Thought experimentStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck pendingjudge pass (gpt-6-luna)

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Entropy difference with unfixed entropy zero

Physical scope

Conditional entropy difference for two positive-mass states on the selected zero-spin, zero-charge Schwarzschild slice of the same connected, strictly subextremal stationary family, under the adopted reversible-heat identification and pointwise reduced-branch temperature matching. It is a model deduction with no measured black-hole entropy, microstate count, general Hawking claim, or selected absolute entropy zero.

Statement

Assume ${A C}_{ω}$ as defined in The Axiom of Countable Choice ( ${A C}_{ω}$ ), inherited from the conditional mathematical spectrum input. Let $m_{*, 1}, m_{*, 2} > 0$ be two physical mass parameters on the zero-angular-momentum, zero-electric-charge Schwarzschild slice of the selected family in Physical KN family work and constancy. Assume the conditional area calibration Conditional area entropy coefficient with free offset with its full hypotheses on the same connected family, including the adopted reversible-heat identification and the pointwise reduced-branch assignment from Restricted reversible horizon entropy identification. Then the entropy difference between these two model states is $S_{H} (m_{*, 2}) - S_{H} (m_{*, 1}) = \frac{4 π k_{B} G_{N}}{ℏ c} (m_{*, 2}^{2} - m_{*, 1}^{2}) .$ The additive entropy constant cancels. This is a conditional model deduction, not an observation or a construction of a statistical microstate count.

Facts & Assumptions

Given: Two positive SI mass parameters on the selected zero-spin, zero-charge family slice and the current restricted entropy calibration.

[F1]

Physical KN family work and constancy: the physical map is $m = G_{N} m_{*} / c^{2}$ , $a = J_{S I} / (m_{*} c)$ , and $q = \sqrt{G_{N} / (4 π ε_{0} c^{4})} Q_{e}$ ; on the selected family $s = \sqrt{m^{2} - a^{2} - q^{2}}$ , $r_{+} = m + s$ , $D = r_{+}^{2} + a^{2}$ , and $A_{H} = 4 π D$ . The joint zero-spin, zero-charge Schwarzschild point is included by its separate exact horizon-data supplier.

[F2]

Conditional area entropy coefficient with free offset: under its actual hypotheses— ${A C}_{ω}$ , the exact SI family differential on the connected, strict-subextremal, zero-magnetic-flux stationary Kerr–Newman family, the adopted $C^{1}$ entropy state function and reversible-heat identity, and for each family point $x$ a reduced-branch instance whose Killing flow and reference normalization match that member's generator and whose positive-gap/positive-rate conditions hold, with constant parameter $κ = κ_{g e o m} (x)$ and $T_{K} (x) = ℏ c κ_{g e o m} (x) / (2 π k_{B})$ —the theorem gives $S_{H} = k_{B} A_{H} / (4 ℓ_{P}^{2}) + s_{*}$ throughout that family for one constant $s_{*}$ , with $ℓ_{P}^{2} = G_{N} ℏ / c^{3}$ . This consumer uses only the formula on its Schwarzschild slice.

[A2]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): every sequence $(X_{n})_{n \in N}$ of nonempty sets has a choice function.

[A1]

$m_{*, 1}, m_{*, 2} > 0$ label two admitted states on that same selected slice and connected family domain to which [F2] applies.

Proof

1.1givenF1A1A2algebra

Under [A2], for either mass $m_{*} > 0$ , set $J_{S I} = Q_{e} = 0$ . The map in [F1] gives $a = q = 0$ and $m = G_{N} m_{*} / c^{2} > 0$ ; hence $s = m$ , $r_{+} = 2 m$ , and $D = 4 m^{2}$ . Therefore $A_{H} (m_{*}) = 4 π D = 16 π m^{2} = \frac{16 π G_{N}^{2} m_{*}^{2}}{c^{4}} .$ This is the selected stationary-family area, not a process or observation.

1.2F2A1A2algebra

Under [A2], apply [F2] at both states in [A1]. The same constant $s_{*}$ cancels, giving $S_{H} (m_{*, 2}) - S_{H} (m_{*, 1}) = \frac{k_{B}}{4 ℓ_{P}^{2}} (A_{H} (m_{*, 2}) - A_{H} (m_{*, 1})) .$

2.1step 1.1step 1.2F2A1A2algebra∎

Substitute the two areas from step 1.1 into step 1.2 and use $ℓ_{P}^{2} = G_{N} ℏ / c^{3}$ . The coefficient simplifies to $4 π k_{B} G_{N} / (ℏ c)$ , which has units $J K^{- 1} k g^{- 2}$ before multiplication by the squared-mass difference; the result has units $J K^{- 1}$ and vanishes when $m_{*, 1} = m_{*, 2}$ .

Remarks

The calculation concerns two hypothetical members of one selected stationary model family. It reports no measurement and assigns no preferred zero of entropy.

Entropy difference with unfixed entropy zero

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources