Curl of gradient is not zero

Question

I have heard that for some functions $T$, if we calculate $\nabla \times (\nabla T )$ in $2$-dimensional polar coordinates, then we get the delta function. Why do we get that result?

Answer 1

Consider $T = \theta$, the angular polar coordinate. This is badly behaved at the origin, and cannot be defined continuously around the origin (although $\nabla \theta$ can be), so we will need some new ideas to make sense of $\nabla \times \nabla \theta$. Let's try!

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One sensible thing we could do is compute the area integral $$ I = \int_{S} {\rm d}^2x \ \nabla \times \nabla \theta$$ using Stokes's Theorem to convert it into a line integral: $$ I = \int_{\partial S} {\rm d} {\bf l} \cdot \nabla \theta$$ Here, $\partial S$ is the boundary of $S$, so it is a circle if $S$ is a disc.

Now this is the integral of a total derivative along a line, and generally that just evaluated to the difference of the function at the start and end points: $$ I = \theta[\mbox{end}] - \theta[\mbox{start}]$$ But the start and end points are the same, because the boundary is a closed loop!

Suppose that the area $S$ did not include the origin. Then $\theta$ is just a smooth continuous function. Hence $I = 0$.

But suppose it did include the origin. Now the loop $\partial S$ goes around the origin! But $\theta$ is discontinuous as you go around a circle. In particular, it is $2\pi$ bigger after going around the origin once. Hence $I = 2\pi$.

Thus $$I = \begin{cases} 2\pi & \mbox{if $S$ contains $\bf 0$} \\ 0 & \mbox{otherwise} \end{cases}$$ and consequently $$\nabla \times \nabla \theta = 2\pi \delta({\bf x})$$

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There are other ways to think about this result, but this is one of the most natural!

Note that the above argument shows that this situation is inherently about non-single-valued functions, with branch cuts. This is very closely related with the fact that the usual 2D Green's function for the Laplacian is proportional to $\log r$, but $\log r$ cannot be extended continuously to the complex plane without a branch cut.

(Indeed, look at $\log (r e^{i\theta}) = \log r + i \theta$. This shows that $\theta$ is the harmonic conjugate of $\log r$. This is why it appears in the solution.)