Problems and Solutions on Thermodynamics and Statistical Mechanics

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-.