91Ó°ÊÓ

We need to show that these vanish in the thermodynamic limit.

We have to start with a proof that equation $8.22$contains in itself 8.18 integral $8.22and8.18$are in the diagram form:

If we want to write$8.22$in the integral form up to the third potential we get:

$$\begin{gathered} 1+\frac{1}{2}\frac{N\left(N−1\right)}{V^2}∫d^3{r}_1{d}^3{r}_2{f}_{12}+\frac{1}{2}â‹\dots \frac{1}{4}\frac{N^2{(N−1)}^2}{V^4}∫d^2{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{f}_{12}{f}_{34} \\ +\frac{1}{6}â‹\dots \frac{1}{8}\frac{N^3{(N−1)}^3}{V^6}∫d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5{d}^3{r}_6{f}_{12}{f}_{34}{f}_{56}+… \end{gathered}$$

Now we can use the trick done for simplest diagram (argument of an exp function in $8.22$):

$$\begin{gathered} \frac{1}{2}\frac{N^2}{V^2}∫d^3{r}_1{d}^3{r}_2{f}_{12}=\frac{1}{2}\frac{N^2}{V^2}∫d^3{r}_1∫d^3{r}_2{f}_{12} \\ \left(\begin{array}{c}{f}_{12}=f\left(r_2−r_1\right)=f\left(r\right)\\ r=r_2−r_1\end{array}\right) \\ =\frac{1}{2}\frac{N^2}{V^2}∫d^3{r}_1∫f\left(r\right)dr \\ =\frac{1}{2}\frac{N^2}{V}∫f\left(r\right)dr \end{gathered}$$

Here we used that $f_{12}$is the function of distance $r_1-r_2$where all are vectors, and thus we can make a change of variable$r$to and$f_{12}\left(r_1−r_2\right)=f\left(r\right)$

Let's do similar with our diagram $r=r_3=r_{2}andr\text{'}=r_2=r_1$.

From the square of the simplest diagram in $8.22$.

Here used thermodynamic approximation of $N(N-1)(N-2)≈N^3$disregarding other terms. Also used,

$∫d^3{r}_1=V$

Now, proving that $8.18$is contained in$8.22$however we have to prove that we can disregard other terms.

To prove that other terms disappear we can note that all terms of

$\frac{N\left(N−1\right)\left(N−2\right)…}{V^m}$

have $\raisebox{1ex}{$N^m$}\!\left/ \!\raisebox{-1ex}{$V^m$}\right.$term that we included, and other terms we rejected. Those other terms can be written proportional to:

$\frac{N^n}{V^m}$

and here $n<m$So now in the thermodynamic limit we have $N→\mathrm{∞}$and $V→\mathrm{∞}$while $N/V=const$,so we can write:

$$\begin{gathered} \frac{N^n}{V^m}={\left[\frac{V^m}{N^n}\right]}^{−1}={\left[{\left(\frac{V}{N}\right)}^n{V}^{m−n}\right]}^{−1}={\left[C^n{V}^{m−n}\right]}^{−1} \\ ={\left[\mathrm{∞}\right]}^{−1}→0 \end{gathered}$$

Where we used$m>n$and we took the infinity limit.