# Derivation of the energy of an alloy

• MathematicalPhysicist
In summary, the conversation discusses a simple model for low-temperature structures, where the energy of nearest-neighbor pairs depends on the type of pair. The model introduces occupation numbers for two sublattices and defines an order parameter. The conversation then makes approximations to calculate the mean energy and entropy. Finally, the model introduces variables that can be expressed in terms of Ising spin variables, and the energy is rewritten in terms of these variables.
MathematicalPhysicist
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## 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:
$$(4.13) \ \ \ \ \ H = \sum_{<ij>} \sigma_i \sigma_j +\frac{q}{4} (e_{AA}-e_{BB})\sum_{i} \sigma_i +q/8 N (e_{AA}+e_{BB}+2e_{AB})$$

## 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.

No one knows?

## 1. What is the energy of an alloy?

The energy of an alloy is the total amount of internal energy stored within the alloy's atoms and molecules.

## 2. How is the energy of an alloy derived?

The energy of an alloy is typically derived through mathematical equations and models that take into account the atomic and molecular structure of the alloy, as well as any external factors such as temperature and pressure.

## 3. Why is it important to understand the energy of an alloy?

Understanding the energy of an alloy is crucial for predicting its behavior and properties, as well as for designing and creating new alloys with desired characteristics.

## 4. Can the energy of an alloy be changed?

Yes, the energy of an alloy can be changed through various methods such as heating, cooling, and applying external forces. These changes can affect the alloy's properties and behavior.

## 5. Are there any practical applications for the derivation of the energy of an alloy?

Yes, the derivation of the energy of an alloy has many practical applications, including in the fields of material science, metallurgy, and engineering. It can help in the development of new and improved alloys for use in various industries such as aerospace, automotive, and construction.

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