# Phase Portraits Given Hamiltonian

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.

Thanks

• 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. – ACuriousMind Feb 9 '16 at 22:06

$$\frac{dq}{dt}=\frac{\partial H}{\partial p}\qquad \frac{dp}{dt}=-\frac{\partial H}{\partial q}$$ 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.