III Theoretical Physics of Soft Condensed Matter

6Model B

6 Model B

We now apply this theory to model some concrete systems. We shall first consider

a simple model of binary fluids. We assume that diffusion happens but without

fluid flow. As before, this is modeled by a scalar composition field

(

), and

evolves under the equations

φ = −∇ ·J

J = −M∇µ +

T MΛ

µ =

δF

δφ

Recall that the system is modelled under the Landau–Ginzburg free energy

F [φ] =



| {z }

(∇φ)



dr.

We then have

µ = aφ + bφ

− κ∇

φ.

As before, the mean field theory for equilibrium looks like

a > 0

a < 0

a > 0

a < 0

Here

±φ

are the spinodals, where the second derivative changes sign. This

gives the phase diagram

−1 1

a(T ) = 0

Here

φ is the global composition, which is a control variable.

In the past, we discussed what the field looks like in each region when we are

at equilibrium. At (1), the system is in a uniform phase that is globally stable.

If we set up our system at (1), and then rapidly change the temperature so that

we lie in (2) or (3), then we know that after the system settles, we will have a

phase separated state. However, how this transition happens is not something

mean field theory tells us. Heuristically, we expect

–

In (2), we have

φ| < φ

, and

(

)

0 implies local instability. The

system rapidly becomes phase separated. This is spinodal behaviour.

–

In (3), we have

< |

φ| < φ

. A uniform phase is locally stable, but

not globally. To reach the phase separated state, we need nucleation and

growth to occur, and requires the contribution of noise.

We now study these in detail.

Regime 1

We know that regime (1) is stable, and we shall see how it responds to perturba-

tion about φ(r) =

φ. Put

φ =

φ +

φ(r).

We can then write

µ =

δF

δφ

∂f

∂φ

− κ∇

φ = f

(

φ) +

φf

(

φ) − κ∇

φ.

Note that the first term is a constant. We then have

φ = −∇ · J

J = −M∇[f

φ − κ∇

φ] +

T MΛ.

We drop the tildes and take the Fourier transform to get

= −Mq

+ κq

)φ

+ iq ·

T MΛ

Compare this with an overdamped particle in a simple harmonic oscillator,

V =

κx

where we have

˙x = −

Mκx +

MΛ.

Indeed, we can think of our system as an infinite family of decoupled harmonic

oscillators, and solve each of them independently.

In the second example sheet, we compute

S(q, t) ≡ hφ

(0)φ

−q

(t)i = S(q)e

−r(q)t

This

(

q, t

) is called the dynamic structure factor , which can be measured by light

scattering. This doesn’t say the fluctuations go away completely — we expect

there to be fluctuations all the time. What this says is that fluctuations at late

times come completely from the random noise, and not the initial fluctuations.

Regime 2

Consider the case where we are in the second regime. As before, we have the

equation

= −Mq

+ κq

)

| {z }

r(q)

+ iq ·

T MΛ

but crucially, now

(

)

0, so it is possible to have

(

)

0. The system is

unstable.

If we ignore the noise by averaging the equation, then we get

i = −r(q)hφ

So if we have a noisy initial state φ

(0), then the perturbation grows as

hφ

(t)i = φ

(0)e

−r(q)t

When

(

)

0, then this amplifies the initial noise. In this world, even if we

start with a perfectly uniform

, noise terms will kick in and get amplified over

time. Moreover, since we have an exponential growth, the earliest noise gets

amplified the most, and at late times, the new perturbations due to the noise

are negligible.

We can plot our r(q) as follows:

r(q)

∗

The maximally unstable mode

∗

is given by the solution to

(

∗

) = 0, which

we can easily check to be given by

∗

−f

2κ

Now consider the equal time non-equilibrium structure factor

(t) = hφ

(t)φ

−q

(t)i ∼ S

(0)e

−2r(q)t

As time evolves, this gets more and more peaked around q = q

∗

(t)

∗

So we see a growth of random structure with scale

L ∼ π/q

∗

. This process is

called spinodal decomposition.

Note that this computation was done on the assumption that

is small, where

we ignored the quadratic terms. At intermediate

, as these phase separated

states grow, the quartic terms are going to kick in. An informal use of variational

theory says we should replace f

, where

= f

(2π)

max

(t) d

This says

is less negative as the fluctuations grow. Since

∗

−

this moves to a smaller

. So

(

)

∼ π/q

∗

(

) starts increasing. This is called

domain growth.

In the late stages, we have large regions of

φ ≈ ±φ

, so it is almost in

equilibrium locally. We are well away from the exponential growth regime, and

the driving force for domain growth is the reduction of interfacial area. We can

estimate the free energy (per unit volume) as

σA(t)

where

(

) is the area of the interface. So by dimensional analysis, this is

∼ σ/L(t). We have calculated the interfacial surface tension σ before to be

σ =



−8κa



1/2

but it doesn’t really matter what this is. The ultimate configuration with

minimal surface area is when we have complete phase separation. The result is

that

L(t) ∼





1/3

We will not derive this yet, because this result is shared with the late time

behaviour of the third regime, and we will discuss this at that point.

Regime 3

Finally, consider the third regime. Suppose we have

−φ

, where

small. The system is locally stable, so

(

)

0 for all

. On the other hand, it is

globally unstable, so phase separation is preferred. To achieve phase separation,

we must overcome a nucleation barrier, and we must rely on noise to do that.

To understand how the process will look like, formally, we can inspect the

path probabilities

P[φ(r, t)] = N exp



−

|J + M∇µ|

dr dt



given by the Langevin equation. We seek to find the most likely trajectory from

the initial to the final state. In field theory, this is called the instanton path,

and in statistical physics, this is called large deviation theory. Instead of doing

this, we use our physical intuition to guide ourselves.

Heuristically, we expect that if we start with a uniform phase

−φ

then at some point, there will be some random small droplet with

= +

small radius

. This is already unlikely, but after this, we need

to increase

until we have full phase separation. The key question is — how unlikely is this

process?

The idea is to consider the cost of having a droplet of radius

. First there is

the cost of having an interface, which is 4

πσR

. However, the fact that we have

areas is energetically favorable, and this grows as the volume. So we get a

cubic term ∼ −R

. If we add these two together, we get a barrier:

F (R)

∗

Once

R > R

∗

, it is then energetically favorable for the radius to continue

increasing, and then we can easily reach phase separation. To reach this, we

must rely on noise to push us over this barrier, and this noise-induced rate is

∼ e

−βF

∗

. To see what happens afterwards, we need to better understand how

droplets work.

Droplet in equilibrium

The mechanics of a droplet is slightly less straightforward than what one might

hope, due to the presence of surface tension that tries to compress the droplet.

The result is that the value of

inside and outside the droplet is not exactly

±φ

, but with a shift.

For simplicity, we shall first consider an equilibrium system with a droplet.

This is achieved by having a large box of fluid with

just slightly above

−φ

Then in the phase separated state, the +

phase will lump together in a droplet

(if we had

φ = 0, then we would have a horizontal interface).

Within each region 1 and 2, the value of

is constant, so the term that contributes

to the free energy is

f(φ) =

We would expect 1 and 2 to respectively be located at

1 2

When we have a spherical interface, 1 and 2 are not exactly at

±φ

. To see this,

Consider the bulk chemical potential

µ =

∂f

∂φ

The thermodynamic pressure is then

Π = µφ − f.

This is the negative of the y-intercept of the tangent line at φ.

If we have a flat interface, which we can think of as the limit

R → ∞

, then

we require

bulk

= µ

bulk

, Π

bulk

= Π

bulk

This means the points 1, 2 have a common tangent

1 2

If we have a droplet, then there is surface tension. Consider an imaginary

interface between the upper and lower interface. Then the pressure difference

tries to move the upper hemisphere up, and contributes a force of (Π

−

)

πR

while the interfacial tension pulls the boundary down by 2

πRσ

. In general, in

dimensions, we require

= Π

(d − 1)

This is called the Laplace pressure.

In static equilibrium, we still require

, since this is the same as saying

∇µ

= 0. So

has the same slope at

and

. However, the two tangent

lines no longer have a common intercept, but they are separated by

(

d −

1).

So it looks like

−Π

To solve for this, we take the approximation that

is small for

decently

large. Then we can write

= f(−φ

+ δ

) ≈

(−φ

)δ

+ f (−φ

)

= f(+φ

+ δ

) ≈

(+φ

)δ

+ f (+φ

So µ

= αδ

, where α = f

(±φ

). So we find that up to first order, δ

= δ

To compute δ, we compute

= µ

− f

= −αδφ

Similarly, we have Π

= +αδφ

. So

− Π

= −2αφ

δ.

Since this equals −(d − 1)

, we have

δ =

d − 1

2αφ

Multiple droplet dynamics

We now move on to understand multiple droplet dynamics. This is relevant

because we expect that noise will produce multiple droplets around the fluid,

which will then interact and combine to a single phase separated state.

The way droplets interact with each other is that once we have a droplet

of large

, then the average

outside of the droplet will decrease. So to begin

understanding this scenario, we first see how a droplet reacts when the relative

density of the outside bath is not what it expects.

So suppose we have a (3D) droplet of radius

inside a bath with

−φ

where

ε 6

(

). This

is called the supersaturation. Note that to have a

droplet of radius

, the value of

inside and immediately outside the droplet

must be

±φ

. Outside of the droplet, the value of

will slowly decay to

−φ

+ ε. Thus, outside of the droplet, we write

φ(r) = −φ

φ(r),

where

φ(∞) = ε and

φ(R

) = δ.

In this situation, unless

happens to be

, we have a gradient of

, hence a

gradient of chemical potential, hence a flux. Again in Model B, we have

φ = −∇ · J, J = −M∇µ = −Mα∇

φ(r),

assuming a weak enough gradient. We assume this has a quasi-static behaviour,

which is reasonable since molecules move quickly relative to how quickly the

droplet changes in size. So to solve for

(

) at any point in time, we set

= 0.

So ∇

φ = 0. We solve this with boundary conditions

φ(∞) = ε,

φ(R

) = δ.

So we have

φ = ε + (δ − ε)

Now if we assume this is what

(

) looks like, then the current just outside the

droplet gives

J(R

) = −M∇µ = −αM

∂

∂r



= αM(δ − ε)



r=R

αM(δ − ε)

Thus, when

and

are not the same, there is a flow of fluid in or out of the

droplet. The discontinuity in

across the boundary is ∆

= 2

. So mass

conservation implies

2φ

R = −J = −

αM(δ − ε)

Thus, we conclude that

R =

2φ



αM

(ε − δ(R))



We can plug in our previous expression for

. Fixing

, we can plot

as follows:

∗

where

∗

αεφ

So if we have a bath containing many droplets, then the big droplets grow and

the small droplets shrink. Indeed, the interfacial tension penalizes small droplets

more heavily than large droplets.

To understand exactly how these grow, we make a scaling ansatz that there

is only one length scale, namely the mean droplet size

R. Then we have

R ≈

2φ

αM

(ε − δ(

R)).

We know that the value of

is also determined by

, so we know

ε − δ

(

) is of

order δ(

R). Hence

R ∼

Mσ

∼

Mσt

So the typical droplet size is ∼ t

1/3

. Likewise, R

∗

∼ t

1/3

, and so ε ∼ t

−1/3

So if we have a sea of droplets, they go into this competitive process, and

we get fewer and fewer droplets of larger and larger size. This is called Ostwald

ripening, and is a diffusive coarsening mechanism.

We have the same scaling for non-droplet geometries, e.g. spinodal decom-

position at late times. In this case, our domains flatten and enlarge, and we

have

L(t) ∼



Mσ



1/3

In practice, we often want to stop this from happening. One way to do so is

to add trapped species insoluble in the continuous phase, e.g. polymers or salt.

If there are

particles inside the droplet exerting ideal gas pressure, then we

have

− Π

2σ

4/3πR

We again have µ

= µ

. This ends up giving a new boundary condition at R

φ(R

) =

αRφ

−

3Nk

8αφ

πR

2αφ

(Π

Lap

− Π

sol

)

The first term is the Laplace pressure just as before, while the second term is

the extra term from the trapped species.

If we put this back into the system of equations before, then we have a new

equation of motion

R =

2φ



αM



ε −

αφ

3Nk

8αφ

πR



We now see that there is a stable fixed point

, and further shrinkage is

prevented by the presence of the trapped species that will be further compressed

by shrinking. Thus, if we manage to produce a system with all droplets of size

< R

∗

, then we end up with a lot of small but finite size droplets R