# Green identity, poisson equation.

1. Jun 5, 2008

### Thaakisfox

Suppose $$\phi$$ is a scalar function: $$R^n\to R$$, and it satisfies the Poisson equation:
$$\nabla^2 \phi=-\dfrac{\rho}{\varepsilon_0}$$

Now I want to calculate the following integral:
$$\int \phi \nabla^2 \phi \,dV$$
So using Greens first identity I get:
$$\int \phi \nabla^2 \phi \,dV = \oint_S \phi \nabla \phi d\vec A -\int |\nabla \phi|^2 dV$$
Where S is some closed surface.
Now when we are calculating the electrostatic potential energy, we have to calculate exactly this integral, but we know the end, that:
$$W=-\dfrac{\varepsilon_0}{2}\int \phi \nabla^2 \phi \,dV =\dfrac{\varepsilon_0}{2}\int |\nabla \phi|^2 dV$$

So what I am wondering is why will, this be true:
$$\oint_S \phi \nabla \phi d\vec A=0$$
If $$\phi$$ satisfies the Poisson equation??

2. Jun 5, 2008

### Thaakisfox

No worries, I figured it out :D

3. Mar 21, 2009

### JANm

Hallo Thaakisfox
I don't and didn't worry, thought you did. Are you able to explain the :D you figured out?
greetings Janm

4. Apr 9, 2009

### Thaakisfox

Hello JANm.

Been a while since I posted this lol

The main thing is that since we are integrating over the whole space the closure surface is at infinity, and since the potential vanishes (this is how we choose it) in infinity the surface integral itself is zero.

5. Apr 9, 2009

### JANm

Hello Thaakisfox
to start with what is a lol. Secondly i'm interested in: aren't you a believer of finite space theories?
greetings Janm

6. Apr 9, 2009

### Thaakisfox

Could you be a bit more clear, on what you mean to ask by that?

A lol means "laughing out loud"...

7. Apr 9, 2009

### JANm

You say closure surface is at infinity. The primordial Hoyle family says that isn't necessary, or yet some sort of impossible. Space is curved it is started at one point at one moment and three seconds later a finite blast fills and is a finite yet expanding universe...

8. Apr 14, 2009

### LuisVela

Hi , I have a question,..in the example im working on i got a potential $$\phi(r)$$
Now, when i calculate the total energy of the system, witht he integral over all space, should i change my result to smth like $$\phi(x,y,z)$$ and then perform the integral from -oo to +oo in all the variables? ..or is it fine to calculate the integral in sferical coordinates...with integral limits of the form (0,+oo)x(0,2$$\pi$$)x(0,$$\pi$$)???
should i cnlude the jacobian (|J|=$$r^{2} sin(\vartheta)$$) in the calculation?

Last edited: Apr 15, 2009
9. Apr 15, 2009

### JANm

Hello LuisVela
Found something for you written by Maxwell: If the potential is a function of r the potential equation becomes d^2V/dr^2+2dV/(rdr)=0 he calculates this with the example of two concentric spheres.
Try to calculate the solution to the diffential equation. When volume integration is needed I think the surface differential looks like:
dS=4 pi r^2 dr
Greetings Janm

10. Apr 15, 2009

### LuisVela

Hello JANm.
I know what you mean by that. But i got the potential using Gauss law, so i actually didnt use Laplace equation, but as i understand, the two methods are equivalent, and when symmetry is present Gauss law is way cheaper to use, than laplace/poisson equation. So i thing i dont need to find the potential again, but thx for the advice
$$W=\dfrac{1}{2}\int_{all space} \rho \phi dV$$
should $$\rho$$ and/or $$\phi$$ be calculated in cartesian coordinates only, and then perform the substitution to spherical using $$dv=4 \pi r^2$$ ?
or should i just argue that ''all space'' is reached by integrating $$\rho(r)\cdot \phi(r)$$ from $$r=0$$ to $$r=\infty$$ .
.......Its quite frustraiting because is kinda complicated to explain.

One more thing. The integral is over all space, or over the boundaries of ''all space''?..i mean, why did you say ''surface '' differential ?

11. Apr 16, 2009

### JANm

Hello Luisvela
Something is wrong with the security on my computer. Written a hell of a reply to you but it vanished...
What is the volume of a sphere with radius a it the integral of 4 * pi *r^2 dr from r=0 until r=a. I hope you can get 4*pi *a^3 /3 out of that? So volume integrating over a radial function no difference in centigrade or lattitude is the easier way to make a volume integral than over x in some interval and y over some interval and z over some interval. That is my point...
greetings Janm