Transcript Ising model in zeroth approximation

Ising model in the
zeroth approximation
Done by
Ghassan M. Masa’deh
Introduction:
In principle the Ising model is not a very
good approximation for any temperature
range. However it has the advantage of
starting directly from the energy levels
and skipping all the steps that lead to
them from the Hamiltonian, in other
methods.
This convenient short cut makes it possible
to concentrate on the details of the statistical
mechanics.
Therefore, the ising model is very widely
used in a variety of other problems, more
than in ferromagnetism for which it was
originally developed.
The Ising model
In 1928 Gorsky attempted a statistical study
of order - disorder transitions in binary
alloys on the basis of the assumption that
the work expended in transferring an atom
from an order position to a disordered one is
directly proportional to the degree of order
prevailing in the system
This idea was further developed by Bragg
and Williams, who, for the first time
introduced the concept of long range order
The basic assumption in the Bragg-Williams
approximation is " the energy of an individual
atom in the given system is determined by
the average degree of order prevailing in the
entire system rather than by the fluctuating
configurations of the neighboring atoms." .

Define: along range parameter 'L' is given by :
L  1 / N  i = N+ - N-/N …….(1)
-1<L<1
Where: σi = +1 for an up spin
= -1 for adown spin
N+ = total number of up spins
N- = total number of down spins
L
Where N = N+ + NSubstitute in (1) we get
L = (2N+/N) -1
So, N+=N/2(1+L)
And
N-= N/2 (1-L)

The magnetization M is given by:
M = (N+- N-)µ =µ NL
; -Nµ<M<Nµ
For Ising model we can write the Hamiltonian
by :
H{σi} = -J ∑σI σ j - µB∑σi ………………(2)
Where : - µB is the potential energy
-Jσi σj is the kinetic energy
H{σi} = -J (1/2 q σ) ∑σi - µB∑σi ……………(3)
We can find the total configurational energy of
the system is given by
E= -1/2 (qJL) NL –( µB) NL ……….(4)
And
2
<E> = U = -1/2 qJNL - µBNL………..(5)
Define: ∆ɛ is the difference between the
over all configurational energy of an up spin
and the over all configurational energy of
down spin , the energy expended in
changing any up spin into a down one is
given by:
∆ɛ = -J(qσ) ∆σ - µB∆σ ……………..(6)
= 2µ (qJσ/(µ+B) )
;
∆σ= -2……….(7)
The relative values of the equilibrium numbers
N+ and N- then follow from the Boltzmann
principle :
N-/N+ = exp (-∆ɛ/KT)
= exp (-2µ(B'+B)/KT) ………(8)
Where B' the internal molecular field and given
by :
B' = qJM/Nµ
2
(1-L)/(1+L) = exp [-2 (qJL+B)/KT] ……..(11)
(qJL+B)/ KT =1/2 ln [(1+L)/(1-L)]
= tanh L
………(12)
L= tanh [(qJL+B)/KT ]
………………(13)
-1
Let B =0 => L0 = tanh [(qJL0)/KT] ……..(14)
This is called the possibility of spontaneous
magnetization
We obtain a Tc below which the system can
acquire a non zero spontaneous
magnetization and above which it can not .
We can identify the Tc with the Curi
temperature. The temperature that marks a
transition from the ferromagnetic to the
paramagnetic behavior of the system or
vice verse.
From equation (14):
1/2
L0(T) ≃{3(1-T/Tc)} ;(T≲Tc,B→0)…(15)
At T→0 => L0 → 1
L0(T)≃ 1-2 exp(-2Tc/T) ; (T/Tc ≪ 1)….(16)
The configrational energy of the system is
given by :
2
U0(T) = -1/2 qJNL0 ……………(17)
And the specific heat is
C0(T) = -qJNL0 dL0/dT
2
2
2
= (NKL0) / [(T/Tc)/(1- L0) - (T/Tc)] ...(18)
At T>Tc => U0(T) = C0(T) = 0
`
The specific heat at the transition
temperature Tc is :
2
C0(T) = lim {(NK* 3x)/[[(1-x)/(1-3x)] – (1-x)]}
= 3/2 NK ……(19)
And at T→ 0
2
C0(T) ≃ 4NK (Tc/T) exp(-2Tc/T)
(20)
Note that the vanishing of the configurational
energy and the specific heat of the system at
temperature above Tc is directly related to the
configurational order prevailing in the system
at lower temperatures is completely wiped out
as T→ Tc .
We note that all the microstate are equally
likely to occur is related to the fact that for
T ≥ Tc there is no configurational order in the
system
Define :
X is the magnetic susceptibility of the system
and given by :
X(B,T) = (dM/dT)T = Nµ(dL/dB)T
2
2
2
=(Nµ /K)[(1-L(B,T))/[T-Tc{1L(B,T)}]]
For L≪ 1 we obtain the Curi – weiss law
2
X0(T) ≃(Nµ /K)/ (T-Tc)
(T≳ Tc ,B→ 0 )
For T <Tc we get.
2
X0(T) ≃ (Nµ /2K)/ (Tc-T)
(T≲ Tc , B→ 0 )
Experimentally the Curi – Weiss law is
satisfied with considerable accuracy except
that the empirical value of Tc thus obtained is
always some what larger than the true
transition temperature of the material.
As T → 0 the law field susceptibility
vanishes in accordance with the formula
X0(T) ≃ (4 Nµ2 /KT) exp (-2Tc/T)
Finally , the relation ship between L and B
at T=Tc and use
-1
3
tanh x ≃ x + x /3 we have :
1/3
L ≃ (3 µB/KT)
(T =Tc ,B→ 0 )
Thank you