Is "detailed balance" equivalent with a continuity equation in state space?

Question

I have a talk tomorrow in which detailed balance is needed and I don't want to bore my audience with elaborate explanations for it so I'm looking for simpler explanations.

As far as I understood it detailed balance means, that for any given time the possibility to be in state $X$ and leave it equals the possibility to be in any state $Y\neq X$ and go from there to $Y$.

So if I draw a circle in (2 dimensional) state space around Y the "flow" in and out of that circle is zero ($\forall X$) - which leads to $div(S)=0$ ($S$ is the state space, also I think there is no need for a changing density ($\rightarrow \rho = const. \rightarrow \frac{\delta \rho}{\delta t} = 0$).

Is this correct? If yes, do you think, this makes understanding detailed balance easier?

If not - where's the mistake?

Answer 1

I could be wrong, but in my understanding, you're describing and justifying steady state, not detailed balance. In thermal equilibrium, steady state is true always, and detailed balance is true sometimes. Detailed balance means that the rate $X \rightarrow Y$ is always the same as the rate $Y \rightarrow X$.

If a system is both in thermal equilibrium and has time-reversal symmetry, you will have detailed balance. If time-reversal symmetry is broken, for example an electrolyte in an external magnetic field, you can have thermal equilibrium but you will not necessarily have detailed balance. The process $X \rightarrow Y$ might be balanced by $Y \rightarrow Z \rightarrow X$, instead of being balanced by $Y \rightarrow X$.

As an explanation of steady-state, your "continuity equation" explanation is fine. But in my opinion you can say the same thing more clearly without using the words "continuity equation" or writing down any math. If you just say that the number of times per second that the system enters state Y equals the number of times per second that the system leaves state Y, I think that's intuitively sensible.