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I am having problems solving a particular problem in my Statistical Mechanics course.

We have a system that is composed of $N$ non-interacting particles each of mass $m$. The particles are bound to move on a 2-dimensional disk of radius $R$. The Hamiltonian function for the single particle is: $$ \mathcal{H}(\vec{p}, \vec{q}) = \frac{p^2}{2m} + Aq^2 $$ Where $q$ and $p$ are the lenght of the vectors $\vec{q}$ and $\vec{p}$ and $A$ is a positive constant.

Assuming that the system is in contact with a Thermal reservoir at temperature $T$ and that Boltzmann's classical statistics can be used calculate the density of probability $p(\epsilon)$ for the energy of a single particle.

I have been trying to solve this but all i could find was that this function $p(\epsilon)$ has to be:

\begin{equation*} p(\epsilon) = \frac{e^{-\beta\epsilon}}{Z_1(T,V)}G(\epsilon) \end{equation*} Where $Z_1$ is tha partition function, which i managed to find quite easily, and $G(\epsilon)$ is the density of states, which i wasn't able to find a way to compute. I thank you in advance for all the help you can give me.

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    $\begingroup$ look for the density of states of harmonic oscillator. For the 3D case, here is a relevant StackExchange post: physics.stackexchange.com/q/185501 $\endgroup$
    – wcc
    Commented Feb 12, 2019 at 17:09

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The density of states of one particle in 2D is $$ G(\varepsilon) = \int\!\!\int d^2\vec{p}\ d^2\vec{q}\ \delta\left(\varepsilon - H(\vec{p},\vec{q})\right) $$ For the given Hamiltonian function this integral can be transformed to the following expression $$ G(\varepsilon) = \frac{2\pi^2m}{A}\int_0^\infty\!\! d\xi \int_0^{AR^2}\!\! d\eta\ \delta(\varepsilon - \xi - \eta). $$ The last one can be calculated in analytic form.

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  • $\begingroup$ Thank You! Can you explain to me which sustitutions you have made to get there. In particular i would lime to understand what those $\xi$ and $\eta$ variables stand for $\endgroup$
    – Defcon97
    Commented Feb 12, 2019 at 19:05
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    $\begingroup$ Here $\xi = p^2/2m$, $\eta = Aq^2$. Also due to the symmetry we have $\int d^2 \vec{p} = 2\pi \int_0^\infty dp p$. $\endgroup$
    – Gec
    Commented Feb 12, 2019 at 19:08

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