Derivation of the energy of an alloy

[MathematicalPhysicist]

Homework Statement

The simplest model that can account for the low-temperature structure is
one in which the energy of nearest-neighbor pairs depends on what kind of pair
it is. We define the quantities $N_{AA}, N_{BB}, N_{AB}$ to be the number of nearest-
neighbor pairs of the Cu-Cu, Zn-Zn, and Cu-Zn type, respectively, and take
the energy of the configuration to be:
$$ (4.1) \ \ \ \ E = N_{AA}e_{AA}+N_{AB}e_{AB}+N_{BB}e_{BB}$$
where $e_{AA}, e_{AB}, \ and\ e_{BB}$ are, respectively, the energies of an AA, AB, and
BB bond.

Let $N$ be the numbers of lattice sites and $N_A$ , $N_B$ the number of Cu and
Zn atoms, respectively. Referring to Figure 4.1, we introduce the occupation
numbers for each of the two simple cubic sublattices: $N_{A1}$ and $N_{B1}$ are the number of atoms of each type on sublattice 1, $N_{A2}$ and $N_{B2}$ the number of atoms of each type on sublattice 2. We have
$$(4.2) \ \ \ \ N_{A1}+N_{A2} = N_{A} = c_A N$$
$$N_{B1}+N_{B2} = N_B = c_B N$$
$$N_{A_1}+N_{B_1} = \frac{1}{2}N$$
$$N_{A_2}+N_{B_2} = \frac{1}{2}N$$

For the sake of definiteness we let $N_A \le N_B$ and define the order parameter:

$$(4.3) \ \ \ \ m= \frac{N_{A1}-N_{A2}}{N_A}$$
$$(4.4) \ \ \ \ N_{A_1} = 1/2 N_A(1+m) \ \ N_{B_1} = 1/2(N_B-N_A m )$$
$$ N_{A_2} = 1/2 N_A(1-m) \ \ N_{B_2} = 1/2(N_B+N_A m)$$

Up to this point our treatment is exact. We now make the crucial approximations:

$$(4.5) \ \ \ \ N_{AA}=\frac{qN_{A1}N_{A2}}{\frac{1}{2}N} \ \ N_{BB} = \frac{qN_{B1}N_{B2}}{\frac{1}{2}N}\ \ \ N_{AB} = q\bigg( \frac{N_{A1}N_{B2}}{\frac{1}{2}N}+\frac{N_{A2}N_{B1}}{\frac{1}{2}N}$$

where $q$ is the number of nearest neighbors surrounding each atomic site. The
mean energy is obtained by substituting (4.5) into (4.1), while the entropy can
be evaluated using the method of Section 3.2. The appropriate free energy is
$$(4.7) \ \ \ \ E = \frac{1}{2}qN( e_{AA}c_A^2+2e_{AB}c_A c_B + e_{BB}c_B^2)-qN\epsilon c_A^2m^2$$

where:
$$(4.8) \ \ \ \ \epsilon = \frac{1}{2} (e_{AA}+e_{BB})-e_{AB}$$

$$\ldots$$

We introduce the variables $n_{iA}$ and $n_{iB}$, where $n_{iA}=1$ if an atom of type $A$ occupies site $i$ otherwise it's zero.
$n_{iB} = 1-n_{iA}$
These variables can be expressed in terms of Ising spin variables:

$$(4.12) \ \ \ n_{iA} = 1/2(1+\sigma_i) , \ \ n_{iB} = 1/2 (1-\sigma_i)$$

with $\sigma_i = \pm 1$.
With $\epsilon = 2J$, the energy (4.1) becomes:
$$(4.13) \ \ \ \ \ H = \sum_{<ij>} \sigma_i \sigma_j +\frac{q}{4} (e_{AA}-e_{BB})\sum_{i} \sigma_i +q/8 N (e_{AA}+e_{BB}+2e_{AB})$$

Homework Equations

The Attempt at a Solution

My question is how to derive $(4.13)$ from the above preceding paragraphs?, I am not sure how achieve these terms.

 

[MathematicalPhysicist]

No one knows?