1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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] 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:


    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted