What Is the Speed of a Neutral Pion Decaying into Two Photons?

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Homework Statement



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.

Homework 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


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.
 
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koab1mjr said:

Homework Statement



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.

Homework 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


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.

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!
 
Thanks for the help!

Solved it
 
Let me take a crack at it. I'll be back