Can we impose Coulomb gauge without using temporal gauge in source-free Maxwell electrodynamics?

Question

Coulomb gauge is: $$\vec{\nabla} \cdot A=0$$ Now, from expression for electric field in terms of potentials: $$\vec{E}=-\vec{\nabla} \phi-\frac{\partial \vec{A}}{\partial t}$$ and Gauss' Law $\vec{\nabla} \cdot \vec{E}=0$, we have: $$\nabla^2 \phi=-\frac{\partial }{\partial t}(\vec{\nabla}\cdot \vec{A})=0$$ whose solution is: $$\phi=0$$

if the scalar potential dies off at infinity.

So, does Coulomb Gauge imply temporal gauge? If so, why is temporal gauge treated differently than Coulomb gauge in most field theory textbooks? When exactly are they the same and when they are not and what are the consequences.

Can somebody discuss something around this dilemma?

Answer 1

It is true that in your case there is no visible difference between the Coulomb gauge and the $\phi=0$ (Weyl) gauge. But this is because you didn't include a source! If you do include it, so $\vec{E} = \rho$, then your equation is ($\epsilon_0 =1$) $$ \nabla^2 \phi + \dfrac{\partial}{\partial t}(\nabla\cdot\vec{A}) = -\rho, $$ which becomes using Coulomb's gauge the famous Poisson equation $$ \nabla^2 \phi = -\rho. $$ This can be "easily" solved, so that $$ \phi(t,\vec{x}) = \dfrac{1}{4\pi}\int d \vec{r} \dfrac{\rho(t,\vec{r})}{|\vec{x}-\vec{r}|} \neq 0. $$ So both gauges are generally different. Also, you can see that $\phi=0$ iff $\rho=0$.

About the uses, Coulomb's gauge is extremely useful whenever you don't care about Lorentz covariance (as this gauge fixing isn't) and is thus used in non-relativistic classical EM and non-relativistic QM, as it simplifies equations a lot.

I don't know much about the uses of Weyl gauge, but maybe someone else can add about it!