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

## Homework Statement

Given $\nabla\frac{1}{r}$, show $\nabla\bullet\nabla\frac{1}{r}$ = -4πδ(r), where δ(r) is the delta dirac function.

## 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.

## Answers and Replies

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Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

Given $\nabla\frac{1}{r}$, show $\nabla\bullet\nabla\frac{1}{r}$ = -4πδ(r), where δ(r) is the delta dirac function.

## 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.
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

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
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.

Ray Vickson
Homework Helper
Dearly Missed
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.
Just read the link; it's all there.

As for "rigour", well, that is another matter altogether, and it concerns the undeniable fact that δ(r) is not really a legitimate function at all---it is a so-called "distribution", or "generalized function". There are lots of different approaches to defining distributions and to proving results about them. The first few pages of http://www.math.umn.edu/~olver/pd_/gf.pdf [Broken]
deal with the basic issues involved in the 1-dimensional case. Some of the same basic ideas carry over to the 3-dimensional case you are dealing with. For more material, Google "generalized functions".

RGV

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