Free-wall pressure on arbitrary van Hove shapes under superstability

Use the half-open unit-cell partition $C_i=i+[0,1{)}^d$ from Poisson Configuration Reference and first take a finite union $\mathcal{R}$ of these cells. Let ${\mathrm{\Pi }}_{\mathcal{R}}$ be the unnormalized positional sector measure whose $N$-point part is $z^N\mathrm{/}N!$ times product Lebesgue measure, and let ${\mu }_{\mathcal{R}}$ have density $e^{-\beta H_{\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{e}}}\mathrm{/}{\mathrm{\Xi }}_{\mathcal{R}}$ relative to ${\mathrm{\Pi }}_{\mathcal{R}}$. The factorial-moment identities from the correlation-density definition in [L1] and Tonelli [F5] give $E{N}_i^2=E{N}_i+E[N_i(N_i-1)]\le \xi \mathrm{\mid }{C}_i\cap \mathcal{R}\mathrm{\mid }+{\xi }^2\mathrm{\mid }{C}_i\cap \mathcal{R}{\mathrm{\mid }}^2\le M:=\xi +{\xi }^2\mathrm{.}$ This gives a uniform second-moment bound for every unit cell in every such region.

For each ${\mathrm{\Omega }}_k$, let ${\mathcal{R}}_k$ be the union of unit cells meeting ${\mathrm{\Omega }}_k$. Then ${\mathrm{\Xi }}_{{\mathrm{\Omega }}_k}\le {\mathrm{\Xi }}_{{\mathcal{R}}_k}$ by domain monotonicity, and $\mathrm{\mid }{\mathcal{R}}_k\mathrm{\mid }\mathrm{/}\mathrm{\mid }{\mathrm{\Omega }}_k\mathrm{\mid }\to 1$ by the van Hove boundary condition. Put $J_0=\lceil R_0+2\rceil$, where $R_0$ is the regular-tail cutoff in Superstable regular pair interactions and configurational ensemble notation, and choose $M_0\ge \max \{2B,\psi (J_0-1)\}$. For $j\in {\mathbb{Z}}^d$, define ${\kappa }_j=M_0$ when $\mathrm{\mid }j{\mathrm{\mid }}_{\mathrm{\infty }}<J_0$ and ${\kappa }_j=\psi (\mathrm{\mid }j{\mathrm{\mid }}_{\mathrm{\infty }}-1)$ otherwise. If two particles occupy cells with offset $j$, their negative pair energy is bounded by ${\kappa }_j$: for near offsets, stability on a two-point configuration gives $\varphi (x)\ge -2B$, and for far offsets the regularity envelope applies. Also ${\sum }_j{\kappa }_j<\mathrm{\infty }$ by [F1]–[F2]. Partition the cells of ${\mathcal{R}}_k$ into boxes $B$ of $\mathrm{\ell }$ unit cells per side and, for each occupied cell $i$, set $e_i={\sum }_{j:B(j)\mathrm{\ne }B(i)}{\kappa }_{i-j}$. The negative cross-box energy obeys $W^{-}\le {\sum }_i{e}_i{N}_i^2$, by $N_i{N}_j\le (N_i^2+N_j^2)\mathrm{/}2$ and summation over ordered cross-box cell pairs. Step 1.1 gives $E{W}^{-}\le M{\sum }_i{e}_i$, so Markov's inequality shows $W^{-}\le 2M{\sum }_i{e}_i$ with probability at least $1\mathrm{/}2$ (with probability one if the sum vanishes). On this event $H_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}\ge H_{\mathrm{i}\mathrm{n}}-2M{\sum }_i{e}_i$, where $H_{\mathrm{i}\mathrm{n}}$ is the sum of within-box energies. Therefore ${\mathrm{\Xi }}_{{\mathcal{R}}_k}\mathrm{/}2\le e^{2\beta M{\sum }_i{e}_i}{\prod }_B{\mathrm{\Xi }}_{{\mathcal{R}}_k\cap B}\mathrm{.}$ Indeed, ${\mathrm{\Xi }}_{{\mathcal{R}}_k}{\mu }_{{\mathcal{R}}_k}(E)={\int }_E{e}^{-\beta H_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}}d{\mathrm{\Pi }}_{{\mathcal{R}}_k}$; bound this by $e^{2\beta M{\sum }_i{e}_i}{\int }_E{e}^{-\beta H_{\mathrm{i}\mathrm{n}}}d{\mathrm{\Pi }}_{{\mathcal{R}}_k}$, then factor the latter integral by boxes. Each unrestricted partial-box partition is at most the full $\mathrm{\ell }$-cube partition. For one full box, put $E_{\mathrm{\ell }}={\sum }_{i\in B,j\mathrm{\notin }B}{\kappa }_{i-j}$. For fixed $R$, cells within distance $R$ of the boundary contribute at most $O_R({\mathrm{\ell }}^{d-1}){\sum }_j{\kappa }_j$, while every other cell contributes at most ${\mathrm{\ell }}^d{\sum }_{\mathrm{\mid }j{\mathrm{\mid }}_{\mathrm{\infty }}>R}{\kappa }_j$. Thus $E_{\mathrm{\ell }}\mathrm{/}{\mathrm{\ell }}^d\to 0$, by first letting $\mathrm{\ell }\to \mathrm{\infty }$ and then $R\to \mathrm{\infty }$. Consequently ${\mathrm{lim\hspace{0.17em}sup}}_k\mathrm{\mid }{\mathrm{\Omega }}_k{\mathrm{\mid }}^{-1}\log {\mathrm{\Xi }}_{{\mathrm{\Omega }}_k}\le {\mathrm{\ell }}^{-d}[\log {\mathrm{\Xi }}_{{\mathrm{\Lambda }}_{\mathrm{\ell }}}+2\beta M{E}_{\mathrm{\ell }}]\mathrm{.}$ Let $\mathrm{\ell }\to \mathrm{\infty }$ and use [F4].

For the lower bound, put $q=z{e}^{\beta B}$ and choose $K>eq+1$. For integer core side $\mathrm{\ell }\ge 1$, let $n_{\mathrm{\ell }}$ be an integer with $K{\mathrm{\ell }}^d<n_{\mathrm{\ell }}\le K{\mathrm{\ell }}^d+1$ and set ${\mathrm{\Xi }}_{\mathrm{\ell }}^{\mathrm{c}\mathrm{u}\mathrm{t}}={\sum }_{N=0}^{n_{\mathrm{\ell }}}{z}^{N}Z({\mathrm{\Lambda }}_{\mathrm{\ell }},N;\beta )$. The stability tail estimate in the proof of [F4] gives ${\mathrm{\Xi }}_{\mathrm{\ell }}^{\mathrm{c}\mathrm{u}\mathrm{t}}\ge (1-e^{-{\mathrm{\ell }}^d}){\mathrm{\Xi }}_{{\mathrm{\Lambda }}_{\mathrm{\ell }}}(z,\beta )$. Fix $g>2R_0$ and retain every complete $(\mathrm{\ell }+g)$-tile whose core lies in ${\mathrm{\Omega }}_k$. By [F2], the retained count $M_k$ divided by $\mathrm{\mid }{\mathrm{\Omega }}_k\mathrm{\mid }$ tends $(\mathrm{\ell }+g{)}^{-d}$. Each core has at most $(K+1){\mathrm{\ell }}^d$ particles. If two tile indices differ by maximum norm $m\ge 1$, their core points are separated by at least $mg$; hence the positive cross energy per core is at most $C_d(K+1{)}^2{\mathrm{\ell }}^{2d}{\sum }_{m\ge 1}{m}^{d-1}\psi (mg)\le C_d^{\mathrm{\prime }}(K+1{)}^2{\mathrm{\ell }}^{2d}{g}^{-d}\eta (g)\mathrm{.}$ Summing over cores, using the exact ordered-label assignment count to cancel the global $N!$, and discarding negative cross energy gives ${\mathrm{\Xi }}_{{\mathrm{\Omega }}_k}(z,\beta )\ge e^{-\beta D_{k,\mathrm{\ell },g}}({\mathrm{\Xi }}_{\mathrm{\ell }}^{\mathrm{c}\mathrm{u}\mathrm{t}}{)}^{M_k},\frac{D_{k,\mathrm{\ell },g}}{\mathrm{\mid }{\mathrm{\Omega }}_k\mathrm{\mid }}\le C_d^{\mathrm{\prime }}(K+1{)}^2{\left(\frac{\mathrm{\ell }}{g}\right)}^d\eta (g)+o_k(1)\mathrm{.}$ First let $k\to \mathrm{\infty }$. Then choose $g=g(\mathrm{\ell })$ from Diagonal gap for a decreasing integrable interaction tail and let integer $\mathrm{\ell }\to \mathrm{\infty }$; $g\mathrm{/}\mathrm{\ell }\to 0$, the packing fraction tends one, $(\mathrm{\ell }\mathrm{/}g{)}^d\eta (g)\to 0$, and ${\mathrm{\ell }}^{-d}\log (1-e^{-{\mathrm{\ell }}^d})\to 0$. The resulting liminf is at least the cube pressure [F4].