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

  • Thread starter bladesong
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Homework Statement


Given [itex]\nabla\frac{1}{r}[/itex], show [itex]\nabla\bullet\nabla\frac{1}{r}[/itex] = -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

  • #2
Ray Vickson
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Homework Statement


Given [itex]\nabla\frac{1}{r}[/itex], show [itex]\nabla\bullet\nabla\frac{1}{r}[/itex] = -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
http://www.deepspace.ucsb.edu/wp-content/uploads/2010/08/033_Chapter-22-Flux-and-Gauss-Law-PML.pdf .

RGV
 
  • #3
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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
http://www.deepspace.ucsb.edu/wp-content/uploads/2010/08/033_Chapter-22-Flux-and-Gauss-Law-PML.pdf .

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.
 
  • #4
Ray Vickson
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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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