# Spacetime translation is not a gauge invariant symmetry in Maxwell's theory

Consider the Action for Electromagnetism,

\begin{align} I=-\frac{1}{4}\int d^4x\, F_{\mu\nu}F^{\mu\nu}, \end{align}

Suppose we perform a space-time translation, $$x^{\mu} \to x^{\mu} + \epsilon ^{\mu}$$ , we can write the variation in $$A_{\mu}$$ as the Lie Derivative.

$$\delta A_{\mu} = -\epsilon ^{\alpha} \partial _{\alpha} A_{\mu}$$

It is mentioned in "A short review on Noether's theorems, gauge symmetries and boundary terms" by Maximo Banados and I.A. Reyes in Page 22 that this is not a gauge invariant symmetry. I have a problem in this statement. Suppose you let $$A_{\mu} \to A_{\mu} + \partial_{\mu} f$$ . Under a spacetime translation, $$\delta f = -\epsilon ^{\alpha} \partial _{\alpha} f$$ . So,

$$\delta (A_{\mu} + \partial_{\mu} f) = \delta A_{\mu} + \partial_{\mu}(\delta f) \implies \delta (A_{\mu} + \partial_{\mu} f) = -\epsilon ^{\alpha} \partial _{\alpha} A_{\mu} -\epsilon ^{\alpha} \partial_{\mu}\partial _{\alpha} f = -\epsilon ^{\alpha} \partial _{\alpha} (A_{\mu} + \partial_{\mu}f)$$

So, $$\delta (A_{\mu} + \partial_{\mu} f) = -\epsilon ^{\alpha} \partial _{\alpha} (A_{\mu} + \partial_{\mu}f)$$ . which looks gauge invariant. I know that the conserved stress energy tensor obtained from this symmetry is not gauge invariant. I seem to be missing some detail here.

My question is, in what sense is the transformation not gauge invariant?

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– Chris
Commented Dec 21, 2019 at 20:39

The expression $$\delta(A_\mu) = \epsilon^\alpha\partial_\alpha A_\mu$$ is not gauge invariant in the usual, straightforward sense. Under a gauge transformation $$A\mapsto A' = A +\mathrm{d}f$$, it transforms exactly as you have shown, $$\delta A'_\mu = \delta(A_\mu + \partial_\mu f) = \epsilon^\alpha\partial_\alpha(A_\mu + \partial_\mu f).$$ Were the expression gauge-invariant, we would have $$\delta A' = \delta A$$, which is clearly not the case.