Posted by Mischa on Wednesday, May 7, 2008 at 6:08pm.
a) follows from conservation of energy:
Photon energy + m = photon energy after collision plus gamma m
(in c = 1 units)
In b) you have to take into account that both the photon after the collision and the electron after the collison are not moving in the same direction the incident photon was moving.
The simplest way to solve this problem is by writing conservation of energy and momentum as a single four-momentum equation:
pf + pe = qf + qe (1)
where
pf = four-mometum of photon before collision
pe = four-momentum of electron before collision
qf = four-mometum of photon after collision
qe = four-momentum of electron after collision
We are not interested in qe, so we want to eliminate it from the equation. What we do is we write (1) as:
qe = pf + pe - qf
We then square both sides:
qe^2 = (pf + pe - qf)^2 =
pf^2 + pe^2 + qf^2 + 2 pf dot pe +
- 2 pf dot qf -2 pe dot qf
Next substitute the energy momentum relation p^2 = m^2, so we have:
qe^2 = m^2
pe^2 = m^2
pf^2 = 0
qf^2 = 0
and the equation simplifies to:
2 pf dot pe - 2 pf dot qf -2 pe dot qf = 0
Pf dot qf contains the angle you want to know. Pe = (m,0,0) because the electron before the collision was at rest, so the Lorentz inner product of the two other terms are just the electron mass times the photon energies.
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