Editing Maxwell–Boltzmann statistics (section)

=== Derivation from canonical ensemble ===
{{Technical|section|date=December 2013}}
In the above discussion, the Boltzmann distribution function was obtained via directly analysing the multiplicities of a system. Alternatively, one can make use of the [[canonical ensemble]]. In a canonical ensemble, a system is in thermal contact with a reservoir.  While energy is free to flow between the system and the reservoir, the reservoir is thought to have infinitely large heat capacity as to maintain constant temperature, ''T'', for the combined system.

In the present context, our system is assumed to have the energy levels <math>\varepsilon _i</math> with degeneracies <math>g_i</math>. As before, we would like to calculate the probability that our system has energy <math>\varepsilon_i</math>.

If our system is in state <math>\; s_1</math>, then there would be a corresponding number of microstates available to the reservoir. Call this number <math>\; \Omega _ R (s_1)</math>. By assumption, the combined system (of the system we are interested in and the reservoir) is isolated, so all microstates are equally probable. Therefore, for instance, if <math> \; \Omega _ R (s_1) = 2 \; \Omega _ R (s_2) </math>, we can conclude that our system is twice as likely to be in state <math>\; s_1</math> than <math>\; s_2</math>. In general, if <math>\; P(s_i)</math> is the probability that our system is in state <math>\; s_i</math>,

:<math>\frac{P(s_1)}{P(s_2)} = \frac{\Omega _ R (s_1)}{\Omega _ R (s_2)}.</math>

Since the [[entropy]] of the reservoir <math>\; S_R = k \ln \Omega _R</math>, the above becomes

:<math>\frac{P(s_1)}{P(s_2)} = \frac{ e^{S_R(s_1)/k} }{ e^{S_R(s_2)/k} } = e^{(S_R (s_1) - S_R (s_2))/k}.</math>

Next we recall the thermodynamic identity (from the [[first law of thermodynamics]]):

:<math>d S_R = \frac{1}{T} (d U_R + P \, d V_R - \mu \, d N_R).</math>

In a canonical ensemble, there is no exchange of particles, so the <math>d N_R</math> term is zero. Similarly, <math>d V_R = 0.</math>  This gives

:<math> S_R (s_1) - S_R (s_2) = \frac{1}{T} (U_R (s_1) - U_R (s_2)) = - \frac{1}{T} (E(s_1) - E(s_2)),</math>

where <math> U_R (s_i) </math> and <math> E(s_i) </math> denote the energies of the reservoir and the system at <math>s_i</math>, respectively. For the second equality we have used the conservation of energy. Substituting into the first equation relating <math>P(s_1), \; P(s_2)</math>:

:<math>
\frac{P(s_1)}{P(s_2)} =  \frac{ e^{ - E(s_1) / kT } }{ e^{ - E(s_2) / kT} },
</math>

which implies, for any state ''s'' of the system

:<math>
P(s) = \frac{1}{Z} e^{- E(s) / kT},
</math>

where ''Z'' is an appropriately chosen "constant" to make total probability 1. (''Z'' is constant provided that the temperature ''T'' is invariant.)

:<math> Z = \sum _s e^{- E(s) / kT}, </math>

where the index ''s'' runs through all microstates of the system. ''Z'' is sometimes called the Boltzmann '''sum over states''' (or "Zustandssumme" in the original German). If we index the summation via the energy eigenvalues instead of all possible states, degeneracy must be taken into account. The probability of our system having energy <math>\varepsilon _i</math> is simply the sum of the probabilities of all corresponding microstates:

:<math>P (\varepsilon _i) = \frac{1}{Z} g_i e^{- \varepsilon_i / kT}</math>

where, with obvious modification,

:<math>Z = \sum _j g_j  e^{- \varepsilon _j / kT},</math>

this is the same result as before.

Comments on this derivation:
*Notice that in this formulation, the initial assumption "... ''suppose the system has total ''N'' particles''..." is dispensed with. Indeed, the number of particles possessed by the system plays no role in arriving at the distribution. Rather, how many particles would occupy states with energy <math>\varepsilon _i</math> follows as an easy consequence.
*What has been presented above is essentially a derivation of the canonical partition function. As one can see by comparing the definitions, the Boltzmann sum over states is equal to the canonical partition function.
*Exactly the same approach can be used to derive [[Fermi–Dirac statistics|Fermi–Dirac]] and [[Bose–Einstein statistics|Bose–Einstein]] statistics. However, there one would replace the canonical ensemble with the [[grand canonical ensemble]], since there is exchange of particles between the system and the reservoir. Also, the system one considers in those cases is a single particle ''state'', not a particle. (In the above discussion, we could have assumed our system to be a single atom.)