Biological Physics and Fluid Dynamics

T H E S C I E N T I S T D O E S N O T S T U DY N AT U R E B E C A U S E I T I S U S E F U L

H E S T U D I E S I T B E C A U S E H E D E L I G H T S I N I T A N D H E D E L I G H T S

I N I T B E C A U S E I T I S B E A U T I F U L I F N AT U R E W E R E N O T B E A U T I F U L

I T W O U L D N O T B E W O R T H K N O W I N G A N D I F N AT U R E W E R E N O T

W O R T H K N O W I N G L I F E W O U L D N O T B E W O R T H L I V I N G

H E N R I P O I N C A R Eacute S C I E N C E A N D M E T H O D 1 9 0 8

R AY M O N D E G O L D S T E I N A N D E R I C L A U G A

B I O L O G I C A L P H Y S I C SA N D F L U I D DY N A M I C S( A G R A D U AT E C O U R S E I N 2 4 L E C T U R E S )

D E PA R T M E N T O F A P P L I E D M AT H E M AT I C S A N DT H E O R E T I C A L P H Y S I C S U N I V E R S I T Y O F C A M B R I D G E

Copyright copy 2021 Raymond E Goldstein and Eric Lauga

published by department of applied mathematics and

theoretical physics university of cambridge

First printing October 2021

Contents

Preface 7

1 Overview (Lecture 1) 9

11 Length scales and methodologies 9

12 Scaling arguments 10

2 Microscopic Physics (Lectures 2 amp 3) 15

21 Review of molecular physics 15

22 Van der Waals interactions 18

221 Interaction of extended objects 20

23 Screened electrostatic interactions 21

231 Electrostatic interactions between surfaces 24

24 Geometrical aspects of screened electrostatics 27

241 Comparative approach 28

242 Perturbative approach 29

3 Fluctuations amp Brownian Motion (Lectures 4 amp 5) 31

4 Biological Filaments (Lectures 6 amp 7) 33

5 Membranes (Lectures 8 amp 9) 35

6 Chemical Kinetics (Lectures 10 amp 11) 37

7 Cellular Motion (Lectures 12-17) 39

8 Chemotaxis (Lecture 18) 41

9 Cytoplasmic Streaming (Lecture 19) 43

10 Pattern Formation (Lectures 20-22) 45

11 Population Dynamics (Lectures 23 amp 24) 47

12 Bibliography 49

Preface

Physics and Biology have been intertwined since the dawn of mod-ern science from Antony van Leeuwenhoekrsquos use in 1700 of ad-vanced optics to reveal the hidden world of microscopic life1 to 1 A van Leeuwenhoek IV Part of a

letter from Mr Antony van Leeuwen-hoek concerning the worms in sheepslivers gnats and animalcula in the ex-crements of frogs Philos Trans R Soc22509ndash18 1700

Bonaventura Cortirsquos 1774 discovery of the persistent fluid motioninside large eukaryotic cells2 Robert Brownrsquos 1828 study of random

2 B Corti Osservazioni Microscopischesulla Tremella e sulla Circolazione del Flu-ido in Una Pianta Acquajuola AppressoGiuseppe Rocchi Lucca Italy 1774

motion at the microscale3 and Theodor Engelmannrsquos determination

3 R Brown XXVII A brief account ofmicroscopical observations made in themonths of June July and August 1827on the particles contained in the pollenof plants and on the general existenceof active molecules in organic and in-organic bodies Philos Mag 4161ndash731828

in 1882 of the wavelength dependence of photosynthetic activity4

4 TW Engelmann Ueber Sauerstof-fausscheidung van Pfalnzenzellen imMikrospektrum Pfluumlger Arch 27485ndash89 1882

Although at the time it might have been difficult to define the precisedisciplines of each of these scientists mdashperhaps they were all lsquonatu-ral philosophersrsquo mdashin hindsight we can see clearly the way in whichtheir discoveries impacted both biology and physics Despite thislong history of discoveries at the boundary between the two fieldsand the innumerable fundamental contributions to both disciplinesover the long arc of time the field of biological physics as a disci-pline within the research enterprise of physics has only risen to greatprominence since the postwar era particularly since the mid 1980sWe are now at the point that most academic physics departmentshave an identifiable group in biological physics alongside those inhigh energy condensed matter atomic and astrophysics

As a subject biological physics is in a much earlier stage of devel-opment than the traditional ones found in physics or applied mathe-matics curricula This makes it an exciting field in which to work butalso poses challenges for students and faculty as there is no univer-sally accepted canon to follow Instead each lecturer has had to cob-ble together concepts and presentations from a number of disparatesources to arrive at a syllabus These lecture notes have been de-veloped very much in that spirit They derive from several masterrsquoslevel (Part III) courses taught by each of us sometimes in collabora-tion with Prof Ulrich Keyser of the Cavendish Laboratory We haveattempted to strike a balance between coverage depth and pedagogyin order to address mdashin the 24 lectures of a Cambridge termmdashtopicsin biological physics that cover length scales from the molecular tothe terrestrial We presume no prior exposure to the subject butassume familiarity with the basic ideas of statistical physics elec-tromagnetism nonlinear dynamics and fluid mechanics As this isprimarily a course for Part III students in the mathematical Triposthe presentation often includes digressions into the applied mathe-matical aspects of the problems at hand

We have organised the topics covered by these notes roughly in or-

der of increasing length scales from the basic molecular forces thatgovern the cohesion and interaction of molecules and supramolecu-lar assembles through the principles of chemical kinetics (with someideas from systems biology) the elastic properties of filaments andmembranes the role of thermal fluctuations in their conformationand interactions on to cellular motion chemotaxis biological pat-tern formation and population dynamics In a course of this lengthand scope it is not possible to be comprehensive We refer the studentto a number of excellent textbooks that have appeared in recent yearscovering various aspects of biological physics5 and physical biology6 5 P Nelson Biological Physics Energy

Information Life Chiliagon SciencePhiladelphia PA student edition 2020

6 R Phillips J Kondev J Theriot HGGarcia and N Orme Physical Biology ofthe Cell Garland Science Boca RatonFL 2nd edition 1998

as well as more specialized works on elasticity7 Much motivation for

7 B Audoly and Y Pomeau Elasticityand Geometry Oxford University PressOxford UK 2010

the physicistrsquos approach to biology comes from DrsquoArcy WentworthThompsonrsquos classic book8 and we recommend highly the classic text

8 DW Thompson On Growth and FormDover Publications Mineola NY thecomplete revised edition 1992

on applied mathematics by Lin and Segel9 Where possible we have

9 CC Lin and LA Segel MathematicsApplied to Deterministic Problems in theNatural Sciences Macmillan PublishingCo Inc New York 1974

included references to the original often old literature in the beliefthat reading such scientific works is highly worthwhile10

10 RE Goldstein Coffee stains cell re-ceptors and time crystals Lessons fromthe old literature Phys Today 7132ndash382018a

Associated with these notes is a set of Example Sheets In manyplaces throughout the text we have deliberately left out certain cal-culational details to encourage the reader to fill them in At thesame time we have taken pains to explain in a fair amount of detailthe ins and outs of various key calculations While we emphasizeanalytical calculations throughout the text there are many contextsin which numerical work is an important adjunct to those calcula-tions whether for something as simple as understanding the detailedshape of a function finding the roots of a transcendental equationor to solve a nonlinear PDE to understand its global behaviour Tothat end we have provided a series of simple Matlab programs forthese various tasks Finally for convenience we have collected to-gether pdfs of all of the references (apart from books) cited in thesenotes

RE Goldstein and E Lauga October 2021

1Overview (Lecture 1)

11 Length scales and methodologies

The field of biological physics covers phenomena on an enormousrange of length scales Historically there has been a great emphasisplaced on the cellular scale of microns1 but as the field broadens 1 HC Berg and EM Purcell Physics of

chemoreception Biophys J 20193ndash2191977

to include problems ranging from neuroscience to development andeven ecology that range now extends to kilometers (Figure 11)

proteins hair

algaecoiled DNAstretched DNAbacteria

lipid vesicles

nm microm mm m km

10-9 m 10-6 m 10-3 m 1 m 103 m

Thermal fluctuations Other forms of noise

neuroscience

paern formation

ecology

spermatozoa

ciliates

Figure 11 The range of length scalesconsidered in these notes and a sam-pling of the living systems of interest

In this course we will study methods to quantify the physics ofliving systems on many of these scales While many of the under-lying properties of the components of cells tissues and organismscan be described by molecular physics and its continuum general-isations life is inherently out of equilibrium and characterized bynoise from many sources At the smallest scales mdashfrom nanome-ters up to hundreds of microns mdashequilibrium thermal fluctuationsmay be dominant but even there and certainly at larger scalesdescriptions must incorporate such features as biochemical fluctu-ations within single organisms and organism-to-organism variabilityin phenotypes Thus the mathematical methods utilized here in-clude not only equilibrium statistical physics but also deterministicand stochastic ordinary and partial differential equations of varioustypes In addition many of the important questions in biologicalphysics involve understanding the time-dependent geometry of liv-ing matter and its constituents In order to make these notes self-

10 biological physics and fluid dynamics

contained we have included where appropriate some backgroundmaterial on differential geometry and geometric evolution and pro-vided references to more complete treatments in the literature

One of the key lessons of applied mathematics is the importance ofproper rescaling of the governing equations to reveal the relevant (di-mensionless) control parameters that determine the behaviour Notonly is this good practice in solving problems it also promotes apoint of view that focuses on the size of the parameter space and thescaling laws that underlie a given phenomenon It serves to makeclear the conclusions of models in interdisciplinary collaborations2 2 RE Goldstein Are theoretical lsquore-

sultsrsquo results eLife 7e40018 2018bA corollary to the above is that in biological physics numbersmatter That is to say gaining an intuitive feel for the typical sizes ofthings energy scales and time scales is very important in buildinga quantitative picture for a given problem But what system of unitsshould we use Many physicists and applied mathematicians havebeen educated to use the MKS system but its use can lead to veryawkward numbers take for example the bending modulus kc of acell membrane (units of energy) which might be 4 times 10minus20 NmiddotmWe shall see that the typical forces on the cellular scale are pN andwith typical sizes of molecular constituents being nanometric theenergy scale pNmiddotnm is much more appropriate for then kc sim 40pNmiddot nm And most importantly with the Boltzmann constant kB =

138times 10minus23NmiddotmK and the absolute temperature T = 300K thermalenergy is sim 4times 10minus21 Nmiddotm or

kBT 4 pN middot nm (11)

So concisely the bending modulus is kc sim 10 kBTFor convenience and to keep these notes self-contained we collect

in Table 11 the first set of physical quantities that will appear inthese notes Other tables will follow as we develop the subject

quantity symboldefinition value alternate value

viscosity of water micro 001 gcmmiddots (1 cP) 0001 kgmmiddotsdensity of water ρ 1 gcm3

103 kgm3

kinematic viscosity of water ν 001 cm2s 106 microm2s

dielectric constant of water ε 80size of a bacterium major axis (1minus 3)times 10minus4 cm 1minus 3 microm

size of unicellular algae radius 5times 10minus4 cm 5 micromsize of multicellular algae radius (100minus 400)times 10minus4 cm 100minus 400 microm

charge of the proton e 16times 10minus19 Cmembrane thickness δ (3minus 5)times 10minus9 m 3-5 nm

membrane elastic modulus kc 4times 10minus20 Nmiddotm 10 kBT

Table 11 Some physical quantitiesused in the early chapters of thesenotes

12 Scaling arguments

To continue this overview we use several examples to illustrate thepower of scaling arguments (sometimes called dimensional analysisas usually attributed to Rayleigh3) to reveal the essential physics in

3 JC Maxwell Remarks on the mathe-matical classification of physical quan-tities Proc London Math Soc s1-32241869

a problem and to point out several particular features of the cellularscale that will become our primary focus

overview (lecture 1) 11

Although quantum phenomena will make only a cursory appear-ance in these notes (when we discuss molecular interactions in Chap-ter 2) the hydrogen atom provides our first example of scaling argu-ments We know that the kinetic energy is p22m while the potentialenergy (in the Gaussian system of units) is minuse2r where r is the sep-aration between electron and proton If we invoke the de Broglierelation p = hλ between the momentum p and the associated par-ticle wavelength λ and the quantization condition λ = 2πrn forsome integer n then p = nhr there is a single length scale r thatgoverns the physics and the total energy is

E(r) sim n2h2

2mr2 minuse2

r (12)

The function E(r) diverges to +infin as r rarr 0 and approaches 0 frombelow as r rarr infin having thus a single minimum We find r by min-imising E(r) yielding the minimiser

rlowastn =n2h2

me2 and En(rlowastn) = minusme4

2n2h2 (13)

Here rlowastn is n2rB where rB is the Bohr radius and En = Ryn2 whereRy is the Rydberg constant These are the exact values for the hy-drogen atom Note that this approach required none of the intricateanalysis of Laguerre polynomials or any other aspects of the fullsolution to the Schrodinger equation4 and even obtains the correct 4 LD Landau and EM Lifshitz Quan-

tum Mechanics Non-relativistic theoryPergamon Press Oxford UK 2nd edi-tion 1965

factor of 2 in the Rydberg constantA second approach to Eq 12 more in keeping with the ethos

of Applied Mathematics5 is to observe that by comparing the two 5 CC Lin and LA Segel MathematicsApplied to Deterministic Problems in theNatural Sciences Macmillan PublishingCo Inc New York 1974

terms on the rhs we have n2h2mr2 sim e2r (ignoring the factor of2) so there is a length rn sim n2h2me2 namely n2 times the Bohrradius If we define a dimensionless radius R and dimensionlessenergy E as

R =rrn

and E =E

me42n2h2 (14)

then we obtain the marvellously simple result

R2 minus2R

which is minimised at R = 1 giving E = minus1 as aboveLet us now turn to an important biophysical question concerning

the role of inertia in the motion of microscopic organisms such asbacteria6 Suppose we have a spherical body of radius a and density 6 EM Purcell Life at low Reynolds

number Am J Phys 451ndash11 1977ρ equal to that of the surrounding fluid whose viscosity is micro If itsposition is x(t) then its equation of motion is

πa3ρd2xdt2 = minus6πmicroa

where we have invoked the Stokes drag force 6πmicroav on a particlemoving with speed v We wish to answer the question if the particleinitially is moving with speed v0 how long will it take to come torest under the action of drag and how far will it travel during that

time Without solving this explicitly we can simply assume there isa characteristic time τ and displacement ` (the coasting length) andbalance terms setting

πa3ρ`

τ2 sim 6πmicroa`

τ(17)

to obtain

τ =29

νand ` = v0τ (18)

where ν = microρ is the kinematic viscosity with units of length2timeFrom the structure of Eq 16 we expect exponential decay of themotion with decay time τ

To get a feel for these quantities let us consider a bacterium witha sim 2 microm from Table 11 we then have τ sim 10minus6 s one microsecondIf the initial speed is say v0 = 10 microms the distance traveled is` sim 10minus5 microm = 10minus2 nm far less than the size of the bacterium andindeed just a small fraction of the Bohr radius (005 nm) We concludethat inertia is completely irrelevant to the motion of micron-sizedobjects in water In this simple picture inertia would only begin tomatter when the coasting length is comparable to the particle radiusSetting these equal leads to the critical radius

ac sim9ν

2v0 (19)

From this we see that even when the speed is sim 500 microms which isamong the highest speeds known for large multicellular algae suchas Volvox carteri shown in Fig 12 7 the critical radius would still be 7 RE Goldstein Green algae as model

organisms for biological fluid dynam-ics Annu Rev Fluid Mech 47343ndash3752015

sim 1 cm an order of magnitude larger than those organisms so thereis a vast range of organisms that swim in the low Reynolds numberworld the description of which will be presented in Chapter 7

daughter colonies

somatic cells

250 μm

Figure 12 The green alga Volvox Itconsists of sim 103 biflagellated somaticcells embedded in a transparent ex-tracellular matrix (ECM) with a smallnumber of daughter colonies that growup inside the spheroid

Using the Stokes drag relation we are in a position to understandthe relationship between thermal energy and diffusion as first doneby Einstein8 and Sutherland9 The basic idea is to consider a suspen- 8 A Einstein Investigations on the the-

ory of the Brownian movement Ann dPhys 17549ndash560 1905

9 W Sutherland LXXV A dynamicaltheory of diffusion for non-electrolytesand the molecular mass of albuminPhil Mag 9781ndash785 1905

sion of microscopic particles at concentration c(x) with diffusionconstant D acted on by a force F(x) = minusdφdx where φ is thepotential energy The flux J of particles is the sum of a Fickrsquos lawcontribution and an advective term with some speed u

J = minusDdcdx

+ uc (110)

where in the viscous regime the speed u is determined by the forcebalance ζu = minusdφdx where ζ = 6πmicroa for spherical particles Todetermine D by the principles of equilibrium statistical physics wemust indeed be in equilibrium so J = 0 Integrating to find theconcentration we obtain

c sim exp(minusφDζ) (111)

If this is to be consistent with the Boltzmann factor of statisticalphysics then Dζ = kBT or

D =kBT

6πmicroa 02

amicrom2s (112)

The second relation holds for water and a is the radius in micronsThis immediately reveals that a small molecule on the order of 2Aring(a sim 2times 10minus4) has D sim 103 microm2s whereas a 2 microm microsphere ofthe kind used to trace flows has D sim 01 microm2s

From the diffusion equation

partcpartt

= Dnabla2c (113)

we see by dimensional analysis that lengths ` and time t are relatedby Dt sim `2 or

t sim `2

D (114)

From the values of D derived above we see that the time it takesa small molecule to diffuse across a bacterium (` sim 1 microm) is 10minus3swhile that to cross a colony of Volvox (` sim 300) is about 103 s sim16 minutes On the other hand consider the cels of the aquatic plantChara corallina (Fig 13) Its cells are some of the largest known to sci-ence they can reach ` sim 10 cm = 105 microm giving a diffusive timescaleof sim 107 s Remembering that there are about 10π megaseconds in ayear (315times 107) this time is about 4 months

Figure 13 The aquatic plant Characorallina

Clearly diffusion is not effective for large eukaryotic cells andsome active transport mechanism is needed This point was em-phasized eloquently by JBS Haldane many years ago in a famousessay10 One such mechanism involves cargo transport by ldquomotor 10 JBS Haldane On Being the Right

Size 1927proteins moving along filaments in the cell or indeed by the fluidentrained by their motion a phenomenon termed cytoplasmic stream-ing that was first discovered in plants such as Chara as long ago as1774

11 At larger scales there are ever more complex vascular sys- 11 B Corti Osservazioni Microscopischesulla Tremella e sulla Circolazione del Flu-ido in Una Pianta Acquajuola AppressoGiuseppe Rocchi Lucca Italy 1774

tems driving by muscular pumps We shall discuss this at greaterlength in Chapter 9

As a final exercise in the use of dimensional analysis we ask aboutthe typical scale of voltages to be expected inside living systemsWe have in mind a situation like that in neurons in which thereare differences in ionic concentrations across membranes Since thechemical potential for a dilute solution of ions at concentration c iskBT ln c the difference in chemical potential between the inside andoutside of a membrane must be kBT ln(cincout) Setting this equal tothe electrostic energy eV for an elementary charge to move throughthe voltage difference across the membrane we find

V sim kBTe

) (115)

From Table 11 we obtain kBTe sim 25 mV which is indeed the basicscale of voltages across neuronal membranes12 as we shall discuss

12 AL Hodgkin and AF Huxley Aquantitative description of membranecurrent and its application to conduc-tance and excitation in nerve J Physiol117500ndash544 1952

in Chapter 10

2Microscopic Physics (Lectures 2 amp 3)

21 Review of molecular physics

At the smallest scales of interest in these lectures we are concernedwith the elementary interactions between molecules and how theydetermine interactions between more extended objects such as mem-branes To that end we review the basics features of van der Walls in-teractions between atoms and of screened electrostatics along withsome basic principles in non-ideal gas theory that serve as the basisof molecular solutions including those that can separate into distinctphases Although it might seem that a discussion of matters relatedto equilibrium phase transitions is somewhat off-topic in a courseon biological physics it is important to note that a rapidly growingbody of recent work has shown that the phenomenon of liquid-liquidphase separation plays a crucial role inside many cells1 1 AA Hyman CA Weber and

F Juumllicher Liquid-liquid phase sepa-ration in biology Annu Rev Cell DevBiol 3039ndash58 2014

The ideal gas law

pV = NkBT orp

kBT= ρ (21)

relates the pressure p volume V and number of molecules N wherekB is the Boltzmann constant T is the absolute temperature and ρ =

NV is the density It is only an approximation At low densitiesreal gases are described by a virial expansion

ρ + B2(T)ρ2 + B3(T)ρ3 + middot middot middot (22)

where the Bi are termed virial coefficients Intuitively the quadraticterm in ρ involves 2-body interactions the cubic term captures 3-body effects and so on Experimental measurements (Fig 21) of thesecond virial coefficient B2 show that it is negative at low tempera-tures (indicating attraction between pairs) and becomes positive (re-pulsive) at high temperatures crossing through zero at the so-calledBoyle point TB

Figure 21 Temperature dependence ofthe second virial coefficient of a gas

One of the great early triumphs of statistical physics was to showhow B2(T) arises from the underlying intermolecular potential Itsderivation gives us an opportunity to review some basic features ofclassical statistical physics that will be needed later For the pur-pose of this deviation it is more convenient to work in the canonicalensemble although the more systematic approach uses the grand

canonical ensemble2 We assume that classical statistical physics is 2 DA McQuarrie Statistical MechanicsHarper amp Row New York 1976valid whereby if the energy of a system of N particles is

E = sumi

2m+ sum

iltju(ri rj) (23)

where u is the intermolecular potential that we will assume is onlya function of the distance rij = |ri minus rj| and the restriction on thesum avoids double-counting The classical partition function for theN-particle system is then

intd3N p

intd3Nr eminusβE (24)

where β = 1kBT and the notation d3N p stands for dp1dp2 etcRecall that the factor of hminus3N is a vestige of the fully quantum ap-proach where one must measure volumes in phase space and servesto render Z dimensionless (recall that the product of a coordinateand a momentum has units of angular momentum (energytimestime)as does h The factor of 1N arises from the indistinguishability ofthe particles

Recognising that the configurational integralint

d3Nr middot middot middot is VN for anoninteracting system (u = 0) where V is the volume of the systemand that the momentum integrals can be done explicitly indepen-dent of the presence of u we can factor out the ideal noninteractingpartion function Z0 leaving the interacting part Q as

︸︷︷︸Z0

intd3Nr prod

iltjeminusβuij

︸︷︷︸Q

where uij = u(rij) and Λ = (2πh2mkBT)12 is the thermal deBroglie wavelength and we have written the exponential of a sumas the product of exponentials

The next step in developing a perturbation theory is to use thefirst important method of mathematical physics (adding zero) andthus to write

eminusβuij = 1 + fij with fij = eminusβuij minus 1 (26)

and observe that fij vanishes for a non-interacting system and servesas a small parameter for an expansion in density The quantityfij is known as the Mayer f -function3 From the exact expression 3 JE Mayer Statistical Mechanics of

Condensing Systems I J Chem Phys567 1937

prodiltj(1 + fij

)we can expand for small f noting that there are N(Nminus

1)2 sim N22 identical terms linear in fij leading to

intd3Nr

(1 + sum

iltjfij + middot middot middot

intd3r f (r) + middot middot middot (28)

where having done N minus 1 integrals there is one remaining factorof V in the denominator From the partition function Z we obtain

microscopic physics (lectures 2 amp 3) 17

the free energy F = minuskBT ln Z and then from F the pressure p =

minus(partFpartV)TN in the form of Eq 22 and thus we identify

B2 = minus12

intd3r(

eminusβu(r) minus 1)

This is an exact expression for B2

Figure 22 Decomposition of molecularinteraction potential (red) into purelyattractive and purely repulsive compo-nents (dashed)

A typical intermolecular potential u(r) has a strong repulsion atshort distances and a long-range attractive tail falling off as rminus6 asr rarr infin due to van der Waals interactions (the details of which are dis-cussed below) It was van der Waals who in his revolutionary 1873

PhD thesis4 had the crucial idea that the thermodynamics of liquids

4 JD van der Waals Over de Continuiteitvan den Gas- en Vloeistoftoestand (On thecontinuity of the gas and liquid state) PhDthesis Leiden University 1873

and gases potential could be understood by partitioning the inter-molecular potential into two parts a purely repulsive part that leadto ldquoexcluded volume around each molecule and a purely attractivepart whose effects on the thermodynamics could be estimated withinperturbation theory We can compare this calculation to our resultson the virial coefficient

In van der Waalsrsquo calculation of the contribution of ua to the en-ergy of a gas we assume the probability distribution of particlesaround a given one (the radial distribution function) is uniformrather than having the characteristic oscillations due to particle im-penetrability Under this assumption the energy Ea is

Ea =12

Nρint

d3rua(r) (210)

where again the factor of 12 avoids double-counting We then de-fine the parameter a as

a = minus12

intd3r ua(r) (211)

so Ea = minusaNρ and the contribution to the pressure isminus(partEapartV)TN =

minusaρ2 Thus a first correction to the equation of state is

p ρkBT minus aρ2 (212)

Further van der Waals realized that effect of the hard-core interac-tions could be accounted for by subtracting from the total volumeV an amount proportional to the number of particles N with aneffective excluded volume per particle b Putting these two effectstogether one has the revised equation of state

(p + aρ2)(V minus Nb) = NkBT (213)

Expanding for small ρ we find pkBT sim ρ + (bminus akBT) ρ2 + so

B2(T) = bminus akBT

This has a form that is qualitatively like that seen in experimentwith saturation at high temperatures to a constant reflecting entropiceffects and a divergence to minusinfin at low temperatures as the attractivepart of the potential dominates A final point to make is that this

result emerges naturally from the exact form of B2 in Eq 29 if weassume the interaction potential is +infin between r = 0 and someradius σ and sufficiently weak beyond σ that the βua(r) 1 Then

B2(T) =2πσ3

3minus a

kBT (215)

where a is as defined in Eq 211 with the integration running from σ

to infin This gives a precise meaning to the excluded volume parameterb σ is twice the hard-core radius of the sphere so b is four times thevolume of the particle

An important lesson from above is that in many contexts interac-tion terms quadratic in a density or concentration variable often mustbe interpreted as free energies involving energy and entropy ratherthan being purely energetic Moreover we see how the integratedstrength of pairwise interactions arise naturally through mean fieldarguments They may also be understood as the zero-momentumvalues of the Fourier transform of the potential but for that trans-form to be well-defined it is necessary to remove the singular repul-sive contribution to the potential as we did in the decomposition ofFig 22

22 Van der Waals interactions

Van der Waalsrsquo name is of course also associated with the fluctuating-dipole interactions between neutral objects To get insight into thephysics responsible for the long-range attraction between neutralatoms or molecules we follow essentially verbatim the very nicederivation by Holstein5 The model employed is two charged har- 5 BR Holstein The van der Waals in-

teraction Am J Phys 69441 2001monic oscillators with positive charges that are fixed in place at someseparation R and whose negative charges can oscillate back and forthunder the action of a spring of constant k as in Figure 23 + _ + _

Figure 23 Geometry of two interactingldquoatoms separated by a distance R eachwith electron and proton connected bya spring of constant k with internal dis-placements between each electron andits nucleus of x12

The artifice of a spring connecting each proton and electron servesto define a natural oscillation frequency ω0 =

radickm for each ldquoatom

where m is the electron mass The total Hamiltonian H is a sum ofthe electron kinetic energy and spring energy H0

H0 =p2

mω20x2

2 (216)

and the electrostatic interactions H1

H1 = e2[

1Rminus x1 + x2

minus 1Rminus x1

minus 1R + x2

] (217)

This is exact but rather cumbersome The physically interesting limitis when the separation R |x1| |x2| where the interaction simplifiesto

H1 asymp minus2e2x1x2

R3 (218)

whose Rminus3 fall-off with distance we recognize as a dipole term Thiscross term can be eliminated through the coordinate transformation

xplusmn =x1 plusmn x2radic

2rArr x1 =

x+ + xminusradic2

x2 =x+ minus xminusradic

2 (219)

The total Hamiltonian then becomes

mω2+x2

mω2minusx2minus (220)

where the new frequencies are

ω2plusmn = ω2

0 plusmn2e2

mR3 (221)

Having diagonalised the Hamiltonian the interaction potential u(r)is just the change to the ground state energy of the system due to theCoulomb interactions namely the shift is zero-point energies

u(r) =12

hω+ +12

hωminus minus 2 middot 12

hω0 (222)

minus12

hω0(e2mω2

R6 + middot middot middot (223)

where we have expanded the frequencies ωplusmn to leading order Ingrouping the terms as shown above we have isolated a characteristicenergy (hω0) that sets the overall scale for the interaction It wouldbe typical of an internal excitation energy from an s state to a p statesince the interaction is fundamentally due to virtual transitions tostates with dipole moments Hence hω0 is typically several electronvolts where 1 eV is 40 kBT We emphasize that this is an attractiveinteraction

Recognizing the Rminus6 dependence we observe that e2mω20 must

be a characteristic volume (remember we are working in Gaussianunits) To check this note that the simplest relationship between aninduced electric dipole moment d and an applied electric field E is

d = αE (224)

where α is the polarisability The dimensions of d are chargetimeslength(Q middot L) and the units of E are QL2 and thus α sim L3 it is a volumeThat it is proper to interpret e2mω2

0 as the polarisability can be seenby placing the charged harmonic oscillators in Fig 23 in an electricfield E0x acting to displace the electrons only (recall that the positivecharges are fixed in place) so

H = H0 + eE0x1 + eE0x2 (225)

A little manipulation shows that this can be rewritten as

mω20z2

2 + middot middot middot z12 = x12 +eE0

where the ellipses represents unimportant constants We thus havea new pair of oscillators whose equilibrium positions are linearlyshifted by the field The induced dipole moment is the electroncharge times that shift namely e2E0mω2

0 and thus the polarisabilityis indeed

α =e2

Putting together all of the pieces we see that the interaction be-tween two fluctuating dipoles can be written in the compact form

u(r) = minus12

hω0α2

r6 (228)

It follows that the typical scale of this energy within a condensedphase of number density n is obtained by setting rminus6 sim n2 giving

u sim hω0 (αn)2 (229)

The quantity αn occupies a very important role in the theory ofmetal-insulator transitions which are predicted to occur when it ex-ceeds an O(1) threshold6 For most dense strongly insulating liquids 6 KF Herzfeld On atomic properties

which make an element a metal PhysRev 29701ndash705 1927

the product αn sim 01minus 02 This is relevant to the calculation of in-teracting extended bodies as we discuss in the next section

221 Interaction of extended objects

We now use the basic results from the previous section to understandthe van der Waals interaction between extended objects A prime ex-ample is the stacks of thylakoid membranes found in chloroplasts ofgreen algae as shown in Fig 247 The balance of forces that de-

7 Y Bussi E Shimoni A WeinerR Kapon D Charuvi R NevoE Efrati and Z Reich Fundamentalhelical geometry consolidates the plantphotosynthetic membrane Proc NatlAcad Sci USA 11622366ndash22375 2019

Figure 24 Thylakoid membranes A10 nm thick slice of lettuce chloroplastimaged with electron microscopy Theso-called grana (G) stacks of disc-shaped thylakoid membranes are inter-spersed with unstacked stroma thy-lakoids (SL) all immersed in water andenclosed in the chloroplast envelope(black arrow) Scale bar is 200 nmFrom Ref 7

termine the spacings observed in these structures has been analysedalong the lines we discuss below8

8 MJ Sculley JT Duniec SW ThorneWS Chow and NK Boardman Thestacking of chloroplast thylakoidsQuantitative analysis of the balance offorces between thylakoid membranes ofchloroplasts and the role of divalentions Arch Bioch Biophys 201339ndash3461980

Figure 25 Geometry of an atom inter-acting with a laterally infinite slab ofthickness δ

To calculate the interaction between slabs we start with the inter-action of a single atom with a laterally infinite slab as indicated inFig 25 Let us denote the interaction between two atoms as

u11(r) = minusCr6 (230)

Working in cylindrical coordinates the interaction u1S between oneatom and the slab is

u1S(z) =int 0

minusδdzprime

int 2π

0dφint infin

0rdrnu11(

radic(zminus zprime)2 + r2) (231)

= minus2πCnint z+δ

zdzprime

int infin

1(zprime2 + r2)3 (232)

= minusπCn6

(1z3 minus

1(z + δ)3

) (233)

where we changed variables from zprime to zprimeprime = zminus zprime in the second line(and then dropped a prime) and n is the number density of particlesin the slab Note that the power law rminus6 of the interparticle potentialhas become zminus3 by integration over three spatial dimensions

Figure 26 Geometry of two slabs ofthickness δ separated by face-to-facedistance d

Now we consider two slabs separated by a distance d With theresult 233 we calculate the slab-slab interaction uSS(d) by integrat-ing u1S(z) over the thickness of the second slab If A is the areaof that slab then there are Andz atoms each per differential thick-ness dz and thus attractive interaction energy per unit area VA(z) equivuSS(z)A is

VA(d) =int d+δ

ddznu1S(z)dz (234)

= minus AH12π

(1d2 minus

2(d + δ)2 +

1(d + 2δ)2

) (235)

where have introduced the Hamaker constant9 9 HC Hamaker The London-van derWaals attraction between spherical par-ticles Physica 41058ndash1072 1937AH = π2n2C (236)

with units of energy Based on our simple picture of the constant Cin Eq 228 we see that AH sim (12)π2hω0(αn)2 so with hω0 sim 3 eVand αn sim 01minus 02 as discussed above we estimate Ah sim 5minus 25 kBT

23 Screened electrostatic interactions

Many of the interesting objects in biology interact both through vander Waals forces and (screened) electrostatic interactions As the for-mer typically decay as an inverse power of distance and the latter(as we shall see) decay exponentially the combined potential is at-tractive at long distances and repulsive at short distances leading toa potential minimum of the kind that sets the spacing of thylakoidmembranes for example The behaviour of systems with van derWaals and screened electrostatic interactions was established in theclassic work of Verwey and Overbeek10 10 EJW Verwey and JThG Overbeek

Theory of the Stability of Lyophobic Col-loids Elsevier Publishing CompanyInc New York 1948

Two elementary charges in vacuum separated by a distance r havean electrostatic energy (in cgs units) of

E = eφ(r) =e2

r (237)

where φ = er is the electrostatic potential Essentially all of liv-ing matter is infused with water whose dielectric constant ε sim 80reduces the energy scale to

εr (238)

While reduced by nearly two orders of magnitude this interactionstill falls off only as an inverse power of distance A sense of the scaleof E is obtained by finding the distance at which it is comparable tothermal energy This Bjerrum length is

λB =e2

εkBTsim 07 nm (239)

Even pure water has small amounts of ionic species Water moleculesare in equilibrium with protons (H+) and hydroxyl ions (OHminus) ac-cording to the reaction scheme

H2O H+ + OHminus (240)

Recall that the pH of a solution is defined as minus log10 [H+] where the

notation [middot middot middot ] means concentration in moleslitre Thus a neutralpH of 7 corresponds to a proton concentration of 10minus7 Mlsim 6 times1013 cmminus3 These and any additional ions present will screen the bareelectrostatic interactions considered above mobile charges oppositein sign to a given charge will cluster around it in a diffuse cloudscreening it from other distant charges The standard formalism tocompute the ionic distribution and resulting interactions is knownas Poisson-Boltzmann theory or in its linearised form as Debye-Huumlckel theory11 They are based on two principles The first is the 11 P Debye and E Huumlckel The the-

ory of electrolytes I Lowering of freez-ing point and related phenomena (En-glish translation) Phys Zeit 24185ndash206 1923

Poisson equation relating the electrostatic potential φ to the chargedensity ρ

nabla2φ = minus4πρ

ε (241)

The second is the Boltzmann distribution relating the ionic concen-trations cs of the species s of charge zse (with zs the valence) to theelectrostatic potential

cs = cs0eminusβzseφ (242)

where cs0 is a background concentration and β = 1kBT as usual

Combining these into a single self-consistent equation we obtain thePoisson-Boltzmann equation

nabla2φ = minus4π

ε sums

zsecs0eminusβzseφ (243)

It is simplest to consider the case of a z z electrolyte in whichthe dissolved neutral species dissociates into two ions of equal andopposite charges such as sodium chloride (1 1 NaClrarr Na++Clminus)or copper sulfate (2 2 CuSO4 rarr Cu2++SO2minus

4 ) Then we have

nabla2φ =8πzec0

εsinh(βzeφ) (244)

This is a very nonlinear equation for which analytical solutions canbe found only in simplified geometries Much of the importantphysics of screened electrostatics can be seen in the weak field limitwhen βzeφ 1 where we keep only the linearised rhs of 244 andobtain the Debye-Huumlckel equation(

nabla2 minus κ2)

φ = 0 (245)

where κminus1 is the Debye-Huckel length

κminus1 =

[εkBT

8πz2e2c0

]12sim 10 nmradic

c0[mM] (246)

Here we have expressed the concentration in the units of millimolar(mM) as typical concentrations of ions such as sodium are in the

range 1minus 200 mM We conclude that the screening length is nano-metric in typical biophysical contexts

The mathematical problem presented by the Debye-Huumlckel equa-tion 245 is the modified Helmholtz equation Let us first solve this inone spatial dimension for the half-space above a surface lying inthe x minus y plane (z = 0) The relevant general solution of 245 in-volves the functions exp(plusmnκz) and the requirement that the solutionbe bounded as z rarr infin leaves only the decaying exponential If thesurface potential is fixed at φ0 say then

φ(z) = φ0 eminusκz (247)

The induced charge density σ0 on the surface can be computed inthe usual way (by considering a Gaussian pillbox that straddles thesurface) It obeys

minusn middotnablaφ

∣∣∣∣z=0

=4πσ0

εrArr σ0 =

4πφ0 (248)

where n is the outward normal to the surface An analogous calcu-lation yields the potential when the surface charge itself is specified

φ(z) =4πσ0

εκeminusκz (249)

Within Debye-Huumlckel theory the surface charge and surface poten-tial are linearly related but in the more general Poisson-Boltzmannsetting this relationship is nonlinear

Observe that the DH equation(nabla2 minus κ2) φ = 0 is the Euler-Lagrange

equation for the functional

intd3r[

12(nablaφ)2 +

κ2φ2]

where we employ the general Euler-Lagrange formula

δGδφ

= minus part

partxpart(middot middot middot )

partφx+

part(middot middot middot )partφ

where (middot middot middot ) is the integrand of the functional We recognize the firstterm in 250 as the electrostatic energy density εE28π The secondis the weak-field approximation of an entropic contribution

If we integrate by parts the term in 250 involving (nablaφ)2 we ob-tain from Greenrsquos first identity a bulk contribution that vanishes bythe DH equation leaving only the surface term

dSσφ (252)

where we have used 248 to express the normal derivative of thepotential in terms of the charge density σ This is the appropriatefree energy for a system in which the surface charge is specified andit is completely determined by those Neumann boundary conditionson the surface There is no free lunch for of course the potential onthe surface must be found from the solution of the DH everywhere

For situations with fixed surface potential rather than fixed chargethe surface free energy must be Legendre transformed which isequivalent to accounting for the work done against the battery thatheld the potential fixed This new free energy F is

F = G minusint

SdSσφ = minus1

dSσφ (253)

The factors of 12 in Eqs 252 and 253 arise from the linear relationbetween surface potential and surface charge In the more generalcase the free energy is expressed as a charging integral of the form

minusint

SdSint φ0

σ(φprime)dφ

prime (254)

For our simple planar examples above

G =2πσ2

A and F = minusεκφ2

A (255)

where A is the total area of the (planar) surface

231 Electrostatic interactions between surfaces

Now we move on to calculate the electrostatic interaction betweentwo surfaces considering both the case of fixed charged densityand the case of fixed potential Although the case of a specifiedcharge density is more biophysically relevant than a specified poten-tial which one would imagine is appropriate to an electrochemicalcontext the comparison between the two serves to illustrate somegeneral features of the way that quadratic field theories determineinteractions between surfaces This takes on added relevance forneutral dipolar membranes which are believed to polarize the in-tervening water molecules and produce a hydration repulsion be-tween them12 measured by clever experiments some years ago13 12 S Marcelja and N Radic Repulsion

of interfaces due to boundary waterChem Phys Lett 42129ndash130 1976

13 LJ Lis M McAlister N Fuller RPRand and VA Parsegian Interactionsbetween neutral phospholipid bilayermembranes Biophys J 37657 1982

For two surfaces held at the same potential φ0 and located atplusmnd2 the potential is the symmetric combination of the fundamentalexponential solutions found previously

φ = φ0cosh(κz)

cosh(κd2) (256)

Using this we find the charge density at the plate at z = d2 Hereminusn middotnabla = ddz thus

σ(d2) =εκφ0

4πtanh(κd2) (257)

At the bottom plate minusn middotnabla = minusddz but with sinh(minusd2λ) =

minus sinh(d2λ) the charge density is the same As the charge andpotential do not vary with position over these flat surfaces the sur-face integration will just give a factor of the surface area A The freeenergy per unit area U(d) equiv F (d)A is

U(d) = minusεκφ2

tanh(κd2) (258)

In the limit drarr infin we recover twice the single-surface result 255Subtracting the self energy of the surfaces we obtain the repul-

sive interaction potential UR(d) = U(d)minusU(infin)

UR(d) =εκφ2

[1minus tanh

)] (259)

At large argument tanh approaches unity from below so this isclearly a repulsion as expected In detail if κd 1 we note thattanh(z) 1minus 2eminus2z + middot middot middot and thus

VR(d) εκφ2

eminusκd (260)

which therefore has the same exponential decay as the electrostaticpotential itself A second important point is that VR(d) is boundedas drarr 0 while the two charged surfaces are brought closer togethertheir induced charges decrease fast enough to avoid a singularity

Now we consider the case of two surfaces with fixed charge den-sity σ0 A simple calculation shows that the required potential is

φ(x) =4πσ0

cosh(κz)sinh(κd2)

with corresponding potential energy per unit area

VR(d) =4πσ2

)minus 1)

The function coth approaches unity from above at large argumentand thus we again have a repulsive interaction However this timethere is a divergence at short distances due to the fixed charge den-sities that are brought ever closer together

hydrocarbon tails

phosphatidylcholine

head group

C16 (palmitic acid)

Figure 27 The chemical structureof the phospholipid DPPC (dipalmi-toylphosphatidylcholine)

Let us estimate the scale of energies involved in the case of chargedlipid membranes Lipid molecules are amphiphilic with a struc-ture that is a fusion of parts that are hydrophilic (affinity for wa-ter) and hydrophobic (repelling water) Figure 27 shows the struc-ture of the lung surfactant dipalmitoylphosphatidylcholine (DPPC)which has two 16-carbon hydrophilic tails and a phosphatidyl-choline head group In diagrams of this type the zig-zag line rep-resents the repeating group CH2 terminated by the methyl groupCH3 with the hydrogen atom labels omitted for clarity These par-ticular hydrocarbon chains are said to be saturated because everycarbon atom is 4-fold coordinated bonded to two carbon atoms andtwo hydrogen atoms (the maximum possible) unlike unsaturatedlipids with one or more CminusminusC double bonds thereby having carbonatoms that have only a single bond to a hydrogen atom

When placed in water lipids self-organize into structures that iso-late the oily hydrocarbon tails from contact with water moleculesleaving the hydrophilic headgroups in contact with water This hy-drophobic effect plays a central role in cellular biology dictating thestructure of lipid assemblies the folding of proteins and localizationof channels in membranes14 The simplest structure is a micelle a

14 C Tanford The hydrophobic effectformation of micelles and biological mem-branes Wiley New York 1973

spherical or cylindrical structure with the tails on the inside and theheads facing outwards

hydrocarbon

head groupswater

Figure 28 Structure of a lipid mem-brane in water

Bilayer membranes have the structure shown in Fig 28 Theyare essentially thin fluid sheets a few nm thick in which typically(but not always) the lipid molecules are free to diffuse laterally Atypical charged head group has a single elementary charge in anarea of sim 1 nm2 If the Debye-Huumlckel screening length is sim 1 nmthen 4πσ2

0 κε sim 40 mNm which can be compared to the surfacetension of water (80 mNm) Expressed another way 40 mNm is40 pNmiddotnmnm2 or about 10 kBTnm2

Returning to the motivational problem of systems such as stacksof thylakoid membranes we now add the van der Waals interaction235 between two membranes to the screened electrostatic repulsion262 A convenient set of nondimensionalisations involves scalingdistances with the membrane thickness δ and the potential energywith the Hamaker constant Thus defining

x =dδ

K = κδ V =12πδ2

AH(VA + VR) (263)

we obtain

V(x) = Γ(

)minus 1)minus 1

(x + 1)2 minus1

(x + 2)2 (264)

Γ =24π2σ2

εκAH (265)

Using our estimates from above and taking a membrane thicknessof 3minus 6 nm we estimate Γ sim 100 and K gt 5 While the large sizeof Γ seems to imply a very large mismatch between the electrostaticand van der Waals scales of energy the exponential attenuation ofthe former allows a competition with the latter at distances beyond afew screening lengths The graph shown in fig 29 reveals an essen-tial feature of this competition namely a minimum in V at values ofx of order unity This would correspond to the equilibrium spacingof the thylakoid membranes Moreover the global minimum of theenergy is at contact (x = 0) where the van der Waals term diverges

x0 05 1 15 2

Figure 29 Interaction energy 264 be-tween membranes for Γ = 30 and vari-ous values of K as shown

As K increases ( corresponding to an increase in salt concentration)the minimum at larger x disappears leaving only the minimum atcontact This implies the well-known salting-out effect in colloidalsuspensions whereby an increase of dissolved salt leads to precipi-tation of the suspended particles

24 Geometrical aspects of screened electrostatics

In the previous section we considered only the simplest planar ge-ometries but many of the interesting structures in biology have non-trivial curvatures As the Debye-Huumlckel screening length is nano-metric while the radius of curvature R of structures within the cellis often much larger it is natural to exploit that separation of lengthscales to develop a systematic perturbative approach using κR asa small parameter Of particular interest historically has been thecontribution of screened electrostatic interactions to the elasticity ofmembranes Intuitively bending a charged surface brings the sur-face charges closer together producing a repulsion that effectivelyincreases the stiffness of the surface

In Chapter 5 we develop the necessary differential geometry ofsurfaces to deal with detailed problems in membrane physics Herewe simply recall that at every point on a surface there are two prin-cipal radii of curvature termed R1 and R2 and from these we canconstruct two important geometric quantities the mean curvature

)(266)

and the Gaussian curvature

R1R2 (267)

In the early 1970s Canham15 and Helfrich16 independently intro-

15 PB Canham The minimum energyof bending as a possible explanation ofthe biconcave shape of the human redblood cell J Theor Biol 2661ndash76 1970

16 W Helfrich Elastic properties oflipid bilayers theory and possible ex-periments Zur Naturforsch C 28693ndash703 1973

duced a model for the elasticity of fluid membranes of fixed areacomposed of lipids that are free to diffusive within the sheet thathave vanishing shear modulus They proposed that the characteris-tic biconcave discoid shapes of red blood cells (erythrocytes) shownin Fig 210 are minimisers of this energy under the constraint of afixed enclosed volume of water less than the equivalent sphere (thesphere having the same surface area as the membrane)

Figure 210 Electron micrograph of redblood cells

This functional is a surface integral of a quadratic form in thecurvatures

E =int

kc(H minus H0)2 +

Here kc and kc are the bending modulus and Gaussian curvaturemodulus of the membrane and H0 is the spontaneous curvature thatincorporates the possibility of a preferred curvature in the groundstate It is often a constant but can vary spatially if the membranecomposition is inhomogeneous The second integral in 268 is us-ing the Gauss-Bonnet theorem a constant that depends only on thetopology of the closed surface of interest

The problem of quantifying the electrostatic contribution to theelastic moduli and H0 can be addressed several ways The first17 is 17 M Winterhalter and W Helfrich Ef-

fect of surface charge on the curvatureelasticity of membranes J Phys Chem926865ndash6867 1988

a method by which comparison of the energies of planes cylindersand spheres yields those quantities directly The second is a bound-ary perturbation method for nearly planar surfaces18 while the third 18 RE Goldstein AI Pesci and

V Romero-Rochiacuten Electric doublelayers near modulated surfaces PhysRev A 415504ndash5515 1990

utilizes a powerful multiple-scattering method19 introduced in the

19 B Duplantier RE GoldsteinV Romero-Rochiacuten and AI PesciGeometrical and topological aspectsof electric double layers near curvedsurfaces Phys Rev Lett 65508ndash5111990

context of the wave equation20 We outline the essential features of

20 R Balian and C Bloch Distribu-tion of eigenfrequencies for the waveequation in a finite domain I Three-dimensional problem with smoothboundary surface Ann Phys (NY) 60401ndash447 1970

the first two approaches leaving the full calculations to the ExampleSheets

241 Comparative approach

For simplicity we focus on the problem of fixed surface charge den-sity σ0 On dimensional grounds we expect the elastic constants(units of energy) to scale as

kc kc sim4πσ2

0εκ3 (269)

and based on the estimate 4πσ20 εκ sim 10 kBTnm2 above Eq 263

and an assumed screening length of 1 nm these moduli are on theorder of 10 kBT Likewise as the screening length is the only lengthscale in the problem we expect H0 sim κminus1 The purpose of thesecalculations is to make these estimates precise

As we have already solved the problem for a single plane wenext consider the cylinder We wish to solve the modified Helmholtzequation 245 inside and outside a cylinder of radius R assumingaxisymmetry The Debye-Huumlckel equation becomes(

r2 part2

partr2 + rpart

partrminus (κr)2

)φ = 0 (270)

As this is homogeneous in powers of r the solution is a function ofκr The two solutions to this are modified Bessel functions K0(κr) for theouter problem (decaying at infinity) and I0(κr) for the inner problem(well-behaved at the origin) We solve the problem separately inthose two domains

Using the identity Iprime0(z) = I1(z) the inner solution is found to be

φ(r) =4πσ

I0(κr)I1(κR)

The (uniform) free energy per unit area of the inner surface is

∣∣∣∣S=

4πσ20

I0(κR)I1(κR)

As outlined above the limit of interest is κR 1 We then appeal tothe asymptotic results for modified Bessel functions [Abramowitz ampStegun21 (henceforth AampS) 971] 21 M Abramowitz and IA Stegun

Handbook of Mathematical Functions withFormulas Graphs and Mathematical Ta-bles Dover Publications Inc MineolaNY 1965

Iν(z) =ezradic

1minus microminus 1

(microminus 1)(microminus 9)2(8z)2 + middot middot middot

where micro = 4ν2 From these results we see that the exponentialprefactors of the asymptotic forms cancel in the ratio 272 leaving asimple expansion in inverse powers of κR

I0(κR)I1(κR)

sim 1 +1

38(κR)2 + middot middot middot (274)

Precise calculations must account for the finite thickness of the mem-brane and the associated charge density difference between the twosides We leave it to the student (Example Sheet 1) to complete thesecalculations

242 Perturbative approach

Next we sketch the basic features of a perturbative approach to find-ing the energetics of electric double layers near a non-flat boundary

Figure 211 A ripply surface held atfixed potential

In the simplest case the surface position is z = h(x) where h isa single-valued height function independent of y and the surfaceis held at a fixed potential φ0 (Fig 211) Thus we must solve theDebye-Huumlckel equation for the potential φ(x z) subject to the bound-ary condition

φ(x h(x)) = φ0 (275)

and φ = 0 at z = infin The mathematical challenge arises from the factthat in general we do not know the Greenrsquos function of the modi-fied Helmholtz operator for the domain bounded by some arbitraryheight function h(x) But we can perturbatively connect the solu-tions at finite h(x) to those at h = 0 where we know the solution

It is convenient to introduce a dimensionless small parameter ε asa counting device and to expand the surface boundary condition ina Taylor series

φ(x εh(x)) φ(x 0) + εh(x)φz(x 0) +12

ε2h(x)2φzz(x 0) + middot middot middot (276)

where subscripts denotes partial differentiation We expect that thesolution itselfmdashin the bulkmdashalso has an expansion in powers of ε

φ(x z)) φ(0)(x z) + εφ(1)(x z) + ε2φ(2)(x z) + middot middot middot (277)

The governing equation does not depend on ε if we substitute277 into 245 at every order we will have

(partxx + partzz minus κ2)φ(n)(x z) = 0 (278)

Merging the expansion of the boundary condition with the expan-sion of the solution we arrive at a sequence of boundary conditionsfor each order of solution At leading order we recover the boundarycondition at a flat surface

O(ε0) φ(0)(x 0) = φ0 (279)

Clearly the solution for all (x z) is

φ(0)(x z) = φ0eminusκz

At first order we find the boundary condition

O(ε1) φ(1)(x 0) = minush(x)φ(0)z (x 0) = κh(x)φ0 (280)

Thus the problem with a boundary condition on a ripply surface hasbecome one with a flat boundary but an inhomogeneous potentialAt quadratic order we have a similar kind of result

O(ε2) φ(2)(x 0) = minus12

h2(x)φ(0)zz (x 0)minus h(x)φ(1)

z (x 0) (281)

A convenient way to solve these boundary-value problems is towork in Fourier space Let us define the Fourier transforms of thepotential and h with respect to x (keeping the z-dependence of φ)

φ(m)(k z) =int

dxeikxφ(m)(x z) and h(q) =int

dxeiqxh(x) (282)

The modified Helmholtz equation (partxx + partzz minus κ2)φ = 0 becomes(partzz minus κ2

)φ(n)(q z) = 0 where κ2

q = κ2 + q2 (283)

and is solved by

φ(n)(q z) = φ(n)(q 0)eminusκqz (284)

To solve the sequence of boundary-value problems we need theFourier transform of φ at z = 0 this can be determined directly fromthe order-by-order boundary conditions For example at order ε1 wehave from 280

φ(1)(q 0) = κφ0h(q) (285)

In addition we must expand the surface normal vector

n = minusminushxex + ezradic1 + h2

in order to compute the surface charge as

σ(x) = minus ε

4πn middotnablaφ(x εh(x))

Details are left for the Example Sheet

3Fluctuations amp Brownian Motion (Lectures 4 amp 5)

4Biological Filaments (Lectures 6 amp 7)

5Membranes (Lectures 8 amp 9)

6Chemical Kinetics (Lectures 10 amp 11)

7Cellular Motion (Lectures 12-17)

8Chemotaxis (Lecture 18)

9Cytoplasmic Streaming (Lecture 19)

10Pattern Formation (Lectures 20-22)

11Population Dynamics (Lectures 23 amp 24)

12Bibliography

M Abramowitz and IA Stegun Handbook of Mathematical Functionswith Formulas Graphs and Mathematical Tables Dover PublicationsInc Mineola NY 1965

B Audoly and Y Pomeau Elasticity and Geometry Oxford UniversityPress Oxford UK 2010

R Balian and C Bloch Distribution of eigenfrequencies for the waveequation in a finite domain I Three-dimensional problem withsmooth boundary surface Ann Phys (NY) 60401ndash447 1970

HC Berg and EM Purcell Physics of chemoreception Biophys J20193ndash219 1977

R Brown XXVII A brief account of microscopical observations madein the months of June July and August 1827 on the particlescontained in the pollen of plants and on the general existence ofactive molecules in organic and inorganic bodies Philos Mag 4161ndash73 1828

Y Bussi E Shimoni A Weiner R Kapon D Charuvi R NevoE Efrati and Z Reich Fundamental helical geometry consolidatesthe plant photosynthetic membrane Proc Natl Acad Sci USA 11622366ndash22375 2019

PB Canham The minimum energy of bending as a possible expla-nation of the biconcave shape of the human red blood cell J TheorBiol 2661ndash76 1970

B Corti Osservazioni Microscopische sulla Tremella e sulla Circolazionedel Fluido in Una Pianta Acquajuola Appresso Giuseppe RocchiLucca Italy 1774

P Debye and E Huumlckel The theory of electrolytes I Lowering offreezing point and related phenomena (English translation) PhysZeit 24185ndash206 1923

B Duplantier RE Goldstein V Romero-Rochiacuten and AI Pesci Ge-ometrical and topological aspects of electric double layers nearcurved surfaces Phys Rev Lett 65508ndash511 1990

A Einstein Investigations on the theory of the Brownian movementAnn d Phys 17549ndash560 1905

TW Engelmann Ueber Sauerstoffausscheidung van Pfalnzenzellenim Mikrospektrum Pfluumlger Arch 27485ndash89 1882

RE Goldstein Green algae as model organisms for biological fluiddynamics Annu Rev Fluid Mech 47343ndash375 2015

RE Goldstein Coffee stains cell receptors and time crystalsLessons from the old literature Phys Today 7132ndash38 2018a

RE Goldstein Are theoretical lsquoresultsrsquo results eLife 7e40018 2018b

RE Goldstein AI Pesci and V Romero-Rochiacuten Electric doublelayers near modulated surfaces Phys Rev A 415504ndash5515 1990

JBS Haldane On Being the Right Size 1927

HC Hamaker The London-van der Waals attraction between spher-ical particles Physica 41058ndash1072 1937

W Helfrich Elastic properties of lipid bilayers theory and possibleexperiments Zur Naturforsch C 28693ndash703 1973

KF Herzfeld On atomic properties which make an element a metalPhys Rev 29701ndash705 1927

AL Hodgkin and AF Huxley A quantitative description of mem-brane current and its application to conductance and excitation innerve J Physiol 117500ndash544 1952

BR Holstein The van der Waals interaction Am J Phys 694412001

AA Hyman CA Weber and F Juumllicher Liquid-liquid phase sepa-ration in biology Annu Rev Cell Dev Biol 3039ndash58 2014

LD Landau and EM Lifshitz Quantum Mechanics Non-relativistictheory Pergamon Press Oxford UK 2nd edition 1965

CC Lin and LA Segel Mathematics Applied to Deterministic Problemsin the Natural Sciences Macmillan Publishing Co Inc New York1974

LJ Lis M McAlister N Fuller RP Rand and VA Parsegian Inter-actions between neutral phospholipid bilayer membranes BiophysJ 37657 1982

S Marcelja and N Radic Repulsion of interfaces due to boundarywater Chem Phys Lett 42129ndash130 1976

JC Maxwell Remarks on the mathematical classification of physicalquantities Proc London Math Soc s1-3224 1869

JE Mayer Statistical Mechanics of Condensing Systems I J ChemPhys 567 1937

bibliography 51

DA McQuarrie Statistical Mechanics Harper amp Row New York1976

P Nelson Biological Physics Energy Information Life Chiliagon Sci-ence Philadelphia PA student edition 2020

R Phillips J Kondev J Theriot HG Garcia and N Orme PhysicalBiology of the Cell Garland Science Boca Raton FL 2nd edition1998

EM Purcell Life at low Reynolds number Am J Phys 451ndash111977

MJ Sculley JT Duniec SW Thorne WS Chow and NK Board-man The stacking of chloroplast thylakoids Quantitative analysisof the balance of forces between thylakoid membranes of chloro-plasts and the role of divalent ions Arch Bioch Biophys 201339ndash346 1980

W Sutherland LXXV A dynamical theory of diffusion for non-electrolytes and the molecular mass of albumin Phil Mag 9781ndash785 1905

C Tanford The hydrophobic effect formation of micelles and biologicalmembranes Wiley New York 1973

DW Thompson On Growth and Form Dover Publications MineolaNY the complete revised edition 1992

JD van der Waals Over de Continuiteit van den Gas- en Vloeistoftoes-tand (On the continuity of the gas and liquid state) PhD thesis LeidenUniversity 1873

A van Leeuwenhoek IV Part of a letter from Mr Antony vanLeeuwenhoek concerning the worms in sheeps livers gnats andanimalcula in the excrements of frogs Philos Trans R Soc 22509ndash18 1700

EJW Verwey and JThG Overbeek Theory of the Stability of LyophobicColloids Elsevier Publishing Company Inc New York 1948

M Winterhalter and W Helfrich Effect of surface charge on thecurvature elasticity of membranes J Phys Chem 926865ndash68671988

Preface

Overview (Lecture 1)

Length scales and methodologies

Scaling arguments

Microscopic Physics (Lectures 2 amp 3)

Review of molecular physics

Van der Waals interactions

Screened electrostatic interactions

Geometrical aspects of screened electrostatics

Fluctuations amp Brownian Motion (Lectures 4 amp 5)

Biological Filaments (Lectures 6 amp 7)

Membranes (Lectures 8 amp 9)

Chemical Kinetics (Lectures 10 amp 11)

Cellular Motion (Lectures 12-17)

Chemotaxis (Lecture 18)

Cytoplasmic Streaming (Lecture 19)

Pattern Formation (Lectures 20-22)

Population Dynamics (Lectures 23 amp 24)

Bibliography

R AY M O N D E G O L D S T E I N A N D E R I C L A U G A

B I O L O G I C A L P H Y S I C SA N D F L U I D DY N A M I C S( A G R A D U AT E C O U R S E I N 2 4 L E C T U R E S )

D E PA R T M E N T O F A P P L I E D M AT H E M AT I C S A N DT H E O R E T I C A L P H Y S I C S U N I V E R S I T Y O F C A M B R I D G E

Contents

Preface 7

12 Bibliography 49

Preface

proteins hair

lipid vesicles

nm microm mm m km

10-9 m 10-6 m 10-3 m 1 m 103 m

neuroscience

paern formation

ecology

spermatozoa

ciliates

103 kgm3

E(r) sim n2h2

2mr2 minuse2

r (12)

rlowastn =n2h2

2n2h2 (13)

R =rrn

and E =E

me42n2h2 (14)

R2 minus2R

πa3ρ`

τ2 sim 6πmicroa`

τ(17)

to obtain

τ =29

νand ` = v0τ (18)

ac sim9ν

2v0 (19)

daughter colonies

somatic cells

250 μm

J = minusDdcdx

+ uc (110)

D =kBT

6πmicroa 02

amicrom2s (112)

partcpartt

= Dnabla2c (113)

t sim `2

D (114)

V sim kBTe

) (115)

in Chapter 10

The ideal gas law

pV = NkBT orp

kBT= ρ (21)

E = sumi

2m+ sum

iltju(ri rj) (23)

intd3N p

︸︷︷︸Z0

intd3Nr prod

iltjeminusβuij

︸︷︷︸Q

prodiltj(1 + fij

intd3Nr

(1 + sum

B2 = minus12

intd3r(

Ea =12

Nρint

d3rua(r) (210)

a = minus12

intd3r ua(r) (211)

B2(T) = bminus akBT

B2(T) =2πσ3

3minus a

kBT (215)

H0 =p2

mω20x2

2 (216)

H1 = e2[

1Rminus x1 + x2

minus 1Rminus x1

minus 1R + x2

] (217)

R3 (218)

2rArr x1 =

x+ + xminusradic2

2 (219)

mω2+x2

ω2plusmn = ω2

0 plusmn2e2

mR3 (221)

u(r) =12

hω+ +12

hω0 (222)

minus12

hω0(e2mω2

d = αE (224)

H = H0 + eE0x1 + eE0x2 (225)

mω20z2

α =e2

u(r) = minus12

hω0α2

r6 (228)

u sim hω0 (αn)2 (229)

u11(r) = minusCr6 (230)

u1S(z) =int 0

minusδdzprime

int 2π

0dφint infin

0rdrnu11(

zdzprime

int infin

1(zprime2 + r2)3 (232)

= minusπCn6

(1z3 minus

1(z + δ)3

) (233)

VA(d) =int d+δ

ddznu1S(z)dz (234)

= minus AH12π

(1d2 minus

2(d + δ)2 +

1(d + 2δ)2

) (235)

E = eφ(r) =e2

r (237)

εr (238)

λB =e2

ε (241)

nabla2φ = minus4π

ε sums

4 ) Then we have

nabla2φ =8πzec0

nabla2 minus κ2)

φ = 0 (245)

κminus1 =

[εkBT

8πz2e2c0

]12sim 10 nmradic

c0[mM] (246)

∣∣∣∣z=0

=4πσ0

εrArr σ0 =

4πφ0 (248)

φ(z) =4πσ0

εκeminusκz (249)

intd3r[

12(nablaφ)2 +

κ2φ2]

δGδφ

= minus part

partφx+

dSσφ (252)

F = G minusint

SdSσφ = minus1

dSσφ (253)

minusint

SdSint φ0

σ(φprime)dφ

prime (254)

G =2πσ2

A (255)

φ = φ0cosh(κz)

cosh(κd2) (256)

σ(d2) =εκφ0

4πtanh(κd2) (257)

U(d) = minusεκφ2

tanh(κd2) (258)

UR(d) =εκφ2

[1minus tanh

)] (259)

VR(d) εκφ2

eminusκd (260)

φ(x) =4πσ0

cosh(κz)sinh(κd2)

VR(d) =4πσ2

)minus 1)

hydrocarbon tails

phosphatidylcholine

head group

C16 (palmitic acid)

hydrocarbon

head groupswater

x =dδ

K = κδ V =12πδ2

AH(VA + VR) (263)

we obtain

V(x) = Γ(

)minus 1)minus 1

(x + 1)2 minus1

(x + 2)2 (264)

Γ =24π2σ2

εκAH (265)

x0 05 1 15 2

)(266)

R1R2 (267)

E =int

kc(H minus H0)2 +

kc kc sim4πσ2

0εκ3 (269)

r2 part2

partr2 + rpart

partrminus (κr)2

)φ = 0 (270)

φ(r) =4πσ

I0(κr)I1(κR)

∣∣∣∣S=

4πσ20

I0(κR)I1(κR)

Iν(z) =ezradic

1minus microminus 1

I0(κR)I1(κR)

sim 1 +1

φ(x h(x)) = φ0 (275)

φ(x εh(x)) φ(x 0) + εh(x)φz(x 0) +12

O(ε0) φ(0)(x 0) = φ0 (279)

O(ε2) φ(2)(x 0) = minus12

z (x 0) (281)

φ(m)(k z) =int

dxeiqxh(x) (282)

q = κ2 + q2 (283)

and is solved by

φ(1)(q 0) = κφ0h(q) (285)

σ(x) = minus ε

12Bibliography

bibliography 51

Preface


Scaling arguments















Bibliography

Contents

Preface 7

12 Bibliography 49

Preface

proteins hair

lipid vesicles

nm microm mm m km

10-9 m 10-6 m 10-3 m 1 m 103 m

neuroscience

paern formation

ecology

spermatozoa

ciliates

103 kgm3

E(r) sim n2h2

2mr2 minuse2

r (12)

rlowastn =n2h2

2n2h2 (13)

R =rrn

and E =E

me42n2h2 (14)

R2 minus2R

πa3ρ`

τ2 sim 6πmicroa`

τ(17)

to obtain

τ =29

νand ` = v0τ (18)

ac sim9ν

2v0 (19)

daughter colonies

somatic cells

250 μm

J = minusDdcdx

+ uc (110)

D =kBT

6πmicroa 02

amicrom2s (112)

partcpartt

= Dnabla2c (113)

t sim `2

D (114)

V sim kBTe

) (115)

in Chapter 10

The ideal gas law

pV = NkBT orp

kBT= ρ (21)

E = sumi

2m+ sum

iltju(ri rj) (23)

intd3N p

︸︷︷︸Z0

intd3Nr prod

iltjeminusβuij

︸︷︷︸Q

prodiltj(1 + fij

intd3Nr

(1 + sum

B2 = minus12

intd3r(

Ea =12

Nρint

d3rua(r) (210)

a = minus12

intd3r ua(r) (211)

B2(T) = bminus akBT

B2(T) =2πσ3

3minus a

kBT (215)

H0 =p2

mω20x2

2 (216)

H1 = e2[

1Rminus x1 + x2

minus 1Rminus x1

minus 1R + x2

] (217)

R3 (218)

2rArr x1 =

x+ + xminusradic2

2 (219)

mω2+x2

ω2plusmn = ω2

0 plusmn2e2

mR3 (221)

u(r) =12

hω+ +12

hω0 (222)

minus12

hω0(e2mω2

d = αE (224)

H = H0 + eE0x1 + eE0x2 (225)

mω20z2

α =e2

u(r) = minus12

hω0α2

r6 (228)

u sim hω0 (αn)2 (229)

u11(r) = minusCr6 (230)

u1S(z) =int 0

minusδdzprime

int 2π

0dφint infin

0rdrnu11(

zdzprime

int infin

1(zprime2 + r2)3 (232)

= minusπCn6

(1z3 minus

1(z + δ)3

) (233)

VA(d) =int d+δ

ddznu1S(z)dz (234)

= minus AH12π

(1d2 minus

2(d + δ)2 +

1(d + 2δ)2

) (235)

E = eφ(r) =e2

r (237)

εr (238)

λB =e2

ε (241)

nabla2φ = minus4π

ε sums

4 ) Then we have

nabla2φ =8πzec0

nabla2 minus κ2)

φ = 0 (245)

κminus1 =

[εkBT

8πz2e2c0

]12sim 10 nmradic

c0[mM] (246)

∣∣∣∣z=0

=4πσ0

εrArr σ0 =

4πφ0 (248)

φ(z) =4πσ0

εκeminusκz (249)

intd3r[

12(nablaφ)2 +

κ2φ2]

δGδφ

= minus part

partφx+

dSσφ (252)

F = G minusint

SdSσφ = minus1

dSσφ (253)

minusint

SdSint φ0

σ(φprime)dφ

prime (254)

G =2πσ2

A (255)

φ = φ0cosh(κz)

cosh(κd2) (256)

σ(d2) =εκφ0

4πtanh(κd2) (257)

U(d) = minusεκφ2

tanh(κd2) (258)

UR(d) =εκφ2

[1minus tanh

)] (259)

VR(d) εκφ2

eminusκd (260)

φ(x) =4πσ0

cosh(κz)sinh(κd2)

VR(d) =4πσ2

)minus 1)

hydrocarbon tails

phosphatidylcholine

head group

C16 (palmitic acid)

hydrocarbon

head groupswater

x =dδ

K = κδ V =12πδ2

AH(VA + VR) (263)

we obtain

V(x) = Γ(

)minus 1)minus 1

(x + 1)2 minus1

(x + 2)2 (264)

Γ =24π2σ2

εκAH (265)

x0 05 1 15 2

)(266)

R1R2 (267)

E =int

kc(H minus H0)2 +

kc kc sim4πσ2

0εκ3 (269)

r2 part2

partr2 + rpart

partrminus (κr)2

)φ = 0 (270)

φ(r) =4πσ

I0(κr)I1(κR)

∣∣∣∣S=

4πσ20

I0(κR)I1(κR)

Iν(z) =ezradic

1minus microminus 1

I0(κR)I1(κR)

sim 1 +1

φ(x h(x)) = φ0 (275)

φ(x εh(x)) φ(x 0) + εh(x)φz(x 0) +12

O(ε0) φ(0)(x 0) = φ0 (279)

O(ε2) φ(2)(x 0) = minus12

z (x 0) (281)

φ(m)(k z) =int

dxeiqxh(x) (282)

q = κ2 + q2 (283)

and is solved by

φ(1)(q 0) = κφ0h(q) (285)

σ(x) = minus ε

12Bibliography

bibliography 51

Preface


Scaling arguments















Bibliography

Contents

Preface 7

12 Bibliography 49

Preface

proteins hair

lipid vesicles

nm microm mm m km

10-9 m 10-6 m 10-3 m 1 m 103 m

neuroscience

paern formation

ecology

spermatozoa

ciliates

103 kgm3

E(r) sim n2h2

2mr2 minuse2

r (12)

rlowastn =n2h2

2n2h2 (13)

R =rrn

and E =E

me42n2h2 (14)

R2 minus2R

πa3ρ`

τ2 sim 6πmicroa`

τ(17)

to obtain

τ =29

νand ` = v0τ (18)

ac sim9ν

2v0 (19)

daughter colonies

somatic cells

250 μm

J = minusDdcdx

+ uc (110)

D =kBT

6πmicroa 02

amicrom2s (112)

partcpartt

= Dnabla2c (113)

t sim `2

D (114)

V sim kBTe

) (115)

in Chapter 10

The ideal gas law

pV = NkBT orp

kBT= ρ (21)

E = sumi

2m+ sum

iltju(ri rj) (23)

intd3N p

︸︷︷︸Z0

intd3Nr prod

iltjeminusβuij

︸︷︷︸Q

prodiltj(1 + fij

intd3Nr

(1 + sum

B2 = minus12

intd3r(

Ea =12

Nρint

d3rua(r) (210)

a = minus12

intd3r ua(r) (211)

B2(T) = bminus akBT

B2(T) =2πσ3

3minus a

kBT (215)

H0 =p2

mω20x2

2 (216)

H1 = e2[

1Rminus x1 + x2

minus 1Rminus x1

minus 1R + x2

] (217)

R3 (218)

2rArr x1 =

x+ + xminusradic2

2 (219)

mω2+x2

ω2plusmn = ω2

0 plusmn2e2

mR3 (221)

u(r) =12

hω+ +12

hω0 (222)

minus12

hω0(e2mω2

d = αE (224)

H = H0 + eE0x1 + eE0x2 (225)

mω20z2

α =e2

u(r) = minus12

hω0α2

r6 (228)

u sim hω0 (αn)2 (229)

u11(r) = minusCr6 (230)

u1S(z) =int 0

minusδdzprime

int 2π

0dφint infin

0rdrnu11(

zdzprime

int infin

1(zprime2 + r2)3 (232)

= minusπCn6

(1z3 minus

1(z + δ)3

) (233)

VA(d) =int d+δ

ddznu1S(z)dz (234)

= minus AH12π

(1d2 minus

2(d + δ)2 +

1(d + 2δ)2

) (235)

E = eφ(r) =e2

r (237)

εr (238)

λB =e2

ε (241)

nabla2φ = minus4π

ε sums

4 ) Then we have

nabla2φ =8πzec0

nabla2 minus κ2)

φ = 0 (245)

κminus1 =

[εkBT

8πz2e2c0

]12sim 10 nmradic

c0[mM] (246)

∣∣∣∣z=0

=4πσ0

εrArr σ0 =

4πφ0 (248)

φ(z) =4πσ0

εκeminusκz (249)

intd3r[

12(nablaφ)2 +

κ2φ2]

δGδφ

= minus part

partφx+

dSσφ (252)

F = G minusint

SdSσφ = minus1

dSσφ (253)

minusint

SdSint φ0

σ(φprime)dφ

prime (254)

G =2πσ2

A (255)

φ = φ0cosh(κz)

cosh(κd2) (256)

σ(d2) =εκφ0

4πtanh(κd2) (257)

U(d) = minusεκφ2

tanh(κd2) (258)

UR(d) =εκφ2

[1minus tanh

)] (259)

VR(d) εκφ2

eminusκd (260)

φ(x) =4πσ0

cosh(κz)sinh(κd2)

VR(d) =4πσ2

)minus 1)

hydrocarbon tails

phosphatidylcholine

head group

C16 (palmitic acid)

hydrocarbon

head groupswater

x =dδ

K = κδ V =12πδ2

AH(VA + VR) (263)

we obtain

V(x) = Γ(

)minus 1)minus 1

(x + 1)2 minus1

(x + 2)2 (264)

Γ =24π2σ2

εκAH (265)

x0 05 1 15 2

)(266)

R1R2 (267)

E =int

kc(H minus H0)2 +

kc kc sim4πσ2

0εκ3 (269)

r2 part2

partr2 + rpart

partrminus (κr)2

)φ = 0 (270)

φ(r) =4πσ

I0(κr)I1(κR)

∣∣∣∣S=

4πσ20

I0(κR)I1(κR)

Iν(z) =ezradic

1minus microminus 1

I0(κR)I1(κR)

sim 1 +1

φ(x h(x)) = φ0 (275)

φ(x εh(x)) φ(x 0) + εh(x)φz(x 0) +12

O(ε0) φ(0)(x 0) = φ0 (279)

O(ε2) φ(2)(x 0) = minus12

z (x 0) (281)

φ(m)(k z) =int

dxeiqxh(x) (282)

q = κ2 + q2 (283)

and is solved by

φ(1)(q 0) = κφ0h(q) (285)

σ(x) = minus ε

12Bibliography

bibliography 51

Preface


Scaling arguments















Bibliography

12 Bibliography 49

Preface

proteins hair

lipid vesicles

nm microm mm m km

10-9 m 10-6 m 10-3 m 1 m 103 m

neuroscience

paern formation

ecology

spermatozoa

ciliates

103 kgm3

E(r) sim n2h2

2mr2 minuse2

r (12)

rlowastn =n2h2

2n2h2 (13)

R =rrn

and E =E

me42n2h2 (14)

R2 minus2R

πa3ρ`

τ2 sim 6πmicroa`

τ(17)

to obtain

τ =29

νand ` = v0τ (18)

ac sim9ν

2v0 (19)

daughter colonies

somatic cells

250 μm

J = minusDdcdx

+ uc (110)

D =kBT

6πmicroa 02

amicrom2s (112)

partcpartt

= Dnabla2c (113)

t sim `2

D (114)

V sim kBTe

) (115)

in Chapter 10

The ideal gas law

pV = NkBT orp

kBT= ρ (21)

E = sumi

2m+ sum

iltju(ri rj) (23)

intd3N p

︸︷︷︸Z0

intd3Nr prod

iltjeminusβuij

︸︷︷︸Q

prodiltj(1 + fij

intd3Nr

(1 + sum

B2 = minus12

intd3r(

Ea =12

Nρint

d3rua(r) (210)

a = minus12

intd3r ua(r) (211)

B2(T) = bminus akBT

B2(T) =2πσ3

3minus a

kBT (215)

H0 =p2

mω20x2

2 (216)

H1 = e2[

1Rminus x1 + x2

minus 1Rminus x1

minus 1R + x2

] (217)

R3 (218)

2rArr x1 =

x+ + xminusradic2

2 (219)

mω2+x2

ω2plusmn = ω2

0 plusmn2e2

mR3 (221)

u(r) =12

hω+ +12

hω0 (222)

minus12

hω0(e2mω2

d = αE (224)

H = H0 + eE0x1 + eE0x2 (225)

mω20z2

α =e2

u(r) = minus12

hω0α2

r6 (228)

u sim hω0 (αn)2 (229)

u11(r) = minusCr6 (230)

u1S(z) =int 0

minusδdzprime

int 2π

0dφint infin

0rdrnu11(

zdzprime

int infin

1(zprime2 + r2)3 (232)

= minusπCn6

(1z3 minus

1(z + δ)3

) (233)

VA(d) =int d+δ

ddznu1S(z)dz (234)

= minus AH12π

(1d2 minus

2(d + δ)2 +

1(d + 2δ)2

) (235)

E = eφ(r) =e2

r (237)

εr (238)

λB =e2

ε (241)

nabla2φ = minus4π

ε sums

4 ) Then we have

nabla2φ =8πzec0

nabla2 minus κ2)

φ = 0 (245)

κminus1 =

[εkBT

8πz2e2c0

]12sim 10 nmradic

c0[mM] (246)

∣∣∣∣z=0

=4πσ0

εrArr σ0 =

4πφ0 (248)

φ(z) =4πσ0

εκeminusκz (249)

intd3r[

12(nablaφ)2 +

κ2φ2]

δGδφ

= minus part

partφx+

dSσφ (252)

F = G minusint

SdSσφ = minus1

dSσφ (253)

minusint

SdSint φ0

σ(φprime)dφ

prime (254)

G =2πσ2

A (255)

φ = φ0cosh(κz)

cosh(κd2) (256)

σ(d2) =εκφ0

4πtanh(κd2) (257)

U(d) = minusεκφ2

tanh(κd2) (258)

UR(d) =εκφ2

[1minus tanh

)] (259)

VR(d) εκφ2

eminusκd (260)

φ(x) =4πσ0

cosh(κz)sinh(κd2)

VR(d) =4πσ2

)minus 1)

hydrocarbon tails

phosphatidylcholine

head group

C16 (palmitic acid)

hydrocarbon

head groupswater

x =dδ

K = κδ V =12πδ2

AH(VA + VR) (263)

we obtain

V(x) = Γ(

)minus 1)minus 1

(x + 1)2 minus1

(x + 2)2 (264)

Γ =24π2σ2

εκAH (265)

x0 05 1 15 2

)(266)

R1R2 (267)

E =int

kc(H minus H0)2 +

kc kc sim4πσ2

0εκ3 (269)

r2 part2

partr2 + rpart

partrminus (κr)2

)φ = 0 (270)

φ(r) =4πσ

I0(κr)I1(κR)

∣∣∣∣S=

4πσ20

I0(κR)I1(κR)

Iν(z) =ezradic

1minus microminus 1

I0(κR)I1(κR)

sim 1 +1

φ(x h(x)) = φ0 (275)

φ(x εh(x)) φ(x 0) + εh(x)φz(x 0) +12

O(ε0) φ(0)(x 0) = φ0 (279)

O(ε2) φ(2)(x 0) = minus12

z (x 0) (281)

φ(m)(k z) =int

dxeiqxh(x) (282)

q = κ2 + q2 (283)

and is solved by

φ(1)(q 0) = κφ0h(q) (285)

σ(x) = minus ε

12Bibliography

bibliography 51

Preface


Scaling arguments















Bibliography

Preface

proteins hair

lipid vesicles

nm microm mm m km

10-9 m 10-6 m 10-3 m 1 m 103 m

neuroscience

paern formation

ecology

spermatozoa

ciliates

103 kgm3

E(r) sim n2h2

2mr2 minuse2

r (12)

rlowastn =n2h2

2n2h2 (13)

R =rrn

and E =E

me42n2h2 (14)

R2 minus2R

πa3ρ`

τ2 sim 6πmicroa`

τ(17)

to obtain

τ =29

νand ` = v0τ (18)

ac sim9ν

2v0 (19)

daughter colonies

somatic cells

250 μm

J = minusDdcdx

+ uc (110)

D =kBT

6πmicroa 02

amicrom2s (112)

partcpartt

= Dnabla2c (113)

t sim `2

D (114)

V sim kBTe

) (115)

in Chapter 10

The ideal gas law

pV = NkBT orp

kBT= ρ (21)

E = sumi

2m+ sum

iltju(ri rj) (23)

intd3N p

︸︷︷︸Z0

intd3Nr prod

iltjeminusβuij

︸︷︷︸Q

prodiltj(1 + fij

intd3Nr

(1 + sum

B2 = minus12

intd3r(

Ea =12

Nρint

d3rua(r) (210)

a = minus12

intd3r ua(r) (211)

B2(T) = bminus akBT

B2(T) =2πσ3

3minus a

kBT (215)

H0 =p2

mω20x2

2 (216)

H1 = e2[

1Rminus x1 + x2

minus 1Rminus x1

minus 1R + x2

] (217)

R3 (218)

2rArr x1 =

x+ + xminusradic2

2 (219)

mω2+x2

ω2plusmn = ω2

0 plusmn2e2

mR3 (221)

u(r) =12

hω+ +12

hω0 (222)

minus12

hω0(e2mω2

d = αE (224)

H = H0 + eE0x1 + eE0x2 (225)

mω20z2

α =e2

u(r) = minus12

hω0α2

r6 (228)

u sim hω0 (αn)2 (229)

u11(r) = minusCr6 (230)

u1S(z) =int 0

minusδdzprime

int 2π

0dφint infin

0rdrnu11(

zdzprime

int infin

1(zprime2 + r2)3 (232)

= minusπCn6

(1z3 minus

1(z + δ)3

) (233)

VA(d) =int d+δ

ddznu1S(z)dz (234)

= minus AH12π

(1d2 minus

2(d + δ)2 +

1(d + 2δ)2

) (235)

E = eφ(r) =e2

r (237)

εr (238)

λB =e2

ε (241)

nabla2φ = minus4π

ε sums

4 ) Then we have

nabla2φ =8πzec0

nabla2 minus κ2)

φ = 0 (245)

κminus1 =

[εkBT

8πz2e2c0

]12sim 10 nmradic

c0[mM] (246)

∣∣∣∣z=0

=4πσ0

εrArr σ0 =

4πφ0 (248)

φ(z) =4πσ0

εκeminusκz (249)

intd3r[

12(nablaφ)2 +

κ2φ2]

δGδφ

= minus part

partφx+

dSσφ (252)

F = G minusint

SdSσφ = minus1

dSσφ (253)

minusint

SdSint φ0

σ(φprime)dφ

prime (254)

G =2πσ2

A (255)

φ = φ0cosh(κz)

cosh(κd2) (256)

σ(d2) =εκφ0

4πtanh(κd2) (257)

U(d) = minusεκφ2

tanh(κd2) (258)

UR(d) =εκφ2

[1minus tanh

)] (259)

VR(d) εκφ2

eminusκd (260)

φ(x) =4πσ0

cosh(κz)sinh(κd2)

VR(d) =4πσ2

)minus 1)

hydrocarbon tails

phosphatidylcholine

head group

C16 (palmitic acid)

hydrocarbon

head groupswater

x =dδ

K = κδ V =12πδ2

AH(VA + VR) (263)

we obtain

V(x) = Γ(

)minus 1)minus 1

(x + 1)2 minus1

(x + 2)2 (264)

Γ =24π2σ2

εκAH (265)

x0 05 1 15 2

)(266)

R1R2 (267)

E =int

kc(H minus H0)2 +

kc kc sim4πσ2

0εκ3 (269)

r2 part2

partr2 + rpart

partrminus (κr)2

)φ = 0 (270)

φ(r) =4πσ

I0(κr)I1(κR)

∣∣∣∣S=

4πσ20

I0(κR)I1(κR)

Iν(z) =ezradic

1minus microminus 1

I0(κR)I1(κR)

sim 1 +1

φ(x h(x)) = φ0 (275)

φ(x εh(x)) φ(x 0) + εh(x)φz(x 0) +12

O(ε0) φ(0)(x 0) = φ0 (279)

O(ε2) φ(2)(x 0) = minus12

z (x 0) (281)

φ(m)(k z) =int

dxeiqxh(x) (282)

q = κ2 + q2 (283)

and is solved by

φ(1)(q 0) = κφ0h(q) (285)

σ(x) = minus ε

12Bibliography

bibliography 51

Preface


Scaling arguments















Bibliography

proteins hair

lipid vesicles

nm microm mm m km

10-9 m 10-6 m 10-3 m 1 m 103 m

neuroscience

paern formation

ecology

spermatozoa

ciliates

103 kgm3

E(r) sim n2h2

2mr2 minuse2

r (12)

rlowastn =n2h2

2n2h2 (13)

R =rrn

and E =E

me42n2h2 (14)

R2 minus2R

πa3ρ`

τ2 sim 6πmicroa`

τ(17)

to obtain

τ =29

νand ` = v0τ (18)

ac sim9ν

2v0 (19)

daughter colonies

somatic cells

250 μm

J = minusDdcdx

+ uc (110)

D =kBT

6πmicroa 02

amicrom2s (112)

partcpartt

= Dnabla2c (113)

t sim `2

D (114)

V sim kBTe

) (115)

in Chapter 10

The ideal gas law

pV = NkBT orp

kBT= ρ (21)

E = sumi

2m+ sum

iltju(ri rj) (23)

intd3N p

︸︷︷︸Z0

intd3Nr prod

iltjeminusβuij

︸︷︷︸Q

prodiltj(1 + fij

intd3Nr

(1 + sum

B2 = minus12

intd3r(

Ea =12

Nρint

d3rua(r) (210)

a = minus12

intd3r ua(r) (211)

B2(T) = bminus akBT

B2(T) =2πσ3

3minus a

kBT (215)

H0 =p2

mω20x2

2 (216)

H1 = e2[

1Rminus x1 + x2

minus 1Rminus x1

minus 1R + x2

] (217)

R3 (218)

2rArr x1 =

x+ + xminusradic2

2 (219)

mω2+x2

ω2plusmn = ω2

0 plusmn2e2

mR3 (221)

u(r) =12

hω+ +12

hω0 (222)

minus12

hω0(e2mω2

d = αE (224)

H = H0 + eE0x1 + eE0x2 (225)

mω20z2

α =e2

u(r) = minus12

hω0α2

r6 (228)

u sim hω0 (αn)2 (229)

u11(r) = minusCr6 (230)

u1S(z) =int 0

minusδdzprime

int 2π

0dφint infin

0rdrnu11(

zdzprime

int infin

1(zprime2 + r2)3 (232)

= minusπCn6

(1z3 minus

1(z + δ)3

) (233)

VA(d) =int d+δ

ddznu1S(z)dz (234)

= minus AH12π

(1d2 minus

2(d + δ)2 +

1(d + 2δ)2

) (235)

E = eφ(r) =e2

r (237)

εr (238)

λB =e2

ε (241)

nabla2φ = minus4π

ε sums

4 ) Then we have

nabla2φ =8πzec0

nabla2 minus κ2)

φ = 0 (245)

κminus1 =

[εkBT

8πz2e2c0

]12sim 10 nmradic

c0[mM] (246)

∣∣∣∣z=0

=4πσ0

εrArr σ0 =

4πφ0 (248)

φ(z) =4πσ0

εκeminusκz (249)

intd3r[

12(nablaφ)2 +

κ2φ2]

δGδφ

= minus part

partφx+

dSσφ (252)

F = G minusint

SdSσφ = minus1

dSσφ (253)

minusint

SdSint φ0

σ(φprime)dφ

prime (254)

G =2πσ2

A (255)

φ = φ0cosh(κz)

cosh(κd2) (256)

σ(d2) =εκφ0

4πtanh(κd2) (257)

U(d) = minusεκφ2

tanh(κd2) (258)

UR(d) =εκφ2

[1minus tanh

)] (259)

VR(d) εκφ2

eminusκd (260)

φ(x) =4πσ0

cosh(κz)sinh(κd2)

VR(d) =4πσ2

)minus 1)

hydrocarbon tails

phosphatidylcholine

head group

C16 (palmitic acid)

hydrocarbon

head groupswater

x =dδ

K = κδ V =12πδ2

AH(VA + VR) (263)

we obtain

V(x) = Γ(

)minus 1)minus 1

(x + 1)2 minus1

(x + 2)2 (264)

Γ =24π2σ2

εκAH (265)

x0 05 1 15 2

)(266)

R1R2 (267)

E =int

kc(H minus H0)2 +

kc kc sim4πσ2

0εκ3 (269)

r2 part2

partr2 + rpart

partrminus (κr)2

)φ = 0 (270)

φ(r) =4πσ

I0(κr)I1(κR)

∣∣∣∣S=

4πσ20

I0(κR)I1(κR)

Iν(z) =ezradic

1minus microminus 1

I0(κR)I1(κR)

sim 1 +1

φ(x h(x)) = φ0 (275)

φ(x εh(x)) φ(x 0) + εh(x)φz(x 0) +12

O(ε0) φ(0)(x 0) = φ0 (279)

O(ε2) φ(2)(x 0) = minus12

z (x 0) (281)

φ(m)(k z) =int

dxeiqxh(x) (282)

q = κ2 + q2 (283)

and is solved by

φ(1)(q 0) = κφ0h(q) (285)

σ(x) = minus ε

12Bibliography

bibliography 51

Preface


Scaling arguments















Bibliography

proteins hair

lipid vesicles

nm microm mm m km

10-9 m 10-6 m 10-3 m 1 m 103 m

neuroscience

paern formation

ecology

spermatozoa

ciliates

103 kgm3

E(r) sim n2h2

2mr2 minuse2

r (12)

rlowastn =n2h2

2n2h2 (13)

R =rrn

and E =E

me42n2h2 (14)

R2 minus2R

πa3ρ`

τ2 sim 6πmicroa`

τ(17)

to obtain

τ =29

νand ` = v0τ (18)

ac sim9ν

2v0 (19)

daughter colonies

somatic cells

250 μm

J = minusDdcdx

+ uc (110)

D =kBT

6πmicroa 02

amicrom2s (112)

partcpartt

= Dnabla2c (113)

t sim `2

D (114)

V sim kBTe

) (115)

in Chapter 10

The ideal gas law

pV = NkBT orp

kBT= ρ (21)

E = sumi

2m+ sum

iltju(ri rj) (23)

intd3N p

︸︷︷︸Z0

intd3Nr prod

iltjeminusβuij

︸︷︷︸Q

prodiltj(1 + fij

intd3Nr

(1 + sum

B2 = minus12

intd3r(

Ea =12

Nρint

d3rua(r) (210)

a = minus12

intd3r ua(r) (211)

B2(T) = bminus akBT

B2(T) =2πσ3

3minus a

kBT (215)

H0 =p2

mω20x2

2 (216)

H1 = e2[

1Rminus x1 + x2

minus 1Rminus x1

minus 1R + x2

] (217)

R3 (218)

2rArr x1 =

x+ + xminusradic2

2 (219)

mω2+x2

ω2plusmn = ω2

0 plusmn2e2

mR3 (221)

u(r) =12

hω+ +12

hω0 (222)

minus12

hω0(e2mω2

d = αE (224)

H = H0 + eE0x1 + eE0x2 (225)

mω20z2

α =e2

u(r) = minus12

hω0α2

r6 (228)

u sim hω0 (αn)2 (229)

u11(r) = minusCr6 (230)

u1S(z) =int 0

minusδdzprime

int 2π

0dφint infin

0rdrnu11(

zdzprime

int infin

1(zprime2 + r2)3 (232)

= minusπCn6

(1z3 minus

1(z + δ)3

) (233)

VA(d) =int d+δ

ddznu1S(z)dz (234)

= minus AH12π

(1d2 minus

2(d + δ)2 +

1(d + 2δ)2

) (235)

E = eφ(r) =e2

r (237)

εr (238)

λB =e2

ε (241)

nabla2φ = minus4π

ε sums

4 ) Then we have

nabla2φ =8πzec0

nabla2 minus κ2)

φ = 0 (245)

κminus1 =

[εkBT

8πz2e2c0

]12sim 10 nmradic

c0[mM] (246)

∣∣∣∣z=0

=4πσ0

εrArr σ0 =

4πφ0 (248)

φ(z) =4πσ0

εκeminusκz (249)

intd3r[

12(nablaφ)2 +

κ2φ2]

δGδφ

= minus part

partφx+

dSσφ (252)

F = G minusint

SdSσφ = minus1

dSσφ (253)

minusint

SdSint φ0

σ(φprime)dφ

prime (254)

G =2πσ2

A (255)

φ = φ0cosh(κz)

cosh(κd2) (256)

σ(d2) =εκφ0

4πtanh(κd2) (257)

U(d) = minusεκφ2

tanh(κd2) (258)

UR(d) =εκφ2

[1minus tanh

)] (259)

VR(d) εκφ2

eminusκd (260)

φ(x) =4πσ0

cosh(κz)sinh(κd2)

VR(d) =4πσ2

)minus 1)

hydrocarbon tails

phosphatidylcholine

head group

C16 (palmitic acid)

hydrocarbon

head groupswater

x =dδ

K = κδ V =12πδ2

AH(VA + VR) (263)

we obtain

V(x) = Γ(

)minus 1)minus 1

(x + 1)2 minus1

(x + 2)2 (264)

Γ =24π2σ2

εκAH (265)

x0 05 1 15 2

)(266)

R1R2 (267)

E =int

kc(H minus H0)2 +

kc kc sim4πσ2

0εκ3 (269)

r2 part2

partr2 + rpart

partrminus (κr)2

)φ = 0 (270)

φ(r) =4πσ

I0(κr)I1(κR)

∣∣∣∣S=

4πσ20

I0(κR)I1(κR)

Iν(z) =ezradic

1minus microminus 1

I0(κR)I1(κR)

sim 1 +1

φ(x h(x)) = φ0 (275)

φ(x εh(x)) φ(x 0) + εh(x)φz(x 0) +12

O(ε0) φ(0)(x 0) = φ0 (279)

O(ε2) φ(2)(x 0) = minus12

z (x 0) (281)

φ(m)(k z) =int

dxeiqxh(x) (282)

q = κ2 + q2 (283)

and is solved by

φ(1)(q 0) = κφ0h(q) (285)

σ(x) = minus ε

12Bibliography

bibliography 51

Preface


Scaling arguments















Bibliography

103 kgm3

E(r) sim n2h2

2mr2 minuse2

r (12)

rlowastn =n2h2

2n2h2 (13)

R =rrn

and E =E

me42n2h2 (14)

R2 minus2R

πa3ρ`

τ2 sim 6πmicroa`

τ(17)

to obtain

τ =29

νand ` = v0τ (18)

ac sim9ν

2v0 (19)

daughter colonies

somatic cells

250 μm

J = minusDdcdx

+ uc (110)

D =kBT

6πmicroa 02

amicrom2s (112)

partcpartt

= Dnabla2c (113)

t sim `2

D (114)

V sim kBTe

) (115)

in Chapter 10

The ideal gas law

pV = NkBT orp

kBT= ρ (21)

E = sumi

2m+ sum

iltju(ri rj) (23)

intd3N p

︸︷︷︸Z0

intd3Nr prod

iltjeminusβuij

︸︷︷︸Q

prodiltj(1 + fij

intd3Nr

(1 + sum

B2 = minus12

intd3r(

Ea =12

Nρint

d3rua(r) (210)

a = minus12

intd3r ua(r) (211)

B2(T) = bminus akBT

B2(T) =2πσ3

3minus a

kBT (215)

H0 =p2

mω20x2

2 (216)

H1 = e2[

1Rminus x1 + x2

minus 1Rminus x1

minus 1R + x2

] (217)

R3 (218)

2rArr x1 =

x+ + xminusradic2

2 (219)

mω2+x2

ω2plusmn = ω2

0 plusmn2e2

mR3 (221)

u(r) =12

hω+ +12

hω0 (222)

minus12

hω0(e2mω2

d = αE (224)

H = H0 + eE0x1 + eE0x2 (225)

mω20z2

α =e2

u(r) = minus12

hω0α2

r6 (228)

u sim hω0 (αn)2 (229)

u11(r) = minusCr6 (230)

u1S(z) =int 0

minusδdzprime

int 2π

0dφint infin

0rdrnu11(

zdzprime

int infin

1(zprime2 + r2)3 (232)

= minusπCn6

(1z3 minus

1(z + δ)3

) (233)

VA(d) =int d+δ

ddznu1S(z)dz (234)

= minus AH12π

(1d2 minus

2(d + δ)2 +

1(d + 2δ)2

) (235)

E = eφ(r) =e2

r (237)

εr (238)

λB =e2

ε (241)

nabla2φ = minus4π

ε sums

4 ) Then we have

nabla2φ =8πzec0

nabla2 minus κ2)

φ = 0 (245)

κminus1 =

[εkBT

8πz2e2c0

]12sim 10 nmradic

c0[mM] (246)

∣∣∣∣z=0

=4πσ0

εrArr σ0 =

4πφ0 (248)

φ(z) =4πσ0

εκeminusκz (249)

intd3r[

12(nablaφ)2 +

κ2φ2]

δGδφ

= minus part

partφx+

dSσφ (252)

F = G minusint

SdSσφ = minus1

dSσφ (253)

minusint

SdSint φ0

σ(φprime)dφ

prime (254)

G =2πσ2

A (255)

φ = φ0cosh(κz)

cosh(κd2) (256)

σ(d2) =εκφ0

4πtanh(κd2) (257)

U(d) = minusεκφ2

tanh(κd2) (258)

UR(d) =εκφ2

[1minus tanh

)] (259)

VR(d) εκφ2

eminusκd (260)

φ(x) =4πσ0

cosh(κz)sinh(κd2)

VR(d) =4πσ2

)minus 1)

hydrocarbon tails

phosphatidylcholine

head group

C16 (palmitic acid)

hydrocarbon

head groupswater

x =dδ

K = κδ V =12πδ2

AH(VA + VR) (263)

we obtain

V(x) = Γ(

)minus 1)minus 1

(x + 1)2 minus1

(x + 2)2 (264)

Γ =24π2σ2

εκAH (265)

x0 05 1 15 2

)(266)

R1R2 (267)

E =int

kc(H minus H0)2 +

kc kc sim4πσ2

0εκ3 (269)

r2 part2

partr2 + rpart

partrminus (κr)2

)φ = 0 (270)

φ(r) =4πσ

I0(κr)I1(κR)

∣∣∣∣S=

4πσ20

I0(κR)I1(κR)

Iν(z) =ezradic

1minus microminus 1

I0(κR)I1(κR)

sim 1 +1

φ(x h(x)) = φ0 (275)

φ(x εh(x)) φ(x 0) + εh(x)φz(x 0) +12

O(ε0) φ(0)(x 0) = φ0 (279)

O(ε2) φ(2)(x 0) = minus12

z (x 0) (281)

φ(m)(k z) =int

dxeiqxh(x) (282)

q = κ2 + q2 (283)

and is solved by

φ(1)(q 0) = κφ0h(q) (285)

σ(x) = minus ε

12Bibliography

bibliography 51

Preface


Scaling arguments















Bibliography

E(r) sim n2h2

2mr2 minuse2

r (12)

rlowastn =n2h2

2n2h2 (13)

R =rrn

and E =E

me42n2h2 (14)

R2 minus2R

πa3ρ`

τ2 sim 6πmicroa`

τ(17)

to obtain

τ =29

νand ` = v0τ (18)

ac sim9ν

2v0 (19)

daughter colonies

somatic cells

250 μm

J = minusDdcdx

+ uc (110)

D =kBT

6πmicroa 02

amicrom2s (112)

partcpartt

= Dnabla2c (113)

t sim `2

D (114)

V sim kBTe

) (115)

in Chapter 10

The ideal gas law

pV = NkBT orp

kBT= ρ (21)

E = sumi

2m+ sum

iltju(ri rj) (23)

intd3N p

︸︷︷︸Z0

intd3Nr prod

iltjeminusβuij

︸︷︷︸Q

prodiltj(1 + fij

intd3Nr

(1 + sum

B2 = minus12

intd3r(

Ea =12

Nρint

d3rua(r) (210)

a = minus12

intd3r ua(r) (211)

B2(T) = bminus akBT

B2(T) =2πσ3

3minus a

kBT (215)

H0 =p2

mω20x2

2 (216)

H1 = e2[

1Rminus x1 + x2

minus 1Rminus x1

minus 1R + x2

] (217)

R3 (218)

2rArr x1 =

x+ + xminusradic2

2 (219)

mω2+x2

ω2plusmn = ω2

0 plusmn2e2

mR3 (221)

u(r) =12

hω+ +12

hω0 (222)

minus12

hω0(e2mω2

d = αE (224)

H = H0 + eE0x1 + eE0x2 (225)

mω20z2

α =e2

u(r) = minus12

hω0α2

r6 (228)

u sim hω0 (αn)2 (229)

u11(r) = minusCr6 (230)

u1S(z) =int 0

minusδdzprime

int 2π

0dφint infin

0rdrnu11(

zdzprime

int infin

1(zprime2 + r2)3 (232)

= minusπCn6

(1z3 minus

1(z + δ)3

) (233)

VA(d) =int d+δ

ddznu1S(z)dz (234)

= minus AH12π

(1d2 minus

2(d + δ)2 +

1(d + 2δ)2

) (235)

E = eφ(r) =e2

r (237)

εr (238)

λB =e2

ε (241)

nabla2φ = minus4π

ε sums

4 ) Then we have

nabla2φ =8πzec0

nabla2 minus κ2)

φ = 0 (245)

κminus1 =

[εkBT

8πz2e2c0

]12sim 10 nmradic

c0[mM] (246)

∣∣∣∣z=0

=4πσ0

εrArr σ0 =

4πφ0 (248)

φ(z) =4πσ0

εκeminusκz (249)

intd3r[

12(nablaφ)2 +

κ2φ2]

δGδφ

= minus part

partφx+

dSσφ (252)

F = G minusint

SdSσφ = minus1

dSσφ (253)

minusint

SdSint φ0

σ(φprime)dφ

prime (254)

G =2πσ2

A (255)

φ = φ0cosh(κz)

cosh(κd2) (256)

σ(d2) =εκφ0

4πtanh(κd2) (257)

U(d) = minusεκφ2

tanh(κd2) (258)

UR(d) =εκφ2

[1minus tanh

)] (259)

VR(d) εκφ2

eminusκd (260)

φ(x) =4πσ0

cosh(κz)sinh(κd2)

VR(d) =4πσ2

)minus 1)

hydrocarbon tails

phosphatidylcholine

head group

C16 (palmitic acid)

hydrocarbon

head groupswater

x =dδ

K = κδ V =12πδ2

AH(VA + VR) (263)

we obtain

V(x) = Γ(

)minus 1)minus 1

(x + 1)2 minus1

(x + 2)2 (264)

Γ =24π2σ2

εκAH (265)

x0 05 1 15 2

)(266)

R1R2 (267)

E =int

kc(H minus H0)2 +

kc kc sim4πσ2

0εκ3 (269)

r2 part2

partr2 + rpart

partrminus (κr)2

)φ = 0 (270)

φ(r) =4πσ

I0(κr)I1(κR)

∣∣∣∣S=

4πσ20

I0(κR)I1(κR)

Iν(z) =ezradic

1minus microminus 1

I0(κR)I1(κR)

sim 1 +1

φ(x h(x)) = φ0 (275)

φ(x εh(x)) φ(x 0) + εh(x)φz(x 0) +12

O(ε0) φ(0)(x 0) = φ0 (279)

O(ε2) φ(2)(x 0) = minus12

z (x 0) (281)

φ(m)(k z) =int

dxeiqxh(x) (282)

q = κ2 + q2 (283)

and is solved by

φ(1)(q 0) = κφ0h(q) (285)

σ(x) = minus ε

12Bibliography

bibliography 51

Preface


Scaling arguments















Bibliography

πa3ρ`

τ2 sim 6πmicroa`

τ(17)

to obtain

τ =29

νand ` = v0τ (18)

ac sim9ν

2v0 (19)

daughter colonies

somatic cells

250 μm

J = minusDdcdx

+ uc (110)

D =kBT

6πmicroa 02

amicrom2s (112)

partcpartt

= Dnabla2c (113)

t sim `2

D (114)

V sim kBTe

) (115)

in Chapter 10

The ideal gas law

pV = NkBT orp

kBT= ρ (21)

E = sumi

2m+ sum

iltju(ri rj) (23)

intd3N p

︸︷︷︸Z0

intd3Nr prod

iltjeminusβuij

︸︷︷︸Q

prodiltj(1 + fij

intd3Nr

(1 + sum

B2 = minus12

intd3r(

Ea =12

Nρint

d3rua(r) (210)

a = minus12

intd3r ua(r) (211)

B2(T) = bminus akBT

B2(T) =2πσ3

3minus a

kBT (215)

H0 =p2

mω20x2

2 (216)

H1 = e2[

1Rminus x1 + x2

minus 1Rminus x1

minus 1R + x2

] (217)

R3 (218)

2rArr x1 =

x+ + xminusradic2

2 (219)

mω2+x2

ω2plusmn = ω2

0 plusmn2e2

mR3 (221)

u(r) =12

hω+ +12

hω0 (222)

minus12

hω0(e2mω2

d = αE (224)

H = H0 + eE0x1 + eE0x2 (225)

mω20z2

α =e2

u(r) = minus12

hω0α2

r6 (228)

u sim hω0 (αn)2 (229)

u11(r) = minusCr6 (230)

u1S(z) =int 0

minusδdzprime

int 2π

0dφint infin

0rdrnu11(

zdzprime

int infin

1(zprime2 + r2)3 (232)

= minusπCn6

(1z3 minus

1(z + δ)3

) (233)

VA(d) =int d+δ

ddznu1S(z)dz (234)

= minus AH12π

(1d2 minus

2(d + δ)2 +

1(d + 2δ)2

) (235)

E = eφ(r) =e2

r (237)

εr (238)

λB =e2

ε (241)

nabla2φ = minus4π

ε sums

4 ) Then we have

nabla2φ =8πzec0

nabla2 minus κ2)

φ = 0 (245)

κminus1 =

[εkBT

8πz2e2c0

]12sim 10 nmradic

c0[mM] (246)

∣∣∣∣z=0

=4πσ0

εrArr σ0 =

4πφ0 (248)

φ(z) =4πσ0

εκeminusκz (249)

intd3r[

12(nablaφ)2 +

κ2φ2]

δGδφ

= minus part

partφx+

dSσφ (252)

F = G minusint

SdSσφ = minus1

dSσφ (253)

minusint

SdSint φ0

σ(φprime)dφ

prime (254)

G =2πσ2

A (255)

φ = φ0cosh(κz)

cosh(κd2) (256)

σ(d2) =εκφ0

4πtanh(κd2) (257)

U(d) = minusεκφ2

tanh(κd2) (258)

UR(d) =εκφ2

[1minus tanh

)] (259)

VR(d) εκφ2

eminusκd (260)

φ(x) =4πσ0

cosh(κz)sinh(κd2)

VR(d) =4πσ2

)minus 1)

hydrocarbon tails

phosphatidylcholine

head group

C16 (palmitic acid)

hydrocarbon

head groupswater

x =dδ

K = κδ V =12πδ2

AH(VA + VR) (263)

we obtain

V(x) = Γ(

)minus 1)minus 1

(x + 1)2 minus1

(x + 2)2 (264)

Γ =24π2σ2

εκAH (265)

x0 05 1 15 2

)(266)

R1R2 (267)

E =int

kc(H minus H0)2 +

kc kc sim4πσ2

0εκ3 (269)

r2 part2

partr2 + rpart

partrminus (κr)2

)φ = 0 (270)

φ(r) =4πσ

I0(κr)I1(κR)

∣∣∣∣S=

4πσ20

I0(κR)I1(κR)

Iν(z) =ezradic

1minus microminus 1

I0(κR)I1(κR)

sim 1 +1

φ(x h(x)) = φ0 (275)

φ(x εh(x)) φ(x 0) + εh(x)φz(x 0) +12

O(ε0) φ(0)(x 0) = φ0 (279)

O(ε2) φ(2)(x 0) = minus12

z (x 0) (281)

φ(m)(k z) =int

dxeiqxh(x) (282)

q = κ2 + q2 (283)

and is solved by

φ(1)(q 0) = κφ0h(q) (285)

σ(x) = minus ε

12Bibliography

bibliography 51

Preface


Scaling arguments















Bibliography

partcpartt

= Dnabla2c (113)

t sim `2

D (114)

V sim kBTe

) (115)

in Chapter 10

The ideal gas law

pV = NkBT orp

kBT= ρ (21)

E = sumi

2m+ sum

iltju(ri rj) (23)

intd3N p

︸︷︷︸Z0

intd3Nr prod

iltjeminusβuij

︸︷︷︸Q

prodiltj(1 + fij

intd3Nr

(1 + sum

B2 = minus12

intd3r(

Ea =12

Nρint

d3rua(r) (210)

a = minus12

intd3r ua(r) (211)

B2(T) = bminus akBT

B2(T) =2πσ3

3minus a

kBT (215)

H0 =p2

mω20x2

2 (216)

H1 = e2[

1Rminus x1 + x2

minus 1Rminus x1

minus 1R + x2

] (217)

R3 (218)

2rArr x1 =

x+ + xminusradic2

2 (219)

mω2+x2

ω2plusmn = ω2

0 plusmn2e2

mR3 (221)

u(r) =12

hω+ +12

hω0 (222)

minus12

hω0(e2mω2

d = αE (224)

H = H0 + eE0x1 + eE0x2 (225)

mω20z2

α =e2

u(r) = minus12

hω0α2

r6 (228)

u sim hω0 (αn)2 (229)

u11(r) = minusCr6 (230)

u1S(z) =int 0

minusδdzprime

int 2π

0dφint infin

0rdrnu11(

zdzprime

int infin

1(zprime2 + r2)3 (232)

= minusπCn6

(1z3 minus

1(z + δ)3

) (233)

VA(d) =int d+δ

ddznu1S(z)dz (234)

= minus AH12π

(1d2 minus

2(d + δ)2 +

1(d + 2δ)2

) (235)

E = eφ(r) =e2

r (237)

εr (238)

λB =e2

ε (241)

nabla2φ = minus4π

ε sums

4 ) Then we have

nabla2φ =8πzec0

nabla2 minus κ2)

φ = 0 (245)

κminus1 =

[εkBT

8πz2e2c0

]12sim 10 nmradic

c0[mM] (246)

∣∣∣∣z=0

=4πσ0

εrArr σ0 =

4πφ0 (248)

φ(z) =4πσ0

εκeminusκz (249)

intd3r[

12(nablaφ)2 +

κ2φ2]

δGδφ

= minus part

partφx+

dSσφ (252)

F = G minusint

SdSσφ = minus1

dSσφ (253)

minusint

SdSint φ0

σ(φprime)dφ

prime (254)

G =2πσ2

A (255)

φ = φ0cosh(κz)

cosh(κd2) (256)

σ(d2) =εκφ0

4πtanh(κd2) (257)

U(d) = minusεκφ2

tanh(κd2) (258)

UR(d) =εκφ2

[1minus tanh

)] (259)

VR(d) εκφ2

eminusκd (260)

φ(x) =4πσ0

cosh(κz)sinh(κd2)

VR(d) =4πσ2

)minus 1)

hydrocarbon tails

phosphatidylcholine

head group

C16 (palmitic acid)

hydrocarbon

head groupswater

x =dδ

K = κδ V =12πδ2

AH(VA + VR) (263)

we obtain

V(x) = Γ(

)minus 1)minus 1

(x + 1)2 minus1

(x + 2)2 (264)

Γ =24π2σ2

εκAH (265)

x0 05 1 15 2

)(266)

R1R2 (267)

E =int

kc(H minus H0)2 +

kc kc sim4πσ2

0εκ3 (269)

r2 part2

partr2 + rpart

partrminus (κr)2

)φ = 0 (270)

φ(r) =4πσ

I0(κr)I1(κR)

∣∣∣∣S=

4πσ20

I0(κR)I1(κR)

Iν(z) =ezradic

1minus microminus 1

I0(κR)I1(κR)

sim 1 +1

φ(x h(x)) = φ0 (275)

φ(x εh(x)) φ(x 0) + εh(x)φz(x 0) +12

O(ε0) φ(0)(x 0) = φ0 (279)

O(ε2) φ(2)(x 0) = minus12

z (x 0) (281)

φ(m)(k z) =int

dxeiqxh(x) (282)

q = κ2 + q2 (283)

and is solved by

φ(1)(q 0) = κφ0h(q) (285)

σ(x) = minus ε

12Bibliography

bibliography 51

Preface


Scaling arguments















Bibliography

The ideal gas law

pV = NkBT orp

kBT= ρ (21)

E = sumi

2m+ sum

iltju(ri rj) (23)

intd3N p

︸︷︷︸Z0

intd3Nr prod

iltjeminusβuij

︸︷︷︸Q

prodiltj(1 + fij

intd3Nr

(1 + sum

B2 = minus12

intd3r(

Ea =12

Nρint

d3rua(r) (210)

a = minus12

intd3r ua(r) (211)

B2(T) = bminus akBT

B2(T) =2πσ3

3minus a

kBT (215)

H0 =p2

mω20x2

2 (216)

H1 = e2[

1Rminus x1 + x2

minus 1Rminus x1

minus 1R + x2

] (217)

R3 (218)

2rArr x1 =

x+ + xminusradic2

2 (219)

mω2+x2

ω2plusmn = ω2

0 plusmn2e2

mR3 (221)

u(r) =12

hω+ +12

hω0 (222)

minus12

hω0(e2mω2

d = αE (224)

H = H0 + eE0x1 + eE0x2 (225)

mω20z2

α =e2

u(r) = minus12

hω0α2

r6 (228)

u sim hω0 (αn)2 (229)

u11(r) = minusCr6 (230)

u1S(z) =int 0

minusδdzprime

int 2π

0dφint infin

0rdrnu11(

zdzprime

int infin

1(zprime2 + r2)3 (232)

= minusπCn6

(1z3 minus

1(z + δ)3

) (233)

VA(d) =int d+δ

ddznu1S(z)dz (234)

= minus AH12π

(1d2 minus

2(d + δ)2 +

1(d + 2δ)2

) (235)

E = eφ(r) =e2

r (237)

εr (238)

λB =e2

ε (241)

nabla2φ = minus4π

ε sums

4 ) Then we have

nabla2φ =8πzec0

nabla2 minus κ2)

φ = 0 (245)

κminus1 =

[εkBT

8πz2e2c0

]12sim 10 nmradic

c0[mM] (246)

∣∣∣∣z=0

=4πσ0

εrArr σ0 =

4πφ0 (248)

φ(z) =4πσ0

εκeminusκz (249)

intd3r[

12(nablaφ)2 +

κ2φ2]

δGδφ

= minus part

partφx+

dSσφ (252)

F = G minusint

SdSσφ = minus1

dSσφ (253)

minusint

SdSint φ0

σ(φprime)dφ

prime (254)

G =2πσ2

A (255)

φ = φ0cosh(κz)

cosh(κd2) (256)

σ(d2) =εκφ0

4πtanh(κd2) (257)

U(d) = minusεκφ2

tanh(κd2) (258)

UR(d) =εκφ2

[1minus tanh

)] (259)

VR(d) εκφ2

eminusκd (260)

φ(x) =4πσ0

cosh(κz)sinh(κd2)

VR(d) =4πσ2

)minus 1)

hydrocarbon tails

phosphatidylcholine

head group

C16 (palmitic acid)

hydrocarbon

head groupswater

x =dδ

K = κδ V =12πδ2

AH(VA + VR) (263)

we obtain

V(x) = Γ(

)minus 1)minus 1

(x + 1)2 minus1

(x + 2)2 (264)

Γ =24π2σ2

εκAH (265)

x0 05 1 15 2

)(266)

R1R2 (267)

E =int

kc(H minus H0)2 +

kc kc sim4πσ2

0εκ3 (269)

r2 part2

partr2 + rpart

partrminus (κr)2

)φ = 0 (270)

φ(r) =4πσ

I0(κr)I1(κR)

∣∣∣∣S=

4πσ20

I0(κR)I1(κR)

Iν(z) =ezradic

1minus microminus 1

I0(κR)I1(κR)

sim 1 +1

φ(x h(x)) = φ0 (275)

φ(x εh(x)) φ(x 0) + εh(x)φz(x 0) +12

O(ε0) φ(0)(x 0) = φ0 (279)

O(ε2) φ(2)(x 0) = minus12

z (x 0) (281)

φ(m)(k z) =int

dxeiqxh(x) (282)

q = κ2 + q2 (283)

and is solved by

φ(1)(q 0) = κφ0h(q) (285)

σ(x) = minus ε

12Bibliography

bibliography 51

Preface


Scaling arguments















Bibliography

E = sumi

2m+ sum

iltju(ri rj) (23)

intd3N p

︸︷︷︸Z0

intd3Nr prod

iltjeminusβuij

︸︷︷︸Q

prodiltj(1 + fij

intd3Nr

(1 + sum

B2 = minus12

intd3r(

Ea =12

Nρint

d3rua(r) (210)

a = minus12

intd3r ua(r) (211)

B2(T) = bminus akBT

B2(T) =2πσ3

3minus a

kBT (215)

H0 =p2

mω20x2

2 (216)

H1 = e2[

1Rminus x1 + x2

minus 1Rminus x1

minus 1R + x2

] (217)

R3 (218)

2rArr x1 =

x+ + xminusradic2

2 (219)

mω2+x2

ω2plusmn = ω2

0 plusmn2e2

mR3 (221)

u(r) =12

hω+ +12

hω0 (222)

minus12

hω0(e2mω2

d = αE (224)

H = H0 + eE0x1 + eE0x2 (225)

mω20z2

α =e2

u(r) = minus12

hω0α2

r6 (228)

u sim hω0 (αn)2 (229)

u11(r) = minusCr6 (230)

u1S(z) =int 0

minusδdzprime

int 2π

0dφint infin

0rdrnu11(

zdzprime

int infin

1(zprime2 + r2)3 (232)

= minusπCn6

(1z3 minus

1(z + δ)3

) (233)

VA(d) =int d+δ

ddznu1S(z)dz (234)

= minus AH12π

(1d2 minus

2(d + δ)2 +

1(d + 2δ)2

) (235)

E = eφ(r) =e2

r (237)

εr (238)

λB =e2

ε (241)

nabla2φ = minus4π

ε sums

4 ) Then we have

nabla2φ =8πzec0

nabla2 minus κ2)

φ = 0 (245)

κminus1 =

[εkBT

8πz2e2c0

]12sim 10 nmradic

c0[mM] (246)

∣∣∣∣z=0

=4πσ0

εrArr σ0 =

4πφ0 (248)

φ(z) =4πσ0

εκeminusκz (249)

intd3r[

12(nablaφ)2 +

κ2φ2]

δGδφ

= minus part

partφx+

dSσφ (252)

F = G minusint

SdSσφ = minus1

dSσφ (253)

minusint

SdSint φ0

σ(φprime)dφ

prime (254)

G =2πσ2

A (255)

φ = φ0cosh(κz)

cosh(κd2) (256)

σ(d2) =εκφ0

4πtanh(κd2) (257)

U(d) = minusεκφ2

tanh(κd2) (258)

UR(d) =εκφ2

[1minus tanh

)] (259)

VR(d) εκφ2

eminusκd (260)

φ(x) =4πσ0

cosh(κz)sinh(κd2)

VR(d) =4πσ2

)minus 1)

hydrocarbon tails

phosphatidylcholine

head group

C16 (palmitic acid)

hydrocarbon

head groupswater

x =dδ

K = κδ V =12πδ2

AH(VA + VR) (263)

we obtain

V(x) = Γ(

)minus 1)minus 1

(x + 1)2 minus1

(x + 2)2 (264)

Γ =24π2σ2

εκAH (265)

x0 05 1 15 2

)(266)

R1R2 (267)

E =int

kc(H minus H0)2 +

kc kc sim4πσ2

0εκ3 (269)

r2 part2

partr2 + rpart

partrminus (κr)2

)φ = 0 (270)

φ(r) =4πσ

I0(κr)I1(κR)

∣∣∣∣S=

4πσ20

I0(κR)I1(κR)

Iν(z) =ezradic

1minus microminus 1

I0(κR)I1(κR)

sim 1 +1

φ(x h(x)) = φ0 (275)

φ(x εh(x)) φ(x 0) + εh(x)φz(x 0) +12

O(ε0) φ(0)(x 0) = φ0 (279)

O(ε2) φ(2)(x 0) = minus12

z (x 0) (281)

φ(m)(k z) =int

dxeiqxh(x) (282)

q = κ2 + q2 (283)

and is solved by

φ(1)(q 0) = κφ0h(q) (285)

σ(x) = minus ε

12Bibliography

bibliography 51

Preface


Scaling arguments















Bibliography

B2 = minus12

intd3r(

Ea =12

Nρint

d3rua(r) (210)

a = minus12

intd3r ua(r) (211)

B2(T) = bminus akBT

B2(T) =2πσ3

3minus a

kBT (215)

H0 =p2

mω20x2

2 (216)

H1 = e2[

1Rminus x1 + x2

minus 1Rminus x1

minus 1R + x2

] (217)

R3 (218)

2rArr x1 =

x+ + xminusradic2

2 (219)

mω2+x2

ω2plusmn = ω2

0 plusmn2e2

mR3 (221)

u(r) =12

hω+ +12

hω0 (222)

minus12

hω0(e2mω2

d = αE (224)

H = H0 + eE0x1 + eE0x2 (225)

mω20z2

α =e2

u(r) = minus12

hω0α2

r6 (228)

u sim hω0 (αn)2 (229)

u11(r) = minusCr6 (230)

u1S(z) =int 0

minusδdzprime

int 2π

0dφint infin

0rdrnu11(

zdzprime

int infin

1(zprime2 + r2)3 (232)

= minusπCn6

(1z3 minus

1(z + δ)3

) (233)

VA(d) =int d+δ

ddznu1S(z)dz (234)

= minus AH12π

(1d2 minus

2(d + δ)2 +

1(d + 2δ)2

) (235)

E = eφ(r) =e2

r (237)

εr (238)

λB =e2

ε (241)

nabla2φ = minus4π

ε sums

4 ) Then we have

nabla2φ =8πzec0

nabla2 minus κ2)

φ = 0 (245)

κminus1 =

[εkBT

8πz2e2c0

]12sim 10 nmradic

c0[mM] (246)

∣∣∣∣z=0

=4πσ0

εrArr σ0 =

4πφ0 (248)

φ(z) =4πσ0

εκeminusκz (249)

intd3r[

12(nablaφ)2 +

κ2φ2]

δGδφ

= minus part

partφx+

dSσφ (252)

F = G minusint

SdSσφ = minus1

dSσφ (253)

minusint

SdSint φ0

σ(φprime)dφ

prime (254)

G =2πσ2

A (255)

φ = φ0cosh(κz)

cosh(κd2) (256)

σ(d2) =εκφ0

4πtanh(κd2) (257)

U(d) = minusεκφ2

tanh(κd2) (258)

UR(d) =εκφ2

[1minus tanh

)] (259)

VR(d) εκφ2

eminusκd (260)

φ(x) =4πσ0

cosh(κz)sinh(κd2)

VR(d) =4πσ2

)minus 1)

hydrocarbon tails

phosphatidylcholine

head group

C16 (palmitic acid)

hydrocarbon

head groupswater

x =dδ

K = κδ V =12πδ2

AH(VA + VR) (263)

we obtain

V(x) = Γ(

)minus 1)minus 1

(x + 1)2 minus1

(x + 2)2 (264)

Γ =24π2σ2

εκAH (265)

x0 05 1 15 2

)(266)

R1R2 (267)

E =int

kc(H minus H0)2 +

kc kc sim4πσ2

0εκ3 (269)

r2 part2

partr2 + rpart

partrminus (κr)2

)φ = 0 (270)

φ(r) =4πσ

I0(κr)I1(κR)

∣∣∣∣S=

4πσ20

I0(κR)I1(κR)

Iν(z) =ezradic

1minus microminus 1

I0(κR)I1(κR)

sim 1 +1

φ(x h(x)) = φ0 (275)

φ(x εh(x)) φ(x 0) + εh(x)φz(x 0) +12

O(ε0) φ(0)(x 0) = φ0 (279)

O(ε2) φ(2)(x 0) = minus12

z (x 0) (281)

φ(m)(k z) =int

dxeiqxh(x) (282)

q = κ2 + q2 (283)

and is solved by

φ(1)(q 0) = κφ0h(q) (285)

σ(x) = minus ε

12Bibliography

bibliography 51

Preface


Scaling arguments















Bibliography

B2(T) =2πσ3

3minus a

kBT (215)

H0 =p2

mω20x2

2 (216)

H1 = e2[

1Rminus x1 + x2

minus 1Rminus x1

minus 1R + x2

] (217)

R3 (218)

2rArr x1 =

x+ + xminusradic2

2 (219)

mω2+x2

ω2plusmn = ω2

0 plusmn2e2

mR3 (221)

u(r) =12

hω+ +12

hω0 (222)

minus12

hω0(e2mω2

d = αE (224)

H = H0 + eE0x1 + eE0x2 (225)

mω20z2

α =e2

u(r) = minus12

hω0α2

r6 (228)

u sim hω0 (αn)2 (229)

u11(r) = minusCr6 (230)

u1S(z) =int 0

minusδdzprime

int 2π

0dφint infin

0rdrnu11(

zdzprime

int infin

1(zprime2 + r2)3 (232)

= minusπCn6

(1z3 minus

1(z + δ)3

) (233)

VA(d) =int d+δ

ddznu1S(z)dz (234)

= minus AH12π

(1d2 minus

2(d + δ)2 +

1(d + 2δ)2

) (235)

E = eφ(r) =e2

r (237)

εr (238)

λB =e2

ε (241)

nabla2φ = minus4π

ε sums

4 ) Then we have

nabla2φ =8πzec0

nabla2 minus κ2)

φ = 0 (245)

κminus1 =

[εkBT

8πz2e2c0

]12sim 10 nmradic

c0[mM] (246)

∣∣∣∣z=0

=4πσ0

εrArr σ0 =

4πφ0 (248)

φ(z) =4πσ0

εκeminusκz (249)

intd3r[

12(nablaφ)2 +

κ2φ2]

δGδφ

= minus part

partφx+

dSσφ (252)

F = G minusint

SdSσφ = minus1

dSσφ (253)

minusint

SdSint φ0

σ(φprime)dφ

prime (254)

G =2πσ2

A (255)

φ = φ0cosh(κz)

cosh(κd2) (256)

σ(d2) =εκφ0

4πtanh(κd2) (257)

U(d) = minusεκφ2

tanh(κd2) (258)

UR(d) =εκφ2

[1minus tanh

)] (259)

VR(d) εκφ2

eminusκd (260)

φ(x) =4πσ0

cosh(κz)sinh(κd2)

VR(d) =4πσ2

)minus 1)

hydrocarbon tails

phosphatidylcholine

head group

C16 (palmitic acid)

hydrocarbon

head groupswater

x =dδ

K = κδ V =12πδ2

AH(VA + VR) (263)

we obtain

V(x) = Γ(

)minus 1)minus 1

(x + 1)2 minus1

(x + 2)2 (264)

Γ =24π2σ2

εκAH (265)

x0 05 1 15 2

)(266)

R1R2 (267)

E =int

kc(H minus H0)2 +

kc kc sim4πσ2

0εκ3 (269)

r2 part2

partr2 + rpart

partrminus (κr)2

)φ = 0 (270)

φ(r) =4πσ

I0(κr)I1(κR)

∣∣∣∣S=

4πσ20

I0(κR)I1(κR)

Iν(z) =ezradic

1minus microminus 1

I0(κR)I1(κR)

sim 1 +1

φ(x h(x)) = φ0 (275)

φ(x εh(x)) φ(x 0) + εh(x)φz(x 0) +12

O(ε0) φ(0)(x 0) = φ0 (279)

O(ε2) φ(2)(x 0) = minus12

z (x 0) (281)

φ(m)(k z) =int

dxeiqxh(x) (282)

q = κ2 + q2 (283)

and is solved by

φ(1)(q 0) = κφ0h(q) (285)

σ(x) = minus ε

12Bibliography

bibliography 51

Preface


Scaling arguments















Bibliography

mω2+x2

ω2plusmn = ω2

0 plusmn2e2

mR3 (221)

u(r) =12

hω+ +12

hω0 (222)

minus12

hω0(e2mω2

d = αE (224)

H = H0 + eE0x1 + eE0x2 (225)

mω20z2

α =e2

u(r) = minus12

hω0α2

r6 (228)

u sim hω0 (αn)2 (229)

u11(r) = minusCr6 (230)

u1S(z) =int 0

minusδdzprime

int 2π

0dφint infin

0rdrnu11(

zdzprime

int infin

1(zprime2 + r2)3 (232)

= minusπCn6

(1z3 minus

1(z + δ)3

) (233)

VA(d) =int d+δ

ddznu1S(z)dz (234)

= minus AH12π

(1d2 minus

2(d + δ)2 +

1(d + 2δ)2

) (235)

E = eφ(r) =e2

r (237)

εr (238)

λB =e2

ε (241)

nabla2φ = minus4π

ε sums

4 ) Then we have

nabla2φ =8πzec0

nabla2 minus κ2)

φ = 0 (245)

κminus1 =

[εkBT

8πz2e2c0

]12sim 10 nmradic

c0[mM] (246)

∣∣∣∣z=0

=4πσ0

εrArr σ0 =

4πφ0 (248)

φ(z) =4πσ0

εκeminusκz (249)

intd3r[

12(nablaφ)2 +

κ2φ2]

δGδφ

= minus part

partφx+

dSσφ (252)

F = G minusint

SdSσφ = minus1

dSσφ (253)

minusint

SdSint φ0

σ(φprime)dφ

prime (254)

G =2πσ2

A (255)

φ = φ0cosh(κz)

cosh(κd2) (256)

σ(d2) =εκφ0

4πtanh(κd2) (257)

U(d) = minusεκφ2

tanh(κd2) (258)

UR(d) =εκφ2

[1minus tanh

)] (259)

VR(d) εκφ2

eminusκd (260)

φ(x) =4πσ0

cosh(κz)sinh(κd2)

VR(d) =4πσ2

)minus 1)

hydrocarbon tails

phosphatidylcholine

head group

C16 (palmitic acid)

hydrocarbon

head groupswater

x =dδ

K = κδ V =12πδ2

AH(VA + VR) (263)

we obtain

V(x) = Γ(

)minus 1)minus 1

(x + 1)2 minus1

(x + 2)2 (264)

Γ =24π2σ2

εκAH (265)

x0 05 1 15 2

)(266)

R1R2 (267)

E =int

kc(H minus H0)2 +

kc kc sim4πσ2

0εκ3 (269)

r2 part2

partr2 + rpart

partrminus (κr)2

)φ = 0 (270)

φ(r) =4πσ

I0(κr)I1(κR)

∣∣∣∣S=

4πσ20

I0(κR)I1(κR)

Iν(z) =ezradic

1minus microminus 1

I0(κR)I1(κR)

sim 1 +1

φ(x h(x)) = φ0 (275)

φ(x εh(x)) φ(x 0) + εh(x)φz(x 0) +12

O(ε0) φ(0)(x 0) = φ0 (279)

O(ε2) φ(2)(x 0) = minus12

z (x 0) (281)

φ(m)(k z) =int

dxeiqxh(x) (282)

q = κ2 + q2 (283)

and is solved by

φ(1)(q 0) = κφ0h(q) (285)

σ(x) = minus ε

12Bibliography

bibliography 51

Preface


Scaling arguments















Bibliography

u(r) = minus12

hω0α2

r6 (228)

u sim hω0 (αn)2 (229)

u11(r) = minusCr6 (230)

u1S(z) =int 0

minusδdzprime

int 2π

0dφint infin

0rdrnu11(

zdzprime

int infin

1(zprime2 + r2)3 (232)

= minusπCn6

(1z3 minus

1(z + δ)3

) (233)

VA(d) =int d+δ

ddznu1S(z)dz (234)

= minus AH12π

(1d2 minus

2(d + δ)2 +

1(d + 2δ)2

) (235)

E = eφ(r) =e2

r (237)

εr (238)

λB =e2

ε (241)

nabla2φ = minus4π

ε sums

4 ) Then we have

nabla2φ =8πzec0

nabla2 minus κ2)

φ = 0 (245)

κminus1 =

[εkBT

8πz2e2c0

]12sim 10 nmradic

c0[mM] (246)

∣∣∣∣z=0

=4πσ0

εrArr σ0 =

4πφ0 (248)

φ(z) =4πσ0

εκeminusκz (249)

intd3r[

12(nablaφ)2 +

κ2φ2]

δGδφ

= minus part

partφx+

dSσφ (252)

F = G minusint

SdSσφ = minus1

dSσφ (253)

minusint

SdSint φ0

σ(φprime)dφ

prime (254)

G =2πσ2

A (255)

φ = φ0cosh(κz)

cosh(κd2) (256)

σ(d2) =εκφ0

4πtanh(κd2) (257)

U(d) = minusεκφ2

tanh(κd2) (258)

UR(d) =εκφ2

[1minus tanh

)] (259)

VR(d) εκφ2

eminusκd (260)

φ(x) =4πσ0

cosh(κz)sinh(κd2)

VR(d) =4πσ2

)minus 1)

hydrocarbon tails

phosphatidylcholine

head group

C16 (palmitic acid)

hydrocarbon

head groupswater

x =dδ

K = κδ V =12πδ2

AH(VA + VR) (263)

we obtain

V(x) = Γ(

)minus 1)minus 1

(x + 1)2 minus1

(x + 2)2 (264)

Γ =24π2σ2

εκAH (265)

x0 05 1 15 2

)(266)

R1R2 (267)

E =int

kc(H minus H0)2 +

kc kc sim4πσ2

0εκ3 (269)

r2 part2

partr2 + rpart

partrminus (κr)2

)φ = 0 (270)

φ(r) =4πσ

I0(κr)I1(κR)

∣∣∣∣S=

4πσ20

I0(κR)I1(κR)

Iν(z) =ezradic

1minus microminus 1

I0(κR)I1(κR)

sim 1 +1

φ(x h(x)) = φ0 (275)

φ(x εh(x)) φ(x 0) + εh(x)φz(x 0) +12

O(ε0) φ(0)(x 0) = φ0 (279)

O(ε2) φ(2)(x 0) = minus12

z (x 0) (281)

φ(m)(k z) =int

dxeiqxh(x) (282)

q = κ2 + q2 (283)

and is solved by

φ(1)(q 0) = κφ0h(q) (285)

σ(x) = minus ε

12Bibliography

bibliography 51

Preface


Scaling arguments















Bibliography

VA(d) =int d+δ

ddznu1S(z)dz (234)

= minus AH12π

(1d2 minus

2(d + δ)2 +

1(d + 2δ)2

) (235)

E = eφ(r) =e2

r (237)

εr (238)

λB =e2

ε (241)

nabla2φ = minus4π

ε sums

4 ) Then we have

nabla2φ =8πzec0

nabla2 minus κ2)

φ = 0 (245)

κminus1 =

[εkBT

8πz2e2c0

]12sim 10 nmradic

c0[mM] (246)

∣∣∣∣z=0

=4πσ0

εrArr σ0 =

4πφ0 (248)

φ(z) =4πσ0

εκeminusκz (249)

intd3r[

12(nablaφ)2 +

κ2φ2]

δGδφ

= minus part

partφx+

dSσφ (252)

F = G minusint

SdSσφ = minus1

dSσφ (253)

minusint

SdSint φ0

σ(φprime)dφ

prime (254)

G =2πσ2

A (255)

φ = φ0cosh(κz)

cosh(κd2) (256)

σ(d2) =εκφ0

4πtanh(κd2) (257)

U(d) = minusεκφ2

tanh(κd2) (258)

UR(d) =εκφ2

[1minus tanh

)] (259)

VR(d) εκφ2

eminusκd (260)

φ(x) =4πσ0

cosh(κz)sinh(κd2)

VR(d) =4πσ2

)minus 1)

hydrocarbon tails

phosphatidylcholine

head group

C16 (palmitic acid)

hydrocarbon

head groupswater

x =dδ

K = κδ V =12πδ2

AH(VA + VR) (263)

we obtain

V(x) = Γ(

)minus 1)minus 1

(x + 1)2 minus1

(x + 2)2 (264)

Γ =24π2σ2

εκAH (265)

x0 05 1 15 2

)(266)

R1R2 (267)

E =int

kc(H minus H0)2 +

kc kc sim4πσ2

0εκ3 (269)

r2 part2

partr2 + rpart

partrminus (κr)2

)φ = 0 (270)

φ(r) =4πσ

I0(κr)I1(κR)

∣∣∣∣S=

4πσ20

I0(κR)I1(κR)

Iν(z) =ezradic

1minus microminus 1

I0(κR)I1(κR)

sim 1 +1

φ(x h(x)) = φ0 (275)

φ(x εh(x)) φ(x 0) + εh(x)φz(x 0) +12

O(ε0) φ(0)(x 0) = φ0 (279)

O(ε2) φ(2)(x 0) = minus12

z (x 0) (281)

φ(m)(k z) =int

dxeiqxh(x) (282)

q = κ2 + q2 (283)

and is solved by

φ(1)(q 0) = κφ0h(q) (285)

σ(x) = minus ε

12Bibliography

bibliography 51

Preface


Scaling arguments















Bibliography

ε (241)

nabla2φ = minus4π

ε sums

4 ) Then we have

nabla2φ =8πzec0

nabla2 minus κ2)

φ = 0 (245)

κminus1 =

[εkBT

8πz2e2c0

]12sim 10 nmradic

c0[mM] (246)

∣∣∣∣z=0

=4πσ0

εrArr σ0 =

4πφ0 (248)

φ(z) =4πσ0

εκeminusκz (249)

intd3r[

12(nablaφ)2 +

κ2φ2]

δGδφ

= minus part

partφx+

dSσφ (252)

F = G minusint

SdSσφ = minus1

dSσφ (253)

minusint

SdSint φ0

σ(φprime)dφ

prime (254)

G =2πσ2

A (255)

φ = φ0cosh(κz)

cosh(κd2) (256)

σ(d2) =εκφ0

4πtanh(κd2) (257)

U(d) = minusεκφ2

tanh(κd2) (258)

UR(d) =εκφ2

[1minus tanh

)] (259)

VR(d) εκφ2

eminusκd (260)

φ(x) =4πσ0

cosh(κz)sinh(κd2)

VR(d) =4πσ2

)minus 1)

hydrocarbon tails

phosphatidylcholine

head group

C16 (palmitic acid)

hydrocarbon

head groupswater

x =dδ

K = κδ V =12πδ2

AH(VA + VR) (263)

we obtain

V(x) = Γ(

)minus 1)minus 1

(x + 1)2 minus1

(x + 2)2 (264)

Γ =24π2σ2

εκAH (265)

x0 05 1 15 2

)(266)

R1R2 (267)

E =int

kc(H minus H0)2 +

kc kc sim4πσ2

0εκ3 (269)

r2 part2

partr2 + rpart

partrminus (κr)2

)φ = 0 (270)

φ(r) =4πσ

I0(κr)I1(κR)

∣∣∣∣S=

4πσ20

I0(κR)I1(κR)

Iν(z) =ezradic

1minus microminus 1

I0(κR)I1(κR)

sim 1 +1

φ(x h(x)) = φ0 (275)

φ(x εh(x)) φ(x 0) + εh(x)φz(x 0) +12

O(ε0) φ(0)(x 0) = φ0 (279)

O(ε2) φ(2)(x 0) = minus12

z (x 0) (281)

φ(m)(k z) =int

dxeiqxh(x) (282)

q = κ2 + q2 (283)

and is solved by

φ(1)(q 0) = κφ0h(q) (285)

σ(x) = minus ε

12Bibliography

bibliography 51

Preface


Scaling arguments















Bibliography

∣∣∣∣z=0

=4πσ0

εrArr σ0 =

4πφ0 (248)

φ(z) =4πσ0

εκeminusκz (249)

intd3r[

12(nablaφ)2 +

κ2φ2]

δGδφ

= minus part

partφx+

dSσφ (252)

F = G minusint

SdSσφ = minus1

dSσφ (253)

minusint

SdSint φ0

σ(φprime)dφ

prime (254)

G =2πσ2

A (255)

φ = φ0cosh(κz)

cosh(κd2) (256)

σ(d2) =εκφ0

4πtanh(κd2) (257)

U(d) = minusεκφ2

tanh(κd2) (258)

UR(d) =εκφ2

[1minus tanh

)] (259)

VR(d) εκφ2

eminusκd (260)

φ(x) =4πσ0

cosh(κz)sinh(κd2)

VR(d) =4πσ2

)minus 1)

hydrocarbon tails

phosphatidylcholine

head group

C16 (palmitic acid)

hydrocarbon

head groupswater

x =dδ

K = κδ V =12πδ2

AH(VA + VR) (263)

we obtain

V(x) = Γ(

)minus 1)minus 1

(x + 1)2 minus1

(x + 2)2 (264)

Γ =24π2σ2

εκAH (265)

x0 05 1 15 2

)(266)

R1R2 (267)

E =int

kc(H minus H0)2 +

kc kc sim4πσ2

0εκ3 (269)

r2 part2

partr2 + rpart

partrminus (κr)2

)φ = 0 (270)

φ(r) =4πσ

I0(κr)I1(κR)

∣∣∣∣S=

4πσ20

I0(κR)I1(κR)

Iν(z) =ezradic

1minus microminus 1

I0(κR)I1(κR)

sim 1 +1

φ(x h(x)) = φ0 (275)

φ(x εh(x)) φ(x 0) + εh(x)φz(x 0) +12

O(ε0) φ(0)(x 0) = φ0 (279)

O(ε2) φ(2)(x 0) = minus12

z (x 0) (281)

φ(m)(k z) =int

dxeiqxh(x) (282)

q = κ2 + q2 (283)

and is solved by

φ(1)(q 0) = κφ0h(q) (285)

σ(x) = minus ε

12Bibliography

bibliography 51

Preface


Scaling arguments















Bibliography

F = G minusint

SdSσφ = minus1

dSσφ (253)

minusint

SdSint φ0

σ(φprime)dφ

prime (254)

G =2πσ2

A (255)

φ = φ0cosh(κz)

cosh(κd2) (256)

σ(d2) =εκφ0

4πtanh(κd2) (257)

U(d) = minusεκφ2

tanh(κd2) (258)

UR(d) =εκφ2

[1minus tanh

)] (259)

VR(d) εκφ2

eminusκd (260)

φ(x) =4πσ0

cosh(κz)sinh(κd2)

VR(d) =4πσ2

)minus 1)

hydrocarbon tails

phosphatidylcholine

head group

C16 (palmitic acid)

hydrocarbon

head groupswater

x =dδ

K = κδ V =12πδ2

AH(VA + VR) (263)

we obtain

V(x) = Γ(

)minus 1)minus 1

(x + 1)2 minus1

(x + 2)2 (264)

Γ =24π2σ2

εκAH (265)

x0 05 1 15 2

)(266)

R1R2 (267)

E =int

kc(H minus H0)2 +

kc kc sim4πσ2

0εκ3 (269)

r2 part2

partr2 + rpart

partrminus (κr)2

)φ = 0 (270)

φ(r) =4πσ

I0(κr)I1(κR)

∣∣∣∣S=

4πσ20

I0(κR)I1(κR)

Iν(z) =ezradic

1minus microminus 1

I0(κR)I1(κR)

sim 1 +1

φ(x h(x)) = φ0 (275)

φ(x εh(x)) φ(x 0) + εh(x)φz(x 0) +12

O(ε0) φ(0)(x 0) = φ0 (279)

O(ε2) φ(2)(x 0) = minus12

z (x 0) (281)

φ(m)(k z) =int

dxeiqxh(x) (282)

q = κ2 + q2 (283)

and is solved by

φ(1)(q 0) = κφ0h(q) (285)

σ(x) = minus ε

12Bibliography

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Bibliography

UR(d) =εκφ2

[1minus tanh

)] (259)

VR(d) εκφ2

eminusκd (260)

φ(x) =4πσ0

cosh(κz)sinh(κd2)

VR(d) =4πσ2

)minus 1)

hydrocarbon tails

phosphatidylcholine

head group

C16 (palmitic acid)

hydrocarbon

head groupswater

x =dδ

K = κδ V =12πδ2

AH(VA + VR) (263)

we obtain

V(x) = Γ(

)minus 1)minus 1

(x + 1)2 minus1

(x + 2)2 (264)

Γ =24π2σ2

εκAH (265)

x0 05 1 15 2

)(266)

R1R2 (267)

E =int

kc(H minus H0)2 +

kc kc sim4πσ2

0εκ3 (269)

r2 part2

partr2 + rpart

partrminus (κr)2

)φ = 0 (270)

φ(r) =4πσ

I0(κr)I1(κR)

∣∣∣∣S=

4πσ20

I0(κR)I1(κR)

Iν(z) =ezradic

1minus microminus 1

I0(κR)I1(κR)

sim 1 +1

φ(x h(x)) = φ0 (275)

φ(x εh(x)) φ(x 0) + εh(x)φz(x 0) +12

O(ε0) φ(0)(x 0) = φ0 (279)

O(ε2) φ(2)(x 0) = minus12

z (x 0) (281)

φ(m)(k z) =int

dxeiqxh(x) (282)

q = κ2 + q2 (283)

and is solved by

φ(1)(q 0) = κφ0h(q) (285)

σ(x) = minus ε

12Bibliography

bibliography 51

Preface


Scaling arguments















Bibliography

hydrocarbon

head groupswater

x =dδ

K = κδ V =12πδ2

AH(VA + VR) (263)

we obtain

V(x) = Γ(

)minus 1)minus 1

(x + 1)2 minus1

(x + 2)2 (264)

Γ =24π2σ2

εκAH (265)

x0 05 1 15 2

)(266)

R1R2 (267)

E =int

kc(H minus H0)2 +

kc kc sim4πσ2

0εκ3 (269)

r2 part2

partr2 + rpart

partrminus (κr)2

)φ = 0 (270)

φ(r) =4πσ

I0(κr)I1(κR)

∣∣∣∣S=

4πσ20

I0(κR)I1(κR)

Iν(z) =ezradic

1minus microminus 1

I0(κR)I1(κR)

sim 1 +1

φ(x h(x)) = φ0 (275)

φ(x εh(x)) φ(x 0) + εh(x)φz(x 0) +12

O(ε0) φ(0)(x 0) = φ0 (279)

O(ε2) φ(2)(x 0) = minus12

z (x 0) (281)

φ(m)(k z) =int

dxeiqxh(x) (282)

q = κ2 + q2 (283)

and is solved by

φ(1)(q 0) = κφ0h(q) (285)

σ(x) = minus ε

12Bibliography

bibliography 51

Preface


Scaling arguments















Bibliography

)(266)

R1R2 (267)

E =int

kc(H minus H0)2 +

kc kc sim4πσ2

0εκ3 (269)

r2 part2

partr2 + rpart

partrminus (κr)2

)φ = 0 (270)

φ(r) =4πσ

I0(κr)I1(κR)

∣∣∣∣S=

4πσ20

I0(κR)I1(κR)

Iν(z) =ezradic

1minus microminus 1

I0(κR)I1(κR)

sim 1 +1

φ(x h(x)) = φ0 (275)

φ(x εh(x)) φ(x 0) + εh(x)φz(x 0) +12

O(ε0) φ(0)(x 0) = φ0 (279)

O(ε2) φ(2)(x 0) = minus12

z (x 0) (281)

φ(m)(k z) =int

dxeiqxh(x) (282)

q = κ2 + q2 (283)

and is solved by

φ(1)(q 0) = κφ0h(q) (285)

σ(x) = minus ε

12Bibliography

bibliography 51

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Scaling arguments















Bibliography

kc kc sim4πσ2

0εκ3 (269)

r2 part2

partr2 + rpart

partrminus (κr)2

)φ = 0 (270)

φ(r) =4πσ

I0(κr)I1(κR)

∣∣∣∣S=

4πσ20

I0(κR)I1(κR)

Iν(z) =ezradic

1minus microminus 1

I0(κR)I1(κR)

sim 1 +1

φ(x h(x)) = φ0 (275)

φ(x εh(x)) φ(x 0) + εh(x)φz(x 0) +12

O(ε0) φ(0)(x 0) = φ0 (279)

O(ε2) φ(2)(x 0) = minus12

z (x 0) (281)

φ(m)(k z) =int

dxeiqxh(x) (282)

q = κ2 + q2 (283)

and is solved by

φ(1)(q 0) = κφ0h(q) (285)

σ(x) = minus ε

12Bibliography

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Scaling arguments















Bibliography

I0(κR)I1(κR)

sim 1 +1

φ(x h(x)) = φ0 (275)

φ(x εh(x)) φ(x 0) + εh(x)φz(x 0) +12

O(ε0) φ(0)(x 0) = φ0 (279)

O(ε2) φ(2)(x 0) = minus12

z (x 0) (281)

φ(m)(k z) =int

dxeiqxh(x) (282)

q = κ2 + q2 (283)

and is solved by

φ(1)(q 0) = κφ0h(q) (285)

σ(x) = minus ε

12Bibliography

bibliography 51

Preface


Scaling arguments















Bibliography

O(ε2) φ(2)(x 0) = minus12

z (x 0) (281)

φ(m)(k z) =int

dxeiqxh(x) (282)

q = κ2 + q2 (283)

and is solved by

φ(1)(q 0) = κφ0h(q) (285)

σ(x) = minus ε

12Bibliography

bibliography 51

Preface


Scaling arguments















Bibliography

12Bibliography

bibliography 51

Preface


Scaling arguments















Bibliography

12Bibliography

bibliography 51

Preface


Scaling arguments















Bibliography

12Bibliography

bibliography 51

Preface


Scaling arguments















Bibliography

12Bibliography

bibliography 51

Preface


Scaling arguments















Bibliography

12Bibliography

bibliography 51

Preface


Scaling arguments















Bibliography

12Bibliography

bibliography 51

Preface


Scaling arguments















Bibliography

12Bibliography

bibliography 51

Preface


Scaling arguments















Bibliography

12Bibliography

bibliography 51

Preface


Scaling arguments















Bibliography

12Bibliography

bibliography 51

Preface


Scaling arguments















Bibliography

12Bibliography

bibliography 51

Preface


Scaling arguments















Bibliography

bibliography 51

Preface


Scaling arguments















Bibliography

bibliography 51

Preface


Scaling arguments















Bibliography

Documents

Biological Physics and Fluid Dynamics