- #1
_Andreas
- 144
- 1
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
[tex]\frac{1}{|r-r'|},[/tex]
where r is the the point where the potential is to be evaluated and r' the location of the source.
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:
[tex]\frac{1}{\sqrt{(r-r')^2+a^2}}[/tex]
Now, what is the Laplacian of this expression? According to Jackson, it's
[tex]\frac{3a^2}{((r-r')^2+a^2)^{\frac{5}{2}}}[/tex]
According to my own calculations, the Laplacian (the second derivative) of the above function with respect to |r-r'| is
[tex]\frac{2|r-r'|-a^2}{((r-r')^2+a^2)^{\frac{5}{2}}}.[/tex]
I've checked this result with the help of wolframalpha, and it's correct. Am I using the wrong variable for differentiation, or what?
[tex]\frac{1}{|r-r'|},[/tex]
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
[tex]\frac{1}{\sqrt{(r-r')^2+a^2}}[/tex]
Now, what is the Laplacian of this expression? According to Jackson, it's
[tex]\frac{3a^2}{((r-r')^2+a^2)^{\frac{5}{2}}}[/tex]
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
[tex]\frac{2|r-r'|-a^2}{((r-r')^2+a^2)^{\frac{5}{2}}}.[/tex]
I've checked this result with the help of wolframalpha, and it's correct. Am I using the wrong variable for differentiation, or what?