# Mean Field Models

Interactions in real matters are complicated and hard to solve exactly since this problem involves many-body interactions. A number of physics-chemical systems which undergo phase transitions have long-range orders. One example is ferromagnetism. Magnetic moments of individual paritcles in a ferromagnet align themselves under a certain critical temperature even without external magnetic field.

The ordering indicates the interactions in these systems ara not completely random. The main idea of **Mean Field Models** is treating the interactions on any one particle as an average interaction [2]. This model effectively turns a many-body problem to a solvable one-body problem. It also provides a theoretical basis for understanding a variaty of phenomena such as ferromagnetism, gas-liquid transitions, and order-disorder transitions in alloys [3].

## Example-Ferromagnetism

In the following example we can see how ferromagnetism arises from Ising model in the zeroth approximaiton [3].
Consider the Ising model on an N-dimensional cubic lattice under an external magnetic field **B**.
The Hamiltonian of the system in configuration {<math>\sigma_1,\sigma_2,...,\sigma_N</math>} is given by

<math>\; H</math>{<math>\sigma_i</math>}=<math>-J \sum_{n.n}\sigma_i\sigma_j-\mu B \sum_i\sigma_i</math>

, where <math>\sigma_i</math>=1 for an "up" spin and -1 for a "down" spin, <math>J</math> is the exchange interaction, and <math>\mu</math> is the magnetic dipole moment. A long-range order parameter <math>L</math> can be defined as

- <math>L=1/N \sum_i\sigma_i = (N_+ - N_-)/N </math>

, where <math>N_+</math> (<math>N_-</math>) is the total number of "up" ("down") spins, and numbers <math>N_+</math> and <math>N_-</math> must satisfy the relation

- <math>N_+ + N_- =N</math>

The magentization <math>M</math> is then given by

- <math>M= (N_+ - N_-) \mu = N \mu L </math>

, the parameter <math>L</math> is a direct measure the magnetizaiton in the system.

In the spirit of mean-field model, the first part of the Hamiltonian can be replaced by the expression <math>-J (1/2q<\sigma>) \sum_i \sigma_i</math>, i.e. for a given <math>\sigma_i</math> with <math>q</math> nearest-neighbors, each of the <math>q \sigma_j</math> is replaced by <<math>\sigma</math>>.

Noting that <math><\sigma>=1/N<\sum_i \sigma_i>= <L></math>, the total confiqurational energy of the system can be written as

<math>E=-1/2 (q J <L>) NL - (\mu B) N L</math>

The expectation valve of <math>E</math> is then given by

<math>U=-1/2 q J N <L>^2 - \mu B N L</math>

The energy expended in changing any "up" spin into a "down" on is given by

<math>\Delta \epsilon = -J (q <\sigma>) \Delta \sigma -\mu B \Delta \sigma =2 \mu (q J/ \mu <\sigma > + B)</math>

, for <math>\Delta \sigma =-2</math> here.

The relative values of the equilibrium numbers <<math>N_+</math>> and <<math>N_-</math>> then follow the Boltzmann distribution,

<math><N_-> / <N_+> = </math> exp<math>( -\Delta \epsilon / k T) = </math> exp<math>(- 2 \mu (B' +B) /k T)</math> , where <math>B'</math> denotes the internal molecular field

<math>B'=q J <\sigma> / \mu = qJ (<M> / N \mu^2)</math>

Given the ratio of <math>N_-</math> and <math>N_+</math> and defnition of order parameter <math>L</math>, we can obtain

<math>(q j <L> + \mu B) / k T = 1/2 </math>ln <math>(1 +<L>)/(1-<L>) = </math> tanh<math>^{-1} <L></math>

To see the possibility of spontaneous magnetization, we can let <math>B \rightarrow 0</math>, which leads to the relationship

<math><L_0>=</math> tanh(<math>q j <L_0> / k T</math>)

The above equatin has solutions when

<math>q J /kT > 1</math>.

This also means spontaneous magnetization can appear when

**<math>T < q j/ k = T_c</math>**

The exact solution of <math><L_0></math> can be solved numerically. Fig. 2 shows <math><L_0></math> as a function of temperature.

## References

[1] http://en.wikipedia.org/wiki/File:Ferromagnetic_ordering.svg

[2] http://en.wikipedia.org/wiki/Mean_field_theory

[3] R. K. Pathria, "Statistical Mechanics", 2nd ed., Butterworth-Heinemann, 1996