# Integral form of Poisson's equation

1. May 20, 2015

### spaghetti3451

Is it true that $V(\textbf{r}) = \frac{1}{4 \pi \epsilon_0} \int \frac{1}{|\textbf{r}-\textbf{r}^{'}|}\ \rho(\textbf{r}^{'})\ d \tau^{'}$ is the integral form of Poission equation $\nabla^{2} V = - \frac{1}{\epsilon_0} \rho$?

Can you trace the steps that lead from one to the other, or at the very least offer some possible way to tackle the proof?

2. May 20, 2015

### vanhees71

Yes, it is. It is a very important insight, because it provides a simple example for a Green's function. I set $\epsilon_0=1$ (Heaviside-Lorentz units). Then you have to solve

\label{1}
\Delta \Phi(\vec{x})=-\rho(\vec{x}).

Now you can think of the field $\Phi$ as of made up by a field, which is due to very many point-like sources, located at places $\vec{r}'$ with charge

\label{2}
\mathrm{d} Q=\mathrm{d}^3 \vec{x}' \rho(\vec{x})'.

You know the solution for that case, because it's simply the Coulomb Law for a charge siting at $\vec{r}'$. This makes a contribution to the field of

\label{3}
\mathrm{d} \Phi(\vec{x})=\mathrm{d}^3 \vec{x}' \rho{\vec{x}'} \frac{1}{4 \pi |\vec{x}-\vec{x}'|}.

Now the Poisson equation is a linear differential equation, and thus the total field is just given by "summing" up all the contributions. But since you've a continuous charge distribution this "sum" turns into an integral over the infinitesimal charge increments (\ref{2}), i.e.,

\label{4}
\Phi(\vec{x})=\int_{V} \mathrm{d}^3 \vec{x}' \frac{\rho(\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|}.

This is the physicists' shortcut to the solution.

It ist worthwhile to study the mathematical formalism behind it. I'll explain this tomorrow (lack of time now :-().

3. May 20, 2015

### spaghetti3451

Thanks!

I'll read up on Green's functions in the meantime!

4. May 20, 2015

### samalkhaiat

Are you familiar with the Dirac delta function? $$\rho ( x ) = \int d^{3} \bar{x} \ \rho ( \bar{x} ) \ \delta^{3} ( x - \bar{x} ) .$$
Can you prove (or do you know) the following relation? $$\nabla^{2} | \frac{1}{x - \bar{x}} | = - 4 \pi \ \delta^{3} ( x - \bar{x} ) ,$$ where $x$ and $\bar{x}$ are vectors in $\mathbb{R}^{3}$. These two relations let you go from the differential representation to the integral one and vice versa.
Sam

5. May 22, 2015

### vanhees71

So here's the promised more mathematical formulation. There are two ways to derive the Green's function for the Laplace operator: You can use Fourier transformation to "momentum space" and then do the back transformation or you integrate the defining equation (see the previous posting):

\label{1}
\Delta G(\vec{x}-\vec{x}')=-\delta^{(3)}(\vec{x}-\vec{x}'),

where I already used the translation invariance of the Green's function to write $G(\vec{x}-\vec{x}')$. So we can as well solve

\label{2}
\Delta G(\vec{x})=-\delta^{(3)}(\vec{x}).

Let's follow the approach to directly integrate the equation. Since everything is radially symmetric around the origin in (\ref{2}), we can introduce spherical coordinates. There's however one caveat: You cannot express the $\delta$ distribution peaked at the origin in these coordinates, because they are singular along the $z$ axis. But for $\vec{x} \neq 0$ you just have

\label{3}
\Delta G(\vec{x})=0, \quad \vec{x} \neq 0.

Now we can look for the radially symmetric solution of this potential equation, i.e., we make the Ansatz $G(\vec{x})=G(r)$ with $r=|\vec{x}|$. Then we use the Laplace operator in spherical coordinates to get

\label{4}
\frac{1}{r} (rG)''=0.

This you can immediately integrate twice. In the first step you get

\label{5}
(r G)'=C_1 =\text{const}.

Integrating again gives

\label{6}
G(r)=C_1+\frac{C_2}{r}.

Now we want $G(r) \rightarrow 0$ for $r \rightarrow \infty$, which makes $C_1=0$.

To determine $C_2$ we must bring in the $\delta$ distribution again. To this end we rewrite $G$ in terms of Cartesian coordinates,

\label{7}
G(\vec{x})=\frac{C_2}{\sqrt{x^2+y^2+z^3}}.

Now we can determine the constant from (\ref{2}). To this end we integrate this equation over a sphere of arbitrary radius $a$ around the origin, using Gauß's integral theorem:

\label{8}
\int_{K_a} \mathrm{d}^3 \vec{x} \Delta G(\vec{x}) = \int_{\partial K_a} \mathrm{d}^2 \vec{F} \cdot \vec{\nabla} G(\vec{x}) \stackrel{!}{=} -1.

Now for $\vec{r} \neq 0$ we have

\label{9}
\vec{\nabla} G(r)=(\vec{\nabla} r) G'(r)=-C_2 \frac{\vec{x}}{r^3}.

In spherical coordinates the normal-surface element of the sphere is

\label{10}
\mathrm{d}^2 \vec{F}=\mathrm{d} \vartheta \mathrm{d} \varphi \vec{e}_r r^2 \sin \vartheta.

Plugging everything into (\ref{8}) gives

-4 \pi C_2=-1 \; \Rightarrow \; C_2=\frac{1}{4 \pi},

leading to the correct result, namely the Coulomb potential for a unit charge at the origin,

G(\vec{x})=\frac{1}{4 \pi |\vec{x}|}.

Last edited: May 22, 2015
6. May 22, 2015

### spaghetti3451

This is some pretty heavy-going mathematics. I'll wait till I study Green's functions from Sadri's textbook and then tackle the proof.