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Speed of the muon

  1. Feb 2, 2006 #1
    The positive pion decays into a muon and a neutrino. The pion has rest mass [tex] m_\pi [/tex], the muon has [tex] m_\mu [/tex] , while the neutrino has [tex] m_v = 0[/tex]. Assuming the original pion was at rest, use conservation of momentum and energy to show that the speed of the muon is given by:

    [tex] \frac{u}{c} = \frac{ (m_\pi/m_\mu)^2 - 1}{ (m_\pi/m_\mu)^2 + 1} [/tex]

    I've tried [tex] m_\pi c^2 = \gamma m_\mu c^2 [/tex]. No success with that. I've also tried the relationship [tex] \beta = pmuc/E [/tex]. Still no. I know it SAYS to use conservation of energy and momentum, but I have yet to put together a correct relationship. Any help?
    Last edited: Feb 2, 2006
  2. jcsd
  3. Feb 2, 2006 #2


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    HINT: Since the neutrino is massless its energy and momentum are related by E = pc.
  4. Feb 2, 2006 #3
    So did I leave out the energy from the neutrino?
    [tex] \pi^+ \rightarrow \mu^+ +v[/tex]

    [tex] m_\pi c^2 = \gamma m_\mu c^2 + p_vc[/tex]
    Last edited: Feb 3, 2006
  5. Feb 3, 2006 #4


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    Yes, you left out the neutrino energy.

    You should arrive at

    [tex]\frac { m_{\pi}} {m_{\mu}} = \gamma (1 +u/c)[/tex]

    after combining the energy and momentum equations and the rest is simple algebra.
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