Integration: Action Variable J

[quantumkiko]

Homework Statement

A particle of mass m moves in one dimension subject to the potential

$$V = \frac{a}{\sin^2(x/x_0)}$$

Obtain an integral expression for Hamilton's characteristic function. Under what conditions can action-angle variables be used? Assuming these are met, find the frequency of oscillation by the action-angle method. (The integral for J can be evaluated by manipulating the integrand so that the square root appears in the denominator.)

2. The relevant equations

I set up a Hamiltonian

$$H = \frac{p^2}{2m} + \frac{a}{\sin^2(x/x_0)}$$.

Since it doesn't depend on time explicitly, then I can express Hamilton's principal function S as

$$S = W(x, \alpha) + \alpha t$$.

The Hamilton-Jacobi equation is then given by

$$\frac{1}{2m} \left( \frac{\partial W}{\partial x} \right)^2 + \frac{a}{\sin^2(x/x_0)} = \alpha$$

The Attempt at a Solution

a.) Obtain an integral solution for Hamilton's characteristic function.

Isolating the partial derivative of the characteristic function W, I get an integral expression

$$W = \int \sqrt{2m\alpha - \frac{2ma}{\sin^2(x/x_0)}} dx$$.

b.) Under what conditions can action-angle variables be used?

The condition that must be met is $$\alpha \geq \frac{a}{\sin^2(x/x_0)}}$$.

c.) Assuming these are met, find the frequency of oscillation by the action-angle method. (The integral for J can be evaluated by manipulating the integrand so that the square root appears in the denominator.)

For finding the frequency of oscillation, I'm having a problem with integrating the action variable J:

$$J = \int p dx = \int \sqrt{2m\alpha - \frac{2ma}{\sin^2(x/x_0)}} dx$$.

I tried to follow the hint, so I got

$$J = \int \frac{2m\alpha - \frac{2ma}{\sin^2(x/x_0)}}{\sqrt{2m\alpha - \frac{2ma}{\sin^2(x/x_0)}}} dx$$.

Still, I'm stuck. How do I integrate this? Thanks!

 

[Jhero]

Thank you for your question. I am happy to assist you in finding a solution to your problem.

To find the frequency of oscillation using the action-angle method, we first need to find the action variable J. In order to integrate the expression you have, we can use the substitution u = x/x0. This will give us the following integral:

J = \int \frac{2m\alpha - \frac{2ma}{\sin^2(u)}}{\sqrt{2m\alpha - \frac{2ma}{\sin^2(u)}}} x0 du.

Next, we can use the trigonometric identity \sin^2(u) = \frac{1}{2}(1 - \cos(2u)) to simplify the integral:

J = \int \frac{2m\alpha - \frac{2ma}{\frac{1}{2}(1 - \cos(2u))}}{\sqrt{2m\alpha - \frac{2ma}{\frac{1}{2}(1 - \cos(2u))}}} x0 du.

Now, we can use the substitution v = \cos(2u) to simplify the integral even further:

J = \int \frac{2m\alpha - \frac{4ma}{1 - v}}{\sqrt{2m\alpha - \frac{4ma}{1 - v}}} x0 \frac{dv}{-2\sin(u)}.

Simplifying this, we get:

J = \int \frac{2m\alpha - \frac{4ma}{1 - v}}{\sqrt{2m\alpha - \frac{4ma}{1 - v}}} x0 \frac{-2dv}{\sqrt{1 - v^2}}.

Using the substitution w = \sqrt{2m\alpha - \frac{4ma}{1 - v}}, we can finally solve the integral:

J = \int \frac{-2x0dw}{\sqrt{w^2 - (2m\alpha - \frac{4ma}{1 - v})}}.

This integral can be evaluated using standard techniques, and once we have the value of J, we can find the frequency of oscillation using the formula:

\omega = \frac{dH}{dJ} = \frac{1}{2m}\frac{d}{dJ