# 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.

Please help me clarify my doubts, is my tutor correct, or are my thoughts correct? Your guidance is much appreciated, thank you!

• 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 }