I have a question in reaction to an article by M. Born and L. Infeld (cf. [1]) concerning the relation between the hodge dual of the electromagnetic tensor and the antisymmetrization of its derivative.
Basically, they state the following equivalence: $$ \partial_{[\nu} f_{\rho\sigma]} = 0 \iff \partial_{\nu} (\sqrt{-g} \star f^{\mu\nu}) = 0 $$ where $f_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$, $\star f^{\mu\nu} = \frac{1}{2\sqrt{-g}} \epsilon^{\rho\sigma\mu\nu} f_{\rho\sigma}$ and $g$ is the determinant of the metric (on a 4-dimensional manifold with signature $(+,-,-,-)$). Has anybody any idea where this equivalence comes from? I have come to the following identities: $$ \partial_\nu (\sqrt{-g} \star f^{\mu\nu}) = \frac{1}{2} \epsilon^{\mu\nu\rho\sigma} \partial_\nu f_{\rho\sigma},\\ \partial_{[i_1} f_{i_2 i_3]} = \frac{1}{3!} \epsilon^{j_1 j_2 j_3}_{i_1 i_2 i_3} \partial_{j_1} f_{j_2 j_3}, $$ but I don't see how the $\mu$ in the first expression drops out.
As a side note, in the article it is frequently assumed that $\partial_\mu \sqrt{-g} = 0$. Was it a convention that partial derivatives are to be considered covariant?
Cheers, Eric
The expression: $$ \partial_\nu (\sqrt{-g} \star f^{\mu\nu}) = \frac{1}{2} \epsilon^{\mu\nu\rho\sigma} \partial_\nu f_{\rho\sigma}, $$
that you found, is exactly what you are trying to prove. The non-zero components of that expressions are: $$ \frac{1}{2} \epsilon^{\mu\nu\rho\sigma} \partial_{[\nu} f_{\rho\sigma]}, $$
where $\mu$ is a free index. The expression is zero for every $\mu$ if and only if: $$ \partial_{[\nu} f_{\rho\sigma]}=0. $$