Alphabeta Math
CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)
How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Equal number density, mean velocity, and scalar energy do not determine covariance stress

Statement refuted

For solutions in the setting of Exact kinetic moment balances and unclosed fluxes, any two phase-space number densities with the same number-density field n, mean-velocity vector u, and internal kinetic-energy density e have the same covariance stress Π.

Facts & Assumptions

Given: a particle mass m>0, number density n0>0, and positive velocity-squared parameters α,β,γ with α≠β.

[F1]

Kinetic density, velocity, covariance stress and heat flux: u=n−1∫vF dv, Π=m∫(v−u)⊗(v−u)F dv, and e=12tr⁡Π.

[F2]

Exact kinetic moment balances and unclosed fluxes: the local moment equations apply to periodic, sufficiently decaying F solving the displayed kinetic equation with velocity-independent acceleration and a source whose number, momentum, and energy moments vanish.

[F3]

Regularity and invariance of the cutoff Gaussian-mixture class: for 0<a<b and Σ∈Ka,b, the normalized Gaussian is γΣ,u(v)=(2π)−3/2(det⁡Σ)−1/2exp⁡(−(v−u)TΣ−1(v−u)/2).

[F5]

Tonelli's theorem for nonnegative measurable functions on a sigma-finite product: nonnegative measurable product integrals may be evaluated iteratively on the stated sigma-finite factors.

Counterexample

1.1F2F3F4F5algebra

Define Σ1:=diag⁡(α,β,γ) and Σ2:=diag⁡(β,α,γ). Set a∗:=min⁡{α,β,γ} and b∗:=max⁡{α,β,γ}. Since α,β,γ>0 and α≠β, one has 0<a∗<b∗ and Σ1,Σ2∈Ka∗,b∗. On any flat three-torus with positive periods and on I=R, take a=0 and Q=0. Define Fj(r,v,t)=n0γΣj,0(v) using [F3]. Each Fj is smooth and rapidly decreasing in velocity, is constant in r,t, and solves the kinetic equation with zero acceleration and zero source. Gaussian normalization follows from [F4] by scaling each coordinate, and [F5] permits the product integration.

2.1F1F4step 1.1algebra∎

Each density has number density n0 and mean-velocity vector u=0 by normalization and oddness. In one coordinate with variance s2, differentiating xexp⁡(−x2/(2s2)) and integrating over R gives ∫x2e−x2/(2s2)dx=s2∫e−x2/(2s2)dx; the boundary term vanishes because the Gaussian decays to zero. Applying this to each coordinate gives covariance matrices Σ1 and Σ2. Thus both have the same scalar internal energy density e=mn0(α+β+γ)/2, while Π1=mn0diag⁡(α,β,γ) and Π2=mn0diag⁡(β,α,γ). Since α≠β, Π1≠Π2, contradicting determination of Π by (n,u,e).

Depends on

Dependency tree · 2 levels