# Is $F^{\mu\nu}F_{\mu\nu}$ equivalent to $A^{\mu}\nabla^{\alpha}\nabla_{\alpha}A_{\mu}$ for $U(1)$ gauge field lagrangian?

The two seem to yield the same equation of motion is why I asked. Where of course the standard notation for exterior forms applies $$dA=F$$.

We all know how the field strength tensor plays into the equations of motion. The Lagrangian density $$A^{\mu}\nabla^{\alpha}\nabla_{\alpha}A_{\mu}$$ seems to have an equation of motion (from varying the $$A$$) of the form:

$$\nabla^{\alpha}\nabla_{\alpha}A_{\mu}+\nabla_{\mu}\nabla_{\alpha}A^{\alpha}=0$$ Under the standard Lorenz gauge $$\nabla_{\mu}A^{\mu}=0$$ this becomes simply:

$$\nabla^{\alpha}\nabla_{\alpha}A_{\mu}=0$$ Now maybe this only works for an abelian field (like the Maxwell)? I'm honestly not sure. If they are equivalent can this be extended to non-abelian fields? I had thought Lagrangians were only equivalent if they're related by a total divergence term, but I'm not seeing it (maybe I'm missing it though).

EDIT

This is what I've tried, but I'm new to exterior calculus:

$$\intop_{M}\left\{ A^{\mu}\nabla^{\alpha}\nabla_{\alpha}A_{\mu}\right\} (dvol)_{M}$$

We can use the self-adjointness of the Laplace-Beltrami operator to write the above as:

$$=\intop_{M}-\langle dA,dA\rangle(dvol)_{M}$$

For some function $$f$$ (0-form) we have that a gauge transformation looks like:

$$A\longrightarrow A+df$$

$$=\intop_{M}-\langle d\left(A+df\right),d\left(A+df\right)\rangle(dvol)_{M}$$

$$=\intop_{M}-\left\{ \langle dA,dA\rangle+\langle d\left(df\right),dA\rangle+\langle dA,d\left(df\right)\rangle+\langle d(df),d(df)\rangle\right\} (dvol)_{M}$$

My understanding is that, $$ddf=0$$ which is like saying that the partial derivatives commute, or that the divergence of a curl is zero (the latter statement only being in three dimensions). Suppose all boundary terms vanish (I'm actually working where there is no boundary), this Leaves us simply with:

$$=\intop_{M}-\left\{ \langle dA,dA\rangle\right\} (dvol)_{M}$$

$$=\intop_{M}\left\{ A^{\mu}\nabla^{\alpha}\nabla_{\alpha}A_{\mu}\right\} (dvol)_{M}$$

I really don't know if that's right (I think I missed something somewhere), I'm trying to teach myself differential forms and exterior calculus. Maybe it's just wishful thinking lol!

• You want to impose Lorenz gauge. But is your proposed Lagrangean $A_\mu\Box A^\mu$ gauge invariant? – Toffomat Dec 5 '18 at 11:55
• @Toffomat Good point, I'll check and come back tomorrow though I believe it was correct before the gauge transformation (which was for clarity sake). I'll come back fresh in the a.m – R. Rankin Dec 5 '18 at 12:40

Your expression $$\intop_{M}-\langle dA,dA\rangle(dvol)_{M}$$ is the usual Lagrangian for electromagnetism written in a highbrow way. You can work this out with Hodge duals and $$dA=F$$ if you've never seen it before. That is why it is no surprise you are showing it is gauge invariant.

Now you can go to $$\intop_{M}-\langle A,d^\dagger dA\rangle(dvol)_{M}$$, but this isn't the Laplacian yet. The Laplacian is $$d^\dagger d+d d^\dagger$$, so you need the gauge condition $$d^\dagger A=0.$$ Again, this is just the Lorentz gauge in fancier language. So your Lagrangian does rely crucially on Lorentz gauge.

You can see all this by the way without using differential forms in flat space easily through integration by parts, by the way. $$-\int dx^4 \partial_{[\mu}A_{\nu]}\partial^{[\mu}A^{\nu]}=-\int dx^4 \partial_{\mu}A_{\nu}\partial^{[\mu}A^{\nu]}=+\int dx^4 A_{\nu}\partial_{\mu}\partial^{[\mu}A^{\nu]}=\int dx^4 A_{\nu}\partial_{\mu}\partial^{\mu}A^{\nu}$$ where the last line I commuted the partial derivatives and used Lorentz gauge.

• gotcha! I was totally neglecting the codifferential $\delta$ in $$\Delta=(d+\delta)^2$$ (or dagger as you used) I think. Thank you! – R. Rankin Dec 6 '18 at 8:52
• I get it, so to be equivalent, I'd have to write my action as(deferring to your notation): $$S=\intop A^{\mu}d^{\dagger}dA_{\mu}\sqrt{|g|}d^{4}x$$ Which is like the "left side" of the laplacian. how to separate that left and right "handed" parts. I need to read up. – R. Rankin Dec 6 '18 at 9:06
• Yes, that form is correct but slightly different from your original Lagrangian since $dA$ is antisymmetrized. It differs by that term with the Ricci tensor J.G. mentioned, which vanishes in flat space of course. – octonion Dec 6 '18 at 9:13

Your alternative lagrangian is not gauge invariant, since if you shift $$A_\mu$$ by the gradient of a scalar function, the expression clearly changes; $$F_{\mu\nu}$$ is antisymmetric in the indices $$\mu \nu$$ so basically any shift in $$A$$ gets cancelled. I think it's just an accident of a gauge choice that you've gotten the right EOM.

• I'm super new to differential forms, but I think it's gauge invariant, but you have to use the self-adjointness of the Laplace-Beltrami operator. Honestly I'm having trouble showing it, but I'm pretty new to differential forms and exterior calculus. – R. Rankin Dec 6 '18 at 6:22
• added an attempt to show gauge invariance emphasis on attempt. – R. Rankin Dec 6 '18 at 7:25
• So I think it's equal to $F^2$ in a certain gauge. But if you change gauge, then it isn't. That's what I mean by saying it's not gauge invariant – user215119 Dec 6 '18 at 18:28
• Of course you are correct!! It was very late when I was thinking about that – R. Rankin Dec 6 '18 at 22:26

Your choice of Lagrangian density is equivalent, up to a total derivative, to $$-(\nabla_\alpha A_\mu)^2$$, while $$F^2_{\alpha\mu}=2\nabla_\alpha A_\mu F^{\alpha\mu}$$, so a $$\nabla_\alpha A_\mu\nabla^\mu A^\alpha$$ term makes them different. Even in your gauge choice, this isn't a total derivative because $$A^\mu\nabla_\alpha\nabla_\mu A^\alpha=A^\mu R_{\mu\beta} A^\beta$$. However, very different-looking Lagrangians can obtain the same EOMs.