Electrostatic energy in annihilation process

[FredMadison]

When an electron and a positron annihilate, they typically produce two gamma rays, each of energy mc^2 plus whatever kinetic energy available before annihilation. I was recently told that it is an experimental fact that the electrostatic energy between the electron and the positron does NOT contribute to the photon energies. The person who told me used this to argue that the extra, not accounted for, energy must be carried away by another medium.

I got curious and browsed around a bit, but I haven't found any mention of this effect.

Does anyone know what the experiments say about this? Is the Coulomb energy accounted for?

 

[mathman]

From wikipedia
http://en.wikipedia.org/wiki/Electron–positron_annihilation

Low energy case

There are only a very limited set of possibilities for the final state. The most possible is the creation of two or more gamma ray photons. Conservation of energy and linear momentum forbid the creation of only one photon. (An exception to this rule can occur for tightly bound atomic electrons.[1]) In the most common case, two photons are created, each with energy equal to the rest energy of the electron or positron (511 keV).[2] A convenient frame of reference is that in which the system has no net linear momentum before the annihilation; thus, after collision, the gamma rays are emitted in opposite directions. It is also common for three to be created, since in some angular momentum states, this is necessary to conserve C parity.[3] It is also possible to create any larger number of photons, but the probability becomes lower with each additional photon because these more complex processes have lower probability amplitudes.

Since neutrinos also have a smaller mass than electrons, it is also possible—but exceedingly unlikely—for the annihilation to produce one or more neutrino–antineutrino pairs. The same would be true for any other particles, which are as light, as long as they share at least one fundamental interaction with electrons and no conservation laws forbid it. However, no other such particles are known.

High energy case

If either the electron or positron, or both, have appreciable kinetic energies, other heavier particles can also be produced (such as D mesons), since there is enough kinetic energy in the relative velocities to provide the rest energies of those particles. It is still possible to produce photons and other light particles, but they will emerge with higher energies.

At energies near and beyond the mass of the carriers of the weak force, the W and Z bosons, the strength of the weak force becomes comparable with electromagnetism.[3] This means that it becomes much easier to produce particles such as neutrinos that interact only weakly.

The heaviest particle pairs yet produced by electron–positron annihilation in particle accelerators are W⁺–W⁻ pairs. The heaviest single particle is the Z boson. The driving motivation for constructing the International Linear Collider is to produce Higgs bosons in this way.

 

[FredMadison]

Yes, but that doesn't really answer my question.

I mean, if the energy balance of a process is out of order, it seems to me that the physics community would be well aware of it, trying to find a solution. Since this debate obviously doesn't exist, I imagine that the person who told me is wrong. But since I cannot find any mention of it, I'm a bit unsure.

 

[sheaf]

isn't it because in the QFT calculation, we take the in-state to be a free electron and positron (free, so no Coulomb interaction), and whatever particles are produced in the out-state (in this case photons) are also free ?

 

[FredMadison]

Yes, that's true, but there are higher-order diagrams contributing in which one or more photons are exchanged before annihilation. So theory can take into account the Coulomb interaction, but the question is whether the energy shows up in experiments.

 

[K^2]

FredMadison, whoever told you this does not understand anything about relativistic electrodynamics.

At a distance, electrostatic energy contributes to electron and positron masses. Some of that energy is emitted when positronium atom forms prior to annihilation, but it is a very small portion of the mc². The rest of it is emitted in the 2+ photons produced in annihilation process.

 

[FredMadison]

K^2:
What do you mean by "at a distance"?

Do I understand you correctly like this:

* Prior to any electron/positron annihilation, a positronium bound system forms.
* This formation releases radiation.
* After formation, annihilation occurs, and the particles are effectively at rest.

This all sounds a bit strange to me. Perhaps I've misunderstood you?

 

[K^2]

The rest mass for electron is defined for an isolated electron. It's basically the same thing as an electron with the positron "infinitely" far away. When electron and positron are close to each other, you have to subtract the decrease in electrostatic potential energy from the rest masses of the two particles. Keep in mind that we are talking a few eV here, compared to electron mass of 511 keV.

Particles are only "at rest" with respect to each other if annihilation occurs from s0 state of the positronium. While that's the most likely scenario (highest branching fraction), it's not the only possible one. Positronium can annihilate from p1 state, producing 3 photons as a product. As you go to higher and higher excited states of positronium, probability of it decaying to lower state prior to annihilation increases.

You can think of initial state of electron and positron "far" from each other as some very high level excited state of positronium. In that case, the energy released as it decays to s0 or whichever state is that "missing" electrostatic energy.