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I have canonical electromagnetic stress-energy-momentum tensor defined as:

$T_{\mu\nu}=\frac{1}{4}\eta^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}-F^{\mu\lambda}F^{\nu}_{\,\,\lambda}-F^{\mu\lambda}\partial_{\lambda}A^{\nu}\qquad (*)$

and I need to show that it is not symmetric.

I tried to calculate:

$T_{\nu\mu}=\frac{1}{4}\eta^{\nu\mu}F_{\alpha\beta}F^{\alpha\beta}-F^{\nu\lambda}F^{\mu}_{\,\,\lambda}-F^{\nu\lambda}\partial_{\lambda}A^{\mu}$ $T_{\nu\mu}=\frac{1}{4}\eta^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}-[(-)F^{\nu}_{\,\,\lambda}(-)F^{\mu\lambda}]-F^{\nu\lambda}\partial_{\lambda}A^{\mu}$ $T_{\nu\mu}=\frac{1}{4}\eta^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}-F^{\mu\lambda}F^{\nu}_{\,\,\lambda}-F^{\nu\lambda}\partial_{\lambda}A^{\mu}$.

The first two parts of last equation looks like in original one $(*)$. So antisymmetry has to be in last term (of course if I didn't make any stupid mistake). I do not know, what to do next, how to end proof.

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I do not know, what to do next, how to end proof.

Assuming that the algebra you presented is correct, I think you are basically done.

You have shown that $$ T^{\mu\nu} - T^{\nu\mu} = F^{\nu\lambda}\partial_{\lambda}A^{\mu} - F^{\mu\lambda}\partial_{\lambda}A^{\nu}\;. $$

As long as the difference $$ T^{\mu\nu} - T^{\nu\mu}\equiv \Delta^{\mu\nu}=F^{\nu\lambda}\partial_{\lambda}A^{\mu} - F^{\mu\lambda}\partial_{\lambda}A^{\nu} $$ is not exactly zero for all $\mu$ and $\nu$ then you have done your job.

You could work out the above $\Delta$ in terms of $A$ to show that the difference is explicitly non-zero, say in the $A^0=0$ gauge, say for $\mu=0$ and $\nu=i$: $$ \Delta^{0i} = \vec E\cdot \vec\nabla A^i \neq 0 $$

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    $\begingroup$ Thank you so much! I got it! $\endgroup$
    – Lilla_mu
    Commented Jan 8 at 7:22