Expression for potential vector of a central field

Question

I know that for the central field $$ {\bf F(x)}=\alpha\cdot\frac{\bf x}{|{\bf x}|^{3}}=\alpha\cdot\left(\frac{x_{1}}{|{\bf x}|^{3}},\frac{x_{2}}{|{\bf x}|^{3}},\frac{x_{3}}{|{\bf x}|^{3}}\right) $$ holds $\nabla\cdot{\bf F(x)}=\delta_{0}$, where $\alpha$ is a constant, $|\bf x|$ is the euclidian norm (i.e. $|{\bf x}|=\sqrt{x_1^2+x_2^2+x_3^2}$ ) and $\delta_{0}$ is the Dirac delta. If one consider a region of the space that doesn't surround the origin, then $\bf F(x)$ should have a potential vector $\bf A$ such that $\nabla \times {\bf A}={\bf F}$, because in this case we can safely say that $\nabla\cdot{\bf F(x)}=0$. But is there an analytical expression for that potential vector? Moreover, my reasonings are correct or am I wrong?

Answer 1

Setting $\alpha = 1$, here is a vector potential for $\mathbf F$. It is defined and smooth off the $z$-axis.

Set $$\mathbf A = \frac1{(x^2+y^2)(x^2+y^2+z^2)^{1/2}}\big(yz,-xz,0\big).$$

The answer is very non-unique. Indeed, you can find this solution, as I did, by looking for $\mathbf A = (P,Q,R)$ with $R=0$. Now find other solutions with, correspondingly, $P=0$ and $Q=0$. Any convex linear combination of these three will be another solution. Of course, the domain of such a vector field will be only the complement of all three coordinate axes.