# Why does $\epsilon_{\mu\nu\rho\sigma}\partial^{\rho}\partial^\mu A^\nu-\epsilon_{\mu\nu\rho\sigma}\partial^{\rho}\partial^\nu A^\mu=0$?

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?

• 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/… Oct 20, 2020 at 11:30
• A different title for this question could be "why $\epsilon_{\mu\nu\rho\sigma}\partial^{\rho}\partial^\mu A^\nu-\epsilon_{\mu\nu\rho\sigma}\partial^{\rho}\partial^\nu A^\mu=0$?" Oct 20, 2020 at 11:36
• 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) Oct 20, 2020 at 11:43

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.