# Curl of gradient is not zero

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?

• Not sure what this has to do with the curl. There are indeed (scalar) functions out there whose Laplacian (the divergence of the gradient) is the delta function. Are you suggesting that that gradient itself is the curl of something? That's possible: it can happen that the divergence of a curl is not zero in the sense of distribution theory, if the domain isn't simply connected. – Ian Mar 28 '18 at 13:43
• Sorry. Edited the orignial post – lminsl Mar 28 '18 at 13:45

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!

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 $$\nabla \times \nabla \theta = 2\pi \delta({\bf x})$$

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.)