Convergence of Cluster and Virial expansions for Repulsive Classical Gases

Abstract

We study the convergence of cluster and virial expansions for systems of particles subject to positive two-body interactions. Our results strengthen and generalize existing lower bounds on the radii of convergence and on the value of the pressure. Our treatment of the cluster coefficients is based on expressing the truncated weights in terms of trees and partition schemes, and generalize to soft repulsions previous approaches for models with hard exclusions. Our main theorem holds in a very general framework that does not require translation invariance and is applicable to models in general measure spaces. Our virial results, stated only for homogeneous single-space systems, rely on an approach due to Ramawadh and Tate. The virial coefficients are computed using Lagrange inversion techniques but only at the level of formal power series, thereby yielding diagrammatic expressions in terms of trees, rather than the doubly connected diagrams traditionally used. We obtain a new criterion that strengthens, for repulsive interactions, the best criterion previously available (proposed by Groeneveld and proven by Ramawadh and Tate). We illustrate our results with a few applications showing noticeable improvements in the lower bound of convergence radii.

Proof of the Identity (4.9)

For completeness, we present the proof of identity (4.9), which plays crucial role in the proof of Theorem 4.5. The proof, reproduced from [31], uses the formalism of species and its associated combinatorial theory exposed, for instance, in [2]. Let us introduce ${\mathfrak {a}}$ the tree graph species and ${\mathfrak {a}}^\bullet$ the rooted tree graph species.

The exponential generating function of the tree graph species is defined as

$$\begin{aligned} {\mathfrak {a}}(X)=\sum _{n=1}^{\infty }\frac{n^{n-2}}{n!}X^n, \end{aligned}$$

(A.1)

and the exponential generating function of the pointed tree graph species is defined by

$$\begin{aligned} {\mathfrak {a}}^\bullet (X)=X\frac{d}{dX}{\mathfrak {a}}(X)=X{\mathfrak {a}}^{\prime }(X), \end{aligned}$$

(A.2)

and it coincides with the exponential generating function of the rooted tree graph species.

The main tool for our proofs will be the following theorem called The Dissymmetry Theorem for Trees which is presented in [2, Chapter 4, Theorem 1].

Theorem A.1

(The Dissymmetry Theorem for Trees) The species of structure, ${\mathfrak {a}}$, of trees, and ${\mathfrak {a}}^\bullet$, of rooted tree, are related by the natural isomorphism

$$\begin{aligned} {\mathfrak {a}}^{\bullet }+E_2({\mathfrak {a}}^\bullet )={\mathfrak {a}} +[{\mathfrak {a}}^\bullet ]^2 \end{aligned}$$

(A.3)

where $E_2$ denotes the species of sets of cardinality 2.

We restate the dissymmetry theorem for trees in an alternative form as the equation below

$$\begin{aligned} {\mathfrak {a}}(X)={\mathfrak {a}}^\bullet (X)-\frac{1}{2}[{\mathfrak {a}}^\bullet (X)]^2. \end{aligned}$$

(A.4)

Note that,

$$\begin{aligned} {\mathfrak {a}}^\bullet (X)=X \mathrm{e}^{{\mathfrak {a}}^\bullet (X)} \end{aligned}$$

(A.5)

see [2, Chapter 3, Example 3]. Moreover, the identity (A.5) can be rewriten in the form

$$\begin{aligned} {\mathfrak {a}}^{\prime }(X)=\mathrm{e}^{X{\mathfrak {a}}^{\prime }(X)} \end{aligned}$$

(A.6)

Lemma A.2

The following equations hold

$$\begin{aligned} {\mathfrak {a}}^{\bullet \bullet }(X)= & {} \frac{{\mathfrak {a}}^\bullet (X)}{1-{\mathfrak {a}}^{\bullet }(X)} \end{aligned}$$

(A.7)

$$\begin{aligned} {\mathfrak {a}}^{\prime \prime }(X)= & {} \frac{[{\mathfrak {a}}^{\prime }(X)]^2}{1-X{\mathfrak {a}}^\prime (X)}\; \end{aligned}$$

(A.8)

where ${\mathfrak {a}}^{\bullet \bullet }$ is defined as

$$\begin{aligned} {\mathfrak {a}}^{\bullet \bullet }(X):= X\frac{d}{dX}{\mathfrak {a}}^{\bullet }(X) \end{aligned}$$

(A.9)

Proof

We differentiate both side of the equation (A.4) and then multiply both sides by X to obtain

$$\begin{aligned} X\frac{d}{dX}{\mathfrak {a}}(X)=X\frac{d}{dX}{\mathfrak {a}}^\bullet (X) -X{\mathfrak {a}}^\bullet (X)\frac{d}{dX}{\mathfrak {a}}^\bullet (X)\;. \end{aligned}$$

(A.10)

This implies

$$\begin{aligned} {\mathfrak {a}}^{\bullet }(X)={\mathfrak {a}}^{\bullet \bullet }(X) -{\mathfrak {a}}^\bullet (X){\mathfrak {a}}^{\bullet \bullet }(X), \end{aligned}$$

(A.11)

which can be manipulated into the form (A.7).

Taking the derivative of both sides of the equation (A.6) we get

$$\begin{aligned} {\mathfrak {a}}^{\prime \prime }(X)={\mathfrak {a}}^{\prime }(X) \mathrm{e}^{X{\mathfrak {a}}^{\prime }(X)}+X{\mathfrak {a}}^{\prime \prime }(X) \mathrm{e}^{X{\mathfrak {a}}^{\prime }(X)} \end{aligned}$$

(A.12)

Equation (A.6) readily implies that

$$\begin{aligned} {\mathfrak {a}}^{\prime \prime }(X)={\mathfrak {a}}^{\prime } (X){\mathfrak {a}}^{\prime }(X)+X{\mathfrak {a}}^{\prime \prime } (X){\mathfrak {a}}^{\prime }(X) \end{aligned}$$

(A.13)

which can be manipulated into the form (A.8). $\square$

Let us introduce the formal series

$$\begin{aligned} {\mathfrak {a}}^{{{\mathrm{Pen}}},l}(X)=\sum _{n=1}^{\infty } \frac{X^n}{n!}\left| {\mathcal {T}}^0_{{\mathrm{Pen}},l}[n] \right| \;. \end{aligned}$$

(A.14)

As

$$\begin{aligned} \sum _{m=1}^{n-1}\left| {\mathcal {T}}^0_{{\mathrm{Pen}},m}[n-1] \right| =\left| {\mathcal {T}}^0[n-1] \right| =n^{n-2}, \end{aligned}$$

(A.15)

by an argument analogous to that used for the proof of Lemma 7.10 we get

$$\begin{aligned} {\mathfrak {a}}^{{\mathrm{Pen}},1}(X)=1-[{\mathfrak {a}}^{\prime }(X)]^{-1}\;. \end{aligned}$$

(A.16)

We differentiate both sides of (A.16) to obtain

$$\begin{aligned} \frac{d}{dX}{\mathfrak {a}}^{{\mathrm{Pen}},1}(X)=\frac{{\mathfrak {a}}^{\prime \prime }(X)}{[{\mathfrak {a}}^{\prime }(X)]^2}\;. \end{aligned}$$

(A.17)

By Lemma A.2 this implies

$$\begin{aligned} \frac{d}{dX}{\mathfrak {a}}^{{\mathrm{Pen}},1}(X)=\frac{1}{1-X{\mathfrak {a}}^\prime (X)}=\frac{1}{1-{\mathfrak {a}}^\bullet (X)}=1+{\mathfrak {a}}^{\bullet \bullet }(X)=1+\sum _{n=1}^{\infty }n^n\frac{X^{n+1}}{(n+1)!}.\quad \end{aligned}$$

(A.18)

Using the fact that $\mathrm {a}^{{\mathrm{Pen}},1}(0)=0$, we integrate both sides of Equation (A.18) and find:

$$\begin{aligned} {\mathfrak {a}}^{{\mathrm{Pen}},1}(X)=X+\sum _{n=2}^{\infty }(n-1)^{(n-1)}\frac{X^n}{n!} \end{aligned}$$

(A.19)

The proof of the identity (4.9) is completed.