Statistical Phyaics 33 1
or
@ +- PJm = tanh-' m
kT kT
With Ifo = 0,m = tanh. Therefore, only when T < T, = PJ/k,
has the above equation a solution m # 0. Thus, the critical temperature is
T, = PJ/k.
(c) When Ho = 0, we have rn = tanh -m. For T --+ T, we can use
(: )
the Taylor expansion and write
112
m - const. (1 - $)
1
2
Hence p = -.
a
ap
(d) From E = - - In z, we obtain
E 1
- = -poHom - -PJm2 ,
N 2
and
am
When Ho = 0,C = -PJm--. When T 2 Tc,m = 0,C = 0. When
aT
T < T,, we have near T,,
112
m = const. (1 - g) ,
T
Tc
m2al--,
PJ
TC
ca-.