Why is the curl of the gradient of the polar angle phi zero, counter to my intuition?

Question

I was trying to test my intuition about curl and divergence by constructing vector functions whose curl and divergence I think I have intuitive expectations about. An idea I came up with was to think of a scalar function with some shape whose gradient I would intuitively expect to have the desired property. I thought, for example, the function $\phi=\arctan(y/x)$ can be geometrically thought of as a staircase of linearly increasing angle winding around the origin. Therefore I thought that $\vec{\nabla}\phi$ would be a vector field whirlpooling around the z-axis (i.e. in the x-y plane), and therefore would have a nonzero curl in the z direction at (0,0). However in this case $\vec{\nabla}\times\vec{\nabla}\phi = \vec{0}$ for all (x,y), contrary to my expectation. What is wrong about my intuition here?

Answer 1

The curl of the gradient is zero away from the origin, but the singular contribution at the origin integrates with the help of Stokes' theorem to $$\int\int_S \hat{z}\cdot(\nabla\times\nabla\phi)\,dxdy=\oint_{\delta S}\nabla\phi\cdot dl=2\pi.$$ Here $\hat{z}$ is a unit vector in the $z$-direction, $\phi$ is the polar angle in the $x$-$y$ plane, $S$ is any surface in the $x$-$y$ plane containing the origin, $\delta S$ is its perimeter and the final integral equals the amount the angle $\phi$ increases as we go once around the perimeter.

This integral is expressed by the identification $$\nabla\times\nabla\phi=2\pi\delta(x)\delta(y)\hat{z}.$$

Answer 2

If $\phi=arctan(y/x)$ is electric scalar potential, I believe that $\phi$ is single valued. If $\phi$ is single valued, the vector $\nabla\phi$ directs in reverse within the small region $(2\pi-\epsilon)<\theta<2\pi$, where $\epsilon$ is some positive small number. Thus the vector $\nabla\phi$ field is not whirlpooling around the z-axis.