Question: I just started in Griffith's Introduction to electrodynamics and I stumbled upon the divergence of $\frac{ \hat r}{r^2} \equiv \frac{{\bf r}}{r^3}$, now from the book, Griffiths says:

[![enter image description here](https://i.stack.imgur.com/cxaLe.jpg)](https://i.stack.imgur.com/cxaLe.jpg)

Now what is the paradox, exactly? Ignoring any physical intuition behind this (point charge at the origin) how are we supposed to believe that the source of $\vec v$ is concentrated at the origin mathematically? Or are we forced to believe that because there was a contradiction with the divergence theorem?

Also how would the situation differ if $\vec v$ was the same vector function but not for a point charge? Or is it impossible?

Answer: 
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> Now what is the paradox, exactly?
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The paradox is that the vector field $\vec{v}$ considered obviously points away from the origin and hence seems to have a non-zero divergence, however, when you actually calculate the divergence, it turns out to be zero.

[![enter image description here](https://i.stack.imgur.com/MlEimm.jpg)](https://i.stack.imgur.com/MlEimm.jpg)

> 
> How are we supposed to believe that the source of $\vec v$ is concentrated at the origin mathematically?
> 
> 
> 

Most important point to observe is that $\nabla \cdot \vec v = 0$ everywhere except at the origin. The diverging lines appearing are from the origin. Our calculations cannot account for that since $\vec v$ blows up at $r = 0$. Moreover, eq. (1.84) is not even valid for $r = 0$. In other words, $\nabla \cdot \vec v \rightarrow \infty$ at that point.

However, if you apply the divergence theorem, you will find $$\int \nabla \cdot \vec v \ \text{d}V = \oint \vec v \cdot \text{d}\vec a = 4 \pi$$
Irrespective of the radius of a sphere centred at the origin, we must obtain the surface integral as $4 \pi$. The only conclusion is that this must be contributed from the point $r = 0$.

This serves as the motivation to define the Dirac delta function: a function which vanishes everywhere except blowing up at a point and has a finite area under the curve.
I am not sure I can answer the question exactly how you meant it, but I can give you some things to think about.

Mathematically, the peculiarity of this situation is caused by the fact that the function is defined on $\mathbb{R}^3- \{0\}$, which is homeomorphic to a sphere, whose second (de Rham) cohomology group is $\mathbb{R}$. Hence you can have closed 2-forms that are not exact. The flux form associated to your vector field is precisely one of these forms. 

Now, you are supposedly in a second year electromagnetism course, I guess? So you probably don’t know the meaning of what I just wrote. Let me put it this way. If you saw complex analysis already, all this is just kind of the residue theorem. If you integrate on a closed loop, you get zero if there’s nothing weird happening inside, or (possibly) non zero if the function diverges somewhere inside the loop, i.e. you have a pole. This is exactly the same thing, but in 3 dimensions, with closed surfaces instead of closed loops, and with flux integrals instead of complex integrals!