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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?

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    $\begingroup$ There are proofs in most E&M books, such as those by Griffiths or Jackson. $\endgroup$ – rob Nov 22 '14 at 18:39
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    $\begingroup$ 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 $r=0$. $\endgroup$ – Fedxa Nov 22 '14 at 18:42
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    $\begingroup$ @identicon: In how many dimensions? $\endgroup$ – Qmechanic Nov 22 '14 at 18:48
  • $\begingroup$ @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. $\endgroup$ – Immortal Player Nov 22 '14 at 19:10
  • $\begingroup$ This probably should be asked on Math.SE, as it's really a mathematics question. $\endgroup$ – Kyle Kanos Nov 23 '14 at 0:59
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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 $r\neq 0$.

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  • $\begingroup$ Welcome to PSE! Jumped from MSE, ha! $\endgroup$ – Immortal Player Nov 22 '14 at 19:11
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This proof from Griffiths book introduction to electrodynamics

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 .

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  • $\begingroup$ Can you explain the proof from griffith $\endgroup$ – identicon Nov 23 '14 at 2:35

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