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Why the renormalization group flow depends only the basic symmetry,but not Lagrangian

  1. Sep 19, 2011 #1
    Please teach me this:
    Why the renormalization group flow and the fix-point depends only on the basic symmetry but not on the Lagrangian form.In general speaking,the physics laws depend only the basic symmetries?By the way,the Klein-Gordon,linear sigma,nonlinear sigma Lagrangian flow to one another under the basic symmetry that is O(N) of scalar field?
    Thank you very much in advance.
     
  2. jcsd
  3. Sep 19, 2011 #2
    Re: Why the renormalization group flow depends only the basic symmetry,but not Lagran

    Knowing the physics laws depend only on symmetry is empirical knowledge or ''theoretical'' knowing?
     
  4. Sep 22, 2011 #3
    Re: Why the renormalization group flow depends only the basic symmetry,but not Lagran

    In the absence of a reply from more competent PFers- I'm not sure that what you've said is actually true. What I think is true is that the RG procedure generates all possible terms in the effective action that are consistent with the symmetries of the original action.

    For example, consider phi^4 theory. By doing loop integrals, you can generate, in perturbation theory, 6-point correlation functions. In the Wilsonian approach, you've "integrated out" degrees of freedom above some momentum scale. These are no longer allowed to contribute to your loop integrals, so you have to add a phi^6 term to your effective lagrangian in order to keep the 6-point amplitude right. However, you can't generate a phi^7 term from phi^4 interactions- no matter how hard you try, you can't draw a Feynman diagram with 7 external lines, using only four-point vertices. Such a term would be inconsistent with the phi---->-(phi) symmetry of the original action, and is thus forbidden.
     
  5. Sep 24, 2011 #4
    Re: Why the renormalization group flow depends only the basic symmetry,but not Lagran

    Thank you very much for the answer.
     
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