I am trying to find the magnetic vector potential a distance of $s$ (cylindrical radial variable) from an infinite wire carrying current $I$. The magnetic field at a distance $s$ from a wire is $$B=\dfrac{\mu_{\circ}I}{2\pi s}\hat{\phi}.$$
Using the fact that $\nabla\times A=B$ and $\nabla\cdot A=0$, I speculated that $A=\dfrac{\mu_\circ I z}{2\pi s}\hat{s}$ satisfies the necessary conditions:
In cylindrical coordinates the curl is just:
\begin{align}
\nabla \times A
& =\left(\dfrac{1}{s}\dfrac{\partial A_z}{\partial \phi}-\dfrac{\partial A_{\phi}}{\partial z}\right)\hat{s}+\left(\dfrac{\partial A_s}{\partial z}-\dfrac{\partial A_z}{\partial s}\right)\hat{\phi}
+\dfrac{1}{s}\left(\dfrac{\partial}{\partial s}(s\,A_{\phi})-\dfrac{\partial A_s}{\partial \phi}\right)\hat{z}
\\ & =\dfrac{\partial A_s}{\partial z}\hat{\phi}=\dfrac{\mu_{\circ}I}{2\pi s}\hat{\phi}
\\ & =B
\end{align}
and the divergence is:
$\nabla \cdot A=\dfrac{1}{s}\dfrac{\partial}{\partial s}(s\,A_s)+\dfrac{1}{s}\dfrac{\partial A_{\phi}}{\partial \phi}+\dfrac{\partial A_z}{\partial z}=\dfrac{1}{s}\dfrac{\partial}{\partial s}\left(\dfrac{\mu_{\circ}Iz}{2\pi}\right)=0$.
So this potential certainly satisfies the necessary requirements, but is different from everything I've looked up. I would have thought that these potentials were unique; I have been staring at this for too long and need another opinion. Is what I have correct, or have I made a hiccup somewhere?
Best Answer
Magnetic potentials are nowhere near unique, as you have conclusively shown; for more details, look up 'gauge freedom' in your favourite EM textbook or in wikipedia. Imposing the Coulomb gauge condition $\nabla\cdot\mathbf A=0$ reduces the gauge freedom but you can still transform the potential to $$\mathbf A\mapsto \mathbf A'=\mathbf A+\nabla\psi$$ for any harmonic $\psi$ such that $\nabla^2\psi$, and obtain a different potential which (i) returns the same magnetic field, and (ii) also satisfies the Coulomb gauge condition.
For regular-enough magnetic fields, you can often introduce additional requirements on the vector potential - regularity conditions, and suitable decay at infinity - which can specify it uniquely, but their feasibility depends on the niceness of the magnetic field.
To make this a bit more explicit, you've shown that $\mathbf A_1=\dfrac{\mu_0Iz}{2\pi s}\hat s$ works as a vector potential. However, it is equally easy to check that $\mathbf A_2=-\dfrac{\mu_0I}{2\pi}\ln(s)\hat z$ works equally well: it returns the same field, and it's also in the Coulomb gauge. What gives? Well, the two gauges are related by the transformation $$ \mathbf A_1 = \mathbf A_2+\nabla\psi = \mathbf A_2+\nabla\left(\dfrac{\mu_0I}{2\pi}z\ln(s)\right), $$ where $\psi\propto z\ln(s)$ obeys the Laplace equation. Which one is preferable? Neither, really - they are both singular at the wire ($\mathbf A_1$ more than $\mathbf A_2$), and while $\mathbf A_2$ grows at infinity, the $\sim 1/s$ decay of $\mathbf A_1$ there is probably too slow to be much help. In this situation, the magnetic field is too singular (infinitely thin wire) and contains too much energy (infinitely long wire) for regularity and boundedness conditions to be much help in specifying a unique vector potential.
In general, gauge freedom is something which you can often fix to a large extent but which will always be there lurking in the background. Moreover, there are simply no universally-optimal ways to fix the gauge (so, for instance, the Coulomb gauge $\nabla\cdot \mathbf A=0$ is not Lorentz invariant but the Lorenz gauge $\nabla\cdot \mathbf A+\frac{1}{c^2}\frac{\partial \varphi}{\partial t}=0$ is awkward for non-relativistic work, and so on) so you always need to think of gauge-dependent constructs as temporary, non-unique, and non-physical. The broader-picture answer is to simply let go of the unicity of the magnetic potential.