# Phase space harmonic oscillator area and probability

I want to find the probability of finding an oscillator between $$x$$ and $$x+dx$$. I calculated the volume $$\frac{8\pi EdE}{\omega^2}$$ enclosed in the phase space for the oscillator with energy between $$E$$ and $$E+dE$$. This is the total volume $$V_{T}$$. Now, I want to get the volume in $$[x,x+dx]\times[E,E+dE]$$ $$V$$. And that should give me the probability of finding the oscillator between $$x$$ and $$x+dx$$ as $$P=\frac{V}{V_{T}}$$. However, I dont know how to calculate $$V$$, the only way I see is by integrating $$\sqrt{1-x^2}$$ which is obtained from the ellipse equation. But that integral gives me arcsines and it gets too complicated. I should get the answer in terms of the energy $$E$$ and the position $$x$$.

To get $$V_{T}$$ I used the fact that the area of an ellipse is $$\pi a b$$ with a and b its semiaxis, and I didnt take into account the quadtratic term $$dE^2$$

If you define it by choosing a random moment in time (uniform distribution over a lon period), and asking for a chance that the oscillator is in a position between $$x$$ and $$x+dx$$, then notice that for an oscillator with a given energy $$E$$ the probability is proportional to the time it takes for it to move from $$x$$ to $$x+dx$$ (or the other way around), that is, inversely proportional to the valocity it has at this position: $$\rho(x|E) \sim \frac{1}{v(x,E)} \sim \frac{1}{\sqrt{x_\text{max}^2(E)-x^2}}$$ $$\rho(x|E)$$ denotes the consitional probability denisty - the probability desnsity that the oscilator is int he point $$x$$ assuming it has energy $$E$$.
To find the normalization constant, we use the fact that $$\int_{-x_\text{max}(E)}^{x_\text{max}(E)} \rho(x|E) dx =1$$ which gives $$\rho(x|E) = \frac{1}{\pi} \frac{1}{\sqrt{x_\text{max}^2(E)-x^2}}$$
If you have some collections of oscilators with some probability density $$\rho(E)$$ of them having a specific energy, you can define the probability density that the osciliator has position $$x$$ AND energy $$E$$ by $$\rho(x,E) = \rho(x|E)\rho(E)$$ or you can average over the collection, obtaing $$\rho(x) = \int\rho(x|E)\rho(E)dE$$
• Yeah, I wouldn't call that equilibrium, but I get the situation now. If they all have the same amplitude and frequency, tand therefore the same total energy, it is the situation I was talking about in the first part of my answer. As I argued, in such case the probability density of any of them having at a particual displacement at given time is inverse proportional to the velocity at the given displacement and equal to $$\rho(x) = \frac{1}{\pi} \frac{1}{\sqrt{x_\text{max}^2-x^2}}$$ where $x_{max}$ is the amplitude of the oscilations. – Adam Latosiński May 6 at 14:03