- #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

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! ##

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