# Deriving the Boltzmann Distribution

I have a basic understanding of thermodynamics, and I came across this derivation of the Boltzmann distribution. I understand all of it except the end and I need some clarification.

At the end, the website claims that the number of particles with energy $E_i$ denoted as $n_i$ is given by: $$n_i=\dfrac{N}{\sum_i e^{-\beta E_i}}e^{-\beta E_i}$$

From what I understand, the Boltzmann Distribution tells me the probability of finding a particle with energy $E_i$, so I simply need $\dfrac{n_i}{N}$?

My last question is that the website abruptly claims that $\beta=1/k_BT$, which apparently comes from applying the zeroth law of Thermodynamics. How can show this is true?

• – higgsss Apr 3 '18 at 4:22
• The point is how the Lagrange multiplier $\beta$ is equal to $\partial \ln \Omega/ \partial E = 1/kT$, where $\Omega$ is the number of microstates, and $T$ is the thermodynamic temperature. In fact, a theorem about constrained extrema (stated and proved in the linked post) guarantees this. "Working with real systems and applying the zeroth law of thermodynamics" is an extraneous statement. – higgsss Apr 3 '18 at 7:12

When $\beta$ is the same for systems A and B as well as for B and C, also A and C will be in thermal equilibrium. That is what the zeroth law is about. For historical reasons, the connection with the thermodynamic temperature is $\beta = 1/kT$. But one could have used coldness $\beta = \frac{1}{\Omega} \frac{d\Omega}{dE}$ instead. For example at room temperature, the coldness is about 4 % per milli-eV. I wrote about that in an earlier answer.