Skip to main content

The Fourth Virial Coefficient for Hard Spheres in Even Dimension


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.

This is a preview of subscription content, access via your institution.

Fig. 1


  1. [14], Sec. 2.519 Eq. (1), at p. 156.

  2. [14], Sec. 2.512 Eq. (4), at p. 152.

  3. [14], Sec. 2.512 Eqs.(1,2), at p. 152.

  4. [14], Sec. 2.511 Eq. (3), at p. 152.

  5. [14], Sec. 2.511 Eq. (2), at p. 152.

  6. [14], Sec. 2.631 Eq. (2), at p. 214.

  7. [14], Sec. 2.631 Eq. (1), at p. 214.


  1. van der Waals, J.D.: Over de Continuiteit van den Gas-en Vloeistoftoestand (on the continuity of the gas and liquid state). PhD Thesis, University of Leiden (1873)

  2. Clisby, Nathan, McCoy, Barry M.: Analytic calculation of B4 for hard spheres in even dimensions. J. Stat. Phys. 114(5), 1343–1361 (2004)

    ADS  Article  Google Scholar 

  3. Lyberg, I.: The fourth virial coefficient of a fluid of hard spheres in odd dimensions. J. Stat. Phys. 119(3), 747–764 (2005)

    ADS  MathSciNet  Article  Google Scholar 

  4. Boltzmann, L.: About the fourth virial coeff of hs in 3d. Versl. Gewone Vergadering Afk. Natuurk. Nederlandse Akad. Wtensch. 7, 484 (1899)

    Google Scholar 

  5. van Laar, J.J.: About the fourth virial coeff. of hs in 3d. Amsterdam Akad. Versl. 7, 350 (1899)

    Google Scholar 

  6. Nairn, J.H., Kilpatrick, J.E.: Van der Waals, Boltzmann, and the fourth virial coefficient of hard spheres. Am. J. Phys. 40(4), 503–515 (1972)

    ADS  Article  Google Scholar 

  7. Kilpatrick, J.E.: The Computation of Virial Coefficients, pp. 39–69. Wiley, Chichester (1971)

    Google Scholar 

  8. Santos, A.: A Concise Course on the Theory of Classical Liquids: Basics and Selected Topics. Lecture Notes in Physics 923. Springer (2016)

  9. Luban, M., Baram, A.: Third and fourth virial coefficients of hard hyperspheres of arbitrary dimensionality. J. Chem. Phys. 76(6), 3233–3241 (1982)

    ADS  Article  Google Scholar 

  10. Rowlinson, J.S.: The virial expansion in two dimensions. Mol. Phys. 7(6), 593–594 (1964)

    ADS  MathSciNet  Article  Google Scholar 

  11. Hemmer, P.C.: Virial coefficients for the hard-core gas in two dimensions. J. Chem. Phys. 42(3), 1116–1118 (1965)

    ADS  MathSciNet  Article  Google Scholar 

  12. Tonks, Lewi: The complete equation of state of one, two and three-dimensional gases of hard elastic spheres. Phys. Rev. 50(10), 955–963 (1936)

    ADS  Article  Google Scholar 

  13. Joslin, C.G.: Third and fourth virial coefficients of hard hyperspheres of arbitrary dimensionality. J. Chem. Phys. 77(5), 2701–2702 (1982)

    ADS  MathSciNet  Article  Google Scholar 

  14. Gradshteyn, I.S., Ryzhik, I.M.: Table of integrals, series, and products. Academic Press, 7th ed edition (2007)

  15. de Boer, J.: Molecular distribution and equation of state of gases. Rep. Prog. Phys. 12(1), 305 (1949)

    ADS  MathSciNet  Article  Google Scholar 

  16. Nijboer, B.R.A., van Hove, L.: Radial distribution function of a gas of hard spheres and the superposition approximation. Phys. Rev. 85(5), 777–783 (1952)

    ADS  MathSciNet  Article  Google Scholar 

  17. Clisby, N., McCoy, B.M.: Negative virial coefficients and the dominance of loose packed diagrams for d-dimensional hard spheres. J. Stat. Phys. 114, 1361–1392 (2004)

    ADS  MathSciNet  Article  Google Scholar 

  18. Zhang, C., Pettitt, B.M.: Computation of high-order virial coefficients in high-dimensional hard-sphere fluids by Mayer sampling. Mol. Phys. 112(9–10), 1427–1447 (2014)

    ADS  Article  Google Scholar 

Download references


I am grateful to an anonymous reviewer for the suggestion of the Eq. (60) that describes the asymptotic behavior of \(B_{4}\) and to the other anonymous reviewer for his comment about the Conjecture done above.

Author information

Authors and Affiliations


Corresponding author

Correspondence to Ignacio Urrutia.

Additional information

Communicated by Eric A. Carle.

Publisher's Note

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.


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

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

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


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


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


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 useFootnote 5

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

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 applyFootnote 6

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

iteratively, to obtain

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

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 useFootnote 7

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

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

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

Rights and permissions

Reprints and Permissions

About this article

Verify currency and authenticity via CrossMark

Cite this article

Urrutia, I. The Fourth Virial Coefficient for Hard Spheres in Even Dimension. J Stat Phys 187, 29 (2022).

Download citation

  • Received:

  • Accepted:

  • Published:

  • DOI: