Past exam of the particle physics course of the University of Oxford / 2015 / 1

Treat the electron and electron antineutrino as effectively massless particles in the centre-of-momentum frame, retain only the charged current, and use the Fermi interaction: $m_e^2\ll s\ll M_W^2$. Take ${\theta }^{\ast }$ between incoming and outgoing electron antineutrino directions. The V-A interaction selects a right-helicity electron antineutrino and a left-helicity electron. Their opposite incoming directions give total angular momentum projection one. The outgoing pair has the same projection along its own axis. The Wigner D-matrix element is $d_{11}^1({\theta }^{\ast })=(1+\cos {\theta }^{\ast })/2$, so the squared helicity amplitude is proportional to $(1+\cos {\theta }^{\ast }{)}^2$. The inactive initial electron helicity is included in the unpolarized normalization supplied by the paper.

The boost from the centre-of-momentum frame to the laboratory frame has $\beta =E/(E+m_e)$, while the incoming electron antineutrino has $E_{\ast }=(s-m_e^2)/(2\sqrt{s})=m_{e}E/\sqrt{s}$ and $s=m_e^2+2m_{e}E$. ThusThe requested identity therefore uses $E\gg m_e$; it is not exact at finite target mass. With $c_{\ast }=\cos {\theta }^{\ast }$, the absolute Jacobian determinant is $|dy/d{c}_{\ast }|=1/2$, givingThe suppression at backward scattering follows from angular momentum conservation: rotating projection $+1$ through $\pi$ cannot leave projection $+1$ along the original axis. Replacing $1+c_{\ast }$ by $2(1-y)$ gives the supplied $(1-y{)}^2$ dependence.

Use the parton model, neglect antiquarks, binding and target mass corrections, quark masses, muon mass, QCD scaling violation and the W boson propagator correction. The struck quark is effectively free during the short interaction and carries parton momentum fraction $x$, giving $\hat{s}\simeq xs$. The charged current reaction is ${\bar{\nu }}_{\mu }{u}_i\to {\mu }^{+}{d}_j$. Its helicity amplitude has the same $(1-y)$ factor as in a. Define the effective parton distribution function by $q(x)={\sum }_{i,j}|V_{ij}{|}^2{u}_i(x)$, summing only kinematically accessible final quarks; for valence-only protons and accessible $d,s$ this is approximately $u(x)$. It is not the unweighted sum $u+d$.Here $y=p\cdot (k-k^{\prime })/(p\cdot k)$ is the inclusive deep inelastic scattering variable, and $x=Q^2/(2p\cdot (k-k^{\prime }))$ in the scaling limit. Integrating $0<y<1$ gives $1/3$, henceThe moment measures the momentum fraction in the active quarks. Approximate linearity is evidence for scattering from pointlike spin-one-half constituents with approximately scale-independent parton distribution functions, rather than coherent scattering from an extended proton. It is an intermediate-energy prediction: thresholds, experimental deep inelastic scattering cuts, QCD scaling violation and eventually the finite W boson propagator modify it.

Strike a target with high-energy protons, collect negative pions and kaons with magnetic horns, and let them decay in a long evacuated decay tunnel: ${\pi }^{-}\to {\mu }^{-}{\bar{\nu }}_{\mu }$ and $K^{-}\to {\mu }^{-}{\bar{\nu }}_{\mu }$. Absorb residual hadrons and range out muons in shielding. Focusing selects the negative parent sign and therefore enriches muon antineutrinos; other parent decays leave a finite contamination. Monitor the parent flux and use a near detector to constrain the unoscillated spectrum.

A massive segmented calorimeter supplies interaction probability and measures the hadronic shower. Alternate dense absorber plates with scintillation counters or tracking planes, and magnetize the iron or add a downstream magnetic muon detector. The signal is a positive muon plus a hadronic shower; the sign rejects much of the muon neutrino contamination. For a contained muon, its range gives energy; for a penetrating track, curvature gives $p_{\perp }[\mathrm{GeV}]=0.2998B[\mathrm{T}]R[\mathrm{m}]$. Multiple measurements and a field map determine the fitted momentum and sign. Summed calibrated shower response estimates the struck quark's visible hadronization energy, not the energy of a free observable quark. Correct for invisible nuclear losses, noncompensating response, leakage and acceptance. Neglecting target recoil corrections, $E_{\nu }\simeq E_{\mu }+E_{\mathrm{had}}$ and $y\simeq E_{\mathrm{had}}/E_{\nu }$.