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

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 same and when they are not and what are the consequences: can somebody discuss something around this dilemma?

• Many gauges will coincide in source-free case. That is not at all saying that they are the same thing. It is only interesting to state the source-ful imposed conditions. Commented Apr 30 at 9:38

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 $$\rho=0$$.