In David J. Griffiths's Introduction to Electrodynamics, the author gave the following problem in an exercise.
Sketch the vector function $$ \vec{v} ~=~ \frac{\hat{r}}{r^2}, $$ and compute its divergence, where $$\hat{r}~:=~ \frac{\vec{r}}{r} , \qquad r~:=~|\vec{r}|.$$ The answer may surprise you. Can you explain it?
I found the divergence of this function as $$ \frac{1}{x^2+y^2+z^2} $$ Please tell me what is the surprising thing here.
Pretty sure the question is about $\frac{\hat{r}}{r^2}$, i.e. the electric field around a point charge. Naively the divergence is zero, but properly taking into account the singularity at the origin gives a delta-distribution.
I have the same book, so I take it you are referring to Problem 1.16, which wants to find the divergence of $\frac{\hat{r}}{r^2}$.
If you look at the front of the book. There is an equation chart, following spherical coordinates, you get $\nabla\cdot\vec{v} = \frac{1}{r^2}\frac{\mathrm{d}}{\mathrm{d}r}\left (r^2 v_r\right) + \text{ extra terms}$. Since the function $\vec{v}$ here has no $v_\theta$ and $v_\phi$ terms the extra terms are zero. Hence $\nabla\cdot\vec{v} = \frac{1}{r^2}\frac{\mathrm{d}}{\mathrm{d}r}\left(r^2 \frac{1}{r^2}\right) = \frac{1}{r^2}\frac{\mathrm{d}}{\mathrm{d}r}\left(1\right) = 0$.
At least this is how I interpret the surprising element of the question.
For me another surprising thing about this question was that the divergence was not negative, seeing as the flow decreases as we move radially outwards. I found an excellent explanation of this here:
