A question regarding particle trajectories in the symplectic manifold formalism

How to solve a free particle on a 2-sphere using symplectic manifold formalism of classical mechanics ?

Is there a way to get coriolis effect directly, without going into Newton mechanics?

And is there a good textbook dealing with classical mechanics problems using symplectic manifolds?

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Concerning textbooks, see e.g. this and this question. – Qmechanic♦ Apr 4 '12 at 11:13

1 Answer

In this special case, the symplectic manifold is the cotangent bundle of the sphere. A cotangent bundle is always a symplectic manifold which can be interpreted as the phase space for a particle wandering around the manifold. The position of the particle is the location on the original manifold, while the momentum of the particle is (a scalar multiple of) the cotangent vector. This defines the phase space for particle motion.

You might wonder why a cotangent vector and not a tangent vector. The reason is that it is the momentum dot the velocity which is an invariant (the change in phase with time in quantum mechanics, the increment of action in time classically). If this is not intuitive, no worries, just keep reading, it is made mathematically apparent below.

To define a symplectic manifold you need a symplectic form, which is an object that takes a the gradient of a scalar hamiltonian function into a phase-space vector (a vector in the tangent bundle of the phase space, the tangent bundle of the cotangent bundle) which tells you how things move around in the phase space in response to the Hamiltonian. In this case, the symplectic form takes a gradient of a scalar energy function on the cotangent bundle (not on the sphere) into a vector field on the cotangent bundle (not on the sphere). This is completely intuitive--- the Hamiltonian is a function of the position and momenta both:

Let the sphere point be coordinatized by s^i, and the cotangent vector by p_i. Given a Hamiltonian function H(s,v), the Hamilton's equations are

$${ds^i\over dt} = {\partial H\over \partial p^i}$$ $${Dp_i\over Dt} = -{\partial H \over \partial s_i}$$

This is a covariant equation--- the derivative of the position is a vector, and the covariant derivative of the covector is a covector, so the left and right hand sides are consistent tensorially. This can be seen from the consistent pattern of upper and lower indices.

From this, you get a mathematical justification of why cotangent bundle-- the derivative with respect to p must give a vector, in other words, something that dots with a covector to make a scalar. A gradient with respect to a covector dots with a covector to make a scalar (this is just a mathematician's way of saying the indices are in the right place).

The reason there is a covariant derivative on p and not s is because the cotangent bundle is flat in the cotangent directions (these are a vector space). The cotangent bundle is only non-flat along the sphere.

If H={1\over 2m} p_i p_j \delta^{ij}, the result is geodesic equation:

$${D\over Dt} {ds^i\over dt} = 0$$

along the sphere with speed ${p^i\over m}$, and you can see this is a great circle with constant velocity by symmetry, or formally by solving the equation after choosing the particle to initially lie on the equator, with its velocity all in the $\phi$ direction in standard spherical coordinates (or any way you prefer).

This problem is a bit too trivial. If you want a case where the cotangent bundle motion is interesting, you should choose a potential function on the sphere, or consider the motion in the group manifold SO(3), whose solution is the spinning top.

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