Inverse Compton scattering in SR

[diazona]

This is actually for a graduate course but it's a basic special relativity problem, i.e. undergraduate-level material, so I'm posting it here...

Homework Statement

Inverse Compton scattering describes the process whereby a photon scatters off a charged particle moving with a speed very nearly that of light. In this problem we analyze an inverse Compton scattering event "geometrically".
(a) An observer, moving with four-velocity U, observes a charged particle traveling with four-velocity V and rest mass m. Describe, in terms of U and V, the condition that the charged particle is moving with a speed very nearly that of light.
(b) The charged particle encounters a photon with four-momentum P. Express, in terms of the appropriate four-vectors, the energy of the photon incident on the charge particle as seen by the observer.
(c) Express, in terms of the appropriate four-vectors, the photon's four-momentum P' following the scattering event.

Homework Equations

$$U^{\mu} = (1, 0, 0, 0)$$ (in the observer's rest frame)
$$V^{\mu} = (\gamma, \gamma \vec{v})$$
$$P^{\mu} = (E, \vec{p})$$

The Attempt at a Solution

I got parts (a) and (b) easily enough by evaluating 4-vector products in the observer's rest frame,
(a) $$\gamma = -U^{\mu} V_{\mu} \gg 1$$
(b) $$E = U^{\mu}P_{\mu}$$
The problem is with part (c). We're supposed to express P' in terms of U, V, and P, but it doesn't seem to be possible without knowing the angle at which the scattered photon exits (or the angle at which the charged particle exits). Am I missing something, or is this actually impossible?

 

[gabbagabbahey]

The problem is with part (c). We're supposed to express P' in terms of U, V, and P, but it doesn't seem to be possible without knowing the angle at which the scattered photon exits (or the angle at which the charged particle exits). Am I missing something, or is this actually impossible?

What quantity is conserved during the scattering event?

 

[diazona]

Momentum, I know... P + mV = P' + mV', where V' is the final four-velocity of the charged particle. But I still don't see how that helps... I don't know V'.

 

[gabbagabbahey]

Why not just express your answer in terms of V'?

 

[diazona]

I thought about that, but it'd be just expressing one unknown in terms of another :/ Anyway, it doesn't matter now, the assignment was due earlier today. In the end even our professor did admit that it was impossible.