# Electric fields of electron and positron that undergo pair annihilation.

1. Jul 10, 2007

### urnchurl

Here is a question, which may seem dumb:

Consider an electron and a positron that annihilate, both of which are non-relativistic (taking the observer's frame of reference to be the center of mass frame of the electron-positron pair) so that we only consider their electric fields for simplicity. Where does the energy in their electric fields go after they annihilate?

Thanks.

Last edited: Jul 10, 2007
2. Jul 10, 2007

### Ariste

Away?

3. Jul 10, 2007

### belliott4488

Well, you can't really analyze this problem in any detail non-relativistically, since e+ e- annihilation is an outcome of relativistic quantum field theory. But ... what you really have to keep in mind is the pair of photons that are produced in this interaction - they carry off all the energy and momentum of the original electron and positron.

- Bruce

4. Jul 10, 2007

### urnchurl

What about the energy of the electric fields of the electron and positron?

5. Jul 10, 2007

### lalbatros

photons take the energy away

6. Jul 10, 2007

### olgranpappy

I think that you have to be a little bit careful about how you define the energy of the fields because you can run into some (rather more formal than physical) divergences. So, here is how I would think of the situation:

Consider an electron and a positron being held in place with magic tweezer a distance of 1 meter apart. Let us define the field energy of this configuration as zero. Thus the total energy is the field energy (zero) plus the rest mass energy of 2*511keV.

Then release your magic tweezers and let the electron and positron move. They will of course be attracted towards each other and as such their kinetic energy and their potential energy $$\frac{-e^2}{r}$$ will decrease but always in exactly the right way such that the total energy remains equal to zero (KE + PE) plus 2*511keV (rest mass).

After the positron and electron annihilate the energy is still zero plus 2*511keV. Thus the photons (necessarily more than one) carry off 2*511keV. Furthermore, since in the frame we have been considering the total momentum is zero we must have each photon carry off 511keV a piece.

Cheers,

7. Jul 10, 2007

### urnchurl

That's right, only differences in energy matter (are measurable). There are no absolute energies. I should have remembered that. But isn't the energy of the electric fields just the energy needed to construct the charges?

In this case the energy needed to construct an electron or positron is infinite or otherwise very large depending on whether you assume they are point particles or not. By taking the field configuration energy to be zero you are subtracting off the divergence. This gets into renormalization theory. What I thought was a dumb question, is really a deep one, and I always end up kicking myself for asking things like this.

Thanks again.

Last edited: Jul 10, 2007
8. Jul 11, 2007

### Parlyne

This bothers me a bit. If you start with the electron and positron a finite distance apart and at rest, you have a bound system, which must have a negative total mechanical energy, meaning that the total energy present must be less than 2*511 keV.

9. Jul 11, 2007

### belliott4488

No, the negative energy of a bound system comes from the convention of setting the potential energy equal to zero at an infinite distance. You can set it to be zero anywhere, since it's only changes in potential that are absolute. You might set it to zero at some finite distance if that distance has other significance for some reason.

10. Jul 12, 2007

### Parlyne

Let's play this reasoning out. Let's say I have an electron and a positron starting an infinite distance apart. But, let's take the convention that you're preferring here and set the potential energy to 0 when they're 1 meter apart. This means that they must have non-zero potential energy when they start. Now, let them accelerate towards each other and annihilate. The conservation of energy demands that the total energy of the photons be 2*511 keV + PE_i. This is clearly not the case. So, there must be something wrong with your assertion that, in talking about particle/anti-particle annihilation, we're free to set the 0 of potential energy anywhere we want.

By the by, if you want a demonstration of binding energy reducing the energy available to be released, just think about how nuclear masses compare to nucleon masses. Certainly, the total energy stored in a nucleus at rest is its mass; but, this is less than the combined masses of the nucleons comprising it.

11. Jul 15, 2007

### triggernum5

Considering the two particles would have opposing velocity vectors and equal masses, and that they are both fermions, couldn't the unexplained energy be put to work on a work function that is proportional to KE to decellerate rapidly, almost like a Fermi billiard ball perfect collision?

12. Jul 16, 2007

### Parlyne

No. The details of the interaction don't matter (and won't unless we care about issues of angular distributions). If belliot4488's reasoning is right, conservation of energy demands my conclusion, no matter how the electron and positron interact. Put simply, there's nowhere for energy to go except into the two photons generated.

I agree that in classical situations, changes in potential energy are all that matter, which leads to the freedom to set zeros wherever we like. However, when annihilation is concerned, things get a bit more subtle. It is, hopefully, obvious that if we start the electron and positron at rest an infinite distance apart, the energy released can be nothing other than the sum of their masses, since their kinetic energies are both zero and the energy of assembly of the system is also zero. Now, the energy released from another initial configuration will be this energy plus the difference in energy between the other configuration and the simple case. In the case of two particles at rest one meter apart, the only difference is in potential energy, which has decreased thereby decreasing the available energy for the subsequent photons.

13. Jul 16, 2007

### triggernum5

So correct interpretation of current theory regarding fermion anhililation doesn't incorporate the Pauli Exclusion principle at all?

14. Jul 16, 2007

### Parlyne

The exclusion principle only applies to identical particles. If two fermions can annihilate, they are not identical.

15. Jul 16, 2007

### triggernum5

But in a formation of a neutron star, wouldn't existing neutrons exert a degeneracy pressure on the protons and electrons (which are not identical particles), while the lower degeneracy pressure between the electrons and protons collapes to form more neutrons..
I realize charge is a major aspect that can't be overlooked, but if its irrelevant to the particular wave-functions, or if another symetry is formed, then the particles would be indestinguishable would they not?

16. Jul 16, 2007

### Parlyne

In a neutron star, the degeneracy pressure directly affect the neutrons. There is no degeneracy pressure between protons and electrons. However, there is quite high pressure in the ordinary classical sense, simply due to how far gravity has compacted the matter. This makes is energetically favorable for protons and electrons to inverse beta-decay into neutrons and neutrinos, and these new neutrons do feel the neutron degeneracy pressure.

Since electron/positron annihilation is an electromagnetic process, nothing you can do will let you ignore charge. Spin directions, though, are not a problem, as spin can flip.

17. Jul 17, 2007

### triggernum5

Thanks, laying in bed last night I realized how wrong I must be.. That explains why..

18. Jul 17, 2007

### urnchurl

My original question about the electron and positron annihilating has no place in classical thoery. One must consider quantum electrodynamics, since afterall, pair annihilation is a consequence of relativistic quantum field theory and is easily handled in QED, as opposed to classical electromagnetic theory which does not predict pair annihilation or antimatter. However, this original thought prompted me to review renormalization theory. I'm glad I could stimulate further thought for myself and everyone else.

19. Jul 19, 2007

### affinefield

Pair Production

Pair production is ALWAYS "RELATIVISTIC"!!!.Think of Feynman.Think of the interpretation of processes in space-time,and reverse the direction.The production of a e(+)/e(-) pair from a photon of energy >2mc squared is EXACTLY THE SAME as the time-reversed process.Everything is conserved.The "change-in-the-number" of field quanta is a built-in feature of quantum field theory.The relativistic theory of photon/charged particle interaction.