Entropy of disks in a 2d box

[ergospherical]

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?

 

[Philip Koeck]

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.

This doesn't look right to me. The dimensions are inconsistent.

Now I see. You probably mean A = (L - 2r)².

 

[ergospherical]

Yes, it should be $A = (L-2r)^2$.
As for the question itself - I was trying to figure out if there's a simple reason the correction is $2\pi r^2$, or if it just comes out that way...

 

[Philip Koeck]

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

I wouldn't say "large N". N should only be large compared to 1, but not too large.

 

[ergospherical]

Yes, $1 \ll N \ll A_{\mathrm{box}}/(\pi r^2)$

 

[Philip Koeck]

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?

I would say A - (N-1) 4π r² is the area available to any given disk.

Is this a model for something discussed in a textbook?
I assume you got this exercise from a book, or?

 

[Philip Koeck]

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?

Did you get any feedback on your result or even a solution?