发表机构
Braude Academic College of Engineering（布雷德工程学院）
机构由 AI 辅助整理，请以论文原文为准。AI 中文总结
该研究从熵为不可约不确定性和视界为完全扰乱器两个假设出发，推导出黑洞熵S=A/4ℓ_P²，每个普朗克元胞携带一个纳特信息，不依赖微观理论。
AI 中文摘要
黑洞视界每个普朗克元胞恰好携带一个纳特的信息，而无需指明黑洞由什么构成。两个输入即可：熵是不可约的不确定性（在普朗克尺度下任何观测都无法分辨的构型，这是一个客观界限，而非无知），且视界是一个完全扰乱器，因此到厄伦费斯特时间其维格纳函数已折叠到地板以下，没有任何元胞能被证明为空。于是单个元胞持有未分辨的弦交叉；一个被穿越n次的元胞允许n!种排序，扰乱会抹去这些排序，因此基数n携带权重1/n!，其计数为Ω₁=∑ₙ1/n!=e，即一个纳特。N=A/ℓ_P²个独立元胞给出Ω=e^N，因此S=N。乘以运动学因子1/4（零相空间中因果可及的部分，另行确立），即得S_BH=A/4ℓ_P²。该推导不涉及场方程、微观理论或假定的希尔伯特空间。
英文摘要
A black hole horizon carries exactly one nat of information per Planck cell, without naming what the black hole is made of. Two inputs suffice: entropy is irreducible uncertainty (the configurations no observation can resolve at the Planck floor, an objective bound, not ignorance), and the horizon is a complete scrambler, so by the Ehrenfest time its Wigner function has folded below the floor and no cell can be certified empty. A single cell then holds unresolved filament crossings; a cell crossed $n$ times admits $n!$ orderings that scrambling erases, so cardinality $n$ carries weight $1/n!$ and its count is $Ω_1=\sum_n 1/n!=e$, one nat. The $N=A/\ell_P^2$ independent cells give $Ω=e^N$, hence $S=N$. Multiplied by the kinematic factor $\tfrac{1}{4}$ (the causally accessible fraction of the null phase space, established separately), this returns $S_{BH}=A/4\ell_P^2$. The derivation invokes no field equations, no microscopic theory, and no posited Hilbert space.
Comments3 pages, published version. EPL 156 (2026) 49002
Journal refEPL 156, 49002 (2026), pp. 49002-p1-p3
