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Why dimensionless coupling in nonlinear sigma model corresponding to temperature?

  1. Sep 9, 2011 #1
    Please teach me this:
    In QTF theory book of Schoeder say:
    In nonlinear sigma model L=[itex]\int[/itex]dx[itex]\frac{1}{2g^{2}}[/itex]([itex]\delta[/itex]
    [itex]_{\mu}[/itex]n[itex]^{\rightarrow}[/itex])(consider2<d<4 where d is the dimension number of spacetime).If we consider the Lagrangian as the Boltzman weight of a partition function,then the dimensionless coupling square(g)M[itex]^{d-2}[/itex] is proportional with temperature.But I do not understand why?
    Thank you very much in advance
  2. jcsd
  3. Sep 10, 2011 #2
    Sorry,I have a mistaking:I have missed the square of the derivative of vector n in the Lagrangian of nonlinear sigma model.By the way I would like to ask another question:Why the critical temperature Tc corresponding with the fix-point T*(where T* is Wilson fix-point of T=g[itex]^{2}[/itex]M[itex]^{d-2}[/itex])(and T is corresponding with temperature as I asked above).
  4. Sep 10, 2011 #3
    I would like to add that the condition pose on n that n[itex]^{2}[/itex]=1 in nonlinear sigma model.
  5. Sep 11, 2011 #4
    It seem to me the critical phenomena would be related with Landau-Ginzbua theory(classical saying).Then I do not understand why the nonlinear sigma model also relate with critical theory?In Landau theory the T-Tc corresponds to mass parameter,but how about the nonlinear model,because in this model where is mass parameter?
  6. Sep 12, 2011 #5
    It seem to me that at critical temperature point the heat capacity has a leap,then the temperature at this point stops to decrease(or increase),so the beta function corresponding the temperature(beta(T))(in nonlinear sigma model) is zero at this point.So that critical temperature Tc coincides with the Wilson fix-point of the temperature.Is that correct?
  7. Sep 13, 2011 #6
    Please teach this is correct or not:
    It seem that in first and second phase transition,at critical point the heat capacity is diverge.
    Thank you very much for any answer.
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