1

I have been given a charge density in 2 dimensions.

$\sigma (\rho, \phi) = \frac{\lambda}{a} \sum_{n = 0}^3 (-1)^n \delta(\rho - a) \delta( \phi - \frac{n \pi}{2}) $

I have also been reminded about a Green function from a previous problem

$G(\rho,\phi;\rho',\phi') = \ln(\rho_{>}^2) + 2 \sum_{m=1}^\infty \frac{1}{m} (\frac{\rho_{<}}{\rho_{>}})^m \cos[m(\phi - \phi')]$

The goal is to try to show the electrostatic potential is

$\phi(\rho,\phi) = \frac{\lambda}{\pi \epsilon_0} \sum_{k=0}^\infty \frac{1}{2k + 1}(\frac{\rho_<}{\rho_>})^{4k+2} \cos[(4k + 2)\phi]$

Ok so here is what I am trying

I recalled that

$\phi(x) = \frac{1}{4 \pi \epsilon_0}\int_v \rho(x') G(x,x')d^3x' + \frac{1}{4 \pi} \oint_s[G(x,x') \frac{\partial \phi}{\partial n'} - \phi(x')\frac{\partial G(x,x')}{\partial n}] da'$

my guess (someone please confirm or refute this guess) from physical reasoning is that this should collapse to

$\phi(x) = \frac{1}{4 \pi \epsilon_0} \oint \sigma(x') G(x,x')da'$

This compels me to write the problem in question to

$\phi(x) = \int \frac{\lambda}{a} \sum_{n = 0}^3 (-1)^n \delta(\rho - a) \delta( \phi - \frac{n \pi}{2}) [\ln(\rho_{>}^2) + 2 \sum_{m=1}^\infty \frac{1}{m} (\frac{\rho_{<}}{\rho_{>}})^m \cos[m(\phi - \phi') ]\rho' d \rho' d \phi' $

or

$\phi(\rho,\phi)=\int \frac{\lambda}{a} \sum_{n = 0}^3 (-1)^n [\ln(\rho_{>}^2) + 2 \sum_{m=1}^\infty \frac{1}{m} (\frac{\rho_{<}}{\rho_{>}})^m \cos[m(\phi - \phi') ]\rho' d \delta(\rho - a) \delta( \phi - \frac{n \pi}{2})\rho' d \phi'$

Now I don't know if I setup the problem correctly. If I did it is left to try to show that it is indeed going to result in what we expect, in which case I am not entirely sure how to proceed. The delta functions make it look promising though.

Kevin Njokom
  • 581
  • 2
  • 12

1 Answers1

2

Note that we have

$$\begin{align} 4\pi \varepsilon_0\Phi(\rho,\phi)&=\langle \sigma,G_{\rho,\phi} \rangle \\\\ &=\lambda \sum_{n=0}^3 (-1)^n \langle \delta_{a,n\pi/2},G_{\rho,\phi}\rangle\\\\ &=\lambda \sum_{n=0}^3 (-1)^n G(\rho,\phi;a,n\pi/2)\\\\ &=2\lambda \sum_{n=0}^3 (-1)^n\sum_{m=1}^\infty \frac1m\left(\frac{\rho_<}{\rho_>}\right)^m \cos(m(\phi-n\pi/2))\\\\ &=2\,\lambda \sum_{m=1}^\infty \frac{1}m \left(\frac{\rho_<}{\rho_>}\right)^m \left(1+(-1)^m\right)\left(\cos(m\phi)-\cos(m(\phi-\pi/2))\right)\\\\ &=2\,\lambda \sum_{m=1}^\infty \frac{1}m \left(\frac{\rho_<}{\rho_>}\right)^{2m} \left(\cos(2m\phi)-\cos(2m(\phi-\pi/2))\right)\\\\ &=2\,\lambda \sum_{m=1}^\infty \frac{1}m \left(\frac{\rho_<}{\rho_>}\right)^{2m}\left(1-(-1)^m\right) \cos(2m\phi)\\\\ &=4\,\lambda \sum_{m=1}^\infty \frac{1}{2m-1} \left(\frac{\rho_<}{\rho_>}\right)^{2(2m-1)} \cos(2(2m-1)\phi)\\\\ &=4\,\lambda \sum_{m=0}^\infty \frac{1}{2m+1} \left(\frac{\rho_<}{\rho_>}\right)^{2(2m+1)} \cos(2(2m+1)\phi)\\\\ \Phi(\rho,\phi)&=\frac\lambda {\pi\varepsilon_0} \sum_{m=0}^\infty \frac{1}{2m+1} \left(\frac{\rho_<}{\rho_>}\right)^{2(2m+1)} \cos(2(2m+1)\phi)\\\\ \end{align}$$

as expected!

Mark Viola
  • 179,405