# Classical Mechanics and the Hamiltonian

This problem has to do with a canonical transformation and Hamiltonian formalism. Below is my attempt at solving it, but I am not too sure about it. Please help!

Let $\theta$ be some parameter.

And $$X_1=x_1\cos \theta-y_2\sin\theta\\ Y_1=y_1\cos \theta+x_2\sin\theta\\ X_2= x_2\cos \theta-y_1\sin\theta\\ Y_2=y_2\cos\theta+x_1\sin \theta$$

Suppose the original Hamiltonian is $$H(x,y)={1\over 2}(x_1^2+y_1^2+x_2^2+y_2^2)$$ I wish to find solve for the motion in terms of the new variables. I am also given the restriction that $X_2=Y_2=0$

Attempt:

I believe we have $$H(X,Y)={1\over 2}(X_1^2+Y_1^2+X_2^2+Y_2^2)$$

Now the normal Hamiltonian formalism would suggest that $$\dot X_i={\partial H\over \partial Y_i }\\ \dot Y_i=-{\partial H\over \partial X_i }$$

Which gives $$\ddot X_1=-X_1\\ \ddot Y_1=-Y_1$$ Therefore, $$X_1(t)=A(\theta)\cos t+B(\theta)\sin t\\ Y_2(t)=C(\theta)\cos t+D(\theta)\sin t$$*Is this form of solutions right?*

We see that the $${\partial X_1\over \partial \theta}=-Y_2=0\\ {\partial Y_1\over \partial \theta}=X_2=0$$ So $A,B,C,D$ must be constants.

Are these arguments right? And can I get a better solution, say by getting a more specific set of $A,B,C,D$, given only the given information?

Thank you.

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You've conflated the problem and the solution; it's not clear where the problem statement ends and your solution attempt begins. –  joriki Dec 30 '12 at 3:06
Dear @joriki , I have edited the question. Hopefully it is of better form now. –  Morgan Dec 30 '12 at 3:19

Almost there. First, you made a typo: you mean to write $Y_1(t)$, not $Y_2(t)$, for the expression $C(\theta)\cos t + D(\theta)\sin t .$
Finally, knowing $\dot X_1 = {\partial H \over \partial Y_i}$, you obtain $$-A \sin t + B \cos t = Y_1 = C \cos t + D \sin t.$$ And the equation for $\dot Y_1$ yields (redundantly) $$-C \sin t + D \cos t = - A \cos t - B \sin t.$$ This means $$X_1(t) = A \cos t + B \sin t, \qquad Y_1(t) = B \cos t - A \sin t.$$ This is a parametrization of a circle with radius $\sqrt{A^2 + B^2}$ in the $(X_1,Y_1)$ plane.