Past exam of the physics 1 paper of the University of Oxford / 2014 resit / 3

In the centre-of-mass frame, the two equal incoming rest masses have opposite momenta and total energy $2E$. The squared total four-momentum fixes the composite invariant mass $\mathcal{M}$:A change of inertial frame, including going to either particle's rest frame, cannot change this invariant mass. It includes the internal kinetic energy and is generally larger than $2M_1$.

There is a terminology convention worth making explicit. The older “relativistic mass” means frame-dependent total energy divided by $c^2$. To stop particle 1 requires a boost with Lorentz factor $\gamma =E/(M_1{c}^2)$, so the system has total energy $2\gamma E=2E^2/(M_1{c}^2)$ in that frame. Under that alternative convention, its relativistic mass would be $2E^2/(M_1{c}^4)$. This is not its invariant mass; both quantities coincide only when $E=M_1{c}^2$ and the incoming particles already have zero relative motion.

Four-momentum conservation fixes the outgoing system's invariant mass to the same $2E/c^2$. In the rest frame of particle 3, let particle 4 have energy $E_4^{\prime }$ and momentum ${\mathbf{p}}_4^{\prime }$. Each outgoing particle has rest mass $M_2$, soConsequentlyHere $E_4^{\prime }$ is the total energy, including rest energy. At the production threshold $E=M_2{c}^2$, it reduces to $M_2{c}^2$, as it should. Production is kinematically possible only for $E\ge M_2{c}^2$. The initial rest mass $M_1$ drops out because the conserved total centre-of-mass energy, rather than the identity of the incoming particles, determines this outgoing relative motion.