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Physics theoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)
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Exact entropy identity in the selected bulk branch

Physical scope

Pointwise extended-entropy identity for actual sufficiently regular solutions of the selected near-equilibrium bulk-only branch, plus one explicitly checked homogeneous Minkowski solution; no general nonlinear existence, causality, stability, universal material, or empirical claim.

Statement

On a nonempty open region of the smooth (−,+,+,+) Lorentzian model of Specified bulk extended entropy and relaxation law, assume an actual solution of its source-free stress and particle balances and corrected relaxation law. Use the state maps and units of Rest fluid state, frame and geometric current definitions and Thermodynamics-GR fields, currents, reference normalization and units, with smooth future unit observer congruence u (the frame, not a chart or tetrad), Δab=gab+uaub, D=ua∇a, θ=∇aua, n,T,c>0, positive C1 coefficient fields ζ,τ, and β=τ/ζ. The state function and map are C2 and u is smooth as in Rest fluid state, frame and geometric current definitions; take Π smooth as required by Corrected bulk relaxation and exact extended entropy, with the positive coefficients C1 as in the postulate. Thus the displayed stress and current and Sa are C1; require the balances and corrected law to hold pointwise. The solution remains in the application's declared open near-equilibrium regime. With Tab=εuaub+(p+Π)Δab,Ja=nua,Sa=(s−βΠ22T)ua, the exact pointwise identity is ∇aSa=Π2ζcT≥0. This is conditional on an actual solution and does not assert general solution existence.

There is also an explicit homogeneous solution under the additional application condition that the declared open regime contains constant zero-gradient reference data with Π=0. Let the selected material's state domain contain (v0,σ0) with n0=1/v0>0 and T0=eσp(v0,σ0)>0, and let the regime contain those constant state data. On M=R4 with global length coordinates (x0,x1,x2,x3)=(ct,x1,x2,x3), Minkowski metric g=−(dx0)2+∑i=13(dxi)2, and future direction ∂x0, choose constant positive ζ0,τ0 and set u=∂x0. For some δ>0 seconds take the open region O=(−cδ,∞)×R3, put t=x0/c, and define Π(t)=Π0e−t/τ0,t>−δ. Choose Π0 sufficiently small that the full field and first-derivative data stay in the declared regime throughout O. Then the source-free balances and corrected law hold, and the entropy production is strictly positive at every finite t≥0 when Π0≠0, and zero when Π0=0.

The argument is a conditional model deduction, not an observation or a general nonlinear existence, causality, or stability theorem. No empirical premise is used.

Facts & Assumptions

Given: An actual solution of the selected bulk-only constitutive postulate on its declared near-equilibrium regime, the C2 state data and smooth u of Rest fluid state, frame and geometric current definitions, smooth Π, positive n,T,c,ζ,τ with ζ,τ∈C1, and the x0=ct conventions.

[A1]

The geometric and tensor construction in the mathematical supplier chain assumes exactly countable choice ACω (The Axiom of Countable Choice (ACω)); no stronger choice principle is assumed.

[A2]

For the explicit witness, the selected material state domain contains a point (v0,σ0) with v0>0 and T0=eσp(v0,σ0)>0, and the application's open regime contains its constant zero-gradient reference data at Π=0.

[F1]

Specified bulk extended entropy and relaxation law: the selected branch adopts Tab=εuaub+(p+Π)Δab, Ja=nua, source-free balances, Sa=[s−βΠ2/(2T)]ua, and βDΠ+Π/(ζc)=−θ−(TΠ/2)∇a(βua/T), with positive C1 ζ,τ and an application-declared near-equilibrium regime.

[F2]

Corrected bulk relaxation and exact extended entropy (current batch-56 mathematical carrier): under smooth (−,+,+,+) state and Lorentzian data, source-free bulk-only balances, n,T,c>0, positive C1 ζ,τ, β=τ/ζ, and the exact corrected law in [F1], every actual smooth solution satisfies ∇aSa=Π2/(ζcT)≥0. Its proof is conditional on its current mathematical suppliers.

[F3]

Rest fluid state, frame and geometric current definitions: on a nonempty open set, a C2 state map into X⊂(0,∞)×R and e∈C2(X) define n=v−1, s=nσp, ε=ne, T=eσp, and p=−ev; u is a smooth future unit timelike observer congruence, and the frame is not identified with a chart or tetrad.

[F4]

Thermodynamics-GR fields, currents, reference normalization and units: with length coordinates x0=ct, c has units m s−1, Π has units Pa, D and θ have units m−1, ∇aSa has units J K−1 m−4, and c∇aSa is the physical production rate per volume.

Proof

1.1givenA1F1F2

The actual solution in the Statement satisfies every input in [F2]: [F1] supplies the same source-free bulk-only stress and current, positive C1 coefficients, entropy current, and corrected law, while the given solution is smooth and remains in the declared regime. The mathematical proof chain carries exactly [A1]. Applying [F2] yields ∇aSa=Π2/(ζcT); since Π2≥0 and ζ,c,T>0, this is nonnegative.

1.2givenA2F1F3F4algebra

For the witness, the constant state gives constant n0,ε0,p0,T0,s0. In the stated Minkowski coordinates, u0=1, ui=0, Δ00=Δ0i=0, and Δij=δij. Thus J0=n0, Ji=0, T00=ε0, T0i=0, and Tij=(p0+Π(t))δij. Since the only nonconstant component depends on time and occurs in spatial slots, ∇aJa=0, ∇aTa0=∂x0ε0=0, and ∇aTai=∂xj[(p0+Π(t))δji]=0 for i=1,2,3.

2.1step 1.1F1F4algebra

In SI length coordinates, [Π]=Pa and [ζcT]=Pa s m s−1 K=Pa m K, so [Π2/(ζcT)]=Pa m−1 K−1=J K−1 m−4, in agreement with [F4]. Multiplication by c gives J K−1 m−3 s−1.

2.2step 1.2givenA2F1algebra

Here θ=0, ∇a(βua/T0)=0, and DΠ=∂x0Π=c−1∂tΠ=−Π/(cτ0). Therefore the left side of the corrected law is −(τ0/ζ0)Π/(cτ0)+Π/(ζ0c)=0, matching its right side. For fixed δ, ∣Π(t)∣ and ∣DΠ(t)∣ are bounded on t>−δ by ∣Π0∣eδ/τ0 and ∣Π0∣eδ/τ0/(cτ0); taking ∣Π0∣ sufficiently small keeps all field and first-derivative data in the stated open regime.

3.1step 1.1step 1.2step 2.2A2F1F3algebra∎

Steps 1.2 and 2.2 verify that the specified fields are an actual smooth solution in the model regime. Applying the conditional identity from step 1.1 pointwise gives ∇aSa=Π(t)2/(ζ0cT0)=Π02e−2t/τ0/(ζ0cT0). Its denominator is positive, hence production is zero for Π0=0 and strictly positive for every finite t≥0 when Π0≠0.

Depends on

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