# Understanding Green's function in the mirror charge method

Initially I want to point out that there are many way as to how we can define the green's function in vacuum or when other objects are taken into consideration, and the same goes for the Potential in the case when we consider conducting objects, or dielectric etc. So I want to make it clear what type of notation I use, in order not to have a confusion down the line.

In vacuum, the potential of a charge distribution is: $$\Phi(\vec r)=\frac 1 {4\pi \epsilon_0}\int_V \frac {\rho(\vec r')}{|\vec r - \vec r'|}dV'.$$

I consider as the Green function of vacuum: $$G_0(\vec r, \vec r')= \frac 1 {4\pi \epsilon_0}\frac {1}{|\vec r - \vec r'|}.$$

The Potential when we consider charge distributions and also other objects in space such as conductors or etc is, as I write it:

$$\Phi(\vec r)=\int_V G(\vec r)\,\rho(\vec r')\,dV' + \Sigma_i\Phi_i\Gamma_i (1)$$

$$\Gamma_i = -\epsilon_0\int_{(V_i)} \frac {\partial G(\vec r,\vec r')}{\partial \vec n'}\,d\vec f'$$ where the normal $$\vec n'$$ to the surface of the object points inwards.

The above expression for the potential, from what I was able to understand can be derived when we use the 2nd Green's identity.

But you can also find the above expressions as:

1. $$G_0(\vec r, \vec r')=\displaystyle -\frac 1 {4\pi}\frac {1}{|\vec r - \vec r'|}$$
2. $$\Phi(\vec r)=-\displaystyle \frac 1 {\epsilon_0}\int_V G(\vec r)\, \rho(\vec r')\,dV' + \Sigma_i\Phi_i\Gamma_i$$

The exercise is the following:

Consider a hollow sphere of radius a with the surface potential U (θ, φ) ≡ ϕ (a, θ, φ) (Dirichlet boundary condition). There are no charges either outside or inside the sphere.

1.Determine the corresponding Green's function G by placing a charge q inside the sphere and placing a mirror charge $$q_b$$ outside the sphere, so that the necessary boundary conditions are met.

2.Show that the potential for $$|\vec r|=r has the following form: (not important for what I want).

So initially I found the with the method of electric mirror charges the green function for this set up:

$$G(\vec r)\,\rho(\vec r') = \frac 1 {4\pi \epsilon_0}\left[\frac 1{|\vec r - \vec r'|} - \frac {a/r'}{|\vec r - (a^2/{r'}^2)\,\vec r'|}\right]$$.

With the other notation you would get:

$$G(\vec r)\,\rho(\vec r') = -\frac 1 {4\pi} \left[\frac 1{|\vec r - \vec r'|} - \frac{a/r'} {|\vec r - (a^2/{r'}^2)\,\vec r'|}\right]$$.

In any case, if you find the potential outside the sphere, you would get the same result.

The problem I am having is solving the 2nd part of this exercise:

In order to find the potential inside the hollow sphere I will use (1). The exercise has a solution and it uses the same equation but the difference is that in the solution it is said:

"Since the interior of the sphere is free of charge, $$\rho(\vec r')=0$$ applies and the first term vanishes." Therefore we have:

$$\Phi(\vec r)=\Sigma_i\Phi_i\Gamma_i=-\epsilon_0\int_{V_i} U (θ,φ)\,\frac {\partial G(\vec r,\vec r')}{\partial \vec n'}\,d\vec f'$$ This is the thing I don't understand. In the exercise it is said that we have no charge inside and outside the hollow sphere, but then in part 1, we put a charge inside the sphere and we also have the mirror charge outside. So how exactly is the interior of the sphere free of charge when in the first part it is clearly said that we put a point charge in the interior. I know it's a lengthy thread, but I hope someone can help me, as I feel this is a fundamental problem that I need to understand.