Why can we always find $\vec A$ such that it satisfies Coulomb (or Lorenz) gauge and Maxwell's equations?

Question

I have a short question about the Coulomb potential.

Let $\vec{E}$ and $\vec{B}$ be the electric field and magnetic field respectively. The electric field and magnetic fields are described by the scalar potential $V$ and vector potential $\vec{A}$ respectively. As we know, there are a lot of scalar potentials and vector potentials describing the same electric field and magnetic field. My question is, can we $\textbf{always}$ find, in this case, the vector potential $\vec{A}$ that describes the given field $\vec{B}$ and also satisfy $$\nabla \cdot \vec{A} = 0~?$$

And is this also the case for Lorenz gauge, $$\nabla \cdot \vec{A} = -\mu_{0}\varepsilon_{0}\frac{\partial V}{\partial t}~?$$

Answer 1

The short answer is yes. The general proof of this does not make any assumptions about the fields and thus is valid in general. In particular, if $\vec{A}_1$ is a vector potential for $\vec{B}$, i.e. a potential such that $\nabla \times \vec{A}_1 = \vec{B}$, then $\vec{A}_2 = \vec{A}_1 + \nabla f$ describes the same magnetic field $\vec{B}$ since $\nabla \times \nabla f = \vec{0}$ for any function $f$. However,$$\nabla \cdot \vec{A}_2 = \nabla\cdot \vec{A}_1 + \nabla^2 f,$$ and thus by appropriately choosing the function $f$ we can set the divergence of $\vec{A}_2$ to anything we may want!