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Is Yang-Mills scale invariant?

  1. Apr 2, 2014 #1
    I remember hearing this, but not sure if it's true.
     
  2. jcsd
  3. Apr 2, 2014 #2
    No, Yang-Mills theory is not scale invariant. Scale invariance is broken by the running of the coupling constant with energy.
     
  4. Apr 2, 2014 #3
    Hm, is there a version of this statement that's true? I remember reading this for 4 dimensions, but I can't see how that would be relevant. Hm, is it perhaps true at tree level? Sorry for the vague questions.

    EDIT: for example "MHV amplitudes", which are basically tree amplitudes for gluons, can be explicitly calculated and shown to be conformally invariant... Maybe this is ruined for loop calculations, compatible with your claim? I'm guessing here...
     
    Last edited: Apr 2, 2014
  5. Apr 2, 2014 #4
    you might have seen conformal invariance of pure yang-mills theory,which happens when d=4.Scale transformation holds for pure yang mills.
     
  6. Apr 2, 2014 #5
    Ah excellent, so in Minkowski spacetmie pure Yang-Mills is scale invariant! Do you have a reference?
     
  7. Apr 2, 2014 #6
    What? As I said above pure Yang-Mills theory is not scale invariant/conformally invariant because of the running of the coupling.

    I think nonequilibrium must be thinking of N=4 supersymmetric Yang-Mills theory.
     
  8. Apr 2, 2014 #7
    Hm, I know about N = 4 SYM and I wasn't thinking of that. The_Duck, what you say makes sense, but how am I to reconcile with the fact that MHV amplitudes are conformally invariant?
     
  9. Apr 2, 2014 #8
    A pure yang mills theory is classical in nature and at the tree level the vanishing of trace of energy momentum tensor guarantees that conformal invariance is maintained if d=4.This is lost very easily when we go to a quantum case.Quantum Yang mills theory involves an intrinsic mass scale(AS in QCD).This scale determines the masses of physical hadronic states.The regularized path integral is not invariant under scale transformation even though the action is and develops a conformal anomaly.Conformal symmetry is broken explicitly by quantum effects.
    @nonequilibrium-You should take a look at 'Quantum fields and strings-a course for mathematicians Vol.1',The section on classical field theory.Probably 4th chapter of this section.
     
  10. Apr 2, 2014 #9
    Excellent thanks :)
     
  11. Apr 2, 2014 #10
    Let me just do a quick summary for future people visiting this thread with the same confusion:

    So the basic statement is that the (free) Yang-Mills action has conformal symmetry in Minkowski space. One way to check this statement is via the vanishing of the trace of the energy-momentum tensor. To see how this is crucial: remember [itex]T_{\mu \nu} = \frac{\partial S}{\partial g^{\mu \nu}}[/itex] hence the *trace* of this should express how the action changes under a scaling of the metric, which is what conformal symmetry is all about.

    For Yang-Mills one can check that [itex]T^\mu_\mu = \left( \frac{g^\mu_\mu}{2} - 1 \right) F^{\mu \nu} F_{\mu \nu} [/itex] (source: the Deligne book that andrien refers to, p200, equation (3.79)). Hence this vanishes if and only if our spacetime is 4D, in which case the trace of the metric is -1+1+1+1 = 2.

    When we do QFT we look at the path-integral [itex]\int e^{iS} \mathcal D A[/itex] and it turns out our measure is not conformally invariant, hence our theory/amplitudes will not be (this is called an anomaly). However, anomaly only enter in loop calculations(?), hence we should still expect Yang-Mills to have conformally invariant tree scattering amplitudes, an example being the so-called MHV amplitudes.
     
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