Does a charged particle at rest (m_{0}>0) have more total energy than an uncharged particle with the same rest mass at rest? Thanks in advance.
The energy of a single particle depends only on its mass. Any EM energy of a charged particle is already included in its mass.
If the EM energy is included into the mass, it contributes to it, right? Then what does contribute to the mass of a neutral particle of the same mass?
Actually, I'm having second thoughts about the validity of my original question. Is the charge (coulombs) on a fundamental particle, say like the electron, energy or force - or rather the sign (+ or -) of a force field?
A single charged particle has (classically) infinite self energy: that is what you get from integrating the energy density over a spherical volume and letting the volume shrink to zero. If you now bring in a second charge at a finite nonzero distance from the first one, there is a finite "interaction energy" for this two-charge system -- it is just the work done in bringing the second charge in from infinity to this finite location, in the field created by the first charge. Of course, the self energy of each charge is still infinite. (You can look upon the infinite self energy of a point charge as the energy required to create a charge out of "free space". We don't know how exactly we could do this, and I don't know if this is really a very good way to look at this idea, but it is at least intuitive classically.) So to summarize, 1. the total energy of the point charge is the sum of its rest energy (by virtue of its mass) and its self energy (which is infinite). 2. for an uncharged particle, the total energy is just the rest energy. (Caveat: an uncharged "mass" can be considered consist of equal positive and negative point charges, each of which should -- by the above argument -- have an infinite amount of self energy. We therefore need more justification for point 2.)
>A single charged particle has (classically) infinite self energy: that is what you get from >integrating the energy density over a spherical volume and letting the volume shrink to >zero. The classical notion of point-like particle is not "fundamental" but illusive: it corresponds to the inclusive picture where different observations (inelastic events) are added up (see my work at arxiv:0806.2635 or Central European Journal of Physics, Volume 7, pp. 1-11, 2009, “Atom as a dressed nucleus”). Concerning the mass renormalizations, see my another work at arxiv:0811.4416, “Reformulation instead of renormalizations”. Quantum mechanically any charge is smeared since it makes part of the EM field oscillators. So there is no problem of the self-energy divergence. Regards, Bob.
Interesting. I haven't checked out your paper yet, but do you mean to say that a mathematical formalism exists, using quantum mechanics to show that such a smearing is just right to ensure that there is 'no problem of the self-energy divergence'? I'd be very eager to learn about that. I've pondered about the divergent self-energy problem throughout my education and I'd be very happy to see a resolution :-) Please let me know.
Yes, I mean exactly that. The formalism is simple and it uses another, quite natural approach. You know, the self-action term jA was introduced by H. Lorentz by analogy with electron interaction with external fields. Lorentz wanted to add some "friction" term for the sake of the energy-momentum conservation. In fact, there is another way of the electron-field coupling preserving the energy-momentum. My papers are accessible at arXiv and they are intentionally made as simple as possible for understanding. Start from "Atom as a dressed nucleus". You will see immediately the quantum mechanical effects of binding. Then read "Reformulation instead of Renormalizations" where the "self-action" ansatz is replaced with the "interaction" ansatz. The latter eliminates the problem of divergences. Regards, Bob.