Homework 3 Problem 8

### Problem:

A mass $m$ moves in the plane, joined to the origin by a linear spring with spring constant $k$ and zero rest length. Assume the position of the mass to be described by its polar coordinates $(r, \theta)$ on the plane excluding the origin. Show that the Lagrangian for this system is hyperregular. Compute the canonical one form $\Theta$, the canonical symplectic form $\Omega$, and the Liouville volume form $\Lambda$ on $T^*Q$. Compute the Lagrangian one form $\Theta_L$ and the Lagrangian two form $\Omega_L$ on $TQ$. Compute the action $A$, the energy $E$, and the Hamiltonian $H$ associated with the Lagrangian $L$. Compute the Lagrangian vector field $Z_L$ on $TQ$ and the Hamiltonian vector field $X_H$ on $T^*Q$.

### Solution:

The Lagrangian for this system is:

$L: TQ \longrightarrow \mathbb{R}: (r, \theta, \dot r, \dot \theta) \longmapsto \frac{1}{2} m(\dot r^2 + r^2 \dot \theta^2) - \frac{1}{2} k r^2$

The fiber derivative is:

$\mathbb{F}L: TQ \longrightarrow T^*Q: (r, \theta, \dot r, \dot \theta) \longmapsto (r, \theta, \frac{\partial L}{\partial \dot r}, \frac{\partial L}{\partial \dot \theta}) = (r, \theta, m \dot r, m r^2 \dot \theta)$, while

$(\mathbb{F}L)^{-1}: T^*Q \longrightarrow TQ: (r, \theta, p_r, p_\theta) \longmapsto (r, \theta, \frac{\partial L}{\partial \dot r}, \frac{\partial L}{\partial \dot \theta}) = (r, \theta, \frac{p_r}{m}, \frac{p_\theta}{m r^2})$

Hyperregularrity is ensured if $\mathbb{F}L$ is a diffeomorpism.
Indeed, $\mathbb{F}L$ is bijective; since $r, \theta, m \dot r, m r^2 \dot \theta, \frac{p_r}{m}$, and $\frac{p_\theta}{m r^2}$ are infinitely differentialble, $\mathbb{F}L$ and $(\mathbb{F}L)^{-1}$ are smooth.
So, the Lagrangian is hyperregular.

The canonical one form is: $\Theta = p_r dr + p_\theta d\theta$, and the canonical symplectic form is: $\Omega = dr \wedge dp_r + d\theta \wedge dp_\theta$

The Liouville volume form on $T^*Q$ is: $\Lambda = dr \wedge d\theta \wedge dp_r \wedge dp_\theta$

The Lagrangian one form is: $\Theta_L = (\mathbb{F}L)^*\Theta = \frac{\partial L}{\partial \dot q^i} dq^i = \frac{\partial L}{\partial \dot r} dr + \frac{\partial L}{\partial \dot \theta} d\theta = m \dot r dr + m r^2 \dot \theta d\theta$

The Lagrangian two form is:

$\Omega_L = -d\Theta_L = -\frac{\partial}{\partial r} [m \dot r dr + m r^2 \dot \theta d\theta] \wedge dr -\frac{\partial}{\partial \theta} [m \dot r dr + m r^2 \dot \theta d\theta] \wedge d\theta -\frac{\partial}{\partial \dot r} [m \dot r dr + m r^2 \dot \theta d\theta] \wedge d\dot r -\frac{\partial}{\partial \dot \theta} [m \dot r dr + m r^2 \dot \theta d\theta] \wedge d\dot \theta$
$= -2m r \dot \theta d\theta \wedge dr - m \ddot r dr \wedge d\dot r - m r^2 \ddot \theta d\theta \wedge d\dot \theta = 2m r \dot \theta dr \wedge d\theta - m \ddot r dr \wedge d\dot r - m r^2 \ddot \theta d\theta \wedge d\dot \theta$

The action is:

$A: TQ \longrightarrow \mathbb{R}: (r, \theta, \dot r, \dot \theta) \longmapsto \dot r \frac{\partial L}{\partial \dot r} + \dot \theta \frac{\partial L}{\partial \dot \theta} = m \dot r^2 + m r^2 \dot \theta^2$

The energy is:

$E: TQ \longrightarrow \mathbb{R}: (r, \theta, \dot r, \dot \theta) \longmapsto A(r, \theta, \dot r, \dot \theta) - L(r, \theta, \dot r, \dot \theta)$
$= (m \dot r^2 + m r^2 \dot \theta^2) - (\frac{1}{2} m(\dot r^2 + r^2 \dot \theta^2) - \frac{1}{2} k r^2) = m \dot r^2 + m r^2 \dot \theta^2 - \frac{1}{2} m \dot r^2 - \frac{1}{2} m r^2 \dot \theta^2 + \frac{1}{2} k r^2$
$= \frac{1}{2} m \dot r^2 + \frac{1}{2} m r^2 \dot \theta^2 + \frac{1}{2} k r^2$

The Hamiltonian is:

$H: T^*Q \longrightarrow \mathbb{R}: (r, \theta, p_r, p_\theta) \longmapsto E \circ (\mathbb{F}L)^{-1}(r, \theta, p_r, p_\theta)$
$= \frac{1}{2} m (\frac{p_r}{m})^2 + \frac{1}{2} m r^2 (\frac{p_\theta}{m r^2})^2 + \frac{1}{2} k r^2 = \frac{p_r^2}{2 m} + \frac{p_\theta^2}{2 m r^2} + \frac{k r^2}{2}$

The Lagrangian vector field $Z_L$ on $TQ$ can be found by solving the equation: $dE = Z_L \hook \Omega_L$

$dE = (m r \dot \theta^2 + k r) dr + 2m \dot r d\dot r + m r^2 \dot \theta d\dot \theta$

$Z_L = Z_{L_r} \frac{\partial}{\partial r} + Z_{L_\theta} \frac{\partial}{\partial \theta} + Z_{L_{\dot r}} \frac{\partial}{\partial \dot r} + Z_{L_{\dot \theta}} \frac{\partial}{\partial \dot \theta}$

$(Z_{L_r} \frac{\partial}{\partial r} + Z_{L_\theta} \frac{\partial}{\partial \theta} + Z_{L_{\dot r}} \frac{\partial}{\partial \dot r} + Z_{L_{\dot \theta}} \frac{\partial}{\partial \dot \theta}) \hook (2m r \dot \theta dr \wedge d\theta - m \ddot r dr \wedge d\dot r - m r^2 \ddot \theta d\theta \wedge d\dot \theta)$
$= 2m r \dot \theta Z_{L_r} d\theta - m \ddot r Z_{L_r} d\dot r - 2m r \dot \theta Z_{L_\theta} dr - m r^2 \ddot \theta Z_{L_\theta} d\dot \theta + m \ddot r Z_{L_{\dot r}} dr + m r^2 \ddot \theta Z_{L_{\dot \theta}} d\theta$
$= (2m r \dot \theta Z_{L_\theta} + m \ddot r Z_{L_{\dot r}}) dr + (2m r \dot \theta Z_{L_r} + m r^2 \ddot \theta Z_{L_{\dot \theta}}) d\theta + (-m \ddot r Z_{L_r}) d\dot r + (-m r^2 \ddot \theta Z_{L_\theta})$

By inspection, we can find the components of $Z_L$.

Then, $Z_L = (\frac{-2\dot r}{\ddot r}, \frac{-\dot \theta}{\ddot \theta}, (\frac{2}{\ddot \theta} + 1) \frac{r \dot \theta^2}{\ddot r} + \frac{k r}{m \ddot r}, \frac{\dot r \dot \theta}{\ddot r \ddot \theta})$

The Hamiltonian vector field $X_H$ on $T^*Q$ can be found by solving the equation: $dH = X_H \hook \Omega$

$dH = (\frac{-p_\theta^2}{m r^3} + k r) dr + \frac{p_r}{m} dp_r + \frac{p_\theta}{m r^2} dp_\theta$

$X_H = X_{H_r} \frac{\partial}{\partial r} + X_{H_\theta} \frac{\partial}{\partial \theta} + X_{H_{p_r}} \frac{\partial}{\partial p_r} + X_{H_{p_\theta}} \frac{\partial}{\partial p_\theta}$

$X_H \hook \Omega = X_{H_{p_r}} dr + X_{H_{p_\theta}} d\theta - X_{H_r} dp_r - X_{H_\theta} dp_\theta$

By inspection, we can find the components of $X_H$.

Then, $X_H = (-\frac{p_r}{m}, -\frac{p_\theta}{m r^2}, \frac{-p_\theta^2}{m r^3} + k r, 0)$

page revision: 3, last edited: 09 May 2007 05:41