hamiltonian formalism - Evaluating limits of action angle problems

I am really troubled with finding the limits in "action-angle integral" problems. It is said that the limit is taken over generalised coordinate $q$ such that we have a complete liberation or rotation in the $p$ vs $q$ space. But how can we get this limit?

considering a particular problem, let's say $V(x)=F|x|$ is given. Then the variable $J$ the is defined as $J= \int_a^bdx({2mE-2mF|x|})^{1/2}$ where E is a constant.

How do I evaluate $a$ and $b$ now? Is there a general scheme that we can use for such problems?

Answer

In general start with

$$E=\frac{p^2}{2m}+V(x)\, . \tag{1}$$ For a given $E$ the turning points of the motion $x_\pm$ are at found when $V(x_\pm)=E$ since, at the turning points, there is no kinetic energy (the momentum $p=0$). The turning points define the boundaries of your motion and thus your integration limits.

Reorganize (1) into

$$p=\pm\sqrt{2m(E-V(x))}\,$$ and integrate. Because of the sign change in $p$ the integration over a full cycle ought to be broken into a part where $p>0$ and a part where $p<0$. It shouldn’t be too hard to justify that $$J=2\int_{x_-}^{x_+} \sqrt{2m(E-V(x))}dx\, ,$$ so it’ s just a job of finding $x_\pm$ for your specific potential.

[Nota: your potential is $k\vert x\vert$ but your integral has instead $F\vert x\vert$. I presume there’s a typo somewhere]