From: terry_pilling@my-deja.com Subject: Re: Yang-Mills Lagrangian and trace Date: Thu, 25 Jan 2001 23:45:15 GMT Newsgroups: sci.physics.research Summary: Derivation of the Lagrangian in Yang-Mills theory In article <94e7rl$3ef$1@nnrp1.deja.com>, semorrison@hotmail.com wrote: > I've been wondering about the usual prescription for the Lagrangian in > Yang-Mills theory, defined in terms of the Lie algebra valued 2-form > curvature. With some coefficient, this is written as the integral of > > Tr (F^{m n} F_{m n}). > > Here indices are lowered or raised using the metric on the base space, > and the repeated indices indicate contraction, as usual. The curvature > form takes values in the Lie algebra, and as we are considering a > particular representation of the relevant group, there is a matrix > representation of the Lie algebra. The product in the expression above > is then taken as the product of these matrices (this is correct, I > hope!) Finally, we take the trace of this matrix. [...] > So - some questions. > 1) Why is it that we take the _trace_? > Certainly we have to form some scalar. Moreover, we have to worry about > gauge transformations, and so some invariant way of forming this scalar > is necessary. But why the trace rather than, say, the determinant? I'm > afraid I have no physical intuition on this one! I can give you the short answer to this part and let someone else cover the other questions. The gauge invariance that you mentioned is the first thing. If you notice the normalization property of the group generators Tr(G^a G^b) = 1/2 \delta^{ab} Then you see that the trace term in the action actually reduces to Tr(F^{mn} F_{mn}) = 1/2 F_a^{mn} F^a_{mn} One important fact about the lagrangian in a gauge theory is that it must be gauge invariant, and writing this term as a trace makes the gauge invariance manifest. (In a non-abelian theory the field strength is not gauge invariant by itself as it is with QED). The reason that it is manifest is because of the cyclic property of the trace, i.e. Under a gauge transformation of the field strenght we have F -> F' = U F U^{-1} and hence Tr( F'_{mn} F'^{mn} ) = Tr( U F U^{-1} U F U^{-1} ) = Tr( U F^{mn} F_{mn} U^{-1} ) = Tr ( U^{-1} U F^{mn} F_{mn} ) = Tr (F^{mn} F_{mn}) so it is gauge invariant. Note that these gauge transformations are U(r) = exp(-i r^a G^a) Okay, now what about the determinant? My `guess' is representation independence: Our action has to be dimensionless and the dimension must be representation independent. In four spacetime dimensions, if the gauge field (say gluon) has a mass dimension, or inverse length then the trace term has dimension L^{-4} and the action is dimensionless. In SU(3) and up in the fundamental representation a determinant term would not have the correct dimension (but this could maybe be fixed?). BUT, in any gauge theory a determinant term would have a dimension that is representation dependent. Which is a bad thing since then higher dimensional representations would not have dimensionless actions. To see this look at the fundamental representation of SU(3). The matrices are 3 dimensional and our determinant term is then det(F^{mn}_a F_{mn}_b G^a G^b) = [F^{mn}_a F_{mn}_b ]^3 det(G^a G^b) and we have terms of L^{-12}. I hope that made some sense. -Ter ============================================================================== From: baez@galaxy.ucr.edu (John Baez) Subject: Re: Yang-Mills Lagrangian and trace Date: Thu, 1 Feb 2001 03:48:14 GMT Newsgroups: sci.physics.research In article , Charles Torre wrote: >semorrison@hotmail.com writes: >> 1) Why is it that we take the _trace_? >[...] the best reason for using the traditional choice >is because it really seems to describe the world around us! Right, and let's not forget that Yang-Mills theory was invented as a generalization of Maxwell's equations. Yang and Mills were following the principle "If it works, generalize it!" They took Maxwell's equations, replaced the vector potential by something taking values in the Lie algebra su(2), and messed around with the equations until they had gauge symmetry. Only later did it become clear that you could get this automatically by replacing the Lagrangian for Maxwell theory, namely F ^ *F, by a slight generalization, namely tr(F ^ *F). >> 2) Does multiplying the two Lie algebra valued forms as if >> they were matrices seem unnatural? Yes, but that's just a lowbrow way of describing what's really going on here. >One often says that in order to build a non-Abelian gauge >theory with the standard type of Lagrangian you need a Lie >group (to form the "gauge group") whose Lie algebra admits >a "group invariant, bilinear form". This just means that >you have a way of forming an invariant scalar from a pair >of Lie algebra elements. Right: that's the highbrow way. >For the Lie groups that one >usually considers (for physical reasons compact and >semi-simple) such a bilinear form exists (via the "Killing >form", I believe), and it reduces to the rule "trace >of the product" when using a matrix representation. >Presumably though, for each such bilinear form you could >get a different theory. (Of course, one obvious bit of >freedom in the specification of the bilinear form is in an >overall numerical factor, which usually gets used to >specify a coupling constant.) Even better, for a compact *simple* gauge group like SU(n) or SO(n), the Killing form is unique up to an overall numerical factor. So you can completely compensate for the representation you take the trace in by messing with the numerical factor out front front. This factor corresponds to the strength of the corresponding force field. In short, taking the trace in a particular representation is just a matter of taste. For a gauge group like SU(3) x SU(2) x U(1) there is not a single factor to worry about, but three, corresponding to the strength of the 3 fundamental fields in the Standard Model. But again, the representation in which you take the trace can be completely compensated for by changing these factor out front in an appropriate way. >Also, keeping in mind your >first question, I suppose higher-rank invariant tensors on >the Lie algebra could lead to other kinds of Lagrangians. Sure, and one can have fun with that, too! >However, from some points of view these new kinds of >Lagrangians probably lead to theories that are somewhat >pathological as I mentioned above. Right, they'll give lousy quantum field theories, except in low dimensions. But it's fun to know that at least classically, we can do stuff like make nonlinear versions of Maxwell's equations without ruining gauge invariance. If we run out of ideas for papers, we can always work on that!