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The close-packing endpoint can have zero canonical weight

Example

In one dimension, let the pair potential be infinite when the separation is less than $a > 0$ , and zero otherwise. For $N \geq 1$ labeled rods in $[0, L]$ , with canonical positional normalization $1 / N!$ , write $S = L - (N - 1) a$ . If $S > 0$ , translating the ordered coordinates by $y_{i} = x_{i} - (i - 1) a$ gives $Z (L, N; β) = \frac{S^{N}}{N!} = \frac{(L - (N - 1) a)^{N}}{N!}, β > 0.$ If $S = 0$ , the allowed configuration set has zero Lebesgue measure and $Z (L, N; β) = 0$ . Along $N \to \infty$ with $L = (N - 1) a$ , the density $N / L \to 1 / a$ while every finite-volume partition is zero. Thus the thermodynamic close-packing endpoint is approached by a sequence of zero-weight finite systems; it is not realized by a positive canonical partition. The interior canonical limit theorem makes no assertion at that endpoint.

Facts & Assumptions

Given: Hard-core distance $a > 0$ and inverse temperature $β > 0$ .

[F1]

Superstable regular pair interactions and configurational ensemble notation: the canonical positional partition integrates over labeled coordinates and divides by $N!$ .

[F2]

Canonical free-cube limit at every interior attainable density: its canonical limit assertion is restricted to interior attainable densities.

[F3]

The product measure of two sigma-finite measure spaces, A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included, For sigma-finite factors, the product measure exists, has the rectangle formula, is sigma-finite, and is unique, and Tonelli's theorem for nonnegative measurable functions on a sigma-finite product: product Lebesgue measure assigns volume $S^{N}$ to the cube $[0, S]^{N}$ and permits iterated integration.

[F4]

The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ : the coordinate-order partition has $N!$ permutation chambers and the canonical normalization divides by $N!$ .

[F5]

Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation: translating a Lebesgue-measurable set preserves its Lebesgue measure.

Verification

Given: $N \geq 1$ and $L \geq (N - 1) a$ .

1.1F1F3F4F5algebra

Order the coordinates $0 \leq x_{1} < \dots < x_{N} \leq L$ . The exclusion constraints are $x_{i + 1} - x_{i} \geq a$ . Apply the coordinatewise translation $x_{i} \mapsto y_{i} = x_{i} - (i - 1) a$ ; each one-coordinate translation preserves Lebesgue measure [F5], so product measure is preserved. The ordered region becomes $0 \leq y_{1} \leq \dots \leq y_{N} \leq S$ , where $S = L - (N - 1) a$ . The cube $[0, S]^{N}$ splits, up to equality sets of zero product measure, into $N!$ coordinate-order regions. Product-measure uniqueness makes these permutation chambers equal in measure, so the ordered region has volume $S^{N} / N!$ . The full labeled allowed volume is $N!$ times this, and canonical division by $N!$ gives $Z (L, N; β) = S^{N} / N!$ .

2.1F1F2F3F4step 1.1algebra∎

If $L = (N - 1) a$ , then $S = 0$ and the transformed region is a null set; its integral is zero. If $L > (N - 1) a$ , then $S > 0$ and the ordered region has positive product volume. Thus the maximal-density endpoint does not carry a positive canonical partition at exact close packing; the interior-only theorem [F2] makes no contrary assertion.

Depends on

Dependency tree · 2 levels

Sources

D. Dereudre, Introduction to the theory of Gibbs point processes