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Interacting Equilibrium Preparations — Examples

1 · Prerequisites

2 · Summary

These leaf examples calculate a finite hard-core positional partition, exhibit a reduced variational model with two minimizers, retain hard-core exclusions in the declared zero-inverse-temperature continuation, and compute the pressure change caused by a constant energy shift throughout an ideal gas volume. Each calculation states its model and boundary scope; none is a reported experiment or a proof that the model describes a particular material.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: AI-generatedVerification: AI-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)Open item page →

Two indistinguishable hard-core particles in an interval

Physical scope

A finite one-dimensional positional toy model with two classical identical particles and a hard-core exclusion; it is a model calculation, not a reported measurement.

Example

Take a dimensionless interval $Λ = [0, L]$ with $0 < a < L$ , two identical particles, and the hard-core pair potential $φ (r) = + \infty$ for $∣ r ∣ < a$ and $0$ otherwise. At any positive temperature the configurational canonical partition function is independent of $T$ and equals $Z_{Λ, 2} = \frac{1}{2} (L - a)^{2} .$ The factor $1 / 2$ is the adopted indistinguishable positional measure. This finite-volume calculation has no momentum variables or experimental data.

Facts & Assumptions

[F1]

Stable regularly tempered classical positional preparation: for two positions the hard-core pair energy is infinite precisely on forbidden separations $∣ x_{1} - x_{2} ∣ < a$ .

[A1]

Indistinguishable-particle positional measure: integrate symmetric positional weights with $d μ_{Λ, 2} = \frac{1}{2} d x_{1} d x_{2}$ .

[A2]

Finite-volume equilibrium ensemble sampling: at exact $N = 2$ , finite-energy pairs carry weight $e^{- β H_{2}}$ and forbidden infinite-energy pairs carry weight zero, integrated against $μ_{Λ, 2}$ .

Verification

Given: The interval, hard-core radius, exact particle number, and the adopted positional measure.

1.1givenF1algebra

The allowed labelled pairs form two disjoint triangles, $x_{2} \geq x_{1} + a$ and $x_{1} \geq x_{2} + a$ . The first has area $\int_{0}^{L - a} (L - a - x_{1}) d x_{1} = (L - a)^{2} / 2$ , and the second has the same area by exchanging coordinates.

2.1givenA1A2step 1.1∎

Their total allowed area is $(L - a)^{2}$ . On it $H_{2} = 0$ and the weight is $1$ ; on the excluded band the adopted finite-energy convention assigns weight zero, without evaluating an exponential at $+ \infty$ . Multiplying the allowed area by the measure factor $1 / 2$ from [A1] gives $Z_{Λ, 2} = (L - a)^{2} / 2$ , as asserted.

ExampleConstruction: AI-generatedVerification: AI-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)Open item page →

Two minimizers in a reduced coexistence model

Physical scope

A dimensionless coarse-grained order-parameter example illustrating a possible failure of a singleton variational-minimizer assumption; it is not a continuum Gibbs-potential calculation.

Example

Consider the reduced coarse-grained variational model with scalar dimensionless order parameter $m \in [- 1, 1]$ and dimensionless free-energy function $g (m) = (m^{2} - 1)^{2}$ . Its global minimizers are $m = - 1$ and $m = 1$ . This model shows why uniqueness must be stated as a parameter-specific assumption. It does not identify these two minimizers with Gibbs measures of a continuum pair potential.

Facts & Assumptions

[F1]

Selected grand-variational phase uniqueness: the selected continuum conclusion is licensed only when the relevant minimizer set on the finite-intensity stationary-law domain is a singleton.

[A1]

Adopt the stated toy variational function $g (m) = (m^{2} - 1)^{2}$ on the closed interval $[- 1, 1]$ ; $m$ and $g$ are dimensionless.

Verification

Given: The domain $[- 1, 1]$ , the explicit function $g$ , and the limited illustrative scope.

1.1givenA1algebra

For every $m \in [- 1, 1]$ , $g (m) = (1 - m^{2})^{2} \geq 0$ . Equality holds exactly when $m^{2} = 1$ , namely at $m = - 1$ or $m = 1$ .

2.1givenF1step 1.1∎

The two equality points are distinct, so this scalar model's global-minimizer set is ${- 1, 1}$ and is not a singleton. This illustrates the need for a separate uniqueness assumption in a variational model. The domain in [F1] consists of stationary configuration laws, whereas this domain is $[- 1, 1]$ ; no map identifying their minimizers has been specified. Thus this calculation neither instantiates nor refutes the continuum assumption [F1].

ExampleConstruction: AI-generatedVerification: AI-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)Open item page →

Hard-core exclusion retained at zero inverse temperature

Physical scope

A finite-volume positional hard-core model under the explicitly defined retained-admissibility beta-zero continuation; no universal infinite-temperature claim is made.

Example

In the dimensionless coordinates of Stable regularly tempered classical positional preparation, take $d = 1$ , an interval $Λ = [0, L]$ , and a hard-core exclusion distance $a$ with $0 < a < L$ . Both $L$ and $a$ are dimensionless; the physical interval length and exclusion distance are $ℓ_{0} L$ and $ℓ_{0} a$ in metres. Set $φ (r) = + \infty$ for $∣ r ∣ < a$ and $0$ otherwise. Under the retained-admissibility zero-inverse-temperature preparation at $z > 0$ , the two-particle weight is zero on $∣ x_{1} - x_{2} ∣ < a$ and equals $z^{2}$ on $∣ x_{1} - x_{2} ∣ \geq a$ . Hence the retained grand law is not the unrestricted Poisson activity law, which assigns positive mass to the forbidden two-particle band. The conclusion concerns the finite positional continuation defined on the A page.

Facts & Assumptions

[F1]

Retained hard-core continuation at zero inverse temperature: at $β = 0$ the adopted weight is $z^{N} 1_{{H_{N} < \infty}}$ .

[F2]

Indistinguishable-particle positional measure: the two-particle reference measure is $\frac{1}{2} d x_{1} d x_{2}$ .

[A1]

The hard-core potential forbids exactly the pairs with $∣ x_{1} - x_{2} ∣ < a$ .

Verification

Given: $0 < a < L$ , the finite interval, and the retained-admissibility convention.

1.1givenF1A1

By [F1] and [A1], every pair in the open band $∣ x_{1} - x_{2} ∣ < a$ has zero retained weight, while every admissible pair has weight $z^{2}$ .

2.1givenF1F2step 1.1algebra∎

The band has positive dimensionless Lebesgue area $L^{2} - (L - a)^{2} = 2 a L - a^{2} > 0$ ; its collision diagonal has area zero. The unrestricted grand normalizer is $\sum_{N \geq 0} (z L)^{N} / N! = e^{z L}$ , so its probability of this two-particle band is $e^{- z L} z^{2} (2 a L - a^{2}) / 2 > 0$ . The retained grand normalizer is between $1$ (the vacuum contribution) and $e^{z L}$ , so its law is defined and gives that band probability zero. Therefore the two probability laws differ.

ExampleConstruction: AI-generatedVerification: AI-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)Open item page →

A volume-filling field changes ideal-gas pressure

Physical scope

A classical ideal positional gas in dimension d at least two with a constant per-particle external energy throughout the volume; a nonzero shift cannot be represented by a uniformly tempered exterior, while the zero shift agrees with the empty exterior. No empirical observation is asserted.

Example

For the ideal positional gas $φ = 0$ in dimension $d \geq 2$ and a bounded measurable region of dimensionless volume $0 \leq V < \infty$ , take $z > 0$ and $T > 0$ , with $β = (k_{B} T)^{- 1} > 0$ in $J^{- 1}$ . The grand partition is $Ξ_{V} (z, β) = e^{z V}$ . Now impose a constant external energy $c \in R$ joules per particle at every point of the volume, adding $c N$ to the configuration energy and using the same grand weighting convention. The new grand partition is $Ξ_{V}^{(c)} = e^{z e^{- β c} V}$ . For $V > 0$ , the dimensionless per-volume log-partition rate changes from $z$ to $z e^{- β c}$ when $c \neq 0$ . At $V = 0$ , both partitions are $1$ and no per-volume rate is assigned. For $c \neq 0$ , this energy shift cannot be represented on the nonempty cubes $Λ_{n}$ by the deterministic exterior interaction used by Qualified van Hove and tempered-exterior thermodynamic limits. For $c = 0$ , it vanishes and agrees with the empty exterior $\emptyset \in Ω (0)$ of Van Hove regions and uniformly tempered exterior preparations. For every $c$ , the ideal potential itself fails the cited theorem's stronger interaction hypotheses.

Facts & Assumptions

[F1]

Stable regularly tempered classical positional preparation: $V$ is dimensionless reference volume, $β$ is inverse energy, and the ideal model has $H_{N} = 0$ .

[A1]

Finite-volume equilibrium ensemble sampling: $Ξ_{V} = \sum_{N \geq 0} z^{N} Z_{V, N}$ with the indistinguishable positional factor $1 / N!$ .

[F2]

Van Hove regions and uniformly tempered exterior preparations: the deterministic exterior class $Ω (t)$ has one fixed square-count bound $t \geq 0$ , and $W_{n} (ω, ζ) = \sum_{x \in ω} \sum_{y \in ζ ∖ Λ_{n}} φ (x - y)$ . Thus $\emptyset \in Ω (0)$ and, for the ideal potential $φ = 0$ , $W_{n} = 0$ for every allowed exterior.

[L1]

Qualified van Hove and tempered-exterior thermodynamic limits: its current conditional statement treats exterior energy only as $W_{n} (ω, ζ)$ for a fixed $ζ \in Ω (t)$ and assumes a superstable regular pair potential with an additional positive divergent core for the exterior conclusion. The ideal potential has no such positive core. The theorem is draft and remains conditional on its mathematical suppliers.

Verification

Given: The finite-volume grand weight, an ideal gas, and the uniform per-particle shift $c$ over the entire region.

1.1givenF1A1algebra

Since $H_{N} = 0$ , the finite-volume canonical positional integral is $Z_{V, N} = V^{N} / N!$ . Summing the adopted grand weights gives $Ξ_{V} = \sum_{N \geq 0} (z V)^{N} / N! = e^{z V}$ .

2.1givenA1step 1.1algebra

Adding $c N$ to the energy multiplies the $N$ -particle weight by $e^{- β c N}$ , so $Ξ_{V}^{(c)} = \sum_{N \geq 0} (z e^{- β c} V)^{N} / N! = e^{z e^{- β c} V}$ . When $V > 0$ , division of the two logarithms by $V$ gives $z$ and $z e^{- β c}$ ; these differ for $c \neq 0$ because $β > 0$ and $z > 0$ . When $V = 0$ , only the vacuum contributes, both partitions equal $1$ , and division by $V$ is undefined.

3.1givenF2L1step 2.1∎

For $φ = 0$ , [F2] gives $W_{n} (ω, ζ) = 0$ for every allowed exterior. If $c \neq 0$ , a singleton $ω = {x}$ in any $Λ_{n}$ has $c N = c \neq 0$ , so no such exterior represents this energy shift. If $c = 0$ , the shift is zero and agrees with $ζ = \emptyset \in Ω (0)$ . In either case the ideal potential lacks the positive divergent core required by [L1], so the calculation in step 2.1 contradicts no qualified exterior conclusion.