What is the connection between detailed balance and thermodynamic equilibrium?

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The detailed balance condition (DBC) is crucial for understanding thermodynamic equilibrium, as it establishes a relationship between the transition rates between states and the equilibrium probability distribution. Specifically, the DBC states that the rate of transitions from state i to j equals the rate from j to i when the system is in equilibrium, ensuring time-reversibility. This implies that the probability distribution remains unchanged over time, which is a hallmark of equilibrium. While the presence of DBC guarantees that the probability distribution is an equilibrium distribution, the converse—that every equilibrium distribution satisfies the DBC—remains uncertain. Thus, the DBC is a key criterion for identifying equilibrium states in Markov processes.
M.B.
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Hey,

I was wondering in what way the detailed balance condition (DBC) has to do with thermodynamic equilibrium.

The DBC is defined as
<br /> W(i\to j)P(i) = W(j\to i)P(j)<br />
with W the rates (probabilities per unit time) to jump from state i to j. P(i) is the equilibrium prob. density to have state i, this could be e.g. a Boltzmann Gibbs distribution in case of canonical equilibrium.
When for a (Markov jump) proces the above equality holds, then the system is time-reversible: it is even likely to follow a path forward in time or backwards in time

My question is:
The above formula is always possible to write down, but how is this related to equilibruim?
 
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It is obvious that the probability distribution must not change by the flows. So you see that your equation states that the same amount of thingamajigs are jumping from i to j as from j to i. Hence the distribution stays the same under time evolution. If you know W then you can derive an equilibrium distribution with the DBC.

If the DBC holds than P is an equilibrium distribution. I am not sure if the inverse direction is also true.
 
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