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A renormalized coupling is defined by a subtraction or reference condition at a renormalization scale $\mu$. Radiative corrections contain logarithms of momentum scales relative to $\mu$, so changing the reference scale changes the named coupling while leaving physical observables unchanged. Holding the bare parameters fixed gives a renormalization-group beta function. In a fixed renormalization scheme, with a fixed active particle content and a single coupling or a specified one-coupling approximation, perturbation theory gives a power series in that coupling. Its coefficients depend on the fields, representations and scheme; beyond the universal leading coefficients they need not be scheme independent. A theory with several couplings generally has coupled beta functions, so the question's scalar equation is schematic rather than literally the form of every quantum field theory. Also $d/d\ln {\mu }^2=\frac{1}{2}d/d\ln \mu$, which fixes factors of two.

For the stated leading equation let $t=\ln ({\mu }^2/{\mu }_0^2)$. Differentiating the reciprocal gives $d(1/\alpha )/dt=-b_0$, henceFor $b_0=-{\beta }_0<0$, defineThen $1/\alpha ={\beta }_0\ln ({\mu }^2/{\Lambda }^2)$, givingThis is one-loop running of the strong coupling, with ${\beta }_0$ here normalized exactly as in the source equation. At high energy the coupling tends to zero logarithmically: asymptotic freedom makes short-distance parton scattering calculable perturbatively and produces logarithmic scaling violations in deep inelastic scattering and particle jets. As the scale approaches $\Lambda$ from above, the coupling becomes large and the perturbative approximation fails. Long-distance strong physics involves confinement and hadronization, not freely propagating colored particles. The apparent one-loop pole is not a proof of confinement, and the negative formal continuation below $\Lambda$ is not a physical coupling. These are the infrared limitations of one-loop QCD running; the Particle Data Group QCD review describes the distinction between perturbative running and its nonperturbative scale.

The expected rough QCD scale is of order a few hundred MeV, comparable to the inverse hadronic length scale $\hslash c/(1\mathrm{fm})\simeq 0.2\mathrm{GeV}$. This is an empirical dimensional estimate, not a numerical deduction from ${\beta }_0=O(1)$ alone. Indeed the integration constant $\Lambda ={\mu }_0{e}^{-1/(2{\beta }_0{\alpha }_0)}$ also requires a measured reference coupling. Its precise value depends on flavor number, perturbative order and renormalization scheme.

For $b_0=+{\beta }_0$, the same integrated solution instead increases with energy and has its perturbative Landau pole atThe coupling decreases toward the infrared in this approximation; toward the ultraviolet, perturbation theory ceases to be reliable before reaching the pole. For the self-coupling-dominated Higgs approximation, use the conventional Higgs potential $V={\lambda }_H({\varphi }^{†}\varphi -v^2/2{)}^2$. ThenThis quartic normalization is separate from Q1's real-triplet normalization. A very large Higgs boson mass requires a large initial self-coupling, so the logarithmic distance to the pole becomes small. Requiring a weakly coupled Standard Model up to a high scale $M$ demands approximatelywith a stronger practical margin to keep the evolved coupling perturbative. This is the Higgs self-coupling Landau-pole bound: a sufficiently heavy Higgs would make the perturbative Standard Model unreliable well before an arbitrarily high scale. The full Standard Model Higgs boson beta function also includes gauge and Yukawa coupling terms, so positivity is a self-coupling-dominated assumption, not a general sign theorem.

For the requested bottom threshold, take the physical color number $N_c=3$. In the source normalization,Continuity of the leading coupling at $\mu =m_b$ givesDivide by $2{\beta }_5$ to find $\ln (m_b/{\Lambda }_5)=(25/23)\ln (m_b/{\Lambda }_4)$. Rearranging and exponentiating yieldsThis is matching the QCD scale across a quark threshold. The named integration constant changes when the beta coefficient changes; the matched coupling itself stays continuous at leading order. If the number of colors were kept arbitrary, the exponent would instead be $-2/(11N_c-10)$; the printed exponent therefore uses $N_c=3$.