Homework 3 Problem 10

Problem

Give an example of a Lagrangian system that has a degenerate Lagrangian $L:T \mathbb{R}^{2}\rightarrow \mathbb{R}$ but a second-order Lagrangian vector field $Z_{L}$.

Melih's Solution:

Let

(1)
\begin{align} L:T\mathbb{R}^{2}\rightarrow \mathbb{R}:(q_{1},q_{2},\dot{q}_{1},\dot{q}_{2})\mapsto \pounds =\frac{1}{2}\dot{q}_{1}^{2}+\frac{1}{2}q_{1}^{2}q_{2}+\lambda \dot{q}_{2} \end{align}

Using this, we can obtain a coordinate expression for the this Lagrangian as $4\times 4$ matrix. Let $B$ be the the skew-symmetrization of $\frac{\partial ^{2}L}{\partial \dot{q}_{i}\partial q_{j}}$.

(2)
\begin{array} {ll} \Omega_L& =\begin{bmatrix} B & \left[ \frac{\partial ^{2}L}{\partial \dot{q}_{i}\partial \dot{q}_{j}}\right] \\ \left[ -\frac{\partial ^{2}L}{\partial \dot{q}_{i}\partial \dot{q}_{j}}\right] & 0 \end{bmatrix}\\~\\ &= \begin{bmatrix}0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{bmatrix} \end{array}

This is singular at every point so the Lagrangian is denerate. Also note that the dynamics of that bead moving in the plane has a constraint on the velocity in $2$-direction.

The Legendre transform; $FL:T\mathbb{R}^{2}\rightarrow T^{\ast }\mathbb{R}^{2}$ $:(q_{i},\dot{q}_{i})\mapsto \left( q_{i},\frac{\partial L}{\partial \dot{q}_{i}}\right)$ is found as,

(3)
\begin{align} FL=(q_{1},q_{2},\dot{q}_{1},\lambda ) \end{align}

The action of $L$; $A:T\mathbb{R}^{2}\rightarrow \mathbb{R}:(q_{1},q_{2}, \dot{q}_{1},\dot{q}_{2})\mapsto A=FL(v)\cdot v$ for $v\in T_{q}\mathbb{R}^{2}\Rightarrow$

(4)
\begin{align} A=\left\langle \frac{\partial L}{\partial \dot{q}_{i}},\dot{q}% _{i}\right\rangle =\dot{q}_{1}^{2}+\lambda \dot{q}_{2}. \end{align}

The energy of $L$;

(5)
\begin{align} E:T\mathbb{R}^{2}\rightarrow \mathbb{R}:(q_{1},q_{2},\dot{q}_{1},\dot{q}_{2})\mapsto E=A-L=(\dot{q}_{1}^{2}+\lambda\dot{q}_{2})-(\frac{1}{2}\dot{q}_{1}^{2}+\frac{1}{2}q_{1}^{2}q_{2}+\lambda\dot{q}_{2})=\frac{1}{2}\dot{q}_{1}^{2}-\frac{1}{2}q_{1}^{2}q_{2}\Rightarrow \end{align}
(6)
\begin{align} dE=\dot{q}_{1}d\dot{q}_{1}-q_{1}q_{2}dq_{1}-\frac{1}{2}q_{1}^{2}dq_{2}. \end{align}

Since, $dE=Z_{L}\lrcorner \Omega _{L}+w$, where

(7)
\begin{align} \Omega _{L}=\frac{\partial^{2}L}{\partial \dot{q}_{i}\partial q_{j}}dq_{i}\wedge dq_{j}+\frac{\partial^{2}L}{\partial \dot{q}_{i}\partial \dot{q}_{j}}dq_{i}\wedge d\dot{q}_{j}=dq_{1}\wedge d\dot{q}_{1} \end{align}

and $w$ is the one form exterior force;

(8)
\begin{align} Z_{L} =(\dot{q}_{1},-q_{1}q_{2}) \end{align}
(9)
\begin{align} w =-\frac{1}{2}q_{1}^{2}dq_{2} \end{align}

i.e. there exists a second order vector field, $Z_{L}$, such that;

(10)
\begin{align} Z_{L_{q_{1}}} &=&\dot{q}_{1} \end{align}
(11)
\begin{align} Z_{L_{\dot{q}_{1}}} &=&-q_{1}q_{2} \end{align}

with the constraint; $\dot{q}_{2}=0$.

Discussion
Please check the Lagrangian I suggested.

page revision: 3, last edited: 21 Apr 2007 21:31