I've read the Green function derivation for Poisson Equation (electrostatics) in this document. There are some points which are not clear for me.
On page 10, the document starts with the Poisson Equation involving the Green Function:
$$\nabla^2 G(\textbf x – \textbf x_0) = \delta(\textbf x – \textbf x_0)$$
where $\textbf x$ and $\textbf x_0$ are the observer and point source position vectors (they may be expressed arbitrarily in carthesian, spherical or cilindrical coordinates).
Now, on page 11, it integrates both members in a sphere with radius R with Volume $V_R$, applies the divergence theorem and gets the final solution.
$$ 1 = \iiint_{V_R} \delta(\textbf x – \textbf x_0)\;dV = \iiint_{V_R} \nabla^2 G(\textbf x – \textbf x_0)\;dV = \iiint_{V_R} \nabla \cdot \nabla G(\textbf x – \textbf x_0)\;dV = \iint_{S_R} \nabla G(\textbf x – \textbf x_0) \cdot \textbf n\;dS = 4 \pi R^2 G'(R)$$
where $\textbf n$ is the unit radial vector.
I have lots of missing points in these steps:
I. Integration Volume. The author integrates in a sphere of radius R. What does R mean? it is the "observation radius", i.e. the radial distance from the observation point? If it is, why does he call it R instead of $\textbf x$?
II. Integration Volume. The resulting Green function is nice and "spherically symmetric" because the author integrates on a Sphere. Why? Is it correct to try to integrate on a cube of volume $V$?
III. $R, r, \textbf x$. The author magically converts $R$ to $r$ and then $r$ to $\textbf x$ (page 11). Can you help me understand this?
Best Answer
The Green's function satisfies $$ \nabla_{\bf x}^2 G({\bf x}) = \delta ({\bf x}) \tag{1} $$ where ${\bf x}=(x,y,z)$ is a 3-vector. The general solution to this equation that vanishes at infinity is given by the Fourier transform $$ G({\bf x}) = \int d{\bf k} \frac{1}{|{\bf k}|^2} e^{ i {\bf k} \cdot {\bf x} } $$ It follows from this that $G({\bf x})$ is rotationally symmetric (prove this yourself!) so we must have $G({\bf x}) = G(r)$ where $r=|{\bf x}|$.
Now, to determine $G(r)$, we can either just calculate the integral above or we can do what the author did. We integrate (1) inside a ball of radius $R$. This $R$ does not mean anything. It is a mathematical tool employed to find out what the function $G(r)$ is. We find $$ \int_B dx dy dz \nabla_{\bf x}^2 G({\bf x}) = \int_B dx dy dz \delta ({\bf x}) = 1 $$ We now move to spherical coordinates $$ 1 = \int_0^R dr r^2 \int_0^\pi d\theta \sin \theta \int_0^{2\pi} d \phi \frac{1}{r^2} \frac{d}{dr} ( r^2 G'(r) ) $$ All the integrals are super easy to evaluate and we find $$ 1 = 4\pi ( r^2 G'(r) ) |_{r=R} = 4\pi R^2 G'(R) . $$ Now, this is true for any value of the the radius $R$ so we can setup another differential equation $$ 4\pi R^2 G'(R) = 1 \implies G(R) = - \frac{1}{4\pi R} + C $$ The constant can be fixed by requiring that $G(R) \to 0$ as $R \to \infty$ so we have $C=0$. In summary, $$ G({\bf x}) = G(r) = - \frac{1}{4\pi r} = - \frac{1}{4\pi |{\bf x} | } . $$