Question

Problem 8.8. Show that the $nth$virial coefficient depends on the diagrams in equation 8.23 that have $n$dots. Write the third virial coefficient, $C\left(T\right),$ in terms of an integral of $f-$functions. Why it would be difficult to carry out this integral?

Short answer

The third virial coefficient is$C\left(T\right)=\frac{1}{3!}{\left(\int f\left(r\right)dr\right)}^2.$

Step-by-step solution

Given Information

Virial coefficient $=nth$

Dots denoted by $=n$

Explanation

Let's begin with the definition of the $nth$Virial Coefficient.

$V_n\left(T\right)=-\frac{1}{n!}{\left(\int d^{3}rf\left(r\right)\right)}^n$

If we consider diagrams in 8.23, we have

We need to have another $d^{3r_i}$for each dot and a $f_{ij}$for each link when translating diagrams into integrals.

In this case, the symmetry factor, $N$value, and $V$value are not relevant since they are in front of the integral.

In this example, we can see how every dot gives us one $d^{3r}$integral:

Find the value of:

localid="1651136353483" $=\frac{1}{2}\frac{N(N-1)(N-2)}{V^3}\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{f}_{12}{f}_{23}$

$\begin{array}{rl}& =\frac{1}{2}\frac{N(N-1)(N-2)}{V^3}\int d^3{r}_1{d}^3{r}_2{f}_{12}\int d^{3}rf\left(r\right)\\ & =\frac{1}{2}\frac{N(N-1)(N-2)}{V^3}\int d^3{r}_1\int d^3{r}^{\mathrm{\prime }}f\left(r^{\mathrm{\prime }}\right)\int d^{3}rf\left(r\right)\end{array}$

Explanation

Similarly to the second example, the third virial coefficient $CT$involves one more $f-$link, and so can be calculated by:

$=\frac{1}{2}\frac{N(N-1)(N-2)}{V^3}\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{f}_{12}{f}_{23}{f}_{13}$

$$\begin{gathered} =\frac{1}{2}\frac{N(N-1)(N-2)}{V^3}\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{f}_{12}{f}_{23}{f}_{13} \\ =\frac{1}{2}\frac{N(N-1)(N-2)}{V^3}\int d^3{r}_1{d}^3{r}_2{f}_{12}{f}_{13}\int f\left(r\right)dr,r=r_2-r_3 \\ =\frac{1}{2}\frac{N(N-1)(N-2)}{V^3}\int d^3{r}_1{f}_{13}\int f\left(r^{\mathrm{\prime }}\right)d{r}^{\mathrm{\prime }}\int f\left(r\right)dr,r^{\mathrm{\prime }}=r_2-r_1 \\ =\frac{1}{2}\frac{N(N-1)(N-2)}{V^3}\int d^3{r}_1{f}_{13}C\left(T\right)3! \end{gathered}$$

Therefore, $C\left(T\right)$is obtained by looking at 8.32 (partial by $V$provides a $+$sign, and$2+3$is a factor out of the equation):

We get:

$C\left(T\right)=\frac{1}{3!}{\left(\int f\left(r\right)dr\right)}^2$