Why does a Lagrangian matter for the standard model?

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Summary:

Why does a Lagrangian matter for the standard model?
Hi,

I don't know much about the standard model but I'm asking out of interest. Why do we actually need a Lagrangian for the standard model? Surely when you apply the relevant Euler-Lagrange equations, you end up with a variety of equations like the Maxwell equations or Dirac equations. Why can't we just have those - why do we have to put them all into a Lagrangian?

For instance, why is it important that we have a Lagrangian for the Maxwell field? Why can't we just have Maxwell's equations?

Thanks!
 

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Orodruin
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First of all, the SM is a quantum field theory. This means that you cannot just use the EL equations of motion, you must somehow quantize your theory. As for the use of writing down the action in terms of the Lagrangian, you do not have to - but it significantly simplifies computations to use the path integral formalism, which is based on the action. In general, choosing between two equivalent descriptions is always coming down to computational simplicity. Why would you use the Lagrange formulation of classical mechanics when you have Newton’s laws? Same idea and same answer: because it is useful. In the case of QFTs, it is so much more useful that we typically define our theories by writing down Lagrangians - including the SM.
 
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dextercioby
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Putting the answer straight to the point: because our current understanding of the theory of the Standard Model is in terms of perturbation theory, very neatly formulated in terms of "classical" Lagrangians actions using Feynman's path integrals.
 
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vanhees71
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Well, with the caveat that classical Lagrangians are not the full truth, but the theory is somehow quantized. The path-integral method is one possibility to do so. Though many properties of the classical Lagrangian are astonishingly stable under quantization, particularly symmetry principles, sometimes they are not. In connection of symmetries there can be anomalies, i.e., quantization of the theory may lead to the explicit breaking of some symmetries the classical theory obeys.

Anomalies can be as well good as bad. An example for a good anomaly is the Adler-Bell-Jakiw anomaly, i.e., the breaking of the axial U(1) symmetry in chiral hadronic models (or rather speaking in a modern way QCD in the chiral limit). This symmetry breaking makes, e.g., the decay rate of neutral pions to two photons consistent with experiment without giving up the (approximate) chiral symmetry which is crucial for a description of the strong interactions of hadrons.

Another example for a good anomaly is the trace anomaly of QCD, i.e., the breaking of scaling symmetry in the massless limit of QCD, which provides an explanation for most of the mass of the matter around us (the Higgs mechanism providing the current quark masses to the (light) quarks is a tiny 1-2% effect on the mass of hadrons). It's quite likely that this anomaly is the main mechanism to dynamically create the hadron masses, while the spontaneous breaking of the (approximate) chiral symmetry only provides the splitting of the masses of chiral partners in the light-quark hadron spectrum.

A bad anomaly is if a local gauge symmetry is anomalously broken (which is also finally the argument that it must be the axial U(1) that must be broken; you are not allowed to choose the vector U(1) to be broken, because then electromagnetic gauge theory would be broken too). Then there's in fact no local gauge symmetry in the quantized model, and the model would simply be garbage. In the Standard Model the high-risk peace is electroweak theory, which is a gauged chiral symmetry. It's gauge group ##\mathrm{SU}(2)_{\text{WISO}} \times \mathrm{U}(1)_{\text{Y}}## is not automatically save from being broken by anomalies, but nature is kind to us, providing precisely the charges to the quarks and leptons (in each generation separately), including the 3 colors of QCD to the quarks, to perfectly cancel the putatively dangerous anomaly.
 
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