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Phase flow is the one-parameter group of transformations

  1. Mar 27, 2010 #1
    The phase flow is the one-parameter group of transformations of phase space

    [tex]g^t:({\bf{p}(0),{\bf{q}(0))\longmapsto({\bf{p}(t),{\bf{q}(t)) [/tex],

    where [tex]{\bf{p}(t)[/tex] and [tex]{\bf{q}}(t)[/tex] are solutions of the Hamilton's system of equations corresponding to initial condition [tex]{\bf{p}}(0) [/tex]and [tex]{\bf{q}}(0)[/tex].

    Show that [tex]\{g^t\}[/tex] is a group.


    Can anyone help me prove the composition?

    [tex]g^t\circ g^s=g^{t+s}[/tex]
     
  2. jcsd
  3. Mar 27, 2010 #2

    quasar987

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    Re: Composition

    On the one hand, [itex]g^{t+s}(p_0,q_0)=(p(t+s),q(t+s))[/itex] where p and q are solutions of the Hamilton's system of equations corresponding to initial condition p_0 and q_0. On the other hand [itex]g^t\circ g^s(p_0,q_0)=g^t(p(s),q(s))=(p'(t),q'(t))[/itex] where p' and q' are solutions of the Hamilton's system of equations corresponding to initial condition p(s) and q(s). To show that (p(t+s),q(t+s))=(p'(t),q'(t)) for all s and t, observe that for s fixed, both sides satisfy the same initial value problem (in the variable t). By uniqueness of the solution to an initial value problem, you get the desired equality.
     
  4. Mar 27, 2010 #3
    Re: Composition


    Can you show that explicitly? I don't follow.

    Associativity:

    [itex]
    ( g^t\circ g^s) \circ g^r = g^{(t+s)} \circ g^r = g^{(t+s) +r}
    [/itex]

    [itex]
    g^t\circ (g^s \circ g^r) = g^{t} \circ g^{(s+r)} = g^{t+(s +r)}
    [/itex]

    Identity,

    at t = 0
    [tex]
    g^t:({\bf{p}(0),{\bf{q}(0))\longmapsto({\bf{p}(0), {\bf{q}(0))
    [/tex]

    Thus
    [itex]
    g^0 \circ g^s = g^{(0+s)} =g^s
    [/itex]


    [itex]
    g^s \circ g^0 = g^{(s+0)} =g^s
    [/itex]


    Invertible

    reverse the sign of Hamiltonian = reverse time, so [tex]g^t[/tex] is invertible and it follows that [tex](g^t)^{-1}=g^{-t}[/tex], so [tex]g^t\circ g^{-1}=g^{t+(-t)} = g^0[/tex] for all [tex]t,s\in\mathbb{R}[/tex].

    Is that correct? I'm not sure of the quantifiers are correct for the composition. Should it be t,s > 0 or t,s in R?
     
    Last edited: Mar 27, 2010
  5. Mar 27, 2010 #4

    quasar987

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    Re: Composition

    Your last post shows that all the group properties follow easily once you have

    [tex]g^t\circ g^s=g^{t+s}[/tex].

    I thought this is what you wer trying to show. Is it not?
     
  6. Mar 27, 2010 #5
    Re: Composition

    Yes.


    Can you show that explicitly? I don't follow.


    The remainder of the post was just to see if there was anything wrong.
     
  7. Mar 28, 2010 #6

    quasar987

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    Re: Composition

    For invertibility, perhaps simpler is this:

    [tex]g^t\circ g^{-t}=g^{-t}\circ g^{t}=g^0=\mathrm{id}[/tex]

    so g^t is invertible with inverse g^{-t}.
     
  8. Mar 28, 2010 #7

    quasar987

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    Re: Composition

    Fix s and set Q(t):=p(t+s), P(t):=p(t+s). Then, by the chain rule and the fact that q,p satisfy Hamilton's equations, we get

    [tex]\frac{dQ}{dt}(t) =\frac{d}{dt}q(t+s)=\frac{dq}{dt}(t+s)=\frac{\partial H}{\partial p}(q(t+s),p(t+s))=\frac{\partial H}{\partial p}(Q(t),P(t))[/tex]
    [tex]-\frac{dP}{dt}(t) =-\frac{d}{dt}p(t+s)=-\frac{dp}{dt}(t+s)=\frac{\partial H}{\partial q}(Q(t),P(t))[/tex]
    [tex]Q(0)=q(0+s)=q(s), \ \ P(0)=p(0+s)=p(s)[/tex]

    And by definition of q',p',

    [tex]\frac{dq'}{dt}(t)=\frac{\partial H}{\partial p}(q'(t),p'(t))[/tex]
    [tex]-\frac{dp'}{dt}(t)=\frac{\partial H}{\partial q}(q'(t),p'(t))[/tex]
    [tex]q'(0)=q(s), \ \ p'(0)=p(s)[/tex]

    So we see that Q,P satisfies the same initial value problem as q',p', so by uniqueness of the solution of such systems, it must be that Q=q' and P=p'. That is, [itex]g^{t+s}(q_0,p_0)=(g^t\circ g^s)(q_0,p_0)[/tex].
     
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