Maxwell–Boltzmann statistics

In statistical mechanics, Maxwell–Boltzmann statistics describes the statistical distribution of material particles over various energy states in thermal equilibrium, when the temperature is high enough and density is low enough to render quantum effects negligible. Maxwell–Boltzmann statistics are therefore applicable to almost any terrestrial phenomena for which the temperature is above a few tens of kelvins.

The expected number of particles with energy $$\epsilon_i$$ for Maxwell–Boltzmann statistics is $$N_i$$ where:



\frac{N_i}{N} = \frac {g_i} {e^{(\epsilon_i-\mu)/kT}} = \frac{g_i e^{-\epsilon_i/kT}}{Z} $$

where:
 * $$N_i$$ is the number of particles in state i
 * $$\epsilon_i$$ is the energy of the i-th state
 * $$g_i$$ is the degeneracy of energy level i, the number of particle's states (excluding the "free particle" state) with energy $$\epsilon_i$$
 * μ is the chemical potential
 * k is Boltzmann's constant
 * T is absolute temperature
 * N is the total number of particles
 * $$N=\sum_i N_i\,$$


 * Z is the partition function
 * $$Z=\sum_i g_i e^{-\epsilon_i/kT}$$


 * e(...) is the exponential function

Equivalently, the distribution is sometimes expressed as



\frac{N_i}{N} = \frac {1} {e^{(\epsilon_i-\mu)/kT}}= \frac{e^{-\epsilon_i/kT}}{Z} $$

where the index i now specifies a particular state rather than the set of all states with energy $$\epsilon_i$$



A derivation of the Maxwell–Boltzmann distribution
In this particular derivation, the Boltzmann distribution will be derived using the assumption of distinguishable particles, even though the ad hoc correction for Boltzmann counting is ignored, the results remain valid.

Suppose we have a number of energy levels, labelled by index i, each level having energy $$\epsilon_i$$ and containing a total of $$N_i$$ particles. To begin with, let's ignore the degeneracy problem. Assume that there is only one way to put $$N_i$$ particles into energy level i.

The number of different ways of performing an ordered selection of one object from $$N$$ objects is obviously $$N$$. The number of different ways of selecting 2 objects from $$N$$ objects, in a particular order, is thus $$N(N-1)$$ and that of selecting $$n$$ objects in a particular order is seen to be $$N!/(N-n)!$$. The number of ways of selecting 2 objects from $$N$$ objects without regard to order is $$N(N-1)$$ divided by the number of ways 2 objects can be ordered, which is 2!. It can be seen that the number of ways of selecting $$n$$ objects from $$N$$ objects without regard to order is the binomial coefficient: $$N!/n!(N-n)!$$. If we have a set of boxes numbered $$1,2, \ldots, k$$, the number of ways of selecting $$N_1$$ objects from $$N$$ objects and placing them in box 1, then selecting $$N_2$$ objects from the remaining $$N-N_1$$ objects and placing them in box 2 etc. is


 * $$W=\left(\frac{N!}{N_1!(N-N_1)!}\right)~\left(\frac{(N-N_1)!}{N_2!(N-N_1-N_2)!}\right)~\ldots

\left(\frac{N_k!}{N_k!0!}\right)$$
 * $$=N!\prod_{i=1}^k (1/N_i!)$$

where the extended product is over all boxes containing one or more objects. If the i-th box has a "degeneracy" of $$g_i$$, that is, it has $$g_i$$ sub-boxes, such that any way of filling the i-th box where the number in the sub-boxes is changed is a distinct way of filling the box, then the number of ways of filling the i-th box must be increased by the number of ways of distributing the $$N_i$$ objects in the $$g_i$$ boxes. The number of ways of placing $$N_i$$ distinguishable objects in $$g_i$$ boxes is $$g_i^{N_i}$$. Thus the number of ways ($$W$$) that $$N$$ atoms can be arranged in energy levels each level $$i$$ having $$g_i$$ distinct states such that the i-th level has $$N_i$$ atoms is:


 * $$W=N!\prod \frac{g_i^{N_i}}{N_i!}$$

For example, suppose we have three particles, $$a$$, $$b$$, and $$c$$, and we have three energy levels with degeneracies 1, 2, and 1 respectively. There are 6 ways to arrange the 3 particles so that $$N_1$$ = 2, $$N_2$$ = 1 and $$N_3$$ = 0.

The six ways are calculated from the formula:


 * $$W=N!\prod \frac{g_i^{N_i}}{N_i!}= 3!

\left(\frac{1^2}{2!}\right) \left(\frac{2^1}{1!}\right) \left(\frac{1^0}{0!}\right)=6 $$

We wish to find the set of $$N_i$$ for which $$W$$ is maximized, subject to the constraint that there be a fixed number of particles, and a fixed energy. The maxima of $$W$$ and $$\ln(W)$$ are achieved by the same values of $$N_i$$ and, since it is easier to accomplish mathematically, we will maximize the latter function instead. We constrain our solution using Lagrange multipliers forming the function:



f(N_1,N_2,...,N_n)=\ln(W)+\alpha(N-\sum N_i)+\beta(E-\sum N_i \epsilon_i) $$

Using Stirling's approximation for the factorials



N! \approx N^N e^{-N},

$$

we obtain:



\ln (N!) = N \ln N - N

$$

Then



\ln W=\ln\left[N!\prod\limits_{i=1}^{n}\frac{g_i^{N_i}}{N_i!}\right]=\ln N!+\sum\limits_{i=1}^n\left(N_i\ln g_i-N_i\ln N_i + N_i\right) $$

Finally



f(N_1,N_2,...,N_n)=N\ln(N)-N+\alpha N +\beta E + \sum\limits_{i=1}^n\left(N_i\ln g_i-N_i\ln N_i + N_i-(\alpha+\beta\epsilon_i) N_i\right) $$

In order to maximize the expression above we apply Fermat's theorem (stationary points), according to which local extrema, if exist, must be at critical points (partial derivatives vanish):



\frac{\partial f}{\partial N_i}=\ln g_i-\ln N_i -(\alpha+\beta\epsilon_i) = 0 $$

By solving the equations above ($$i=1\ldots n$$) we arrive to an expression for $$N_i$$:



N_i = \frac{g_i}{e^{\alpha+\beta \epsilon_i}} $$

It can be shown thermodynamically that β = 1/kT where $$k$$ is Boltzmann's constant and T is the temperature, and that α = -μ/kT where μ is the chemical potential, so that finally:



N_i = \frac{g_i}{e^{(\epsilon_i-\mu)/kT}} $$

Note that the above formula is sometimes written:



N_i = \frac{g_i}{e^{\epsilon_i/kT}/z} $$

where $$z=exp(\mu/kT)$$ is the absolute activity.

Alternatively, we may use the fact that


 * $$\sum_i N_i=N\,$$

to obtain the population numbers as



N_i = N\frac{g_i e^{-\epsilon_i/kT}}{Z} $$

where $$Z$$ is the partition function defined by:



Z = \sum_i g_i e^{-\epsilon_i/kT} $$

Another derivation
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 be have energy levels $$\epsilon _i$$ with degeneracies $$g_i$$. As before, we would like to calculate the probability that our system has energy $$\epsilon_i$$.

If our system is in state $$\; s_1$$, then there would be a corresponding number of microstates available to the reservoir. Call this number $$\; \Omega _ R (s_1)$$. 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 $$ \; \Omega _ R (s_1) = 2 \; \Omega _ R (s_2) $$, we can conclude that our system is twice as likely to be in state $$\; s_1$$ than $$\; s_2$$. In general, if $$\; P(s_i)$$ is the probability that our system is in state $$\; s_i$$,


 * $$\frac{P(s_1)}{P(s_2)} = \frac{\Omega _ R (s_1)}{\Omega _ R (s_2)}.$$

Since the entropy of the reservoir $$\; S_R = k \ln \Omega _R$$, the above becomes


 * $$\frac{P(s_1)}{P(s_2)} = \frac{ e^{S_R(s_1)/k} }{ e^{S_R(s_2)/k} } = e^{(S_R (s_1) - S_R (s_2))/k}.$$

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


 * $$d S_R = \frac{1}{T} (d U_R + P d V_R - \mu d N_R)$$.

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


 * $$(S_R (s_1) - S_R (s_2)) = \frac{1}{T} (U_R (s_1) - U_R (s_2)) = - \frac{1}{T} (E(s_1) - E(s_2))$$,

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



\frac{P(s_1)}{P(s_2)} = \frac{ e^{ - E(s_1) / kT } }{ e^{ - E(s_2) / kT} } $$,

which implies, for any state s of the system



P(s) = \frac{1}{Z} e^{- E(s) / kT} $$,

where Z is an appropriately chosen "constant" to make total probability 1. (Z is constant provided that the temperature T is invariant.) It is obvious that


 * $$\; Z = \sum _s e^{- E(s) / kT} $$,

where the index s runs through all microstates of the system. 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 $$\epsilon _i$$ is simply the sum of the probabilities of all corresponding microstates:


 * $$P (\epsilon _i) = \frac{1}{Z} g_i e^{- \epsilon_i / kT}$$

where, with obvious modification,

$$Z = \sum _j g_j e^{- \epsilon _j / kT}$$,

this is the same result as before.

Comments

 * 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 $$\epsilon _i$$ follows as an easy consequence.


 * What has been presented above is essentially a derivation of the canonical partition function. As one can tell by comparing the definitions, the Boltzman sum over states is really no different from the canonical partition function.


 * Exactly the same approach can be used to derive Fermi–Dirac and 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.)

Limits of applicability
The Bose–Einstein and Fermi–Dirac distributions may be written:



N_i = \frac{g_i}{e^{(\epsilon_i-\mu)/kT}\pm 1} $$

Assuming the minimum value of $$\epsilon_i$$ is small, it can be seen that the condition under which the Maxwell–Boltzmann distribution is valid is when


 * $$e^{-\mu/kT} \gg 1$$

For an ideal gas, we can calculate the chemical potential using the development in the Sackur–Tetrode article to show that:


 * $$\mu=\left(\frac{\partial E}{\partial N}\right)_{S,V}=-kT\ln\left(\frac{V}{N\Lambda^3}\right)$$

where $$E$$ is the total internal energy, $$S$$ is the entropy, $$V$$ is the volume, and  $$\Lambda$$ is the thermal de Broglie wavelength. The condition for the applicability of the Maxwell–Boltzmann distribution for an ideal gas is again shown to be


 * $$\frac{V}{N\Lambda^3}\gg 1.$$