calculushomework-and-exercisesVector Fields
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.
Both identities are correct. The point is that they refer to different bases. The former uses the decomposition, $$\vec{V} = V^r \partial_r + V^\theta \partial_\theta + V^z \partial_z\:,$$ the latter uses the decomposition, $$\vec{V} = V'^r \vec{e}_r + V'^\theta \vec{e}_\theta + V'^z \vec{e}_z\:,$$ where $\vec{e}_i = \frac{1}{\sqrt{g_{ii}}} \partial_{x^i}$ are unit vectors, so that $V'^i = \sqrt{g_{ii}} V^i$ (there is no sum over the repeated index $i$).
Now what is the paradox, exactly?
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.
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.
Best Answer
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.