Phase flow is the one-parameter group of transformations

[Nusc]

The phase flow is the one-parameter group of transformations of phase space

g^t:({\bf{p}(0),{\bf{q}(0))\longmapsto({\bf{p}(t),{\bf{q}(t)),

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

Show that \{g^t\} is a group.


Can anyone help me prove the composition?

g^t\circ g^s=g^{t+s}

 

[quasar987]

On the one hand, g^{t+s}(p_0,q_0)=(p(t+s),q(t+s)) 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 g^t\circ g^s(p_0,q_0)=g^t(p(s),q(s))=(p'(t),q'(t)) 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.

 

[Nusc]

] 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.

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

Associativity:

<br /> ( g^t\circ g^s) \circ g^r = g^{(t+s)} \circ g^r = g^{(t+s) +r} <br />

<br /> g^t\circ (g^s \circ g^r) = g^{t} \circ g^{(s+r)} = g^{t+(s +r)} <br />

Identity,

at t = 0
<br /> g^t:({\bf{p}(0),{\bf{q}(0))\longmapsto({\bf{p}(0), {\bf{q}(0)) <br />

Thus
<br /> g^0 \circ g^s = g^{(0+s)} =g^s<br /> <br /> g^s \circ g^0 = g^{(s+0)} =g^s<br />Invertible

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

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?

 

[quasar987]

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

g^t\circ g^s=g^{t+s}.

I thought this is what you wer trying to show. Is it not?

 

[Nusc]

Yes.

] 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.

Can you show that explicitly? I don't follow.The remainder of the post was just to see if there was anything wrong.

 

[quasar987]

For invertibility, perhaps simpler is this:

g^t\circ g^{-t}=g^{-t}\circ g^{t}=g^0=\mathrm{id}

so g^t is invertible with inverse g^{-t}.

 

[quasar987]

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

\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))
-\frac{dP}{dt}(t) =-\frac{d}{dt}p(t+s)=-\frac{dp}{dt}(t+s)=\frac{\partial H}{\partial q}(Q(t),P(t))
Q(0)=q(0+s)=q(s), \ \ P(0)=p(0+s)=p(s)

And by definition of q',p',

\frac{dq&#039;}{dt}(t)=\frac{\partial H}{\partial p}(q&#039;(t),p&#039;(t))
-\frac{dp&#039;}{dt}(t)=\frac{\partial H}{\partial q}(q&#039;(t),p&#039;(t))
q&#039;(0)=q(s), \ \ p&#039;(0)=p(s)

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, g^{t+s}(q_0,p_0)=(g^t\circ g^s)(q_0,p_0)[/tex].