Consider we have been provided a RGE $$ \mu^2\frac{d}{d\mu^2}\ln Z=\frac{\epsilon}{2}+\beta $$ where $\beta=-\beta_0\alpha_s^2$, upto the $\mathcal{O}(\alpha_s^2)$ and $Z$ is the renormalization constant for $\alpha_s$. I want to solve the RGE to find out $Z$ in terms of $\beta_0$ and $\alpha_s$. I am trying to solve this as following: we rewrite $$ \mu^2\frac{d}{d\mu^2}Z=Z\left(\frac{\epsilon}{2}+\beta\right) $$ $$ or,\qquad Z=\frac{2}{\epsilon}\frac{1}{(1-\frac{2}{\epsilon}\beta_0\alpha_s^2)}\mu^2\frac{d}{d\mu^2}Z $$ Now the $NLO$ of the LHS can be calculated with the $LO$ of the RHS and so on. But I do not know (at least here) how to calculate the $LO$ of $\frac{d}{d\mu^2}Z$. The answer should be $Z=1+\frac{1}{\epsilon}2\beta_0\alpha_s$. Any hint is appreciated.
$\begingroup$
$\endgroup$
2
-
$\begingroup$ You are supposed to solve the differential equation, which gives $Z$ as a function of $\mu$. The right-hand side of the RGE is a constant, let's call it $c=\epsilon/2+\beta$, so it's just $d/d\mu^2 \ln Z= c/\mu^2$. Integrate twice, you should get $\ln Z=-c\ln \mu+a\mu+b$, where $a$ and $b$ are undetermined constants and has to be fixed by initial conditions. $\endgroup$– Meng ChengCommented Mar 9 at 13:42
-
$\begingroup$ It looks like $\mu^2$ as a whole is the variable, so they need only integrate once. $\endgroup$– bbrinkCommented Mar 9 at 14:29
Add a comment
|