# Why is $\nabla\cdot(\hat{\bf r}/r^2)$ giving 0 as answer? [closed]

While I was reading I encountered the statement $\nabla\cdot(\hat{\bf r}/r^2)$ (r cap divided by $r$ square) is 0. Can anyone explain proof of the statement why is it giving 0?

• There are proofs in most E&M books, such as those by Griffiths or Jackson. – rob Nov 22 '14 at 18:39
• Try to write it explicitly in $x,y,z$ components and calculate. It is very instructive and you'll get 0 everywhere, except for badly divergent quantity for $r=0$. – Fedxa Nov 22 '14 at 18:42
• @identicon: In how many dimensions? – Qmechanic Nov 22 '14 at 18:48
• @identicon: Remember you are not doing dot product over there (it seems to be a dot product)! [citation required]. While solving just see whether you are multiplying the magnitudes of del and other vector or just differentiating. Good luck. – Immortal Player Nov 22 '14 at 19:10
• This probably should be asked on Math.SE, as it's really a mathematics question. – Kyle Kanos Nov 23 '14 at 0:59

Indeed the answer is not zero but $-4\pi\delta(r)$ (Dirac delta function). The formula of divergence can be found in any standard textbook on mathematical physics, for example chapter 2 of Mathematical methods for physicists by Arfken. But since this function is singular at $r=0$ we must be careful. At any other points is easy to calculate it. It is proportional to $\frac{\partial}{\partial r} (r^2 V_r)$ where $V_r$‌ is the component of vector $\vec V$ along $\hat r$. So the answer is zero for $r\neq 0$.
Consider the vector function $$\vec{a}=\frac{1}{r^2}\hat{r}$$ At every location $\vec{a}$ is directed radially outward ; if ever there was a function that ought to have a large positive divergence, this is it. and yet, when you actually calculate the divergence, you will get $$\nabla.\vec{a}=\frac{1}{r^2}\frac{d}{dr}\left(r^2 \frac{1}{r^2}\right)=\frac{1}{r^2}\frac{d}{dr}(1)=0$$
note: $\nabla$ is in spherical coordinate .