Maxwell–Boltzmann statistics can be derived in various
statistical mechanical thermodynamic ensembles: • The
grand canonical ensemble, exactly. • The
canonical ensemble, exactly. • The
microcanonical ensemble, but only in the thermodynamic limit. In each case it is necessary to assume that the particles are non-interacting, and that multiple particles can occupy the same state and do so independently.
Derivation from microcanonical ensemble Suppose we have a container with a huge number of very small particles all with identical physical characteristics (such as mass, charge, etc.). Let's refer to this as the
system. Assume that though the particles have identical properties, they are distinguishable. For example, we might identify each particle by continually observing their trajectories, or by placing a marking on each one, e.g., drawing a different number on each one as is done with
lottery balls. The particles are moving inside that container in all directions with great speed. Because the particles are speeding around, they possess some energy. The Maxwell–Boltzmann distribution is a mathematical function that describes about how many particles in the container have a certain energy. More precisely, the Maxwell–Boltzmann distribution gives the non-normalized probability (this means that the probabilities do not add up to 1) that the state corresponding to a particular energy is occupied. In general, there may be many particles with the same amount of energy \varepsilon. Let the number of particles with the same energy \varepsilon_1 be N_1, the number of particles possessing another energy \varepsilon_2 be N_2, and so forth for all the possible energies \{ \varepsilon_i \mid i = 1, 2, 3, \ldots\}. To describe this situation, we say that N_i is the
occupation number of the
energy level i. If we know all the occupation numbers \{N_i \mid i=1,2,3,\ldots\}, then we know the total energy of the system. However, because we can distinguish between
which particles are occupying each energy level, the set of occupation numbers \{ N_i \mid i=1,2,3,\ldots\} does not completely describe the state of the system. To completely describe the state of the system, or the
microstate, we must specify exactly which particles are in each energy level. Thus when we count the number of possible states of the system, we must count each and every microstate, and not just the possible sets of occupation numbers. To begin with, assume that there is only one state at each energy level i (there is no degeneracy). What follows next is a bit of combinatorial thinking which has little to do in accurately describing the reservoir of particles. For instance, let's say there is a total of k boxes labelled a,b,\ldots,k. With the concept of
combination, we could calculate how many ways there are to arrange N into the set of boxes, where the order of balls within each box isn’t tracked. First, we select N_a balls from a total of N balls to place into box a, and continue to select for each box from the remaining balls, ensuring that every ball is placed in one of the boxes. The total number of ways that the balls can be arranged is \begin{align} W & = \frac{N!}{N_a!\cancel{(N-N_a)!}} \times \frac{\cancel{(N-N_a)!}}{N_b!\cancel{(N-N_a-N_b)!}} \times \frac{\cancel{(N-N_a-N_b)!}}{N_c!\cancel{(N-N_a-N_b-N_c)!}} \times \cdots \times \frac{\cancel{(N-\cdots-N_\ell)!}}{N_k!(N-\cdots-N_\ell-N_k)!} \\[8pt] & = \frac{N!}{N_a!N_b!N_c!\cdots N_k!(N-N_a-\cdots-N_\ell-N_k)!} \end{align} As every ball has been placed into a box, (N - N_a - N_b - \cdots - N_k)! = 0! = 1, and we simplify the expression as W = N!\prod_{\ell=a,b,\ldots}^k \frac{1}{N_\ell!} This is just the
multinomial coefficient, the number of ways of arranging
N items into
k boxes, the
lth box holding
Nl items, ignoring the permutation of items in each box. Now, consider the case where there is more than one way to put N_i particles in the box i (i.e. taking the degeneracy problem into consideration). If the ith box has a "degeneracy" of g_i, that is, it has g_i "sub-boxes" (g_i boxes with the same energy \varepsilon_i. These states/boxes with the same energy are called degenerate states.), such that any way of filling the ith 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
ith box must be increased by the number of ways of distributing the N_i objects in the g_i "sub-boxes". The number of ways of placing N_i distinguishable objects in g_i "sub-boxes" is g_i^{N_i} (the first object can go into any of the g_i boxes, the second object can also go into any of the g_i boxes, and so on). Thus the number of ways W that a total of N particles can be classified into energy levels according to their energies, while each level i having g_i distinct states such that the
ith level accommodates N_i particles is: W=N!\prod_{i}\frac{g_i^{N_i}}{N_i!} This is the form for
W first derived by
Boltzmann. Boltzmann's fundamental equation S=k_\text{B}\,\ln W relates the thermodynamic
entropy S to the number of microstates
W, where
kB is the
Boltzmann constant. It was pointed out by
Gibbs however, that the above expression for
W does not yield an
extensive entropy, and is therefore faulty. This problem is known as the
Gibbs paradox. The problem is that the particles considered by the above equation are not
indistinguishable. In other words, for two particles (
A and
B) in two energy sublevels the population represented by [
A,
B] is considered distinct from the population [
B,
A] while for indistinguishable particles, they are not. If we carry out the argument for indistinguishable particles, we are led to the
Bose–Einstein expression for
W: W=\prod_i \frac{(N_i+g_i-1)!}{N_i!(g_i-1)!} The Maxwell–Boltzmann distribution follows from this Bose–Einstein distribution for temperatures well above absolute zero, implying that g_i\gg 1. The Maxwell–Boltzmann distribution also requires low density, implying that g_i\gg N_i. Under these conditions, we may use
Stirling's approximation for the factorial: N! \approx N^N e^{-N}, to write: \begin{align} W &=\prod_i \frac{g_i}{N_i+g_i}\frac{(N_i+g_i)!}{N_i!g_i!} \approx\prod_i \frac{(N_i+g_i)!}{N_i!g_i!}\\ &\approx \prod_i \frac{(N_i+g_i)^{N_i+g_i}e^{-N_i-g_i}}{N_i!g_i^{g_i}e^{-g_i}}=\prod_i \frac{g_i^{N_i}(1+N_i/g_i)^{N_i+g_i}e^{-N_i}}{N_i!} \end{align} Using the fact that (1+N_i/g_i)^{N_i+g_i}\approx e^{N_i} for g_i\gg N_i we get: W\approx\prod_i \frac{g_i^{N_i}}{N_i!} This is essentially a division by
N! of Boltzmann's original expression for
W, and this correction is referred to as ''''''. We wish to find the N_i for which the function W is maximized, while considering the constraint that there is a fixed number of particles \left(N=\sum N_i\right) and a fixed energy \left(E=\sum N_i \varepsilon_i\right) in the container. 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,\ldots,N_n) = \textstyle \ln(W)+\alpha(N - \sum_i N_i) + \beta(E - \sum_i N_i \varepsilon_i) \ln W=\ln\left[\prod_{i=1}^{n}\frac{g_i^{N_i}}{N_i!}\right] \approx \sum_{i=1}^n\left(N_i\ln g_i-N_i\ln N_i + N_i\right) Finally f(N_1,N_2,\ldots,N_n)=\alpha N +\beta E + \sum_{i=1}^n\left[N_i\ln g_i-N_i\ln N_i + N_i - \left(\alpha+\beta\varepsilon_i\right) 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\varepsilon_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 \varepsilon_i}} Substituting this expression for N_i into the equation for \ln W and assuming that N \gg 1 yields: \ln W = (\alpha+1) N+\beta E\, or, rearranging: E=\frac{\ln W}{\beta}-\frac{N}{\beta}-\frac{\alpha N}{\beta} Boltzmann realized that this is just an expression of the
Euler-integrated fundamental equation of thermodynamics. Identifying
E as the internal energy, the Euler-integrated fundamental equation states that : E=TS-PV+\mu N where
T is the
temperature,
P is pressure,
V is
volume, and
μ is the
chemical potential. Boltzmann's equation S=k_\text{B} \ln W is the realization that the entropy is proportional to \ln W with the constant of proportionality being the
Boltzmann constant. Using the ideal gas equation of state (
PV =
NkB
T), It follows immediately that \beta=1/k_\text{B}T and \alpha=-\mu/k_\text{B}T so that the populations may now be written: N_i = \frac{g_i}{e^{(\varepsilon_i-\mu)/(k_\text{B}T)}} Note that the above formula is sometimes written: N_i = \frac{g_i}{e^{\varepsilon_i/k_\text{B}T}/z} where z=\exp(\mu/k_\text{B}T) 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^{-\varepsilon_i/k_\text{B}T}}{Z} where
Z is the
partition function defined by: Z = \sum_i g_i e^{-\varepsilon_i/k_\text{B}T} In an approximation where
εi is considered to be a continuous variable, the
Thomas–Fermi approximation yields a continuous degeneracy
g proportional to \sqrt{\varepsilon} so that: \frac{\sqrt{\varepsilon}\,e^{-\varepsilon/k T}}{\int_0^\infty\sqrt{\varepsilon}\,e^{-\varepsilon/k T}} = \frac{\sqrt{\varepsilon}\,e^{-\varepsilon/k T}}{\frac{\sqrt{\pi}}{2}(k_\text{B}T)^{3/2}} = \frac{2\sqrt{\varepsilon}\,e^{-\varepsilon/k T}}{\sqrt{\pi(k_\text{B}T)^3}} which is just the
Maxwell–Boltzmann distribution for the energy.
Derivation from canonical ensemble 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 \varepsilon _i with degeneracies g_i. As before, we would like to calculate the probability that our system has energy \varepsilon_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_\text{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_\text{R} (s_1) = 2 \; \Omega_\text{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_\text{R} (s_1)}{\Omega_\text{R} (s_2)}. Since the
entropy of the reservoir S_\text{R} = k \ln \Omega_\text{R}, the above becomes \frac{P(s_1)}{P(s_2)} = \frac{ e^{S_\text{R}(s_1)/k} }{ e^{S_\text{R}(s_2)/k} } = e^{(S_\text{R} (s_1) - S_\text{R} (s_2))/k}. Next we recall the thermodynamic identity (from the
first law of thermodynamics and
second law of thermodynamics): d S_\text{R} = \frac{1}{T} (d U_\text{R} + P \, d V_\text{R} - \mu \, d N_\text{R}). In a canonical ensemble, there is no exchange of particles, so the d N_\text{R} term is zero. Similarly, d V_\text{R} = 0. This gives S_\text{R} (s_1) - S_\text{R} (s_2) = \frac{1}{T} (U_\text{R} (s_1) - U_\text{R} (s_2)) = - \frac{1}{T} (E(s_1) - E(s_2)), where U_\text{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) / k_\text{B}T } }{ e^{ - E(s_2) / k_\text{B}T} }, which implies, for any state
s of the system P(s) = \frac{1}{Z} e^{- E(s) / k_\text{B}T}, where
Z is an appropriately chosen "constant" to make total probability 1. (
Z is constant provided that the temperature
T is invariant.) Z = \sum _s e^{- E(s) / k_\text{B}T}, 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 \varepsilon _i is simply the sum of the probabilities of all corresponding microstates: P (\varepsilon _i) = \frac{1}{Z} g_i e^{- \varepsilon_i / k_\text{B}T} where, with obvious modification, Z = \sum _j g_j e^{- \varepsilon _j / k_\text{B}T}, 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 \varepsilon _i 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 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.) == Derivation from canonical ensemble ==