Physical degree of freedom and gauge fixing?

I'm confused with the gauge fixing in the Higgs mechanism.

So if we have an action like $$S=\int |D\phi|-\frac{1}{4}F^2 -V(\phi) ~ ,\tag{1}$$ then expand around some non-trivial vacuum, then we have terms like $$\int \partial_\mu \phi\partial^\mu \phi -2 m A_\mu \partial^\mu \phi +m^2 A_\mu A^\mu +\dots ~,\tag{2}$$ which can be written as $$\int m^2(A_\mu -\frac{1}{m}\partial_\mu \phi)(A^\mu -\frac{1}{m}\partial^\mu \phi)+\dots ~.\tag{3}$$ Then since the original action (1) has a gauge symmetry $$A\rightarrow A-\partial \alpha$$ for any function $$\alpha$$, we can fix the gauge (by choosing some $$\alpha$$), in this case $$\alpha=\frac{1}{m}\phi$$. Then above actions looks like $$\int m^2 B_\mu B^\mu+\dots ~,$$ where $$B\equiv A-\frac{1}{m}\partial \phi$$. I think this is how Higgs mechanism works and essencially you are removing the "unphysical degree of freedom" (in this case $$\phi$$)

My question is what if we take $$m=0$$? Then Eq.(2) looks like $$\int \partial_\mu \phi\partial^\mu \phi -2 m A_\mu \partial^\mu \phi +\dots ~.$$ So we can't rewrite the action in nice forms like Eq.(3). Then what kind of gauge should I choose to see the physical degree of freedom? I'm guessing I can't choose a gauge like $$A\rightarrow A-\partial \alpha A$$ where $$\partial \alpha = A$$.

• What does your standard summary say? – Cosmas Zachos Aug 12 at 16:48
• When $m=0$ then you don't have a Higgs particle. Also, why is there a $m$ in the last term in the last equation- there aren't $2$ different masses - the last term should vanishes too. – Cinaed Simson Aug 12 at 18:42
• @CinaedSimson Very true. Sorry did't even notice. Thanks – user239970 Aug 13 at 10:03