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Relativistic decay of a particle

  1. Jul 24, 2012 #1
    1. The problem statement, all variables and given/known data

    Pions can decay via the reaction π+ → μ+ νμ. Show that the energy of the neu- trino in the rest frame of the pion is given by

    [tex] E_v = \frac{m^2_∏-m^2_μ}{2m_∏} [/tex]



    Pions with energy Eπ in the laboratory frame (Eπ >> mπc2) decay via the above reaction. Show that the maximum energy of the neutrinos in the laboratory frame is given by


    [tex] E^{max}_v = E_∏\frac{m^2_∏-m^2_μ}{m^2_∏} c^2[/tex]


    2. Relevant equations

    Lorentz Transform

    [tex] \left(\begin{array}{cc}E'\\p'c\end{array}\right) = \left(\begin{array}{cc}\gamma&\beta\gamma\\-\beta\gamma&\gamma\end{array}\right) * \left(\begin{array}{cc}E\\pc\end{array}\right) [/tex]



    3. The attempt at a solution

    I am able to get the first result, the problem I am having is changing frames from the first result to the second result. I take the first result and use the Lorentz transform ont he energy, where prime denotes the lab frame and unprimed is the rest frame of the pion.
    this give me:


    [tex] E_v' = \frac{m^2_∏-m^2_μ}{2m_∏} \gamma(1-\beta) c^2 [/tex]

    i then use the fact that E'_∏ = gamma mc^2 to find gamma, giving me:


    [tex] E_v' = \frac{m^2_∏-m^2_μ}{2m^2_∏} E_∏(1-\beta) c^2 [/tex]

    However I have done something wrong as this means I need beta =-1 to give me the right result. Any help would be greatly appreciated.
     
  2. jcsd
  3. Jul 24, 2012 #2
    You've done something wrong with your LT matrix. Both [itex]\beta \gamma[/itex] terms should have the same sign.
     
  4. Jul 25, 2012 #3
    Ahh yes, however that was only a mistake I made when typing the problem up to pf, my calculations had both with the -ve sign to get the result below it.
     
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