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Must all symmetries have consequences?

We know that transnational invariance, for example, leads to momentum conservation, etc, cf. Noether's Theorem.

Is it possible for a theory or a model to have a symmetry of some kind with no physical consequences at all for that symmetry?

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Noether's theorem states: for each symmetry, there is a conserved quantity. It's true, because it rhymes ;) See math.ucr.edu/home/baez/noether.html –  Alex Nelson Aug 1 '12 at 4:31
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3 Answers 3

Noether's theorem is overzealously applied--- it only applies to theories with a Lagrangian formulation, or to quantum mechanics. This is true of fundamental systems, but for non-fundamental systems, you can have classical equations which are symmetrical the symmetry does not imply a conservation law.

The symmetry does not come with no consequence, however, it comes with the consequence of symmetrical solutions! If the equations are symmetrical under a transformation, the solutions must come in families that turn into each other under the symmetry. For classical systems, this is not a particularly profound consequence. So I will consider systems where this is the only consequence.

For a stupid example, consider Newton's laws for an object free-falling in a gravitational field. The acceleration is uniform in the z direction, but z momentum is not conserved. The reason is that the Lagrangian is not invariant in the z direction. But you wouldn't know it from looking at the equations of motion.

For a less stupid example, consider Newton's laws for a particle with constant force and a $v^3$ friction law, say:

$$ {dv \over dt} = a + v^3 $$

And there is a symmetry in time translation and for translations in x. But aside from telling you the trivial fact that solutions are translatable in x and in t, it doesn't tell you anything more.

These problems are sort of silly, so I'll give the granddaddy of all examples--- the incompressible Navier stokes equations, with hyperviscosity (so that what I am saying is definitely true). Here you have a time-dependent diffeomorphism $X(x)$ from n-dimensional space whose general point is called x, to itself.

The time derivative of X is the velocity field v, and v obeys the equation

$$ {\partial_t} v = v\cdot \nabla v + \nabla P + \nu \nabla^2 v + \epsilon \nabla^4 v $$

Where the $\epsilon$ term is introduced to make sure that the equation has a unique and smooth initial value problem, so that the X diffeomorhism makes sense. This equation is translationally invariant to t translations, and completely diffeomorphism invariant--- composing X with a diffeomorphism takes a solution to a solution, but it doesn't have a conserved energy (although formally the limit $\nu=\epsilon=0$ does), nor does it have any conserved quantity corresponding to the diffeomorphism invariance. The diff invariance is a gauge redundancy in the X description.

These symmetries still have the trivial consequence of symmetrical solutions--- so you can translate any solution in time, and do a diffeomorphism on the initial positions. It just doesn't mean anything to studying the equation.

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What do you mean by "consequences"?

Classically, topological numbers might fit that bill. In quantum field theory, that depends upon whether it is sensible to consider superpositions of topological sectors. If there is an S-duality, it might possibly be.

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How are topological numbers a "symmetry"? Same for S-duality. This reads like nonsense superficially. –  Ron Maimon Aug 1 '12 at 8:39
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If there is a continuous symmetry of the action, it is necessary to take the quotient by the symmetry--gauge fix it--when quantizing. One way to see this is to just consider perturbation theory. The quadratic coupling in the Lagrangian will be constant along the flow of the symmetry, so it will be have degenerate derivative in the tangent space directions along that flow and hence not be invertible. Thus, we won't be able to find a propagator.

Abstractly, perturbation theory works when the classical theory one is quantizing is integrable. When showing a 2n dimensional system is integrable, one must specify n Poisson-commuting integrals of motion. The continuous symmetry gives one by Noether's theorem, and n is the most possible in 2n dimensions, so one has to come up with n integrals of motion which combine linearly into the Noether charge of the symmetry. In other words, one eventually includes the symmetry in the solution regardless of whether he meant to.

I feel like there might be weirder things that happen nonperturbatively, but I am not sure. Certainly when one wants all observables to also obey the symmetry, one can cook up a BRST operator. The existence of this operator implies that some states in the "big" Hilbert space that doesn't take account of the symmetry necessarily have zero norm. Can one do this in all cases?

It is possible on the other hand to lose symmetries when one quantizes. Usually this is the fault of the regularization scheme, but in certain cases one can prove that no regularization scheme is invariant under the symmetry. An example is the 2+1 dimensional Chern Simons theory with Wilson loops. Even though the Lagrangian is fully topologically invariant, one needs to pick a framing of loops to regularize self-intersections. Similarly there are theories with manifestly conformal Lagrangians but which require breaking scale invariance in regularization.

Discrete symmetries are of course of a much different flavor.

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This is false. You only need to take the quotient when doing the path integral over a locally infinite volume gauge invariance. You most definitely do not take the quotient for a single free particle which has a continuous symmetry which takes all points to the origin! Then you couldn't have free particle motion. –  Ron Maimon Oct 31 '12 at 1:47
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