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Suppose $K=1+a+a^2[C-b\ln (q^2/{\mu }^2)]$, with $da/d\ln {\mu }^2=-b{a}^2$. Differentiation gives $dK/d\ln {\mu }^2=-2b{a}^3[C-b\ln (q^2/{\mu }^2)]$. For nonzero $a,b$, the principle of minimal sensitivity therefore gives $\boxed{{\mu }_{\ast }^2=q^2{e}^{-C/b}}$. The derivative vanishes identically through order $a^2$: this prescription retains the residual $a^3$ term generated by one-loop running of the truncated expression. Higher-order running and coefficients can shift the stationary scale. For $b=0$ or $a=0$, no unique scale is selected.

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.

Set $a={\alpha }_s({\mu }^2)/\pi$, $b={\beta }_0/4$, $C=1.99-0.11n_f$ and $L=\ln {\mu }^2$, treating $n_f$ as fixed. The notation ${\alpha }_s({\mu }^2)$ means the coupling evaluated at energy scale $\mu$, using the squared-scale label. The one-loop running of the strong coupling gives $da/dL=-b{a}^2$. Write $\ell =\ln (q^2/{\mu }^2)$, so $d\ell /dL=-1$ andDifferentiating both explicit and implicit scale dependence givesFor nonzero coupling and ${\beta }_0\ne 0$, the principle of minimal sensitivity applied to this truncated expression with one-loop running therefore selectsThis is the one-loop stationary scale for the hadronic annihilation correction. For five active flavours it is approximately ${\mu }_{\ast }^2=0.472q^2$, or ${\mu }_{\ast }=0.687\sqrt{q^2}$. It is meaningful within a fixed-flavour perturbative region; crossing a threshold requires the appropriate matched coupling.

The derivative cancels identically through order $a^2$. The prescription retains the residual order-$a^3$ term generated by differentiating the truncated observable with the one-loop running. If one instead discards every order-$a^3$ contribution, no unique scale is determined: every scale is stationary to the retained accuracy. Two-loop running and uncomputed higher observable coefficients can shift the optimum. For ${\beta }_0=0$, or the trivial $a=0$ limit, this leading approximation likewise selects no unique scale.