 Solution

Neglecting the electron mass in the centre-of-mass kinematics, take $k=(E^{\ast },0,0,E^{\ast })$, $p=(E^{\ast },0,0,-E^{\ast })$, and $k^{\prime }=(E^{\ast },E^{\ast }\sin {\theta }^{\star },0,E^{\ast }\cos {\theta }^{\star })$. Then $p\cdot k=2E^{\ast 2}$ and $p\cdot k^{\prime }=E^{\ast 2}(1+\cos {\theta }^{\star })$. HenceThe constant Jacobian $|d\cos {\theta }^{\star }/dy|=2$ givesIncluding both Standard Model chiral channels in the contact approximation instead gives $d\sigma /dy\propto g_L^2(1-y{)}^2+g_R^2$. The right-electron term is flat in $y$.

In the electron-rest laboratory frame $p=(m_e,0)$, the invariant definition immediately givesThis last equality is exact. With electron mass retained in the centre-of-mass relation, the angular formula becomes $y=(s-m_e^2)(1-\cos {\theta }^{\star })/(2s)$, approaching the boxed massless relation at high energy.