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A particle discharge a photon

  1. May 12, 2013 #1
    1. The problem statement, all variables and given/known data
    A particle of mass m is at rest in the lab frame.
    The particle discharge a photon with energy 1/2mc^2 to the direction of x+.
    A spaceship is moving at v = 0.8c in the direction x+ (in the same direction the photon is moving).
    What is the total energy of the particle after it discharge the photon relativity to the spaceship frame?


    2. Relevant equations
    [itex]E = m{\gamma}c^2[/itex]
    [itex]P = m{\gamma}v[/itex]
    [itex]E_{photon} = pc [/itex]

    3. The attempt at a solution
    Let's mark the new mass after the discharge: [itex]m'[/itex]
    And the new [itex]\gamma[/itex] after the discharge : [itex]\gamma'[/itex]
    And the energy of the particle after the discharge relative to the lab: [itex]E_{particle}[/itex]


    The energy before the discharge:
    [itex]E_i = mc^2[/itex]

    The momentum before the discharge:
    [itex]P_i = 0[/itex]

    After the discharge the energy and the momentum are same, so:

    [itex]E_{photon} = 1/2mc^2[/itex]
    [itex]E_f = mc^2 = E_{photon} + E_{particle} = m'{\gamma'}c^2 + 1/2mc^2[/itex]
    so:
    [itex]1/2m = m'{\gamma'}[/itex]

    since [itex]E_{photon} = pc [/itex] we can say that:
    [itex]P_{photon} = 1/2mc[/itex]
    and:
    [itex]0 = P_i = P_f = 1/2mc + m'{\gamma'}v[/itex].
    but
    [itex]m'{\gamma'} = 1/2m[/itex]
    so:
    [itex]0 = P_i = P_f = 1/2mc + 1/2mv[/itex].
    so:
    [itex]v = -c[/itex].
    So the particle is seen from all frames with the same velocity and there for
    the energy is the same that observed in the lab frame:
    [itex]Ans = mc^2-1/2mc^2=1/2mc^2[/itex].

    However this is not correct (by my reference), does anyone know why?
    By my reference the right answer is:
    [itex]Ans = {3/2}mc^2[/itex].
    Thanks,
    Shai.
     
  2. jcsd
  3. May 12, 2013 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    The particle has to be massless, I agree (which is a bit strange, but possible). This gives a speed of c in all reference frames, but the energy is still frame-dependent.

    By the way, I never heard "discharge" for those processes. Usually, they are called "emission", if the result is another massive particle. If a particle decays to two photons (as in this example), I would simply call it "decay".
     
  4. May 12, 2013 #3
    You are right!!
    After I apply

    [itex]E'={\gamma}(E-pv)[/itex]
    I get the correct result!
    Thank.
    Regarding the name of the process you are probably right, I translated this question from Hebrew so...
     
  5. May 12, 2013 #4

    tiny-tim

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    Science Advisor
    Homework Helper

    <h2><strong>welcome to pf!</strong></h2>

    hi buc030! welcome to pf! :smile:

    a more direct way would be to use the red-shift formula to calculate the energy of the photon, and the initial energy of the particle, in the .8c frame, and just subtract :wink:
     
    Last edited: May 12, 2013
  6. May 13, 2013 #5
    Yeah you have a point there :)
     
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