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# Ising model in the zeroth approximation Done by Ghassan M. Masa’deh - PowerPoint PPT Presentation

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.

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### Ising model in the zeroth approximationDone by Ghassan M. Masa’deh

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 tothem 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 of other problems, more than in ferromagnetism for which it was originally developed

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 : " 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." .

= N+ - N-/N …….(1) -1<L<1

Where: σi = +1 for an up spin

N+ = total number of up spins

N- = total number of down spins

Where " 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." .N = N+ + N-

Substitute 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: " 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." .

M = (N+- N-)µ =µ NL ; -Nµ<M<Nµ

For Ising model we can write the Hamiltonian by :

H{σi} = -J ∑σI σj - µB∑σi ………………(2)

Where : - µB is the potential energy

-Jσiσj is the kinetic energy

H{ " 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." .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

<E> = U = -1/2 qJNL - µBNL………..(5)

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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

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• (1-L)/(1+L) = exp [-2 (qJL+B)/KT] ……..(11) then follow from the Boltzmann principle :

• (qJL+B)/ KT =1/2 ln [(1+L)/(1-L)]

= tanh L ………(12)

L= tanh [(qJL+B)/KT ] ………………(13)

• Let B =0 => L0 = tanh [(qJL0)/KT] ……..(14)

• This is called the possibility of spontaneous magnetization

-1

We obtain a T then follow from the Boltzmann principle :c 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): then follow from the Boltzmann principle :

• 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)

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The configrational energy of the system is given by : then follow from the Boltzmann principle :

U0(T) = -1/2 qJNL0 ……………(17)

And the specific heat is

C0(T) = -qJNL0 dL0/dT

= (NKL0) / [(T/Tc)/(1- L0) - (T/Tc)] ...(18)

At T>Tc => U0(T) = C0(T) = 0

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The specific heat at the transition temperature Tc is : then follow from the Boltzmann principle :

C0(T) = lim {(NK* 3x)/[[(1-x)/(1-3x)] – (1-x)]} = 3/2 NK ……(19)

And at T→ 0

C0(T) ≃ 4NK (Tc/T) exp(-2Tc/T) (20)

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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 : the specific heat of the system at temperature above T

X is the magnetic susceptibility of the system and given by :

X(B,T) = (dM/dT)T = Nµ(dL/dB)T

=(Nµ /K)[(1-L(B,T))/[T-Tc{1-L(B,T)}]]

For L≪ 1 we obtain the Curi – weiss law

X0(T) ≃(Nµ /K)/ (T-Tc) (T≳ Tc ,B→ 0 )

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For T <T the specific heat of the system at temperature above Tc we get.

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.

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As T → 0 the law field susceptibility vanishes in accordance with the formula

X0(T) ≃ (4 Nµ /KT) exp (-2Tc/T)

Finally , the relation ship between L and B at T=Tc and use

tanh x ≃ x + x /3 we have :

L ≃ (3 µB/KT) (T =Tc ,B→ 0 )

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3

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Thank you accordance with the formula