Silly question about pair annihilation

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 2K views
Einj
Messages
464
Reaction score
59
Hi all. I have a (probably stupid) question about e[itex]^{+}[/itex]e[itex]^{-}[/itex] annihilation. I was reading about the emission of two 511keV photons which is the electron's mass. But, is this energy fixed every time e[itex]^{+}[/itex]e[itex]^{-}[/itex] annihilates in a two photons or is just the energy of the positronium dacay?

My idea is that in e[itex]^{+}[/itex]e[itex]^{-}[/itex] → γ γ the energy of the two photons must be equal, and equal to one half the total energy in the initial state. But who tells me that this initial energy is just the sum of electron e positron masses?? Don't we have a contribution from their momentum??
 
Physics news on Phys.org
Apart from electron-positron colliders, annihilation usually occurs with slow particles - if you shoot a positron in matter, it usually slows down before it annihilates, as the cross-section increases with decreasing kinetic energy. There, 511keV per photon is a good approximation.

In the center of mass frame of electron+positron, the photons have equal energy. In other frames (like the lab), the energy can be different.
 
Understood. Thank you very much! :biggrin: