A Correction to a Remark in a Paper by Procacci and Yuhjtman: New Lower Bounds for the Convergence Radius of the Virial Series

Abstract

In this note we deduce a new lower bound for the convergence radius of the Virial series of a continuous system of classical particles interacting via a stable and tempered pair potential using the estimates on the Mayer coefficients obtained in the recent paper by Procacci and Yuhjtman (Lett Math Phys 107:31–46, 2017). This corrects the wrongly optimistic lower bound for the same radius claimed (but not proved) in the above cited paper (in Remark 2 below Theorem 1). The lower bound for the convergence radius of the Virial series provided here represents a strong improvement on the classical estimate given by Lebowitz and Penrose in 1964.

Appendices

Appendix A: Proof of (2.7)

Estimate (2.7) is basically the bound (1.2) proved in [7] with the unique difference that the factor $e^{\beta \bar{B}(n-1)}$ replaces the factor $e^{\beta B n}$. Of course, since $\bar{B}\ge B$, this is a bad deal for someone interested in an upper bound the Mayer series. On the other hand, as shown in Sec. 3, the use (2.7) instead of (1.2) happens to be a good deal towards an upper bound the Virial series. Inequality (2.7) relies on two lemmas originally proved in [7] (see there Lemma 1 and Lemma 2). For completeness we report these lemmas and their proofs here below.

The first of these two lemmas involves the concept of partition scheme, which, we remind, is a map $\varvec{M}$ from the set $T_n$ of the labeled trees in [n] to the set $G_n$ of the connected graphs in [n] such that $G_n=\biguplus _{\tau \in {T_n}}[\tau ,{\varvec{M}}(\tau )]$ with $\biguplus$ disjoint union and $[\tau ,\varvec{M}(\tau )]=\{g\in G_n: \tau \subseteq g\subseteq \varvec{M}(\tau )\}$.

Lemma 1

For fixed V and $(x_1,\dots , x_n)\in \mathbb {R}^{dn}$, choose a total order $\succ$ in the set of edges $E_n$ of the complete graph $K_n$ in such a way that $\{i,j\}\succ \{k,l\}\Longrightarrow V(x_i-x_j)\ge V(x_k-x_l)$ and let $\varvec{T}:G_n\rightarrow T_n$ be the map that associates to $g\in G_n$ the tree $\varvec{T}(g)\in T_n$ constructed by starting from $\emptyset$ and keeping adding the lowest edge in g that does not create a cycle (Kruskal algorithm).

Let $\varvec{M}: T_n\rightarrow G_n$ be the map that associates to $\tau \in T_n$ the graph $\varvec{M}( \tau )\in G_n$ whose edges are the $\{i,j\}\in E_n$ such that $\{i,j\} \succeq \{k,l\}$ for every edge $\{k,l\} \in E_\tau$ belonging to the path from i to j through $\tau$.

Then $\varvec{T}^{-1}(\tau )=\{g \in {\mathcal G}_n :\, \tau \subseteq g \subseteq \varvec{M}(\tau )\}$ and therefore $\varvec{M}$ is a partition scheme in ${G}_n$.

Proof

Assume first that $g \in \varvec{T}^{-1}(\tau )$. Then $\tau =\varvec{T}(g) \subset g$. Now take $\{i,j\} \in E_g$, and let $e \in E_\tau$ be any edge belonging to the path from i to j in $\tau$. Consider the tree $\tau '$ obtained from $\tau$ after replacing the edge e by $\{i,j\}$. By minimality of $\tau$ we must have $\{i,j\}\succ e$, i.e. $\{i,j\} \in E_{\varvec{M}(\tau )}$, whence $g \subset \varvec{M}(\tau )$. Conversely, let $\tau \subset g \subset \varvec{M}(\tau )$. We must show $\varvec{T}(g)=\tau$. By contradiction, take $\{i,j\} \in E_{\varvec{T}(g)} \setminus E_\tau$. Consider the path $p^\tau (\{i,j\})$ in $\tau$ joining i with j. Since $\varvec{T}(g) \subset \varvec{M}(\tau )$, $\{i,j\}$ is greater (w.r.t. $\succ$) than any edge in the path $p^\tau (\{i,j\})$. If we remove $\{i,j\}$ from $\varvec{T}(g)$, the tree $\varvec{T}(g)$ splits into two trees. Necessarily, one of the edges in the path $p^\tau (\{i,j\})$ joins a vertex of one tree with a vertex of the other. Thus, by adding this edge we get a new tree which contradicts the minimality of $\varvec{T}(g)$. $\square$

Remark

The proof of Lemma 1 above, as well as the definition of the partition scheme $\varvec{M}$, are identical to those given in the homonymous lemma of [7], but the map $\varvec{T}: G_n\rightarrow T_n$ appearing in the enunciate, based on Kruskal’s algorithm, replaces a similar (but more involved) map constructed in [7] via so-called admissible functions with values in a tomonoid. The idea to use the Kruskal’s algorithm, which eases the definition of the “minimal tree” map $\varvec{T}$, has been suggested during an Oberwolfach meeting by David Brydges, Tyler Helmuth and Daniel Ueltschi (see [10]).

Using the partition scheme $\varvec{M}$ defined in Lemma 1 we now state and prove the second key lemma of [7].

Lemma 2

Let V be stable with Basuev stability constant $\bar{B}$, and let $\tau \in { T}_n$. Let $(x_1,...,x_n) \in {\mathbb R}^{dn}$ and let $\varvec{M}$ be the partition scheme given above, then

$$\begin{aligned} \sum _{\{i,j\} \in \varvec{E}_{M(\tau )} \setminus E_\tau ^+} v(x_i-x_j) \ge -\bar{B}(n-1) \end{aligned}$$

(2.20)

Proof

. The set of edges $E_\tau \setminus E_\tau ^+$ forms the forest $\{\tau _1,...,\tau _k\}$. Let us denote ${ V}_{\tau _s}$ the vertex set of the tree $\tau _s$ from the forest. Assume $i \in { V}_{\tau _a}$, $j \in {V}_{\tau _b}$. If $a \ne b$, the path from i to j through $\tau$ involves an edge $\{k,l\}$ in $E_\tau ^+$. Thus, if in addition $\{i.j\} \in E_{\varvec{M}(\tau )}$, we have $\{i,j\}\succeq \{k,l\}$ and therefore $v(x_i-x_j) \ge v(x_k-x_l) \ge 0$. If $a=b$, the path from i to j through $\tau$ is contained in $\tau _a$. Thus, if in addition $\{i,j\} \notin E_{\varvec{M}(\tau )}$, there must be at least one edge $\{r,s\}$ in that path such that $\{i,j\}\prec \{r,s\}$ and therefore $v(x_i-x_j) \le v(x_r-x_s) < 0$. This allows to bound:

$$\begin{aligned} \sum _{\{i,j\} \in E_{\varvec{M}(\tau )} \setminus E_\tau ^+}v(x_i-x_j) \ge \sum _{s=1}^k \sum _{\{i,j\} \subset { V}_{\tau _s}} v(x_i-x_j) \ge \sum _{s=1}^k -| (V_{\tau _s}|-1)\bar{B} \ge -(n-1)\bar{B} \end{aligned}$$

where to get the last inequality we have used (2.4). $\square$

We are now ready to conclude the proof of (2.7). By the so-called Penrose tree-graph identity [6], given a pair potential V and $(x_1,...,x_n) \in {\mathbb R}^{dn}$, one has that

$$\begin{aligned} \sum _{g\in G_{n}}~ \prod _{\{i,j\}\in E_g}\left[ e^{ -\beta V(x_i-x_j)} -1\right] = \sum _{\tau \in T_n} e^{-\beta \sum _{\{i,j\}\in E_{\varvec{M}(\tau )}\backslash E_\tau }V(x_i-x_j)} \prod _{\{i,j\}\in E_\tau }\left( e^{- \beta V(x_i-x_j)}-1\right) \end{aligned}$$

(2.21)

where $\varvec{M}: T_n\rightarrow G_n$ is any map such that $G_n=\biguplus _{\tau \in {T_n}}[\tau ,{\varvec{M}}(\tau )]$ with $\biguplus$ disjoint union and $[\tau ,\varvec{M}(\tau )]=\{g\in G_n: \tau \subseteq g\subseteq \varvec{M}(\tau )\}$ (partition scheme) and ${ T}_n$ is the set of all trees with vertex set [n].

Set now $E_\tau ^+=\{\{i,j\}\in E_\tau :V(x_i-x_j)\ge 0\}_{}$, then from (2.21) one immediately gets

$$\begin{aligned}&\left| \sum \limits _{g\in G_{n}}~ \prod \limits _{\{i,j\}\in E_g}\left[ e^{ -\beta V(x_i-x_j)} -1\right] \right| \nonumber \\&\quad ~\le ~ \sum _{\tau \in T_n} e^{-\beta \sum \limits _{\{i,j\}\in E_{\varvec{M}(\tau )}\backslash E^+_\tau }V(x_i-x_j)} \prod _{\{i,j\}\in E_\tau }\left( 1-e^{- \beta |V(x_i-x_j)|}\right) \end{aligned}$$

(2.22)

Let us now choose the partition scheme defined in Lemma 1. Then from Lemma 2, inserting (2.20) into (2.22), one obtains, for any $n\ge 2$ and any $(x_1,\dots ,x_n)\in \mathbb {R}^{dn}$, the following inequality

$$\begin{aligned} \left| \sum _{g \in {G}_n} \prod _{\{i,j\} \in E_g} (e^{-\beta V(x_i-x_j)}-1)\right| ~\le ~ e^{\beta \bar{B} (n-1)}\sum _{\tau \in {T}_n} \prod _{\{i,j\} \in E_\tau } (1-e^{-\beta | V(x_i-x_j)|}) \end{aligned}$$

(2.23)

Now (2.7) follows easily from (2.23) recalling that $|T_n|=n^{n-2}$ and observing that, for any $\tau \in T_n$, it holds

$$\begin{aligned} \int _{{\Lambda } }d{x}_1\dots \int _{{\Lambda } }d{x}_{n} \prod _{\{i,j\}\in E_{\tau }} \left( 1-e^{-\beta |v(x_i-x_j)|}\right) \le |{\Lambda } |\left[ \tilde{C}(\beta )\right] ^{n-1} \end{aligned}$$

(2.24)

Appendix B: Proof of (2.5) and (2.6)

Proof of (2.5)

Let assume that V(|x|) is a stable and tempered pair potential in d dimensions with stability constant B and Basuev stability constant $\bar{B}$. Let

$$\begin{aligned} B_n =\sup _{(x_1,\dots ,x_n)\in \mathbb {R}^{dn}}\Big \{-{1\over n}\sum _{1\le i<j\le n}V(x_i-x_j)\Big \} \end{aligned}$$

and

$$\begin{aligned} \bar{B}_n =\sup _{(x_1,\dots ,x_n)\in \mathbb {R}^{dn}}\Big \{-{1\over n- 1}\sum _{1\le i<j\le n}V(x_i-x_j)\Big \} \end{aligned}$$

so that $B=\sup _{n\in \mathbb {N}} B_n$ and $\bar{B}=\sup _{n\in \mathbb {N}} \bar{B}_n$.

Let us first prove that if

$$\begin{aligned} \bar{B}= \limsup _{n\rightarrow \infty } \bar{B}_n \end{aligned}$$

(2.25)

then

$$\begin{aligned} \bar{B} = B \end{aligned}$$

Suppose by contradiction that $\bar{B}- B=\delta >0$. If this holds then necessarily there exists a finite $m \ge 2$ such that $B=B_m$ otherwise if $B=\limsup _{n\rightarrow \infty } B_n$ then $B=\bar{B}$. Now, due to the hypothesis (2.25), for all $\varepsilon >0$ there exists $n_0$ such that for infinitely many $n>n_0$

$$\begin{aligned} \bar{B}_n> \bar{B}-\varepsilon ~~~~~~~~~~\Longrightarrow ~~~~~~ B_n> {n-1\over n}(\bar{B}-\varepsilon )= {n-1\over n}(B+\delta -\varepsilon )~ \end{aligned}$$

Choose $\varepsilon ={\delta \over 2}$, then for infinitely many n we have that

$$\begin{aligned} B_n> {n-1\over n}(B+{\delta \over 2})> B~~~~~~~~~~~~ \text{ as } \text{ soon } \text{ as } n> {2B\over \delta }+1 \end{aligned}$$

in contradiction with the assumption that $B=B_m$. Hence we must have $B=\limsup B_n= \limsup \bar{B}_n=\bar{B}$.

So let us suppose that the $\sup \bar{B}_n$ is reached at some finite integer m, i.e. $\bar{B}= \bar{B}_m$.

Since V is stable, it is bounded from below and since V is tempered $\inf V$ cannot be positive. Let $\inf _{x\in \mathbb {R}^d}V(x)=-C$ with $C\ge 0$. Then for any $\varepsilon >0$ there exists $r_\varepsilon$ such that $V(r_\varepsilon )< -(C -\varepsilon )$. Take the configuration $(x_1,x_2,\dots , x_{d+1})\in (\mathbb {R}^d)^{d+1}$ such that $x_1$, $x_2,\dots ,x_{d+1}$ are vertices of a d-dimensional hypertetrahedron with sides of length $r_\varepsilon$. Recall that a d-dimensional hypertetrahedron has $d+1$ vertices and $d(d+1)/2$ sides. Then $U(x_1,x_2,\dots , x_{d+1})< -{d(d+1)\over 2}(C-\varepsilon )$, which implies that $B> {d\over 2}(C-\varepsilon )$ and by the arbitrariness of $\varepsilon$ we get $B\ge {d\over 2}C$. On the other hand we also have, for any $(x_1,\dots ,x_m)\in \mathbb {R}^{dm}$ that $U(x_1,\dots ,x_m)\ge -{m(m-1)C/2}$ which implies that $\bar{B}_m=\bar{B}\le {m\over 2}C$. Hence we can write ${d\over 2}C\le B\le \bar{B}=\bar{B}_m\le {m\over 2}C$ which implies $m\ge d+1$ and so $\bar{B}= \bar{B}_m= {m\over m-1}B_m\le {m\over m-1}B\le {d+1\over d}B$. $\square$

Proof of (2.6)

Let us assume that V(|x|) is a stable pair potential in d dimensions with stability constant B and Basuev stability constant $\bar{B}$ and that V(|x|) reaches its negative minimum $-C$ at some $|x|=r_0$ and it is negative for all $|x|>r_0$. We want to prove that the inequalities (2.6) hold.

First note that $d(d-1)C$ is always a lower bound for B when $d\ge 3$. Just consider a configuration in which n particle (with n as large as we want) are arranged in close-packed configuration at the sites of a d-dimensional face-centered cubic lattice with step $r_0$. The energy of such configuration is (asymptotically as $n\rightarrow \infty$) less than or equal to $-d(d-1)Cn$ since in a d-dimensional face-centered cubic lattice each site has $2d(d-1)$ neighbors (see e.g. [9]) and so there are (asymptotically) $d(d-1)n$ pairs of neighbors in the configuration. On the other hand, for any n-particle configuration $(x_1,\dots ,x_n)\in \mathbb {R}^{dn}$, it holds that $U(x_1,\dots ,x_n)\ge -n(n-1)C/2$, i.e. $\bar{B}_n\le nC/2$. Now, if $\bar{B}= \sup _n\bar{B}_n$ is attained at $n\rightarrow \infty$ then, as previously seen, we have $\bar{B}=B$. So let us suppose that $\bar{B}= \sup _n\bar{B}_n$ is attained at a certain finite m. Then we must have that $d(d-1)C\le B\le \bar{B}\le mC/2$ and so $m \ge 2d(d-1)$. Now if $m=2d(d-1)$ then $\bar{B}=B$. So when $\bar{B}>B$ then $m > 2d(d-1)$ and so we have $\bar{B}=\frac{m}{m-1}B_m \le {2d(d-1)+1\over 2d(d-1)} B$. The case $d=1$ and $d=2$ are treated analogously by just observing that the close-packed arrangement in $d=1$ is simply the cubic lattice with 2 neighbors for each site while for $d=2$ is the triangular lattice with 6 neighbors for each site. So $d(d-1)$ must be replaced by 1 for $d=1$ and by 3 for $d=2$ yielding $m>2$ and $m>6$ for $d=1$ and $d=2$ respectively. $\square$