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Pion decay question

  1. Jun 18, 2010 #1
    Forgive my question for being so elementry, but I'm jumping a bit ahead of my curriculum...

    I've attached a decay process for the negative pion. I actually spliced together the initial pion decay and then the subsequent muon decay, assuming the entire process is correct.

    1) the first W- boson decays to a muon, the second W- boson decays to an electron. I assume that the greater energy for the first W- decay is due to it's greater momentum - given it's mass is set at about 80Gev, I assume this is due to greater velocity than the second W- boson decay to an electron???? Is that correct??

    2) If the first W- boson has greater velocity, is it shorter lived? or is the life cycle of the W- boson set? Meaning it would travel a greater distance before decaying....

    any insights, feedback, etc would be greatly appreciated...



    Attached Files:

  2. jcsd
  3. Jun 18, 2010 #2

    It also seems that the electron anti-neutrino would have to carry off more momenta than the muon anti-neutrino....???
  4. Oct 16, 2010 #3
    Weak interaction is a V-A theory. Through calculation, pion decays into muon nearly 100 percents. It could decay into electron, but the decaying rate is very small. Then muon decays into electron.
  5. Oct 16, 2010 #4
    As far as I know, it is impossible to detect the W bosons directly so far. In pion's or muon's decay, it could only exist for a very short time. You can estimate it according to the uncertainty relation.
  6. Oct 16, 2010 #5
    In fact, if you check you'll find that conservation of energy and momentum at each vertex forbids the Ws from having mass anywhere near 80 GeV, given that the initial pion has a mass of only about 150 MeV. This means that the Ws are both highly off-shell, which greatly suppresses the rate of these processes.

    The muon, on the other hand, is on-shell and can survive for macroscopic times and distances, which is why these are usually drawn as separate processes. The muon has time to interact with matter around it, making the muon decay process decohere from the pion decay.

    Back to your questions. Since velocity is a frame-dependent quantity, we can't unambiguously talk about any particle in this process, whether on or off-shell. So, we will consider the process to occur in the rest frame of the pion. Since the entire pion transmutes into a W in the process of decaying, the first W must, in fact, be at rest as well. The muon will have momentum
    [tex]p = \frac{m_\pi^2-m_\mu^2}{2m_\pi}[/tex]
    and (thus) energy
    [tex]E = \frac{m_\pi^2+m_\mu^2}{2m_\pi}[/tex].
    This means it has velocity
    [tex]v = \frac{m_\pi^2-m_\mu^2}{m_\pi^2+m_\mu^2}.[/tex]

    The kinematics of the muon decay will depend on the angles between its velocity and those of the decay products, as well as the angles among those products. However, we can see that the second W has no particular reason to be at rest. But, you could use relativistic kinematics to put limits on the allowed velocities (remembering that its effective mass will be the invariant mass of the electron/electron-neutrino pair).
    Last edited: Oct 16, 2010
  7. Oct 16, 2010 #6
    You probably should analyze the pi-μ-e decay from the standpoint of the pion rest frame (as stated by Parlyne). In this frame, the muon kinetic energy is ~ 4.12 MeV. So it is nearly at rest. So in the pion rest frame, nearly all of the electron kinetic energy comes from the Q of the μ-e decay. All the rest of the energy and angular distribution (excepting parity conservation issues) comes from the Lorentz transformation into the moving pion frame.

    Bob S
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