What is the Laplacian of the scalar potential with an extra term?

[_Andreas]

On page 35 of Jackson's Classical Electrodynamics, he calculates the Laplacian of a scalar potential due to a continuous charge distribution. In the expression for the potential, the operand of the Laplacian is

\frac{1}{|r-r'|},

where r is the the point where the potential is to be evaluated and r' the location of the source.

Homework Statement

Jackson says that, "Because it turns out that the resulting
integrand is singular, we invoke a limiting procedure." He thus adds an extra term, a^2, in the following way:

Homework Equations

\frac{1}{\sqrt{(r-r')^2+a^2}}

Now, what is the Laplacian of this expression? According to Jackson, it's

\frac{3a^2}{((r-r')^2+a^2)^{\frac{5}{2}}}

The Attempt at a Solution

According to my own calculations, the Laplacian (the second derivative) of the above function with respect to |r-r'| is

\frac{2|r-r'|-a^2}{((r-r')^2+a^2)^{\frac{5}{2}}}.

I've checked this result with the help of wolframalpha, and it's correct. Am I using the wrong variable for differentiation, or what?

 

[henry_m]

Hi,
Careful: the Laplacian in spherical coordinates is not simply the sum of second derivatives. Check http://en.wikipedia.org/wiki/Laplace_operator" , up to a sign.

 

[_Andreas]

Hi,
Careful: the Laplacian in spherical coordinates is not simply the sum of second derivatives. Check http://en.wikipedia.org/wiki/Laplace_operator" , up to a sign.

Ah, yes, of course he's using spherical coordinates! *facepalm*

Thanks, henry!

 

[_Andreas]

Actually, there are some things I still don't understand here. Isn't the term r in the Laplacian given in spherical coordinates the same as the term r showing up in the expression for the infinitesimal volume element when you go from cartesian to spherical coordinates? And isn't that term r the distance from the origin to the volume element, rather than the distance between the field point and source point?

 

[Dick]

Actually, there are some things I still don't understand here. Isn't the term r in the Laplacian given in spherical coordinates the same as the term r showing up in the expression for the infinitesimal volume element when you go from cartesian to spherical coordinates? And isn't that term r the distance from the origin to the volume element, rather than the distance between the field point and source point?

Yes. r and r' are not intended to be spherical coordinate r values. Put r=(x,y,z) and r'=(x',y',z') and evaluate the laplacian in the usual way.

 

[_Andreas]

Yes. r and r' are not intended to be spherical coordinate r values. Put r=(x,y,z) and r'=(x',y',z') and evaluate the laplacian in the usual way.

I don't quite see what you mean. Should I not use the Laplacian expressed in spherical coordinates after all? Could you elaborate?

 

[Dick]

I don't quite see what you mean. Should I not use the Laplacian expressed in spherical coordinates after all? Could you elaborate?

I'm saying that you can use any coordinates you like. Put one point at the origin. Now your spherical expression is 1/sqrt(r^2+a^2). You could also write it as 1/sqrt(x^2+y^2+z^2+a^2) in cartesian coordinates.

 

[_Andreas]

I'm saying that you can use any coordinates you like. Put one point at the origin. Now your spherical expression is 1/sqrt(r^2+a^2). You could also write it as 1/sqrt(x^2+y^2+z^2+a^2) in cartesian coordinates.

Ok, I think I finally get it. I've been mixing up the variable of integration and the variable of differentiation completely*. Thanks.

*"With respect to |r-r'|", as I wrote in the OP, is wrong. The differentiation is with respect to r.