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Are the Maxwell equations written with the derivatives of the EM field strength tensor not generally covariant? I can't seem to prove that is.

The Maxwell equations in 4-tensor form:

$\partial_{\mu}F_{\alpha\beta}+\partial_{\alpha}F_{\beta\mu}+\partial_{\beta}F_{\mu\alpha}=0\tag{1}$

Transforms like this:

$J^{\theta}_{\mu}\partial_{\theta}\left(J^{\sigma}_{\alpha}J^{\gamma}_{\beta} F_{\sigma\gamma}\right)+J^{\theta}_{\alpha}\partial_{\theta}\left(J^{\sigma}_{\beta}J^{\gamma}_{\mu} F_{\sigma\gamma}\right)+J^{\theta}_{\beta}\partial_{\theta}\left(J^{\sigma}_{\mu}J^{\gamma}_{\alpha} F_{\sigma\gamma}\right)=0\tag{2}$

With the product rule this can be broken up into 9 terms, 6 of which have to cancel out in order for LHS of $(1)$ to transform like a tensor.

These six should then give:

$J^{\theta}_{\mu}J^{\sigma}_{\alpha}F_{\sigma\gamma}J^{\gamma}_{\theta\beta}+J^{\theta}_{\mu}J^{\sigma}_{\beta}F_{\sigma\gamma}J^{\sigma}_{\theta\alpha}+J^{\theta}_{\alpha}J^{\sigma}_{\beta}F_{\sigma\gamma}J^{\gamma}_{\theta\mu}+J^{\theta}_{\alpha}J^{\gamma}_{\mu}F_{\sigma\gamma}J^{\sigma}_{\theta\beta}+J^{\theta}_{\beta}J^{\sigma}_{\mu}F_{\sigma\gamma}J^{\gamma}_{\theta\alpha}+J^{\theta}_{\beta}J^{\gamma}_{\alpha}F_{\sigma\gamma}J^{\sigma}_{\theta\mu}=0\tag{3}$

These we should group in three pairs that have to cancel, one such pair should give:

$J^{\theta}_{\mu}J^{\sigma}_{\alpha}F_{\sigma\gamma}J^{\gamma}_{\theta\beta}+J^{\theta}_{\alpha}J^{\gamma}_{\mu}F_{\sigma\gamma}J^{\sigma}_{\theta\beta}=0\tag{4}$

But switching the indices of the second, antisymmetric, EM field strength tensor:

$J^{\theta}_{\mu}J^{\sigma}_{\alpha}F_{\sigma\gamma}J^{\gamma}_{\theta\beta}-J^{\theta}_{\alpha}J^{\gamma}_{\mu}F_{\gamma\sigma}J^{\sigma}_{\theta\beta}=0\tag{5}$

Renaming $\gamma$ as $\sigma$ and vice versa in the second term:

$J^{\theta}_{\mu}J^{\sigma}_{\alpha}F_{\sigma\gamma}J^{\gamma}_{\theta\beta}-J^{\theta}_{\alpha}J^{\sigma}_{\mu}F_{\sigma\gamma}J^{\gamma}_{\theta\beta}=0\tag{6}$

We seem to be stuck...

Am I missing something?

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  • $\begingroup$ You could rewrite (1) using covariant derivative and antisymetry property of $F_{\mu\nu}$. In such form general coordinate invariance is obvious. $\endgroup$ – Nikita Oct 8 '20 at 0:52
  • $\begingroup$ Yeah it would definitely be covariant with the covariant derivative, but I was expecting this to work out. maybe that just wasn't justified though.... $\endgroup$ – Stijn Boshoven Oct 8 '20 at 1:01
  • $\begingroup$ What is the definition of $J^a_b$? More precisely: how exactly is $J^a_b$ related to the new and old coordinate systems (which are not clearly distinguished from each other by the $\partial$ notation in equations (1) and (2))? $\endgroup$ – Chiral Anomaly Oct 8 '20 at 12:06
  • $\begingroup$ They're the elements of the Jacobian, they are all from the same Jacobian, there's no mixing of terms from the Jacobian and inverse Jacobian because all the terms $(1)$ are covariant. $\endgroup$ – Stijn Boshoven Oct 8 '20 at 13:41
  • $\begingroup$ You need also use antisymetry property of equation 1 $\endgroup$ – Nikita Oct 10 '20 at 20:58
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Hint:

$$ F_{\mu\nu} =2 D_{[\mu} A_{\nu]} = 2 \partial_{[\mu} A_{\nu]} $$

After Lorentz transformation we will have:

$$ F^\prime_{\mu\nu} = J_\mu^{\;\rho}J_\nu^{\;\rho} F_{\rho\sigma} + J_{[\mu}^{\;\rho} \partial_\rho J_{\nu]}^{\;\sigma} A_{\sigma} $$

Why

$$ J_{[\mu}^{\;\rho} \partial_\rho J_{\nu]}^{\;\sigma} =0? $$

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