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Versions of this question have been asked on this site before but have not directly addressed by concern. In the $(+---)$ convention the EM lagrangian in the presence of charge sources is

$$ \mathcal{L} = -\frac{\sqrt{-g}}{4} F^{\mu \nu} F_{\mu \nu} + \sqrt{-g} j^\mu A_\mu. $$

Using the equation $$ T_{\mu \nu} = \frac{2}{\sqrt{-g}} \frac{\partial \mathcal{L} }{\partial g^{\mu \nu} } $$

we get $$ T_{\mu \nu} = - F_{\mu}^{\: \, \alpha} F_{\nu \alpha} + g_{\mu \nu} \frac{1}{4} F^{\alpha \beta} F_{\alpha \beta} + g_{\mu \nu} j^\alpha A_\alpha $$

The first two terms correspond to the stress-energy tensor of a free $(j = 0)$ EM field. The third term, however, is not gauge invariant. Isn't this a problem? The right hand side of $G_{\mu \nu} = 8 \pi G T_{\mu \nu}$ should not depend on gauge. Isn't there an ambiguity when you have charged matter in a gravitational field?

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    $\begingroup$ "The third term, however, is not gauge invariant. Isn't this a problem?" Yes, it is a problem. You skipped the matter theory this term makes gauge invariant through its gauge variation. This matter term needs the third term just as the third term needs it. $\endgroup$ – Cosmas Zachos Feb 9 at 17:49
  • $\begingroup$ What if our matter theory is just a bunch of point charges, with the matter action just being the proper time of the particles? It seems there is no $A_\mu$ in that part of the action so there is no other term that could possibly compensate to make the total $T_{\mu \nu}$ gauge invariant. $\endgroup$ – user1379857 Feb 9 at 17:53
  • $\begingroup$ The derivatives on the part of the charges contribute to the gauge variation. The third term is effectively their gauge covariant completion. $\endgroup$ – Cosmas Zachos Feb 9 at 18:39
  • $\begingroup$ Which term are you referring to for the derivatives on the part of the charges? I'm having a difficult time understanding which term. Are you also referring to varying with respect to $g^{\mu \nu}$ (to get $T_{\mu \nu}$) or $A_\mu$? $\endgroup$ – user1379857 Feb 9 at 18:47
  • $\begingroup$ The missing term of evolving matter: the one you skipped. It too contributes to the stress energy tensor. Fix gauge invariance term and bother with metrics and geometry later. $\endgroup$ – Cosmas Zachos Feb 9 at 19:12
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Our charge coupling term is of the form

\begin{equation} S = \int_U j^\alpha A_\alpha \end{equation}

Now take the gauge transformation. We have

\begin{equation} A_\alpha \to A_\alpha + \partial_\alpha f \end{equation}

giving us the difference

\begin{equation} \delta S = \int_U (j^\alpha \partial_\alpha f) \end{equation}

As a Noether current, the electric current obeys the relation

\begin{equation} \partial_\alpha j^\alpha = 0 \end{equation}

Therefore, we can integrate by parts to obtain

\begin{eqnarray} \delta S &=& \int_{\partial U} f n_\alpha j^\alpha - \int_U ((\partial _\alpha j^\alpha) \alpha f)\\ &=& \int_{\partial U} f n_\alpha j^\alpha \end{eqnarray}

with $n$ the normal vector to the boundary of integration. The action then only differs by a surface term. In general, we don't consider boundary terms for the EM action (we just pick $M$ as the volume of integration which has no boundary), but things will get more complicated if we allow our volume have a boundary.

But in terms of the stress-energy tensor, which is a local quantity for which we can always pick the appropriate volume around it, this isn't an issue. The difference from the gauge vanishes and therefore its variation does as well.

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Okay, I understand the answer. I was really thinking about this in terms of point particles coupled to the gauge field. So the interaction term is

$$ \mathcal{L}_{\rm int} = j^\mu A_\mu = q \int d\lambda \frac{dx^\mu}{d \lambda}A_\mu (x(\lambda)). $$ where $\lambda$ parameterizes the path.

Note that there is actually no metric dependence in this term. Therefore $$ \frac{\partial \mathcal{L}_{\rm int}}{\partial g^{\mu \nu}} = 0 $$ and there is therefore no contribution from this term to $T_{\mu \nu}$. What confused me was that, in my original question, when I wrote $$ \mathcal{L}_{\rm int} = \sqrt{-g} J^\mu A_\mu $$ I assumed that $J$ was independent of $g^{\mu \nu}$ which is actually not the case. In fact,

$$ J^\mu(x)= \frac{j^\mu(x)}{\sqrt{-g}} = \frac{q}{\sqrt{-g}} \int d\lambda \frac{d x^\mu(\lambda)}{d \lambda}\delta^4(x - x(\lambda) ). $$

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