Phase behaviour, sedimentation and stacking in colloidal

mixtures of thick and thin hard rods

Patrick HooijerUtrecht University

MSc. thesis

Supervisor:Prof. Dr. Rene van Roij

Institute for Theoretical PhysicsUtrecht University

June 30, 2015

Abstract

In this thesis we take a look at a binary mixture of thick and thin hard rods using Onsagertheory, and look at their high pressure expansion to study the nematic phase more accurately.We will numerically solve the non-linear integral equations leading to the orientation distri-bution functions of the system for several diameter ratios, concentrations and compositions.Using these solutions, we will draw the phase diagrams summarizing the possible phases of thesystem, and find isotropic-nematic, nematic-nematic as well as isotropic-isotropic demixing,with corresponding isotropic-nematic-nematic and isotropic-isotropic-nematic triple points.In a few cases, a nematic-nematic or isotropic-isotropic critical point will appear.

Then we will look at sedimentation profiles by including the effects of gravity in our calcu-lations, and calculate the paths these mixtures can take in various phase diagrams as functionof the height in a test tube. Here we find many configurations of the contents of the test tube,for example an isotropic phase between two nematic phases, and the composition variable ofthe heavier particles that is not increasing as function of the height. We will summarize thesedifferent configurations in a phase stacking diagram, and see that this diagram is not trivial,even for the simple case where there is only isotropic-nematic phase coexistence.

1

1 Introduction

Colloids are particles with a typical size of 10-1000 nm, and appear in many shapes andsizes. Colloids, along with other particles like polymers and surfactants are called mesoscopicparticles, because they are much heavier and larger than microscopic particles like atoms,but much lighter and smaller than macroscopic particles like desks and houses. Because ofthe size of colloids, they are typically classical objects instead of quantummechanical. Whilesimultaneously having a low enough binding energy, of the order of the thermal energy, thatthermal fluctuations and Brownian motion become major components in describing them.Classical statistical mechanics is usually the framework used to describe colloids.

1.1 Liquid Crystals

The gas, liquid and crystalline states of matter are well understand, and that the distinguishingfeature between these states is the positional ordering of the molecules between different states.While gases and liquids do have translational and rotational symmetry, these symmetries arefully broken in the crystalline phase. This makes these phase transitions interesting to study.Since many molecules are approximately spherical, most studies involving these substancesinvolves modelling these molecules by spheres.

For almost a century, there has been an increasing interest in other shapes as models formolecules: most commonly rod-like and disc-like colloids. The study of the phase behaviourof these kinds of molecules date back to at least the 1920s, where Zocher [1] discoveredexperimentally that a solution of rod-like colloids undergoes a phase transitions from anisomorphic phase to a phase where the rods are oriented along a specific direction.

Rod- and disc-like colloids exhibit additional forms of ordering compared to sphericalparticles: in addition to positional ordering there is also orientational ordering. The mostdisordered phase is the isotropic phase, where neither positions nor orientations are ordered.When all particles point in approximately the same direction, the fluid is in a nematic phase.Other phases include the smectic phase, where the particles are in addition to being orderedin one orientational direction are also ordered in one positional direction; and the crystallinephase, where the system is ordered in all three orientational and all three positional directions.Schematic configurations of these phases can be seen in Fig. 1. Because these additionalphases have optical properties of a crystal, while having the mechanical properties of a liquid,these systems are called Liquid Crystals. These combinations of properties of the traditionalfluid and crystalline phases makes them useful in multiple applications, for example in LiquidCrystalline Displays (LCDs).

Figure 1: Typical configurations of rod-like particles in the isotropic, nematic, smectic and crys-talline phases respectively. The nematic and smectic phases are examples of liquid crystal phasesnot appearing in a system with spherical particles; they have often the mechanical properties of afluid, but optical properties of a crystal. This thesis focusses on the first two phases.

2

2 Onsager Theory for Monodisperse Systems

To study a system consisting of N colloidal particles, we have to study its Hamiltonian. Intheoretical models of particles it is often assumed that the interaction between the particles isthe sum over pairwise terms, each of them characterized by a pair potential φ(~ri−~rj , ~Ωi− ~Ωj)that depends on the relative coordinates of the particles ~ri − ~rj and the relative orientationof the particles ~Ωi− ~Ωj . For these pairwise-additive particles, the Hamiltonian is given by [5]

H(~pN , ~rN , ~ΩN ) =

N∑i=1

~p2i2m

+

N∑i<j

φ(~ri − ~rj , ~Ωi − ~Ωj). (1)

In the 1940’s Onsager [2] successfully explained the isotropic to nematic phase transitiontheoretically in terms of the interparticle interactions. He modelled rod-like particles withrigid, infinitely long rods, where the only interaction between the rods is a short-range hard-core interaction

φ(~ri − ~rj , ~Ωi − ~Ωj) =

∞ : particle i overlaps with particle j

0 : otherwise. (2)

The hard-core potential does not contain any attracting force. This is present in real particlesdue to to Van der Waals forces, and describes the steeply repulsive interaction when twoparticles are very close to each other crudely. However, the hard-core potential is of greattheoretical importance because it is a zeroth-order approximation in the perturbation theoryof liquids. Moreover, experimental realizations of hard-core systems can be made due toadvances in the synthesis of colloidal particles in the form of colloidal suspensions. Onsagerused this simpel model to explain qualitatively the experimental observed interaction byZocher [1] between the particles.

2.1 Virial Expansion

The classic ideal-gas law pV = NkBT , where p is the pressure, V the volume, N the numberof particles, kB the Boltzmann constant and T the temperature; does not hold for real gassesat finite density ρ = N/V because of the interactions between the particles. Kamerlingh-Onnes proposed the virial expansion of the pressure as correction to this law for orientationalindependent particles by fitting correction terms to experimental data:

p(ρ, T ) = kBT (ρ+B2(T )ρ2 +B3(T )ρ3 + ...) (3)

where Bn(T ) are the virial coefficients. The second and third virial coefficient are respectively[5]

B2(T ) =− 1

2V

∫d3~ri

∫d3~rj

(exp(−βφ(~rij))− 1

)= −1

2

∫d3~rf(~r) and (4)

B3(T ) =− 1

3V

∫d3~ri

∫d3~rj

∫d3~rkf(~rij)f(~rjk)f(~rki). (5)

Here ~rij = ~ri − ~rj , β = 1/kBT is the inverse temperature and f(~r) = exp(−βφ(~r)) − 1 isthe Mayer function. We are however interested in orientational dependent particles and thusreplace the volume element with

d3~r → d3~r d3~Ω∫d3~Ω

. (6)

The virial coefficients become

B2(T ) =− 1

2V (∫

d3~Ω)2

∫d3~r

∫d3~Ωf(~r, ~Ω), (7)

B3(T ) =− 1

3V (∫

d3~Ω)3

∫d3~ri

∫d3~Ωi

∫d3~rj

∫d3~Ωj

∫d3~rk

∫d3~Ωk

× f(~rij , ~Ωij)f(~rjk, ~Ωjk)f(~rki, ~Ωki),

(8)

3

with ~Ωij = ~Ωi − ~Ωj . When filling in the hard core potential of Eq. (2), the exponentialterm exp(−βφ(rij ,Ωij) is either equal to 1 (if φ(rij ,Ωij) = 0) or 0 (if φ(rij ,Ωij) =∞). Thisremoves all dependencies on β, and thus the virial coefficients of hard-core interactions areindependent on the temperature T . Thus the interactions and dynamics of the system withhard-core potentials are driven purely by the entropy of the system.

The second virial coefficient for hard rod shaped particles has been calculated by Onsagerin a rather extensive calculation [2] and is equal to the excluded volume of two identicalparticles close to each other:

B2,HR =π

2

(D3| sin γ|+ LD2(1 + | cos γ|+ E(| sin γ|))

)+ 2L2D| sin γ|, (9)

where L is the length of the rod, D is the diameter of the rod, γ is the relative angle betweenthe rods and E(x) is the elliptic integral of the second kind

E(x) =

∫ π/2

0

dϕ

√1− x2 sin2 ϕ. (10)

In the infinite needle limit, where L D, the last term of Eq. (9) is the dominating term

B2,HR|LD = 2L2D| sin γ| (11)

In the isotropic phase, where all particles are even likely to point in any direction, we cantake the average of | sin γ|, which is equal to 〈| sin γ|〉iso = 1/(π)

∫ π0

dγ| sin γ| = 2/π. So thesecond virial coefficient in the infinite needle limit in the isotropic phase is equal to

B2,INL,iso =4

πL2D (12)

Onsager showed that it is plausible that the third and higher virial coefficients are negligible,and subsequent numerical work supports this claim. [4]

2.2 Helmontz Free Energy

Since the dynamics of the system are driven only by entropy, the internal energy dU =T dS − pdV + µ dN , where T is the temperature, S is the entropy, p is the pressure, V is thevolume and µ is the chemical potential; is not an interesting quantity to calculate becauseof its independence of the entropy S. We can however perform a Legendre transformationto transform the internal energy to the Helmontz free energy F = U − TS, such that dF =−S dT −p dV +µ dN . This is dependent on the entropy, and thus suited to study this systemof infinitely long rods.

The Helmontz free energy of an ideal gas is equal to F = NkBT (log(ρΛ3)− 1), [5] whereΛ = h/

√2πmkBT with h Planck’s constant is the thermal (De Broglie) wavelength of the

particle. For interacting liquids, we can use a virial expansion similar to Eq. (3) and write

F (T, V,N) = NkBT (−1 + log

(N

VΛ3

)+B2

N

V+

1

3B3

(N

V

)2

+ . . . ) (13)

Generalizing this expression for s species of Ns particles with N =∑iNi, Eq. (13) becomes

F (T, V,N1, N2, . . . , Ns) = kBT (−N +

s∑i=1

Ni log

(NiV

Λ3

)+

1

V

s∑i,j=1

B2(i, j)NiNj + . . . ) (14)

where B2(i, j) = − 12V

∫d3~r

∫d3~Ω

(exp(−βφij(~r, ~Ω))− 1

)is the second virial coefficient term

due to interactions between particles of species i and j. When looking at a large number ofparticles occupying all the directions Ω, we can replace the sum with an integral

F (T,N, V ) = kBT (−N +

∫dΩN(Ω) log

(N(Ω)

VΛ3

)+

1

V

∫dΩ

∫dΩ′B2(Ω,Ω′)N(Ω)N(Ω′))

(15)where we omitted the terms of B3 and higher order. Because rods have an axial symmetry, itis convenient to write dΩ in spherical coordinates dθ dϕ sin θ. Next we replace N(Ω) with the

4

orientation distribution function (ODF) ψ(Ω), which gives the fraction of particles pointingin the Ω direction: N(Ω) = Nψ(Ω). Filling in B2 from Eq. (11) we arrive at

F (T,N, V ) = NkBT(− 1 + 2π log

(N

VΛ3

)+ 2π

∫ π

0

dθ sin θψ(θ) log(ψ(θ))

+2

πL2D

N

V

∫ π

0

dθ

∫ 2π

0

dϕ

∫ π

0

dθ′∫ 2π

0

dϕ′ sin θ sin θ′| sin γ(θ, ϕ, θ′, ϕ′)|ψ(θ)ψ(θ′)).

(16)

We can write | sin γ(θ, θ′)| explicitly in terms of θ and θ′

| sin γ(θ, ϕ, θ′, ϕ′)| =√

1− (cos θ cos θ′ + sin θ sin θ′ cos(ϕ− ϕ′))2. (17)

Because this is the only term dependent on ϕ, we can move the ϕ integrals inside Eq. (17)and define the azimuthally averaged kernel K(θ, θ′):

K(θ, θ′)

2π=

1

2π

∫ 2π

0

dϕ

∫ 2π

0

dϕ′| sin γ| =∫ 2π

0

dϕ√

1− (cos θ cos θ′ + sin θ sin θ′ cosϕ)2. (18)

Because you can add or subtract any constant to the Helmontz free energy, we subtract theterms −1 and log(Λ3). By defining the dimensionless number density c = B2,IHR,isoN/V andthe Helmontz free energy per particle per kBT , we arrive at

f(c) ≡ F (T,N, V )

NkBT= log(c) + 4π

∫ π/2

0

dθ sin θψ(θ) log(ψ(θ))

+ 32c

∫ π/2

0

dθ

∫ π/2

0

dθ′ sin θ sin θ′K(θ, θ′)ψ(θ)ψ(θ′),

(19)

where we used the up-down symmetry for both isotopic and nematic systems ψ(θ) = ψ(π−θ).This free energy consists of an orientational entropy term for, which is part of the ideal gasterm log c+ for, and an excess term with respect to the ideal gas due to the excluded volumeof the particles fex.

One can minimize this free energy with respect to the ODF by using the Euler-Lagrangeequation δf/δψ(θ) = 0, which gives

λ = logψ(θ) +8c

π

∫ π

0

dθ′ sin θ′K(θ, θ′)ψ(θ′), (20)

where λ is the Lagrange multiplier that fixes the normalization 4π∫ π/20

dθ sin θψ(θ) = 1.This equation is a non-linear integral equation, and has not been solved analytically. It

can however easily be checked that the isotropic phase ψiso(θ) = 1/(4π) is a solution of Eq.(20) for any c. The nematic solution can be calculated using numeric methods.

Given a solution ψ(θ) of Eq. (20), we can calculate the pressure, chemical potential andorder parameter of the particles

p ≡ βpB2 = B2

(β∂F

∂V

)N

=∂f

∂cc2, (21)

µ ≡ βµ =

(β∂F

∂N

)V

= f +∂f

∂cc, (22)

S = 4π

∫ π/2

0

dθ sin θψ(θ)3 cos2 θ − 1

2. (23)

where p and µ are dimensionless versions of the pressure p and the chemical potential µrespectively. Having the expression for the pressure in Eq. (21), we can perform anotherLegendre transformation to transform the Helmontz free energy into the Gibbs free energy:G(T, p, N) = F + pV . Thus the Gibbs free energy is

dG = dF + d(pV ) = −S dT + V dp+ µ dN (24)

g(p) = G/(NkBT ) = f + βpB2/c. (25)

The Gibbs free energy is useful for studying phase equilibria, since the pressure is constantalong the interface between two phases.

5

0

5

10

15

20

25

30

0 0.1 0.2 0.3 0.4 0.5

ψ(θ)

θ

Nematic ODF solutions for various dimensionless number densities c

c=1 (isotropic)c=5c=10c=15

Figure 2: Nematic solutions of Eq. 20 for various dimensionless number densities c. The ODFsbecome more and more peaked for higher c, while having a constant surface due to the normal-ization. No nematic solution was found for c = 1, so the isotropic solution ψ(θ) = 1/(4π) ≈ 0.796is calculated instead. The nematic solutions exponentially decay. The ODFs are only plotted forθ ∈ [0, 5], while the actual domain of θ is [0, π/2].

-6

-4

-2

0

2

4

6

0 1 2 3 4 5 6 7

f(c)

c

Dimensionless Helmontz free energy as function of c

c∗ ' 3.539

fnem(c)fiso(c)

•

Figure 3: Dimensionless Helmontz free energy as function of the dimensionless number density.The isotropic and nematic solution have equal f for c∗ ' 3.539. For c < c∗, the isotropic phase isfavourable, while for c > c∗ the nematic phase is favourable.

6

2.3 Numerical Methods

The non-linear integral equation of Eq. (20) can be solved using an iterative scheme, aprocedure very common in the theory of many-body systems. This method is used andexplained by van Roij in [6]. Starting with an initial guess ψ0(θ), we calculate the next stepψn+1(θ) by filling in the result from the previous step ψn(θ) in Eq. (20):

ψn+1(θ) =1

Zexp

(−8c

π

∫ π

0

dθ′ sin θ′K(θ, θ′)ψn(θ′)

)(26)

where Z is a normalization constant such that the normalization 14π

∫ π/20

sin θψn+1(θ) = 1 issatisfied. Self-consistency is obtained by applying this iterative scheme multiple times untilthe desired accuracy is reached.

The ODF ψ(θ) will be determined on a discrete grid of a finite set of N polar angles θk inthe interval [0, π/2], with k = 1, 2, . . . , N . For an equidistant grid, θk is equal to

θk =π

2

k

N + 1, k = 1, 2, . . . , N, (27)

however, other grids that are finer close to θ = 0 are used , because the ODFs become verypeaked for higher densities (see Fig. 2). Because dθ sin θ = − d(cos θ), we can approximatefor any arbitrary function f(θ)∫ π/2

0

dθ sin θf(θ) ≈N∑k=1

∆kf(θk), (28)

where we define ∆k as

∆k ≡

1− (cos θk + cos θk+1)/2 : k = 1,

(cos θk−1 − cos θk+1)/2 : k = 2, 3, . . . , N − 1,

(cos θk + cos θk−1)/2 : k = N.

(29)

This choice of ∆k guarantees the correct normalization∑Nk=1 ∆k = 1 for any θ-grid for N ≥ 2.

To calculate the azimuthally averaged kernel K(θ, θ′), defined in Eq. (18), a equidistant gridof the Nϕ azimuthal angles ϕj with

ϕj = 2πj

Nϕ + 1, j = 1, 2, . . . , Nϕ. (30)

Using this grid, the values of K(θ, θ′) on this grid are equal to

Kkl ≡ K(θk, θl) =2π

Nϕ + 1

3

2| sin γkl(ϕ1)|+

Nϕ−1∑j=2

| sin γkl(ϕj)|+3

2| sin γkl(ϕNϕ)|

(31)

with | sin γkl(ϕ)| the discrete version of Eq. (17):

| sin γkl(ϕ)| =√

1− (cos θk cos θl + sin θk sin θl cosϕ)2. (32)

When calculating ψ(θ) for a wide range of cm, m = 1, . . . ,M , it is more efficient to usethe result of the previous step as your first initial guess: ψ0(θ)|cm = ψ(θ)|cm−1 . Because theODFs in Fig. 2 are very similar to the Gaussian distribution, but not identical, it is convenientto choose our first guess ψ0(θ)|c0 = (c0/π)2 exp(−2c20θ

2/π). This ensures quick convergencefor small enough cm − cm−1, except around the density where fiso(c∗) = fnem(c∗), where theODFs jump from a peaked function around θ to a constant function. Because Eq. (31) isindependent of c, it is advised to calculate and store Kkl before applying the iterative schemeand load them when calculating the ODFs.

7

2.4 Phase separation

The dimensionless Helmontz free energy f as function of the dimensionless number density cof the isotropic and nematic phase are equal to each other at c∗ ' 3.539, see Fig. 3. Thus onecould argue that a phase transition appears at c = c∗, with an isotropic phase at c < c∗ and anematic phase at c > c∗. However, the system can lower its energy further at a given densityc ∈ (cI , cN ) by splitting up into a subvolume in the isotropic phase with density cI , and asubvolume in the nematic phase with density cN . Such a coexistence requires mechanical anddiffusive equilibrium between the two phases, which are only true if the pressure and chemicalpotential between the two phases are the same:

piso(cI) = pnem(cN ) µiso(cI) = µnem(cN ). (33)

These two equations with two variables can easily be solved using numerical root findingmethods. Another way of finding the two coexisting pressures cI and cN is by finding thecommon tangents of φ(c) ≡ f(c)c, this is called the Maxwell construction. Finding a commenttangent of a function φ(c) is equivalent to solving

dφ(c)

dc

∣∣∣∣cI

=dφ(c)

dc

∣∣∣∣cN

, φ(cN ) = φ(cI) +dφ(c)

dc

∣∣∣∣cI

(cN − cI), (34)

for variables cI and cN . These equations are equivalent to the equations in (33):

p(c) =df(c)

dcc2 =

dφ(c)

dcc− φ(c) → φ(cN )− dφ(c)

dc

∣∣∣∣cN

cN = φ(cI)−dφ(c)

dc

∣∣∣∣cI

cI , (35)

µ(c) = f(c) +df(c)

dcc =

dφ(c)

dc→ dφ(c)

dc

∣∣∣∣cI

=dφ(c)

dc

∣∣∣∣cN

, (36)

where we used Eqs. (21) and (22). This construction can be seen in Fig. 4, where the rightpicture is the same as the left picture minus the linear term cµIN . Since φ(c) − cµ(c) =φ(c)−cdφ/dc = −p(c), the common tangent in the right picture is a horizontal line at −pIN .

The physical quantities at this coexistence are cI ' 3.290, cN ' 4.191, µIN ' 5.241,pIN ' 14.12, SI = 0 and SN ' 0.7922, where we used the definitions of Eqs. (21)-(23).

8

Common tangent of φ(c)

0

2

4

6

8

10

12

14

16

2.5 3 3.5 4 4.5 5

φ(c)

c

c∗cI cN

φiso(c)φnem(c)

•N

N

-14.4

-14.2

-14

-13.8

-13.6

2.5 3 3.5 4 4.5 5

c

c∗cI cN

−pIN

φ(c)−cµIN

φiso(c)φnem(c)

•N N

Figure 4: On the left: common tangent of φ(c) = cf(c). Because the common tangent is hard tosee, we subtract in the figure right the linear term cµIN , where µIN is the chemical potential atthe interface. Since φ(c)− cµ(c) = −p(c), the coexistence pressure −pIN can be read off the rightpicture.

Physical quantities of the solutions of Eq. (20) as function of c.

0

5

10

15

20

25

30

35

40

0 1 2 3 4 5 6 7

p(c)

c

Pressure

cI cN

pIN

pnem(c)piso(c)

N N

-5

0

5

10

15

0 1 2 3 4 5 6 7

µ(c)

c

Chemical potential

cI cN

µIN

µnem(c)µiso(c)

N N

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 1 2 3 4 5 6 7

S(c)

c

Order parameter

cI cN

SI

SN

Snem(c)Siso(c)

N

N

Figure 5: Graphs of the pressure, chemical potential and order parameter respectively for theisotropic and nematic phases, with the coexistence values highlighted.

9

3 Binary Mixtures

Onsager theory can easily be extended for mixtures of different particles. The orientationalfree energy is replaced by a sum of the orientational free energy of each species; the excess freeenergy, which consist of the interactions between particles, now has separate terms for everypair of species; and an extra mixing free energy term is added, which favours the system tospread the particles of different species evenly.

Lekkerkerker et al. argued in 1984 [7] that the same isotropic-nematic demixing happensin a binary mixture of long and short hard rods, but with different concentrations of the longerrods in the nematic phase compared to the isotropic phase. This is because when the thickrods are in the nematic phase and thus are aligned, there is little room for the shorter rods topoint in every orientation with equal probability. It costs less energy to split the system in anematic phase, where the longer rods are aligned and thus are efficient with the volume, andan isotropic phase, where the shorter rods can freely explore all orientations without beingconstricted by the aligned longer rods.

It was not until 1993 that Vroege and Lekkerkerker [8] argued that there should alsoappear a nematic-nematic demixing when the longer rods are three times larger than theshorter rods. This is similar to the isotropic-nematic demixing for binary mixtures, where theshorter rods are less aligned (but are still in a nematic phase) than the longer rods. In 1995,Sear and Jackson [9] showed that a mixture of hard rods can also have an isotropic-isotropicdemixing. They showed it for a system of thick and thin hard rods, where the thicker rodsare more than five times thicker than the thin rods. It has been argued that this demixingis due to the depletion effect, where the excluded volume per particle around the thick rodswhere thin rods can appear is smaller if the thicker rods have overlapping excluded volumes,thus favouring a thick rod rich phase and a thin rod rich phase.

It is also possible to extend Onsager theory to include charged rods [10] and chiral particles[11].

3.1 Extension to Onsager theory

Next we consider a system with Nσ hard rods of species σ = 1, 2 with diameters Dσ andidentical length L in a macroscopic volume V , and define the diameter ratio d = D2/D1 > 1.We characterize the thermodynamic state by the dimensionless number density c = (N1 +N2)B2/V with respect to the second virial coefficient of the thinner rods in the isotropic phaseB2 = (π/4)L2D1; and the composition variable x = N2/(N1 +N2), which is the mole fractionof the thick rods. We also consider separate ODFs ψσ(θ) for the two different species, suchthat ψ(θ) = (1− x)ψ1(θ) + xψ2(θ). Filling in this substitution, the orientational term of Eq.(19) becomes

4π

∫ π/2

0

dθ sin θψ(θ) log(ψ(θ))

= 4π

∫ π/2

0

dθ sin θ ((1− x)ψ1(θ) + xψ2(θ)) log((1− x)ψ1(θ) + xψ2(θ))

= 4π(

(1− x) log(1− x)

∫ π/2

0

dθ sin θψ1(θ) + x log x

∫ π/2

0

dθ sin θψ2(θ)

+

∫ π/2

0

dθ sin θ ((1− x)ψ1(θ) log(ψ1(θ)) + xψ2(θ) log(ψ2(θ))))

= fmix(x) + (1− x)for,1 + xfor,2

(37)

where for,σ is the orientational term of the free energy of the monodisperse system of speciesσ, replacing ψ with ψσ:

for(x) = (1− x)for,1 + xfor,2

= 4π

∫ π/2

0

dθ sin θ ((1− x)ψ1(θ) log(ψ1(θ)) + xψ2(θ) log(ψ2(θ)))(38)

Using the identity 4π∫ π/20

dθ sin θψσ(θ) = 1, we can write the mixing term of the freeenergy as

fmix(x) = (1− x) log(1− x) + x log x. (39)

10

The excess term fex of Eq. (19) is replaced in the same way. However, we have totake notice that the virial coefficients of the two species are not equal, from Eq. (11) isB2,INL,σ = 2L2Dσ| sin γ|. However, B2,INL,2 = B2,INL,1d, thus the excess term of the freeenergy for a binary mixture is equal to

fex(c, x, d) = 32c

∫ π/2

0

dθ

∫ π/2

0

dθ′ sin θ sin θ′K(θ, θ′)

×((1− x)2ψ1(θ)ψ1(θ′) + x(1− x)(1 + d)ψ1(θ)ψ2(θ′) + x2dψ2(θ)ψ2(θ′)

),

(40)

so the total free energy of the binary mixture is

f(c, x, d) = log(c) + fmix(x) + for(x) + fex(c, x) (41)

where fmix, for and fex are defined in Eqs. (38)-(40).Using the Euler-Lagrange equations, we can minimize f with respect to the ODFs by

solving δf/δψσ(θ) = 0, which gives

µ1 = logψ1(θ) +8c

π

∫ π/2

0

dθ′ sin θ′K(θ, θ′)(2(1− x)ψ1(θ′) + x(1 + d)ψ2(θ′)

),

µ2 = logψ2(θ) +8c

π

∫ π/2

0

dθ′ sin θ′K(θ, θ′)((1− x)(1 + d)ψ1(θ′) + 2xdψ2(θ′)

),

(42)

where we fix both Lagrange multipliers µσ by normalizing the ODFs: 4π∫ π/20

dθ sin θψσ(θ) =1. Again it is easy to show that ψiso,σ(θ) = 1/(4π) are solutions to Eq. (42).

Once a solution for (42) has been found, one can calculate f by plugging in the resultingODFs in (41). By knowing f , one can calculate multiple physical quantities related to thesystem. The physical quantities we are interested in are the pressure p, chemical potentialsµ1 and µ2, the order parameters Sσ and the Gibbs free energy G. It is convenient to lookat the dimensionless pressure βpB2, dimensionless chemical potentials βµ1 and βµ2, and theGibbs free energy per particle per kBT g, which are defined as

p ≡ βpB2 =

(B2β∂F

∂V

)N1,N2

=

(∂f

∂c

)x

c2, (43)

µ1 ≡ βµ1 =

(β∂F

∂N1

)V,N2

= f −(∂f

∂x

)c

x+

(∂f

∂c

)x

c, (44)

µ2 ≡ βµ2 =

(β∂F

∂N2

)V,N2

= f +

(∂f

∂x

)c

(1− x) +

(∂f

∂c

)x

c. (45)

Sσ = 4π

∫ π/2

0

dθ sin θψσ(θ)3 cos2 θ − 1

2(46)

dG = −S dT + V dp+ µ1 dN1 + µ2 dN2 (47)

g =F + pV

NkBT= f +

βpB2

c(48)

Starting from here, p and µσ refer to the dimensionless versions of the pressure andchemical potentials respectively.

3.2 High pressure limit

To study the nematic-nematic coexistence, we have to look at large concentrations c, and thuslarge pressures p. However, the resulting ODFs from Eq. (42) are extremely peaked aroundθ = 0, causing even very fine θ-grids to be too crude for accurate results. It is more efficientto scale the ODFs with respect to the concentration c, and then expand the free energy f forlarge c [14] [15].

First we introduce the reduced ODFs

φσ(θ) =ψσ(θ)

ψσ(0)(49)

11

0

2

4

6

8

10

0 0.1 0.2 0.3 0.4 0.5

ψ(θ)

θ

ODFs for d=4, c=3 and x=0.5

ψiso,1ψiso,2ψnem,1ψnem,2

Figure 6: Solutions of the equations in (42) for diameter ratio d = 4, dimensionless number densityc = 3 and composition variable x = 0.5. ψiso,σ = 1/(4π) are the solutions of the isotropic phase,and ψnem,σ are the solutions of the nematic phase. It shows that in the nematic case the thickerrods are more alligned to the nematic director than the thinner rods. The ODFs for differentparameters yield similar shapes of the nematic solutions, but with different width and heights.

which is equal to 1 at θ = 0 by construction. Next we introduced the scaled polar anglet = cθ. Using this definition, we can expand the terms in (41) for large c:

K(θ, θ′) = cK0(t, t′) +K2(t, t′)/c+O(1/c3),

dθ sin θ = (dtt)/c2 × (1− t2/6c2 +O(1/c4)),

φσ(θ) = φσ0(t) + φσ1(t)/c+ φσ2(t)/c2 +O(1/c3).

(50)

where the kernels K0 and K2 are equal to

K0(t, t′) = 4(t+ t′)E

(4tt′

(t+ t′)2

)K2(t, t′) =

1

3(t+ t′)

((t− t′)2K

(4tt′

(t+ t′)2

)− 3(t2 + t′2)E

(4tt′

(t+ t′)2

)) (51)

where K(x) and E(x) are the elliptic integrals of the first and second kind, respectively.Note that Kn for odd n are equal to 0.

Substituting these expansions in Eq. (42), and using the short-hand notations ∆Kn(t, t′) =Kn(t, t′)−Kn(0, t′) and Ξσρ =

∫dttφσρ(t), we get

log φ1,0(t) = − 2

π2

∫ πc/2

0

dt′t′∆K0(t, t′)

(2(1− x)

φ10(t′)

Ξ10+ x(1 + d)

φ20(t′)

Ξ20

),

log φ1,0(t) = − 2

π2

∫ πc/2

0

dt′t′∆K0(t, t′)

((1− x)(1 + d)

φ10(t′)

Ξ10+ 2xd

φ20(t′)

Ξ20

) (52)

12

φ11(t) = − 2

π2φ10(t)

∫ πc/2

0

dt′t′∆K0(t, t′)

×(

2(1− x)

(φ11(t′)

Ξ10− φ10(t′)Ξ11

Ξ210

)+ x(1 + d)

(φ21(t′)

Ξ20− φ20(t′)Ξ21

Ξ220

)),

φ21(t) = − 2

π2φ20(t)

∫ πc/2

0

dt′t′∆K0(t, t′)

×(

(1− x)(1 + d)

(φ11(t′)

Ξ10− φ10(t′)Ξ11

Ξ210

)+ 2xd

(φ21(t′)

Ξ20− φ20(t′)Ξ21

Ξ220

)),

(53)

φ12(t) = − 2

π2φ10(t) (2(1− x)v1(t) + x(1 + d)v2(t)) +

φ211(t)

2φ210(t)

,

φ22(t) = − 2

π2φ20(t) ((1− x)(1 + d)v1(t) + 2xdv2(t)) +

φ221(t)

2φ220(t)

,

(54)

with

vσ(t) =1

Ξσ0

∫ πc/2

0

dt′t′(− t′2

6∆K0(t, t′)φσ0(t′) + ∆K2(t, t′)φσ0(t′) + ∆K0(t, t′)φσ2(t′)

+ ∆K0(t, t′)φσ1Ξσ1Ξσ0

+ ∆K0(t, t′)φσ0(t′)Ξ2σ1 − Ξσ0

∫dt′′t′′(φσ2(t′′)− t′′2

6)

Ξ2σ0

).

(55)

Note however that the equations in (53) are linear integral equations with respect to φσ1without a source term, and thus are equal to 0. Therefore we can remove Eq. (53) andsimplify (55) by

vσ(t) =1

Ξσ0

∫ πc/2

0

dt′t′(− t′2

6∆K0(t, t′)φσ0(t′) + ∆K2(t, t′)φσ0(t′)

+ ∆K0(t, t′)φσ2(t′)−∆K0(t, t′)φσ0(t′)

∫dt′′t′′(φσ2(t′′)− t′′2

6)

Ξσ0

).

(56)

The φσρ terms still are dependent on c: they appear in the integration limits. However, theODFs decay exponentially to 0 (see Fig. 7). For any integrable function h(t) the followingholds: ∫ πc/2

0

dth(t) =

∫ ∞0

dth(t)−∫ ∞πc/2

dth(t). (57)

If h(t) is an exponentially decaying function h(t) ∼ exp(−λt), then the last term of Eq. (57)is equal to ∫ ∞

πc/2

dth(t) ∼ 1

λexp(−λπc/2)

c→∞−−−→ 0. (58)

Thus we can replace∫ πc/20

with∫∞0

in Eqs. (52)-(56), meaning these equations are indepen-dent of c.

Using this expansion, we can expand the free energy too

f(c, x) = 3 log c+ f0(x) + f2(x)/c2 +O(1/c4). (59)

We will not show the derivation of f0 and f2 explicitly, because the calculations are long butstraightforward.

After calculating f(c, x), we can calculate the physical quantities in the same way as Eqs.(43)-(45) using Eq. (59)

p =∂f

∂cc2 = 3c− 2

cf2(x) (60)

µ1 = f − ∂f

∂xx+

∂f

∂cc = 3 log c+ f0(x)− xf ′0(x) + 3− 1

c2(f2(x)− xf ′2(x)

)(61)

µ2 = f +∂f

∂x(1− x) +

∂f

∂cc = 3 log c+ f0(x) + (1− x)f ′0(x) + 3− 1

c2(f2(x) + (1− x)f ′2(x)

)(62)

13

1e-60

1e-50

1e-40

1e-30

1e-20

1e-10

1

0 2 4 6 8 10 12 14

φ(t)

t

ODFs of the high pressure expansion for d = 4. and x = 0.5.

φ10(t)φ20(t)φ12(t)φ22(t)

Figure 7: Logarithmic plot of the solutions of Eqs. (52) and (54), showing the exponential decayrequired to perform the limit in Eq. (58).

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

φ(t)

t

Comparison the numeric solution and the zeroth order approximation of species 1 for d = 4 and x = 0.5

φ10(t)ψ1(cθ)/ψ(0) for c = 2ψ1(cθ)/ψ(0) for c = 5ψ1(cθ)/ψ(0) for c = 10ψ1(cθ)/ψ(0) for c = 20

Figure 8: Comparison between the zeroth order term of the high pressure expansion, which is thesolution of Eq. (53); and the solution of Eq. (20), for species 1. For c = 2, there is large differencebetween the zeroth order approximation and the numerical solution, because the next order in theexpansion scales with 1/22 = 1/4. For c = 5, the two curves lie much closer, since the next orderin the expansion scales with 1/52 = 1/25.

14

Gibbs free energy of the isotropic phase for d = 10 and p = 2, with its common tangent.

0

1

2

3

4

5

6

0.2 0.4 0.6 0.8

g(p, x)

x

xI1 xI2

giso(2, x)Common tangent

•

•

-0.04

-0.02

0

0.02

0.04

0.06

0.08

0.1

0.2 0.4 0.6 0.8

x

xI1 xI2

giso(2, x)−linear termCommon tangent

• •

Figure 9: On the left: g(p = 2, x), which is in the isotropic phase, and its common tangent withitself. Because the common tangent is hard to see, we subtract in the figure right a linear term.The common tangent touches g(p, x) at the points xI1 ' 0.0257 and xI2 ' 0.576, thus the volumesplits into two subvolumes with isotropic phases with compositions xI1 and xI2 .

where f ′n(x) = dfn(x)/dx.

3.3 Phase coexistence

Once we have solved Eq. (42) and calculated the related physical quantities (43)-(46) for fixedd and variable c and x, we can compare the (Gibbs) free energy for both the isotropic and thenematic solutions. The isotropic phase is more energetic favourable for x . 0.568 while thenematic phase is favoured for x & 0.568. However, an even more energetic favourable state isby phase seperating the isotropic from the nematic part. Such a phase coexistence can onlyhappen if we find two solutions of Eq. (42) where the chemical potentials and pressures arethe same:

µ1(c1, x1) = µ1(c2, x2), µ2(c1, x1) = µ2(c2, x2), p(c1, x1) = p(c2, x2) (63)

where p and µσ are defined in Eqs. (43) - (45). Note that these are 3 equations with 4unknowns, so we add a fourth equation fixing the pressure: p(c1, x1) = p. This equation isequal to finding the common tangent between the Gibbs free energies of the isotropic and thenematic phase, as demonstrated in Fig. 9.

Isotropic-nematic phase coexistence is not the only type of coexistence found in binarysystems of thin and thick hard rods: for high pressure (p & 14) and high diameter ratios(d & 4), a nematic-nematic phase coexistence is energetic favourable, where one nematicsystem has significantly more colloids of species 1. For even greater diameter ratios (d & 8)and low pressures (p . 2.5), an isotropic-isotropic phase coexistence appears.

Because the equations at (63) are three equations with four unknowns, a solution curveappears. By adding a fourth constraint fixing the pressure p(c1, x1) = p, we can visualize thissolution space in a phase diagram. Phase diagrams in the p-x plane are shown in Fig. 10.

These phase diagrams can also be represented in different planes, for example in the c1−c2plane, where c1 = (1 − x)c and c2 = xc are the dimensionless number densities of species 1and 2 respectively; or the µ1 − µ2 plane, which is of main interest in sedimentation profiles.

15

2

4

6

8

10

12

14

0 0.2 0.4 0.6 0.8 1

p

x

Phase diagram for d=3.5

I

I-N

N

5

10

15

20

25

30

0 0.2 0.4 0.6 0.8 1

p

x

Phase diagram for d=4.2

I

I-N

N2N1

N1-N2

I-N1-N2 Triple point

N1-N2 Critical Point

N

N

5

10

15

20

25

30

0 0.2 0.4 0.6 0.8 1

p

x

Phase diagram for d=4.3

I

I-N

N2

N1

N1-N2

Triple Point

1

1.5

2

2.5

3

0 0.2 0.4 0.6 0.8 1

p

x

Phase diagram for d=10

I1 I2

N2

I1-I2I2-N2

I1-N2

I1-I2-N2 Triple Point

I1-I2 Critical Point

H

H

Figure 10: Phase diagrams in the p-x plane for d = 3.5, 4.2, 4.3 and 10.

16

Phase diagram in the µ1 − µ2 plane for diameter ratios d = 3.5 and d = 10.

4.5

5

5.5

6

6.5

7

7.5

8

8.5

-4 -2 0 2 4 6 8

µ2

µ1

N

I

0

2

4

6

8

10

12

14

-2 0 2 4 6 8

µ2

µ1

N2

I2 I1

N1

Triple Points

I1-I2 Critical Point

N

N

Figure 11: Phase diagrams in the µ1 - µ2 representation for diameter ratios d = 3.5 and d = 10respectively, with the critical and triple points highlighted.

Phase diagram in the c1 − c2 plane for diameter ratios d = 3.5 and d = 10.

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5

c2

c1

I

N

0

1

2

3

4

5

0 1 2 3 4 5

c2

c1

N2

I2I1 N1

N1 −N2

I1 −N2

Figure 12: Phase diagrams in the c1 - c2 representation for diameter ratios d = 3.5 and d = 10respectively. Coexistence are dotted in the picture on the left. All phase lines stop at the c1axis at the densities of the monodisperse system, while at the c2 axis, these endpoints multipliedwith a factor 1/d. In the picture on the right, the triangles represent the triple points, with theI1 − I2 −N2 triple point on the left, and the I1 −N1 −N2 triple point on the right.

17

4 Sedimentation

Experimental research on binary mixtures is often more challenging than studying monodis-perse systems because of the effects of gravity which are often unavoidable. By updating thefree energy with height-dependent density profiles caused by gravity, one can calculate thegrand chemical potential and minimize this to study the effects of gravity in binary mixtures.While this calculation is straightforward, many interesting phenomena happen when puttinga binary mixture in a test tube. For example, an isotropic phase can appear between twonematic phases. Even for relatively simple systems with low diameter ratios, where there isno demixing except for the isotropic-nematic demixing, multiple layers of alternating isotropicand nematic phases appear. Another interesting phenomena is that the fraction of the heavierparticles is not always decreasing higher in the tube, it is quite possible that a heavy particlerich layer appears between two phases dominated by lighter particles. [16] [17] The theory tomodel this behaviour is inspired by [18] [19].

The grand potential functional as function of the density is given by

Ω[ρ1(~x, θ), ρ2(~x, θ)] = F [ρ1(~x, θ), ρ2(~x, θ)]

+

∫d3~x

∫dθ sin θ

(ρ1(~x, θ)(Vext,1(~x, θ)− µ1) + ρ2(~x, θ)(Vext,2(~x, θ)− µ2)

) (64)

where F [ρ1, ρ2] is the free energy functional and Vext,σ is an external potential on species σ.We can approximate the free energy functional using a local density approximation because thegravitational length of the particles, which is often in the order of a millimetre, is much largerthan the size of the colloids, which is of the order of nanometres. Using this approximation,the free energy functional can be approximated as

F [ρ(z, θ)] ≈∫

d3~xfb(ρ1(~x), ρ2(~x)) (65)

where fb(ρ) = Fb(N,V, T )/(V kBT ) is the bulk free energy per volume per kBT , as defined inEq. (41). Using gravity as our external potential Vext,σ = mσgz, Eq. (64) reduces to

Ω[ρ1(z), ρ2(z)] = A

∫dz(fb(ρ1(z), ρ2(z)) + ρ1(z)(m1gz − µ1) + ρ2(z)(m2gz − µ2)

), (66)

where A is a macroscopic area perpendicular to z. We minimize this grand potential usingthe Euler-Lagrange equations

δΩ[ρ1(z), ρ2(z)]

δρ1(z)= 0 = A

(∂f(ρ1, ρ2)

∂ρ1+m1gz + µ1

)= A(µ1b(ρ1) +m1gz − µ1)

δΩ[ρ1(z), ρ2(z)]

δρ2(z)= 0 = A

(∂f(ρ1, ρ2)

∂ρ2+m2gz + µ2

)= A(µ2b(ρ2) +m2gz − µ2)

(67)

and thus we want to solve the equations

µ1b(ρ1(z), ρ2(z)) = µ1 −m1gz µ2b(ρ1(z), ρ2(z)) = µ2 −m2gz (68)

where µσb is the bulk chemical potential and µσ is the chemical potential in absence of gravity.We can rewrite Eq. (68) as

µ2b(µ1b) =m2

m1(µ1b − µ1) + µ2 = sµ1b + a (69)

with

s = m2/m1 a = µ2 − µ1m2/m1 (70)

the mass ratio between the two species and the scaled chemical potential difference respec-tively. Note that Eq. (69) is a linear equation with respect to µ1b, and therefore are straightlines in the µ1 − µ2 phase diagrams like those in Fig. 11.

These straight lines in the µ1-µ2 with slope s and point of crossing the µ2-axis a correspondto the path the chemical potentials of the mixtures would take in a test tube with gravity. Eachcrossing with the phase-diagram corresponds to an interface between two phases, resulting inmany different stacking sequences for varying s and a.

18

2

2.2

2.4

2.6

2.8

3

0 2 4 6 8 10

µ2

µ1

A

B

Asymptotic terminal line

Terminal line

Sedimentation binodal

Critical point•

Figure 13: Example of a sedimentation binodal, terminal line and asymptotic terminal line for anexample phase diagram between phases A and B. Sedimentation binodals are equal to the tangentlines of the phase diagram, terminal lines go through critical or triple points, and asymptoticterminal lines are parallel to the asymptote.

4.1 Phase stacking diagram

We can summarize all possible stacking sequences for different mass ratios s and scaled chem-ical potential differences a in a phase stacking diagram. By calculating when the straightline changes the amount of crossings with a line of the phase diagram in the s− a parameterspace, we can calculate the boundaries of different stacking sequences. There are three typesof boundaries: sedimentation binodals, which correspond to the points in s − a space wherethe line is tangent to a phase diagram line; terminal lines, which are lines that touch certainterminal points like critical and triple points; and asymptotic terminal lines, which appeardue to the phase diagram having asymptotes. These boundaries are visualized in Fig. 13.

4.1.1 Sedimentation binodals

A straight line can cross a curve multiple times. When considering a one-dimensional curvef(x) with an infinite domain (−∞,∞), a variable straight line a + sx where a and s varychanges its amount of intersections with f(x) when it crosses a tangent of f(x). By calculatingthese tangents and their corresponding a and s, we can draw a curve in the s − a planerepresenting all tangent lines of f(x). These curves in the s− a plane correspond to a changein the amount of intersections the line a+sx has with f(x), and thus is the boundary betweentwo different stacking sequences.

By calculating the curves µ1α : [µ2,α, µ2α′ ] → I with µ1α(µ2), corresponding to the nlines in the µ1 − µ2 phase diagram like those in Fig. 11, we can parametrize the curvebα(µ2) = (s(µ2), a(µ2)) in the s− a plane, which is equal to

bs,α(µ2) =

(dµ1α

dµ2, µ1α(µ2)− dµ1α

dµ2µ2

)(71)

where µ2 ranges from [µ2,α, µ2α′ ]. This results in n boundaries in the s − a plane, with atone side a phase stacking . . . A . . . and on the other side . . . ABA . . . , where A and B are thetwo phases of the µ1α(µ2) boundary in the µ1−µ2 phase diagram, and . . . represent differentphases due to other boundaries.

19

4.1.2 Terminal lines

When the curve µ1α(µ2) does not have an infinite domain, there is a terminal point (µ1α(µ2α), µ2α)where the curve reaches a triple point (there curve splits into different lines µ1β(µ2) andµ1γ(µ2)) or reaches a critical point, where the curve stops. Straight lines in the µ1−µ2 planechange the amount of crossings with line α when crossing this terminal point. For a terminalpoint (µ1α, µ2α), all lines going through this point have the relation

bt,α(s) = (s , µ2α − µ1αs) (72)

where s ranges from (−∞,∞). These are infinite straight lines in the s−a plane. For a triplepoint, this is the boundary between phase stacking . . . AB . . . and phase stacking . . . ACB . . . .For a critical point, this is the boundary between . . . A . . . and . . . A1A2 . . . , where A1 and A2

are phases which are able to continuous transform into each other via the supercritical phaseA, without having to undergo a phase transition.

4.1.3 Asymptotic terminal lines

For very low concentrations cσ of species σ, the chemical potential µσ goes to −∞, while theother chemical potentials µγ with ρ 6= γ go asymptotically to a certain value.

limρσ→0

µσ(ρ) = −∞ limρσ→0

µγ(ρ) = µγ,as (73)

This is because at very low chemical potentials, the amount of particles of species σ Nσ goeslower, resulting in a very low ρσ. If species σ undergo a phase transition, the other particlesγ will barely notice the change in structure of species σ due to the low amount of particles ofspecies σ.

In the s − a plane, lines parallel to the asymptote cross the asymptote if they have aslightly different slope s, thus a boundary between two phases in the phase stacking diagramis

ba,α(a) = (s, a) (74)

where s is the slope of the asymptote, and a ranges from (−∞,∞). These lines correspondto vertical lines at s in the s− a plane. Asymptotic lines where µσ(µ2)→ finite if µ2 →∞do not appear in the phase stacking diagram, because they correspond to lines with infiniteslope s =∞.

4.1.4 Results

Combining the results of Eq. (71), (72) and (74) in a single plot in the s − a plane gives usthe phase stacking diagram of this mixture. Diagrams for diameter ratios d = 3.5 and d = 10are shown in Figs. (14) and (15) respectively.

4.2 Paths in other planes

For experimental purposes, the µ1−µ2 plane is not a very interesting plane due to the difficultyof measuring the chemical potentials. However, it is possible to transform the straight linesin the chemical potentials plane to different planes.

4.2.1 Pressure-composition plane

The pressure-composition (p− x) plane is an interesting plane to study because phase coex-istences appear as horizontal lines, while giving information about the jump in compositionalong the coexistence surface. For a given diameter ratio d and mass ratio s, we can calculatethe scaled chemical potential difference a using Eqs. (70), (44) and (45)

a(c, x) = µ2(c, x)− sµ1(c, x) = (1− s)f + (1− x− sx)df

dx+ (1− s)cdf

dc. (75)

With a coordinate transformation c → p(c, x) using Eq. (43), we can transform a(c, x) toa(p, x), and thus calculate for every point in the p− x the corresponding point in the µ1−µ2

plane, and since we have fixed s, the corresponding a associated with that point. Connecting

20

-0.4 -0.2 0.0 0.2 0.4 0.6 0.8 1.0m1m2

4

5

6

7

8

9

a

Phase Stacking Diagram for d=3.5

IN

IN

ININ

N

NIN

Figure 14: Phase stacking diagram for diameter ratio d = 3.5. This diagram consists of onesedimentation binodal and one asymptotic terminal line on the vertical line s = m1/m2 = 0.

-1 0 1 2 3 4m1m2

3.5

4.0

4.5

5.0

5.5

6.0

a

Phase Stacking Diagram for d=10

I2N2

I1I2N2I2I1N2

I N2

I N1

I2I1N1

I2N2I1N2

I2N2I1N1

I2N2I2I1N1

N2I1N1

N2I2I1N1

N2I N1

Figure 15: Phase stacking diagram for diameter ratio d = 10.. This diagram consists of fivesedimentation binodals, one for each line segment in Fig. 11; three terminal lines, correspondingto the two triple points and one critical point; and one asymptotic terminal line related to theasymptote of the I2−N2 line segment in the µ1−µ2 representation. The three thin regions arounds = 1.1 are from top to bottom: I2N2I1N2N1N2, I2I1N2N1N2 and IN2N1N2.

21

Phase stacking lines in the p− x plane and µ1 − µ2 plane for d = 3.5 and s = m1/m2 = 0.2.

0.0 0.2 0.4 0.6 0.8 1.0

4

6

8

10

12

14

x

p

Figure 16: Phase stacking lines in the pressure-composition plane and the chemical potentialsplane for diameter ratio d = 3.5 and mass ratio m1/m2 = 0.2. The thick black lines are the phasediagrams taken from Figs. 10 and 11 respectively. The thin lines correspond to lines of equalscaled chemical potential difference a, where we highlighted several of these lines with a colour.The coloured lines in the left picture are the same as those in the right picture, both showing thatfor example the blue line has a I −N − I −N phase stacking.

the points with equal a gives us the curve the fluid obeys when in a tube with gravity. Thesecurves are the think black lines in Figs. 16

When drawing the contours of a in the p − x plane, we notice a discontinuity from thetop left at x = 0 and p = pcrit, the monodisperse solution of thin rods, to the bottom rightat x = 1 and p = pcrit/d, the monodisperse solution of the thick rods. This discontinuitycorresponds to the points where the nematic phase is more favourable than the isotropic phase(but still less favourable than a coexistence of both phases), while the (Gibbs) free energymay be continuous at this line, other physical quantities do not have to be. This is illustratedin Fig. 5, where the chemical potential µ is discontinuous at the concentration c where theisotropic and nematic phases have equal free energy. Since a is dependent on the chemicalpotentials, we don’t expect a to be continuous at this line in the p− x plane.

When drawing the phase diagram on top of the contours of a(p, x), the endpoints of thehorizontal coexistence lines (p, x1) and (p, x2) have equal scaled chemical potential differencea(p, x1) = a(p, x2). This is to be expected, since the horizontal coexistence lines in the p− xplane correspond to a single point in the µ1 − µ2 phase diagram. For a isotropic-isotropicor nematic-nematic demixing, the contours of a associated with this demixing cross certainvalues of p multiple times. If there were no isotropic-isotropic or nematic-nematic demixing,the system would follow the a contour, and would have an interval with decreasing pressurewhen going down the tube, something non-physical.

4.2.2 Other chemical potential planes

Planes consisting of one of the chemical potentials are interesting because the difference inchemical potentials is equal (up to a constant) to the difference in height. Starting from Eq.(68), we can rewrite these equations as

∆µσ = −mσg∆z (76)

where ∆µσ is the difference between two points on the same line µ1(µ2) = a + sµ2 and∆z the corresponding height difference. By introducing the gravitational length of species

22

Phase stacking lines in the p− x plane and µ1 − µ2 plane for d = 10.

0.0 0.2 0.4 0.6 0.8 1.0

5

10

15

x

p

0.0 0.2 0.4 0.6 0.8 1.0

1.5

2.0

2.5

3.0

x

p

0.0 0.2 0.4 0.6 0.8 1.0

1.5

2.0

2.5

x

p

Figure 17: Phase stacking lines in the p−x plane and the µ1−µ2 plane for diameter ratio d = 10.All but the bottom right picture have mass ratio s = 1, while the bottom right picture has massratio 3.5. The thick black lines are the phase diagrams taken from Figs. 10 and 11. The topright picture is the same as the top left pictures, but zoomed in at the low pressure regime. Thethin lines correspond to lines of equal scaled chemical potential difference a, where we highlightedseveral of these lines with a colour. The coloured lines in the s = 1 pictures are the same phasestacking, showing for example that the red line has phase stacking I2 −N1 − I1 −N2.

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Composition and order parameter as function of the height at d = 10 and s = 1.

0.2 0.4 0.6 0.8x

- 7

- 6

- 5

- 4

- 3

z g

I 2

I1

0.2 0.4 0.6 0.8S

-7

-6

-5

-4

-3

zg

I2

I1

0.2 0.4 0.6 0.8x

- 7

- 6

- 5

- 4

- 3

z g

I 2

N2

I1

0.2 0.4 0.6 0.8S

- 7

- 6

- 5

- 4

- 3

z g

I 2

N2

I1

Figure 18: Composition x (blue) and order parameters S1 (red) and S2 (purple) as function of theheight per gravitational length z/`g, with the different phases coloured. The two ’tubes’ on theleft are at a = µ1 − sµ2 = 4., and the two tubes on the right for a = 4.48.

Composition and order parameter as function of the height at d = 10 and s = 3.5.

0.2 0.4 0.6 0.8x

- 6

- 5

- 4

- 3

- 2

z g

I1

I 2

N2

0.2 0.4 0.6 0.8S

- 6

- 5

- 4

- 3

- 2

z g

I1

I 2

N2

0.2 0.4 0.6 0.8x

- 6

- 5

- 4

- 3

- 2

z g

I1

N2

0.2 0.4 0.6 0.8S

- 6

- 5

- 4

- 3

- 2

z g

I1

N2

Figure 19: Composition x (blue) and order parameters S1 (red) and S2 (purple) as function of theheight per gravitational length z/`g, with the different phases coloured. The two ’tubes’ on theleft are at a = µ1 − sµ2 = 4.08, and the two tubes on the right for a = 4.42.

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Line segment in the µ1 − µ2 plane

Figure 20: Line segment in the µ1 − µ2 plane. With a known diameter ratio d, mass ratio s,scaled chemical potential difference a and chemical potential of species 1 at height z µ1(z), onecan calculate the location of the line segment using Eqs. (69) and (78).

1 `g = kBT/(m1g) we can relate the height difference to the non-dimensionless chemicalpotential difference

∆z = `g∆µ1. (77)

Thus, calculating the lines of equal scaled chemical potential a in a plane with both µ1 andanother physical quantity gives us this physical quantity as function of the height in a tubeconsisting of a mixture of these two species. In Figs. 18 and 19, the composition x and theorder parameters S1 and S2 are shown as function of the height of such a finite tube.

4.3 Finite size effects

All these calculations assume the test tube has an infinite length in both direction, however,in Eq. (77) the chemical potential is proportional to the height in the test tube. Thus bymeasuring the chemical potential µ1(z) at a certain height z ∈ [0, h], where h is the height ofthe mixture in the test tube, the chemical potentials at the bottom and the top of the tubeare

µ1(0) = µ1(z)− z/`g, µ1(h) = µ1(z) + (h− z)/`g. (78)

This construction is visualized in Fig. 20. However, the chemical potential is difficult tomeasure. But by measuring the dimensionless number density c and composition x at acertain height z, one can calculate the chemical potential in Onsager theory using Eqs. (41),(42) and (44).

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5 Thanks

This thesis could not be made without the help of certain people. First, I want to thank Prof.Dr. Rene van Roij for supervising this project and giving me countless tips and tricks on howto tackle this subject. Next I want to thank Tara Drwenski for helping and listening to mewhen I had any problems. I also want to thank Darius Keijdener for supporting me to getthrough the difficult parts of the thesis, as well as my family for supporting me all the time.There are many other people who supported me directly or indirectly during my thesis, andI would like to thank them as well.

References

[1] H. Zocher, Zeitschrift fur anorganische und allgemeine Chemie 147, 91 (1925)

[2] L. Onsager, Annals of the New York Academy of Sciences 51, 627 (1949)

[3] J.D. Parsons, Physical Review A 19, 1225 (1979)

[4] D. Frenkel, The Journal of Physical Chemistry 91, 4912 (1987)

[5] D. Chandler, Introduction to Modern Statistical Mechanics (1987)

[6] R.H.H.G. van Roij, European Journal of Physics 26, S57 (2005)

[7] H.N.W. Lekkerkerker, Ph. Coulon, R. van der Haegen and R. Deblieck, The Journal ofChemical Physics 80, 3427 (1984)

[8] G.J. Vroege and H.N.W. Lekkerkerker, The Journal of Chemical Physics 97, 3601 (1993)

[9] R.P. Sear and G. Jackson, The Journal of Chemical Physics 103, 8684 (1995)

[10] E. Eggen, M. Dijkstra and R.H.H.G. van Roij, Physical Review E 79, 041401 (2009)

[11] S. Belli, S. Dussi, M. Dijkstra and R.H.H.G. van Roij, Physical Review E 90, 020503(2014)

[12] R.H.H.G. van Roij, B. Mulder and M. Dijkstra, Physica A 261, 374 (1998)

[13] R.H.H.G. van Roij and B. Mulder, Physical Review E 54, 6430 (1996)

[14] R.H.H.G. van Roij and B. Mulder, The Journal of Chemical Physics 105, 11237 (1996)

[15] R.H.H.G. van Roij and B. Mulder, Europhysics Letters 34, 201 (1996)

[16] D. de las Heras, N. Doshi, T. Cosgrove, J. Phipps, D.I. Gittins, J.S. van Dijneveldt andM. Schmidt, Scientific Reports 2, 789 (2012)

[17] K.R. Purdy, S. Varga, A. Galindo, G. Jackson and S. Fraden, Physical Review Letters94, 057801 (2005)

[18] D. de las Heras and M. Schmidt, Soft Matter 9, 8636 (2013)

[19] D. de las Heras and M. Schmidt, Journal of Physics: Condensed Matter 27, 194115(2015)

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