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Special Rel. 4 momenta, particle decay.

  1. May 1, 2013 #1
    1. The problem statement, all variables and given/known data

    A particle X with rest mass mx is travelling with speed vx = 0.8c along
    the x axis in the lab frame.
    (a) Write down the four-vector momentum of the particle in the lab frame
    in terms of vx and mx.
    The particle decays to two particles, A and B, each with mass mx/4.
    Calculate the energy of these particles, in terms of mx, in the rest frame of
    X. Calculate their speed in this frame.

    2. Relevant equations

    E=γmc[itex]^{2}[/itex]
    E=mc[itex]^{2}[/itex]+pc
    note: four mom. vector = [px,py,pz,E/c] here.

    3. The attempt at a solution

    Alright, so I have the 4 momenta which is p = [E=γmx0.8c, 0, 0, E=γmxc]. so γ=5/3 using the velocity given?
    I am having trouble finding the energy and speed of the particles in the rest frame.
    I think I have mxc[itex]^{2}[/itex]=Ea+Eb. Ea=Eb does it not? In that case
    Ea+Eb=E=γ(m[itex]_{x}[/itex]/2)c[itex]^{2}[/itex]
    because Ea = E=γ(m[itex]_{x}[/itex]/4)c[itex]^{2}[/itex].
    I think I have some bad logic somewhere. I get γ=2 from this, so β=v/c=[itex]\sqrt{3}[/itex]/2?
    If this is correct, is this v/c value for both particles? As they are traveling in opposite directions. It's mostly the intuition which confuses me.

    Thanks :)
     
  2. jcsd
  3. May 2, 2013 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Yes. Good.

    Yes, but you shouldn't be hesitant on this point! Remember, 3-momentum must also be conserved and the created particles have equal mass.

    I'd say your logic is great here!

    Each final particle has β=## \frac{\sqrt{3}}{2}## in this frame of reference.
     
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