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Calculation of the running coupling

  1. Jul 22, 2015 #1
    The beta function is defined as:

    [tex]\beta(\lambda)=M\frac{d}{dM}\lambda[/tex]

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

    [tex]\beta(\lambda)=-\frac{d}{dt}\lambda[/tex]

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

    [tex]\beta(e)=\frac{e^3}{12\pi^3}[/tex]

    and for [itex]e(p=M)=e_0[/itex] the result is

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

    which is clearly false.

    What am I missing?
     
  2. jcsd
  3. Jul 22, 2015 #2

    fzero

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    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.
     
  4. Jul 23, 2015 #3
    I've made some typing errors.

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

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

    which yields the final result

    [tex]e(p)^2 = \frac{ e(M)^2}{1 + \frac{e(M)^2}{6\pi^2} \ln \frac{p}{M}}[/tex]

    The only disagreement between the two results that I was referring to in my original post is the wrong sign in the denominator.
     
  5. Jul 23, 2015 #4

    fzero

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    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
    $$ I = - \int_0^{\ln p/M} dt.$$
     
  6. Aug 12, 2015 #5
    You are right, I was thinking about p as the integral variable, but it was M' instead. Thanks for clearing this out!
     
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