Canonical Transformation (two degrees of freedom)

[sayebms]

Homework Statement

Point transformation in a system with 2 degrees of freedom is: $$Q_1=q_1^2\\Q_2=q_q+q_2$$
a) find the most general $P_1$ and $P_2$ such that overall transformation is canonical
b) Show that for some $P_1$ and $P_2$ the hamiltonain $$H=\frac{a}{2}(\frac{p_1-p_2}{wq_1})^2+bp_2+c(q_1+q_2)^2$$
(Note: a,b,c, constant) can be transformed in a way that $Q_1$ and $Q_2$ could be ignored.
c) Finally solve the problem and find equations for $q_1,q_2,p_1$ and $p_2$ in terms of t and their initial values.

Homework Equations

-Symplictic approach

The Attempt at a Solution

a) in order to find the transformation for $P$'s I use the symplectic approach:
in this approach we have the following condition for the canonical transformations $$\tilde{M}JM=J$$
where M is the matrix:
$$M=\begin{bmatrix} \frac{\partial Q_1}{\partial q_1}&\frac{\partial Q_1}{\partial q_2}&\frac{\partial Q_1}{\partial p_1}&\frac{\partial Q_1}{\partial p_2}\\\frac{\partial Q_2}{\partial q_1}&\frac{\partial Q_2}{\partial q_2}&\frac{\partial Q_2}{\partial p_1}&\frac{\partial Q_2}{\partial p_2}\\\frac{\partial P_1}{\partial q_1}&\frac{\partial P_1}{\partial q_2}&\frac{\partial P_1}{\partial p_1}&\frac{\partial P_1}{\partial p_2}\\\frac{\partial P_2}{\partial q_1}&\frac{\partial P_2}{\partial q_2}&\frac{\partial
P_2}{\partial p_1}&\frac{\partial P_2}{\partial p_2}\end{bmatrix}$$
and $$J=\begin{bmatrix} 0&0&1&0\\0&0&0&1\\-1&0&0&0\\0&-1&0&0\end{bmatrix}$$
so applying the symplectic methods conditions I get the following equations:$$\frac{\partial P_2}{\partial p_2}=1\\\frac{\partial P_2}{\partial p_1}=0\\\frac{\partial P_1}{\partial p_2}=\frac{-1}{2q_1}\\\frac{\partial P_1}{\partial p_1}=\frac{1}{2q_1}$$
which then gives me the following most general transformations for $P_1$ and $P_2$ :
$$P_1=\frac{1}{2q_1}(p_1-p_2)+g(q_1,q_2)\\P_2=p_2+f(q_1,q_2)\\2q_1 \frac{\partial g}{\partial q_2}+\frac{\partial f}{\partial q_2}-\frac{\partial f}{\partial q_1}$$
b) this part is very simple if we choose $$P_1=\frac{p_1-p_2}{2q_1}\\g(q_1,q_2)=0$$and $$P_2=p_2+\frac{c}{b}(q_1+q_2)^2\\f(q_1,q_2)=\frac{c}{b}(q_1+q_2)^2$$
we see that with these choices the three equations we found earlier are satisfied. and the Hamiltonian becomes:$$H=\frac{a}{2}P_1^2+bP_2$$
But I don't know how I can proceed for solving part c). Any help is appreciated. Thank you very much.[/B]

 

[Orodruin]

You have the Hamiltonian, so you can just write down the equations of motion and solve them.

 

[sayebms]

so using the Hamiltonian equations of motion I have: $$\dot{Q_1}=\frac{\partial H}{\partial P_1}=aP_1 \Longrightarrow \frac{d(q_1^2)}{dt}=2q_1\dot{q_1}=aP_1\\\dot{Q_2}=\frac{\partial H}{\partial P_2}=b \Longrightarrow \dot{q_1}+\dot{q_2}=b\\\dot{P_1}=0\\\dot{P_2}=0$$ So I guess these are the equations that I am supposed to solve. But there is one more thing that I don't get; the problem hasn't given us anything on initial conditions but still wants the equations of q,p,t in terms of t and Initial values.Thank you very much again.

 

[Johnny O]

$$P_1$$ is a constant of motion, so you can integrate the right side to get $$Q_1 = aP_1 \cdot t$$.