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Shannon entropy decreases in a biased detailed-balance bath

Physical scope

An exact two-state finite-rate calculation distinguishing Shannon entropy from relative entropy to a biased stationary law; no empirical claim.

Example

Let $S = {0, 1}$ , with rates $q_{01} = 1$ and $q_{10} = 9$ seconds $^{- 1}$ . The positive detailed-balance law is $π = (0.9, 0.1)$ , and take initial law $p_{1} (0) = 1 / 2$ .

Facts & Assumptions

Given: The finite detailed-balance rate model and natural-log Shannon entropy $H (p) = - p \log p - (1 - p) \log (1 - p)$ .

[F1]

Finite detailed-balance rate and transition models: the master equation is ${\dot{p}}_{1} = (1 - p_{1}) q_{01} - p_{1} q_{10}$ .

[F2]

Relative-entropy decrease and convergence for a finite detailed-balance chain: the theorem controls $D (p ∥ π)$ ; it does not claim that $H (p)$ increases.

[F3]

The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t and Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm: $(\log x)^{'} = 1 / x$ for $x > 0$ , and the natural logarithm is strictly increasing.

Verification

Given: $q_{01} = 1$ , $q_{10} = 9$ , and $p_{1} (0) = 1 / 2$ .

1.1givenF1algebra

Solving ${\dot{p}}_{1} = 1 - 10 p_{1}$ gives $p_{1} (t) = 0.1 + 0.4 e^{- 10 t}$ , which lies in $(0.1, 0.5)$ and strictly decreases for every $t > 0$ .

2.1F3step 1.1algebra

For $0 < p < 1 / 2$ , $H^{'} (p) = \log ((1 - p) / p) > 0$ by [F3]. Therefore $d H (p_{1} (t)) / d t = H^{'} (p_{1} (t)) p_{1}^{'} (t) < 0$ for every $t > 0$ . The Shannon entropy decreases while detailed balance holds.

3.1F1F2step 1.1step 2.1∎

The two-state rate graph is connected, both components of $p (0)$ and $π$ are positive, and detailed balance holds. Thus [F2] applies to the relative entropy $D (p (t) ∥ π)$ at the same time that step 2.1 shows the Shannon entropy decreases.

Depends on

Dependency tree · 2 levels

Sources

U. Seifert, Stochastic thermodynamics, fluctuation theorems, and molecular machines