- #1

ergospherical

- 966

- 1,276

- Homework Statement
- - 2d ideal gas of ##N## identical, hard disks (radius ##r##)

- inside a box of area ##A_{\mathrm{box}} = L^2##

- dilute: ##N \pi r^2 \ll A_{\mathrm{box}}##

- what is the entropy?

- Relevant Equations
- N/A

I'd like to check if my reasoning is right here and that the numerical factors in the final result are correct. The disks occupy an effective area ##A = (A_{\mathrm{box}}-2r)^2##, excluding the region of width ##r## at the boundary. The area available to the ##n##th disk is then ##A_n = A - 4\pi (n-1) r^2##, i.e. excluding the region of width ##r## around each of the ##(n-1)## existing disks.

Consider the zero-energy configuration, in which case the volume of phase space is then$$\Gamma = \frac{1}{N!} A_1 A_2 \dots A_N$$where ##1/N!## arises due to indistinguishability. Making use of Stirling's approximation & using the Taylor expansion of ##\log(1+x) \approx x## a few times,\begin{align*}

\log \Gamma &\approx -(N\log N - N) + \sum_{n=1}^N \log(A - 4\pi(n-1)r^2) \\

&\approx -(N\log N - N) + \sum_{n=1}^N \left\{ \log(A) - \frac{4\pi(n-1)r^2}{A} \right\} \\

&\approx N\left(1 + \log \left(\frac{A}{N}\right) - \frac{2\pi r^2}{A}(N-1) \right) \\

&\approx N\left(1 + \log \left( \frac{A}{N} -2\pi r^2 \frac{N-1}{N} \right)\right)

\end{align*}In the limit of large ##N##, then ##S \approx k_B N \left( 1 + \log\left( \frac{A}{N} - 2\pi r^2\right) \right)##.

About the term ##\frac{A}{N} - 2\pi r^2##: it seems to represent the effective area available to any given particle, with the ##2\pi r^2## an excluded area due to other disks. Is that factor right?

Consider the zero-energy configuration, in which case the volume of phase space is then$$\Gamma = \frac{1}{N!} A_1 A_2 \dots A_N$$where ##1/N!## arises due to indistinguishability. Making use of Stirling's approximation & using the Taylor expansion of ##\log(1+x) \approx x## a few times,\begin{align*}

\log \Gamma &\approx -(N\log N - N) + \sum_{n=1}^N \log(A - 4\pi(n-1)r^2) \\

&\approx -(N\log N - N) + \sum_{n=1}^N \left\{ \log(A) - \frac{4\pi(n-1)r^2}{A} \right\} \\

&\approx N\left(1 + \log \left(\frac{A}{N}\right) - \frac{2\pi r^2}{A}(N-1) \right) \\

&\approx N\left(1 + \log \left( \frac{A}{N} -2\pi r^2 \frac{N-1}{N} \right)\right)

\end{align*}In the limit of large ##N##, then ##S \approx k_B N \left( 1 + \log\left( \frac{A}{N} - 2\pi r^2\right) \right)##.

About the term ##\frac{A}{N} - 2\pi r^2##: it seems to represent the effective area available to any given particle, with the ##2\pi r^2## an excluded area due to other disks. Is that factor right?