Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 356 / 1 / g

Table of contents
- Solution g

Solution

For the oriented edge

i \to j = i + 1 (mod 3)

,

J_{i} = α P_{i} - β P_{j}

. Substitution in the Markov-chain entropy production gives

\dot{S}_{tot} = \sum_{i} J_{i} lo g \frac{α P _{i}}{β P _{j}} .

Since

\sum_{i} J_{i} = α - β

,

\dot{S}_{tot} = (α - β) lo g \frac{α}{β} + i \sum [α P_{i} - β P_{j}] lo g \frac{P _{i}}{P _{j}} .

The second term is relaxational and vanishes for the uniform stationary distribution. The steady entropy exported to the environment is therefore

\dot{S}_{neq} = (α - β) lo g (α / β) \geq 0.

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