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Consider a bidimensional harmonic oscillator.

Ref. 1 says that, when the frequencies are commensurable,

separating the variables in cartesian or polar coordinates leads to different action-angle variables

(and ultimately to different invariant tori)...but I can not figure how. Could someone offer an example?

References:

  1. V.I. Arnold, Mathematical Methods of Classical Mechanics, 1989; $\S51$ p. 291 Example 1. Read it here or here.
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  • $\begingroup$ Where does Arnold talk about commensurable frequencies? $\endgroup$
    – Qmechanic
    Jan 16, 2019 at 12:54

1 Answer 1

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  1. In polar coordinates $(r,\theta)$ and in rectangular coordinates $$(x,y)~=~r(\cos\theta,\sin\theta)$$ the 2D harmonic oscillator has Lagrangian $$L ~=~\frac{1}{2}(\dot{r}^2+r^2\dot{\theta}^2-r^2)~=~L_x+L_y, $$ $$L_x~=~\frac{1}{2}(\dot{x}^2-x^2), \qquad L_y~=~\frac{1}{2}(\dot{y}^2-y^2), $$ and Hamiltonian $$H ~=~\frac{1}{2}(p_r^2+\frac{p_{\theta}^2}{r^2}+r^2)~=~H_x+H_y,$$ $$ H_x~=~\frac{1}{2}(p_x^2+x^2), \qquad H_y~=~\frac{1}{2}(p_y^2+y^2), $$ where $$r^2~=~x^2+y^2, \qquad p_r^2~=~p_x^2+p_y^2, \qquad p_{\theta}~=~xp_y-yp_x. $$

  2. The angle-action variables are $$ (\varphi_r,\theta, H, p_{\theta})\qquad\text{and}\qquad(\varphi_x,\varphi_y, H_x, H_y),$$ respectively, where $$ p_r+ir~=~\sqrt{2H}e^{i\varphi_r}, \qquad p_x+ix~=~\sqrt{2H_x}e^{i\varphi_x}, \qquad p_y+iy~=~\sqrt{2H_y}e^{i\varphi_y}. $$

References:

  1. V.I. Arnold, Mathematical Methods of Classical Mechanics, 1989; $\S$51 p. 291 Example 1.
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  • $\begingroup$ I know that's unnecessary...but thank you!! $\endgroup$
    – Lo Scrondo
    Jan 16, 2019 at 13:33

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