# Fix temporal gauge $A_0=f$ using an appropriate gauge transformation

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?

• Are you sure $A^0$ can be set to any function $f$? Dec 9 '20 at 5:50
• 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 Dec 9 '20 at 10:22

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 $$\partial_0\alpha =0$$.