Electron-Positron Annihilation, low energy vs high energy mode- do both end in 511 ke

Aug 2019
2
0
Electron-Positron Annihilation, low energy vs high energy mode- do both end in 511 keV Annihilation?

Normal room temperature electron-positron annihilation is a well documented process, used in industrial, medical and metroloogy fields, among others.

There are two modes of e-p annihilation, the room temperature or so-called low energy model, and the high energy model (think storage Rings, accelerators, colliders).

In the latter category, do the e-p particles annihilate and produce 511 keV opposed photons like in a normal annihilation, or is there a threshold where e-p annihilation can be forced by kinetic energy alone?

Citations/ references if you please, not personal opinions, I need to prove this to someone, one way or the other.

Thank you for your time.

George Dowell
 

topsquark

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Apr 2008
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Electron-Positron Annihilation, low energy vs high energy mode- do both end in 511 keV Annihilation?

Normal room temperature electron-positron annihilation is a well documented process, used in industrial, medical and metroloogy fields, among others.

There are two modes of e-p annihilation, the room temperature or so-called low energy model, and the high energy model (think storage Rings, accelerators, colliders).

In the latter category, do the e-p particles annihilate and produce 511 keV opposed photons like in a normal annihilation, or is there a threshold where e-p annihilation can be forced by kinetic energy alone?

Citations/ references if you please, not personal opinions, I need to prove this to someone, one way or the other.

Thank you for your time.

George Dowell
2 times 511 keV is a minimum, for when both particles have no momentum. If either (or both) particles are moving we get a relativistic "bump" in final energy content fron the momenta. The most likely channel ends up with photons but high energy collisions can use other channels, like pions or any of the weak vector bosons as an example.

As a source I can't point to any papers. They were probably written in the 40's (at the latest). I don't have access to any journals so I can't give you anything better than to say you can look for them. I can point you to a few texts that you can look for in a library or online. I have an old edition of Kane's which is pretty decent and you can probably find it in a good Science library. There is a more modern version you can find here if you are interesting in purchasing it. There is a good explanation of the Relativity aspects in "Introduction to Elementary Particles" by David Griffiths. (From 1987. I think there are a couple of later edictions of this one.)

-Dan
 
Aug 2019
2
0
Asked and answered- thanks

2 times 511 keV is a minimum, for when both particles have no momentum. If either (or both) particles are moving we get a relativistic "bump" in final energy content fron the momenta. The most likely channel ends up with photons but high energy collisions can use other channels, like pions or any of the weak vector bosons as an example.

As a source I can't point to any papers. They were probably written in the 40's (at the latest). I don't have access to any journals so I can't give you anything better than to say you can look for them. I can point you to a few texts that you can look for in a library or online. I have an old edition of Kane's which is pretty decent and you can probably find it in a good Science library. There is a more modern version you can find here if you are interesting in purchasing it. There is a good explanation of the Relativity aspects in "Introduction to Elementary Particles" by David Griffiths. (From 1987. I think there are a couple of later edictions of this one.)

-Dan
Thanks for the thoughtful reply Dan. Actually I did find the answer at Lawrence Berkley Lab Particle site :The Particle Adventure | Particle decays and annihilations | Electron / positron annhiliation

the formulais e+e- > D+D-

been looking for that for a long time, just never used the right search I guess.

Nice meeting you. I'm into nuclear radiation detection.

George