6.1 Classical

Back

Theorem 6.1. Stokes Let S be a piecewise-smooth, orientable surface in

R3 with boundary @S. Let n be any normal vector to the surface, and let this

induce the derived orientation on @S.

Let F be a smooth vector field on R3.

Then Z

@S

F q dr =

Z

S

curl F q dS

where dS is the normal vector to the surface element with the same orientation

as n and dr is the tangent to the length element having the derived

orientation.

Remark 6.1.1. This is the standard form of Stokes’ Theorem and is in a

form which Stokes might almost have recognised. The next job is to explain

what some of the words mean.

Remark 6.1.2. If U is a smooth surface in R3 and if ˆn(x) is a unit length

normal vector to x 2 U then there is a map

ˆn : U −! R3

x   ˆn(x)

99

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

Figure 6.1: A non-orientable surface

Well, there is always a choice of directions; there are always two unit normal

vectors to a smooth surface at a point. One is the negative of the other.

Picking one of them is equivalent to deciding which way is up.

This ought to depend on the surface being smooth; if it looked like the roof

of a house then there would be ridges without a normal vector. If however

there were two linearly independent tangent vectors to the surface at a point,

the cross product would give me a normal vector. So if the surface were the

graph of a differentiable function at a point there certainly ought to be a

normal vector, in fact a normal line.

Suppose we make a choice of which of two unit normals to take at some

particular point x 2 S.

Now I take a path in U. I can ensure that I make “the same” choice of

unit normal along the path. What I mean by this is that I ensure that the

function ˆn is continuous. Small changes in the position x on the surface

will make small changes in ˆn(x). This makes ˆn : U −! R3 continuous, the

vectors ˆn(x) will change as we move about, but not too drastically if U is

smooth.

Now you have agreed to this as blindingly obvious, look at the M¨obius strip

of figure 6.1. You can carry the normal vector all the way around a loop and

on returning to your starting point, the vector is pointing in the opposite

direction.

6.1. CLASSICAL 101

Figure 6.2: The derived orientation

Oops.

We get around this by calling the surfaces where this doesn’t happen orientable,

and things like the mobius strip are non-orientable. Then we

simply ignore the non-orientable ones.

Definition 6.1. Induced Orientation If we have an orientable surface,

we choose an orientation which is equivalent to making a choice of normal

vector, n. Now I can move this about over the surface in such a way that it

changes continuously with x

Move it towards the boundary of the surface (if it has one).

At the boundary, take a vector normal to the boundary and also to the vector

n pointing away from the surface call it w. This is shown in figure 6.2

Then there are two tangent vectors. Choose t so that (w, t,n) is a righthanded

system like

2

4

2

4

1

0

0

3

5

2

4

0

1

0

3

5

2

4

0

0

1

3

5

3

5, otherwise known as (i, j, k)

Then the direction of t is called the induced orientation on the boundary.

Remark 6.1.3. It may not have been too obvious from the statement of

Stokes’ Theorem, but the idea is the same as Green’s Theorem. All we do

is to go from a vector field in R2 to one in R3, and instead of having a flat

surface sitting in R2 it is still a bit of surface, sitting curved in R3. Then if

you integrate the spin of the vector field over the surface, you must get the

same result as if you take the path integral around the boundary.

In figure 6.3 I sketch the picture this should evoke:

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

Figure 6.3: Green’s Theorem in R3

Remark 6.1.4. The only problem is to get clear the business of integrating

the curl of F over the surface. We do this by taking the curl of the vector

field to be, as in the last chapter, a vector normal to the plane of rotation

of the vector field F. We take the vector dS to be normal to the surface

and of length the ‘infinitesimal area element’. The dot product of these two

vectors gives the amount of spin in the tangent to the surface. Integrating

this dot product over the surface ought to give us the same result as it does

in Green’s Theorem, and for the same reason.

“Proof” of Stokes Theorem (Classical Version) (For the case where S

is g(I2) for g a piecewise smooth embedding.)

Partition I2 into little rectangles side 4u,4v and map them into g(I2) by

g : I2 −! R3

_

u

v

_

 

2

4

x(u, v)

y(u, v)

z(u, v)

3

5

I show a picture of this in figure 6.4

Let

_

a

b

_

be the centre of one such square. The curl of F at the point g

_

a

b

_

is the vector r × F.

It should be clear that

@g

@u

____

a

b

6.1. CLASSICAL 103

Figure 6.4: Parametrising a surface in R3

is a tangent curve in the surface g(I2) at the point g

_

a

b

_

, and that

@g

@v

____

a

b

is another one. (After all if you keep one one of two variables fixed you are

putting a curve in R3 which lies in the surface.)

Then a normal to the surface at g

_

a

b

_

is

n(a, b) =

@g

@u

×

@g

@v

=

2

64

@x

@u

@y

@u

@z

@u

3

75

×

2

64

@x

@v

@y

@v

@z

@v

3

75

all partial derivatives being evaluated at

_

a

b

_

. This is because the cross

product is always orthogonal to the other two vectors, both of which are

tangent to the surface.

The amount of ”twist” in the plane tangent to U at g

_

a

b

_

is the dot product

[r × F] q ˆn which is the projection on ˆn where

ˆn =

n

knk

is the unit normal.

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

The “area element” dS is the vector ˆn multiplied by the area of the ‘infinitesimal’

area that 4u4v has become after being stretched by g0. This turns out

by what would look slightly miraculous if one were disposed to think that

way, to be the length of the vector

2

64

@x

@u

@y

@u

@z

@u

3

75

×

2

64

@x

@v

@y

@v

@z

@v

3

75

So the amount of twist that curl(F) exerts on the surface at the point g

_

a

b

_

multiplied by the infintesimal area element is

[r × F] q

0

B@

2

64

@x

@u

@y

@u

@z

@u

3

75

×

2

64

@x

@v

@y

@v

@z

@v

3

75

1

CA

du dv

where [r × F] is evaluated at g

_

a

b

_

and the partial derivatives are all

evaluated at

_

a

b

_

Integrating this over the surface gives the spin part of Stokes Theorem.

But this just pulls back the integral to I2; taking limits as 4u ! 0 4v ! 0

we get Z 1

u=0

Z 1

v=0

([r × F]) q

_

@g

@u

×

@g

@v

_

du dv

Now

[r × F] q

_

@g

@u

×

@g

@v

_

at g

_

a

b

_

defines a spin field on I2 multiplied by the area stretch g0 does,

and by Green’s Theorem for a square this is equal to the integral of the

corresponding vector field around the boundary:

Z

@I2

F _ g

if we again take the component tangent to the boundary. Which is what we

get if we take Z

@g(I2)

F q g0 =

Z

@g(I2)

F q dr

6.2. MODERN 105

_

Remark 6.1.5. In other words, Stokes’ theorem is just Green’s theorem

after we pull it back to I2 properly.

Remark 6.1.6. Stokes’ Theorem (updated version) (dates from about 1870)

and was not proved by Stokes. In fact the original statement is in a letter

from Sir William Thomson, later known as Lord Kelvin, to Stokes dated

1850. Stokes set the problem of proving it in an examination set for the

top mathematics students at Cambridge in 1854, but we don’t know if anyone

proved it. Probably not. Kelvin himself proved it. Maxwell proved

it in Electricity and Magnetism. Stokes was rather lucky to have his name

perpetuated by a theorem he may never have proved.

Навигация