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That the two gamma rays are only one possible decay mode?

  1. Jun 15, 2005 #1
    Hi, I'm new. I read aboot how an electron, and a positron colliding is supposed to produce two seperate gamma rays in completely opposite directions 180 deg away from each other. I understand that the gamma rays are supposed to correspond perfectly to the mass of the particles involved.

    I am wondering aboot any information regarding this. Especially if sometimes more or less than two gamma rays are produced?...That the two gamma rays are only one possible decay mode? Or maybe that sometimes they give off lesser particles plus electromagnetic radiation?
  2. jcsd
  3. Jun 15, 2005 #2
    The angle (as observed in the laboratory) will depend on the initial velocities of the electron and positron.

    Anywho, a colliding electron and positron will annihalate to 'pure' energy. From this (and simplifying greatly), essentially any particles with total rest mass less than or equal to the colliding energy can be created. Certain conservation laws must hold, however. Laws such as charge conservation (e-e+ collision has charge 0; the created products must have sum charge 0), angular momentum conservation (most importantly spin - the number of gamma rays produced will depend of the spin of the annihalating particles. Electrons have spin ±1/2, photons have spin ±1).

    The gamma photons are indeed only one decay mode; many other decay channels are available, what is allowed depends on the conservation laws. In basic terms, the more energy, the more 'interesting' stuff can happen (ie creation of W and Z bosons which can decay in sorts of ways).

    Do a google search for relativistic kinematics to see what's going on...
  4. Jun 16, 2005 #3
    Thanks for the quick response, I shall check out google.
  5. Jun 16, 2005 #4
    As to the number of gamma rays generated, it must always be a multiple of two. This is because of conservation of momentum. If an electron-positron pair collides head on, the resulting products must have a momentum in the x-direction of 0(because they have the same mass, same velocity) and a momentum in the y-direction of 0(amount of y-momentum possessed by the pair)
  6. Jun 16, 2005 #5
    You don't have to have photons released in multiples of two for momentum conservation - you could quite happily imagine a singular photon flying in the +y direction, and two photons going -y -x and -y +x respectivaly. Assuming this pair are of equal momentum, the angle between these two and the y axis would have to be the same, let's call it [itex]\theta[/itex]. Suppose the momentum of the photon in the +y direction has momentum [itex]p_0[/itex] and the momentum of each of the other photons is [itex]p_1[/itex] we then have the conidition (for conservation of momentum in the y axis):


    It's easy to generalise to any photon momenta.
  7. Jun 16, 2005 #6
    You could have single photons generated, as long there are other particles. For example: e+e- -> e+e- gamma.
  8. Jun 16, 2005 #7
    It's bloody annoying when that happens though :)
  9. Jun 16, 2005 #8
    Wow, I love this forum! I did try to google relativistic kinematics, but that is a little above me right now, but thanks to you I shall be posting here quite often...I had been wondering if 2 x gamma photons were the only decay mode for electron/positron pair anihilation for quite some time. You sir have renewed my interest in this mind-boggling thing called quantum mechanics. Thanks! -Mike
  10. Jun 17, 2005 #9


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    In the standard model, it's quite possible for three photons to emerge from an electron / positron collision. I know I'm just a construction worker so you shouldn't take my word for it, so instead read chapter 4 of this link:

    Less than two photons is impossible, unless there is something else produced. If the "something else" decays into neutrinos, then it could look like just one photon, as the above article mentions (if I recall). Any number of photons more than one is possible. The above article goes on to talk about four photons being produced.

    But more and more photons make the process less and less likely.

    Basically, one photon is forbidden because the two particles coming in have very well established momenta and energies (they are in momentum eigenstates, or as the physicists say, they are on their "mass shell"), and it is impossible to make momentum and energies of the photon (also assumed to be on its mass shell) equal that of the incoming particles. But with any other number, it is possible.

    There is no violation of angular momentum conservation (if it applies, that is, if the incoming particles are in eigenstates of the appropriate operators) because of a bunch of complicated reasons that have to do with stuff that is beyond the scope of this thread.

    Last edited: Jun 17, 2005
  11. Jun 17, 2005 #10
    Hello redwraith94,

    when a positron and electron collide, a so-called positronium can be generated. Positronium is like a hydrogen atom, but there the electron and and positron move around their common center of mass.


    There are two different kinds of positronium:
    (i) para-positronium: the positronium decays to two gamma photons,
    each photon with an energy of 511keV. And the photons decay into opposite directions.

    (ii) ortho-positronium: this positronium has a total spin of [itex]1\hbar[/itex], so the two photon decay is forbidden.
    Instead, ortho-positronium decays to three gamma photons.
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