Take the 2-minute tour ×  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.  share|improve this question WHat
is $\Omega$? For uniform distribution $p_i=1/n$ the entropy is
$ln n$. –  Andyk Jan 30 '12 at 17:39  1 Answer 1  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 $$S
= - \sum_{i=1}^\Omega\ p_i \ln p_i=- \sum_{i=1}^\Omega\ p_i \ln
\frac{1}{\Omega}=\left(-\ln \frac{1}{\Omega}\right)
\sum_{i=1}^\Omega\ p_i=\ln\Omega.$$  share|improve this answer
Well, that's exact for microcanonical ensemble . The thing is
that $\ln \Omega$ is used for canonical ens