# Electromagnetism Bianchi Identity

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.

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 :

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

• Yes! This is more along the lines of what I wanted to ask, but how do you get the first equation here (the "from the Bianchi Identity")? I understand that it's from the Bianchi identity, but how can we say it's true for a general antisymmetric tensor? Apr 12 '20 at 7:49
• @user132849 A simple way to see this is to check that the Bianchi Identity is symmetric in all its indices i.e. totally symmetric whereas the Levi Civita symbol is totally antisymmetric in its indices and the product of a symmetric quantity with antisymmetric one is always zero as far as I have seen. Apr 12 '20 at 8:48
• @user132849 A more mathematical way would be to say that we have written $x_i =0$ for all i in this form : $A x_{i1} + B x_{i2} + ... = 0$ with A,B etc being constant coefficients corresponding to $\epsilon$ values (notice that the presence of $\epsilon$ there implies that the indices are summed and you have more than three terms there). Apr 12 '20 at 8:57

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.

• I'm only vaguely familiar with the exterior derivative. Would you mind showing how it applies here to $F_{\mu \nu}$? Edit: nevermind, I get it now. Thanks! Apr 12 '20 at 8:58
• Well, actually, just to make sure, when you day that it only applies as F is the form of an exterior derivative, is this the same as saying that F contains partial derivatives with switched indices in this case? $F_{\mu \nu} = \partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}$? Apr 12 '20 at 9:04
• @user132849 I have made edits. Hope this is more helpful. Apr 12 '20 at 9:17
• @user132849 glad to help. I just added another couple more new lines, this might make it clearer. Apr 12 '20 at 9:21
• Okay thanks for the references by the way! Apr 12 '20 at 9:26

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

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.

• This is not quite what I was asking. But thanks for the answer though! Apr 12 '20 at 7:50