# Conversion between $\beta (e)$ and $\beta (\alpha)$

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

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

• +1 for this question; I was always puzzled by this when I first did QFT, but brushed it aside as unimportant. – JamalS May 22 '17 at 10:04
• Why did you not ask your teacher who taught you QFT? – Shen May 22 '17 at 10:14
• 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. – JamalS May 22 '17 at 10:27

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 $\alpha$.