Yang-Mills Lagrangian and trace
semor...@hotmail.com
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.
Incidentally, we can write the same expression as Tr ( F ^ *F ), where
^ is the usual exterior multiplication of forms (remembering we are
actually multiplying matrices however), and * is the Hodge star. This
might be more or less familiar to various people! It looks
more 'geometrical' this way, but if we're not afraid of abstract
indices it makes little difference.
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!
2) Does multiplying the two Lie algebra valued forms as if they were
matrices seem unnatural?
There is no problem with actually doing this - because we are in a
particular representation the elements of the Lie algebra are in fact
matrices. On the other hand the product won't generally be in the Lie
algebra any more, which seems a little peculiar. (Of course, if we took
the Lie algebra product instead things would go horribly wrong - U(1)
Yang-Mills, electromagnetism, would become trivial!)
Thanks, Scott Morrison
semor...@hotmail.com
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terry_...@my-deja.com
semor...@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
Robert C. Helling
>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!
>
>2) Does multiplying the two Lie algebra valued forms as if they were
>matrices seem unnatural?
>There is no problem with actually doing this - because we are in a
>particular representation the elements of the Lie algebra are in fact
>matrices. On the other hand the product won't generally be in the Lie
>algebra any more, which seems a little peculiar. (Of course, if we took
>the Lie algebra product instead things would go horribly wrong - U(1)
>Yang-Mills, electromagnetism, would become trivial!)
Let me answer both questions at once: As you noted, we want to form
invariants of the Lie algebra to get an gauge invariant Lagrangian.
Those are classified for all Lie algebras, the simplest one being the
quadratic Casimir which is just the trace in a matrix representation (but
there is a more abstract notion if you are afraight of concrete matrices,
both boil down to the same thing), but there are others, details depending on
the algebra under study.
The physical input is that you want something quadratic in the field strengths
because that leads to second order PDEs for the classical equations of
motion for the A's. This comes handy with describing a configuration by
fields and momenta. If higher derivatives were present, you would have to
specify more constants of integration. There is also a quantum analogue
of this ("unitarity"). But note, that this equation
second derivatives <=> unitary theory
is just a rule of thumb and there are counter examples (string theory for
example).
Robert
--
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Robert C. Helling Institut fuer Physik
Humboldt-Universitaet zu Berlin
print "Just another Fon +49 30 2093 7964
stupid .sig\n"; http://www.aei-potsdam.mpg.de/~helling
Charles Torre
[Concerning the Lagrangian for non-Abelian gauge theory.]
I do not have quick or definitive answers to your questions,
but I have a number of ideas.
> 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!
With this choice, the field equations are (modulo gauge),
hyperbolic equations - nonlinear wave equations - so
they have nice properties. I would think that higher
powers of the field strength in the Lagrangian, such
as you would get with, say, a determinant, would lead to
PDEs with less desirable qualities from both the classical
point of view and, more importantly for physics, from the
quantum point of view. I say this, not because the field
equations would be higher than second order, which need not
be the case (e.g. with a determinant or something), but
because the symbol of the PDEs (which has a lot to say
about the Cauchy problem, propagators, etc.) would now be
dangerously field dependent. My guess is that the second
most important reason for the traditional choice is that
other choices most likely spoil key properties of the quantum
field theory relative to the traditional choice. I am
thinking of unitarity, renormalizability and asymptotic
freedom here. Finally, the best reason for using the traditional choice
is because it really seems to describe the world around us!
>
> 2) Does multiplying the two Lie algebra valued forms as if they were
> matrices seem unnatural?
> There is no problem with actually doing this - because we are in a
> particular representation the elements of the Lie algebra are in fact
> matrices. On the other hand the product won't generally be in the Lie
> algebra any more, which seems a little peculiar. (Of course, if we took
> the Lie algebra product instead things would go horribly wrong - U(1)
> Yang-Mills, electromagnetism, would become trivial!)
>
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. 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.) Also, keeping in mind your
first question, I suppose higher-rank invariant tensors on
the Lie algebra could lead to other kinds of Lagrangians.
However, from some points of view these new kinds of
Lagrangians probably lead to theories that are somewhat
pathological as I mentioned above.
-charlie
Laurence Yaffe
>semor...@hotmail.com writes:
>[Concerning the Lagrangian for non-Abelian gauge theory.]
>I do not have quick or definitive answers to your questions,
>but I have a number of ideas.
>> 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!
Charles Torre already gave some reasons. Other possible answers include:
1. The trace of the square of the field strength (or tr(F^2))
is the lowest dimension gauge invariant, Lorentz invariant scalar
one can form. In four dimensions, this is the only action for
a (parity invariant) quantized non-Abelian theory you can use
which has decent short distance behavior. [That is, adding any
higher dimension invariant will render your theory non-renormalizable.]
2. tr(F^2) gives you equations of motion which reduce to (multiple copies of)
the usual Maxwell equations for weak fields. This is necessary for
good short distance behavior (``asymptotic freedom'').
3. tr(F^2) works --- it yields a sensible theory of strong interactions (QCD)
which agrees with experiment.
John Baez
>semor...@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!