Question 3.4: Consider a group with Cj = 20, Nj = 10. Using Eqs. 3.6 and 3...
Consider a group with C_{j} = 20, N_{j} = 10. Using Eqs. 3.6 and 3.8 we find that W_{BE} = 2 × 10^{7} and W_{FD} = 2 × 10^{5}. Thus, at least for this specific example, there are significant differences in the results obtained for the degree of randomness for bosons and fermions.
Equations 3.6 and 3.8 provide the number of combinations for a particular group. For a given distribution of particles across energy groups, N_{j}, the total number of microstates for a system is then the product over all j groups:
(W_{BE} )_{j}=\frac{(N_{j}+ C_{j} – 1)! }{(N_{j})!(C_{j}- 1 )! } (3.6)
(W_{FD} )_{j}=\frac{(c_{j})! }{(C_{j}- N_{j} )!(N_{j} )!} (3.8)
BE: W(N_{j} )=\prod\limits_{j}{\frac{(N_{j}+ C_{j}- 1 )! }{(N_{j} )!(C_{j}- 1 )!} } (3.9)
FD: W(N_{j} )= \prod\limits_{j}{\frac{(C_{j} )!}{(C_{j}- N_{j} )!(N_{j} )! } } (3.10)
where W(N_{j} ) indicates the result for a particular macrostate N_{j} . The final result for the total number of microstates of the system is obtained from
\Omega =\sum{W(N_{j} )} (3.11)
such that \sum\limits_{j}{N_{j} }=N and \sum\limits_{j}{N_{j} \epsilon _{j} }=E .
We next consider part (1) of the overall counting process.