Lagrangian formalism and Contact Bundles

Question

In his Applied Differential Geometry book, William Burke says the following after telling that the action should be the integral of a function $L$:

A line integral makes geometric sense only if it's integrand is a 1-form. Is $Ldt$ a 1-form? Well, that's the wrong question. The correct question is: On what space is $Ldt$ a 1-form? It is not a 1-form on configuration space, the space of positions, because it can have a non-linear dependence on velocities. A 1-form must be a linear operator on the tangent vectors. The correct space for $L dt$ is the line-element contact bundle of the configuration space.

Now, why intuitively the correct setting for lagrangian mechanics is on the contact bundle? I understand the contact bundle as pairs $(p,[v])$ where $p$ is a point in configuration space and $[v]$ is an equivalence class of vectors, explicitly $v\sim kv$.

Thinking not on all that arguments for selecting the space on which $Ldt$ is a $1$-form, physically, how can we intuit that the contact bundle is useful for that? I mean, is there some observation in classical mechanics that guides us in building the theory on that space?

This post imported from StackExchange Physics at 2014-04-01 12:39 (UCT), posted by SE-user user1620696

Answer 1

The special feature of contact geometry is the contact 1-form $\lambda$, which satisfies $\lambda\wedge d\lambda\ne0$ (let's restrict to 3-dimensions). In our Lagrangian mechanics example, $\lambda = dq-vdt$. You want this to pull-back to zero on the ``permissible'' curves in phase space -- these curves represent the motions of your system.

For a more detailed but tangential explanation, see:
http://mathoverflow.net/questions/72498/what-is-the-role-of-contact-geometry-in-the-hamiltonian-mechanics

This post imported from StackExchange Physics at 2014-04-01 12:39 (UCT), posted by SE-user Chris Gerig