Gradient Diver Curl

Chapter 1

Gradient, Divergence & Curl

Scalar and vector fields 

Imagine a cooling system of a reactor which is using fluid as the cooler medium vb

va

Fluid Tc Td

2

FIELD is a description of how a physical quantity varies from one point to another in the region of the field (and with time). (a) Scalar fields Ex: Depth of a lake, d(x, y) Temperature in a room, T(x, y, z) Depicted graphically by constant magnitude ycontours or surfaces. d3

d1 d2 x

3

At any point P, we can measure the temperature T.  The temperature will depend upon whereabouts in the reactor we take the measurement. Of course, the temperature will be higher close to the radiator than the opening valve.  Clearly the temperature T is a function of the position of the point. If we label the point by its Cartesian coordinates ( x, y, z ) , then T will be a function of x, y and z, i.e. 

T = T ( x, y , z )

.  This is an example of a scalar field since

4

•

A field is a quantity which can be specified everywhere in space as a function of position.

•

The quantity that is specified may be a scalar or a vector.

•

For instance, we can specify the temperature at every point in a room.

•

The room may, therefore, be said to be a region of “temperature field” which is a scalar field because the temperature T (x, y, z) is a scalar function of the position.

•

An example of a scalar field in electromagnetism is the electric potential. 5

Meanwhile, at each point, the fluid will be moving with a certain speed in a certain direction  That is, each small fluid element has a particular velocity and direction, depending upon whereabouts in the fluid it is.  This is an example of a vector field since velocity is a vector. The velocity can be expressed as a vector function, i.e. 

v = v ( x, y, z ) = v1 ( x, y, z )i + v2 ( x, y, z ) j + v3 ( x, y, z )k v1 , v2 and v3

where functions.

will each be scalar 6

Example: Linear velocity vector field of points on a rotating disk 1-7

7

Physical examples of scalar fields:

+

Electric potential around a charge

Temperature near a heated wall

(The darker region representing higher values ) 8

Physical examples of vector fields: +

−

Electric field surrounding a positive and a negative charge.

Hurricane

Magnetic field lines shown by iron filings

The flow field around an airplane

9

Gradient of Scalar Fields The gradient of a scalar field is a vector field,  which points in the direction of the greatest rate of increase of the scalar field, and whose magnitude is the greatest rate of change. 

10

  x, y, z   ctt uuu r dr grad 

uuu r Physical meaning:grad  dr is the local variation of Φ along dr. Particularly, grad Φ is perpendicular to the line Φ = ctt. 11

Gradient operator Suppose , we have a function of three variables- say, the temperature T(x, y, z) in a room. For the temperature distribution we see how a scalar would vary as we moved off in an arbitrary direction. Now a derivative is supposed to tell us how fast the function varies, if we move a little distance.

12

,

If T(r) is a scalar field, its gradient is defined in Cartesian coordinates by

∂T ∂T ∂T grad T = i+ j+ k ∂x ∂y ∂z It is usual to define the vector operator ∇ which is called “del”. We can write

.

 ∂ ∂ ∂ ∇ = i + j + k  ∂y ∂z   ∂x

grad T ≡ ∇T

Without thinking too hard, notice that grad T tends to point in the direction of greatest change of the scalar 13 field T.

The significance of the gradient: A theorem on partial derivatives states that  ∂T   ∂T   ∂T    dT =  dy +  dx +  dz   ∂x   ∂z   ∂y  This rule tells us how T changes when we alter all three variables by the infinitesimal amounts dx, dy, dz. Change in T can be written as,  ∂T ∂T ∂T  dT =  i+ j+ k  • ( dx i + dy j + dz k ) ∂y ∂z   ∂x

 →   → =  ∇T  ⋅ d l      The conclusion is that, the RHS of above equation is the small change in temperature T when we move by dl. 14

If we divide the above eq. by dl We get →   dT  →   d l  =  ∇T  ⋅  dl    dl     → → d l  but  is a unit vector in the direction of d l .   dl   

So , we can conclude that, grad T has the property that the rate of change of T w.r.t. distance in any direction â is the projection of grad T onto that direction â. That is

dT dl

  in direction of 

 → ^ a  = ∇T ⋅ a  ^

15

the quantity

dT dl

is called a directional derivative.

In general, •

a directional derivative had a different value for each direction,

•

has no meaning untill you specify the direction

16

Gradient Perpendicular to T constant surfaces If we move a tiny amount within the surface, that is in any tangential direction, there is no change in T , so dT = 0. dl

Surface of constant T, These are called level surfaces. →

Surfaces of constant T →

dl dl in the surface ∇T ⋅ =0 So for any dl dl Conclusion is that; grad T is normal to a surface of constant T. 17

Geometrical Interpretation of the Gradient (1.5)

Like any vector, a gradient has magnitude and direction. To determine its geometrical meaning, lets rewrite the dot product In its abstract form:

dT = ∇T • dl = ∇Tdl cos θ where θ is the angle between ∇T and dl. Now, if we fix the magnitude dl and search around in various directions (that is, vary θ), the maximum change in T evidently occurs when θ =0 (for then cos θ = 1). That is for a fixed distance dl, dT is greatest when I move in the same direction as ∇ T .

18

In the above two images, the scalar field is in black and white, black representing higher values, and its corresponding gradient is represented by blue arrows. 19

Example 1 If φ(x,y,z) = 3x2y– y2z2, find grad φ and ∇φ at the point (1,2,−1).

20

Solution ∂φ ∂φ ∂φ grad φ = ∇φ = i + j +k ∂x ∂y ∂z = 6 xyi + (3 x − 2 yz ) j + (−2 y z )k 2

2

2

At the point (1,2,−1), ∇φ = 6(1)(2)i + [3(1) 2 − 2(2)(−1) 2 ]j − 2(2) 2 (−1)k = 12i − j + 8k ∴ ∇φ (1, 2, −1) = 12i − j + 8k = 12 + (−1) + 8 = 209 2

2

2

21

Example 2 

If

r = x 2 +y 2 +z 2 Find out ∇ r =? →

^

^

^

here r =x i +y j +z k Q.Show that (a) ∇ ( r 2 ) =2r ^

r 1  (b) ∇   =− 2 r r 

Example 3 Find φ ( x, y ) , if ∇φ = y cos x i + (sin x + e y ) j

Given φ (0,0) = 0.

23

Solution y ∇ φ = y cos x i + (sin x + e ) j , we have Since

∂φ = y cos x .....(1) ∂x

∂φ = sin x + e y .....(2) ∂y

Integrating (1) and (2) w.r.t. x and y respectively, we obtain φ = ∫ y cos xdx = y sin x + f ( y ) .....(3) φ = ∫ (sin x + e y )dy = y sin x + e y + g ( x) .....(4) 24

Comparing (3) and (4), we can conclude that f ( y ) = e y + C and g ( x) = C where C is an arbitrary constant of integration y φ ( x , y ) = y sin x + e +C Hence, To find constant C, use φ (0,0) = 0.

φ (0,0) = 0 sin 0 + e 0 + C = 0 1+ C = 0 ∴ C = −1

Therefore, φ ( x, y ) = y sin x + e − 1 y

♣

25

Example 4 Find φ ( x, y, z ) if 2 3 2 3 3 2 2 ∇φ = ( y − 2 xyz )i + (3 + 2 xy − x z ) j + (4 z − 3 x yz )k and φ (0,0,0) = −2 .

26

Solution ∂φ = y 2 − 2 xyz 3 .....(1) ∂x ∂φ 3 2 2 ∂φ 2 3 = 4 z − 3 x yz .....(3) = 3 + 2 xy − x z .....(2) ∂z ∂y

We have

Integrating (1), (2) and (3) w.r.t. x, y and z respectively, we obtain φ = ∫ ( y − 2 xyz )dx = xy − x yz + f ( y, z ) .....(4) 2

3

2

2

3

φ = ∫ (3 + 2 xy − x 2 z 3 )dy = 3 y + xy 2 − x 2 yz 3 + g ( x, z ) .....(5) φ = ∫ (4 z 3 − 3 x 2 yz 2 )dz = z 4 − x 2 yz 3 + h( x, y ) .....(6)

27

Comparing (4) with (5) and (6) we get f ( y, z ) = 3 y + z 4 + C

Therefore 2 2 3 4 φ = xy − x yz + 3 y + z + C To find constant C, use φ (0,0,0) = −2 ∴φ = xy − x yz + 3 y + z − 2 2

2

3

4

28

Problem 5 →

Show that ∇φis a vector perpendicular to the surface φ( x, y , z ) = k where k is const.

φ =φ( x, y , z ) = k ∂φ ∂φ ∂φ dφ = dx + dy + dz = o ∂x

∂y

∂z

^   ∂φ ∂φ   ∂φ i +j +k •(idx + jdy +kdz )   ∂y ∂z   ∂x   →

→

∇φ.d r = 0

29

Application of gradient: Surface normal vector 

A normal, n to a flat surface is a vector which is perpendicular to that surface.



A normal, n to a non-flat surface at a point P on the surface is a vector perpendicular to the tangent plane to that surface at P. n z = f ( x, y ) 30



Therefore, for a non-flat surface, the normal vector is different, depending at the point P where the normal vector is located. n

z = f ( x, y )



Unit vector normal isndefined as nˆ =

n 31



To find the unit vector normal to the surface

z = f ( x, y ) we follow thezfollowing = f ( x, y )steps: (i) Rewrite the as φ ( xfunction , y, z ) = k , k any constant

n = ∇φ

(ii) Find the normal vector that is nvector ∇φis (iii) Then, the unit normal ˆ

n=

n

=

∇φ

(iv) Hence, P ( x0 , y0the , z0 )unit normal vector at a point

∇φ ( x0 , y0 , z0 ) nˆ = ∇φ ( x0 , y0 , z0 ) is

32

Example 5 Find the unit normal vector of the surface at the indicated point. (a) z = 6 − x 2 − y 2 at (−1,3,2) y 3 2 xe + y = z (b) at (1,0,−1)

33

Solution 2 2 2 x + y + z = 14 (a) Rewrite z = 14 − x − y as 2

2

2 2 2 φ ( x , y , z ) = x + y + z Thus, we obtain

Then, ∇φ = 2 xi + 2 yj + 2 zk = 2( xi + yj + zk ) At the point (−1,3,2), ∇φ = 2(−i + 3 j + 2k ) and ∇φ = 2 (−1) 2 + 32 + 2 2 = 2 14 ∇φ − i + 3 j + 2k The unit normal vector is nˆ = = ∇φ 14

34

(b) Rewrite xe + y = z y

3

2

as xe + y − z = 0 y

3

Thus, we obtain φ ( x, y, z ) = xe + y − z y

3

2

2

Then, ∇φ = e i + ( xe + 3 y ) j − 2 zk At the point (1,0,−1), y

y

2

∇φ = e 0 i + [1e 0 + 3(0) 2 ]j − 2(−1)k = i + j + 2k

and

∇φ = 12 + 12 + 2 2 = 6

∇φ i + j + 2k The unit normal vector is nˆ = = ∇φ 6 35

Divergence of Vector Fields The divergence is an operator that measures the magnitude of a vector field's source or sink at a given point  The divergence of a vector field is a scalar 

uu r V(x, y, z)

x

uu r V(x  dx, y, z)

x+dx

36



The divergence of a vector field F ( x, y, z ) = F1 ( x, y, z )i + F2 ( x, y, z ) j + F3 ( x, y, z )k

is defined as div F = ∇ ⋅ F  ∂ ∂ ∂ =  i + j + k  ⋅ ( F1i + F2 j + F3k ) ∂y ∂z   ∂x ∂F1 ∂F2 ∂F3 = + + ∂x ∂y ∂z 37

 Divergence uu r V(x, y, z)

x

uu r V(x  dx, y, z)

ur v(x, y, z)

is a differentiable vector field

x+dx

ur ur v vy vz x div v =  v     uvu x y z 2 – Physical meaning

ur div v ur v

is associated to local conservation laws: for example, we will show how that if the mass of fluid (or of charge) outcoming ur from a domain is equal to the mass entering, then is the fluid velocity (or the current) vectorfield

div v  0

38

Geometrical Interpretation. The name divergence is well chosen, for ∇ . F is a measure of how much the vector F spreads out (diverges) from the point in question. The vector function has a large (positive) divergence at the point P; it is spreading out. (If the arrows pointed in, it would be a large negative divergence.) P

NOTE:

P=electric field due to charge (+ ve or – ve)

39

On the other hand, the function has zero divergence at P; it is not spreading out at all.

P

So, for example, if the divergence is positive at a point, it means that, overall, that the tendency is for fluid to move away from that point (expansion); if the divergence is negative, then the fluid is tending to move towards that point (compression). 40

Fundamental theorem of divergence The fundamental theorem for divergences states that:

( ) ∇ • F d τ = F • da ∫ ∫

volume

surface

This theorem has at least three special names: Gauss’s theorem, Green’s theorem, or, simply, the dτ divergence theorem. dτ is function at the boundary element of volume (in

Cartesian coordinates, dτ = dx, dy, dz), and The volume integration is really a triple integral. 41



da represents an infinitesimal element of area; it is a vector , whose magnitude is the area of the element and whose direction is perpendicular ( normal ) to the surfaces, pointing outward.

On the front face of the cube, a surface element is

da1 = ( dy dz ) iˆ

42

on the right face, it would be

da 2 = ( dz dx ) ˆj whereas for the bottom it is

( )

da3 = ( dx dy ) − kˆ

43

Problem 1 

Q. Calculate the divergence of the following vector functions?

(a ) v1 = x i + 3 xz j − 2 xzk 2

2

(b) v2 = xyi + 2 yzj + 3zxk

Problem 2 

Check the divergence theorem using the function ∧

∧

∧

v = y i + (2 xy + z ) j + (2 yz ) k 2

2

And the unit cube is situated at the origin. Z

Y X

45

Curl of Vector Fields 

Curl is a vector operator that shows a vector field's rate of rotation, i.e. the direction of the axis of rotation and the magnitude of the rotation.

∇×v = 0

ur curl v  0 46

47



The curl of a vector field F ( x, y, z ) = F1 ( x, y, z )i + F2 ( x, y, z ) j + F3 ( x, y, z )k

is defined as i ∂ curl F = ∇ × F = ∂x F1  ∂F3 ∂F2   − = i − ∂z   ∂y

j ∂ ∂y F2

k ∂ ∂z F3

 ∂F3 ∂F1   ∂F2 ∂F1   j − −  + k   ∂x ∂z   ∂x ∂y  48

Problem 1 Z



Find the curl of

∇ × v3 =

i ∂ ∂x −y

v3 = − yi + xj j ∂ ∂y x

k ∂ ∂z 0

y

The curl of v3 points in the z-direction X

Curl To find a possible interpretation of the curl, let us consider a body rotating with uniform angular speed ω about and axis l. Let us define the vector angular velocity to be a vector of length ω extending along l in the direction Take the point O as the origin of coordinates we can write R = xi + yj + zk

the radius at which P rotates is |R||sinθ| Hence, the linear speed if P is v = ω|R||sinθ| = Ω|R||sinθ|

If we take the curl of V, we therefore have

that is

Expanding this, remembering that Ω is a vector, we find

Conclusion: The angular velocity of a uniform rotating body is thus equal to one-half the curl of the linear velocity of any point of the body.

Example: For velocity field,

(

) ( ^

)

^

u = x + e sin ( yz ) i + x + e cos( yz ) j x

^

x

^

i ∂ ω = ∇×u = ∂x u

, find the angular velocity ω.

^

j ∂ ∂y v

k ∂ ∂z w

For the field, u = ( x + e sin ( yz ) ) i + ( x + e cos( yz ) ) j ^

x

^

x

, we obtain: ^

^

i ∂ ω = ∇×u = ∂x x x + e sin ( yz ) =−

^

j ∂ ∂y x x + e cos( yz )

k ∂ ∂z 0

^ ^ ∂ ∂ ∂ ∂ x + e x cos( yz ) i + x + e x sin ( yz ) j +  x + e x cos( yz ) − x + e x sin ( yz ) ∂z ∂z ∂y  ∂x

(

)

^

(

^

)

{

(

)

}

^

= e y sin ( yz ) i + e y cos( yz ) j + 1 + e cos( yz ) − ze cos( yz ) k x

x

x

x

(

) k ^



Fundamental theorem of curl The fundamental theorem for curls, which goes by the special name of Stokes’ theorem, states that

∫ ( ∇ × v ) ⋅ da = ∫ v.dl

surface

boundary line

The integral of a curl over a region (a patch of surface) is equal to the value of the function at the boundary (the perimeter of the patch).

54

Example 6 Find both div F and curl F at the point (2,0,3) if

F( x, y, z ) = ze i + 2 xz cos yj + ( x + 2 y )k 2 xy

55

Solution

∂F1 ∂F2 ∂F3 div F = ∇ ⋅ F = + + ∂x ∂y ∂z ∂ ∂ ∂ 2 xy = ( ze ) + (2 xz cos y ) + ( x + 2 y ) ∂x ∂y ∂z = 2 yze 2 xy − 2 xz sin y

[Notice that div F is a scalar!] At the point (2,0,3), ∇ ⋅ F = 2(0)(3)e 2( 2)( 0) − 2(2)(3) sin 0 = 0

56

i ∂ curl F = ∇ × F = ∂x 2 xy ze

j ∂ ∂y 2 xz cos y

k ∂ ∂z x + 2y

∂  ∂ = i  ( x + 2 y ) − (2 xz cos y ) ∂z  ∂y  ∂ ∂ 2 xy  − j ( x + 2 y ) − ( ze ) ∂z  ∂x  ∂ ∂ 2 xy  + k  (2 xz cos y ) − ( ze ) ∂y  ∂x  = (2 − 2 x cos y )i − (1 − e 2 xy ) j + (2 z cos y − 2 xze 2 xy )k 57

[Notice that curl F is also a vector] At the point (2,0,3), ∇ × F = [2 − 2(2) cos 0]i − [1 − e 2 ( 0)(3) ]j + [2(3) cos 0 − 2(2)(3)e 2 ( 2 )( 0 ) ]k = −2i − 6k

58

Properties of Del If F(x,y,z) and G(x,y,z) are differentiable vector functions φ(x,y,z) and ψ(x,y,z) are differentiable scalar functions, then ∇(φ ±ψ ) = ∇φ ± ∇ψ (i) (ii) ∇(φψ ) = φ∇ψ +ψ∇φ (iii)

φ ∇ ψ

 ψ∇φ − φ∇ψ  = 2 ψ  59

(iv) ∇ ⋅ (F ± G ) = ∇ ⋅ F ± ∇ ⋅ G (v) ∇ × (F ± G ) = ∇ × F ± ∇ × G (vi)

2 2 2   ∂ ∂ ∂ 2 ∇ ⋅ (∇φ ) ≡ ∇ φ ≡  2 + 2 + 2 φ ∂y ∂z   ∂x 2 2 2 ∂φ ∂φ ∂φ = 2+ 2+ 2 ∂x ∂y ∂z

(vii) ∇ × (∇φ ) = 0 or curl grad φ = 0 (viii) ∇ ⋅ (∇ × F) = 0 or div curl F = 0 *Notes: In (vi), ∇ 2 is called the Laplacian operator 60