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Anit-matter Pro-annihilation

  1. Sep 5, 2006 #1
    We know an electron colliding into a positron annihilate one another and 2 recoiling gamma rays are produced moving in opposite directions...
    Is the process of pair-production, where an electron and positron are created, due to 2 colliding gamma rays annihilating eachother? Furthermore, can a single gamma ray, of sufficient energy, spontaneously annihilate itself to create the same electron and positron pair? Thirdly, is there any basis in experiment or theory that allows for a single gamma ray to interact with the quantum field in a manner than could create an electron/ poistron pair? And finally, is there a version of string theory that would allow for a single gamma ray to interact with tiny dimesions whereupon an electron/ positron pair could be created?
  2. jcsd
  3. Sep 5, 2006 #2
    Nope, it's due to a single > 1022 keV gamma ray photon decaying.

    Yup, see above. :smile:

    Yup, QED explains it theoretically and we have particle accelerators to show it in action. See this image for example.
  4. Sep 5, 2006 #3

    Meir Achuz

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    1. It is difficult to collide two gammas. It does occur vitually, but at such a high energy that just a single e-p (I'll use p for positron.)
    pair is rare.
    2. A single gamma cannot-->e-p. It is forbidden by E and p conservation.
    Pair production usually occurs in the electric field of a nearby nucleus.
    3. No, as in 2.
    4. String theory also conserves E and p, if little else.
  5. Sep 5, 2006 #4


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    Pair production occurs with the interaction of a gamma-ray of at least 1.022 MeV (1022 keV), which is the threshold based on the rest masses of electron/positron, with the nucleus of an atom, or perhaps a heavy particle. Otherwise, photons scatter off electrons - Compton effect - or are absorbed - photoelectric effect.
  6. Sep 5, 2006 #5
  7. Sep 5, 2006 #6

    First off, very nice post. It was well-formatted, concise and insightful, so thank you. Secondly, your link to the bubble chamber picture puzzles me as an explanation for single gamma -->e-p. I'm far from an expert, but I can see the e-p diverge in opposite directions from a single S->North line, BUT this line continues even after the e-p pair is created. So IF this line is the gamma (which I believe it is?) how can it continue beyond the e-p divergence? Is the S->North line consisting of many gammas?

  8. Sep 5, 2006 #7


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    Last edited: Sep 5, 2006
  9. Sep 6, 2006 #8
  10. Sep 6, 2006 #9
    Yes, something like that. I meant that you don't actually need two photons to have an e-p pair produced; a single gamma photon can "kick off" from an atomic nucleus, for example, to produce a electron-positron pair.

    I bet it is a neutron, because it sure doesn't seem to have a charge (it doesn't steer to any specific direction in the magnetic field of the bubble chamber). So I think the image is actually showing a single gamma photon "kicking off" from a neutron to produce the e-p pair. Maybe, maybe not. It could be a high-energy atomic nucleus also.
  11. Sep 6, 2006 #10
    Just a comment on CarlB's proof. There is a much easier and more general way to prove that a photon cannot decay into electron anti electron pair.

    Let P be the fourmomentum of the photon. let P1 and P2 be the 4momenta of the resulting electron and anti electron respectively. the photon deacying is the statement:

    P = P1 + P2 => P^2 = 0 = m^2 + m^2 + 2P1*P2

    now go to the rest frame of the electron, P1 = (m,0,0,0)

    and you get the contradiction that 0 > 2m^2 >0.

    thus a photon (or any other massless particle) can never split into two massive particles, atleast in the absence of external fields.
    Last edited: Sep 6, 2006
  12. Sep 9, 2006 #11
    What is the heaviest known pair of particles produced from gamma annihilation?
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