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I understand the concept, but I'm having a hard time applying the consequences of conservation (energy/momentum). For example:

A proton with kinetic energy 437 MeV hits a proton at rest elastically and as result both protons rebound with equal energies. What is the included angle between them?

Attempt at a solution: Since the collision is elastic the masses remain the same. I'm also assuming that the velocities of the protons form the same angle with the incident proton in order to preserve momentum in the direction perpendicular to the incident proton. Writing up the four-vectors for the system (particle 2 being the target, primes are post event, $\phi$ is the angle from the incident proton's path, T is the kinetic energy) $$ P_1 = (mc^2+T_1,|\vec{p_1}|,0,0) \hspace{4 mm} P_2 = (mc^2,0,0,0) \\ P'_1 = (E'/2c,|\vec{p'_1}|\cos{\phi}, |\vec{p'_1}|\sin{\phi},0) \\ P'_2 = (E'/2c, |\vec{p'_1}|\cos{\phi}, -|\vec{p'_1}|\sin{\phi}, 0) $$ If I postulate that ${P_1 + *{P_2}*= P'_1+P'_2}$ then this yields $$ 2|\vec{p'_1}|\cos{\phi} = |\vec{p_1}| = T_1/c $$ but this is where things go pearshaped (or at least when I realize that they have done so much earlier).

(** are edited sections)

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You'll save some minor headaches by working in $c = 1$ units (i.e. masses, energies, momenta all in eV). –  dmckee Mar 20 '13 at 20:05
    
You write "If I postulate that $P_1=P'_1+P′_2$ [...]" which is wrong. Notice that these are four-momenta and that $P_2$ is not zero. As a result the second equality in your next line is also wrong. –  dmckee Mar 20 '13 at 20:15
    
yes, this was a copyerror from my notes but of no significance since I only use the momentum parts which is 0 for $P_2$ –  Hippie_Eater Mar 20 '13 at 20:37
    
OK. I see where you are going with this. You've still made an error is identifing $p_1$ with $T_1$. Recall that $E^2 = p^2 + m^2$ with those inconvenient squares in there, and you regime is not highly relativistic, so you have to care. –  dmckee Mar 20 '13 at 20:42
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