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 $m_b$ gives
$$
\frac{25}{3}\ln\frac{m_b^2}{\Lambda_4^2}=\frac{23}{3}\ln\frac{m_b^2}{\Lambda_5^2},\qquad \ln\frac{m_b}{\Lambda_5}=\frac{25}{23}\ln\frac{m_b}{\Lambda_4}.
$$
Thus <matching the QCD scale across a quark threshold> yields
$$
\boxed{\Lambda_5=m_b\left(\frac{\Lambda_4}{m_b}\right)^{25/23}=\Lambda_4\left(\frac{m_b}{\Lambda_4}\right)^{-2/23},\qquad p=2,\quad r=23.}
$$
The 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.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 305 / 4 / c / Solution

At a heavy-quark mass $m$, leading-order continuity gives $\beta_- \ln(m/\Lambda_-)=\beta_+\ln(m/\Lambda_+)$. Therefore $\boxed{\Lambda_+=m(\Lambda_-/m)^{\beta_-/\beta_+}}$. This is a change of the <strong-coupling scale> parametrization across a change of active flavour number, not a discontinuity of the matched <QCD coupling>. For four to five active flavours in $SU(3)$, $\beta_-=25/3$ and $\beta_+=23/3$, hence $\Lambda_5=\Lambda_4(m_b/\Lambda_4)^{-2/23}$.

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