General formulation

Recall the expression for the configurational partition function:

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

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

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

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

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

Then, we may express $Z_N$ as

$$Z_N = {Z_N{^{(0)}}\over Z_N{^{(0)}}}\int {d{\bf r}_1\cdots d{\bf r}_N}e^{-\beta U_0({{\bf r}_1,...,{\bf r}_N})} e^{-\beta U_1({{\bf r}_1,...,{\bf r}_N})}$$

$$= Z_N{^{(0)}}\langle e^{-\beta U_1({{\bf r}_1,...,{\bf r}_N})}\rangle_0$$

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 $U_1$:

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

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

The free energy is given by

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

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 $\omega_k$ 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

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

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

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

$$\sum_{k=1}^{\infty}(-1)^{k-1}{1 \over k} \left(\sum_{l=1}^{\... ...\right)^k = \sum_{k=1}^{\infty} (-\beta)^k {\omega_k \over k!}$$

This must be solved for each of the undetermined parameters $\omega_k$, 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 $l=1$ 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:

$${\rm Left\ Side}: -{\beta^3 \over 6}\langle U_1^3 \rangle + ... ... \rangle_0 + {1 \over 2}\beta^2 \langle U_1^2 \rangle\right)^2$$

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

$$A = A{^{(0)}}+ \omega_1 - {\beta \over 2}\omega_2 + {\beta^2 \over 6}\omega_3 \cdots$$

$$= -{1 \over \beta}\ln \left({Z_N{^{(0)}}\over N! \lambda^{3N}}\righ... ...1 \rangle_0 \langle U_1^2 \rangle_0 + 2\langle U_1 \rangle_0^3 \right) + \cdots$$

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,

$$\langle U_1 \rangle_0 = {1 \over Z_N{^{(0)}}} \int {d{\bf r}_1\cdots d{\bf r}_N}{1 \over ... ... j}u_1(\vert{\bf r}_i-{\bf r}_j\vert) e^{-\beta U_0({{\bf r}_1,...,{\bf r}_N})}$$

$$= {N(N-1) \over 2 Z_N{^{(0)}}} \int d{\bf r}_1 d{\bf r}_2 u_1(\vert... ...ert) \int d{\bf r}_3\cdots d{\bf r}_N e^{-\beta U_0({{\bf r}_1,...,{\bf r}_N})}$$

$$= {N^2 \over 2V^2} \int d{\bf r}_1 d{\bf r}_2 u_1(\vert{\bf r}_1-{\bf r}_2\vert) g_0^{(2)}({\bf r}_1,{\bf r}_2)$$

$$= {\rho^2 V \over 2}\int_0^{\infty} 4\pi r^2 u_1(r)g_0(r)dr$$

The free energy is therefore given by

$$A(N,V,T) = -{1 \over \beta}\ln \left({Z_N{^{(0)}}\over N! \l... ...langle U_1^2 \rangle_0 - \langle U_1 \rangle_0^2\right) \cdots$$