3161: Microstructural Dynamics I
Peter W. Voorhees and Kenneth R. ShullDepartment of Materials Science and EngineeringNorthwestern University
1 Catalog Description (3161,2)
Principles underlying development of microstructures. Defects, diffusion, phase transformations, nucleation and growth, thermal and mechanical treatment of materials. Lectures, laboratory. Prerequisite: 315 or equivalent.
2 Course Outcomes
At the conclusion of 3161 students will be able to:
 Describe the Kirkendall effect, diffusion in ternary systems, and the importance of shortcircuit diffusion.
 Describe the structure of various types of interfaces and the effects these structures have on interfacial energy.
 Apply concepts of mathematics and physics to imperfections, diffusion and phase transformations.
 Use basic concepts of dislocation theory: topology and energetics of dislocations in crystalline materials.
 Exhibit a good understanding of dislocations as related to their type (edge, screw, mixed), stress fields, energies, geometry (bowing, kinks, jogs) and interaction.
 Correlate dislocation motion to plastic flow.
 Describe how the grain size of a material can be controlled by mechanical and thermal processing of materials
 Demonstrate laboratory skills in structural and thermal processing of materials.
3 Diffusion
3.1 Review of the Basic Equations
The diffusion equation describes the evolution of the composition profile with time as the individual components diffuse within a sample. These components can be either atoms or molecules, but for our purposes we'll assume that the diffusing species are atoms (as in a metallic sample). For a binary system of A and B components, we can use either ${C}_{A}$ or ${C}_{B}$ (the respective concentrations of A and B species in atoms/volume) to describe the composition. These compositions sum to the total atomic concentration, ${C}_{0}$:
$$\begin{array}{cc}{C}_{0}={C}_{A}+{C}_{B}& (3.1)\end{array}$$
If the molar volumes of A and B are equal to one another, then ${C}_{0}$ is fixed, so that the following conditions hold:
$$\begin{array}{cc}\frac{\partial {C}_{A}}{\partial z}=\frac{\partial {C}_{B}}{\partial z}& (3.2)\end{array}$$
$$\begin{array}{cc}\frac{\partial {C}_{A}}{\partial t}=\frac{\partial {C}_{B}}{\partial t}& (3.3)\end{array}$$
Note that we are using
$z$ as our spatial variable. For a binary A/B alloy we can use either
${C}_{A}$ or
${C}_{B}$ to describe the overall composition of the alloy. The flux of atoms is given by
:
If the diffusion coefficient is independent of concentration (and hence, independent of $x$ as well) then the diffusion equation can be written as follows:
The diffusion equation involves a first derivative with respect to time an a second derivative with respect to distance, so in general we need an initial condition and two boundary conditions. Consider for example the following situation:
 Boundary conditions: ${C}_{A}={C}_{1}$ for $z\to \infty $ and ${C}_{A}={C}_{2}$ for $z\to \infty $
 Initial condition: The concentration jumps discontinuously from ${C}_{1}$ to ${C}_{2}$ at $x=0$
Here $w$ is the following diffusion length, which enters into all diffusion problems:
$$\begin{array}{cc}w\left(t\right)=2\sqrt{{D}_{A}t}& (3.9)\end{array}$$
$$\begin{array}{cc}erf\left(x\right)=\frac{2}{\sqrt{\pi}}{\int}_{0}^{x}{e}^{{t}^{2}}dt& (3.10)\end{array}$$
Note that erf(
$x$) transitions from 1 to large negative values of
$x$ to +1 for high positive values of
$x$, as shown in Figure
3.1.
The solution to Eq. 3.8 is shown in Figure
3.2. To show how the concentration profile evolves with time, we have included values of
$w$ in the plot.
figure
figformat % set some defaults so the figures look pretty
z=linspace(1,1,200); % These are the z points
w=[0.2,0.4,0.6]; % these are the three values of the normalized diffusion length that we will include in our calculations
c=@(z,w) erf(z/w); % define a function of two variables, z and w
col={[1,0,0],[0,0.5,0],[0,0,1]}; % these are the three colors (rgb format)
linetype={'','','.'}; % these are the three line types we well used (plain, dashed and dashdot)
axes
hold on
for i=1:3
plot(z,c(z,w(i)),'color',col{i},'linestyle',linetype{i})
legendtext{i}=['$w/L$=' num2str(w(i))];
end
legend(legendtext,'location','best','interpreter','latex')
ylabel('C_{a}')
xlabel('z/L')
ylim([1.2 1.2])
set(gca,'ytick',[1,1])
set(gca,'yticklabel',[]) % turn off the y axis tick labls by making 'yticklable' an empty vector
text(1.15, 1, 'C_{1}','fontsize',16)
text(1.15, 1, 'C_{2}','fontsize',16)
print(gcf,'../figures/erfsolution.eps','depsc2') % save as an eps file
We can also consider the situation where we have layer of material at
$z=0$, which diffuses in the positive and negative directions into the bulk material. In this case the initial and boundary conditions are as follows:
 Boundary conditions: ${C}_{A}=0$ for $z=\pm \infty $
 Initial condition: All of the A component is confined to a very layer at $z=0$, with a surface coverage (atoms/area) of ${C}_{s}$
 Normalization condition: The total amount of material in the sample must be conserved, so if we integrate the concentration profile we must end up with ${C}_{s}$:
$$\begin{array}{cc}{\int}_{\infty}^{\infty}{C}_{A}\left(z\right)dz={C}_{s}& (3.11)\end{array}$$
In this case the following solution to the diffusion equation is obtained:
Eq. 3.12 is plotted in Figure
3.3 for three different time points.
3.2 Mole Fractions and Volume Fractions
An assumption that we make throughout this text is that the atomic volumes of different chemical species are all identical, equal to ${V}_{0}\mathrm{.}$ In reality, this is almost never exactly true. Fortunately, it doesn't really matter when thinking about diffusion because we can always work with volume fractions instead of mole fractions. In a generalized formulation the molecular volumes of the A and B molecules are given by multiplying the reference volume by a factor of $N$, which is not necessarily the same for each molecule:
$$\begin{array}{cc}\begin{array}{c}{V}_{A}={N}_{A}{V}_{0}\\ {V}_{B}={N}_{B}{V}_{0}\end{array}& (3.13)\end{array}$$
We can relate concentrations to mole fractions and volume fractions by considering a binary A/B system with $n$ total atoms. Of these, $n{X}_{A}$ are A atoms and $n{X}_{B}$ are B atoms. Multiplying by the the atomic volume gives the total volume of each component. The total volume of A atoms is $n{X}_{A}{N}_{A}{V}_{0}$ and the total volume of B atoms is $n{X}_{B}{N}_{B}{V}_{0}$. From these expressions we obtain the following for ${\phi}_{A}$, the volume fraction of A atoms in the system:
$$\begin{array}{cc}\begin{array}{c}{\phi}_{A}=\frac{n{X}_{A}{N}_{A}{V}_{0}}{{V}_{tot}}={C}_{A}{N}_{A}{V}_{0}\\ {\phi}_{B}=\frac{n{X}_{A}{N}_{A}{V}_{0}}{{V}_{tot}}={C}_{B}{N}_{B}{V}_{0}\end{array}& (3.14)\end{array}$$
where ${V}_{tot}$ is the total volume of the system. Note that we have used ${C}_{A}=n{X}_{A}/{V}_{tot}$ and ${C}_{B}=n{X}_{B}/{V}_{tot}$. Throughout the rest of this text we generally assume that ${N}_{A}={N}_{B}=1.$ In the case where ${N}_{A}$ and/or ${N}_{B}$ are not equal to one we can define renormalized concentrations, ${C}_{A}^{0}$ and ${C}_{B}^{0}$ that describe the concentration of subunits of volume ${V}_{0}\mathrm{.}$ These fluxes are related to the atomic fluxes, ${C}_{A}$ and ${C}_{B}$ by multiplying by the appropriate value of $N$:
The renormalized fluxes, ${J}_{A}^{0}$ and ${J}_{B}^{0}$ are obtained by a similar normalization:
$$\begin{array}{cc}\begin{array}{c}{J}_{A}^{0}={J}_{A}{N}_{A}\\ {J}_{B}^{0}={J}_{B}{N}_{B}\end{array}& (3.16)\end{array}$$
$$\begin{array}{cc}\frac{\partial {\phi}_{i}}{\partial t}=\frac{\partial}{\partial x}{D}_{i}\left(\frac{\partial {\phi}_{i}}{\partial z}\right)& (3.17)\end{array}$$
The bottom line of all this is that Fick's second law still applies, with same diffusion coefficient used for the case where the atomic volumes are equal, provided that we simply replace concentrations with volume fractions.
3.3 Vacancy Diffusion Mechanism
Figure
3.5 shows the output of a vacancy diffusion simulation of the interdiffusion between two materials. Vacancies move when an atom from an adjacent site moves into the vacancy. The resulting net motion of the atoms provides a means for diffusive mixing across an interface, and this is the process being illustrated in Figure
3.5 If the probability of hopping into a vacancy is different for A and B atoms, then
$\left{J}_{A}\right\ne \left{J}_{B}\right$, and we need to consider additional effects. These are described below in our discussion of the Kirkendall effect.
tic % start a time so that we can see how long the program takes to run
n=30; % set the number of boxes across the square grid
vfrac=0.01; % vacancy fraction
matrix=ones(n);
map=[1,1,1;1,0,0;0,0,1]; % define 3 colors: white, red, blue
figure
colormap(map) % set the mapping of values in 'matrix' to a specific color
caxis([0 2]) % range of values in matrix goes from 0 (vacancy) to 2
% the previous three commands set things up so a 0 will be white, a 1 will
% be red and a 2 sill be blue
matrix(:,n/2+1:n)=2; % set the right half of the matrix to 'blue'
i=round(n/2); % put one vacancy in the middle
j=round(n/2);
matrix(i,j)=0;
imagesc(matrix); % this is the command that takes the matrix and turns it into a plot
t=0;
times=[1e4,2e4,5e4,1e5];
showallimages=1; % set to zero if you want to speed things up by not showing images, set to 1 if you want to show all the images during the simulation
%% now we start to move things around
vacancydiff.matrices={}; % makea blank cell array
while t<max(times)
t=t+1;
dir=randi([1 4], 1, 1);
if dir==1
in=i+1;
jn=j;
if in==n+1; in=1; end
elseif dir==2
in=i1;
jn=j;
if in==0; in=n; end
elseif dir==3
in=i;
jn=j+1;
if jn>n; jn=n; end
elseif dir==4
in=i;
jn=j1;
if jn==0; jn=1; end
end
% now we need to make switch
neighborix=sub2ind([n n],in,jn);
vacix=sub2ind([n n],i,j);
matrix([vacix neighborix])=matrix([neighborix vacix]);
if showallimages
imagesc(matrix);
drawnow
end
if ismember(t,times)
vacancydiff.matrices=[vacancydiff.matrices {matrix}]; % append matrix to output file
imagesc(matrix);
set(gcf,'paperposition',[0 0 5 5])
set(gcf,'papersize',[5 5])
print(gcf,['vacdiff' num2str(t) '.eps'],'depsc2')
end
i=in;
j=jn;
end
vacancydiff.times=times;
vacancydiff.n=n;
save('vacancydiff.mat','vacancydiff') % writes the vacancydiff structure to a .mat file that we can read in later
toc
3.4 Kirkendall Effect
The geometry of the Kirkendall experiment (1947) is shown in Figure 3.6 [
7]. In the experiment a small block of brass (70% copper, 30% Zn) was surrounded by inert, Molybdenum (Mo) wires. The sample was then coated with copper, and heated to a high temperature to allow atoms within the material to diffuse. In the measurement, the distance,
$w$, between the Mo markers decreased as a function of time. This result implies that the flux of Zn out of the brass portion of the sample is larger than the copper flux back into the brass from the outside.
Diffusion does not need to occur by a vacancy motion in order for the Kirkendall effect to be observed, all that is needed is an asymmetry in the diffusion coefficients of the individual components in the material. However, for our purposes we will assume for now that diffusion occurs by a vacancy hopping mechanism. This assumption is valid for the original Kirkendall experiment, and it also enables us to make a connection to the relevant microscopic diffusion mechanisms. It is an excellent example of the structure/property relationships that define the field of materials science.
Our starting point is to assume that the vacancy concentration remains at equilibrium, so that the total number of lattice sites (including vacant sites) remains constant. A consequence of this assumption is that the fluxes of of A atoms, B atoms and vacancies must sum to zero:
$$\begin{array}{cc}{J}_{A}+{J}_{B}+{J}_{v}=0& (3.18)\end{array}$$
$$\begin{array}{cc}{J}_{v}=\left({D}_{A}{D}_{B}\right)\frac{\partial {C}_{A}}{\partial z}& (3.19)\end{array}$$
The situation for
${D}_{A}{D}_{B}>0$ is illustrated in Figure
3.7. In this case the net vacancy flux is negative (to the left), and has a maximum magnitude at the point where the concentration gradient is the largest. Because the vacancy flux varies with position, there will be a time dependent increase or decrease in the local vacancy concentration that can be obtained from a site conservation equation similar to Eq.
3.5:
$$\begin{array}{cc}\frac{\partial {C}_{v}}{\partial t}=\frac{\partial {J}_{v}}{\partial z}& (3.20)\end{array}$$
This results in a net depletion of vacancies in some regions of the sample (the right in Figure
3.7c) and a net supersaturation of the vacancy concentration in other regions of the sample (the left in Figure
3.7c). In most cases processes exist that enable these concentration variations to be eliminated, by the creation of vacancies at the right portion of the sample and the destruction of vacancies at the left portion of the sample. Typically, these processes involve the addition or removal of vacancies to the core of a dislocation.
3.5 The Interdiffusion Coefficient
In general, a material flux, $J$, within a material can be related to a velocity, $v$. This velocity is obtained simply by multiplying $J$ by the reference volume, ${V}_{0}$, that is used to define the diffusive flux: (${V}_{0}=1/{C}_{0}$):
We will often find it useful to use mole fractions instead of concentrations in our expressions, so we need to keep the following relationship in mind:
This is the velocity that individual planes are moving with respect to a fixed position in the sample that is far from the interface (the ends of the sample, for example). The fluxes obtained from Fick's first law are defined in terms of a reference plane that is moving with a velocity ${v}_{\ell}$. We can also define fluxes of A and B atoms across stationary planes, and we refer to these fluxes as ${J}_{A}^{\prime}$ and ${J}_{B}^{\prime}$. We can get ${J}_{A}^{\prime}$ by adding ${v}_{\ell}{C}_{A}$ to ${J}_{A}$, where ${v}_{\ell}{C}_{A}$ is the net flux of A atoms across a fixed plane in space due to the lattice plane velocity:
$$\begin{array}{cc}{J}_{A}^{\prime}={D}_{A}\frac{\partial {C}_{A}}{\partial z}+{v}_{\ell}{C}_{A}& (3.26)\end{array}$$
We can combine this expression with Eq. for ${v}_{\ell}$ to get:
$$\begin{array}{cc}{J}_{A}^{\prime}={D}_{A}\frac{\partial {C}_{A}}{\partial z}+\left({D}_{A}{D}_{B}\right){X}_{A}\frac{\partial {C}_{A}}{\partial z}& (3.28)\end{array}$$
After a bit of algebra, keeping in mind that ${X}_{A}+{X}_{B}=1$, we obtain the following:
$$\begin{array}{cc}{J}_{A}^{\prime}=\left[{D}_{A}{X}_{B}+{X}_{A}{D}_{B}\right]\frac{\partial {C}_{A}}{\partial z}& (3.29)\end{array}$$
with this definition we have:
$$\begin{array}{cc}{J}_{A}^{\prime}=\tilde{D}\frac{\partial {C}_{A}}{\partial z}& (3.31)\end{array}$$
3.6 Connection to thermodynamics
$$\begin{array}{cc}{v}_{B}={M}_{B}\frac{\partial {\mu}_{B}}{\partial z}& (3.33)\end{array}$$
Note that ${M}_{B}$ must be positive. Atoms always move down a chemical potential potential gradient, although in some cases diffusion may take place up a concentration gradient (more on this in 3162). We can use the previous expression for ${v}_{B}$ to obtain the following expression for the diffusive flux of B atoms:
$$\begin{array}{cc}{J}_{B}={C}_{B}{v}_{B}={M}_{B}{C}_{B}\frac{\partial {\mu}_{B}}{\partial {X}_{B}}\frac{\partial {X}_{B}}{\partial z}& (3.34)\end{array}$$
$$\begin{array}{cc}{J}_{B}={M}_{B}{X}_{B}\frac{\partial {\mu}_{B}}{\partial {X}_{B}}\frac{\partial {C}_{B}}{\partial z}& (3.35)\end{array}$$
Comparing to Fick's first law (Eq.
3.4), we obtain the following for
${D}_{B}$:
We see the this intrinsic diffusion coefficients involves purely kinetic parameter (the mobility, ${M}_{B}$), and a thermodynamic parameter (the derivative of ${\mu}_{B}$ with concentration). As discussed in more detail in 315 (see the sectional on Type II (binary) phase diagrams), chemical potentials are most commonly expressed in terms of activity coefficients in the following way:
$$\begin{array}{cc}{\mu}_{B}={\mu}_{B}^{0}+RTln{a}_{B}& (3.37)\end{array}$$
Here ${\mu}_{B}^{0}$ is the standard state, which is generally defined to be zero for a pure material at thermodynamic equilibrium. This equation can be used to write the chemical potential derivative in the following way:
3.7 Tracer Diffusion Coefficients
 Like mobilities, tracer diffusion coefficients are purely kinetic parameters.
 Tracer diffusion coefficients will depend on the local composition (the relative amount of A and B atoms in a binary alloy, for example).
 A binary AB alloy there are two, independent, compositiondependent tracer diffusion coefficients. If the tracer species is chemically identical to atom A, then we refer to the tracer diffusion coefficient as ${D}_{A}^{*}$. If the tracer species is chemically identical to atom B, then we refer to the tracer diffusion coefficient as ${D}_{B}^{*}$ .
By definition, tracer diffusion coefficients are are defined in the dilute limit, where the activity increases linearly with concentration in a way that is given by the Henry's law coefficient, $H$. We'll illustrate things by assuming that the tracer is chemically identical to B. In this case we express Henry's law in the following way:
$$\begin{array}{cc}{a}_{B}={H}_{B}{X}_{B}& (3.39)\end{array}$$
From Eq. 3.38 we obtain the following expression for the chemical potential derivative in the dilute (Henry's law) regime.
$$\begin{array}{cc}\frac{\partial {\mu}_{B}}{\partial {X}_{b}}=\frac{RT}{{X}_{B}}& (3.40)\end{array}$$
The tracer diffusion coefficient for the B atoms is related to the mobility by the following expression:
$$\begin{array}{cc}{D}_{B}=\frac{{D}_{B}^{*}{X}_{B}}{RT}\frac{\partial {\mu}_{B}}{\partial {X}_{B}}=\frac{{D}_{B}^{*}{X}_{B}}{a}\frac{\partial {a}_{B}}{\partial {X}_{B}}& (3.42)\end{array}$$
As a result, ${D}_{B}={D}_{B}^{*}$ whenever ${a}_{B}$ is proportional to ${X}_{B}$. This is the case when ${X}_{B}$ is very small (Henry's law regime), and when ${X}_{B}$ is close to 1 (Rault's law regime), but it is not necessarily true or intermediate values of ${X}_{B}$.
3.8 Summary of Diffusion in a Binary System
We have defined three interrelated types diffusion coefficients: $\tilde{D}$, ${D}_{i}$ and ${D}_{i}^{*}$. Here we provide a brief summary of these different diffusion coefficients and the relationships between them.
 $\tilde{D}$: The interdiffusion coefficient (often referred to as the mutual diffusion coefficient). If you are interested in the timedependent evolution of the composition profile, this is the diffusion coefficient that you use when you are solving the diffusion equation:
$$\frac{\partial {C}_{B}}{\partial t}=\frac{\partial}{\partial x}\tilde{D}\left({C}_{B}\right)\left(\frac{\partial {C}_{B}}{\partial z}\right)$$
note that for a binary system, we only need to specify one of the compositions, since ${C}_{A}={C}_{0}{C}_{B}$. Also note that in general, $\tilde{D}$ depends on the composition, so cannot be treated as a constant.
 ${D}_{A}$ and ${D}_{B}$: The intrinsic diffusion coefficients for the individual components. These are important for two reasons. First, they are needed if you want to describe motions of atomic planes relative to the external boundaries of the sample (the Kirkendall effect). This motion was determined from the the atomic fluxes relative to atomic planes, as opposed to fixed points in space. These fluxes are determined by the appropriate intrinsic diffusion coefficient. For example, for the B component, we have:
$${J}_{B}={D}_{B}\frac{\partial {C}_{B}}{\partial z}$$
Also, predictive models of interdiffusion are generally based on the relationship between these intrinsic diffusion coefficients and the interdiffusion coefficient through the following expression:
$$\begin{array}{cc}\tilde{D}& =\left[{X}_{B}{D}_{A}+{X}_{A}{D}_{B}\right]\end{array}$$
 ${D}_{A}^{*}$ and ${D}_{B}^{*}$: The tracer diffusion coefficients for the individual components. Imagine a single atom in a homogeneous material. The tracer diffusion coefficient describes the probability that the atom has diffused a certain distance in a given period of time. These diffusion coefficients are purely kinetic parameters, and can be expressed in terms of a mobilty:
$${D}_{B}^{*}=RT{M}_{B}$$
Unlike the interdiffusion and intrinsic diffusion coefficients, they are not affected by the thermodynamics of the system. In general, values of the tracer diffusion coefficients will depend on the concentration of the material in which the tracer atoms are diffusing. The special cases of ${D}_{A}^{*}$ at ${X}_{A}=1$ and ${D}_{B}^{*}$ at ${X}_{B}=1$ are self diffusion coefficients. The intrinsic diffusion coefficients are related to the tracer diffusion coefficients through the following relationship:
$${D}_{B}=\frac{{D}_{B}^{*}{X}_{B}}{RT}\frac{\partial {\mu}_{B}}{\partial {X}_{B}}$$
3.9 Diffusion in Ternary Systems
Atomic diffusion in ternary systems is driven by chemical potential gradients, just as it is does in binary systems. In systems with more than two components, however, the composition is no longer specified by a single composition variable. Some interesting effects can be observed in this case, as exemplified by carbon diffusion in FeSiC ternary alloys . The carbon chemical potential is now a function of the concentration of both the silicon and carbon in the alloy:
$$\begin{array}{cc}{\mu}_{c}=f\left({X}_{Si},{X}_{C}\right)& (3.43)\end{array}$$
The diffusion coefficient of carbon is much larger than the diffusion coefficient for silicon (
${D}_{c}\gg {D}_{Si}$), so we can assume that the silicon remains stationary during a diffusion experiment, as shown in Figure
3.10. Silicon and carbon have a unfavorable thermodynamic interaction within the alloy, so
${\mu}_{c}$ increases with increasing silicon content,
${X}_{Si}$. In order for the carbon chemical potential to remain constant across and interface between two regions of differing Si content, the carbon concentration in the region with low Si content needs to be smaller than the carbon concentration in the region with high Si content. This chemical potential discontinuity at the interface is eliminated by the jump of carbon atoms from the left (high Si side) to the right (low Si side) of the interface. Diffusion then continues from left to right, down the carbon potential gradient that has been established.
3.10 Crystal Defects and High Diffusivity Paths
“Crystals are like people. It is the defects in them which tend to make them interesting.”  Colin Humphreys
Real crystals are never perfect, and they always contain some sort of defects. These defects can be classified into four categories, based on their dimension:
 0dimensional (point) defects: These include missing atoms (vacancies), or atoms in location where they would not be in a perfect crystal structure (interstitials or substitutional impurities. From purely thermodynamic considerations we know that point defects must exist at some finite concentration for temperatures above 0K.
 1dimensional (line) defects: These are dislocations.
 2dimensional (planar) defects: These include grain boundaries, which are internal interfaces between regions of different crystalline orientation, and the external surfaces of a material.
 3dimensional (volume) defects: These are geometric imperfections in a material, like pores and cracks. We don't consider these types of defects in this class, but they become very important when we discuss the fracture properties of bulk, brittle materials in subsequent courses.
As illustrated in Figure 3.11, dislocation, grain boundaries and surfaces are associated with a more open structure. As a result diffusion along these defects is much faster than in the bulk of the material.
4 Dislocations
where $\phi $ is the angle between the tensile axis and the slip plane normal, $\overrightarrow{n}$, and $\lambda $ is the angle between the tensile axis and the slip direction, $\overrightarrow{d}$.
.
Why is the force to deform a single crystal so low? We'll start by considering what we would expect for the critical resolved shear stress if the shear deformation were to occur by the sliding of atomic planes over one another, as shown conceptually in Figure
4.3. We refer to the stress required to slide these planes over one another as the dislocationfree critical resolved shear stress,
${\tau}_{crss}^{0}\mathrm{.}$
$$\begin{array}{cc}{e}_{xy}=\frac{u}{d}& (4.2)\end{array}$$
For a linearly elastic material, the shear stress, $\tau $ is proportional to ${e}_{xy}$, with the shear modulus $G$ defined as the ratio of shear stress over shear strain:
In Figure 4.5 show a schematic representation of the stress as a function of displacement for the atomic planes shown in Figure
4.3. The stress function has the following features:
 The stress is a periodic function, with the stress repeating every time the displacement is increased by an amount equal to $b$, the distance between atoms along the slip direction.
 The stress is equal to zero at the stable equilibrium positions at $u=0,\phantom{\rule{6px}{0ex}}b,\phantom{\rule{6px}{0ex}}2b$, etc.
 For $u<b/2$ the stress is positive because we need to apply a stress to move the atoms out of their stable equilibrium positions.
 At $u=b/2$ the system is at an unstable equilibrium. The stress is also equal to zero at this position, but the equilibrium is unstable because any slight perturbation in the displacement will cause the atomic plane to fall back into an equilibrium position at $u=0$ or $u=b$.
 The maximum stress is at $u=b/4$ . The stress actually reverses sign for $u>b/2$, since a stress must be applied to avoid having the atoms fall into the equilibrium position at $u=b$.
The simplest mathematical expression for the shear stress that has the right periodicity is a sinusoidal function:
$$\begin{array}{cc}\tau =\frac{bG}{2\pi d}sin\left(\frac{2\pi u}{b}\right)& (4.6)\end{array}$$
The critical resolved shear stress in this picture corresponds to the maximum value of
$\tau $, equal to
$bG/2\pi d$. The interplanar spacing,
$d$ is comparable to
$b$. (We're not going to worry about the exact numerical factor here, since we're just aiming to get an approximate expression for
${\tau}_{crss}$). We take
$b\approx d$ and
$2\pi \approx 6$ to end up with the following expression for the ideal critical resolved shear stress,
${\tau}_{crss}^{0}$, which is the value of the critical resolved shear stress we would expect to have if dis:
$$\begin{array}{cc}{\tau}_{crss}^{0}\approx G/6.& (4.7)\end{array}$$
4.1 Edge Dislocations
Motion of an edge dislocation is illustrated in response to an applied shear stress is illustrated in Figure 4.9. Note that for every atom moving away from its equilibrium on one side of the dislocation core, there is an equivalent atom moving toward an equilibrium position on the other side of the dislocation core. In energetic terms, for every atom that must be forced out of its lowest energy position, there is atom moving toward its lowest energy position. As a result the energy changes cancel (or very nearly so), and the energy barrier to moving a dislocation is much less than the barrier to slide surfaces across one another. As a result the net force to move a dislocation is very small. The stress needed to move a dislocation is generally much less than
$G/6$, and is as low or lower than the observed critical resolved shear stress for single crystals.
Figure 4.13 illustrates the the motion of a screw dislocation through a crystal. In this case the dislocation moves from the front of the crystal to the back of the crystal. The net effect of this motion is for the top and bottom halves of the crystal to be displaced to the right, by an amount and in the direction given by the Burgers vector. This figure illustrates the following:
 When a dislocation line travels through a material, the motion of the line traces out a plane.
 The relative displacement between the material on either side of this plane is given by the Burgers vector $\overrightarrow{b}$.
Note that this is true for ANY dislocation (edge, screw, or mixed).
4.3 The Burgers Circuit
 Draw a circuit around the dislocation line that starts end ends at the same point. A 'right handed' convention is typically used to describe the direction that we take the circuit. (Clockwise looking along the direction of $\stackrel{\u02c6}{s}$, counterclockwise if $\stackrel{\u02c6}{s}$ is pointed at you).
 Repeat the procedure, using the same numbers of atomic steps in each direction in a perfect crystal.
 The Burgers vector is the vector connecting the start and end positions for the circuit drawn in the perfect crystal.
 Move four steps down (a to b)
 Move three steps to the right (b to c)
 Move four steps up (c to d)
 Move four steps to the left (d back to a)
When this same procedure is repeated in the perfect crystal we end up at point e, which is one step to the left of our starting point at point a. Our convention is to define the
$\overrightarrow{b}$ as the vector starting at point a and ending at point b. When the procedure is repeated for a dislocation where the half plane is in the bottom half of the crystal we end up with the Burgers vector pointing in the opposite direction, as shown in Figure
4.15.
In Figure 4.16 we repeat the same process for a screw dislocation. In this example we have defined the direction of
$\stackrel{\u02c6}{s}$ so that the dislocation is pointed toward the bottom of the figure. The procedure for determining
$\overrightarrow{b}$ is as follows:
 Draw a circuit in the clockwise direction (viewed from the top, so we are looking in the direction of $\stackrel{\u02c6}{s}$) around the dislocation line. The circuit begins and ends at point a.
 Repeat the circuit in a perfect part of the crystal. The circuit begins at point $s$ and ends at point $f$.
 The Burgers vector is obtained as the vector that starts at $s$ and ends at $f\mathrm{.}$
Note that
$\overrightarrow{b}$ is parallel to
$\stackrel{\u02c6}{s}$, as it must be for a screw dislocation, but that
$\overrightarrow{b}$ and
$\stackrel{\u02c6}{s}$ are pointed in opposite directions,
i.e., they are antiparallel. With our convention of drawing the
$\overrightarrow{b}$ from the starting point to the ending point of the Burgers circuit in the perfect crystal, right handed screw dislocations have negative Burgers vectors and left handed screw dislocations have positive Burgers vectors. The left handed version of the dislocation shown in Figure
4.16 is shown in Figure
4.17.
4.4 The $\overrightarrow{b}\times \stackrel{\u02c6}{s}$ cross product
Unfortunately, the ambiguity introduced by our definition of the direction of
$\stackrel{\u02c6}{s}$ along the dislocation line is impossible to remove. In figure
4.16 we defined
$\stackrel{\u02c6}{s}$ so that it points along the negative z direction, but there's no reason that we couldn't have defined
$\overrightarrow{s}$ so that it is directed in the positive z direction instead. We end up with a Burgers vector that points in one of two opposite directions, depending on how we define
$\stackrel{\u02c6}{s}$ in the first place. The good news is that
${\overrightarrow{n}}_{d}$, the vector cross product of
$\stackrel{\u02c6}{s}$ and
$\overrightarrow{b}$ is independent of our convention for defining the direction of
$\overrightarrow{s}$. As a reminder, the vector cross product between vectors
$\stackrel{\u02c6}{s}$ and
$\overrightarrow{b}$ is defined as follows, as illustrated in Figure
4.18:[
#_cross_2014]
Here ${\stackrel{\u02c6}{n}}_{d}$ is a unit vector in the direction perpendicular to the plane containing $\stackrel{\u02c6}{s}$ and $\overrightarrow{b}$. It's orientation is defined using the right hand rule: We place our right hand along $\stackrel{\u02c6}{s}$, with our fingers oriented in the positive $\theta $ direction. Our right thumb is then pointed along ${\stackrel{\u02c6}{n}}_{d}$.
When defined in this way, ${\overrightarrow{n}}_{d}$ has the following properties:
 Because redefining $\overrightarrow{s}$ to have the opposite orientation also changes the orientation of $\overrightarrow{b}$, the negative signs cancel and we end up with a value for ${\overrightarrow{n}}_{d}$ that is independent of the way that we choose to define $\overrightarrow{s}$.
 For a pure screw dislocation, $\theta =0$ or $\theta =18{0}^{\u25cb}$ . In either case, ${\overrightarrow{n}}_{d}$ = 0.
 For an edge dislocation, the magnitude of ${\overrightarrow{n}}_{d}$ is equal to the $b$, the magnitude of Burgers vector. In addition, ${\overrightarrow{n}}_{d}$ points toward the extra half plane.
With our convention for using the Burgers circuit to obtain $\overrightarrow{b}$ (Righthandrule, start to finish), we have the following relationships between $\stackrel{\u02c6}{s}$ and $\overrightarrow{b}$:
 Righthanded screw dislocation: $\stackrel{\u02c6}{s}$ and $\overrightarrow{b}$ point in opposite directions.
 Lefthanded screw dislocation: $\stackrel{\u02c6}{s}$ and $\overrightarrow{b}$ point in the same direction.
 Edge dislocation: $\stackrel{\u02c6}{s}$ perpendicular to $\overrightarrow{b}$, $\overrightarrow{b}\times \stackrel{\u02c6}{s}$ points to the extra half plane.
4.5 Connection to the Crystal Structure
The Burgers vector must correspond to an atomic repeat distance in the crystal structure. As we show below, the energy of a dislocation is proportional to the square of the magnitude of the Burgers vector. For this reason the Burgers vector will correspond to closest atomic distance in crystal structure. As shown in Figure
4.19, the Burgers vector is half the unit cell diagonal for the BCC structure, and half the face diagonal of the unit cell in the FCC structure.
4.6 Dislocation loops
Exercise: What happens to the shape of the crystal in Figure
4.20 if the loop contracts to nothing and disappears?
Solution: The dislocation just disappears, and a perfect crystal (at least in this region) is recovered.
4.7 Dislocation Density
 Total line length of dislocations per volume.
 The number of intersections that the dislocations make with a plane of unit area.
Both definitions give dislocation densities with units of 1/area, and are equivalent if the dislocations are straight. Typical dislocation densities are as follows:
 A well annealed metal: $1{0}^{6}1{0}^{8}$/cm${}^{2}$.
 Plastically deformed metal: can be as high as $5x1{0}^{11}$/cm${}^{2}$.
 Ceramics: Much lower, typically 10/cm${}^{2}$.
 Si used in microelectronics: dislocation density of zero! Macroscopic single crystals are typically grown without a single dislocation. The down side of this is that Si is very brittle, since there is no plastic deformation mechanism.
4.8 Dislocation Motion
4.8.1 Dislocation Glide
4.8.2 Dislocation Climb
4.9 Dislocation Energy
Dislocation disrupts the regularity of the lattice, and introduces strain into the sample. The strain field that results from the presence of a dislocation has a very long range, and can easily be more than 100 times the unit cell dimension. As a result the total strain energy is very large as well. This strain field and the energy associated with it is important because it provides a mechanism for dislocations to interact with one another over long distances. In essence, dislocations 'talk' to each other through these strain fields.
4.9.1 Screw Dislocations
For a screw dislocation we can use some simple concepts to calculate this strain energy, so we'll start with this example. Our starting point is that the material surrounding a screw dislocation is in a state of pure shear, with shear deformation as defined in Figure
4.4. We see this by considering a cylindrical portion of the material around a screw dislocation, using the illustration in Figure
4.24. The displacement applied across the dislocation is given by the Burgers vector,
$\overrightarrow{b}$ (Figure
4.24a). (When referring to the magnitude of the Burgers vector we'll drop the vector symbol and just use the
$b$). If we unwrap the circumference of the cylinder at a distance
$r$ from the dislocation line (Figure
4.24b) we see that the shear displacement of
$b$ is applied over a distance of
$2\pi r$. The cylinder has been unwrapped in the circumferential direction,
i.e. the
$\theta $ direction, and the displacement is along the
$z$ direction so we have the following for th shear strain,
${e}_{z\theta}$:
The distortion is pure shear, with a shear strain at a radius of $r$ given by the following:
Note that as $r\to 0,$ ${e}_{z\theta}\to \infty $. The strain can't really go to infinity, so we have a problem here that we're going to have to deal with eventually. The elastic stress is obtained by multiplying by the shear modulus:
The elastic strain energy per unit volume, ${E}_{v}$ is obtained from the following expression:
Dimensionally this makes sense, since $G$ has units of force/area, or energy/volume.
Because the strain energy is radially symmetric, we can get the energy per unit length of the dislocation (${E}_{s}/h$) by integrating all values of $r$. Because the area element in radial coordinates is $2\pi rdr$, we have:
$$\begin{array}{cc}{E}_{s}/h=2\pi {\int}_{0}^{{r}_{max}}r{E}_{v}\left(r\right)dr& (4.13)\end{array}$$
$$\begin{array}{cc}{E}_{s}=\frac{G{b}^{2}h}{4\pi}{\displaystyle {\displaystyle {\int}_{0}^{{r}_{max}}}\frac{1}{r}dr}& (4.14)\end{array}$$
After integration we obtain:
$$\begin{array}{cc}{E}_{s}=\frac{G{b}^{2}h}{4\pi}{\displaystyle \left[ln{r}_{max}ln0\right]}& (4.15)\end{array}$$
We have a problem here because $ln0=\infty $. This is because for $r\to 0$, the shear strain goes to infinity and we can no longer use the simple, continuum picture of linear elasticity to describe what is going on. Instead what we typically do is separate out a core energy, ${E}_{s}^{core}$ that corresponds to the strain energy inside some small core radius, ${r}_{0}$. We do this simply be adding ${E}_{s}^{core}$ to ${E}_{s}$ and replacing the lower bound on the integration from 0 to ${r}_{o}$.
$$\begin{array}{cc}{E}_{s}={E}_{s}^{core}+\frac{G{b}^{2}h}{4\pi}{\displaystyle ln\left(\frac{{r}_{max}}{{r}_{0}}\right)}& (4.16)\end{array}$$
We can generally choose ${r}_{0}$ so that it is large enough so that our assumption of linear elasticity holds for $r>{r}_{0}$, yet it is small enough so that ${E}_{s}^{core}$ is a relatively small fraction of the overall dislocation energy. In this case we can ignore the core energy and approximate the dislocation energy as follows:
4.9.2 Edge Dislocations
 In the slip plane itself the material is in a state of pure shear.
 Above the slip plane there is compressive component to the strain field.
 Below the slip plane there is a tensile component to the strain field.
A more detailed calculation shows that the strain still decays as $1/r$, with an expression for the edge dislocation energy per length that is similar to the expression obtained for a screw dislocation:
Here, $\nu $ is Poisson's ratio. Typically $\approx 0.3$ for most metals.
The conceptual picture shown in Figure 4.25 is useful, but we can do a little bit better by reminding ourselves of some definitions pertaining to a stress state. A twodimensional stress state in the xy plane has three independent components of the stress: the shear stress,
$\tau $, and two normal stresses,
${\sigma}_{xx}$ and
${\sigma}_{xy}$, as shown in Figure
4.26. In Figure
4.27a the regions around an edge dislocation where stress components have different signs are illustrated. Figure
4.27b has similar information, but in this case we plot contours of equal stress for each of the three stress components,
${\sigma}_{xx}$,
${\sigma}_{yy}$ and
${\sigma}_{xy}$.
4.9.3 General Comments
The following general comments are valid for edge, screw and mixed dislocations:
 Elastic strain energy scales with $lnr$ so it has a very long range.
 The boundary conditions matter, so the energy depends on the shape of the sample. A small crystal with low value of ${r}_{max}$ will have a lower dislocation energy than a large crystal with a very large value of ${r}_{max}$.
 Energy scales as ${b}^{2}$. Dislocations with small values of $b$ are therefore preferred, which is why the Burgers vector in a material corresponds to the smallest interatomic spacing in the material.
 Energy scales as $h$. Energy is proportional to the length of the dislocation. This means the strain energy will decreases as the line length decreases.
This last point seems trivial at first, but it has some important consequences. Consider for example a dislocation loop. If the radius of a circular loop decreases, the energy associated with the loop will decrease as well. There's a line tension acting on the loop causing it to contract. This tension is like the tension in a rubber band that once to squeeze things inward, and can be viewed as a driving force for the dislocation loop to shrink in size. An applied stress can cause a dislocation loop to grow instead of shrink, and this will be considered later.
$$\begin{array}{cc}{E}_{s}\approx \frac{G{b}^{3}}{4\pi}{\displaystyle ln\left(\frac{{r}_{max}}{{r}_{0}}\right)=\frac{\left(3x1{0}^{10}\phantom{\rule{6px}{0ex}}Pa\right){\left(3x1{0}^{10}\phantom{\rule{6px}{0ex}}m\right)}^{3}}{4\pi}ln\left(\frac{1000\phantom{\rule{6px}{0ex}}nm}{1\phantom{\rule{6px}{0ex}}nm}\right)=4.5x1{0}^{19}\phantom{\rule{6px}{0ex}}J}& (4.19)\end{array}$$
A more convenient energy scale on an atomic basis is the electron volt, which we obtain from the energy in Joules by dividing by the electron charge (1.6x10${}^{19}$ C). In these units the energy per atom along the dislocation line is 2.8 eV. This energy of comparable magnitude to typical vacancy formation energies of $\approx $ 1eV, but is actually larger because of the nature of the longrange strain field that is produced around a dislocation.
4.10 Dislocation Line Tension
An energy per unit length has dimensions of a force. The dislocation energy per unit length is therefore equivalent to a force, or tension, exerted by the dislocation. We refer to this line tension as ${T}_{s}$:
$$\begin{array}{cc}{F}_{s}^{r}2\pi rdr={T}_{s}2\pi dr& (4.21)\end{array}$$Rearrangement gives the following expression for ${F}_{s}^{r}$:
$$\begin{array}{cc}{F}_{s}^{r}=\frac{{T}_{s}}{r}& (4.22)\end{array}$$
4.11 Effect of Applied Stress
This work goes into moving the dislocation, and must be equal to the force applied to the dislocation multiplied by the distance the dislocation moves as it translates across the sample. In our notation this distance is the sample width, $w$, so we have:
Equating these two expressions for the work gives:
So the force per unit length acting on the dislocation is simply the shear stress multiplied by the magnitude of the Burgers vector.
The only real assumption in Eq. 4.25 is that
$\tau $ is the component of the shear stress oriented along the direction of the Burgers vector. The resulting force is perpendicular to the dislocation line itself, regardless of the specific orientation of the dislocation line. This point as an important one that is not completely obvious, so we illustrate it for a screw dislocation in Figure
4.30. The orientation of the Burgers vector is identical to that of the Burgers vector for the edge dislocation in Figure
4.29, but the dislocation is now a screw dislocation oriented along the y direction that propagates in the negative x direction as the dislocation moves through the crystal. Because the final state of the crystal is the same as for the edge dislocation in Figure
4.29, the work done by the applied stress is still given by
$w\ell \tau b$,
i.e. Eq.
$\mathrm{[Unknown\; [string\; eq:external\; work\; on\; dislocation\; mathalpha]\; ]}$ still applies. The dislocation moves a distance
$\ell $ in this case. Because the length of the dislocation is
$w$, the total force applied to the dislocation is
${F}_{s}^{\tau}w$, and the energy required to translate it by a distance
$\ell $ is
${F}_{s}^{\tau}w\ell $. So we see that Eq.
4.24 still applies as well. The net result is that the
${F}_{s}^{\tau}$ is still given by
${\tau}_{rss}b$, just as it was for an edge dislocation. It can be shown that the same must be true for a mixed dislocation as well.
$$\begin{array}{cc}{F}_{s}={F}_{s}^{\tau}{F}_{s}^{r}={\tau}_{rss}b\frac{{T}_{s}}{r}& (4.26)\end{array}$$
At equilibrium the net force acting on the dislocation is zero (${F}_{s}=0$). This occurs when the applied stress is equal to a critical value that we refer to as ${\tau}^{*}$:
If ${\tau}_{rss}>{\tau}^{*}$ the dislocation loop expands, and if ${\tau}_{rss}<{\tau}^{*}$ the dislocation shrinks and disappears altogether.
So why do precipitates strengthen a material? The answer is connected to Eq. 4.27. Consider a dislocation that is moving toward two precipitates. The applied stress results in a force per unit length,
${F}_{s}^{\tau}$ that moves the dislocation. The pinning of the dislocation between the precipitates results in a curvature,
$r$, with an associated stress
${\tau}^{*}$ that must be applied in order for the dislocation to move. The maximum value of
${\tau}^{*}$ corresponds to the minimum value of the dislocation curvature
$r$, which is equal to half the interparticle spacing. In this example,
${\tau}^{*}$ is the critical resolved shear stress for the material. For optimum strengthening, what we want very small precipitates with a correspondingly small interparticle spacing.
4.12 Dislocation Multiplication
Where do these dislocations come from in the first place? Shape change associated with the emergence of dislocation to the exterior of the crystal must be decreasing their density. A typical dislocation density of
$\approx 1{0}^{7}/c{m}^{2}$ is way too small to give the experimentally measured plastic strain observed in a typical metal. So there must be some mechanism of creating new dislocations. One possibility we can consider is that the applied stress is itself sufficient to nucleate a dislocation loop. To figure out if this makes sense, we can calculate the shear stress required to expand a relatively small dislocation loop with a radius,
$r$, of 10
$b$. We'll assume
${r}_{0}=b$ and
${r}_{max}=10b$ and estimate the dislocation line tension From Eqs.
4.17 and
4.20:
$$\begin{array}{cc}{T}_{s}\approx \frac{G{b}^{2}}{4\pi}{\displaystyle ln\left(10\right)}\approx \frac{G{b}^{2}}{8}& (4.28)\end{array}$$
The resolved shear stress,
${\tau}^{*}$, required to expand the loop is given by Eq.
4.27:
$$\begin{array}{cc}{\tau}^{*}=\frac{{T}_{s}}{rb}=\frac{G}{80}& (4.29)\end{array}$$
Actual values of the critical resolved shear stress are
$\approx G/1{0}^{4}$ (see Table
1), so there must be some other mechanism operating at a lower stress that enables new dislocations to be created. This mechanism is the
.
The process by which new dislocations are produced by a FrankRead source is illustrated in Figure 4.31. It is based on the behavior of a dislocation segment that is pinned between two points (precipitate particles for example), labeled A and B in Figure
4.31. In the absence of an applied shear stress, this dislocation is a straight line between points A and B, (line 1 in Figure
4.31). As a shear stress is applied to the material the dislocation expands outward in a series of arcs, labeled as 2, 3, 4 and 5 in Figure
4.31. Because
${F}_{s}^{\tau}$ is always acting normal to the dislocation line, pushing it outward, the dislocation bends around the pinning points. Eventually, two segments of the dislocation with opposite
$\stackrel{\u02c6}{s}$ are in close proximity to each other (arc 4 in Figure
4.31). These segments of the dislocation annihilate each other, and the dislocation breaks into two separate arcs, both of which are now labeled as 5. The larger of these arcs is a dislocation loop that continues to expand, and the smaller of the arcs repeats the process as it expands in response to the stress. In this way an unlimited number of dislocation loops can be created by the original segment of the dislocation.
We can also use this argument to obtain values for the critical resolved shear stress in the system. Because the shear stress needed to expand the dislocation is inversely proportional to the dislocation radius of curvature, $r$ (from Eq. 4.27), the largest stress corresponds to the smallest radius of curvature for the dislocation line. In its original unstressed configuration (line 1 in Figure
4.31), the dislocation is a straight line, with
$r=\infty $. Then the radius of curvature decreases as the dislocation begins to grown in response to the applied stress. The minimum radius of curvature is
$d/2$, where
$d$ is the distance between the pinning points of the dislocation. This corresponds to line 1 in Figure
4.31. This corresponds to the maximum applied stress, which for
$r=d/2$ is
$2{T}_{s}/db$. This is the critical resolved shear stress,
${\tau}_{CRSS}$, for the system, if dislocation pinning is the strengthening mechanism in the material. If we estimate
${T}_{s}$ as
$G{b}^{2}/8$ as we did above, we obtain:
$$\begin{array}{cc}{\tau}_{crss}\approx \frac{Gb}{4d}& (4.30)\end{array}$$
Precipitation strengthening of a material is based on the introduction of very closely spaced nanoscale precipitates, giving the smallest possible value of $d$, and hence the maximum ${\tau}_{CRSS}$.
5 Thermodynamics of Interfaces
5.1 A Brief Review of the Thermodynamic Potentials
This expression is used frequently in the calculations in the following sections.
5.1.1 Internal Energy
The variation in the internal energy, $U$, for a multicomponent system at equilibrium is given by the following expression:
Here ${\mu}_{i}$is the chemical potential of component $i$ and ${n}_{i}$ is the number of atoms of this component. Note that $\delta U=0$ for fixed entropy, volume and number of atoms ($\delta S=\delta V=\delta {n}_{i}=0)\mathrm{.}$ This means that at equilibrium under conditions of fixed entropy, volume total amount of each component, the internal energy is minimized at equilibrium.
5.1.2 Helmholtz Free Energy
The Helmholtz free energy, $F$, is defined in the following way:
The variation of $F$ is:
So that $\delta F=0$ for fixed $V$, $T$, ${n}_{i}$. This means that at equilibrium under conditions of fixed temperature, volume and total amount of each component, the Helmholtz free energy is minimized.
5.1.3 Gibbs Free Energy
The Gibbs free energy is defined in a very similar manner, but in this case we replace the internal energy, $U$, with the enthalpy, $H$:
$$\begin{array}{cc}G\equiv HTS=U+PVTS& (5.6)\end{array}$$
Here the enthalpy is given by the following espression. :
$$\begin{array}{cc}H=U+PV& (5.7)\end{array}$$
$$\begin{array}{cc}G=F+PV& (5.8)\end{array}$$
In differential form
So that $\delta G=0$ for fixed $P$, $T$, ${n}_{i}$. This means that at equilibrium under conditions of fixed temperature, pressure and amount of each species, the Gibbs free energy is minimized.
5.1.4 Chemical Potential Expressions
One useful thing that emerges from all of these expressions is that we can get some useful, equivalent expressions for the chemical potential. The chemical potential of component $i$ is always given by the derivative of some thermodynamic potential with respect to ${n}_{i}$. The thermodynamic potential to use just depends on what we are holding constant during the differentiation: $S$, $V$ if we use $U$; $T$, $V$ if we use $F$; $P$, $T$ if we use $G\mathrm{.}$ In mathematical terms we have:
$$\begin{array}{cc}{\mu}_{i}={.\frac{\partial U}{\partial {n}_{i}}}_{S,V}={.\frac{\partial F}{\partial {n}_{i}}}_{T,V}={.\frac{\partial G}{\partial {n}_{i}}}_{P,V}& (5.11)\end{array}$$
In this class we are going to be working primarily with the Gibbs free energy. A couple other statements about $G$ and its relationship to the chemical potentials is useful here. The first is that the chemical potential is equivalent to the partial molar free energy. So writing the chemical potential as a derivative of the free energy, we can sum up the potentials to get the free energy:
$$\begin{array}{cc}G={\sum}_{i}{\mu}_{i}{n}_{i}& (5.12)\end{array}$$
In differential form, we have:
$$dG={\sum}_{i}\left({\mu}_{i}d{n}_{i}+{n}_{i}d\right)$$
At equilibrium the chemical potentials must be equal. This is true even if the pressure is not uniform throughout the system, a situation that is nearly always true in multiphase systems because of interfacial energy effects, as we see below. In that case the appropriate thermodynamic potential is the Gibbs free energy, because we need to be able to calculate the pressuredependence of the chemical potentials.
5.1.5 Grand Canonical Potential
$$\Omega =UTS{\mu}_{A}{n}_{A}{\mu}_{B}{n}_{B}$$.
5.2 Interfacial Free Energy and the Dividing Surface
Interfaces have an energy associated with them. This is easiest to see in the case where there is a big structural change across the interface (a solidvapor interface, for example). In the simple example illustrated in Figure
5.1 the atoms at the surface have fewer bonds than the atoms in the bulk of the material. The lower number of bonds implies that there is an excess energy associated with atoms near the surface. In the simple nearest neighbor picture only those atoms at the surface are affected. In most cases, however, many atoms near the surface are affected, especially in cases where the density and/or structure of the phases are very similar. For example, in a liquid/liquid system like the interface between oil and water, structural changes across the interface are more subtle, and the interface can be very wide on an atomic scale. If we plot the concentration of one of the components across the interface between
$\alpha $ and
$\beta $ phases as shown schematically in Figure
5.2, we see that it transitions from
${C}^{\alpha}$ to
${C}^{\beta}$ over an interfacial region that can in some cases be many atomic dimensions wide.
The change in density across the interface means that the energy in the transition zone is different than the energy in either of the bulk phases. Even if the structure is the same, for example a coherent interface between alloys of the same crystal structure, the change in composition across the interface will lead to a region of the material with a different energy.
How can we develop a generalized description of interfacial thermodynamics that is valid for all types of interface (crystalline, amorphous, narrow, broad, etc.)? Fortunately, this was done by Gibbs even before atoms were discovered (c. 1880)! Our basic assumption is that all quantities vary across the interface in a continuous manner, like the density plot shown in Figure 5.2. We need to develop the corresponding condition for the inhomogeneous interfacial region of finite width. We begin by considering the interface between two phases,
$\alpha $ and
$\beta $. As shown in Figure
5.3 we can separate the system in to three regions:
$\alpha $ and
$\beta $ bulk phase regions where the properties are completely uniform, and an interfacial region
$I$, where the properties (
$S,\phantom{\rule{6px}{0ex}}{n}_{i},$ etc. are nonuniform).
At a planar interface, we still get the usual thermodynamic condition that the temperature and chemical potentials are uniform everywhere at equilibrium. But what about pressure effects? What if $\delta {V}^{\alpha}\ne 0$? To answer these questions we will make a relatively large conceptual leap and replace the real system where the interfacial has some finite width with an equivalent model system where the the interface is a true surface with no volume. This model system is obtained by extending bulk phase properties all the way up to the fictitious location of the dividing surface, $\Sigma $, where we have the two phases $\alpha $ and $\beta $ in our example directly in contact with one another.
Once we specify the precise location of the dividing surface we can determine the number of atoms that are associated with the interface. Once we know where the dividing surface is, we also know the volumes of each phase, ${V}^{\alpha}$ and ${V}^{\beta}\mathrm{.}$ Multiplying by the bulk phase concentration gives the total number of atoms in each phase:
$$\begin{array}{cc}{n}_{i}^{\Sigma}={n}_{i}{n}_{i}^{\alpha}{n}_{i}^{\beta}& (5.14)\end{array}$$
We commonly divide by the interfacial area, $A$, to get an interfacial excess of component $i$ per area, which we define as ${\Gamma}_{i}$:
We can also define an interfacial energy (${U}^{\Sigma}$) and an interfacial entropy (${S}^{\Sigma})$ in a similar way:
$$\begin{array}{cc}{U}^{\Sigma}=U{U}^{\alpha}{U}^{\beta}& (5.16)\end{array}$$
$$\begin{array}{cc}{F}_{a}^{\Sigma}={U}_{a}^{\Sigma}T{S}_{a}^{\alpha \beta}& (5.17)\end{array}$$
$$\begin{array}{cc}{G}_{a}^{\Sigma}={U}_{a}^{\Sigma}+P{V}^{\Sigma}T{S}_{a}^{\Sigma}& (5.18)\end{array}$$
Because the dividing surface is defined so that it has no volume (${V}^{\Sigma}=0)$, these two versions of the interfacial free energy are equal to one another. We define them as the interfacial free energy, ${\gamma}_{\alpha \beta}$:
$$\begin{array}{cc}{\gamma}_{\alpha \beta}={F}_{a}^{\Sigma}={G}_{a}^{\Sigma}={U}_{a}^{\Sigma}T{S}_{a}^{\Sigma}& (5.19)\end{array}$$
The interfacial contribution to the interfacial energy obeys the following version of Eq.
5.10:
$$\begin{array}{cc}d{G}^{\Sigma}=Sd{T}^{\Sigma}+\sum {\mu}_{i}\delta {n}_{i}^{\Sigma}+{\gamma}_{\alpha \beta}\delta {A}^{\Sigma}& (5.21)\end{array}$$
5.3 Equilibrium Condition for a System with an Interface
We are interested in the effects of an interface on the equilibrium conditions in a binary alloy, twophase system. We shall assume that the bulk phases far from the interface are uniform (no stress, gravity, electric field...). We can thus use the dividing surface construction wherein the phases are taken uniform up to a dividing surface. Thus the total number of atoms of components $A$ and $B$, ${n}_{A}$, ${n}_{B}$ entropy, $S$ and energy, $U$ are given by the summing the contributions arising from the individual phases and from the interface:
$$\begin{array}{cc}\begin{array}{c}U={U}^{\alpha}+{U}^{\beta}+{U}^{\Sigma}\\ {n}_{a}={n}_{a}^{\alpha}+{n}_{a}^{\beta}+{n}_{a}^{\Sigma}\\ {n}_{b}={n}_{b}^{\alpha}+{n}_{b}^{\beta}+{n}_{b}^{\Sigma}\\ S={S}^{\alpha}+{S}^{\beta}+{S}^{\Sigma}\\ V={V}^{\alpha}+{V}^{\beta}\end{array}& (5.22)\end{array}$$
In the last equation there is no ${V}^{\Sigma}$ since the dividing surface is of zero thickness.
So, we need to minimize, $$\begin{array}{cc}{f}^{*}=f\lambda \left({x}^{2}+{y}^{2}1\right)& (5.29)\end{array}$$The first variation of ${f}^{*}$ is, $$\begin{array}{cc}\delta {f}^{*}=\delta f\lambda \left(2x\delta x+2y\delta y\right)=\delta x+\delta y\lambda \left(2x\delta x+2y\delta y\right)& (5.30)\end{array}$$Since we are looking for an extremum, $$\begin{array}{cc}\left(12\lambda x\right)\delta x+\left(12\lambda y\right)\delta y=0& (5.31)\end{array}$$For this to hold for any variations in $x$ and $y$ (i.e. for any values of $\delta x$ and $\delta y$), the following conditions must be met:
$$\begin{array}{cc}\begin{array}{c}12\lambda x=0\\ 12\lambda y=0\end{array}& (5.32)\end{array}$$which implies
5.5 Determining the Equilibrium
5.6 No Change in Location or Shape of the Interface
$$\begin{array}{cc}\delta {U}^{*}=\begin{array}{c}\left[\left({T}^{\alpha}{\lambda}_{S}\right)\delta {S}_{v}^{\alpha}+\left({\mu}_{A}^{\alpha}{\lambda}_{{n}_{A}}\right)\delta {C}_{A}+\left({\mu}_{B}^{\alpha}{\lambda}_{{n}_{B}}\right)\delta {C}_{B}\right]{V}^{\alpha}+\\ \left[\left({T}^{\beta}{\lambda}_{S}\right)\delta {S}_{v}^{\beta}+\left({\mu}_{A}^{\beta}{\lambda}_{{n}_{A}}\right)\delta {C}_{A}+\left({\mu}_{B}^{\beta}{\lambda}_{{n}_{B}}\right)\delta {C}_{B}\right]{V}^{\beta}+\\ \left[\left({T}^{\Sigma}{\lambda}_{S}\right)\delta {S}_{a}^{\Sigma}+\left({\mu}_{A}^{\Sigma}{\lambda}_{{n}_{A}}\right)\delta {\Gamma}_{A}+\left({\mu}_{B}^{\Sigma}{\lambda}_{{n}_{B}}\right)\delta {\Gamma}_{B}\right]A\end{array}& (5.40)\end{array}$$
At equilibrium, $\delta {U}^{*}=0$ for all potential variations in ${C}_{A}$, ${C}_{B},$ ${\Gamma}_{A}$, ${\Gamma}_{B}$, ${S}_{v}^{\alpha}$, ${S}_{v}^{\beta}$ and ${S}_{a}^{\Sigma}$. This is only possible when the following equilibrium conditions are satisfied.
$$\begin{array}{cc}\begin{array}{c}{\lambda}_{S}={T}^{\alpha}={T}^{\beta}={T}^{\Sigma}\\ {\lambda}_{{n}_{a}}={\mu}_{a}^{\alpha}={\mu}_{a}^{\beta}={\mu}_{a}^{\Sigma}\\ {\lambda}_{{n}_{b}}={\mu}_{b}^{\alpha}={\mu}_{b}^{\beta}={\mu}_{b}^{\Sigma}\end{array}& (5.41)\end{array}$$In this way we obtain the usual equilibrium conditions of constant temperature and chemical potential for a system at equilibrium.
5.7 Changes in Location or Shape of the Interface
Now we examine the cases where we let the volumes of
$\alpha $ and
$\beta $ change. Since the case we just did will hold in this case too, we just have to examine the terms that involve the variations in the volumes and area of the interface. Thus, from
$$\begin{array}{cccc}\delta U& =& {U}_{v}^{\alpha}\delta {V}^{\alpha}+{U}_{v}^{\beta}\delta {V}^{\beta}+{U}_{A}^{\Sigma}\delta A& (5.42)\\ \delta S& =& {S}_{v}^{\alpha}\delta {V}^{\alpha}+{S}_{v}^{\beta}\delta {V}^{\beta}+{S}_{A}^{\Sigma}\delta A& \\ \delta {n}_{a}& =& {\rho}_{a}^{\alpha}\delta {V}^{\alpha}+{\rho}_{a}^{\beta}\delta {V}^{\beta}+{\Gamma}_{a}\delta A& \\ \delta {n}_{b}& =& {\rho}_{b}^{\alpha}\delta {V}^{\alpha}+{\rho}_{b}^{\beta}\delta {V}^{\beta}+{\Gamma}_{b}\delta A& \end{array}$$Thus we must minimize ${U}^{*}$, $$\begin{array}{cc}\delta {U}^{*}=\delta U{\lambda}_{T}\delta S{\lambda}_{{n}_{a}}\delta {n}_{a}{\lambda}_{{n}_{b}}\delta {n}_{b}& (5.43)\end{array}$$Using the values of the Lagrange multipliers, $$\begin{array}{cc}\delta {U}^{*}=\delta UT\delta S{\mu}_{a}\delta {n}_{a}{\mu}_{b}\delta {n}_{b}& (5.44)\end{array}$$Using Eq. 5.42,
$$\begin{array}{cccc}\delta {U}^{*}& =& \left({U}_{v}^{\alpha}T{S}_{v}^{\alpha}{\mu}_{a}{\rho}_{a}^{\alpha}{\mu}_{b}{\rho}_{b}^{\alpha}\right)\delta {V}^{\alpha}+& \\ & & \left({U}_{v}^{\beta}T{S}_{v}^{\beta}{\mu}_{a}{\rho}_{a}^{\beta}{\mu}_{b}{\rho}_{b}^{\beta}\right)\delta {V}^{\beta}+& \\ & & \left({U}_{A}^{\Sigma}T{S}_{a}^{\Sigma}{\mu}_{a}{\Gamma}_{a}^{\Sigma}{\mu}_{b}{\Gamma}_{b}^{\Sigma}\right)\delta A& (5.45)\end{array}$$The terms in the brackets is a less well known energy function, the Grand Canonical free energy. The Grand Canonical free energy,
$\Omega =UTS{\mu}_{A}{n}_{A}{\mu}_{B}{n}_{B}$. Since
$U=TS{\mu}_{A}{n}_{A}{\mu}_{B}{n}_{B}PV$,
$\Omega =PV$, so on per volume basis,
${\Omega}_{v}=P$. Since
${\Omega}_{v}={U}_{v}T{S}_{v}{\mu}_{A}{\rho}_{A}{\mu}_{B}{\rho}_{B}$,
$$\begin{array}{cc}\delta {U}^{*}=\left({\Omega}_{v}^{\alpha}\right)\delta {V}^{\alpha}+\left({\Omega}_{v}^{\beta}\right)\delta {V}^{\beta}+\left(\sigma \right)\delta A& (5.46)\end{array}$$where the
${\Omega}_{A}^{\Sigma}=\sigma $. So the interfacial energy is the excess Grand canonical energy per area associated with the interface. The variations in the volumes and areas shown in are not independent.
5.7.1 Application to a Planar Interface
Consider a planar interface, illustrated in Figure
5.6. If the volume of
$\alpha $ increases the volume of
$\beta $ decreases. So, for a planar interface,
$\delta A=0$, and
$\delta {V}^{\beta}=\delta {V}^{\alpha}$.
$$\begin{array}{cc}\delta {U}^{*}=\left({\Omega}_{v}^{\alpha}{\Omega}_{v}^{\beta}\right)\delta {V}^{\beta}=0& (5.47)\end{array}$$Thus at equilibrium,
${\Omega}_{v}^{\alpha}={\Omega}_{v}^{\beta}$, or
${P}^{\alpha}={P}^{\beta}$.
5.7.2 Application to a Curved Interface
$$\begin{array}{cc}{V}^{\beta}=\frac{4}{3}\pi {R}^{3}& (5.48)\end{array}$$
$$\begin{array}{cc}{A}^{\Sigma}=4\pi {R}^{2}& (5.49)\end{array}$$
If $R$ changes by a small amount $\delta R$, this leads to the following for $\delta {V}^{\beta}$ and $\delta {A}^{\Sigma}$:
$$\begin{array}{cc}\delta {V}^{\beta}=4\pi {R}^{2}\delta R& (5.50)\end{array}$$
$$\begin{array}{cc}\delta {A}^{\Sigma}=8\pi R\delta R& (5.51)\end{array}$$
We are working in a system with total fixed, i.e., $\delta V=\delta {V}^{\alpha}+\delta {V}^{\beta}=0.$ From this we obtain:
$$\begin{array}{cc}\delta {V}^{\alpha}=\delta {V}^{\beta}=4\pi {R}^{2}\delta R& (5.52)\end{array}$$
$$\begin{array}{cc}{P}^{\alpha}4\pi {R}^{2}\delta R{P}^{\beta}4\pi {R}^{2}\delta R+{\gamma}_{\alpha \beta}8\pi R\delta R=0& (5.53)\end{array}$$
Note that this pressure equation is an additional equilibrium condition, in addition to those already obtained (constant temperature and constant chemical potential). Note that at equilibrium the chemical potentials are uniform everywhere, even in conditions where the pressure is nonuniform. For systems with curved interfaces we need to account of the effect of pressure (and hence, the interface curvature) on the chemical potential. These pressureinduced chemical potential differences drive a variety of important processes in microstructure development in materials, including coarsening and grain growth.
5.7.3 Chemical Potential Expressions
5.7.4 A practical example
Example: Magnitude of the Laplace pressure
How large is the Laplace pressure
5.7.5 Effects of Interfacial Curvature on the Melting Transition
The enthalpy of melting, $\Delta {H}_{m}$, is a more intuitive quantity than the entropy of melting, and is more directly measured experimentally. At equilibrium for an interface between solid and liquid with a planar interface (so the pressure is the same in both phases), the free energies of the solid and liquid are equal to one another. This fact is generally used to write thermodynamic quantities in terms of $\Delta {H}_{m}$ instead of $\Delta {S}_{m}$. We begin by recognizing that the free energy change between solid and liquid phases is zero at $T={T}_{m}:$
$$\begin{array}{cc}\Delta {G}_{m}\left({T}_{m}\right)=\Delta {H}_{m}{T}_{m}\Delta {S}_{m}=0& (5.62)\end{array}$$
We can use this equation to write the entropy of mixing in terms of the enthalpy of mixing and the equilibrium melting temperature:
Finally, with ${P}^{s}{P}^{\ell}=2{\gamma}_{s\ell}/r$ we have:
So how large is this effect? To understand this, we need to put in some real numbers. Let's consider the case for gold with a particle radius, $R$ of 5 nm:
 $\gamma $ (solid/liquid interfacial free energy for gold): 0.177 J/m${}^{2}$.
 ${V}_{m}^{s}$ (molar volume of solid gold): $\frac{197\phantom{\rule{6px}{0ex}}g}{mole}\cdot \frac{c{m}^{3}}{17.3\phantom{\rule{6px}{0ex}}g}\cdot \frac{1{0}^{6}\phantom{\rule{6px}{0ex}}{m}^{3}}{c{m}^{3}}=1.14x1{0}^{5}\phantom{\rule{6px}{0ex}}{m}^{3}$
 $\Delta {H}_{m}$ (molar heat of fusion): $1.25x1{0}^{4}\phantom{\rule{6px}{0ex}}J/mole$
 ${T}_{m}$ (equilibrium melting temperature): 1064 ${}^{\u25cb}$C (1337 K)
 $R$ (droplet radius): $5x1{0}^{9}$ m
We conclude this section with a useful graphical interpretation of the effect of pressure.
By taking only the first term in the Taylor expansion, we are assuming that the plots of ${G}_{m}$ vs $T$ are straight, $i\mathrm{.}e\mathrm{.}$ we are neglecting any temperature dependence of the entropy. We are also assuming that $\partial {G}_{m}/\partial P$ is constant, which means we are saying that the molar volume is independent of the pressure (the system is assume to be incompressible). This is generally a reasonable approximation for most solid and liquid materials, but will fail miserably if one of the phases is a gas.
5.7.6 Sizedependent solubility
Another consequence of the increased pressure within a small precipitate is that small precipitates are more soluble in their surroundings than large precipitates.
5.7.6.1 General Concepts
At temperatures below the eutectic temperature, solid $\alpha $ and solid $\beta $ are in equilibrium with one another. For flat interfaces, ($R=\infty $) the phase compositions are given by the solvus lines, and are equal to ${X}_{B}^{\alpha}\left(\infty \right)$ and ${X}_{B}^{\beta}\left(\infty \right)$. How does this change if the interface is curved? Suppose we are in the Arich portion of the phase diagram, where small $\beta $ precipitates of radius $r$ exist in a matrix of $\alpha $.
$$\begin{array}{cc}\frac{\partial {G}_{m}^{\beta}}{\partial P}={V}_{m}^{\beta}& (5.66)\end{array}$$
If we assume that $\beta $ is incompressible, then ${V}_{m}^{\beta}$ does not change with pressure and we have:
$$\begin{array}{cc}{G}_{m}^{\beta}\left({P}^{\beta}\right)={G}_{m}^{\beta}\left({P}^{\alpha}\right)+{V}_{m}^{\beta}\left({P}^{\beta}{P}^{\alpha}\right)& (5.67)\end{array}$$
With ${P}^{\beta}{P}^{\alpha}=2{\gamma}_{\alpha \beta}/R$, where ${\gamma}_{\alpha \beta}$ is the interfacial free energy for the interface between $\alpha $ and $\beta $ phases, we have:
$$\begin{array}{cc}{G}_{m}^{\beta}\left({P}^{\beta}\right)={G}_{m}^{\beta}\left({P}^{\alpha}\right)+\frac{2{V}_{m}^{\beta}{\gamma}_{\alpha \beta}}{R}& (5.68)\end{array}$$
$$\begin{array}{cc}\begin{array}{c}{X}_{B}^{\alpha}\left(r\right)>{X}_{B}^{\alpha}\left(\infty \right)\\ {X}_{B}^{\beta}\left(r\right)>{X}_{B}^{\beta}\left(\infty \right)\end{array}& (5.69)\end{array}$$
To develop expressions for ${X}_{B}^{\alpha}\left(r\right)$ and ${X}_{B}^{\beta}\left(r\right)$, we just need to equate the chemical potentials for $A$ and $B$ atoms in each phase:
where ${P}^{\beta}={P}^{\alpha}+2{\gamma}_{\alpha \beta}/r$ . In general Eq. 5.70 is a set of two, nonlinear equations (obtained by setting
$i$ to
$A$ or
$B$) that must be solved numerically in order to obtain
${X}_{B}^{\alpha}$ and
${X}_{B}^{\beta}$, the compositions of the
$\alpha $ and
$\beta $ phases that re in equilibrium with one another. Note that because
${X}_{A}=1{X}_{B}$ we can use the single composition variable,
${X}_{B}$ to describe the compositions.
5.7.6.2 Activity Coefficients
In general the chemical potential of species $i$ is related to its activity coefficient, ${a}_{i}$:
$$\begin{array}{cc}{\mu}_{i}={\mu}_{i}^{0}+RTln\left({a}_{i}\right)& (5.71)\end{array}$$
Here $R$ is the gas constant (8.314 J/mole$\cdot $K) $T$ is the absolute temperature and ${\mu}_{i}^{0}$ is the chemical potential in it's standard state (which we'll take at a pressure of ${P}^{\alpha}$). We'll define the standard state chemical potentials as zero, so the chemical potentials for $P={P}^{\alpha}$ are simply:
$$\begin{array}{cc}{\mu}_{i}\left({P}^{\alpha}\right)=RTln\left({a}_{i}\right)& (5.72)\end{array}$$
5.7.6.3 Effect of Pressure
In the beta phase, we need to account for the fact that pressure in the $\beta $ phase, ${P}^{\beta}$, is no longer equal to the reference pressure, ${P}^{\alpha}$:
$$\begin{array}{cc}{\mu}_{i}\left({P}^{\beta}\right)=RTln{a}_{i}+\frac{\partial {\mu}_{i}^{\beta}}{\partial P}\left({P}^{\beta}{P}^{\alpha}\right)& (5.73)\end{array}$$
The pressure derivatives appearing in these equations can be replaced by the partial molar volumes of the A and B components in the $\beta $ phase, defined as follows:
where ${\stackrel{\u203e}{V}}_{i}$ is the partial molar volume of component $i$. We can therefore right the chemical potential in the following generalized form, which accounts for its dependence on both composition (${X}_{B}$) and pressure $\left({P}^{\beta}\right)$. and pressure concentration and pressure dependence:
5.7.6.4 Expression for ${X}_{B}^{\alpha}\left(r\right)$ in the dilute regime.
$$\begin{array}{cc}{a}_{i}={H}_{i}{X}_{i}\phantom{\rule{6px}{0ex}}\phantom{\rule{6px}{0ex}}\phantom{\rule{6px}{0ex}}\left(Rault\text{'}s\phantom{\rule{6px}{0ex}}Law\right)& (5.76)\end{array}$$
Flat Interface $R\to \infty $:
We know that ${\mu}_{B}^{\beta}$ is close to zero because ${X}_{B}^{\beta}\left(\infty \right)$ is close to one. Component B is dilute in the the alpha phase, so we are in the Henry's law regime. The chemical potential is given as follows:
Curved interface  finite $R$:
The chemical potentials are now modified by the pressure contribution to the molar free energy of the beta phase, and are no longer zero. One consequence of this is that the equilibrium compositions are changed. Because the $\alpha $ phase pressure is taken as our reference pressure, the equations for the chemical potentials in this phase are unchanged. We just need to specify that $r$ is no longer infinite:
$${\mu}_{B}^{\alpha}\left(r\right)=RT\left[ln{H}_{B}+ln{X}_{B}^{\alpha}\left(r\right)\right]$$
Now we can develop some analytic expressions that are useful in the dilute limit. We'll Eq. 5.74, along with the fact that
${P}^{\beta}{P}^{\alpha}=2{\gamma}_{\alpha \beta}/r$ to simplify some of the expressions. We're specifically interested in the increase in the minority phase fraction when
$r$ becomes very small. In other words, how large is
${X}_{B}^{\alpha}\left(r\right)$ compared to
${X}_{B}^{\alpha}\left(\infty \right)$? Requiring that
${\mu}_{B}^{\beta}\left(r\right)={\mu}_{B}^{\alpha}\left(r\right)$ gives:
$$\begin{array}{cc}RT\left[ln{H}_{B}+ln{X}_{B}^{\alpha}\left(r\right)\right]=RTln{X}_{B}^{\beta}\left(r\right)+\frac{2{\gamma}_{\alpha \beta}{\stackrel{\u203e}{V}}_{B}}{r}& (5.81)\end{array}$$
Eq.
5.80 holds in the alpha phase, which is assumed to be nearly pure A, so we can replace
$ln{H}_{B}$ with
$ln{X}_{B}^{\alpha}\left(\infty \right)$. Similarly, the
$\beta $ phase is assumed to be nearly pure B, so
$ln\left({X}_{B}^{\beta}\right)\approx 0$ and we have:
The molar volume of the $\beta $ precipitate is related to the partial molar volumes in the following way:
$$\begin{array}{cc}{V}_{m}^{\beta}={X}_{A}^{\beta}{\stackrel{\u203e}{V}}_{A}^{\beta}+{X}_{B}^{\beta}{\stackrel{\u203e}{V}}_{B}^{\beta}& (5.83)\end{array}$$
$$\begin{array}{cc}{X}_{B}^{\alpha}\left(r\right)={X}_{B}^{\alpha}\left(\infty \right)exp\left[\frac{2{\gamma}_{\alpha \beta}{V}_{m}^{\beta}}{RTr}\right]& (5.84)\end{array}$$
For small $x$, $exp\left(x\right)\approx 1+x$. If $r$ is not too small (generally the case) we can use this expansion for the exponential function to write ${X}_{B}^{\alpha}\left(r\right)$ as follows:
$$\begin{array}{cc}{X}_{B}^{\alpha}\left(r\right)={X}_{B}^{\alpha}\left(\infty \right)\left[1+\frac{2{\gamma}_{\alpha \beta}{V}_{m}^{\beta}}{RTr}\right]& (5.85)\end{array}$$
We see that the surface free energy term makes small precipitates more soluble in the matrix than larger precipitates. This increased solubility drives the coarsening of the microstructure over time, giving larger precipitates over time. We're not going to do much more with this specific equation in 3161, but it is very important when we start talking more about the evolution of microstructure in 3162. It is given here largely as an illustration of the importance of the interfacial free energy.
6 Two Dimensional Defects in Crystals: Surfaces and grain boundaries.
Dislocations are one dimensional defects in a crystalline structure. We now consider crystal interfaces, which can be viewed as twodimensional crystal defects. We'll consider three kinds of interfaces:
 Free surfaces
 Grain boundaries
 Interphase interfaces
All of these interfaces have an interfacial energy, given by the energy required to create extra interfacial area:
$$\begin{array}{cc}\gamma =\frac{\partial U}{\partial A}& (6.1)\end{array}$$
One of the unique features of crystalline materials is that
$\gamma $ is no longer isotropic. Certain crystal facets have a lower interfacial energy than other facets. This is why natural crystals like quartz (see Figure
6.1) have beautiful shapes and are not boring solid blobs. In the following sections we'll investigate some of the features that give rise to the anisotropy in
$\gamma $, and will see how this anisotropy determines the equilibrium crystal shape.
6.1 Surface Energy of a ClosePacked Plane
We can use the 'missing bond' picture of crystalline surfaces to estimate the surface energy for a close packed plane of atoms. Consider, for example a {111} surface in the FCC crystal structure. The FCC crystal structure consists of ABC stacking of these closepacked planes, as shown in Figure
6.3.
$$\begin{array}{cc}{L}_{m}=12{N}_{av}\frac{\epsilon}{2}=6{N}_{av}\epsilon & (6.3)\end{array}$$
Rearrangement of this equation gives:
so the surface energy per atom is ${L}_{m}/4{N}_{av}$. There is also an excess entropy associated with the surface due to changes in the vibrations of the surface atoms, configuration entropy due to surface vacancies, but this is typically a small contribution to the overall surface free energy and is ignored in our treatment. The surface energy is obtained by multiplying the excess surface energy per atom by the surface density of atoms on the plane of interest, ${\Sigma}_{s}$:
where $r$ is the atomic radius.
$$\begin{array}{cc}{\gamma}_{sv}=\frac{{L}_{m}}{8{r}^{2}\sqrt{3}{N}_{av}}& (6.7)\end{array}$$
In addition to having a higher value of
${\gamma}_{sv}$, we also expect that materials with a higher value of
${L}_{m}$ will have higher melting points (
${T}_{m}$). Values for
${T}_{m}$ and
${\gamma}_{sv}$ are listed in Table
3.
6.2 Orientation Dependence of the Surface Energy
$$\begin{array}{cc}{\Sigma}_{s}=\frac{1}{{a}^{2}}\left(cos\theta +sin\theta \right)& (6.8)\end{array}$$
The surface energy is obtained multiplying by the energy per bond, $\epsilon /2:$
Equation 6.9 is valid for values of
$\theta $ between 0 and 90
${}^{\u25cb}$, where
$sin\theta $ and
$cos\theta $ are both positive. These sin and cos terms came from the projected length of the tilted surface along the [100] and [010] directions. A
(Note that the equation here is approximate because we have neglected entropic contributions to surface free energy). There's no derivative at $\theta =0$, so the free energy function must have a cusp. (show plot). How can we represent ${\gamma}_{sv}$ as a function of $\theta $? Describe in terms of ${\theta}_{n}$ (angle of normal of a plane with respect to the x axis). So we can plot ${\gamma}_{sv}$on a polar plot. In MATLAB we use the 'polar' command to do this. We'll give an example when we illustrate the Wulff construction in the following section.
6.3 Equilibrium Shape of Crystals
We know that $\gamma $ is a function of the angle, but what are the implications on the equilibrium shape of the crystal? We need to minimize the total surface energy subject to volume conservation.
$${\int}_{A}^{A}\gamma ds$$ If $\gamma $ is a constant (independent of the angle), then we just need to minimize the overall surface area for a fixed volume. We get a sphere in this case. Now we have $\gamma =f\left(\theta \right)$. Suppose we have two facets with surface free energies of ${\gamma}_{1}$ and ${\gamma}_{2}$.
$${G}_{t}={\gamma}_{1}{A}_{1}+{\gamma}_{2}{A}_{2}$$
 Draw $\gamma =\gamma \left(\theta \right)$
 Draw line from origin to $\gamma $ curve for a given value of $\theta $
 Draw perpendicular to this line.
 Repeat for all values of $\theta $.
 Inner envelope is equilibrium shape.
close all
gamma=@(alpha) abs(cos(alpha))+abs(sin(alpha));
r=@(theta,alpha) gamma(alpha)/cos(thetaalpha);
alpha=linspace(0,2*pi,200);
polar(alpha,gamma(alpha),'r');
title('\gamma=cos\theta+sin\theta', 'fontsize', 16)
hold on % plot all subsequent curves on existing axes
for alpha=linspace(0,2*pi,17) % this is the loop that draws all the lines
theta(1)=alpha+2*pi/5; % specify two angles on either side of alpha
theta(2)=alpha2*pi/5;
rvals(1)=r(theta(1),alpha); % use the equation provided to get r for each
% of the specified angles
rvals(2)=r(theta(2),alpha);
polar(theta,rvals) % plot lines connecting the two points we just defined
end
set(gcf,'paperposition',[0 0 5 5],'papersize',[5 5])
print(gcf,'../figures/matlabwulffenergyexample.eps', 'depsc2') % save the eps file
7 Grain Boundaries
In the two dimensional Wulff construction the interface surface of interest is specified by a single variable, $\theta $. For a true, threedimensional crystal the Wulff circle becomes a Wulff sphere, and we need two different angles to specify the the orientation on this sphere. In other words, we have two degrees of freedom in specifying a specific surface of a threedimensional crystal. We commonly specify a surface by using the normal vector, $\stackrel{\u02c6}{n}$, that is perpendicular to that surface. This surface has three components, ${n}_{1}$, ${n}_{2}$ and ${n}_{3}$, in the x, y and z directions, respectively:
$$\begin{array}{cc}\stackrel{\u02c6}{n}=\left({n}_{1}\stackrel{\u02c6}{x}+{n}_{2}\stackrel{\u02c6}{y}+{n}_{3}\stackrel{\u02c6}{z}\right)& (7.1)\end{array}$$
Because $\stackrel{\u02c6}{n}$ is a unit vector with ${n}_{1}^{2}+{n}_{2}^{2}+{n}_{3}^{2}=1$, only two of the three components of $\stackrel{\u02c6}{n}$ are independent, so we again come to the conclusion that there are two degrees of freedom associated with the specification of a crystal surface.
In order to fully describe a grain boundary between two crystals we need to specify three additional degrees of freedom, so there are five degrees of freedom altogether. In order to illustrate these degrees of freedom, we can consider the following conceptual procedure for producing a grain boundary.
 Cut the crystal along a plane specified by the unit normal to the plane, $\stackrel{\u02c6}{n}$. Two degrees of freedom are associated with the specification of $\stackrel{\u02c6}{n}$.
 Rotate one of the two halves of the crystal by $\theta $ about an axis directed along the unit normal, $\stackrel{\u02c6}{c}$.
Two additional degrees of freedom are used in the specification of $\stackrel{\u02c6}{c}$, just as we use two degrees of freedom to specify $\stackrel{\u02c6}{n}$. The fifth and final degree of freedom is the the rotation angle, $\theta $.
The fact that 5 different parameters are needed to specify a grain boundary within a given crystal means that it is impossible for us to be exhaustive in our treatment of the different possibilities. Instead, we'll consider the following three cases:
 :
Twist boundaries correspond to rotation about an axis that is perpendicular to the plane. In terms of $\stackrel{\u02c6}{c}$ and $\stackrel{\u02c6}{n}$, they correspond to the case where these unit vectors are parallel to one another:
$$\stackrel{\u02c6}{c}\parallel \stackrel{\u02c6}{n}$$
 :
Tilt boundaries correspond to the opposite limiting case, where $\stackrel{\u02c6}{c}$ and $\stackrel{\u02c6}{n}$ are perpendicular to one another:
$$\stackrel{\u02c6}{c}\perp \stackrel{\u02c6}{n}$$

. This is a special type of low energy tilt boundary, where lattice planes on either side of the boundary are in registry with one another.
7.1 Tilt Boundaries
$$\begin{array}{cc}tan\left(\theta /2\right)=\frac{b/2}{2sin\left(\theta /2\right)}& (7.2)\end{array}$$
For very small values of $\theta $, we can assume $tan\left(\theta /2\right)\approx \theta /2$, so we have:
The energy per unit length of a dislocation is ${T}_{s}$. For a low angle grain boundary consisting of dislocations separated by a distance b, we expect the following for the grain boundary energy, ${\gamma}_{gb}$:
$$\begin{array}{cc}{\gamma}_{gb}=\frac{{T}_{s}}{d}\approx \frac{{T}_{s}\theta}{b}& (7.4)\end{array}$$
$$\begin{array}{cc}{\gamma}_{gb}=\frac{Gb\theta}{4\pi \left(1\nu \right)}ln\left({r}_{max}/{r}_{0}\right)& (7.5)\end{array}$$
The simplest thing to do here is to let the upper cutoff, ${r}_{max}$, correspond to the dislocation spacing $d,$ which in our case is equal to $b/\theta $. From this we get:
where we have defined ${\gamma}_{0}$ in the following way:
$$\begin{array}{cc}{\gamma}_{0}=\frac{Gb}{4\pi \left(1\nu \right)}& (7.7)\end{array}$$
A more detailed treatment (the ReadSchockley model of low angle tilt boundaries) gives the following very similar form:
$$\begin{array}{cc}{\gamma}_{gb}={\gamma}_{0}\theta \left(Bln\theta \right)& (7.8)\end{array}$$
7.2 Twin Boundaries
 The third layer goes above position 'A', so that the repeating structure of the stacking is ABABAB... This results in the FCC structure.
 The third layer goes above position 'C', so that the repeating structure of the stacking is ABCABCABC... This stacking produces the FCC structure.
Now we can talk about twin boundaries in an FCC structure. In an untwinned structure this is a regular, uninterupted of the stacking of the 'A', 'B' and 'C' closepacked planes:
At a twin boundary this stacking gets interrupted, as in the following example.
 Twinned: ABCABCBACBA (twin indicated)
The twin plane is the 'C' plane in the middle of this sequence. Note that the sequences on either side of the twin boundary are mirror images of one another. The sequence of planes working out from the twin plane is 'BACBACBAC...' in both cases. Twin boundaries have very low energies because there are no broken bonds, dislocations, step edges, etc. The energy only comes from the small unfavorable energy associted from second nearest neighbor interactions, as described above.
7.3 Twist Boundaries
7.4 Grain Boundary Junctions
7.5 Thermally Activated Migration of Grain Boundaries
(General treatment is much more general than just grain boundaries)
Assume isotropic grain boundary properties.
In general, grain boundaries are not flat. As a result the boundaries are subject to a force. Very similar to force exerted on a line by the line tension.
Recall the pressure dependence of the chemical potential:
$$\begin{array}{cc}\mu \left({P}^{2}\right)=\mu \left({P}^{1}\right)+\frac{2{V}_{m}\gamma}{r}& (7.10)\end{array}$$
$$\begin{array}{cc}\frac{\Delta \mu}{{V}_{m}}=\Delta P=\frac{force}{area}& (7.11)\end{array}$$
Boundaries move toward their center of curvature.
Now we know the driving force  this comes from thermodynamics
Need to study the kinetics to understand if the grain will actually move in response to this driving force.
Now plot free energy as a function of position. Draw at equilibrium. An activation barrier exists that has a high of $\Delta {G}_{a}$.
What if the interface is curved so that the curvature is toward grain 1 (grain 1 is smaller).
Redraw curve  now grain one has higher energy than grain 2.
This decreases $\mu $ a bit smaller than it was before. Also, have a negative $\mu $ for atoms going from grain 1 to grain 2. Now the fluxes in the two directions are different. It's clear now that there is a net flux of atoms from grain 1 to grain 2. The actual flux from 1 to 2 is:
$$\begin{array}{cc}{J}_{1\to 2}={A}_{2}{n}_{1}{\nu}_{1}exp\left(\Delta {\mu}^{*}/RT\right)& (7.12)\end{array}$$
${A}_{2}$ is the probability that the atom is accommodated in grain 2.
${n}_{1}$ the number of atoms that are able to make the jump (in molar units).
${\nu}_{1}$= vibrational frequency.
The flux in the backward direction (from 2 to 1) is given by:
$$\begin{array}{cc}{J}_{2\to 1}={A}_{1}{n}_{2}{\nu}_{2}exp\left(\left(\Delta {\mu}^{*}+\Delta \mu \right)/RT\right)& \\ ={A}_{1}{n}_{2}{\nu}_{2}exp\left(\Delta {\mu}^{*}/RT\right)exp\left(\Delta \mu /RT\right)& (7.13)\end{array}$$
If $\Delta \mu =0$, we have equilibrium and ${J}_{1\to 2}={J}_{2\to 1}$. A consequence of this is that the following ${A}_{2}{n}_{1}{\nu}_{1}={A}_{1}{n}_{2}{v}_{2}$.
If $\Delta \mu >0$ the net flux is given by the difference:
$$\begin{array}{cc}{J}_{net}={J}_{1\to 2}{J}_{2\to 1}={A}_{2}{n}_{1}vexp\left(\frac{\Delta {\mu}^{*}}{RT}\right)\left(1exp\left(\Delta \mu /RT\right)\right)& (7.14)\end{array}$$
Assume that the $\Delta \mu \ll RT$, so we can use the approximation that $exp\left(x\right)\approx 1+x$ for small x. The expression for ${J}_{net}$ then reduces to the following:
$$\begin{array}{cc}{J}_{net}={J}_{1\to 2}{J}_{2\to 1}={A}_{2}{n}_{1}vexp\left(\frac{\Delta {\mu}^{*}}{RT}\right)\frac{\Delta \mu}{RT}& (7.15)\end{array}$$
Now we can get an expression for the velocity of the grain boundary:
$$\begin{array}{cc}v={J}_{net}{V}_{m}={A}_{2}{n}_{1}{V}_{m}vexp\left(\frac{\Delta {\mu}^{*}}{RT}\right)\frac{\Delta \mu}{RT}& (7.16)\end{array}$$
Can substitute this and get an expression for $v$. (expand exponential for small argument).
Now we define an interface mobility in the following way:
$$\begin{array}{cc}v=\frac{M\Delta \mu}{{V}_{m}}& (7.17)\end{array}$$
Break things down into enthalpy and entropy:
$$\begin{array}{cc}\Delta {\mu}^{*}=\Delta {H}^{*}T\Delta {S}^{*}& (7.18)\end{array}$$
$$\begin{array}{cc}M=\frac{{A}_{2}{n}_{1}{\nu}_{1}{V}_{m}^{2}}{RT}exp\frac{\Delta {S}^{*}}{R}exp\left[\frac{\Delta {H}^{*}}{RT}\right]& (7.19)\end{array}$$
What are the factors affecting M:
 Temperature
 Structure of the boundary (low angle vs. high angle boundary)
 Impurities (alloying elements)
Show Fig. 3.27
Impurities have a huge effect on grain boundary mobility. Grain boundaries are like garbage dumps for impurities. Structure also plays a role. Tricks used in the heat treatment of high temperature superconductors.
Effect of impurities  LangmuirMclean model for grain boundary segregation.
${X}_{gB}$= fraction of a monolayer adsorbed on the boundary
$${\left(\frac{X}{1X}\right)}_{gb}={\left(\frac{X}{1X}\right)}_{bulk}exp\left[\frac{\Delta {G}_{b}}{RT}\right]$$
Here $\Delta {G}_{B}>0$ for an element that adsorbs to the boundary.
 Strongly temperature dependent
 more dilute elements adsorb more (show figure)
 segregation affects the mobility of the boundaries
Show grain boundary composition vs. atomic solid solubility.
7.5.1 Transformation kinetics (crystallization, recrystallization):
$$\begin{array}{cc}f\left(t\right)=\frac{fraction\phantom{\rule{6px}{0ex}}transformed\phantom{\rule{6px}{0ex}}at\phantom{\rule{6px}{0ex}}time\phantom{\rule{6px}{0ex}}t}{final\phantom{\rule{6px}{0ex}}fraction\phantom{\rule{6px}{0ex}}transformed}& (7.20)\end{array}$$
We can derive an expression for $f\left(t\right)$ by assuming that nucleation occurs at a uniform rate, ${\u1e44}_{v}$ (nuclei formed per volume per time) that does not depend on $t$. The volume of a single crystalline sphere of radius $r$ is $4\pi {r}^{3}/3$. If the sphere forms at $t=0$ and $r$ increases linearly with a growth velocity of $v$, we have:
$$\begin{array}{cc}Vol\left(t\right)=\frac{4}{3}\pi {r}^{3}=\frac{4}{3}\pi (vt{)}^{3}& (7.21)\end{array}$$
If nucleation does not occur until $t=\tau $ we have:
$$\begin{array}{cc}f\left(t\right)={\u1e44}_{v}{\int}_{0}^{t}Vol\left(t,\tau \right)d\tau ={\u1e44}_{v}\frac{4}{3}\pi {v}^{3}{\int}_{0}^{t}(t\tau {)}^{3}d\tau ={\u1e44}_{v}\frac{\pi}{3}{v}^{3}{t}^{4}& (7.23)\end{array}$$
This equation is only valid for short times, since it neglects the fact that individual crystalline regions stop growing once they impinge on one another. In reality, $f\left(t\right)$ must reach an asymptotic value of 1 for $t=\infty $. A more detailed solution to the problem gives the following expression:
$$\begin{array}{cc}f\left(t\right)=1exp\left(\frac{\pi}{3}{\u1e44}_{v}{v}^{3}{t}^{4}\right)& (7.24)\end{array}$$This is a specific example of the following more general expression, referred to as the
JohnsonMehlAvramiKolmogorov (JMAK) equation
:
$$\begin{array}{cc}f\left(t\right)=1exp\left(k{t}^{n}\right)& (7.25)\end{array}$$
where $n$ is an empirical constant obtained from experimental data that is found to vary between 1 and 4. This is the simplest equation that has the basic behavior observed experimentally.
7.5.2 Relationship to Material Strength
$$\begin{array}{cc}{\sigma}_{y}={\sigma}_{0}+\frac{{k}_{y}}{\sqrt{d}}& (7.26)\end{array}$$
Two regions of space meet at a plane, and three regions of space meet at a line. These lines are important in a variety of problems in materials science. In Figure 8.1 we consider the most general case, where the 3 regions of space are labeled 1, 2 and 3. The junction between these three regions may correspond to three different material phases, or they may correspond to grain boundaries within a single phase region. In either case we refer to 1, 2 and 3 as 'phases', and refer to the line at which they meet as a '3phase contact line'. The way in which the three phases meet at this contact line are specified by two angles. These two angles can be defined in a variety of ways, but we use the angles defined in Figure
8.1 as
${\theta}_{1}$ and
${\theta}_{2}$, which give the orientation of the 1/3 and 2/3 boundaries with respect to the 1/2 boundary.
At equilibrium ${\theta}_{1}$ and ${\theta}_{2}$ are related to the interfacial free energies of the 3 phases that meet at the contact line. Because interfaces have a contribution to the free energy that is associated with them, there is a thermodynamic driving force for any interface to shrink in area. As a result an interface exerts a force on the contact line along the direction of the interface, with a force per length of equal to the relevant interfacial free energy. At the three phase contact line three different forces, ${\gamma}_{12}$, ${\gamma}_{23}$ and ${\gamma}_{13}$, are pulling on the contact line. At equilibrium the net force on the contact line is zero. We can obtain ${\theta}_{1}$ and ${\theta}_{2}$ by considering separate force balances in the directions parallel and perpendicular to the 1/2 interface:
 Horizontal force balance (x direction):$$\begin{array}{cc}{\gamma}_{13}sin{\theta}_{1}={\gamma}_{23}sin{\theta}_{2}& (8.1)\end{array}$$
 Vertical force balance (y direction):
$$\begin{array}{cc}{\gamma}_{12}={\gamma}_{13}cos{\theta}_{1}+{\gamma}_{23}cos{\theta}_{2}& (8.2)\end{array}$$
These are coupled, nonlinear equations that generally need to be solved numerically. An example procedure using MATLAB is given below in the section on wetting.
8.1 Wetting
 ${\gamma}_{w}$: the surface free energy of water
 ${\gamma}_{o}$: the surface free energy of the oil
 ${\gamma}_{ow}$: the interfacial free energy between oil and water
Note that we refer to ${\gamma}_{w}$ and ${\gamma}_{o}$ as 'surface energies' and not 'interfacial energies' because one of the contact phases is air. We drop the subscript for air in this case, which is the convention that is commonly used. The horizontal force balance in this case can be written as follows:
go=30; gow=50; gw=72; % specify the different interfacial energies
verticalforce=@(theta) go*sind(theta(1))gow*sind(theta(2)); % this is the net force in the vertical direction
horizontalforce=@(theta) gwgo*cosd(theta(1))gow*cosd(theta(2));
ftosolve=@(theta) [verticalforce(theta), horizontalforce(theta)]; % write the function so that it returns the two components of the net force that both must be zero
thetaguess=[10,10]; % initial guess for theta1 and theta2
thetasolution=fsolve(ftosolve, thetaguess); % returns the solution as thetasolution
$$\begin{array}{cc}{\gamma}_{o}+{\gamma}_{ow}<{\gamma}_{w}& (8.5)\end{array}$$
8.2 Grain Boundary Junctions
In this case the three phase contact line is actually a junction between grain boundaries, as opposed to a place where three distinct phases come into contact with one another. The important points are the following:
 If the three grain boundary energies for the boundaries meeting at the contact line are all equal to one another, ${\theta}_{1}={\theta}_{2}=6{0}^{\u25cb}$. In other words, the interior angles between the different grains are all 120${}^{\u25cb}$.
 As a corollary to the point above, the boundaries of grains with fewer than 6 sides will be curved outward and the grain will tend to shrink, whereas grains with more than 6 sides will have boundaries that are curved and the grains will tend to grow.
8.3 Liquid Drop on a Solid Surface
$$\begin{array}{cc}{\gamma}_{\ell}={\gamma}_{sl}+{\gamma}_{\ell}cos\theta & (8.6)\end{array}$$
Here ${\gamma}_{\ell}$ is the liquid surface energy, ${\gamma}_{sl}$is the solid/liquid interfacial energy and ${\gamma}_{s}$ is the solid surface energy.
9 Interphase Interfaces
Interfaces between two coexisting phases can have three types of interfaces: coherent, semicoherent and incoherent. Here we briefly describe these three types of interface.
9.1 Coherent Interfaces
$$\begin{array}{cc}{\gamma}_{coherent}={\gamma}_{ch}& (9.1)\end{array}$$
For interfaces between FCC and HCP crystal structures, only certain planes are coherent. For all planes to be coherent, both phases have to have the same crystal structure. However, they don't have to have the same lattice parameter. In this case elastic strains are generated.
9.2 Semicoherent Interfaces
Typical values for ${\gamma}_{semicoherent}$ are in the range of 0.10.5 mJ/m${}^{2}$. Note that
$${\gamma}_{st}\propto dislocation\phantom{\rule{6px}{0ex}}density$$
9.3 Incoherent interfaces
If the lattice mismatch becomes too late, we the energy associated with all of the required dislocations to make the interface at least partially coherent is too high. Instead the interface becomes incoherent as shown in Figure the dislocation cores begin to overlap. The interface becomes incoherent, with ${\gamma}_{incoherent}\approx 5001000\phantom{\rule{6px}{0ex}}mJ/{m}^{2}$. ${\gamma}_{coherent}$ is relatively isotropic.
9.4 Case Study I: the SiGe system
$$\begin{array}{cc}\overrightarrow{b}=\frac{a}{2}\u27e8110\u27e9& (9.3)\end{array}$$
It makes sense that Si and Ge would have the same crystal structure, since the properties of these elements are very similar, with Ge residing just below Si on the periodic table. The lattice parameters are different however, and are given as follows:
Si: ${a}_{Si}$=5.431 Å
Ge: $a{}_{Ge}$=5.658 Å
This lattice parameter mismatch corresponds to a mismatch strain, ${\delta}_{a}$, of 0.04 in this case, which we obtain from the relative difference between the lattice parameters:$$\begin{array}{cc}{\delta}_{a}=\frac{{a}_{Ge}{a}_{Si}}{{a}_{Si}}& (9.4)\end{array}$$
$${d}_{200}^{Si}=\frac{{a}_{Si}}{\left({h}^{2}+{k}^{2}+{\ell}^{2}\right)}=\frac{{a}_{Si}}{2}$$
Similarly, we have ${d}_{200}^{Ge}={a}_{Ge}/2=$. The fractional difference in the interatomic spacings is the same as the fractional difference in the lattice parameters:
Define lattice misfit, $\delta $:
$$\begin{array}{cc}{\delta}_{200}=\left({d}_{200}^{Ge}{d}_{200}^{Si}\right)/{d}_{200}^{Si}={\delta}_{a}& (9.5)\end{array}$$
We can rearrange this expression to get the following for ${d}_{200}^{Ge}$
$$\begin{array}{cc}{d}_{200}^{Ge}={d}_{200}^{Si}+{\delta}_{a}{d}_{200}^{Si}& (9.6)\end{array}$$
No we introduce a quantity ${D}_{200}$ which is the average distance between dislocations in the x direction. Within this distance there are $n$ (200) Ge planes but there are $n+1$ (200) Si planes, so we have:
$$\begin{array}{cc}{D}_{200}=n{d}_{200}^{Ge}=\left(n+1\right){d}_{200}^{Si}& (9.7)\end{array}$$
From these two equations we obtain $n=1/{\delta}_{a}$.
For
${\delta}_{a}=0.04$ and
${d}_{200}^{Ge}=2.82$ Å , we have
$n=25$ and
${D}_{200}=70.5$ Å. We also have to account for the misfit in the other surface direction (the z direction in our case). The same argument holds in this direction as well, so we'll end up with dislocation spaced by
${D}_{002}$ in this direction with
${D}_{002}={D}_{200}$. Overall, we get a grid of dislocations at the interface, as shown in Figure
9.6.
9.5 Case Study II: The Cu/Al system
9.6 Second Phase Shape
9.7 Elastic Effects
$$\begin{array}{cc}{\gamma}_{\alpha \beta}={\gamma}_{ch}& (9.8)\end{array}$$
If the lattice parameters of the $\alpha $ and $\beta $ phases do not match exactly, which will almost certainly be the case for any real system, there will be a positive elastic strain energy, ${W}_{el}$ that we need to consider. For a spherical, coherent precipitate in an elastically isotropic medium, ${W}_{el}$ is given by the following expression:
${K}^{\beta}$= bulk modulus of $\beta $ phase: ${K}^{\beta}=V\frac{\partial P}{\partial {V}^{\beta}}$
${G}^{\alpha}$= shear modulus of $\alpha $ phase: $G=\frac{shear\phantom{\rule{6px}{0ex}}stress}{shear\phantom{\rule{6px}{0ex}}strain}$
$\delta $= misfit: $\delta =\frac{1}{3}\left[\frac{{V}_{m}^{\beta}{V}_{m}^{\alpha}}{{V}_{m}^{\alpha}}\right]$
For cubic systems and for the small values of $\delta $ that are generally relevant here, $\delta $ is also equal to the fractional mismatch in the lattice parameter:
$$\begin{array}{cc}\delta =\left(\frac{{a}^{\beta}{a}^{\alpha}}{{a}^{\alpha}}\right)& (9.10)\end{array}$$
 compressible precipitate in a rigid matrix: ${G}^{\alpha}>>{K}^{\beta}$
$$\begin{array}{cc}\frac{{W}_{el}}{{V}^{\beta}}\approx \frac{9{\delta}^{2}{K}^{\beta}}{2}& (9.11)\end{array}$$
 rigid precipitate in a deformable matrix: ${K}^{\beta}>>{G}^{\alpha}$
$$\begin{array}{cc}\frac{{W}_{el}}{{V}^{\beta}}\approx 6{\delta}^{2}{G}^{\alpha}& (9.12)\end{array}$$
 Precipitate and matrix with the same elastic properties.
An isotropic material is characterized by just two independent elastic constants. It's convenient to express $K$ in terms of $G$ and $\nu $, using the following expression (note that the Wikipedia page has a very useful summary of the relationships between different elastic constants for an isotropic material)
$$\begin{array}{cc}K=\frac{2G\nu}{12\nu}& (9.13)\end{array}$$
As an example, we can take $\nu =1/3,$ in which case we get $K=2G,$ and ${W}_{el}$ is given by the following expression. $$\begin{array}{cc}\frac{{W}_{el}}{{V}^{\beta}}=3.6{\delta}^{2}G& (9.14)\end{array}$$
In each of these cases the most important points to keep in mind are the following:
 The elastic strain energy is proportional to the volume of the precipitate.
 The elastic strain energy is proportional to the square of the lattice mismatch.
We are now in a position to compare the overall free energy of coherent and incoherent precipitates, and to see how each of these depend on the precipitate size. For a coherent precipitate we just need to add the elastic strain energy to the chemical contribution to the interfacial free energy:
Incoherent precipitates have a larger value of ${\gamma}_{\alpha \beta}$ because we also need to account for the structural component of the interfacial free energy that arises for the dislocations that are present at the interface between the $\alpha $ and $\beta $ phases. However, the strain energy in the bulk is reduced to zero, so have the following for ${W}_{inc}$, the total excess free energy of an incoherent precipitate:
$$\begin{array}{cc}{r}_{crit}=3{\gamma}_{st}\frac{{V}^{\beta}}{{W}_{el}}& (9.17)\end{array}$$
9.8 Effects of Elastic Anisotropy
in reality, no crystalline material is completely isotropic. An FCC crystal, for example, is generally stiffest along the [110] directions and softest along [100] directions. This is because the linear density of atoms is highest along the [110] direction, where in a hard sphere model of the crystal structure the atoms are in contact with one another. As a result coherent precipitates end up with facets perpendicular to the 'soft' [100] directions. Faceting becomes more important as the precipitates grow (assuming they stay coherent), since the elastic contribution to the energy scales with the volume of the precipitate, whereas the total surface area scales with the 2/3 power of the volume.
10 Surfactants
Diblock copolymer molecules can act as macromolecular 'surfactants', segregating preferentially to the interface between the corresponding homopolymers. In the schematic illustration below, A/B diblock copolymer molecules segregate preferentially to the interface between A and B phases, thereby limiting the ability of the morphology to coarsen by coalescence of the B domains. An actual example of this for polystyrene/poly(methyl methacrylate) (PS/PMMA) system is illustrated here.
 ${\gamma}_{Si}$: The surface free energy of the Si substrate
 ${\gamma}_{Ge}$: The surface free energy of Ge
 ${\gamma}_{Ch}$: The chemical contribution to the free energy of the Si/Ge interface
 ${\gamma}_{St}$: The structural component to the free energy of the Si/Ge interface
 ${W}_{el}/A$: The strain energy per unit area within the Ge film.
The full thickness dependence of the free energy can be understood by investigating the way that these contributions contribute to the overall free energy as the Ge film thickness increases:
 For ${t}_{Ge}=0$ the free energy is simply ${\gamma}_{Si}$, the surface free energy of the silicon substrate.
 For very thin Ge films the elastic strain energy within the Ge film does not contribute significantly to the strain energy, so the overall value of $W/A$ is the sum of the energies of the Si/Ge and Ge/vapor interfaces. Because the Si/Ge interface is fully coherent for sufficiently thin Ge films, its interfacial free energy is just the chemical part, ${\gamma}_{ch}$, so the overall free energy for very thin Ge films is ${\gamma}_{Ge}$ + ${\gamma}_{Ch}$. This free energy is less than ${\gamma}_{Si}$, so the system is in the wetting regime.
 As the film thickness increases the Ge film remains coherent, but $W/A$ increases linearly with thickness, according to the thickness dependence of ${W}_{el}$.
 When the elastic energy exceeds the structural component of the Si/Ge interfacial free energy associated with the loss of full coherence, the Ge film becomes incoherent. The elastic energy ${W}_{el}$ is now large enough so that it is energetically favorable for the Si/Ge interface to be less coherent.
For film thicknesses larger than the thickness for which
${W}_{el}$ is a minimum, a thin Ge layer with a thickness corresponding to the thickness at the free energy minimum will coexist with Ge droplets that are much thicker  the 'islands' shown in Figure
11.1.
12 Review Questions
12.1 Diffusion
 What are the tracer, interdiffusion and intrinsic diffusion coefficients? Which are purely kinetic quantities, and which involve thermodynamics? How are these diffusion coefficients related to one another?
 What is the Kirkendall effect? How can you figure out the direction of vacancy motion?
 What is the mechanism by which vacancies are either created or destroyed?
 Under what conditions will voids form, and where will they form (Lab 1)?
 Where must dislocations be created or destroyed to maintain an equilibrium vacancy concentration?
12.2 Dislocations
 Explain the physical origin of shear bands observed on the surface of a plastically deformed metal.
 What is the critical resolved shear stress? How is it calculated for a tensile experiment?
 What is the value of the ratio of the theoretical critical resolved shear stress (in the absence of dislocations) to a typical experimental value of this same quantity.
 Define an edge and a screw dislocation in terms of their Burgers vectors and the sense vectors.
 What are the Burgers vectors of perfect dislocations in the simple cubic, facecentered cubic and bodycentered cubic lattices?: use a drawing to illustrate the Burgers vectors. What are the magnitudes of these vectors.
 Explain how to make pure edge or screw dislocations by cutting and slipping operations.
 Explain how to make a curved dislocation by cutting and slipping operations. Demonstrate that its character varies from pure screw to pure edge as one moves along the curved dislocation line.
 Demonstrate carefully how the Burgers vector of an edge or a screw dislocation is determined employing a Burgers circuit.
 What is the difference between a righthand and a lefthand screw dislocation?
 Given $\overrightarrow{b}$ and $\overrightarrow{s}$, how do you know where the slip plane is, and what direction the dislocation will move in response to an applied shear stress.
 Explain why a pure screw dislocation does not have a unique glide or slip plane.
 Explain why a pure edge dislocation has a unique glide or slip plane.
 Explain the main differences between glide motion of a dislocation and climb motion?
 How does the temperature dependence of glide differ from the temperature dependence of climb? Which is more important at lower temperatures, and why?
 Explain how climb of an edge dislocation can relieve a super or subsaturation of vacancies or selfinterstitial atoms.
 Do pure screw dislocations crossslip?
 What are the physical origins of the energy of a dislocation line?
 If no external stress is supplied to a dislocation loop, why does the loop shrink until it disappears from a crystal?
 Describe qualitatively the state of stress associated with a pure edge dislocation and compare it with the state of stress associated with a pure screw dislocation.
 What is the physical significance of ${F}_{s}^{\tau}$ and ${F}_{s}^{r}$?
 Describe qualitatively the state of stress associated with a pure edge dislocation and compare it with the state of stress associated with a pure screw dislocation.
 When do parallel edge dislocations move toward each other? Under what conditions do they move away from each other?
 When do parallel screw dislocations move toward or away from each other?
 How does a FrankRead source work?
 How does the stress need to be oriented to either expand or contract a dislocation loop?
 How does precipitation hardening work? What is the role of the precipitate spacing and of FrankRead sources? Why does the precipitate spacing matter?
12.3 Solid/Liquid and Solid/Vapor Interfaces
 Why does the surface energy of a crystal depend on the orientation?
 How does the interfacial energy affect the melting temperature for a small droplet?
 How does the interfacial free energy affect precipitate solubility?
 How can the surface energy be approximated from the crystal structure and thermodynamic data?
 What is the Wulff construction and how is it used?
12.4 Grain Boundaries
 What are the 5 parameters needed to fully characterize a grain boundary?
 What special relationship exists between these parameters for twist and tilt boundaries?
 How are dislocations arranged for low angle twist and tilt boundaries?
 How does the force balance at the triple junctions of grains affect grain shape?
 What happens to grains with different numbers of sides during grain growth?
 What is the role of curvature in grain growth?
 Derive the expected time dependence of the grain size for grain growth driven by curvature.
 What is a twin boundary? For what crystal structures is it observed?
12.5 Interphase Interfaces
 What is the general condition governing the equilibrium shape of a precipitate when there is no contribution from the elastic energy?
 (a) What is the expression for the total elastic strain energy of a precipitate, if the matrix is elastically isotropic? (b) Explain the physical significance of each term in this equation.
 How does the shape of a misfitting precipitate in an elastically anisotropic system vary with particle size? Why is this variation observed?
 (a) Derive an approximate expression for the critical radius at which a spherical precipitate loses its coherency and become semicoherent or incoherent. (b) Explain why precipitates often exceed this calculated critical value.
 Describe the nature of a solidliquid interface and how it differs from a solidsolid interface.
 What is the significance of the chemical and structural components to the interfacial free energy between solids?
 Why do Ge films on Si form islands on top of a thin wetting layer?
12.6 Crystallization or Recrystallization
 Make a plot of the volume fraction, $f$, transformed, 0 to 1.0, as a function of time for a general phase transformation occurring by nucleation and growth.
 Derive for $f\left(t\right)\ll 1$ the JohnsonMehlAvramiKolomogorov (JMAK) equation for the volume fraction transformed as a function of time, $f\left(t\right)$, under the assumption that a specimen contains a number $n$ of effective point heterogeneities per unit volume and nucleation occurs at all of these points very quickly and that the nuclei have a spherical shape. State any and all assumptions made.
 Derive a JMAK equation, for $f\left(t\right)\ll 1$ , for the case where all the nuclei do not form at time $t=0$, but rather form randomly throughout a specimen at a constant rate, which is N nuclei formed per unit volume per unit time of untransformed material. State any and all assumptions made.
12.7 MATLAB
 How do I make plots suitable for publication when those plots generated by Excel just aren't good enough anymore?
 How do I write arbitrary functions that can be plotted or compared with experimental data?
 How do I fit a userdefined function to experimental data?
 How do I use fsolve to solve a system of coupled equations?
 How can I run a write and run a simple simulation in MATLAB (like the vacancy diffusion simulation)
 What is a polar plot and how can I generate one?
 How do I solve the Wulff construction numerically?
13 3161 Problems
Introduction
13.0.0.1
 Anything about yourself (why you are interested in MSE, previous work experience, etc., outside interests apart from MSE) that will help me get to know you a bit (feel free to be brief  any info here is fine).
 Your level of experience and comfort level with MATLAB. Be honest about your assessment (love it, hate it, don't understand it, etc.).
 Let us know if you have NOT taken 314 or 315 for some reason.
Diffusion
13.0.0.2
Consider a diffusion couple with composition ${C}_{1}$ as $z\to \infty $ and ${C}_{2}$ as $z\to \infty $. The solution to the diffusion equation is:
$$C\left(z,t\right)=\frac{{C}_{1}+{C}_{2}}{2}\frac{{C}_{1}{C}_{2}}{2}erf\left(\frac{z}{2\sqrt{Dt}}\right)$$where $erf\left(y\right)=\frac{2}{\pi}{\int}_{0}^{y}{e}^{{t}^{2}}dt$. Note that in the definition of the error function t is a dummy variable of integration, thus the error function is a function of y. Also, erf(0)=0, and erf( $\infty $)=1. You will determine if these boundary conditions are correct.
 Show that the boundary conditions at $z=\pm \infty $ are satisfied by the solution.
 Does the composition at $z=0$ vary with time? If not, what is its value? Why do you think this is the case?
 Write the solution in terms of $\eta =z/{t}^{1/2}$.
 Show that the solution satisfies the following diffusion equation that is written in terms of $\eta $:
$$D\frac{{d}^{2}C}{d{\eta}^{2}}+\frac{\eta}{2}\frac{dC}{d\eta}=0$$
You will needed to take a derivative of the error function. Leibniz’s formula for the differentiation of integrals will be helpful:
$$\frac{d}{dz}{\int}_{h\left(z\right)}^{g\left(z\right)}f\left(t\right)dt=\frac{dg\left(z\right)}{dz}f\left(g\left(z\right)\right)\frac{dh\left(z\right)}{dz}f\left(h\left(z\right)\right)$$
13.0.0.3
A diffusion couple including inert wires was made by plating pure copper on to a block of
$\alpha $brass with
${X}_{Zn}=0.3$, as shown in Figure
13.1. After 56 days at 785
${}^{\u25cb}$C the marker velocity was 2.6x10
${}^{8}$ mm/s, with a composition at the markers of
${X}_{Zn}=0.22$, and a composition gradient,
$\partial {X}_{Zn}/\partial z$ of 0.089 mm
${}^{1}$. A detailed analysis of the data gives
$\tilde{D}=4.5x1{0}^{13}\phantom{\rule{6px}{0ex}}{m}^{2}/s$ for
${X}_{Zn}=0.22$. Use these data to calculate
${D}_{Zn}$ and
${D}_{Cu}$ for
${X}_{Zn}=0.22$. How would you expect
${D}_{Zn}$,
${D}_{Cu}$ and
$\tilde{D}$ to vary as a function of composition?
13.0.0.4
In class we developed an expressions for ${J}_{a}^{\prime}$. Show that ${J}_{a}^{\prime}={J}_{b}^{\prime}$. (Recall that these primed fluxes correspond to fluxes in the laboratory frame of reference).
13.0.0.5
Consider two binary alloys with compositions ${X}_{b}={X}_{1}$ and ${X}_{b}={X}_{2}$, shown in Figure13.2 along with the free energy curves for
$\alpha $ and
$\beta $ phases formed by this alloy. Draw the composition profile across the interface shortly after the two alloys are brought into contact with one another, assuming that the interface is in “local equilibrium”, i.e. the interface compositions are given by the equilibrium phase diagram. Describe the direction in which you expect the B atoms to diffuse on each side of the interface.
13.0.0.6
The following MATLAB script runs the vacancy simulation shown in class. It saves the data into a 'structure' called output, which can be loaded into MATLAB later. The file can be downloaded from this link:
tic % start a time so that we can see how long the program takes to run
n=30; % set the number of boxes across the square grid
vfrac=0.01; % vacancy fraction
matrix=ones(n);
map=[1,1,1;1,0,0;0,0,1]; % define 3 colors: white, red, blue
figure
colormap(map) % set the mapping of values in 'matrix' to a specific color
caxis([0 2]) % range of values in matrix goes from 0 (vacancy) to 2
% the previous three commands set things up so a 0 will be white, a 1 will
% be red and a 2 sill be blue
matrix(:,n/2+1:n)=2; % set the right half of the matrix to 'blue'
i=round(n/2); % put one vacancy in the middle
j=round(n/2);
matrix(i,j)=0;
imagesc(matrix); % this is the command that takes the matrix and turns it into a plot
t=0;
times=[1e4,2e4,5e4,1e5];
showallimages=1; % set to zero if you want to speed things up by not showing images, set to 1 if you want to show all the images during the simulation
%% now we start to move things around
vacancydiff.matrices={}; % makea blank cell array
while t<max(times)
t=t+1;
dir=randi([1 4], 1, 1);
if dir==1
in=i+1;
jn=j;
if in==n+1; in=1; end
elseif dir==2
in=i1;
jn=j;
if in==0; in=n; end
elseif dir==3
in=i;
jn=j+1;
if jn>n; jn=n; end
elseif dir==4
in=i;
jn=j1;
if jn==0; jn=1; end
end
% now we need to make switch
neighborix=sub2ind([n n],in,jn);
vacix=sub2ind([n n],i,j);
matrix([vacix neighborix])=matrix([neighborix vacix]);
if showallimages
imagesc(matrix);
drawnow
end
if ismember(t,times)
vacancydiff.matrices=[vacancydiff.matrices {matrix}]; % append matrix to output file
imagesc(matrix);
set(gcf,'paperposition',[0 0 5 5])
set(gcf,'papersize',[5 5])
print(gcf,['vacdiff' num2str(t) '.eps'],'depsc2')
end
i=in;
j=jn;
end
vacancydiff.times=times;
vacancydiff.n=n;
save('vacancydiff.mat','vacancydiff') % writes the vacancydiff structure to a .mat file that we can read in later
toc
 Run the vacancy diffusion script, and include in your homework the .jpg files generated for time steps of 1e4, 2e4, 4e4 and 1e5.
 For the longest time step, develop a plot of average composition along the horizontal direction.
Here is the MATLAB script that I used to do this (available at
http://msecore.northwestern.edu/3161/matlab/vacancyplot.m):
load vacancydiff % load the previously saved output.mat file
figure
figformat % not necessary, this is the standard initialization script I use to standardize what my plots look like
n=vacancydiff.n;
matrix=vacancydiff.matrices{4};
matrixsum=sum(matrix,1); % sum of each column in the matrix
plot(1:n,matrixsum/n,'+b')
xlabel ('z')
ylabel ('C')
print(gcf,'../figures/vacancyplot.eps','depsc2') % this creates an .eps file, which I use for the coursenotes but which may not be as useful for many of you as the jpg file
% saveas(gcf,'vacancyplot.jpg') % this is what to do if you just want to save a .jpg file
set(0,'defaultaxesbox', 'on') % draw the axes box (including the top and right axes)
set(0,'defaultlinelinewidth',2)
set(0,'defaultaxesfontsize',16)
set(0,'defaultfigurepaperposition',[0,0,7,5])
set(0,'defaultfigurepapersize',[7,5]')
 In the previous problem set we obtained concentration profiles from the MATLAB. Now we'll take these concentration profiles and see if they are consistent with the solution to the diffusion equation.
 For each of the 4 time points used in the simulation, plot the concentration profile and fit it to the error function to the diffusion equation, using the interfacial width, $w$, ($w=2\sqrt{Dt}$) as a fitting parameter:
$$C\left(z,t\right)=\frac{{C}_{1}+{C}_{2}}{2}\frac{{C}_{1}{C}_{2}}{2}erf\left(\frac{z}{w}\right)$$
Note: This problem is a curve fitting exercise in MATLAB. The most frustrating part is getting all the syntax right, but once you know the proper format for the MATLAB code, it's pretty straightforward. Take a look at the section entitled 'Fitting a Function to a Data Set' in the MSE MATLAB help file:
This section includes a MATLAB script that you can download and modify as needed.
 Plot ${w}^{2}$ as a function of the time (expressed here as the number of time steps in the simulation). Obtain the slope of a line drawn through the origin that best fits the data.
 When diffusion occurs by a vacancy hopping mechanism in a 2dimensional system like the one used in our simulation, the diffusion coefficient is given by the following expression:
$$D=K{X}_{v}\Gamma {a}^{2}$$
Here is the average hop frequency for any given vacancy and $a$ is the hopping distance. From the the slope of the curve of $w$ vs. the total number of jumps, extract an estimated value for $K$.
13.0.0.7
A region of material with a different composition is created in an infinitely long bar. The following plot shows the mole fraction of component A as a function of position. Assume that the intrinsic diffusion coefficient of the A atoms is twice as large as the intrinsic diffusion coefficient for the B atoms.
 Plot the flux of A and the flux of B relative to the lattice as a function of position in the graph above.
 Plot the vacancy creation rate as a function of position in the graph above.
 Plot the flux of A and B in the lab frame as a function of position in the graph above.
 Plot the lattice velocity as a function of position in the graph below. What are the physical implications of this plot?
13.0.0.8
The values for the intrinsic diffusion coefficients for Cu and Ni in a binary Cu/Ni alloy are shown below on the left (note that Cu and Ni are completely miscible in the solid state). A diffusion couple is made with the geometry shown below on the right.
 What is the value of the interdiffusion coefficient $\tilde{D}$, for an alloy consisting of nearly pure Nickel?
 Will the markers placed initially at the Cu/Ni interface move toward the copper end of the sample, the nickel end of the sample, or stay at exactly the same location during the diffusion experiment.
 The copper concentration across the sample is sketched below after diffusion has occurred for some time.
 Sketch the fluxes of Copper, Nickel and vacancies, defining positive fluxes as those moving to the right.
 Now sketch the rate at which vacancies are created or destroyed within the sample in order to maintain a constant overall vacancy concentration throughout.
13.0.0.9
An experiment is performed to determine the tracer diffusion coefficient of metal A in a matrix of metal B. This is done by depositing a very thin film of metal A onto the surface of metal B and measuring the concentration profile of metal A into the depth of the material at different times. The concentration profiles in the left figure below are obtained at two times, ${t}_{1}$ and ${t}_{2}$:
 Estimate the ratio ${t}_{2}/{t}_{1}$
 Now suppose we measure the self diffusion coefficients of A and B. Performing measurements at the same time and temperature gives the concentration profiles shown in the figure above to the right. Which element (A or B) do you expect has the highest melting temperature, and why?
 Now we'll make a diffusion couple with element A on the right half and element B on the left half. Assume that A and B are miscible at the diffusion temperature, and form a one phase alloy. Mark up the following diagram as directed on the next page:
 Put an arrow labeled 'M' on the diagram indicating the direction that inert markers placed originally at the interface will move.
 Put an arrow labeled 'V' on the diagram indicating the the net vacancy flux due to diffusion in the sample.
 Put a 'C' on the region of the sample where you expect vacancies to be created, and a 'D' on the sample where you expect vacancies to be destroyed, assuming that the total vacancy concentration must remain at equilibrium.
 Two edge dislocations are also indicated in the diagram. Place arrows on top of each dislocation to illustrate he directions you expect these dislocations to move in order to create or destroy the vacancies from part iii.
Stress and Strain
13.0.0.10
A tensile stress, $\sigma $, is applied to a single crystal of zinc, which has an HCP structure. The close packed planes of atoms (the slip plane for an HCP material) is oriented with its surface normal in the plane of the paper, inclined to the tensile axis by an angle $\phi $ as shown below, with $\phi =3{0}^{\u25cb}$. Assume that the critical resolved shear stress for motion of the dislocation is 50 MPa (5x10 ${}^{7}$ Pa). The shear modulus of Zn is 43 GPa (4.3x10 ${}^{10}$ Pa) and its atomic radius is 0.13 nm.
 Is this an edge dislocation, a screw dislocation, or a mixed dislocation, and how do you know?
 Put an arrow on the drawing above to indicate the direction in which the dislocation moves under an applied tensile stress.
 Calculate the tensile yield stress for this sample.
 Suppose that the slip plane is oriented so that $\overrightarrow{b}$ is still in the plane of the paper, but that $\phi $ is increased to $6{0}^{\u25cb}$. Will the yield stress increase, decrease or stay the same.
 Suppose that the dislocation is impeded by pinning points (precipitates, for example), that are uniformly spaced and separated by 1 $\mu $m (10 ${}^{6}$ m). The resolved shear stress is determined by the stress required to move the dislocation around these pinning points. Use the information given in this problem to determine the energy per length of the dislocation. Compare this to the expressions given for the energies of edge and screw dislocations to see if it makes sense.
Dislocation Structure
13.0.0.11
A right handed screw dislocation initially located in the middle of the front face of the sample shown below moves toward the back of the sample in response to an applied shear stress on the sample.
 Sketch the shape of the sample after the dislocation has propagated halfway through the sample, and again when it has propagated all the way through the sample. Use arrows to specify the shear force that is being applied.
 Repeat part a for a lefthanded screw dislocation.
13.0.0.12
Draw an edge dislocation and on the same figure dot in the positions of the atoms after the dislocation has shifted by $\overrightarrow{b}\mathrm{.}$
13.0.0.13
How can two edge dislocations with opposite Burgers vectors meet to form a row of vacancies? How can they meet to form a row of interstitials? Draw pictures of both situations.
13.0.0.14
Given a crystal containing a dislocation loop as shown in the following figure:
Let the loop be moved (at constant radius) toward a corner until threefourths of the loop runs out of the crystal. This leaves a loop segment that goes in one face and comes out the orthogonal face. Sketch the resultant shape of the crystal, both above and below the slip plane.
13.0.0.15
Given a loop with a Burger’s vector that is perpendicular everywhere to the dislocation line, determine the resulting surface morphology after the loop propagates out of the crystal. Assume that the loop moves only by glide.
13.0.0.16
Show that it is impossible to make a dislocation loop all of whose segments are pure screw dislocations, but that it is possible with edge dislocations. For the case of the pure edge dislocation loop, describe the orientation of the extra half plane with respect to the dislocation loop.
13.0.0.17
Draw the compressive and tensile regions surrounding an edge dislocation.
13.0.0.18
Consider the dislocation loop shown below:
 Circle the drawing below that corresponds to the shape of the material after the dislocation has expanded and moved out outside the crystal.
 Indicate in the spaces below the locations (a, b, c, or d) where the dislocation has the following characteristics:
 It is a right handed screw dislocation:_____
 It is a left handed screw dislocation:_____
 It is an edge dislocation with the extra half plane above the plane of the loop:_____
 It is an edge dislocation with the extra half plane below the plane of the loop:_____
 Add arrows to the illustration of the dislocation loop to show the orientation of the shear stress that will most efficiently cause the dislocation to loop to grow.
Dislocation Interactions
13.0.0.19
If edge dislocations with opposite signs of the Burger’s vectors meet, does the energy of the crystal increase or decrease? Defend your answer.
13.0.0.20
A nanowire is grown such that it is free of dislocations. Why would the stress required to deform the nanowire be larger than a bulk material?
13.0.0.21
If an anisotropic alloy system has a nearly zero dislocation line tension, would you expect the precipitate spacing to have a large effect on the yield stress of the alloy? Explain your reasoning
13.0.0.22
Given an edge dislocation in a crystal, whose top twothirds is under a compressive stress $\sigma $ acting along the glide plane (see figure below):
 If diffusion occurs, which way will thee dislocation move? Explain why and tell where the atoms go that leave the dislocation.
 Derive an equation relating the stress, $\sigma $ to $b$ and the force tending to make the dislocation move in the vertical plane.
 If the edge dislocation is replaced by a screw dislocation, which which way will the dislocation tend to move?
13.0.0.23
Construct a plot of the interaction energy vs. dislocation separation distance for two identical parallel edge dislocations that continue to lie one above the other as climb occurs. Justify your plot qualitatively by explaining how the strain energy changes with vertical separation.
13.0.0.24
Repeat the previous problem for edge dislocations of opposite sign.
13.0.0.25
On the following sketch of a dislocation, indicate the direction that it must move in order for vacancies to be created.
13.0.0.26
Consider an isolated righthanded screw dislocation. Suppose a shear force is applied parallel to the dislocation line, as illustrated below.
 What is the direction of the force, ${F}_{s}^{\tau}$, that is applied to the dislocation as a result of the applied stress.
 Suppose the screw dislocation is replaced by a dislocation loop with the same Burgers vector as the dislocation from part a, as shown below. Use arrows to indicate the direction ${F}_{s}^{\tau}$ at different points along the dislocation loop. (The direction of ${F}_{s}^{\tau}$has already been indicated at the right edge of the dislocation).
 Describe how the magnitude of ${F}_{s}^{\tau}$ changes (if at all) for different locations along the dislocation loop.
 What to you expect to happen to the dislocation loop if you remove the external applied stress (will the loop grow, shrink or stay the same size)?
 Suppose the straight screw dislocation from is pinned by obstacles that are separated by a distance $d$, as illustrated in the following figure. Sketch the shape of the dislocation for an applied shear stress that is just large enough for dislocation to pass around the obstacles.
 What do you expect to happen to the critical resolved shear stress of the material if $d$ is decreased by a factor of 2. (Will the critical resolved shear stress increase, decrease or stay the same).
Interfacial Thermodynamics
13.0.0.27
Consider the following:
 Is the molar latent heat positive or negative?
 Is the melting temperature $,$ $T$ , for a very small particle greater to or less than the equilibrium value of ${T}_{m}$ for a bulk material?
 Must this always be the case?
 For metals, what is the typical value of $r$ for which a change in melting temperature of 10K is observed. What about a change of 1K?
13.0.0.28
The molar enthalpy of a phase varies with temperature as
$${H}_{m}\left(T\right){H}_{m}\left({T}_{0}\right)+{\displaystyle {\int}_{{T}_{0}}^{T}}{C}_{p}\left(T\right)dT$$where ${C}_{p}$ is the molar heat capacity. Given this, at what temperature is the latent heat appearing in expression for the melting point reduction evaluated?
13.0.0.29
Consider the case of a pure liquid spherical droplet embedded in a pure solid. Create a graphical construction plotting the temperature dependence of the free energy of the solid and liquid phases(similar to Figure
5.11) for this case, and use it to determine if the melting point above or below the bulk melting temperature.
13.0.0.30
Consider the CoCu phase diagram shown below:
 Plot the equilibrium activity of Cobalt as a function of composition across the entire phase diagram at 900ºC.
 From the phase diagram, estimate the solubility limit of Co in Cu at 900 ${}^{\u25cb}$C. Suppose the interfacial free energy for the Cu/Co interface is $300\phantom{\rule{6px}{0ex}}mJ/{m}^{2}$. For what radius of a Co precipitate will this solubility limit be increased by 10%?
Surface and Interface Structure
13.0.0.31
Look up values for heats of sublimation for any of the materials in Table 6.1 that have closepacked crystal structures (FCC or HCP). Compare the estimated values of the surface free energy that you obtain from these heats of sublimation to the tabulated values in Table 6.1.
13.0.0.32
Determine the equilibrium shape of a crystal. This should be done using a computer and your favorite program or language (most likely MATLAB). The equation of a straight line in polar coordinates drawn from the origin of the polar coordinate system is
$rcos\left(\theta \alpha \right)=d$, where
$\left(r,\phantom{\rule{6px}{0ex}}\theta \right)$ locate the points on the line,
$d$ is the perpendicular distance from the origin to the line and
$\alpha $ is the angle between the perpendicular to the line and the xaxis (see Figure
13.3).
 Determine the equilibrium shape of a crystal where the surface energy is given by $\gamma =1$ J/m ${}^{2}$ (independent of $\alpha $).
 Determine the equilibrium shape of a crystal where the surface energy is given by $\gamma =1+0.05cos\left(4\alpha \right)$ J/m ${}^{2}$ ( $\alpha $ in radians). Are there any corners on the equilibrium shape?
 Determine the equilibrium shape of a crystal where the surface energy is given by $\gamma =1+0.07cos\left(4\alpha \right)$ J/m ${}^{2}$. Are there any corners on the equilibrium shape?
 Determine the equilibrium shape of a crystal where the surface energy is given by $\gamma =1+0.6cos\left(4\alpha \right)$ J/m ${}^{2}$. Are there any corners on the equilibrium shape? How is the shape shown in (c) different from that in (d), and why (argue on the basis of the physics of the problem)?
close all
A=[0,0.05,0.07,0.6]; % these are the 4 values of A defined in the problem
% define a function where the radius d is the surface energy and alpha
% is the angle
d=@(A,alpha) 1+A*cos(4*alpha);
figure
for k=1:4
alpha=linspace(0,2*pi,200);
subplot(2,2,k) % this makes a 2 by 2 grid of plots
polar(alpha,d(A(k),alpha),'r'); % poloar is the command to make a polar plot
title(['A=' num2str(A(k))],'fontsize',20) % label each subplot
end
% adjust the print command as necessary to change the format, filename,
% etc.
print(gcf,'../figures/matlabwulffenergy.eps', 'depsc2') % save the eps file
This generates the following polar plots for the four different functions that are given (with $A$ defined so that $\gamma =1+Acos\left(4\alpha \right)$).
13.0.0.33
Assume a simple cubic crystal structure with nearest neighbor interactions. Calculate the ratio of the surface energies for the {110} and {100} surfaces.
13.0.0.34
The octahedral particles of FCC golds shown below were created by controlling the growth rates of the different crystal facets. For these crystals, were the growth rates fastest in the $\u27e8100\u27e9$ directions or in the $\u27e8111\u27e9$ directions? Provide a brief explanation of your answer.
13.0.0.35
The relationship between the the interfacial energy between $\alpha $ and $\beta $ phases and the pressure difference across a curved interface is obtained from the following expression:
$${P}^{\alpha}\delta {V}^{\alpha}{P}^{\beta}\delta {V}^{\beta}+{\gamma}_{\alpha \beta}\delta {A}^{\Sigma}=0$$
 Use this expression to obtain the pressure difference between a cylinder of $\beta $ phase with a radius $r$ and a surrounding $\alpha $ phase.
 Repeat the calculation for a cube where the length of each side is $a$. Assume that the surface energy of each of the cube faces is the same.
Wetting and Contact Angles
13.0.0.36
Consider the an oil droplet that forms on the surface of water, as shown schematically in the following Figure:
Determine ${\theta}_{1}$ and ${\theta}_{2}$ if the air/water interfacial free energy is 72 mJ/m ${}^{2}$, the air/oil interfacial free energy is 30 mJ/m ${}^{2}$ and the oil/water interfacial free energy is 50 mJ/m ${}^{2}$.
13.0.0.37
Suppose a, hemispherical liquid Au droplet with a radius of curvature of $r$ is in contact with solid Si cylinder with the same radius as shown below. Derive a relationship between the three interfacial energies that must be valid in order for the equilibrium shape of the Au/Si interface to be flat, as drawn in the picture.
Grain Boundaries
13.0.0.38
The surface energy of the interface between nickel and its vapor is 1.580 J/m ${}^{2}$ at 1100K. The average dihedral angle measured for grain boundaries intersecting the free surface is 168 ${}^{\u25cb}$. Thoria dispersed nickel alloys are made by dispersing fine particles of ThO ${}_{2}$ in nickel powder and consolidating the aggregate. The particles are left at the grain boundaries in the nickel matrix. Prolonged heating at elevated temperatures gives the particles their equilibrium shape. The average dihedral angle measured inside the particle is 145 ${}^{\u25cb}$. Estimate the interfacial energy of the thorianickel interface. Assume the interfacial energies are isotropic.
13.0.0.39
Consider a gold line deposited on a silicon substrate. The grain boundaries run laterally completely across the line, giving a “bamboo” structure as shown in the figure below. The grain boundary energy of gold at 600K is 0.42 J/m ${}^{2}$ and the surface energy is 1.44 J/m ${}^{2}$. Assume all the interfacial energies are isotropic.
 Compute the dihedral angle ( $\theta $ in the diagram above) where a grain boundary meets the external surface.
 Find the critical grain boundary spacing ${\ell}_{c}$ for which the equilibrium grain shape produces a hole in the film, assuming $h=1\phantom{\rule{6px}{0ex}}\mu m$. Note that for a spherical cap, $\ell ,$ h and $\phi $ are related to each other by the following expression: $tan\left(\phi /2\right)=2h/\ell $.
13.0.0.40
Why does the velocity of a grain boundary depend on temperature? Assume that the driving force for grain boundary motion is independent of temperature.
13.0.0.41
Consider the following junction between three grains. Suppose that the grain boundary free energy between grains 1 and 2, and between 1 and 3, is 0.5 J/m ${}^{2}$. What is the grain boundary energy between grains 2 and 3?
13.0.0.42
Consider the following image from the grain growth simulation:
 The boundary marked with an 'X' separates grains 1 and 2. Do you expect this boundary to move toward grain 1 or grain 2 during the process of grain growth?
 Suppose that the interface marked above is the cross section through a grain boundary in aluminum, and that this section of the grain boundary has a spherical shape with a radius of curvature of $1$ $\mu $m. Assuming a grain boundary energy of 0.25 J/m ${}^{2}$, calculate the chemical potential difference, $\Delta \mu $ between Al atoms on the '1' and '2' sides of the grain boundary.
 On the schematic below, indicate which grain is grain 1 and which one is grain 2.
 Suppose ${J}_{1\to 2}$ is the rate at which Al atoms hop from grain 1 to grain 2, and ${J}_{2\to 1}$ is the rate at which atoms hop from grain 2 to grain 1. Calculate the ratio, ${J}_{1\to 2}/{J}_{2\to 1}$ at $T=500$K.
Transformation Kinetics
13.0.0.43
Does the time to 50% transformed increase or decrease with an increase in nucleation rate? Defend your answer without using any equations.
Interphase Interfaces
13.0.0.44
Consider a material with the orientational dependence of the surface energy shown in each of the 3 plots below. For each of these three materials, sketch the equilibrium shape that you would expect to obtain. On each drawing, indicate any interfaces that you expect to be coherent.
13.0.0.45
Consider the shapes of the particles in the simulations below of misfitting particles in an elastically anisotropic system. The left column is the entire system, whereas the right column is a magnification of a small region of the figure in the left column. These are snapshots taken as function of time while the particles are growing. Are these cuboidal shapes due to elastic stress, an anisotropic interfacial energy, or both?
13.0.0.46
Explain the structure and energies of coherent, semicoherent and incoherent interfaces, paying particular attention to the role of orientation relationships and misfit.
13.0.0.47
Explain why fully coherent precipitates tend to lose coherency as they grow.
13.0.0.48
Why do very small precipitates tend to have coherent interfaces?
13.0.0.49
A thin film of Zn with an HCP crystal structure is deposited on a Ni FCC substrate with a {111} orientation. Which plane of the HCP crystal would you expect to contact the {111} Ni surface?
13.0.0.50
Given an example of an interface between two crystals that that displays a very large change in free energy with a change in the orientation of the interface.
13.0.0.51
Consider an FCC metal (metal A) with a surface energy of 1 $J/{m}^{2}\mathrm{.}$ An HCP metal (metal B) with a surface energy of 0.7 $J/{m}^{2}$ is deposited onto the {111} surface of metal A. Assume that the atomic diameter of the HCP metal is 3% larger than the atomic diameter of the FCC metal, and that the chemical component of the interfacial energy between the two metals is 0.2 $J/{m}^{2}\mathrm{.}$
 For B layers that are sufficiently thin, do you expect that a coherent interface will form between the A and B materials? Justify your answer.
 How do you expect the interface between the A and B metals to change as the thickness of the B layer increases?
 Do you expect thick films to remain continuous, or will isolated drops of B be formed on the surface. Describe any assumptions that you make.
13.0.0.52
Consider the vacancy shown below, for a simulation of 'red' and 'blue' atoms that are undergoing phase separation. Is the vacancy more likely to move to the right or to the left? Justify your answer.
13.0.0.53
Consider the tilt boundary shown in the image to the left. On the axes on the right, sketch the relationship between the grain boundary free energy and the tilt angle that you expect to observe for values of theta between 0 and 10 ${}^{\u25cb}$.
(5
13.0.0.54
Suppose you need to apply a coating to a surface, and you want the coating to spread as a smooth uniform film for all thicknesses. You have a choice of three different coatings, which have the thicknessdependent free energies shown below. Which material to you choose, and why?
14 3161 Simulation Exercise: Monte Carlo Simulation of Decomposition in a Binary Alloy
14.1 Background
14.1.1 Scientific problem
 We mix together the same number of moles of elements A and B to obtain a homogeneous alloy at some temperature above ${T}_{c}$.
 The temperature is reduced to ${T}_{0}$.
 The temperature is held fixed at ${T}_{0}$, and the system evolves to form two different phases, with compositions ${X}_{B}^{\alpha}$ and ${X}_{B}^{\beta}$.
14.1.2 Atomistic Monte Carlo Model
In this section, we introduce the Atomistic Monte Carlo model that we will use to model the decomposition of the AB alloy.
14.1.4 Monte Carlo model
The thermodynamic evolution of the alloy is modeled with a Monte Carlo process. The principle of Monte Carlo simulations is to model the AB alloy evolution in a statistic way. To understand this model we can consider individual jumps of a vacancy into one of the $z$ nearest neighbor positions. Within a certain specified time step, $\Delta t$, these different possible jumps occur with a probability ${\mathrm{varGamma}}_{\mu}$ where $\mu $ is an index that indicates which direction the vacancy will move. In a simple cubic lattice, for example, $z$ = 6, and the 6 values of $\mu $ correspond to jumps in the positive and negative x, y and z directions. The sum over all possible jump probabilities in the statistical time must sum to 1:
To figure out which direction the vacancy moves, we draw a random number ${r}_{n}$ between 0 and 1. The jump performed by the system during the time $\Delta t$ is the ${k}^{th}$ one such that the following condition holds:
Probabilities of transitions ${\mathrm{varGamma}}_{\mu}$ are related to the energetic barrier associated with vacancy motion, which we refer to as $\Delta {E}_{\mu}$. Because vacancy hopping is a thermally activated process, we can use an Arrhenius rate expression:
$$\begin{array}{cc}{\mathrm{varGamma}}_{\mu}={\mathrm{varGamma}}_{0}exp\left(\frac{\Delta {E}_{\mu}}{{k}_{B}{T}_{0}}\right)& (14.3)\end{array}$$
where ${\mathrm{varGamma}}_{0}$ is a constant, ${k}_{B}$ is Boltzmann's constant and ${T}_{0}$ is the temperature of the system.
The energy barrier is the difference between the maximum energy of the system during the jump (the position of the migrating atom at this maximum energy is called the saddle point) and the energy of the system before the jump.
$$\begin{array}{cc}\Delta {E}_{\mu}={E}^{SP}{E}^{ini}& (14.4)\end{array}$$
To compute the energy barriers of the different possible jumps $\Delta {E}_{\mu}$, we have to use an energetic model. In Monte Carlo simulations, we usually use an Ising model or Broken bond model. In this energetic model, we assume that the total energy of the system is equal to the sum of interaction energies ${\u03f5}_{ij}$ between the different elements (atoms of type A and B and vacancies V) placed on the lattice sites.
$$\begin{array}{cc}{E}_{\nu}={\displaystyle {\mathrm{varSigma}}_{ij}{\u03f5}_{ij}}& (14.5)\end{array}$$
With this energetic model, the migration barrier of an exchange between an element $X$ and the vacancy $V$ becomes:
where ${\u03f5}_{Ak}^{SP}$ are interaction energies between the atom migrating and its neighbors at the saddle point, ${\u03f5}_{Ai}$ are interaction energies between the atom migrating and its neighbors before the jump and ${\u03f5}_{Vj}$ are interaction energies between the vacancy and its neighbors before the jump. The indices i, j and k indicate the following neighbors:
Index

Meaning

i

nearest neighbors of the migrating atom before the jump

j

nearest neighbors of the vacancy before the jump

k

nearest neighbors of migrating atom at the saddle point

In theory, the range of interaction distances between elements are unlimited. In practice, we usually restrict these interactions to first and sometimes second nearest neighbors.
 3 AV interactions
 1 BV interaction
 3 AB interactions
 9 AA interactions
 12 BB interactions
Therefore,
${E}_{\nu}=3{\u03f5}_{AV}+1{\u03f5}_{BV}+3{\u03f5}_{AB}+9{\u03f5}_{AA}+12{\u03f5}_{BB}$. If we suppose that the vacancy exchange its position with the B atom on its left side, the configuration of the system at the saddle point is the one presented in figure
14.5.
In this configuration, the system has:
 2 BB interactions at the saddle point
 2 AB ineractions at the saddle point
 3 AB interactions
 9 AA interactions
 9 BB interactions
so ${E}^{SP}=2{\u03f5}_{BB}^{SP}+2{\u03f5}_{BA}^{SP}+3{\u03f5}_{AB}+9{\u03f5}_{AA}+9{\u03f5}_{BB}$. The migration barrier of this jump is therefore:
$$\Delta {E}_{\nu}=2{\u03f5}_{BB}^{SP}+2{\u03f5}_{BA}^{SP}3{\u03f5}_{AV}1{\u03f5}_{BV}3{\u03f5}_{BB}$$
Here we assume that the two elements A and B have the same simple cubic lattice. We model the AB alloy as a matrix in 2D with $nx$ rows and $ny$ columns and with periodic boundary conditions on its edges. To simplify the problem, we introduce only one vacancy in the lattice (so 1 vacancy for $nx\times ny$ sites), initially located in the middle of the matrix. As we only interest ourselves to the thermodynamic evolution of the system (and not to its kinetic evolution), we assume that the alloy evolves with normalized time steps of 1 until a maximum time ${t}_{max}$. At each time step, the vacancy exchanges its position with one of its neighbors.
To simplify the energetic model we suppose that the sum of interaction energies between the atom migrating and its neighbors at the saddle point ${\sum}_{k}}{\u03f5}_{Xk}^{SP$ is a constant equal to $3\phantom{\rule{6px}{0ex}}eV$. In addition, we suppose that ${\u03f5}_{AA}={\u03f5}_{BB}={\u03f5}_{AV}={\u03f5}_{BV}=0\phantom{\rule{6px}{0ex}}eV$. The only interaction which can be different from zero is thus ${\u03f5}_{AB}$.
The free enthalpy of the alloy is expressed by
$$\begin{array}{cc}\Delta {G}_{mix}=\Omega {X}_{A}{X}_{B}T\Delta {S}_{mix}& (14.7)\end{array}$$
with $\Omega $ the ordering energy of the alloy and $\Delta {S}_{mix}$ the configurational entropy of mixing of the alloy given by :
$$\begin{array}{cc}\Delta {S}_{mix}={k}_{B}\left[{X}_{A}ln{X}_{A}+{X}_{B}ln{X}_{B}\right]& (14.8)\end{array}$$
For a symmetrical miscibility gap, the ordering energy is
where ${T}_{C}$ is the critical temperature of the miscibility gap (${T}_{C}=1000\phantom{\rule{6px}{0ex}}K$ in this study). In broken bond models with only first nearest neighbors interactions we have:
where $z$ is the number of first nearest neighbors for a given site.
In this section we translate the problem described previously in an algorithm scheme. As we are modeling an evolution according to time, our code will contain an initial state and an incremental loop on time which will start from the initial time (${t}_{0}$) and finish at a final time (${t}_{max}$). During the time loop (for example between time ${t}_{n}$ and ${t}_{n+1}$), the code will repeat the same operations which will make the matrix go from the configuration at ${t}_{n}$ to the one at ${t}_{n+1}$. In this code we suggest that the system evolves with the following steps in the time loop:
 Evolution of time from ${t}_{n}$ and ${t}_{n+1}$
 Computation of jump frequencies of all possible jumps ${\mathrm{varGamma}}_{\mu}$
 Drawing of a random number ${r}_{n}$ and choice of a jump according to Eq. 14.2.
 Completion of chosen jump: exchange of position between vacancy and nearest neighbor chosen.
14.2 Exercise
Random walks
In this first work, we model the evolution of the system if the equilibrium configuration of the alloy is an homogenized state. As we only interest ourselves to the thermodynamic evolution of the system (and not to its kinetic evolution), we assume that the alloy evolves with normalized time steps of 1 until a maximum time ${t}_{max}$. At each time step, the vacancy exchanges its position with one of its neighbors. The vacancy can exchange its position with all its first nearest neighbors $X$ (and only its first nearest neighbors). The difference is that in this section we suppose that all exchanges have the same jump frequency ${\mathrm{varGamma}}_{XV}$. This is called a“ random walk”.
14.2.1 Preliminary work
 Consider a vacancy located on the lattice site $\left(xv,yv\right)$ as in Figure (14.6). In this figure, identify the first nearest neighbors of the vacancy by numbers and give the coordinates of these neighbors according to $\left(xv,yv\right)$.
 Suppose that all exchanges of the vacancy with its first nearest neighbors have the same jump frequency. Using equation (14.1), give the probability of a given jump ${\mathrm{varGamma}}_{\mu}$.
14.2.2 Simulation
 Create a folder for this MATLAB project. Open a new script in Matlab and save it in your folder as “part1.m”.
 We first write the initial state of the system in the file part1.m. Save the matrix given in the file called 'matini.mat' available from the following link:
https://www.dropbox.com/s/y4o2q3v53ffwinw/matini.mat?dl=0
Load this matrix in part1.m as “matrix”. Define $nx$ and $ny$ as the number of rows and column respectively of matrix. In this matrix, elements A are identified by a number 1 and elements B are identified by a number 2. Place a vacancy (identified by a 0) in the middle of the matrix:
 define in part1.m the coordinates $\left(xv,yv\right)$ (where $xv$ is the row and $yv$ the column of the vacancy position) as the coordinates at the middle of the matrix.
 Place a 0 in the matrix at the appropriate coordinates $\left(xv,yv\right)$ .
Initialize time $t$ to 0.
 If the matrix has the configuration of figure 14.6, what does the matrix in Matlab look like (with the numbers)? (Include a printout of the matrix).
 Create a loop on time $t$ where time evolves by steps of 1 as long as $t$ remains lower than ${t}_{max}$. Place ${t}_{max}=10$. We now have the part 1 in the algorithm (see section 14.1.7). Attach part1.m that includes all of the steps so far.
 We now have to create the next part of the algorithm: the computation of the jump frequency of all possible jumps. (Remark: in this random walk program, this part could be placed outside of the time loop since all jumps have the same frequency. However, we include it in the time loop to prepare the second part of the problem where we will have to compute the ${\mathrm{varGamma}}_{XV}$ according to the environment). In the program, we call $Gamma$ the vector such that $Gamma\left(i\right)$ is the jump frequency of the exchange $i$. Use a “for” loop to compute the values of the different $Gamma\left(i\right)$ components.
 We now have to choose a jump amount the different possibilities. For this, we suggest the MATLAB code shown below  just a single line that results in a random integer between 1 and 4:
njump = randi([1 4], 1, 1)
Run this command 5 times and write down the numbers you get for njump. Does this make sense?
 For the chosen jump, identify in your code by $\left(xn,yn\right)$ the coordinates of the corresponding nearest neighbor according to $\left(xv,yv\right)$. For this, we suggest you to define a matrix ($2\times z$) of the different possible evolutions (for example $\left(\begin{array}{c}+1\\ 0\end{array}\right)$ or $\left(\begin{array}{c}0\\ 1\end{array}\right)$ ) and to write $\left(xn,yn\right)$ according to $\left(xv,yv\right)$ and the column of the matrix corresponding to the jump.
 We use periodic boundary conditions in this model (see part 14.1.3). For a site $\left(x,y\right)$, verify that the following function enables to apply boundary conditions presented in figure (14.2) $$\begin{array}{c}x=mod\left(x1,\phantom{\rule{6px}{0ex}}nx\right)+1\\ y=mod\left(y1,\phantom{\rule{6px}{0ex}}ny\right)+1\end{array}$$For this, respond to the following questions: what is the value of $x$ returned by this function if the $x$ in input is between 1 and $nx$ ? equal to 0? equal to $nx+1$? Apply this function to $xn$ and $yn$.
 Exchange types of elements corresponding to the vacancy an the neighbor migrating in the matrix.
 Update the vacancy coordinates to its new site.
 In this random walk model, what is the equilibrium state of the system? (Help: the fact that all the $Gamma\left(i\right)$ are equal induces that the migration barriers for all possible jumps $\Delta {E}_{\mu}$ are equal. From equation (14.6) it induces that all saddle point interactions ${\u03f5}_{Ak}^{SP}$ and ${\u03f5}_{Bk}^{SP}$ are equal, all atomatom interations are equal and ${\u03f5}_{AV}={\u03f5}_{BV}$. What is thus the value of the ordering energy $\Omega $ in equation (14.10)? And the value of ${T}_{C}$ ? So at any temperature, what is the equilibrium state of the system?)
 Test: Replace the initial matrix by a matrix of same size with all A atoms on the half left side and all B elements on the half right side. Print an image of this initial matrix. Make the code run until ${t}_{max}=1{0}^{6}$. What do you observe? Print an image of the final matrix.
14.2.3 Introduction of alloy thermodynamic properties
We now have to introduce the alloy thermodynamic properties in the code. We thus have to compute the jump frequency of possible exchanges between the vacancy and its neighbors according to the alloy thermodynamic properties.
 We recall here that ${\u03f5}_{AA}={\u03f5}_{BB}=0\phantom{\rule{6px}{0ex}}eV$. Express ${\u03f5}_{AB}$ according to the ordering energy $\Omega $ and then to the critical temperature ${T}_{C}$. Give a numerical value of ${\u03f5}_{AB}$ in eV if ${T}_{C}=1000\phantom{\rule{6px}{0ex}}K$.
 We analyze the migration barrier of an exchange between a vacancy V and one of its nearest neighbors $X$. We note $NA$ the number of $X$ first nearest neighbors of type A and $NB$ the number of $X$ first nearest neighbors of type B. How many first nearest neighbors does X have (we do not count the vacancy)? Express equation (14.6) according to $NA$, $NB$ and ${\u03f5}_{XA}$and ${\u03f5}_{XB}$. Using that ${\sum}_{k}}{\u03f5}_{Xk}^{SP}=3\phantom{\rule{6px}{0ex}}eV$ and that ${\u03f5}_{AA}={\u03f5}_{BB}={\u03f5}_{AV}={\u03f5}_{BV}=0\phantom{\rule{6px}{0ex}}eV$, simplify the equation obtained if X is an element A. Same question if X is an element B. We observe from these calculations that, to compute the migration barrier of a jump, we need to know the type of the element of the exchange (so the type of $X$) and the type of all $X$ first nearest neighbors (to compute $NA$ and $NB$).
 For a given vacancy position, we want to compute the jump frequency of the jump $i$ (so $Gamma\left(i\right)$). We note $X$ the vacancy neighbor corresponding to this jump. We start by computing $NA$ and $NB$ (the number of $X$ first nearest neighbors of type A and B). We note $\left(xn,yn\right)$ the position of $X$ and $\left(xnk,ynk\right)$ the coordinates of $X$ first nearest neighbor $k$ ($k$ goes from 1 to 3, the vacancy position is excluded from these nearest neighbors). We write $$\left(\begin{array}{c}xnk\\ ynk\end{array}\right)=\left(\begin{array}{c}xn\\ yn\end{array}\right)+nvec\left(k\right)$$where $nvec\left(k\right)$ is the column $k$ of the $2\times 3$ matrix of relative position of $\left(xnk,\phantom{\rule{6px}{0ex}}ynk\right)$ compared to $\left(xn,\phantom{\rule{6px}{0ex}}yn\right)$. Graph 14.7 gives the position of neighbors $X$ compared to the vacancy.$nvec=\left(\begin{array}{cccc}1& 0& 1& 0\\ 0& 1& 0& 1\end{array}\right)$
0For each of these jumps, associate the matrix $nveci$ of the relative position of $X$ first nearest neighbors.
$$\begin{array}{cc}\left(1\right)\phantom{\rule{6px}{0ex}}nneigh=\left(\begin{array}{ccc}0& 0& 1\\ +1& 1& 0\end{array}\right)& \left(2\right)\phantom{\rule{6px}{0ex}}nneigh=\left(\begin{array}{ccc}1& 1& 0\\ 0& 0& 1\end{array}\right)\\ \left(3\right)\phantom{\rule{10px}{0ex}}nneigh=\left(\begin{array}{ccc}1& 1& 0\\ 0& 0& 1\end{array}\right)& \left(4\right)\phantom{\rule{10px}{0ex}}nneigh=\left(\begin{array}{ccc}0& 0& +1\\ +1& 1& 0\end{array}\right)\end{array}$$
 Inside the loop to compute $Gamma\left(i\right)$ coefficients write the following steps:
 define by $\left(xn,yn\right)$ the vacancy neighbor corresponding to $i$ (use the $nvec$ matrix). Apply periodic conditions to $\left(xn,yn\right)$.
 Initialize $NA$ and $NB$ to zero. Compute $NA$ and $NB$ of the exchange by analyzing the type of element on all $\left(xnk,ynk\right)$ sites. To define the $nveci$ matrix corresponding to the jump, you can distinguish the different cases with ifstatements or you can use a structure with all the $nveci$ matrices and load the one corresponding to the jump. Don't forget to apply boundary conditions to $\left(xnk,ynk\right)$.
 Express the migration barrier of each jump depending on the type of the neighbor $X$ (located on $\left(xn,yn\right)$) and $NA$ and $NB$. Compute the jump frequency associated to this migration barrier (place the temperature to an arbitrary valuedon't forget to define ${\u03f5}_{AB}$ in your code).
 Normalize the $Gamma$ vector to 1 so that the sum of $Gamma\left(i\right)$ is equal to 1.
 Analytic calculation: Suppose that for a given position, the vacancy can exchange it's position with either of 2 different A atoms. One on them is in a local configuration with NB=0 (the jump frequency of this exchange is noted ${\mathrm{varGamma}}_{NB=0}$) and the other one is in a local configuration with NB=3 (the jump frequency of this exchange is noted ${\mathrm{varGamma}}_{NB=3}$). Compute ${\mathrm{varGamma}}_{NB=3}/{\mathrm{varGamma}}_{NB=0}$ for T=100K and for T=2000K. Explain why these ratios are consistent with the alloy phase diagram.
 Place the temperature to 100K. Run the simulation until ${t}_{max}=1{0}^{6}$. What do you observe?
Nomenclature
 $C_{0}$
 Overall concentration of atoms
 $C_{i}$
 Concentration of component i
 $D_{i}$
 Intrinsic diffusion coefficient for component i (m$^{2}$/s)
 $D_{i}^{*}$
 Tracer diffusion coefficient for component i (m$^{2}$/s)
 $E$
 Energy (J)
 $E_{s}$
 Dislocation Energy (J)
 $F_{s}$
 Force per unit length acting on a dislocation (N)
 $F_{s}^{\tau}$
 Stressinduced force per unit length acting on a dislocation (N/m)
 $F_{s}^{r}$
 Curvatureinduced force per unit length acting on a dislocation (N/m)
 $G$
 Shear modulus (Pa)
 $H_{i}$
 Henry' law coefficient for component i (dimensionless)
 $J_{i}$
 Diffusive flux of component i with respect to a cooridnate system fixed to the lattice planes (atoms/m$^{2}$/s)
 $J_{i}^{'}$
 Diffusive flux of component i with respect to a cooridnate system fixed to the external dimensions of the sample (atoms/m$^{2}$/s)
 $L_{m}$
 Molar heat of sublimation (Joules)
 $M_{i}$
 Mobility of component i (units of velocity/force)
 $R$
 Gas constant (8.314 J/mole$\cdot$ K)
 $S_{m}^{L}$
 Molar entropy of the liquid phase
 $S_{m}^{S}$
 Molar entropy of the critical nucleus
 $T$
 Absolute temperature (K)
 $T_{s}$
 Dislocation line tension (N or J/m)
 $V$
 Volume (m$^{3}$)
 $V_{m}^{S}$
 Molar volume of the solid phase
 $\ell_{i}^{D}$
 Diffusion length for component i (m)
 $\gamma_{\alpha\beta}$
 Interfacial free energy between $\alpha$ and $\beta$ phases (J/m$^{2}$)
 $\hat{s}$
 Unit vector directed along a dislocation core (dimensionless)
 $\mu_{i}$
 Chemical potential of component i (J/mole)
 $\sigma$
 Tensile stress (Pa)
 $\tau$
 Shear stress (Pa)
 $\tau_{crss}$
 Critical resolved shear stress (Pa)
 $\tau_{crss}^0$
 critical resloved shear stress in the absence of dislocations
 $\tau_{rss}$
 Resolved shear stress (Pa)
 $\tilde{D}$
 Interdiffusion coefficient (m$^{2}$/s)
 $\vec{b}$
 Burgers vector (m)
 $\vec{n}_{d}$
 Vector cross product of $\vec{s}$ and $\vec{b}$ (m)
 $a_{i}$
 Activity coeffiienct for component i (dimensionless)
 $b$
 Magnitude of $\vec{b}$ (m)
 $e_{xy}$
 Shear strain in the xy plane
 $k_{B}$
 Boltzmann's constant (1.38x10$^{23}$ J/K)
 $n_{i}$
 Total number of atoms of component i.
 $r$
 Particle radius (m)
 $v_{\ell}$
 Velocity of the lattice planes with respect to a laboratory coordinate system (m/s)
 F
 Helmholtz free energy
 G
 Gibbs free energy
Index
 Burgers circuit: 1
 Burgers vector: 1
 Climb (of a dislocation): 1
 Contact angle: 1
 Critical resolved shear stress: 1
 limiting value in the absence of dislocations: 1
 Cross slip: 1
 Diffusion equation: 1
 Dislocation
 Dislocation density: 1
 Dislocations
 $\overrightarrow{b}\times \overrightarrow{s}$ cross product: 1
 Screw: 1
 Sense vector: 1
 Edge dislocation: 1
 error function: 1
 Fick's first law: 1
 FrankRead source: 1
 Gibbs free energy: 1
 Glide: 1
 Glide plane: 1
 HallPetch relationship: 1
 Heat of sublimation: 1
 Helmholtz free energy: 1
 Henry's law: 1
 Henry's law coefficient: 1
 Heteroepitaxy: 1
 Interdiffusion coefficient: 1
 Intrinsic diffusion coefficient: 1
 JohnsonMehlAvramiKolmogorov (JMAK) equation: 1
 Lagrange Multipliers: 1
 Laplace pressure equation: 1
 LaplaceYoung equation: 1
 MATLAB
 fsolve: 1
 Function plot: 1
 loading and plotting data from .mat file: 1
 polar plot: 1
 setting plotting defaults: 1
 Vacancy diffusion simulation: 1
 Wulff Construction: 1
 Miller indices: 1
 mobility coefficient: 1
 Partial wetting: 1
 Resolved shear stress: 1
 Screw Dislocations: 1
 Sense vector: 1
 Slip Plane: 1
 Thin Film Growth: 1
 Three phase contact lines: 1
 Tilt boundary: 1
 Twin boundaries: 1
 Twin boundary: 1
 Twist boundary: 1
 Wulff construction: 1
15 3161 Labs
15.1 Laboratory 1: Diffusion in Substitutional CuNi Alloys
15.1.1 Objectives
 To observe diffusion in a CuNi diffusion couple.
 To determine if these observations are consistent with a compositiondependent interdiffusion coefficient, expected for diffusion in substitutional alloys.
 To begin to model the diffusion process using MATLAB.
15.1.2 Introduction
In the case of packcarburization, we were able to make the assumption that diffusivity of carbon in iron was independent of composition. For substitutional alloys, this is not the case. The interdiffusion coefficient in this case is composition dependent and related to the intrinsic diffusion coefficients as follows:
$$\tilde{D}={X}_{a}{D}_{b}+{X}_{b}{D}_{a}$$
In addition, in situations where ${D}_{a}$ and ${D}_{b}$ differ from one another, there will be a net vacancy flux in the material, giving rise to the motion of an inert set of markers that can be observed experimentally.
15.1.3 Samples
Samples have been prepared using two techniques:
 electroplating of nickel layers onto copper, and
 welding NiCu sandwich layers.
In both cases, Mo wires were placed at the interface, to mark the position of the original interface; however in the case of the electroplated samples, these wires sometimes shifted away from the surface during plating. After electroplating/ welding, the samples were sealed in evacuated quartz tubes to prevent oxidation, and annealed at 1000°C for 4, 16, and 72 hours.
15.1.3.1 Laboratory Procedure
Refer to the class notes in addition to the paper describing the background and history of the Kirkendall effect [
6]. Look at the CuNi samples (annealed at 4, 16 and 72 hours at 1000 °C) under the optical microscope. Note that there are two types of sample: 1) copper strips wound with Mo wire which were nickelelectroplated and 2) a welded "sandwich" of nickel with outer copper layers and rolled molybdenum "marker wires" at the interface. Note that in (1) the Mo was not secured to the copper strip wellenough to mark the original interface (this will be obvious in your observations). In (2) you will find enough pairs of wires that are nearly across from each other to measure the distance between markers as a function of time at elevated temperature. (Unfortunately, the weld broke on the unannealed (time = 0) samples; but you should be able to assess the three remaining samples quantitatively or at least semiquantitatively. Include these measurements with your other observations, as well as a discussion of what you expected. Discuss whether or not your observations and measurements are consistent with the Kirkendall effect. In future exercises we will be comparing these diffusion profiles to what we would expect from published values of the relevant diffusion coefficients. For now document your inclass observations, including well labeled sketches and micrographs.
15.2 Laboratory 2: Recovery, Recrystallization and Grain Growth in Cold Worked 70/30 Brass
15.2.1 Objectives
To observe the phenomena of recovery, recrystallization and grain growth. To understand the effect of processing on microstructure, specifically the effect of amount of coldwork on recrystallization and final grain size. To understand the time dependence of grain growth. To understand the predictions of the HallPetch relationship.
15.2.2 General Procedure: Week 1
You will be provided with brass (70%Cu, 30% Zn) that has been heated to 700° C for six hours, from the asreceived state and then rolled to reductions of ~ 15% and ~ 30%, as well as some brass that has not yet been rolled. Your groups will coldroll samples to similar reductions for the next group. The specified amount of coldwork will be introduced using the rolling mill.
 Measure the thickness and the Rockwell hardness of your asreceived and rolled samples. Choose an appropriate Rockwell scale over which you can anticipate measuring your sample after it is rolled – then subsequently annealed. Always check to make sure the load and indenter size correspond to the correct scale. Use a standard to check the tester.
 As a group, roll two samples, using the rolling mill, one to a reduction of ~ 3040%, a second to a lesser reduction, e.g. 1520%. Anticipate the target thickness before you begin rolling. Calculate target thicknesses for each reduction, assuming width does not change with rolling. Percent reduction (or percent coldwork) is defined as:
$$\begin{array}{cc}\%CW=\frac{{A}_{0}{A}_{d}}{{A}_{d}}\times 100& (15.1)\end{array}$$
which may be rewritten for this lab:
$$\begin{array}{cc}\%CW=\frac{{t}_{0}{t}_{d}}{{t}_{d}}\times 100& (15.2)\end{array}$$
where ${t}_{0}$ is the starting thickness and ${t}_{d}$ is the final thickness. Set aside for the next group.
 Remeasure hardness after rolling. (Make sure to measure a flat region. The sample should not deflect when the indenter is applied.)
 Section the rolled samples into about 8 pieces (~ 1cm long). Note that we will be interested in observing the transverse sections, defined in the figure below. Set aside a time = 0 sample; each of the other 1 cm long “coupons” will be annealed at a specified time at the temperature assigned to your group
 Record the temperature assigned to your group. T=___ degrees C.
 samples that have been annealed from 2 minutes, 8 minutes, 32 minutes….up to a week. You will be measuring and recording Rockwell hardness on each of these samples, then mounting them for polishing and etching.
 After reserving the time = 0 sample, place the remaining samples in the furnace assigned to your lab group. (All samples of both reductions, except t=0, should be annealed at the SAME TEMPERATURE
***Suggest (the entire group’s) annealing conditions by reviewing information available in the Metals Handbook, and by discussion with your lab mates & instructor. You want to achieve conditions under which you will observe partial to total recrystallization. Consider how you will need to vary the conditions to test the JohnsonAvramiMehl equation.
15.2.2.1 General Procedure: Week 2
 Make sure you have measured the Rockwell hardness of each annealed sample. Note that you should try to take all your hardness readings on the same scale.
 Mount transverse crosssections of each of the annealed samples, along with an unannealed piece in an acrylic mount for polishing. Follow the instructions for the autopolisher. Wash your sample carefully and ultrasonic between each step to avoid contaminating the wheels. (These are soft samples; it will be difficult to remove the scratches that are introduced by such contamination!)
 Etch to reveal grains. (Be careful; the different reductions and different temperatures of annealing may result in different etch rates.) Record a photomicrograph of each sample at an appropriate magnification.
 From your micrographs, calculate the volume fraction of recrystallized material, and the grain size of samples that are completely recrystallized.
 Measure the Vickers hardness of each sample (three indents, minimum, on each sample.)
15.2.3 InLab Questions DUE at the Beginning of Week 2:
Rolling, hardness testing and cutting will take some time. If you are waiting you may use time in lab to answer the following. Make sure you define all terms and cite sources:
 What equation describes the rate of grain growth?
 Refer to Chapter 3 of Shewmon and summarize the “Engineering Laws of Recrystallization” relevant to this experiment. (You may summarize all – then determine which you might be able to test vs. not able to test.)
 What equation describes the volume fraction of material recrystallized with time?
 How can the rate of recrystallization at a given temperature be determined?
 What is the HallPetch equation? Discuss the equation and any limitations.
15.2.4 Final Deliverable  Group PowerPoint Presentation
Your presentation will be judged on content, delivery (presentation style), neatness, completeness. You must submit a hardcopy of your presentation slides. Imagine you are presenting this to Prof. Voorhees and other MSE students who were not in lab; they are familiar with terms like grain size and hardness, but do not know the details of your sample preparation and what you are testing (i.e. which of the Engineering laws of Recrystallization you were able to test.) Length: 12 minutes. Each group member must participate.
Due: one week after completing inclass measurements.
 Refer to Chapter 3 of Shewmon; discuss whether or not the class data substantiates the “Engineering Laws of Recrystallization,” i.e. how do hardness, grain size, volume fraction of recrystallized material vary with the amount of coldrolling, and time of anneal? Plot hardness (Rockwell is OK, here) as a function of annealing time for both reductions, including time = 0 values. Explain changes in hardness by comparison with micrographs.
 Estimate the recrystallization rate for your group’s annealing temperature: Rate = 1/(time for volume fraction transformed = 0.5).
Note: We will try to use the information from different groups to compare recrystallization as a function of temperature. If you have enough points (this is unlikely), you may be able to fit the Avrami (JMAK) equation:
$$\begin{array}{cc}y\phantom{\rule{6px}{0ex}}\left(fraction\phantom{\rule{6px}{0ex}}recrystallized\right)=1exp\left(k{t}^{n}\right)& (15.3)\end{array}$$
 Make sure you use actual – not target reductions – when discussing your results. Doublecheck that the reduction is, for example, 40%, not 70%.
 For samples in which complete recrystallization was observed – does the HallPetch relationship hold? Assume that hardness is proportional to yield strength (see next page). The Hall Petch equation states that the yield stress, ${\sigma}_{y}$, is increases linearly with ${d}^{1/2}$, where $d$ is the average grain size:
where${\sigma}_{0}$ and ${k}_{y}$ are constants for a given material. Note that you do not have to confine comparisons to a single recrystallization; use all the samples available that have recrystallized. (It tends not to be valid for very large or very small grains.)
 For completely recrystallized samples, is normal grain growth observed? Measure grain sizes for recrystallized material at a given reduction and determine the exponent for grain growth as a function of annealing time at a given temperature: $$\begin{array}{cc}{d}^{n}{d}_{0}^{n}=Kt& (15.5)\end{array}$$
Solve to see if $n$ is greater than or equal to 2, as expected. Note that at the start of recrystallization, the grain size is infinitesimally small.
 Estimate the average grain size by counting, on a micrograph, screen or the specimen itself, the number of grains intercepted by one or more straight lines sufficiently long to yield at least 50 intercepts. Select the magnification such that this can be done in a single field.
 Make counts on 35 blindly selected, widely separated fields.
 Use a factor of 1.5 to determine the average grain size from the lineal intercept length.
15.2.4.2 Hall–Petch determination:
 Measure Vickers hardness.
 Use hardness and grain size to determine if the HallPetch relationship holds true for your data. (Plot HV vs. $1/\sqrt{d}$)
 You can use Vickers hardness to calculate the Yield strength of brass. Assume 1/3 of the applied load in a Vickers Hardness test plastically deforms the sample and use the appropriate conversion factor ($CF$) to convert to MPa:
$${\sigma}_{y}=\frac{HV\phantom{\rule{6px}{0ex}}\left(kg/m{m}^{3}\right)}{3}\times CF$$
Q – Are your values of yield strength within a reasonable range? Compare to typical values (Metals Handbook)
15.2.4.3 Empirical relationship between Rockwell B and Vickers hardness (kg/mm${}^{2}$).
Note that it is best to measure the Vickers hardness directly. The following relationship between the Vickers hardness ($HV)$ and Rockwell B hardness (${R}_{b}$) is obtained from ASTM Standard E140 (table 4, Conversion data for Cartridge brass), Annual Book of ASTM Standards, volume 3.01, 1989:
$$\begin{array}{cc}HV=0.002{R}_{b}^{3}0.0092{R}_{b}^{2}+0.8163{R}_{b}+52.865& (15.6)\end{array}$$
15.3 Laboratory 3: Surface Energy and Contact Angles
15.3.1 Objectives
 To understand what aspects of liquid behavior are determined by surface and interfacial energies.
 To understand how contact angles are used to characterize material surfaces.
15.3.2 Introduction
The properties of solid surfaces are often probed by measuring the ability of liquids to spread over the surface of a material. The relevant property is the contact angle,
$\theta $, illustrated in Figure
15.1a. If the droplet is small enough so that it is not affected by gravity, the radius of curvature,
$R$, of the droplet is uniform, and shape of the droplet is a spherical cap,
i.e., the portion of a sphere that exists above a specified plane. The relationship between the droplet height,
$h$, the basal radius of the droplet,
$r$, and the contact angle in this situation is as follows:
It is often useful to rewrite Eq. 15.8 in terms of the thermodynamic work of adhesion,
${W}_{adh}$, which describes the energy required to remove the liquid from the solid surface, replacing the solid/liquid interface with a liquid/air interface and a air/solid interface (see Figure
15.2):
Equation 15.10 indicates that we can know the quantitative interaction between the liquid and the solid if we are able to measure the liquid surface energy,
${\gamma}_{\ell}$ and the equilibrium contact angle,
${\theta}_{e}$. This purpose of this lab is to measure both of these quantities in some model systems and to show how these quantities can be easily modified. Before we do that, we need to talk about two important issues:
 The actual contact angle you will measure is almost certainly not going to be the equilibrium contact angle.
 Liquid surface energies are often measured by understanding the effect of gravity on a relatively small drop.
15.3.2.1 Nonequilibrium effects:
In reality, the situation is more complicated than is implied by Eq.
15.8, and the contact angles you will measure depend on a whole bunch of factors, in addition to the surface and interfacial energies. Factors like surface roughness and surface inhomogeneities on the nanometer scale cause the measured contact angles differ from the
${\theta}_{e}$, and to depend on the details of the way the experiment is done. When a droplet is originally applied to the materials surface and the droplet volume is increasing with time, the contact angle is referred to an advancing contact angle,
${\theta}_{a}$. The receding contact angle,
${\theta}_{r}$, corresponds to the opposite situation, where the droplet size is shrinking. The advancing contact angle is larger than
${\theta}_{e}$ and the receding contact angle will be less than
${\theta}_{e}$:
$$\begin{array}{cc}{\theta}_{r}<{\theta}_{e}<{\theta}_{a}& (15.11)\end{array}$$
Generally you'll want to report both advancing and receding angles in your work. The difference between ${\theta}_{a}$ and ${\theta}_{r}$ is an important parameter referred to as the contact angle hysteresis, and controls the tendency of droplets to stick to an inclined surface.
15.3.2.2 The Effect of Gravity and the Measurement of ${\gamma}_{\ell}:$
We know from experience that Eq.
15.7 can't work for very large droplets. Eventually, gravity flattens the droplet and the drop height,
$h$, no longer continues to increase as
$r$ gets larger and larger. This situation is as shown in the left part of Figure
15.3, where we show the behavior of small and large droplets sitting on a surface (sessile drops). The obvious question to ask here is 'how small is small'? and what controls the maximum value of
$h$ that can be obtained? The answer to this question is the capillary length,
${\lambda}_{c}$, which can be viewed as the radius of the spherical droplet for which the Laplace pressure inside the drop (
$2\gamma /R)$ is equal to the gravitational hydrostatic pressure at the bottom of the drop (
$2\rho gR$, where
$g$ is the gravitational acceleration and
$\rho $ is the liquid density). These pressures are equal to one another for
$R={\lambda}_{c}$, where
${\lambda}_{c}$ is given by the following:
A reasonable estimate of the surface free energy can be obtained by continuously injecting liquid through the syringe needle with the pendant drop geometry and measuring the critical droplet volume, ${V}_{c}$, where a droplet detaches and a new one is formed. Droplet detachment happens when the force corresponding to the surface tension around the perimeter of the droplet ($2\pi {R}_{m}{\gamma}_{\ell}$, where ${R}_{m}$ is the inner radius of the capillary) is equal to the gravitational force exerted by the droplet ($\rho g{V}_{c})$. By equating these two forces we get the following approximate expression for ${\gamma}_{\ell}$:
15.3.3 Samples
The following materials will be provided:
 Clean water
 A soap solution that can be added to the water to reduce it's surface energy
 A variety of materials expected to have different contact angles with water
 Access to a UVozone cleaner for surface modification
15.3.3.1 Laboratory Procedure and Writeup
References
1, "Error Function", Wikipedia (2017).
2, "Lagrange Multiplier", Wikipedia (2018).
3Daniel Carvajal, Evan J. Laprade, Kevin J. Henderson, and Kenneth R. Shull, "Mechanics of Pendant Drops and Axisymmetric Membranes", Soft Matter 7 (2011), pp. 10508.
4 E04 Committee, "Test Methods for Determining Average Grain Size", ASTM International (2013).
5Maochang Liu, Dengwei Jing, Zhaohui Zhou, and Liejin Guo, "TwinInduced OneDimensional Homojunctions Yield High Quantum Efficiency for Solar Hydrogen Generation", Nat Commun 4 (2013).
6Hideo Nakajima, "The Discovery and Acceptance of the Kirkendall Effect: The Result of a Short Research Career", JOM 49, 6 (1997), pp. 1519.
7Ad Smigelskas and Eo Kirkendall, "Zinc Diffusion in AlphaBrass", Transactions of the American Institute of Mining and Metallurgical Engineers 171 (1947), pp. 130142.
Index
 Burgers circuit: 1
 Burgers vector: 1
 Climb (of a dislocation): 1
 Contact angle: 1
 Critical resolved shear stress: 1
 limiting value in the absence of dislocations: 1
 Cross slip: 1
 Diffusion equation: 1
 Dislocation
 Dislocation density: 1
 Dislocations
 $\overrightarrow{b}\times \overrightarrow{s}$ cross product: 1
 Screw: 1
 Sense vector: 1
 Edge dislocation: 1
 error function: 1
 Fick's first law: 1
 FrankRead source: 1
 Gibbs free energy: 1
 Glide: 1
 Glide plane: 1
 HallPetch relationship: 1
 Heat of sublimation: 1
 Helmholtz free energy: 1
 Henry's law: 1
 Henry's law coefficient: 1
 Heteroepitaxy: 1
 Interdiffusion coefficient: 1
 Intrinsic diffusion coefficient: 1
 JohnsonMehlAvramiKolmogorov (JMAK) equation: 1
 Lagrange Multipliers: 1
 Laplace pressure equation: 1
 LaplaceYoung equation: 1
 MATLAB
 fsolve: 1
 Function plot: 1
 loading and plotting data from .mat file: 1
 polar plot: 1
 setting plotting defaults: 1
 Vacancy diffusion simulation: 1
 Wulff Construction: 1
 Miller indices: 1
 mobility coefficient: 1
 Partial wetting: 1
 Resolved shear stress: 1
 Screw Dislocations: 1
 Sense vector: 1
 Slip Plane: 1
 Thin Film Growth: 1
 Three phase contact lines: 1
 Tilt boundary: 1
 Twin boundaries: 1
 Twin boundary: 1
 Twist boundary: 1
 Wulff construction: 1