# Derivation of Covariant Maxwell's Equations

I am trying to derive the covariant formulation of Maxwell's equations.

I understand that all four of Maxwell's equations can be written compactly as

$$\partial_{\mu}F^{\mu\nu} - j^{\mu} = 0 \;, \tag{1}$$

and

$$\partial_{[\mu}F_{\alpha\beta]} = 0\;. \tag{2}$$

However, the second of these equations is often expressed in terms of the dual $$\tilde{F}^{\mu\nu}$$ as $$\partial_{\mu}\tilde{F}^{\mu\nu} = 0\;.$$ I am trying to derive eqn. (2) from this one. Until now, I have

$$\partial_{\mu}\tilde{F}^{\mu\nu} = \epsilon^{\mu\nu\alpha\beta}\partial_{\mu}F_{\alpha\beta}.$$ (where $$\epsilon^{\mu\nu\alpha\beta}$$ is the 4-dimensional Levi-Civita symbol)

I don't know how to prove that the RHS is the same as eqn.(2). Any help is greatly appreciated!

Since the Levi-Civita symbol is antisymmetric, contraction with it extracts the antisymmetric part, $$\epsilon^{\mu\nu\alpha\beta} \partial_\mu F_{\alpha \beta} = \epsilon^{\mu\nu\alpha\beta} \partial_{[\mu} F_{\alpha \beta]} = 0.$$ However, the Levi-Civita symbol is also invertible, i.e. we can remove it by contracting with another, $$\epsilon_{\mu' \nu \alpha' \beta'} \epsilon^{\mu\nu\alpha\beta} \partial_{[\mu} F_{\alpha \beta]} = 3! \, \partial_{[\mu'} F_{\alpha' \beta']}.$$ We know the left-hand side is zero, so the right-hand side is zero, recovering the result.
• @Sankarshana Fix a pair of indices, say $T_{ab \ldots}$. You can always write the tensor as $T_{ab \ldots} = T_{(ab) \ldots} + T_{[ab] \ldots}$. Then it's easy to see contracting with the Levi-Civita kills the first term, because it is antisymmetric in any two indices. Since this holds for every pair of indices, the only surviving part is antisymmetric in all indices. – knzhou Dec 19 '18 at 15:17