6.3 Divergence

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Remark 6.3.1. It will have occurred to the more reflective of you that we

ought to be able to take the exterior derivative of a 2-form on R3 and get a

differential 3-form.

Definition 6.4. If

  , E(x, y, z) dx ^ dy + F(x, y, z) dx ^ dz + G(x, y, z) dy ^ dz

is a smooth differential 2-form on R3, the exterior derivative applied to it

gives:

@E

@z

dz ^ dx ^ dy +

@F

@y

dy ^ dx ^ dz +

@G

@x

dx ^ dy ^ dz

Where I followed the same rules as before and didn’t bother to get the zero

terms. Collecting up:

d  =

_

@E

@x

−

@F

@y

+

@G

@x

_

dx ^ dy ^ dz

Definition 6.5. Classical Notation If

F = Pi + Qj + Rk

is a smooth vector field on R3 then

divF =

@P

@x

+

@Q

@y

+

@R

@z

We memorise this by writing:

divF = r q F

Definition 6.6.

I3 ,

8<

:

2

4

x

y

z

3

5 2 R3 : 0 _ x _ 1, 0 _ y _ 1, 0 _ z _ 1

9=

;

6.3. DIVERGENCE 119

Theorem 6.4. Divergence: Classical Form If V is a subset of R3 that

is the image of a smooth embedding a.e. of I3 and F is a smooth vector field

on an open neighbourhood of V then

ZZ

@V

F q dS =

ZZZ

V

div F

where dS is the area element times the unit outward normal to the surface

@V .

Theorem 6.5. Modern Form If V is a subset of R3 that is the image of

a smooth embedding a.e. of I3 and ! is a differential 2-form on an open

neighbourhood of V then Z

@V

! =

Z

V

d!

Proof: I shall prove it for the special case of the cube I3. The case for the

embedding of a cube then follows by the usual argument, the only complicated

bit being the load of partial derivatives in the higher dimensional part

showing

g_d! = dg_!

which is just another computation.

To add variety I shall prove it for the cube using the classical notation and

pretending the 2-form is a vector field.

Look at one face of the cube and observe that over this face,

Z

F q dS

has meaning the amount of flow (flux was used in the seventeenth century

and still is in some quarters) coming out of the surface. See figure 6.5. If we

subdivide the cube into six subcubes and add up

R

F q dS for each subcube

we must get the same result as

Z

F q dS

for the whole cube. This is because the flow out of any one interior cube face

is counted twice, one in each direction so cancels out in the sum. The same

process can be repeated indefinitely over progressively small subcubes.

120 CHAPTER 6. STOKES’ THEOREM (CLASSICAL AND MODERN)

Figure 6.5: Flow from a face

Now look at a very small subcube centred at a point

2

4

a

b

c

3

5

and having edge 24

The flow out of the face with centre

2

4

a +4

b

c

3

5

is only in the i direction so is approximately

_

P +

@P

@x

4

_

× 442

(which is the field at the centre of the face in the i direction multiplied by

the surface area.)

6.3. DIVERGENCE 121

The flow out of the opposite face is

−(P −

@P

@x

4) × 442

with a minus sign because the normal is pointing in the opposite direction.

For the other four faces of the cube we get the corresponding terms with

_

Q +

@Y

@y

4

_

× 442

and its opposite and _

R +

@R

@z

4

_

× 442

and its opposite. Adding these up we get

_

@P

@x

+

@Q

@y

+

@R

@z

_

× 843

which is div F multiplied by the volume of the little cube.

Integrating this function over the cube is done by approximating by Riemann

sums and taking limits so we get, for the cube,

ZZ

@V

F q dS =

ZZZ

V

div F

Saying this in new-fangled language we get

Z

@(I3)

! =

Z

I3

d!

The rest is a straightforward calculation to make it work for regions which

are the images of smooth embeddings a.e. of cubes in R3, and also images

by maps which are smooth embeddings except on a set of area zero. _

Remark 6.3.2. It is good clean fun to show that the sets which are smooth

emebeddings a.e. of a cube includes the unit ball,

D3 =

8<

:

2

4

x

y

z

3

5 2 R3 : x2 + y2 + z2 _ 1

9=

;

Try it. It will make you better and purer people. More like me.

122 CHAPTER 6. STOKES’ THEOREM (CLASSICAL AND MODERN)

Example 6.3.1. Find the amount of flow of the vector field zi + xj + yk

through the unit sphere S2.

Solution The normal vector to a point

8<

:

2

4

x

y

z

3

5

9=

; 2 S2

is just 8

< 

:

2

4

x

y

z

3

5

9=

;

The projection of the pseudovector zi + xj + yk on this is (xz + xy + yz)

which is a straightforward function and we want to integrate this over S2

This we know how to do, we parametrise by

g : [0, 2_] × [−1, 1] −! R3

u, v 

2

4

p

1 − v2 p cos(u)

1 − v2 sin(u)

v

3

5

And we have to put in the area stretching term which is

 

@g

@u

×

@g

@v

 

to get

Z 1

v=−1

Z 2_

u=0

_

v

p

1 − v2 cos(u) + (1 − v2) cos(u) sin(u) + v

p

1 − v2 sin(u)

_

×(the area stretching factor) du dv

This is clearly doable but would make the bravest heart sink a bit. Even

typing it correctly into Mathematica would take a while.

We have Gauss galloping to the rescue this time (he invented the Divergence

Theorem) telling us that the result is the same as if we integrate div F over

the interior of the ball.

Now div Fis:

r q F =

@z

@x

+

@x

@y

+

@y

@z

= 0

6.3. DIVERGENCE 123

Integrating the zero function over the solid ball D3 gives zero. You have to

allow that this is the easy way to do it.

Knowing the right answer, you can see that this is correct because the sphere

has eight octants which are moved into each other by reversing the sign of

one or more axes, and the symmetry in the field causes the total to cancel

out. But this is being wise after the event.

Example 6.3.2. Prove that for any region U which is the image by a smooth

embedding a.e. of a cube in R3 and any vector field F

Z

@U

curl F = 0

Proof: I claim that d2 = 0 Hence div(curl(F)) = 0. So the integral over U

of div curl is zero, so the surface intagral of curl(F) is zero.

All I have to do is a simple calculation to show that going from 1-forms

to 2-forms by d and then from 2-forms to three forms by d gives me zero.

Starting with

P dx + Q dy + R dz

I get

_

@Q

@x

−

@P

@y

_

dx ^ dy +

_

@R

@x

−

@P

@z

_

dx ^ dz +

_

@R

@y

−

@Q

@z

_

dy ^ dz

and taking the exterior derivative of this I get:

_

@

@x

_

@Q

@x

−

@P

@y

_

−

@

@y

_

@R

@x

−

@P

@z

_

+

@

@z

_

@R

@y

−

@Q

@z

__

dz ^ dy ^ dz

You can see that the terms cancel pairwise to give zero, so the result is

proved.

Remark 6.3.3. There are a large number of applications of these ideas in

electromagnetism and fluid mechanics. Alas, I have no time to cover them

but you should be in a good position to understand them when they are used

in Physics.

124 CHAPTER 6. STOKES’ THEOREM (CLASSICAL AND MODERN)

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