# Problem with finding the density of states of an $N$-body system

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.

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

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.
• 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 – Defcon97 Feb 12 at 19:05
• Here $\xi = p^2/2m$, $\eta = Aq^2$. Also due to the symmetry we have $\int d^2 \vec{p} = 2\pi \int_0^\infty dp p$. – Gec Feb 12 at 19:08