What is the connection between detailed balance and thermodynamic equilibrium?

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SUMMARY

The detailed balance condition (DBC) is crucial for understanding thermodynamic equilibrium in Markov jump processes. It is mathematically expressed as W(i→j)P(i) = W(j→i)P(j), where W represents transition rates and P denotes the probability density of states. When DBC is satisfied, the system exhibits time-reversibility, ensuring that the probability distribution remains unchanged over time. Consequently, if DBC holds, P is confirmed as an equilibrium distribution, although the converse relationship requires further investigation.

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  • Understanding of Markov jump processes
  • Familiarity with probability density functions
  • Knowledge of Boltzmann Gibbs distribution
  • Concept of time-reversibility in statistical mechanics
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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
[tex] W(i\to j)P(i) = W(j\to i)P(j)[/tex]
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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