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Muon momentum

  1. Nov 24, 2007 #1
    1. The problem statement, all variables and given/known data

    in the decay process, [tex] \pi ^{+} -> \mu^{+} + \nu_{\mu} [/tex]

    show that for a neutrino of finite (but small) mass, compared with the case of the massless neutrino, the muon momentum would be reduced by the fraction:

    [tex]\frac{p'}{p}= - \frac{m_{\nu}^2 (m_{\pi}^2+m_{\mu}^2)}{(m_{\pi}^2-m_{\mu}^2)^2} [/tex]

    Calculation of the muon's momentum with massless neutrino:

    [tex] p_{\mu}=p_{\pi}-p_{\nu}[/tex]
    [tex] p_{\mu}^2=p_{\pi}^2+p_{\nu}^2-2m_{\pi}E_{\nu}[/tex]
    We know:
    [tex]p_{\mu}=m_{\mu}^2c^2[/tex]
    [tex]p_{\pi}=m_{\pi}^2c^2[/tex]
    [tex]p_{\nu}=0[/tex]

    [tex] m_{\nu}^2c^2=m_{\pi}^2c^2-2m_{\pi}E_{\nu}[/tex]

    But: [tex] E_{\nu}=|p_{\nu}|c=|p_{\mu}|c [/tex]

    [tex] 2 m_{\pi} |p_{\mu}|=(m_{\pi}^2-m_{\mu}^2)c [/tex]
    [tex]|p_{\mu}|=\frac{m_{\pi}^2-m_{\mu}^2}{2m_{\pi}}c [/tex]

    Using the same method, but with [tex]p_{\nu}=m_{\nu}^2c^2[/tex]

    we get:
    [tex] 2 m_{\pi} |p_{\nu}|=(m_{\pi}^2-m_{\mu}^2-m_{\nu}^2)c [/tex]
    [tex]|p'_{\mu}|=\frac{m_{\pi}^2-m_{\mu}^2-m_{\nu}^2}{2m_{\pi}}c [/tex]

    The fraction I get is:

    [tex]\frac{p'}{p}=\frac{m_{\pi}^2-m_{\mu}^2-m_{\nu}^2}{m_{\pi}^2-m_{\mu}^2}[/tex]

    If i multiply by [tex]\frac{m_{\pi}^2-m_{\mu}^2}{m_{\pi}^2-m_{\mu}^2}[/tex] I do not get the above result. Any idea where i went wrong?
     
    Last edited: Nov 24, 2007
  2. jcsd
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