Inverse-Square vector fields have both a divergence and curl of $0$?

Question

Consider an inverse- square vector field

$$ \vec{F} = \frac{x}{r^3}\hat{x} + \frac{y}{r^3}\hat{y} + \frac{z}{r^3}\hat{z} = \frac{\hat{r}}{r^2}$$

where $r = \sqrt{x^2 + y^2 + z^2}$. The curl $\nabla \times \vec{F} = \vec{0}$, therefore we might go looking for a potential $V$. I find that $V = -1/r$ works and therefore one can say that $\vec{F} = \nabla V$ is derivable from a potential function $V$. I'll point out right now that $\vec{F}$ is undefined at the origin.

The divergence $\nabla \cdot \vec{F} = 0$. Therefore, we might go looking for a vector potential $\vec{A}$ such that $\vec{F} = \nabla \times \vec{A}$. One would say that $\vec{F}$ is derivable from a vector potential $\vec{A}$. But I'm having trouble seeing that an inverse-square vector field is derivable from both a vector potential and a scalar potential. So I know we have a trouble point at the origin. Yet this trouble point doesn't really seem to affect the "conservativeness" or path-independence of the vector field. But this trouble point does seem to affect the surface-independence of the vector field. As long as the surface doesn't wrap around the origin, I'd expect the inverse-square vector field to be surface-independent for a given boundary curve.

Can an inverse-square vector field be derivable from both a scalar potential and a separate a vector potential? (Helmholtz theorem comes to mind. But the question I'm asking involves two separate equations. One $\nabla V$ gives $\vec{F}$ and another $\nabla \times \vec{A}$ gives $\vec{F}$ as well).

Answer 1

If all you want is an inverse-square vector field in a topologically trivial region, then the Dirac string vector potential does the trick: $$ \vec{A} = \frac{1 - \cos \theta}{r \sin \theta} \hat{\phi} = \frac{\tan (\theta/2)}{r} \hat{\phi}. $$ It is not too hard to show that the curl of this vector field is $$ \vec{\nabla} \times \vec{A} = \frac{1}{r^2} \hat{r}. $$ However, $\vec{A}$ is ill-defined when $\theta = \pi$, which means that it can't be extended to all of space. As noted in my previous answer, there are topological obstructions to extending any such vector field over all of space.

Answer 2

This is an old question but I find the comments and the other answer are a little evasive. There is a simple, clear answer to your question: NO, your field $F$ does not possess a vector potential everywhere it is defined. It only possesses a vector potential on regions away from the origin, but none of these vector potentials can be extended to all of space minus the origin.

You seem to think $F$ should have a vector potential everywhere because it has zero divergence. But an arbitrary divergence-free field is only guaranteed to have a vector potential on a region with trivial second de Rham cohomology. In this case, the domain of $F$ is space minus the origin, which has nontrivial second de Rham cohomology. As a result, knowing that $F$ has zero divergence is not enough to conclude that it has a vector potential: it could go either way. (There are some divergence-free fields on space minus a point that do have vector potentials; others don’t.) To check, you need to ask whether the integral of $F$ vanishes over every closed surface contained in the domain. If you can find just one closed surface where the integral fails to vanish, then $F$ does not have a vector potential.

That is what happens in this case. (In fact, your $F$ is the classic, canonical example of a divergence-free field without a potential.) You can see that $F$ fails the test I gave in the last paragraph by observing that its integral over the unit sphere is nonzero (this is just Gauss’s law!).