Dirac definitions...

[sanctus]

I have once again a little black out: why is it that (in three dimensions)

[math]

\nabla^2 (\frac{1}{\vert \vec{r}\vert})=-4\pi\delta(\vec r)

[/math]

?

 

I mean it is easily verifiable that one actually gets zero if r has a norm different from zero and tends to infinity if the norm is zero but where dos the [imath] -4\pi[/imath] come from (the minus mainly, since I think the 4 \pi comes from the definition via the fourier transform) ? I know it is something simple and quite evident, but don't find it atm.

 

Thanks

[Essay]

I suppose a negative sigma would be too simple?

Oh well....

 

ScienceDirect - Physics Letters B : Hyperkähler metrics from (4,0) superspace*1

 

Abstract

We construct (4,0) supersymmetric non-linear sigma models using the superspace formalism. An explicit expression for hyperkähler metrics in terms of quaternionic structures results. These metrics are vacuum solutions to Einstein's equation, satisfy the tree level equations of motion of the superstring and are analogs of the Calabi-Yau manifolds favored for superstring compactification.

 

Keep on Spinoring

:hihi:

[sanctus]

Although, if I could read more than the abstract, it would be very interesting, it doesn't really help me...

 

I also do not understand why you connect to the sigma, since I ask about the delta-dirac...

[Erasmus00]

I have once again a little black out: why is it that (in three dimensions)

[math]

nabla^2 (frac{1}{vert vec{r}vert})=-4pidelta(vec r)

[/math]

?

 

I mean it is easily verifiable that one actually gets zero if r has a norm different from zero and tends to infinity if the norm is zero but where dos the [imath] -4pi[/imath] come from (the minus mainly, since I think the 4 pi comes from the definition via the fourier transform) ? I know it is something simple and quite evident, but don't find it atm.

 

If you take the fourier transform [imath] \nabla \to ik[/imath]. Two i s and you get a negative sign to match the right side.

 

Its also useful to think of the left hand side as Poisson's equation for a point charge, and the right hand side as the charge density of a point charge.

-Will

[sanctus]

Thanks Erasmus! All clear now.