 Entropy production rate of a Markov chain

For a finite Continuous-time Markov chain, let $w_{ij}$ be the rate from $i$ to $j$, and let $P_i$ be its current state probability. Define forward traffic $a_{ij}=P_i{w}_{ij}$ and current $J_{ij}=a_{ij}-a_{ji}$. The total entropy production rate, in dimensionless entropy units $k_B=1$ with natural logarithms, isThe factor $1/2$ avoids counting an unordered pair twice. The expression is the same as ${\sum }_{i\ne j}{a}_{ij}\log (a_{ij}/a_{ji})$ after pairing opposite directions.^[ref]

Nonnegativity is elementary: for $a,b>0$, the signs of $a-b$ and $\log a-\log b$ agree. A pair with both flows zero contributes zero. If exactly one flow is positive, its contribution is $+\infty$, the corresponding one-sided limiting convention. The rate vanishes exactly when every pair has equal opposing traffic at that instant, which is Detailed balance for that Probability distribution.

For a stationary three-state cycle with clockwise rate $r>0$ and anticlockwise rate $\ell >0$, the stationary law is uniform. Each edge has current $(r-\ell )/3$, so ${\dot{S}}_{\mathrm{tot}}=(r-\ell )\log (r/\ell )$. It is zero at equal rates and positive otherwise, even though the state probabilities no longer change. Stationarity balances total flow at each state; it need not eliminate circulating currents.