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Ideal photon reservoir and active-area mass balance — Examples

1 · Prerequisites

2 · Summary

This companion gives one exact half-mass cutoff calculation inside the stipulated fixed-reference effective-emitter model. It is a conditional thought experiment, not a reported observation; the positive cutoff avoids any conclusion about the zero-mass endpoint.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

Thought experimentStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Ideal active-area half-mass cutoff interval

Physical scope

Conditional fixed-asymptotic-reference ideal emitter calculation with eta=1 and a positive half-mass cutoff. The effective area is not a material null-horizon surface, the exported entropy is coarse Gibbs/ray entropy, and the example is not an observation or a duration prediction for an actual evaporating black hole.

Statement

Assume ${A C}_{ω}$ and the conditional model conclusion of Conditional ideal Planck emission and entropy accounting. Choose $η = 1$ and $M_{\min} = M_{0} / 2$ for $M_{0} > 0$ , with $G_{N}, c, ℏ, k_{B}$ as in that theorem. Define the model's algebraic zero-mass time scale

t_{*} : = \frac{M_{0}^{3}}{3 α} = \frac{5120 π G_{N}^{2} M_{0}^{3}}{ℏ c^{4}} .

The stipulated model stops earlier, at

τ = \frac{7}{8} t_{*} = \frac{4480 π G_{N}^{2} M_{0}^{3}}{ℏ c^{4}},

when its mass parameter first reaches $M_{0} / 2 > 0$ . It exports energy $c^{2} M_{0} / 2$ joules. Its stationary-family entropy drop, coarse photon entropy export, and total coarse entropy gain are respectively

Δ S_{H} = \frac{3 π k_{B} G_{N} M_{0}^{2}}{ℏ c}, S_{r a d} (τ) = \frac{4 π k_{B} G_{N} M_{0}^{2}}{ℏ c}, Δ (S_{H} + S_{r a d}) = \frac{π k_{B} G_{N} M_{0}^{2}}{ℏ c} .

For any $ϵ_{0} > 0$ , if $M_{0}^{2} \geq ℏ c / (1920 π G_{N} ϵ_{0})$ , then the model's maximum diagnostic on the closed interval $[0, τ]$ , including its positive cutoff endpoint, satisfies $ϵ_{q s} \leq ϵ_{0}$ . This is an exact ideal-model calculation, not a measured evaporation time or an error estimate for a gravitational backreaction solution.

Facts & Assumptions

Given: ${A C}_{ω}$ , the conditional model conclusion of Conditional ideal Planck emission and entropy accounting, $M_{0} > 0$ , and the choices $η = 1$ , $M_{\min} = M_{0} / 2$ .

[F1]

Conditional ideal Planck emission and entropy accounting: for the fixed-reference model, $M (t_{\infty}) = (M_{0}^{3} - 3 α t_{\infty})^{1 / 3}$ , $τ = (M_{0}^{3} - M_{\min}^{}) / (3 α)$ , energy export is $c^{2} (M_{0} - M_{\min})$ , $S_{r a d} = (4 / 3) Δ S_{H}$ , and $ϵ_{q s} = 2 G_{N} α / (c^{3} M^{2})$ . These are conditional on the theorem's stated model and ${A C}_{ω}$ .

[A1]

The thought experiment selects $η = 1$ and $M_{\min} = M_{0} / 2$ ; $M_{0}$ remains an arbitrary positive mass parameter, not a measured value.

[A2]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): ${A C}_{ω}$ is the exact countable-choice principle; no full Axiom of Choice is used.

Proof

Given: The model formulas of [F1] and the exact parameter choices [A1].

1.1givenF1A1A2algebra

With $η = 1$ , [F1] gives $α = ℏ c^{4} / (15360 π G_{N}^{2})$ . Hence $t_{*} = M_{0}^{3} / (3 α) = 5120 π G_{N}^{2} M_{0}^{3} / (ℏ c^{4})$ , and

τ = \frac{M_{0}^{3} - (M_{0} / 2)^{3}}{3 α} = (1 - \frac{1}{8}) t_{*} = \frac{7}{8} t_{*} .

The cutoff remains positive, $M (τ) = M_{0} / 2$ , so the model does not reach its algebraic zero-mass time.

1.2F1A1A2algebra

The energy export in [F1] is $c^{2} (M_{0} - M_{\min})$ . Substituting [A1] gives $c^{2} M_{0} / 2$ joules.

1.3givenF1A1A2algebra

By [F1], $S_{H} (M) = 4 π k_{B} G_{N} M^{2} / (ℏ c) + s_{*}$ on the selected branch, with an unfixed constant $s_{*}$ . Subtracting its values at $M_{0}$ and $M_{0} / 2$ cancels $s_{*}$ and gives $Δ S_{H} = 4 π k_{B} G_{N} (3 M_{0}^{2} / 4) / (ℏ c) = 3 π k_{B} G_{N} M_{0}^{2} / (ℏ c)$ . The entropy export is $4 Δ S_{H} / 3 = 4 π k_{B} G_{N} M_{0}^{2} / (ℏ c)$ , and the total gain is $Δ S_{H} / 3 = π k_{B} G_{N} M_{0}^{2} / (ℏ c)$ .

2.1F1A1A2algebra∎

The maximum diagnostic occurs at $M = M_{\min} = M_{0} / 2$ , so [F1] gives

ϵ_{q s, m a x} = \frac{2 G_{N} α}{c^{3} (M_{0} / 2)^{2}} = \frac{8 G_{N} α}{c^{3} M_{0}^{2}} = \frac{ℏ c}{1920 π G_{N} M_{0}^{2}} .

The displayed lower bound on $M_{0}^{2}$ is equivalent, after multiplication by the positive $1920 π G_{N} M_{0}^{2} ϵ_{0} / (ℏ c)$ , to $ϵ_{q s, m a x} \leq ϵ_{0}$ . This remains an internal diagnostic for the stipulated ODE, not a comparison with a coupled spacetime solution.