I'm wondering about how to show that $A_a\rightarrow A_a+\alpha\partial_0A_a$, with $\alpha$ infinitesimal, is an infinitesimal symmetry of $\mathcal L=-\frac14F_{ab}F^{ab}$.

\begin{equation} F_{ab}\rightarrow F_{ab}+\partial_a(\alpha\partial_0A_b)-\partial_b(\alpha\partial_0A_a)=F_{ab}+\partial_0(\alpha F_{ab}). \end{equation}

\begin{equation} \Rightarrow \mathcal L\rightarrow\mathcal L-\frac12F_{ab}\partial_0(\alpha F^{ab})=\mathcal L-\frac14\partial_0(\alpha F_{ab}F^{ab}). \end{equation}

We require $\delta\mathcal L=0$ for it to be called "a symmetry of $\mathcal L$", right? Or do we only require $\delta\mathcal L=$ a total derivative? Either way, I don't seem to get what we want.


Comments to the question (v2):

It seems that OP assumes that $\alpha$ is independent of $x$, i.e., OP considers a global quasisymmetry

$$\delta A_{\mu}=\alpha \dot{A}_{\mu}.$$

The corresponding conserved quantity is energy, cf. Example 1 on the Wikipedia page for Noether's (first) theorem.

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We may approach the problem via differential forms, or ordinary tensor calculus:

  1. Differential Forms: The field strength tensor $F$ is a differential form given by the exterior derivative of the 1-form $A$, i.e. $F=\mathrm{d}A$ which in components is $\partial_{[\mu}A_{\nu]}$. To add a total derivative to the form $A$ is equivalent to adding the exterior derivative of a 0-form, i.e. $A \to A + \mathrm{d}\alpha$. But the new $F' = \mathrm{d}(A + \mathrm{d}\alpha) = \mathrm{d}A$ because $\mathrm{d}^2\alpha=0$. So it is a symmetry.
  2. Tensor Calculus: The field strength tensor is given by $F_{\mu\nu}=\partial_\mu A_\nu - \partial_\nu A_\mu$. Consider now a change of $A$, i.e. $A_\nu \to A_\nu + \partial_{\nu} \alpha$. When we plug this into the field strength tensor, we obtain $F'_{\mu \nu} = \partial_{\mu}(A_\nu + \partial_{\nu} \alpha) - \partial_\nu (A_\mu + \partial_{\mu} \alpha) = \partial_\mu A_\nu - \partial_\nu A_\mu$ because $\partial_\mu \partial_\nu \alpha = \partial_\nu \partial_\mu \alpha$, and hence the terms cancel in the expression for $F'_{\mu \nu}$.
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  • $\begingroup$ Thank you! My main confusion arises because this is not about the usual transformation $A_\nu\rightarrow A_\nu+\partial_\nu\Lambda$. It is $A_\nu\rightarrow A_\nu+\alpha\partial_0A_\nu$. I don't understand how this $\partial_0$ makes things work. $\endgroup$ – LorentzNoether Mar 23 '14 at 10:40
  • $\begingroup$ If you only add the time derivative of some scalar to the 1-form, then I don't think it is a (quasi)symmetry. $\endgroup$ – JamalS Mar 23 '14 at 10:42
  • $\begingroup$ Well, it's the time derivative of the 1-form (with an infinitesimal parameter scaling it) added to the 1-form. Does that make you think otherwise?... $\endgroup$ – LorentzNoether Mar 23 '14 at 10:55
  • $\begingroup$ I think it still wouldn't be a symmetry, even though it may leave the Lagrangian invariant for certain $\partial_0 A_a$, e.g. when $A_a = \mathrm{constant}$, or $\partial_0 A_a = \partial_a \alpha(x)$. $\endgroup$ – JamalS Mar 23 '14 at 11:13

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