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Coarse Shannon entropy rises and falls in a reversible rotor

Statement refuted

For every reversible Hamiltonian flow, every finite measurable partition, and every initial density, the partition Shannon entropy is nondecreasing in time.

Facts & Assumptions

Given: Let $θ \in R / Z$ be a position on the circle, $I \in R$ an action coordinate, and choose $ν > 0$ , $I_{0} > 0$ , and $a_{0} > 0$ .

[F1]

Hamiltonian phase, complete-domain flow and reference measure: on $T^{*} S^{1}$ the canonical symplectic form is $d θ \land d I$ ; a complete invariant Borel domain carries the induced reference measure $m_{0} = ∣ d θ \land d I ∣ / a_{0}$ .

[A1]

Adopt countable choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )) for the reference measure and the transport supplier [F2]; no regular-shell premise is used.

[A2]

A flow on a symplectic manifold is reversible here if there is a smooth involution $J$ (so $J^{2} = id$ ) with $J^{*} ω = - ω$ and $J \circ Φ_{t} \circ J = Φ_{- t}$ for every real $t$ .

[F2]

Liouville probability transport and constant fine entropy: on an open complete invariant domain, a finite-valued normalized Borel $L^{1} (m_{0})$ density is transported by $ρ_{t} = ρ_{0} \circ Φ_{- t}$ and preserves the fine $- k \int ρ \log ρ d m_{0}$ whenever the absolute entropy integral is finite.

Counterexample

Given: [A1], the convention [A2], [F1]–[F2], and the positive parameters above.

1.1A2F1givenconstructalgebra

Use the Hamiltonian $H (θ, I) = ν I$ , with $θ$ dimensionless, $I$ in J s, and $ν$ in s $^{- 1}$ , so $H$ is in J. The convention in [F1] gives $\dot{θ} = ν$ and $\dot{I} = 0$ . Thus $Φ_{t} (θ, I) = (θ + ν t m o d 1, I)$ is complete. The smooth involution $J (θ, I) = (- θ m o d 1, I)$ satisfies $J^{*} (d θ \land d I) = - d θ \land d I$ and $J Φ_{t} J (θ, I) = (θ - ν t m o d 1, I) = Φ_{- t} (θ, I)$ , proving reversibility in [A2]. Use the open invariant domain $R = S^{1} \times (I_{0}, I_{0} + a_{0}) \subset T^{*} S^{1}$ ; it is preserved by $J$ , complete under the flow, and its reference measure has total mass $a_{0}^{- 1} \int_{0}^{1} d θ \int_{I_{0}}^{I_{0} + a_{0}} d I = 1$ . Define on this open domain the density $ρ_{0} = 4 1_{[0, 1 / 4) \times (I_{0}, I_{0} + a_{0})}$ relative to $m_{0} ∣_{R}$ . Its integral is $4 \cdot (1 / 4) \cdot (a_{0} / a_{0}) = 1$ ; excluding the two action endpoints changes no reference-measure mass.

2.1F1step 1.1algebra

Partition the open domain into $C_{1} = [0, 1 / 2) \times (I_{0}, I_{0} + a_{0})$ and $C_{2} = [1 / 2, 1) \times (I_{0}, I_{0} + a_{0})$ . The transported density has angular support $[ν t, ν t + 1 / 4)$ modulo one. At $t = 0$ its cell probabilities are $(1, 0)$ . At $t = 3 / (8 ν)$ its support is $[3 / 8, 5 / 8)$ , so each cell receives angular length $1 / 8$ and the probabilities are $(1 / 2, 1 / 2)$ . At $t = 1 / (2 ν)$ its support is $[1 / 2, 3 / 4)$ , so the probabilities are $(0, 1)$ .

3.1step 2.1algebra

For any fixed $k > 0$ , the coarse Shannon entropy $S_{C} (t) = - k \sum_{j = 1}^{2} p_{j} (t) \log p_{j} (t)$ , with $0 \log 0 = 0$ , takes the values $0$ , $k \log 2$ , and $0$ at those three times. It therefore decreases between $3 / (8 ν)$ and $1 / (2 ν)$ , contradicting the proposed monotonicity. Here $k$ is a mathematical scale; choosing $k = k_{B}$ gives the usual entropy units J/K without changing the calculation.

4.1A1F2step 1.1algebra∎

The initial density satisfies $\int_{R} ∣ ρ_{0} \log ρ_{0} ∣ d m_{0} = 4 \log 4 \cdot m_{0} (supp ρ_{0}) = \log 4 < \infty$ . By [F2] its fine reference-measure entropy remains constant, even though the displayed coarse partition entropy changes.

Remarks

The angular coordinate is periodic; no global real-valued angle function is assumed. The calculation concerns an explicit mathematical coarse partition, not an observed entropy trace or a universal thermodynamic law.

Coarse Shannon entropy rises and falls in a reversible rotor

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Depends on

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