# Isn't Stress-Energy Tensor of Maxwell field in presence of charges gauge variant?

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?

• "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. – Cosmas Zachos Feb 9 at 17:49
• 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. – user1379857 Feb 9 at 17:53
• The derivatives on the part of the charges contribute to the gauge variation. The third term is effectively their gauge covariant completion. – Cosmas Zachos Feb 9 at 18:39
• 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 $A_\mu$? – user1379857 Feb 9 at 18:47
• 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. – Cosmas Zachos Feb 9 at 19:12

Our charge coupling term is of the form

$$$$S = \int_U j^\alpha A_\alpha$$$$

Now take the gauge transformation. We have

$$$$A_\alpha \to A_\alpha + \partial_\alpha f$$$$

giving us the difference

$$$$\delta S = \int_U (j^\alpha \partial_\alpha f)$$$$

As a Noether current, the electric current obeys the relation

$$$$\partial_\alpha j^\alpha = 0$$$$

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.

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,

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