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?

  • $\begingroup$ Minor comment to the post (v2): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$
    – Qmechanic
    Jan 20 at 9:13

2 Answers 2


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, $$ G({\bf x}) = G(r) = - \frac{1}{4\pi r} = - \frac{1}{4\pi |{\bf x} | } . $$

  • 1
    $\begingroup$ I think you are missing some terms in your second equation, no? I.e. inside the integral, before the exponential, there should be $(1/|\mathbf k|^2+c_1 \delta(\mathbf k) + \mathbf c_2 \cdot \nabla \delta (\mathbf k))$, no? I.e. if you take the Fourier of the Poisson equation, you get $\hat G(\mathbf k) |\mathbf k|^2 = 1$, which you have to invert in the distributional sense. $\endgroup$ Jan 20 at 9:55
  • 1
    $\begingroup$ @TobiasFünke - I should have said, "The general solution to this equation that vanishes at infinity". That fixes the other constants. Fixed! $\endgroup$
    – Prahar
    Jan 20 at 9:56

Short answer

The results come from:

  • the assumption of homogeneity (no preferential point in space) and isotropy (no preferential orientation) of space,
  • a reasonable choice of parameters to describe such a space (spherical coordinates in 3D space, cylindrical in 2D space),
  • an application of divergence theorem to proof the result does not depend on the domain of integration.

Let's answer the points in your question.

Poisson equation, active/passive points

But before focusing on these points, let's inspect Poisson equation with Dirac's delta force $\delta(\mathbf{x} - \mathbf{x}_0)$ \begin{equation} -\nabla^2 G(\mathbf{x}, \mathbf{x_0}) = \delta(\mathbf{x} - \mathbf{x}_0) \ , \end{equation} and give a meaning to $\mathbf{x}$ and $\mathbf{x}_0$ points in space:

  • point $\mathbf{r}_0$ can be interpreted as the point where we can find the "singularity" of the Dirac delta, and it can be interpreted as a parameter of the problem; since it's the location of the forcing, let's call it active point;
  • point $\mathbf{x}$ can be interpreted as the independent variable of the problem, i.e. the set of point in space $\mathbf{x} \in V$ where we want to evaluate the function $G$; let's call it passive point.


Following the assumption of space homogeneity and isotropy, it's reasonable to look for a solution with spherical symmetry around the active point $\mathbf{x}_0$. Let's thus:

  • introduce a set of spherical coordinates with origin in $\mathbf{x}_0$, so that
    • each point in space can be written as $\mathbf{x} = \mathbf{x}_0 + \mathbf{r}$, being $\mathbf{r}$ the vector connecting $\mathbf{x}$ and $\mathbf{x_0}$, the origin of the spherical coordinate system,
    • the radial coordinate of a point $\mathbf{x}$ reads $r = |\mathbf{r}| = |\mathbf{x} - \mathbf{x}_0|$
  • assume that the solution of the problem, the Green's function, only depends on $r$ (and thus on $|\mathbf{r}| = |\mathbf{x} - \mathbf{x}_0|$), and not on the angles $\phi$, $\theta$.

I. Integration volume (I). Performing an integration over a spheres and its surface of radius $R$, the author identifies all the points of the sphere $\mathbf{x}$ with their distance from the origin: 2-dimensional surface in 3d-space only needs an equation (in simple cases, like a sphere) to be determined; thus, instead of writing $\mathbf{x} \in S_R$, he identifies this points with the distance from the origin, i.e. the radial coordinate $r=R$.

II. Integration volume (II). The result doesn't depend on the choice of the domain of integration, but only on the assumptions of homogeneity and isotropy. You can prove it using divergence theorem on a "shell" volume $\Omega$ that doesn't contain point $\mathbf{r}_0$, so that Poisson equation in $\Omega$ is homogeneous \begin{equation} -\nabla^2 \mathbf{G}(\mathbf{x},\mathbf{x}_0) = 0 \ . \end{equation}

\begin{equation} \begin{aligned} 0 & = \int_{\Omega} \nabla^2 \mathbf{G} = \\ & = \oint_{\partial \Omega} \nabla G \cdot \mathbf{\hat{n}} = 0 \\ & = \oint_{S_R} \nabla G \cdot \mathbf{\hat{n}} + \oint_{\tilde{S}} \nabla G \cdot \mathbf{\hat{n}}^- \ , \end{aligned}\end{equation} where surfaces $S_R$ and the arbitrary surface $\tilde{S}$ are the outer and the inner surface of the "shell" volume $\Omega$. Here, surface $\tilde{S}$ is supposed to be the inner surface of the shell, and thus with reversed unit normal $\mathbf{\hat{n}}^- = \mathbf{\hat{n}}$.

Thus, it's easy to realize that \begin{equation} \oint_{S_R} \nabla G \cdot \mathbf{\hat{n}} = \oint_{\tilde{S}} \nabla G \cdot \mathbf{\hat{n}} \ . \end{equation}

A spherical domain is usually chosen, in order to get integral to evaluate as simple as possible.

III. Space and parametrization, $\mathbf{r}$, $\mathbf{x}$, $r$, $R$. Once you got the result for all the points in space lying on the sphere of radius $R$ centered in the Dirac's delta location, and determined by the radial coordinates $r=R$, \begin{equation} G(R) = - \dfrac{1}{4\pi R} \ , \end{equation} this solution holds for all the points in space, and thus for every value of the radial coordinate $r$, \begin{equation} G = -\dfrac{1}{4\pi r} = -\dfrac{1}{4 \pi |\mathbf{r}|} = -\dfrac{1}{4\pi |\mathbf{x} - \mathbf{x}_0|} \ . \end{equation}

Last remark on notation

To be honest and very precise, when you change the independent variables, you should also change the name or the letter you use to indicate the functions, since mathematically these functions different relations (roughly speaking, a function is nothing more than a relation between variables) between independent and dependent variables, i.e. \begin{equation} G(r) = -\dfrac{1}{4 \pi r} = G(|\mathbf{r}|) = -\dfrac{1}{4 \pi |\mathbf{r}|} = G(|\mathbf{x}-\mathbf{x}_0|) = -\dfrac{1}{4 \pi |\mathbf{x}-\mathbf{x}_0|} = \tilde{G}(\mathbf{x}, \mathbf{x}_0) \ , \end{equation} being the function $\tilde{G}(\mathbf{a}, \mathbf{b})$ the composition of the function $N(\mathbf{a}, \mathbf{b}) = |\mathbf{a} - \mathbf{b}|$ and $G(r)$,

\begin{equation} \tilde{G}(\mathbf{a}, \mathbf{b}) = G(|\mathbf{a} - \mathbf{b}|) = G(N(\mathbf{a}, \mathbf{b})) \ . \end{equation}


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