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Homework Help: Counting multiplicities of a particle lattice

  1. Feb 20, 2013 #1
    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, [itex]W_{a}[/itex] of the system in state A?


    (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 [itex]W_{b}[/itex] 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


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

    For [itex]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 [itex]2^{2n}[/itex] in the numerator of [itex]W_{b}[/itex], so [itex]\frac{W_{a}}{W_{b}} = \frac{1}{2^{2n}}[/itex]

    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 [itex]{{N}\choose{n}}^{2} < {{2N}\choose{2n}}[/itex].

    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:

    [itex]W_{c} ≈ \frac{(2N)^{2N}}{(2n)^{2n}(2(N-n))^{2(N-n)}}[/itex]

    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 [itex]W_{a} = {{N}\choose{n}}^{2} \\
    W_{c} = {{2N}\choose{2n}} \\
    We have [itex] {{N}\choose{n}}^{2} = {{2N}\choose{2n}} [/itex] for N sufficiently large.

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

    This implies that I've miscounted either [itex]W_{a}[/itex] or [itex]W_{c}[/itex] - could someone show me the way?
    Last edited: Feb 21, 2013
  2. jcsd
  3. Feb 22, 2013 #2


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    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}##
  4. Feb 24, 2013 #3
    Thank you!
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