Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

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

In particular, a symplectic integrator to solve:

$$\ddot{\theta} + \dfrac{g}{l} \sin(\theta) = 0.$$

I'm currently using velocity verlet - by realizing that

$$\ddot{\theta} = -\nabla (-cos(\theta)) = A(\theta(t)),$$

ie. letting $$x = \theta$$

$$v = d \theta/dt$$

$$a = d^2 \theta /dt^2.$$

Is it safe to apply Verlet integration to generalized coordinates? In particular, does this hold true for a generalized coordinate theta:

$$\theta_{t+dt} \approx \theta_t + \dot{\theta}_t dt + \frac{1}{2} \ddot{\theta}_t (dt)^2?$$

share|cite|improve this question

The canonical momentum corresponding to the generalizaed coordinate $\theta$ is:

$p_{\theta} = \frac{\partial L}{\dot{\partial\theta}} = ml^2\dot{\theta} = ml^2 v$

where $L$ is the Lagrangian:

$ L = T-V = \frac{1}{2}ml^2\dot{\theta}^2+mgl(1-cos\theta)$

Thus the symplectic form is

$\omega = dp_{\theta} \wedge d\theta = ml^2 dv \wedge d\theta$

This is the symplectic form of a particle moving on a circle (symplectic form of $T^*S^1$). It looks locally just like the canonical symplectic form of a particle moving on a line. This is the reason why the Verlet integration formula will provide a symplectic integrator for this problem. This is provided that it is implemented just in the form written in the Wikipedia page:

$v(t+\frac{1}{2}\Delta t) = v(t)+ \frac{1}{2}A(t) \Delta t$

$x(t+\Delta t) = x(t)+v(t+\frac{1}{2}\Delta t) \Delta t$

$v(t+\Delta t) = v(t+\frac{1}{2}\Delta t))+\frac{1}{2}A(t+\Delta t) \Delta t$

share|cite|improve this answer

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.