Given a Hamiltonian say $$ H = 5p^2 $$ What is the correct procedure for producing a phase portrait.

My initial thoughts were to solve the system of equations $\frac{dq}{dt} = 0$ and $\frac{dp}{dt} = 10p $ which yields: $$p[t] -> e^{(10 t)} C[1]$$ $$ q[t] -> C[2]$$

How am I then supposed to proceed? I don't know any further information, so cannot solve for the constants.

I do have Mathematica at my disposal for this but I would like to understand the theory.


  • $\begingroup$ Don't you produce a phase portrait by just plotting a few trajectories? That is, just choose some initial conditions at (semi-)random, and plot the trajectories you get. $\endgroup$ – ACuriousMind Feb 9 '16 at 22:06

Your solutions are wrong. As

$$ \frac{dq}{dt}=\frac{\partial H}{\partial p}\qquad \frac{dp}{dt}=-\frac{\partial H}{\partial q} $$ you get $$ \frac{dq}{dt}=10\,p\qquad \frac{dp}{dt}=0 $$ i.e. $$ q(t)=10\,p_{0}\,t+q_{0}\qquad p=p_{0} $$ The $q$ coordinate flows in time in straight lines, while the $p$ coordinate doesn't change in time. So each phase trajectory lies on a different $p=p_{0}$ line. For $p_{0}>0$, they flow in the direction of increasing $q$. For $p_{0}<0$, they flow in the direction of decreasing $q$. For $p_{0}=0$, the trajectories are just points: $(q_{0},0)$. When $p_{0}\neq 0$, $q$'s initial conditions cannot be represented, unless you mark the starting point with a dot or something like that. Remember that a phase portrait is a collection of phase trajectories which differ by their initial conditions.

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  • $\begingroup$ Okay, now I understand, so to produce a mathematica plot I could just Streamplot[{10p_0 t + 0, p_0}, {t, 0, 5}, {p_0, 0, 5}] ? $\endgroup$ – user Feb 9 '16 at 22:26
  • $\begingroup$ I guess so, I should check whether that is the right command, but it should be so. $\endgroup$ – Giorgio Comitini Feb 9 '16 at 22:36

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