Homework 4 Problem 4

#### Problem

Recall the “extensible barbell” example from class. The system's symmetry with respect to the group action

(1)
\begin{align} \Phi: \text{SE}(2) \times Q \to Q: ((a, b, \phi), (d, x, y, \theta)) \mapsto (d, a + x \cos \phi - y \sin \phi, b + x \sin \phi + y \cos \phi, \theta + \phi) \end{align}

corresponded, via Noether's theorem, to a conservation law for the vector quantity $\mathbb{J} = [2 m \dot x \quad 2 m \dot y \quad 2 m \dot y x - 2 m \dot x y + \frac{1}{2} m d^2 \dot \theta]$.

The system is also symmetric with respect to the group action

(2)
\begin{align} \Psi: (\mathbb{R}^3, +) \times Q \to Q: ((\alpha, \beta, \gamma), (d, x, y, \theta)) \mapsto (d, x + \alpha, y + \beta, \theta + \gamma) \end{align}

What's the corresponding conservation law in this case?

#### Solution

Our first step is to calculate the infinitesimal generator. Let $\xi$ be an element of $\mathfrak{g}$. Because $(\mathbb{R}^3, +)$ is vector group, we know that $T_e\mathbb{R}^3 = \mathbb{R}^3$ and $\exp(t\xi)=t\xi$ (Marsden pg. 274). Let $x$ be in $Q$. In coordinates then

(3)
\begin{array} {lcr} \xi_Q(x) & = & \left.\frac{d}{dt}\right|_{t=0}\Psi(exp(t\xi),x)\\ ~\\ & = &\left.\frac{d}{dt}\right|_{t=0}\Psi((t\xi_x,t\xi_y,t\xi_\theta),(d,x,y,\theta))\\ ~\\ & = & \left.\frac{d}{dt}\right|_{t=0}(d,x+t\xi_x,y+t\xi_y,\theta+t\xi_\theta) ~\\ & = & (0,\xi_x,\xi_y,\xi_\theta) \end{array}

Finally we determine the value of $\mathbb{J}(z)$ where $z = (d,x,y,\theta,p_d,p_x,p_y,p_\theta)$

(4)
\begin{align} \left\langle \mathbb{J}(z),\xi\right\rangle = \left\langle z,\xi_Q(q)\right\rangle \end{align}

Writing both sided in coordinates

(5)
\begin{align} \mathbb{J}_1(z) \xi_x + \mathbb{J}_2(z) \xi_y +\mathbb{J}_3(z) \xi_\theta = p_d\cdot 0 + p_x\xi_x + p_y\xi_y + p_\theta\xi_\theta \end{align}

From this, we see that the conserved quantities are

(6)
\begin{array} {lcr} \mathbb{J}_1(z)&=&p_x \\ \mathbb{J}_2(z)&=& p_y\\ \mathbb{J}_3(z)&=&p_\theta\\ \end{array}

Equivalently, we could have carried this out on the Lagrangian side. Let $u = (d,x,y,\theta,\dot{d},\dot{x},\dot{y},\dot{\theta})$. In this case we would have found

(7)
\begin{align} \left\langle \mathbb{J}(u),\xi\right\rangle = \left\langle \mathbb{F}L(u),\xi_Q(q)\right\rangle \end{align}

In coordinates

(8)
\begin{align} \mathbb{J}_1(u) \xi_x + \mathbb{J}_2(u) \xi_y +\mathbb{J}_3(u) \xi_\theta = \frac{1}{2}m\dot{d}\cdot0 + 2m\dot{x}\xi_x + 2m\dot{y}\xi_y + \frac{1}{2}md^2\dot{\theta}\xi_\theta \end{align}

Again, we see that the conserved quantities are

(9)
\begin{array} {lcr} \mathbb{J}_1(u)&=& 2m\dot{x}\\ \mathbb{J}_2(u)&=& 2m\dot{y}\\ \mathbb{J}_3(u)&=&\frac{1}{2}md^2\dot{\theta}\\ \end{array}
page revision: 4, last edited: 07 May 2007 00:29