 Solution

A Lorentz invariant has the same value when the same physical system is described in any inertial frame. The components of its energy-momentum four-vector change, but their Minkowski norm does not. In particular the relativistic energy-momentum relation iswhere $M$ is the rest mass. Individual total energy and momentum are frame-dependent; the combination on the left is invariant.

For the fixed-target annihilation, the initial total laboratory energy is $E+M{c}^2$ and its total momentum is the incident particle's $\mathbf{p}$, with $c^2{p}^2=E^2-M^2{c}^4$. The centre-of-mass energy available for new-particle creation is thereforeIn the centre-of-mass frame the total momentum vanishes, so $W$ can supply the new particles' total rest energy at threshold. One does not subtract the annihilated particles' original rest energy. The excess laboratory energy beyond $W$ is associated with motion of the whole final system in the laboratory frame.

For the subsequent elastic collision, the equal and opposite transverse components of final momentum imply equal final momentum magnitudes, and hence equal final energies, for the two identical particles. By conservation of relativistic energy, each hasTheir final momentum magnitude $p_f$ and the incident momentum magnitude $p_i$ obeyLongitudinal momentum conservation gives $p_i=2p_f\cos \theta$. It follows that the equal-angle relativistic elastic scattering relation isThis derivation assumes nonzero incident kinetic energy and the stated symmetric outgoing paths. The angle between those paths is $2\theta$, which is evaluated in the two following parts.