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

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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

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

Answer 4

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, $$ \begin{aligned} \nabla\cdot\left(\frac{1}{r^{2}} \hat{\mathbf{r}}\right) &=4\pi\delta^3({\bf r}) \, . \end{aligned} $$

Answer 5

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

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

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