PART - A
L(q,\dot{q},t)=m/2[\dot{q}^2\sin^2\omegat+\dot{q}q\omega\sin(2\omegat)+q^2\omega^2]......(EQ0)
Step 1: First calculate the generalized momentum-
p\equiv(\frac{\delL}{\del\dot{q}}) = m/2[2\dot{q}\sin^2\omegat+qw\sin(2\omegat)]...... (EQ1)
Step 2: Next, invert (1) and write \dot{q} in terms of p and q -
\dot{q}(p,q) = [p/(m\sin^2\omegat)-(q\omega)/(\tan\omegat)] ...... (EQ2)
For later convenience we also find expressions for \dot{q}\sin\omegat and \dot{q}^2\sin^2\omegat.
\dot{q}\sin\omegat = [ p/(m\sin\omegat)-q\omega\cos\omegat ]......(EQ3)
\dot{q}^2\sin^2\omegat = [p^2/(m^2\sin^2\omegat) + q^2\omega^2\cos^2\omegat] ......(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 \dot{q} with p,
L(q,\dot{q},t) = m/2[(\dot{q}\sin\omegat)^2+2\omegaq(\dot{q}\sin\omegat)\cos\omegat+q^2\omega^2]
Substitute EQ3 and EQ4 for the terms in parentheses in the above equation, simplify and rearrange -
L(q,p,t) = m/2[p^2/(m^2\sin^2\omegat) + q^2\omega^2\sin^2\omegat] ...... (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(q,p,t) = [p^2/(2m\sin^2\omegat) - (pq\omega)/(\tan\omegat) - (m\omega^2q^2)/2\sin^2\omegat] ......(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
\frac{\delH}{\delt}=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(q,\dot{q},t)=m/2[\dot{q}^2\sin^2\omegat+\dot{q}q\omega\sin(2\omegat)+q^2\omega^2]......(EQ0)
Introduce a new coordinate \quad Q = q\sin\omegat;
\dot{Q} = \dot{q}\sin\omegat+q\omega\cos\omegat...... (EQ6)
\dot{Q}^2 = \dot{q}^2\sin^2\omegat+\dot{q}q\omega\sin(2\omegat) + q^2\omega^2\cos^2\omegat
\dot{q}^2\sin^2\omegat+\dot{q}q\omega\sin(2\omegat) = \dot{Q}^2 - q^2\omega^2\cos^2\omegat
\dot{q}^2\sin^2\omegat+\dot{q}q\omega\sin(2\omegat) = \dot{Q}^2 - q^2\omega^2(1-\sin^2\omegat)
\dot{q}^2\sin^2\omegat+\dot{q}q\omega\sin(2\omegat) + q^2\omega^2= \dot{Q}^2+ q^2\omega^2\sin^2\omegat
\dot{q}^2\sin^2\omegat+\dot{q}q\omega\sin(2\omegat) + q^2\omega^2= \dot{Q}^2+ \omega^2Q^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, \dot{Q} as -
L(Q, \dot{Q},t) = m/2[\dot{Q}^2+\omega^2Q^2] ...... (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 \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(Q,P,t)\equiv P\dot{Q}(Q,P)-L(Q,P,t)
Simplify the above expression
H(Q,P,t) = P(P/m)-[P^2/(2m)+(m\omega^2)/2Q^2]
H(Q,P,t) = [P^2/(2m) - (m\omega^2)/2Q^2] ...... (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.
