So, our professor introduced the Bose-Einstein statistics by deriving the Grand Canonical Partition function of a boson system associated to a single energy state $\epsilon_r$. So the formula is: $$\mathcal{Z} = \sum_{n = 0}^\infty \exp(-\beta(n\epsilon_r - \mu n)) = \frac{1}{1-\exp(-\beta(\epsilon_r-\mu))}$$ Then I know that we can derive the Maxwell-Boltzmann statistics of the same system (now consisting of distinguishable particles) through its Canonical Partition function. Let's assume that we have $N$ particles and $m$ different energy states. Then the formula is: $$Z = \sum_{n = 0}^N \frac{N!}{n_1!n_2!\dots n_m!} \exp(-\beta n_1 \epsilon_1)\dots\exp(-\beta n_m \epsilon_m) = (\exp(-\beta \epsilon_1)+\dots+\exp(-\beta \epsilon_m))^N$$ Now if I want to derive again the Maxwell-Boltzmann statistics, but in the Grand Canonical framework like I did for the Bose-Einstein, it does not add up. Given a single energy state $\epsilon_r$ (that I assume being connected to a particle reservoir with chemical potential $\mu$) the Grand Canonical Partition function becomes: $$\mathcal{Z} = \sum_{n = 0}^N \frac{N!}{n! (N-n)!} \exp(-\beta(n \epsilon_r - \mu n)) = (\exp(-\beta(\epsilon_r - \mu))+1)^N$$ By multiplying by the binomial coefficient I'm taking care of all possible distinguishable states (which are the number of possible ways to pick $n$ particles from the existing $N$). But when I try to evaluate the expected number of particles of this energy state, the èresult is different from the Canonical prediction. $$<N_{\epsilon_r}> = \frac{1}{\beta} \frac{\partial \log(\mathcal{Z})}{\partial \mu} = N \frac{\exp(-\beta(\epsilon_r-\mu))}{\exp(-\beta(\epsilon_r - \mu))+1}$$ and from this point I'm not able to get to the actual correct prediction: $$<N_{\epsilon_r}> = N \frac{\exp(-\beta\epsilon_r)}{\sum_r \exp(-\beta\epsilon_r)}$$ Why?

  • $\begingroup$ It seems to me that Maxwell-Boltzmann statistics cannot be derived consistently from the first principles of statistical physics and it is approximation for Bose-Einstein and Fermi- Dirac statistics for the case when the average numbers of particles in one quantum state (ocupation numbers) are much less than unity. $\endgroup$ – Aleksey Druggist Jun 29 at 13:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.