Boltzmann Entropy
- Date:
28 July 2019
Dependency
What is the problem ?
The Key of Statistical Mechanics
Link between Thermodynamics (macroscopic) And Statistical Mechanics (microscopic)
Derivation
Step 1 Fundamental equation in the Entropy Representation
From Thermodynamic Potentials, we have
\[\begin{split}\begin{align*}
&dS ={1\over T}dE
- \sum\limits_{i=1}^r {F_i \over T} dx_i
- \sum\limits_{j=1}^s {\mu_j \over T} dN_j \\
&\Rightarrow S=S(E,{x_i},{N_j})
\end{align*}\end{split}\]
Step 2 Previously, we treated \(S\) purely as a theoretical macroscopic variable. But Boltzmann figured out the exact microscopic expression (origin) of it!
\[\begin{split}\begin{align*}
& \boxed{ S(E,{x_i},{N_j}) = k_B \ln \Omega(E,{x_i},{N_j}) }\\
& where\\
& k_B: \text{ Boltzmann Constant }\\
& \Omega(E,{x_i},{N_j}): \text{ Number of Microstates corresponding to } (E,{x_i},{N_j})
\end{align*}\end{split}\]
Example 1: Two State System
\(N\) identical particles, with two energy levels \(\epsilon_n = (-1)^n \epsilon, \;\; n = 1,2\)
Combination : The number of ways of picking \(k\) unordered outcomes from \(n\) possibilities
Stirling Approximation: \(\ln N! \approx N\ln N -N\)
\[\begin{split}\begin{align*}
& \begin{cases}
N=N_1+N_2\\
E=-\epsilon N_1+ \epsilon N_2
\end{cases}
\Rightarrow \begin{cases}
N_1 = {N\over 2}(1-{E\over N \epsilon})\\
N_2 = {N\over 2}(1+{E\over N \epsilon})
\end{cases}\\
\Omega(E,N;N_2)& = {N!\over N_2!(N-N_2)!}\;\; (\text{Combination})\\
S(E,N;N_2)& = k_B \ln \Omega=...(\text{Stirling Approximation})\\
& = -k_B N\left[ {N_2\over N}\ln {N_2\over N}+ {N-N_2\over N}\ln {N-N_2\over N}\right]\\
& \text{Next just substitute } N_2 \text{ we get } S(E,N)
\end{align*}\end{split}\]