Past exam of the symmetry and relativity course of the University of Oxford / 2016 / 2 / c

Each photon has energy $\epsilon =\hslash \omega$. Their total four-momentum isIts squared invariant is $P^2=2{\epsilon }^2(1-\cos \theta )/c^2$. Therefore the system's rest energy isTwo individually massless particles can thus have positive total invariant mass. Back-to-back photons give rest energy $2\hslash \omega$, whereas exactly parallel photons have zero invariant mass and no massive-system rest frame.

The center-of-momentum frame moves with velocity $\mathbf{V}=c^2{\mathbf{p}}_{\mathrm{tot}}/E_{\mathrm{tot}}$. Using the total four-momentum in solution givesFor $0<\theta \le \pi$, its direction bisects the two photon directions, and its speed decreases from the limiting value $c$ to zero. At $\theta =\pi$, the lab already is the center-of-momentum frame. At $\theta =0$, the formal speed is $c$ and no inertial rest frame exists; this endpoint is a limit, not an allowable boost.