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I'm looking at the first part of question 7 here (I'm a mathematician trying to self teach some physics, this isn't a homework assignment so I'm just in need of hints)! I'm struggling to make sense of the set-up, but will explain what I've done.

The number of particles in the system is equal to: \begin{equation*} \int_{p_1}^{p_2}\int_{x_1}^{x_2}f(x,p,t)~dx~dp = \int_{p_1}^{p_2}\int_{x_1}^{x_2}f_1~dx~dp = N \end{equation*} where $[p_1,p_2]$ is the range of values the momentum of each particle takes, and $[x_1,x_2]$ is the range of values the position of each particle takes. To work out these ranges, I compute the equation of motion for a particle in the system - I did this using Hamilton's equations but I'll skip the details because I think it's common knowledge this is simple harmonic motion: \begin{equation*} \ddot{x}=-\omega_1^2x \end{equation*}

which has solution $x=A\cos(\omega_1t)+B\sin(\omega_1t)$ where $A$, $B$ are integration constants. But in order to work out the range of $x$ and $p$ I need to know the value of these integration constants, but don't have enough information, so I'm not sure what to do.

Supposing for a second that I did know $A$ and $B$, would the following answer be correct: $x$ would then take values in $[-\sqrt{A^2+B^2},\sqrt{A^2+B^2}]$, and using $p=m\dot{x}$, we'd have:

\begin{equation*} p=Bm\omega_1\cos{\omega_1t}-Am\omega_1\sin{\omega_1t} \end{equation*}

and hence $p$ would take values in $[-m\omega_1\sqrt{A^2+B^2},m\omega_1\sqrt{A^2+B^2}]$. Substituting these into the integral above and using the fact that $f_1$ is constant gives: \begin{equation*} N=f_1(2\sqrt{A^2+B^2})(2m\omega_1\sqrt{A^2+B^2}) = 4f_1m\omega_1(A^2+B^2) \end{equation*}

I feel this can't be right as $E_1$ doesn't come into it, and given the lecture notes I should probably be using Liouville's Theorem, I'm just not sure where.

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  • $\begingroup$ Typically, $A$ and $B$ are material/problem dependent quantities (e.g., mass of oscillating bob & the spring constant for the simple harmonic oscillator). You probably need to define some (boundary|initial) conditions for your problem. $\endgroup$ – Kyle Kanos Sep 18 '15 at 11:51
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The energy $E$ of an oscillator is given by

$$E=\frac{p^2}{2m}+\frac12m\omega_1^2x^2 $$

This defines an ellipse in phase space! So now, when $E=E_1$ everything within the ellipse defined by $E_1$ will have energy less than $E_1$. To proceed with finding the limits of of integration, we consider the cases when the particles' have all kinetic or all potential energy. So, the maximum momentum $P$ is defined by, $$E_1=\frac{P^2}{2m}$$ and the maximum position $X$ will be defined by $$E_1=\frac12\omega_1^2X^2.$$ This is enough to define an ellipse with its major and minor radii. So, doing a coordinate change will let you integrate. But, the area of an ellipse is well known so maybe you won't even have to integrate because $f_1$ is constant.

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