1549380323-Statistical Mechanics Theory and Molecular Simulation

620 Critical phenomena

∆(N,h,T)≈

∑

σ 1 =± 1

∑

σ 2 =± 1

···

∑

σN=± 1

exp

{

−β

[

1

2

NJm^2 −(Jm+h)

∑

i

σi

]}

= e−βNJm

(^2) / 2 ∑
σ 1 =± 1

∑

σ 2 =± 1

···

∑

σN=± 1

exp

[

β(Jm+h)

∑

i

σi

]

= e−βNJm

(^2) / 2 ∑
σ 1 =± 1
exp [β(Jm+h)σ 1 ]×···×

∑

σN=± 1

exp [β(Jm+h)σN]

= e−βNJm

(^2) / 2

(

∑

σ=± 1

eβ(Jm+h)σ

)N

= e−βNJm

(^2) / 2 (
eβ(Jm+h)+ e−β(Jm+h)

)N

= e−βNJm

(^2) / 2
(2coshβ(Jm+h))N. (16.5.7)
From eqn. (16.5.7), the Gibbs free energyG(N,h,T) can be calculated according to
G(N,h,T) =−

1

β

ln ∆(N,h,T) =

1

2

NJm^2 −

N

β

ln [2coshβ(Jm+h)]. (16.5.8)

The average magnetization isM=−(∂G/∂h), which means that the average mag-
netization per spinm=M/Ncan be expressed asm=−(∂g/∂h), whereg(h,T) =
G(N,h,T)/Nis the Gibbs free energy per spin:

g(h,T) =

1

2

Jm^2 −

1

β

ln [2coshβ(Jm+h)]. (16.5.9)

Thus, the average magnetization per spin is given by

m=−

∂g

∂h

= tanhβ(Jm+h). (16.5.10)

Notice, however, that sincemwas introduced into the Hamiltonian, the result of this
derivative is an implicit relation formthat takes the form of a transcendental equation.
We now ask if an ordered phase exists at zero field. Settingh= 0 in eqn. (16.5.10),
the transcendental equation becomesm= tanh(βJm). Of course,m= 0 is a trivial
solution to this equation; however, we seek solutions for finitem, which we can obtain
by solving the equation graphically. That is, we plot the two functionsf 1 (m) =mand
f 2 (m) = tanh(βJm) on the same graph and then look for points at which the two
curves intersect for different values ofkT= 1/β. The plot is shown in Fig. 16.6. We
see that depending on the value ofT, the curves intersect at either three points or one
point. Excluding the trivial casem= 0 we see that for small enoughT(βJ >1), there
are two other solutions, which we labelm 0 and−m 0. These solutions correspond to
magnetizations at zero field aligned along the positive and negativez-axis, respectively.