Statistical Mechanics: Boltzmann partition function

Question

I’m a high school student and was recently studying basic statistical mechanics (for use in physical chemistry). In the derivation of the Boltzmann derivation or the partition function we arrive at a factor $e^{-\beta T}$. So I searched for the connection of $\beta$ to temperature and got one proof for that on the wikipedia site. But it seems to make to make the assumptions that

1. The energy of a system in equilibrium depends only on the temperature.

2. At equilibrium the number of microstates is maximized.

3. It uses the formula $S=k\ln(W)$.

   So I just wanted to know that which of the assumptions is an experimental fact and whether we can arrive at the equation $S=k\ln(W)$ without already using $\beta=1/(k_BT)$ otherwise the proof would become circular.

To improvise, we can use the property of Lagrange multipliers to obain that $\beta=\frac{d(ln(W))}{dT}$ but I still don’t see how can we reach to the value of $\beta$ from here.

P.S. I am just a beginner so please use simple language if possible.

Answer 1

From (2) and $\Omega = \Omega_A \times \Omega_B$ one can see that the fractional change of the number of microstates with energy $\frac{1}{\Omega}\frac{d\Omega}{dE}$ must be equal for two systems in thermal equilibrium. This is the thermodynamic beta $\beta.$ At room temperature it is about 4 % per meV.

Now consider a tiny system that is in thermal contact with a large system. The $\beta$ of that large system will not change much when the tiny system absorbs some energy. But its number of microstates will become smaller. The probability will be lower because of that.

For simplicity, assume one microstate per energy for the small system, maybe just one harmonic oscillator. Rewrite the expression for $\beta = \frac{1}{\Omega}\frac{d\Omega}{dE}$ to a differential equation for the large system $\frac{d\Omega}{dE} = - \beta \Omega$. (The minus sign appears because the energy $dE$ is transferred to the small system.)

Finally solve to obtain the Boltzmann factor: $P(E) \propto \exp[-\beta E]$.

The expression is not valid when particles are indistinguishable and Bose- or Fermi-statistics should be used. The assumption that this derivation relies on is the distinguishability of the small harmonic oscillator.