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Canonical Transformation (two degrees of freedom)
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[QUOTE="sayebms, post: 5309111, member: 300413"] [h2]Homework Statement [/h2] 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. [h2]Homework Equations[/h2] -Symplictic approach [h2]The Attempt at a Solution[/h2] 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] [/QUOTE]
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Canonical Transformation (two degrees of freedom)
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