Kepler Problem and SO(4) Momentum Map

A complete solution of the Kepler problem is given; the SO(4) action on (TS,Ω4) is discussed; and the incomplete Kepler hamiltonian vector field is regularized by the LS map, embedding it into (TS, ω̃4). [2] did excellent work on these topics. I read through them and write this report. O Introduction Kepler problem is a famous and classical problem in both mathematics and physics. Since Kepler, there are various solutions. In Section 1, we are going to solve it using symplectic geometry. The hamiltonian vector field in the solution to the Kepler problem is not complete, and we will regularize it by embedding it into a complete flow. The LS map provides a nice way to regularize the negative energy Keplerian orbits all at once. This is discussed in section 3. In the regularization, the LS map sends negative energy set (Σ− = {(q, p) ∈ TR|H(q, p) < 0}, ω3|Σ−) to (TS, ω̃4), which is SO(4) symmetric. To see the connections between these two vector fields, we devote section 2 to the discussion of some properties of the SO(4) momentum map. 1 The Kepler Problem Kepler Problem: Consider two particles in R. One is fixed at the origin, and the other one moves under the influence of gravitational field of the fixed particle. The problem, named after Kepler who first gave the solution, is to describe the motion of the second particle. In this section, we will give a complete solution using symplectic geometry. This is done in three steps: 1.1 We define the Kepler Hamiltonian system (H,TR0, ω3) and discuss some properties of the Kepler Hamiltonian vector field XH . 1.2 On (Σ−, ω̃3 = ω3|Σ−) (Σ− is the open subset of (TR0, ω3) where the energy H is negative), we define a representation of so(4) in the space of Hamiltonian vector fields which has a momentum map G. And by studying G , we get the characterization of the orbits of XH . 1 1.3 Kepler’s equation. There is a similar discussion in [1]II.3. The idea of the proof of our theorems just comes from it. 1.1 Kepler Hamiltonian system (H,TR0, ω3) On the phase space TR0 = (R−{0})×R with coordinates (q,p) and ω3 = ∑3 i=1 dqi∧dpi the symplectic form, consider the Kepler Hamiltonian H : TR0 → R : (q, p) 7−→ 1 2 < p, p > − μ |q| . (Here <,> is the Euclidean inner product on R and |q| is the length of vector q. As we consider the case where the force is attractive, μ is supposed to be > 0.) This is what we call Kepler Hamiltonian system. The integral curves of the Hamiltonian vector field XH on TR0 satisfy the equations  q̇ = p ṗ = −μ q |q|3 . In physics, we know that energy h, angular momentum J and Runge-Lenz vector e, so called eccentricity vector, are conserved quantities in a two-body system. Thus we get some integrals of the Kepler vector field XH . Define: h def = 12 < p, p > − μ |q| ; J def = (J1, J2, J3) = q × p; e def = (e1, e2, e3) = − q |q| + 1 μp× (q × p). Proposition 1. h,J,e are integrals of XH . Proof: d dth =< p, d dtp > + μ |q|3 < q, d dtq > = − < p, μ |q|3 q > + μ |q|3 < q, p >= 0 d dtJ = d dtq × p+ q × d dtp = p× p− μ |q|3 q × q = 0 d dte = − d dt q |q| + 1 μ d dtp× J = 1 |q|3 < d dtq, q > q − 1 |q| d dtq + 1 μ d dtp× J = 1 |q|3 (< p, q > q− < q, q > p− q × J) = 1 |q|3 (q × (q × p)− q × (q × p)) = 0. 2 Proposition 2. If the energy h is negative, then the image of each integral curve of the Kepler vector field under the bundle projection τ : TR0 → R : (q, p) 7−→ q is bounded.