Dimitri D. Vvedensky The Blackett Laboratory, Imperial ...

Solutions to Exercises inTransformations of Materials

Dimitri D. VvedenskyThe Blackett Laboratory,

Imperial College London, London SW7 2AZ

Copyright c©2019 by Dimitri D. Vvedensky

Contents

Preface 1

1 Overview of Thermodynamics 3

2 Brownian Motion, Random Walks, and the Diffusion Equation 17

3 Atomic Diffusion in Solids 33

4 Spinodal Decomposition 45

5 Nucleation and Growth 63

6 Instabilities of Solidification Fronts 79

7 Diffusionless Transformations 85

i

ii

Preface

This book contains solutions to all the exercises in the accompanying book, Transformations ofMaterials. Problem sets with these exercises were assigned each week during the 8-9 weeks ofthe course (3 lectures per week). Each problem set consisted of 5-8 exercises, which were one ofthree types: (i) examples of the methods and concepts presented in class, (i) extensions of resultsobtained in class, or (iii) the derivation of new results. Approximately half of the problems ineach set were designated for “rapid feedback”. These were marked by a teaching assistant, whopresented the solutions to the class at weekly sessions and handed out the solutions to all exercises.In most cases, there is more detail in these solutions than that expected from the students, in thatvery few steps are omitted, and diagrams are provided in support of the solutions. In effect, thesesolutions provided a separate learning stream for the course.

A word about usage. The exercises formed an integral part of the course and their solutionsare provided as an aid to self-study. They are a key component of the learning process, but shouldbe referred to only after an attempt has been made to solve the particular problem. Many of theproblems can be easily modified by an instructor to suit the needs of a particular course – in thatsense, they provide a framework for the learning outcomes of an introductory course on the kineticsof materials.

The exercises in this book were developed over the five-year period (2009–2013) I taught thiscourse in the Centre for Doctoral Training in the Theory and Simulation of Materials at Imperial.I am grateful to the many students and teaching assistants who found algebraic and typographicalerrors and made suggestions for the improvement of particular exercises. Special thanks go toHannes Jonsson for permission to use Exercise 3.2 and its solution from his unpublished lecturenotes.

Equations and figures in this book are labelled and referred to by the prefix S, while equationsand figures in the accompanying book are referred to by their labels in that book.

Dimitri D VvedenskyApril 2019

1

2

Chapter 1

Overview of Thermodynamics

Exercise 1.1

First Law: You can’t win. The First Law is a statement about the conservation of energy.Not being able to win means that you cannot get more energy out of a system than you putin to that system.

Second Law: You can’t break even. The Second Law is a statement about entropy pro-duction during a processes. In particular, the Second Law states that converting betweendifferent forms of energy in a real, i.e. irreversible, process results in a loss of usable energythrough the production of entropy. Reversible processes, being an unattainable idealization,fall outside the scope of physical processes. Hence, in any real process, you always ‘lose’energy.

Third Law: You can’t get out of the game. There are three forms of the third law. The mostgermane for the statement here mandates the impossibility of reaching absolute zero, whereentropy differences vanish and one could ‘break even,’ for example, by having a perfectlyefficient heat engine. Hence, one cannot ‘get out of the game’ in which entropy alwaysincreases, which results in the loss of usable energy.

Exercise 1.2

(a) We have two equal amounts of substances A and B that are initially thermally isolated andhave temperatures TA and TB. These systems are then brought into thermal contact and al-lowed to equilibrate. We assume that the two systems together are thermally isolated. Nowork is done in the process and, since the total system is itself isolated, the total energy can-not change during this process, nor can there be any heat exchanged with the surroundings.Thus,

QA +QB = 0, (S1.1)

3

4 Solutions to Exercises in Transformations of Materials

where QA and QB represent the heat transfers of A and B during equilibration. These quanti-ties can be calculated from the constant-volume specific heats as

QA =∫ Tf

TA

CA dT =CA(Tf −TA) ,

(S1.2)

QB =∫ Tf

TB

CB dT =CB(Tf −TB) ,

since CA and CB are assumed to be independent of temperature. By substituting these ex-pressions into (S1.1),

CA(Tf −TA)+CB(Tf −TB) = 0, (S1.3)

and solving for Tf , we obtain

Tf =CATA +CBTB

CA +CB. (S1.4)

(b) The entropy changes of A and B can now be calculated as

∆SA = SA(Tf )−SA(TA) =CA

∫ Tf

TA

dTT

=CA ln(

Tf

TA

),

(S1.5)

∆SB = SB(Tf )−SB(TB) =CB

∫ Tf

TB

dTT

=CB ln(

Tf

TB

).

Since the entropy is a function of state, we can calculate the entropy difference betweenthe initial and final states of A and B even though the process is not quasi-static. The totalentropy change during the process is thus given by

∆S = ∆SA +∆SB =CA ln(

Tf

TA

)+CB ln

(Tf

TB

). (S1.6)

By using the fact that lnx≤ x−1 (Fig. S1.1), we have

ln(

1x

)=− lnx≤ 1

x−1, (S1.7)

orlnx≥ 1− 1

x. (S1.8)

Using this inequality, we can write equation (S1.6) as

∆S =CA ln(

Tf

TA

)+CB ln

(Tf

TB

)≥CA

(1− TA

Tf

)+CB

(1− TB

Tf

)=

1Tf

[CA(Tf −TA)+CB(Tf −TB)

]= 0, (S1.9)

where we have used equation (S1.3) in the last step. Thus, ∆S≥ 0.

Overview of Thermodynamics 5

0.0 0.5 1.0 1.5 2.0-1.0

-0.5

0.0

0.5

1.0

x

lnx,

x1

lnx

x1

1

Figure S1.1: Comparison of the functions lnx and x−1, showing that lnx ≤ x−1 for all x exceptx = 0, where the two functions are equal. The shaded region indicates the difference between thetwo functions, and the dot at x = 0 corresponds to the point where the two functions are equal.

Exercise 1.3The Legendre transformation φ of a function y = f (x) is defined by

φ(m) = f (x)−mx , (S1.10)

wherem =

d fdx

. (S1.11)

Whereas the representation y = f (x) of the function is a curve in the x-y plane, the Legendretransformation provides an alternative representation as the envelope of the family of tangent linesin which φ(m) is the y-intercept of the line with slope m, i.e.

y = mx+φ(m) . (S1.12)

As an explicit example of the Legendre transformation, consider the function f = x2. Sincem = 2x, we have that x = 1

2m, so the Legendre transformation (S1.10) of x2 is

φ(m) =(1

2m)2−m

(12m)=−1

4m2 . (S1.13)

The original and Legendre representations of x2 are shown in Fig. S1.2. The left panel showsthe representation of y = x2 as a curve in the x-y plane. The right panel shows the Legendretransformation of x2 as the envelope of the family of straight lines. Each of these lines intersectsthe curve x2 once and only once, so the information provided by the entire family is the same asthat of the original curve.


-2 -1 0 1 20

1

2

3

4

x

y

-2 -1 0 1 20

1

2

3

4

x

y

1

Figure S1.2: (Left panel) The curve y = x2 in the x-y plane. (Right panel) the Legendre transfor-mation of x2 as the envelope of the family of straight lines.

Exercise 1.4To understand the different level of information contained in U(T ,V ) and U(S,V ), we begin withU(T ,V ), which for a monatomic ideal gas, is given by:

U(T ,V ) =32

NkBT . (S1.14)

To express this form of U as a function of S and V , we observe from the fundamental equation ofthermodynamics,

dU = T dS−PdV , (S1.15)

that T is given as a function of S and V through the relation

T (S,V ) =

(∂U∂S

)V

. (S1.16)

Thus, substituting this expression for T into the equation of state in (S1.14), yields

U(S,V ) =U [T (S,V ),V ] =3NkB

2

(∂U∂S

)V

, (S1.17)

which is a linear first-order differential equation for U(S,V ). This is a separable equation, so it canbe rearranged as

23NkB

dS =dUU

. (S1.18)

By integrating this equation from S0 to S, we obtain

23NkB

(S−S0) = ln[

U(S,V )

U(S0,V )

], (S1.19)


which yields

U(S,V ) =U(S0,V )exp[

23NkB

(S−S0)

]. (S1.20)

The function U(S0,V ) can be obtained by integrating the fundamental equation (S1.15) at constantentropy. By using (S1.14) and the equation of state, we have

dU =−PdV =−NkBTdVV

=−2U3

dVV

, (S1.21)

which can be rearranged into a separable differential equation:

dUU

=−23

dVV

. (S1.22)

Integrating this equation, ∫ U

U0

dU ′

U ′=−2

3

∫ V

V0

dV ′

V ′, (S1.23)

which yields

ln(

UU0

)=−2

3ln(

VV0

), (S1.24)

or,

U(S0,V ) =U0

(VV0

)− 23

, (S1.25)

where U0 ≡U(S0,V0). Thus, after substitution into (S1.20), we obtain

U(S,V ) =U0

(VV0

)− 23

exp[

23NkB

(S−S0)

]. (S1.26)

The comparison between equations (S1.14) and (S1.26) demonstrates that the equation of statein (S1.14) contains less thermodynamic information than the solution (S1.26) of the fundamentalequation.

Exercise 1.5

(a) The fundamental equation for the internal energy U(S,V ,N) is

dU = T dS−PdV +µ dN , (S1.27)

This internal energy is an exact differential, so the three equations are determined by thepartial derivatives with respect to each independent variable:(

∂U∂S

)V ,N

= T ,(

∂U∂V

)S,N

=−P ,(

∂U∂N

)S,V

= µ . (S1.28)


(b) As U , S, V and N are extensive variables, they transform under the scaling of the system sizeby λ as

U → λU , S→ λS , V → λV , N→ λN . (S1.29)

Thus, the transformation of the internal energy is

λU(S,V ,N) =U(λS,λV ,λN) . (S1.30)

(c) The derivative of both sides of (S1.30) with respect to λ is

U =

(∂U

∂ (λS)

)V ,N

d(λS)dλ

+

(∂U

∂ (λV )

)S,N

d(λV )

dλ+

(∂U

∂ (λN)

)S,V

d(λN)

dλ

=

(∂U

∂ (λS)

)V ,N

S+(

∂U∂ (λV )

)S.N

V +

(∂U

∂ (λN)

)S,V

N . (S1.31)

Since this equation is valid for all positive λ , we may set λ = 1 to obtain

U =

(∂U∂S

)V ,N

S+(

∂U∂V

)S.N

V +

(∂U∂N

)S,V

N = T S−PV +µN , (S1.32)

where we have used the equations of state in (S1.28).

(d) Equation (S1.32) is the integrated form of the fundamental equation in (S1.27). The inde-pendent variables are S, V , and N, and the dependent variable is, of course, U . The intensivevariables T , P, and µ are expressed as partial derivatives of U as equations of state, as ex-plained in the solution to (a). For the solution (S1.26) of the fundamental equation for anideal monatomic gas, we have the following derivatives:

T =

(∂U∂S

)V ,N

=2U

3NkB,

P =−(

∂U∂V

)S,N

=2U3V

, (S1.33)

µ =

(∂U∂N

)S,V

=5U3N− 2SU

3N2kB.

The first equation yields the internal energy, U = 32NkBT , the second the equation of state,

PV = 23U = NkBT , and the third the chemical potential, µN = 5

2NkBT −T S. Substitutingthese expressions into (S1.32),

U = T S−PV +µN

=

(2U

3kBT

)S−(

2U3V

)V +

5U3−(

2U3kBT

)S =U (S1.34)

which verifies this equation for the monatomic ideal gas.


(e) For the Gibbs free energy, there is only one variable N that is extensive. Hence, the transfor-mation equation is

λG(T ,P,N) = G(T ,P,λN) . (S1.35)

Differentiating this equation with respect to λ ,

G =∂G

∂ (λN)

d(λN)

dλ=

∂G∂ (λN)

N , (S1.36)

and setting λ = 1, yields G = µN, where we have used(∂G∂N

)T ,P

= µ (S1.37)

from the fundamental equation for G in (1.40). That the result obtained is consistent withthe fundamental equation for U can be seen from

G =U−T S+PV = (T S−PV +µN)−T S+PV = µN . (S1.38)

Exercise 1.6(a) We begin by writing

dU = T dS−PdV = T dS−PdV −V dP+V dP

= T dS−d (PV )+V dP . (S1.39)

Bringing the total differentials to the left-hand side of this equation,

d(U +PV ) = T dS+V dP , (S1.40)

shows that the independent variables are now the entropy S and the pressure P. The quantitywhose differential is the right-hand side of this equation is called the enthalpy, or the heatfunction, and is denoted by H:

H =U +PV . (S1.41)

(b) Since the right-hand of (S1.41) is comprised of state functions, the enthalpy is also a statefunction. The mathematical consequences of this are obtained by first taking the differentialof H(S,P),

dH =

(∂H∂S

)P

dS+(

∂H∂P

)S

dP , (S1.42)

and comparing withdH = T dS+V dP , (S1.43)

which allows us to make the following identifications for the partial derivatives of the en-thalpy: (

∂H∂S

)P= T ,

(∂H∂P

)S=V . (S1.44)


(c) The change of the enthalpy during an reversible isobaric process is equal to the heat flowduring that process. To see this, suppose a system in contact with a pressure reservoir un-dergoes an infinitesimal reversible transformation. We have from the fundamental law ofthermodynamics that

dU = T dS−PdV . (S1.45)

Since the pressure is assumed constant, this can be written as

dU = T dS−PdV −V dP , (S1.46)

which can be rearranged as

T dS = dU +d(PV ) = d(U +PV ) . (S1.47)

Utilizing Clausius’ equality (for a reversible process) and the definition of the enthalpy in(S1.41), we obtain d Q = dH. Thus, the heat added to a system at constant pressure ismanifested as an increase in enthalpy. An important consequence of this equation is that thechange in enthalpy during a chemical reaction under constant external pressure is equal tothe heat of reaction. This is the situation that normally occurs, since most chemical reactionstake place at atmospheric pressure.

(d) In Fig. 1.1(d) the system is maintained under constant pressure and is thermally isolated. Ifwe consider reversible processes, then the first law of thermodynamics reduces to

dU =−PdV , (S1.48)

since d Q = 0 because of the adiabatic walls surrounding the system. Using the fact thatdP = 0, we can rewrite (S1.48) as

dU +PdV +V dP = d(U +PV ) = 0 (S1.49)

so for this experiment the pressure P and the quantity U +PV are constant.

Exercise 1.7

(a) There is only a single component in the phase diagram, so c−1 and Gibbs phase rule reducesto f = 3− p. In regions with only a single phase, p = 1, so f = 2. The phase diagram indeedshows that there are extended planar regions where independent variations in P and T do notchange the phase of water.

(b) Along a coexistence line, there are two phases in equilibrium, so p = 2. Hence, f = 1,which is to be expected for a line. This means that any change in, say, P, must have aspecific (i.e. not independent) compensating change in T to remain on the coexistence line.


(c) At the triple point, there are three phases in equilibrium, so p = 3. Hence f = 0, which is tobe expected for a point, i.e. any change in either p or T takes us away from this point.

(d) As for the triple point, where the pressure and temperature are specifically defines, there areno degrees of freedom associated with the critical point.

(e) The Gibbs phase rule for a one-component system is f = 3− p. If we set f = 0, then p = 3,which is clearly incorrect, and if we set p = 2, then f = 1, which is also incorrect.

(f) The calculation in (a) remains valid, but the calculation in (b) is amended to c(p− 1)+ 1because of the equality of the densities of two phases. The number f of degrees of freedomobtained is

f =[p(c−1)+2

]−[c(p−1)+1

]= c− p+1. (S1.50)

Now, with c = 1, we have f = 2− p, so with p = 2 (liquid and vapor), we have f = 0, asrequired.

Exercise 1.8

(a) The latent heat is a measure of the extent to which bonds are disrupted or broken, whileentropy is a measure of the configurational disorder, Thus, in the solid-to-liquid transi-tion, the bonds are not actually broken, but disrupted by the additional mobility of thewater molecules. In the liquid-to-vapor transition, however, bonds must be broken for themolecules to escape from the liquid. In the solid-to-vapor transition, the bond breaking iseven more severe, as the bonds in the ordered solid must be broken for the molecules to jointhe vapor. Thus, we expect that

λ (s→ v)> λ (`→ v) λ (s→ `) , (S1.51)

which is consistent with the corresponding entries in the Table.

To assess the entropy changes, we first observe that the solid is ordered and has the lowestentropy, with the liquid having a somewhat higher entropy because of the additional configu-rational freedom of the molecules. The vapor has the highest entropy because the moleculesoccupy the greatest volume and have trajectories that are ballistic between collisions withother molecules in the vapor. We therefore expect that

∆s(s→ v)> ∆s(`→ v) ∆s(s→ `) , (S1.52)

which is again consistent with the corresponding entries in the Table.

(b) When expressed in terms of the natural logarithm, the Clausius–Clapeyron equation reads

d lnPdT

=1P

dPdT

=λ

PT (vb− va). (S1.53)


solid

liquid

vapor

T

lnP

Figure S1.3: The slopes of the coexistence lines of water near the triple point as determined by theClausius-Clapeyron equation.

For the solid–vapor transition, we have

d lnPdT

∣∣∣∣s→v≈ 50×103

6×102×3×102×2×10−2 =50036≈ 14 K−1 , (S1.54)

which is close to the exact value of 13.85 K−1. For the liquid–vapor transition, we find

d lnPdT

∣∣∣∣`→v≈ 45×103

6×102×3×102×2×10−2 =45036≈ 12.5 K−1 , (S1.55)

which is to be compared with the exact value of 12.2 K−1. Finally, for the solid–liquidtransition, we have

d lnPdT

∣∣∣∣s→`

≈− 6×103

6×102×3×102×1.5×10−6 =−6×105

27≈−2.22×104 K−1 , (S1.56)

and the exact value of −2.18× 104 K−1. A qualitative sketch of these slopes is shown inFig. S1.3.

Exercise 1.9Beginning with the Clausius-Clapeyron equation,

dPdT

=λ

T (vv− v`), (S1.57)

we use the following observations to enable this equation to be integrated:


(a) Particles in a vapor are much further apart that those in a liquid, as evidenced, for example,by the long mean-free path of vapor particles. Hence, the specific volume vv of the vapor isexpected to be greater than the specific volume v` of the liquid. We may thereby make theapproximation vv− v` ≈ vv.

(b) In the limit of low densities and/or high temperatures, the vapor may described by the equa-tion of state of an ideal gas: PV = NvkBT , which we rearrange as

vv =VNv

=kBT

P. (S1.58)

(c) Since λ is observed to vary slowly with temperature, we regard this quantity as a constant.

We are now able to approximate the Clausius-Clapeyron equation by

dPdT

=λP

kBT 2 . (S1.59)

Separating variables,dPP

=λdTkBT 2 , (S1.60)

and integrating, ∫ P

P0

dP′

P′=

λkB

∫ T

T0

dT ′

T ′2, (S1.61)

yields

lnP′∣∣∣∣PP0

= ln(

PP0

)=− λ

kB

1T ′

∣∣∣∣TT0

=− λkBT

+λ

kBT0, (S1.62)

or, after solving for P,

P = P0 exp[− λ

kB

(1T− 1

T0

)]. (S1.63)

Exercise 1.10

(a) The transformation shown has a positive Gibbs free energy change, so does not occur sponta-neously. However, the reverse transformation, from diamond to graphite, has a negative freeenergy change, and so does occur spontaneously. Although the free energy change is quitesmall, this transformation proceeds so slowly because the steric barriers for the movementof atoms required for the structural change are so high, ultimately because of the strengthof the C–C bonds in diamond. Thus, for this transformation, slow kinetics pre-empt thethermodynamic driving force.


(b) At constant temperature T0, the fundamental equation for G reduces to dG = V (T0,P)dP.Integrating this equation from pressure P0 to P yields

∫ P

P0

dG = G(T0,P)−G(T0,P0)+∫ P

P0

V (T0,P′)dP′ , (S1.64)

or, when solved for G(T0,P),

G(T0,P) = G(T0,P0)+∫ P

P0

V (T0,P′)dP′ . (S1.65)

Using the corresponding equation for graphite and diamond and taking the difference pro-duces the change ∆G(T0,P) for the transformation of graphite to diamond:

∆G(T0,P) = Gd(T0,P)−Gg(T0,P)

= Gd(T0,P0)−Gg(T0,P))+∫ P

P0

[Vd(T0,P′)−Vg(T0,P′)

]dP′

= ∆G+∫ P

P0

∆V (T0,P′)dP′ . (S1.66)

(c) Given that ∆V (T0,P0) can be taken as independent of pressure, the coexistence pressureof graphite and diamond is determined by the equality of their Gibbs free energies, thatis, ∆G(T0,P) = 0:

∆P = P−P0 =∆G

∆V (T0,P). (S1.67)

Hence, the coexistence pressure P at room temperature is

P = 105 Pa+2.9J/mol×103

1.9×10−6 m3/mol

= 105 Pa+1.53×109 Pa≈ 1.53×109 Pa, (S1.68)

which is about 15,000 atmospheres.

Exercise 1.11

(a) We can use the Gibbs free energy to specify the temperature T , the pressure P, and thequantities of material of each species in each phase: Nα

i , for i= 1,2, . . . ,c and α = 1,2, . . . , p.As there are cp of the Nα

i , the total number of thermodynamic variables in the system iscp+2.


(b) The chemical equilibrium between each species requires that

µ1i (T ,P) = µ2

i (T ,P) = · · ·= µαi (T ,P) . (S1.69)

which represents p−1 constraints for each i. To see this, take, say, µ1i . Then, this relation is

equivalent to

µ1i (T ,P) = µ2

i (T ,P) , µ1i (T ,P) = µ3

i (T ,P) , . . . , µ1i (T ,P) = µα

i (T ,P) , (S1.70)

which is p−1 equations. For c components, we have a total of c(p−1) constraints.

(c) Since the total amount of each component is fixed, we have

p

∑α=1

Nαi = Ni (S1.71)

for each i. There are clearly c such constraints.

(d) To determine the degrees of freedom f in this scenario, we subtract the total number ofconstraints from the initial number variables:

cp+2− c(p−1)− c = cp+2− cp+ c− c = 2. (S1.72)


Chapter 2

Brownian Motion, Random Walks, and theDiffusion Equation

Exercise 2.1The recursion relation for a biased random walk is

p(na,kτ) = rp((n−1)a,(k−1)τ)+ `p((n+1)a,(k−1)τ) , (S2.1)

where p(na,kτ) is the probability density of the walker occupying site na at time kτ , the probabilityof movement to the left is ` and to the right is r such that `+ r = 1. We proceed as for the case ofthe unbiased random walk and take the limits a→ 0 and τ → subject to additional constraints thatwill be discussed below. In this limit, na becomes the continuous variable x and kτ becomes thecontinuous variable t, and the two terms on the right-hand side of (S2.1) can be expanded as

p((n±1)a,(k−1)τ) = p(x±a, t− τ)

= p(x, t)+∂ p∂x

(±a)+∂ p∂ t

(−τ)+12

∂ 2 p∂x2 (±a)2 +

∂ 2 p∂x∂ t

(±a)(−τ)+12

∂ 2 p∂ t2 (−τ)2 + · · · . (S2.2)

Substitution of (S2.2) into (S2.1) produces

p(x, t) = r[

p(x, t)− ∂ p∂x

a− ∂ p∂ t

τ +12

∂ 2 p∂x2 a2− ∂ 2 p

∂x∂ taτ +

12

∂ 2 p∂ t2 τ2 + · · ·

]

+ `

[p(x, t)+

∂ p∂x

a− ∂ p∂ t

τ +12

∂ 2 p∂x2 a2 +

∂ 2 p∂x∂ t

aτ +12

∂ 2 p∂ t2 τ2 + · · ·

]

= (r+ `)p(x, t)− (r− `)a∂ p∂x− (r+ `)τ

∂ p∂ t

+12(r+ `)a2 ∂ 2x

∂x2

− (r− `)aτ∂ 2 p

∂x∂τ+

12(r+ `)τ2 ∂ 2 p

∂τ2 + · · · . (S2.3)

17


Since r+ `= 1, we see that all of the terms in the unbiased random walk are present on the right-hand side but, because r−` 6= 0, terms with odd-order derivatives with respect to x are also present.A rearrangement of (S2.3) yields

τ∂ p∂ t

+τ2

2∂ 2 p∂ t2 + · · ·=−(r− l)a

∂ p∂x

+a2

2∂ 2 p∂x2 − (r− `)aτ

∂ 2 p∂x∂τ

+ · · · ,

which, after dividing both sides by τ , becomes

∂ p∂ t

+τ2

∂ 2 p∂ t2 + · · ·=−(r− l)a

τ∂ p∂x

+a2

2τ∂ 2 p∂x2 − (r− `)a

∂ 2 p∂x∂τ

+ · · · . (S2.4)

We now examine the limits a→ 0 and τ → 0 more closely. In our discussion of the unbiasedrandom walk, we found that the appropriate limiting forms of these quantities were a(ε) = aε andτ(ε) = τε2 where ε → 0, whereupon

limε→0

[a2(ε)2τ(ε)

]=

a2

2τ≡ D , (S2.5)

where D is the diffusion constant. Consider the first term on the right-hand side of (S2.4):

−(r− l)aτ

∂ p∂x

, (S2.6)

which appears to become infinite as ε → 0. This problem can be avoided by letting the quantity(r− `) go to zero as (r− `)ε , in which case we obtain

limε→0

[(r− `)ε]aε

τε2

=

(r− l)aτ

≡ u . (S2.7)

The quantity u has the dimensions of speed and represents the net ‘drift’ to the right (if r > `) orto the left (if r < `).

Before taking the limit ε→ 0, we examine the effect of our limits on higher-order terms. Thereare two types of such terms: those with even-order derivatives with respect to x, and those withodd-order x derivatives. For even-order derivatives, we have, after dividing by τ , as in the passageto (S2.4),

1τ

∂ 2r+s p∂x2r∂ ts a2rτs =

∂ 2r+s p∂x2r∂ ts a2rτs−1→ ∂ 2r+s p

∂x2r∂ ts a2rτs−1ε2r+2s−2 , (S2.8)

in which r = 0,1,2, . . . and s = 0,1,2, . . . , such that not both are equal to zero, and we have usedthe limiting forms of a and τ . This term is non-zero as ε → 0 only if 2r+ 2s = 2, which has thesolutions r = 1 and s = 0, i.e. pxx, and r = 0 and s = 1, i.e. pt . All other terms vanish as ε → 0.

The terms with odd-order spatial derivatives are, after dividing by τ ,

r− `

τ∂ 2r+1+s p∂x2r+1∂ ts a2r+1τs = (r− `)

∂ 2r+s p∂x2r∂ ts a2r+1τs−1

→ (r− `)∂ 2r+s p∂x2r∂ ts a2r+1τs−1ε2r+2s , (S2.9)

Brownian Motion, Random Walks, and the Diffusion Equation 19

in which r = 0,1,2, . . . and s = 0,1,2, . . . , where both r and s can be equal to zero, and we haveused the limiting forms of a, τ , and (r− `). This term is non-zero as ε → 0 only if 2r+ 2s = 0,which has the solution r = 0 and s = 0, i.e. px. All other terms vanish as ε → 0. Hence, the onlysurviving terms of the limiting process are

∂ p∂ t

=−u∂ p∂x

+D∂ 2 p∂x2 . (S2.10)

The result of unequal left and right hopping probabilities is the term −upx. This equation is calledthe diffusion equation with drift.

Exercise 2.2Beginning with (S2.10), we consider the effect of changing variables from (x, t) to (ξ = x+at, t)where a is a constant to be determined. Then the required partial derivatives in (S2.10) are

∂ p∂ t

=∂ p∂ξ

∂ξ∂ t

+∂ p∂ t

= a∂ p∂ξ

+∂ p∂ t

,

∂ p∂x

=∂ p∂ξ

∂ξ∂x

=∂ p∂ξ

,

∂ 2 p∂x2 =

∂ 2 p∂ξ 2

∂ξ∂x

=∂ 2 p∂ξ 2 .

(S2.11)

Substitution of these expressions into (S2.10) yields

a∂ p∂ξ

+∂ p∂ t

=−u∂ p∂ξ

+D∂ 2 p∂ξ 2 . (S2.12)

By choosing a =−u, the terms with pξ cancel, and this equation reduces to the diffusion equationin the variables (ξ , t):

∂ p∂ t

= D∂ 2 p∂ξ 2 . (S2.13)

Hence, if we have a solution p(x, t) of the diffusion equation, the corresponding solution of thediffusion equation with drift is p(x−ut, t).

Exercise 2.3If c(x, t) is a solution of the diffusion equation,

∂c∂ t

= D∂ 2c∂x2 , (S2.14)


then the time-derivative of the entropy associated with this solution,

S(t) =−kB

∫c(x, t) ln

[c(x, t)

]dx , (S2.15)

isdSdt

=−kB

∫ ∂c∂ t

ln[c(x, t)

]dx− kB

∫ ∂c∂ t

dx . (S2.16)

We now use the fact that c(x, t) is a solution of (S2.14) to write

dSdt

=−DkB

∫ ∂ 2c∂x2 ln

[c(x, t)

]dx− kBD

∫ ∂ 2c∂x2 dx . (S2.17)

The second term on the right-hand side can be evaluated immediately:

−D∫ ∂ 2c

∂x2 dx =−D∂c∂x

∣∣∣∣∞−∞

= 0 (S2.18)

because the diffusion current vanishes as x→±∞. The first term on the right-hand side of (S2.17)is evaluated by an integration by parts:

−D∫ ∂ 2c

∂x2 ln[c(x, t)

]dx =−D

∂c∂x

ln[c(x, t)

]∣∣∣∣∞−∞

+D∫ 1

c

(∂c∂x

)2

dx . (S2.19)

The first term on the right-hand side vanishes because the diffusion current decays to zero asx→±∞, typically as an exponential of x2. The second term is positive, because c≥ 0 and (cx)

2≥ 0everywhere. Hence,

dSdt

= kBD∫ 1

c

(∂c∂x

)2

dx≥ 0. (S2.20)

The equality sign is obtained only for a uniform distribution, in which case cx = 0. In all othercases, St > 0, so a non-uniform initial distribution has an increasing entropy under the action ofdiffusion. This can be understood from the standpoint of information. A non-uniform distributioncontains information about where the diffusing species is more likely to be found. This is a low-entropy condition. With a uniform distribution, however, all of this information is lost by the actionof diffusion, which corresponds to a maximum entropy condition. The general solution solution ofthe one-dimensional diffusion equation is a straight line whose intercept and slope are determinedby the boundary conditions.

Exercise 2.4We are given that c(x, t) is a solution of the one-dimensional equation,

∂c∂ t

= D∂ 2c∂x2 , (S2.21)


for −∞ < x < ∞ with the initial condition c(x,0) = 0. Then, beginning with the definition,

〈x(t)〉=∫

∞

−∞

xc(x, t)dx , (S2.22)

we have that

d〈x〉dt

=∫

∞

−∞

x∂c∂ t

dx = D∫

∞

−∞

x∂ 2c∂x2 dx

= D(

x∂c∂x

∣∣∣∣∞−∞

−∫

∞

−∞

∂c∂x

dx)= D

(x

∂c∂x

∣∣∣∣∞−∞

− c(x, t)∣∣∣∣∞−∞

). (S2.23)

Both the solutions and its derivatives must vanish exponentially at ±∞ so the right-hand side ofthis equation vanishes, leaving

d〈x〉dt

= 0, (S2.24)

so 〈x(t)〉= constant. At t = 0, x(0) = 0, so 〈x(t)〉= 0. We also have that

〈(x(t)−〈x(t)〉)2〉= 〈x2(t)〉 . (S2.25)

Beginning with the definition,

〈x2(t)〉=∫

∞

−∞

x2c(x, t)dx , (S2.26)

and proceeding as above, yields

d〈x2〉dt

=∫

∞

−∞

x∂c∂ t

dx = D∫

∞

−∞

x2 ∂ 2c∂x2 dx

= D(

x2 ∂c∂x

∣∣∣∣∞−∞

−2∫

∞

−∞

x∂c∂x

)= D

(x2 ∂c

∂x

∣∣∣∣∞−∞

−2xc(x, t)∣∣∣∣∞−∞

+2∫

∞

−∞

c(x, t)dx)

. (S2.27)

The vanishing of the solution and its derivatives exponentially at ±∞ eliminates the surface terms.The diffusion equation is a conservation equation, as the derivation in Sec. 2.2 showed, so theintegral over the concentration is unity. We are left with

d〈x2〉dt

= 2D , (S2.28)

whose solution is 〈x2〉= 2Dt+constant. For an initial condition localized at the origin, the constantvanishes and we obtain 〈x2〉= 2Dt.


Exercise 2.5Substitution of the Fourier transforms f (k) and g(k) into the expression given, yields

12π

∫∞

−∞

f (k)g(k)e−ikxdk

=1

2π

∫∞

−∞

[∫∞

−∞

f (x′)eikx′dx′][∫

∞

−∞

g(x′′)eikx′′dx′′]

e−ikxdx

=1

2π

∫∞

−∞

dx′ f (x′)∫

∞

−∞

dx′′g(x′′)∫

∞

−∞

dk eik(x′+x′′−x) . (S2.29)

The integral over k can be evaluated by using (2.41):∫∞

−∞

dk eik(x′+x′′−x) = 2πδ (x′+ x′′− x) , (S2.30)

in which δ (x) is the Dirac delta-function. Then, using the property of δ (x) that∫ a

−af (x)δ (x)dx = f (0) , (S2.31)

for any a > 0, (S2.29) reduces to

12π

∫∞

−∞

f (k)g(k)e−ikxdk =∫

∞

−∞

dx′ f (x′)∫

∞

−∞

dx′′g(x′′)δ (x′+ x′′− x) (S2.32)

=∫

∞

−∞

f (x′)g(x− x′)dx′ , (S2.33)

which is the convolution theorem.By examining (2.30),

c(x, t) =1

2π

∫∞

−∞

c0(k)e−Dk2t e−ikx dk , (S2.34)

we identify c0(k) with f (k) and e−Dk2t with g(k). Then, since

c0(k) =∫

∞

−∞

c0(x)eikxdx , e−Dk2t =1√

4πDt

∫∞

−∞

e−x2/4Dteikx dx , (S2.35)

the convolution theorem yields

c(x, t) =1√

4πDt

∫∞

−∞

c0(x′)e−(x−x′)2/4Dt dx′ (S2.36)

as the solution of the initial-value problem of the diffusion equation for −∞≤ x≤ ∞.


Exercise 2.6

(a) For solutions of the diffusion equation,

∂c∂ t

= D∂ 2c∂x2 , (S2.37)

on the semi-infinite interval 0≤ x≤ ∞, the Fourier transform of c(x, t) is

c(k, t) =∫

∞

0c(x, t)eikx dx

=∫

∞

0c(x, t)coskxdx+ i

∫∞

0c(x, t)sinkxdx

≡ cr(k, t)+ i ci(k, t) , (S2.38)

in which we have defined the real cr(k, t) and imaginary ci(k, t) parts of the Fourier transformof the concentration. Proceeding as in Sec. 2.3.1, we obtain

∂ c(k, t)∂ t

= D∂c∂x

eikx∣∣∣∣∞0+ iDkceikx

∣∣∣∣∞0−Dk2

∫∞

0c(x, t)eikx dx . (S2.39)

We still require that

limx→∞

[c(x, t)

]= 0, lim

x→∞

(∂c∂x

)= 0, (S2.40)

so the evaluation of the right-hand side of (S2.39) yields

∂ c(k, t)∂ t

=−D∂c∂x

∣∣∣∣x=0

+ iDkc∣∣∣∣x=0−Dk2c(k, t) . (S2.41)

(b) Both the concentration and the current at x = 0 appear on the right-hand side of (S2.41).These cannot both be independently specified, so we proceed by expressing this equation interms of its real and imaginary parts. By using the definitions in (S2.38), we have

∂ cr

∂ t+ i

∂ ci

∂ t=−D

∂c∂x

∣∣∣∣x=0

+ iDkc∣∣∣∣x=0−Dk2[cr(k, t)+ ici(k, t)

], (S2.42)

so equating the real and imaginary parts separately yields

∂ cr

∂ t=−D

∂c∂x

∣∣∣∣x=0−Dk2cr(k, t) , (S2.43)

∂ ci

∂ t= Dkc

∣∣∣∣x=0−Dk2ci(k, t) .) (S2.44)


Suppose that we impose a reflecting boundary condition at x = 0: cx(0, t) = 0. Then (S2.43)reduces to

∂ cr

∂ t=−Dk2cr(k, t) , (S2.45)

which has the solutioncr(k, t) = cr(k,0)e−Dk2t . (S2.46)

However, equation (S2.44) remains unspecified. This problematic situation can be allevi-ated by working with the cosine Fourier transform, for which there is no imaginary part of(S2.38).

Similarly, if we have an reflecting boundary at x = 0, cx(0, t) = 0, then (S2.44) becomes

∂ ci

∂ t=−Dk2ci(k, t) , (S2.47)

which has the solutionci(k, t) = ci(k,0)e−Dk2t . (S2.48)

Equation (S2.43) can then be eliminated by working with the Fourier sine transform, forwhich there is no real part of (S2.38).

(c) In the case of an absorbing boundary, the solution c(x, t) is expressed by the Fourier cosinetransform

c(x, t) =2π

∫∞

0cr(k,0)e−Dk2t coskxdk . (S2.49)

We take as the initial condition a delta-function at a point x′ > 0: c(x,0) = δ (x− x′), theFourier cosine transform of which is

cr(k,0) =∫

∞

0δ (x− x′)coskxdx = coskx′ . (S2.50)

The solution in (S2.49) becomes

c(x, t) =2π

∫∞

0e−Dk2t coskx′ coskxdk . (S2.51)

The trigonometric identity

cosacosb = 12

[cos(a−b)+ cos(a+b)

], (S2.52)

can be used to write the solution as

c(x, t) =1π

∫∞

0e−Dk2t cos

[k(x− x′)

]dk+

1π

∫∞

0e−Dk2t cos

[k(x+ x′)

]dk . (S2.53)


The integrals on the right-hand side can be evaluated by expressing the cosine functions interms of complex exponentials,

c(x, t) =1

2π

∫∞

0e−Dk2t

[eik(x−x′)+ e−ik(x−x′)

]dk+

12π

∫∞

0e−Dk2t

[eik(x+x′)+ e−ik(x+x′)

]dk

=1

2π

∫∞

−∞

e−Dk2t+ik(x−x′) dk+1

2π

∫∞

−∞

e−Dk2t+ik(x+x′) dk , (S2.54)

and using the integral evaluated in Sec. 2.3.2:∫∞

−∞

e−(Dk2t±ikξ ) dk =√

π√Dt

exp(− ξ 2

4Dt

). (S2.55)

We obtain

c(x, t) =1√

4πDtexp[−(x− x′)2

4Dt

]+

1√4πDt

exp[−(x+ x′)2

4Dt

]. (S2.56)

An analogous procedure can be applied to obtain the solution for an absorbing boundary atx = 0:

c(x, t) =1√


4Dt

]− 1√

4πDtexp[−(x+ x′)2

4Dt

]. (S2.57)

These expressions for reflecting and absorbing boundaries are analogous to the method ofimage used in electrostatics, where fictitious solutions are strategically placed to ensure thatthe boundary conditions are satisfied at all times. The linearity if the diffusion equation (andof Poisson’s’s equation in electrostatics) is why this method works.

Exercise 2.7The region where there is a non-zero concentration of the A species is 0 ≤ x ≤ d, where thisconcentration is unity:

c(x,0) =

1, if 0≤ x≤ d;

0, if d ≤ x ,(S2.58)

The boundary condition on the concentration of A atoms states that no atoms enter or leave atx = 0, which is a reflecting boundary condition. Hence, the solution for the concentration of A is

c(x, t) =∫

∞

0c(x,0)GR(x,x′; t)dx′ =

∫ d

0Gr(x,x′; t)dx′ (S2.59)

=1√

4πDt

[∫ d

0e(x−x′)2/4Dt dx′+

∫ d

0e(x+x′)2/4Dt dx′

]. (S2.60)


In the first integral on the right-hand side, we change the integration variable to

s =x− x′√

4Dt, (S2.61)

which means that x′ = x− s√

4Dt. Making the appropriate changes to the integration element, theintegrand, and the limits of integration yields

1√4πDt

∫ d

0e(x−x′)2/

√4Dt dx′ =

1√π

∫ x/√

4Dt

(x−d)/√

4Dte−s2

ds . (S2.62)

In the second integral, we change the integration variable to

s =x+ x′√4πDt

, (S2.63)

which means that x′ = −x+ s√

4πDt. Again making the appropriate changes to the integrationelement, the integrand, and the limits of integration yields

1√4πDt

∫ d

0e(x+x′)2/

√4Dt dx′ =

1√π

∫ (x+d)/√

4Dt

x/√

4Dte−s2

ds . (S2.64)

Combining the two integrals yields

c(x, t) =1√π

∫ 0

(x−d)/√

4Dte−s2

ds+1√π

∫ x/√

4Dt

0e−s2

ds

+1√π

∫ 0

x/√

4Dte−s2

ds+1√π

∫ (x+d)/√

4Dt

0e−s2

ds

=− 1√π

∫ (x−d)/√

4Dt

0e−s2

ds+1√π

∫ (x+d)/√

4Dt

0e−s2

ds

=12

[erf(

x+d√4Dt

)− erf

(x−d√

4Dt

)]. (S2.65)

Exercise 2.8We seek a solution of the diffusion equation,

∂c∂ t

= D∂ 2c∂x2 , (S2.66)

over the range 0≤ x < ∞ subject to the initial and boundary conditions

c(x,0) = 0, c(0, t) = c0 , (S2.67)


in which c0 is a constant. We define the function u(x, t) = c(x, t)− c0 and, upon substitution into(S2.66) and (S2.67), obtain

∂u∂ t

= D∂ 2u∂x2 , (S2.68)

with the initial and boundary conditions

u(x,0) =−c0 , u(0, t) = 0. (S2.69)

We now have an absorbing boundary at x = 0, with the initial condition u(x,0) = −c0, for x > 0.The appropriate extension of the fundamental solution in the presence of the absorbing boundaryis given in (S2.57):

G(x,x′; t) =1√


4Dt

]− 1√

4πDtexp[−(x+ x′)2

4Dt

]. (S2.70)

in terms of which the solution to (S2.66) and (S2.67) for u is

u(x, t) =∫

∞

0u(x′,0)G(x,x′; t)dx′

=− c0√4πDt

∫∞

0

exp[−(x− x′)2

4Dt

]dx′+

c0√4πDt

∫∞

0exp[−(x+ x′)2

4Dt

]dx′ . (S2.71)

The integrals on the right-hand of this equations can be evaluated by making the substitution

s =x′± x√

4Dt←→ x′ = s

√4Dt∓ x , (S2.72)

which yields

u(x, t) =− c0√π

∫∞

−x/√

4Dte−s2

ds+c0√

π

∫∞

x/√

4Dte−s2

ds

=− c0√π

∫ x/√

4Dt

−x/√

4Dte−s2

ds =− 2c0√π

∫ x/√

4Dt

0e−s2

ds

=−c0 erf(

x√4Dt

). (S2.73)

Thus, the solution to the original problem in (S2.66) and (S2.67) is

c(x, t) = u(x, t)+ c0 = c0

[1− erf

(x√4Dt

)]. (S2.74)

The evolution of this solution is shown in Fig. S2.1.


0 1 2 3 4 50.0

0.2

0.4

0.6

0.8

1.0

x

c c 0

Dt = 0

Dt = 0.5

Dt = 5

Dt = 50

1

Figure S2.1: The solution c/c0 in (S2.74) for the indicated times Dt. The initial sharp drop in theconcentration is gradually smoothed out over an expanding range of x. For large enough x at anyfinite time, the solution decays exponentially to zero.

Exercise 2.9

(a) The required derivatives for substitution into the diffusion equation are

∂c∂ t

=− 12t√

tφ − x

2t2dφdξ

,

∂c∂x

=1t

dφdξ

, (S2.75)

∂ 2c∂x2 =

1t√

td2φdξ 2 .

Upon substitution of these expressions into the diffusion equation, ct = Dcxx, we obtain

− 12t√

tφ − x

2t2dφdξ

=D

t√

td2φdξ 2 , (S2.76)

which, after multiplying both sides by 2t√

t and bringing all terms to the same side of theequation, becomes

2Dd2φdξ 2 +ξ

dφdξ

+φ = 0, (S2.77)

Since the last two terms can be written as a derivative, ξ φ ′+φ = (ξ φ)′, the equation for φcan be written as

ddξ

(2D

dφdξ

+ξ φ)= 0. (S2.78)


(b) Equation (S2.78) can be integrated directly to obtain

2Ddφdξ

+ξ φ = A , (S2.79)

in which A is a constant of integration. This constant must be the same for all values of ξ ,so we may use the fact that at any finite time, the concentration and diffusion current vanishas x→±∞ to deduce that A = 0. The equation for φ therefore reduces to

2Ddφdξ

+ξ φ = 0. (S2.80)

This is a separable equation whose solution may be represented as,∫ φ

φ0

dφ ′

φ ′=− 1

2D

∫ ξ

ξ0

ξ ′ dξ ′ , (S2.81)

which yieldsφ(ξ ) = φ0eξ0/4De−ξ 2/4D ≡ Be−ξ 2/4D , (S2.82)

which defines the integration constant B. By imposing normalization,

B√t

∫∞

−∞

e−x2/4Dt dx = B√

4D∫

∞

−∞

e−s2ds︸︷︷︸√

π

= 1, (S2.83)

so B = 1/√

4Dπ , we obtain

c(x, t) =1√

4πDtexp(− x2

4Dt

)(S2.84)

for our normalized solution, which is the fundamental solution of the diffusion equation.

Returning to (S2.79), if we do not suppose that A = 0, we can find a solution of this equationby first solving the corresponding homogeneous equation. Neglecting, for the moment thequestion of normalization, the solution φh of this equation obtained in (c) is

φh(ξ ) = e−ξ 2/4D . (S2.85)

Hence, the solution of (S2.78) with A 6= 0 has the form φ(ξ ) = φh(ξ ) f (ξ ), where f is to bedetermined. Substitution of this solution into (S2.78),

2D(− ξ

2De−ξ 2/4D

)f +2De−ξ 2/4D d f

dξ+ξ f e−ξ 2/4D = A , (S2.86)

yields, after cancelling terms and a simple rearrangement, a differential equation for f :

d fdξ

=A

2Deξ 2/4D . (S2.87)


Hence, f is given by

f (ξ ) =A

2D

∫ ξes2/4D ds+C , (S2.88)

where C is a second constant of integration. The solution of (S2.78) is therefore given by

φ(ξ ) =A

2De−ξ 2/4D

∫ ξes2/4D ds+Ce−ξ 2/4D , (S2.89)

from which we see immediately that, if A 6= 0, the solution is not normalizable. Since we areassuming that the initial distribution is contains a finite amount of diffusing material, and thediffusion equation is a conservation equation, we must choose A = 0.

Exercise 2.10The required derivatives of

c(r, t) =1t

φ(

r√t

)(S2.90)

for substitution into the two-dimensional isotropic diffusion equation are:

∂c∂ t

=− 1t2 φ − r

2t2√

tdφdξ

,

∂c∂ r

=1

t√

tdφdξ

, (S2.91)

∂ 2c∂ r2 =

1t2

d2φdξ 2 .

Substitution of these expressions into the diffusion equation yields,

− 1t2 φ − r

2t2√

tdφdξ

=Dr

1t√

tdφdξ

+Dt2

d2φdξ 2 . (S2.92)

By multiplying both sides of this equation by t2,

−φ − r2√

tdφdξ

=D√

tr

dφdξ

+Dd2φdξ 2 , (S2.93)

and then multiplying by 2r/√

t = 2ξ , we obtain

2Dξd2φdξ 2 +2D

dφdξ

+ξ 2 dφdξ

+2ξ φ = 0. (S2.94)


We can write this equation as a first-order equation by observing that

2Dξd2φdξ 2 +2D

dφdξ

=d

dξ

(2Dξ

dφdξ

),

(S2.95)

ξ 2 dφdξ

+2ξ φ =d

dξ(ξ 2φ) ,

so that

2Dξd2φdξ 2 +2D

dφdξ

+ξ 2 dφdξ

+2ξ φ =d

dξ

(2Dξ

dφdξ

+ξ 2φ)= 0. (S2.96)

The solution to this equation is

2Dξdφdξ

+ξ 2φ = constant . (S2.97)

As in Exercise 2.9, this constant is zero (for the same reasons), with the resulting equation for φreducing to

2Ddφdξ

+ξ φ = 0, (S2.98)

which is the same as that in (S2.80). The solution is

φ(ξ ) = Ae−ξ 2/4D . (S2.99)

Imposing normalization,

2πAt

∫∞

0e−r2/4Dtr dr =

2πAt

(4Dt)∫

∞

0e−s2

sds

= 8πDA(−1

2e−s2

)∣∣∣∣∞0= 4πDA = 1. (S2.100)

Hence, A = 1/(4πD), and we obtain the fundamental solution

c(r, t) =1

4πDtexp(− r2

4Dt

). (S2.101)

Exercise 2.11The fundamental solutions along the x- and y-directions are

c(x, t) =1√

4πDtexp(− x2

4Dt

), c(y, t) =

1√4πDt

exp(− y2

4Dt

). (S2.102)

The product of these solutions,

c(x, t)c(y, t) =1

4πDtexp(−x2 + y2

4Dt

)=

14πDt

exp(− r2

4Dt

), (S2.103)


with x2 + y2 = r2, is the same as that in (S2.101). We conclude from this that an isotropic randomwalk in two dimensions is equivalent to independent random walks along the x- and y- directions.In d spatial dimensions, we have that the fundamental solution is

c(r, t) =1

(4πDt)d/2 exp(− r2

4Dt

). (S2.104)

The spatial dimension enters only in the prefactor, so in light of the first part of this problem, weconclude that this expression is an n-fold product of one-dimensional random walks. The prefactorindicates that the decay of the solution decreases for the same elapsed time with increasing dimen-sion. This is due to the fact that there is simply more space for the random to explore as the spatialdimension increases.

Chapter 3

Atomic Diffusion in Solids

Exercise 3.1

(a) The number of ways of arranging n interstitial atoms on N sites is determined by first notingthat there N sites for the first interstitial atom, N−1 for the second, N−2 for the third, andso on, until we reach the nth interstitial, for which there are N−n+1 available sites. As theorder in which we place the vacancies is immaterial, the number of ways Ωi of arranging theinterstitials is

Ωi =N(N−1) · · ·(N−n+1)

n!=

N!(N−n)!n!

, (S3.1)

where n! represents the number of permutations of the n interstitials. The same procedurecan be used to count the number Ωv of distinct ways of creating n vacancies:

Ωv =N!

(N−n)!n!, (S3.2)

Since these numbers are independent of one another, the total number Ω of ways of creatinga Frenkel defect is

Ω = ΩiΩv =

[N!

(N−n)!n!

]2

. (S3.3)

(b) The entropy S = k lnΩ. By using Stirling’s approximation to evaluate the factorials, weobtain

S = 2kB[lnN!− ln(N−n)!− lnn!

]= 2kB

[N lnN−N− (N−n) ln(N−n)+(N−n)−n lnn+n

]= 2kB

[N lnN− (N−n) ln(N−n)−n lnn

]. (S3.4)

The energy of n noninteracting defects is nEv, so the Helmholtz free energy of the defects is

F =U−T S = nEv−2kBT[N lnN− (N−n) ln(N−n)−n lnn

]. (S3.5)

33


(c) The minimum of F is obtained by differentiating with respect to n,

∂F∂n

= Ev−2kBT[ln(N−n)+1− lnn−1

]= Ev−2kBT

[ln(N−n)− lnn

], (S3.6)

setting the result equal to zero,

Ev = 2kBT ln(

N−nn

), (S3.7)

and solving for n,

n =Ne−Ev/2kBT

1+ e−Ev/2kBT. (S3.8)

This result is valid only if n N, i.e. for temperatures that are not too high in the sense thate−Ev/kBT 1, in which case we obtain

n = N exp(− Ev

2kBT

). (S3.9)

Exercise 3.2

(a) Surface and contour plots of the function

V (x,y,z) =Vs

[e−cos(2πx/b)−cos(2πy/b)−2az−2e−az

], (S3.10)

in Fig. S3.1 show that the minima are located at

(x,y,z) = (mb,nb,zmin) , (S3.11)

in which m and n are integers and zmin will be determined below by minimizing V (0,0,z).The saddle points are of two types: (i) a maximum with respect to variations along thex-direction, but a minimum with respect to variations along the y- and z-axes, and (ii) amaximum with respect to variations along the y-direction, but a minimum with respect tovariations along the x- and z-axes. These are located at

(i): (x,y,z)=[(

m+ 12

)b,nb,zSP

], (S3.12)

(ii): (x,y,z)=[mb,(n+ 1

2

)b,zSP

], (S3.13)

where m and n are integers. The quantities zmin and zSP are determined by minimizing thepotential at the appropriate positions. At (mb,nb,z),

Vmin(mb,nb,z) =Vs(e−2−2αz−2e−αz) . (S3.14)

Atomic Diffusion in Solids 35

x

y

V

0 2 4 6 80

2

4

6

8

x

y

(a) (b)

1

Figure S3.1: (a) Surface and (b) contour plots of the potential energy in (S3.10) for Vs = 0.2 eV,b = 3A, and α = 2A−1 at z = 0.

Differentiating this expression with respect to z,dVmin

dz=−2αVse−αz(e−2−αz−1

), (S3.15)

and setting the result equal to zero yields

zmin =−2α

=−1 A . (S3.16)

Similarly, at[(

m+ 12

)b,nb,z

],

VSP((

m+ 12

)b,nb,z

)=Vs

(e−2αz−2e−αz) . (S3.17)

Differentiating with respect to z,dVSP

dz=−2αVse−αz(e−αz−1

), (S3.18)

and setting the result equal to zero yields

zSP = 0. (S3.19)

(b) Differentiating (S3.10) in turn with respect to x, y, and z yields

∂V∂x

=2πVs

bsin(

2πxb

)e−cos(2πx/b)−cos(2πy/b)−2az ,

∂V∂y

=2πVs

bsin(

2πyb

)e−cos(2πx/b)−cos(2πy/b)−2az , (S3.20)

∂V∂ z

=−2αe−αzVs

[e−cos(2πx/b)−cos(2πy/b)−az−1

].


The extreme values of V are found by setting these expressions equal to zero. For Vx and Vy,this necessitates setting the sine factors equal to zero:

sin(

2πxb

)= 0, sin

(2πy

b

)= 0. (S3.21)

The solutions of these equations are

2πxb

= nπ ,2πx

b= mπ , (S3.22)

for any integers m and n, i.e.

x =nb2

, y =mb2

. (S3.23)

If n is even, then n = 2p, and if n is odd, then n = 2p+1, for integer p, and similarly for m.This yields the two types of solutions for the (x,y) coordinates for minima and saddle-points.The corresponding values for z are obtained as in (a).

The potentials at Vmin and VSP are determined as

Vmin =V (mb,nb,zmin) =Vs(e2−2e2) =−Vse2 =−1.48 eV, (S3.24)

andVSP =V

(mb,(n+ 1

2

)b,zSP

)=−Vs =−0.2 eV. (S3.25)

Hence, the activation energy for diffusion is

VSP−Vmin =−0.2 eV+1.48 eV = 1.28 eV. (S3.26)

(c) We can choose any minimum and any saddle point about which to perform the expansion ofthe potential to second order. For the expansion about the minimum, we take (0,0,−1 A)as our reference point. Expanding (S3.10) in Taylor series and using the fact that all firstderivatives vanish, we have the following terms

V(

0,0,− 2α

)=Vs

(e2−2e2)=−Vse2 ,

Vxx

(0,0,− 2

α

)=

(2πb

)2

Vse2 ,

Vxy

(0,0,− 2

α

)=Vxz

(0,0,− 2

α

)= 0,

(S3.27)

Vyy

(0,0,− 2

α

)=

(2πb

)2

Vse2 ,

Vyx

(0,0,− 2

α

)=Vyz

(0,0,− 2

α

)= 0,

Vzz

(0,0,− 2

α

)= 2α2Vse2 .


Then, upon setting b = 3 A, we obtain

V (x,y,z) =−Vse2 +Vse2 2π2

9x2 +Vse2 2π2

9y2 +Vse24(z+1)2

=Vse2[−1+

2π2

9x2 +

2π2

9y2 +4(z+1)2

]. (S3.28)

For the expansion about the saddle point, we take(1

2b,0,0)

as a reference point. We obtain

V(

b2

,0,0)=−Vs ,

Vxx

(b2

,0,0)=−

(2πb

)2

Vs ,

Vxy

(b2

,0,0)=Vxz

(b2

,0,0)= 0,

(S3.29)

Vyy

(b2

,0,0)=

(2πb

)2

Vs ,

Vyx

(b2

,0,0)=Vyz

(b2

,0,0)= 0,

Vzz

(b2

,0,0)= 2α2Vs .

Then, by setting b = 3 A and α = 2 A−1, we obtain

V (x,y,z) =−Vs−Vs2π2

9(x−1.5)2 +Vs

2π2

9y2 +4Vsz2

=Vs

[−1− 2π2

9(x−1.5)2 +

2π2

9y2 +4z2

]. (S3.30)

The expansions in (S3.28) and (S3.30) are already in normal mode coordinates, so we canidentify the corresponding values of the effective spring constants with the second derivativewith respect to the appropriate variable. In the units we are using, we must multiply thenumerical value by the conversion factor

ω ≡[

J/eV(m/A)mH

]1/2

= 9.82×1013 s−1 , (S3.31)


in which mH is the mass of the hydrogen atom in kilograms. Hence, at the minimum, wehave

ω I,x = ω[(

2πb

)2

Vse2]1/2

= 2.50×1014 s−1 ,

ω I,y = ω[(

2πb

)2

Vse2]1/2

= 2.50×1014 s−1 , (S3.32)

ω I,z = ω(2α2Vse2)1/2

= 3.38×1014 s−1 .

At the saddle point, we have only two stable modes:

ωD,y = ω[(

2πb

)2

Vs

]1/2

= 9.20×1013 s−1 ,

(S3.33)ωD,z =

[2α2Vs

]1/2= 1.24×1014 s−1 .

Hence, the prefactor in the harmonic approximation to transition-state theory is

∏3i=1 νI,i

∏2i=1 νD,i

=1

2π∏

3i=1 ωI,i

∏2i=1 ωD,i

=1

2πωI,xωI,yωI,z

ωD,yωD,z

=

(2.50×1014)2 (3.38×1014)

2π(9.2×1013

)(1.24×1014

) = 2.95×1014 . (S3.34)

The complete transition rate K is therefore

K(T ) = 2.95×1014 exp(−1.28 eV

kBT

). (S3.35)

(d) The transition rate (S3.35) at T = 300 K, 400 K, and 500 K is

K(300 K) = 9.25×10−8 s−1 ,

K(400 K) = 0.022 s−1 , (S3.36)

K(500 K) = 36 s−1 ,

from which we obtain the corresponding average residence times τ(T ) = 1/K(T ):

τ(300 K) = 1.08×107 s ,

τ(400 K) = 45 s , (S3.37)

τ(500 K) = 0.027 s .


Exercise 3.3The energy of the one-dimensional harmonic oscillator is

E(q, p) =p2

2m+

kq2

2, (S3.38)

from which we calculate the classical partition function as

Z =1h

∫∞

−∞

d p∫

∞

−∞

dqe−βE(q,p)

=1h

∫∞

−∞

d pexp(− p2

2mkBT

)∫∞

−∞

dqexp(− kq2

2kBT

). (S3.39)

By changing variables tos =

p√2mkBT

(S3.40)

in the integral over p, and

t = q

√k

2kBT(S3.41)

in the integral over q, we obtain

Z =2kBT

h

√mk

∫∞

−∞

e−s2ds︸︷︷︸√

π

∫∞

−∞

e−t2dt︸︷︷︸√

π

=2πkBT

h

√mk

. (S3.42)

The natural frequency ν = 2πω of the oscillator is

ν = 2π√

km

, (S3.43)

in terms of which we can write the partition function as

Z =kBThν

. (S3.44)

The Helmholtz free energy F =−kBT lnZ for the harmonic oscillator is

F =−kBT ln(

kBThν

). (S3.45)

Since dF =−SdT −PdV , the entropy is calculated as

S =−∂F∂T

= kB ln(

kBThν

)+ kB , (S3.46)


which when solved for the frequency, yields

ν =kBT

hexp(− S

kB+1)

. (S3.47)

The transition rate calculated within the harmonic approximation of transition-state theory isgiven in (3.40):

K =∏

3Ni=1 νI,i

∏3N−1i=1 νD,i

e−(VSP−Vmin)/kBT . (S3.48)

Each frequency factor in this expression can be expressed as in (S3.47):

νI,i =kBT

hexp(−SI,i

kB+1)

,

(S3.49)

νD,i =kBT

hexp(−SD,i

kB+1)

.

Hence, the numerator in (S3.48) becomes

3N

∏i=1

νI,i =

(kBT

h

)3N

e3N3N

∏i=1

exp(−SI,i

kB

)

=

(kBT

h

)3N

e3N exp(− 1

kB

3N

∑i=1

SI,i

)

=

(kBT

h

)3N

e3N exp(− 1

kBSmin

), (S3.50)

where Smin is the total entropy of the minimum energy state. Similarly, for the denominator in(S3.48), we obtain

3N−1

∏i=1

νD,i =

(kBT

h

)3N−1

e3N−1 exp(− 1

kBSSP

), (S3.51)

where SSP is the total entropy of the saddle point. Hence, the frequency prefactor in (S3.48) is

∏3Ni=1 νI,i

∏3N−1i=1 νD,i

=

(kBT

he)

exp[

1kB

(SSP−SImin)]

. (S3.52)

Before proceeding, we comment on the factor of e in the frequency prefactor. This originatesin the additive factor of kB in the entropy (S3.46) of the harmonic oscillator. Since hν is theenergy difference between successive states of the quantum harmonic oscillator, then for classicalstatistical mechanics to provide a valid description of this oscillator, we must have that kBT >> hν ,in which case we can neglect the addition factor of kB in comparison to the logarithm. With thisproviso, the frequency prefactor in (S3.48) reduces to

∏3Ni=1 νI,i

∏3N−1i=1 νD,i

=kBT

hexp[

1kB

(SSP−Smin)

], (S3.53)


and the transition rate in (S3.48) becomes

K =kBT

hexp[− 1

kBT(∆V −T ∆S)

], (S3.54)

in which∆V =VSP−Vmin , ∆S = SSP−Smin . (S3.55)

How do we interpret the prefactor in (S3.54)? If we imagine the moving particle as a harmonicoscillator in the original potential energy minimum, then according to the equipartition theorem,we have that the average energy 〈E〉= kBT , i.e. 1

2kBT each for the position and momentum degreesof freedom. If we equate this to the energy of an effective frequency hν(T ), we obtain

ν(T ) =kBT

h, (S3.56)

which is the prefactor in (S3.54).

Exercise 3.4Given that cA + cB = constant, we have that

∂cA

∂ t+

∂cB

∂ t= 0, (S3.57)

and∂cA

∂x+

∂cB

∂x= 0. (S3.58)

Thus,∂cB

∂ t=−∂cA

∂ t,

∂cB

∂x=−∂cA

∂x. (S3.59)

Hence, the diffusion equation for B,

∂cB

∂ t=

∂∂x

(DB

∂cB

∂x

), (S3.60)

can be written as∂cA

∂ t=

∂∂x

(DB

∂cA

∂x

). (S3.61)

Subtracting this equation from that for cA,

∂cA

∂ t=

∂∂x

(DA

∂cA

∂x

), (S3.62)

yields∂∂x

(DA

∂cA

∂x

)− ∂

∂x

(DB

∂cA

∂x

)=

∂∂x

[(DA−DB)

∂cA

∂x

]= 0, (S3.63)

which implies that DB = DA. Hence, the diffusion of cA and cB are governed by the same equation.


Exercise 3.5Using the solution in Exercise 2.8, we first set c0 equal to the concentration of P at the surface,c0 = 1021 atoms/cm2. Then by setting the value of the solution to the concentration of B, i.e. 1016

atoms/cm2,

c(x, t) = 1021[

1− erf(

x√4Dt

)]= 1016 , (S3.64)

yields, after a simple rearrangement,

erf(

x√4Dt

)= 1− 1016

1021 . (S3.65)

Solving for the value of x/√

4Dt produces

x√4Dt

= 3.12 , (S3.66)

which, in a form solved for t, is

t =x2

4D(3.12)2 . (S3.67)

With x given by the junction depth of 1 µm = 10−4 cm and D = 1.28×10−14 cm2/s, we obtain

t =(10−4)2

4×1.28×10−14× (3.12)2 = 20064 s = 5.6 hr . (S3.68)

Exercise 3.6

(a) The concentration in the steel initially is c(x,0) = c0 and the boundary conditions for thecarbon concentration at the surface of the product is c(0, t) = cs.

(b) By shifting the concentration according to c(x, t) = u(x, t)+ cs, we have that u also satisfiesthe diffusion equation, but with the initial condition u(x,0) = c0− cs and the boundary con-dition u(0, t) = 0, i.e. the problem for u is expressed in terms of an absorbing boundary atx = 0. Hence, the solution to the boundary-value problem for u is

u(x, t) =c0− cs√

4πDt

∫∞

0exp[−(x− x′)2

4Dt

]dx′− c0− cs√

4πDt

∫∞

0exp[−(x+ x′)2

4Dt

]dx′ , (S3.69)

where the superposition of fundamental solutions guarantees that the boundary condition atx = 0 is satisfied at all times.

(c) The integrals in (b) can be evaluated with the variable transformation

s =x′± x√

4Dt←→ x′ = s

√4Dt∓ x , (S3.70)


which yields

u(x, t) =(c0− cs)√

π

∫∞

−x/√

4Dte−s2

ds− (c0− cs)√π

∫∞

x/√

4Dte−s2

ds

=(c0− cs)√

π

∫ x/√

4Dt

−x/√

4Dte−s2

ds =2(c0− cs)√

π

∫ x/√

4Dt

0e−s2

ds

= (c0− cs)erf(

x√4Dt

). (S3.71)

Thus, the solution for c is

c(x, t) = u(x, t)+ cs = cs +(c0− cs)erf(

x√4Dt

). (S3.72)

(d) The definition of xs yields

c(xs, t) = cs +(c0− cs)erf(

xs√4Dt

)=

c0 + cs

2, (S3.73)

which may be rearranged as

erf(

xs√4Dt

)=

12

. (S3.74)

Thus,xs√4Dt≈ 1

2, (S3.75)

so xs ≈√

Dt.


Chapter 4

Spinodal Decomposition

Exercise 4.1(a) Beginning with the free energy F ,

F = N(EAB−EAA

)cAcB +NkBT

(cA lncA + cB lncB

), (S4.1)

we make the substitutionscA = 1

2 + c , cB = 12 − c , (S4.2)

in which −12 ≤ c≤ 1

2 , to obtain

F = N(EAB−EAA

)(12 + c

)(12 − c

)+NkBT

[(12 + c

)ln(1

2 + c)+(1

2 − c)

ln(1

2 − c)]

. (S4.3)

Since,ln(1

2 ± c)= ln

(12

)+ ln(1±2c) =− ln2+ ln(1±2c) , (S4.4)

we can use the Taylor series of the natural logarithm,

ln(1+ x) = x− 12x2 + 1

3x3− 14x4 , (S4.5)

to obtain(12 + c

)ln(1

2 + c)+(1

2 − c)

ln(1

2 − c)

=(1

2 + c)[− ln2+ ln(1+2c)

]+(1

2 − c)[

ln(1

2

)+ ln(1−2c)

]=− ln2+

(12 + c

)ln(1+2c)+

(12 − c

)ln(1−2c)

=− ln2+(1

2 + c)(

2c−2c2 + 83c3−4c4)+ (1

2 − c)(−2c−2c2− 8

3c3−4c4)=− ln2+

(−2c2−4c4)+ (4c2 + 16

3 c4)=− ln2+2c2 + 4

3c4 . (S4.6)

45


Notice that, because of the symmetry of the original free energy under the interchange of cAand cB, only even powers of c appear in this expansion. The quartic approximation to thefree energy is

F = N(EAB−EAA

)(14 − c2)+NkBT

[− ln2+2c2 + 4

3c4]= 1

4N(EAB−EAA

)−NkBT ln2+N

[2kBT − (EAB−EAA

)]c2 + 4

3NkBT c4 . (S4.7)

This expression can be simplified by proceeding as in the main text and dividing both sidesof the equation by N

(EAB−EAA

)and defining the dimensionless temperature

t ≡ kBTEAB−EAA

, (S4.8)

whereupon we obtain

f ≡ FN(EAB−EAA)

= 14 − t ln2+(2t−1)c2 + 4

3tc4 . (S4.9)

The quartic polynomial approximation to the free energy, indicated by a solid line, togetherwith the original expression in the form

FN(EAB−EAA)

=(1

2 + c)(1

2 − c)+ t[(1

2 + c)

ln(1

2 + c)+(1

2 − c)

ln(1

2 − c)]

, (S4.10)

indicated by a broken line, are shown in Fig. S4.1. The two expressions for the free energyagree near c = 0, but discrepancies appear at the extreme points that become more apparentwith increasing t. This is because the energy is reproduced exactly by the quartic expression,but the entropy is treated only approximately. The deviations between the full and approx-imate expressions are most apparent near the inflection points near c = ±1

2 , where, withincreasing t, the entropy becomes increasingly important. But the overall trends are repro-duced by the polynomial approximation and the two expression agree near t = 1

2), which iswhere is expansion is made.

(b) The phase diagram produced by the free energy in (S4.9) requires expressions for the bi-modal and spinodal lines. The bimodal line is obtained from the points where fc = 0. Usingthe expression in (S4.9), the required derivative is

∂ f∂c

= 2(2t−1)c+ 163 tc3 = 2c

[(2t−1)+ 8

3tc2] . (S4.11)

Setting this equal to zero yields three solutions:

c = 0, c =±[

3(1−2t)8t

]1/2

. (S4.12)

Spinodal Decomposition 47

= 0

= 0.2

= 0.3

= 0.4

= 0.5

= 0.6

t

t

t

t

t

t

-0.4 -0.2 0.0 0.2 0.4-0.2

-0.1

0.0

0.1

0.2

0.3

c

F

NDE

Figure S4.1: Comparison of the free energy (S4.10) obtained from the Bragg–Williams model(broken line) withe the approximate expression (S4.9) obtained from a quartic expansion of thelogarithms.

The latter two solutions, which are real only for t < 12 , are always minima which, for t = 1

2coincide with the extreme point at c = 0. From the free energy diagram in Fig. S4.1, thispoint is a maximum for t < 1

2 , and a minimum for t > 12 .

The spinodal line is determined from the points where the second derivative vanishes. Wehave

∂ 2 f∂c2 = 2(2t−1)+16tc2 = 0, (S4.13)

which yields

c =±[

1−2t8t

]1/2

. (S4.14)

Notice that this expression also has real solutions only for t < 12 .

The phase diagram of the binary solution produced by these solutions is shown in Fig. S4.2,with the stable, metastable, and unstable regions marks as ‘s’, ‘m,’ and ‘u,’ respectively.The basic topology of this diagram is the same as that in Fig. S4.1, but the details are quitedifferent. Most apparent is the expanded unstable region at the expense of the metastableregions. The position of the critical point and the neighboring regions are represented quitewell by the approximation, which is expected, as this is the reference point for the expansion.


Show@%12, Graphics@8Disk@80, 0.5<, 80.02, 0.01<D,Text@Style@"critical point", FontFamily Ø TimesD, 80., 0.525<D,Text@

Style@"s", FontFamily Ø Times, FontSlant -> Italic, FontSize Ø 16D, 8-0.4, 0.5<D,Text@Style@"s", FontFamily Ø Times, FontSlant -> Italic, FontSize Ø 16D,

80.4, 0.5<D,Text@Style@"m", FontFamily Ø Times, FontSlant -> Italic, FontSize Ø 16D,

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80., 0.2<D<DD

critical points s

m m

u

-0.5 -0.25 0. 0.25 0.50.0

0.1

0.2

0.3

0.4

0.5

c

t

PS4.2.nb 3

Figure S4.2: The phase diagram for the Bragg–Williams model obtained from the quartic approx-imation (S4.9) to the free energy.

Exercise 4.2(a) For a binary mixture,

µα = µ0α + kBT lnaα = µ0

α + kBT ln(γαxα) , (S4.15)

where α = A,B , aα is the activity of species α and γα is the corresponding activity coeffi-cient, and

xα =cα

cA + cB. (S4.16)

Thus,dµαdx

=dµαdxα

dxαdx

, (S4.17)

and, since cA + cB = 1, so c′A + c′B = 0 (the prime indicates the derivative), we have

dxαdx

=ddx

(cα

cA + cB

)=

c′αcA + cB

. (S4.18)

Hence,

Jα = cαmα

(−dµα

dx

)=−cαmα

(1

cA + cB

dµA

dxα

dcαdx

)=−mα

(xα

dµαdxα

dcαdx

)=−mα

dµαd lnxα

dcαdx

. (S4.19)


Referring to (S4.15), we have

dµαd lnxα

= kBT(

d lnγαdlnxα

+1)

. (S4.20)

Substitution of this result into (S4.19) yields

Jα =−mαkBT(

1+d lnγαd lnxα

)dcαdx

. (S4.21)

(b) Beginning with G = NAµA +NBµB, where NA and NB are the numbers of A and B atoms,respectively, the Gibbs free energy g per atom in a binary system is

g =GN

=NAµA

N+

NBµB

N= xAµA + xBµB . (S4.22)

Hence, focussing on component A,

dgdxA

= µA +µBdxB

dxA= µA +µB

d(1− xA)

xA= µA−µB (S4.23)

d2gdx2

A=

dµA

dxA− dµB

dxB

dxB

dxA=

dµA

dxA+

dµB

dxB. (S4.24)

The Gibbs–Duhem equation is a consequence of G being a state function. The fundamentalequation for g is given in (1.39):

dg =−sdT + vdP+µAdxA +µBdxB , (S4.25)

where quantities in lower case indicate the corresponding thermodynamic variable per atom:s = S/N and v =V/N. As G is a function of state, (S4.22) and (S4.25) must be equal, so atconstant temperature and pressure, we obtain

xAdµA + xBdµB = 0, (S4.26)

which is the Gibbs–Duhem equation. For our purposes, this equation is used to write

xAdµA

xAdxA + xB

dµB

dxB

dxB

dxAdxA =

(xA

dµA

xA− xB

dµB

dxB

)dxA = 0, (S4.27)

which can be rearranged asdµB

dxB=

xA

xB

dµA

dxA. (S4.28)

By substituting this result into (S4.25) and using (S4.15), we obtain

d2gdx2

A=

dµA

dxA+

xA

xB

dµA

dxA=

1xB

dµA

xA

=1xB

(kBT

d lnγA

dxA+

kBTxA

)=

kBTxAxB

(1+ xA

d lnγA

dxA

)=

kBTxAxB

(1+

d lnγA

d lnxA

). (S4.29)


Comparing this result with (S4.21), we can write

JA =−(

xAxBmAd2gdx2

A

)dcA

dx≡−DA

dcA

dx. (S4.30)

so the sign of Dα depends on the sign of the second derivative of the Gibbs free energy. Inthe case of uphill diffusion DA < 0.

Exercise 4.3Combining the continuity equation for a concentration c,

∂c∂ t

+∇∇∇·JJJ = 0, (S4.31)

the generalized statement of Fick’s first law,

JJJ =−K(c)∇∇∇µ , (S4.32)

in which K(c) is a transport coefficient which we take as a constant, and µ is the chemical potential,with

µ =−γ∇∇∇2c , (S4.33)

where γ is a surface tension, yields

∂c∂ t

=−∇∇∇·JJJ = K∇∇∇2µ =−γK∇∇∇

4c , (S4.34)

which is the Mullins–Herring equation:

∂c∂ t

=−γK∂ 4c∂x4 . (S4.35)

The minus sign on the right-hand side of this equation is the stable sign, i.e. initial disturbances ofany wave number k > 0 decay with time, as we will now show.

Fourier transforming this equation and neglecting boundary terms yields,

∂∂ t

∫∞

−∞

c(x, t)eikx dx =−γK∫

∞

−∞

∂ 4c∂x4 eikx dx

= iγKk∫

∞

−∞

∂ 3c∂x3 eikx dx = γKk2

∫∞

−∞

∂ 2c∂x2 eikx dx

=−iγKk3∫

∞

−∞

∂c∂x

eikx dx =−γKk4∫

∞

−∞

c(x, t)eikx dx . (S4.36)


k2

k4

0 1 20

1

2

3

4

k

k2 ,k4

Figure S4.3: Comparison between k2 (dotted curve) and k4 (solid curve). For k < 1, k2 is greaterthan k4, while, for k > 1, k4 is greater than k2.

In terms of the Fourier transform

c(k, t) =∫

∞

−∞

c(x, t)eikx dx , (S4.37)

this equation reads∂ c∂ t

=−γKk2c , (S4.38)

the solution to which isc(k, t) = c(k,0)e−γKk4t . (S4.39)

The fourth power of the wave number in the exponential means that the relaxation of sharp(i.e. large k) concentration profiles is much faster than that of the diffusion equation, which hasonly a quadratic dependence on k. However, the relaxation of small-k profiles is slower. The dia-gram below shows k2 and k4, both with unit coefficients. The crossover point that separates whereeach is the dominant term is k = 1..

Exercise 4.4

(a) The free energy obtained in Exercise 4.1 is

f =F

N(EAB−EAA)= 1

4 − t ln2+(2t−1)c2 + 43 tc4 , (S4.40)


where t = kBT/(EAB−EAA). Substitution of this expression into the general form of theCahn–Hilliard equation, and assuming a constant K, yields

∂c∂ t

= K∂ 2

∂x2

[−2(1−2t)+

16t3

c2−2κ∂c∂x2

]

= 2K∂ 2

∂x2

[−(1−2t)c+

8t3

c3−κ∂ 2c∂x2

]. (S4.41)

(b) The Cahn–Hilliard equation (S4.41) can be written in dimensionless form (i.e. where theconstants are eliminated), by introducing rescaled variables u, ξ , and τ defined by

c = Au , ξ = Bx , τ =Ct , (S4.42)

in which A, B, and C are constants to be determined. By calculating the derivatives using thechain rule, we obtain

∂∂ t

=∂τ∂ t

∂∂τ

=C∂

∂τ, (S4.43)

∂ n

∂xn =

(∂ξ∂x

)n ∂ n

∂ξ n = Bn ∂ n

∂ξ n , (S4.44)

since B is a constant. Substitution of these transformations into (S4.41) yields

AC∂u∂τ

= 2KB2 ∂ 2

∂ξ 2

[−(1−2t)Au+

8t3

A3u3−κAB2 ∂ 2u∂ξ 2

], (S4.45)

which can be written as

∂u∂τ

=−(1−2t)2KB2

C∂ 2u∂ξ 2 +2K

B2A2

C8t3

u3−2κKB4

C∂ 4u∂ξ 4 . (S4.46)

Setting each coefficient equal to one, yields three equations:

(1−2t)2KB2

C= 1, 2K

B2A2

C8t3= 1 2κK

B4

C. (S4.47)

Dividing the second equation by the first yields

A2 =3(1−2t)

8t, (S4.48)

or

A =

[3(1−2t)

8t

] 12

. (S4.49)


Similarly, dividing the first equation by the third yields

B2 =1−2t

κ, (S4.50)

or,

B =

[1−2t

κ

] 12

. (S4.51)

Then, solving the first equation for C and using the solution for B yields

C = 2K(1−2t)B2 =2Kκ

(1−2t)2 . (S4.52)

Hence, we obtain∂u∂τ

=∂ 2

∂ξ 2

(−u+u3− ∂ 2u

∂ξ 2

). (S4.53)

as the dimensionless form of the Cahn–Hilliard equation.

(c) We attempt a trial stationary solution to (S4.53) of the form u = a tanh(bξ ), a and b are tobe determined. Rather than substituting this form directly into the Cahn–Hilliard equation,we first observe that a stationary solution of this equation satisfies

d2

dξ 2

(−u+u3− d2u

dξ 2

)= 0, (S4.54)

the general solution to which is

−u+u3− d2udξ 2 = Aξ +B , (S4.55)

where A and B are constants of integration. We can use what know about the behaviorof the hyperbolic tangent function to determine the values of these constants. At ξ = 0,tanh(bξ ) = 0 (because tanhx is an odd function), and the curvature vanishes there as well(because this function is linear near the origin). Hence, B = 0. To determine A, we use thefact that

limξ→∞

a tanhbξ = a , limξ→∞

(a

d tanhbξdξ

)= 0, (S4.56)

to take the limits of both sides of (S4.64) as ξ → ∞. The left-hand side approaches −a+a3,which is finite, while Aξ becomes unbounded. Hence, we must choose A = 0. Thus, theequation that our trial solution must solve is, in fact, a second-order equation:

−u+u3− d2udξ 2 = 0. (S4.57)


The required derivative is

d2udξ 2 =−2ab2 (sechbξ )2 tanhbξ , (S4.58)

substitution of which into (S4.57) yields,

−a tanhbξ +(a tanhbξ )3 +2ab2 (sechbξ )2 tanhbξ

=−a tanhbξ[1− (a tanhbξ )2−2b2 (sechbξ )2

]= 0. (S4.59)

We can use standard identities for hyperbolic functions,

tanhx =sinhxcoshx

, sechx =1

coshx, (S4.60)

to write this expression as

− a tanhbξcosh2 bξ

[cosh2 bξ −a2 sinh2 bξ −2b2

]=−a tanhbξ

cosh2 bξ

[1−2b2 +(1−a2)sinh2 bξ

]= 0, (S4.61)

which can be made to vanish by choosing a and b to satisfy 1−2b2 = 0 and 1−a2 = 0, i.e.

a =±1, b =1√2

. (S4.62)

Choosing a = 1 (the choice a =−1 yields a symmetrically equivalent solution), the solutionto (S4.53) is therefore,

u(ξ ) = tanh(

ξ√2

), (S4.63)

or, in the original variables, the solution to (S4.41) is

c(x) =[

3(1−2t)8t

]1/2

tanh[

(1−2t)2κ

]1/2

x

. (S4.64)

(d) The solution smoothly interpolates between the concentrations

c =±[

3(1−2t)8t

]1/2

, (S4.65)

where fc = 0, which are the ‘bimodal’ points. Hence, this solution describes the stationaryinterface between the two stable compositions of the alloy at temperature t < 1

2 . The quantityκ determines the width of the interface, which becomes sharper as κ increases.


At these compositions fcc > 0, so small compositional changes cause the free energy to in-crease. This system is therefore locally stable with respect to local compositional fluctuationsbut, as the stationary solution to the Cahn–Hilliard equation shows, it is globally unstablewith respect to separation into two coexisting phases. There is a free energy barrier that mustbe surmounted to attain the global minimum of the phase-separated state. Consequently, thiscomposition is metastable, that is, not in equilibrium, but with a long lifetime.

Exercise 4.5A variation on the Cahn–Hilliard equation is

∂u∂τ

=∂ 2

∂ξ 2

(−u+ ru2 +u3− ∂ 2u

∂ξ 2

), (S4.66)

where r is a constant. By writing this equation as

∂u∂τ

=∂ 2

∂ξ 2

(d fdu− ∂ 2u

∂ξ 2

), (S4.67)

the free energy corresponding to this equation is

f =−u2

2+

ru3

3+

u4

4. (S4.68)

This free energy is plotted in Fig. S4.4.We consider the stationary form of (S4.66):

d2

dξ 2

(−u+ ru2 +u3− d2u

dξ 2

)=− d2u

dξ 2 + rd2u2

dξ 2 +d2u3

dξ 2 −d4udξ 4 = 0. (S4.69)

-2 -1 0 1 2-0.4-0.20.00.20.40.60.81.0-2 -1 0 1 2

-0.4-0.20.00.20.40.60.81.0

-2 -1 0 1 2-0.4-0.20.00.20.40.60.81.0-2 -1 0 1 2

-0.4-0.20.00.20.40.60.81.0

-2 -1 0 1 2-0.4-0.20.00.20.40.60.81.0-2 -1 0 1 2

-0.4-0.20.00.20.40.60.81.0

2 2 21 1 10 0 01 1 12 2 2

0.4 0.4

0 0

0.4 0.4

0.8 0.8

u u u

f

1

Figure S4.4: The free energy (S4.68) for r =−14 (left), r = 0 (center), and r = 1

4 (right). The quan-tity r is clearly seen to be an asymmetry parameter that extends the capability of the polynomialapproximation to situations where EAA 6= EBB.


For a trial solution of the formu(ξ ) = a+b tanhcξ , (S4.70)

in which a, b, and c are constants to be determined, the required derivatives are

d2udξ 2 =−2bc2(sechcξ )2 tanhcξ ,

d4udξ 4 = 16bc4(sechcξ )4 tanhcξ −8bc4(sechcξ )2(tanhcξ )3 ,

(S4.71)d2(u2)

dξ 2 = 2b2c2(sechcξ )4−4abc2(sechcξ )2 tanhcξ −4b2c2(sechcξ )2(tanhcξ )2 ,

d(u3)

dξ 2 = 6ab2c2(sechcξ )4 +6b3c2(sechcξ )4 tanhcξ −6a2bc2(sechcξ )2 tanhcξ

− 12ab2c2(sechcξ )2(tanhcξ )2−6b3c2(sechcξ )2(tanhcξ )3 .

Substituting these expressions into (S4.66) yields

− d2udξ 2 + r

d2(u2)

dξ 2 +d2(u3)

dξ 2 −d4udξ 4

= 2bc2(sechcξ )2 tanhcξ +2rb2c2(sechcξ )4−4rabc2(sechcξ )2 tanhcξ

− 4rb2c2(sechcξ )2(tanhcξ )2 +6ab2c2(sechcξ )4 +6b3c2(sechcξ )4 tanhcξ

− 6a2bc2(sechcξ )2 tanhcξ −12ab2c2(sechcξ )2(tanhcξ )2

− 6b3c2(sechcξ )2(tanhcξ )3−16bc4(sechcξ )4 tanhcξ +8bc4(sechcξ )2(tanhcξ )3

= 2bc2(sechcξ )4[

tanhcξ(sechcξ )2 + rb−2ra

tanhcξ(sechcξ )2 −2rb

(tanhcξsechcξ

)2

+3ab

+ 3b2 tanhcξ −3a2 tanhcξ(sechcξ )2 −6ab

(tanhcξsechcξ

)2

−3b2 (tanhcξ )3

(sechcξ )2

− 8c2 tanhcξ +4c2 (tanhcξ )3

(sechcξ )2

]= 2bc2(sechcξ )4

[(r+3a)b+(1−2ra−3a2)coshcξ sinhcξ

− 2(r+3a)b(sinhcξ )2 +(3b2−8c2) tanhcξ +(4c2−3b2)(sinhcξ )3

coshcξ

]


= 2bc2(sechcξ )4[(r+3a)b+(1−2ra−3a2 +4c2−3b2)coshcξ sinhcξ

−2(r+3a)(sinhcξ )2−6(b2−2c2) tanhcξ]

. (S4.72)

This expression can be made to vanish by choosing a, b, and c to satisfy

r+3a = 0, 1−2ra−3a2 +4c2−3b2 = 0, b2−2c2 = 0. (S4.73)

The first equation immediately yields a = −13r and the third equation required that b2 = 2c2.

Substituting these results into the second equation yields

1+2r2

3− r2

3+2b2−3b2 = 0, (S4.74)

which yieldsb2 = 1+ 1

3r2 , (S4.75)

or,

b =±√

1+ 13r2 , (S4.76)

whereupon we obtainc2 = 1

2

(1+ 1

3r2) , (S4.77)

or

c =±

√1+ 1

3r2

√2

. (S4.78)

The solution of the modified Cahn–Hilliard equation in (S4.66) is therefore, given by

u(ξ ) =− r3±(

1+r2

3

)1/2

tanh[(

1+r2

3

)1/2 ξ√2

]. (S4.79)

Exercise 4.6

(a) Given the free energy functional,

F =∫ [

ϕ2(1−ϕ)2 + 12ε(∇∇∇ϕ)2] dxxx , (S4.80)

the equation of motion for ϕ is

∂ϕ∂ t

=−δFδϕ

=− ∂∂ϕ[ϕ2(1−ϕ)2]+ ε∇∇∇

2ϕ

=−2ϕ +6ϕ2−4ϕ3 + ε∇∇∇2ϕ . (S4.81)


(b) In one spatial dimension, (S4.81) reads

∂ϕ∂ t

= ε∂ 2ϕ∂x2 −2ϕ +6ϕ2−4ϕ3 . (S4.82)

Based on the methods in Exercise 4.4 and Exercise 4.5, we attempt a trial solution of theform

ϕ(x) = a+b tanhcξ , (S4.83)

where a, b, and c are constants to be determined. The required expressions are

d2ϕdx2 =−2bc2(sechcξ )2 tanhcξ ,

φ 2 = a2 +2ab tanhcξ +b2(tanhcξ )2 , (S4.84)

φ 3 = a3 +3a2b tanhcξ +3ab2(tanhcξ )2 +b2(tanhcξ )3 .

Substitution into the stationary form of (S4.82) yields

εd2ϕdx2 −2ϕ +6ϕ2−4ϕ3

=−2εbc2(sechcξ )2 tanhcξ −2a−2b tanhcξ +6a2 +12ab tanhcξ

+ 6b2(tanhcξ )2−4a3−12a2b tanhcξ −12ab2(tanhcξ )2−4b3(tanhcξ )3

=−2a+6a2−4a3 +(12ab−2b−12ab2) tanhcξ +(6b2−12ab2)(tanhcξ )2

− 4b3(tanhcξ )3−2εbc2(sechcξ )2 tanhcξ

=−2a+6a2−4a3 +(12ab−2b−12ab2−4b3) tanhcξ

+ (6b2−12ab2)(tanhcξ )2 +(4b3−2εbc2)(sechcξ )2 tanhcξ

=−2a(1−3a+2a2)+2b(6a−1−6ab−2b2) tanhcξ +6b2(1−2a)(tanhcξ )2

+ 2b(2b2− εc2)(sechcξ )2 tanhcξ . (S4.85)

We must first of all require that 2b2− εc2 = 0 and then that 1− 2a = 0, i.e. a = 12 . This

choice of a also yields

1−3a+2a2 = 1− 32+

24= 0. (S4.86)

That leaves only the coefficient of tanhcξ :

6a−1−6ab−2b2 = 2−3b−2b2 = 0, (S4.87)


which has two solutions: b = −2 and b = 12 . Only the positive solution is ‘physical,’ in

that the phase field must interpolate between zero and one. This choice also determinesc: εc2 = 1

2 , i.e.

c =1√2ε

. (S4.88)

Hence, our solution for the stationary phase field is

u(x) =12

[1+ tanh

(x√2ε

)]. (S4.89)

Note that, as x→ ∞, u→ 1 and, as x→ −∞, u→ 0. Moreover, since ε is an effectivediffusion constant, the crossover region between these asymptotic values becomes sharperas ε decreases.

Exercise 4.7

(a) For the order parameter written as NA = a+ bW . When W = 1, all the atoms are on theirown sublattice (perfect long-range order), so NA = 1

2N = a+b. When W =−1, all atoms areon the opposite sublattice, which also corresponds to perfect long-range order, so a−b = 0.Finally, when W = 0, there is no long-range order, with the occupancy of each lattice siteassigned randomly. We therefore have NA = 1

4N = a. Hence, we have a = b = 14N, which

yieldsNA = 1

4N(1+W ) . (S4.90)

(b) To determine the equilibrium of the alloy, we must calculate the Helmholtz free energyF = U − T S as a function of W . For the entropy, note that each sublattice is a randomsolution of 1

2N sites

c =NA12N

= 12(1+W ) , (S4.91)

so that1− c = 1

2(1−W ) . (S4.92)

The total entropy of the system is twice that for each sublattice which, by following the samesteps leading to (4.5) yields

S =−NkB

[(1−W

2

)ln(

1−W2

)+

(1+W

2

)ln(

1+W2

)]. (S4.93)

(c) We now turn to the internal energy. As in Sec. 4.1, the exact internal energy is determined bythe numbers of pairs NAA, NAB and NBB pairs of AA, AB, and BB atoms, since we are againrestricting ourselves to nearest-neighbor interactions. Each atomic site is surrounded by 8nearest neighbors of the other sublattice. The number nAA of AA bonds formed by an A atom


on an A sublattice site is 8 times the fraction of A atoms that occupy these sites of the othersublattice. If we again adopt a mean-field approach and assume that this fraction is given bythe concentration of A atoms on the other sublattice, we obtain

nAA = 8× 12(1−W ) = 4(1−W ) . (S4.94)

Thus, NAA is determined by the product of nAA and the number of A atoms in the A sublattice:

NAA = nAA×NA = 4(1−W )× 14N(1+W ) = N(1−W 2) . (S4.95)

The same argument applied to B atoms yields

NBB = N(1−W 2) . (S4.96)

The determination of NAB is most expediently carried out by by observing that the total num-ber of nearest-neighbor pairs is given by the product of number 1

2N sites on one sublatticeand the number of nearest-neighbor sites on the other sublattice: 1

2N×8 = 4N. Hence,

NAB = 4N−NAA−NBB = 4N−2N(1−W 2) = 2N(1+W 2) . (S4.97)

Thus, the internal energy U is

U = NAAEAA +NBBEBB +NABEAB

= N(1−W 2)EAA +N(1−W 2)EBB +2N(1+W 2)EAB

= N(EAA +EBB +2EAB)+NW 2∆E , (S4.98)

where we have defined ∆E ≡ 2EAB−EAA−EBB < 0.

(d) The Helmholtz free energy F =U−T S is given as a function of W as

F = N(EAA +EBB +2EAB)+NW 2∆E

+NkBT[(

1−W2

)ln(

1−W2

)+

(1+W

2

)ln(

1+W2

)]. (S4.99)

The equilibrium condition is, therefore, determined by

dFdW

= 2NW∆E +NkBT[−1

2ln(

1−W2

)− 1

2+

12

ln(

1+W2

)+

12

]= 2NW∆E +

NkBT2

ln(

1+W1−W

)= 0. (S4.100)

Using the fact that ln(1+ x) ≈ x for small x, we see that as W → 0 (that is, as the alloybecomes increasingly disordered),

ln(

1+W1−W

)= ln(1+W )− ln(1−W )≈ 2W , (S4.101)


0.0 0.2 0.4 0.6 0.8 1.0 1.2-1.0

-0.5

0.0

0.5

1.0

T

Tc

W

Figure S4.5: The degree of order W as a function of the reduced temperature T/Tc for the Bragg–Williams model of the order-disorder transition.

so the equilibrium condition becomes (after cancelling the factor of N),

2W∆E + kBTW = 0. (S4.102)

Solving for T yields the critical temperature kBTc = −2∆E, which is, of course, positive,since ∆E < 0. Substitution of this expression into the equilibrium condition and rearrangingyields, finally,

Tc

T=

12W

ln(

1+W1−W

), (S4.103)

The solution of this equation for the degree of order W is shown in Fig. S4.5. At lowtemperatures, W is close to±1, indicating that the order is nearly perfect. As the temperatureis raised, the long-range order begins to be lost, slowly at first, but more rapidly as the criticaltemperature is approached and W→ 0. This behavior is caused by the fact that, in a perfectly-ordered alloy, the movement of atoms between sublattices is energetically unfavorable, as theinternal energy dominates the entropy in the Helmholtz free energy. But as the temperatureis raised, and more atoms exchange sublattices, the ordered energy becomes less effective,and the effect of entropy begins to be important. Finally, at the critical point and beyond,there is no long-range order at all.

The phase transition at Tc is of second-order, as there is no latent heat. The transition issmooth, in that there no abrupt change in structure, and the two phases, corresponding toW = 1 and W = −1 are indistinguishable at T = Tc, This is the type of transition for CuZnand other systems based on the body-centered cubic structure. However, systems basedon the face-centered cubic structure, such as Cu3Au, have a latent heat accompanying theorder-disorder transition, and are, therefore, of first order.


Chapter 5

Nucleation and Growth

Exercise 5.1

(a) The coordinates of the vertices of the equilateral triangle at z = 0 are

1:(0, 1

3

√3,0)

, 2:(−1

2 ,−16

√3,0)

, 3:(1

2 ,−16

√3,0)

, (S5.1)

which, at z, become

1′:(

0,

√3

3

(1− z

h

),z)

,

2′:(−1

2

(1− z

h

),−√

36

(1− z

h

),z)

, (S5.2)

3′:(

12

(1− z

h

),−√

36

(1− z

h

),z)

,

for 0≤ z≤ h. The distance between each pair of vertices is

|1′−2′|=[

14

(1− z

h

)2

+34

(1− z

h

)2

+0]1/2

= 1− zh

,

|1′−3′|=[

14

(1− z

h

)2

+34

(1− z

h

)2

+0]1/2

= 1− zh

, (S5.3)

|2′−3′|=[(

1− zh+0+0

)2]1/2

= 1− zh

.

Thus, with the base b and height h of the triangle given by

b(z) = 1− zh

, h(z) =

√3

2

(1− z

h

), (S5.4)

63


the area A(z) = 12h(z)b(z) of the triangle at z is

A(z) =

√3

4

(1− z

h

). (S5.5)

Hence, the volume V of the right equilateral pyramid is obtained by integrating the volumeelement A(z)dz over the height of the pyramid:

V =∫ h

0A(z)dz =

√3

4

∫ h

0

(1− z

h

)2

dz . (S5.6)

This integral can be easily evaluated by making the variable change z = hs:

V =h√

34

∫ 1

0(1− s2)ds =−h

(1− s)3

3

∣∣∣∣10=

h√

312

=h3

(√3

4

), (S5.7)

where 14

√3 is the area of the base triangle (i.e. the triangle at z = 0).

(b) Consider now an oblique equilateral triangular pyramid whose apex lies at the point (xh,yh,h).Then any point (x0,y0,0) on the perimeter of the base triangle is mapped to(

xhzh+

(1− z

h

)x0,yh

zh+

(1− z

h

)y0,z

), (S5.8)

for 0≤ z≤ h. In particular, the coordinates of the vertices 1, 2, and 3 are transformed to

1′′:(

xhzh

,yhzh+

√3

3

(1− z

h

),z)

,

2′′:(

xhzh− 1

2

(1− z

h

),yh

zh−√

36

(1− z

h

),z)

, (S5.9)

3′′:(

xhzh+

12

(1− z

h

),yh

zh−√

36

(1− z

h

),z)

.

The distance between each of vertices is unchanged from that in (a) because the x- andy-components each differ by the same additive factor and he z-component is unchanged.Hence, the volumeV of the oblique triangular with base area A and height h is again givenby V = 1

3hA.

(c) Suppose we have a simple closed curve in the x-y plane whose area is A0. We construct apyramid with this area as a base by choosing a vertex at a point (xh,yh,z). Then, because aunit element of area in the base plane is transformed to the area (1− z/h)2 at z, the area A(z)at z is

A(z) = A0

(1− z

h

)2

. (S5.10)

Nucleation and Growth 65

Since the volume element dV of this pyramid is dV = A(z)dz, we can proceed as in (a) toobtain the volume V this pyramid as

V =13

hA . (S5.11)

Exercise 5.2As the following figure shows schematically in two dimensions,

for a solid bounded by n polygonal faces, any interior point can be chosen as the common vertexfor n pyramids whose bases are the areas Σi of the bounding faces and whose heights are theperpendicular distances hi from these bases to the vertex. The volume Vc of such a crystal is, fromExercise 5.1(c) obtained by summing over the volumes of each of the pyramids:

Vc =n

∑i=1

hiΣi . (S5.12)

Exercise 5.3

(a) Beginning withdF =−Pv dVv−Pc dVc +∑

iσi dΣi = 0, (S5.13)

we use the fact that V =Vv+Vc = constant to write dV = dVv+dVc = 0, i.e. dVv =−dVc, toobtain

−(Pc−Pv)dVc +∑i

σi dΣi = 0. (S5.14)

(b) Taking the differential of (S5.12) yields

dVc =13

n

∑i=1

(Σi dhi +hi dΣi

). (S5.15)


Alternatively, the approach taken in Exercise 5.1 implies that a change in the volume of theparticle is given in terms of the changes to the heights hi:

dVc =n

∑i=1

Σi dhi . (S5.16)

Equating these two expressions for dVc,

13

n

∑i=1

(Σi dhi +hi dΣi

)=

n

∑i=1

Σi dhi , (S5.17)

yieldsn

∑i=1

hi dΣ = 2n

∑i=1

Σi dhi , (S5.18)

which, with (S5.16), implies that

dVc =12

n

∑i=1

hi dΣi . (S5.19)

Upon substituting this result into (S5.14),

− 12(Pc−Pv)

n

∑i=1

hi dΣi +n

∑i=1

σi dΣi

=n

∑i=1

[−1

2(Pc−Pv)hi +σi

]dΣi = 0, (S5.20)

and noting that equality must hold for any independent variations of the Σi, we deduce thatthe coefficient of each dΣi must separately vanish, which yields

Pc−Pv = 2σi

hi, (S5.21)

for each i.

(c) If the pressure difference Pc−Pv is independent of the crystallographic orientation,that is,independent of i, then

σi

hi=

Pc−Pv

2= constant . (S5.22)

(d) Given that(Pc−Pv)vc = ∆µ , (S5.23)

then (S5.21) implies that

Pc−Pv =∆µvc

= 2σi

hi, (S5.24)

or∆µ =

2σivc

hi. (S5.25)


Exercise 5.4The critical Gibbs free energy ∆G∗ for a crystalline nucleus of (equilibrium) volume V ∗ is

∆G∗ =−V ∗

vc∆µ +∑

iσiΣi , (S5.26)

where vc is the volume of a basic unit of the crystal. From Problem 2, we can write

V ∗ =13

n

∑i=1

hiΣi , (S5.27)

and, from (S5.25), we can solve for the hi as

hi =2σivc

∆µ(S5.28)

to obtain

V ∗ =2vc

3∆µ

n

∑i=1

σiΣi . (S5.29)

Hence, upon substitution into (S5.26), we obtain

∆G∗ =−23

n

∑i=1

σiΣi +n

∑i=1

σiΣi =13

n

∑i=1

σiΣi , (S5.30)

which generalizes the result obtained in Sec. 5.1.2 for a spherical particle.

Exercise 5.5(a) Referring to Fig. 5.3, the free energy change ∆G associated with the formation of the nucleus

of phase β from phase α on a surface S is the sum of four terms: (i) the free energy changefrom the volume Vβ of the nucleus, ∆GαβVβ , where ∆Gαβ is the difference in free energiesbetween the α and β phases, (ii) the free energy gain from the β -S interface, γβSAβS, whereASβ is the area of the interface, (iii) the free energy gain from the α-β interface, γαβ Aαβ ,where Aαβ is the area of the interface, and (iv) the free energy reduction from the α-Sinterface by the formation of the nucleus, γαSAαS. Hence,

∆G = (µβ −µα)Vβ + γβSAβS + γαβ Aαβ − γαSAβS . (S5.31)

The equilibrium of forces at the edge of the nucleus, where the α and β phases and thesurface meet, requires that

γαS = γβS + γαβ cosθ . (S5.32)

Combining this equation with (S5.31), yields

∆G = ∆GαβVβ + γβSAβS + γαβ Aαβ − (γβS + γαβ cosθ)AβS

= ∆GαβVβ + γαβ Aαβ − γαβ AβS cosθ . (S5.33)


ar

x

Figure S5.1: Three-dimensional construction of the spherical cap on the surface (left), cut away toshow the planar construction in the right panel, which will be the basis of our calculations. Thespherical cap in the plane of the cut is shown in bold.

(b) To calculate the volumes and areas in (S5.33), we refer to Fig. S5.1.

The key quantities are

x =√

r2−a2 , cosθ =ar

, sinθ =xr=

√1−(

ar

)2

. (S5.34)

The area AβS between the nucleus and the surface is

AβS = πx2 = πr2(

x2

r2

)= πr2 sin2 θ . (S5.35)

The volume Vβ of the spherical cap is calculated by integrating πx2(z):

Vβ = π∫ r

a

(r2− z2)dz = πr2

∫ r

adz−π

∫ r

az2 dz

= πr2z∣∣∣∣ra− πz3

3

∣∣∣∣ra= πr2(r−a)− π

3(r3−a3)

= πr3(

1− ar

)− πr3

3

[1−(

ar

)3]

= πr3(1− cosθ)− πr3

3(1− cos3 θ) =

πr3

3(2−3cosθ + cos3 θ

). (S5.36)


The area Aαβ is calculated by using the radius x = r sinϕ , where 0≤ ϕ ≤ θ :

Aαβ = 2πr∫ θ

0sinϕ (r dϕ) =−2πr2 cosϕ

∣∣∣∣θ0

= 2πr2−2πr2 cosθ = 2πr2(1− cosθ) . (S5.37)

By substituting (S5.35), (S5.36), and (S5.37) into (S5.33), we obtain

∆G =πr3

3∆Gαβ (2−3cosθ + cos3 θ)+2πr2γαβ (1− cosθ)−πr2γαβ cosθ sin2 θ

=πr3

3∆Gαβ (2−3cosθ + cos3 θ)+πr2γαβ (2−2cosθ − cosθ + cos3 θ)

=

(−4πr3

3∆Gαβ +4πr2γαβ

)(12− 3cosθ

4+

cos3 θ4

). (S5.38)

The first factor on the right-hand side is that same as that in (5.5) for homogeneous nucle-ation. The second factor is the reduction of this free energy caused by the presence of thesurface.

(c) The factor

f (θ)≡ 12− 3cosθ

4+

cos3 θ4

(S5.39)

in (S5.38) represents the reduction in the free energy associated with the formation of anucleus on a surface. This function is plotted in Fig. S5.2 over the range 0≤ θ ≤ π .

Exercise 5.6Beginning with

dVtr =

(1− Vtr

V

)dVex , (S5.40)

we divide both sides of the equation by the total volume V to obtain

dVtr

V= d(

Vtr

V

)=

(1− Vtr

V

)dVex

V=

(1− Vtr

V

)d(

Vex

V

), (S5.41)

since V is a constant. Using the notation

f =Vtr

V, fex =

Vex

V, (S5.42)

equation (S5.41) can be written in a more concise form as

d f = (1− f )d fex . (S5.43)


This is a separable differential equation whose solution may be represented as∫ f

0

d f ′

1− f ′=∫ fex

0d f ′ex . (S5.44)

which incorporates the initial conditions, i.e. the lower limit of each integral has been set equal tozero because when there is no transformed volume ( f = 0), there can be no extended volume either( fex = 0). Evaluating the integrals in (S5.44) is a straightforward matter:

− ln(1− f ′)∣∣∣∣ f

0=− ln(1− f ) = fex , (S5.45)

or, after a solving for f ,f = 1− e− fex . (S5.46)

This is equivalent to the solution (5.31).

Exercise 5.7The KJMA equation is

f = 1− exp[−4π

3

∫ t

0n(τ)R3(t− τ)dτ

]. (S5.47)

In[1]:= f@t_D := H1 ê 2L - HH3 * Cos@tDL ê 4L + HHCos@tDL^3 ê 4L

In[2]:= Plot@f@tD, 8t, 0, Pi<, Frame Ø True, PlotStyle Ø 8Black, [email protected]<,FrameTicks Ø 880, Pi ê 4, Pi ê 2, 3 Pi ê 4, Pi<, Automatic, Automatic, Automatic<,BaseStyle Ø 8FontFamily Ø "Times", FontSize Ø 16<,FrameLabel Ø 8"q", RowBox@8"f", "H", "q", "L"<D êê DisplayForm<D

Out[2]=

0 p4

p2

3 p4

p0.0

0.2

0.4

0.6

0.8

1.0

q

fHqL

Figure S5.2: The reduction (S5.39) in the free energy associated with the formation of a nucleuson a surface. For θ < π , the change in the Gibbs free energy is lower for heterogeneous nucleationcompared to the homogeneous case. Nuclei corresponding to several values of θ are shown inFig. S5.3. For small θ , f (θ) is also small, and the droplet wets the surface, indicating that thechange in the Gibbs free energy is also small. At the other extreme, as θ → π , the sphere isimmersed in the α phase, and the situation is the same as for homogeneous nucleation, with thesame change in Gibbs free energy.


qq

(a) (b)

q q

(c) (d)

Figure S5.3: Spherical caps on surface corresponding to several values of θ . This shows that therange of θ is 0 ≤ θ ≤ π . With increasing θ , the reduction in the Gibbs free energy compared tothe homogeneous case is reduced, withe the difference vanishing altogether as θ approaches π .

There are two cases to consider: (i) constant continuous nucleation, where n(τ) = n, independentof τ , and (ii) a sudden burst of nucleation at τ = 0: n(τ) = n(τ)δ (τ). For the growth of linearcircular and spherical particles at a constant rate G, we have

d = 1: V (t) = 2Gt ,

d = 2: V (t) = π(Gt)2 , (S5.48)

d = 3: V (t) = 43π(Gt)3 .

These cases can be subsumed into a single expression by writing V (t) =Vdtd . Thus, for a suddenburst of nucleation at τ = 0, we find

f = 1− exp[−n(0)Vd td] , (S5.49)

so the Avrami exponent n = d. For constant continuous nucleation,

f = 1− exp(−nVd

∫ t

0τd dτ

)= 1− exp

(− nVd

d +1td+1

), (S5.50)

so the Avrami exponent n = d +1.


Exercise 5.8

(a) The origin of the system is arbitrary, so we could take any particle position as the origin, inwhich case w(r) corresponds to the distribution of nearest neighbors.

(b) The definition of 〈r〉 is

〈r〉=∫

∞

04πr3n0 exp

(−4πr3n0

3

)dr . (S5.51)

The calculation of this integral proceeds by transforming the integration variable to a newvariable s according to 4

3πr3n0 = s, which implies that

r3 =3s

4πn0, (S5.52)

so the differential is

3r2 dr =3ds4πn0

. (S5.53)

Since

r2 =(r3)2/3

=

(3

4πn0

)2/3

s2/3 , (S5.54)

the integration element transforms to

3dr =(

34πn0

)1/3

s−2/3 ds . (S5.55)

The limits of integration are unchanged, so the integral for 〈r〉 becomes

〈r〉=∫

∞

0

4πr3n0

3exp(

4πr3n0

3

)3dr

=∫

∞

0se−s

(3

4πn0

)1/3

s−2/3 ds =(

4πn0

3

)−1/3 ∫ ∞

0s1/3e−s ds . (S5.56)

The definition of the Gamma function is

Γ(x) =∫

∞

0sx−1 e−s ds , (S5.57)

so that

〈r〉=(

4πn0

3

)−1/3

Γ(4

3

)= 0.554n−1/3

0 . (S5.58)


(c) Consider a simple cubic lattice with a lattice constant a, which is the length of a side of thecube, as shown below:

There are 8 atoms in the basic unit cell, one at each vertex of the cube. Each of these atomsis shared by eight adjacent cubes, so the atomic density n0 of the simple cubic lattice a−3,i.e. one atom per cube. Given that the nearest neighbor distance is a, we have that the averagenearest-neighbor distance rSC of a simple cubic lattice can be expressed as rSC = n−1/3

0 .

The body-centered cubic lattice, shown below,

has an additional atom in the center of the fundamental unit cell and the nearest neighbordistance is 1

2

√3a, half the distance along the main diagonal. The atomic density is n0 = 2a−3,


so

rBCC =

√3

2a =

√3

2

(2n0

)1/3

=

√3

22/3 n−1/30 = 1.09n−1/3

0 . (S5.59)

The face-centered cubic lattice, which is shown below,

has 8 atoms at the corners of a cube and 6 atoms in the center of the faces, so there are8× 1

8 + 6× 12 = 1+ 3 = 4 atoms per volume a3. The nearest-neighbor distance is 1

2

√2a,

along the diagonal on a face, so

rFCC =

√2

2a =

√2

2

(4n0

)1/3

= 21/6n−1/30 = 1.12n−1/3 . (S5.60)

(d) To understand the effects of clustering, suppose that we have a large number N of particlesthat we distribute randomly within a large volume. The resulting average nearest-neighbordistance is that calculated in (b). Now suppose that we take a number nN of these particlesand distribute them randomly within a volume vV , while the remaining N−n particles aredistributed randomly within the entire volume V . The average distance between neighborsmust decrease, since the N− n particles have the nearest neighbor distance in (b), but then particles within v will have much small nearest neighbor distances between themselves,while having approximately the random value with the remaining N− n particles. Hence,since this mimics the effect of clustering, we expect that any tendency toward clustering willbe reflected in a reduced nearest-neighbor distance compared with the random distribution.

(e) The results of (b) and (d) show that, for a random distribution, 〈r〉 ≈ 0.554n−10 , while for a

regular three-dimensional array, 〈r〉 ≥ n−1/30 . Moreover, we argued in (d) that any tendency

toward clustering yields 〈r〉 < 0.554n−1/30 . Thus, a measurement of 〈r〉 and n0 in the form

〈r〉= mn−1/30 reveals the degree of randomness in an ensemble of nuclei (or, for that matter,


any set of points). A value of m > 0.554 indicates a tendency toward ordering, while a valuem < 0.554 indicates a tendency toward clustering.

Exercise 5.9For a nucleation rate that is a decreasing function of time,

n(t) =Atα , (S5.61)

where A is a constant and 0≤ α < 1, we substitute into the KJMA expression, with V (t) =Vdtd asin Exercise 5.7, to obtain

f = 1− exp[−VdA

∫ t

0

(t− τ)d

τα dτ]

. (S5.62)

For d = 1, we perform an integration by parts∫ t

0(t− τ)τ−α dτ = (t− τ)

τ1−α

1−α

∣∣∣∣t0︸︷︷︸

= 0

+1

1−α

∫ t

0τ1−α dτ (S5.63)

=τ2−α

(1−α)(2−α)(S5.64)

which yields an Avrami exponent n = 2−α . For d = 2, we again perform an integration by partsto obtain ∫ t

0(t− τ)2τ−α dτ = (t− τ)2 τ1−α

1−α

∣∣∣∣t0︸︷︷︸

= 0

+2

1−α

∫ t

0(t− τ)τ1−α dτ . (S5.65)

The integral on the right-hand side of this equation is that same as that for d = 1 with the replace-ment α 7→ α−1. Hence, ∫ t

0(t− τ)2τ−α dτ =

2τ3−α

(1−α)(2−α)(3−α), (S5.66)

so the Avrami exponent n = 3−α . Finally, for d = 3, we again begin with an integration by partsand use the (suitably modified) result for d = 2:∫ t

0(t− τ)3τ−α dτ = (t− τ)3 τ1−α

1−α

∣∣∣∣t0︸︷︷︸

= 0

+3

1−α

∫ t

0(t− τ)2τ1−α dτ

=6τ4−α

(1−α)(2−α)(3−α)(4−α), (S5.67)


so the Avrami exponent n = 4−α .We can summarize the results here and in Exercise 5.7 as follows. For a sudden burst of

nucleation in d dimensions, n= d, for constant continuous nucleation n= d+1, and for continuousnucleation n(τ)∼ τ−α , n = d +1−α .

Exercise 5.10

(a) By dividing both sides of the first equation by V ,

dVi

V

(1− V1 +V2 + · · ·+Vn

V

)dV ex

iV

, (S5.68)

we obtain, using the definitions given

dvi = (1− z)dvexi . (S5.69)

Then, summing over i and using the definitions given yields

dz = (1− z)dzex . (S5.70)

This is a separable differential equation whose solution can be represented as∫ z

0

ds1− s

=∫ zex

0ds , (S5.71)

which, upon integration and a simple rearrangement, yields z = 1− e−zex.

(b) Dividing equation (S5.69) by (S5.70),

dvi

dz=

dvexi

dzex , (S5.72)

rearranging as

dvi =dvex

idzex dz , (S5.73)

using (S5.70) again,

dvi =dvex

idzex (1− z)dzex , (S5.74)

and, finally, invoking the solution for z, yields

dvi =dvex

idzex e−zex

dzex . (S5.75)

The integral of this equation yields

vi =∫ dvex

idzex e−zex

dzex . (S5.76)


(c) Setting Bvex1 −C = 0 yields vex

1 =C/B. Hence, since vex2 ≥ 0,

zex = vex1 + vex

2 =

vex

1 , if 0≤ vexi ≤C/B;

vex1 (1+B)−C , if vex

1 >C/B .(S5.77)

Thus, with dvexi /dzex = 1, (S5.76) can be integrated for 0≤ vex

1 ≤C/B to obtain

v1 =∫ vex

1

0e−s ds =−e−s

∣∣∣∣vex1

0= 1− e−vex

1 , (S5.78)

with v2 = 0.

(d) For vex1 >C/B, we have

vex1 =

zex +C1+B

, (S5.79)

sodvex

1dzex =

11+B

, dzex = (1+B)dvex1 . (S5.80)

Thus,

v1 =∫ C/B

0e−s ds+

∫ vex1

C/B

11+B

[e−(1+B)s+C

](1+B)ds

= 1− e−C/B +

[− 1

1+Be−(1+B)s+C

]∣∣∣∣vex1

C/B

= 1− e−C/B− 11+B

e−(1+B)vex1 +C +

11+B

e−C/B

= 1− B1+B

[e−C/B− e−(1+B)vex

1 +C]

, (S5.81)

Since zex = vex1 (1+B)−C, we have that

zex =(vex

2 +C)1+B

B−C = vex

21+B

B+

CB

. (S5.82)

Hence,

v2 =∫ vex

2

0exp(−s

1+BB−C

B

)ds

= e−C/B[− B

1+Bexp(−s

1+BB

)]∣∣∣∣vex2

0

=Be−C/B

1+B

[1− exp

(−vex

21+B

B

)], (S5.83)

which, as vex2 = Bvex

1 −C, can be expressed in terms of vex1 as

v2 =B

B+1

[e−C/B− e−vex

1 (1+B)+C]

. (S5.84)


Chapter 6

Instabilities of Solidification Fronts

Exercise 6.1Small particles have a larger surface to volume ratio than larger particles. Thus, a correspondinglylarger fraction of the atoms are surface atoms in larger clusters. As these atoms have fewer bondsthan atoms within the cluster, they can detach more quickly, leading to a lower melting temperature.As the size increases and the fraction of surface atoms decreases, the melting temperature of thecluster approaches that of the bulk.

The same principle can be used to explain Ostwald ripening. Smaller clusters dissolve morequickly than larger ones, so the atoms from smaller clusters are made available for capture by largerclusters. Thus, larger clusters grow at the expense of smaller clusters, which is Ostwald ripening.

Exercise 6.2To determine how the Euler equation

ax2 d2ydx2 +bx

dydx

+ cy = 0

behaves under the change of variable x = et (or t = lnx), we first need to determine how thederivatives are transformed. This is done by applying the chain rule:

ddx

=dtdx

ddt

=1x

ddt

= e−t ddt

d2

dx2 =d2tdx2

ddt

+

(dtdx

)2 d2

dt2

=− 1x2

ddt

+1x2

d2

dt2 =−e−2t ddt

+ e−2t d2

dt2 .

79


Substituting these expressions into Euler’s equation, yields

ae2t(− e−2t dy

dt+ e−2t d2y

dt2

)+bet

(e−t dy

dt

)+ cy

= ad2ydt2 +(b−a)

dydt

+ cy = 0.

which is a second-order equation with constant coefficients. These are equations are solved with atrial solution y = emt , where t is determined by requiring y to be a solution. By substituting y intothe equation, [

am2 +(b−a)m+ c]emt = 0,

we obtain the characteristic equation: am2 +(b−a)m+ c = 0.For the Euler equation

r2 d2δul

dr2 +2rdδul

dr− l(l +1)δul = 0,

we have that a = 1, b = 2, and c =−l(l +1). Thus, the characteristic equation is

m2 +m− l(l +1) = 0,

whose solutions are

m =12

[−1±

√1+4l(l +1)

]=

12[−1± (2l +1)

]= l,−(l +1) .

Hence, our solution isδul(t) = al e−(l+1)t +bl elt ,

or, in the original variable t = lnr,

δul(r) = al e−(l+1) lnr +blel lnr =al

rl+1 +blrl .

Exercise 6.3

(a) We determine the solution for a planar solidification front moving along the positive z-direction with speed vby first transforming to a coordinate system moving with the interface:

(x,y,z, t) 7→ (x,y,z− vt, t) , (S6.1)

whereupon u(x,y,z, t) 7→ u(x,y,z− vt, t). Using the notation ζ = z− vt, we have

∂u∂ t7→ ∂u

∂ t+

∂u∂ζ

∂ζ∂ t

=∂u∂ζ− v

∂u∂ζ

, (S6.2)

∇∇∇2u =

∂ 2u∂x2 +

∂ 2u∂y2 +

∂ 2u∂ z2 7→

∂ 2u∂x2 +

∂ 2u∂y2 +

∂ 2u∂ζ 2 . (S6.3)

Instabilities of Solidification Fronts 81

Hence, under the transformation (S6.1), the diffusion equations in the liquid and solid phasesbecome1

∂u∂ t− v

∂u∂ z

= K`∇∇∇2u , z≥ 0(liquid) , (S6.4)

∂u∂ t− v

∂u∂ z

= Ks∇∇∇2u z≤ 0(solid) . (S6.5)

The temperature variations in this system are only along the z-axis and, since the velocityof the plane along this direction is constant, there are no remaining temporal changes to thetemperature field, i.e. we can confine ourselves to steady-state diffusion equations:

d2udz2 +

1ξ

dudz

= 0, z≥ 0(liquid) , (S6.6)

d2udz2 +

1ξ ′

dudz

= 0, z≤ 0(solid) , (S6.7)

where ξ = K`/v and ξ ′ = Ks/v. These equations have the same form on either side of theinterface, so their general solutions can be determined together. By writing, say, the firstequation as

d2udz2 +

1ξ

dudz

=ddz

(dudz

+uz

)= 0, (S6.8)

we immediately see that the quantity within the parentheses must be a constant, independentof z:

dudz

+uξ= constant≡ A . (S6.9)

The constant A can be eliminated by shifting u, leaving a separable first-order differentialequation whose general solution can be written as

u(z) =

Be−z/ξ +Aξ , z≥ 0 (liquid) ,

B′e−z/ξ ′+A′ξ ′ , z≤ 0 (solid) .(S6.10)

In the solid region, where z ≤ 0, so we must set B′ = 0 to obtain a physical solution, sou=A′ξ ′, which is a constant. We now consider the value uI of u on the solid-liquid boundary.Since κ = 0 for a plane, uI = 0, so u = 0 on the solid side of the interface. Applying thisboundary condition to the solution on the liquid side of the interface yields

u(0) = B+Aξ = 0, (S6.11)

1For notational and conceptual clarity, we have retained the notation z for the coordinate along the growth velocitydirection in the knowledge that this coordinate in the moving frame is not the same that in the laboratory frame usedoriginally.


so Aξ = −B. Substitution of these solutions into the conservation equation yields, withvn = v,

v = K`

[− d

dz

(Be−z/ξ −B

)∣∣∣∣z=0

]=

K`Bξ

= Bv , (S6.12)

because ξ = K`/v. Thus, we find that B = 1, so the complete solution for the temperaturefield is

u0(z) =

e−z/ξ −1, z≥ 0 (liquid) ;

0 , z≤ 0 (solid) .(S6.13)

(b) The decay length ξ = K`/v of the temperature field is determined by the ratio of the thermaldiffusion constant, which determines the transport of heat away from the interface, to thevelocity v of the front, which determines the transport of heat to the front.

(c) The temperature fields for both planar and spherical growth fronts decay to the undercoolingtemperature in the liquid phase, albeit at different rates. The main difference between thetwo cases is the temperature within the solid phase, which is zero in the planar case, butnonzero in the spherical case. This is ultimately an effect of the curvature of the growingspherical ‘droplet:’ the Gibbs–Thomson relation requires that the temperature of the surfaceis lower than TM. Since the temperature in the droplet must be a constant, this constant mustbe the Gibbs–Thomson value −2d0/R0 since, for a sphere of radius R0, κ = 2/R0.

(d) This steady-state solution exists for any v > 0, but requires an undercooling to u = −1 asz→ ∞, i.e. to T = TM −L/cP,`, so the undercooled liquid must be at a temperature L/cP,`below the melting temperature. This is a direct consequence of the conservation of energy.For the plane to move at constant speed, the net effect of the moving interface is to replacea volume of liquid at temperature T∞ by a volume of solid at temperature TM, while the heatgenerated per unit volume is L. Equating L to the heat cP,`(TM−T∞) yields

T∞ = TM−L

cP,`, (S6.14)

which is the asymptotic undercooling in (S6.13).

Exercise 6.4

(a) The dispersion relationω = v

[k− (1+β )d0ξ k3] , (S6.15)

in which is the wave vector of the small sinusoidal perturbation, v is the velocity of the front,d0 is the capillary length, and ξ = K`/v, is shown below:

Instabilities of Solidification Fronts 83

k

!(k

)

kmax

For small k, ω increases linearly with slope v. All modes with ω > 0, i.e. for

k < k0 ≡1√

(1+β )d0ξ, (S6.16)

have positive growth rates and are therefore unstable. The maximum growth rate is forkmax = k0/

√3. Hence, and initially planar interface would exhibit small protrusions on

a length scale λmax = 2π/kmax that would grow outward. At some point, the interface isno longer described by the linear equations we have derived, whereupon a full nonlinearanalysis is required. Even in this regime, λmax remains an important length scale, but otherparameters enter the description of the nonlinear evolution of the interface.

(b) Increasing either ξ or d0 increases kmax because both quantities facilitate the transfer of heataway from the a small protrusion (cf. Fig. 6.6).

Exercise 6.5Referring to the diagram in the main text, which is reproduced on the next page, when the solid firstforms, the liquid is enriched by the solute because lower-concentration material is removed fromthe liquid. This higher-composition liquid then solidifies to form a solid whose solute compositionhas increased. We denote the volume fraction of material that has solidified by vs. Then, ignoringany volume difference between the liquid and solid, the volume fraction of liquid is 1− vs. Amass balance between the solute increase in the fluid upon solidification, (c`− cs)dvs and thecorresponding increase in the solute concentration in the liquid, (1− vs)dc`, yields

(c`− cs)dvs = (1− vs)dc` . (S6.17)

The initial concentration in the liquid is c0 and in the solid kc0, where k = cs/c` because we haveapproximated the liquidus and solidus by straight lines. Equation (S6.17) is a separable differential

84 Solutions to Exercises in Transformations of MaterialsIn[14]:= r = 0.008;

Show@%9, Graphics@8Line@880, 0.55<, 80.357143, 0.55<<D,[email protected], 0.55<, rD,[email protected], 0.55<, rD<DD

Out[15]=

Figure6.7.nb 3

Liquid

a Liquid+a

c0c00 c000

cs c`

Concentration

Tem

pera

ture

T0

T

T1

equation,dvs

1− vs=

dc`c`− cs

=dc`

c`− kc`=

dc`(1− k)c`

. (S6.18)

The initial volume fraction is zero, and initial concentration of the liquid is c0. Hence, integratingboth sides of this equation, ∫ vs

0

dv′s1− v′s

=1

(1− k)

∫ c`

c0

dc′`c′`

, (S6.19)

yields

− ln(1− v′s)∣∣∣∣vs

0=− ln(1− vs) =

1(1− k)

lnc′`

∣∣∣∣c`c0

=1

(1− k)ln(

c`c0

), (S6.20)

which, when solved for c`, isc` = c0(1− vs)

k−1 = c0vk−1` . (S6.21)

Equivalently, for the concentration of the solid,

cs = kc0(1− vs)k−1 , (S6.22)

which is Scheil’s equation.

Chapter 7

Diffusionless Transformations

Exercise 7.1

(a) The free energy f (e, t) proposed by Falk is, in terms of the dimensionless shear strain e andtemperature t, given by

f (e, t) = f0(t)+(t + 14)e

2− e4 + e6 , (S7.1)

Differentiating the free energy with respect to e,

∂ f∂e

= 2(t + 14)e−4e3 +6e5 , (S7.2)

and setting the result equal to zero, yields the extrema of f , while the sign of the secondderivative,

∂ 2 f∂e2 = 2(t + 1

4)−12e2 +30e4 , (S7.3)

determines their stability. Setting (S7.2) equal to zero yields

e[3e4−2e2 +(t + 1

4)]= 0. (S7.4)

the solutions to this equation are e = 0 and, from the quadratic formula,

e2± =

16

[2±√

4−12(t + 14)

]=

16

(2±√

1−12 t)

. (S7.5)

The discriminant is equal to zero, that is, when t = 112 . At high temperatures, where t > 1

12 ,there is a single extreme point at e = 0 which, according to (S7.3), is a local minimum. Thisextreme point remains a minimum with decreasing temperature for t >−1

4 ; for t <−14 , this

point becomes a local maximum.

85


To assess the stability of e±, we substitute (S7.5) into (S7.3):

∂ 2 f∂e2

∣∣∣∣e±

= 2(

t +14

)− 12

6

(2±√

1−12 t)+

3036

(2±√

1−12 t)2

= 2t +12−4∓2

√1−12 t +

56

(4±4

√1−12 t +1−12 t

)=

(12−4+

103+

56

)+(2−10)t±

(−2+

103

)√1−12 t

=23−8 t± 4

3

√1−12 t

=23(1−12 t)± 4

3

√1−12 t

=23

√1−12 t

(√1−12 t±2

). (S7.6)

As in (S7.5), for solutions of (S7.2) to exist in the first place, we must have that 1− 12 t ≥0, that is, t < 1

12 . With this condition, the upper sign in (S7.6) is always positive with,according to (S7.6), a positive second derivative, which means that the points±e+ are alwayslocal minima. If the solution corresponding to the lower sign in (S7.5) is negative, we haveno solution, since e− must be real. If it is positive, then (S7.6) indicates that the secondderivative at that point is negative, which means that the points ±e− are local maxima.Hence, we have the following extrema of f over the indicated ranges of temperatures:

t > 112 , e = 0 (global minimum) ,

−14 < t < 1

12 , e = 0 (local minimum),

± e+ =±[1

6

(2+√

1−12 t)]1/2

(local minima) ,

± e− =±[1

6

(2−√

1−12 t)]1/2

(local maxima) ,

t <−14 , e = 0 (local maximum) ,

± e+ =±[1

6

(2+√

1−12 t)]1/2

(global minima) ,

where the local minima determine the stable states as a function of temperature.

Plots of the free energy (S7.1) are sown in the figure below. We see explicitly that, at hightemperatures (t = 1), there is only a single minimum at e = 0. With decreasing temperature,the profile near e = 0 begins to flatter and, for t < 1

12 , there are three shallow local minima,at e = 0 and at e =±e+. Finally, low temperatures, see the development of deep minima ate =±e+. The local maxima at e =±e− are seen only over a restricted temperature range.

Diffusionless Transformations 87

-1.0 -0.5 0.0 0.5 1.0

-0.5

0.

0.5

-1.0 -0.5 0.0 0.5 1.0

-0.5

0.

0.5

Shear strain (e) Shear strain (e)

Free

ener

gy(f

)

Free

ener

gy(f

)

t = 1 t = 14

-1.0 -0.5 0.0 0.5 1.0

-0.5

0.

0.5

-1.0 -0.5 0.0 0.5 1.0

-0.5

0.

0.5

Shear strain (e) Shear strain (e)

Free

ener

gy(f

)

Free

ener

gy(f

)

t = 116 t = 3

4

(b) The unstrained state is identified as austenite. This is the only stable state for t > 112 . For

−14 < t < 1

12 , there are additional locally stable points, which are identified as twin marten-sitic phases M±. These are indicated in the diagram below:

Show@%11, %12D

Show@%13, Graphics@[email protected],Arrow@88Sqrt@2D ê 4, f@Sqrt@2D ê 4, 2 ê 16D<,

8Sqrt@2D ê 4 - 0.02, f@Sqrt@2D ê 4 - 0.02, 2 ê 16D<<D<DD

6 Figure.ex7.1.1.nb

f

M A M+ e

1

23

tt+


The shape memory effect in Falk’s free energy model is based on the two types of freeenergy curves shown in the figure. Beginning with the high-temperature (t < 1

12 ) austenitephase, the material is cooled to 1

12 > t >−14 . The austenite remains stable and undeformed

(e = 0). Although the model does not specify so, this cooling corresponds to the austenite-twinned martensite transition in Fig. 7.3, as both are stable at e = 0. For the path 1, aload is applied to the austenite, which produces a transition to the deformed martensite M+

phase, which is stable for e 6= 0. Upon heating along path 2 to the original temperature, theabsence of martensitic minima means that the material becomes unstable and strain drivesthe material to the only thermodynamic minimum (path 3), which is at e = 0, that is, theoriginal undeformed state.

Exercise 7.2The general isokinetic equation is

dxdt

= g(x)h(T ) . (S7.7)

This equation is separable,

dt =dx

g(x)h(T ), (S7.8)

and, under isothermal conditions, integrates to∫ ta(T )

0dt = ta(T ) =

1h(T )

∫ xa

0

dxg(x)

. (S7.9)

For a thermal history T (t), additivity reads∫ τ

0

dtta(T (t))

=∫ τ

0

dxta(T ))

dtdx

=∫ xa

0

dx

ta(T ))dxdt

, (S7.10)

where we have changed the integration variable from t to x, and made the concomitant changes tothe upper limit of integration. Substitution of (S7.7) and (S7.9) into the right-hand side, yields∫ τ

0

dtta(T (t))

=1∫ xa

0

dxg(x)

∫ xa

0

dxg(x)

= 1, (S7.11)

so the general isokinetic equation (S7.7) reduces additivity to an identity.

Exercise 7.3The time derivative of a KJMA equation of the form

f (t) = 1− exp[−k(T )tn] , (S7.12)

Diffusionless Transformations 89

where n is a constant (i.e. independent of temperature), is

d fdt

= k(T )ntn−1 exp[−k(T )tn] . (S7.13)

We must now express the right-hand side in terms of f . We first have that

exp[−k(T )tn]= 1− f . (S7.14)

By solving this equation for t, we obtain

t = k(T )−1n[− ln(1− f )

] 1n . (S7.15)

Substituting (S7.14) and (S7.15) into (S7.13) produces

d fdt

= nk(T )k(T )−n−1

n[− ln(1− f )

] n−1n (1− f )

= nk(T )1− n−1n[− ln(1− f )

] n−1n (1− f ) , (S7.16)

which can be written concisely as

d fdt

= nk(T )1−γ[− ln(1− f )]γ(1− f ) , (S7.17)

where γ = (n−1)/n. Thus, (S7.13) has the isokinetic form (S7.7), with

g(x) =[− ln(1− f )

]γ(1− f ) ,

h(T ) = nk(T )1−γ . (S7.18)

Exercise 7.4The time derivative of

f (t) = 1− exp[−k(T )tn(T )] , (S7.19)

is determined by proceeding as in Exercise 7.3. Using the fact that

dtn(T )

dt=

ddt

[en(T ) ln t]= n(T )

ten(T ) ln t

=n(T )

ttn(T ) = n(T )tn(T )−1 , (S7.20)

we obtaind fdt

= k(T )n(T )tn(T )−1 exp[−k(T )tn(T )] , (S7.21)


which is the same as making the replacement n 7→ n(T ) in (S7.13). We thereby arrive at theanalogue of (S7.22):

d fdt

= n(T )k(T )1−γ(T )[− ln(1− f )]γ(T )(1− f ) , (S7.22)

where

γ(T ) =n(T )−1

n(T ), (S7.23)

which does not have the isokinetic form (S7.7).

Dimitri D. Vvedensky The Blackett Laboratory, Imperial ...

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