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Ising model in the zeroth approximation Done by Ghassan M. Masa’deh.

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Presentation on theme: "Ising model in the zeroth approximation Done by Ghassan M. Masa’deh."— Presentation transcript:

1 Ising model in the zeroth approximation Done by Ghassan M. Masa’deh

2 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.

3 This convenient short cut makes it possible to concentrate on the details of the statistical mechanics.

4 Therefore, the ising model is very widely used in a variety of other problems, more than in ferromagnetism for which it was originally developed.

5 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

6 This idea was further developed by Bragg and Williams, who, for the first time introduced the concept of long range order

7 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.".

8 Define: along range parameter 'L' is given by : = N + - N-/N …….(1) -1

9 Where N = N + + N- Substitute in (1) we get L = (2N+/N) -1 So, N + =N/2(1+L) And N-= N/2 (1-L)

10 The magnetization M is given by: M = (N + - N-)µ =µ NL ; -Nµ

11 H{σ i } = -J (1/2 q σ) ∑σ i - µB∑σ i ……………(3) We can find the total configurational energy of the system is given by E= -1/2 (qJL) NL –( µB) NL ……….(4) And = U = -1/2 qJNL - µBNL………..(5) 2

12 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)

13 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

14 (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) Let B =0 => L 0 = tanh [(qJL 0 )/KT] ……..(14) This is called the possibility of spontaneous magnetization

15 We obtain a T c below which the system can acquire a non zero spontaneous magnetization and above which it can not. We can identify the T c with the Curi temperature. The temperature that marks a transition from the ferromagnetic to the paramagnetic behavior of the system or vice verse.

16 From equation (14):  L 0 (T) ≃ {3(1-T/Tc)} ;(T ≲ T c,B→0)…(15) At T→0 => L 0 → 1 L 0 (T) ≃ 1-2 exp(-2T c /T) ; (T/T c ≪ 1)….(16) 1/2

17 The configrational energy of the system is given by : U 0 (T) = -1/2 qJNL 0 ……………(17) And the specific heat is C 0 (T) = -qJNL 0 dL 0 /dT = (NKL 0 ) / [(T/Tc)/(1- L 0 ) - (T/Tc)]...(18) At T>T c => U 0 (T) = C 0 (T) = 0 `

18 The specific heat at the transition temperature Tc is : C 0 (T) = lim {(NK* 3x)/[[(1-x)/(1-3x)] – (1-x)]} = 3/2 NK ……(19) And at T→ 0 C 0 (T) ≃ 4NK (T c /T) exp(-2T c /T) (20) 2 2

19 Note that the vanishing of the configurational energy and the specific heat of the system at temperature above T c is directly related to the configurational order prevailing in the system at lower temperatures is completely wiped out as T→ T c. We note that all the microstate are equally likely to occur is related to the fact that for T ≥ T c there is no configurational order in the system

20 Define : 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-T c {1-L(B,T)}]] For L ≪ 1 we obtain the Curi – weiss law X 0 (T) ≃ (Nµ /K)/ (T-T c ) (T ≳ T c,B→ 0 ) 222 2

21 For T

22 As T → 0 the law field susceptibility vanishes in accordance with the formula X 0 (T) ≃ (4 Nµ /KT) exp (-2T c /T) Finally, the relation ship between L and B at T=T c and use tanh x ≃ x + x /3 we have : L ≃ (3 µB/KT) (T =T c,B→ 0 ) 2 3 1/3

23 Thank you


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