= Two-loop asymptotic running coupling
{title2=$\frac{4\pi}{\alpha_\lambda}=\beta_0L+\frac{\beta_1}{\beta_0}\log L+O(\log L/L)$}
For $\beta(\lambda)=-\beta_0\lambda^3/(4\pi)^2-\beta_1\lambda^5/(4\pi)^4+O(\lambda^7)$, define $y=(4\pi)^2/\lambda^2$ and $L=\log(\mu^2/\Lambda^2)$. Then $dy/dL=\beta_0+\beta_1/y+O(y^{-2})$. In the asymptotically free regime $\beta_0>0$, successive substitution gives the displayed expansion. If the <beta function (physics)> is exactly truncated to two loops, then $y-\frac{\beta_1}{\beta_0}\log[(y+\beta_1/\beta_0)/\beta_0]=\beta_0L$, after choosing the integration constant. Its next term is $\beta_1^2(\log L+1)/(\beta_0^3L)$. Thus the generic remainder after the first two terms is not $O(1/L)$ when $\beta_1\ne0$. Higher-loop terms change the constant coefficient of $1/L$, but not the forced $\log L/L$ term.
Back to article page
