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Decay of muon

  1. Nov 21, 2007 #1
    the mc^2 for a pion and muon are 139.57 MeV and 105.66 MeV respectively. Find the kinetic energy of the muon in its decay from [tex] \pi ^+ -> \mu^+ + \nu_{\mu} [/tex] assuming the neutrino is massless. Here's what I did:

    Since [tex]E^2=p^2c^2+m^2c^4[/tex] and that c=1, then E, p and m have same units.

    [tex]E^2 = p^2 +m^2[/tex]
    [tex](139.57 MeV)^2 - (105.66MeV)^2 =p^2[/tex]

    Also consider the case where there is a small neutrino mass:

    [tex]E^2 = p^2 +m^2[/tex]
    [tex](m_{\pi})^2 - (m_{\mu}+m_{\nu})^2 =p^2[/tex]
    [tex]p=\sqrt{(m_{\pi})^2 - (m_{\mu}+m_{\nu})^2}[/tex]

    I feel like there is ill logic here. Comments on my work would be appreciated.
  2. jcsd
  3. Nov 21, 2007 #2


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    There is ill logic. You are dealing with a three body problem. Think four vectors. (E_pion,p_pion)=(E_muon,p_muon)+(E_neutrino,p_neutrino). You can only apply E^2=p^2+m^2 to each individual vector, not somehow magically to the whole system.
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