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

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.

Treat a nucleon as incoherent collinear spin-half quarks and antiquarks carrying fractions $x$ of its momentum. Neglect their masses and transverse motion, use a charged-current four-fermion interaction, and take $Q^2\ll m_W^2$. The subprocess energy is $\hat{s}\simeq xs$. Antineutrino scattering on a quark has the same selected-current helicity factor $(1-y{)}^2$ as in part a; crossing to an antiquark changes the allowed spin pairing and gives a flat distribution. Multiplying the elementary strength by the respective parton densities givesHere $q,\bar{q}$ are the effective weak flavour combinations with the isoscalar proton-neutron average and CKM weights absorbed into their definitions. With only light flavours and charge symmetry, the relevant up and down combinations coincide on an isoscalar target. Rescaling those density definitions requires rescaling the prefactor consistently.

The equation is a leading-order scaling result: it neglects logarithmic QCD evolution, heavy-quark thresholds, higher-twist and target-mass corrections, nuclear modifications, and weak-propagator suppression. The current produces ${\mu }^{+}$ for an incident muon antineutrino. At higher momentum transfer the weak propagator supplies a factor approximately $[m_W^2/(m_W^2+Q^2){]}^2$, and at low energy resonance and quasielastic scattering invalidate a pure DIS description.

For a nucleon at rest and beam energy well above its mass scales, $s\simeq 2m_{N}E$. If the leading parton distributions are approximately independent of beam energy, integration over $x,y$ therefore makes both total cross sections approximately proportional to $E$. The plotted plateaus are evidence for scattering on approximately pointlike constituents in this range, with the departures expected from QCD evolution and threshold effects.

Define the momentum-weighted contents $Q={\int }_0^{1}xq(x)dx$ and $\bar{Q}={\int }_0^{1}x\bar{q}(x)dx$. Since ${\int }_0^1(1-y{)}^{2}dy=1/3$, the two cross sections obeyRead the original high-energy plateaus as approximately $0.67$ and $0.33$ in the plotted $10^{-38}\mathrm{c}{\mathrm{m}}^2/GeV$ units. With $R={\sigma }_{\bar{\nu }}/{\sigma }_{\nu }\simeq 1/2$ and $r=\bar{Q}/Q$,Using the literal readings $0.33/0.67$ gives $r=0.1905$, so twenty percent is an appropriate graphical estimate, not a precision fit. A purely valence target would have $R=1/3$, while a fully symmetric sea would give $R=1$.

The neutrino-antineutrino DIS sea momentum ratio is a ratio of momentum fractions in the effective flavour combinations, not an unweighted count of antiquarks. The cross-section integral contains $xq(x)$: it cannot determine $\int \bar{q}/\int q$ without additional assumptions about their shapes. A distribution growing at small $x$ can have a large number integral while its momentum integral remains finite.