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How to derive an expression for entropy in form of

$S = \ln \Omega$

from the form

$\displaystyle{S = - \sum_i \; p_i \ln p_i}$ ?

That is the last formula taken as a definition of entropy.

Just a reference will do. The backwards derivation (probably with some assumptions) is given in Landau and Lifshitz.

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WHat is $\Omega$? For uniform distribution $p_i=1/n$ the entropy is $ln n$. –  Andyk Jan 30 '12 at 17:39
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1 Answer

up vote 3 down vote accepted

If there are $\Omega$ states and $p_i$ is constant by the fundamental postulate of statistical mechanics, then you have

$$1=\sum_{i=1}^\Omega\ p_i=\Omega\ p_i\ \ \Longrightarrow\ \ p_i=\frac{1}{\Omega},$$ and consequently $$S = - \sum_{i=1}^\Omega\ p_i \ln p_i=- \sum_{i=1}^\Omega\ p_i \ln \frac{1}{\Omega}=\left(-\ln \frac{1}{\Omega}\right) \sum_{i=1}^\Omega\ p_i=\ln\Omega.$$

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Well, that's exact for microcanonical ensemble . The thing is that $\ln \Omega$ is used for canonical ensemble too. After a bit of thinking I think I got why it is valid and what does $\Omega$ mean for a canonical ensemble. Anyway I don't need it anymore, so I won't check my guess. –  Yrogirg Jan 31 '12 at 15:27
    
@Yrogirg: Wait. In the canonical ensamble, don't you just compute the partition function with $p_i=e^{-H/T}$ and take the logarithm, so you get the free energy and compute an expression for the entropy from there? –  NiftyKitty95 Jan 31 '12 at 15:56
    
The question is not how do you compute entropy (anyway one cannot generally compute the partition function), but how do you define entropy in statistical physics. I prefer the definition as $S = - \sum_i \; p_i \ln p_i$ ($- \int \rho \ln \rho \; d \Gamma$, $\text{Tr} \hat \rho \ln \hat \rho$). But following Boltzmann some use $\ln \Omega$ definition. I don't quite get that definition, at least in ensembles other than micro canonical. So I've asked how to derive one from the other partly because I wanted to understand that $\ln \Omega$ definition. –  Yrogirg Jan 31 '12 at 16:10
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