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I had to find the canonical energy-momentum tensor defined by this Lagrangian density $ \mathcal{L} = - {1 \over 4} F_{\mu \nu} F^{\mu \nu}$ and I got the result of $ T^{\mu \nu} = - F^{\mu \lambda} \partial^{\nu} A_{\lambda} + {1 \over 4} \eta^{\mu \nu} F^{\rho \sigma} F_{\rho \sigma}$ In order to make it a symmetric tensor, I had to add total derivative term and knowing from equation of motion that $ \partial_{\lambda} F^{\mu \lambda}=0$, I got new tensor $$ \hat{T}^{\mu \nu} = F^{\mu \lambda}F^{\nu}_{\lambda} + {1 \over 4} \eta^{\mu \nu} F^{\rho \sigma} F_{\rho \sigma}$$ which is symmetric. The question is - what is the trace of $\hat{T}^{\mu \nu}$?

I know that, as far as it comes to the stress-energy tensor $T^{\mu \nu}$, we can write the trace as $T^{\mu}_{\mu}$ and using the formula for the energy-momentum tensor, do we get sth like that $T^{\mu}_{\mu} = {{\partial \mathcal{L}} \over {\partial (\partial_{\mu} A_{\lambda})}} \partial_{\nu} A_{\lambda} - \delta^{\mu}_{\nu} \mathcal{L} $? Honestly, I can't see what would the result for $\hat{T}^{\mu \nu}$ be. Would it be that the canonical stress-energy tensor is traceless? But how actually prove it?

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    $\begingroup$ $T^{\mu}_{\mu} = {{\partial \mathcal{L}} \over {\partial (\partial_{\mu} A_{\lambda})}} \partial_{\nu} A_{\lambda} - \delta^{\mu}_{\nu} \mathcal{L} $ This equation doesn’t make sense because the indices are inconsistent. You have no free (i.e., uncontracted) indices on the left, but two free indices ($\mu$ and $\nu$) on the right. $\endgroup$
    – G. Smith
    Dec 20, 2020 at 18:18

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You can find the trance of $T^{\mu \nu}$ like this,

$T^\mu_{\;\; \mu} = \eta_{\mu \nu} T^{\mu \nu} = \eta_ {\mu \nu} \Big(F^{\mu \rho}F^{\nu}_{\rho} + \frac{1}{4} \eta^{\mu \nu} F^{\alpha \beta}F_{\alpha \beta} \Big)$

I leave the rest of the calculation to yourself.

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  • $\begingroup$ I actually meant the trace of this improved tensor, namely $\hat{T}^{\mu \nu}$, not this $T^{\mu \nu}$. Perhaps it wasn't clear. $\endgroup$
    – Chakalaka
    Dec 20, 2020 at 21:05
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    $\begingroup$ @Chakalaka Please observe that Kian did write the expression for your $\hat T^{\mu\nu}$, they simply did not include the hats. Besides this, the definition of the trace of a tensor is independent of the tensor's definition. $\endgroup$ Dec 21, 2020 at 1:46

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