# Relativistic kinetic energy - particle decay

1. Mar 25, 2015

### Phynos

1. The problem statement, all variables and given/known data

A pion at rest (mπ = 273me) decays into a muon (mμ = 207me) and an antineutrino (mn ≈ 0).

Find (a) the kinetic energy of the muon and (b) the energy of the antineutrino in electron volts.

2. Relevant equations

K = (γ-1)mc2

E = γmc2

ER = mc2

E2 = p2c2 + (mc2)2

I didn't use all of these, but I'm assuming a solution requires one or more of them.

3. The attempt at a solution

Not really sure how to go about this when the mass of the neutrino is so close to zero. If I assume the total energy after the reaction of the neutrino is essentially zero, then the missing energy goes into the motion of the muon.

Kμ ≈ E - E
= c2 (mπ - mμ)
= c2 (0.511*106eV/c2)(273-207)
= 3.37*107 eV
= 33.7 MeV

The answer in the book is 4.08MeV so either my assumption that Kn ≈ 0 is a bad one, or I've made a mistake somewhere here.

2. Mar 25, 2015

### Orodruin

Staff Emeritus
Yes, your assumption is bad. In fact, in a two body decay, the lighter decay product will carry most of the kinetic energy.

You need to fulfill the conservation laws involved, which are they?

3. Mar 25, 2015

### Phynos

So momentum needs to be conserved, however the momentum is proportional to mass just like kinetic energy, which is what lead me to conclude the kinetic energy must be close to zero. How do I go about calculating the momentum of something for which I do not know the mass?

4. Mar 25, 2015

### Orodruin

Staff Emeritus
Momentum is only proportional to mass for fixed velocity < c. You can only draw the conclusion that a light particle has smaller energy or momentum than a heavy if they have the same speed.

You already have a relation between the momentum and total energy of a particle. This relation is true also for massless particles.