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Classical mechanics exercise, pion decay!

  1. Sep 15, 2013 #1
    1. The problem statement, all variables and given/known data

    If anyone could help me with this classical mechanics exercise I would be very grateful! The exercise is as follows:

    The muon (μ) is a particle with mass mμ=207me, with me being the electron mass. The pion (∏) has a mass of m=273me. The pion can decay into a muon plus a massless neutrino, v, in the reaction  ∏ → μ + v. Find the kinetic energy of the muon when a pion decays at rest. Hint: Use conservation of both energy and momentum.


    2. Relevant equations

    The momentum conservation looks like this:

    Pbefore=Pafter

    Where the momentum is given by:

    P=mμγ(u)

    The energy conservation is:

    Ekin, before=Ekin, after

    The kinetic energy is given by:

    Ekin=mc^2(γ(u)-1)

    quantities we know:

    mμ=207 me

    m=273 me

    mv=0

    u=0 (the speed of the pion)

    3. The attempt at a solution

    I have tried to write the equations, please correct me if I'm wrong:

    Ekin, before = m c^2 (γ(u)-1)

    Ekin, after = Ekin, μ + Ekin, v = mμ c^2 (γ(u)-1) + mv c^2 (γ(u)-1)

    Pbefore=P=m u γ(u)=0

    Pafter=Pμ+Pv=mμ uμ γ(u) + mv uv γ(u) = mμ uμ γ(u) + [itex]\frac{E}{c}[/itex] (because a massless particle can still have momentum)

    Now I'm a bit stuck. What is the kinetic energy of the neutrino with zero mass, zero? How do I find the energy, E in the momentum expression, is that the kinetic energy? It seems to me that there are too many unknowns, but I'm pretty sure I'm wrong here!


    If anyone could help me, I would I will be very grateful!
    Thanks in advance.
    mr. bean.
     
  2. jcsd
  3. Sep 15, 2013 #2
    The pion is initially at rest so the total energy is equal to the rest mass of the pion, there is not kinetic energy before so get rid of the γ. Start there and then imagine that when the particle decays in order to make it so that the total momentum is still zero the muon and the neutrino must travel in opposite directions with equal momentum. From this you can solve for E.
     
  4. Sep 15, 2013 #3

    vela

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    A few suggestions:
    1. Avoid using ##\gamma## and any velocities. It just makes the algebra more painful. Work only with energy, momentum, and mass instead. For example, the kinetic energy is given ##E-mc^2##.
    2. Use the relationship ##(mc^2)^2 = E^2 - (pc)^2##.
    Your equation for conservation of energy is incorrect. Kinetic energy obviously is not conserved. Initially there is no kinetic energy because the pion is at rest whereas the muon and neutrino do have kinetic energy. It's total energy E that's conserved.
     
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