Dynamics of ﬁlaments and membranes in a …Dynamics of ﬁlaments and membranes in a viscous...

Dynamics of filaments and membranes in a viscous fluid

Thomas R. Powers*

Division of Engineering, Brown University, Providence, Rhode Island 02912, USA

Published 19 May 2010; corrected 3 June 2010

Motivated by the motion of biopolymers and membranes in solution, this article presents aformulation of the equations of motion for curves and surfaces in a viscous fluid. The focus is ongeometrical aspects and simple variational methods for calculating internal stresses and forces, and thefull nonlinear equations of motion are derived. In the case of membranes, particular attention is paidto the formulation of the equations of hydrodynamics on a curved, deforming surface. The formalismis illustrated by two simple case studies: 1 the twirling instability of straight elastic rod rotating in aviscous fluid and 2 the pearling and buckling instabilities of a tubular liposome or polymersome.

DOI: 10.1103/RevModPhys.82.1607 PACS numbers: 02.40.Hw

CONTENTS

I. Introduction 1607

II. Characteristic Length and Time Scales 1608

A. Thermal fluctuations and the persistence length 1608

B. Time scales for stretching, bending, and twisting of

filaments 1609

C. Time scales for stretching and bending of

membranes 1610

III. Filaments 1612

A. Rod kinematics 1612

B. Energies, forces, and moments 1613

C. Compatibility relations for rods 1615

D. Viscous forces: Slender-body theory and

resistive-force theory 1616

E. Equations of motion for curves 1617

F. Illustrative example: Twirling and whirling of elastic

filaments 1617

IV. Membranes 1618

A. Geometry of a stationary surface 1619

B. Energies and forces 1622

1. Soap film 1623

2. Fluid membrane 1623

C. Moving surfaces: Kinematics 1624

1. Coordinates on deforming surfaces with

flow 1624

2. Strain rate 1625

3. Equation of continuity 1625

D. Dynamical equations for an incompressible

membrane 1626

E. Illustrative example: Buckling and pearling

instabilities of tubular polymersomes 1627

V. Outlook 1628

Acknowledgments 1629

Appendix A: Coordinate Transformations 1629

Appendix B: Green’s Theorem 1629

References 1630

I. INTRODUCTION

By definition, soft matter is easy to deform. Some ma-terials are soft because they have a high degree of sym-metry. For example, the absence of long-range transla-tional order in a liquid means that there is no resistanceto shear in static equilibrium. Other materials, such ascolloidal crystals, have low symmetry but are soft sincethe fundamental constituents are large and the interac-tions between these constituents are of entropic origin,and therefore weak. For example, to estimate the shearmodulus G of a colloidal crystal, we assume a character-istic length scale a1 m and a characteristic energyscale of about ten times the thermal energy at roomtemperature, and find G10kBT /a3410−2 Pa. Thisestimate is far smaller than the typical elastic modulusfor a molecular crystal, where covalent bonds and mo-lecular sizes determine the size of the shear modulus.Finally, many objects are soft since they are thin in oneor two dimensions. It is a matter of common experiencethat bending a thin rod or plate is much easier thanstretching. Examples of thin objects at the micrometerscale abound: bacterial flagella, actin filaments, and cellmembranes readily deform in response to thermal fluc-tuations or viscous forces arising in fluid flow. The key tounderstanding the physics of these systems often hingeson understanding the dynamics of their shapes. Thus,the student of soft matter is led directly to the problemof describing how curves and surfaces deform under ex-ternal forces. Differential geometry is the natural toolfor this task. In this review, we give a self-contained in-troduction to differential geometry of curves and sur-faces that evolve in time. Our intended audience issomeone who wants to use these equations to elucidatephysical phenomena. Therefore, although our main fo-cus is on mathematics, we give physical interpretationsof the mathematics whenever possible.

It is natural to ask why we need another review ofdifferential geometry. There are introductory books of apurely mathematical nature that are good starting pointsfor students Coxeter, 1969; Millman and Parker, 1977;Struik, 1988; Morgan, 1998. There are useful books with*[email protected]

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http://dx.doi.org/10.1103/RevModPhys.82.1607

a broader mathematical scope, such as that of Nakahara2003. Finally, there are books and reviews that intro-duce differential geometry in the context of solid me-chanics Green and Zerna, 1968, fluid mechanics Aris,1989, and soft matter David, 1989; Kamien, 2002; vanHemmen and Leibold, 2007. Our review is closest inspirit to these last four, but it is complementary since itemphasizes the variational point of view for derivingforces due to internal stresses acting on a stiff polymeror membrane. Like the review of Kamien 2002, wegive a unified treatment of curves and surfaces from the“extrinsic” point of view, in which distances along curvesand surfaces sitting in three-dimensional space are mea-sured using the notion of distance in that three-dimensional space. The alternative “intrinsic” approachto the differential geometry of surfaces is to work onlywith quantities that can be measured by a two-dimensional being living on the surface. For example,the curvature K of a surface can be defined by compar-ing the circumference C of a circle on the surface to theradius R of the circle, as measured by the inhabitant ofthe surface. See Fig. 1; the formula relating K, C, and Rdisplayed in the figure can be readily confirmed for thecase of a sphere. Since polymers and membranes inter-act and bend in three-dimensional space, it is impossibleto formulate their equations of motions in terms of in-trinsic quantities only, and we will not hesitate to char-acterize curves and surfaces with extrinsic quantitiessuch as length in three dimensions.

In addition to the tension between the extrinsic andintrinsic points of view, the issue of choosing coordinatesalways arises when studying curves and surfaces. Sincegeometric quantities such as vector fields and curvaturesexist independently of the choice of coordinates, it isnatural to define these quantities without invoking coor-dinates at all Misner et al., 1973. Although this ap-proach is elegant, it requires too much mathematicalmachinery for our purposes. In the case of curves, thebasic geometrical ideas can be captured with vector cal-culus. Surfaces are more complicated and therefore re-quire us to introduce some of the elements of the calcu-

lus of tensors, which we attempt to do with a minimalamount of formalism.

Another reason not to insist on a coordinate-free ap-proach is that physical considerations often single out acoordinate system or a class of coordinate systems asspecial. For example, we will see that arclength param-etrization is natural for inextensible rods, and that con-vected coordinates—coordinates that are carried alongby material particles—are natural for fluid membranes.A second point related to choice of coordinates is thedistinction between an interface between two differentphases and the physical surfaces that we study. For ex-ample, since a solidification front Langer, 1980 ad-vances through a substance while the material points ofthe substance remain at rest, we are free to use choosecoordinates such that the interface velocity is alwaysnormal to the tangent plane Brower et al., 1984. Incontrast, the velocity of a material point in a polymer ormembrane may have a component along the tangentialdirection. For motion of a polymer or membrane arisingfrom deformation, this velocity component has physicalconsequences and cannot be removed by a change ofcoordinates.

II. CHARACTERISTIC LENGTH AND TIME SCALES

A. Thermal fluctuations and the persistence length

For brevity, we limit the scope of the review by disre-garding thermal fluctuations. This assumption is a vastsimplification, since the flexibility of polymers and mem-branes ensures that thermal undulations are alwayspresent. However, there are many shape-evolution prob-lems in which these effects are small. Furthermore, theformalism described in this review is also useful even forproblems where thermal effects play an important role,such as the dynamics of tension in semiflexible polymersHallatschek et al., 2007 and the surface tension of anundulating fluid membrane Fournier and Barbetta,2008. In this section we delineate the conditions underwhich thermal fluctuations are important.

First consider a filament in solution. The filamentcould be a semiflexible polymer such as DNA or actin,or the flagellar filament of a bacterium Boal, 2002.These filaments resist bending, and may be described byan energy

E =A

2 1

R2ds , 1

where A is the bending modulus with dimensions of en-ergy times length, s is the distance measured along thefilament, and R is the local radius of curvature of thefilament. We describe the curvature of a filament in Sec.III.A; for now one can simply imagine that the thermalmotion of the molecules of the solution causes the fila-ment to flex back and forth between arcs of constantcurvature. If we imagine lengthening the filament, thenour experience with macroscopic rods tells us that thefilament is easier to bend. That is, for fixed filament stiff-ness A, the same force applied perpendicular to the end

C ≈ 2πR(1 − KR2/6)

2R

C

FIG. 1. A two-dimensional being after Burger and Starbird,2005 considers the relation between the circumference andradius of a circle in his curved world; K is a number he candetermine by making careful measurements of length.

1608 Thomas R. Powers: Dynamics of filaments and membranes in a …

Rev. Mod. Phys., Vol. 82, No. 2, April–June 2010

of the filament leads to a greater deflection as the fila-ment increases in length. Thus, the longer the filament,the more it will bend spontaneously due to thermal fluc-tuations. We can estimate the length p of a filament atwhich thermal fluctuations become important by equat-ing thermal energy kBT with the bending energy of afilament bent into an arc of one radian: kBTA /p, orpkBT /A. The length p is called the persistencelength. At room temperature, the persistence length ofDNA is about 50 nm; p is of the order of micrometersfor actin, and millimeters for microtubules and bacterialflagellar filaments Boal, 2002. The effects of thermalfluctuations are small when the length L of a filament issmall compared to p, Lp.

Unlike interfaces which are simply the boundary be-tween two phases, membranes are comprised of mol-ecules that resist bending. Therefore, the shapes ofmembranes are also governed by a bending energy. Aswe review in Sec. IV.A, there are two radii of curvatureat any point of a surface. Thus, the bending energy of afluid membrane is

E =

2 1

R1+

1

R22

dS , 2

where is the bending modulus or stiffness, with dimen-sions of energy, and dS is the element of the surfacearea. Just as for filaments we expect that as a membranegets larger and larger, the thermal motion of the solventis more and more effective at bending the membrane.However, since the units of do not contain a length, itis not immediately obvious how to estimate the persis-tence length using dimensional analysis. Therefore, wemust use a more detailed approach.

Consider a membrane that is weakly perturbed bythermal motion and spanning a frame with dimensionsLL. Since the membrane is almost flat, it can be de-scribed as the graph of a height function h over the x-yplane. In Sec. IV.A we will see that the energy in Eq. 2may be given as

E

2 2h2dxdy . 3

Our strategy for quantifying the effect of thermal fluc-tuations will be to estimate the scale L at which fluctua-tions in the normal to the membrane, n z−h, becomeof the order of unity, or in other words large enough tospoil the assumption that the membrane is almost flat.First, use the equipartition theorem to assign to eachmode of the quadratic energy function 3 the contribu-tion kBT /2 toward the average energy. The simplest wayto apply the equipartition theorem is to assume periodicboundary conditions and write the height as a Fourierseries, hx ,y=q expiq · x ,yhq /L2, where the sum isover all wave vectors q= 2m /L ,2n /L with the inte-gers m and n less than a maximum value mmax. Thewavelength corresponding to the maximum is the small-est length at which the continuum theory applies, and isa molecular scale mol such as the thickness of the mem-

brane. Thus, the mean-square amplitude of the fluctua-tions of the height is

hqhq = L2q+q,0kBT

q4 . 4

Longer wavelength ripples are easier to excite. We cannow calculate the fluctuations in the normal direction n.Since the membrane is almost flat, it is convenient tocalculate the mean-square fluctuations of the deviationn= n− z−h Chaikin and Lubensky, 1995

n · n = qq

q · q

L2 hqhq 5

d2q

22

kBT

q2 . 6

The integral in Eq. 6 runs over wave vectors rangingfrom the short-distance cutoff qmin2 /mol to the long-distance cutoff given by the scale of the frame, qmax2 /L. Assuming for simplicity a disk-shaped domainof integration, we find n ·nkBT /2lnL /mol.The scale at which the thermal fluctuations lead to sig-nificant deformation of the membrane is the length scaleL=p for which n ·n1, or

p mol exp2

kBT . 7

For typical values mol1 nm and 10kBT, the persis-tence length is extremely large. Thus, thermal fluctua-tions are often disregarded in studies of the shape dy-namics of vesicles.

B. Time scales for stretching, bending, and twisting offilaments

To understand the physical mechanisms that governthe dynamics of filaments, we describe the characteristictime scales for stretching, bending, and twisting. To getthe orders of magnitude of these time scales, it is suffi-cient to use scaling arguments and work in the limit ofsmall deflection from a straight filament. The precisenonlinear equations of motion require the developmentof the geometric tools presented in the later sections ofthis review.

We begin by recalling some basic definitions from themechanics of rods Landau and Lifshitz, 1986 and vis-cous fluids Landau and Lifshitz, 1987. Consider astraight rod of radius a, with long axis aligned along zand under longitudinal deformation. The stress zz, orforce per unit area acting through a cross section at z, isproportional to the strain for small deformations: zz=Yzu, where u is the displacement of the cross sectionof the rod originally at z and Y is the Young’s moduluswith units of energy per volume. The net elastic forceper length acting on the element of size dz is thereforegiven by a2Yz

2udz. Now suppose the rod is allowed torelax, with the only resistance to the relaxation being theviscous drag of the solvent.

1609Thomas R. Powers: Dynamics of filaments and membranes in a …


To estimate the viscous resistance, we first argue thatat the small scales of polymer and cells the typical vis-cous forces are much larger than the forces that are re-quired to change the velocity of the fluid. Viscousstresses are given by the product of the viscosity andthe strain rate. Thus, the viscous force on an object ofsize is given by

Fv v/2 = v , 8

where is the viscosity and v is a characteristic speed ofthe flow. The viscosity of water is 10−3 N s/m2. Theproduct of mass and acceleration for an element of size of a fluid of density is given by

Fi 3v2/ = v22. 9

For micrometer-sized objects in water moving at speedsof the order of micrometers per second, the inertialforce Fi is about a 106 times smaller than the viscousforce Fv. This regime is the regime of “low Reynoldsnumber,” in which inertia is irrelevant Happel andBrenner, 1965; Hinch, 1988; Kim and Karrila, 1991.Thus, we may estimate the viscous drag on the elementof the rod of length dz to be of the order of vdz. InSec. III we address the question of how the length scalea enters the drag.

We now have the ingredients required to estimate therelaxation time for longitudinal extension and compres-sion of a filament. Assuming that the inertia of the rodas well as the fluid is negligible, we balance the viscousdrag with the elastic force, identify v with tu, and dropnumerical factors such as to find tua2Yz

2u. Thus,the characteristic relaxation time for longitudinalstretching or compression of a rod of length L in a vis-cous fluid is

s

YL

a2

. 10

To estimate the relaxation time for bending deforma-tions, note that the bending modulus A for a thin fila-ment of radius a is related to the Young’s modulus byAYa4 Landau and Lifshitz, 1986. Just as we did formembranes, we can describe small deflections of a fila-ment by a height function hz, and the elastic energy isgiven as EA /2 z

2h2dz. By the principle of virtualwork, the elastic force per unit length due to bending isgiven by the functional derivative −E /h−Az

4h.Once again disregarding numerical factors, we approxi-mate the resistance to transverse motion as thdz, andbalance the elastic and viscous forces to obtain the bend-ing relaxation time

b L4

A

YL

a4

. 11

The ratio of the stretch relaxation time to the bend re-laxation time involves two powers of the aspect ratio,s /ba /L2. This aspect ratio is typically very small,ranging from 1/50 for a 50 nm segment of DNA to 10−3

for a typical bacterial flagellum Boal, 2002. Therefore,

modulations in the extension relax much faster thanbending deformations for polymer filaments. When thefocus of interest is bending dynamics, thin filaments aretypically assumed to be inextensible.

To estimate the relaxation time for twisting deforma-tions, suppose the filament is straight but twisted. Weprecisely define twist for a filament with arbitrary con-tour in Sec. III.A, but for a straight filament define theangle z through which the cross section at z has ro-tated due to twist. The torque applied through a crosssection is Cz, and the net elastic torque on an elementof length dz due to internal stresses is Cz

2dz.The viscous torque resisting rotation at rate t is the

product of the viscous stress t, the surface area of theelement adz, and the moment arm a. Thus, the twistdegree of freedom obeys the diffusion equation a2t=Cz

2, and the twist relaxation time is

t a2L2

C. 12

For many materials, CA Landau and Lifshitz, 1986,and thus

t

YL

a2

. 13

The twist relaxation time is much smaller than the bendrelaxation time. Interestingly, the twist relaxation time iscomparable to the stretch relaxation time. Nevertheless,it is common to consider the dynamics of inextensibleelastic rods with both bending and twisting degrees offreedom, since twisting and stretching are usually drivenby different physical processes. A simple example is pro-vided by an elastic filament, initially straight and twirlingabout its long axis Sec. III.F. Since the rotation inducestwist but no stretching, it is convenient to treat the fila-ment as inextensible.

C. Time scales for stretching and bending of membranes

In this section we review the basic mechanics of mem-branes Seifert, 1997. We confine our attention to fluidmembranes, in which there is no long-range order in themolecules making up the membrane. These membraneswrap up into closed surfaces known as vesicles—liposomes in the case of lipid molecules, and polymer-somes in the case of block copolymers Yu and Eisen-berg, 1998; Discher and Ahmed, 2006. Both the lipidsand the block copolymers are amphiphilic, consisting oftwo parts with different affinities for water. The hydro-phobic effect drives the molecules to form a bilayer, withthe oily chains with the least affinity for water inside thebilayer.

We now estimate the relaxation time for membranestretching. Consistent with the discussion of Sec. II.A,we suppose the membrane is in the “high-tension”regime of stretching in which the resistance arisesfrom elasticity rather than entropic effects Evans andRawicz, 1990. Disregarding the bilayer structure for themoment, we may view the membrane as a two-



dimensional fluid. Unlike the filaments considered inSec. II.B, the membrane is a liquid, and to describe theresistance of the membrane to stretching and compres-sion we use the density rather than a displacement vari-able. Let denote the number of molecules per unitarea, and 0 denote the preferred number density. Forsmall deformations, the elastic energy of stretching is

Es =k

2

0− 12

dS , 14

where k is the compression modulus with units of en-ergy. Since the modulus k is usually large enough thatthe density is close to 0, work in terms of =−0.From dimensional analysis, we deduce that the two-dimensional pressure in the membrane, with units offorce per length, is p2D=k /0. Gradients in the two-dimensional pressure lead to tangential forces in themembrane, which drive regions of nonuniform densityto relax to constant density. The balance of these tan-gential forces with dissipative forces determines thestretch relaxation time. We consider the relaxation of adensity modulation with wave number q.

First we suppose that the dominant dissipative forcesarise from the viscous forces the solvent exerts on themembrane. For a characteristic flow field v near themembrane, the stress is vqv. Thus the balanceof the gradient of pressure with the viscous stress leadsto

qk/0 qv . 15

Assuming the no-slip boundary condition between thesolvent velocity and the lipid velocity holds, we may usev to represent the lipid velocity as well. The rate ofchange of the density is related to the lipid flow by thecontinuity equation expressing conservation of particles.In Sec. IV.D we will see how to formulate the continuityequation on a moving curved surface, but for our scalingargument it is sufficient to assert that tv, or

/s q0v , 16

where s is the stretching relaxation time of the mem-brane. Combining Eqs. 15 and 16 leads to

s

kq, 17

or taking the smallest wave number to be set by the sizeL of the vesicle, sL /k.

To estimate the relaxation time for bending deforma-tions with wave number q, we balance the bending elas-tic force per unit area with viscous stress or tractionfrom the solvent. We give the precise form of the elasticforce per area on a deformed membrane in Sec. IV.B.2;but working in analogy with our expression for thebending force per unit length on a filament, we have f4hq4h. Again invoking the no-slip condition toidentify th with the velocity in the solvent near themembrane, we estimate the viscous traction on themembrane as qh /b, where b is the bending relaxation

time scale. Balancing the elastic force per area with thetraction leads to Brochard and Lennon, 1975

b

q3 . 18

Since the smallest q scales with vesicle size L, we mayalso write bL3 /. A simple scaling estimate for themagnitude of the bending stiffness is kmol

2 , wheremol is a molecular scale such as the thickness of themembrane Seifert, 1997. Thus, the ratio of the stretchrelaxation time to the bend relaxation time for mem-branes scale as s /bmol/L2, and just as in the caseof filaments we expect density modulations to relaxmuch faster than bending modulations. To estimate thetime scales directly from the measured parameters forthe lipid stearoyloleoyl phosphatidylcholine, we use k0.2 N/m and 10−19 J Dimova et al., 2002 whichyields for a vesicle with L=10 m a stretch relaxationtime of s10−7 s and a bending relaxation time of 10 s.1 Note that the bending relaxation time dependsstrongly on length scale; a ripple with a 1 m wave-length relaxes 1000 times faster than a ripple with a10 m wavelength. Thus, the time scale for bending dy-namics of lipid membranes can easily be of the order ofseconds.

In addition to the viscous forces from the solvent,there are two other dissipative forces acting on themembrane: viscous forces due the surface viscosity ofthe membrane itself, and friction forces arising from theslipping of the two monolayers past each other. The rela-tive importance of these forces changes with scale, withbilayer friction and then membrane viscosity becomingdominant as the length scale decreases Seifert andLanger, 1993; Seifert, 1997. The implications of bilayerstructure, both for dynamics and for morphology, aredescribed in Seifert 1997 and will not be discussed inthis review. However, we will treat intrinsic membraneviscosity, since it is even more important than solventviscosity in the case of polymersomes.

Surface viscosity relates the viscous force per unitlength fv to the shear rate, fvv. Thus, the ratio ofsurface and bulk viscosity is a length, SD= /, termedthe “Saffman-Delbrück length” Henle et al., 2008 inhonor of Saffman and Delbrück’s study of the mobilityof particles embedded in a thin liquid film atop a deepliquid subphase Saffman and Delbrück, 1975. Since theforce per unit area acting on an element of membrane isgiven by a derivative of the force per length, the surfaceviscous force per unit area scales as q2v compared toqv for the traction from the solvent. Therefore, surfaceviscosity effects become dominant when SDq 1. Sur-face viscosities for lipid membranes are of the order of

1In principle, the relaxation dynamics of the compressionmodes could be so fast that inertial effects become relevant,invalidating our estimates for viscous friction. This applies toboth filaments and membranes, but does not change the con-clusion that the compression modes relax much faster thanbending modes Seifert and Langer, 1993.



10−9 N s/m Dimova et al., 1999, 2000, 2006, whereasthe surface viscosity of diblock copolymer vesicles canbe about 500 times higher Dimova et al., 2002. TheSaffman-Delbrück length is SD1 m for liposomesand SD1 mm for polymersomes. Therefore, surfaceviscosity effects dominate the dynamics of polymer-somes. However, we will see in Sec. IV.D that solventviscosity cannot be neglected in the dynamics of almostflat membranes. Nevertheless, the bending relaxationtime for a polymersome may be obtained by the sameargument that led to Eq. 18 with q replacing , lead-ing to

b

q2 , 19

or bL2 / for a vesicle of size L. Since surface vis-cous stresses will resist the relaxation of both compres-sion and bending modes, the relation sb still holdsfor polymersomes.

III. FILAMENTS

In this section we derive the equations of motion forflexible filaments in a viscous liquid Doi and Edwards,1986. We begin with kinematics and review how to com-pactly describe bending and twisting using a body-fixedframe of directors. Then we use variational arguments toderive the elastic forces per unit length acting on anelement of a rod in an arbitrary configuration. After adiscussion of compatibility relations for rod strain andangular velocity, we conclude the section with an over-view of the viscous forces acting on slender rods in theoverdamped regime of Stokes flow.

A. Rod kinematics

The centerline of a rod is given by r, the position ofthe point with coordinate . Since is fixed for a givenmaterial point, we call a material coordinate. Alterna-tively, is sometimes called a convected coordinate sincethe label is carried along with the material point as thebody deforms. As mentioned, it is much easier to bend arod than it is to stretch it. Thus, we consider inextensiblerods. Note that is a material or convected coordinate;it is a label for material points of the centerline of therod. It is convenient to take the parameter to run from0 to 1 from one end of the rod to the other. The distanceds between two nearby points on the rod centerline isgiven by the Pythagorean theorem and the chain rule,

ds2 = dr2 =r

·r

d2. 20

The vector r=r / is tangent to the centerline; nor-malizing this vector leads to the unit tangent vector rsr.

Our characterization of the configuration of a rod iscompleted by specifying the orientation of the rod cross

section at each s. Let d1 and d2 lie along the principaldirections of the rod cross section see Fig. 2. Then de-

fine d3= d1 d2. For simplicity, we consider unshearablerods, for which the cross section always remains perpen-

dicular to the rod centerline, or d3=r Antman, 1995.Note that if the cross section is circular, then we must

arbitrarily choose a direction d1 at each point s. For thecase of a rod which is straight in the absence of external

forces, a sensible choice is to make d1s constant.Let the subscripts a ,b , . . . =1, 2, or 3 label the mem-

bers of orthonormal frames. The configuration of therod is given by the orientation of the frame for all s.Instead of specifying the orientation of the frame, it ismuch more convenient to describe the configuration ofthe rod in terms of how fast the orthonormal frame ro-tates as s increases. The frames for any two nearbypoints differ by a small rotation,

da = da, 21

where in the body-fixed frame of directors, =1d1

+2d2+3d3. For brevity we use the Einstein conven-

tion of summing repeating indices =ada. The infini-tesimal rotations a are uniquely defined once the direc-

tors are chosen, since 1= d2 · d3, and 2 and 3 aredetermined by cyclically permuting the indices. Thequantities 1 and 2 are the components of the curva-ture of the centerline, and 3 is the rate at which the

cross section twists around d3 as s increases. For com-parison, it is useful to recall the Serret-Frenet frame

curve of a space curve, T ,N , B, where T=r is the unit

tangent vector, and the normal N and the binormal Bare defined by

N = B − T , 22

B = − N , 23

T = N . 24

The curvature is the rate of rotation of T about B, and

the torsion is the rate of rotation of N about T. It is

d1

d1

d2

d2

d3

d3

FIG. 2. Color online The directors for a rod are chosen alongthe principal axes of the cross section of the rod.



conventional to take 0 to make N point to the centerof curvature. Comparison of Eqs. 21–24 reveals that2=1

2+22 and 3=+, where is the angle between

N and d, d2 ·N=cos Fig. 3. Note that the directions ofthe normal and binormal become ambiguous when therod is straight, =0. This ambiguity can be avoided usingthe so-called natural frame, which is well defined even atpoints where =0 Bishop, 1975; Goldstein et al., 1998;Wolgemuth et al., 2004. However, there is no ambiguity

about d1 and d2 when the rod is straight; note furtherthat 1 and 2 can have either sign. In this review, weconsider rods which are straight in absence of externalforces. Thus, a may be interpreted as strains associatedwith bending and twisting. Note that a rod with

=0d2 is an arc of a circle, and a rod with =0d2

+0d3 is a helix. Also note that all three variables amust be given to specify the shape of a rod; specifyingonly the curvature and twist does not determine theshape. For example,

circle =1

Rcoss/Rd1 +

1

Rsins/Rd2 +

1

Rd3 25

describes a rod bent into a circle with curvature 1/R andtwist 1 /R, whereas

helix =1

Rd2 +

1

Rd3 26

describes a helix that also has curvature 1/R and twist1 /R see Fig. 4.

B. Energies, forces, and moments

Our next task is to find the force fds exerted on anelement of the rod of length ds by the rest of the rod.This force equals the force the element exerts on thesurrounding medium. It is useful to first consider asimple but classic question: What is the force per unitlength that a curve in the plane exerts on its surround-ings when its energy function is E= ds? Suppose alsothat the curve encloses a fixed area. Thus our problemmodels a two-dimensional droplet of incompressiblefluid. To find the force for an arbitrary configuration rsof the curve, we impose an external force per unit lengthequal and opposite to fs to put the instantaneous con-figuration in mechanical equilibrium. Thus, f is given bythe principle of virtual work, or the condition Etot=0,where

Etot = E + fs · rsds − pA , 27

and p is the Lagrange multiplier enforcing the constraintof fixed enclosed area A. To calculate Etot, we must findhow the length ds changes under a variation of the curveFig. 5

ds = r + r2ds − ds = r · rds , 28

where the second line of Eq. 28 is accurate to first

order in r. Note that r= d3 is the unit tangent vector.Our formula correctly shows that the length element isinvariant under rigid motions, such as the translationsr=a and the rotations r=r, where a and are con-stant vectors. Using ds to calculate Etot yields

Etot = r · rds + f − pN · rds

= − r + f − pN · rds , 29

where we integrated by parts and used the fact thatrL=r0 for a closed curve. Since r is an arbitraryvariation, the force is given by

f = + pN . 30

Note that the curve can only support normal forces perunit length. In equilibrium, f=0, and we recover theYoung-Laplace law, with the Lagrange multiplier p iden-tified as the difference in the pressure outside the curveand inside the curve. Since p is constant, is constant,and the length-minimizing curve is a circle.

d1

d2

Ω2d1

N

Bψ

κN−Ω1d2

FIG. 3. The basis vectors of the material and Serret-Frenetframes in the plane of the rod cross section.

FIG. 4. Two rods with the same curvature and the same twist3. For each rod, the black line traces the path of the tip of d1.Note that the fact that one rod is closed and one is open isimmaterial for determining the twist.

r(s1)r(s2)

r(s2) + δr(s2)r(s1) + δr(s1)

FIG. 5. The variation of a curve stretches the curve.



We now apply this approach to inextensible rods thatresist bending Hallatschek et al., 2007 and twist. Unlikethe previous example of a one-dimensional interface inthe plane, an elastic rod has internal structure with itsstate defined by the strains a. Suppose the elastic en-ergy of the rod is given by

E = E1,2,3ds . 31

For example, the energy density

E =A

21

2 +22 +

C

23

2 =A

22 +

C

23

2 32

describes a rod with bend modulus A, twist modulus C,and a straight shape in the unstressed configuration. Ourformalism is easily generalized to handle models inwhich E depends on derivatives of the a Kan andWolgemuth, 2007, such as models for polymorphicphase transformations in bacterial flagella Goldstein etal., 2000; Coombs et al., 2002; Srigiriraju and Powers,2005, 2006. However, we disregard such “strain-gradient” terms for simplicity.

It is natural to suppose that the ends of the rod aresubject to external forces Fext0 and FextL, and mo-ments Mext0 and MextL. To hold an arbitrary configu-ration in equilibrium with these boundary conditions, wemust apply an external force per unit length −fs. Since

the rod can twist about the tangent d3, we must also

apply an external moment per unit length −msd3. As in

the closed curve example, fs and msd3 are the forcesand torques per unit length that the rod exerts on themedium for a given configuration. These are again givenby the condition Etot=0, where Etot includes the workdone by all external forces. Note that we are not consid-ering the most general form of the external torque perunit length. A torque per unit length parallel to the cen-terline of the rod arises naturally when considering ro-tation of a filament in a viscous liquid see Sec. III.F,but our formalism can be easily modified to handle ex-ternal torques per unit length that are perpendicular tothe centerline.

Since the rod is inextensible, care must be taken whenperforming the variation, since most variations rsstretch the rod Fig. 5. The three components of r arenot independent for a variation that conserves the dis-tance between material points on the rod. It is conve-nient to introduce a Lagrange multiplier function en-forcing r ·r=1 and then vary the energy with respect tothe three components of r independently Goldsteinand Langer, 1995. Instead we follow the less commonprocedure of dispensing with the Lagrange multiplierfunction and constraining the variation Hallatschek etal., 2007. Since the constrained variation does notstretch the rod, arclength is conserved,

ds = 0. 33

Therefore, the order of applying the variation and de-rivatives with respect to s is immaterial,

s = s . 34

We may drop the distinction between r and r,and simply write r.

Since the energy depends on the a only, the configu-ration of the rod is completely determined by the orien-

tation of the directors in space, da. Thus, we may varythe configuration without stretching the rod by subject-ing the orthonormal frame of directors at each point s to

an infinitesimal rotation, da= da, where ada.The infinitesimal angles a are related to the directorsvia

1 = d2 · d3, 35

with 2 and 3 determined by cyclically permuting theindices in Eq. 35. Thus, under a variation, the ortho-

normal frame vectors d2 and d3 rotate about d1 throughthe angle 1 Fig. 6. The two angles 1 and 2 cor-respond to the two degrees of freedom of r under theconstraint r ·r=1,

1 = − d2 · r, 36

2 = d1 · r. 37

In general, there are no globally defined angles a whosevariations correspond to the infinitesimal angles a.The exceptions are cases when one of the directors is

constant for the whole rod. For example, we may take d1to be normal to the plane containing a planar curve;then 1 may be defined as the angle between the localtangent vector and a fixed direction in the plane, and 1

is the curvature. Likewise, the director d3 has a constantdirection for a straight, twisted rod, with 3 the cumula-

tive angle through which d1 has rotated, and 3 the twist3.

The variation in the strain is found by varying Eq.

21, using Eq. 34, da= da, and ada, and fi-

nally noting that V da=0 for all a implies V=0,

d1

d2

δχ1

d2 + δd2

d3 + δd3d3

δχ1

FIG. 6. Rotation of directors under a variation with 10,2=3=0.



= + . 38

It is also useful to write the variation in the followingcomponent form:

1 = 1 +23 −32, 39

2 = 2 +31 −13, 40

3 = 3 +12 −21. 41

The components of and not appear on theright-hand sides of Eqs. 39–41, since the basis vectors

da in and must be varied and differentiated, re-spectively. To interpret these formulas, note that a varia-

tion with =3sd3 changes the twist 3 and the prin-cipal curvatures 1 and 2, but not =1

2+22. A

deformation of the centerline =1sd1+2sd2changes the twist if the rod is curved. This fact underliesthe coupling of twist and writhe Celugereanu, 1959;White, 1969; Fuller, 1971; Goldstein et al., 1998; Kamien,1998; Dennis and Hannay, 2005.

We now have all the necessary ingredients to find theforce per unit length f. We demand Etot=0, where

Etot = E + f · r + m3ds − FextL · rL

− Fext0 · r0 − MextL · L

− Mext0 · 0 . 42

Since arclength is conserved 33 and the energy is afunction of a only, we have

E = Maads , 43

where MaE /a. The formulas for the variation ofthe strain, Eqs. 39–41, imply that

Maa = Maa + M · 44

=Maa − Maa − · M 45

=Maa − M · . 46

Likewise, if we introduce the force Fs acting on thecross section of the rod at s by F= f, we have

f · r = F · r 47

=F · r − F · r 48

=F · r − · d3 F . 49

Combining Eqs. 46, 49, and 42, and demandingEtot=0 for arbitrary inextensible r yieldsFL=FextL, F0=−Fext0, ML=MextL, M0=−Mext0, and

md3 = M + d3 F . 50

The boundary conditions on M lead us to identify Msas the moment exerted through the cross section at s bythe part of the rod with arclength greater than s on thepart with arclength less than s. Note that when m=0, Eq.50 is the equilibrium moment balance equation for arod Landau and Lifshitz, 1986.

For the rod energy density 32, the relation Ma

=E /a yields M=A1d1+2d2+C3d3, or

M = Ad3 d3 + C3d3. 51

The transverse part of Eq. 50 yields

F = − Ar + C3r r + r. 52

Our variational procedure does not determine the func-tion . The value of is chosen to enforce the con-straint r ·r=1, but from Eq. 52 we can see that it maybe interpreted as a tension see Deutsch 1988 or Gold-stein and Langer 1995, for example. Note that −Arhas a tangential part.

To summarize this section, we have used the principleof virtual work to derive the moment M and force Facting through any given cross section of a rod with en-ergy density 32. The constitutive relation Ma=E /aarises naturally from the variation. Likewise, the condi-tions of force balance F=0 and moment balance M

+ d3F=0 in equilibrium, when f=0 and m=0, are notimposed but are implied by the principle of virtual work.These statements can be generalized to arbitrary energydensities along a curve Starostin and van der Heijden,2007, 2009.

C. Compatibility relations for rods

The variational formula 38 leads to important rela-tions between the angular velocity and strain. The localangular velocity vector is given by = /t as t→0,where the variation is interpreted as the change of the

rod shape over a time interval t. Thus da= da im-plies

da

t= da. 53

Likewise, since we may similarly define /t= /t ast→0, Eq. 38 implies

t= + . 54

These compatibility relations are also conveniently de-rived using the rotation matrices that act on the directorframes Wolgemuth et al., 2004. A third way of derivingthe compatibility relations is to recognize that since sand t are independent variables, the order of taking de-rivatives does not matter, and thus we must have

2da /ts=2da /st. We will see below that the thirdcomponent of Eq. 54,



3

t=

3

s+12 −21, 55

is required to obtain the evolution equation for twist.Once again, it is useful to note that contains deriva-tives of a and derivatives of the directors when compar-ing Eqs. 55 and 54; compare with the discussion afterEqs. 39–41.

We may use Eqs. 21 and 53 and the fact that de-rivatives with respect to arclength and time commute torecast Eq. 55 in terms of the shape of the filament r,

t3 = s3 + r r · tr. 56

This relation may be viewed as a conservation law fortwist density 3, with 3 the twist current, and the sec-ond term on the right-hand side acting as a source fortwist Klapper and Tabor, 1994; Goldstein et al., 1998;Kamien, 1998. The first term may be understood byobserving that the only way the twist in a straight rodcan change is if the rate of rotation 3 is not uniformalong the rod. The second term reflects the fact thatmotions of the backbone that do not involve a rotationalvelocity 3 can also change the twist. Such motions re-quire a change in the writhe, where the writhe of aclosed curve is computed by counting the number ofself-crossing of the curve in a planar projection, andthen averaging over all projections see Dennis andHannay 2005, and references therein. Althoughwrithe is defined for a closed curve, its variation has alocal form and can be considered for an open curveKlapper and Tabor, 1994; Goldstein et al., 1998; Ka-mien, 1998.

D. Viscous forces: Slender-body theory and resistive-forcetheory

In this section we describe how to approximately cal-culate the forces acting on a slender body immersed in aviscous fluid, and give a slightly more quantitative dis-cussion of viscous fluid dynamics than presented in Sec.II.B; see Lauga and Powers 2009, and referencestherein for a more detailed treatment. The stress in anincompressible viscous fluid is related to the strain rateby ij=−pij+iuj+jui, where p is the pressure, isthe viscosity, and u is the flow velocity Landau and Lif-shitz, 1987. Note that we use the indices i , j , . . . for theCartesian coordinates. As argued in Sec. II.B, viscousforces dominate at the small scales of polymer and cells.Thus we are permitted to disregard inertia. In this limit,the flow is governed by the Stokes equations

2u − p = 0 , 57

· u = 0. 58

Note that the pressure p plays the same role as did inour discussion of inextensible polymers, Eq. 52: thevalue of p is determined to enforce the constraint ofincompressibility, ·u=0. The boundary conditions forthe Stokes equations are that the velocity of the fluid at

a solid boundary must match the velocity of the bound-ary.

Ideally, we want to calculate the viscous forces actingon a moving and deforming filament. Since no time de-rivatives appear in the Stokes equations 57 and 58,the flow u and thus the forces acting on the filament at agiven instant depend only on the geometry of thefilament—its velocity and shape at that instant. How-ever, it is generally impossible to find analytical expres-sions for the forces for anything but the simplest shapes.The problem can be simplified by developing an ap-proximation scheme that exploits the smallness of theratio of the rod diameter to its length; this approach isknown as slender-body theory Lighthill, 1976. Slender-body theory relies on the linearity of the Stokes equa-tions to construct solutions for the flow around a movingor deforming filament by superposition of singular solu-tions. These singular solutions are the flows induced by apoint force or a dipole source, and are analogous to thesolutions used in the method of images in electrostatics.The end result is that slender-body theory provides amethod, typically numerical, for computing the hydrody-namic forces distributed along a filament. Representa-tive examples include work by Shelley and Ueda 2000and Cortez 2001.

In the limit of an asymptotically thin filament, slender-body theory simplifies to a local theory known asresistive-force theory Gray and Hancock, 1955; Chwangand Wu, 1971; Lighthill, 1976. Resistive-force theory islocal in the sense that it describes the hydrodynamicforce per unit length acting at a point on a filament interms of the velocity of the filament at that point, incontrast with slender-body theory, which accounts forthe fact that the motion of every point along the fila-ment induces a flow that affects the hydrodynamicforces at any given point on the filament. Thus, it isoften stated that resistive-force theory does not properlyaccount for hydrodynamic interactions. Nevertheless,since resistive-force theory is much simpler to imple-ment that slender-body theory, and because it can giveaccurate results, we use resistive-force theory to formu-late the equations of motion of filaments.

In resistive-force theory, the fluid is treated as a pas-sive background material that does not respond to themotion of the slender object. The viscous force per unitlength acting on the rod is anisotropic, and given by

fv = − v − d3 · vd3 − d3 · vd3, 59

where v is the velocity of the rod relative to the localvelocity of the fluid. In the limit of small a /, the frictioncoefficients in Eq. 59 are defined by

4/ln/a , 60

2/ln/a , 61

where a is the filament diameter and is a cutoff repre-senting either the rod length or the characteristic radiusof curvature of the rod Keller and Rubinow, 1976.Note that replacing by 2 in Eqs. 60 and 61 merely



leads to subdominant additive terms in the denominator.Equation 59 becomes accurate when a becomes muchsmaller than the rod’s length or typical radius of curva-ture Keller and Rubinow, 1976. Macroscopic scaleexperiments show that resistive-force theory does sur-prisingly well in describing the shape of driven elasticfilaments in highly viscous fluid Koehler and Powers,2000; Yu et al., 2006; Qian et al., 2008. Numerical calcu-lations show that resistive-force theory is inaccurate fortightly coiled helices in flow, where hydrodynamic inter-actions are expected to be more important Kim andPowers, 2005. Even in a quiescent fluid, resistive-forcetheory can give qualitatively wrong predictions for thesense of rotation of a towed superhelix Jung et al.,2007. Thus, resistive-force theory is usually justified in afirst approach to a new problem, but its predictions mustalways be viewed with a critical eye.

In some situations, such as the rotation of a nearlystraight rod about its axis Wolgemuth et al., 2000, it isnecessary to include the viscous moment per unit length

for rotation about the local tangent vector d3. The ap-propriate local approximation for this moment is the vis-cous torque per unit length on a rotating cylinder,

mv = − rd3 · d3, 62

where r=4a2 is the rotational drag coefficient Lan-dau and Lifshitz, 1987. In other situations, such as therotation of helix, the contribution to the torque due tothe rigid-body motion of the centerline dominates thecontributions from Eq. 62.

E. Equations of motion for curves

We are ready to combine all the parts we have intro-duced and arrive at the coupled equations for the evo-lution of a rod. Since we work in the highly overdampedregime, all forces must sum to zero,

F + fv = 0 . 63

Suppose that there is no imposed flow in the fluid; thusthe velocity of the rod relative to the fluid is v=tr. Us-ing the elastic force on a cross section 52 and the vis-cous force per unit length 59, we find

tr + tr = − Ar + C3r r + r,

64

where tr=tr− d3 · trd3 and tr = d3 · trd3. Todetermine , write v in terms of the mobilities

–1 and

–1 and the internal force per unit length f=F,

v =f

+f

, 65

and use 0=tr ·r=2tr ·r=2sv ·r to find

sf · d3 + 1 −

f · sd3 = 0. 66

Using Eq. 52 to calculate f=F yields the desired dif-ferential equation for .

The elastic and viscous moments must also sum tozero,

M + d3 F + mv = 0 . 67

The only nontrivial terms in Eq. 67 are the tangentialcomponents, which simplify to

C3 = r3. 68

To find the evolution equation for twist, combine Eqs.68 and 56 to find

rt3 = C3 + rr r · tr, 69

where tr must be deduced from Eq. 64.We close this section by recalling the characteristic

time scales for the motion of filaments in a viscous fluiddescribed in Sec. II.B. Examination of Eq. 64 for anearly straight filament with no twist reveals that bend-ing modes have a characteristic relaxation time bL4 /A. Likewise, Eq. 69 applied to a straight buttwisted rod implies that twist deformations have a char-acteristic relaxation time t=rL

2 /C.

F. Illustrative example: Twirling and whirling of elasticfilaments

To illustrate the use of the equations of motion, weconsider a fundamental problem: the whirling instabilityof a rotating elastic rod in a viscous fluid Wolgemuth etal., 2000; Lim and Peskin, 2004; Wada and Netz, 2006.A motor at s=0 rotates the rod about its long axis withangular speed 0 Fig. 7. Initially, the rod is straight.Viscous forces cause twist to build up. For sufficientlyhigh rotation speed, the straight state is unstable, andthe rod writhes. Study of this problem will provide uswith intuition that will help us understand more compli-cated phenomena in rotating filaments, such as thepropagation of polymorphic instabilities in bacterial fla-gella Hotani, 1982; Coombs et al., 2002 or the chiraldynamics of beating cilia in embryonic developmentNonaka et al., 1998; Hilfinger and Jülicher, 2008.

As mentioned, in this review we disregard the effectsof thermal fluctuations. We consider a rod with lengthmuch smaller than the bend persistence length and twistpersistence length. We also suppose the length is muchsmaller than the twist persistence length kBT /C, which isdefined by an argument similar to the one in Sec. II.A.

ω0

x

y

η

z

FIG. 7. An elastic filament immersed in a liquid with viscosity, and twirled about the z axis with angular speed 0. Thestraight state is unstable above a critical angular speed.



See Wada and Netz 2006 for a study of the whirlinginstability for rods so long that thermal effects are im-portant.

Since we assume the rod is thin, it is natural to use thelocal drag formulas 59 and 62 to describe the viscousforces and moments acting on the rod. Dropping thesubscript 3 to save space, the balance of torques aboutthe tangential direction 68 becomes C=r. In thebase state, the rod is straight with r=zz, the tension vanishes, and the rod undergoes rigid-body rotationabout the z axis with speed =0, leading to a constantgradient in twist. At the free end, s=L, the torque van-ishes: C=0. Thus,

= r0s − L/C . 70

To estimate when the distributed viscous drag causes therod to twist so much that the straight state becomes un-stable, we may consider the related problem of a rodtwisted by external moments acting at either end, in theabsence of viscous forces. In this simpler problem, thestraight state of the rod becomes unstable when the twistmoment is roughly equal to the characteristic bendingmoment of a rod of length L: CcritA /L Landau andLifshitz, 1986. Since CA for ordinary materials Lan-dau and Lifshitz, 1986, the rod writhes into a nonplanarshape once it is twisted by about one turn. In our prob-lem with distributed external moments, the speed atwhich the typical twist torque balances the characteristicbending torque is given by rcritLA /L, or

crit A/rL2. 71

Since CA, the critical rotation rate is the same orderof magnitude as the relaxation rate for twist.

To find the precise value of the critical speed, linearizethe equation of motion about the base state, Eq. 64,

tr = − Ar + Cz r , 72

where r= x ,y ,0 and the twist is given by Eq. 70. Toleading order, the deflection of the rod is perpendicularto the z axis. We therefore expect that the tension issecond order in x and y, which may be confirmed byexpanding Eq. 66 in powers of x and y.

Perturbations to the twist are governed by the diffu-sion equation with diffusion constant C /r. Thus, pertur-bations in shape and twist may be studied independently.Since perturbations to twist are not destabilizing, we fo-cus on Eq. 72, which governs the shape of the filament.As in the static twist instability, this equation is mostconveniently analyzed using complex notation =x+ iyLandau and Lifshitz, 1986. In these variables, Eq. 72becomes

t = − A + ir0s − L. 73

For boundary conditions, we assume the end at z=0 isheld fixed, 0, t=0, and clamped, 0, t=0. At the farend the conditions are vanishing force, L , t=0, andvanishing bending moment L , t=0. The solutions toEq. 73 are proportional to expit. When the imagi-nary part of is negative, the base state is unstable; the

real part of is the rate of rigid-body rotation of thefilament about the z axis. Equation 73 has nontrivialsolutions only for special values of 0; to determine thevalue for which the straight state first becomes unstable,we assume s , t=expit, where is real, and use theboundary conditions to determine the critical 0. Thisprocedure is best done numerically. Wolgemuth et al.2000 found

crit 8.9A

rL2 , 74

in agreement with our rough estimate Eq. 71. The mo-tion of the filament can be described in terms of twodifferent rotations. One is the rapid local rotation ofeach element of the rod about the local tangent vector;this rotation has rate 0. But the centerline of the fila-ment undergoes rigid-body motion at a different, muchsmaller rate , which at =crit takes the value2

22.9A

L4 . 75

The rigid-body rotation rate is of the same order ofmagnitude as the relaxation time for bending. The ratioof crit to is of order L2 /a2 since the ratio of the cor-responding resistance coefficients /r is of order 1/a2.

Numerical methods applied to the full equations ofmotion are required to find the whirling state for 0 crit. Wolgemuth et al. 2000 attempted a weakly non-linear analysis by applying a numerical pseudospectralanalysis Goldstein, Muraki, and Petrich, 1996 to theequations of motion truncated at second order in dis-placement, and found a small-amplitude steady-statewhirling state for just above crit. However, subse-quent work using the immersed boundary method Limand Peskin, 2004, as well as simulations that include thefull coupling of thermal, elastic, and hydrodynamicforces Wada and Netz, 2006, showed that there is nosmall-amplitude whirling state, and that the weakly non-linear analysis yields incorrect results. Wada and Netz2006 found the same value for the critical frequencythat we quote here; Wolgemuth 2009 also found thisvalue as well as the large-amplitude whirling state usingthe pseudospectral methods employed in Wolgemuth etal. 2000. Lim and Peskin 2004 found a much lowervalue for the critical frequency. The reasons for the dis-crepancy remain unclear.

IV. MEMBRANES

In this section we follow the same format as our dis-cussion of curve dynamics, but we consider fluid mem-branes Boal, 2002. We begin with the geometry of afixed surface in space. This material is standard, but weinclude it for completeness. Then we turn to the calcu-lation of membrane forces per unit area as a function of

2Wolgemuth 2009 provided a more accurate value than theone stated in Wolgemuth et al. 2000.



membrane shape. Since traditional approaches to thiscalculation are somewhat tedious, we review a recentidea of Guven that embodies economy of effort Guven,2004. We then revisit the basic geometric concepts for amoving surface, taking care to distinguish between timedependence arising from flow or surface deformationand time dependence arising from time-dependent coor-dinates. Finally we conclude with the equations of mo-tion for an incompressible fluid membrane and applythem to the instability of a cylindrical membrane tubeunder pressure.

A. Geometry of a stationary surface

The intrinsic geometry of a space curve is Euclidean: aone-dimensional being confined to a space curve cannottell if the curve bends and twists in the ambient three-dimensional space. This fact is why we can alwayschoose arclength coordinates for a curve. In contrast,curved surfaces are non-Euclidean. Anyone who has at-tempted to flatten an orange peel onto a tabletop knowsthat it is impossible to construct a Cartesian coordinatesystem on a sphere. Thus we are forced to use curvilin-ear coordinates when a surface has curvature, and thedescription of the geometry of surfaces is necessarilymore complicated than for curves. A key theme is thatgeometrical and physical concepts do not depend onthe choice of coordinate system. Thus, we will graduallybe led to the basic ideas of tensor calculus, which willallow us to construct geometrical quantities such as cur-vature in terms of any curvilinear coordinate system.

An additional complication arises for fluid surfacesthat was not present for solid rods: even for a stationaryshape, there may be a complex flow pattern on the sur-face. We return to this complication later; first we con-sider stationary surfaces with no flow and review thegeometry necessary to calculate the force per unit areadue to internal stresses as a function of membraneshape.

Denote the coordinates of a stationary surface as1 ,2. Greek letters , , . . . refer to the two-dimensional coordinates, such as , and latin lettersi , j , . . . continue to refer to Cartesian three-dimensionalcoordinates. Thus the position of a point in space on thesurface is given by r1 ,2. The distance between twopoints on the surface is given by the Pythagorean theo-rem

ds2 = dr2 =r

·r

dd. 76

The right-hand side of Eq. 76 is known as the firstfundamental form. The quantity

g = r · r 77

is known as the metric tensor, because it converts dis-tances d1 ,d2 measured in coordinate space to dis-tance measured in real, three-dimensional space. Sincedistances can be measured by a being on a two-

dimensional surface, the metric tensor is an intrinsicquantity.

A surface has two independent tangent vectors at anypoint, which may be taken to be tr along the direc-tion of increasing , for =1,2. In general, the basist1 , t2 is not an orthonormal basis. The tangent vectorst are not geometric objects, since they depend on thechoice of coordinates, but a tangent vector field W=Wt is a geometric object independent of the choiceof coordinates. The unit normal to the surface is givenby the cross product of the two coordinate tangent vec-tors, suitably normalized,

n =t1 t2

t1 t2. 78

Note that there are two unit normal vectors at everypoint, n and −n. Unless there is an asymmetry such asdifferent fluids on either side of the membrane, there isno reason to prefer n over −n. The element of area on asurface is given by the parallelogram formed by the in-finitesimal vectors t1d1 and t2d2, which has area

dS = n · t1d1 t2d2 = gd1d2, 79

where g=g11g22−g12g21 is the determinant of the metrictensor.

The curvature of a space curve is the tangential com-ponent of the rate of change of the normal to the curveEq. 22. Since a surface has two tangential directions,curvature of a surface is described by a bilinear formknown as the second fundamental form,

dn · dr = − Kdd, 80

where

K = − n · t. 81

Note that since n · t=0, we have

n · t = − n · t = − n · t = n · t, 82

and therefore conclude that the second fundamentalform is symmetric: K=K.

To make a stronger connection between our discus-sion of curves and the second fundamental form, con-sider a curve on a surface, parametrized by arclength,xs=r„1s ,2s…, and define the normal curvature n

by n= n ·N= n ·dT /ds, or

n = n ·d

ds

d

dst 83

=K

d

ds

d

ds. 84

The normal curvature is the ratio of the second and firstfundamental forms Struik, 1988,

n =Kd

d

gdd

. 85

Equation 85 implies that at a given point all curveswith the same direction d2 /d1 and parametrized by



arclength have the same normal curvature.It is useful to characterize the curvature of a surface

by the maximum and minimum values of the normalcurvature at a point. If v=vt is a unit vector, then theextreme values of n are given by the critical points of

f = vKv − vgv − 1 , 86

where is a Lagrange multiplier. Demanding thatf /v=0 yields

Kv − gv

= 0. 87

Note that multiplying Eq. 87 by v and summing over, together with the definition of normal curvature 85,implies that =n. Since v0, the condition for Eq.87 to have a solution is detK−ng=0. If we useg to represent the inverse of the metric tensor,

gg = , 88

then what we have shown is that the maximum andminimum values of the normal curvature are given bythe eigenvalues of the matrix

K gK. 89

These extreme values are known as the principal curva-tures of the surface. Note that since K and g aresymmetric, the principal directions v corresponding tothe two principal curvatures are orthogonal. Also notethat Eq. 81 may be written in terms of K

to yield theWeingarten equations,

n = − Kt. 90

As an aside, we emphasize the importance of whether ornot an index is raised or lowered. The convention is touse the metric tensor to lower indices, and the inverse ofthe metric tensor to raise indices. Quantities with upperand lower indices transform oppositely from one an-other under coordinate changes; for example, W and ttransform under coordinate changes in such as way thatthe vector field W=Wt remains invariant. These trans-formation rules are reviewed in Appendix A.

Returning now to the discussion of curvature, we notethat it is common to describe the curvature of a surfacein terms of the mean curvature, H=K

/2, and theGaussian curvature, K=det K

. The mean curvature His the average of the sum of the two principal curvaturesof a surface. It is a measure of how a surface bends inthree-dimensional space; H is analogous to the curva-ture of a space curve. A plane has vanishing meancurvature, a cylinder has mean curvature equal to halfthe inverse of its radius, and a sphere has mean curva-ture equal to the inverse of its radius. The Gauss curva-ture K is the product of the two principal curvatures of asurface. It is an intrinsic quantity; K can be determinedby two-dimensional beings making local measurementsof length on the surface. For example, by choosing ap-propriate coordinates, one can show that the relationbetween the circumference C and radius R of a smallcircle on a surface involves K only do Carmo, 1976:

C2R1−KR2 /6 Fig. 1. Below we use direct calcu-lation to see that K is an intrinsic quantity. Note that aplane and a cylinder both have vanishing Gauss curva-ture, which is consistent with the fact that local measure-ments confined to the surface cannot distinguish a planefrom a cylinder.

In our study of curves we found it useful to expressthe derivative of directors in terms of the directors.Likewise, it is useful to examine the derivative of thecoordinate tangent vector fields t in terms of t and n.Since the coordinate tangent vectors on a surface neednot have unit length, the derivative of a tangent vectorhas both tangential and normal components,

t = Kn + t, 91

where the normal components have been identified us-ing the Weingarten equations 90 and n · t=0, and theas yet unknown quantities

are called the Christoffelsymbols. Note that

= since r=r and K

=K. The formula K= n ·r also suggests anotherinterpretation of the second fundamental form. Choosea point p on the surface and introduce Cartesian coordi-nates x ,y ,z in three-dimensional space such that z isparallel to n. Near p, we may choose coordinates1 ,2x ,y. The height of the surface over the tan-gent plane is given by the second fundamental form Fig.8.

To see how the Christoffel symbols are related to theshape, observe that

g = t · t = g +

g. 92

By cyclically permuting , , and in Eq. 92 and thenadding and subtracting appropriately, we can solve forthe Christoffel symbols to find

= g =

g

2g + g − g . 93

Since the Christoffel symbols depend only on the metricand its derivatives, they are intrinsic quantities.

Equations 91 are sometimes called the Gauss equa-tions or the partial differential equations of surfaces

y

x

z

h

∆x

∆y

FIG. 8. Color online The height near the origin given by thesecond fundamental form: hK, where 1=x and2=y.



Struik, 1988. These equations are indispensable for for-mulating the equations of motion for fluids on a deform-ing surface. In particular, we need to take derivatives ofthe velocity field, which has tangential and normal com-ponents. However, we limit this section to a simplifieddiscussion in which we consider the derivative of apurely tangential vector field, W=Wt,

W = Wt + W t + WKn . 94

Once the direction t of the derivative is chosen, thetangential and normal components of W are uniquelydefined, independent of the choice of coordinates. Thesequantities contain all the information about how Wchanges along the direction t. In contrast, the quantitiesW

do not contain complete information about thechange in W. For example, we can imagine a vector fieldon small region of a curved surface in which the compo-nents W are constants. Then W

=0, but W is notconstant, since the surface is curved. Therefore, we areled to define the covariant derivative of the vector fieldW as the tangential components of W,

W = W +

W. 95

The covariant derivative of a tangent vector field is anintrinsic quantity since the Christoffel symbols are in-trinsic.

The covariant derivative is readily generalized toother quantities defined on a surface. For scalar func-tions, f=f. Since each Cartesian component of avector field may be considered a scalar function, we mayalso write W=W. The Leibnitz rule is used to definethe covariant derivative of higher-order tensors. For ex-ample, we may write the dot product of two vector fieldsU and W as U ·W=UW

, where U=gU. The Leib-

nitz rule then implies

UW = UW + UW. 96

Since UW is a scalar function, UW

=UW

and solving Eq. 96 for W yields

W = W − W. 97

Since the Cartesian coordinates of a vector are scalarfunctions, we may write

t = t − t = Kn . 98

At first sight this relation may seem confusing since wedefined the covariant derivative of W by removing thenormal part from W, and here we have normal com-ponents in the covariant derivative of t. The key pointis that the two-dimensional indices determine the natureof the covariant derivative. The rule 98 ensures thatthe Leibnitz rule holds for the components and basisvectors of a vector: W=W= Wt+Wt.

The general rule is that the covariant derivative of atensor has a term with a Christoffel symbol added forevery upper index, and a term with a Christoffel symbolsubtracted for every lower index. For example,

K = K

+ K

− K

. 99

Since the metric tensor g is derived from the Euclid-ean metric of the ambient three-dimensional space, thecovariant derivative of the metric tensor vanishes,

g = g − g −

g = 0. 100

This fact may be deduced by writing Eq. 96 in terms ofU, W, and g. The metric tensor and its inverse is freeto pass through the covariant derivative.

To formulate the generalization of the Stokes equa-tions 57 and 58 for membrane flow on a curved sur-face we need to consider multiple covariant derivativesacting on the velocity field. Special care must be takensince covariant derivatives do not commute when actingon a vector field. The Gauss-Codazzi equations, whichwe now derive, determine the relation betweenW and W. Since the tangent vector fieldW is three scalar functions on the surface, covariant de-rivatives commute when acting on W,

− W = 0. 101

Resolving this formula into tangential and normal com-ponents yields the Gauss-Codazzi equations,

− W = KK − K

KW, 102

K = K. 103

The right-hand side of Eq. 102 can be simplified bynoting that K

K −K

K is asymmetric in , and

,, and therefore proportional to the determinant ofK. The only independent component is =1, =2, and

we may write

12 − 21W1 = KW2, 104

12 − 21W2 = − KW1. 105

The Gaussian curvature, originally defined in terms ofthe extrinsic quantity K= n ·r, is intrinsic. The factthat K may be determined by a two-dimensional beingliving on a surface is known as Gauss’s TheoremaEgregium Morgan, 1998.

We close this section with a discussion of the covariantform of Green’s theorem for a two-dimensional surface.This relation will be required when we derive the stresstensor for a membrane. First note that Eq. 93 impliesthat the covariant divergence of a two-dimensional vec-tor field has a simple form

V =1g

gV . 106

This formula is useful when integrating by parts, sincethe area element dS=gd2 depends on ,

VfdS = − VfdS + VfdS . 107

In Appendix B we show that the second term of Eq.107 can be written as an integral over the boundary ofthe surface



D

BdS =

DBpds , 108

where p=r /s n is the unit tangent vector perpen-dicular to the boundary D.

B. Energies and forces

In this section we find the force exerted on a patch ofa membrane by the rest of the membrane, or, equiva-lently, the force the patch exerts on the surrounding me-dium. We follow the geometric approach of Guven2004; see also Capovilla and Guven 2002, 2004. Forconcreteness, we focus on fluid membranes; however,the approach may also be applied to membranes withinternal degrees of freedom such as tilt order Müller etal., 2005. The elastic energy for a fluid membrane takesthe form

E = 2

2H2 + KdS , 109

where and are moduli with units of energy Canham,1970; Helfrich, 1973. For a closed vesicle, the Gauss-Bonnet theorem implies that the integral of the Gauss-ian curvature K over the surface is a topological invari-ant Kamien, 2002, and therefore does not changeunder a variation of the shape. Henceforth we drop thisterm.

To find the force fdS on a patch for an arbitrary shape,we invoke the principle of virtual work as we did withfilaments. We suppose the shape is held in equilibriumby an external force per area −f1 ,2 and examine thecondition Etot=0, where

Etot = E + f · rdS . 110

Finding the variation of E for surfaces is harder then it isfor curves. Since E= Eg ,KdS, we must find gand K in terms of r. This direct approach is straight-forward but tedious Zhong-can and Helfrich, 1989.Guven 2004 suggested introducing Lagrange multiplierfunctions which allow the variations r, g, and K to betaken independently,

W = E + f · rdS + F · r − tdS

+ t · n + nn2 − 1dS

+ K + t · n

− T/2g − t · tdS . 111

The energy density E is now regarded not as function ofr but instead as a function of g and K. The conditionof equilibrium is that W=0 for the variations r, t,n, K, and g. Since the variables appear quadrati-

cally in the constraints, the variations are easy to com-pute. For example, the variation with respect to r impliesthat

F = f . 112

Thus the Lagrange multiplier function F is the two-dimensional stress tensor. To see why the stress is ofmixed character, it is useful to apply Green’s theorem108,

D

FdS =

DFpds . 113

The force per unit length acting through a contour in thesurface is a three-dimensional vector, and the normal pto the contour lies in the local tangent plane and istherefore a two-dimensional vector.

It is convenient to define the functional derivativeW /h, where h is a generic mixed tensor, as

W = Wh

· hdS . 114

The condition W /t=0 yields the stress in terms ofLagrange multipliers,

F = T − Kt +

n . 115

Likewise, the condition W /n=0 along with Eq. 98gives

and n in terms of ,

=

, 116

2n = K. 117

Finally, W /K=0 implies

= − E, 118

where E=E /K.We need a few identities to help us find the variation

of E with respect to the metric tensor. For example, tofind the variation of the determinant of the metric ten-sor, we use the identity

expln det g = exptr ln g , 119

which yields

g = ggg. 120

Thus dS=ggdS /2. Note also that Eq. A5 implies

g = − ggg. 121

With these formulas we see that W /g=0 implies

T = gE + 2E

g. 122

The two-dimensional stress tensor may also be writtenas



T =2g

ggE . 123

Note that the contributions from varying dS in the con-straint terms vanish since the constraints are invokedafter taking the variation. To summarize, we have founda general formula for the stress tensor of a membrane interms of E=E /K, the curvature tensor K, andthe two-dimensional stress T Guven, 2004,

F = T + EKt − En . 124

The variational approach with auxiliary variables canalso be used to find the moments per length actingacross any infinitesimal line segment in the membraneCapovilla and Guven, 2002; Müller et al., 2007. Thefollowing two examples help illustrate this formalism.

1. Soap film

The energy of a soap film is given by the interfacialtension times the area. Thus, the energy density E=,and E=0. The stress depends only on the two-dimensional tensor T=g and

Ffilm = gt. 125

Ffilm is in the tangent direction as expected. The force

per unit area acting on the film is ffilm=Ffilm or

ffilm = 2Hn + gt. 126

The tangential component of Eq. 126 corresponds to aMarangoni stress and arises when surfactant concentra-tion varies over the film. For soap film on a frame inequilibrium, is uniform and f=0 is the minimal surfaceequation H=0. If the film forms a closed surface withfixed enclosed volume, we introduce another Lagrangemultiplier as in Eq. 27 and find that the mean curva-ture must be constant.

2. Fluid membrane

Consider the fluid membrane energy 109 with =0.The energy density is E= /2gK2. To find T, wemust differentiate E with respect to g while holdingK fixed. Note that Eq. 121 implies

g

g= −

12

gg −12

gg, 127

where we have used the symmetry g=g. A short cal-culation reveals that the bending stress in a fluid mem-brane is

Fbend = 2gH2 − HKt − Hn . 128

The force per unit area acting on the membrane due tointernal bending stresses is given by the divergence ofthe stress tensor Fbend

,

fbend = − 22H + 2H3 − 2HKn . 129

We used Eq. 103 to show that the tangential compo-nents of fbend vanish for constant . There is no tangen-

tial component of the force per unit area because thebending energy is invariant under changes in coordi-nates, and a small change in coordinates corresponds toa deformation of the surface along tangent directions atevery point,

r + r + t. 130

Equation 129 for fbend may be derived directly from f=−E /r with considerably more effort Zhong-can andHelfrich, 1989.

Although we argued that membranes in the tense re-gime are approximately incompressible, it is instructiveto examine the stress and force per unit area arisingfrom the stretching energy, Es= EsdS, where Es= k /2 /0−12. First we work by analogy with a two-dimensional liquid and identify the two-dimensionalpressure p2D by considering a situation with uniformdensity Cai and Lubensky, 1995. In that case, thetwo-dimensional pressure is p2D=−Es /A, where A isthe area and the derivative is taken at fixed number A.Carrying out the differentiation leads to p2D=−Es

−Es, where Es is regarded as a function of .To apply the principle of virtual work to find the stress

and elastic force, we must keep the identity of the par-ticles fixed under the variation. This constraint arises be-cause the principle of virtual work is derived by assum-ing mechanical equilibrium of a fixed set of particles.Therefore, the variation satisfies g=0, which im-plies =−1/2gg and

Es = − 12

ggp2DdS . 131

From Eqs. 111 and 131, the corresponding two-dimensional stress tensor is Ts

=−p2Dg and we seethat the two-dimensional pressure acts like a negativeinterfacial tension compare with Eq. 125. Thus, Fs

=−p2Dgt and fs=−2p2DHn− p2Dgt.We can make a further simplification by observing

that the large modulus k implies that the density will beclose to its preferred value: =0+1, where 10.Working to first order in 1 leads to p2D=k1 /0. Thestress arising from stretching is therefore

Fs = − k1/0gt = + gt, 132

where =−p2D. In general, the stress Fs cannot be dis-

regarded in a membrane, since the smallness of thestretch 1 is offset by the largeness of the modulus k.Note that in the absence of flow the tangential stressesacting on the membrane vanish, and therefore p2D and1 are uniform. Often membrane shapes are studied un-der the assumption of incompressibility, k→, in whichcase =−p2D is treated as a constant Lagrange multi-plier, and Fs

is the associated force per area enforcingthe constraint. For dynamics, the Lagrange multipliermay have spatial dependence if there are tangentialstresses.

The formulas for the stress tensors Fbend Eq. 128

and Fs Eq. 132 give insight into the mechanics of



membranes. Consider a tubule pulled out of a vesiclewith a point force Fig. 9 Evans and Yeung, 1994; Fy-genson et al., 1997. It is common to regard the sphericalpart of the vesicle in the right-most frame of Fig. 9 as areservoir of lipid, and to suppose there is an interfacialtension that remains constant during the deformation.Equivalently, we may regard the membrane as undertension, with =−p2D 0 and constant. Since the spheri-cal part of the vesicle is much bigger than the tubule, wesuppose that p2D is independent of tubule length oncethe tubule is formed. Thus, E=+ /22H2. The cylin-drical part of the vesicle is easy to analyze, since K=0and H=−1/ 2R, where R is the tubule radius. We maydisregard the pressure jump across the membrane sincethe large radius of the spherical part of the vesicle en-sures that the pressure jump is small. The total force onan infinitesimal patch of the cylinder therefore vanishes,f= ffilm+ fbend=0, or

− /R + /2R3 = 0. 133

In equilibrium, R= / 2.Closer inspection of the equilibrium configuration

may seem to reveal a paradox Waugh and Hochmuth,1987; Powers et al., 2002; Fournier, 2007. With the usualcylindrical coordinates 1 ,2= ,z, we expect that theinterfacial tension will lead to a force per unit lengthRF= along the edge AB of Fig. 10, with a force perunit length of equal magnitude along the other edge,CD. But since there is no pressure jump across the sur-face, there are no other forces to balance these forces.To resolve this apparent paradox, examine the stressF=Ffilm

+Fbend . In cylindrical coordinates, the first fun-

damental form is ds2=R2d2+dz2, and the second fun-damental form is Kd

d=−Rd2. Along a circularcontour of constant z, Fz=2z. Thus, the force required

to hold the tether in equilibrium is 4z; twice what itwould be for a cylinder with no bending stiffness. How-ever, along a line of longitude, constant F= /R− / 2R3=0; the contributions to the stress from thebending energy cancel the contributions from the inter-facial energy, as they must since there is no pressuredifference.

C. Moving surfaces: Kinematics

1. Coordinates on deforming surfaces with flow

The description of a moving surface which may haveinternal flows requires careful thought. Curves are muchsimpler. We saw that we can always use arclength for thespatial coordinate, and in the case of an inextensible fila-ment, the arclength coordinate of a material point doesnot change. There is no analog of arclength parametri-zation for surfaces, and instead we must choose coordi-nates that are most appropriate for the situation. Thereare several common choices, including convective ormaterial coordinates, surface-fixed coordinates, and gen-eral coordinates. Material and surface-fixed coordinatesare useful for formulating equations of motion Aris,1989, while general coordinates are useful when onemust solve for the evolution of a shape. In this sectionwe describe important aspects of each class of coordi-nates.

To define material or convective coordinates, param-etrize the surface at time t=0 with coordinates u

= uI ,uII. We will use roman numerals or upper-casegreek letters for material coordinates. Since we disre-gard thermal effects such as Brownian motion of themolecules making up the membrane, the motion of thematerial points defines a smooth flow velocity. The co-ordinates u are labels that stay with the material par-ticles as time evolves, with the trajectory of the materialpoint uI ,uII given by ruI ,uII , t. The velocity of a ma-terial point is given by

V = rt

uI,uII= Vt + Vnn . 134

As mentioned, convected coordinates are useful for for-mulating the equations of motion. For example, we willsee in Sec. IV.D that the Reynolds transport theorem forconservation of particles is conveniently derived usingconvective coordinates Aris, 1989. However, since themotion of a surface is typically something to be solvedfor, ruI ,uII , t is usually unknown at the outset of anyproblem.

An alternate approach is to attempt to define a coor-dinate system that is fixed to the surface, independent ofthe flows of the molecules making up the surface. Such achoice is easy enough for a surface that does not changeshape, but is impossible for a deforming surface. Thebest we can do is to choose coordinates 1 ,2 such thatthe velocity of a point with fixed coordinate—not neces-sarily a material point—is purely normal to the surface,

10m

icro

ns

FIG. 9. Equilibrium shapes of a vesicle subject deformed withan optical tweezer. The magnitude of the force increases fromleft to right. From Fygenson et al., 1997.

Fϕ

Fz

ϕxz

y

A

BC

D

FIG. 10. Color online A section of a tubule, showing theforces per unit length Fz and RF.



t ·rt1,2

= 0. 135

Thus,

V =drdt

uI,uII= tV

+ Vnn , 136

where V=d /dt, t is the trajectory of a particlewith constant u, and Vnn=r /t. Note that r /t ispurely normal. Surface-fixed coordinates are conceptu-ally useful when formulating equations of motion ofmoving surfaces. For example, we will see below howthe condition 135 leads to alternate expression of thecontinuity equation that commonly appears in the litera-ture Aris, 1989; Youhei, 1994; Dörries and Foltin, 1996;Fujitani, 1997; Buzza, 2002. But like material coordi-nates, surface-fixed coordinates also require knowledgeof the shape evolution which is usually unknown untilthe complete problem is solved.

Therefore, it is common to use general coordinateswhen one is solving for the shape of a membrane as afunction of time. By general coordinates we mean coor-dinates in the Eulerian point of view Chaikin andLubensky, 1995, which are not constrained by condi-tions such as Eq. 135 or the Lagrangian label of con-vected coordinates. For example, we use cylindrical co-ordinates r ,z , t=hz , tr+zz in our example ofaxisymmetric surfaces. Here r is a unit vector in theradial direction in cylindrical coordinates. The velocityin general coordinates is

V =drdt

uI,uII=

ru

du

dt+

rt

, 137

where r /t has both normal and tangential components.

2. Strain rate

To formulate the hydrodynamics of fluid membranes,we need to measure the strain rate, defined as rate thatthe distance between two nearby points on the surfacechanges with time. If dr is the vector connecting the twonearby points at a given time, then after a time intervaldt the two points are connected by dr=dr+ dVdt. Thestrain rate tensor S is then defined by

ds2 = ds2 + 2dtddS, 138

where ds2=dr ·dr and ds2=dr ·dr. Writing ds2 interms of dr and V leads to

S = 12 V · t + V · t . 139

Equation 139 holds for convective, surface-fixed, orgeneral coordinates. Note that S is unchanged by ro-tation with constant, uniform rate , V=r, as re-quired by the definition of strain Dörries and Foltin,1996. Likewise, the strain is invariant under translationsrr+a and Galilean boosts VV+V0, where a and V0are constant and uniform Dörries and Foltin, 1996.Applying the Gauss-Weingarten equations 91 and 90to Eq. 139 yields an alternate form,

S = 12 V + V − 2KVn . 140

The first two terms on the right-hand side of Eq. 140are the covariant generalization of the strain rate of atwo-dimensional material. The third term represents thestrain that takes place, for example, when a sphere in-flates uniformly—even though there are no velocity gra-dients, there is strain whenever there is curvature and anormal velocity Vn. Note that gradients in Vn do notlead to strain at first order in dt.

3. Equation of continuity

It is useful to formulate the covariant version of theconservation of particles, or, more generally, to formu-late the covariant version of the Reynolds transporttheorem Aris, 1989. To this end, consider a patch ofmaterial points St on a fluid membrane, and a scalarquantity , which for concreteness can be taken as thenumber per unit area. We want to calculate the rate ofchange of the total number of particles in the patch St,assuming that the particles never leave the membrane toenter the solvent. To eliminate the time dependence ofthe patch region, use convected coordinates,

0 =d

dt

StdS =

d

dt

S0

„u,t,t…GduIduII,

141

where S0=St=0 and G is the determinant of the metricin the material coordinates uI ,uII. The total time de-rivative of is given by the chain rule,

d

dt= t + V , 142

where V is the tangential component of the velocityV=Vt+Vnn. To evaluate the time derivative ofG, note that Eq. 120 implies dG /dt=GGdG /dt /2 and dG /dt=V · t+ t ·V.Reverting to general coordinates, we have

GdG

dt= gV · t + V · t . 143

We may now use the Gauss-Weingarten equations 91and 90 to deduce

d

dt

StdS =

Std

dt+ V

− 2HVndS 144

or

d

dt+ V

− 2HVn = 0. 145

The continuity equation 145 is sometimes expressed interms of derivatives of the metric tensor using surface-fixed coordinates Aris, 1989; Youhei, 1994; Dörries andFoltin, 1996; Fujitani, 1997; Buzza, 2002, in which thecondition 135 can be used to show



1g

tg + V = 0. 146

However, since Eq. 146 only holds for surface-fixed co-ordinates, and not, for example, in cylindrical coordi-nates we use to solve our illustrative problem in Sec.IV.E, we will use Eq. 145 in this review.

For an incompressible membrane, is constant, andthe continuity equation simplifies to

V = 2HVn. 147

In terms of the strain rate, incompressibility implies S=0.

D. Dynamical equations for an incompressible membrane

There are several complementary approaches to thedynamics of membranes from the points of view of fluidmechanics and interfaces Scriven, 1960; Waxman, 1984;Kralchevsky et al., 1994; Arroyo and DeSimone, 2009,statistical mechanics Cai and Lubensky, 1994, 1995;Miao et al., 2002, and the director approach analogousto our treatment of rods Hu et al., 2007. Instead ofpresenting the most general models, we discuss thesimple case of an incompressible fluid membrane. Wedisregard the bilayer nature of the membrane and there-fore do not consider effects such as bilayer frictionSeifert and Langer, 1993; Yeung and Evans, 1995.

The form of the two-dimensional viscous stress tensoris similarly derived as the stress for three-dimensionalfluid mechanics in flat space Aris, 1989. Disregardingthe possibility of viscoelastic effects Rey, 2006, thestress must be a symmetric tensor linear in the velocitygradients,

Tvis = Ag + BS, 148

with A linearly proportional to S and B independent ofS. Since S=0 for an incompressible membrane, onlyB enters Tvis

. The symmetries of the stress and rateof strain tensors imply that B is symmetric underexchange of and , and also and . In principle, wecould build B out of combinations of both of thetensors K and g. However, we work to leading orderin an expansion in curvature, in which case

B = gg + gg + gg. 149

The constant is the dilational viscosity, which does notenter the stress for an incompressible membrane sinceS=0. Thus, Tvis

=2S, and the total membrane vis-cous stress is

Fvis = V + V − 2KVnt. 150

The force per unit area acting on the membrane due tothe internal viscous stress is fvis=Fvis

or

fvis = 2V + KV − 2VnH

+ 2Hg − KVnt+ 2K

V − Vn4H2 − 2Kn , 151

where the Gauss-Codazzi equations 104 have beenused to simplify the final form. Various limits and formsof fvis have appeared in the literature Scriven, 1960;Waxman, 1984; Aris, 1989; Youhei, 1994; Dörries andFoltin, 1996; Fujitani, 1997; Arroyo and DeSimone,2009. Note that Eq. 151 leads to a qualitative changein the equations for fluid flow even in the simplest caseof flow on a fixed sphere, for which fvis=2V

+KVt, where K is a constant Henle et al., 2008.It is natural to ask if there are any other terms of the

same order in gradients of velocity that should be in-cluded in the surface viscous stress. For example, theterm Fn

=2nVn with “transverse” viscosity 2 has of-ten been used in the theory of capillary waves onsurfactant-laden interfaces Goodrich, 1961. However,this term has been shown to be unphysical by Buzza2002. Buzza’s argument is that Fn

does not obey frameinvariance. Frame invariance is a fundamental assump-tion of continuum mechanics that states that the relationbetween stress and strain for a small element of materialcannot depend on the motion of the element Oldroyd,1950; Larson, 1988, since at the scale of the microstruc-ture inertial effects are small for typical shear rates. Inparticular, the relation between stress and strain for asmall element must remain the same if the element issubject to a time-dependent rotation. Removing the spu-rious term Fn

eliminates unphysical results such as nega-tive measured dilatational viscosities that have plaguedthe interpretation of quasielastic light scattering experi-ments for many years Buzza, 2002; Cicuta and Hopkin-son, 2004.

To see why Fn is unphysical, one can show that in a

rotating frame the force per unit volume depends on therate of rotation for a rigid-body rotation Buzza, 2002.Perhaps a more direct argument that something is amisswith Fn

is that it predicts a positive dissipation rate forrigid-body rotation of a flat membrane Buzza, 2002. Ifwe include the 1 term, then the dissipation rate

W = V · Fvis dS , 152

has a term in the integrand proportional to2VnVn, which is nonzero for a rotation aboutan axis lying in the plane of the membrane. Thus weconclude that 1=0.

The membrane is also subject to tractions from theviscous solvent. The stress tensor for a viscous fluid is

ij = − pij + iuj + jui . 153

For membranes and interfaces, there is no systematicexpansion that leads to simple approximations analo-gous to slender-body theory, although an approximationsimilar to resistive-force theory is sometimes used forillustrative purposes Cai and Lubensky, 1995. Balanc-ing all the forces on a patch of membranes, we find

niij+ − ij

− + Ftot j = 0, 154

where ij± is the three-dimensional viscous stress evalu-

ated at the membrane surface with n the outward-



pointing normal, and Ftot is the elastic and viscous

stresses of Eqs. 128, 132, and 150.We close this section by writing out the equations of

motion for a membrane which is almost flat. We work tolowest order in deflection, and study the evolution of aripple with wavelength 2 /q. The membrane shape isgiven by rx ,y , t= „x ,y ,zx ,y , t…, where zx ,y , t=htexpiqx, and complex notation is used since wework to linear order in h. The large value of the mem-brane bulk modulus implies that tangential flow in themembrane is small, which is why there is no motion inthe x-y plane in the expression for rx ,y , t. The equa-tions of motion for the membrane simplify greatly sincethe membrane flow vanishes, V=0, and many of the geo-metrical quantities such as K are second order in h. Themean curvature is H2z /2, where to leading order22 /x2+2 /y2. The normal component of force bal-ance on the membrane is thus

zz+ − zz

− z=0 − 4z = 0. 155

To find the stresses z±, solve the Stokes equations 57

and 58 subject to the no-slip boundary conditions ux

=0 and uz= h expiqx at z=0,

ux± = − iqzheiqx qz, 156

uz± = h1 ± qzeiqx qz, 157

p± = ± 2qheiqx qz. 158

Thus, zz+ −zz

− z=0=−4qh expiqx, which togetherwith Eq. 155 implies

h = −q3

4h . 159

The ripple relaxes with the bending relaxation time 18Brochard and Lennon, 1975. Note that even if qSD!1, the membrane viscosity does not enter since thereare no tangential flows due to the high bulk modulus.However, curvature can lead to tangential flows for in-compressible membranes, as we show in the next sec-tion.

E. Illustrative example: Buckling and pearling instabilities oftubular polymersomes

In this section we discuss two instabilities of a tubularvesicle: pearling and buckling. Pearling arises when themembrane is subject to tension, and buckling ariseswhen the membrane is under compression. Both insta-bilities consist of a cylinder deforming into a series ofbulges, but the physical mechanisms are different. Pearl-ing can be induced by applying a laser tweezer to a tu-bular liposome Bar-Ziv and Moses, 1994; Bar-Ziv et al.,1998. The laser tweezer induces a tension, and if thetension is sufficiently high, the cylinder can lower its en-ergy by deforming into series of bulges. This instabilityis similar to the Rayleigh-Plateau instability of cylindri-

cal interfaces Rayleigh, 1892; Tomotika, 1935, and ithas been extensively studied in liposomes Granek andOlami, 1995; Nelson et al., 1995; Goldstein, Nelson, etal., 1996. Buckling of a membrane is analogous to thebuckling of a rod under tension Landau and Lifshitz,1986. Buckling of a spherical membrane is well studiedDeuling and Helfrich, 1976a, 1976b; Jenkins, 1977;Zhong-can and Helfrich, 1987; Peterson, 1988, 1989, butthe buckling of a cylindrical membrane has received lessattention.

Consider a cylindrical polymersome of radius a. If a isof the order of microns, then SD/a!1, and surfacemembrane viscosity cannot be disregarded. We study theinstability of the cylinder as a function of the pressure pinside the vesicle, with the pressure outside taken to bezero. Increasing the pressure increases the tension in themembrane and leads to pearling; decreasing the pres-sure compresses the membrane and leads to buckling.Suppose that in the initial state that there are justenough molecules to enclose a cylindrical volume of ra-dius a without stretching the membrane. Thus, =0,=−p2D=0, and there is no stretching stress in the mem-brane, Fs

=0 Eq. 132. However, because the mem-brane is curved, there is a nonzero bending stress Fbend

and a corresponding bending force per unit area Eq.129, which must balance the internal pressure: pn+ fbend=0. Since H=−1/ 2a, normal force balance in theinitial equilibrium state yields p=p0=− / 2a3.

Now imagine changing the pressure to a value pp0.The cylinder stretches or compresses, depending onwhether p 0 or p"0. Since the stretch modulus k ishigh, ka2 /!1, there is little stretch: 1 /0= −0 /01. However, the tension is not small cf. dis-cussion after Eq. 132, and

/a = p + /2a3 . 160

Our task is to determine if sufficiently large positive ornegative tension in the membrane destabilizes the cy-lindrical shape.

To proceed, parametrize the vesicle in cylindrical co-ordinates, r ,z , t= a+hz , t+ zz+wz , t. Notethat the deformed shape is assumed axisymmetric. Tofirst order in the displacements h and w, the mean cur-vature is H−1/a+h /a2+h /2, and the Gaussian cur-vature is K−h /a. The evolution of the shape of themembrane is governed by normal force balance,

n · fvis + fbend + fs + nn+ − nn

− =a = 0, 161

where nn± =a= niij

±=a ,z , tnj is the normal-normalcomponent of the stress tensor for the solvent outside# and inside $ the vesicle evaluated at the mem-brane, to leading order in h.

We suppose the deformation is sinusoidal, hz , t%expiqz. Again, since ka2 /!1, the change in den-sity is small and we may use the condition of continuityfor an incompressible membrane, Eq. 147, to relate theshape to the flow Vz= w in the membrane. To linear or-der,



iqVz = − h/a . 162

The unperturbed solvent pressure and internal forcesper unit area acting on the membrane are thus

p + n · fvis + fbend + fs − 2h/a2 − Eqh , 163

where Eq=q4a2+ pq2a2+1− p /a4, and we have intro-duced the dimensionless pressure p=pa3 /.

The solvent forces acting on the membrane due to thedeformation are determined by solving Stokes equations57 and 58 for the flow u± inside and outside the tube,and then calculating the stress nn

± . Note that the im-posed pressure p enters the stress tensor through thedynamic pressure P: ij

±=−P±ij+iuj±+jui

±, where

P−=p when h=0, and P+=0 when h=0. Since the Stokesequations are linear, the stress at the membrane has the

form nn+ −nn

− =a=−−1h+p, where the “dynamicalfactor” depends only on the wave number q and ra-dius a. The Stokes equations in an axisymmetric geom-etry are readily solved using the stream function ±,where u±= ±z Happel and Brenner, 1965. Herewe only quote the result and refer the interested readerto Goldstein, Nelson, et al. 1996 for the details. Withthe no-slip boundary conditions at the membrane,

u±a= h and uz

±a=Vz, the dynamical factor is

= −a

2

qaI02 − I1

2 − 2I0I1qaK02 − K1

2 + 2K0K1I1K1 + I0K0 + qaI1K0 − I0K1

,

164

where In and Kn are modified Bessel functions of ordern evaluated at qa. Note that this expression corrects anerror in Eq. D.9 of Goldstein, Nelson, et al. 1996,which gave a dynamical factor which is too small.

Balancing internal membrane forces and the solvent

force leads to h=h, where

= −

a2

q4a4 + pq2a2 + 1 − p

2 + a/a/SD. 165

The sign of is determined by the sign of Eq=q4a4

+ pq2a2+1− p /a4. When p /a3, E can be negative, andthe cylinder is unstable. By Eq. 160, this pressure cor-responds to a tension =3 / 2a2, the critical tension forthe pearling instability Goldstein, Nelson, et al., 1996.Figure 11 shows the growth rate for the pearling insta-bility for various pressures and SD/a=1000. Also shownfor comparison is the growth rate for p=3 /a3 andSD/a=. Note how , which arises from the dynamicsof the incompressible solvent, forces the growth rate tovanish at q=0, even when SD/a!1.

When p"−21+2 /a3−4.828 /a3, the cylinder isalso unstable. Since negative pressure corresponds tocompressive stress, this instability is analogous to theEuler buckling instability of a rod. Figure 12 shows thegrowth rate for the buckling instability. Note the con-trast with the pearling instability. In the buckling insta-bility, when the pressure is just sufficient to induce the

instability, only a narrow band of wave number centeredabout a nonzero wave number qa2= 1+2 has posi-tive growth rate. In the pearling instability, the band ex-tends from q=0 to a finite value, and therefore there isno limit to the wavelength of the unstable modes.

V. OUTLOOK

The aim of this review has been to give a practicaloverview of the geometrical tools necessary to formulatethe equations of motion of polymers and membranes.Equilibrium and dynamic properties of thin filamentsand membranes are best described using geometricalideas such as curvature. We have also seen how naturallythese geometric ideas fit with the variational point ofview. In the case of filaments, the variation of the energyled directly to the constitutive relation between mo-ments and the rate of rotation of the directors, or strain.In the case of membranes, variation using auxiliary vari-ables provided a quick route to the elastic forces perarea, with the added bonus of the explicit form of thestress tensor of the membrane. Geometric ideas are alsocrucial for understanding viscous flow on a deformingcurved surface. This study of polymers and membranesremains an active area of research, and the purely me-

0.2 0.4 0.6 0.8 1.0 1.2

-2.0

-1.5

-1.0

-0.5

0.5

1.0

λµa2/κ

qap = 1 p = 2

p = 3

FIG. 11. Dimensionless growth rate a2 / vs dimensionlesswave number qa for the pearling instability of a tubular poly-mersome, in which membrane surface viscosity dominates thedissipation. The solid lines have SD/a=1000 and the dashedline has SD/a=.

0.5 1.0 1.5 2.0 2.5 3.0

-12.5

-10.0

-7.5

-5.0

-2.5

2.5

qa

λµa2/κ

0.0

p = −8

p = −2(1 +√

2)

p = −6

FIG. 12. Dimensionless growth rate a2 / vs dimensionlesswave number qa for the buckling instability of a tubular poly-mersome, in which membrane surface viscosity dominates thedissipation. The solid lines have SD/a=1000 and the dashedline has SD/a=.



chanical problems considered here provide a solid foun-dation for probing the dynamics of fluctuating polymersand membranes.

ACKNOWLEDGMENTS

I would like to thank Ray Goldstein, Greg Huber,Randy Kamien, Phil Nelson, and Charles Wolgemuthfor many discussions. I am grateful to Charles Wolge-muth for the code that produced Fig. 4, and DeborahFygenson for Fig. 9. Some of this review was written atthe Aspen Center for Physics. This work was supportedin part by National Science Foundation under GrantsNo. NIRT-0404031, No. DMS-0615919, and No. CBET-0854108.

APPENDIX A: COORDINATE TRANSFORMATIONS

In this appendix we describe how the components oftensors transform under a change of coordinates. In gen-eral, a tensor is a quantity that transforms linearly undera change of coordinates. We begin with the metric tensoras a concrete example.

Consider two systems of coordinates related by

= 1 ,2. From the chain rule, the metric tensor g inthe new coordinate system is related to metric in the oldcoordinates by a linear transformation law,

g =r

·

r

=

g. A1

The differentials have the opposite transformation law,

d =

d, A2

which ensures that

gdd = gd

d. A3

Quantities that transform linearly under a change of co-ordinates are tensors. The tensor transformation law forg reflects the fact that the distance ds2 is a geometricquantity, independent of the choice we make for coordi-nates.

It is traditional to use upper or lower indices to indi-cate what kind of transformation law a tensor has. Thus,a tensor T12¯

12¯ transforms homogeneously under achange of coordinates as

T12¯12¯ =

1

1

2

2¯

1

1

2

2¯ T12¯

12¯. A4

Tensors with upper indices are “contravariant,” and ten-sors with lower indices are “covariant.”

As mentioned in Sec. IV, we use raised indices to de-note the inverse of the metric tensor

gg = . A5

Viewing Eq. A1 as a matrix equation and invertingyields

g =

g; A6

that is, g transforms as a contravariant tensor. Usingthe metric tensor, we can create a covariant tensor froma contravariant one, and vice versa. For example, W

=gW and W=gW. Geometrically, we can think of

W as the components of the vector field W=Wt withrespect to a dual basis t defined by

t · t = . A7

In other words, W=Wt=Wt. The dual basis vectorst are like reciprocal lattice vectors, and also obey therule about raising and lowering indices,

t = gt. A8

Recall that the covariant derivative was defined in Sec.IV to be a geometric object, independent of coordinatesonce the direction t is chosen. The quantity V

= t ·V is a tensor since V and t are both tensors.Note, however, that the Christoffel symbol

is not atensor.

The importance of these linear transformation laws isthat they allow us to easily construct coordinate-invariant expressions. A quantity is coordinate invariantwhen there is an equal number of upper and lower indi-ces that are summed over. For example, the mean cur-vature H=K

and the divergence V are both invari-

ant under coordinate changes. Likewise, the Gaussiancurvature det K

is invariant under coordinate changes,since the upper and lower indices have associated withthem transformation matrices that are inverses. On theother hand, g=det g is not coordinate invariant since ithas two lower indices and no upper indices, and there-fore gets two factors of the Jacobian matrix

1 ,2 /1 , 2 under a coordinate change. These fac-tors are precisely what is needed to make dS=gd1d2

invariant under coordinate change.

APPENDIX B: GREEN’S THEOREM

Consider Stokes theorem for a tangent vector fieldA=At on a region D of a surface,

D

n · AdS = D

A · ds , B1

where ds is the infinitesimal tangent vector along bound-ary and denotes the three-dimensional gradient inCartesian coordinates. Note that is the covariant de-rivative for Euclidean space in standard coordinates.Recall that the normal component of the curl has deriva-tives along the tangent direction only,



n · A = n · A = t1 t2

g · A

=1g

t2t1 · − t1t2 · · At

=1g

1A2 − 2A1 =1g

1A2 − 2A1 .

B2

Since A ·ds=Ad, we have shown that

AdS = Ad, B3

where we have introduced the two-dimensional antisym-metric tensor with components 12=−21=1/g and 11

=22=0. Defining =gg note that 12=g and

A=B, we also have

BdS = Bd. B4

A short calculation shows that Bd=B · pds, whereB=Bt; p is the outward-pointing unit vector normalto the curve D defining the region of integration andlying in the tangent plane, p= td n /gd

d

=r /s n; and ds=gdd is measure of arclength

along the curve. Therefore we may write

D

BdS =

DBpds , B5

where pds=d.

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Dynamics of ﬁlaments and membranes in a …Dynamics of ﬁlaments and membranes in a viscous...

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