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Elastic helices Comprehensive exam report (revised version) Bojan Durickovic Advisor: Alain Goriely April 14, 2008 1

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Page 1: Elastic helices Comprehensive exam report

Elastic helices

Comprehensive exam report

(revised version)

Bojan Durickovic

Advisor: Alain Goriely

April 14, 2008

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Abstract

This report reviews the classification of infinite helical equilibria ofinextensible, unshearable uniform rods with quadratic strain-energy den-sity, presented in the paper Helices by Nadia Chouaieb, Alain Goriely,and John H. Maddocks [1]. A more detailed presentation of the basicunderlying theory of Kirchhoff equations is included.

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Contents

Introduction 5

1 General equations 71.1 Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.1.1 Director basis. Strips . . . . . . . . . . . . . . . . . . . . 71.2 Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.2.1 Spin vector . . . . . . . . . . . . . . . . . . . . . . . . . . 101.2.2 Compatibility relations . . . . . . . . . . . . . . . . . . . 10

1.3 Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.3.1 Conservation of linear momentum . . . . . . . . . . . . . 101.3.2 Conservation of angular momentum . . . . . . . . . . . . 11

1.4 Elasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.4.1 Linear elasticity . . . . . . . . . . . . . . . . . . . . . . . 13

1.5 Kirchhoff equations . . . . . . . . . . . . . . . . . . . . . . . . . . 141.6 Scaling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.6.1 Diagonal linear constitutive relation . . . . . . . . . . . . 151.6.2 General constitutive relation . . . . . . . . . . . . . . . . 16

2 Static equations 162.1 Static equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.2 Kirchhoff top analogy . . . . . . . . . . . . . . . . . . . . . . . . 172.3 Static equations in the director basis . . . . . . . . . . . . . . . . 18

2.3.1 Force equation . . . . . . . . . . . . . . . . . . . . . . . . 182.3.2 Moment equation . . . . . . . . . . . . . . . . . . . . . . . 182.3.3 Constitutive relation . . . . . . . . . . . . . . . . . . . . . 19

2.4 First integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.4.1 Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.4.2 Projection of moment onto the force . . . . . . . . . . . . 202.4.3 Tangential component of the moment in isotropic rods . . 202.4.4 Energy integral . . . . . . . . . . . . . . . . . . . . . . . . 20

2.5 Noncanonical Hamiltonian formulation . . . . . . . . . . . . . . . 212.6 Variational characterization of relative equilibria . . . . . . . . . 22

3 Helices 233.1 Equations for uniform helices . . . . . . . . . . . . . . . . . . . . 243.2 First integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243.3 Surfaces in twist-space . . . . . . . . . . . . . . . . . . . . . . . . 25

3.3.1 Helix hyperboloid . . . . . . . . . . . . . . . . . . . . . . 253.3.2 Integral surfaces . . . . . . . . . . . . . . . . . . . . . . . 25

3.4 Relative equilibria as solutions of the dual variational problem . 263.5 Discussion of helical equilibria . . . . . . . . . . . . . . . . . . . . 27

4 Conclusion and further work 27

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Bibliography 30

Appendices 30

Appendix A Geometry of curves 30A.1 Parametric form of a curve . . . . . . . . . . . . . . . . . . . . . 30A.2 Tangent vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30A.3 Principal normal, curvature . . . . . . . . . . . . . . . . . . . . . 30A.4 Binormal, Frenet basis . . . . . . . . . . . . . . . . . . . . . . . . 31A.5 Torsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31A.6 Frenet-Serret equations . . . . . . . . . . . . . . . . . . . . . . . 33

Appendix B Geometry of helices 34Index37

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Introduction

Over the course of the past two decades, experimental techniques for manipu-lation of individual DNA molecules have been perfected [2–7]. In these experi-ments, it is possible to apply a force and/or moment parallel to the helical axisof a DNA molecule, and to measure the elastic response in terms of elongationand angle of twisting about the helical axis. (See an example of an experimen-tal setup in Figure 1.) An abundance of data has been collected for variousexperimental scenarios.

Figure 1: A schematic of an experimental setup (from [8]).

The worm-like chain model (WLC) is predominantly used in these papers.Although this model does capture the entropic elasticity effects at large lengthscales, it breaks down at scales smaller than the persistance length1 (∼ 50 nm ∼150 base pairs ∼ 20 helical turns) [9].

DNA molecules have the well-known double helix structure, the degree ofcomplexity of which is sufficiently high that the derivation of the elastic proper-ties of the molecule from first principles (molecular forces) remains out of reach.Simpler models are needed to capture DNA’s mechanical properties. One of thesimplest models we can think of is to neglect the discrete structure of the basemolecules, and the double helix structure, and use a continuous elastic (single)helix. This simplified model does capture qualitatively certain aspects of elasticproperties of DNA molecules [10, 11].

Although inspired by the experiments referred to above, the questions raisedbelow are by no means restricted to DNA molecules.

Stability Certain instabilities of helices have been well studied, including su-percoiling [11] and the perversion instability [12], as well as the stability of

1 The persistance length is the length over which the correlation in the orientation of themolecule is lost.

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certain helical configurations [13, 14]. Work has also been done in dynamicinstabilities of helices [15]. However, a systematic study and classification ofhelix instabilities seems to remain an open research topic. Which region of thecurvature-torsion plane correspond to stable helical states, and which part ofthe boundary of that region correspond to what type of instability?

Inverse spring problem With the knowledge of elastic parameters of thespring, the elastic response (strains) can be calculated for a given wrench (stress)applied (or vice versa) in experiments of type described above. A valid questionto ask is: how to determine the elastic parameters from measurements of bothstresses and strains?

Even within the realm of linear elasticity, there are several ways to model thissystem: the isotropic diagonal case (no preferred bending direction, no bend-twist coupling; 2 scalar parameters), the non-isotropic diagonal case (two differ-ent bending stiffnesses, no bend-twist coupling; 3 scalar parameters), isotropicwith bend-twist coupling (no preferred bending direction, with bend-twist cou-pling; 3 scalar parameters), non-isotropic with bend-twist coupling (two differ-ent bending stiffnesses, with bend-twist coupling; 5 scalar parameters). Thefour models are listed in increasing number of scalar parameters, and each newparameter presents a new degree of freedom in fitting the data. What measure-ments does it take to fit each of these models to the data? Can all four be fittedwith simple wrench-type experiments? How much information about the elasticparameters can be extracted without complete knowledge of the stresses andstrains (the moment being difficult to measure)? How do the fits for variousmodels compare in real systems? Does more scalar parameters translate intobetter fits?

In order to tackle these and other questions regarding systems consisting ofa helical spring with a wrench applied along the helical axis, we lay down thebasic theory of elastic rods, with an emphasis on application to static helicalrods.

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1 General equations

This section outlines the basic concepts underlying the Kirchhoff theory of rods,culminating in time-dependent Kirchhoff equations, which describe the dynam-ics of inextensible and unshearable rods. For a rigorous derivation with arbitrarystrain-energy density, see [16]. A reduction of the obtained equations will beused in the following section in the study of static solutions.

1.1 Geometry

Within the realm of Kirchhoff equations, a full description of a rod is givenby a curve connecting the centroids of the rod cross sections, and an angularparameter that captures the orientation of the material cross sections at eachpoint of the curve. We will refer to this curve as the centerline, and to theangular parameter as the register2. As we assume that the rod is inextensible,we parametrize the centerline by the arc length s. For a brief overview of thegeometry of curves, see Appendix Section A.

We note here a useful geometric fact: the centerline is prescribed by twoscalar functions: curvature (cf. Subsection A.3), and torsion (cf. Subsection A.5),as shown by the following theorem.

Theorem 1 (Fundamental theorem). A curve in space is uniquely determinedup to a rigid body motion by the curvature and the torsion.

In other words, knowing two scalar functions κ(s) and τ(s), given the initialconditions for the Frenet basis (ν,β, τ ) (i.e. the orientation of the curve), it ispossible to reconstruct the Frenet basis (ν,β, τ ) at every point of the curve.Then, the tangent vector τ alone is then sufficient to recover the parametricform of the curve x(s) (cf. equation (82)). For proof, see e.g. [18, pp.56].

The advantages of using the curvature and torsion (κ(s), τ(s)) to representthe centerline, rather than e.g. a parametric representation x(s), are: (1) twoscalar parameters are used instead of three coordinates, (2) it gives a representa-tion independent from a reference frame (i.e. rigid body motion is automaticallyeliminated), and (3) the values of the curvature and torsion at a given point pro-vide information about the local behavior of the curve.

1.1.1 Director basis. Strips

When dealing with material rods, instead of using the Frenet basis, which isdefined by the geometry of curve, it is more convenient to use a basis that isfixed with reference to the material of the rod, for example in the direction of theprincipal axes of inertia. While the tangent vector is still useful in this picture,the two remaining basis vectors should be allowed to rotate in the (ν,β) planeas the rod is twisted.3 An additional degree of freedom (angle ϕ) thus is required

2 Following [17].3 Note that a rod that is twisted (in a material sense) can still have the same geometric

shape, even though the material description is different.

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for a full description of a material rod — this is the angle that describes the(material) twisting of the rod about its centerline. A geometric object specifiedby a curve and an additional angular parameter (represented parametrically by(x(s), ϕ(s))) is called a strip. We can therefore say that, within the frameworkof the Kirchhoff equations, a strip is a geometrical representation of a physicalrod.

The director basis (d1,d2,d3) is a generalization of the Frenet basis (cf. Sub-section A.4) for strips. This basis is defined so that d3 coincides with the tangentvector, while the remaining two vectors are rotated in the normal-binormal plane(ν,β) by the register ϕ:

d1 := cosϕν + sinϕβ (1a)

d2 := − sinϕν + cosϕβ (1b)

d3 := τ (1c)

The coordinates of a vector in the director basis are sometimes referred to aslocal coordinates, and we denote them using sans serif fonts.4

Expressed in the local director basis, the coordinates of the Darboux vector(cf. Subsection A.6) are

Ω1 = κ sinϕ (2a)

Ω2 = κ cosϕ (2b)

Ω3 = τ . (2c)

The vector that plays the role of the Darboux vector for the director basis (in thesense that it captures the evolution of the basis along the strip by an analogueof the Frenet-Serret equations (95)) needs to take into account the change of theregister ϕ = ϕ(s) along the strip. As the rotation in the (ν,β)-plane is describedby the third (tangential) component, there will be an extra term ϕ′(s) in thisdirection. The generalization of the Darboux vector is called the twist vector5

u, whose director basis coordinates are thus

u1 = κ sinϕ (3a)

u2 = κ cosϕ (3b)

u3 = τ + ϕ′ . (3c)

We will refer to the tangential coordinate of the twist vector u3 as the twistdensity, and to ϕ′(s) as the excess twist (thus twist density = torsion +excess twist), although these terms are not used in a consistent way elsewhere.6

4 The reason to distinguish them so clearly from coordinates expressed in an external(laboratory) system of reference is that the director basis changes orientation both in timeand in space, and care must be exerted when relating the derivative of vector with its localcoordinates.

5 Some authors use the term Darboux vector for both.6 Other terms used for twist density include: material torsion, physical torsion [19], and

physical twist [20], and for torsion: geometric torsion, and mathematical torsion [19].

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A strip with zero excess twist (ϕ′ ≡ 0) is said to be uniform (also called aFrenet strip).

Equations (3) are easily inverted:

ϕ = arctanu1

u2(4a)

κ =√

u12 + u2

2 (4b)

τ = u3 +u2′u1 − u1

′u2

u12 + u2

2. (4c)

We denote the column-matrix [u1, u2, u3]T (i.e. the matrix representation of thetwist vector u in the director basis) by u (in sans-serif font). The coordinates of uin the normal-binormal plane (d1,d2) plane, u1, u2 are called the componentsof curvature.

The generalized Frenet-Serret equations (also called the twist equa-tions) describe the evolution of the director basis along the curve:

(di)′ = u× di , i = 1, 2, 3 , (5)

where u is the twist vector defined in (3).7 This form of the equations is imposedby the requirement that the director basis remains orthonormal for all values ofthe parameter s.

Proposition 2 (Fundamental theorem for strips). A strip is uniquely deter-mined up to rigid body motion by the director basis components of the twistvector u(s).

Proof. The curvature κ and angle ϕ represent the polar representation of (u1, u2).The torsion is then obtained as τ = u3 − ϕ′, and the claim follows from Theo-rem 1.

Note that the key for the proof lies in the fact that director basis com-ponents are used. If this was not the case, additional information about theangular parameter would be required, without which not even the curve couldbe reconstructed, since the torsion τ could not be extracted from the twistdensity u3.

1.2 Kinematics

So far, the only parameter used was s. Now, we allow all variables to exhibittime dependence as well, meaning that the curve is moving and changing itsshape in time.

di = di(s, t) , i = 1, 2, 3 .

7 Note the intricate interrelation between the twist vector, which is defined in terms ofits director basis coordinates, and the director basis vectors, whose rate of change along thecurve is given by the twist vector.

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1.2.1 Spin vector

If there is no shear deformation (this is our assumption), the director basiswill preserve its orthonormality in time. This yields equations analogous to thegeneralized Frenet-Serret equations (5) for time dependence, which we will referto as the spin equations:

di = w × di , i = 1, 2, 3 , (6)

where the time derivative is denoted with a dot,

˙( · ) :=∂

∂t( · ) ,

and the vector w is called the spin vector.

1.2.2 Compatibility relations

The twist vector and the spin vector are related in the following way. Differen-tiating (5) with respect to time, and (6) with respect to arc length s,8 we obtainthe following constraint

w′ − u = u×w . (7)

1.3 Mechanics

A rod is a physical system having the geometry of a strip and physical prop-erties (such as mass density, moments of inertia, Young modulus) that we nowconsider. A material point in the rod is described by the radius vector

x = x + X1d1 + X2d2 ,

where x = x(s, t) is the position of the center of the corresponding cross section,and X1,X2 are the coordinates of the point in the rod in the local director basiswith reference to the center x(s, t).

1.3.1 Conservation of linear momentum

The (infinitesimal) force acting on an infinitesimally thin slice of the rod ofwidth ds is

dn(s, t) +

∫∫A(s,t)

F(s, t) dX1 dX2 ds =

∫∫A(s,t)

ρ¨x dX1 dX2 ds ,

where ρ(s) is the mass density of the rod, and F is the body force per unitvolume element. Integrating over the cross section area A, in the left-hand sidewe obtain a per-unit-length body force f , while in the right-hand side, the X1

and X2 terms dissapear by the center of mass property, and we have

n′ + f = ρAx . (8)

8 and assuming the order of differentiation can be interchanged

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Differentiating with respect to s and noting that x′ = d3,9 this equation becomes

n′′ + f ′ = ρAd3 . (9)

We will refer to 9 (and equations derived therefrom) as the force equation.The tension in the rod is the projection of the force onto the tangent vector,

n ·d3 = n3.

1.3.2 Conservation of angular momentum

Similarly, the conservation of angular momentum yields the following equation

m′ + x′ × n +

∫∫A

L dX1 dX2 . =

∫∫Aρ r× x dX1 dX2 ,

where L is the body couple per unit volume. This yields

m′ + d3 × n + ` = ρI2 d1 × d1 + ρI1 d2 × d2 , (10)

where ` is the body couple per unit length, and

I1 =

∫∫A

X12 dX1 dX2 , I2 =

∫∫A

X22 dX1 dX2 (11)

are the two principal moments of inertia in the cross section plane.10 Wewill refer to 10 (and equations derived therefrom) as the moment equation.

1.4 Elasticity

The equations presented thus far all express general principles, and the elasticproperties of the rod have not been yet been taken into account.

Intrinsic twist The intrinsic coordinates of the rod are the coordinateswith no stress applied. We denote them with hatted variables. In the unstressedstate, a rod has intrinsic twist vector u, the components of which encapsulatethe intrinsic curvature κ, intrinsic torsion τ , and intrinsic register ϕthrough (the hatted version of) equation (4).

Constitutive relation The system of equations obtained so far is closed (seethe equation and variable count in the next subsection) by the constitutiverelation, which relates stresses (m) with strains (u- u):

m = f(u− u) (12)

9 and assuming that we can interchange the order of integration with respect to s and t10 These are “geometric” moments of inertia, which are the “physical” moments of inertia

of a homogenous body, divided by the (constant) mass density.

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Strain-energy density A rod is called hyperelastic if the right hand sidecan be written as a gradient of a scalar function W : R3 → R+

m =∂W

∂u(u− u) , (13)

whereW (u) is the strain-energy density function. We assume that the strain-energy desity is convex, coercive function with W (0) = 0.

For inextensible, unshearable rods—which we assume throughout—the strain-energy density W is a function of the strains u only, while the force is a reactiveparameter to be determined from the equations.

The strain-energy density function is said to be isotropic11 if

W (u cosϕ, u sinϕ, v) (14)

is independent of ϕ for all u and v. Physically, this corresponds to isotropyof elastic response in the cross section plane of the rod. Generally, however,the rod will have a preferred bending direction, and this is where it becomesapparent that the argument of W in (13) must be in terms of coordinates in abasis attached to the rod. The strain-energy density is therefore a function ofthree scalars—director basis coordinates of the twist vector, W = W (u1, u2, u3).We abbreviate this notation to W = W (u), where u is the column matrix withcoordinates of the twist vector in the director basis. The gradient must also betaken with respect to the director basis, so we denote it accordingly12

∂W

∂u(u−u) ≡Wu(u−u) ≡

(d1

∂u1+ d2

∂u2+ d3

∂u3

)W (u1−u1, u2−u2, u3−u3)

The rod is said to be isotropic if the strain-energy density is isotropic andu1 = 0 = u2.

Legendre transform Because the strain-energy density is a convex and co-ercive function, it can be inverted

u =∂W ∗

∂m(m) + u , (15)

where W ∗ is the Legendre transform of W

W ∗(y) = supx∈R3

y · x−W (x) . (16)

11 Strictly speaking, the accurate term would be transversely isotropic, since we only con-sider isotropy in the cross section plane. However, as the rod has one dimension distinguishedfrom the other two, this is the most isotropy that can be achieved in such a system.

12 Note that although the arguments of W are dependent on the choice of director basis, thegradient Wu is a vector (independent of coordinate representation), and, to be formally correct,the “vectors” on the two sides of the equation m = Wu(u) should be marked in different fonts(left hand side: vector; arguments in right hand side: director basis coordinates).

However, we will also use the looser notation m = Wu(u), by which we mean a directorbasis coordinate representation of the equation above. (The right hand sides are denoted thesame way, but represent different types of objects: a vector and a column matrix of directorbasis coordinates, respectively. The interpretation is obvious from the context, i.e. from theleft hand side.)

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1.4.1 Linear elasticity

When the strain-energy density is a quadratic function,

W (u) =1

2uTKu (17)

where K is a symmetric positive definite matrix,13 the constitutive relation islinear:

m = K(u− u) . (18)

As the director basis vectors d1,d2 can always be oriented along principalaxes of the rod cross section, we can assume without loss of generality that themost general form for the matrix K is

K =

K1 0 K13

0 K2 K23

K13 K23 K3

(19)

The coefficients K1 and K2 are called the principal bending stiffnesses, andK3 — the torsional stiffness. Off-diagonal elements K13 and K23 are thebend-twist coupling coefficients.

Moreover, we can always point d1 in the direction of the principal axis withthe smaller bending stiffness, so we can assume without loss of generality that

K1 ≤ K2 . (20)

Diagonal case In the simplest—diagonal—case, we have

W (u) =EI1

2u1

2 +EI2

2u2

2 +µJ

2u3

2 , (21)

for which the constitutive relation reads

m = EI1(u1 − u1)d1 + EI2(u2 − u2)d2 + µJ(u3 − u3)d3 . (22)

Here, E is the Young modulus, µ is the shear modulus14, J is a constantthat depends on the shape of the cross section15, and I1, I2 are the principalmoments of inertia (11). Here, EI1 and EI2 are the principal bending stiffnesses,and µJ is the torsional stiffness. A symmetric rod (I1 = I2 =: I) has a singlebending stiffness EI.

13 As K only has meaning through the quadratic form of the strain energy density, thematrix K can always be chosen as symmetric.

14 Although we consider the case with no shear deformation, the shear modulus comes intoplay in the torsional deformation.

15 For a circular cross section J = 2I, I := I1 = I2.

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1.5 Kirchhoff equations

The conservation laws (9) and (10), together with the constitutive relation (13)and the twist equations (5) and spin equations (6) are collectively called theKirchhoff equations. When writing these equations as a system, we oftenwrite down only the force equation, the moment equation and the constitutiverelation:

n′′ + f ′ = ρAd3 (23a)

m′ + d3 × n + ` = ρI2d1 × d1 + ρI1d2 × d2 (23b)

m = Wu(u− u) , (23c)

and leave out the twist (5) and spin (6) equations as understood.

Equation and variable count In terms of scalar equations, the number ofequations is:

• 9 twist equations: d′i = u× di

• 9 spin equations: di = w × di

• −6 constraints: d1,d2,d3 orthonormal,

• 3 force equations: n′′(s, t) + f ′ = ρAd3

• 3 moment equations: m′ + d3 × n + ` = ρI2 d1 × d1 + ρI1 d2 × d2

• 3 constitutive relations: m = Wu

which is a total of 21 equations, for 21 variables: (n,m,u,w,d1,d2,d3).Note that all equations are PDEs except the constitutive relation, which is

an algebraic relation that can be used to eliminate the moment from the momentequation, thus reducing the system to 18 PDEs for 18 variables (n,u,w,d1,d2,d3).

Even though the position vector x(s) of the rod centerline does not figurein the system explicitly, it can be obtained from the defining equation for thetangent vector (82) by integration:

x(s) = x(0) +

∫ s

0

τ (s′) ds′ = x(0) +

∫ s

0

d3(s′) ds′ . (24)

1.6 Scaling

Let us reduce the Kirchhoff equations (23) to dimensionless form by rescalingthe material coordinate s, time and the force

s = Ls =⇒ ∂

∂s=

1

L

∂s,

t = T t =⇒ ∂

∂t=

1

T

∂t,

n = φn .

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The new force and length scales will give the following scales for the moment offorce ([m] = [ns]) and twist ([u] = 1/[s]):

m = φLm

u =1

Lu .

We set the body force and body couple to zero.16 The rescaled Kirchhoff equa-tions are

φ

L2n′′ = ρA

1

T 2d3

φL

Lm′ + φd3 × n = ρI1

1

T 2

(I2I1

d1 × d1 + d2 × d2

)φLm = LWu(u/L) .

Equating the coefficients for the first two equations,

φT 2 = ρAL2

φT 2 = ρI1 ,

we see that the force and moment equations alone determine the length scale:

L =

√I1A. (25)

Thus, the natural length scale L is independent of the constitutive relation; it de-pends only on the geometry of the rod, not on its mechanical properties. On theother hand, the time scale T cannot be set without specifying the constitutiverelation — the strain energy density W fixes the time scale.

1.6.1 Diagonal linear constitutive relation

For a quadratic strain energy density, the rescaled constitutive relation is

φLm = EI11

Lu1d1 + EI2

1

Lu2d2 + µJ

1

Lu3d3 .

We can set the coefficient of the first term in the constitutive relation to unityby imposing the following relation between coefficients:

φL2 = EI1 .

Dividing the moment equation by this constitutive relation, we obtain the timescale:

T =

√ρI1AE

(26)

16There is no loss in generality in doing this: we can always scale them by the obtainedforce and moment scales, and reinsert them in the end.

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Finally, using these two scales, the moment equation gives the force scale:

φ = AE . (27)

The dimensionless rescaled Kirchhoff equations are

n′′ = d3 (28a)

m′ + d3 × n = ad1 × d1 + d2 × d2 (28b)

m = (u1 − u1) d1 + a (u2 − u2) d2 + Γ (u3 − u3) d3 , (28c)

where

a =I2I1, Γ =

µJ

EI1(29)

are constants:

a is the asymmetry coefficient, which is a geometric property17 of the rod(measure of asymmetry of the cross section). Note that by the choice (20),a ≥ 1.

Γ is the twist-to-bend stiffness ratio, which is an elastic property of therod (ratio of the torsional and bending stiffnesses).18

1.6.2 General constitutive relation

The equations for the quadratic strain-energy density (28) are easily generalizedto an arbitrary strain-energy density:

n′′ = d3 (30a)

m′ + d3 × n = ad1 × d1 + d2 × d2 (30b)

m = Wu(u− u) , (30c)

However, it should be remembered that the time and force scales cannot be fixeduntil a concrete form of the strain energy density is specified. The right-handside of the constitutive relation can be scaled so as to set the coefficient of one ofits terms to unity, while the remaining terms will generally have dimensionlesscoefficients (as a and Γ in (28c)).

2 Static equations

In this section, we present the general theory of static Kirchhoff equations,including a Hamiltonian formulation. The results will be applied to helices inthe following section.

17 We assume here that the rod is made of transversely isotropic material.18 In the ideal incompressible case Γ = 2/3. Typical values for macroscopic rods are

23< Γ < 1. For DNA, Γ ≈ 1.4.

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2.1 Static equations

Static equations follow from (23) by simply setting all time derivatives to zero.However, note that the force equation (23a) was obtained by differentiating (8)with respect to s. Therefore, we can use (8) instead and thus reduce the orderof the static force equation.19 The static system then reads

n′ = 0 (31a)

m′ + d3 × n = 0 (31b)

m = Wu(u− u) . (31c)

In the static case w ≡ 0, and (after elimination of the moment using theconstitutive relation) the system is reduced to 15 equations for 15 variables(n,u,d1,d2,d3). In the typical process of solving the system, the force is elimi-nated and the equations are expressed in terms of the director basis coordinates,yielding 3 scalar odes for the three coordinates of the twist vector, (u1, u2, u3).For that reason, and due to Theorem 2 and the fact that all other variablescan be reconstructed from the twist vector coordinates, we will refer to u as thesolution of the static system (31), although strictly speaking the actual solutioncomprises 4 other vectors as well.

2.2 Kirchhoff top analogy

The analogy between the static equations (31) and the heavy spinning top wasfirst discovered by Gustav Kirchhoff. For an intrinsically straight rod (u = 0),the formal analogy of the equations is complete, and the correspondance ofvarious variables is listed in Table 1. However, the analogy cannot be pushedas far as one between two problems, since the spinning top is an initial valueproblem, while the elastic rod is a boundary value problem. Imposing a positionand velocity of the top at two points in time does not make physical sense.

Table 1: Top analogy

quantity elastic rod spinning tops arc length time

(d1,d2,d3) basis attached to rod basis attached to rigid bodyd3 unit tangent vector unit vector from fixed point to center of massn force −mgm moment of force angular momentumu twist vector angular velocity vector

EI1, EI2 principal bending stiffnesses principal moments of inertia ⊥ d3

µJ torsional stiffness principal moment of inertia about d3

W strain-energy density angular kinetic energym = Wu constitutive relation angular momentum definition

19 In other words, a comparison of equations (23a) and (8) implies that integrating theformer with respect to s, the integration constant is always zero.

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2.3 Static equations in the director basis

We now write down the coordinate equations in the director basis. As with thetwist vector director basis coordinates u, we use sans-serif fonts to denote thecolumn vectors of force and moment coordinates in the director basis

n :=

n1

n2

n3

, m :=

m1

m2

m3

(32)

2.3.1 Force equation

We can express the force equation in terms of its director basis coordinates. 20

n′ =∑i

(ni di)′

=∑i

ni′ di + ni di

=∑i

ni′ di + u×

∑i

nidi = n′ + u× n

The force equation is thus:21

n′ + u× n = 0 (33)

Written by coordinates, this is

n1′ = u3n2 − u2n3 (34a)

n2′ = u1n3 − u3n1 (34b)

n3′ = u2n1 − u1n2 . (34c)

2.3.2 Moment equation

We analogously expand the derivative of the moment using the twist equa-tions (5):

m′ = m′ + u×m (35)

and the moment equation in director basis coordinates becomes

m′ + u×m + d3 × n = 0 (36)

Noting that

d3 × n =

−n2

n1

0

, (37)

20 Note that, as the basis is not constant, the coordinate derivatives notation v′1 for a vectorv is ambiguous, since (v1)′ 6= (v′)1. This is where a different font for director basis coordinatescomes in handy: by v′1 we always mean (v ·d1)′.

21 This can be thought of as follows: the force vector is constant, but the coordinate systemwe are using is rotating with angular velocity u. Looking from the moving frame, the forcevector, along with the whole fixed frame is rotating by −u, and so (n1d1 + n2d2 + n3d3)′ =−u× n.

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this equation written in coordinate form is

m1′ = u3m2 − u2m3 + n2 (38a)

m2′ = u1m3 − u3m1 − n1 (38b)

m3′ = u2m1 − u1m2 . (38c)

2.3.3 Constitutive relation

The constitutive relation (13) is naturally expressed in the director basis:

mi =∂W

∂ui(u− u) , i = 1, 2, 3 (39)

and can be used to eliminate the moment coordinates from the remaining equa-tions.

2.4 First integrals

We now look for the integrals of motion of the static system (31).

2.4.1 Force

The force equation (31a) is readily integrated

n = const (40)

We orient the z-axis of the fixed reference frame along the constant force vector,and denote the magnitude of the force by F :

n = Fe3 , F = const , (41)

Even though we have obtained a vector integral, it can be shown [20] that,when expressed in terms of director basis coordinates, only one scalar integralremains:

I1 := F 2 = n12 + n2

2 + n32 = const . (42)

The moment equation Using the constant force vector, the moment equa-tion can be read to give the direction of the rate of change of the moment:

m′ = −d3 × I1e3 = I1(e3 × d3) .

The rate of change of the moment is orthogonal to both e3 and d3. The mag-nitude of this rate of change is proportional to I1 and to the sine of the anglebetween e3 and d3.22

22 For helices, this is the cosine of the pitch angle.

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2.4.2 Projection of moment onto the force

Another first integral can be obtained by projecting the moment equation (31b)onto the force

m′ ·n + (d3 × n) ·n = 0 .

The mixed product in the second term vanishes, while the first term can betransformed noting that the force vector is constant (40), m′ ·n = (m ·n)′.Thus, projection of the moment onto the force is a first integral

I2 := m ·n = const . (43)

An equivalent way of expressing this first integral is using the fixed z-axis,pointed in the direction of the force vector (cf. (41)). Dividing m ·n by theconstant magnitude of the force I1 yields the z coordinate of the moment, whichwe label as an alternative first integral to I2:

mz =I2I1

=: I2 = const . (44)

Note that both first integrals I1 and I2 are constitutive-independent.

2.4.3 Tangential component of the moment in isotropic rods

In the special case of an isotropic rod (cf. Subsection 1.4), another first integralexists: the tangential coordinate of the moment, m3. This can be verified byprojecting the moment equation (31b) onto the tangent d3:

m′ ·d3 + (d3 × n) ·d3 = 0 (45)

The mixed product vanishes, and the first term yields

m′ ·d3 = (m ·d3)′ −m · (u× d3) (46)

For an isotropic rod, the last term on the right hand side vanishes, yielding anintegral:

I3 := m ·d3 = m3 = const . (47)

2.4.4 Energy integral

Let us look for another constant of motion by dotting the moment equationwith u, the force equation with d3, and adding them together:

u ·m′ + u · (d3 × n) + n′ ·d3 = 0

u ·m′ + n ·d3′ + n′ ·d3 = 0

u ·m′ + (n ·d3)′ = 0 .

Now we use the constitutive relation (31c)

u · (Wu(u− u))′+ (n ·d3)′ = 0 . (48)

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Rewriting the first term as

u · (Wu(u− u))′

= (u ·Wu(u− u))′ − u′ ·Wu(u− u)

= (u ·Wu(u− u))′ −∑k

∂uk∂s

∂W

∂uk(u− u)

= (u ·Wu(u− u))′ −W ′(u− u) ,

the equation yields an integral

IH := u ·Wu(u− u)−W (u− u) + n3

= u ·m−W (u− u) + n3 = const , (49)

where n3 = Fe3 ·d3 is the (generally nonconstant) tension (d3 coordinate of theconstant force vector).

2.5 Noncanonical Hamiltonian formulation

The static Kirchhoff equations as expressed in the director basis ((33), (36), (39))can be cast as a noncanonical Hamiltonian system, with associated Hamiltonian:

H(m,n) := W ∗(m) + m · u + n ·d3 , (50)

where W ∗ is the Legendre transform of W (cf. (16)), and d3 is the director basisrepresentation of the tangent vector, d3 = [0, 0, 1]T . The inverted constitutiverelation (15) translates into

∂H

∂m(m,n) = u , (51)

while the inextensibility and unshearability constraint is formally captured by

∂H

∂n(m,n) = d3 . (52)

The Hamiltonian structure is given by[mn

]′= J(m,n)∇H(m,n) (53)

where J is the 6× 6 skew-symmetric matrix

J(m,n) =

[m× n×

n× 0

](54)

Note that when n 6= 0, J has a two-dimensinal null space spanned by[nm

]and

[0n

](55)

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The first integrals I1 and I2 expressed in terms of the Hamiltonian systemvariables (i.e. director basis coordinates) are

I1(m,n) = n ·n , I2(m,n) = m ·n, . (56)

The third integral, IH , is the Hamiltonian itself.The integrals I1 and I2 are called Casimir functions, because their gradients

lie in the null space of the structure matrix J :

∇I1 =

[0n

], ∇I2 =

[nm

]. (57)

2.6 Variational characterization of relative equilibria

A relative equilibrium is a solution to the variational problem

Minimize H(m,n) (58)

subject to the constraints

I1 = n ·n = C12 , I2 = m ·n = C2C1 , (59)

where C1 and C2 are constants. We now look for a dual problem in terms ofthe strains u.

A point z = [m,n]T in phase space is a relative equilibrium of an autonomousHamiltonian system if

∇H(z) =

N∑i=1

λi∇Ii(z) , (60)

where H(z) is the Hamiltonian, and Ii(z) are the other integrals of the system.A non-isotropic rod has only two integrals I1 and I2. With (51), (52) and (57),the equilibrium criterion (60) becomes:[

ud3

]=

[λ2I 0λ1I λ2I

] [nm

](61)

Note that the right hand side is linear due to the fact that both I1 and I2 arequadratic in the Hamiltonian variables m and n. Equation (61) is a first-ordernecessary condition for the problem (58).

The case λ2 = 0 corresponds to u = 0 and d3 = λ1n, which is a configura-tion with a straight centerline, tangential force, constant moment, and constantdirector frame.

For λ2 6= 0, (61) can be inverted[nm

]=

[µ2I 0µ1I µ2I

] [ud3

], (62)

where

µ1 = − λ1

λ22 , µ2 =

1

λ2. (63)

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The force part of equation (62),

n = µ2u , (64)

states that the force and twist vectors are parallel, while the moment part yields

m ≡ ∂W

∂u(u− u) = µ1u + µ2d3 . (65)

The case µ2 = 0 corresponds to n = 0, and moment parallel to the twistvector.

Equation (62) is a first-order necessary condition for the dual problem:

Minimize W (u− u) (66)

subject to constraints

u ·u = η12 , u ·d3 = η1η2 , (67)

where the relation between the constants η1, η2 and C1, C2 is found from (62)to be [

λ2 0λ1 λ2

] [C1

2

C1C2

]= µ2

[η1

2

η1η2

]. (68)

Note that the coefficients µ1 and µ2 are Lagrange multipliers associated withconstraints (67).

For more details, see [20].

3 Helices

A helix is a curve of constant curvature and torsion. Hence, any helix is de-scribed by a twist vector of the form

u =

κ cosϕκ sinϕτ + ϕ′

, κ = const, τ = const, ϕ = ϕ(s) (69)

Helices with constant register ϕ are called uniform, and are described by apoint in u-space.

In the non-uniform case, the orientation of the director basis changes withrespect to the Frenet frame (cf. (1)), and this is what induces the changingdirector basis coordinates of the twist vector in (69). The twist vector u is stillconstant, as in the uniform case, but the helix is no longer described by a pointin u-space. Instead, the director basis coordinates of the twist vector will followa trajectory that lies on a cylinder of radius κ, whose axis is on the u3-axis.

In this section, we focus on uniform helices. The justification for such arestriction can be found in [1, Supplementary material], where it is shown thatnon-uniform helical solutions are highly exceptional.

We will say that a helix is isotropic if its strain-energy density is isotropic.

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3.1 Equations for uniform helices

For uniform helices, the director basis coordinates of all variables are constant:First, it follows the constitutive relation that u = const implies m = const.Then, (38a) and (38b) yield n2 = 0 and n1 = 0, respectively. Finally, it followsfrom the force integral (42) that the tangential force coordinate is also constant.

The static system of ODEs (34), (38) becomes a system of algebraic equa-tions:

u3n2 − u2n3 = 0 (70a)

u1n3 − u3n1 = 0 (70b)

u2n1 − u1n2 = 0 (70c)

u3m2 − u2m3 + n2 = 0 (70d)

u1m3 − u3m1 − n1 = 0 (70e)

u2m1 − u1m2 = 0 , (70f)

which can be used to solve for the force, and express the first integrals in termsof the twist and moment vectors only.

3.2 First integrals

Solving (70a) (70b) and (70d) for the force coordinates yields:

n1 =u1

u2(m3u2 −m2u3) (71a)

n2 = m3u2 −m2u3 (71b)

n3 =u3

u2(m3u2 −m2u3) . (71c)

We have thus determined the Lagrange multiplier µ2 in (64), which is the con-stant of proportionality between the twist vector and the force:

µ2 =m3u2 −m2u3

u2(72)

Now, we can now write down the expression for the integral I1 in terms ofthe director basis coordinates of moment and twist:

I1 ≡ n ·n = µ22 u ·u

=

(m3u2 −m2u3

u2

)2 (u1

2 + u22 + u3

2), (73)

and similarly for the projection of the moment onto the force:

I2 ≡ m ·n = µ2 m ·u

=m3u2 −m2u3

u2(m1u1 + m2u2 + m3u3) . (74)

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Equations (73) and (74) with a linear constitutive relation represent a vari-ant of the “spring formula,” known since the 19th century ([21, art.271] is afrequently cited reference).

Equations (73) and (74), along with the constraint (70f) and the constitutiverelation yield a complete set of equations. Eliminating the moment coordinatesusing the constitutive relation, we have three scalar equations for the three co-ordinates of the twist vector, with two parameters I1 and I2. We have thusconverted the constraints (59) of the variational problem (58) to u-space. Note,however, that the results in this subsection were derived directly from the Kirch-hoff equations, not from the variational formulation.

3.3 Surfaces in twist-space

3.3.1 Helix hyperboloid

Note that (70f) does not involve the first integrals I1 or I2; it is a universalconstraint on u for a given constitutive relation.

With a linear constitutive relation m = K(u− u), (70f) becomes an equationof a hyperboloid in u-space. Using the general form for the matrix K (19), theequation reads:

(K1−K2)u1u2−K23u1u3+K13u2u3+(K2u2+K23u3)u1−(K1u1+K13u3)u2 = 0(75)

Note that the hyperboloid is independent of the torsional stiffness K3; it isentirely given by the bending stiffnesses and the bend-twist coupling coefficients.

In the diagonal isotropic case (K1 = K2, K23 = 0 = K13), the hyperboloidreduces to:

u2u1 = u1u2 . (76)

If the rod is intrinsically straight (u1 = 0 = u2), this relation is trivially satisfied,meaning that the entire u-space corresponds to relative equilibria. If the rod isintrinsically curved, (76) is the plane spanned by the intrinsic twist u, and theu3-axis. Not only are the solutions uniform (constant register), but all points inthe plane share the same register ϕ = arctan u1

u2= arctan u1

u2= ϕ.

In the diagonal non-isotropic case (K23 = 0 = K13, K1 6= K2), the hyper-boloid exhibits translational symmetry along u3.

3.3.2 Integral surfaces

The first integrals (73) and (74) provide two more (constitutive-dependent) sur-faces in u-space. These surfaces correspond to the constraints in the variationalproblem (58). In the case of a linear constitutive relation, the I1 surface is asextic, and the I2 a quartic. The shape of these surfaces is not particularlyilluminating for illustration purposes.

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Figure 2: Helix hyperboloid with K1 = 1, K2 = 32 , K13 = K23 = 1

2 , u1 = 1,u2 = 0, u3 = 0. Solutions for various points on the hyperboloid are shown.The inset shows the intersection of the hyperboloid with the horizontal planeu3 = 3

2 . (From [1].)

3.4 Relative equilibria as solutions of the dual variationalproblem

Next, we consider the dual problem of minimizing the strain-energy densityW (u− u) (66) subject to constraints (67), rewritten below:

u ·u = η12 (77a)

u3 = η1η2 (77b)

In u-space, (77a) corresponds to a sphere of radius η1, and (77b) to a planeparallel to the (u1, u2)-plane, intersecting the u3-axis at η1η2. The resultingconstraint set (intersection of the above two sets) is a circle.

As it was pointed out previously, a first-order necessary condition for mini-mizing W (u− u) is

m ≡ ∂W

∂u(u− u) = µ1u + µ2d3 . (78)

The Lagrange multipliers µ1, µ2 are associated with the constraints (77), andcan be eliminated by noting that the constraint set is a circle with normalξ := (u2,−u1, 0) so that the first-order necessary condition (78) can be writtenexplicitly:

ξ · ∂W∂u

(u− u) = 0 . (79)

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Note that this condition is identical to (70f), which was shown in the previoussubsection to correspond to a hyperboloid in u-space for a quadratic strain-energy density W . Therefore, intersecting the constraint set (circle) (77) withthe hyperboloid provides the local extrema of W (u− u) on the circle. The insetin Figure 2 illustrates the fact that for different values of η1, η2, the circle canintersect the hyperboloid in two or four points (half of them are local minimaof W (u− u)).

3.5 Discussion of helical equilibria

As noted in Subsection 2.6, there are two exceptional families of solutions, u = 0and n = 0. The former represents a degenerate helix—a straight line, wherethe moment is given by the constitutive relation, and the force is parallel to thehelix centerline with arbitrary magnitude. The latter consists of helices withmoment and twist vectors parallel, both directed along the helical axis. Thisfamily contains the absolute energy minimizer u = u, for which m = 0. The twofamilies intersect at the unique uniform helical equilibrium for which u = 0 = n,which is a helix untwisted into a straight line by the moment m = Wu(−u) withno tension applied.

All other uniform helical solutions are found directly from the constraints(77) and the strain-energy density minimization necessary condition (79): Forany given curvature κ =

√u1

2 + u22 and torsion τ = u3, there are two or more

uniform helical equilibria, found by intersecting the constraint circle (77),

η1 =√κ2 + τ2 , η2 =

τ√κ2 + τ2

, (80)

with the helix hyperboloid (75). In the case of an isotropic rod, W has aconstant value on the constraint circle, and the entire circle represents uniformequilibria.

4 Conclusion and further work

The problem of finding helical equilibria was reduced to a geometric problem inu-space. Two approaches were seen, which correspond to dual variational prob-lems. One is expressed in terms of the first inegrals, which are simply relatedto the wrench (force and moment) applied to the helical rod in the experimentsreferred to in the introduction, and the dual problem uses constraints on thehelix configuration space (twist-space, or u-space) variables (essentially curva-ture and torsion), which are easily measured helix parameters. The two dualstress/strain formulations complement each other nicely and provide a morecomplete picture of uniform helical solutions.

Research topics that build on the theory presented in this report, which Ihave been and/or plan to work on in the future, include:

Instabilities of helices Finding regions in u-space that correspond to stable

27

Page 28: Elastic helices Comprehensive exam report

and unstable helical configurations, and classifying the types of instabili-ties on the boundary between the two.

Inverse spring problem Extracting elastic parameters from stress-strain mea-surements carried out in types of experiments described in the introduc-tion.

Modes of vibration of helices Finding eigenmodes of vibration of finite he-lices.

Compact support solutions Having found solutions with compact supportfor two-dimensional elastic rods with nonlinear strain-energy density, thequestion arises whether such qualitative behavior can be reproduced on ahelix.

References

[1] Nadia Chouaieb, Alain Goriely, and John H. Maddocks. Helices. Proc.Natl. Acad. Sci. USA, 103(25):9398–9403 (electronic), June 2006. ISSN1091-6490. doi: 10.1073/pnas.0508370103.

[2] Carlos J. Bustamante, Laura Finzi, Page E. Sebring, and Steven B. Smith.Manipulation of single-DNA molecules and measurements of their elasticproperties under an optical microscope. Proceedings of SPIE, 1435:179–187,July 1991. doi: 10.1117/12.44242.

[3] S. B. Smith, L. Finzi, and C. Bustamante. Direct mechanical measurementsof the elasticity of single DNA molecules by using magnetic beads. Science,258(5085):1122–1126, November 1992.

[4] Steven B. Smith, Yujia Cui, and Carlos Bustamante. Overstretchingb-DNA: the elastic response of individual double-stranded and single-stranded dna molecules. Science, 271(5250):795–799, February 1996. doi:10.1126/science.271.5250.795.

[5] Michelle D. Wang, Hong Yin, Robert Landick, Jeff Gelles, and Steven M.Block. Stretching DNA with optical tweezers. Biophysical Journal, 72:1335–1346, March 1997.

[6] C. Bustamante, S. B. Smith, J. Liphardt, and D. Smith. Single moleculestudies of DNA mechanics. Curr. Opin. Struct. Biol., 10:279–285, 2000.

[7] M. Salomo, K. Kegler, C. Gutsche, M. Struhalla, J. Reinmuth, W. Skokow,U. Hahn, and F. Kremer. The elastic properties of single double-strandedDNA chains of different lengths as measured with optical tweezers. Col-loid and Polymer Science, 284(11):1325–1331, August 2006. doi: 10.1007/s00396-006-1517-4.

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Page 29: Elastic helices Comprehensive exam report

[8] Jeff Gore, Zev Bryant, Marcelo Nollmann, Mai U. Le, Nicholas R. Coz-zarelli, and Carlos Bustamante. DNA overwinds when stretched. Nature,442(7104):836–839, August 2006. doi: 10.1038/nature04974.

[9] Brendan Maher. Physics in the cell: Spring theory. Nature, 448:984–986,August 2007. doi: 10.1038/448984a.

[10] Alain Goriely and John H. Maddocks. Overwinding DNA: All you need isLove. Unpublished, 2007.

[11] Nicolas Clauvelin, Basile Audoly, and Sebastien Neukirch. Analyticalresults for the plectonemic response of supercoiled DNA. Journal ofComputer-Aided Materials Design, 14(Supplement 1):95–101, December2007. doi: 10.1007/s10820-007-9072-y.

[12] T. McMillen and A. Goriely. Tendril perversion in intrinsically curved rods.J. Nonlinear Sci., 12:241–281, 2002. doi: 10.1007/s00332-002-0493-1.

[13] Alain Goriely and Patrick Shipman. Dynamics of helical strips. Phys.Rev. E (3), 61(4, part B):4508–4517, 2000. ISSN 1539-3755. doi: 10.1103/PhysRevE.61.4508.

[14] Alain Goriely and Michael Tabor. The nonlinear dynamics of filaments.Nonlinear Dynam., 21(1):101–133, January 2000. ISSN 0924-090X. doi:nonlineardynamicsoffilaments. The theme of solitary waves and localizationphenomena in elastic structures.

[15] Alain Goriely and Michael Tabor. Nonlinear dynamics of filaments. II.Nonlinear analysis. Phys. D, 105(1-3):45–61, 1997. ISSN 0167-2789.

[16] Stuart S. Antman. Nonlinear Problems of Elasticity. Springer, 2 edition,2004.

[17] Sanjay R. Sanghani, Krystyna Zakrzewska, Stephen C. Harvey, andRichard Lavery. Molecular modelling of (A4T4NN)n and (T4A4NN)n: se-quence elements responsible for curvature. Nucleic Acids Research, 24(9):1632–1638, March 1996.

[18] Erwin Kreyszig. Differential Geometry. Dover, 1991.

[19] W. Helfrich. Helical bilayer structures due to spontaneous torsion of theedges. J. Chem. Phys., 85(2):1085–1087, July 1986. orals.

[20] Nadia Chouaieb and John H. Maddocks. Kirchhoff’s problem of helicalequilibria of uniform rods. Journal of Elasticity, 77:221–247, 2004. doi:10.1007/s10659-005-0931-z.

[21] Augustus Edward Hough Love. A Treatise on the Mathematical Theory ofElasticity. Cambridge University Press, 2 edition, 1906.

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Appendices

A Geometry of curves

A.1 Parametric form of a curve

The simplest representation of a curve is the parametric one: x = x(σ), whereσ ∈ R is an arbitrary parameter. In the inextensible case, a practical andmeaningful choice for the parameter, which we will be using exclusively, is thearc length s:

s =

∫ σ

0

√dx

dσ(ζ) · dx

dσ(ζ) dζ , (81)

We denote the derivatives with respect to the parameter s with primes:

( · )′ :=d

ds( · ) .

A.2 Tangent vector

With the arc length as the parameter, the tangent vector τ = τ (s),

τ :=dx

ds= x′ (82)

is a unit vector. This can be easily verified:

|τ |2 = τ · τ =dx

ds· dx

ds= 1 ,

sincedx ·dx = ds2 .

The tangent vector points in the direction of increasing values of the param-eter s, and thus depends on the parametrization.

A.3 Principal normal, curvature

Since the tangent vector τ (s) is a unit vector for all values of the parameter s,its change can only be in the perpendicular direction:

τ · τ = 1 =⇒ τ · τ ′ = 0

=⇒ τ ′ = 0 or τ ′ ⊥ τ .

The first case gives τ = const, i.e. a straight line. Otherwise, we define thecurvature κ = κ(s) as the magnitude of the derivative of the tangent vector:

κ := |τ ′| = |x′′| . (83)

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The principal normal vector23 (short: normal) ν = ν(s) is defined asthe unit vector in the direction of τ ′:

ν :=1

κτ ′ =

τ ′

|τ ′|=

x′′

|x′′|. (84)

Unlike the tangent vector, the normal vector is independent of parametrization,and is directed towards the local center of curvature24.

The osculating plane is the plane spanned by the tangent and normalvectors.

A.4 Binormal, Frenet basis

The binormal vector β = β(s) is normal to the osculating plane, and isdefined as

β := τ × ν . (85)

Since τ and ν are normalized, so is β.The triad (τ ,ν,β) composed of the tangent, principal normal, and binor-

mal vector forms a right orthonormal system and is called the Frenet basis(Figure 3).

A.5 Torsion

If β′ = 0, then β = const, and the osculating plane is the same for all values ofthe parameter s, i.e. the curve lies in a plane. It is reasonable to expect that β′

measures the magnitude and sense of deviation of the curve from the osculatingplane.

Since β is normalized, its derivative (if non-zero) must be orthogonal to it:

β ·β = 1 =⇒ β ·β′ = 0

=⇒ β′ = 0 or β′ ⊥ β .

On the other hand, since β ⊥ τ by definition, and τ ′ = κν,

β · τ = 0 =⇒ β′ · τ + β · τ ′ = 0

=⇒ β′ · τ = −β · τ ′ = −κβ ·ν = 0

=⇒ β′ ⊥ τ .

Now, since β′ is perpendicular both to β and τ , it must be in the direction ofthe principal normal ν. The coefficient of proportionality between β′ and ν iscalled the torsion τ = τ(s):

β′ = −τν . (86)

23 Also called the curvature vector [19].24 The point that is the center of the circle of radius ρ which locally approximates the curve.

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ΤΝ

Β

Figure 3: Frenet basis.

The torsion is therefore the magnitude of β′ with an additional sign.25 Asthe vector ν is independent of the parametrization, this sign has a geometricalsignificance, and it was chosen26 so that a right helix has a positive torsion.

Multiplying the above defining equation by ν, we have

τ = −β′ ·ν . (87)

This expression can be rewritten entirely in terms of the tangent vector τ (s):

τ = −ν ·β′ = −ν · d

ds(τ × ν) = −ν · (τ ′ × ν + τ × ν′)

= −ν · (τ × ν′) = −τ′

κ·

(τ ×

(τ ′

κ

)′)=

1

κ2τ · (τ ′ × τ ′′) ,

and even further in terms of x:

τ =1

κ2x′ · (x′′ × x′′′) =

x′ · (x′′ × x′′′)

x′′ ·x′′. (88)

25 Unlike the curvature κ which is always positive, the torsion τ can be positive or negative.26 The negative sign in the definition!

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From this expression it is clear that τ is invariant with respect to the inver-sion s → −s. Equation (88) also implies that a curve with continous torsion τhas a C3 parametric representation x(s).

Taylor expansion. Expanding x(s) in a Taylor series, the curvature and thetorsion appear as correcting factors of second and third order, respectively:

x(s+ h) = x(s) + h x′(s)︸ ︷︷ ︸=τ (s)

+h2

2!x′′(s)︸ ︷︷ ︸

=κ(s)ν(s)

+h3

3!x′′′(s)︸ ︷︷ ︸∝τ(s)β(s)

+o(h4) .

Radius of curvature, radius of torsion. The radius of curvature andthe radius of torsion are defined as the reciprocicals of the curvature andtorsion, respectively:

ρ :=1

κ, σ :=

1

τ. (89)

A.6 Frenet-Serret equations

The Frenet-Serret equations relate the derivatives τ ′, ν′, β′ with the basis vec-tors τ , ν, β. The equations for τ ′ and β′ are already known from the definitionsof curvature and torsion. The missing equation is the one expressing ν′ in thebasis (τ ,ν,β). Since ν preserves unit norm, ν′ cannot have a component in thedirection of ν, therefore

ν′ = aτ + cβ .

The coefficients a = ν′ · τ and c = ν′ ·β are found from:

ν · τ = 0 =⇒ ν′ · τ + ν · τ ′ = 0

=⇒ ν′ · τ = −ν · τ ′ = −κν ·ν = −κ

and

ν ·β = 0 =⇒ ν′ ·β + ν ·β′ = 0

=⇒ ν′ ·β = −ν ·β′ = τ .

Thereforeν′ = −κτ + τβ . (90)

The Frenet-Serret equations are thus

ν′ = τβ − κτ (91a)

β′ = −τν (91b)

τ ′ = κν , (91c)

or in matrix form: νβτ

′ =

0 τ −κ−τ 0 0κ 0 0

νβτ

. (92)

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Note that this is a vector/scalar relation: the multiplication between the sqarematrix and the column matrix is the hybrid (scalar · vector) multiplication.

When the Frenet basis vectors are written out as column-vectors, the relationis written as [

ν β τ]′

=[ν β τ

] 0 −τ κτ 0 0−κ 0 0

. (93)

(All matrices here are 3× 3 with over the field of scalars.)

Darboux vector Denoting

Ω = κβ + ττ , (94)

the Frenet-Serret equations can also be written as

ν′ = Ω× ν (95a)

β′ = Ω× β (95b)

τ ′ = Ω× τ (95c)

The vector Ω is called the Darboux vector. The component of Ω in thetangential direction is the torsion τ , and its binormal component is the cur-vature κ.27 Note that in (94), κ and τ are scalar quantities independent ofparametrization, while β and τ are vectors that flip direction under the inversions 7→ −s. Therefore, the Darboux vector is also dependent on parametrization:Darboux vectors Ω, Ω corresponding to opposite parametrizations are relatedby Ω = −Ω.

B Geometry of helices

A helix is a curve of constant curvature and torsion. Theorem 1 thus impliesthat the entire (infinite) helix curve is specified (up to rigid body motion) by twoconstant scalars, κ and τ . Alternatively, various other pairs of scalar parametersmay be used:

• radius of curvature ρ and radius of torsion σ (89)

• helix radius28 r and pitch29 2πp

• radius r and pitch angle α := arcsin(e3 ·d3) (angle between the tangentand the plane perpendicular to the helix axis)

27 Note that by construction the Darboux vector has no normal component.28 The helix radius r is not to be confused with the (generally larger) radius of curvature

ρ.29 We define p as the “reduced pitch” to abbreviate the formulae. The actual pitch—distance

per full helical turn along the helix axis—is 2πp.

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• twist magnitude u :=√κ2 + τ2 and helix angle θ := arccos(e3 ·d3)

(angle between the helix axis and the tangent, complement of the pitchangle: θ = π

2 − α)

These pairs of parameters are interconnected as follows.

κ, τ ↔ u, θ:

u =√κ2 + τ2 , θ = arctan

κ

τ(96)

κ = u sin θ , τ = u cos θ (97)

r, p↔ κ, τ :

r =κ

κ2 + τ2, p =

τ

κ2 + τ2(98)

κ =r

r2 + p2, τ =

p

r2 + p2(99)

r, p↔ ρ, σ:

r =ρ

1 + ρ2

σ2

, p =σ

1 + σ2

ρ2

(100)

ρ =r2 + p2

r, σ =

r2 + p2

p(101)

r, p↔ u, θ:

u =1√

r2 + p2, θ = arctan

r

p(102)

r =1

usin θ , p =

1

ucos θ (103)

κ, τ ↔ r, α:

κ =cos2 α

r, τ =

sinα cosα

r(104)

r =κ

κ2 + τ2, α = arctan

τ

κ(105)

Parametric form of the helix

r(s) = r cos

(s√

r2 + p2

)e1 + r sin

(s√

r2 + p2

)e2 + p

(s√

r2 + p2

)e3 (106)

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Helix height For a finite helix, the height of a helix30 of contour length Lis found by multiplying the number of turns N = L

2π√r2+p2

by the pitch 2πp:

h = N2πp = Lp√

r2 + p2= L cos θ = L sinα (107)

30By height we mean the length of the projection of the helix onto the helix axis.

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Index

arc length, 30asymmetry coefficient, 16

bend-twist coupling coefficients, 13bending stiffness, 13binormal vector, 31

centerline, 7components of curvature, 9constitutive relation, 11

hyperelastic rod, 12linear, 13

curvature, 30vector, 31

Darboux vector, 8, 34director basis, 8

excess twist, 8

first integrals, 19–21, 22force equation, 11Frenet basis, 31Frenet strip, 9Frenet-Serret equations, 33

generalized, 9Fundamental theorem, 7

for strips, 9

helix, 23angle, 35height, 36isotropic, 23radius, 34uniform, 23

hyperelastic, 12

intrinsiccoordinates, 11curvature, 11register, 11torsion, 11twist vector, 11

Kirchhoff equations

static, 17in director basis, 18–19

time dependent, 14

local coordinates, 8

moment equation, 11

normal vector, 31

osculating plane, 31

pitch, 34angle, 34

principal bending stiffnesses, 13principal moments of inertia, 11principal normal vector, 31

radiusof curvature, 33of torsion, 33

register, 7, 8rod, 10

isotropic, 12

shear modulus, 13spin

equations, 10vector, 10

strain-energy density, 12isotropic, 12

strip, 8uniform, 9

tangent vector, 30tension, 11top analogy, 17torsion, 31

geometric, 8material, 8mathematical, 8physical, 8

torsional stiffness, 13twist

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density, 8equations, 9magnitude, 35physical, 8vector, 8

twist-to-bend stiffness ratio, 16

Young modulus, 13

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