- #1
Mentz114
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- 292
Starting from the definition of energy levels ##e_n## and occupations ##a_n## and
the conditions ##\sum_n a_n = N## (2.2) and ##\sum_n a_n e_n = E## (2.3) where ##N## and ##E## are fixed I'm trying to find the distribution which extremizes the Shannon entropy.
Using the frequency ##f_n=a_n/N## and ##S=\sum f_n\log(f_n)## I need to solve the variation ( ##\lambda## and ##\mu## are Lagrange multipliers) ##\delta\left(S+\lambda(\sum_n f_n-1) + \mu(\sum_n e_n f_n -E/N\right)=0##.
[tex]
\begin{align*}
\delta(S) &= \sum \delta f_n\log(f_n) + \sum f_n \delta(\log(f_n) )\\
&\sum f_n \delta(\log(f_n) ) = \sum f_n (\delta f_n)/f_n=\sum \delta f_n
\end{align*}
[/tex]
and the total variation is
[tex]
\begin{align*}
\delta &= (1+\lambda)\sum \delta f_n + \sum \delta f_n\log(f_n) + \mu\sum e_n\delta f_n
\end{align*}
[/tex]
Setting ##\lambda=-1## we are left with ##\sum \delta f_n\log(f_n)=- \mu\sum e_n\delta f_n ## from which follows ##a_n=Ne^{-\mu e_n}##.
This is the first time I've attempted a constrained variation so I'm not entirely confident that this is right. I hope it is because it is more satisfactory than methods** that use Stirlings formula. Also ##\lambda## can be trivially eliminated.
The interpetation can now be made that the state derived has the greatest entropy/disorder of all possible states that satisfy the constraints. There is a physical connection which is absent if a probability is extremised, because probability has no physical analog, unlike entropy.
** This method extremises the log of ##P=N!/\Pi a_n! ##
the conditions ##\sum_n a_n = N## (2.2) and ##\sum_n a_n e_n = E## (2.3) where ##N## and ##E## are fixed I'm trying to find the distribution which extremizes the Shannon entropy.
Using the frequency ##f_n=a_n/N## and ##S=\sum f_n\log(f_n)## I need to solve the variation ( ##\lambda## and ##\mu## are Lagrange multipliers) ##\delta\left(S+\lambda(\sum_n f_n-1) + \mu(\sum_n e_n f_n -E/N\right)=0##.
[tex]
\begin{align*}
\delta(S) &= \sum \delta f_n\log(f_n) + \sum f_n \delta(\log(f_n) )\\
&\sum f_n \delta(\log(f_n) ) = \sum f_n (\delta f_n)/f_n=\sum \delta f_n
\end{align*}
[/tex]
and the total variation is
[tex]
\begin{align*}
\delta &= (1+\lambda)\sum \delta f_n + \sum \delta f_n\log(f_n) + \mu\sum e_n\delta f_n
\end{align*}
[/tex]
Setting ##\lambda=-1## we are left with ##\sum \delta f_n\log(f_n)=- \mu\sum e_n\delta f_n ## from which follows ##a_n=Ne^{-\mu e_n}##.
This is the first time I've attempted a constrained variation so I'm not entirely confident that this is right. I hope it is because it is more satisfactory than methods** that use Stirlings formula. Also ##\lambda## can be trivially eliminated.
The interpetation can now be made that the state derived has the greatest entropy/disorder of all possible states that satisfy the constraints. There is a physical connection which is absent if a probability is extremised, because probability has no physical analog, unlike entropy.
** This method extremises the log of ##P=N!/\Pi a_n! ##
Last edited: