Simple Example: Particle on the Surface of a Cylinder

Consider a particle constrained to move on the surface of a cylinder of radius $R$, with a force toward the origin $\vec{F}=-k\vec{r}$. We will work the problem in cylindrical coordinates, $(R,\theta,z)$ with $R$ fixed. It is rather simple to work this problem from Newton's laws for from the Lagrangian so the point of this exercise is only to see that the Hamiltonian formalism has a lot in common with the methods we know.

Of course the potential energy is

$$U={1\over 2}kr^2={1\over 2}k(R^2+z^2)$$

and the kinetic energy is

$$T={1\over 2}m(\dot{z}^2+R^2\dot{\theta}^2) .$$

Since the potential does not depend on the velocities,

$$H=T+U={1\over 2}m(\dot{z}^2+R^2\dot{\theta}^2)+{1\over 2}k(R^2+z^2) .$$

We need to write this as a function of the coordinates and momenta. So first find the momenta conjugate to $z$ and $\theta$.

$$p_i={\partial L\over \partial q_i} = {\partial T\over \partial q_i}$$

since the potential doesn't depend on the velocities.

$$p_z = {\partial T\over \partial \dot{z}}=m\dot{z}$$

$$p_\theta = {\partial T\over \partial \dot{\theta}}=mR^2\dot{\theta}$$

(Eventually, you should know the momenta conjugate to the standard coordinates.)

Next we write the Hamiltonian in terms of the momenta.

$$H={1\over 2m}\left(p_z^2+{p_\theta^2\over R^2}\right)+{1\over 2}k(R^2+z^2) .$$

Now we look at Hamilton's equations. The equations for $\dot{q}_i$ give us the momentum in terms of $\dot{q}$ again and so there is nothing new there.

$$\dot{z} ={\partial H\over\partial p_z}={p_z\over m}$$

$$\dot{\theta} ={\partial H\over\partial p_\theta}={p_\theta\over mR^2}$$

The equations of motion can be found in the equations for $\dot{p}_i$

$$\dot{p}_z =-{\partial H\over\partial z } = -kz$$

$$\dot{p}_\theta =-{\partial H\over\partial \theta } = 0$$

The $\theta$ momentum (angular momentum) is conserved while $z$ executes harmonic motion.

$$\dot{p}_z = -kz$$

$$m\ddot{z} = -kz$$