 Solution

At the stated order the renormalization-group beta function is separable:ConsequentlyFor a positive initial running coupling, this expression is used on the connected interval containing ${\mu }_i$ where its denominator is positive and perturbation theory remains valid.

If $b_0<0$, the running coupling decreases at high renormalization scale and tends to zero: this is asymptotic freedom. Decreasing the renormalization scale increases the running coupling, with a formal infrared pole at $\Lambda ={\mu }_i\exp [1/(b_0{\alpha }_i)]$. Near $\Lambda$, perturbation theory fails; the pole alone is not a proof of confinement. Both QCD and the SU(2) weak gauge coupling have this sign for the Standard Model particle content above the relevant thresholds.

If $b_0>0$, the running coupling decreases towards low renormalization scale and grows towards high renormalization scale. The formal ultraviolet Landau pole is ${\mu }_L={\mu }_i\exp [1/(b_0{\alpha }_i)]$. The pole signals failure of this perturbative evolution, rather than a controlled physical infinity. The hypercharge gauge coupling and ordinary QED with charged matter have this sign.

If $b_0=0$, the running coupling is constant at this order; higher-order terms may still make it run. No Standard Model gauge factor has a vanishing one-loop order coefficient for the usual particle content.

For precise normalization, take ${\alpha }_i=g_i^2/(4\pi )$ and $Q=T_3+Y$. The Standard Model one-loop gauge coefficients then giveThese coefficients use $d/d\ln \mu$, not $d/d\ln {\mu }^2$. Threshold matching changes the active particle content, so the single constant-$b_0$ expression is applied piecewise across thresholds.