Question. Does the vector field $\vec{F}(r, \phi) = \frac{1}{r} \hat{\phi}$ have something like an associated potential function?
Context. I know that $\vec{F}$ is not conservative, and so I should not expect $\vec{F}$ to have a potential function that is defined everywhere. But could $\vec{F}$ nevertheless have a potential function that is defined almost everywhere? Specifically, I know that taking line integrals around closed paths through $\vec{F}$ leads to a result of either 0 or $2\pi$, and this depends on whether the closed path encircles the origin or not. So it seems like $\vec{F}$ acts like a conservative vector field when you're away from the origin. I also find it strangely suspicious that $\vec{F}$ itself will be undefined when $r = 0$, i.e. at the origin. So I'm wondering if maybe there is an associated potential function that is defined everywhere except the origin, and has an essential singularity at the origin, so that integrating around the singularity leads to nonzero results?
Best Answer
I figured out the answer, plus a little bit more. I was forgetting that if $f(r, \phi)$ is defined in polar coordinates, then $\nabla f(r, \phi) = \frac{\partial f}{\partial r} \hat{r} + \frac{1}{r} \frac{\partial f}{\partial \phi} \hat{\phi}$. So if I let $f(r, \phi) = \phi$, then I obtain $\nabla f(r, \phi) = \frac{1}{r} \hat{\phi}$ as desired.
So I wanted to know what this "potential" function might mean. First of all, $f$ is clearly not defined at the origin since $\phi$ is undefined at the origin. Worse still, since multiple values of $\phi$ correspond to each point in space, the "function" is multi-valued, I believe, and therefore is not a function at all. Second, if one imposes certain restrictions on the range of the function so that it is single valued, one could imagine taking a circular path integral around the origin, and attempting to apply something like the fundamental theorem of line integrals. Well, as the $\phi$ coordinate increases due to the circular path, the value of $f(r, \phi) = \phi$ would increase. And if one took a circular path starting at $\phi = 0$ and ending at $\phi = 2\pi$, i.e. a complete revolution, one would obtain $2\pi - 0 = 2\pi$, which is the same result one obtains when taking an actual line integral around the origin.
Furthermore, I wanted to "see" what this all would look like, so I graphed this:
Clearly, the path has the property of starting at a height 0 and ending at a height 2$\pi$, but then there is a huge discontinuity. I'm not sure I can describe why all of this is explicitly problematic, but it seems like it is clearly problematic!