Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 66 / 4 / b / Solution

SetThe chain rule givesFor the ultraviolet expansion take ${\beta }_0>0$ and $L=\log ({\mu }^2/{\Lambda }^2)\gg 1$. The leading solution is $y\sim {\beta }_{0}L$. Substituting it in the next term gives $dy/dL={\beta }_0+{\beta }_1/({\beta }_{0}L)+O(\log L/L^2)$, whose integration yieldsChoosing $\Lambda$ removes the additive constant. This proves the two displayed leading terms of the two-loop asymptotic running coupling.

The printed remainder is too small in general. This can be checked even for an exactly two-loop renormalization-group beta function, so it does not depend on unknown higher-loop effects. Put $c={\beta }_1/{\beta }_0$. Separating variables and fixing the integration constant givesExpanding this identity one order further givesThe $\log L/L$ term cannot be absorbed into a fixed change of $\Lambda$. For example, with ${\beta }_0={\beta }_1=1$, the exact implicit relation is $y-\log (y+1)=L$, and the next term is $(\log L+1)/L$, which is not $O(1/L)$. The unspecified $O(y^{-2})$ contribution to the differential equation changes the constant coefficient of $1/L$, while the coefficient ${\beta }_1^2/{\beta }_0^3$ of $\log L/L$ is already fixed. Only special cases such as ${\beta }_1=0$ avoid this particular obstruction.

The asymptotic formula also presupposes an asymptotically free ultraviolet regime. If ${\beta }_0=0$, division by it is invalid and the leading scaling changes. If ${\beta }_0<0$, the weak-coupling asymptotic direction is generally toward the infrared and the logarithmic variable must be chosen accordingly. These qualifications distinguish the valid perturbative solution from an unconditional formula.