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.

  • $\begingroup$ 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, 2015 at 11:51

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.