The magnetic bending angle is $\varphi \simeq |q|BL/p$. For a momentum estimate based on this angle, propagation of uncertainty gives $\sigma (p)/p=\sigma (\varphi )/\varphi$. Comparing the total Multiple Coulomb scattering angle with the bend therefore gives the commonly intended characteristic estimateIts independence of $p$ is the key result for relativistic particles: the scattering angle and magnetic bend both scale as $1/p$.

A precise standard deviation requires specifying how tracks are measured. If $\varphi$ is obtained from independently measured entrance and exit directions and all the material lies between them, the bending-plane projected root mean square is $\mathrm{RMS}(\theta )/\sqrt{2}$. For that estimator the equality isA three-point sagitta fit weights scatter positions differently again. The paper does not specify these measurement details, so it fixes the scaling and characteristic estimate, but not one universal geometric coefficient for all track fits.

For a thickness $x$ with number density ${\rho }_n$ of scattering centres, the mean count is $N={\rho }_{n}x\sigma$. Using the total cross section in i and the variance addition in ii,ThusThe stated assumptions make the prefactor independent of $x$ and approximately independent of $p$. The lower-cutoff factors cancel except inside the slowly varying logarithm. Treating $N$ as a Poisson random variable rather than a fixed count gives the same result after replacing $N$ by its mean.

Let $\ell =\ln ({\theta }_{\max }/{\theta }_{\min })$. The supplied second moment gives the single-scatter root mean square angleFor independent scatters with uniform azimuth, the small angular-deflection vectors have zero mean. Their cross terms average to zero, so their variances add. After $N$ scatters,This is the root mean square magnitude of the two-dimensional angular deflection. Its projection into one specified plane has a variance smaller by a factor of two, hence a root mean square smaller by $\sqrt{2}$.

For azimuthally symmetric Rutherford scattering, integrate over the solid angle, $d\Omega =2\pi \sin \theta d\theta$. Substituting $u=\sin (\theta /2)$ gives $\sin \theta d\theta =4udu$, soThe small lower cutoff dominates when ${\theta }_{\max }\gg {\theta }_{\min }$ and ${\theta }_{\min }\ll 1$. Using $\sin ({\theta }_{\min }/2)\simeq {\theta }_{\min }/2$, the leading total cross section isA nonzero cutoff is essential: the unscreened forward cross section otherwise diverges.

The sagitta is the maximum distance between the circular track and its chord. The perpendicular bisector of a chord of length $L$ gives a right triangle with hypotenuse $R$ and sides $L/2$ and $R-s$. HenceAt fixed electric charge, magnetic field and lever arm, $p=|q|B{L}^2/(8s)$, so differentiation and propagation of uncertainty giveThus a fixed spatial measurement uncertainty produces a relative momentum uncertainty growing linearly with $p$. If momentum is expressed in $\mathrm{GeV}/c$, $B$ in tesla, $L,s$ in metres and $z=q/e$, restore the usual conversion $p=0.2998|z|BR$.

The Lorentz force is $d\mathbf{p}/dt=q\mathbf{v}\times \mathbf{B}$. Because it is perpendicular to the velocity, it changes the direction of momentum without changing its magnitude. For a charged particle in a uniform magnetic field, the radius of curvature is $R=p/(|q|B)$. Over the short track, the bending angle is $\varphi \simeq L/R$. Equivalently, integrate the force for a time $L/v$ while neglecting the small rotation in its direction. The magnitude of the acquired momentum component perpendicular to the incident direction isIts direction follows $q\mathbf{v}\times \mathbf{B}$; a signed component can therefore be written $qBL$ after fixing an oriented bending axis. Chord length and arc length agree at this order in $L/R$.

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.

Choose $W^{+}\to {\mu }^{+}{\nu }_{\mu }$. Important backgrounds include quantum chromodynamics production of heavy-flavour hadrons with semileptonic decays, in-flight pion or kaon decays, and hadrons penetrating the calorimeters. These generally have accompanying jet activity; semileptonic decays can contain real neutrinos, while jet mismeasurement can fake missing transverse momentum. Heavy-flavour tracks can originate at a displaced decay vertex. Require a high-$p_T$ isolated muon, agreement between inner tracking and the external muon system, and substantial missing transverse momentum inconsistent with detector noise or a mismeasured jet.

The genuine electroweak backgrounds include $Z\to {\mu }^{+}{\mu }^{-}$ with one muon lost and $W\to \tau \nu \to \mu \nu \nu \nu$. Veto a second reconstructed opposite-sign muon and use the dimuon invariant mass when both are present to reject the Z boson channel. The tau-mediated spectrum is softer and contains extra missing particles; template fits and control samples estimate its residual contribution. Detector acceptance determines how much of the lost-muon background survives.

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 $M_W$.

Initial-state radiation of gluons gives the W boson nonzero transverse momentum. Its transverse Lorentz boost then changes the charged-lepton $p_T$, washing out the sharp $M_W/2$ edge and permitting events above it. The recoil must be modelled with quantum chromodynamics and constrained by measured hadronic activity. Independently, the finite W boson decay width spreads its actual invariant mass, also smearing the fixed-mass endpoint. Either is a theoretical broadening mechanism; neither is a detector-resolution effect.

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 $c$.

Let ${\theta }^{\star }$ be the angle from the incoming up quark, along the proton direction, to the outgoing charged antilepton in the W boson rest frame. The initial and final spin projections in ii are $-1$ and $+1$. The Wigner small-d matrix therefore givesAfter squaring and normalizing over $c=\cos {\theta }^{\star }$, the charged-lepton angular distribution isIts integral is one since ${\int }_{-1}^1(1-c{)}^{2}dc=8/3$. The charged antilepton is preferentially emitted toward the antiproton. Measuring the angle from the incoming down antiquark instead replaces $c$ by $-c$ and produces $(1+c{)}^2$; the axis convention is part of the answer. Per unit solid angle the normalized density is $3(1-\cos {\theta }^{\star }{)}^2/(16\pi )$.

Neglect the participating quark and lepton masses, so that the weak charged current selects a left-helicity $u$ and a right-helicity $\bar{d}$. Choose $+z$ along the proton beam in the partonic centre-of-momentum frame. Then the up quark moves along $+z$ with spin projection $-1/2$, while the down antiquark moves along $-z$ with helicity $+1/2$ and also has spin projection $-1/2$ along $+z$. ThusThe collinear two-body initial state has no orbital projection about its collision axis. In this approximation the produced W boson therefore has the indicated transverse polarization; its longitudinal spin state is absent. The unpolarized hadron beams contain other constituent spin states, but these do not participate in this massless charged-current subprocess.

In the final state the Positron has helicity $+1/2$ and the electron neutrino has helicity $-1/2$. Along the outgoing charged-lepton direction their combined angular-momentum projection is $+1$, because the oppositely moving neutrino has spin projection $+1/2$ on that axis. Finite masses, noncollinear production and sea contributions modify the ideal configuration.

A leading valence-level process is $u\bar{d}\to W^{+}\to e^{+}{\nu }_e$, with the up quark supplied by the proton and the down antiquark by the antiproton. The same Feynman diagram works with $e^{+}{\nu }_e$ replaced by ${\mu }^{+}{\nu }_{\mu }$. Both vertices are weak charged current vertices; the production scattering amplitude contains the CKM matrix element $V_{ud}$ or its conjugate according to the chosen fermion-flow convention. The intermediate W boson is produced near its mass shell.

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 an incoming ${\mu }^{-}$, a putative $e^{-}{\mu }^{-}\to {\nu }_e{\nu }_{\mu }$ process would change total electric charge from $-2$ to zero. It is forbidden: both negative charged leptons would have to emit a $W^{-}$, and a single exchanged charged boson cannot connect those two emission vertices. Ordinary elastic $e^{-}{\mu }^{-}\to e^{-}{\mu }^{-}$ scattering remains possible through $t$-channel photon or $Z$ exchange, shown in the figure, but this is not the charged-current reaction of part d. The neutral-current weak contribution is distinct from the electromagnetic one.

Interference CP violation concerns the relative phase between direct decay and decay after mixing. For a common CP eigenstate final state, it can occur with both $|q/p|=1$ and $|{\bar{A}}_f/A_f|=1$, so neither of the first two mechanisms is needed. In that limit $|\lambda |=1$, butproduces different interference terms for $\lambda$ and ${\lambda }^{-1}={\lambda }^{\ast }$. With $\Delta m=m_h-m_{\ell }$,The term proportional to $\mathrm{Im}\lambda \sin (\Delta mt)$ consequently reverses sign. In the CP-conserving limit, $\lambda$ is the real CP parity of the final state in a consistent phase convention. For non-CP-eigenstate final states, compare the full coefficients in CP-conjugate rates rather than imposing this simplified criterion.

CP violation in mixing occurs whenThe flavour-transition probabilities are proportional to $|q/p{|}^2|g_{-}{|}^2$ for $K^0\to {\bar{K}}^0$, and $|p/q{|}^2|g_{-}{|}^2$ for the reverse. Their inequality remains when decay probability amplitudes themselves conserve CP. The coefficients multiplying the mixed-decay terms in the rate equations encode this asymmetry.

Direct CP violation means unequal magnitudes of CP-conjugate decay probability amplitudes:It changes decay probabilities even at $t=0$, before any mixing. For a common CP eigenstate final state, this reduces to $|A_f|\ne |{\bar{A}}_f|$. Usually it requires interfering decay contributions with different weak phases and different strong phases; it is not equivalent to a nonzero weak phase in one isolated probability amplitude.

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.