Why is the divergence of $\widehat{r}/r^2$ equal to $0$?

Question

I have read that $\nabla\cdot\dfrac{\widehat{r}}{r^2}$ is equal to $0$.

But I cannot understand why. I tried but I cannot solve it. Can anyone explain it please?

Answer 1

If you put it into spherical coordinates (link to Wikipedia article "Del in cylindrical and spherical coordinates"), $$ \nabla\cdot{\hat{r}\over r^2}=1/r^2{\partial\over\partial r} {r^2\hat{r}\over r^2} $$ $$ =1/r^2{\partial\over\partial r} \hat{r} $$ Assuming $r\neq 0$, this is zero because $\hat{r}$ doesn't change with $r$.

Answer 2

This is not zero everywhere. Use the integral definition of divergence. Let $M$ be a volume centered on a point $p$. Let $V$ be the volume of this region.

$$\left. \nabla \cdot \frac{\hat r}{r^2} \right|_p = \lim_{V \to 0} \frac{1}{V} \oint_{p \in M} \frac{\hat r}{r^2} \cdot \hat n \, dS$$

In particular, when $p$ is the origin, use spherical coordinates and a spherical integration volume. Then $\hat n = \hat r$ and $dS = r^2 \sin \theta \, d\theta \, d\phi$, and we get

$$\left . \nabla \cdot \frac{\hat r}{r^2} \right|_\vec{0} = \lim_{R \to 0} \frac{4\pi R}{4\pi R^3/3} $$

which diverges. This only happens at the origin. At all other points, you can use the differential form of the divergence and get the right answer--which is zero.

Answer 3

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{}$ \begin{align} \color{#66f}{\large\nabla\cdot\pars{\hat{r} \over r^{2}}}& =\nabla\cdot\pars{{1 \over r^{3}}\,\vec{r}} =\nabla\pars{1 \over r^{3}}\cdot\vec{r} + {1 \over r^{3}}\,\nabla\cdot\vec{r} =\bracks{\hat{r}\,\totald{}{r}\pars{1 \over r^{3}}}\cdot\vec{r} + {1 \over r^{3}}\times 3 \\[5mm]&=\bracks{{\vec{r} \over r}\pars{-\,{3 \over r^{4}}}}\cdot\vec{r} +{3 \over r^{3}} =-\,{3 \over r^{3}}+ {3 \over r^{3}}=\color{#66f}{\Large 0}\,,\qquad \vec{r} \not= 0 \end{align}