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Relativity - photon 'decay'

  1. May 27, 2010 #1
    1. The problem statement, all variables and given/known data

    A photon decays into a positron and an electron - show why this is impossible. Under what conditions could this be possible?

    2. Relevant equations
    conservation of linear momentum:
    [tex] P_{PHOTON} = P_{POSITRON} + P_{ELECTRON} [/tex]

    conservation of energy:
    [tex] E_{PHOTON} = E_{POSITRON} + E_{ELECTRON} [/tex]


    3. The attempt at a solution

    [tex] P_{PHOTON} = P_{POSITRON} + P_{ELECTRON} [/tex]
    [tex] \frac{E_{PHOTON}}{c} = \gamma mu + \gamma mu = 2\gamma mu [/tex]


    [tex] E_{PHOTON} = E_{POSITRON} + E_{ELECTRON} [/tex]
    [tex] E_{PHOTON} = \gamma mc^{2} + \gamma mc^{2} = 2\gamma mc^{2} [/tex]

    [tex] \frac{E_{PHOTON}}{c} = 2\gamma mu = \frac{2\gamma mc^{2}}{c} [/tex]

    therefore:

    [tex] u = c [/tex] …which would be impossible. is this right? I'm a bit confused on how you apply conservation of energy and momentum in relativity. for instance i haven't considered any reference frames here and i'm not sure if it requires some kind of transformation.

    any help is appreciated, thank you!

    btw I've assumed that the product particles have equal rest mass and share the energy equally. and u is the velocity of the positron/ electron

    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: May 27, 2010
  2. jcsd
  3. May 27, 2010 #2

    kuruman

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    Your momentum conservation equation assumes that the electron and positron have the same speed and direction of travel. This is not necessarily the case. Actually the argument for this is quite simple. I will point you in the right direction. Consider the center-of-mass frame of the two particles. In that frame, what is the momentum after the "decay" of the single photon? What about before the "decay"?
     
  4. May 27, 2010 #3
    To complement this argument, consider a photon in any frame of reference, what is its momentum?
     
  5. May 28, 2010 #4
    thanks, so it's just a case of showing that in the c.o.m frame the process doesn't conserve momentum, e.g. the sum of momenta of products is zero, which doesn't tally with momentum of the photon (always > 0).

    hi, not sure how to apply this to all reference frames. surely in some momentum is conserved?

    also, when solving particle production problems etc. is it generally best to work in the c.o.m frame to ensure linear momentum is conserved? perhaps i've misunderstood something...

    thanks again!
     
    Last edited: May 28, 2010
  6. May 28, 2010 #5
    Remember the two postulates of Special Relativity. What do they say about the speed of light?

    The center of mass frame is useful in certain situations. Since total linear momentum is zero, some of the equations are simpler to deal with. However, there are cases in which there is an easier solution that doesn't involve considering the center of mass frame.
     
  7. May 28, 2010 #6

    vela

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    Momentum is conserved in every reference frame.
     
  8. May 29, 2010 #7
    You've hit the nail on the head, but didn't notice it. :)

    Whenever you have two material particles (Not photons, meaning that their velocity is confined to be lower than the speed of light), you can always find a frame of reference in which the total momentum is 0.

    In this frame of reference, the total momentum before the decay, would be solely that of the photon, but that would have to be 0, which is impossible, because as you stated, a photon always has non-zero momentum in every possible frame of reference (Ones moving at the speed of light are senseless)
     
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