# Question b44f9

Dec 8, 2017

(A)$\setminus \quad H \left(q , p , t\right) = \left[{p}^{2} / \left(2 m \setminus {\sin}^{2} \setminus \omega t\right) - \frac{p q \setminus \omega}{\setminus \tan \setminus \omega t} - \frac{m \setminus {\omega}^{2} {q}^{2}}{2} \setminus {\sin}^{2} \setminus \omega t\right]$

(B)\quad H(Q,P,t) = [P^2/(2m) - (m\omega^2)/2Q^2]; H is conserved.

#### Explanation:

PART - A

$L \left(q , \setminus \dot{q} , t\right) = \frac{m}{2} \left[\setminus {\dot{q}}^{2} \setminus {\sin}^{2} \setminus \omega t + \setminus \dot{q} q \setminus \omega \setminus \sin \left(2 \setminus \omega t\right) + {q}^{2} \setminus {\omega}^{2}\right]$......(EQ0)

Step 1: First calculate the generalized momentum-
$p \setminus \equiv \left(\setminus \frac{\setminus \partial L}{\setminus \partial \setminus \dot{q}}\right) = \frac{m}{2} \left[2 \setminus \dot{q} \setminus {\sin}^{2} \setminus \omega t + q w \setminus \sin \left(2 \setminus \omega t\right)\right]$...... (EQ1)

Step 2: Next, invert (1) and write $\setminus \dot{q}$ in terms of $p$ and $q$ -

$\setminus \dot{q} \left(p , q\right) = \left[\frac{p}{m \setminus {\sin}^{2} \setminus \omega t} - \frac{q \setminus \omega}{\setminus \tan \setminus \omega t}\right]$ ...... (EQ2)

For later convenience we also find expressions for $\setminus \dot{q} \setminus \sin \setminus \omega t$ and $\setminus {\dot{q}}^{2} \setminus {\sin}^{2} \setminus \omega t$.

$\setminus \dot{q} \setminus \sin \setminus \omega t = \left[\frac{p}{m \setminus \sin \setminus \omega t} - q \setminus \omega \setminus \cos \setminus \omega t\right]$......(EQ3)

$\setminus {\dot{q}}^{2} \setminus {\sin}^{2} \setminus \omega t = \left[{p}^{2} / \left({m}^{2} \setminus {\sin}^{2} \setminus \omega t\right) + {q}^{2} \setminus {\omega}^{2} \setminus {\cos}^{2} \setminus \omega t\right]$ ......(EQ4)

Step 3: Rewrite the Lagrangian as a function of $q$, $p$ and $t$,

Let us rewrite the Lagrangian in a form that would make it easy to replace $\setminus \dot{q}$ with $p$,

$L \left(q , \setminus \dot{q} , t\right) = \frac{m}{2} \left[{\left(\setminus \dot{q} \setminus \sin \setminus \omega t\right)}^{2} + 2 \setminus \omega q \left(\setminus \dot{q} \setminus \sin \setminus \omega t\right) \setminus \cos \setminus \omega t + {q}^{2} \setminus {\omega}^{2}\right]$

Substitute EQ3 and EQ4 for the terms in parentheses in the above equation, simplify and rearrange -

$L \left(q , p , t\right) = \frac{m}{2} \left[{p}^{2} / \left({m}^{2} \setminus {\sin}^{2} \setminus \omega t\right) + {q}^{2} \setminus {\omega}^{2} \setminus {\sin}^{2} \setminus \omega t\right]$ ...... (EQ4)

Step 4: Perform a Legendre transformation and construct the Hamiltonian,

H(q,p,t) \equiv p\dot{q}(q,p)-L(q,p,t);

Substituting EQ2 and EQ4 in the above relation, simplifying and rearranging,

$H \left(q , p , t\right) = \left[{p}^{2} / \left(2 m \setminus {\sin}^{2} \setminus \omega t\right) - \frac{p q \setminus \omega}{\setminus \tan \setminus \omega t} - \frac{m \setminus {\omega}^{2} {q}^{2}}{2} \setminus {\sin}^{2} \setminus \omega t\right]$ ......(EQ5)

Is the Hamiltonian conserved?: The Hamiltonian is conserved if it is not an explicit function of time. Though the above Hamiltonian seems to have some explicit terms involving time, it is not clear if
$\setminus \frac{\setminus \partial H}{\setminus \partial t} = 0$.

I tried hard to verify if it were so. But it grows monstrous to the point where I had to give up and start wondering if the second part of the problem intends to teach us something regarding this ....

PART - B

$L \left(q , \setminus \dot{q} , t\right) = \frac{m}{2} \left[\setminus {\dot{q}}^{2} \setminus {\sin}^{2} \setminus \omega t + \setminus \dot{q} q \setminus \omega \setminus \sin \left(2 \setminus \omega t\right) + {q}^{2} \setminus {\omega}^{2}\right]$......(EQ0)

Introduce a new coordinate \quad Q = q\sin\omegat;
$\setminus \dot{Q} = \setminus \dot{q} \setminus \sin \setminus \omega t + q \setminus \omega \setminus \cos \setminus \omega t$...... (EQ6)
$\setminus {\dot{Q}}^{2} = \setminus {\dot{q}}^{2} \setminus {\sin}^{2} \setminus \omega t + \setminus \dot{q} q \setminus \omega \setminus \sin \left(2 \setminus \omega t\right) + {q}^{2} \setminus {\omega}^{2} \setminus {\cos}^{2} \setminus \omega t$
$\setminus {\dot{q}}^{2} \setminus {\sin}^{2} \setminus \omega t + \setminus \dot{q} q \setminus \omega \setminus \sin \left(2 \setminus \omega t\right) = \setminus {\dot{Q}}^{2} - {q}^{2} \setminus {\omega}^{2} \setminus {\cos}^{2} \setminus \omega t$
$\setminus {\dot{q}}^{2} \setminus {\sin}^{2} \setminus \omega t + \setminus \dot{q} q \setminus \omega \setminus \sin \left(2 \setminus \omega t\right) = \setminus {\dot{Q}}^{2} - {q}^{2} \setminus {\omega}^{2} \left(1 - \setminus {\sin}^{2} \setminus \omega t\right)$
$\setminus {\dot{q}}^{2} \setminus {\sin}^{2} \setminus \omega t + \setminus \dot{q} q \setminus \omega \setminus \sin \left(2 \setminus \omega t\right) + {q}^{2} \setminus {\omega}^{2} = \setminus {\dot{Q}}^{2} + {q}^{2} \setminus {\omega}^{2} \setminus {\sin}^{2} \setminus \omega t$
$\setminus {\dot{q}}^{2} \setminus {\sin}^{2} \setminus \omega t + \setminus \dot{q} q \setminus \omega \setminus \sin \left(2 \setminus \omega t\right) + {q}^{2} \setminus {\omega}^{2} = \setminus {\dot{Q}}^{2} + \setminus {\omega}^{2} {Q}^{2}$

Observe that the left-hand-side of the above expression if the term inside the square bracket of the Lagrangian expression. Thus we rewrite the Lagrangian in terms of the new variable $Q$ and its time derivative, $\setminus \dot{Q}$ as -

$L \left(Q , \setminus \dot{Q} , t\right) = \frac{m}{2} \left[\setminus {\dot{Q}}^{2} + \setminus {\omega}^{2} {Q}^{2}\right]$ ...... (EQ8)

Step 1: First calculate the generalized momentum-

P\equiv(\frac{\delL}{\del\dot{Q}})=m\dot{Q};......(EQ9)

Step 2: Invert the above expression to write $\setminus \dot{Q}$ in terms of $P$ and also write the Lagrangian in terms of $Q$, $P$ and $t$,

\dot{Q} = P/m;\qquad L(Q, P, t) =[P^2/(2m)+(m\omega^2)/2Q^2]#

Step 3: Perform a Legendre transformation and construct the Hamiltonian,

$H \left(Q , P , t\right) \setminus \equiv P \setminus \dot{Q} \left(Q , P\right) - L \left(Q , P , t\right)$

Simplify the above expression
$H \left(Q , P , t\right) = P \left(\frac{P}{m}\right) - \left[{P}^{2} / \left(2 m\right) + \frac{m \setminus {\omega}^{2}}{2} {Q}^{2}\right]$

$H \left(Q , P , t\right) = \left[{P}^{2} / \left(2 m\right) - \frac{m \setminus {\omega}^{2}}{2} {Q}^{2}\right]$ ...... (EQ10)

This is the desired Hamiltonian function. In this case it is obvious that the Hamiltonian is purely a function of $Q$ and $P$ and is not explicitly dependent on $t$. So the Hamiltonian is conserved.