- #1

spaghetti3451

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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?

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In summary, the integral form of Poisson's equation, ##\nabla^{2} V = - \frac{1}{\epsilon_0} \rho##, can be derived from the differential form, ##\Delta V = -\rho##, by considering the field as a sum of contributions from point-like sources. This leads to the equation ##V(\textbf{r}) = \frac{1}{4 \pi \epsilon_0} \int \frac{1}{|\textbf{r}-\textbf{r}^{'}|}\ \rho(\textbf{r}^{'})\ d \tau^{'}##, which is a shortcut for solving the Poisson equation

- #1

spaghetti3451

- 1,344

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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?

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- #2

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\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 :-().

- #3

spaghetti3451

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Thanks!

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

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

- #4

samalkhaiat

Science Advisor

- 1,802

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failexam said:

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?

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

- #5

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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}

\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:

- #6

spaghetti3451

- 1,344

- 34

The integral form of Poisson's equation is an alternative way of expressing the relationship between a scalar function (such as electric potential or temperature) and its sources (such as charge density or heat generation rate). It is given by the integral of the function over a volume, equal to the integral of the sources over the same volume plus the integral of the normal derivative of the function over the boundary of the volume.

The integral form is a global representation of the equation, while the differential form is a local representation. This means that the integral form considers the entire volume and its boundary, while the differential form only considers a small region around a specific point. Additionally, the integral form does not require the function to be differentiable, making it more applicable to a wider range of problems.

Poisson's equation is a fundamental equation in physics and engineering, as it describes the relationship between a scalar field and its sources. It is used in a wide range of applications, including electrostatics, heat transfer, fluid dynamics, and more. It also serves as the basis for other important equations, such as Laplace's equation and the Helmholtz equation.

To solve Poisson's equation in its integral form, one must first determine the sources and boundary conditions for the problem. Then, the function is integrated over the volume and its boundary, and the resulting equation is solved for the function. This may involve using mathematical techniques such as Green's functions or Fourier transforms, depending on the problem at hand.

The integral form of Poisson's equation may not be applicable to all problems, as it requires the function to be integrable over the volume and its boundary. Additionally, it may be difficult to solve analytically for more complex geometries or sources. In these cases, the differential form may be a more suitable approach.

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