91Ó°ÊÓ

We need to find the necessary to rewrite the series in exponential form.

In this problem we need to write size of the diagrams in equation , the diagrams are shown in the following figure, for simplicity I will indicate each equation by its number in the figure. The integral for these diagrams are:

$$\begin{gathered} \left(1\right)=\frac{1}{2}\frac{N\left(N−1\right)}{V^2}∫d^3{r}_1{d}^3{r}_2{f}_{12} \\ \left(2\right)=\frac{1}{2}\frac{N\left(N−1\right)\left(N−2\right)}{V^3}∫d^3{r}_1{d}^3{r}_2{d}^3{r}_3{f}_{12}{f}_{23} \\ \left(3\right)=\frac{1}{8}\frac{N\left(N−1\right)\left(N−2\right)\left(N−3\right)}{V^4}∫d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{f}_{12}{f}_{34} \\ \left(4\right)=\frac{1}{6}\frac{N\left(N−1\right)\left(N−2\right)}{V^3}∫d^3{r}_1{d}^3{r}_2{d}^3{r}_3{f}_{12}{f}_{23}{f}_{31} \\ \left(5\right)=\frac{1}{6}\frac{N\left(N−1\right)\left(N−2\right)\left(N−3\right)}{V^4}∫d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{f}_{14}{f}_{24}{f}_{34} \\ \left(6\right)=\frac{1}{4}\frac{N\left(N−1\right)\left(N−2\right)\left(N−3\right)\left(N−4\right)}{V^5}∫d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5{f}_{12}{f}_{34}{f}_{45} \\ \left(7\right)=\frac{1}{8}\frac{N\left(N−1\right)\left(N−2\right)\left(N−3\right)\left(N−4\right)\left(N\right)}{V^6} \\ ×∫d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5{d}^3{r}_6{f}_{12}{f}_{34}{f}_{56} \\ \left(8\right)=\frac{1}{6}\frac{N\left(N−1\right)\left(N−2\right)\left(N−3\right)}{V^4}∫d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{f}_{13}{f}_{12}{f}_{24} \end{gathered}$$

according to the above equation, we can write (realizing that $f\left(r\right)$is in order of $1$out of the distance about the diameter of the molecule and $f≈0$beyond that):

$\left(1\right)≈\frac{Nv}{V}$

what I have done here is just approximate $N≈N-1$, and since we have only one factor of we have one factor of $f$where is the sized volume of the molecule and the integration over the renaming factor of $r$gives the volume so the volume in the denominator reduced by one order. For the second diagram we get:

$\left(2\right)≈\frac{N^3{v}^2}{V^2}$

and so on, we can write:

$\left(3\right)≈\frac{N^4{v}^3}{V^3}$

For the fourth diagram, we need to change the coordinates of the diagram as follow:

$$\begin{gathered} {\stackrel{→}{r}}_a={\stackrel{→}{r}}_1−\stackrel{→}{r} \\ {\stackrel{→}{r}}_b={\stackrel{→}{r}}_1−{\stackrel{→}{r}}_3 \end{gathered}$$

so we can write:

$$\begin{gathered} \left(4\right)=\frac{1}{6}\frac{N^3}{V^2}∫d^3{r}_a{d}^3{r}_{b}f\left(r_a\right)f\left(r_b\right)f\left({\stackrel{→}{r}}_a−{\stackrel{→}{r}}_b\right) \\ \left(4\right)≈\frac{N^3{v}^2}{V^2} \end{gathered}$$

Following the same method above, we can write the size for the rest of the diagram as:

$$\begin{gathered} \left(5\right)≈\frac{N^4{v}^3}{V^3}\left(6\right)≈\frac{N^5{v}^3}{V^3} \\ \left(7\right)≈\frac{N^6{v}^2}{V^2}\left(8\right)≈\frac{N^4{v}^3}{V^3} \end{gathered}$$