Nonlinear Mechanics - Physics at Oregon State University

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38 CHAPTER 3. ABSTRACT TRANSFORMATION THEORY 3.2 Geometry in n Dimensions: The Hairy Ball Have another look at equation (3.2). Let’s call ˙η a velocity field. By this I mean that it associates a complete set of ˙q’s and ˙p’s with each point in phase space. In what direction does ˙η point? This is an easy question in one dimension; ˙η evaluated at the point P points in the direction tangent to the trajectory through P . Since trajectories can’t cross in phase space, there is only one trajectory through P , and the direction is unambiguous. If we use action-angle variables, the trajectory is a circle, and ˙η is what we would call in Ph211, a tangent velocity. The same is true, no doubt, for n > 1, but how do these circles fit together? How does one visualize this in higher dimensions? The answer, as I have mentioned before, is that the trajectories all lie on the surface of an n dimensional torus imbedded in 2n dimensional space. Your ordinary breakfast donut is a two dimensional torus imbedded in three dimensional space. 2 This is easy to visualize, so let’s limit the discussion to two degrees of freedom for the time being. The step from one degree of freedom to two involves some profound new ideas. The step from two to higher dimension is mostly a matter of mathematical generalization. Since we are dealing with conservative systems, the trajectories are limited by the conservation of energy, i.e. H(q1, q2; p1, p2) = E is an equation of constraint. The trajectories move on a manifold with three independent variables. Now the gradient of a function has a well defined geometrical significance: at the point P , ∇f points in a direction perpendicular to the surface or contour of constant f through P . In this case ∇H is a four component vector perpendicular to the surface of constant energy. Unfortunately, ˙η points in the direction of J · ∇H. What direction is that? Well, (∇H) T · J · ∇H = [H, H] = 0 Consequently, J · ∇H points in a direction perpendicular to ∇H, which is perpendicular to the plane of constant H, i.e. ˙η lies somewhere on the three dimensional surface of constant H. We could have guessed that ahead of time, of course, but we can take the argument further. H is probably not the only constant of motion. Suppose there are others; call them F , G, etc. For each of these constants we can 2 The word “dimension” gets used in two different ways. When we talk about physical systems, Lagrangians, Hamiltonians, etc., the dimension is equal to the number of degrees of freedom. Here I am using dimension to mean the number of independent variables required to describe the system.
3.2. GEOMETRY IN N DIMENSIONS: THE HAIRY BALL 39 construct a vector field using (2). ˙ηF = J · ∇F ˙ηG = J · ∇G · · · etc. · · · How many such fields can we construct that are independent of one another? To put it another way, how many independent constants of motion are there? That’s a good question – what do you mean by “independent”? The answer comes from differential geometry. I’m afraid I can only give a hand-waving introduction to it. There are two related requirements: 1. Suppose F and G are independent constants of motion. Take any trajectory from the manifold of constant F and another from constant G. There is no continuous canonical transformation that maps the one trajectory into another. 2. For each point P in space there must be one unique trajectory that lies in the plane of constant F and simultaneously in the plane of constant G. Think about this last requirement in the case where there are two degrees of freedom and two independent constants of motion. The trajectories must lie on a two- dimensional surface. If we use action-angle variables, the trajectories are circles. This sounds like a globe of the earth. Trajectories with constant ϕ are called longitudes, lines of constant θ are latitudes. But wait! We have a serious problem at the poles. The north and south poles have all possible longitudes. Requirement 2 is violated. Could you rearrange the lines so that this problem doesn’t occur? It turns out that this is not possible. This deep result is known in mathematical circles a the Poincare- Hopf theorem. In the sort of less exalted company we keep, it’s the Hairy Ball Theorem. The idea is this: try to comb the hair on a hairy ball so that there is no bald spot. It can’t be done. So long as you really use a comb, i.e. so long as the trajectories don’t cross, you will always be left with one hair standing straight up! This is not a proof, of course, but it is a vivid way of visualizing the content of the theorem. It is easy to see, however, that what is impossible on a sphere is trivially easy on the surface of a donut. It can be done in an infinite variety of ways. The simplest is to choose your “longitudes” so they go around the donut the long way. Latitudes go around the short way. This also satisfies requirement 1. You can’t deform a latitude into a longitude without cutting through the donut.
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