- #1

fluidistic

Gold Member

- 3,948

- 263

## Homework Statement

Consider a harmonic oscillator with generalized coordinates q and p with a frequency omega and mass m.

Let the transformation (p,q) -> (Q,P) be such that [itex]F_2(q,P,t)=\frac{qP}{\cos \theta }-\frac{m\omega }{2}(q^2+P^2)\tan \theta[/itex].

1)Find [itex]K(Q,P)[/itex] where [itex]\theta[/itex] is a function of time and K is the Hamiltonian in function of Q and P.

2)For which values of [itex]\theta (t)[/itex] does [itex]K(Q,P)[/itex] vanishes?

## Homework Equations

[itex]H=\frac{p^2}{2m}+\frac{\omega m }{2}[/itex].

[itex]p=\frac{\partial F_2}{\partial q}, Q=\frac{\partial F_2}{\partial P}[/itex]

[itex]K=H+\frac{\partial F_2}{\partial t}[/itex]

## The Attempt at a Solution

I guess the main idea is to calculate [itex]\frac{\partial F_2}{\partial t}[/itex] and express p and q in terms of P and Q.

Playing with the relevant equations I get that [itex]q=Q\cos \theta +m \omega P \sin \theta[/itex], [itex]p=\frac{P}{\cos \theta }-m\omega \tan \theta (Q\cos \theta + m\omega P \sin \theta )[/itex].

Also, [itex]\frac{\partial F_2}{\partial t}=qP\dot \theta \sin \theta - \frac{m\omega }{2}(q^2+P^2)\left ( \frac{\dot \theta }{\sin ^2 \theta } \right ) [/itex].

This gave me [itex]K=\frac{1}{2m} \left [ \frac{P}{\cos \theta }-m\omega (Q\cos \theta+m \omega \sin \theta ) \tan \theta \right ] ^2+\frac{m\omega}{2}(Q\cos \theta + m \omega P \sin \theta )^2+ (Q \cos \theta + m \omega P \sin \theta )P\dot \theta \sin \theta -\frac{m\omega }{2} [(Q\cos \theta + m\omega P \sin \theta)^2+P^2]\left ( \frac{\dot \theta }{\sin ^2 \theta } \right )[/itex].

The new Hamiltonian does indeed depends on the variables it should, but it looks so horrible that I cannot believe I made things right. What am I missing?