The Laplacian of the potential q*exp(-r)/r

[Elder1994]

Hello, I have a problem where I'm supposed to calculate the charge distribution ρ. I need to calculate it by applying the Laplacian operator to the potential Θ. The potential is the function: q*exp(-αr)/r
I found on the internet that for this type of potentials I cannot just apply the Laplacian operator because the function is not differentiable as r approaches to 0. So I'm not sure what I'm supposed to do with this.

 

[jasonRF]

You have a potential that is the product of two functions, $f = \exp(-\alpha r)$ and $g = \frac{1}{r}$, and you need to calculate $\nabla^2 f g$. I would expand $\nabla^2 f g$, then plug in your specific $f$ and $g$.

 

[vanhees71]

Also carefully think about what $\Delta(1/r)$ is (of course, there's only a problem at the origin $r=0$). Also note that to answer this question, you should be very careful and NOT use spherical but Cartesian coordinates in your calculations since spherical coordinates are singular along the polar axis including the origin $r=0$!

 

[jasonRF]

vanhees71,

When I solved this, I used knowledge of electrostatics to know what $\nabla^2 \frac{1}{r}$ is, but used spherical coordinates for the remaining terms and it certainly yielded what I was expecting (this is the potential of a point charge in a hot plasma so we know the equation it should satisfy). The other terms include
$$\left( \nabla \frac{1}{r}\right) \cdot \nabla e^{-\alpha r}$$
and
$$ \nabla^2 e^{-\alpha r}$$
What is wrong with using spherical coordinates for those?

 

[vanhees71]

There's nothin a priori wrong. If you are lucky the coordinate singularities of spherical coordinates don't do any harm, but you can't express
$$\Delta \frac{1}{r}=-4 \pi \delta^{(3)}(\vec{x})$$
in terms of spherical coordinates, because there the coordinate singularities hit with maximal impact ;-)). I've seen this done wrong in otherwise respectable textbooks!

 

[jasonRF]

Well, the three dimensional delta function at the origin can certainly be expressed in Cartesian or spherical coordinates. Using your notation (assuming I understand it correctly), I get,
$$\delta^{(3)}(\overrightarrow{x}) = \delta(x)\delta(y)\delta(z) = \frac{\delta(r)}{4\pi r^2}$$

Perhaps we are getting off-top from the OP...