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E'lir Kramer
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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?
[itex]{{N}\choose{n}}{{M}\choose{m}}[/itex]
(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
[itex]\frac{(N+M)!}{n!m!(N+M-n-m)!}[/itex]
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)}} \\
[/itex]
[itex]
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)}} \\
[/itex]
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:
[itex]
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}
[/itex]
But since [itex]W_{a} = {{N}\choose{n}}^{2} \\
W_{c} = {{2N}\choose{2n}} \\
[/itex]
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?
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?
[itex]{{N}\choose{n}}{{M}\choose{m}}[/itex]
(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
[itex]\frac{(N+M)!}{n!m!(N+M-n-m)!}[/itex]
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)}} \\
[/itex]
[itex]
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)}} \\
[/itex]
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:
[itex]
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}
[/itex]
But since [itex]W_{a} = {{N}\choose{n}}^{2} \\
W_{c} = {{2N}\choose{2n}} \\
[/itex]
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?
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