ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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A constant potential has no canonical Gibbs probability on the line

Example

On $R$ , take $V (q) = 0$ , any fixed mass $m > 0$ , friction $γ > 0$ , and inverse temperature $β > 0$ . The configurational partition integral diverges, so the canonical Gibbs expression does not define a probability. The momentum factor is normalizable but cannot repair the infinite configurational factor.

Facts & Assumptions

Given: $V (q) = 0$ for every $q \in R$ , $m > 0$ , $γ > 0$ , $β > 0$ , and the canonical normalizer in Underdamped Langevin equation and canonical Gibbs density.

[F1]

Underdamped Langevin equation and canonical Gibbs density: a canonical Gibbs probability requires a finite positive normalizing integral $Z = \int e^{- β H (q, p)} d q d p$ .

[A1]

The Axiom of Choice: AC is inherited from the model-definition interface [F1]; the finite-rectangle lower bound uses no further choice.

Verification

Given: $V (q) = 0$ for every $q \in R$ , $m > 0$ , $γ > 0$ , $β > 0$ , and the canonical normalizer in Underdamped Langevin equation and canonical Gibbs density.

[F1] Underdamped Langevin equation and canonical Gibbs density: a canonical Gibbs probability requires a finite positive normalizing integral $Z = \int e^{- β H (q, p)} d q d p$ .

[A1] The Axiom of Choice: AC is inherited from the model-definition interface [F1]; the finite-rectangle lower bound uses no further choice.

Given: The data in the Example.

1.1givenalgebra

For every $R > 0$ , $\int_{- R}^{R} e^{- β V (q)} d q = \int_{- R}^{R} 1 d q = 2 R$ . These integrals are unbounded as $R \to \infty$ , so $\int_{R} e^{- β V (q)} d q = \infty$ .

2.1A1F1givenalgebra∎

On the finite rectangle $[- R, R] \times [- 1, 1]$ the phase-space integrand is at least $e^{- β / (2 m)}$ , so its integral is at least $4 R e^{- β / (2 m)}$ . These lower bounds diverge as $R \to \infty$ , hence the full normalizer $Z$ is infinite and the expression cannot be normalized to a probability.

Depends on

Used by

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Sources

Jean-Noël Roux, Stéphane Rodts, and Gabriel Stoltz, Introduction à la physique statistique et à la physique quantique, §4.1 and §5.2.2 (standard reference, not scraped)
Assyr Abdulle, Gilles Vilmart, and Konstantinos C. Zygalakis, Long time accuracy of Lie-Trotter splitting methods for Langevin dynamics, §2.1 (standard reference, not scraped)