# Contour Integration: Action Angle Variables (Mechanics)

1. Oct 27, 2015

### Xyius

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

1. The problem statement, all variables and given/known data

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 $(\phi,J)$

2. Relevant equations
$$J=\frac{1}{2\pi}\oint p dq$$

3. 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!

Last edited: Oct 27, 2015
2. Oct 27, 2015

### TSny

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.

3. Oct 27, 2015

### Xyius

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...

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