The tensor $F^{\mu\nu}$ is defined as $\partial^\mu A^\nu-\partial^\nu A^\mu$. Why is the equation $$\epsilon_{\mu\nu\rho\sigma}\partial^{\rho} F^{\mu\nu} = 0$$ identically satisfied by $F^{\mu\nu}=\partial^\mu A^\nu-\partial^\nu A^\mu$?

We have $$\epsilon_{\mu\nu\rho\sigma}\partial^{\rho} (\partial^\mu A^\nu-\partial^\nu A^\mu)=\epsilon_{\mu\nu\rho\sigma}\partial^{\rho}\partial^\mu A^\nu-\epsilon_{\mu\nu\rho\sigma}\partial^{\rho}\partial^\nu A^\mu$$ I am told that since $\epsilon$ is antisymmetric and $\partial\partial$ is symmetric (no doubts about it), the product $(\text{antisymmetric})(\text{symmetric})=0$. Here is my attempt at understanding this last statement:

$$\epsilon_{\mu\nu\rho\sigma}\partial^{\rho}\partial^\mu A^\nu=\epsilon_{\mu\nu\rho\sigma}\partial^{\mu}\partial^\rho A^\nu=-\epsilon_{\rho\nu\mu\sigma}\partial^{\mu}\partial^\rho A^\nu=-\epsilon_{\mu\nu\rho\sigma}\partial^{\rho}\partial^\mu A^\nu$$

  1. Step 1: symmetry of $\partial^\rho\partial^\mu$
  2. Step 2: antisymmetry of $\epsilon$
  3. Step 3: I call $\mu$ $\rho$ and viceversa, since they are to be summed over

Then I got $\epsilon_{\mu\nu\rho\sigma}\partial^{\rho}\partial^\mu A^\nu=-\epsilon_{\mu\nu\rho\sigma}\partial^{\rho}\partial^\mu A^\nu=0$. Are these steps right?

  • $\begingroup$ The title question is a little bit obscure for me. The sourceless Maxwell equation is an applied form of the Bianchi identity. See physics.stackexchange.com/questions/296164/… $\endgroup$ Oct 20, 2020 at 11:30
  • $\begingroup$ A different title for this question could be "why $\epsilon_{\mu\nu\rho\sigma}\partial^{\rho}\partial^\mu A^\nu-\epsilon_{\mu\nu\rho\sigma}\partial^{\rho}\partial^\nu A^\mu=0$?" $\endgroup$
    – Lorenzo B.
    Oct 20, 2020 at 11:36
  • $\begingroup$ Nice. Change it :) Did the link answer your question? (and also for the queston : it is simply because of the negative sign caused by tbe index-interchange in antisymmetric tensor. So you're right) $\endgroup$ Oct 20, 2020 at 11:43

1 Answer 1


You got it a little bit wrong, but the main ideas are here. Starting from $\epsilon_{\mu \nu\rho\sigma}\partial^\rho\partial^\mu A^\nu$, you commute $\partial^\rho$ with $\partial^\mu$ without changing anything. Then, you use anti symmetry of $\epsilon$ to exchange the two indices $\mu$ and $\rho$. At this point: $$\epsilon_{\mu \nu\rho\sigma}\partial^\rho\partial^\mu A^\nu=-\epsilon_{\rho \nu\mu\sigma}\partial^\mu\partial^\rho A^\nu$$ And since $\mu$ and $\rho$ are dummy indices, you can exchange them in the right hand side:$$\epsilon_{\mu \nu\rho\sigma}\partial^\rho\partial^\mu A^\nu=-\epsilon_{\mu\nu\rho\sigma}\partial^\rho\partial^\mu A^\nu$$ Since that thing is equal to its opposite, it should be zero indeed.

  • $\begingroup$ isn't that what I did? $\endgroup$
    – Lorenzo B.
    Oct 20, 2020 at 11:31
  • $\begingroup$ I had made some typos on the indices, now I think it is equal to your calculation $\endgroup$
    – Lorenzo B.
    Oct 20, 2020 at 11:34
  • $\begingroup$ Yep, without the typos on the indices our two calculations match :) $\endgroup$ Oct 20, 2020 at 11:36
  • $\begingroup$ perfect! thank you $\endgroup$
    – Lorenzo B.
    Oct 20, 2020 at 11:39
  • $\begingroup$ You are welcome $\endgroup$ Oct 20, 2020 at 19:33

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