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?)
  • $\begingroup$ 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. $\endgroup$ – Cinaed Simson Aug 21 at 5:45

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