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: 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


    User Avatar
    Science Advisor
    Homework Helper

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