8 Basic ideas
Step I. The mechanicalproblemistobesolvedtogive the possible states ofone
particle. We take the solution to be states of (non-degenerate) energies 0,ε,2ε,3ε,...
For conveniencewelabelthese statesj= 0 ,1,2,...withεεj=jε.
Step II. Defining thedistribution numbers as{nnj},withj= 0 ,1,2,...,we note that
anyallowable distributions must satisfy
∑
nnj=4;∑
nnjεεj= 4 ε(i.e.∑
nnjjjj= 4 )There are thus five possible distributions:
Distribution n 0 n 1 n 2 n 3 n 4 n 5 ...
1 300010...
2 210100...
3 2 0 2 000 ...
4 121000...
5 040000 ...
Step III. Amicrostate specifies the state ofeachofthefour particles. We needto
count the number of microstates to each of the five distributions. To take distribution
1 as an example, we can identify four possible microstates:
(i)A is in statej= 4 ; B, C and D are in statej= 0
(ii)Bisin statej= 4 ,theothers arein statej= 0
(iii)C is in statej=4,the others are in statej= 0
(iv)D is in statej= 4 , the others are in statej= 0
Hencet(^1 )= 4 .Similarly one can workout (an exercisefor the reader?) the numbers
ofmicrostatesfor theotherfourdistributions. The answers aret(^2 )=12,t(^3 )=
6,t(^4 )= 12 ,t(^5 )= 1. It is significant that the distributions which spread the particles
between the states as muchas possiblehave the most microstates, andthus (Step
IV) are the mostprobable. The total number of microstatesis equal to the sum
t(^1 )+t(^2 )+t(^3 )+t(^4 )+t(^5 ),i.e.= 35 in this example.
(((Aside.A general formula for this situation is developed in Appendix A and we shall
make muchuse ofitinChapter 2. The resultis
t({nnj})=N!/∏
jnnj! (1.4)where the denominator
∏
nnj! represents the extended product
n 0 !n 1 !n 2 !...nnj!....The resultfort(^1 )followsfrom (1.4) as 4!/3!, when one substitutes
0! = 1 and 1! = 1. The othertvalues follow similarly.)