lightarrow said:
But to evaluate the electric field's energy and compare it with the energy you give to the system to establish if the mass is in the field or in the particles, you should (also) know the electron's shape and dimensions. Do you know them?
Hello lightarrow, I didn't have energy to answer this back then when the thread was active, and then I forgot this. But anyway, yes, the energy of the interaction is precisely the energy of the field. Point charges bring a little problem of infinite rest energies, but they can be ignored and the interaction energy can be solved. You don't need to know the shape of the charges really. The Coulomb's potential is valid, if the distance between two charges is much larger than the size of the charges. Here's the calculation for point charges (in the physicist way).
If we have a charge q at location x', then the electric field is
<br />
E(x) = \frac{q}{4\pi \epsilon_0}\frac{x-x'}{|x-x'|^3}<br />
The energy density of the electric field is
<br />
\frac{1}{2}\epsilon_0 |E|^2<br />
So the total energy of the field around the point charge is
<br />
\int d^3x\; \frac{q^2}{32\pi^2 \epsilon_0}\frac{1}{|x-x'|^4} = \infty<br />
This is the already mentioned infinite rest energy. Now suppose we have two charges, q_1 and q_2, at locations x_1 and x_2. The the electric field is
<br />
E(x) = \frac{1}{4\pi\epsilon_0}\Big(\frac{q_1(x-x_1)}{|x-x_1|^3} + \frac{q_2(x-x_2)}{|x-x_2|^3}\Big)<br />
The energy of the electric field is now
<br />
\int\frac{d^3x}{32\pi^2\epsilon_0} \Big(\frac{q_1^2}{|x-x_1|^4} + \frac{q_2^2}{|x-x_2|^4} + \frac{2q_1q_2(x-x_1)\cdot(x-x_2)}{|x-x_1|^3 |x-x_2|^3}\Big)<br />
The first two terms give the same infinities that are the rest energies of the both particles, so we can ignore them. Only the last term contributes to the interaction. Substituting
<br />
\nabla\frac{1}{|x-x_2|} = -\frac{x-x_2}{|x-x_2|^3}<br />
makes the interaction energy to be
<br />
\int\frac{d^3x\; q_1q_2}{16\pi^2\epsilon_0} \frac{x-x_1}{|x-x_1|^3}\cdot \nabla\frac{-1}{|x-x_2|}<br />
The nabla can be moved onto the first factor using integration by parts, and this gives
<br />
\int \frac{d^3x\;q_1q_2}{16\pi^2\epsilon_0}\Big( \nabla\cdot\frac{x-x_1}{|x-x_1|^3}\Big)\frac{1}{|x-x_2|}<br />
Now using an delta function representation
<br />
\nabla\cdot\frac{x-x_1}{|x-x_1|^3} = 4\pi\delta^3(x-x_1)<br />
the interaction energy becomes
<br />
\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{|x_1-x_2|}<br />
which is just the Coulomb's potential. So when you speak about the potential energy given by the Coulomb's potential formula, you are speaking about the energy in the electric field. And when the mass of the system is changing as result of changing potential energy, it is the field that is carrying the changing mass.
I asked about this some years ago from some lecturer, and in the next lecture he had brought Jackson's book with him, and he showed some calculations where Coulomb's potential was derived. I didn't copy them back then, but when I later tried to go through the details, I was able to do it on my own (because I had seen it once before and I already new something about delta functions).
