# Functional analysis and diff. forms

1. Mar 15, 2010

### Tomsk

Hi PF,

I am currently trying to teach myself the rudiments of differential forms, in particular their application to physics, and there's something I'd like to ask.

It seems like diff forms can be used to express all kinds of physics, but the area I haven't been able to figure out is stuff related to functional analysis. Basically, what I want to know is, can we make sense of something like $\delta F / \delta \alpha$ where F is a functional of the differential form alpha, i.e.$F[\alpha]=\int f(\alpha)$. I've been trying to use the Gateaux derivative, something like $\frac{d}{dt}F[\alpha+t\tau]|_{t=0}$ where tau is a test form (of the same degree as alpha). I think this would be related to the functional derivative by $$\int \left(\frac{\delta F}{\delta \alpha}\right)\wedge\tau = \frac{d}{dt}F[\alpha+t\tau]\Big |_{t=0}$$,
which seems to nearly work, but not quite.

Now is probably a good time for an example. I've been mainly trying to apply this to electromagnetism. Using diff forms we can express Maxwell's equations as dF=0 and d*F=J, where F is the curvature of something, and F=dA. The action can be written:
$$S[A] = \int F \wedge \star F +A \wedge J$$
Since we already know F is exact (because of the Bianchi identity), we just need to derive d*F=J using the principle of least action ($\delta S / \delta A = 0$). Using the Gateaux derivative I get as far as
$$\frac{d}{dt}S[A+t\tau]\Big |_{t=0} = \frac{d}{dt}\left(\int (F+td\tau)\wedge(\star F+t \star d\tau) + (A+t\tau)\wedge J \right)\Big |_{t=0}$$
$$=\int F \wedge \star d\tau + d\tau \wedge \star F +\tau \wedge J$$
$$=\int F \wedge \star d\tau - \tau \wedge d\star F +\tau \wedge J$$
if we assume boundary terms vanish. The only problem is that first term, I don't know how to get rid of it. So... what's going on? Am I even allowed to use the Gateaux derivative like that? I feel like I'm on the right track, but it would be great to hear from someone else, I need a nudge in the right direction (I'm mainly just learning from wikipedia and stuff). Maybe there's a reason not to send *F to *(F+t dtau)? That would be nice!

Thanks :)

2. Mar 16, 2010

### Ben Niehoff

hmm...one issue is that your action integral is not an integral over time unless you pull back the forms to the worldline. You can't really do this for the pure Maxwell part, so I don't think your method is quite going to work. Also, you're missing a factor of -1/2. I would do it like this:

$$S = -\frac12 \int F \wedge * F + \int A \wedge *J$$

In my notation J is a one-form rather than a three-form. Then the notation is consistent. Note that each integral is simply an inner product on the space of n-forms. Then

$$\delta S = -\frac12 \int \left( \delta F \wedge * F + F \wedge * \delta F \right) + \int \delta A \wedge *J$$

Note that $\delta$ commutes with the Hodge dual because the metric is fixed. This is important if you want to try this in curved space...things get more complicated in that case. Now, we note that the inner product $\int \alpha \wedge * \beta$ is symmetric, so we may write

$$\delta S = -\int \delta F \wedge * F + \int \delta A \wedge *J$$

Now, we use the definition F = dA:

$$\delta S = -\int d\delta A \wedge * dA + \int \delta A \wedge *J$$

and use the identity

$$\int_R d(\alpha \wedge \beta) = \int_{\partial R} \alpha \wedge \beta = \int_R d\alpha \wedge \beta + (-1)^p \int_R \alpha \wedge d\beta$$

where in our case, the boundary integral vanishes because the variation $\delta A$ vanishes (sufficiently fast) at infinity. Since A is a 1-form, we have p=1, and so we get no sign change when we integrate by parts. Then we have

$$\delta S = -\int \delta A \wedge d * dA + \int \delta A \wedge *J$$

Now, for the action to be stationary, we must have $\delta S = 0$ for all possible variations, and so

$$d*dA = *J$$

as desired.

Edit:

On second thought, it looks like your method actually works fine, except for the missing -1/2, and some mistakes here:

There are two problems. First, the second term on the second line should not have a minus sign. As explained above, you do not get a minus sign when integrating by parts on odd-degree forms.

Second, you can take care of the first term by using the symmetry of the inner product. A little algebra should convince you that

$$\alpha \wedge * \beta = \beta \wedge * \alpha$$

in general, where alpha and beta are forms of the same degree. After doing that and integrating by parts, you have the answer you want (provided you also insert the -1/2 where it goes).

Last edited: Mar 16, 2010
3. Mar 17, 2010

### Tomsk

Thanks for your reply. I hadn't realised that about the inner product. I am always missing minus signs and factors of 1/2, typical! I can't quite match the 1/2 to what I already know though. I learned the lagrangian as $-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$ (oops, the minus should have been there!), and the 2-form F is $F = \frac{1}{2}F_{\mu\nu}dx^\mu \wedge dx^\nu$, so I thought that gave the 1/4. In fact now that I think of it the hodge star introduces another 1/2, which would make 1/4 with the one you introduced, but then I have 1/8 overall... Then again there is a hidden half in d*F=*J. hmm.

The other thing is, what about deriving the stress energy tensor? I gather from here that I am looking for a vector valued 3-form. [I am not sure what a vector valued k-form even is, can I think of it like a (1,k)-tensor that's antisymmetric in its k lower indices?] I haven't even been able to express it in terms of F (I'm just focusing on the part associated with the field). As a tensor it is:
$$T_\mu^\nu = -(F_{\mu\alpha}F_{\beta\gamma}g^{\alpha\beta}g^{\gamma\nu}+\frac{1}{4}\delta_\mu^\nu F_{\sigma\alpha}F_{\beta\rho} g^{\alpha\beta} g^{\rho\sigma})$$
And I gather the form we're looking for is something like
$$\star T = T_\mu^\nu \sqrt{-g} g^{\rho\mu}\frac{1}{3!}\epsilon_{\rho\alpha\beta\gamma} e^\alpha \wedge e^\beta \wedge e^\gamma \otimes e_\nu$$
(I think?!)
But can I write that in terms of F? I am thinking it won't be possible to write it in a nice neat and tidy way. And can I derive it from the lagrangian by differentiating with respect to the metric? It seems to be harder because the metric is kind of hidden inside the hodge star. I'm looking at doing it by another method, by working out the Noether current associated with spatial translations. That lead me on to things I don't really understand like flows and Lie derivatives.

I am thinking this is all quite far beyond what I know...

4. Mar 18, 2010

### Ben Niehoff

When you write out the indices,

$$\frac12 F \wedge *F = \frac12 \left( \frac12 F_{\mu\nu} dx^\mu \wedge dx^\nu \right) \wedge \left( \frac12 F^{\alpha \beta} \frac1{4!} \epsilon_{\alpha \beta \rho \sigma} dx^\rho \wedge dx^\sigma \right) = \frac18 \cdot \frac1{4!} F_{\mu\nu} F^{\alpha \beta} \epsilon_{\alpha \beta \rho \sigma} dx^{\mu} \wedge dx^{\nu} \wedge dx^{\rho} \wedge dx^{\sigma}$$

Now, note that alpha and beta must always be different from rho and sigma, which means they must be the same as mu and nu. There are two possibilities: alpha=mu, beta=nu, or alpha=nu, beta=mu. Both possibilities give the same term in the sum, so we can simply choose one option and multiply by 2:

$$= \frac14 \cdot \frac1{4!} F_{\alpha \beta} F^{\alpha \beta} \epsilon_{\mu \nu \rho \sigma} dx^{\mu} \wedge dx^{\nu} \wedge dx^{\rho} \wedge dx^{\sigma}$$

Now, recognizing the volume form, we can write

$$= \frac14 F_{\alpha \beta} F^{\alpha \beta} d^4x$$

I don't have a good answer for your bit about the energy-momentum tensor. I will try to think of one.

5. Mar 19, 2010

### Tomsk

Thanks, I get it now.

About the stress energy tensor, I'm not sure diff forms are the best language to describe it. I've been reading about geometric algebra too, which might be more relevant. Apparently T(n) = -1/2 FnF, where F is now a bivector, and n is an arbitrary vector, but I'm not sure about the derivation.