# Pion decaying into two photons

1. Sep 19, 2009

### koab1mjr

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

A neutral pion traveling along the x axis decays into two photons, one being ejected exactly forward and the other exactly backward. The first photon has three times the energy of the second. Prove that the original pion had speed 0.5c.

2. Relevant equations

for m=0, E=p*c
conservation of Energy E^2=(c*p)^2+(m*c^2)^2
gamma=1/sqrt(1-Beta^2)
Beta = v/c
p=gamma*m*v
E=gamma*m*c^2

3. The attempt at a solution

I know momentum and energy are conserved giving me the following equations. Epion = 3*Ephoton1 + Ephoton2 and Ppion = Photon1 - Pphoton2. I know the mass of a pion at 140MeV/C^2. For a massless particles E = pc and for the pion E = gamma*m*c^2. I want to find expressions for Ephoton1 + Ephoton2 that do not have P or E in it to solve for the velocity. Though I cannot seem to make any progress with this approach. I need a hint.

2. Sep 20, 2009

### Drew7918

I had troubles on this one too.

First and foremost is that the frequency of either photon is not given, so right away, conservation of energy gives nothing. However, if you solve for the frequency of one of the photons, say photon(a) by using conservation of momentum, you will get:

f(a)=(3(gamma)mvc/2h)

Then, by plugging that number into f(a) for conservation of energy, Doctor Fenstermacher made it very easy to cancel out all variables that we don't need.

See you in class!

3. Sep 21, 2009

Staff Emeritus
Use E2 - p2 = m2. The use the relationship between E and m, or p and m, to give you v.

Note that you don't need the pion mass.

4. Sep 21, 2009

### koab1mjr

Thanks for the help!!!!

Solved it

5. Sep 21, 2009