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M e m b e r o f t h e H e l m h o l t z A s s o c i a t i o n Solutions of Exercises in An Introduction to Dynamics of Colloids

Jan Karel George Dhont, Kyongok Kang

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Schriften des Forschungszentrums JlichReihe Schlsseltechnologien / Key Technologies Band / Volume 65

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Forschungszentrum Jlich GmbHInstitute of Complex Systems (ICS)

Soft Condensed Matter (ICS-3)

Solutions of Exercises inAn Introduction to Dynamics of Colloids

Jan Karel George Dhont, Kyongok Kang

Schriften des Forschungszentrums Jlich

Reihe Schlsseltechnologien / Key Technologies Band / Volume 65ISSN 1866-1807 ISBN 978-3-89336-882-2

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Bibliographic information published by the Deutsche Nationalbibliothek.The Deutsche Nationalbibliothek lists this publication in the DeutscheNationalbibliograe; detailed bibliographic data are available in the

Internet at http://dnb.d-nb.de.

Publisher and Forschungszentrum Jlich GmbHDistributor: Zentralbibliothek 52425 Jlich

Tel: +49 2461 61-5368Fax: +49 2461 61-6103 Email: [email protected] www.fz-juelich.de/zb Cover Design: Grasche Medien, Forschungszentrum Jlich GmbH

Printer: Grasche Medien, Forschungszentrum Jlich GmbH

Copyright: Forschungszentrum Jlich 2013

Schriften des Forschungszentrums JlichReihe Schlsseltechnologien / Key Technologies, Band / Volume 65

ISSN 1866-1807ISBN 978-3-89336-882-2

The complete volume is freely available on the Internet on the Jlicher Open Access Server (JUWEL)at www.fz-juelich.de/zb/juwel

Neither this book nor any part of it may be reproduced or transmitted in any form or by anymeans, electronic or mechanical, including photocopying, microlming, and recording, or by anyinformation storage and retrieval system, without permission in writing from the publisher.

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PREFACE

This solution book provides solutions to exercises of the book of J.K.G. Dhont An

Introduction to Dynamics of Colloids as published in the Interface Science Series,

vol II, Elsevier press, 1996. As mentioned in the preface in the book, colloid science

is the domain of both chemists and (theoretical) physicists. The first chapter in the

book is therefore aimed to provide a minimum mathematical background for those

who did not have a sufficient mathematical training. A single chapter can of course

not fully cover this gap in the mathematical background between experimentalists

and theoreticians, and might render the exercises still problematic for those who are

not used to work with equations.

The aim of this solution book is to bridge this gap, by providing the mathematics

needed to solve the exercises in detail. We tried to give the solutions in a way that it

is also accessible for those who are less trained in mathematics. Every step is

worked out in detail. Some times we added remarks on the interpretation and

physical implications of results. In this solution book, exercises are selected that are

not purely concerned with mathematics, but where rather the understanding ofphysics is the aim. We hope that this exercise book will lower a possible activation

energy concerned with the mathematics involved in exercises.

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The content of the book is focused on equations of motion for Brownian systems (in

particular the Smoluchowski equation), hydrodynamics, light scattering, diffusion,

sedimentation, critical phenomena, and phase separation kinetics. The effects of

simple shear flow on various phenomena are presented throughout the various

chapters.

Some of the original figures in the book are re-arranged for illustration whenever

helpful within the context of an exercise. We also added some text to further explain

the physics on an intuitive level, and mentioned some typos in the book whenever

relevant within an exercise.

We sincerely hope that you enjoy solving the exercises with the help of this solution

book.

Kyongok Kang ( [email protected] )

Jan Karel George Dhont ( [email protected] )

June-July, 2013

Forschungszentrum Juelich,

Juelich, Germany

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Solutions of Exercises in An Introduction to Dynamics of Colloids

Exercises Chapter 1: INTRODUCTION

Looking down view of the surroundings at Les-Houches, France

1.1 The sedimentation velocity of a colloidal sphere in very diluted suspensions

is equal to,

where is the shear viscosity of the solvent, and the external force acting onthe colloidal sphere with a radius . The gravitational external force, corrected for

buoyancy, is equal to (with the specific mass of the colloidal particle, and thatof the solvent fluid)

The question is what the maximum size of a colloidal particle (silica) in water can be, such

that the particle displacement due to sedimentation is not larger than its own radius, duringan experiment of 1 sec . For the calculation, use that the viscosity of water is 0.001 ,the specific mass of water is 1.0 and that of amorphous silica particle is ~ 1.8 .

The gravitational force accelerat ion constant is .

Substitution gives,

0

0

16

ext S v F a

34 ( )3

ext p f F g a

0 ext F a

p f

2/ Ns m / g ml / g ml

29.8 / g m s

0 30 0

1 1 4

6 6 3ext

S p f v F g aa a

C ol l oi d a l S y s t e m s : M a t h e m a t i c a l P r e l i mi n a r i e s : S t a t i s t i c a l M

e c h a ni c s

1

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1.9 Interaction of two charged colloidal spheres (depicted in Fig. 1.1(a))

(a) Consider a small charged colloidal particle, located at the origin, in a solvent that containsfree ions. The electrostatic potential is related to the free charge density byPoissons equation as

where is the dielectric constant of the solvent, which, for simplicity, is assumed to be thesame as the colloidal particle.

The charge density is composed of two terms; one is from the colloidal particle at the originand the other is from the solvent that may have unequal locally unequal ion

concentrations .

Then

For simplicity, the colloidal particle is regarded as a point-like, as mathematically described by the delta function. The ion concentrations are related to the electrostatic potential by(assuming of point-like ions)

2 ( )( ) r

r

2 ( )( ) ( )S r Q

r r

0( ) exp ( ) j j jr ez r

( )r ( )r

( )Q r

( )S r

j

Fig. 1.1: The most common kinds of pair-interaction potentials for sphericalcolloidal particles: (a) the screened Coulomb potential (which will be treated inthis exercise), that is the DLVO potential with negligible van der Waals attraction,(b) an almost ideal hard-core interaction, (c) steric repulsion of long polymers in agood solvent, where the polymers are grafted to the particles surface, (d) short-ranged attraction of colloids grafted with polymers in a marginal solvent.

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with the number density of that species outside the double layer, where the electrostatic potential is zero. The local electrostatic energy of ions of species j is equal to, which is the energy in the electrostatic field generated by the remaining ions and thecolloidal particle.

The resulting non-linear Poisson-Boltzmann equation

cannot be solved analytically. For sufficiently small potentials (where is smallcompared to ), however, we can linearize the above equation.

Note that this is always true for large distances from the colloid, and is only true for all distances whenever the surface potential is sufficiently small.

Also due to electroneutrality, for large volumes of the system.

Linearization leads to the so-called linear Poisson-Boltzmann equation

This can be seen as follows

which is valid for

2 01( ) exp ( ) ( ) j j j j

Qr ez ez r r

2 2

2 2 0

( ) ( ) ( ),

,

exp( )

4

j j j

B

Qr r r

e z

k T

r Qr

r

0 0exp ( ) 1 ( ) s j j j j j j j j

ez ez r ez ez r

0

0 0 02 2

1 ( )

( ) 0 ( )

j s j j

j

j j j j j j

j j j

ez ez r

ez ez r ez r

0 j

( ) jez r

Bk T

0 0 j j j

Qez

V

( )r

( )( ) 1 j j

B

ez r ez r

k T

( ) jez r

4

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It is convenient to use Fourier Transformation to solve the linearized Poisson- Boltzmann equation (see exercise 1.5)

Fourier inversion thus leads to

Using the identity, where the angular integration are performed explicitly,

it is found that

From the residue theorem it is found that

so that it is finally found that the electrostatic potential is equal to

with

the inverse Debye-screening length.

(b) The Helmholtz free energy of a system of two particles and the free ions in the solvent isthe colloid-colloid pair-interaction potential. The pair-interaction force between the twocolloidal particles is , where U is the potential energy and S the entropy offree ions in solution. Within the linearization approximation discussed in (a), for particles witha given, fixed charge , the total electrostatic potential is the sum of each of the separatecolloidal particles

where are the position coordinates of the colloidal particles.

2 2 Qk k k

2 1

00 1

2 sin 2ik r ik r ikrx ikr ikr dk e d d e dx e e eikr

2 2 28ikr ikr Q k r dk e e

ir k

ikr r z dz e i e

z i z i

4

r Q er

r

2 2 0 j j

j

B

e z

k T

2

3 32 2 2 20

2 2

ik r ik r Q e Q k r dk dk dk e

k k

F U T S

Q

1 2( )t r r R r R

1,2 R

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The local electrostatic energy density is equal to , so that the total electrostatic

energy can be written equally both in real space and Fourier space as (see Exercise 1.4b)

For the non-interacting ions, the entropy is

where the probability distribution function (pdf) is equal to

with the configurational partition function equal to

Expansion the entropy up to quadratic order in the assumed small quantity

thus leads to

This can be seen from (here is not the colloid charge, but the partition function)

by using the expansions

22 23

1( ) ( )

2 2(2 )t t U dr r dkk k

11

1

exp ( )

( , ..., )( ,..., , , )

M

j t j j

M m

ez r

P r r Q N N V T

21 ( )2 t

r

2 2 21 1 1exp 1 , 1 , ln 12 1 2

x x x x x and x x x x

Q

1 1 1... ( ,..., ) ln ( ,..., ) B M M M S k dr dr P r r P r r

1 11

( ,..., , , ) ... exp ( )M

m M j t j

Q N N V T dr dr ez r

( ) j

B

ez r

k T

2

12

11

1 1ln ,

2 2

... ( ) , 1, 2

M M BM M

iM

i M j t j j

k I S V V I

V V

I dr dr ez r i

1 1 1

1

2 21

... ( ,..., ) ln ( ,..., )

... ln ,

1, ...

2

B M M M

B M

M M M

S k dr dr P r r P r r

ek dr dr QQ

Q V dR dR V dr dr

6

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Since we are interested in the change of the entropy with the relative position of the colloidal particles, i.e., , the thermodynamic entropy term can be omitted.

Furthermore since is a constant, it is also does not contribute to the pair-interaction force.Therefore the relevant entropy is

The pair-interaction potential is thus equal to

From the Fourier transform of the linearized solution of the Poisson-Boltzmann equation in (a), it is found that

The pair-interaction potential for electrostatic interactions through double-layer overlap isequal to

Where the integration is performed using the residue theorem. This is the screened Coulombor Yukawa potential.

1 2 R R

( ) j t jdr r

lnM M V V

1 2

1 2

2

1 2 3 2 2

2

1 2

1(2 )

4

ik R R

R R

Q eV R R dk

k

Q e

R R

2 2223

1( ) ( )

2 2(2 ) t t T S dr r dk k

2

2 2

1 2 3 ( )2 (2 ) t V R R U T S dk k k

1 2

1 2

0

2 2

1 1( )

2

1

ik R ik R r ikr ikr t

ik R ik R

Qk e e dre e e

k i

Qe e

k

Note that we assumed that the charge of the colloids is independent of their separation. This is strictly true when the degree of ionization of the chemicalgroups on the surfaces of the colloidal particles is close to 100%. For partial

ionization, the local electrostatic potential affects the ionization equilibrium andthereby the charge on the colloidal particles. Then the appropriate condition is a

constant surface potential rather than a constant charge.

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1.11 The effective interaction potential

The effective interaction potential is defined, for isotropic and homogeneoussystems, which are both translational and rotational invariant, as

where is the pair-correlation function. In this exercise, it is shown that the gradient ofthis effective potential with respect the distance r between two colloids is equal to theinteraction force between the two colloids, averaged over the positions of the remainingcolloidal particles.

From the definition of the pair-correlation function of

and where

is the one-particle pdf for a homogeneous system.

Then the pair-correlation function becomes

Now the effective potential can be expressed in terms of the volume and partition function as

where the 1st and 2nd terms are independent of the position and .

Thus with a space differentiation with respect to gives

( ) exp ( )eff g r V r

1 2 1 2 1 2 3 1 2 3, , , , , , N N N P r r P r P r g r r dr dr P r r r r

1 2 3, , , , N N e

P r r r r Q

( )eff V r

( ) g r

( ) g r

11

P r V

21 2 3 1 2 3

32

, , , , ,

eff

N N N

N V

g r r V dr dr P r r r r

dr dr eV e

Q

1 2 3( , ) 2 ln ln lneff N V r r V Q dr dr e

1r 2r

1 2 3

1 2 3

, , , ,3 1

, , , ,3

N

N

r r r r N

r r r r N

dr dr e

dr dr e

1 2 3, , , ,1 1 2 1 3( , ) ln N r r r r eff B N V r r k T dr dr e

1r

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Dividing both the numerator and denominator with the partition function gives

Here the is so-called the conditional probability distribution functionof the particle coordinates , for given positions of particles, 1 and 2.

Therefore is the force that particle 2 exerts on particle 1, averaged over the position coordinates of all other particles, 3, , N.

Q

1 2 3

1 2 3

, , , ,3 1

1 1 2 , , , ,3

/( , )

/

N

N

r r r r N eff

r r r r N

dr dr e QV r r

dr dr e Q

3 1 1 2 3

1 2

, , , ,

, N N N dr dr P r r r r

P r r

1 2 33 1

1 2

3 1 3 4 1 2

, , , ,,

, , , ,

N N N

N c N

P r r r r dr dr P r r

dr dr P r r r r r

3 4 1 2, , , ,c N P r r r r r

3 4, , , N r r r

1eff V

9

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1.12 The pair-correlation function for hard spheres

The hard sphere pair-interaction potential is defined as with a , radius of the hard core.

Using the definition of the Mayer-function

it can be verified that

( )( ) 1 0, 2 ,

1, 2 .

hsV r hs f r e for r a

for r a

0, 2( )

, 2hsr a

V r r a

( ) g r

0.1

( )hsV r

1cos / 42 2

23 1 3 2 3

0 0 ( / 2)cos

2 sinr a a

r

dr f r r f r r d d dR R

31 22 3

/ 4 / 2

4 14 8 3 , 4

3 16

a

r a r x

r r dx dR R a for r a

a a

Fig. 1.10 (a) The pair-correlation function to first order in concentration for hard-spheres,

with , (b) a sketch for hard-spheres at larger concentrations, and (c) for charged spheres with a long-ranged repulsive pair-interaction potential.

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To see this, we can use the spherical coordinate system as depicted above

Then the pair-correlation function for hard spheres for low concentrations is (see eqn. (1.56))

This function is displayed and plotted in the above figure, Fig. 1.10 (a) from the book.

2a

1cos / 4r a

1cos / 42 2 / 4 22 2

0 0 ( / 2)cos 1 / 2

221 33 3

3/ 2 2/ 4

33

2 sin 4

41 4 1 14 8 2 1

3 3 8 2

4 18 3

3 16

r a a r a a

r r x

R a

R r xr a

d d dR R dx dR R

ar r dx R a

a a r

r r a

a a

3

0 1

1, 4

1( ) ( ) ( ) 1 8 3 , 2 4

16

0, 2

hs

r a

r r g r g r g r a r a

a a

r a

11

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1.13 Number density fluctuations

A measure for the amplitude of the fluctuations of the microscopic density is its standarddeviation

where is the microscopic number density.

To show that

use that the microscopic density is defined as

First of all, notice that, for identical particles

with the macroscopic density. Similarly, by definition

Using that

for , while

for , and

and similarly for the other term, and using that the pdf for all position coordinates isnormalized, it is found that

2 ( , ') ( ) ( ) ( ') ( ')m m m mr r r r r r

1( ) ( , , )m N r r r r

2 ( , ') ( ) ( ') ( ) ( ') ( , ') 1r r r r r r r g r r

1( ) ( )

N

m j jr r r

1 1 11( ) ( ) ( , , ) ( ) ( )

N

m N j n jr dr dr r r P r r N P r r

21 , 1

1 1

1

( , ') ( ' ) ( )

( ) ( ' ) ( ') ( )

( ) ( ') ( , , )

[

]

N

N i ji j

N N

m i m ji j

m m n

r r dr dr r r r r

r r r r r r

r r P r r

1 1 2( ' ) ( ) ( , , ) ( , ') N i j ndr dr r r r r P r r P r r i j

1 1 1( ' ) ( ) ( , , ) ( ) ( ') N i j ndr dr r r r r P r r P r r r i j

1 1 11( ) ( ' ) ( , , ) ( ) ( ')

N

N m i n midr dr r r r P r r N r P r

( )r

22 1

1 1

( , ') ( 1) ( , ') ( ) ( ')

( ) ( ') ( ') ( ) ( ) ( ')m m m m

r r N N P r r NP r r r

N r P r N r P r r r

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Substitution of the definition (1.52) in the book of the pair-correlation function immediatelyleads to

provided that N is large.

Now define the phase function as

which is the number of particles contained in the volume V . Suppose that the linear dimensionof the volume V is much larger than the distance over which the pair-correlation functionattains its limiting value of 1.

Integration of the defining equation for the standard deviation gives

where N is now understood to be the above defined phase function. Hence by integration of

the above expression for the standard deviation

Since by assumption V is large as compared to the range over which the pair-correlationfunction tends to unity, and for a homogeneous system, thisleads to

where is so-called total correlation function and is the averagedensity. The total correlation function thus measures the amplitude of number fluctuations inlarge volumes.

Note that the relative standard deviation tends to zero for large systems.

1 1( , , ) ( , , ) ( ) N N mV V

N r r dr r r r dr r

2

2' ( , ') 1

V V

N N N N dr dr g r r

V

2 2/ N N N

221 4 ( )

N N dR R h R

N

1h g

1( , , ) N N r r

2

2

' ( , ') ( ) ( ) ' ( ') ( ')m m m mV V V V

dr dr r r dr r r dr r r

N N

2 ( , ') ( ) ( ') ( ) ( ') ( , ') 1r r r r r r r g r r

( , ') ( ' ) ( ) g r r g r r g R

/ N V

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1.14 This exercise is related to Chapter 4, where equations of motion are discussed.

The pdf for the position coordinate of a non-interacting Brownian particle at time t satisfies the following equation of motion (EOM),

The initial condition is , which specifies that the particle is located at theorigin at time .

In this exercise we evaluate the collective dynamic structure factor, defined as

The time-evolution operator for the particular case of non-interacting spheres underconsideration here, according to the above equation of motion, is equalto .

For , we have

since

where the last line defines the function , which function, for large volumes, becomesequal to the delta function, which is zero for . The structure factor for non-interacting

particles thus reduces to

For interacting particles, this will be defined later as the self-dynamic structure factor.For non-interacting particles the collective and self-dynamic structure factors are the same.

Written in terms of the Cartesian coordinates x, y and z, the time-evolution operator reads

20( , ) ( , ) P r t D P r t t

( , 0) ( ) P r t r

0 01( , ) * ( ) ( )cS k t t X t k X t k N

0, 1

1exp ( ) ( )

N

i ji j

ik r t r t N

exp ( 0) ( ) exp ( 0) exp ( ) 0i j i jik r t r t ik r t ik r t

1 1( , ) exp ( 0) ( )cS k t ik r t r t

2 2 2

0 2 2 2 x y z i k x k y k z ik r L e D e

x y z

( , ) P r t

0t

20

L D

i j

31 (2 )exp ( ) exp ( ) j j j V

V

ik r t dr ik r k V V

( )V k 0k

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It is readily verified that and similarly for y and z .

Hence

Repeating the operation n times leads to

so that

Hence from eqn. (1.67) with , and , we find that

where it is used that the pdf for non-interacting particles is equal to 1/V.

Here is a summary of the equations and results of this exercise:

Remember that all this refers to a very dilute suspension of spheres, where inter-colloidalinteractions are neglected.

20 exp expnn L ik r D k ik r

20 200 0

! !

n nn D k t Lt ik r n ik r ik r ik r

n n

t t e e L e D k e e en n

2 2 20 0 01 1

1 1 D k t D k t D k t ik r ik r sS dX e e e e dX eV V

20

20

21 1 0

( , ) ( , )

exp exp exp exp

( , ) exp ( 0) ( ) expS

P r t D P r t t

Lt ik r D k t ik r

S k t ik r t r t D k t

1ik r f e

1ik r g e

2

22

x y z x y z i k x k y k z i k x k y k z xe k e x

20 exp exp L ik r D k ik r

1 X r

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1.15 For non-interacting particles, the static structure factor is identically equal to 1for .

.We will first show this from the defining equation

of the structure factor. Since for

as already shown in exercise 1.14, only the terms with survive, which immediatelyleads to for .

The same result is recovered from the middle expression in eqn. (1.72)

Since

where is the delta function, it follows that for , again, .

Note that the last equation in eqn.(1.72) is obtained from

where the last integral (between the square brackets) ranges over all orientations of (thatis, the spherical angular coordinates), and is the unit vector along , and

, 1

1( ) exp

N

i ji j

S k ik r r N

( ) 1 ( ) ik RS k dR g R e

2 1cos0 0 1

sin sin 2 2 4ikR ikR

ik R ikR ikRx kRe edR e d d e dx eikR kR

32 0ik RdR e k

0k

i j

i j

exp exp exp 0i j i jik r r ik r ik r

0k

( ) 1 g R ( ) 1S k

( ) 1S k

( )k

2

0

( ) ( )ik R ik R RdR g R e dR R g R dR e

/ R R R

R R

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Exercises Chapter 2:BROWNIAN MOTION OF NON-INTERACTING PARTICLES

A dragonfly with a drag force in the oily solvent, taken at Juelich in Germany

2.1 Newtons equation of motion for a spherical Brownian particle is

Let us first integrate once to obtain an expression for the momentum in terms ofthe fluctuating force . In the first step, consider the equation of motion withoutthe random force

the solution of which is a single exponential

where is an integration constant. The method of variation of constant is based onmaking this constant a function of time (the constant is thus assumed to vary with time),in such a way that the full equation of motion is satisfied. Hence

( )dp p

f t dt M

( ) f t ( ) p t

dp pdt M

( )t

M p t Ae

A

( ) t M dp p dA t edt M dt

Di f f u s i v e T i m

e S c a l e : T r a n s l a t i on

&R

o t a t i on a l M o t i on s

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so that

and hence

This now immediately leads to

Since , a second integration is necessary to obtain an explicit expressionfor the position coordinate . Integration of the above expression for the momentumcoordinate leads to an expression for the position coordinate involving the double integral

The integration range in the plane is indicated in Fig.2.7 by the dashed area. The

vertical lines in the figure below indicate the new integration directions.

Fig. 2.7 The integration range in the (t, t) -plane

The integration ranges can be read-off the above figure.

"( " ')

0 0

" ' ( ')t t

t t M dt dt f t e

r t

( ')

0

( ) ( 0) ' ( ')t t t t M M p t p t e dt f t e

( )( )

t M dA t f t e

dt

'

0

( ) ( 0) ' ( ')t

t M A t A t dt f t e

( ' , ")t t

( ) ( ) / p t M dr t dt

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It follows from this figure that an interchange of the order of integration leads to

The final expression for the position coordinates in terms of the fluctuating force istherefore

This is the expression that is used in the book to calculate the mean squared displacement.

/ / ( ')0

0 10 1 ' ( ') 1

t M t M t t p t r t r t e dt f t e

" "( " ') ( " ')

0 0 0 '

( ')

0

" ' ( ') ' '' ( ')

' ( ') 1

t t t t t t t t t

M M

t

t t t

M

dt dt f t e dt dt f t e

M dt f t e

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2.3 A spherical Brownian particle with a radius of 100 nm and a mass density of 1.8is immersed in water, with a viscosity of 0.001 .

Use that the friction coefficient of a macroscopically large sphere is equal to ,to calculate the momentum relaxation time constant and its corresponding diffusivelength scale . Also calculate the time at which the mean squared displacement (MSD) isequal to the square of the radius of the Brownian sphere.

Plug into the numerical values of the given quantities

The temperature is taken equal to 300 K , and the value of the Boltzmann constant is.

Also, the specific mass is corrected for that of water, which is 1.0 .

Since the mean squared displacement is given by

so that the time at which the MSD is equal to the squared radius of the colloidis equal to

This numerical example illustrates the time-scale separation that was mentioned in the maintext of the book.

23

3 20

32

0

4 4(1.8 1.0) / * 0.8 / * * 100

3 3~ ~ 1.86 6 * 1*10 /

3 3~ 1.1 106

B B D

g ml a g ml nmM

t nsa Ns m

Mk T Mk T l a a a

2

0

( ) ( 0) 6 Bk T

MSD r t r t Dt t a

3

0 76

B

at ms

k T

/ g ml 2/ N s m

06 a /M

Dl

231.38 10 / Nm K

/ g ml

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2. 4 Brownian motion in an external force field

A constant force field is applied to a spherical Brownian particle, such as a gravitationalforce. The Langevin equation for this case reads

On the diffusive time scale, where , we canneglect the inertial force, so that

The average, stationary velocity resulting from the force field is equal to ,thus the Langevin equation can be rewritten as

This is precisely the original Langevin equation in the absence of a field, where now themomentum is taken relative to the drift velocity. Since in this co-moving frame the equi-

partition theorem holds, exactly the same analysis as without a field leads to

It follows that also in the present case with a constant force field

In order to find the probability density function (pdf) for the position coordinate, usingChandrasekhars theorem, the above Langevin equation on the diffusive time scale isintegrated once

Comparing this with eqn.(2.29) we have the identification with the quantities defined insection 2.4 on Chandrasekhars theorem

lim ( ( ) ) ( ( ) )t

M p t p p t p I

0

2' '

f t

f t f t I t t

F

2

2

d r dr M F f

dt dt

1dr F f

dt

F /v F

d p p p p f dt M

1 1 2( ) ' X t r t t F t t t H I

0

1 1( ) ' ( ')

t

r t F t dt f t

/t M

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2.5 Brownian motion in shear flow

We will calculate the mean position and the mean squared displacementfor a Brownian particle in a simple shear flow from the Langevin equation. The particle hasan arbitrary initial position .

The local shear flow velocity is equal to

The Langevin equation reads

Exactly as in exercise 2.4, it is shown that the strength of the fluctuating force is unaffected bythe flow, so that, again (see also the discussion in section 2.7)

On the diffusive time scale, in the overdamped limit, the Langevin equation reduces to

This differential equation is solved by the method of variation of constants, which wasalready discussed in detail in exercise 2.1. Integration with the neglect of the fluctuating forceGives

where is an integration constant, and where the exponent of the matrix is defined as (seeeqn.(2.58))

00 1 0

0 0 0

0 0 0

u r r

dp pr f

dt M

( )r t ( ) ( )r t r t

0r

1dr r f

dt

0

2' '

f t

f t f t I t t

0

1exp{ }

!n n

n

t t n

0( ) exp{ }r t r t A

A

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From the definition of it is easily verified that , and hencefor all . It thus follows that

Hence

We now turn the vector into a function of time, such that the full Langevin equation,including the fluctuating term, is satisfied. Substitution into the Langevin equation gives(use again that )

The inverse of the matrix is equal to . It follows that

and hence

Finally, the position coordinate is found to be equal to

The average position coordinate is thus equal to

The interpretation of this result is that, on average, the particle is dragged along by the flowwith a velocity equal to the local flow velocity

The mean squared displacement is equal to

0

0n

1n

exp{ }t I t

0( ) [ ]r t r I t A

0 0( )r t r r t

A

0

( ) 1[ ] ( )dA t

I t r f t dt

0

[ ] I t [ ] I t

0( ) 1 ( )

dA t r I t f t

dt

00

1 ( ) ' ' ( ')t

A t r t dt I t f t

0 00 0

1 1( ) ' ( ') ' ( ') ( ')

t t

r t r r t dt f t dt t t f t

20 0 0 0 2

0 0

20 0

20 0

1( ( ) ) ( ( ) ) ( ) ( ) ' '' ( ') ( '')

1' '' ( ') ( '') ( ') ( ') ( '')

1' '' ( ') ( '') ( ') ( '')

t t

t t T

t t T

r t r r t r r r t dt dt f t f t

dt dt t t f t f t f t f t

dt dt t t t t f t f t

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Using that the fluctuating force is delta-correlated it follows that the mean squareddisplacement in the co-moving frame is equal to

where is the Einstein diffusion coefficient, and

In matrix notation this reads

with the mean squared displacement in the absence of flow.

The effect of flow is measured by the difference , which is thematrix that is written explicitly in the above expression. All components of whereone of the indices refers to the z-direction are zero. This is the vorticity direction, which is

both perpendicular to the flow and gradient direction. Diffusion in the vorticity direction isthus unaffected, which is intuitively expected. Also the yy-components of isunaffected by flow.

The probability to move upwards in the gradient direction is equal to that for a displacementdownwards, giving on average a zero net effect. That the xy- and yx-components are affected

by flow can be understood as follows.

As depicted in the figure, if the particle moves downward, to lower values of its position y inthe gradient direction, the flow will induce an additional velocity to the left. When the particlemoves upwards, it attains an equal change in velocity in the opposite direction.

The product of these two displacements is equal, so that there is a non-zero contribution. Thatthe xx-component of is non-zero is easily understood: when the particle happens to

be displaced in the y-direction, the displacement in the x-direction is very much enhancedsince the particle is taken along with the flow, precisely as was already depicted in the figure.

0 0 0

2 30 0 0

( ) ( ( ) ) ( ( ) )

2 2 23

cW t r t r r t r t r r t

ID t ED t UD t

12

T T T E U

( )cW t

2 2

0 0

1 10

3 21

( ) 2 0 02

0 0 0

c

t t

W t W t D t t

0 1 / D

0 0( ) 2W t I D t

0( ) ( ) ( )cW t W t W t

( )W t

( )W t

( )W t

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2.6 In this exercise we consider a Brownian particle when one can occupy only discrete positions that are indexed by integers.

Suppose that the probability per unit time for a single step to the left or the right side is equalto . Let denote the conditional pdf for the position of the Brownian particle,given that at time the position was equal to .

For simplicity of notation, we shall take , and refrain from the explicit notation of thecondition in the argument of the pdf. The equation of motion (EOM) is

The interpretation of the various terms on the right is as follows.

If a particle resides at position , the probability that it jumps per unit time to theneighboring position is , multiplied by the probability that the particleis at position . The product is thus the rate of increase ofdue to jumps from to .

This explains the first term on the right hand side, and similarly the second terms account for jumps from to . The last term of the right hand side accounts for jumps away from position , to the left or right, giving rise to the factor 2.

To calculate the average displacement , both sides of the EOM are multiplied by ,and then a summation over all positions is performed.

The left hand side gives,

The first term on the right hand side gives,

and similarly for the remaining terms. This leads to

Since at time , time integration gives

To calculate the mean squared displacement, multiply the EOM by and sum over all .The first term on the right hand side of the EOM, as an example, is equal to

, ) 1, 1, 2 , P n t P n t P n t P n t t

n

, 3, 2, 1, 0,1, 2, 3,n

, P n t 0t 0n

1n n 1, P n t

1, P n t 1n

1n

n

nn

,n

d d n P n t n

d t d t

1, ( 1) , 1n m

n P n t m P m t n

0 0n

2 0d

n n n nd t

0n 0t

0n

2n n

n

, P n t

n

1n

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The following EOM for the mean squared displacement is found

Integration, using that the mean squared displacement is zero at time zero finally gives

The mean squared displacement is again found to be a linear function of time.

Comparing this with eqn. (2.21), the diffusion coefficient is thus equal to .

Note that the factor of 2 in the above equation and that in eqn. (2.21) both relate todiffusion in a single dimension.

2 2n t

2 2

2 2

1, ( 1) ,

( 2 1) , 2 1

n m

m

n P n t m P m t

m m P m t n n

2 2 2 2( 2 1) ( 2 1) 2 2d n n n n n nd t

D

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2.8 The diffusive angular time scale

The thermally averaged Langevin equation (2.129) reads,

where is the rotational velocity. The solution of this differential equation is

Consider a rod that at time zero is aligned along the x-direction and the angular velocity attime zero is along the z-axis, as shown in the figure below.

The direction of the angular velocity remains along the z-direction, and only its magnitudedecreases with time (as can be seen from the above equation), due to friction with the solvent.

The orientation of the rod is therefore always within the xy-plane. Hence we can write

with the angle of the orientation with the y-axis (see the figure). Hence

where is unit vector along the z-axis. Thus

Integration thus leads to

2112 r

d ML

dt

2

01

exp /12 r

MLt

0

cos( ) sin( ) x yu u

( ) y x z x y z du dudu d t

u e u u edt dt dt dt

z e

0( )

exp{ / }d t

t dt

0 0( ) 1 exp{ / }t t t

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2.9 For a fixed orientation of the rod along the z-direction , the Langevin equations(2.86,87), read, on the diffusive time scale where can be set to zero

The friction tensor is equal to (see eqn.( 2.91))

the inverse of which is,

so that the above over-damped Langevin equation reads

where it is used that

Integration thus gives

0

1 1( ) (0) ' ( ') ( ')

t

r t r dt f t f t

( )( ) ( ) ( ) f

dr t f t f t f t

dt

/dp dt

[ ] f uu I uu

1 1 1 [ ] f uu I uu

( ) 1 1( ) ( )

dr t f t f t

dt

0 [ ]

[ ] 0

uu f I uu f f

uu f f I uu f

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For the x-component it is thus found that

with the x-component of the perpendicular fluctuating force.

Hence, from eqns. (2.107, 112)

Note that we used in eqn. (2.107), that we only consider here the x-component of the force,giving rise to the factor 2 instead of 4 in the fluctuation strength. The two other mean squareddisplacements follow similarly.

,0

1( ) (0) ' ( ')

t

x x t x dt f t

, ( ') x f t

2, ,2 2

0 0

21( ( ) (0)) ' '' ( ') ( '') 2

t t B

x x

k T t x t x dt dt f t f t D t

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3.3 In the evaluation of the scattered intensity , the following integral is encountered

In order to perform the spherical integration, we rewrite this as

where with the unit vector along . The last integral is with respect to theorientation of the wave vector, that is, the two spherical angles. Since the integral isindependent of the direction of the wave vector, the angular integral can be written as

Hence

Since the integrand is an even function of k , we can replace the integral from zero to infinity by half of the integral from minus infinity to plus infinity. Re-expressing the sinus as a sumof exponential, we thus arrive at

Consider the first integral, which can be evaluated by means of the residue theorem by closingin the upper part of complex half plane, as depicted in the left of figure 3.3.

There is just a single first order pole at

22 0

exp "( )

ik r r I k dk

k k i

2 22

0 0

1 ( ) exp " I k dkk dk ikk r r k k i

k k k k k

2

0 0

1

1

exp " sin{ }exp cos{ } "

2 exp "

exp " exp "2

"

sin{ " }4

"

dk ikk r r d d ik r r

dx ik x r r

ik r r ik r r

ik r r

k r r

k r r

2

2 200

sin{ " }( ) 4

( ) "

k r r k I k dk

k k i k r r

2 20

exp{ " } exp{ " }( )

( ) " "

i k r r i k r r k I k dk

i k k i r r r r

0 .k k i

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Fig. 3.3 The integration contours for the evaluation of the k-integral in eqn. (3.36).

According to the residue theorem we thus arrive (for ) at

The second integral is evaluated similarly, and turns out to be equal to the firstintegral. Hence

Now use that (with )

A second differentiation gives

Performing the differentiation with respect to R and adding the various contributions leads toeqn. (3.37).

0 0

0 0

exp{ " exp{ }"" "

"" 1 exp "

"

ik r r ik Rr r d r r r r dR R

r r ik r r ik r r

r r

200

2 20 0

02

exp{ " } exp{ ( ) " }2

( ) " 2( ) "

exp{ " }

"

i k r r i k i r r k ik dk i

i k k i k r r i k i r r

i k r r

r r

0

02 exp "( ) 2 "

ik r r I k r r

" R r r

0 0 03

003

exp{ "" 1 exp "

" "

1( ") ( ")exp{ }

"

ik r r I ik r r ik r r

r r r r

ik Rr r r r d ik R

r r dR R

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3.4 The scattered electric field strength is calculated for a fixed configuration of Brownian particles (see Fig.3.1), which is a valid procedure when the phase change of the scattered lightdue to Brownian motion is small during the time that light needs to propagate through thecuvette. The time required for light to travel over a distance of the typical size of 1 cm of thecuvette, with water as the solvent (for which the refractive index is 1.3 ), is equal to

A typical diffusion coefficient for a colloid is

The displacement of a single colloidal particle during 0.04 [ns] is therefore

A appropriate upper limit for the relative displacement of two colloids is therefore2l=0.03 nm. A typical value for the scattering vector is , where

is the wavelength of the light in vacuum.

The change in phase shift is therefore

This phase shift is very small, and can be therefore neglected. This validates the calculation ofthe phases as if the colloidal-particle configuration does not change during the time needed

for light to propagate through the system.

Fig. 3.1: A schematic representation of the scattering of light by an assembly of point-like particles. Each of the Brownian particles can comprise many of the point-like scatterers.

1[ ] /(300.000 /1.3) [ / ] 0.04[ ]t cm km s ns

12 210 [ / ]. D m s

6 0.015l D t nm

04 ( 1.3) /n 0 600 nm

4 00.03[ ] 4 1.3 / 600[ ] 8 10 [ ] 0.05nm nm rad

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3.6 Expressing the static structure factor in terms of the total pair-correlation function ,the integral

with the scattering volume, is replaced by the integral

for non-zero wave vectors. It is thus assumed that

We calculate the integral for two scattering geometries: (i) the scattering volume is arectangular box, where the incident intensity within the box is constant, and is zero outsidethe box, and (ii) the more realistic case, where the incident intensity is Gaussian.

(i) The integral for the rectangular box is equal to

where l is the linear dimension of the box and is a typical value of the wavevector. Substitution of the typical values and shows that thisintegral is quite large: . The above claimed smallness of the integral is thus completelyfalse for the rectangular scattering volume.

(ii) In this case, the incident intensity reads , where r is the radialdistance from the center of the scattering volume, and measures the overall (constant)intensity of the incident light. The integral is now equal to (see section 1.3.4 in the book)

We now find that the integral is extremely small.

The conclusion is that an unrealistically sharp edge of the scattering volume gives rise to verylarge Fourier components: the actual scattering pattern now contains very intensive scatteringrings. For the realistic case of smooth edges, the above calculation justifies the neglect of theintegral.

S V

exp{ } 1S V

dr ik r

20( ) exp{ ( / ) } I r I r l

0.5l mm

2 23 3

0

( )exp{ } (2 ) exp

2 I r k l

dr ik r l I

7 1~ 2 * 10k m19 310 m

( ) exp{ } sV

dr g r ik r

[ ( ) 1]exp{ } sV

dr g r ik r

3 3/ 2/ 23

/ 2/ 2

exp{ } sin{ / 2}exp{ }

/ 2

l l

l l

ikx kl dx ik x l

ik kl

2210

0 I

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3.7 Small size polydispersity and static light scattering

(a) The polydisperse form factor is defined as the intensity normalized to unity at zerowavevector,

where

is the measured polydisperse Rayleigh ratio for a dilute suspension. Here a is the colloidalradius and is the size-distribution function.

For optically homogeneous particles,

where

is the optical contrast.

Fig. 3.8: The instantaneous speckle pattern of the scattered light.The circular hole in the screen is the detector pinhole.

( ) ( ) / ( 0) pol pol pol P k R k R k

00

( ) ( ) ( , ) pol R k da P a R k a

2

6 63

cos sin( , ) * ( ) * 3

ka ka ka R k a K a P k K a

ka

24 40 0*9 9

p f

f

k k K C

0 P

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In order to expand around the average radius , we use,

where is the form factor of a sphere with the average radius

The standard deviation in size is defined as

A Taylor expansion also gives

Using the above expansions we have

Further expansion in terms of the small quantity leads to

( , ) P k a a

00

( )a da P a a

22 00

( )da P a a a

226

2

65 6

2 6 24 5 6

2 2

( ) 1 ( ), ( ) ( )

2

, ( , )6 ,

, ( , ) ( , )30 , 12

df a d f aa P k a f a f a a a a a

da dad a P k a dP k a

a P k a ada da

d a P k a dP k a d P k aa P k a a a

da da da

a a

/ a

26 6 5 46 15a a a a a a a a

60

0

60

0

226 4 5 6

0 20

26 40

0

( ) ( , )( ) ,

( )

1 ( , ) ( , )( ) ( , ) 30 , 12

2

( ) 15

pol

da P a a P k a P k

da P a a

dP k a d P k ada P a a P k a a a a P k a a a

da da

da P a a a a a

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This polydisperse form factor is plotted in Fig.3.11 as a function of for various relativestandard deviations . As a result of polydispersity, the minima in the scattering curves

become less pronounced.

Fig. 3.11: The logarithm of the polydisperse form factor versus

for various values of the relative standard deviation of , as indicated by the numbers attached to the different curves

2 2

6 2 4 5 66 2 2

1 1 ( , ) ( , )( ) ( , ) 30 , 12 1 15

2 pol dP k a d P k a P k a P k a a P k a a a

a da da a

2 26 2 4 5 6 6

6 2 2

2 22

2 2

2 62

2 4 2

1 1 ( , ) ( , )( , ) 30 , 12 15 ( , )

2

1 ( , ) ( , )( , ) 12

2

( , )1( , ) 15 ( , )2

dP k a d P k aa P k a a P k a a a a P k a

a da da a

dP k a d P k a P k a a a

a da da

d a P k a P k a P k aa a da

/ a 0ka

0ka

( ) ( ) / ( 0) pol pol pol P k R k R k / a

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(b) For small wavevectors, i.e. , the polydisperse radius is defined as

which is the radius that is experimentally obtained, assuming monodisperse opticallyhomogeneous Brownian particles. In the following we will show that to leading order in the

polydispersity

By using the result for of (a), and

it is found that

Thus, for sufficiently small wave vectors, the form factor is equal to

1ka

2 22 21 1( ) exp 15 5

pol pol pol P k k a k a

2 213

1 13 12

pol a a aa a

2 21( , ) 15

P k a k a

22

22 2 2 2 2

2

22 2 2 2 2 2

2

22 2 2 2

2

22 2 2 2

2

1( ) 1

51 1 1 1

1 12 2 25 2 5 5

1 1 1 11 24 2

5 2 5 5

1 1 261

5 2 5

1 131

5 5

pol pol P k k a

k a a k a a k a

k a k a k aa

k a k aa

k a k aa

222

1( ) 1 1 13 ; 15

pol P k ka kaa

pol a

pol P

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3.8 Small polydispersity and dynamic light scattering: second cumulant analysis

The measured (or polydisperse) EACF is equal to (see eqn.(3.105))

where is the scattering amplitude and is the size probability density function.For narrow size-distribution functions, the a-dependent functions (other than ) can beexpanded around the average value of . First rewrite the above expression as

where the polydisperse diffusion coefficient will be defined later. For radii close to theaverage value, the polydisperse diffusion coefficient is close to . Hence we canexpand

so that

provided that the polydisperse diffusion coefficient is defined as

2 20 00

20

0

( ) ( , ) exp ( ) ( , )

( ) ( , )

pol E

da P a B k a D a k t g k t

da P a B k a

0 ( ) P a( , ) B k a0 ( ) P a

a a

2 20 0 0

2 00

20

0

( ) ( , ) exp ( ) ( , ) exp

( ) ( , )

pol

pol pol E

da P a B k a D D a k t g k t D k t

da P a B k a

0 ( ) D a

222 2 20 0 0 0 0 01exp ( ) 1 ( ) ( )2 pol pol pol D D a k t D D a k t D D a k t

2 22 2 20 0 0 0 0

2 00

20

0

( , )

1( ) ( , ) 1 ( ) ( )

2exp

( ) ( , )

pol E

pol pol

pol

g k t

da P a B k a D D a k t D D a k t D k t

da P a B k a

2 22 20 0 0

2 00

2

00

1( ) ( , ) 1 ( )

2exp

( ) ( , )

pol

pol

da P a B k a D D a k t D k t

da P a B k a

20 0

00

20

0

( ) ( , ) ( )( )

( ) ( , )

pol

da P a B k a D a D k

da P a B k a

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The standard deviation in the diffusion coefficient is now defined as

so that the above expression for the correlation function can be written as

In order to express the polydisperse diffusion coefficient in terms of the size-standarddeviation, we employ the Taylor expansion

The term vanishes by definition upon integration. A similar expansion must bemade for the denominator in the expression for . Hence, to leading order in polydispersity

To leading order in the standard deviation , only the term ,resulting after substitution of the above result into the defining expression forcontributes. Since , we have

from which it thus follows that

2 4 2 2 2 4 2 20 01 1 ( , ) exp 1 exp2 2

pol pol pol E D D g k t D k t k t D k t k t

220 0 02 0

20

0

( ) ( , ) ( ) ( )

( ) ( , )

pol

D

daP a B k a D a D k

daP a B k a

20

222 2 2

0 0 02

( , ) ( )

1( , ) ( ) ( , ) ( ) ( , ) ( )

2

B k a D a

d d B k a D a a a B k a D a a a B k a D a

da da

a a

22 2 2

0 02

0 22 2 2

2 2

2 2 2 22 2

0 0 02 2 2

1( , ) ( ) ( , ) ( )

2( )1

( , ) 1 ( , )2 ( , )

( ) ( , ) ( ) ( ) ( , )2 ( , )

pol

d B k a D a B k a D a

da D k d

B k a B k a B k a da

a d d D a B k a D a D a B k a

a B k a da da

0 pol D

0 0( ) ( )dD a D ada a

0 ( ) D

D a a

20 0( ) ( ) D a D a2

D

2

0 ( ) 1/ D a a

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Using that for small wave vectors , we have andis a constant, independent of the average radius, the above expression forleads to

Performing the differentiations finally gives

Putting things together, we thus arrive at an expression for the measured, polydispersecorrelation function in terms of the standard deviation in the radius

This result enables the determination of the average radius and its standard deviation from aso-called second-cumulant fit, where a term in the exponent is included in the fit toexperimental data.

Fig. 3.12: The polydisperse diffusion coefficient, relative to the monodisperse diffusioncoefficient , versus for various values of the relative standard deviation ,as indicated by the numbers attached to the different curves. The radius is related to ,as discussed in section 3.9.1 (see especially eqn.(3.102)).

2 2

22 2 20 0

1 ( 0, ) exp ( ) 1 5 ( )2

pol E g k t D a k t D a k t a a

1/ 2ka

2 2 2 2

6 60 0 0 06 2 2

1( ) ( ) ( ) ( )

2 pol a d d D k D a a aD a D a a

a a da a da

3( , ) B k a a 0 ( )aD a

0 pol D

2

0 0( ) ( ) 1 5 pol D k D a

a

4 2k t

0 ( ) D a 0ka / a 0a a

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3.11 Form factor of a thick rod

For static light scattering by thin rods, where the integral ranging over a cylinderwith its geometrical center at the origin

was calculated in subsection 3.10.2. Now suppose that is not small. Here we evaluatethe form factor for that case where the rod has an arbitrary orientation of the rod.

The wave vector is now decomposed in its component parallel to the rod, andits component perpendicular to the rods long axis. We can now write

where is the unit vector along . Note that the form factor only depends on thedirection of the wave vector through its angle with the z -axis (see the figure below).

The integral thus reads, in terms of cylindrical coordinates

The z -integral is relatively simple

0

1exp

V

I dr ik r V

2

1 / yk k k u k

0.2kD

kD

z k k u

2 1 ( ) ( )k r k y k u k z k u

2 / 2 / 22

0 / 2 0

1 exp ( ) sin{ } 1 ( ) L D

L

I d dz d i k z k u k k uV

/k k k k

/ 2

0/ 2

1 sin ( )12 exp ( ) ( )

1 2 ( )2

L

L

Lk k udz ik z k u L L j Lk k u

Lk k u

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where the last identity defines the spherical Bessel function of zeroth order.

For n=0 we have

so that

Now using that

it follows that

Since the volume of the cylinder is equal to

it is finally found that

2

1

2

1 12 1 / sin2 2

11 1 /22

J kD k u k Lk u I

Lk ukD k u k

2

00 0

1 1( ) cos sin exp sin

2 J x d x d ix

1 0( ) ( )d

xJ x xJ xdx

/ 2 2 / 2

2 20

0 0 0

exp sin{ } 1 ( ) 2 1 ( ) D D

d d ik k u d J k k u

2

2

1 1 ( )/ 2 2

20 02

20 0

1 1 ( )2

122 0

21

2

1 1 ( ) ( ) 1 ( )

1( ) 1 ( )

/ 2 1 1 ( )2 1 ( )

Dk k u D

Dk k u

d J k k u dx xJ xk k u

d dx xJ xdxk k u

D J Dk k u

k k u

2

4V D L

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(b) For , that is, small scattering angles, the integrand in the integral in exercise (a)can be expanded as

so that

Re-exponentiation thus leads to the scattered intensity being approximately equal to

This is the equivalent of the Guinier approximation as discussed in detail in section 3.8.1 for

spherical colloids. The slope of the Guinier plot for thin rods is thus equal to . 2136

L

1kL

2sin 116

z z

z

2 2/ 2 / 22 2 4

0 0

2 sin 2 1 11 1 ( ) (( ) )6 36

kL kL z dz dz z kL kLkL z kL

2 21exp36

I L k

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3.13 Heterodyne dynamic light scattering

When the scattered light is mixed with incident light directed towards the detector, thedetected electric field strength is

where is the field scattered by the particles, and is so-called local oscillatorfield strength from the incident field that is mixed with the scattered light. The instantaneousdetected intensity is now

The IACF for heterodyne light scattering is therefore equal to

where is the intensity of the light that is mixed in with the scattered light. Sincethe average of the oscillating field strength is zero

and the average scattered intensity is equal to ,while (see page 133-134 inthe book)

and

with the scattered intensity, this expression reduces to

Note that for this heterodyne IACF is approximately equal to the homodyneEACF.

( ) ( )het loc S E t E E t

22 2 2 ( , ) 2 2 Re ( , ) ( , )het loc loc loc I E E g k t I I I I I I g k t I g k t

50loc I I

* ** * *

( ) ~ ( ) * ( )

( ) ( ) ( ) ( )

loc locS S

loc loc locS S S S

i t E E t E E t

I E t E E E t E t E t

2loc loc I E

( ) s E t loc E

* * *

* * *

( ) (0) ( ) ( ) ( ) ( )

* (0) (0) (0) (0)

loc loc locS S S S

loc loc locS S S S

i t i I E t E E E t E t E t

I E E E E E E

2 * * *

2* * *

* *

(0) (0) (0) (0)

( ) (0) ( ) (0) ( ) (0)

( ) ( ) (0) (0)

loc loc loc loc loc locS S S S

loc loc locS S S S S S

locS S S S

I I E E I E E I E E

E t E E I E t E I E t E

I I E t E t E E

*(0) 0 (0)S S E E

*(0) (0)S S I E E

2* * 2 *( ) ( ) (0) (0) (0) ( )S S S S S S E t E t E E I E E t

( ) (0) 0S S E t E

I

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3.14 For a very dilute system of Brownian spheres , where interactions between thecolloids can be neglected, the pdf in the presence of a constant external field isequal to (see exercise 2.4)

where is the displacement during the time t . In case , according to the previousexercise, the heterodyne correlation function becomes equal to

The normalized electric field auto correlation function (EACF) is equal to

Introducing the new integration variable , we arrive at

where is the stationary velocity that the Brownian spheres attain under the actionof the external force. According to the appendix in chapter 1, page 49, on integration ofGaussian functions, this leads to

the real part of which is equal to

The measured heterodyne IACF is thus equal to

This correlation function is a damped exponential as depicted in the figure.

2 202 1 cos exphet loc loc I g I I I k v t D k t

( , ) ( ', ) exp ' E g k t dr P r t ik r

2

03/ 2

0

1', exp ' / 4

4

F P r t r t D t

D t

/v F

loc I I

F

2 ( , ) 2 1 Re ( , )het loc loc I E g k t I I I g k t

'r

' ( / ) R r F t

2 03/ 2

0

1 ( , ) exp exp / 4 exp

4 E g k t ik vt dR R D t ik R

D t

20 ( , ) exp exp E g k t ik vt D k t

20 ( , ) cos exp E g k t k vt D k t

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Setting in the equation for the IACF in the previous exercise equal to zero, it isimmediately found that the homodyne correlation function is not affected by the velocityof the Brownian particles (since ).

Heterodyne light scattering is required in order to be able to measure particle velocities.

Fig.: A sketch of the heterodyne IACF

loc I

exp 1i k v t

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Exercises Chapter 4: FUNDAMENTAL EQUATIONS OF MOTION

4.2 The Brownian oscillatorTwo identical Brownian spheres are connected to each other with a spring, corresponding to a

potential energy

where and are the position coordinates of the two spheres, and C is the spring constant.

Define the distance between the spheres and the center-of-mass position . Now, for example,

where is the matrix with components i and j equal to the i-th component of andthe j-th component of , and similarly for . A similar calculation for the derivativewith respect to thus gives

2

1 2

12

C r r

1 2 R r r

1r 2r

2( ) / 2r r r

1

1 1

1 2 1 2( , ( ) / 2)

1( , ) ( , ) ( , )2

r

r R r r R r

f R r r r r r

R f R r r f R r f R r

1r R

1r

R 1r r 2r

1 21 1

,2 2r R r r R r

Smoluchowski Equation: Free Diffusion

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The Smoluchowski equation in (4.40, 41) for two particles (with the neglect of hydrodynamicinteractions) thus reads in terms of the new coordinates

as

Now substitute the separation variables to obtain

Dividing both sides with and equalizing the independent terms depending onlyon and gives

The center-of-mass thus diffuses as a free single Brownian sphere with a diffusion coefficient

equal to half of that of a single sphere. The interesting part of this result is the equation ofmotion for . Comparing this equation of motion with eqn. (4.59) we have theidentification

and . The corresponding equations of motion in (4.58, 59) are

where, as inferred at the top of page189,

The initial conditions are

( , , ) ( , ) ( , ) P R r t P R t P r t

, ( , ), , ( , ) ( , ) P r t P R t P R r t P R t P r t t t t

2 20 1( , ) , ( , ) 2 ( , ) 2 ( , )2 r R R D P R t P r t P r t C RP R t P R t

20, 2 ( , ) 2 ( , ) R R P R t

D C RP R t P R t t

20

, 1( , )

2 r P r t

D P r t t

0

0

22

A D C I

B D I

0

0 0

2 ,

4 4

d m D C m

dt d

M D I D C M dt

r

X R

2 20 1, , 2 2 2 R R r P R r t D C RP P P t

( , ) ( , ) P R t P r t

R

( , ) P R t

( ) ( )

( ) ( )

m t R t

M t R m R m t

0( 0) , ( 0) 0m t R M t

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4.3 Diffusion in an inhomogeneous solvent

For very dilute homogeneous suspensions, the diffusion coefficient is equal to

Now consider an inhomogeneous solvent, so that the diffusion coefficient is different at each position. Since there is now a direction that is associated with the inhomogeneities, thediffusion coefficient is not a scalar but rather a tensorial quantity. The flux is now equal to

, so that the Smoluchowski equation is

The average velocity is

The integral can be rewritten, using Gausss integral theorem, with the neglect of surfacecontributions

Applying Gausss integral theorem once more gives

Hence, the velocity induced by the inhomogeneous solvent is equal to

As an example, consider a solvent that consists of a mixture of water and ethanol, where thecomposition changes in the x-direction, say. This leads to a spatial change of the viscosity. Inthis case

where is the unit vector along the x-direction. To leading order in spatial gradients, in theevaluation of the average, the pdf can simply be taken as a constant, so that the velocity isequal to

Note that the particle moves from regions of high viscosity to regions of low viscosity.

00

16

Bk T Da

0( , ) ( ) ( , )r r P r t D r P r t t

0 ( )T

r d

r D r dt

0 0

0 0

( ) ( , ) ( ) ( , )

( ) ( , ) ( ) ( , )

r r r r

r r

Int dr r D r P r t dr D r P r t r

dr D r P r t I dr D r P r t

0 0( , ) ( ) ( )T T

r r Int dr P r t D r D r

0 ( ) ( , )r D r P r t

0, ( ) ( , )r d r

dr r P r t dr r D r P r t dt t

0 00 02

0 0

( ) ln ( ) ( ) ( )

6 ( ) 6 ( )T B B

r x x x

d x d xk T k T d D r e e e D x

dx x a x a dx dx

xe

00

ln ( )( ) ( ) x

d xv x e D x

dx

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(a) For spherical particles, the hydrodynamic torques are zero in the absence of an externalfield. From eqn.(4.128)

Since the torque on the sphere is zero, this implies that

and hence

,

From the last equation it follows that the orientational velocities are equal to

Substitution into the above equation for the forces gives

11

11

h

h TT TR N N

h RT RR

h N N

v F

v F

1 1

10

0

h

h TT TR N N

RT RR

N

F v

F v

1 1 1h

TT TR

h N N N

F v

F v

1 10

0

RT RR

N N

v

v

1 1

1 RR RT

N N

v

v

1 1 1 1 1

1

h

TT TR TT TR RR RT

h N N N N N

F v v v

F v v v

1

1TT TR RR RT

N

v

v

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4.5 The direct torque on a rod

Consider a very thin and long rod, the core of which we approximate as a line. Let be thecontour variable, . The position of a line element on the core is given by ,where is the orientation of the rod.

Let be the force on line elements (either due to an external field and/or interactionswith other rods). When the orientation of the rod is changed by a small amount , theaccompanied change in potential energy is (see the figure)

Using that and it is easilyverified that,

where is the torque, which is equal to

The first equation defines the torque (with the volume occupied by the core), while in thesecond equation its approximation for the very long and thin rod is given.

On the other hand we have , where is the gradient operator with respect tothe Cartesian coordinates of . Comparing to the above result we thus have

Since is an arbitrary vector, but always lies in the plane perpendicular to , theconclusion is that the components of and in that plane are equal.

The vectors are thus equal when they do not have a component along . For this isimmediately clear, since it is perpendicular to . That is also perpendicular tofollows from , since is constrained to have a fixed length of unity.Hence

Taking the outer product of both sides, noting that is perpendicular to , and using theabove relation for an outer product of three vectors, leads to

with the rotational operator. This is the rotational analogue of the translational resultfor the force.

/ 2

/ 2

( ) L

L

dl f lu l u

/ 2 / 2

/ 2 / 2

( ) ( )

L L

L L

dl l f lu u u u u u dl l f lu u

u u u u

0

/ 2

/ 2

( ) [ ( )] L

V L

dr r f r u dl lu f lu

uu

lu

u

0V

( ) f r lu

/ 2 / 2 L l L l

u

( ) ( ),a b c c a b

( ) ( ) ( ) , ,a b c a c b a b c u u

u

u

( )uu u u

u u

/ 0uu d d u

u uu u

uu

uu u

uu R

R F

u

u

'l l u l u l u u

l

'u

u

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4.6 To evaluate the following steps can be made

The components of the rotational operator are, by definition

where the partials denote differentiation, , that is, is the differentiation withrespect to the component of . Since

and

which follows from the above component-wise representation of the rotationoperator, we have

From the same component-wise representation, it also follows that

so that

The other components are calculated similarly, leading to .

23 3

22

1 1r i j i j i nj j ni

n nn n

r r r r r r r r

3

1in nj jn ni

n

3

1

2 2in nj ijn

2 3 3 2

3 1 1 3

1 2 2 1

u

u u

R u u u

u u

j u j

3 3

2

1 1

[ ] [ ]u un n

R u R R u u u u

2 3 3 2

1 3 1 1 3 1 3

1 2 2 1 2

0

u u

R u u u u u

u u u

3

21 1 1 2 3 3 2

1

0n nn

R u R R u R R u R u

2 3 3 1 1 3 3 1

3 2 1 2 2 1 2 1

R u u u u u R u u u u u

21 1 1 1

2 R u u u u

2r r r

jth j u

2 2 R u u

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For an arbitrary vector the component of is

For the different values of i , this is, according to the component-wise notation of the rotationoperator

so that

This concludes the proof of the three identities

These and other identities are necessary for calculations of ensemble averages

a

3

1

i u i n u inn

a Ru a u u a u u

1 2 3 3 2 2 3 3 2

1 3 2 3 1 3 1 1 3

1 2 2 1 3 1 2 2 1

1; 0 ( ) ( )

2; 0 ( )

3; 0 ( )

i a a u a u a u a u

i a u a a u a u a ui a u a u a a u a u

2 3 3 2

3 1 1 3

1 2 2 1

a u a u

a R u a u a u a u

a u a u

a Ruthi

2

2

2 2

r r I

R u u

a R u a u

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4.7 Small angle depolarized time resolved static light scattering by rods

In this exercise we consider a dilute suspension of a rigid, rod like Brownian particles whichare strongly aligned in the z-direction by means of an external field. The external field isturned off at time . After a long time the rods attain an isotropic distribution. We areconsidering here the kinetics of relaxation to the isotropic state after turning off the externalfield. Experimentally, the rotational relaxation kinetics can be measured by means of depolarized light scattering. The polarizat ion direction of the incident light is chosen in the

z -direction, which is along the direction of alignment of the rods at time zero. The polarizationdirection of the detected light is chosen in the x-direction.

We consider small angle light scattering such that , where is the wave vector andis the length of the rods. The ensemble averaged scattered intensity is given by eqn (3.131) forthe anisotropic structure factor

For , the Bessel functions are essentially equal to unity; while for non-interacting

rods only the term where survive. Hence

The scattered intensity is thus a strong function of the orientations. Since these change withtime, the scattered intensity R is time dependent, which characterizes the relaxation oforientational order.

The time dependence of this depolarized small angle scattered intensity is calculated form theSmoluchowski equation

Since the quantity is independent of the position coordinates of the rods, according toGausss integral theorem, only the rotational contribution in the Smoluchowskioperator contributes. Multiplying both sides

1kL

0 0

( , ) 0 0

, 1

20

1 11 ( ) 2 2 exp1 2

s i s j i j

N a a i j

i j i j

n u n u n u n u

j Lk u j Lk u R S k N ik r r

j Lk u

0t

k L

sn

2 2 ~ z x R u u

0n

1kL

i j

0

0 2 2

, , , , ,

1 3r

S

S r r r

P r u t L P r u t t

L D D D uu I

2r D

0S L

2 2 z xu u

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of the Smoluchowski equation by , and integrating, we thus arrive at the followingexpression for the time dependence of the scattered intensity

where the integral ranges over all orientations of , and we replaced the indices z by 3 and x by 1 (and later we will replace y by 2). In the last line we used Stokess theorem in the form

Using the same steps as in exercise 4.6, we find with some effort that

The details of the derivation of this result are given at the end of this exercise. Here, is thegradient operator with respect to the component of . It is thus found that

In order to explicitly solve this equation, we need an expression for . This can beobtained similarly as the above expression, by multiplying both sides of the Smoluchowskiequation by , and integrate

Again using the above expression for , it is found that

This equation can be integrated by means of variation of constants (see exercise 2.1; details

are also given at the end of exercise)

Substitution into the equation of motion for , which can be integrated by variation ofconstants, to give (details are again given later in this exercise)

u

2 2 2 2 2 2 2 2 2 22 3 1 1 3 2 1 2 3

2 3 2 3 1 3 1 3 1 2 1 2

1 1 2 2 3 3

2

2

u u u u u u

u u u u u u

u u u

22u

2 22 2 2 6r

d u D u

dt

6 202 23 1

1 1 4 ~15 21 35

r r D t D t R u u e e

2 2 2 2 2 23 1 3 1

( , ) ( , ) ( )r r d

R D du u u P u t D du P u t u ud t

2 2 ( ) ( ) ( ) ( )du f u g u dS g u f u

2 2 z xu u

jth j u

2 2 2 2 22 2 2 , ,r r d

u D du u P u t D du P u t udt

2 2 2 2 23 1 3 1 2 20 2 1r d u u D u u udt

22u

2

622 1 13r D t u e

2 23 1 u u

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The homogeneous equation reads

the solution of which reads

The integration constant A is now considered a function of time (hence the name variation ofconstants), such that it satisfies the full equation of motion. One finds after substitution

Hence

Since at time zero we have it follows that the initial value of A is zero: . Thus

Substitution of this result in the equation of the motion of gives

Again, first the homogeneous equation is solved

which gives

The integration constant is again considered now a function of time such that it satisfies thefull equation of motion. Substitution into the equation of motion

2 22 2 6 r

d u D u

dt

622 r

D t u Ae

6( ) 2 r D t r d

A t D edt

60 1( ) 13r D t A t A e

0( 0) 0 A t A

622 1 13

r D t u e

22 0u

2 23 1 u u

2 2 2 2 23 1 3 1 262 2

3 1

20 2 1

4 1 20 13 2

r

r

D t r r

d u u D u u u

dt

D u u D e

2 2 2 23 1 3 1 20 r

d u u D u u

dt

202 23 1 r

D t u u B e

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Exercises Chapter 5: HYDRODYNAMICS

San Francisco, CA, USA

5.4 The effective viscosity

On a length scale that is large in comparison to the size of the Brownian particles, a flowingsuspension can be described as an effective fluid (see the figure). The Navier-Stokes onsuch a coarsened length scale is that of a mono-component fluid, where the viscosity is aneffective viscosity. The stress tensor is thus written as

where is the coarse-grained suspension velocity and P the pressure.

The effective viscosity depends on the volume fraction of colloids and the type of inter-colloidal interactions. In the present exercise we calculate the effective viscosity to leadingorder in concentration.

( ) ( ) ( ) T eff eff r U r U r P r I

eff

U

F r e e Di f f u

s i on

S m

ol u

c h o w s k i E

q u a t i on : F r e e Di f f u

s i on

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Fig. 5.10: Left: The flowing suspension on a length scale large comparedto the size of a Brownian particle. Right: A blow up of a fictitious

volume element, showing the flow on a length scale smaller than the size of the Brownian particles. The dotted straight line indicates the flow velocity gradient pertaining to the effective flow.

For non-interacting Brownian particles, the stress tensor at position is equal to

where is the position coordinate of colloidal particle j, and is the stressgenerated by a single Brownian particle. For non-interacting colloids, the coarse-grainedstress tensor is equal to

The stress tensor within the solvent is equal to

Here, and are the solvent velocity and pressure on the small length scale, much smallerthan the size of the colloidal spheres. At this point, we have to make the distinction betweenvolume elements within the solvent and the core. Let denote the volume occupied by asingle colloidal sphere with its center at the origin. The effective stress tensor is now writtenas

1 2 01

, , , ' 'n

N j j

r r r r r r

0 ' jr r

'r

jr

0( ) ' 'eff V

N r dr r

V

0 0 0 0 0 ( ) ( ') ( ') 'T r u r u r p r I

0u 0 p

0 0

0 0\

( ) ' ' ' 'eff

V V V

N r dr r dr r

V

0V

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The volume is occupied by solvent, for which the above formula for the stress tensoris valid. The integral over the volume occupied by the core can be cast into an integralover the surface area of the core, just inside the solvent. To this end we use the mathematicalidentity (summation over n is understood)

From Gausss integral theorem we thus have (with the surface of )

The body force is zero in a stationary state, or more generally, on a time scale that islarge to the elastic relaxation time of the material of the colloidal core. The last integraltherefore vanishes, so that, apart from pressure contributions

The coarse-grained values of gradients of the suspension velocity are, similarly to thecoarse-grained stress tensor, defined as a spatial average of the corresponding microscopicflow velocity

As before, is the solvent velocity. Hence

where is the unit normal vector pointing out of .

0\V V 0V

0, 0, 0,( ') ' ( ') ' ' ( ') 'ij n nj i

fzj-2013-04149.pdf

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