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Physics theoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)
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Entropy transition in a specified two-candidate model

Physical scope

A conditional prediction for a boundary region whose prescribed candidate class consists of exactly two quantum-stationary surfaces with finite, continuously differentiable generalized-entropy branches on an open boundary time interval. The cusp is a property of the adopted minimum rule, not a deduction about an exact finite-cutoff microscopic entropy or about which saddle contributes in a different gravitational path integral.

Statement

Let I⊂R be an open interval of boundary time, measured in seconds, and let t0∈I. Suppose the prescribed candidate class for the adopted entropy dictionary consists, near t0, of exactly two admissible surfaces X1(t) and X2(t). Let their finite generalized entropies be S1,S2∈C1(I,R), with values in J K−1; here C1 means continuously differentiable. Assume S1(t0)=S2(t0), that S1(t)≠S2(t) for all sufficiently close t≠t0, and that S1′(t0)≠S2′(t0). Then the adopted semiclassical boundary-entropy prediction S∂model(t)=min⁡{S1(t),S2(t)} is continuous at t0 and is not differentiable there. This conclusion concerns the selected two-candidate minimum. The hypotheses do not identify it with an exact finite-cutoff microscopic entropy, which need not have this cusp.

Facts & Assumptions

Given: The open time interval, the two prescribed admissible candidates, their finite C1 generalized-entropy functions, the isolated crossing, and the unequal derivatives specified in the Statement.

[F1]

Restricted semiclassical entropy dictionary: for a specified candidate class with finite values and an attained minimum, the adopted boundary-entropy prediction is the minimum of the scheme-fixed generalized entropies; the postulate does not identify this prediction with the exact finite-cutoff entropy.

Proof

1.1givenF1algebra

For real numbers x,y, min⁡{x,y}=(x+y−∣x−y∣)/2. Therefore S∂model(t)=(S1(t)+S2(t)−∣S1(t)−S2(t)∣)/2. Since S1 and S2 are continuous, and absolute value is continuous, S∂model is continuous at t0. By [F1], this minimum is the model's semiclassical prediction.

1.2givenalgebra

Put D(t)=S1(t)−S2(t) and m=D′(t0)≠0. Since D(t0)=0, the definition of derivative gives D(t0+h)=mh+r(h) with r(h)/h→0 as h→0. Choose δ>0 small enough that ∣r(h)∣<∣m∣∣h∣/2 whenever 0<∣h∣<δ. Then D(t0+h) has the sign of mh on each side of t0. Hence the smaller branch on the left is different from the smaller branch on the right.

2.1givenstep 1.2algebra

On each side the minimum equals the selected C1 branch from step 1.2. Its left derivative at t0 is one of S1′(t0),S2′(t0) and its right derivative is the other. These numbers are unequal by hypothesis, so the two-sided derivative of S∂model does not exist.

3.1givenF1step 2.1algebra∎

Supplement the same model data with a regulated boundary code space C2 and constant boundary-state family ρ∂(t)=I2/2 for t∈I. This is a density operator supported on that code space. Its eigenvalues are 1/2,1/2, so Sexact(t)=−kBtr⁡(ρ∂(t)log⁡ρ∂(t))=kBlog⁡2 is constant and smooth. The postulate does not impose Sexact=S∂model, so this assignment is compatible with the two-branch premises and the cusp established in step 2.1. Thus those premises do not force a cusp in exact finite-cutoff entropy.

Remarks

For the symmetric branches S1=S0+a(t−t0) and S2=S0−a(t−t0), with a>0 measured in J K−1 s−1, the plot below shows the crossing and the lower envelope. It is illustrative only; the proof above establishes the result for all functions satisfying the stated hypotheses.

t¡t0(s)S¡S0(JK¡1)S1S2min(S1;S2)t0

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