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Slow widening of a constriction approximately conserves the transverse mode action, reducing transverse momentum as the width $W$ increases. A lowering band bottom also increases longitudinal kinetic energy. Together they can narrow the angular distribution of an emitted electron beam. An abrupt narrow aperture alone can instead diffract strongly.

Expanding an angular distribution in multipoles converts the Boltzmann equation into a hierarchy in which free streaming couples neighboring multipoles and collisions damp or source selected moments.

Emission is the production of electromagnetic radiation by a physical system. Mechanisms include accelerated charges, transitions between quantum states and thermal fluctuations.

For an isolated atomic transition, a photon carries energy $h\nu =E_i-E_f$. In thermal emission, a macroscopic distribution of microscopic motions and states produces a spectrum controlled by temperature and material response. Emission must be distinguished from scattering of incident radiation, though a measured signal can contain both contributions. The process and observation geometry determine the relevant spectrum and angular distribution.

Two adjacent half-slits of width h with opposite field transmission have centres separated by h. The aperture transform is $h\mathrm{sinc}(qh/2)[e^{-iqh/2}-e^{iqh/2}]$. Thus the phase-reversed pair factor multiplies the finite-half-slit envelope, giving the displayed normalized intensity with $s=qh/2$ and $\mathrm{sinc}s=\sin s/s$. The on-axis field cancels. Positive side maxima satisfy $\tan s=2s$, and total integrated transmission is unchanged by a lossless phase plate even though the angular distribution changes.

An optical concentrator redirects light from an entrance area $A_{\mathrm{in}}$ to a smaller receiver area $A_{\mathrm{out}}$. Its geometric concentration ratio is $C=A_{\mathrm{in}}/A_{\mathrm{out}}$. Specify the admitted angular distribution and losses before interpreting $C$ as an irradiance gain: if a fraction $\eta$ of incident radiant flux reaches the receiver, the ratio of mean receiver to mean entrance irradiance is $\eta C$.

Include the Z boson amplitude and its quantum interference with the photon amplitude. Define $s_W=\sin {\theta }_W$, $c_W=\cos {\theta }_W$, $e=g{s}_W$, $g_L^f=T_3^f-Q_f{s}_W^2$, $g_R^f=-Q_f{s}_W^2$, andFor the same unpolarized and massless approximations as b, the complete tree-level angular distribution isThe absolute squares include the pure terms and their interference. Put $A=|C_{LL}{|}^2+|C_{RR}{|}^2$ and $B=|C_{LR}{|}^2+|C_{RL}{|}^2$. The even part is $(A+B)(1+c_{\ast }^2)$ and the odd part is $2(A-B)c_{\ast }$. Thus the forward-backward asymmetry is $A_{\mathrm{FB}}=3(A-B)/(4(A+B))$.

At the Z boson pole, neglect the small photon contribution and use real tree-level couplings. Thenwhere $g_V=g_L+g_R$, $g_A=g_L-g_R$. For charged leptons, $g_V/g_A=1-4{\sin }^2{\theta }_W$. Fitting the forward/backward counts, after detector and radiation corrections, therefore measures the weak mixing angle, conventionally an effective leptonic angle once electroweak corrections are included. For illustration ${\sin }^2{\theta }_W=0.231$ gives ${\mathcal{A}}_{\ell }\simeq 0.151$ and $A_{\mathrm{FB}}^0\simeq 0.0171$. This illustrative parameter choice is not extracted from this paper's data. The real tree-level photon-Z boson interference vanishes exactly at the pole but matters away from it.

Work in the W boson rest frame, taking the spin-quantization axis along its laboratory motion. Neglect charged-lepton and neutrino masses. The weak charged current produces a negative-helicity ${\ell }^{-}$ and a positive-helicity antineutrino. Their momenta are opposite, so both spin Vectors point opposite to the charged lepton's direction $\hat{\mathbf{n}}$; the final total spin projection along $\hat{\mathbf{n}}$ is $-1$.

Let ${\lambda }_W=+1,0,-1$ denote the parent spin projection along $z$. In the given probability amplitude, use $j=1$ and the rotation element $d_{-1,{\lambda }_W}^1(\theta )$. The supplied spin-one rotation elements giveThe transverse states have spin projections $\pm 1$; the longitudinal state has projection zero, not total spin zero. The original diagram below distinguishes momenta from spins and shows the corresponding normalized angular profiles.

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 $3(1-\cos {\theta }^{\star }{)}^2/(16\pi )$.

Doubling the number of identical-angle bends halves each angle. The minimum emittance scales as ${\theta }^3$, so it becomes $\boxed{0.240\mathrm{nm}\mathrm{rad}}$. At unchanged energy and unchanged individual magnet length, the bending radius doubles and field halves, giving $B=0.70$ T and $\boxed{E_c\simeq 4.19\mathrm{keV}}$.

At fixed stored current the loss per turn, proportional to $E^4\oint ds/{\rho }^2$, halves: twice as many equal-length bends each have one quarter of the original contribution. Total radiation power therefore halves. The spectrum softens by the factor-two critical-energy change. The characteristic high-energy vertical opening angle remains of order $1/\gamma$ at fixed energy, while each magnet's horizontal sweep halves. At a fixed absolute photon energy the normalized spectral position changes, so a claim that every angular distribution is identical would be too strong. Lower emittance also reduces the electron source size/divergence subject to the optical functions.

For a resting W boson with real spatial polarization vector along $z$, the fermion spin sum gives $\sum |\mathcal{M}{|}^2=g^2{M}_W^2{\sin }^2\theta /2$ for massless charged lepton and antineutrino momenta. The two-body Lorentz-invariant phase space gives the displayed angular decay rate. Since $\int {\sin }^2\theta d\Omega =8\pi /3$, the integrated width is $g^2{M}_W/(48\pi )$, equal to the unpolarized width. The massive-vector spin average changes the angular distribution, not the total width. No initial spin average is taken for a specified pure polarization.

The visible angular distribution and laboratory energy in ${\tau }^{-}\to h^{-}{\nu }_{\tau }$ carry information about the parent's spin. For fixed hadronic invariant mass, $E_h=\gamma (E_h^{\ast }+\beta p_h^{\ast }\cos \vartheta )$. A spin-zero recoil has analyzing coefficient ${\alpha }_h=1$. For a spin-one recoil coupled through the V-A interaction, longitudinal and transverse partial rates have weights $m_{\tau }^2:2m_h^2$ and opposite slopes, giving ${\alpha }_h=(m_{\tau }^2-2m_h^2)/(m_{\tau }^2+2m_h^2)$. Thus the harder-helicity ordering can reverse when the recoil mass exceeds $m_{\tau }/\sqrt{2}$. A mean invariant mass without spin-dependent spectral weights cannot fix the sign for a mixed sample.