Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Some notes on magnetic flows F. Pasquotto Hamiltonian Dynamics, Fall 2015 0.1 The Hamiltonian formulation Let (Q, g) be a Riemannian manifold of dimension n. Denote by π : T ∗ Q → Q its cotangent bundle. With respect to local coordinates (q1 , . . . , qn ) on Q and corresponding cotangent coordinates (p1 , . . . , pn ), the canonical symplectic form Pn on T ∗ Q is given by ω0 = i=1 dpi ∧ dqi . Let σ be a closed 2-form on Q: then ωσ = ω0 + π ∗ σ is a symplectic form on T ∗ Q, called a twisted symplectic form. Consider the Hamiltonian function 1 H : T ∗ Q → R, H(q, p) = gq∗ (p, p), 2 where g ∗ denotes the pullback of the metric g to the cotangent bundle via the metric isomorphism. The Hamiltonian flow associated to H on (T ∗ Q, ωσ ), that is, the flow of the Hamiltonian vector field X defined by iX ωσ = −dH, is called a magnetic flow, while σ is called the magnetic form. In local coordinates, the vector field X has the form n X ∂H ∂H ∂ ∂H ∂ , σij (q) Xi = − + ∂pi ∂qi ∂qi j=1 ∂pj ∂pi Pn where σ = 12 i,j=1 σij (q)dqi ∧dqj and σij = −σji. Via the metric isomorphism, we can pull-back the twisted magnetic form ωσ to a symplectic form Ωσ on the tangent bundle T Q. The magnetic flow of the pair (g, σ) on T Q is the flow associated to the Hamiltonian 1 E : T Q → R, E(q, v) = gq (v, v) 2 with respect to Ωσ . Define the Lorentz force to be the bundle endomorphism Y : T Q → T Q defined by gq (Yq (u), v) = σq (u, v) for all u, v ∈ Tq Q and all q ∈ Q. Then a curve t 7→ (γ(t), γ̇(t)) is an orbit of the magnetic flow on T Q if and only if it satisfies D γ̇(t) = Yγ(t) (γ̇(t)) dt for all t. If this holds, the curve γ : R → Q is called a magnetic geodesic. In the above identity, D/dt denotes the covariant derivative along γ. As a geometric D object, dt γ̇(t) is again a vector field along γ and in local coordinates it is given by expressions of the form γ̈k (t) + Γkij (γ(t))γ̇i (t)γ̇j (t), where Γkij are the Christoffel symbols of the metric g. Notice that if our Riemannian manifold is R2n with the standard Euclidean metric, these symbols vanish and the covariant derivative along γ is just the usual accelleration vector field. 0.2 Exact magnetic flows: Lagrangian formulation and the horocycle flow on H. Now suppose the form σ is not only closed, but exact. That is, there exists a 1-form θ on Q such that dθ = σ. Then the magnetic geodesics we encountered in the previous sections are the solutions of the Euler-Lagrange equation associated to the Lagrangian L : T Q → R, L(q, v) = 1 gq (v, v) + θq (v). 2 Consider the hyperbolic upper half-plane H = {z = x + iy ∈ C | y > 0}, equipped with the Riemannian metric g = Euler-Lagrange flow associated to L : T H → R, L(z, v) = dx2 +dy 2 . y2 We want to study the 1 gz (v, v) + ηz (v), 2 where ηz = dx y . The energy function is E(z, v) = gz (v, v), and the associated D Euler-Lagrange equation is dt ż(t) = J ż(t), where J is the standard almost complex structure, that is, counterclockwise rotation by π2 . In local coordinates the Lagrangian can be written as L(z, v) = L(x, y, vx , vy ) = 1 vx (v 2 + vy2 ) + , 2y 2 x y and the Euler-Lagrange equation becomes vx2 + vy2 d vx 1 d vy vx + = 0, − 2. = − 2 2 3 dt y y dt y y y From this we see that ẋ = −y and ẏ = 0 are solutions. These are the horizontal horospheres associated to the vertical geodesics of H. The energy of these solutions is 1 1 E(x, v) = k(−y, 0)k2 = . 2 2 In fact, the solutions of the Euler-Lagrange equation at the energy level 12 are the horospheres, parametrized by arc-length. This can be shown in the following steps: • isometries of H preserve the property of being a minimizer of the Lagrangian action functional; • horizontal horospheres are solutions o the EL equation and isometries of H act transitively on T 1 H, so all horospheres parametrized by arc-length are solutions; • any solution with energy 12 must be a horosphere: in fact, for any (z, v) ∈ T 1 H, there is a unique horosphere through (z, v), so a non-horosphere solution would contradict uniqueness of solutions. 2