each drawing, the occupation numbers N 0 ,N 1 ,N 2 , and N 3 are listed. By ap-
plying equation 17.7, we can see thatW(1, 0, 1, 1)
1!0!3
!
1!1!
6
W(0, 1, 2, 0)
0!1!3
!
2!0!
3
W(0, 2, 0, 1)
0!2!3
!
0!1!
3
Figure 17.6 shows that there are 6, 3, and 3 ways of distributing the particles
among the possible energy levels so that the total energy equals 5.Referring to Example 17.2, we see that there are 12 possible arrangements
of the particles in three distinct ways, each of which has multiple possibilities.
Which of these 12 ways is preferred? Putting the question another way, if we
have an ensemble of microstates for a system that has 5 energy units, which of
the arrangements in Figure 17.6 will be favored? Statistical thermodynamics
assumes that none of the arrangements is preferred over the others, that all
possible arrangements are equally probable. This is known as the principle of
equal a priori probabilities.
Does this mean that each set of occupation numbers (1, 0, 1, 1), (0, 1, 2, 0),
and (0, 2, 0, 1) will make up a third of the overall system? No, it doesn’t, be-
cause there are six ways of making the occupation numbers (1, 0, 1, 1) and only
three ways each of making (0, 1, 2, 0) and (0, 2, 0, 1). It does mean that 112 of592 CHAPTER 17 Statistical Thermodynamics: Introduction
(N 0 , N 1 , N 2 , N 3 ) (1, 0, 1, 1)3
2
1
0
(1, 0, 1, 1)3
2
1
0
(1, 0, 1, 1)3
2
1
0(N 0 , N 1 , N 2 , N 3 ) (1, 0, 1, 1)3
2
1
0
(1, 0, 1, 1)3
2
1
0
(1, 0, 1, 1)3
2
1
0(N 0 , N 1 , N 2 , N 3 ) (0, 1, 2, 0)3
2
1
0
(0, 1, 2, 0)3
2
1
0
(0, 1, 2, 0)3
2
1
0(N 0 , N 1 , N 2 , N 3 ) (0, 2, 0, 1)3
2
1
0
(0, 2, 0, 1)3
2
1
0
(0, 2, 0, 1)3
2
1
0Figure 17.6 Refer to Example 17.2. These are the only ways of distributing the three particles
(shown as a left, a middle, and a right dot) to get 5 energy units in the system. Notice that there
are six possible combinations of the occupation number set (1, 0, 1, 1) but only three possible
combinations of the occupation number sets (0, 1, 2, 0) and (0, 2, 0, 1).