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I have a question about understanding the Lagrangian of standard model, should we view it as a "fundamental" or effective theory? The "fundamental" theory here means the theory with physical cutoff (maybe Planck scale) $\Lambda$; the effective theory here means the theory given by lowering the cutoff $\Lambda'<\Lambda$, (Wilsonian renormalization), and we pick up the leading terms as we saw in the action.

The reason to think the standard model Lagrangian is an effective action

We haven't tested the physics up to the physical cutoff (Planck scale?), this is an extra assumption that the standard model is valid up to physical cutoff. To the contrary, in the effective theory, the cutoff is much more reasonable (e.g. LHC scale).

The reason to think the standard model Lagrangian is a "fundamental" action

If we follow the renormalization flow, the coupling constant of QCD will grow in low energy scale, then the perturbation breaks down. We know from lattice calculations of hardron mass, that even in low energy, QCD works. The Wilsonian renormalization group transformation is formulated from perturbative expansion (may be this is just my limited knownledge, correct me if I am wrong). Since QCD agree with experiment beyond the regiom of perturbation, it has to be a "fundamental" action, than effective action. Similar attitude applies to the standard model.

It seems the question is rather a mental setup. Since standard model works very well so far at LHC, it seems hard to distinguish these two opinions: whether viewing standard model action as "fundamental" or effective.

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    $\begingroup$ The Standard Model has "input data" (free parameters), so it is not a fundamental theory. $\endgroup$ – Trimok Sep 16 '13 at 18:21
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    $\begingroup$ "fundamental" here means the theory with a physical cutoff $\Lambda$, I follows the terminology from my course of QFT, maybe that's a misnomer... I edited the post with "" on fundamental $\endgroup$ – user26143 Sep 16 '13 at 18:26
  • $\begingroup$ Well there are neutrino masses so there is a dimension 5 term already. And that's not even talking about dark matter, dark energy, gravity... $\endgroup$ – Michael Brown Sep 17 '13 at 0:26
  • $\begingroup$ how to see neutrino mass term has dimension 5? thx $\endgroup$ – user26143 Sep 17 '13 at 0:34
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    $\begingroup$ It's not true that the Wilsonian renormalization group transformation is defined using the perturbative expansion. It can be done on the lattice. $\endgroup$ – user1504 Dec 3 '13 at 20:59
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The Standard Model isn't consistent up to arbitrarily high energy scales – because of the Landau pole for the $U(1)$ part of the gauge group at an exponentially high scale – which is one of the reasons why it can't be a fundamental description of Nature.

(If one is satisfied with up-to-all-orders perturbative approximation of the Standard Model, it's consistent at all energies because the Standard Model is perturbatively renormalizable. So neglecting nonperturbative effects, the Standard Model may be considered effective as well as fundamental.)

Other reasons why it's not fundamental is the absence of gravity and the presence of unexplained parameters, the coupling constants and masses, especially the unexplained smallness of the Higgs mass (although if one denies that gravity exists, there's no high scale to compare the electroweak scale with). Another reason why it can't be fundamental is that the Higgs potential becomes unstable for the value of the Higgs mass measured in 2012 – unstable at a high enough scale.

On the other hand, the behavior of QCD at extremely low energies doesn't affect the "fundamental vs effective" debate at all. QCD obviously does work at arbitrarily low energies or long distances, whether its action is a fundamental action or an effective action. The "divergence" of the coupling constant at low energies is a hint that the behavior qualitatively changes. When one investigates how it changes, he finds out that QCD is confining and has a gap – all of its excitations are massive so the spectrum beneath a certain QCD-like scale contains no excitations at all.

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  • $\begingroup$ Thanks for your answer. How to regard QCD as an effective theory in low energy regiom? What I learned the effective theory is lower the cutoff in the perturbativa expansion, if the perturbative expansion breaks down, in this case, how to define effective theory? $\endgroup$ – user26143 Sep 17 '13 at 17:24

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