Thought experimentStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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Entropy-impermeable finite-slab prediction

Physical scope

Conditional ideal boundary; growth equals integrated production and zero production gives equal endpoint charge; no observations/GSL.

Statement

Take the finite chart slab, current and local-admissibility assumptions of Conditional finite entropy balance with explicit boundary flux. Call the side boundary entropy-impermeable in this setup when the pullback of the entropy-current flux three-form vanishes on every side face; this implies $Φ_{S} (B) = 0$ . Then, conditional on the current draft supplier chain recorded by Conditional finite entropy balance with explicit boundary flux,

$Q_{S} (x_{+}^{0}) - Q_{S} (x_{-}^{0}) = \int_{K} Π {v o l}_{g} \geq 0.$

If $Π = 0$ throughout the slab, the endpoint charges are equal. This is an idealized model prediction, not a reported experiment. It does not identify this finite material entropy with horizon area or exterior quantum entropy, and does not assert a global generalized second law.

Facts & Assumptions

Given: The conditional finite-slab theorem Conditional finite entropy balance with explicit boundary flux, its smooth local entropy-current premise, and the same finite chart slab. The side boundary is entropy-impermeable in the pointwise sense defined in the Statement.

[L1]

Conditional finite entropy balance with explicit boundary flux states, conditionally on the draft mathematical finite-current balance, $Q_{S} (x_{+}^{0}) - Q_{S} (x_{-}^{0}) + Φ_{S} (B) = \int_{K} Π {v o l}_{g} \geq 0$ .

[A1]

For the hypothetical boundary in this item, the entropy-current flux three-form vanishes on every side face, so the defined net outward flux is $Φ_{S} (B) = 0$ .

[A2]

In the zero-production case only, assume $Π = 0$ at every point of $K$ .

Proof

technique · direct substitution of the ideal boundary condition into the conditional finite-slab theorem

1.1givenA1

By [A1], the pullback flux vanishes on each side face, so its finite face integral and hence the net outward flux satisfy $Φ_{S} (B) = 0$ .

2.1L1step 1.1

Substitute this equality into [L1]. The finite material slice-entropy change equals the integrated local production and is nonnegative under the theorem's adopted local premise.

3.1A2step 2.1algebra∎

Under the additional zero-production case [A2], the integral in step 2.1 is zero, so the endpoint charges are equal. This remains a conditional model deduction and supplies no measured outcome or horizon-entropy identification.

Depends on

Conditional finite entropy balance with explicit boundary flux

Entropy-impermeable finite-slab prediction

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources