For the oriented edge $i\to j=i+1\pmod3$, $J_i=\alpha P_i-\beta P_j$. Substitution in the Markov-chain entropy production gives
$$
\dot S_{\rm tot}=\sum_iJ_i\log\frac{\alpha P_i}{\beta P_j}.
$$
Since $\sum_iJ_i=\alpha-\beta$,
$$
\boxed{\dot S_{\rm tot}=(\alpha-\beta)\log\frac\alpha\beta
+\sum_i[\alpha P_i-\beta P_j]\log\frac{P_i}{P_j}.}
$$
The second term is relaxational and vanishes for the uniform stationary distribution. The steady entropy exported to the environment is therefore
$$
\boxed{\dot S_{\rm neq}=(\alpha-\beta)\log(\alpha/\beta)\geq0.}
$$


Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-356/1/g/solution