# Entropy of disks in a 2d box

• ergospherical
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?

ergospherical said:
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.

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...

ergospherical said:
&\approx N\left(1 + \log \left( \frac{A}{N} -2\pi r^2 \frac{N-1}{N} \right)\right)
\end{align*}

ergospherical said:
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.

Last edited:
Yes, ##1 \ll N \ll A_{\mathrm{box}}/(\pi r^2)##

ergospherical said:
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?

ergospherical said:
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?

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