Homework 3 Problem 7

Problem

For the "falling chain" problem described in class, show that the Lagrangian 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 (revisiting an earlier homework problem) 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 $L$ for this problem was determined by Andy for HW1:

ยต = mass per unit length
l = length of the chain
x = displacement of the chain from the x-axis
g = gravity

(1)
\begin{align} L = T - V = 2\mu x\dot x^2 - \mu g\left( {x^2 - \frac{{l^2 }}{2}} \right) \\ \end{align}

In order to show that the $L$ is hyperregular, one needs to check that the fiber derivative, $\mathbb F L$, is a diffeomorphism ($\mathbb F L$ is bijective and $\mathbb F L$ and $\mathbb F L^{-1}$ are both smooth).

First make sure that $\mathbb {F} L$ is locally invertible:

(2)
\begin{align} \left[\frac{\partial^{2}L}{\partial{\dot q^i}\partial \dot q^j}\right] \neq 0 \rightarrow \left[\frac{\partial^{2}L}{\partial{\dot x^2}}\right] = 4\mu x \end{align}

Clearly, this is non-zero for all x not equal to zero. For this problem, $\mathbb F L$ and $\mathbb F L^{-1}$ are found to be:

(3)
\begin{align} \mathbb F L: TQ \rightarrow T^*Q:\left (x,\dot x \right) \mapsto \left(x,4 \mu x \dot x \right) \ \end{align}
(4)
\begin{align} \mathbb F L^{-1}: T^*Q \rightarrow TQ:\left (x,p_x \right) \mapsto \left(x,\frac {p_x}{ 4\mu x} \right) \end{align}

Clearly, $\mathbb F L$ is bijective. However, $\mathbb F L^{-1}$ is not smooth at $x = 0$. Therefore, I feel like $L$ is not hyperregular for this problem.

Moving on, the canonical one form $\Theta$, the canonical symplectic form $\Omega$, and the Liouville volume form $\Lambda$ on $T^*Q$ are computed as follows:

(5)
\begin{align} \Theta = p_x dx \end{align}
(6)
\begin{align} \Omega = dx \wedge dp_x \end{align}
(7)
\begin{align} \Lambda = dx \wedge dp_x \end{align}

The Lagrangian one form $\Theta_L$ and the Lagrangian two form $\Omega_L$ on $TQ$ are:

(8)
\begin{align} \Theta_L = \left(\mathbb {F}L\right)^*\left(\Theta\right) = \frac {\partial L }{\partial \dot x}dx = 4\mu x \dot x dx \end{align}
(9)
\begin{align} \Omega_L = - d\Theta_L = -\frac{\partial}{\partial x}\left[4\mu x \dot x \right] dx \wedge dx \\ -\frac{\partial}{\partial x}\left[4\mu x \dot x \right] d \dot x \wedge dx = 4 \mu x \ dx \wedge d\dot x \end{align}

Next, the action $A$, the energy $E$, and the Hamiltonian $H$ associated with the Lagrangian $L$are:

(10)
\begin{align} A = \dot q^i \frac {\partial L}{\partial \dot q^i} = \dot x \frac {\partial L}{\partial \dot x}= 4 \mu x \dot x^2 \end{align}
(11)
\begin{align} E = A - L = 4 \mu x \dot x^2 - 2\mu x \dot x^2 + \mu g \left(x^2 - \frac {l^2}{2}\right) \\ = 2 \mu x \dot x^2 + \mu g \left(x^2 - \frac {l^2}{2}\right) \end{align}
(12)
\begin{align} H = E \cdot \left(\mathbb {F}L^{-1}\right)(x,p_x) = 2 \mu x \left(\frac{p_{x}^2}{16 \mu ^2 x^2}\right) \\ +\mu g \left(x^2 - \frac{l^2}{2}\right) = \left(\frac{p_{x}^2}{8 \mu x}\right) +\mu g \left(x^2 - \frac{l^2}{2}\right) \end{align}

As expected, the Hamiltonian found using this method matches what was found by Andy in HW1-2.

Now for the Hamiltonian vector field $X_H$ on $T^*Q$:

(13)
\begin{align} dH(q,p)\cdot \left(V_x\frac{\partial}{\partial x}+V_p_x \frac{\partial}{\partial p_x}\right)=\Omega(x,p_x)\left(X_{H_x}\frac{\partial}{\partial x}+X_{H_p}\frac{\partial}{\partial p_x},V_x\frac{\partial}{\partial x}+V_p_x \frac{\partial}{\partial p_x}\right) \end{align}
(14)
\begin{align} \rightarrow X_H = \left( \frac{\partial H}{\partial p_x}, - \frac{\partial H}{\partial x} \right) = \left(\frac{p_x}{4\mu x}, \frac{-p^{2}_{x}}{8\mu x^2}+2\mu gx \right) \end{align}

Now for the Lagrangian vector field $Z_L$ on $TQ$:

(15)
\begin{align} dE = \left(2\mu \dot x^2 + 2 \mu gx\right)dx +4 \mu x \dot x d \dot x \end{align}
(16)
\begin{align} dE = Z_L \lrcorner \Omega_L = \left(Z_{L_x}\frac{\partial}{\partial x}+Z_{L_{\dot x}} \frac {\partial}{\partial \dot x}\right) \lrcorner \left(4 \mu x dx \wedge d\dot x\right) = 4 \mu x \left(Z_{L_{x}} d \dot x - Z_{L_{\dot x}} dx\right) \end{align}

Therefore, by inspection $Z_L$ is found to be:

(17)
\begin{align} Z_L = \left(\dot x, -\left(\frac{\dot x^2}{2x} + \frac{g}{2} \right ) \right) \end{align}

Discussion

I am not sure if my conclusion that $L$ is not hyperregular is correct because usually when a problem says show something, if you show that it is not something, you screwed up…

Also, I'm not 100% sure about the Liouville volume form, but I think that it is just equal to $\Omega$ since n = 1 for this problem.

Lastly, I'm a little confused about $Z_L$ and $X_H$. While looking up Hamiltonian Vector field, I found the relation that I used to solve for it $\left(X_H = \left( \frac{\partial H}{\partial p_x}, - \frac{\partial H}{\partial x} \right)\right)$. I assumed that $Z_L$ would be similar, but w/ $L, x, \dot x$ replacing $H, x, p_x$. However, I could not find confirmation of this, so I worked it out the way we did it in class. After finishing the solution, I noticed that if I did not pull the $4 \mu x$ out, $Z_L$ would look like:

(18)
\begin{align} Z_L = \left( \frac{\partial L}{\partial \dot x}, - \frac{\partial L}{\partial x} \right) \end{align}

which is what I expected based on the method I found/used for $X_H$. Anyone have any thoughts on this?

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