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Simple cons. momentum question

  1. Mar 3, 2013 #1
    Hi, I've been struggling with this question for a while, can't help but think I'm missing something obvious
    Let's say I have a particle of rest mass 2m. In the frame S, it moves with velocity v. S' is the rest frame of the particle. In S, the momentum of the particle (c=1)
    [itex]p=\frac{2mv}{\sqrt{1-v^2}}[/itex]

    Now, in S', the particle splits into two particles of mass m with velocities v and -v. The particles with -v will be stationary in S, the particle with v will have velocity vnew
    [itex]v_{new}=\frac{v+v}{1+v^2}=\frac{2v}{1+v^2}[/itex]
    by the velocity addition formula

    So the momentum in S
    [itex]p=\frac{mv_{new}}{\sqrt{1-v_{new}^2}}[/itex]

    This doesn't appear to be equal to the p before (just by trying v=0.5), so something's clearly gone wrong. Can someone help?
    Thanks
     
  2. jcsd
  3. Mar 3, 2013 #2

    Bill_K

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    Science Advisor

    This step does not conserve energy. In S' before the split the total energy was 2m. But after the split each particle will have energy greater than m, since it has rest energy m plus kinetic energy.
     
  4. Mar 3, 2013 #3
    Ah I see. Thank you
     
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