# Statistical Mechanics: Boltzmann partition function

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.

• Please provide a more descriptive title. Some people who might answer will not even read the question. – garyp Jun 17 '18 at 14:32
• You may want to consider emphasizing the question(s). I think the question is about the assumptions for the proof? – user191954 Jun 17 '18 at 15:36
• yes @Chair i want to know wether S=kln(W) came first or B=1/kt – DHYEY Jun 17 '18 at 15:55
• The historical development of thermodynamics and statistical mechanics is not often that enlightening. In my answer below, I avoided entropy. It does not clarify things at highschool level. – user137289 Jun 17 '18 at 16:01

## 1 Answer

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.

• thanks. but my main confusion is that how do we know that b=1/kT. – DHYEY Jun 17 '18 at 15:57
• @DHYEY What do you mean with $T$? Is it the ideal gas temperature or the thermodynamic Carnot temperature that you want to connect to statistical mechanics? In my answer, I had interpreted your question as asking about the connection between the Boltzmann factor and beta, which is a measure of temperature. And temperature is most easily understood as a function of beta, the relative change of the number of microstates with energy. – user137289 Jun 17 '18 at 16:14
• I wanted a connection between the thermodyanamic temperature and beta.. – DHYEY Jun 21 '18 at 11:44
• BTW just as a curiosity do the thermodynamic and ideal gas temperatures differ anywhere except in their way of derivation – DHYEY Jun 21 '18 at 11:46