Statistical physics
9.2 Maxwell-Boltzmann distribution
To describe the behavior of a classical gas we can use the following assumptions:
• the particles are distinguishable1 (i.e. these are classical particles)
• the probability of the occupation of every energy level Ei is equal
• one level may be occupied by any number of particles (no exclusion principle)
• the probability of a distribution (or partition) {ni} is proportional to the number of particle configurations that can realize it.
• thermal equilibrium corresponds to the maximum probability distribution E1 is occupied by n1 particles. This may happen in nN
1
different ways2. For each of these there are N−nn21
possibilities to fill E2. For every possible configuration on E1
1See section 6.7for a short introduction between distinguishable and indistinguishable particles.
2 N m
≡m! (NN−m)!! . Also observe that if a level is unoccupied thenni= 0, and 0! = 1.
and E2 there are N−nn1−n2
3
possibilities to fill E3, etc. The total number of possible configurations therefore3 When the probabilities of occupation are different for different Eis, i.g. an energy level may be compatible with more different angular momentum states then the others, then it is more likely to be occupied, this formula should be modified to become4
w= where the gi(≥1) numbers give the degeneracy of each energy level.
We calculated so far the number of possible ways the given partition can occur. If we are interested in the probability of this partition5 then we must divide it with the number of all possible particle permutations, i.e. with N!. So the probability of a given partition in the general case:
P({ni}) =Y
i
gini
ni! (9.2.3)
The next step is to calculate the maximum of this probability. However we are looking for a maximum with the following additional conditions: the total number of particles Ntot = P
i
ni is constant and the total energy Etot = P
i
Eini is also constant. This is called a conditional maximum problem. The result6:
3If we write the product of binomial coefficients using factorials we can observe that the denominator of a factor in this product is the same as the numerator of the next factor, so these cancels each other out. This leads to (9.2.1).
We should have arrived to the same formula notingN particles may be orderedN! times (permuta-tions), and because we are not interested in the order of the particles on a given level this number must be divided by the product of all ni!.
4If levelEi hasgi possible sub-states and we haveni particles on this level then each particle may be put into any of thesegisub-states with the same probability, therefore the number of possible sub-states isgnii.
5probability of an event (e.g. of a partition) ≈number of ways it can happen divided by the total number of possible outcomes.
6Mathematical details are in an appendix: Appendix 22.13.
Important 9.2.1.
and kB = 1.38 10−23 J/K is the Boltzmann constant, T is the absolute temperature (in K) and Z is called the partition function.
The factore−βEi =e−
Ei
kB T that gives the (unnormalised) relative probability of a state (i.e. the statistical weight) is called the Boltzmann factor.
Knowing the probability of a given macrostate we can easily calculate theexpectation value of ni ; i.e. the value we expect as the average of the results of many successive measurements of ni at a givenEi
hnii=N · Pmax,i= N
Z gie−βEi (9.2.7)
This is the Maxwell-Boltzmann statistics.
Important 9.2.2. For any energy dependent physical quantity F(E) the average value is the ni weighted average of possible F(Ei) values :
Note that in this case the sum is the negative derivative of the (9.2.6) partition function with respect to β:
Furthermore d
d β = d d T
d T
d β =−kBT2 d d T, so U =N kBT2 d
d T ln Z (9.2.11)
The average energy of a particle is
Eaver =kBT2 d
d T ln Z (9.2.12)
therefore the temperature of a system is determined by the average energy per particle and the structure of the energy levels of the system described in Z.
Example 9.1. In a system with equidistant energy levels how many ways can you dis-tribute 9 units of energy among 6 identical. distinguishable particles? The energy of the ground state (i=0) is 0, and the levels are one unit of energy distant from each other.
Solution In this case the observable different macrostates give the number of particles on every level, while the microstates are the possible ways to achieve a given macrostate.
Because we must distribute 9 units of energy among the particles and the energy of the ground state is 0, we have to use 10 energy levels.
The number of the macrostates are few so they can easily be counted in this case and the result is 26. The figure shows all macrostates with a total energy of 9 units, together with the number of the microstates that correspond to the same macrostate. The first macrostate in the first row have 6!
6! = 6 microstates, the second one 6!
4! 1! 1! = 30, while the first one in the second row have 6!
2! 2! 1! 1! = 180, etc. The total number of microstates is 2002.
Example 9.2. Graph the distribution for Problem 9.1and compare it with the Maxwell-Boltzmann distribution function!
Solution We have to graph the ni vs E discreet function of Problem 9.1. The average occupation numbers for the levels are
hnii=
26
P
n=1
wi(n)ni(n)
26
P
n
wi
where the summation goes from 1 to the number of all possible macrostates, and wi(n)is the number of the microstates that results in then-th macrostate.
The denominator is the total number of microstates, which is 2002 as we have shown previously in Problem 9.1. So for instance for i= 0
n0 =6·5 + 4·30·4 + (3·120 + 3·60 + 20)·3 + (2·60 + 4·180)·2
2002 +
(30 + 120 + 60 + 180 + 30)·1 + (30 + 6 + 30 + 20)·0
2002 = 2.143
The average occupation numbers or average population of the levels:
Energy level 0 1 2 3 4 5
hnii 2.143 1.484 0.989 0.629 0.378 0.210
Energy level 6 7 8 9
hnii 0.104 0.045 0.015 0.003
while in the figure you can see the results compared to that of the continuous Maxwell-Boltzmann distribution function.
As you can see the distribution for even as few as 6 particles closely ap-proximates the Maxwell-Boltzmann distribution function.
9.2.1 Application of the Maxwell-Boltzmann statistics to the ideal gas
Even though we are talking about a classical ideal gas molecular excitations are quantum mechanical processes with transitions between discreet energy levels. The “classical”
nature here refers to the molecules themselves, which at not extremely large pressures and at not extremely low temperatures are distinguishable particles.
According to (9.2.4) the equilibrium ratio of the (average) number of particles on energy levels Ei and Ej is
ni nj = Pi
Pj = gi gj e−
Ej−Ei
kB T (9.2.13)
which depends exponentially on ∆E ≡ Ej − Ei. Table 9.1 summarizes this:
The kinetic energy levels of gas molecules in a container, whose size is very large compared to the size of the molecules are so very close to each other at normal temper-atures and pressures, that we cannot observe the discreetness of the levels and consider
∆E[eV] 100 K 300 K 1000 K rotational 10−4 0.989 0.996 0.999 vibrational 5·10−2 3·10−3 0.150 0.560 electronic 3 13·10−164 8·10−49 8·10−16
Table 9.1: Excitation energies (∆E) and excitation probabilities of rotational, vibrational and electronic transitions in molecules of an ideal gas at different temperatures.
the kinetic energy as a continuous quantity.
For this case let us introduce a continuous fM B(E) function, which gives the proba-bility of occupation, so that the number ∆n of particles in a ∆E energy interval around E is N ·fM B(E)g(E) ∆E = ∆n, or equivalently: fM B(E) = N g(E)1 dn(E)dE .
where g(E)dE is the number of possible states in the energy interval dE around E. The average value of any energy dependent physical quantity F(E) for one particle is:
hF i= 1
Again using F =E we find that (9.2.12) is still valid:
Eaver ≡ hEi=kBT2 d d T ln Z
For a classical ideal gas enclosed in a large container of volume V the degeneracy of the energy states according to (3.5.20) (where we called it the density of states):
g(E) = 4π V √ 2m3 h3
√E
The partition function in (9.2.15): Therefore the internal energy of the gas is
U =NEaver =N kBT2 d
For classical monatomic ideal gases the energy is only kinetic, that is independent of the mass of the molecule, can be expressed with the velocity of the molecules E =
1
2m v2. Substituting into the Maxwell-Boltzmann distribution formula gives the number of particles in a unit energy interval as a function of the magnitude of the velocity:
d N(E) = g(E)fM B(E)dE =d N(v) = g(E(v))fM B(v)d v
This Maxwell-Boltzmann velocity distribution function is shown in Fig. 9.1.