Past exam of the mathematics course of the University of Cambridge / 2012 / ia / Paper 4 / 12B / b / Solution

For $u>0$, take the initial electron direction as the longitudinal axis and put $\beta =u/c$. Incoming four-momentum conservation givesThe transverse components must cancel. In the noncollinear case the two transverse directions are opposite, and ${\nu }_1\sin {\theta }_1={\nu }_2\sin {\theta }_2$. Hence the ratio of longitudinal momentum times $c$ to energy isWriting $s=({\theta }_1+{\theta }_2)/2$ and $d=({\theta }_1-{\theta }_2)/2$, the last ratio is $\cos s/\cos d$. It also equals $(1+\cos ({\theta }_1+{\theta }_2))/(\cos {\theta }_1+\cos {\theta }_2)$ whenever that denominator is nonzero. Since ${\gamma }^2{\beta }^2={\gamma }^2-1$, the angular relation for two-photon electron-positron annihilation isThe positive root follows from $u>0$. For noncollinear emission the positive momentum-to-energy ratio ensures $\cos s>0$, and $|d|<\pi /2$ ensures the stated denominator is positive.

The printed quotient needs a qualification for exactly collinear photons. The allowed configuration ${\theta }_1=0$, ${\theta }_2=\pi$ has positive frequencieswhich obey both conservation equations, but the printed left side is $0/0$. Thus the quotient holds for noncollinear angles, or as its appropriate noncollinear limit. The undivided identityremains true in the collinear case, where both sides vanish. The limit $u=0$ similarly requires care because the initial direction is then undefined and the photons are antiparallel.