TransWikia.com

Correct expression for multipole expansion

Physics Asked on August 18, 2021

Panofsky and Philips state the multipole expansion of a potential due to a volume charge distribution $rho$ in a finite volume $V$ given by $phi(mathbf R)=frac{1}{4piepsilon_0}int_Vfrac{rho(mathbf r’)}{|mathbf R-mathbf r’|} d^3mathbf r’$ as
$$
rho (mathbf R) = frac{1}{4piepsilon_0}left{ frac{1}{R} int_Vrho(mathbf r’)d^3mathbf r’
+ left[ frac{partial}{partial x’_alpha}left(frac{1}{r}right) right]_{r=R} int_Vx’_alpharho(mathbf r’)d^3mathbf r’
+frac{1}{2!}left[ frac{partial^2}{partial x’_alpha partial x’_beta}left(frac{1}{r}right) right]_{r=R} int_Vx’_alpha x’_betarho(mathbf r’)d^3mathbf r’+cdots
right},
$$

where by $r$ they mean $|mathbf{R-r’}|$.

However, I don’t understand their notation. So I tried to write an expression on my own. I get the following expansion.
$$
rho(mathbf R)=frac{1}{4piepsilon_0}
left{
frac{1}{R}int_Vrho(mathbf r’)d^3mathbf r’
+
frac{partial f}{partial x_alpha} (mathbf R) int_Vx’_alpharho(mathbf r’)d^3mathbf r’
+
frac{1}{2!}frac{partial^2 f}{partial x’_alpha partial x’_beta}(mathbf R) int_Vx’_alpha x’_betarho(mathbf r’)d^3mathbf r’+cdots
right},
$$

where $f(mathbf r):=1/|mathbf r|$.

I just to to confirm if I’m correct.

Add your own answers!

Ask a Question

Get help from others!

© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP