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## Homework Statement

Verify that ##\partial_{\mu}F_{\nu\lambda}+\partial_{\nu}F_{\lambda\mu}+\partial_{\lambda}F_{\mu\nu}## is equivalent to ##\partial_{[\mu}F_{\nu\lambda]}=0##,

and that they are both equivalent to ##\tilde{\epsilon}^{ijk}\partial_{j}E_{k}+\partial_{0}B^{i}=0## and ##\partial_{i}B^{i}=0##.

## Homework Equations

## The Attempt at a Solution

##\partial_{[\mu}F_{\nu\lambda]}=0##

##\implies \frac{1}{3!}(\partial_{\mu}F_{\nu\lambda}-\partial_{\mu}F_{\lambda\nu}-\partial_{\nu}F_{\mu\lambda}+\partial_{\nu}F_{\lambda\mu}-\partial_{\lambda}F_{\nu\mu}+\partial_{\lambda}F_{\mu\nu})=0##, using the definition of antisymmetrization of a tensor

##\implies \partial_{\mu}F_{\nu\lambda}-\partial_{\mu}F_{\lambda\nu}-\partial_{\nu}F_{\mu\lambda}+\partial_{\nu}F_{\lambda\mu}-\partial_{\lambda}F_{\nu\mu}+\partial_{\lambda}F_{\mu\nu}=0##

##\implies \partial_{\mu}F_{\nu\lambda}+\partial_{\mu}F_{\nu\lambda}+\partial_{\nu}F_{\lambda\mu}+\partial_{\nu}F_{\lambda\mu}+\partial_{\lambda}F_{\mu\nu}+\partial_{\lambda}F_{\mu\nu}=0##

##\implies 2(\partial_{\mu}F_{\nu\lambda}+\partial_{\nu}F_{\lambda\mu}+\partial_{\lambda}F_{\mu\nu})=0##

##\implies \partial_{\mu}F_{\nu\lambda}+\partial_{\nu}F_{\lambda\mu}+\partial_{\lambda}F_{\mu\nu}=0##

Is my working so far correct?