2
$\begingroup$

Under the entry Beta function (physics) in Wikipedia, the one-loop beta function in electrodynamics (QED) is given by \begin{equation} \beta(e) = \frac{e^{3}}{12\pi^{2}} ,\tag{1} \end{equation} or, equivalently, \begin{equation} \beta(\alpha) = \frac{2\alpha ^{2}}{3\pi} ,\tag{2} \end{equation} where \begin{equation} \alpha = \frac{e^2}{4\pi} .\tag{3} \end{equation}

I am puzzled by the conversion between (1) and (2). I plug (3) into (2) and get \begin{equation} \beta (e) = \frac{e^4}{24\pi ^{3}} \end{equation} But this is not right. How should I make the conversion?

$\endgroup$
  • $\begingroup$ +1 for this question; I was always puzzled by this when I first did QFT, but brushed it aside as unimportant. $\endgroup$ – JamalS May 22 '17 at 10:04
  • $\begingroup$ Why did you not ask your teacher who taught you QFT? $\endgroup$ – Shen May 22 '17 at 10:14
  • $\begingroup$ I taught myself QFT when I was around 16 from Peskin and Schroeder - I'm at university now but back then had nobody to ask. $\endgroup$ – JamalS May 22 '17 at 10:27
5
$\begingroup$

The "argument" of $\beta$ here is actually referring to the variable which gets differentiated, i.e. $$\begin{align} \beta(e) &= \frac{\partial e}{\partial \ln\mu} & \beta(\alpha) &= \frac{\partial \alpha}{\partial \ln\mu} \end{align}$$ In this way, it's not a traditional function argument; it's actually an operator argument, and some of the usual logic about function composition doesn't apply. In more explicit notation you might want to write $\beta[e](\mu)$ and $\beta[\alpha](\mu)$.

To demonstrate the relation that the Wikipedia article shows, you can use the fact that $$\frac{\partial \alpha}{\partial \ln\mu} = \frac{e}{2\pi} \frac{\partial e}{\partial \ln\mu}$$ which follows from the definition of $\alpha$.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.