Electromagnetism Bianchi Identity

Question

I am currently studying a book about the Lorentz covariant formulation of electromagnetism. At the point I am at in the book, the author has just introduced the covariant form of Maxwell's equations (with (+---)):

$$ \partial_\beta F^{\alpha \beta} = -\frac{1}{c} j^\alpha $$ and $$ \partial_\rho F_{\nu \sigma} + \partial_\nu F_{\sigma \rho} + \partial_\sigma F_{\rho \nu} = 0 $$

My question is about this second equation. I understand where this comes from in regards to the electrodynamics of the problem, but when introducing it in the text, the author states the following:

"...for any antisymmetric tensor $F_{\mu \nu}$ satisfies the identity: $$ \epsilon^{\mu \nu \sigma \rho} \partial_\rho F_{\nu \sigma} = 0$$ "

Where $\epsilon$ is the Levi-Civita symbol here. I've had a little difficulty with this statement because I've seen other people appeal to the same argument regarding the general properties of antisymmetric tensors, but I am having a difficult time proving it myself. I can understand if we have an $F$ of the form: $$F_{\mu \nu} = \partial_{\mu}A_{\nu} - \partial_{\nu} A_{\mu} $$ That's a rather trivial proof, but it seems that the author (and others that I've seen) appeal to this as a general property of antisymmetric tensors. So, if anyone would want to show me how to prove why this would be a general property of antisymmetric tensors, I would be very grateful.

Answer 1

It's not true for a general antisymmetric tensor. The equation you write is a bianchi identity, $$d(dA)=0$$. This is true only because $F$ is of the form $dA$(an exterior derivative). It didn't have to be, ofcourse.

As an analogy, the Reimann tensor satisfies a Bianchi identity too. It is certainly a special tensor-not any good old tensor will describe curvature.

EDIT: In response to comments. Roughly, an exterior derivative is a map between differential forms-it maps a $k$-form to a $k+1$ form. It's an extension of the notion of successive differentiation. For smooth functions($0$-forms) $f$, it is the regular derivative.

These have the defining property(for the wedge product $\wedge$, an extension of cross products)-$$d(x\wedge y)=dx\wedge y+(-1)^p( y\wedge dx)$$, for a $p$-form $x$. In this case, $x$ corresponds to $\partial_\mu$, which we know is a $1$-form('dual' vector/covector), so $p=1$ and it's easy to see how the definition of the Maxwell tensor is an exterior derivative. For such derivatives, $d^2=0$ holds as an operator identity-the Bianchi identity.

In a coordinate basis, it can be shown that for a $1$-form $A$, the components of the exterior derivative $dA$(which is a two form-note that $A_\mu$ and $F_{\mu\nu}$ are one forms and two forms respectively) are-$$(dA)_{ij}=\partial_iA_j-\partial_jA_i$$.

Wikipedia is a good reference. MTW-Gravity has a dedicated chapter to this, if you're interested.

Answer 2

If I understand it correctly, you want to prove that $ \epsilon^{\mu\nu\sigma\rho} \partial_\rho F_{\nu\sigma} = 0 $ for a general anti-symmetric $F_{\nu\sigma}$.

From Bianchi Identity we have : $$ \epsilon^{\mu\nu\sigma\rho} (\partial_\rho F_{\nu \sigma} + \partial_\nu F_{\sigma \rho} + \partial_\sigma F_{\rho \nu}) = 0 \\ \epsilon^{\mu\nu\sigma\rho}\partial_\rho F_{\nu \sigma} + \epsilon^{\mu\nu\sigma\rho}\partial_\nu F_{\sigma \rho} + \epsilon^{\mu\nu\sigma\rho}\partial_\sigma F_{\rho \nu} = 0 \\ $$

Now realise that in the all the terms the indices are contracted (hence can be replaced by other indices) and write all the terms such that the indices on $\partial$ and F are same in each term so that the $\epsilon$ indices get modified :

$$ \epsilon^{\mu\nu\sigma\rho}\partial_\rho F_{\nu \sigma} + \epsilon^{\mu\rho\nu\sigma}\partial_\rho F_{\nu \sigma} + \epsilon^{\mu\sigma\rho\nu}\partial_\rho F_{\nu \sigma} = 0 \\ (\epsilon^{\mu\nu\sigma\rho}+\epsilon^{\mu\rho\nu\sigma}+\epsilon^{\mu\sigma\rho\nu}) \partial_\rho F_{\nu \sigma} = 0 $$

Now, use the fact that $\epsilon$ is fully anti-symmetric in its indices so that :

$$ \epsilon^{\mu\nu\sigma\rho} = \epsilon^{\mu\rho\nu\sigma} =\epsilon^{\mu\sigma\rho\nu} $$

So, we get :

$$ \epsilon^{\mu\sigma\rho\nu}\partial_\rho F_{\nu \sigma} =0 $$

Also, again exchanging indices we get :

$$ \epsilon^{\mu\nu\sigma\rho}\partial_\rho F_{\nu \sigma} =0 $$

Answer 3

$$F_{\mu \nu} = \partial_{\mu}A_{\nu} - \partial_{\nu} A_{\mu} $$

is the definition of the Faraday tensor from the electromagnetic potential $A(x)$. A proof would only mean a proof of the consistency of the definition, meaning that what is actually required is to prove the existence of the potential, $A$, from Maxwell's equations.