Equal-angle relativistic elastic scattering (${\cos }^2\theta =(T+2M{c}^2)/(T+4M{c}^2)$)

For two identical particles of rest mass $M$, one initially stationary and the other with kinetic energy $T$, symmetric final paths at $\pm \theta$ force equal final energies $M{c}^2+T/2$. The relativistic energy-momentum relation and longitudinal momentum conservation give the title formula. The angle between the paths is $2\theta$, tending to $90^{\circ }$ at low kinetic energy and to zero at high kinetic energy.