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Symplectic structure vs. metric structure |
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| Jun21-07, 05:00 AM | #1 |
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Symplectic structure vs. metric structure
Symplectic structure vs. metric structure
A question about the relationship between the phase space of the Hamiltonian formulation of classical mechanics and of the Lagrangian formulation; that is, between the cotangent bundle of configuration space, T*Q, which has a natural symplectic structure given by the canonical symplectic form, and the tangent bundle, TQ, which has a natural metric structure given by the Riemannian line element, viz. a quadratic differential form of the q dots. It seems there's a sense in which the former has *less structure* than the latter. Hamiltonian phase space has symplectic structure (this determines a volume element), not metric structure. And in general, metric structure determines, or presupposes, a volume structure, but not the other way around. A metric would then add another level of structure to what's needed for the Hamiltonian equations of motion. (``Level of structure'' in the sense that, starting with a set of points, we can define mathematical objects on it, some of which presuppose others; in this sense a topological space has more structure than a set of points, a metric space has more structure than a topological space--a metric induces a topology-- and so on.) It also seems the Lagrangian formulation needs this metric structure, for the quadratic differential form is the invariant quantity of the Lagrangian transformations, and a symplectic manifold is ``floppy'', having no local notion of curvature that would distinguish one symplectic manifold from another locally (from Darboux's theorem). This is because of the two sets of generalized coordinates used by the Lagrangian as opposed to the Hamiltonian formulation: for the Hamiltonian, the canonically conjugate q's and p's, treated as independent variables on the phase space (so that the energy function is linear in each); for the Lagrangian, the generalized coordinates and their first time derivatives, the generalized velocities (giving rise to the quadratic differential form of the q dots). (Would a more coordinate-free version of Lagrangian mechanics be able to get by without the Riemannian metric structure on the base manifold? I would've thought not, given the above, but I'm not sure.) It seems this should mean that not every symplectic manifold (similarly, not every cotangent bundle) is isomorphic to a Riemannian manifold (tangent bundle). However, there is a natural isomorphism between T*Q and TQ, given by the Legendre transformation. On the other hand, the isomorphism is non-canonical (there is no basis- independent isomorphism). More generally, then, how can we compare the structures of these two spaces, given that they are, after all, *different* spaces? Is there a structure-preserving map between the two? How should one go about trying to find such a thing? Any and all feedback or references would be extremely helpful! I am in philosophy of physics, struggling with a paper, and having trouble figuring out an answer on the basis of the books and online references I've seen so far. Profuse apologies if I have simply gotten myself rather mixed up. |
| Jun22-07, 05:01 AM | #2 |
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On 2007-06-21, jillnorth@hotmail.com <jillnorth@hotmail.com> wrote:
> It seems this should mean that not every symplectic manifold > (similarly, not every cotangent bundle) is isomorphic to a Riemannian > manifold (tangent bundle). However, there is a natural isomorphism > between T*Q and TQ, given by the Legendre transformation. On the > other hand, the isomorphism is non-canonical (there is no basis- > independent isomorphism). More generally, then, how can we compare > the structures of these two spaces, given that they are, after all, > *different* spaces? Is there a structure-preserving map between the > two? How should one go about trying to find such a thing? The Legendre transform, strictlly speaking, is not a map between tangent and cotangent space. Rather, it is a map between functions on a vector space and its dual (in particular the tangent space at a point, and its dual). Specifically, it invertibly maps bounded (from below), convex functions on the tangent space to bounded, convex functions on the cotangent space. As such, it is independent of the choice of basis in either vector space. The map from the cotangent space to the tangent space is contained in one of Hamilton's equations: dq/dt = @H/@p, where @ denotes partial derivative. The reverse map, p = @L/@(dq/dt), is defined with the help of the Lagrangian. As long as both L and H obey the above boundedness and convexity restrictions (as functions of dq/dt and p, respectively), then both maps are well defined and basis independent. However, they require (the Legendre dual) L and H to be provided. Note, however, that for the usual Legendre tansform to be defined, you need to work with vector spaces. So, if your symplectic manifold is not at least a vector bundle (where momenta live in vector space fibers above position points), you'll have trouble getting an equivalent Lagrangian formulation. On the topic of comparing structures on T*Q and TQ. As you've noticed, T*Q has a natural symplectic form on it. On the other hand TQ has a natural prolongation structure. What I mean by this is that given a (parametrized) curve in Q, it can be lifted to a curve in TQ simply by attaching the curve's tangent vector to every point. While in T*Q, the prolongation only exists for curves that solve Hamilton's equations of motion. In a general symplectic manifold it's not even possible to meaninfully separate the position and momentum coordinates, as they can be freely mixed by symplectic transformations. In the end, you can always restrict treat the phase space of a mechanical system as the set of the solutions of its equations of motion. Then both TQ and T*Q are just particular ways of parametrizing it. Thus, if you have a symplectic form in any one parametrization, you can always pull it back to different parametrization through a diffeomorphism (like the one I described above). > Any and all feedback or references would be extremely helpful! I am > in philosophy of physics, struggling with a paper, and having trouble > figuring out an answer on the basis of the books and online > references I've seen so far. Profuse apologies if I have simply > gotten myself rather mixed up. For anyone trying to master mechanics from the symplectic point of view, I would first recommend V.I. Arnold's book _Mathematical Methods of Classical Mechanics_. Another book on this subject that I know by reputation is Abraham's and Marsden's _Foundations of Mechanics_. I don't know if these books will help answer your questions, but they should contain all the basics that relate mechanics and symplectic manifolds. Hope this helps. Igor |
| Jun24-07, 09:11 AM | #3 |
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Hi Jill,
if M is a manifold and t(M) (t*(M)) its tangent (cotangent) bundle then the partial differentials d/dx_{j} and the differentials dx_{j} form a dual basis in the tangent- (cotangent-) vector spaces T_{x}M(T*_{x}(M)). A dual basis is a somehow orthogonal basis. That means the partial differentials and the differentials are somehow orthogonal on each other. Think of it as the dual of the electromagnetical field-tensor, where the electrical and magnetical field are simply exchanged. I must admit, that I didn't really understand your point, when you were talking about the metric, the Lagrangian and the Poisson brackets. Are you wondering, were the metric is hidden in the Hamilton formalism ? The metric is incorporated in Einsteins variational formalism of the Hamilton function. Check any book, f.e. his original work from 1915 collected in "The Principle of Relativity" (Dover). Regards Kay |
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