# Boltzmann's Entropy and Statistical Entropy, are they equal? [duplicate]

• 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.

• Oct 12, 2022 at 21:58
• You will find there is aa lot of confusion on this site regarding entropy, and this is a reflection of the even-greater confusion of the world at large regarding entropy. It is not helped by the fact that we use the word "entropy" to mean a lot of different things.
– hft
Oct 12, 2022 at 22:37
• Have a look at Eq (18) and the surrounding explanation from this paper: informationphilosopher.com/solutions/scientists/jaynes/…
– hft
Oct 12, 2022 at 22:37
• – hft
Oct 12, 2022 at 22:52