# Relativistic energy/momentum, massless particles

zippideedoda

## Homework Statement

A pion at rest decays into a muon and an antineutrino. The mass of the antineutrino is zero, find the energies and momenta of the muon and antineutrino. Mass of the pion is 139.57 MeV/c^2 and the mass of the muon is 105.66 MeV/c^2

## Homework Equations

pion -> muon + antineutrino

(1) E=mc^2
(2) E=pc
(3) E^2 = (pc)^2 + (mc^2)^2

## The Attempt at a Solution

Conservation of energy: E(pion) = E(muon) + E(antineutrino)
Using equation 1 and the given masses:
E(pion) = 139.57 MeV
E(muon) = 105.66 MeV (?)
so E(antineutrino) = E(pion) - E(muon) = 33.91 MeV
Use equation 2 for the massless antineutrino and get p = 33.91 MeV/c

since momentum is conserved and p(pion) = 0 (at rest),
p(muon) = -p(antineutrino) = -33.91 MeV/c

I don't think I'm right because if the muon has momentum it is moving and thus I can't find its energy by simply plugging its mass into equation 1.

Thanks for any help.

Last edited:

## Answers and Replies

You are correct in that you can't use Equation 1 for the energy of the muon. You have to use Equation 3. Otherwise you are on target.

zippideedoda

p(muon) = p(antineutrino) = p
E(antineutrino) = pc
E(muon) = sqrt[(pc)^2+(mc^2)^2]

plug these into conservation of E and solve for p
after some math I get

p = 29.79 MeV/c
so E(antineutrino) = 29.79 MeV
E(muon) = 109.78 MeV

The two E's add up to equal E(pion), 139.57