# General covariance of the Maxwell equations in 4-tensor form

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?

• You could rewrite (1) using covariant derivative and antisymetry property of $F_{\mu\nu}$. In such form general coordinate invariance is obvious. Oct 8, 2020 at 0:52
• 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.... Oct 8, 2020 at 1:01
• 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))? Oct 8, 2020 at 12:06
• 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. Oct 8, 2020 at 13:41
• You need also use antisymetry property of equation 1 Oct 10, 2020 at 20:58

$$F_{\mu\nu} =2 D_{[\mu} A_{\nu]} = 2 \partial_{[\mu} A_{\nu]}$$
$$F^\prime_{\mu\nu} = J_\mu^{\;\rho}J_\nu^{\;\rho} F_{\rho\sigma} + J_{[\mu}^{\;\rho} \partial_\rho J_{\nu]}^{\;\sigma} A_{\sigma}$$
$$J_{[\mu}^{\;\rho} \partial_\rho J_{\nu]}^{\;\sigma} =0?$$