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https://en.wikipedia.org/wiki/Boltzmann%27s_entropy_formula#Generalization https://en.wikipedia.org/wiki/Boltzmann%27s_entropy_formula#... According to Wikiped

https://en.wikipedia.org/wiki/Boltzmann%27s_entropy_formula#Generalization https://en.wikipedia.org/wiki/Boltzmann%27s_entropy_formula#... According to Wikipedia, if you start with the Gibbs entropy (which is the same as Shannon entropy), and then assume all microstate probabilities are equal (which Boltzmann does), you get the Boltzmann entropy formula. It also says Boltzmann himself used a p ln(p) formulation. So aren't they the same, perhaps up to a constant factor?

hexxiiiz 6y ago
If you count the number of microstates for a given macrostate you get a hyper geometric number N!/(n_1!n_2!...) The log of this is the Boltzmann entropy. However, if you consider N to be very large or infinite, you can show using the Stirling approximation that this ends up being the Gibbs/Shannon entropy in that case. So, in general, no.