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My textbook gives the RG equation: $$ \frac{d \bar{m}(\mu)}{d \ln \mu}=\gamma_{m} \bar{m}(\mu) ; \quad \gamma_{m}=-\frac{3 \alpha}{2 \pi} $$ And then says this is easy to solve and the solution is: $$ \begin{aligned} \bar{m}(\mu) &=\bar{m}\left(\mu_{0}\right) \exp \left[\int_{\ln \mu_{0}}^{\ln \mu} d \ln \mu^{\prime} \gamma_{m}\right] \\ &=\bar{m}\left(\mu_{0}\right) \exp \left[\gamma_{m} \ln \frac{\mu}{\mu_{0}}\right]=\bar{m}\left(\mu_{0}\right)\left(\frac{\mu}{\mu_{0}}\right)^{\gamma_{m}} \end{aligned} $$ Though I can't see how they make this jump to solve the equation, and there seems to be quite little about solving RG equations online. Please could someone break down how they have got to this solution?

EDIT: so I have got as far as: $$ \bar{m}(\mu) = \gamma_{m} \int \bar{m}(\mu) d\ln{\mu} $$ I am not sure how they got the form in the first line given from where I am currently.

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  • $\begingroup$ Which theory? Which textbook? Which page? $\endgroup$
    – Qmechanic
    Commented Apr 9, 2022 at 16:34
  • $\begingroup$ You tried integrating $d\ln f=\gamma dt$ for $t=\ln \mu$ already? Show your work. $\endgroup$ Commented Apr 9, 2022 at 19:36
  • $\begingroup$ @CosmasZachos I have edited the question to show where I am currently. $\endgroup$
    – John
    Commented Apr 9, 2022 at 20:12
  • $\begingroup$ @CosmasZachos Integrating the expression you gave: $ ln{f} = \gamma \int dt = \gamma t$? I could then impose limits $\ln{\mu_{0}}$ to $ \ln{\mu} $ on t, is this correct? This would give $ \gamma \ln{\frac{\mu}{\mu_{0}}} $? $\endgroup$
    – John
    Commented Apr 9, 2022 at 20:42
  • $\begingroup$ So $ f = \frac{\mu}{\mu_0} \exp{\gamma} $, but I am not trying to integrate $d\ln{f}$, I am trying to integrate $d\bar{m}(\mu)$, so how does this solution relate to my problem, and why isn't $\bar{m}(\mu_0)$ involved in the integral - it is a function of $\mu$ yet it just becomes $\bar{m}(\mu_0)$, and lastly, why does $\frac{\mu}{\mu_0}$ go to the power of $\gamma_{m}$, when it was $\exp{\gamma}$? Thanks for the help - I will update my question to be more precise in light of our discussion. $\endgroup$
    – John
    Commented Apr 9, 2022 at 20:45

1 Answer 1

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You are integrating the trivial ODE $\frac{d\ln \bar m(\mu)}{d\ln \mu}= \gamma$, a constant.

So the solution is just $$ \ln \frac{ \bar m(\mu)}{ \bar m(\mu_0)}= \gamma \ln \frac{\mu}{\mu_0} , \leadsto \\ \frac{ \bar m(\mu)}{ \bar m(\mu_0)}= \left (\frac{\mu}{\mu_0} \right )^ \gamma , $$ upon exponentiation.

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  • $\begingroup$ Why is it that $ \frac{d \bar{m}(\mu)}{d \ln \mu}=\gamma_{m} \bar{m}(\mu) \rightarrow \frac{d\ln{\bar{m}(\mu)}}{d\ln{\mu}} = \gamma $ ? $\endgroup$
    – John
    Commented Apr 10, 2022 at 20:30
  • $\begingroup$ Calculus identity ${dy\over dx} =\gamma y\leftrightarrow {d\ln y \over dx} =\gamma$. $\endgroup$ Commented Apr 10, 2022 at 20:48

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