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The squared invariant amplitude weights each accessible final configuration. The final on-shell momentum measure counts the available states, and the four-dimensional delta function imposes energy and momentum conservation:To prove its Lorentz invariance for proper orthochronous transformations, writeThe four-volume measure has unit Lorentz Jacobian, $p^2$ is invariant, and the future mass shell is preserved. Likewise ${\delta }^{(4)}(\Lambda Q)={\delta }^{(4)}(Q)$ because $|\det \Lambda |=1$. The product therefore proves that $d{\Phi }_n$ is invariant. Spin averages and sums belong in the amplitude as appropriate; identical final particles require a symmetry divisor if the integration covers a fully labelled phase space.

The initial $1/(2E_i)$ factors come from relativistic normalization of the incoming states. They are not individually invariant, so the displayed coordinate-time transition rate must not be called a Lorentz scalar merely because its final phase space is invariant. A one-particle decay has proper width ${\Gamma }_0=(2M_0{)}^{-1}\int |\mathcal{M}{|}^{2}d{\Phi }_n$ in its rest frame, while its laboratory rate is ${\Gamma }_{\mathrm{lab}}=(M_0/E_0){\Gamma }_0$. A scattering cross section instead uses the invariant incident flux $4\sqrt{(p_1\cdot p_2{)}^2-m_1^2{m}_2^2}$. This distinguishes invariance of phase space from transformation of a measured rate.

Write $s_{ij}=M_{ij}^2$ and factor the three-body phase space through the pair $q=p_1+p_3$, with $s=q^2$:A scalar parent permits integration over the overall orientation. In its rest frame let the pair have momentum magnitude $p$, and in the pair rest frame let particle 3 have magnitude $p^{\ast }$. After integrating the azimuth of its direction, the two-body factors giveIn that pair rest frame, particle 2 has momentum magnitude $M_{0}p/\sqrt{s}$, so $|d{s}_{23}/d\cos {\theta }^{\ast }|=2M_{0}p{p}^{\ast }/\sqrt{s}$. The Jacobian cancels both momentum factors, leavingThe physical domain is bounded by the requirement that the relative cosine lie between $-1$ and $1$. The phase-space density is uniform within that domain; the event density is uniform only for a constant squared amplitude. Scalar parent spin alone does not forbid resonance-dependent amplitudes. A factor $1/(s_{13}-m_R^2+i{m}_R{\Gamma }_R)$ is a concrete nonuniform example. For two identical pions, a conventional overall $1/2!$ compensates integration over both labelings without changing the local flatness of phase space.^[ref]

The three on-shell spatial momenta have nine components. Four conservation constraints leave five, and an unobserved overall spatial rotation removes three for the scalar decay, leaving two internal kinematic variables. The pair invariants obeyThus two determine the third. They also determine each rest-frame daughter energy,and hence all momentum magnitudes and pairwise angles. The momentum triangle is fixed up to overall orientation. The Dalitz plot describes this internal kinematics, rather than claiming to record three independently oriented Four-vectors. Polarized parents or additional measured spin information can require orientation variables beyond the two invariant axes.

Nuclear-field pair creation is $\gamma Z\to e^{-}{e}^{+}Z$. A virtual photon exchanged with the nucleus supplies recoil momentum, while the real photon and virtual photon attach to the electron line joining the outgoing pair. The figure shows a representative graph. The second attachment ordering must be included in the full amplitude. A real photon cannot create a free massive pair in vacuum: its invariant mass is zero, whereas $(p_{e^{-}}+p_{e^{+}}{)}^2\ge 4m_e^2$. Nuclear recoil resolves that kinematic obstruction and permits energy-momentum conservation.

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]

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.

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/6/Physics%20Papers%202017.pdf?cookie_passthrough=1. The held cover identifies C4 on 14 June 2018; the annual bundle's filename does not determine the year. Natural units $\hslash =c=1$ are used except in the silicon-detector calculations, where SI units and the momentum conversion are stated explicitly. All six questions are solved.

With relativistically normalized one-particle states, Fermi's golden rule becomesIn the parent rest frame, the three-momentum delta function fixes ${\mathbf{p}}_2=-{\mathbf{p}}_1$. The energy delta function fixes the common magnitude $p^{\ast }$; its radial Jacobian converts the remaining phase-space factor intoOnly one momentum magnitude remains because the two final momenta are constrained, not independent. Explicitly,For leptonic W boson decay both final masses are small compared with $m_W$, including the tau mass at the precision requested, so $p^{\ast }\simeq m_W/2$.

At a hadron collider, select $W\to \ell \nu$ events and reconstruct the transverse massusing missing transverse momentum for the neutrino. Fit the shape, especially the region above the nominal Jacobian edge, with finite-width resonance templates convolved with detector resolution, recoil modelling and backgrounds. The Breit-Wigner width controls this off-shell tail. Calibrated momentum and missing-momentum resolutions must be separated from the physical width. This is a lineshape measurement; the lifetime is far too short for a direct flight-distance measurement.

The valence quark distributions describe net quark content: $\int u_v(x)dx=2$ and $\int d_v(x)dx=1$ for a proton. Their momentum densities $x{u}_v,x{d}_v$ are small at very small $x$, peak at moderate momentum fractions, and vanish toward $x=1$, where one parton carrying almost all proton momentum is unlikely. The larger $u$ distribution reflects two valence up quarks and its comparatively harder shape; its pointwise value need not be exactly twice the down distribution.

The sea quark distribution is generated largely by gluon radiation and $g\to q\bar{q}$ splitting. It grows strongly toward small $x$ and falls at large $x$. The plotted sea curve is scaled by $0.05$, so it must be multiplied by twenty before numerical use. The total up-quark PDF isusing the valence-up distribution and the up-sea distribution. The summed sea curve alone does not fix $\bar{u}$ without a flavour assumption. Under the later stated four-flavour symmetry, $\bar{u}=\Sigma /8$.

At $y=0$, both fractions equal $x=0.0402$. Reading the original plot near this value gives representative momentum densitiesThese are approximate graphical readings, not a precision PDF fit. Undo the sea-curve scale and divide by eight flavours/antiflavours:All products have the same $1/x^2$ factor, which cancels in the fractions. The three relative weights areTheir sum is $0.3128$, givingReading the central curves more finely changes these percentages by about a percentage point, well within what the plot supports. Omitting its factor $0.05$ would substantially underestimate the sea contribution.

Put $r=p/(E+m)$ and assume $p>0$. The common plane wave has unit modulus. The component norm isFor unit spinor probability density choosewhere $E^2=p^2+m^2$ was used. This is spinor normalization, not square normalization of a plane wave over infinite space; a box or wavepacket gives the corresponding spatial normalization. If the convention is instead ${\psi }^{†}\psi =2E$, use $N=\sqrt{(E+m)/10}$.

For momentum along $+z$, the positive-helicity components are the first and third, while the second and fourth have negative helicity. HenceBoth are independent of speed. In the source's chiral block order the exact left-chiral weight is $1/2+3p/(10E)$, tending to $4/5$ at high energy; helicity and chirality probabilities agree only in that limit.

A proton contains up quarks on which the charged current can act:through $W^{-}$ exchange, weighted by $V_{ud}$. Spectator quarks complete the hadronic final state. At the exclusive nucleon level the charge assignment is $e^{-}p\to {\nu }_{e}n$; at the high momentum transfers specified, the generally dominant description is inclusive $e^{-}p\to {\nu }_{e}X$ with proton structure and hadronic breakup, not an unsuppressed pointlike proton-to-neutron transition. The quark-level diagram in the figure displays this allowed charge flow. The electron's left-chiral selection is the same as in part d.

For binary strip readout, report the centre of the hit strip. With uniform impact position across pitch $w$, the position error is uniformly distributed on $[-w/2,w/2]$, soThis assumes no analogue charge sharing between strips.

The magnetic force is $d\mathbf{p}/dt=q\mathbf{v}\times \mathbf{B}$. Since relativistic momentum is parallel to velocity, $d(p^2)/dt=2\mathbf{p}\cdot q(\mathbf{v}\times \mathbf{B})=0$. A uniform transverse field therefore changes momentum direction but not magnitude. If the trajectory turns at angular speed $v/R$, then $|d\mathbf{p}/dt|=p_{T}v/R$. Equating force magnitudes yieldsThis is the positive radius. The sign of $q$ determines the direction of curvature; the source's signed formula can describe an oriented radius. In practical tracking units,

The source labels $L$ as the full chord from $P_1$ to $P_3$, with $L/2$ on each side of its midpoint. The Right triangle between the Circle centre, chord midpoint and endpoint givesFor fixed geometry and field, the inverse transverse momentum isIf the uncertainty is dominated by the central point, ${\sigma }_s\simeq {\sigma }_x$. Linear propagation of uncertainty givesUsing $|q/e|=1$, $B=2\mathrm{T}$, $L=0.6\mathrm{m}$ and ${\sigma }_x=28.87\mu \mathrm{m}$,The inverse-momentum resolution is approximately momentum independent in this spatial-error-only model; the relative momentum resolution grows as $p_T\sigma (1/p_T)$. Endpoint errors and multiple scattering are excluded by the stated assumptions.

For negligible fermion masses, the neutral-current vertex connects a fermion and antifermion of opposite helicity. The two incoming choices are $e_L^{-}{e}_R^{+}$ and $e_R^{-}{e}_L^{+}$; the two outgoing choices are $f_L{\bar{f}}_R$ and $f_R{\bar{f}}_L$. Their combinations give four channels, denoted $LL,LR,RL,RR$, where each letter refers to the fermion rather than antifermion.

Take the incoming electron momentum along $+z$ and Positron momentum along $-z$. The initial $L$ pair has both spins along $-z$, giving total projection $-1$; the $R$ pair has both along $+z$, giving projection $+1$. If the outgoing fermion moves along $\hat{\mathbf{n}}$, an outgoing $L$ pair has both spins along $-\hat{\mathbf{n}}$ and an $R$ pair both along $+\hat{\mathbf{n}}$. The figure shows momentum and spin arrows separately for all four channels.

Take an on-shell W boson of fixed mass $M_W$, zero transverse recoil, negligible charged-lepton mass, and unrestricted angular acceptance. In its rest frame the charged lepton has momentum magnitude $p_0=M_W/2$. A longitudinal Lorentz boost to the laboratory preserves transverse momentum, soBoth angular roots contribute to the same $p_T$. Set $x=p_T/p_0$ and $a=\sqrt{1-x^2}$. The change-of-variable Jacobian determinant has $|dc/d{p}_T|=x/(p_{0}a)$, and the two-root angular weight is $\frac{3}{8}[(1-a{)}^2+(1+a{)}^2]=\frac{3}{4}(2-x^2)$. ConsequentlyThis is the ideal W-decay Jacobian peak. The divergence at the endpoint is integrable; it is a change-of-variable effect, not an infinite decay probability. Reversing the polar axis does not change this transverse momentum density, because the sum of the two roots removes the term odd in $c$.

For the selected muon channel, finite curvature resolution in the magnetic tracking system broadens the reconstructed transverse momentum spectrum. A high-$p_T$ track has small sagitta, so spatial measurement errors, alignment errors and magnetic-field calibration errors become important; at high momentum an approximately fixed uncertainty in $1/p_T$ gives $\delta p_T\propto p_T^2$. Multiple Coulomb scattering contributes at lower momenta. If the transverse mass is fitted, finite hadronic-recoil resolution additionally smears the inferred missing transverse momentum. These are experimental effects, distinct from the physical recoil and width in ii.

The useful modes are $\boxed{W^{+}\to e^{+}{\nu }_e\text{and}{W}^{+}\to {\mu }^{+}{\nu }_{\mu }}$ and their charge conjugates. An electron gives an isolated track and localized electromagnetic shower; a muon gives an isolated penetrating track. Both allow a well-calibrated charged-lepton transverse momentum and a missing transverse momentum measurement for the neutrino.

The tau lepton decays before detection, usually adding further neutrinos. Its visible momentum is only part of the original tau momentum, so the simple two-body endpoint is lost. For the direct electron and muon modes, the transverse massalso has an ideal endpoint at $M_W$ and avoids needing the unmeasured neutrino longitudinal momentum. Fits to this distribution or the charged-lepton $p_T$ spectrum determine $M_W$.