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A little confused about photon decay

  1. Dec 7, 2006 #1
    I have this question.

    Give relativistic expressions for the energy E and the magnitude of the momentum p of a particle of mass m moving at speed v. The expression E^2 − p^2c^2 is an invariant in the Special Theory of Relativity. Explain what this means and state the value of the invariant for a single particle.

    A particle of mass m moving at speed 3/5c collides with an identical particle at rest, and forms a new particle of mass M which moves off at speed v. Show that v = 1/3c and find an expression for M.

    The particle of mass M subsequently decays into two photons, one travelling
    in the direction of incidence of the mass m, and the other travelling in the opposite direction. Determine the energy of each photon.

    If instead the decay produces photons of equal energies, what are their
    momenta and directions?


    I have no problem in answering the first two parts of the question. the last part I know how to solve too: by saying photons are moving in opposite directions along the y-axis in the zero momentum frame. It is the third part where photons are moving parallel to x-axis which I don't get. The way it asks is implying that the two photons have different energies. Well fine if it is not a massless particle, in which case the speed of each particle in the stationary frame will have different speed and thus different kinetic/total enery. But with photons and such their speed is fixed, and they don't have mass, so in the zero momentum frame the two photons will have equal magnitude of momentum but in opposite directions and this will be the same in the stationary frame, since there is no change in velocity. So because E=p/c for photons, than they will have the same energy!? I am confused, could somebody sort me out please. thanks a million.
     
  2. jcsd
  3. Dec 7, 2006 #2

    OlderDan

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    What do you know about the Doppler effect for light?
     
  4. Dec 7, 2006 #3
    oh i see, why didn't i not realize that before! thanks a lot. So I say that in the zero momentum frame, each photon has same amount of energy i.e. half of the original energy, so use this energy the original emitted frequency can be calculated, than use the Doppler effect for light calculate the observed frequency for each photon, and thus the energy. Is my logic right? thanks ^_^
     
  5. Dec 7, 2006 #4

    OlderDan

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    That is one way to approach the problem. I don't think it would be difficult for two photons to do the calculation directly in the lab frame using conservation of energy and momentum, which would be another approach.
     
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