 Solution

For $0<T\ll M{c}^2$, the ratio in 9 tends to $1/2$, givingThis recovers the nonrelativistic equal-mass elastic collision result: momentum conservation gives ${\mathbf{v}}_i={\mathbf{v}}_1+{\mathbf{v}}_2$, while conservation of kinetic energy gives $v_i^2=v_1^2+v_2^2$. Subtracting the squared momentum relation leaves ${\mathbf{v}}_1\cdot {\mathbf{v}}_2=0$. Exactly at $T=0$ there are no outgoing nonzero velocities defining such angles; the result is the limit through positive kinetic energies.