Пресс-релиз популярных книг

Авторы: 111 А Б В Г Д Е Ж З И Й К Л М Н О П Р С Т У Ф Х Ц Ч Ш Щ Э Ю Я

Книги: 164 А Б В Г Д Е Ж З И Й К Л М Н О П Р С Т У Ф Х Ц Ч Ш Щ Э Ю Я

На сайте 111 авторов, 92 книг, 72 статей, 5913 глав.

6.2 Modern Theorem

Back

6.2. Let g : I2 −! U _ R3 be a smooth embedding a.e. and ! a

smooth differential 1−form on U.

Then Z

@g(I2)

! =

g(I2)

d!.

Remark 6.2.1. This looks like Greens Theorem with only the dimension of

U changed from 2 to 3. This is right. The proof is the same too.

Recall that we had for

g : I2 −! U _ R2

and if ! is a 1−form on U, ! = Pdx + Qdy, g_! was the 1−form on I2

defined by

(g_!)

= P

dx + Q

and _

= g0

# _

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

And if ! = Pdx ^ dy is a 2−form on R2, g_! is the 2−form on I2 defined by

(g_!)

= P

dx ^ dy

where dx ^ dy is obtained by

dx =

du +

dy =

du +

dx ^ dy =

du +

−

du ^ dv.

Remark 6.2.2. This extends without fuss to g : I2 −! U _ R3.

Definition 6.2. If ! is a differential 1−form on U, ! , Pdx + Qdy + Rdz

on U

g_!

, P

dx + Q

dy + R

and

g0 =

664

775

so 2

5 =

664

775

ie.

dx =

du +

dy =

du +

dz =

du +

6.2. MODERN 107

g_(!) = (P _ g)

du + (P _ g)

+ (Q _ g)

du + (Q _ g)

+ (R _ g)

du + (R _ g)

(P _ g)

+ (Q _ g)

+ (R _ g)

(P _ g)

+ (Q _ g)

+ (R _ g)

Remark 6.2.3. The idea is the same: evaluate the pullback to I2 by using

composition with g to get the value and composition with g0 to get the

differential part.

Remark 6.2.4. Now we do it for a 2-form on U _ R3.

Definition 6.3. If ! , Pdx ^ dy + Qdx ^ dz + Rdy ^ dz on U _ R3 and

g : I2 −! U is smooth, g_(!) is the 2-form on I2 given by:

P _ g

dx ^ dy + Q _ g

dx ^ dz + R _ g

dy ^ dz

and

dx =

du +

dv, dy =

du +

dx ^ dy =

du +

−

du ^ dv

dx ^ dz =

−

du ^ dv

dy ^ dz =

−

du ^ dv

which gives g_! as a differential 2-form on U _ R3. We get:

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

g_(!) =

(P _ g)

−

+ (Q _ g)

_ _

−

+(R _ g)

_ _

−

du ^ dv

R _ g

−Q _ g

P _ g

5 q

; du ^ dv

You will recognise the second term as part of the “area stretching factor” for

g : I2 −! U _ R3 at a point.

Note again that this comes out of the computation of the induced 2-form

quite automatically.

Proposition 6.2.1. If ! is a smooth differential 1-form on U _ R3 and

g : I2 −! U is a smooth embedding a.e.,

@g(I2)

! =

@I2

g_!

Proof We have for each edge of (I2), g defines a parametric curve in U _ R3

which is part of the boundary of g(I2) and g_! is constructed to take care of

the length stretch automatically:

g(I)

! =

Z 1

+ Q

+ R

g_!

Proposition 6.2.2. If ! is a 1-form on I2,

@I2

! =

Proof This is just Green’s Theorem for I2. _

Proposition 6.2.3. If ! is a smooth differential 1-form on U _ R3 and

g : I2 −! U is a smooth embedding,

then dg_! = g_d!.

6.2. MODERN 109

Proof If ! = Pdx + Qdy + Rdz we have from definition 6.2

g_! =

P _ g

+ Q _ g

+ R _ g

du +

P _ g

+ Q _ g

+ R _ g

dg_! =

−

du ^ dv

where

A =

(P _ g)

+ (Q _ g)

+ (R _ g)

B =

(P _ g)

+ (Q _ g)

+ (R _ g)

ie.

dg_! =

@(P _ g)

+ (P _ g)

@2u

@u@v

@(Q _ g)

+ (Q _ g)

@2y

@u@v

@(R _ g)

+ (R _ g)

@2z

@u@v

−

@(P _ g)

+ (P _ g)

@2x

@v@u

@(Q _ g)

+ (Q _ g)

@2y

@v@u

@(R _ g)

+ (R _ g)

@2z

@z@u

du ^ dv

@(P _ g)

−

@(P _ g)

@(Q _ g)

−

@(Q _ g)

@(R _ g)

−

@(R _ g)

du ^ dv

but

@(P _ g)

and similarly for Q _ g and R _ g.

dg_! =

8>< >:

@v + @P

−@P

@u − @P

+ similar terms for Q and R

9>=

du ^ dv

( h

_ @y

@v − @y

+ @P

_ @z

@v − @z

+ similar terms for Q and R

)

du ^ dv

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

From definition 6.3, g_d! is

8>>>>>><

>>>>>>:

@x − @P

_ g

_ _ @x

@v − @x

_@R

@x − @P

_ g

_ _ @x

@v − @x

@y − @Q

_ g

_ _ @y

@v − @y

9>>>>>>=

>>>>>>;

du ^ dv.

The P terms are

__ _

−

__ _

−

and the Q and R terms can be collected in the same way, to establish that

dg_! = g_d! _.

Remark 6.2.5. It has to be said that by your standards this is a nasty

calculation, but all it requires of you is lots of partial differentiating.

Proposition 6.2.4. If ! is a smooth differential 2-form on U _ R3 and

g : I2 −! U is a smooth embedding a.e.,

g(I2)

! =

g_!

Proof g parametrises the surface in U and if ! = Pdx ^ dy + Qdx ^ dz +

Rdy ^ dz then g_! has P _ g, Q _ g and R _ g for the function part and Dg

acts on du and dv to give us du ^ dv on the differential part. We have again

5 =

664

775

We need to verify that we get the correct “area stretching” formula out.

From definition 6.3 we had, recall,

g_(!) =

P _ g

_ _

−

+ Q _ g

_ _

−

+R _ g

_ _

−

du ^ dv

6.2. MODERN 111

This is 2

R _ g

−Q _ g

P _ g

5 q

__3

5du ^ dv

(The permutation of the P, Q,R (and the minus sign) come from the way

the dx ^ dy acts on a piece of surface normal to the (d)z direction.)

We can rewrite this as

R _ g

−Q _ g

P _ g

5 q ˆn[u, v]

5 du ^ dv

where ˆn[u, v] is the unit normal to the surface at g

, and

is the “area stretching factor”.

We have that Z

g(I2)

is the limit of the sums of values of ! on small elements of the surface g(I2).

Suppose g takes a rectangle 4u×4v in I2 to a (small) piece of the surface.

! at g

is, say,

Pdx ^ dy + Qdx ^ dz + Rdy ^ dz

and the unit normal to the surface is ˆn[u, v] (located at g

Write ˆn[u, v] as 2

ˆnx

ˆny

ˆnz

The dx^dy of the 2-form affects only the ˆnz component and does so linearly,

likewise the dx ^ dz is a rotation in the plane orthogonal to ˆny. The sum of

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

the three components is easily seen to be.

66666664

R _ g

−Q _ g

P _ g

77777775

ˆnx

ˆny

ˆnz

5 =

R _ g

−Q _ g

P _ g

5 q ˆn

multiplying by the “area stretching factor” we obtain

g(I2)

! =

g_!

Theorem 6.3. [Stokes] Let g : I2 −! U _ R3 be a smooth embedding a.e.

and ! a smooth differentiable 1−form on U. Then

@g(I2)

! =

g(I2)

Proof By taking the pieces separately and summing the results we can

assume without loss of generality that g is smooth. Then

@g(I2)

@I2

g_! by proposition 6.2.1

dg_! by Green’s Theorem for a square

g_d! by proposition 6.2.3

g(I2)

d! by proposition 6.2.4

Remark 6.2.6. It looks believable that if

g : Im −! U _ Rn

6.2. MODERN 113

is a smooth embedding a.e.(hence m _ n) and ! is a smooth m−form on

Rn, then Z

gIm

! =

g_!

and this is indeed the case. This is the general ‘differential measure-stretching’

change of variables rule.

It looks also believable that on Im, m > 1 if ! is an m − 1 form, and

g : Im −! U _ Rn is a smooth embedding,

dg_! = g_d!

which is also the case.

It also looks plausible that for any m − 1 form ! on Im,

@Im

! =

This is also the case.

Hence we have for any dimension, by copying out the proof of Theorem 6.3,

if ! is a smooth k − 1 form on Rn and g : Ik −! U _ Rn is a smooth

embedding a.e., Z

@g(Ik)

! =

g(Ik)

In this or more general forms, this is now known (to the well informed) as

Stokes’ Theorem. It is fair to say that Stokes would have needed to do some

work to recognise it.

It includes the case when n = 1, k = 0 when it says:

Z b

f(x)dx = F(b) − F(a)

when

f =

For this reason Stokes’ Theorem is sometimes known as the Fundamental

Theorem of Calculus. Please note that this is not as your textbook author

appears to think an analogy, it is simply a consequence of correct generalisation.

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

Remark 6.2.7. Since we are integrating, we can neglect the failure of g to

be smooth or 1 − 1 on sets of measure (length, area, volume · · · ) zero. So

Stokes Theorem works for lots of sets including spheres and things which

have no boundary. In the exercises you will find that this can be pushed a

lot further than such restricted cases.

Example 6.2.1. Let U be the hemisphere

U =

5 2 R3 : x2 + y2 + z2 = 1, z _ 0

;

And

! = xyz dx + x dy + y dz

be a differential 1-form on R3.

Calculate Z

and I

and show they are the same.

Solution 1:

The line integral looks easier so I do that first. I parametrise S1 = @U by

x = cos(t), y = sin(t), z = 0; 0 _ t _ 2_

Then I want Z 2_

cos(t) sin(t)0 dx + cos(t) dy + sin(t)0

since z=0 and hence dz = 0 Substituting for dy (dy = dy/dtdt),this becomes

Z 2_

cos(t) cos(t)dt

Z 2_

cos(2t) + 1

dt = _

Now for the surface integral over the hemisphere. First I parametrise the

hemisphere with

x =

1 − v2 cos(u), y =

1 − v2 sin(u), z = v

6.2. MODERN 115

This gives:

dx =

du +

and similarly for dy and dz. Working it out:

dx = −sin(u)

1 − v2 du −

v cos(u)

1 − v2

dy = +cos(u)

1 − v2 du −

v sin(u)

1 − v2

dz = 0 du + 1 dv

Now we calculate d! where:

! = xyz dx + x dy + y dz

d! = (1 − xz) dx ^ dy + (0 − xy) dx ^ dz + (1 − 0) dy ^ dz

This gives:

d! = (1−v

1 − v2 cos(u)) dx ^ dy −(1−v2) cos(u) sin(u) dx ^ dz +dy ^ dz

Now we have to calculate the ^ terms: dx ^ dy =

(−sin(u)

1 − v2 du −

v cos(u)

1 − v2

dv) ^ (+ cos(u)

1 − v2 du −

v sin(u)

1 − v2

dv)

= v sin2(u) − (−v cos2(u)) du ^ dv

= v du ^ dv

And

dx ^ dz = (−sin(u)

1 − v2 du −

v cos(u)

1 − v2

dv) ^ (0du + 1dv)

= −sin(u)

1 − v2 du ^ dv

and finally:

dy ^ dz = cos(u)

1 − v2 du ^ dv

Substituting in the integral gives:

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

(1 − v

1 − v2 cos(u))(v) +

((1 − v2) sin(u) cos(u))(sin(u)

1 − v2) + 1(cos(u)

1 − v2 )

du ^ dv

This is

ZZ n

(v − v2

1 − v2 cos(u) + (1 − v2)3/2 cos(u) sin2(u) + cos(u)

1 − v2

du^dv

All that remains is to ensure that the orientation is correct: at u = 0, v = 0

we ought to get the outward normal: it is enough to verify that dy ^ dz = k

is pointing out. This is cos(u)

1 − v2 which is +1 which is correct.

Now we can leave the wedge out and perform the double integral. Using the

mathematica expression:

Integrate[(v - v^2*Sqrt[1 - v^2]*Cos[u]) +

Cos[u]*(Sin[u])^2*Sqrt[1 - v^2]*(1 - v^2) +

Cos[u]*Sqrt[1 - v^2], {u, 0, 2Pi}, {v, 0, 1}]

we get the result _ again.

Solution 2 Now I do it all again but using the old fashioned physicist’s

notation.

Now we write the 1-form as a vector field:

F = xyz i + x j + y k

and

curl(F) = r × F = 1 i + xy j + (1 − xz) k

For the path integral around the unit circle at z = 0 we have again

x = cos(t), y = sin(t), z = 0

so the path integral becomes

0 + cos(t)

+ sin(t)

cos2(t) dt

6.2. MODERN 117

as before.

For the integral over the surface we again need a parametrisation g:

x =

1 − v2 cos(u), y =

1 − v2 sin(u), z = v

for

0 _ v _ 1, 0 _ u _ 2_

and I need to calculate

This is: 2

−sin(u)

1 − v2

cos(u)

1 − v2

5 ×

664

cos(u)(−v) p

1−v2

sin(u)(−v) p

1−v2

775

Evaluating the cross product gives:

dS =

cos(u)

1 − v2

sin(u)

1 − v2

We have

curlF =

1 − xz

5 =

(1 − v2) sin(u) cos(u)

1 − v

1 − v2 cos(u)

Now we integrate

(1 − v2) sin(u) cos(u)

1 − v

1 − v2 cos(u)

5 q

cos(u)

1 − v2

sin(u)

1 − v2

with respect to u, v over the region 0 _ v _ 1, 0 _ u _ 2_

This comes out as

ZZ n

(v − v2

1 − v2 cos(u) + (1 − v2)3/2 cos(u) sin2(u) + cos(u)

1 − v2

dudv

as before.

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

Remark 6.2.8. As you can see, it is worth being able to decide if a 2-form is

derived from a 1-form: in the above case the double integral is much harder

than the single one around the circle. Given the double integral of d! it

would have been worth looking for an ! to integrate around the boundary.

Пресс-релиз популярных книг

6.2 Modern Theorem

Популярные книги

Популярные статьи

Навигация