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Differentiating the strong coupling constant ${\alpha }_s=g^2/(4\pi )$ givesIntegrate this constant derivative and express the integration constant through the strong-coupling scale $\Lambda$ at which the one-loop inverse coupling vanishes. The one-loop running of the strong coupling isThe positive-coupling branch has $\mu >\Lambda$ and is perturbatively reliable only when the logarithm is sufficiently large. The pole at $\Lambda$ signals failure of the one-loop expansion, not a prediction of a physically infinite observable. Equivalently, $\Lambda ={\mu }_0\exp [-2\pi /({\beta }_0{\alpha }_s({\mu }_0))]$ is scale-independent to this order; this is dimensional transmutation.

For the Quantum chromodynamics gauge group $SU(3)$, the coefficients on the two sides of the bottom threshold are ${\beta }_0^{(4)}=25/3$ and ${\beta }_0^{(5)}=23/3$. Continuity of the one-loop running of the strong coupling at $m_b$ givesThus matching the QCD scale across a quark threshold yieldsThe scale parameter changes because the beta-function coefficient changes, even though the matched coupling is continuous. This is leading-order matching with a fixed flavour number within each region; the stated four-flavour region is the effective region above the charm threshold, not a literal claim that four quarks remain active down to arbitrarily small scales.

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.