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rak576
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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