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?

  • 2
    $\begingroup$ It is derived in every statistical physics textbook. $\endgroup$
    – Roger V.
    Sep 13, 2021 at 18:35
  • $\begingroup$ @Roger Vadim What is you favourite example of such a book for simplicity of derivation with no skipped steps or assumed knowledge? $\endgroup$
    – Connor
    Sep 13, 2021 at 18:36
  • $\begingroup$ The normalization simply comes from the fact that $N = \sum_i N_i$ $\endgroup$ Sep 13, 2021 at 18:39
  • $\begingroup$ @BySymmetry Okay, that sort of makes sense, could you explain how one uses that to move from the upper relation, to the lower equation? Or are the two not as related as the Wikipedia article implies? In what sense is the Boltzmann constant related to the normalising factor? Thanks! $\endgroup$
    – Connor
    Sep 13, 2021 at 18:48
  • $\begingroup$ Every demonstration will skip some steps, that you are assumed to know if you are reading at that level. Otherwise books would become extremely long, and boring too $\endgroup$
    – user65081
    Sep 13, 2021 at 19:00

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

  • $\begingroup$ That's great thankyou, but where does the $k_B$ come from? Other than that though, nicely derived! $\endgroup$
    – Connor
    Sep 14, 2021 at 16:39
  • $\begingroup$ Short answer: The exponent must be a dimensionless quantity, the denominator must be added a constant to have a unit of energy. Long answer: This is the hard part to derive the Boltzmann distribution from micro-canonical formulation. $\endgroup$
    – ytlu
    Sep 15, 2021 at 2:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.