How do you derive the Maxwell-Boltzmann distribution?

Question

I have searched for a reasonable derivation online, but so far have been unable to find one which doesn't skip steps or presume prior knowledge.

I found a derivation on this Wikipedia article which gives the following relation as motivation for the rest of the equations:

$$-\log\Big(\frac{N_i}{N}\Big) \propto \frac{E_i}{T}$$

where $N_i$ is the expected number of particles in the single-particle microstate $i$, $N$ is the total number of particles in the system, $E_i$ is the energy of microstate $i$, and $T$ is the equilibrium temperature of the system.

This part intuitively makes sense to me, however, the next part of the derivation says you can gain the following by introducing a normalising factor:

$$\frac{N_i}{N} = \frac{\exp\big(\frac{-E_i}{k_BT}\big)}{\sum_j \exp\big(\frac{-E_j}{k_BT}\big)}$$

where $E_j$ is the energy of microstate $j$, and $k_B$ is the Boltzmann constant. What's the meaning behind this normalising factor, and how does its meaning motivate its use? What relation does it have to the Boltzmann constant?

Answer 1

Staring from the first part that make sense to you (this part is considered rather difficult to me!!!) : \begin{align} -\log\Big(\frac{N_i}{N}\Big) &\propto \frac{E_i}{K_BT}\\ \frac{N_i}{N} &\propto \exp\left( -\frac{E_i}{K_BT}\right)\\ \frac{N_i}{N} &= C \exp\left( -\frac{E_i}{K_BT}\right) \end{align} Where $C$ is a constant to be determined.

Then find the constant $C$ from the condition $\sum \frac{N_i}{N} = \frac{N}{N} = 1$: \begin{align} 1 &= \sum_i \frac{N_i}{N}\\ &= \sum_i C \exp\left( -\frac{E_i}{K_BT}\right)\\ &= C \sum_i \exp\left( -\frac{E_i}{K_BT}\right) \end{align}

Therefore $$ C= \frac{1}{\sum_j \exp\big(\frac{-E_j}{k_BT}\big)}$$

And $$ \frac{N_i}{N} = \frac{ \exp\left( -\frac{E_i}{K_BT}\right) }{ \sum_j \exp\big(\frac{-E_j}{k_BT}\big) } $$