Boltzmann Entropy
=================================

:Date: 28 July 2019

Dependency
-------------

:doc:`Thermodynamic-Potentials`


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 :doc:`Thermodynamic-Potentials`, we have

.. math::
	\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*}

**Step 2** Previously, we treated :math:`S` purely as a theoretical macroscopic variable. But Boltzmann figured out the exact microscopic expression (origin) of it! 

.. math::
	\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*}

Example 1: Two State System
---------------------------------

:math:`N` identical particles, with two energy levels :math:`\epsilon_n = (-1)^n \epsilon, \;\; n = 1,2`

- Combination_ : The number of ways of picking :math:`k` unordered outcomes from :math:`n` possibilities
- Stirling Approximation: :math:`\ln N! \approx N\ln N -N`

.. _Combination: http://mathworld.wolfram.com/Combination.html

.. math::
	\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*}