Physics Mechanics - Lecture Notes

Physics 1. Mechanics Math0. Useful mathematical formulae Here you will find a (limited) number of useful mathematical formulae which you must know in order to understand physics. • Let f = f (x) and g = g(x) and f 0 = df /dx, . . . . Then (f + g)0 = f 0 + g 0 ,

(f g)0 = f 0 g + f g 0

• Let f = f (y), and y = y(x), then df = dx • Following two previous rules



df dy



dy dx



 0 f gf 0 − f g 0 = g g2

• Differential:

 df =

df dx

 dx

• Taylor expansion  f (x) = f (x0 ) + (x − x0 )

df dx

 + |x=x0

1 (x 2

2

− x0 )



d2f dx2

 + ... |x=x0

or for the infinite series  n  ∞ X 1 d f n f (x) = f (x0 ) + (x − x0 ) n! dxn |x=x i=1

0

• Useful approximations for |x|  1 sinx ≈ x,

cosx ≈ 1 − x2 /2,

ex ≈ 1 + x.

ln(1 + x) ≈ x,

(1 + x)α ≈ 1 + αx

• Let f = f (x, y). Partial derivative ∂f /∂x is the derivative with respect to x while y is assumed constant. 1

Physics 1. Mechanics

Math0

• Total differential

 df =

∂f ∂x



 dx +

∂f ∂y

 dy

• First term of the Taylor expansion  f (x, y) = f (x0 , y0 ) +

∂f ∂x



 (x − x0 ) + |x=x0 ,y=y0

∂f ∂y

 (y − y0 ) |x=x0 ,y=y0

• Features of mixed derivatives for normal functions f x, y: ∂2f ∂2f = ∂x∂y ∂y∂x • Indefinite integral Z

• Definite integral

x2

Z

f 0 dx = f + C

f 0 dx = f (x2 ) − f (x1 )

x1

• Substitution. Let f = f (y) and y = y(x) then Z

Z f (y)dy =

f (y(x))

dy dx dx

• Integration by parts Z

0

f g dx = f g −

Z

gf 0 dx

2

Physics 1. Mechanics Lecture 0. Partial derivatives

1

Functions

Let x be some variable (like time), and for each x we measure another variable y which depends on x (like a body temperature). The dependence of y on x is described by a function y = y(x). It is essential that x be continuous, and that for each x from some interval x1 < x < x2 there exist a corresponding y.

2

Derivative

Let the argument x change from x1 to x2 , ∆x ≡ x2 − x1 . The function change is ∆y = y(x2 ) − y(x1 ). The ratio ∆y/∆x shows how fast the function changes on average when the argument changes on the interval (x1 , x2 ). Alternatively, let us choose a point x and shift from this point ∆x. The above ratio then will be written as

y(x + ∆x) − y(x) ∆y = ∆x ∆x

as depends on x and ∆x as well. If we now require ∆x → 0 (infinitesimal change) then the ratio (if it exists at all) will show the local rate of change of the function y in the point x, y0 ≡

dy ∆y = lim , dx ∆x→0 ∆x

(1)

and is called a derivative. In this notation dy/dx is not a ratio but a new function. However, it is obtained by calculating the ratio of two infinitesimal values. Starting with the obvious expression  ∆y =

∆y ∆x

 ∆x

we can extrapolate this onto infinitesimal changes and obtain a differential : 

0

dy = y dx =

dy dx

 dx,

which is the change of y when x changes by infinitesimal dx → 0 (but dx 6= 0 !). Example: Let y(x) = xn , then (dy/dx) = nxn−1 , dy = nxn−1 dx. Example: The function y = |x| does not have a derivative at x = 0.

1

(2)

Physics 1. Mechanics

3

Lecture 0

Partial derivatives

Let now y be a function of more than one variable xi , i = 1, 2, . . . , n (here n is the number of independent variables). We define a partial derivative (∂y/∂xl ) of y with respect to the variable xl as a regular derivative, taken as if y were function of xl only while other variables remain constant. In this case dy is produced by the change of each xi independently, so that one has:  dy =

∂y ∂x1



 dx1 +

∂y ∂x2

 dx2 + . . . =

X  ∂y  i

∂xi

dxi .

(3)

Example: Let y = x21 + x1 x2 , then (∂y/∂x1 ) = 2x1 + x2 , (∂y/∂x2 ) = x1 , dy = (2x1 + x2 )dx1 + x1 dx2 . Q: Partially differentiate and write down the full differential for the following functions: y = x21 + 2x1 x22 + x52 , y=

exp(−x21

−

x22 )

(4)

· [sin x1 + 2 cos(2x1 x2 )].

(5)

Each partial derivative is itself a function.

4 4.1

Some applications Differentiation along a curve

Let y = y(x1 , x2 , . . .), and each argument is, in turn, a function of a single argument t: xi = xi (t). Eventually, y is a function of t too, and one may ask what is the derivative (dy/dt). One way is to substitute all xi (t) into the expression for y and, after an explicit form of y(t) is obtained, differentiate it as usual. This is not always convenient. Instead, let us write the full differential dy in two forms: 

 dy dy = dt, dt  X ∂y  dy = dxi ∂xi

(6) (7)

i

However,  dxi =

dxi dt

 dt

(8)

and therefore 

dy dt

 dt =

X  ∂y 

dxi ∂xi X  ∂y   dxi  =[ ]dt ⇒ ∂xi dt i   X   ∂y dxi dy = dt ∂xi dt

(9)

i

(10) (11)

i

2

Physics 1. Mechanics

4.2

Lecture 0

Implicit function

Sometimes it is impossible to give an explicit form of a function y = y(x1 , x2 , . . .). Instead, and implicit function is given in the form of an equation F (y, x1 , x2 , . . .) = 0, which is difficult or impossible to solve with respect to y. In order to find derivatives of y we use again the full differential:  X  ∂F  ∂F dF = 0 = dy + dxi ⇒ ∂y ∂xi i   X  ∂F  ∂F dy = −[ dxi ]/ ⇒ ∂xi ∂y i     ∂y ∂F ∂F =− / ∂xi ∂xi ∂y 

5

(12) (13) (14)

Taylor expansion

We use the full differential to approximate a function variation:  ∆y =

∆y ∆x



 ∆x ≈

dy dx

 ∆x

(15)

or, in a more appropriate form,  y(x) = y(x0 ) +

dy dx

 |x=x0 (x − x0 ) + . . .

(16)

Here |x=x0 means that the derivative is evaluated in the point x = x0 , and . . . means that the expression is not exact and (hopefully) smaller terms should be added. For a function of many variables one has, respectively y(xi ) = y(xi,0 ) +

X  ∂y  i

∂xi

|xi =xi,0 (xi − xi,0 )

(17)

3

Physics 1. Mechanics Lecture 1. Coordinates

1

Coordinates. General principles

We start with the description of the motion of a point mass (point or particle). First thing to do is to learn how to describe the position of the point. In order to do that one needs coordinates. In the simplest case of a motion along a line (not necessarily straight line) one needs to choose an origin O and a method of assigning a coordinate, that is, a rule to establish a correspondence of each point of the curve to a real number (positive of negative). In order to be useful the coordinate should change continuously and, if possible, different points have to have different coordinates and each point has to have only one coordinate.

O

Figure 1: Coordinate on a curve: one-dimensional space.

1.1 1.1.1

Examples Straight line

Any point can be taken as the origin O. Let us choose the positive direction to the right (see Fig. 2), and assign to each point a number corresponding to the distance from O with the sign + if the point is to the right of O, and − if it is to the left. Let us denote this coordinate as x, then −∞ < x < ∞. Q: Is this choice unique ? Give an example of another choice. Try to find a coordinate x′ which would change in the range −1 ≤ x′ ≤ 1 and yet cover the whole line. 1

O

x

Figure 2: Coordinates on a straight line

ϕ

O

Figure 3: Coordinates on a circle. 1.1.2

Circle

A simple choice of the coordinate would be the angle ϕ measured counterclockwise from some radius (see Fig. 3). In order to make this coordinate unique we have to require, for example, 0 ≤ ϕ < 2π. However, in this case the coordinate is not continuous at ϕ = 0. On the other hand, in many applications it is more desirable to have the coordinate continuous. To do that we may say that the coordinates ϕ and ϕ + 2πn (where n is an arbitrary integer) describe the same point. 1.1.3

Closed curve with no intersections

Let us consider a simple closed curve which does not intersect itself. A simple coordinate choice is the angle ϕ measured counterclockwise from some straight line intersecting the curve. In some cases this coordinate may be ambiguous, and a better coordinate would be the curve length measured from one chosen point in a chosen direction. In this case the coordinate is always positive.

2

Proceeding further

In the above examples we needed only one coordinate to describe a point position. In this case it is said that the space is one-dimensional. If two-coordinates are necessary, the space is called two-dimensional and if n coordinates are needed, the space is called n-dimensional of nD. The space we live in is three-dimensional, that is, in general three coordinates are needed to describe a point position. In what follows we shall denote a point as P and its coordinates as (xi ), where i = 1, . . . , n. 2

θ O

Figure 4: Coordinates on a closed curve with not self-intersections.

x2

x1

Figure 5: General 2D coordinates.

2.1

Examples of 2D coordinates

In the 1D case we had a single curve. In the 2D case we build two family of “parallel” curves (see Fig. 5), one family for each coordinate. “Parallel” means that the curves from the same family do not intersect. For each curve from one family the second coordinate is constant. 2.1.1

Cartesian coordinates

The two families are straight lines intersecting and 90◦ . The coordinates (x, y) are chosen along each line as in the 1D case of a straight line (distance with sign). The curves of x = const are straight lines parallel to y-axis. The curves of y = const are straight lines parallel to x-axis. 2.1.2

Polar coordinates

One family is the radii coming out of the origin, the other is the circles with the origin as a center. One coordinate, r, is a distance from the origin (nonnegative), the second is the angle ϕ as in the 3

P (x, y)

y

x Figure 6: 2D Cartesian coordinates.

P (r, ϕ) r

ϕ

Figure 7: Polar coordinates. case of a circle (see above). The curves of r = const are circles. The curves of ϕ = const are radii.

2.2

3D examples

For 3D case xi = const gives a surface. 2.2.1

3D Cartesian coordinates

Similar to the 2D case. x = x0 = const corresponds to a plane (2D space !) parallel to the y − z plane and crossing the x axis at x = x0 . 2.2.2

Cylindrical coordinates

The coordinates (r, ϕ, z) are the polar coordinates (r, ϕ) with the addition of the “height” z. r = const are cylindrical surfaces around z-axis. z = const are planes parallel to x − y plane. Q: What are ϕ = const surfaces ? 4

z P (x, y, z)

y

x

Figure 8: 3D Cartesian coordinates.

z P (r, ϕ, z)

ϕ r Figure 9: Cylindrical coordinates.

5

P (R, θ, ϕ) R θ ϕ

Figure 10: Spherical coordinates. 2.2.3

Spherical coordinates

The coordinates (R, θ, ϕ) are the distance from the origin, the angle measured downward from z-axis, and the polar angle ϕ measured in the projection on the plane x − y counterclockwise from x-axis. Q: What are R = const, θ = const, and ϕ = const surfaces, respectively ? What are the dimensions of the (sub)space r = const and ϕ = const ? What is this space ?

2.3

Relation between different coordinate systems

The same point can be assigned coordinates with the help of different coordinate systems, and it is necessary to know the relation between those coordinates. In the following we assume that the coordinate origin is the same for all chosen systems. 2.3.1

2D Cartesian - polar

The two systems are shown in the figure. It is accepted to choose x-axis as the base of the polar coordinates. The following relation holds: x = r cos ϕ, y = r sin ϕ. 2.3.2

(1)

3D

The three coordinate systems are shown in the same figure. One has

6

P

y r

ϕ x Figure 11: 2D Cartesian - relation.

z P R y

θ

x

ϕ r Figure 12: 3D relation.

7

x = r cos ϕ = R sin θ cos ϕ, y = r sin ϕ = R sin θ sin ϕ,

(2)

z = z = R cos θ.

2.4

Distance element

Let P (xi ) and P ′(xi + dxi ), i = 1, 2, 3, be two (infinitesimally) close points. We shall denote the distance between these points as ds and call it distance element. If the coordinates are cartesian, the distance element is given by ds2 = dx2 + dy 2 + dz 2 ,

(3)

and such space is called Euclidean. The expression for the distance element will be different in different coordinate systems. As an example we derive the distance element for spherical coordinates. Let P (r, θ, ϕ) and P ′(r + dr, θ + dθ, ϕ + dϕ) be to close points. If we knew dx, dy, dz in the cartesian coordinates the distance element would be obtained from (3). However, in this case (r, θ, ϕ) are independent variables, while (x, y, z) are functions of these variables. Thus, we have to derive the full differentials, according to the rule (??) and using the relations (2): dx =



∂x ∂R



dR +



∂x ∂θ



dθ +



∂x ∂ϕ



dϕ

(4)

= sin θ cos ϕdR + R cos θ cos ϕdθ − R sin θ sin ϕdϕ, and similarly for dy and dz. After substitution into (3) one gets ds2 = dR2 + R2 dθ2 + R2 sin2 θdϕ2 .

(5)

Q: Derive ds2 in cylindrical coordinates.

3

Curvilinear orthogonal coordinates

In all cases above the distance element had the form ds2 =

X

h2i dx2i

(6)

i=1,2,3

Indeed, let us take the cylindrical coordinates with x1 = r, x2 = ϕ, and x3 = z. One has ds2 = dr 2 + r 2 dϕ2 + dz 2

(7)

that is, h1 ≡ hr = 1,

h2 = hϕ = r,

h3 ≡ hz = 1

(8) 8

For spherical coordinates x1 = R, x2 = θ, x3 = ϕ one has ds2 = dR2 + R2 dθ2 + R2 sin θ2 dϕ2

(9)

and h1 ≡ hR = 1,

h2 ≡ hθ = R,

h3 ≡ hϕ = R sin θ

(10)

The coordinates where the distance element takes the form (6) are called orthogonal coordinates. Cartesian coordinates with h1 = h2 = h3 = 1 are the special case of orthogonal coordinates. If at least one of hi 6= 1 then the coordinates are called curvilinear coordinates. Let us assume that we have previously introduced Cartesian coordinates (x1 , x2 , x3 ) = (x, y, z) but it would be more convenient to use some other orthogonal coordinates (X1 , X2 , X3 ). We have to derive the distance element with the use of these new coordinates. In order to do that we start with ds2 = dx2 + dy 2 + dz 2

(11)

Unless we are given the relation between the new and old coordinates we cannot do anything further. Thus, we assume that the functions x = x(X1 , X2 , X3 ),

y = y(X1 , X2 , X3 ),

z = z(X1 , X2 , X3 )

(12)

are given. If so, we have ∂x dX1 + ∂X1 ∂y dX1 + dy = ∂X1 ∂z dx = dX1 + ∂X1 dx =

∂x dX2 + ∂X2 ∂y dX2 + ∂X2 ∂z dX2 + ∂X2

∂x dX3 , ∂X3 ∂y dX3 , ∂X3 ∂z dX3 , ∂X3

(13) (14) (15)

9

and substituting into (11) one has  ∂x 2 ∂y 2 ∂z 2 ds = ( ) +( ) +( ) dX12 ∂X1 ∂X1 ∂X1   ∂x 2 ∂y 2 ∂z 2 + ( ) +( ) +( ) dX22 ∂X2 ∂X2 ∂X2   ∂y 2 ∂z 2 ∂x 2 ) +( ) +( ) dX32 + ( ∂X3 ∂X3 ∂X3   ∂x ∂x ∂y ∂y ∂z ∂z +2 ( )( )+( )( )+( )( ) dX1 dX2 ∂X1 ∂X2 ∂X1 ∂X2 ∂X1 ∂X2   ∂x ∂x ∂y ∂y ∂z ∂z +2 ( )( )+( )( )+( )( ) dX1 dX3 ∂X1 ∂X3 ∂X1 ∂X3 ∂X1 ∂X3   ∂x ∂y ∂y ∂z ∂z ∂x )( )+( )( )+( )( ) dX2 dX3 +2 ( ∂X2 ∂X3 ∂X2 ∂X3 ∂X2 ∂X3 2



which means 1/2 ∂x 2 ∂y 2 ∂z 2 h1 = ( , ) +( ) +( ) ∂X1 ∂X1 ∂X1  1/2 ∂x 2 ∂y 2 ∂z 2 h2 = ( ) +( ) +( ) , ∂X2 ∂X2 ∂X2  1/2 ∂x 2 ∂y 2 ∂z 2 h2 = ( ) +( ) +( ) ∂X3 ∂X3 ∂X3 

and also gives the conditions which have to be fulfilled for (X1 , X2 , X3 ) to be orthogonal coordinates:  ∂x ∂y ∂y ∂z ∂z ∂x )( )+( )( )+( )( ) = 0, ( ∂X1 ∂X2 ∂X1 ∂X2 ∂X1 ∂X2   ∂x ∂x ∂y ∂y ∂z ∂z ( )( )+( )( )+( )( ) = 0, ∂X1 ∂X3 ∂X1 ∂X3 ∂X1 ∂X3   ∂x ∂y ∂y ∂z ∂z ∂x )( )+( )( )+( )( ) =0 ( ∂X2 ∂X3 ∂X2 ∂X3 ∂X2 ∂X3



10

Physics 1. Mechanics Advanced 1. Curvilinear coordinates

1

Advanced material

Let us assume that we are interested in describing the same n-dimensional space with two different P coordinate systems, xi and x0i . For simplicity we assume that xi are cartesian, that is, ds2 = i dx2i . Of course, there should be a relation between the two systems, that is, xi = xi (x0j ), i = 1, . . . , n, P 2 j = 1, . . . , n. We are interested in the distance element which is ds2 = i dxi in the cartesian coordinates. Now dxi is a full differential: dxi =

X  ∂xi  ∂x0j

j

dx0j ,

(1)

so that ds2 =

X

dx2i

i

!

X X  ∂xi 

X  ∂xi 

dx0j · 0 ∂x ∂x0k j i j k " # X X  ∂xi   ∂xi  = · dx0j dx0k 0 0 ∂xj ∂xk i j,k X = gjk dx0j dx0k =

! dx0k (2)

jk

Now we can forget the initial cartesian coordinates and conclude that the most general form for the distance element is X ds2 = gij dxi dxj , (3) i,j

where gij = gji . The construction gij is called a metric tensor. The metric tensor is symmetric gij = gji , which means that it has 6 independent components in 3-dimensional space. Q: How many independent components has gij in n-dimensional space ? Q: Derive metric tensor for x = x0 − y 0 , y = y 0 . Q: Derive metric tensor for elliptical coordinates ρ, ϕ, where ρ2 = x2 /a2 + y 2 /b2 , tan ϕ = y/x.

1

Physics 1. Mechanics

Advanced 1

In cartesian coordinates gij = δij , where  1 if i = j δij = 0 if i 6= j

(4)

In cylindrical coordinates (1 = r, 2 = ϕ, 3 = z) one has g11 = 1,

g22 = r2 ,

g33 = 1,

(5)

g12 = g13 = g23 = 0. In spherical coordinates (1 = R, 2 = θ, 3 = ϕ) g11 = 1,

g22 = R2 ,

g33 = R2 sin2 θ,

(6)

g12 = g13 = g23 = 0. Coordinates, for which gij = 0 if i 6= j are called orthogonal. The space where it is possible to choose global (that is, which cover the whole space) cartesian coordinates, such that gij = δij everywhere, is called Euclidean. General remarks. The space of special relativity is pseudo-Euclidean, since it is possible to choose coordinates (1 = x, 2 = y, 3 = z, 4 = ct - time) so that g11 = g22 = g33 = 1, g44 = −1: ds2 = dx2 + dy 2 + dz 2 − c2 dt2 . This space is called Minkowsky space. General relativity states that the space is locally Minkowskian.

2

Physics 1. Mechanics Lecture 2. Vectors

1

A simplest vector: Displacement

Let us assume that a particle starts moving from the point P1 (x1 , y1 ) and proceeds subsequently to the point P2 (x2 , y2 ) and further to P (x3 , y3 ) (we shall work in the two-dimensional space for simplicity of graphical representation, see Fig. 1). As a result of the first move the coordinates

y

P1(x1, y1 )

− P1−P→ 2

P2(x2, y2) 3 −−→ P 2P

2 −−→ P 1P

P3(x3, y3) x Figure 1: Displacement vector.

change by (x2 − x1 , y2 − y1 ). The same can be described by saying that the particle moved by p the distance (x2 − x1 )2 + y2 − y1 )2 in certain direction. The two parameters (two coordinates or distance and direction) completely determine the particle displacement from the initial point. We −−→ shall say that a displacement vector P1 P2 is defined either as the two components (x2 − x1 , y2 − y1 ) or the distance+direction (see Fig. 2). The numbers x2 − x1 and y2 − y1 are called the vector x and y components, respectively. In order to assign any physical meaning to this “displacement vector” we have to say what can be done −−→ −−→ with it. First, let us notice that P2 P3 = (x3 − x2 , y3 − y2 ) and P1 P3 = (x3 − x1 , y3 − y1 ) are also −−→ −−→ −−→ displacement vectors, and P1 P3 = P1 P2 + P2 P3 if we define the vector summation as the summation 1

y

P2

y2 y1 P1

x1

x2

x

Figure 2: Displacement vector and coordinates. of the corresponding components: x3 −x1 = (x2 −x1 )+(x3 −x2 ), and similarly for y. This means that moving by some distance in some direction and afterwards by another distance to another direction we as a result move by some third distance to some third direction, and the rule to find this resulting move is known (and independent of the order of the two movements). −−→ Let us denote the distance between P1 and P2 as |P1 P2 | and call that the magnitude of the vector. −−→ If we want to move from the point P1 in the same direction but by a distance λ|P1 P2 |, where λ is some real number, we shall say that the movement is in the same direction, if λ > 0, and in the opposite −−→ direction, if λ < 0. It is easy to see that the same can said as follows: λP1 P2 = (λ(x2 −x1 ), λ(y2 −y1 )), that is, multiplication of a vector by a number is multiplication of all its components by this number. These two operations, vector summation and multiplication by a number, make the construction “something with direction” meaningful. In fact, a construction may be called a vector only if these two operations are properly defined.

2

Vectors more generally

In analogy with the displacement vector we shall call an object A a vector if a) it has several components (3 in the three-dimensional space), b) the summation is defined, and c) multiplication by a real number is defined. We shall write A = (Ax , Ay , Az ) or A = (A1 , A2 , A3 ) or A = (Ai , i = 1, 2, 3). Then, for each A and B, the vector sum C = A + B means that Ci = Ai + Bi , i = 1, 2, 3. For each A and real α, D = αA means Di = αAi (we shall omit i = 1, 2, 3 where this is obvious). The

2

y P3

y3 P2

y2 y1 P1

x1

x2

x3

x

Figure 3: Multiplication by a number important features of the summation and multiplication are the following: A + B = B + A,

(1)

α(A + B) = αA + αB,

(2)

(α + β)A = αA + βA.

(3)

The graphical representation of vectors would be the same as for the case of the displacement vector. Fig. 4 shows the relation between the vector components (Ax , Ay ) (REMEMBER!: it is the c between the vector and x-axis for same as (A1 , A2 ) !), the vector magnitude |A| and the angle Ax the two-dimensional case: c c Ax = |A| cos(Ax), Ay = |A| sin(Ax). (4)

It is also clear that |A| =

p

A2x + A2y .

Geometrically vector summation is shown in Fig. 5.

3

Scalar product

Given two vectors A = (Ai ) and B = (Bi ) (see Fig. 6) one can construct a scalar A·B =

X

Ai Bi .

(5)

i

This construction is called scalar product or dot-product. Scalar is an object which does not depend on the choice of coordinates. Let us check whether the construction (5) is indeed coordinate independent. 3

y

Ay A

x

Ax Figure 4: Vector and components.

c A2 = |A| sin(Ax), c For simplicity we consider a two-dimensional case, for which A1 = |A| cos(Ax), d and B2 = |B| sin(Bx). d Substituting into (5), we have B1 = |B| cos(Bx), c − Bx), d A · B = |A| · |B| · cos(Ax

(6)

d A · B = |A| · |B| · cos(AB).

(7)

c − Bx d = AB d is the angle between A and B, one gets or, noticing that Ax

In the right hand side of the last expression the vector magnitudes and the angle between the two vectors are completely independent of the coordinate choice (coordinates or vector components do not even appear there) which means that the scalar product is also coordinate independent, that is, invariant. This is true not only in the two-dimensional space but in any dimensions. Properties of the scalar product. From the very definition it is clear that A · B = B · A,

(8)

(αA) · B = α(A · B),

(9)

A · (B + C) = A · B + A · C. Vector magnitude (length). From the definition A · A = |A| =

√

A · A.

P

i

Ai Ai =

(10) P

i

A2i = |A|2, so that (11)

This property allows to define a unit vector e such that e · e = 1: for any nonzero vector A/|A| 4

C =A+B

C

A

C

A

B

B Figure 5: Vector summation. will have a unit length. Of special use are the unit vectors along the axes of the cartesian coordinate system ei , i = 1, 2, 3 (or i = x, y, z - the same !). With the help of these unit vectors we can represent any vector as follows (see Fig. 8): A=

X

(12)

Ai ei .

i

Now we have several equivalent representations of a vector: A = (Ai , i = 1, 2, 3) =

X

Ai ei .

(13)

i

It is important to understand that all these representations mean exactly the same. Use the one which is most convenient. Unit vectors and vector decomposition. We saw above that with the use of ex , ey , and ez , a vector A can be decomposed as A = Ax ex + Ay ey + Az ez . The choice of the unit vectors ei in the cartesian coordinates is global, that is, their direction does not depend on the point. However, we can choose three unit vectors in each point of the space independently and decompose a vector in the same way. The three unit vectors do not even have to be orthogonal, that is, the condition ei · ej = 0, if i 6= j, is not necessary (it is very convenient though and we shall use it in what follows). The only necessary condition is that they are independent, that is, if a1 e1 +a2 ej +a3 e3 = 0, then a1 = a2 = a3 = 0. Thus, in general, we can choose three unit vectors ei which are not position dependent (that is, their directions are different in different points) and 5

A

d (AB)

B

Figure 6: Two vectors: notation. decompose a vector as in (13). In practice, such unit vectors are usually related to some coordinate frame, that is, the direction is chosen along the tangential to the curves building the frame. Example - spherical coordinates. In spherical coordinates a curve θ = const, ϕ = const, is a radius, which means that we choose the unit vector e1 ≡ eR parallel to the radius-vector r: eR =

r |r|

= (sin θ cos ϕ, sin θ sin ϕ, cos θ)

(14)

= sin θ cos ϕex + sin θ sin ϕey + cos θez . The vector e2 ≡ eθ is tangential to R = const and ϕ = const, and the vector e3 ≡ eϕ is tangential to R = const and θ = const. They can be found from geometrical considerations but it is instructive to derive these two from the orthogonality conditions. We shall start with eϕ which can be written as eϕ = a1 ex + a2 ey (it is always in a plane parallel to x − y plane since θ = const. The two conditions are eϕ · eR = a1 sin θ cos ϕ + a2 sin θ sin ϕ = 0,

(15)

eϕ · eϕ = a21 + a22 = 1.

(16)

The solution is a1 = − sin ϕ, a2 = cos ϕ, or a1 = sin ϕ, a2 = − cos ϕ. In the special case eR = ex (θ = π/2, ϕ = 0) one has eϕ = ey , which shows that we have to choose the first set: eϕ = − sin ϕex + cos ϕey .

6

z

eˆz eˆx

y

eˆy

x Figure 7: Unit vectors eˆi . The third unit vector can be expressed as eθ = b1 ex + b2 ey + b3 ez and there are three conditions: eθ · eR = b1 sin θ cos ϕ + b2 sin θ sin ϕ + b3 cos θ = 0,

(17)

eθ · eϕ = −b1 sin ϕ + b2 cos ϕ = 0,

(18)

eθ · eθ = 1.

(19)

The second condition gives b2 = b1 sin ϕ/ cos ϕ. Substituting into the first equation, one gets b1 sin θ/ cos ϕ + b3 cos θ = 0, and after substituting into the third condition one would get b3 = − sin θ,

b1 = cos θ cos ϕ,

b2 = cos θ sin ϕ.

Q: Why the sign of b3 is chosen as above and not b3 = sin θ ? Now eθ = cos θ cos ϕex + cos θ sin ϕey − sin θez . With the use of these unit vectors the radius-vector can be written as r = ReR . Exercise. Find er , eϕ , and ez for cylindrical coordinates. Scalar product - scientific notation. We can use (12) and (8)-(10) to write the scalar product as

7

A = Ax eˆx + Ay eˆy

A

y

eˆy x

eˆx

Figure 8: Vector, components, and unit vectors. follows: X X A·B =( Ai ei ) · ( Bj ej ) i

j

X

Ai Bj (ei · ej )

=

X

Ai Bj δij

=

X

Ai Bi .

=

i,j

(20)

i,j

i

Here we used the properties of the unit vectors ei : ei · ej = 1 is i = j and zero otherwise, and have defined the famous Kronecker delta-symbol :  1, if i = j, δij = 0, if i = 6 j.

(21)

Properties of δij . The delta-symbol is fully symmetric: δij = δji . The delta-symbol is used to filter indices: X δij Aj = Ai . (22) j

8

Useful formulae: X

δii = 3,

(23)

X

δij δjk = δik .

(24)

i

j

d one Scalar product - angle between two vectors. From the relation A · B = |A| · |B| cos(AB)

immediately finds:

d =√ cos(AB)

A·B √ . A·A B·B

(25)

Thus, if the components of the two vectors A and B are known it is straightforward to find the angle. Cosine theorem. Let there is a triangle built on two vectors running from the same point, A and B. The third vector is then C = B − A. The length of C can be found using the scalar product: C 2 = (B − A)2 = A2 + B 2 − 2A · B.

3.1

Decomposition for two vectors

Let there are two vectors, A and B, and we want to represent the vector B as a sum B = Bk + B⊥ , where = Bk k A, and = B⊥ ⊥ A (see figure).

B B⊥

Bk

A Figure 9: Decomposition for two vectors.

9

4

Vector product

Besides the above described scalar product, another object can be constructed which can be shown to be a vector. This object is called a vector product C = A × B and is derived according to the following rule C1 = A2 B3 − A3 B2 , C2 = A3 B1 − A1 B3 ,

(26)

C3 = A1 B2 − A2 B1 . From the definition we can see that such vector product can be constructed only in three-dimensional space. From (26) one has e1 × e2 = −e2 × e1 = e3 , e2 × e3 = −e3 × e2 = e1 ,

(27)

e3 × e1 = −e1 × e3 e2 , and using A =

P

i

Ai ei and similar representation for B it is possible to write

e1 e2 e3



C = A × B = det A1 A2 A3 .

B1 B2 B3

(28)

Q: Prove (27) and (28). Properties of the vector product. The following properties of the vector product follow from the definition: A × B = −B × A, (αA) × B = α(A × B),

(29)

A × (B + C) = A × B + A × C. Magnitude and direction. Let for simplicity A and B be in x − y plane, that is, A3 = 0, B3 = 0. We shall also write c A1 = |A| cos(Ax), d B1 = |B| cos(Bx),

Substituting (30) into (28) one finds (see Fig. 10):

c A2 = |A| sin(Ax), d B2 = |B| sin(Bx).

(30)

10

A×B

90◦

B A Figure 10: Direction of the vector product. C = A×B

Q: Prove (31)

d 3. = |A||B| sin(AB)e

(31)

Q: Show that the length of A × B is equal to the area of the parallelogram built on the two vectors. Q: Show that C ⊥ A and C ⊥ B, and A, B, and C obey the right hand rule. Vector product -scientific notation. One most important construction is defined usually: the Levy-Chivita symbol    1, if ijk = 123, or 231, or 312   εijk = −1, if ijk = 132, or 321, or 213 (32)    0, if at least two from ijk are equal.

With this symbol the vector product C = A × B can be written as follows: Ci =

X

εijk Aj Bk .

(33)

j,k

Q: Prove (33) Another, probably even better definition is ε123 = 1 and it is completely antisymmetric: εijk = −εjik = −εikj . In other words, each transposition of two indices multiplies by −1. Q: How many independent components has εijk ?

11

4.1

EXERCISES.

The below exercises with the indices and symbols are very important for better understanding. P P 1. Calculate i δii and ij δij . P 2. Calculate ij (Ai Bj − Aj Bi )δij . P 3. Calculate ijk εijk . P 4. Calculate jk εijk δjk . P 5. Calculate ijk εijk εijk . P 6∗ . Calculate m εmij εmkl . HINT: try X

εmij εmkl = K(δik δjl − δil δjk )

m

and find K. P 7∗ . Calculate mn εimn εjmn . HINT: try X

εimn εjmn = Xδij

mn

and find X.

5

More complex operations with vectors

Let A, B, and C are three arbitrary vectors. We know how to build a scalar from two vectors: A · B. This scalar can be multiplied by the third vector to give another vector (A · B)C which is parallel to C. On the other hand, we can build a vector A × B. This vector can be multiplied by the third vector either to give a scalar: C · (A × B) or to give a new vector C × (A × B) Q: Show that |C · (A × B)| is a volume of the parallelepiped built on the three vectors. Q: Show that C · (A × B) = (C × A) · B

12

Q∗ : Derive the formula: A × (B × C) = B(A · C) − C(A · B).

(34)

ATTENTION: this expressions are very important and useful and we will need them in future.

6

Displacement vector and curvilinear coordinates

Displacement vector may be used for derivation of the scale factors hi in curvilinear coordinates. Indeed, dr 2 = ds2 , thus calculating 2  ∂r ∂r ∂r )dX1 + ( )dX2 + ( )dX3 dr = ( ∂X1 ∂X2 ∂X3 ∂r 2 2 ∂r 2 2 ∂r 2 2 ) dX1 + ( ) dX2 + ( ) dX3 =( ∂X1 ∂X2 ∂X3 ∂r ∂r ∂r ∂r ∂r ∂r + 2( )·( )dX1 dX2 + 2( )·( )dX1 dX3 + 2( )·( )dX2 dX3 ∂X1 ∂X2 ∂X1 ∂X3 ∂X2 ∂X3 2

we have ∂r 2 ∂r 2 ∂r 2 ) , h22 = ( ) , h23 = ( ) , ∂X1 ∂X2 ∂X3 ∂r ∂r ∂r ∂r ∂r ∂r )·( )=( )·( )=( )·( )=0 ( ∂X1 ∂X2 ∂X1 ∂X3 ∂X3 ∂X3

h21 = (

This becomes even more obvious if we write r ∂r ∂X1 ∂r ∂X2 ∂r ∂X3 ∂r 2 ) ( ∂X1

ˆ + y yˆ + z z, ˆ = xx ∂y ∂z ∂x ˆ+ ˆ x yˆ + z, = ∂X1 ∂X1 ∂X1 ∂y ∂z ∂x ˆ+ ˆ x yˆ + z, = ∂X2 ∂X2 ∂X2 ∂y ∂z ∂x ˆ+ ˆ x yˆ + z, = ∂X3 ∂X3 ∂X3 ∂x 2 ∂y 2 ∂z 2 =( ) +( ) +( ) , ∂X1 ∂X1 ∂X1 ...

and compare the obtained expressions with Lecture 1.

13

Physics 1. Mechanics Advanced 2. Coordinate transformation and vectors, tensors

1

Advanced material

Let us consider two cartesian coordinate frames (x, y) (“old”) and (x
, y
) (“new”), which are related by a rotation to the angle θ (see Fig. 1). Let A is a vector in this plane. This vector can be represented by its components in both frames A = (Ax , Ay ) = (A
x , A
y ). Obviously, the should be some relations between the old and new coordinates, old and new vector components, and old and new unit vectors (ex , ey ) and (e
x , e
y ). From the figure one can find: x = x
cos θ − y
sin θ,

y = x
sin θ + y
cos θ,

Ax = A
x cos θ − A
y sin θ, ex = e
x cos θ − e
y sin θ,

Ay = A
x sin θ + A
y cos θ, ey = e
x sin θ + e
y cos θ.

We see that the vector components transform similarly to the coordinate transformation.

Q: Show that i Ai ei = i A
i e
i .

(1) (2) (3)

This is a basic feature of vectors and another, in many cases better, definition of a vector. In a more general way, let x
i = x
i (xj ) in an n-dimensional space. The differentials dx
i depend on the

Figure 1: Two coordinate systems - same vector.

1

Physics 1. Mechanics

Advanced 2

differentials dxj as follows: dx
i

=

  ∂x
 i

∂xj

j

dxj ,

(4)

which is just the rule for the full differential. Then a vector Ai , i = 1, . . . , n, is an object which transforms like the differentials: A
i

=

  ∂x
 i

∂xj

j

Aj .

(5)

A vector in n-dimensional space has n components. It is possible to define a more general construction which is called tensor. For example, a 2-rank tensor is a two-index object Tij (n2 components) which transforms as follows: Tij


=

  ∂x
  ∂x
j  i

∂xk

kl

∂xl

Tkl .

(6)

In previous lecture we introduced the metric tensor gij . Let us check whether it is indeed a tensor.
The distance element ds2 = ij gij dxi dxj is invariant under coordinate transformations, that is, ds2 is independent of the coordinate choice. Therefore, writing the same distance element in two different coordinate frames, one has 

gij dxi dxj =

ij




gkm dx
k dx
m ,

km

and using dxi =

  ∂xi  k

we obtain

dxj =




km

∂x
k

ij

  ∂xj  m

    ∂xi   ∂xj  km

Comparing this with

∂x
k

dx
k ,

∂x
m

∂x
m

dx
m ,



gij dx
k dx
m .


gkm dx
k dx
m we find that
gkm

=

  ∂xi   ∂xj  ∂x
k

ij

∂x
m

gij ,

(7)

which does not look exactly as (6). Nevertheless, gij is a tensor. In fact, in curvilinear coordinates exist tensors (and vectors too) of two kinds (they are not completely independent but related through gij ). We shall call them up and down (these are not real names which we will not use now). For example, up-vector transforms as (5), while down-vector transforms as follows: A
i

=

  ∂xj  j

∂x
i

Aj .

(8)

2

Physics 1. Mechanics

Advanced 2

ATTENTION: index i in both expressions goes with the primed component A
i and coordinate x
i . Example - two-dimensional cartesian coordinates. For convenience let us introduce new notation: Lij =



∂x
i ∂xj



,

L
ji

=



∂xj ∂x
i



.

(9)

Then the up- and down-vectors transform as follows: up: A
i =



down A
i =

Lij Aj ,

j



L
ji Aj .

(10)

j

As in the beginning we consider two cartesian frames related by rotation to the angle θ: x1 = x
1 cos θ − x
2 sin θ,

x2 = x
1 sin θ + x
2 cos θ,

(11)

x
1 = x1 cos θ + x2 sin θ,

x
2 = −x1 sin θ + x2 cos θ.

(12)

Using these relations, we obtain L11 = L22 = L
11 = L
22 = cos θ,

(13)

L12 = −L21 = −L
12 = L
21 = − sin θ. The up- and down-vectors transform as (see (10)) A
1 = L11 A1 + L12 A2 ,

A
2 = L21 A1 + L22 A2 ,

up

(14)

A
1 = L
11 A1 + L
21 A2 ,

A
2 = L
12 A1 + L
22 A2 ,

down

(15)

and, because of (13), the two transformation laws are the same. Thus, there is no difference between up- and down-vectors in this case. Transformations, for which Lij = L
ji , that is, (∂x
i /∂xj ) = (∂xj /∂xi ), are called orthogonal. Rotations are orthogonal transformations.

3

Physics 1. Mechanics Lecture 3. Motion

1

General concepts

Any change of the position with time is called motion. This “definition” implies that we have a good method to measure time everywhere and always. For the time being let us assume that indeed such method exists so that we can observe xi (t), that is, coordinates of a particle are functions of time. If the coordinates do not change we say that the particle is in the rest and consider that as a special case of motion. By providing xi (t) we describe the motion completely, since we know where is the particle at each moment of the time. However, sometimes we want to know some particular things about the motion, for example, how quickly the particle position is changing. For the description of this “quickly” we define the velocity vi ≡ (dxi /dt). It is important to understand that, in general, the velocity is defined using coordinates and can differ from our usual perception of the speed. Thus, in cartesian coordinates one has the following components of the velocity: (dx/dt), (dy/dt), and (dz/dt). In spherical coordinates the components are (dr/dt), (dθ/dt), and (dϕ/dt), that is, even do not have the same dimensions. If necessary, we shall emphasize this by calling this velocity a coordinate velocity, and the other one (see below) a physical velocity. In general relativity, however, it is not always possible to define globally the physical velocity as we understand it in our everyday life (measured in km/h, for example), and one has to use the velocity introduced above. In a similar way one introduces acceleration ai ≡ (d2 xi /dt2 ) which shows how quickly the velocity is changing. One might go further and introduce higher derivatives. However, this is useful only in very few cases, because of the structure of classical mechanics, where external influence (force) determines acceleration. Once we know xi (t), a simple procedure (derivation) gives us subsequently the velocity and acceleration. The inverse procedure is a little bit more complicated but still remains only a technical one. Namely, if we know vi (t) and the initial conditions xi (t = t0 ) = xi0 , one immediately gets: xi (t) = xi0 +




t

t0

vi (t
)dt
.

(1)

ATTENTION: for not to make a mistake one has to make clear the distinction between the moment of time t in which we would like to have xi (t), and the time variable t
which is the integration 1

Physics 1. Mechanics

Lecture 3

variable only. Similarly, if we know the acceleration ai (t) and the initial conditions vi (t = t0 ) = vi0 , we have: vi (t) = vi0 +




t

t0

ai (t
)dt
.

(2)

Calculation of the distance is a little more complicated since we have to know how to write the distance element ds. We shall delay the discussion of this until the next session where our ordinary space is considered (in which it is always possible to choose cartesian coordinates). It has to be mentioned, however, that both velocity and acceleration are vectors. All points which are passed by the particle in space constitute a trajectory (or path or orbit). Examples of trajectories are shown in Fig. 1

y

x Figure 1: Trajectories in two-dimensional (left) and one-dimensional (right) cases.

2

Alternative approach

An alternative approach is usually used in the case where cartesian coordinates are possible (but not necessarily chosen). In our three-dimensional space we can define a displacement vector (see previous lecture). We can go even further and determine a particle position by the displacement vector from the origin of the coordinates to the point where the particle is now. Following general principles this vector is P0 P , where P = (x, y, z) (in cartesian coordinates) and P0 = (0, 0, 0) (origin !). Thus, P0 P = (x, y, z). This vector is denoted r and is called a position vector or radius-vector. Using this vector, we can define velocity and acceleration in a more concise way: v=

dr , dt

a=

dv d2 r = 2. dt dt

(3)

2

Physics 1. Mechanics

Lecture 3

Now (1)-(2) will take the following shape: r(t) = r0 +




v(t) = v0 +

t t0




v(t
)dt
, t

t0

(4)

a(t
)dt
,

(5)

It is now even sufficiently easy to calculate the distance. Indeed, from ds2 = coordinates), one has  1/2   dxi 2  ds = =( vi2 )1/2 = (v · v)1/2 = |v|, dt dt i i so that s=




t

t0

|v|(t
)dt
.



i

dx2i (in cartesian

(6)

(7)

In practice, in order to calculate the distance one always has to substitute |v| =



dx dt

2

+



dy dt

2

+



dz dt

2 1/2

.

(8)

In spherical coordinates one has to use |v| =



dr dt

2

+ r2



dθ dt

2

+ r 2 sin2 θ



dϕ dt

2 1/2

.

(9)

Q: Write |v| in cylindrical coordinates.

2.1

Usage of unit vectors

The velocity vector can be written as v = xe ˙ x + ye ˙ y + ze ˙ z. Here we use the short notation A˙ ≡ dA/dt for any A. Instead of using the cartesian unit vectors one could use another set of those, similarly as we did in Lecture 2. We shall show how it is done on the example of two-dimensional polar coordinates. We define er = r/r as a unit vector along the radius-vector (tangential to the curve ϕ = const, and eϕ as a unit vector tangential to r = const (there is no z - two-dimensional case). It is easy to find er = cos ϕex + sin ϕey ,

eϕ = − sin ϕex + cos ϕey .

(10)

3

Physics 1. Mechanics

Lecture 3

From these expressions we find ex = cos ϕer − sin ϕeϕ ,

ey = sin ϕex + cos ϕey .

(11)

Now the velocity v=

dr = xe ˙ x + ye ˙ y, dt

(12)

where x = r cos ϕ,

y = sin ϕ,

(13)

and x˙ = r˙ cos ϕ − r sin ϕϕ, ˙

(14)

y˙ = r˙ sin ϕ + r cos ϕϕ. ˙ Substituting (11) and (14) into (12) we have ˙ ϕ. v = re ˙ r + r ϕe

(15)

Q: Express v in terms of eR , eθ , and eϕ (spherical coordinates).

3

Motion - examples

3.1

Constant acceleration

Let a = const. This means that neither the magnitude nor the direction of the acceleration do not change. From (4)-(5) one has v = v0 + a(t − t0 ),

(16)

r = r0 + v0 (t − t0 ) + 12 a(t − t0 )2 .

(17)

Q: Prove these expressions. The most famous example of such motion is the motion near the earth surface, where the free-fall acceleration g = const. Let us choose the coordinate system as in Fig. 2. Then g = (0, −g) and one has for the components and coordinates vx = v0 cos α,

(18)

vy = v0 sin α − g(t − t0 ),

(19)

x = x0 + v0 cos α(t − t0 ),

(20)

y = y0 + v0 sin α(t − t0 ) − 12 g(t − t0 )2 .

(21) 4

Physics 1. Mechanics

Lecture 3

y g

v0 α

x Figure 2: Motion near the earth surface.

Trajectory Trajectory (path) is the curve along which the particle moves. The equation for the trajectory can be obtained by excluding time from the equation for x and further substituting into the equation for y: g(x − x0 )2 . (22) y = y0 + (x − x0 ) tan α − 2 2v0 cos2 α Q: Derive (22). Distance In order to calculate the distance along the path (trajectory) we have to integrate s= =




t

t
0t t0

3.2

|v(t
)|dt
 v02 cos2 α + (v0 sin α − g(t
− t0 ))2 dt
.

(23)

Circular motion

This motion is in a plane. It is natural to use polar coordinates. In this coordinates r = const, while the angle ϕ = ϕ(t) is time dependent. The derivative ω = dϕ/dt is (naturally) called angular velocity (if the radius r is changing, dr/dt is called radial velocity). The velocity magnitude is found from |v| = (dr/dt)2 + r 2 (dϕ/dt)2 = ωr. In most cases, however, this motion is described in terms of r and v. In order to do that let us write down the relation between the cartesian and cylindrical coordinates of the particle: x = r cos ϕ,

y = r sin ϕ,

z = z,

(24)

5

Physics 1. Mechanics

Lecture 3

from which (with the use of simple differentiation) one gets vx = −ωr sin ϕ,

vy = ωr cos ϕ,

vz = 0.

(25)

Now we introduce the angular velocity vector ω = ωez and notice that (25) can be written as v = ω × r.

(26)

The physical sense of this vector is that it shows the rate of the angle change and also the direction of the rotation axis. Further differentiating (26) with respect to time one has dω × r + ω × (dr/dt) dt dω = ×r+ω×v dt dω = × r + ω × (ω × r) dt

a=

(27)

Let us now write r = r + r⊥ , where r = zez  ω, r⊥ = xex + yey ⊥ ω. Then application of the formulae of the vector algebra gives a=

dω × r − ω 2 r⊥ . dt

(28)

Q: Prove (28). In the special case when the angular velocity does not change ω = const the centripetal acceleration is a = −ω 2 r⊥ . Let us now consider a more general case when ω = ω(t)ez . In the polar coordinates r⊥ = rer , and ez × er = eϕ , so that (28) will give a = ωe ˙ ϕ − ω 2rer .

(29)

The first term in the fight hand side of (29) is called tangential acceleration, the second term is called centripetal acceleration.

4

General

A particle motion is given by r(t) and the velocity is v = dr/dt. From the very definition the velocity is tangent to the trajectory. What about the acceleration ? The acceleration is defined as a = dv/dt, and, in general, it is neither parallel nor perpendicular to the velocity. However, we can decompose

6

Physics 1. Mechanics

Lecture 3

it similarly to what has been done in the case of the circular motion. For this to do let us write v = vˆ v,

(30)

ˆ = v/v is the unit vector in the direction of the velocity. where v = |v| is the velocity magnitude, and v Of course, so far we have not done anything new by writing the identity (30). Now, however, the acceleration can be written as follows: a=

d(vˆ v) dv dˆ v ˆ+v . = v dt dt dt

(31)

ˆ  v. The second term is perpendicular to the We see that the first term in (31) is parallel to v ˆ·v ˆ = 1 one has velocity. Indeed, since v d dˆ v ˆ ) = 0 = 2ˆ (ˆ v·v v· . dt dt ˆ . Thus, we have decomposed the acceleration into the tangential and which means that dˆ v/dt ⊥ v normal components: dv ˆ, v dt dˆ v an = v . dt

at =

(32) (33)

The first one shows the change in the velocity magnitude, the second one gives the change of the direction. The centripetal acceleration is the special case of the normal acceleration when a particle moves along a circle. In the circular motion the magnitudes of the velocity and centripetal acceleration are related through the circle radius: |a| = v 2 /R. We can define similarly the curvature radius of the trajectory in the general case as R = v 2 /|an |, so that R=

1 dˆ v . v dt

Let us now pay attention that the vector dˆ v is a unit vector, while the vector dˆ v/dt is not, in general. We can build a new unit vector v/dt ˆ = dˆ , N |dˆ v/dt|

(34)

which is called the vector of the normal. In order to complete these two vectors we have to add a ˆ=v ˆ ˆ × N. third one, which can be done as b

7

Physics 1. Mechanics Advanced 3. Motion in curvilinear coordinates

1 1.1

Advanced material General coordinates

In general coordinates vi = dxi /dt and ds2 =

P

ij

dxi dxj . Therefore,

X dxi dxj ds =( gij )1/2 dt dt dt ij X =( gij vi vj )1/2

(1)

qP Thus, the velocity magnitude is |v| = ij gij vi vj . However, is velocity a vector ? From the previous lecture we know what should be the transformation law when we change coordinates. Indeed, vi0

dx0i X ∂x0i dxj X ∂x0i = = = vj , dt ∂xj dt ∂xj j j

provided that the time t does not change when we change the coordinates. This means that the velocity defined as vi = dxi /dt is a vector only if t is invariant under coordinate transformation. In nonrelativistic classical mechanics t is an absolute time, which is the same for any coordinate choice, so that vi is a vector. If xi = xi (t) this defines a one-dimensional curve in the space. Is the choice of time the only method to define a curve ? Obviously, any substitution t = t(τ ) is also good, if there is one-to-one correspondence. Since s(t) is a monotonic function, s can be used for the curve parametrization with equal success. Let us say that s is the new “time” and see what happens if we define new velocity as dxi /ds. The only effect of our choice would be in the magnitude of the velocity: v·v =

X

gij

ij

1

dxi dxj = 1. ds ds

Physics 1. Mechanics

1.2

Advanced 3

Differentiation along the curve

In section 1.1 we encountered the derivative of the kind v(dA/dt) (where A is some vector). Such derivative is called the derivative along the curve. Using the definition of the curve length ds/dt = v we can rewrite such derivative in terms of derivation with respect to the path: v

d d ˆ . =v dt ds

With this definition the curvature radius takes the form dˆ v −1 R = . ds

(2)

(3)

Since we live in a three-dimensional space, any curve is three-dimensional, in general, and one curvature radius is not sufficient to describe the curve behavior. Indeed, a second curvature is defined related to the twisting db ˆ (4) T = ds ˆ and N). ˆ (see section 1.1 for the definition of b The first and second curvatures are defined as C1 = 1/R, C2 = T , and the total curvature is C 2 = C12 + C22 . ˆ ×v ˆ =b ˆ and differentiating with respect to s show that Exercise: Using N dN ˆ C= . ds

2

Physics 1. Mechanics Lecture 4. Relativity

1

Reference frames

A reference frame is an observer with a coordinate system. This implies that a) some coordinates are chosen to describe the position of any particle, b) there is some physical body relative to which the motion is considered (it is possible to perform physically meaningful measurements), c) there is a method of time measurements. The last means that a clock should exist in each point of the space, and all these clocks should be synchronized. They are synchronized by sending some signal from some reference clock to all others, telling them what is the correct time in the reference point. Such procedure requires that the clocks in all points of the space be identical. In Newtonian mechanics the signal speed is assumed to be infinite, so that the synchronization is straightforward and easy. Let us now assume that there are two frames, S and S
, so that S
moves with the velocity V relative to S. We also assume that both observers put themselves in the coordinate origin of each of these systems, and both choose cartesian coordinates building the axes so that x
is parallel to x, y
is parallel to y, and z
is parallel to z. If the signal velocity is infinite it is possible to synchronize the clocks in S and S
as well so that the two observers measure the same time: t = t
. On the other hand, it is clear (see Fig. 1) that for each point P , r = r
+ R, where r is the radius-vector of P measured by the observer S, r
is the radius-vector of P measured by the observer S
, and R is the radius-vector of S
(coordinate origin) measured by the observer S. Thus, one has the Galileo transformation in the following form: t = t
,

r = r
+ R.

(1)

Differentiating (1) with respect to time t and taking into account that t = t
, we immediately find v = v
+ V,

(2)

where v = dr/dt is the velocity of P measured by S, v
= dr
/dt
is the velocity of P measured by S
, and V = dR/dt is the velocity of S measured by S
, that is, the relative velocity of the two frames. The two frames, S and S
, are said to be related by the Galileo transformation if V = const. In Newtonian mechanics the following Galileo relativity principle holds: all physics laws have the same shape in two systems related by the Galileo transformation. In particular, this means that the second Newton law looks identical, F = ma and F
= m
a
, in both frames. Indeed, mass m is a scalar 1

Physics 1. Mechanics

Lecture 4

V y
y

r
r x


R x Figure 1: Galilean transformations.

and the same in both frames. The acceleration can be obtained by differentiating (2) with respect to time,which gives a = a
(if V = const). From this we derive the law of the force transformation F = F
. In what follows we assume that the frame S is inertial. This, in particular, means that the force in the second Newton law is caused by other physical bodies. In the absence of such bodies there is no any force and a = 0. The frame S
is also inertial if V = const. What happens if the frame S
is non-inertial, that is V is time-dependent ? Direct differentiation of (2) gives a = a
+ A,

(3)

where A is the acceleration of the frame S
measured by S. Substituting in the Newton law, we have Fext = ma = ma
+ mA, where the index ext emphasizes the nature of the force: it is produced by other physical bodies. If we want to have the Newton law i in the accelerated frame S
similar to its usual form, that is, F
= ma
, we have to define F
= Fext − mA. The second term in this expression is the inertia force which is not caused by any physical body but is due to the the acceleration of the frame.

2

Physics 1. Mechanics

Lecture 4

a y


F = −ma

y

r
r x


R x

Figure 2: Inertial force in the accelerated system. d This expression can be also considered as the transformation rule for the force.

1.1

Examples of inertial forces

Hanging body. Let a body hang on a rope from a ceiling (see Fig. 3) and let the ceiling move

θ −ma mg

Figure 3: Hanging body in non-inertial frame. with constant acceleration a. What would be the angle θ between the rope with the vertical ? The simplest (and most effective) way is to consider the situation in the non-inertial accelerated frame, where the position of the body is determined by the balance of the rope tension T (directed along the rope), vertical gravity force mg and the inertia force −ma. This means that the vector sum g + (−a) should be directed along the rope. It is clear now that tan θ = |a|/|g|. Sliding down accelerated slope. What should be the acceleration of the slope to prevent sliding of the body (see Fig. 4)? In the accelerated frame the vector sum of the gravity force mg and inertia 3

Physics 1. Mechanics

Lecture 4

−ma

mg α

Figure 4: Sliding down accelerated slope. force −ma should be normal to the slope, which gives |a| = |g| tan α.

2

Rotating frames

A special and one of the most important case of non-inertial frames is a rotating frame. Let S be again an inertial (“standing”) frame and S
rotates with the constant angular velocity ω around the axis z (z
= z), so that the coordinate origins in S and S
are the same (see Fig. 5). According to

y y
x


x Figure 5: Standing x, y, z and rotating x
, y
.z
frames. our definition of the angular velocity, ω = ωez . Let in the moment t the angle between x and x
axes is ϕ. Let r and r
be the radii-vector of the particle P as measured in S and S
, respectively. Attention: although both are shown by the same “arrow” in the figure, the vectors cannot be said to be the same since the components are different,

4

Physics 1. Mechanics

Lecture 4

instead we have to say that r = r
r , where the index r means “rotated from S
into S”. In the following we shall not emphasize this difference which should, nevertheless, be kept in mind. Our objective is to establish the relation between the velocities and accelerations measured in both frames. Let us start with the simplest case where the particle P does not move at all in the rotating frame, v
= 0. In this case, from the point of view of S, the particle P simple experiences circular motion with the constant angular velocity. We know that the observer S would measure the velocity v = ω × r, and the acceleration a = ω × v, The nature of this acceleration is simply the rotation of the velocity vector. The radius-vector r and the velocity vector v rotate by the angle ωdt during the infinitesimal time dt. Since the rotation of the both vectors is the same, the relation between the velocity change dv and v is the same as between dr and r. The latter can be found from what we already know: dr = vdt and v = ω × r, so that dr = ω × rdt. We conclude that for the velocity vector one would have dv = ω × vdt. Moreover, for any vector A rotating with S
we will have in the standing frame S: dA = ω × Adt. In a more general way, if a vector A is changing in the rotating frame then dA
dA = +ω×A (4) dt dt where (dA/dt) is the rate of the vector change as measured by the standing observed and (dA
/dt) is the rate of the vector change as measured by the rotating observer and rotated into the standing frame. Let now the particle P move in the frame S
with the velocity v
in the moment t. The observer S will measure in the same moment the velocity v = v
+ ω × r. During dt this velocity changes by dv
in the frame S
. The observer S will measure the velocity change dv which is due to a) the change dv
, b) the change due to the rotation of the velocity vector ω × vdt = ω × (v
+ ω × r), and c) due to the change of the rotational term d(ω × r)m = ω × dr = ω × dr
= ω × v
dt =. The last contribution simply says that the particle is on another circler in the moment t + dt, so that its rotation velocity is not the same as before. Combining all these contributions together, dividing by dt and taking into account that dv/dt = a, dv
/dt = a
, we finally obtain a = a
+ 2ω × v
+ ω × (ω × r).

(5)

The last term is already known to us: it is the centripetal acceleration and can be written as −ω 2 r. The second term is called Coriolis acceleration. If now we wish to write the second Newton law in the usual form F
= ma
, we have to write F
= Fext − 2mω × v
+ mω 2 r
,

(6)

where the first term is the force cause by other bodies, the second term is called the Coriolis force, and the last one is the centrifugal force. The last two are inertia forces and are caused by the frame rotation. 5

Physics 1. Mechanics

Lecture 4

The below figures show once again the change of the velocity in the rotating and nonrotating frame.

y


y


t

t + dt

v


v


r


r
x


x


Figure 6: Velocity change in the rotating frame. The left figure shows the position r
(blue) and velocity v
(red) of the particle in the moment t. The right panel shows what happens in the moment t + dt. The particle moved into the new position r
+ v
dt and has the new velocity v
+ ∆v
.




y

y

t

y

t + dt




y v

v




x




r

x

r

x

x

Figure 7: Velocity change in the nonrotating frame. In order to find the velocity in the nonrotating frame we have to rotate the velocity in the rotating frame by the angle ϕ and add ω × r. The left figure shows the velocity in the moment t: the velocity v
(red) is rotated by the angle ϕ, the position r
(blue) is rotated by the same angle (and gives r) and ω × r (green) is added. The total velocity v is shown in magenta. The right figure shows the velocity in the moment t + dt: v
+ dv
and r
+ dr
are rotated by the angle ϕ + dϕ, and ω × (r + dr) is added.

6

Physics 1. Mechanics Advanced 4. Rotations, general noninertial frames

1

Advanced material: Rotations

Two cartesian coordinate systems S(x, y, z) and S 0 (x0 , y 0 , z 0 ) with the common origin can be transformed one into another by a rotation. Let us start with the rotation around z axis by the angle ϕz . Then the transformation reads x0 = x cos ϕz + y sin ϕz , y 0 = −x sin ϕz + y cos ϕz ,

(1)

z 0 = z. It is convenient (and not only convenient !) to represent this transformation in a matrix form. Let us define row   ˜r = x y z , and column   x   r = y  , z and similarly for S 0 . The transformation (1) can be now written as follows:  0    x cos ϕz − sin ϕz 0 x  0    y  =  sin ϕz cos ϕz 0 y  z0 0 0 1 z

(2)

or in the following symbolic form r0 = Rz (ϕz )r,   cos ϕz − sin ϕz 0   Rz (ϕz ) =  sin ϕz cos ϕz 0 . 0 0 1 1

(3) (4)

Physics 1. Mechanics

Advanced 4

The construction Rz (ϕz ) is called an operator, and (4) is the matrix form of the rotation operator. In a similar way we can find the operators (matrices) for rotations around x and y axes:  1 0  Rx (ϕx ) = 0 cos ϕx 0 sin ϕx  cos ϕy  Ry (ϕy ) =  0 − sin ϕy

 0  − sin ϕx  cos ϕx  0 sin ϕy  1 0  0 cos ϕy

(5)

(6)

If we want to rotate, say, first around z by ϕz and after that around x by ϕx , the result will be r00 = Rx (ϕx )r0 = Rx (ϕx )Rz (ϕz )r.

(7)

In the component form (3) is written as follows: x0i =

X

Rij xj .

(8)

j

For the rows one would have ˜ r˜0 = ˜rR,

(9)

X

(10)

(remember matrix multiplication !) or x0i =

˜ ji . xj R

j

Thus, ˜ ji = Rij R

→

˜ = RT , R

(11)

T where RT is the transposed matrix: Rij = Rji . P 0 0 P For any rotation i xi xi = k xk xk , which gives

X k

x0k x0k =

X

Rki xi Rkj xj =

ijk

X

T Rkjxi xj = Rik

ijk

X

δij xi xj .

(12)

ij

Thus, X

T Rik Rkj = δij ,

(13)

RT R = I,

(14)

k

or

where I is the unit matrix (all diagonal elements equal to one, all nondiagonal equal to zero).

2

Physics 1. Mechanics

Advanced 4

˜ such that RR ˜ = I, the matrix R ˜ is called the Definition. If for a matrix R exists a matrix R inverse matrix and denoted R−1 . Thus, for rotations we have R−1 = RT . Since the matrix product is noncommutative, that is, for two matrices A and B, in general, AB 6= BA, the result of two successive rotations depends on their order. However, if we consider infinitesimal rotations ϕ → dϕ, the situation is different. Using sin(dϕ) → ϕ, cos(dϕ) → 1, we obtain Ri (dϕ) = I + dϕJi , (15) where 

0  J x = 0 0  0  J y = 0 1  0  J z = 0 0

 0 0  0 1 −1 0  0 −1  0 0 0 0  1 0  −1 0 0 0

(16)

(17)

(18)

Q: Prove that Ri (dϕi )Rj (dϕj ) = Rj (dϕj )Ri (dϕi ). Now we can define the infinitesimal rotation angle vector as dϕ = (dϕx , dϕy , dϕz ) and a vector of matrices R = (Rx , Ry , Rz ) (and similarly for J ), so that the general infinitesimal rotation will be r0 = r + (J · dϕ)r,

(19)

where J · dϕ is an operator which acts on r. Q: Prove that (19) can be written as dr = dϕ × r.

2

Advanced material: noninertial frames

The Galilean relativity (Newtonian mechanics) state that there is a class of special frames (inertial frames) in which the motion laws have the same shape. What happens in noninertial frames and why they are worse than inertial ones ? Actually, if we want that the physical laws be universal we should find a way to describe these laws in any frame. We already know how to include inertia forces in a uniformly accelerated frame and in a uniformly rotating frame. Now let us try to do the same in an arbitrary frame. In order to do that we shall write down the second Newton law in an inertial

3

Physics 1. Mechanics

Advanced 4

frame S(xi ) (where xi are cartesian coordinates) in the form m

d2 r = F, dt2

(20)

where F is an external (“true”) force, caused by other bodies. Let the observer in a noninertial frame S 0 also chooses cartesian coordinates x0i , and we know the relation r(r0 , t). The dependence on t shows that this is not a simple coordinate transformation but the relation between the coordinates in two frames with some relative motion. The only transformation between the two frames, which conserves the scale, is the time dependent translation (this is not precise, we shall look into it more closely later) and rotation: r = X + Rr0 , (21) where X(t) is the radius-vector of the coordinate origin of S 0 according to the observer S. Now one has dR 0 r + Rv0 , (22) v=V+ dt where V is the relative velocity of the two frames as measured by S. Differentiating once again with respect to time we get d2 R dR a = A + 2 r0 + 2 v0 + Ra0 , (23) dt dt where A = dV/dt. Left multiplying by R−1 and taking into account (20) we have finally ma0 = R−1 Fext − mR−1 A − mR−1

d2 R 0 dR r − 2mR−1 v0 . 2 dt dt

(24)

The first term in the right hand side of (24) is the true force measured in S 0 , the second term is the generalization of the inertia force due to the relative motion of the coordinate origin of the two frames, the third term is the generalized centrifugal force, and the last term is the generalized Coriolis force. We see that there no other inertia forces. Exercise: Derive the usual centrifugal and Coriolis force for A = 0 and 

 cos(ωt) − sin(ωt) 0   R =  sin(ωt) cos(ωt) 0 0 0 1

(25)

4

Physics 1. Mechanics Lecture 5. Particle dynamics I

1

Newton’s laws, momentum, kinetic energy

This section is to remind the basics of the particle dynamics. We shall proceed in an inertial frame, so that there are no inertia forces. The second Newton law says that the equation is proportional to the force F = ma.

(1)

p = mv,

(2)

Momentum p is defined as follows:

which immediately gives the second Newton law in a more general form: F=

dp . dt

(3)

Now the momentum change ∆p = p2 − p1 = Kinetic energy is a scalar: K=




t2 t1

F(t)dt.

mv 2 mv · v = . 2 2

(4)

(5)

The rate of change of the kinetic energy dv dK = mv · = mv · a = (ma) · v = F · v. dt dt

(6)

During the time dt the energy change is dK = F · vdt = F · dr = dW.

(7)

The scalar quantity dW is called the infinitesimal work. If the particle moves from r1 to r2 the kinetic energy change ∆K is equal to the work which is done along the trajectory: ∆K = W =




r2

r1

1

F · dr.

(8)

Physics 1. Mechanics

Lecture 5

The last integral is taken along the trajectory. It, and therefore, the work depends, in general, on the path (and not only on the initial and final points r1 and r2 ). The work produced in unit time P = dW/dt is called power. For our case it is P = F · v.

(9)

Third Newton law states that if two particles interact then F1→2 = −F2→1 . Here F1→2 is the force which particle 1 exerts on particle 2. This has to be completed by the superposition principle: if there are a number of particles acting on particle i the total force which particle i experiences is the vectorial sum of the forces from all other particles: Fi =



Fj→i .

(10)

j

Two additional important concepts are the concept of a physical system which may include a number of particles, and the concept of the resulting force acting on the whole system, which is the vectorial sum of all the forces acting on all parts of the system. If the system consists of a number of particles, the resulting force would be F=



(Fi,ext +

i



Fj→i ),

(11)

j

where Fi,ext is the external force (that is, not by any particle from the same system) on particle i, and Fj→i is the force exerted by particle j on particle i (both i and j belong to the physical system under consideration). Using the third Newton law Fi→j = −Fj→i we obtain F=



Fi,ext ,

(12)

i

that is, the internal forces do not contribute to the resulting forced acting on the system as a whole.

2

Work as a path integral

Work is calculated differently from the integrals we have seen before. In the expression W =




2 1

F · dr

the integral is done along the curve. The force can be a function of r and t, in general, F = F(r, t) and in this case the work can be calculated only if we know r(t) as follows W =




t2

t1

F(r(t), t) · v(t)dt,

(13)

2

Physics 1. Mechanics

Lecture 5

so that the integrand becomes a function of only one variable t. In many cases, however, F = F(r) and is independent of t. If this is the case, the path (trajectory) should be expressed in a parametric form.

2.1

Example: friction

We assume the a body is moving in a horizontal plane so that the friction force magnitude is constant while it is always directed against the velocity F = −f v/v. Now F · vdt = −f vdt = −f ds, where ds is the length of the path. Thus, W = −f s, where s is the length of the whole path made by the particle.

2.2

Example: constant force

For a constant force F one has W =




1

2

F · dr = F ·




1

2

dr = F · (r2 − r1 )

and does not depend on the path but only on the initial and final positions. This is an example of a conservative force for which work does not depend on the path. In other words, a conservative force does not produce work along any closed path.

3

Momentum conservation, center of mass

The total momentum of the system is defined as the vectorial sum of the momenta of all its parts: P=

 i

pi =



mvi .

(14)

i

The change of the total momentum with time is given by dP/dt: d  P = pi dt dt i  dpi  = = Fi = Fext , dt i i

(15)

where Fext is the sum of the forces caused by the bodies which do not belong to the system (external forces). If this resulting external force Fext = 0 the total system momentum does not change dP/dt = 0, that is, the momentum is conserved.

3

Physics 1. Mechanics

Lecture 5

Let us now define the center-of-mass (CM) as the point with the radius-vector Rcm

  mi ri mi ri i , =  = i M i mi

(16)

where M is the total mass of the system. There does not have to be any particle in this point - it is completely imaginary. Differentiating (16) with respect to time we find that CM is moving with the velocity 

mi vi P = . (17) M M Thus, the physical sense of the center-of-mass is that this is a point which may be assigned the total momentum and the total mass of the system. From (15) we find the acceleration of the center-of-mass Vcm =

as follows acm = Fext /M.

(18)

In the absence of external forces the center-of-mass moves with constant velocity. If it was in the rest in the beginning it would remain in the rest if only internal forces are present. While the total momentum can be attributed to the motion of the center-of-mass, the kinetic energy cannot. Indeed, the total kinetic energy K=

 i

Ki =

 mi

 mi i

2

vi2

[(vi − Vcm ) + Vcm ]2 2 i  mi   mi 2 = mi (vi − Vcm ) · Vcm + . (vi − Vcm )2 + Vcm 2 2 i i i =

(19)

The second term in the last expression of (19) vanishes (SHOW THAT !), and the first term can be written in terms of the velocity relative to the center-of-mass ui = vi − Vcm , so that we get finally: K=

 mi M 2 Vcm + u2i . 2 2 i

(20)

Thus, the total kinetic energy of the system is the sum of the kinetic energy of the center-of-mass and the kinetic energy of the relative motion of the system parts. The last contribution is called internal energy.

4

Center-of-mass: continuous systems

Most of the systems we deal with do not consist of point-like particles, but are continuous bodies of finite size. The generalization onto such systems is straightforward, although not always simple for calculations. We divide the whole body onto small (infinitesimal) parts with the mass dm, 4

Physics 1. Mechanics

Lecture 5

radius-vector r , and velocity v , so that the sums are substituted by integrals: M=




P=




Rcm

dm,

(21)

v dm,
= r dm/M.

(22) (23)

The integrals should cover all material, they are written in a symbolical form. In practice, one has always to express the integrals in terms of coordinates.

4.1

Center-of-mass: examples

A rod 1. A zero thickness rod 0 < x < l of the mass m (see Fig. 1).

O

x

dx

Figure 1: Center of mass for a rod. In the integral (23) dm = (m/l)dx (λ = dm/dx = m/l is the linear density), and we have Xcm

1 = m




0

l

l xmdx = , l 2

(24)

as could be expected. 2. Same but the linear density is not constant λ = ax. Then l

xλdx 2l Xcm = 0 l = . 3 λdx 0

(25)

5

Physics 1. Mechanics

Lecture 5

A triangle A uniform density triangle has the vertices (0, 0), (l, 0), and (l, h), and the mass m. The surface density σ ≡ dm/dA = m/A, where A = lh/2 is the triangle area. We divide the triangle onto infinitesimal dxdy (see Fig. 2) so that for each of which dm = σdxdy. We have (for any general

y

dy

O

x

dx

Figure 2: Center of mass for triangle. surface density σ(x, y)  xσdxdy Xcm =   , σdxdy  yσdxdy , Ycm =   σdxdy

(26) (27)

where the integral limits should be chosen according to the conditions. In our case



(. . .)dxdy →


l 
0

0

hx/l



dy(. . .) dx,

(28)

so that we have  l  hx/l ( xσdy)dx = 2l/3 Xcm = 0 l 0 hx/l ( σdy)dx 0 0  l  hx/l ( yσdy)dx = h/3. Ycm = 0 l 0 hx/l ( σdy)dx 0 0

(29) (30)

A half-disk Given a uniform thin half-disk with the radius a (see Fig. 3). It is obvious that

6

Physics 1. Mechanics

Lecture 5

Ycm = 0. For Xcm we use (26): Xcm

5

 a  √a2 −x2 ( √ 2 2 xσdy)dx 4a = = 0 −√ a2 −x2 . a a −x 3π √ ( σdy)dx − a2 −x2 0

(31)

Reactive motion

From the general form of the second Newton law F = dp/dt and p = mv it follows that the velocity can change if the mass changes. A rocket is based on this principle. Let us consider one-dimensional motion of a rocket which ejects backwards gas with the velocity u relative to the rocket. Let v be the velocity of the rocket in the moment t and v + dv is its velocity in the moment t + dt. Let also m be its mass in the moment t and m + dm is its mass in the moment t + dt (dm < 0 since the rocket ejects gas). The mass of the ejected gas is −dm (so that the total mass (m + dm) + (−dm) = m does not change), while the velocity of the gas is v − u (here v > 0 and u > 0). The momentum conservation (no external force) gives mv = (m + dm)(v + dv) + (−dm)(v − u)

→

dv = −udm/m.

(32)

Differentiating the last equation with respect to time one gets the rocket acceleration a = µu/m,

(33)

where µ = −dm/dt is the rate of the gas ejection. On the other hand, integrating (32) one gets vf − vi = u ln

mi , mf

(34)

where i and f mean initial and final, respectively.

7

Physics 1. Mechanics

Lecture 5

y √ y = a2 − x2

dx O

x

√ y = − a2 − x2

Figure 3: Center of mass for halfdisk.

8

Physics 1. Mechanics Math1. Complex numbers

In this lecture we bring (without proof) the basics of the usage of complex numbers, which is necessary for a physicist. A complex number is a construction (it is widely accepted use z for general complex numbers) z = x + iy, where x and y are real numbers, and i is a special number such that i2 = −1. For any two complex numbers z1 = x1 + iy1 and z2 = x2 + iy2 we can find z1 + z2 and z1 · z2 as follows: z1 + z + 1 = (x1 + x2 ) + i(y1 + y2 ),

(1)

z1 z2 = (x1 x2 − y1 y2 ) + i(x1 y2 + x2 y1 ).

(2)

The absolute value of the complex number z = x + iy is |z| =

p x2 + y 2 .

The number x is called the real part and the number y is called the imaginary part: x = Re z,

y = Im z.

p Thus, z = Re z +i Im z, and |z| = (Re z)2 + (Im z)2 . A complex conjugate (cc) number z ∗ is defined as z ∗ = Re z − Im z. It is easy to see that |z|2 = zz ∗ . Now the division of the complex numbers if defined as follows: z1 z2∗ z1 z2∗ z1 = = z2 z2 z2∗ |z2 |2 Exercise: Express z1 /z2 in terms of x1 , y1 , x2 , y2 , i. We can also write Re z = (z + z ∗ )/2, Im z = (z − z ∗ )/2i. Any complex number can be represented as a point in the complex plane, where x axis is for Re z and y axis is for Im z. The same point in the x − y plane can be represented by its polar coordinates r and ϕ, so that x = r cos ϕ, y = r sin ϕ. With this one can write z = x + iy = |z|(cos ϕ + i sin ϕ).

1

(3)

Physics 1. Mechanics

Math1

This is completed with the (most important) Euler formula cos ϕ + i sin ϕ = exp(iϕ) = eiϕ .

(4)

With this one can write z = |z| exp(iϕ), z ∗ = |z| exp(−iϕ), 1/z = (1/|z|) exp(−iϕ), z1 z2 = |z1 ||z2 | exp[i(ϕ1 + ϕ2 )], z n = |z|2 exp(inϕ), and so on.

2

Physics 1. Mechanics Lecture 6. Particle dynamics II

In this lecture we shall study several important cases of particle dynamics in electric and magnetic fields.

1

Field

We say that an electric field E exists in the space if we can put an electric charge q (charged particle) in any point and it will experience the force F = qE. The concept of the field is one of the central concepts in physics. Since we shall deal with fields a lot in future it is worth to devote some tome to this concept now. We start with the example of the Earth and satellites around it. It is known that each two masses are attracted to each other. For the time being the precise shape of the force is not important. Let us say that the two bodies have masses M and m, M > m, and are in the positions R and r, respectively. Then the force acting on, say m, is proportional to m and M and depends on R and r. The same force but with the opposite direction acts on M. Although the magnitude of the force is the same, the magnitudes of the accelerations are inversely proportional to masses, so that the larger mass is less affected by this interaction. Let us assume for simplicity that M
m, so that the effect of the interaction on M can be neglected. Then we can regard the large mass M as a “source” of the force acting on the small mass m. Thus, we can take this small mass (which is called test particle or test mass), put it in an arbitrary point of the space and measure the force acting on it. Since we fixed the position and the mass of the large body, the only meaningful dependence of the measured force would be on the mass of the test particle and its position. If we put different test particles in the same point, each would feel the force proportional to its mass, that is, the force can be written as F→m = mE, where E does not depend at all on the test particle mass m. If we take the same test particle in two different points, say 1 and 2, it will experience forces F1 = mE1 , and F2 = mE2 , which are, in general, different. That means that the vector E depends, in general, on the position, E = E(r). If the large body moves somehow, this vector can depend on time too, E = E(r, t). Let us now forget about the body which causes E and look at everything from the point of view of the test particle. The test particle does know anything about what causes the forces, the only think it knows is that in each point of space r and every moment t it would feel the force F = mE(r, t). The test panicle (and we too) concludes that the vector E(r, t) exists in space even without presence of 1

Physics 1. Mechanics

Lecture 6

this test particle. Indeed, any test particle could be placed in any position at any time, and the relation F = mE(r, t) does not change because of our bringing the test particle into consideration. Thus, E(r, t) is real, depends only on other bodies (source) and is independent of test measurements. We say that there is a field E(r, t). This field exists in space in time and produced by other bodies which are not of interest to us and the motion of which is usually assumed to be known. The mass m which, being multiplied by the field, gives the force acting on the test particle, is called, in general, a charge. When certain field is considered, the characteristic word should be added to make clear what is studied. Thus, the field produced by the gravitational forces is called gravitational field and the mass is the gravitational charge. In this lecture we will study the motion in the electric and magnetic fields acting on the electric charge. Fields in general Fields are not necessarily related to forces. In general, we say that there is a field in space if some quantity can be measured at different positions and times, but the very existence of which is not determined by such measurement. In the above consideration what was measure was the force vector, thus the field was a vector field. A well-known example of another field is temperature, which can be measured at any position and any time. Temperature is a scalar field which exists even if we do not measure it. A “test particle” would be in this case a thermometer.

2

Motion in a constant electric field

When we consider a motion in a constant electric field we assume that the field vector E does not depend on r and t, at least in that part of the space where our test particles moves. The electric force is F = qE, where q is the electric charge of the particle (we omit the word “test” for brevity). This case is not especially interesting since the second Newton law states F = ma → ma = qE → a = qE/m = const, that is, the particle moves with constant acceleration, the case we studied earlier.

3

Motion in a uniform oscillating field

A field is called uniform is it does not depend on r. It can still depend on time. In the section we shall try the field of the form E = E0 sin(ωt),

(1)

where E0 = const is a constant vector (both the magnitude and direction does not change), and ω ˆ in the direction of is the frequency (or angular frequency). It is convenient to define a unit vector e

2

Physics 1. Mechanics

Lecture 6

ˆ. The equation of motion reads ˆ = E0 /|E0 |, so that E0 = E0 e E0 , e d2 r dv = 2 = a= dt dt




qE0 m



ˆ sin(ωt). e

(2)

It is convenient (this procedure will be widely used in future) to split the position vector r = xˆ e + r⊥ , ˆ. Then the equation of motion (2) splits into where r⊥ ⊥ e d2 r⊥ = 0, dt2
 qE0 d2 x = sin(ωt). dt2 m

(3) (4)

Eq. (3) describes the motion with zero acceleration (constant velocity), that is, we immediately know the answer: r⊥ = r⊥,0 + v⊥,0 (t − t0 ).

(5)

ˆ. Thus, the particle moves with the constant velocity v⊥,0 in the direction perpendicular Here v⊥,0 ⊥ e to the electric field direction (!!! it is only a part of the motion !!!). Frame transformation. Here we introduce a very important method of studying particle dynamics. We should remember that it is always possible to switch to another inertial frame, and the second Newton law will not change the form. If we now switch to the frame moving with the velocity v⊥,0 with respect to our original frame, we will see the particle not moving in the direction perpendicular to the electric field. In fact, any motion with constant velocity can be excluded from the consideration by switching to the corresponding moving inertial frame. (WARNING: we have to know, in general, how the fields transform with the frame transformation. For the time being, we shall leave this issue aside, assuming that our electric field remains the same.) Thus, in our case we may say that the particle does not move in the perpendicular direction in the properly chosen reference frame. With all above, the only equation of motion to be solved is the equation (4) for single coordinate in the direction of the electric field. This equation is easily integrated to give qE0 (1 − cos(ωt)) , vx = vx0 +
mω  qE0 qE0 x = x0 + vx0 t + sin(ωt), t− mω mω 2

(6) (7)

where we added the initial conditions x = x0 , vx = vx0 at t = 0. We see that the motion in x direction is the motion with the constant velocity vd = vx0 + qE0 /mω and periodic oscillations. Again we can switch to the reference frame moving in x direction with the velocity vd . In this frame the particle only oscillates with the frequency ω. 3

Physics 1. Mechanics

4

Lecture 6

Constant magnetic field

The electric and magnetic field are the two parts of the electromagnetic field, and act on electric charges. However, while the electric field acts on any charges, the magnetic field acts only on moving charges. The magnetic force looks as follows (SI units) F = qv × B,

(8)

where B is the magnetic field and v is the particle velocity. [In the Gaussian (CGS) system of units magnetic and electric fields are measured in the same units and the force is written as F = qv × B/c, where c is the speed of light. In our course we shall use SI units throughout.] The force (8) is perpendicular both to the velocity and magnetic field direction. As an obvious consequence, the magnetic force cannot change the kinetic energy of the particle, that is, the magnetic force does not produce work: P = dW/dt = F · v = q(v × B) · v = 0. Since the kinetic energy K = mv 2 /2 does not change, the magnitude of the particle velocity v = |v| remains constant. Let is now follow the principles of the previous section and split r = r⊥ + r (and similarly for the velocity) where the indices ⊥ and  mean perpendicular and parallel to the magnetic field (remember: the direction of the magnetic field is constant). Exercise: Show that the equations of motion will take the form dv = 0, dt
 dv⊥ qB = − × v⊥ . dt m

(9) (10)

Eq. (9) shows that the velocity in the direction of magnetic field does not change at all. We already know that it is possible to move to another inertial reference frame in which this motion is absent. Let us concentrate on the motion in the plane perpendicular to the magnetic field, which is described by (10). Because of the importance of the case we shall do that in two ways.

4.1

Analogy

This is simple. We have only to remember the equation for circular motion with constant angular velocity which reads (dv/dt) = ω × v. Comparing with (10) we see that the motion is the circular motion, gyration, with the angular velocity ω = −qB/m, called gyrofrequency. Respectively, the radius of gyration, gyroradius, is r⊥ = mv⊥ /|qB|. 4

Physics 1. Mechanics

4.2

Lecture 6

Coordinate representation

Let B = Bˆez , than (10) will be written as (we use the widely accepted notation X˙ ≡ dX/dt for any variable X): v˙x = ωvy ,

(11)

v˙y = −ωvx ,

(12)

ω = qB/m.

(13)

In order to proceed we shall use complex numbers. See mathematical appendix Math1 as a separate lecture. Proceed only after having learnt the material. It is convenient to define vs = vx + isvy , where s = ±1. Multiplying the equation for vy by is and adding to the equation for vx we get v˙s = −isωvs ,

(14)

vs = v⊥ exp(isωt + isφ),

(15)

which has the solution in the form

where v⊥ = const and φ = const, so that vx = v⊥ cos φ and vy = v⊥ sin φ at t = 0 . In order to get vx and vy we have to use the Euler formula: vs = v⊥ cos(ωt + φ) + isv⊥ sin(ωt + φ),

(16)

or vx = v⊥ cos(ωt + φ),

5

vy = v⊥ sin(ωt + φ).

(17)

Electric and magnetic field together

Here we consider the case where there are both constant electric E and magnetic B fields. The total force (Lorentz force) is then F = qE + qv ×B. As before we split all vectors onto components parallel (index ) and perpendicular (index ⊥) to the magnetic field. Let b = B/B be the unit vector in the direction of the magnetic field, then we shall write r = zb + r⊥ , v = v b + v⊥ , and E = E b + E⊥ . The equation or motion reads q (E + v × B) → m q q v˙  b + v˙ ⊥ = E b + [E⊥ + v⊥ × B]. m m

v˙ =

(18)

5

Physics 1. Mechanics

Lecture 6

Separating parallel and perpendicular components we obtain q E , m q v˙ ⊥ = (E⊥ + v⊥ × B. m v˙  =

(19) (20)

We see that (19) contains only parallel components, while (20) contains only perpendicular ones. This allows to study separately the motion along () and across (⊥) the magnetic field. The motion along the magnetic field is the simple constant acceleration motion (see (19)) and the analysis remains as an exercise for the reader. In order to study the motion across the magnetic field, eq. (20) it is convenient and instructive to switch to another frame which is moving with some velocity Vd ⊥ B. We shall choose this frame so that the equations of motion reduce to what we already know. Let us denote the particle velocity in the new frame as u. According to the rules of Galilean transformation, v⊥ = Vd + u. Substituting into (20) and taking into account that Vd is constant, we obtain u˙ =

q (E⊥ + Vd × B) + ω × u, m

(21)

where ω = −qB/m. The first term in parentheses is constant. If it were absent we would obtain the equation identical to the equation of motion for a particle in the magnetic field, without any electric field. Now we can choose the new frame velocity to eliminate this term. We require E ⊥ + Vd × B = 0

(22)

from which it immediately follows

E⊥ × B . (23) B2 This can be done only if E⊥ /B < c where c is the speed of light !. The obtained velocity is called the drift velocity. In the frame moving with this velocity, the particle is gyrating around the magnetic field, as if the electric field was absent (in the course Physics 2 we shall see that the electric and Vd =

magnetic fields change when switching frames, and the electric field indeed vanished for the frame transformation that we did). Thus, we can conclude that the particle gyrates around the magnetic field and in addition drifts with the constant velocity Vd . This drift is perpendicular to the electric and magnetic field as well.

6

Physics 1. Mechanics

Lecture 6

Figure 1: Gyration, drift, and motion along the magnetic field.

7

Physics 1. Mechanics Advanced 6. Perturbation theory in particle dynamics

Not all problems in physics can be solved exactly. In fact, only very few problems have been solved exactly and not many can be solved exactly. Thus, approximate methods play a very important role in physics. Here we shall study two simple examples of such methods.

1

General principles of perturbation theory

Let us assume that the equation of motion can be written in the form m¨ x = F0 (x, t) + F1 (x, t),

(1)

where  is a symbolic small parameter, that is, shows that the addition F1 is small relative to the main force F0 . Let us also assume that we know the solution x = x0 (t) of the unperturbed equation m¨ x = F0 .

(2)

We proceed assuming that the perturbation F1 perturbs the solution only weakly, that is, that the solution of the complete equation (1) can be presented as x = x0 (t)+x1 (t), where  again symbolizes the smallness of the change. It is important that it is the same  as in the force perturbation. Substituting into (1) we get m(¨ x0 + ¨ x1 ) = F0 (x0 + x1 , t) + F1 (x0 + x1 , t) dF0 + F1 (x0 (t), t) ≈ F0 (x0 (t), t) + x1 dx |x=x0 (t) and therefore m¨ x1 = x1

dF0 + F1 (x0 (t), t). dx |x=x0 (t)

(3)

(4)

The new equation is usually easier to solve than the complete one. Although the above is written for a one-dimensional motion for simplicity, it can be two- and three-dimensional as well.

1

Physics 1. Mechanics

2

Advanced 6

Example 1: Magnetic field

Let us start with the particle gyration in the constant magnetic field mv˙ x = qvy Bz ,

(5)

mv˙ y = −qvx Bz ,

(6)

v0,x = v⊥ cos(ωt),

(7)

v0,y = −v⊥ sin(ωt),

(8)

with the solution

where ω = qBz /m (Bz is assumed constant). Now let us add perturbation in the form of the electric force F1,x = 0, F1,y = qEy . Following the method outlined above we write vx = v0,x + v1,x and vy = v0,y + v1,y . Substituting this into the equations of motion we obtain mv˙ 1,x = qv1,y Bz , mv˙ 1,y = −qv1,x Bz + qEy ,

(9) (10)

which has the special solution of the form v1,x = Ey /Bz , v1,y = 0. We rediscovered the E × B drift. This example is simple but not very informative since we know how to solve the equations exactly in this case.

3

Example 2: Wave field

Let us now solve the one-dimensional equation of motion for the force of the kind F = F1 cos(kx), where k is a constant parameters. (Actually, in this case the problem can be solved exactly but the situation changes if there are two forces of the kind Fi cos(ωi t − ki x), i = 1, 2, where ω2 /k2 6= ω1 /k1 . The method of solution in this case is the same as the method we are going to learn now.) The solution of the unperturbed equation x¨ = 0 is x0 (t) = v0 t (for simplicity we ignore the initial coordinate). Substituting this into the force we have for the perturbation (x = x0 + x1 ) mx¨1 = F1 cos(kv0 t)

(11)

and the solution can be found by direct integration: F1 sin(kv0 t), mkv0 F1 x1 = − 2 2 cos(kv0 t). mk v0

v1 =

(12) (13)

2

Physics 1. Mechanics

Advanced 6

This approximation is valid if the perturbation is weak, that is, max|v1 |  v0 , which gives F1 1 mkv02 Let us now notice that for the potential force F1 cos(kx) the potential U = −(F1 /k) sin(kx) therefore (F1 /k) has the meaning of the maximum potential energy. More physically speaking, F1 /k is of the order of the typical potential energy F1 /k ∼ U1 (this is not a precise statement but an estimate which is one of the most powerful instruments of a physicist). In the same sense mv02 ∼ K, thus the applicability criterion reads U1 /K  1, where U1 is the potential energy of the particle in the perturbed state and K is the kinetic energy of the particle in the unperturbed state. The small parameter  becomes now a quite real parameter  ∼ U1 /K.

4

Averaging

Let us consider a particle moving in the x − y plane perpendicular to the magnetic field, only in this case the magnetic field will be allowed to be weakly inhomogeneous, Bz = Bz (x). Later we shall see what does it mean: “weakly”. If the magnetic field was constant the motion would be a gyration, circular motion around the magnetic field, vx = v⊥ cos(ωt),

(14)

vy = −v⊥ sin(ωt),

(15)

x = x0 + r⊥ sin(ωt),

(16)

y = y0 + r⊥ cos(ωt),

(17)

where the gyrofrequency ω = qBz /m, and gyroradius r⊥ = v⊥ /ω. Because of the magnetic field inhomogeneity the particle “feels” different magnetic fields in different points of its circular trajectory which eventually would distort the circle. If the magnetic field does not change substantially from one point of the circle to another, that is, |Bz (x0 + r⊥ ) − Bz (x0 − r⊥ )|  |Bz (x0 )| we may Taylor expand to obtain dBz . (18) Bz (x) = Bz (x0 ) + (x − x0 ) dx |x=x0 Now we see that the condition for weak inhomogeneity is r⊥

dBz  Bz , dx

(19)

that is, the magnetic field does not change significantly on the gyroradius. If we estimate dBz /dx ∼ Bz /L, where L is the typical length of the magnetic field inhomogeneity then the condition will be L  r⊥ . 3

Physics 1. Mechanics

Advanced 6

Coming back to the equations of motion we shall now add the velocity perturbation v → v + v1 and write the magnetic field as Bz = B0 + (x − x0 )(dBz /dx) where  ∼ r⊥ /L is the small parameter of the problem. The equations of motion read mv˙ 1,x = qv1,y B0 + qvy (x − x0 )

dBz dx

= qv1,y B0 − qv⊥ sin(ωt)r⊥ sin(ωt) mv˙ 1,y = −qv1,x B0 − qvx (x − x0 )

(20) dBz , dx

dBz dx

= −qv1,x B0 − qv⊥ cos(ωt)r⊥ sin(ωt)

(21) (22)

dBz . dx

(23)

Instead of trying to solve these equations we shall try to find out if there is some average velocity changes, that it, whether there is a velocity addition which is not periodic with the period T = 2π/ω. In order to do that we perform averaging according to the recipe 1 v¯x = T

Z

t+T

vx (t0 )dt0

(24)

t

¯ = 0) which immediately gives (obviously, dv/dt v¯1,y =

2 v⊥ 1 dBz , 2ω Bz dx

v¯1,x = 0.

(25)

This is the so called gradient drift which is perpendicular to the direction of the magnetic field (z) and to the direction of the magnetic field change (x). We may regard the particle motion as a rapid gyration with local gyrofrequency ω = qBz (x) and local gyroradius, while the center of the circle slowly drifts along y direction with the velocity given by (25).

4

Physics 1. Mechanics Lecture 7. Particle dynamics III

1

Potential energy

Let F(r) be a conservative force. This means that for each two points r1 and r2 the work W =


2 1

F·dr

does not depend on the path but only on the initial and final points themselves. The minimum work that we have to do against this force to move a particle from r1 to r2 is Wext = −W = −



r2

r1

F · dr.

Let us now fix some point r0 as a reference point and consider the minimum work we have to do to move a particle from this reference point to any other point r. This work depends only on r0 and r so that we can write  r U(r) − U(r0 ) = −

r0

F · dr.

(1)

The function U(r) is called a potential energy, and the difference U(r) − U(r0 ) what work the force does when moving a particle from r to r0 or what minimum work an external force must do to move particle from r0 to r. From this analysis it is clear that there is no physical meaning to the value of the potential energy in the reference point, since only the difference is meaningful. The reference point is usually chosen from the conveniency arguments. For example, is a particle is very far from bodies the interaction is negligible, and it is natural to choose U0 = 0. Let us now consider two close points, r and r + dr. According to (1) we have dU = U(r + dr) − U(r) = −F · dr = −(Fx dx + Fy dy + Fz dz).

(2)

Thus, if we know the potential energy as a function of position r we can find the force at r as follows: ∂U , ∂x ∂U Fy = − , ∂y ∂U . Fz = − ∂z

Fx = −

1

(3)

Physics 1. Mechanics

Lecture 7

For brevity this is usually written in a symbolical form as follows: ∂U , ∂r

(4)

F = −∇U,

(5)

F=− or

where the special symbol ∇ = ex

∂ ∂ ∂ + ey + ez ∂x ∂y ∂z

(6)

denotes the differential operator, that is, the partial derivatives act on everything which stands to the right or them. The construction ∇U is called a gradient of U.

1.1

Examples

Here we provide several simple examples showing how force is obtained from potential energy. 1. Potential energy related to a constant vector. U = f · r → F = −∂U/∂r = −f, that is, the force is constant. 2. General dependence on distance. U = U(R), where R = |r − r0 |). Following the same procedure, we find ∂U ∂R . ∂R ∂r The last can be easily found it we take into account that F=−

R2 = (r − r0 ) · (r − r0 ). Differentiating the last expression with respect to r we have R or

∂R = r − r0 , ∂r

r − r0 r − r0 ∂R = = = (r
− r0 ), ∂r R |r − r0 |

where (r
− r0 ) is the unit vector in the direction of r − r0 . Thus, F=−

∂U
(r − r0 ). ∂R

2

Physics 1. Mechanics

Lecture 7

3. Coulomb potential. The function U = A/R is called Coulomb potential. Following the previous derivation we find A(r − r0 ) . F=− |r − r0 |3 4. Harmonic oscillator. The function U = k1 x2 /2 + k2 y 2 /2 + k3 z 2 /2 describes three-dimensional harmonic oscillations. The corresponding force is given by (3): F = −k1 xex − k2 y ey − k3 z ez .

2

Conservative (potential) forces

We have seen that if a force is conservative (work does not depend on path) a potential energy can be defined. Thus, we can call such forces potential forces too. Let F(r) be some force (more precisely, force field since it depends on r). In order to find out whether this force is potential or not, it is not necessary to calculate integrals to show that they are independent of trajectory. Instead it is possible to use the relations (3) and the identities ∂ ∂xi



∂ ∂x



∂U ∂xj



∂ = ∂xj



∂ = ∂y



∂U ∂xi



for any i and j. For example, from ∂U ∂y



∂U ∂x



we find ∂ ∂ Fy = Fx ∂x ∂y

(7)

and similarly ∂ Fz = ∂y ∂ Fx = ∂z

∂ Fy ∂z ∂ Fz ∂x

(8) (9)

3

Physics 1. Mechanics Advanced7. Differential operators

0.1

Advanced material

In a more compact form we can write ∂U , ∂xi

(1)

X ∂Fk ∂U =− = 0. εijk ∂xj ∂xj ∂xk jk

(2)

Fi = − and the condition of potentiality looks as follows: X jk

εijk

Using the vector operator ∇ the right hand side of (2) can be written as ∇ × F and the whole relation takes the form ∇ × F = ∇ × (−∇u) = −(∇ × ∇)U = 0. (3) The last relation becomes obvious since for any vector A the vector product A × A = 0. This shows that ∇ can be treated as an ordinary vector. One has to remember, though, that it is an operator acting on everything on the right, so that the order is important. Let us consider, for example, the construction ∇ × (B × C) If ∇ was an ordinary vector, it could written as follows (INCORRECT !): ∇ × (B × C) = B(∇ · C) − C(∇ · B). However, in the left hand side ∇ always acts on B and C, while in the right hand side it is not so. Therefore, the correct form will be ∇ × (B × C) = (∇ · C)B − (∇ · B)C, where ∇ further acts as a derivative. Taking into account (xy)0 = x0 y + xy 0 we find eventually ∇ × (B × C) = B(∇ · C) + (C · ∇)B − C(∇ · B) − (B · ∇)C, where B · ∇ =

P

i

Bi (∂/∂xi ). 1

Physics 1. Mechanics

1

Advanced7

Differential operators: More systematic approach

We have defined above ∇ as a differential operator. Depending to what it is applied and how this operator has different names. When applied to a scalar field U (r) it gives a vector field called gradient A = grad U = ∇U =

∂U ∂U ∂U ˆ+ ˆ+ x y zˆ ∂x ∂y ∂z

(4)

When it is applied to a vector field it can give a scalar field called divergence: V = div A =

∂Ax ∂Ay ∂Az + + ∂x ∂y ∂z

(5)

When applied to a vector field it can give a vector field called rotor or curl : 

 ∂Az ∂Ay ˆ B = rot A = − x ∂y ∂z   ∂Ax ∂Az ˆ − y + ∂z ∂x   ∂Ay ∂Ax + − zˆ ∂x ∂y

(6)

Differential operators can be applied more than once. Indeed, if we apply grad to a scalar field U we get a vector field to which div or rot can be applied, which we can write as div grad U . Or rot can be applied giving rot grad U . If we apply div to a vector field A we get a scalar field to which grad can be applied giving grad div A. Finally, if rot is applied to a vector field A we get again a vector field an div can be applied to give div rot A. Exercise: Prove div grad U =

∂2U ∂2U ∂2U + + ≡ ∆U, ∂x2 ∂y 2 ∂y 2

(7)

div rot A = 0,

(8)

rot grad U = 0.

(9)

2

Physics 1. Mechanics Lecture 8. Conservation laws

1 1.1

Energy conservation Simple derivation

We already know that the change of the kinetic energy K = mv 2 /2 is caused by work, that is Z K2 − K1 = W =

2

F · dr.

(1)

1

If the force is conservative, the right hand side can be written as U1 − U2 , and we have E = K1 + U1 = K2 + U2 = const,

(2)

which is the simplest form of the energy conservation. The sum E = K + U is called mechanical energy.

1.2

More complicated (but more informative) derivation

Let us assume that the force is conservative (potential ) and can be written as F = −∂U/∂r, where the potential energy U (r, t) can, in principle, explicitly depend on time t and must depend on r. The rate of change of the kinetic energy K = mv 2 /2 along the particle trajectory is dv dr ∂U dr dK = mv · =F· =− . dt dt dt ∂r dt

(3)

Let us now notice that the total time derivative of U is dU ∂U dr ∂U = + , dt ∂r dt ∂t

(4)

and substitute into (3): d ∂U (K + U ) = (5) dt ∂t We see that the mechanical energy E = K + U is conserved, dE/dt = 0, if the potential energy does not depend on time, ∂U/∂t = 0. 1

Physics 1. Mechanics

1.3

Lecture 8

Nonconservative forces

If the total force can be decomposed into the conservative and nonconservative components, F=−

∂U + Fn ∂r

the energy equation takes the following form Z

2

Fn · dr,

E2 − E1 =

(6)

1

where E = K + U , or, in the local form dE = Fn · v dt

2

(7)

Momentum conservation

A particle momentum is not conserver if there is a potential energy U (r), since the conservation requires F = −(∂U/∂r) = 0.

3

Angular momentum conservation

Angular momentum is defined as follows: J=r×p

(8)

Let us assume that the acting forces are potential, then the rate of change of angular momentum is dJ =v×p+r×F=r×F dt

(9)

since the first term vanishes identically. The construction N = r × F is called a momentum of force or torque. Thus, in order that the angular momentum be conserved, dJ/dt = 0, the force should be always parallel to r. In the previous lecture we have seen that ∂U/∂r k r is the potential energy depends only on r = |r|. It is possible to show that this is the sufficient and necessary condition for angular momentum conservation.

4

Interaction

In all above we considerer a particle moving under influence of a external force, that is, all other bodies were excluded from our consideration. It is time to consider the interaction between bodies, 2

Physics 1. Mechanics

Lecture 8

so that all they are included in the consideration. It appears that it is sufficient to study a twoparticle system, since the superposition principle states that interactions in a many body system can be reduced to the interactions between all possible couples, that is, a many body system “consists” of two-particle systems. Thus, let us consider a system consisting of two particles, 1 and 2, with no other forces acting on these particles. If particle 1 exerts a force F1→2 on particle 2, then particle 2 also exerts a force F2→1 = −F1→2 on particle 1. We say that the two particles interact. If one of the forces is conservative, the other one is conservative too. From the point of view of particle 1 it has a potential energy U (r1 ), where r1 is its position vector. If the distance between the particles is infinite they should not feel each other, and the potential energy should vanish. Thus, the potential energy of the particle 1 can be calculated by fixing the particle 2 in the point r2 and bringing the particle 1 from infinity into the point r1 . On the other hand, from the point of view of particle 2 it has a potential energy U (r2 ), and this potential energy can be calculated by fixing the particle 1 in the point r1 and bringing the particle 2 from infinity into the point r2 . In both cases we come to the same configuration (particle 1 in r1 and particle 2 in r2 ) so that the result should be the same. With all this and noticing that this potential energy does not exist when one of the particles is absent we come to the conclusion that the potential energy is the energy of interaction and belongs to the system of the two particles, so that the total mechanical energy is E = K1 + K2 + U12 (r1 , r2 ). This, in particular, means that F2 ≡ F1→2 = −

∂U , ∂r2

F1 ≡ F2→1 = −

∂U ∂r1

Let us check the infinitesimal change of the mechanical energy E during time dt: dE = dK1 + dK2 + dU12 = mv1 · dv1 + mv2 · dv2 +

∂U ∂U · dr1 + · dr2 ∂r1 ∂r2

(10)

= mv1 · a1 dt + mv2 · a2 dt − F1 · dr1 − F2 · dr2 = F1 · (v1 dt) + F2 · (v2 dt) − F1 · dr1 − F2 · dr2 = 0, that is, the total mechanical energy is conserved. It is easy to generalize this derivation onto systems P with arbitrary number of particles. We conclude that total mechanical energy E = i Ki + U is conserved. Here U is the potential energy of the system and depends on all ri . In this derivation we assumed that the system is closed, that is, there are no other bodies which interact with the parts of out system and which do not belong to the system.

3

Physics 1. Mechanics

5

Lecture 8

Momentum conservation

It is possible to show that the third Newton law means that the potential energy of the interaction of two particles depends only on the difference r2 − r1 . This is the manifestation of the homogeneity of the space. In other words, it means that we can arbitrarily shift the coordinate origin, so that r01 = r1 + R, and r02 = r2 + R, where R is some constant vector, and the potential energy will not change (since r02 − r01 = r2 − r1 ). The inverse statement is also correct: if the choice of the coordinate origin does not affect the potential energy, it must depend only on the difference. Thus, U = U (mbr2 − r1 ), so that F1 = −(∂U/∂r1 ) = (∂U/∂r2 ) = −F2 . The change of the total momentum is dP = dp1 + dp2 = F1 dt + F2 dt = 0, that is, the total momentum is conserved. In this derivation the system is also assumed to be closed.

6

Angular momentum

The vector product J = r × p is called angular momentum. The importance of this quantity is that it is also conserved, if the potential is isotropic, that is, the interaction does not depend on the direction. For two particles this means that the potential energy U = U (X), where X = |r2 − r1 |, that is, depends only on the distance between the two particles, and does not depend on the direction of the vector r2 − r1 . Indeed, dJ = dJ1 + dJ2 = dr1 × p1 + r1 × dp1 + dr2 × p2 + r2 × dp2 [dr = vdt||p → dr × p = 0]

(11)

= r1 × F1 dt + r2 × F2 dt   ∂U ∂U = −r1 − r2 dt. ∂r1 ∂r2 Now we use ∂U ∂U r1 − r2 = , ∂r1 ∂X X ∂U ∂U r2 − r1 = . ∂r2 ∂X X

(12)

4

Physics 1. Mechanics

Lecture 8

Substituting (12) into (11) we obtain 

 r1 − r2 r2 − r1 ∂U dJ = −r1 × − r2 × dt X X ∂X (r1 − r2 ) × (r1 − r2 ) ∂U dt = 0. =− X ∂X

(13)

Thus, angular momentum of a closed system is conserved.

7

Open systems

Open are systems which are under influence of some external bodies. In such systems energy, momentum, and angular momentum are not conserved and their changes should be determined by the external forces.

7.1

Single particle

Single particle under the influence of external force is an open system. We already know that dp = Fext dt and

dK = Fext · v. dt

If the external force Fext is conservative, that is, Fext = − ∂U (U does not depend on time) then the ∂r second relation can be written as dE/dt = 0 where E = K + U is the particle energy. For the angular momentum we get dr dp dJ = ×p+r× =v×p+r×F dt dt dt Since v k p only the last term survives so that dJ = r × Fext ≡ N dt where the right hand side is called momentum of force or torque.

5

Physics 1. Mechanics

7.2

Lecture 8

Many particle system

We have shown that the total momentum of the system can be expressed in terms of the total mass and the velocity of the center-of-mass: P=

X

pi = M Vcm

i

We have already shown that the total energy if the kinetic energy of the center-of-mass plus the internal energy 2 X mi X M Vcm + (vi − Vcm )2 . K= Ki = 2 2 i i In a similar way for the angular momentum we get J=

X

Ji =

X

i

=

ri × pi

i

X

[(r − Rcm ) + Rcm ] × pi

(14)

i

=

X

(r − Rcm ) × pi + Rcm × P.

i

Thus, the angular momentum of the system is the angular momentum of the center of mass Rcm × P plus the internal angular momentum, that is, the angular momentum of the parts of the system with respect to the center-of-mass. This part of the angular momentum is often called spin: Jint =

X

(r − Rcm ) × pi

i

Now for a system of particles one has X dP = Fext = Fi,ext . dt i For energy we would obtain d dt and



2 M Vcm 2

 = Fext · Vcm

dEint X = Fi,ext · (vi − Vcm ) dt i

where Eint is the internal energy, which is now the sum of the kinetic energy of the motion of the system parts relative to the center-of-mass and the potential energy of the interaction between the particles of the system.

6

Physics 1. Mechanics

Lecture 8

For angular momentum we obtain d (Rcm × P) = Rcm × Fext dt and

X X d Jint = (vi − Vcm ) × pi + (ri − Rcm ) × Fi . dt i i

Q: Show that the first term in the right hand side is identically zero, so that X d Jint = (ri − Rcm ) × Fi ≡ Ncm . dt i Here Ncm is the torque relative to the center of mass.

8

Collisions

A collision is a kind of interaction where the time of interaction is negligible, that is, what is of interest is the initial and final states. During a collision the momentum and angular momentum are conserved (since only internal forces are working), while the kinetic energy may be converted into the internal energy and vice versa. The collision where the kinetic energy is conserved is called an elastic collision. If the kinetic energy is not conserved the collision is called inelastic. During inelastic collisions the kinetic energy often dissipates to plastic deformations. An example of the collision with the conversion of the internal energy into kinetic energy is explosion. Here we shall consider a special case of the elastic collision of two particles. We shall call the lab frame the reference frame where we see one particle, m1 , v1 , colliding with another particle m2 at rest, v2 = 0. In this case the collision is often called scattering of the particle 1 on the particle 2. The momentum and energy conservation give (primed quantities are after the collision) p1 + p2 = p01 + p02 , p0 21

(15) p0 22

p2 p21 + 2 = + . 2m1 2m2 2m1 2m2

(16)

Our objective is to find as much information about p0 1 and p0 2 if we know p1 and p2 . Let me remind you that in the lab frame p2 = 0. It is extremely convenient to work with two frames, the lab frame and CM frame (center-of-mass frame). The latter moves with the velocity Vcm = (p1 + p2 )/(m1 + m2 ) = p1 /(m1 + m2 ) relative to the lab frame. In what follows we denote the particle velocities in the lab frame as v1 , v2 , v0 1 , v0 2 , and in CM frame as u1 , u2 , u0 1 , u0 2 . According to the velocity transformation rule we have v = u + Vcm and similarly for all other velocities in both frames. Therefore, if we know the initial and final velocities in CM we can easily find them in the lab frame. 7

Physics 1. Mechanics

Lecture 8

Physics 1. Mechanics

Lecture 8

Figure 1: Comparison of what is measured in the laboratory frame (left) and center-of-mass frame

Figure 1: Comparison of what is measured in the laboratory frame (left) and center-of-mass frame (right). (right). Attention: p2 6= 0 in this figure for better visualization, despite our definition of the laboratory system. In CM frame the total momentum should vanish (according to the definition of CM frame),

so that m1u1 + m2u2 = m1u! 1 + m2 u!2 . Energy conservation requires that m1u2 /2 + m2u2 /2 =

1 In CM frame the total momentum should vanish (according to the definition of CM2 frame), m1 u!21 /2 + m2u! 22 /2. It is easy to see that the only way to satisfy this two conditions is to rotate so that m1 u1 + m2 u2 = m1 u0 1 + m2 u0 2 . Energy conservation requires that m1 u21 /2 + m2 u22 /2 = both vectors u1 and u2 by the same angle, while their magnitudes do not change. Moreover, all four m1 u0 21 /2 + m2 u0 22 /2. It is easy to see that the only way to satisfy this two conditions is to rotate vectors u, the vector Vcm , and the vectors v remain in the same plain. For convenience we choose bothcartesian vectors u while their magnitudes do not alleˆfour 1 and u2 by coordinates so the thatsame v1 = angle, v1 eˆx (remember that v2 = 0). Then Vcmchange. = (m1v1Moreover, /(m1 + m2)) x, vectors u, the vector V , and the vectors v remain in the same plain. For convenience we choose and u1 = v1 − Vcm =cm (m2v1/(m1 + m2))eˆx . Respectively, u2 = −m1 u1/m2 = −(m1v1/(m1 + m2))eˆx . cartesian thatvelocity v1 = v1vectors eˆx (remember that rotate v2 = 0). Then Vcm 1 vwe 1 /(m 1 + m2 ))eˆx , If aftercoordinates the collisionsoboth in CM frame by the angle θcm=, (m then have and u1 = v1 − Vcm = (m2 v1 /(m1 + m2 ))eˆx . Respectively, u2 = −m1 u1 /m2 = −(m1 v1 /(m1 + m2 ))eˆx . m2 v1 u!1x in = CM frame cosrotate θcm , by the angle θcm , then we have(17) If after the collision both velocity vectors m +m 1

2

m2 v1 = m2 v1 sin θcm , u1x = m1 + m2 cos θcm , m1 +mm 1 v21 u!2x = −m cos θcm , 21v1+ m2 0 m u1y = m1v1 sin θcm , u!1y =m −1 + m2 sin θcm . v1m2 mm 1 1+ 0 u! 0 1y

Using v! = u + Vcm we find

Using v0 = u + Vcm we find

(18)

(17)

(19)

(18) (20)

u2x = −

(19)

u01y

(20)

cos θcm , m1 + m2 m1 v1 =− sin θcm . m1 + m2

m1 + m2 cos θcm v1 , m1 + m2 m2 sin θcm ! v1y = v1 , +m mm 1 1+ m2 2cos θcm 0 v1x! = m2 − m1 cos θcm v1 , m1 + m2 v2x = v1 , m1θ + m2 m sin 2 cm 0 v1y v1 , m1 sin θcm != v1y =m − 1 + m2 v 1 . m1 + m2 ! v1x =

m2 − m1 cos θcm v1 , m1 + m2 m1 sin θcm =− v1 . m1 + m2

(21) (22)

(21)

(23)

(22)

(24)

0 v2x =

(23)

0 v1y

(24) 8

8

Physics 1. Mechanics

Lecture 8

Now the angle between the initial velocity v1 and final velocity v2 is given by tan θ =

0 v1y m2 sin θcm = . 0 v1x m1 + m2 cos θcm

(25)

The obtained expressions fully describe the collision (scattering) with the use of only one free parameter θcm . It is impossible to find this parameter from the conservation laws, and this angle is arbitrary, unless some additional conditions are imposed.

9

Physics 1. Mechanics Lecture 9. Motion in a potential, I. 1D and central forces

1

Potential energy, symmetries, and conservation

In what follows we shall study the motion of a particle under the potential force, that is, the case when the force can be written as F = −∇U = − grad U = −

∂U ∂r

(all forms are the same and we shall use the one which is more convenient). The mechanical energy of the particle, E = K + U = mv 2 /2 + U . In general, the potential energy is a function of all three coordinates R = (x, y, z) and time t and the energy E, momentum p = mv and the angular momentum J = r × p are not conserved. However, if the potential does not depend on one or several coordinates and/or time, some of the mentioned quantities are conserved. Let us assume that U does not depend on t. Then ∂E dv ∂E dr ∂E dE = · + · + dt ∂v dt ∂r dt ∂t ∂U = mv · a − F · v + = 0, ∂t

(1)

that is, the energy is conserved. In what follows we restrict ourselves with the cases where the potential energy is time-independent, that is, energy is conserved, E = const. Let us assume now, that the potential energy (or, for brevity, potential ) does not depend on one coordinate, say x, that is, ∂U/∂x = 0. Then ∂U dpx = Fx = − =0 dt ∂x which means that momentum component in x direction is conserved, px = const. This, in turn, means that the velocity in this direction vx = px /m = const and the motion along x-axis is trivial so that we can ignore this coordinate at all (such coordinate is called ignorable). Let us assume now that the potential energy depends only on the length r = |r|, U = U (r). We

1

Physics 1. Mechanics

Lecture 9

have seen earlier, that the force can be written as follows F=−

∂U dU ∂r dU =− = − ˆr ∂r dr ∂r dr

where ˆr = r/r is the unit vector in the direction of r and we substituted ∂U/∂r → dU/dr since U now depends only on one variable and there is no sense in partial derivatives. Then for the angular momentum we have dJ =r×F=0 dt so that the angular momentum J = const. Presence of conserved quantities (so called integrals of motion) greatly simplifies analysis of particle motion.

2

One-dimensional potential

In this section we consider the potentials of the form U = U (x). According to what is found above, y and z are ignorable coordinates (py and pz do not change and we can ignore them at all) and the energy E = mvx2 /2 + U (x) = const. The straightforward approach is to write down the equation of motion d2 x dU m 2 =− dt dx (since U depends only on one variable) and to try to solve it. However, it is much more simple and physically meaningful to write down the energy conservation as follows m 2



dx dt

2 + U (x) = E,

(2)

and solve it with respect to dx/dt taking into account that E is some constant parameter:  1/2 2 dx =± (E − U (x)) dt m

(3)

and from here Z

x

± x0

dx0 p

(2/m)(E − U (x0 ))

Z

t

= t − t0

=

(4)

t0

where x = x0 at t = t0 . Even if we cannot perform the integration practically, the solution is known in principle (only technical problems remain) and the motion is said to be integrable. The above expressions require that E ≥ U , which means that all points x where U (x) > E cannot be reached by the particle. In the point x∗ where U (x∗) = E the particle velocity vx = 0 and its sign changes,

2

Physics 1. Mechanics

Lecture 9

that is, this point is the turning point. Three possible types of particle trajectories exist: a) if U ≤ E for all points −∞ < x < ∞ the trajectory is infinite and unbound, the particle velocity does not change sign and the particle crosses the whole space; b) if U (x) ≤ E only for x0 ≤ x < ∞ (or −∞ < x ≤ x0 ) then the trajectory is infinite but bound from one side: the particle comes from ∞ (−∞ in the second case) turns at x = x0 and returns to ∞ (−∞); c) if U (x) ≤ E for x1 ≤ x ≤ x2 than the particle trajectory is finite and bound, the particle moves back and forth between the two turning points. In the last case the motion becomes periodic and the period is Z

x2

T =2 x1

dx0 p

(2/m)(E − U (x0 ))

.

Attention: The last statement is correct only if in both turning points dU/dx 6= 0. If U (x∗) = E and (dU/dx)x=x∗ = 0 simultaneously, the time which is necessary for the particle to reach the point x∗ becomes infinite.

u 0.6 0.5 0.4 0.3 0.2 0.1 -7.5

-5

-2.5

2.5

5

7.5

x

Figure 1: Potential energy U (x) is shown by a thick black line. Upper thin black line - infinite unbound motion. Green line - infinite bound motion (two cases), vertical green lines show the coordinates of turning points. Blue line, right, - bound infinite. Blue, left, - bound, infinite, requires infinite time to reach the turning point. Blue, middle, - bound, infinite period. Red, left, right, bound infinite. Red, middle, - bound finite, periodic.

3

Central forces

Central forces correspond to the potential energy of the form U = U (r) where r = |r|. As we have found already in this case the angular momentum J = r×p is conserved. This means that the vectors 3

Physics 1. Mechanics

Lecture 9

r and p are always in the the plane perpendicular to the constant vector J so that the motion is two-dimensional. Let the particle move in the x − y plane. It is convenient to use polar coordinates ˆr = r/r = cos ϕˆ x = r cos ϕ, y = r sin ϕ. We shall also introduce the unit vectors e ex + sin ϕˆ ey ˆϕ = − sin ϕˆ and e ex + cos ϕˆ ey . It should be understood that x, y, r, ϕ are coordinates of the moving ˆr and e ˆϕ . Using this notation particle and therefore all depend on time, as well as the unit vectors e we can write v = r˙ = xˆ ˙ ex + yˆ ˙ ey (r˙ cos ϕ − r sin ϕϕ)ˆ ˙ ex + (r˙ sin ϕ + r cos ϕϕ)ˆ ˙ ey

(5)

= rˆ ˙ er + rϕˆ ˙ eϕ The angular momentum has only one component J = (0, 0, J), J = xpy − ypx = m(xy˙ − y x) ˙ = m[r cos ϕ(r˙ sin ϕ + r cos ϕϕ) ˙ − r sin ϕ(r˙ cos ϕ − r sin ϕϕ)] ˙ = mr2 ϕ˙ and this immediately gives us J . mr2 Now we shall use the energy conservation in the form ϕ˙ =

(6)

m 2 v + U (r) = E = const 2 Substituting (5) we have

m 2 (r˙ + r2 ϕ˙ 2 ) + U (r) = E 2

or finally m dr 2 ( ) = E − Uef f (r), 2 dt

(7)

where

J2 (8) 2mr2 is the effective potential. With the use of the effective potential the problem is reduced to onedimensional analyzed earlier with the only difference that 0 ≤ r < ∞ (and not −∞ < x < ∞). Uef f = U (r) +

4

Physics 1. Mechanics

Lecture 9

u 40 20

1

2

3

4

5

r

-20 -40 -60 Figure 2: Potential U = −a2 /r (blue) and effective potential (red). It is of interest to analyze possible trajectories for the potentials of the type U (r) = A/rα , where α > 0. If A > 0 the potential is repelling: a particle cannot come close to r = 0 even in the case of J = 0. The centrifugal part J 2 /2mr2 is always repelling and prevents the particle from coming close to the center r = 0. Thus, for A > 0 the only possible trajectories are infinite trajectories bound from lower r, and the energy is always positive E > 0. If A < 0 the potential is attractive. If J = 0 the particle would fall onto the center for all initial radial velocities (that is, all energies). Presence of the centrifugal part may prevent such fall. If α < 2 the effective potential has a minimum at r∗ such that (dUef f /dr)r=r∗ = 0, that is, at r∗ = (−J 2 /αmA)1/(2−α) (A < 0). The effective potential Uef f → ∞ when r → 0 and Uef f → 0 (while remaining negative) when r → ∞. As a result, all trajectories with E > 0 are infinite and bound only from the lower r side. All trajectories with E < 0 are finite and bound, that is, the motion occurs between the two radii r1 < r < r2 , such that Uef f (r1 ) = Uef f (r2 ) = E < 0. Exercise: Show that if α > 2 a particle can fall onto the center even if J 6= 0. What happens when α = 2 ? The solution of (7) is Z r dr0 p . (9) t − t0 = ± (2/m)(E − Uef f ) r0 p Using (dϕ/dt) = J/mr2 and dr/dt = ± (2/m)(E − Uef f ) we find dϕ J =± p 2 dr r 2m(E − Uef f )

(10)

5

Physics 1. Mechanics

Lecture 9

and further

Z

r

ϕ − ϕ0 = ± r0

3.1

Jdr0 p . r0 2 2m(E − Uef f )

(11)

Example: Coulomb potential

The potential energy of the form U = −k/r is called Coulomb potential. It describes a charged particle in the field of a point charge or a point mass in the gravitational field of a spherical body. In what follows we consider the gravitational field which is attractive, so that k > 0. Exercise: Using substitution u = 1/r and the integral Z

√

du 1 2cu − b = √ arccos( √ ) 2 c a + bu − cu b2 + 4ac

find the solution of (11) in the form r = p/[1 +  cos(ϕ − ϕ0 )] where p = J 2 /mk, and  = p 1 + 2EJ 2 /mk 2 . The trajectory found in the last exercise is an ellipse with the major semiaxis a = (rmax +rmin )/2 = p/(1 − 2 ) = −k/2E, if  < 1. It is obvious that the orbit is circular if  = 0, that is, E = −mk 2 /2J 2 (ATTENTION: finite orbits correspond to negative values of energy E !). When  = 1 the orbit is an infinite parabola, and when  > 1 the orbit is a hyperbola (also infinite). In both cases the energy is positive.

6

Physics 1. Mechanics

Lecture 9

y 4

2

2

4

6

8

x

-2

-4 Figure 3: Trajectories in Coulomb potential. Black - circular trajectory,  = 0, red - ellipse, 0 <  < 1, green - parabola,  = 1, blue - hyperbola,  > 1. Angular momentum J is the same for all orbits, energy E changes.

7

Physics 1. Mechanics Advanced 9. Orbits in a central potential

1

Epicyclic approximation

As we have seen earlier a particle orbit in a central potential U (r) is determined by two conservation laws. The angular momentum conservation gives J mr2

(1)

2 (E − Uef f ), m

(2)

ϕ˙ = where J = const. The energy conservation gives r˙ 2 = 2

J where E = const and Uef f = U + 2mr 2 . (The conserved quantities J and E are also called integrals of motion.) Although (2) can be formally solved in the form of an integral (is fully integrable), in most cases the integral cannot be calculated in terms of elementary functions which makes it difficult to understand the motion. In many cases approximate methods provide more clear information. The simplest orbit is, of course, a circular orbit. In this case r˙ = 0 which means that the orbit radius should be find as a minimum of Uef f and E and J are not independent at the orbit. Let r0 be the minimum, that is, dUef f /dr = 0 at r = r0 . In other words, dU/dr − J 2 /mr03 = 0. The energy and angular momentum are relate by E = Uef f (r0 ) = U (r0 ) + J 2 /2mr02 . The angular velocity on this J trajectory is ϕ˙ = mr 2 0 In order to move away from the circular trajectory we have to break the relation between E and J. It is sufficient to keep J constant and change E. We are interested in the trajectories which are close to the found circular orbit, so that we shall assume that E = E0 + E1 = Uef f (r0 ) + E1 and E1 is sufficiently small. As usual, the condition of the smallness will be found after we find the new trajectory. It is worth noting that E1 > 0 since we were already in the minimum. Now r = r0 + r1 , where r1 is the perturbation. Taylor expanding the potential near the minimum we have

Uef f (r) = Uef f (r0 ) + 12 r12

d2 Uef f dr2 |r=r0

(3)

To make the notation more compact let us denote Uef f (r0 ) = U¯0 , U (r0 ) = U0 , (d2 U/dr2 )|r=r0 = U 00 , 1

Physics 1. Mechanics

Advanced 9

(d2 Uef f /dr2 )|r=r0 = (U 00 + 3J 2 /mr04 ) = mκ2 . Then the energy conservation will give r˙12 + κ2 r12 =

2E1 m

(4)

The solution of this equation is 1 r1 = κ

r

2E1 cos(κt − φ). m

(5)

For ϕ we have J 2J − r1 2 mr0 mr03 r J 2J 2E1 = − cos(κt − φ) mr02 κmr03 m

ϕ˙ =

so that J 2J ϕ= t− 2 3 2 mr0 κ mr0

r

2E1 sin(κt − φ). m

(6)

(7)

This is the so called epicyclic approximation. It represents the motion as the “drift” on the circle on the radius r0 and “rotation” on the epicycles (perturbations).

2

Keplerian potential

The potential U = −A/r is called Keplerian (or Coulomb). We know that all closed orbits in this potential are ellipses. Let us apply the above method to the analysis of ellipses with small eccentricity (deviations from circular shape). We have Uef f = −

J2 A + r 2mr2

and

dUef f A J2 = 2− =0 dr r mr3 gives the radius of the circular orbit as r0 = J 2 /mA. Let A = mG (as for the gravitation). Then it is appropriate to use the normalized variable l = J/m and e = E/m, so that r0 = l2 /G. The angular velocity of the circular motion is Ω = ϕ˙ = l/r02 = G2 /l3 , while the total normalized energy is e0 = −G/2r0 = −G2 /2l2 . Now 2

κ =



2G 3l2 − 3 + 4 r0 r0



= Ω2

2

Physics 1. Mechanics

Advanced 9

Thus, the Keplerian ellipse is represented with the epicyclic motion with the equal orbital Ω and epicyclic κ angular velocities.

3

Physics 1. Mechanics Lecture 10. Motion in a potential, II, Harmonic oscillations

1

Importance of the oscillator potential

Let us consider a particle moving in an one-dimensional potential field, that is, having potential energy U(x) depending only on one coordinate. As we already know we can ignore two other coordinates (y and z) and concentrate on x only. The equation of motion is m

d2 x dU =− , 2 dt dx

(1)

where partial derivative ∂U/∂x → dU/dx since only one coordinate is changing. The energy conservation is now 1 mx˙ 2 2

+ U(x) = E = const.

(2)

Let us assume that the particle is in an equilibrium at the point x = x0 , that is, x¨ = 0, x˙ = 0. The first condition (zero acceleration) immediately means that the force (dU/dx)x=x0 = 0 . If we perturb a little the particle it will no longer be in the equilibrium but will move so that (hopefully) its coordinate is close to x = x0 . If the difference |x − x0 | is sufficiently small (which should be checked separately in each case) then we can write approximately U(x) = U(x0 ) +

dU d2 U (x − x0 ) + 21 2 (x − x0 )2 , dx |x=x0 dx |x=x0

(3)

which is nothing but the Taylor expansion of the potential energy near the equilibrium point x = x0 . In (3) the derivatives are taken in the equilibrium point, and, therefore, are simply constants. Moreover, in the equilibrium point (dU/dx)x=x0 = 0. Let us denote U0 ≡ U(x0 ), k ≡ (d2 U/dx2 )|x=x0 , and for simplicity put x0 = 0 and U0 = 0 (we can always do that), so that U(x) = U0 +

1

kx2 . 2

(4)

Physics 1. Mechanics

Lecture 10 U

8

6

4

2

-2

-1

1

2

x

Figure 1: Approximation of the potential energy near stable equilibrium by (4). Now the equation of motion and energy conservation take the form: m¨ x = −kx,

(5)

1 mx˙ 2 2

(6)

+ 21 kx2 = E = const.

Equations (5) and (6) describe the so called harmonic oscillations. Note that the derivation of these equations is quite general, which means that every stable system harmonically oscillates when perturbed from equilibrium.

2

Harmonic oscillator: Direct solution

The solution of the equations for motion for a harmonic oscillator can be obtained directly from the energy conservation (as we did in the general one-dimensional case) by solving for x: ˙ p x˙ = ± (2E − kx2 )/m and integrating t − t0 = ±

Z

x

x0

dx p . (2E − kx2 )/m

Q: Show that the solution is of the form x = A cos(ωt − φ) and find A, ω, and φ.

2

Physics 1. Mechanics

3

Lecture 10

Harmonic oscillator: General method

The above method is not good if there are more than one degrees of freedom (motion along more than one coordinate) or there are additional nonconservative forces which break the energy conservation (we shall add such in future). There is another, a more general and very powerful method which is based on solving linear differential equations with constant coefficients. We start with (5) and assume that a solution can be written in the form x = C exp(pt),

(7)

where p and C are to be found upon substitution of (7) into (5). In fact, this is not an assumption: any solution of (5) must be of the form x = C exp(pt). Substituting, we obtain p2 = −(k/m),

(8)

while C is arbitrary (in fact, it has to be determined from initial conditions, we’ll do that later). p The solution of (8) is p = ±iω, ω = k/m. Thus, there are two possibilities, x = C1 exp(iωt) and x = C2 exp(−ωt). Since we have two initial conditions, x(t = 0) = x0 and vx (t = 0) = v0 we have to combine the two into the general solution x = C1 exp(iωt) + C2 exp(−iωt).

(9)

ATTENTION: C1 and C2 are complex constants, in general, while x must be real. From (9) we find the velocity in the form vx = x˙ = iω[C1 exp(iωt) − C2 exp(−iωt)],

(10)

Now, taking into account the initial conditions, we have C1 + C2 = x0 ,

iω(C1 − C2 ) = v0

3

Physics 1. Mechanics

Lecture 10

and eventually v0 v0 x = 12 (x0 + ) exp(iωt) + 21 (x0 − ) exp(−iωt)  iω    iω v0 exp(iωt) − exp(−iωt) exp(iωt) + exp(−iωt) + = x0 2 ω 2i v0 = x0 cos(ωt) + sin(ωt) ω = A cos(ωt − φ), q A = x20 + (v0 /ω)2 , x0 = A cos φ,

(11) (12) (13) (14) (15)

v0 /ω = A sin φ

(16)

A is called the amplitude of the oscillations and φ is called the phase. The amplitude shows the maximum displacement from the equilibrium A = |x|max . The phase gives the relation between the initial velocity and initial displacement.

x 1

0.5

2

4

6

8

10

12

14

t

-0.5

-1 Figure 2: Coordinate (red solid line) and velocity (blue dashed line ) as functions of time. It is easy to show that the energy of the particle is E=

4

m 2 k 2 kA2 v + x = = const 2 x 2 2

(17)

Damping oscillations

Here we are going to add a friction force of special kind, F = −γvx . When friction is present the energy is no longer conserved and we cannot use the energy conservation but have to solve the 4

Physics 1. Mechanics

Lecture 10

equation of motion which is now m¨ x = −kx − γvx

(18)

x¨ + Γx˙ + ω 2 x = 0,

(19)

or, in a more widely used form,

where Γ = γ/m, and ω 2 = k/m. Equation (19) is also a linear differential equation with constant coefficients and is also solved using substitution x = C exp(pt), which gives p2 + Γp + ω 2 = 0

(20)

so that

Γ p 2 (21) ± Γ /4 − ω 2 . 2 Since there are two different solutions p1 6= p2 the general solution will again have the form x = C1 exp(p1 t) + C2 exp(p2 t), where C1 and C2 are determined from the initial conditions. Let us first consider the case of strong friction, Γ/2 > ω. In this case both pi < 0, so that exp(pi t) p1,2 = −

are monotonically decreasing functions of time. The total energy m k E = x˙ 2 + x2 2 2 m 2 = [x˙ + ω 2x2 ] 2 m = [(p1 C1 exp(p1 t) + p2 C2 exp(p2 t))2 + ω 2(C1 exp(p1 t) + C2 exp(p2 t))2 ] 2 m 2 = [(p1 + ω 2 )C12 exp(2p1 t) + (p22 + ω 2)C22 exp(2p2 t) + 2(p1 p2 + ω 2 )C1 C2 exp((p1 + p2 )t)] 2

(22)

Each term in the last expression is positive and decreases with time, which means that the total energy monotonically decreases. The motion is strongly damped. Let us now consider the case of weak friction, Γ/2 < ω. In this case p1,2 = −

Γ ± i˜ ω 2

(23)

where ω ˜=

p

ω 2 − (Γ/2)2 ,

(24)

and the general solution can be written as follows: x = [C1 exp(i˜ ω t) + C2 exp(−i˜ ω t)] exp(−Γt/2).

(25)

We have seen earlier that the expression in brackets can be written as follows: C1 exp(i˜ ω t) + C2 exp(−i˜ ω t) = A cos(˜ ω t − φ) 5

Physics 1. Mechanics

Lecture 10

and describes harmonic oscillations. Now (25) can be rewritten as x = A(t) cos(˜ ω t − φ)

(26)

where now the “amplitude” A(t) = A0 exp(−Γt/2) monotonically decreases with time (A0 is the amplitude at t = 0). The cos term is periodic in time and describes oscillations, while exp(−Γt/2) describes damping. Thus, altogether (26) describes damped oscillations. During these oscillations the displacement crosses x = 0 each T = 2π/˜ ω , that is, infinite number of times, while the maximum displacement during each period T monotonically decreases.

x 0.8 0.6 0.4 0.2 2

4

6

8

10

12

14

t

-0.2 -0.4 Figure 3: Coordinate (red solid line) and velocity (blued dashed line) in the case of weak damping. The power of the friction force P = F · v = −γvx2 ≤ 0 so that the total mechanical energy E = K + U should decrease monotonically. Using the expression (26) we find the velocity in the form Γ vx = x˙ = − A(t) cos(˜ ω t − φ) − ω ˜ A(t) sin(˜ ω t − φ). (27) 2 Substituting (26) and (27) into E = mvx2 /2 + kx2 /2 = (m/2)(vx2 + ω 2x2 ) we find  m Γ2 E(t) = cos2 (˜ ω t − φ) + ω ˜ 2 sin2 (˜ ω t − φ) 2 4  +Γ˜ ω cos(˜ ω t − φ) sin(˜ ω t − φ) + ω 2 cos2 (˜ ω t − φ) A20 exp(−Γt)   m 2 Γ2 Γ˜ ω = ω + cos(2˜ ω t − 2φ) + sin(2˜ ω t − 2φ) A20 exp(−Γt) 2 4 2

(28)

It can be shown that the energy E(t) decreases but we get more physical information when speaking 6

Physics 1. Mechanics

Lecture 10

about the average energy which is defined as ¯ = 1 E(t) T

Z

t+T

E(t)dt

(29)

t

Doing such averaging in our case we find ¯ = k [A(t)]2 = k A2 exp(−Γt) E(t) 2 2 0 We see that

(30)

¯ + 1/Γ) E(t 1 , = ¯ e E(t)

thus the time τ = 1/Γ describes the rate of the change of the average oscillation energy. It is called damping or decay time.

e 1 0.8 0.6 0.4 0.2

2

4

6

8

10

12

14

t

Figure 4: Energy (red solid line) and period average energy (blue dashed line) for the weak damping case.

5

Forced oscillations

Oscillations which occur in the system when slightly perturbed from the equilibrium and no other force is present are called natural oscillations or a normal mode and occur at the natural frequency p ω = k/m. The friction force of the kind Fx = −γvx changes the oscillation frequency to ω ¯ = p ω 2 − (Γ/2)2 and brings about damping. In this case we shall call the oscillations damping natural oscillations or a damping normal mode. Now we are going to consider the case where such system is under the influence of an external force of the kind Fext = F0 cos(˜ ω t). That means that we have to 7

Physics 1. Mechanics

Lecture 10

solve the equation of the form x¨ + Γx˙ + ω 2x = (F0 /m) cos(˜ ω t)

(31)

We shall try the solution of the form x = A cos(˜ ω t − φ0 ). Then x = A[cos(˜ ω t) cos φ0 − sin(˜ ω t) sin φ0 ],

(32)

x˙ = ω ˜ A[− sin(˜ ω t) cos φ0 − cos(˜ ω t) sin φ0 ],

(33)

x¨ = −˜ ω 2 A[cos(˜ ω t) cos φ0 − sin(˜ ω t) sin φ0 ]

(34)

Substituting this into (31) and collecting all terms with cos(˜ ω t) and with sin(˜ ω t) separately, we arrive at two equations [(ω 2 − ω ˜ 2 ) cos φ0 − Γ˜ ω sin φ0 ]A = (F0 /m),

(35)

(ω 2 − ω ˜ 2 ) sin φ0 + Γ˜ ω cos φ0 = 0

(36)

From the second equation we have ω2 − ω ˜2 , cos φ0 = [(ω 2 − ω ˜ 2)2 + Γ2 ω ˜ 2 ]1/2 Γ˜ ω sin φ0 = − 2 , [(ω − ω ˜ 2 )2 + Γ2 ω ˜ 2 ]1/2

(37) (38)

and substituting into (35) we find eventually A=

1 F0 . 2 2 2 m [(ω − ω ˜ ) + Γ2 ω ˜ 2 ]1/2

(39)

The dependence of the amplitude A on ω ˜ /ω for several ratios Γ/ω is shown in Fig 5. When Γ/ω  1 the amplitude reaches its maximum at ω ˜ ≈ ω (not exactly but very close) and the maximum is very sharp. This phenomenon is called a resonance.

8

Physics 1. Mechanics

Lecture 10

A - amplitude

20

15

10

5

0.5

1

1.5

2

ω ˜ /ω

Figure 5: Amplitude of forced oscillations as a function of ω ˜ /ω for various values of γ/ω. For lower γ/ω the maximum is higher and more narrow.

6

Oscillations in physical systems

Whenever motion of a physical system is described by equation (5) or its energy can be written similarly to (6) it experiences harmonic oscillations.

6.1

A mass on a spring

In this case the energy is exactly as (6) so that the derived expressions can be directly applied to the mass on a spring.

6.2

A pendulum

When a pendulum deviates by the angle θ its energy is E=

m 2 ˙2 l θ + mgl(1 − cos θ) 2

For small θ  1 we use the Taylor expansion cos θ = 1 − θ2 /2, so that E=

ml2 ˙2 mgl 2 θ + θ 2 2

9

Physics 1. Mechanics

Lecture 10

Comparing with (6) we see that m → ml2 , k → mgl, and x → θ. We, therefore, immediately conclude that the motion can be described as follows: θ = A cos(ωt − φ), where ω =

p g/l.

10

Physics 1. Mechanics Advanced 10. Oscillations, several degrees of freedom

1

Two degrees of freedom

We shall learn now how to deal with the oscillator-like problems when there is more than one degree of freedom. In this section we restrict ourselves with two degrees of freedom, that is, the potential U = U (x1 , x2 ). Respectively, we shall write the equations of motion in the form x˙1 = v1 ,

(1)

x˙2 = v2 ,

(2)

∂U , ∂x1 ∂U mv˙2 = − . ∂x2

mv˙1 = −

(3) (4)

The equations are deliberately written as a system of the first order differential equations. We assume that there is an equilibrium x1 = X1 , x2 = X2 ,where x˙i = 0, i = 1, 2. This is possible only if (∂U/∂xi )|xi =Xi = 0. Now let us, as usual, perturb the system from the equilibrium: xi (t) = Xi + x˜i (t), vi (t) = v˜i (t). If we remain (at least in the beginning) near the equilibrium point, the potential U (x1 , x2 ) can be Taylor expanded (we need to expand up to the second order since the first derivatives are zero): ∂U ∂U + x˜2 ∂x1 |xi =Xi ∂x2 |xi =Xi ∂2U ∂2U ∂2U + 12 x˜21 2 + x˜1 x˜2 + 12 x˜22 2 ∂x1 |xi =Xi ∂x1 ∂x2 |xi =Xi ∂x2 |xi =Xi X ∂2U = U (X1 , X2 ) + 21 x˜i x˜j ∂xi ∂xj i,j X = U0 + 12 kij x˜i x˜j .

U (x1 , x2 ) = U (X1 , X2 ) + x˜1

ij

1

(5)

Physics 1. Mechanics

Advanced 10

The equations of motion will be rewritten as follows ˙i = v˜i , ˜x X ˜v˙i = − κij x˜j ,

(6) (7)

j

where κij = kij /m. A more formal approach would be to defined a column vector   x˜1   x˜2   r=  v˜   1 v˜2 and a matrix



0 0   0 0 T= −κ  11 −κ12 −κ21 −κ22

(8)

1 0 0 0

 0  1  0  0

(9)

so that the equations of motion will be written as r˙ = Tr,

(10)

Now we start looking for a solution of the kind x˜i = Ai exp(pt), v˜i = Bi exp(pt), or in terms of column vectors   A1    A2   r(t) =  (11) B  exp(pt) = r0 exp(pt)  1 B2 Substituting this into (10) we have pr = Tr → (T − pI)r = 0

(12)

The last equation has nonzero solutions only if det ||T − pI|| = 0..

(13)

Eq. (13) is a fourth order equation with respect to p. In our particular case it looks as follows: (p2 + κ11 )(p2 + κ22 ) − κ212 = 0

(14)

2

Physics 1. Mechanics

Advanced 10

(since κ12 = κ21 ), and has the solutions p21 p22

κ11 + κ22 − =− 2 κ11 + κ22 =− + 2

s s

κ11 − κ22 2

2

κ11 − κ22 2

2

+ κ212 ,

(15)

+ κ212 .

(16)

One solution, p21 < 0, which means that we have two possibilities: r ∝ exp(iω1 t) and r ∝ exp(−iω1 t) where ω12 = −p21 > 0. These are oscillating solutions. The second option, p22 , is negative is κ11 κ22 > κ212 and is positive otherwise. 1.0.1

p22 negative

In this case the other options are ∝ exp(±iω2 t), where ω22 = −p22 > 0. The general solution would be r = r1 exp(iω1 t) + r2 exp(−iω1 t) + r3 exp(iω2 t) + r4 exp(−iω2 t) (17) where ri , i = 1, . . . , 4 are constant column vectors. In a more general way, we say that there are four frequencies - normal modes, and the general solution is a superposition of these normal modes. 1.0.2

p22 positive

In this case we have one of the solutions of the type ∝ exp(p2 t), where p2 > 0 (the other one with p2 < 0 tends to zero as t → ∞). This solution is called unstable.

2

General approach

Systems can consist of a larger number of particles, like, for example, several masses connected with springs, so that there may be a rather large number of degrees of freedom. In general, such systems can be represented near the equilibrium by the equation of motion similar to (10) only the size of the vector r and matrix T corresponds to twice the number of degrees of freedom. The procedure which leads to this system is called linearization since the obtained equations are linear. We always search for solutions ∝ exp(pt), and possible p are found from the equation (13). Each imaginary p corresponds to a normal mode of the system. If Re(p) < 0 the mode is said to be damping. If Re(p) > 0 the mode is unstable.

3

Physics 1. Mechanics Lecture 11. Rigid body rotation I, Theory

1

Rotation of a rigid body

A rigid body is a body in which the distance between each two points |r(1) − r(2) | does not change during its motion. Let us fix a point in the body, say P0 (r(0) ), and consider the motion of all other points relative to this reference point. Since for each point r(i) the distance |r(i) − r(0) | does not change, that means that at every moment the point i moves on a circle with the radius |r(i) − r(0) | around the reference point.

Figure 1: Rotation of a rigid body. As we already know such motion is described by an angular velocity vector ω (i) (which, in general, depends on time) and its velocity (relative to the reference point) is v(i) = ω (i) × (r(i) − r(0) ). Indeed, dr(i) d|r(i) − r(0) |2 = 2(r(i) − r(0) ) · =0 dt dt This is right for all points. Moreover, the distance |r(i) − r(j) | should not change for any i and j, 1

Physics 1. Mechanics

Lecture 11

which gives, as above d|r(i) − r(j) |2 dr(i) dr(j) = 2(r(i) − r(j) ) · ( −( ) dt dt dt = 2(r(i) − r(j) ) · (v(i) − v(j) ) = 2(r(i) − r(j) ) · [ω (i) × (r(i) − r(0) ) − ω (j) (r(j) − r(0) )] = 0 It is easy to see that in order that the last equality be satisfied ω (i) = ω (j) , that is, all point of the body rotate with the same angular velocity ω. Thus, at any time t each point of the body has the velocity (relative to r(0) ) v(i) = ω × r0 (i) , where we used the notation r0 (i) = r(i) − r(0) . Let us calculate the angular momentum and kinetic energy of this motion, assuming that the point r(0) is at rest. X

J=

r0 (i) × p(i) =

(i)

=

X

X

r0 (i) × m(i) (ω × r0 (i) )

(1)

(i) 2

m(i) [r 0 (i) ω − r0 (i) (r0 (i) · ω)]

(2)

X

(3)

(i)

K=

1 2

m(i) [ω × r0 (i) ]2

(i)

=

1 2

X

2

m(i) [r 0 (i) ω 2 − (r0 (i) · ω)2 ]

(4)

(i)

One important thing we can see immediately: J is not, in general, parallel to ω. It is more instructive to write down the angular momentum and kinetic energy in components. In what follows the indices a = 1, 2, 3 and b = 1, 2, 3 denote coordinates: r0 = (x01 , x02 , x03 ). Q: Show that the above expressions can be written as follows: Ja =

XX

m(i) (x0 (i),b x0 (i),b ωa − x0 (i),a x0 (i),b ωb )

b

(i)

XX 2 = [ m(i) (r 0 (i) δab − x0 (i),a x0 (i),b )]ωb b

=

(5) (6)

(i)

X

Iab ωb ,

(7)

b

Iab =

X

2

(8)

Iab ωa ωb

(9)

m(i) (r 0 (i) δab − x0 (i),a x0 (i),b )

(i)

K=

1 2

X a,b

The construction Iab which stands in the expressions for the angular momentum and kinetic energy is called tensor of inertia.

2

Physics 1. Mechanics

2

Lecture 11

Rotation around a fixed axis

If the axis of rotation is fixed the above expressions can be simplified. Let the body rotate around z axis, that is, ω = ωˆ z = (0, 0, ω). Then (7) and (9) give Jz = Izz ω, X X 2 2 2 Izz = m(i) (x0 (i) + y 0 (i) ) = m(i) r0 ⊥ , (i)

K=

(10) (11)

(i)

1 I ω2 2 zz

(12)

The (single) quantity Izz is called the moment of inertia with respect to z-axis. We shall write down immediately also the equations for the change of the angular momentum and kinetic energy. The P first one is dJ/dt = N = (i) r0 (i) × F(i) which gives for the z component X dJz dω X 0 0 = Izz = r i,⊥ × F(i) = (x0(i) F(i),y − y(i) F(i),x = Nz dt dt (i)

(13)

(i)

For the kinetic energy one has dK X = F(i) · v(i) dt (i) X = F(i) · (ω × r0 (i) ) (14)

(i)

=

X

ω · r0 i,⊥ × F(i)

(i)

= Nz ω We see the complete analogy with the one-dimensional particle motion:

3

Physics 1. Mechanics

Lecture 11

Particle

Rigid body

Mass m

Moment of inertia Izz

Coordinate x

Angle of rotation θ

Velocity vx = x˙

Angular velocity ω = θ˙

Acceleration ax = v˙x

Angular acceleration α = ω˙

Force Fx

Torque Nz

Momentum px = mvx

Angular momentum Jz = Izz ω

Kinetic energy K = mvx2 /2 Kinetic energy K = Izz ω 2 /2 Thus, the motion of the rigid body rotating around a fixed axis is described by a single coordinate (or a single equation of motion). It is said to have one degree of freedom, as well as a particle moving in a one-dimensional space.

3

Center-of-mass in action

In the previous sections we derived the equations of motion for a rigid body rotating around an arbitrary axis. This axis passes through some point r(0) of the body, which means that we can choose each point as a momentary reference point. However, in many cases it is physically meaningful to use the special point, the center-of-mass, as the reference point. Indeed, we know that M V˙cm = Fext

(15)

where Fext is the vector sum of all external forces acting on the body, M is the total mass, and Vcm the velocity of the center-of-mass. We also know that d (Mr × Vcm ) = Rcm × Fext dt

(16)

and J˙ int = N =

X

(r(i) − Rcm ) × F(i)

(17)

(i)

Thus, a rigid body motion can be always represented as a motion of the center-of-mass and the rotation around the center of mass. The importance of this representation is that the tensor of inertia Icm,ab (or moment of inertia) relative to the center-of-mass is well-defined and is a quantity 4

Physics 1. Mechanics

Lecture 11

which depends only on the mass distribution in the body and can be calculated once and forever for each given body. The kinetic energy of such body K=

2 X MVcm m(i) (v(i) − Vcm )2 + 12 2

(18)

(i)

where the second term (internal kinetic energy) was shown above to be equal (1/2) so that finally K=

P

2 X MVcm + 12 Icm,ab ωa ωb 2 a,b

a,b Icm,ab ωa ωb ,

(19)

A question can be asked: what is the relation between the tensor of inertia Iab calculated relative to some arbitrary point r0 and the tensor of inertia Icm,ab . In order to establish the relation we just write down the definition X Iab = m(i) [(r(i) − r0 ) · (r(i) − r0 )δab − (x(i),a − x0,a )(x(i),b − x0,b )] (i)

=

X

m(i) [((r(i) − rcm ) + (rcm − r0 )) · ((r(i) − rcm ) + (rcm − r0 ))δab

(i)

− ((x(i),a − xcm,a ) + (xcm,a − x0,a ))((x(i),a − xcm,a ) + (xcm,a − x0,a ))] X = m(i) [(r(i) − rcm ) · (r(i) − rcm )δab − (x(i),a − xcm,a )(x(i),b − xcm,b )] (i)

+

X

m(i) [(rcm − r0 ) · (rcm − r0 )δab − (xcm,a − x0,a )(xcm,b − x0,b )]

(i)

+

X

m(i) [2(r(i) − rcm ) · (r0 − rcm ) − (x(i),a − xcm,a )(x0,b − xcm,b) − (x(i),b − xcm,b )(x0,a − xcm,a )]

(i)

= Icm,ab + M[(rcm − r0 ) · (rcm − r0 )δab − (xcm,a − x0,a )(xcm,b − x0,b )] (20) P since all terms like (i) m(i) (x(i),a − xcm,a ) = 0 because of the definition of the center-of-mass. Thus, we arrive at the generalized Steiner theorem: Iab = Icm,ab + M[(rcm − r0 ) · (rcm − r0 )δab − (xcm,a − x0,a )(xcm,b − x0,b )]

(21)

For the case when the direction of the angular velocity (rotation axis) does not change (let it be in z direction) the above reduces to the more familiar Steiner theorem Izz = Icm,zz + Ml2 ,

(22)

p where l = (x0 − xcm )2 + (y0 − ycm )2 is the distance between the rotation axis and a parallel axis passing through the center-of-mass. 5

Physics 1. Mechanics

4

Lecture 11

Meaning of the tensor of inertia

P Since Ja = b Iab ωb it means that, in general, the direction of the angular momentum J does not coincide with the direction of the angular velocity ω.

Figure 2: Directions of angular velocity (red) and angular momentum (blue) do not coincide, in general. Let, for example, ω = ωˆz. Then Jx = Ixz ω,

Jy = Iyz ω,

Jz = Izz ω

(23)

The components Jx and Jy vanish only if Ixz = Iyz = 0. However, even if they vanish at some moment they can become nonzero in other moment. Indeed, look at the definition of the tensor of inertia. It includes summation over the distribution of masses. When a body rotates this distribution changes with time in the inertial frame, which means that the tensor of inertia remains constant and is good for use only in the frame rotating with the body. Let us assume that the torque N = 0 in the inertial frame. Then the angular momentum J should be constant, that is dJ/dt = 0. Writing this in components we have X J˙a = [I˙ab ωb + Iab ω˙ b ] = 0 (24) b

and since I˙ab 6= 0, in general, for a rotating body in the inertial frame, the angular velocity ω also 6

Physics 1. Mechanics

Lecture 11

changes with time, dωa /dt 6= 0. There is important exclusion from this general rule. It can be shown that for any (even extremely irregular) body it is possible to choose cartesian coordinates so that Ixy = Iyx = Ixz = Izx = Iyz = Izy = 0 (in the reference frame related to the body). Let us denote I1 = Ixx , I2 = Iyy m I3 = Izz in these coordinates. If the initial angular velocity is along one of these coordinate axes (say, z), then the angular momentum is also along this axis, J = I3 ωˆz and remains in the same direction as long as the angular momentum is conserved. In this case I˙3 = 0 and ω = const.

Figure 3: Directions of angular velocity do not coincide, of angular velocity is not directed along one of the principal axes (left). If angular velocity is directed along on of the principal axes, the angular momentum points in the same direction.

5

Degrees of freedom

So far we did not mention degrees of freedom. Number of degrees of freedom is simply a number of coordinates which are necessary to describe the motion, or number of equations of motion which have to be solved, or three components of (generalized) momenta. Single particle in a three-dimensional space has three degrees of freedom. A system of N independently moving particles has 3N degrees of freedom. A rigid body has six degrees of freedom. Three of these are related to the motion of the center-of-mass (like a particle) and three other are related to three independent rotations around three axes. Otherwise it can be said that the six degrees of freedom correspond to three components of the momentum and three component of the angular momentum.

7

Physics 1. Mechanics Lecture 12. Rigid body rotation II, Applications

1

Calculation of moment of inertia

In the following examples we usually restrict ourselves with the calculation of the moment of inertia relative to the axis passing through the center-of-mass. Moment of inertia relative to any other parallel axis can be found from the Steiner theorem I = Icm + Ml2 where l is the distance between the axes.

Figure 1: Steiner theorem Calculation of moment of inertia for a system of particles is straightforward and will not be 1

Physics 1. Mechanics

Lecture 12

considered here. The general expression reads Iab =

X

2 m(i) (r(i) δab − x(i),a x(i),b )

(1)

(i)

for the tensor of inertia, or X

Izz ==

2 m(i) (x2(i) + y(i) )

(2)

(i)

for the moment of inertia relative to the rotation axis parallel to z-axis. In both cases the coordinate origin is on the rotation axis. For a rigid body summation is substituted by integral as follows Iab =

Z

(r 2 δab − xa xb )dm,

(3)

Izz =

Z

(x2 + y 2 )dm,

(4)

where the integral should be taken over all masses. (Compare with the calculation of the center-ofmass position.)

1.1

Thin circle

A homogeneous circle (radius R, mass m) is in x − y axis. Let us find Ixx , Iyy , and Izz . Equation (4) gives Izz =

Z

2

2

(x + y )dm =

Z

2

R dm = R

2

Z

dm = MR2 .

(5)

Similarly, Ixx =

Z

Iyy =

Z

(y + z )dm = /z = 0/ =

Z

y 2dm

(6)

(x2 + z 2 )dm = /z = 0/ =

Z

x2 dm

(7)

2

2

Straightforward calculation would be more lengthy, but we can notice that (1/2)Izz , so that Ixx = Iyy = MR2 /2.

R

x2 dm =

R

y 2 dm =

2

Physics 1. Mechanics

Lecture 12

Figure 2: Thin circle: moments of inertia. This is a special case of the perpendicular axis theorem for bodies with no thickness: Ixx + Iyy =

1.2

Z

2

y dm +

Z

2

x dm =

Z

(x2 + y 2 )dm = Izz

Disk

Let a homogeneous disk (mass M and radius R) lie in the x − y plane. We shall calculate Izz and Ixx = Iyy (the rotation axis always goes through the center of the disk). In order to do so we have to divide the disk onto small (infinitesimal) parts whose position is given by suitably chosen coordinates, and the mass dm is expressed in terms of density and coordinate differentials.

3

Physics 1. Mechanics

Lecture 12

Figure 3: Disk It is convenient to use polar coordinates x = r cos ϕ, y = r sin ϕ. With the use of the surface density σ = M/πR2 the mass element would be dm = σrdrdϕ, so that R

2π

πR4 σ MR2 Izz = (x + y )dm = ( σr rdϕ)dr = = , 2 2 0 0 Z Z R Z 2π MR2 2 2 Ixx = (y + z )dm = ( σr 2 sin2 ϕrdϕ)dr = 4 0 0 Z

1.3

2

2

Z

Z

2

(8) (9)

Sphere

Here we assume that all mass M is homogeneously distributed on the surface of the sphere of the radius R, so that the surface density is σ = M/4πR2 . Because of the symmetry Ixx = Iyy = Izz so that we need to calculate only one. We shall calculate Izz using spherical coordinates x = R sin θ cos ϕ, y = R sin θ sin ϕ (we do not need z !). The infinitesimal surface element on the sphere surface has the area dA = R2 dθ sin θdϕ and dm = σdA.

4

Physics 1. Mechanics

Lecture 12

Figure 4: Sphere. Now Izz =

1.4

Z

2

2

(x + y )dm =

Z

π

( 0

Z

0

2π

2 8 R2 sin2 θσR2 sin θdϕ)dθ = πσR4 = MR2 3 3

(10)

Filled sphere

This time the mass M is homogeneously distributed in the volume of the sphere of the radius R, so that the density is ρ = M/(4πR3 /3).

5

Physics 1. Mechanics

Lecture 12

Figure 5: Filled sphere. The only difference with the previous case is that we need the volume element dV = r 2 sin θdrdθdϕ, so that dm = ρdV and the integration goes in the following way:

2

Izz =

Z

=

Z

2

2

(x + y )dm =

Z

0

R 0

R

(

Z

0

π

(

Z

2π

r 2 sin2 θρr 2 sin θdϕ)dθ)dr

0

8 8 2 πρr 4 dr = πρR5 = MR2 3 15 5

(11)

Rolling

Let a body with a spherical or cylindrical symmetry moves on a surface with friction so that the point(s) which are in contact with surface at always at rest. Then this body rotates along the axis which is parallel to the surface but perpendicular to the velocity. Such motion is called rolling. This motion can be viewed from the two points of view. First, we can say that the center of mass moves with some velocity, and the body rotates around the axis passing thought the center of mass. If the distance between the point of contact and the rotation axis is R then the center-of-mass velocity v and the angular velocity of rotation ω are related by v = ωR in order that the contact point not slide relative to the surface. In this case the kinetic energy is K = Mv 2 /2 + Icm ω 2 /2 = (M + Icm /R2 )v 2 /2. 6

Physics 1. Mechanics

Lecture 12

Figure 6: Rolling On the other hand, we can regards the body at each moment as rotating around the instantaneous rotation axis, passing thought the contact point(s). Since the velocity of the center-of-mass is v and it rotates on the radius R the corresponding angular velocity is ω = v/R, that is, the same as in the previous method of consideration. The kinetic energy is K = Iω 2/2 = (Icm + MR2 )ω 2 /2 = (Icm + MR2 )v 2 /2R2 . Since we obtained the same as in the previous method, both are consistent. Let us consider, for example, a cylinder on a slope, and assume that friction is sufficiently strong to ensure rolling (and prevent sliding). If we consider the cylinder as rotating around the instantaneous axis of rotation passing through the contact point, the following equation of motion holds: I ω˙ = N = MgR sin α.

(12)

˙ = v/R On the other hand, ω˙ = v/R ˙ = a/R so that we immediately have a=

2.1

1 MR2 g sin α = g sin α. I 1 + Icm /MR2

(13)

Physical pendulum

Let a body can rotate around the point O as shown in Fig. 8. If the center-of-mass is at the distance l from the axis of rotation, the energy can be written as follows: E = mgl(1 − cos θ) + 21 I θ˙2

7

Physics 1. Mechanics

Lecture 12

Figure 7: Cylinder rolling down a slope. and for small θ  1 (cos θ ≈ 1 − θ2 /2) we have E = 12 mglθ2 + 12 I θ˙2 . The last expression has the oscillatory form, and the oscillation frequency can immediately written p as ω = mgl/I.

8

Physics 1. Mechanics

Lecture 12

Figure 8: Physical pendulum.

9

Physics 1. Mechanics Advanced 12. Rigid body rotation

1

General

Let us look at the body rotation from the point of view of the nonrotating S and rotating S 0 frames. The rotating frame rotates with the instantaneous angular velocity of the body ω. The vectors measure by the two observers are related by a rotation A = RA0 or A0 = R−1 A. Let us write down the equation of motion J˙ = N

(1)

ot expressing with the use of the tensor of inertia I d (Iω) = N. dt

(2)

Because of the rotation of the body the tensor of inertia is time dependent in the nonrotating frame dI/dt 6= 0 but it should be constant in the frame related to the body itself, dI0 /dt = 0 We have to establish the rule of transformation for I. Kinetic energy is invariant under rotations. Remembering matrix notations it can be written as follows’ K = 21 ω T Iω

(3)

where the superscript T means transposed matrix, in particular, it makes a row from a column. Let me remind also the feature of rotations RT = R−1 . Now, using ω T = ω 0 T RT we have T

K = 21 ω 0 (RT IR)ω 0 T

= K 0 = 12 ω 0 I0 ω 0

(4)

from which we find I0 = RT IR = R−1 IR

(5)

I = RI0 R−1 .

(6)

or

1

Physics 1. Mechanics

Advanced 12

Now, substituting all this into (2) we have d d d (Iω) = (RI0 R−1 Rω 0 ) = (RI0 ω 0 ) dt dt  dt  d d R I0 ω 0 = RN0 = RI0 ω 0 + dt dt

(7) (8)

Multiplying by R−1 from the left we finally obtain ˙ 0 ω 0 = N0 I0 ω˙ 0 + R−1 RI

(9)

which is the equation of motion in the rotating frame where I0 is constant. On the other hand, we have seen earlier that ˙ = RA˙ 0 + ω × A A (10) Since a vector product ω × A transforms as a vector, we can write also ˙ = R(A˙ 0 + ω 0 × A0 ) A

(11)

˙ 0 = ω 0 × A0 R−1 RA

(12)

I0 ω˙ 0 + ω 0 × (I0 ω 0 ) = N0 .

(13)

We immediately find that

and further

In order to proceed we shall use the property of the tensor of inertia I0 which says that it is possible 0 0 to choose cartesian coordinates (x01 , x02 , x03 ) (rotating with the body) so that I11 = I1 , I22 = I2 , and 0 0 0 I33 = I3 and all other components are zero. In other words Iij = Ii δij . With this Ji = Ii ωi0 and substitution into (13) gives X Ii ω˙ 0 i + εijk ωj0 Ik ωk0 = Ni0 (14) jk

or in the extended form I1 ω˙ 0 1 + (I3 − I2 )ω20 ω30 = N10 ,

(15)

I2 ω˙ 0 2 + (I1 − I3 )ω30 ω10 = N20 ,

(16)

I3 ω˙ 0 3 + (I2 − I1 )ω10 ω20 = N30 ,

(17)

It is of interest to consider first the case where there is external torque, that is, N0 = 0. In this case ω 0 does not remain constant but changes according to I1 ω˙ 0 1 = (I2 − I3 )ω20 ω30 ,

(18)

I2 ω˙ 0 2 = (I3 − I1 )ω30 ω10 ,

(19) 2

Physics 1. Mechanics

Advanced 12 I3 ω˙ 0 3 = (I1 − I2 )ω10 ω20 ,

(20)

It is easily seen that it ω 0 is directed along one of the principal axes (x01 , x02 , x03 ) (that is, only one component is nonzero in the beginning) it remains directed along this axis and is constant. This happens in the rotating frame. The observed in the nonrotating frame would see the angular velocity vector ω rotating as ω ×ω = 0. Thus, in the nonrotating frame the angular velocity remains constant also. Let as assume now that ω 0 = ω10 eˆ1 + ω˜0 , where ω˜0 is a small perturbation. We are already familiar with the linearization procedure to derive the following equations for the perturbations: ˙ I1 ω˜10 = 0, ˙ I2 ω˜20 = (I3 − I1 )ω˜30 ω10 , ˙ I3 ω˜30 = (I1 − I2 )ω10 ω˜20 .

(21) (22) (23)

Following the standard procedure, we substitute ω˜0 ∝ exp(pt) and get 2

p2 = −(I1 − I2 )(I1 − I3 )ω 0 1 /I2 I3 ,

(24)

which is negative (stable equilibrium) if (I1 − I2 )(I1 − I3 ) > 0 (I1 is either the largest or the smallest), and positive (unstable) if I1 is between I2 and I3 . Thus, the rotation around the largest and smallest moments of inertia (principal axes) is stable, the rotation around the intermediate is unstable. It is also easy to see that if N0 = 0 the following quantity is conserved 1 1 (I ω 0 2 1 1

2

2

+ I2 ω 0 2 + I3 ω 0 3 ) = const

(25)

which is nothing but the kinetic energy.

2

Precession etc.

Let us assume that a body is symmetrical, so that I2 = I3 and it rotates around I1 with angular velocity ω10 . Let us also assume that N2 = const 6= 0 while N1 = N3 = 0. The equations of motion read I1 ω˙ 0 1 = 0,

(26)

I2 ω˙ 0 2 = (I3 − I1 )ω10 ω30 + N2 ,

(27)

I3 ω˙ 0 3 = (I1 − I2 )ω10 ω20 .

(28)

3

Physics 1. Mechanics

Advanced 12

We see that ω10 = const. There is a solution with ω20 = 0, ω30 = N2 /(I1 − I3 )ω10 = const, which corresponds to the rotation of the angular velocity vector (in the non-rotating frame). This motion is called precession.

4

Physics 1. Mechanics Lecture 13. Gravitation

1

Basic law

Every two point masses, m1 and m2 , positioned in r1 and r2 interact (attract each other) with the force F1→2 = −

Gm1 m2 (r2 − r1 ) . |r2 − r1 |3

(1)

Comparing with what is already known to us we find that this force is conservative and can be derived from the potential energy U = U(r) = −

2

Gm1 m2 , r

r ≡ |r2 − r1 |.

(2)

Two-body problem

Let there are two gravitationally interacting particles without any additional external forces. The equations of motion can be written as follows: Gm1 m2 (r1 − r2 ) , |r2 − r1 |3 Gm1 m2 (r2 − r1 ) . m2¨r2 = − |r2 − r1 |3 m1¨r1 = −

(3) (4)

Since the forces are internal, the center-of-mass Rcm = (m1 r1 + m2 r2 )/(m1 + m2 ) is in the rest all the time (this can be checked once again by adding equations (3) and (4). On the other hand, dividing (3) by m1 , (4) by m2 and subtracting the first from the second, we have ¨r = −

G(m1 + m2 )r , r3

(5)

where r = r2 − r1 . In order to bring the force to the same expression as above it is widely accepted to introduce the reduced mass µ = m1 m2 /(m1 + m2 ). With the help of the reduced mass the equation of motion is written as follows: µ¨r = −

Gm1 m2 r r3 1

(6)

Physics 1. Mechanics

Lecture 13

Thus the two-body problem is reduced to the motion of a single particle with the reduced mass µ in the same potential energy U = −Gm1 m2 /r. This problem has been solved already. Once r is found the position vectors of the the two bodies are immediately obtained from r1 = −m2 r/(m1 +m2 )+Rcm and r2 = m1 r/(m1 + m2 ) + Rcm .

3

Gravitational attraction by spherically symmetric distributions

Let us assume that a mass M is homogeneously distributed on the surface of a thin (zero thickness) spherical shell of the radius R. Let us put a test mass m at the distance r from the center of the shell and calculate the potential energy of the interaction of the test mass with the whole shell. If we divide the sphere onto small parts dM the potential energy of the interaction of the test mass with this part is dU = −GmdM/|r − r |, where r is the position vector of m and r is the position vector of dM. The total potential energy is then U =−




GmdM |r − r |

(7)

In order to calculate this integral let us choose the coordinate system so that the origin is in the center of the sphere and the test mass is on z axis, that is, r = (0, 0, r). It is convenient to use spherical co ordinates where r = (R sin θ cos ϕ, R sin θ sin ϕ, R cos θ). Then |r − r | = R2 sin2 θ + (r − R cos θ)2 , and dM = σR2 sin θdθdϕ. Here σ = M/4πR2 . With all this U = −GmσR

2




0

π

(




0

2π

sin θ 

R2 sin2 θ + (r − R cos θ)2

dϕ)dθ

(8)

and substituting z = cos θ, k = r/R
dz GmM 1  U =− 2R (1 + k 2 ) − 2kz −1 GmM , if r > R =− r GmM , if r < R. =− R

(9) (10) (11)

A particle outside the shell has the potential energy which is the same as it would be if the whole mass of the shell be in the center. The potential energy of a particle inside the shell is constant. Since F = −(∂U/∂r), the particle outside the shell experiences the force as if the whole mass of the shell is in the center. There is not force on the particle inside the shell. This can be easily generalized onto any spherical distribution of mass, since it can be divided onto many thin shells. Thus, let us

2

Physics 1. Mechanics

Lecture 13

assume that the mass distributed so that the density depends on the radius only ρ = ρ(r). Let us out a test particle m at some radius R, then the above analysis shows that only the mass inside the sphere r < R affects the particle as if the whole mass M(R) =




R

4πρ(r)r 2dr

0

is in the center, so that GM(R)m . R2 The material which is outside the sphere, r > R, does not affect the test particle at all. |F| =

3

Physics 1. Mechanics Lecture 14. Basics of special relativity

1

Observer and event

Special relativity heavily uses two concepts: observer and event. An observer is a physical body which can server as a basis for a reference frame. Since the velocity of each body cannot exceed the speed of light, no observer can exist which moves with a superluminal speed. An event is something which happens in the world. Unless the time duration of the event is of special importance, it is assumed to be zero. That is, all time intervals should be measured between two events, as well as spatial separation.

2

Basic principles

The following principles are the cornerstones of special relativity. 1. Inertial frames remain the specially chosen frames where the laws of physics are formulated. 2. No physical body can move at a velocity higher than the light speed in the vacuum. As a result the synchronizing signals propagate with the speed of light and not with infinite velocity. 3. Speed of light is the same in all frames. 4. As a result, each observer measures his own coordinates and time. Time becomes a fourth coordinate. Our space is a four-dimensional space-time. 5. Let S and S
be two inertial frames (observers) and let the two observers measure two events, A1 and A2 . According to S the event A1 occurred in r1 in the moment t1 , the event A2 occurred in r2 in the moment t2 , the corresponding measurements by S
are denoted by
. Then (r2 − r1 )2 − c2 (t2 − t1 )2 = (r
2 − r
1 )2 − c2 (t
2 − t
1 )2

1

(1)

Physics 1. Mechanics

3

Lecture 14

Lorentz transformations

Let S
move relative to S in x direction with the velocity v0 , both coordinate frames are cartesian and the axes are parallel, and the origins coincide initially. If the event coordinates are (x, y, z, t) and (x
, y
, z
, t
), then they are related as follows y
= y,

z
= z,

(2)

x
= γ0 (x − v0 t),

(3)

t
= γ0 (t − v0 x/c2 ),
γ0 = 1/ 1 − v02 /c2 .

(4) (5)

 For any velocity v the quantity γ = 1/ 1 − v 2 /c2 is called a Lorentz-factor. The inverse transformation reads y = y
,

3.1

z = z
,

(6)

x = γ0 (x
+ v0 t
),

(7)

t = γ0 (t
+ v0 x
/c2 ),
γ0 = 1/ 1 − v02 /c2 .

(8) (9)

Simultaneous events

Let two events be observed by the moving observer S
simultaneously, t
1 = t
2 but at different positions x
1
= x
2 . The observer in rest S will see this events at t1
= t2 , and t2 − t1 = γ0 v0 (x
2 − x
1 )/c2 . This means that two events may be simultaneous in one frame and not simultaneous in another. The spatial separation between the two events in S is x2 − x1 = γ0 (x
2 − x
1 ). Let two events be observed by S
in the same position, x
1 = x
2 but at different times, t
1
= t
2 . They are observed by S at different positions, x2 − x1 = γ0 v0 (t
2 − t
1 ). The time interval between the two events in S is t2 − t1 = γ0 (t
2 − t
1 ).

3.2

World lines

In what follows we restrict ourselves with x and t only and forget about y and z for simplicity. A world line of a particle is the line in the x − t plane, that is, x(t) (or x
(t
) in the moving frame S
). Let us consider now two particles which are at rest in the frame S
, that is, x
1 (t
) = const and x
2 (t
) = const. The distance between the particles which the moving observer S
measures is l
= |x
2 − x
1 |. This distance is measured so that the two coordinates are measured simultaneously by S
. If we want to know what distance S measures we have to require that the coordinate measurements in S are simultaneous. Thus, writing x
1 = γ0 (x1 − v0 t1 ), x
2 = γ0 (x2 − v0 t2 ) and t1 = t2 we get 2

Physics 1. Mechanics

Lecture 14

l = |x2 − x1 | = |x
2 − x
1 |/γ0 = l
/γ0 , that is the observer at rest measure small length then the observer moving with the particles. Example 3.1.

L et a rigid ruler move with the velocity v0 = (v0 , 0, 0) and let its length, as

measured in its rest frame S
(moving with the same velocity) be l
. What is it length measured in the non-moving frame S if it lies along x axis ? Let us denote the coordinates of the two ends of the ruler in its rest frame S
as x
1 and x
2 . These coordinates remain constant and the ruler length in this frame is simply l
= x
2 − x
1 . In order to measure the ruler length in S we have to measure the coordinates of the both ends simultaneously, that is, t1 = t2 . Then we have x
2 = γ0 (x2 − v0 t1 ),

(10)

x
1 = γ0 (x1 − v0 t1 ),

l
= x
2 − x
1 = γ0 (x2 − x1 ) = γ0 l

(11)

y
S

v0




y S

l
x
l x Example 3.2.

L et the same rigid ruler lie at an angle θ
in S
. What would be the angle

in S ? According to the above we have lx = lx
/γ0 , and ly
= ly . Since tan θ = ly /lx we have tan θ = ly
/(lx
/γ0 ) = γ0 tan θ
.

3

Physics 1. Mechanics

Lecture 14

y
S




v0

y S

x


x Let is now assume that certain event occurs in some point x
of S
so that is starts at t
1 and ends at t
2 . The observer S
measures the duration of the event T
= |t
2 − t
1 |. The observer S will the event starting at the moment t1 = γ0 (t
1 + v0 x
/c2 ) and ending at t2 = γ0 (t
2 + v0 x
/c2 ), so that the event duration, according to S, is T = |t − 2 − t1 | = γ0 |t
2 − t
1 | = γ0 T
.

3.3

Velocity transformation

According to the velocity definition, dx dy dz , vy = , vz = , dt dt dt

dx dy dz
vx
=
, vy
=
, vz
=
. dt dt dt

vx =

(12) (13)

From the Lorentz transformations we have dy
= dy,

Example 3.3.

dz
= dz,

(14)

dx
= γ0 (dx − v0 dt),

(15)

dt
= γ0 (dt − v0 dx/c2 ), vy , vy
= γ0 (1 − v0 vx /c2 ) vz vz
= , γ0 (1 − v0 vx /c2 ) vx − v0 . vx
= 1 − v0 vx /c2

(16) (17) (18) (19)

L et one body move with the velocity v1 = (v, 0, 0) and another with v2 = (−v, 0, 0).

4

Physics 1. Mechanics

Lecture 14

What is the relative velocity ? Using the above expression we find vrel = (v−(−v))/(1−v·(−v)/c2 ) = 2v/(1 + v 2 /c2 ). If we notice that vx v0 = v · v0 , v = vx eˆx  v0 , and v⊥ = vy hatey + vz hatez ⊥ v0 , the above expressions can be written as follows: v − v0 , 1 − v · v0 /c2 v⊥ . = γ(1 − v · v0 /c2 )

v
=

(20)

v
⊥

(21)

These expressions are useful for finding relative velocity. Example 3.4.

L et v1 = v1 (cos 30◦ , sin 30◦ , 0), v2 = v2 (cos 30◦ , − sin 30◦ , 0). We shall rewrite the

above
expression substituting v0 → v1 , v → v2 , and v
→ vrel . Now v = v2 · v1 /v1 = v2 cos 60◦ , v⊥ = v22 − v2 = v2 sin 60◦ , and we get v2 cos 60◦ − v1 , 1 − v1 v2 cos 60◦ /c2 v2 sin 60◦ = γ(1 − v1 v2 cos 60◦ /c2 )


vrel, =

(22)


vrel,⊥

(23)

where γ = (1 − v12 /c2 )−1/2 . Now, using vrel =

vrel

3.4




2 2 vrel, + vrel,⊥ we eventually find

 v12 + v22 − 2v1 v2 cos 60◦ − v12 v22 sin2 60◦ /c2 = 1 − v1 v2 cos 60◦ /c2

(24)

Aberration

Let a particle move with the velocity v = v(cos θ, sin θ, 0) in S (v
= v
(cos θ
, sin θ
, 0) in S
, respectively). From the velocity transformation vy
v sin θ , tan θ =
= vx γ0 (v cos θ − v0 )


(25)

that is, θ

= θ, in general. Of particular interest is v = c, that is, when the particle is a photon, emitted in S at the angle θ to x axis. The above expression gives the angle θ
at which the photon propagates in S
: sin θ tan θ
= . (26) γ0 (cos θ − v0 /c) This effect is called aberration.

5

Physics 1. Mechanics

3.5

Lecture 14

Doppler effect

Let an electromagnetic emitter (at (x, y) at present) move with the velocity v = (v, 0, 0) and emits a signal (electromagnetic wave pulse or photon) each T
according to its proper time (as measured by the clock moving with the emitter). What is the time T between the two successive pulses received by an observer at the coordinate origin of S ? Let us denote by A the event of the first photon emission in (x, y) at the time tA = t
A = 0, by A
the event of the first photon reception in (0, 0) at tA
, by B the event of the second photon emission in (x + ∆x, y) (the emitter is moving !) at tB , and by B
the event of the second photon reception in (0, 0) at tB
. The time we are looking for is the time between two successive receptions T = tB
− tA
. tA


If the first photon is emitted in the point (x, y) at t = 0 it arrives at the coordinate origin at  = lA /c, where lA = x2 + y 2 is the distance between the emitter and receiver in the emission

time t = 0. If the second photon is emitted at the time t
B = T
(by the clock of the emitter - this is the proper time), the time of the second emission, by the clock of the observer S is tB = γT
. By that time the moving emitter will be in the point (x + ∆x, y), ∆x = vγT
, so that at the time  of the second emission the distance between the emitter and the receiver is lB = (x + ∆x)2 + y 2. The second photon will come to the receiver at the time tB
= tB + lB /c, and the difference we are

looking for is T = tB
− tA
= tB + (lB − lA )/c. For small T
the difference lB − lA can be found as  a full differential: lB − lA = (x/ x2 + y 2 )∆x = cos θvγT
, where θ is the angle between the line of sight and the emitter velocity, that is, between r and v. Using all this we find T = γ(1 + v cos θ/c). T
For a periodic process such as a monochromatic wave T
and T are periods in the emitter rest frame and the receiver rest frame, respectively. This period is related to the frequency ω as follows ω = 2π/T , and to the wavelength λ as follows: λ = cT , so that we have λ ω
= γ(1 + v cos θ/c). = λ
ω

(27)

This is called Doppler effect. The case θ = 0 corresponds to the emitter moving away from the receiver along the line of sight.  The relation (27) gives λ/λ
= (1 + v/c)/(1 − v/c) > 1 (redshift). The case θ = π corresponds to  the emitter approaching the receiver along the line of sight, and λ/λ
= (1 − v/c)/(1 + v/c) < 1 (blueshift). The case θ = π/2 corresponds to the emitter moving perpendicularly to the line of sight, so that the distance between the emitter and observer does not change. The relation (27) gives λ/λ
= γ > 1 (redshift again). This redshift occurs only because of the time dilation ∆t = γ∆t
. Example 3.5.

A star is moving on a circular orbit of the radius R around another star. The

6

Physics 1. Mechanics

Lecture 14

angular velocity is ω. The moving star emits electromagnetic radiation with the wavelength λ
in its rest frame. What wavelength would measure a distant observer which is in the plane of the rotation ? Since the observer is very far from the start system we can ignore the change of the line of sight due to the finite orbit radius, and choose x axis as the line of sight. Then star velocity is v0 = ωR, while the angle between the velocity and x axis changes as θ = ωt. Substituting this in the expression for the Doppler effect we have 1 λ  = [1 + ωR cos(ωt)/c] λ
1 − ω 2 R2 /c2

4

(28)

Momentum and energy

The proper expressions for the momentum and kinetic energy of a particle moving with the velocity v are  γ = 1/ 1 − v 2 /c2 . (29) p = mvγ, E = mc2 γ, Momentum and energy transformation: p
y = py ,

p
z = pz ,

(30)

p
x = γ0 (px − v0 E/c2 ),

(31)

E
= γ0 (E − v0 px ).

(32)

It is easy to see that the combination E 2 − p2 c2 = m2 c2 is frame independent. If the energy is much higher than the rest energy, E  mc2 , it is not convenient to use the velocity which is almost equal to c. Instead, it is widely accepted to use γ. Example 4.1.

W hat is the particle velocity is γ = 10 ? Since γ = (1 − v 2 /c2 )−1/2 we have

 v = c 1 − 1/γ 2 ≈ 0.995c. Some particles (photon, neutrino) have a zero rest mass, which means that for these particles E = pc. Example 4.2.

T wo photons with p1 = −p2 collide and disappear producing an electron-positron

pair. This reaction can be written as γ + γ → e− + e+ (here γs stand for photons !). The rest masses of electron and positron, m, are identical. What is the minimal photon energy for this to occur ? Since the initial momentum (of the two photons) is zero the final momentum (of the electron and positron) is also zero. Let p be the electron momentum. Then the total energy after the collision  will be 2 p2 c2 + m2 c4 . Since the energy is conserved (there is no internal structure so that it is  impossible to get of loose energy), we have Eγ = p2 c2 + m2 c4 , and the minimum is achieved when 7

Physics 1. Mechanics

Lecture 14

p = 0: Emin = mc2 .

4.1

Aberration and Doppler-effect (another approach)

A photon (electromagnetic wave) is a massless particle, so that E = pc. A photon energy is related to its frequency as E = hν, so that p = hν/c. Let a photon with the frequency ν
is emitted by S
at an angle θ
to x
axis, so that p
x = hν
cos θ
/c, p
y = hν
sin θ
/c. From the momentum-energy transformation we have px = hν cos θ/c = γ0 (hν
cos θ
/c + v0 hν
/c2 ) = γ0 (cos θ
+ v0 /c)hν
/c,

(33)

py = hν sin θ/c = hν
sin θ
/c,

(34)

E = hν = γ0 (hν
+ v0 hν
cos θ
/c),

(35)

so that ν = γ0 (1 + v0 cos θ
/c), ν
cos θ
+ v0 /c . cos θ = 1 + v0 cos θ
/c

(36) (37)

The relation (36) describes the change of the frequency depending on the relative velocity of the emitter (S) and receiver (S
). The frequency decreases when the emitter moves away from the receiver, and increases when the emitter approaches the receiver. This is called Doppler-effect. The relation (37) describes the phenomenon which is called aberration: the angle at which a photon is emitted is different from he angle at which it is received by a moving receiver. We can invert these relations to have ν
= γ0 (1 − v0 cos θ/c), ν cos θ − v0 /c . cos θ
= 1 − v0 cos θ/c

(38) (39)

so that by measuring the frequency and angle in the receiver frame we can know what are the frequency and the angle at the emitter.

8