Consider an infinitely thin, infinitely long wire,
Using the standard formula for the magnetic vector potential, we obtain the result that the magnetic vector potential is infinity,( with no variables)
And takes the form that $A$ is proportional to
$$ \ln|\sec(\theta)+\tan(\theta)| $$ evaluated at $-\pi/2$ to $\pi/2$
This is because we have specifically used the gauge choice $\nabla \cdot \vec{A} = 0$ in order to find the standard formula.
I have also seen that the formula
$$\vec A=-\frac{\mu_0I}{2\pi} \ln(r) \hat e_z$$
Would ALSO work as a valid potential function.
So my question is this:
What gauge choice do I need to evoke in order to obtain a valid formula for the magnetic vector potential that does NOT diverge for an infinitely thin, and infinitely long wire(or how to find it another way)
I also don't understand why there isn't at least SOME variables that I can take the curl of, to find the correct magnetic field, As it technically IS a perfect solution to the equation
$$\nabla(\nabla \cdot \vec{A}) – \nabla^2 \vec{A} = -\mu_{0} \vec{J}$$
Best Answer
The equation we're trying to solve is $$\nabla(\nabla \cdot \mathbf A) - \nabla^2 \mathbf A = -\mu_0 \mathbf J$$ From a mathematical standpoint, an infinitely long, infinitely thin wire does not constitute a well-defined current density in the usual sense - rather, it would take the form of a delta distribution $\mathbf J = I \delta(x) \delta(y)\hat z$ in cartesian coordinates.
The easiest way to handle this is to use cylindrical coordinates $(r,\theta,z)$ and solve the equation $$-\nabla(\nabla \cdot \mathbf A) - \nabla^2 \mathbf A = 0$$ on the domain $(r,\theta,z)\in (0,\infty)\times[0,2\pi) \times (-\infty,\infty)$ subject to some boundary conditions that we'll get to in a moment. Choosing the Coulomb gauge condition $\nabla \cdot \mathbf A=0$ yields the familiar Laplace equation, and taking advantage of the cylindrical symmetry of the problem (so $\partial_\theta,\partial_z \rightarrow 0$) yields the fairly straightforward equation $$\frac{1}{r}\frac{\partial}{\partial r} \left(r \frac{\partial \mathbf A}{\partial r}\right)=0 \implies r \frac{\mathbf A}{\partial r} = \mathbf c \implies \mathbf A = \mathbf c \ln(r/r_0)$$ for some unknown vector $\mathbf c$. $r_0$ is simply an integration constant, and can be chosen to be any value you wish. The subsequent magnetic field is given by $$\mathbf B= \nabla \times \mathbf A = -\frac{c_z}{r} \hat \theta + \frac{c_\phi}{r}\left(\ln(r/r_0) + 1\right)\hat z$$
The aforementioned boundary condition is that if consider an arbitrary curve $\mathbf r(t)$ which encloses the wire and compute $\oint \mathbf B \cdot \mathrm d\mathbf r$, then we will obtain $\mu_0 I$. After a bit of work, we find that in order for this to be true for arbitrary curves, $c_\phi=0$ (otherwise the result of the line integral would be curve-dependent) and $c_z = - \frac{\mu_0 I}{2\pi}$, and so we finally obtain
$$\mathbf A = -\frac{\mu_o I}{2\pi} \ln(r/r_0) \hat z \implies \mathbf B = \frac{\mu_0 I}{2\pi r} \hat \theta \tag{$\star$}$$
The "standard solution" referenced by the OP and others is a special case of a Green's function solution. Briefly, the underlying idea goes as follows. Consider a differential equation for an unknown function $u(\mathbf r)$ of the form $$L\big[u\big](\mathbf r) = f(\mathbf r)$$ (where $L$ is a linear differential operator) on some region $D$ with boundary $\partial D$, where $u$ is subject to e.g. Dirichlet boundary conditions on $\partial D$. If we first solve the equation $$L[G](\mathbf r) = \delta(\mathbf r-\mathbf r')$$ where $G$ is subject to the same boundary conditions, then we find that the correct solution to the original equation is given by $$u(\mathbf r) = \int_D \mathrm d^n r' \ f(\mathbf r') G(\mathbf r;\mathbf r')$$
As an example, one might consider the 3D Laplacian with $D$ the ball of radius $R$ centered at the origin, and Dirichlet boundary conditions on $\partial D$. It turns out that the correct solution is $$G_R(\mathbf r,\mathbf r') = \frac{1}{4\pi|\mathbf r-\mathbf r'|} - \frac{R/r'}{4\pi|\mathbf r - \frac{R^2}{r'^2}\mathbf r'|}$$ which can be seen to vanish when $|\mathbf r|=R$. Furthermore, this $G_R$ has a well-defined limit as $R\rightarrow \infty$, and taking that limit yields the so-called free space Green's function $G_\infty(\mathbf r,\mathbf r')= 1/4\pi|\mathbf r- \mathbf r'|$. Note that there is no reason to expect this limit to exist in general.
Since the problem of an infinite line charge is mathematically equivalent to that of a 2D point charge, consider the 2D version of the example above. In this case, it turns out that the Green's function is
$$G_R(\mathbf r,\mathbf r') = \frac{1}{2\pi}\log|\mathbf r-\mathbf r'| - \frac{1}{2\pi}\log\left(\frac{r'}{R}|\mathbf r-\frac{R^2}{r'^2} \mathbf r'|\right)= \frac{1}{2\pi}\log\left(\sqrt{\frac{r^2-2\mathbf r\cdot \mathbf r' + r'^2}{\frac{r'^2}{R^2}r^2 - 2\mathbf r \cdot \mathbf r' + R^2}}\right)$$
Observe that $G_R$ does not have a well-defined limit as $R\rightarrow \infty$. As a result, it is not possible to solve $\nabla^2G = \delta(\mathbf r-\mathbf r')$ subject to the boundary condition that $G\rightarrow 0$ as $|\mathbf r|\rightarrow \infty$; you know this already, in the form of the fact that the electrostatic potential due to an infinite line charge diverges logarithmically as $|\mathbf r|\rightarrow \infty$. Instead, we are forced to solve this equation on a finite domain - say, a disk of radius $R$. If apply this solution to your problem, we find that
$$\mathbf A = \int_D \mathrm d^2 r' [\mu_0 I \delta(\mathbf r')\hat z] G_R(\mathbf r,\mathbf r') = \frac{\mu_0 I}{2\pi} \ln\left(\frac{r}{R}\right)\hat z$$
This gives us a nice interpretation for the integration constant $r_0$ in $(\star)$. If we want to apply Dirichlet boundary conditions to the equation $\nabla^2 \mathbf A = \mu_0 I\delta(x)\delta(y)\hat z$, then we cannot do so at infinity like we can in the case of a 3D point charge. Instead, we pick a finite disk of radius $r_0$ and impose our Dirichlet conditions on the boundary of the disk, and that $r_0$ appears in the solution to enforce this condition.