1549380323-Statistical Mechanics Theory and Molecular Simulation

Mean-field theory 621

tanh( )bJm

f(m)=m

m

J < 1

J = 1

bJ > 1

b

b

m 0

m 0

Fig. 16.6 Graphical solution of the transcendental equation tanh(βJm) =m.

When the temperature is too high, that is, whenβJ <1, only them= 0 solution exists.
Since there is no magnetization at zero field,there is no spontaneousordering. The
caseβJ= 1 just separates these two regimes and corresponds, therefore, to a critical
isotherm. The conditionβJ= 1 can be used to determine the critical temperature:

βJ=

J

kT

= 1 ⇒ kTc=J. (16.5.11)

In order to clarify further the behavior of the system near the critical point, consider
expanding the free energy aboutm= 0 at zero field for temperatures near the critical
temperature. If the expansion is carried out up to quadratic order inm, we obtain

g(0,T)≈c 1 +J(1−βJ)m^2 +c 2 m^4 =a(T), (16.5.12)

wherec 1 andc 2 are constants withc 2 >0. Note that at zero field, the Gibbs free
energy per spin becomes the Helmholtz free energy per spina(T). IfβJ >1, the
sign of the quadratic term is negative, and a plot of the free energyas a function of
mis shown in Fig. 16.7(a). We can see from the figure that the free energy has two
minima atm=±m 0 and a maximum atm= 0, indicating that the ordered states,
predicted by solving for the magnetizationm, are thermodynamically stable while the
disordered state withm= 0 is thermodynamically unstable. ForβJ <1, the sign of
the quadratic term is positive, and the free energy plot, shown in Fig. 16.7(b), possesses
a single minimum atm= 0, indicating that there are no solutions corresponding to
ordered states.