# What does the probability $p_r$ for Boltzmann's distribution represent?

In F.Mandl's book Statisical Physics. He derived Boltsmann's Distribution for a canonical system of N particles to be:

$$P_r = \frac{e^{-\beta E_r}}{Z}$$

Where Z is the partition function of the system, and $\beta = \frac{1}{kT}$.

This is my uncertainty: My tutor told me that this probability is the probability of one of the mircrostates of Energy = $E_r$ in other words $P_r = \frac{1}{\Omega(E_r)} = \frac{1}{all \hspace{2mm} arrangements \hspace{2mm} with \hspace{2mm} energy \hspace{2mm} E_r}$. Therefore if the canonical system was made up of N subsystems instead of N particles, the partition function would be simply powered by N. Thus the new boltzmann's distribution is given by: $P_r = \frac{e^{-\beta E_r}}{Z^N}$.

However to me, from the derivation, $P_r$ looks more like the probability of finding the state of the system of interest to have energy $E_r$. I guessed this was so, since we started the derivation of $P_r$ in Mandl's textbook to be given by the ratio of statistical weight of the heatbath to have energy corresponding to total energy minus energy of system of interest, $E_{0}-E_r$, to the sum of all statistical weights of all energies.

• If the number of microstates of energy $E_r$ is $\Omega(E_r)$, then the probability that the system has energy $E_r$ in the canonical ensemble at inverse temperature $\beta$ is $\Omega(E_r) P_r$. The probability of observing one specific microstate of energy $E_r$ is $P_r$. – Yvan Velenik Mar 4 '17 at 13:13

Let $P_i$ be,

$$P_i := \frac{1}{Z}\exp (-\beta E_i).$$

Then $P_i$ is the probability that the system described by the partition function $Z$, occupies the microstate with energy $E_i$. Thus, for example, the probability that the system is in the state with energy $E_4$ or the state with energy $E_5$ is,

$$P(E_4 \lor E_5) = \frac{1}{Z} \left[\exp(-\beta E_4) + \exp(-\beta E_5) \right]$$

since the probabilities are summed. Now suppose we have two identical copies of the same system. If we want to compute the probability that a system is in a state with energy $E_1$ and another being in the state with energy $E_2$, we have,

$$P(\mathrm{subsytem\, in \, state \, E_1} \land\mathrm{subsytem\, in \, state \, E_2}) = \frac{2}{Z^2} \exp(-\beta(E_1+E_2))$$

since we multiply probabilities since we want the system to be in state $E_1$ and another in $E_2$. Notice since they are indistinguishable, it is achievable two ways, hence the factor of $2$. This is equivalent to the definition of the partition function of $N$ indistinguishable subsystems:

$$Z_{\mathrm{total}} = \frac{Z^N}{N!}.$$

in micro canonical ensemble all states has an equal probability to reach.then based on your tutor, the probability of each micro state is

P_r=\frac{1}{Ω(Er)}=\frac{1}{all arrangements with energy E_r}.


but in canonical ensemble based on your system Temperature, you can reach to one state.then you cannot take all states with equal probability and this means that z^N isn't the normalization factor. because each state based on the ratio of it's energy to k_BT has a probability of finding.

in fact, in canonical ensemble the probability of finding system with energy E_r in temperature T is

p_r={all arrangements with energy E_r }/{all possible state with boltzman }