Contour Integration: Action Angle Variables (Mechanics)

[Xyius]

I only really need help evaluating the contour integral. I added more detail for completeness.

1. Homework Statement
A particle of mass $m$ is constrained to move on the $x$ axis subject to the following potential.
V(x)=a \sec^2\left( \frac{x}{l} \right)

Find the action angle variables $(\phi,J)$

Homework Equations

J=\frac{1}{2\pi}\oint p dq

The Attempt at a Solution

So I really only need help with the evaluation of the contour integral for J. My background in complex analysis is limited and I believe I must use the residue theorem. I will walk you through what I have.

So first find the lagrangian
$L=T-V=\frac{1}{2}m\dot{x}^2-a\sec^2\left( \frac{x}{l} \right)$
Then find the Hamiltonian using the energy function definition and writing everything in terms of momentum.
$H=\frac{p^2}{2m}+a\sec^2\left( \frac{x}{l} \right)$

Next, use the Hamilton-Jacobi equation to find the generating function (I won't go through this as it is not the focus of my question). From here You can find that $p=\sqrt{2m(X_0-a\sec^2\left( \frac{x}{l} \right))}$ where $X_0$ is the space canonical transformation (acting as a constant for the purposes of the integral next).

So using the equation in section 2 from above, I have the following.

J=\frac{\sqrt{2m}}{2\pi}\oint \sqrt{X_0-a\sec^2\left( \frac{x}{l} \right)}dx

And this is where I am stuck. I believe I need to use the residue theorem, but again my background in complex analysis is limited. So one thing I notice is that because the secant squared is inside the square root, it cannot go to infinity, otherwise the square root becomes complex. $p$ will equal zero when $x=l \arccos{\sqrt{\frac{a}{x}}}$. So if I plot $p$ I can sort of see it's motion but I am not sure. The curve starts on the Y (or p) axis and then dips down and touches the x axis, then it goes imaginary is is periodic from here. I Believe the path in phase space is such that this dip is mirrored about the y-axis and forms an upside-down parabolic looking shape. From this I can clearly see a closed loop path. Two curves segments making the parabolic-looking shape, and a straight segment connection the two when p=0.

If anyone can offer any help I would appreciate it!

 

[TSny]

From here You can find that $p=\sqrt{2m(X_0-a\sec^2\left( \frac{x}{l} \right))}$ where $X_0$

When taking the square root you should allow for plus or minus: $\pm \sqrt{}$.

The p-x trajectory should be qualitatively similar to that for a simple harmonic potential. Both potentials are "bowl" shaped. So, the particle should just oscillate back and forth along the x-axis and the p-x graphs will be qualitatively similar.

For half the cycle, p is positive and for the other half p is negative. There will not be a straight line segment along p = 0 in the p-x graph. A point where p = 0 corresponds to a "turning point" where x is max or min and the momentum is changing from one sign to the other.

 

[Xyius]

When taking the square root you should allow for plus or minus: $\pm \sqrt{}$.

The p-x trajectory should be qualitatively similar to that for a simple harmonic potential. Both potentials are "bowl" shaped. So, the particle should just oscillate back and forth along the x-axis and the p-x graphs will be qualitatively similar.

For half the cycle, p is positive and for the other half p is negative. There will not be a straight line segment along p = 0 in the p-x graph. A point where p = 0 corresponds to a "turning point" where x is max or min and the momentum is changing from one sign to the other.

Ah that makes perfect sense! So now that the path in phase space is determined, how do I do the integration? I have attached a plot I made in MATLAB defining the loop the the particles motion defines in phase space. It would just be the center loop.

My professor said I have to look at the branch points. Not having background in complex analysis, I have done some more reading on branch points and found that a branch point is a point such that when you take a loop around the point, you don't end up in the same location you started at. The example they used was the logarithm and writing it as $\log(R e^{i\theta})=\log(R)+i\theta$. So when $\theta=2\pi$ (corresponding to a full revolution of the path around the point) you don't get the same value. The question is, how can I apply that to this problem?? This is where I am confused...