# Finding action-angle variable of given system

I need to find the action variable of a particle of mass $m$ moves in a one-dimensional potential $V = V_o \tan^2{\frac{πq}{2a}}$, where $V_o$ and a are potential parameters, and $q$ is the particle co_ordinate. Also what will be the period of motion when $E>>V_o$ & $E<< Vo$.

I have started as below, \begin{align} E&=\frac{p^2}{2m}+V_o\tan^2\frac{q\pi}{2a}\, ,\\ p^2 &=2mE-2mV_o\tan^2\frac{q\pi}{2a}\qquad\Rightarrow \qquad p=\sqrt{2mE}\sqrt{1-\frac{V_o\sin^2\frac{q\pi}{2a}}{E \cos^2\frac{q\pi}{2a}}} \end{align} Now, $$I= \frac{1}{2\pi}\int{\sqrt{2mE}\sqrt{1-\frac{V_o\sin^2\frac{q\pi}{2a}}{E \cos^2\frac{q\pi}{2a}}}dq}$$ I am stuck with this. How to choose limits? Can we write, \begin{align} &\frac{1}{2\pi}\int{\sqrt{2mE}\sqrt{1-\frac{V_o\sin^2\frac{q\pi}{2a}}{E \cos^2\frac{q\pi}{2a}}}dq}\\ &=\frac{\sqrt{2mE}}{2\pi}\int {\frac{1+\frac{V_o^2}{E^2}}{\sqrt{1-\frac{V_o\sin^2\frac{q\pi}{2a}}{E \cos^2\frac{q\pi}{2a}}}}dq}-\frac{\sqrt{2mE}}{2\pi}\int{\frac{\frac{V_o^2}{E^2\cos^2\frac{q\pi}{2a}}}{\sqrt{1-\frac{V_o\sin^2\frac{q\pi}{2a}}{E \cos^2\frac{q\pi}{2a}}}}dq} \end{align}

The limits of integration are the turning points of the motion, i.e. you need to integrate between $q_1$ and $q_2$ such that $$V_0\tan^2\frac{\pi q_1}{2a}=V_0\tan^2\frac{\pi q_2}{2a}=E\, .$$ The substitution needed to complete the integral is of the type $$q=\arctan\left(\frac{\nu}{\sqrt{A(1+\nu^2)}}\right)\, .$$
• @Icchyamoy I'm sorry but I'm not sure I understand your question. The action integral is $\int pdq$ so you need to find the range of $q$ for the motion, and the motion is bounded by the turning points. To find the turning points set $p=0$ in the expression of the energy (since there is no kinetic energy at the turning points). For given $E$ the turning points $\{q_1,q_2\}$ are solutions to $V_0\tan^2\frac{\pi q}{2a}=E$. The integral then goes $\int_{q_1}^{q_2}$. Oct 24, 2017 at 15:12