Should the energy-momentum tensor be invariant under gauge transformations?

For example, consider the electromagnetic theory given by \begin{align} I=-\frac{1}{4}\int d^4x\, F_{\mu\nu}F^{\mu\nu}, \end{align} where $$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$$. The action has a symmetry given by the space-time translations \begin{align} \delta A_\mu=-\epsilon^\alpha\partial_\alpha A_\mu, \end{align} with $$\epsilon^\alpha$$ the transformation parameter. In this case the energy-momentum tensor will be given by \begin{align} T^{\alpha}_\nu=F^{\mu\alpha}\partial_\nu A_\mu-\frac{1}{4}\delta^{\alpha}_\nu F_{\lambda\rho}F^{\lambda\rho}. \end{align} Everything seems to be ok, but note $$T^{\alpha}_\nu$$ will not be gauge-invariant and this is because $$\delta A_\mu=-\epsilon^\alpha\partial_\alpha A_\mu$$ is not gauge-invariant. This seems to be a problem, so one can do some sleight-of-hand and do an 'improved translation', which is a space-time translation + a particular gauge transformation \begin{align} \delta A_\mu=-\epsilon^{\alpha}\partial_\alpha A_\mu+\partial_\mu(\epsilon^{\alpha}A_\alpha)=F_{\mu\alpha}\epsilon^\alpha. \end{align} Note the transformation will be gauge-invariant. For completeness, the (gauge-invariant) energy-momentum tensor will be \begin{align} T^{\alpha}_\nu=-F^{\alpha\beta}F_{\nu\beta}+\frac{1}{4}\delta^{\alpha}_\nu F^{\gamma\delta}F_{\gamma\delta}. \end{align} So this was a happy example.

My questions are:

1. What would happen if I can't do the 'improved translation' always?
2. Energy-momentum tensor should always be gauge-invariant? What does imply this? (Furthermore, what would be the consequences at the quantum theory?)
3. There's some examples of not gauge-invariant energy-momentum tensor? (or perhaps some loss of - some more (or less) "dramatic" - symmetry?)
• First, you're missing the $J^{\mu}A_{\mu}$ in the Lagrangian. Also, the Lorentz gauge has already been fixed, i.e., the tensors are antisymmetric. There's only a harmonic function left, i.e., $A_{\mu} \rightarrow A_{\mu}+\partial_{\mu}f$ - see the Lorentz gauge on wikipedia. And you may be confusing Noether’s theorem which is based on global symmetries with gauge fixing which is local. Typically, one use gauge fixing to simplify the equations. – Cinaed Simson Aug 21 '19 at 5:45
• If you have an answer to your question you should post it in the Answers section, not add it as an edit to the question. – PM 2Ring Dec 21 '19 at 11:55
• This thread (v4) does not conform with the Q&A format. Answers should be in answers, not part of the question. Also on Phys.SE we don't write SOLVED in the title when it's answered. – Qmechanic Dec 21 '19 at 11:55
• Additionally, you don't add "solved" to the title. The way to do it on this Q&A site is to accept the answer that answers your question. – Kyle Kanos Dec 21 '19 at 11:56
• This is a Q&A site, which means that you should post the answer separately, not as an edit. I have removed that section and posted it separately in accordance with that. You're obviously free to post that text as an answer yourself, in which case I'll delete the answer quoting your text. Similarly, if you want to mark the question as resolved, do not restore the text on the title - use the green checkmarks on the answer to mark it as accepted, instead. – Emilio Pisanty Dec 21 '19 at 13:41

"Energy-momentum tensor should always be gauge-invariant? What does imply this? (Furthermore, what would be the consequences at the quantum theory?)". We want it to be gauge-invariant because energy and momentum will be a measure and it must be the ideal case, nevertheless, it'll not always be gauge-invariant and it's fine. Consider the gravitational wave theory, whose action is the linearised general relativity action. The energy-momentum tensor $$t_{\mu\nu}$$ will not be gauge invariant. As a consequence we will consider the mean value of it $$\langle t_{\mu\nu}\rangle$$ in order to obtain a dependence of the physical modes $$h_{ij}^{TT}$$ [1]. $$\langle t_{\mu\nu}\rangle$$ will be the observable indeed, and not $$t_{\mu\nu}$$.