Past exam of the particle physics course of the University of Oxford / 2019 / 6 / b / Solution

For the stationary target, $p=(m_N,0)$. The invariant definition immediately givesNo massless approximation is needed for this laboratory identity when binding and target recoil conventions are neglected.

For the angular relation, specify the neutrino-parton centre-of-momentum frame and neglect masses there. The incoming and outgoing leptons then each have energy $\sqrt{\hat{s}}/2$. Their momentum transfer satisfiesSince $Q^2=xys$ and $\hat{s}=xs$, cancellation givesThis identifies the antiquark weight $(1-y{)}^2$ with the corresponding squared angular factor. The relevant massless condition is sufficient energy in the lepton-parton subprocess; $E_{\nu }\gg m_{\mu }$ alone does not control every endpoint configuration.

The frame qualifier matters: the complete neutrino-nucleon centre-of-mass final state contains a variable-mass hadronic system. In that frame, still neglecting masses but retaining $x$, direct two-body kinematics for $\mu +X$ gives insteadFor example $x=1/2$ and ${\theta }_{\nu N}=\pi /2$ give $y=2/3$, not $1/2$. Thus the simple angle-only relation implicitly uses the partonic frame.