Divergence of $ \frac{1}{s}\hat{s}$ in cylindrical coordinates

Question

In Griffiths' electrodynamics, the divergence of $\frac{1}{r^{2}}\hat{r}$ is evaluated in spherical coordinates to be $4\pi\delta(r)$. I encounter the same problem in case of $\frac{1}{s}\hat{s}$ where $s$ is the radius in cylindrical coordinates. I think that it is $2\pi\delta(s)$ but I am not sure.

If I put this in the expression for divergence in cylindrical coordinates I get zero to be the answer but it is only intuitive that there should be some divergence at the origin as in the case of $\frac{1}{r^{2}}\hat{r}$. If I compute the flux through a closed surface say a cylinder the flux through the top and bottom end is zero while flux through the lateral surface is $2\pi h$,where h is the height of cylinder .In this case it is dependent on the surface features(as opposed to the case of $\frac{1}{r^{2}}\hat{r}$).This violates the divergence theorem.What is the correct divergence of this vector field?

Could someone please help me with this problem?

Answer 1

So $1/s$ does not have a divergence. It cannot have a divergence. Divergences are a property of vector fields, it is a scalar field.

I think you may have meant $\vec F(s, \varphi, z) = \hat s / s = (x\hat x + y \hat y)/(x^2 + y^2)$, which has the nice property that it has zero divergence if $s\ne 0.$ If so, then I think this is an an answerable question.

But first I also have to correct a misconception (you may not have it but many students do): the divergence of $\hat r/r^2$ is not $4\pi \delta(r)$. It is, rather, $4\pi ~\delta_3\big(\vec r\big).$ This is a 3D Dirac $\delta$-function which looks for whether a volume includes the origin point $\vec 0$ in it, so we can write this in Cartesian coordinates as $\delta(x)~\delta(y)~\delta(z)$. This is distinct from the 1D Dirac $\delta$ function applied to the spherical coordinate $r$, $\delta(r).$ For one reason why this matters, it matters because we have a mathematical ambiguity about what $\int_0^X dx~\delta(x)$ should be, should it be 0 or 1 or maybe $1/2$ in between? The starting point of the integral is not clear on whether it contains that point at $x=0$ completely or halfway or not at all. And this is important here because all of our radial integrals start at 0 so if you choose $1/2$ as a nice intermediary then you would find that because the angular integral gives you $4\pi$ of solid angle already, the divergence of $\hat r / r^2$ is actually something like $2 \delta(r)$ when understood in that way. But it's just not a great way to understand the problem.

Now this misconception becomes important for us, because you are about to encounter a 2D Dirac $\delta$-function in 3D space and then it becomes very hard to denote what even we're talking about. To be clear, we are talking about a scalar field that in Cartesian coordinates would be $2\pi ~\delta(x)~\delta(y).$ You have the right prefactor. But as for how you would clearly denote that you want a 2D $\delta$ function within the 3D space, I am not sure... Possibly you could write $$\nabla\cdot \vec F = 2\pi~\int_{-\infty}^{\infty} dz_0~\delta_3\big(\vec r - z_0~\hat z\big),$$ using superposition to give a clear sense of “here is exactly what I mean.” Because if you were to multiply that by some $f(\vec r)$ and then $\iiint d^3 r$ over some volume, that sort of expression should ultimately reduce it down to $\int_a^b dz_0~f(0, 0, z_0)$ as needed when all of the commutations are done. Or maybe we just invent the non-square matrix $$\mathbf P_z = \begin{bmatrix}1&0&0\\0&1&0\end{bmatrix}$$ and simply write $\nabla\cdot\vec F = 2\pi~\delta_2\big(\mathbf P_z~\vec r \big)$ for it, or so.

So if you are very careful about how you write $\delta_2$ then what you have said is correct, but if you are not very careful and you just write it as $\delta$ then people may misread you as talking about a 1D Dirac $\delta$ rather than a 2D Dirac $\delta$ and they would be very confused.