Special Relativity - energy-momentum conservation

[rak576]

Homework Statement

A positron of rest mass me, kinetic energy equal to its rest mass-energy, strikes an electron at rest. They annihilate, creating two high energy photons a and b. The photon a is emitted at the angle of 90 degress with respect to the direction of the incident positron.

(a) Split 4-vector law of energy-momentum conservation into energy and momentum conservation laws in the rest frame of the electron.

(b) Show that the total energy of the emitted photons Ea+Eb=3me*c^2 and that Eb^2 = Ea^2 + (p^2)*(c^2) where p is 3momentum of positron and Ea, Eb are energies of photons a and b.

(c) Use these results and identity E^2 = (p^2)*(c^2) + m0^2*c^4 to show Eb = 2me*c^2 and Ea = me*c^2. Find the direction of motion of photon b. In particular show that the angle between its direction and the direction of the positron is theta = arcsin(1/2)

Homework Equations

The Attempt at a Solution

Qp, Qe = 4momenta of positron and electron
Pa, Pb = 4momenta of photon a and b.

Qp = (2me*c, me*v) v=velocity of positron
Qe = (me*c, 0)
Pa = (Ea/c, Ea/c n) n=direction of photon
Pe = (Eb/c, Eb/c n)

This is far as I can get! Can anybody please help? Have I split up the conservation law okay?
Thanks

 

[kuruman]

A conservation law is usually represented as an equation. Here, a "split" would be to say

Momentum before = Momentum after
Energy before = Energy after

You have not done that.

 

[rak576]

Sorry, I didn't put that in.

Qp + Qe = Pa + Pb

 

[kuruman]

That's only one equation. You need two plus you need to put in symbols for the rest masses, speed of positron, etc. in these two equations.