ∂x ∂xi ∂ajk ∂aki ∂aij + + = 0. components of a second order tensor. (b) Show that ∂xk ∂xi ∂xj
42. Show that xi = K eijk Aj Bk , with K = 0 and arbitrary, is a general solution of the system of equations


41.

Ai xi = 0, Bi xi = 0, i = 1, 2, 3. Give a geometric interpretation of this result in terms of vectors.
= y
Given the vector A e1 + z
e2 + x
e3 where
e1 ,
e2 ,
e3 denote a set of unit basis vectors which
2 = 4

3 =

1 = 3
e1 + 4
e2 , E e1 + 7
e2 and E e3 denote a set of define a set of orthogonal x, y, z axes. Let E


43.

basis vectors which define a set of u, v, w axes. (a) Find the coordinate transformation between these two
3, E
3 . (c) Calculate the covariant components of A.

1, E sets of axes. (b) Find a set of reciprocal vectors E
(d) Calculate the contravariant components of A.
44.

Let A = Aij
ei
ej denote a dyadic. Show that A : Ac = A11 A11 + A12 A21 + A13 A31 + A21 A12 + A22 A22 + A23 A32 + A31 A13 + A32 A23 + A23 A33


45.


= Ai

= Bi

= Ci

= Di

B,
ψ =C
D
denote ei , B ei , C ei , D ei denote vectors and let φ = A Let A

dyadics which are the outer products involving the above vectors. Show that the double dot product satisfies
B
:C
D
= (A
· C)(
B
· D)
φ:ψ=A
46.

Show that if aij is a symmetric tensor in one coordinate system, then it is symmetric in all coordinate

systems.
47.

Write the transformation laws for the given tensors. (a)


48.

Show that if Ai = Aj

and unbarred systems.

Akij

(b)

Aij k

(c)

Aijk m

∂xj ∂xj i , then Ai = Aj ∂xi . Note that this is equivalent to interchanging the bar ∂x


49. (a) Show that under the linear homogeneous transformation x1 =a11 x1 + a21 x2 x2 =a12 x1 + a22 x2 the quadratic form Q(x1 , x2 ) = g11 (x1 )2 + 2g12 x1 x2 + g22 (x2 )2

becomes

Q(x1 , x2 ) = g11 (x1 )2 + 2g12 x1 x2 + g 22 (x2 )2

where g ij = g11 aj1 ai1 + g12 (ai1 aj2 + aj1 ai2 ) + g22 ai2 aj2 . (b) Show F = g11 g22 − (g12 )2 is a relative invariant of weight 2 of the quadratic form Q(x1 , x2 ) with respect to the group of linear homogeneous transformations. i.e. Show that F = ∆2 F where F = g 11 g22 −(g12 )2 and ∆ = (a11 a22 − a21 a12 ).

64
50.

Let ai and bi for i = 1, . . . , n denote arbitrary vectors and form the dyadic Φ = a1 b1 + a2 b2 + · · · + an bn .

By definition the first scalar invariant of Φ is φ1 = a1 · b1 + a2 · b2 + · · · + an · bn where a dot product operator has been placed between the vectors. The first vector invariant of Φ is defined
= a1 × b1 + a2 × b2 + · · · + an × bn φ where a vector cross product operator has been placed between the vectors. (a) Show that the first scalar and vector invariant of Φ=
e1
e2 +
e3 +
e3 e2
e3
are respectively 1 and
e1 +
e3 . e1 + f 2
e2 + f 3
e3 one can form the dyadic ∇f having the matrix components (b) From the vector f = f1
 ∂f1 ∂f2 ∂f3  ∇f = 

∂x ∂f1 ∂y ∂f1 ∂z

∂x ∂f2 ∂y ∂f2 ∂z

∂x ∂f3 ∂y ∂f3 ∂z

.

Show that this dyadic has the first scalar and vector invariants given by ∂f2 ∂f3 ∂f1 + + ∇·f = ∂x ∂y ∂z       ∂f1 ∂f2 ∂f3 ∂f2 ∂f3 ∂f1


− − − e1 + e2 + e3 ∇×f = ∂y ∂z ∂z ∂x ∂x ∂y
51.

Let Φ denote the dyadic given in problem 50. The dyadic Φ2 defined by 1 Φ2 = ai × aj bi × bj 2 i,j

is called the Gibbs second dyadic of Φ, where the summation is taken over all permutations of i and j. When i = j the dyad vanishes. Note that the permutations i, j and j, i give the same dyad and so occurs twice in the final sum. The factor 1/2 removes this doubling. Associated with the Gibbs dyad Φ2 are the scalar invariants

1 (ai × aj ) · (bi × bj ) 2 i,j 1 φ3 = (ai × aj · ak )(bi × bj · bk ) 6

φ2 =

i,j,k

Show that the dyad Φ = as + tq + cu has

the first scalar invariant φ1 = a · s + b · t + c · u
= a×s+b×t+c×u the first vector invariant φ Gibbs second dyad

Φ2 = b × ct × u + c × au × s + a × bs × t

second scalar of Φ φ2 = (b × c) · (t · u) + (c × a) · (u × s) + (a × b) · (s × t) third scalar of Φ φ3 = (a × b · c)(s × t · u)

65
52. (Spherical Trigonometry) Construct a spherical triangle ABC on the surface of a unit sphere with sides and angles less than 180 degrees. Denote by a,b c the unit vectors from the origin of the sphere to the vertices A,B and C. Make the construction such that a · (b × c) is positive with a, b, c forming a right-handed system. Let α, β, γ denote the angles between these unit vectors such that a · b = cos γ

c · a = cos β

b · c = cos α.

(1)

The great circles through the vertices A,B,C then make up the sides of the spherical triangle where side α is opposite vertex A, side β is opposite vertex B and side γ is opposite the vertex C. The angles A,B and C between the various planes formed by the vectors a, b and c are called the interior dihedral angles of the spherical triangle. Note that the cross products a × b = sin γ c

b × c = sin α a

c × a = sin β b

(2)

define unit vectors a, b and c perpendicular to the planes determined by the unit vectors a, b and c. The dot products a · b = cos γ

b · c = cos α

c · a = cos β

(3)

define the angles α,β and γ which are called the exterior dihedral angles at the vertices A,B and C and are such that α=π−A

β =π−B

γ = π − C.

(4)

(a) Using appropriate scaling, show that the vectors a, b, c and a, b, c form a reciprocal set. (b) Show that a · (b × c) = sin α a · a = sin β b · b = sin γ c · c (c) Show that a · (b × c) = sin α a · a = sin β b · b = sin γ c · c (d) Using parts (b) and (c) show that sin β sin γ sin α = = sin α sin γ sin β (e) Use the results from equation (4) to derive the law of sines for spherical triangles sin β sin γ sin α = = sin A sin B sin C (f) Using the equations (2) show that sin β sin γb · c = (c × a) · (a × b) = (c · a)(a · b) − b · c and hence show that cos α = cos β cos γ − sin β sin γ cos α. In a similar manner show also that cos α = cos β cos γ − sin β sin γ cos α. (g) Using part (f) derive the law of cosines for spherical triangles cos α = cos β cos γ + sin β sin γ cos A cos A = − cos B cos C + sin B sin C cos α A cyclic permutation of the symbols produces similar results involving the other angles and sides of the spherical triangle.

65 §1.3 SPECIAL TENSORS Knowing how tensors are defined and recognizing a tensor when it pops up in front of you are two different things. Some quantities, which are tensors, frequently arise in applied problems and you should learn to recognize these special tensors when they occur. In this section some important tensor quantities are defined. We also consider how these special tensors can in turn be used to define other tensors. Metric Tensor Define y i , i = 1, . . . , N as independent coordinates in an N dimensional orthogonal Cartesian coordinate system. The distance squared between two points y i

and y i + dy i ,

i = 1, . . . , N is defined by the

expression ds2 = dy m dy m = (dy 1 )2 + (dy 2 )2 + · · · + (dy N )2 .

(1.3.1)

Assume that the coordinates y i are related to a set of independent generalized coordinates xi , i = 1, . . . , N by a set of transformation equations y i = y i (x1 , x2 , . . . , xN ),

i = 1, . . . , N.

(1.3.2)

To emphasize that each y i depends upon the x coordinates we sometimes use the notation y i = y i (x), for i = 1, . . . , N. The differential of each coordinate can be written as dy m =

∂y m j dx , ∂xj

m = 1, . . . , N,

(1.3.3)

and consequently in the x-generalized coordinates the distance squared, found from the equation (1.3.1), becomes a quadratic form. Substituting equation (1.3.3) into equation (1.3.1) we find ds2 =

∂y m ∂y m i j dx dx = gij dxi dxj ∂xi ∂xj

where gij =

∂y m ∂y m , ∂xi ∂xj

i, j = 1, . . . , N

(1.3.4)

(1.3.5)

are called the metrices of the space defined by the coordinates xi , i = 1, . . . , N. Here the gij are functions of the x coordinates and is sometimes written as gij = gij (x). Further, the metrices gij are symmetric in the indices i and j so that gij = gji for all values of i and j over the range of the indices. If we transform to another coordinate system, say xi , i = 1, . . . , N , then the element of arc length squared is expressed in terms of the barred coordinates and ds2 = g ij dxi dxj , where gij = g ij (x) is a function of the barred coordinates. The following example demonstrates that these metrices are second order covariant tensors.

66 EXAMPLE 1.3-1.

Show the metric components gij are covariant tensors of the second order.

Solution: In a coordinate system xi , i = 1, . . . , N the element of arc length squared is ds2 = gij dxi dxj

(1.3.6)

while in a coordinate system xi , i = 1, . . . , N the element of arc length squared is represented in the form ds2 = g mn dxm dxn .

(1.3.7)

The element of arc length squared is to be an invariant and so we require that gmn dxm dxn = gij dxi dxj

(1.3.8)

Here it is assumed that there exists a coordinate transformation of the form defined by equation (1.2.30) together with an inverse transformation, as in equation (1.2.32), which relates the barred and unbarred coordinates. In general, if xi = xi (x), then for i = 1, . . . , N we have dxi =

∂xi dxm ∂xm

and dxj =

∂xj dxn ∂xn

(1.3.9)

Substituting these differentials in equation (1.3.8) gives us the result g mn dxm dxn = gij

∂xi ∂xj dxm dxn ∂xm ∂xn

or

  ∂xi ∂xj g mn − gij m n dxm dxn = 0 ∂x ∂x

For arbitrary changes in dxm this equation implies that g mn = gij

∂xi ∂xj and consequently gij transforms ∂xm ∂xn

as a second order absolute covariant tensor.

EXAMPLE 1.3-2. (Curvilinear coordinates) Consider a set of general transformation equations from rectangular coordinates (x, y, z) to curvilinear coordinates (u, v, w). These transformation equations and the corresponding inverse transformations are represented

Here y 1 = x, y 2 = y, y 3 = z

x = x(u, v, w)

u = u(x, y, z)

y = y(u, v, w)

v = v(x, y, z)

z = z(u, v, w).

w = w(x, y, z)

(1.3.10)

and x1 = u, x2 = v, x3 = w are the Cartesian and generalized coordinates

and N = 3. The intersection of the coordinate surfaces u = c1 ,v = c2 and w = c3 define coordinate curves of the curvilinear coordinate system. The substitution of the given transformation equations (1.3.10) into the position vector
r = x
e1 + y
e2 + z
e3 produces the position vector which is a function of the generalized coordinates and e2 + z(u, v, w)
e3
r =
r(u, v, w) = x(u, v, w)
e1 + y(u, v, w)


67 and consequently d
r =

∂
r ∂
r ∂
r du + dv + dw, where ∂u ∂v ∂w
1 = ∂
r = E ∂u
2 = ∂
r = E ∂v ∂

3 = r = E ∂w

∂x ∂y ∂z


e1 + e2 + e3 ∂u ∂u ∂u ∂x ∂y ∂z


e1 + e2 + e3 ∂v ∂v ∂v ∂x ∂y ∂z


e1 + e2 + e3 . ∂w ∂w ∂w

(1.3.11)

are tangent vectors to the coordinate curves. The element of arc length in the curvilinear coordinates is ∂
r ∂
r ∂
r ∂
r ∂
r ∂
r · dudu + · dudv + · dudw ∂u ∂u ∂u ∂v ∂u ∂w ∂
r ∂
r ∂
r ∂
r ∂
r ∂
r · dvdu + · dvdv + · dvdw + ∂v ∂u ∂v ∂v ∂v ∂w ∂
r ∂
r ∂
r ∂
r ∂
r ∂
r + · dwdu + · dwdv + · dwdw. ∂w ∂u ∂w ∂v ∂w ∂w

ds2 = d
r · d
r =

(1.3.12)

Utilizing the summation convention, the above can be expressed in the index notation. Define the quantities

x2 = v,

∂
r ∂
r · ∂u ∂v ∂
r ∂
r · = ∂v ∂v ∂
r ∂
r · = ∂w ∂v

g13 =

g21

g22

g23

g32

g33

x3 = w. Then the above element of arc length can be expressed as
i · E
j dxi dxj = gij dxi dxj , ds2 = E

where

∂
r ∂
r · ∂u ∂w ∂
r ∂
r · = ∂v ∂w ∂
r ∂
r · = ∂w ∂w

g12 =

g31 and let x1 = u,

∂
r ∂
r · ∂u ∂u ∂
r ∂
r · = ∂v ∂u ∂
r ∂
r · = ∂w ∂u

g11 =

m m
i · E
j = ∂
r · ∂
r = ∂y ∂y , gij = E ∂xi ∂xj ∂xi ∂xj

i, j = 1, 2, 3

i, j free indices

(1.3.13)

are called the metric components of the curvilinear coordinate system. The metric components may be thought of as the elements of a symmetric matrix, since gij = gji . In the rectangular coordinate system x, y, z, the element of arc length squared is ds2 = dx2 + dy 2 + dz 2 . In this space the metric components are 

1 0 gij =  0 1 0 0

 0 0. 1

68 EXAMPLE 1.3-3. (Cylindrical coordinates (r, θ, z)) The transformation equations from rectangular coordinates to cylindrical coordinates can be expressed as x = r cos θ,

y = r sin θ,

z = z. Here y 1 = x, y 2 = y, y 3 = z

and x1 = r, x2 = θ, x3 = z, and the

e2 + z
e3 . The derivatives of this position position vector can be expressed
r =
r(r, θ, z) = r cos θ
e1 + r sin θ
vector are calculated and we find
1 = ∂
r = cos θ
E e1 + sin θ
e2 , ∂r


2 = ∂
r = −r sin θ
e1 + r cos θ
E e2 , ∂θ


3 = ∂
r =
e3 . E ∂z

From the results in equation (1.3.13), the metric components of this space are 

1 0 gij =  0 r2 0 0

 0 0. 1

We note that since gij = 0 when i = j, the coordinate system is orthogonal.

Given a set of transformations of the form found in equation (1.3.10), one can readily determine the metric components associated with the generalized coordinates. For future reference we list several different coordinate systems together with their metric components. Each of the listed coordinate systems are orthogonal and so gij = 0 for i = j. The metric components of these orthogonal systems have the form 

h21  gij = 0 0

0 h22 0

 0 0  h23

and the element of arc length squared is ds2 = h21 (dx1 )2 + h22 (dx2 )2 + h23 (dx3 )2 .

1. Cartesian coordinates (x, y, z) x=x

h1 = 1

y=y

h2 = 1

z=z

h3 = 1

The coordinate curves are formed by the intersection of the coordinate surfaces x =Constant, y =Constant and z =Constant.

69

Figure 1.3-1. Cylindrical coordinates.

2. Cylindrical coordinates (r, θ, z) x = r cos θ

r≥0

h1 = 1

y = r sin θ

0 ≤ θ ≤ 2π

h2 = r

z=z

−∞
h3 = 1

The coordinate curves, illustrated in the figure 1.3-1, are formed by the intersection of the coordinate surfaces

x2 + y 2 = r2 , y/x = tan θ

Cylinders Planes

z = Constant

Planes.

3. Spherical coordinates (ρ, θ, φ) x = ρ sin θ cos φ

ρ≥0

h1 = 1

y = ρ sin θ sin φ

0≤θ≤π

h2 = ρ

z = ρ cos θ

0 ≤ φ ≤ 2π

h3 = ρ sin θ

The coordinate curves, illustrated in the figure 1.3-2, are formed by the intersection of the coordinate surfaces

x2 + y 2 + z 2 = ρ2

Spheres

x2 + y 2 = tan2 θ z 2

Cones

y = x tan φ Planes. 4. Parabolic cylindrical coordinates (ξ, η, z) x = ξη 1 y = (ξ 2 − η 2 ) 2 z=z

−∞
 ξ 2 + η2  h2 = ξ 2 + η 2

η≥0

h3 = 1

−∞<ξ <∞

h1 =

70

Figure 1.3-2. Spherical coordinates.

The coordinate curves, illustrated in the figure 1.3-3, are formed by the intersection of the coordinate surfaces x2 = −2ξ 2 (y −

ξ2 ) 2

Parabolic cylinders

η2 ) Parabolic cylinders 2 z = Constant Planes.

x2 = 2η 2 (y +

Figure 1.3-3. Parabolic cylindrical coordinates in plane z = 0.

5. Parabolic coordinates (ξ, η, φ) x = ξη cos φ

ξ≥0

y = ξη sin φ 1 z = (ξ 2 − η 2 ) 2

η≥0

 ξ 2 + η2  h2 = ξ 2 + η 2

0 < φ < 2π

h3 = ξη

h1 =

71 The coordinate curves, illustrated in the figure 1.3-4, are formed by the intersection of the coordinate surfaces x2 + y 2 = −2ξ 2 (z −

ξ2 ) 2

Paraboloids

η2 ) Paraboloids 2 y = x tan φ Planes.

x2 + y 2 = 2η 2 (z +

Figure 1.3-4. Parabolic coordinates, φ = π/4.

6. Elliptic cylindrical coordinates (ξ, η, z) x = cosh ξ cos η

ξ≥0

h1 =

sinh2 ξ + sin2 η

y = sinh ξ sin η

0 ≤ η ≤ 2π

h2 =

sinh2 ξ + sin2 η

z=z

−∞
h3 = 1

The coordinate curves, illustrated in the figure 1.3-5, are formed by the intersection of the coordinate surfaces

y2 x2 + =1 cosh2 ξ sinh2 ξ y2 x2 − =1 cos2 η sin2 η

Elliptic cylinders Hyperbolic cylinders

z = Constant

Planes.

72

Figure 1.3-5. Elliptic cylindrical coordinates in the plane z = 0.

7. Elliptic coordinates (ξ, η, φ) !  x = (1 − η 2 )(ξ 2 − 1) cos φ  y = (1 − η 2 )(ξ 2 − 1) sin φ

1≤ξ<∞ −1≤η ≤1 0 ≤ φ < 2π

z = ξη

h1 = ! h2 = h3 =

ξ 2 − η2 ξ2 − 1 ξ 2 − η2 1 − η2

 (1 − η 2 )(ξ 2 − 1)

The coordinate curves, illustrated in the figure 1.3-6, are formed by the intersection of the coordinate surfaces

x2 y2 z2 + + =1 Prolate ellipsoid ξ2 − 1 ξ2 − 1 ξ2 z2 x2 y2 − − =1 Two-sheeted hyperboloid η2 1 − η2 1 − η2 y = x tan φ Planes

8. Bipolar coordinates (u, v, z) a sinh v , 0 ≤ u < 2π cosh v − cos u a sin u , −∞ < v < ∞ y= cosh v − cos u z=z −∞
x=

h21 = h22 h22 =

a2 (cosh v − cos u)2

h23 = 1

73

Figure 1.3-6. Elliptic coordinates φ = π/4.

Figure 1.3-7. Bipolar coordinates. The coordinate curves, illustrated in the figure 1.3-7, are formed by the intersection of the coordinate surfaces

a2 Cylinders sinh2 v a2 x2 + (y − a cot u)2 = Cylinders sin2 u z = Constant Planes.

(x − a coth v)2 + y 2 =

74 9. Conical coordinates (u, v, w) uvw , b 2 > v 2 > a2 > w 2 , ab " u (v 2 − a2 )(w2 − a2 ) y= a a2 − b 2 " 2 u (v − b2 )(w2 − b2 ) z= b b 2 − a2

x=

u≥0

h21 = 1 u2 (v 2 − w2 ) − a2 )(b2 − v 2 ) u2 (v 2 − w2 ) h23 = 2 (w − a2 )(w2 − b2 )

h22 =

(v 2

The coordinate curves, illustrated in the figure 1.3-8, are formed by the intersection of the coordinate surfaces

x2 + y 2 + z 2 = u2 2

2

Spheres

2

y z x + 2 + 2 =0, Cones v2 v − a2 v − b2 x2 y2 z2 + + = 0, Cones. w2 w 2 − a2 w 2 − b2

Figure 1.3-8. Conical coordinates.

10. Prolate spheroidal coordinates (u, v, φ) x = a sinh u sin v cos φ,

u≥0

h21 = h22

y = a sinh u sin v sin φ,

0≤v≤π

h22 = a2 (sinh2 u + sin2 v)

z = a cosh u cos v,

0 ≤ φ < 2π

h23 = a2 sinh2 u sin2 v

The coordinate curves, illustrated in the figure 1.3-9, are formed by the intersection of the coordinate surfaces

x2 y2 z2 + + = 1, 2 2 (a sinh u) (a sinh u) (a cosh u)2 z2 x2 y2 − − = 1, 2 2 (a cos v) (a sin v) (a sin v)2

Prolate ellipsoids Two-sheeted hyperboloid

y = x tan φ,

Planes.

75

Figure 1.3-9. Prolate spheroidal coordinates

11. Oblate spheroidal coordinates (ξ, η, φ) ξ≥0 π π y = a cosh ξ cos η sin φ, − ≤η≤ 2 2 z = a sinh ξ sin η, 0 ≤ φ ≤ 2π

x = a cosh ξ cos η cos φ,

h21 = h22 h22 = a2 (sinh2 ξ + sin2 η) h23 = a2 cosh2 ξ cos2 η

The coordinate curves, illustrated in the figure 1.3-10, are formed by the intersection of the coordinate surfaces

x2 y2 z2 + + = 1, 2 2 (a cosh ξ) (a cosh ξ) (a sinh ξ)2 x2 y2 z2 + − = 1, 2 2 (a cos η) (a cos η) (a sin η)2

Oblate ellipsoids One-sheet hyperboloids

y = x tan φ,

Planes.

12. Toroidal coordinates (u, v, φ) a sinh v cos φ , cosh v − cos u a sinh v sin φ , y= cosh v − cos u a sin u , z= cosh v − cos u

x=

0 ≤ u < 2π −∞ < v < ∞ 0 ≤ φ < 2π

h21 = h22 h22 =

a2 (cosh v − cos u)2

h23 =

a2 sinh2 v (cosh v − cos u)2

The coordinate curves, illustrated in the figure 1.3-11, are formed by the intersection of the coordinate surfaces

 a cos u 2 a2 x2 + y 2 + z − , = sin u sin2 u  2 cosh v a2 x2 + y 2 − a + z2 = , sinh v sinh2 v y = x tan φ,

Spheres Torus planes

76

Figure 1.3-10. Oblate spheroidal coordinates

Figure 1.3-11. Toroidal coordinates EXAMPLE 1.3-4. Show the Kronecker delta δji is a mixed second order tensor. Solution: Assume we have a coordinate transformation xi = xi (x), i = 1, . . . , N of the form (1.2.30) and i

possessing an inverse transformation of the form (1.2.32). Let δ j and δji denote the Kronecker delta in the barred and unbarred system of coordinates. By definition the Kronecker delta is defined i δj

 =

δji

=

0,

if

i = j

1,

if

i=j

.

77 Employing the chain rule we write ∂xm ∂xi ∂xm ∂xk i ∂xm = δ n = n ∂x ∂xi ∂x ∂xi ∂xn k By hypothesis, the xi , i = 1, . . . , N are independent coordinates and therefore we have

(1.3.14) ∂xm ∂xn

m

= δ n and (1.3.14)

simplifies to m

δ n = δki

∂xm ∂xk . ∂xi ∂xn

Therefore, the Kronecker delta transforms as a mixed second order tensor.

Conjugate Metric Tensor Let g denote the determinant of the matrix having the metric tensor gij , i, j = 1, . . . , N as its elements. In our study of cofactor elements of a matrix we have shown that cof (g1j )g1k + cof (g2j )g2k + . . . + cof (gN j )gN k = gδkj .

(1.3.15)

We can use this fact to find the elements in the inverse matrix associated with the matrix having the components gij . The elements of this inverse matrix are g ij =

1 cof (gij ) g

(1.3.16)

and are called the conjugate metric components. We examine the summation g ij gik and find: g ij gik = g 1j g1k + g 2j g2k + . . . + g N j gN k 1 = [cof (g1j )g1k + cof (g2j )g2k + . . . + cof (gN j )gN k ] g 1  j gδk = δkj = g The equation g ij gik = δkj

(1.3.17)

is an example where we can use the quotient law to show g ij is a second order contravariant tensor. Because of the symmetry of g ij and gij the equation (1.3.17) can be represented in other forms. EXAMPLE 1.3-5. Let Ai and Ai denote respectively the covariant and contravariant components of a
Show these components are related by the equations vector A. Ai = gij Aj k

jk

A = g Aj where gij and g ij are the metric and conjugate metric components of the space.

(1.3.18) (1.3.19)

78 Solution: We multiply the equation (1.3.18) by g im (inner product) and use equation (1.3.17) to simplify the results. This produces the equation g im Ai = g im gij Aj = δjm Aj = Am . Changing indices produces the result given in equation (1.3.19). Conversely, if we start with equation (1.3.19) and multiply by gkm (inner j Aj = Am which is another form of the equation (1.3.18) with product) we obtain gkm Ak = gkm g jk Aj = δm

the indices changed. Notice the consequences of what the equations (1.3.18) and (1.3.19) imply when we are in an orthogonal Cartesian coordinate system where 

1 gij =  0 0

 0 0 1 0 0 1



and g ij

1 = 0 0

 0 0 1 0. 0 1

In this special case, we have A1 = g11 A1 + g12 A2 + g13 A3 = A1 A2 = g21 A1 + g22 A2 + g23 A3 = A2 A3 = g31 A1 + g32 A2 + g33 A3 = A3 . These equations tell us that in a Cartesian coordinate system the contravariant and covariant components are identically the same.

EXAMPLE 1.3-6. We have previously shown that if Ai is a covariant tensor of rank 1 its components in a barred system of coordinates are Ai = Aj

∂xj . ∂xi

(1.3.20)

Solve for the Aj in terms of the Aj . (i.e. find the inverse transformation). Solution: Multiply equation (1.3.20) by

∂xi ∂xm

Ai

(inner product) and obtain

∂xi ∂xj ∂xi = A . j ∂xm ∂xi ∂xm

(1.3.21)

∂xj ∂xi ∂xj j = = δm since xj and xm are assumed to be independent ∂xm ∂xi ∂xm coordinates. This reduces equation (1.3.21) to the form In the above product we have

Ai

∂xi j = Aj δm = Am ∂xm

(1.3.22)

which is the desired inverse transformation. This result can be obtained in another way. Examine the transformation equation (1.3.20) and ask the question, “When we have two coordinate systems, say a barred and an unbarred system, does it matter which system we call the barred system?” With some thought it should be obvious that it doesn’t matter which system you label as the barred system. Therefore, we can interchange the barred and unbarred symbols in ∂xj equation (1.3.20) and obtain the result Ai = Aj i which is the same form as equation (1.3.22), but with ∂x a different set of indices.

79 Associated Tensors Associated tensors can be constructed by taking the inner product of known tensors with either the metric or conjugate metric tensor.

Definition: (Associated tensor) Any tensor constructed by multiplying (inner product) a given tensor with the metric or conjugate metric tensor is called an associated tensor.

Associated tensors are different ways of representing a tensor. The multiplication of a tensor by the metric or conjugate metric tensor has the effect of lowering or raising indices. For example the covariant and contravariant components of a vector are different representations of the same vector in different forms. These forms are associated with one another by way of the metric and conjugate metric tensor and g ij Ai = Aj

gij Aj = Ai .

EXAMPLE 1.3-7. The following are some examples of associated tensors. Aj = g ij Ai mi Am Aijk .jk = g mk nj A.nm g Aijk i.. = g

Aj = gij Ai ijk Ai.k m = gmj A

Amjk = gim Ai.jk

Sometimes ‘dots’are used as indices in order to represent the location of the index that was raised or lowered. If a tensor is symmetric, the position of the index is immaterial and so a dot is not needed. For example, if Amn is a symmetric tensor, then it is easy to show that An.m and A.n m are equal and therefore can be written as Anm without confusion. Higher order tensors are similarly related. For example, if we find a fourth order covariant tensor Tijkm we can then construct the fourth order contravariant tensor T pqrs from the relation T pqrs = g pi g qj g rk g sm Tijkm . This fourth order tensor can also be expressed as a mixed tensor. Some mixed tensors associated with the given fourth order covariant tensor are: p T.jkm = g pi Tijkm ,

pq p T..km = g qj T.jkm .

80 Riemann Space VN A Riemannian space VN is said to exist if the element of arc length squared has the form ds2 = gij dxi dxj

(1.3.23)

where the metrices gij = gij (x1 , x2 , . . . , xN ) are continuous functions of the coordinates and are different from constants. In the special case gij = δij the Riemannian space VN reduces to a Euclidean space EN . The element of arc length squared defined by equation (1.3.23) is called the Riemannian metric and any geometry which results by using this metric is called a Riemannian geometry. A space VN is called flat if it is possible to find a coordinate transformation where the element of arclength squared is ds2 = Fi (dxi )2 where each Fi is either +1 or −1. A space which is not flat is called curved. Geometry in VN
i and B
j , then their dot product can be represented
= Ai E
= Bj E Given two vectors A
i · E
·B
= Ai B j E
j = gij Ai B j = Aj B j = Ai Bi = g ij Aj Bi = |A||
B|
cos θ. A

(1.3.24)


and B
Consequently, in an N dimensional Riemannian space VN the dot or inner product of two vectors A is defined: gij Ai B j = Aj B j = Ai Bi = g ij Aj Bi = AB cos θ.

(1.3.25)

In this definition A is the magnitude of the vector Ai , the quantity B is the magnitude of the vector Bi and θ is the angle between the vectors when their origins are made to coincide. In the special case that θ = 90◦ we have gij Ai B j = 0 as the condition that must be satisfied in order that the given vectors Ai and B i are orthogonal to one another. Consider also the special case of equation (1.3.25) when Ai = B i and θ = 0. In this case the equations (1.3.25) inform us that g in An Ai = Ai Ai = gin Ai An = (A)2 .

(1.3.26)

From this equation one can determine the magnitude of the vector Ai . The magnitudes A and B can be 1

written A = (gin Ai An ) 2

1

and B = (gpq B p B q ) 2 and so we can express equation (1.3.24) in the form cos θ =

gij Ai B j 1

1

(gmn Am An ) 2 (gpq B p B q ) 2

.

(1.3.27)

An import application of the above concepts arises in the dynamics of rigid body motion. Note that if a vector Ai has constant magnitude and the magnitude of i

dA dt

dAi dt

is different from zero, then the vectors Ai and j

must be orthogonal to one another due to the fact that gij Ai dA dt = 0. As an example, consider the unit

vectors
e1 ,
e2 and
e3 on a rotating system of Cartesian axes. We have for constants ci , i = 1, 6 that d
e1 e 2 + c2
e3 = c1
dt

d
e2 e 3 + c4
e1 = c3
dt

d
e3 e 1 + c6
e2 = c5
dt

because the derivative of any
ei (i fixed) constant vector must lie in a plane containing the vectors
ej and
ek , (j = i , k = i and j = k), since any vector in this plane must be perpendicular to
ei .

81 The above definition of a dot product in VN can be used to define unit vectors in VN .

Definition: (Unit vector)

Whenever the magnitude of a vec-

i

tor A is unity, the vector is called a unit vector. In this case we have gij Ai Aj = 1.

(1.3.28)

EXAMPLE 1.3-8. (Unit vectors) In VN the element of arc length squared is expressed ds2 = gij dxi dxj which can be expressed in the dxi dxi dxj . This equation states that the vector , i = 1, . . . , N is a unit vector. One application form 1 = gij ds ds ds of this equation is to consider a particle moving along a curve in VN which is described by the parametric equations xi = xi (t), for i = 1, . . . , N. The vector V i =

dxi dt ,

i = 1, . . . , N represents a velocity vector of the

particle. By chain rule differentiation we have Vi = where V =

ds dt

dxi ds dxi dxi = =V , dt ds dt ds

is the scalar speed of the particle and

dxi ds

(1.3.29)

is a unit tangent vector to the curve. The equation

(1.3.29) shows that the velocity is directed along the tangent to the curve and has a magnitude V. That is 

ds dt

2 = (V )2 = gij V i V j .

EXAMPLE 1.3-9. (Curvilinear coordinates) Find an expression for the cosine of the angles between the coordinate curves associated with the transformation equations x = x(u, v, w),

y = y(u, v, w),

z = z(u, v, w).

82

Figure 1.3-12. Angles between curvilinear coordinates. Solution: Let y 1 = x, y 2 = y, y 3 = z and x1 = u, x2 = v, x3 = w denote the Cartesian and curvilinear coordinates respectively. With reference to the figure 1.3-12 we can interpret the intersection of the surfaces v = c2 and w = c3 as the curve
r =
r(u, c2 , c3 ) which is a function of the parameter u. By moving only along ∂
r du and consequently this curve we have d
r = ∂u ∂
r ∂
r ds2 = d
r · d
r = · dudu = g11 (dx1 )2 , ∂u ∂u or 1= This equation shows that the vector be represented by

tr(1)

=

dx1 ds

=

d
r d
r · = g11 ds ds √1 g11



dx1 ds

2 .

is a unit vector along this curve. This tangent vector can

r √ 1 δ1 . g11

The curve which is defined by the intersection of the surfaces u = c1 and w = c3 has the unit tangent vector tr(2) =

r √ 1 δ2 . g22

Similarly, the curve which is defined as the intersection of the surfaces u = c1 and

v = c2 has the unit tangent vector tr(3) =

r √ 1 δ3 . g33

The cosine of the angle θ12 , which is the angle between the

unit vectors tr(1) and tr(2) , is obtained from the result of equation (1.3.25). We find g12 1 1 cos θ12 = gpq tp(1) tq(2) = gpq √ δ1p √ δ2q = √ √ . g11 g22 g11 g22 For θ13 the angle between the directions ti(1) and ti(3) we find g13 cos θ13 = √ √ . g11 g33 Finally, for θ23 the angle between the directions ti(2) and ti(3) we find g23 cos θ23 = √ √ . g22 g33 When θ13 = θ12 = θ23 = 90◦ , we have g12 = g13 = g23 = 0 and the coordinate curves which make up the curvilinear coordinate system are orthogonal to one another. In an orthogonal coordinate system we adopt the notation g11 = (h1 )2 ,

g22 = (h2 )2 ,

g33 = (h3 )2

and

gij = 0, i = j.

83 Epsilon Permutation Symbol Associated with the e−permutation symbols there are the epsilon permutation symbols defined by the relations Fijk =

√ geijk

and

1 Fijk = √ eijk g

(1.3.30)

where g is the determinant of the metrices gij . It can be demonstrated that the eijk permutation symbol is a relative tensor of weight −1 whereas the Fijk permutation symbol is an absolute tensor. Similarly, the eijk permutation symbol is a relative tensor of weight +1 and the corresponding Fijk permutation symbol is an absolute tensor. EXAMPLE 1.3-10. (F permutation symbol) Show that eijk is a relative tensor of weight −1 and the corresponding Fijk permutation symbol is an absolute tensor. Solution: Examine the Jacobian

 1  ∂x1  ∂x  2 J =  ∂x  ∂x13 x  ∂x1 x

∂x

and make the substitution aij =

∂x1 ∂x2 ∂x2 ∂x2 ∂x3 ∂x2

∂x1 ∂x3 ∂x2 ∂x3 ∂x3 ∂x3

      

∂xi , i, j = 1, 2, 3. ∂xj

From the definition of a determinant we may write x eijk aim ajn akp = J( )emnp . x

(1.3.31)

By definition, emnp = emnp in all coordinate systems and hence equation (1.3.31) can be expressed in the form

 x −1 ∂xi ∂xj ∂xk J( ) eijk m n p = emnp x ∂x ∂x ∂x

(1.3.32)

which demonstrates that eijk transforms as a relative tensor of weight −1. We have previously shown the metric tensor gij is a second order covariant tensor and transforms ∂xm ∂xn . Taking the determinant of this result we find according to the rule g ij = gmn ∂xi ∂xj  m 2  x 2  ∂x  g = |gij | = |gmn |  i  = g J( ) (1.3.33) x ∂x where g is the determinant of (gij ) and g is the determinant of (g ij ). This result demonstrates that g is a scalar invariant of weight +2. Taking the square root of this result we find that  x √ g = gJ( ). x

(1.3.34)

√ Consequently, we call g a scalar invariant of weight +1. Now multiply both sides of equation (1.3.32) by √ g and use (1.3.34) to verify the relation  ∂xi ∂xj ∂xk √ g eijk m n p = g emnp . ∂x ∂x ∂x √ This equation demonstrates that the quantity Fijk = g eijk transforms like an absolute tensor.

(1.3.35)

84

Figure 1.3-14. Translation followed by rotation of axes In a similar manner one can show eijk is a relative tensor of weight +1 and Fijk =

√1 eijk g

is an absolute

tensor. This is left as an exercise. Another exercise found at the end of this section is to show that a generalization of the e − δ identity is the epsilon identity g ij Fipt Fjrs = gpr gts − gps gtr .

(1.3.36)

Cartesian Tensors Consider the motion of a rigid rod in two dimensions. No matter how complicated the movement of the rod is we can describe the motion as a translation followed by a rotation. Consider the rigid rod AB illustrated in the figure 1.3-13.

Figure 1.3-13. Motion of rigid rod

In this figure there is a before and after picture of the rod’s position. By moving the point B to B  we have a translation. This is then followed by a rotation holding B fixed.

85

Figure 1.3-15. Rotation of axes

A similar situation exists in three dimensions. Consider two sets of Cartesian axes, say a barred and unbarred system as illustrated in the figure 1.3-14. Let us translate the origin 0 to 0 and then rotate the (x, y, z) axes until they coincide with the (x, y, z) axes. We consider first the rotation of axes when the origins 0 and 0 coincide as the translational distance can be represented by a vector bk , k = 1, 2, 3. When the origin 0 is translated to 0 we have the situation illustrated in the figure 1.3-15, where the barred axes can be thought of as a transformation due to rotation. Let
r = x
e1 + y
e2 + z
e3

(1.3.37)

denote the position vector of a variable point P with coordinates (x, y, z) with respect to the origin 0 and the e2 ,
e3 . This same point, when referenced with respect to the origin 0 and the unit vectors unit vectors
e1 ,
ˆ2 , e ˆ3 , has the representation ˆ1 , e e ˆ1 + y e ˆ2 + z e ˆ3 .
r = x e

(1.3.38)

By considering the projections of
r upon the barred and unbarred axes we can construct the transformation equations relating the barred and unbarred axes. We calculate the projections of
r onto the x, y and z axes and find:

ˆ1 ·
ˆ2 ·
ˆ3 ·

r ·
e1 = x = x( e e1 ) + y( e e1 ) + z( e e1 ) ˆ1 ·
ˆ2 ·
ˆ3 ·

r ·
e2 = y = x( e e2 ) + y( e e2 ) + z( e e2 )

(1.3.39)

ˆ1 ·
ˆ2 ·
ˆ3 ·

r ·
e3 = z = x( e e3 ) + y( e e3 ) + z( e e3 ). We also calculate the projection of
r onto the x, y, z axes and find: ˆ1 ) + y(
ˆ1 ) + z(
ˆ1 ) ˆ1 = x = x(
e1 · e e2 · e e3 · e
r · e ˆ2 = y = x(
ˆ2 ) + y(
ˆ2 ) + z(
ˆ2 )
r · e e1 · e e2 · e e3 · e

(1.3.40)

ˆ3 ) + y(
ˆ3 ) + z(
ˆ3 ). ˆ3 = z = x(
e1 · e e2 · e e3 · e
r · e By introducing the notation (y1 , y2 , y3 ) = (x, y, z) (y 1 , y2 , y3 ) = (x, y, z) and defining θij as the angle ˆ between the unit vectors
ei and ej , we can represent the above transformation equations in a more concise

86 form. We observe that the direction cosines can be written as ˆ1 = cos θ11 711 =
e1 · e

ˆ2 = cos θ12 712 =
e1 · e

ˆ3 = cos θ13 713 =
e1 · e

ˆ1 = cos θ21 721 =
e2 · e

ˆ2 = cos θ22 722 =
e2 · e

ˆ3 = cos θ23 723 =
e2 · e

ˆ1 = cos θ31 731 =
e3 · e

ˆ2 = cos θ32 732 =
e3 · e

ˆ3 = cos θ33 733 =
e3 · e

(1.3.41)

which enables us to write the equations (1.3.39) and (1.3.40) in the form yi = 7ij y j

and

y i = 7ji yj .

(1.3.42)

ˆr
ep = 7pr e

(1.3.43)

Using the index notation we represent the unit vectors as: ˆr = 7pr
ep e

or

where 7pr are the direction cosines. In both the barred and unbarred system the unit vectors are orthogonal and consequently we must have the dot products ˆr · e ˆp = δrp e

and


em ·
en = δmn

(1.3.44)

where δij is the Kronecker delta. Substituting equation (1.3.43) into equation (1.3.44) we find the direction cosines 7ij must satisfy the relations: ˆr · e ˆs = 7pr
e em = 7pr 7ms δpm = 7mr 7ms = δrs ep · 7ms
em = 7pr 7ms
ep ·
and

ˆn = 7rm 7sn δmn = 7rm 7sm = δrs . ˆm · 7sn e ˆn = 7rm 7sn e ˆm · e
er ·
es = 7rm e

The relations 7mr 7ms = δrs

and

7rm 7sm = δrs ,

(1.3.45)

with summation index m, are important relations which are satisfied by the direction cosines associated with a rotation of axes. Combining the rotation and translation equations we find yi = 7ij y j + # $% & rotation

bi #$%&

.

(1.3.46)

translation

We multiply this equation by 7ik and make use of the relations (1.3.45) to find the inverse transformation yk = 7ik (yi − bi ).

(1.3.47)

These transformations are called linear or affine transformations. Consider the xi axes as fixed, while the xi axes are rotating with respect to the xi axes where both sets
= Ai
ei denote a vector fixed in and rotating with the xi axes. We of axes have a common origin. Let A 

dA  dA 
with respect to the fixed (f) and rotating (r) axes. We can denote by and the derivatives of A dt  dt  f

r

87  i
 d
ei d
ei dA  = dA
ei + Ai . Note that is the derivative of a write, with respect to the fixed axes, that dt f dt dt dt vector with constant magnitude. Therefore there exists constants ωi , i = 1, . . . , 6 such that d
e1 e2 − ω 2
e3 = ω3
dt

d
e2 e3 − ω 4
e1 = ω1
dt

d
e3 e1 − ω 6
e2 = ω5
dt


e2 + d
e1 ·
i.e. see page 80. From the dot product
e1 ·
e2 = 0 we obtain by differentiation
e1 · ddt e2 = 0 dt which implies ω4 = ω3 . Similarly, from the dot products
e1 ·
e3 and
e2 ·
e3 we obtain by differentiation the
additional relations ω5 = ω2 and ω6 = ω1 . The derivative of A with respect to the fixed axes can now be represented   i

 dA dA  = dA
 +
ω × A



ei + (ω2 A3 − ω3 A2 ) e1 + (ω3 A1 − ω1 A3 ) e2 + (ω1 A2 − ω2 A1 ) e3 = dt f dt dt r
represents the where
ω = ωi
ei is called an angular velocity vector of the rotating system. The term
ω × A i
 dA  = dA
velocity of the rotating system relative to the fixed system and ei represents the derivative with dt r dt respect to the rotating system. Employing the special transformation equations (1.3.46) let us examine how tensor quantities transform when subjected to a translation and rotation of axes. These are our special transformation laws for Cartesian tensors. We examine only the transformation laws for first and second order Cartesian tensor as higher order transformation laws are easily discerned. We have previously shown that in general the first and second order tensor quantities satisfy the transformation laws: ∂yj ∂y i ∂y i A = Aj i ∂yj ∂y ∂y mn A = Aij m n ∂yi ∂yj ∂yi ∂yj Amn = Aij ∂y m ∂yn ∂y ∂yj m An = Aij m ∂yi ∂y n Ai = Aj

(1.3.48) (1.3.49) (1.3.50) (1.3.51) (1.3.52)

For the special case of Cartesian tensors we assume that yi and y i , i = 1, 2, 3 are linearly independent. We differentiate the equations (1.3.46) and (1.3.47) and find ∂yj ∂yi = 7ij = 7ij δjk = 7ik , ∂yk ∂yk

and

∂yi ∂y k = 7ik = 7ik δim = 7mk . ∂ym ∂ym

Substituting these derivatives into the transformation equations (1.3.48) through (1.3.52) we produce the transformation equations Ai = Aj 7ji i

A = Aj 7ji mn

A

= Aij 7im 7jn

Amn = Aij 7im 7jn m

An = Aij 7im 7jn .

88

Figure 1.3-16. Transformation to curvilinear coordinates

These are the transformation laws when moving from one orthogonal system to another. In this case the direction cosines 7im are constants and satisfy the relations given in equation (1.3.45). The transformation laws for higher ordered tensors are similar in nature to those given above. In the unbarred system (y1 , y2 , y3 ) the metric tensor and conjugate metric tensor are: gij = δij

and

g ij = δij

where δij is the Kronecker delta. In the barred system of coordinates, which is also orthogonal, we have g ij =

∂ym ∂ym . ∂y i ∂yj

From the orthogonality relations (1.3.45) we find g ij = 7mi 7mj = δij

and

gij = δij .

We examine the associated tensors Ai = g ij Aj Aij = g im g jn Amn Ain = g im Amn

Ai = gij Aj Amn = gmi gnj Aij Ain = gnj Aij

and find that the contravariant and covariant components are identical to one another. This holds also in the barred system of coordinates. Also note that these special circumstances allow the representation of contractions using subscript quantities only. This type of a contraction is not allowed for general tensors. It is left as an exercise to try a contraction on a general tensor using only subscripts to see what happens. Note that such a contraction does not produce a tensor. These special situations are considered in the exercises. Physical Components
can be represented in many forms depending upon We have previously shown an arbitrary vector A the coordinate system and basis vectors selected. For example, consider the figure 1.3-16 which illustrates a Cartesian coordinate system and a curvilinear coordinate system.

89

Figure 1.3-17. Physical components
as In the Cartesian coordinate system we can represent a vector A
= Ax
e1 + Ay
e2 + Az
e3 A where (
e1 ,
e2 ,
e3 ) are the basis vectors. Consider a coordinate transformation to a more general coordinate 1
can be represented with contravariant components as system, say (x , x2 , x3 ). The vector A
= A1 E
1 + A2 E
2 + A3 E
3 A

(1.3.53)


1, E
2, E
3 ). Alternatively, the same vector A
can be represented with respect to the tangential basis vectors (E in the form
= A1 E
1 + A2 E
2 + A3 E
3 A

(1.3.54)


1, E
2, E
3 ). These equations are having covariant components with respect to the gradient basis vectors (E just different ways of representing the same vector. In the above representations the basis vectors need not be orthogonal and they need not be unit vectors. In general, the physical dimensions of the components Ai and Aj are not the same.
in a direction is defined as the projection of A
upon a unit The physical components of the vector A
in the direction E
1 is vector in the desired direction. For example, the physical component of A

· E1 = A1 = projection of A
on E
1. A
1|
1| |E |E

(1.3.58)


in the direction E
1 is Similarly, the physical component of A 1
1
· E = A = projection of A
on E
1. A 1 1

|E | |E |

(1.3.59)

EXAMPLE 1.3-11. (Physical components) Let α, β, γ denote nonzero positive constants such that the product relation αγ = 1 is satisfied. Consider the nonorthogonal basis vectors
1 = α
E e1 , illustrated in the figure 1.3-17.


2 = β
E e1 + γ
e2 ,


3 =
E e3

90 It is readily verified that the reciprocal basis is
1 = γ
E e1 − β
e2 ,


2 = α
E e2 ,


3 =
E e3 .


= Ax
e1 + Ay
e2 in the contravariant vector form Consider the problem of representing the vector A
1 + A2 E
2
= A1 E A

or tensor form Ai , i = 1, 2.

This vector has the contravariant components
·E
1 = γAx − βAy A1 = A

and


·E
2 = αAy . A2 = A

Alternatively, this same vector can be represented as the covariant vector
= A1 E
1 + A2 E
2 A

which has the tensor form

Ai , i = 1, 2.

The covariant components are found from the relations
·E
1 = αAx A1 = A


·E
2 = βAx + γAy . A2 = A


2 are found to be:
in the directions E
1 and E The physical components of A 1
1 x − βAy
· E = A = γA  A = A(1) 1 1

|E | |E | γ2 + β2 2
2
· E = A = αAy = Ay = A(2). A
2|
2| α |E |E


Note that these same results are obtained from the dot product relations using either form of the vector A. For example, we can write

and


1
1
1
2
1
· E = A1 (E · E ) + A2 (E · E ) = A(1) A
1|
1| |E |E
2
1
2
2
2
· E = A1 (E · E ) + A2 (E · E ) = A(2). A
2|
2| |E |E


in a direction of a unit vector λi is the generalized In general, the physical components of a vector A dot product in VN . This dot product is an invariant and can be expressed
in direction of λi gij Ai λj = Ai λi = Ai λi = projection of A

91 Physical Components For Orthogonal Coordinates In orthogonal coordinates observe the element of arc length squared in V3 is ds2 = gij dxi dxj = (h1 )2 (dx1 )2 + (h2 )2 (dx2 )2 + (h3 )2 (dx3 )2 

where

(h1 )2  gij = 0 0

0 (h2 )2 0

 0 0 . (h3 )2

(1.3.60)

In this case the curvilinear coordinates are orthogonal and h2(i) = g(i)(i)

i not summed and

gij = 0, i = j.

At an arbitrary point in this coordinate system we take λi , i = 1, 2, 3 as a unit vector in the direction of the coordinate x1 . We then obtain λ1 =

dx1 , ds

λ2 = 0,

λ3 = 0.

This is a unit vector since 1 = gij λi λj = g11 λ1 λ1 = h21 (λ1 )2 or λ1 =

1 h1 .

Here the curvilinear coordinate system is orthogonal and in this case the physical component

of a vector Ai , in the direction xi , is the projection of Ai on λi in V3 . The projection in the x1 direction is determined from A(1) = gij Ai λj = g11 A1 λ1 = h21 A1

1 = h1 A1 . h1

Similarly, we choose unit vectors µi and ν i , i = 1, 2, 3 in the x2 and x3 directions. These unit vectors can be represented 1 dx2 = , ds h2 ν 2 =0,

µ1 =0,

µ2 =

ν 1 =0,

µ3 =0 ν3 =

1 dx3 = ds h3

and the physical components of the vector Ai in these directions are calculated as A(2) = h2 A2

and

A(3) = h3 A3 .

In summary, we can say that in an orthogonal coordinate system the physical components of a contravariant tensor of order one can be determined from the equations A(i) = h(i) A(i) =

√ g(i)(i) A(i) ,

i = 1, 2 or 3 no summation on i,

which is a short hand notation for the physical components (h1 A1 , h2 A2 , h3 A3 ). In an orthogonal coordinate system the nonzero conjugate metric components are g (i)(i) =

1 , g(i)(i)

i = 1, 2, or 3 no summation on i.

92 These components are needed to calculate the physical components associated with a covariant tensor of order one. For example, in the x1 −direction, we have the covariant components λ1 = g11 λ1 = h21

1 = h1 , h1

λ2 = 0,

λ3 = 0

and consequently the projection in V3 can be represented gij Ai λj = gij Ai g jm λm = Aj g jm λm = A1 λ1 g 11 = A1 h1

1 A1 = = A(1). h21 h1

In a similar manner we calculate the relations A(2) =

A2 h2

and

A(3) =

A3 h3

for the other physical components in the directions x2 and x3 . These physical components can be represented in the short hand notation A(i) =

A(i) A(i) =√ , h(i) g(i)(i)

i = 1, 2 or 3

no summation on i.

In an orthogonal coordinate system the physical components associated with both the contravariant and covariant components are the same. To show this we note that when Ai gij = Aj is summed on i we obtain A1 g1j + A2 g2j + A3 g3j = Aj . Since gij = 0 for i = j this equation reduces to A(i) g(i)(i) = A(i) ,

i not summed.

Another form for this equation is A(i) √ A(i) = A(i) g(i)(i) = √ g(i)(i)

i not summed,

which demonstrates that the physical components associated with the contravariant and covariant components are identical. NOTATION The physical components are sometimes expressed by symbols with subscripts which represent the coordinate curve along which the projection is taken. For example, let H i denote the contravariant components of a first order tensor. The following are some examples of the representation of the physical components of H i in various coordinate systems: orthogonal

coordinate

tensor

physical

coordinates

system

components

components

general

(x1 , x2 , x3 )

Hi

H(1), H(2), H(3)

rectangular cylindrical spherical general

(x, y, z) (r, θ, z) (ρ, θ, φ) (u, v, w)

H

i

Hx , Hy , Hz

H

i

Hr , Hθ , Hz

H

i

Hρ , Hθ , Hφ

H

i

Hu , Hv , Hw

93 Higher Order Tensors The physical components associated with higher ordered tensors are defined by projections in VN just like the case with first order tensors. For an nth ordered tensor Tij...k we can select n unit vectors λi , µi , . . . , ν i and form the inner product (projection) Tij...k λi µj . . . ν k . When projecting the tensor components onto the coordinate curves, there are N choices for each of the unit vectors. This produces N n physical components. The above inner product represents the physical component of the tensor Tij...k along the directions of the unit vectors λi , µi , . . . , ν i . The selected unit vectors may or may not be orthogonal. In the cases where the selected unit vectors are all orthogonal to one another, the calculation of the physical components is greatly simplified. By relabeling the unit vectors λi(m) , λi(n) , . . . , λi(p) where (m), (n), ..., (p) represent one of the N directions, the physical components of a general nth order tensor is represented T (m n . . . p) = Tij...k λi(m) λj(n) . . . λk(p) EXAMPLE 1.3-12. (Physical components) In an orthogonal curvilinear coordinate system V3 with metric gij , i, j = 1, 2, 3, find the physical components of (i) the second order tensor Aij . (ii) the second order tensor Aij . (iii) the second order tensor Aij . Solution: The physical components of Amn , m, n = 1, 2, 3 along the directions of two unit vectors λi and µi is defined as the inner product in V3 . These physical components can be expressed n A(ij) = Amn λm (i) µ(j)

i, j = 1, 2, 3,

where the subscripts (i) and (j) represent one of the coordinate directions. Dropping the subscripts (i) and (j), we make the observation that in an orthogonal curvilinear coordinate system there are three choices for the direction of the unit vector λi and also three choices for the direction of the unit vector µi . These three choices represent the directions along the x1 , x2 or x3 coordinate curves which emanate from a point of the curvilinear coordinate system. This produces a total of nine possible physical components associated with the tensor Amn . For example, we can obtain the components of the unit vector λi , i = 1, 2, 3 in the x1 direction directly from an examination of the element of arc length squared ds2 = (h1 )2 (dx1 )2 + (h2 )2 (dx2 )2 + (h3 )2 (dx3 )2 . By setting dx2 = dx3 = 0, we find 1 dx1 = = λ1 , ds h1

λ2 = 0,

λ3 = 0.

This is the vector λi(1) , i = 1, 2, 3. Similarly, if we choose to select the unit vector λi , i = 1, 2, 3 in the x2 direction, we set dx1 = dx3 = 0 in the element of arc length squared and find the components λ1 = 0,

λ2 =

1 dx2 = , ds h2

λ3 = 0.

94 This is the vector λi(2) , i = 1, 2, 3. Finally, if we select λi , i = 1, 2, 3 in the x3 direction, we set dx1 = dx2 = 0 in the element of arc length squared and determine the unit vector λ1 = 0,

λ2 = 0,

λ3 =

1 dx3 = . ds h3

This is the vector λi(3) , i = 1, 2, 3. Similarly, the unit vector µi can be selected as one of the above three directions. Examining all nine possible combinations for selecting the unit vectors, we calculate the physical components in an orthogonal coordinate system as: A11 h1 h1 A21 A(21) = h1 h2 A31 A(31) = h3 h1

A12 h1 h2 A22 A(22) = h2 h2 A32 A(32) = h3 h2

A(11) =

A(12) =

A13 h1 h3 A23 A(23) = h2 h3 A33 A(33) = h3 h3 A(13) =

These results can be written in the more compact form A(ij) =

A(i)(j) h(i) h(j)

no summation on i or j .

(1.3.61)

Aij = g im Amj = g i1 A1j + g i2 A2j + g i3 A3j .

(1.3.62)

For mixed tensors we have

From the fact g ij = 0 for i = j, together with the physical components from equation (1.3.61), the equation (1.3.62) reduces to (i)

A(j) = g (i)(i) A(i)(j) =

1 · h(i) h(j) A(ij) no summation on i and i, j = 1, 2 or 3. h2(i)

This can also be written in the form (i)

A(ij) = A(j)

h(i) h(j)

no summation on i or j.

(1.3.63)

Hence, the physical components associated with the mixed tensor Aij in an orthogonal coordinate system can be expressed as A(11) = A11 h2 h1 h3 A(31) = A31 h1 A(21) = A21

A(12) = A12

h1 h2

A(22) = A22 A(32) = A32

h3 h2

h1 h3 h2 A(23) = A23 h3 A(33) = A33 . A(13) = A13

For second order contravariant tensors we may write Aij gjm = Aim = Ai1 g1m + Ai2 g2m + Ai3 g3m .

95 We use the fact gij = 0 for i = j together with the physical components from equation (1.3.63) to reduce the (i)

above equation to the form A(m) = A(i)(m) g(m)(m)

no summation on m . In terms of physical components

we have h(m) A(im) = A(i)(m) h2(m) h(i)

or A(im) = A(i)(m) h(i) h(m) . no summation i, m = 1, 2, 3

(1.3.64)

Examining the results from equation (1.3.64) we find that the physical components associated with the contravariant tensor Aij , in an orthogonal coordinate system, can be written as: A(11) = A11 h1 h1

A(12) = A12 h1 h2

A(13) = A13 h1 h3

A(21) = A21 h2 h1

A(22) = A22 h2 h2

A(23) = A23 h2 h3

A(31) = A31 h3 h1

A(32) = A32 h3 h2

A(33) = A33 h3 h3 .

Physical Components in General In an orthogonal curvilinear coordinate system, the physical components associated with the nth order tensor Tij...kl along the curvilinear coordinate directions can be represented: T (ij . . . kl) =

T(i)(j)...(k)(l) h(i) h(j) . . . h(k) h(l)

no summations.

These physical components can be related to the various tensors associated with Tij...kl . For example, in ij...m can be an orthogonal coordinate system, the physical components associated with the mixed tensor Tn...kl

expressed as: (i)(j)...(m) h(i) h(j)

T (ij . . . m n . . . kl) = T(n)...(k)(l)

. . . h(m) h(n) . . . h(k) h(l)

no summations.

(1.3.65)

EXAMPLE 1.3-13. (Physical components) Let xi = xi (t), i = 1, 2, 3 denote the position vector of a particle which moves as a function of time t. Assume there exists a coordinate transformation xi = xi (x), for i = 1, 2, 3, of the form given by equations (1.2.33). The position of the particle when referenced with respect to the barred system of coordinates can be found by substitution. The generalized velocity of the particle in the unbarred system is a vector with components vi =

dxi , i = 1, 2, 3. dt

The generalized velocity components of the same particle in the barred system is obtained from the chain rule. We find this velocity is represented by vi =

dxi ∂xi dxj ∂xi j v . = = j dt ∂x dt ∂xj

This equation implies that the contravariant quantities (v 1 , v 2 , v 3 ) = (

dx1 dx2 dx3 , , ) dt dt dt

96 are tensor quantities. These quantities are called the components of the generalized velocity. The coordinates x1 , x2 , x3 are generalized coordinates. This means we can select any set of three independent variables for the representation of the motion. The variables selected might not have the same dimensions. For example, in cylindrical coordinates we let (x1 = r, x2 = θ, x3 = z). Here x1 and x3 have dimensions of distance but x2 has dimensions of angular displacement. The generalized velocities are v1 =

dr dx1 = , dt dt

v2 =

dθ dx2 = , dt dt

v3 =

dz dx3 = . dt dt

Here v 1 and v 3 have units of length divided by time while v 2 has the units of angular velocity or angular change divided by time. Clearly, these dimensions are not all the same. Let us examine the physical components of the generalized velocities. We find in cylindrical coordinates h1 = 1, h2 = r, h3 = 1 and the physical components of the velocity have the forms: vr = v(1) = v 1 h1 =

dr , dt

vθ = v(2) = v 2 h2 = r

dθ , dt

vz = v(3) = v 3 h3 =

dz . dt

Now the physical components of the velocity all have the same units of length divided by time.

Additional examples of the use of physical components are considered later. For the time being, just remember that when tensor equations are derived, the equations are valid in any generalized coordinate system. In particular, we are interested in the representation of physical laws which are to be invariant and independent of the coordinate system used to represent these laws. Once a tensor equation is derived, we can chose any type of generalized coordinates and expand the tensor equations. Before using any expanded tensor equations we must replace all the tensor components by their corresponding physical components in order that the equations are dimensionally homogeneous. It is these expanded equations, expressed in terms of the physical components, which are used to solve applied problems. Tensors and Multilinear Forms Tensors can be thought of as being created by multilinear forms defined on some vector space V. Let us define on a vector space V a linear form, a bilinear form and a general multilinear form. We can then illustrate how tensors are created from these forms.

Definition: (Linear form)

Let V denote a vector space which

contains vectors
x,
x1 ,
x2 , . . . . A linear form in
x is a scalar function ϕ(
x) having a single vector argument
x which satisfies the linearity properties: (i) (ii)

ϕ(
x1 +
x2 ) = ϕ(
x1 ) + ϕ(
x2 )

(1.3.66)

ϕ(µ
x1 ) = µϕ(
x1 )

for all arbitrary vectors
x1 ,
x2 in V and all real numbers µ.

97 An example of a linear form is the dot product relation
·
x ϕ(
x) = A

(1.3.67)


is a constant vector and
x is an arbitrary vector belonging to the vector space V. where A Note that a linear form in
x can be expressed in terms of the components of the vector
x and the base e2 ,
e3 ) used to represent
x. To show this, we write the vector
x in the component form vectors (
e1 ,
ei = x1
e1 + x2
e2 + x3
e3 ,
x = xi
where xi , i = 1, 2, 3 are the components of
x with respect to the basis vectors (
e1 ,
e2 ,
e3 ). By the linearity property of ϕ we can write ei ) = ϕ(x1
e1 + x2
e2 + x3
e3 ) ϕ(
x) = ϕ(xi
e1 ) + ϕ(x2
e2 ) + ϕ(x3
e3 ) = ϕ(x1
= x1 ϕ(
e1 ) + x2 ϕ(
e2 ) + x3 ϕ(
e3 ) = xi ϕ(
ei ) Thus we can write ϕ(
x) = xi ϕ(
ei ) and by defining the quantity ϕ(
ei ) = ai as a tensor we obtain ϕ(
x) = xi ai .
1, E
2, E
3 ) then the components of
x also must change. e2 ,
e3 ) to (E Note that if we change basis from (
e1 ,
Letting xi denote the components of
x with respect to the new basis, we would have
i
x = xi E


i ) = xi ϕ(E
i ). and ϕ(
x) = ϕ(xi E


i ) so that ϕ(
x) = xi ai . Whenever there is a definite relation The linear form ϕ defines a new tensor ai = ϕ(E
1, E
2, E
3 ), say, e2 ,
e3 ) and (E between the basis vectors (
e1 ,
j
i = ∂x
E ej , ∂xi

then there exists a definite relation between the tensors ai and ai . This relation is
i ) = ϕ( ai = ϕ(E

∂xj ∂xj ∂xj

aj . i ej ) = i ϕ( ej ) = ∂x ∂x ∂xi

This is the transformation law for an absolute covariant tensor of rank or order one. The above idea is now extended to higher order tensors.

Definition: ( Bilinear form)

A bilinear form in
x and
y is a

scalar function ϕ(
x,
y) with two vector arguments, which satisfies the linearity properties: (i) ϕ(
x1 +
x2 ,
y1 ) = ϕ(
x1 ,
y1 ) + ϕ(
x2 ,
y1 ) (ii) ϕ(
x1 ,
y1 +
y2 ) = ϕ(
x1 ,
y1 ) + ϕ(
x1 ,
y2 )

(1.3.68)

(iii) ϕ(µ
x1 ,
y1 ) = µϕ(
x1 ,
y1 ) (iv) ϕ(
x1 , µ
y1 ) = µϕ(
x1 ,
y1 ) for arbitrary vectors
x1 ,
x2 ,
y1 ,
y2 in the vector space V and for all real numbers µ.

98 Note in the definition of a bilinear form that the scalar function ϕ is linear in both the arguments
x and
y . An example of a bilinear form is the dot product relation ϕ(
x,
y) =
x ·
y

(1.3.69)

where both
x and
y belong to the same vector space V. The definition of a bilinear form suggests how multilinear forms can be defined.

Definition: (Multilinear forms) A multilinear form of degree M or a M degree linear form in the vector arguments
x1 ,
x2 , . . . ,
xM is a scalar function ϕ(
x1 ,
x2 , . . . ,
xM ) of M vector arguments which satisfies the property that it is a linear form in each of its arguments. That is, ϕ must satisfy for each j = 1, 2, . . . , M the properties: (i) ϕ(
x1 , . . . ,
xj1 +
xj2 , . . .
xM ) = ϕ(
x1 , . . . ,
xj1 , . . . ,
xM ) + ϕ(
x1 , . . . ,
xj2 , . . . ,
xM ) (ii)

ϕ(
x1 , . . . , µ
xj , . . . ,
xM ) = µϕ(
x1 , . . . ,
xj , . . . ,
xM ) (1.3.70)

for all arbitrary vectors
x1 , . . . ,
xM in the vector space V and all real numbers µ.

An example of a third degree multilinear form or trilinear form is the triple scalar product ϕ(
x,
y ,
z) =
x · (
y ×
z).

(1.3.71)

Note that multilinear forms are independent of the coordinate system selected and depend only upon the e2 ,
e3 ) and represent vector arguments. In a three dimensional vector space we select the basis vectors (
e1 ,
all vectors with respect to this basis set. For example, if
x,
y ,
z are three vectors we can represent these vectors in the component forms ei ,
x = xi


ej ,
y = y j


ek
z = z k


(1.3.72)

where we have employed the summation convention on the repeated indices i, j and k. Substituting equations (1.3.72) into equation (1.3.71) we obtain ei , y j
ej , z k
ek ) = xi y j z k ϕ(
ei ,
ej ,
ek ), ϕ(xi


(1.3.73)

since ϕ is linear in all its arguments. By defining the tensor quantity ej ,
ek ) = eijk ϕ(
ei ,


(1.3.74)

99 (See exercise 1.1, problem 15) the trilinear form, given by equation (1.3.71), with vectors from equations (1.3.72), can be expressed as ϕ(
x,
y ,
z) = eijk xi y j z k ,

i, j, k = 1, 2, 3.

(1.3.75)

The coefficients eijk of the trilinear form is called a third order tensor. It is the familiar permutation symbol considered earlier. In a multilinear form of degree M , ϕ(
x,
y , . . . ,
z), the M arguments can be represented in a component e2 ,
e3 ). Let these vectors have components xi , y i , z i , i = 1, 2, 3 form with respect to a set of basis vectors (
e1 ,
with respect to the selected basis vectors. We then can write ei ,
x = xi


ej ,
y = y j


ek .
z = z k


Substituting these vectors into the M degree multilinear form produces ei , y j
ej , . . . , z k
ek ) = xi y j · · · z k ϕ(
ϕ(xi
ei ,
ej , . . . ,
ek ).

(1.3.76)

Consequently, the multilinear form defines a set of coefficients aij...k = ϕ(
ei ,
ej , . . . ,
ek )

(1.3.77)

which are referred to as the components of a tensor of order M. The tensor is thus created by the multilinear form and has M indices if ϕ is of degree M.
2, E
3 ) the multilinear form defines
1, E Note that if we change to a different set of basis vectors, say, (E a new tensor
i, E
j, . . . , E
k ). aij...k = ϕ(E

(1.3.78)

This new tensor has a bar over it to distinguish it from the previous tensor. A definite relation exists between the new and old basis vectors and consequently there exists a definite relation between the components of the barred and unbarred tensors components. Recall that if we are given a set of transformation equations y i = y i (x1 , x2 , x3 ), i = 1, 2, 3,

(1.3.79)

from rectangular to generalized curvilinear coordinates, we can express the basis vectors in the new system by the equations j
i = ∂y
ej , i = 1, 2, 3. (1.3.80) E ∂xi For example, see equations (1.3.11) with y 1 = x, y 2 = y, y 3 = z, x1 = u, x2 = v, x3 = w. Substituting

equations (1.3.80) into equations (1.3.78) we obtain ∂y α ∂y β ∂y γ
eα , j
eβ , . . . , k
eγ ). i ∂x ∂x ∂x By the linearity property of ϕ, this equation is expressible in the form aij...k = ϕ(

∂y α ∂y β ∂y γ . . . k ϕ(
eα ,
eβ , . . . ,
eγ ) i j ∂x ∂x ∂x ∂y α ∂y β ∂y γ aij...k = . . . aαβ...γ ∂xi ∂xj ∂xk This is the familiar transformation law for a covariant tensor of degree M. By selecting reciprocal basis aij...k =

vectors the corresponding transformation laws for contravariant vectors can be determined. The above examples illustrate that tensors can be considered as quantities derivable from multilinear forms defined on some vector space.

100 Dual Tensors The e-permutation symbol is often used to generate new tensors from given tensors. For Ti1 i2 ...im a skew-symmetric tensor, we define the tensor 1 j1 j2 ...jn−m i1 i2 ...im Tˆ j1 j2 ...jn−m = e Ti1 i2 ...im m!

m≤n

(1.3.81)

as the dual tensor associated with Ti1 i2 ...im . Note that the e-permutation symbol or alternating tensor has a weight of +1 and consequently the dual tensor will have a higher weight than the original tensor. The e-permutation symbol has the following properties ei1 i2 ...iN ei1 i2 ...iN = N ! N ei1 i2 ...iN ej1 j2 ...jN = δji11 ij22...i ...jN m ek1 k2 ...km i1 i2 ...iN −m ej1 j2 ...jm i1 i2 ...iN −m = (N − m)!δkj11jk22...j ...km

(1.3.82)

m δkj11jk22...j ...km Tj1 j2 ...jm = m!Tk1 k2 ...km .

Using the above properties we can solve for the skew-symmetric tensor in terms of the dual tensor. We find Ti1 i2 ...im =

1 Tˆ j1 j2 ...jn−m . ei i ...i j j ...j (n − m)! 1 2 m 1 2 n−m

(1.3.83)

For example, if Aij i, j = 1, 2, 3 is a skew-symmetric tensor, we may associate with it the dual tensor Vi =

1 ijk e Ajk , 2!

which is a first order tensor or vector. Note that Aij has the components   0 A12 A13  −A12 0 A23  −A13 −A23 0

(1.3.84)


are and consequently, the components of the vector V (V 1 , V 2 , V 3 ) = (A23 , A31 , A12 ).

(1.3.85)

Note that the vector components have a cyclic order to the indices which comes from the cyclic properties of the e-permutation symbol. As another example, consider the fourth order skew-symmetric tensor Aijkl , i, j, k, l = 1, . . . , n. We can associate with this tensor any of the dual tensor quantities 1 ijkl e Aijkl 4! 1 V i = eijklm Ajklm 4! 1 V ij = eijklmn Aklmn 4! 1 V ijk = eijklmnp Almnp 4! 1 V ijkl = eijklmnpr Amnpr 4! V =

Applications of dual tensors can be found in section 2.2.

(1.3.86)

101 EXERCISE 1.3
1. (a) From the transformation law for the second order tensor g ij = gab solve for the gab in terms of gij .

∂xa ∂xb ∂xi ∂xj

(b) Show that if gij is symmetric in one coordinate system it is symmetric in all coordinate systems.  x √ (c) Let g = det(gij ) and g = det(gij ) and show that g = gJ 2 ( xx ) and consequently g = gJ( ). This x √ shows that g is a scalar invariant of weight 2 and g is a scalar invariant of weight 1.
2.

For gij =


3.

∂y m ∂y m ∂xi ∂xj

show that g ij =

∂xi ∂xj ∂y m ∂y m

Show that in a curvilinear coordinate system which is orthogonal we have: (a)

g = det(gij ) = g11 g22 g33

(b)

gmn = g mn = 0 for m = n 1 gN N = for N = 1, 2, 3 (no summation on N) gN N

(c)


4.

 i 2  ∂y  Show that g = det(gij ) =  j  = J 2 , where J is the Jacobian. ∂x


5.

Define the quantities h1 = hu = |

∂
r |, ∂u

h2 = hv = |

∂
r |, ∂v

h3 = hw = |

∂
r | and construct the unit ∂w

vectors 1 ∂
r 1 ∂
r 1 ∂
r

, ev = , ew = . h1 ∂u h2 ∂v h3 ∂w (a) Assume the coordinate system is orthogonal and show that
eu =

 g11 = h21 = 

g22 g33

∂x ∂u



2 +



2 +

∂z ∂u

2 ,

2  2 ∂y ∂z = = + + , ∂v ∂v  2  2  2 ∂x ∂y ∂z = h23 = + + . ∂w ∂w ∂w h22

∂x ∂v

2



∂y ∂u

eu du + h2
ev dv + h3
ew dw. (b) Show that d
r can be expressed in the form d
r = h1
(c) Show that the volume of the elemental parallelepiped having d
r as diagonal can be represented dτ = Hint:

√ ∂(x, y, z) dudvdw. g dudvdw = J dudvdw = ∂(u, v, w)   A1 


|A · (B × C)| =  B1  C1

A2 B2 C2

 A3  B3  C3 

102

Figure 1.3-18 Oblique cylindrical coordinates.
6.

For the change d
r given in problem 5, show the elemental parallelepiped with diagonal d
r has:

(a) the element of area dS1 = (b) The element of area dS2 = (c) the element of area dS3 =

2 dvdw in the u =constant surface. g22 g33 − g23 2 dudw in the v =constant surface. g33 g11 − g13 2 dudv in the w =constant surface. g11 g22 − g12

(d) What do the above elements of area reduce to in the special case the curvilinear coordinates are orthog
× B)
· (A
× B)

× B|
= (A |A . onal? Hint:
· A)(
B
· B)
− (A
· B)(
A
· B)
= (A
7.

In Cartesian coordinates you are given the affine transformation. xi = 7ij xj where x1 =

1 (5x1 − 14x2 + 2x3 ), 15

1 x2 = − (2x1 + x2 + 2x3 ), 3

x3 =

1 (10x1 + 2x2 − 11x3 ) 15

(a) Show the transformation is orthogonal.
1 , x2 , x3 ) in the unbarred system has the components (b) A vector A(x A1 = (x1 )2 ,

A2 = (x2 )2

A3 = (x3 )2 .

Find the components of this vector in the barred system of coordinates.
8.

Calculate the metric and conjugate metric tensors in cylindrical coordinates (r, θ, z).


9.

Calculate the metric and conjugate metric tensors in spherical coordinates (ρ, θ, φ).


10.

Calculate the metric and conjugate metric tensors in parabolic cylindrical coordinates (ξ, η, z).


11.

Calculate the metric and conjugate metric components in elliptic cylindrical coordinates (ξ, η, z).


12.

Calculate the metric and conjugate metric components for the oblique cylindrical coordinates (r, φ, η),

illustrated in figure 1.3-18, where x = r cos φ, 0<α≤

π 2.

Note: When α =

π 2

y = r sin φ + η cos α,

cylindrical coordinates result.

z = η sin α and α is a parameter

103
13.

Calculate the metric and conjugate metric tensor associated with the toroidal surface coordinates

(ξ, η) illustrated in the figure 1.3-19, where x = (a + b cos ξ) cos η

a>b>0

y = (a + b cos ξ) sin η

0 < ξ < 2π

z = b sin ξ

0 < η < 2π

Figure 1.3-19. Toroidal surface coordinates
14.

Calculate the metric and conjugate metric tensor associated with the spherical surface coordinates

(θ, φ), illustrated in the figure 1.3-20, where x = a sin θ cos φ

a>0

y = a sin θ sin φ

0 < φ < 2π π 0<θ< 2

z = a cos θ

is constant


15.

Consider gij , i, j = 1, 2 g22 −g12 g11 , g 12 = g 21 = , g 22 = where ∆ = g11 g22 − g12 g21 . (a) Show that g 11 = ∆ ∆ ∆ ik k (b) Use the results in part (a) and verify that gij g = δj , i, j, k = 1, 2.


16.

Let Ax , Ay , Az denote the constant components of a vector in Cartesian coordinates. Using the

transformation laws (1.2.42) and (1.2.47) to find the contravariant and covariant components of this vector upon changing to (a) cylindrical coordinates (r, θ, z). (b) spherical coordinates (ρ, θ, φ) and (c) Parabolic cylindrical coordinates.
17.

Find the relationship which exists between the given associated tensors. (a) Apqk r. (b) Ap.mrs

and Apq rs and Apq ..rs

(c)

Ai.j. .l.m

and A.s.p r.t.

(d)

Amnk

and Aij ..k

104

Figure 1.3-20. Spherical surface coordinates
18.

Given the fourth order tensor Cikmp = λδik δmp + µ(δim δkp + δip δkm ) + ν(δim δkp − δip δkm ) where λ, µ

and ν are scalars and δij is the Kronecker delta. Show that under an orthogonal transformation of rotation of axes with xi = 7ij xj

where 7rs 7is = 7mr 7mi = δri the components of the above tensor are unaltered. Any

tensor whose components are unaltered under an orthogonal transformation is called an ‘isotropic’ tensor. Another way of stating this problem is to say “Show Cikmp is an isotropic tensor.”
19.

Assume Aijl is a third order covariant tensor and B pqmn is a fourth order contravariant tensor. Prove

that Aikl B klmn is a mixed tensor of order three, with one covariant and two contravariant indices.
20.

Assume that Tmnrs is an absolute tensor. Show that if Tijkl + Tijlk = 0 in the coordinate system xr

then T ijkl + T ijlk = 0 in any other coordinate system xr .
21.

Show that Fijk Frst

  gir  =  gjr  gkr

gis gjs gks

 git  gjt  gkt 

Hint: See problem 38, Exercise 1.1
22.

Determine if the tensor equation Fmnp Fmij + Fmnj Fmpi = Fmni Fmpj is true or false. Justify your answer.


23.

Prove the epsilon identity g ij Fipt Fjrs = gpr gts − gps gtr . Hint: See problem 38, Exercise 1.1


24. Frmn

1 Let Ars denote a skew-symmetric contravariant tensor and let cr = Frmn Amn where 2 √ = germn . Show that cr are the components of a covariant tensor. Write out all the components.

1 rmn 1 F Amn where Frmn = √ ermn . 2 g Show that cr are the components of a contravariant tensor. Write out all the components.


25.

Let Ars denote a skew-symmetric covariant tensor and let cr =

105
26.

s s Let Apq Brqs = Cpr where Brqs is a relative tensor of weight ω1 and Cpr is a relative tensor of weight

ω2 . Prove that Apq is a relative tensor of weight (ω2 − ω1 ).
27.

When Aij is an absolute tensor prove that

√ i gAj is a relative tensor of weight +1.


28.

When Aij is an absolute tensor prove that

√1 Ai g j

is a relative tensor of weight −1.


29. (a) Show eijk is a relative tensor of weight +1. (b) Show Fijk =
30.

√1 eijk g

is an absolute tensor. Hint: See example 1.1-25.

The equation of a surface can be represented by an equation of the form Φ(x1 , x2 , x3 ) = constant.

Show that a unit normal vector to the surface can be represented by the vector ni =

∂Φ g ij ∂x j

1

∂Φ ∂Φ 2 (g mn ∂x m ∂xn )

.


31.

Assume that g ij = λgij with λ a nonzero constant. Find and calculate g ij in terms of g ij .


32.

Determine if the following tensor equation is true. Justify your answer. Frjk Ari + Firk Arj + Fijr Ark = Fijk Arr .

Hint: See problem 21, Exercise 1.1.
33.

Show that for Ci and C i associated tensors, and C i = Fijk Aj Bk , then Ci = Fijk Aj B k


34.

Prove that Fijk and Fijk are associated tensors.


35.

Show Fijk Ai Bj Ck = Fijk Ai B j C k .


36.

Let Tji , i, j = 1, 2, 3 denote a second order mixed tensor. Show that the given quantities are scalar

invariants.

Hint: Consider the determinant of gij .

(i) I1 = Tii  1 i 2 i (Ti ) − Tm (ii) I2 = Tim 2 (iii) I3 = det|Tji |


37. (a) Assume Aij and B ij , i, j = 1, 2, 3 are absolute contravariant tensors, and determine if the inner product C ik = Aij B jk is an absolute tensor? ∂xj ∂xj = δnm is satisfied, and determine whether the inner product in (b) Assume that the condition ∂xn ∂xm part (a) is a tensor? (c) Consider only transformations which are a rotation and translation of axes y i = 7ij yj + bi , where 7ij are ∂y j ∂yj direction cosines for the rotation of axes. Show that = δnm ∂yn ∂ym

106
38.

For Aijk a Cartesian tensor, determine if a contraction on the indices i and j is allowed. That

is, determine if the quantity Ak = Aiik ,

(summation on i) is a tensor. Hint: See part(c) of the previous

problem.
39.
40.

j k k δn − δnj δm . Prove the e-δ identity eijk eimn = δm

Consider the vector Vk , k = 1, 2, 3 and define the matrix (aij ) having the elements aij = eijk Vk ,

where eijk is the e−permutation symbol. (a) Solve for Vi in terms of amn by multiplying both sides of the given equation by eijl and note the e − δ identity allows us to simplify the result. (b) Sum the given expression on k and then assign values to the free indices (i,j=1,2,3) and compare your results with part (a). (c) Is aij symmetric, skew-symmetric, or neither?
41.

It can be shown that the continuity equation of fluid dynamics can be expressed in the tensor form 1 ∂ √ ∂L = 0, ( gLV r ) + √ r g ∂x ∂t

where L is the density of the fluid, t is time, V r , with r = 1, 2, 3 are the velocity components and g = |gij | is the determinant of the metric tensor. Employing the summation convention and replacing the tensor components of velocity by their physical components, express the continuity equation in (a) Cartesian coordinates (x, y, z) with physical components Vx , Vy , Vz . (b) Cylindrical coordinates (r, θ, z) with physical components Vr , Vθ , Vz . (c) Spherical coordinates (ρ, θ, φ) with physical components Vρ , Vθ , Vφ .
42.

Let x1 , x2 , x3 denote a set of skewed coordinates with respect to the Cartesian coordinates y 1 , y 2 , y 3 .
2, E
3 are unit vectors in the directions of the x1 , x2 and x3 axes respectively. If the unit
1, E Assume that E

vectors satisfy the relations
1 · E
1 = 1 E


2 = cos θ12
1 · E E


2 · E
2 = 1 E


1 · E
3 = cos θ13 E


3 · E
3 = 1 E


2 · E
3 = cos θ23 , E

then calculate the metrices gij and conjugate metrices g ij .
43.

Let Aij , i, j = 1, 2, 3, 4 denote the skew-symmetric second rank tensor 

0 a b d  −a 0 Aij =  −b −d 0 −c −e −f

 c e , f 0

where a, b, c, d, e, f are complex constants. Calculate the components of the dual tensor V ij =

1 ijkl e Akl . 2

107
44.

In Cartesian coordinates the vorticity tensor at a point in a fluid medium is defined   1 ∂Vj ∂Vi − ωij = 2 ∂xi ∂xj where Vi are the velocity components of the fluid at the point. The vorticity vector at a point in a fluid 1 medium in Cartesian coordinates is defined by ω i = eijk ωjk . Show that these tensors are dual tensors. 2


45.

Write out the relation between each of the components of the dual tensors 1 Tˆ ij = eijkl Tkl i, j, k, l = 1, 2, 3, 4 2 and show that if ijkl is an even permutation of 1234, then Tˆ ij = Tkl .


46.

Consider the general affine transformation x ¯i = aij xj where (x1 , x2 , x3 ) = (x, y, z) with inverse

transformation xi = bij x¯j . Determine (a) the image of the plane Ax + By + Cz + D = 0 under this transformation and (b) the image of a second degree conic section Ax2 + 2Bxy + Cy 2 + Dx + Ey + F = 0.
47.

Using a multilinear form of degree M, derive the transformation law for a contravariant vector of

degree M.
48.

Let g denote the determinant of gij and show that

∂g ∂gij = gg ij k . ∂xk ∂x


49.

We have shown that for a rotation of xyz axes with respect to a set of fixed x ¯y¯z¯ axes, the derivative
of a vector A with respect to an observer on the barred axes is given by  

 dA
 = dA  + ω
× A. dt f dt r Introduce the operators


 dA  = derivative in fixed system dt f 

= dA  = derivative in rotating system Dr A dt r
= (Dr +

(a) Show that Df A ω ×)A.
is the position vector
r. Show that Df
r = (Dr +
ω×)
r (b) Consider the A  special  case that the vector    
 = V
 +

 represents the velocity of a particle relative to the fixed system produces V ω ×
r where V    f r f   and V
 represents the velocity of a particle with respect to the rotating system of coordinates. r         (c) Show that
a =
a +
ω × (
ω ×
r) where
a represents the acceleration of a particle relative to the f f r  fixed system and
a represents the acceleration of a particle with respect to the rotating system.
= Df A

r

(d) Show in the special case
ω is a constant that  
+ω
a = 2
ω × V
× (
ω ×
r) f


is the velocity of the particle relative to the rotating system. The term 2
ω × V
is referred to where V as the Coriolis acceleration and the term
ω × (
ω ×
r) is referred to as the centripetal acceleration.

108 §1.4 DERIVATIVE OF A TENSOR In this section we develop some additional operations associated with tensors. Historically, one of the basic problems of the tensor calculus was to try and find a tensor quantity which is a function of the metric ∂gij ∂ 2 gij tensor gij and some of its derivatives , , . . . . A solution of this problem is the fourth order ∂xm ∂xm ∂xn Riemann Christoffel tensor Rijkl to be developed shortly. In order to understand how this tensor was arrived at, we must first develop some preliminary relationships involving Christoffel symbols. Christoffel Symbols Let us consider the metric tensor gij which we know satisfies the transformation law g αβ = gab

∂xa ∂xb . ∂xα ∂xβ

Define the quantity (α, β, γ) =

∂gαβ ∂ 2 xa ∂xb ∂xa ∂ 2 xb ∂gab ∂xc ∂xa ∂xb + g + g ab ab γ = γ α α γ ∂x ∂xc ∂x ∂x ∂xβ ∂x ∂x ∂xβ ∂xα ∂xβ ∂xγ

1 [(α, β, γ) + (β, γ, α) − (γ, α, β)] to obtain the result 2    ∂g γα ∂xb ∂ 2 xa ∂gbc ∂gca ∂xa ∂xb ∂xc 1 ∂gab − + − + g . = ab 2 ∂xc ∂xa ∂xb ∂xα ∂xβ ∂xγ ∂xβ ∂xβ ∂xα ∂xγ

and form the combination of terms  ∂gβγ 1 ∂gαβ + 2 ∂xγ ∂xα

(1.4.1)

In this equation the combination of derivatives occurring inside the brackets is called a Christoffel symbol of the first kind and is defined by the notation [ac, b] = [ca, b] =

  1 ∂gab ∂gbc ∂gac + − . 2 ∂xc ∂xa ∂xb

(1.4.2)

The equation (1.4.1) defines the transformation for a Christoffel symbol of the first kind and can be expressed as [α γ, β] = [ac, b]

∂xa ∂xb ∂xc ∂ 2 xa ∂xb + g . ab ∂xα ∂xβ ∂xγ ∂xα ∂xγ ∂xβ

(1.4.3)

Observe that the Christoffel symbol of the first kind [ac, b] does not transform like a tensor. However, it is symmetric in the indices a and c. At this time it is convenient to use the equation (1.4.3) to develop an expression for the second derivative term which occurs in that equation as this second derivative term arises in some of our future considerations. ∂xβ de To solve for this second derivative we can multiply equation (1.4.3) by g and simplify the result to the ∂xd form ∂xa ∂xc ∂xβ de ∂ 2 xe de = −g [ac, d] + [α γ, β] g . (1.4.4) ∂xα ∂xγ ∂xα ∂xγ ∂xd The transformation g de = g λµ

∂xd ∂xe allows us to express the equation (1.4.4) in the form ∂xλ ∂xµ

∂ 2 xe ∂xa ∂xc ∂xe de βµ [α γ, β] µ . α γ = −g [ac, d] α γ +g ∂x ∂x ∂x ∂x ∂x

(1.4.5)

109 Define the Christoffel symbol of the second kind as 

i jk

'

 =

i kj

'

1 = g [jk, α] = g iα 2



iα

∂gkα ∂gjα ∂gjk + − ∂xj ∂xk ∂xα

 .

(1.4.6)

This Christoffel symbol of the second kind is symmetric in the indices j and k and from equation (1.4.5) we see that it satisfies the transformation law 

µ αγ

'

∂xe = ∂xµ



e ac

'

∂xa ∂xc ∂ 2 xe . α γ + ∂x ∂x ∂xα ∂xγ

(1.4.7)

Observe that the Christoffel symbol of the second kind does not transform like a tensor quantity. We can use the relation defined by equation (1.4.7) to express the second derivative of the transformation equations in terms of the Christoffel symbols of the second kind. At times it will be convenient to represent the Christoffel symbols with a subscript to indicate the metric from which they are calculated. Thus, an alternative notation  '  ' i i for j k is the notation j k . g

EXAMPLE 1.4-1. (Christoffel symbols) Solve for the Christoffel symbol of the first kind in terms of the Christoffel symbol of the second kind. Solution: By the definition from equation (1.4.6) we have 

i jk

' = g iα [jk, α].

We multiply this equation by gβi and find  gβi 

and so [jk, α] = gαi

i jk

i jk

' = δβα [jk, α] = [jk, β]

'

 = gα1

1 jk

'

 + · · · + gαN

N jk

' .

EXAMPLE 1.4-2. (Christoffel symbols of first kind) Derive formulas to find the Christoffel symbols of the first kind in a generalized orthogonal coordinate system with metric coefficients gij = 0

for

i = j

and

g(i)(i) = h2(i) ,

i = 1, 2, 3

where i is not summed. Solution: In an orthogonal coordinate system where gij = 0 for i = j we observe that 1 [ab, c] = 2



∂gac ∂gbc ∂gab + − b a ∂x ∂x ∂xc

 .

Here there are 33 = 27 quantities to calculate. We consider the following cases:

(1.4.8)

110 CASE I Let a = b = c = i, then the equation (1.4.8) simplifies to [ab, c] = [ii, i] =

1 ∂gii 2 ∂xi

(no summation on i).

(1.4.9)

From this equation we can calculate any of the Christoffel symbols [11, 1],

[22, 2],

or [33, 3].

CASE II Let a = b = i = c, then the equation (1.4.8) simplifies to the form [ab, c] = [ii, c] = −

1 ∂gii 2 ∂xc

(no summation on i and i = c).

(1.4.10)

since, gic = 0 for i = c. This equation shows how we may calculate any of the six Christoffel symbols [11, 2],

[11, 3],

[22, 1],

[22, 3],

[33, 1],

[33, 2].

CASE III Let a = c = i = b, and noting that gib = 0 for i = b, it can be verified that the equation (1.4.8) simplifies to the form [ab, c] = [ib, i] = [bi, i] =

1 ∂gii 2 ∂xb

(no summation on i and i = b).

(1.4.11)

From this equation we can calculate any of the twelve Christoffel symbols [12, 1] = [21, 1]

[31, 3] = [13, 3]

[32, 3] = [23, 3]

[21, 2] = [12, 2]

[13, 1] = [31, 1]

[23, 2] = [32, 2]

CASE IV Let a = b = c and show that the equation (1.4.8) reduces to [ab, c] = 0,

(a = b = c.)

This represents the six Christoffel symbols [12, 3] = [21, 3] = [23, 1] = [32, 1] = [31, 2] = [13, 2] = 0. From the Cases I,II,III,IV all twenty seven Christoffel symbols of the first kind can be determined. In practice, only the nonzero Christoffel symbols are listed.

EXAMPLE 1.4-3. (Christoffel symbols of the first kind)Find the nonzero Christoffel symbols of the first kind in cylindrical coordinates. Solution: From the results of example 1.4-2 we find that for x1 = r, g11 = 1,

g22 = (x1 )2 = r2 ,

x2 = θ,

g33 = 1

the nonzero Christoffel symbols of the first kind in cylindrical coordinates are: 1 ∂g22 = −x1 = −r 2 ∂x1 1 ∂g22 = x1 = r. [21, 2] = [12, 2] = 2 ∂x1 [22, 1] = −

x3 = z and

111 EXAMPLE 1.4-4. (Christoffel symbols of the second kind) Find formulas for the calculation of the Christoffel symbols of the second kind in a generalized orthogonal coordinate system with metric coefficients gij = 0

for

i = j

and

g(i)(i) = h2(i) ,

i = 1, 2, 3

where i is not summed. Solution: By definition we have 

i jk

' = g im [jk, m] = g i1 [jk, 1] + g i2 [jk, 2] + g i3 [jk, 3]

(1.4.12)

By hypothesis the coordinate system is orthogonal and so g ij = 0 for

i = j

and g ii =

1 gii

i not summed.

The only nonzero term in the equation (1.4.12) occurs when m = i and consequently 

i jk

' = g ii [jk, i] =

[jk, i] gii

no summation on i.

(1.4.13)

We can now consider the four cases considered in the example 1.4-2. CASE I Let j = k = i and show 

i ii

' =

[ii, i] 1 ∂gii 1 ∂ = = ln gii gii 2gii ∂xi 2 ∂xi

no summation on i.

(1.4.14)

CASE II Let k = j = i and show 

i jj

' =

−1 ∂gjj [jj, i] = gii 2gii ∂xi

no summation on i or j.

(1.4.15)

CASE III Let i = j = k and verify that 

j jk

'

 =

j kj

' =

[jk, j] 1 ∂gjj 1 ∂ = = ln gjj gjj 2gjj ∂xk 2 ∂xk

no summation on i or j.

CASE IV For the case i = j = k we find 

i jk

' =

[jk, i] = 0, gii

The above cases represent all 27 terms.

i = j = k

no summation on i.

(1.4.16)

112 EXAMPLE 1.4-5.

(Notation) In the case of cylindrical coordinates we can use the above relations and

find the nonzero Christoffel symbols of the second kind:  ' 1 1 ∂g22 = −x1 = −r =− 22 2g11 ∂x1  '  ' 2 2 1 1 1 ∂g22 = 1 = = = 12 21 2g22 ∂x1 x r Note 1: The notation for the above Christoffel symbols are based upon the assumption that x1 = r, x2 = θ and x3 = z. However, in tensor calculus the choice of the coordinates can be arbitrary. We could just as well have defined x1 = z, x2 = r and x3 = θ. In this latter case, the numbering system of the Christoffel symbols changes. To avoid confusion, an alternate method of writing the Christoffel symbols is to use coordinates in place of the integers 1,2 and 3. For example, in cylindrical coordinates we can write  '  '  ' r θ θ 1 and = −r. = = θθ rθ θr r If we define x1 = r, x2 = θ, x3 = z, then the nonzero Christoffel symbols are written as  '  '  ' 1 2 2 1 and = −r. = = 22 12 21 r In contrast, if we define x1 = z, x2 = r, x3 = θ, then the nonzero Christoffel symbols are written  '  '  ' 2 3 3 1 and = −r. = = 33 23 32 r Note 2: Some textbooks use the notation Γa,bc for Christoffel symbols of the first kind and Γdbc = g da Γa,bc for Christoffel symbols of the second kind. This notation is not used in these notes since the notation suggests that the Christoffel symbols are third order tensors, which is not true. The Christoffel symbols of the first and second kind are not tensors. This fact is clearly illustrated by the transformation equations (1.4.3) and (1.4.7).

Covariant Differentiation Let Ai denote a covariant tensor of rank 1 which obeys the transformation law Aα = Ai

∂xi . ∂xα

(1.4.17)

Differentiate this relation with respect to xβ and show ∂Aα ∂ 2 xi ∂Ai ∂xj ∂xi = Ai α β + . β ∂xj ∂xβ ∂xα ∂x ∂x ∂x

(1.4.18)

Now use the relation from equation (1.4.7) to eliminate the second derivative term from (1.4.18) and express it in the form ∂Aα = Ai ∂xβ

(

σ αβ

'

∂xi − ∂xσ



i jk

'

∂xj ∂xk ∂xα ∂xβ

) +

∂Ai ∂xj ∂xi . ∂xj ∂xβ ∂xα

(1.4.19)

113 Employing the equation (1.4.17), with α replaced by σ, the equation (1.4.19) is expressible in the form  '  ' σ i ∂xj ∂xk ∂Aα ∂Aj ∂xj ∂xk − A − A = σ i α β β k αβ j k ∂xα ∂xβ ∂x ∂x ∂x ∂x or alternatively

(

 ')   ' σ ∂Aj i ∂xj ∂xk ∂Aα − A − A . = σ i αβ jk ∂xk ∂xα ∂xβ ∂xβ

Define the quantity Aj,k

∂Aj = − Ai ∂xk



(1.4.20)

(1.4.21)

'

i jk

(1.4.22)

as the covariant derivative of Aj with respect to xk . The equation (1.4.21) demonstrates that the covariant derivative of a covariant tensor produces a second order tensor which satisfies the transformation law Aα,β = Aj,k

∂xj ∂xk . ∂xα ∂xβ

(1.4.23)

Other notations frequently used to denote the covariant derivative are: Aj,k = Aj;k = Aj/k = ∇k Aj = Aj |k .

(1.4.24)

In the special case where gij are constants the Christoffel symbols of the second kind are zero, and conse∂Aj . That is, under the special circumstances where the quently the covariant derivative reduces to Aj,k = ∂xk Christoffel symbols of the second kind are zero, the covariant derivative reduces to an ordinary derivative. Covariant Derivative of Contravariant Tensor i

A contravariant tensor Ai obeys the transformation law A = Aα form α

Ai = A

∂xi which can be expressed in the ∂xα

∂xi ∂xα

(1.4.24)

by interchanging the barred and unbarred quantities. We write the transformation law in the form of equation (1.4.24) in order to make use of the second derivative relation from the previously derived equation (1.4.7). Differentiate equation (1.4.24) with respect to xj to obtain the relation α

2 i ∂A ∂xβ ∂xi ∂xβ ∂Ai α ∂ x = A + . ∂xj ∂xα ∂xβ ∂xj ∂xβ ∂xj ∂xα

(1.4.25)

Changing the indices in equation (1.4.25) and substituting for the second derivative term, using the relation from equation (1.4.7), produces the equation ∂Ai α =A j ∂x

(

σ αβ

'

∂xi − ∂xσ



i mk

'

∂xm ∂xk ∂xα ∂xβ

)

α

∂A ∂xβ ∂xi ∂xβ + . ∂xj ∂xβ ∂xj ∂xα

Applying the relation found in equation (1.4.24), with i replaced by m, together with the relation ∂xβ ∂xk = δjk , ∂xj ∂xβ

(1.4.26)

114 we simplify equation (1.4.26) to the form 

∂Ai + ∂xj



i mj

'

 m

A

(

σ

∂A + = ∂xβ

Define the quantity i

A

,j

∂Ai = + ∂xj





' ) β i σ α ∂x ∂x A σ. j αβ ∂x ∂x

' i Am mj

(1.4.27)

(1.4.28)

as the covariant derivative of the contravariant tensor Ai . The equation (1.4.27) demonstrates that a covariant derivative of a contravariant tensor will transform like a mixed second order tensor and β σ ∂x ,β j

Ai ,j = A

∂xi . ∂x ∂xσ

(1.4.29)

∂Ai and the ∂xj covariant derivative of a contravariant tensor reduces to an ordinary derivative in this special case.

Again it should be observed that for the condition where gij are constants we have Ai ,j =

In a similar manner the covariant derivative of second rank tensors can be derived. We find these derivatives have the forms: Aij,k Aij ,k Aij ,k

 '  ' σ σ ∂Aij = − Aσj − Aiσ k ik jk ∂x  '  ' ∂Aij i σ = + Aσj − Aiσ σk jk ∂xk  '  ' ∂Aij i j σj iσ = +A +A . σk σk ∂xk

(1.4.30)

In general, the covariant derivative of a mixed tensor Aij...k lm...p of rank n has the form Aij...k lm...p,q =

∂Aij...k lm...p ∂xq



'  '  ' i j k ij...σ + Aiσ...k + · · · + A lm...p lm...p σq σq σq  '  '  ' σ σ σ ij...k ij...k − Aij...k − A − · · · − A σm...p lσ...p lm...σ lq mq pq + Aσj...k lm...p

(1.4.31)

and this derivative is a tensor of rank n + 1. Note the pattern of the + signs for the contravariant indices and the − signs for the covariant indices. Observe that the covariant derivative of an nth order tensor produces an n+ 1st order tensor, the indices of these higher order tensors can also be raised and lowered by multiplication by the metric or conjugate metric tensor. For example we can write g im Ajk |m = Ajk |i

and g im Ajk |m = Ajk |i

115 Rules for Covariant Differentiation The rules for covariant differentiation are the same as for ordinary differentiation. That is: (i) The covariant derivative of a sum is the sum of the covariant derivatives. (ii) The covariant derivative of a product of tensors is the first times the covariant derivative of the second plus the second times the covariant derivative of the first. (iii) Higher derivatives are defined as derivatives of derivatives. Be careful in calculating higher order derivatives as in general Ai,jk = Ai,kj . EXAMPLE 1.4-6. (Covariant differentiation)

Calculate the second covariant derivative Ai,jk .

Solution: The covariant derivative of Ai is Ai,j

 ' σ ∂Ai = − Aσ . ij ∂xj

By definition, the second covariant derivative is the covariant derivative of a covariant derivative and hence Ai,jk = (Ai,j ) ,k =

∂ ∂xk



 '  '  ' ∂Ai σ m m − A − A − A . σ m,j i,m ij ik jk ∂xj

Simplifying this expression one obtains  '  ' σ ∂ ∂ 2 Ai ∂Aσ σ − − A σ ∂xj ∂xk ∂xk i j ∂xk i j   '  '   '  ' ∂Am σ m ∂Ai σ m − − . − A − A σ σ mj ik im jk ∂xj ∂xm

Ai,jk =

Rearranging terms, the second covariant derivative can be expressed in the form  '  '  ' ∂ 2 Ai ∂Aσ σ ∂Am m ∂Ai m − − − ∂xj ∂xk ∂xk i j ∂xj i k ∂xm j k   '  ' '  ' ' ∂ σ σ m m σ − Aσ − − . im jk ik mj ∂xk i j

Ai,jk =

(1.4.32)

116 Riemann Christoffel Tensor Utilizing the equation (1.4.32), it is left as an exercise to show that σ Ai,jk − Ai,kj = Aσ Rijk

where σ Rijk =

∂ ∂xj



σ ik

' −

∂ ∂xk



σ ij

'

 +

m ik

'

σ mj

'

 −

m ij

'

σ mk

' (1.4.33)

is called the Riemann Christoffel tensor. The covariant form of this tensor is i Rhjkl = gih Rjkl .

(1.4.34)

It is an easy exercise to show that this covariant form can be expressed in either of the forms  '  ' s s ∂ ∂ [nk, i] − [nj, i] + [ik, s] − [ij, s] nj nk ∂xj ∂xk   2 ∂ gil 1 ∂ 2 gjl ∂ 2 gik ∂ 2 gjk = − i k − j l + i l + g αβ ([jk, β][il, α] − [jl, β][ik, α]) . 2 ∂xj ∂xk ∂x ∂x ∂x ∂x ∂x ∂x

Rinjk = or

Rijkl

From these forms we find that the Riemann Christoffel tensor is skew symmetric in the first two indices and the last two indices as well as being symmetric in the interchange of the first pair and last pairs of indices and consequently Rjikl = −Rijkl

Rijlk = −Rijkl

Rklij = Rijkl .

In a two dimensional space there are only four components of the Riemann Christoffel tensor to consider. These four components are either +R1212 or −R1212 since they are all related by R1212 = −R2112 = R2121 = −R1221 . In a Cartesian coordinate system Rhijk = 0. The Riemann Christoffel tensor is important because it occurs in differential geometry and relativity which are two areas of interest to be considered later. Additional properties of this tensor are found in the exercises of section 1.5.

117 Physical Interpretation of Covariant Differentiation
1, E
2, E
3 ). These In a system of generalized coordinates (x1 , x2 , x3 ) we can construct the basis vectors (E basis vectors change with position. That is, each basis vector is a function of the coordinates at which they are evaluated. We can emphasize this dependence by writing
i = E
i (x1 , x2 , x3 ) = ∂
r E ∂xi

i = 1, 2, 3.

Associated with these basis vectors we have the reciprocal basis vectors
i = E
i (x1 , x2 , x3 ), E

i = 1, 2, 3


can be represented in terms of contravariant components as which are also functions of position. A vector A
= A1 E
1 + A2 E
2 + A3 E
3 = Aj E
j A

(1.4.35)

or it can be represented in terms of covariant components as
1 + A2 E
2 + A3 E
3 = Aj E
j.
= A1 E A

(1.4.36)


is represented as A change in the vector A
= dA


∂A dxk ∂xk

where from equation (1.4.35) we find

∂Aj
∂A j ∂ Ej Ej = A + ∂xk ∂xk ∂xk

(1.4.37)

or alternatively from equation (1.4.36) we may write
j
∂E ∂Aj
j ∂A E . = A + j k k ∂x ∂x ∂xk

(1.4.38)

We define the covariant derivative of the covariant components as Ai,k =


j
∂A
i = ∂Ai + Aj ∂ E · E
i. ·E k k ∂x ∂x ∂xk

(1.4.39)

The covariant derivative of the contravariant components are defined by the relation Ai ,k =

i

∂A
i = ∂A + Aj ∂ Ej · E
i. ·E k k ∂x ∂x ∂xk

(1.4.40)

Introduce the notation
j ∂E = ∂xk We then have



' m
Em jk



i · ∂ Ej = E ∂xk

and 

 '
j j ∂E
m. E =− mk ∂xk

'  '  ' m
i
i = m δi = Em · E jk jk m jk

(1.4.41)

(1.4.42)

118 and

   ' ' '
j
i · ∂E = − j E
m · E
i = − j δim = − j . E ∂xk mk mk ik

(1.4.43)

Then equations (1.4.39) and (1.4.40) become ' j Aj ik  ' i ∂Ai = + Aj , k jk ∂x ∂Ai = − ∂xk

Ai,k Ai ,k



which is consistent with our earlier definitions from equations (1.4.22) and (1.4.28). Here the first term of the covariant derivative represents the rate of change of the tensor field as we move along a coordinate curve. The second term in the covariant derivative represents the change in the local basis vectors as we move along the coordinate curves. This is the physical interpretation associated with the Christoffel symbols of the second kind. We make the observation that the derivatives of the basis vectors in equations (1.4.39) and (1.4.40) are related since
i · E
j = δj E i and consequently
j
∂

j
i · ∂ E + ∂ Ei · E
j = 0 (Ei · E ) = E k k ∂x ∂x ∂xk
j

i · ∂ E = −E
j · ∂ Ei or E k ∂x ∂xk Hence we can express equation (1.4.39) in the form Ai,k =


∂Ai
j · ∂ Ei . − Aj E k ∂x ∂xk

(1.4.44)

We write the first equation in (1.4.41) in the form
j ∂E = ∂xk



' m
i = [jk, i]E
i gim E jk

(1.4.45)

and consequently

 '  '  '
j i m ∂E m
i · E
m = i E
m = · E δ = i k jk jk jk ∂x (1.4.46)
∂ Ej
i i
·E
m = [jk, i]δ = [jk, m]. and · Em =[jk, i]E m ∂xk These results also reduce the equations (1.4.40) and (1.4.44) to our previous forms for the covariant derivatives. The equations (1.4.41) are representations of the vectors

'i ∂E ∂xk

and

'j ∂E ∂xk

in terms of the basis vectors and

reciprocal basis vectors of the space. The covariant derivative relations then take into account how these vectors change with position and affect changes in the tensor field. The Christoffel symbols in equations (1.4.46) are symmetric in the indices j and k since
j ∂E ∂ = ∂xk ∂xk



∂
r ∂xj

 =

∂ ∂xj



∂
r ∂xk

 =


k ∂E . ∂xj

(1.4.47)

119 The equations (1.4.46) and (1.4.47) enable us to write ) (
j
j
k ∂ E 1 ∂ E ∂ E
m ·
m ·
m · = +E E [jk, m] =E ∂xk 2 ∂xk ∂xj ) (  
m
m ∂ 
1 ∂E ∂E ∂ 




= − Ek · Em · Ej + j Em · Ek − Ej · 2 ∂xk ∂x ∂xk ∂xj ( )   
k
j ∂ ∂ 
1 ∂ E ∂ E
m · E
j ·
j +
k − E
k · Em · E E = −E 2 ∂xk ∂xj ∂xm ∂xm       1 ∂ 

m · E
j · E
j + ∂ E
k − ∂
k E E · E = m 2 ∂xk ∂xj ∂xm   ∂gmk ∂gjk 1 ∂gmj + − m = [kj, m] = k j 2 ∂x ∂x ∂x which again agrees with our previous result.
j,
is represented in the form A
= Aj E For future reference we make the observation that if the vector A involving contravariant components, then we may write  

∂Aj
∂A k j ∂ Ej
Ej + A dxk dx = dA = ∂xk ∂xk ∂xk  j  '  ∂A
i
j Ei dxk Ej + A = k jk ∂x  j  '  ∂A j m
j dxk = Aj dxk E
j. E = + A ,k mk ∂xk

(1.4.48)


is represented in the form A
= Aj E
j involving covariant components it is left as Similarly, if the vector A an exercise to show that
j
= Aj,k dxk E dA

(1.4.49)

Ricci’s Theorem Ricci’s theorem states that the covariant derivative of the metric tensor vanishes and gik,l = 0. Proof: We have gik,l gik,l gik,l

 '  ' m m ∂gik = − gim − gmk kl il ∂xl ∂gik = − [kl, i] − [il, k] ∂xl     ∂gik 1 ∂gik ∂gil ∂gkl ∂gkl ∂gil 1 ∂gik = − + k − + − k = 0. − ∂xl 2 ∂xl ∂x ∂xi 2 ∂xl ∂xi ∂x

Because of Ricci’s theorem the components of the metric tensor can be regarded as constants during covariant differentiation. i EXAMPLE 1.4-7. (Covariant differentiation) Show that δj,k = 0.

Solution i δj,k

 '  '  '  ' ∂δji i σ i i σ i = + δj − δσ = − = 0. σk jk jk jk ∂xk

120 EXAMPLE 1.4-8. (Covariant differentiation) Show that g ij,k = 0. Solution: Since gij g jk = δik we take the covariant derivative of this expression and find k =0 (gij g jk ),l = δi,l

gij g jk,l + gij,l g jk = 0. But gij,l = 0 by Ricci’s theorem and hence gij g jk,l = 0. We multiply this expression by g im and obtain g im gij g jk,l = δjm g jk,l = g mk ,l = 0 which demonstrates that the covariant derivative of the conjugate metric tensor is also zero.

EXAMPLE 1.4-9. (Covariant differentiation) Some additional examples of covariant differentiation are:

(i) (gil Al ),k = gil Al ,k = Ai,k (ii) (gim gjn Aij ) ,k = gim gjn Aij,k = Amn,k

Intrinsic or Absolute Differentiation The intrinsic or absolute derivative of a covariant vector Ai taken along a curve xi = xi (t), i = 1, . . . , N is defined as the inner product of the covariant derivative with the tangent vector to the curve. The intrinsic derivative is represented dxj δAi = Ai,j δt dt   ' j ∂Ai α dx δAi = − Aα ij δt ∂xj dt  ' j dAi δAi α dx = − Aα . i j dt δt dt

(1.4.50)

Similarly, the absolute or intrinsic derivative of a contravariant tensor Ai is represented δAi dAi dxj = Ai ,j = + δt dt dt



' i dxj Ak . jk dt

The intrinsic or absolute derivative is used to differentiate sums and products in the same manner as used in ordinary differentiation. Also if the coordinate system is Cartesian the intrinsic derivative becomes an ordinary derivative. The intrinsic derivative of higher order tensors is similarly defined as an inner product of the covariant derivative with the tangent vector to the given curve. For example, δAij dxp klm = Aij klm,p δt dt is the intrinsic derivative of the fifth order mixed tensor Aij klm .

121 EXAMPLE 1.4-10. (Generalized velocity and acceleration) Let t denote time and let xi = xi (t) for i = 1, . . . , N , denote the position vector of a particle in the generalized coordinates (x1 , . . . , xN ). From the transformation equations (1.2.30), the position vector of the same particle in the barred system of coordinates, (x1 , x2 , . . . , xN ), is xi = xi (x1 (t), x2 (t), . . . , xN (t)) = xi (t), The generalized velocity is v i =

dxi dt ,

i = 1, . . . , N.

i = 1, . . . , N. The quantity v i transforms as a tensor since by definition vi =

∂xi dxj ∂xi j dxi = = v . dt ∂xj dt ∂xj

(1.4.51)

Let us now find an expression for the generalized acceleration. Write equation (1.4.51) in the form vj = v i

∂xj ∂xi

(1.4.52)

and differentiate with respect to time to obtain dv i ∂xj dv j ∂ 2 xj dxk = vi i k + dt dt ∂xi ∂x ∂x dt The equation (1.4.53) demonstrates that

dv i dt

(1.4.53)

does not transform like a tensor. From the equation (1.4.7)

previously derived, we change indices and write equation (1.4.53) in the form dv j dxk = vi dt dt

(

σ ik

'

∂xj − ∂xσ



) ' j ∂xa ∂xc ∂xj dv i . + i k a c ∂x ∂x ∂xi dt

Rearranging terms we find ∂v j dxk + ∂xk dt



j ac

'

 c k  ' ∂xa i ∂x dx σ ∂xj dxk ∂xj ∂v i dxk + v vi σ = i i k dt k dt ik ∂x dt ∂x ∂x ∂x ∂x (  j  '  k  ' ) k σ j ∂v j σ ∂v a dx i dx ∂x = + v + v ak ik ∂xk dt dt ∂xσ ∂xk

or

δv σ ∂xj δv j . = δt δt ∂xσ The above equation illustrates that the intrinsic derivative of the velocity is a tensor quantity. This derivative is called the generalized acceleration and is denoted dv i dxj δv i = v i,j = + f = δt dt dt i



'  ' m n i i dx dx d2 xi m n v v = , + 2 mn m n dt dt dt

i = 1, . . . , N

To summarize, we have shown that if xi = xi (t), i

i = 1, . . . , N

is the generalized position vector, then

dx , i = 1, . . . , N is the generalized velocity, and dt i dxj δv = v i,j , i = 1, . . . , N is the generalized acceleration. fi = δt dt vi =

(1.4.54)

122

Parallel Vector Fields Let y i = y i (t), i = 1, 2, 3 denote a space curve C in a Cartesian coordinate system and let Y i define a constant vector in this system. Construct at each point of the curve C the vector Y i . This produces a field of parallel vectors along the curve C. What happens to the curve and the field of parallel vectors when we transform to an arbitrary coordinate system using the transformation equations y i = y i (x1 , x2 , x3 ),

i = 1, 2, 3

xi = xi (y 1 , y 2 , y 3 ),

i = 1, 2, 3?

with inverse transformation

The space curve C in the new coordinates is obtained directly from the transformation equations and can be written xi = xi (y 1 (t), y 2 (t), y 3 (t)) = xi (t),

i = 1, 2, 3.

The field of parallel vectors Y i become X i in the new coordinates where Y i = Xj

∂y i . ∂xj

(1.4.55)

Since the components of Y i are constants, their derivatives will be zero and consequently we obtain by differentiating the equation (1.4.55), with respect to the parameter t, that the field of parallel vectors X i must satisfy the differential equation dY i ∂ 2 y i dxm dX j ∂y i = = 0. + Xj j m j dt ∂x ∂x ∂x dt dt

(1.4.56)

Changing symbols in the equation (1.4.7) and setting the Christoffel symbol to zero in the Cartesian system of coordinates, we represent equation (1.4.7) in the form ∂ 2yi = ∂xj ∂xm



' i α ∂y j m ∂xα

and consequently, the equation (1.4.56) can be reduced to the form dX j δX j = + δt dt



' j dxm Xk = 0. km dt

(1.4.57)

The equation (1.4.57) is the differential equation which must be satisfied by a parallel field of vectors X i along an arbitrary curve xi (t).

123 EXERCISE 1.4
1. 1

Find the nonzero Christoffel symbols of the first and second kind in cylindrical coordinates

(x , x , x3 ) = (r, θ, z), where x = r cos θ,
2. 1

2

y = r sin θ,

z = z.

Find the nonzero Christoffel symbols of the first and second kind in spherical coordinates 2

(x , x , x3 ) = (ρ, θ, φ), where x = ρ sin θ cos φ,

y = ρ sin θ sin φ,

z = ρ cos θ.


3.

Find the nonzero Christoffel symbols of the first and second kind in parabolic cylindrical coordinates 1 (x , x , x3 ) = (ξ, η, z), where x = ξη, y = (ξ 2 − η 2 ), z = z. 2 1

2


4.

Find the nonzero Christoffel symbols of the first and second kind in parabolic coordinates 1 (x , x , x3 ) = (ξ, η, φ), where x = ξη cos φ, y = ξη sin φ, z = (ξ 2 − η 2 ). 2 1


5. 1

2

Find the nonzero Christoffel symbols of the first and second kind in elliptic cylindrical coordinates

(x , x , x3 ) = (ξ, η, z), where x = cosh ξ cos η,
6.

2

y = sinh ξ sin η,

z = z.

Find the nonzero Christoffel symbols of the first and second kind for the oblique cylindrical coordinates

1

(x , x2 , x3 ) = (r, φ, η), where x = r cos φ,

y = r sin φ+η cos α,

z = η sin α with 0 < α <

π 2

and α constant.

Hint: See figure 1.3-18 and exercise 1.3, problem 12.
7.
8.

Show [ij, k] + [kj, i] =

∂gik . ∂xj



' r (a) Let = g ri [st, i] and solve for the Christoffel symbol of the first kind in terms of the Christoffel st symbol of the secondkind. ' n (b) Assume [st, i] = gni and solve for the Christoffel symbol of the second kind in terms of the st Christoffel symbol of the first kind.


9. (a) Write down the transformation law satisfied by the fourth order tensor Fijk,m . (b) Show that Fijk,m = 0 in all coordinate systems. √ (c) Show that ( g),k = 0.
10.

Show Fijk ,m = 0.


11.

Calculate the second covariant derivative Ai ,kj .

∂φ . ∂xi (a) Find the physical components associated with the covariant components φ ,i Ai φ,i dφ = (b) Show the directional derivative of φ in a direction Ai is . dA (gmn Am An )1/2


12.


i The gradient of a scalar field φ(x1 , x2 , x3 ) is the vector grad φ = E

124
13. (a) Show

√ g is a relative scalar of weight +1.

(b) Use the results from 9(c) and problem 44, Exercise 1.4, to show that  problem ' √ ∂ g m √ √ − g = 0. ( g),k = k ∂x  ' km m ∂ 1 ∂g √ (c) Show that = ln( g) = . k km ∂x 2g ∂xk 
14.

Use the result from problem 9(b) to show Hint: Expand the covariant derivative Frst,p

multiplication with
15.

rst e√ g

' m ∂ 1 ∂g √ = ln( g) = . km ∂xk 2g ∂xk√ and then substitute Frst = gerst . Simplify by inner

and note the Exercise 1.1, problem 26.

Calculate the covariant derivative Ai,m and then contract on m and i to show that 1 ∂ √ i

gA . Ai,i = √ g ∂xi 

'


16.

1 ∂ √ ij

Show √ gg + g ∂xj


17.

Prove that the covariant derivative of a sum equals the sum of the covariant derivatives.

i pq

g pq = 0. Hint: See problem 14.

Hint: Assume Ci = Ai + Bi and write out the covariant derivative for Ci,j .
18.

Let Cji = Ai Bj and prove that the covariant derivative of a product equals the first term times the

covariant derivative of the second term plus the second term times the covariant derivative of the first term. α

β

∂x ∂x and take an ordinary derivative of both sides Start with the transformation law A¯ij = Aαβ i ∂x ¯ ∂x ¯j k with respect to x ¯ and hence derive the relation for Aij,k given in (1.4.30).


19.

∂xi ∂xj and take an ordinary derivative of both sides ∂x ¯α ∂ x ¯β with respect to xk and hence derive the relation for Aij,k given in (1.4.30).


20.

Start with the transformation law Aij = A¯α β


21.

Find the covariant derivatives of (a) Aijk


22.

(b) Aijk

(c) Aijk

Find the intrinsic derivative along the curve xi = xi (t), (a) Aijk

(b) Aijk

(c) Aijk


23.
i and show that dA
i.
= Ai dxk E
= Ai E (a) Assume A ,k i k
and show that dA
i.
= Ai,k dx E
= Ai E (b) Assume A

(d)

Aijk

i = 1, . . . , N for (d)

Aijk

125
24.

(parallel vector field) Imagine a vector field Ai = Ai (x1 , x2 , x3 ) which is a function of position.

Assume that at all points along a curve xi = xi (t), i = 1, 2, 3 the vector field points in the same direction,
is a constant, then we would then have a parallel vector field or homogeneous vector field. Assume A
= dA

' ∂A ∂xk

dxk = 0. Show that for a parallel vector field the condition Ai,k = 0 must be satisfied. 

σ ik

'



' σ . ik


25.

∂ ∂[ik, n] Show that = gnσ j ∂xj ∂x


26.

Show Ar,s − As,r =


27.

In cylindrical coordinates you are given the contravariant vector components

+ ([nj, σ] + [σj, n])

∂Ar ∂As − . ∂xs ∂xr

A1 = r

(a) Find the physical components Ar ,

A2 = cos θ

A3 = z sin θ

Aθ , and Az . Arr

Arθ

Arz

(b) Denote the physical components of Ai,j , i, j = 1, 2, 3, by Aθr

Aθθ

Aθz

Azr

Azθ

Azz .

Find these physical components.
28.

Find the covariant form of the contravariant tensor C i = Fijk Ak,j .

Express your answer in terms of Ak,j . 1 In Cartesian coordinates let x denote the magnitude of the position vector xi . Show that (a) x ,j = xj x 1 1 1 2 −δij 3xi xj (b) x ,ij = δij − 3 xi xj (c) x ,ii = . (d) LetU = , x = 0, and show that U ,ij = + and 3 x x x x x x5 U ,ii = 0.


29.


30.

Consider a two dimensional space with element of arc length squared  g11 2 1 2 2 2 ds = g11 (du ) + g22 (du ) and metric gij = 0

0 g22



where u1 , u2 are surface coordinates. (a) Find formulas to calculate the Christoffel symbols of the first kind. (b) Find formulas to calculate the Christoffel symbols of the second kind.
31.

Find the metric tensor and Christoffel symbols of the first and second kind associated with the

two dimensional space describing points on a cylinder of radius a. Let u1 = θ and u2 = z denote surface coordinates where

x = a cos θ = a cos u1 y = a sin θ = a sin u1 z = z = u2

126
32.

Find the metric tensor and Christoffel symbols of the first and second kind associated with the

two dimensional space describing points on a sphere of radius a. Let u1 = θ and u2 = φ denote surface coordinates where

x = a sin θ cos φ = a sin u1 cos u2 y = a sin θ sin φ = a sin u1 sin u2 z = a cos θ = a cos u1


33.

Find the metric tensor and Christoffel symbols of the first and second kind associated with the

two dimensional space describing points on a torus having the parameters a and b and surface coordinates u1 = ξ,

u2 = η. illustrated in the figure 1.3-19. The points on the surface of the torus are given in terms

of the surface coordinates by the equations x = (a + b cos ξ) cos η y = (a + b cos ξ) sin η z = b sin ξ
34.

Prove that eijk am bj ck ui,m + eijk ai bm ck uj,m + eijk ai bj cm uk,m = ur,r eijk ai bj ck . Hint: See Exercise 1.3,

problem 32 and Exercise 1.1, problem 21.
35.

Calculate the second covariant derivative Ai,jk .


36.

1 ∂ √ ij

Show that σ ij,j = √ gσ + σ mn g ∂xj


37.

Find the contravariant, covariant and physical components of velocity and acceleration in (a) Cartesian



i mn

'

coordinates and (b) cylindrical coordinates.
38.

Find the contravariant, covariant and physical components of velocity and acceleration in spherical

coordinates.
39.

In spherical coordinates (ρ, θ, φ) show that the acceleration components can be represented in terms

of the velocity components as fρ = v˙ ρ −

vθ2 + vφ2 , ρ

fθ = v˙ θ +

vφ2 vρ vθ − , ρ ρ tan θ

fφ = v˙ φ +

vθ vφ vρ vφ + ρ ρ tan θ

Hint: Calculate v˙ ρ , v˙ θ , v˙ φ .
40.

The divergence of a vector Ai is Ai,i . That is, perform a contraction on the covariant derivative

Ai,j to obtain Ai,i . Calculate the divergence in (a) Cartesian coordinates (b) cylindrical coordinates and (c) spherical coordinates.
41.

If S is a scalar invariant of weight one and Aijk is a third order relative tensor of weight W , show

that S −W Aijk is an absolute tensor.

127
42.

Let Y¯ i ,i = 1, 2, 3 denote the components of a field of parallel vectors along the curve C defined by yi the equations y i = y¯i (t), i = 1, 2, 3 in a space with metric tensor g¯ij , i, j = 1, 2, 3. Assume that Y¯ i and d¯ dt are unit vectors such that at each point of the curve C¯ we have g¯ij Y¯ i

d¯ yj = cos θ = Constant. dt

¯ (i.e. The field of parallel vectors makes a constant angle θ with the tangent to each point of the curve C.) i i i i 1 2 3 Show that if Y¯ and y¯ (t) undergo a transformation x = x (¯ y , y¯ , y¯ ), i = 1, 2, 3 then the transformed m ∂x m i vector X = Y¯ j makes a constant angle with the tangent vector to the transformed curve C given by ∂y ¯

y 1 (t), y¯2 (t), y¯3 (t)). xi = xi (¯
43.

Let J denote the Jacobian determinant |

∂xi |. Differentiate J with respect to xm and show that ∂xj

 '  ' ∂J α ∂xp r = J − J . α p ∂xm rm ∂xm Hint: See Exercise 1.1, problem 27 and (1.4.7).
44.

Assume that φ is a relative scalar of weight W so that φ = J W φ. Differentiate this relation with

respect to xk . Use the result from problem 43 to obtain the transformation law: (  '  m  ' )  r ∂x α ∂φ ∂φ W −W φ =J −W φ . mr αk ∂xm ∂xk ∂xk The quantity inside the brackets is called the covariant derivative of a relative scalar of weight W. The covariant derivative of a relative scalar of weight W is defined as φ ,k

∂φ = −W ∂xk



' r φ kr

and this definition has an extra term involving the weight. It can be shown that similar results hold for relative tensors of weight W. For example, the covariant derivative of first and second order relative tensors of weight W have the forms i T ,k

Tji ,k

 '  ' i r ∂T i m = + T −W Ti km kr ∂xk  '  '  ' ∂Tji i σ r σ i = + Tj − Tσ − W Ti k kσ jk kr j ∂x

When the weight term is zero these covariant derivatives reduce to the results given in our previous definitions.
45.

Let

dxi dt

= v i denote a generalized velocity and define the scalar function of kinetic energy T of a

particle with mass m as T =

1 1 m gij v i v j = m gij x˙ i x˙ j . 2 2

Show that the intrinsic derivative of T is the same as an ordinary derivative of T. (i.e. Show that

δT δT

=

dT dt

.)

128
46. Verify the relations

∂gij = −gmj gni ∂xk ∂g in = −g mn g ij ∂xk

∂g nm ∂xk ∂gjm ∂xk

1 ∂ √ ijk

a tensor? Justify Assume that B ijk is an absolute tensor. Is the quantity T jk = √ gB g ∂xi your answer. If your answer is “no”, explain your answer and determine if there any conditions you can


47.

impose upon B ijk such that the above quantity will be a tensor?
48.

The e-permutation symbol can be used to define various vector products. Let Ai , Bi , Ci , Di

i = 1, . . . , N denote vectors, then expand and verify the following products: (a) In two dimensions R =eij Ai Bj Ri =eij Aj

a scalar determinant.

a vector (rotation).

(b) In three dimensions S =eijk Ai Bj Ck Si =eijk Bj Ck Sij =eijk Ck

a scalar determinant.

a vector cross product.

a skew-symmetric matrix

(c) In four dimensions T =eijkm Ai Bj Ck Dm Ti =eijkm Bj Ck Dm Tij =eijkm Ck Dm Tijk =eikm Dm

Expand the curl operator for: (a) Two dimensions B = eij Aj,i (b) Three dimensions Bi = eijk Ak,j (c) Four dimensions Bij = eijkm Am,k

4-dimensional cross product.

skew-symmetric matrix.

skew-symmetric tensor.

with similar products in higher dimensions.
49.

a scalar determinant.

129 §1.5 DIFFERENTIAL GEOMETRY AND RELATIVITY In this section we will examine some fundamental properties of curves and surfaces. In particular, at each point of a space curve we can construct a moving coordinate system consisting of a tangent vector, a normal vector and a binormal vector which is perpendicular to both the tangent and normal vectors. How these vectors change as we move along the space curve brings up the subjects of curvature and torsion associated with a space curve. The curvature is a measure of how the tangent vector to the curve is changing and the torsion is a measure of the twisting of the curve out of a plane. We will find that straight lines have zero curvature and plane curves have zero torsion. In a similar fashion, associated with every smooth surface there are two coordinate surface curves and a normal surface vector through each point on the surface. The coordinate surface curves have tangent vectors which together with the normal surface vectors create a set of basis vectors. These vectors can be used to define such things as a two dimensional surface metric and a second order curvature tensor. The coordinate curves have tangent vectors which together with the surface normal form a coordinate system at each point of the surface. How these surface vectors change brings into consideration two different curvatures. A normal curvature and a tangential curvature (geodesic curvature). How these curvatures are related to the curvature tensor and to the Riemann Christoffel tensor, introduced in the last section, as well as other interesting relationships between the various surface vectors and curvatures, is the subject area of differential geometry. Also presented in this section is a brief introduction to relativity where again the Riemann Christoffel tensor will occur. Properties of this important tensor are developed in the exercises of this section. Space Curves and Curvature For xi = xi (s),i = 1, 2, 3, a 3-dimensional space curve in a Riemannian space Vn with metric tensor gij , and arc length parameter s, the vector T i =

dxi ds

represents a tangent vector to the curve at a point P on

i

the curve. The vector T is a unit vector because gij T i T j = gij

dxi dxj = 1. ds ds

(1.5.1)

Differentiate intrinsically, with respect to arc length, the relation (1.5.1) and verify that gij T i

δT j δT i j + gij T = 0, δs δs

(1.5.2)

δT i = 0. δs

(1.5.3)

which implies that gij T j Hence, the vector

δT i δs

is perpendicular to the tangent vector T i . Define the unit normal vector N i to the

space curve to be in the same direction as the vector Ni =

δT i δs

and write

1 δT i κ δs

(1.5.4)

where κ is a scale factor, called the curvature, and is selected such that gij N i N j = 1

which implies

gij

δT i δT j = κ2 . δs δs

(1.5.5)

130 The reciprocal of curvature is called the radius of curvature. The curvature measures the rate of change of the tangent vector to the curve as the arc length varies. By differentiating intrinsically, with respect to arc length s, the relation gij T i N j = 0 we find that gij T i

δN j δT i j + gij N = 0. δs δs

(1.5.6)

Consequently, the curvature κ can be determined from the relation gij T i

δN j δT i j = −gij N = −gij κN i N j = −κ δs δs

(1.5.7)

which defines the sign of the curvature. In a similar fashion we differentiate the relation (1.5.5) and find that gij N i This later equation indicates that the vector

δN j δs

δN j = 0. δs

(1.5.8)

is perpendicular to the unit normal N i . The equation

(1.5.3) indicates that T i is also perpendicular to N i and hence any linear combination of these vectors will also be perpendicular to N i . The unit binormal vector is defined by selecting the linear combination δN j + κT j δs

(1.5.9)

and then scaling it into a unit vector by defining 1 B = τ j



δN j + κT j δs

 (1.5.10)

where τ is a scalar called the torsion. The sign of τ is selected such that the vectors T i , N i and B i form a right handed system with Fijk T i N j B k = 1 and the magnitude of τ is selected such that B i is a unit vector satisfying gij B i B j = 1.

(1.5.11)

The triad of vectors T i , N i , B i at a point on the curve form three planes. The plane containing T i and B i is called the rectifying plane. The plane containing N i and B i is called the normal plane. The plane containing T i and N i is called the osculating plane. The reciprocal of the torsion is called the radius of torsion. The torsion measures the rate of change of the osculating plane. The vectors T i , N i and B i form a right-handed orthogonal system at a point on the space curve and satisfy the relation B i = Fijk Tj Nk .

(1.5.12)

By using the equation (1.5.10) it can be shown that B i is perpendicular to both the vectors T i and N i since gij B i T j = 0

and gij B i N j = 0.

It is left as an exercise to show that the binormal vector B i satisfies the relation relations

δT i = κN i δs δN i = τ B i − κT i δs δB i = −τ N i δs

δB i δs

= −τ N i . The three

(1.5.13)

131 are known as the Frenet-Serret formulas of differential geometry. Surfaces and Curvature Let us examine surfaces in a Cartesian frame of reference and then later we can generalize our results to other coordinate systems. A surface in Euclidean 3-dimensional space can be defined in several different ways. Explicitly, z = f (x, y), implicitly, F (x, y, z) = 0 or parametrically by defining a set of parametric equations of the form x = x(u, v),

y = y(u, v),

z = z(u, v)

which contain two independent parameters u, v called surface coordinates. For example, the equations x = a sin θ cos φ,

y = a sin θ sin φ,

z = a cos θ

are the parametric equations which define a spherical surface of radius a with parameters u = θ and v = φ. See for example figure 1.3-20 in section 1.3. By eliminating the parameters u, v one can derive the implicit form of the surface and by solving for z one obtains the explicit form of the surface. Using the parametric form of a surface we can define the position vector to a point on the surface which is then represented in terms of the parameters u, v as
r =
r(u, v) = x(u, v)
e1 + y(u, v)
e2 + z(u, v)
e3 .

(1.5.14)

The coordinates (u, v) are called the curvilinear coordinates of a point on the surface. The functions x(u, v), y(u, v), z(u, v) are assumed to be real and differentiable such that
r(u, c2 )

and

∂' r ∂u

×

∂' r ∂v

= 0. The curves


r(c1 , v)

(1.5.15)

with c1 , c2 constants, then define two surface curves called coordinate curves, which intersect at the surface coordinates (c1 , c2 ). The family of curves defined by equations (1.5.15) with equally spaced constant values ci , ci + ∆ci , ci + 2∆ci , . . . define a surface coordinate grid system. The vectors

∂' r ∂u

∂' r and ∂v evaluated at the

surface coordinates (c1 , c2 ) on the surface, are tangent vectors to the coordinate curves through the point and are basis vectors for any vector lying in the surface. Letting (x, y, z) = (y 1 , y 2 , y 3 ) and (u, v) = (u1 , u2 ) and utilizing the summation convention, we can write the position vector in the form
r =
r(u1 , u2 ) = y i (u1 , u2 )
ei .

(1.5.16)

The tangent vectors to the coordinate curves at a point P can then be represented as the basis vectors i
α = ∂
r = ∂y
ei , E α α ∂u ∂u

α = 1, 2

(1.5.17)

where the partial derivatives are to be evaluated at the point P where the coordinate curves on the surface intersect. From these basis vectors we construct a unit normal vector to the surface at the point P by calculating the cross product of the tangent vector
ru = n
=n
(u, v) =

∂' r ∂u

and
rv =

∂' r ∂v .


1 × E
2 E
ru ×
rv =

|
ru ×
rv | |E1 × E2 |

A unit normal is then (1.5.18)

132
1, E
2 and n and is such that the vectors E
form a right-handed system of coordinates. If we transform from one set of curvilinear coordinates (u, v) to another set (¯ u, v¯), which are determined by a set of transformation laws u = u(¯ u, v¯),

v = v(¯ u, v¯),

the equation of the surface becomes u, v¯), v(¯ u, v¯))
e2 + z(u(¯ u, v¯), v(¯ u, v¯))
e3
r =
r(¯ u, v¯) = x(u(¯ u, v¯), v(¯ u, v¯))
e1 + y(u(¯ and the tangent vectors to the new coordinate curves are ∂
r ∂u ∂
r ∂v ∂
r = + ∂u ¯ ∂u ∂ u ¯ ∂v ∂ u ¯

and

∂
r ∂
r ∂u ∂
r ∂v = + . ∂¯ v ∂u ∂¯ v ∂v ∂¯ v

Using the indicial notation this result can be represented as ∂y i ∂y i ∂uβ = . ∂u ¯α ∂uβ ∂ u ¯α This is the transformation law connecting the two systems of basis vectors on the surface. A curve on the surface is defined by a relation f (u, v) = 0 between the curvilinear coordinates. Another way to represent a curve on the surface is to represent it in a parametric form where u = u(t) and v = v(t), where t is a parameter. The vector ∂
r du ∂
r dv d
r = + dt ∂u dt ∂v dt is tangent to the curve on the surface. An element of arc length with respect to the surface coordinates is represented by ds2 = d
r · d
r = where aαβ =

∂' r ∂uα

·

∂' r ∂uβ

∂
r ∂
r · duα duβ = aαβ duα duβ α ∂u ∂uβ

(1.5.19)

with α, β = 1, 2 defines a surface metric. This element of arc length on the surface is

often written as the quadratic form A = ds2 = E(du)2 + 2F du dv + G(dv)2 =

1 EG − F 2 2 (E du + F dv)2 + dv E E

(1.5.20)

and called the first fundamental form of the surface. Observe that for ds2 to be positive definite the quantities E and EG − F 2 must be positive. The surface metric associated with the two dimensional surface is defined by
α · E
β = aαβ = E

∂
r ∂
r ∂y i ∂y i · = , α β ∂u ∂u ∂uα ∂uβ

α, β = 1, 2

(1.5.21)

with conjugate metric tensor aαβ defined such that aαβ aβγ = δγα . Here the surface is embedded in a three dimensional space with metric gij and aαβ is the two dimensional surface metric. In the equation (1.5.20) the quantities E, F, G are functions of the surface coordinates u, v and are determined from the relations ∂y i ∂y i ∂
r ∂
r · = ∂u ∂u ∂u1 ∂u1 ∂y i ∂y i ∂
r ∂
r · = = ∂u ∂v ∂u1 ∂u2 ∂y i ∂y i ∂
r ∂
r · = = ∂v ∂v ∂u2 ∂u2

E =a11 = F =a12 G =a22

(1.5.22)

133 Here and throughout the remainder of this section, we adopt the convention that Greek letters have the range 1,2, while Latin letters have the range 1,2,3. Construct at a general point P on the surface the unit normal vector n
at this point. Also construct a plane which contains this unit surface normal vector n
. Observe that there are an infinite number of planes which contain this unit surface normal. For now, select one of these planes, then later on we will consider all such planes. Let
r =
r(s) denote the position vector defining a curve C which is the intersection of the selected plane with the surface, where s is the arc length along the curve, which is measured from some fixed point on the curve. Let us find the curvature of this curve of intersection. The vector T
= d'r , evaluated ds

at the point P, is a unit tangent vector to the curve C and lies in the tangent plane to the surface at the point P. Here we are using ordinary differentiation rather than intrinsic differentiation because we are in a Cartesian system of coordinates. Differentiating the relation T
· T
= 1, with respect to arc length s we find that T
· dT
= 0 which implies that the vector dT
is perpendicular to the tangent vector T
. Since the ds

ds

coordinate system is Cartesian we can treat the curve of intersection C as a space curve, then the vector
= dT
, evaluated at point P, is defined as the curvature vector with curvature |K|
= κ and radius of K ds
to the space curve is taken in the same direction as dT
so that the curvature R = 1/κ. A unit normal N ds dT


. Consider the geometry of figure 1.5-1 curvature will always be positive. We can then write K = κN = ds and define on the surface a unit vector u
=n
× T
which is perpendicular to both the surface tangent vector T
and the surface normal vector n
, such that the vectors T i ,ui and ni forms a right-handed system.

Figure 1.5-1 Surface curve with tangent plane and a normal plane.

134
1 and E
2 . Note that The direction of u
in relation to T
is in the same sense as the surface tangents E
the vector ddsT
is perpendicular to the tangent vector T
and lies in the plane which contains the vectors n
and u
. We can therefore write the curvature vector K in the component form

= dT = κ(n) n
n +K
g K
+ κ(g) u
=K ds

(1.5.23)

where κ(n) is called the normal curvature and κ(g) is called the geodesic curvature. The subscripts are not indices. These curvatures can be calculated as follows. From the orthogonality condition n
· T
= 0 we obtain
d
n dT + T
· = 0. Consequently, the normal by differentiation with respect to arc length s the result n
· ds ds curvature is determined from the dot product relation d
r d
n n
= κ(n) = −T
· d
n
·K =− · . ds ds ds

(1.5.24)

By taking the dot product of u
with equation (1.5.23) we find that the geodesic curvature is determined from the triple scalar product relation κ(g) = u
·

dT
dT
= (
n × T
) · . ds ds

(1.5.25)

Normal Curvature The equation (1.5.24) can be expressed in terms of a quadratic form by writing κ(n) ds2 = −d
r · d
n.

(1.5.26)

The unit normal to the surface n
and position vector
r are functions of the surface coordinates u, v with d
r =

∂
r ∂
r du + dv ∂u ∂v

and d
n=

∂
n ∂
n du + dv. ∂u ∂v

(1.5.27)

We define the quadratic form  B = −d
r · d
n=−

   ∂
n ∂
r ∂
r ∂
n du + dv · du + dv ∂u ∂v ∂u ∂v

2

2

α

B = e(du) + 2f du dv + g(dv) = bαβ du du 

where n ∂
r ∂
· , e=− ∂u ∂u and bαβ

2f = −

∂
r ∂
n ∂
n ∂
r · + · ∂u ∂v ∂u ∂v

 ,

(1.5.28)

β

g=−

n ∂
r ∂
· ∂v ∂v

(1.5.29)

α, β = 1, 2 is called the curvature tensor and aαγ bαβ = bγβ is an associated curvature tensor.

The quadratic form of equation (1.5.28) is called the second fundamental form of the surface. Alternative methods for calculating the coefficients of this quadratic form result from the following considerations. The unit surface normal is perpendicular to the tangent vectors to the coordinate curves at the point P and therefore we have the orthogonality relationships ∂
r ·n
=0 ∂u

and

∂
r ·n
= 0. ∂v

(1.5.30)

135 Observe that by differentiating the relations in equation (1.5.30), with respect to both u and v, one can derive the results

∂ 2
r n ∂
r ∂
· = b11 ·n
=− 2 ∂u ∂u ∂u ∂
r ∂
n ∂
n ∂
r ∂ 2
r ·n
=− · =− · = b21 = b12 f= ∂u∂v ∂u ∂v ∂u ∂v n ∂ 2
r ∂
r ∂
· = b22 g= ·n
=− 2 ∂v ∂v ∂v and consequently the curvature tensor can be expressed as e=

bαβ = −

(1.5.31)

∂
r ∂
n · . α ∂u ∂uβ

(1.5.32)

The quadratic forms from equations (1.5.20) and (1.5.28) enable us to represent the normal curvature in the form of a ratio of quadratic forms. We find from equation (1.5.26) that the normal curvature in the direction

du dv

is κ(n) =

e(du)2 + 2f du dv + g(dv)2 B = . A E(du)2 + 2F du dv + G(dv)2

(1.5.33)

r ∂' r duα If we write the unit tangent vector to the curve in the form T
= d' ds = ∂uα ds and express the derivative n ∂
n duβ of the unit surface normal with respect to arc length as dds
= ∂u β ds , then the normal curvature can be

expressed in the form κ(n)

d
n =− = −T
· ds



∂
r ∂
n · ∂uα ∂uβ



duα duβ ds ds

(1.5.34)

bαβ duα duβ bαβ duα duβ = = . ds2 aαβ duα duβ Observe that the curvature tensor is a second order symmetric tensor.

In the previous discussions, the plane containing the unit normal vector was arbitrary. Let us now consider all such planes that pass through this unit surface normal. As we vary the plane containing the unit surface normal n
at P we get different curves of intersection with the surface. Each curve has a curvature associated with it. By examining all such planes we can find the maximum and minimum normal curvatures associated with the surface. We write equation (1.5.33) in the form κ(n) = where λ =

dv du .

e + 2f λ + gλ2 E + 2F λ + Gλ2

(1.5.35)

From the theory of proportions we can also write this equation in the form κ(n) =

f + gλ e + fλ (e + f λ) + λ(f + gλ) = = . (E + F λ) + λ(F + Gλ) F + Gλ E + Fλ

(1.5.36)

Consequently, the curvature κ will satisfy the differential equations (e − κE)du + (f − κF )dv = 0

and (f − κF )du + (g − κG)dv = 0.

The maximum and minimum curvatures occur in those directions λ where

dκ(n) dλ

= 0. Calculating the deriva-

tive of κ(n) with respect to λ and setting the derivative to zero we obtain a quadratic equation in λ (F g − Gf )λ2 + (Eg − Ge)λ + (Ef − F e) = 0,

(1.5.37)

(F g − Gf ) = 0.

136 This equation has two roots λ1 and λ2 which satisfy λ1 + λ2 = −

Eg − Ge F g − Gf

and

λ1 λ2 =

Ef − F e , F g − Gf

(1.5.38)

where F g − Gf = 0. The curvatures κ(1) ,κ(2) corresponding to the roots λ1 and λ2 are called the principal curvatures at the point P. Several quantities of interest that are related to κ(1) and κ(2) are: (1) the principal radii of curvature Ri = 1/κi ,i = 1, 2; (2) H =

1 2 (κ(1)

+ κ(2) ) called the mean curvature and K = κ(1) κ(2)

called the total curvature or Gaussian curvature of the surface. Observe that the roots λ1 and λ2 determine two directions on the surface d
r1 ∂
r ∂
r = + λ1 du ∂u ∂v

and

∂
r ∂
r d
r2 = + λ2 . du ∂u ∂v

If these directions are orthogonal we will have d
r1 d
r2 ∂
r ∂
r ∂
r ∂
r · =( + λ1 )( + λ2 ) = 0. du du ∂u ∂v ∂u ∂v This requires that Gλ1 λ2 + F (λ1 + λ2 ) + E = 0.

(1.5.39)

It is left as an exercise to verify that this is indeed the case and so the directions determined by the principal curvatures must be orthogonal. In the case where F g − Gf = 0 we have that F = 0 and f = 0 because the coordinate curves are orthogonal and G must be positive. In this special case there are still two directions determined by the differential equations (1.5.37) with dv = 0, du arbitrary, and du = 0, dv arbitrary. From the differential equations (1.5.37) we find these directions correspond to κ(1) = We let λα =

duα ds

e E

and

κ(2) =

g . G

denote a unit vector on the surface satisfying aαβ λα λβ = 1. Then the equation (1.5.34)

can be written as κ(n) = bαβ λα λβ or we can write (bαβ − κ(n) aαβ )λα λβ = 0. The maximum and minimum normal curvature occurs in those directions λα where (bαβ − κ(n) aαβ )λα = 0 and so κ(n) must be a root of the determinant equation |bαβ − κ(n) aαβ | = 0 or |a

αγ

bαβ −

κ(n) δβγ |

 1 b −κ =  1 2 (n) b1

  b12  = κ2 − bαβ aαβ κ(n) + b = 0. (n) b22 − κ(n)  a

(1.5.40)

This is a quadratic equation in κ(n) of the form κ2(n) − (κ(1) + κ(2) )κ(n) + κ(1) κ(2) = 0. In other words the principal curvatures κ(1) and κ(2) are the eigenvalues of the matrix with elements bγβ = aαγ bαβ . Observe that from the determinant equation in κ(n) we can directly find the total curvature or Gaussian curvature which αγ is an invariant given by K = κ(1) κ(2) = |bα β | = |a bγβ | = b/a. The mean curvature is also an invariant

obtained from H =

1 2 (κ(1)

+ κ(2) ) =

1 αβ bαβ , 2a

where a = a11 a22 − a12 a21 and b = b11 b22 − b12 b21 are the

determinants formed from the surface metric tensor and curvature tensor components.

137 The equations of Gauss, Weingarten and Codazzi
and unit binormal At each point on a space curve we can construct a unit tangent T
, a unit normal N
The derivatives of these vectors, with respect to arc length, can also be represented as linear combinations B.
, B.
See for example the Frenet-Serret formulas from equations (1.5.13). In a similar of the base vectors T
, N fashion the surface vectors
ru ,
rv , n
form a basis and the derivatives of these basis vectors with respect to the surface coordinates u, v can also be expressed as linear combinations of the basis vectors
ru ,
rv , n
. For
. We can write example, the derivatives
ruu ,
ruv ,
rvv can be expressed as linear combinations of
ru ,
rv , n

ruu = c1
ru + c2
rv + c3 n
ruv = c4
ru + c5
rv + c6 n


(1.5.41)



rvv = c7
ru + c8
rv + c9 n where c1 , . . . , c9 are constants to be determined. It is an easy exercise (see exercise 1.5, problem 8) to show that these equations can be written in the indicial notation as  ' γ ∂
r ∂ 2
r = + bαβ n
. α β α β ∂uγ ∂u ∂u

(1.5.42)

These equations are known as the Gauss equations. In a similar fashion the derivatives of the normal vector can be represented as linear combinations of the surface basis vectors. If we write ∂
n = c1
ru + c2
rv ∂u ∂
n = c3
ru + c4
rv ∂v

∂
r ∂
n ∂
n = c∗1 + c∗2 ∂u ∂u ∂v ∂
n ∂
n ∂
r = c∗3 + c∗4 ∂v ∂u ∂v

or

(1.5.43)

where c1 , . . . , c4 and c∗1 , . . . , c∗4 are constants. These equations are known as the Weingarten equations. It is easily demonstrated (see exercise 1.5, problem 9) that the Weingarten equations can be written in the indicial form

where bβα = aβγ bγα

∂
r ∂
n = −bβα β ∂uα ∂u is the mixed second order form of the curvature tensor.

(1.5.44)

The equations of Gauss produce a system of partial differential equations defining the surface coordinates i

x as a function of the curvilinear coordinates u and v. The equations are not independent as certain compatibility conditions must be satisfied. In particular, it is required that the mixed partial derivatives must satisfy ∂ 3
r ∂uα ∂uβ ∂uδ

=

∂ 3
r ∂uα ∂uδ ∂uβ '



We calculate ∂ 3
r = ∂uα ∂uβ ∂uδ



γ αβ

'

∂ 2
r + ∂uγ ∂uδ

∂

.

γ αβ

∂uδ

∂bαβ ∂
r ∂
n + bαβ δ + n
∂uγ ∂u ∂uδ

and use the equations of Gauss and Weingarten to express this derivative in the form   '  ω  '  '  '  ∂ αβ   r γ ω γ ∂ 3
r ∂bαβ ω  ∂
 = + − b b + b + n
. αβ δ  γδ  ∂uδ αβ γδ αβ ∂uα ∂uβ ∂uδ ∂uω ∂uδ

138 Forming the difference ∂ 3
r ∂uα ∂uβ ∂uδ

−

∂ 3
r ∂uα ∂uδ ∂uβ

we find that the coefficients of the independent vectors n
and

∂' r ∂uω

=0 must be zero. Setting the coefficient of n


equal to zero produces the Codazzi equations 

γ αβ

'

 bγδ −

γ αδ

' bγβ +

∂bαβ ∂bαδ − = 0. ∂uδ ∂uβ

(1.5.45)

These equations are sometimes referred to as the Mainardi-Codazzi equations. Equating to zero the coefficient of

∂' r ∂uω

we find that Rδαγβ = bαβ bδγ − bαγ bδβ or changing indices we have the covariant form aωδ Rδαβγ = Rωαβγ = bωβ bαγ − bωγ bαβ ,

where Rδαγβ =

∂ ∂uγ



δ αβ

' −

∂ ∂uβ



δ αγ

'

 +

ω αβ

'

δ ωγ

'

(1.5.46) 

−

ω αγ

'

δ ωβ

' (1.5.47)

is the mixed Riemann curvature tensor. EXAMPLE 1.5-1 Show that the Gaussian or total curvature K = κ(1) κ(2) depends only upon the metric aαβ and is R1212 where a = det(aαβ ). K= a Solution: Utilizing the two-dimensional alternating tensor eαβ and the property of determinants we can write eγδ K = eαβ bγα bδβ

where from page 137, K = |bγβ | = |aαγ bαβ |. Now multiply by eγζ and then contract on

ζ and δ to obtain

But eγδ aγµ aδν = aeµν so that

eγδ eγδ K = eγδ eαβ bγα bδβ = 2K

2K = eγδ eαβ (aγµ bαu ) aδν bβν √ 2K = eαβ a eµν bαµ bβν . Using aeµν = Fµν we have 2K = Fµν Fαβ bαµ bβν .

Interchanging indices we can write 2K = Fβγ Fωα bωβ bαγ

and 2K = Fγβ Fωα bωγ bαβ .

Adding these last two results we find that 4K = Fβγ Fωγ (bωβ bαγ − bωγ bαβ ) = Fβγ Fωγ Rωαβγ . Now multiply βγ ωα both sides by Fστ Fλν to obtain 4KFστ Fλν = δστ δλν Rωαβγ . From exercise 1.5, problem 16, the Riemann

curvature tensor Rijkl is skew symmetric in the (i, j), (k, l) as well as being symmetric in the (ij), (kl) pair βγ ωα δλν Rωαβγ = 4Rλνστ and hence Rλνστ = KFστ Fλν and we have the special case of indices. Consequently, δστ √ √ b R1212 . A much simpler way to obtain this result is to observe K = where K ae12 ae12 = R1212 or K = a a (bottom of page 137) and note from equation (1.5.46) that R1212 = b11 b22 − b12 b21 = b.

Note that on a surface ds2 = aαβ duα duβ where aαβ are the metrices for the surface. This metric is a ∂uα ∂uβ and by taking determinants we find tensor and satisfies a ¯γδ = aαβ γ ∂u ¯ ∂u ¯δ   ∂uα  ∂uβ      a ¯ = ¯ aγδ  γ  δ  = aJ 2 ∂u ¯ ∂u ¯

139 where J is the Jacobian of the surface coordinate transformation. Here the curvature tensor for the surface Rαβγδ has only one independent component since R1212 = R2121 = −R1221 = −R2112 (See exercises 20,21). From the transformation law

α β γ δ ¯ #ηλµ = Rαβγδ ∂u ∂u ∂u ∂u R ∂u ¯# ∂ u ¯η ∂ u ¯λ ∂ u ¯µ ¯ 1212 = R1212 J 2 and consequently one can sum over the repeated indices and show that R

¯ 1212 R R1212 = =K a ¯ a which shows that the Gaussian curvature is a scalar invariant in V2 . Geodesic Curvature
associated with this curve, is For C an arbitrary curve on a given surface the curvature vector K,
and geodesic curvature κ(g) u
and lies in a plane which the vector sum of the normal curvature κ(n) n is perpendicular to the tangent vector to the given curve on the surface. The geodesic curvature κ(g) is obtained from the equation (1.5.25) and can be represented κ(g)

dT
dT

=u = (
n × T
) · = =u
·K
· ds ds



dT
T
× ds

 ·n
.

Substituting into this expression the vectors d
r du dv T
= =
ru +
rv ds ds ds
dT
=
ruu (u  )2 + 2
ruv u  v  +
rvv (v  )2 +
ru u  +
rv v  , =K ds where



=

d ds ,

and by utilizing the results from problem 10 of the exercises following this section, we find

that the geodesic curvature can be represented as 

κ(g)

'   '  ' 2 2 1  3 = (u ) + 2 − (u  )2 v  + 11 12 11  '   '  ' 1 2 1 (v  )3 + (u  v  − u  v  ) EG − F 2 . −2 u  (v  )2 − 22 22 12

(1.5.48)

This equation indicates that the geodesic curvature is only a function of the surface metrices E, F, G and the derivatives u  , v  , u  , v  . When the geodesic curvature is zero the curve is called a geodesic curve. Such curves are often times, but not always, the lines of shortest distance between two points on a surface. For example, the great circle on a sphere which passes through two given points on the sphere is a geodesic curve. If you erase that part of the circle which represents the shortest distance between two points on the circle you are left with a geodesic curve connecting the two points, however, the path is not the shortest distance between the two points. For plane curves we let u = x and v = y so that the geodesic curvature reduces to kg = u  v  − u  v  =

dφ ds

140 where φ is the angle between the tangent T
to the curve and the unit vector
e1 . Geodesics are curves on the surface where the geodesic curvature is zero. Since kg = 0 along a geodesic
to the curve will be in the same surface curve, then at every point on this surface curve the normal N
= 0 and
rv · n
= 0 which reduces to direction as the normal n
to the surface. In this case, we have
ru · n dT
·
ru = 0 ds since the vectors n
and

' dT ds

and

dT
·
rv = 0, ds

(1.5.49)

have the same direction. In particular, we may write d
r ∂
r du ∂
r dv T
= = + =
ru u +
rv v  ds ∂u ds ∂v ds dT
=
ruu (u  )2 + 2
ruv u  v  +
rvv (v  )2 +
ru u  +
rv v  ds

Consequently, the equations (1.5.49) become dT
·
ru = (
ruu ·
ru ) (u  )2 + 2(
ruv ·
ru ) u  v  + (
rvv ·
ru ) (v  )2 + Eu  + F v  = 0 ds . dT
 2    2   ·
rv = (
ruu ·
rv ) (u ) + 2(
ruv ·
rv ) u v + (
rvv ·
rv ) (v ) + F u + Gv = 0. ds

(1.5.50)

Utilizing the results from exercise 1.5,(See problems 4,5 and 6), we can eliminate v  from the equations (1.5.50) to obtain d2 u + ds2



1 11

'

du ds



2 +2

1 12

'

du dv + ds ds



1 22

'

dv ds

2 =0

and eliminating u  from the equations (1.5.50) produces the equation d2 v + ds2



2 11

'

du ds



2 +2

2 12

'

du dv + ds ds



2 22

'

dv ds

2 = 0.

In tensor form, these last two equations are written d2 uα + ds2



α βγ

' a

duβ duγ = 0, ds ds

α, β, γ = 1, 2

(1.5.51)

where u = u1 and v = u2 . The equations (1.5.51) are the differential equations defining a geodesic curve on a surface. We will find that these same type of equations arise in considering the shortest distance between two points in a generalized coordinate system. See for example problem 18 in exercise 2.2.

141 Tensor Derivatives Let uα = uα (t) denote the parametric equations of a curve on the surface defined by the parametric equations xi = xi (u1 , u2 ). We can then represent the surface curve in the spatial geometry since the surface curve can be represented in the spatial coordinates through the representation xi = xi (u1 (t), u2 (t)) = xi (t). Recall that for xi = xi (t) a given curve C , the intrinsic derivative of a vector field Ai along C is defined as the inner product of the covariant derivative of the vector field with the tangent vector to the curve. This intrinsic derivative is written ( )  ' j j i δAi ∂Ai i dx k dx = A ,j = + A j jk g δt dt ∂x dt or



dAi δAi = + δt dt

'

i jk

Ak g

dxj dt

where the subscript g indicates that the Christoffel symbol is formed from the spatial metric gij . If Aα is a surface vector defined along the curve C, the intrinsic derivative is represented  α  '  β ∂A α δAα duβ γ du = Aα,β = + A βγ a δt dt ∂uβ dt or



δAα dAα = + δt dt

α βγ

' Aγ a

duβ dt

where the subscript a denotes that the Christoffel is formed from the surface metric aαβ . Similarly, the formulas for the intrinsic derivative of a covariant spatial vector Ai or covariant surface vector Aα are given by dAi δAi = − δt dt and δAα dAα = − δt dt Consider a mixed tensor

Tαi





k ij

'

γ αβ

Ak g

dxj dt

' Aα a

duβ . dt

which is contravariant with respect to a transformation of space coordinates

i

x and covariant with respect to a transformation of surface coordinates uα . For Tαi defined over the surface curve C, which can also be viewed as a space curve C, define the scalar invariant Ψ = Ψ(t) = Tαi Ai B α where Ai is a parallel vector field along the curve C when it is viewed as a space curve and B α is also a parallel vector field along the curve C when it is viewed as a surface curve. Recall that these parallel vector fields must satisfy the differential equations dAi δAi = − δt dt



k ij

' Ak g

dxj = 0 and dt

δB α dB α = + δt dt



α βγ

' Bγ a

duβ = 0. dt

(1.5.52)

The scalar invariant Ψ is a function of the parameter t of the space curve since both the tensor and the parallel vector fields are to be evaluated along the curve C. By differentiating the function Ψ with respect to the parameter t there results dΨ dT i dAi α dB α = α Ai B α + Tαi B + Tαi Ai . dt dt dt dt

(1.5.53)

142 But the vectors Ai and B α are parallel vector fields and must satisfy the relations given by equations (1.5.52). This implies that equation (1.5.53) can be written in the form ( )  '  ' j β i γ dΨ dTαi k dx i du = + − Ai B α . T T k j g α dt β α a γ dt dt dt

(1.5.54)

The quantity inside the brackets of equation (1.5.54) is defined as the intrinsic tensor derivative with respect to the parameter t along the curve C. This intrinsic tensor derivative is written  '  ' j δTαi dTαi i γ duβ k dx Tα Tγi = + − . dt dt kj g dt βα a dt

(1.5.55)

The spatial representation of the curve C is related to the surface representation of the curve C through the defining equations. Therefore, we can express the equation (1.5.55) in the form ( )  '  ' j β i γ ∂Tαi δTαi k ∂x i du = + T − T k j g α ∂uβ β α a γ dt dt ∂uβ

(1.5.56)

The quantity inside the brackets is a mixed tensor which is defined as the tensor derivative of Tαi with respect to the surface coordinates uβ . The tensor derivative of the mixed tensor Tαi with respect to the surface coordinates uβ is written i Tα,β

∂Tαi = + ∂uβ



i kj

' Tαk g

∂xj − ∂uβ



γ βα

' Tγi . a

i...j In general, given a mixed tensor Tα...β which is contravariant with respect to transformations of the

space coordinates and covariant with respect to transformations of the surface coordinates, then we can define the scalar field along the surface curve C as i...j Ai · · · Aj B α · · · B β Ψ(t) = Tα...β

(1.5.57)

where Ai , . . . , Aj and B α , . . . , B β are parallel vector fields along the curve C. The intrinsic tensor derivative is then derived by differentiating the equation (1.5.57) with respect to the parameter t. Tensor derivatives of the metric tensors gij , aαβ and the alternating tensors Fijk , Fαβ and their associated tensors are all zero. Hence, they can be treated as constants during the tensor differentiation process. Generalizations In a Riemannian space Vn with metric gij and curvilinear coordinates xi , i = 1, 2, 3, the equations of a surface can be written in the parametric form xi = xi (u1 , u2 ) where uα , α = 1, 2 are called the curvilinear coordinates of the surface. Since dxi =

∂xi α du ∂uα

(1.5.58)

then a small change duα on the surface results in change dxi in the space coordinates. Hence an element of arc length on the surface can be represented in terms of the curvilinear coordinates of the surface. This same element of arc length can also be represented in terms of the curvilinear coordinates of the space. Thus, an element of arc length squared in terms of the surface coordinates is represented ds2 = aαβ duα duβ

(1.5.59)

143 where aαβ is the metric of the surface. This same element when viewed as a spatial element is represented ds2 = gij dxi dxj .

(1.5.60)

By equating the equations (1.5.59) and (1.5.60) we find that gij dxi dxj = gij

∂xi ∂xj α β du du = aαβ duα duβ . ∂uα ∂uβ

(1.5.61)

The equation (1.5.61) shows that the surface metric is related to the spatial metric and can be calculated ∂xi ∂xj from the relation aαβ = gij α β . This equation reduces to the equation (1.5.21) in the special case of ∂u ∂u Cartesian coordinates. In the surface coordinates we define the quadratic form A = aαβ duα duβ as the first fundamental form of the surface. The tangent vector to the coordinate curves defining the surface are given by

∂xi ∂uα

and can be viewed as either a covariant surface vector or a contravariant spatial vector. We define

this vector as xiα =

∂xi , ∂uα

i = 1, 2, 3,

α = 1, 2.

(1.5.62)

Any vector which is a linear combination of the tangent vectors to the coordinate curves is called a surface vector. A surface vector Aα can also be viewed as a spatial vector Ai . The relation between the spatial representation and surface representation is Ai = Aα xiα . The surface representation Aα , α = 1, 2 and the spatial representation Ai , i = 1, 2, 3 define the same direction and magnitude since gij Ai Aj = gij Aα xiα Aβ xjβ = gij xiα xjβ Aα Aβ = aαβ Aα Aβ . Consider any two surface vectors Aα and B α and their spatial representations Ai and B i where Ai = Aα xiα

and

B i = B α xiα .

(1.5.63)

These vectors are tangent to the surface and so a unit normal vector to the surface can be defined from the cross product relation ni AB sin θ = Fijk Aj B k

(1.5.64)

where A, B are the magnitudes of Ai , B i and θ is the angle between the vectors when their origins are made to coincide. Substituting equations (1.5.63) into the equation (1.5.64) we find ni AB sin θ = Fijk Aα xjα B β xkβ .

(1.5.65)

In terms of the surface metric we have AB sin θ = Fαβ Aα B β so that equation (1.5.65) can be written in the form (ni Fαβ − Fijk xjα xkβ )Aα B β = 0

(1.5.66)

which for arbitrary surface vectors implies ni Fαβ = Fijk xjα xkβ

or ni =

1 αβ F Fijk xjα xkβ . 2

(1.5.67)

The equation (1.5.67) defines a unit normal vector to the surface in terms of the tangent vectors to the coordinate curves. This unit normal vector is related to the covariant derivative of the surface tangents as

144 is now demonstrated. By using the results from equation (1.5.50), the tensor derivative of equation (1.5.59), with respect to the surface coordinates, produces xiα,β

∂ 2 xi = + ∂uα ∂uβ



i pq

'

 xpα xqβ g

−

σ αβ

' xiσ

(1.5.68)

a

where the subscripts on the Christoffel symbols refer to the metric from which they are calculated. Also the tensor derivative of the equation (1.5.57) produces the result gij xiα,γ xjβ + gij xiα xjβ,γ = aαβ,γ = 0.

(1.5.69)

Interchanging the indices α, β, γ cyclically in the equation (1.5.69) one can verify that gij xiα,β xjγ = 0.

(1.5.70)

The equation (1.5.70) indicates that in terms of the space coordinates, the vector xiα,β is perpendicular to the surface tangent vector xiγ and so must have the same direction as the unit surface normal ni . Therefore, there must exist a second order tensor bαβ such that bαβ ni = xiα,β .

(1.5.71)

By using the relation gij ni nj = 1 we can transform equation (1.5.71) to the form bαβ = gij nj xiα,β =

1 γδ F Fijk xiα,β xjγ xkδ . 2

(1.5.72)

The second order symmetric tensor bαβ is called the curvature tensor and the quadratic form B = bαβ duα duβ

(1.5.73)

is called the second fundamental form of the surface. Consider also the tensor derivative with respect to the surface coordinates of the unit normal vector to the surface. This derivative is ni, α =

∂ni + ∂uα



i jk

' nj xkα .

(1.5.74)

g

Taking the tensor derivative of gij ni nj = 1 with respect to the surface coordinates produces the result gij ni nj,α = 0 which shows that the vector nj,α is perpendicular to ni and must lie in the tangent plane to the surface. It can therefore be expressed as a linear combination of the surface tangent vectors xiα and written in the form ni,α = ηαβ xiβ

(1.5.75)

where the coefficients ηαβ can be written in terms of the surface metric components aαβ and the curvature components bαβ as follows. The unit vector ni is normal to the surface so that gij ni xjα = 0.

(1.5.76)

145 The tensor derivative of this equation with respect to the surface coordinates gives gij niβ xjα + gij ni xjα,β = 0.

(1.5.77)

Substitute into equation (1.5.77) the relations from equations (1.5.57), (1.5.71) and (1.5.75) and show that bαβ = −aαγ ηβγ .

(1.5.78)

Solving the equation (1.5.78) for the coefficients ηβγ we find ηβγ = −aαγ bαβ .

(1.5.79)

Now substituting equation (1.5.79) into the equation (1.5.75) produces the Weingarten formula ni,α = −aγβ bγα xiβ .

(1.5.80)

This is a relation for the derivative of the unit normal in terms of the surface metric, curvature tensor and surface tangents. A third fundamental form of the surface is given by the quadratic form C = cαβ duα duβ

(1.5.81)

where cαβ is defined as the symmetric surface tensor cαβ = gij ni,α nj,β .

(1.5.82)

By using the Weingarten formula in the equation (1.5.81) one can verify that cαβ = aγδ bαγ bβδ .

(1.5.83)

Geodesic Coordinates In a Cartesian coordinate system the metric tensor gij is a constant and consequently the Christoffel symbols are zero at all points of the space. This is because the Christoffel symbols are dependent upon the derivatives of the metric tensor which is constant. If the space VN is not Cartesian then the Christoffel symbols do not vanish at all points of the space. However, it is possible to find a coordinate system where the Christoffel symbols will all vanish at a given point P of the space. Such coordinates are called geodesic coordinates of the point P. Consider a two dimensional surface with surface coordinates uα and surface metric aαβ . If we transform to some other two dimensional coordinate system, say u ¯α with metric a ¯αβ , where the two coordinates are related by transformation equations of the form u 1, u ¯ 2 ), uα = uα (¯

α = 1, 2,

(1.5.84)

146 then from the transformation equation (1.4.7) we can write, after changing symbols,  '  ' α δ ∂uα ∂uδ ∂u# ∂ 2 uα = + . δ F a ∂u β γ a¯ ∂ u ¯δ ¯ β ∂u ¯γ ∂u ¯ β ∂u ¯γ

(1.5.85) 

This is a relationship between the Christoffel symbols in the two coordinate systems. If a point P , then for that particular point the equation (1.5.85) reduces to  ' α ∂ 2 uα ∂uδ ∂u# = − β γ δ F a ∂u ∂u ¯ ∂u ¯ ¯ β ∂u ¯γ

' δ βγ

vanishes at a ¯

(1.5.86)

where all terms are evaluated the point P. Conversely, if the equation (1.5.86) is satisfied at the point P,  at ' δ then the Christoffel symbol β γ must be zero at this point. Consider the special coordinate transformaa ¯

tion

uα = uα ¯α − 0 +u

 ' 1 α u¯ β u ¯α 2 βγ a

(1.5.87)

where uα 0 are the surface coordinates of the point P. The point P in the new coordinates is given by u ¯ α = 0. We now differentiate the relation (1.5.87) to see if it satisfies the equation (1.5.86). We calculate the derivatives

 '  '  1 α 1 α ∂uα α β γ  = δ − u ¯ − u ¯  α τ ∂u ¯τ 2 βτ a 2 τγ a u =0

(1.5.88)

 '  α ∂ 2 uα  =−  τ σ a uα =0 ∂u ¯ τ ∂u ¯σ

(1.5.89)

and

where these derivative are evaluated at u¯ α = 0. We find the derivative equations (1.5.88) and (1.5.89) do satisfy the equation (1.5.86) locally at the point P. Hence, the Christoffel symbols will all be zero at this particular point. The new coordinates can then be called geodesic coordinates. Riemann Christoffel Tensor Consider the Riemann Christoffel tensor defined by the equation (1.4.33). Various properties of this tensor are derived in the exercises at the end of this section. We will be particularly interested in the Riemann Christoffel tensor in a two dimensional space with metric aαβ and coordinates uα . We find the Riemann Christoffel tensor has the form  '  '  ' '  ' ' δ δ τ δ τ δ ∂ ∂ δ R. αβγ = − + − αγ βτ αβ γτ ∂uβ α γ ∂uγ α β

(1.5.90)

where the Christoffel symbols are evaluated with respect to the surface metric. The above tensor has the associated tensor Rσαβγ = aσδ R.δαβγ

(1.5.91)

which is skew-symmetric in the indices (σ, α) and (β, γ) such that Rσαβγ = −Rασβγ

and

Rσαβγ = −Rσαγβ .

(1.5.92)

The two dimensional alternating tensor is used to define the constant K=

1 αβ γδ F F Rαβγδ 4

(1.5.93)

147 (see example 1.5-1) which is an invariant of the surface and called the Gaussian curvature or total curvature. In the exercises following this section it is shown that the Riemann Christoffel tensor of the surface can be expressed in terms of the total curvature and the alternating tensors as Rαβγδ = KFαβ Fγδ .

(1.5.94)

Consider the second tensor derivative of xrα which is given by xrα,βγ =

∂xrα,β + ∂uγ



r mn

'

 xrα,β xnγ −

g

δ αγ

'

 xrδ,β − a

δ βγ

' xrα,γ

(1.5.95)

a

which can be shown to satisfy the relation xrα,βγ − xrα,γβ = Rδ.αβγ xrδ .

(1.5.96)

Using the relation (1.5.96) we can now derive some interesting properties relating to the tensors aαβ , bαβ , cαβ ,

Rαβγδ , the mean curvature H and the total curvature K. Consider the tensor derivative of the equation (1.5.71) which can be written xiα,βγ = bαβ,γ ni + bαβ ni,γ

where bαβ,γ =

∂bαβ − ∂uα



σ αγ

'

 bσβ −

a

σ βγ

(1.5.97) ' bασ .

(1.5.98)

a

By using the Weingarten formula, given in equation (1.5.80), the equation (1.5.97) can be expressed in the form xiα,βγ = bαβ,γ ni − bαβ aτ σ bτ γ xiσ

(1.5.99)

and by using the equations (1.5.98) and (1.5.99) it can be established that xrα,βγ − xrα,γβ = (bαβ,γ − bαγ,β )nr − aτ δ (bαβ bτ γ − bαγ bτ β )xrδ .

(1.5.100)

Now by equating the results from the equations (1.5.96) and (1.5.100) we arrive at the relation Rδ.αβγ xrδ = (bαβ,γ − bαγ,β )nr − aτ δ (bαβ bτ γ − bαγ bτ β )xrδ .

(1.5.101)

Multiplying the equation (1.5.101) by nr and using the results from the equation (1.5.76) there results the Codazzi equations bαβ,γ − bαγ,β = 0.

(1.5.102)

Multiplying the equation (1.5.101) by grm xm σ and simplifying one can derive the Gauss equations of the surface Rσαβγ = bαγ bσβ − bαβ bσγ .

(1.5.103)

By using the Gauss equations (1.5.103) the equation (1.5.94) can be written as KFσα Fβγ = bαγ bσβ − bαβ bσγ .

(1.5.104)

148 Another form of equation (1.5.104) is obtained by using the equation (1.5.83) together with the relation aαβ = −aσγ Fσα Fβγ . It is left as an exercise to verify the resulting form −Kaαβ = cαβ − aσγ bσγ bαβ .

(1.5.106)

Define the quantity H=

1 σγ a bσγ 2

(1.5.107)

as the mean curvature of the surface, then the equation (1.5.106) can be written in the form cαβ − 2H bαβ + K aαβ = 0.

(1.5.108)

By multiplying the equation (1.5.108) by duα duβ and summing, we find C − 2H B + K A = 0

(1.5.109)

is a relation connecting the first, second and third fundamental forms. EXAMPLE 1.5-2 In a two dimensional space the Riemann Christoffel tensor has only one nonzero independent component R1212 . ( See Exercise 1.5, problem number 21.) Consequently, the equation (1.5.104) can be written in the √ √ form K ae12 ae12 = b22 b11 − b21 b12 and solving for the Gaussian curvature K we find K=

R1212 b b22 b11 − b12 b21 . = = a11 a22 − a12 a21 a a

(1.5.110)

Surface Curvature For a surface curve uα = uα (s),α = 1, 2 lying upon a surface xi = xi (u1 , u2 ),i = 1, 2, 3, we have a two duα is a unit tangent vector to dimensional space embedded in a three dimensional space. Thus, if tα = ds α β du du the surface curve then aαβ = aαβ tα tβ = 1. This same vector can be represented as the unit tangent ds ds dxi dxj dxi vector to the space curve xi = xi (u1 (s), u2 (s)) with T i = . That is we will have gij = gij T i T j = 1. ds ds ds The surface vector tα and the space vector T i are related by ∂xi duα = xiα tα . ∂uα ds

Ti =

(1.5.111)

The surface vector tα is a unit vector so that aαβ tα tβ = 1. If we differentiate this equation intrinsically with β

respect to the parameter s, we find that aαβ tα δtδs = 0. This shows that the surface vector α

δtα δs

is perpendicular

α

to the surface vector t . Let u denote a unit normal vector in the surface plane which is orthogonal to the tangent vector tα . The direction of uα is selected such that Fαβ tα uβ = 1. Therefore, there exists a scalar κ(g) such that

δtα = κ(g) uα δs

(1.5.112)

149 where κ(g) is called the geodesic curvature of the curve. In a similar manner it can be shown that is a surface vector orthogonal to tα . Let

δuα δs

δuα δs

= αtα where α is a scalar constant to be determined. By

differentiating the relation aαβ tα uβ = 0 intrinsically and simplifying we find that α = −κ(g) and therefore δuα = −κ(g) tα . δs

(1.5.113)

The equations (1.5.112) and (1.5.113) are sometimes referred to as the Frenet-Serret formula for a curve relative to a surface. Taking the intrinsic derivative of equation (1.5.111), with respect to the parameter s, we find that δtα duβ α δT i = xiα + xiα,β t . δs δs ds

(1.5.114)

Treating the curve as a space curve we use the Frenet formulas (1.5.13). If we treat the curve as a surface curve, then we use the Frenet formulas (1.5.112) and (1.5.113). In this way the equation (1.5.114) can be written in the form κN i = xiα κ(g) uα + xiα,β tβ tα .

(1.5.115)

By using the results from equation (1.5.71) in equation (1.5.115) we obtain κN i = κ(g) ui + bαβ ni tα tβ

(1.5.116)

where ui is the space vector counterpart of the surface vector uα . Let θ denote the angle between the surface normal ni and the principal normal N i , then we have that cos θ = ni N i . Hence, by multiplying the equation (1.5.116) by ni we obtain κ cos θ = bαβ tα tβ .

(1.5.117)

Consequently, for all curves on the surface with the same tangent vector tα , the quantity κ cos θ will remain constant. This result is known as Meusnier’s theorem. Note also that κ cos θ = κ(n) is the normal component of the curvature and κ sin θ = κ(g) is the geodesic component of the curvature. Therefore, we write the equation (1.5.117) as κ(n) = bαβ tα tβ

(1.5.118)

which represents the normal curvature of the surface in the direction tα . The equation (1.5.118) can also be written in the form κ(n) = bαβ

B duα duβ = ds ds A

(1.5.119)

which is a ratio of quadratic forms. The surface directions for which κ(n) has a maximum or minimum value is determined from the equation (1.5.119) which is written as (bαβ − κ(n) aαβ )λα λβ = 0.

(1.5.120)

The direction giving a maximum or minimum value to κ(n) must then satisfy (bαβ − κ(n) aαβ )λβ = 0

(1.5.121)

150 so that κ(n) must be a root of the determinant equation det(bαβ − κ(n) aαβ ) = 0.

(1.5.122)

The expanded form of equation (1.5.122) can be written as κ2(n) − aαβ bαβ κ(n) +

b =0 a

(1.5.123)

where a = a11 a22 − a12 a21 and b = b11 b22 − b12 b21 . Using the definition given in equation (1.5.107) and using the result from equation (1.5.110), the equation (1.5.123) can be expressed in the form κ2(n) − 2H κ(n) + K = 0.

(1.5.124)

The roots κ(1) and κ(2) of the equation (1.5.124) then satisfy the relations H=

1 (κ(1) + κ(2) ) 2

(1.5.125)

and K = κ(1) κ(2) .

(1.5.126)

Here H is the mean value of the principal curvatures and K is the Gaussian or total curvature which is the product of the principal curvatures. It is readily verified that H=

Eg − 2f F + eG 2(EG − F 2 )

and K =

eg − f 2 EG − F 2

are invariants obtained from the surface metric and curvature tensor. Relativity Sir Isaac Newton and Albert Einstein viewed the world differently when it came to describing gravity and the motion of the planets. In this brief introduction to relativity we will compare the Newtonian equations with the relativistic equations in describing planetary motion. We begin with an examination of Newtonian systems. Newton’s viewpoint of planetary motion is a multiple bodied problem, but for simplicity we consider only a two body problem, say the sun and some planet where the motion takes place in a plane. Newton’s law of gravitation states that two masses m and M are attracted toward each other with a force of magnitude GmM ρ2 ,

where G is a constant, ρ is the distance between the masses, m is the mass of the planet and M is the

mass of the sun. One can construct an x, y plane containing the two masses with the origin located at the e1 + sin φ
e2 denote a unit vector at the origin of this coordinate center of mass of the sun. Let
eρ = cos φ
system and pointing in the direction of the mass m. The vector force of attraction of mass M on mass m is given by the relation

−GmM
F
= eρ . ρ2

(1.5.127)

151

Figure 1.5-2. Parabolic and elliptic conic sections The equation of motion of mass m with respect to mass M is obtained from Newton’s second law. Let ρ
= ρ
eρ denote the position vector of mass m with respect to the origin. Newton’s second law can then be written in any of the forms
−GmM −GmM
d2 ρ dV
F
= e = = m = m ρ
ρ ρ2 dt2 dt ρ3

(1.5.128)

and from this equation we can show that the motion of the mass m can be described as a conic section. Recall that a conic section is defined as a locus of points p(x, y) such that the distance of p from a fixed point (or points), called a focus (foci), is proportional to the distance of the point p from a fixed line, called a directrix, that does not contain the fixed point. The constant of proportionality is called the eccentricity and is denoted by the symbol F. For F = 1 a parabola results; for 0 ≤ F ≤ 1 an ellipse results; for F > 1 a hyperbola results; and if F = 0 the conic section is a circle. With reference to figure 1.5-2, a conic section is defined in terms of the ratio

FP PD

= F where F P = ρ and

P D = 2q − ρ cos φ. From the F ratio we solve for ρ and obtain the polar representation for the conic section ρ=

p 1 + F cos φ

(1.5.129)

152 where p = 2qF and the angle φ is known as the true anomaly associated with the orbit. The quantity p is called the semi-parameter of the conic section. (Note that when φ =

π 2,

then ρ = p.) A more general form

of the above equation is ρ=

p 1 + F cos(φ − φ0 )

or u =

1 = A[1 + F cos(φ − φ0 )], ρ

(1.5.130)

where φ0 is an arbitrary starting anomaly. An additional symbol a, known as the semi-major axes of an elliptical orbit can be introduced where q, p, F, a are related by p = q = a(1 − F) 1+F

or p = a(1 − F2 ).

(1.5.131)

To show that the equation (1.5.128) produces a conic section for the motion of mass m with respect to mass M we will show that one form of the solution of equation (1.5.128) is given by the equation (1.5.129). To verify this we use the following vector identities: ρ
×
eρ =0   d
ρ d2 ρ d
ρ
× =
ρ× 2 dt dt dt d
eρ
eρ · =0 dt   d
eρ d
eρ
eρ ×
eρ × . =− dt dt

(1.5.132)

From the equation (1.5.128) we find that d dt

 
GM d
ρ d2 ρ
×
eρ =
0 ρ
× =ρ
× 2 =− 2 ρ dt dt ρ

(1.5.133)

so that an integration of equation (1.5.133) produces ρ
×

d
ρ
= h = constant. dt

(1.5.134)

ρ

= ρ
= ρ The quantity H
× mV
× m d' dt is the angular momentum of the mass m so that the quantity h represents the angular momentum per unit mass. The equation (1.5.134) tells us that
h is a constant for our

two body system. Note that because
h is constant we have  
GM d
ρ d 

 dV
eρ × ρ V ×h = ×h=− 2

× dt dt ρ dt dρ d
eρ GM
eρ × [
+ eρ )] ρ
eρ × (ρ =− 2
ρ dt dt GM d
eρ 2 d
eρ =− 2
eρ × (
)ρ = GM eρ × ρ dt dt and consequently an integration produces
×
h = GM

V eρ + C

153
is a vector constant of integration. The triple scalar product formula gives us where C d
ρ
×
h) =
h · (

ρ
· (V ρ × ) = h2 = GM ρ
·
eρ + ρ
· C dt or h2 = GM ρ + Cρ cos φ

(1.5.135)


and ρ
. From the equation (1.5.135) we find that where φ is the angle between the vectors C ρ=

p 1 + F cos φ

(1.5.136)

where p = h2 /GM and F = C/GM. This result is known as Kepler’s first law and implies that when F < 1 the mass m describes an elliptical orbit with the sun at one focus. We present now an alternate derivation of equation (1.5.130) for later use. From the equation (1.5.128) we have 2


d d
ρ d2 ρ · 2 = dt dt dt



d
ρ d
ρ · dt dt

 = −2

GM d d
ρ GM =− 3 (
ρ·ρ
) . ρ
· 3 ρ dt ρ dt

(1.5.137) dρ dt , ρ sin θ dφ dt

Consider the equation (1.5.137) in spherical coordinates ρ, θ, φ. The tensor velocity components are V 1 = V2 =

dθ dt ,

V3 =

dφ dt

and the physical components of velocity are given by Vρ =

dρ dt ,

Vθ = ρ dθ dt , Vφ =

so that the velocity can be written d
ρ dρ dθ dφ

V
= = eρ + ρ
eθ + ρ sin θ eφ . dt dt dt dt

(1.5.138)

Substituting equation (1.5.138) into equation (1.5.137) gives the result (   2 )  2   2 dρ dφ d dθ d 1 GM d 2 2GM dρ 2 2 2 (ρ ) = − 2 = 2GM =− 3 +ρ + ρ sin θ dt dt dt dt ρ dt ρ dt dt ρ which can be integrated directly to give 

dρ dt



2 + ρ2

dθ dt



2 + ρ2 sin2 θ

dφ dt

2 =

2GM −E ρ

where −E is a constant of integration. In the special case of a planar orbit we set θ =

(1.5.139) π 2

constant so that

the equation (1.5.139) reduces to  

dρ dt

dρ dφ dφ dt



2 +ρ

2

+ρ

2



2

dφ dt dφ dt

2 = 2

2GM −E ρ

(1.5.140)

2GM − E. = ρ

Also for this special case of planar motion we have |
ρ× By eliminating

dφ dt

d
ρ dφ | = ρ2 = h. dt dt

(1.5.141)

from the equation (1.5.140) we obtain the result 

dρ dφ

2 + ρ2 =

2GM 3 E ρ − 2 ρ4 . h2 h

(1.5.142)

154

Figure 1.5-3. Relative motion of two inertial systems. The substitution ρ =

1 u

can be used to represent the equation (1.5.142) in the form 

du dφ

2

+ u2 −

2GM E u+ 2 =0 2 h h

(1.5.143)

which is a form we will return to later in this section. Note that we can separate the variables in equations (1.5.142) or (1.5.143). The results can then be integrate to produce the equation (1.5.130). Newton also considered the relative motion of two inertial systems, say S and S. Consider two such systems as depicted in the figure 1.5-3 where the S system is moving in the x−direction with speed v relative to the system S. For a Newtonian system, if at time t = 0 we have clocks in both systems which coincide, than at time t a point P (x, y, z) in the S system can be described by the transformation equations x =x − vt

x =x + vt y =y

y =y or

(1.5.144)

z =z

z =z

t =t

t =t.

These are the transformation equation of Newton’s relativity sometimes referred to as a Galilean transformation. Before Einstein the principle of relativity required that velocities be additive and obey Galileo’s velocity addition rule VP/R = VP/Q + VQ/R .

(1.5.145)

155 That is, the velocity of P with respect to R equals the velocity of P with respect to Q plus the velocity of Q with respect to R. For example, a person (P ) running north at 3 km/hr on a train (Q) moving north at 60 km/hr with respect to the ground (R) has a velocity of 63 km/hr with respect to the ground. What happens when (P ) is a light wave moving on a train (Q) which is moving with velocity V relative to the ground? Are the velocities still additive? This type of question led to the famous Michelson-Morley experiment which has been labeled as the starting point for relativity. Einstein’s answer to the above question was ”NO” and required that VP/R = VP/Q = c =speed of light be a universal constant. In contrast to the Newtonian equations, Einstein considered the motion of light from the origins 0 and 0 of the systems S and S. If the S system moves with velocity v relative to the S system and at time t = 0 a light signal is sent from the S system to the S system, then this light signal will move out in a spherical wave front and lie on the sphere x2 + y 2 + z 2 = c2 t2

(1.5.146)

where c is the speed of light. Conversely, if a light signal is sent out from the S system at time t = 0, it will lie on the spherical wave front 2

x2 + y 2 + z 2 = c2 t .

(1.5.147)

Observe that the Newtonian equations (1.5.144) do not satisfy the equations (1.5.146) and (1.5.147) identically. If y = y and z = z then the space variables (x, x) and time variables (t, t) must somehow be related. Einstein suggested the following transformation equations between these variables x = γ(x − vt) and x = γ(x + vt)

(1.5.148)

where γ is a constant to be determined. The differentials of equations (1.5.148) produce dx = γ(dx − vdt)

and dx = γ(dx + vdt)

(1.5.149)

from which we obtain the ratios dx γ(dx − v dt) = dx γ(dx + v dt)

or

1 γ(1 +

v

dx )

= γ(1 −

dt

When

v dx dt

).

(1.5.150)

dx dx = = c, the speed of light, the equation (1.5.150) requires that dt dt γ 2 = (1 −

v 2 −1 ) c2

or γ = (1 −

v 2 −1/2 ) . c2

(1.5.151)

From the equations (1.5.148) we eliminate x and find t = γ(t −

v x). c2

(1.5.152)

We can now replace the Newtonian equations (1.5.144) by the relativistic transformation equations x =γ(x − vt)

x =γ(x + vt) y =y

y =y or

z =z t =γ(t +

v x) c2

(1.5.153)

z =z t =γ(t −

v x) c2

156 where γ is given by equation (1.5.151). These equations are also known as the Lorentz transformation. v Note that for v << c, then 2 ≈ 0, γ ≈ 1 , then the equations (1.5.153) closely approximate the equations c (1.5.144). The equations (1.5.153) also satisfy the equations (1.5.146) and (1.5.147) identically as can be readily verified by substitution. Further, by using chain rule differentiation we obtain from the relations (1.5.148) that dx = dt

dx dt

1+

+v dx dt

.

(1.5.154)

v c c

The equation (1.5.154) is the Einstein relative velocity addition rule which replaces the previous Newtonian rule given by equation (1.5.145). We can rewrite equation (1.5.154) in the notation of equation (1.5.145) as VP/R =

VP/Q + VQ/R 1+

VP/Q VQ/R c c

.

(1.5.155)

Observe that when VP/Q << c and VQ/R << c then equation (1.5.155) approximates closely the equation (1.5.145). Also as VP/Q and VQ/R approach the speed of light we have lim

VP/Q + VQ/R

VP/Q →C VQ/R →C

VP/Q VQ/R c c

1+

=c

(1.5.156)

which agrees with Einstein’s hypothesis that the speed of light is an invariant. Let us return now to the viewpoint of what gravitation is. Einstein thought of space and time as being related and viewed the motion of the planets as being that of geodesic paths in a space-time continuum. Recall the equations of geodesics are given by d2 xi + ds2



i jk

'

dxj dxk = 0, ds ds

(1.5.157)

where s is arc length. These equations are to be associated with a 4-dimensional space-time metric gij where the indices i, j take on the values 1, 2, 3, 4 and the xi are generalized coordinates. Einstein asked the question, ”Can one introduce a space-time metric gij such that the equations (1.5.157) can somehow reproduce the law of gravitational attraction

d2 ρ ' dt2

+

GM
ρ3 ρ

= 0?” Then the motion of the planets can be

viewed as optimized motion in a space-time continuum where the metrices of the space simulate the law of gravitational attraction. Einstein thought that this motion should be related to the curvature of the space which can be obtained from the Riemann-Christoffel tensor Rijkl . The metric we desire gij , i, j = 1, 2, 3, 4 has 16 components. The conjugate metric tensor g ij is defined such that g ij gjk = δki and an element of arc length squared is given by ds2 = gij dxi dxj . Einstein thought that the metrices should come from the Riemann-Christoffel curvature tensor which, for n = 4 has 256 components, but only 20 of these are linearly independent. This seems like a large number of equations from which to obtain the law of gravitational attraction and so Einstein considered the contracted tensor  '  '  ' '  ' ' n n m n m n ∂ ∂ Gij = Rtijt = − + − . in mj ij mn ∂xj i n ∂xn i j Spherical coordinates (ρ, θ, φ) suggests a metric similar to ds2 = −(dρ)2 − ρ2 (dθ)2 − ρ2 sin2 θ(dφ)2 + c2 (dt)2

(1.5.158)

157 where g11 = −1, g22 = −ρ2 , g33 = −ρ2 sin2 θ, g44 = c2 and gij = 0 for i = j. The negative signs are 2 = c2 − v 2 is positive when v < c and the velocity is not greater than c. However, introduced so that ds dt this metric will not work since the curvature tensor vanishes. The spherical symmetry of the problem suggest that g11 and g44 change while g22 and g33 remain fixed. Let (x1 , x2 , x3 , x4 ) = (ρ, θ, φ, t) and assume g11 = −eu ,

g22 = −ρ2 ,

g33 = −ρ2 sin2 θ,

g44 = ev

(1.5.159)

where u and v are unknown functions of ρ to be determined. This gives the conjugate metric tensor g 11 = −e−u ,

g 22 =

−1 , ρ2

g 33 =

ρ2

−1 , sin2 θ

g 44 = e−v

(1.5.160)

and g ij = 0 for i = j. This choice of a metric produces ds2 = −eu (dρ)2 − ρ2 (dθ)2 − ρ2 sin2 θ(dφ)2 + ev (dt)2

(1.5.161)

together with the nonzero Christoffel symbols    

1 11 1 22 1 33 1 44

' ' ' '



1 du 2 dρ



= − ρe−u



= − ρe−u sin2 θ



=

dv 1 = ev−u 2 dr

2 12 2 21 2 33

'

1 ρ



1 = ρ



= − sin θ cos θ



= ' '

3 13 3 23 3 31 3 32

' =

1 ρ

=

cos θ sin θ

=

1 ρ

=

cos θ sin θ

' ' '

 

4 14 4 41

' =

1 dv 2 dρ

=

1 dv . 2 dρ

'

(1.5.162)

The equation (1.5.158) is used to calculate the nonzero Gij and we find that  2 1 du 1 d2 v 1 dv 1 du dv − + − 2 2 dρ 4 dρ 4 dρ dρ ρ dρ   1 dv 1 du =e−u 1 + ρ − ρ − eu 2 dρ 2 dρ   dv 1 du 1 − ρ − eu sin2 θ =e−u 1 + ρ 2 dρ 2 dρ    2 1 dv 1 dv 1 d2 v 1 du dv v−u + =−e − + 2 dρ2 4 dρ dρ 4 dρ ρ dρ

G11 = G22 G33 G44

(1.5.163)

and Gij = 0 for i = j. The assumption that Gij = 0 for all i, j leads to the differential equations d2 v 1 + dρ2 2



dv dρ

2

2 du 1 du dv − =0 2 dρ dρ ρ dρ 1 du 1 dv − ρ − eu =0 1+ ρ 2 dρ 2 dρ  2 d2 v 2 dv 1 dv 1 du dv + =0. + − 2 dρ 2 dρ 2 dρ dρ ρ dρ −

(1.5.164)

158 Subtracting the first equation from the third equation gives du dv + =0 dρ dρ

or u + v = c1 = constant.

(1.5.165)

du = 1 − eu dρ

(1.5.166)

The second equation in (1.5.164) then becomes ρ

Separate the variables in equation (1.5.166) and integrate to obtain the result eu =

1 1 − cρ2

(1.5.167)

where c2 is a constant of integration and consequently v

e =e

c1 −u

  c2 =e 1− . ρ c1

(1.5.168)

The constant c1 is selected such that g44 approaches c2 as ρ increases without bound. This produces the metrices g11 =

−1 , 1 − cρ2

g22 = −ρ2 ,

g33 = −ρ2 sin2 θ,

g44 = c2 (1 −

c2 ) ρ

(1.5.169)

where c2 is a constant still to be determined. The metrices given by equation (1.5.169) are now used to expand the equations (1.5.157) representing the geodesics in this four dimensional space. The differential equations representing the geodesics are found to be 1 du d2 ρ + ds2 2 dρ



dρ ds

2

− ρe−u

d2 θ 2 dθ dρ + − sin θ cos θ ds2 ρ ds ds





dθ ds

dφ ds

2

2

− ρe−u sin2 θ



dφ ds

2

+

1 v−u dv e 2 dρ



dt ds

2

=0

=0

(1.5.170) (1.5.171)

d2 φ 2 dφ dρ cos θ dφ dθ + +2 =0 ds2 ρ ds ds sin θ ds ds

(1.5.172)

d2 t dv dt dρ + = 0. ds2 dρ ds ds

(1.5.173)

The equation (1.5.171) is identically satisfied if we examine planar orbits where θ =

π 2

is a constant. This

value of θ also simplifies the equations (1.5.170) and (1.5.172). The equation (1.5.172) becomes an exact differential equation d ds

 ρ2

dφ ds

 =0

dφ = c4 , ds

(1.5.174)

dt v e = c5 , ds

(1.5.175)

or ρ2

and the equation (1.5.173) also becomes an exact differential d ds



dt v e ds

 =0

or

where c4 and c5 are constants of integration. This leaves the equation (1.5.170) which determines ρ. Substituting the results from equations (1.5.174) and (1.5.175), together with the relation (1.5.161), the equation (1.5.170) reduces to d2 ρ c2 c2 c24 c2 c24 ) = 0. + + − (1 − ds2 2ρ2 2ρ4 ρ ρ3

(1.5.176)

159 By the chain rule we have d2 ρ d2 ρ = 2 2 ds dφ



dφ ds

2 +

dρ d2 φ d2 ρ c24 = + 2 dφ ds dφ2 ρ4



dρ dφ

2 

−2c24 ρ5



and so equation (1.5.176) can be written in the form d2 ρ 2 − 2 dφ ρ The substitution ρ =

1 u



dρ dφ

2

  c2 c2 ρ 2 c2 − 1− + + ρ = 0. 2 c24 2 ρ

(1.5.177)

reduces the equation (1.5.177) to the form d2 u c2 3 + u − 2 = c2 u 2 . dφ2 2c4 2

(1.5.178)

du Multiply the equation (1.5.178) by 2 dφ and integrate with respect to φ to obtain



du dφ

2 + u2 −

c2 u = c2 u 3 + c6 . c24

(1.5.179)

where c6 is a constant of integration. To determine the constant c6 we write the equation (1.5.161) in the special case θ =

π 2

and use the substitutions from the equations (1.5.174) and (1.5.175) to obtain  e

or

The substitution ρ =

u

dρ ds



1 u



2 =e

dρ dφ

2

u

dρ dφ dφ ds



2 =1−ρ

2

dφ ds



2 +e

v

dt ds

    c2 ρ 4 c2 c2 − 52 + 1− = 0. ρ2 + 1 − ρ ρ c c24

2

(1.5.180)

reduces the equation (1.5.180) to the form 

du dφ

2 + u 2 − c2 u 3 +

1 c2 c2 − 2 u − 2 5 2 = 0. 2 c4 c4 c c4

(1.5.181)

Now comparing the equations (1.5.181) and (1.5.179) we select  2  c5 1 −1 2 c6 = c2 c4 so that the equation (1.5.179) takes on the form 

du dφ

2

  c2 c25 1 + u − 2u + 1 − 2 = c2 u 3 c4 c c24 2

(1.5.182)

Now we can compare our relativistic equation (1.5.182) with our Newtonian equation (1.5.143). In order that the two equations almost agree we select the constants c2 , c4 , c5 so that c2 2GM = 2 c4 h2

c2

and

1 − c52 E = 2. 2 c4 h

(1.5.183)

The equations (1.5.183) are only two equations in three unknowns and so we use the additional equation lim ρ2

ρ→∞

dφ dφ ds = lim ρ2 =h ρ→∞ dt ds dt

(1.5.184)

160 which is obtained from equation (1.5.141). Substituting equations (1.5.174) and (1.5.175) into equation (1.5.184), rearranging terms and taking the limit we find that c4 c2 = h. c5

(1.5.185)

From equations (1.5.183) and (1.5.185) we obtain the results that c25 =

c2 , 1 + cE2

c2 =

2GM c2



1 1 + E/c2



h c4 =  c 1 + E/c2

,

(1.5.186)

These values substituted into equation (1.5.181) produce the differential equation 

Let α =

c2 c24

=

2GM h2

du dφ

2

and β = c2 =

2GM E 2GM +u − u+ 2 = 2 h h c2



2

2GM 1 c2 ( 1+E/c2 )

1 1 + E/c2

 u3 .

(1.5.187)

then the differential equation (1.5.178) can be written as

3 α d2 u + u − = βu2 . dφ2 2 2

(1.5.188)

We know the solution to equation (1.5.143) is given by u=

1 = A(1 + F cos(φ − φ0 )) ρ

(1.5.189)

and so we assume a solution to equation (1.5.188) of this same general form. We know that A is small and so we make the assumption that the solution of equation (1.5.188) given by equation (1.5.189) is such that φ0 is approximately constant and varies slowly as a function of Aφ. Observe that if φ0 = φ0 (Aφ), then and

2

d φ0 dφ2

dφ0 dφ

= φ0 A

= φ0 A2 , where primes denote differentiation with respect to the argument of the function. (i.e.

Aφ for this problem.) The derivatives of equation (1.5.189) produce du = − FA sin(φ − φ0 )(1 − φ0 A) dφ d2 u =FA3 sin(φ − φ0 )φ0 − FA cos(φ − φ0 )(1 − 2Aφ0 + A2 (φ0 )2 ) dφ2 = − FA cos(φ − φ0 ) + 2FA2 φ0 cos(φ − φ0 ) + O(A3 ). Substituting these derivatives into the differential equation (1.5.188) produces the equations 2FA2 φ0 cos(φ − φ0 ) + A −

3β 2 α = A + 2FA2 cos(φ − φ0 ) + F2 A2 cos2 (φ − φ0 ) + O(A3 ). 2 2

Now A is small so that terms O(A3 ) can be neglected. Equating the constant terms and the coefficient of the cos(φ − φ0 ) terms we obtain the equations A−

3β 2 α = A 2 2

2FA2 φ0 = 3βFA2 +

3β 2 2 F A cos(φ − φ0 ). 2

Treating φ0 as essentially constant, the above system has the approximate solutions A≈

α 2

φ0 ≈

3β 3β Aφ + AF sin(φ − φ0 ) 2 4

(1.5.190)

161 The solutions given by equations (1.5.190) tells us that φ0 varies slowly with time. For F less than 1, the elliptical motion is affected by this change in φ0 . It causes the semi-major axis of the ellipse to slowly rotate at a rate given by

dφ0 dt .

Using the following values for the planet Mercury G =6.67(10−8) dyne cm2 /g2 M =1.99(1033 ) g a =5.78(1012 ) cm F =0.206 c =3(1010 ) cm/sec 2GM β ≈ 2 = 2.95(105) cm c h ≈ GM a(1 − F2 ) = 2.71(1019) cm2 /sec  1/2 GM dφ ≈ sec−1 Kepler’s third law dt a3

(1.5.191)

we calculate the slow rate of rotation of the semi-major axis to be approximately dφ0 dφ0 dφ 3 dφ = ≈ βA ≈3 dt dφ dt 2 dt



GM ch

2 

GM a3

1/2

=6.628(10−14) rad/sec

(1.5.192)

=43.01 seconds of arc per century. This slow variation in Mercury’s semi-major axis has been observed and measured and is in agreement with the above value. Newtonian mechanics could not account for the changes in Mercury’s semi-major axis, but Einstein’s theory of relativity does give this prediction. The resulting solution of equation (1.5.188) can be viewed as being caused by the curvature of the space-time continuum. The contracted curvature tensor Gij set equal to zero is just one of many conditions that can be assumed in order to arrive at a metric for the space-time continuum. Any assumption on the value of Gij relates to imposing some kind of curvature on the space. Within the large expanse of our universe only our imaginations limit us as to how space, time and matter interact. You can also imagine the existence of other tensor metrics in higher dimensional spaces where the geodesics within the space-time continuum give rise to the motion of other physical quantities. This short introduction to relativity is concluded with a quote from the NASA [email protected] news release, spring 1998, Release:98-51. “An international team of NASA and university researchers has found the first direct evidence of a phenomenon predicted 80 years ago using Einstein’s theory of general relativity– that the Earth is dragging space and time around itself as it rotates.”The news release explains that the effect is known as frame dragging and goes on to say “Frame dragging is like what happens if a bowling ball spins in a thick fluid such as molasses. As the ball spins, it pulls the molasses around itself. Anything stuck in the molasses will also move around the ball. Similarly, as the Earth rotates it pulls space-time in its vicinity around itself. This will shift the orbits of satellites near the Earth.”This research is reported in the journal Science.

162 EXERCISE 1.5 ' δT δs


and τ = δN' · B.
Assume in turn that each of the intrinsic derivatives of T
, N
,B
are ·N δs
,B
and hence derive the Frenet-Serret formulas of differential geometry. some linear combination of T
, N


1.

Let κ =


2.

Determine the given surfaces. Describe and sketch the curvilinear coordinates upon each surface.

(a)
r(u, v) = u
e1 + v
e2
3.

(b)
r(u, v) = u cos v
e1 + u sin v
e2

(c)
r(u, v) =

2uv 2 2u2 v

e e2 . + 1 u2 + v 2 u2 + v 2

Determine the given surfaces and describe the curvilinear coordinates upon the surface. Use some

graphics package to plot the surface and illustrate the coordinate curves on the surface. Find element of area dS in terms of u and v. e2 + c cos u
e3 a, b, c constants 0 ≤ u, v ≤ 2π (a)
r(u, v) = a sin u cos v
e1 + b sin u sin v
u u u e1 + (4 + v sin ) sin u
e2 + v cos
(b)
r(u, v) = (4 + v sin ) cos u
e3 − 1 ≤ v ≤ 1, 0 ≤ u ≤ 2π 2 2 2 (c)
r(u, v) = au cos v
e1 + bu sin v
e2 + cu
e3 (d)
r(u, v) = u cos v
e1 + u sin v
e2 + αv
e3 (e)
r(u, v) = a cos v
e1 + b sin v
e2 + u
e3

α constant a, b constant

e3 e2 + u
(f )
r(u, v) = u cos v
e1 + u sin v
2



 E F . Assume that the surface is F G described by equations of the form y i = y i (u, v) and that any point on the surface is given by the position


4.

Consider a two dimensional space with metric tensor (aαβ ) =

vector
r =
r(u, v) = y i
ei . Show that the metrices E, F, G are functions of the parameters u, v and are given by E =
ru ·
ru ,
5.

F =
ru ·
rv ,

G =
rv ·
rv

where
ru =

∂
r ∂u

and
rv =

∂
r . ∂v

For the metric given in problem 4 show that the Christoffel symbols of the first kind are given by [1 1, 1] =
ru ·
ruu

[1 2, 1] = [2 1, 1] =
ru ·
ruv

[2 2, 1] =
ru ·
rvv

[1 1, 2] =
rv ·
ruu

[1 2, 2] = [2 1, 2] =
rv ·
ruv

[2 2, 2] =
rv ·
rvv

∂ 2
r ∂
r · , α, β, γ = 1, 2. ∂uα ∂uβ ∂uγ Show that the results in problem 5 can also be written in the form

which can be represented [α β, γ] =
6.

1 Ev 2 1 1 [1 1, 2] = Fu − Ev [1 2, 2] = [2 1, 2] = Gu 2 2 where the subscripts indicate partial differentiation. [1 1, 1] =

1 Eu 2

[1 2, 1] = [2 1, 1] =


7.

1 [2 2, 1] = Fv − Gu 2 1 [2 2, 2] = Gv 2

For the metricgiven'in problem 4, show that the Christoffel symbols of the second kind can be γ expressed in the form = aγδ [α β, δ], α, β, γ = 1, 2 and produce the results αβ  '  '  '  ' 1 1 1 2 GEu − 2F Fu + F Ev GEv − F Gu 2EFu − EEv − F Eu = = = = 11 12 21 11 2(EG − F 2 ) 2(EG − F 2 ) 2(EG − F 2 )  '  '  '  ' 1 2 2 2 2GFv − GGu − F Gv EGu − F Ev EGv − 2F Fv + F Gu = = = = 2 2 22 12 22 21 2(EG − F ) 2(EG − F ) 2(EG − F 2 ) where the subscripts indicate partial differentiation.

163
8.

Derive the Gauss equations by assuming that
ruu = c1
ru + c2
rv + c3 n
,


ruv = c4
ru + c5
rv + c6 n
,


rvv = c7
ru + c8
rv + c9 n


with the vectors
ru ,
rv , where c1 , . . . , c9 are constants  'determined  by ' taking dot products  of ' the above  vectors ' 1 2 1 2 and n
. Show that c1 = , c2 = , c3 = e, c4 = , c5 = , c6 = f, 11 12 12  '  '1 1  ' 2 1 2 γ ∂
r ∂
r c7 = , c8 = , c9 = g Show the Gauss equations can be written = + bαβ n
. α β 22 22 α β ∂uγ ∂u ∂u
9. Derive the Weingarten equations n
u = c1
ru + c2
rv n
v = c3
ru + c4
rv and show

f F − eG EG − F 2 eF − f E c2 = EG − F 2

c1 =

gF − f G EG − F 2 f F − gE c4 = EG − F 2 c3 =

and


ru = c∗1 n
u + c∗2 n
v
rv = c∗3 n
u + c∗4 n
v f F − gE eg − f 2 f E − eF c∗2 = eg − f 2 c∗1 =

f G − gF eg − f 2 f F − eG c∗4 = eg − f 2 c∗3 =

The constants in the above equations are determined in a manner similar to that suggested in problem 8. Show that the Weingarten equations can be written in the form ∂
n ∂
r = −bβα β . ∂uα ∂u
ru ×
rv , the results from exercise 1.1, problem 9(a), and the results from problem 5, Using n
= √ EG − F 2 verify that


10.



' (
ru ×
ruu ) · n
= EG − F 2  ' 2 (
ru ×
ruv ) · n
= EG − F 2 12  ' 1 (
rv ×
ruu ) · n
=− EG − F 2 11  ' 2 (
ru ×
rvv ) · n
= EG − F 2 22 2 11



' EG − F 2  ' 1 (
rv ×
rvv ) · n
=− EG − F 2 22  (
ru ×
rv ) · n
= EG − F 2

(
rv ×
ruv ) · n
=−

1 21

and then derive the formula for the geodesic curvature given by equation (1.5.48).  ' α dT
dT
αδ

= (T × )·n
and a ]β γ, δ] = . Hint:(
n × T) · βγ ds ds

164
11.

Verify the equation (1.5.39) which shows that the normal curvature directions are orthogonal. i.e.

verify that Gλ1 λ2 + F (λ1 + λ2 ) + E = 0.
12.

βγ ωα Verify that δστ δλν Rωαβγ = 4Rλνστ .


13.

Find the first fundamental form and unit normal to the surface defined by z = f (x, y).


14.

Verify σ Ai,jk − Ai,kj = Aσ R.ijk

where σ = R.ijk

∂ ∂xj



σ ik

' −

∂ ∂xk



σ ij

'

 +

n ik

'

σ nj

'

 −

n ij

'

' σ . nk

which is sometimes written Rinjk
15.

   '  '  ∂   s  s ∂     j k     ∂x ∂x n j n k = +   [nj, k] [nk, i]   [ij, s] [ik, s] 

σ For Rijkl = giσ R.jkl show

Rinjk

 '  ' ∂ ∂ s s = [nk, i] − k [nj, i] + [ik, s] − [ij, s] nj nk ∂xj ∂x

which is sometimes written

σ R.ijk


16.

  '   n     ik  ' +  '   σ σ    nk ik  ∂ ∂xk



'     '  σ   nj n ij

Show Rijkl


17.

  ∂   ∂xj =   '  σ  ij

1 = 2



∂ 2 gjl ∂ 2 gik ∂ 2 gjk ∂ 2 gil − − + ∂xj ∂xk ∂xi ∂xk ∂xj ∂xl ∂xi ∂xl

 + g αβ ([jk, β][il, α] − [jl, β][ik, α]) .

Use the results from problem 15 to show (i) Rjikl = −Rijkl ,

(ii) Rijlk = −Rijkl ,

(iii) Rklij = Rijkl

Hence, the tensor Rijkl is skew-symmetric in the indices i, j and k, l. Also the tensor Rijkl is symmetric with respect to the (ij) and (kl) pair of indices.
18.

Verify the following cyclic properties of the Riemann Christoffel symbol: (i) Rnijk + Rnjki + Rnkij = 0 (ii) Rinjk + Rjnki + Rknij = 0 (iii) Rijnk + Rjkni + Rkinj = 0 (iv)


19. Riijk ,

Rikjn + Rkjin + Rjikn = 0

first index fixed second index fixed third index fixed fourth index fixed

By employing the results from the previous problems, show all components of the form: Rinjj ,

Riijj ,

Riiii ,

(no summation on i or j) must be zero.

165
20.

Find the number of independent components associated with the Riemann Christoffel tensor

Rijkm ,

i, j, k, m = 1, 2, . . . , N. There are N 4 components to examine in an N −dimensional space. Many of

these components are zero and many of the nonzero components are related to one another by symmetries or the cyclic properties. Verify the following cases: CASE I We examine components of the form Rinin ,

i = n

with no summation of i or n. The first index

can be chosen in N ways and therefore with i = n the second index can be chosen in N − 1 ways. Observe that Rinin = Rnini , leaves M1 =

1 2 N (N

(no summation on i or n) and so one half of the total combinations are repeated. This − 1) components of the form Rinin . The quantity M1 can also be thought of as the

number of distinct pairs of indices (i, n). CASE II We next examine components of the form Rinji ,

i = n = j where there is no summation on

the index i. We have previously shown that the first pair of indices can be chosen in M1 ways. Therefore, the third index can be selected in N − 2 ways and consequently there are M2 = 12 N (N − 1)(N − 2) distinct components of the form Rinji with i = n = j. CASE III Next examine components of the form Rinjk where i = n = j = k. From CASE I the first pairs of indices (i, n) can be chosen in M1 ways. Taking into account symmetries, it can be shown that the second pair of indices can be chosen in 12 (N − 2)(N − 3) ways. This implies that there are 14 N (N − 1)(N − 2)(N − 3) ways of choosing the indices i, n, j and k with i = n = j = k. By symmetry the pairs (i, n) and (j, k) can be interchanged and therefore only one half of these combinations are distinct. This leaves 1 N (N − 1)(N − 2)(N − 3) 8 distinct pairs of indices. Also from the cyclic relations we find that only two thirds of the above components are distinct. This produces M3 =

N (N − 1)(N − 2)(N − 3) 12

distinct components of the form Rinjk with i = n = j = k. Adding the above components from each case we find there are M4 = M1 + M2 + M3 =

N 2 (N 2 − 1) 12

distinct and independent components. Verify the entries in the following table: Dimension of space N

1

2

3

4

5

4

1

16

81

256

625

M4 = Independent components of Rijkm

0

1

6

20

50

Number of components N

Note 1: A one dimensional space can not be curved and all one dimensional spaces are Euclidean. (i.e. if we have an element of arc length squared given by ds2 = f (x)(dx)2 , we can make the coordinate transformation  f (x)dx = du and reduce the arc length squared to the form ds2 = du2 .) Note 2: In a two dimensional space, the indices can only take on the values 1 and 2. In this special case there are 16 possible components. It can be shown that the only nonvanishing components are: R1212 = −R1221 = −R2112 = R2121 .

166 For these nonvanishing components only one independent component exists. By convention, the component R1212 is selected as the single independent component and all other nonzero components are expressed in terms of this component. Find the nonvanishing independent components Rijkl for i, j, k, l = 1, 2, 3, 4 and show that R1212

R3434

R2142

R4124

R1313

R1231

R2342

R4314

R2323

R1421

R3213

R4234

R1414

R1341

R3243

R1324

R2424

R2132

R3143

R1432

can be selected as the twenty independent components.
21. (a) For N = 2 show R1212 is the only nonzero independent component and R1212 = R2121 = −R1221 = −R2112 . (b) Show that on the surface of a sphere of radius r0 we have R1212 = r02 sin2 θ.
22.

Show for N = 2 that R1212


23.

 2  ∂x  = R1212 J = R1212   ∂x 2

s Define Rij = R.ijs as the Ricci tensor and Gij = Rji − 12 δji R as the Einstein tensor, where Rji = g ik Rkj

and R = Rii . Show that Rjk = g ab Rjabk  '  '  ' ' √ √ ∂ 2 log g b ∂ log g a b a ∂ (b) Rij = − − a + i j b ij ia jb ∂x ∂x ∂x ∂x i j

(a)

(c)
24.

i Rijk =0

By employing the results from the previous problem show that in the case N = 2 we have R22 R12 R1212 R11 = = =− g11 g22 g12 g

where g is the determinant of gij .
25.

Consider the case N = 2 where we have g12 = g21 = 0 and show that (a)

R12 = R21 = 0

(c)

(b)

R11 g22 = R22 g11 = R1221

(d)

2R1221 g11 g22 1 Rij = Rgij , 2 R=

where

R = g ij Rij

The scalar invariant R is known as the Einstein curvature of the surface and the tensor Gij = Rji − 12 δji R is known as the Einstein tensor.
26.

For N = 3 show that R1212 , R1313 , R2323 , R1213 , R2123 , R3132 are independent components of the

Riemann Christoffel tensor.

167 
27.

For N = 2 and aαβ =

K=

For N = 2 and aαβ = 1 K= √ 2 a



0 a22

 show that

     1 1 ∂a22 ∂ 1 ∂a11 R1212 ∂ √ √ =− √ + . a 2 a ∂u1 a ∂u1 ∂u2 a ∂u2 


28.

a11 0

a11 a21

a12 a22

 show that

   ' a12 ∂a11 2 ∂a12 ∂ 1 ∂a22 1 ∂a11 a12 ∂a11 ∂ √ √ √ −√ −√ − + . ∂u1 a11 a ∂u2 a ∂u1 ∂u2 a ∂u1 a ∂u2 a11 a ∂u1

Check your results by setting a12 = a21 = 0 and comparing this answer with that given in the problem 27.
29.

Write out the Frenet-Serret formulas (1.5.112)(1.5.113) for surface curves in terms of Christoffel

symbols of the second kind.
30. (a) Use the fact that for n = 2 we have R1212 = R2121 = −R2112 = −R1221 together with eαβ , eαβ the two dimensional alternating tensors to show that the equation (1.5.110) can be written as Rαβγδ = KFαβ Fγδ

where

Fαβ =

√ aeαβ

1 and Fαβ = √ eαβ a

are the corresponding epsilon tensors. 1 Rαβγδ Fαβ Fγδ = K. 4 Hint: See equations (1.3.82),(1.5.93) and (1.5.94).

(b) Show that from the result in part (a) we obtain
31.

Verify the result given by the equation (1.5.100).


32.

Show that aαβ cαβ = 4H 2 − 2K.


33.

Find equations for the principal curvatures associated with the surface x = u,

y = v,

z = f (u, v).


34. Geodesics on a sphere Let (θ, φ) denote the surface coordinates of the sphere of radius ρ defined by the parametric equations x = ρ sin θ cos φ, y = ρ sin θ sin φ, z = ρ cos θ.

(1)

Consider also a plane which passes through the origin with normal having the direction numbers (n1 , n2 , n3 ). This plane is represented by n1 x + n2 y + n3 z = 0 and intersects the sphere in a great circle which is described by the relation n1 sin θ cos φ + n2 sin θ sin φ + n3 cos θ = 0.

(2)

This is an implicit relation between the surface coordinates θ, φ which describes the great circle lying on the sphere. We can write this later equation in the form n1 cos φ + n2 sin φ =

−n3 tan θ

(3)

168 and in the special case where n1 = cos β, n2 = sin β,n3 = − tan α is expressible in the form   tan α tan α or φ − β = cos−1 cos(φ − β) = . tan θ tan θ

(4)

The above equation defines an explicit relationship between the surface coordinates which defines a great circle on the sphere. The arc length squared relation satisfied by the surface coordinates together with the equation obtained by differentiating equation (4) with respect to arc length s gives the relations sin2 θ

dφ = ds

tan α 1−

tan2 α tan2 θ

dθ ds

(5)

ds2 = ρ2 dθ2 + ρ2 sin2 θ dφ2

(6)

The above equations (1)-(6) are needed to consider the following problem. (a) Show that the differential equations defining the geodesics on the surface of a sphere (equations (1.5.51)) are  2 dφ d2 θ − sin θ cos θ =0 ds2 ds d2 φ dθ dφ =0 + 2 cot θ ds2 ds ds

(7) (8)

(b) Multiply equation (8) by sin2 θ and integrate to obtain sin2 θ

dφ = c1 ds

(9)

where c1 is a constant of integration. (c) Multiply equation (7) by

dθ ds

and use the result of equation (9) to show that an integration produces 

dθ ds

2 =

−c21 + c22 sin2 θ

(10)

where c22 is a constant of integration. (d) Use the equations (5)(6) to show that c2 = 1/ρ and c1 =

sin α ρ .

(e) Show that equations (9) and (10) imply that tan α dφ = dθ tan2 θ and making the substitution u =

tan α tan θ

sec2 θ 1−

tan2 α tan2 θ

this equation can be integrated to obtain the equation (4). We

can now expand the equation (4) and express the results in terms of x, y, z to obtain the equation (3). This produces a plane which intersects the sphere in a great circle. Consequently, the geodesics on a sphere are great circles.

169
35.
36.

Find the differential equations defining the geodesics on the surface of a cylinder. Find the differential equations defining the geodesics on the surface of a torus. (See problem 13,

Exercise 1.3)
37.

Find the differential equations defining the geodesics on the surface of revolution x = r cos φ,

y = r sin φ,

z = f (r).

Note the curve z = f (x) gives a profile of the surface. The curves r = Constant are the parallels, while the curves φ = Constant are the meridians of the surface and ds2 = (1 + f 2 ) dr2 + r2 dφ2 .


38.

Find the unit normal and tangent plane to an arbitrary point on the right circular cone x = u sin α cos φ,

y = u sin α sin φ,

z = u cos α.

This is a surface of revolution with r = u sin α and f (r) = r cot α with α constant.
39.

Let s denote arc length and assume the position vector
r(s) is analytic about a point s0 . Show that h2 h3 the Taylor series
r(s) =
r(s0 ) + h
r  (s0 ) +
r  (s0 ) +
r  (s0 ) + · · · about the point s0 , with h = s − s0 is 2! 3!
+ 1 h3 (−κ2 T
+ κ N
+ κτ B)
+ · · · which is obtained by differentiating given by
r(s) =
r(s0 ) + hT
+ 12 κh2 N 6

the Frenet formulas.
40. (a) Show that the circular helix defined by x = a cos t,

y = a sin t,

z = bt with a, b constants, has the

property that any tangent to the curve makes a constant angle with the line defining the z-axis. (i.e. T
·
e3 = cos α = constant.)
·
e3 is parallel to the rectifying plane, which implies that (b) Show also that N e3 = 0 and consequently

sin α.
e3 = T
cos α + B (c) Differentiate the result in part (b) and show that κ/τ = tan α is a constant.
41.

Consider a space curve xi = xi (s) in Cartesian coordinates.  dT
    (a) Show that κ =   = xi xi ds

(b) Show that τ =

1 eijk xi xj x r  ·
r  ×
r  k . Hint: Consider
κ2


42. (a) Find the direction cosines of a normal to a surface z = f (x, y). (b) Find the direction cosines of a normal to a surface F (x, y, z) = 0. (c) Find the direction cosines of a normal to a surface x = x(u, v), y = y(u, v), z = z(u, v).
43.

Show that for a smooth surface z = f (x, y) the Gaussian curvature at a point on the surface is given

by K=

2 fxx fyy − fxy . (fx2 + fy2 + 1)2

170
44.

Show that for a smooth surface z = f (x, y) the mean curvature at a point on the surface is given by H=

(1 + fy2 )fxx − 2fx fy fxy + (1 + fx2 )fyy . 2(fx2 + fy2 + 1)3/2


45.

Express the Frenet-Serret formulas (1.5.13) in terms of Christoffel symbols of the second kind.


46.

Verify the relation (1.5.106).


47.

In Vn assume that Rij = ρgij and show that ρ =

R n

where R = g ij Rij . This result is known as

Einstein’s gravitational equation at points where matter is present. It is analogous to the Poisson equation ∇2 V = ρ from the Newtonian theory of gravitation.
48.

In Vn assume that Rijkl = K(gik gjl − gil gjk ) and show that R = Kn(1 − n). (Hint: See problem 23.)


49.

Assume gij = 0 for i = j and verify the following.

(a) Rhijk = 0 for h = i = j = k  2√  √ √ √ √ ∂ gii ∂ log ghh ∂ gii ∂ log gkk ∂ gii √ (b) Rhiik = gii − − for h, i, k unequal. ∂xh∂xk ∂xh ∂xk ∂xk ∂xh     √  √ √ √ n  ∂ gii ∂ ghh  1 ∂ gii ∂ 1 ∂ ghh √ √  ∂ (c) Rhiih = gii ghh  h √ + i √ +  where h = i. ∂x ghh ∂xh ∂x gii ∂xi ∂xm ∂xm m=1 m=h m=i


50.

Consider a surface of revolution where x = r cos θ, y = r sin θ and z = f (r) is a given function of r.

(a) Show in this V2 we have ds2 = (1 + (f  )2 )dr2 + r2 dθ2 where  =

d ds .

(b) Show the geodesic equations in this V2 are  2  2 dr dθ f  f  r d2 r + − =0 2  2  2 ds 1 + (f ) ds 1 + (f ) ds d2 θ 2 dθ dr =0 + ds2 r ds ds a dθ = 2 . Substitute this result for ds in part (a) to show (c) Solve the second equation in part (b) to obtain ds r  a 1 + (f  )2 dθ = ± √ dr which theoretically can be integrated. r r 2 − a2

171 PART 2: INTRODUCTION TO CONTINUUM MECHANICS

In the following sections we develop some applications of tensor calculus in the areas of dynamics, elasticity, fluids and electricity and magnetism. We begin by first developing generalized expressions for the vector operations of gradient, divergence, and curl. Also generalized expressions for other vector operators are considered in order that tensor equations can be converted to vector equations. We construct a table to aid in the translating of generalized tensor equations to vector form and vice versa. The basic equations of continuum mechanics are developed in the later sections. These equations are developed in both Cartesian and generalized tensor form and then converted to vector form. §2.1 TENSOR NOTATION FOR SCALAR AND VECTOR QUANTITIES We consider the tensor representation of some vector expressions. Our goal is to develop the ability to convert vector equations to tensor form as well as being able to represent tensor equations in vector form. In this section the basic equations of continuum mechanics are represented using both a vector notation and the indicial notation which focuses attention on the tensor components. In order to move back and forth between these notations, the representation of vector quantities in tensor form is now considered. Gradient For Φ = Φ(x1 , x2 , . . . , xN ) a scalar function of the coordinates xi , i = 1, . . . , N , the gradient of Φ is defined as the covariant vector Φ,i =

∂Φ , ∂xi

i = 1, . . . , N.

(2.1.1)

The contravariant form of the gradient is g im Φ,m .

(2.1.2)

Note, if C i = g im Φ,m , i = 1, 2, 3 are the tensor components of the gradient then in an orthogonal coordinate system we will have C 1 = g 11 Φ,1 ,

C 2 = g 22 Φ,2 ,

C 3 = g 33 Φ,3 .

We note that in an orthogonal coordinate system that g ii = 1/h2i , (no sum on i), i = 1, 2, 3 and hence replacing the tensor components by their equivalent physical components there results the equations 1 ∂Φ C(1) = 2 1, h1 h1 ∂x

C(2) 1 ∂Φ = 2 2, h2 h2 ∂x

C(3) 1 ∂Φ = 2 3. h3 h3 ∂x

Simplifying, we find the physical components of the gradient are C(1) =

1 ∂Φ , h1 ∂x1

C(2) =

1 ∂Φ , h2 ∂x2

C(3) =

1 ∂Φ . h3 ∂x3

These results are only valid when the coordinate system is orthogonal and gij = 0 for i = j and gii = h2i , with i = 1, 2, 3, and where i is not summed.

172 Divergence The divergence of a contravariant tensor Ar is obtained by taking the covariant derivative with respect to xk and then performing a contraction. This produces div Ar = Ar,r .

(2.1.3)

Still another form for the divergence is obtained by simplifying the expression (2.1.3). The covariant derivative can be represented Ar,k

∂Ar = + ∂xk



' r Am . mk

Upon contracting the indices r and k and using the result from Exercise 1.4, problem 13, we obtain √ ∂Ar 1 ∂( g) m + A √ ∂xr g ∂xm  √  1 √ ∂Ar r∂ g = √ g r +A g ∂x ∂xr 1 ∂ √ r = √ ( gA ) . g ∂xr

Ar,r = Ar,r Ar,r

(2.1.4)

EXAMPLE 2.1-1. (Divergence) Find the representation of the divergence of a vector Ar in spherical coordinates (ρ, θ, φ). Solution: x1 = ρ,

In spherical coordinates we have x2 = θ,

g11 = h21 = 1,

x3 = φ with

gij = 0 for

g22 = h22 = ρ2 ,

The determinant of gij is g = |gij | = ρ4 sin2 θ and

√

i = j

and

g33 = h23 = ρ2 sin2 θ.

g = ρ2 sin θ. Employing the relation (2.1.4) we find

  ∂ √ 1 1 ∂ √ 2 ∂ √ 3 div A = √ ( gA ) + 2 ( gA ) + 3 ( gA ) . g ∂x1 ∂x ∂x r

In terms of the physical components this equation becomes   ∂ √ A(1) 1 ∂ √ A(2) ∂ √ A(3) ( g ( g ( g )+ )+ ) . div A = √ g ∂ρ h1 ∂θ h2 ∂φ h3 r

By using the notation A(1) = Aρ ,

A(2) = Aθ ,

A(3) = Aφ

for the physical components, the divergence can be expressed in either of the forms:   ∂ 2 1 ∂ 2 Aθ Aφ ∂ 2 (ρ (ρ ) + (ρ ) sin θA ) + sin θ sin θ ρ ρ2 sin θ ∂ρ ∂θ ρ ∂φ ρ sin θ 1 1 1 ∂ ∂A ∂ φ div Ar = 2 (ρ2 Aρ ) + (sin θAθ ) + . ρ ∂ρ ρ sin θ ∂θ ρ sin θ ∂φ

div Ar =

or

173 Curl
= curl A
are represented The contravariant components of the vector C C i = Fijk Ak,j .

(2.1.5)

In expanded form this representation becomes: 1 C =√ g



1

1 C =√ g



2

1 C3 = √ g



∂A3 ∂A2 − 2 ∂x ∂x3 ∂A3 ∂A1 − 3 ∂x ∂x1 ∂A2 ∂A1 − 1 ∂x ∂x2

  (2.1.6)  .


in spherical coordinates EXAMPLE 2.1-2. (Curl) Find the representation for the components of curl A (ρ, θ, φ). Solution:

In spherical coordinates we have :x1 = ρ, g11 = h21 = 1,

x2 = θ,

g22 = h22 = ρ2 ,

x3 = φ with gij = 0 for i = j and

g33 = h23 = ρ2 sin2 θ.

√ The determinant of gij is g = |gij | = ρ4 sin2 θ with g = ρ2 sin θ. The relations (2.1.6) are tensor equations
To find the components of curl A
in spherical components representing the components of the vector curl A. we write the equations (2.1.6) in terms of their physical components. These equations take on the form:   ∂ C(1) 1 ∂ (h3 A(3)) − (h2 A(2)) =√ h1 g ∂θ ∂φ   C(2) ∂ 1 ∂ (h1 A(1)) − (h3 A(3)) =√ h2 g ∂φ ∂ρ   ∂ C(3) 1 ∂ (h2 A(2)) − (h1 A(1)) . =√ h3 g ∂ρ ∂θ

(2.1.7)

We employ the notations C(1) = Cρ ,

C(2) = Cθ ,

C(3) = Cφ ,

A(1) = Aρ ,

A(2) = Aθ ,

A(3) = Aφ


in spherical coordinates, to denote the physical components, and find the components of the vector curl A, are expressible in the form:

  ∂ 1 ∂ Cρ = 2 (ρ sin θAφ ) − (ρAθ ) ρ sin θ ∂θ ∂φ   ∂ 1 ∂ (Aρ ) − (ρ sin θAφ ) Cθ = ρ sin θ ∂φ ∂ρ   1 ∂ ∂ Cφ = (ρAθ ) − (Aρ ) . ρ ∂ρ ∂θ

(2.1.8)

174 Laplacian The Laplacian ∇2 U has the contravariant form ∇ U = g U,ij 2

ij

  ij ∂U = (g U,i ),j = g . ∂xi ,j ij

(2.1.9)

Expanding this expression produces the equations:    ' j ∂ ij ∂U im ∂U g + g ∂xj ∂xi ∂xi m j   √ ∂ 1 ∂ g ij ∂U ij ∂U ∇2 U = g g + √ ∂xj ∂xi g ∂xj ∂xi    √  ∂U ∂U ∂ g 1 √ ∂ ∇2 U = √ g j g ij i + g ij i g ∂x ∂x ∂x ∂xj   1 ∂ √ ij ∂U gg ∇2 U = √ . g ∂xj ∂xi ∇2 U =

(2.1.10)

In orthogonal coordinates we have g ij = 0 for i = j and g11 = h21 ,

g22 = h22 ,

g33 = h23

and so (2.1.10) when expanded reduces to the form        ∂ h2 h3 ∂U h1 h3 ∂U h1 h2 ∂U 1 ∂ ∂ ∇ U= + 2 + 3 . h1 h2 h3 ∂x1 h1 ∂x1 ∂x h2 ∂x2 ∂x h3 ∂x3 2

(2.1.11)

This representation is only valid in an orthogonal system of coordinates. EXAMPLE 2.1-3. (Laplacian)

Find the Laplacian in spherical coordinates.

Solution: Utilizing the results given in the previous example we find the Laplacian in spherical coordinates has the form

       ∂ 1 ∂U 1 ∂U ∂ ∂U ∂ 2 ∇ U= 2 ρ sin θ + sin θ + . ρ sin θ ∂ρ ∂ρ ∂θ ∂θ ∂φ sin θ ∂φ 2

(2.1.12)

This simplifies to ∇2 U =

1 ∂ 2U 1 ∂2U ∂2U 2 ∂U cot θ ∂U + + + + . 2 ∂ρ2 ρ ∂ρ ρ2 ∂θ2 ρ2 ∂θ ρ2 sin θ ∂φ2

The table 1 gives the vector and tensor representation for various quantities of interest.

(2.1.13)

175

VECTOR

GENERAL TENSOR
A


·B
A

CARTESIAN TENSOR

Ai or Ai

Ai

Ai Bi = gij Ai B j = Ai B i Ai Bi = g ij Ai Bj

Ai Bi


=A
×B
C

1 C i = √ eijk Aj Bk g

Ci = eijk Aj Bk

∇ Φ = grad Φ

g im Φ,m

Φ,i =

∂Φ ∂xi


= div A
∇·A

1 ∂ √ r g mn Am,n = Ar,r = √ ( gA ) g ∂xr

Ai,i =

∂Ai ∂xi

C i = Fijk Ak,j

Ci = eijk


=C
= curl A
∇×A ∇2 U

1 ∂ g mn U ,mn = √ g ∂xj

∂Ak ∂xj   ∂ ∂U ∂xi ∂xi

  √ ij ∂U gg ∂xi


= (A
· ∇)B
C

C i = Am B i,m

Ci = Am


= A(∇

C · B)

C i = Ai B j,j

Ci = Ai


= ∇2 A
C  
·∇ φ A  
∇ ∇·A  
∇× ∇×A

C i = g jm Ai ,mj

or Ci = g jm Ai,mj

g im Ai φ ,m

g im Ar,r ,m

Fijk g jm Fkst At,s ,m

Table 1 Vector and tensor representations.

Ci =

∂ ∂xm

∂Bi ∂xm

∂Bm ∂xm   ∂Ai ∂xm Ai φ,i ∂ 2 Ar ∂xi ∂xr

∂ 2 Aj ∂ 2 Ai − ∂xj ∂xi ∂xj ∂xj

176 EXAMPLE 2.1-4. (Maxwell’s equations) In the study of electrodynamics there arises the following vectors and scalars:


=Electric force vector, [E]
= Newton/coulomb E
=Magnetic force vector, [B]
= Weber/m2 B
=Displacement vector, [D]
= coulomb/m2 D
=Auxilary magnetic force vector, [H]
= ampere/m H
= ampere/m2 J
=Free current density, [J] L =free charge density, [L] = coulomb/m3

The above quantities arise in the representation of the following laws: Faraday’s Law This law states the line integral of the electromagnetic force around a loop is proportional to the rate of flux of magnetic induction through the loop. This gives rise to the first electromagnetic field equation:

= − ∂B ∇×E ∂t Ampere’s Law

Fijk Ek,j = −

or

∂B i . ∂t

(2.1.15)

This law states the line integral of the magnetic force vector around a closed loop is

proportional to the sum of the current through the loop and the rate of flux of the displacement vector through the loop. This produces the second electromagnetic field equation:
= J
+ ∇×H


∂D ∂t

Fijk Hk,j = J i +

or

∂Di . ∂t

(2.1.16)

Gauss’s Law for Electricity This law states that the flux of the electric force vector through a closed surface is proportional to the total charge enclosed by the surface. This results in the third electromagnetic field equation:
=L ∇·D

or

1 ∂ √ i

gD = L. √ g ∂xi

(2.1.17)

Gauss’s Law for Magnetism This law states the magnetic flux through any closed volume is zero. This produces the fourth electromagnetic field equation:
=0 ∇·B

or

1 ∂ √ i

gB = 0. √ g ∂xi

(2.1.18)

The four electromagnetic field equations are referred to as Maxwell’s equations. These equations arise in the study of electrodynamics and can be represented in other forms. These other forms will depend upon such things as the material assumptions and units of measurements used. Note that the tensor equations (2.1.15) through (2.1.18) are representations of Maxwell’s equations in a form which is independent of the coordinate system chosen. In applications, the tensor quantities must be expressed in terms of their physical components. In a general orthogonal curvilinear coordinate system we will have g11 = h21 , This produces the result

g22 = h22 ,

g33 = h23 ,

and gij = 0

for i = j.

√ g = h1 h2 h3 . Further, if we represent the physical components of Di , Bi , Ei , Hi

by D(i), B(i), E(i), and H(i)

177 the Maxwell equations can be represented by the equations in table 2. The tables 3, 4 and 5 are the representation of Maxwell’s equations in rectangular, cylindrical, and spherical coordinates. These latter tables are special cases associated with the more general table 2.

1 h1 h2 h3 1 h1 h2 h3 1 h1 h2 h3 1 h1 h2 h3 1 h1 h2 h3 1 h1 h2 h3

 ∂ ∂ 1 ∂B(1) (h3 E(3)) − 3 (h2 E(2)) = − 2 ∂x ∂x h1 ∂t   ∂ ∂ 1 ∂B(2) (h1 E(1)) − 1 (h3 E(3)) = − 3 ∂x ∂x h2 ∂t   ∂ ∂ 1 ∂B(3) (h2 E(2)) − 2 (h1 E(1)) = − 1 ∂x ∂x h3 ∂t 

 ∂ ∂ J(1) 1 ∂D(1) (h3 H(3)) − 3 (h2 H(2)) = + 2 ∂x ∂x h1 h1 ∂t   ∂ ∂ J(2) 1 ∂D(2) (h1 H(1)) − 1 (h3 H(3)) = + 3 ∂x ∂x h2 h2 ∂t   ∂ ∂ J(3) 1 ∂D(3) (h2 H(2)) − 2 (h1 H(1)) = + ∂x1 ∂x h3 h3 ∂t 

       1 ∂ D(1) D(2) D(3) ∂ ∂ h1 h2 h3 + 2 h1 h2 h3 + 3 h1 h2 h3 =L h1 h2 h3 ∂x1 h1 ∂x h2 ∂x h3        ∂ B(1) B(2) B(3) ∂ ∂ 1 h1 h2 h3 + 2 h1 h2 h3 + 3 h1 h2 h3 =0 h1 h2 h3 ∂x1 h1 ∂x h2 ∂x h3 Table 2 Maxwell’s equations in generalized orthogonal coordinates. Note that all the tensor components have been replaced by their physical components.

178

∂Ez ∂Ey ∂Bx − =− ∂y ∂z ∂t ∂Ex ∂Ez ∂By − =− ∂z ∂x ∂t ∂Ex ∂Bz ∂Ey − =− ∂x ∂y ∂t

∂Hy ∂Hz − = Jx + ∂y ∂z ∂Hx ∂Hz − = Jy + ∂z ∂x ∂Hx ∂Hy − = Jz + ∂x ∂y

∂Dx ∂t ∂Dy ∂t ∂Dz ∂t

∂Dy ∂Dz ∂Dx + + =L ∂x ∂y ∂z ∂By ∂Bz ∂Bx + + =0 ∂x ∂y ∂z

Here we have introduced the notations:

with x1 = x,

Dx = D(1)

Bx = B(1)

Hx = H(1)

Jx = J(1)

Ex = E(1)

Dy = D(2)

By = B(2)

Hy = H(2)

Jy = J(2)

Ey = E(2)

Dz = D(3)

Bz = B(3)

Hz = H(3)

Jz = J(3)

Ez = E(3)

x2 = y,

x3 = z,

h1 = h2 = h3 = 1

Table 3 Maxwell’s equations Cartesian coordinates

∂Eθ ∂Br 1 ∂Ez − =− r ∂θ ∂z ∂t ∂Ez ∂Bθ ∂Er − =− ∂z ∂r ∂t ∂Bz 1 ∂ 1 ∂Er (rEθ ) − =− r ∂r r ∂θ ∂t ∂Dz 1 ∂Dθ 1 ∂ (rDr ) + + =L r ∂r r ∂θ ∂z

∂Hθ ∂Dr 1 ∂Hz − = Jr + r ∂θ ∂z ∂t ∂Hz ∂Hr ∂Dθ − = Jθ + ∂z ∂r ∂t 1 ∂ 1 ∂Hr ∂Dz (rHθ ) − = Jz + r ∂r r ∂θ ∂t 1 ∂ ∂Bz 1 ∂Bθ (rBr ) + + =0 r ∂r r ∂θ ∂z

Here we have introduced the notations:

with x1 = r,

Dr = D(1)

Br = B(1)

Hr = H(1)

Jr = J(1)

Er = E(1)

Dθ = D(2)

Bθ = B(2)

Hθ = H(2)

Jθ = J(2)

Eθ = E(2)

Dz = D(3)

Bz = B(3)

Hz = H(3)

Jz = J(3)

Ez = E(3)

x2 = θ,

x3 = z,

h1 = 1,

h2 = r,

h3 = 1.

Table 4 Maxwell’s equations in cylindrical coordinates.

179

  ∂ 1 ∂Eθ ∂Bρ (sin θEφ ) − =− ρ sin θ ∂θ ∂φ ∂t 1 ∂ 1 ∂Eρ ∂Bθ − (ρEφ ) = − ρ sin θ ∂φ ρ ∂ρ ∂t 1 ∂ ∂Bφ 1 ∂Eρ (ρEθ ) − =− ρ ∂ρ ρ ∂θ ∂t

  1 ∂ ∂Hθ ∂Dρ (sin θHφ ) − = Jρ + ρ sin θ ∂θ ∂φ ∂t 1 ∂ 1 ∂Hρ ∂Dθ − (ρHφ ) = Jθ + ρ sin θ ∂φ ρ ∂ρ ∂t 1 ∂ 1 ∂Hρ ∂Dφ (ρHθ ) − = Jφ + ρ ∂ρ ρ ∂θ ∂t

∂ 1 ∂ 2 1 1 ∂Dφ (ρ Dρ ) + (sin θDθ ) + =L ρ2 ∂ρ ρ sin θ ∂θ ρ sin θ ∂φ 1 ∂ 2 ∂ 1 1 ∂Bφ (ρ Bρ ) + (sin θBθ ) + =0 2 ρ ∂ρ ρ sin θ ∂θ ρ sin θ ∂φ Here we have introduced the notations:

with x1 = ρ,

Dρ = D(1)

Bρ = B(1)

Hρ = H(1)

Jρ = J(1)

Eρ = E(1)

Dθ = D(2)

Bθ = B(2)

Hθ = H(2)

Jθ = J(2)

Eθ = E(2)

Dφ = D(3)

Bφ = B(3)

Hφ = H(3)

Jφ = J(3)

Eφ = E(3)

x2 = θ,

x3 = φ,

h1 = 1,

h2 = ρ,

h3 = ρ sin θ

Table 5 Maxwell’s equations spherical coordinates.

Eigenvalues and Eigenvectors of Symmetric Tensors Consider the equation Tij Aj = λAi ,

i, j = 1, 2, 3,

(2.1.19)

where Tij = Tji is symmetric, Ai are the components of a vector and λ is a scalar. Any nonzero solution Ai of equation (2.1.19) is called an eigenvector of the tensor Tij and the associated scalar λ is called an eigenvalue. When expanded these equations have the form (T11 − λ)A1 +

T12 A2 +

T13 A3 = 0

T21 A1 + (T22 − λ)A2 +

T23 A3 = 0

T31 A1 +

T32 A2 + (T33 − λ)A3 = 0.

The condition for equation (2.1.19) to have a nonzero solution Ai is that the characteristic equation should be zero. This equation is found from the determinant equation    T11 − λ T12 T13   f (λ) =  T21 T22 − λ T23  = 0,  T31 T32 T33 − λ 

(2.1.20)

180 which when expanded is a cubic equation of the form f (λ) = −λ3 + I1 λ2 − I2 λ + I3 = 0,

(2.1.21)

where I1 , I2 and I3 are invariants defined by the relations I1 = Tii 1 1 I2 = Tii Tjj − Tij Tij 2 2 I3 = eijk Ti1 Tj2 Tk3 .

(2.1.22)

When Tij is subjected to an orthogonal transformation, where T¯mn = Tij 7im 7jn , then 7im 7jn (Tmn − λ δmn ) = T¯ij − λ δij

and

det (Tmn − λ δmn ) = det T¯ij − λ δij .

Hence, the eigenvalues of a second order tensor remain invariant under an orthogonal transformation. If Tij is real and symmetric then • the eigenvalues of Tij will be real, and • the eigenvectors corresponding to distinct eigenvalues will be orthogonal. Proof: To show a quantity is real we show that the conjugate of the quantity equals the given quantity. If (2.1.19) is satisfied, we multiply by the conjugate Ai and obtain Ai Tij Aj = λAi Ai .

(2.1.25)

The right hand side of this equation has the inner product Ai Ai which is real. It remains to show the left hand side of equation (2.1.25) is also real. Consider the conjugate of this left hand side and write Ai Tij Aj = Ai T ij Aj = Ai Tji Aj = Ai Tij Aj . Consequently, the left hand side of equation (2.1.25) is real and the eigenvalue λ can be represented as the ratio of two real quantities. ˆ 1 and L ˆ2 Assume that λ(1) and λ(2) are two distinct eigenvalues which produce the unit eigenvectors L with components 7i1 and 7i2 , i = 1, 2, 3 respectively. We then have Tij 7j1 = λ(1) 7i1

and

Tij 7j2 = λ(2) 7i2 .

(2.1.26)

Consider the products λ(1) 7i1 7i2 = Tij 7j1 7i2 ,

(2.1.27)

λ(2) 7i1 7i2 = 7i1 Tij 7j2 = 7j1 Tji 7i2 . and subtract these equations. We find that [λ(1) − λ(2) ]7i1 7i2 = 0.

(2.1.28)

By hypothesis, λ(1) is different from λ(2) and consequently the inner product 7i1 7i2 must be zero. Therefore, the eigenvectors corresponding to distinct eigenvalues are orthogonal.

181 Therefore, associated with distinct eigenvalues λ(i) , i = 1, 2, 3 there are unit eigenvectors ˆ (i) = 7i1 e ˆ1 + 7i2 e ˆ2 + 7i3 e ˆ3 L with components 7im , m = 1, 2, 3 which are direction cosines and satisfy 7in 7im = δmn

and

7ij 7jm = δim .

(2.1.23)

The unit eigenvectors satisfy the relations Tij 7j1 = λ(1) 7i1

Tij 7j2 = λ(2) 7i2

Tij 7j3 = λ(3) 7i3

and can be written as the single equation Tij 7jm = λ(m) 7im ,

m = 1, 2, or 3

m not summed.

Consider the transformation xi = 7ij xj

or

xm = 7mj xj

which represents a rotation of axes, where 7ij are the direction cosines from the eigenvectors of Tij . This is a linear transformation where the 7ij satisfy equation (2.1.23). Such a transformation is called an orthogonal transformation. In the new x coordinate system, called principal axes, we have T mn = Tij

∂xi ∂xj = Tij 7im 7jn = λ(n) 7in 7im = λ(n) δmn ∂xm ∂xn

(no sum on n).

(2.1.24)

This equation shows that in the barred coordinate system there are the components  λ(1)

T mn =  0 0

0 λ(2) 0

 0 0 . λ(3)

That is, along the principal axes the tensor components Tij are transformed to the components T ij where T ij = 0 for i = j. The elements T (i)(i) , i not summed, represent the eigenvalues of the transformation (2.1.19).

182 EXERCISE 2.1
1. In cylindrical coordinates (r, θ, z) with f = f (r, θ, z) find the gradient of f.
= A(r,
θ, z) find div A.

2. In cylindrical coordinates (r, θ, z) with A
= A(r,
θ, z) find curl A.

3. In cylindrical coordinates (r, θ, z) for A
4. In cylindrical coordinates (r, θ, z) for f = f (r, θ, z) find ∇2 f.
5. In spherical coordinates (ρ, θ, φ) with f = f (ρ, θ, φ) find the gradient of f.
= A(ρ,
θ, φ) find div A.

6. In spherical coordinates (ρ, θ, φ) with A
= A(ρ,
θ, φ) find curl A.

7. In spherical coordinates (ρ, θ, φ) for A
8. In spherical coordinates (ρ, θ, φ) for f = f (ρ, θ, φ) find ∇2 f. ˆ1 + y e ˆ2 + z e ˆ3 denote the position vector of a variable point (x, y, z) in Cartesian coordinates.
9. Let
r = x e Let r = |
r| denote the distance of this point from the origin. Find in terms of
r and r: (a) grad (r)

(b)

grad (rm )

(c)

1 grad ( ) r

(d) grad (ln r)

(e)

grad (φ)

where φ = φ(r) is an arbitrary function of r. ˆ1 +y e ˆ2 +z e ˆ3 denote the position vector of a variable point (x, y, z) in Cartesian coordinates.
10. Let
r = x e Let r = |
r| denote the distance of this point from the origin. Find: (a)

div (
r) (b) div (rm
r) (c)

div (r−3
r) (d)

div (φ
r)

where φ = φ(r) is an arbitrary function or r.
11.

ˆ1 + y e ˆ2 + z e ˆ3 denote the position vector of a variable point (x, y, z) in Cartesian Let
r = x e

coordinates. Let r = |
r| denote the distance of this point from the origin. Find: (a)

curl
r

where φ = φ(r) is an arbitrary function of r.

12. Expand and simplify the representation for curl (curl A).
13. Show that the curl of the gradient is zero in generalized coordinates.
14. Write out the physical components associated with the gradient of φ = φ(x1 , x2 , x3 ).
15. Show that

 1 ∂ √ im 1 ∂ √ i  g im Ai,m = √ gg Am = Ai,i = √ gA . i g ∂x g ∂xi

(b) curl (φ
r)

183
16. Let r = (
r ·
r)1/2 =
17.

 x2 + y 2 + z 2 ) and calculate (a) ∇2 (r)

Given the tensor equations Dij =

1 2 (vi,j

+ vj,i ),

(b) ∇2 (1/r) (c) ∇2 (r2 )

(d) ∇2 (1/r2 )

i, j = 1, 2, 3. Let v(1), v(2), v(3) denote the

physical components of v1 , v2 , v3 and let D(ij) denote the physical components associated with Dij . Assume the coordinate system (x1 , x2 , x3 ) is orthogonal with metric coefficients g(i)(i) = h2i , i = 1, 2, 3 and gij = 0 for i = j. (a) Find expressions for the physical components D(11), D(22) and D(33) in terms of the physical compo1 ∂V (i)  V (j) ∂hi + no sum on i. nents v(i), i = 1, 2, 3. Answer: D(ii) = hi ∂xi hi hj ∂xj j =i

(b) Find expressions for the physical components D(12),  D(13)  and D(23)  in terms  of the physical compo V (i) V (j) hj ∂ 1 hi ∂ + nents v(i), i = 1, 2, 3. Answer: D(ij) = 2 hj ∂xj hi hi ∂xi hj
18. Write out the tensor equations in problem 17 in Cartesian coordinates.
19. Write out the tensor equations in problem 17 in cylindrical coordinates.
20. Write out the tensor equations in problem 17 in spherical coordinates.
21. Express the vector equation (λ + 2µ)∇Φ − 2µ∇ ×
ω + F
=
0 in tensor form.
22. Write out the equations in problem 21 for a generalized orthogonal coordinate system in terms of physical components.
23. Write out the equations in problem 22 for cylindrical coordinates.
24. Write out the equations in problem 22 for spherical coordinates.
25. Use equation (2.1.4) to represent the divergence in parabolic cylindrical coordinates (ξ, η, z).
26. Use equation (2.1.4) to represent the divergence in parabolic coordinates (ξ, η, φ).
27. Use equation (2.1.4) to represent the divergence in elliptic cylindrical coordinates (ξ, η, z). Change the given equations from a vector notation to a tensor notation.
28.
29.
30.
31.
32.


=
v ∇ · A
+ (∇ ·
v ) A
B


d


= dA · (B
× C)
+A
· ( dB × C)
+A
· (B
× dC ) [A · (B × C)] dt dt dt dt ∂
v d
v = + (
v · ∇)
v dt ∂t
1 ∂H
= −curl E c ∂t
dB
· ∇)
v + B(∇
− (B ·
v ) =
0 dt

184 Change the given equations from a tensor notation to a vector notation.
33.

Fijk Bk,j + F i = 0


34.

gij Fjkl Bl,k + Fi = 0 ∂L + (Lvi ), i = 0 ∂t ∂vi ∂P ∂ 2 vi ∂vi + vm m ) = − i + µ m m + Fi L( ∂t ∂x ∂x ∂x ∂x


35.
36.

 
37. The moment of inertia of an area or second moment of area is defined by Iij =

(ym ym δij −yi yj ) dA A

where dA is an element of area. Calculate of inertia Iij , i, j = 1, 2 for the triangle illustrated in  1 the3 moment  1 2 2 bh − b h 12 24 the figure 2.1-1 and show that Iij = . 1 2 2 1 3 − 24 b h 12 b h

Figure 2.1-1 Moments of inertia for a triangle


38. Use the results from problem 37 and rotate the axes in figure 2.1-1 through an angle θ to a barred system of coordinates. (a) Show that in the barred system of coordinates 

   I11 + I22 I11 − I22 + cos 2θ + I12 sin 2θ 2 2   I11 − I22 =− sin 2θ + I12 cos 2θ 2     I11 + I22 I11 − I22 = − cos 2θ − I12 sin 2θ 2 2

I 11 = I 12 = I 21 I 22

(b) For what value of θ will I 11 have a maximum value? (c) Show that when I 11 is a maximum, we will have I 22 a minimum and I 12 = I 21 = 0.

185

Figure 2.1-2 Mohr’s circle
39. Otto Mohr1 gave the following physical interpretation to the results obtained in problem 38: • Plot the points A(I11 , I12 ) and B(I22 , −I12 ) as illustrated in the figure 2.1-2 • Draw the line AB and calculate the point C where this line intersects the I axes. Show the point C has the coordinates I11 + I22 , 0) 2 • Calculate the radius of the circle with center at the point C and with diagonal AB and show this (

! 

radius is r=

I11 − I22 2

2 2 + I12

• Show the maximum and minimum values of I occur where the constructed circle intersects the I axes. I11 + I22 I11 + I22 Show that Imax = I 11 = +r Imin = I 22 = − r. 2 2   I11 I12
40. Show directly that the eigenvalues of the symmetric matrix Iij = are λ1 = Imax and I21 I22 λ2 = Imin where Imax and Imin are given in problem 39.
41. Find the principal axes and moments of inertia for the triangle given in problem 37 and summarize your results from problems 37,38,39, and 40.
42. Verify for orthogonal coordinates the relations 

3   e(i)jk ∂(h(k) A(k))
· eˆ(i) = ∇×A h(i) h1 h2 h3 ∂xj k=1

or 1
= ∇×A h1 h2 h3

  h1 e ˆ1  ∂  ∂x1   h1 A(1)

 ˆ3  h3 e ∂ ∂ . ∂x2 ∂x3  h2 A(2) h3 A(3)  ˆ2 h2 e


43. Verify for orthogonal coordinates the relation 

1

3   h(i) ∂
· eˆ(i) = e(i)jr ersm ∇ × (∇ × A) h1 h2 h3 ∂xj m=1

Christian Otto Mohr (1835-1918) German civil engineer.

(

h2(r) ∂(h(m) A(m)) h1 h2 h3 ∂xs

)

186
44. Verify for orthogonal coordinates the relation   '    1 ∂(h2 h3 A(1)) ∂(h1 h3 A(2)) ∂(h1 h2 A(3)) ∂ 1
∇ ∇·A · eˆ(i) = + + h(i) ∂x(i) h1 h2 h3 ∂x1 ∂x2 ∂x3
45. Verify the relation   3    ∂h(i) ∂hk A(k) ∂B(i)  B(k)
· ∇)B
·e ˆ(i) = (A + − A(k) A(i) h(k) ∂xk hk h(i) ∂xk ∂x(i) k=1

k =i


46. The Gauss divergence theorem is written    1 

∂F 2 ∂F 3 ∂F + + n1 F 1 + n2 F 2 + n3 F 3 dσ dτ = ∂x ∂y ∂z V S where V is the volume within a simple closed surface S. Here it is assumed that F i = F i (x, y, z) are continuous functions with continuous first order derivatives throughout V and ni are the direction cosines of the outward normal to S, dτ is an element of volume and dσ is an element of surface area. (a) Show that in a Cartesian coordinate system ∂F 2 ∂F 3 ∂F 1 + + ∂x ∂y ∂z   and that the tensor form of this theorem is F,ii dτ = F i ni dσ. F,ii =

V

S

(b) Write the vector form of this theorem. (c) Show that if we define ∂u ∂v , vr = ∂xr ∂xr = g im (uvi,m + um vi ) ur =

then F,ii = g im Fi,m

and Fr = grm F m = uvr

(d) Show that another form of the Gauss divergence theorem is    im m g um vi dτ = uvm n dσ − ug im vi,m dτ V

S

Write out the above equation in Cartesian coordinates.
47. Show
48. Show
49. Show

V



1 Find the eigenvalues and eigenvectors associated with the matrix A =  1 2 that the eigenvectors are orthogonal.  1 Find the eigenvalues and eigenvectors associated with the matrix A =  2 1 that the eigenvectors are orthogonal.  1 Find the eigenvalues and eigenvectors associated with the matrix A =  1 0 that the eigenvectors are orthogonal.

 1 2 2 1. 1 1  2 1 1 0. 0 1  1 0 1 1. 1 1


50. The harmonic and biharmonic functions or potential functions occur in the mathematical modeling of many physical problems. Any solution of Laplace’s equation ∇2 Φ = 0 is called a harmonic function and any solution of the biharmonic equation ∇4 Φ = 0 is called a biharmonic function. (a) Expand the Laplace equation in Cartesian, cylindrical and spherical coordinates. (b) Expand the biharmonic equation in two dimensional Cartesian and polar coordinates. Hint: Consider ∇4 Φ = ∇2 (∇2 Φ). In Cartesian coordinates ∇2 Φ = Φ,ii and ∇4 Φ = Φ,iijj .

187 §2.2 DYNAMICS Dynamics is concerned with studying the motion of particles and rigid bodies. By studying the motion of a single hypothetical particle, one can discern the motion of a system of particles. This in turn leads to the study of the motion of individual points in a continuous deformable medium. Particle Movement The trajectory of a particle in a generalized coordinate system is described by the parametric equations xi = xi (t),

i = 1, . . . , N

(2.2.1)

where t is a time parameter. If the coordinates are changed to a barred system by introducing a coordinate transformation xi = xi (x1 , x2 , . . . , xN ),

i = 1, . . . , N

then the trajectory of the particle in the barred system of coordinates is xi = xi (x1 (t), x2 (t), . . . , xN (t)),

i = 1, . . . , N.

(2.2.2)

The generalized velocity of the particle in the unbarred system is defined by vi =

dxi , dt

i = 1, . . . , N.

(2.2.3)

By the chain rule differentiation of the transformation equations (2.2.2) one can verify that the velocity in the barred system is vr =

∂xr dxj ∂xr j dxr = = v , j dt ∂x dt ∂xj

r = 1, . . . , N.

(2.2.4)

Consequently, the generalized velocity v i is a first order contravariant tensor. The speed of the particle is obtained from the magnitude of the velocity and is v 2 = gij v i v j . The generalized acceleration f i of the particle is defined as the intrinsic derivative of the generalized velocity. The generalized acceleration has the form n dv i δv i i dx = v,n = + f = δt dt dt i



'  ' m n i d2 xi i dx dx m n v v = + mn m n dt dt dt2

and the magnitude of the acceleration is f 2 = gij f i f j .

(2.2.5)

188

Figure 2.2-1 Tangent, normal and binormal to point P on curve. Frenet-Serret Formulas The parametric equations (2.2.1) describe a curve in our generalized space. With reference to the figure 2.2-1 we wish to define at each point P of the curve the following orthogonal unit vectors: T i = unit tangent vector at each point P. N i = unit normal vector at each point P. B i = unit binormal vector at each point P. These vectors define the osculating, normal and rectifying planes illustrated in the figure 2.2-1. In the generalized coordinates the arc length squared is ds2 = gij dxi dxj . Define T i =

dxi ds

as the tangent vector to the parametric curve defined by equation (2.2.1). This vector is a

unit tangent vector because if we write the element of arc length squared in the form 1 = gij

dxi dxj = gij T i T j , ds ds

(2.2.6)

we obtain the generalized dot product for T i . This generalized dot product implies that the tangent vector is a unit vector. Differentiating the equation (2.2.6) intrinsically with respect to arc length s along the curve produces gmn

δT m n δT n T + gmn T m = 0, δs δs

which simplifies to gmn T n

δT m = 0. δs

(2.2.7)

189 The equation (2.2.7) is a statement that the vector vector is defined as Ni =

1 δT i κ δs

δT m δs

is orthogonal to the vector T m . The unit normal

or

Ni =

1 δTi , κ δs

(2.2.8)

where κ is a scalar called the curvature and is chosen such that the magnitude of N i is unity. The reciprocal of the curvature is R =

1 κ,

which is called the radius of curvature. The curvature of a straight line is zero

while the curvature of a circle is a constant. The curvature measures the rate of change of the tangent vector as the arc length varies. The equation (2.2.7) can be expressed in the form gij T i N j = 0.

(2.2.9)

Taking the intrinsic derivative of equation (2.2.9) with respect to the arc length s produces gij T i or gij T i

δN j δT i j + gij N =0 δs δs

δN j δT i j = −gij N = −κgij N i N j = −κ. δs δs

(2.2.10)

The generalized dot product can be written gij T i T j = 1, and consequently we can express equation (2.2.10) in the form gij T

i δN

δs

j

 = −κgij T T i

j

or

gij T

i

δN j + κT j δs

 = 0.

(2.2.11)

Consequently, the vector δN j + κT j δs

(2.2.12)

is orthogonal to T i . In a similar manner, we can use the relation gij N i N j = 1 and differentiate intrinsically with respect to the arc length s to show that gij N i

δN j = 0. δs

This in turn can be expressed in the form  gij N

i

δN j + κT j δs

 = 0.

This form of the equation implies that the vector represented in equation (2.2.12) is also orthogonal to the unit normal N i . We define the unit binormal vector as   1 δN i i i B = + κT or τ δs

1 Bi = τ



δNi + κTi δs

 (2.2.13)

where τ is a scalar called the torsion. The torsion is chosen such that the binormal vector is a unit vector. The torsion measures the rate of change of the osculating plane and consequently, the torsion τ is a measure

190 of the twisting of the curve out of a plane. The value τ = 0 corresponds to a plane curve. The vectors T i , N i , B i , i = 1, 2, 3 satisfy the cross product relation B i = Fijk Tj Nk . If we differentiate this relation intrinsically with respect to arc length s we find   δTj δB i δNk = Fijk Tj + Nk δs δs δs = Fijk [Tj (τ Bk − κTk ) + κNj Nk ]

(2.2.14)

= τ Fijk Tj Bk = −τ Fikj Bk Tj = −τ N i . The relations (2.2.8),(2.2.13) and (2.2.14) are now summarized and written δT i = κN i δs δN i = τ B i − κT i δs δB i = −τ N i . δs

(2.2.15)

These equations are known as the Frenet-Serret formulas of differential geometry. Velocity and Acceleration Chain rule differentiation of the generalized velocity is expressible in the form vi = where v =

ds dt

dxi ds dxi = = T i v, dt ds dt

(2.2.16)

is the speed of the particle and is the magnitude of v i . The vector T i is the unit tangent vector

to the trajectory curve at the time t. The equation (2.2.16) is a statement of the fact that the velocity of a particle is always in the direction of the tangent vector to the curve and has the speed v. By chain rule differentiation, the generalized acceleration is expressible in the form fr =

dv r δv r δT r = T +v δt dt δt dv r δT r ds T +v = dt δs dt dv r T + κv 2 N r . = dt

(2.2.17)

The equation (2.2.17) states that the acceleration lies in the osculating plane. Further, the equation (2.2.17) indicates that the tangential component of the acceleration is 2

eration is κv .

dv dt ,

while the normal component of the accel-

191 Work and Potential Energy Define M as the constant mass of the particle as it moves along the curve defined by equation (2.2.1). Also let Qr denote the components of a force vector (in appropriate units of measurements) which acts upon the particle. Newton’s second law of motion can then be expressed in the form Qr = M f r

or

Qr = M fr .

(2.2.18)

The work done W in moving a particle from a point P0 to a point P1 along a curve xr = xr (t), r = 1, 2, 3, with parameter t, is represented by a summation of the tangential components of the forces acting along the path and is defined as the line integral 

P1

W = P0

dxr ds = Qr ds



P1

 r

t1

Qr dx = P0

t0

dxr dt = Qr dt



t1

Qr v r dt

(2.2.19)

t0

where Qr = grs Qs is the covariant form of the force vector, t is the time parameter and s is arc length along the curve. Conservative Systems If the force vector is conservative it means that the force is derivable from a scalar potential function V = V (x1 , x2 , . . . , xN )

such that

Qr = −V ,r = −

∂V , ∂xr

r = 1, . . . , N.

(2.2.20)

In this case the equation (2.2.19) can be integrated and we find that to within an additive constant we will have V = −W. The potential function V is called the potential energy of the particle and the work done becomes the change in potential energy between the starting and end points and is independent of the path connecting the points. Lagrange’s Equations of Motion The kinetic energy T of the particle is defined as one half the mass times the velocity squared and can be expressed in any of the forms 1 T = M 2



ds dt

2 =

1 1 1 M v 2 = M gmn v m v n = M gmn x˙ m x˙ n , 2 2 2

(2.2.21)

where the dot notation denotes differentiation with respect to time. It is an easy exercise to calculate the derivatives

∂T = M grmx˙ m r ∂ x ˙     d ∂T ∂grm n m m x ¨ + x ˙ x ˙ = M g rm dt ∂ x˙ r ∂xn ∂T 1 ∂gmn m n = M x˙ x˙ , r ∂x 2 ∂xr

(2.2.22)

and thereby verify the relation d dt



∂T ∂ x˙ r

 −

∂T = M fr = Qr , ∂xr

r = 1, . . . , N.

(2.2.23)

192 This equation is called the Lagrange’s form of the equations of motion. EXAMPLE 2.2-1. (Equations of motion in spherical coordinates)

Find the Lagrange’s form of

the equations of motion in spherical coordinates. Solution: Let x1 = ρ, x2 = θ, x3 = φ then the element of arc length squared in spherical coordinates has the form ds2 = (dρ)2 + ρ2 (dθ)2 + ρ2 sin2 θ(dφ)2 . The element of arc length squared can be used to construct the kinetic energy. For example,  2  ds 1 1  2 ˙ 2 + ρ2 sin2 θ(φ) ˙ 2 . ˙ + ρ2 (θ) T = M = M (ρ) 2 dt 2 The Lagrange form of the equations of motion of a particle are found from the relations (2.2.23) and are calculated to be:

  d ∂T − dt ∂ ρ˙   d ∂T − M f2 = Q2 = dt ∂ θ˙   d ∂T M f3 = Q3 = − dt ∂ φ˙

M f1 = Q1 =

  ∂T ˙ 2 − ρ sin2 θ(φ) ˙ 2 = M ρ¨ − ρ(θ) ∂ρ     d ∂T ˙ 2 =M ρ2 θ˙ − ρ2 sin θ cos θ(φ) ∂θ dt    d ∂T =M ρ2 sin2 θφ˙ . ∂φ dt

In terms of physical components we have   ˙ 2 − ρ sin2 θ(φ) ˙ 2 Qρ = M ρ¨ − ρ(θ)   M d  2 ˙ ˙ 2 ρ θ − ρ2 sin θ cos θ(φ) Qθ = ρ dt    d M ρ2 sin2 θφ˙ . Qφ = ρ sin θ dt

Euler-Lagrange Equations of Motion Starting with the Lagrange’s form of the equations of motion from equation (2.2.23), we assume that the external force Qr is derivable from a potential function V as specified by the equation (2.2.20). That is, we assume the system is conservative and express the equations of motion in the form   ∂V d ∂T ∂T − r = − r = Qr , r = 1, . . . , N dt ∂ x˙ r ∂x ∂x

(2.2.24)

The Lagrangian is defined by the equation L = T − V = T (x1 , . . . , xN , x˙ 1 , . . . , x˙ N ) − V (x1 , . . . , xN ) = L(xi , x˙ i ).

(2.2.25)

Employing the defining equation (2.2.25), it is readily verified that the equations of motion are expressible in the form d dt



∂L ∂ x˙ r

 −

∂L = 0, ∂xr

r = 1, . . . , N,

which are called the Euler-Lagrange form for the equations of motion.

(2.2.26)

193

Figure 2.2-2 Simply pulley system EXAMPLE 2.2-2. (Simple pulley system) Find the equation of motion for the simply pulley system illustrated in the figure 2.2-2. Solution: The given system has only one degree of freedom, say y1 . It is assumed that y1 + y2 = 7 = a constant. The kinetic energy of the system is T =

1 (m1 + m2 )y˙ 12 . 2

Let y1 increase by an amount dy1 and show the work done by gravity can be expressed as dW = m1 g dy1 + m2 g dy2 dW = m1 g dy1 − m2 g dy1 dW = (m1 − m2 )g dy1 = Q1 dy1 . Here Q1 = (m1 − m2 )g is the external force acting on the system where g is the acceleration of gravity. The Lagrange equation of motion is d dt



∂T ∂ y˙ 1

 −

∂T = Q1 ∂y1

or y1 = (m1 − m2 )g. (m1 + m2 )¨ Initial conditions must be applied to y1 and y˙ 1 before this equation can be solved.

194 EXAMPLE 2.2-3. (Simple pendulum) Find the equation of motion for the pendulum system illustrated in the figure 2.2-3. Solution: Choose the angle θ illustrated in the figure 2.2-3 as the generalized coordinate. If the pendulum is moved from a vertical position through an angle θ, we observe that the mass m moves up a distance h = 7 − 7 cos θ. The work done in moving this mass a vertical distance h is W = −mgh = −mg7(1 − cos θ), since the force is −mg in this coordinate system. In moving the pendulum through an angle θ, the arc length s swept out by the mass m is s = 7θ. This implies that the kinetic energy can be expressed  2 ds 1 1  2 1 ˙ 2. T = m = m 7θ˙ = m72 (θ) 2 dt 2 2

Figure 2.2-3 Simple pendulum system

The Lagrangian of the system is L=T −V =

1 2 ˙ 2 m7 (θ) − mg7(1 − cos θ) 2

and from this we find the equation of motion   d ∂L ∂L =0 or − dt ∂ θ˙ ∂θ

d  2 ˙ m7 θ − mg7(− sin θ) = 0. dt

This in turn simplifies to the equation g θ¨ + sin θ = 0. 7 This equation together with a set of initial conditions for θ and θ˙ represents the nonlinear differential equation which describes the motion of a pendulum without damping.

195 EXAMPLE 2.2-4. (Compound pendulum)

Find the equations of motion for the compound pendulum

illustrated in the figure 2.2-4. Solution: Choose for the generalized coordinates the angles x1 = θ1 and x2 = θ2 illustrated in the figure 2.2-4. To find the potential function V for this system we consider the work done as the masses m1 and m2 are moved. Consider independent motions of the angles θ1 and θ2 . Imagine the compound pendulum initially in the vertical position as illustrated in the figure 2.2-4(a). Now let m1 be displaced due to a change in θ1 and obtain the figure 2.2-4(b). The work done to achieve this position is W1 = −(m1 + m2 )gh1 = −(m1 + m2 )gL1 (1 − cos θ1 ). Starting from the position in figure 2.2-4(b) we now let θ2 undergo a displacement and achieve the configuration in the figure 2.2-4(c).

Figure 2.2-4 Compound pendulum

The work done due to the displacement θ2 can be represented W2 = −m2 gh2 = −m2 gL2 (1 − cos θ2 ). Since the potential energy V satisfies V = −W to within an additive constant, we can write V = −W = −W1 − W2 = −(m1 + m2 )gL1 cos θ1 − m2 gL2 cos θ2 + constant, where the constant term in the potential energy has been neglected since it does not contribute anything to the equations of motion. (i.e. the derivative of a constant is zero.) The kinetic energy term for this system can be represented   2 2 1 ds1 ds2 1 + m2 m1 2 dt 2 dt 1 1 T = m1 (x˙ 21 + y˙ 12 ) + m2 (x˙ 22 + y˙ 22 ), 2 2 T =

(2.2.27)

196 where

(x1 , y1 ) = (L1 sin θ1 , −L1 cos θ1 ) (x2 , y2 ) = (L1 sin θ1 + L2 sin θ2 , −L1 cos θ1 − L2 cos θ2 )

(2.2.28)

are the coordinates of the masses m1 and m2 respectively. Substituting the equations (2.2.28) into equation (2.2.27) and simplifying produces the kinetic energy expression T =

1 1 (m1 + m2 )L21 θ˙12 + m2 L1 L2 θ˙1 θ˙2 cos(θ1 − θ2 ) + m2 L22 θ˙22 . 2 2

(2.2.29)

Writing the Lagrangian as L = T − V , the equations describing the motion of the compound pendulum are obtained from the Lagrangian equations d dt



∂L ∂ θ˙1



∂L =0 − ∂θ1

d dt

and



∂L ∂ θ˙2

 −

∂L = 0. ∂θ2

Calculating the necessary derivatives, substituting them into the Lagrangian equations of motion and then simplifying we derive the equations of motion L1 θ¨1 +

m2 m2 L2 θ¨2 cos(θ1 − θ2 ) + L2 (θ˙2 )2 sin(θ1 − θ2 ) + g sin θ1 = 0 m1 + m2 m1 + m2 L1 θ¨1 cos(θ1 − θ2 ) + L2 θ¨2 − L1 (θ˙1 )2 sin(θ1 − θ2 ) + g sin θ2 = 0.

These equations are a set of coupled, second order nonlinear ordinary differential equations. These equations are subject to initial conditions being imposed upon the angular displacements (θ1 , θ2 ) and the angular velocities (θ˙1 , θ˙2 ).

Alternative Derivation of Lagrange’s Equations of Motion Let c denote a given curve represented in the parametric form xi = xi (t),

i = 1, . . . , N,

t0 ≤ t ≤ t 1

and let P0 , P1 denote two points on this curve corresponding to the parameter values t0 and t1 respectively. Let c denote another curve which also passes through the two points P0 and P1 as illustrated in the figure 2.2-5. The curve c is represented in the parametric form xi = xi (t) = xi (t) + Fη i (t),

i = 1, . . . , N,

t0 ≤ t ≤ t1

in terms of a parameter F. In this representation the function η i (t) must satisfy the end conditions η i (t0 ) = 0 and η i (t1 ) = 0

i = 1, . . . , N

since the curve c is assumed to pass through the end points P0 and P1 . Consider the line integral



t1

I(F) = t0

L(t, xi + Fη i , x˙ i + Fη˙ i ) dt,

(2.2.30)

197

Figure 2.2-5. Motion along curves c and c where i L = T − V = L(t, xi , x˙ )

is the Lagrangian evaluated along the curve c. We ask the question, “What conditions must be satisfied by the curve c in order that the integral I(F) have an extremum value when F is zero?”If the integral I(F) has a minimum value when F is zero it follows that its derivative with respect to F will be zero at this value and  dI(F)  = 0. dF #=0

we will have

Employing the definition

 I(F) − I(0) dI  = I  (0) = 0 = lim  dF #=0 #→0 F

we expand the Lagrangian in equation (2.2.30) in a series about the point F = 0. Substituting the expansion 

 ∂L i ∂L i L(t, x + Fη , x˙ + Fη˙ ) = L(t, x , x˙ ) + F η + i η˙ + F2 [ ] + · · · ∂xi ∂ x˙ i

i

i

i

i

i

into equation (2.2.30) we calculate the derivative I(F) − I(0) = lim I (0) = lim #→0 #→0 F 



t1

t0



 ∂L i ∂L i η (t) + i η˙ (t) dt + F [ ] + · · · = 0, ∂xi ∂ x˙

where we have neglected higher order powers of F since F is approaching zero. Analysis of this equation informs us that the integral I has a minimum value at F = 0 provided that the integral 

t1

δI = t0



 ∂L i ∂L i η (t) + η ˙ (t) dt = 0 ∂xi ∂ x˙ i

(2.2.31)

198 is satisfied. Integrating the second term of this integral by parts we find 

t1

δI = t0

 t1  t1   ∂L i ∂L i d ∂L η dt + η (t) − η i (t) dt = 0. ∂xi ∂ x˙ i ∂ x˙ i t0 dt t0

(2.2.32)

The end condition on η i (t) makes the middle term in equation (2.2.32) vanish and we are left with the integral



t1

δI =

 η i (t)

t0

∂L d − ∂xi dt



∂L ∂ x˙ i

 dt = 0,

(2.2.33)

which must equal zero for all η i (t). Since η i (t) is arbitrary, the only way the integral in equation (2.2.33) can be zero for all η i (t) is for the term inside the brackets to vanish. This produces the result that the integral of the Lagrangian is an extremum when the Euler-Lagrange equations d dt





∂L ∂ x˙ i

∂L = 0, ∂xi

−

i = 1, . . . , N

(2.2.34)

are satisfied. This is a necessary condition for the integral I(F) to have a minimum value. In general, any line integral of the form 

t1

I=

φ(t, xi , x˙ i ) dt

(2.2.35)

t0

has an extremum value if the curve c defined by xi = xi (t), i = 1, . . . , N satisfies the Euler-Lagrange equations d dt





∂φ ∂ x˙ i

−

∂φ = 0, ∂xi

i = 1, . . . , N.

(2.2.36)

The above derivation is a special case of (2.2.36) when φ = L. Note that the equations of motion equations (2.2.34) are just another form of the equations (2.2.24). Note also that δ δT = δt δt



1 mgij v i v j 2

 = mgij v i f j = mfi v i = mfi x˙ i

∂V and if we assume that the force Qi is derivable from a potential function V , then mfi = Qi = − i , so ∂x δ ∂V i δV δT i i = mfi x˙ = Qi x˙ = − i x˙ = − or (T + V ) = 0 or T + V = h = constant called the energy that δt ∂x δt δt constant of the system. Action Integral The equations of motion (2.2.34) or (2.2.24) are interpreted as describing geodesics in a space whose line-element is ds2 = 2m(h − V )gjk dxj dxk where V is the potential function for the force system and T + V = h is the energy constant of the motion. The integral of ds along a curve C between two points P1 and P2 is called an action integral and is  √ A = 2m

P2

P1

 '1/2 dxj dxk dτ (h − V )gjk dτ dτ

199 where τ is a parameter used to describe the curve C. The principle of stationary action states that of all curves through the points P1 and P2 the one which makes the action an extremum is the curve specified by Newton’s second law. The extremum is usually a minimum. To show this let φ=

√

 '1/2 dxj dxk 2m (h − V )gjk dτ dτ

in equation (2.2.36). Using the notation x˙ k =

dxk dτ

we find that

∂φ 2m (h − V )gik x˙ k = i ∂ x˙ φ ∂gjk j k 2m ∂V ∂φ 2m (h − V ) = x˙ x˙ − gjk x˙ j x˙ k . ∂xi 2φ ∂xi 2φ ∂xi The equation (2.2.36) which describe the extremum trajectories are found to be   ∂gjk j k 2m ∂V d 2m 2m (h − V )gik x˙ k − (h − V ) x˙ x˙ + gjk x˙ j x˙ k = 0. dt φ 2φ ∂xi φ ∂xi By changing variables from τ to t where satisfy the equation m

d dt

dt dτ

=

√

√

mφ 2(h−V )

we find that the trajectory for an extremum must

  ∂V dxk m ∂gjk dxj dxk + i =0 gik − i dt 2 ∂x dt dt ∂x

which are the same equations as (2.2.24). (i.e. See also the equations (2.2.22).) Dynamics of Rigid Body Motion Let us derive the equations of motion of a rigid body which is rotating due to external forces acting upon it. We neglect any translational motion of the body since this type of motion can be discerned using our knowledge of particle dynamics. The derivation of the equations of motion is restricted to Cartesian tensors and rotational motion. Consider a system of N particles rotating with angular velocity ωi , i = 1, 2, 3, about a line L through the center of mass of the system. Let V
(α) denote the velocity of the αth particle which has mass m(α) and (α)

position xi , i = 1, 2, 3 with respect to an origin on the line L. Without loss of generality we can assume that the origin of the coordinate system is also at the center of mass of the system of particles, as this choice of an origin simplifies the derivation. The velocity components for each particle is obtained by taking cross products and we can write
(α) =
V ω ×
r (α)

or

(α)

Vi

(α)

= eijk ωj xk .

(2.2.37)

The kinetic energy of the system of particles is written as the sum of the kinetic energies of each individual particle and is N N 1 1 (α) (α) (α) T = m(α) Vi Vi = m(α) eijk ωj xk eimn ωm x(α) n . 2 α=1 2 α=1

(2.2.38)

200 Employing the e − δ identity the equation (2.2.38) can be simplified to the form T =

N   1 (α) (α) (α) . m(α) ωm ωm xk xk − ωn ωk xk x(α) n 2 α=1

Define the second moments and products of inertia by the equation Iij =

N 

  (α) (α) (α) (α) m(α) xk xk δij − xi xj

(2.2.39)

α=1

and write the kinetic energy in the form T =

1 Iij ωi ωj . 2

(2.2.40)

Similarly, the angular momentum of the system of particles can also be represented in terms of the second moments and products of inertia. The angular momentum of a system of particles is defined as a summation of the moments of the linear momentum of each individual particle and is Hi =

N 

(α) (α)

m(α) eijk xj vk

=

α=1

N 

(α)

m(α) eijk xj ekmn ωm x(α) n .

(2.2.41)

α=1

The e − δ identity simplifies the equation (2.2.41) to the form Hi = ωj

N 

  (α) (α) (α) = ωj Iji . m(α) x(α) n xn δij − xj xi

(2.2.42)

α=1

The equations of motion of a rigid body is obtained by applying Newton’s second law of motion to the system of N particles. The equation of motion of the αth particle is written (α)

m(α) x ¨i

(α)

= Fi

.

(2.2.43)

Summing equation (2.2.43) over all particles gives the result N 

(α)

m(α) x ¨i

=

α=1

N 

(α)

Fi

.

(2.2.44)

α=1

This represents the translational equations of motion of the rigid body. The equation (2.2.44) represents the rate of change of linear momentum being equal to the total external force acting upon the system. Taking (α)

the cross product of equation (2.2.43) with the position vector xj (α)

produces

(α)

¨t erst x(α) = erst x(α) m(α) x s s Ft and summing over all particles we find the equation N  α=1

(α) m(α) erst x(α) ¨t s x

=

N  α=1

(α)

erst x(α) s Ft

.

(2.2.45)

201 The equations (2.2.44) and (2.2.45) represent the conservation of linear and angular momentum and can be 

written in the forms d dt 

and d dt

 m(α) x˙ (α) r

=

α=1

N 

Fr(α)

(2.2.46)

α=1

 (α) m(α) erst x(α) ˙t s x

=

α=1

N 

(α)

erst x(α) s Ft

.

(2.2.47)

α=1

0 (α) representing the linear momentum, Fr = Fr the total force 0 (α) (α) acting on the system of particles, Hr = m(α) erst xs x˙ t is the angular momentum of the system relative 0 (α) (α) to the origin, and Mr = erst xs Ft is the total moment of the system relative to the origin. We can By definition we have Gr =

0

N 

N 

(α)

m(α) x˙ r

therefore express the equations (2.2.46) and (2.2.47) in the form dGr = Fr dt

(2.2.48)

dHr = Mr . dt

(2.2.49)

and

The equation (2.2.49) expresses the fact that the rate of change of angular momentum is equal to the moment of the external forces about the origin. These equations show that the motion of a system of particles can be studied by considering the motion of the center of mass of the system (translational motion) and simultaneously considering the motion of points about the center of mass (rotational motion). We now develop some relations in order to express the equations (2.2.49) in an alternate form. Toward this purpose we consider first the concepts of relative motion and angular velocity. Relative Motion and Angular Velocity Consider two different reference frames denoted by S and S. Both reference frames are Cartesian coordinates with axes xi and xi , i = 1, 2, 3, respectively. The reference frame S is fixed in space and is called an inertial reference frame or space-fixed reference system of axes. The reference frame S is fixed to and rotates with the rigid body and is called a body-fixed system of axes. Again, for convenience, it is assumed that the origins of both reference systems are fixed at the center of mass of the rigid body. ei , i = 1, 2, 3, while the reference system S has the basis Further, we let the system S have the basis vectors
ˆi , i = 1, 2, 3. The transformation equations between the two sets of reference axes are the affine vectors e transformations xi = 7ji xj

and

xi = 7ij xj

(2.2.50)

where 7ij = 7ij (t) are direction cosines which are functions of time t (i.e. the 7ij are the cosines of the angles between the barred and unbarred axes where the barred axes are rotating relative to the space-fixed unbarred axes.) The direction cosines satisfy the relations 7ij 7ik = δjk

and

7ij 7kj = δik .

(2.2.51)

202 EXAMPLE 2.2-5. (Euler angles φ, θ, ψ)

Consider the following sequence of transformations which

are used in celestial mechanics. First a rotation about the x3 axis taking the xi axes to the yi axes      y1 cos φ sin φ 0 x1  y2  =  − sin φ cos φ 0   x2  0 0 1 y3 x3 where the rotation angle φ is called the longitude of the ascending node. Second, a rotation about the y1 axis taking the yi axes to the yi axes      1 0 0 y1 y1  y2  =  0 cos θ sin θ   y2  y3 y3 0 − sin θ cos θ where the rotation angle θ is called the angle of inclination of the orbital plane. Finally, a rotation about the y3 axis taking the yi axes to the x ¯i axes       cos ψ sin ψ 0 y1 x ¯1 x ¯2  =  − sin ψ cos ψ 0   y2  0 0 1 x ¯3 y3 where the rotation angle ψ is called the argument of perigee. The Euler angle θ is the angle x ¯3 0x3 , the angle x1 . These angles are illustrated in the figure 2.2-6. Note also that φ is the angle x1 0y1 and ψ is the angle y1 0¯ ˙ θ, ˙ ψ˙ in the directions the rotation vectors associated with these transformations are vectors of magnitude φ, indicated in the figure 2.2-6.

Figure 2.2-6. Euler angles. By combining the above transformations there results the transformation equations (2.2.50)      cos ψ cos φ − cos θ sin φ sin ψ cos ψ sin φ + cos θ cos φ sin ψ sin ψ sin θ x1 x ¯1 x ¯2  =  − sin ψ cos φ − cos θ sin φ cos ψ − sin ψ sin φ + cos θ cos φ cos ψ cos ψ sin θ   x2  . sin θ sin φ − sin θ cos φ cos θ x ¯3 x3 It is left as an exercise to verify that the transformation matrix is orthogonal and the components 7ji satisfy the relations (2.2.51).

203 Consider the velocity of a point which is rotating with the rigid body. Denote by vi = vi (S), for i = 1, 2, 3, the velocity components relative to the S reference frame and by v i = v i (S), i = 1, 2, 3 the velocity components of the same point relative to the body-fixed axes. In terms of the basis vectors we can write dxi
= v1 (S) e ˆi ˆ1 + v2 (S) e ˆ2 + v3 (S) e ˆ3 = e V dt

(2.2.52)

as the velocity in the S reference frame. Similarly, we write dxi

= v (S)
V e1 + v 2 (S)
e2 + v 3 (S)
e3 = ei 1 dt

(2.2.53)

as the velocity components relative to the body-fixed reference frame. There are occasions when it is desirable
in the S frame of reference and V
in the S frame of reference. In these instances we can write to represent V
= v1 (S)
V e1 + v2 (S)
e2 + v3 (S)
e3

(2.2.54)


= v (S) e ˆ1 + v 2 (S) e ˆ2 + v 3 (S) e ˆ3 . V 1

(2.2.55)

and

Here we have adopted the notation that vi (S) are the velocity components relative to the S reference frame and vi (S) are the same velocity components relative to the S reference frame. Similarly, v i (S) denotes the velocity components relative to the S reference frame, while v i (S) denotes the same velocity components relative to the S reference frame.
are vectors and so their components are first order tensors and satisfy the transfor
and V Here both V mation laws v i (S) = 7ji vj (S) = 7ji x˙ j

and

vi (S) = 7ij v j (S) = 7ij x˙ j .

(2.2.56)

The equations (2.2.56) define the relative velocity components as functions of time t. By differentiating the equations (2.2.50) we obtain dxi = v i (S) = 7ji x˙ j + 7˙ji xj dt

(2.2.57)

dxi = vi (S) = 7ij x˙ j + 7˙ij xj . dt

(2.2.58)

and

Multiply the equation (2.2.57) by 7mi and multiply the equation (2.2.58) by 7im and derive the relations vm (S) = vm (S) + 7mi 7˙ji xj

(2.2.59)

v m (S) = v m (S) + 7im 7˙ij xj .

(2.2.60)

and

The equations (2.2.59) and (2.2.60) describe the transformation laws of the velocity components upon changing from the S to the S reference frame. These equations can be expressed in terms of the angular velocity by making certain substitutions which are now defined. The first order angular velocity vector ωi is related to the second order skew-symmetric angular velocity tensor ωij by the defining equation ωmn = eimn ωi .

(2.2.61)

204 The equation (2.2.61) implies that ωi and ωij are dual tensors and ωi =

1 eijk ωjk . 2

Also the velocity of a point which is rotating about the origin relative to the S frame of reference is vi (S) = eijk ωj xk which can also be written in the form vm (S) = −ωmk xk . Since the barred axes rotate with the rigid body, then a particle in the barred reference frame will have vm (S) = 0, since the coordinates of a point in the rigid body will be constants with respect to this reference frame. Consequently, we write equation (2.2.59) in the form 0 = vm (S) + 7mi 7˙ji xj which implies that vm (S) = −7mi 7˙ji xj = −ωmk xk

or ωmj = ωmj (S, S) = 7mi 7˙ji .

This equation is interpreted as describing the angular velocity tensor of S relative to S. Since ωij is a tensor, it can be represented in the barred system by ω mn (S, S) = 7im 7jn ωij (S, S) = 7im 7jn 7is 7˙js = δms 7jn 7˙js

(2.2.62)

= 7jn 7˙jm By differentiating the equations (2.2.51) it is an easy exercise to show that ωij is skew-symmetric. The second order angular velocity tensor can be used to write the equations (2.2.59) and (2.2.60) in the forms vm (S) = vm (S) + ωmj (S, S)xj

(2.2.63)

v m (S) = v m (S) + ωjm (S, S)xj The above relations are now employed to derive the celebrated Euler’s equations of motion of a rigid body. Euler’s Equations of Motion We desire to find the equations of motion of a rigid body which is subjected to external forces. These equations are the formulas (2.2.49), and we now proceed to write these equations in a slightly different form. Similar to the introduction of the angular velocity tensor, given in equation (2.2.61), we now introduce the following tensors 1. The fourth order moment of inertia tensor Imnst which is related to the second order moment of inertia tensor Iij by the equations Imnst =

1 ejmn eist Iij 2

or Iij =

1 Ipqrs eipq ejrs 2

(2.2.64)

2. The second order angular momentum tensor Hjk which is related to the angular momentum vector Hi by the equation Hi =

1 eijk Hjk 2

or Hjk = eijk Hi

(2.2.65)

3. The second order moment tensor Mjk which is related to the moment Mi by the relation Mi =

1 eijk Mjk 2

or Mjk = eijk Mi .

(2.2.66)

205 Now if we multiply equation (2.2.49) by erjk , then it can be written in the form dHij = Mij . (2.2.67) dt Similarly, if we multiply the equation (2.2.42) by eimn , then it can be expressed in the alternate form Hmn = eimn ωj Iji = Imnst ωst and because of this relation the equation (2.2.67) can be expressed as d (Iijst ωst ) = Mij . (2.2.68) dt We write this equation in the barred system of coordinates where I pqrs will be a constant and consequently its derivative will be zero. We employ the transformation equations Iijst = 7ip 7jq 7sr 7tk I pqrk ω ij = 7si 7tj ωst M pq = 7ip 7jq Mij and then multiply the equation (2.2.68) by 7ip 7jq and simplify to obtain

d 7iα 7jβ I αβrk ωrk = M pq . dt Expand all terms in this equation and take note that the derivative of the I αβrk is zero. The expanded 7ip 7jq

equation then simplifies to dω rk + (δαu δpv δβq + δpα δβu δqv ) I αβrk ω rk ωuv = M pq . (2.2.69) dt Substitute into equation (2.2.69) the relations from equations (2.2.61),(2.2.64) and (2.2.66), and then multiply I pqrk

by empq and simplify to obtain the Euler’s equations of motion dω i − etmj I ij ω i ωt = M m . (2.2.70) dt Dropping the bar notation and performing the indicated summations over the range 1,2,3 we find the I im

Euler equations have the form dω1 dω2 dω3 + I21 + I31 dt dt dt dω1 dω2 dω3 + I22 + I32 I12 dt dt dt dω1 dω2 dω3 + I23 + I33 I13 dt dt dt In the special case where I11

+ (I13 ω1 + I23 ω2 + I33 ω3 ) ω2 − (I12 ω1 + I22 ω2 + I32 ω3 ) ω3 = M1 + (I11 ω1 + I21 ω2 + I31 ω3 ) ω3 − (I13 ω1 + I23 ω2 + I33 ω3 ) ω1 = M2

(2.2.71)

+ (I12 ω1 + I22 ω2 + I32 ω3 ) ω1 − (I11 ω1 + I21 ω2 + I31 ω3 ) ω2 = M3 . the barred axes are principal axes, then Iij = 0 for i = j and the Euler’s

equations reduces to the system of nonlinear differential equations dω1 + (I33 − I22 )ω2 ω3 = M1 dt dω2 (2.2.72) + (I11 − I33 )ω3 ω1 = M2 I22 dt dω3 + (I22 − I11 )ω1 ω2 = M3 . I33 dt In the case of constant coefficients and constant moments the solutions of the above differential equations I11

can be expressed in terms of Jacobi elliptic functions.

206 EXERCISE 2.2
1. Find a set of parametric equations for the straight line which passes through the points P1 (1, 1, 1) and P2 (2, 3, 4). Find the unit tangent vector to any point on this line.
2. Consider the space curve x =

1 2

sin2 t, y = 12 t − 14 sin 2t, z = sin t where t is a parameter. Find the unit

vectors T i , B i , N i , i = 1, 2, 3 at the point where t = π.
3. A claim has been made that the space curve x = t, y = t2 , z = t3 intersects the plane 11x-6y+z=6 in three distinct points. Determine if this claim is true or false. Justify your answer and find the three points of intersection if they exist.
4. Find a set of parametric equations xi = xi (s1 , s2 ), i = 1, 2, 3 for the plane which passes through the points P1 (3, 0, 0), P2 (0, 4, 0) and P3 (0, 0, 5). Find a unit normal to this plane. 2 t find the equation of the tangent plane to the curve at the π point where t = π/4. Find the equation of the tangent line to the curve at the point where t = π/4.


5. For the helix x = sin t y = cos t z =

∂T = M grm x˙ m . ∂ x˙ r     d ∂T ∂grm n m m
7. Verify the derivative ¨ + x˙ x˙ = M grm x . dt ∂ x˙ r ∂xn


6. Verify the derivative


8. Verify the derivative

∂T 1 ∂gmn m n = M x˙ x˙ . ∂xr 2 ∂xr


9. Use the results from problems 6,7 and 8 to derive the Lagrange’s form for the equations of motion defined by equation (2.2.23).
10. Express the generalized velocity and acceleration in cylindrical coordinates (x1 , x2 , x3 ) = (r, θ, z) and show

 2 dθ d2 r δV 1 = 2 −r f = δt dt dt 2 2 θ d dr dθ δV 2 = 2 + f2 = δt dt r dt dt d2 z δV 3 3 = 2 f = δt dt Find the physical components of velocity and acceleration in cylindrical coordinates and show dr dx1 = V = dt dt dθ dx2 2 = V = dt dt 3 dz dx = V3 = dt dt 1

dr Vr = dt dθ Vθ =r dt dz Vz = dt

1

 2 dθ d2 r fr = 2 − r dt dt 2 d θ dr dθ fθ =r 2 + 2 dt dt dt d2 z fz = 2 dt

207
11. Express the generalized velocity and acceleration in spherical coordinates (x2 , x2 , x3 ) = (ρ, θ, φ) and show dρ dx1 = V = dt dt 2 dθ dx = V2 = dt dt 3 dφ dx = V3 = dt dt 1

f1 =

d2 ρ δV 1 = 2 −ρ δt dt



dθ dt

2

 − ρ sin2 θ

dφ dt

2

 2 dφ d2 θ δV 2 2 dρ dθ = 2 − sin θ cos θ f = + δt dt dt ρ dt dt 3 2 d dρ dφ dθ dφ φ δV 2 = 2 + + 2 cot θ f3 = δt dt ρ dt dt dt dt 2

Find the physical components of velocity and acceleration in spherical coordinates and show d2 ρ fρ = 2 − ρ dt

dρ Vρ = dt dθ Vθ =ρ dt Vφ =ρ sin θ



dθ dt



2 − ρ sin θ 2

dφ dt

2

 2 d2 θ dφ dρ dθ − ρ sin θ cos θ +2 2 dt dt dt dt 2 dθ dφ d φ dρ dφ + 2ρ cos θ fφ =ρ sin θ 2 + 2 sin θ dt dt dt dt dt fθ =ρ

dφ dt


12. Expand equation (2.2.39) and write out all the components of the moment of inertia tensor Iij .
13. For ρ the density of a continuous material and dτ an element of volume inside a region R where the material is situated, we write ρdτ as an element of mass inside R. Find an equation which describes the center of mass of the region R.
14. Use the equation (2.2.68) to derive the equation (2.2.69).
15. Drop the bar notation and expand the equation (2.2.70) and derive the equations (2.2.71).
16. Verify the Euler transformation, given in example 2.2-5, is orthogonal.
17. For the pulley and mass system illustrated in the figure 2.2-7 let a = the radius of each pulley. 71 = the length of the upper chord. 72 = the length of the lower chord. Neglect the weight of the pulley and find the equations of motion for the pulley mass system.

208

Figure 2.2-7. Pulley and mass system
18. Let φ =

ds dt ,

where s is the arc length between two points on a curve in generalized coordinates.  (a) Write the arc length in general coordinates as ds = gmn x˙ m x˙ n dt and show the integral I, defined by equation (2.2.35), represents the distance between two points on a curve.

(b) Using the Euler-Lagrange equations (2.2.36) show that the shortest distance between two points in a  ' d2 s i 2 x˙ j x˙ k = x˙ i dt generalized space is the curve defined by the equations: x¨i + ds jk dt  ' i dxj dxk d2 xi = 0, for (c) Show in the special case t = s the equations in part (b) reduce to + j k ds ds ds2 i = 1, . . . , N. An examination of equation (1.5.51) shows that the above curves are geodesic curves. (d) Show that the shortest distance between two points in a plane is a straight line. (e) Consider two points on the surface of a cylinder of radius a. Let u1 = θ and u2 = z denote surface coordinates in the two dimensional space defined by the surface of the cylinder. Show that the shortest  distance between the points where θ = 0, z = 0 and θ = π, z = H is L = a2 π 2 + H 2 .
19. For T = 12 mgij v i v j the kinetic energy of a particle and V the potential energy of the particle show that T + V = constant. Hint:

∂V mfi = Qi = − ∂x i,

i = 1, 2, 3 and

dxi dt

= x˙ i = v i , i = 1, 2, 3.


20. Define H = T + V as the sum of the kinetic energy and potential energy of a particle. The quantity H = H(xr , pr ) is called the Hamiltonian of the particle and it is expressed in terms of: • the particle position xi and • the particle momentum pi = mvi = mgij x˙ j . Here xr and pr are treated as independent variables. (a) Show that the particle momentum is a covariant tensor of rank 1. (b) Express the kinetic energy T in terms of the particle momentum. ∂T . (c) Show that pi = ∂ x˙ i

209

Figure 2.2-8. Compound pendulum dpi dxi ∂H ∂H and = = − i . These are a set of differential equations describing the dt ∂pi dt ∂x position change and momentum change of the particle and are known as Hamilton’s equations of motion

(d) Show that

for a particle.
21.

Let

δT i δs

= κN i and

B i = Fijk Tj Nk and find

δB i δs

δN i δs

= τ B i − κT i and calculate the intrinsic derivative of the cross product

in terms of the unit normal vector.


22. For T the kinetic energy of a particle and V the potential energy of a particle, define the Lagrangian 1 L = L(xi , x˙ i ) = T − V = M gij x˙ i x˙ j − V as a function of the independent variables xi , x˙ i . Define the 2 1 ij g pi pj + V, as a function of the independent variables xi , pi , Hamiltonian H = H(xi , pi ) = T + V = 2M where pi is the momentum vector of the particle and M is the mass of the particle. ∂T (a) Show that pi = . ∂ x˙ i ∂L ∂H =− i (b) Show that i ∂x ∂x
23.

When the Euler angles, figure 2.2-6, are applied to the motion of rotating objects, θ is the angle

of nutation, φ is the angle of precession and ψ is the angle of spin. Take projections and show that the time derivative of the Euler angles are related to the angular velocity vector components ωx , ωy , ωz by the relations

ωx = θ˙ cos ψ + φ˙ sin θ sin ψ ωy = −θ˙ sin ψ + φ˙ sin θ cos ψ ωz = ψ˙ + φ˙ cos θ

where ωx , ωy , ωz are the angular velocity components along the x1 , x2 , x3 axes.
24. Find the equations of motion for the compound pendulum illustrated in the figure 2.2-8.

210 GM m
25. Let F
= − 3
r denote the inverse square law force of attraction between the earth and sun, with r G a universal constant, M the mass of the sun, m the mass of the earth and 'rr a unit vector from origin at the center of the sun pointing toward the earth. (a) Write down Newton’s second law, in both vector d (
r ×
v ) =
0 and and tensor form, which describes the motion of the earth about the sun. (b) Show that dt r
consequently
r ×
v =
r × d' dt = h = a constant.
26. Construct a set of axes fixed and attached to an airplane. Let the x axis be a longitudinal axis running from the rear to the front of the plane along its center line. Let the y axis run between the wing tips and let the z axis form a right-handed system of coordinates. The y axis is called a lateral axis and the z axis is called a normal axis. Define pitch as any angular motion about the lateral axis. Define roll as any angular motion about the longitudinal axis. Define yaw as any angular motion about the normal axis. Consider two sets of axes. One set is the x, y, z axes attached to and moving with the aircraft. The other set of axes is denoted X, Y, Z and is fixed in space ( an inertial set of axes). Describe the pitch, roll and yaw of an aircraft with respect to the inertial set of axes. Show the transformation is orthogonal. Hint: Consider pitch with respect to the fixed axes, then consider roll with respect to the pitch axes and finally consider yaw with respect to the roll axes. This produces three separate transformation matrices which can then be combined to describe the motions of pitch, roll and yaw of an aircraft. i i
27. In Cartesian coordinates let Fi = Fi (x1 , x2 , x3 ) denote a force field  andlet x2=  x (t) denote a curve dxi d 1 dxi C. (a) Show Newton’s second law implies that along the curve C m = Fi (x1 , x2 , x3 ) dt 2 dt dt (no summation on i) and hence (   2  2 2  3 2 )  dx1 dx dx d 1 dx1 dx2 dx3 d 1 2 m mv = F1 + F2 + F3 + + = dt 2 dt dt dt dt 2 dt dt dt

(b) Consider two points on the curve C, say point A, xi (tA ) and point B, xi (tB ) and show that the work done in moving from A to B in the force field Fi is tB  B 1 2 mv = Fi dx1 + F2 dx2 + F3 dx3 2 A tA where the right hand side is a line integral along the path C from A to B. (c) Show that if the force field is derivable from a potential function U (x1 , x2 , x3 ) by taking the gradient, then the work done is independent of the path C and depends only upon the end points A and B.
28. Find the Lagrangian equations of motion of a spherical pendulum which consists of a bob of mass m suspended at the end of a wire of length 7, which is free to swing in any direction subject to the constraint that the wire length is constant. Neglect the weight of the wire and show that for the wire attached to the origin of a right handed x, y, z coordinate system, with the z axis downward, φ the angle between the wire and the z axis and θ the angle of rotation of the bob from the y axis, that there results the equations of  2   dθ d dθ g d2 φ 2 motion − sin φ cos φ + sin φ = 0 sin φ =0 and dt dt dt2 dt 7

211
29. In Cartesian coordinates show the Frenet formulas can be written dT
=
δ × T
, ds


dN
=
δ × N, ds


dB
=
δ × B ds


where
δ is the Darboux vector and is defined
δ = τ T
+ κB.
30. Consider the following two cases for rigid body rotation. Case 1: Rigid body rotation about a fixed line which is called the fixed axis of rotation. Select a point 0 ˆR on this fixed axis and denote by
e a unit vector from 0 in the direction of the fixed line and denote by e a unit vector which is perpendicular to the fixed axis of rotation. The position vector of a general point ˆR in the rigid body can then be represented by a position vector from the point 0 given by
r = h
e + r0 e

(a) (b)

(a) (b) (c) (d) (e)

ˆR is fixed in and rotating with the rigid body. where h, r0 and
e are all constants and the vector e dθ ˆR as it rotates about the scalar angular change with respect to time of the vector e Denote by ω = dt dθ d
e) = e where θ
e is defined as the the fixed line and define the vector angular velocity as
ω = (θ
dt dt vector angle of rotation. ˆR de ˆR . =
e× e Show that dθ ˆR ˆR dθ de
= d
r = r0 d e ˆR ) =
ω × (h
ˆR ) =
ω ×
r. Show that V = r0 =
ω × (r0 e e + r0 e dt dt dθ dt ˆ1 which is Case 2: Rigid body rotation about a fixed point 0. Construct at point 0 the unit vector e ˆ1 de fixed in and rotating with the rigid body. From pages 80,87 we know that must be perpendicular dt ˆ1 de ˆ1 and so we can define the vector e ˆ2 as a unit vector which is in the direction of to e such that dt ˆ1 de ˆ2 for some constant α. We can then define the unit vector e ˆ3 from e ˆ3 = e ˆ1 × e ˆ2 . = αe dt ˆ3 de ˆ3 , is also perpendicular to e ˆ1 . , which must be perpendicular to e Show that dt ˆ3 ˆ3 de de ˆ2 for some constant β. Show that can be written as =βe dt dt ˆ2 de ˆ3 − β e ˆ3 × e ˆ1 show that ˆ1 ) × e ˆ2 ˆ2 = e = (α e From e dt ˆ1 ˆ2 ˆ3 de de de ˆ1 , ˆ2 , ˆ3 ˆ3 − β e ˆ1 and show that =
ω × e =ω
×e =
ω × e Define
ω = αe dt dt dt ˆ1 + y e ˆ2 + z e ˆ3 denote an arbitrary point within the rigid body with respect to the point 0. Let
r = x e d
r Show that =
ω ×
r. dt ˆ3 and e ˆ1 are constantly changing. Note that in Case 2 the direction of ω
is not fixed as the unit vectors e In this case the direction
ω is called an instantaneous axis of rotation and
ω , which also can change in magnitude and direction, is called the instantaneous angular velocity.

211 §2.3 BASIC EQUATIONS OF CONTINUUM MECHANICS Continuum mechanics is the study of how materials behave when subjected to external influences. External influences which affect the properties of a substance are such things as forces, temperature, chemical reactions, and electric phenomena. Examples of forces are gravitational forces, electromagnetic forces, and mechanical forces. Solids deform under external forces and so deformations are studied. Fluids move under external forces and so the velocity of the fluid is studied. A material is considered to be a continuous media which is a collection of material points interconnected by internal forces (forces between the atoms making up the material). We concentrate upon the macroscopic properties rather than the microscopic properties of the material. We treat the material as a body which is homogeneous and continuous in its makeup. In this introduction we will only consider solid media and liquid media. In general, most of the ideas and concepts developed in this section can be applied to any type of material which is assumed to be a collection of material points held together by some kind of internal forces. An elastic material is one which deforms under applied forces in such a way that it will return to its original unloaded state when the applied forces are removed. When a linear relation exists between the applied forces and material displacements, then the material is called a linear elastic material. In contrast, a plastic material is one which deforms under applied forces in such a way that it does not return to its original state after removal of the applied forces. Plastic materials will always exhibit some permanent deformation after removal of the applied forces. An elastic material is called homogeneous if it has the same properties throughout. An isotropic material has the same properties, at a point, in all directions about the point. In this introduction we develop the basic mathematical equations which describe how a continuum behaves when subjected to external forces. We shall discover that there exists a set of basic equations associated with all continuous material media. These basic equations are developed for linear elastic materials and applied to solids and fluids in later sections. Introduction to Elasticity Take a rubber band, which has a rectangular cross section, and mark on it a parallelepiped having a length 7, a width w and a height h, as illustrated in the figure 2.3-1. Now apply a force F to both ends of the parallelepiped cross section on the rubber band and examine what happens to the parallelepiped. You will see that: 1.

7 increases by an amount ∆7.

2.

w decreases by an amount ∆w.

3.

h decreases by an amount ∆h.

There are many materials which behave in a manner very similar to the rubber band. Most materials, when subjected to tension forces will break if the change ∆7 is only one or two percent of the original length. The above example introduces us to several concepts which arise in the study of materials when they are subjected to external forces. The first concept is that of strain which is defined as strain =

change in length , original length

(dimensionless).

212

Figure 2.3-1. Section of a rubber band When the force F is applied to our rubber band example there arises the strains ∆7 , 7

∆w , w

∆h . h

The second concept introduced by our simple example is stress. Stress is defined as a force per unit area. In particular, Force force , with dimension of . Area over which force acts unit area We will be interested in studying stress and strain in homogeneous, isotropic materials which are in equilibstress =

rium with respect to the force system acting on the material. Hooke’s Law For linear elastic materials, where the forces are all one dimensional, the stress and strains are related by Hooke’s law which has two parts. The Hooke’s law, part one, states that stress is proportional to strain in the stretch direction, where the Young’s modulus E is the proportionality constant. This is written   ∆7 F =E Hooke’s law part 1 . (2.3.1) A 7 A graph of stress vs strain is a straight line with slope E in the linear elastic range of the material. The Hooke’s law, part two, involves the fact that there is a strain contraction perpendicular to the stretch direction. The strain contraction is the same for both the width and height and is proportional to the strain in the stretch direction. The proportionality constant being the Poisson’s ratio ν. Hooke’s law part 2

∆h ∆7 ∆w = = −ν , w h 7

0<ν<

1 . 2

(2.3.2)

The proportionality constants E and ν depend upon the material being considered. The constant ν is called the Poisson’s ratio and it is always a positive number which is less than one half. Some representative values for E and ν are as follows. Various types of steel Various types of aluminium

28 (10)6 psi ≤ E ≤ 30 (10)6 psi 9.0 (10)6 psi ≤ E ≤ 11.0 (10)6 psi

0.26 ≤ ν ≤ 0.31 0.3 ≤ ν ≤ 0.35

213

Figure 2.3-2. Typical Stress-strain curve. Consider a typical stress-strain curve, such as the one illustrated in the figure 2.3-2, which is obtained by placing a material in the shape of a rod or wire in a machine capable of performing tensile straining at a low rate. The engineering stress is the tensile force F divided by the original cross sectional area A0 . Note that during a tensile straining the cross sectional area A of the sample is continually changing and getting smaller so that the actual stress will be larger than the engineering stress. Observe in the figure 2.3-2 that the stress-strain relation remains linear up to a point labeled the proportional limit. For stress-strain points in this linear region the Hooke’s law holds and the material will return to its original shape when the loading is removed. For points beyond the proportional limit, but less than the yield point, the material no longer obeys Hooke’s law. In this nonlinear region the material still returns to its original shape when the loading is removed. The region beyond the yield point is called the plastic region. At the yield point and beyond, there is a great deal of material deformation while the loading undergoes only small changes. For points in this plastic region, the material undergoes a permanent deformation and does not return to its original shape when the loading is removed. In the plastic region there usually occurs deformation due to slipping of atomic planes within the material. In this introductory section we will restrict our discussions of material stress-strain properties to the linear region. EXAMPLE 2.3-1. (One dimensional elasticity)

Consider a circular rod with cross sectional area A

which is subjected to an external force F applied to both ends. The figure 2.3-3 illustrates what happens to the rod after the tension force F is applied. Consider two neighboring points P and Q on the rod, where P is at the point x and Q is at the point x + ∆x. When the force F is applied to the rod it is stretched and P moves to P  and Q moves to Q . We assume that when F is applied to the rod there is a displacement function u = u(x, t) which describes how each point in the rod moves as a function of time t. If we know the displacement function u = u(x, t) we would then be able to calculate the following distances in terms of the displacement function P P  = u(x, t),

0P  = x + u(x, t),

QQ = u(x + ∆x, t)

0Q = x + ∆x + u(x + ∆x, t).

214

Figure 2.3-3. One dimensional rod subjected to tension force The strain associated with the distance 7 = ∆x = P Q is ∆7 P  Q − P Q (0Q − 0P  ) − (0Q − 0P ) = = 7 PQ PQ [x + ∆x + u(x + ∆x, t) − (x + u(x, t))] − [(x + ∆x) − x] e= ∆x u(x + ∆x, t) − u(x, t) e= . ∆x e=

Use the Hooke’s law part(i) and write F u(x + ∆x, t) − u(x, t) =E . A ∆x Taking the limit as ∆x approaches zero we find that F ∂u(x, t) =E . A ∂x Hence, the stress is proportional to the spatial derivative of the displacement function.

Normal and Shearing Stresses Let us consider a more general situation in which we have some material which can be described as having a surface area S which encloses a volume V. Assume that the density of the material is L and the material is homogeneous and isotropic. Further assume that the material is subjected to the forces
b and
t (n) where
b is a body force per unit mass [f orce/mass], and
t (n) is a surface traction per unit area [f orce/area]. The superscript (n) on the vector is to remind you that we will only be interested in the normal component of the surface forces. We will neglect body couples, surface couples, and concentrated forces or couples that act at a single point. If the forces described above are everywhere continuous we can calculate the resultant
acting on the material by constructing various surface and volume integrals force F
and resultant moment M which sum the forces acting upon the material. In particular, the resultant force F
acting on our material can be described by the surface and volume integrals:  
t (n) dS + F
= L
b dτ S

V

(2.3.3)

215

Figure 2.3-4. Stress vectors acting upon an element of volume which is a summation of all the body forces and surface tractions acting upon our material. Here L is the density of the material, dS is an element of surface area, and dτ is an element of volume.
about the origin is similarly expressed as The resultant moment M 


r ×
t (n) dS +


= M

L(
r ×
b) dτ.

S

(2.3.4)

V

The global motion of the material is governed by the Euler equations of motion. • The time rate of change of linear momentum equals the resultant force or d dt

  
t (n) dS + L
v dτ = F
= L
b dτ.

   V

S

(2.3.5)

V

This is a statement concerning the conservation of linear momentum. • The time rate of change of angular momentum equals the resultant moment or d dt

  






= L
r ×
v dτ = M V


r ×
t (n) dS +

S

L(
r ×
b) dτ.

(2.3.6)

V

This is a statement concerning conservation of angular momentum. The Stress Tensor Define the stress vectors


t 1 = σ 11 e ˆ1 + σ 12 e ˆ2 + σ 13 e ˆ3
t 2 = σ 21 e ˆ1 + σ 22 e ˆ2 + σ 23 e ˆ3

(2.3.7)


t 3 = σ 31 e ˆ1 + σ 32 e ˆ2 + σ 33 e ˆ3 , where σ ij , i, j = 1, 2, 3 is the stress tensor acting at each point of the material. The index i indicates the coordinate surface xi = a constant, upon which
t i acts. The second index j denotes the direction associated with the components of
t i .

216

Figure 2.3-5. Stress distribution at a point For i = 1, 2, 3 we adopt the convention of sketching the components of
t i in the positive directions if the exterior normal to the surface xi = constant also points in the positive direction. This gives rise to the figure 2.3-4 which illustrates the stress vectors acting upon an element of volume in rectangular Cartesian coordinates. The components σ 11 , σ 22 , σ 33 are called normal stresses while the components σ ij , i = j are called shearing stresses. The equations (2.3.7) can be written in the more compact form using the indicial notation as
t i = σ ij e ˆj ,

i, j = 1, 2, 3.

(2.3.8)

If we know the stress distribution at three orthogonal interfaces at a point P in a solid body, we can then determine the stress at the point P with respect to any plane passing through the point P. With reference to the figure 2.3-5, consider an arbitrary plane passing through the point P which lies within the material body being considered. Construct the elemental tetrahedron with orthogonal axes parallel to the x1 = x, x2 = y and x3 = z axes. In this figure we have the following surface tractions: −
t 1

on the surface 0BC

−
t 2

on the surface 0AC

−
t 3

on the surface 0AB


(n)

on the surface ABC

t

The superscript parenthesis n is to remind you that this surface traction depends upon the orientation of the plane ABC which is determined by a unit normal vector having the direction cosines n1 , n2 and n3 .

217 Let ∆S1 = the surface area 0BC ∆S2 = the surface area 0AC ∆S3 = the surface area 0AB ∆S = the surface area ABC . These surface areas are related by the relations ∆S1 = n1 ∆S,

∆S2 = n2 ∆S,

∆S3 = n3 ∆S

(2.3.9)

which can be thought of as projections of ∆S upon the planes xi =constant for i = 1, 2, 3. Cauchy Stress Law Let tj (n) denote the components of the surface traction on the surface ABC. That is, we let
t (n) = t1 (n) e ˆ1 + t2 (n) e ˆ2 + t3 (n) e ˆ3 = tj (n) e ˆj .

(2.3.10)

It will be demonstrated that the components tj (n) of the surface traction forces
t (n) associated with a plane through P and having the unit normal with direction cosines n1 , n2 and n3 , must satisfy the relations tj (n) = ni σ ij ,

i, j = 1, 2, 3.

(2.3.11)

This relation is known as the Cauchy stress law. Proof: Sum the forces acting on the elemental tetrahedron in the figure 2.3-5. If the body is in equilibrium, then the sum of these forces must equal zero or 1 (−
t ∆S1 ) + (−
t 2 ∆S2 ) + (−
t 3 ∆S3 ) +
t (n) ∆S = 0.

(2.3.12)

The relations in the equations (2.3.9) are used to simplify the sum of forces in the equation (2.3.12). It is readily verified that the sum of forces simplifies to
t (n) = n1
t 1 + n2
t 2 + n3
t 3 = ni
t i .

(2.3.13)

Substituting in the relations from equation (2.3.8) we find
t (n) = tj (n) e ˆj = ni σ ij e ˆj ,

i, j = 1, 2, 3

(2.3.14)

or in component form tj (n) = ni σ ij which is the Cauchy stress law.

(2.3.15)

218 Conservation of Linear Momentum Let R denote a region in space where there exists a material volume with density L having surface tractions and body forces acting upon it. Let v i denote the velocity of the material volume and use Newton’s second law to set the time rate of change of linear momentum equal to the forces acting upon the volume as in (2.3.5). We find δ δt



  



j

 σ ij ni dS +

Lv dτ = R

S

Lbj dτ. R

Here dτ is an element of volume, dS is an element of surface area, bj are body forces per unit mass, and σ ij are the stresses. Employing the Gauss divergence theorem, the surface integral term is replaced by a volume integral and Newton’s second law is expressed in the form   j  Lf − Lbj − σ ij ,i dτ = 0,

(2.3.16)

R

where f j is the acceleration from equation (1.4.54). Since R is an arbitrary region, the equation (2.3.16) implies that σ ij ,i + Lbj = Lf j .

(2.3.17)

This equation arises from a balance of linear momentum and represents the equations of motion for material in a continuum. If there is no velocity term, then equation (2.3.17) reduces to an equilibrium equation which can be written σ ij ,i + Lbj = 0.

(2.3.18)

This equation can also be written in the covariant form g si σms,i + Lbm = 0, which reduces to σij,j + Lbi = 0 in Cartesian coordinates. The equation (2.3.18) is an equilibrium equation and is one of our fundamental equations describing a continuum. Conservation of Angular Momentum The conservation of angular momentum equation (2.3.6) has the Cartesian tensors representation d dt

   Leijk xj vk dτ = eijk xj σpk np dS + Leijk xj bk dτ.

   R

S

(2.3.19)

R

Employing the Gauss divergence theorem, the surface integral term is replaced by a volume integral to obtain   R

'  d ∂ dτ = 0. eijk L (xj vk ) − eijk Lxj bk + p (xj σpk ) dt ∂x

Since equation (2.3.20) must hold for all arbitrary volumes R we conclude that  ' ∂σpk d eijk L (xj vk ) = eijk Lxj bk + xj + σjk dt ∂xp

(2.3.20)

219

Figure 2.3-6. Shearing parallel to the y axis which can be rewritten in the form   ∂σpk dvk eijk σjk + xj ( p + Lbk − L ) − Lvj vk = 0. ∂x dt

(2.3.21)

In the equation (2.3.21) the middle term is zero because of the equation (2.3.17). Also the last term in (2.3.21) is zero because eijk vj vk represents the cross product of a vector with itself. The equation (2.3.21) therefore reduces to eijk σjk = 0,

(2.3.22)

which implies (see exercise 1.1, problem 22) that σij = σji for all i and j. Thus, the conservation of angular momentum requires that the stress tensor be symmetric. Consequently, there are only 6 independent stress components to be determined. This is another fundamental law for a continuum. Strain in Two Dimensions Consider the matrix equation

   x 1 = y β

0 1

  x y

(2.3.23)

which can be used to transform points (x, y) to points (x, y). When this transformation is applied to the unit square illustrated in the figure 2.3-6(a) we obtain the geometry illustrated in the figure 2.3-6(b) which represents a shearing parallel to the y axis. If β is very small, we can use the approximation tan β ≈ β and then this transformation can be thought of as a rotation of the element P1 P2 through an angle β to the position P1 P2 when the barred axes are placed atop the unbarred axes. Similarly, the matrix equation

   x 1 = y 0

α 1

  x y

(2.3.24)

can be used to represent a shearing of the unit square parallel to the x axis as illustrated in the figure 2.3-7(b).

220

Figure 2.3-7. Shearing parallel to the x axis

Figure 2.3-8. Shearing parallel to x and y axes Again, if α is very small, we may use the approximation tan α ≈ α and interpret α as an angular rotation of the element P1 P4 to the position P1 P4 . Now let us multiply the matrices given in equations (2.3.23) and (2.3.24). Note that the order of multiplication is important as can be seen by an examination of the products          1 0 1 α x 1 α x x = = y β 1 0 1 y β 1 + αβ y          x 1 α 1 0 x 1 + αβ α x = = . y 0 1 β 1 y β 1 y

(2.3.25)

In equation (2.3.25) we will assume that the product αβ is very, very small and can be neglected. Then the order of matrix multiplication will be immaterial and the transformation equation (2.3.25) will reduce to    x 1 = y β

α 1

  x . y

(2.3.26)

Applying this transformation to our unit square we obtain the simultaneous shearing parallel to both the x and y axes as illustrated in the figure 2.3-8. This transformation can then be interpreted as the superposition of the two shearing elements depicted in the figure 2.3-9. For comparison, we consider also the transformation equation      x 1 0 x = y −α 1 y

(2.3.27)

221

Figure 2.3-9. Superposition of shearing elements

Figure 2.3-10. Rotation of element P1 P2 where α is very small. Applying this transformation to the unit square previously considered we obtain the results illustrated in the figure 2.3-10. Note the difference in the direction of shearing associated with the transformation equations (2.3.27) and (2.3.23) illustrated in the figures 2.3-6 and 2.3-10. If the matrices appearing in the equations (2.3.24) and (2.3.27) are multiplied and we neglect product terms because α is assumed to be very small, we obtain the matrix equation       x 1 α x 1 = = y −α 1 y 0 #

     0 α x 0 x . + −α 0 y 1 y $% & # $% & rotation identity

(2.3.28)

This can be interpreted as a superposition of the transformation equations (2.3.24) and (2.3.27) which represents a rotation of the unit square as illustrated in the figure 2.3-11. The matrix on the right-hand side of equation (2.3.28) is referred to as a rotation matrix. The ideas illustrated by the above simple transformations will appear again when we consider the transformation of an arbitrary small element in a continuum when it under goes a strain. In particular, we will be interested in extracting the rigid body rotation from a deformed element and treating this rotation separately from the strain displacement.

222

Figure 2.3-11. Rotation of unit square Transformation of an Arbitrary Element In two dimensions, we consider a rectangular element ABCD as illustrated in the figure 2.3-12. Let the points ABCD have the coordinates A(x, y),

B(x + ∆x, y),

C(x, y + ∆y),

D(x + ∆x, y + ∆y)

(2.3.29)

and denote by u = u(x, y),

v = v(x, y)

the displacement field associated with each of the points in the material continuum when it undergoes a deformation. Assume that the deformation of the element ABCD in figure 2.3-12 can be represented by the matrix equation

   x b11 = y b21

b12 b22

  x y

(2.3.30)

where the coefficients bij , i, j = 1, 2, 3 are to be determined. Let us define u = u(x, y) as the horizontal displacement of the point (x, y) and v = v(x, y) as the vertical displacement of the same point. We can now express the displacement of each of the points A, B, C and D in terms of the displacement field u = u(x, y) and v = v(x, y). Consider first the displacement of the point A to A . Here the coordinates (x, y) deform to the new coordinates x = x + u,

y = y + v.

That is, the coefficients bij must be chosen such that the equation 

x+u y+v



 =

b11 b21

b12 b22

  x y

(2.3.31)

is satisfied. We next examine the displacement of the point B to B  . This displacement is described by the coordinates (x + ∆x, y) transforming to (x, y), where x = x + ∆x + u(x + ∆x, y),

y = y + v(x + ∆x, y).

(2.3.32)

223

Figure 2.3-12. Displacement of element ABCD to A B  C  D Expanding u and v in (2.3.32) in Taylor series about the point (x, y) we find x = x + ∆x + u + y =y+v+

∂u ∆x + h.o.t. ∂x

(2.3.33)

∂v ∆x + h.o.t., ∂x

where h.o.t. denotes higher order terms which have been neglected. The equations (2.3.33) require that the coefficients bij satisfy the matrix equation 

x + u + ∆x + ∂u ∂x ∆x ∂v y + v + ∂x ∆x



 =

b11 b21

b12 b22



x + ∆x y

 .

(2.3.34)

224 The displacement of the point C to C  is described by the coordinates (x, y + ∆y) transforming to (x, y) where x = x + u(x, y + ∆y),

y = y + ∆y + v(x, y + ∆y).

(2.3.35)

Again we expand the displacement field components u and v in a Taylor series about the point (x, y) and find

∂u ∆y + h.o.t. ∂y ∂v ∆y + h.o.t. y = y + ∆y + v + ∂y x = x+u+

(2.3.36)

This equation implies that the coefficients bij must be chosen such that 

x + u + ∂u ∂y ∆y y + v + ∆y + ∂v ∂y ∆y



 =

b12 b22

b11 b21



x y + ∆y

 .

(2.3.37)

Finally, it can be verified that the point D with coordinates (x + ∆x, y + ∆y) moves to the point D  with coordinates x = x + ∆x + u(x + ∆x, y + ∆y),

y = y + ∆y + v(x + ∆x, y + ∆y).

(2.3.38)

Expanding u and v in a Taylor series about the point (x, y) we find the coefficients bij must be chosen to satisfy the matrix equation 

x + ∆x + u + y + ∆y + v +

∂u ∂x ∆x + ∂v ∂x ∆x +

 ∂u ∂y ∆y ∂v ∂y ∆y

 =

b11 b21

b12 b22



x + ∆x y + ∆y

 .

(2.3.39)

The equations (2.3.31),(2.3.34),(2.3.37) and (2.3.39) give rise to the simultaneous equations b11 x + b12 y = x + u b21 x + b22 y = y + v b11 (x + ∆x) + b12 y = x + u + ∆x +

∂u ∆x ∂x

∂v ∆x ∂x ∂u ∆y b11 x + b12 (y + ∆y) = x + u + ∂y ∂v b21 x + b22 (y + ∆y) = y + v + ∆y + ∆y ∂y ∂u ∂u b11 (x + ∆x) + b12 (y + ∆y) = x + ∆x + u + ∆x + ∆y ∂x ∂y ∂v ∂v b21 (x + ∆x) + b22 (y + ∆y) = y + ∆y + v + ∆x + ∆y. ∂x ∂y b21 (x + ∆x) + b22 y = y + v +

(2.3.40)

It is readily verified that the system of equations (2.3.40) has the solution b11 = 1 + b21

∂v = ∂x

∂u ∂x

b12 = b22

∂u ∂y

∂v . =1+ ∂y

(2.3.41)

225

Figure 2.3-13. Change in 45◦ line Hence the transformation equation (2.3.30) can be written as      ∂u 1 + ∂u x x ∂x ∂y . = ∂v ∂v y y 1 + ∂x ∂y

(2.3.42)

A physical interpretation associated with this transformation is obtained by writing it in the form:            e11 e12 ω11 ω12 1 0 x x x x + + , (2.3.43) = y 0 1 y e21 e22 ω21 ω22 y y # $% & # $% & # $% & strain matrix

identity

rotation matrix

  1 ∂u ∂v + e21 = e11 2 ∂y ∂x ∂v e12 e22 = ∂y are the elements of a symmetric matrix called the strain matrix and   1 ∂u ∂v ω11 = 0 − = ω 12   2 ∂y ∂x ∂u 1 ∂v ω21 = − ω22 = 0 2 ∂x ∂y where

∂u = ∂x  ∂u 1 ∂v + = 2 ∂x ∂y

(2.3.44)

(2.3.45)

are the elements of a skew symmetric matrix called the rotation matrix. The strain per unit length in the x-direction associated with the point A in the figure 2.3-12 is e11 =

∆x +

∂u ∂x ∆x

− ∆x

=

∆x and the strain per unit length of the point A in the y direction is ∆y +

∂v ∂y ∆y

− ∆y

∂u ∂x

(2.3.46)

∂v . (2.3.47) ∆y ∂y These are the terms along the main diagonal in the strain matrix. The geometry of the figure 2.3-12 implies e22 =

=

that tan β =

∂v ∂x ∆x , ∆x + ∂u ∂x ∆x

and

tan α =

∂u ∂y ∆y . ∂v ∆y + ∂y ∆y

(2.3.48)

For small derivatives associated with the displacements u and v it is assumed that the angles α and β are small and the equations (2.3.48) therefore reduce to the approximate equations ∂u ∂v tan α ≈ α = . (2.3.49) ∂x ∂y For a physical interpretation of these terms we consider the deformation of a small rectangular element which tan β ≈ β =

undergoes a shearing as illustrated in the figure 2.3-13.

226

Figure 2.3-14. Displacement field due to state of strain The quantity

 α+β =

∂u ∂v + ∂y ∂x

 = 2e12 = 2e21

(2.3.50)

is the change from a ninety degree angle due to the deformation and hence we can write 12 (α + β) = e12 = e21

as representing a change from a 45◦ angle due to the deformation. The quantities e21 , e12 are called the shear strains and the quantity γ12 = 2e12

(2.3.51)

is called the shear angle. In the equation (2.3.45), the quantities ω21 = −ω12 are the elements of the rigid body rotation matrix and are interpreted as angles associated with a rotation. The situation is analogous to the transformations and figures for the deformation of the unit square which was considered earlier. Strain in Three Dimensions The development of strain in three dimensions is approached from two different viewpoints. The first approach considers the derivation using Cartesian tensors and the second approach considers the derivation of strain using generalized tensors. Cartesian Tensor Derivation of Strain. Consider a material which is subjected to external forces such that all the points in the material undergo a deformation. Let (y1 , y2 , y3 ) denote a set of orthogonal Cartesian coordinates, fixed in space, which is used to describe the deformations within the material. Further, let ui = ui (y1 , y2 , y3 ), i = 1, 2, 3 denote a displacement field which describes the displacement of each point within the material. With reference to the figure 2.3-14 let P and Q denote two neighboring points within the material while it is in an unstrained state. These points move to the points P  and Q when the material is in a state of strain. We let yi , i = 1, 2, 3 represent the position vector to the general point P in the material, which is in an unstrained state, and denote by yi + ui , i = 1, 2, 3 the position vector of the point P  when the material is in a state of strain.

227 For Q a neighboring point of P which moves to Q when the material is in a state of strain, we have from the figure 2.3-14 the following vectors: position of P : yi ,

i = 1, 2, 3



position of P : yi + ui (y1 , y2 , y3 ), position of Q : yi + ∆yi ,

i = 1, 2, 3 (2.3.52)

i = 1, 2, 3

position of Q : yi + ∆yi + ui (y1 + ∆y1 , y2 + ∆y2 , y3 + ∆y3 ),

i = 1, 2, 3

Employing our earlier one dimensional definition of strain, we define the strain associated with the point P L − L0 in the direction P Q as e = , where L0 = P Q and L = P  Q . To calculate the strain we need to first L0 calculate the distances L0 and L. The quantities L20 and L2 are easily calculated by considering dot products of vectors. For example, we have L20 = ∆yi ∆yi , and the distance L = P  Q is the magnitude of the vector yi + ∆yi + ui (y1 + ∆y1 , y2 + ∆y2 , y3 + ∆y3 ) − (yi + ui (y1 , y2 , y3 )),

i = 1, 2, 3.

Expanding the quantity ui (y1 + ∆y1 , y2 + ∆y2 , y3 + ∆y3 ) in a Taylor series about the point P and neglecting higher order terms of the expansion we find that L2 = (∆yi +

∂ui ∂ui ∆ym )(∆yi + ∆yn ). ∂ym ∂yn

Expanding the terms in this expression produces the equation L2 = ∆yi ∆yi +

∂ui ∂ui ∂ui ∂ui ∆yi ∆yn + ∆ym ∆yi + ∆ym ∆yn . ∂yn ∂ym ∂ym ∂yn

Note that L and L0 are very small and so we express the difference L2 − L20 in terms of the strain e. We can write L2 − L20 = (L + L0 )(L − L0 ) = (L − L0 + 2L0 )(L − L0 ) = (e + 2)eL20 . Now for e very small, and e2 negligible, the above equation produces the approximation   1 ∂um L2 − L20 ∂un ∂ur ∂ur eL20 ≈ = + + ∆ym ∆yn . 2 2 ∂yn ∂ym ∂ym ∂yn The quantities emn =

  1 ∂um ∂un ∂ur ∂ur + + 2 ∂yn ∂ym ∂ym ∂yn

(2.3.53)

is called the Green strain tensor or Lagrangian strain tensor. To show that eij is indeed a tensor, we consider the transformation yi = 7ij yj +bi , where 7ji 7ki = δjk = 7ij 7ik . Note that from the derivative relation

∂yi ∂y j

= 7ij

and the transformation equations ui = 7ij uj , i = 1, 2, 3 we can express the strain in the barred system of coordinates. Performing the necessary calculations produces   1 ∂ui ∂uj ∂ur ∂ur eij = + + 2 ∂yj ∂yi ∂y i ∂yj   ∂yn ∂ ∂ym ∂ ∂yk ∂ ∂yt 1 ∂ (7ik uk ) + (7jk uk ) + (7rs us ) (7rm um ) = 2 ∂yn ∂yj ∂ym ∂yi ∂yk ∂y i ∂yt ∂y j   ∂um ∂uk ∂us ∂up 1 + 7jk 7mi + 7rs 7rp 7ki 7tj = 7im 7nj 2 ∂yn ∂ym ∂yk ∂yt   1 ∂um ∂un ∂us ∂us = + + 7im 7nj 2 ∂yn ∂ym ∂ym ∂yn or

eij = emn 7im 7nj .Consequently, the strain eij transforms like a second order Cartesian tensor.

228 Lagrangian and Eulerian Systems Let xi denote the initial position of a material particle in a continuum. Assume that at a later time the particle has moved to another point whose coordinates are xi . Both sets of coordinates are referred to the same coordinate system. When the final position can be expressed as a function of the initial position and time we can write xi = xi (x1 , x2 , x3 , t). Whenever the changes of any physical quantity is represented in terms of its initial position and time, the representation is referred to as a Lagrangian or material representation of the quantity. This can be thought of as a transformation of the coordinates. When the Jacobian J( xx ) of this transformation is different from zero, the above set of equations have a unique inverse xi = xi (x1 , x2 , x3 , t), where the position of the particle is now expressed in terms of its instantaneous position and time. Such a representation is referred to as an Eulerian or spatial description of the motion. Let (x1 , x2 , x3 ) denote the initial position of a particle whose motion is described by xi = xi (x1 , x2 , x3 , t), then ui = xi − xi denotes the displacement vector which can by represented in a Lagrangian or Eulerian form. For example, if

x1 = 2(x1 − x2 )(et − 1) + (x2 − x1 )(e−t − 1) + x1 x2 = (x1 − x2 )(et − 1) + (x2 − x1 )(e−t − 1) + x2 x3 = x3

then the displacement vector can be represented in the Lagrangian form u1 = 2(x1 − x2 )(et − 1) + (x2 − x1 )(e−t − 1) u2 = (x1 − x2 )(et − 1) + (x2 − x1 )(e−t − 1) u3 = 0 or the Eulerian form u1 = x1 − (2x2 − x1 )(1 − e−t ) − (x1 − x2 )(e−2t − e−t ) − x1 e−t u2 = x2 − (2x2 − x1 )(1 − e−t ) − (x2 − x1 )(e−2t − e−t ) − x2 e−t u3 = 0. Note that in the Lagrangian system the displacements are expressed in terms of the initial position and time, while in the Eulerian system the independent variables are the position coordinates and time. Euler equations describe, as a function of time, how such things as density, pressure, and fluid velocity change at a fixed point in the medium. In contrast, the Lagrangian viewpoint follows the time history of a moving individual fluid particle as it moves through the medium.

229 General Tensor Derivation of Strain. With reference to the figure 2.3-15 consider the deformation of a point P within a continuum. Let 1

(y , y 2 , y 3 ) denote a Cartesian coordinate system which is fixed in space. We can introduce a coordinate transformation y i = y i (x1 , x2 , x3 ),

i = 1, 2, 3 and represent all points within the continuum with respect

to a set of generalized coordinates (x1 , x2 , x3 ). Let P denote a general point in the continuum while it is in an unstrained state and assume that this point gets transformed to a point P  when the continuum experiences external forces. If P moves to P  , then all points Q which are near P will move to points Q near P  . We can imagine that in the unstrained state all the points of the continuum are referenced with respect to the set of generalized coordinates (x1 , x2 , x3 ). After the strain occurs, we can imagine that it will be convenient to represent all points of the continuum with respect to a new barred system of coordinates (x1 , x2 , x3 ). We call the original set of coordinates the Lagrangian system of coordinates and the new set of barred coordinates the Eulerian coordinates. The Eulerian coordinates are assumed to be described by a set of coordinate transformation equations xi = xi (x1 , x2 , x3 ), i

i

1

2

3

x = x (x , x , x ),

i = 1, 2, 3 with inverse transformations

i = 1, 2, 3, which are assumed to exist. The barred and unbarred coordinates can

be related to a fixed set of Cartesian coordinates y i , i = 1, 2, 3, and we may assume that there exists transformation equations y i = y i (x1 , x2 , x3 ),

i = 1, 2, 3 and y i = y i (x1 , x2 , x3 ),

i = 1, 2, 3

which relate the barred and unbarred coordinates to the Cartesian axes. In the discussion that follows be sure to note whether there is a bar over a symbol, as we will be jumping back and forth between the Lagrangian and Eulerian reference frames.

Figure 2.3-15. Strain in generalized coordinates
i = ∂
r which produce In the Lagrangian system of unbarred coordinates we have the basis vectors E ∂xi

the metrices gij = Ei · Ej . Similarly, in the Eulerian system of barred coordinates we have the basis vectors
·E
. These basis vectors are illustrated in the figure 2.3-15.
= ∂
r which produces the metrices G = E E i ij i j ∂xi

230 We assume that an element of arc length squared ds2 in the unstrained state is deformed to the element of arc length squared ds2 in the strained state. An element of arc length squared can be expressed in terms of the barred or unbarred coordinates. For example, in the Lagrangian system, let d
r = P Q so that L20 = d
r · d
r = ds2 = gij dxi dxj ,

(2.3.54)

where gij are the metrices in the Lagrangian coordinate system. This same element of arc length squared can be expressed in the barred system by L20 = ds2 = gij dxi dxj ,

where g ij = gmn

∂xm ∂xn . ∂xi ∂xj

(2.3.55)

Similarly, in the Eulerian system of coordinates the deformed arc length squared is L2 = d
r · d
r = ds2 = Gij dxi dxj ,

(2.3.56)

where Gij are the metrices in the Eulerian system of coordinates. This same element of arc length squared can be expressed in the Lagrangian system by the relation L2 = ds2 = Gij dxi dxj ,

where Gij = Gmn

∂xm ∂xn . ∂xi ∂xj

(2.3.57)

In the Lagrangian system we have ds2 − ds2 = (Gij − gij )dxi dxj = 2eij dxi dxj where eij =

1 (Gij − gij ) 2

(2.3.58)

is called the Green strain tensor or Lagrangian strain tensor. Alternatively, in the Eulerian system of coordinates we may write

ds2 − ds2 = Gij − g ij dxi dxj = 2eij dxi dxj

where eij =

1 Gij − gij 2

is called the Almansi strain tensor or Eulerian strain tensor.

(2.3.59)

231 Note also in the figure 2.3-15 there is the displacement vector
u. This vector can be represented in any of the following forms:
i
u = ui E

contravariant, Lagrangian basis


i
u = ui E

u = ui E

covariant, Lagrangian reciprocal basis

i


i
u = ui E

contravariant, Eulerian basis covariant, Eulerian reciprocal basis.

By vector addition we have
r +
u =
r and consequently d
r + d
u = d
r . In the Lagrangian frame of reference
i and write d
r in the form d
r = dxi E
i . By at the point P we represent
u in the contravariant form
u = ui E
i . These substitutions produce the use of the equation (1.4.48) we can express d
u in the form d
u = ui,k dxk E
i in the Lagrangian coordinate system. We can then express ds2 in the representation d
r = (dxi + ui,k dxk )E Lagrangian system. We find
i · (dxj + uj dxm )E
j d
r · d
r = ds2 = (dxi + ui,k dxk )E ,m = (dxi dxj + uj,m dxm dxi + ui,k dxk dxj + ui,k uj,m dxk dxm )gij and consequently from the relation (2.3.58) we derive the representation eij =

1 ui,j + uj,i + um,i um ,j . 2

(2.3.60)

This is the representation of the Lagrangian strain tensor in any system of coordinates. The strain tensor eij is symmetric. We will restrict our study to small deformations and neglect the product terms in equation (2.3.60). Under these conditions the equation (2.3.60) reduces to eij = 12 (ui,j + uj,i ). If instead, we chose to represent the displacement
u with respect to the Eulerian basis, then we can write

u = ui E i

with


. d
u = ui,k dxk E i

These relations imply that
. d
r = d
r − d
u = (dxi − ui,k dxk )E i This representation of d
r in the Eulerian frame of reference can be used to calculate the strain eij from the relation ds2 − ds2 . It is left as an exercise to show that there results eij =

1 ui,j + uj,i − um,i um ,j . 2

(2.3.61)

The equation (2.3.61) is the representation of the Eulerian strain tensor in any system of coordinates. Under conditions of small deformations both the equations (2.3.60) and (2.3.61) reduce to the linearized Lagrangian and Eulerian strain tensor eij = 12 (ui,j + uj,i ). In the case of large deformations the equations (2.3.60) and (2.3.61) describe the strains. In the case of linear elasticity, where the deformations are very small, the product terms in equations (2.3.60) and (2.3.61) are neglected and the Lagrangian and Eulerian strains reduce to their linearized forms eij =

1 [ui,j + uj,i ] 2

eij =

1 [ui,j + uj,i ] . 2

(2.3.62)

232

Figure 2.3-16. Displacement due to strain Compressible and Incompressible Material With reference to figure 2.3-16, let xi , i = 1, 2, 3 denote the position vector of an arbitrary point P in a continuum before there is a state of strain. Let Q be a neighboring point of P with position vector xi + dxi , i = 1, 2, 3. Also in the figure 2.3-16 there is the displacement vector
u. Here it is assumed that
u =
u(x1 , x2 , x3 ) denotes the displacement field when the continuum is in a state of strain. The figure 2.3-16 illustrates that in a state of strain P moves to P  and Q moves to Q . Let us find a relationship between the distance P Q before the strain and the distance P  Q when
2, E
3 basis functions constructed at P we have previously
1, E the continuum is in a state of strain. For E shown that if
i
u(x1 , x2 , x3 ) = ui E

then


i. d
u = ui,j dxj E

Now for
u + d
u the displacement of the point Q we may use vector addition and write P Q +
u + d
u =
u + P  Q .

(2.3.63)


i = ai E
i denote an arbitrary small change in the continuum. This arbitrary displacement Let P Q = dxi E
i due to the state of strain in the continuum. Employing the equation (2.3.63) gets deformed to P  Q = Ai E we write dxi + ui,j dxj = ai + ui,j aj = Ai which can be written in the form δai = Ai − ai = ui,j aj

where

dxi = ai , i = 1, 2, 3

(2.3.64)

denotes an arbitrary small change. The tensor ui,j and the associated tensor ui,j = git ut,j are in general not symmetric tensors. However, we know we can express ui,j as the sum of a symmetric (eij ) and skewsymmetric(ωij ) tensor. We therefore write ui,j = eij + ωij

or ui,j = eij + ω ij ,

where eij =

1 1 (ui,j + uj,i ) = (gim um ,j + gjm um ,i ) 2 2

and

ωij =

1 1 (ui,j − uj,i ) = (gim um ,j − gjm um ,i ) . 2 2

The deformation of a small quantity ai can therefore be represented by a pure strain Ai − ai = eis as followed by a rotation Ai − ai = ωsi as .

233 Consider now a small element of volume inside a material medium. With reference to the figure 2.317(a) we let
a,
b,
c denote three small arbitrary independent vectors constructed at a general point P within the material before any external forces are applied. We imagine
a,
b,
c as representing the sides of a small parallelepiped before any deformation has occurred. When the material is placed in a state of strain the
B,
C
as illustrated in point P will move to P  and the vectors
a,
b,
c will become deformed to the vectors A,
B,
C
represent the sides of the parallelepiped after the deformation. the figure 2.3-17(b). The vectors A,

Figure 2.3-17. Deformation of a parallelepiped Let ∆V denote the volume of the parallelepiped with sides
a,
b,
c at P before the strain and let ∆V 
B,
C
at the denote the volume of the deformed parallelepiped after the strain, when it then has sides A, point P  . We define the ratio of the change in volume due to the strain divided by the original volume as the dilatation at the point P. The dilatation is thus expressed as Θ=

∆V  − ∆V = dilatation. ∆V

(2.3.65)

Since ui , i = 1, 2, 3 represents the displacement field due to the strain, we use the result from equation
B,
C
in the form (2.3.64) and represent the displaced vectors A, Ai = ai + ui,j aj B i = bi + ui,j bj i

i

C =c +

(2.3.66)

ui,j cj

where
a,
b,
c are arbitrary small vectors emanating from the point P in the unstrained state. The element of volume ∆V, before the strain, is calculated from the triple scalar product relation ∆V =
a · (
b ×
c) = eijk ai bj ck . The element of volume ∆V  , which occurs due to the strain, is calculated from the triple scalar product
· (B
× C)
= eijk Ai B j C k . ∆V  = A

234 Substituting the relations from the equations (2.3.66) into the triple scalar product gives ∆V  = eijk (ai + ui,m am )(bj + uj,n bn )(ck + uk,p cp ). Expanding the triple scalar product and employing the result from Exercise 1.4, problem 34, we find the simplified result gives us the dilatation Θ=

∆V  − ∆V = ur,r = div (
u). ∆V

(2.3.67)

That is, the dilatation is the divergence of the displacement field. If the divergence of the displacement field is zero, there is no volume change and the material is said to be incompressible. If the divergence of the displacement field is different from zero, the material is said to be compressible. Note that the strain eij is expressible in terms of the displacement field by the relation eij =

1 (ui,j + uj,i ), 2

and consequently g mn emn = ur,r .

(2.3.68)

Hence, for an orthogonal system of coordinates the dilatation can be expressed in terms of the strain elements along the main diagonal. Conservation of Mass Consider the material in an arbitrary region R of a continuum. Let L = L(x, y, z, t) denote the density of the material within the region. Assume that the dimension of the density L is gm/cm3 in the cgs system of units. We shall assume that the region R is bounded by a closed surface S with exterior unit normal
n defined everywhere on the surface. Further, we let
v =
v (x, y, z, t) denote a velocity field associated with all points within the continuum. The velocity field has units of cm/sec in the cgs system of units. Neglecting sources and sinks, the law of conservation of mass examines    all the material entering and leaving a region R. Enclosed within R is the material mass m where m =

L dτ with dimensions of gm in the cgs system of R

units. Here dτ denotes an element of volume inside the region R. The change of mass with time is obtained by differentiating the above relation. Differentiating the mass produces the equation  ∂L ∂m = dτ ∂t R ∂t

(2.3.69)

and has the dimensions of gm/sec. 

Consider also the surface integral

L
v · n ˆ dσ

I=

(2.3.70)

S

where dσ is an element of surface area on the surface S which encloses R and n ˆ is the exterior unit normal vector to the surface S. The dimensions of the integral I is determined by examining the dimensions of each term in the integrand of I. We find that [I] =

gm gm cm · cm2 = · 3 cm sec sec

and so the dimension of I is the same as the dimensions for the change of mass within the region R. The surface integral I is the flux rate of material crossing the surface of R and represents the change of mass

235 entering the region if
v · n ˆ is negative and the change of mass leaving the region if
v · n ˆ is positive, as n ˆ is always an exterior unit normal vector. Equating the relations from equations (2.3.69) and (2.3.70) we obtain a mathematical statement for mass conservation   ∂m ∂L = dτ = − L
v ·
n dσ. ∂t R ∂t S

(2.3.71)

The equation (2.3.71) implies that the rate at which the mass contained in R increases must equal the rate at which the mass flows into R through the surface S. The negative sign changes the direction of the exterior normal so that we consider flow of material into the region. Employing the Gauss divergence theorem, the surface integral in equation (2.3.71) can be replaced by a volume integral and the law of conservation of mass is then expressible in the form   R

 ∂L + div (L
v ) dτ = 0. ∂t

(2.3.72)

Since the region R is an arbitrary volume we conclude that the term inside the brackets must equal zero. This gives us the continuity equation ∂L + div (L
v ) = 0 ∂t

(2.3.73)

which represents the mass conservation law in terms of velocity components. This is the Eulerian representation of continuity of mass flow. Equivalent forms of the continuity equation are: ∂L +
v · grad L + L div
v = 0 ∂t ∂L ∂L ∂vi + vi i + L i = 0 ∂t ∂x ∂x ∂vi DL +L i =0 Dt ∂x ∂L ∂L dxi ∂L ∂L DL = + = + vi is called the material derivative of the density L. Note that the Dt ∂t ∂xi dt ∂t ∂xi ∂E material derivative contains the expression ∂x i vi which is known as the convective or advection term. If the where

density L = L(x, y, z, t) is a constant we have DL ∂L ∂L dx ∂L dy ∂L dz ∂L ∂L dxi = + + + = + i =0 Dt ∂t ∂x dt ∂y dt ∂z dt ∂t ∂x dt

(2.3.74)

and hence the continuity equation reduces to div (
v ) = 0. Thus, if div (
v ) is zero, then the material is incompressible. EXAMPLE 2.3-2. (Continuity Equation) Find the Lagrangian representation of mass conservation. Solution: Let (X, Y, Z) denote the initial position of a fluid particle and denote the density of the fluid by L(X, Y, Z, t) so that L(X, Y, Z, 0) denotes the density at the time t = 0. Consider a simple closed region in our continuum and denote this region by R(0) at time t = 0 and by R(t) at some later time t. That is, all the points   in  R(0) move in a one-to-one fashion to points in R(t). Initially the mass of material in R(0) is m(0) = L(X, Y, Z, 0) dτ (0) where dτ (0) = dXdY dZ is an element of volume in R(0). We have after a R(0)

236  time t has elapsed the mass of material in the region R(t) given by m(t) =

L(X, Y, Z, t) dτ (t) where  x,y,z dτ (t) = dxdydz is a deformed element of volume related to the dτ (0) by dτ (t) = J X,Y,Z dτ (0) where J is R(t)

the Jacobian of the Eulerian (x, y, z) variables with respect to the Lagrangian (X, Y, Z) representation. For mass conservation we require that m(t) = m(0) for all t. This implies that L(X, Y, Z, t)J = L(X, Y, Z, 0)

(2.3.75)

for all time, since the initial region R(0) is arbitrary. The right hand side of equation (2.3.75) is independent of time and so d (L(X, Y, Z, t)J) = 0. dt

(2.3.76)

This is the Lagrangian form of the continuity equation which expresses mass conservation. Using the result dJ = Jdiv V
, (see problem 28, Exercise 2.3), the equation (2.3.76) can be expanded and written in the that dt form DL
=0 + L div V (2.3.77) Dt where

DE Dt

is from equation (2.3.74). The form of the continuity equation (2.3.77) is one of the Eulerian forms

previously developed. In the Eulerian coordinates the continuity equation is written system the continuity equation is written

d(E J) dt

∂E ∂t

+ div (L
v ) = 0, while in the Lagrangian

= 0. Note that the velocity carries the Lagrangian axes and

the density change grad L. This is reflective of the advection term
v · grad L. Thus, in order for mass to be conserved it need not remain stationary. The mass can flow and the density can change. The material derivative is a transport rule depicting the relation between the Eulerian and Lagrangian viewpoints. In general, from a Lagrangian viewpoint, any quantity Q(x, y, z, t) which is a function of both position and time is seen as being transported by the fluid velocity (v1 , v2 , v3 ) to Q(x + v1 dt, y + v2 dt, z + v3 dt, t + dt). Then the time derivative of Q contains both

∂Q ∂t

and the advection term
v · ∇Q. In terms of mass flow, the

Eulerian viewpoint sees flow into and out of a fixed volume in space, as depicted by the equation (2.3.71), In contrast, the Lagrangian viewpoint sees the same volume moving with the fluid and consequently    D ρ dτ = 0, Dt R(t) where R(t) represents the volume moving with the fluid. Both viewpoints produce the same continuity equation reflecting the conservation of mass. Summary of Basic Equations Let us summarize the basic equations which are valid for all types of a continuum. We have derived: • Conservation of mass (continuity equation) ∂L + (Lv i ),i = 0 ∂t

237 • Conservation of linear momentum sometimes called the Cauchy equation of motion. σ ij ,i + Lbj = Lf j ,

j = 1, 2, 3.

• Conservation of angular momentum σij = σji • Strain tensor for linear elasticity eij =

1 (ui,j + uj,i ). 2

If we assume that the continuum is in equilibrium, and there is no motion, then the velocity and acceleration terms above will be zero. The continuity equation then implies that the density is a constant. The conservation of angular momentum equation requires that the stress tensor be symmetric and we need find only six stresses. The remaining equations reduce to a set of nine equations in the fifteen unknowns: 3 displacements u1 , u2 , u3 6 strains

e11 , e12 , e13 , e22 , e23 , e33

6 stresses σ11 , σ12 , σ13 , σ22 , σ23 , σ33 Consequently, we still need additional information if we desire to determine these unknowns. Note that the above equations do not involve any equations describing the material properties of the continuum. We would expect solid materials to act differently from liquid material when subjected to external forces. An equation or equations which describe the material properties are called constitutive equations. In the following sections we will investigate constitutive equations for solids and liquids. We will restrict our study to linear elastic materials over a range where there is a linear relationship between the stress and strain. We will not consider plastic or viscoelastic materials. Viscoelastic materials have the property that the stress is not only a function of strain but also a function of the rates of change of the stresses and strains and consequently properties of these materials are time dependent.

238 EXERCISE 2.3
1.

Assume an orthogonal coordinate system with metric tensor gij = 0 for i = j and g(i)(i) = h2i (no

summation on i). Use the definition of strain ers =

1 1 (ur,s + us,r ) = grt ut,s + gst ut,r 2 2

and show that in terms of the physical components e(ij) =

eij hi hj

u(i) = hi ui

no summation on i or j no summation on i

there results the equations:  '  t ∂ut m eii = git + u no summation on i mi ∂xi ∂ut ∂ut 2eij = git j + gjt i , i = j ∂x ∂x   3 u(i) ∂ 1  u(m) ∂ 2

h no summation on i e(ii) = + ∂xi hi 2h2i m=1 hm ∂xm i     u(i) u(j) hi ∂ hj ∂ 2e(ij) = + , no summation on i or j, i = j. hj ∂xj hi hi ∂xi hj 


2. Use the results from problem 1 to write out all components of the strain tensor in Cartesian coordinates. Use the notation u(1) = u,u(2) = v,u(3) = w and e(11) = exx ,

e(22) = eyy ,

e(33) = ezz ,

e(12) = exy ,

to verify the relations: exx eyy ezz

∂u = ∂x ∂v = ∂y ∂w = ∂z

e(13) = exz ,

e(23) = eyz

  ∂u 1 ∂v + = 2 ∂x ∂y   1 ∂u ∂w + = 2 ∂z ∂x   1 ∂w ∂v + = 2 ∂y ∂z

exy exz ezy


3. Use the results from problem 1 to write out all components of the strain tensor in cylindrical coordinates. Use the notation u(1) = ur , u(2) = uθ , u(3) = uz and e(11) = err ,

e(22) = eθθ ,

e(33) = ezz ,

e(12) = erθ ,

e(13) = erz ,

to verify the relations: err eθθ ezz

∂ur = ∂r ur 1 ∂uθ + = r ∂θ r ∂uz = ∂z

  ∂uθ uθ 1 1 ∂ur + − 2 r ∂θ ∂r r   ∂ur 1 ∂uz + = 2 ∂r ∂z   1 ∂uz 1 ∂uθ + = 2 ∂z r ∂θ

erθ = erz eθz

e(23) = eθz

239
4. Use the results from problem 1 to write out all components of the strain tensor in spherical coordinates. Use the notation u(1) = uρ ,u(2) = uθ ,u(3) = uφ and e(11) = eρρ ,

e(22) = eθθ ,

e(33) = eφφ ,

e(12) = eρθ ,

e(13) = eρφ ,

e(23) = eθφ

to verify the relations eρρ eθθ eφφ

∂uρ = ∂ρ uρ 1 ∂uθ + = ρ ∂θ ρ uθ 1 ∂uφ uρ + + cot θ = ρ sin θ ∂φ ρ ρ

eρθ eρφ eθφ

  uθ ∂uθ 1 1 ∂uρ − + = 2 ρ ∂θ ρ ∂ρ   1 ∂uρ uφ ∂uφ 1 − + = 2 ρ sin θ ∂φ ρ ∂ρ   uφ 1 ∂uθ 1 1 ∂uφ − cot θ + = 2 ρ ∂θ ρ ρ sin θ ∂φ


5. Expand equation (2.3.67) and find the dilatation in terms of the physical components of an orthogonal system and verify that Θ=

  ∂(h2 h3 u(1)) ∂(h1 h3 u(2)) ∂(h1 h2 u(3)) 1 + + h1 h2 h3 ∂x1 ∂x2 ∂x3


6. Verify that the dilatation in Cartesian coordinates is Θ = exx + eyy + ezz =

∂u ∂v ∂w + + . ∂x ∂y ∂z


7. Verify that the dilatation in cylindrical coordinates is Θ = err + eθθ + ezz =

1 ∂uθ 1 ∂ur ∂uz + + ur + . ∂r r ∂θ r ∂z


8. Verify that the dilatation in spherical coordinates is Θ = eρρ + eθθ + eφφ =

1 ∂uθ 2 uθ cot θ ∂uρ 1 ∂uφ + + uρ + + . ∂ρ ρ ∂θ ρ ρ sin θ ∂φ ρ


9. Show that in an orthogonal set of coordinates the rotation tensor ωij can be written in terms of physical components in the form 1 ω(ij) = 2hi hj



 ∂(hi u(i)) ∂(hj u(j)) − , ∂xj ∂xi

no summations

Hint: See problem 1.
10. Use the result from problem 9 to verify that in Cartesian coordinates   1 ∂v ∂u − ωyx = 2 ∂x ∂y   1 ∂u ∂w − ωxz = 2 ∂z ∂x   1 ∂w ∂v − ωzy = 2 ∂y ∂z

240
11. Use the results from problem 9 to verify that in cylindrical coordinates   1 ∂(ruθ ) ∂ur − ωθr = 2r ∂r ∂θ   ∂uz 1 ∂ur − ωrz = 2 ∂z ∂r   ∂uθ 1 1 ∂uz − ωzθ = 2 r ∂θ ∂z
12. Use the results from problem 9 to verify that in spherical coordinates   1 ∂(ρuθ ) ∂uρ − ωθρ = 2ρ ∂ρ ∂θ   1 ∂uρ ∂(ρuφ ) 1 − ωρφ = 2ρ sin θ ∂φ ∂ρ   ∂(uφ sin θ) ∂uθ 1 − ωφθ = 2ρ sin θ ∂θ ∂φ
13. The conditions for static equilibrium in a linear elastic material are determined from the conservation law σij ,j + Lbi = 0,

i, j = 1, 2, 3,

where σji are the stress tensor components, bi are the external body forces per unit mass and L is the density of the material. Assume an orthogonal coordinate system and verify the following results. (a) Show that

1 ∂ √ j σij ,j = √ ( gσi ) − [ij, m]σ mj g ∂xj

(b) Use the substitutions σ(ij) = σij

hj hi

no summation on i or j

bi no summation on i hi σ(ij) = σ ij hi hj no summation on i or j b(i) =

and express the equilibrium equations in terms of physical components and verify the relations √  3 3  ghi σ(ij) 1 ∂ 1  σ(jj) ∂(h2j ) + hi Lb(i) = 0, − √ hj 2 j=1 h2j ∂xi g ∂xj j=1 where there is no summation on i.
14. Use the results from problem 13 and verify that the equilibrium equations in Cartesian coordinates can be expressed ∂σxx ∂σxy ∂σxz + + + Lbx = 0 ∂x ∂y ∂z ∂σyy ∂σyz ∂σyx + + + Lby = 0 ∂x ∂y ∂z ∂σzy ∂σzz ∂σzx + + + Lbz = 0 ∂x ∂y ∂z

241
15. Use the results from problem 13 and verify that the equilibrium equations in cylindrical coordinates can be expressed 1 ∂σrθ ∂σrz 1 ∂σrr + + + (σrr − σθθ ) + Lbr = 0 ∂r r ∂θ ∂z r 1 ∂σθθ ∂σθz 2 ∂σθr + + + σθr + Lbθ = 0 ∂r r ∂θ ∂z r 1 ∂σzθ ∂σzz 1 ∂σzr + + + σzr + Lbz = 0 ∂r r ∂θ ∂z r
16. Use the results from problem 13 and verify that the equilibrium equations in spherical coordinates can be expressed 1 ∂σρθ 1 ∂σρφ 1 ∂σρρ + + + (2σρρ − σθθ − σφφ + σρθ cot θ) + Lbρ = 0 ∂ρ ρ ∂θ ρ sin θ ∂φ ρ 1 ∂σθθ 1 ∂σθφ 1 ∂σθρ + + + (3σρθ + [σθθ − σφφ ] cot θ) + Lbθ = 0 ∂ρ ρ ∂θ ρ sin θ ∂φ ρ 1 ∂σφθ 1 ∂σφφ 1 ∂σφρ + + + (3σρφ + 2σθφ cot θ) + Lbφ = 0 ∂ρ ρ ∂θ ρ sin θ ∂φ ρ
17. Derive the result for the Lagrangian strain defined by the equation (2.3.60).
18. Derive the result for the Eulerian strain defined by equation (2.3.61).
19.

The equation δai = ui,j aj , describes the deformation in an elastic solid subjected to forces. The

quantity δai denotes the difference vector Ai − ai between the undeformed and deformed states. (a) Let |a| denote the magnitude of the vector ai and show that the strain e in the direction ai can be represented δ|a| = eij e= |a|



ai |a|



aj |a|

 = eij λi λj ,

where λi is a unit vector in the direction ai . (b) Show that for λ1 = 1, λ2 = 0, λ3 = 0 there results e = e11 , with similar results applying to vectors λi in the y and z directions. Hint: Consider the magnitude squared |a|2 = gij ai aj . ˆ1 +
20. At the point (1, 2, 3) of an elastic solid construct the small vector
a = F( 23 e

2 3

ˆ2 + e

1 3

ˆ3 ), where e

F > 0 is a small positive quantity. The solid is subjected to forces such that the following displacement field results. ˆ2 + xz e ˆ3 ) × 10−2 ˆ1 + yz e
u = (xy e
after the displacement field has been imposed. Calculate the deformed vector A
21. For the displacement field ˆ1 + (xy + z 2 ) e ˆ2 + xyz e ˆ3
u = (x2 + yz) e (a) Calculate the strain matrix at the point (1, 2, 3). (b) Calculate the rotation matrix at the point (1, 2, 3).

242
22. Show that for an orthogonal coordinate system the ith component of the convective operator can be written
· ∇) A]
i= [(V

  3 3  V (m) ∂A(i)  A(m) ∂hm ∂hi + V (i) m − V (m) i hm ∂xm hm hi ∂x ∂x m=1 m=1 m=i


23.

Consider a parallelepiped with dimensions 7, w, h which has a uniform pressure P applied to each

face. Show that the volume strain can be expressed as ∆V ∆7 ∆w ∆h −3P (1 − 2ν) = + + = . V 7 w h E The quantity k = E/3(1 − 2ν) is called the bulk modulus of elasticity.
24. Show in Cartesian coordinates the continuity equation is ∂L ∂(Lu) ∂(Lv) ∂(Lw) + + + = 0, ∂t ∂x ∂y ∂z where (u, v, w) are the velocity components.
25. Show in cylindrical coordinates the continuity equation is ∂L 1 ∂(rLVr ) 1 ∂(LVθ ) ∂(LVz ) + + + =0 ∂t r ∂r r ∂θ ∂z where Vr , Vθ , Vz are the velocity components.
26. Show in spherical coordinates the continuity equation is ∂L 1 ∂(ρ2 LVρ ) 1 ∂(LVθ sin θ) 1 ∂(LVφ ) + 2 + + =0 ∂t ρ ∂ρ ρ sin θ ∂θ ρ sin θ ∂φ where Vρ , Vθ , Vφ are the velocity components.
27. (a) Apply a stress σyy to both ends of a square element in a x, y continuum. Illustrate and label all changes that occur due to this stress. (b) Apply a stress σxx to both ends of a square element in a x, y continuum. Illustrate and label all changes that occur due to this stress. (c) Use superposition of your results in parts (a) and (b) and explain each term in the relations exx =

σyy σxx −ν E E

and

eyy = 


28. Show that the time derivative of the Jacobian J = J div V
=

∂V1 ∂V2 ∂V3 + + ∂x ∂y ∂z

and V1 =

σxx σyy −ν . E E

x, y, z X, Y, Z

dx , dt



V2 =

satisfies dy , dt

dJ
where = J div V dt

V3 =

dz . dt

Hint: Let (x, y, z) = (x1 , x2 , x3 ) and (X, Y, Z) = (X1 , X2 , X3 ), then note that eijk

∂V1 ∂x2 ∂x3 ∂V1 ∂xm ∂x2 ∂x3 ∂x1 ∂x2 ∂x3 ∂V1 = eijk = eijk , ∂Xi ∂Xj ∂Xk ∂xm ∂Xi ∂Xj ∂Xk ∂Xi ∂Xj ∂Xk ∂x1

etc.

243 §2.4 CONTINUUM MECHANICS (SOLIDS) In this introduction to continuum mechanics we consider the basic equations describing the physical effects created by external forces acting upon solids and fluids. In addition to the basic equations that are applicable to all continua, there are equations which are constructed to take into account material characteristics. These equations are called constitutive equations. For example, in the study of solids the constitutive equations for a linear elastic material is a set of relations between stress and strain. In the study of fluids, the constitutive equations consists of a set of relations between stress and rate of strain. Constitutive equations are usually constructed from some basic axioms. The resulting equations have unknown material parameters which can be determined from experimental investigations. One of the basic axioms, used in the study of elastic solids, is that of material invariance. This axiom requires that certain symmetry conditions of solids are to remain invariant under a set of orthogonal transformations and translations. This axiom is employed in the next section to simplify the constitutive equations for elasticity. We begin our study of continuum mechanics by investigating the development of constitutive equations for linear elastic solids. Generalized Hooke’s Law If the continuum material is a linear elastic material, we introduce the generalized Hooke’s law in Cartesian coordinates σij = cijkl ekl ,

i, j, k, l = 1, 2, 3.

(2.4.1)

The Hooke’s law is a statement that the stress is proportional to the gradient of the deformation occurring in the material. These equations assume a linear relationship exists between the components of the stress tensor and strain tensor and we say stress is a linear function of strain. Such relations are referred to as a set of constitutive equations. Constitutive equations serve to describe the material properties of the medium when it is subjected to external forces. Constitutive Equations The equations (2.4.1) are constitutive equations which are applicable for materials exhibiting small deformations when subjected to external forces. The 81 constants cijkl are called the elastic stiffness of the material. The above relations can also be expressed in the form eij = sijkl σkl ,

i, j, k, l = 1, 2, 3

(2.4.2)

where sijkl are constants called the elastic compliance of the material. Since the stress σij and strain eij have been shown to be tensors we can conclude that both the elastic stiffness cijkl and elastic compliance sijkl are fourth order tensors. Due to the symmetry of the stress and strain tensors we find that the elastic stiffness and elastic compliance tensor must satisfy the relations cijkl = cjikl = cijlk = cjilk

(2.4.3)

sijkl = sjikl = sijlk = sjilk and consequently only 36 of the 81 constants are actually independent. If all 36 of the material (crystal) constants are independent the material is called triclinic and there are no material symmetries.

244 Restrictions on Elastic Constants due to Symmetry The equations (2.4.1) and (2.4.2) can be replaced by an equivalent set of equations which are easier to analyze. This is accomplished by defining the quantities

where

e1 ,

e2 ,

e3 ,

σ1 ,

σ2 ,

σ3 ,



e4 e2 e6

e1  e4 e5 

and

σ4 σ2 σ6

σ1  σ4 σ5

e4 , σ4 ,

e5 ,

e6

σ5 ,

σ6

  e5 e11 e6  =  e21 e3 e31

e12 e22 e32

 e13 e23  e33

  σ5 σ11 σ6  =  σ21 σ3 σ31

σ12 σ22 σ32

 σ13 σ23  . σ33

Then the generalized Hooke’s law from the equations (2.4.1) and (2.4.2) can be represented in either of the forms σi = cij ej

or ei = sij σj

where i, j = 1, . . . , 6

(2.4.4)

where cij are constants related to the elastic stiffness and sij are constants related to the elastic compliance. These constants satisfy the relation smi cij = δmj 

Here eij = and similarly

 σij =

where

i, m, j = 1, . . . , 6

ei , e1+i+j ,

i = j = 1, 2, 3 i = j, and i = 1, or, 2

σi , σ1+i+j ,

i = j = 1, 2, 3 i = j, and i = 1, or, 2.

(2.4.5)

These relations show that the constants cij are related to the elastic stiffness coefficients cpqrs by the relations cm1 = cij11

cm4 = 2cij12

cm2 = cij22

cm5 = 2cij13

cm3 = cij33

cm6 = 2cij23



where m=

i, 1 + i + j,

if i = j = 1, 2, or 3 if i = j and i = 1 or 2.

A similar type relation holds for the constants sij and spqrs . The above relations can be verified by expanding the equations (2.4.1) and (2.4.2) and comparing like terms with the expanded form of the equation (2.4.4).

245 The generalized Hooke’s law can now be expressed in a form where the 36 independent constants can be examined in more detail under special material symmetries. We will examine the form      s11 s12 s13 s14 s15 s16 σ1 e1  e2   s21 s22 s23 s24 s25 s26   σ2        e3   s31 s32 s33 s34 s35 s36   σ3   =  .  e4   s41 s42 s43 s44 s45 s46   σ4       e5 s51 s52 s53 s54 s55 s56 σ5 e6 s61 s62 s63 s64 s65 s66 σ6 Alternatively, in the arguments that follow, one can examine    σ1 c11 c12 c13 c14 c15  σ2   c21 c22 c23 c24 c25     σ3   c31 c32 c33 c34 c35  =  σ4   c41 c42 c43 c44 c45    σ5 c51 c52 c53 c54 c55 σ6 c61 c62 c63 c64 c65

(2.4.6)

the equivalent form   e1 c16 c26   e2    c36   e3   . c46   e4    c56 e5 c66 e6

Material Symmetries A material (crystal) with one plane of symmetry is called an aelotropic material. If we let the x1 x2 plane be a plane of symmetry then the equations (2.4.6) must remain invariant under the coordinate transformation



  1 0 x1  x2  =  0 1 x3 0 0

  0 x1 0   x2  x3 −1

(2.4.7)

which represents an inversion of the x3 axis. That is, if the x1 -x2 plane is a plane of symmetry we should be able to replace x3 by −x3 and the equations (2.4.6) should remain unchanged. This is equivalent to saying that a transformation of the type from equation (2.4.7) changes the Hooke’s law to the form ei = sij σ j where the sij remain unaltered because it is the same material. Employing the transformation equations x1 = x1 ,

x2 = x2 ,

x3 = −x3

(2.4.8)

we examine the stress and strain transformation equations σ ij = σpq

∂xp ∂xq ∂xi ∂xj

and

eij = epq

∂xp ∂xq . ∂xi ∂xj

(2.4.9)

If we expand both of the equations (2.4.9) and substitute in the nonzero derivatives ∂x1 = 1, ∂x1

∂x2 = 1, ∂x2

∂x3 = −1, ∂x3

(2.4.10)

we obtain the relations σ 11 = σ11

e11 = e11

σ 22 = σ22

e22 = e22

σ 33 = σ33

e33 = e33

σ 21 = σ21

e21 = e21

σ 31 = −σ31

e31 = −e31

σ 23 = −σ23

e23 = −e23 .

(2.4.11)

246 We conclude that if the material undergoes a strain, with the x1 -x2 plane as a plane of symmetry then e5 and e6 change sign upon reversal of the x3 axis and e1 , e2 , e3 , e4 remain unchanged. Similarly, we find σ5 and σ6 change sign while σ1 , σ2 , σ3 , σ4 remain unchanged. The equation (2.4.6) then becomes   s11 e1  e2   s21     e3   s31 =   e4   s41    −e5 s51 −e6 s61 

s12 s22 s32 s42 s52 s62

s13 s23 s33 s43 s53 s63

s14 s24 s34 s44 s54 s64

s15 s25 s35 s45 s55 s65

  σ1 s16 s26   σ2    s36   σ3   . s46   σ4    s56 −σ5 s66 −σ6

(2.4.12)

If the stress strain relation for the new orientation of the x3 axis is to have the same form as the old orientation, then the equations (2.4.6) and (2.4.12) must give the same results. Comparison of these equations we find that s15 = s16 = 0 s25 = s26 = 0 s35 = s36 = 0

(2.4.13)

s45 = s46 = 0 s51 = s52 = s53 = s54 = 0 s61 = s62 = s63 = s64 = 0. In summary, from an examination of the equations (2.4.6) and (2.4.12) we find that for an aelotropic material (crystal), with one plane of symmetry, the 36 constants sij reduce to 20 constants and the generalized Hooke’s law (constitutive equation) has the form   e1 s11  e2   s21     e3   s31  =  e4   s41    e5 0 e6 0 

s12 s22 s32 s42 0 0

s13 s23 s33 s43 0 0

s14 s24 s34 s44 0 0

0 0 0 0 s55 s65

  0 σ1 0   σ2    0   σ3   . 0   σ4    s56 σ5 s66 σ6

Alternatively, the Hooke’s law can be represented in the form   c11 σ1  σ2   c21     σ3   c31  =  σ4   c41    σ5 0 σ6 0 

c12 c22 c32 c42 0 0

c13 c23 c33 c43 0 0

c14 c24 c34 c44 0 0

0 0 0 0 c55 c65

  e1 0 0   e2    0   e3   . 0   e4    c56 e5 c66 e6

(2.4.14)

247 Additional Symmetries If the material (crystal) is such that there is an additional plane of symmetry, say the x2 -x3 plane, then reversal of the x1 axis should leave the equations (2.4.14) unaltered. If there are two planes of symmetry then there will automatically be a third plane of symmetry. Such a material (crystal) is called orthotropic. Introducing the additional transformation x1 = −x1 ,

x2 = x2 ,

x3 = x3

which represents the reversal of the x1 axes, the expanded form of equations (2.4.9) are used to calculate the effect of such a transformation upon the stress and strain tensor. We find σ1 , σ2 , σ3 , σ6 , e1 , e2 , e3 , e6 remain unchanged while σ4 , σ5 , e4 , e5 change sign. The equation (2.4.14) then becomes 

  e1 s11  e2   s21     e3   s31  =  −e4   s41    0 −e5 e6 0

s12 s22 s32 s42 0 0

s13 s23 s33 s43 0 0

s14 s24 s34 s44 0 0

0 0 0 0 s55 s65

  0 σ1 0   σ2    0   σ3   . 0   −σ4    s56 −σ5 s66 σ6

(2.4.15)

Note that if the constitutive equations (2.4.14) and (2.4.15) are to produce the same results upon reversal of the x1 axes, then we require that the following coefficients be equated to zero: s14 = s24 = s34 = 0 s41 = s42 = s43 = 0 s56 = s65 = 0. This then produces the constitutive equation   s11 e1  e2   s21     e3   s31  =  e4   0    e5 0 0 e6

s12 s22 s32 0 0 0

s13 s23 s33 0 0 0

0 0 0 s44 0 0

0 0 0 0 s55 0

  σ1 0 0   σ2    0   σ3    0   σ4    0 σ5 s66 σ6

  c11 σ1  σ2   c21     σ3   c31  =  σ4   0    0 σ5 0 σ6

c12 c22 c32 0 0 0

c13 c23 c33 0 0 0

0 0 0 c44 0 0

0 0 0 0 c55 0

  e1 0 0   e2    0   e3    0   e4    0 e5 c66 e6



or its equivalent form



(2.4.16)

and the original 36 constants have been reduced to 12 constants. This is the constitutive equation for orthotropic material (crystals).

248 Axis of Symmetry If in addition to three planes of symmetry there is an axis of symmetry then the material (crystal) is termed hexagonal. Assume that the x1 axis is an axis of symmetry and consider the effect of the transformation x1 = x1 ,

x2 = x3

x3 = −x2

upon the constitutive equations. It is left as an exercise to verify that the constitutive equations reduce to the form where there are 7 independent constants having either of the forms   s11 e1  e2   s21     e3   s21  =  e4   0    0 e5 0 e6 

or



  σ1 c11  σ2   c21     σ3   c21  =  σ4   0    0 σ5 σ6 0

s12 s22 s23 0 0 0

s12 s23 s22 0 0 0

0 0 0 s44 0 0

0 0 0 0 s44 0

  σ1 0 0   σ2    0   σ3    0   σ4    0 σ5 s66 σ6

c12 c22 c23 0 0 0

c12 c23 c22 0 0 0

0 0 0 c44 0 0

0 0 0 0 c44 0

  0 e1 0   e2    0   e3   . 0   e4    0 e5 e6 c66

Finally, if the material is completely symmetric, the x2 axis is also an axis of symmetry and we can consider the effect of the transformation x1 = −x3 ,

x2 = x2 ,

x3 = x1

upon the constitutive equations. It can be verified that these transformations reduce the Hooke’s law constitutive equation to the form 

  e1 s11 s e  2   12     e3   s12  =  e4   0    0 e5 e6 0

s12 s11 s12 0 0 0

s12 s12 s11 0 0 0

0 0 0 s44 0 0

0 0 0 0 s44 0

  0 σ1 0   σ2    0   σ3   . 0   σ4    0 σ5 σ6 s44

(2.4.17)

Materials (crystals) with atomic arrangements that exhibit the above symmetries are called isotropic materials. An equivalent form of (2.4.17) is the relation   σ1 c11  σ2   c12     σ3   c12  =  σ4   0    0 σ5 σ6 0 

c12 c11 c12 0 0 0

c12 c12 c11 0 0 0

0 0 0 c44 0 0

0 0 0 0 c44 0

  0 e1 0   e2    0   e3   . 0   e4    0 e5 e6 c44

The figure 2.4-1 lists values for the elastic stiffness associated with some metals which are isotropic1 1

Additional constants are given in “International Tables of Selected Constants”, Metals: Thermal and

Mechanical Data, Vol. 16, Edited by S. Allard, Pergamon Press, 1969.

249 Metal Na Pb Cu Ni Cr Mo W

c11 0.074 0.495 1.684 2.508 3.500 4.630 5.233

c12 0.062 0.423 1.214 1.500 0.678 1.610 2.045

c44 0.042 0.149 0.754 1.235 1.008 1.090 1.607

Figure 2.4-1. Elastic stiffness coefficients for some metals which are cubic. Constants are given in units of 1012 dynes/cm2 Under these conditions the stress strain constitutive relations can be written as σ1 = σ11 = (c11 − c12 )e11 + c12 (e11 + e22 + e33 ) σ2 = σ22 = (c11 − c12 )e22 + c12 (e11 + e22 + e33 ) σ3 = σ33 = (c11 − c12 )e33 + c12 (e11 + e22 + e33 )

(2.4.18)

σ4 = σ12 = c44 e12 σ5 = σ13 = c44 e13 σ6 = σ23 = c44 e23 . Isotropic Material Materials (crystals) which are elastically the same in all directions are called isotropic. We have shown that for a cubic material which exhibits symmetry with respect to all axes and planes, the constitutive stress-strain relation reduces to the form found in equation (2.4.17). Define the quantities s11 =

1 , E

s12 = −

ν , E

s44 =

1 2µ

where E is the Young’s Modulus of elasticity, ν is the Poisson’s ratio, and µ is the shear or rigidity modulus. For isotropic materials the three constants E, ν, µ are not independent as the following example demonstrates. EXAMPLE 2.4-1. (Elastic constants) For an isotropic material, consider a cross section of material in the x1 -x2 plane which is subjected to pure shearing so that σ4 = σ12 is the only nonzero stress as illustrated in the figure 2.4-2. For the above conditions, the equation (2.4.17) reduces to the single equation e4 = e12 = s44 σ4 = s44 σ12

or

µ=

σ12 γ12

and so the shear modulus is the ratio of the shear stress to the shear angle. Now rotate the axes through a 45 degree angle to a barred system of coordinates where x1 = x1 cos α − x2 sin α

x2 = x1 sin α + x2 cos α

250

Figure 2.4-2. Element subjected to pure shearing where α =

π 4.

Expanding the transformation equations (2.4.9) we find that σ 1 = σ 11 = cos α sin α σ12 + sin α cos α σ21 = σ12 = σ4 σ 2 = σ 22 = − sin α cos α σ12 − sin α cos α σ21 = −σ12 = −σ4 ,

and similarly e2 = e22 = −e4 .

e1 = e11 = e4 , In the barred system, the Hooke’s law becomes

e1 = s11 σ 1 + s12 σ 2

or

e4 = s11 σ4 − s12 σ4 = s44 σ4 . Hence, the constants s11 , s12 , s44 are related by the relation s11 − s12 = s44

or

ν 1 1 + = . E E 2µ

(2.4.19)

This is an important relation connecting the elastic constants associated with isotropic materials. The above transformation can also be applied to triclinic, aelotropic, orthotropic, and hexagonal materials to find relationships between the elastic constants. Observe also that some texts postulate the existence of a strain energy function U ∗ which has the property that σij = ∗

∂U ∗ ∂eij .

In this case the strain energy function, in the single index notation, is written

U = cij ei ej where cij and consequently sij are symmetric. In this case the previous discussed symmetries give the following results for the nonzero elastic compliances sij : 13 nonzero constants instead of 20 for aelotropic material, 9 nonzero constants instead of 12 for orthotropic material, and 6 nonzero constants instead of 7 for hexagonal material. This is because of the additional property that sij = sji be symmetric.

251 The previous discussion has shown that for an isotropic material the generalized Hooke’s law (constitutive equations) have the form e11 = e22 = e33 = e21 = e12 = e32 = e23 = e31 = e13 =

1 [σ11 − ν(σ22 + σ33 )] E 1 [σ22 − ν(σ33 + σ11 )] E 1 [σ33 − ν(σ11 + σ22 )] E , 1+ν σ12 E 1+ν σ23 E 1+ν σ13 E

(2.4.20)

where equation (2.4.19) holds. These equations can be expressed in the indicial notation and have the form eij =

1+ν ν σij − σkk δij , E E

(2.4.21)

where σkk = σ11 + σ22 + σ33 is a stress invariant and δij is the Kronecker delta. We can solve for the stress in terms of the strain by performing a contraction on i and j in equation (2.4.21). This gives the dilatation eii =

1+ν 3ν 1 − 2ν σii − σkk = σkk . E E E

Note that from the result in equation (2.4.21) we are now able to solve for the stress in terms of the strain. We find 1+ν ν σij − ekk δij E 1 − 2ν E νE eij = σij − ekk δij 1+ν (1 + ν)(1 − 2ν) E νE or σij = eij + ekk δij . 1+ν (1 + ν)(1 − 2ν) eij =

The tensor equation (2.4.22) represents the six scalar equations E [(1 − ν)e11 + ν(e22 + e33 )] (1 + ν)(1 − 2ν) E [(1 − ν)e22 + ν(e33 + e11 )] = (1 + ν)(1 − 2ν) E [(1 − ν)e33 + ν(e22 + e11 )] = (1 + ν)(1 − 2ν)

E e12 1+ν E e13 = 1+ν E e23 . = 1+ν

σ11 =

σ12 =

σ22

σ13

σ33

σ23

(2.4.22)

252 Alternative Approach to Constitutive Equations The constitutive equation defined by Hooke’s generalized law for isotropic materials can be approached from another point of view. Consider the generalized Hooke’s law σij = cijkl ekl ,

i, j, k, l = 1, 2, 3.

If we transform to a barred system of coordinates, we will have the new Hooke’s law σ ij = cijkl ekl ,

i, j, k, l = 1, 2, 3.

For an isotropic material we require that cijkl = cijkl . Tensors whose components are the same in all coordinate systems are called isotropic tensors. We have previously shown in Exercise 1.3, problem 18, that cpqrs = λδpq δrs + µ(δpr δqs + δps δqr ) + κ(δpr δqs − δps δqr ) is an isotropic tensor when we consider affine type transformations. If we further require the symmetry conditions found in equations (2.4.3) be satisfied, we find that κ = 0 and consequently the generalized Hooke’s law must have the form σpq = cpqrs ers = [λδpq δrs + µ(δpr δqs + δps δqr )] ers σpq = λδpq err + µ(epq + eqp ) or

(2.4.23)

σpq = 2µepq + λerr δpq ,

where err = e11 + e22 + e33 = Θ is the dilatation. The constants λ and µ are called Lame’s constants. Comparing the equation (2.4.22) with equation (2.4.23) we find that the constants λ and µ satisfy the relations µ=

E 2(1 + ν)

λ=

νE . (1 + ν)(1 − 2ν)

(2.4.24)

In addition to the constants E, ν, µ, λ, it is sometimes convenient to introduce the constant k, called the bulk modulus of elasticity, (Exercise 2.3, problem 23), defined by k=

E . 3(1 − 2ν)

(2.4.25)

The stress-strain constitutive equation (2.4.23) was derived using Cartesian tensors. To generalize the equation (2.4.23) we consider a transformation from a Cartesian coordinate system y i , i = 1, 2, 3 to a general coordinate system xi , i = 1, 2, 3. We employ the relations g ij = and σ mn = σij

∂y i ∂y j , ∂xm ∂xn

∂y m ∂y m , ∂xi ∂xj

g ij =

emn = eij

∂y i ∂y j , ∂xm ∂xn

∂xi ∂xj ∂y m ∂y m

or

erq = eij

∂xi ∂xj ∂y r ∂y q

253 and convert equation (2.4.23) to a more generalized form. Multiply equation (2.4.23) by the result σ mn = λ

∂y p ∂y q and verify ∂xm ∂xn

∂y q ∂y q err + µ (emn + enm ) , ∂xm ∂xn

which can be simplified to the form σ mn = λg mn eij g ij + µ (emn + enm ) . Dropping the bar notation, we have σmn = λgmn g ij eij + µ (emn + enm ) . The contravariant form of this equation is σ sr = λg sr g ij eij + µ (g ms g nr + g ns g mr ) emn . Employing the equations (2.4.24) the above result can also be expressed in the form σ

rs

E = 2(1 + ν)

 g ms g nr + g ns g mr +

 2ν sr mn g g emn . 1 − 2ν

(2.4.26)

This is a more general form for the stress-strain constitutive equations which is valid in all coordinate systems. Multiplying by gsk and employing the use of associative tensors, one can verify σji or

E = 1+ν

 eij +

ν em δ i 1 − 2ν m j



i σji = 2µeij + λem m δj ,

are alternate forms for the equation (2.4.26). As an exercise, solve for the strains in terms of the stresses and show that m i δj . Eeij = (1 + ν)σji − νσm

EXAMPLE 2.4-2.

(Hooke’s law)

Let us construct a simple example to test the results we have

developed so far. Consider the tension in a cylindrical bar illustrated in the figure 2.4-3.

Figure 2.4-3. Stress in a cylindrical bar

254 F

Assume that

A

σij =  0 0

0 0 0

 0 0 0

where F is the constant applied force and A is the cross sectional area of the cylinder. Consequently, the generalized Hooke’s law (2.4.21) produces the nonzero strains 1+ν ν σ11 σ11 − (σ11 + σ22 + σ33 ) = E E E −ν σ11 = E −ν σ11 = E

e11 = e22 e33 From these equations we obtain: The first part of Hooke’s law

σ11 = Ee11 or

F = Ee11 . A

The second part of Hooke’s law −e22 −e33 lateral contraction = = = ν = Poisson’s ratio. longitudinal extension e11 e11 This example demonstrates that the generalized Hooke’s law for homogeneous and isotropic materials reduces to our previous one dimensional result given in (2.3.1) and (2.3.2).

Basic Equations of Elasticity Assuming the density L is constant, the basic equations of elasticity reduce to the equations representing conservation of linear momentum and angular momentum together with the strain-displacement relations and constitutive equations. In these equations the body forces are assumed known. These basic equations produce 15 equations in 15 unknowns and are a formidable set of equations to solve. Methods for solving these simultaneous equations are: 1) Express the linear momentum equations in terms of the displacements ui and obtain a system of partial differential equations. Solve the system of partial differential equations for the displacements ui and then calculate the corresponding strains. The strains can be used to calculate the stresses from the constitutive equations. 2) Solve for the stresses and from the stresses calculate the strains and from the strains calculate the displacements. This converse problem requires some additional considerations which will be addressed shortly.

255 Basic Equations of Linear Elasticity • Conservation of linear momentum. σ ij,i + Lbj = Lf j

j = 1, 2, 3.

(2.4.27(a))

where σ ij is the stress tensor, bj is the body force per unit mass and f j is the acceleration. If there is no motion, then f j = 0 and these equations reduce to the equilibrium equations σ ij,i + Lbj = 0 • Conservation of angular momentum. • Strain tensor. eij =

j = 1, 2, 3.

(2.4.27(b))

σij = σji

1 (ui,j + uj,i ) 2

(2.4.28)

where ui denotes the displacement field. • Constitutive equation. For a linear elastic isotropic material we have σji =

E i E e + ek δ i 1 + ν j (1 + ν)(1 − 2ν) k j

i, j = 1, 2, 3

(2.4.29(a))

or its equivalent form σji = 2µeij + λerr δji

i, j = 1, 2, 3,

(2.4.29(b))

where err is the dilatation. This produces 15 equations for the 15 unknowns u1 , u2 , u3 , σ11 , σ12 , σ13 , σ22 , σ23 , σ33 , e11 , e12 , e13 , e22 , e23 , e33 , which represents 3 displacements, 6 strains and 6 stresses. In the above equations it is assumed that the body forces are known.

Navier’s Equations The equations (2.4.27) through (2.4.29) can be combined and written as one set of equations. The resulting equations are known as Navier’s equations for the displacements ui over the range i = 1, 2, 3. To derive the Navier’s equations in Cartesian coordinates, we write the equations (2.4.27),(2.4.28) and (2.4.29) in Cartesian coordinates. We then calculate σij,j in terms of the displacements ui and substitute the results into the momentum equation (2.4.27(a)). Differentiation of the constitutive equations (2.4.29(b)) produces σij,j = 2µeij,j + λekk,j δij .

(2.4.30)

256 A contraction of the strain produces the dilatation err =

1 (ur,r + ur,r ) = ur,r 2

(2.4.31)

From the dilatation we calculate the covariant derivative ekk,j = uk,kj .

(2.4.32)

Employing the strain relation from equation (2.4.28), we calculate the covariant derivative eij,j =

1 (ui,jj + uj,ij ). 2

(2.4.33)

These results allow us to express the covariant derivative of the stress in terms of the displacement field. We find σij,j = µ [ui,jj + uj,ij ] + λδij uk,kj or

(2.4.34)

σij,j = (λ + µ)uk,ki + µui,jj .

Substituting equation (2.4.34) into the linear momentum equation produces the Navier equations: (λ + µ)uk,ki + µui,jj + Lbi = Lfi ,

i = 1, 2, 3.

(2.4.35)

In vector form these equations can be expressed (λ + µ)∇ (∇ ·
u) + µ∇2
u + L
b = Lf
,

(2.4.36)

where
u is the displacement vector,
b is the body force per unit mass and f
is the acceleration. In Cartesian coordinates these equations have the form:  (λ + µ)

∂ 2 u1 ∂ 2 u2 ∂ 2 u3 + + ∂x1 ∂xi ∂x2 ∂xi ∂x3 ∂xi

 + µ∇2 ui + Lbi = L

∂ 2 ui , ∂t2

for i = 1, 2, 3, where ∇2 ui =

∂ 2 ui ∂ 2 ui ∂ 2 ui + + . ∂x1 2 ∂x2 2 ∂x3 2

The Navier equations must be satisfied by a set of functions ui = ui (x1 , x2 , x3 ) which represent the displacement at each point inside some prescribed region R. Knowing the displacement field we can calculate the strain field directly using the equation (2.4.28). Knowledge of the strain field enables us to construct the corresponding stress field from the constitutive equations. In the absence of body forces, such as gravity, the solution to equation (2.4.36) can be represented in the form
u =
u (1) +
u (2) , where
u (1) satisfies div
u (1) = ∇ ·
u (1) = 0 and the vector
u (2) satisfies curl
u (2) = ∇ ×
u (2) = 0. The vector field
u (1) is called a solenoidal field, while the vector field
u (2) is called an irrotational field. Substituting
u into the equation (2.4.36) and setting
b = 0, we find in Cartesian coordinates that  L

∂ 2
u (1) ∂ 2
u (2) + 2 ∂t ∂t2



  = (λ + µ)∇ ∇ ·
u (2) + µ∇2
u (1) + µ∇2
u (2) .

(2.4.37)

257 The vector field
u (1) can be eliminated from equation (2.4.37) by taking the divergence of both sides of the equation. This produces L

∂ 2 ∇ ·
u (2) = (λ + µ)∇2 (∇ ·
u (2) ) + µ∇ · ∇2
u (2) . ∂t2

The displacement field is assumed to be continuous and so we can interchange the order of the operators ∇2 and ∇ and write



∂ 2
u (2) ∇· L − (λ + 2µ)∇2
u (2) ∂t2

 = 0.

This last equation implies that L

∂ 2
u (2) = (λ + 2µ)∇2
u(2) ∂t2

and consequently,
u (2) is a vector wave which moves with the speed

 (λ + 2µ)/L. Similarly, when the vector

field
u (2) is eliminated from the equation (2.4.37), by taking the curl of both sides, we find the vector
u (1) also satisfies a wave equation having the form L This later wave moves with the speed

∂ 2
u (1) = µ∇2
u (1) . ∂t2

 µ/L. The vector
u (2) is a compressive wave, while the wave u (1) is

a shearing wave. The exercises 30 through 38 enable us to write the Navier’s equations in Cartesian, cylindrical or spherical coordinates. In particular, we have for cartesian coordinates ∂2w ∂2u ∂2u ∂2u ∂2u ∂2v ∂2u + ) + µ( 2 + 2 + 2 ) + Lbx =L 2 + 2 ∂x ∂x∂y ∂x∂z ∂x ∂y ∂z ∂t ∂2v ∂2v ∂2u ∂2w ∂2v ∂2v ∂2v + ) + µ( 2 + 2 + 2 ) + Lby =L 2 (λ + µ)( + ∂x∂y ∂y 2 ∂y∂z ∂x ∂y ∂z ∂t ∂2v ∂2w ∂2u ∂2w ∂2w ∂2w ∂2w + + (λ + µ)( ) + µ( 2 + + ) + Lbz =L 2 ∂x∂z ∂y∂z ∂z 2 ∂x ∂y 2 ∂z 2 ∂t (λ + µ)(

and in cylindrical coordinates   ∂uz ∂ 1 ∂ 1 ∂uθ (rur ) + + + ∂r r ∂r r ∂θ ∂z 1 ∂ 2 ur 1 ∂ur ∂ 2 ur ur 2 ∂uθ ∂ 2 ur ∂ 2 ur + 2 ) + Lbr =L 2 + − 2 − 2 µ( 2 + 2 2 ∂r r ∂r r ∂θ r r ∂θ ∂t  ∂z  ∂uz 1 ∂ 1 ∂ 1 ∂uθ (rur ) + + (λ + µ) + r ∂θ r ∂r r ∂θ ∂z 1 ∂ 2 uθ uθ 1 ∂uθ ∂ 2 uθ 2 ∂ur ∂ 2 uθ ∂ 2 uθ + 2 − + + ) + Lb =L µ( 2 + θ ∂r r ∂r r ∂θ2  ∂z 2 r2 ∂θ r2 ∂t2  ∂uz ∂ 1 ∂ 1 ∂uθ (rur ) + + (λ + µ) + ∂z r ∂r r ∂θ ∂z 1 ∂ 2 uz 1 ∂uz ∂ 2 uz ∂ 2 uz ∂ 2 uz + 2 + ) + Lb =L µ( 2 + z ∂r r ∂r r ∂θ2 ∂z 2 ∂t2 (λ + µ)

258 and in spherical coordinates  1 ∂ 2 ∂ 1 1 ∂uφ (ρ (u u ) + sin θ) + + ρ θ ρ2 ∂ρ ρ sin θ ∂θ ρ sin θ ∂φ 2uθ cot θ 2 2 ∂uθ 2 ∂uφ ∂ 2 uρ − ) + Lbρ =L 2 − 2 µ(∇2 uρ − 2 uρ − 2 2 ρ ρ ∂θ ρ ρ sin θ ∂φ ∂t   1 ∂ 2 ∂ 1 ∂ 1 1 ∂uφ (ρ uρ ) + (uθ sin θ) + (λ + µ) + ρ ∂θ ρ2 ∂ρ ρ sin θ ∂θ ρ sin θ ∂φ uθ 2 ∂uρ ∂ 2 uθ 2 cos θ ∂uφ − 2 2 − 2 ) + Lbθ =L 2 µ(∇2 uθ + 2 2 ρ ∂θ ∂t ρ sin θ ρ sin θ ∂φ   ∂ 1 ∂ 2 ∂ 1 1 1 ∂uφ (ρ uρ ) + (uθ sin θ) + (λ + µ) + ρ sin θ ∂φ ρ2 ∂ρ ρ sin θ ∂θ ρ sin θ ∂φ 2 cos θ ∂uθ 1 2 ∂uρ ∂ 2 uφ + 2 2 ) + Lbφ =L 2 µ(∇2 uφ − 2 2 uφ + 2 ρ sin θ ∂φ ∂t ρ sin θ ρ sin θ ∂φ (λ + µ)

∂ ∂ρ



where ∇2 is determined from either equation (2.1.12) or (2.1.13). Boundary Conditions In elasticity the body forces per unit mass (bi , i = 1, 2, 3) are assumed known. In addition one of the following type of boundary conditions is usually prescribed: • The displacements ui ,

i = 1, 2, 3 are prescribed on the boundary of the region R over which a solution

is desired. • The stresses (surface tractions) are prescribed on the boundary of the region R over which a solution is desired. • The displacements ui , i = 1, 2, 3 are given over one portion of the boundary and stresses (surface tractions) are specified over the remaining portion of the boundary. This type of boundary condition is known as a mixed boundary condition. General Solution of Navier’s Equations There has been derived a general solution to the Navier’s equations. It is known as the Papkovich-Neuber solution. In the case of a solid in equilibrium one must solve the equilibrium equations (λ + µ)∇ (∇ ·
u) + µ∇2
u + L
b = 0 or L 1 1 ∇(∇ ·
u) +
b = 0 (ν =  ) ∇2
u + 1 − 2ν µ 2

(2.4.38)

259 THEOREM


A general elastostatic solution of the equation (2.4.38) in terms of harmonic potentials φ,ψ

is
− 4(1 − ν)ψ

u = grad (φ +
r · ψ)

(2.4.39)


are continuous solutions of the equations where φ and ψ ∇2 φ =

−L
r ·
b 4µ(1 − ν)

and


= ∇2 ψ

L
b 4µ(1 − ν)

(2.4.40)

ˆ1 + y e ˆ2 + z e ˆ3 a position vector to a general point (x, y, z) within the continuum. with
r = x e Proof: First we write equation (2.4.38) in the tensor form ui,kk +

1 L (uj,j ) ,i + bi = 0 1 − 2ν µ

(2.4.41)

Now our problem is to show that equation (2.4.39), in tensor form, ui = φ,i + (xj ψj ),i − 4(1 − ν)ψi

(2.4.42)

is a solution of equation (2.4.41). Toward this purpose, we differentiate equation (2.4.42) ui,k = φ,ik + (xj ψj ),ik − 4(1 − ν)ψi,k

(2.4.43)

and then contract on i and k giving ui,i = φ,ii + (xj ψj ),ii − 4(1 − ν)ψi,i .

(2.4.44)

Employing the identity (xj ψj ),ii = 2ψi,i + xi ψi,kk the equation (2.4.44) becomes ui,i = φ,ii + 2ψi,i + xi ψi,kk − 4(1 − ν)ψi,i .

(2.4.45)

By differentiating equation (2.4.43) we establish that ui,kk = φ,ikk + (xj ψj ),ikk − 4(1 − ν)ψi,kk = (φ,kk ),i + ((xj ψj ),kk ),i − 4(1 − ν)ψi,kk

(2.4.46)

= [φ,kk + 2ψj,j + xj ψj,kk ],i − 4(1 − ν)ψi,kk . We use the hypothesis φ,kk =

−Lxj Fj 4µ(1 − ν)

and

ψj,kk =

LFj , 4µ(1 − ν)

and simplify the equation (2.4.46) to the form ui,kk = 2ψj,ji − 4(1 − ν)ψi,kk .

(2.4.47)

Also by differentiating (2.4.45) one can establish that uj,ji = (φ,jj ),i + 2ψj,ji + (xj ψj,kk ),i − 4(1 − ν)ψj,ji     −Lxj Fj Lxj Fj = + 2ψj,ji + − 4(1 − ν)ψj,ji 4µ(1 − ν) ,i 4µ(1 − ν) ,i = −2(1 − 2ν)ψj,ji .

(2.4.48)

260 Finally, from the equations (2.4.47) and (2.4.48) we obtain the desired result that ui,kk +

1 LFi uj,ji + = 0. 1 − 2ν µ

Consequently, the equation (2.4.39) is a solution of equation (2.4.38). As a special case of the above theorem, note that when the body forces are zero, the equations (2.4.40) become ∇2 φ = 0

and


=
0. ∇2 ψ

In this case, we find that equation (2.4.39) is a solution of equation (2.4.38) provided φ and each component of
are harmonic functions. The Papkovich-Neuber potentials are used together with complex variable theory ψ to solve various two-dimensional elastostatic problems of elasticity. Note also that the Papkovich-Neuber
can produce the same value for
u. potentials are not unique as different combinations of φ and ψ Compatibility Equations If we know or can derive the displacement field ui , i = 1, 2, 3 we can then calculate the components of the strain tensor eij =

1 (ui,j + uj,i ). 2

(2.4.49)

Knowing the strain components, the stress is found using the constitutive relations. Consider the converse problem where the strain tensor is given or implied due to the assigned stress field and we are asked to determine the displacement field ui , i = 1, 2, 3. Is this a realistic request? Is it even possible to solve for three displacements given six strain components? It turns out that certain mathematical restrictions must be placed upon the strain components in order that the inverse problem have a solution. These mathematical restrictions are known as compatibility equations. That is, we cannot arbitrarily assign six strain components eij and expect to find a displacement field ui , i = 1, 2, 3 with three components which satisfies the strain relation as given in equation (2.4.49). EXAMPLE 2.4-3. Suppose we are given the two partial differential equations, ∂u =x+y ∂x

and

∂u = x3 . ∂y

Can we solve for u = u(x, y)? The answer to this question is “no”, because the given equations are inconsistent. The inconsistency is illustrated if we calculate the mixed second derivatives from each equation. We ∂2u ∂2u = 1 and from the second equation we calculate = 3x2 . These find from the first equation that ∂x∂y ∂y∂x √ mixed second partial derivatives are unequal for all x different from 3/3. In general, if we have two first ∂u ∂u order partial differential equations = f (x, y) and = g(x, y), then for consistency (integrability of ∂x ∂y the equations) we require that the mixed partial derivatives ∂f ∂2u ∂g ∂2u = = = ∂x∂y ∂y ∂y∂x ∂x be equal to one another for all x and y values over the domain for which the solution is desired. This is an example of a compatibility equation.

261 A similar situation occurs in two dimensions for a material in a state of strain where ezz = ezx = ezy = 0, called plane strain. In this case, are we allowed to arbitrarily assign values to the strains exx , eyy and exy and from these strains determine the displacement field u = u(x, y) and v = v(x, y) in the x− and y−directions? Let us try to answer this question. Assume a state of plane strain where ezz = ezx = ezy = 0. Further, let us assign 3 arbitrary functional values f, g, h such that exx =

∂u = f (x, y), ∂x

exy =

1 2



∂u ∂v + ∂y ∂x

 = g(x, y),

eyy =

∂v = h(x, y). ∂y

We must now decide whether these equations are consistent. That is, will we be able to solve for the displacement field u = u(x, y) and v = v(x, y)? To answer this question, let us derive a compatibility equation (integrability condition). From the given equations we can calculate the following partial derivatives ∂ 2 exx ∂3u ∂2f = = 2 2 ∂y ∂x∂y ∂y 2 ∂ 2 eyy ∂3v ∂2h = = 2 2 ∂x ∂y∂x ∂x2 2 3 ∂ u ∂ exy ∂3v ∂2g = . 2 + = 2 ∂x∂y ∂x∂y 2 ∂y∂x2 ∂x∂y This last equation gives us the compatibility equation 2

∂ 2 exx ∂ 2 eyy ∂ 2 exy = + ∂x∂y ∂y 2 ∂x2

or the functions g, f, h must satisfy the relation 2

∂2f ∂2h ∂2g = + . ∂x∂y ∂y 2 ∂x2

Cartesian Derivation of Compatibility Equations If the displacement field ui , i = 1, 2, 3 is known we can derive the strain and rotation tensors eij =

1 (ui,j + uj,i ) 2

and

ωij =

1 (ui,j − uj,i ). 2

(2.4.50)

Now work backwards. Assume the strain and rotation tensors are given and ask the question, “Is it possible to solve for the displacement field ui , i = 1, 2, 3?” If we view the equation (2.4.50) as a system of equations with unknowns eij , ωij and ui and if by some means we can eliminate the unknowns ωij and ui then we will be left with equations which must be satisfied by the strains eij . These equations are known as the compatibility equations and they represent conditions which the strain components must satisfy in order that a displacement function exist and the equations (2.4.37) are satisfied. Let us see if we can operate upon the equations (2.4.50) to eliminate the quantities ui and ωij and hence derive the compatibility equations. Addition of the equations (2.4.50) produces ui,j =

∂ui = eij + ωij . ∂xj

(2.4.51)

262 Differentiate this expression with respect to xk and verify the result ∂ 2 ui ∂eij ∂ωij = + . ∂xj ∂xk ∂xk ∂xk

(2.4.52)

We further assume that the displacement field is continuous so that the mixed partial derivatives are equal and

∂ 2 ui ∂ 2 ui = . ∂xj ∂xk ∂xk ∂xj

(2.4.53)

Interchanging j and k in equation (2.4.52) gives us ∂ 2 ui ∂eik ∂ωik = + . ∂xk ∂xj ∂xj ∂xj

(2.4.54)

Equating the second derivatives from equations (2.4.54) and (2.4.52) and rearranging terms produces the result

Making the observation that ωij

∂eij ∂eik ∂ωik ∂ωij − = − (2.4.55) ∂xk ∂xj ∂xj ∂xk ∂ωik ∂ωij ∂ωjk satisfies − = , the equation (2.4.55) simplifies to the ∂xj ∂xk ∂xi

form

The term involving ωjk

∂eik ∂ωjk ∂eij − = . ∂xk ∂xj ∂xi can be eliminated by using the mixed partial derivative relation

(2.4.56)

∂ 2 ωjk ∂ 2 ωjk = . (2.4.57) ∂xi ∂xm ∂xm ∂xi To derive the compatibility equations we differentiate equation (2.4.56) with respect to xm and then interchanging the indices i and m and substitute the results into equation (2.4.57). This will produce the compatibility equations ∂ 2 eij ∂ 2 emk ∂ 2 eik ∂ 2 emj + − − = 0. (2.4.58) ∂xm ∂xk ∂xi ∂xj ∂xm ∂xj ∂xi ∂xk This is a set of 81 partial differential equations which must be satisfied by the strain components. Fortunately, due to symmetry considerations only 6 of these 81 equations are distinct. These 6 distinct equations are known as the St. Venant’s compatibility equations and can be written as ∂ 2 e11 ∂ 2 e12 ∂ 2 e23 ∂ 2 e31 = − + 2 ∂x2 ∂x3 ∂x1 ∂x3 ∂x1 ∂x1 ∂x2 2 2 2 ∂ e22 ∂ e23 ∂ e31 ∂ 2 e12 = − + 2 ∂x1 ∂x3 ∂x2 ∂x1 ∂x2 ∂x2 ∂x3 ∂ 2 e33 ∂ 2 e31 ∂ 2 e12 ∂ 2 e23 = − + ∂x1 ∂x2 ∂x3 ∂x2 ∂x3 2 ∂x3 ∂x1 2 2 2 ∂ e12 ∂ e11 ∂ e22 2 = + ∂x1 ∂x2 ∂x2 2 ∂x1 2 2 2 ∂ e23 ∂ e22 ∂ 2 e33 2 = + ∂x2 ∂x3 ∂x3 2 ∂x2 2 2 2 ∂ e31 ∂ e33 ∂ 2 e11 2 = + . 2 ∂x3 ∂x1 ∂x1 ∂x3 2 Observe that the fourth compatibility equation is the same as that derived in the example 2.4-3.

(2.4.59)

These compatibility equations can also be expressed in the indicial form eij,km + emk,ji − eik,jm − emj,ki = 0.

(2.4.60)

263 Compatibility Equations in Terms of Stress In the generalized Hooke’s law, equation (2.4.29), we can solve for the strain in terms of stress. This in turn will give rise to a representation of the compatibility equations in terms of stress. The resulting equations are known as the Beltrami-Michell equations. Utilizing the strain-stress relation eij =

1+ν ν σij − σkk δij E E

we substitute for the strain in the equations (2.4.60) and rearrange terms to produce the result σij,km + σmk,ji − σik,jm − σmj,ki = ν [δij σnn,km + δmk σnn,ji − δik σnn,jm − δmj σnn,ki ] . 1+ν

(2.4.61)

Now only 6 of these 81 equations are linearly independent. It can be shown that the 6 linearly independent equations are equivalent to the equations obtained by setting k = m and summing over the repeated indices. We then obtain the equations σij,mm + σmm,ij − (σim,m ),j − (σmj,m ),i =

ν [δij σnn,mm + σnn,ij ] . 1+ν

Employing the equilibrium equation σij,i + Lbj = 0 the above result can be written in the form σij,mm +

1 ν σkk,ij − δij σnn,mm = −(Lbi ),j − (Lbj ),i 1+ν 1+ν

∇2 σij +

1 ν σkk,ij − δij σnn,mm = −(Lbi ),j − (Lbj ),i . 1+ν 1+ν

or

This result can be further simplified by observing that a contraction on the indices k and i in equation (2.4.61) followed by a contraction on the indices m and j produces the result σij,ij =

1−ν σnn,jj . 1+ν

Consequently, the Beltrami-Michell equations can be written in the form ∇2 σij +

1 ν σpp,ij = − δij (Lbk ) ,k − (Lbi ) ,j − (Lbj ) ,i . 1+ν 1−ν

(2.4.62)

Their derivation is left as an exercise. The Beltrami-Michell equations together with the linear momentum (equilibrium) equations σij,i + Lbj = 0 represent 9 equations in six unknown stresses. This combinations of equations is difficult to handle. An easier combination of equations in terms of stress functions will be developed shortly. The Navier equations with boundary conditions are difficult to solve in general. Let us take the momentum equations (2.4.27(a)), the strain relations (2.4.28) and constitutive equations (Hooke’s law) (2.4.29) and make simplifying assumptions so that a more tractable systems results.

264 Plane Strain The plane strain assumption usually is applied in situations where there is a cylindrical shaped body whose axis is parallel to the z axis and loads are applied along the z−direction. In any x-y plane we assume that the surface tractions and body forces are independent of z. We set all strains with a subscript z equal to zero. Further, all solutions for the stresses, strains and displacements are assumed to be only functions of x and y and independent of z. Note that in plane strain the stress σzz is different from zero. In Cartesian coordinates the strain tensor is expressible in terms of its physical components which can be represented in the matrix form 

e11  e21 e31

e12 e22 e32

  e13 exx e23  =  eyx e33 ezx

exy eyy ezy

 exz eyz  . ezz

If we assume that all strains which contain a subscript z are zero and the remaining strain components are functions of only x and y, we obtain a state of plane strain. For a state of plane strain, the stress components are obtained from the constitutive equations. The condition of plane strain reduces the constitutive equations to the form: exx = eyy = 0= exy = eyx = ezy = eyz = ezx = exz = where σxx ,

σyy ,

1 [σxx − ν(σyy + σzz )] E 1 [σyy − ν(σzz + σxx )] E 1 [σzz − ν(σxx + σyy )] E 1+ν σxy E 1+ν σyz = 0 E 1+ν σxz = 0 E

σzz ,

σxy ,

σxz ,

E [(1 − ν)exx + νeyy ] (1 + ν)(1 − 2ν) E [(1 − ν)eyy + νexx ] = (1 + ν)(1 − 2ν) E [ν(eyy + exx )] = (1 + ν)(1 − 2ν) E exy = 1+ν =0

σxx = σyy σzz σxy σxz

(2.4.63)

σyz = 0

σyz are the physical components of the stress. The above constitutive

equations imply that for a state of plane strain we will have σzz = ν(σxx + σyy ) 1+ν exx = [(1 − ν)σxx − νσyy ] E 1+ν [(1 − ν)σyy − νσxx ] eyy = E 1+ν σxy . exy = E Also under these conditions the compatibility equations reduce to ∂ 2 exx ∂ 2 eyy ∂ 2 exy . + =2 2 2 ∂y ∂x ∂x∂y

265 Plane Stress An assumption of plane stress is usually applied to thin flat plates. The plate thinness is assumed to be in the z−direction and loads are applied perpendicular to z. Under these conditions all stress components with a subscript z are assumed to be zero. The remaining stress components are then treated as functions of x and y. In Cartesian coordinates the stress tensor is expressible in terms of its physical components and can be represented by the matrix



σ11  σ21 σ31

σ12 σ22 σ32

  σxx σ13 σ23  =  σyx σ33 σzx

σxy σyy σzy

 σxz σyz  . σzz

If we assume that all the stresses with a subscript z are zero and the remaining stresses are only functions of x and y we obtain a state of plane stress. The constitutive equations simplify if we assume a state of plane stress. These simplified equations are 1 ν σxx − σyy E E 1 ν = σyy − σxx E E ν = − (σxx + σyy ) E 1+ν σxy = E =0

E [exx + νeyy ] 1 − ν2 E = [eyy + νexx ] 1 − ν2 = 0 = (1 − ν)ezz + ν(exx + eyy ) E exy = 1+ν =0

exx =

σxx =

eyy

σyy

ezz exy exz

σzz σxy σyz

(2.4.64)

σxz = 0

eyz = 0.

For a state of plane stress the compatibility equations reduce to ∂ 2 exx ∂ 2 eyy ∂ 2 exy + =2 2 2 ∂y ∂x ∂x∂y

(2.4.65)

and the three additional equations ∂ 2 ezz = 0, ∂x2

∂ 2 ezz = 0, ∂y 2

∂ 2 ezz = 0. ∂x∂y

These three additional equations complicate the plane stress problem. Airy Stress Function In Cartesian coordinates we examine the equilibrium equations (2.4.25(b)) under the conditions of plane strain. In terms of physical components we find that these equations reduce to ∂σxx ∂σxy + + Lbx = 0, ∂x ∂y

∂σyy ∂σyx + + Lby = 0, ∂x ∂y

∂σzz = 0. ∂z

The last equation is satisfied since σzz is a function of x and y. If we further assume that the body forces are conservative and derivable from a potential function V by the operation L
b = −grad V or Lbi = −V ,i we can express the above equilibrium equations in the form: ∂σxx ∂σxy ∂V + − =0 ∂x ∂y ∂x ∂σyx ∂σyy ∂V + − =0 ∂x ∂y ∂y

(2.4.66)

266 We will consider these equations together with the compatibility equations (2.4.65). The equations (2.4.66) will be automatically satisfied if we introduce a scalar function φ = φ(x, y) and assume that the stresses are derivable from this function and the potential function V according to the rules: σxx =

∂2φ +V ∂y 2

σxy = −

∂2φ ∂x∂y

σyy =

∂2φ + V. ∂x2

(2.4.67)

The function φ = φ(x, y) is called the Airy stress function after the English astronomer and mathematician Sir George Airy (1801–1892). Since the equations (2.4.67) satisfy the equilibrium equations we need only consider the compatibility equation(s). For a state of plane strain we substitute the relations (2.4.63) into the compatibility equation (2.4.65) and write the compatibility equation in terms of stresses. We then substitute the relations (2.4.67) and express the compatibility equation in terms of the Airy stress function φ. These substitutions are left as exercises. After all these substitutions the compatibility equation, for a state of plane strain, reduces to the form

∂4φ ∂4φ ∂ 4 φ 1 − 2ν +2 2 2 + 4 + 4 ∂x ∂x ∂y ∂y 1−ν



∂2V ∂2V + 2 ∂x ∂y 2

 = 0.

(2.4.68)

In the special case where there are no body forces we have V = 0 and equation (2.4.68) is further simplified to the biharmonic equation. ∂4φ ∂4φ ∂4φ + 2 + = 0. ∂x4 ∂x2 ∂y 2 ∂y 4 In polar coordinates the biharmonic equation is written  2  2  ∂ ∂ φ 1 ∂φ 1 ∂2 1 ∂2φ 1 ∂ ∇4 φ = ∇2 (∇2 φ) = + + + + = 0. ∂r2 r ∂r r2 ∂θ2 ∂r2 r ∂r r2 ∂θ2 ∇4 φ =

(2.4.69)

For conditions of plane stress, we can again introduce an Airy stress function using the equations (2.4.67). However, an exact solution of the plane stress problem which satisfies all the compatibility equations is difficult to obtain. By removing the assumptions that σxx , σyy , σxy are independent of z, and neglecting body forces, it can be shown that for symmetrically distributed external loads the stress function φ can be represented in the form φ=ψ−

νz 2 ∇2 ψ 2(1 + ν)

(2.4.70)

where ψ is a solution of the biharmonic equation ∇4 ψ = 0. Observe that if z is very small, (the condition of a thin plate), then equation (2.4.70) gives the approximation φ ≈ ψ. Under these conditions, we obtain the approximate solution by using only the compatibility equation (2.4.65) together with the stress function defined by equations (2.4.67) with V = 0. Note that the solution we obtain from equation (2.4.69) does not satisfy all the compatibility equations, however, it does give an excellent first approximation to the solution in the case where the plate is very thin. In general, for plane strain or plane stress problems, the equation (2.4.68) or (2.4.69) must be solved for the Airy stress function φ which is defined over some region R. In addition to specifying a region of the x, y plane, there are certain boundary conditions which must be satisfied. The boundary conditions specified for the stress will translate through the equations (2.4.67) to boundary conditions being specified for φ. In the special case where there are no body forces, both the problems for plane stress and plane strain are governed by the biharmonic differential equation with appropriate boundary conditions.

267 EXAMPLE 2.4-4 Assume there exist a state of plane strain with zero body forces. For F11 , F12 , F22 constants, consider the function defined by φ = φ(x, y) =

1 F22 x2 − 2F12 xy + F11 y 2 . 2

This function is an Airy stress function because it satisfies the biharmonic equation ∇4 φ = 0. The resulting stress field is σxx =

∂2φ = F11 ∂y 2

σyy =

∂ 2φ = F22 ∂x2

σxy = −

∂2φ = F12 . ∂x∂y

This example, corresponds to stresses on an infinite flat plate and illustrates a situation where all the stress components are constants for all values of x and y. In this case, we have σzz = ν(F11 +F22 ). The corresponding strain field is obtained from the constitutive equations. We find these strains are exx =

1+ν [(1 − ν)F11 − νF22 ] E

eyy =

1+ν [(1 − ν)F22 − νF11 ] E

exy =

1+ν F12 . E

The displacement field is found to be   1+ν 1+ν u = u(x, y) = [(1 − ν)F11 − νF22 ] x + F12 y + c1 y + c2 E E   1+ν 1+ν [(1 − ν)F22 − νF11 ] y + v = v(x, y) = F12 x − c1 x + c3 , E E with c1 , c2 , c3 constants, and is obtained by integrating the strain displacement equations given in Exercise 2.3, problem 2. EXAMPLE 2.4-5. A special case from the previous example is obtained by setting F22 = F12 = 0. This is the situation of an infinite plate with only tension in the x−direction. In this special case we have φ = 12 F11 y 2 . Changing to polar coordinates we write φ = φ(r, θ) =

F11 2 F11 2 2 r sin θ = r (1 − cos 2θ). 2 4

The Exercise 2.4, problem 20, suggests we utilize the Airy equations in polar coordinates and calculate the stresses

1 ∂2φ 1 ∂φ F11 + 2 2 = F11 cos2 θ = (1 + cos 2θ) r ∂r r ∂θ 2 2 ∂ φ F11 (1 − cos 2θ) = = F11 sin2 θ = 2 ∂r 2 F11 1 ∂φ 1 ∂ 2 φ − =− sin 2θ. = 2 r ∂θ r ∂r∂θ 2

σrr = σθθ σrθ

268 EXAMPLE 2.4-6. We now consider an infinite plate with a circular hole x2 + y 2 = a2 which is traction free. Assume the plate has boundary conditions at infinity defined by σxx = F11 ,

σyy = 0,

σxy = 0. Find

the stress field. Solution: The traction boundary condition at r = a is ti = σmi nm or t1 = σ11 n1 + σ12 n2

and

t2 = σ12 n1 + σ22 n2 .

For polar coordinates we have n1 = nr = 1, n2 = nθ = 0 and so the traction free boundary conditions at the surface of the hole are written σrr |r=a = 0 and σrθ |r=a = 0. The results from the previous example are used as the boundary conditions at infinity. Our problem is now to solve for the Airy stress function φ = φ(r, θ) which is a solution of the biharmonic equation. The previous example 2.4-5 and the form of the boundary conditions at infinity suggests that we assume a solution to the biharmonic equation which has the form φ = φ(r, θ) = f1 (r) + f2 (r) cos 2θ, where f1 , f2 are unknown functions to be determined. Substituting the assumed solution into the biharmonic equation produces the equation 

d2 1 d + dr2 r dr

  2    d 4 1 1 d 1  f2  − f + + − 4 f cos 2θ = 0. f1 + f1 + 2 r dr2 r dr r2 r 2 r2

We therefore require that f1 , f2 be chosen to satisfy the equations 

or

d2 1 d + 2 dr r dr (iv)

r 4 f1

  1 f1 + f1 = 0 r

+ 2r3 f1 − r2 f1 + rf1 = 0



d2 4 1 d − 2 + 2 dr r dr r (iv)

r 4 f2

  1 f2 f2 + f2 − 4 2 = 0 r r

+ 2r3 f2 − 9r2 f2 + 9rf2 = 0

These equations are Cauchy type equations. Their solutions are obtained by assuming a solution of the form f1 = rλ and f2 = rm and then solving for the constants λ and m. We find the general solutions of the above equations are f1 = c1 r2 ln r + c2 r2 + c3 ln r + c4

and f2 = c5 r2 + c6 r4 +

c7 + c8 . r2

The constants ci , i = 1, . . . , 8 are now determined from the boundary conditions. The constant c4 can be arbitrary since the derivative of a constant is zero. The remaining constants are determined from the stress conditions. Using the results from Exercise 2.4, problem 20, we calculate the stresses c3  c7 c8  cos 2θ − 2c + 6 + 4 5 r2  r4 r2  c3 c7 = c1 (3 + 2 ln r) + 2c2 − 2 + 2c5 + 12c6 r2 + 6 4 cos 2θ r r  c7 c8  = 2c5 + 6c6 r2 − 6 4 − 2 2 sin 2θ. r r

σrr = c1 (1 + 2 ln r) + 2c2 + σθθ σrθ

269 The stresses are to remain bounded for all values of r and consequently we require c1 and c6 to be zero to avoid infinite stresses for large values of r. The stress σrr |r=a = 0 requires that 2c2 +

c3 =0 a2

and 2c5 + 6

The stress σrθ |r=a = 0 requires that 2c5 − 6

c7 c8 + 4 2 = 0. 4 a a

c7 c8 − 2 2 = 0. 4 a a

In the limit as r → ∞ we require that the stresses must satisfy the boundary conditions from the previous F11 F11 example 2.4-5. This leads to the equations 2c2 = and 2c5 = − . Solving the above system of equations 2 2 produces the Airy stress function φ = φ(r, θ) =

F11 2 a2 F11 + r − F11 ln r + c4 + 4 4 2



F11 a2 F11 2 F11 a4 − r − 2 4 4r2

 cos 2θ

and the corresponding stress field is σrr σrθ σθθ

    F11 a2 F11 a2 a4 = 1− 2 + 1 + 3 4 − 4 2 cos 2θ 2 r 2 r r   F11 a2 a4 =− 1 − 3 4 + 2 2 sin 2θ 2 r r     F11 F11 a2 a4 = 1+ 2 − 1 + 3 4 cos 2θ. 2 r 2 r

There is a maximum stress σθθ = 3F11 at θ = π/2, 3π/2 and a minimum stress σθθ = −F11 at θ = 0, π. The effect of the circular hole has been to magnify the applied stress. The factor of 3 is known as a stress concentration factor. In general, sharp corners and unusually shaped boundaries produce much higher stress concentration factors than rounded boundaries. EXAMPLE 2.4-7. Consider an infinite cylindrical tube, with inner radius R1 and the outer radius R0 , which is subjected to an internal pressure P1 and an external pressure P0 as illustrated in the figure 2.4-7. Find the stress and displacement fields. Solution: Let ur , uθ , uz denote the displacement field. We assume that uθ = 0 and uz = 0 since the cylindrical surface r equal to a constant does not move in the θ or z directions. The displacement ur = ur (r) is assumed to depend only upon the radial distance r. Under these conditions the Navier equations become d (λ + 2µ) dr This equation has the solution ur = c1 err =

dur , dr



 1 d (rur ) = 0. r dr

c2 r + and the strain components are found from the relations 2 r eθθ =

ur , r

ezz = erθ = erz = ezθ = 0.

The stresses are determined from Hooke’s law (the constitutive equations) and we write σij = λδij Θ + 2µeij ,

270 where Θ=

ur 1 ∂ ∂ur + = (rur ) ∂r r r ∂r

is the dilatation. These stresses are found to be σrr = (λ + µ)c1 −

2µ c2 r2

σθθ = (λ + µ)c1 +

We now apply the boundary conditions   2µ σrr |r=R1 nr = − (λ + µ)c1 − 2 c2 = +P1 R1

2µ c2 r2

σzz = λc1

σrθ = σrz = σzθ = 0.



 2µ and σrr |r=R0 nr = (λ + µ)c1 − 2 c2 = −P0 . R0

Solving for the constants c1 and c2 we find c1 =

R12 P1 − R02 P0 , (λ + µ)(R02 − R12 )

c2 =

R12 R02 (P1 − P0 ) . 2µ(R02 − R12 )

This produces the displacement field     r R02 r R12 R12 P1 R02 P0 ur = + + − , 2(R02 − R12 ) λ + µ µr 2(R02 − R12 ) λ + µ µr and stress fields

uθ = 0,

uz = 0,

    R12 P1 R02 R02 P0 R12 1 − − 1 − R02 − R12 r2 R02 − R12 r2     R2 P1 R2 R2 P0 R2 = 2 1 2 1 + 20 − 2 0 2 1 + 21 R − R1 r R0 − R1 r 0  2 2 λ R1 P1 − R0 P0 = λ+µ R02 − R12

σrr = σθθ σzz

σrz = σzθ = σrθ = 0 EXAMPLE 2.4-8. By making simplifying assumptions the Navier equations can be reduced to a more tractable form. For example, we can reduce the Navier equations to a one dimensional problem by making the following assumptions 1. Cartesian coordinates x1 = x, 2. u1 = u1 (x, t),

x2 = y,

x3 = z

u2 = u3 = 0.

3. There are no body forces. ∂u1 (x, 0) =0 ∂t 5. Boundary conditions of the displacement type u1 (0, t) = f (t), 4. Initial conditions of

u1 (x, 0) = 0 and

where f (t) is a specified function. These assumptions reduce the Navier equations to the single one dimensional wave equation ∂ 2 u1 ∂ 2 u1 = α2 , 2 ∂t ∂x2

α2 =

λ + 2µ . ρ

The solution of this equation is  u1 (x, t) =

f (t − x/α), 0,

x ≤ αt . x > αt

271 The solution represents a longitudinal elastic wave propagating in the x−direction with speed α. The stress wave associated with this displacement is determined from the constitutive equations. We find σxx = (λ + µ)exx = (λ + µ)

∂u1 . ∂x

This produces the stress wave  σxx =

 − (λ+µ) α f (t − x/α),

x ≤ αt

0,

x > αt

.

Here there is a discontinuity in the stress wave front at x = αt. Summary of Basic Equations of Elasticity The equilibrium equations for a continuum have been shown to have the form σ ij,j + Lbi = 0, where bi are the body forces per unit mass and σ ij is the stress tensor. In addition to the above equations we have the constitutive equations σij = λekk δij + 2µeij which is a generalized Hooke’s law relating stress to strain for a linear elastic isotropic material. The strain tensor is related to the displacement field ui by 1 the strain equations eij = (ui,j + uj,i ) . These equations can be combined to obtain the Navier equations 2 µui,jj + (λ + µ)uj,ji + Lbi = 0. The above equations must be satisfied at all interior points of the material body. A boundary value problem results when conditions on the displacement of the boundary are specified. That is, the Navier equations must be solved subject to the prescribed displacement boundary conditions. If conditions on the stress at the boundary are specified, then these prescribed stresses are called surface tractions and must satisfy the relations ti (n) = σ ij nj , where ni is a unit outward normal vector to the boundary. For surface tractions, we need to use the compatibility equations combined with the constitutive equations and equilibrium equations. This gives rise to the Beltrami-Michell equations of compatibility σij,kk +

1 ν σkk,ij + L(bi,j + bj,i ) + Lbk,k = 0. 1+ν 1−ν

Here we must solve for the stress components throughout the continuum where the above equations hold subject to the surface traction boundary conditions. Note that if an elasticity problem is formed in terms of the displacement functions, then the compatibility equations can be ignored. For mixed boundary value problems we must solve a system of equations consisting of the equilibrium equations, constitutive equations, and strain displacement equations. We must solve these equations subject to conditions where the displacements ui are prescribed on some portion(s) of the boundary and stresses are prescribed on the remaining portion(s) of the boundary. Mixed boundary value problems are more difficult to solve. For elastodynamic problems, the equilibrium equations are replaced by equations of motion. In this case we need a set of initial conditions as well as boundary conditions before attempting to solve our basic system of equations.

272 EXERCISE 2.4
1. Verify the generalized Hooke’s law constitutive equations for hexagonal materials. In the following problems the Young’s modulus E, Poisson’s ratio ν, the shear modulus or modulus of rigidity µ (sometimes denoted by G in Engineering texts), Lame’s constant λ and the bulk modulus of elasticity k are assumed to satisfy the equations (2.4.19), (2.4.24) and (2.4.25). Show that these relations imply the additional relations given in the problems 2 through 6.
2.

µ(3λ + 2µ) µ+λ λ(1 + ν)(1 − 2ν) E= ν

E=


3.


4.


5.


6.

E=

3k(1 − 2ν) µ= 2(1 + ν) 3Ek µ= 9k − E

3kν 1+ν µ(2µ − E) λ= E − 3µ

9kµ µ + 3k

E = 3(1 − 2ν)k

E = 2µ(1 + ν)

 (E + λ)2 + 8λ2 + (E + 3λ) k= 6 2µ + 3λ k= 3

λ=

E=

 (E + λ)2 + 8λ2 − (E + λ) ν= 4λ 3k − 2µ ν= 2(µ + 3k)

3k − E ν= 6k λ ν= 2(µ + λ)

3(k − λ) µ= 2 λ(1 − 2ν) µ= 2ν

9k(k − λ) 3k − λ

E 3(1 − 2ν) µE k= 3(3µ − E)

k=

E − 2µ 2µ λ ν= 3k − λ ν=

2µ(1 + ν) 3(1 − 2ν) λ(1 + ν) k= 3ν

k=

 (E + λ)2 + 8λ2 + (E − 3λ) µ= 4 E µ= 2(1 + ν)

3k − 2µ 3 3k(3k − E) λ= 9k − E

λ=

νE (1 + ν)(1 − 2ν) 2µν λ= 1 − 2ν

λ=


7. The previous exercises 2 through 6 imply that the generalized Hooke’s law σij = 2µeij + λδij ekk is expressible in a variety of forms. From the set of constants (µ,λ,ν,E,k) we can select any two constants and then express Hooke’s law in terms of these constants. (a) Express the above Hooke’s law in terms of the constants E and ν. (b) Express the above Hooke’s law in terms of the constants k and E. (c) Express the above Hooke’s law in terms of physical components. Hint: The quantity ekk is an invariant hence all you need to know is how second order tensors are represented in terms of physical components. See also problems 10,11,12.

273
8. Verify the equations defining the stress for plane strain in Cartesian coordinates are E [(1 − ν)exx + νeyy ] (1 + ν)(1 − 2ν) E [(1 − ν)eyy + νexx ] = (1 + ν)(1 − 2ν) Eν [exx + eyy ] = (1 + ν)(1 − 2ν) E exy = 1+ν =0

σxx = σyy σzz σxy σyz = σxz


9. Verify the equations defining the stress for plane strain in polar coordinates are E [(1 − ν)err + νeθθ ] (1 + ν)(1 − 2ν) E [(1 − ν)eθθ + νerr ] = (1 + ν)(1 − 2ν) νE [err + eθθ ] = (1 + ν)(1 − 2ν) E erθ = 1+ν =0

σrr = σθθ σzz σrθ σrz = σθz


10. Write out the independent components of Hooke’s generalized law for strain in terms of stress, and stress in terms of strain, in Cartesian coordinates. Express your results using the parameters ν and E. (Assume a linear elastic, homogeneous, isotropic material.)
11. Write out the independent components of Hooke’s generalized law for strain in terms of stress, and stress in terms of strain, in cylindrical coordinates. Express your results using the parameters ν and E. (Assume a linear elastic, homogeneous, isotropic material.)
12. Write out the independent components of Hooke’s generalized law for strain in terms of stress, and stress in terms of strain in spherical coordinates. Express your results using the parameters ν and E. (Assume a linear elastic, homogeneous, isotropic material.)
13. For a linear elastic, homogeneous, isotropic material assume there exists a state of plane strain in Cartesian coordinates. Verify the equilibrium equations are ∂σxx ∂σxy + + Lbx = 0 ∂x ∂y ∂σyy ∂σyx + + Lby = 0 ∂x ∂y ∂σzz + Lbz = 0 ∂z Hint: See problem 14, Exercise 2.3.

274
14 . For a linear elastic, homogeneous, isotropic material assume there exists a state of plane strain in polar coordinates. Verify the equilibrium equations are 1 ∂σrθ 1 ∂σrr + + (σrr − σθθ ) + Lbr = 0 ∂r r ∂θ r 1 ∂σθθ 2 ∂σrθ + + σrθ + Lbθ = 0 ∂r r ∂θ r ∂σzz + Lbz = 0 ∂z Hint: See problem 15, Exercise 2.3.
15. For a linear elastic, homogeneous, isotropic material assume there exists a state of plane stress in Cartesian coordinates. Verify the equilibrium equations are ∂σxx ∂σxy + + Lbx = 0 ∂x ∂y ∂σyy ∂σyx + + Lby = 0 ∂x ∂y
16. Determine the compatibility equations in terms of the Airy stress function φ when there exists a state of plane stress. Assume the body forces are derivable from a potential function V.
17. For a linear elastic, homogeneous, isotropic material assume there exists a state of plane stress in polar coordinates. Verify the equilibrium equations are ∂σrr 1 ∂σrθ 1 + + (σrr − σθθ ) + Lbr = 0 ∂r r ∂θ r 1 ∂σθθ 2 ∂σrθ + + σrθ + Lbθ = 0 ∂r r ∂θ r

275
18. Figure 2.4-4 illustrates the state of equilibrium on an element in polar coordinates assumed to be of unit length in the z-direction. Verify the stresses given in the figure and then sum the forces in the r and θ directions to derive the same equilibrium laws developed in the previous exercise.

Figure 2.4-4. Polar element in equilibrium.

Hint: Resolve the stresses into components in the r and θ directions. Use the results that sin dθ 2 ≈ cos

dθ 2

dθ 2

and

≈ 1 for small values of dθ. Sum forces and then divide by rdr dθ and take the limit as dr → 0 and

dθ → 0.
19.

Express each of the physical components of plane stress in polar coordinates, σrr , σθθ , and σrθ

in terms of the physical components of stress in Cartesian coordinates σxx , σyy , σxy . Hint: Consider the ∂xa ∂xb . transformation law σ ij = σab i ∂x ∂xj
20. Use the results from problem 19 and assume the stresses are derivable from the relations σxx = V +

∂2φ , ∂y 2

σxy = −

∂2φ , ∂x∂y

σyy = V +

∂2φ ∂x2

where V is a potential function and φ is the Airy stress function. Show that upon changing to polar coordinates the Airy equations for stress become σrr = V +

1 ∂2φ 1 ∂φ + 2 2, r ∂r r ∂θ

σrθ =

1 ∂φ 1 ∂ 2 φ − , r2 ∂θ r ∂r∂θ

σθθ = V +

∂2φ . ∂r2


21. Verify that the Airy stress equations in polar coordinates, given in problem 20, satisfy the equilibrium equations in polar coordinates derived in problem 17.

276
22.

In Cartesian coordinates show that the traction boundary conditions, equations (2.3.11), can be

written in terms of the constants λ and µ as      ∂u1 ∂u1 ∂u1 ∂u2 ∂u3 + + T1 = λn1 ekk + µ 2n1 1 + n2 + n 3 ∂x ∂x2 ∂x1 ∂x3 ∂x1      ∂u2 ∂u2 ∂u2 ∂u1 ∂u3 + + n + + 2n T2 = λn2 ekk + µ n1 2 3 ∂x1 ∂x2 ∂x2 ∂x3 ∂x2       ∂u3 ∂u3 ∂u3 ∂u1 ∂u2 T3 = λn3 ekk + µ n1 + + + n + 2n 2 3 ∂x1 ∂x3 ∂x2 ∂x3 ∂x3 where (n1 , n2 , n3 ) are the direction cosines of the unit normal to the surface, u1 , u2 , u3 are the components of the displacements and T1 , T2 , T3 are the surface tractions.
23. Consider an infinite plane subject to tension in the x−direction only. Assume a state of plane strain and let σxx = T with σxy = σyy = 0. Find the strain components exx , eyy and exy . Also find the displacement field u = u(x, y) and v = v(x, y).
24. Consider an infinite plane subject to tension in the y-direction only. Assume a state of plane strain and let σyy = T with σxx = σxy = 0. Find the strain components exx , eyy and exy . Also find the displacement field u = u(x, y) and v = v(x, y).
25. Consider an infinite plane subject to tension in both the x and y directions. Assume a state of plane strain and let σxx = T , σyy = T and σxy = 0. Find the strain components exx , eyy and exy . Also find the displacement field u = u(x, y) and v = v(x, y).
26. An infinite cylindrical rod of radius R0 has an external pressure P0 as illustrated in figure 2.5-5. Find the stress and displacement fields.

Figure 2.4-5. External pressure on a rod.

277

Figure 2.4-6. Internal pressure on circular hole.

Figure 2.4-7. Tube with internal and external pressure.
27. An infinite plane has a circular hole of radius R1 with an internal pressure P1 as illustrated in the figure 2.4-6. Find the stress and displacement fields.
28. A tube of inner radius R1 and outer radius R0 has an internal pressure of P1 and an external pressure of P0 as illustrated in the figure 2.4-7. Verify the stress and displacement fields derived in example 2.4-7.
29. Use Cartesian tensors and combine the equations of equilibrium σij,j + Lbi = 0, Hooke’s law σij = 1 λekk δij + 2µeij and the strain tensor eij = (ui,j + uj,i ) and derive the Navier equations of equilibrium 2 σij,j + Lbi = (λ + µ)

∂Θ ∂ 2 ui + µ + Lbi = 0, ∂xi ∂xk ∂xk

where Θ = e11 + e22 + e33 is the dilatation.
30. Show the Navier equations in problem 29 can be written in the tensor form µui,jj + (λ + µ)uj,ji + Lbi = 0 or the vector form µ∇2
u + (λ + µ)∇ (∇ ·
u) + L
b =
0.

278
31. Show that in an orthogonal coordinate system the components of ∇(∇ ·
u) can be expressed in terms of physical components by the relation   ' 1 ∂(h2 h3 u(1)) ∂(h1 h3 u(2)) ∂(h1 h2 u(3)) 1 ∂ [∇ (∇ ·
u)]i = + + hi ∂xi h1 h2 h3 ∂x1 ∂x2 ∂x3
32. Show that in orthogonal coordinates the components of ∇2
u can be written  2  ∇
u i = g jk ui,jk = Ai and in terms of physical components one can write  ' ' 3 3  3    m ∂(hm u(m))  m ∂(hi u(i)) 1  ∂ 2 (hi u(i)) hi A(i) = − 2 − ij jj h2 ∂xj ∂xj ∂xj ∂xm m=1 m=1 j=1 j    '  ' '  '  ' 3 3  3   m m p m p ∂  − − − hm u(m) j i j i p j j j p i j ∂x m=1 p=1 p=1
33. Use the results in problem 32 to show in Cartesian coordinates the physical components of [∇2
u]i = Ai can be represented

 ∂2u ∂2u ∂2u ˆ1 = A(1) = + 2 + 2 ∇2
u · e ∂x2 ∂y ∂z 2 2  2  ∂ v ∂ v ∂2v ˆ2 = A(2) = ∇
u · e + + ∂x2 ∂y 2 ∂z 2 2 2  2  ∂ w ∂ w ∂2w ˆ3 = A(3) = ∇
u · e + + ∂x2 ∂y 2 ∂z 2



where (u, v, w) are the components of the displacement vector
u.
34. Use the results in problem 32 to show in cylindrical coordinates the physical components of [∇2
u]i = Ai can be represented



 1 2 ∂uθ ˆr = A(1) = ∇2 ur − 2 ur − 2 ∇2
u · e r r ∂θ  2  1 2 ∂ur 2 ˆθ = A(2) = ∇ uθ + 2 − 2 uθ ∇
u · e r ∂θ r  2  ˆz = A(3) = ∇2 uz ∇
u · e

1 ∂2α ∂2α ∂ 2 α 1 ∂α + + + ∂r2 r ∂r r2 ∂θ2 ∂z 2
35. Use the results in problem 32 to show in spherical coordinates the physical components of [∇2
u]i = Ai where ur , uθ , uz are the physical components of
u and ∇2 α =

can be represented  2 cot θ 2 2 ∂uθ 2 ∂uφ ˆρ = A(1) = ∇2 uρ − 2 uρ − 2 − uθ − 2 ∇2
u · e 2 ρ ρ ∂θ ρ ρ sin θ ∂φ  2  ∂uθ 1 ∂u 2 2 cos θ ρ ˆθ = A(2) = ∇2 uθ + 2 ∇
u · e − 2 uθ − 2 2 ρ ∂θ ρ sin θ ρ sin θ ∂φ  2  1 2 ∂uρ 2 cos θ ∂uθ ˆφ = A(3) = ∇2 uφ − 2 2 uφ + 2 + 2 2 ∇
u · e ρ sin θ ∂φ ρ sin θ ρ sin θ ∂φ 

where uρ , uθ , uφ are the physical components of
u and where ∇2 α =

∂2α 1 ∂ 2 α cot θ ∂α 1 ∂ 2 α 2 ∂α + + + + ∂ρ2 ρ ∂ρ ρ2 ∂θ2 ρ2 ∂θ ρ2 sin2 θ ∂φ2

279
36. Combine the results from problems 30,31,32 and 33 and write the Navier equations of equilibrium in Cartesian coordinates. Alternatively, write the stress-strain relations (2.4.29(b)) in terms of physical components and then use these results, together with the results from Exercise 2.3, problems 2 and 14, to derive the Navier equations.
37. Combine the results from problems 30,31,32 and 34 and write the Navier equations of equilibrium in cylindrical coordinates. Alternatively, write the stress-strain relations (2.4.29(b)) in terms of physical components and then use these results, together with the results from Exercise 2.3, problems 3 and 15, to derive the Navier equations.
38. Combine the results from problems 30,31,32 and 35 and write the Navier equations of equilibrium in spherical coordinates. Alternatively, write the stress-strain relations (2.4.29(b)) in terms of physical components and then use these results, together with the results from Exercise 2.3, problems 4 and 16, to derive the Navier equations.
39. Assume L
b = −grad V and let φ denote the Airy stress function defined by ∂2φ σxx =V + 2 ∂y ∂2φ σyy =V + ∂x2 ∂2φ σxy = − ∂x∂y (a) Show that for conditions of plane strain the equilibrium equations in two dimensions are satisfied by the above definitions. (b) Express the compatibility equation ∂ 2 exx ∂ 2 eyy ∂ 2 exy + = 2 ∂y 2 ∂x2 ∂x∂y in terms of φ and V and show that 1 − 2ν 2 ∇ V = 0. ∇4 φ + 1−ν
40. Consider the case where the body forces are conservative and derivable from a scalar potential function such that Lbi = −V,i . Show that under conditions of plane strain in rectangular Cartesian coordinates the 1 ∇2 V compatibility equation e11,22 + e22,11 = 2e12,12 can be reduced to the form ∇2 σii = , i = 1, 2 1−ν involving the stresses and the potential. Hint: Differentiate the equilibrium equations. i
41. Use the relation σji = 2µeij + λem m δj and solve for the strain in terms of the stress.


42. Derive the equation (2.4.26) from the equation (2.4.23).
43.

In two dimensions assume that the body forces are derivable from a potential function V and

Lbi = −g ij V ,j . Also assume that the stress is derivable from the Airy stress function and the potential function by employing the relations σ ij = Fim Fjn um,n + g ij V

i, j, m, n = 1, 2 where um = φ ,m and

Fpq is the two dimensional epsilon permutation symbol and all indices have the range 1,2. (a) Show that Fim Fjn (φm ) ,nj = 0. (b) Show that σ ij,j = −Lbi . (c) Verify the stress laws for cylindrical and Cartesian coordinates given in problem 20 by using the above expression for σ ij . Hint: Expand the contravariant derivative and convert all terms to physical components. Also recall that Fij =

√1 eij . g

280
44. Consider a material with body forces per unit volume ρF i , i = 1, 2, 3 and surface tractions denoted by σ r = σ rj nj , where nj is a unit surface normal. Further, let δui denote a small displacement vector associated with a small variation in the strain δeij .

 (a) Show the work done during a small variation in strain is δW = δWB + δWS where δWB = ρF i δui dτ V  σ r δur dS is a surface is a volume integral representing the work done by the body forces and δWS = S

integral representing the work done by the surface forces.

(b) Using the Gauss divergence theorem show that the work done can be represented as   1 1 δW = cijmn δ[emn eij ] dτ or W = σ ij eij dτ. 2 V 2 V The scalar quantity 12 σ ij eij is called the strain energy density or strain energy per unit volume. 1 Hint: Interchange subscripts, add terms and calculate 2W = V σ ij [δui,j + δuj,i ] dτ.
45. Consider a spherical shell subjected to an internal pressure pi and external pressure po . Let a denote the inner radius and b the outer radius of the spherical shell. Find the displacement and stress fields in spherical coordinates (ρ, θ, φ). Hint: Assume symmetry in the θ and φ directions and let the physical components of displacements satisfy the relations uρ = uρ (ρ),
46.

uθ = uφ = 0.

(a) Verify the average normal stress is proportional to the dilatation, where the proportionality

constant is the bulk modulus of elasticity. i.e. Show that 13 σii =

E 1 i 1−2ν 3 ei

= keii where k is the bulk modulus

of elasticity. (b) Define the quantities of strain deviation and stress deviation in terms of the average normal stress s = 13 σii and average cubic dilatation e = 13 eii as follows strain deviator

εij = eij − eδji

stress deviator

sij = σji − sδji

Show that zero results when a contraction is performed on the stress and strain deviators. (The above definitions are used to split the strain tensor into two parts. One part represents pure dilatation and the other part represents pure distortion.) (c) Show that (1 − 2ν)s = Ee

or s = (3λ + 2µ)e

(d) Express Hooke’s law in terms of the strain and stress deviator and show E(εij + eδji ) = (1 + ν)sij + (1 − 2ν)sδji which simplifies to sij = 2µεij .
47. Show the strain energy density (problem 44) can be written in terms of the stress and strain deviators (problem 46) and 1 W = 2

 V

and from Hooke’s law W =

1 σ eij dτ = 2



ij

3 2

(3se + sij εij ) dτ V

 ((3λ + 2µ)e2 + V

2µ ij ε εij ) dτ. 3

281
48. Find the stress σrr ,σrθ and σθθ in an infinite plate with a small circular hole, which is traction free, when the plate is subjected to a pure shearing force F12 . Determine the maximum stress.
49. Show that in terms of E and ν C1111 =

E(1 − ν) (1 + ν)(1 − 2ν)

C1122 =

Eν (1 + ν)(1 − 2ν)

C1212 =

E 2(1 + ν)


50. Show that in Cartesian coordinates the quantity S = σxx σyy + σyy σzz + σzz σxx − (σxy )2 − (σyz )2 − (σxz )2 1 (σii σjj − σij σij ). 2
51. Show that in Cartesian coordinates for a state of plane strain where the displacements are given by is a stress invariant. Hint: First verify that in tensor form S =

u = u(x, y),v = v(x, y) and w = 0, the stress components must satisfy the equations ∂σxy ∂σxx + + Lbx =0 ∂x ∂y ∂σyy ∂σyx + + Lby =0 ∂x ∂y −L ∇ (σxx + σyy ) = 1−ν 2



∂bx ∂by + ∂x ∂y




52. Show that in Cartesian coordinates for a state of plane stress where σxx = σxx (x, y), σyy = σyy (x, y), σxy = σxy (x, y) and σxz = σyz = σzz = 0 the stress components must satisfy ∂σxx ∂σxy + + Lbx =0 ∂x ∂y ∂σyy ∂σyx + + Lby =0 ∂x ∂y ∇ (σxx + σyy ) = − L(ν + 1) 2



∂bx ∂by + ∂x ∂y



282 §2.5 CONTINUUM MECHANICS (FLUIDS) Let us consider a fluid medium and use Cartesian tensors to derive the mathematical equations that describe how a fluid behaves. A fluid continuum, like a solid continuum, is characterized by equations describing: 1. Conservation of linear momentum σij,j + Lbi = Lv˙ i

(2.5.1)

2. Conservation of angular momentum σij = σji . 3. Conservation of mass (continuity equation) ∂L ∂L ∂vi + vi + L =0 ∂t ∂xi ∂xi

or

DL
= 0. + L∇ · V Dt

(2.5.2)

In the above equations vi , i = 1, 2, 3 is a velocity field, L is the density of the fluid, σij is the stress tensor and bj is an external force per unit mass. In the cgs system of units of measurement, the above quantities have dimensions [v˙ j ] = cm/sec2 ,

[σij ] = dyne/cm2 ,

[bj ] = dynes/g,

[L] = g/cm3 .

(2.5.3)

The displacement field ui , i = 1, 2, 3 can be represented in terms of the velocity field vi , i = 1, 2, 3, by 

the relation ui =

t

vi dt.

(2.5.4)

0

The strain tensor components of the medium can then be represented in terms of the velocity field as  t  t 1 1 eij = (ui,j + uj,i ) = (vi,j + vj,i ) dt = Dij dt, (2.5.5) 2 0 2 0 where 1 (vi,j + vj,i ) 2 is called the rate of deformation tensor , velocity strain tensor, or rate of strain tensor. Dij =

(2.5.6)

Note the difference in the equations describing a solid continuum compared with those for a fluid continuum. In describing a solid continuum we were primarily interested in calculating the displacement field ui , i = 1, 2, 3 when the continuum was subjected to external forces. In describing a fluid medium, we calculate the velocity field vi , i = 1, 2, 3 when the continuum is subjected to external forces. We therefore replace the strain tensor relations by the velocity strain tensor relations in all future considerations concerning the study of fluid motion. Constitutive Equations for Fluids In addition to the above basic equations, we will need a set of constitutive equations which describe the material properties of the fluid. Toward this purpose consider an arbitrary point within the fluid medium and pass an imaginary plane through the point. The orientation of the plane is determined by a unit normal (n)

ni , i = 1, 2, 3 to the planar surface. For a fluid at rest we wish to determine the stress vector ti on the plane element passing through the selected point P. We desire to express

(n) ti

acting

in terms of the stress

tensor σij . The superscript (n) on the stress vector is to remind you that the stress acting on the planar element depends upon the orientation of the plane through the point.

283 (n)

We make the assumption that ti

is colinear with the normal vector to the surface passing through

the selected point. It is also assumed that for fluid elements at rest, there are no shear forces acting on the planar element through an arbitrary point and therefore the stress tensor σij should be independent of the orientation of the plane. That is, we desire for the stress vector σij to be an isotropic tensor. This requires σij to have a specific form. To find this specific form we let σij denote the stress components in a general coordinate system xi , i = 1, 2, 3 and let σ ij denote the components of stress in a barred coordinate system xi , i = 1, 2, 3. Since σij is a tensor, it must satisfy the transformation law σ mn = σij

∂xi ∂xj , ∂xm ∂xn

i, j, m, n = 1, 2, 3.

(2.5.7)

We desire for the stress tensor σij to be an invariant under an arbitrary rotation of axes. Consider therefore the special coordinate transformations illustrated in the figures 2.5-1(a) and (b).

Figure 2.5-1. Coordinate transformations due to rotations For the transformation equations given in figure 2.5-1(a), the stress tensor in the barred system of coordinates is σ 11 = σ22

σ 21 = σ32

σ 31 = σ12

σ 12 = σ23

σ 22 = σ33

σ 32 = σ13

σ 13 = σ21

σ 23 = σ31

σ 33 = σ11 .

(2.5.8)

If σij is to be isotropic, we desire that σ 11 = σ11 , σ 22 = σ22 and σ 33 = σ33 . If the equations (2.5.8) are to produce these results, we require that σ11 , σ22 and σ33 must be equal. We denote these common values by (−p). In particular, the equations (2.5.8) show that if σ 11 = σ11 , σ 22 = σ22 and σ 33 = σ33 , then we must require that σ11 = σ22 = σ33 = −p. If σ 12 = σ12 and σ 23 = σ23 , then we also require that σ12 = σ23 = σ31 . We note that if σ 13 = σ13 and σ 32 = σ32 , then we require that σ21 = σ32 = σ13 . If the equations (2.5.7) are expanded using the transformation given in figure 2.5-1(b), we obtain the additional requirements that σ 11 = σ22 σ 12 = −σ21 σ 13 = σ23

σ 21 = −σ12 σ 22 = σ11 σ 23 = −σ13

σ 31 = σ32 σ 32 = −σ31 σ 33 = σ33 .

(2.5.9)

284 Analysis of these equations implies that if σij is to be isotropic, then σ 21 = σ21 = −σ12 = −σ21 or σ21 = 0 which implies

σ12 = σ23 = σ31 = σ21 = σ32 = σ13 = 0.

(2.5.10)

The above analysis demonstrates that if the stress tensor σij is to be isotropic, it must have the form σij = −pδij .

(2.5.11)

Use the traction condition (2.3.11), and express the stress vector as (n)

tj

= σij ni = −pnj .

(2.5.12)

This equation is interpreted as representing the stress vector at a point on a surface with outward unit normal ni , where p is the pressure (hydrostatic pressure) stress magnitude assumed to be positive. The negative sign in equation (2.5.12) denotes a compressive stress. Imagine a submerged object in a fluid medium. We further imagine the object to be covered with unit normal vectors emanating from each point on its surface. The equation (2.5.12) shows that the hydrostatic pressure always acts on the object in a compressive manner. A force results from the stress vector acting on the object. The direction of the force is opposite to the direction of the unit outward normal vectors. It is a compressive force at each point on the surface of the object. The above considerations were for a fluid at rest (hydrostatics). For a fluid in motion (hydrodynamics) a different set of assumptions must be made. Hydrodynamical experiments show that the shear stress components are not zero and so we assume a stress tensor having the form σij = −pδij + τij ,

i, j = 1, 2, 3,

(2.5.13)

where τij is called the viscous stress tensor. Note that all real fluids are both viscous and compressible. Definition: (Viscous/inviscid fluid) If the viscous stress tensor τij is zero for all i, j, then the fluid is called an inviscid, nonviscous, ideal or perfect fluid. The fluid is called viscous when τij is different from zero. In these notes it is assumed that the equation (2.5.13) represents the basic form for constitutive equations describing fluid motion.

285

Figure 2.5-2. Viscosity experiment. Viscosity Most fluids are characterized by the fact that they cannot resist shearing stresses. That is, if you put a shearing stress on the fluid, the fluid gives way and flows. Consider the experiment illustrated in the figure 2.5-2 which illustrates a fluid moving between two parallel plane surfaces. Let S denote the distance between the two planes. Now keep the lower surface fixed or stationary and move the upper surface parallel to the lower surface with a constant velocity V
0 . If you measure the force F required to maintain the constant velocity of the upper surface, you discover that the force F varies directly as the area A of the surface and the ratio V0 /S. This is expressed in the form F V0 = µ∗ . A S

(2.5.14)

The constant µ∗ is a proportionality constant called the coefficient of viscosity. The viscosity usually depends upon temperature, but throughout our discussions we will assume the temperature is constant. A dimensional analysis of the equation (2.5.14) implies that the basic dimension of the viscosity is [µ∗ ] = M L−1 T −1 . For example, [µ∗ ] = gm/(cm sec) in the cgs system of units. The viscosity is usually measured in units of centipoise where one centipoise represents one-hundredth of a poise, where the unit of 1 poise= 1 gram per centimeter per second. The result of the above experiment shows that the stress is proportional to the change in velocity with change in distance or gradient of the velocity. Linear Viscous Fluids The above experiment with viscosity suggest that the viscous stress tensor τij is dependent upon both the gradient of the fluid velocity and the density of the fluid. In Cartesian coordinates, the simplest model suggested by the above experiment is that the viscous stress tensor τij is proportional to the velocity gradient vi,j and so we write τik = cikmp vm,p ,

(2.5.15)

where cikmp is a proportionality constant which is dependent upon the fluid density. The viscous stress tensor must be independent of any reference frame, and hence we assume that the proportionality constants cikmp can be represented by an isotropic tensor. Recall that an isotropic tensor has the basic form cikmp = λ∗ δik δmp + µ∗ (δim δkp + δip δkm ) + ν ∗ (δim δkp − δip δkm )

(2.5.16)

286 where λ∗ , µ∗ and ν ∗ are constants. Examining the results from equations (2.5.11) and (2.5.13) we find that if the viscous stress is symmetric, then τij = τji . This requires ν ∗ be chosen as zero. Consequently, the viscous stress tensor reduces to the form τik = λ∗ δik vp,p + µ∗ (vk,i + vi,k ).

(2.5.17)

The coefficient µ∗ is called the first coefficient of viscosity and the coefficient λ∗ is called the second coefficient of viscosity. Sometimes it is convenient to define 2 ζ = λ∗ + µ∗ 3

(2.5.18)

as “another second coefficient of viscosity,” or “bulk coefficient of viscosity.” The condition of zero bulk viscosity is known as Stokes hypothesis. Many fluids problems assume the Stoke’s hypothesis. This requires that the bulk coefficient be zero or very small. Under these circumstances the second coefficient of viscosity is related to the first coefficient of viscosity by the relation λ∗ = − 32 µ∗ . In the study of shock waves and acoustic waves the Stoke’s hypothesis is not applicable. There are many tables and empirical formulas where the viscosity of different types of fluids or gases can be obtained. For example, in the study of the kinetic theory of gases the viscosity can be calculated C1 gT 3/2 from the Sutherland formula µ∗ = where C1 , C2 are constants for a specific gas. These constants T + C2 can be found in certain tables. The quantity g is the gravitational constant and T is the temperature in degrees Rankine (o R = 460 + o F ). Many other empirical formulas like the above exist. Also many graphs and tabular values of viscosity can be found. The table 5.1 lists the approximate values of the viscosity of some selected fluids and gases.

Table 5.1 Substance Water Alcohol Ethyl Alcohol Glycol Mercury Air Helium Nitrogen

Viscosity of selected fluids and gases gram in units of cm−sec = Poise at Atmospheric Pressure.

0◦ C 0.01798 0.01773

0.017 1.708(10−4) 1.86(10−4) 1.658(10−4)

20◦ C 0.01002

60◦ C 0.00469

0.012 0.199 0.0157

0.00592 0.0495 0.013

1.94(10−4 ) 1.74(10−4 )

1.92(10−4)

100◦ C 0.00284

0.0199 0.0100 2.175(10−4) 2.28(10−4) 2.09(10−4)

The viscous stress tensor given in equation (2.5.17) may also be expressed in terms of the rate of deformation tensor defined by equation (2.5.6). This representation is τij = λ∗ δij Dkk + 2µ∗ Dij ,

(2.5.19)

where 2Dij = vi,j + vj,i and Dkk = D11 + D22 + D33 = v1,1 + v2,2 + v3,3 = vi,i = Θ is the rate of change of the dilatation considered earlier. In Cartesian form, with velocity components u, v, w, the viscous stress

287 tensor components are

τyy τzz





∂u ∂v ∂w + λ∗ + ∂x ∂y ∂z   ∂u ∂w ∗ ∗ ∂v ∗ =(λ + 2µ ) +λ + ∂y ∂x ∂z  ∂u ∂v  ∂w + λ∗ + =(λ∗ + 2µ∗ ) ∂z ∂x ∂y

τxx =(λ∗ + 2µ∗ )

τzx = τxz τzy = τyz

 ∂u



∂v ∂y ∂x   ∂w ∂u =µ∗ + ∂x ∂z   ∂w ∂v =µ∗ + ∂z ∂y

τyx = τxy =µ∗

+

In cylindrical form, with velocity components vr , vθ , vz , the viscous stess tensor components are ∂vr  + λ∗ ∇ · V ∂r  1 ∂vθ vr  τθθ =2µ∗ + λ∗ ∇ · V + r ∂θ r ∂vz  + λ∗ ∇ · V τzz =2µ∗ ∂z  = 1 ∂ (rvr ) + 1 ∂vθ + ∂vz ∇·V r ∂r r ∂θ ∂z τrr =2µ∗

where

τrz = τzr τzθ = τθz



∂vθ vθ 1 ∂vr + − r ∂θ ∂r r   ∂vr ∂vz ∗ =µ + ∂z ∂r   ∂vθ 1 ∂vz ∗ =µ + r ∂θ ∂z

τθr = τrθ =µ∗



In spherical coordinates, with velocity components vρ , vθ , vφ , the viscous stress tensor components have the form ∂vρ  + λ∗ ∇ · V ∂ρ   1 ∂vθ vρ  τθθ =2µ∗ + λ∗ ∇ · V + ρ ∂θ ρ   vρ vθ cot θ 1 ∂vφ  + λ∗ ∇ · V + + τφφ =2µ∗ ρ sin θ ∂φ ρ ρ

∂ 1 ∂vφ  = 1 ∂ ρ2 vρ + 1 (sin θvθ ) + ∇·V 2 ρ ∂ρ ρ sin θ ∂θ ρ sin θ ∂φ τρρ =2µ∗

where



τφρ = τρφ τθφ = τφθ







∂ 1 ∂vρ vθ + ∂ρ ρ ρ ∂θ    ∂ 1 ∂vr vθ =µ∗ +ρ ρ sin θ ∂φ ∂ρ ρ   v   1 ∂vθ sin θ ∂ φ =µ∗ + ρ ∂θ sin θ ρ sin θ ∂φ

τρθ = τθρ =µ∗ ρ

Note that the viscous stress tensor is a linear function of the rate of deformation tensor Dij . Such a fluid is called a Newtonian fluid. In cases where the viscous stress tensor is a nonlinear function of Dij the fluid is called non-Newtonian. Definition: (Newtonian Fluid) If the viscous stress tensor τij is expressible as a linear function of the rate of deformation tensor Dij , the fluid is called a Newtonian fluid. Otherwise, the fluid is called a non-Newtonian fluid. Important note: Do not assume an arbitrary form for the constitutive equations unless there is experimental evidence to support your assumption. A constitutive equation is a very important step in the modeling processes as it describes the material you are working with. One cannot arbitrarily assign a form to the viscous stress and expect the mathematical equations to describe the correct fluid behavior. The form of the viscous stress is an important part of the modeling process and by assigning different forms to the viscous stress tensor then various types of materials can be modeled. We restrict our study in these notes to Newtonian fluids. In Cartesian coordinates the rate of deformation-stress constitutive equations for a Newtonian fluid can be written as σij = −pδij + λ∗ δij Dkk + 2µ∗ Dij

(2.5.20)

288 which can also be written in the alternative form σij = −pδij + λ∗ δij vk,k + µ∗ (vi,j + vj,i )

(2.5.21)

involving the gradient of the velocity. Upon transforming from a Cartesian coordinate system y i , i = 1, 2, 3 to a more general system of coordinates xi , i = 1, 2, 3, we write σ mn = σij

∂y i ∂y j . ∂xm ∂xn

(2.5.22)

Now using the divergence from equation (2.1.3) and substituting equation (2.5.21) into equation (2.5.22) we obtain a more general expression for the constitutive equation. Performing the indicated substitutions there results

  ∂y i ∂y j σ mn = −pδij + λ∗ δij v k,k + µ∗ (vi,j + vj,i ) ∂xm ∂xn ∗ k ∗ σ mn = −pgmn + λ gmn v ,k + µ (v m,n + v n,m ).

Dropping the bar notation, the stress-velocity strain relationships in the general coordinates xi , i = 1, 2, 3, is σmn = −pgmn + λ∗ gmn g ik vi,k + µ∗ (vm,n + vn,m ).

(2.5.23)

Summary The basic equations which describe the motion of a Newtonian fluid are : Continuity equation (Conservation of mass) ∂L i

+ Lv ,i = 0, ∂t

or

DL
=0 + L∇ · V Dt

Conservation of linear momentum σ ij,j + Lbi = Lv˙ i ,

1 equation.

(2.5.24)

3 equations

DV
= L
b + ∇ · σ = L
b − ∇p + ∇ · τ (2.5.25) Dt 03 03 03 03 = i=1 j=1 (−pδij + τij ) eˆi eˆj and τ = i=1 j=1 τij eˆi eˆj are second order tensors. Conseror in vector form L

where

σ

vation of angular momentum σ ij = σ ji ,

(Reduces the set of equations (2.5.23) to 6 equations.) Rate of

deformation tensor (Velocity strain tensor) Dij =

1 (vi,j + vj,i ) , 2

6 equations.

(2.5.26)

Constitutive equations σmn = −pgmn + λ∗ gmn g ik vi,k + µ∗ (vm,n + vn,m ),

6 equations.

(2.5.27)

289 In the cgs system of units the above quantities have the following units of measurements in Cartesian coordinates is the velocity field , i = 1, 2, 3,

vi σij

is the stress tensor, i, j = 1, 2, 3, L

bi

is the fluid density

is the external body forces per unit mass Dij

is the rate of deformation tensor p is the pressure

λ∗ , µ∗

are coefficients of viscosity

[vi ] = cm/sec [σij ] = dyne/cm2 [L] = gm/cm3 [bi ] = dyne/gm [Dij ] = sec−1 [p] = dyne/cm2

[λ∗ ] = [µ∗ ] = Poise

where 1 Poise = 1gm/cm sec If we assume the external body forces per unit mass are known, then the equations (2.5.24), (2.5.25), (2.5.26), and (2.5.27) represent 16 equations in the 16 unknowns L, v1 , v2 , v3 , σ11 , σ12 , σ13 , σ22 , σ23 , σ33 , D11 , D12 , D13 , D22 , D23 , D33 .

Navier-Stokes-Duhem Equations of Fluid Motion Substituting the stress tensor from equation (2.5.27) into the linear momentum equation (2.5.25), and assuming that the viscosity coefficients are constants, we obtain the Navier-Stokes-Duhem equations for fluid motion. In Cartesian coordinates these equations can be represented in any of the equivalent forms Lv˙ i = Lbi − p,j δij + (λ∗ + µ∗ )vk,ki + µ∗ vi,jj L

∂vi + Lvj vi,j = Lbi + (−pδij + τij ) ,j ∂t ∂Lvi + (Lvi vj + pδij − τij ) ,j = Lbi ∂t D
v = L
b − ∇ p + (λ∗ + µ∗ )∇ (∇ ·
v ) + µ∗ ∇2
v L Dt

(2.5.28)

∂
v D
v = + (
v · ∇)
v is the material derivative, substantial derivative or convective derivative. This Dt ∂t derivative is represented as where

v˙ i =

∂vi dxj ∂vi ∂vi ∂vi ∂vi + j = + j vj = + vi,j v j . ∂t ∂x dt ∂t ∂x ∂t

(2.5.29)

In the vector form of equations (2.5.28), the terms on the right-hand side of the equation represent force terms. The term L
b represents external body forces per unit volume. If these forces are derivable from a potential function φ, then the external forces are conservative and can be represented in the form −L∇ φ. The term −∇ p is the gradient of the pressure and represents a force per unit volume due to hydrostatic pressure. The above statement is verified in the exercises that follow this section. The remaining terms can be written f
viscous = (λ∗ + µ∗ )∇ (∇ ·
v ) + µ∗ ∇2
v

(2.5.30)

290 and are given the physical interpretation of an internal force per unit volume. These internal forces arise from the shearing stresses in the moving fluid. If f
viscous is zero the vector equation in (2.5.28) is called Euler’s equation. If the viscosity coefficients are nonconstant, then the Navier-Stokes equations can be written in the Cartesian form    ∂vi ∂vi ∂vi ∂vk ∂ ∂vj ∗ ∗ L[ + vj ] =Lbi + +µ + −pδij + λ δij ∂t ∂xj ∂xj ∂xk ∂xj ∂xi      ∂v ∂v ∂p ∂ ∂vj ∂ k i ∗ ∗ =Lbi − + + λ + j µ ∂xi ∂xi ∂xk ∂x ∂xj ∂xi which can also be written in terms of the bulk coefficient of viscosity ζ = λ∗ + 23 µ∗ as L[

     ∂vi ∂vi ∂p ∂ ∂vk ∂vj ∂vi 2 ∂ + vj ] =Lbi − + + (ζ − µ∗ ) + j µ∗ ∂t ∂xj ∂xi ∂xi 3 ∂xk ∂x ∂xj ∂xi      ∂vi ∂p ∂ ∂vj 2 ∂vk ∂vk ∂ =Lbi − + + − δij ζ + j µ∗ ∂xi ∂xi ∂xk ∂x ∂xj ∂xi 3 ∂xk

These equations form the basics of viscous flow theory. In the case of orthogonal coordinates, where g(i)(i) = h2i (no summation) and gij = 0 for i = j, general expressions for the Navier-Stokes equations in terms of the physical components v(1), v(2), v(3) are: Navier-Stokes-Duhem equations for compressible fluid in terms of physical components: (i = j = k)  

v(2) ∂v(i) v(3) ∂v(i) ∂v(i) v(1) ∂v(i) + + + ∂t h1 ∂x1 h2 ∂x2 h3 ∂x3 −



v(j) hi hj



v(j)

∂hj ∂hi − v(i) ∂xi ∂xj



+

∗

1 ∂p 1 ∂ ∗ b(i)  + µ − + λ ∇·V hi hi ∂xi hi ∂xi hi hj

+

µ∗ hi hk

−

2µ∗ hi hk

+

∂ ∂xj







hi ∂ hk ∂xk



v(i) hi



+

hk ∂ hi ∂xi



v(k) hk





1 ∂v(k) v(i) ∂hk v(k) ∂hk + + hk ∂xk hi hk ∂xi hk hj ∂xi

µ∗ hi hk



hj ∂ hi ∂xi



v(j) hj



+



v(k) hi hk

hi ∂ hj ∂xj

v(i)

hj ∂ hi ∂xi

∂hi ∂hk − v(k) ∂xk ∂xi



v(j) hj

2µ∗ ∂hi − ∂xk hi hj







+

v(i) hi

'

+

hi ∂ hj ∂xj

=



v(i) hi



∂hi ∂hj

v(k) ∂hj v(i) ∂hj 1 ∂v(j) + + hj ∂xj hj hk ∂xk hi hj ∂xi

∂hk 1 + ∂xi hi hj hk







∂ ∂xk

∂ ∂xi

2





2µ∗ hj hk

µ∗ hi hj





1 ∂v(i) v(j) ∂hi v(k) ∂hi + + hi ∂xi hi hj ∂hj hi hk ∂xk

hi ∂ hk ∂xk



v(i) hi



+

hk ∂ hi ∂xi



v(k) hk

'

3 (2.5.31)

where ∇ ·
v is found in equation (2.1.4). In the above equation, cyclic values are assigned to i, j and k. That is, for the x1 components assign the values i = 1, j = 2, k = 3; for the x2 components assign the values i = 2, j = 3, k = 1; and for the x3 components assign the values i = 3, j = 1, k = 2. The tables 5.2, 5.3 and 5.4 show the expanded form of the Navier-Stokes equations in Cartesian, cylindrical and spherical coordinates respectively.

291





DVx ∂p ∂Vx ∂  2µ∗ =bx − + + λ∗ ∇ · V Dt ∂x ∂x ∂x



DVy ∂p ∂ µ∗ =by − + Dt ∂y ∂x



DVz ∂p ∂ µ∗ =bz − + Dt ∂z ∂x











∂Vy ∂Vx + ∂x ∂y ∂Vz ∂Vx + ∂x ∂z





+



∂ µ∗ ∂y



∂Vx ∂Vy + ∂y ∂x





+

∂Vy ∂  2µ∗ + λ∗ ∇ · V ∂y ∂y

+

∂ µ∗ ∂y





∂Vz ∂Vy + ∂y ∂z













+

∂ µ∗ ∂z

+

∂ µ∗ ∂z

+

∂Vx ∂Vz + ∂z ∂x ∂Vy ∂Vz + ∂z ∂y



 

∂ ∂Vz  2µ∗ + λ∗ ∇ · V ∂z ∂z



D ∂( ) ∂( ) ∂( ) ∂( ) () = + Vx + Vy + Vz Dt ∂t ∂x ∂y ∂z

where

and ∇ · V =

∂Vx ∂Vy ∂Vz + + ∂x ∂y ∂z

(2.5.31a) Table 5.2 Navier-Stokes equations for compressible fluids in Cartesian coordinates.

 

V2 DVr − θ Dt r

 =br − +

 

DVθ Vr Vθ + Dt r





∂p ∂Vr ∂  2µ∗ + + λ∗ ∇ · V ∂r ∂r ∂r



∂ µ∗ ∂z

 ∂V

r

∂z

+

∂Vz ∂r





where

+

2µ∗ r



1 ∂ µ∗ r ∂θ

+

 ∂V

r

∂r

−



1 ∂Vr ∂Vθ Vθ + − r ∂θ ∂r r

1 ∂Vθ Vr − r ∂θ r





DVz ∂p 1 ∂ µ∗ r =bz − + Dt ∂z r ∂r



∂Vr ∂Vz + ∂z ∂r



+



1 ∂ µ∗ r ∂θ







1 ∂Vr 1 ∂p 1 ∂ ∂ ∂Vθ Vθ µ∗ + 2µ∗ + + − r ∂θ ∂r r ∂θ ∂r r r ∂θ      1 ∂Vz ∂Vθ ∂Vθ Vθ ∂ 2µ∗ 1 ∂Vr + + − + µ∗ + ∂z r ∂θ ∂z r r ∂θ ∂r r

=bθ −











1 ∂Vθ Vr + r ∂θ r

1 ∂Vz ∂Vθ + r ∂θ ∂z



+



 + λ∗ ∇ · V





∂Vz ∂  2µ∗ + λ∗ ∇ · V ∂z ∂z



D ∂( ) ∂( ) ∂( ) Vθ ∂( ) () = + Vr + + Vz Dt ∂t ∂r r ∂θ ∂z

and ∇ · V =

1 ∂(rVr ) 1 ∂Vθ ∂Vz + + r ∂r r ∂θ ∂z

Table 5.3 Navier-Stokes equations for compressible fluids in cylindrical coordinates.

(2.5.31b)

292 Observe that for incompressible flow

DE Dt

= 0 which implies ∇ · V
= 0. Therefore, the assumptions

of constant viscosity and incompressibility of the flow will simplify the above equations. If on the other hand the viscosity is temperature dependent and the flow is compressible, then one should add to the above equations the continuity equation, an energy equation and an equation of state. The energy equation comes from the first law of thermodynamics applied to a control volume within the fluid and will be considered in the sections ahead. The equation of state is a relation between thermodynamic variables which is added so that the number of equations equals the number of unknowns. Such a system of equations is known as a closed system. An example of an equation of state is the ideal gas law where pressure p is related to gas density L and temperature T by the relation p = LRT where R is the universal gas constant.



Vθ2 + Vφ2 DVρ  − Dt ρ

 = bρ −





∂p ∂Vρ ∂  2µ∗ + + λ∗ ∇ · V ∂ρ ∂ρ ∂ρ





+





1 ∂ ∂ µ∗ ρ ρ ∂θ ∂ρ

Vφ µ∗ ∂Vρ ∂ 1 ∂ + µ∗ ρ ρ sin θ ∂φ ρ sin θ ∂φ ∂ρ ρ  ∂V 2 ∂Vθ 4Vρ 2 ∂Vφ 2Vθ cot θ ∂ µ∗ ρ − − − − + ρ cot θ + 4 ρ ∂ρ ρ ∂θ ρ ρ sin θ ∂φ ρ ∂ρ +

 

Vφ2 cot θ DVθ Vρ Vθ + − Dt ρ ρ



 = bθ −







1 ∂p ∂ ∂ µ∗ ρ + ρ ∂θ ∂ρ ∂ρ





Vθ ρ



+

µ∗ ∂Vρ ρ ∂θ

1 ∂ 2µ∗ ∂Vθ  + Vρ + λ∗ ∇ · V ρ ∂θ ρ ∂θ   V   ∂ µ∗ sin θ ∂ 1 µ∗ ∂Vθ φ + + ρ sin θ ∂φ ρ ∂θ sin θ ρ sin θ ∂φ   1 ∂V   ∂ 1 ∂Vφ Vθ cot θ µ∗ θ − − + 2 cot θ +3 ρ ρ ρ ∂θ ρ sin θ ∂φ ρ ∂ρ +



where

 DV

φ

Dt

+





ρ

Vθ ρ

V  θ

ρ

+

1 ∂Vρ ρ ∂θ

+

Vθ Vφ cot θ

= bφ −



+

µ∗ ∂Vρ ρ ∂θ

+

cot θ ∂Vρ ρ ∂θ





 V 

1 ∂p ∂ µ∗ ∂Vρ ∂ + + µ∗ ρ ρ ρ ρ sin θ ∂φ ∂ρ ρ sin θ ∂φ ∂ρ   V   1 ∂ µ∗ sin θ ∂ µ∗ ∂Vθ φ + + ρ ∂θ ρ ∂θ sin θ ρ sin θ ∂φ     1 ∂Vφ 1 ∂ 2µ∗  + + Vρ + Vθ cot θ + λ∗ ∇ · V ρ sin θ ∂φ ρ sin θ ∂φ  V    V  µ∗ 1 3 ∂Vρ sin θ ∂ ∂ φ φ + + 2 cot θ + + 3ρ ρ ρ sin θ ∂φ ∂ρ ρ ρ ∂θ sin θ ρ sin θ Vθ Vφ





V  θ



φ

ρ

∂Vθ ∂φ



Vφ ∂( ) D ∂( ) ∂( ) Vθ ∂( ) () = + Vρ + + Dt ∂t ∂ρ ρ ∂θ ρ sin θ ∂φ

and ∇ · V =

1 ∂(ρ2 Vρ ) 1 ∂Vθ sin θ 1 ∂Vφ + + ρ2 ∂ρ ρ sin θ ∂θ ρ sin θ ∂φ

(2.5.31c) Table 5.4 Navier-Stokes equations for compressible fluids in spherical coordinates.

293 We now consider various special cases of the Navier-Stokes-Duhem equations. Special Case 1: Assume that
b is a conservative force such that
b = −∇ φ. Also assume that the viscous v force terms are zero. Consider steady flow ( ∂' ∂t = 0) and show that equation (2.5.28) reduces to the equation

(
v · ∇)
v =

−1 ∇ p − ∇ φ L is constant. L

(2.5.32)

Employing the vector identity 1 (
v · ∇)
v = (∇ ×
v ) ×
v + ∇(
v ·
v ), 2

(2.5.33)

we take the dot product of equation (2.5.32) with the vector
v . Noting that
v · [(∇ ×
v ) ×
v ] =
0 we obtain   p 1 2 (2.5.34)
v · ∇ + φ + v = 0. L 2 This equation shows that for steady flow we will have p 1 + φ + v 2 = constant L 2

(2.5.35)

along a streamline. This result is known as Bernoulli’s theorem. In the special case where φ = gh is a v2 p + gh = constant. This equation is known as force due to gravity, the equation (2.5.35) reduces to + L 2 Bernoulli’s equation. It is a conservation of energy statement which has many applications in fluids. Special Case 2: Assume that
b = −∇ φ is conservative and define the quantity Ω by
= ∇ ×
v = curl
v Ω

ω=

1 Ω 2

(2.5.36)

as the vorticity vector associated with the fluid flow and observe that its magnitude is equivalent to twice the angular velocity of a fluid particle. Then using the identity from equation (2.5.33) we can write the Navier-Stokes-Duhem equations in terms of the vorticity vector. We obtain the hydrodynamic equations ∂
v
1 1 1 + Ω ×
v + ∇ v 2 = − ∇ p − ∇ φ + f
viscous , ∂t 2 L L

(2.5.37)

where f
viscous is defined by equation (2.5.30). In the special case of nonviscous flow this further reduces to the Euler equation 1 1 ∂
v
+ Ω ×
v + ∇ v 2 = − ∇ p − ∇ φ. ∂t 2 L If the density L is a function of the pressure only it is customary to introduce the function  p dP 1 dp so that ∇P = ∇p = ∇p P = L dp L c then the Euler equation becomes ∂
v
1 + Ω ×
v = −∇(P + φ + v 2 ). ∂t 2 Some examples of vorticies are smoke rings, hurricanes, tornadoes, and some sun spots. You can create a vortex by letting water stand in a sink and then remove the plug. Watch the water and you will see that a rotation or vortex begins to occur. Vortices are associated with circulating motion.

294 Pick 4 an arbitrary simple closed curve C and place it in the fluid flow and define the line integral
v · eˆt ds, where ds is an element of arc length along the curve C,
v is the vector field defining the K = C

velocity, and eˆt is a unit tangent vector to the curve C. The integral K is called the circulation of the fluid around the closed curve C. The circulation is the summation of the tangential components of the velocity field along the curve C. The local vorticity at a point is defined as the limit lim Area→0

Circulation around C = circulation per unit area. Area inside C

By Stokes theorem, if curl
v =
0, then the fluid is called irrotational and the circulation is zero. Otherwise the fluid is rotational and possesses vorticity. If we are only interested in the velocity field we can eliminate the pressure by taking the curl of both sides of the equation (2.5.37). If we further assume that the fluid is incompressible we obtain the special equations

∇ ·
v = 0

Incompressible fluid, L is constant.


= curl
v Ω ∗
∂Ω

×
v ) = µ ∇2 Ω + ∇ × (Ω ∂t L

Definition of vorticity vector.

(2.5.38)

Results because curl of gradient is zero.

Note that when Ω is identically zero, we have irrotational motion and the above equations reduce to the
×
v ) is neglected, then the last equation in Cauchy-Riemann equations. Note also that if the term ∇ × (Ω equation (2.5.38) reduces to a diffusion equation. This suggests that the vorticity diffuses through the fluid once it is created. Vorticity can be caused by a rigid rotation or by shear flow. For example, in cylindrical coordinates let

= ∇×V
= 2ω eˆz , which shows the V = rω eˆθ , with r, ω constants, denote a rotational motion, then curl V vorticity is twice the rotation vector. Shear can also produce vorticity. For example, consider the velocity
= y eˆ1 with y ≥ 0. Observe that this type of flow produces shear because |V
| increases as y increases. field V
= − eˆ3 . The right-hand rule tells us that if an imaginary paddle For this flow field we have curl V
= ∇ × V wheel is placed in the flow it would rotate clockwise because of the shear effects. Scaled Variables In the Navier-Stokes-Duhem equations for fluid flow we make the assumption that the external body forces are derivable from a potential function φ and write
b = −∇ φ [dyne/gm] We also want to write the Navier-Stokes equations in terms of scaled variables
v =
v v0 p p= p0

L L0 t t= τ

L=

φ , gL x x= L

φ=

y L z z= L

y=

which can be referred to as the barred system of dimensionless variables. Dimensionless variables are introduced by scaling each variable associated with a set of equations by an appropriate constant term called a characteristic constant associated with that variable. Usually the characteristic constants are chosen from various parameters used in the formulation of the set of equations. The characteristic constants assigned to each variable are not unique and so problems can be scaled in a variety of ways. The characteristic constants

295 assigned to each variable are scales, of the appropriate dimension, which act as reference quantities which reflect the order of magnitude changes expected of that variable over a certain range or area of interest associated with the problem. An inappropriate magnitude selected for a characteristic constant can result in a scaling where significant information concerning the problem can be lost. This is analogous to selecting an inappropriate mesh size in a numerical method. The numerical method might give you an answer but details of the answer might be lost. In the above scaling of the variables occurring in the Navier-Stokes equations we let v0 denote some characteristic speed, p0 a characteristic pressure, L0 a characteristic density, L a characteristic length, g the acceleration of gravity and τ a characteristic time (for example τ = L/v0 ), then the barred variables v, p, L,φ, t, x, y and z are dimensionless. Define the barred gradient operator by ∇=

∂ ∂ ∂ eˆ1 + eˆ2 + eˆ3 ∂x ∂y ∂z

where all derivatives are with respect to the barred variables. The above change of variables reduces the Navier-Stokes-Duhem equations L

to the form

∂
v + L(
v · ∇)
v = −L∇φ − ∇ p + (λ∗ + µ∗ )∇ (∇ ·
v ) + µ∗ ∇2
v , ∂t

 L v  ∂
v  L v 2  p 

0 0 0 0 0 + L L
v · ∇
v = −L0 gL∇ φ − ∇p τ L L ∂t  ∗ 

µ v0 (λ∗ + µ∗ ) 2
v0 ∇ ∇ · v + ∇
v. + 2 2 L L

(2.5.39)

(2.5.40)

Now if each term in the equation (2.5.40) is divided by the coefficient L0 v02 /L, we obtain the equation SL



∂
v −1 L∇ φ − E∇p + + L
v · ∇
v = F ∂t



λ∗ +1 µ∗



1
1 2
∇ ∇·v + ∇ v R R

(2.5.41)

which has the dimensionless coefficients p0 = Euler number L0 v02 v2 F = 0 = Froude number, g is acceleration of gravity gL

E=

L0 V0 L = Reynolds number µ∗ L = Strouhal number. S= τ v0

R=

Dropping the bars over the symbols, we write the dimensionless equation using the above coefficients. The scaled equation is found to have the form SL

1 ∂
v + L(
v · ∇)
v = − L∇φ − E∇p + ∂t F



λ∗ +1 µ∗



1 1 ∇ (∇ ·
v ) + ∇2
v R R

(2.5.42)

296 Boundary Conditions Fluids problems can be classified as internal flows or external flows. An example of an internal flow problem is that of fluid moving through a converging-diverging nozzle. An example of an external flow problem is fluid flow around the boundary of an aircraft. For both types of problems there is some sort of boundary which influences how the fluid behaves. In these types of problems the fluid is assumed to adhere to a boundary. Let
rb denote the position vector to a point on a boundary associated with a moving fluid, and let
r denote the position vector to a general point in the fluid. Define
v (
r) as the velocity of the fluid at the point
r and define
v (
rb ) as the known velocity of the boundary. The boundary might be moving within the fluid or it could be fixed in which case the velocity at all points on the boundary is zero. We define the boundary condition associated with a moving fluid as an adherence boundary condition. Definition: (Adherence Boundary Condition) An adherence boundary condition associated with a fluid in motion is defined as the limit lim
v (
r) =
v (
rb ) where
rb is the position ' r →' rb

vector to a point on the boundary. Sometimes, when no finite boundaries are present, it is necessary to impose conditions on the components of the velocity far from the origin. Such conditions are referred to as boundary conditions at infinity. Summary and Additional Considerations Throughout the development of the basic equations of continuum mechanics we have neglected thermodynamical and electromagnetic effects. The inclusion of thermodynamics and electromagnetic fields adds additional terms to the basic equations of a continua. These basic equations describing a continuum are: Conservation of mass The conservation of mass is a statement that the total mass of a body is unchanged during its motion. This is represented by the continuity equation ∂L + (Lv k ),k = 0 or ∂t

DL
=0 + L∇ · V Dt

where L is the mass density and v k is the velocity. Conservation of linear momentum The conservation of linear momentum requires that the time rate of change of linear momentum equal the resultant of all forces acting on the body. In symbols, we write D Dt where

Dv i Dt

=

∂v i ∂t

+

∂v i ∂xk







Lv i dτ = V

S

i F(s) ni dS +

V

i LF(b) dτ +

n 

i F(α)

(2.5.43)

α=1

i i v k is the material derivative, F(s) are the surface forces per unit area, F(b) are the

i represents isolated external forces. Here S represents the surface and body forces per unit mass and F(α)

V represents the volume of the control volume. The right-hand side of this conservation law represents the resultant force coming from the consideration of all surface forces and body forces acting on a control volume.

297 Surface forces acting upon the control volume involve such things as pressures and viscous forces, while body forces are due to such things as gravitational, magnetic and electric fields. Conservation of angular momentum The conservation of angular momentum states that the time rate of change of angular momentum (moment of linear momentum) must equal the total moment of all forces and couples acting upon the body. In symbols, D Dt



 j k

V

Leijk x v dτ =

 j

S

eijk x

k F(s)

j

dS + V

Leijk x

k F(b)

dτ +

n 

k i (eijk xj(α) F(α) + M(α) )

(2.5.44)

α=1

i k represents concentrated couples and F(α) represents isolated forces. where M(α)

Conservation of energy The conservation of energy law requires that the time rate of change of kinetic energy plus internal energies is equal to the sum of the rate of work from all forces and couples plus a summation of all external energies that enter or leave a control volume per unit of time. The energy equation results from the first law of thermodynamics and can be written D ˙ + Q˙ h (E + K) = W Dt

(2.5.45)

˙ is the rate of work associated with surface and where E is the internal energy, K is the kinetic energy, W body forces, and Q˙ h is the input heat rate from surface and internal effects.  Let e denote the internal specific energy density within a control volume, then E = Le dτ represents V

the total  internal energy of the control volume. The kinetic energy of the control volume is expressed as 1 Lgij v i v j dτ where v i is the velocity, L is the density and dτ is a volume element. The energy (rate K= 2 V of work) associated with the body and surface forces is represented  ˙ = W S

 i gij F(s) vj

i Lgij F(b) vj

dS + V

dτ +

n 

i i (gij F(α) v j + gij M(α) ωj )

α=1

i i where ω j is the angular velocity of the point xi(α) , F(α) are isolated forces, and M(α) are isolated couples.

Two external energy sources due to thermal sources are heat flow q i and rate of internal heat production

∂Q ∂t

per unit volume. The conservation of energy can thus be represented D Dt



1 L(e + gij v i v j ) dτ = 2 V

 S

+

 i (gij F(s) v j − qi ni ) dS + n 

V

i (Lgij F(b) vj +

∂Q ) dτ ∂t (2.5.46)

i i (gij F(α) v j + gij M(α) ω j + U(α) )

α=1

where U(α) represents all other energies resulting from thermal, mechanical, electric, magnetic or chemical sources which influx the control volume and D/Dt is the material derivative. In equation (2.5.46) the left hand side is the material derivative of an integral of the total energy et = L(e + 12 gij v i v j ) over the control volume. Material derivatives are not like ordinary derivatives and so

298 we cannot interchange the order of differentiation and integration in this term. Here we must use the result  



that D Dt

V

et dτ =

V

 ∂et + ∇ · (et V
) dτ. ∂t

To prove this result we consider a more general problem. Let A denote the amount of some quantity per unit mass. The quantity A can be a scalar, vector or tensor. The total amount of this quantity inside the 1 control volume is A = V LA dτ and therefore the rate of change of this quantity is ∂A = ∂t



D ∂(LA) dτ = ∂t Dt

V



 V

LA dτ −

S


·n LAV ˆ dS,

which represents the rate of change of material within the control volume plus the influx into the control volume. The minus sign is because n ˆ is always a unit outward normal. By converting the surface integral to a volume integral, by the Gauss divergence theorem, and rearranging terms we find that     D ∂(LA)
+ ∇ · (LAV ) dτ. LA dτ = Dt V ∂t V i i In equation (2.5.46) we neglect all isolated external forces and substitute F(s) = σ ij nj , F(b) = bi where

σij = −pδij + τij . We then replace all surface integrals by volume integrals and find that the conservation of energy can be represented in the form ∂et
) = ∇(σ · V
) − ∇ ·
q + L
b · V
+ ∂Q + ∇ · (et V (2.5.47) ∂t ∂t 03 03 where et = Le + L(v12 + v22 + v32 )/2 is the total energy and σ = i=1 j=1 σij eˆi eˆj is the second order stress tensor. Here

σ · V



+ = −pV

3  j=1

∗

τ1j vj eˆ1 +

3 

τ2j vj eˆ2 +

j=1

3 


+τ ·V
τ3j vj eˆ3 = −pV

j=1

∗

and τij = µ (vi,j + vj,i ) + λ δij vk,k is the viscous stress tensor. Using the identities L

∂et D(et /L) = + ∇ · (et V
) Dt ∂t

and L

D(et /L) De D(V 2 /2) =L +L Dt Dt Dt


as together with the momentum equation (2.5.25) dotted with V L

DV


− ∇p · V
+ (∇ · τ ) · V
· V = L
b · V Dt

the energy equation (2.5.47) can then be represented in the form L

De
) = −∇ ·
q + ∂Q + Φ + p(∇ · V Dt ∂t

(2.5.48)

where Φ is the dissipation function and can be represented
. Φ = (τij vi ) ,j − vi τij,j = ∇ · (τ · V
) − (∇ · τ ) · V As an exercise it can be shown that the dissipation function can also be represented as Φ = 2µ∗ Dij Dij +λ∗ Θ2 where Θ is the dilatation. The heat flow vector is determined from the Fourier law of heat conduction in

299 terms of the temperature T as
q = −κ∇ T , where κ is the thermal conductivity. Consequently, the energy equation can be written as L

De
) = ∂Q + Φ + ∇(k∇T ). + p(∇ · V Dt ∂t

(2.5.49)

In Cartesian coordinates (x, y, z) we use D ∂ ∂ ∂ ∂ = + Vx + Vy + Vz Dt ∂t ∂x ∂y ∂z ∂V ∂V ∂V x y z
= ∇·V + + ∂x ∂y ∂z       ∂ ∂T ∂ ∂T ∂ ∂T ∇ · (κ∇T ) = κ + κ + κ ∂x ∂x ∂y ∂y ∂z ∂z In cylindrical coordinates (r, θ, z) ∂ Vθ ∂ ∂ ∂ D = + Vr + + Vz Dt ∂t ∂r r ∂θ ∂z ∂ ∂V 1 ∂V 1 θ z
= (rVr ) + 2 + ∇·V r ∂r  r ∂θ ∂z      1 ∂ ∂T 1 ∂ ∂T ∂ ∂T ∇ · (κ∇T ) = rκ + 2 κ + κ r ∂r ∂r r ∂θ ∂θ ∂z ∂z and in spherical coordinates (ρ, θ, φ) ∂ Vθ ∂ Vφ ∂ ∂ D = + Vρ + Dt ∂t ∂ρ ρ ∂θ ρ sin θ ∂φ 1 ∂ ∂ 1 1 ∂Vφ ∇ · V
= 2 (ρVρ ) + (Vθ sin θ) + ρ ∂ρ ρ sin θ ∂θ ρ sin θ ∂φ       ∂ 1 ∂ ∂ ∂T 1 ∂T 1 ∂T 2 ∇ · (κ∇T ) = 2 ρ κ + 2 κ sin θ + 2 2 κ ρ ∂ρ ∂ρ ρ sin θ ∂θ ∂θ ∂φ ρ sin θ ∂φ The combination of terms h = e + p/L is known as enthalpy and at times is used to express the energy equation in the form L

D p ∂Q Dh = + − ∇ ·
q + Φ. Dt Dt ∂t

The derivation of this equation is left as an exercise. Conservative Systems Let Q denote some physical quantity per unit volume. Here Q can be either a scalar, vector or tensor field. Place within this field an imaginary simple closed surface S which encloses a volume V. The total 111 amount of Q within the surface is given by V Q dτ and the rate of change of this amount with respect 111 ∂ Q dτ. The total amount of Q within S changes due to sources (or sinks) within the volume to time is ∂t
called current, which represents a and by transport processes. Transport processes introduce a quantity J, 11 flow per unit area across the surface S. The inward flux of material into the volume is denoted S −J
· n ˆ dσ (ˆ n is a unit outward normal.) The sources (or sinks) SQ denotes a generation (or loss) of material per unit 111 S dτ denotes addition (or loss) of material to the volume. For a fixed volume we then volume so that V Q have the material balance

 V

∂Q dτ = − ∂t



 J
· n ˆ dσ + S

SQ dτ. V

300 Using the divergence theorem of Gauss one can derive the general conservation law ∂Q + ∇ · J
= SQ ∂t

(2.5.50)

The continuity equation and energy equations are examples of a scalar conservation law in the special case where SQ = 0. In Cartesian coordinates, we can represent the continuity equation by letting Q=L


= L(Vx eˆ1 + Vy eˆ2 + Vz eˆ3 ) and J
= LV

(2.5.51)

The energy equation conservation law is represented by selecting Q = et and neglecting the rate of internal heat energy we let

( J
= (et + p)v1 −

3 

) vi τxi + qx eˆ1 +

i=1

( (et + p)v2 −

3 

) vi τyi + qy eˆ2 +

i=1

( (et + p)v3 −

3 

(2.5.52)

) eˆ3 .

vi τzi + qz

i=1

In a general orthogonal system of coordinates (x1 , x2 , x3 ) the equation (2.5.50) is written ∂ ∂ ∂ ∂ ((h1 h2 h3 Q)) + ((h2 h3 J1 )) + ((h1 h3 J2 )) + ((h1 h2 J3 )) = 0, ∂t ∂x1 ∂x2 ∂x3 where h1 , h2 , h3 are scale factors obtained from the transformation equations to the general orthogonal coordinates. The momentum equations are examples of a vector conservation law having the form ∂
a + ∇ · (T ) = L
b ∂t where
a is a vector and T is a second order symmetric tensor T =

(2.5.53) 3 3  

Tjk eˆj eˆk . In Cartesian coordinates

k=1 j=1

we let
a = L(Vx eˆ1 + Vy eˆ2 + Vz eˆ3 ) and Tij = Lvi vj + pδij − τij . In general coordinates (x1 , x2 , x3 ) the
and Tij = Lvi vj + pδij − τij . In a general orthogonal system momentum equations result by selecting
a = LV the conservation law (2.5.53) has the general form    ∂  ∂  ∂  ∂ (h2 h3 T · eˆ1 ) + (h1 h3 T · eˆ2 ) + (h1 h2 T · eˆ3 ) = L
b. ((h1 h2 h3
a)) + ∂t ∂x1 ∂x2 ∂x3

(2.5.54)

Neglecting body forces and internal heat production, the continuity, momentum and energy equations can be expressed in the strong conservative form

where

∂U ∂E ∂F ∂G + + + =0 ∂t ∂x ∂y ∂z

(2.5.55)

 ρ  ρVx    U =  ρVy    ρVz et

(2.5.56)



301 

 ρVx ρVx2 + p − τxx     E= ρVx Vy − τxy    ρVx Vz − τxz (et + p)Vx − Vx τxx − Vy τxy − Vz τxz + qx   ρVy ρVx Vy − τxy     F = ρVy2 + p − τyy    ρVy Vz − τyz (et + p)Vy − Vx τyx − Vy τyy − Vz τyz + qy   ρVz ρVx Vz − τxz     G= ρVy Vz − τyz    2 ρVz + p − τzz (et + p)Vz − Vx τzx − Vy τzy − Vz τzz + qz

(2.5.57)

(2.5.58)

(2.5.59)

where the shear stresses are τij = µ∗ (Vi,j + Vj,i ) + δij λ∗ Vk,k for i, j, k = 1, 2, 3. Computational Coordinates To transform the conservative system (2.5.55) from a physical (x, y, z) domain to a computational (ξ, η, ζ) domain requires that a general change of variables take place. Consider the following general transformation of the independent variables ξ = ξ(x, y, z)

η = η(x, y, z)

ζ = ζ(x, y, z)

(2.5.60)

with Jacobian different from zero. The chain rule for changing variables in equation (2.5.55) requires the operators ∂( ) ∂( ) ∂( ) ∂( ) = ξx + ηx + ζx ∂x ∂ξ ∂η ∂ζ ∂( ) ∂( ) ∂( ) ∂( ) = ξy + ηy + ζy ∂y ∂ξ ∂η ∂ζ ∂( ) ∂( ) ∂( ) ∂( ) = ξz + ηz + ζz ∂z ∂ξ ∂η ∂ζ

(2.5.61)

The partial derivatives in these equations occur in the differential expressions 

dξ =ξx dx + ξy dy + ξz dz dη =ηx dx + ηy dy + ηz dz

or

dζ =ζx dx + ζy dy + ζz dz

  dξ ξx  dη  =  ηx ζx dζ

ξy ηy ζy

  ξz dx ηz   dy  ζz dz

(2.5.62)

In a similar mannaer from the inverse transformation equations x = x(ξ, η, ζ)

y = y(ξ, η, ζ)

z = z(ξ, η, ζ)

(2.5.63)

we can write the differentials 

dx =xξ dξ + xη dη + xζ dζ dy =yξ dξ + yη dη + yζ dζ dz =zξ dξ + zζ dζ + zζ dζ

or

  dx xξ  dy  =  yξ zξ dz

xη yη zη

  dξ xζ yζ   dη  zζ dζ

(2.5.64)

302 The transformations (2.5.62) and (2.5.64) are inverses of each other and so we can write 

ξx  ηx ζx

ξy ηy ζy

  xξ ξz ηz  =  yξ ζz zξ 

xη yη zη

−1 xζ yζ  zζ

 −(xη zζ − xζ zη ) xη yζ − xζ yη yη zζ − yζ zη =J  −(yξ zζ − yζ zξ ) xξ zζ − xζ zξ −(xξ yζ − xζ yξ )  yξ zη − yη zξ −(xξ zη − xη zξ ) xξ yη − xη yξ

(2.5.65)

By comparing like elements in equation (2.5.65) we obtain the relations ξx =J(yη zζ − yζ zη )

ηx = − J(yξ zζ − yζ zξ )

ζx =J(yξ zη − yη zξ )

ξy = − J(xη zζ − xζ zη )

ηy =J(xξ zζ − zζ zξ )

ζy = − J(xξ zη − xη zξ )

ξz =J(xη yζ − xζ yη )

ηz = − J(xξ yζ − xζ yξ )

ζz =J(xξ yη − xη yξ )

(2.5.66)

The equations (2.5.55) can now be written in terms of the new variables (ξ, η, ζ) as ∂U ∂E ∂E ∂E ∂F ∂F ∂F ∂G ∂G ∂G + ξx + ηx + ζx + ξy + ηy + ζy + ξz + ηz + ζz = 0 ∂t ∂ξ ∂η ∂ζ ∂ξ ∂η ∂ζ ∂ξ ∂η ∂ζ

(2.5.67)

Now divide each term by the Jacobian J and write the equation (2.5.67) in the form ∂ ∂t



U J



  ∂ Eξx + F ξy + Gξz ∂ξ J   ∂ Eηx + F ηy + Gηz + ∂η J   ∂ Eζx + F ζy + Gζz + ∂ζ J  '    ∂ ζx ∂ ξx ∂  ηx  + + −E ∂ξ J ∂η J ∂ζ J     '   ∂ ξy ∂ ζy ∂ ηy + −F + ∂ξ J ∂η J ∂ζ J     '   ∂ ξz ∂ ζz ∂ ηz + −G + =0 ∂ξ J ∂η J ∂ζ J

+

(2.5.68)

Using the relations given in equation (2.5.66) one can show that the curly bracketed terms above are all zero and so the transformed equations (2.5.55) can also be written in the conservative form

where


∂ F
∂ G

∂E ∂U + + + =0 ∂t ∂ξ ∂η ∂ζ

(2.5.69)


=U U J Eξ
= x + F ξy + Gξz E J Eη + F ηy + Gηz x F
= J
= Eζx + F ζy + Gζz G J

(2.5.70)

303 Fourier law of heat conduction The Fourier law of heat conduction can be written qi = −κT,i for isotropic material and qi = −κij T,j for anisotropic material. The Prandtl number is a nondimensional constant defined as P r =

cp µ∗ κ

so that

the heat flow terms can be represented in Cartesian coordinates as qx = −

cp µ∗ ∂T P r ∂x

qy = −

cp µ∗ ∂T P r ∂y

qz = −

cp µ∗ ∂T P r ∂z

Now one can employ the equation of state relations P = Le(γ − 1), cp = above equations in the alternate forms     ∂ γP ∂ γP µ∗ µ∗ qx = − qy = − P r(γ − 1) ∂x L P r(γ − 1) ∂y L ! The speed of sound is given by a =

γR γ−1

, cp T =

γRT γ−1

∂ µ∗ qz = − P r(γ − 1) ∂z

and write the 

γP L



 γP γP = γRT and so one can substitute a2 in place of the ratio L L

in the above equations. Equilibrium and Nonequilibrium Thermodynamics High temperature gas flows require special considerations. In particular, the specific heat for monotonic and diatomic gases are different and are in general a function of temperature. The energy of a gas can be written as e = et + er + ev + ee + en where et represents translational energy, er is rotational energy, ev is vibrational energy, ee is electronic energy, and en is nuclear energy. The gases follow a Boltzmann distribution for each degree of freedom and consequently at very high temperatures the rotational, translational and vibrational degrees of freedom can each have their own temperature. Under these conditions the gas is said to be in a state of nonequilibrium. In such a situation one needs additional energy equations. The energy equation developed in these notes is for equilibrium thermodynamics where the rotational, translational and vibrational temperatures are the same. Equation of state It is assumed that an equation of state such as the universal gas law or perfect gas law pV = nRT holds which relates pressure p [N/m2 ], volume V [m3 ], amount of gas n [mol],and temperature T [K] where R [J/mol − K] is the universal molar gas constant. If the ideal gas law is represented in the form p = LRT where L [Kg/m3 ] is the gas density, then the universal gas constant must be expressed in units of [J/Kg − K] (See Appendix A). Many gases deviate from this ideal behavior. In order to account for the intermolecular forces associated with high density gases, an empirical equation of state of the form p = ρRT +

M1 

2

βn ρn+r1 + e−γ1 ρ−γ2 ρ

n=1

M2 

cn ρn+r2

n=1

involving constants M1 , M2 , βn , cn , r1 , r2 , γ1 , γ2 is often used. For a perfect gas the relations e = cv T

γ=

cp cv

cv =

R γ−1

cp =

γR γ−1

h = cp T

hold, where R is the universal gas constant, cv is the specific heat at constant volume, cp is the specific heat at constant pressure, γ is the ratio of specific heats and h is the enthalpy. For cv and cp constants the relations p = (γ − 1)Le and RT = (γ − 1)e can be verified.

304 EXAMPLE 2.5-1. (One-dimensional fluid flow) Construct an x-axis running along the center line of a long cylinder with cross sectional area A. Consider the motion of a gas driven by a piston and moving with velocity v1 = u in the x-direction. From an Eulerian point of view we imagine a control volume fixed within the cylinder and assume zero body forces. We require the following equations be satisfied. ∂L
) = 0 which in one-dimension reduces to ∂L + ∂ (Lu) = 0. + div(LV Conservation of mass ∂t ∂t ∂x ∂ 2 ∂p ∂ (Lu) + Lu + = 0. Conservation of momentum, equation (2.5.28) reduces to ∂t ∂x ∂x Conservation of energy, equation (2.5.48) in the absence of heat flow and internal heat production,  ∂e ∂e ∂u becomes in one dimension L +u +p = 0. Using the conservation of mass relation this ∂t ∂x ∂x ∂ ∂ ∂u equation can be written in the form (Le) + (Leu) + p = 0. ∂t ∂x ∂x In contrast, from a Lagrangian point of view we let the control volume move with the flow and consider advection terms. This gives the following three equations which can then be compared with the above Eulerian equations of motion. DL ∂u d +L = 0. Conservation of mass (LJ) = 0 which in one-dimension is equivalent to dt Dt ∂x Du ∂p + = 0. Conservation of momentum, equation (2.5.25) in one-dimension L Dt ∂x ∂u De +p = 0. In the above equations Conservation of energy, equation (2.5.48) in one-dimension L Dt ∂x D( ) ∂ ∂ Dt = ∂t ( ) + u ∂x ( ). The Lagrangian viewpoint gives three equations in the three unknowns ρ, u, e. In both the Eulerian and Lagrangian equations the pressure p represents the total pressure p = pg + pv where pg is the gas pressure and pv is the viscous pressure which causes loss of kinetic energy. The gas pressure c

is a function of L, e and is determined from the ideal gas law pg = LRT = L(cp − cv )T = L( cpv − 1)cv T or pg = L(γ − 1)e. Some kind of assumption is usually made to represent the viscous pressure pv as a function of e, u. The above equations are then subjected to boundary and initial conditions and are usually solved numerically.

Entropy inequality Energy transfer is not always reversible. Many energy transfer processes are irreversible. The second law of thermodynamics allows energy transfer to be reversible only in special circumstances. In general, the second law of thermodynamics can be written as an entropy inequality, known as the Clausius-Duhem inequality. This inequality states that the time rate of change of the total entropy is greater than or equal to the total entropy change occurring across the surface and within the body of a control volume. The ClausiusDuhem inequality places restrictions on the constitutive equations. This inequality can be expressed in the form

   n  D i Ls dτ s ni dS + ρb dτ + B(α) ≥ Dt V S V α=1 $% & # # $% & Rate of entropy increase Entropy input rate into control volume

where s is the specific entropy density, si is an entropy flux, b is an entropy source and B(α) are isolated entropy sources. Irreversible processes are characterized by the use of the inequality sign while for reversible

305

Figure 2.5-3. Interaction of various fields. processes the equality sign holds. The Clausius-Duhem inequality is assumed to hold for all independent thermodynamical processes. If in addition there are electric and magnetic fields to consider, then these fields place additional forces upon the material continuum and we must add all forces and moments due to these effects. In particular we must add the following equations Gauss’s law for magnetism

Gauss’s law for electricity

1 ∂ √ i ( gB ) = 0. √ g ∂xi


= Le ∇·D

1 ∂ √ i ( gD ) = Le . √ g ∂xi


=− ∇×E

Faraday’s law

Ampere’s law


=0 ∇·B


∂B ∂t



= J
+ ∂ D ∇×H ∂t

Fijk Ek,j = −

∂B i . ∂t

Fijk Hk,j = J i +

∂Di . ∂t

where Le is the charge density, J i is the current density, Di = Fji Ej + Pi is the electric displacement vector, Hi is the magnetic field, Bi = µji Hj + Mi is the magnetic induction, Ei is the electric field, Mi is the magnetization vector and Pi is the polarization vector. Taking the divergence of Ampere’s law produces the law of conservation of charge which requires that ∂Le + ∇ · J
= 0 ∂t

∂Le 1 ∂ √ i +√ ( gJ ) = 0. ∂t g ∂xi

The figure 2.5-3 is constructed to suggest some of the interactions that can occur between various variables which define the continuum. Pyroelectric effects occur when a change in temperature causes changes in the electrical properties of a material. Temperature changes can also change the mechanical properties of materials. Similarly, piezoelectric effects occur when a change in either stress or strain causes changes in the electrical properties of materials. Photoelectric effects are said to occur if changes in electric or mechanical properties effect the refractive index of a material. Such changes can be studied by modifying the constitutive equations to include the effects being considered. From figure 2.5-3 we see that there can exist a relationship between the displacement field Di and electric field Ei . When this relationship is linear we can write Di = Fji Ej and Ej = βjn Dn , where Fji are

306 dielectric constants and βjn are dielectric impermabilities. Similarly, when linear piezoelectric effects exist we can write linear relations between stress and electric fields such as σij = −gkij Ek and Ei = −eijk σjk , where gkij and eijk are called piezoelectric constants. If there is a linear relation between strain and an electric fields, this is another type of piezoelectric effect whereby eij = dijk Ek and Ek = −hijk ejk , where dijk and hijk are another set of piezoelectric constants. Similarly, entropy changes can cause pyroelectric effects. Piezooptical effects (photoelasticity) occurs when mechanical stresses change the optical properties of the material. Electrical and heat effects can also change the optical properties of materials. Piezoresistivity occurs when mechanical stresses change the electric resistivity of materials. Electric field changes can cause variations in temperature, another pyroelectric effect. When temperature effects the entropy of a material this is known as a heat capacity effect. When stresses effect the entropy in a material this is called a piezocaloric effect. Some examples of the representation of these additional effects are as follows. The piezoelectric effects are represented by equations of the form σij = −hmij Dm

Di = dijk σjk

eij = gkij Dk

Di = eijk ejk

where hmij , dijk , gkij and eijk are piezoelectric constants. Knowledge of the material or electric interaction can be used to help modify the constitutive equations. For example, the constitutive equations can be modified to included temperature effects by expressing the constitutive equations in the form σij = cijkl ekl − βij ∆T

and

eij = sijkl σkl + αij ∆T

where for isotropic materials the coefficients αij and βij are constants. As another example, if the strain is modified by both temperature and an electric field, then the constitutive equations would take on the form eij = sijkl σkl + αij ∆T + dmij Em . Note that these additional effects are additive under conditions of small changes. That is, we may use the principal of superposition to calculate these additive effects. If the electric field and electric displacement are replaced by a magnetic field and magnetic flux, then piezomagnetic relations can be found to exist between the variables involved. One should consult a handbook to determine the order of magnitude of the various piezoelectric and piezomagnetic effects. For a large majority of materials these effects are small and can be neglected when the field strengths are weak. The Boltzmann Transport Equation The modeling of the transport of particle beams through matter, such as the motion of energetic protons or neutrons through bulk material, can be approached using ideas from the classical kinetic theory of gases. Kinetic theory is widely used to explain phenomena in such areas as: statistical mechanics, fluids, plasma physics, biological response to high-energy radiation, high-energy ion transport and various types of radiation shielding. The problem is basically one of describing the behavior of a system of interacting particles and their distribution in space, time and energy. The average particle behavior can be described by the Boltzmann equation which is essentially a continuity equation in a six-dimensional phase space (x, y, z, Vx , Vy , Vz ). We

307 will be interested in examining how the particles in a volume element of phase space change with time. We introduce the following notation: (i)
r the position vector of a typical particle of phase space and dτ = dxdydz the corresponding spatial volume element at this position.
the velocity vector associated with a typical particle of phase space and dτv = dVx dVy dVz the (ii) V corresponding velocity volume element.
a unit vector in the direction of the velocity V
= v Ω.
(iii) Ω (iv) E = 12 mv 2 kinetic energy of particle.
is a solid angle about the direction Ω
and dτ dE dΩ
is a volume element of phase space involving the (v) dΩ solid angle about the direction Ω.
t) the number of particles in phase space per unit volume at position
r per unit velocity (vi) n = n(
r, E, Ω,
at time t and N = N (
r, E, Ω,
t) = vn(
r, E, Ω,
t) at position V
per unit energy in the solid angle dΩ
at time t. The quantity the number of particles per unit volume per unit energy in the solid angle dΩ
t)dτ dE dΩ
represents the number of particles in a volume element around the position
r with N (
r, E, Ω,
in the solid angle dΩ
at time t. energy between E and E + dE having direction Ω
t) = vN (
r, E, Ω,
t) is the particle flux (number of particles/cm2 − Mev − sec). (vii) φ(
r, E, Ω,
 → Ω)
a scattering cross-section which represents the fraction of particles with energy E  (viii) Σ(E  → E, Ω
 that scatter into the energy range between E and E + dE having direction Ω
in the and direction Ω
per particle flux. solid angle dΩ (ix) Σs (E,
r) fractional number of particles scattered out of volume element of phase space per unit volume per flux. (x) Σa (E,
r) fractional number of particles absorbed in a unit volume of phase space per unit volume per flux. Consider a particle at time t having a position
r in phase space as illustrated in the figure 2.5-4. This
in a direction Ω
and has an energy E. In terms of dτ = dx dy dz, Ω
and E an particle has a velocity V
where dΩ
= dΩ(θ,
element of volume of phase space can be denoted dτ dEdΩ, ψ) = sin θdθdψ is a solid angle
about the direction Ω. The Boltzmann transport equation represents the rate of change of particle density in a volume element
of phase space and is written dτ dE dΩ d
t) dτ dE dΩ
= DC N (
r, E, Ω,
t) N (
r, E, Ω, dt

(2.5.71)

where DC is a collision operator representing gains and losses of particles to the volume element of phase space due to scattering and absorption processes. The gains to the volume element are due to any sources
t) per unit volume of phase space, with units of number of particles/sec per volume of phase space, S(
r, E, Ω, together with any scattering of particles into the volume element of phase space. That is particles entering the volume element of phase space with energy E, which experience a collision, leave with some energy E − ∆E and thus will be lost from our volume element. Particles entering with energies E  > E may,

308

Figure 2.5-4. Volume element and solid angle about position
r. depending upon the cross-sections, exit with energy E  − ∆E = E and thus will contribute a gain to the volume element. In terms of the flux φ the gains due to scattering into the volume element are denoted by  
 → Ω)φ(


t) dτ dE dΩ

 dE  Σ(E  → E, Ω r , E  , Ω, dΩ and represents the particles at position
r experiencing a scattering collision with a particle of energy E  and
 which causes the particle to end up with energy between E and E + dE and direction Ω
in dΩ.
direction Ω The summations are over all possible initial energies. In terms of φ the losses are due to those particles leaving the volume element because of scattering and are
t)dτ dE dΩ.
Σs (E,
r)φ(
r, E, Ω, The particles which are lost due to absorption processes are
t) dτ dE dΩ.
Σa (E,
r)φ(
r, E, Ω, The total change to the number of particles in an element of phase space per unit of time is obtained by summing all gains and losses. This total change is   dN
 dE  Σ(E  → E, Ω
 → Ω)φ(


t) dτ dE dΩ
dτ dE dΩ = dΩ r , E  , Ω, dt
t)dτ dE dΩ − Σs (E,
r)φ(
r, E, Ω,
t) dτ dE dΩ
− Σa (E,
r)φ(
r, E, Ω,
t)dτ dE dΩ.
+ S(
r, E, Ω, The rate of change

dN dt

on the left-hand side of equation (2.5.72) expands to ∂N ∂N dx ∂N dy ∂N dz dN = + + + dt ∂t ∂x dt ∂y dt ∂z dt ∂N dVx ∂N dVy ∂N dVz + + + ∂Vx dt ∂Vy dt ∂Vz dt

(2.5.72)

309 which can be written as

where

' dV dt

=

' F m


∂N dN
· ∇'r N + F · ∇ ' N = +V V dt ∂t m

(2.5.73)

represents any forces acting upon the particles. The Boltzmann equation can then be

expressed as
∂N
· ∇'r N + F · ∇ ' N = Gains − Losses. +V V ∂t m

(2.5.74)

If the right-hand side of the equation (2.5.74) is zero, the equation is known as the Liouville equation. In the special case where the velocities are constant and do not change with time the above equation (2.5.74) can be written in terms of the flux φ and has the form 

 1 ∂

t) = DC φ + Ω · ∇'r + Σs (E,
r) + Σa (E,
r) φ(
r, E, Ω, v ∂t 

where DC φ =


 dΩ



(2.5.75)


 → Ω)φ(


 , t) + S(
r, E, Ω,
t). dE  Σ(E  → E, Ω r, E , Ω

The above equation represents the Boltzmann transport equation in the case where all the particles are the same. In the case of atomic collisions of particles one must take into consideration the generation of secondary particles resulting from the collisions. Let there be a number of particles of type j in a volume element of phase space. For example j = p (protons) and j = n (neutrons). We consider steady state conditions and define the quantities
as the flux of the particles of type j. (i) φj (
r, E, Ω)
Ω
 , E, E  ) the collision cross-section representing processes where particles of type k moving in (ii) σjk (Ω,
with energy E.
 with energy E  produce a type j particle moving in the direction Ω direction Ω (iii) σj (E) = Σs (E,
r) + Σa (E,
r) the cross-section for type j particles. The steady state form of the equation (2.5.64) can then be written as
· ∇φj (
r, E, Ω)+σ

Ω r , E, Ω) j (E)φj (
 
Ω
 , E, E  )φk (
r, E  , Ω
 )dΩ
 dE  σjk (Ω, =

(2.5.76)

k

where the summation is over all particles k = j. The Boltzmann transport equation can be represented in many different forms. These various forms are dependent upon the assumptions made during the derivation, the type of particles, and collision crosssections. In general the collision cross-sections are dependent upon three components. (1) Elastic collisions. Here the nucleus is not excited by the collision but energy is transferred by projectile recoil. (2) Inelastic collisions. Here some particles are raised to a higher energy state but the excitation energy is not sufficient to produce any particle emissions due to the collision. (3) Non-elastic collisions. Here the nucleus is left in an excited state due to the collision processes and some of its nucleons (protons or neutrons) are ejected. The remaining nucleons interact to form a stable structure and usually produce a distribution of low energy particles which is isotropic in character.

310 Various assumptions can be made concerning the particle flux. The resulting form of Boltzmann’s equation must be modified to reflect these additional assumptions. As an example, we consider modifications to Boltzmann’s equation in order to describe the motion of a massive ion moving into a region filled with a homogeneous material. Here it is assumed that the mean-free path for nuclear collisions is large in comparison with the mean-free path for ion interaction with electrons. In addition, the following assumptions are made (i) All collision interactions are non-elastic. (ii) The secondary particles produced have the same direction as the original particle. This is called the straight-ahead approximation. (iii) Secondary particles never have kinetic energies greater than the original projectile that produced them. (iv) A charged particle will eventually transfer all of its kinetic energy and stop in the media. This stopping distance is called the range of the projectile. The stopping power Sj (E) =

dE dx

represents the energy dR

loss per unit length traveled in the media and determines the range by the relation dEj = Sj 1(E) or 1E  1 Rj (E) = 0 SjdE (E  ) . Using the above assumptions Wilson, et.al. show that the steady state linearized Boltzmann equation for homogeneous materials takes on the form
· ∇φj (
r, E, Ω)
− 1 ∂ (Sj (E)φj (
r, E, Ω))
+ σj (E)φj (
r, E, Ω)
Ω Aj ∂E 
 σjk (Ω,
Ω
 , E, E  )φk (
r, E  , Ω
) = dE  dΩ

(2.5.77)

k =j


is the flux of ions of type j moving in where Aj is the atomic mass of the ion of type j and φj (
r, E, Ω)
with energy E. the direction Ω Observe that in most cases the left-hand side of the Boltzmann equation represents the time rate of change of a distribution type function in a phase space while the right-hand side of the Boltzmann equation represents the time rate of change of this distribution function within a volume element of phase space due to scattering and absorption collision processes. Boltzmann Equation for gases
, t) which can be Consider the Boltzmann equation in terms of a particle distribution function f (
r, V written as




∂
· ∇'r + F · ∇ ' +V V ∂t m


, t)
, t) = DC f (
r, V f (
r, V

(2.5.78)

for a single species of gas particles where there is only scattering and no absorption of the particles. An element of volume in phase space (x, y, z, Vx , Vy , Vz ) can be thought of as a volume element dτ = dxdydz for the spatial elements together with a volume element dτv = dVx dVy dVz for the velocity elements. These elements are centered at position
r and velocity V
at time t. In phase space a constant velocity V1 can be thought of as a sphere since V12 = Vx2 + Vy2 + Vz2 . The phase space volume element dτ dτv changes with time
change with time. The position vector
r changes because of velocity since the position
r and velocity V 1

John W. Wilson, Lawrence W. Townsend, Walter Schimmerling, Govind S. Khandelwal, Ferdous Kahn,

John E. Nealy, Francis A. Cucinotta, Lisa C. Simonsen, Judy L. Shinn, and John W. Norbury, Transport Methods and Interactions for Space Radiations, NASA Reference Publication 1257, December 1991.

311 and the velocity vector changes because of the acceleration

' F m.


, t)dτ dτv represents the expected Here f (
r, V

number of particles in the phase space element dτ dτv at time t. Assume there are no collisions, then each of the gas particles in a volume element of phase space centered
1 move during a time interval dt to a phase space element centered at position at position
r and velocity V ' F
1 + dt. If there were no loss or gains of particles, then the number of particles must be
r + V
1 dt and V m

conserved and so these gas particles must move smoothly from one element of phase space to another without any gains or losses of particles. Because of scattering collisions in dτ many of the gas particles move into or
1 + dV
1 . These collision scattering processes are denoted by the collision
1 to V out of the velocity range V
, t) in the Boltzmann equation. operator DC f (
r, V Consider two identical gas particles which experience a binary collision. Imagine that particle 1 with
2 . Denote by σ(V
1 → V
1 , V
2 → V
2 ) dτV1 dτV2 the velocity V
1 collides with particle 2 having velocity V
 and V
 + dV
 and the
1 to between V conditional probability that particle 1 is scattered from velocity V 1 1 1
 + dV
 . We will be interested in collisions
2 to between V
 and V struck particle 2 is scattered from velocity V 2 2 2
 ) → (V
1 , V
2 ) for a fixed value of V
1 as this would represent the number of particles scattered
, V of the type (V 1

2


1 , V
2 ) → (V
1 , V
2 ) for a fixed value V
1 as this represents into dτV1 . Also of interest are collisions of the type (V particles scattered out of dτV1 . Imagine a gas particle in dτ with velocity V
1 subjected to a beam of particles
1 − V
2 |f (
r, V
2 , t)dτV  and hence
2 . The incident flux on the element dτ dτV  is |V with velocities V 1 2
1 → V
1 , V
2 → V
2 ) dτV1 dτV2 dt |V
1 − V
2 |f (
r, V
2 , t) dτV  σ(V 2

(2.5.79)


1 and V
1 + dV
1 represents the number of collisions, in the time interval dt, which scatter from V
1 to between V
 to between V
2 and V
2 + dV
2 . Multiply equation (2.5.79) by the density of particles as well as scattering V 2


1 ,V
2 and final velocities V
2 not equal in the element dτ dτV1 and integrate over all possible initial velocities V to V
1 . This gives the number of particles in dτ which are scattered into dτV1 dt as  
 →V
1 , V
 →V
2 )|V
 − V
 |f (
r, V
 , t)f (
r, V
 , t). N sin = dτ dτV1 dt dτV2 dτV2 dτV1 σ(V (2.5.80) 1 2 1 2 1 2 In a similar manner the number of particles in dτ which are scattered out of dτV1 dt is   
1 , t) dτV2 dτV  dτV  σ(V
 →V
1 , V
 →V
2 )|V
2 − V
1 |f (
r, V
2 , t). N sout = dτ dτV1 dtf (
r, V 1 2 2 1

(2.5.81)

Let
 →V
1 , V
 →V
2 ) = |V
1 − V
2 | σ(V
 →V
1 , V
 → V
2 ) W (V 1 2 1 2

(2.5.82)


, t) = N sin − N sout to represent the define a symmetric scattering kernel and use the relation DC f (
r, V Boltzmann equation for gas particles in the form   

 ∂  · ∇r + F · ∇  +V V ∂t m



dτV  1

1 , t) = f (r , V



dτV  2





(2.5.83)

1 → V  , V 2 → V   ) f (r , V   , t)f (r , V   , t) − f (r , V 1 , t)f (r , V 2 , t) . dτV2 W (V 1 2 1 2


1 ). That Take the moment of the Boltzmann equation (2.5.83) with respect to an arbitrary function φ(V
1 ) and then integrate over all elements of velocity space dτV1 . Define is, multiply equation (2.5.83) by φ(V the following averages and terminology:

312 • The particle density per unit volume  n = n(
r, t) =

+∞ 
, t) = dτV f (
r, V


, t)dVx dVy dVz f (
r, V

(2.5.84)

−∞

where ρ = nm is the mass density. • The mean velocity
1 = V
= 1 V n

+∞ 
1 f (
r, V
1 , t)dV1x dV1y dV1z V −∞


1 ) define the barred quantity For any quantity Q = Q(V 1 Q = Q(
r, t) = n(
r, t) Further, assume that

' F m




)f (
r, V
, t) dτV = 1 Q(V n

+∞ 
)f (
r, V
, t)dVx dVy dVz . Q(V

(2.5.85)

−∞

is independent of V
, then the moment of equation (2.5.83) produces the result 3 3 

Fi ∂φ ∂  ∂ nφ + nV φ − n =0 1i i ∂t ∂x m ∂V1i i=1 i=1

(2.5.86)

known as the Maxwell transfer equation. The first term in equation (2.5.86) follows from the integrals 


1 , t) ∂f (
r, V
1 )dτV1 = ∂
1 , t)φ(V
1 ) dτV1 = ∂ (nφ) φ(V f (
r, V ∂t ∂t ∂t

(2.5.87)

where differentiation and integration have been interchanged. The second term in equation (2.5.86) follows from the integral



  3

∂f φ dτV1 ∂xi i=1   3  ∂ = V1i φf dτV1 ∂xi i=1


1 )dτV1 = V
1 ∇'r f φ(V

=

V1i

(2.5.88)

3 

∂ nV1i φ . i ∂x i=1

The third term in equation (2.5.86) is obtained from the following integral where integration by parts is employed

 
  3  F Fi ∂f ∇V'1 f φ dτV1 = φ dτV1 m m ∂V1i i=1   +∞   3 Fi ∂f = φ dV1x dV1y dV1y m ∂V1i −∞ i=1    Fi ∂ φ f dτV1 =− ∂V1i m   Fi Fi ∂φ ∂ φ =− = −n ∂V1i m m ∂V1i

(2.5.89)

313
1 and f (
r, V
, t) equals zero for Vi equal to ±∞. The right-hand side of since Fi does not depend upon V equation (2.5.86) represents the integral of (DC f )φ over velocity space. This integral is zero because of the symmetries associated with the right-hand side of equation (2.5.83). Physically, the integral of (Dc f )φ over velocity space must be zero since collisions with only scattering terms cannot increase or decrease the number of particles per cubic centimeter in any element of phase space. In equation (2.5.86) we write the velocities V1i in terms of the mean velocities (u, v, w) and random velocities (Ur , Vr , Wr ) with V11 = Ur + u,

V12 = Vr + v,

V13 = Wr + w

(2.5.90)


1 = V
r + V
r = 0 (i.e. the average random velocity is zero.) For
r + V
with V
= V
since V or V
1 = V future reference we write equation (2.5.86) in terms of these random velocities and the material derivative. Substitution of the velocities from equation (2.5.90) in equation (2.5.86) gives 3     ∂  Fi ∂φ ∂(nφ) ∂  ∂  + n(Vr + v)φ + n(Wr + w)φ − n n(Ur + u)φ + =0 ∂t ∂x ∂y ∂z m ∂V1i i=1

or

(2.5.91)





∂(nφ) ∂ ∂ ∂ + nuφ + nvφ + nwφ ∂t ∂x ∂y ∂z 3 



∂ Fi ∂φ ∂ ∂ + nUr φ + nVr φ + nWr φ − n = 0. ∂x ∂y ∂z m ∂V1i i=1

Observe that

(2.5.92)

+∞ 
, t)dVx dVy dVz = nuφ uφf (
r, V

nuφ =

(2.5.93)

−∞

and similarly nvφ = nvφ, nwφ = nwφ. This enables the equation (2.5.92) to be written in the form n

∂φ ∂φ ∂φ ∂φ + nu + nv + nw ∂t ∂x ∂y ∂z   ∂n ∂ ∂ ∂ +φ + (nu) + (nv) + (nw) ∂t ∂x ∂y ∂z +

(2.5.94)

3 



∂ Fi ∂φ ∂ ∂ nUr φ + nVr φ + nWr φ − n = 0. ∂x ∂y ∂z m ∂V1i i=1

The middle bracketed sum in equation (2.5.94) is recognized as the continuity equation when multiplied by m and hence is zero. The moment equation (2.5.86) now has the form n

3 



Dφ ∂ Fi ∂φ ∂ ∂ + nUr φ + nVr φ + nWr φ − n = 0. Dt ∂x ∂y ∂z m ∂V1i i=1

(2.5.95)

Note that from the equations (2.5.86) or (2.5.95) one can derive the basic equations of fluid flow from continuum mechanics developed earlier. We consider the following special cases of the Maxwell transfer equation.

314 (i) In the special case φ = m the equation (2.5.86) reduces to the continuity equation for fluids. That is, equation (2.5.86) becomes ∂
1 ) = 0 (nm) + ∇ · (nmV ∂t

(2.5.96)

∂ρ
)=0 + ∇ · (ρV ∂t

(2.5.97)

which is the continuity equation


is the mean velocity defined earlier. where ρ is the mass density and V
1 is momentum, the equation (2.5.86) reduces to the momentum equation (ii) In the special case φ = mV
1 V
1 in the form for fluids. To show this, we write equation (2.5.86) in terms of the dyadic V

or

Let

σ = −ρV
r V
r

 ∂ 
1 + ∇ · (nmV
1 V
1 ) − nF
= 0 nmV ∂t

(2.5.98)

 ∂ 

) + ∇ · (ρ(V
r + V
)(V
r + V
)) − nF
= 0. ρ(Vr + V ∂t

(2.5.99)

denote a stress tensor which is due to the random motions of the gas particles and

write equation (2.5.99) in the form
∂ρ ∂V
(∇ · V
) + V
(∇ · (ρV
)) − ∇ · σ − nF
= 0. + V
+ ρV (2.5.100) ∂t ∂t 
) = 0 because of the continuity equation and so equation (2.5.100) reduces + ∇ · (ρV ρ


The term V



∂ρ ∂t

to the momentum equation

 ρ


∂V
∇·V
+V ∂t



= nF
+ ∇ · σ .

(2.5.101)


+ qV
×B
+ m
b, where q is charge, E
and B
are electric and magnetic fields, and
b is a For F
= q E body force per unit mass, together with

σ=

3  3 

(−pδij + τij )
ei e
j

(2.5.102)

i=1 j=1

the equation (2.5.101) becomes the momentum equation ρ

DV

+V
× B).
= ρ
b − ∇p + ∇ · τ + nq(E Dt

(2.5.103)


and B
vanish, the equation (2.5.103) reduces to the previous momentum In the special case were E equation (2.5.25) . (iii) In the special case φ =

m

2 V1 · V1

=

m 2 2 2 2 (V11 + V12 + V13 )

is the particle kinetic energy, the equation (2.5.86)

simplifies to the energy equation of fluid mechanics. To show this we substitute φ into equation (2.5.95) and simplify. Note that

 m (Ur + u)2 + (Vr + v)2 + (Wr + w)2 2  m 2 φ= Ur + Vr2 + Wr2 + u2 + v 2 + w2 2 φ=

(2.5.104)

315 since uUr = vVr = wWr = 0. Let V 2 = u2 + v 2 + w2 and Cr2 = Ur2 + Vr2 + Wr2 and write equation (2.5.104) in the form φ=

 m 2 Cr + V 2 . 2

(2.5.105)

Also note that  nm  Ur (Ur + u)2 + Ur (Vr + v)2 + Ur (Wr + w)2 2 ( ) nm Ur Cr2 + uUr2 + vUr Vr + wUr Wr = 2 2

nUr φ =

(2.5.106)

and that  nm  Vr Cr2 + uVr Ur + vVr2 + wVr Wr 2  nm  nWr φ = Wr Cr2 + uWr Ur + vWr Vr + wWr2 2

(2.5.107)

nVr φ =

(2.5.108)

are similar results. We use

∂ ∂V1i

(φ) = mV1i together with the previous results substituted into the equation (2.5.95), and

find that the Maxwell transport equation can be expressed in the form    V2 ∂  D Cr2 + =− ρ[uUr2 + vUr Vr + wUr Wr ] ρ Dt 2 2 ∂x  ∂  − ρ[uVr Ur + vVr2 + wVr Wr ] ∂y  ∂  ρ[uWr Ur + vWr Vr + wWr2 ] − ∂z       ∂ ∂ Ur Cr2 Vr Cr2 Wr Cr2 ∂
. ρ − ρ − ρ + nF
· V − ∂x 2 ∂y 2 ∂z 2 Compare the equation (2.5.109) with the energy equation (2.5.48)   De D V2
) − ∇ ·
q + ρ
b · V
ρ +ρ = ∇(σ · V Dt Dt 2

(2.5.109)

(2.5.110)

C2

where the internal heat energy has been set equal to zero. Let e = 2r denote the internal energy due to random motion of the gas particles, F
= m
b, and let       ∂ ∂ ∂ Ur Cr2 Vr Cr2 Wr Cr2 ρ − ρ − ρ ∇·
q =− ∂x 2 ∂y 2 ∂z 2 (2.5.111)       ∂ ∂T ∂ ∂T ∂ ∂T =− k − k − k ∂x ∂x ∂y ∂y ∂z ∂z represent the heat conduction terms due to the transport of particle energy

mCr2 2

by way of the random

particle motion. The remaining terms are related to the rate of change of work and surface stresses giving  ∂ ∂  ρ[uUr2 + vUr Vr + wUr Wr ] = (uσxx + vσxy + wσxz ) − ∂x ∂x  ∂ ∂  ρ[uVr Ur + vVr2 + wVr Wr ] = (uσyx + vσyy + wσyz ) − (2.5.112) ∂y ∂y  ∂ ∂  ρ[uWr Ur + vWr Vr + wWr2 ] = (uσzx + vσzy + wσzz ) . − ∂z ∂z

316 This gives the stress relations due to random particle motion σxx = − ρUr2

σyx = − ρVr Ur

σzx = − ρWr Ur

σxy = − ρUr Vr

σyy = − ρVr2

σzy = − ρWr Vr

σxz = − ρUr Wr

σyz = − ρVr Wr

σzz = − ρWr2 .

(2.5.113)

The Boltzmann equation is a basic macroscopic model used for the study of individual particle motion where one takes into account the distribution of particles in both space, time and energy. The Boltzmann equation for gases assumes only binary collisions as three-body or multi-body collisions are assumed to rarely occur. Another assumption used in the development of the Boltzmann equation is that the actual time of collision is thought to be small in comparison with the time between collisions. The basic problem associated with the Boltzmann equation is to find a velocity distribution, subject to either boundary and/or initial conditions, which describes a given gas flow. The continuum equations involve trying to obtain the macroscopic variables of density, mean velocity, stress, temperature and pressure which occur in the basic equations of continuum mechanics considered earlier. Note that the moments of the Boltzmann equation, derived for gases, also produced these same continuum equations and so they are valid for gases as well as liquids. In certain situations one can assume that the gases approximate a Maxwellian distribution f (
r, V
, t) ≈ n(
r, t)

 m 3/2   m
·V
V exp − 2πkT 2kT

(2.5.114)

thereby enabling the calculation of the pressure tensor and temperature from statistical considerations. In general, one can say that the Boltzmann integral-differential equation and the Maxwell transfer equation are two important formulations in the kinetic theory of gases. The Maxwell transfer equation depends upon some gas-particle property φ which is assumed to be a function of the gas-particle velocity. The Boltzmann equation depends upon a gas-particle velocity distribution function f which depends upon
and time t. These formulations represent two distinct and important viewpoints position
r, velocity V considered in the kinetic theory of gases.

317 EXERCISE 2.5
1.

Let p = p(x, y, z), [dyne/cm2 ] denote the pressure at a point (x, y, z) in a fluid medium at rest

(hydrostatics), and let ∆V denote an element of fluid volume situated at this point as illustrated in the figure 2.5-5.

Figure 2.5-5. Pressure acting on a volume element. (a) Show that the force acting on the face ABCD is p(x, y, z)∆y∆z eˆ1 . (b) Show that the force acting on the face EF GH is   ∂ 2 p (∆x)2 ∂p −p(x + ∆x, y, z)∆y∆z eˆ1 = − p(x, y, z) + ∆x + 2 + · · · ∆y∆z eˆ1 . ∂x ∂x 2! (c) In part (b) neglect terms with powers of ∆x greater than or equal to 2 and show that the resultant force ∂p in the x-direction is − ∆x∆y∆z eˆ1 . ∂x (d) What has been done in the x-direction can also be done in the y and z-directions. Show that the ∂p ∂p resultant forces in these directions are − ∆x∆y∆z eˆ2 and − ∆x∆y∆z eˆ3 . (e) Show that −∇p = ∂y ∂z   ∂p ∂p ∂p eˆ1 + eˆ2 + eˆ3 is the force per unit volume acting at the point (x, y, z) of the fluid medium. − ∂x ∂y ∂z
2. Follow the example of exercise 1 above but use cylindrical coordinates and find the force per unit volume at a point (r, θ, z). Hint: An element of volume in cylindrical coordinates is given by ∆V = r∆r∆θ∆z.
3. Follow the example of exercise 1 above but use spherical coordinates and find the force per unit volume at a point (ρ, θ, φ). Hint: An element of volume in spherical coordinates is ∆V = ρ2 sin θ∆ρ∆θ∆φ.
4. Show that if the density L = L(x, y, z, t) is a constant, then v r,r = 0.
5. Assume that λ∗ and µ∗ are zero. Such a fluid is called a nonviscous or perfect fluid. (a) Show the Cartesian equations describing conservation of linear momentum are ∂u ∂u ∂u ∂u 1 ∂p +u +v +w = bx − ∂t ∂x ∂y ∂z L ∂x ∂v ∂v ∂v ∂v 1 ∂p +u +v +w = by − ∂t ∂x ∂y ∂z L ∂y ∂w ∂w ∂w ∂w 1 ∂p +u +v +w = bz − ∂t ∂x ∂y ∂z L ∂z where (u, v, w) are the physical components of the fluid velocity. (b) Show that the continuity equation can be written ∂L ∂ ∂ ∂ + (Lu) + (Lv) + (Lw) = 0 ∂t ∂x ∂y ∂z

318
6. Assume λ∗ = µ∗ = 0 so that the fluid is ideal or nonviscous. Use the results given in problem 5 and make the following additional assumptions: • The density is constant and so the fluid is incompressible. • The body forces are zero. • Steady state flow exists. • Only two dimensional flow in the x-yplane is considered such that u = u(x, y), v = v(x, y) and w = 0. (a) Employ the above assumptions and simplify the equations in problem 5 and verify the results

∂u 1 ∂p ∂u +v + =0 ∂x ∂y L ∂x ∂v ∂v 1 ∂p u +v + =0 ∂x ∂y L ∂y ∂u ∂v + =0 ∂x ∂y

u

(b) Make the additional assumption that the flow is irrotational and show that this assumption produces the results ∂v ∂u − =0 ∂x ∂y

and

1 1 2 u + v 2 + p = constant. 2 L

(c) Point out the Cauchy-Riemann equations and Bernoulli’s equation in the above set of equations.
7. Assume the body forces are derivable from a potential function φ such that bi = −φ,i . Show that for an ideal fluid with constant density the equations of fluid motion can be written in either of the forms 1 ∂v r + v r,s v s = − g rm p,m − g rm φ,m ∂t L

or

∂vr 1 + vr,s v s = − p,r − φ,r ∂t L

1 ∇ (
v ·
v ) −
v × (∇ ×
v ) are 2 used to express the Navier-Stokes-Duhem equations in alternate forms involving the vorticity Ω = ∇ ×
v .


8. The vector identities ∇2
v = ∇ (∇ ·
v ) − ∇ × (∇ ×
v )

and

(
v · ∇)
v =

(a) Use Cartesian tensor notation and derive the above identities. (b) Show the second identity can be written ∂v 2 mj k in generalized coordinates as v j v m v vk,j − Fmnp Fijk gpi vn vk,j . Hint: Show that = 2v k vk,j . ,j = g ∂xj
9. Use problem 8 and show that the results in problem 7 can be written

or

  p ∂v r v2 ∂ − Frnp Ωp vn = −g rm m +φ+ ∂t ∂x L 2   2 p v ∂ ∂vi − Fijk v j Ωk = − i +φ+ ∂t ∂x L 2


10. In terms of physical components, show that in generalized coordinates,  orthogonal    for i = j, the rate v(i) v(j) 1 hi ∂ hj ∂ of deformation tensor Dij can be written D(ij) = + , no summations 2 hj ∂xj hi hi ∂xi hj 3 ∂hi 1 ∂v(i) v(i) ∂hi  1 − + v(k) k , no summations. (Hint: See and for i = j there results D(ii) = 2 i i hi ∂x hi ∂x hi hk ∂x k=1 Problem 17 Exercise 2.1.)

319

Figure 2.5-6. Plane Couette flow
11. Find the physical components of the rate of deformation tensor Dij in Cartesian coordinates. (Hint: See problem 10.)
12. Find the physical components of the rate of deformation tensor in cylindrical coordinates. (Hint: See problem 10.)
13. (Plane Couette flow) Assume a viscous fluid with constant density is between two plates as illustrated in the figure 2.5-6. (a) Define ν =

µ∗ E

as the kinematic viscosity and show the equations of fluid motion can be written ∂v i 1 + v i,s v s = − g im p,m + νg jm v i,mj + g ij bj , ∂t L

i = 1, 2, 3

(b) Let
v = (u, v, w) denote the physical components of the fluid flow and make the following assumptions • u = u(y), v = w = 0 • Steady state flow exists • The top plate, with area A, is a distance 7 above the bottom plate. The bottom plate is fixed and a constant force F is applied to the top plate to keep it moving with a velocity u0 = u(7). • p and L are constants • The body force components are zero. Find the velocity u = u(y) u0 F = σ21 = σxy = σyx = µ∗ . This A 7 example illustrates that the stress is proportional to u0 and inversely proportional to 7.

(c) Show the tangential stress exerted by the moving fluid is


14. In the continuity equation make the change of variables t=

t , τ

L=

L , L0


v =
v , v0

x=

x , L

y=

y , L

z=

z L

and write the continuity equation in terms of the barred variables and the Strouhal parameter.
15. (Plane Poiseuille flow)

Consider two flat plates parallel to one another as illustrated in the figure

2.5-7. One plate is at y = 0 and the other plate is at y = 27. Let
v = (u, v, w) denote the physical components of the fluid velocity and make the following assumptions concerning the flow The body forces are zero. The ∂p ∂p ∂p = −p0 is a constant and = = 0. The velocity in the x-direction is a function of y only derivative ∂x ∂y ∂z

320

Figure 2.5-7. Plane Poiseuille flow with u = u(y) and v = w = 0 with boundary values u(0) = u(27) = 0. The density is constant and ν = µ∗ /L is the kinematic viscosity. d2 u p0 + = 0, u(0) = u(27) = 0 dy 2 L (b) Find the velocity u = u(y) and find the maximum velocity in the x-direction. (c) Let M denote the (a) Show the equation of fluid motion is ν

mass flow rate across the plane x = x0 = constant, , where 0 ≤ y ≤ 27, and 0 ≤ z ≤ 1. 2 Show that M = ∗ Lp0 73 . Note that as µ∗ increases, M decreases. 3µ ∂(δcu) , where c is the ∂t 3 specific heat [cal/gm C], δ is the volume density [gm/cm ], H is the rate of heat generation [cal/sec cm3 ], u


16. The heat equation (or diffusion equation) can be expressed div ( k grad u)+ H =

is the temperature [C], k is the thermal conductivity [cal/sec cm C]. Assume constant thermal conductivity, volume density and specific heat and express the boundary value problem ∂2u ∂u = δc , 0 < x < L ∂x2 ∂t u(L, t) = u1 , u(x, 0) = f (x) k

u(0, t) = 0,

in a form where all the variables are dimensionless. Assume u1 is constant.
17. Simplify the Navier-Stokes-Duhem equations using the assumption that there is incompressible flow.
18. (Rayleigh impulsive flow)

The figure 2.5-8 illustrates fluid motion in the plane where y > 0 above a

plate located along the axis where y = 0. The plate along y = 0 has zero velocity for all negative time and at time t = 0 the plate is given an instantaneous velocity u0 in the positive x-direction. Assume the physical components of the velocity are
v = (u, v, w) which satisfy u = u(y, t), v = w = 0. Assume that the density of the fluid is constant, the gradient of the pressure is zero, and the body forces are zero. (a) Show that the velocity in the x-direction is governed by the differential equation ∂ 2u ∂u = ν 2, ∂t ∂y

with

ν=

µ∗ . L

Assume u satisfies the initial condition u(0, t) = u0 H(t) where H is the Heaviside step function. Also assume there exist a condition at infinity limy→∞ u(y, t). This latter condition requires a bounded velocity at infinity. (b) Use any method to show the velocity is  u(y, t) = u0 − u0 erf

y √ 2 νt



 = u0 erfc

y √ 2 νt



321

Figure 2.5-8. Rayleigh impulsive flow where erf and erfc are the error function and complimentary error function respectively. Pick a point on the √ line y = y0 = 2 ν and plot the velocity as a function of time. How does the viscosity effect the velocity of the fluid along the line y = y0 ?
19. Simplify the Navier-Stokes-Duhem equations using the assumption that there is incompressible and irrotational flow.
20. Let ζ = λ∗ + 23 µ∗ and show the constitutive equations (2.5.21) for fluid motion can be written in the   2 σij = −pδij + µ∗ vi,j + vj,i − δij vk,k + ζδij vk,k . 3

form


21. (a) Write out the Navier-Stokes-Duhem equation for two dimensional flow in the x-y direction under the assumptions that • λ∗ + 23 µ∗ = 0

(This condition is referred to as Stoke’s flow.)

• The fluid is incompressible • There is a gravitational force
b = −g∇ h Hint: Express your answer as two scalar equations involving the variables v1 , v2 , h, g, L, p, t, µ∗ plus the continuity equation. (b) In part (a) eliminate the pressure and body force terms by cross differentiation and subtraction. (i.e. take the derivative of one equation with respect to x and take the derivative of the other equationwith respect to y ∂v1 1 ∂v2 − and then eliminate any common terms.) (c) Assume that
ω = ω eˆ3 where ω = and 2 ∂x ∂y derive the vorticity-transport equation dω = ν∇2 ω dt

where

∂ω ∂ω ∂ω dω = + v1 + v2 . dt ∂t ∂x ∂y

Hint: The continuity equation makes certain terms zero. (d) Define a stream function ψ = ψ(x, y) ∂ψ ∂ψ satisfying v1 = and v2 = − and show the continuity equation is identically satisfied. ∂y ∂x Show also that ω = − 21 ∇2 ψ and that ∇4 ψ =

  1 ∂∇2 ψ ∂ψ ∂∇2 ψ ∂ψ ∂∇2 ψ + − . ν ∂t ∂y ∂x ∂x ∂y

If ν is very large, show that ∇4 ψ ≈ 0.

322
22. In generalized orthogonal coordinates, show that the physical components of the rate of deformation stress can be written, for i = j σ(ij) = µ∗



hi ∂ hj ∂xj



v(i) hi

 +

hj ∂ hi ∂xi



v(j) hj

 ,

no summation,

and for i = j = k   1 ∂v(i) 1 ∂hi 1 ∂hi + v(j) + v(k) σ(ii) = −p + 2µ∗ hi ∂xi hi hj ∂xj hi hk ∂xk   ∂ λ∗ ∂ ∂ + {h h v(1)} + {h h v(2)} + {h h v(3)} , 2 3 1 3 1 2 h1 h2 h3 ∂x1 ∂x2 ∂x3

no summation


23. Find the physical components for the rate of deformation stress in Cartesian coordinates. Hint: See problem 22.
24. Find the physical components for the rate of deformations stress in cylindrical coordinates. Hint: See problem 22. 1
25. Verify the Navier-Stokes equations for an incompressible fluid can be written v˙ i = − p,i + νvi,mm + bi L ∗ where ν = µE is called the kinematic viscosity.
26. Verify the Navier-Stokes equations for a compressible fluid with zero bulk viscosity can be written ∗ 1 ν v˙ i = − p,i + vm,mi + νvi,mm + bi with ν = µE the kinematic viscosity. L 3
27. The constitutive equation for a certain non-Newtonian Stokesian fluid is σij = −pδij +βDij +γDik Dkj . Assume that β and γ are constants (a) Verify that σij,j = −p,i + βDij,j + γ(Dik Dkj,j + Dik,j Dkj ) (b) Write out the Cauchy equations of motion in Cartesian coordinates. (See page 236).
28. Let the constitutive equations relating stress and strain for a solid material take into account thermal 1+ν ν σij − σkk δij +α T δij stresses due to a temperature T . The constitutive equations have the form eij = E E where α is a coefficient of linear expansion for the material and T is the absolute temperature. Solve for the stress in terms of strains.
29. Derive equation (2.5.53) and then show that when the bulk coefficient of viscosity is zero, the NavierStokes equations, in Cartesian coordinates, can be written in the conservation form ∂(Lu) ∂(Lu2 + p − τxx ) ∂(Luv − τxy ) ∂(Luw − τxz ) + + + = Lbx ∂t ∂x ∂y ∂z ∂(Lv) ∂(Luv − τxy ) ∂(Lv 2 + p − τyy ) ∂(Lvw − τyz ) + + + = Lby ∂t ∂x ∂y ∂z ∂(Lw) ∂(Luw − τxz ) ∂(Lvw − τyz ) ∂(Lw2 + p − τzz ) + + + = Lbz ∂t ∂x ∂y ∂z 2 where v1 = u,v2 = v,v3 = w and τij = µ∗ (vi,j + vj,i − δij vk,k ). Hint: Alternatively, consider 2.5.29 and use 3 the continuity equation.

323
30. Show that for a perfect gas, where λ∗ = − 32 µ∗ and η = µ∗ is a function of position, the vector form of equation (2.5.25) is L

4 D
v = L
b − ∇p + ∇(η∇ ·
v ) + ∇(
v · ∇η) −
v ∇2 η + (∇η) × (∇ ×
v ) − (∇ ·
v )∇η − ∇ × (∇ × (η
v )) Dt 3

D p ∂Q Dh = + − ∇ ·
q + Φ. Hint: Use the continuity equation. Dt Dt ∂t
32. Show that in Cartesian coordinates the Navier-Stokes equations of motion for a compressible fluid
31. Derive the energy equation L can be written       Du ∂p ∂ ∂u
+ ∂ µ∗ ( ∂u + ∂v ) + ∂ µ∗ ( ∂w + ∂u ) =ρbx − + 2µ∗ + λ∗ ∇ · V Dt ∂x ∂x ∂x ∂y ∂y ∂x ∂z ∂x ∂z       ∂ ∂w ∂ ∂w Dv ∂v ∂p ∂ ∂v ∂w ∗ ∗ ∗ ∗
=ρby − + +λ ∇·V + + ) + + ) ρ 2µ µ ( µ ( Dt ∂y ∂y ∂y ∂z ∂z ∂y ∂x ∂y ∂x       Dv ∂ ∂u ∂ ∂w ∂p ∂ ∗ ∂w ∗ ∗ ∂w ∗ ∂v
ρ =ρbz − + +λ ∇·V + + ) + + ) 2µ µ ( µ ( Dt ∂z ∂z ∂z ∂x ∂x ∂z ∂y ∂z ∂y

ρ

where (Vx , Vy , Vz ) = (u, v, w).
33. Show that in cylindrical coordinates the Navier-Stokes equations of motion for a compressible fluid can be written       DVr V2 ∂ ∂Vr ∂p
+ 1 ∂ µ∗ ( 1 ∂Vr + ∂Vθ − Vθ ) − θ =Lbr − + + λ∗ ∇ · V L 2µ∗ Dt r ∂r ∂r ∂r r ∂θ r ∂θ ∂r r   ∗ ∂ ∂V 2µ ∂V 1 ∂V V ∂V r z r θ r + + ) + ( − − ) µ∗ ( ∂z ∂z ∂r r ∂r r ∂θ r       DVθ Vr Vθ Vr 1 ∂p 1 ∂ 1 ∂Vθ
+ ∂ µ∗ ( 1 ∂Vz + ∂Vθ ) + + + ) + λ∗ ∇ · V L =Lbθ − 2µ∗ ( Dt r r ∂θ r ∂θ r ∂θ r ∂z r ∂θ ∂z   ∗ ∂ ∂V ∂V V 2µ 1 ∂V ∂V V 1 r θ θ r θ θ + + − ) + ( + − ) µ∗ ( ∂r r ∂θ ∂r r r r ∂θ ∂r r     DVz ∂ ∂p ∂Vz
+ 1 ∂ µ∗ r( ∂Vr + ∂Vz ) L =Lbz − + + λ∗ ∇ · V 2µ∗ Dt ∂z ∂z ∂z r ∂r ∂z ∂r   1 ∂ ∂V ∂V 1 z θ + + ) µ∗ ( r ∂θ r ∂θ ∂z
34. Show that the dissipation function Φ can be written as Φ = 2µ∗ Dij Dij + λ∗ Θ2 .
35. Verify the identities: (a)

L

D ∂et
) (et /L) = + ∇ · (et V Dt ∂t

(b) L

D D 2

De (et /L) = L +L V /2 . Dt Dt Dt


36. Show that the conservation law for heat flow is given by ∂T + ∇ · (T
v − κ∇T ) = SQ ∂t where κ is the thermal conductivity of the material, T is the temperature, J
advection = T
v, J
conduction = −κ∇T and SQ is a source term. Note that in a solid material there is no flow and so
v = 0 and

324 the above equation reduces to the heat equation. Assign units of measurements to each term in the above equation and make sure the equation is dimensionally homogeneous.
37. Show that in spherical coordinates the Navier-Stokes equations of motion for a compressible fluid can be written

    2 2 ∂Vρ ∂ ∂p DVρ Vθ + Vφ
+ 1 ∂ µ∗ (ρ ∂ (Vθ /ρ) + 1 ∂Vρ ) − ) = Lbρ − + + λ∗ ∇ · V 2µ∗ Dt ρ ∂ρ ∂ρ ∂ρ ρ ∂θ ∂ρ ρ ∂θ   1 ∂ ∂ 1 ∂Vρ + +ρ (Vφ /ρ)) µ∗ ( ρ sin θ ∂φ ρ sin θ ∂φ ∂ρ µ∗ ∂Vρ 2 ∂Vθ 4Vρ 2 ∂Vφ 2Vθ cot θ ∂ cot θ ∂Vρ + (4 − − − − + ρ cot θ (Vθ /ρ) + ) ρ ∂ρ ρ ∂θ ρ ρ sin θ ∂φ ρ ∂ρ ρ ∂θ   DVθ Vρ Vθ Vφ2 cot θ 1 ∂p 1 ∂ 2µ∗ ∂Vθ
L( + − ) = Lbθ − + ( + Vρ ) + λ∗ ∇ · V Dt ρ ρ ρ ∂θ ρ ∂θ ρ ∂θ     ∂ sin θ ∂ 1 ∂Vθ ∂ 1 ∂Vρ 1 ∂ (Vφ / sin θ) + ) + (Vθ /ρ) + ) + µ∗ ( µ∗ (ρ ρ sin θ ∂φ ρ ∂θ ρ sin θ ∂φ ∂ρ ∂ρ ρ ∂θ      ∗ µ 1 ∂Vθ 1 ∂Vφ Vθ cot θ 1 ∂Vρ ∂ + − − (Vθ /ρ) + 2 cot θ + 3 ρ ρ ρ ∂θ ρ sin θ ∂φ ρ ∂ρ ρ ∂θ      1 ∂Vρ 1 ∂p Vφ Vρ Vθ Vφ cot θ ∂ ∂ DVφ + + = Lbφ − + µ∗ +ρ (Vφ /ρ) L Dt ρ ρ ρ sin θ ∂φ ∂ρ ρ sin θ ∂φ ∂ρ     ∂ 2µ∗ 1 ∂Vφ 1
+ Vρ + Vθ cot θ + λ∗ ∇ · V + ρ sin θ ∂φ ρ sin θ ∂φ    sin θ ∂ 1 ∂ 1 ∂Vθ + (Vφ / sin θ) + µ∗ ρ ∂θ ρ ∂θ ρ sin θ ∂φ      sin θ ∂ 1 ∂Vρ ∂ µ∗ 1 ∂Vθ +ρ (Vφ /ρ) + 2 cot θ (Vφ / sin θ) + + 3 ρ ρ sin θ ∂φ ∂ρ ρ ∂θ ρ sin θ ∂φ L(


38. Verify all the equations (2.5.28).
39. Use the conservation of energy equation (2.5.47) together with the momentum equation (2.5.25) to derive the equation (2.5.48).
40. Verify the equation (2.5.55).
41. Consider nonviscous flow and write the 3 linear momentum equations and the continuity equation and make the following assumptions: (i) The density L is constant. (ii) Body forces are zero. (iii) Steady state flow only. (iv) Consider only two dimensional flow with non-zero velocity components u = u(x, y) and v = v(x, y). Show that there results the system of equations u

∂u ∂u 1 ∂P +v + = 0, ∂x ∂y L ∂x

u

∂v ∂v 1 ∂P +v + = 0, ∂x ∂y L ∂y

∂u ∂v + = 0. ∂x ∂y

Recognize that the last equation in the above set as one of the Cauchy-Riemann equations that f (z) = u − iv be an analytic function of a complex variable. Further assume that the fluid flow is irrotational so that

P 1 2 ∂v ∂u − = 0. Show that this implies that u + v 2 + = Constant. If in addition u and v are derivable ∂x ∂y 2 L and v = ∂φ from a potential function φ(x, y), such that u = ∂φ ∂x ∂y , then show that φ is a harmonic function. By constructing the conjugate harmonic function ψ(x, y) the complex potential F (z) = φ(x, y) + iψ(x, y) is such that F  (z) = u(x, y) − iv(x, y) and F  (z) gives the velocity. The family of curves φ(x, y) =constant are called equipotential curves and the family of curves ψ(x, y) = constant are called streamlines. Show that these families are an orthogonal family of curves.

325 §2.6 ELECTRIC AND MAGNETIC FIELDS Introduction In electromagnetic theory the mks system of units and the Gaussian system of units are the ones most often encountered. In this section the equations will be given in the mks system of units. If you want the equations in the Gaussian system of units make the replacements given in the column 3 of Table 1. Table 1. MKS AND GAUSSIAN UNITS

volt/m

Replacement symbol
E

statvolt/cm


(Magnetic field) B

weber/m2

' B c

gauss


(Displacement field) D

coulomb/m2

' D 4π

statcoulomb/cm2


(Auxiliary Magnetic field) H

ampere/m

' cH 4π

oersted

J
(Current density)

ampere/m2

J


statampere/cm2


(Vector potential) A

weber/m

' A c

gauss-cm

V (Electric potential)

volt

V

statvolt

MKS symbol

MKS units


(Electric field) E

F (Dielectric constant)

# 4π

µ (Magnetic permeability)

4πµ c2

GAUSSIAN units

Electrostatics A basic problem in electrostatic theory is to determine the force F
on a charge Q placed a distance r from another charge q. The solution to this problem is Coulomb’s law 1 qQ
er F
= 4πF0 r2

(2.6.1)

where q, Q are measured in coulombs, F0 = 8.85 × 10−12 coulomb2 /N · m2 is called the permittivity in a vacuum, r is in meters, [F
] has units of Newtons and
er is a unit vector pointing from q to Q if q, Q have
= F
/Q is called the the same sign or pointing from Q to q if q, Q are of opposite sign. The quantity E
= F
and so Q = 1 is called electric field produced by the charges. In the special case Q = 1, we have E a test charge. This tells us that the electric field at a point P can be viewed as the force per unit charge exerted on a test charge Q placed at the point P. The test charge Q is always positive and so is repulsed if q is positive and attracted if q is negative. The electric field associated with many charges is obtained by the principal of superposition. For example, let q1 , q2 , . . . , qn denote n-charges having respectively the distances r1 , r2 , . . . , rn from a test charge Q placed at a point P. The force exerted on Q is F
=F
1 + F
2 + · · · + F
n   1 q1 Q q2 Q qn Q


F
= e e e + + · · · + r r r 1 2 n 4πF0 r12 r22 rn2 n  qi

= E(P
) =F = 1
er or E Q 4πF0 i=1 ri2 i

(2.6.2)

326
= E(P
) is the electric field associated with the system of charges. The equation (2.6.2) can be genwhere E eralized to other situations by defining other types of charge distributions. We introduce a line charge density λ∗ , (coulomb/m), a surface charge density µ∗ , (coulomb/m2 ), a volume charge density ρ∗ , (coulomb/m3 ), then we can calculate the electric field associated with these other types of charge distributions. For example, if there is a charge distribution λ∗ = λ∗ (s) along a curve C, where s is an arc length parameter, then we would have
)= E(P

1 4πF0




er ∗ λ ds r2

C

(2.6.3)

as the electric field at a point P due to this charge distribution. The integral in equation (2.6.3) being a line integral along the curve C and where ds is an element of arc length. Here equation (2.6.3) represents a continuous summation of the charges along the curve C. For a continuous charge distribution over a surface S, the electric field at a point P is
)= E(P

 

1 4πF0


er ∗ µ dσ r2

S

(2.6.4)

where dσ represents an element of surface area on S. Similarly, if ρ∗ represents a continuous charge distribution throughout a volume V , then the electric field is represented
)= E(P

1 4πF0

  


er ∗ ρ dτ r2

V

(2.6.5)

where dτ is an element of volume. In the equations (2.6.3), (2.6.4), (2.6.5) we let (x, y, z) denote the position of the test charge and let (x , y  , z  ) denote a point on the line, on the surface or within the volume, then e1 + (y − y  )
e2 + (z − z  )
e3
r = (x − x )


(2.6.6)


r er = . represents the distance from the point P to an element of charge λ∗ ds, µ∗ dσ or ρ∗ dτ with r = |
r| and
r
= 0, and so it is derivable from a potential function V If the electric field is conservative, then ∇ × E by taking the negative of the gradient of V and
= −∇V. E

(2.6.7)


· d
r is an exact differential so that the potential function can For these conditions note that ∇V · d
r = −E be represented by the line integral



P

V = V(P ) = −


· d
r E

(2.6.8)

α

where α is some reference point (usually infinity, where V(∞) = 0). For a conservative electric field the line integral will be independent of the path connecting any two points a and b so that 

b

V(b) − V(a) = − α

 
E · d
r − −



a


· d
r E

α

 =−



b


· d
r = E a

b

∇V · d
r.

(2.6.9)

a

Let α = ∞ in equation (2.6.8), then the potential function associated with a point charge moving in the radial direction
er is  V(r) = −


· d
r = −q E 4πF0 ∞ r



r

∞

1 q 1 r q | = . dr = r2 4πF0 r ∞ 4πF0 r

327 By superposition, ofcharges is given by   the ∗potential at a point P for a continuous volume distribution  ρ µ∗ 1 1 dτ and for a surface distribution of charges V(P ) = dσ and for a line V(P ) = 4πF0 4πF0 V r S r  ∗ 1 λ distribution of charges V(P ) = ds; and for a discrete distribution of point charges 4πF0 C r N 1  qi V(P ) = . When the potential functions are defined from a common reference point, then the 4πF0 i=1 ri principal of superposition applies. The potential function V is related to the work done W in moving a charge within the electric field. The work done in moving a test charge Q from point a to point b is an integral of the force times distance
and so the force F
= −QE
is in opposition to this moved. The electric force on a test charge Q is F
= QE force as you move the test charge. The work done is 



b

F
· d
r =

W = a



b


· d
r = Q −QE a

b

∇V · d
r = Q[V(b) − V(a)].

(2.6.10)

a

The work done is independent of the path joining the two points and depends only on the end points and the change in the potential. If one moves Q from infinity to point b, then the above becomes W = QV (b).
= E(P
) is a vector field which can be represented graphically by constructing vectors An electric field E at various selected points in the space. Such a plot is called a vector field plot. A field line associated with a vector field is a curve such that the tangent vector to a point on the curve has the same direction as the vector field at that point. Field lines are used as an aid for visualization of an electric field and vector fields
at that point. in general. The tangent to a field line at a point has the same direction as the vector field E For example, in two dimensions let
r = x
e1 + y
e2 denote the position vector to a point on a field line. The
= E(x,
e2 . If E y) = −N (x, y)
e1 + M (x, y)
e2 tangent vector to this point has the direction d
r = dx
e1 + dy

and d
r must be colinear. Thus, for each point (x, y) is the vector field constructed at the same point, then E
for some constant K. Equating like components we find that the on a field line we require that d
r = K E field lines must satisfy the differential relation. dx dy = =K −N (x, y) M (x, y) or

(2.6.11)

M (x, y) dx + N (x, y) dy =0.

In two dimensions, the family of equipotential curves V(x, y) = C1 =constant, are orthogonal to the family of field lines and are described by solutions of the differential equation N (x, y) dx − M (x, y) dy = 0 obtained from equation (2.6.11) by taking the negative reciprocal of the slope. The field lines are perpendicular to the equipotential curves because at each point on the curve V = C1 we have ∇V being perpendicular
at this same point. Field lines associated with electric to the curve V = C1 and so it is colinear with E fields are called electric lines of force. The density of the field lines drawn per unit cross sectional area are proportional to the magnitude of the vector field through that area.

328

Figure 2.6-1. Electric forces due to a positive charge at (−a, 0) and negative charge at (a, 0). EXAMPLE 2.6-1. Find the field lines and equipotential curves associated with a positive charge q located at the point (−a, 0) and a negative charge −q located at the point (a, 0).
on a test charge Q = 1 place Solution: With reference to the figure 2.6-1, the total electric force E at a general point (x, y) is, by superposition, the sum of the forces from each of the isolated charges and is
=E
1 + E
2 . The electric force vectors due to each individual charge are E e2 e1 + kqy

1 = kq(x + a)
E with r12 = (x + a)2 + y 2 3 r1 e2 e1 − kqy

2 = −kq(x − a)
E with r22 = (x − a)2 + y 2 r23 where k =

(2.6.12)

1 is a constant. This gives 4πF0     kq(x + a) kq(x − a) kqy kqy



e1 + e2 . E = E1 + E2 = − − 3
r13 r23 r13 r2

This determines the differential equation of the field lines dx kq(x+a) r13

−

kq(x−a) r23

=

kqy r13

dy . − kqy r3

(2.6.13)

2

To solve this differential equation we make the substitutions cos θ1 =

x+a r1

and

cos θ2 =

x−a r2

(2.6.14)

329

Figure 2.6-2. Lines of electric force between two opposite sign charges. as suggested by the geometry from figure 2.6-1. From the equations (2.6.12) and (2.6.14) we obtain the relations − sin θ1 dθ1 =

r1 dx − (x + a) dr1 r12

2r1 dr1 =2(x + a) dx + 2ydy − sin θ2 dθ2 =

r2 dx − (x − a)dr2 r22

2r2 dr2 =2(x − a) dx + 2y dy which implies that (x + a)y dy y 2 dx + 3 r13 r1 2 (x − a)y dy y dx − sin θ2 dθ2 = − + 3 r23 r2

− sin θ1 dθ1 = −

(2.6.15)

Now compare the results from equation (2.6.15) with the differential equation (2.6.13) and determine that y is an integrating factor of equation (2.6.13) . This shows that the differential equation (2.6.13) can be written in the much simpler form of the exact differential equation − sin θ1 dθ1 + sin θ2 dθ2 = 0

(2.6.16)

in terms of the variables θ1 and θ2 . The equation (2.6.16) is easily integrated to obtain cos θ1 − cos θ2 = C

(2.6.17)

where C is a constant of integration. In terms of x, y the solution can be written x−a x+a  − = C. 2 2 (x + a) + y (x − a)2 + y 2 These field lines are illustrated in the figure 2.6-2.

(2.6.18)

330 The differential equation for the equipotential curves is obtained by taking the negative reciprocal of the slope of the field lines. This gives dy = dx

kq(x−a) r23 kqy r13

− −

kq(x+a) r13 kqy r23

.

This result can be written in the form     (x + a)dx + ydy (x − a)dx + ydy − + =0 r13 r23 which simplifies to the easily integrable form −

dr1 dr2 + 2 =0 r12 r2

in terms of the new variables r1 and r2 . An integration produces the equipotential curves

or

1 1 − =C2 r1 r2 1 1  − =C2 . (x + a)2 + y 2 (x − a)2 + y 2

The potential function for this problem can be interpreted as a superposition of the potential functions kq kq V1 = − and V2 = associated with the isolated point charges at the points (−a, 0) and (a, 0). r1 r2 Observe that the electric lines of force move from positive charges to negative charges and they do not cross one another. Where field lines are close together the field is strong and where the lines are far apart the field is weak. If the field lines are almost parallel and equidistant from one another the field is said to be
If one moves along a field uniform. The arrows on the field lines show the direction of the electric field E. line in the direction of the arrows the electric potential is decreasing and they cross the equipotential curves
= 0. at right angles. Also, when the electric field is conservative we will have ∇ × E In three dimensions the situation is analogous to what has been done in two dimensions. If the electric
= E(x,
e2 + R(x, y, z)
e3 and
r = x
e1 + y
e2 + z
e3 is the position field is E y, z) = P (x, y, z)
e1 + Q(x, y, z)

must be colinear so that vector to a variable point (x, y, z) on a field line, then at this point d
r and E
for some constant K. Equating like coefficients gives the system of equations d
r = K E dx dy dz = = = K. P (x, y, z) Q(x, y, z) R(x, y, z)

(2.6.19)

From this system of equations one must try to obtain two independent integrals, call them u1 (x, y, z) = c1 and u2 (x, y, z) = c2 . These integrals represent one-parameter families of surfaces. When any two of these
These surfaces intersect, the result is a curve which represents a field line associated with the vector field E. type of field lines in three dimensions are more difficult to illustrate.
over a surface S is defined as the summation of the normal The electric flux φE of an electric field E
over the surface and is represented component of E 
·n ˆ dσ E

φE = S

with units of

N m2 C

(2.6.20)

331 ˆ is a unit normal to the surface. The flux φE can be thought of as being proportional to the number where n of electric field lines passing through an element of surface area. If the surface is a closed surface we have by the divergence theorem of Gauss




·n ˆ dσ E


dτ = ∇·E

φE = V

S

where V is the volume enclosed by S. Gauss Law Let dσ denote an element of surface area on a surface S. A cone is formed if all points on the boundary of dσ are connected by straight lines to the origin. The cone need not be a right circular cone. The situation is illustrated in the figure 2.6-3.

Figure 2.6-3. Solid angle subtended by element of area. ˆ denote a We let
r denote a position vector from the origin to a point on the boundary of dσ and let n ˆ ·
r = r cos θ where r = |
r| and θ is the unit outward normal to the surface at this point. We then have n ˆ and
r. Construct a sphere, centered at the origin, having radius r. This sphere angle between the vectors n dΩ intersects the cone in an element of area dΩ. The solid angle subtended by dσ is defined as dω = 2 . Note r that this is equivalent to constructing a unit sphere at the origin which intersect the cone in an element of area dω. Solid angles are measured in steradians. The total solid angle about a point equals the area of the sphere divided by its radius squared or 4π steradians. The element of area dΩ is the projection of dσ on the ˆ ·
r ˆ ·
r n n dΩ dσ so that dω = 3 dσ = 2 . Observe that sometimes the constructed sphere and dΩ = dσ cos θ = r r r ˆ ·
r is negative, the sign depending upon which of the normals to the surface is constructed. dot product n (i.e. the inner or outer normal.) The Gauss law for electrostatics in a vacuum states that the flux through any surface enclosing many charges is the total charge enclosed by the surface divided by F0 . The Gauss law is written  Qe  for charges inside S
·n ˆ dσ = #0 E 0 for charges outside S S

(2.6.21)

332 ˆ the unit outward normal to the surface. where Qe represents the total charge enclosed by the surface S with n The proof of Gauss’s theorem follows. Consider a single charge q within the closed surface S. The electric
= 1 q
er and so the flux field at a point on the surface S due to the charge q within S is represented E 4πF0 r2 integral is    ˆ er · n q
dΩ q q
·n ˆ dσ = E φE = dσ = = (2.6.22) 2 2 4πF r 4πF r F 0 0 0 S S S 
ˆ er · n cos θ dσ dΩ since = = = dω and dω = 4π. By superposition of the charges, we obtain a similar r2 r2 r2 S n  result for each of the charges within the surface. Adding these results gives Qe = qi . For a continuous i=1  distribution of charge inside the volume we can write Qe = ρ∗ dτ , where ρ∗ is the charge distribution V

per unit volume. Note that charges outside of the closed surface do not contribute to the total flux across the surface. This is because the field lines go in one side of the surface and go out the other side. In this 
ˆ dσ = 0 for charges outside the surface. Also the position of the charge or charges within the E·n case S

volume does not effect the Gauss law. The equation (2.6.21) is the Gauss law in integral form. We can put this law in differential form as follows. Using the Gauss divergence theorem we can write for an arbitrary volume that    ∗  Qe 1 ρ

ˆ dσ = E·n ∇ · E dτ = dτ = = ρ∗ dτ F0 F0 S V V F0 V which for an arbitrary volume implies

∗
= ρ . ∇·E F0

(2.6.23)

The equations (2.6.23) and (2.6.7) can be combined so that the Gauss law can also be written in the form ρ∗ which is called Poisson’s equation. ∇2 V = − F0 EXAMPLE 2.6-2 Find the electric field associated with an infinite plane sheet of positive charge. Solution: Assume there exists a uniform surface charge µ∗ and draw a circle at some point on the plane surface. Now move the circle perpendicular to the surface to form a small cylinder which extends equal distances above and below the plane surface. We calculate the electric flux over this small cylinder in the limit as the height of the cylinder goes to zero. The charge inside the cylinder is µ∗ A where A is the area of the circle. We find that the Gauss law requires that  Qe µ∗ A
·n ˆ dσ = E = F0 F0 S

(2.6.24)

ˆ is the outward normal to the cylinder as we move over the surface S. By the symmetry of the where n situation the electric force vector is uniform and must point away from both sides to the plane surface in the en and direction of the normals to both sides of the surface. Denote the plane surface normals by
en and −

= −β

= β
en on the other side of the surface for some assume that E en on one side of the surface and E constant β. Substituting this result into the equation (2.6.24) produces 
·n ˆ dσ = 2βA E S

(2.6.25)

333 since only the ends of the cylinder contribute to the above surface integral. On the sides of the cylinder we ˆ · ±
will have n en = 0 and so the surface integral over the sides of the cylinder is zero. By equating the µ∗ and consequently we can write results from equations (2.6.24) and (2.6.25) we obtain the result that β = 2F0 ∗
= µ
en where
E en represents one of the normals to the surface. 2F0 Note an electric field will always undergo a jump discontinuity when crossing a surface charge µ∗ . As in ∗ ∗
down = − µ

up = µ
en and E en so that the difference is the above example we have E 2F0 2F ∗
down = µ

up − E en E F0

(1)

E i ni

or

(2)

+ E i ni

+

µ∗ = 0. F0

(2.6.26)

It is this difference which causes the jump discontinuity. EXAMPLE 2.6-3. Calculate the electric field associated with a uniformly charged sphere of radius a. Solution: We proceed as in the previous example. Let µ∗ denote the uniform charge distribution over the surface er denote the unit normal to the sphere. The total charge then is written as   of the sphere and let
∗ 2 ∗ µ dσ = 4πa µ . If we construct a sphere of radius r > a around the charged sphere, then we have q= Sa



by the Gauss theorem

Qe q
·
E er dσ = = . F0 F0 Sr

(2.6.27)


and assume that it points radially outward in the direction of the Again, we can assume symmetry for E
= β

into the er for some constant β. Substituting this value for E surface normal
er and has the form E equation (2.6.27) we find that 


·
E er dσ = β

Sr

dσ = 4πβr2 = Sr

q . F0

(2.6.28)

1 q
er where
er is the outward normal to the sphere. This shows that the electric field 4πF0 r2 outside the sphere is the same as if all the charge were situated at the origin.
= This gives E

For S a piecewise closed surface enclosing a volume V and F i = F i (x1 , x2 , x3 ) i = 1, 2, 3, a continuous vector field with continuous derivatives the Gauss divergence theorem enables us to replace a flux integral of F i over S by a volume integral of the divergence of F i over the volume V such that     ˆ dσ = F
· n F i ni dσ = F i,i dτ or div F
dτ. S

V

S

(2.6.29)

V

If V contains a simple closed surface Σ where F i is discontinuous we must modify the above Gauss divergence theorem. EXAMPLE 2.6-4. We examine the modification of the Gauss divergence theorem for spheres in order to illustrate the concepts. Let V have surface area S which encloses a surface Σ. Consider the figure 2.6-4 where the volume V enclosed by S and containing Σ has been cut in half.

334

Figure 2.6-4. Sphere S containing sphere Σ. Applying the Gauss divergence theorem to the top half of figure 2.6-4 gives     i T F i nTi dσ + F i nbi T dσ + F i nΣ dσ = F ,i dτ i ST

Sb1

ΣT

(2.6.30)

VT

where the ni are the unit outward normals to the respective surfaces ST , Sb1 and ΣT . Applying the Gauss divergence theorem to the bottom half of the sphere in figure 2.6-4 gives     i bB i ΣB F i nB dσ + F n dσ + F n dσ = i i i SB

Sb2

ΣB

F i,i dτ

(2.6.31)

VB

Observe that the unit normals to the surfaces Sb1 and Sb2 are equal and opposite in sign so that adding the equations (2.6.30) and (2.6.31) we obtain 

 F ni dσ + S

 (1) F i ni

i

Σ

i F ,i dτ

dσ = VT +VB

(2.6.32)

335 where S = ST + SB is the total surface area of the outside sphere and Σ = ΣT + ΣB is the total surface area (1)

of the inside sphere, and ni

is the inward normal to the sphere Σ when the top and bottom volumes are

combined. Applying the Gauss divergence theorem to just the isolated small sphere Σ we find   (2) F i ni dσ = F i,i dτ Σ (2)

where ni

(2.6.33)

VΣ

is the outward normal to Σ. By adding the equations (2.6.33) and (2.6.32) we find that 

 

i

F ni dσ + S

(1) F i ni

+

(2) F i ni





F i,i dτ

(2.6.34)

   (1) (2) F i ni + F i ni dσ.

(2.6.35)

dσ =

Σ

V

where V = VT + VB + VΣ . The equation (2.6.34) can also be written as 



i F ,i dτ −

F i ni dσ = S

V

Σ

In the case that V contains a surface Σ the total electric charge inside S is   ∗ Qe = ρ dτ + µ∗ dσ V

(2.6.36)

Σ

where µ∗ is the surface charge density on Σ and ρ∗ is the volume charge density throughout V. The Gauss theorem requires that



Qe 1 E ni dσ = = F F 0 0 S



1 ρ dτ + F0 ∗

i

V



µ∗ dσ.

(2.6.37)

Σ

In the case of a jump discontinuity across the surface Σ we use the results of equation (2.6.34) and write 

 E i,i dτ −

E i ni dσ = S

 

V

(1)

E i ni

(2)

+ E i ni

 dσ.

(2.6.38)

Σ

Subtracting the equation (2.6.37) from the equation (2.6.38) gives       ρ∗ µ∗ i i (1) i (2) E ,i − dτ − E ni + E ni + dσ = 0. F0 F0 V Σ

(2.6.39)

For arbitrary surfaces S and Σ, this equation implies the differential form of the Gauss law E i,i =

ρ∗ . F0

(2.6.40)

Further, on the surface Σ, where there is a surface charge distribution we have (1)

E i ni

(2)

+ E i ni

+

µ∗ =0 F0

which shows the electric field undergoes a discontinuity when you cross a surface charge µ∗ .

(2.6.41)

336 Electrostatic Fields in Materials When charges are introduced into materials it spreads itself throughout the material. Materials in which the spreading occurs quickly are called conductors, while materials in which the spreading takes a long time are called nonconductors or dielectrics. Another electrical property of materials is the ability to hold local charges which do not come into contact with other charges. This property is called induction. For example, consider a single atom within the material. It has a positively charged nucleus and negatively
the negative cloud charged electron cloud surrounding it. When this atom experiences an electric field E
while the positively charged nucleus moves in the direction of E.
If E
is large enough it moves opposite to E can ionize the atom by pulling the electrons away from the nucleus. For moderately sized electric fields the atom achieves an equilibrium position where the positive and negative charges are offset. In this situation the atom is said to be polarized and have a dipole moment p
. Definition: When a pair of charges +q and −q are separated by a distance 2d
the electric dipole
where p
has dimensions of [C m]. moment is defined by p
= 2dq,
and the material is symmetric we say that p
In the special case where d
has the same direction as E
and write p
= αE,
where α is called the atomic polarizability. If in a material subject is proportional to E to an electric field their results many such dipoles throughout the material then the dielectric is said to be polarized. The vector quantity P
is introduced to represent this effect. The vector P
is called the polarization vector having units of [C/m2 ], and represents an average dipole moment per unit volume of material. The vectors Pi and Ei are related through the displacement vector Di such that Pi = Di − F0 Ei .

(2.6.42)

For an anisotropic material (crystal) Di = Fji Ej

and

Pi = αji Ej

(2.6.43)

where Fji is called the dielectric tensor and αji is called the electric susceptibility tensor. Consequently, Pi = αji Ej = Fji Ej − F0 Ei = (Fji − F0 δij )Ej

so that

αji = Fji − F0 δij .

(2.6.44)

A dielectric material is called homogeneous if the electric force and displacement vector are the same for any two points within the medium. This requires that the electric force and displacement vectors be constant parallel vector fields. It is left as an exercise to show that the condition for homogeneity is that Fji,k = 0. A dielectric material is called isotropic if the electric force vector and displacement vector have the same direction. This requires that Fji = Fδji where δji is the Kronecker delta. The term F = F0 Ke is called the dielectric constant of the medium. The constant F0 = 8.85(10)−12 coul2 /N · m2 is the permittivity of free space and the quantity ke =

# #0

is called the relative dielectric constant (relative to F0 ). For free space ke = 1.

Similarly for an isotropic material we have αji = F0 αe δij where αe is called the electric susceptibility. For a
and E
are related by linear medium the vectors P
, D Di = F0 Ei + Pi = F0 Ei + F0 αe Ei = F0 (1 + αe )Ei = F0 Ke Ei = FEi

(2.6.45)

337 where Ke = 1 + αe is the relative dielectric constant. The equation (2.6.45) are constitutive equations for dielectric materials. The effect of polarization is to produce regions of bound charges ρb within the material and bound surface charges µb together with free charges ρf which are not a result of the polarization. Within dielectrics we have ∇ · P
= ρb for bound volume charges and P
·
en = µb for bound surface charges, where
en is a unit normal to the bounding surface of the volume. In these circumstances the expression for the potential function is written V=

1 4πF0

 V

1 ρb dτ + r 4πF0

 S

µb dσ r

(2.6.46)

and the Gauss law becomes
= ρ∗ = ρb + ρf = −∇ · P
+ ρf F0 ∇ · E

or


+ P
) = ρf . ∇(F0 E

(2.6.47)


+ P
the Gauss law can also be written in the form
= F0 E Since D
= ρf ∇·D

or

D,ii = ρf .

(2.6.48)

When no confusion arises we replace ρf by ρ. In integral form the Gauss law for dielectrics is written 
·n ˆ dσ = Qf e D (2.6.49) S

where Qf e is the total free charge density within the enclosing surface. Magnetostatics
while a moving charge generates a magnetic field B.
A stationary charge generates an electric field E Magnetic field lines associated with a steady current moving in a wire form closed loops as illustrated in the figure 2.6-5.

Figure 2.6-5. Magnetic field lines. The direction of the magnetic force is determined by the right hand rule where the thumb of the right hand points in the direction of the current flow and the fingers of the right hand curl around in the direction
The force on a test charge Q moving with velocity V
in a magnetic field is of the magnetic field B.
× B).
F
m = Q(V

(2.6.50)

The total electromagnetic force acting on Q is the electric force plus the magnetic force and is  
+ (V
× B)
F
= Q E

(2.6.51)

338 which is known as the Lorentz force law. The magnetic force due to a line charge density λ∗ moving along 

a curve C is the line integral


× B)
= λ∗ ds(V

F
mag = C


I
× Bds.

(2.6.52)

C

Similarly, for a moving surface charge density moving on a surface  
×B
dσ
× B)
= F
mag = K µ∗ dσ(V S

(2.6.53)

S

and for a moving volume charge density  
dτ
× B)
= F
mag = J
× B ρ∗ dτ (V V

(2.6.54)

V


, K
= µ∗ V
and J
= ρ∗ V
are respectively the current, the current per unit where the quantities I
= λ∗ V length, and current per unit area. A conductor is any material where the charge is free to move. The flow of charge is governed by Ohm’s law. Ohm’s law states that the current density vector Ji is a linear function of the electric intensity or Ji = σim Em , where σim is the conductivity tensor of the material. For homogeneous, isotropic conductors σim = σδim so that Ji = σEi where σ is the conductivity and 1/σ is called the resistivity. Surround a charge density ρ∗ with an arbitrary simple closed surface S having volume V and calculate the flux of the current density across the surface. We find by the divergence theorem   ˆ dσ = J
· n ∇ · J
dτ. S

(2.6.55)

V

If charge is to be conserved, the current flow out of the volume through the surface must equal the loss due to the time rate of change of charge within the surface which implies     ∂ρ∗ d ˆ dσ = J
· n dτ ∇ · J
dτ = − ρ∗ dτ = − dt S V V V ∂t  

or

∇ · J
+ V

∂ρ∗ ∂t

(2.6.56)

 dτ = 0.

(2.6.57)

This implies that for an arbitrary volume we must have ∇ · J
= −

∂ρ∗ . ∂t

(2.6.58)

Note that equation (2.6.58) has the same form as the continuity equation (2.3.73) for mass conservation and so it is also called a continuity equation for charge conservation. For magnetostatics there exists steady line ∗ currents or stationary current so ∂ρ = 0. This requires that ∇ · J
= 0. ∂t

339

Figure 2.6-6. Magnetic field around wire. Biot-Savart Law The Biot-Savart law for magnetostatics describes the magnetic field at a point P due to a steady line current moving along a curve C and is
) = µ0 B(P 4π


I×
er ds 2 r C

(2.6.59)

with units [N/amp · m] and where the integration is in the direction of the current flow. In the Biot-Savart law we have the constant µ0 = 4π × 10−7 N/amp2 which is called the permeability of free space, I
= I
et is er is a unit vector directed the current flowing in the direction of the unit tangent vector
et to the curve C,
from a point on the curve C toward the point P and r is the distance from a point on the curve to the general point P. Note that for a steady current to exist along the curve the magnitude of I
must be the same everywhere along the curve. Hence, this term can be brought out in front of the integral. For surface
and volume currents J
the Biot-Savart law is written currents K 


×
K er dσ 2 r S 
J×
er
) = µ0 dτ. B(P 2 4π r V
) = µ0 B(P 4π

and EXAMPLE 2.6-5.


a distance h perpendicular to a wire carrying a constant current I.
Calculate the magnetic field B Solution: The magnetic field circles around the wire. For the geometry of the figure 2.6-6, the magnetic field points out of the page. We can write I
×
er = I
et ×
er = Iˆ e sin α ˆ is a unit vector tangent to the circle of radius h which encircles the wire and cuts the wire perpenwhere e dicularly.

340 For this problem the Biot-Savart law is 


) = µ0 I B(P 4π

ˆ e ds. r2

In terms of θ we find from the geometry of figure 2.6-6 tan θ =

s h

ds = h sec2 θ dθ

with

Therefore,
) = µ0 B(P π



θ2

θ1

and

cos θ =

h . r

Iˆ e sin α h sec2 θ dθ. h2 / cos2 θ

But, α = π/2 + θ so that sin α = cos θ and consequently e
) = µ0 Iˆ B(P 4πh



θ2

cos θ dθ = θ1

e µ0 Iˆ (sin θ2 − sin θ1 ). 4πh

e
) = µ0 Iˆ . For a long straight wire θ1 → −π/2 and θ2 → π/2 to give the magnetic field B(P 2πh For volume currents the Biot-Savart law is
) = µ0 B(P 4π

 V

J
×
er dτ r2

(2.6.60)

and consequently (see exercises)
= 0. ∇·B

(2.6.61)

∗


= ρ is known as the Gauss’s law for electric fields and so Recall the divergence of an electric field is ∇ · E #0
= 0 is sometimes referred to as Gauss’s law for magnetic fields. If ∇ · B
= 0, in analogy the divergence ∇ · B
such that B
= ∇ × A.
The vector field A
is called the vector potential of then there exists a vector field A
Note that ∇ · B
= ∇ · (∇ × A)
= 0. Also the vector potential A
is not unique since B
is also derivable B.
+ ∇φ where φ is an arbitrary continuous and differentiable scalar. from the vector potential A Ampere’s Law Ampere’s law is associated with the work done in moving around a simple closed path. For example,
around a circular path of radius h consider the previous example 2.6-5. In this example the integral of B which is centered at some point on the wire can be associated with the work done in moving around this path. The summation of force times distance is    µ0 I
· d
r =
B
·e ˆ ds =
B
ds = µ0 I 2πh C C C

(2.6.62)

 ˆ ds is a tangent vector to the circle encircling the wire and
ds = 2πh is the distance where now d
r = e C

around this circle. The equation (2.6.62) holds not only for circles, but for any simple closed curve around the wire. Using the Stoke’s theorem we have   
· d
r =
·
en dσ
B (∇ × B) en dσ = µ0 I = µ0 J
·
C

S

S

(2.6.63)

341  J
·
en dσ is the total flux (current) passing through the surface which is created by encircling

where S

some curve about the wire. Equating like terms in equation (2.6.63) gives the differential form of Ampere’s law

= µ0 J. ∇×B

(2.6.64)

Magnetostatics in Materials Similar to what happens when charges are introduced into materials we have magnetic fields whenever there are moving charges within materials. For example, when electrons move around an atom tiny current loops are formed. These current loops create what are called magnetic dipole moments m
throughout the
material. When a magnetic field B is applied to a material medium there is a net alignment of the magnetic
, called the magnetization vector is introduced. Here M
is associated with a dipoles. The quantity M dielectric medium and has the units [amp/m] and represents an average magnetic dipole moment per unit
volume and is analogous to the polarization vector P
used in electrostatics. The magnetization vector M acts a lot like the previous polarization vector in that it produces bound volume currents J
b and surface
= J
b is a volume current density throughout some volume and M
×

b is a
b where ∇ × M en = K currents K surface current on the boundary of this volume. '

From electrostatics note that the time derivative of F0 ∂∂tE has the same units as current density. The ' total current in a magnetized material is then J
t = J
b + J
f + F0 ∂ E where J
b is the bound current, J
f is the ∂t

'

free current and F0 ∂∂tE is the induced current. Ampere’s law, equation (2.6.64), in magnetized materials then becomes
= µ0 J
t = µ0 (J
b + J
f + F0 ∇×B



∂E ∂E ) = µ0 J
+ µ0 F0 ∂t ∂t

(2.6.65)

' where J
= J
b + J
f . The term F0 ∂∂tE is referred to as a displacement current or as a Maxwell correction to

the field equation. This term implies that a changing electric field induces a magnetic field.
defined by An auxiliary magnet field H Hi =

1 Bi − Mi µ0

(2.6.66)


and magnetization vector M
. This is another conis introduced which relates the magnetic force vector B stitutive equation which describes material properties. For an anisotropic material (crystal) Bi = µji Hj

Mi = χji Hj

and

(2.6.67)

where µji is called the magnetic permeability tensor and χji is called the magnetic permeability tensor. Both of these quantities are dimensionless. For an isotropic material µji = µδij

where

µ = µ0 km .

Here µ0 = 4π × 10−7 N/amp2 is the permeability of free space and km = coefficient. Similarly, for an isotropic material we have

χji

=

χm δij

(2.6.68) µ µ0

is the relative permeability

where χm is called the magnetic sus-

ceptibility coefficient and is dimensionless. The magnetic susceptibility coefficient has positive values for

342 materials called paramagnets and negative values for materials called diamagnets. For a linear medium the
M
and H
are related by quantities B, Bi = µ0 (Hi + Mi ) = µ0 Hi + µ0 χm Hi = µ0 (1 + χm )Hi = µ0 km Hi = µHi

(2.6.69)

where µ = µ0 km = µ0 (1 + χm ) is called the permeability of the material.
for magnetostatics in materials plays a role similar to the Note: The auxiliary magnetic vector H
for electrostatics in materials. Be careful in using electromagnetic equations from displacement vector D
and H.
Some authors call H
the magnetic field. different texts as many authors interchange the roles of B
should be the fundamental quantity.1 However, the quantity B Electrodynamics In the nonstatic case of electrodynamics there is an additional quantity J
p = current which satisfies

∂ ∂ρb ∂ P
= ∇ · P
= − ∇ · J
p = ∇ · ∂t ∂t ∂t

' ∂P ∂t

called the polarization (2.6.70)

and the current density has three parts

+ J
f + ∂ P J
= J
b + J
f + J
p = ∇ × M ∂t

(2.6.71)

consisting of bound, free and polarization currents. Faraday’s law states that a changing magnetic field creates an electric field. In particular, the electromagnetic force induced in a closed loop circuit C is proportional to the rate of change of flux of the magnetic field associated with any surface S connected with C. Faraday’s law states   ∂
·

B en dσ.
E · d
r = − ∂t S C Using the Stoke’s theorem, we find 


·
(∇ × E) en dσ = −

S

S


∂B ·
en dσ. ∂t

The above equation must hold for an arbitrary surface and loop. Equating like terms we obtain the differential form of Faraday’s law
=− ∇×E


∂B . ∂t

(2.6.72)

This is the first electromagnetic field equation of Maxwell. Ampere’s law, equation (2.6.65), written in terms of the total current from equation (2.6.71) , becomes


= µ0 (∇ × M
+ J
f + ∂ P ) + µ0 F0 ∂ E ∇×B ∂t ∂t which can also be written as ∇×( 1

1

) = J
f + ∂ (P
+ F0 E)
B−M µ0 ∂t

D.J. Griffiths, Introduction to Electrodynamics, Prentice Hall, 1981. P.232.

(2.6.73)

343 or
= J
f + ∇×H


∂D . ∂t

(2.6.74)

This is Maxwell’s second electromagnetic field equation. To the equations (2.6.74) and (2.6.73) we add the Gauss’s law for magnetization, equation (2.6.61) and Gauss’s law for electrostatics, equation (2.6.48). These four equations produce the Maxwell’s equations of electrodynamics and are now summarized. The general form of Maxwell’s equations involve the quantities Ei , Electric force vector, [Ei ] = Newton/coulomb Bi , Magnetic force vector, [Bi ] = Weber/m2 Hi , Auxilary magnetic force vector, [Hi ] = ampere/m Di , Displacement vector, [Di ] = coulomb/m2 Ji , Free current density, [Ji ] = ampere/m2 Pi , Polarization vector, [Pi ] = coulomb/m2 Mi , Magnetization vector, [Mi ] = ampere/m for i = 1, 2, 3. There are also the quantities L, representing the free charge density, with units [L] = coulomb/m3 F0 , Permittivity of free space, [F0 ] = farads/m or coulomb2 /Newton · m2 . µ0 , Permeability of free space, [µ0 ] = henrys/m or kg · m/coulomb2 In addition, there arises the material parameters: µij , magnetic permeability tensor, which is dimensionless Fij , dielectric tensor, which is dimensionless αij , electric susceptibility tensor, which is dimensionless χij , magnetic susceptibility tensor, which is dimensionless These parameters are used to express variations in the electric field Ei and magnetic field Bi when acting in a material medium. In particular, Pi , Di , Mi and Hi are defined from the equations Di =Fji Ej = F0 Ei + Pi

Fij = F0 δji + αji

Bi =µji Hj = µ0 Hi + µ0 Mi , Pi =αji Ej ,

and

µij = µ0 (δji + χij )

Mi = χji Hj

for i = 1, 2, 3.

The above quantities obey the following laws: Faraday’s Law

This law states the line integral of the electromagnetic force around a loop is proportional

to the rate of flux of magnetic induction through the loop. This gives rise to the first electromagnetic field equation:

= − ∂B ∇×E ∂t

or

Fijk Ek,j = −

∂B i . ∂t

(2.6.75)

344 Ampere’s Law

This law states the line integral of the magnetic force vector around a closed loop is

proportional to the sum of the current through the loop and the rate of flux of the displacement vector through the loop. This produces the second electromagnetic field equation:

= J
f + ∂ D ∇×H ∂t

Fijk Hk,j = Jfi +

or

∂Di . ∂t

(2.6.76)

Gauss’s Law for Electricity This law states that the flux of the electric force vector through a closed surface is proportional to the total charge enclosed by the surface. This results in the third electromagnetic field equation:
= ρf ∇·D

D,ii = ρf

or

or

1 ∂ √ i

gD = ρf . √ g ∂xi

(2.6.77)

Gauss’s Law for Magnetism This law states the magnetic flux through any closed volume is zero. This produces the fourth electromagnetic field equation:
=0 ∇·B

B,ii = 0

or

or

1 ∂ √ i

gB = 0. √ g ∂xi

(2.6.78)

When no confusion arises it is convenient to drop the subscript f from the above Maxwell equations. Special expanded forms of the above Maxwell equations are given on the pages 176 to 179. Electromagnetic Stress and Energy Let V denote the volume of some simple closed surface S. Let us calculate the rate at which electromagnetic energy is lost from this volume. This represents the energy flow per unit volume. Begin with the first two Maxwell’s equations in Cartesian form ∂Bi ∂t ∂Di . =Ji + ∂t

Fijk Ek,j = −

(2.6.79)

Fijk Hk,j

(2.6.80)

Now multiply equation (2.6.79) by Hi and equation (2.6.80) by Ei . This gives two terms with dimensions of energy per unit volume per unit of time which we write ∂Bi Hi ∂t ∂Di Ei . Fijk Hk,j Ei =Ji Ei + ∂t

Fijk Ek,j Hi = −

(2.6.81) (2.6.82)

Subtracting equation (2.6.82) from equation (2.6.81) we find ∂Di Ei − ∂t ∂Di Ei − + Hi,j Ek ] = − Ji Ei − ∂t

Fijk (Ek,j Hi − Hk,j Ei ) = − Ji Ei − Fijk [(Ek Hi ),j − Ek Hi,j

∂Bi Hi ∂t ∂Bi Hi ∂t

Observe that Fjki (Ek Hi ),j is the same as Fijk (Ej Hk ),i so that the above simplifies to Fijk (Ej Hk ),i + Ji Ei = −

∂Di ∂Bi Ei − Hi . ∂t ∂t

(2.6.83)

345 Now integrate equation (2.6.83) over a volume and apply Gauss’s divergence theorem to obtain     ∂Di ∂Bi Ei + Hi ) dτ. Fijk Ej Hk ni dσ + Ji Ei dτ = − ( ∂t ∂t S V V

(2.6.84)

The first term in equation (2.6.84) represents the outward flow of energy across the surface enclosing the volume. The second term in equation (2.6.84) represents the loss by Joule heating and the right-hand side is the rate of decrease of stored electric and magnetic energy. The equation (2.6.84) is known as Poynting’s theorem and can be written in the vector form  


· ∂B − E
· J)
dτ.
× H)
·n
· ∂D − H ˆ dσ = (E (−E ∂t ∂t S V

(2.6.85)

For later use we define the quantity Si = Fijk Ej Hk


=E
×H
or S

[Watts/m2 ]

(2.6.86)

as Poynting’s energy flux vector and note that Si is perpendicular to both Ei and Hi and represents units of energy density per unit time which crosses a unit surface area within the electromagnetic field. Electromagnetic Stress Tensor Instead of calculating energy flow per unit volume, let us calculate force per unit volume. Consider a region containing charges and currents but is free from dielectrics and magnetic materials. To obtain terms with units of force per unit volume we take the cross product of equation (2.6.79) with Di and the cross product of equation (2.6.80) with Bi and subtract to obtain  −Firs Fijk (Ek,j Ds + Hk,j Bs ) = Fris Ji Bs + Fris

∂Di ∂Bs Bs + Di ∂t ∂t



which simplifies using the e − δ identity to −(δrj δsk − δrk δsj )(Ek,j Ds + Hk,j Bs ) = Fris Ji Bs + Fris

∂ (Di Bs ) ∂t

which further simplifies to −Es,r Ds + Er,s Ds − Hs,r Bs + Hr,s Bs = Fris Ji Bs +

∂ (Fris Di Bs ). ∂t

Observe that the first two terms in the equation (2.6.87) can be written Er,s Ds − Es,r Ds =Er,s Ds − F0 Es,r Es 1 =(Er Ds ),s − Er Ds,s − F0 ( Es Es ),r 2 1 =(Er Ds ),s − ρEr − (Ej Dj δsr ),s 2 1 =(Er Ds − Ej Dj δrs ),s − ρEr 2 which can be expressed in the form E − ρEr Er,s Ds − Es,r Ds = Trs,s

(2.6.87)

346 where 1 E = Er Ds − Ej Dj δrs Trs 2

(2.6.88)

is called the electric stress tensor. In matrix form the stress tensor is written   E1 D2 E1 D3 E1 D1 − 12 Ej Dj E . Trs = E2 D1 E2 D2 − 12 Ej Dj E2 D3 E3 D1 E3 D2 E3 D3 − 12 Ej Dj

(2.6.89)

By performing similar calculations we can transform the third and fourth terms in the equation (2.6.87) and obtain M Hr,s Bs − Hs,r Bs = Trs,s

(2.6.90)

1 M = Hr BS − Hj Bj δrs Trs 2

(2.6.91)

where

is the magnetic stress tensor. In matrix form the magnetic stress tensor is written 

M Trs

B1 H1 − 12 Bj Hj = B2 H1 B3 H1

B1 H2 B2 H2 − 12 Bj Hj B3 H2

 B1 H3 . B2 H3 B3 H3 − 12 Bj Hj

(2.6.92)

The total electromagnetic stress tensor is E M Trs = Trs + Trs .

(2.6.93)

Then the equation (2.6.87) can be written in the form Trs,s − ρEr = Fris Ji Bs +

∂ (Fris Di Bs ) ∂t

ρEr + Fris Ji BS = Trs,s −

∂ (Fris Di Bs ). ∂t

or (2.6.94)

For free space Di = F0 Ei and Bi = µ0 Hi so that the last term of equation (2.6.94) can be written in terms of the Poynting vector as µ0 F0

∂ ∂Sr = (Fris Di Bs ). ∂t ∂t

(2.6.95)

Now integrate the equation (2.6.94) over the volume to obtain the total electromagnetic force 

 ρEr dτ +

V



V

 Trs,s dτ − µ0 F0

Fris Ji Bs dτ = V

V

∂Sr dτ. ∂t

Applying the divergence theorem of Gauss gives     ∂Sr dτ. ρEr dτ + Fris Ji Bs dτ = Trs ns dσ − µ0 F0 V V S V ∂t

(2.6.96)

The left side of the equation (2.6.96) represents the forces acting on charges and currents contained within the volume element. If the electric and magnetic fields do not vary with time, then the last term on the right is zero. In this case the forces can be expressed as an integral of the electromagnetic stress tensor.

347 EXERCISE 2.6
1.

Find the field lines and equipotential curves associated with a positive charge q located at (−a, 0) and

a positive charge q located at (a, 0). The field lines are illustrated in the figure 2.6-7.

Figure 2.6-7. Lines of electric force between two charges of the same sign.
2. Calculate the lines of force and equipotential curves associated with the electric field
= E(x,
E y) = 2y
e1 + 2x
e2 . Sketch the lines of force and equipotential curves. Put arrows on the lines of force to show direction of the field lines.
3.

A right circular cone is defined by x = u sin θ0 cos φ,

y = u sin θ0 sin φ,

z = u cos θ0

with 0 ≤ φ ≤ 2π and u ≥ 0. Show the solid angle subtended by this cone is Ω =

A r2

= 2π(1 − cos θ0 ).


4.

A charge +q is located at the point (0, a) and a charge −q is located at the point (0, −a). Show that −2aq
at the position (x, 0), where x > a is E
= 1
e2 . the electric force E 4πF0 (a2 + x2 )3/2

Let the circle x2 + y 2 = a2 carry a line charge λ∗ . Show the electric field at the point (0, 0, z) is ∗ e3
= 1 λ az(2π)
E . 2 2 3/2 4πF0 (a + z )


5.


6.

Use superposition to find the electric field associated with two infinite parallel plane sheets each

carrying an equal but opposite sign surface charge density µ∗ . Find the field between the planes and outside ∗

µ and perpendicular to plates. of each plane. Hint: Fields are of magnitude ± 2# 0 
J×
er µ0


= 0.
7. For a volume current J the Biot-Savart law gives B = dτ. Show that ∇ · B 2 4π r V
r
r Hint: Let
er = and consider ∇ · (J
× 3 ). Then use numbers 13 and 10 of the appendix C. Also note that r r ∇ × J
= 0 because J
does not depend upon position.

348
8.

A homogeneous dielectric is defined by Di and Ei having parallel vector fields. Show that for a

homogeneous dielectric Fji,k = 0.
9.
10.

Show that for a homogeneous, isotropic dielectric medium that F is a constant. Show that for a homogeneous, isotropic linear dielectric in Cartesian coordinates Pi,i =


11.

αe ρf . 1 + αe

Verify the Maxwell’s equations in Gaussian units for a charge free isotropic homogeneous dielectric.
=0
=1∇ · D ∇·E F
=µ∇H
=0 ∇·B


12.



1 ∂B µ ∂H =− c ∂t c ∂t

1 ∂ D 4π F ∂E 4π

= ∇×H + J
= + σE c ∂t c c ∂t c
=− ∇×E

Verify the Maxwell’s equations in Gaussian units for an isotropic homogeneous dielectric with a

charge.
=4πρ ∇·D
=0 ∇·B
13.


1 ∂B c ∂t
4π
= J
+ 1 ∂ D ∇×H c c ∂t
=− ∇×E

For a volume charge ρ in an element of volume dτ located at a point (ξ, η, ζ) Coulombs law is  1 ρ

e dτ E(x, y, z) = 2 r 4πF0 r V

(a) Show that r2 = (x − ξ)2 + (y − η)2 + (z − ζ)2 . 1 e1 + (y − η)
e2 + (z − ζ)
e3 ) . (b) Show that
er = ((x − ξ)
r (c) Show that    
1 er (x − ξ)
e1 + (y − η)
e2 + (z − ζ)
e3 1
E(x, y, z) = ρ dξdηdζ = ∇ ρ dξdηdζ 2 2 2 3/2 4πF0 4πF0 r2 V [(x − ξ) + (y − η) + (z − ζ) V   ] ρ(ξ, η, ζ)
is V = 1 dξdηdζ (d) Show that the potential function for E 2 2 2 1/2 4πF0 V [(x − ξ) + (y − η) + (z − ζ) ]
= −∇V. (e) Show that E ρ (f) Show that ∇2 V = − Hint: Note that the integrand is zero everywhere except at the point where F (ξ, η, ζ) = (x, y, z). Consider the integral split into two regions. One region being a small sphere about the  point  (x, y, z) in the  limit  as the radius of this sphere approaches zero. Observe the identity

er er = −∇(ξ, η, ζ) enables one to employ the Gauss divergence theorem to obtain a ∇(x,y,z) r2 r2 
er ρ ρ surface integral. Use a mean value theorem to show − ·n ˆ dS = 4π since n ˆ = −
er . 2 4πF0 r 4πF 0 S
14.

Show that for a point charge in space ρ∗ = qδ(x − x0 )δ(y − y0 )δ(z − z0 ), where δ is the Dirac delta

function, the equation (2.6.5) can be reduced to the equation (2.6.1).
15.
er is irrotational. Here
er = 'rr is a unit vector in the direction of r.
= −∇V which satisfies V(r0 ) = 0 for r0 > 0. (b) Find the potential function V such that E
= (a) Show the electric field E

1 r2

349
16.
is a conservative electric field such that E
= −∇V, then show that E
is irrotational and satisfies (a) If E
= curl E
= 0. ∇×E
= curl E
= 0, show that E
is conservative. (i.e. Show E
= −∇V.) (b) If ∇ × E Hint: The work done on a test charge Q = 1 along the straight line segments from (x0 , y0 , z0 ) to (x, y0 , z0 ) and then from (x, y0 , z0 ) to (x, y, z0 ) and finally from (x, y, z0 ) to (x, y, z) can be written 



x



y

E1 (x, y0 , z0 ) dx −

V = V(x, y, z) = − x0

y0

Now note that ∂V = −E2 (x, y, z0 ) − ∂y

z

E2 (x, y, z0 ) dy − 

E3 (x, y, z) dz. z0

z

z0

∂E3 (x, y, z) dz ∂y


= 0 we find ∂E3 = ∂E2 , which implies ∂V = −E2 (x, y, z). Similar results are obtained and from ∇ × E ∂y ∂z ∂y ∂V ∂V
and . Hence show −∇V = E. for ∂x ∂z
17.
= 0, then there exists some vector field A
such that B
= ∇ × A.
(a) Show that if ∇ · B
is called the vector potential of B.
The vector field A  1
y, z) =
Hint: Let A(x, sB(sx, sy, sz) ×
r ds where
r = x
e1 + y
e2 + z
e3 0  1 dBi 2 s ds by parts. and integrate ds 0
= 0. (b) Show that ∇ · (∇ × A)
18.

Use Faraday’s law and Ampere’s law to show g im (E j,j ),m − g jm E i,mj = −µ0


19.

  ∂ ∂E i J i + F0 ∂t ∂t


where σ is the conductivity. Show that for ρ = 0 Maxwell’s equations produce Assume that J
= σ E

∂2E ∂E
+ µ0 F0 2 =∇2 E ∂t ∂t

∂2B ∂B
+ µ0 F0 2 =∇2 B. µ0 σ ∂t ∂t µ0 σ

and


and B
satisfy the same equation which is known as the telegrapher’s equation. Here both E
20.

Show that Maxwell’s equations (2.6.75) through (2.6.78) for the electric field under electrostatic

conditions reduce to


=0 ∇×E
=ρf ∇·D


is irrotational so that E
= −∇V. Show that ∇2 V = − ρf . Now E F

350
21.

Show that Maxwell’s equations (2.6.75) through (2.6.78) for the magnetic field under magnetostatic
= J
and ∇ · B
= 0. The divergence of B
being zero implies B
can be derived conditions reduce to ∇ × H
such that B
= ∇ × A.
Here A
is not unique, see problem 24. If we select from a vector potential function A


such that ∇ · A
= 0 then show for a homogeneous, isotropic material, free of any permanent magnets, that A
= −µJ.
∇2 A
22.

Show that under nonsteady state conditions of electrodynamics the Faraday law from Maxwell’s
= −∇V. Why is this? Observe that equations (2.6.75) through (2.6.78) does not allow one to set E
= 0 so we can write B
= ∇×A

∇·B  vector potential A. Using this vector potential show that  for some

+ ∂ A = 0. This shows that the quantity inside the parenthesis is Faraday’s law can be written ∇ × E ∂t

+ ∂ A = −∇V for some scalar potential V. The representation conservative and so we can write E ∂t

= −∇V − ∂ A E ∂t is a more general representation of the electric potential. Observe that for steady state conditions

' ∂A ∂t

=0

so that this potential representation reduces to the previous one for electrostatics.

= −∇V − ∂ A derived in problem 22, show that in a vacuum Using the potential formulation E ∂t
ρ ∂∇ · A =− (a) Gauss law can be written ∇2 V + ∂t F0 (b) Ampere’s law can be written    
∂V ∂2A

∇ × ∇ × A = µ0 J − µ0 F0 ∇ − µ0 F0 2 ∂t ∂t


23.

(c) Show the result in part (b) can also be expressed in the form    
∂ A ∂V 2

− ∇ ∇ · A + µ0 F0 ∇ A − µ0 F0 = −µ0 J
∂t ∂t
24.

The Maxwell equations in a vacuum have the form

= ∂ D + ρ V
∇×H ∂t



= − ∂B ∇×E ∂t


=ρ ∇·D


=0 ∇·B



= F0 E, where D


= µ0 H
with F0 and µ0 constants satisfying F0 µ0 = 1/c2 where c is the speed of light. B

and scalar potential V defined by B
= ∇×A
and E
= − ∂ A − ∇ V. Introduce the vector potential A ∂t Note that the vector potential is not unique. For example, given ψ as a scalar potential we can write
= ∇×A
= ∇ × (A
+ ∇ ψ), since the curl of a gradient is zero. Therefore, it is customary to impose some B
and B
are kind of additional requirement on the potentials. These additional conditions are such that E 1 ∂V
and V satisfy ∇ · A
+ not changed. One such condition is that A = 0. This relation is known as the c2 ∂t
and V and show Lorentz relation or Lorentz gauge. Find the Maxwell’s equations in a vacuum in terms of A that



1 ∂2 ∇ − 2 2 c ∂t 2



ρ V=− F0

 and

1 ∂2 ∇ − 2 2 c ∂t 2


= −µ0 ρV
. A

351
25.


and B
satisfy In a vacuum show that E
= ∇2 E


1 ∂2E c2 ∂t2


= ∇2 B


1 ∂2B c2 ∂t2


=0 ∇·E


=0 ∇B


26. (a) Show that the wave equations in problem 25 have solutions in the form of waves traveling in the x- direction given by
= E(x,

0 ei(kx±ωt) E t) = E

and


= B(x,

0 ei(kx±ωt) B t) = B


0 and B
0 are constants. Note that wave functions of the form u = Aei(kx±ωt) are called plane where E harmonic waves. Sometimes they are called monochromatic waves. Here i2 = −1 is an imaginary unit. Euler’s identity shows that the real and imaginary parts of these type wave functions have the form A cos(kx ± ωt)

and

A sin(kx ± ωt).

These represent plane waves. The constant A is the amplitude of the wave , ω is the angular frequency, and k/2π is called the wave number. The motion is a simple harmonic motion both in time and space. That is, at a fixed point x the motion is simple harmonic in time and at a fixed time t, the motion is harmonic in space. By examining each term in the sine and cosine terms we find that x has dimensions of length, k has dimension of reciprocal length, t has dimensions of time and ω has dimensions of reciprocal time or angular velocity. The quantity c = ω/k is the wave velocity. The value λ = 2π/k has dimension of length and is called the wavelength and 1/λ is called the wave number. The wave number represents the number of waves per unit of distance along the x-axis. The period of the wave is T = λ/c = 2π/ω and the frequency is f = 1/T. The frequency represents the number of waves which pass a fixed point in a unit of time. (b) Show that ω = 2πf (c) Show that c = f λ (d) Is the wave motion u = sin(kx − ωt) + sin(kx + ωt) a traveling wave? Explain. 1 ∂2φ (e) Show that in general the wave equation ∇2 φ = 2 2 have solutions in the form of waves traveling in c ∂t either the +x or −x direction given by φ = φ(x, t) = f (x + ct) + g(x − ct) where f and g are arbitrary twice differentiable functions. (f) Assume a plane electromagnetic wave is moving in the +x direction. Show that the electric field is in the xy−plane and the magnetic field is in the xz−plane. Hint: Assume solutions Ex = g1 (x − ct),

Ey = g2 (x − ct), Ez = g3 (x − ct), Bx = g4 (x − ct),

By = g5 (x − ct), Bz = g6 (x − ct) where gi ,i = 1, ..., 6 are arbitrary functions. Then show that Ex
= 0 which implies g1 must be independent of x and so not a wave function. Do does not satisfy ∇ · E
Since both ∇ · E
= ∇·B
= 0 then Ex = Bx = 0. Such waves the same for the components of B. are called transverse waves because the electric and magnetic fields are perpendicular to the direction
and B
waves must be in phase and be mutually of propagation. Faraday’s law implies that the E perpendicular to each other.

352 BIBLIOGRAPHY • Abramowitz, M. and Stegun, I.A., Handbook of Mathematical Functions, 10th ed, New York:Dover, 1972. • Akivis, M.A., Goldberg, V.V., An Introduction to Linear Algebra and Tensors, New York:Dover, 1972. • Aris, Rutherford, Vectors, Tensors, and the Basic Equations of Fluid Mechanics, Englewood Cliffs, N.J.:Prentice-Hall, 1962. • Atkin, R.J., Fox, N., An Introduction to the Theory of Elasticity, London:Longman Group Limited, 1980. • Bishop, R.L., Goldberg, S.I.,Tensor Analysis on Manifolds, New York:Dover, 1968. • Borisenko, A.I., Tarapov, I.E., Vector and Tensor Analysis with Applications, New York:Dover, 1968. • Chorlton, F., Vector and Tensor Methods, Chichester,England:Ellis Horwood Ltd, 1976. • Dodson, C.T.J., Poston, T., Tensor Geometry, London:Pittman Publishing Co., 1979. • Eisenhart, L.P., Riemannian Geometry, Princeton, N.J.:Univ. Princeton Press, 1960. • Eringen, A.C., Mechanics of Continua, Huntington, N.Y.:Robert E. Krieger, 1980. • D.J. Griffiths, Introduction to Electrodynamics, Prentice Hall, 1981. • Fl¨ ugge, W., Tensor Analysis and Continuum Mechanics, New York:Springer-Verlag, 1972. • Fung, Y.C., A First Course in Continuum Mechanics, Englewood Cliffs,N.J.:Prentice-Hall, 1969. • Goodbody, A.M., Cartesian Tensors, Chichester, England:Ellis Horwood Ltd, 1982. • Hay, G.E., Vector and Tensor Analysis, New York:Dover, 1953. • Hughes, W.F., Gaylord, E.W., Basic Equations of Engineering Science, New York:McGraw-Hill, 1964. • Jeffreys, H., Cartesian Tensors, Cambridge, England:Cambridge Univ. Press, 1974. • Lass, H., Vector and Tensor Analysis, New York:McGraw-Hill, 1950. • Levi-Civita, T., The Absolute Differential Calculus, London:Blackie and Son Limited, 1954. • Lovelock, D., Rund, H. ,Tensors, Differential Forms, and Variational Principles, New York:Dover, 1989. • Malvern, L.E., Introduction to the Mechanics of a Continuous Media, Englewood Cliffs, N.J.:Prentice-Hall, 1969. • McConnell, A.J., Application of Tensor Analysis, New York:Dover, 1947. • Newell, H.E., Vector Analysis, New York:McGraw Hill, 1955. • Schouten, J.A., Tensor Analysis for Physicists,New York:Dover, 1989. • Scipio, L.A., Principles of Continua with Applications, New York:John Wiley and Sons, 1967. • Sokolnikoff, I.S., Tensor Analysis, New York:John Wiley and Sons, 1958. • Spiegel, M.R., Vector Analysis, New York:Schaum Outline Series, 1959. • Synge, J.L., Schild, A., Tensor Calculus, Toronto:Univ. Toronto Press, 1956.

Bibliography

353 APPENDIX A UNITS OF MEASUREMENT The following units, abbreviations and prefixes are from the Syst`eme International d’Unit`es

(designated SI in all Languages.)

Prefixes. Abreviations Multiplication factor 1012 109 106 103 102 10 10−1 10−2 10−3 10−6 10−9 10−12

Symbol T G M K h da d c m µ n p

Basic units of measurement Name Length meter Mass kilogram Time second Electric current ampere Temperature degree Kelvin Luminous intensity candela

Symbol m kg s A ◦ K cd

Prefix tera giga mega kilo hecto deka deci centi milli micro nano pico

Basic Units. Unit

Unit Plane angle Solid angle

Supplementary units Name radian steradian

Symbol rad sr

354 Name Area Volume Frequency Density Velocity Angular velocity Acceleration Angular acceleration Force Pressure Kinematic viscosity Dynamic viscosity Work, energy, quantity of heat Power Electric charge Voltage, Potential difference Electromotive force Electric force field Electric resistance Electric capacitance Magnetic flux Inductance Magnetic flux density Magnetic field strength Magnetomotive force

DERIVED UNITS Units square meter cubic meter hertz kilogram per cubic meter meter per second radian per second meter per second squared radian per second squared newton newton per square meter square meter per second newton second per square meter joule watt coulomb volt volt volt per meter ohm farad weber henry tesla ampere per meter ampere

Symbol m2 m3 −1 Hz (s ) kg/m3 m/s rad/s m/s2 rad/s2 N (kg · m/s2 ) N/m2 m2 /s N · s/m2 J (N · m) W (J/s) C (A · s) V (W/A) V (W/A) V/m Ω (V/A) F (A · s/V) Wb (V · s) H (V · s/A) T (Wb/m2 ) A/m A

Physical constants. 4 arctan 1 = π = 3.14159 26535 89793 23846 2643 . . .  n 1 lim 1 + = e = 2.71828 18284 59045 23536 0287 . . . n→∞ n Euler’s constant γ = 0.57721 56649 01532 86060 6512 . . .   1 1 1 γ = lim 1 + + + · · · + − log n n→∞ 2 3 n speed of light in vacuum = 2.997925(10)8 m s−1 electron charge = 1.60210(10)−19 C Avogadro’s constant = 6.02252(10)23 mol−1 Plank’s constant = 6.6256(10)−34 J s Universal gas constant = 8.3143 J K −1 mol−1 = 8314.3 J Kg −1 K −1 Boltzmann constant = 1.38054(10)−23 J K −1 Stefan–Boltzmann constant = 5.6697(10)−8 W m−2 K −4 Gravitational constant = 6.67(10)−11 N m2 kg −2

355 APPENDIX B CHRISTOFFEL SYMBOLS OF SECOND KIND 1. Cylindrical coordinates (r, θ, z) = (x1 , x2 , x3 ) x = r cos θ

r≥0

h1 = 1

y = r sin θ

0 ≤ θ ≤ 2π

h2 = r

z=z

−∞
h3 = 1

The coordinate curves are formed by the intersection of the coordinate surfaces x2 + y 2 = r2 ,

Cylinders

y/x = tan θ

Planes

z = Constant 

1 22

'

 = −r

2 12

Planes. '

 =

2 21

' =

1 r

2. Spherical coordinates (ρ, θ, φ) = (x1 , x2 , x3 ) x = ρ sin θ cos φ

ρ≥0

h1 = 1

y = ρ sin θ sin φ

0≤θ≤π

h2 = ρ

z = ρ cos θ

0 ≤ φ ≤ 2π

h3 = ρ sin θ

The coordinate curves are formed by the intersection of the coordinate surfaces x2 + y 2 + z 2 = ρ2 2

2

Spheres 2

x + y = tan θ z y = x tan φ 

' 1 = −ρ 22  ' 1 = −ρ sin2 θ 33  ' 2 = − sin θ cos θ 33



Cones Planes.

'  ' 2 2 1 = = ρ 12 21  '  ' 3 3 1 = = ρ 13 31  '  ' 3 3 = = cot θ 32 23

356 3. Parabolic cylindrical coordinates (ξ, η, z) = (x1 , x2 , x3 ) x = ξη 1 y = (ξ 2 − η 2 ) 2 z=z

−∞
 ξ 2 + η2  h2 = ξ 2 + η 2

η≥0

h3 = 1

−∞<ξ <∞

h1 =

The coordinate curves are formed by the intersection of the coordinate surfaces x2 = −2ξ 2 (y −

ξ2 ) 2

Parabolic cylinders

η2 ) 2 z = Constant

x2 = 2η 2 (y +

Parabolic cylinders Planes.



' 1 ξ = 2 11 ξ + η2  ' 2 η = 2 22 ξ + η2  ' 2 −η = 2 11 ξ + η2



' 1 −ξ = 2 22 ξ + η2  '  ' 1 1 η = = 2 12 21 ξ + η2  '  ' 2 2 ξ = = 2 21 12 ξ + η2

4. Parabolic coordinates (ξ, η, φ) = (x1 , x2 , x3 ) x = ξη cos φ

ξ≥0

y = ξη sin φ 1 z = (ξ 2 − η 2 ) 2

η≥0

 ξ 2 + η2  h2 = ξ 2 + η 2

0 < φ < 2π

h3 = ξη

h1 =

The coordinate curves are formed by the intersection of the coordinate surfaces x2 + y 2 = −2ξ 2 (z − x2 + y 2 = 2η 2 (z + y = x tan φ 

' 1 = 11  ' 2 = 22  ' 1 = 22  ' 2 = 11  ' 2 = 33

ξ 2 ξ + η2 η 2 ξ + η2 −ξ 2 ξ + η2 −η 2 ξ + η2 −ηξ 2 ξ 2 + η2

ξ2 ) 2

η2 ) 2

Paraboloids Paraboloids Planes. 

' 1 = 33  '  ' 1 1 = = 21 21  '  ' 2 2 = = 21 12  '  ' 3 3 = = 32 23  '  ' 3 3 = = 13 31

−ξη 2 ξ 2 + η2 η 2 ξ + η2 ξ 2 ξ + η2 1 η 1 ξ

357 5. Elliptic cylindrical coordinates (ξ, η, z) = (x1 , x2 , x3 ) x = cosh ξ cos η

ξ≥0

h1 =

sinh2 ξ + sin2 η

y = sinh ξ sin η

0 ≤ η ≤ 2π

h2 =

sinh2 ξ + sin2 η

z=z

−∞
h3 = 1

The coordinate curves are formed by the intersection of the coordinate surfaces x2 y2 + =1 cosh2 ξ sinh2 ξ y2 x2 − =1 cos2 η sin2 η

Elliptic cylinders Hyperbolic cylinders

z = Constant 

' 1 = 11  ' 1 = 22  '  ' 1 1 = = 12 21

Planes. 

' 2 sin η cos η = 22 sinh2 ξ + sin2 η  ' 2 − sin η cos η = 11 sinh2 ξ + sin2 η  '  ' 2 2 sinh ξ cosh ξ = = 12 21 sinh2 ξ + sin2 η

sinh ξ cosh ξ sinh2 ξ + sin2 η − sinh ξ cosh ξ sinh2 ξ + sin2 η sin η cos η sinh2 ξ + sin2 η

6. Elliptic coordinates (ξ, η, φ) = (x1 , x2 , x3 ) !  x = (1 − η 2 )(ξ 2 − 1) cos φ  y = (1 − η 2 )(ξ 2 − 1) sin φ z = ξη

h1 =

1≤ξ<∞

!

−1≤η ≤1

h2 =

0 ≤ φ < 2π

h3 =

ξ 2 − η2 ξ2 − 1 ξ 2 − η2 1 − η2

 (1 − η 2 )(ξ 2 − 1)

The coordinate curves are formed by the intersection of the coordinate surfaces y2 z2 x2 + + =1 ξ2 − 1 ξ2 − 1 ξ2 z2 x2 y2 − − =1 η2 1 − η2 1 − η2 y = x tan φ 

' 1 ξ ξ + 2 =− 11 −1 + ξ 2 ξ − η2  ' 2 η η − 2 = 2 22 1−η ξ − η2

 ' ξ −1 + ξ 2 1 =− 22 (1 − η 2 ) (ξ 2 − η 2 )



 ' ξ −1 + ξ 2 1 − η 2 1 =− 33 ξ 2 − η2

 ' η 1 − η2 2 = 11 (−1 + ξ 2 ) (ξ 2 − η 2 )

Prolate ellipsoid Two-sheeted hyperboloid Planes



' −1 + ξ 2 η 1 − η 2 2 = ξ 2 − η2 33  ' 1 η =− 2 12 ξ − η2  ' 2 ξ = 2 21 ξ − η2  ' 3 ξ = 31 −1 + ξ 2  ' 3 η =− 32 1 − η2



358 7. Bipolar coordinates (u, v, z) = (x1 , x2 , x3 ) a sinh v , 0 ≤ u < 2π cosh v − cos u a sin u , −∞ < v < ∞ y= cosh v − cos u z=z −∞
h21 = h22

x=

h22 =

a2 (cosh v − cos u)2

h23 = 1

The coordinate curves are formed by the intersection of the coordinate surfaces a2 sinh2 v a2 x2 + (y − a cot u)2 = sin2 u z = Constant

(x − a coth v)2 + y 2 =



' 1 sin u = 11 cos u − cosh v  ' 2 sinh v = cos u − cosh v 22  ' 1 sin u = 22 − cos u + cosh v

Cylinders Cylinders Planes.



' 2 sinh v = 11 − cos u + cosh v  ' 1 sinh v = cos u − cosh v 12  ' 2 sin u = 21 cos u − cosh v

8. Conical coordinates (u, v, w) = (x1 , x2 , x3 ) uvw , b 2 > v 2 > a2 > w 2 , ab " u (v 2 − a2 )(w2 − a2 ) y= a a2 − b 2 " v (v 2 − b2 )(w2 − b2 ) z= b b 2 − a2

x=

u≥0

h21 = 1 u2 (v 2 − w2 ) − a2 )(b2 − v 2 ) u2 (v 2 − w2 ) h23 = 2 (w − a2 )(w2 − b2 ) h22 =

(v 2

The coordinate curves are formed by the intersection of the coordinate surfaces x2 + y 2 + z 2 = u2 2

2

Spheres

2

x y z + 2 + 2 = 0, v2 v − a2 v − b2 x2 y2 z2 + 2 + 2 = 0, 2 2 w w −a w − b2 

' 2 v v v = 2 − + 2 22 b − v2 −a2 + v 2 v − w2  ' 3 w w w − − =− 2 33 v − w2 −a2 + w2 −b2 + w2

 ' u v 2 − w2 1 =− 2 22 (b − v 2 ) (−a2 + v 2 )

 ' u v 2 − w2 1 =− (−a2 + w2 ) (−b2 + w2 ) 33



 ' v b2 − v 2 −a2 + v 2 2 =− 2 33 (v − w2 ) (−a2 + w2 ) (−b2 + w2 )

Cones Cones.



' w −a2 + w2 −b2 + w2 3 = 2 22 (b − v 2 ) (−a2 + v 2 ) (v 2 − w2 )  ' 2 1 = 21 u  ' 2 w =− 2 v − w2 23  ' 3 1 = 31 u  ' 3 v = 2 v − w2 32 

359 9. Prolate spheroidal coordinates (u, v, φ) = (x1 , x2 , x3 ) x = a sinh u sin v cos φ,

u≥0

h21 = h22

y = a sinh u sin v sin φ,

0≤v≤π

h22 = a2 (sinh2 u + sin2 v)

z = a cosh u cos v,

0 ≤ φ < 2π

h23 = a2 sinh2 u sin2 v

The coordinate curves are formed by the intersection of the coordinate surfaces x2 y2 z2 + + = 1, 2 2 (a sinh u) a sinh u) a cosh u)2 y2 z2 x2 − − = 1, (a cos v)2 (a sin v)2 (a cos v)2

Prolate ellipsoids Two-sheeted hyperpoloid

y = x tan φ, 

' 1 cosh u sinh u = 11 sin2 v + sinh2 u  ' 2 cos v sin v = 22 sin2 v + sinh2 u  ' 1 cosh u sinh u =− 2 22 sin v + sinh2 u  ' 1 sin2 v cosh u sinh u =− 33 sin2 v + sinh2 u  ' 2 cos v sin v =− 2 11 sin v + sinh2 u

Planes.



' 2 cos v sin vsinh2 u =− 33 sin2 v + sinh2 u  ' 1 cos v sin v = 12 sin2 v + sinh2 u  ' 2 cosh u sinh u = 21 sin2 v + sinh2 u  ' 3 cosh u = sinh u 31  ' 3 cos v = 32 sin v

10. Oblate spheroidal coordinates (ξ, η, φ) = (x1 , x2 , x3 ) x = a cosh ξ cos η cos φ, y = a cosh ξ cos η sin φ, z = a sinh ξ sin η,

ξ≥0 π π − ≤η≤ 2 2 0 ≤ φ ≤ 2π

h21 = h22 h22 = a2 (sinh2 ξ + sin2 η) h23 = a2 cosh2 ξ cos2 η

The coordinate curves are formed by the intersection of the coordinate surfaces x2 y2 z2 + + = 1, (a cosh ξ)2 (a cosh ξ)2 (a sinh ξ)2 y2 z2 x2 + − = 1, (a cos η)2 (a cos η)2 (a sin η)2 y = x tan φ, 

' 1 cosh ξ sinh ξ = 11 sin2 η + sinh2 ξ  ' 2 cos η sin η = 22 sin2 η + sinh2 ξ  ' 1 cosh ξ sinh ξ =− 2 22 sin η + sinh2 ξ  ' 1 cos2 η cosh ξ sinh ξ =− 33 sin2 η + sinh2 ξ  ' 2 cos η sin η =− 2 11 sin η + sinh2 ξ



Oblate ellipsoids One-sheet hyperboloids Planes.

' 2 cos η sin ηcosh2 ξ = 33 sin2 η + sinh2 ξ  ' 1 cos η sin η = 12 sin2 η + sinh2 ξ  ' 2 cosh ξ sinh ξ = 21 sin2 η + sinh2 ξ  ' 3 sinh ξ = 31 cosh ξ  ' 3 sin η =− 32 cos η

360 11. Toroidal coordinates (u, v, φ) = (x1 , x2 , x3 ) a sinh v cos φ , cosh v − cos u a sinh v sin φ y= , cosh v − cos u a sin u z= , cosh v − cos u

x=

0 ≤ u < 2π −∞ < v < ∞ 0 ≤ φ < 2π

h21 = h22 h22 =

a2 (cosh v − cos u)2

h23 =

a2 sinh2 v (cosh v − cos u)2

The coordinate curves are formed by the intersection of the coordinate surfaces  a cos u 2 a2 x2 + y 2 + z − , = sin u sin2 u  2 cosh v a2 , x2 + y 2 − a + z2 = sinh v sinh2 v y = x tan φ, 

' 1 = 11  ' 2 = 22  ' 1 = 22  ' 1 = 33  ' 2 = 11

sin u cos u − cosh v sinh v cos u − cosh v sin u − cos u + cosh v sin usinh v 2 − cos u + cosh v sinh v − cos u + cosh v



Spheres Tores planes

' 2 sinh v (cos u cosh v − 1) =− cos u − cosh v 33  ' 1 sinh v = cos u − cosh v 12  ' 2 sin u = 21 cos u − cosh v  ' 3 sin u = cos u − cosh v 31  ' 3 cos u cosh v − 1 = 32 cos u sinh v − cosh v sinh v

361 12. Confocal ellipsoidal coordinates (u, v, w) = (x1 , x2 , x3 ) (a2 − u)(a2 − v)(a2 − w) , (a2 − b2 )(a2 − c2 ) (b2 − u)(b2 − v)(b2 − w) , y2 = (b2 − a2 )(b2 − c2 ) (c2 − u)(c2 − v)(c2 − w) , z2 = (c2 − a2 )(c2 − b2 )

x2 =

u < c2 < b 2 < a2 c2 < v < b 2 < a 2 c2 < b 2 < v < a2

(u − v)(u − w) 4(a2 − u)(b2 − u)(c2 − u) (v − u)(v − w) h22 = 4(a2 − v)(b2 − v)(c2 − v) (w − u)(w − v) h23 = 4(a2 − w)(b2 − w)(c2 − w)

h21 =



' 1 1 1 1 1 1 + + + + = 2 2 2 11 2 (a − u) 2 (b − u) 2 (c − u) 2 (u − v) 2 (u − w)  ' 1 1 1 1 2 1 + + + + = 2 2 2 2 (a − v) 2 (b − v) 2 (c − v) 2 (−u + v) 2 (v − w) 22  ' 3 1 1 1 1 1 + + + + = 2 2 2 33 2 (a − w) 2 (b − w) 2 (c − w) 2 (−u + w) 2 (−v + w) 2





 '  ' a − u b2 − u c2 − u (v − w) 1 1 −1 = = 22 2 (a2 − v) (b2 − v) (c2 − v) (u − v) (u − w) 2 (u − v) 12





2  '  ' a − u b2 − u c2 − u (−v + w) 1 1 −1 = = 33 2 (u − v) (a2 − w) (b2 − w) (c2 − w) (u − w) 13 2 (u − w) 2





 '  ' a − v b2 − v c2 − v (u − w) 2 2 −1 = = 2 2 2 11 2 (a − u) (b − u) (c − u) (−u + v) (v − w) 2 (−u + v) 21  '

2

2

2  ' 2 −1 − v b − v c − v (−u + w) a 2 = = 2 2 2 23 2 (v − w) 33 2 (−u + v) (a − w) (b − w) (c − w) (v − w)  ' 2

2

2

 ' 3 −1 (u − v) a − w b − w c − w 3 = = 2 (−u + w) 31 11 2 (a2 − u) (b2 − u) (c2 − u) (−u + w) (−v + w)  ' 2

2

2

 ' 3 −1 (−u + v) a − w b − w c − w 3 = = 32 2 (−v + w) 22 2 (a2 − v) (b2 − v) (c2 − v) (−u + w) (−v + w)

362 APPENDIX C VECTOR IDENTITIES
B,
C,
D
are differentiable vector functions of position while The following identities assume that A, f, f1 , f2 are differentiable scalar functions of position.

1.


· (B
× C)
=B
· (C
× A)
=C
· (A
× B)
A

2.


× (B
× C)
= B(
A
· C)
− C(
A
· B)
A

3.


× B)
· (C
× D)
= (A
· C)(
B
· D)
− (A
· D)(
B
· C)
(A

4.


× (B
× C)
+B
× (C
× A)
+C
× (A
× B)
=
0 A

5.


× B)
× (C
× D)
= B(
A
·C
× D)
− A(
B
·C
× D)
(A
A
·B
× C)
− D(
A
·B
× C)
= C(

6.


× B)
· (B
× C)
× (C
× A)
= (A
·B
× C)
2 (A

7.

∇(f1 + f2 ) = ∇f1 + ∇f2

8.


+ B)
=∇·A
+∇·B
∇ · (A

9.


+ B)
=∇×A
+∇×B
∇ × (A

10.


= (∇f ) · A
+ f∇ · A
∇(f A)

11.

∇(f1 f2 ) = f1 ∇f2 + f2 ∇f1

12.


=)∇f ) × A
+ f (∇ × A)
∇ × (f A)

13.


× B)
=B
· (∇ × A)
−A
· (∇ × B)
∇ · (A  
2
× (∇ × A)

· ∇)A
= ∇ |A| −A (A 2

14. 15.


· B)
= (B
· ∇)A
+ (A
· ∇)B
+B
× (∇ × A)
+A
× (∇ × B)
∇(A

16.


× B)
= (B
· ∇)A
− B(∇

− (A
· ∇)B
+ A(∇
· B)
∇ × (A · A)

17.

∇ · (∇f ) = ∇2 f

18.

∇ × (∇f ) =
0

19.


=0 ∇ · (∇ × A)

20.



= ∇(∇ · A)
− ∇2 A ∇ × (∇ × A)

363

INDEX A Absolute differentiation 120 Absolute scalar field 43 Absolute tensor 45,46,47,48 Acceleration 121, 190, 192 Action integral 198 Addition of systems 6, 51 Addition of tensors 6, 51 Adherence boundary condition 294 Aelotropic material 245 Affine transformation 86, 107 Airy stress function 264 Almansi strain tensor 229 Alternating tensor 6,7 Ampere’s law 176,301,337,341 Angle between vectors 80, 82 Angular momentum 218, 287 Angular velocity 86,87,201,203 Arc length 60, 67, 133 Associated tensors 79 Auxiliary Magnetic field 338 Axis of symmetry 247

Cauchy stress law 216 Cauchy-Riemann equations 293,321 Charge density 323 Christoffel symbols 108,110,111 Circulation 293 Codazzi equations 139 Coefficient of viscosity 285 Cofactors 25, 26, 32 Compatibility equations 259, 260, 262 Completely skew symmetric system 31 Compound pendulum 195,209 Compressible material 231 Conic sections 151 Conical coordinates 74 Conjugate dyad 49 Conjugate metric tensor 36, 77 Conservation of angular momentum 218, 295 Conservation of energy 295 Conservation of linear momentum 217, 295 Conservation of mass 233, 295 Conservative system 191, 298 Conservative electric field 323

B Basic equations elasticity 236, 253, 270 Basic equations for a continuum 236 Basic equations of fluids 281, 287 Basis vectors 1,2,37,48 Beltrami 262 Bernoulli’s Theorem 292 Biharmonic equation 186, 265 Bilinear form 97 Binormal vector 130 Biot-Savart law 336 Bipolar coordinates 73 Boltzmann equation 302,306 Boundary conditions 257, 294 Bulk modulus 251 Bulk coefficient of viscosity 285 C Cartesian coordinates 19,20,42, 67, 83 Cartesian tensors 84, 87, 226

Constitutive equations 242, 251,281, 287 Continuity equation 106,234, 287, 335 Contraction 6, 52 Contravariant components 36, 44 Contravariant tensor 45 Coordinate curves 37, 67 Coordinate surfaces 37, 67 Coordinate transformations 37 Coulomb law 322 Covariant components 36, 47 Covariant differentiation 113,114,117 Covariant tensor 46 Cross product 11 Curl 21, 173 Curvature 130, 131, 134, 149 Curvature tensor 134, 145 Curvilinear coordinates 66, 81 Cylindrical coordinates 18, 42, 69

364

INDEX D Deformation 222 Derivative of tensor 108 Derivatives and indicial notation 18, 31 Determinant 10, 25, 32, 33 Dielectric tensor 333 Differential geometry 129 Diffusion equation 303 Dilatation 232 Direction cosines 85 Displacement vector 333 Dissipation function 297

Equilibrium equations 273,300 Elastic constants 243,248 Equipotential curves 325 Euler number 294 Euler-Lagrange equations 192 Eulerian angles 201, 209 Eulerian form 287 Eulerian system 227 Eulers equations of motion 204 F

Distribution function 302

Faraday’s law 176,301, 340

Divergence 21, 172

Field lines 324, 327

Divergence theorem 24

Field electric 322

Dot product 5

First fundamental form 133,143

Double dot product 50, 62

Fourier law 297, 299

Dual tensor 100

Free indices 3

Dummy index 4, 5

Frenet-Serret formulas 131, 188

Dyads 48,62,63

Froude number 294

Dynamics 187

Fluids 281

E

G

e Permutation symbol 6, 7, 12

Gas law 300

e-δ identity 12

Gauss divergence theorem 24, 330

Eigenvalues 179,189

Gauss equations 138

Eigenvectors 179,186

Gauss’s law for electricity 176,301,328

Einstein tensor 156

Gauss’s law for magnetism 176,301,341

Elastic constants 248

Gaussian curvature 137,139, 149

Elastic stiffness 242

Geodesics 140, 146

Elasticity 211,213

Geodesic curvature 135, 140

Electrostatic field 322,333

General tensor 48

Electric flux 327

Generalized e − δ identity 84, 104

Electric units 322

Generalized Hooke’s law 242

Electrodynamics 339

Generalized Kronecker delta 13, 31

Electromagnetic energy 341

Generalized stress strain 242

Electromagnetic stress 341,342

Geometry in Riemannian Space 80

Elliptic coordinates 72

Gradient 20, 171

Elliptical cylindrical coordinates 71

Gradient basis 37

Enthalpy 298

Green’s theorem 24

Entropy 300

Group properties 41, 54

Epsilon permutation symbol 83

Generalized velocity 121

Equation of state 300

Generalized acceleration 121

365

INDEX H Hamiltonian 208 Heat equation 316 Hexagonal material 247 Higher order tensors 47, 93 Hooke’s law 212, 242, 252 Hydrodynamic equations 283 I

M Magnitude of vector 80 Magnetostatics 334,338 Magnetic field 334 Magnetization vector 337 Magnetic permeability 337 Material derivative 234, 288 Material symmetry 244, 246

Ideal fluid 283

Maxwell equations 176, 339

Idemfactor 50

Maxwell transfer equation 308

Incompressible material 231

Maximum, minimum curvature 130, 140

Index notation 1, 2, 14

Mean curvature 137, 148

Indicial notation 1, 2, 14,24

Metric tensor 36, 65

Inner product 52

Meusnier’s Theorem 150

Inertia 30

Mixed tensor 49

Integral theorems 24

Mohr’s circle 185

Intrinsic derivative 120

Moment of inertia 30, 184, 200

Invariant 43

Momentum 217, 218

Inviscid fluid 283

Multilinear forms 96, 98

Isotropic material 248

Multiplication of tensors 6, 51

Isotropic tensor 104 N J Navier’s equations 254, 257 Jacobian 17, 30, 40, 101, 127 Jump discontinuity 330 K

Navier-Stokes equations 288, 290 Newtonian fluids 286 Nonviscous fluid 283 Normal curvature 135, 136

Kronecker delta 3, 8, 13, 31, 76

Normal plane 188

Kinetic energy 201

Normal stress 214

Kinematic viscosity 302

Normal vector 130, 132

L Lagrange’s equation of motion 191, 196

Notation for physical components 92 O

Lagrangian 209 Laplacian 174 Linear form 96 Linear momentum 209, 287 Linear transformation 86 Linear viscous fluids 284 Lorentz transformation 57 Lame’s constants 251

Oblate Spheroidal coordinates 75 Oblique coordinates 60 Oblique cylindrical coordinates 102 Order 2 Orthogonal coordinates 78, 86 Orthotropic material 246 Outer product 6, 51 Osculating plane 188

366

INDEX P Parallel vector field 122

Rayleigh implusive flow 317

Pappovich-Neuber solution 263

Reciprocal basis 35, 38

Parabolic coordinates 70

Relative scalar 127

Parabolic cylindrical coordinates 69

Relative tensor 50, 121

Particle motion 190

Relative motion 202

Pendulum system 197, 210

Relativity 151

Perfect gas 283, 299

Relative motion 155

Permutations 6

Reynolds number 294

Phase space 302

Ricci’s theorem 119

Physical components 88, 91,93

Riemann Christoffel tensor 116, 129,139, 147

Piezoelectric 300

Riemann space 80

Pitch,roll, Yaw 209

Rectifying plane 188

Plane Couette flow 315

Rigid body rotation 199

Plane Poiseuille flow 316

Rotation of axes 85, 87, 107

Plane strain 263

Rules for indices 2

Plane stress 264 Poisson’s equation 329

S

Poisson’s ratio 212 Polar element 273

Scalar 40, 43

Polarization vector 333

Scalar invariant 43, 62, 105

Polyads 48

Scalar potential 191

Potential energy 191

Scaled variables 293

Potential function 323

Second fundamental form 135, 145

Poynting’s vector 341

Second order tensor 47

Pressure 283

Shearing stresses 214

Principal axes 183

Simple pulley system 193

Projection 35

Simple pendulum 194

Prolated Spheroidal coordinates 74

Skew symmetric system 3, 31

Pully system 194, 207

Skewed coordinates 60, 102 Solid angle 328

Q

Space curves 130 Special tensors 65

Quotient law 53 R

Spherical coordinates 18, 43, 56, 69, 103,194 Stokes flow 318 Stokes hypothesis 285

Radius of curvature 130, 136

Stokes theorem 24

Range convention 2, 3

Straight line 60

Rate of deformation 281, 286

Strain 218, 225, 228

Rate of strain 281

Strain deviator 279

367

INDEX

Stress 214

U

Stress deviator 279 Strong conservative form 298

Unit binormal 131, 192

Strouhal number 294

Unit normal 131, 191

St Venant 258

Unit tangent 131, 191

Subscripts 2

Unit vector 81, 105

Subtraction of tensors 51, 62 Summation convention 4, 9

V

Superscripts 2

Vector identities 15, 20, 315

Surface 62, 131

Vector transformation 45, 47

Surface area 59

Vector operators 20, 175

Surface curvature 149

Vector potential 188

Surface metric 125, 133

Velocity 95, 121, 190, 193

Susceptibility tensor 333

Velocity strain tensor 281

Sutherland formula 285

Viscosity 285

Symmetric system 3, 31, 51, 101

Viscosity table 285

Symmetry 243

Viscous fluid 283

System 2, 31

Viscous forces 288

T Tangential basis 37 Tangent vector 130

Viscous stress tensor 285 Vorticity 107, 292 W

Tensor and vector forms 40, 150

Wave equation 255, 269

Tensor derivative 141

Weighted tensor 48, 127

Tensor general 48

Weingarten’s equation 138, 153

Tensor notation 92, 160

Work 191, 279

Tensor operations 6, 51, 175

Work done 324

Test charge 322 Thermodynamics 299

Y

Third fundamental form 146 Third order systems 31 Toroidal coordinates 75, 103 Torus 124 Transformation equations 17, 37, 86 Transitive property 45,46 Translation of coordinates 84 Transport equation 302 Transposition 6 Triad 50 Trilinear form 98 Triple scalar product 15

Young’s modulus 212