Calculation of the running coupling

[Trifis]

The beta function is defined as:

$$\beta(\lambda)=M\frac{d}{dM}\lambda$$

If we make the substitution $t=ln(p/M)$ the above equation becomes:

$$\beta(\lambda)=-\frac{d}{dt}\lambda$$

Now if we use e.g. the QED beta function

$$\beta(e)=\frac{e^3}{12\pi^3}$$

and for $e(p=M)=e_0$ the result is

$$e=\frac{e_0}{1+(3e_0/16\pi^2)log(p/M)}$$

which is clearly false.

What am I missing?

 

[fzero]

I think there must be some mistake in your integration. I would compute
$$ \int_{e(p)}^{e(M)} \frac{ de}{e^3} = \int_p^M \frac{1}{12\pi^2} \frac{dM}{M}$$
to find
$$ e(p)^2 = \frac{ e(M)^2}{1 + \frac{e(M)^2}{6\pi^2} \ln \frac{M}{p}},$$
which seems to have the behavior we ordinarily associate with QED.

 

[Trifis]

I've made some typing errors.

The integration after introducing the new variable $t=\ln{(p/M)}$ is:

$$\frac{de}{dt}=-\frac{e^3}{12\pi^2}$$
$$\int_{e(p)}^{e(M)} \frac{ de}{e^3} = -\frac{1}{12\pi^2} \int_{\ln{(p/M)}}^0 dt$$

which yields the final result

$$e(p)^2 = \frac{ e(M)^2}{1 + \frac{e(M)^2}{6\pi^2} \ln \frac{p}{M}}$$

The only disagreement between the two results that I was referring to in my original post is the wrong sign in the denominator.

 

[fzero]

It took me a couple of tries to realize it, but the problem is in the bounds on your integral. We start with
$$ I = \int_{M'=p}^{M' = M} \frac{dM'}{M'},$$
so when you change variables to $t= \ln p/M'$, then
$$ I = - \int_0^{\ln p/M} dt.$$

 

[Trifis]

You are right, I was thinking about p as the integral variable, but it was M' instead. Thanks for clearing this out!