Proving that two integrals of potential energy are equal

[Buffu]

I want to prove ##\displaystyle U = {1\over 8\pi}\int \vec E \cdot \vec E dV## and ##\displaystyle U = \frac12 \int \phi \rho dV## are equal.

I started with ##\nabla \cdot (\phi \nabla\phi) =(\nabla \phi)^2 + \phi \nabla^2 \phi##'

Then

##\displaystyle {1\over 8\pi}\int \vec E \cdot \vec E dV = {1\over 8\pi}\int \nabla \phi \cdot \nabla \phi dV = {1\over 8\pi}\int \nabla \cdot (\phi \nabla\phi) - \phi \nabla^2 \phi dV = {1\over 8\pi}\int \nabla \cdot (\phi \nabla\phi) dV - {1\over 8\pi}\int \phi \nabla^2 \phi dV ##

By poisson's equation, ##\nabla^2 \phi = 4\pi \rho##

So, ##\displaystyle {1\over 2}\int \rho \phi dV##

So I guess I need to prove ##\displaystyle {1\over 8\pi}\int \nabla \cdot (\phi \nabla\phi) dV## is zero

Using divergence theorem,

## \displaystyle {1\over 8\pi}\int \nabla \cdot (\phi \nabla\phi) dV = {1\over 8\pi}\int (\phi \nabla \phi)\cdot d\vec a##

Now I don't have any clue what to do, I guess I need to use ##-\nabla \phi = \vec E## but I don't know how.

 

[haruspex]

Haven't you flipped a sign?
You have ##-\displaystyle {1\over 2}\int \rho \phi dV##, but you want ##+\displaystyle {1\over 2}\int \rho \phi dV##

 

[Buffu]

Haven't you flipped a sign?
You have ##-\displaystyle {1\over 2}\int \rho \phi dV##, but you want ##+\displaystyle {1\over 2}\int \rho \phi dV##

I made a typo in poisson's equation, it is ##\nabla^2 \phi = -4\pi\rho## so they got cancelled.

 

[vanhees71]

First note that you want to find the total energy of the electric field,
$$U=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \frac{\vec{E}^2}{8 \pi}.$$
It's safe to assume that you have a your charge distribution in a finite volume (it's most realistic ;-)). Then from the result I also guess you deal with stationary states, i.e., time-independent fields, where ##\vec{E}## has a potential, i.e.,
$$\vec{E}=-\vec{\nabla} \phi.$$
Then you can use
$$\vec{E}^2=-\vec{E} \cdot \vec{\nabla} \phi = - E_j \partial_j \phi=-[\partial_j (E_j \phi)-\phi \partial_{j} E_j]=-\vec{\nabla} \cdot (\phi \vec{E})+\phi \vec{\nabla} \cdot \vec{E}.$$
For the 2nd term you use Gauss's Law (which is one of Maxwell's Equations)
$$\vec{\nabla} \cdot \vec{E}=4 \pi \rho.$$
Then we get
$$\vec{E}^2=-\vec{\nabla} \cdot(\phi \vec{E})+4 \pi \phi \rho.$$
Now we have to integrate that. Let's start with the first term. Because it's a divergence, we can use Gauss's Integral Theorem. So let's assume some large volume ##V## with a boundary surface ##\partial V##. Then we get
$$\int_V \mathrm{d}^3 \vec{x} \vec{\nabla} \cdot (\phi \vec{E})=\int_{\partial V} \mathrm{d}^2 \vec{f} \cdot (\phi \vec{E}).$$
Now take ##V## to be a sphere of radius ##R## around the origin. Since our charge distribution is assumed to occupy only some finite volume, and we can also assume that this is located around the origin of our coordinate system, we know that ##\phi \sim 1/r## and ##\vec{E} \sim 1/r^2## for large ##r##. This implies that the surface integral goes like ##1/R##. So in the limit ##R \rightarrow \infty## this surface contribution vanishes, and we are left with the 2nd part. So finally we indeed get
$$U=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \frac{1}{2} \rho \phi.$$

 

[Buffu]

Now take ##V## to be a sphere of radius ##R## around the origin. Since our charge distribution is assumed to occupy only some finite volume, and we can also assume that this is located around the origin of our coordinate system, we know that ##\phi \sim 1/r## and ##\vec{E} \sim 1/r^2## for large ##r##. This implies that the surface integral goes like ##1/R##. So in the limit ##R \rightarrow \infty## this surface contribution vanishes, and we are left with the 2nd part. So finally we indeed get
$$U=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \frac{1}{2} \rho \phi.$$

Two things,

One, Is the equivalence of these two integrals only when the surface is very large ? I think so because you used ##R \to \infty##.

Second, Why you write ##d (...)## before the actual function ? :)

 

[vanhees71]

It's the total field energy. So you have to integrate over all space, and thus the surface of the volume is at infinity.

I use the physicists' notation for integrals, where the integral sign together with the ##\mathrm{d}^3 \vec{x}## acts as an operator on the integrand. I think it's a very clear notation, because you see at once wrt. to which variables you integrate. That's particularly convenient when you have nested integrals.