I am working on a problem in Jackson that asks me to show explicitly by separation of variables in polar coordinates that the green function can be expressed as Fourier series in azimuthal coordinate,
$G = \frac{1}{2 \pi} \sum_{-{\infty}}^{\infty} e^{i m (\phi - \phi')} g_m(\rho,\rho')$
where the radial green functions satisfy $\frac{1}{\rho'}\frac{\partial}{\partial \rho'}(\rho' \frac{\partial g_m}{\partial \rho'}) - \frac{m^2}{\rho'^2}g_m = -4 \pi \frac{\delta(\rho-\rho')}{\rho}$
I gather the problem wants me to literally do some kind of separation of variables.
Here is where I am taking this:
$\nabla^2 = \frac{1}{\rho'} \frac{\partial}{\partial \rho'}(\rho' \frac{\partial}{\partial \rho'}) - \frac{1}{\rho'^2} \frac{\partial}{\partial \phi'^2}$ in polar coordinates
Ok then I think the delta function is something like $\frac{1}{\rho}\delta(\rho -\rho')\delta{(\phi - \phi')}$
Ok so setting it up I want to guess that it should be something like
$\nabla^2 G = (\frac{1}{\rho'} \frac{\partial}{\partial \rho'}(\rho' \frac{\partial}{\partial \rho'}) - \frac{1}{\rho'^2} \frac{\partial}{\partial \phi'^2}) G = \frac{1}{\rho}\delta(\rho -\rho')\delta{(\phi - \phi')}$
where $G = \mathbf{R}(\rho)\Phi(\phi)$
It seems I would want to find a superposition of the homogeneous part and the part with a discontinuity.
At this point focusing on the homogeneous radial part, I want to set the radial part to a constant say $\nu^2$ as is usually done.
One gets something like
$\frac{1}{\rho'}\frac{\partial}{\partial \rho'}(\rho' \frac{dG}{d\rho'}) = \frac{m^2}{\rho'^2}$
$\frac{\rho}{R}\frac{\partial}{\partial \rho'}(\rho' \frac{dR}{d\rho'}) = \nu^2$
Ok I am thinking to myself that one needs to then do next is to try to find
$\frac{1}{\rho}\frac{\partial}{\partial \rho'}(\rho' \frac{dG}{d\rho'}) -\frac{m^2}{\rho'^2}G = \frac{1}{\rho}\delta(\rho -\rho')$
At which point I feel like throwing an integration sign on both sides
$\int \frac{1}{\rho}\frac{\partial}{\partial \rho'}(\rho' \frac{dG}{d\rho'}) -\frac{m^2}{\rho'^2}G) d \rho' = \int \frac{1}{\rho}\delta(\rho -\rho') d \rho'$
My hope was to try and get
i) the form $\frac{1}{\rho'}\frac{\partial}{\partial \rho'}(\rho' \frac{\partial g_m}{\partial \rho'}) - \frac{m^2}{\rho'^2}g_m = -4 \pi \frac{\delta(\rho-\rho')}{\rho}$ How does one get this form?
ii)Also, now that I think about it after all this typing it seems like $G = \frac{1}{2 \pi} \sum_{-{\infty}}^{\infty} e^{i m (\phi - \phi')} g_m(\rho,\rho')$ here is already of the form $G = \mathbf{R}(\rho)\Phi(\phi)$ and I just need to plug in and show that it works. Is that the case?
