Homework 4 Problem 5


The scalar quantity $2 m (\dot y x - \dot x y)$ is also conserved in the “extensible barbell” example. What group action corresponds to this conservation law?


Let $G$ be $S^1$ with the group action addition modulo $2\pi$. Define the left action of $S^1$ on $Q$ by

\begin{align} \Phi: S^1 \times Q \to Q: \phi, (d, x, y, \theta)) \mapsto (d, x \cos \phi - y \sin \phi, x \sin \phi + y \cos \phi, \theta) \end{align}

The tangent lift of this action is represented by the matrix

\begin{bmatrix} 1&0&0&0\\ 0&\cos\phi&-\sin\phi&0\\ 0&\sin\phi&\cos\phi&0\\ 0&0&0&1 \end{bmatrix}

From this we can see that the tangent lifted action has the form

\begin{array} {l} \Phi: S^1 \times TQ \to TQ: \phi, (d, x, y, \theta,\dot{d},\dot{x},\dot{y},\dot{\theta}) \mapsto \\ \quad (d, x \cos \phi - y \sin \phi, x \sin \phi + y \cos \phi, \theta ,\dot{d}, \dot{x} \cos \phi - \dot{y} \sin \phi, \dot{x} \sin \phi + \dot{y} \cos \phi, \dot{\theta}) \end{array}

Note that the Lagrangian is invariant under this action. Let $\mathbf{q}$ be the two vector $(x,y)$ and $A_\phi$ be the two by two block of the above matrix. Note that $A_\phi$ is orthogonal. Then

\begin{align} L(\Phi_\phi(d,\mathbf{q},\theta,\dot{d},\dot{\mathbf{q}},\dot{\theta})) = m||A_\phi\dot{\mathbf{q}}||^2+ \frac{1}{2}d^2\dot{\theta}}^2 = m||\dot{\mathbf{q}}||^2 +\frac{1}{2}d^2\dot{\theta}}^2 = L(d,\mathbf{q},\theta,\dot{d},\dot{\mathbf{q}},\dot{\theta}) \end{align}

Therefore the Lagrangian is invariant under the lifted action.

Now to determine the infinitesimal generator. Note that $\mathfrak{s}^1 \cong \mathbb{R}$. Let $\phi,\eta$ be in $S^1$ and $\xi$ be in $\mathfrak{s}^1$. In coordinates, $L_\phi\eta = \phi+\eta$. Thus, in coordinates, $T_eL_\phi\xi = \xi$, In coordinates, the left invariant vector field $X_\xi$ corresponding to $\xi$ is

\begin{align} X_\xi(\phi) = \xi \frac{\partial}{\partial \phi} \end{align}

From this we can read off that $$\exp(t\xi) = t\xi$. Now we are ready to compute $\xi_Q$

\begin{array} {lcr} \xi_Q(d,x,y,\theta) &=&\left.\frac{d}{dt}\right|_{t=0} \Phi(\exp(t\xi),(d,x,y,\theta))\\ &=&\left.\frac{d}{dt}\right|_{t=0} \Phi(t\xi,(d,x,y,\theta))\\ &=&\left.\frac{d}{dt}\right|_{t=0} (d, x \cos t\xi - y \sin t\xi, x \sin t\xi + y \cos t\xi, \theta)\\ &=&(0, - y\xi, x\xi,0)\\ \end{array}

Now we are ready to calculate $\mathbb{J}(z)$ where $z=(d, x, y, \theta,\dot{d},\dot{x},\dot{y},\dot{\theta})$

\begin{array} {lcr} \left\langle\mathbb{J}(z),\xi\right\rangle &=&\left\langle\mathbb{F}L(z),\xi_Q(d,x,y,\theta) \right\rangle \\ \mathbb{J}_1(z)\xi&=&md\dot{\theta}^2 \cdot 0 + 2m\dot{x}\cdot*(-y\xi) + 2\dot{y}\cdot(x\xi) + \frac{1}{2}md^2\dot{\theta} \cdot 0\\ &=& 2m (\dot{y}x-\dot{x}y)\xi \end{array}

From this we can read off the conserved quantity $\mathbb{J}_1(z) = 2m(\dot{y}x - \dot{x}y)$

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