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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.

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.