That's what I was trying to do in my second post, but I couldn't come to the conclusion that the two angles have to be equal. I'll try to make it clearer what I did there so you can point out what I'm missing.
Say the photon is moving in the +x direction and the resultant particles move in the x-y plane, for simplicity (so no z-component of momenta). Let's say the electron is scattered at an angle \theta_1 and the positron at \theta_2. By the conservation of momentum in the y direction,
\gamma_1mv_1sin\theta_1=\gamma_2mv_2sin\theta_2
\gamma_1v_1sin\theta_1=\gamma_2v_2sin\theta_2The condition for the x-components of the momenta to be different is:
\gamma_1mv_1cos\theta_1\neq\gamma_2mv_2cos\theta_2
\frac{\gamma_1v_1}{\gamma_2v_2}\neq\frac{cos\theta_2}{cos\theta_1}
which is satisfied as long as the two angles are different because:
\frac{\gamma_1v_1}{\gamma_2v_2}=\frac{sin\theta_2}{sin\theta_1}\neq\frac{cos\theta_2}{cos\theta_1}Also, some of the momentum of the photon will be transferred to a nucleus, so the initial momentum of the photon will be greater that the sum of the x-components of the final momenta, but as far as I know, we can't write an exact equality for it:
\frac{hν}{c}>\gamma_1mv_1+\gamma_2mv_2
hν>\gamma_1mv_1c+\gamma_2mv_2cAnd for the energy conservation, the energy transferred to the nucleus will be insignificant compared to the momentum transferred, so:
hν=\gamma_1mc^2+\gamma_2mc^2 (approximately)
It's clear that
hν=\gamma_1mc^2+\gamma_2mc^2>\gamma_1mv_1c+\gamma_2mv_2c
so there is no contradiction with what we've found using the energy conservation and the momentum conservation in the x direction.I don't see anything here that violates one of the conservation laws if the angles are not equal. It is even easier to see that it should be allowed if we let the speeds be much less than c, so \gamma_1=\gamma_2=1. That way, there will be a linear relationship between the speeds, and you can easily determine the speeds for a given ν and two angles you pick, as I explained in my second post.
tom.stoer said:
or 2)
Start in the c.o.m. frame with vanishing total momentum. The initial momenta (in x and -x direction) are carried by photon and the target and add up to zero. So the total momentum in x-direction is zero. Now make a Lorentz transformation from c.o.m to rest frame of M; the Lorentz trf. introduces one velocity so both particles must have same momentum in x direction.
In that case, in the CoM frame, the electron and the positron could move in any two opposite directions, since the total momentum is zero and any two opposite directions would be equally good. So, the x components of their momenta would add up to 0. Let's call the electron's speed in the x direction (in the CoM frame) u' and the positron's -u'.
Now the CoM is moving in the +x direction with respect to the frame where M is at rest. Let's call this speed v. Then using the Lorentz transformation, the speeds in the x direction u_1 (of the electron) and u_2 (of the positron) in the rest frame are:
u_1=\frac{u'+v}{1+\frac{u'v}{c^2}}
u_2=\frac{-u'+v}{1-\frac{u'v}{c^2}}
which cannot be equal unless u'=0 or v=c. u'=0 is the case where the particles' scattering angles are the same in the rest frame.