electrostatics - Method of image charge for cylindrical conductor

I am simply puzzled that only for spherical and planar conducting surfaces the method of images is applied. Is it (really) impossible to find image charge or charge distribution which can simulate the behaviour of potential in the volume of interest. Is there any method which may be used to find the image charge/charge distribution ?

Answer

The reason why the method of the images is easily applicable in the case of the sphere or the plane is that it uses the symmetries of the Laplace operator

$$\Delta\Phi=\frac{\partial^2\Phi}{\partial x^2}+\frac{\partial^2\Phi}{\partial y^2}+\frac{\partial^2\Phi}{\partial z^2}$$ or in the spherical coordinates $$\Delta\Phi=\frac{1}{r}\frac{\partial^2}{\partial r^2}(r\Phi)+\frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial\Phi}{\partial\theta}\right)+\frac{1}{r^2\sin^2\theta}\frac{\partial^2\Phi}{\partial\phi^2}$$

which appears in the equation on the potential

$$\Delta\Phi=-4\pi\rho$$

First you have the rotations, shifts and reflections. The common trait of all these tranformations are that they conserve the metric $ds^2=dx^2+dy^2+dz^2$. For them we have the following property,

$$\Delta \Phi({\bf x})=-4\pi\rho({\bf x})\Rightarrow \Delta \Phi({\bf x}')=-4\pi\rho({\bf x}')$$ where ${\bf x}'$ is a transformed coordinates as a function of non-transformed $\bf{x}$. So for example if you know the potential $\Phi(x,y,z)$ of the charge distribution $\rho(x,y,z)$ you also know that the potential $\Phi(x,y,-z)$ corresponds to the charge distribution $\rho(x,y,-z)$.

However we may expand this class including more general mapping - conformal transformations. They don't conserve the metric however they only multiply it on some function,

$$ds^2=dx^2+dy^2+dz^2\mapsto (ds')^2=g(x',y',z')^2\left((dx')^2+(dy')^2+(dz')^2\right)$$ So they don't conserve the distances but conserve the angles. They obviously include the rotations, shifts and reflections and in fact can all be represented as their combination with extra transformation called the inversion, $$r\mapsto \frac{R^2}{r}$$ Now for this tranformation the Laplace equation changes a bit trickier, $$\Delta\Phi(r,\theta,\phi)=-4\pi\rho(r,\theta,\phi)\Rightarrow \Delta\tilde{\Phi}(r,\theta,\phi)=-4\pi\tilde{\rho}(r,\theta,\phi),$$ where $$\tilde{\Phi}(r,\theta,\phi)=\frac{R}{r}\Phi\left(\frac{R^2}{r},\theta,\phi\right), \tilde{\rho}(r,\theta,\phi)=\left(\frac{R}{r}\right)^5\rho\left(\frac{R^2}{r},\theta,\phi\right)$$ For point charge at $r=a$ that means (thanks to the properties of the delta-function) that $\tilde{q}=q\frac{R}{a}$

Now how the image charge method works. You need to perform the conformal transformation that will exchange one side with another while conserving the form of the surface AND the potential on it. For plate it's easy - just use reflection. For sphere it's also easy - just use inversion. Hovewer for cylinder there's no such conformal transformation that will not rescale the $z$ coordinate! That means that there's no trivial way to obtain the charge distribution reproducing the field of the point charge inside the cylinder.

The only way I see is to compute it directly using the expansion of the Green function in terms of the Bessel function

$$\frac{1}{|\bf{x}-\bf{x}'|}=\frac{2}{\pi}\sum_{m=-\infty}^{+\infty}\int_0^{+\infty}dk e^{im(\phi-\phi')}\cos[k(z-z')]I_m(k\rho_<)K_m(k\rho_>)$$ However playing with it a bit now I couldn't reach any good answer. It will be good if you will be able to obtain the representation of $\rho$ as a series without any integrals. As result it's much easier just to compute the field of the point charge inside the cylinder without thinking about image charges at all.

On the other hand for the homogeneous line in the cylinder the field doesn't depend on $z$ so you can just use the inversion in the $xy$ plane without worrying about the rescaling of the $z$. That's why it's easy to find the image for the line but not for the point charge in the case of the cylinder.