# Why are $S = -k_B\sum_i P_i \ln P_i$ and $S = k_B \ln\Omega$ equivalent?

This might be a silly question, but I don't see the equivalence relation between these two equations. Could somebody explain to me how to derive one from the other? Thanks in advance!

• Actually if $P_i$ simply means the probability of each microstate within each macrostate, isn't is just simply $P_i = 1/\Omega$, which makes them equivalent, and then each $P_i$ would be the same. Then why bother writing it in the way on the left? Jan 31, 2016 at 3:00

Since $$P_i = 1/\Omega$$, obviously $$\ln P_i =-\ln\Omega$$ and since $$P_i$$ is probability, then the $$\Sigma_i P_i=1$$
• Oh got it! But I don't understand why writing it as a summation? Is there something revealing about writing it in that way? What I feel is that the left way does not show anything new, and it makes me feel as if the $P_i's$ are not the same. Jan 31, 2016 at 3:04
• Oh I see! So the form $S = k_B \ln \Omega$ is not the general form but only applies to the situation where the fundamental postulate of thermodynamics holds (mostly likely the system is isolated and at equilibrium). (I thought the assumption always holds) Jan 31, 2016 at 3:21
• This answer is far from complete. The task is to show not just the agreement in the case $P_i = 1/\Omega$ but also to show that different ways of treating ensembles give mutually consistent results. e.g. take a huge set of microstates and divide it into parts which contain various fractions of the whole, etc. Oct 14, 2022 at 17:32