Divergence of 1/r^2; delta dirac's role

1. Sep 26, 2012

1. The problem statement, all variables and given/known data
Given $\nabla\frac{1}{r}$, show $\nabla\bullet\nabla\frac{1}{r}$ = -4πδ(r), where δ(r) is the delta dirac function.

3. The attempt at a solution
I've used divergence theorem and also solved the equation itself, so I know that outright solving is zero and the divergence theorem gives -4π. But I'm not sure how to show the presence, or rather, how I get the delta dirac function. I understand its role and but not necessarily why it needs to be there. Any help on direction on this one would be greatly appreciated.

2. Sep 26, 2012

Ray Vickson

Think of $\nabla \frac{1}{r}$ as an electrical field $\mathbf{E}$, so that $\nabla \cdot \mathbf{E}$ is its divergence. Now use Gauss' Law

RGV

3. Sep 26, 2012

Thank you for your prompt reply - this also makes it a little clearer. Seeing as this is a general mathematical property, is it simply enough to state this or is there some more rigorous proof needed?

i.e., I see now why at 0 it has the delta dirac, but not why ithe -4pi portion. This is the result of the integral - how does it fit back into the original divergence? Please pardon my ignorance - it's been a while.

4. Sep 26, 2012