The question is:

A pion decays into a muon and a neutrino (which is almost massless, so we’ll take it to have a mass of zero). If the pion is initially at rest, find the energy of the outgoing muon.

Final Answer: $$E_μ=(m_π^2+m_μ^2)c^2/2m_π$$

I used conservation of energy and so the initial energy is just the rest energy of the pion which equals $m_πc^2$ and this should equal the final energy.

Getting the final energy is what I am having trouble with, the mass of the neutrino is approx. 0 so I used the $$E=\sqrt{p^2c^2 + m^2c^4}$$ and crossed out the term with m leaving $E_\nu=pc$. Then, for the muon there is rest energy and kinetic energy I eventually get $$E_μ=m_μc^2/\sqrt{1-u_μ^2/c^2}.$$

However, the final answer neither includes a $p$ nor a $u$, so I think that the problem is in finding the final energy and isolating for the $E_μ$.

  • 1
    $\begingroup$ Hi and welcome to physics SE. Please, use laTex notation for formulae. It's about writing them in between of dollar symbols, and laTex commands inside. See here: math.meta.stackexchange.com/questions/5020/… $\endgroup$
    – FGSUZ
    Jan 29 '19 at 0:44
  • $\begingroup$ You are simply forgetting an equation: conservation of momentum $\endgroup$ Jan 29 '19 at 0:48
  • $\begingroup$ @GabrielGolfetti That's more like an answer than like a comment. $\endgroup$
    – rob
    Jan 29 '19 at 1:24

Try to use conservation of momentum to get a new equation. It should eliminate the velocity dependence of the answer.

  • $\begingroup$ OK thank you, I also realized that for the momentum, I need to isolate pv from the Ev=pv*c equation $\endgroup$ Jan 29 '19 at 3:19
  • $\begingroup$ note that $-p_{\mu} = p_{\nu} = E_{\nu} = m_{\pi} - E_{\mu} $ $\endgroup$
    – JEB
    Jan 29 '19 at 3:27
  • $\begingroup$ @JEB can you explain that more? $\endgroup$ Jan 29 '19 at 4:29
  • $\begingroup$ To the first part: Note that $-p_{\mu} = p_{\nu}$ because you can consider the process in the centre-of mass frame. To the middle part: $p_{\nu} = E{\nu}$ because the neutrino is assumed to be mass-less (remember $\left|\vec p\right| = E\left|\vec \beta\right|$). To the last part: That I don't know yet .. $\endgroup$
    – user248824
    Nov 23 '20 at 18:46

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