I am working on the dual Lagrangian as given by $$\mathcal{L}_D=F_{\mu\nu}\tilde{F}^{\mu\nu}$$ In literature, this term is often written as $$\boxed{\mathcal{L}_D=2\partial^\mu(\varepsilon_{\mu\nu\rho\sigma}A^\nu\partial^\rho A^\sigma)}$$ I - with my limited knowledge on field theory - would like to reproduce this result. So far, I've computed $$\mathcal{L}_D=\frac{1}{2}(\varepsilon_{\mu\nu\rho\sigma}\partial^\mu A^\nu\partial^\rho A^\sigma-\varepsilon_{\mu\nu\rho\sigma}\partial^\mu A^\nu \partial^\sigma A^\rho - \varepsilon_{\mu\nu\rho\sigma} \partial^\nu A^\mu \partial^\rho A^\sigma + \varepsilon_{\mu\nu\rho\sigma}\partial^\nu A^\mu \partial^\sigma A^\rho)$$ By carefully evaluating a few combinations of the Levi-Civita symbol, I was able to deduce that $$\mathcal{L}_D=2\varepsilon_{\mu\nu\rho\sigma}(\partial^\mu A^\nu)(\partial^\rho A^\sigma)$$ To go from here to the boxed term, my reference states "One can write out the product rule for differentiation. All terms other than the boxed one are zero."

Here I do not really see how the product rule should be invoked and why terms drop out. Naively, I would say $$\mathcal{L}_D=2\varepsilon_{\mu\nu\rho\sigma}\left[\partial^\mu A^\nu \partial^\rho A^\sigma\right]=2\varepsilon_{\mu\nu\rho\sigma}\left[\partial^\mu(A^\nu)\partial^\rho A^\sigma + A^\nu \partial^\mu(\partial^\rho A^\sigma) \right]$$

  • $\begingroup$ The partial derivatives in the last term commute, so you get two identical terms by swapping their indices. But the Levi-Civita tensor gives a minus sign for swapping indices, so you get pairs of identical terms with opposite signs. $\endgroup$
    – ragnar
    Commented Jun 24, 2019 at 16:25
  • $\begingroup$ That makes perfect sense, thank you! I was puzzled by the boxed equation as it still looks like something that needs to be expanded, but the extra term there will drop out as well then $\endgroup$
    – DaanMusic
    Commented Jun 24, 2019 at 16:27

1 Answer 1


The question has been answered by ragnar yesterday: $\epsilon^{\mu\nu\rho\sigma}(\partial_\mu\partial_\nu(\cdots))=0$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.