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Running couplings

  1. Aug 14, 2008 #1
    Hi there!

    I'm struggling a bit with running couplings. Srednicki introduces dimensional regularization and the [itex]\overline{\mathrm{MS}}[/itex] scheme, then calculates a squared transition amplitude for some reaction in [itex]\varphi^3[/itex] theory. Eventually, he calculates the beta function for the coupling and solves the renormalization group equation. Clear so far.

    In the transition amplitude, a factor [itex]\ln(s/\mu^2)[/itex] occurs, where [itex]s[/itex] is a Mandelstam variable and [itex]\mu[/itex] is the factor arising from dimensional regularization. Now he says, to avoid large logarithms, we should put [itex]\mu^2 \sim s[/itex] (which we can do since physics must be independent of [itex]\mu[/itex]). Then, according to the beta function, the coupling runs with the involved momenta, the well-known behavior.

    Now, [itex]\mu^2 \sim s[/itex] certainly is a convenient choice, but what if I choose [itex]\mu[/itex] to be constant (or something else), which I am free to do? Then, the coupling doesn't run at all. It seems to me that the running coupling is just a way of viewing things (here in particular by using the [itex]\overline{\mathrm{MS}}[/itex] scheme and [itex]\mu^2 \sim s[/itex])?

    Another quick question: I read everywhere that renormalization is related to the behavior of the theory at small distances/large momenta. How are small distances and large momenta related? And what has renormalization to do with it?
     
  2. jcsd
  3. Aug 14, 2008 #2

    Avodyne

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    Correct. If you use a fixed [itex]\mu[/itex], then the coupling doesn't run. The expansion parameter governing perturbation theory then appears to be [itex]\alpha(\mu)\ln(s/\mu^2)[/itex], and if this is large, perturbation theory breaks down. The advantage of using the running coupling is that we see that the expansion parameter is really [itex]\alpha(s^{1/2})[/itex], which might be small even if [itex]\alpha(\mu)\ln(s/\mu^2)[/itex] is not. Indeed, this is the case at high energies in QCD.

    Fourier transform.

    With small distances = large momenta, nothing. As for the connection between behavior at small distances and renormalization, see section 29 of Srednicki.
     
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