# Finding period from action-angle variable in one dimensional potential [closed]

I want to calculate the period from the action-angle variable for a particle in a one dimensional potential $$V = V_0 \tan^2(q \pi/2a)$$. After doing some algebra I get $$I = \frac{\sqrt{2mE}}{2\pi} \int_{q_1}^{q_2}\sqrt{1-\frac{V_0}{E}\tan^2{\frac{q\pi}{2a}}}dq.$$ Solving which gives me $$I = \frac{a}{\pi}\sqrt{2mE}\left(\sqrt{1+ \frac{V_0}{E}} - \sqrt{\frac{V_0}{E}}\right).$$ How can I get the period of the motion and how will it behave when $$E\gg V_0$$ and vice versa? Also how does this picture change if I change $$a$$ adiabatically with time?

The trick is to re-express the Hamiltonian in terms of the action variables. In your case this simply means to invert the expression $$I(E)$$ into an expression $$E(I) \equiv H(I)$$. This is then your Hamiltonian expressed in terms of action-angle coordinates. Finally, you can compute the evolution of the angle $$\phi$$ canonically conjugate to $$I$$ from Hamilton's equations of motion $$\dot{\phi} = \frac{\partial H}{\partial I}$$ Notice that $$\dot{I} = -\partial H/\partial \phi = 0$$, $$I$$ is constant, and thus also the frequency $$\dot{\phi}$$ is constant for any trajectory.
You are also asking about the behaviour when $$E \ll V_0$$ as opposed to $$E \gg V_0$$. This is respectively equivalent to $$I \ll a\sqrt{m E}$$ and $$I \gg a\sqrt{m E}$$, as you should directly verify. You will then see that in the leading-order limit of a small action, the rate $$\dot{\phi}$$ is independent of $$I$$ - the system behaves like a harmonic oscillator. However, for large actions (large energies) the frequency will have a steep dependence on $$I$$.
• thanks for your answer. Would you be able to comment on how the energy of the system changes when I change $a$ adiabatically ? – 112358 Apr 29 '19 at 14:33
• @112358 Certain variables will adiabatic invariants. You can assume that $\dot{a}/a\ll \dot{\phi}$. How large will be $\dot{E},\dot{I}$ in that case? – Void Apr 29 '19 at 14:53