{disambiguate=physics}
{title2=$\beta(g)=\mu\,dg/d\mu$}
{wiki}

The beta function of a coupling $g$ is
$$
\beta(g)=\mu\frac{dg}{d\mu}=\frac{dg}{d\log\mu}.
$$
It describes how the <running coupling> changes when the <renormalization scale> changes.

Beta function

Beta function (physics)

Regard $\lambda$ as a <running coupling> $\lambda(M)$. At fixed $s,t,u$ and to order $\lambda^2$, differentiating the one-loop answer with respect to $\log M$ gives
$$
0=\frac{d\Gamma_4}{d\log M}
=\frac{d\lambda}{d\log M}
-\frac{\lambda^2}{32\pi^2}\int_0^1 6\,dx+O(\lambda^3),
$$
because $m\ll M$ implies $F(M^2)\simeq-x(1-x)M^2$. Thus the <beta function (physics)> is
$$
\beta(\lambda)=\frac{d\lambda}{d\log M}
=\frac{3\lambda^2}{16\pi^2}+O(\lambda^3).
$$
Solving this <ordinary differential equation> with $\lambda(M_0)=\lambda_0$ gives
$$
\frac1{\lambda(M)}=\frac1{\lambda_0}
-\frac3{16\pi^2}\log\frac{M}{M_0},
\qquad
\lambda(M)=\frac{\lambda_0}{1-\dfrac{3\lambda_0}{16\pi^2}\log(M/M_0)}
$$
to leading-logarithmic order.

Solved by gpt-5.6-sol high.

Solution

Paper 304 / 1 / c / Solution

{wiki}

A running coupling is a renormalized coupling regarded as a function of the <renormalization scale>. Its scale derivative is its <beta function (physics)>.

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 Incoming links: Running coupling