Darwin–Fowler method

Classical statistics

For ${\displaystyle N=\sum _i{n}_i}$ independent elements with ${\displaystyle n_i}$ on level with energy ${\displaystyle {\epsilon }_i}$ and ${\displaystyle E=\sum _i{n}_i{\epsilon }_i}$ for a canonical system in a heat bath with temperature ${\displaystyle T}$ we set

${\displaystyle Z=\sum _{\text{arrangements}}{e}^{-E/kT}=\sum _{\text{arrangements}}\prod _i{z}_i^{n_i},z_i=e^{-{\epsilon }_i/kT}.}$

The average over all arrangements is the mean occupation number

${\displaystyle (n_i{)}_{\text{av}}=\frac{\sum _j{n}_{j}Z}{Z}=z_j\frac{\mathrm{\partial }}{\mathrm{\partial }{z}_j}\ln Z.}$

Insert a selector variable ${\displaystyle \omega }$ by setting

${\displaystyle Z_{\omega }=\sum \prod _i(\omega z_i{)}^{n_i}.}$

In classical statistics the ${\displaystyle N}$ elements are (a) distinguishable and can be arranged with packets of ${\displaystyle n_i}$ elements on level ${\displaystyle {\epsilon }_i}$ whose number is

${\displaystyle \frac{N!}{\prod _i{n}_i!},}$

so that in this case

${\displaystyle Z_{\omega }=N!\sum _{n_i}\prod _i\frac{(\omega z_i{)}^{n_i}}{n_i!}.}$

Allowing for (b) the degeneracy ${\displaystyle g_i}$ of level ${\displaystyle {\epsilon }_i}$ this expression becomes

${\displaystyle Z_{\omega }=N!\prod _{i=1}^{\mathrm{\infty }}{\left(\sum _{n_i=0,1,2,\dots }\frac{(\omega z_i{)}^{n_i}}{n_i!}\right)}^{g_i}=N!e^{\omega \sum _i{g}_i{z}_i}.}$

The selector variable ${\displaystyle \omega }$ allows one to pick out the coefficient of ${\displaystyle {\omega }^N}$ which is ${\displaystyle Z}$. Thus

${\displaystyle Z={\left(\sum _i{g}_i{z}_i\right)}^N,}$

and hence

${\displaystyle (n_j{)}_{\text{av}}=z_j\frac{\mathrm{\partial }}{\mathrm{\partial }{z}_j}\ln Z=N\frac{g_j{e}^{-{\epsilon }_j/kT}}{\sum _i{g}_i{e}^{-{\epsilon }_i/kT}}.}$

This result which agrees with the most probable value obtained by maximization does not involve a single approximation and is therefore exact, and thus demonstrates the power of this Darwin–Fowler method.

Quantum statistics

We have as above

${\displaystyle Z_{\omega }=\sum \prod (\omega z_i{)}^{n_i},z_i=e^{-{\epsilon }_i/kT},}$

where ${\displaystyle n_i}$ is the number of elements in energy level ${\displaystyle {\epsilon }_i}$. Since in quantum statistics elements are indistinguishable no preliminary calculation of the number of ways of dividing elements into packets ${\displaystyle n_1,n_2,n_3,...}$ is required. Therefore the sum ${\displaystyle \sum }$ refers only to the sum over possible values of ${\displaystyle n_i}$.

In the case of Fermi–Dirac statistics we have

${\displaystyle n_i=0}$ or ${\displaystyle n_i=1}$

per state. There are ${\displaystyle g_i}$ states for energy level ${\displaystyle {\epsilon }_i}$. Hence we have

${\displaystyle Z_{\omega }=(1+\omega z_1{)}^{g_1}(1+\omega z_2{)}^{g_2}\cdots =\prod (1+\omega z_i{)}^{g_i}.}$

In the case of Bose–Einstein statistics we have

${\displaystyle n_i=0,1,2,3,\dots \mathrm{\infty }.}$

By the same procedure as before we obtain in the present case

${\displaystyle Z_{\omega }=(1+\omega z_1+(\omega z_1{)}^2+(\omega z_1{)}^3+\cdots {)}^{g_1}(1+\omega z_2+(\omega z_2{)}^2+\cdots {)}^{g_2}\cdots .}$

But

${\displaystyle 1+\omega z_1+(\omega z_1{)}^2+\cdots =\frac{1}{(1-\omega z_1)}.}$

Therefore

${\displaystyle Z_{\omega }=\prod _i(1-\omega z_i{)}^{-g_i}.}$

Summarizing both cases and recalling the definition of ${\displaystyle Z}$, we have that ${\displaystyle Z}$ is the coefficient of ${\displaystyle {\omega }^N}$ in

${\displaystyle Z_{\omega }=\prod _i(1\pm \omega z_i{)}^{\pm g_i},}$

where the upper signs apply to Fermi–Dirac statistics, and the lower signs to Bose–Einstein statistics.

Next we have to evaluate the coefficient of ${\displaystyle {\omega }^N}$ in ${\displaystyle Z_{\omega }.}$ In the case of a function ${\displaystyle \varphi (\omega )}$ which can be expanded as

${\displaystyle \varphi (\omega )=a_0+a_1\omega +a_2{\omega }^2+\cdots ,}$

the coefficient of ${\displaystyle {\omega }^N}$ is, with the help of the residue theorem of Cauchy,

${\displaystyle a_N=\frac{1}{2\pi i}\oint \frac{\varphi (\omega )d\omega }{{\omega }^{N+1}}.}$

We note that similarly the coefficient ${\displaystyle Z}$ in the above can be obtained as

${\displaystyle Z=\frac{1}{2\pi i}\oint \frac{Z_{\omega }}{{\omega }^{N+1}}d\omega \equiv \frac{1}{2\pi i}\int e^{f(\omega )}d\omega ,}$

where

${\displaystyle f(\omega )=\pm \sum _i{g}_i\ln (1\pm \omega z_i)-(N+1)\ln \omega .}$

Differentiating one obtains

${\displaystyle f^{\prime }(\omega )=\frac{1}{\omega }\left[\sum _i\frac{g_i}{(\omega z_i{)}^{-1}\pm 1}-(N+1)\right],}$

and

${\displaystyle f^{″}(\omega )=\frac{N+1}{{\omega }^2}\mp \frac{1}{{\omega }^2}\sum _i\frac{g_i}{[(\omega z_i{)}^{-1}\pm 1{]}^2}.}$

One now evaluates the first and second derivatives of ${\displaystyle f(\omega )}$ at the stationary point ${\displaystyle {\omega }_0}$ at which ${\displaystyle f^{\prime }({\omega }_0)=0.}$. This method of evaluation of ${\displaystyle Z}$ around the saddle point ${\displaystyle {\omega }_0}$ is known as the method of steepest descent. One then obtains

${\displaystyle Z=\frac{e^{f({\omega }_0)}}{\sqrt{2\pi f^{″}({\omega }_0)}}.}$

We have ${\displaystyle f^{\prime }({\omega }_0)=0}$ and hence

${\displaystyle (N+1)=\sum _i\frac{g_i}{({\omega }_0{z}_i{)}^{-1}\pm 1}}$

(the +1 being negligible since ${\displaystyle N}$ is large). We shall see in a moment that this last relation is simply the formula

${\displaystyle N=\sum _i{n}_i.}$

We obtain the mean occupation number ${\displaystyle (n_i{)}_{av}}$ by evaluating

${\displaystyle (n_j{)}_{av}=z_j\frac{d}{d{z}_j}\ln Z=\frac{g_j}{({\omega }_0{z}_j{)}^{-1}\pm 1}=\frac{g_j}{e^{({\epsilon }_j-\mu )/kT}\pm 1},e^{\mu /kT}={\omega }_0.}$

This expression gives the mean number of elements of the total of ${\displaystyle N}$ in the volume ${\displaystyle V}$ which occupy at temperature ${\displaystyle T}$ the 1-particle level ${\displaystyle {\epsilon }_j}$ with degeneracy ${\displaystyle g_j}$ (see e.g. a priori probability). For the relation to be reliable one should check that higher order contributions are initially decreasing in magnitude so that the expansion around the saddle point does indeed yield an asymptotic expansion.