Laplacian of 1/r in Darwin term

[NanakiXIII]

The http://en.wikipedia.org/wiki/Fine_structure#Darwin_term" contains a (3D-)delta function as a result of taking the Laplacian of the Coulomb potential. I'm trying to find out why. I've been searching, and I've so far come across different views of the Laplacian of 1/r at the origin. Either it's considered zero, or

$$\nabla^2\frac 1 r = -\,\frac{\delta(r)}{r^2},$$

but I can't find any reference that says it's $\delta^3(r)$. It has the right units, but that's about all I can say about it. Could someone clarify this?

 

[samalkhaiat]

The relation

$$\nabla^{2}|\frac{1}{\vec{r}}| = -4\pi \delta^{3}(\vec{r})$$

is proved in post #10 of;

www.physicsforums.com/showthread.php?t=200580

In spherical coordinates, the delta function can be written as

$$\delta^{3}(\vec{r}) = \frac{1}{r^{2}} \delta (r) \delta (\phi) \delta (\cos \theta )$$

regards

sam

 

[NanakiXIII]

Ah, I see. I assumed $\delta(r)$ and $\delta(\vec{r})$ were pretty much the same thing, but I guess not. Thanks. Is there actually any point in including those delta functions in $\phi$ and $\theta$? The delta function in $r$ narrows things down to a single point.

 

[samalkhaiat]

I assumed $\delta(r)$ and $\delta(\vec{r})$ were pretty much the same thing,

Yes, thse are the same thing! However, $\delta(r)$ is different from $\delta^{3}(r)$.

Is there actually any point in including those delta functions in $\phi$ and $\theta$?

Well, the theta and the phi are there! you can't just leave them out. For certain potential you can integrate or average over the angular dependence. But, in general, potentials do depend on theta and phi.

sam