Coulomb gauge does not fix $\vec A$ uniquely. Is the solution to $\nabla^2{\vec A} = -\mu_0{\vec J}$ given in Griffiths unique?

Question

With the Coulomb gauge $\nabla\cdot{\vec A}=0$, and $\nabla\times\vec A=\vec B$, the vector potential satisfies the Poisson's equation, $$\nabla^2{\vec A} = -\mu_0{\vec J},$$ which for $\vec J\to0$ at infinity, leads to the solution (Griffiths' electrodynamics, Eq. 5.65, (see [1]): $${\vec A}({\vec r})= \frac{\mu_0}{4\pi}\int\frac{{\vec J}({\vec r}\,')\,d^3r'}{|{\vec r}-{\vec r}\,'|}.$$ Since the Coulomb gauge does not fix ${\vec A}$ uniquely (see here), is the above solution unique? I think that the solution is actually $${\vec A}({\vec r})= \frac{\mu_0}{4\pi}\int\frac{{\vec J}({\vec r}')d^3r'}{|{\vec r}-{\vec r}'|}+\nabla\phi(\vec r)$$ where $\phi(\vec r)$ is a solution of $\nabla^2\phi=0$. I think that to set $\phi=0$, we need additional conditions on $\vec A$. Please explain what condition do we need to invoke, such that we can get Griffiths' solution.

Answer 1

The Helmholtz Theorem says that a vector field $\mathbf{A}$ is uniquely determined by ${\rm div}\,\mathbf{A}$, ${\rm curl}\,\mathbf{A}$, and $\mathbf{A}\rightarrow0$ as $r\rightarrow\infty$. So, when dealing with a localized source, as described in the question, there is only a single transverse $\mathbf{A}$ that will vanish at spatial infinity.

This can be seen directly from the harmonic property of the $\phi$ in the question. If $\nabla\phi$ is to be added to $\mathbf{A}$ without changing $\nabla\cdot\mathbf{A}$, then, as pointed out, $\phi$ must satisfy $\nabla^{2}\phi=0$. However, if we are to have $\nabla\phi\rightarrow0$ as $r\rightarrow\infty$, then $\phi$ goes to a constant at spatial infinity. Since $\phi$ is a solution of Laplace's Equation, it has no local extrema. If $\phi$ goes to a constant value in every direction for $r\rightarrow\infty$, the only way it can have no local maxima or minima is if it is constant; thus, the $\nabla\phi$ added to $\mathbf{A}$ vanishes.

However, there are many reasonable source configurations for which the source $\mathbf{J}$ does not vanish at infinity, and so neither does $\mathbf{A}$. The simplest example is an infinite wire. There is no vector potential $\mathbf{A}$ for an infinite wire than vanishes as $r\rightarrow\infty$, since the source extends to spatial infinity. Thus, a point can be near to the source for arbitrarily large $r$. There are also situations where we may be given not $\mathbf{J}$ but $\mathbf{B}$. For example, a constant background magnetic field $\mathbf{B}_{0}$ everywhere cannot be written as the curl of a $\mathbf{A}$ that vanishes at infinity. It is because we want to account for the kinds of configurations discussed in this paragraph that we consider the Coulomb gauge not to be a complete gauge.