The Fourth Virial Coefficient for Hard Spheres in Even Dimension

Abstract

The fourth virial coefficient is calculated exactly for a fluid of hard spheres in even dimensions. For this purpose the complete star cluster integral is expressed as the sum of two three-folded integrals only involving spherical angular coordinates. These integrals are solved analytically for any even dimension d, and working with existing expressions for the other terms of the fourth cluster integral, we obtain an expression for the fourth virial coefficient $B_{4}(d)$ for even d. It reduces to the sum of a finite number of simple terms that increases with d.

Appendices

Appendix A: Explicit Form of 

To evaluate this Mayer diagram we transform Eq. (8) as

with $H=\intop _{0}^{\arccos x}(\sin \gamma )^{2\nu }d\gamma$, and also

$$\begin{aligned} H=\frac{(2\nu -1)!!}{2^{\nu }\nu !}\arccos x-\frac{x}{2\nu \sqrt{1-x^{2}}} \left[ \left( 1-x^{2}\right) ^{\nu }+\sum _{k=1}^{\nu -1} \frac{\prod _{i=1}^{k}(2\nu -2i+1)}{2^{k}\prod _{i=1}^{k} (\nu -i)}\left( 1-x^{2}\right) ^{\nu -k}\right] , \end{aligned}$$

where we used Eq. (C1). We integrate by parts to obtain

$$\begin{aligned} \int _{0}^{\frac{1}{2}}x^{2\nu -1}H^{2}dx=\left. \frac{x^{2\nu }}{2\nu } H^{2}\right| _{0}^{\frac{1}{2}}+\int _{0}^{\frac{1}{2}}\frac{x^{2\nu }}{\nu }\left( 1-x^{2}\right) ^{\nu -\frac{1}{2}}H\,dx, \end{aligned}$$

(A1)

where the first term on the right gives $\frac{2^{-2\nu }}{2\nu }\left\{ \frac{(2\nu -1)!!}{2^{\nu }\nu !}\frac{\pi }{3}-\frac{\sqrt{3}}{6\nu } \left[ \left( \frac{3}{4}\right) ^{\nu }+\sum _{k=1}^{\nu -1} \frac{\prod _{i=1}^{k}(2\nu -2i+1)}{2^{k}\prod _{i=1}^{k}(\nu -i)} \left( \frac{3}{4}\right) ^{\nu -k}\right] \right\} ^{2}$. To solve the remaining integral on the right of Eq. (A1) we change variable by replacing $x=\cos y$ to obtain

$$\begin{aligned}&\intop _{0}^{\frac{1}{2}}\frac{x^{2\nu }}{\nu }\left( 1-x^{2}\right) ^{\nu -\frac{1}{2}}H\,dx\\&= \frac{1}{\nu }\intop _{\frac{\pi }{3}}^{\frac{\pi }{2}}\left( \sin y\cos y\right) ^{2\nu }\left\{ \frac{(2\nu -1)!!}{2^{\nu }\nu !}y-\frac{\cos y}{2\nu \sin y}\left[ \left( \sin y\right) ^{2\nu }+\sum _{k=1}^{\nu -1}\frac{\prod _{i=1}^{k}(2\nu -2i+1)}{2^{k} \prod _{i=1}^{k}(\nu -i)}\left( \sin y\right) ^{2(\nu -k)}\right] \right\} dy. \end{aligned}$$

Here, there are terms that can be written using $D\left( 2l+1,2k+1\right)$ analyzed in Eq. (39), and a term $\intop _{\frac{\pi }{3}}^{\frac{\pi }{2}} y\left( \sin y\cos y\right) ^{2\nu }dy$ $=2^{-2\nu -2}\intop _{\frac{2\pi }{3}}^{\pi }y\left( \sin y\right) ^{2\nu }dy$. This last integral is solved using Eq. (C2) and gives

$$\begin{aligned} \intop _{\frac{2\pi }{3}}^{\pi }y\left( \sin y\right) ^{2\nu }dy= & {} \frac{5\pi ^{2}}{18}\frac{(2\nu -1)\text {!!}}{(2\nu )!!}-\frac{\pi \sqrt{3}}{9} \left[ \frac{\left( \frac{3}{4}\right) ^{\nu }}{\nu }+\sum _{k=1}^{\nu -1} \frac{\left( \frac{3}{4}\right) ^{\nu -k}\prod _{i=0}^{k-1}(2\nu -2i-1)}{2^{k} (\nu -k)\prod _{i=0}^{k-1}(\nu -i)}\right] \\&-\frac{1}{4}\left[ \frac{\left( \frac{3}{4}\right) ^{\nu }}{\nu ^{2}}+ \sum _{k=1}^{\nu -1}\frac{\left( \frac{3}{4}\right) ^{\nu -k}\prod _{i=0}^{k-1} (2\nu -2i-1)}{2^{k}(\nu -k)^{2}\prod _{i=0}^{k-1}(\nu -i)}\right] \,. \end{aligned}$$

Thus, one finds

$$\begin{aligned} \intop _{0}^{\frac{1}{2}}x^{2\nu -1}H^{2}dx= & {} \frac{2^{-2\nu }}{2\nu }\biggl \{\frac{(2\nu -1)!!}{(2\nu )!!}\frac{\pi }{3}-\frac{\sqrt{3}}{6\nu }\biggl [\left( \frac{3}{4}\right) ^{\nu }+\sum _{k=1}^{\nu -1}\frac{\prod _{i=1}^{k}(2\nu -2i+1)}{2^{k}\prod _{i=1}^{k}(\nu -i)}\left( \frac{3}{4}\right) ^{\nu -k}\biggr ]\biggr \}^{2}\\&+\frac{(2\nu -1)!!}{2^{2(\nu +1)}\nu (2\nu )!!}\biggl \{\frac{5\pi ^{2}(2\nu -1)\text {!!}}{18(2\nu )!!}-\frac{\pi \sqrt{3}}{9}\biggl [\frac{\left( \frac{3}{4}\right) ^{\nu }}{\nu }\\&+\sum _{k=1}^{\nu -1}\frac{\left( \frac{3}{4}\right) ^{\nu -k}\prod _{i=0}^{k-1}(2\nu -2i-1)}{2^{k}(\nu -k)\prod _{i=0}^{k-1}(\nu -i)}\biggr ]\\&-\frac{1}{4}\biggl [\frac{\left( \frac{3}{4}\right) ^{\nu }}{\nu ^{2}}+\sum _{k=1}^{\nu -1}\frac{\left( \frac{3}{4}\right) ^{\nu -k}\prod _{i=0}^{k-1}(2\nu -2i-1)}{2^{k}(\nu -k)^{2}\prod _{i=0}^{k-1}(\nu -i)}\biggr ]\biggr \}\\&+\frac{1}{2\nu ^{2}}\biggl \{\frac{(2\nu -1)!\,l!}{2((2\nu -1)+l+1)!}+\sum _{k=1}^{\nu -1}\frac{\prod _{i=1}^{k}(2(2\nu -1)-2i+1)}{2^{k}\prod _{i=1}^{k}((2\nu -i)}\\&+D\left( 2\nu +1,4\nu -1\right) +\sum _{k=1}^{\nu -1}\frac{\prod _{i=1}^{k}(2\nu -2i+1)}{2^{k}\prod _{i=1}^{k}(\nu -i)}D\left( 2\nu +1,4\nu -2k-1\right) \biggr \}\,. \end{aligned}$$

The final result is given in Eq. (11).

Appendix B: Origin of $\cos \varphi$ Term

Here we derive the identity Eq. (21) showing the origin of the $\cos \varphi$ term. Let us start from Eq. (17)

we apply the derivative $\frac{\partial }{\partial \sigma }$ to both sides, to obtain

(B1)

Here the dashed bond corresponds to $\delta '(\sigma -r)$, being $f''=\frac{\partial ^{2}f}{\partial ^{2}r}=-\delta '(\sigma -r)$. Some terms on the left are readily integrated on, and . For the diagram with a dashed bond we have

(B2)

We replace Eq. (B2) in Eq. (B1) to obtain

(B3)

Appendix C: Integration of the Different Terms in L

Here we solve the split terms of L (see Eq. (35)), the four integrals: $\left( \sqrt{3}K+\frac{4\pi }{3}\right) \int _{0}^{\frac{\pi }{3}} \sin \left( \varphi \right) ^{2m+2}d\varphi$, $-\left( \frac{\sqrt{3}}{2}K+\frac{2\pi }{3}\right) \int _{0}^{\frac{\pi }{3}}\cos \varphi \left( \sin \varphi \right) ^{2m+2}d\varphi$, $-2\int _{0}^{\frac{\pi }{3}}\varphi \left( \sin \varphi \right) ^{2m+2}d\varphi$, and $\int _{0}^{\frac{\pi }{3}}\varphi \cos \varphi \left( \sin \varphi \right) ^{2m+2}d\varphi$. To solve the first one we use⁵

$$\begin{aligned} \int _{0}^{u}\left( \sin \varphi \right) ^{2l}d\varphi =\frac{(2l-1)!!}{2^{l}l!} u-\frac{\cos u}{2l}\biggl [\left( \sin u\right) ^{2l-1}+\sum _{k=1}^{l-1}\frac{\prod _{i=1}^{k}(2l-2i+1)}{2^{k} \prod _{i=1}^{k}(l-i)}\left( \sin u\right) ^{2l-2k-1}\biggr ], \end{aligned}$$

(C1)

to obtain

$$\begin{aligned} \int _{0}^{\frac{\pi }{3}}\left( \sin \varphi \right) ^{2m+2}d\varphi= & {} \frac{\frac{\pi }{3}(2m+1)!!}{2^{m+1}(m+1)!}-\frac{\sqrt{3}}{8(m+1)} \biggl [\left( \frac{3}{4}\right) ^{m}+\sum _{k=1}^{m}\frac{\prod _{i=0}^{k-1} (2m-2i+1)}{2^{k}\prod _{i=0}^{k-1}(m-i)}\left( \frac{3}{4}\right) ^{m-k}\biggr ]\,. \end{aligned}$$

The second one is straightforward, it gives

$$\begin{aligned} \int _{0}^{\frac{\pi }{3}}\cos \varphi \left( \sin \varphi \right) ^{2m+2}d \varphi =\frac{\sqrt{3}}{4m+6}\left( \frac{3}{4}\right) ^{m+1}\,. \end{aligned}$$

To solve the third one, $-2\int _{0}^{\frac{\pi }{3}}\varphi \left( \sin \varphi \right) ^{2m+2}d\varphi$, we apply⁶

$$\begin{aligned}&\int _{0}^{u}x^{l}\left( \sin x\right) ^{n}dx=\frac{u^{l-1} \left( \sin u\right) ^{n-1}}{n^{2}}\left( l\sin u-n\,u\cos u\right) \nonumber \\&\quad +\frac{n-1}{n}\int _{0}^{u}x^{l}\left( \sin x\right) ^{n-2}dx-\frac{l(l-1)}{n^{2}}\int _{0}^{u}x^{l-2}\left( \sin x\right) ^{n}dx\,, \end{aligned}$$

(C2)

iteratively, to obtain

$$\begin{aligned} \int _{0}^{u}x\left( \sin x\right) ^{2n}dx= & {} \frac{u^{2}(2n-1) \text {!!}}{2\left( 2^{n}n!\right) }+\frac{1}{4}\biggl [\frac{(\sin u)^{2n}}{n^{2}}+ \sum _{k=1}^{n-1}\frac{\prod _{i=0}^{k-1}(2n-2i-1)}{2^{k}(n-k)^{2}\prod _{i=0}^{k-1}(n-i)} (\sin u)^{2(n-k)}\biggr ]\nonumber \\&-\frac{u\cos u}{2\sin u}\biggl [\frac{(\sin u)^{2n}}{n}+\sum _{k=1}^{n-1}\frac{\prod _{i=0}^{k-1}(2n-2i-1)}{2^{k}(n-k) \prod _{i=0}^{k-1}(n-i)}(\sin u)^{2(n-k)}\biggr ], \end{aligned}$$

(C3)

and finally

$$\begin{aligned} \int _{0}^{\frac{\pi }{3}}\varphi \left( \sin \varphi \right) ^{2m+2} d\varphi= & {} \frac{\pi ^{2}(2m+1)\text {!!}}{18(2m+2)!!}+\frac{1}{4} \biggl [\frac{\left( \frac{3}{4}\right) ^{m+1}}{(m+1)^{2}}+\sum _{k=1}^{m} \frac{\left( \frac{3}{4}\right) ^{m-k+1}\prod _{i=0}^{k-1}(2m-2i+1)}{2^{k} (m-k+1)^{2}\prod _{i=0}^{k-1}(m-i+1)}\biggr ]\\&-\frac{\pi \sqrt{3}}{18}\biggl [\frac{\left( \frac{3}{4}\right) ^{m+1}}{m+1}+\sum _{k=1}^{m}\frac{\left( \frac{3}{4}\right) ^{m-k+1}\prod _{i=0}^{k-1} (2m-2i+1)}{2^{k}(m-k+1)\prod _{i=0}^{k-1}(m-i+1)}\biggr ]\,. \end{aligned}$$

To solve the fourth one we use⁷

$$\begin{aligned}&\int _{0}^{u}x\left( \sin x\right) ^{p}\left( \cos x\right) ^{q}\,dx\\&\quad = \frac{1}{\left( p+q\right) ^{2}}\left[ \left( p+q\right) u\left( \sin u\right) ^{p+1} \left( \cos x\right) ^{q-1}+\left( \sin u\right) ^{p}\left( \cos x\right) ^{q}\right. \,,\\&\qquad \left. -p\int _{0}^{u}\left( \sin x\right) ^{p-1}\left( \cos x\right) ^{q-1}dx +\left( q-1\right) \left( p+q\right) \int _{0}^{u}\left( \sin x\right) ^{p} \left( \cos x\right) ^{q-2}dx\right] \,. \end{aligned}$$

In particular, for $q=1$ it reduces to

$$\begin{aligned} \int _{0}^{u}x\left( \sin x\right) ^{p}\cos x\,dx=\frac{1}{(p+1)^{2}} \left[ (p+1)u\left( \sin u\right) ^{p+1}+\left( \sin u\right) ^{p}\cos u-p\int _{0}^{u}\left( \sin x\right) ^{p-1}dx\right] \,, \end{aligned}$$

which gives the result

$$\begin{aligned} \int _{0}^{\frac{\pi }{3}}\varphi \cos \varphi \left( \sin \varphi \right) ^{2m+2} d\varphi =\frac{\pi \sqrt{3}\left( \frac{3}{4}\right) ^{m+1}}{6(2m+3)}+ \frac{\left( \frac{3}{4}\right) ^{m+1}}{2(2m+3)^{2}}-\frac{2(m+1)}{(2m+3)^{2}} \int _{0}^{\pi /3}\left( \sin x\right) ^{2m+1}dx\,. \end{aligned}$$

To solve the last integral we used Eq. (38) (with $q=0$) to obtain

$$\begin{aligned} \int _{0}^{\frac{\pi }{3}}\left( \sin x\right) ^{2n+1}dx=\frac{2^{n}n!}{\left( 2n+1\right) !!}-\frac{1}{2(2n+1)} \biggl [\left( \frac{3}{4}\right) ^{n}+\sum _{k=1}^{n}\frac{2^{k} \prod _{i=0}^{k-1}(n-i)}{\prod _{i=0}^{k-1}(2n-2i-1)} \left( \frac{3}{4}\right) ^{n-k}\biggr ]\,. \end{aligned}$$