# General formulation of Distribution Functions

Recall the expression for the configurational partition function:

$Z_N = \int d{\textbf r}_1\cdots d{\textbf r}_N e^{-\beta U(r_1,...,r_N)}$

Suppose that the potential $$U$$ can be written as a sum of two contributions

$U({{\textbf r}_1,...,{\textbf r}_N}) = U_0({{\textbf r}_1,...,{\textbf r}_N}) + U_1({{\textbf r}_1,...,{\textbf r}_N})$

where $$U_1$$ is, in some sense, small compared to $$U_0$$. An extra bonus can be had if the partition function for $$U_0$$ can be evaluated analytically.

Let

$Z_N{^{(0)}}= \int {d{\textbf r}_1\cdots d{\textbf r}_N}e^{-\beta U_0({r_1,...,r_N})}$

Then, we may express $$Z_N$$ as

$Z_N= {Z_N{^{(0)}}\over Z_N{^{(0)}}}\int d{\textbf r}_1\cdots d{\textbf r}_Ne^{-\beta U_0(r_1,...,r_N)}e^{-\beta U_1(r_1,...,r_N)}$

$= Z_N{^{(0)}}\langle e^{-\beta U_1(r_1,...,r_N)}\rangle_0$

where $$\langle \cdots \rangle _0$$ means average with respect to $$U_0$$ only. If $$U_1$$ is small, then the average can be expanded in powers of $$U_1$$:

$\langle e^{-\beta U_1}\rangle_0 = \underline { 1 - \beta \langle U_1\rangle_0 +{\beta^2 \over 2!} \langle U_1^2 \rangle_0 - {\beta^3 \over 3!}\langle U_1^3 \rangle_0 +\cdots}$

$=\underline { \sum_{k=0}^{\infty} {(-\beta)^k \over k!}\langle U_1^k \rangle_0}$

The free energy is given by

$A(N,V,T) = -{1 \over \beta}\ln \left({Z_N \over N!\lambda^{3N}}\right) = -{1 \over \beta}\ln \left({Z_N^{(0)} \over N!\lambda^{3N}}\right)-- {1 \over \beta}\ln \langle e^{-\beta U_1}\rangle_0$

Separating $$A$$ into two contributions, we have

$A(N,V,T) = A{^{(0)}}(N,V,T) + A{^{(1)}}(N,V,T)$

where $$A^{(0)}$$ is independent of $$U_1$$ and is given by

$A{^{(0)}}(N,V,T) = -{1 \over \beta}\ln \left({Z_N{^{(0)}}\over N!\lambda^{3N}}\right)$

and

 $$A{^{(1)}}(N,V,T)$$ $$-{1 \over \beta}\ln \langle e^{-\beta U_1}\rangle_0$$ $$-{1 \over \beta}\ln \langle \sum_{k=0}^{\infty}{(-\beta)^k \over k!}\langle U_1^k \rangle_0$$

We wish to develop an expansion for $$A^{(1)}$$ of the general form

$A{^{(1)}}= \sum_{k=1}^{\infty} {(-\beta)^{k-1} \over k!}\omega_k$

where $$\underline {\omega _k}$$ are a set of expansion coefficients that are determined by the condition that such an expansion be consistent with $$\ln\langle \sum_{k=0}^{\infty} (-\beta)^k \langle U_1^k\rangle_0 /k!$$.

Using the fact that

$\ln(1+x) = \sum_{k=1}^{\infty} (-1)^{k-1} {x^k \over k}$

we have that

 $$-{1 \over \beta}\ln \left(\sum_{k=0}^{\infty}{(-\beta)^k \over k!} \langle U_1^k \rangle_0\right)$$ $$-{1 \over \beta}\ln \left(1 + \sum_{k=0}^{\infty}{(-\beta)^k \over k!} \langle U_1^k \rangle_0\right)$$ $$\underline { -{1 \over \beta}\sum_{k=1}^{\infty}(-1)^{k-1}{1 \over k}\left(\sum_{l=1}^{\infty}{(-\beta)^l \over l!}\langle U_1^l\rangle_0\right)^k }$$

Equating this expansion to the proposed expansion for $$A^{(1)}$$, we obtain

$\sum_{k=1}^{\infty}(-1)^{k-1}{1 \over k}\left(\sum_{l=1}^{\infty} {(-\beta)^l \over l!} \langle U^l_1 \rangle _0\right)^k = \sum_{k=1}^{\infty} (-\beta)^k {\omega_k \over k!}$
This must be solved for each of the undetermined parameters $$\underline {\omega_k}$$, which can be done by equating like powers of $$\beta$$ on both sides of the equation. Thus, from the $$\beta ^1$$ term, we find, from the right side:

${\rm Right\ Side}:\;\;\;-{\beta \omega_1 \over 1!}$

and from the left side, the $$j = 1$$ and $$k = 1$$ term contributes:

${\rm Left\ Side}:\;\;\;-{\beta \langle U_1 \rangle_0 \over 1!}$

from which it can be easily seen that

$\omega_1 = \langle U_1 \rangle_0$

Likewise, from the $$\beta ^2$$ term,

${\rm Right\ Side}:\;\;\; {\beta^2 \over 2!}\omega_2$

and from the left side, we see that the $$l = 1, k = 2$$ and $$l = 2, k = 1$$ terms contribute:

${\rm Left\ Side}:\;\;\; {\beta^2 \over 2}\left(\langle U_1^2 \rangle_0- \langle U_1 \rangle_0^2\right)$

Thus,

$\omega_2 = \langle U_1^2 \rangle_0 -\langle U_1\rangle_0^2$

For $$\beta ^3$$, the right sides gives:

${\rm Right\ Side}:\;\;\; -{\beta^3 \over 3!}\omega_3$

the left side contributes the $$l = 1, k = 3, k = 2, l = 2$$ and $$l = 3, k = 1$$ terms:

${\rm Left\ Side}: -{\beta^3 \over 6}\langle U_1^3 \rangle + (-1)^2 {1 \over 3}(-\beta \langle U_1\rangle _0 )^3 - {1 \over 2} \left ( -\beta \langle U_1 \rangle _0 + {1 \over 2}\beta^2\langle U_1^2 \rangle\right)^2$

Thus,

$\omega_3 = \langle U_1^3 \rangle_0 + 2\langle U_1 \rangle_0^3- 3\langle U_1 \rangle_0\langle U_1^2 \rangle_0$

Now, the free energy, up to the third order term is given by

 $$A$$ $$\underline { A{^{(0)}}+ \omega_1 - {\beta \over 2}\omega_2 + {\beta^2 \over 6}\omega_3 \cdots}$$ $$-{1 \over \beta}\ln \left({Z_N{^{(0)}}\over N! \lambda^{3N}}\right) + \langle U_1 \rangle_0 - {\beta \over 2} \left \langle U_1^2 \rangle_0 - \langle U_1\rangle _0^2 \right ) + {\beta^2 \over 6} \left (\langle U_1^3 \rangle - 3 \langle U_1 \rangle _0\langle U_1^2 \rangle_0 + 2\langle U_1 \rangle_0^3 \right)+ \cdots$$

In order to evaluate $$\langle U_1 \rangle _0$$, suppose that $$U_1$$ is given by a pair potential

$U_1({{\bf r}_1,...,{\bf r}_N}) = {1 \over 2}\sum_{i\neq j}u_1(\vert{\bf r}_i - {\bf r}_j\vert)$

Then,

 $$\langle U_1 \rangle_0$$ $${1 \over Z_N{^{(0)}}}\int {d{\textbf r}_1\cdots d{\textbf r}_N}{1 \over 2} \sum_{i \ne j} u_1(\vert{\textbf r}_i-{\textbf r}_j\vert)e^{-\beta U_0( r_1,...,r_N)}$$ $$\underline { {N(N-1) \over 2 Z_N{^{(0)}}}\int d{\textbf r}_1 d{text\bf r}_2 u_1(\vert r_1 - r_2 \vert)\int d{\textbf r}_3\cdots d{\textbf r}_Ne^{-\beta U_0({{\bf r}_1,...,{\bf r}_N})} }$$ $$\underline { {N^2 \over 2V^2}\int d{\textbf r}_1 d{\textbf r}_2 u_1(\vert {\textbf r}_1-{\textbf r}_2\vert) g_0^{(2)}({\textbf r}_1,{\textbf r}_2)}$$ $$\underline { {\rho^2 V \over 2}\int_0^{\infty}4\pi r^2 u_1(r)g_0(r)dr }$$

The free energy is therefore given by

$A(N,V,T) = -{1 \over \beta}\ln\left({Z_N^{(0)} \over N! \lambda ^{3N} } \right ) + {1 \over 2} \rho ^2 V \int _0^{\infty} 4 \pi r^2 u_1 (r) g_0 (r) dr - {\beta \over 2} \left ( \langle U_1^2 \rangle_0 - \langle U_1 \rangle_0^2\right)\cdots$