In electrodynamics, the scar potential $\phi$ and the vector potential $\textbf{A}$ satisfy the equations $$\frac{\partial}{\partial t}(\boldsymbol{\nabla}\cdot\textbf{A})+\nabla^2\phi=-\frac{\rho}{\epsilon_0}$$ and $$\boldsymbol{\nabla}\Big((\boldsymbol{\nabla}\cdot\textbf{A})+\frac{1}{c^2}\frac{\partial\phi}{\partial t}\Big)+\frac{1}{c^2}\frac{\partial^2\textbf{A}}{\partial t^2}-\nabla^2\textbf{A}=\mu_0\textbf{J}.$$ Radiation gauge implies that one can change from $(\phi,\textbf{A})\to(\phi^\prime,\textbf{A}^\prime)$ such that $\phi^\prime=0$ as well as $\nabla\cdot\textbf{A}^\prime=0$. If $\nabla\cdot\textbf{A}\neq 0$, we can choose a function $\chi(\textbf{r},t)$ such that $$\nabla^2\chi=-\nabla\cdot\textbf{A},\tag{1}$$ we get, $\nabla\cdot\textbf{A}^\prime=0$ because $\textbf{A}^\prime=\textbf{A}+\nabla\chi$.
If one is in vacuum, $\rho=0$. In that case only, one can choose $\phi^\prime=0$, by demanding $$\frac{\partial\chi}{\partial t}=\phi\tag{2}$$ since $\phi^\prime=\phi-\frac{\partial\chi}{\partial t}$.
For radiation gauge, we require both $\nabla\cdot\textbf{A}=\phi=0$. But for that choice of gauge, (1) and (2) has to be satisfied simultaneously. What is the guarantee that we can always find a $\chi(\textbf{r},t)$ satisfying both (1) and (2)?
Best Answer
$\nabla\cdot \mathbf{A} = 0$ is known as the Coulomb (or transverse) gauge. It's also called the radiation gauge. Note that, for boundary conditions at infinity, the Coulomb gauge is complete - there is no ambiguity left for making further modifications to the gauge fields that don't violate $\lim_{r\rightarrow \infty} A^\mu(r, t) = 0$ (because any function that satisfies $\nabla^2 f(\mathbf{x})=0$ and $\lim_{r\rightarrow \infty} f(\mathbf{x}) = 0$ is identically zero).
$\phi = 0$ is a different gauge called the Weyl gauge. The residual gauge symmetries aren't enough to make it compatible with the Coulomb gauge. This gauge is unspoilt by any gauge transformation that satisfies $\frac{\partial f}{\partial t} = 0$, leaving $$\mathbf{A}'(\mathbf{x}, t) = \mathbf{A}(\mathbf{x}, t) + \nabla f(\mathbf{x}).$$ True, it might be possible to use the residual freedom in the Weyl gauge to satisfy the Coulomb gauge condition at some particular time, but it will only be satisfied for that instant, and not in general.