# Divergence of $\vec{F} = \frac{\hat{\mathrm{r}}}{r^{2}}$

Consider the vector field $$\vec{\mathrm{F}} = \frac{\hat{\mathrm{r}}}{r^{2}},$$ then the divergence of this field is: $$\vec{\nabla}\cdot\left(\frac{\hat{\mathrm{r}}}{r^{2}}\right) = 4\pi\delta^{3}(\mathrm{\vec{r}})$$

What is the proof of this relation?

• use the fact that $\text{div } F= \frac 1V \int F.n \, dS$
– abel
Jun 22, 2015 at 23:45
• @MohammedSalamaIbrah I provided a more rigorous version of the "proof" - one based on a common method used in the theory of distributions (Generalized Functions). Please let me know how I can improve the answer. I really just want to give you the best answer I can. Jun 23, 2015 at 4:00

A common way to show that $\nabla \cdot \left(\frac{\hat r}{r^2}\right)=4\pi \delta (\vec r)$ is to regularize the function $\left(\frac{\hat r}{r^2}\right)$ in terms of a parameter, say $a$. To that end, let $\vec \psi$ be the regularized function given by

$$\vec \psi(\vec r;a)=\frac{\vec r}{(r^2+a^2)^{3/2}} \tag 1$$

Taking the divergence of $(1)$ reveals that

$$\nabla \cdot \vec \psi(\vec r; a)=\frac{3a^2}{(r^2+a^2)^{5/2}}$$

Now, for any sufficiently smooth test function $\phi$, we have that

\begin{align} \lim_{a \to 0}\int_V \nabla \cdot \vec \psi(\vec r; a)\phi(\vec r)dV&=\lim_{a \to 0}\int_V \frac{3a^2}{(r^2+a^2)^{5/2}}\phi(\vec r)dV\\\\ &=0 \end{align}

if $V$ does not include the origin.

Now, suppose that $V$ does include the origin. Then, we have

\begin{align} \lim_{a \to 0}\int_V \nabla \cdot \vec \psi(\vec r; a)\phi(\vec r)dV&=\lim_{a\to 0}\int_{V-V_{\delta}} \frac{3a^2}{(r^2+a^2)^{5/2}}\phi(\vec r)dV+\lim_{a\to 0}\int_{V_{\delta}} \frac{3a^2}{(r^2+a^2)^{5/2}}\phi(\vec r)dV\\\\ &=\lim_{a\to 0}\int_{V_{\delta}} \frac{3a^2}{(r^2+a^2)^{5/2}}\phi(\vec r)dV \end{align}

where $V_{\delta}$ is a spherical region centered at $\vec r=0$ with radius $\delta$. For any $\epsilon>0$, take $\delta>0$ such that $|\phi(\vec r)-\phi(0)|\le \epsilon/(4\pi)$ whenever $0<|\vec r|< \delta$. Then, we have

\begin{align} \lim_{a \to 0}\left|\int_V \nabla \cdot \vec \psi(\vec r; a)(\phi(\vec r)-\phi(0))\,dV\right|&\le \lim_{a\to 0} \int_{V_{\delta}} \left|\phi(\vec r)-\phi(0)\right|\frac{3a^2}{(r^2+a^2)^{5/2}}dV\\\\ &\le \left(\frac{\epsilon}{4\pi}\,4\pi\right)\,\lim_{a \to 0}\int_{0}^{\infty}\frac{3a^2}{(r^2+a^2)^{5/2}}r^2\,dr\\\\ &\le \epsilon \end{align}

Thus, we have for any test function $\phi$,

\begin{align} \lim_{a \to 0}\int_V \nabla \cdot \vec \psi(\vec r; a)\phi(\vec r)\,dV&=4\pi \phi(0) \end{align}

and it is in this sense that

$$\bbox[5px,border:2px solid #C0A000]{\lim_{a\to 0} \nabla \cdot \vec \psi(\vec r;a)=4\pi \delta(\vec r)}$$

• @MichaelLevy What additional steps? If you have a specific questtion, I'd be happy to answer. Dec 1, 2023 at 14:54
• @MichaelLevy $\left|\int_V f(\vec r)\,dV\right|\le \int_V |f(\vec r)|\,dV\le \int_{V_\delta}| |f(\vec r)|\,dV$$Dec 1, 2023 at 19:46 • @MichaelLevy Is the sentence "where V_{\delta} is a spherical region centered at \vec r=0 with radius \delta." not explicit? Dec 2, 2023 at 16:40 What does "4\pi\delta^3(r)" mean? It means if you integrate any compactly supported test function f against it, the result will be 4\pi f(0). So what you really want to show is that for any compactly supported test function f, you have$$ \int \bigg(\nabla\cdot \frac{\hat{r}}{r^2}\bigg)f(r)\ dr = 4\pi f(0).$$Integrate by parts and take a limit around the singularity at$0$to get the answer. Can't really give "the" proof if I don't know what you're allowed to assume. But, to suit my personal taste I would start with the well known identity$ \int d^3x \mathbf{\nabla\cdot F} = \oint d^2x \mathbf{\hat{n}}\cdot\mathbf{F}$Insert$\mathbf{F}=\mathbf{\hat{r}}/r^2\$ and take it from there.

The below is a more detailed version of bob.sacamento's answer. I will use boldface for vector quantities.

Suppose we integrate over a sphere of radius $$R$$ centered at the origin. The surface integral is \begin{aligned} \oint \mathbf{F} \cdot d \mathbf{a} &=\int\left(\frac{1}{R^{2}} \hat{\mathbf{r}}\right) \cdot\left(R^{2} \sin \theta d \theta d \phi \hat{\mathbf{r}}\right) \\ &=\left(\int_{0}^{\pi} \sin \theta d \theta\right)\left(\int_{0}^{2 \pi} d \phi\right)=4 \pi \end{aligned} Using the divergence theorem: \begin{aligned} \oint \mathbf{F} \cdot d \mathbf{a} &=\int \nabla\cdot{\bf F}\, d{\mathbf{\tau}} \end{aligned} where $${\mathbf{\tau}}$$ is the volume differential. The above two equations imply that \begin{aligned} \int \nabla\cdot{\bf F}\, d{\mathbf{\tau}} &=4\pi\, \end{aligned} Therefore, \begin{aligned} \nabla\cdot\left(\frac{1}{r^{2}} \hat{\mathbf{r}}\right) &=4\pi\delta^3({\bf r}) \, . \end{aligned}

IMHO I think that all the other answers of this question are excessively complex. A very simple explanation, just based in the generalized Stokes theorem, is as follow: suppose that $$R$$ is an open, bounded and convex region in $$\mathbb{R}^3$$ and let $$\mathbf{F}:=\frac{\mathbf{\hat r}}{r^2}$$, then it follows from Stokes theorem that if $$\mathbf{0}\notin R$$ then $$\operatorname{div}\mathbf{F}$$ is well-defined and equal to zero at every point of $$R$$, then from Stokes theorem we have that

$$\int_{\partial R}\mathbf{F}\cdot d\mathbf{S}=\int_{R}\operatorname{div} \mathbf{F}\, dV=\int_{R}0\, dV=0\tag1$$

Now, to handle the case when $$\mathbf{0}\in R$$ we can exploit (1) and the simple result

$$\int_{\partial \mathbb{B}(\mathbf 0,\epsilon )}\mathbf{F} \cdot d\mathbf{S}=4\pi\tag2$$

what follows easily using spherical coordinates in $$\mathbb{B}(\mathbf 0,\epsilon )$$. Then as $$R$$ is open by assumption then there exists some $$\epsilon >0$$ such that $$\mathbb{B}(\mathbf{0},\epsilon )\subset R$$, therefore

$$0=\int_{R\setminus \mathbb{B}(\mathbf{0},\epsilon )}\operatorname{div} \mathbf{F}\,dV=\int_{\partial (R\setminus \mathbb{B}(\mathbf 0,\epsilon ))}\mathbf{F}\cdot d\mathbf{S}=\int_{\partial R}\mathbf{F}\cdot d\mathbf{S}-\int_{\partial \mathbb{B}(\mathbf 0,\epsilon )}\mathbf{F}\cdot d\mathbf{S}\\ \therefore\quad \int_{\partial R}\mathbf{F}\cdot d\mathbf{S}=4\pi\tag3$$

Then, to extend the Stokes theorem to the case where $$\mathbf{0}\in R$$ we can set $$\operatorname{div}\left(\frac{\mathbf{\hat r}}{r^2}\right):=4\pi \delta ^3$$ and with this definition we have that

$$\int_{\partial R}\mathbf{F}\cdot d\mathbf{S}=4\pi=\int_{R}\operatorname{div}\mathbf{F}\,dV\tag4$$