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!
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$\begingroup$ 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? $\endgroup$– snsunxJan 31, 2016 at 3:00
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Citing Wikipedia here,
In what has been called the fundamental assumption of statistical thermodynamics or the fundamental postulate in statistical mechanics, the occupation of any microstate is assumed to be equally probable (i.e. Pi = 1/Ω, where Ω is the number of microstates); this assumption is usually justified for an isolated system in equilibrium.
Since $P_i = 1/\Omega$, obviously $\ln P_i =-\ln\Omega$ and since $P_i$ is probability, then the $\Sigma_i P_i=1$
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$\begingroup$ 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. $\endgroup$– snsunxJan 31, 2016 at 3:04
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$\begingroup$ @chemicaholic I've expanded the citation from wikipedia and given the link for further reading - the expression with the probabilities is the general expression - the probabilities might in fact not be equal at all. We assume that they are in some conditions, since we have no reason to think otherwise. $\endgroup$ Jan 31, 2016 at 3:16
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$\begingroup$ 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) $\endgroup$– snsunxJan 31, 2016 at 3:21
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$\begingroup$ 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. $\endgroup$ Oct 14, 2022 at 17:32