Classical Canonical Partition Function in Two Dimensions

[Diracobama2181]

Homework Statement

Consider a classical gas in two dimensions trapped inside a circle of radius $R$ and a potential well given by
$$U= \begin{cases}
-U_0 & r<r_0 \\
0 & r\geq r_0 \\
\end{cases}
$$
Assume $$r_0<<R$$.
Find the canonical partition function for this system. Then find the average number of particles in the well and the average potential energy of the particles.

Relevant Equations

$$Z=\frac{1}{h^2}\int e^{-\beta H(p,q)}d^2pd^2q$$
$$H=\frac{P^2}{2m}+U(r)$$

For a single particle,
$$Z=\frac{1}{h^2}\int_{-\infty}^{\infty} e^{-\beta \frac{P^2}{2m}}d^2p \int e^{-U(r)}drd\theta=
\frac{1}{h^2}(\frac{2\pi m}{\beta}) 2\pi [\int_{0}^{r_0}e^{U_0}dr+\int_{r_0}^{R}dr]$$
$$
=\frac{1}{h^2}(\frac{2\pi m}{\beta}) 2\pi [e^{U_0}(r_0)+(R-r_0)]=\frac{\pi R^2}{h^2}(\frac{2\pi m}{\beta})2\pi [e^{U_0}\frac{(r_0)}{\pi R^2}+(\frac{R}{\pi R^2}-\frac{r_0}{\pi R^2})] \approx \frac{1}{h^2}(\frac{2\pi m}{\beta})2\pi R
$$
Thus,
$$Z_{tot}=(\frac{1}{\hbar^2}(\frac{m}{\beta})R)^N$$.
For the average number of particles in the well, I would assume it would be
$$<n>=\frac{N\frac{1}{h^2}\int e^{-\beta(\frac{P^2}{2m}-U_0)}d^2 p d^2q} {Z}=\frac{N\frac{1}{h^2}(\frac{2\pi m}{\beta}) 2 \pi \int_0^{r_0} e^{\beta U_0}dr} {Z}$$
and
$$<U>=\frac{N (-U_0)\frac{1}{h^2}(\frac{2\pi m}{\beta}) 2 \pi \int_0^{r_0} e^{\beta U_0}dr} {Z}$$
Would appreciate if anyone could confirm whether I'm on the right track. Thanks!

 

[Vitani1]

If the potential is zero for r > ro wouldn't that mean that your second integral from ro to R is zero? You integrated "1" when the potential states zero in this interval. Other than this everything else seems fine.

 

[TSny]

For a single particle,
$$Z=\frac{1}{h^2}\int_{-\infty}^{\infty} e^{-\beta \frac{P^2}{2m}}d^2p \int e^{-U(r)}drd\theta$$

You wrote the area element $d^2q$ as $dr d\theta$. But $dr d\theta$ doesn't have the dimensions of area.

Otherwise, I think you're on the right track.

 

[TSny]

If the potential is zero for r > ro wouldn't that mean that your second integral from ro to R is zero? You integrated "1" when the potential states zero in this interval. Other than this everything else seems fine.

Are you referring to the last integral in the following expression for $Z$?

$$Z=\frac{1}{h^2}\int_{-\infty}^{\infty} e^{-\beta \frac{P^2}{2m}}d^2p \int e^{-U(r)}drd\theta=
\frac{1}{h^2}(\frac{2\pi m}{\beta}) 2\pi [\int_{0}^{r_0}e^{U_0}dr+\int_{r_0}^{R}dr]$$

The "1" looks Ok to me. $e^0 = 1$.

 

[Vitani1]

My apologies.