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Consider the Lagrangian \begin{equation} \mathcal{L}= -\frac{1}{4} F_{\mu \nu}F^{\mu \nu} - A_{\mu}J^{\mu} \ \ \ \ \text{ with } \ \ \ \ F_{\mu \nu}=\partial_\mu A_\nu - \partial_{\nu}A_{\mu}. \end{equation} The current $J^{\mu}$ is conserved and hence it satisfies $\partial_{\mu}J^{\mu}=0$.

I showed action $S$ is invariant under the gauge transformation \begin{equation} A_{\mu} \to A_{\mu} + \partial_{\mu} \alpha, \end{equation} where $\alpha$ is an arbitrary function that vanishes at infinity. Moreover I know that the equations of motions are $$\partial^{\mu}F_{\mu \nu} = J_{\nu}.$$

Question: Show that one can choose the function $\alpha$ so that after an appropriate gauge transformation one can set $A_0=f$, where $f$ is a given function.

I think I'm missing some fundamental stuff about gauge theory since e.g. I don't know why the coulomb gauge or the lorenz gauge are allowed.
Anyways..

My idea: Say $\alpha = \int (f- A_0) dt$, then we get under the gauge transformation . $$A_0 \to A_0' = A_0 + \partial_{0} \int (f- A_0) dt = A_0 +f - A_0 = f.$$ And then we redefine $A_0'$ as $A_0$ which is possible by gauge invariance and therefore obtain $A_0=f$.

Is what I'm doing correct, or am I doing something fundamentally wrong?
Moreover, what would the residual gauge transformation be?

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  • $\begingroup$ Are you sure $A^0$ can be set to any function $f$? $\endgroup$ Dec 9 '20 at 5:50
  • $\begingroup$ well that's the question. Of course in physics we don't consider non-integrable functions e.g. everything is always smooth. or is that not the point $\endgroup$ Dec 9 '20 at 10:22
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Suppose that the gauge field $A'_{\mu}$ does not satisfy the desired property, so let's say instead $A'_{0} = g$ for some function $g$. Choose $\alpha$ such that $$ \partial_0 \alpha = f-g, $$ which is guaranteed to have a solution. We apply a gauge transformation $A'_{\mu} \to A_{\mu} = A'_{\mu} + \partial_{\mu} \alpha $ and obtain that $$ A_0 = A'_0 + \partial_0 \alpha = g + f-g =f. $$ Therefore after an appropriate gauge transformation one can set $A_0=f$. Of course if we'd apply another gauge transformation such that $\partial_0 \alpha =0$ and we'd still have that $A_0=f$. So the residual gauge symmetry is $\partial_0\alpha =0$.

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