MAXWELL-BOLTZMANN STATISTICS
INTRODUCTION
This part of the course is on Statistical Mechanics. The purpose of the course is to
come up with a macroscopic picture of a collection of particles by considering the
distribution of the particles at different energy levels. Throughout this part of the course, I
shall refer to the whole collection of particles under consideration as an “assembly” and
individual particles as “systems”. These terms are adopted in quite a number of textbooks.
Roughly speaking, there are two kinds of quantum statistics, namely, the BoseEinstein (BE) statistics and the Fermi-Dirac (FD) statistics. The former is valid for particles
with zero or integral spins (e.g. photons in EM waves and phonons in crystal vibration). The
latter is specifically for odd half spins, e.g. electrons with ½ spin.
Strictly speaking, these are the only two classes of particles in Statistical Mechanics.
When you find a particle in nature, it is either a boson obeying BE statistics or a fermion
obeying FD statistics. In practice, the mathematics of these two forms of statistics is very
hard to handle. Most integrals involved require very difficult treatments or even tedious
numerical analysis.
Fortunately, Many BE or FD assemblies do behave classically in dilute systems where
particles do not see each other behaving as those they were independent classical particles.
When the dilute conditions prevail, one would obtain the theoretical results in two ways.
Firstly, one can start with the correct BE or FD quantum statistics and then let your
expressions tend to the classical limits. This procedure is not too dissimilar to the
correspondence principle in Quantum Mechanics. Secondly, one can start from scratch and
device a classical statistics for independent classical particles. In fact, the two main
approaches here give exactly the same results.
The second approach is called the Maxwell-Boltzmann (MB) statistics. The present
lecture notes will describe this classical MB statistics, i.e. second approach. This MB
approach has a number of advantages. The mathematics involved is much more manageable.
Secondly, it is easier to get to the underlying physical principles with this technique. With
this, we shall be able to derive macroscopic thermodynamical functions such as entropy S,
the free energy F and the internal energy U. We shall find that we can obtain much physical
insight into the subject matter with this classical approach. For more condensed systems in
which the quantum nature of the particles involved becomes important, the BE and FD
statistics will be described in separate lecture notes.
MATHEMATICAL TOOLS
1. Stirling’s approximation (see appendix 2 of supplementary notes)
In this subject, we often need to handle the function log(n!). This function may be
written as
log(n!) = n.log(n) - n + 1
(1)
If the variable n >> 1, which is always the case in Statistical Mechanics, the last term
of the rhs may be neglected. Further, if n is large, nlogn is larger still, thus
log(n!)  nlogn - n  nlogn
(2)
The purpose of using (2) is that the approximated value on the rhs is much easier to
handle, e.g. when differentiating with respect to (wrt) n.
2. Change of variables
Suppose we have a function g(p) expressed in the space (variable) p and we need to
express the same function in another space (variable) e. We do however know the
relationship between the two variables p and e given by
e = f(p)
(3)
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where f is a function for e and has nothing to do with the first function g. (3) also
gives us the derivative of e
de/dp = df(p)/dp = f’ say
(4)
Let us now consider the area under the curve g(p) in the g-p plot. Specifically, let us
consider a small area under this curve between p and p + dp. This area is given by
A = g(p)dp
(5)
We now consider a similar and corresponding element of area in the e-space under the
curve g(e),
A’ = g(e)de
(6)
To transform from one space to another, we impose an important condition such that
the two areas A and A’ are equal. This is achieved by adjusting the width de, thus
g(p)dp = g(e)de
(7)
multiplying the lhs of (7) by de/de, we get
g(e)de = g(p).(dp/de)de
(8)
From the derivative in (4) and dividing throughout by de, we have
g(e) = g(p)/f’
(9)
(9) will prove to be very useful. Thus, to change the function from p-space to another
e-space, we need both a direct substitution from p to e using (3) and the knowledge of the
derivative
de/dp = f’
(10)
This procedure will be invoked for the density of states function g(p) and g(e) in the
momentum and energy space respectively in both Statistical Mechanics and Solid State
Physics.
THE WEIGHT OF A CONFIGURATION
Let us consider a no. of energy levels as in fig. 1. The energies are shown as e s for
level s etc. The corresponding population and degeneracy are respectively ns and gs. The gs
is sometimes called the weight of the level s and is the no. of states for level s.
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Fig. 1.
For MB classical statistics, there is no restriction on the number of particles going to a
given state. each particle (each of the ns) will have the gs states to go to. The total no. of
such arrangements is
vs = gsns
(11)
Again, note that (11) is classical as it allows any no. of particles in any given one
state.
For simplicity, let us consider only the lowest three levels 1 to 3. The total no. n is
n = n1 + n2 + n3
(12)
The task is to find the no. of combinations for choosing from the total no. n, n 1 for
level 1, n2 for level 2 etc.
Given that it does not matter how the systems n1 are arranged among themselves at
level 1, (i.e. interchanging any two particles on the same energy level will not change the no.
of combinations) the problem for level 1 is a problem of combination rather than
permutation, i.e.
n
Cn1 = n!/[n1!(n - n1)!]
(13)
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As for level 2, a similar formula may be obtained except that n in (13) should be
replaced by (n - n1) being the remaining no. from which we can choose for populating level 2,
hence
(n-n1)
Cn2 = (n - n1)!/[n2!(n - n1 - n2)!]
(14)
If level 3 is the last level to be populated, the combination is just
n3
Cn3 = 1
(15)
where n3 = n - n1 - n2
(16)
This is because there is only one way for placing the remaining n3 in the final single
energy level. The total combination is just the product of (13) to (15) and thus, using (16),
we have
w’ = n!/[n1!n2!n3!]
(17)
In finding the total weight w, we must include (11) for all the levels involved, hence
w = v1v2v3.w’ = n!v1v2v3/[n1!n2!n3!]
(18)
Basically (18) shows us that w involves n! and two products containing respectively v
and n for all the three levels. We can therefore, by deduction, extend this derivation to r
levels. For simplicity, we use a product notation  for multiplying from s = 1 to s = r, the
general weight becomes
w = n!.svs/ns! = n!.sgsns/ns!
(19)
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MOST PROBABLE WEIGHT
(19) is a general expression involving different populations for different energy levels.
We have no information about the likely values of ns. In nature, a configuration (which is
defined to be a certain set of ns and es) is unstable until the weight w for that configuration
has reached the most probable value. This we shall need to evaluate now.
First, we have to modify (19) before finding the maximum weight wm as w is in
product form which is mathematically difficult to handle. Instead, we want to handle log(w),
mindful that maximising log(w) is the same as maximising w. From (19)
W = log(w) = logn! + s[nsloggs - logns!]
(20)
Using Stirling’s formula in (2), we have
W  nlogn + s[nsloggs - nslogns]
(21)
For simplicity, we shall drop the subscript s for the summation. In the following
partial differentiation, the first term will disappear as n is a constant.
W/ns = 0 + log(gs) - ns(1/ns) - log(ns) = log(gs/ns) - 1
 log(gs/ns)
(22)
All the other terms under the summation disappear on partial differentiation. The
term -1 has also been dropped as it is negligible compared to log(gs/ns).
THE LAGRANGE UNDETERMINED MULTIPLIERS TECHNIQUE
In this case, as often is the case in Physics, we need other constraints (limiting
conditions) in order to find the maximum. Here we have the energy and system number
constraints. In other words, the Langrange technique is to take into account the energy and
population factors. The total energy is
e = nses = constant
(23)
This is fixing the total energy
de = (nsdes + esdns) = esdns = 0
(24)
We have put des = 0 as the energy levels are not allowed to move for this exercise.
We only allow the populations to re-distribute, i.e., dns is finite, hence the remaining term in
(24).
Also, the total n is constant so that
n = ns = constant
(25)
dn = dns = 0
(26)
Note that although dn = 0, the individual dns are allowed to be finite. This means that
while the total is fixed, we allow individual populations to re-distribute.
Now we have the three conditions
For maximum of weight
dW = dlogw = 0
(27)
For fixing energy and total number, we have
de = 0
(28)
and dn = 0
(29)
These are simultaneous conditions and the Lagrange technique thus have the
following
dW + adn + bde = 0
(30)
where a and b are the Lagrange undetermined multipliers which we shall evaluate in
due course. It should be mentioned that (30) is strictly speaking for finding the turning point
of W. However, it can be shown that W has no minimum and this procedure is effectively for
finding the maximum of W or w. Students can later prove that the turning point here is in
fact a maximum as required.
In terms of the ns variables, we have
dW = (W/ns)dns = 0
(31)
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(cf (1) of the Thermodynamics notes).
Thus, looking at (31), (24) and (26), all the three terms in (30) have the same
summation for s, we thus have
[(W/ns) + a + bes]dns = 0
(32)
We know that dns is not 0 as we need to vary the populations to maximise. Thus,
each bracket under the summation must vanish at all times, i.e. all the coefficients of dns must
be 0. Thus
(W/ns) + a + bes = 0
(33)
In anticipation, we have already evaluated the first term of this in (22), thus
log(gs/ns) + a + bes = 0
(34)
ns = gsexp(a + bes)
(35)
Essentially (35) is the MB distribution. It tells us how all the populations n1, n2…
should be valued so as to maximise W while satisfying all the conditions
dW = dn = de = 0
(36)
The MB statistics in (35) does have another form though. Suppose we only consider
states i (each of the gs) rather than levels s, then
ni = (ns/gs) = exp(a + bes) = exp(a + bei)
(37)
Here ni is the probability that any given gs state is occupied. Note that (37) can be
fractional (<1) as ns << gs. This means that we must go through all the steps leading up to
(35)with ns>>1 so that we are allowed to use Stirling’s approximation.
It is possible to show that (35) gives a maximum for w rather than a minimum. Fig 2
shows how sharp w looks with fractional concentration variations of the order of 10-10 and n
= 1020. (see supplementary notes).
Fig. 2. Rapid decrease of W with very small deviations from the most probable distribution.
THE VALUE OF b
For convenience, let us re-write (35) as
ns = Ags.exp(bes)
(38)
where A = exp(a)
(39)
The total energy is
e = nses = Agses.exp(bes)
(40)
n = ns = Agsexp(bes)
(41)
If b was positive, both e and n in (40) and (41) would blow up to infinity making it
impossible to conserve either energy or system number.
If b is negative, this problem is solved as the population nr at infinite energy = 0.
Thus, we conclude that b must be a negative number.
Let us now consider two sub-assemblies as in fig. 3, with W’, n’, e’ and W”, n”, e”
respectively.
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Fig. 3.
Suppose we establish a heat link between these as shown while preventing an
exchange of systems. Both sub-assemblies are otherwise isolated from the outside world.
Energy is allowed to transfer and eventually, the two sub-assemblies are in thermal
equilibrium with each other, namely, they have the same thermodynamical parameter known
as the temperature T. We now have
w = w’.w”
(42)
W = logw = log(w’.w”) = W’ + W”
(43)
In maximising W, we also have
dn’ = dn” = de = 0
(44)
Note that we do not have de’ = de” = 0 as e’ and e” are not fixed quantities. (43) and
(44) now have four conditions and thus we need three Lagrange undetermined multipliers
similar to (30) with only one assembly.
dW + a’dn’ + a”dn” + bde = 0
(45)
Given that
dn’ = dn’s
dn” = dn”s
de = e’sdn’s + e”sdn”s
(46)
Using the same procedures as in (32) and (33), (45) now gives
W’/n’s + a’ + be’s = 0
W”/n”s + a” + be”s = 0
(47)
Now in thermal equilibrium, (47) shows that the only quantity in common between
the two sub-assemblies is b. In terms of Thermodynamics, the only thing in common is T,
therefore
b = f(T)
(48)
Thus we have discovered two features of b, namely, b is negative and is a function of
T only.
ENTROPY S AND THE WEIGHT w
Let us look at (30) again
dW + adn + bde = 0
(49)
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In equilibrium, all the three terms are all 0 hence giving a total of 0 on the rhs of this
equation. Do not forget that this was the condition giving rise to the MB distribution in (35).
The first law gives
de = dq - PdV
(50)
Let us now consider a process which does no work but puts in dq amount of heat, then
de = dq
(51)
Suppose that, from this equilibrium position, we now disturb this assembly by
supplying the heat dq and let it reach a new equilibrium (i.e. rhs = 0 again), (49) becomes
dW + adn + bdq = 0
(52)
But dn = 0
Thus dW = dlog(w) = -bdq
(53)
This means that in supplying dq, there will be a finite dw or w will increase until it is
maximised again.
We know that dW is a perfect differential, and according to (53), so is (-bdq). In
Thermodynamics, we know that dq is not a perfect differential but it can become a perfect
differential by way of its relationship with the change of entropy, i.e. an integrating factor
(1/T) can be found so that dq becomes a perfect differential,
dS = dq/T
(54)
Comparing (53) and (54), dS and dlogw must only differ from each other by a
constant of proportionality. Thus
dS = kdlogw = k(-bdq) = dq/T
(55)
or -bk = 1/T
(56)
b = -1/kT
(57)
The MB distribution now becomes
ns = Agsexp(-es/kT)
(58)
k is known as the Boltzmann constant.
(55) now gives us an alternative definition of the entropy
dS = kdW = kdlog(w)
(59)
or S = kW = klogw
(60)
w is often called the disorder. We know that S tends to a maximum. (59) says that
the weight or disorder also tends to a maximum. Thus, an assembly is in thermal equilibrium
if it is as chaotic as possible. (60) is quite consistent with itself. Let us consider two subassemblies at the same temperature and pressure. The total weight is
w = w’.w”
(61)
The total entropy is
S = S’ + S” = klogw’ + klogw” = klog(w’.w”) = klogw
(62)
(58) shows that the population of the lower levels are larger. As T increases, more
upper levels will get populated while keeping the total number n constant. The ratio of
populations between any two levels (not necessarily adjacent) separated by an energy e will
depend on the Boltzmann factor exp(-e/kT) with a negative argument (-e/kT). On this
model, therefore, as T tends to 0, the Boltzmann factor tends to 0 for all finite energy
separation e meaning that all the n particles will go to the lowest possible energy level at
0K.
Last updated
22 January 2015
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