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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 $dE^2$

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The probability doesn't have to be uniform in the phase space. You need to be careful about how you deine it.

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 $$ \rho(x) = \int\rho(x|E)\rho(E)dE$$

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  • $\begingroup$ I'm allowed to suppose an uniform distribution on the phase because the system is isolated and in equilibrium, so I'm using the principle of equal probabilities a priori $\endgroup$ – Juan Pablo Arcila May 5 at 22:55
  • $\begingroup$ The concept of quilibrium doesn't make sense if the don't interact with each other. Do all of them have the same energy, or is this system of oscillators in constant temperature, or is that something else? What kind of equilibrium is it? $\endgroup$ – Adam Latosiński May 6 at 5:16
  • $\begingroup$ well, because all of them move with the same amplitude and frequency, their energy must be the same, in that sense, they would be in equilibrium? I hadn't asked to myself what kind of equilibrium was it $\endgroup$ – Juan Pablo Arcila May 6 at 13:52
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    $\begingroup$ 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. $\endgroup$ – Adam Latosiński May 6 at 14:03

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