exterior derivative


by Terilien
Tags: derivative, exterior
Terilien
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#1
Apr15-07, 01:51 PM
P: 140
What exactly is the exterior derivative? What is its motivation? how do you compute it? Most importantly why is that how you copute it?
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mathwonk
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#2
Apr15-07, 03:37 PM
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well physicists understand this better than i do, but heres a try. in one variable, we know that F'(x) is the limit of [F(b)-F(a)]/(b-a) as a,b, approach x. this is by the ftc. or equivalently, the mvt.

this also holds true on the plane for a function of one variable. i.e. the value of the gradient on a tangent vector is this same limit as we let a,b, denoite the ends of a parametrized segment with the given tangent vector as velocity vector at x.

if F is a one form in 2 variables, by the ftc. or green, or whatever you call it, then dF at a point x, is the limit of the quotient formed by the integral of F over a loop centered at x, by the area of the inside of that loop.

thus it measures the tendency of the vector field dual to the one form in terms of a dot product, to rotate around the point x. so they call it the curl.

up one dimension it measures the tendency of these vectors to radiate through the surface of a sphere centered at x, hence it is called the divergence.

so to evaluate the exterior derivative of a k form on a k+1 vector at a point, you put a k+1 dimensional parametrized surface through that point, with its k+1 directional tangent vectors spanning the given k+1 vector, then you integrate the k form over the boundary of the surface, and divide by the K+1 dimensional volume of the surface.

take the limit, and by stokes theorem that is the value of the exterior derivative.

put another way, it is the thing that makes stokes theorem work. i.e. it is the "adjoint" of the boundary operator on surfaces, wrt the inner product defined by integration.

take this with a grain of salt. or maybe read bachmans book on this.
Terilien
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#3
Apr16-07, 02:45 PM
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Well could someone explain why we impose the condition d(da)=0. I think I understand but would still like an explanation...

Hurkyl
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Apr16-07, 05:29 PM
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exterior derivative


Quote Quote by Terilien View Post
Well could someone explain why we impose the condition d(da)=0. I think I understand but would still like an explanation...
Well, if you followed mathwonk's description that we use the exterior derivative because it makes Stoke's theorem work... then apply Stoke's theorem!

[tex]\int_R d(da) = \int_{\delta R} da = \int_{\delta(\delta R)} a[/tex]

If you take some shape (e.g. a disk), then look at its boundary (a circle)... what is the boundary of that?
coalquay404
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#5
Apr16-07, 06:05 PM
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Quote Quote by Terilien View Post
Well could someone explain why we impose the condition d(da)=0, and it's not something to be "imposed" but is instead a consequence of the manner in which the exterior derivative is defined. I think I understand but would still like an explanation...
This condition is referred to as "nilpotency" of the exterior derivative. Suppose that you have some coordinate basis [itex]\{dx^a\}[/itex] for the cotangent space at a point in your manifold [itex]M[/itex], and denote the space of [itex]p[/itex]-forms at a point [itex]q\in M[/itex] by [itex]\Lambda^p_q(M)[/itex]. The existence of this coordinate basis means that we can write a [itex]p[/itex]-form [itex]\alpha\in\Lambda^p_q(M)[/itex] as

[itex]\alpha = \frac{1}{p!}\alpha_{a_1a_2\ldots a_p}\wedge_{i=1}^p dx^{a_i}[/itex]

where I have defined

[itex]\wedge_{i=1}^p dx^{a_i} \equiv dx^{a_1}\wedge dx^{a_2}\wedge \ldots dx^{a_p}[/itex]

Now the exterior derivative is defined to be a map [itex]d:\Lambda^p_q(M)\to\Lambda^{p+1}_q(M)[/itex], [itex]d:\alpha\mapsto d\alpha[/itex]. The exterior derivative of [itex]\alpha[/itex] can be expressed in terms of its components in the coordinate basis as

[tex]d\alpha = \frac{1}{p!}\partial_{a_1}\alpha_{a_2a_3\ldots a_{p+1}}\wedge_{i=2}^{p+1}dx^{a_i}[/tex]

If we apply the exterior derivative a second time we find that we obtain a [itex](p+2)[/itex]-form that can be written as

[tex]d(d\alpha) = d^2\alpha
= \frac{1}{p!}\partial_{a_1}\partial_{a_2}\alpha_{a_3a_4\ldots a_{p+2}}\wedge_{i=3}^{p+2}dx^{a_i}[/tex]

However, since partial derivatives are commutative and since [itex]\wedge_{i=3}^{p+2}dx^{a_i}[/itex] is totally antisymmetric it shouldn't take a great deal of thought to convince yourself that [itex]d^2\alpha=0[/itex]. See if you can figure this out - it's important that you understand it.
Terilien
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#6
Apr16-07, 08:05 PM
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So I did understand it. One last thing, I'm not sure if I understand the modified leibniz rule very well. could someone prove it rigorously?

I'm talking about the liebniz rule between wedge products. I don't quite know how to prove it...:(
Hurkyl
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Apr16-07, 09:17 PM
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Quote Quote by Terilien View Post
So I did understand it. One last thing, I'm not sure if I understand the modified leibniz rule very well. could someone prove it rigorously?

I'm talking about the liebniz rule between wedge products. I don't quite know how to prove it...:(
What's modified about it? It says d(f g) = (df) g + f (dg), which is the same as the ordinary product rule.

Hrm -- are you simply thinking about the simplifications you can make by applying the identities (x and y are scalar fields):
d(df) = 0,
(dx)(dx) = 0,
(dx)(dy) = -(dy)(dx)?
Terilien
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#8
Apr16-07, 09:27 PM
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I was talking about the exterior derivative of a wedge product.

It's suppose to be something like, p^dq +-1^p (q^dp) or something along those lines. how do we get that?>

i know its silly but i really don't know how its proven.
Hurkyl
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Apr16-07, 10:10 PM
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Quote Quote by Terilien View Post
I was talking about the exterior derivative of a wedge product.

It's suppose to be something like, p^dq +-1^p (q^dp) or something along those lines. how do we get that?>

i know its silly but i really don't know how its proven.
When doing exterior algebra, I'm very much used to writing the product as
ab
rather than
a/\b.

What did you mean by -1^p? I assume you meant (-1)^(the rank of p * the rank of q)


Anyways, it's quite straightforward -- if you write down any example at all, it's fairly obvious what's happening. At least, if you apply the identities I mentioned. Algebraically, they are the defining property of the exterior algebra, so you need to take them to heart! (And they are actually all the same identity)


Or... did you mean to ask why d(fg) should be f dg + df g in the first place?
Terilien
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Apr16-07, 10:16 PM
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Oh no. it's simply the wedge product thing. anyway could you write out an example because i don' think I got that(unless i arranged the terms wrong which i rpobably did without noticing).

I know this is stupid, but please do this for me. I know its weird and stupid...

i'm awful with details and miss them all the time.
Hurkyl
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Apr16-07, 10:21 PM
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Well, what did you do? Incidentally, ignore the fact a derivative is involved. You're just looking at the basic properties of the product: how pq and qp are related, for arbitrary p and q.

(This might help with constructing a simpler example to understand what's going on)
Terilien
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#12
Apr16-07, 10:23 PM
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what i did was (p+dp)^(q+dq) -p^q. i evaluated that and got + p^dq +dp^q +dp^dq

what do we do with that? Wher do the other parts come in?

Again i'm sorry for my stupidity.
Hurkyl
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Apr16-07, 10:28 PM
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Quote Quote by Terilien View Post
what i did was (p+dp)^(q+dq) -p^q. i evaluated that and got + p^dq +dp^q +dp^dq

what do we do with that?
That (generally) shouldn't be equal to d(p /\ q). So it's a good thing you didn't wind up with p /\ dq - q /\ dp.
Terilien
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Apr16-07, 10:30 PM
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so where does the other thing come into play?
Hurkyl
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Apr16-07, 10:32 PM
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Quote Quote by Terilien View Post
so where does the other thing come into play?
Well, I'm not really sure what you're asking anymore. And it's past my bedtime, so I can't help anymore today.
Super_Leunam
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Nov4-09, 06:16 AM
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This website should clarify all your doubts:

http://www.maths.adelaide.edu.au/mic...ns/node26.html
wofsy
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Nov4-09, 03:17 PM
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Quote Quote by Terilien View Post
What exactly is the exterior derivative? What is its motivation? how do you compute it? Most importantly why is that how you copute it?
The exterior derivative allows the fundamental theorem of calculus to be generalized to integrals over areas and volumes and higher dimensional analogues.

The Fundamental Theorem of calculus says Integral from a to b of dF = F(a)-F(b). This means that the value of F on the boundary of the interval equals the integral of dF over the interior of the interval.

Stokes' Theorem says the same thing for integrating a form - its integral over the boundary of a region equals the integral of its exterior derivative over the interior of the region. You should try to prove this for yourself in a simple region. You will find that it is just the Fundamental Theorem all over again and the reason for its method of calculation will become obvious.

In three dimensions exterior derivatives can be thought of as describing properties of vector fields. These are the usual vector calculus pictures that are used in Physics.

A vector field has two key properties - its curl and its divergence. The curl represents the field's tendency to circulate The divergence detects the presence of sources of the field.

For instance in magnetostatics Gauss's Law says that the divergence of the magnetic field is zero. This means it has no sources and so must follow closed loops (in S3).

This law can be expressed in terms of differential forms as follows. If the magnetic field has components (a,b,c)the the exterior derivative of the 2 form ady^dz - bdx^dz + cdx^dy will be zero.

The curl of (a,b,c) can be thought of as the exterior derivative of the 1 form adx + bdy + cdz. By Ampere's Law, the integral of this 2 form over a piece of a surface will measure the total electric current passing through the region and this is the same as the circulation of the field around the boundary of the region.

I have found it helpful to learn all of these Physics pictures. There are many more.


One more picture. A 2 form can be thought of as dual to a small plane at each point and its integral over this plane determines an area measure. At each different point one gets a different area measure. But if the form is closed these measures are related. The kernel of the 2 form is 1 dimensional and thus can be integrated to determine a flow. If the form is closed then the area measure is preserved by the flow - Stokes Theorem again.
haushofer
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Nov5-09, 02:59 AM
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Quote Quote by Terilien View Post
I was talking about the exterior derivative of a wedge product.

It's suppose to be something like, p^dq +-1^p (q^dp) or something along those lines. how do we get that?>

i know its silly but i really don't know how its proven.
Just write it out, like a lot of exercises :) I will do it for a one and a two form; you should generalize the argument

[tex]
\alpha = \alpha_{a}dx^a, \ \ \ \ \ \beta = \beta_{de}dx^d \wedge dx^e, \ \ \ \ \ \ \alpha\wedge\beta = (\alpha_a \wedge \beta_{de})dx^a \wedge dx^d \wedge dx^e
[/tex]

Then


[tex]
d(\alpha\wedge\beta) = [(\partial_{c}\alpha_a)\beta_{de} + \alpha_a \partial_c(\beta_{de})]dx^c \wedge dx^a \wedge dx^d \wedge dx^e
[/tex]

Look at the contraction of the components with the basisvectors: the indices of the first term on the LHS are in the same order as the basis indices; {cade}. But the order of the second term on the LHS are {acde}. So in the second term we exchange the {c} and {a} index, and we pick up a minus sign. Notice that this depends on the position of the partial derivative in the second term on the LHS, which is important for your generalization; it depends on what kind of form beta is. So in the end we obtain

[tex]
d(\alpha\wedge\beta) = [(\partial_{c}\alpha_a)\beta_{de} - \alpha_c \partial_a(\beta_{de})]dx^c \wedge dx^a \wedge dx^d \wedge dx^e
[/tex]

Hope this helps :)


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