Master equation

For transition rate $w_{ij}$ from state $i$ to state $j$, the Master equation is

Use the master-equation convention that $w_{ij}$ is the transition rate from $i$ to $j$. In a stationary law $\pi$, the mean directed transition count per unit time isThe departure-state occupation multiplies the departure-to-destination rate. The reversed rate index would describe a different transition. A physical event transporting signed charge $q_{\mathrm{tr}}$ in the forward direction has mean charge current $q_{\mathrm{tr}}{J}_{ij}$.^[ref]

For a closed finite chain whose complete transition graph is a tree, every stationary edge current is zero. Removing one edge separates the states into two sets; stationarity makes total probability flow across that cut vanish, and the removed edge is its only crossing. Both directed counts can nevertheless be positive. On a three-state path with symmetric adjacent rates $a,b>0$, the stationary law is uniform and the two edge traffic rates are $a/3$ and $b/3$ in each direction.

Different physical transition mechanisms between the same states should be kept as separate edges of a multigraph when their currents matter. Summing their rates produces an aggregated edge. Opposing mechanism currents can cancel in that aggregate, so zero aggregate current does not force every mechanism's current to vanish.

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.