Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I don't get how to use Hamilton's equations in mechanics, for example let's take the simple pendulum with $$H=\frac{p^2}{2mR^2}+mgR(1-\cos\theta)$$ Now Hamilton's equations will be: $$\dot p=-mgR\sin\theta$$ $$\dot\theta=\frac{p}{mR^2}$$ I know one of the points of Hamiltonian formalism is to get first order diff. equations instead of second order that Lagrangian formalism gives you, but how can I proceed from here without just derivating again wrt. $\dot\theta$ and substituting $\dot p$ to get the same equation that I get with the Lagrangian formulation? Or is that the way to do it? And how could I get the path of the system on the phase space with those equations?

share|cite|improve this question
up vote 3 down vote accepted

Generally both formulations (Largangian and Hamiltonian) are equivalent, but in your case, if $\theta$ is small, you have a simplified equation for $p$ and you can use a solution ansatz like $e^{i\omega t}$ for both $p$ and $\theta$.

To draw a path in the phase space, you have to solve the equations and/or manage to express $p(\theta)$ or $\theta(p)$.

share|cite|improve this answer
Thanks for the info, so I will get to the same second order differential equation? And about the phase space? – MyUserIsThis Jan 22 '13 at 16:41
Not obligatory, you can solve $\dot{p}\propto \theta$ and $\dot{\theta}\propto p$ by the ansatz. – Vladimir Kalitvianski Jan 22 '13 at 16:48
To add to what Vladimir said, if you consider a system with $n$ generalized coordinates (in this case $n=1$ since your system is described by the coordinate $\theta$), then you will obtain $2n$ first order differential equations, $n$ for the coordinates and $n$ for their corresponding canonical momenta. To get the path in phase space, you can do what Vladimir suggests; this will allow you to obtain $\theta(t)$ and $p(t)$, and then you simply plot these functions as a parametric curve on the $\theta$-$p$ plane. Cheers! – joshphysics Jan 22 '13 at 16:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.