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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.

Assume unpolarized beams, $s\gg m_{\mu }^2$, a single exchanged photon, full azimuthal coverage and no radiative corrections. Take $c_{\ast }=\cos {\theta }^{\ast }$ between incoming $e^{-}$ and outgoing ${\mu }^{-}$. The four nonzero helicity amplitudes have magnitudes $e^2(1+c_{\ast })$ for equal incoming/outgoing negative-particle helicities, and $e^2(1-c_{\ast })$ for opposite ones. Averaging four initial spin states and summing the final ones givesThe massless two-body phase space givesThe even distribution reflects the equal left/right photon couplings and is symmetric under interchange of forward and backward directions.

The four allowed two-real-photon states are ${\eta }_c$, ${\eta }_c(2S)$, ${\chi }_{c0}$ and ${\chi }_{c2}$. Two photons have $C=+1$, excluding $J/\psi$, $\psi (2S)$ and $h_c$; the Landau-Yang theorem excludes ${\chi }_{c1}$ because $J=1$. Both scalar and pseudoscalar two-photon amplitudes exist, so $0^{++}$ and $0^{-+}$ are allowed. The J/psi meson was discovered first because its $1^{--}$ assignment permits direct, resonantly enhanced single-photon formation in $e^{+}{e}^{-}$ and a clean dilepton peak in hadron production. The lighter ${\eta }_c$ cannot be formed by one photon; it needs less favorable production channels and has a broader two-gluon decay.

A spin-zero ${\eta }_c$ at rest has no preferred axis, so $d\Gamma /d{\Omega }_p$ is isotropic after summing final spins. A J/psi meson formed in unpolarized high-energy $e^{+}{e}^{-}$ has transverse $m=\pm 1$ alignment. Its $p\bar{p}$ distribution can be writtenwhere $a,b$ contain the two independent helicity amplitudes. It is generally anisotropic; the approximate massless, helicity-conserving limit has $b\to 0$. The tabulated branching fraction does not fix $a/b$. An ensemble equally populated in all three spin projections would instead be isotropic: alignment of the production ensemble matters.

Neglect radiation, take $E_{\tau }=M_Z/2\simeq 45.6\mathrm{GeV}$, $m_{\tau }\simeq 1.777\mathrm{GeV}$, and treat a fixed hadronic invariant mass $m_h=1.5\mathrm{GeV}$ as a single recoil system in ${\tau }^{-}\to h^{-}{\nu }_{\tau }$. With negligible neutrino mass,Thus $E_h^{\min ,\max }=\gamma (E_h^{\ast }\mp \beta p_h^{\ast })$, numerically approximately $32.50$ to $45.60\mathrm{GeV}$. In the ultrarelativistic limit the range is $E_{\tau }{m}_h^2/m_{\tau }^2\le E_h\le E_{\tau }$. A spread of hadronic invariant masses broadens these endpoints; an average mass alone is not a strict endpoint specification for the whole sample.

Let $P_{\tau }=+1$ denote a positive-helicity ${\tau }^{-}$, and measure $\vartheta$ from its laboratory flight direction in its rest frame. For a spin-zero hadronic recoil, the left-helicity tau neutrino and angular momentum conservation giveFor example a tau lepton with spin along $+z$ can emit a left-helicity neutrino along $-z$, carrying spin along $+z$, with the spin-zero recoil along $+z$. A positive-helicity tau lepton then gives a harder visible recoil. If $x_h=E_h/E_{\tau }$, $r=m_h^2/m_{\tau }^2$ and $\beta \simeq 1$, the more general tau hadronic polarimetry formula isFor spin-zero recoil ${\alpha }_h=1$. For a spin-one system produced by the V-A interaction, the longitudinal and transverse weights are $m_{\tau }^2:2m_h^2$, with opposite angular slopes; summing gives ${\alpha }_h=(m_{\tau }^2-2m_h^2)/(m_{\tau }^2+2m_h^2)$. At $m_h=1.5\mathrm{GeV}$ this is about $-0.175$, so the negative-helicity tau lepton has the higher mean total hadronic energy, about $39.43$ versus $38.67\mathrm{GeV}$, within that spin-one model. ^[ref]

The supplied mean invariant mass does not specify the recoil's spin composition or its helicity amplitudes. Consequently the last sign cannot be universal: positive helicity is harder for a spin-zero recoil, negative helicity is harder for the specified $1.5\mathrm{GeV}$ spin-one recoil, and a mixture requires its measured spectral weights. Polarization of an intermediate vector or axial-vector meson also changes the division of energy among its daughter hadrons; that information is not fixed by the total jet invariant mass alone.

The intermediate photon carries spin $J=1$. In the massless limit, a back-to-back fermion-antifermion pair coupled to its vector current has opposite helicities, so the initial and final projections along their respective axes are $m,m^{\prime }=\pm 1$. A rotation between these axes contributes the Wigner D-matrix elements $d_{m^{\prime }m}^1(\theta )$. The relevant elements have magnitudes $(1+\cos \theta )/2$ and $(1-\cos \theta )/2$. Summing the four squared helicity amplitudes gives the angular factorThe common coupling and normalization factors recover the result in d. The $m=0$ terms do not contribute: the massless external vector currents eliminate the corresponding helicity configurations.