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5.3 Spin fields and Differential 2-forms

Back

The idea of a vector field having an amount of twist or spin at each point

turned out to make sense. Now I want to consider something a bit wilder.

Suppose I have a physical system which has, for each point in the plane, an

amount of twist or spin associated with it. We do not need to assume this

comes from a vector field, although it might. I could call such a thing a

‘twist field’ or ‘spin field’ on R2. If I did I would be the only person doing

so, but there is a proper name for the idea, it is called3 a differential 2-form.

To signal the fact that there is a number associated with each point of the

plane and it matters which orientation the plane has, we write

R(x, y) dx ^ dy

for this ‘spin field’. The dx ^ dy tells us the positive direction, from x to y.

If we reverse the order we reverse the sign:

dx ^ dy = −dy ^ dx

3 A differential 2-form exists to represent anything that is associated with an oriented

plane in a space, not necessarily spin. We could use it for describing pressure in a fluid.

But I can almost imagine a spin field, and so I shall pretend that these are the things for

which 2-forms were invented, which is close.

5.3. SPIN FIELDS AND DIFFERENTIAL 2-FORMS 61

Remark 5.3.1. The idea makes sense, and we now have a spin field and

we can ask, does it come from a vector field? Or more simply, we have a

differential 2-form and we would like to know if it comes from a differential

1-form.

Given R(x, y) dx ^ dy, is there always a P dx + Q dy that gives

R(x, y) =

−

It is easy to see that there are lots of them.

Example 5.3.1. Let

= xy sin(y) dx ^ dy

be a “spin field” on R2. Is it derived from some 1-form ! = P dx + Q dy?

Solution: Yes, put P = 0 and Q = x2y sin(y)/2 Then

−

= xy sin(y)

All I did was to set P to zero and integrate Q with respect to x. This is a

bit too easy to be interesting. It stops being so silly if we do it on R3, as we

shall see later.

Remark 5.3.2. It should be obvious that just as we had the derivative taking

0-forms to 1-forms, so we have a process for getting 2-forms from 1-forms.

The process is called the exterior derivative, written d and on R2 is is defined

by:

Definition 5.1. If ! , P dx+Q dy is a 1-form on R2 the exterior derivative

of !, d!, is defined by

d! ,

−

dx ^ dy

Remark 5.3.3. Although I have been doing all this on R2, it all goes over

to R3 and indeed Rn for any larger n. It is particularly important in R3, so

I shall go through this case separately.

Remark 5.3.4. If you can believe in a ‘spin field’ in R2 you can probably

believe in one on R3. Again, you can see that a little paddle wheel in a vector

field flow on R3 could turn around as a result of different amounts of push

on different sides. Now this time the paddle wheel could be chosen to be in

62 CHAPTER 5. GREEN’S THEOREM

any plane through the point, and the amount of twist would depend on the

point and the plane chosen. If you think of time as a fourth dimension, you

can see that it makes just as much sense to have a spin field on R4. In both

cases, there is a point and a preferred plane and there has to be a number

associated with the point and the plane. After all, twists occur in planes.

This is exactly why 2-forms were invented. Another thing about them: if

you kept the point fixed and varied the plane continuously until you had the

same plane, only upside down, you would get the negative of the answer you

got with the plane the other way up. In R3 the paddle wheel stick would be

pointing in the opposite direction.

We can in fact specify the amount of spin in three separate planes, the x−y

plane, the x − z plane, and the y − z plane, and this is enough to be able to

calculate it for any plane. This looks as though we are really doing Linear

Algebra, and indeed we are.

Definition 5.2. 2-forms on R3 A smooth differential 2-form on R3 is written

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

where the functions E, F,G are all smooth.

Remark 5.3.5. If you think of this as a spin field on R3 with E(x, y, z)

giving the amount of twist in the x − y plane, and similarly for F,G, you

won’t go wrong. This is a useful way to visualise a differential 2-form on R3.

Remark 5.3.6. It might occur to you that I have told you how we write a

differential 2-form, and I have indicated that it can be used for talking about

spin fields, and told you how to visualise the spin fields and hence differential

2-forms. What I have not done is to give a formal definition of what one is.

Patience, I’m coming to this.

Example 5.3.2. Suppose the plane x + y + z = 0 in R3 is being rotated in

the positive direction when viewed from the point

5, at a constant rate

of one unit. Express the rotation in terms of its projection on the x−y, x−z

and y − z planes.

Solution If you imagine the plane rotating away and casting a shadow on

the x − y (z = 0, but remember the orientation!) plane, clearly there would

be some ‘shadow’ twist, but not the full quantity. Likewise the projections

on the other planes.

5.3. SPIN FIELDS AND DIFFERENTIAL 2-FORMS 63

Take a basis for the plane consisting of two orthogonal vectors in the plane

of length one. There are an infinite number of choices: I pick

u =

−1/

5, v =

−1/

(I got these by noticing that

−1

5 was in the plane and then I took the

cross product with the normal to the plane to get

−1

finally I scaled them to have length one.)

Now I write

du =

−

dv =

−1

dx +

dy −

which I got by telling myself that the projection onto the basis vector u should

be called du, and that it would in fact send [x, y, z]T to 1/

2 i − 1/

2 k

which is the mixture

−

And similarly for the expression for dv.

Last, I write the spin as

1 du ^ dv =

dx −

−1

dx +

dy −

Expanding this and using dx ^ dx = 0 and dz ^ dx = −dx ^ dz I get

dx ^ dy −

dx ^ dz +

dy ^ dz

dx ^ dy −

dx ^ dz +

dy ^ dz

64 CHAPTER 5. GREEN’S THEOREM

This shows equal amounts of spin on each plane, and a negative twist on the

x − z plane, which is right. (Think about it!)

Note that the sum of the squares of the coefficients is 1. This is the amount

of spin we started with. Note also that the sums are easy although they

introduce the ^ as if it is a sort of multiplication. I shall not try to justify

this here. At this point I shall feel happy if you are in good shape to do the

sums we have coming up.

Remark 5.3.7. It would actually make good sense to write it out using

dz ^ dx instead of dx ^ dz Then the plane x + y + z = 0 with the positive

orientation can be written as

dx ^ dy +

dz ^ dx +

dy ^ dz

In this form it is rather strikingly similar to the unit normal vector to the

plane. Putting dx ^ dy = k and so on, is rather tempting. It is a temptation

to which physicists have succumbed rather often.

Example 5.3.3. The spin field 2 dx ^ dy + 3dz ^ dx + 4 dy ^ dz on R3 is

examined by inserting a probe at the origin so that the oriented plane is

again x + y + z = 0 with positive orientation seen from the point i + j + k.

What is the amount of spin in this plane?

Solution 1 Project the vector 2 dx^dy +3dz ^dx+4 dy ^dz on the vector

dx ^ dy +

dz ^ dx +

dy ^ dz

to get

dx ^ dy +

dz ^ dx +

dy ^ dz

= 3 dx ^ dy + 3 dz ^ dx + 3 dy ^ dz

= 3

3 du ^ dv

Solution 2 (Physicist’s solution) Write the spin as a vector 4i+3j+2k and

the normal to the oriented plane as

i +

j +

Now take the dot product to get the length of the spin (pseudo)vector:

3 = 3

3. The whole vector is therefore

3 i + 3 j + 3 k

This is 3

3 times the unit normal to the plane x + y + z = 0.

5.3. SPIN FIELDS AND DIFFERENTIAL 2-FORMS 65

Remark 5.3.8. It should be clear that in R3 it is largely a matter of taste

as to which system, vectors or 2-forms, you use. The advantage of using

2-forms is partly that they generalise to higher dimensions. You could solve

a problem similar to the above but in four dimensions if you used 2-forms,

while the physicist would be stuck. So differential forms have been used in

the past for terrorising physicists, which takes a bit of doing. The modern

physicists are, of course, quite comfortable with them.

Remark 5.3.9. Richard Feynman in his famous “Lecture Notes in Physics”

points out that vectors used in this way, to represent rotations, are not

really the same as ordinary vectors such as are used to describe force fields.

He calls them ‘pseudovectors’ and makes the observation that we can only

confuse them with vectors because we live in a three dimensional space, and

in R4 we would have six kinds of rotation.

Exercise 5.3.1. Confirm that Feynman knows what he is talking about and

that six is indeed the right number.

5.3.1 The Exterior Derivative

Remark 5.3.10. Now I tell you how to do the exterior derivative from 1-

forms to 2-forms on R3. Watch carefully!

Definition 5.3. If

! , P(x, y, z)dx + Q(x, y, z)dy + R(x, y, z)dz

is a smooth 1-form on R3 then the exterior derivative applied to it gives the

2-form:

d! ,

−

dx^dy+

−

dx^dz+

−

dy^dz

Remark 5.3.11. This is not so hard to remember as you might think and I

will now give some simple rules for working it out. Just to make sure you can

do it on R4 I give it in horrible generality. (Actually I have a better reason

than this which will emerge later.)

Definition 5.4. If

! , P1 dx1 + P2 dx2 + · · · + Pn dxn

66 CHAPTER 5. GREEN’S THEOREM

is a differential 1-form on Rn, the exterior derivative of !, d! is the differential

2-form

@P2

@x1

−

@P1

@x2

dx1 ^ dx2 +

@P3

@x1

−

@P1

@x3

dx1 ^ dx3 + · · ·

· · · +

@P3

@x2

−

@P2

@x3

dx2 ^ dx3 + · · · +

@Pn

@xn−1

−

@Pn−1

@xn

dxn−1 ^ dxn

This looks frightful but is actually easily worked out:

Rule 1: Partially differentiate every function Pj by every variable xi. This

gives n2 terms.

Rule 2 When you differentiate Pj dxj with respect to xi, write the new

differential bit as:

@Pj

@xi dxi ^ dxj

Rule 3: Remember that dxi ^ dxj = −dxj ^ dxi. Hence dxi ^ dxi = 0 So we

throw away n terms leaving n(n − 1), and collect them in matching pairs.

Rule 4 Bearing in mind Rule 3, collect up terms in increasing alphabetical

order so if i < j, we get a term for dxi ^ dxj .

Remark 5.3.12. It is obvious that if you have a 1-form on Rn, the derived

2-form has n(n − 1)/2 terms in it.

Proposition 5.3.1. When n = 2, this gives

d! =

−

dx ^ dy

Proof

Start with:

! , P dx + Q dy

Following rule 1 we differentiate everything in sight and put du^ in front of

the differential already there when we differentiate with respect to u, where

5.3. SPIN FIELDS AND DIFFERENTIAL 2-FORMS 67

u is x or y. This gives us:

d! =

dx ^ dx +

dy ^ dx +

dx ^ dy +

dy ^ dy

Now we apply rule 3 and throw out the first and last term to get

d! =

dy ^ dx +

dx ^ dy

and finally we apply rules 3 and 4 which has dx^dy as the preferred (alphabetic)

ordering so we get:

d! =

−

dx ^ dy

as required. _

Example 5.3.4. Now I do it for R3 and you can see how easy it is to get

the complicated expression for d! there:

I shall take the exterior derivative of

P(x, y, z) dx + Q(x, y, z) dy + R(x, y, z) dz

which is a 1-form on R3:

Rules 1, 2 give me

dx ^ dx +

dy ^ dx +

dz ^ dx

when I do the P term.

The Q term gives me:

dx ^ dy +

dy ^ dy +

dz ^ dy

and finally the R term gives me:

dx ^ dz +

dy ^ dz +

dz ^ dz

Rule 3 tells me that of these nine terms, three are zero, the dx ^ dx, the

dy ^ dy and dz ^ dz terms. I could have saved a bit of time by not even

writing them down.

68 CHAPTER 5. GREEN’S THEOREM

It also tells me that the remaining six come in pairs. I collect them up in

accordance with Rule 4 to get:

−

dx ^ dy +

−

dx ^ dz +

−

dy ^ dz

Remark 5.3.13. Not so terrible, was it? All you have to remember really is

to put du^ in front of the old differential when you partially differentiate with

respect to u, and do it for every term and every variable. Then remember

du ^ dv = −dv ^ du and so du ^ du = 0 and collect up the matching pairs.

After some practice you can do them as fast as you can write them down.

Remark 5.3.14. Just as in two dimensions, the exterior derivative applied

to a 1-form or vector field gives us a 2-form or spin-field. Only now it has

three components, which is reasonable.

Remark 5.3.15. If you are an old fashioned physicist who is frightened of

2-forms, you will want to pretend dy ^dz = i, dz ^dx = j and dx^dy = k.

Which means that, as pointed out earlier, spin fields on R3 can be confused

with vector fields by representing a spin in a plane by a vector orthogonal to

the plane of the spin and having length the amount of the spin.

This doesn’t work on R4.

You will therefore write that if

F = P i + Q j + R k

is a vector field on R3, there is a derived vector field which measures the spin

of F It is called the curl and is defined by:

Definition 5.5.

curl(F) ,

BB@ @

@y − @Q

@z − @R

@x − @P

CCA

Remark 5.3.16. This is just our formula for the exterior derivative with

dy ^ dx put equal to i, dx ^ dz put equal to −j and dx ^ dy put equal to k.

It is a problem to remember this, since the old fashioned physicists couldn’t

easily work it out, so instead of remembering the simple rules for the exterior

derivative, they wrote:

r =

5.3. SPIN FIELDS AND DIFFERENTIAL 2-FORMS 69

This they pretended was a vector in R3 and then :

Definition 5.6.

curl(F) = r × F

where × is the cross product in R3.

Exercise 5.3.2. Confirm that the two notations are equivalent.

Example 5.3.5. Find the 2-form d! where

! , xz dx + xyz dy − y2 dz

Solution:

Just writing down the terms following the four rules but skipping the zero

terms I get:

(yz) dx ^ dy + (−x) dx ^ dz + (−2y − xy) dy ^ dz

Doing it using the curl rule I get:

curl(!) = r ×

xyz

−y2

5 =

−2y − xy

These are obviously the same under the convention of writing dx ^ dy = k

and so on.

I know which I prefer, but you can make up your own minds. Be sure you

can use both systems with reasonable skill.

Remark 5.3.17. Again, the old fashioned physicists had no satisfactory way

of handling problems in Rn for n > 3. The exterior derivative d! looks a

fairly straightforward thing compared with r × F. Of course, you can get

used to any notation if you use it often enough. Some of the more inflexible

minds, having spent a lot of time mastering one notation scream in horror

at the thought of having to learn another. The question of which is better

doesn’t come into it. The well known equation

NEW = EVIL

is invoked.

Remark 5.3.18. We can generalise 2-forms to Rn with minimal fuss:

70 CHAPTER 5. GREEN’S THEOREM

Definition 5.7. Differential 2-forms on Rn

A smooth differential 2-form on Rn is written as

1_i<j_n

Fi,j(x) dxi ^ dxj

where the Fi,j are smooth functions from Rn to R.

Remark 5.3.19. We can again ask, this time for R3, the question, is any

spin field (differential 2-form) on R3 in fact the exterior derivative (derived

spin field) of a vector field (differential 1-form)? Algebraically, if

! = E(x) dx ^ dy + F(x) dx ^ dz + G(x) dY ^ dz

is there a 1-form

= P dx + Q dy + R dz

such that

d = !?

that is, can we find P, Q,R such that

−

= E

and _

−

= F

and _

−

= G

Remark 5.3.20. You might like to try this out for simple cases. Experiment,

explore, it is a more interesting world if you do. I shall come back to this

later.

Remark 5.3.21. Another question we might ask (based on simple curiosity

and trying to push things from two dimensions to three) is: If ! is a smooth

1-form on R3 and d! = 0, is it the case that ! = df for some 0-form f?

Or, saying it in the old fashioned language, if the curl of a smooth vector

field is zero, is it conservative? Note that if the curl of a vector field exists,

it has to be a vector field on at least a subset of R3, since this is the only

place where we have a cross product to be able to compute the curl.

The answer to both questions is yes. On the other hand it can fail to be true

if the vector field/1-form is defined on only a subset of R3 which has holes

in it.

5.3. SPIN FIELDS AND DIFFERENTIAL 2-FORMS 71

Exercise 5.3.3. Define “has no holes in it” for subsets of R3

Proposition 5.3.2. If curlF = 0 on all of R3, then F is conservative.

Recall:

Definition 5.8. Exact A 1-form ! is exact iff it there is a 0-form f such

that ! = df.

Remark 5.3.22. This was stated for forms on R2 but since the dimension

isn’t mentioned, it must work for R3 as well.

Proposition 5.3.3. If ! is a 1-form and d! = 0 then ! is exact.

Proof: Later.

Example 5.3.6.

! = 2xy3z4 dx + 3x2y2z4 dy + 4x2y3z3 dz

Is this 1-form exact? That is, is ! = df for some 0-form f on R3?

Applying the claim (as yet unproven) above:

d! = (6xy2z4 − 6xy2z4) dx ^ dy + (8xy3z3 − 8xy3z3) dx ^ dz

+ (12x2y2z3 − 12x2y2z3) dy ^ dz = 0

This tells us that provided the form is defined on all of R3, which it is, then

it must be exact, and the corresponding vector field is conservative. So there

must be a 0-form f. We must have

= 2xy3z4

so integrating with respect to x we get

x2y3z4 + u(y, z)

for some unknown function of y and z only. Similar integration for the other

two functions leads us to the conclusion that u = 0 and

f(x, y, z) = x2y3z4

Remark 5.3.23. There is nothing much new here over the two dimensional

case and you could do it for 1-forms on R4 in exactly the same way.

72 CHAPTER 5. GREEN’S THEOREM

5.3.2 For the Pure Mathematicians.

Remark 5.3.24. You may be feeling uneasy that we have not given a formal

definition of a differential 2-form on U _ Rn, for any n. Instead I have just

told you how to write them down and how to derive (some of) them from

1-forms. In this respect your expereince of them is just like your experience

of cats. You know how to recognise one, and you know and what to do with

one when you meet it. In Mathematics, if not in real life, it is possible to do

better.

I told you that a differential 0-form on U was a smooth map

f : U −! R

and that a differential 1-form on U _ Rn was a smooth map

! : U −! Rn_

In the first case we attached a number to each point of U, in the second we

attach a covector, barely distinguishable from a vector.

You are within your rights to expect me to tell you that a differential 2-form

on U is a map from U to some vector space of thingies which can be used

to represent torques. Certainly these ‘thingies’ have to have some sort of

association with oriented planes at the very least.

We can actually do this: we find it is a map from U to a vector space of things

called alternating 2-tensors written 2(Rn) and the dxi^dxj are basis vectors

of it.

You can see that the dimension of this space will be n(n−1)/2 because that

is how many dxi ^ dxj there are. So a point in this vector space will be

1_i<j_n

ai,j dxi ^ dxj

for some collection of n(n − 1)/2 numbers ai,j .

A differential 2-form on U attaches such things to points of the space U: the

ai,j vary as we move about in U. We are now attaching alternating tensors,

which gives us a tensor field. A differential form is an example of a tensor

field: 1-forms and vector fields are the easiest cases, 2-forms a bit harder.

Remark 5.3.25. The actual definition of 2(Rn) may come as a bit of an

anticlimax.

5.3. SPIN FIELDS AND DIFFERENTIAL 2-FORMS 73

I shall do it by defining the basis vectors of the space, the dxi^dxj , for i < j.

Each dxi is a projection from Rn to R so it is not surprising that dxi ^ dxj

is a map:

dxi ^ dxj : Rn × Rn −! R

6664

...

7775

6664

...

7775

xiyj − xjyi

In other words, it is just the determinant of the i and j rows. It is obvious

that there are Cn

2 = n(n − 1)/2 ways of picking two rows from n, and any

two choices are different maps. If you reverse the order of the rows you get

dxj ^ dxi = −dxi ^ dxj which is right.

Definition 5.9. 2(Rn) is the vector space of maps from Rn × Rn to R

spanned by the above maps.

Remark 5.3.26. It is now easy to prove that it really is a vector space with

the right dimension. This is not the only way to define it, or even the best

way, but it is an easy way which is why I have picked it.

Remark 5.3.27. So now you know what a differential 2-form on U _ Rn

really is, and you have met a definition of an alternating tensor field. Crikey,

life doesn’t get much better than this.

Remark 5.3.28. For the record, for possible future needs, but not for your

present needs, a differential k-form on U _ Rn is still an alternating tensor

field, i.e a differentiable map from Rn to k(Rn). The space k(Rn) is defined

as a space of maps

Rn × Rn · · · × Rn

| {z }

k copies

−! R

A basis vector in this space makes a choice of k different rows and calculates

the determinant of the resulting k × k matrix. There are obviously Cn

different basis vectors and the span of them all is k(Rn).

Remark 5.3.29. The exterior derivative generalises to an exterior derivative

d that takes k-forms to k + 1 forms. If you have a lot of terms of the form

Fi1,i2,···ikdxi1 ^ dxi2 · · · ^ dxik

you just differentiate all of them with respect to everything as in rule 1.

74 CHAPTER 5. GREEN’S THEOREM

You put du^ in front of the differential part when you have differentiated

the function part with respect to u, just as for Rule 2.

You remember that if any two terms get swapped among all those

dxi1 ^ dx12 ^ · · ·

the sign is changed, so if any are the same you put the term to be zero.

Then you collect up in alphabetic order. And that does it.

Remark 5.3.30. It requires, perhaps, rather more (multi-)linear algebra

than you have met so far for you to feel altogether happy about this. If you

are prepared to take my word for it that I can write any oriented plane in

R3 as so much dx ^ dy plus some amount of dx ^ dz added to a quantity

of dy ^ dz, then you can proceed without further worry. If you feel insecure

without formal definitions, they are in the appendix.

5.3.3 Return to the (relatively) mundane.

Remark 5.3.31. It is easy to see how to write differential 2-forms on R4:

On R4 with

664

775

specifying the four components.

We would have

Pdx ^ dy + Qdx ^ dz + Rdx ^ dw + Sdy ^ dz + Tdy ^ dw + Udz ^ dw

as a differential 2-form. This agrees with Richard Feynman which is cheering.

Remark 5.3.32. You will perhaps be even more cheered to note that we

don’t go beyond 3-forms on R3.

Proposition 5.3.4. We have 0-forms d −! 1-forms d −! 2-forms on R2

and d2 = 0.

Proof: If f is the 0-form,

df =

dx +

5.4. MORE ON DIFFERENTIAL STRETCHING 75

is the 1-form and

d2f =

@2f

@x@y

−

@2f

@y@x

dx ^ dy = 0

is the derived 2-form. _

Remark 5.3.33. This also holds on R3:

Proposition 5.3.5. For 0-forms on R3, d2 = 0.

Proof: Given f : R3 −! R is a 0-form,

df =

dx +

dy +

and

d2f =

@2f

@y@x

−

@2f

@x@y

dx^dy+

@2f

@z@x

−

@2f

@x@z

dx^dz+

@2f

@z@y

−

@2f

@y@z

dy^dz

= 0

Remark 5.3.34. We can turn this into old-fashioned language by writing:

curl(r(f)) = 0

r × r(f) = 0

Remark 5.3.35. Since there is no new idea in this and it is just the preceding

subject rewritten in the old fashioned notation, I leave it to you to verify it.

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