- Boltmann's entropy: $$S_B=k_B ln(\Omega)$$ where $k_B$ is the Boltzmann constant, and $\Omega$ is the number of accessible microstates of the system.
- Statistical entropy:$$S_s=-k\sum_i P_i ln(P_i)$$ where k is a positive constant, and $P_i$ is the probability to be in the accessible $i$ microstate .
Observation: if $k=k_B$ and $P_i=1/\Omega$, the statistical entropy becomes equals to the Boltzmann entropy, $S_s=S_b$. So it seems that the statistical entropy equals the Boltzmann entropy only if the probability distribution is the uniform distribution $P_i=1/\Omega$.
How is it possible that, in canonical distribution, at equilibrium, $S_B=S_s$ and, at the same time, $P_i \neq 1/\Omega$?
The only possible answer that I came up with is that $S_B=S_s$ is true only for isolated systems, because for an isolated system the $P_i$ distribution is indeed $1/\Omega$. However, reading online, it seems that $S_B=S_s$ is true for every equilibrium state, so I don't know.
