Electrostatics – Question About the Green’s Function for a Conducting Sphere

electrostaticsgreens-functionspotential

In Jackson's Classical Electrodynamics, he gives the Green's function for a conducting sphere with Dirichlet boundary conditions as
$$ G(\mathbf{x},\mathbf{x}^\prime) = \frac{1}{|\mathbf{x} – \mathbf{x}^\prime|} – \frac{a}{x^\prime\left|\mathbf{x} – \frac{a^2}{x^{\prime2}}\mathbf{x}^\prime\right|}. $$
The way this Green function was obtained was from using the method of images on a grounded (zero potential at the surface) conducting sphere with a point charge outside the sphere at position $\mathbf{x}^\prime$. The second term then corresponds to the image charge.

So this Green function was obtained using a specific case, i.e. a conducting sphere with a point charge outside, and with zero potential at the surface. What I understand is that the Green function is specific to the boundary conditions imposed. So I would imagine that if a different boundary condition at the surface of the sphere is imposed (e.g. a constant non-zero potential at the surface of the sphere), a different Green function would correspond to it, in particular one that would correspond to the potential solved by Jackson using the method of images for the case of a conducting sphere held at a constant potential $V$:
$$\Phi(\mathbf{x}) = \frac{1}{4\pi\epsilon_0}\left[\frac{q}{|\mathbf{x}-\mathbf{y}|}-\frac{aq}{y\left|\mathbf{x}-\frac{a^2}{y^2}\mathbf{y}\right|}\right] +\frac{Va}{|\mathbf{x}|},$$
where $a$ is the radius of the sphere and $\mathbf{y}$ is the position of the point charge.

However, Jackson argues that the Green function above (1st equation) would correspond to a conducting sphere for any Dirichlet boundary condition. Why is this the case?

Best Answer

What I understand is that the Green function is specific to the boundary conditions imposed.

No. The Green function is independent of the specific boundary conditions of the problem you are trying to solve. In fact, the Green function only depends on the volume where you want the solution to Poisson's equation. The process is:

  1. You want to solve $\nabla^2V=\displaystyle-\frac{\rho}{\epsilon_0}$ in a certain volume $\varOmega$.

  2. You define the Green function as one which is a solution of ${\nabla^{\prime}}^2G(\mathbf{r},\mathbf{r^\prime})=-4\pi\delta(\mathbf{r}-\mathbf{r'})$ with null Dirichlet boundary conditions, i.e. $G(\mathbf{r},\mathbf{r'})=0$ for all $\mathbf{r'\in\partial\varOmega}$.

    NB: Be careful, in Jackson the Green function has a $4\pi$ floating around in the definition, but in other texts the $4\pi$ is absorbed in the Green function and the definition is just ${\nabla^{\prime}}^2G(\mathbf{r},\mathbf{r^\prime})=-\delta(\mathbf{r}-\mathbf{r'})$.

  3. Then construct the solution to your problem as $$V(\mathbf{r})=\frac{1}{4\pi\varepsilon_0}\int_\varOmega G(\mathbf{r},\mathbf{r}^\prime)\rho(\mathbf{r^\prime})dV'-\frac{1}{4\pi}\int_{\partial\varOmega} V(\mathbf{r'})\frac{\partial G}{\partial n'}dS'$$ where in the second integral you are adding the specific boundary conditions of your problem.

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