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Hamiltonian flow question

  1. Jun 5, 2013 #1
    1. The problem statement, all variables and given/known data

    At what time does the particle reach infinity given that H(p,x)=(1/2)p^2 -(1/2)x^4. And initial values are x(0)=1 and p(0)=1

    2. Relevant equations


    The hamiltonian equations i believe are given by the partial derivatives let d mean partial derivative so x'=dH/dp and p'=-dH/dx
    3. The attempt at a solution well so far i have that x'=p and p'=2x^3 but at this point, im confused over how to impose the initial values... This is my attempt. By seperation of variables, dx/dt=p so (1/p)x=t+ c by integration and now i impose initial value of x(0)=1which gives me c=1/p so i get t=(1/p)x-(1/p) and for dp/dt i got t=(1/2x^3)p - (1/2x^3) . At this point im very confused what to do next or if im doing this right in the first place... How can i tell when the particle reaches infinity with two times??
     
    Last edited: Jun 5, 2013
  2. jcsd
  3. Jun 5, 2013 #2

    Ray Vickson

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    You wrote H(p,x)=(1/2)x^2 -(1/2)x^4. Did you mean H(p,x) = (1/2)p^2 - (1/2)x^4?
     
  4. Jun 5, 2013 #3
    Yes thanks for that :)
     
  5. Jun 5, 2013 #4

    haruspex

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    p' = 2x3?
    Doesn't look valid to me. Try looking at x''.
     
  6. Jun 5, 2013 #5

    Ray Vickson

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    If I were dong this question I would avoid the dynamical equations and use instead conservation of energy (i.e., constant H).
     
  7. Jun 5, 2013 #6
    Im just confused on how to use two initial conditions . Ray Vickson , what do you mean by the conservation of energy method?
     
  8. Jun 6, 2013 #7

    Ray Vickson

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    Because of the form of the Hamiltonian, it is constant over time; that is, ##{\cal{H}}(t) \equiv H(p(t),x(t))## is constant. That means that for any t we have
    [tex] p^2(t) - x^4(t) = p^2(0) - x^4(0) = 0.[/tex] Since the mass is 1 (from the form of H) we have
    [tex] p(t) = \frac{d x(t)}{dt},[/tex]
    so we get immediately a first-order DE for ##x(t)##.
     
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