# Mean Field Models

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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.

Fig. 1, Ferromagnetic ordering [1]

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 {$\sigma_1,\sigma_2,...,\sigma_N$} is given by

$\; H${$\sigma_i$}=$-J \sum_{n.n}\sigma_i\sigma_j-\mu B \sum_i\sigma_i$

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

$L=1/N \sum_i\sigma_i = (N_+ - N_-)/N$

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

$N_+ + N_- =N$

The magentization $M$ is then given by

$M= (N_+ - N_-) \mu = N \mu L$

, the parameter $L$ 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 $-J (1/2q<\sigma>) \sum_i \sigma_i$, i.e. for a given $\sigma_i$ with $q$ nearest-neighbors, each of the $q \sigma_j$ is replaced by <$\sigma$>.

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

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

The expectation valve of $E$ is then given by

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

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

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

, for $\Delta \sigma =-2$ here.

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

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

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

Given the ratio of $N_-$ and $N_+$ and defnition of order parameter $L$, we can obtain

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

Fig. 2, The spontaneous magnetization of a Weiss ferromagnet as function of temperature. The experimetal points are for iron (x), nickel (o), cobal($\Delta$), and magnetite (+) [3]

To see the possibility of spontaneous magnetization, we can let $B \rightarrow 0$, which leads to the relationship

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

The above equatin has solutions when

$q J /kT > 1$.

This also means spontaneous magnetization can appear when

$T < q j/ k = T_c$

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

## References

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