= Solution
A regulator and a <renormalization condition> introduce the reference mass scale $\mu$, even when the classical theory has no mass. Loop amplitudes contain dimensionless logarithms of momentum or distance ratios involving $\mu$. The resulting <running coupling> and field normalization compensate changes of this arbitrary reference scale.
Let $\phi_0=Z_\phi^{1/2}\phi$ and $G_n=Z_\phi^{-n/2}G_{n,0}$. Hold the bare parameters fixed and define $\beta(g)=\mu\,dg/d\mu$ and $\gamma(g)=\frac12\mu\,d\log Z_\phi/d\mu$. Assuming multiplicative field <renormalization> and no mixing or additive contact terms for the correlator, differentiating gives the <Callan-Symanzik equation>
$$
\boxed{(\mu\partial_\mu+\beta(g)\partial_g+n\gamma(g))G_n=0.}
$$
For the dimensionless two-point factor put $r=p^2/\mu^2$. Its equation is $(-2r\partial_r+\beta\partial_g+2\gamma)C=0$. Let
$$
\frac{dg(t)}{dt}=\beta(g(t)),\quad g(0)=g,\qquad f(t)=\exp\left(2\int_0^t\gamma(g(u))\,du\right).
$$
The <characteristic solution of the multiplicative Callan-Symanzik equation> is
$$
\boxed{C(e^{2t}r,g)=f(t)C(r,g(t)).}
$$
To check the sign, $\partial_tC(e^{2t}r,g)=2e^{2t}r\partial_rC$ equals $(\beta\partial_g+2\gamma)C$. The characteristic flow and its accumulated multiplier give precisely this evolution. Changing $\mu$ changes the dimensionless momentum and renormalized $g$; the same bare <two-point correlation function> is recovered after the compensating field normalization. An unnormalized renormalized correlator need not remain numerically identical under that change, but physical predictions do.
With a mass, write $v=m/\mu$ and define its <running mass> by $m'(t)=\delta(g(t))m(t)$, $m(0)=m$. The dimensionless equation becomes
$$
[-2r\partial_r+\beta\partial_g+(\delta-1)v\partial_v+2\gamma]C=0.
$$
Its flow is therefore
$$
\boxed{C(e^{2t}r,m/\mu,g)=f(t)C(r,e^{-t}m(t)/\mu,g(t)),\quad m(t)=m\exp\left(\int_0^t\delta(g(u))\,du\right).}
$$
The <renormalization-group mass suppression criterion> is $\int_0^t(\delta(g(u))-1)du\to-\infty$, for example an eventual bound $\delta\le1-\eta$ with $\eta>0$. Then the mass argument on the right tends to zero. A regular <massless limit>, uniform along the limiting coupling trajectory, makes the mass negligible at high energies. Merely calling $\delta$ small without controlling this integrated exponent is insufficient.
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