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I am taking a course in classical electrodynamics and I am facing this problem

Prove that $A_\mu A^\mu=0$ is a good gauge fixing or not. If it is, there are residual gauge freedom?

I know that, for any $A_\mu$, I have to find a function $\alpha$ that make $(A_\mu + \partial_\mu \alpha )( A^\mu + \partial^\mu \alpha)=0$. I also know that I can not make $(A_\mu + \partial_\mu \alpha )=0$ (as a tensor), so I have tried to expand the product, but I haven't managed to get an explicit expression for $\alpha$. Is this a good approach or should I try something else?

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  • $\begingroup$ Check this. When you expand, you get the original term PLUS a surface term PLUS a Lorentz Gauge term PLUS a massless Klein Gordon term. (I am loose with the definition here as this is a comment and not answer.) So, my conclusion is that within the Lorentz gauge, the condition $A^\mu A_\mu =0$ is good. $\endgroup$ – negligible_singularity Jun 29 '16 at 6:49
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You can find out $\dot{\alpha}=\sqrt{\sum_i(A^i+\partial^i\alpha)^2}-A^0$, thus, you can pose a Cauchy problem for $\alpha$. $\alpha_{,i}$ at $t=t_0$ is arbitrary ($i=1,2,3$).

EDIT(6/29/16): A better wording would probably be that $\alpha$ at $t=t_0$ is arbitrary (but smooth enough).

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