Past exam of the particle physics course of the University of Oxford / 2017 / 5 / a

Assume first an ultrarelativistic charged-current interaction with negligible electron mass in the centre-of-mass spin analysis. Choose the incident antineutrino along $+z$. Its right helicity gives spin $+z$; the interacting left-helicity electron moves along $-z$ and also has spin $+z$. The total initial projection is $+1$. The final antineutrino and selected electron likewise have total spin $+1$ along the outgoing antineutrino direction. The spin-one rotation therefore givesAn unpolarized electron target changes the normalization through the spin average, not this selected-current shape. For electron antineutrinos the charged-current diagram is the $s$-channel $W^{-}$ graph. A contact approximation further requires $m_e^2\ll s\ll m_W^2,m_Z^2$.

The full Standard Model also has neutral-current $Z$ exchange. In the contact limit it supplies a right-electron contribution as well as interference in the left-electron channel. With $g_L=1/2+{\sin }^2{\theta }_W$, $g_R={\sin }^2{\theta }_W$, the full high-energy shape is proportional to $g_L^2(1+\cos {\theta }^{\star }{)}^2/4+g_R^2$. Thus the pure squared angular form is an explicitly stated charged-current approximation, not an exact prediction for all Standard Model contributions. At momentum transfers comparable to the weak-boson masses, propagators must also be retained.^[ref]

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.