# Pion decaying energy

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 $$E_μ$$.

• 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/… Jan 29 '19 at 0:44
• You are simply forgetting an equation: conservation of momentum Jan 29 '19 at 0:48
• @GabrielGolfetti That's more like an answer than like a comment.
– rob
Jan 29 '19 at 1:24

## 1 Answer

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

• OK thank you, I also realized that for the momentum, I need to isolate pv from the Ev=pv*c equation Jan 29 '19 at 3:19
• note that $-p_{\mu} = p_{\nu} = E_{\nu} = m_{\pi} - E_{\mu}$
– JEB
Jan 29 '19 at 3:27
• @JEB can you explain that more? Jan 29 '19 at 4:29
• 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 ..
– user248824
Nov 23 '20 at 18:46