Homework 3 Problem 5


Recall the bead on a rotating hoop described in class. Show that the canonical Hamiltonian equations obtained using the total energy in the system as the Hamiltonian disagree with the canonical Euler-Lagrange equations obtained using the difference between the kinetic energy and the potential energy as the Lagrangian.


To refresh your memories:

  • $m$ is the mass of the bead.
  • $R$ is the radius of the hoop.
  • $\omega$ is the rate at which the hoop is rotating.
  • $\theta$ is the position of the bead on the hoop, with $\theta = 0$ corresponding to the bottom of the hoop.
  • $g$ is gravity.
  • The middle of the hoop is the level at which $PE = 0$.

We start by writing expressions for kinetic and potential energy.

\begin{align} KE = \frac{1}{2}m\left[\left(\omega R \sin \theta \right)^2+R^2 \dot \theta^2\right] \par \end{align}
\begin{align} PE = -mgR \cos \theta \end{align}

And then the Lagrangian:

\begin{align} L = KE-PE = \frac{1}{2}m\left[\left(\omega R \sin \theta \right)^2+R^2 \dot \theta^2\right]+mgR \cos \theta \end{align}

The canonical Euler-Lagrange equations, with L=KE-PE

\begin{align} \frac{\partial L}{\partial\theta} = m\omega^2 R^2 \sin\theta \cos\theta - mgR\sin\theta \end{align}
\begin{align} \frac{\partial L}{\partial\dot\theta} = mR^2\dot\theta \end{align}
\begin{align} \frac{d}{dt}\frac{\partial L}{\partial\dot\theta} = mR^2\ddot\theta \end{align}
\begin{align} \frac{\partial L}{\partial\theta} - \frac{d}{dt}\frac{\partial L}{\partial\dot\theta} = m\omega^2 R^2 \sin\theta \cos\theta - mgR\sin\theta - mR^2\ddot\theta = 0 \end{align}

Simplifying this yields the canonical Euler-Lagrange equation:

\begin{align} R \left( \omega^2 \sin\theta \cos\theta - \ddot\theta \right) - g\sin\theta = 0 \end{align}

The canonical Hamiltonian equations, with H=KE+PE

To find the canonical Hamiltonian equations using the total energy in the system as the Hamiltonian, we should find expressions for some more stuff first…

\begin{align} \mathbb{F}L:(\theta, \dot \theta) \mapsto \left(\theta,\frac{\partial L}{\partial \dot \theta}\right) = \left(\theta,mR^2\dot\theta\right) \end{align}
\begin{align} \left(\mathbb{F}L\right)^{-1} : (\theta, p) \mapsto \left(\theta,\frac{p}{mR^2}\right) \end{align}
\begin{align} \Omega = d\theta\wedge d p \end{align}

Now we write the Hamiltonian as the sum of kinetic and potential energy, but we replace $\dot \theta$ with $\frac{p}{mR^2}$:

\begin{align} H = KE + PE = \frac{1}{2}m\left[\left(\omega R \sin \theta \right)^2+R^2 \left(\frac{p}{mR^2}\right)^2\right] - mgR\cos\theta \end{align}


\begin{align} dH = \left(m\omega^2 R^2 \sin\theta \cos\theta + mgR\sin\theta \right) d\theta + \left(\frac{p}{mR^2}\right)\right) dp \end{align}

Using a generic Hamiltonian vector field

\begin{align} $X_H = X_{H\theta}\frac{\partial}{\partial\omega} + X_{Hp}\frac{\partial}{\partial p} \medspace, \end{align}

we find that

\begin{align} X_H \overset\lrcorner\,\, \Omega = \left( X_{H\theta}\frac{\partial}{\partial\theta}+X_{Hp}\frac{\partial}{\partial p} \right) \overset\lrcorner\,\ \left( d\theta\wedge d p \right) = X_{H\theta}d p - X_{Hp} d\theta \end{align}

Now using the relation $dH = X_H \overset\lrcorner\,\, \Omega$, we compare Eq.(13) and Eq.(15) to find

\begin{align} X_{H\theta} = \frac{p}{mR^2} \end{align}


\begin{align} X_{Hp} = m\omega^2 R^2 \sin\theta \cos\theta + mgR\sin\theta \end{align}

This gives us the canonical Hamiltonian equations

\begin{align} \dot\theta = \frac{p}{mR^2} \end{align}


\begin{align} \dot p = m\omega^2 R^2 \sin\theta \cos\theta + mgR\sin\theta \end{align}

However, these can be combined into one equation as follows:

Through rearranging and differentiating Eq.(18), we get

\begin{align} \dot p = mR^2\ddot\theta \end{align}

Then we equate Eq.(19) and Eq.(20) and rearrange to get

\begin{align} R \left( \omega^2 \sin\theta \cos\theta - \ddot\theta \right) + g\sin\theta = 0 \end{align}

Comparison of the Euler-Lagrange equation 8 and Hamilton's equation 21 reveals that they are in fact different!

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