General formulation of Distribution Functions
- Page ID
- 5245
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
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
\[ A(N,V,T) = -{1 \over \beta}\ln\left({Z_N^{(0)} \over N! \lambda ^{3N} } \right ) + {1 \over 2} \rho ^2 V \int _0^{\infty} 4 \pi r^2 u_1 (r) g_0 (r) dr - {\beta \over 2} \left ( \langle U_1^2 \rangle_0 - \langle U_1 \rangle_0^2\right)\cdots \]