Chapter 7
Solid-Solid Transformations
We now turn to a discussion of solid-solid transformations. As an example, we
consider a single phase solid which is at a temperature and composition where
there would be two phases in equilibrium. An example is a precipitation
reaction α → α + β where α is the super saturated α-phase obtained from
quenching from a higher temperature, as shown in Fig. 7.1 a). A second
example would be a eutectoid transformation γ → α + β as shown in Fig. 7.1
b).
The approach to equilibrium can occur by at least two mechanisms:
• Nucleation and growth
• Spinodal decomposition
These processes are shown schematically in Fig. 7.2. In nucleation and
growth, thermal fluctuations form a nucleus of β-phase with a sharp interface
between it and the surrounding α matrix. As the β-phase particle grows, it
depletes the matrix of B atoms, and the growth of the β-phase particle is
limited by diffusion of B atoms down the concentration gradient in the surrounding matrix. In spinodal decomposition, small local fluctuations develop
in the super saturated phase. These fluctuations increase in amplitude as B
atoms diffuse uphill toward higher B atom concentration levels. This is the
consequence of a negative diffusion coefficient D̃ which, as you may recall, is
given by:
d ln γB
D̃ = (xA DB + xB DA ) 1 +
d ln xB
229
230
CHAPTER 7. SOLID-SOLID TRANSFORMATIONS
a)
β
α
α+ β
cα
cβ
b)
γ
β
α
α+ β
cα
cβ
Figure 7.1: Phase diagrams for solid-solid transformations. a) Precipitation
reaction α → α + β. b) Eutectoid reaction γ → α + β.
231
where DA and DB are the tracer diffusivities of the species A and B, and γB
is the activity coefficient of B in the A-B solution. For the case of a regular
solution in the quasi-chemical approximation we found:
D̃ = (xA DB + xB DA ) 1 −
∆Hmix
kT
where ∆Hmix is the heat of mixing of the solution. If this is large enough,
that is, if unlike bonds are much less favorable than like bonds, then the
chemical diffusion coefficient can be negative, and uphill diffusion can occur.
We will explore this further in future sections and problems.
a)
cβ
t1
t2
t3
t1
t2
t3
c0
cα
b)
cβ
c0
cα
Figure 7.2: Schematic of a) nucleation and growth and b) spinodal decomposition. In nucleation and growth the β-phase particle starts out small and
grows as B atoms diffuse down the concentration gradient to the α-β interface. In spinodal decomposition a small composition fluctuation forms and
increases as B atoms diffuse up the composition gradient by a process of
uphill diffusion.
232
7.1
CHAPTER 7. SOLID-SOLID TRANSFORMATIONS
Chemical Concentration of the Precipitate
Consider the situation of a β phase precipitate in an α phase matrix as shown
in Fig. 7.3 which is a schematic of the free energy for the two phases α and
β as a function of the atomic fraction of B atoms, x. The equilibrium concentrations for the α and β phases are xαβ and xβα . We are interested in
the composition of the β phase precipitate which initially precipitates from
a supersaturated α-phase of composition xα0 . We consider the metastable
equilibrium between the the matrix of composition xα1 and precipitate of
composition xβ1 . We allow the system to seek its lowest energy with respect to the composition of the matrix and precipitate, while fixing the total
number of atoms in the precipitate, N β , and matrix, N α . A change in precipitate composition from xβ1 to xβ2 and the corresponding change in matrix
composition from xα1 to xα2 will have the change in energy:
*
+
δG = [g α (xα2 ) − g α (xα1 )] N α + g β (xβ2 ) − g β (xβ1 ) N β
(7.1)
where g α (x) and g β (x) are the free energy per atom for α and β phases. For
infinitesimal changes, we can write the difference in free energies as:
dg α
(xα2 − xα1 ) =
dx
dg β
g β (xβ2 ) − g β (xβ1 ) =
(xβ2 − xβ1 ) =
dx
g α (xα2 ) − g α (xα1 ) =
dg α
δxα
dx
dg β
δxβ
dx
(7.2)
We conserve the total number of atoms so that the number of B-atoms
added to the β phase is equal to the number of B-atoms leaving the α-phase
matrix.
δNBα = −δNBβ
(7.3)
Since xα = NBα /N α we can see that:
δNBα = N α δxα
δNBβ = N β δxβ
(7.4)
(Note that we can see from Eqn. 7.4 that the change in composition in the
matrix is much smaller than that of the precipitate, since most of the atoms
7.1. CHEMICAL CONCENTRATION OF THE PRECIPITATE
β
α
G
xαβ
A
233
xα xα xα
x
xβα xβ xβ
B
Figure 7.3: Schematic of free energy of A-B binary system.
are still in the matrix, NBα NBβ .) Inserting Eqns. 7.2, 7.3 and 7.4 into
Eqn. 7.1, we find:
dg β dg α
δG =
δNBβ
(7.5)
−
dx
dx
From Eqn. 7.5 we see that unless dg α /dx = dg β /dx, the system can always
lower its energy by changing the composition of the precipitate. For example
if:
dg β
dg α
>
(7.6)
dx
dx
then the system can lower its energy by transferring B atoms from the βphase precipitate to the α-phase matrix. This will continue until moving
down on the β-phase free energy curve causes condition 7.6 to no longer be
true. Thus the composition which minimizes the energy is the one which
234
CHAPTER 7. SOLID-SOLID TRANSFORMATIONS
corresponds to the condition:
dg β
dg α
=
dx
dx
(7.7)
This is the so-called “tangent construction.”
7.2
Nucleation Energy Contributions
We consider the formation of a nucleus of β-phase in a matrix of super
saturated α -phase. We have found in Section 7.1 that the concentration of
the precipitate will be found by the tangent construction. Now we turn to
consideration of the free energy change associated with formation of a nucleus
containing N β atoms of this composition. For a solid-solid transformation,
the free energy can be written:
∆GT = N β (−∆gA + ∆gel ) + ηγ
Nβ
2/3
(7.8)
where ∆gA is the difference in free energy per atom between the supersaturated α phase and the β phase, ∆gel is the elastic strain energy per atom, η
is the shape factor defined by:
η=
A
(N β )2/3
as in our treatment of nucleation in solidification, and γ is the interface free
energy. We first turn to the free energy of the transformation.
7.2.1
Transformation Free Energy
The free energy difference between the α matrix and the matrix with a βphase precipitate is the driving force for the transformation. We refer to the
free energy schematic shown in Fig. 7.4 which shows the free energies of the
α and β phases as a function of atomic fraction x of B atoms. The total free
energy drop associated with the transformation is shown as an arrow from
the free energy of the original supersaturated α -phase at atomic fraction x0
to the common tangent line connecting the free energy curves of the α and
β phases at atomic fractions xαβ and xβα .
7.2. NUCLEATION ENERGY CONTRIBUTIONS
235
µBα
β
α
g
µAαβ
µαβ
B
µAα
xαβ
A
xα xα
x
xβα xβ
B
Figure 7.4: Schematic of free energy for precipitation reaction.
236
CHAPTER 7. SOLID-SOLID TRANSFORMATIONS
We consider the formation of a β-phase nucleus having N β total atoms,
where the atomic fraction xβ1 in the precipitate satisfies the condition:
dg α dg β =
dx xβ1
dx xα1
where xα1 is the atomic fraction in the α phase matrix after formation of
the β-phase precipitate. This is the tangent construction which determines
the atomic fraction of the precipitate which minimizes the free energy. The
transformation free energy associated with the nucleus is:
∆Gtr ≡ N β ∆gA
=
α
N +N
β
*
(7.9)
g (x0 ) − N g (xβ1 ) − N g (xα1 )
α
β β
α α
+
= N β g α (x0 ) − g β (xβ1 ) + N α [g α (x0 ) − g α (xα1 )]
*
+
dg α
(7.10)
dx
where g α (x) and g β (x) are the free energy of the α and β phases as a function
of atomic fraction of B atoms, and N α is the number of atoms remaining in
the α-phase matrix after forming the β-phase precipitate. Consider the last
term in Eqn. 7.10. In order to identify ∆gA , we would like to transform this
into a product of N β and some other quantity. We can do this by observing
that:
N α x0 = (N − N β )x0 = NB − N β x0
≈ N β g α (x0 ) − g β (xβ1 ) + N α (x0 − xα1 )
and
N α xα1 = NB − N β xβ1
since the number of B atoms in the α-phase plus the number of B atoms
in the β-phase must equal the total number NB of B atoms. Using these
relationships we find:
N α (x0 − xα1 ) = N β (xβ1 − x0 )
Inserting this into Eqn. 7.10 we find:
β
N ∆gA = N
β
dg α
g (x0 ) − g (xβ1 ) + (xβ1 − x0 )
dx
α
β
Thus we identify ∆gA as the quantity in brackets. This is shown schematically in Fig. 7.4. Hence, given the free energy functions g α and g β , we can
find the composition of the precipitate, and the free energy per precipitate
atom associated with the transformation.
7.2. NUCLEATION ENERGY CONTRIBUTIONS
7.2.2
237
Strain Energy
One aspect of a solid-solid nucleation which is different from liquid-solid nucleation, is the possibility of elastic strain resulting from a change in volume
or shape during the α → α + β transition. We consider a simple treatment
of the strain energy ∆gel in Eqn. 7.8, and see that both nucleation rate and
nuclei shape can be affected.
The strain energy associated with this transformation can be examined
by the following sequence of imaginary operations.
• Remove a small volume of α and transform it to β.
• Apply surface tractions to this β phase particle to allow it to return to
its original size and shape.
• Stick it back in its original location and weld surfaces together.
• Allow entire assembly to relax.
This places the sample of matrix and nuclei into a state of self stress, and
the strain energy must be taken into account in the free energy associated
with formation of this transformed region.
Nabarro has shown that if the interfaces are incoherent and all the strains
are in the matrix, the strain energy for a spheroidal precipitate with semi
axes c parallel to the rotation axis and b perpendicular, is given by1 :
∆gel =
2 µα ∆V 2
φ (c/b)
3 Vβ
where µα is the shear modulus of the matrix phase α, ∆V is the difference in
atomic volumes of the two phases, ∆V = Vβ − Vα , and φ (c/b) is a function
of the shape of the particle which has the following limits:



1
if c/b = 1 (sphere)
if c/b 1 (rod)
φ (c/b) = 3/4


3πc/4b if c/b 1 (disk)
This is shown schematically in Fig. 7.5.
1
F.R.N. Nabarro, Proc. Roy. Soc. A 1175, 519, (1940)
238
CHAPTER 7. SOLID-SOLID TRANSFORMATIONS
sphere
φ(c/b)
1
0.75
rod
plate
c/b
1
Figure 7.5: Schematic of function φ (c/b).
7.3
Nucleation Barrier
To see how this can affect nucleation, lets examine a simple case of diskshaped precipitates with large semiaxis b, and small semiaxis c. The volume
and surface area are then:
4 2
πb c
3
A ≈ 2πb2
V
=
The shape factor η is found by:
η ≡
A
(N β )2/3
Vβ
= A
V
= 2πb
2
2/3
Vβ
4πb2 c/3
= 2π
where s = c/b.
1/3
3Vβ
4s
2/3
2/3
7.4. GROWTH IN SOLID-SOLID TRANSFORMATIONS
239
The free energy associated with formation of a transformed region can
then be written as:
∆GT = N
β
3Vβ
πµα (∆V )2
−∆gA +
s + 2π 1/3
2Vβ
4s
By setting:
2/3
γ (N β )2/3
∂∆GT =0
∂N β N β =N β∗
we can find the number of atoms in a critical nuclei, as before, but now we
must also minimize the energy with respect to the shape of the particle, that
is, we must set:
∂∆GT =0
∂s s=s∗
and solve for s∗ . After some algebra we find:
Nβ
∗
s∗
∆G∗
32π 3 (∆V )4 γ 3 µ2α
3∆gA5
∆gA Vβ
=
πµα (∆V )2
8π 3 γ 3 µ2α (∆V )4
=
3
∆gA4
=
(7.11)
Eqn. 7.11 shows that now the nucleation barrier has ∆gA to the fourth power
in the denominator, which gives a much more sudden onset of nucleation
than the (∆gA )2 in the expression for homogeneous nucleation with no strain
energy effect. We also see that a large ∆V will favor flat disk-shaped nuclei
which can minimize the strain energy.
7.4
Growth in Solid-Solid Transformations
Growth in the solid phase has many characteristics of growth of solids from a
liquid. We still must have atoms crossing the interface and attaching to the
growing phase, so the atom attachment mechanisms discussed previously are
important. Transformation kinetics are usually limited by the relatively slow
atomic jump rates; it is rare for heat removal to be the limiting factor as it is
in some solidification processes. Therefore, we discuss isothermal solid-solid
transformations.
240
CHAPTER 7. SOLID-SOLID TRANSFORMATIONS
Under the condition of constant undercooling, there are two possible processes which can dominate the growth rate: interface attachment kinetics and
diffusion. In interface attachment limited growth, the growth rate is limited
by the rate at which atoms can jump from the matrix to the growing phase.
The rate at which atoms are supplied to the interface region is sufficiently
fast so that this process does not limit growth. In this regime, the interface
moves with a constant velocity given by:
ν0 ∆gT
−∆GM
v=a
exp
kB T
kB T
for uniform growth or:
a2 x∗k ν0 ∆gT
−∆GM
v=
exp
ys kB T
kB T
for nonuniform growth, where a is the distance grown per atomic layer, ν0 is
the vibration frequency, ∆gT is the driving force per atom for atomic jumps,
∆GM is the jump (migration) activation energy, x∗k is the fraction of step
sites which are kinks, and ys is the spacing between steps. These equations
are the same as those for solid growth into a liquid.
Interface attachment is likely to limit growth in phase transitions where
there is no composition change, as in crystallization of a metallic glass into
a compound of the same composition. However, in the more common case,
where the growing phase has a composition which differs from the matrix,
diffusion of atomic species is likely to limit the growth rate. The growing
phase can grow only as fast as allowed by the flux of atoms to the interface.
Earlier in the course, we examined an example of this diffusion limited growth
as a homework problem which dealt with a compound forming at the interface
of a diffusion couple. We found that the width of the growing phase was linear
with the square root of time. This dependence is typical of diffusion limited
growth and contrasts with interface attachment limited growth where the
interface position moves linearly with time (constant interface velocity).
As a second example of diffusion limited growth, we consider growth of a
β-phase spherical nuclei of composition cβα growing into an α-phase matrix
of composition c0 (Fig. 7.6). The composition in the α-phase at the interface
is cαβ , and the interface movement is governed by the rate that the flux in the
α-phase can provide atoms to drive the composition change at the interface:
∂cα dR (cβα − cαβ ) = −Jαβ dt = Dα
dt
∂r αβ
7.5. ISOTHERMAL TRANSFORMATION KINETICS
241
where R is the radius of the precipitate particle, Dα is the diffusion coefficient
in the α-phase, and cα is the composition in the α-phase as a function of radial
coordinate r. Hence, the rate of change of the particle radius is:
v=
(∂cα /∂r)|αβ
dR
= Dα
dt
cβα − cαβ
With the spherical symmetry of this problem it is natural to work in spherical
coordinates. For this symmetric case, the diffusion equation in spherical
coordinates is:
2
∂cα
2
∂
c
∂c
α
α
+
= Dα ∇2 cα = Dα
∂t
∂r2
r ∂r
and for a steady-state with the present boundary conditions, we find:
cα = c 0 −
Hence we find:
R
(c0 − cαβ )
r
∂cα c0 − cαβ
=
∂r αβ
R
and the interface velocity becomes:
v=
Integrating we find:
R=
dR
Dα (c0 − cαβ )
=
dt
R(cβα − cαβ )
.
.
/2D
c0 − cαβ
αt
cβα − cαβ
Thus for diffusion limited growth, we again find that the interface position
is proportional to the square root of time.
7.5
Isothermal Transformation Kinetics
Having considered separately the processes of nucleation and growth, we now
wish to describe the kinetics for a complete transformation which occurs by
a process of isothermal nucleation and growth. We consider a transformation
from an undercooled α phase to β phase. During this process, β phase nuclei
will form in regions of α phase. These β phase particles will grow, consuming
242
CHAPTER 7. SOLID-SOLID TRANSFORMATIONS
cβα
c
cαβ
0
2
4
6
8
10
r/R
Figure 7.6: Composition as a function of position for spherical precipitate.
the α phase matrix. The number of β phase particles which nucleate between
time t = τ and t = τ + dτ is given by:
IV α dτ
where I is the nucleation rate per unit volume and V α is the volume of
untransformed α phase at time τ . If we assume that the growth is interface
limited, the volume, Vτ , of a β phase region which nucleated at time t = τ
is:
4π 3
v (t − τ )3 if t > τ
3
Vτ =
0
if t < τ
where v is the growth rate. Actually, this assumes that the particle we are
considering does not impinge on other growing particles, since this would
halt new growth at the boundary between the two impinging particles. This
will only be the case at short times, where the average size of the growing
particle is small compared to the interparticle spacing. Within this short
time assumption, we can find the total volume, Vstβ , of transformed region by
integrating over the nucleation times, τ .
4πV t 3
Iv (t − τ )3 dτ
3 0
where we have assumed that the volume of the transformed region is small,
so that V α = V , the entire sample volume. Assuming a constant nucleation
Vstβ =
7.5. ISOTHERMAL TRANSFORMATION KINETICS
243
rate, we find for the transformed fraction, F :
F ≡
Vstβ
π
= Iv 3 t4
V
3
(7.12)
This is good for short times only. As the particles grow they impinge on
each other, resulting in mutual interference. This presents a messy geometrical problem. In addition, as the transformation continues, the volume of
untransformed region is decreased, so that the region where nucleation can
occur is less than the sample volume, V .
The volume which we calculate by ignoring these effects is known as the
extended volume, Veβ . It is larger than the actual transformed volume since
it includes:
• Nuclei which form in already transformed region - these are known as
phantom nuclei.
• Growth occurring in previously transformed regions, as growing particles impinge on each other.
The extended volume is what we could calculate to be the transformed volume if we:
• Removed each nuclei from the sample as soon as it nucleates, filling the
hole left behind with untransformed α.
• Place the nucleated β phase particle in an infinite chunk of untransformed α phase where no nucleation is occurring, thus letting it grow
unimpeded by other β phase particles.
The expression for the extended volume is the same as that for the short
time volume found previously:
Veβ =
4πV t 3
v I(t − τ )3 dτ
3 0
(7.13)
As time goes on, the extended volume can get to be larger than the sample
volume!
How does the ability to calculate this extended volume help us? Well,
we can relate the extended volume to the actual transformed volume. In
a time interval the extended volume will increase by dVeβ while the actual
244
CHAPTER 7. SOLID-SOLID TRANSFORMATIONS
transformed volume will increase by dV β . If we assume randomly located
nucleation sites, a fraction, V β /V , of the increase in extended volume will
occur in previously transformed material, while a fraction, (1−V β /V ), occurs
in untransformed material. Only the fraction which occurs in untransformed
material will contribute to dV β , hence:
Vβ
dV = 1 −
V
β
Integrating we find:
dVeβ
Veβ
Vβ
= −V ln 1 −
V
So that for the transformed fraction we find:
Vβ
Vβ
= 1 − exp − e
F =
V
V
For the case considered previously, where the nucleation rate is constant and
we have interface limited growth:
−πIv 3 t4
F = 1 − exp
3
(7.14)
Note that for short times, we recover Eqn. 7.12.
Lets consider a different case, where there are a fixed number of randomly
distributed heterogeneous nucleation sites. In this case, the nucleation rate
is not constant. If we say that NN is the number of nucleation sites per unit
volume, then:
dNN = −NN νN dt
where νN is the nucleation rate of a given site. We can then find that:
NN = NN 0 exp (−νN t)
where NN 0 is the initial number of nucleation sites. The nucleation rate, I,
will be given by:
dNN
I=−
= NN 0 νN e−νN t
dt
Inserting this expression for nucleation rate at a given time τ into Eqn. 7.13
we find:
t
4π 3
Veβ
e−νN τ (t − τ )3 dτ
=
v NN 0 νN
V
3
0
7.5. ISOTHERMAL TRANSFORMATION KINETICS
245
Integration by parts three times yields:
νN2 t2 νN3 t3
8πv 3 NN 0 −νN t
Veβ
e
−
1
+
ν
t
−
=
+
N
V
νN3
2
6
(7.15)
Lets consider some limiting cases:
• Slow Nucleation Rate (νN t 1) We first note that in this case:
I = NN 0 νN e−νN t ≈ NN 0 νN = constant
so that we should get the same result as for the constant nucleation
case, i.e. Eqn. 7.14. Expanding the exponential in the right hand side
of Eqn. 7.15 we get:
e−νN t ≈ 1 − νN t +
νN2 t2 νN3 t3 νN4 t4
−
+
2
6
24
(7.16)
The first four terms of Eqn. 7.16 are canceled by the other terms in the
brackets of Eqn. 7.15, so that we have:
Veβ
V
so that:
8πv 3 NN 0 νN4 t4
=
νN3
24
3 4
πIv t
=
3
πIv 3 t4
F = 1 − exp −
3
as we found previously in Eqn. 7.14 as we expected.
• Rapid Nucleation Rate (νN t 1) In this case the nuclei will be used
up quickly. In the brackets of Eqn. 7.15, we keep only the term which
is highest order in νN t and find:
8πNN 0 v 3
Veβ
=
V
νN3
so that:
νN3 t3
6
= NN 0
4π 3 3
v t
3
4πNN 0 3 3
v t
F = 1 − exp −
3
246
CHAPTER 7. SOLID-SOLID TRANSFORMATIONS
In this case all the nucleation occurs very early and so the volume transformed
is just due to the growing precipitates.
Avrami has proposed that we use as a general expression:
F = 1 − exp (−(kt)n )
(7.17)
where 3 ≤ n ≤ 4, with n = 3 corresponding to a nucleation rate which
decreases with time, and n = 4 corresponding to a constant nucleation rate.
As final examples we consider 2-dimensional and 1-dimensional growth.
If the particles of the β phase grow as disks of thickness d, we have for the
volume of a particle nucleated at time t = τ :
Vτ = πdv 2 (t − τ )2
so that for the transformed fraction we have the same expression as in
Eqn. 7.17 but with:
2≤n≤3
For particles which grow only in one direction we find the transformed fraction again given by Eqn. 7.17, but with:
1≤n≤2
Eqn. 7.17 is a fairly general expression for the transformed fraction as a
function of time for a nucleation and growth process. A typical example is
shown in Fig. 7.7.
7.6
Time-Temperature-Transformation Diagrams
We have a general expression for the volume transformed as a funcition of
time:
−Veβ
= 1 − exp[−(kt)n ]
F = 1 − exp
V
where the rate constant k and Avarmi exponent n depend on the exact
transformation mechanism. For example for a constant nucleation rate and
interface limited growth, we have:
n = 4
k =
πIv 3
3
1/4
7.6. TIME-TEMPERATURE-TRANSFORMATION DIAGRAMS
247
1.0
n
0.6
1
2
3
4
5
F
0.8
0.4
0.2
0.0
0.0
0.5
1.0
1.5
2.0
2.5
3.0
tk
Figure 7.7: Fraction transformed as predicted by isothermal transformation
kinetics where the fraction transformed is given by F = 1 − exp[−(kt)n ].
Shown are behaviors for several different values of n.
where I is the nucleation rate and v is the constant interface velocity. Similar
rate constants can be found for other nucleation and growth situations. We
have seen that the nucleation rate rises and then falls as a function of the
undercooling, being dominated first by its increasing ability to overcome the
nucleation barrier resulting from the surface energy, and then by its decreasing ability to overcome the activation energy for atomic motion. Similarly,
the growth velocity v first increases with increased uncercooling, as the driving force increases, and then decreases as atomic jumps atomic jumps are
frozen out. Hence, the reaction constant k will increase and then decrease
as a function of undercooling. At temperatures close to the transition temperature, the driving force will be small and rate of transformation will be
low. As the temperature decreases, the driving force and the transformation
rate will increase. As the temperature decreases further, the transformation
rate will again decrease as atomic jumps become unlikely due to the lack of
thermal energy.
This behavior can be summarized in a time-temperature-transformation
(TTT) graph which plots the temperature required for a given transformed
fraction as a function of the time required. This is shown schematically in
Fig. 7.8, where the boundary representing a given transformed fraction is
plotted on temperature-time plot. A given thermal history can be traced
248
CHAPTER 7. SOLID-SOLID TRANSFORMATIONS
on this transformation, for example an anneal where the temperature is increased at a steady rate and then held constant for a given amount of time
and then cooled rapidly to an temperature where the transformation is inactive.
T
Te
ln t
Figure 7.8: Time-temperature-transformation (TTT) diagram for a solidsolid transformation showing annealing paths.
7.7
7.7.1
Spinodal Decomposition
Spinodal Instability
So far we have considered transformations which occur by nucleation and
growth. This type of transition always occurs when the two phases have a
difference in symmetry and even sometimes when the two phases differ only
by composition. However, there is a second type of transformation which can
occur if the two phases differ only in composition. This is known as spinodal
decomposition. Shown in Fig. 7.9 is the free energy versus composition for
a system where the two terminal phases have the same structure, but there
is a large positive heat of mixing between the constituents. Note that the
equilibrium configuration as given by the common tangent rule will be a two
phase mixture of distinct phases with compositions c1 and c2 . Between these
compositions there is a one phase mixture which is unstable with respect to
decomposition.
Consider the region between the compositions marked cs1 and cs2 . Within
7.7. SPINODAL DECOMPOSITION
249
A)
g
c
A
s
s
c
B)
c
c
c
B
T
T
Spinodal
Boundary
α
α
α + α
A
c
s
s
c
c
c
c
B
Figure 7.9: A) Schematic of free energy per volume versus composition for
a system where the terminal phases are the same structure, but there is a
large positive heat of mixing. B) Schematic of phase diagram exhibiting a
spinodal region.
250
CHAPTER 7. SOLID-SOLID TRANSFORMATIONS
this region, the curvature of the free energy is negative, that is:
∂2g
<0
∂c2
where g is the free energy per volume, and c is the composition. The
compositions,cs1 and cs2 , delineate this region, so, at these compositions:
∂ 2 g ∂ 2 g =
=0
∂c2 c=cs
∂c2 c=cs
1
2
As shown schematically in Fig. 7.9, within this region, a single phase
sample is unstable with respect to small fluctuations in composition. That
is, inside this region, a small fluctuation in composition will lower the free
energy, while outside this region a small fluctuation in composition will raise
the free energy. This region, where the curvature of the free energy is negative, is known as the spinodal region. A phase diagram is obtained from
making free energy curves like that shown in Fig. 7.9 for many temperatures,
and plotting the locust of points c1 (T ), c2 (T ), cs1 (T ), and cs2 (T ). Depending
on the relative position of the free energy of the liquid and solid, this can
result in either a solubility gap or a eutectic phase diagram. In either case,
the locust of points cs1 (T ), and cs2 (T ) define the spinodal region of the phase
diagram. This is shown schematically in Fig. 7.9.
Up to this point, we have always described the free energy in terms of
variables such as the composition, temperature, pressure and so forth, and
in the case of multiphase samples we have included terms in the free energy
corresponding to surfaces or interfaces. Now however, we have the possibility
of having composition fluctuations with no discrete interfaces. We must
include the effect on the free energy of composition gradients. Ignoring for
the time the effects of changes in the molar volume with composition, and
expanding the dependence of the free energy on the composition gradients,
we write for the free energy per volume:
∂g (∇c)2 ∂ 2 g g(c, ∇c) = g(c, 0) + ∇c
+
+ ···
∂∇c ∇c=0
2 ∂(∇c)2 ∇c=0
(7.18)
where g(c, 0) is the free energy per volume of a homogeneous system of composition c. We know that the free energy of the system cannot depend on the
sign of the composition gradient, so the second term on the right of Eqn. 7.18
must be zero, and writing g (c) = g(c, 0) and k = (∂ 2 g/∂(∇c)2 )/2 we have:
g = g + k(∇c)2
(7.19)
7.7. SPINODAL DECOMPOSITION
251
The free energy, G, of the entire sample is found by integrating Eqn. 7.19
over the sample volume, V :
G=
*
+
g + k(∇c)2 dV
(7.20)
V
We can further expand g (c) about some composition c0 to find:
dg (c − c0 )2 d2 g +
+ ···
g (c) = g (c0 ) + (c − c0 )
dc c=c0
2
dc2 c=c0
(7.21)
We assume a composition of the form:
c − c0 = A cos βz
(7.22)
where A is the composition wave amplitude, and β is the spatial frequency of
the composition wave. This is a general assumption, since any composition
wave can be represented by a series of sinusoidal waves such as Eqn. 7.22. We
then plug Eqn. 7.22 into Eqn. 7.21 and perform the integration of Eqn. 7.20.
We find for the free energy difference between a system with composition
wave given by Eqn. 7.22 and a homogeneous system:
A2 d2 g ∆G = V
+ 2kβ 2
4 dc2
(7.23)
If ∆G < 0 then the system is unstable with respect to sinusoidal fluctuations
with wavelength of 2π/β. In fact, whenever
d2 g <0
dc2 c=c0
the system is unstable with respect to sinusoidal composition fluctuations
with a wavelength greater than some critical wavelength, λc , found by setting
the term in the brackets in Eqn. 7.23 equal to 0.
2π
λc =
=
βc
−8π 2 k
d2 g /dc2
1/2
(7.24)
As can be seen from Eqn. 7.24, λc → ∞ as the spinodal boundary is approached.
252
7.7.2
CHAPTER 7. SOLID-SOLID TRANSFORMATIONS
Estimation of the Gradient Energy Term
Before proceeding it is interesting to estimate the relative size of the terms
involved in spinodal decomposition. One can show that in a simple nearest
neighbor bond counting model a coherent interface between two phases with
atomic fraction difference ∆x across the interface has an enthalpy per area
given by
∆H s = −zFBB σ
(∆x)2
where z is the number of nearest neighbors, FBB is the fraction which are
across the given interface, σ is the atomic density at the interface (atoms/area)
and is the bond enthalpy difference
= HAB −
1
(HAA + HBB )
2
where Hij is the ij bond enthalpy (assumed to be positive for a stable bond).
The enthalpy per volume is then just
∆H =
∆H s
d
where d is the interface width, or atomic plane spacing across the interface.
Recognizing that
1
σ
=
d
VA
where VA is the atomic volume, we find
−zFBB (∆x)2
∆H =
= −zFBB (∆c)2 VA
VA
where c is the concentration in atoms per volume and is related to x through
c = x/VA .
For a material with a composition gradient ∇c, the composition difference
across an interface will be given by
∆c ≈ d∇c
Hence a composition gradient will have associated with it an enthalpy per
volume of
∆H = −zFBB VA d2 (∇c)2
7.7. SPINODAL DECOMPOSITION
253
This allows us to identify the gradient energy parameter in the Cahn-Hilliard
spinodal decomposition theory as
k = −zFBB VA d2
Materials with a positive heat of mixing will have weaker unlike bonds and
hence a negative and positive gradient energy term.
It is interesting to compare this with the term d2 g /dc2 in the expression
A2
∆G = V
4
d2 g + 2kβ 2
dc2
which is the free energy change associated with a sinusoidal composition
modulation of magnitude A and wavenumber β. Using the regular solution
model for g we find
d2 g = VA 2z
+ kB T
dc2
1
1
+
1−x x
!
Hence
d2 g + 2kβ 2 = VA 2z
(1 − FBB d2 β 2 ) + kB T
dc2
1
1
+
1−x x
!
This leads to a critical wavelength given by
-
λc .
4π 2 FBB
.
=/
d
1 + kB T 1 + 1
2z
1−x
x
Note that this has a minimum value of
λmin = 2πd FBB
The behavior as a function of scaled bond parameter is shown in Figure 7.10.
7.7.3
Strain Effect on Spinodal Instability
If the molar volume is a function of composition, the composition wave induces strain. This strain has an associated elastic energy, and thus raises
the free energy cost of the composition fluctuation. This acts to stabilize the
homogeneous system, and to shrink the spinodal region.
254
CHAPTER 7. SOLID-SOLID TRANSFORMATIONS
10
λc/λmin
8
6
4
2
0
1
2
3
4
5
-z
/2kBT
Figure 7.10: Regular solution prediction of critical wavelength as a √
function
of scaled bond parameter. The critical wavelength is scaled by 2πd FBB .
We define to be the linear expansion per unit composition change, and
we can write the molar volume as:
V (c) = V0 [1 + 3
(c − c0 )] = V0 (1 + 3A
cos βz)
Let’s examine the components of the stress free strain:
fxx = fyy = fzz = A
cos βz
If we impose coherency on the system, then there can be no change in the
lattice parameter measured perpendicular to the composition wave, as a function of distance along the composition wave. The total strain, t is given by:
tii = fii + eii
where e is the elastic strain. In the case where we impose coherency, and
the composition wave is along the z direction, we have that:
txx = tyy = 0
so that:
exx = eyy = −A
cos βz
7.7. SPINODAL DECOMPOSITION
255
From isotropic elasticity analysis we find:
exx = [σxx − ν (σyy + σzz )] /E
eyy = [σyy − ν (σxx + σzz )] /E
ezz = [σzz − ν (σyy + σxx )] /E
where ν is Poisson’s ration and E is Young’s modulus. Simple manipulation
yields:
2ν
cos βz
1−ν
−A
cos βz
=
E
1−ν
−A
cos βz
E
=
1−ν
= 0
ezz = A
σxx
σyy
σzz
Having all the elastic strain and stress components allows us to find the local
strain energy, E e (x):
E e (x) =
1
A2 2 E
σii eii =
cos2 βz
2 i
1−ν
The average elastic energy per volume, E e is then:
1 1
σii eii dV
V V 2 i
E e =
A2 2 E
2(1 − ν)
=
which is independent of wavelength.
If we include this term in the free energy we find:
G=
V
and:
2 E
g (c) +
(c − c0 )2 + k(∇c)2 dV
1−ν
∆G
2
2 E
A2 d2 g +
=
+ 2kβ 2
2
V
4 dc
1−ν
256
CHAPTER 7. SOLID-SOLID TRANSFORMATIONS
Now the region of instability is given by:
2
2 E
d2 g +
=0
dc2
1−ν
A larger negative curvature is necessary to make the homogeneous sample
unstable with respect to composition fluctuations. In other words, strain stabilizes the homogeneous solution. The reduced spinodal region is sometimes
called the strain spinodal (Fig. 7.11).
Spinodal
Boundary
T
Solubility
Limit
Strain
Spinodal
c
A
B
Figure 7.11: Schematic of strain stabilized spinodal.
The critical wavelength is now given by:
λc =
−8π 2 k
d2 g /dc2 + 2
2 E/(1 − ν)
1/2
(7.25)
So that we see that strain causes the smallest stable wavelength to become
larger.
7.7.4
Spinodal Decomposition Rate
Consider the change, δG, in free energy associated with a small composition
change, δc:
δG =
V
dg dk
2
2 E
+
(c − c0 ) + (∇c)2 δc + 2k∇cδ(∇c) dV
dc
1−ν
dc
7.7. SPINODAL DECOMPOSITION
257
Integration by parts yields:
δG =
V
dk
dg 2
2 E
+
(c − c0 ) − (∇c)2 − 2k∇2 c δcdV
dc
1−ν
dc
The term in the brackets is the chemical potential, and its gradient will drive
a diffusive flux:
dg dk
2
2 E
2
2
J = −M ∇
+
(c − c0 ) − (∇c) − 2k∇ c
dc
1−ν
dc
where M is the atomic mobility. Applying the continuity equation:
∂c
= −∇ · J
∂t
and keeping only terms which are linear in c or its gradients, we find:
∂c
=M
∂t
2
2 E
d2 g +
∇2 c − 2M k∇4 c
dc2
1−ν
(7.26)
We try a solution to Eqn. 7.26 of the form:
c − c0 = A(β, t) cos βz
and find:
∂A
2
2 E
d2 g +
β 2 A − 2M kβ 4 A
= −M
∂t
dc2
1−ν
We now try a solution for A of the form:
A = A(β, 0) exp [R(β)t]
and we find for the spinodal rate, R:
R(β) = −M β 2
2
2 E
d2 g 2
+
2β
k
+
dc2
1−ν
(7.27)
The behavior of a system can be divided into two regimes:
• If R < 0 then composition fluctuations decay exponentially with time.
This is the case if λ < λc , and we have a homogeneous system which
is outside the strain spinodal and hence stable relative to composition fluctuations. If the composition is still between the c1 and c2 in
Fig. 7.9, then the two phased system is lower in free energy than the homogeneous alloy, but the approach to equilibrium cannon take place by
spinodal decomposition, and must take place by a process of nucleation
and growth.
258
CHAPTER 7. SOLID-SOLID TRANSFORMATIONS
• If R(β) > 0 then composition fluctuations grow exponentially with
time. This will be the case inside the strain spinodal, where λ > λc . A
composition fluctuation will grow and eventually lead to a two phase
sample. The fastest growing composition fluctuation wavelength, λmax ,
can be found by setting:
∂R(β) =0
∂β β=βmax
and we find:
λmax =
√
2λc
that is, the fastest growing wavelength is
wavelength.
√
2 times the shortest stable
References
J.W. Cahn, Acta Met. 9, 795-801 (1961).
J.W. Cahn, Acta Met. 10, 179-83 (1962). boundary