Alphabeta Math
LemmaStatement: AI-adaptedProof: 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.

Corrected bulk relaxation and exact extended entropy

Statement

Use the smooth (−,+,+,+) Lorentzian data, local state, and conventions of Perfect stress and particle balances with entropy identity and Frame-qualified dissipative entropy cancellation, with n>0, T>0, c>0, x0=ct, and smooth future unit u. On the nonempty open spacetime domain O of those suppliers, let ζ,τ∈C1(O;(0,∞)) be scalar coefficient fields, set β:=τ/ζ∈C1(O;(0,∞)), and let Π:O→R be smooth. In the bulk-only branch take

Tab=ϵuaub+(p+Π)Δab,Ja=nua,∇aTab=0,∇aJa=0,

where Δab=gab+uaub, D=ua∇a, θ=∇aua, and aaccb=Dub. Adopt the corrected relaxation law

βDΠ+Πζc=−θ−TΠ2∇a ⁣(βuaT)

and the extended entropy current Sa=[s−βΠ2/(2T)]ua. Then the stress and particle equations imply

Dϵ+(h+Π)θ=0,(h+Π)aaccb+Δba∇a(p+Π)=0,Dn+nθ=0,

and every actual smooth solution of the stated corrected law satisfies

∇aSa=Π2ζcT≥0.

Here β has units Pa−1, Π,ϵ,p,h have units Pa, D,θ,aacc have inverse-length components, and the correction βΠ2/(2T) has the units of entropy density. The law is an explicit model assumption; dropping its derivative term does not yield the stated sign identity in general.

There is an exact homogeneous solution on the explicitly specified Minkowski spacetime M=R4 with global length coordinates (x0,x1,x2,x3)=(ct,x1,x2,x3), metric η=diag⁡(−1,1,1,1), and future direction ∂x0; its Levi-Civita derivative is ordinary coordinate differentiation. Fix a state point (v0,σp0) in the supplied local state domain with v0>0 and T0=eσp(v0,σp0)>0, and take the constant state map (v,σp)(x)=(v0,σp0), so n0=1/v0, ϵ0=n0e(v0,σp0), and p0=−ev(v0,σp0) are constant. For constant positive coefficients ζ0,τ0, let u=∂x0=(1,0,0,0), T=T0, and define the smooth scalar field Π(x)=Π0e−t/τ0 on all of M, where t=x0/c. The stress and particle balances and the corrected law hold, with strict entropy production wherever Π(x)≠0. This mathematical witness is not a general nonlinear existence theorem or a claim of physical applicability in any separately selected near-equilibrium regime.

Facts & Assumptions

Given: The smooth state and metric data specified above, coefficient fields ζ,τ∈C1(O;(0,∞)), smooth bulk scalar Π:O→R, the displayed stress and particle balances, and the displayed corrected relaxation law. All derivatives are classical; no solution existence is assumed beyond any particular smooth field set under discussion.

[A1]

Countable choice ACω is the only choice strength in the geometric supplier chain (The Axiom of Countable Choice (ACω)).

[F1]

Perfect stress and particle balances with entropy identity (current batch-55 draft): for a smooth fixed-composition state and arbitrary smooth force f, the perfect stress T0ab=ϵuaub+pΔab and Ja=nua satisfy Dϵ+hθ=−ubfb, haaccb+Δba∇ap=Δbcfc, and ∇aJa=0  ⟺  Dn+nθ=0. This use is conditional on the current draft state and Gibbs carriers; no independent proof acceptance is inferred.

[F2]

Frame-qualified dissipative entropy cancellation (current batch-55 draft): for the smooth dissipative stress with bulk pressure Π, zero heat flux, zero particle diffusion, zero shear, force f=0, and ∇aJa=0, the entropy current su obeys ∇a(sua)=−Πθ/T. Its proof body is complete conditional on the same current draft state/Gibbs interfaces; authoring readiness does not establish proof acceptance.

Proof

1.1A1F1givenalgebra

Put T0ab=ϵuaub+pΔab and define fb:=∇aT0ab=−∇a(ΠΔab). Since ∇g=0, ∇aΔab=θub+aaccb, so ∇a(ΠΔab)=Δba∇aΠ+Π(θub+aaccb). Apply [F1] to T0 with this force: contracting the force term with ub yields Dϵ+hθ=−Πθ, while its spatial projection gives haaccb+Δba∇ap=−Δba∇aΠ−Πaaccb. The particle balance is unchanged. Rearrangement gives the three projected equations in the Statement.

1.2A1F2givenalgebra

Apply [F2] to the full bulk-only stress and particle balance with zero force; it gives ∇a(sua)=−Πθ/T. Product differentiation of the correction gives ∇a[(βΠ2/(2T))ua]=βΠDΠ/T+(Π2/2)∇a(βua/T). Therefore ∇aSa=−(Π/T)[θ+βDΠ+(TΠ/2)∇a(βua/T)]. The corrected law makes the bracket −Π/(ζc), proving ∇aSa=Π2/(ζcT)≥0 because ζ,c,T>0.

2.1givenstep 1.2algebra∎

For the homogeneous witness, constant n,σp,T,ϵ,p,u give ∇aJa=0 and ∇aTa0=0, ∇aTai=∂i(p+Π)=0. Also DΠ=c−1∂tΠ=−Π/(cτ), θ=0, and ∇a(βua/T)=0, so the two terms on the left of the corrected law cancel and its right side vanishes. The entropy formula in step 1.2 is strict when Π0≠0. If a chosen near-equilibrium domain bounds ∣Π∣, any sufficiently small ∣Π0∣ remains in it for t≥0.

Depends on

Dependency tree · 2 levels

Sources