# Homework Help: Counting multiplicities of a particle lattice

1. Feb 20, 2013

### E'lir Kramer

This is from Molecular Driving Forces, 2nd Ed.

5.3: Calculating the entropy of mixing. Consider a lattice with N sites and n green particles, and another lattice with M sites and m red particles. These lattices cannot exchange particles. This is state A.

(a) What is the total number of configurations, $W_{a}$ of the system in state A?

${{N}\choose{n}}{{M}\choose{m}}$

(b) Now assume that all of the N+M sites are available to all the green and red particles. The particles remain distinguishable by their color. This is state B. What is the total configurations $W_{b}$ of the system?

I'm counting it like this: we have three species: red balls, green balls, and empty spaces, with a population of n, m, and (N+M-n-m) respectively. There are N+M total sites, so we have

$\frac{(N+M)!}{n!m!(N+M-n-m)!}$

Now take N =M and n =m for the following two problems:
(c) Using Stirling's approximation, what is the ratio of $\frac{W_{a}}{W_{b}}$?

For $W_{a} = {{N}\choose{n}}{{N}\choose{n}} \\ = \left(\frac{N!}{n!(N-n)!}\right)^{2} \\ ≈ \left(\frac{\left(\frac{N}{e}\right)^{N}}{\left( \frac{n}{e}\right)^{n}\left(\frac{N-n}{e}\right)^{(N-n)}}\right)^{2} \\ = \left(\frac{N^{N}}{n^{n} (N-n)^{(N-n)}}\right)^{2} \\ = \frac{N^{2N}}{n^{2n} (N-n)^{2(N-n)}} \\$

$W_{b} = \frac{(2N)!}{n!n!((2(N-n))!} \\ ≈ \frac{(2N)^{2N}}{n^{2n}(2(N-n))^{2(N-n)}} \\ = \frac{2^{2N}N^{2N}}{2^{2(N-n)}n^{2n}(N-n)^{2(N-n)}} \\ = \frac{2^{2n}N^{2N}}{n^{2n}(N-n)^{2(N-n)}} \\$

These are the same except for a factor of $2^{2n}$ in the numerator of $W_{b}$, so $\frac{W_{a}}{W_{b}} = \frac{1}{2^{2n}}$

Now while I was trying to figure out how to count a system of red balls and green balls, I started by trying to count a simpler model: a system of size 2N with 2n indistinguishable balls. I assumed that this system would have a higher multiplicity than two systems of size N with n balls each.

In other words, I assumed that ${{N}\choose{n}}^{2} < {{2N}\choose{2n}}$.

I reasoned that the multiplicity of my new system of 2n indistinguishable balls, call it state C, would be approximated by Sterling's approximation like this:

$W_{c} ≈ \frac{(2N)^{2N}}{(2n)^{2n}(2(N-n))^{2(N-n)}}$

But after manipulation, I found that:

$W_{c} ≈ \frac{(2N)^{2N}}{(2n)^{2n}(2(N-n))^{2(N-n)}} \\ = \frac{2^{2N}N^{2N}}{2^{2n}2^{2(N-n)}n^{2n}(N-n)^{2(N-n)}} \\ = \frac{N^{2N}}{n^{2n}(N-n)^{2(N-n)}} \\ = W_{a}$

But since $W_{a} = {{N}\choose{n}}^{2} \\ W_{c} = {{2N}\choose{2n}} \\$
We have ${{N}\choose{n}}^{2} = {{2N}\choose{2n}}$ for N sufficiently large.

I have confirmed with code that $\frac{lim}{N \to ∞} ({{N}\choose{\frac{N}{2}}}^{2} / {{2N}\choose{N}} ) = 0$, so ${{N}\choose{\frac{N}{2}}}^{2} ≠ {{2N}\choose{N}}$

This implies that I've miscounted either $W_{a}$ or $W_{c}$ - could someone show me the way?

Last edited: Feb 21, 2013
2. Feb 22, 2013

### haruspex

You have to use a more accurate approximation for n!. You left out the √(2πn) part.
Suppose we write p = n/N. Then ${{N}\choose{n}} ≈ \left(p^p\left(1-p\right)^{1-p}\right)^{-N-\frac12} \left(2\pi N\right)^{-\frac12}$
Writing $r = p^p\left(1-p\right)^{1-p}$, ${{N}\choose{n}} ≈ r^{-N} \left(2\pi r N\right)^{-\frac12}$
So ${{N}\choose{n}}^2 ≈ r^{-2N} \left(2\pi r N\right)^{-1}$, ${{2N}\choose{2n}} ≈ r^{-2N} \left(4\pi r N\right)^{-\frac12}$. The ratio of these is $\left(\pi r N\right)^{\frac12}$

3. Feb 24, 2013

Thank you!