1. May 29, 2010

### cepheid

Staff Emeritus
My book points out that the combined rest mass of an isolated proton + an isolated electron is greater than the mass of a hydrogen atom:

mH < mp + me

The book then goes on to claim that if the electron is captured into the ground state in order to form a hydrogen atom, the rest mass difference will correspond to exactly 13.6 eV, the ionization potential of hydrogen. Their argument seems to be that if an electron loses 13.6 eV, emitting a photon of that energy in the process, then where did that energy go? My standard response would be that, "it came from the electric potential energy that existed due to the fact that these two opposite charges were initially unbound and separated." Of course, the problem with my viewpoint is that the amount of energy liberated would depend upon initial separations and kinetic energies of the proton and electron, and the electron, once captured, would spiral in, continuously emitting light of ever increasing frequency. We know that this doesn't happen. So I guess my problem stems from the fact that I am using classical physics to try and understand something that should be described by quantum mechanics + special relativity. But in spite of some knowledge of those fields, I cannot really understand the role of "rest energy" and why it is reduced.

The book later gives an example of 4 protons undergoing a fusion reaction to create a 4He nucleus. In this case, I am more willing to believe that the 26.731 MeV liberated by this nuclear reaction in the form of photons + neutrinos come at the expense of a reduction in "rest energy." Maybe it's because I know that this reaction involves the strong force, which is something that I really don't understand, but I have this notion that somehow the binding energy of the strong force and "rest energy" are related (since "rest energy" seems to involve how much energy is tied up in things that are made of quarks).

Can you lose "rest energy" through electromagnetic interactions? If so, how/why does it differ from the classical picture of something in the presence of some Coulomb potential?

I can solve Schrodinger's Equation with the best of them (for simple, 1D, totally unphysical Hamiltonians, LOL), but the more I delve into physics, the less I realize I know.

2. May 29, 2010

### JesseM

Well, in relativity E=mc^2 still works when m is the inertial rest mass of a bound system composed of multiple particles (with inertial rest mass defined in terms of resistance to acceleration in the system's rest frame, proportional to what the system would weigh if placed on a scale), and E is the total energy of the bound system, including the rest masses of all the individual particles and their potential energy and their kinetic energies (the fact that it includes potential energy explains why an atom has a slightly smaller rest mass than the sum of the rest masses of the particles that make it up, and the fact that it includes kinetic energy explains why heating an object would increase its rest mass slightly). There's nothing specifically quantum-mechanical about this, it should be true for a classical bound system that respects Lorentz-invariant laws like Maxwell's equations. I don't know if that helps, if not could you sum up what you're asking exactly?

3. Jun 1, 2010

### cepheid

Staff Emeritus
Helpful, thank you. I will keep that in mind.

Unfortunately no, not really. My thoughts are not too organized at this point, but if I do come up with a more precise question I'll be sure to ask.

4. Jun 2, 2010

### Major_Energy

It almost sounds like the question is not where did the energy go, but where did it come from?

If the electron is captured into the ground state in order to form a hydrogen atom, the rest mass difference will correspond to exactly 13.6 eV.

Which makes sense, but to be "captured" total energy would have to be conserved, which would be any kinetic energy+rest mass. Et=K+Eo

So for the electron, if you give it 13.6 eV of K, and it has a rest mass mec2 it makes sense that you need just enough K energy for it to be captured and what you have left is the rest mass.

It "sounds like" what they are saying is if an electron is launched and it emits a photon of 13.6 eV, then that means its captured by the Proton, or, it just shed its K. If it had more than 13.6 eV, it would just zoom by the Proton, shed a smaler photon, but still have some K left which allows it not to be captured.

Basically it sounds like they didn't explain conservation of Energy and/or Momentum very well.

5. Jun 3, 2010

### closet mathemetician

I've got some confusion about rest energy as well:

4. Mc^2 is the rest energy of a massive object. Presumably, this represents the sum of the following types of energies:
a. Nuclear binding energy between particles in the nuclei of the molecules/atoms of the object,
b. Electromagnetic binding energy between the electrons and nuclei of the molecules/atoms,
c. Average kinetic energy of molecules – temperature

Each of these kinds of energy are associated with very complex mathematical theories (Yang-Mills, QCD, Statistical Mechanics, etc ..). How is it that the total rest energy of the object, as calculated using all these complex mathematical theories, is captured by the simple statement “m*c^2”? Could this be verified? If I took a particle, calculated the energies due to a, b and c above using the above mentioned theories, would the sum of these energies equal mc^2?