CounterexampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)

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A volume-sized external field changes ideal-gas pressure

Statement refuted

At fixed activity $z > 0$ and inverse temperature $β > 0$ , adding any spatially constant one-particle external energy throughout a continuum ideal-gas region leaves the thermodynamic-limit log grand partition sum per volume unchanged, as would a negligible boundary perturbation.

Facts & Assumptions

Given: $V > 0$ , $z > 0$ , $β > 0$ , constant $c \in R$ , and the $1 / N!$ -normalized positional grand sum.

[A1]

The model has zero pair energy and assigns each $N$ -point configuration external energy $c N$ .

[F1]

The product measure of two sigma-finite measure spaces, 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: the product coordinate integral of the constant one over $Ω^{N}$ is $V^{N}$ .

[F2]

The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ , The real exponential function and the number $e$ by a power series, and The exponential addition formula $\exp (x + y) = \exp (x) \exp (y)$ : the nonnegative exponential series satisfies $\sum_{N \geq 0} x^{N} / N! = e^{x}$ .

Counterexample

Given: $V > 0$ , $z > 0$ , $β > 0$ , constant $c \in R$ , and the $1 / N!$ -normalized positional grand sum.

For the continuum ideal gas in a bounded measurable region $Ω$ of volume $V$ , use activity $z > 0$ , inverse temperature $β > 0$ , and no pair interaction. Add a constant external one-particle energy $c$ throughout $Ω$ . This volume-sized field changes the pressure per volume whenever $c \neq 0$ , so it is not a negligible boundary perturbation.

1.1givenA1F1F2algebra

The $N$ -particle term without the field is $z^{N} V^{N} / N!$ , by the product volume [F1] and the $1 / N!$ convention [given]. Summing over $N \geq 0$ gives $Ξ_{0} = \sum_{N} (z V)^{N} / N! = e^{z V}$ by [F2].

2.1givenA1F2step 1.1algebra∎

With the field, each term is multiplied by $e^{- β c N}$ , so $Ξ_{c} = \sum_{N} (z e^{- β c} V)^{N} / N! = e^{z e^{- β c} V}$ by [F2]. Therefore $V^{- 1} \log Ξ_{0} = z$ and $V^{- 1} \log Ξ_{c} = z e^{- β c}$ . If $c \neq 0$ , these differ because $z, β > 0$ .

Depends on

Dependency tree · 2 levels

Sources

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