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At tree level, the Feynman diagram has an electron-positron current and a muon-antimuon current joined by one virtual photon. In the massless limit, the spin average and final-state spin sum give $\overline{|\mathcal{M}{|}^2}=e^4(1+{\cos }^2\theta )$, and the integrated relativistic scattering cross-section is $4\pi {\alpha }^2/(3s)$.

When the full Lorentz-invariant phase-space measure treats $n$ identical final particles as labeled, each physical final configuration is counted $n!$ times. The unlabelled relativistic scattering cross-section therefore includes $1/n!$. For two identical scalars, this divides the full-angle result by two; equivalently, integrate over a region containing only one representative of each exchanged pair.

With relativistically normalized external states, $\mathcal{M}$ is the reduced S-matrix element after removing the four-momentum conservation distribution. A relativistic scattering cross-section combines $|\mathcal{M}{|}^2$ with the Lorentz-invariant phase-space measure and the invariant flux factor. Identical-leg factorials in Feynman vertices and the identical final-state symmetry factor in the event integral have different roles.

In the static charge density model, define the normalized Fourier transformHere $\mathbf{q}=(\mathbf{p}-{\mathbf{p}}^{\prime })/\hslash$ is the spatial momentum transfer divided by the reduced Planck constant: $q=|\mathbf{q}|$ has units of inverse length. Physical momentum transfer is $\hslash \mathbf{q}$. We keep $\hslash$ and the speed of light $c$ explicit; compute $q=|\mathbf{p}-{\mathbf{p}}^{\prime }|c/(\hslash c)$, expressing the numerator in $\mathrm{MeV}$ and $\hslash c$ in $\mathrm{MeV}\mathrm{fm}$, to obtain $q$ in inverse femtometres. The spatial $q$ is distinct from the invariant four-momentum transfer used in relativistic scattering cross-sections.

For spherical symmetry, take the polar axis along $\mathbf{q}$ and use spherical coordinates. With $\mu$ the cosine of the polar angle,Every radial shell contributes an amplitude. At $q=0$ all shells add with the same sign; at larger $q$ their contributions can cancel. This cancellation is the origin of a diffraction minimum in the nuclear charge form factor, rather than an absence of electric charge in the nucleus.

The static Fourier transform interpretation is appropriate for the nuclear model and kinematic regime specified here. A fully relativistic composite system requires frame and current conventions; its electric form factor is not universally the transform of a frame-independent three-dimensional charge density.

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 the Feynman vertex $-i\lambda$ for the scalar Yukawa coupling, the internal scalar propagator $i/(k^2-{\mu }^2+i0)$, incoming factors $u_r(p)$ and outgoing Dirac adjoints ${\bar{u}}_{r^{\prime }}(p^{\prime })$. Define $i\mathcal{M}$ as the coefficient in the connected S-matrix element after removing its overall momentum-conserving Dirac delta function. A direct pairing contributesThe crossed pairing interchanges the identical outgoing fermions and therefore has a relative fermionic sign of minus. With $t=(p-p^{\prime }{)}^2$ and $u=(p-q^{\prime }{)}^2$, the result isIt changes sign under exchange of the two outgoing momenta and spin labels, as required for identical fermions. No fermion spin sum or identical-particle phase-space factor belongs in this amplitude; those enter a relativistic scattering cross-section calculation.

Work in natural units $\hslash =c=k_B=1$, at temperatures far below the $W$-boson mass but of order or above the electron mass. The Fermi interaction then supplies a local four-fermion amplitude. The Fermi constant has dimension energy${}^{-2}$. In scattering of a thermal lepton from a nonrelativistic nucleon, the nucleon current matrix element has size $m_N$, while the lepton current has size $T$. Thus $\mathcal{M}\sim G_F{m}_{N}T$. The two-body relativistic scattering cross-section contains the kinematic factor $s^{-1}\sim m_N^{-2}$; when the final and initial lepton energies are comparable, their momentum ratio is of order one. Consequently $\sigma \sim |\mathcal{M}{|}^2/m_N^2\sim G_F^2{T}^2$. This estimates temperatures above the electron mass and mass-splitting thresholds; those scales only affect the dimensionless rate function near freeze-out.

The light leptons in the thermal bath have number density $n\sim T^3$ and speed of order unity. Their conversion rate per nucleon is thereforeFor example, $n+{\nu }_e\leftrightarrow p+e^{-}$ and $n+e^{+}\leftrightarrow p+{\bar{\nu }}_e$ change the neutron-to-proton ratio. The bath density here is the lepton density, not the much smaller baryon density. Dimensionless weak couplings, phase-space integrals and the several reaction channels set a numerical coefficient. Electron masses and the neutron-proton splitting also become important for a precision rate near cosmological weak freeze-out.

For an unpolarized initial fermion, average over its two spin states and sum over the unobserved final spin:The initial scalar has only one spin state. One can evaluate this sum directly from normalized Dirac spinors, or use consistent fermion spin sums to express it as a trace. It is the squared sum of both tree scattering amplitudes, not the sum of their separate squares.

The relativistic scattering cross-section is obtained by integrating the Lorentz-invariant phase-space measure and dividing by the invariant flux factor:There is no identical-final-particle factor, because the outgoing scalar and fermion are distinct. All energies are positive, with $E_q^2=|\mathbf{q}{|}^2+M^2$ and $E_l^2=|\mathbf{l}{|}^2+m^2$.

Equivalently, in the centre-of-momentum frame set $s=-(p+k{)}^2$. Integrating the energy delta function in the relativistic two-body phase space givesFor this elastic process $k_f=k_i$, so integrate $\overline{|T{|}^2}/(64{\pi }^{2}s)$ over the full solid angle. This supplies the requested prescription without evaluating the angular integral.

There are exactly two tree Feynman diagrams: the incident scalar can be absorbed before the final scalar is emitted, or after it. The internal fermion four-momenta are respectively $r_s=p+k=q+l$ and $r_u=p-l=q-k$. A scalar exchange diagram would require an absent scalar self-interaction, so there is no additional tree channel.

Define the S-matrix convention $S_{fi}={\delta }_{fi}+i(2\pi {)}^4{\delta }^{(4)}(p+k-q-l)T_{s^{\prime }s}$. The two tree scattering amplitudes areUsing the Dirac propagator from the preceding Feynman rules,The two terms add with the same relative sign: neither diagram exchanges identical external fermions. Their interference must be retained when squaring the complete scattering amplitude. In the phase convention above, ${\gamma }^5(M-i/r){\gamma }^5=-(M+i/r)$, which gives an equivalent simplified numerator. An overall phase depends on the S-matrix convention and does not change a relativistic scattering cross-section.

Keep the mostly-plus Minkowski metric and the Fourier transform $e^{ip\cdot x}$. Write $/p={\gamma }^a{p}_a$, with $\{{\gamma }^a,{\gamma }^b\}=2{\eta }^{ab}$. To fix the otherwise unspecified phase of ${\gamma }^5$ and the Dirac adjoint, takewhere ${\Gamma }^a$ are ordinary mostly-minus gamma matrices. Then $({\gamma }^5{)}^2=-1$, $\{{\gamma }^5,{\gamma }^a\}=0$, and the fermionic time-derivative term is $i{\psi }^{†}{\partial }_t\psi$. In these conventions the interaction with real $g$ is Hermitian: in mostly-minus notation it is $ig{\bar{\psi }}_{\mathrm{std}}{\Gamma }^5\psi \varphi$. If one instead calls $i{\gamma }^5$ the square-one chirality matrix, its coefficient must be $-ig$ to represent the same interaction. These phase choices leave physical relativistic scattering cross-sections unchanged.

Expanding $e^{i{S}_{\mathrm{int}}}$ yields the following Feynman rules with relativistically normalized external states:

Strip the overall four-momentum conservation delta function when defining the scattering amplitude. There are no further bare interaction vertices, no gauge fixing and no Faddeev-Popov ghost fields in this theory. Renormalized higher-order calculations add the required counterterms; these are additional to the rules of the displayed classical Lagrangian density.

Substitute the preceding spin average into the supplied differential scattering cross-section. Since $e^4=16{\pi }^2{\alpha }^2$,The angular integral is $2\pi {\int }_{-1}^1(1+x^2)dx=16\pi /3$. Consequently the relativistic scattering cross-section for the massless electron-positron annihilation into a muon pair isThe outgoing particles are distinct, so there is no identical-particle factor. Comparing with the requested monomial form givesThe integrated answer depends only on the incoming invariant $s$; the angular variables $t,u$ have been integrated out.

In four spacetime dimensions with $\hslash =c=1$, a relativistic scattering cross-section has mass dimension $-2$, since it is an area. The action is dimensionless, so the Lagrangian density has dimension four. The scalar kinetic term then gives $[\varphi ]=1$, and the cubic term gives $[\lambda ]=1$.

Each term of the four-point tree amplitude in phi cubed theory has dimension $[{\lambda }^2]-[s-m^2]=2-2=0$. Hence $[|\mathcal{M}{|}^2]=0$. In the final integral,while the numerical factors $X,Y$ are dimensionless. ThereforeIncluding the identical-particle factor changes only a dimensionless constant and preserves this check.

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.