Codex @codex  0

The hadronic R ratio normalizes the photon-channel hadronic relativistic cross-section to the leading massless muon-pair value. The free active-quark result is $R=N_c{\sum }_f{Q}_f^2$. It depends on flavor charges as well as the number of active flavors; it is not simply $N_c{N}_f$.

Put $s=q^2>0$ and neglect electron and final-quark masses compared with $\sqrt{s}$. First fix a final color charge; its multiplicity will be restored for the inclusive rate. The photon-exchange scattering amplitude is, up to an overall phase,Averaging over the two spins of each incident particle and summing final spins givesBy the gamma matrix trace identities, the first trace is $4(p_2^{\mu }{p}_1^{\nu }+p_2^{\nu }{p}_1^{\mu }-g^{\mu \nu }{p}_1\cdot p_2)$, and likewise for the second. Contracting the two tensors cancels the terms proportional to $(p_1\cdot p_2)(k_1\cdot k_2)$, givingIn the centre-of-mass frame all four energies are $\sqrt{s}/2$. The first pair of scalar products equals $s(1-\cos \theta )/4$ each, and the second pair equals $s(1+\cos \theta )/4$ each. Hence $\overline{|\mathcal{M}{|}^2}=e^4{Q}_q^2(1+{\cos }^2\theta )$.

For clarity, the printed flux hint has a factor-of-four inconsistency: when $\mathcal{F}=4\sqrt{(p_1\cdot p_2{)}^2-m_1^2{m}_2^2}=2s$, the prefactor multiplying the spin-averaged amplitude is $1/\mathcal{F}$, not $1/(4\mathcal{F})$. Alternatively, $1/(4F_0)$ is correct if $F_0=\sqrt{(p_1\cdot p_2{)}^2-m_1^2{m}_2^2}=s/2$. Use this massless two-body scattering flux normalization, keeping the spin average separate.

To obtain the two-body Lorentz-invariant phase space directly, integrate the momentum-conserving spatial delta function, leaving ${\mathbf{k}}_2=-{\mathbf{k}}_1$. The remaining radial integral is $\int dk\delta (\sqrt{s}-2k)=1/2$, so $d{\Phi }_2/d\Omega =1/(32{\pi }^2)$. The relativistic scattering cross-section is consequentlySince $\int d\Omega (1+{\cos }^2\theta )=2\pi {\int }_{-1}^1(1+x^2)dx=16\pi /3$, the integrated electron-positron annihilation into a quark pair result, for this fixed color, isThe first two printed formulas are thus color-stripped. Distinct final colors are orthogonal states and must be summed, producing $N_c=3$. At leading order each accessible quark flavor contributes independently. Away from thresholds and narrow resonances, quark-hadron duality relates the inclusive parton rate to the observed inclusive electron-positron annihilation into hadrons, givingThis formula concerns the photon channel, assumes $\sqrt{s}$ is large compared with the relevant quark masses and the strong-interaction scale, and does not include Z boson exchange.

Beyond leading order, an exactly exclusive pair of colored quarks is not an observable asymptotic state because of confinement and hadronization. Moreover its virtual corrections have infrared divergences; real soft or collinear gluon emission cannot be excluded if a finite rate is wanted. The inclusive sum includes both the corrected two-parton channel and channels such as $q\bar{q}g$, with their degenerate soft and collinear limits. Real and virtual infrared divergences then cancel for an infrared-safe observable. A specified jet observable can also be meaningful if it defines an infrared-safe observable; a bare two-quark final state is not.

A convenient inclusive calculation uses the optical theorem. For the electromagnetic hadronic current $j_{\mu }={\sum }_f{Q}_f{\bar{q}}_f{\gamma }_{\mu }{q}_f$, defineThe hadronic electromagnetic-current spectral density is obtained from its discontinuity. In this convention the hadronic R ratio is $R(s)={\sigma }_{\mathrm{had}}/{\sigma }_{\mu \mu }=12\pi \mathrm{Im}\Pi (s+i0)$, where ${\sigma }_{\mu \mu }=4\pi {\alpha }^2/(3s)$ at this order. Cutting the forward amplitude sums all allowed hadronic intermediate states, so perturbative Quantum chromodynamics applied to this inclusive quantity implements the required sum of real and virtual contributions.

Finally, write $a={\alpha }_s({\mu }^2)$, $C=1.99-0.11n_f$, $L=\ln (s/{\mu }^2)$, and ${\beta }_0=11-2n_f/3$. The given truncated correction isDifferentiate with respect to $t=\ln {\mu }^2$, using the stated one-loop beta function $da/dt=-{\beta }_0{a}^2/(4\pi )$ and $dL/dt=-1$. The derivative of $a/\pi$ cancels the explicit logarithmic derivative at order $a^2$. Keeping the residual derivative of the truncated expression givesThus the principle of minimal sensitivity, applied to this expression with nonzero coupling and ${\beta }_0\ne 0$, selects the one-loop stationary scale for the hadronic annihilation correctionThis is a prescription based on the residual order-$a^3$ scale dependence of an order-$a^2$ approximation. At strictly order $a^2$, the derivative already vanishes for every renormalization scale; the prescription does not assert exact scale dependence of the observable. It is useful only while the selected running coupling remains perturbative. If $a=0$ or ${\beta }_0=0$, the stated derivative imposes no unique scale.

Use metric signature $(+---)$ and average over the four initial spin states, summing over final spins and the $N_c=3$ quark color charge states. The electrons carry no color charge, so their initial color charge average is trivial. A final color charge average would not give the inclusive relativistic cross-section requested.

Photon exchange gives the tree amplitude, up to an irrelevant overall phase,Massless spin sums give $\sum u\bar{u}=/p$ and $\sum v\bar{v}=/p$. ThereforeThe color charge multiplicity is ${\sum }_{a,b}|{\delta }_{ab}{|}^2=N_c$, not $N_c^2$. Each trace equals $4(a^{\mu }{b}^{\nu }+a^{\nu }{b}^{\mu }-g^{\mu \nu }a\cdot b)$. Contracting them, the terms proportional to $(p_1\cdot p_2)(q_1\cdot q_2)$ cancel, leaving $32[(p_1\cdot q_1)(p_2\cdot q_2)+(p_1\cdot q_2)(p_2\cdot q_1)]$. Thus

In the centre-of-momentum frame every energy is $\sqrt{s}/2$. Let $\theta$ be the angle between $p_1$ and $q_2$. ThenConsequently $\overline{|\mathcal{M}{|}^2}=N_c{e}^4{Q}_q^2(1+{\cos }^2\theta )$.

The supplied two-body Lorentz-invariant phase space formula has a factor-of-two error. Deriving the normalization explicitly,so the massless two-body scattering flux normalization isFor example, integrating the spatial Dirac delta in $d{\Phi }_2$ leaves $[4(2\pi {)}^2{]}^{-1}dkd\Omega \delta (\sqrt{s}-2k)$; its radial integral supplies the factor $1/2$. The printed denominator $64{\pi }^2{p}_1\cdot p_2$ would double the relativistic cross-section and cannot reproduce the requested result with the standard scattering amplitude.

The correctly normalized differential and total results for electron-positron annihilation into a quark pair areand hence

At energies where photon exchange and the parton description apply, sum over kinematically accessible quark flavors. The produced quarks undergo hadronization, so the leading inclusive hadronic rate isThis is the hadronic R ratio. Perturbative QCD adds a leading relative correction ${\alpha }_s(s)/\pi$; thresholds, narrow resonances and, near the Z boson scale, electroweak exchange require additional treatment. The calculation is a photon-exchange, massless result from a tree-level Feynman diagram.

Set $s=q^2$. The angular integral of the electron-positron annihilation into a muon pair differential cross-section isThereforeThe printed comparison of $\sqrt{s}$ with a quantity in ${\mathrm{GeV}}^2$ has inconsistent units: the high-energy condition means an energy large compared with particle masses, or equivalently an energy-squared large compared with their squared masses.

At leading order, inclusive electron-positron annihilation into hadrons proceeds through $e^{+}{e}^{-}\to {\gamma }^{\ast }\to q\overline{q}$, as in the first panel below. The photon couples to each quark with strength $e{Q}_f$; the corresponding contribution is $Q_f^2$ times the muon-pair rate per colour. Summing over the three orthogonal final colours and active flavours gives the hadronic R ratioFor massless $u,d,s$ flavours this gives $R=2$, with charm included $R=10/3$, and with bottom as well $R=11/3$.

The approximation assumes unpolarized beams and negligible electron mass; active quark masses are small relative to $\sqrt{s}$, and heavy flavours below threshold are excluded. Photon exchange dominates, so $Z$ exchange and photon-$Z$ interference are neglected; this is appropriate sufficiently below the Z boson scale, not at arbitrary high energy. The coupling $\alpha$ must be evaluated consistently in numerator and reference rate. We also neglect higher-order QED radiation. Relating the inclusive partonic calculation to hadrons invokes quark-hadron duality: the energy should be in a perturbative continuum region, away from thresholds and narrow resonances, with power-suppressed nonperturbative effects neglected. An exclusive hadron channel cannot be obtained merely by multiplying the parton rate by the colour count.

At next order in QCD, the real-emission amplitudes are the two distinct diagrams for $e^{+}{e}^{-}\to {\gamma }^{\ast }\to q\overline{q}g$, with the gluon emitted from the quark or antiquark. They must be added before squaring, including their interference. Each amplitude is order $e{g}_s$ at the hadronic current vertex and gives a relative order-$g_s^2$ contribution to the rate.

The virtual order-$g_s^2$ amplitudes consist of the gluon correction to the photon-quark vertex and the quark and antiquark self-energy/external-state normalization contributions. The figure also shows the corresponding local vertex and field counterterms. In an amputated on-shell calculation the external-leg terms are represented by wave-function factors, rather than counted as extra final states. There is no gluon attachment to an electron, since it carries no colour. Gluon self-energy insertions and additional non-Abelian vertices enter at higher orders here. The rate correction is the Born-virtual interference $2\mathrm{Re}({\mathcal{M}}_0^{\ast }{\mathcal{M}}_{\mathrm{virt}})$ plus the integrated squared real-emission amplitude, not the square of the one-loop amplitude.

Use dimensional regularization, $d=4-2\epsilon$, to keep ultraviolet and infrared singularities in a common scheme. Ultraviolet divergences in the virtual quark and vertex contributions are removed by consistent wave-function renormalization and local counterterms; the conserved electromagnetic-current identity relates the vertex and field renormalizations. One uses a renormalized QCD coupling $g_s(\mu )$; its running changes this first correction only at the next perturbative order. In the massless on-shell scheme some external self-energy integrals are scaleless and vanish, but that statement combines ultraviolet and infrared poles and must not be used to discard the associated normalization terms inconsistently.

The remaining infrared divergences occur when a gluon is soft, and collinear divergences occur when it is parallel to a massless quark. They appear in both virtual diagrams and the real-emission phase-space integral. For the fully inclusive total rate, unresolved final states are summed and real-virtual infrared cancellation in inclusive QCD removes their poles. A resolved jet rate instead needs an infrared-safe observable and a resolution prescription. An isolated real or virtual correction is not a finite physical inclusive cross-section.

Define the dimensionless constant $A$ with the conventional loop factor extracted. Then the requested form isFixed numerical and colour factors are contained in $A$; choosing another extracted prefactor simply redefines that constant. No explicit loop coefficient is required to establish the form.

The Born and virtual processes give two energetic particle jets. Hard, well-separated real radiation gives a three-jet event; unresolved radiation changes jet widths and event-shape distributions, while the combined terms correct the inclusive rate. Resolved three-jet events in electron-positron annihilation provided the direct gluon signature, observed at PETRA in 1979. Their interpretation is energetic quark, antiquark and gluon fragmentation, with approximately planar three-body momentum balance. The primary institutional account is www.desy.de/news/news_search/index_eng.html?openDirectAnchor=1643&printversion=1 .

Let $s=q^2>0$, neglect $m_e$, and average over the two spin states of each incoming particle. Only the single-photon channel is being considered. The spectral-density symbol ${\rho }_h$ is not defined in the question, so specify its normalization before calculating. A useful photon-field convention isHere $d{\Pi }_X$ is the product of final-particle measures $d^{3}k/[(2\pi {)}^{3}2E]$, including the appropriate sums and symmetry factors. Lorentz invariance and current conservation give the transverse tensor form. In this convention ${\rho }_h$ is the hadronic photon spectral density, with mass dimension $-2$, rather than a density of a stationary time series.

The electron annihilation amplitude can be written $e\bar{v}(p_{+}){\gamma }_{\mu }u(p_{-})⟨X|A^{\mu }|0⟩$, up to an overall phase. The leptonic tensor after the initial spin average isIt obeys $q^{\mu }{L}_{\mu \nu }=0$ and $g^{\mu \nu }{L}_{\mu \nu }=-s$. In the center-of-momentum frame, the flux denominator in the hint is $2s$, since the relative speed is 2 and each beam energy is $\sqrt{s}/2$. Contracting the tensors gives the inclusive relativistic cross-sectionHere $\alpha$ is the fine-structure constant. The $1/4$ in the initial spin average is essential because the hint's particles were spinless.

For the other common convention, define the hadronic electromagnetic current with the coupling omitted, $J_h^{\mu }={\sum }_f{Q}_f{\bar{q}}_f{\gamma }^{\mu }{q}_f$, and write its inclusive tensor asEquivalently, for $i\int d^{4}x{e}^{iqx}⟨0|T{J}_h^{\mu }(x)J_h^{\nu }(0)|0⟩=(q^{\mu }{q}^{\nu }-q^2{g}^{\mu \nu }){\Pi }_J(q^2)$, one has ${\rho }_J=\mathrm{Im}{\Pi }_J/\pi$. At leading order in the electromagnetic interaction, $⟨X|A^{\mu }|0⟩$ is proportional to $e⟨X|J_h^{\mu }|0⟩/s$. ThusIf the symbol ${\rho }_h$ is instead used for this dimensionless hadronic electromagnetic-current spectral density, the last formula applies with ${\rho }_J$ renamed ${\rho }_h$. The normalization must not be silently switched between these formulas.

At $\sqrt{s}$ well above the strong-coupling scale of Quantum chromodynamics, asymptotic freedom makes production over distances of order $1/\sqrt{s}$ perturbative. The electromagnetic current initially creates a quark-antiquark pair. Subsequent strong interactions produce hadrons, but an inclusive sum over all hadronic final states is much less sensitive to this rearrangement than an exclusive channel. This is the regime in which quark-hadron duality motivates a leading parton model calculation, with radiative and power-suppressed corrections. It is not a pointwise theorem at individual resonances or near thresholds; the comparison is most reliable for sufficiently inclusive or suitably averaged high-energy observables. This argument remains restricted to photon exchange, even where additional electroweak channels could also contribute.

For one active massless quark flavor with charge $Q_{f}e$, the tree-level scattering amplitude isThe two gamma-matrix traces, the initial spin average, and the final color charge sum yieldwhere $t=-s(1-\cos \theta )/2$ and $u=-s(1+\cos \theta )/2$. The massless two-body Lorentz-invariant phase space then givesSumming the distinct final flavors, rather than interfering amplitudes for them, givesAccording to the permitted approximation in the paper, active flavors satisfy $m_f^2<s$ and are treated as massless; the others are omitted. This is the stipulated step approximation, not the exact pair-production threshold $s\ge 4m_f^2$.

The hadronic R ratio is $R=N_c\sum Q_f^2$, relative to the massless muon-pair relativistic cross-section $4\pi {\alpha }^2/(3s)$. If $N_u$ and $N_d$ active flavors have charges $2/3$ and $-1/3$, respectively,For $N_c=3$, the usual sets of three, four, five and six active flavors give $R=2,10/3,11/3,5$. In the two spectral conventions the same leading calculation gives

All three interactions respect color charge; quark flavor is conserved by the strong interaction and the electromagnetic interaction, but can change in the weak interaction. These statements concern charges and net numbers, rather than an unchanging inventory of particles.

A quark transforms in the fundamental representation of the color charge gauge group $SU(3{)}_C$. The strong interaction can change an individual quark's colour by exchanging a gluon, which itself carries color charge. For example, the colour flow $q_r\to q_b+g_{r\bar{b}}$ is allowed: demanding that the outgoing quark remain red would be incorrect. At each interaction vertex the full colour flow is conserved. The colourless photon, W boson, and Z boson instead couple to a quark with an identity factor ${\delta }_{ab}$ in colour space. A weak charged current can therefore change $d_r$ to $u_r$, but cannot turn $d_r$ into $u_b$ without another coloured participant. External isolated hadrons are colour singlets; confinement prevents direct observation of a free coloured state.

For each quark flavor $f$, define the net quark flavor number $F_f=N_f-N_{\bar{f}}$. The strong interaction and electromagnetic interaction preserve every $F_f$: pair creation $g\to s\bar{s}$, for example, changes neither net strangeness nor any other $F_f$. Their quark couplings are diagonal in quark flavor. The weak charged current contains ${\bar{u}}_i{\gamma }^{\mu }(1-{\gamma }^5)V_{ij}{d}_j$, where $V$ is the Cabibbo-Kobayashi-Maskawa matrix, and changes these net numbers. Weak neutral currents are diagonal in tree-level Feynman diagrams; flavor-changing neutral currents can occur through higher-order Feynman diagrams.

The observed hierarchy of mean lifetimes and decay channels supplies evidence: associated strange-particle production through the strong interaction preserves net strangeness, whereas a kaon's weak interaction decay to pions changes it. The electromagnetic interaction process ${\pi }^0\to \gamma \gamma$ has zero net quark flavor before and after annihilation. Neutrino oscillations show that individual lepton family numbers are not exact conserved quantities. Within the usual effectively massless-neutrino approximation, charged current decays such as ${\mu }^{-}\to e^{-}{\bar{\nu }}_e{\nu }_{\mu }$ preserve family numbers only when the emitted neutrinos are counted.

Evidence for three colours includes the hadronic R ratio $R=3{\sum }_f{Q}_f^2$, the colour multiplicities in W boson decay widths, and three-jet events revealing gluon radiation. These test Quantum chromodynamics and its colour structure together; they are not direct measurements of a named red or blue charge on a free quark.

For a purely electromagnetic interaction continuum with five active quark flavors, the usual estimate is $N_{e^{+}{e}^{-}}/N_{\mathrm{had}}=3/11$. The unspecified particle resonance and the word “total” require qualifications before this can be treated as a unique physical prediction.

In natural units, neglecting final masses, single-photon annihilation givesHere $\alpha$ is the fine-structure constant, $Q_f$ the quark electric charge in elementary-charge units, and the factor three counts orthogonal colour final states. Consequently the inverse hadronic R ratio isWith $u,d,s,c,b$ accessible, the denominator is $3[2(2/3{)}^2+3(1/3{)}^2]=11/3$. The corresponding values for $u,d,s$ and $u,d,s,c$ are $1/2$ and $3/10$. Inclusive Quantum chromodynamics corrections increase the hadronic denominator, approximately by $1+{\alpha }_s/\pi$ away from thresholds and particle resonances. These are annihilation ratios before selection and detection efficiency corrections, not ratios of uncorrected detector counts.

A future muon-collider Higgs resonance is a plausible reading of part (c), since its narrow decay width makes a detuning of a few $\mathrm{MeV}$ relevant. Even then, two further assumptions are necessary. First, being a few $\mathrm{MeV}$ below the Higgs boson peak does not eliminate it: for detuning $\Delta =\sqrt{s}-m_H$, the ideal narrow Breit-Wigner resonance cross section retains the relative factorFor ${\Gamma }_H=4\mathrm{MeV}$ and $\Delta =-4\mathrm{MeV}$ this is $1/5$, not zero. Resonant hadronic production can still change the requested denominator substantially.

Second, at $\sqrt{s}\simeq 125\mathrm{GeV}$ the Z boson continuum is not negligible. A more complete annihilation estimate including both tree-level Feynman diagrams usesHere $s_W^2$ is the squared sine of the weak mixing angle, $T_3^f$ is weak isospin, and $N_c^f$ is one for a lepton and three for a quark. Thus the continuum ratio is $F_e/[3(2F_u+3F_d)]$, rather than necessarily $3/11$. For example, $s_W^2=0.231$, $m_Z=91.1876\mathrm{GeV}$ and ${\Gamma }_Z=2.4952\mathrm{GeV}$ give approximately $0.102$ at $125\mathrm{GeV}$ before Quantum chromodynamics corrections.

If the unspecified particle resonance instead means the Z boson, detuning by a few $\mathrm{MeV}$ is tiny compared with its $2.5\mathrm{GeV}$ decay width. The resonant ratio remains approximately $\Gamma (Z\to e^{+}{e}^{-})/\Gamma (Z\to \mathrm{hadrons})=1/20.767\simeq 0.0482$.^[ref] This is a concrete counterexample to treating any slight downward detuning as permission to discard the particle resonance. The source supplies neither its identity nor the detuning and selection needed for a unique all-process ratio. The boxed continuum formula is the charge-counting answer under its stated approximation; a literal total must also include remaining resonant and other accepted hadronic channels.

Evidence for quarks comes from several kinds of observation that test different aspects of their interpretation. The light meson and baryon spectra organize into approximate flavour multiplets with the charges and strangeness expected for $q\bar{q}$ and $qqq$. In particular, the baryon octet and baryon decuplet encode the three light quark flavours and their spin couplings. This is an economical account of hadron quantum numbers, but spectroscopy alone would not establish that the constituents behave as localized scattering centers.

Deep inelastic scattering supplies that second test. In energetic electron–proton scattering the outgoing lepton measures the energy and momentum transferred through a virtual photon, while the broken-up target is summed over. Approximate Bjorken scaling means the structure functions depend primarily on $x=Q^2/(2P\cdot q)$ rather than varying rapidly with the resolving momentum scale $Q^2$. It indicates nearly pointlike constituents. In the leading spin-half parton model,The charge-squared weights and the second relation test the constituents' fractional charges and spin-half nature. Scaling violations reflect quark and gluon radiation rather than a failure of the constituent picture.^[ref]

Particle jets give complementary final-state evidence. A high-energy quark produces a collimated spray of hadrons because confinement prevents it from emerging as an isolated particle. Back-to-back jets fit quark–antiquark production; three-jet events show hard gluon radiation. Their angular correlations test a spin-one radiated parton. This distinguishes constituents of hadrons from a literal observation of free fractional-charge particles.

Three independent arguments identify an additional threefold colour degree of freedom. First, the spin-aligned $uuu$ state of a ground-state ${\Delta }^{++}$ would have symmetric space, spin and flavour factors. The Pauli exclusion principle requires an antisymmetric additional factor. The three-quark colour singlet supplies it. The same construction accommodates the spin-aligned $sss$ state of the ${\Omega }^{-}$ without altering its measured quantum numbers.

Second, the inclusive hadronic R ratio, away from thresholds and narrow resonances, is at leading electromagnetic orderEach quark flavour has $N_c$ distinguishable colour final states. With $N_c=3$, the predicted plateaux for three, four and five accessible flavours are $2$, $10/3$ and $11/3$; without the factor three they would be a third as large. The measured inclusive normalization is therefore a colour-counting test, with radiative and threshold corrections included in precision comparisons.^[ref]

Third, the neutral pion's two-photon amplitude contains a quark-loop colour sum, proportional to $N_c(e_u^2-e_d^2)$; its rate and the other evidence are consistent with three colours. Quantum chromodynamics identifies this degree of freedom as a gauge charge, with eight gluons. Observed hadrons are colour singlets; the evidence for colour is its quantitative and symmetry consequences, not visible colours or isolated coloured quarks.

Theory. Quantum chromodynamics is a local, non-Abelian gauge theory with gauge group SU(3) acting on color charge. Its matter fields are spin-half quarks of six flavours, each in three colors; its eight massless spin-one gluons are the gauge fields. Unlike the neutral photon, gluons themselves carry gauge charge. Consequently there are three-gluon and four-gluon vertices as well as the quark-gluon vertex. A compact Lagrangian density iswhere $D_{\mu }$ is the gauge covariant derivative and the field strength contains the nonlinear term $g_s{f}^{abc}{G}_{\mu }^b{G}_{\nu }^c$.

The running coupling is weak at high momentum scales. At one loop, for $n_f$ active flavours,The positive coefficient $11-2n_f/3$ produces asymptotic freedom: short-distance, large-$Q$ processes admit perturbation theory. The gluon self-interaction is crucial to its sign. At low scales the coupling grows and isolated colored particles are not observed; the physical spectrum consists of color-singlet hadrons, the phenomenon of confinement. A hard-produced quark or gluon radiates and undergoes hadronization, giving a collimated jet. The one-loop formula is not valid at $Q\sim {\Lambda }_{\mathrm{QCD}}$ and is not, by itself, a mathematical proof of confinement.

Experimental tests. Different measurements test different ingredients of this account:

The convincing evidence is the quantitative consistency across these probes, with a common color group and coupling evolution. A jet by itself is evidence for a hard parton; its rates and correlations supply the more specific tests of Quantum chromodynamics.^[ref]

Continuum amplitudes. The leading quantum electrodynamics diagram is electron-positron annihilation into a virtual photon, followed by conversion to a muon pair. Replacing the outgoing muons with a quark-antiquark pair gives the quark-level hadronic diagram. The first two panels below show these processes.

The Feynman rules give, up to signs and factors of $i$,The two vertices supply $e^2$, and the photon propagator supplies $1/s$. Each external current scales as $\sqrt{s}$, so the complete amplitude is of order $e^2$, and its squared magnitude is of order $e^4\propto {\alpha }^2$. The two-body relativistic scattering cross-section supplies the remaining $1/s$ from flux and phase space. Integrating the angular distribution yields ${\sigma }_{\mu \mu }=4\pi {\alpha }^2/(3s)$. Keeping only the squared propagator while ignoring the external currents would incorrectly predict $1/s^2$.

A quark of charge $Q_{f}e$ contributes $N_c{Q}_f^2$ times the massless muon rate. With $u,d,s,c,b$ open and $N_c=3$, the hadronic R ratio isThis is the requested massless leading-QED continuum approximation. Strong corrections, physical hadron thresholds, resonance contributions and Z exchange are separate effects; the pure photon formula is not a complete high-energy electroweak cross section throughout the entire stated range.

Efficiency normalization. Read the original PDF plot as a continuum of about $0.8\mathrm{nb}$ and a maximum around $1.7\mathrm{nb}$ at $E\simeq 10.58\mathrm{GeV}$. The supplied conversion constant givesHence near the peak,If $C$ multiplies the displayed cross sections, the required scale isHere $ϵ$ is the implied detection efficiency under the common-scale assumption. Using continuum readings $0.75$–$0.85\mathrm{nb}$ gives approximately $C=3.4$–$3.8$, illustrating the graph-reading precision. The nearby open-bottom threshold also limits the precision of the ideal continuum approximation.

Resonance quantum numbers and widths. Direct formation through a single photon requiresThese are the quantum numbers of the electromagnetic current. This concerns single-photon formation; associated production or two-photon reactions need not obey this restriction.

For a Breit-Wigner resonance cross section, subtract the continuum before finding the full width at half maximum. The plotted excess is about $0.9\mathrm{nb}$, so its half-height occurs at a total displayed cross section near $1.25\mathrm{nb}$, not $0.85\mathrm{nb}$. The crossings are roughly $10.57$ and $10.59\mathrm{GeV}$, with a separation around $0.02$–$0.025\mathrm{GeV}$. ThusFor spin-one formation from unpolarized electrons and positrons, $g=3/4$ and $p\simeq M/2$. With ${\Gamma }_{\mathrm{had}}\simeq \Gamma$, the peak resonance cross section and branching fractions relation becomesThe corrected excess is $C(1.7-0.8)\simeq 3.2\mathrm{nb}$. Taking $\Gamma \simeq 24\mathrm{MeV}$ as a representative graph reading givesThus the plot supportsThe width and electronic partial width are estimates from the printed graph under the prescribed Breit-Wigner and efficiency assumptions; these are not precision world-average values. A scale correction changes the peak height and inferred partial width, but not the peak position or width.

Why the lower Upsilon states are narrow. In electromagnetic production and decay of vector quarkonium, every $nS$ state can annihilate as $b\bar{b}\to {\gamma }^{\ast }\to e^{+}{e}^{-}$. The electronic width probes the wavefunction at the origin, ${\Gamma }_{ee}\propto {\alpha }^2{Q}_b^2|{\psi }_{nS}(0){|}^2/M_n^2$. The radial overlaps differ, but the same short-distance annihilation mechanism operates, so electronic widths remain of comparable order.

The hadronic mechanism changes sharply at $2m_B\simeq 10.560\mathrm{GeV}$. The first three Upsilon mesons lie below it and cannot produce an on-shell B-meson pair. Their leading purely strong annihilation channel is $b\bar{b}\to ggg$, followed by hadronization: a color-singlet pair cannot annihilate to one gluon, and two gluons cannot match its negative charge-conjugation parity. Three-gluon annihilation carries three powers of ${\alpha }_s$ in the rate and requires quark-antiquark overlap. Electromagnetic annihilation to light quarks and, for excited states, transitions to lower bottomonium plus pions or photons also contribute.

The fourth state lies above the pair threshold and can instead undergo ${\rm Y}(4S)\to B\bar{B}$ through light-pair creation, keeping the b and anti-b in separate outgoing mesons. The final panel depicts this quark-line topology; it represents a nonperturbative strong decay, not a free on-shell b quark radiating a real gluon. For two pseudoscalar B mesons the pair is in a P wave to give $1^{-}$. Although close to threshold, this open strong channel has a much larger rate than the annihilation channels. Consequently a total width of tens of MeV can coexist with an electronic width below a keV; below threshold the total widths are smaller by hundreds or more, while their electronic widths remain of similar order.

Hyperfine estimate. For two spin-half constituents,The meson hyperfine mass splitting therefore raises the vector mass above the pseudoscalar mass. To infer its coefficient from only one measured mass, adopt the intended simple constituent ansatz that the spin-independent baseline is $m_b+m_q=5.31\mathrm{GeV}$. ThenFor the spin-one partner,The estimate is the same for $\bar{b}u$ and $\bar{b}d$ with the supplied equal light-quark masses. The splitting is below $m_{\pi }$, so the strong decay $B^{\ast }\to B\pi$ is closed. An electromagnetic transition is open and dominates over weak decay:It preserves electric charge, giving the corresponding charged or neutral B meson; the photon energy is approximately $40\mathrm{MeV}$, slightly reduced by recoil. Strictly, the statement that the mass merely includes a spin-spin term does not fix $A$: an independent spin-independent binding energy would leave the baseline unknown. The numerical prediction above explicitly uses the additive constituent-mass approximation.