Statiatical Phyaica 169(b) As the particles are distinguishable,N!
= n!(N - n)! *N!
n!(N - n)!Hence S = lcln
We note that S(n = 0) = S(n = N) = 0, and we expect S,,, to appear
= S(n).
at n = N/2 (to be proved in (d) below). The graph of S(n) is shown in
Fig. 2.2.
n
(c) Inn! = C 1nm M lnzdz = nlnn - n + 1 M nlnn - n, (for
m= 1
large n).
S N n
(d) kmNln--nln-
N-n N-n
dS- = 0 gives
dn
n
1-Inn- - + ln(N - n) = 0.
-_ N
N-n N-n
Therefore, S = S,,,,, when n = N/2.
N/2 N
Fig. 2.2.1 1 as
(e) As E = nc, S = S,,,,, when E = -NE. When E > -NE, - < 0
2 2 aE
1 dS
T dE’(see Fig. 2.2). Because - = - we have T < 0 when E > NE/~.
(f) The reason is that here the energy level of a single particle has an
upper limit. For a gas system, the energy level of a single particle does not
have an upper limit, and the entropy is an increasing function of E; hence
negative temperature cannot occur.
From the point of view of energy, we can say that a system with nega-
tive temperature is “hotter” than any system with a positive temperature.