A Statistical View of Entropy — Question 7

Question 7

Binary composition entropy has a calculable maximum

A large collection has $N$ distinguishable sites, with $Nx$ occupied by species A and $N(1-x)$ by B. For fixed counts, $\Omega=N!/[(Nx)!(N(1-x))!]$. Use the leading Stirling approximation $\ln m!\simeq m\ln m-m$, with both species counts large.

1. Derive the leading configurational entropy per site and evaluate it at $x=.200$ and $x=.500$.

2. Differentiate to locate and classify the maximum. Explain why changing the species fraction changes the number of arrangements at fixed $N$.

Question 7 - Solution

Binary composition entropy has a calculable maximum

(a) Method and calculation

$$\frac{S}{k_B}=\ln N!-\ln(Nx)!-\ln[N(1-x)]!.$$$$\frac{S}{Nk_B}\simeq-x\ln x-(1-x)\ln(1-x)\equiv h(x).$$$$h(.200)\approx.500402,\quad h(.500)=\ln2\approx.693147.$$

(b) Interpretation and further calculation

$$h\prime(x)=\ln\frac{1-x}{x}=0\Rightarrow x=1/2.$$$$h\prime\prime(x)=-1/x-1/(1-x)<0.$$At equal composition there are the most ways to assign the two species to sites. This is a configurational count; any energy differences would require additional statistical weights.

Physical check

The approximation neglects subleading terms in $\ln N$; it is not an exact finite-$N$ entropy formula.