I'm reading this article and am struggling with some of the terminology. What is the flow map for a Hamiltonian system? I'm looking for a rigorous definition really!

Many thanks in advance.


1 Answer 1


Usually flow maps are integrals of vector fields. In other words, solutions to dr(t)/dt = v(r). If you pick one specific value for the parameter t then you get one specific flow map that takes r(0) to r(t) as a (bijective) map on the manifold.

A hamiltonian flow is one that is generated as by a symplectic form S and a hamiltonian potential H, as in v = S(dH).

Edit: I think an example might help. Consider the hamiltonian of a harmonic oscillator H = p^2 + x^2. If we take the gradient in phase space we get (dH/dx,dH/dp) = (2x,2p), which points radially away from the coordinate origin with a magnitude that is proportional to the distance from there. Applying the standard symplectic form on phase space gives (dH/dp,-dH/dx) = (2p,-2x) for the hamiltonian vector field. This field points tangentially around the origin and its integral lines are circles. The proportionality of the magnitude to the distance from the origin makes sure that the flow transports points at a constant angular velocity around the origin. The flow maps are therefore rotations, or in other words, the hamiltonian above generates rotations on phase space. Generally if you look at the flow equation in the symplectic vector field dr/dt = v you get (dx/dt,dp/dt) = (dH/dp,-dH/dx), which are just the hamilton equations of motion. So you see this is just a different way to look at the hamilton formalism.

  • $\begingroup$ Thanks - I understand your first paragraph perfectly. Could you expand your second one? I haven't come across symplectic forms before, so could you flesh out the explanation? Many thanks! $\endgroup$ Sep 24, 2012 at 14:56
  • $\begingroup$ You are surely familiar with the gradient as an operation that creates a vector field from a potential. Getting a symplectic flow is very similar. You first take the gradient of the potential, but then you apply the symplectic form which takes the gradient to a vector that is orthogonal to the gradient, but with the same length. In R^2 you can represent the canonical symplectic form as the matrix [ [0 1] [-1 0] ]. $\endgroup$
    – A.O.Tell
    Sep 24, 2012 at 15:08

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