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Integral form of Poisson's equation

  1. May 20, 2015 #1
    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. jcsd
  3. May 20, 2015 #2

    vanhees71

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    2016 Award

    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
    \begin{equation}
    \label{1}
    \Delta \Phi(\vec{x})=-\rho(\vec{x}).
    \end{equation}
    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
    \begin{equation}
    \label{2}
    \mathrm{d} Q=\mathrm{d}^3 \vec{x}' \rho(\vec{x})'.
    \end{equation}
    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
    \begin{equation}
    \label{3}
    \mathrm{d} \Phi(\vec{x})=\mathrm{d}^3 \vec{x}' \rho{\vec{x}'} \frac{1}{4 \pi |\vec{x}-\vec{x}'|}.
    \end{equation}
    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.,
    \begin{equation}
    \label{4}
    \Phi(\vec{x})=\int_{V} \mathrm{d}^3 \vec{x}' \frac{\rho(\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|}.
    \end{equation}
    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 :-().
     
  4. May 20, 2015 #3
    Thanks!

    I'll read up on Green's functions in the meantime!
     
  5. May 20, 2015 #4

    samalkhaiat

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    Are you familiar with the Dirac delta function? [tex]\rho ( x ) = \int d^{3} \bar{x} \ \rho ( \bar{x} ) \ \delta^{3} ( x - \bar{x} ) .[/tex]
    Can you prove (or do you know) the following relation? [tex]\nabla^{2} | \frac{1}{x - \bar{x}} | = - 4 \pi \ \delta^{3} ( x - \bar{x} ) ,[/tex] where [itex]x[/itex] and [itex]\bar{x}[/itex] are vectors in [itex]\mathbb{R}^{3}[/itex]. These two relations let you go from the differential representation to the integral one and vice versa.
    Sam
     
  6. May 22, 2015 #5

    vanhees71

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    2016 Award

    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):

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

    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
    \begin{equation}
    \label{2}
    \Delta G(\vec{x})=-\delta^{(3)}(\vec{x}).
    \end{equation}
    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
    \begin{equation}
    \label{3}
    \Delta G(\vec{x})=0, \quad \vec{x} \neq 0.
    \end{equation}
    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
    \begin{equation}
    \label{4}
    \frac{1}{r} (rG)''=0.
    \end{equation}
    This you can immediately integrate twice. In the first step you get
    \begin{equation}
    \label{5}
    (r G)'=C_1 =\text{const}.
    \end{equation}
    Integrating again gives
    \begin{equation}
    \label{6}
    G(r)=C_1+\frac{C_2}{r}.
    \end{equation}
    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,
    \begin{equation}
    \label{7}
    G(\vec{x})=\frac{C_2}{\sqrt{x^2+y^2+z^3}}.
    \end{equation}
    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:
    \begin{equation}
    \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.
    \end{equation}
    Now for ##\vec{r} \neq 0## we have
    \begin{equation}
    \label{9}
    \vec{\nabla} G(r)=(\vec{\nabla} r) G'(r)=-C_2 \frac{\vec{x}}{r^3}.
    \end{equation}
    In spherical coordinates the normal-surface element of the sphere is
    \begin{equation}
    \label{10}
    \mathrm{d}^2 \vec{F}=\mathrm{d} \vartheta \mathrm{d} \varphi \vec{e}_r r^2 \sin \vartheta.
    \end{equation}
    Plugging everything into (\ref{8}) gives
    \begin{equation}
    -4 \pi C_2=-1 \; \Rightarrow \; C_2=\frac{1}{4 \pi},
    \end{equation}
    leading to the correct result, namely the Coulomb potential for a unit charge at the origin,
    \begin{equation}
    G(\vec{x})=\frac{1}{4 \pi |\vec{x}|}.
    \end{equation}
     
    Last edited: May 22, 2015
  7. May 22, 2015 #6
    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.
     
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