I am a bit confused about the difference between macrostate and microstate in the microcanonical ensemble. So I have read that for the microcanonical ensemble, the probabilities of each microstate are equal $$ p = 1/\Omega $$ where $\Omega$ is the number of microstates. For a given number of particles $N$ the number of microstates of particles distributed over discreet energy levels is given by

$$\Omega = \frac{N!}{\prod_i n_i} $$

where $n_i$ is the number of particles in the $i$th energy level. Maximising $\Omega$ given the constraint that the particle number is constant $\sum_i n_i = N $ and the energy is constant $\sum_i \varepsilon_i n_i = E$ gives

$$ p_i \propto e^{-\varepsilon_i/kT}$$

This gives a probability for a particle to be in the $i$th energy level. I thought if we were in the microcanonical ensemble all probabilities are equal?


  • $\begingroup$ What is equal is the probability of each microstate, that is, a given distribution of particles with total energi E. Each microstate will have particles with different energies with some distribution. Then, after averaging across all posible microstates, you get the probability $p_i$ $\endgroup$ – user126422 Apr 21 '17 at 1:32
  • $\begingroup$ Firstly , you lost the factorial in the formula for $\Omega$: $n_i!$. Secondly , $\ p_i$ is not the probability of a microstate of the microcanonical ensemble $\endgroup$ – Aleksey Druggist Jun 27 '17 at 9:33

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.