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Comprehensive Examwith Extended Derivations

T. N. Shendruk

April 17, 2011

Contents1 Flory-Huggins Theory 4

1.1 Entropy of Mixing . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Flory Parameter . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.3 Free Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.4 Chemical Potential . . . . . . . . . . . . . . . . . . . . . . . . . . 91.5 Osmotic Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.6 Partial Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.7 Phases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2 The Colloidal State 152.1 Colloidal Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2.1.1 Thermal Forces and Constant External Forces (gravity) . 152.1.2 London-van der Waals Forces . . . . . . . . . . . . . . . . 152.1.3 Electrostatic Forces . . . . . . . . . . . . . . . . . . . . . 162.1.4 DLVO Theory . . . . . . . . . . . . . . . . . . . . . . . . 17

2.2 Colloidal Aggregates . . . . . . . . . . . . . . . . . . . . . . . . . 172.3 Thermodynamics of Colloidal Systems . . . . . . . . . . . . . . . 192.4 Surfactants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.4.1 Weak Adsorption . . . . . . . . . . . . . . . . . . . . . . . 212.4.2 Flocculation and Stabilization . . . . . . . . . . . . . . . . 232.4.3 Graft Polymers . . . . . . . . . . . . . . . . . . . . . . . . 242.4.4 Non-Adsorped Polymers . . . . . . . . . . . . . . . . . . . 24

3 Field-Flow Fractionation 253.1 Concentration Profile . . . . . . . . . . . . . . . . . . . . . . . . . 253.2 Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283.3 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283.4 Zone Spreading . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

4 Field-Flow Fractionation Specifics 334.1 Flow Field-Flow Fractionation . . . . . . . . . . . . . . . . . . . . 334.2 Thermal Field-Flow Fractionation . . . . . . . . . . . . . . . . . 36

1

5 Introduction to Tethered Polymer Microstructures 385.1 Flat Solid Interfaces . . . . . . . . . . . . . . . . . . . . . . . . . 385.2 Alexander Brush Model . . . . . . . . . . . . . . . . . . . . . . . 39

5.2.1 Free Energy Cost of Compression Mode . . . . . . . . . . 415.3 Curved Interfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 445.4 Polydisperse Grafting . . . . . . . . . . . . . . . . . . . . . . . . 47

6 Aggregation of Block Copolymers 486.1 Microphases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 486.2 Micelles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

6.2.1 Hairy Micelle . . . . . . . . . . . . . . . . . . . . . . . . . 496.2.2 Crewcut Micelle . . . . . . . . . . . . . . . . . . . . . . . 51

6.3 Adsorbed Copolymer Layers . . . . . . . . . . . . . . . . . . . . . 51

7 Deformation of Single Chains 537.1 Freely Jointed Chain . . . . . . . . . . . . . . . . . . . . . . . . . 537.2 Bead Spring Chain . . . . . . . . . . . . . . . . . . . . . . . . . . 537.3 Correlations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 557.4 Deformation Due to Tension . . . . . . . . . . . . . . . . . . . . . 56

7.4.1 Stress Ensemble: Blob Theory . . . . . . . . . . . . . . . 567.4.2 Stress Ensemble: FJC . . . . . . . . . . . . . . . . . . . . 577.4.3 Stress Ensemble: FJC vs Blobs . . . . . . . . . . . . . . . 59

7.5 Strain Ensemble . . . . . . . . . . . . . . . . . . . . . . . . . . . 597.6 Deformation Due to Flow . . . . . . . . . . . . . . . . . . . . . . 60

7.6.1 Deformation in a Good Solvent . . . . . . . . . . . . . . . 607.6.2 Deformation in an Ideal Solvent . . . . . . . . . . . . . . 65

7.7 Tethered Polymers in Shear Flow . . . . . . . . . . . . . . . . . . 667.7.1 Good Solvent Horn . . . . . . . . . . . . . . . . . . . . . . 687.7.2 Ideal Solvent - Tethered Chain . . . . . . . . . . . . . . . 697.7.3 Fluctuations of Tethered Polymers in Shear Flows . . . . 70

7.8 Polymer Stretching over a Potential Barrier . . . . . . . . . . . . 707.9 Polymer-Obstacle Collisions . . . . . . . . . . . . . . . . . . . . . 72

7.9.1 Force Induced Collision . . . . . . . . . . . . . . . . . . . 737.9.2 Flow Induced Collision . . . . . . . . . . . . . . . . . . . . 76

7.10 Polymer-Polymer Collision . . . . . . . . . . . . . . . . . . . . . . 77

8 Microfluidics: Dimensionless Numbers 788.1 Reynolds Number: Inertial vs Viscous . . . . . . . . . . . . . . . 788.2 Péclet number: Convection vs Diffusion . . . . . . . . . . . . . . 798.3 Capillary Number: Viscous vs Interfacial . . . . . . . . . . . . . . 808.4 Deborah number: Relaxation vs Flow . . . . . . . . . . . . . . . 818.5 Weissenberg number: Relaxation vs Shear Rate . . . . . . . . . . 828.6 Elasticity number: Elastic vs Inertial . . . . . . . . . . . . . . . . 828.7 Rayleigh and Grashof numbers . . . . . . . . . . . . . . . . . . . 828.8 Schmidt number . . . . . . . . . . . . . . . . . . . . . . . . . . . 838.9 Knudsen number . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

9 Microfluidics: Driving Flows at Boundaries 85

2

10 Attractive Interactions Between Colloids 8610.1 Intermolecular Attractions . . . . . . . . . . . . . . . . . . . . . . 86

10.1.1 Keesom Interaction . . . . . . . . . . . . . . . . . . . . . . 8610.1.2 Debye Interaction . . . . . . . . . . . . . . . . . . . . . . 8610.1.3 London Interaction . . . . . . . . . . . . . . . . . . . . . . 86

10.2 London Constant . . . . . . . . . . . . . . . . . . . . . . . . . . . 8710.2.1 Polarizability . . . . . . . . . . . . . . . . . . . . . . . . . 88

10.3 Dispersion Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . 8810.3.1 Plates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8910.3.2 Spheres . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9010.3.3 Cylinders . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

10.4 Hamaker Constant . . . . . . . . . . . . . . . . . . . . . . . . . . 9010.5 Depletion Interaction . . . . . . . . . . . . . . . . . . . . . . . . . 91

11 Replusive Interactions Between Colloids 9311.1 Born Repulsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9311.2 Retardation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9311.3 Electric Double Layer . . . . . . . . . . . . . . . . . . . . . . . . 93

11.3.1 Origins of Surface Charge . . . . . . . . . . . . . . . . . . 9311.3.2 Electric Double Layer . . . . . . . . . . . . . . . . . . . . 9411.3.3 Debye-Hückel Approximation . . . . . . . . . . . . . . . . 9511.3.4 Debye-Hückel Planar . . . . . . . . . . . . . . . . . . . . . 9611.3.5 Debye-Hückel Spherical . . . . . . . . . . . . . . . . . . . 97

11.4 Diffuse Double Layer Interactions . . . . . . . . . . . . . . . . . . 9711.4.1 Diffuse Double Layer: Plates . . . . . . . . . . . . . . . . 9811.4.2 Diffuse Double Layer: Spheres . . . . . . . . . . . . . . . 98

11.5 Steric Replusion . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

12 Stability of Colloidal Dispersions 102

3

(a) Lattice sites.

• •• •

? ? ?

? ? ?

? ? ?

(b) Pure solvent, packed obsticles and poly-mer melt.

•? ? • ? ? ?

? ? ? ?

•

(c) Solution of solvent, obsti-cles and polymers

Figure 1: Schematic of Flory-Huggins theory for ternary mixture.

1 Flory-Huggins TheoryConsider a lattice where each site can be occupied by a solvent molecule, a partof an obsticle or a monomer of a polymer of length Np. The we wish to considerthe difference in entropy between pure materials and the mixed solution. Fig. 1schematically represents the model. In general, we can think of a set of speciesof type i who each occupy Ni lattice sites. The number of i-type constituentsis ni. If each site has some volume ν and there are a total of n sites then thetotal volume is

V = nν. (1.1)

We also want to specify that every lattice site is ocupied so

V = Vs + Vo + Vp

= nsν + noNoν + npNpν.

Generally,

V =∑i

niNiν (1.2a)

n =∑i

niNi (1.2b)

where Ni is the number of units that make up the molecule and ni is the numberof i-type molecules. So then the volume fraction of each species is

φi =ViV

=niNin

. (1.3)

4

1.1 Entropy of MixingBefore mixing the substances, they are in a pure as Fig. 1b tries to represent.They might be a pure solvent, a hard packed colloids or a melt of polymers(Henk’s case). Prior to mixing then, each pure substance has some entropy.In this mean-field approximation (called Flory-Huggins Theory), the number ofstates of each species is simply the number of lattice sites the substance occupies:

Ωpi = niNi = nφi (1.4a)

where the superscript p denotes the pure state. That’s a gross miscountingbut that’s mean-field approximation for you. Ωpi is really the number of statesaccessible when placing a single molecule on nφi sites.

To continue our terrible counting exercise, if we instead look at the entirevolume of the solution then we will place the molecule on one of the n sites.Therefore the number of states is

Ωsi = n (1.4b)

where the superscript s denotes the fact that we are now forming a solution.The difference in entropy between these two cases for species i is

∆Si = −kB ln Ωpi − (−kB ln Ωsi )

= −kB ln

(ΩpiΩsi

)= −kB ln

(nφin

)

∆Si = −kB lnφi (1.5)

which is an increase in entropy since φi < 1.The total entropy of mixing is the weighted sum of the entropy change of

each molecule in the system

∆Smix =∑i

ni∆Si

=∑i

nφiNi

∆Si

= −nkB∑i

φiNi

lnφi

∆smix = −kB∑i

φiNi

lnφi (1.6)

where we’ve defined the specific change in entropy due to mixing ∆smix =∆Smix/n (the entropy of mixing per lattice site) since it’s an intrinsic quantity.

5

?

pp

ss

?pp

?

pp

pp? ss

ss

ss

? (a) Interaction in pure substances. Each neighbour is of thesame species.

×

??

×

??

× ???

??

??× × ??

??

??×

× ×(b) Interaction after mixing.

Figure 2: Schematic of nearest-neighbour contacts.

As an example, if Henk’s solvent molecules each take a single lattice site, hisobsticles occupy No sites, and his macromelcules have a polymerization of Npthen the entropy of mixing per lattice site is

∆smix = −kB[φs lnφs +

φoNo

lnφo +φpNp

lnφp

].

1.2 Flory ParameterHenk’s system is a thermal - there is no there is no interaction energy differencebetween neighbours. If this wasn’t the case then we would have interactionsbetween nearest neighbours, say εij = εss, εps, εpp, εso, εpo, εoo.

Consider a polymer solution. Prior to mixing there are z s-s contacts inthe pure solvent and z p-p contacts in the polymer melt where z is the contactnumber. So then the interaction energy between eight sites as shown in Fig. 2ais

Up = zεpp + zεss.

Then the total interaction energy before mixing must be

Up =z

2

∑i

niεii

=z

2

∑i

nφiεii

up =Up

n=z

2

∑i

φiεii (1.7a)

6

After mixing the two substances into a solution the interaction energy changesas shown in Fig. 2b . If we look at any site (the centre site in Fig. 2b ) thereis a probability φi that it is occupied by an i type molecule. Now considerit’s neighbours: in the mean field approximation there is a φj chance that it isoccupied by a j type molecule. So then the average interaction energy of thecentre particle and its z neighbours

Us = z∑i

φi

∑j

φjεij

.Then the interaction energy of the solution is

us =z

2

∑i,j

φiφjεij (1.7b)

Notice that the mean field approximation has crept in here again. For a polymer,two of any monomer’s neighbours must be it’s neighbouring monomers in thechain and it’s highly probable that some other monomer is the chain occupies oneof it’s other neighbours. This is all ignored in Flory-Huggins theory (recall themonomers are all unconnected). Under this assumption, we have our estimatethe probability of a lattice being occupied by some species i is φi and so we canestimate the probability that it’s neighbour is also of type i to be φi×φi or it’sprobability that it’s neighbour is of some other type j 6= i which is 2φiφj .

The difference between the energies ∆u = us − up is then written

∆umix =z

2

∑i

φi

∑j

φjεij

− εii . (1.7c)

In the case of a binary solution of polymer and solvent of volume fractionφs = 1− φp the energies simply become

up =z

2

∑i

nφiεii

= z [φpεpp + (1− φp)εss]

us =z

2

∑i,j

φiφjεij

=z

2

[φ2pεpp + 2φp(1− φp)εps + (1− φp)2εss

]∆umix =

z

2

∑i

φi

∑j

φjεij − εii

=z

2

φp∑

j

φjεpj − εpp

+ (1− φp)

∑j

φjεsj − εss

=z

2(φp [(φpεpp + (1− φp)εps)− εpp] + (1− φp) [(φpεsp + (1− φp)εss)− εss])

=z

2(φp [− (1− φp) εpp + (1− φp)εps] + (1− φp) [φpεsp − φpεss])

=z

2φp (1− φp) (−εpp + εps + [εsp − εss])

= χφp (1− φp)

7

where

χ ≡ z

2

(2εps − εpp − εss

kBT

). (1.8)

The difference between the energies (scaled by the thermal energy) is called theFlory parameter, χ.

The Flory parameter only makes sense to me for a binary solution. Forinstance, consider Henk’s ternary solution for which

∆umix =z

2

∑i

φi

∑j

φjεij − εii

=z

2

(∑i

φi [φpεip + φsεis + φpεio − εii]

)=z

2(φp [φpεpp + φsεps + φpεpo − εpp]

+φs [φpεsp + φsεss + φpεso − εss]+φo [φpεop + φsεos + φpεoo − εoo])

= z

([φpφsεps + φpφoεpo + φoφsεos]

−φp (1− φp) εpp + φs (1− φs) εss + φo (1− φo) εoo2

)If we insisted on forming a Flory parameter it would have to be a function ofφo or something which undermines the point.

1.3 Free EnergySince there is no work being done on the system, the thermodynamic potentialsHelmholtz free energy F and Gibbs free energy G are equivalent. Eventuallywhen we look at changes in concentration, we will have a chemical potentialdifference and we’ll be using the grand potential. In any case, the Helmholtzfree energy of mixing is

dF = dE − d(TS)

∆Fmix = ∆Umix − T∆Smix

which is generally

∆fmix

kBT=

z

2kBT

∑i

φi

∑j

φjεij

− εii+

∑i

φiNi

lnφi (1.9)

and for an athermal solvent is

∆fmix

kBT=∑i

φiNi

lnφi (1.10)

where ∆fmix = ∆Fmix/n is the specific free energy.

8

1.4 Chemical PotentialThe free energy is important because it is directly related to the chemical po-tential. The chemical potential of the ith species is either

µi =∂F

∂ni

∣∣∣T,V,nj 6=i

(1.11a)

µi =∂G

∂ni

∣∣∣T,P,nj 6=i

. (1.11b)

The difference is just whether we hold volume or pressure fixed. The importantthing is that the chemical potential is the change in the thermodynamic potentialwith number of molecules. It’s the work needed to add additional particles tothe system or it’s the free energy capacity. We’re interested in the chemicalpotential difference between the pure substances and the solution.

We’ll need some standard δij stuff like

∂nj∂ni

= δij∑i

xiδij = xj .

Furthermore, we’ll use that

∂φk∂ni

∣∣∣nj 6=i

=∂

∂ni

(nφkn

)nj 6=i

=∂

∂ni

(Nknk∑j Njnj

)nj 6=i

=

(1∑

j Njnj

)∂

∂ni(Nknk)− Nknk(∑

j Njnj

)2

∂

∂ni

∑k

Nknk

=

(1

n

)Niδik −

Nknkn2

Ni

=φiniδik − φk

φini

=φini

(δik − φk)

and (using the above result) also that

∂

∂ni

∑k

xkφk

∣∣∣∣∣nj 6=i

=∑k

xk∂φk∂ni

∣∣∣nj 6=i

=∑k

xk

(φini

(δik − φk)

)=φini

∑k

xk (δik − φk)

=φini

(xi −

∑k

xkφk

)

9

if xk is some constant specific to each k. Plus I repeat for convinience nφi =niNi.

We are now armed to find the chemical potential difference between thespecies in the solution and the pure substance from Eq. (1.11a)

∆µi

kBT=∂∆Fmix

∂ni

∣∣∣T,V,nj 6=i

=∂

∂ni

∑j

nφj

Njlnφj +

z

2kBT

∑j

nφj

[(∑k

φkεjk

)− εjj

]=

∂

∂ni

∑j

nj lnφj +z

2kBT

∑j

njNj

[(∑k

φkεjk

)− εjj

]=∑j

lnφj∂nj

∂ni+∑j

nj∂ lnφj

∂ni+

z

2kBT

∑j

[(∑k

φkεjk

)− εjj

]∂njNj

∂ni

+∑j

Njnj∂

∂ni

[(∑k

φkεjk

)− εjj

]

=∑j

δij lnφj +∑j

nj

φj

∂φj

∂ni+

z

2kBT

∑j

δij

[(∑k

φkεjk

)− εjj

]

+∑j

Njnj

[∂

∂ni

(∑k

φkεjk

)− 0

]= lnφi +

∑j

nj

φj

φi

ni(δij − φj) +

z

2kBT

(∑k

φkεik

)− εii

+∑j

Njnj∂

∂ni

(∑k

φkεjk

)= lnφi + 1

−∑j

njφi

ni+

z

2kBT

∑k

φkεik − εii +∑j

Njnjφi

ni

(εji −

∑k

εjkφk

)= 1 + lnφi

−Ni∑j

nj

n+

z

2kBT

∑j

φjεij − εii +Ni∑j

φj

(εij −

∑k

εjkφk

)

∆µi

kBT= 1 + lnφi −Ni

∑j

φj

Nj+

z

2kBT

−εii +∑j

φjεij +Ni∑j

φj

(εij −

∑k

εjkφk

) .

(1.12)

That’s quite the formula but it’s pretty general and useful. As a coupleof examples let’s consider the difference in chemical potential for the solute

10

molecules in a binary solution solution (i.e. Ns = 1 and φs = (1− φp):∆µs

kBT= 1 + lnφs −Ns

∑j

φj

Nj

+z

2kBT

∑j

φjεsj − εss +Ns

∑j

φj

(εsj −

∑k

εjkφk

)= 1 + lnφs −Ns

(φs

Ns+φp

Np

)

+z

2kBT

(φsεss + φpεsp)− εss +Ns

∑j

φj (εsj − [εjsφs + εjpφp])

= 1 + ln(1− φp)−

(φs +

φp

Np

)+

z

2kBT(1− φp) εss + φpεsp − εss

+φs (εss − [εssφs + εspφp]) + φp (εsp − [εpsφs + εppφp])

= 1 + ln(1− φp)−(

(1− φp) +φp

Np

)+

z

2kBT−φpεss + φpεsp

+(φsεss − εssφ2s − εspφsφp

)+(φpεsp − εpsφpφs − εppφ2p

)= ln(1− φp) + φp

(1−

1

Np

)+

z

2kBT−φpεss + φsεss

+φpεsp − εspφsφp + φpεsp − εspφpφs − εppφ2p − εssφ2s

= ln(1− φp) + φp

(1−

1

Np

)+

z

2kBT−φpεss + (1− φp)εss

+φpεsp − εsp(1− φp)φp + φpεsp − εspφp(1− φp)− εppφ2p − εssφ2s

= ln(1− φp) + φp

(1−

1

Np

)+

z

2kBT

−φpεss + (1− φp)εss + 2φ2pεsp − εppφ2p − εss(1− 2φp + φ2p)

= ln(1− φp) + φp

(1−

1

Np

)+

z

2kBT

2φ2pεsp − εppφ2p − εssφ2p

= ln(1− φp) + φp

(1−

1

Np

)+ φ2pχ

That was alot of work but we got a standard formula in polymer books from avery general formula.

An easier thing to do is look at the changes in chemical potential in anathermal solvent. In Henk’s system the chemical potential differences for each

11

species are

∆µpkBT

= 1 + lnφp −Np(φpNp

+φoNo

+ φs

)∆µokBT

= 1 + lnφo −No(φpNp

+φoNo

+ φs

)∆µskBT

= 1 + lnφs −(φpNp

+φoNo

+ φs

)1.5 Osmotic PressureImagine a semipermeable membrane that prevents the passage of some moleculesbut not others. This causes a pressure difference called osmotic pressure. Os-motic pressure is the rate of change of total free energy of the system withrespect to volume

Π = −∂∆Fmix

∂V

∣∣∣ni

(1.13)

where i are all the fixed species. The derivative with respect to V can be turnedinto a derivative with respect to lattice number since V = nν. We found thesederiviatives in the last section. When a membrane separates a pure solvent anda binary solvent-polymer solution such that only the solute molecules can crossthe osmotic pressure of the macromolecules on the solvent in the solution is

Πs = −kBT∆µsν

. (1.14)

After getting Eq. (1.14) , de Gennes talks about “an absurd concept”: an osmoticpressure of the solvent molecules on the macromolecules. To get this one needsa semipermeable membrane that allows chains to pass but not solvent. So wehave a semipermeable membrane between solution and polymer melt that onlyallows polymers to cross. The osmotic pressure of the solvent on the monomersis

ν

kBTΠp = −∆µp. (1.15)

But both of these are for a binary system. Again having a ternary solutionadds difficulty. If only a single species is allowed to cross the barrier then there isstill an osmotic pressure. Say the solvent molecules can cross but the polymersand the obsticles are trapped on their side then the osmotic pressure of thepolymers and the obsticles is clearly

ν

kBTΠs = −∆µs.

Likewise if one side (say side 1) is made up of some concentration of polymersand obsticles but the other side (side 2) is a solution of a different concentrationof polymers and obsticles with a semipermeable membrane between them thenthey have unequal differences in chemical potential compared to pure solventand the osmotic pressure is (towards side 1) is

νΠs

kBT= ∆µs,1 −∆µs,2 (1.16)

12

Notice if the concentration of polymers and obsticles on side 1 is zero, then∆µs,1 which is the difference between the chemical potential of the solutionand the pure solvent is zero and we recover the previous result. In general,when two solutions of different concentrations are separated by a semipermeablemembrane which selectively filters some species but not others, then osmosis willoccur meaning there will be flow of material from the less concentrated to themore concentrated side. Henk’s polymers must over come this - not are drivenby it.

That’s all neat enough but that’s not Henk’s situation. In Henk’s systemonly the obsticles must stay on their side and keep their concentration. Both thesolvent and the polymer are allowed to cross the fictitious membrane. In thatcase the concept of osmotic pressure isn’t directly meaningful. Rather there isa difference in partial pressures.

1.6 Partial PressureWhen several particles are involved, equalibrium occurs when∑

i

µi = 0

at constant pressure and temperature. This comes out of

dG =∂G

∂TdT +

∂G

∂pdp+

∑i

∂G

∂nidni

= 0 + 0 +∑i

µidni.

Perhaps this sort of explains the origin of the osmotic pressure better: on side1 of the membrane the chemical potentials summed to some value, on side twothey summed to a different value. Therefore dG between them couldn’t be zero.A dp arose to balance the situation making the sides have different pressuresand thus equalibriating the system.

The partial pressure π of species i in the mixture is the pressure that itwould have if you took all the other substances away and let it alone occupy theentire volume. The partial pressures sum to give the mixture’s total pressure

p =∑i

πi. (1.17)

If the pressure difference has arisen because of a difference in chemical po-tentials between sides (although I’m not sure if this general definition is trulyosmotic pressure, I’m still going to refer to this as osmotic pressure Π) then it’svalue is the difference between sums of the partial pressures. The total osmoticpressure towards side 1 is

νΠi

kBT=∑i

µ1,i − µ2,i

νΠi

kBT=∑i

∆µ1,i −∆µ2,i (1.18)

13

where i is the set of species who can transverse the membrane (who are havingthe osmotic pressure on them) and

∑i is specifically over those species.

And we are finally ready to answer Henk’s question. He has an athermalsystem but the volume fraction of obsticles is different causing the volume frac-tion of solvent to be different. So then the osmotic pressure on the polymer andthe solvent is

νΠp,s

kBT= ∆µ1,s −∆µ2,s + ∆µ1,p −∆µ2,p

=

(1 + lnφ1,s −

[φ1,p

Np+φ1,o

No+ φ1,s

])−(

1 + lnφ2,s −[φ2,p

Np+φ2,o

No+ φ2,s

])+

(1 + lnφ1,p −Np

[φ1,p

Np+φ1,o

No+ φ1,s

])+

(1 + lnφ2,p −Np

[φ2,p

Np+φ2,o

No+ φ2,s

])= ln

(φ1,sφ1,p

φ2,sφ2,p

)− (1 +Np)

[φ1,p − φ2,p

Np+φ1,o − φ2,o

No+ φ1,s − φ2,s

]The conformational entropy difference due to crowding must over come thisosmotic pressure. Also note that with a long enough polymer the free energywould balance once a certain number of monomers is established on each sideand an equalibrium would be setup.

1.7 PhasesIn binary solutions the solution is stable at all concentrations when χ is smallenough. That’s because ∆µrep = ∆µp − Np∆µs, the cost in free energy of re-moving Np solvent molecules and replacing them with a chain, is monotonicallyincreasing with φp. But at χ > 0.605 there is a region where ∆µrep unphysicallydecreases meaning that the solution is unstable and will phase separate.

Henk doesn’t have a χ but he does have a term that depends on obsticleconcentration in both ∆µp and ∆µs. Could that cause an instability? No. Theycancel out. This is what we expect and if we need we can see it explicitly from

∆µrep = ∆µp −Np∆µs

= 1 + lnφp −Np(φpNp

+φoNo

+ φs

)−Np

(1 + lnφs −

(φpNp

+φoNo

+ φs

))= 1−Np + lnφp −Np lnφs + 0

= 1−Np + ln

(φp

φNps

)

which never goes down with φp.

14

2 The Colloidal StateThe term colloidal (from the greek work for glue kolla) is a touch ambiguous.It was initially coined in reference to diffusive properties. We now refer to itmore as reflecting the degree of physical division. The size these objects isloosely between 1 nm and 1000 nm but it’s really the intermediate mesoscalebetween atomic length scales and microscopic scales. There must be a largesurface-to-volume ratio such that the fraction of atoms residing at the interfaceincreases often causing physicochemical properties to change dramatically. Aswe shall see, the surfaces play a very significant role in determining interparticleforces and it is the interplay being these forces and the effects of the continuousmedium in which the objects are dispersed that causes the system to be colloidal.

2.1 Colloidal Forces2.1.1 Thermal Forces and Constant External Forces (gravity)

Thermal agitation opposes sedimentation (the authors J. C. Daniel and R. Au-debert define sedimentation as occuring for particles of higher density than thesuspension fluid and creaming as occuring for particles of lower density that thefluid matrix whereas I’m going to generally refer to both situations as sedimen-tation).

Consider a test tube of height H containing a solution of colloids. It isintuitive that the concentration profile is a decay of the cencentration at thebase of the tube C0. The concentration profile is set up to create an equalibriumbetween the thermal energy kBT and the gravitiational potential mgh

C(h) = C0e−mgh/kBT (2.1)

where m is the boyant mass m = 43πr

3∆ρ defined by the difference in densitybetween the dispersed colloids and the suspension fluid, ∆ρ. Although this isjust stated it is derived in § 3 .

The rate at which equallibrium is arrived at is dictated by the sedimentationspeed of the dispersed phase (just the terminal velocity of the colloids as theymove through the fluid of viscosity ν. It is assumed that although this mayoccur quickly their small size makes them follow Stokes’ law)

vterm =mg

6πηr

=2

9

r2

η∆ρg

which is a balance of their surface area to the viscosity of the medium. Polymersare often used to thicken the solution increasing the viscosity.

2.1.2 London-van der Waals Forces

A perminant dipole can induce a dipole in a neighbour the resulting force iscalled the van der Waals force. But the locations of electrons within atomsfluctuate giving rise to an instantaneous dipole. The instantaneous dipole canthen induce a dipole in a neighbour. This special case of the van der Waalsforce is called called the London dispersion force. Together they are grouped

15

as London-van der Waals forces. Since dispersion force follows the Lennard-Jones potential, the energy coupling between two colloids (modelled as identicalspheres of radius r) is proportional to a Hamaker constant.

EvdW = − A12

r

s(2.2)

where s is the interparticle distance. The authors don’t say why it’s such asimple linear relation and the later plot EvdW according to the Lennard-Jonesform

V (s) = 4ε

[(σs

)12

−(σs

)6]. (2.3)

Anyway, assuming Eq. (2.2) is true, it applies if the spheres are in vacuum andalso if they are immersed in some medium. The Hamaker constant A changesin a complicated way depending on both the medium and the material of thedispersed spheres.

The van der Waals forces are attractive and typical larger than but the sameorder of magnitude as kBT . Whereas thermal agitation stabilized the colloidalsolution against sedimentation, the thermal agitiation causes collisions betweenspheres and so the dipole-dipole coupling energy EvdW would often be instan-taneously high causing the rapid formation of aggregates. If destabilization dueto London dispersion forces is not to occur there must be some repulsive forcebetween colloids.

2.1.3 Electrostatic Forces

Electrostatic forces are really the key to why colloids are best defined as largesurface-to-volume objects. Surface charges provide the primary repulsive in-terparticle interaction between particles. Surface charges originate from threemechanisms:

Structural - Impurity ions within the crystal lattice must be balanced out bycations. Sometimes the available cations can’t fit in the crystal structureand must reside at the surface.

Ionizable Groups - A particle could be a pure crystal except at the surfacewhere it reacts with the environment (ex. oxidizes). Those surface specificgroups may be ionizable in certain circumstances.

Selective Adsorption - Ions from the medium may be selectively adsorptedto the surface. Both charges can build up this way but anions are morefrequent.

The presence of surface charges does not indicate global charge, just an inhomo-geneous distribution of charge. Co-ions (like charge) are repelled and counter-ions (opposite charge) concentrate near the surface. It’s not really most usefulto discuss the concentration of ions. Rather, we’d like to discuss the concentra-tion of charge. The ionic strength scales the concentration of species i by thecharge Zi squared for all ions

I =1

2

∑i

CiZ2i . (2.4)

16

The ionic strength at any point will be determined by the overall concentrationand the electric potential, Ψ.

In the same manner as the authors handle sedimentation, they handle aplanar surface in univalent salt. If Ψ0 is the potential at the surface then

Ψ = Ψ0e−s/κ.

Presumably, this comes out of the Poisson-Boltzmann equation but they don’tmention it. The important thing here is that the potential is controlled by aparameter κ. The parameter must have units of length and is known as theDebye length. The Debye length is a measure of the range of the electrostaticforces and scales with the square root of ionic strength

κ ∝ I1/2. (2.5)

The Debye length is applicable to many geometries. For instance, it enters intothe electrostatic energy for two spheres of radius r separated by s:

Ee = 2πεΨ2

0

κe−s/κ (2.6)

2.1.4 DLVO Theory

When the particles are small enough to neglect sedimentation and gravitationalpotential then the stability of the solution is determined by the sum of theLondon-van der Waals and the electrostatic energy. Balancing the interparticleattraction and repulsion is called Deryaguin-Landau-Vervey-Overbeck(DLVO)Theory in which E = EvdW + Ee.

The electrostatic energy is a fairly steady exponential decay from an initialvalue as the particles move further apart. On the other hand, the Lennard-Jonespotential is a wall at very short distances, a strong attractive well at moderatelyclose distances and virtually nothing when the particles are separated by greatdistances.

Short range EvdW well means that the thermodynamic equalibrium uccurswhen all the particles are in contact and at the bottom of the energy well. Theposition of the energy minimum Emin is called the aggregation distance, smin.However, the electrostatic repulsion forms a barrier in the net energy. If twoparticles are separated by more than smax (the position of the local maximumEmax) and do not have enough thermal or kinetic energy they will not be ableto get beyond the repulsive barrier and reach the attractive global minimum.They are thus in a metastable state.

Since the deep potential well is presumably fairly insensitive to ionic strengthof the liquid, adding salt or otherwise increasing the ionic strength (and there-fore the Debye length by Eq. (2.5) ) lowers Ee. If the barrier is lowered enougha metastable state will no longer be stable (at the given temperature) and floc-culation (or coagulation or agglomeration) occurs (i.e. the particle aggregate).

2.2 Colloidal AggregatesImagine that we have a set of primary particles which if they ever collide sticktogether and form a secondary doublet particle. The rate at which primary

17

particles disapper (equal and opposite to rate that doublets form; therefore,called aggregation rate) is

−dN1

dt= k11N

21 (2.7)

where N1 is the number density. Notice this is different than say radioactivedecay since the aggregation rate is proportional to the square of the number ofparticles which I interpret to be becauase the decay of the particle is a randomprocess of each particle whereas here the aggregation is a random process forboth the particles.

This can be extended to the rate of formation of an aggregate containing mprimary particles. The rate of aggregation must now have two terms:

Source A cluster of m particles will be formed anytime that a cluster of iparticles and a cluster of j particles collide if i+ j = m.

Sink Whenever a cluster of m particles collides with any other cluster, it formsa larger cluster and so an m aggregate has dissapeared.

If Nm is the number density of m aggregates than the total rate of formation is

dNmdt

=1

2kij

∑i,j:i+j=m

NiNj − kmj∞∑j=1

NmNj . (2.8)

These kinetics are extremelly sensitive to the values kij which is called theinterparticle affinity constant. So far we have implicitly been considering hardcore potentials: there is no interaction energy until they collide. In this case,the kinetics are determined by diffusion and the authors state

k11 =8kBT

3η

which using the Einstein relation

D =kBT

6πηr(2.9)

is equivalent to

k11 =8kBT

3η

=8kBT

3η

D

kBT/6πηr

= D8kBT

3η

6πηr

kBT

= 16πDr.

What about if the there are long range interactions. As stated in § 2.1.4 ,the total energy E has local maximum Emax which creates energy threshholdthat must be overcome for aggregation to occur. The aggregation rate when

18

there is a threshold energy (dNm/dt)Emaxis apparently observed to be related

to the aggregation rate when there is no interaction energy (dNm/dt)0 as(dNmdt

)Emax

= K

(dNmdt

)0

=2r

κe−Emax/kBT

(dNmdt

)0

(2.10)

which (although they don’t say it) simply means that the interparticle affinityconstant should be modified to

k11

↓k′11 = k11K

=8kBT

3η

2r

κe−Emax/kBT

= 16πDr2r

κe−Emax/kBT

= 32πDr2

κe−Emax/kBT .

2.3 Thermodynamics of Colloidal SystemsWhen considering the the thermodynamics of a colloidal system, it is once againthe large surface-to-volume ratio that drives the novel behaviour. In colloidalsystems one can no longer assume that the energy due to surface tension isneglible. Surface tension arise from the idea that inside a droplet the meanglobal attraction of a molecule to it’s neighbours is zero because the medium isisotropic. But molecules residing at the surface feel unbalanced van der Waalsforces. The schematic Fig. 3 tries to represent this idea. The shape of thedroplet tries to minimize the energy by minimizing the surface area (creatinga sphere). It costs energy ∆E to deform the droplet’s surface by an area ∆Awhereas

∆E = γ∆A

where γ is the interfacial tension (or surface tension). The surface tension is aconstant material property of the droplet and the surrounding medium.

If we cut a droplet in half the surface area goes up and so energy must besupplied to do this. In a colloidal system enough energy is stored in the divisionof matter that γ must be included in the energy as a thermodynamic force (withsurface area as the corresponding coordinate). In general, we might write theenergy E, the enthalpy H, the Helmholtz free energy F and the Gibbs freeenergy G as

dE = TdS + J · dx (2.11a)dH = TdS − x · dJ (2.11b)dF = −SdT + J · dx (2.11c)dG = −SdT − x · dJ (2.11d)

19

• •

• • • •

• • • • • •

• • • • • • • •

• • • • • • • • • •

• • • • • • • • • •

• • • • • • • •

• • • • • •

• • • •

• •

Figure 3: Schematic of van der Waals forces within a droplet.

where J are any thermodynamic forces of interest and x are their coordinates.For colloidal systems we would write

dE = TdS − PdV + γdA+∑i

µidNi (2.12a)

dH = TdS + V dP −Adγ −∑i

Nidµi (2.12b)

dF = −SdT − PdV + γdA+∑i

µidNi (2.12c)

dG = −SdT + V dP −Adγ −∑i

Nidµi. (2.12d)

This is why the thermal dynamic properties like solubility ϕ (rate of changeof N with pressure), vapour pressure (dido but just difference in phase), etc.depend on particle size. To demonstrate this, the authors state that

ln

(ϕrϕ0

)=

2γ12VmrRT

(2.13)

where the subscripts 1, 2 label the substance, Vm is the molar volume and R isthe ideal gas constant. The subscripts 0 indicates the value in a macroscopic

20

state while the subscript r denotes it’s value when the material has been sub-divided into many small particles af radius r. Eq. (2.13) is called Kelvin’s lawand is a balance of the amount of energy need to transfer a molecule from thebulk state to a droplet (which of course increases it’s surface area).

2.4 SurfactantsSince the surface play such a major role in determining the thermodynamics ofthe system through γ as seen in § 2.3 and establishing stability as seen in § 2.1.4it should be of no surprise that any modification of the interface may have pro-found effects. The most spectacular results are obtained when macromolecules(maybe asymmetric with a hydrophilic head and a long hydrophobic tail) actto reduce the surface tension. As long as at least one end has an affinity to thesurface, these sufactant molecules will organize at the interface.

2.4.1 Weak Adsorption

However, molecules that fix themselves at interfaces loose degrees of freedomdecreasing the entropy. Let’s consider an energy balance that favours adsorptionto the colloidal surfaces. The conformation of the chain at the surface will reflectthis balance of enthalpic attraction and entropic reduction of degrees of freedomsuch that only some fraction of the monomers will actually be in contact withthe colloid. This number and the thickness of the coating is determined by ascaling argument:

• Consider a polymer solution near a planar, weakly adsorping surface. Letthe energy gain per monomer in contact with the surface be δkBT (weaklymeans 0 < δ < 1). The thickness ξads defines the adsorption blob. Theadsorption blob is the length scale over which the interaction energy withthe surface is on the order of kBT . So below that length scale the polymerdoesn’t see the surface - it’s conformation is unperturbed. To estimateξads we need to first estimate the number of monomers in contact with thesurface.

Let ξads contain g monomers each of size b and let the Flory exponent beν then since below the length scales of ξads the statistics are unperturbed,we must have the blob size

ξads = bgν (2.14)

Then the volume fraction of the blobs must be

φads ≈b3g

ξ3ads

=b3g

b3g3ν

= g1−3ν

=

(b

ξads

) 1−3νν

21

So then the number density is φads/b3 which means that the number ofthat are within the shell of thickness b at the surface of the blob (shellvolume ξ2

adsb) is

φadsb3

ξ2adsb =

b

ξadsg

=b

ξads

(ξadsb

)1/ν

=

(ξadsb

)1/ν−1

We already said the energy gain per blob is ≈ kBT and the energy gainper monomer is δkBT which means that

δkBT

(ξadsb

)1/ν−1

≈ kBT

δ

(ξadsb

) 1−νν

= 1

ξ1−νν

ads =b

1−νν

δ

ξads =b

δν/(1−ν)(2.15)

But that’s the dilute limit where every chain is isolated. What if the con-centration is increased? Eventually they begin to perturb each other and forma sort of semidilute solution. The authors don’t do this but let us considermultiple chains adsorping to the surface with another scaling argument:

• The chains are still attracted to the surface in the same manner as de-scribed above but now crowding by the surrounding chains repulse eachother. At the surface, we must still have adsorption blobs of size ξads asgiven by Eq. (2.15) but sections of the chains further away don’t gain fromthe attractive energy and are just decaying from the high concentrationat the surface φads to the background solution concentration φ0.

In a good solvent, the chains form three length scales of blobs:

1. There’s a thermal blob size (the chain is ideal random walk belowthis scale).

2. The chains that make up the correlation blobs are self-avoidingwalks of thermal blobs

3. Beyond the correlation length, the excluded volume is screened andthe chain is a random walk of correlation blobs.

If the solvent is good, the size of the correlation blob is

ξcorr = b( vb3

)2ν−1

gν . (2.16)

22

In other words, the number of monomers in a correlation blob is

g =

(ξcorrb

)1/ν (b3

v

)(2ν−1)/ν

.

If we remember from the above that the concentration is

φ ≈ g b3

ξ3corr

=

(ξcorrb

)1/ν (b3

v

)(2ν−1)/ν (b

ξcorr

)3

=

(b3

v

)(2ν−1)/ν (ξcorrb

)−(3ν−1)/ν

‘

ξcorr = b

(b3

v

)(2ν−1)/(3ν−1)

φ−ν/(3ν−1)

If the solvent is athermal then the excluded volume v = b3. In that case,

ξcorr = bφ−ν/(3ν−1). (2.17)

When the polymers are near the surface the blob size must be determinedby the distance from the surface s so we say s = ξcorr and rearrangeEq. (2.17) to find

φ =(zb

)−(3ν−1)/ν

. (2.18)

This concentration profile is called the de Gennes self-similar carpet.

2.4.2 Flocculation and Stabilization

Consider a metastable solution of colloids. If two colloids are able come closerthan the Debye length, they will likely aggregate. Now let’s add some polymer tothe solution. We assume that it has a large polymerization such that the radiusof gyration at lease twice the Debye length, R > 2κ, and that the polymer isnot a polyelectrolyte and so does not cause a large change in charge distributionabout the colloid.

• If just a little polymer is added to the solution, those polymers that areadsorbed will have a thickness ξads that is less than κ.

• If a little more is added and the thickness of the layer is beyond the κsomething a little surprising happens. Since the polymer adsorption isn’tsaturated on either surface, the polymer will adsorp to both. The twoparticles are now joined and and form an aggregate but the electrostaticbarrier is not overcome and the particles aren’t actually in contact. Thisis called bridging flocculation.

• When there is an excess of polymer added, the particles are stericallyrepelled due to polymer contact long before the electrostatic repulsionis even relavent. Steric repulsion is a combination of two independentmechanisms:

23

1. The monomers surrounding the other particle restrict the volumeavailiable to the chains adsorped to the surface (they don’t want tooverlap). When particles approach each other this causes a lose ofentropy and therefore an increase in free energy which is thermody-namically unfavourable and causes the particles to move apart.

2. When the layers do overlap, there is a supersaturation of polymersand the change in the number of particles causes an osmotic effectwhich pushes the particles apart.

The authors suggest that a great thing to do is to use very low polymer-izations so that the electrostatic repulsion acts at large separations while stericrepulsion comes into play if the energy barrier is overcome. In this way, thereis no risk of bridging flocculation. It is possible to make the system truly stablein this manner. The authors state that the a third energy Ep for the polymersshould then be added to the DLVO theory but don’t suggest a form for it.

2.4.3 Graft Polymers

If the polymers are not just adsorped to the surface but rather are graft, theyform more of a brush than a carpet. Simply speaking, the concentration is astep function rather than a continouous profile. Now the size of the correlationblobs is determined by the distance between grafting sites. Let σ be the graftingdensity (number of sites per unit area) such that

ξcorr =1√σ. (2.19)

Again we assume an athermal solvent and the number of monomers per blob isexactly as before

g =

(ξ

b

)1/ν

=(√σb)−1/ν

.

Then the number of blobs is N/g where N is the polymerization and the heightof the brush is

H = ξcorrN

g= Nb1/νσ(1−ν)/(2ν). (2.20)

Brushes provide better steric repulsion than carpets since the density is stepfunction-ish.

2.4.4 Non-Adsorped Polymers

Polymers do not need to be on the surface of the colloid to have an impact on thethermodynamics. If the polymer does not adsorb then it’s centre of mass can notcome with in a distance R of the surface. This creates concentration dip at thewall which we might call a depletion layer. As long as the polymer concentrationis low, this has little effect on the stability. But at high concentrations thesystem must try to reduce the forbidden space that the polymers can’t access.If the colloids aggregate the depletion volume is reduced and withit the polymerconcentration. This is called depletion flocculation.

24

Figure 4: Separation mechanism of two components (Normal mode of separa-tion).

3 Field-Flow FractionationField-Flow Fractionation (FFF) is a simple concept: we attempt to separatea mixture of different species by placing them in a flow and then applying afield perpendicular to that. The perpendicular field (which can be gravitation,centrifugal, magnetic, thermal, or even a cross flow of fluids) pushes the sam-ple against the accumulation wall but diffusion due to the concentration profilecompetes with the field. An exponential concentration gradient results whenequalibrium is reached. Each of the different species will have a different con-centration profile. BUT that means that the mean position of each is at adifferent place in the flow field and therefore their mean velocities are different.Separation results.

Maybe I’ve been reading Giddings too much but instead of just definingField-Flow Fractionation, I’d like to define it relative to another method, chro-matography. In chromotography there is a mobile phase and a stationary phase(let’s call these velocity phases for my allegory): either the solute is movingwith some velocity in the mobile phase or else it’s sorbed (either adsorbed orabsorbed) to something that isn’t moving at all. The fraction of time that eachspecies spends in the mobile phase determines the separation. In FFF the solutedoesn’t have two velocity phases but rather a continuous distribution of velocityphases (determined by the flow profile) and the time it spends at any velocityis determined by the concentration gradient set up by the perpendicular field.

3.1 Concentration ProfileLet’s consider the concentration gradient established by the perpendicular field.We must assume that the flow is slow enough that an equalibrium concentrationprofile can be set up. In general, the density flux J in a given direction y canbe written

J = (U + u) c−Ddc

dy(3.1)

where U is any velocity given to the solute particles by external fields, u is anyflow velocity in the y-direction (i.e. not the flow but flow in the perpendicularfield direction) and c is the concentration. Since

dc

dt= −dJ

dy, (3.2)

25

we note that the the nonequalibrium transport equation in the given directiony is

dc

dt= −W dc

dy+D

d2c

dy2. (3.3)

Here we defined W = U + u to be all the nondiffusive velocities.The diffusion coefficient D is the total sum of all the effective diffusion co-

efficients, not just the molecular diffusivity. Other random mechanisms for anapparent diffusivity could include

1. Multiple Paths - if the solution flows through a packed-bed that makes upmany different possible paths then any random path choice will cause afluctuation in velocity

2. Capture - if there are obsticles that form traps then the random stop-and-go process will increase the apparent diffusion coefficient

3. Random State - if the solute can randomly change it’s properties (say thenumber of ions an polyelectrolyte has in electrophoresis) then there willbe further spread.

In the steady state, dc/dt = 0 which means dJ/dy = 0 or J = J0 is aconstant. This means that Eq. (3.1) is a linear, first order ODE and we cansolve for the concentration profile. Letting J = J0 and c′ ≡ c−J0/W , Eq. (3.1)becomes

J = (U + u) c−Ddc

dy

J0 = (W ) c−Ddc

dy

c− J0

W=D

W

dc

dy

c′ =D

W

dc′

dy

Let’s do the integration from a reference where y0 ≡ 0 and the concentrationthere is c′0. Then

ln

(c′

c′0

)=

∫ y

0

W

Ddy.

We need to know the form of W/D. It’s usually reasonable to assume that Dis constant. The flows we are interested in are things like:

1. No flow which would mean W is constant with y.

2. Shear flow for which W ∝ y

3. Parabolic flow with W ∝ y2.

So let’s say

W ≡ −ayn (3.4)

26

and chose the appropriate n later on. In that fairly general case,

ln

(c′

c′0

)=

∫ y

0

W

Ddy

ln

(c′

c′0

)= − 1

D

ayn+1

n+ 1

c− J0W

c0 − J0W

= exp

(− ayn+1

D (n+ 1)

)c− J0

W=

(c0 −

J0

W

)exp

(Wy

D (n+ 1)

)

c =J0

W+

(c0 −

J0

W

)exp

(Wy

D (n+ 1)

)(3.5a)

or

c =J0

W+

(c0 −

J0

W

)exp

(− ayn+1

D (n+ 1)

). (3.5b)

So the concentration profile forms some sort of exponential distribution maybesimple exponential decay or a Gaussian. In particular, in FFF a thin steady-state distribution is setup. So the flux is J0 = 0 and the diffusion coefficientshould be approximately the molecular diffusivity. Plus, for FFFW = U+u (thenondiffusive velocities) are pointed down towards the wall and held constant.Since it’s constant n = 0 and because it’s down we explicitely write W = a →−|W |. So then the concentration is just

c = c0 exp

(−|W | y

D

)(3.6)

= c0 exp(−y`

)(3.7)

where ` is the effective mean layer thickness

` ≡ D

|W |=

D

|U + u|. (3.8)

The layer thickness resulting from the field W and the diffusion D is exactlythe quantity that determines the average velocity of the species and so resultsin separation.

Of course, this is exactly what we expected and what we stated in § 2.1.1for sedimentation. If D = kBT/6πηr and W = U + 0 = vterm = mg/6πηr then

` =D

|W |

=kBT

6πνr

6πηr

mg

=kBT

mg

c = c0 exp(−y`

)= c0 exp

(−mghkBT

).

27

3.2 FlowConsider a cross-sectional area of flow. If we discretize the area into n smallregions each with some velocity vi. The solute concentration profile can also bediscretized to ci. So then the average velocity of solute through the plane is

V =

∑ni civi∑ni ci

. (3.9)

Taking the continuous limit

V =

∫cvdA∫cdA

. (3.10)

One should compare this to the cross-sectional average velocity of the solution

〈v〉 =1

n

n∑i

vi (3.11)

↓

〈v〉 =1

A

∫vdA. (3.12)

If we want the different solutes to separate, it is required that their averagevelocities V are different. When comparing experiments, we don’t want to worryabout who’s solution is flowing faster or anything so a retention ratio is definedas

R =V〈v〉

(3.13)

=〈cv〉〈c〉〈v〉

(3.14)

=A∫cvdA∫

cdA∫vdA

. (3.15)

Since V is proportional to 〈v〉, the retention ratio is independent of flow rate.

3.3 ExampleAs an example, let’s consider an exponential concentration gradient (Eq. (3.7)with a mean layer thickness ` = D/ |W |) in parabolic flow in a channel. Thesituation is shown in Fig. 5 .

In a channel of width w, the velocity profile is simply

v =∆pw2

2Lη

(y

w− y2

w2

)(3.16)

= 6 〈v〉(y

w− y2

w2

). (3.17)

28

Figure 5: Parabolic flow and exponential concentration profile in a channel.

This means that the retention ratio is

R =V〈v〉

=

∫cvdA∫cdA

1

〈v〉

=

∫c0 exp

(−y`)

6 〈v〉(yw −

y2

w2

)dA∫

c0 exp(−y`)dA

1

〈v〉

=c0 〈v〉 zdepth

∫ w0

exp(−y`)

6(yw −

y2

w2

)dy

c0 〈v〉 zdepth∫ w

0exp

(−y`)dy

=6

w

∫ w0ye−y/`dy +

∫ w0

y2

w e−y/`dy∫ w

0e−y/`dy

=−6[(λ+ 1) e−1/λ − λ

]+ 6

[e−1/λ + 2λ (λ+ 1) e−1/λ − 2λ2

]1− e−1/λ

= 6−λe−1/λ − e−1/λ + λ+ e−1/λ + 2λ2e−1/λ + 2λe−1/λ − 2λ2

1− e−1/λ

= 6λ

(1− e−1/λ

)− 2λ

(1− e−1/λ

)+ 2e−1/λ

1− e−1/λ

= 6λ(1− 2λ)

(1− e−1/λ

)+ 2e−1/λ

1− e−1/λ

= 6λ

[1− 2λ+

2

e1/λ − 1

]= 6λ

[(1 +

2

e1/λ − 1

)− 2λ

]= 6λ

[coth

(1

2λ

)− 2λ

](3.18)

where we’ve used a dimensionless form of ` called the retention parameter λ

29

defined as

λ =`

w. (3.19)

For small retention parameters (λ 1) the hyperbolic cotangent in ap-proaches unity

limλ1

R ≈ 6λ (1− 2λ) (3.20)

or for even smaller values

limλ→0

R ≈ 6λ. (3.21)

3.4 Zone SpreadingBut we’ve only considered the mean velocities so far. However, looking back atFig. 5 , the zone of solute is bound to spread. Furthermore, diffusion has playedan essential and helpful role in the y-direction but it will play a detrimental rolein the x-direction.

Let’s start by considering the diffusive part. Consider a region of high con-centration that is a Gaussian distribution and call that region a zone. The zonewill spread due to the apparent diffusion as

σ2 = 2Dt. (3.22)

If the zone is translating with some velocity V then it goes a distance L in timet and over that distance the zone has spread

σ2 =

(2D

V

)L (3.23)

= HL (3.24)

The coefficient

H =2D

V(3.25)

=2D

R 〈v〉(3.26)

expresses the rate of growth of the variance along the separation path. Theparameter H is called the plate height for historical reasons. It’s the plateheight that really tells the separation power of a method. When H is large thespreading is large relative to the distance L over which separation occurs. WhenH is small, the zones are relatively narrow and so easier to resolve.

As a tiny detail note that the number of theoretical plates is clearly the totalmigration distance over the place height

N =w

H=Vw2D

(3.27)

Now let’s consider the zone spreading due to the shearing of the flow. Con-sider a zone moving at some (mean) velocity V. Now not only do the particles

30

diffuse along the flow (x-direction) but also up and down it (y-direction). So saya particle was right at the zone centre but then diffuses up or down to anothervelocity state v1. Since it’s diffusing let’s say it spends an average time of teqmoving at v1. During that time it gains a distance

l = (v1 − V) teq = ∆Vteq (3.28)

with respect to the zone centre. Since the transfer between velocity states arerandom, we can imagine that we have a random walk of steps l. The varianceof such a random walk is just

σ2 = l2n (3.29)

where n is the number of steps in the walk. We can estimate the number ofsteps as the total time of the process t = L/V over the exchange time teq:

n =t

teq=

L

Vteq. (3.30)

The variance of the zone due to shear is then

σ2 = l2n

= ∆V2t2eqL

Vteq

=

(∆VV

)2

VteqL. (3.31)

So then by Eq. (3.24) the plate height is

H =

(∆VV

)2

Vteq. (3.32)

The exchange time teq is a pretty abstract variable. Let’s replace it with thedistance diffused in time d =

√2Dteq so that the plate height reads

H =

(∆VV

)2

V d2

2D. (3.33)

Remember d and ∆V are intimately tied together.If we think specifically about FFF, then from the accumulation wall to the

effective mean layer thickness d = ` the velocity has gone from zero to V bydefinition. Going towards the centre of the channel the change in velocity isless and so it takes a larger distance d to get the same ∆V = V. Presumablyone can actually find this distance but we reasonably estimate that taking intoaccount both sides ∆V ∼ V for d ∼ 2`. Therefore, the plate height in FFF isapproximately

H =

(∆VV

)2

V d2

2D

= (1)2 V 4`2

2D

=2V`2

D. (3.34)

31

That’s not a bad estimate. According to Giddings (maybe in Giddings J. Chem.Educ., 50, 667 (1973).) a more rigorous (nonequalibrium) treatment than justa random walk model leads to a factor of 2:

H =4V`2

D. (3.35)

As a final note, this can be written in terms of the retention parameter λ = `/wand the average solvent velocity by defining

χ = 24λ3 (3.36)

such that

H = χ 〈v〉 w2

D. (3.37)

It’s interesting to see that if Eq. (3.25) is the working definition of the theoreticalplate height then the zone spreading due to shear has an effective diffusioncoefficient that is inversely proportional to the actual diffusion coefficient:

H =2Deff

R 〈v〉

= χ 〈v〉 w2

D

Deff = χR 〈v〉2 w2

2D∝ 1

D.

The total theoretical plate height is the sum of both the diffusive and theshearing contributions

H =2D

R 〈v〉+ χ 〈v〉 w

2

D(3.38)

=2D

R 〈v〉

[1 +

χR

2

(〈v〉wD

)2]

(3.39)

32

4 Field-Flow Fractionation SpecificsThe point of retention ratios R and plate heights H are to predict the resolutionbetween two zones of different species. The resolution between two peaks atposition xi with width σi is defined as

Rs =x2 − x1

2 (σ1 + σ2). (4.1)

Since we saw in the previous section that the zone broadening was propertydependant on the solute, let’s assume that for two zones that need resolutionσ1 ≈ σ2 = σ and so estimate

Rs =∆x

4σ. (4.2)

If we would rather the plate height then we use

σ =√Hx.

Also we’ve been dealing with velocity so far so then

∆x = t∆V =x

〈V〉∆V

where we used t = x/V. Substituting both of these into the resolution gives

Rs =∆x

4σ

=x

〈V〉∆V 1

4√Hx

=1

4

(∆V〈V〉

)( xH

)(4.3)

4.1 Flow Field-Flow FractionationFlow Field-Flow Fractionation (fl-FFF) is quite nearly universally applicable. Ithas been able to separate particles as small as ∼ 1 nm and as large as ∼ 0.1 mm

The basic idea is that as the flow parallel to the channel walls carries thesolution over the channel length L a cross flow acts as the field perpendicu-lar to the accumulation wall. The flow exits the channel through a filtrationmembrane. Very generally there are two methods of fl-FFF:

Symmetric Flow Field-Flow Fractionation (sfl-FFF): The cross-flow en-ters the channel through a porous top wall and then exits the channelthrough a filtration membrane over a porous bottom wall.

Asymmetric Flow Field-Flow Fractionation (afl-FFF): Only the accumu-lation wall is permeable to the solvent. The inlet flow is split into channelflow and cross-flow. Notice that since there is a continual lose of cross-flow fluid out the bottom, the volumetric flow rate is reduced as the flowtravels down the channel. This leads to a channel flow velocity gradientbut there are also advantages (to be discussed) that outway this problem.

33

The zone thickness parameter λ can estimated for both of these by sayingthat the velocity in y-direction can be related to the volumetric flow rate vc ofthe cross-flow. If the volume of the channel is V (V = wbL for a simple channel)and the area of the accumulation wall A = V/w (A = bL for a simple channel),then we have

λ =`

w

=D

wW

=D

wU

=D

w (vc/ (V/w))

=DV

vcw2(4.4)

We can introduce another set of useful terms here: We’ve already discussedthe retention ratio R by Eq. (3.15) but in all practicality how is it measured?Since the solute and the solvent both must travel the length of the channel, theratio of the solute velocity to the solvent velocity is a ratio of times

R =V〈v〉

=t0tr

(4.5)

where the void time t0 is the time it takes the unretained solvent to transversethe channel. The other time tr is the retention time, the time it takes the soluteto flow the length of the channel. It is tr that is experimentally measured.

For sfl-FFF,

t0 =V

v(4.6)

(that’s v, the volumetric flow rate along the channel not vc, the volumetricflow rate across the channel). Assuming small λ the retention time is

tr =t0R

≈ t06λ

=w2

6D

t0vcV

(4.7)

=w2

6D

(vcv

)(4.8)

Using either the void time t0 or flow rate v the diffusion coefficient D (andtherefore the hydrodynamic radius) can be found as a function of the retentiontime tr.

As mentioned above, afl-FFF has a complication. The cross-flow is a functionof y, how close to the accumulation wall the particle is. Giddings estimated thecross flow for a channel having one permeable wall to be

U (y) = − |u0|(

1− 3y2

w2+ 2

y3

w3

)(4.9)

34

where u0 = U (y = 0) is the cross-flow velocity at the accumulation wall. Thisfunction does not vary along the longitudinal length of the channel.

But what does vary is the solvent velocity. It’s still assumed to be parabolic.From the same mass conservation considerations that led to Eq. (4.9) , Giddingsfound the gradient of the mean velocity 〈v〉 along the channel (x-direction) tobe linear

〈v〉 (x) = 〈v〉0 −x

wu0. (4.10)

If Eq. (4.9) is substituted into Eq. (3.1) and the concentration is integratedagain then we find

ln

(c′

c′0

)=

∫ y

0

W

Ddy

ln

(c′

c′0

)=

∫ y

0

−|u0|D

(1− 3

y2

w2+ 2

y3

w3

)dy

ln

(c′

c′0

)= −|u0| y

D

(1− y2

w2+

y3

2w3

)c− J0

W

c0 − J0W

= exp

(−|u0| y

D

[1− y2

w2+

y3

2w3

])c (y) = c0 exp

(−|u0| y

D

[1− y2

w2+

y3

2w3

]). (4.11)

The Eq. (4.11) and Eq. (4.10) can be put into Eq. (3.15) to find R but theintegrals apparently lack an analytic solution. If the concentration is primarilynear the accumulation wall (y < w) then both the concentration profile and theretention ratio are approximated by their simple sedimentation values (Eq. (3.7)and Eq. (3.18) ). We’ll say λ is small and use Eq. (3.21) for the rest of this.

The last thing to do is predict the retention time tr = t0/R which under theλ→ 0 assumption comes down to estimating the void time in afl-FFF. We saidthat 〈v〉 was linearly reducing so in terms of the volumetric flow rate v. To firstorder, we still expect Eq. (4.6) but more specifically

t0 =

∫ L

0

dx

〈v〉

=

∫ L

0

dx

〈v〉0 −xwu0

=

[−

ln(〈v〉0 −

xwu0

)u0/w

]L0

(4.12)

which we can transform into volumetric flow rates by u0 = vc/bL, 〈v〉0 = vin/bw

35

and vout = vin − vc. The void time is then

t0 = − wu0

[ln

(〈v〉0 −

L

wu0

)− ln 〈v〉0

]=Lbw

vc

[− ln

(vinbw− L

w

vcbL

)+ ln

(vinbw

)]=V

vc

[− ln

(vout + vc

bw− vcbw

)+ ln

(vout + vc

bw

)]=V

vcln

(vout + vcvout

)=V

vcln

(1 +

vcvout

). (4.13)

Now we can finally say that the retention time is

tr =t0R

=w2

6D

vcVt0

=w2

6D

vcV

V

vcln

(1 +

vcvout

)=w2

6Dln

(1 +

vcvout

). (4.14)

And that’s assuming a simple rectangular channel. Often trapezoidal are usedsince a uniform cross-flow is very difficult to achieve and a trapezoidal channelcan compensate. In fact, trapezoidal channels are used almost always.

There are only two other details that I’d like to mention about fl-FFF:

1. When eluting polymers, high temperatures often improve the solubilityand so diffusivity is increased. When discussing zone broadening by shear,we demonstrated that large D will decrease zone spreading. When sepa-rating polymers by FFF, the higher the temperature the better.

2. Focusing in afl-FFF is often done. By having a second input stream lo-cated some distance far from the initial injection pointing directionly downtowards the accumulation wall, one will get a back flow. The backflow andthe injection flow meet at some position called the focusing point and forma sort of pressure well. The sample is focused where the streams meet whilethe fluid exits through the membrane. The zone width is shrunk duringthis focusing step. Once the focusing flow is stopped the sample the solutewill carry on it’s way but the overall band broadening has been reduced.

4.2 Thermal Field-Flow FractionationThe field perpendicular to the accumulation wall can also be a thermal gradient.The accumulation wall is kept at some cool temperature while the other wall isheated such that the temperature drop across the channel width w is ∆T . Thelarger the temperature gradient, the stronger the field and so high temperaturesare required. For this reason, the channel is usually pressurized to ∼ 10 bars

36

which increases the boiling temperature and allows experiments to reach highertemperatures. The particles drift velocity in the temperature gradient is

U = DT∂T

∂y

≈ DT∆T

w(4.15)

where DT is the thermal diffusion coefficient. The thermal diffusion coefficient(unlike D) is effected by the chemical composition of the particle and the carriersolvent which suggestes the possibility of separating particles of similar sizesbut composed of differing material. The drift velocity immediately tells us thedimensionless zone thickness

λ =D

Ww

=D

DT∆T. (4.16)

The ratio of diffusion coefficients is called the Soret coefficient. The Soret coeffi-cient determines the separation in Thermal Field-Flow Fractionation (th-FFF).While it is great that separation can result from compositional differences, sur-prisingly the theoretical understanding of thermal diffusion is lacking and thereis no predictive theory for DT .

Naively, knowing λ tells us the retention ratio by Eq. (3.18) but there isa minor subtlety that results in some non-neglible effects. The properties likeviscosity and thermal conductivity of a fluid are temperature dependent andso in a strong thermal gradient they have a nonhomogeneous profile across thechannel. This results in diviation from parabolic flow. The maximum velocity ismoved above the centre line of the channel skewing the flow profile upwards. Asbefore we don’t solve it but the idea is that the viscosity is no longer constantin the Navier-Stokes equation so it becomes

d

dy

(ηd

dyv

)=

d

dyp. (4.17)

The viscosity is assumed to be expandable as a series in T

η ≈ a0 + a1T + a2T2 + . . . (4.18)

Furthermore, the temperature which we assumed linear in Eq. (4.15) is a func-tion of the now varying thermal conductivity κ. The temperature profile turnsout to be quite complicated and is best estimated as a Taylor expansion

T ≈ T0 + y

[dT

dy

]0

+y2

2!

[d2T

dy2

]0

+ . . . (4.19)

From all this a series solution for the velocity can be found and then by Eq. (3.15)the retention ratio can be found to be

R = 6λ [1 + α (1− 6λ)]

[coth

(1

2λ

)− 2λ

](4.20)

where α is a velocity distortion factor . For limα→0R reduces to Eq. (3.18) .

37

5 Introduction to Tethered Polymer Microstruc-tures

Common features connect many of the structures formed by tethered polymers.Chains can be tethered at a point (forming a star polymer), a line (forminga centepede), a hard surface (forming a brush), a liquid-liquid interfaces, etc.They can be grafted, absorbed or aggregated.

And yet, despite all this diversity we can handle all of these comnpositemicrostructures in a coherent way. We’ll begin by discussing a foundationalcase for tethered polymers, grafted Alexander brush, and then we’ll extend theideas to other geometries.

5.1 Flat Solid InterfacesWhen polymers are grafted to a solid interface with some grafting density σthey are forced to assume elongated conformations due to the presence of theirneighbours but this elongation has a entropic cost associated with it. Thustheir equalibrium structure is dictated by a minimization of an interaction andan elastic free energy

F = Fint + Fel. (5.1)

We can estimate each of these terms through one of two ways:

1. a Flory Approximation

2. Blob Scaling

Because it is a mean field theory, the Flory Approximations neglect all cor-relations whereas blob scaling theory (although far from perfect) does includeexcluded volume correlations and so is more accurate (even though it is lessintensive). In blob theory (which by the way was already used in § 2 ) we modela layer of polymers as a semidilute solution that is characterized by some cor-relation length ξ. Although we can often cheat and identify the blob size as the“only length scale in the problem” but it will always fall out of the analysis. It ismore precisely defined as the length scale over which the interaction free energyis kBT (i.e. the blob size is defined such that the interaction free energy densityis kBT/ξ3). In this way, the polymer layer can be viewed as a close-packed arrayof blobs.

If the number of monomers (of size b) in a blob is g then the blob size is

ξ = gνb (5.2)

just as we always say the radius of gyration is R0 = Nνb. With the idea ofblobs being the correlation length we can imagine the radius of gyration to bea random walk of N/g blobs:

R20 =

N

gξ2 =

N

gb2g2ν = Nb2g2ν−1. (5.3)

At this point we can determine the free energies. As already said, the inter-action free energy is defined as kBT per blob and there are N/g blobs therefore

Fint =N

gkBT. (5.4)

38

The free energy of stretching the random walk of blobs to a length h is then

FelkBT

=

(h

R0

)2

(5.5)

where the power of 2 results because the blobs form an ideal random walk. Sothen the free energy is

F =N

g+

(h

R0

)2

. (5.6)

We want to minimize the free energy in terms of stretching length h to find theequalibrium height H. Unfortunately, although it might look like there is onlya single h2 term, there are actually hs buried in almost every variable. In thenext section, we go through alot of work to pull them all out and minimize thefree energy to find the equalibrium height, the equalibrium free energy and thefree energy of compression in terms of the equalibrium values. The next sectionshould be treated as an appendix. The lazy-man’s answer is that there are N/gblobs of size ξ therefore the height of the brush is

H =N

gξ (5.7)

and simplify.

5.2 Alexander Brush ModelConsider the blob Alexander model for a brush. We assume that the concen-tration of monomers forms a step-function i.e. is constant within the brush andthe brush abruptly ends. An unperturbed brush has a height H that is a resultof a balance between the excluded volume interaction energy and the elasticenergy. The free energy per chain is

F = Fint + Fel. (5.8)

For a brush of grafting density σ the blob size is ξ = bgν ≈ 1/√σ. Each of these

blobs has a repulsive energy on the order of ∼ kBT and is made of g monomers.There are N/g blobs per chain. That means that the whole chains free energydue to excluded volume interactions is

FintkBT

≈ N

g. (5.9)

The elastic component of the free energy results from the stretching of thecoil by it’s neighbours. The polymer would prefer to have an unperturbed sizeR0 that is the size of a random walk of N/g blobs. The free energy of stretchingto a length h is then

FelkBT

=

(h

R0

)2

(5.10)

39

where the power of 2 results because the blobs form an ideal random walk andso the undeformed size is

R20 =

N

gξ2 =

N

g(bgν)

2

= Nb2g2ν−1. (5.11)

Now the free energy is specified by the number of monomers per blob. We canremove the g dependance by two simple observations:

1. The volume fraction in a blob is φ = gb3/ξ3 and

2. The volume fraction in the brush is φ = Nb3σ/h.

By our primary assumption of an Alexander brush the concentration is homoge-neous in the brush and therefore setting the two values of φ equal to one anotherwe find that

gb3

ξ3=Nb3σ

h

gb3

(bgν)3 =

Nb3σ

h

g1−3ν =Nb3σ

h

g =

(Nb3σ

h

)1/(1−3ν)

So then putting g into Eq. (5.9) -Eq. (5.10) , the total free energy per chain ina brush is

F

kBT=FintkBT

+FelkBT

=N

g+

h2

Nb2g2ν−1

=N(

Nb3σh

)1/(1−3ν)+

h2

Nb21((

Nb3σh

)1/(1−3ν))2ν−1

= N

(h

Nb3σ

)1/(1−3ν)

+h2

Nb2

(Nb3σ

h

)(1−2ν)/(1−3ν)

= N

(Nb3σ

h

)1/(3ν−1)

+h(1−4ν)/(1−3ν)

Nb2(Nb3σ

)(1−2ν)/(1−3ν)

=N

(Nb3σ)1/(1−3ν)

h1/(1−3ν) +1

Nb2 (Nb3σ)(2ν−1)/(1−3ν)

h(1−4ν)/(1−3ν)

(5.12)

40

The equalibrium heightH is the height at which the free energy is minimized:

∂F

∂h=

N

(Nb3σ)1/(1−3ν)

1

1− 3νh1/(1−3ν)−1 +

1

Nb2 (Nb3σ)(2ν−1)/(1−3ν)

1− 4ν

1− 3νh(1−4ν)/(1−3ν)−1

0 = H3ν/(1−3ν) +

(1

Nb3σ

)(2ν−2)/(1−3ν)1− 4ν

N2b2H−ν/(1−3ν)

H3ν/(1−3ν)

H−ν/(1−3ν)=

4ν − 1

σ(2ν−2)/(1−3ν)

(1

N

)4ν/(3ν−1)(1

b

)4/(3ν−1)

H4ν/(1−3ν) = (4ν − 1)σ2(1−ν)/(1−3ν)N4ν/(1−3ν)b4/(1−3ν)

H = (4ν − 1)(1−3ν)/4ν

σ(1−ν)/2νNb1/ν

H ≈ σ(1−ν)/2νb1/νN (5.13)

= σ(1−ν)/2νR1/ν0 (5.14)

which if we put back into the free energy Eq. (5.8) gives the free energy perchain in a brush at equalibrium to be

F0

kBT≈ H√σ +

(H

bNν

)1/(1−ν)

≈ σ(1−ν)/2νb1/νNσ1/2 +

(σ(1−ν)/2νb1/νN

bNν

)1/(1−ν)

= σ1/2νb1/νN + σ1/2νNb1/ν

≈ σ1/2νR1/ν0

= σ1/2ν H

σ(1−ν)/2ν

= H√σ (5.15)

5.2.1 Free Energy Cost of Compression Mode

Compression of the brush lowers the brush height h away from it’s equalib-rium value H and so we would like to rewrite the free energy per chain out ofequalibrium in terms of it’s equalibrium values H and F0.

It’s convinient to use the brushes volume fraction φ so we notice

φ = g

(b

ξ

)3

=

(ξ

b

)1/ν (b

ξ

)3

=

(ξ

b

)−(3ν−1)/ν

. (5.16)

41

Using the above we can find g in terms of φ

g = φ

(ξ

b

)3

= φ(φ−3ν/(3ν−1)

)= φ−1/(3ν−1). (5.17)

But we can also give φ in terms of real values

φ = gb3

ξ3

=b3

ξ2

g

ξ

=b3

ξ2

N

h

=Nb3σ

h(5.18)

Also the equalibrium height and the unperturbed polymer size are relatedthrough Eq. (5.14) to be

H = σ(1−ν)/2νR1/ν0

R0 =Hν

σ(1−ν)/2(5.19)

42

Using the above equations we are ready to go. Let’s start with Fint:

Fint = h√σ

=h

ξ

= N

(1

g

)= N

(1

φ−1/(3ν−1)

)from Eq. (5.16)

= Nφ1/(3ν−1)

= N

(Nb3σ

h

)1/(3ν−1)

from Eq. (5.18)

= N

(Nb3σ

h

)1/(3ν−1)

= N

((H

σ(1−ν)/2νb1/ν

)b3σ

h

)1/(3ν−1)

from Eq. (5.14)

= N

(H

hb3−1/νσ1−(1−ν)/2ν

)1/(3ν−1)

=

(H

h

)1/(3ν−1)

Nb1/νσ1/2ν

= H√σ

(H

h

)1/(3ν−1)

from Eq. (5.14) (5.20)

And doing the same sort of thing for the free energy of stretching:

Fel =

(h

R0

)2

=h(1−4ν)/(1−3ν)

Nb2 (Nb3σ)(2ν−1)/(1−3ν)

(5.21)

=H(1−4ν)/(1−3ν)

Nb2 (Nb3σ)(2ν−1)/(1−3ν)

(h

H

)(1−4ν)/(1−3ν)

=σ

1−ν2ν

(1−4ν)(1−3ν) b

1ν

(1−4ν)(1−3ν)N

(1−4ν)(1−3ν)

Nb2 (Nb3σ)(2ν−1)/(1−3ν)

(h

H

)(1−4ν)/(1−3ν)

= σ1/2νb1/νN

(h

H

)(1−4ν)/(1−3ν)

= H√σ

(h

H

)(1−4ν)/(1−3ν)

(5.22)

So then putting these two together and using Eq. (5.15) , we get the free

43

energy of compression as a function of the brush height to be

F = Fint + Fel

≈ F0

[(H

h

)1/(3ν−1)

+

(h

H

)(1−4ν)/(1−3ν)]. (5.23)

which is more general than but in agreement with references. . . I think.

5.3 Curved InterfacesThe Alexander model of § 5.2 and § 5.2.1 works because the concentration isassumed to be step like. If we wish to discuss spherical and cylindrical geome-tries, we have to account for the fact that the volume accessible to each chaingrows with the distance from the grafting site. What we say is that each layerof blobs is made up of blobs of size ξ but ξ grows with each layer.

Consider f polymers grafted to a surface area S then the grafting density isσ = f/S and we keep our cheating concept from § 5.2 that the blob size mustbe

ξ =1√σ

=

(Sf

)1/2

. (5.24)

To be absolutely explicit:

S ∼

r2 sphericalLr cylindricalconstant planar

(5.25)

ξ ∼

r

f1/2 spherical(Lrf

)1/2

cylindrical1

σ1/2 planar.

(5.26)

And really once you know the blob size every thing else must come naturally.Specifically, we want the volume fraction, the size of the brush and the freeenergy.

The volume fraction is the volume taken up by all the monomers in a blobper volume of a blob or if we recall the definition ξ = gνb then the volumefraction in terms of blob size is

φ ∼ gb3

ξ3

=

(ξb

)1/ν

b3

ξ3

=

(b

ξ

)(3ν−1)/ν

44

which to keep being annoyingly explicit means

φ ∼

(b2fr2

)(3ν−1)/2ν

spherical(b2fLr

)(3ν−1)/2ν

cylindrical(b2σ)(3ν−1)/2ν planar.

(5.27)

We can estimate the average thickness of the polymer layer because we knowthat if we add up all the monomers in all the blobs we must get all the monomersi.e.

N =

∫ Ri+H

Ri

g

(dr

ξ

)where Ri is the radial grafting point and dr/ξ is the number of blobs. So if weput in the number of monomers per blob in terms of the blob size we get

N =

∫ Ri+H

Ri

(ξ

b

)1/ν1

ξdr

=1

b1/ν

∫ Ri+H

Ri

ξ(1−ν)/νdr (5.28)

We do this integral for the three geometries and dump all the numericalpieces to get the scaling:

Planar

Nb1/ν =

∫ H

0

ξ(1−ν)/νdz =

(1

σ1/2

)(1−ν)/ν ∫ H

0

dz

H = Nb1/νσ(1−ν)/2ν

Cylindrical

Nb1/ν =

∫ Ri+H

Ri

ξ(1−ν)/νdz =

∫ Ri+H

Ri

(√Lr

f

)(1−ν)/ν

dr

Nb1/ν(f

L

)(1−ν)/2ν

=

∫ Ri+H

Ri

r(1−ν)/νdr =

[1

1−ν2ν + 1

r1−ν2ν +1

]Ri+HRi

=

(2ν

1 + ν

)[(Ri +H)

(1+ν)/2ν −R(1+ν)/2νi

]∼ H(1+ν)/2ν

H ∼ N2ν/(1+ν)b2ν/ν(1+ν)

(f

L

) (1−ν)2ν

2ν(1+ν)

= N2ν/(1+ν)b2/(1+ν)

(f

L

)(1−ν)/(1+ν)

45

Spherical

Nb1/ν =

∫ Ri+H

Ri

(r

f1/2

)(1−ν)/ν

dr

Nb1/νf (1−ν)/2ν =

∫ Ri+H

Ri

r(1−ν)/νdr

=1

1−νν + 1

[r

1−νν +1

]Ri+HRi

= ν [rν ]Ri+HRi

ν[(Ri +H)

1/ν −R1/νi

]∼ H

H ∼ Nνbf (1−ν)/2

So then to recap, the scaling for each of the geometries is

H ∼

Nνbf (1−ν)/2 spherical

N2ν/(1+ν)b2/(1+ν)(fL

)(1−ν)/(1+ν)

cylindrical

Nb1/νσ(1−ν)/2ν planar

(5.29)

Finally and perhaps most importantly, we can estimate the free energy. Westill have that the free energy per blob is kBT so then the free energy is

F

kBT=

∫ Ri+H

Ri

ξ−1dr. (5.30)

which means that for each geometry we have

Spherical

F

kBT=

∫ Ri+H

Ri

ξ−1dr =

∫ Ri+H

Ri

f1/2

rdr = f1/2 [ln r]

Ri+HRi

= f1/2 ln

(Ri +H

Ri

)Cylindrical

F

kBT=

∫ Ri+H

Ri

ξ−1dr =

∫ Ri+H

Ri

f1/2

L1/2r1/2dr =

(f

L

)1/2 [2r1/2

]Ri+HRi

=

(f

L

)1/2 [√Ri +H −

√Ri

]∼(f

L

)1/2

H1/2

∼

(f

LN2ν/(1+ν)b2/(1+ν)

(f

L

) 1−ν1+ν

)1/2

=

(Nνbf

L

)1/(1+ν)

Planar

F

kBT=

∫ Ri+H

Ri

ξ−1dr =

∫ Ri+H

Ri

σ1/2dr = σ1/2 (Ri +H −Ri) = σ1/2H

= σ1/2Nb1/νσ(1−ν)/2ν = Nb1/νσ1/2ν

46

So in summary

F

kBT∼

f1/2 ln

(Ri+HRi

)spherical(

NνbfL

)1/(1+ν)

cylindrical

Nb1/νσ1/2ν planar

(5.31)

5.4 Polydisperse GraftingImagine that we’ve grafted a mixture of two chain lenths to a flat interface. Let’stry to handle this situation via the Alexander model. Remember the assumptionof the Alexander model is uniform density which causes sharp interfaces. Forthis reason, we can imagine two brushes: a short inner brush of height Hi andan outer brush of height Ho. The extra chain length can’t penetrate the denserinner brush. The longer chains have a degree of polymerization NL while theshorter ones have NS . If the distance between grafting points (of any chain) isd = 1/

√σ then from Eq. (5.29) the equalibrium height of the inner brush is

Hi ∼ NSb1/ν1

d(1−ν)/ν. (5.32)

If the longer chains have a grafting distance of do then the outer brush formedby the left over NL −NS monomers has a height

Ho ∼ [NL −NS ] b1/ν1

d(1−ν)/νo

. (5.33)

Which means that the thickness of the mixed brush is

H = Hi +Ho ∼ NSb1/ν1

d(1−ν)/ν+ [NL −NS ] b1/ν

1

d(1−ν)/νo

=NSb

1/ν

d(1−ν)/ν

[1 +

(NLNS− 1

)(d

do

)(1−ν)/ν]. (5.34)

47

6 Aggregation of Block CopolymersWe’ve yet to discuss how the tethering actually happens. Up to this point it’sbeen implicitly assumed that the chains are grafted to solid interfaces. Butthat’s not the only way that tethering can occur. In this section, we will discussmicrostructures formed by block copolymers.

6.1 MicrophasesConsider block copolymers in a melt that phase separate into layers into layers.In this case the grafting density isn’t arbitrary but rather arises from the equal-ibrium of the self-assembly. Now the interaction energy Fint is dicated by theinterfacial energy. This will still be balaced by an elastic term Fel. Let’s assumea sharp interface with an interfacial energy per unit area of γkBT . Assume thatthe solution of has some number density ρ of monomers. Then the number ofchains per unit area at the interface is hρ/N where h is the height of the struc-ture. That means that the interfacial energy per chain is Fint = γkBTN/(Hρ)and the free energy is

F

kBT=γN

hρ+

h2

N2νb2. (6.1)

Minimization with respect to h gives the equalibrium structure height H to be

H = N (2ν+1)/3

(γb2

ρ

)1/3

. (6.2)

There is really no dependence on NA or NB just N .

6.2 MicellesIf the AB block polymers aggregate into spherical micelles instead then we havea slightly different story. The free energy is composed of three terms:

1. the interfacial, Fint, which favours growth in order to reduce the curvatureof the interface;

2. the core, Fcore, which favours shrinking to keep the melt density fixed;

3. and the corona, Fcorona, which limits growth to increase the accessiblevolume available to the outer chains.

We label the corona block monomers NA and the core NB and consider twolimits. When NA NB the corona dominates and the micelle is almost star-polymer-ish but when NA NB the core dominates. We will rely heavily onthe results of § 5.3 .

The free energy of the interface per chain is specified by a surface energydensity γkBT such that

Fint = γkBTA

f

=γkBT

fπR2

core

48

where it’s important to recall f is the number of chains. The core is assumedto be much denser than the corona. It’s made up of fNB monomers of size b3.So if they are packed in then

Rcore ∼ b (fNB)1/3 (6.3)

which means that

FintkBT

∼γb2N

2/3B

f1/3. (6.4)

We assume that the core is a dense packing resulting in a melt. Since it’s amelt there are no correlations and the chains are random walks (which meanswe set ν = 1/2 in the core necessarily i.e. R0,core = N

1/2B b). The free energy of

the core is due to the deformation of the chains needed to keep the core densityconstant even if Rcore > R0.

FcorekBT

=

(Rcore

R0,core

)2

∼

(b (fNB)

1/3

N1/2B b

)2

∼ f2/3N−1/3B (6.5)

The free energy of the corona was essentially found in § 5.3 since the core isconsidered so much denser we use Eq. (5.31) as the free energy of the corona:

Fcorona ∼ f1/2 ln

(R

Rcore

)(6.6)

where R is the micelle size.Having found the pertenent free energies we can consider the two limits:

6.2.1 Hairy Micelle

In this case, the corona dominates over the core and so the review approximates

F ≈ Fint + Fcorona

F

kBT∼γb2N

2/3B

f1/3+ f1/2. (6.7)

Equalibrium is established by the number of copolymers aggregating into themicelle so we minimize with respect to f to find

f ∼(γb2)6/5

N4/5B . (6.8)

We can understand what this approximation is a touch better if we choose

49

not to make it. A better approximation of the free energy is

F ≈ Fint + Fcorona

F

kBT∼γb2N

2/3B

f1/3+ f1/2 ln

(R

Rcore

)F

kBT=γb2N

2/3B

f1/3+ f1/2 ln

(NνAbf

(1−ν)/2

b (fNB)1/3

)F

kBT=γb2N

2/3B

f1/3+ f1/2 ln

(NνA

N1/3B

)+ f1/2 ln

(f (1−3ν)/6

)F

kBT∼γb2N

2/3B

f1/3+ f1/2 ln

(NνA

N1/3B

)+ f1/2 ln (f)

In which case minimization with respect to f gives

0 = −γb2N

2/3B

3f4/3+

ln (f) + ln

(NνAN

1/3B

)+ 2

2f1/2

2

3γb2N

2/3B =

f4/3

f1/2

(2 + ln

(fNνA

N1/3B

))

γb2N2/3B ∼ f5/6

(1 + ln

(fNνA

N1/3B

))

f

(1 + ln

(fNνA

N1/3B

))6/5

∼(γb2)6/5

N4/5B .

So he effectively drops 1 + ln(fNν

A/N1/3B

)= ln f + ln

(eNνAN

1/3B

). That’s con-

venient but is it reasonable? The number of copolymers forming the micelle fshould be greater than 1. Only if f is small is ln f ≈ 1− f valid. In that case,of small f then f(1 + ln f)6/5 ≈ f(2− f)1 ≈ 2f − f2 ≈ 2f ∼ f and we get backthe review’s result. Conclusion: it’s pretty suspect.

There are a couple of things worth mentioning about this result. Firstly,there is no solvent quality dependance on the number of copolymers making upa micelle, and second, the number of monomers in the core not the number ofmonomers in the corona matters. Since we know the number of polymers ineach micelle, we can now find the size of the core to be

Rcore = b (fNB)1/3 ∼ b

((γb2)6/5

N4/5B NB

)1/3

∼ γ2/5b9/5N3/5B (6.9)

which by the assumption of our limiting case is much smaller than the coronawhich we estimate to be the height of the brush in spherical coordinates fromEq. (5.29)

R ≈ NνAbf

(1−ν)/2 (6.10)

50

which we could put Eq. (6.8) into at any time to find

R ≈ NνAb[(γb2)6/5

N4/5B

](1−ν)/2

∼ NνAN

2(1−ν)/5B b(11−6ν)/5γ3(1−ν)/5 (6.11)

6.2.2 Crewcut Micelle

The other limit is that NA NB so that the corona is very thin and the coremakes up the bulk of the micelle. In this case,

F ≈ Fint + Fcore

F

kBT∼γb2N

2/3B

f1/3+ f2/3N

−1/3B (6.12)

which we again minimize with respect to f and now find that the number ofcopolymers making up the micelle is approximately

f ∼ γb2NB . (6.13)

That means that the core is

Rcore = b (fNB)1/3 ∼ b

(γb2NBNB

)1/3∼ γ1/3b5/3N

2/3B (6.14)

and since the core is assumed much greater than the corona R ≈ Rcore.

6.3 Adsorbed Copolymer LayersNow let’s envision the blobk copolymers to be near a selective surface whichhas an attractive adsorbing energy for B-type monomers. We suppose that theanchoring layer is dense with φB → 1. The polymers have some average spacingd and the dense anchoring layer is ∆ thick so then the volume fraction of theanchoring layer is

φB =NBb

3

d2∆(6.15)

limφB→1

∆ ≈ NBb3

d2(6.16)

The free energy of adsorbtion is from the van der Waals attraction which it issaid goes as

Fabs =A

12π∆2(6.17)

where A is a Hamaker constant. I’m not sure about this. Why is it ∼ ∆−2?In any case, this free energy must be balanced against the stretching of the NAmonomers. From Eq. (5.31) , we know the free energy per unit area

FelkBT

∼ NA(b

d

)1/ν1

d2(6.18)

51

(where we replaced√σ = d and we multiplied by 1/d2 to get the free energy

per unit area rather than per chain). So then the total free energy

F

kBT=

Ad4

12πN2Bb

6+NAb2

(b

d

)(1+2ν)/ν

(6.19)

must be minimized with respect to the separation between adsorbtion points tofind the equalibrium separation

d ∼(kBT

A

)ν/(5ν+2) (NAN

2B

)ν/(5ν+2)b(1+6ν)/(2+5ν) (6.20)

which is a really weak function of kBT/A and even polymerization.

52

7 Deformation of Single Chains

7.1 Freely Jointed ChainA freely-jointed chain (FJC) of N links of length b forms a random walk. There-fore, the mean square end-to-end distance is⟨

h2⟩

= Nb2. (7.1)

Also just like a random walk, the probability distribution of the end-to-enddistance is

p(h) =

(3

2π 〈h2〉

)3/2

exp

(− 3h2

2 〈h2〉

)(7.2)

which we must admit is only true at large N .It one is only interested in h then we can discuss a FJC as a single entropic

spring of spring constant k. If we assume it’s a linear spring such that

fs(h) ≈ kh (7.3)

Us(h) ≈ 1

2kh2 (7.4)

then we have to ask what k recovers Eq. (7.2) ? The total energy of the springis zero and the potential energy is due to thermal fluctuations so the probabilityof having a U corresponding to a h is

p(h) = exp

(− U

kBT

)= exp

(−1

2

kh2

kBT

). (7.5)

Comparison between Eq. (7.2) and Eq. (7.5) gives the spring constant to be

k =3kBT

〈h2〉. (7.6)

7.2 Bead Spring ChainThe end of § 7.1 leads directly from the FJC model to the bead spring chain(BSC). Why not replace each section of n links with a spring of average length⟨h2⟩

= nb2? then we have M = N/n springs with spring constants kn =

3kBT/nb2 and the total end-to-end distance is still

⟨h2⟩

= M(nb2)

= Nb2 justlike the FJC chain.

But what benefit do we gain in switching from FJC to BSC? The BSC formsthe basis of the Rouse model which allows us to consider dynamics. Whenbrownian forces move a bead, the rest of the chain must respond. What’s theequation of motion? If we assume the inertia term is neglible (i.e. overdampedwith ma = 0) and that each bead only interacts with it’s nearest neighboursthen

f = ma = 0 = Φn − ξ0vn + k (rn−1 − rn) + k (rn − rn+1) (7.7)

53

where Φn is the brownian force on bead n and ξ0 is the friction coefficient of amonomer. This is equivalent to

ξ0∂rn∂t

= k (rn−1 + rn+1 − 2rn) + Φn

ξ0∂rn∂t

= k∂2rn∂n2

+ Φn

ξ0∂R

∂t= AR + Φn (7.8)

where we have first gone to the continuous limit and then defined R ≡ r1, r2 . . . rNand the N ×N tridiagonal force matrix A. Obviously, the index based equationdoesn’t strictly apply for the first and last beads. We can make decouple all themodes of this equation by thinking about it in one of two equivalent ways:

1. We can do a Fourier transformation to normal coordinates

qi =1

N

N∑n=1

cos

(inπ

N

)rn (7.9)

2. We can diagonalize A to get eigenvalues

λi = 4 sin2

(iπ

2N

)≈(iπ

N

)2

(7.10)

Using the eigenvalues if we define the ξ = Nξ0 (the fiction coefficient of thechain when no hydrodynamic interactions are included) then we can define anormal coordinate spring constant

κi =ξ

ξ0k

(iπ

N

)2

= N3kBT

b2

(iπ

N

)2

=3kBT

Nb2π2i2 (7.11)

which allows us to write the equation of motion in normal coordinates as

dqidt

= −κiξ

qi +giξ

= − 1

τiqi +

1

ξgi (7.12)

where gi is the transformed random force and τi is the relaxation time of modei (i.e. describing the motion of blobs which contain N/i monomers which isto say mode i = 1 descibes the motion of the whole chain and mode i = Ndescribes the motion of each monomer) which is given by

τi =ξ0N

2b2

3π2kBT

1

i2. (7.13)

54

7.3 CorrelationsWhereas the FJC allowed us to only look at the average end-to-end distance, wecan now look at correlations of the end-to-end vector describing the relaxationof the chain

C =〈h(t) · h(0)〉〈h2〉

. (7.14)

We use the formal solution of Eq. (7.12) which is

qi(t) =1

ξ

∫ t

−∞gi(t) exp [(t′ − t) /τi]dt′ (7.15)

to find the correlation between any two monomer positions in normal coordi-nates. The important thing in doing this is remembering that the properties ofthe white noise that forms the brownian force don’t change in move to normalcoordinates:

〈gi(t)〉 = 0 (7.16a)〈gi(t) · gi(t′)〉 = 6δijξikBTδ (t− t′) . (7.16b)

Therefore, the cross correlation is

〈qi(t) · qj(0)〉 =1

ξi

∫ t

−∞dt1 exp [(t1 − t) /τi]×

1

ξj

∫ 0

−∞dt2 exp [(t2 − 0) /τj ]× 〈gi(t1) · gi(t2)〉

=6δijkBT

ξj

∫ t

−∞dt1 exp [(t1 − t) /τi]

∫ 0

−∞dt2 exp [t2/τj ]δ (t1 − t2)

=6δijkBT

ξj

∫ 0

−∞dt2 exp [(t2 − t) /τj ] exp [t2/τj ]

=6δijkBT

ξje−t/τj

∫ 0

−∞dt2 exp [2t2/τj ]

=6δijkBT

ξje−t/τj

[τj2− 0]

=3kBT

kje−t/τjδij (7.17)

Now the autocorrelation of the end-to-end distance h(t) can be found. First,we note that in normal coordinates the instantaneous end-to-end vector is h(t)is

h(t) = rN−1 − r1

=

q0 + 2

N−1∑i=1

[cos

(i (N − 1)π

N

)qi

]+ (−1)

N−1qN

−

q0 + 2

N−1∑i=1

[cos

(iπ

N

)qi

]+ (−1)

1qN

= 2

N−1∑i=1

[cos

(i (N − 1)π

N

)qi − cos

(iπ

N

)qi

]

≈ 2

N−1∑i=1

[(−1)

i − 1]

qi

= 4∑i,odd

qi

55

That’s a pretty awesome result because now the autocorrelation of the end-to-end distance can be found using Eq. (7.17) , Eq. (7.18) and Eq. (7.11) tobe

C =〈h(t) · h(0)〉〈h2〉

=1

〈h2〉16∑i,odd

〈qi(t) · qi(0)〉

=16

〈h2〉∑i,odd

3kBT

κiexp (−t/τi)

=16

〈h2〉∑i,odd

Nb2

2π2i2exp (−t/τi)

=8

π2

∑i,odd

exp (−t/τi)i2

. (7.18)

Notice how quickly the sum dies out. By the second term the i−2 has alreadygone down to about a tenth the size of the first term and then there’s also theexponential term! Since 8/π2 ≈ 1 a quick estimate is

C ≈ exp (−t/τ1) . (7.19)

7.4 Deformation Due to TensionNow that we have a fairly good understanding of our polymer. Let’s pull on itwith a fixed force. We shall treat this stress ensemble in two different ways:

1. by blob theory

2. seriously.

7.4.1 Stress Ensemble: Blob Theory

We do this very quickly. The blob size is

ξ = bgν (7.20)

where g is the number of monomers in a blob. Then the number of blobs in thechain is N/g and the end-to-end distance is just

〈h〉 ∼ N

gξ =

N

ξ1/ν/b1/νξ =

Nb1/ν

ξ1/ν−1

ξ1/ν−1 =Nb1/ν

〈h〉

ξ =

(Nb1/ν

〈h〉

)1/(1/ν−1)

=

(R

1/ν0

〈h〉

)ν/(1−ν)

.

56

By definition, the free energy per blob is kBT so then the free energy of thestretched chain is

F ∼ kBTN

g= kBT

〈h〉ξ

= kBT 〈h〉

(〈h〉R

1/ν0

)ν/(1−ν)

= kBT

(〈h〉1/ν

R1/ν0

)ν/(1−ν)

= kBT

(〈h〉R0

)1/(1−ν)

which means that the constant force required to extend the chain to an averagelength 〈h〉 is

f =∂F

∂ 〈h〉∼ kBT

(1

R0

)1/(1−ν)∂

∂ 〈h〉〈h〉1/(1−ν)

= kBT

(1

R0

)1/(1−ν)

〈h〉ν/(1−ν)

=kBT

R0

(〈h〉R0

)ν/(1−ν)

.

When ν = 1/2 the solvent is ideal and the chain is a hookian spring (〈h〉 ∝ f)but this is just scaling. Can we do better?

7.4.2 Stress Ensemble: FJC

What is the probability of a chain with a fixed number of particles N in a fixedtemperature bath ]β = 1/kBT pulled on with a fixed force f has an end-to-endlength h?

p(h) =exp [−βH+ βf · h]

Z(7.21)

where Z is the Gibbs partition function. Let’s say H = 0 indicating that thereis no interaction between neighbours but rather there is a mean field of energyf · h acting on each. Then the partition function is

Z =∑µ

exp [βf · h] .

We take the number of states to the continuous limit so∑µ →

∫µand say that

f · h =

N∑i=0

f · bi (7.22)

where b is the vector connecting monomer i to i+1. Then the partition functionbecomes

Z =

∫µ

exp [βf · h]

=

∫µ

exp

[N∑i=0

βf · bi

]

Since it’s a FJC each of the monomers i+1 can be placed anywhere on a sphereof radius b surrounding monomer i so the integral over the states is an integral

57

over the surface of the sphere for each monomer but the other property of FJCsis that each monomer is independent of every other so the N integrals are allexactly the same

Z =

∫ 2π

θi=0

∫ π

φi=0

exp

[N∑i=0

βf · bi

]sin θidθidφi

=

∫ 2π

θi=0

∫ π

φi=0

exp

[N∑i=0

βfb cos θi

]sin θidθidφi

=

N∏i=0

∫ 2π

θi=0

∫ π

φi=0

exp [βfb cos θi] sin θidθidφi

=

[∫ 2π

θ=0

∫ π

φ=0

exp [βfb cos θ] sin θdθdφ

]N=

[π

∫ 2π

θ=0

exp [βfb cos θ] sin θdθ

]N.

By letting y = cos θ such that dy = sin θdθ the integral is easily doable

Z =

[π

∫ 1

y=−1

exp [βfby] dy

]N

=

[π

βfbexp [βfby]

∣∣∣∣1−1

]N

=

[π

βfb(exp [βfb]− exp [−βfby])

]N=

[π

sinh (βfb)

βfb

]N(7.23)

Amazing! The partition function is given very simply interms of the dimension-less force βfb and once the partition function is found everything else comessimply. In particular,

F = −kBT lnZ

= −kBTN ln

(π

sinh (βfb)

βfb

)and

〈h〉 = −∂F∂f

= kBTN∂

∂fln

(π

sinh (βfb)

βfb

)= kBTN

∂

∂fln sinh (βfb)− ∂

∂fln (βfb)

= kBTN

cosh (βfb)

sinh (βfb)βb− 1

f

= Nb

coth (βfb)− 1

βfb

.

58

Since since Nb is the contour length hmax

〈h〉hmax

= L (βfb) = coth (βfb)− 1

βfb(7.24)

where L() is the Langevin function.

7.4.3 Stress Ensemble: FJC vs Blobs

For small forces βfb 1 so that the hyperbolic cotangent can expand to

coth(x) ≈ 1

x+x

3− x3

45+ . . .

then

〈h〉hmax

≈ 1

βfb+βfb

3− 1

βfb

≈ βfb

3(7.25)

which is the hookean answer from blob theory.But in the opposite limit limx→∞ coth(x) = 1. Even coth(10) = 1±4×10−7.

So if the tension energy is at all larger than thermal energy then

〈h〉hmax

≈ 1− 1

βfb(7.26)

which is markedly different behaviour from hookian.

7.5 Strain EnsembleIn § 7.4.2 gives a dimensionless average length h = 〈h〉 /hmax for a finitelyextensible chain that results from pulling on both ends of the chain with aconstant dimensionless force f = βfb:

h = L(f)≈

13 f −

145 f

3 + 2945 f

5 −O(f7)

f 1

1− 1f

f 1(7.27)

which can be inverted to give the constant force reqired to get an average ex-tension h

f = L−1(h)≈

3h+ 95 h

3 + 297175 h

5 +O(h7)

h 1

11−h 1− h 1

. (7.28)

This forms the basis of things like the FENE springs but notice something. Thisis the stress ensemble i.e. the constant force to give an average extension. Whatis really desired is the strain ensemble i.e. a fixed length. In the Rouse modelwe actually want to know the mean force

⟨f⟩giving rise to the instantaneous

59

end-to-end distance h. This is not an easy question but there is an approximateseries solution:

⟨f⟩≈

(− 3N + 3

)h+

(− 99

25N + 95

)h3 +

(− 837

175N + 297175

)h5 +O

(h7)

h 1

1Nh

+ 1−2/N

1−h 1− h 1.

(7.29)

Look at that! Eq. (7.29) only differs from Eq. (7.28) by factors on the orderof ∼ N−1. No big deal.

7.6 Deformation Due to FlowPreviously, we discussed pulling on the ends of a polymer but what if we onlypull on one end? The polymer will move through the fluid. Of course, this isequivalent to pinning one end of the polymer that is in a homogeneous flow ofvelocity v.

7.6.1 Deformation in a Good Solvent

At very small forces f , the polymer will be unperturbed and Stokes law willapply to the roughly spherical polymer of size R0 (this is assuming Zimmian-like dynamics)

f ∼ ηR0v (7.30)

What is a “very small force” though? The natural force scale is the thermalforce. So a small force should be on the order of

f <kBT

R0(7.31a)

which through Eq. (7.30) means that a small velocity is

v <kBT

ηR20

(7.31b)

The critical force fc,1 for transferring to new behaviour should be once theforce on the one end is on the same scale as the thermal forces

fc,1 ∼R0

kBT(7.32a)

vc,1 ∼kBT

ηR20

=R0

τz(7.32b)

Of course! τz = ηR30/kBT is the Zimm relaxation time. If the time scale of the

polymer’s motion is less than the Zimm relaxation time, the polymer will nothave a chance to relax and deformation will result.

To consider this new regime, let’s start the index from the free end and letmonomer N be fixed with a flow of v > vc,1. Monomer N is being tugged on by

60

all the blobs behind it. On the other hand, monomer 1 is free −→ the tensionincreases as you near the fixed end. From blob theory, we know

f =kBT

R0

(〈h〉R0

)ν/(1−ν)

〈h〉R0

=

(fR0

kBT

)(1−ν)/ν

〈h〉N

= bNν−1

(fR0

kBT

)(1−ν)/ν

= b

(fR0

NνkBT

)(1−ν)/ν

= b

(fb

kBT

)(1−ν)/ν

= b(f)(1−ν)/ν

Hey our dimensionless force is back.Anyway, we don’t have to consider the whole chain in the blob model. Let’s

just consider ∆n monomers that are stretched to a length ∆x under the force f

∆x

∆n= b

(f)(1−ν)/ν

which can be taken to the continuous limit such that the local deformation isgiven by

dx

dn= b

(f)(1−ν)/ν

. (7.33)

If the force on each piece follows Stokes law then we can substitute in for f

dx

x(1−ν)/ν= b

(ηvb

kBT

)(1−ν)/ν

dn∫dxx(ν−1)/ν = b

(ηvb

kBT

)(1−ν)/ν

n

ν

2ν − 1x(2ν−1)/ν = nb

(ηvb

kBT

)(1−ν)/ν

x(n) =

(2ν − 1

ν

)ν/(2ν−1)

(nb)ν/(2ν−1)

(ηvb

kBT

)(1−ν)/(2ν−1)

(7.34)

61

This result means that the total length of chain is

〈h〉 = x(N) ∼ (Nb)ν/(2ν−1)

(ηvb

kBT

)(1−ν)/(2ν−1)

= Nν/(2ν−1)b1/(2ν−1)

(ηv

kBT

)(1−ν)/(2ν−1)

= R1/(2ν−1)0

(ηv

kBT

)(1−ν)/(2ν−1)

= R1/(2ν−1)0

(τzv

R30

)(1−ν)/(2ν−1)

=R

1/(2ν−1)0

R2(1−ν)/(2ν−1)0

(τzv

R0

)(1−ν)/(2ν−1)

= R0

(τzv

R0

)(1−ν)/(2ν−1)

= R0

(v

vc,1

)(1−ν)/(2ν−1)

(7.35)

Keeping in mind that x is measured from the free end (i.e. from monomer1) the size of the blob at x is, by definition ξ(h) = kBT/f . The blob size givesus the perpendicular y profile of the chain

ξ (h) =kBT

f(7.36)

↓

y (x) =kBT

ηxv(7.37)

The size is going down as the inverse of the distance from the free end. Thisinverse shape is why this regime is called the trumpet regime.

The last blob (at n = 1) has to drag only it’s self so f = ηvξ1 which meansthe blob size is

ξ1 =kBT

f

=kBT

ηvξ1

ξ1 =

√kBT

ηv(7.38)

At the other extreme, the smallest blob (at n = N) has to drag the whole chain

62

of total length 〈h〉 so the smallest blob is then

ξN =kBT

f

=kBT

ηvx(N)

=kBT

ηv 〈h〉

=kBT

ηvR0

(τzvR0

)(1−ν)/(2ν−1)

= R20

kBT

ηvR30

(R0

τzv

)(1−ν)/(2ν−1)

= R0R0

τzv

(R0

τzv

)(1−ν)/(2ν−1)

= R0

(vc,1v

)ν/(2ν−1)

. (7.39)

That’s interesting in itself but ξN also gives the limit on this so calledtrumpet regime. The above analysis assumes that at all points along the chaina blob theory holds. But consider that if the smallest blob size ξN is smallerthan a monomer the theory will not hold. The transition happens at a secondcritical velocity vc,2 which can be seen to be

b < ξN

b < R0

(vc,1v

)ν/(2ν−1)

b < bNν(vc,1v

)ν/(2ν−1)

(v

vc,1

)ν/(2ν−1)

< Nν

v

vc,1< N (2ν−1)

v < vc,1N(2ν−1)

v < vc,2.

So the trumpet regime is only valid in the regime vc,1 < v < vc,2 where (torepeat for clarity) vc,1 = R0/τz and vc,2 = vc,1N

(2ν−1).At velocities greater than vc,2 part of the chain is fully(-ish) elongated and

has a diameter comparible to the monomer size (the stem). The rest of the n∗monomers still form a trumpet (the flower) of length x(n∗) = x∗. This is thestem and flower regime. We can calculate the number of monomers in the flowerportion by realizing that because x∗ is the point where the flower turns into thestem we have ξn∗ = b which means the length of the flower is

ξn∗ = b =kBT

f=kBT

ηvx∗

x∗ =kBT

ηvb

63

but we can still apply Eq. (7.34) to the flower portion which gives

x∗ = x(n∗) ≈(

2ν − 1

ν

)ν/(2ν−1)

(n∗b)ν/(2ν−1)

(ηvb

kBT

)(1−ν)/(2ν−1)

which if we equate the two (dropping the numerical factor in front) gives thenumber of monomers in the flower to be

kBT

ηvb=

(2ν − 1

ν

)ν/(2ν−1)

(n∗b)ν/(2ν−1)

(ηvb

kBT

)(1−ν)/(2ν−1)

(n∗b)ν/(2ν−1) ∼

(kBT

ηvb

)(ν)/(2ν−1)

n∗b =kBT

ηvb

n∗ =1

b21

v

kBT

η

n∗ =1

b21

v

R30

τz

n∗ =vc,1v

R20

b2

n∗ =vc,1v

b2N2ν

b2

n∗ =vc,1vN2ν

n∗ =vc,2vN (7.40)

Of course, that means the number of monomers in the stem is N − n∗ =N (1− vc,2/v). This also means that with a bit of work we can write the length

64

of the flower

x∗ ∼ (n∗b)ν/(2ν−1)

(ηvb

kBT

)(1−ν)/(2ν−1)

= (n∗b)ν/(2ν−1)

(ηvR3

0

kBT

)(1−ν)/(2ν−1)(b

R30

)(1−ν)/(2ν−1)

= (n∗b)ν/(2ν−1)

(vτz)(1−ν)/(2ν−1)

(b

R30

)(1−ν)/(2ν−1)

= (n∗b)ν/(2ν−1)

(v

vc,1

)(1−ν)/(2ν−1)(b

R20

)(1−ν)/(2ν−1)

= (n∗)ν/(2ν−1)

b1/(2ν−1)

(v

Nvc,1N2ν−1

)(1−ν)/(2ν−1)(N ×N2ν−1

R20

)(1−ν)/(2ν−1)

= (n∗)ν/(2ν−1)

b1/(2ν−1)

(v

Nvc,2

)(1−ν)/(2ν−1)(N2ν

b2N2ν

)(1−ν)/(2ν−1)

= (n∗)ν/(2ν−1)

(1

n∗

)(1−ν)/(2ν−1)(b1/(ν−1)

b2

)(1−ν)/(2ν−1)

= n∗(b(2ν−1)/(ν−1)

)(1−ν)/(2ν−1)

= n∗b (7.41a)

Remember this is from scaling and there is definately a numerical coefficient ofless than 1 in front of this result. The book suggests it’s [(2ν − 1) /ν]

ν/(2ν−1)=

1/3 if ν = 3/5:

x∗ =bn∗

3. (7.41b)

Therefore, an estimate of the entire length of extended polymer is

〈h〉 = hstem + hflower

= (N − n∗) b+ x∗ = (N − n∗) b+bn∗

3

=

(N − 2

3n∗)b = Nb

(1− 2

3

vc,2v

)(7.42)

7.6.2 Deformation in an Ideal Solvent

Those who read the last section too carefully may have noticed that keeping thesolvent quality ν as a variable was dumb. Everything is perfectly fine until theintegration of Eq. (7.33) to get Eq. (7.37) . If we set ν = 1/2 we find

dx

x(1−ν)/ν= b

(ηvb

kBT

)(1−ν)/ν

dn

dx

x= b

(ηvb

kBT

)dn

lnx =

(ηv

kBT

)nb2

65

Clearly, the deformation is enhanced. The stretching will increase exponentiallywith the velocity! We can write the total length in terms of the Zimm relaxationtime

〈h〉 = exp

(ηv

kBTNb2

)= exp

(vτzR3

0

R20

)= exp

(vτzR0

). (7.43)

7.7 Tethered Polymers in Shear FlowImagine a polymer grafted to a wall and subject to a shearing flow. The velocityprofile is given by vx(y) = γy where γ is the shearing rate. If the polymer inthis flow is made up of blobs of size ξi then the friction coefficient of the ith blobis given by Stokes law to be f = 6πηξiv ∼ ηξv. The total friction is the sum ofall the blobs. Now in a shearing flow the blobs stay relatively close to the wallsuch that ξ → y and the friction becomes

f ∼∑

ηξivi

=∑

ηyiγyi

=

∫ηγy2

(dx

y

)= ηγyx.

But by again comparing this to the definition of

kBT

f= ξ = y (7.44)

f =kBT

y(7.45)

we can see that polymer’s profile must be

ηγyx =kBT

y

y2 =kBT

ηγx

y(x) =

√kBT

ηγx(7.46)

which has been called a horn profile. The function y(x) ∝ x−1/2 for a tetheredpolmyer in a flow given by Eq. (7.46) should be compared to Eq. (7.37) whichgives y(x) ∝ x−1 for a polymer being pulled by one end through a viscoussolvent (or equivalently a pinned polymer in a uniform flow).

66

The largest blob ξ1 is given by setting x = y = ξ1 in Eq. (7.46) which means

ξ1 =

√kBT

ηγξ1

ξ31 =

kBT

ηγ

ξ1 =

(kBT

ηγ

)1/3

. (7.47)

This result allows us to say when we get deformations. If ξ1 = R0 the poly-mer is not deformed but when ξ1 < R0 the polymer is significantly deformed.This can be expressed as

1 <R0

ξ1

1 < R0

(ηγ

kBT

)1/3

1 <

(ηR3

0γ

kBT

)1/3

13 <ηR3

0γ

kBT

1 < τz γ

1 <Wi (7.48)Wic,1 = 1 (7.49)

where we have defined the dimensionless number Wi = τ γ called the Weis-senberg number which balances the time scale required by the polymer to relaxto the time scale of the deforming flow. There is no deformation for Wic,1 < 1but above that first critical value we get a horn profile.

Again Eq. (7.33) gives the local deformation (very good formula) and ite-gration over the entire chain gives the total chain length

dx

dn= b

(f)(1−ν)/ν

dx

dn= b

(fb

kBT

)(1−ν)/ν

dx

dn= b

(b

y(x)

)(1−ν)/ν

dx

dn= b

(b

√ηγx

kBT

)(1−ν)/ν

(7.50)

67

7.7.1 Good Solvent Horn

Taking ν = 3/5 Eq. (7.50) becomes

dx

dn= b

(b

√ηγx

kBT

)(1−ν)/ν

dx

dn= b5/3

(ηγx

kBT

)1/3

x−1/3dx = b5/3(ηγ

kBT

)1/3

dn

2

3x2/3 = b5/3

(ηγ

kBT

)1/3

n

which for n→ N becomes

〈h〉 =

(3

2

)3/2

b5/2(ηγ

kBT

)1/2

N3/2

∼ b5/2N3/2

R3/20

(τz γ)1/2

=b5/2N3/2

b3/2N3/4

√τz γ

= bN1/2√τz γ

= R0

√τz γ

= R0

√Wi (7.51)

That’s good information in itself but it also allows us to find the stem flowerregime for tethered polymers. Just like when we found the largest blob ξ1 byEq. (7.46) , now we find the smallest blob ξN and note that when ξN → b weno longer have a profile of blobs but rather a stem and lower regime.

y(x) =

√kBT

ηγx

ξN =

√kBT

ηγ 〈h〉=

√kBT

ηγR0

√τz γ

=

√R2

0

τz γ√τz γ

↓

b =

√R2

0

τz γc,2√τz γc,2

=

(R2

0

(τz γc,2)3/2

)1/2

(τz γc,2)3/2

=R2

0

b2

τz γc,2 =

(N2νb2

b2

)2/3

Wic,2 = N4ν/3 (7.52)

68

7.7.2 Ideal Solvent - Tethered Chain

Again, deformation is going to be enhanced in an ideal solvent but not as muchas it was for a pinned polymer. Eq. (7.33) still holds

dx

dn= b

(b

√ηγx

kBT

)(1−ν)/ν

= b2√ηγx

kBT

dx√x

= b2√

ηγ

kBTdn

Stop!Unlike in § 7.6.2 , integration produces a power of x so in § 7.7.1 , we could

have kept ν as a variable and found the answer for both good and ideal solutions.We make up for our mistake and do so now:

dx

dn= b

(b

√ηγx

kBT

)(1−ν)/ν

dx

x(1−ν)/ν= b

(b

√ηγ

kBT

)(1−ν)/ν

dn

(3ν − 1

2ν

)x(3ν−1)/2ν = b1/ν

(ηγ

kBT

)(1−ν)/2ν

n

x(3ν−1)/2ν =

(2ν

3ν − 1

)nb1/ν

(ηγ

kBT

)(1−ν)/2ν

x =

(2ν

3ν − 1

)2ν/(3ν−1) (nb1/ν

)2ν/(3ν−1)(ηγ

kBT

) 1−ν2ν

2ν3ν−1

x =

(2ν

3ν − 1

)2ν/(3ν−1)

(nνb)2/(3ν−1)

(ηγ

kBT

)(1−ν)/(3ν−1)

(7.53)

From this result, we can find the total length of the tethered polymer in theshear flow to be

〈h〉 = x(N) =

(2ν

3ν − 1

)2ν/(3ν−1)

(Nνb)2/(3ν−1)

(ηγ

kBT

)(1−ν)/(3ν−1)

=

(2ν

3ν − 1

)2ν/(3ν−1)

R2/(3ν−1)0 (τz γ)

(1−ν)/(3ν−1) 1

R3(1−ν)/(3ν−1)0

=

(2ν

3ν − 1

)2ν/(3ν−1)

R0Wi(1−ν)/(3ν−1)

If ν = 3/5 then Eq. (7.54) recovers Eq. (7.51) . If we set ν = 1/2 instead

69

then we get

〈h〉 =

(2ν

3ν − 1

)2ν/(3ν−1)

R0Wi(1−ν)/(3ν−1)

= 22R0Wi

∼ 22R0Wi. (7.54)

7.7.3 Fluctuations of Tethered Polymers in Shear Flows

Recall our regimes for a tethered polymer in shear flow:

1. (Wic,1 = 1) < Wi: When the Weissenberg number is less than one thepolymer is not deformed.

2. (Wic,1 = 1) < Wi <(Wic,2 = N4ν/3

): The polymer deforms slightly into

a horn with a y ∝ x−1/2 profile.

3. Wi >(Wic,2 = N4ν/3

): The polymer is quite deformed with a thin stem

and a flower at the end.

These descriptions are the static conformations of the chain; however, there aredynamic fluctuations that can be characterized as well. When the deformation issmall (Wi < 1) there fluctuations in the y profile are small as well. In the secondregime (they say 1 .Wi & 20 which agrees with Wic,2 = N4ν/3 if 10 . N & 100which is likely for simulations like theirs) the polymer extends into the horn andthe fluctuations about the mean extension 〈h〉 increase substantially but whenWi > 20 the polymer is approaching it’s full extension (and most likely in thestem-flower regime) the fluctuations are reduced!

The reason for this fluctuation in extension is a fluctuation in angle fromthe wall. This occurs cyclicly. Consider a partially elongaged chain. Eventuallysome particularily large fluctuation will bring the chain away from the wall (andtherefore, into stronger flow). The stronger shear in turn further extends thechain. But the extension also results in an increased torque which rotates thechain back towards the wall (and therefore, into the slower flow) allowing thechain to relax into it’s less extended state until another large fluctuation onceagain moves the chain away from the wall.

7.8 Polymer Stretching over a Potential BarrierConsider a polymer in a trap. It must over come a potential barrier. The barriermay be entropic, like a pore or a Craighead device. It doesn’t really matter.Foundamentally we want to know how a polymer crosses a potential barrier. Solet’s consider a double-well potential that’s pretty mundane.

Our Rouse model equation of motion from Eq. (7.8) which we repeat forconvinience

ξ0∂R

∂t= AR + Φn.

The force operator A still includes the spring interactions with nearest neigh-bours k∂2()/∂n2 but now also includes the potential gradient −∇V () so that in

70

this case Eq. (7.8) becomes

ξ0∂R

∂t=∂2R

∂n2−∇V (R) + Φn. (7.55)

To consider each of the monomers at all times is pretty untractable so insteadwe only consider the average of the equalibrium i.e.

〈Φ〉 = 0 (7.56a)〈R〉 = x(n) (7.56b)⟨∂x

∂t

⟩= 0 (7.56c)

where we have defined the length of the chain to have crossed the barrier to bex. Now the Rouse model reduces to

0 = kd2

dn2x(n)− d

dxV (x) + 0

kd2x

dn2=dV

dx. (7.57)

This steady equalibrium can happen in a couple of ways. A stable (trivial) caseis when the entire chain is in one of the wells but there is an unstable solutionwhere the chain straddles the barrier. Although we will only really cover an endover the barrier there must also be a solution for a hernia over the barrier andso on. But it the single end case will give us an idea of the activation energyrequired for translocation. It is pointed out in the review that Eq. (7.57) looksexactly like a particle in a potential with f = ma = mx where k m andn t.

The important thing is that you can find the energy by integrating the springforce

Uspring =

∫kd2x

dn2dx =

∫kd2x

dn

dx

dn=

∫k

(dx

dn

)d

(dx

dn

)=k

2

(dx

dn

)2

(7.58a)

U = Uspring − V

=k

2

(dx

dn

)2

− V (7.58b)

In equalibrium the two forces balance and U = 0 so that

V =k

2

(dx

dn

)2

(7.58c)

or

dx

dn=

√2V

k(7.58d)

You can also find the total free energy F . The total is the free energy fromeach bead at it’s potential V and it’s corresponding stretching. So replacing the

71

summation over every bead with integration gives the free energy

F =

∫ N

0

Uspring + V dn

=

∫ N

0

k

2

(dx

dn

)2

+ V dn.

If we assume equalibrium then we can insert Eq. (7.58d) to get

F =

∫ N

0

k

2

(√2V

k

)2

+ V dn =

∫ N

0

2V (x)dn

=

∫ x(N)

x(0)

2V (x)

(dn

dx

)dx =

∫ x(N)

x(0)

2V (x)

(√k

2V

)dx

=

∫ x(N)

x(0)

√2kV dx. (7.59)

From here we can find the activation energy ∆Fa. This is the free energyrequired to extend the end of the polymer some critical distance xc past thebarrier. We assume that the first polymer x(0) is at the very minimum of thefirst well x0 so

∆Fa =

∫ xc

x0

√2kV dx (7.60)

This is the result. . . and it’s kind of vague and dissappointing. At this point youneed a form for V (x) and you only get the activation free energy as a functionof xc which requires solving (it can be found by setting V (xc) = V (x0) = 0 andhaving the local maximum of the barrier be ∆V ) but the important conclusion isto compare this to how much energy is required to move N unattached particles(i.e. not a Rouse chain) over the barrier ∆V . That costs N∆V . On theother hand, the Rouse chain only needs to extend the head of the chain acrossthe barrier to some critical xc and then the translocation is guaranteed i.e.stretching greatly reduces the effective barrier height.

Notice that for a hernia the free energy is just twice that of an end since it’sexactly twice the free energy and twice the stretched springs.

7.9 Polymer-Obstacle CollisionsIn § 7.8 we considered in a very deterministic (read equalibrium/averaged)way how a polymer translocates a potential barrier. In this section we treat apolymer colliding with an obstacle considering more of the dynamics.

We can think of two ways which are subtly different in which a polymer canbe moved to collide with an obstacle:

1. apply an external force to each monomer

2. put the polymer in a flow.

We’ll consider these in two separate subsections.

72

7.9.1 Force Induced Collision

There are four stages to polymer-obstacle collision:

1. Free-Flow Approach

2. Deformation

3. Escape

4. Relaxation.

We treat them one at a time.

1. Free-Flow Approach Let each monomer feel a force f0 and so the totalforce on the polymer is Nf0. Assuming that the polymer is in an over-damped/viscous fluid it will be moving at a constant velocity v0 which isgiven by Stokes law

v0 =Nf0

ζ(7.61)

where ζ is the friciton coefficient and is different for the Rouse and Zimmmodels. To be repetative the Rouse model treats each monomer as havinga friciton coefficient ζ0 = 6πηb ∼ ηb and since the Rouse model neglectshydrodynamic interaction the total friction coefficient is ζR ∼ Nζ0. Onthe other hand, the Zimm model treats the polymer as a single solid objectof radius Rg so that the friciton coefficient is ζZ ∼ ηRg.

2. Deformation In the free-flow approach the polymer is roughly sphericalbut as it’s forced over the obstacle eventually it takes a pulley-like confor-mation. Of course, it doesn’t need to be two linear threads. It can be twotangles but for simplicity let’s say that we have a completely stretchedpolymer. If the polymer’s contour length is L then we say that one armhas a length `(t) and the other has a length L− `(t) at some time t wherewe pick `(t = 0) = `0 < L/2.

The question here is “How long does this deformation take?” Well, thepolymer is moving at a velocity v0 so each monomer is moving at approx-imately v0 and so if we assume that on collision the ends continue movingat ≈ v0 for the entire deformation then the time scale of deformation isapproximately

τd ∼L− `0v0

(7.62)

where we used L − `0 since it is the longer of the two arms. τd is mostdefinately less than the time scale of escape which we talk about next.

3. Escape The two arms are each in a field and they are competing. theequation of motion is the balance of the forces on the two arms. One armhas length ` (or is made up of `/b monomers). The second arm is longerand has length L− `. The force on the first ` of this longer arm balancesthe shorter arm. The force on the extra L − 2` is the net force (which is

73

n = (L− 2`) /b monomers. We use Eq. (7.61) to give us an idea of themotion:

v = −d`dt

=nf0

ζ∩=f0

ζ

L− 2`(t)

b. (7.63)

At this point the question becomes “What’s the friction coefficient of then monomers in the pulley conformation, ζ∩?”

Rouse Since each monomer feels the drag independently, in the Rousemodel the friciton coefficient doesn’t change with conformation so isstill the drag of all the monomers

ζ∩R = ζR = Nζ0. (7.64a)

Zimm Consider the chain to be a solid object. It’s clearly not a spheresince the chain is so extended but perhaps modelling it as a cylinderisn’t a bad idea. The hydrodynamic radius RH of a cylinder (alongit’s axis) of length L and radius b is

RH(L, b) =23L

ln (2L/b)− 0.5.

It’s pretty clear why we set the the radius of the cylinder to b butit’s a little bit trickier to justify the length of the cylinder to beL since the chain is bent around the obstacle. What we’ve doneis assumed that the arms are far enough apart (the obstacle is bigenough) that there is little hydrodynamic coupling between the armsas the chain moves with velocity d`/dt. When this assumption ismade it is sensical to sum the friction terms (haha the chains ofmonomers hydrodynamicalling interact forming one body but thearms do not allowing us to just sum the friciton coefficients) whichgives approximately the hydrodynamic radius of a cylinder of lengthL. So the friction coefficient is

ζ∩Z = ζZ ≈ 6πηRH =4πηL

ln (2L/b)− 0.5(7.64b)

which we approximate as

ζ∩Z =4πηL

ln (2L/b)− 0.5≈ ηL

ln (L/b)=ηNb

lnN=

Nζ0ln (N)

=ζR

lnN(7.64c)

Now we can return to Eq. (7.63) and solve for `(t)

d`

dt= − f0

ζ∩L− 2`

b

d`

L− 2`= − f0

ζ ∩ bdt

−1

2ln (L− 2`) = − f0t

ζ∩b+ C0

`(t) =L

2− C1 exp

(2f0t

ζ∩b

)

74

The integration constant is given interms of `(t = 0) = `0 to be

`0 =L

2− C1e

0

C1 =L

2− `0

so that the length of the initial shorter arm at time t is

`(t) =L

2−(L

2− `0

)exp

(2f0t

ζ∩b

). (7.65)

But the really important value is an estimate of the escape time τe. Theescape time is the time at which ` = 0. Therefore, from Eq. (7.65) theescape time is

0 =L

2−(L

2− `0

)exp

(2f0τeζ∩b

)2f0τeζ∩b

= ln

(L/2

L/2− `0

)τe =

ζ∩b

2f0ln

(1

1− 2`0/L

). (7.66)

But of course `0 is different for every collision so there is a distribution ofescape times. If we average `0 over 0 to L/2 we find the average escapetime to be

〈τe〉 =

∫ L/20

ζ∩b2f0

ln(

11−2`0/L

)d`

L/2− 0

=ζ∩b

Lf0

∫ L/2

0

ln

(1

1− 2`0/L

)d`

=ζ∩b

Lf0

L

2=ζ∩b

2f0

=ζ∩

2f0

L

N. (7.67)

4. Relaxation At τe the particle is no longer hooked on the pulley but itis in a highly extended state. So as it drifts away from the obstacle itspeeds back up to v0 while it relaxes to it’s nondeformed equalibrium state.Apparently, the relaxation after collision is still not very well understoodbut I would suspect that this must happen on the relaxation time scale

τp =

τR Rouse ModelτZ Zimm Model

see Eq. (8.14a) -Eq. (8.14c) for the form.

75

7.9.2 Flow Induced Collision

When the motion of the chain is due to a flow field we still have the same fourstages to polymer-obstacle collision:

1. Free-Flow Approach

2. Deformation

3. Escape

4. Relaxation.

Free-Flow Approach Now the polymer is carried along by solvent flow withvelocity vs so

v0 = vs

Deformation The deformation time is still approximated by

τd ≈L− `0vs

Escape Here is where the difference lies. With an external force the frictioncoefficient was either Rousian (unchanged) or estimated by a cylinder aslong as the contour length of the polymer. Now we must estimate itdifferently.

Just as before the unbalanced force results from the L − 2` segment butnow that force is due to drag f+ = v0ζ(L−2`) where the friciton coefficentis a cylinder of hydrodynamic radius RH(L−2`, b) as given by Eq. (7.64b)textbfbut and (here’s the kicker) this drag force must work against thedrag force of the entire chain that is moving with a velocity d`/dt i.e.f− = −ζ(L)d`/dt. So then using Eq. (7.64c) the equation of motion is

f− = f+

−ζ(L)d`

dt= −v0ζ(L− 2`)

d`

dt= v0

ζ(L− 2`)

ζ(L)

= v0η (L− 2`) / ln (L− 2`)

ηL/ ln (L)= v0

(1− 2`

L

)ln

(L

b− L− 2`

b

)= v0

(1− 2`

L

)ln

(2`

b

)which we integrate to find the length at anytime... actually I failed atintegrating this and the review only gives the average escape time:

〈τe〉 =

(L

2vs

)(2 lnN − 1

2 lnN + 1

). (7.68)

Relaxation The polymer again accelerates to v0 = vs and relaxes as it movesaway from the obstacle.

76

7.10 Polymer-Polymer CollisionOne last complication: instead of considering a moving polymer (moving dueto an external force on each monomer) colliding with a stationary obstacle let’sconsider a moving polymer colliding with a initialy stationary second polymer.This will produce a double pulley plus the collision will most likely cause acentre of mass motion.

So let’s say that polymer A has a force Nf0 acting on it but polymer Bdoesn’t. Polymer B is being dragged by the effective post created by polymer Aand a post moving through stagnant fluid is indistinguishable from flow over afixed post. Therefore, this combines the effects from § 7.9.1 and § 7.9.2 . Onlynow, the contact point has a velocity vcontact. The total friction coefficient isapproximately

ζT ≈ ζA + ζB . (7.69)

The force can be broken into two components:

1. fcontact whcih generates the contact point velocity which is therefore givenby

vcontact =fcontactζT

=fcontactζA + ζB

. (7.70)

2. The rest of the force is used to unhook the two polymers which is

funhook = f0LA − `A(t)

b(7.71)

Gary doesn’t go any further. He only states that “This contact pointvelocity is defined self-consistently through the solution of the equationsof motion for the escape of the two polymers, because it depends on theinstantaneous lengths of short legs of the molecules. . . ”

77

8 Microfluidics: Dimensionless NumbersWhen Newton’s second law F = ma per unit volume is applied to a continuum(a fluid of density ρ) then the resulting velocity field of a Newtonian fluid uobeys the Navier-Stokes equation

ρDu

Dt= ∇ · σ + f (8.1a)

where D/Dt is the material derivative which gives the rate of change as thefluid particle moves both in time and along a it’s path through the field. Thematerial derivative is made up of the temperal term ∂u/∂t and also the spatialrate of change which is the convective term u · ∇u. The motion is a result ofthe force per unit volume acting on the fluid particle. The external force perunit volume is f and the stress is σ which includes both the pressure p and theviscous dissipation η∇2u (for a noncompressible, Newtonian fluid) which meansthat the Navier-Stokes equation becomes

ρ

(∂u

∂t+ u · ∇u

)= −∇p+ η∇2u + f . (8.1b)

One should notice that the stress can include sources other than viscousity suchas interfacial stress σc ∼ γ/R (where γ is surface tension and R is curvature).

The Navier-Stokes equation is the fulfillment of conservation of momentum ina continuum. The conservation of mass is manifested by the continuity equation

∂ρ

∂t+∇ · (ρu) = 0. (8.2a)

When the fluid is incompressible ρ is constant and the continuity equation re-duces to

∇ · u = 0. (8.2b)

The Navier-Stokes equation is not an easy equation to deal with but oneoften doesn’t need to consider the whole thing. By simply comparing terms,one can get an idea of the type of behaviour that will be exhibited or simplifythe governing equations making them more tractable.

8.1 Reynolds Number: Inertial vs ViscousConsider Eq. (8.1b) :

inertia︷︸︸︷ρ

∂u

∂t︸︷︷︸unsteadyacceleration

+ u · ∇u︸︷︷︸convection

=

stress︷︸︸︷−∇p︸︷︷︸pressure

+ η∇2u︸︷︷︸viscous

+ f︸︷︷︸body

(8.3)

fi ∼ fp + fν + f

78

The idea behind dimensionless numbers is to estimate a set of forces basedoff of characteristic distances, times, velocities, etc. and compare couples offorces to determine if any are irrelavent.

In microfluidics, the inertial force is nearly always irrelavent. Let the fluid bemoving with a characteristic velocity U in a space characterized by the lengthscale L. The steady inertial force density is

fi = ρu · ∇u

fi ∼ ρUU

L

≈ ρU2

L. (8.4a)

The inertial force should be compared to the viscous force since both are alwayspresent in flows:

fν = η∇2u

fν ∼ ηU

L2(8.4b)

The ratio of the inertial to the viscous force densities is called the Reynoldsnumber

Re =fifν∼ ρUL

η. (8.5)

When Re 1 the nonlinear terms in Eq. (8.1b) can be neglected and theNavier-Stokes equation reduces to the linear Stokes equation

ρ∂u

∂t= −∇p+ η∇2u + f . (8.6)

This is obviously a very useful simplification. Notice that we called the said“inertial force density” but then used the convective foce density ρu · ∇u anddid not include the temperal term. We did this because we wanted to be ableto handle unsteady flows. If we consider low Reynolds numbers then we canestimate the inertial time scale by balancing the unsteady inertial force densitywith the viscous force density. The inertia time scale τi is not nesassarily theconvective time scale τc ∼ L/U but rather is the time required for the viscousforces to dissipate the inertia

ρU

τi∼ ηU

L2

τi ∼ρ

ηL2. (8.7)

The form of Eq. (8.7) suggests what this time actually is: It is the time requiredfor the vorticity to diffuse a distance L with a diffusion coefficient η/ρ.

8.2 Péclet number: Convection vs DiffusionIn the last section we weighed forces. The Péclet number is defined as the ratioof the rate of convection to the rate of diffusion. By the random walk nature ofdiffusion

τD ∼ L2/D (8.8a)

79

and the convective motion is

τc ∼ L/U. (8.8b)

The ratio of the two times is the Péclet number

Pe =τDτc∼ LU

D. (8.9)

Clearly, this is a very important number in separation science since it gives theralative importance of convection to diffusion. Notice for instance, that in § 3the equation (Eq. (3.27) ) for the theoretical number of plate heights N is trulya Péclet number:

N =w

H=Vw2D

=UNLN

2D∼ PeN .

In fact, so is the retention parameter λ:

1

λ=w

`=

w

D/ |W |=w |W |D

=UλLλD

∼ Peλ.

The important thing to keep in mind here is that they are Péclet numbers fordifferent things. The number of theoretical plates is a Péclet number for thesolute moving in the flow parallel to the plate while the retention parameter isthe (inverse) Péclet number for the particles in the field moving perpendicularto the wall. Each process has a different velocity scale U , length scale L andpotential diffusion coefficient D (if shear spreading of the zone occurs, see § 3 ).

8.3 Capillary Number: Viscous vs InterfacialNot all fluids are immiscible. When they aren’t there is a surface tension γacting on the interface. The surface tension γ is the energy per unit area orequivalently the force per unit length. In microfluidic devices where surfaces aresmall interfacial forces can be quite large. If the length scale of the surface is Lthen the interfacial force density is

fγ =γ

S

fγ ∼γ

L2(8.10a)

almost by definition of the surface tension. Whether or not the surface tensionplays a significant role is primarily determined by it’s strength relative to thedissipating viscous force density

fν = η∇2u

fν ∼ηU

L2.

The ratio of the two is called the capillary number

Ca =fνfγ∼ ηU

γ. (8.11)

80

This is almost always equivalent to the ratio of the length scales that the vis-cosity and the surface tension act over. If surface tension creates droplets ofradius ∼ R in a channel of length z then

Ca ∼ z

R(8.12a)

or if a long droplet of length L is moving from a hydrophobic to a hydrophilicsurface (with a difference ∆γ) in a tube of width w then the capillary numberis

Ca ∼ w

L. (8.12b)

8.4 Deborah number: Relaxation vs FlowThe next two dimensionless numbers give an idea of the importance of elasticeffects. Consider an elastic object in a flow, say a polymer. The polymer willhave some intrinsic relaxation time τp (in fact it will have many different modesas discussed in § 7.2 but we’re talking about the longest one here). The longestrelaxation time of polymers is the time scale required for the polymer to diffuseover it’s own size with the diffusion coefficient is given by the Stokes-Einsteinequation to be

D ∼ kBT

ζ(8.13)

where ζ is the friction coefficient i.e. the low Re drag is f = ζv = 6πηRv.The relaxation time is thus

τp ∼R2

D∼ ζR2

kBT. (8.14a)

Rousian dynamics considers the friction coefficient of the polymer to be thesum of the friction coefficient on each bead ζ = Nζ0 which means that theRouse relaxation time is

τR ∼ Nζ0R2

kBT∼ NηbNb

2

kBT=

ηb3

kBTN2 (8.14b)

we’ve taken ν = 1/2 for the solvent quality because hydrodynamics interactionsare being neglected (or screened more likely).

On the other hand, Zimmian dynamics considers the friction coefficient ofthe polymer to be the friction coefficient for a hard sphere the same size as thepolymer ζ = ηR which makes the relaxation time

τZ ∼ζR2

kBT∼ ηR3

kBT. (8.14c)

The time scale of the fluid’s motion is simply the convective time scale thatwe considered earlier

τc = L/U

81

and the Deborah number is the ratio of the two

De =τpτc∼ ζR2U

kBTL(8.15)

where τp can be either Eq. (8.14b) or Eq. (8.14c) depending on which is moreappropriate.

If De < 1 then the time scale of the polymer is short compared to the flowand the polymer can relax i.e. the polymer is not deformed by the flow. IfDe > 1 then the flow may deform the configuration of the polymer in the timescale of the experiment. In general, for De 1 means that the material flowson the time scale under consideration while materials with De 1 should beconsidered rigid.

8.5 Weissenberg number: Relaxation vs Shear RateIn many ways the Weissenberg number is a specific example of a Deborah num-ber, although people do like to fight about this. In simple, linear shear flowthe time scale of the fluid must be the inverse of the shear rate γ therefore thedimensionless comparison between the relaxation time scale of the polymer andthe convective time scale becomes what is called the Weissenberg number

Wi = τpγ. (8.16)

We make use of the Weissenberg number extensively when dealing with tetheredpolymers in a deforming shear flow in § 7.7 .

8.6 Elasticity number: Elastic vs InertialThe Deborah number increases linearly with the velocity scale indicating thatthe elastic effects are becoming more important. However, the inertial effectsare also increasing. In fact, the Reynolds number also increases linearly withU . So the elasticity should be measured by an Elasticity number

El =DeRe

=τp/τcρUL/η

=τpη

LU ρUL

=τpη

ρL2(8.17)

which gives the relative importance of elasticity to inertia. Notice that it’sindependent of flow rate but rather determined by the geometry i.e. L−2.

8.7 Rayleigh and Grashof numbersThese numbers are kind of dumb but they have historical importance so theyeach get their own name. Consider a tracer driven flow i.e. you have twofluids with differing densities where the density difference is completely due toa concentration of particles in one of the fluids

∆ρ = ρ0α∆c (8.18)

where ρ0 is the density of the particle free liquid ∆c is the solute concentrationfield and α is a coefficient. The exaqct same system can be considered with atemperature gradient ∆T instead and then α becomes the coefficient of thermalexpansion.

82

What is the Péclet number for this gravity driven situation? We rememberthat

Pe =τDτc∼ L2/D

L/U=LU

D(8.19)

But because it’s gravity driven we know a little more information about U . Itis the quasisteady velocity that is achieved by balancing buoyency and viscousdrag:

fν ≈ fbηUgL2∼ ∆ρg

Ug ∼∆ρgL2

η=ρ0α∆cgL2

η

When the terminal velocity is used then the Péclet number becomes the Rayleighnumber

Ra ∼ τDτc,g∼ LUg

D=ρ0α∆cgL3

ηD. (8.20)

So Ra is useful for comparing convective and diffusive effects in situationswhere variation in density give information about the velocity scale. But if youare unsure if the convective effects or the viscous effects are dominant you needanother number i.e. a Reynolds number specific for varied density. This numberis called the Grashof number

Gr =fifν∼ ρ0UgL

η=ρ2

0

η2α∆cgL3. (8.21)

8.8 Schmidt numberIn the last section, I noted that the Rayleigh and the Grashof numbers are littlemore than specific cases of the Péclet and the Reyonlds numbers respectively.But if we think hard about what a Reynolds number truly is we realize thatit’s pretty much a Péclet number too. It tells you how the inertia dissipates inthe viscous media so you can think of ν = η/ρ (the kinetmatic viscosity) as adiffusion coefficient for vorticity.

Pe =UL

D→ Re =

ρUL

η=UL

ν(8.22)

The ratio of the two characterizes the dissipation by diffusion and viscosity.The ratio is called the Schmidt number

Sc =PeRe

=RaGr

=ν

D. (8.23)

8.9 Knudsen numberWe have been considering the flowing material in the microfluidic device to bea fluid (or a gas, although we didn’t allowed for compression in Eq. (8.1b) ) but

83

as the size scale of the device gets smaller and smaller the number of moleculesin device goes down too. At some point we must question whether the fluid canactually be said to be a continuum or whether it must be viewed as an ensembleof individuals. Clearly, the size scale is the primary concern here. The size scaleof the particle-particle interactions should be on the order of the mean free pathlength λf ∼ na2 where n is the number density of the molecules with radius a.The ratio of the two

Kn =λfL

(8.24)

is called the Knudsen number.As the Knudsen number becomes larger, noncontinuum effects become more

prevalent. Perhaps the most important effect is the violation of the no-slipcondition. The no-slip condition states that fluid in contact with a surface willmove with that surface. This enjoys statement has almost universal applicationas a axiom of fluid mechanics. The reason for this is a touch unclear. Obviouslythere is friction due to surface roughness. Furthermore, it’s been suggested thatmolecules of the fluid can adsorb to the surface creating an immobile layer ofthe fluid molecules but that can hardly be universal. What it must come downto is whether the reflection of the molecules comprising the fluid is specular(reflection) or reverse (bounce back) reflection.

What is clear is that as the we approach the wall, the length scale L decreasesand the fluid can only be said to truly behave as a continuum a distance on theorder of λf from the wall. Near the wall it is often a question of wetting (cohesionvs adhesion). Wetting surfaces (hydrophilic) tend to obey the no-slip conditionwhile clean, non-wetting (hydrophobic) surfaces often exhibit some slip.

If β is the slip-length (of order λf ) the boundary condition becomes

u|0 = βdu

dn

∣∣∣∣0

. (8.25)

We can decide if this complication is worth while by an analogous Knudsennumber

Kn =β

L. (8.26)

84

9 Microfluidics: Driving Flows at BoundariesWe finished § 8 by considering the no-slip condition in the context of the dimen-sionless Knudsen number. Since flows have such long-range, the importance ofboundaries on flows can not be overstated. The boundary conditions determinethe flow. And in microfluidics, the boundaries are never far away. The followinggives some examples of boundary effects driving the flow.

85

10 Attractive Interactions Between ColloidsIn this section we discuss forces of attraction between colloids. In the nextsection (§ 11 ) we will discuss repulsive forces and in § 12 .

10.1 Intermolecular AttractionsWe consider the types of attractions that can occur between molecules. Theyare all dipolar in origin.

10.1.1 Keesom Interaction

Perminant-Dipole/Perminant-Dipole Attraction Two perminant dipoleshave an attractive force and a pair potential

uK(r) = −CKr6

(10.1)

where r is the intermolecular separation and CK is a constant which goesas

CK ∝ µ4

if both molecules have the same dipole moment µ.

10.1.2 Debye Interaction

Perminant-Dipole/Induced-Dipole Attraction One molecule is polar andthe other is non-polar but the one perminant dipole molecule polarizes theelectron cloud of the non-polar molecule inducing a moment and caussingan attractive pair potential

uD(r) = −CDr6

(10.2)

where the interaction constant is now a function of the dipole moments µand the polarizability α by

CD ∝ α1µ22 + α2µ

21

10.1.3 London Interaction

Induced-Dipole/Induced-Dipole In non-polar molecules the electron cloudfluctuates creating an instantaneous dipole that fluctuates. If two atomsare near each other the fluctuations tend to be coupled leading to tempo-rary dipole alignment and a pair potential

uL(r) = −CLr6. (10.3)

It turns out that for most colloids the London (or Dispersion) interactionis far more significant than the Keesom or Debye attractions. It is notexplained why but it is stated that water is the exception.

86

10.2 London ConstantSince Eq. (10.3) is going to be our primary molecular attraction let’s considerthe constant CL in some detail.

If it is appropriate to assume that the polarizability α occurs only at theionization energy/frequency νI then the London constant goes as

CL ∝ h(

νI,1νI,2νI,1 + νI,2

)α1α2 (10.4)

where h is the Planck constant.But we know that this is often a bad approximation since the polarizability

is truly some complex function of frequency, α(iν).Now I can’t do it but the answer can be found as follows:The first dipole has a potential:

ψ1(r) = − 1

4πε0p1 · ∇r−1 (10.5a)

so that if the second dipole moment p2 is placed at r from the first it has aninteraction energy

U = p2 · ∇ψ1. (10.5b)

But the first dipole moment fluctuates and so has an instantaneous value

p1 = pinst =1

2π

∫ ∞−∞

p(ω) exp (−iωt) dω (10.5c)

The second dipole moment is induced and related to the first through it’s po-larizability and the electric field

p2 = αε0E = αε0∇ψ1

= αε0∇(− 1

4πε0p1 · ∇r−1

)= − 1

4π∇(αp1 · ∇r−1

)= − 1

4π∇(α

[1

2π

∫ ∞−∞

p(ω) exp (−iωt) dω]· ∇r−1

)= − 1

8π2

∫ ∞−∞

α(ω)p(ω) · ∇∇r−1 exp (−iωt) dω (10.5d)

where I’m not sure how that last step worked but ok.So I guess the interaction energy is

U = p2 · ∇ψ1 = p2 · ∇(− 1

4πε0p1 · ∇r−1

)= − 1

32π3ε0

∫ ∞−∞

α(ω)p(ω) · ∇∇r−1 · ∇∇r−1 · p(−ω)dω (10.5e)

apparently. And if you thought that this was a suckie derivation so far just youwait for the next bit. The only useful thing we can do is look at Eq. (10.5e) we

87

can already see the u(r) ∝ r−6 dependence which is of key importance. Weneed to know the magnitude of fluctuations in point dipoles p so he invokes thefluctuation-dissipation theorem to get a coth() magically which he mysteriouslyintegrates by the theory of residues to get a summation solution. When we allowfor the two molecules to be different and to be in a medium the pair potentialis

u(r) = − 1

r6

6kBT

(4πε0)2

∞∑n=0

′α1(iν)α2(iν)

ε23(iν)

(10.6)

where the prime over the summation denotes that the n = 0 term is divided inhalf and the summation indicies are n = νnh/ (2πkBT ). Comparing Eq. (10.6)to Eq. (10.3) , we see that the London constant is

CL =6kBT

(4πε0)2

∞∑n=0

′α1(iν)α2(iν)

ε23(iν)

(10.7)

where α1 and α2 are the polarizabilities of molecule 1 and 2 and ε3 is thepermiativity of the medium and it is a dielectric function of frequency. It’s clearthat Eq. (10.4) would be nicer to use if possible.

10.2.1 Polarizability

What kind of function is α(ν)? It will have a perminant alignment term αa whichwill depend on the dipole moment and the rotational frequency of the moleculeνR plus an electric term αe which will depend on the ionization frequency νI .The book proposes

α(ν) = αa(ν) + αe(ν) =µ2

3kBT (1− iν/νR)+

αe

1− (ν/νI)2 (10.8a)

which of course gives

α(iνn) =µ2

3kBT (1 + νn/νR)+

αe

1 + (νn/νI)2 . (10.8b)

The book substitutes α(iνn) into Eq. (10.6) then recalling n = νnh/ (2πkBT )replaces the summation with integration by dn = (h/2πkBT ) dν then does theintegration. In the end, CL is still just a constant.

10.3 Dispersion ForcesSo far we have only considered the pair potential between two point particles. Torecapitualate the pair potential in colloids is dominated by the London potentialuL(r) = −CLr−6 where the London constant can be a complicated function ofperminent dipole moment µ, the dielectric properties of the medium ε3 andthe polarizability α of both the molecules which itself can be a function of allfrequencies. But in the end, it’s still just a constant.

And that’s the saving grace. This fact will allow us to find the interactionpotential of groups of molecules. We call the interaction potential of groupsof molecules Φ. To start consider a single molecule interacting with a slab.

88

Throughout this, we will only consider interacttions to be pairwise as usuallymust be done. The total interaction energy is the sum of all the pairwise inter-actions

Φ =∑i

u(ri) (10.9)

but to do this we would need to know ri for every particle. Since this is impos-sible instead we take a semi-continuum approach and assume that the slab hasa uniform number density ρN . Then we can say the interaction potential is

Φ =

∫Vslab

ρNu(r)dV (10.10)

i.e. we integrate over the volume of the slab.If instead we have two generic volumes then the total energy is the sum over

all moluecules in volume two of the sum over all the molecules in volume oneoff all pairwise interactions or

Φ =

∫V1

∫V2

ρN,1ρN,2u(r)dx1dx2 (10.11)

= −ρN,1ρN,2CL∫V1

∫V2

dx1dx2

r6

= −A132

π2

∫V1

∫V2

dx1dx2

r6(10.12)

where we have bunch all of the material constants of body 1, body 2 andthe medium 3 into a constant called the Hamaker constant. We’ll discuss theHamaker constant in a bit but first we find the dispersion interaction energy forsome important geometries. Until then we’ll assume it’s all in a vacuum andwrite A12. Also I’m not doing the integral so I’ll give the general forms andwe’ll find specific cases from them.

10.3.1 Plates

Two plates of thickness t1 and t2 that are separated by a distance H has adispersion potential per unit area of

Φ

area= −A12

12π

[1

H2+

1

(H + t1 + t2)2 −

1

(H + t1)2 −

1

(H + t2)2

]. (10.13a)

If we take the plates to be the same thickness then

Φ

area= −A12

12π

[1

H2+

1

(H + 2t)2 −

2

(H + t)2

]. (10.13b)

and if we take the plates to be semi-infinite slabs then

Φ

area= −A12

12π

[1

H2+ 0− 0

]= −A12

12π

1

H2(10.13c)

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10.3.2 Spheres

If two spheres of radii a1 and a2 have a centre-to-centre distance r then theirsurfaces are separated by H = r − 2a and the dispersion interaction is

Φ = −A12

6

[2a1a2

r2 − (a1 + a2)2 +

2a1a2

r2 − (a1 − a2)2 + ln

(r2 − (a1 + a2)

2

r2 − (a1 − a2)2

)](10.14a)

If the spheres are the same size a then the potential reduces to

Φ = −A12

6

[2a2

r2 − 4a2+

2a2

r2+ ln

(r2 − 4a2

r2

)](10.14b)

Russel and I get this but Goodwin get’s something that doesn’t make any sense(it doesn’t have the same units for instance). Another thing he says is thatwhen spheres are close (H a) then the dispersion energy is

Φ ≈ −A12

12

a

r − 2aH a. (10.14c)

I don’t really see this but it’s probably correct.Furthermore, and I don’t know if he does solves Eq. (10.12) again or if he

takes the limit of a large particle but he says that the interaction energy betweena sphere of radius a and a plane is

Φ = −A12

6

a

H(10.14d)

where the Goodwin doesn’t say it but I suppose that H is the distance betweenthe surfaces (H = r − a I guess).

10.3.3 Cylinders

There’s also answers for cylinders but they’re not as essential as Eq. (10.13a)and Eq. (10.14a) .

10.4 Hamaker ConstantThe Hamaker constant is just a constant but it does hold a wealth of information.From Eq. (10.12) it is clear that

A132 = π2ρN,1ρN,2CL =3kBT

8

ρN,1ρN,2ε2

0

∞∑n=0

′α1(iν)α2(iν)

ε23(iν)

(10.15)

but we don’t want to calculate this everytime. In fact, most of the time we justwant a reasonable value that we can pull out of a table of materials.

The problem with that is tabulated values are for material-material pairs(i.e. 1 and 2) and don’t take the medium into account. The medium redicesthe net interaction. To make up for this there is a certain rule of thumb forestimating an effective Hamaker constant. The effective Hamaker constant is

90

j&%'$

&%'$&%'$

Figure 6: Schematics explaining excluded volume.

the sum of the particle-particle and the medium-medium interactions in vacuumminus the cross terms i.e.

A132 = A12 +A33 −A13 −A23. (10.16)

But to make things even worse often tables don’t even have cross terms (there’sjust so many permutations) so an estimate for cross Hamaker constants is

Aij ≈√AiiAjj (10.17)

Therefore putting Eq. (10.17) into Eq. (10.16) an estimate of the effectiveHamaker constant is

A132 ≈(√

A11 −√A33

)(√A22 −

√A33

). (10.18)

10.5 Depletion InteractionBefore moving onto repulsive interactions, let’s talk about a relatively weak butfairly common force. Consider a solution of two colloids. The larger colloidshave a radius a and the smaller colloids have a radius of Rg (because they arealmost always polymers). We model both colloids as hard spheres. Imaginethat two of the larger colloids have a centre-to-centre distance of r = 2a−H andthat H > 2Rg. There is a volume between the big spheres that is inaccessibleto the smaller particle.

Now think a moment about osmotic pressure. If you have a solution of largecolloids with a membrane that only they can cross splitting the volume in halfand on one half there are smaller colloids too then the small colloids producean osmotic pressure on the larger colloids to move towards the region wherethe smaller particles can not be. The same thing happens here. The depletedregion between the two close spheres forms a solution without polymers andtherefore there is an osmotic force pushing the colloids together. When can findan effective pair potential through the osmotic pressure

Φd = −∫H

ΠA0 = −ΠV0 (10.19)

where V0 is the volume of liquid that is not accessable to the polymer-colloidand Π is the osmotic pressue. The osmotic pressure is the number density ofsmaller spheres ρp,1 time the thermal energy

Π = ρp,1kBT. (10.20)

What is the excluded volume? Well, the small colloid can’t get within RG ofthe large colloid. So there’s a sphere of radius a+Rg as shown in Fig. 6 . When

91

two of these spheres come together the form a lens which is twice the volumeof a spherical cap. A spherical cap of height h from a sphere of radius R has avolume of

Vcap =πh2

3(3R− h) (10.21)

(for use R = a+Rg). The only question is what’s the height of the cap. If youdraw in the large colloid radius a, the small colloid radius Rg, the centre-to-centre separation r and the cap height h in Fig. 6 (it’s too small for me to doit) then it is clear that

r = (a+Rg − h) + (−h+ a+Rg)

h = a+Rg −r

2.

So the excluded volume is

V0 = 2Vcap =2π

3

(a+Rg −

r

2

)2 [3 (a+Rg)− a−Rg +

r

2

]=

2π

3

(a+Rg −

r

2

)2 [2a+ 2Rg +

r

2

](10.22)

which means that the depletion potential is

Φd = −ΠV0

= −ρp,1kBT2π

3

(a+Rg −

r

2

)2 [2a+ 2Rg +

r

2

]. (10.23)

This can be writen in terms of the volume fraction/concentration of the smallerparticles and it can also be found for a sphere-plate and plate-plate geometries.

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11 Replusive Interactions Between ColloidsThe primary repulsive force is due to electrostatic interactions and we’ll discussthat in a moment but first I’d like to mention two other repulsions that aremore akin to the dispersive forces discussed in § 10

11.1 Born RepulsionWhen molecules get extremely close their electron orbitals begin to overlapcausing a strong repulsion called Born repulsion. The Born potential goes asuB(r) ∝ r−12. According to the text, this is a “found” result although I hadalways thought of it as somewhat arbitrary.

11.2 RetardationConsider the dispersion forces one last time. As the distance r between particlesgrows the time scale of field propogation becomes comparable to the time scaleof the oscillations i.e. we can no longer consider interaction to be instantaneous.This cause the correlation to degrade and thus free energy decreases at a greaterrate than expected if we continued assume interactions were instantaneous.

This isn’t really a repulsion but a weakening of the dispersion forces.

11.3 Electric Double LayerThe primary repulsion is electrostatic since most colloids have surface charge.But ions in the medium cause the interaction to not be Coulombic.

Here is a list of some assumptions and some vocabulary that we will be using:

• Uniformily charged surface

• System is neutral

• When a medium contains free ions it’s called an electrolyte

• Ions of the same charge as the surface are called co-ions

• Ions of the opposite charge as the surface are called counter-ions

• Some ions may interact with the surface and therefore vary the surfacecharge. These are called potential-determining ions

• Other ions have no direct charging role and are refered to as indiferentions.

11.3.1 Origins of Surface Charge

You may ask why are so many colloidal surfaces charged. The answer seems tobe because there are just so many ways for surface charge to originate that it’salmost bound to happen. Here are some examples.

Adsorbed Layers Surfactants coating the colloid are often ionic with the hy-drophobic heads at the interface and the ionic groups at the outer tail.

93

Ionogenic Surfaces Chemical moieties reside on the surface and can be easilyionized.

Isomorphous Substitution The colloid atoms are on a lattice but if someimpurity atom ends up on the lattice that has a different charge then someanion coordination will be incomplete resulting in a net charge. This canbe called entrapment of ions.

Dissolution The surface maybe contain ions that are slightly more soluablethan the other components and so will have an affinity for dissolving.

11.3.2 Electric Double Layer

However the surface charge arises, it sets up a potential which acts on themobile ions which of course changes the distribution of ions until the potential,diffusion and the counter-ions are in equalibrium. There are two main layers tpthe solution charge near the surface before it get’s to it’s bulk value.

Stern Layer This is the inner region: a monolayer (-ish) of counter-ions thathave adsorbed to a fraction of avalliable sites that changes the surfacecharge density σ0 to a lower value creating an effective surface density ofσδ. Of course, that changes the potential at the surface from Ψ0 to Ψδ.The Stern layer has a thickness of δ. It seems sort of ill-defined.

At some point slightly past δ the fluid molecules are again mobile. Thisis called the slipping plane. The electric potential at the slipping plane iscalled the zeta-potential (ζ-potential) and it is often said that

Ψ(δ) = Ψδ ≈ ζ (11.1)

since the ζ-potential is easier to measure.

Diffuse Layer Outside of the compact Stern layer the electrostatics must com-pete with diffusion until equalibrium is established. The potential is foundthrough the Gouy-Chapman model. This assumes that the ions are pointcharges of type-i each with a number density ni,0 in bulk and a valencyzi. The ion density of each type in the diffuse layer is estimated by aBoltzmann distribution to be

ni(x) = ni,0 exp

(−eziΨ(x)

kBT

). (11.2)

That means that the total charge density is

ρ =∑i

ezini(x). (11.3)

An important case is a symmetrical electrolyte i.e. z− = −z+ = −z andn−,0 = n+,0 = n0. In this case, the charge density takes a particularily

94

elegant solution

ρ(x) =∑i

ezini(x) = ez+n+ + ez−n−

= ez [n+ − n−]

= ezn0

[exp

(−ezΨ(x)

kBT

)− exp

(+ezΨ(x)

kBT

)]= −2ezn0 sinh

(ezΨ(x)

kBT

)(11.4)

which allows use to find the potential Ψ through the Poisson equation

∇2Ψ = − ρ

ε0εr. (11.5)

Substituting ρ into the Poisson equation gives

∇2Ψ =2ezn0

ε0εrsinh

(ezΨ(x)

kBT

)(11.6)

11.3.3 Debye-Hückel Approximation

The solution Eq. (11.6) of the Gouy-Chapman model (not to mention Eq. (11.2)and Eq. (11.3) ) is a complete discribtion of the diffuse layer but it’s not veryeasy to deal with. For small potentials the hyperbolic sine function can beexpanded as sinh(x) ≈ x to give the linear, second order differential equation

∇2Ψ =2ezn0

ε0εr

(ezΨ(x)

kBT

)= κ2Ψ(x). (11.7)

This is called the Debye-Hückel approximation and we have made the definition

κ ≡

√√√√∑i

n0 (ezi)2

ε0εrkBT. (11.8)

The inverse of this parameter κ−1 is called the Debye length of the double layer.We’ll look at two geometries for the Debye-Hückel equation.

As for boundary conditions at the interface to the Stern layer the potentialmust be

Ψ(x = δ) = Ψδ ≈ ζ (11.9a)

and far from the surface the potential must eventually go to zero

Ψ(x→∞)→ 0 (11.9b)

just like the slope

d

dxΨ(x→∞)→ 0. (11.9c)

95

11.3.4 Debye-Hückel Planar

For planar geometry Eq. (11.6) becomes

d2Ψ

dx2=

2ezn0

ε0εrsinh

(ezΨ(x)

kBT

)(11.10)

which actually has a solution which we’ll talk about in a second but first weconsider the Deybe-Hückel approximation (Eq. (11.7) ) for a plane

d2Ψ

dx2= κ2Ψ(x) (11.11)

which has the solution

Ψ(x) = A exp (−κx)

to which we apply the boundary condition Eq. (11.9a) to get

A = Ψδ exp (κδ)

and

Ψ(x) = Ψδ exp [−κ (x− δ)]= Ψδ exp (−κx) (11.12)

where we recast x− δ → x to be the distance from the Stern layer rather thanthe distance from the surface (I don’t know why).

It turns out to be possible to get a solution for a plane without making theDebye-Hückel approximation. This solution is

Ψ(x) =2kBT

zeln

1 + exp (−κx) tanh(zeΨδ4kBT

)1− exp (−κx) tanh

(zeΨδ4kBT

) (11.13)

The nice thing about this is that even for strong potentials we can get a sim-ple solution for long distances from the surface. Then the logarithm can beexpanded as

ln (1− x) = −∞∑n=1

xn

n≈ −x

ln (1 + x) =

∞∑n=1

(−1)n+1 x

n

n≈ x

to get (for high potentials tanh(x)→ 1)

Ψ(x) =4kBT

zeexp (−κx) tanh

(zeΨδ

4kBT

)≈ 4kBT

zeexp (−κx) (11.14)

which is exactly the same answer as Eq. (11.12) with an effective sur-face/Stern potential of Ψδ = 4kBT/ze. Cool.

96

With all of this information, we can predict the effective surface chargedensity σδ

σδ = −∫ ∞δ

ρdx = −∫ ∞δ

ε0εr∇2Ψdx

= −∫ ∞δ

ε0εrd2Ψ

dx2dx = −

∫ ∞δ

ε0εrd

(dΨ

dx

)dx

dx

= −ε0εrdΨ

dx

∣∣∣∣∞δ

= −ε0εr [−κΨδ exp (−κδ)]

= ε0εrκΨδ exp (−κδ) ≈ ε0εrκΨδ

≈ ε0εrκζ (11.15)

where we used the Debye-Hückel solution Eq. (11.12) , the boundary conditionEq. (11.9c) and the approximation Eq. (11.9a) . We could have just as easilysubstituted Eq. (11.4) with Eq. (11.12) in directly.

11.3.5 Debye-Hückel Spherical

We repeat the last section but for spherical colloid of radius a. The Poissonequation Eq. (11.6) in spherical coordinates using the Debye-Hückel approxi-mation is

∇2Ψ(r) =2zen

ε0εrsinh

(ezΨ(r)

kBT

)1

r2

∂

∂rr2 ∂

∂rΨ ≈ κ2Ψ(r) (11.16)

which has the solution (if we assume a thin Stern layer δ a)

Ψ(r) = Ψδa

rexp [−κ (r − a)] (11.17)

11.4 Diffuse Double Layer InteractionsIn the absense of a medium, the electrostatics would have been a Coulombicrepulsion between charged surfaces but the electrolyte interaction has screededthe ions in the double layer. Therefore, there is no repulsion directly due toCoulombic forces. But the concentration distribution of ions results in an os-motic pressure between the surfaces. We estimate the osmotic pressure as thepressure at the mid-point between the surfaces. The osmotic pressure is then

Π = kBT [(n+ − n0) + (n− − n0)] = kBT [n+ + n− − 2n0]

which using the Boltzmann equation (Eq. (11.2) ) becomes

Π = kBT

[n0 exp

(− zeΨkBT

)+ n0 exp

(+zeΨ

kBT

)− 2n0

]= 2kBTn0

[cosh

(zeΨ

kBT

)− 1

](11.18)

97

We apply the Debye-Hückel approximation which amounts to cosh(x) ≈ 1 +x2/2 + . . . to find the osmotic pressure at the midpoint between two plates inthe low field approximation to be

Π ≈ kBTn0

[zeΨ

kBT

]2

=(ze)2n0

kBTΨ2 = ε0εrκ

2Ψ2.

The potential Ψ should be the sum of the two independent potentials.

11.4.1 Diffuse Double Layer: Plates

So when we have two plates Ψ is given by Eq. (11.12) (notice we already usedthe Debye-Hückel approximation so we can go ahead and use it again) and theosmotic pressure becomes

Π = ε0εrκ2 [2Ψ(H/2)]

2= 4ε0εrκ

2 [Ψδ exp (−κH/2)]2

= 4ε0εrκ2Ψ2

δ exp (−κH)

The potential between the surfaces is the energy required to overcome the pres-sure in bringing the plates from infinity to H or

ΦDDL = −∫ H

∞Π(H)dH

= −4ε0εrκ2Ψ2

δ

[− 1

κexp (−κH)

]H∞

= 4ε0εrκΨ2δ exp (−κH) (11.19)

11.4.2 Diffuse Double Layer: Spheres

Presumably, it is quite easy to do the same thing for two spheres: We wouldonly need to substitute Eq. (11.17) into the osmotic pressure in Eq. (11.19). Unfortunately, the integral isn’t do-able (there is a series solution, I think).Instead, the review gives a could of limitting cases:

Thin Diffuse Layer κa > 10. When the diffuse layer (characterized by thelength scale κ−1) is small compared to the particle radius then the poten-tial (for constant potential) is

ΦR ≈ 2πε0εraΨ2δ ln [1 + exp (−κH)] (11.20)

This assumes the potential is held constant (which isn’t strictly true).He could have also assume the charge density stays constant (can’t haveboth). This is a pretty good approximation down to about κa > 2 whichoverlaps nicely with the next limit.

Comparable Diffuse Layer κa < 5. When the diffuse layer is a similar mag-nitude to the particle radius then if we still use the Debye-Hückel approx-imation the potential is

ΦR ≈ 2πε0εraΨ2δ exp (−κH) . (11.21)

Here (as with the plates) there is no difference between the constant Sternlayer potential and constant charge density results.

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11.5 Steric ReplusionImagine a surface with a polymer brush grafted on to it. Two such objectswill sterically repulse each other. In this section we will assume a mean fieldtype interaction based in a Flory-Huggins model. Notice that this is in markedcontrast to our discussions of polymer microstructures in § 5 and § 6 . In thosesections, we were quite strict about the fact that brushes have uniform densityby close packing of blobs which does not allow for interpenetrability. See inparticular § 5.4 . That only allows for the compression of interacting brushes.

Here, on the other hand, we do not consider compression at all but say thatpenetration is perfect, by which we mean that the polymer layer will overlapwith another, increasing the concentration and therefore creating a osmoticpressure. We use Flory-Huggins theory to estimate the osmotic pressure and soreview it very quickly here.

There is some set of species of type i each constituent of which takes up Nisites. The number of i like particles is ni. Each site has a volume v and thereare a total of n sites. The entropy of mixing per site is

∆s = −kB∑i

φiNi

lnφi (11.22)

where we sum over all the different i-types of molecules being placed on thelattice. Therefore, if we are only considering chains of polymerization N and avolume fraction φ and a fluid which then has a volume fraction 1− φ then theentropy of mixing is

∆s = −kB[φ

Nln (φ) + (1− φ) ln (1− φ)

]. (11.23)

The energy of mixing is the probability that they are beside one another∏φi =

φ (1− φ) (unconnected mean field theory) times some energy numner χkBT or

∆u = kBTχφ (1− φ) . (11.24)

We know that the Helmholtz free energy is

∆F = ∆U − T∆S = n∆U − nT∆s

= kBTχnφ (1− φ) + nkBTφ

Nln (φ) + nkBT (1− φ) ln (1− φ)

∆F

kBT= χnsφ+ np ln (φ) + ns ln (1− φ) (11.25)

where ns = nφ and np = nφ/N are the total number of solvent particles andtotal number of chains.

We can find the osmotic pressure on the polymers as either Π = −∂∆F/δVor since Π = −kBT∆µs/v since µs = ∂∆G/∂ns. Notice to take derivative mustreplace φ = Nnp/ (Nnp + ns) but anyway get

∆µskBT

= ln (1− φ) +

(1− 1

N

)φ+ χφ2 (11.26)

99

which means that the osmotic pressure is

Π = −kBT∆µsv

(11.27)

=kBT

v

[φ

N− φ− χφ2 − ln (1− φ)

](11.28)

≈ kBT

v

[φ

N− φ− χφ2 +

(φ+

φ2

2+φ3

3+ . . .

)]=

φ

N+

(1

2− χ

)φ2 +

1

3φ3 + . . . (11.29)

This can be done in terms of concentration of polymer c and molecular weightM :

Π =kBTc

M+ kBTB2c

2 + . . . (11.30)

where B2 is the second virial coefficient.Now that we have an estimate of the osmotic pressure (mean field/Flory-

Huggins approximation) in terms of a concentration or the volume fraction wecan estimate the interaction potential due to surfactants just like we did in § 10.5. Say the coating thickness is δ and the surfaces are separated some distance Hthen the surfactant potential is

Φs = −∫ h

2δ

ΠeAdx (11.31)

where A is the area and the subscript e on the osmotic pressure denotes that weare considering the excess osmotic pressure i.e. the concentration is c when theparticles are apart (h ≥ 2δ) and 2c when the surfactants overlap (h < frm−eδ)which means

Πe = −Πoverlap + Πseparate1 + Πseparate2 = 2Πseparate −Πoverlap

= 2Π(c)−Π(2c) (11.32)

Since it’s a mean field approximation the concentration is fixed and so theosmotic pressure is too. That means we only need to consider the volume in theintegration V0. So then

Φs = −∫ H

2δ

ΠeAdx = −V0Πe

= V0 [Π(2c)− 2Π(c)] (11.33)

We then apply Eq. (11.30) to get

Φs = V0

[kBT2c

M+ kBTB24c2 − 2

kBTc

M− 2kBTB2c

2

]= 2kBTB2c

2V0 (11.34)

and just like the last time we dealt with osmotic pair potential the questionbecomes one of what is the volume V0. Let’s assume the height δ of the coating

100

on both is the same. For two plates it is simply the unit area time the thicknessof the overlap h = 2δ − H. For a plane and a sphere it is a cap of heighth = 2δ −H and for two spheres it is a lens which is equivalent to two times acap of height h = a+ δ − r/2 = δ −H/2. As a reminder to § 10.5 the volumeof a cap of height h made from a sphere of radius R is

Vcap =πh2

3[3R− h] . (11.35)

101

12 Stability of Colloidal Dispersions

102

Indexχ parameter, see Flory’s Parameter

Agglomeration, see FlocculationAggregation

Micelles, 46Crewcut, 49Hairy, 47

Microphases, 46Aggregation distance, 15Aggregation rate, 16Alexander Brush, 37, 39

Bead Spring Chain, 51Blob theory, 19, 20, 22, 36, 37, 42, 54, 57,

59, 61Boltzmann distribution, 92Born repulsion, 91

Capillary number, 78Chemical potential, 7, 8chemical potential, 8Chromatography, 23Coagulation, see FlocculationColloids, 13Concentration Profile, 23Continuity equation, 76Contour length, 71Correlation Coefficient, 53Coulombic Interaction, 95Craighead device, 68Creaming, see Sedimentation, 13

Deborah number, 80Debye Interaction, 84Debye length, 15, 21Debye-Hückel approximation, 93–96Depletion interaction, 89, 90Depletion layer, 22Deybe-Hückel approximation, 94de Gennes self-similar carpet, 21Diffusion Coefficient

Apparent, 24Dispersion Forces, 84, 86–88Dispersion forces, 91Dispersion Retardation, 91DLVO theory, 15, 22Double layer

Diffuse layer, 92, 93Gouy-Chapman model, 92Stern layer, 92

Double-well potential, 68

Elasticity number, 80Electrolyte, 91

Co-ions, 91Counter-ions, 91Indifferent ions, 91Potential-determining ions, 91

Electrostatic forces, 14, 15Electrostatic interactions, 91Enthalpy, 17

FENE spring, 57Field-Flow Fractionation, 23

Flow Field-Flow FractionationAsymmetric, 31, 32Symmetric, 31, 32

Sedimentation Field-Flow Fractiona-tion, 26

Flocculation, 15Bridging, 21Depletion, 22

FloculationBridging, 22

Flory’s Parameter, 4, 6Flory-Huggins theory, 2, 3, 97Flow Field-Flow Fractionation, 31Freely Jointed Chain, 51, 55

Gibbs free energy, 6, 17, 55, 56Gibbs partition function, see Partition func-

tionGouy-Chapman model, 93Grand potential, 6Grashof number, 81

Hamaker constant, 14, 87, 88Helmholtz free energy, 6, 17Horn regime, 64, 68

Interparticle affinity constant, 16Ionic strength, 14

Keesom Interaction, 84Kelvin’s law, 19Knudsen number, 82

Langevin function, 57Lennard-Jones potential, 14, 15London Interaction, see Dispersion ForcesLondon-van der Waals forces, 14, 15

Navier-Stokes equation, 76, 77No-Slip condition, 82, 83

103

Normal Coordinates, 52

Osmotic effect, 22Osmotic pressure, 10, 11, 95

Péclet number, 77, 81Partial pressure, 11Partition function, 55Poisson equation, 93, 95Polarizability, 85, 86Polydisperse grafting, 45Polymer adsorption, 19

Rayleigh number, 81Retention Parameter, 27Retention Ratio, 26, 32Retention Time, 32, 33Reynolds number, 76, 77, 80, 81Rouse relaxation time, 52, 79Rousian Dynamics, 51, 68, 69, 79Roussian Dynamics, 71

Schmidt number, 81Sedimentation, 13Sedimentation speed, 13Shear flow, 64Slipping plane, 92Soret Coefficient, 35Stem and flower regime, 61, 66, 68Steric repulsion, 21Stern layer, 93, 94Stokes equation, 77Stokes law, 58, 59, 64, 71Stokes’ law, 13Stokes-Einstein relation, 16, 79Strain ensemble, 57Stress Ensemble, 54Surface tension, 17Surfactants, 19

Theoretical Plate Height, 29, 30Theoretical plates

Height, 28Number, 28, 78

Thermal Field-Flow Fractionation, 34Trumpet regime, 58

Velocity Distortion Factor, 35Void Time, 32, 33Vorticity, 77

Weissenberg number, 65, 68, 80Wetting, 82

Zeta-potential, 92

Zimm relaxation time, 58, 61, 79Zimmian Dynamics, 71, 79Zone Spreading, 28

Diffusive, 28Shear, 28, 34

104

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