By variating the Maxwell Lagrangian we get the equation of motion. The remaining two Maxwell equations can be written as $$\epsilon_{\mu\nu\rho\sigma}\partial^{\rho} F^{\mu\nu} = 0.$$ I have also seen it written as the Bianchi identity: $$\partial_{[\lambda}F_{\mu\nu]} = 0.$$ Why are these two forms equivalent?


closed as off-topic by user108787, user36790, ACuriousMind, Jon Custer, Gert Sep 21 '16 at 2:15

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  • $\begingroup$ Please see edit. It's important to use understandable punctuation etc. so that others can understand the question. $\endgroup$ – DanielSank Sep 17 '16 at 14:06
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    $\begingroup$ Have you tried plugging in the components of $F_{\mu \nu}$? Both forms are equivalent to the same two Maxwell equations, thus they are obviously mutually equivalent. $\endgroup$ – Prof. Legolasov Sep 17 '16 at 14:35
  • $\begingroup$ VTC due to insufficient research, sorry. $\endgroup$ – user108787 Sep 17 '16 at 15:11
  • $\begingroup$ I know they are equivalent by plugging the components of $F_{\mu\nu}$, But can we prove it by pure mathematicas only with $F_{\mu\nu}$ is antisymmetric. $\endgroup$ – Xian-Hui Sep 18 '16 at 10:20
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    $\begingroup$ In both expressions everything is explicitly antisymmetrized, so I think you don't need antisymmetry of F. $\endgroup$ – Simon Sep 18 '16 at 23:59

It's basically just a duality relation analogous to the cross product in three dimensions. But if you want to do some work to show the equivalence, then:

Going from the second equation to the first is easy, just hit it with $\epsilon_{\mu\nu\rho\sigma}$.

Going from the first to the second equation, is a little trickier and relies on knowing how to evaluate the products of Levi-Civita symbols. The basic idea is that you should contact the first equation with $\epsilon^{\mu'\nu'\lambda'\sigma}$ and compare the resulting antisymmetric combination of $\delta$s with the antisymmetrization of the indices in the second equation.


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