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In classical textbooks, like "Introduction to Electrodynamics" by J.D. Griffiths, it is given that $$\nabla\cdot\left(\frac{\widehat{r}}{r^2}\right)=4\pi\delta^3(R).$$

To prove this equality, Griffiths first evaluates the divergence and finds it $0$ for $r \neq 0$ and undefined for $r=0$. Then evaluates the integral of the given divergence and finds it $4\pi$. Then, he concludes that the above equality should hold since the integral has a constant value containing origin.

Now, I am asked about the following equality :

$$\nabla \cdot\left(\frac{\widehat{\rho}}{\rho}\right)=2\pi\delta^2(\rho)$$

(note: the equation is claimed to be valid in cylindrical coordinates).

I was unable to work out the integral properly. Any help would be appreciated.

Arnaud D.
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Buddy
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1 Answers1

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If one integrates the line integral of the normal component of $\frac{\hat \rho}{\rho}$ along the circle $\rho=\epsilon$, then one finds that

$$\oint_{\rho=\epsilon}\left(\frac{\hat \rho}{\rho}\right)\cdot d\vec\ell=\int_0^{2\pi}\left(\frac{\hat \rho}{\epsilon}\right)\cdot \hat \rho\,\epsilon\,d\phi=2\pi$$

Heuristically, we apply naively the divergence theorem (the conditions of the divergence theorem are not satisfied due to the singularity at $\rho=0$) and find for a smooth function $\phi(\vec \rho)$

$$\begin{align} \lim_{\epsilon\to 0}\int_0^{2\pi}\int_0^\epsilon \nabla\cdot\left(\frac{\hat\rho}{\rho}\right)\phi(\vec\rho)\,dS&\approx \phi(0) \lim_{\epsilon\to 0}\int_0^{2\pi}\int_0^\epsilon \nabla\cdot\left(\frac{\hat\rho}{\rho}\right)\,dS\\\\ &=\phi(0)\lim_{\epsilon\to0}\oint_{\rho=\epsilon}\left(\frac{\hat \rho}{\rho}\right)\cdot d\vec\ell\\\\ &=2\pi\phi(0) \end{align}$$

Hence, in distribution, we can write

$$\nabla\cdot\left(\frac{\hat \rho}{\rho}\right)\sim 2\pi \delta(\vec\rho)$$

To be precise, we begin with the regularization

$$\begin{align} \delta_a(\vec \rho)&=\nabla \cdot \left(\frac{\vec \rho}{\rho^2+a^2}\right)\\\\ &=\frac{2a^2}{(\rho^2+a^2)^2} \end{align}$$

Then, for any test function $\phi(\vec \rho)$ and domain $S\in \mathbb{R}^2$ we have

$$\lim_{a\to0}\int_S \phi(\vec \rho)\delta_a(\vec\rho)\,dS=\begin{cases}2\pi \phi(0)&,\vec\rho\in S\\\\0&,\vec\rho\notin S\end{cases}\tag1$$

To show that $(1)$ holds, we used an analogous approach of the analysis in THIS ANSWER for the $3$-D Dirac Delta.

Thus, it is in this sense that

$$\lim_{a\to0}\delta_a(\vec\rho)=2\pi\delta(\vec\rho)$$

Mark Viola
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