Kinetic Energy of Muon in Decay of Pi+ -> Mu+ + Nu_mu

[indigojoker]

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 $$\pi ^+ -> \mu^+ + \nu_{\mu}$$ assuming the neutrino is massless. Here's what I did:

Since $$E^2=p^2c^2+m^2c^4$$ and that c=1, then E, p and m have same units.

$$E^2 = p^2 +m^2$$
$$(139.57 MeV)^2 - (105.66MeV)^2 =p^2$$
$$p=91.19MeV$$

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

$$E^2 = p^2 +m^2$$
$$(m_{\pi})^2 - (m_{\mu}+m_{\nu})^2 =p^2$$
$$p=\sqrt{(m_{\pi})^2 - (m_{\mu}+m_{\nu})^2}$$

I feel like there is ill logic here. Comments on my work would be appreciated.

 

[Dick]

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.