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Is any Hamiltonian system integrable?

  1. Jul 11, 2014 #1
    This is related to classical Hamiltonian mechanics. There is something wrong
    in the following argument but I cannot pinpoint where exactly the pitfall is:
    Consider an arbitrary (smooth) Hamiltonian (let us assume conservative) and 2n
    phase space coordinates (q,p). The Hamiltonian flow gives the evolution
    starting the initial conditions (q(q0,p0,t),p(q0,p0,t)). Now, clearly the
    inverse function (q0(q,p,t),p0(q,p,t)) provides 2n constants of motion: for
    any point (q,p) in the orbit and t one can reconstruct the initial values of
    the coordinates, obtaining always the same values. In addition the
    transformation from (q,p) to (q0,p0) is a canonical one, so the Poisson
    brackets between the various q0 vanish. I would say that obviously the various
    q0(q,p,t) are functionally independent. So apparently, any Hamiltonian system
    is integrable in the sense of Liouville (the fact that these n functions
    q0(q,p,t) are very complicated is a different matter). Can any one point out
    to me where is the fallacy? Thanks.
  2. jcsd
  3. Jul 12, 2014 #2


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    Gold Member

    Is there a reason you think there must be a fallacy? Do you think not all Hamiltonian systems are intrgrable?
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