Towards a classiflcation of Euler-Kirchhofi fllaments

Towards a classification of Euler-Kirchhoff filaments

Michel Nizette∗ and Alain Goriely††University of Arizona, Department of Mathematics and

Program in Applied Mathematics, Bldg #89, Tucson, AZ85721and ∗Universite Libre de Bruxelles

Departement de Physique Statistique CP231e-mail: mnizette@ ulb.ac.be; goriely@ math.arizona.edu

Abstract

Euler-Kirchhoff filaments are solutions of the static Kirchhoff equations for elastic rods withcircular cross-sections. These equations are known to be formally equivalent to the Euler equa-tions for spinning tops. This equivalence is used to provide a classification of the different shapesa filament can assume. Explicit formulas for the different possible configurations and specificresults for interesting particular cases are given. In particular, conditions for which the filamenthas points of self-intersection, self-tangency, vanishing curvature or when it is closed or localizedin space are provided. The average properties of generic filaments are also studied. They areshown to be equivalent to helical filaments on long length scales.

KEYWORDS: Rods, Spinning tops, Kirchhoff equations, Elliptic functions.

PACS: 46.30.Cn, 03.20.+i

I Introduction

The study of elastic deformations in rods has a long tradition in mathematics, physics and engineer-ing, dating back to Euler and Lagrange. Engineers have been confronted to the problem of coilingin sub-oceanic cables and have tried to understand the process of loop formation in twisted wires[1, 2, 3]. In chemistry and biology, increasing interest is taken in the elastic character of filamentarystructures such as polymers [4, 5, 6, 7] (such as DNA molecules [8, 9, 10, 11, 12, 13, 14, 15]) andbacterial fibers [16, 17, 18], for which the macroscopic theory of rods provides an idealized model.Long, twisted structures also play an important role in hydrodynamic models [19] such as scrollwave propagation [20], vortex tube motion [21], or sun spots formation and solar corona heating[22, 23].

The Kirchhoff model (1859) [24] provides the basic framework for the theory of elastic filaments.A remarkable feature of this model, known as the Kirchhoff kinetic analogy, is that the equationsgoverning the static phenomena are formally equivalent to the Euler equations describing the motionof a rigid body with a fixed point under an external force field. The statics of rods is thus intimatelyconnected to the dynamics of spinning tops, a problem to which innumerable work has been devoted.For instance, the most studied case where the filament has a circular cross-section is shown tocorrespond to a top having an axis of revolution, in which case the equations are fully integrable.There is a rich mathematical literature on the statics of rods (see for instance [26]). In the particularcase of circular cross-sections and linear elasticity, various researchers have considered particularequilibrium filament shapes (helices [25, 27], rings [28, 3], localizing buckling modes [1, 29], solutions

1

having points with vanishing curvature [30], supercoiled helices [31], see also [32] for an Hamiltonianformulation and [33] for a group theory approach). Recently, departing from the traditional Eulerangles approach, Shi and Hearst (1994) have obtained a closed form of the general solution ofthe static Kirchhoff equations for circular cross-sections [34]. Despite this achievement, it remainshighly nontrivial to obtain a global picture of all possible static filament shapes. A step towards ageneral geometric classification of the equilibrium solutions is provided here by considering in moredetail the analogy between filaments and spinning tops. In Section III, we rederive Shi and Hearst’sgeneral solution, keeping an explicit dependence in the constants of the motion of the spinning top,and we discuss in detail various filament configurations on the basis of the corresponding orbits ofthe top, recovering the aforementioned shapes as particular cases.

In this paper, we provide new results on the different configurations of Kirchhoff filaments in 2Dand 3D. We give explicit formulas for the centerline coordinate and find conditions for self-tangencyand self-intersection of planar filaments, conditions for vanishing curvature and average behavior oflong filaments. Explicit formulas for periodic or localized (homoclinic) filaments in space are alsogiven

II The Kirchhoff Model

We first introduce the Kirchhoff model. It accounts for the dynamics of a thin elastic filamentsubject to internal stresses and boundary constraints. A filament is a unidimensional piece ofelastic material which can be mathematically modelized by a curve in space, together with extrainformation about its twist, that is, how longitudinal material lines on the edge of the filamentwind around it. This curve-plus-twist concept is formalized in the notion of ribbon discussed inSection II.1 as a preliminary.

II.1 Space Curves and Ribbons

We define a dynamical space curve R(s, t) as a smooth function mapping R2 into the physical spaceR3, and taking as variables the arc length s and the time t. For every s and t we define the Frenetbasis (n,b, t) to be the normal, binormal and tangent vectors to the curves. These vectors formthe Frenet basis. The tangent vector is a unit vector given by: t = ∂R

∂s . The curvature κ of thecurve at the point s is then given by:

κ =∣∣∣∣∂t∂s

∣∣∣∣ . (1)

At points where the curvature does not vanish, the normal vector is defined by:

∂t∂s

= κ n. (2)

The third unit vector b is:

b = t× n. (3)

Therefore, the Frenet basis is a right-handed orthonormal basis on the space curve R. As aconsequence, one has:

∂n∂s

= τ b−κ t. (4)

2

This relation defines the torsion τ , which measures the amount of rotation of the Frenet triadaround the tangent t as the arc length increases. Finally the derivative of the binormal b is givenin terms of the normal and tangent vector:

∂b∂s

= −τ n. (5)

The coupled equations (2), (4) and (5) are the Frenet-Serret equations. If the curvature κ and thetorsion τ are known for all s, the Frenet triad (n,b, t) can be obtained as the unique solution ofthe Frenet-Serret equations. It is then possible to reconstruct the space curve R by integrating thetangent vector t.

A ribbon is a space curve R(s, t) supplied with a smooth unit vector field d2(s, t) orthogonalto the curve. Let d3 = t be the unit vector tangent. A third unit vector field d1 = d2 × d3 isintroduced so that the triad (d1,d2,d3) forms a right-handed orthonormal basis. This basis is ageneralization of the Frenet triad (n,b, t).

The components of the derivatives of the local basis vectors d1, d2 and d3 with respect to arclength s and the time t expressed in the local basis form the twist vector κκκ(s, t) = κ1d1+κ2d2+κ3d3

and the spin vector ωωω(s, t) = ω1d1 + ω2d2 + ω3d3, defined as follows:

∂di∂s

= κκκ× di, (6.a)

∂di∂t

= ωωω × di. (6.b)

The equations (6.a) constitute the generalization of the Frenet-Serret equations for the ribbon.These equations can also be expressed in terms of the twist matrix K(s, t) and the spin matrixW(s, t), which we define as follows:

K =

0 −κ3 κ2

κ3 0 −κ1

−κ2 κ1 0

, W =

0 −ω3 ω2

ω3 0 −ω1

−ω2 ω1 0

. (7)

The Frenet-Serret equations then read:(∂d1

∂s

∂d2

∂s

∂d3

∂s

)=(

d1 d2 d3

)K, (8)

(∂d1

∂t

∂d2

∂t

∂d3

∂t

)=(

d1 d2 d3

)W. (9)

II.2 The Kirchhoff Equations

II.2.1 Main Assumptions and Derivation of the Kirchhoff Equations

A thin filament, or rod, can be modelized by a ribbon constituted of a space curve R joining theloci of the centroids of the cross-sections, together with a vector field d2 attached to the filamentmaterial. The space curve R is referred to as the centerline of the rod.

The Kirchhoff equations describe the dynamical evolution of the filament under the effect ofinternal elastic stresses and boundary constraints, in the absence of external force fields such asgravity. Also, only local interactions, between adjacent cross-sections, are considered, ignoring thepossibility for two remote segments of the filament to intersect with each other. Let F(s) andM(s) be the total force and total moment exerted on the back side of a cylinder C(s) by the

3

cylinder C(s+ ds) whose cross-section shape is defined by S, the set of all the values of the couple(x1, x2) corresponding to a material point inside a given cross-section. The conservation of linearand angular momentum yields [35]:

∂F∂s

= ρA∂2R∂t2

, (10.a)

∂M∂s

+ d3 × F = ρ

(I2d1 ×

∂2d1

∂t2+ I1d2 ×

∂2d2

∂t2

), (10.b)

where A denotes the area of the cross-section and the quantities I1 and I2 are the principal momentsof inertia of the cross-section:

I1 =∫Sdx1dx2 x

22, I2 =

∫Sdx1dx2 x

21. (11)

The equations (10) are closed by using the constitutive relation of linear elasticity relating thetorque M to the twist vector κκκ:

M = EI1κ1d1 + EI2κ2d2 + µJκ3d3, (12)

where E is Young’s modulus, µ is the shear modulus and J is a function of the shape S of thecross-section. In the particular case of a circular cross-section, one has:

I1 = I2 =J

2=πR4

4, (13)

where R is the radius of the cross-section. The combinations EI1 and EI2 are called the principalbending stiffnesses of the rod and measure how strong an applied torque must be in order to bendit, whereas the combination µJ is called the torsional stiffness and measures how large an appliedtorsional moment must be in order to twist the rod.

The coupled equations (10) and (12) constitute the dynamical Kirchhoff equations. These arethree vectorial equations involving the local basis (d1,d2,d3) and its derivatives, the tension F andthe torque M, which add up to nine degrees of freedom, hence the system is closed. In the staticcase (if the time dependencies dropped), the term in R vanishes from equation (10.a):

F′ = 0, (14.a)M′ + d3 × F = 0, (14.b)M = EI1κ1d1 + EI2κ2d2 + µJκ3d3. (14.c)

II.2.2 The Kirchhoff Equations in Scaled Form

In order to restrict the number of independent constants in (14),we scale the variables, by choosingcombination of the length [L], time [T ] and mass [M ] units in the following way:

[M ] = ρ√AI1,

[M ] [L]3

[T ]2= EI1, (15)

with:

a =I2

I1, b =

µJ

EI1=

J

2I1(1 + σ), (16)

4

where σ denotes the Poisson ratio. The constant a measures the asymmetry of the cross-section.Our convention is to orient the vector fields d1 and d2 such that I1 and I2 are, respectively, thelarger and smaller bending stiffnesses. In this case, we have:

0 < a ≤ 1, (17)

the value 1 being reached in the symmetric case where the moments of inertia are identical. Theconstant b is the scaled torsional stiffness. It involves the constant 1

1+σ , which ranges from 23 ,

corresponding to incompressible media (if the volume is unchanged as the material is stretched),to 1, corresponding to hyperelasticity (if there is no striction as the material is stretched). In theparticular case of a circular cross-section, one has

b =1

1 + σ∈[

23, 1]

. (18)

The scaled system reads now:

F′ = 0, (19.a)M′ + d3 × F = 0, (19.b)M = κ1d1 + aκ2d2 + bκ3d3. (19.c)

The quantities involved still have a dimension. From (19), we see that the scaled moment M has thedimension of the inverse of a length, and that the scaled tension F has the dimension of the inverseof a squared length, hence every variable involved is a length to a power. However the systemcannot be further simplified by choosing the length unit [L], because the remaining constants aand b are already dimensionless. It is still possible to choose a convenient length scale for a givenproblem. For example, if we consider a finite rod, a natural choice for the length unit will be thelength of the rod. In Section III, we choose the length unit [L] such that the norm of F has a givenvalue.

The fact that the length unit [L] is undetermined has yet another implication. Consider-ing the static system (14) or (19), we see that every known solution actually determines a one-parameter family of solutions. More precisely, if {F(s),M(s),κκκ(s)} is a solution of the system, then{λ−2F(λs), λ−1M(λs), λ−1κκκ(λs)

}is another solution of the system for every real non vanishing λ.

That is, the system is scale-invariant. Furthermore, if such a transformation is performed on thesolution together with a rescaling of the length unit [L] by a factor λ−1, the solution remains un-changed, although the rod thickness is modified by a factor λ−1. Hence, the statics of a filament,in the limit of the Kirchhoff model, is independent on the rod thickness.

II.2.3 Integrability of the Static Kirchhoff Equations

The static Kirchhoff equations (19) are formally identical to the Euler equations describing themotion of a rigid body with a fixed point under gravity, in the particular case where the axis joiningthe fixed point to the center of mass lies along a principal direction of inertia (the correspondencebetween rigid body variables and rod variables is shown on Table 1). Therefore, they are fullyintegrable in two cases of interest, namely, the Lagrange and Kowalevskaya cases1.

1Actually, the Euler equations are integrable in three cases. The third one corresponds to the motion of a rigidbody with arbitrary shape in free fall and is known as the Euler case. However, one cannot speak of full integrabilityin the context of filaments, since this case corresponds to a vanishing tension vector F and represents only particularboundary conditions, hence a subset of all possible configurations which can be adopted by a filament for given valuesof the material parameters a and b.

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Symbol Meaning for rods Meaning for rigid bodiesd3 Unit tangent vector Unit vector joining the fixed point

to the center of mass(d1,d2,d3) Basis attached to the rod Basis attached to the solid bodys Arc length TimeF Tension Force equal and opposite to gravityM Moment Angular momentumκκκ Twist vector Angular velocity vectorEI1, EI2 Principal bending stiffnesses Principal moments of inertia in

directions orthogonal to d3

µJ Torsional stiffness Principal moment of inertia along d3

a Bending stiffnesses ratio Ratio of the moments of inertia indirections orthogonal to d3

b Scaled torsional stiffness Scaled moment of inertia along d3

Table 1: Analogy between rigid bodies and static filaments.

In the Lagrange case, the rigid body has identical moments of inertia along every directionperpendicular to the axis joining the fixed point to the center of mass, as for a symmetric top.The corresponding rods have identical bending stiffnesses in every direction, so that a = 1. Thisis realized, for instance, in the very common situation where the filament cross-section is circular,showing the importance of the Lagrange case for filaments. Much work has been devoted to it(Euler classified the planar solutions of the equations, see Figure 1), and recently, Shi and Hearst(1994) [34] have obtained a closed form for the general solution. Their results are rederived withspecial emphasis on the correspondence between rod and rigid body variables and developed indetail in Section III.

In the Kowalevskaya case, there exists an axis D, originating from the fixed point, along whichthe moment of inertia is half as large as the moments of inertia along directions perpendicular toD. In addition, the center of mass lies in the plane perpendicular to D. In the context of rods,this corresponds to asymmetric cross-sections with a = 1

2 and b = 1. However, it turns out thatthe Kowalevskaya case, having a very high torsional stiffness, lies far beyond the region covered bythe possible physical values of the parameters. For instance, an elliptic cross-section with a = 1

2and b = 1 would have a Poisson ratio σ = −1

3 . This value describes a material which inflatestransversally as it is stretched in length, and although this is not precluded in theory, it is notrealized in practice.

III Symmetric Rods and Lagrange Tops

III.1 Generalities

The static Kirchhoff equations, written in terms of an appropriate set of variables, are formallyequivalent to the Euler equations describing the dynamics of a heavy top. To every possible motionof the top, one can associate a particular static solution of the Kirchhoff equations. Here wefocus exclusively on the analogy between tops and static rods, in one of the three cases where theequations are fully integrable: the Lagrange case, where the top has two identical principal momentsof inertia. This condition is satisfied if it has a symmetry of revolution. The corresponding filamenthas identical bending stiffnesses in all directions, that is, we must set a = 1 in the scaled form ofthe Kirchhoff equations (19).

The solutions of the Euler equations in the Lagrange case are well known and can be written as

6

Figure 1: Euler’s drawings of planar filaments.

7

combinations of elliptic functions [36, 37, 38]. In order to obtain the centerline R of the correspond-ing static filament, we must identify the tangent vector d3 to a unit vector lying along the axis ofrevolution of the top. The centerline is then obtained by integrating d3 over the arc length s which,in the context of tops, corresponds to time. An obvious difficulty arises: in order to describe thebehavior of real filaments under given external constraints, the space curve R must be available in aform that allows for boundary conditions at two distinct points. That is, it is essential to explicitlycarry the integration of the tangent vector. Although it is not obvious that this integration can beperformed, Shi and Hearst (1994) [34] recently obtained expressions for the centerline R in cylin-dric coordinates in a closed analytic form involving elliptic functions. Despite this achievement, theproblem is still partially unsolved. Indeed, explicit forms of the solutions are by no means sufficientto get a global insight on the large variety of possible filament shapes. Furthermore, the detailedanalysis of the correspondence between spinning Lagrange tops and static symmetric rods providesa way of establishing an exhaustive geometric classification of the solutions. The aim of this paperis to provide a first step towards such a classification.

Shi & Hearst obtained their solutions by first solving the Kirchhoff equations for the curvatureand torsion and then solving the Frenet-Serret equations to obtain the centerline. The resultingexpressions depend on integration constants which do not have a clear meaning. Here, we departfrom their approach and work consistently with variables and integration constants relevant to thetop. Namely, in Section III.3 , we express the Kirchhoff-Euler equations in terms of the Eulerangles. Then we compute the centerline R of the filament as a function of the constants of themotion for the spinning top. Once the expressions for the centerline have been obtained, we studyvarious classes of shapes of the rod and their correspondence with the motion of the top. Thisanalysis can be thought of an extension of Euler’s work, who classified planar shapes of filaments.

In this Section, we show illustrations of the spinning top orbits together with the correspondingfilament shapes. A top orbit is displayed as a curve on the unit sphere which represents theextremity of the unit vector d3. The filament shapes are represented with a circular cross-section,hence b must be set to a value lying between 2

3 and 1. We have chosen b = 34 . The radius of the

cross-section and the zoom factor vary from one figure to another and are chosen for clarity (asseen in Section II, the radius is arbitrary in the static case).

In the remaining of this section, we write down the static Kirchhoff equations (19) in the casea = 1, and identify a set of three first integrals necessary to guarantee the integrability of thesystem. With a = 1, the system (19) reads:

F′ = 0, (20.a)M′ + d3 × F = 0, (20.b)M = κ1d1 + κ2d2 + bκ3d3. (20.c)

The first equation expresses the fact that the tension F is constant. We choose it oriented alongthe third vector of the fixed basis:

F = FeZ. (21)

In terms of spinning tops, the tension corresponds to the opposite of the top weight, −mg, whichdescribes an external force field. In the same spirit, we consider the tension as having a fixed value.As a consequence, we consider F as a parameter rather than a first integral. Inserting (21) into(20.b) and projecting along the fixed basis vector eZ, we have:

M′ · eZ = 0. (22)

The basis vector eZ being independent of s, we can extend the derivative in (22) to take effect overthe whole left-hand side, leading to:

M ′Z = 0, (23)

8

where MZ denotes the component of the moment along eZ. It is a first integral that represents thevertical component of the angular momentum of a spinning top. By projecting (20.b) along d3, weobtain:

M′ · d3 = (M · d3)′ −M · d′3 = 0. (24)

Using the fact that d′3 = κκκ×d3 = M×d3, we see that the second term of (24) vanishes identically,leading to:

M ′3 = 0. (25)

That is, the torsional moment M3 is another first integral. From (20.c), we see that it correspondsto a constant twist density. For the spinning top, M3 represents the component of the angularmomentum along the axis of revolution of the top.

Finally, taking the dot product of both sides of (20.b) with κκκ, we have:

M′ · κκκ+ (d3 × F) · κκκ = M′ · κκκ+ F · (κκκ× d3) = 0. (26)

Using (20.c) and the expressions (6.a) of the derivatives of the local basis (d1,d2,d3) in terms ofκκκ, (26) reduces to:

κ1κ′1 + κ2κ

′2 + F · d3

′ =(

12M · κκκ+F · d3

)′= 0, (27)

which provides the last first integral:

12M · κκκ+ F · d3 = H. (28)

This constant quantity H is the total elastic-plus-strain energy density of the filament and corre-sponds to the total kinetic-plus-potential mechanical energy of the spinning top. We now proceedto analyze different solutions of the system (20).

III.2 Helical Filaments

III.2.1 The General Helical Solution

Helical solutions have constant Frenet curvature κ and torsion τ . Taking into account the fact thatthe twist density κ3 is also constant, we introduce the following definitions.

• A Frenet helix is a helix with pure torsion, that is, κ3 = τ .

• An overtwisted helix is a helix such that (κ3 − τ) has the same sign as τ .

• An undertwisted helix is a helix such that (κ3 − τ) and τ have opposite signs.

These three types of helices are represented on Figures 3, 2 and 4 respectively. The undertwistedand overtwisted helices can be distinguished by the relation between the handedness of the helixitself and the handedness of the apparent twist pattern on the helix. In the case of an overtwistedhelix, both hands are identical, whereas an undertwisted helix has opposite hands.

It is convenient to treat the cases of helical, circular (ring-like) and straight solutions of theKirchhoff equations independently. The rings and the helical rods are easily identified as corre-sponding to spinning tops with the extremity describing a circle (centered at the fixed point inthe case of rings), while the straight solutions correspond to the cases where the extremity of the

9

N

Figure 2: An overtwisted helix (κ = 3, τ = 1, κ3 = 6).

top is at rest (the so-called sleeping tops): pointing upwards in the case of positive tension F3 anddownwards in the case of negative tension. Illustrations of these top orbits are given on Figures 2to 7. For the sake of simplicity, helical, circular and straight filaments will be referred to as helicalfilaments.

We do not introduce Euler angles for the helical solutions. Instead, we introduce a fixed Frenetcurvature κ and a fixed torsion τ into the expressions for the twist vector:

κ1 = κ sin [(κ3 − τ)(s− s0)] , (29.a)κ2 = κ cos [(κ3 − τ)(s− s0)] . (29.b)

Substituting this into (20), we obtain one single nontrivial vector condition:

F = (bκ3 − τ) [κκκ− (κ3 − τ) d3], (30)

which simply gives the tension in the local basis as a constant combination of the other quantities.Hence, there exist helical solutions with arbitrary curvature κ, torsion τ and twist density κ3. Thisis unique to the case of symmetric rods. Indeed, in the regular case a 6= 1 the only possible helicesare Frenet helices with either κ1 = 0 or κ2 = 0 [39].

Three more facts can be noticed from (30). First, in the case of rings (τ = 0), the tension hasno longitudinal component (F · d3 = 0). Second, in the case of a non circular filament (τ 6= 0), avanishing longitudinal tension implies bκ3 = τ , hence the tension vector F itself vanishes. Finally,in the case of circular cross-section, one has b ≤ 1. This means that the Frenet helices (κ3 = τ)have a negative longitudinal tension, and that the helices with null tension are overtwisted.

For straight solutions (κ = 0), (30) does not hold and is replaced by:

F = F3d3, (31)

where F3 is constant and arbitrary. Hence, there exist straight solutions with arbitrary longitudinaltension and twist density. The torsion has no meaning for straight rods. Figures 2 to 7 showrepresentations of helical, ring-like and straight solutions.

10

N

Figure 3: A Frenet helix (κ = 3, τ = −1, κ3 = −1).

N

Figure 4: An undertwisted helix (κ = 3, τ = −1, κ3 = 4).

N

Figure 5: A twisted ring (κ = 1, τ = 0, κ3 = 3).

11

N

Figure 6: A straight rod subject to extensive tension (κ = 0, κ3 = 1, F3 = 1).

S

Figure 7: A straight rod subject to compressive tension (κ = 0, κ3 = 1, F3 = −1).

12

III.2.2 The General Solution with Null Tension.

Substituting F = 0 into (20.b), we see that the moment M is constant. Differentiating (20.c) withrespect to s and projecting the resulting equation successively along d1, d2 and d3, we obtain:

κ′1 + (b− 1)κ3κ2 = 0, (32.a)κ′2 + (1− b)κ3κ1 = 0, (32.b)0 = 0, (32.c)

which, taking into account the fact that κ3 is constant, can be integrated:

κ1 = κ sin [(1− b)κ3(s− s0)] , (33.a)κ2 = κ cos [(1− b)κ3(s− s0)] , (33.b)

where κ and s0 are integration constants. We see that (33) assumes the form (29), with τ = bκ3.In other words, (33) is a helix.

We conclude that all the solutions with vanishing tension are helices. In the following sections,we consider non-helical filaments. Therefore, from now on, we assume F 6= 0. This gives a precisesense to the vertical unit vector eZ.

III.3 General Solution for the Local Basis

III.3.1 Equations for the Euler Angles

We now introduce the Euler angles (ϕ, θ, ψ) for the local basis (d1,d2,d3). The angles ϕ, θ andψ denote, respectively, the precession, nutation and self-rotation angles (see Figure 8). In matrixform, the local basis is obtained from the fixed trihedron (eX, eY, eZ) as follows:(

d1 d2 d3

)=(

eX eY eZ

)E, (34)

where the general rotation matrix E reads:

E =

cosϕ cos θ cosψ − sinϕ sinψ − cosϕ cos θ sinψ − sinϕ cosψ cosϕ sin θsinϕ cos θ cosψ + cosϕ sinψ − sinϕ cos θ sinψ + cosϕ cosψ sinϕ sin θ

− sin θ cosψ sin θ sinψ cos θ

.

(35)

From (34) and (35) we extract the fixed vector eZ as a function of the local basis:

eZ = sin θ(sinψ d2 − cosψ d1) + cos θ d3. (36)

We now express the twist matrix K in terms of the rotation matrix E. Differentiating (34) yields:

∂

∂s

(d1 d2 d3

)=

(eX eY eZ

) ∂

∂sE

=(

d1 d2 d3

)E>E′, (37)

hence, from the definition (7) of the twist matrix, we see that:

K = E>E′. (38)

13

θ

θ

ϕϕ

ψ

ψ

d3

d1

d2

eX

eY

eZ

Figure 8: The Euler Angles.

Therefore, we can express the twist vector in terms of Euler angles:

κ1 = θ′ sinψ − ϕ′ sin θ cosψ, (39.a)κ2 = θ′ cosψ + ϕ′ sin θ sinψ, (39.b)κ3 = ϕ′ cos θ + ψ′. (39.c)

Using (36) and (39), we now write the three first integrals in terms of the Euler angles:

MZ = ϕ′[1 + (b− 1) cos2 θ] + bψ′ cos θ, (40.a)M3 = b(ϕ′ cos θ + ψ′), (40.b)

H =12

(θ′2 + ϕ′2 sin2 θ +

M23

b

)+ F cos θ. (40.c)

It is convenient to introduce the following auxiliary constants:

h =1F

(H − M2

3

2b

), (41.a)

h =1F

(H − M2

3

2b+M2

3 −M2Z

2

), (41.b)

which are well defined since F 6= 0. We also carry out the following change of variable in equations(40.a) to (40.c):

z = cos θ. (42)

Note that a solution with constant z corresponds to a helical rod excluded in this discussion, hencez is not constant. We can solve (40.a) to (40.c) for ϕ′, θ′ and z′ to obtain the final form of ourequations in the Euler angles:

ϕ′ =MZ −M3z

1− z2, (43.a)

ψ′ =(

1b− 1

)M3 +

M3 −MZz

1− z2, (43.b)

14

z′2 = 2F (h− z)(1− z2)− (MZ −M3z)2 (44.a)⇔ z′2 = 2F (h− z)(1− z2)− (M3 −MZz)2. (44.b)

III.3.2 A Better Choice of First Integrals

The differential equation (44.a) or (44.b) is an identity between z′2 and a cubic polynomial in z.In order to solve this equation, we must know the roots z1, z2, z3 of this polynomial. Rather thancomputing the roots in terms of the constants MZ, M3, h or h, the classical approach consists inconsidering the roots z1, z2 and z3 as three independent first integrals, and then expressing theconstants MZ, M3, h and h as functions of z1, z2 and z3. This is achieved by rewriting the cubicpolynomial in (44.a) and (44.b) as a product of three factors involving z1, z2 and z3:

z′2 = 2F (h− z)(1− z2)− (MZ −M3z)2 (45.a)= 2F (h− z)(1− z2)− (M3 −MZz)2 (45.b)= 2F (z − z1)(z2 − z)(z3 − z). (45.c)

By setting z = ±1 in (45.a) or (45.b), the right-hand side assumes a non positive value. Further-more, if we choose F to be positive (which we can always do by defining adequately the verticalunit vector eZ), we see that the right-hand side of (45.a) or (45.b) tends to +∞ as z → +∞. Thismeans that one of the roots (conventionally, z3) lies in the interval [1,+∞[. Finally, in order toobtain solutions with real values of θ, we require that z′2 be positive for z ranging in some intervalcontained in [−1, 1]. Hence the other two roots of the polynomial must be real and lie between −1and 1. Conventionally, we choose z1 ≤ z2. We conclude that the physical values of our independentconstants z1, z2 and z3 must satisfy:

−1 ≤ z1 ≤ z2 ≤ 1 ≤ z3. (46)

In the previous Section, we showed that in order to obtain the scaled form of the Kirchhoff equationsin the static case, it was not necessary to perform a complete scaling. Namely, at this level, everyvariable involved in the equations is a length raised to a given power, and we still have the freedomto choose an arbitrary length unit [L]. In the following, it is convenient to choose the length unitto be:

[L] =

√2

F (z3 − z1), (47)

which is equivalent to the substitution:

F =2

z3 − z1. (48)

The right-hand side of (48) is well-defined as long as z1 6= z3. The condition z1 = z3 impliesz1 = z2 = z3 = 1, which corresponds to a straight rod with z = 1, a case excluded from thisdiscussion.

Expressions for the constants MZ, M3, h and h in terms of the roots can be obtained byidentifying the coefficients of the powers of z in (45.a) to (45.c), or equivalently, by considering theequalities between the right-hand sides of (45.a) to (45.c) for three well-chosen values of z. Settingz = ±1 leads, together with (48), to:

−(MZ ∓M3)2 =4

z3 − z1(±1− z1)(z2 ∓ 1)(z3 ∓ 1). (49)

15

The equations (49) give MZ −M3 and MZ + M3 up to sign determination. A third equality oftype (45) corresponding to another value of z would lead to an equation involving h or h and couldnot provide additional knowledge on MZ or M3 for themselves. We conclude that given values ofz1, z2 and z3 do not yield unique values of MZ and M3. Instead they give these two constantswith a complete sign indetermination. By performing a mirror reflection in space, we can map anyfilament with MZ +M3 < 0 onto a filament with MZ +M3 > 0. Hence, we restrict our analysis tothe case MZ +M3 ≥ 0. Nevertheless, we still have to supply the values of z1, z2 and z3 with extrainformation, namely, the sign S of MZ −M3:

S = + or S = −. (50)

Now, if we set:

M+ =MZ +M3

2≥ 0, (51.a)

M− =|MZ −M3|

2, (51.b)

we have:

MZ = M+ + SM−, (52.a)M3 = M+ − SM−, (52.b)

with

M± =

√(1± z1)(1± z2)(z3 ± 1)

z3 − z1. (53)

The constants h and h in (44.a) and (44.b) are obtained from suitable combinations of equalitiesbetween the coefficients of z in (45.a) to (45.c):

h =12

[z1 + z2 + z3 − z1z2z3 + S(z3 − z1)M+M−] , (54.a)

h =12

[z1 + z2 + z3 − z1z2z3 − S(z3 − z1)M+M−] . (54.b)

III.3.3 General Solution for the Euler Angles

The solutions of equations (43.a)-(44) involve elliptic functions (see Appendix A). With a suitableorigin for the arc length s, the solution of equation (44) assumes the form:

z = z1 + (z2 − z1)sn2(s | k), (55)

where the modulus k ranges between 0 and 1 and is given by:

k2 =z2 − z1

z3 − z1. (56)

We can use the fact that z = cos θ to obtain θ as a function of s. In the generic case where z1 6= −1and z2 6= 1, the cosine never reaches its extreme values ±1, and there is a bijective correspondencebetween z and θ. In this case, we can take θ = arccos z. In the degenerate cases where z1 = −1 orz2 = 1, we must take care of the behavior of θ as cos θ reaches its limiting values. This is achievedby substituting z = cos θ in (44) and examining the behavior of θ′ around z = −1 or z = 1. Theresults are as follows:

16

• If z1 = −1 and z2 6= 1, θ′ has a non vanishing limit as z → −1, hence the sign of θ−π changesat z = −1.

• If z1 6= −1, z2 = 1 and z3 6= 1, θ′ has a non vanishing limit as z → 1, hence the sign of θchanges at z = 1.

• If z1 6= −1, z2 = 1 and z3 = 1, z never reaches its extreme value 1, hence we can takeθ = arccos z.

• If z1 = −1, z2 = 1 and z3 6= 1, θ′ has a non vanishing limit in both cases z → ±1, hence θ isa monotonous function of s. In this case, MZ = M3 = 0, that is, the top behaves like a planependulum. This corresponds to planar rods studied by Euler.

• If z1 = −1, z2 = 1 and z3 = 1, θ′ has a non vanishing limit as z → −1 and z never reaches itsextreme value 1, hence z assumes the value −1 only at the single point s = 0, and θ coversthe interval ]0, 2π[ crossing every value only once. This corresponds to the homoclinic orbitof the plane pendulum.

In order to obtain ϕ and ψ as functions of s, we must integrate (43.a) and (43.b), which, using(52.a) and (52.b), can be rewritten as:

ϕ′ = M+1

1 + z+ SM−

11− z , (57.a)

ψ′ =(

1b− 1

)M3 +M+

11 + z

− SM−1

1− z . (57.b)

Next, we define:

n± = ∓z2 − z1

1± z1. (58)

In the cases where n± have a non vanishing denominator, we can express (57.a) and (57.b) using(55) as:

ϕ′ =M+

1 + z1

11− n+sn2(s | k)

+ SM−

1− z1

11− n−sn2(s | k)

, (59.a)

ψ′ =(

1b− 1

)M3 +

M+

1 + z1

11− n+sn2(s | k)

− S M−1− z1

11− n−sn2(s | k)

. (59.b)

Notice that we have the freedom to perform global rotations of the local basis around d3 and ofthe fixed trihedron around eZ, in such a way that ϕ = 0 and ψ = 0 for s = 0. The equations (59.a)and (59.b) can be integrated to yield:

ϕ =M+

1 + z1Π(s | n+, k) + S

M−1− z1

Π(s | n−, k), (60.a)

ψ =(

1b− 1

)M3s+

M+

1 + z1Π(s | n+, k)− S M−

1− z1Π(s | n−, k), (60.b)

where Π is the incomplete elliptic integral of the third kind in “practical” form, as defined inAppendix A.

In the degenerate cases where n+ or n− has a vanishing denominator, the correct limits of (60.a)and (60.b) are obtained by setting the term involving the ill-defined quantity to zero.

The expressions (60.a) and (60.b) together with (55) constitute the general solution of thespinning symmetric top problem. We can then use (34) to obtain the non-fixed basis (d1,d2,d3)as a function of the Euler angles.

17

III.3.4 Particular Orbits of the Spinning Top

The generic orbits of the spinning top are those for which the extremity of d3 oscillates verticallybetween two parallels on the unit sphere, while it revolves horizontally, either monotonously as onFigures 21, 20, 22 and 23, or making loops as on Figures 16 and 24. The looping orbits arise ifS = − and z3 >

1−z1z2z2−z1 . This condition is obtained by allowing ϕ′ to vanish for some value of z.

The degenerate case S = −, z3 = 1−z1z2z2−z1 separates looping orbits from monotonously precessing

orbits. This corresponds to trajectories which present turn-back points, as shown on Figure 19.We shall see in the following section that the condition z3 = 1−z1z2

z2−z1 has a clear meaning in termsof rods for both values of S.

The case z1 = −1 with z2 6= 1 and z3 6= 1 corresponds to orbits which cross periodically thesouth pole of the unit sphere, as shown on Figure 18, while the case z2 = 1 with z1 6= −1 andz3 6= 1 corresponds to orbits which cross the north pole, as shown on Figure 17.

The case z2 = z3 = 1 with arbitrary z1 represents homoclinic orbits such as those shown onFigures 13 to 15.

Next, there are the orbits for which MZ = M3 = 0, that is, for which the top behaves like aplane pendulum (the extremity of d3 is restricted to a vertical grand circle on the unit sphere).They correspond to z1 = −1 and either z2 = 1 or z3 = 1. The case z2 = 1 describes an oscillatingpendulum, whereas the case z3 = 1 describes a revolving pendulum. The case z2 = z3 = 1corresponds to the homoclinic orbit of the pendulum.

Finally, there are orbits with constant z which we took apart from our preceding analysis, andwhich correspond to the helical, circular and straight rods examined in Section III.2. See Figures2 to 7.

III.4 Centerline in Cylindrical Coordinates, Curvature and Torsion.

III.4.1 Polar Coordinates, Complex Curvature and Complex Centerline Radius

Following Shi & Hearst (1994), we introduce cylindrical coordinates R, Φ, Z for the filamentcenterline R:

R = R cos Φ eX +R sin Φ eY + Z eZ. (61)

Rather than adopting the method of Shi & Hearst to obtain R, Φ and Z as functions of the arclength, we lead the calculations in a way which highlights the remarkable correspondence betweenthe expressions for the radius R and the Frenet curvature κ, as well as between the polar angle Φand the angle ζ =

∫ds (κ3− τ) giving the orientation of the local basis (d1,d2,d3) with respect to

the Frenet triad (n,b,d3). The computations are quite analogous and can be led in parallel.First, we introduce the complex centerline radius R, the complex curvature κ and the complex

horizontal component of the moment M :

R = R exp iΦ, (62.a)κ = κ1 + iκ2, (62.b)M = MX + iMY, (62.c)

where MX and MY are the components of the moment along eX and eY. Using (34), (35) and(39.a), κ and M can be expressed in Euler angles as:

κ = −MZ −M3z + iz′

±√

1− z2exp−iψ, (63.a)

M =M3 −MZz − iz′±√

1− z2exp iϕ. (63.b)

18

The correct sign to put in front of the root in (63.a) and (63.b) is the one of sin θ. The key toobtain the complex centerline R is the moment equation (20.b). Taking into account the fact thatF = FeZ is constant and that R =

∫ds d3, we can integrate both sides of this equation, leading

to:

M + F R× eZ = MZeZ (64)

for an appropriate choice of origin in the tridimensional space. Taking the dot product of (64) witheX + ieY, we obtain:

R =−iFM =

−iF

M3 −MZz − iz′±√

1− z2exp iϕ. (65)

III.4.2 Frenet Curvature and Centerline Radius

Using (63.a) and (65), the Frenet curvature κ and the radius R are:

κ2 = |κ|2 =(MZ −M3z)2 + z′2

1− z2, (66.a)

R2 =∣∣∣R∣∣∣2 =

(M3 −MZz)2 + z′2

F 2(1− z2). (66.b)

We now substitute the expressions (45.a) and (45.b) for z′2 respectively in (66.a) and (66.b) toyield the final forms of κ and R. They depend on s only through the variable z:

κ2 = 2F (h− z), (67.a)

R2 =2F

(h− z). (67.b)

Notice that the left-hand sides of these equations are positive for all z, hence h and h are bothgreater than or equal to z2. Also, ignoring the case of straight rods, (67.a) and (67.b) show thatκ and R can only vanish at isolated points where z = z2. A natural question is: for which valuesof z1, z2, z3 and S do the equalities h = z2 or h = z2 hold? Using expressions (48) for F and(54.a)-(54.b) for h and h, one can easily obtain the following results:

h = z2 ⇔ S = − and z3 =1− z1z2

z2 − z1, (68.a)

h = z2 ⇔ S = + and z3 =1− z1z2

z2 − z1. (68.b)

Condition (68.a) is necessary and sufficient for the curvature to vanish at some isolated points.Remarkably, it is identical to the condition for the orbit of the spinning top to present turn-backpoints. In the same way, (68.b) is a necessary and sufficient condition for the radius R to vanishat isolated points.

In order to have a continuous dependence in s for the polar angle Φ across these isolated pointswhere the radius R vanishes, we must change the sign of R. In the same way, the sign of κ mustchange wherever κ vanishes in order for the angle ζ to be a continuous function of s.

III.4.3 Frenet Torsion and Polar Angle of the Centerline

The argument of the right-hand side of (65) can be used as an expression for the polar angle Φ.However, this is not convenient. A more tractable expression can be obtained by first computing

19

the derivative of the polar angle from (65):

Φ′ =∂

∂sarg R =

∂

∂sarctan

−z′M3 −MZz

+ ϕ′. (69)

Similarly, the Frenet torsion τ can be computed:

τ = κ3 +∂

∂sarg κ =

M3

b+

∂

∂sarctan

z′

MZ −M3z− ψ′. (70)

Taking the derivatives in (69) and (70) leads to expressions for Φ′ and τ involving z, z′2, z′′, ϕ′ andψ′. Using the expressions (43.a)-(44) to express everything in terms of z only, we obtain:

Φ′ =12

(MZ +

M3 −MZh

h− z

), (71.a)

τ =12

(M3 +

M3h−MZ

h− z

). (71.b)

In the case of a non constant z, we see from (71) that the condition for a constant Φ is identicalto the condition for the rod to be planar (τ = 0): MZ and M3 must both vanish. This means thatnon circular planar filaments correspond to the case where the spinning top behaves like a planependulum (z1 = −1 and either z2 = 1 or z3 = 1).

For non-planar rods, we define:

n =z2 − z1

h− z1, (72.a)

n =z2 − z1

h− z1

. (72.b)

We can then integrate (71.a) and (71.b) to yield:

Φ =12

(MZs+

M3 −MZh

h− z1

Π(s | n, k)

)− π

2, (73.a)

ζ =M3

bs− 1

2

(M3s+

M3h−MZ

h− z1Π(s | n, k)

)− π

2. (73.b)

The integration constant −π2 in (73.a) has been determined from the complex radius R (65) in the

limit s → 0. The integration constant −π2 in (73.b) is determined from the fact that d1 and the

binormal b are opposite when ψ = 0.

III.4.4 Vertical Cartesian Coordinate of the Centerline

To complete the filament description, we need an expression for Z, the vertical coordinate of thecenterline. It is given by:

Z ′ = z, (74)

which, using (55) and (56), can be written as:

Z ′ = z3 − (z3 − z1)[1− k2sn2(s | k)] (75)

The last expression is readily integrated to yield, for an appropriate choice of origin of the verticalcoordinate:

Z = z3s− (z3 − z1)E(s | k), (76)

20

where E is the incomplete elliptic integral of the second kind in “practical” form, as defined inAppendix A.

For very large z3, Z is expressed as a difference between two large quantities, although Z itselfis finite for z3 →∞. Hence, (76) could be tricky to handle numerically for large z3.

Finally, we note that the expression for the radius R is bounded while, in general, the expressionfor Z is not. As a consequence, the vector d3, averaged over all s, has no component in the (eX, eY)plane. This means that, on average, a spinning top is vertical whatever the constants of the motionare.

III.5 Filament Shapes

III.5.1 Planar Shapes

This section is dedicated to planar filaments other than the circular and straight ones. This isessentially a modern re-statement of Euler’s results. As seen in the previous section, the non-circular and non-straight planar filaments correspond to values of the constants z1, z2 and z3 forwhich the top behaves like a plane pendulum. In this case, the sign variable S is meaningless. Thereare two one-parameter families of planar solutions. The first one is obtained by setting z1 = −1and z3 = 1 and keeping z2 arbitrary; it corresponds to oscillating orbits of the pendulum. Thesecond one is obtained by setting z1 = −1 and z2 = 1 and keeping z3 arbitrary; it corresponds torevolving orbits of the pendulum. In both cases, it is convenient to adopt the modulus k as thearbitrary parameter.

The two families of pendulum orbits are displayed on the phase portrait shown on Figure 9.Each curve represents a different orbit with a given value of k in the (θ, θ′) space. The closed orbitscorrespond to oscillating states of the pendulum, while the open orbits correspond to revolvingstates of the pendulum. The corresponding filament shapes are represented on Figures 10a to 12f.Since the scaling defined in equation (47) depends on k through z3− z1, so does the length unit forthe filament, or the time unit for the pendulum. This is inadequate to draw a phase portrait, hence,on Figure 9, we have exceptionally chosen the scaled force unit [L]−2 to be F . This is identical tothe scaling defined in (47) in the oscillating case.

We introduce the following notations for the complete elliptic integrals of the first and secondkinds defined in Appendix A:

K = K(k), (77.a)E = E(k). (77.b)

Oscillating orbits of the pendulum: In the case z3 = 1, the expressions for the relevantvariables z, R and Z, reduce to:

z = 2k2sn2(s | k)− 1, (78.a)R = 2kcn(s | k), (78.b)Z = s− 2E(s | k). (78.c)

In the following, we define:

f = FA(

1√2k| k)

, (79.a)

e = EA(

1√2k| k)

, (79.b)

21

-1

0

1

2

3

1 2 3 4 5 6

θ’

θ

k = 0.9092

k = 0.7071

k = 0.3333

k = 0.6103

k = 0.6596

k = 0.7227

k = 0.8063

k = 0.9145

k = 1

Figure 9: Phase portrait for the plane pendulum. The closed curves represent oscillating orbits,while the open curves represent revolving orbits. The homoclinic orbit corresponds to k = 1.

where FA and EA are the incomplete elliptic integrals of the first and second kinds in algebraicform (see Appendix A).

The extrema of Z are given by the condition z = 0, which is satisfied for:

sn2(s | k) =1

2k2⇔ s = ±f + 2mK, (80)

where m is an arbitrary integer. We see from (79.a) that f is real only if k2 ≥ 12 . Hence, for

k2 < 12 , Z is a decreasing function of s (see Figure 10a). The limiting case k2 = 1

2 corresponds toan oscillating pendulum which has just enough energy to reach the horizontal position, as shownof Figure 10b. Above this critical value of k2, Z is not monotonous, and for high enough valuesof k, the filaments can have points of self-tangency or self-intersection (see Figures 10d to 10f).The cases of self-tangency are obtained by imposing the equality of Z at different points where zvanishes. This leads to the condition:

f − 2e+ 2m(2E −K) = 0, (81)

where m is a positive integer which denotes the number of pendulum oscillations before a self-tangency occurs. The solutions km of the equation (81) for the lowest values of m are given in thefirst column of Table 2.

Setting m =∞ in equation (81) yields:

k∞ = 0.9089, (82)

for which the filament shape is the lemniscate represented on Figure 11a. Every value of k < k∞corresponds to a solution Z which, on average, is a decreasing function of s, while for k > k∞,Z is, on average, an increasing function of s. The increasing solutions can also present pointsof intersection and self-tangency for given values of k (see Figures 11b to 11e). The condition ofself-tangency in the case k > k∞ also assumes the form (81), provided that we set m to be negative.The positive integer −m counts the number of pendulum oscillations before a self-tangency occurs.The solutions km of the equation (81) for the lowest values of −m are given in the second columnof Table 2.

22

Homoclinic orbit of the pendulum: The homoclinic orbit is obtained by taking the limit k → 1in either the oscillating case or the revolving case. It is shown on Figure 11f. The expressions forz, R and Z then reduce to:

z = 1− 2sech2s, (83.a)R = 2sechs, (83.b)Z = s− 2 tanh s. (83.c)

Revolving orbits of the pendulum: In this case, we set z1 = −1 and z2 = 1. The expressionsfor θ, R and Z read:

θ = π − 2am(s | k), (84.a)ρ = 2k−2dn(s | k), (84.b)Z = (2k−2 − 1)s− 2k−2E(s | k). (84.c)

Next, we define:

f = F(

1√2| k)

, (85.a)

e = E(

1√2| k)

. (85.b)

The extrema of Z correspond to the values of s for which θ is an integer multiple of π, henceam(s | k) is an integer multiple of π

2 , or

s = ±2f + 2mK, (86)

where m is an integer. Notice that f and e are real for all k. As in the case of oscillating orbits,we can find a condition for the existence of points of self tangency:

(2− k2)f − 2e+m[(2− k2)K − 2E] = 0, (87)

where the positive integer m counts the number of pendulum revolutions before a self-tangencyoccurs. The solutions km of the equation (87) for the lowest values of m are given in the thirdcolumn of Table 2. The sequence (km) has a vanishing limit for m→∞ and, in the limit k → 0, thefilament is a vertical ring. Some filaments corresponding to revolving orbits are shown on Figure12.

III.5.2 Non-planar Localizing Solutions

The homoclinic orbits constitute a one-parameter family of solutions which are obtained by settingz2 = z3 = 1 while z1 is kept arbitrary. Here again, the sign parameter S is meaningless. In terms ofrods, these orbits correspond to the localizing solutions studied by Coyne (1990) [1]. They connectcontinuously the straight state (z1 = 1) to the planar loop (z1 = −1). These solutions have constanttorsion:

τ =

√1 + z1

1− z1. (88)

23

a) k = 0.3333 b) k = 0.7071

c) k = 0.8000 d) k = 0.8551

e) k = 0.8750 f) k = 0.8858

Figure 10: Planar filaments corresponding to low amplitude oscillating orbits of the plane pendu-lum.

24

a) k = 0.9089

c) k = 0.9320

e) k = 0.9700

b) k = 0.9270

d) k = 0.9414

f) k = 1

Figure 11: Planar filaments corresponding to high amplitude oscillating orbits of the plane pendu-lum. The homoclinic orbit is reached in the limit k = 1.

25

a) k = 0.9700 b) k = 0.9185

c) k = 0.8500 d) k = 0.8063

e) k = 0.7600 f) k = 0.7227

Figure 12: Planar filaments corresponding to revolving orbits of the plane pendulum.

26

k1 = 0.8551 k−1 = 0.9414 k1 = 0.9145k2 = 0.8858 k−2 = 0.9270 k2 = 0.8063k3 = 0.8942 k−3 = 0.9214 k3 = 0.7227k4 = 0.8981 k−4 = 0.9185 k4 = 0.6596k5 = 0.9004 k−5 = 0.9167 k5 = 0.6103... ... ...k∞ = 0.9089 k−∞ = 0.9089 k∞ = 0

Table 2: Values of k corresponding to self-tangency.

We see that τ is bijective in z1 and can assume any nonnegative value, from 0 in the case z1 = −1(plane pendulum) to infinity2 in the limit z1 → 1. In the following, we adopt the torsion τ ratherthan z1 as the arbitrary parameter.

The top variables ϕ, z and ψ assume the form:

ϕ = arctan(

1τ

tanh s)

+ τs, (89.a)

z = 1− 21 + τ2

sech2s, (89.b)

ψ = arctan(

1τ

tanh s)

+(

3− 2b

)τs. (89.c)

Notice that the boundary values 23 and 1 for the variable b in the case of circular cross-section

take a new sense in view of the expressions (89). The value b = 23 makes the self-rotation angle ψ

bounded, while b = 1 is the value for which ϕ = ψ.The centerline variables R, Φ and Z read:

R =2

1 + τ2sechs, (90.a)

Φ = τs− π

2, (90.b)

Z = s− 21 + τ2

tanh s. (90.c)

Three typical non-planar (τ 6= 0) homoclinic orbits of the spinning top together with the corre-sponding filament shapes are shown on Figures 13 to 15.

III.5.3 Generic Filament Shapes

In the most generic case, the spinning top oscillates between two parallels z1 and z2, while itprecesses either monotonously or with backward-and-forward motion. The corresponding filamentcenterline R behaves, on average, as a helix around which it is wound.

Average helix: We define the mean helical centerline 〈R〉 in the following way: We introducecylindrical coordinates 〈R〉, 〈Φ〉, 〈Z〉 for the helix 〈R〉, and we take the mean vertical coordinate〈Z〉 to be a linear function of s, namely:

〈Z〉 = 〈z〉 s, (91)

2Actually, for fixed tension, the torsion cannot be arbitrarily large. One must keep in mind that the parameter τused here is the scaled torsion and that the scaling defined in (47) in turn depends on τ through z1. As a consequence,

the unscaled torsion τ is not simply proportional to τ , but instead is given by τ2 = FEI1(1+τ−2)

, where F is the unscaled

tension. Hence, the torsion has actually an upper bound proportional to the square root of the tension.

27

N

Figure 13: A locally buckled filament with low torsion and high curvature (τ = 12).

N

Figure 14: A locally buckled filament with intermediate torsion and curvature (τ = 1).

N

Figure 15: A locally buckled filament with high torsion and low curvature (τ = 2).

28

where 〈z〉 is the average of z over a period, 2K(k). Using (76), we have:

〈z〉 =1

2K(k)

∫ 2K(k)

0z ds = z3 − (z3 − z1)

E(k)K(k)

. (92)

Now, consider the expression (65) for the complex centerline radius R. The polar angle Φ beingthe argument of R, we have:

Φ = arg ρ+ ϕ− π

2, (93)

with

ρ =1F

M3 −MZz − iz′±√

1− z2. (94)

The function ρ depends on s through z and z′ only, hence it describes a closed curve in the complexplane. If ρ does not vanish anywhere, as s increases by a period 2K(k), the argument of ρ increasesby 2πm, where m is some integer. Moreover, the real and imaginary parts of the numerator in (94)being decreasing functions of z and z′ respectively, the curve defined by ρ has no self-crossing (andis parameterized clockwise by s). Hence, it turns around the origin at most once clockwise overa period 2K(k), so that m is restricted to the values 0 and −1. The domain in the (z1, z2, z3, S)space where either value of m holds is delimited by the condition (68.b) ensuring the existence ofpoints where ρ vanishes. We find that m = −1 if S = + and z3 >

1−z1z2z2−z1 , and m = 0 otherwise.

As a consequence, we have:

ϕ (s+ 2K(k))− ϕ(s) = Φ (s+ 2K(k))− Φ(s) +

2π if S = + and z3 >

1− z1z2

z2 − z1,

0 if S = − or z3 <1− z1z2

z2 − z1.

(95)

Hence, over a period of z, the precession angle ϕ of the top covers either the same angular distanceas the polar angle Φ of the rod, or the same angular distance plus one complete revolution.

Now, using (60.a) and (73.a), we define the mean angular velocities 〈ϕ′〉 and 〈Φ′〉 for thecorresponding angles ϕ and Φ as:

⟨ϕ′⟩

=1

2K(k)

∫ 2K(k)

0ds ϕ′ =

M+

1 + z1

Π(n+, k)K(k)

+ SM−

1− z1

Π(n−, k)K(k)

, (96.a)

⟨Φ′⟩

=1

2K(k)

∫ 2K(k)

0ds Φ′ =

12

(MZ +

M3 −MZh

h− z1

Π(n, k)K(k)

). (96.b)

As a consequence, these two quantities either are equal or differ from πK(k) . The question arises

then: which mean angular velocity, 〈ϕ′〉 or 〈Φ′〉, should we use to define the polar angle 〈Φ〉 of ouraverage helix? The most natural choice seems to take 〈Φ〉 = 〈Φ′〉 s. However, the expression (96.b)is not continuous through the boundary (68.b), making the definition of the mean polar angle 〈Φ〉ambiguous at that point. Therefore, we define:

〈Φ〉 =⟨ϕ′⟩s− π

2for S = −, (97.a)

〈Φ〉 =(⟨ϕ′⟩− π

K(k)

)s− π

2for S = +. (97.b)

These expressions reduce to 〈Φ′〉 s for sufficiently large z3, and are continuous across the boundary(68.b). As we shall see, this definition is the most adequate in the case where the filament shape is

29

a supercoiled helix (this case has a great importance in biochemical applications, in particular forDNA supercoiling [31, 40, 41]). The supercoiled helices are defined and discussed below. It remainsnow to define the mean helix radius 〈R〉. A definition consistent with (97) is:

〈R〉 =1

2K(k)

∫ 2K(k)

0ds R exp −i 〈Φ〉 . (98)

Therefore, the generic filament behaves on average like an helical filament.

Supercoiled helices: A supercoiled helix is a curve which looks like a helix on short lengthscales, with the central axis itself shaped like a helix on large length scales. The condition for thecenterline R to be a supercoiled helix, in terms of spinning tops, is that the vector d3 describesslowly precessing, nearly circular, oblique loops on the unit sphere. Two such examples are shownon Figures 16 and 17. If any of these three conditions (slow precession, near-circularity and oblicityof the top orbit) is not fulfilled, the filament shape will not look like a supercoiled helix, but ratherlike a deformed helix, as shown on Figures 18 to 21.

The supercoiled helices can be studied systematically as solutions close to the oblique circularorbits of the spinning top. There are two ways to obtain circular orbits. The first one consists insetting z1 = z2, in which case the orbit is horizontal, and not oblique. The second one is to takethe limit z3 → ∞. Indeed, in view of (48), the tension vanishes as z3 grows without bound, and,as we mentioned in Section III.2, every filament with null tension is a helix, hence every top orbitwith F = 0 is a circle. Furthermore, these asymptotic circular orbits can be arbitrarily orientedsince the vertical direction eZ cannot be distinguished from the other directions in the limit F = 0.As a consequence, we can redefine a supercoiled helix as a solution with a large (“close to infinity”)value of z3. In practice, however, z3 does not need to be very large, so that the threshold valueappearing in (95) can be reached with the centerline R still reasonably looking like a supercoiledhelix.

In the limit of large z3, the vertical axis joining the poles of the unit sphere is interior toprecessing circular orbit in the case S = +, and exterior to it in the case S = −. As a consequence,one passes continuously from the supercoiled helices with S = − to the supercoiled helices withS = + by enlarging the slowly precessing orbit so that it passes through one of the poles, as onFigure 17.

We mentioned above that our definitions of the average helix coordinates 〈R〉, 〈Φ〉, 〈Z〉 are welladapted to supercoiled helices, in the sense that they describe consistently the large scale helicalbehavior of the axis around which the centerline is wound. Namely, in the limit z3 → ∞, thesupercoiled helix tends towards an (infinitely remote) ordinary helix, with a straight axis given bythe limit of the average helix 〈R〉.

Deformed helices: In many cases, although the centerline R winds around the average helix〈R〉, it does not quite look like a supercoiled helix. This happens if the criterion discussed aboveis not fulfilled, which can result in the following.

• The short scale spatial period 2K(k) and the large scale spatial period 2π〈Φ〉′ can be too close

to each other, in which case the two orders of helicity cannot be clearly distinguished. Sucha situation is shown on Figure 18.

• If the top orbit is too far away from a circle, the short scale pattern does not look like a helix.As an example, Figure 19 shows a large-scale helix with curvature varying on short scale.

30

• For some values of the constants, the top orbit can be periodic although quite different froma circle. In this case, the filament shape is periodic in space, but different from a helix (seethe “oblique helix” on Figure 20).

In addition, there is the possibility for the amplitude of the short scale pattern to be largeenough for two consecutive turns of the large-scale helix to overlap each other. In this case, thetopology of the solution is different from the topology of an ordinary helix. Such a “knotted helix”is shown on Figure 21.

III.5.4 Bounded and Closed Filament Shapes

The filament shapes discussed so far are all unbounded in space, except for the twisted ring shownon Figure 5 and the lemniscate shown on Figure 11a. In general, bounded shapes are obtained byimposing the coordinate Z to be a periodic function of s, that is, Z = 0 for s = 2K(k). Using (76),the condition for boundedness reads:

z3K(k)− (z3 − z1)E(k) = 0. (99)

The condition for the space curve R to be closed is obtained by requiring, in addition, that theperiods of the variables Z and Φ be in a ratio of integer numbers. Namely, using (73.a):

MZK(k) +M3 −MZh

h− z1

Π(n, k) = 2πmΦ

mZ, (100)

where the integers mΦ and mZ denote, respectively, the number of periods of the variables Z andΦ in a complete covering of the centerline R.

Finally, one can impose the ribbon associated to the filament to be closed by requiring that theperiods of the variables Z and ζ be in a ratio of integer numbers, or, using (73.b):

2M3

bK(k)−

(M3K(k) +

M3h−MZ

h− z1Π(n, k)

)= 2π

mζ

mZ, (101)

where mζ denotes the number of periods of ζ in a complete covering of the centerline R. Thecondition (101) is useful if one has, for instance, an octogonal cross-section, in which case mζ mustbe set to an integer multiple of 1

8 , in order for the octogons at s = 0 and at s = 2mZK(k) to matcheach other.

This together makes three conditions from which the constants z1, z2 and z3 can be determined.However, the conditions (100) and (101) are not very tractable, because the left-hand side of (100)is proportional to the average angular velocity 〈Φ′〉 which, as we mentioned, is not a continuousfunction of the constants z1, z2 and z3. This holds too for the left-hand side of (101). As aconsequence, numerical root solvers are inefficient in solving the system (99)-(101). In practice, itis well advised to replace the equations (100) and (101) by equivalent conditions on the Euler anglesϕ and ψ. Remember that, over a period of Z, the angles ϕ and Φ cover angular distances whichdiffer by integer multiples of 2π. A similar relation holds for the angles ψ and ζ. The conditionson ϕ and ψ analogous to (100) and (101) read:

M+

1 + z1Π(n+, k) + S

M−1− z1

Π(n−, k) = πmϕ

mZ, (102.a)(

1b− 1

)M3K(k) +

M+

1 + z1Π(n+, k)− S M−

1− z1Π(n−, k) = π

mψ

mZ, (102.b)

where mϕ and mψ differ from mΦ and mζ respectively by an integer.Examples of closed filaments are displayed on Figures 22 to 24. Figures 22 and 23 show torus

knots, that is, knotted curves which are topologically equivalent to closed curves lying on a torus.Figure 24 shows a supercoiled ring (notice the large value of z3).

31

N

Figure 16: A slowly precessing, nearly circular top orbit corresponds to a supercoiled helix (z1 = −12 ,

z2 = 34 , z3 = 4, S = −).

N

Figure 17: A top orbit crossing periodically the north pole. The corresponding filament has aperiodically vertical tangent. (z1 = 1

2 , z2 = 1, z3 = 3).

S

Figure 18: A top orbit crossing periodically the south pole. The corresponding filament has aperiodically vertical tangent. (z1 = −1, z2 = −1

2 , z3 = 1410 , S = −).

32

N

Figure 19: A top orbit presenting turn-back points corresponds to a filament with periodicallyvanishing curvature (z1 = 0, z2 = 1

2 , z3 = 2, S = −).

N

Figure 20: A periodic top orbit results in a spatially periodic filament shape (z1 = −0.7822,z2 = 0.8782, z3 = 1, S = −).

N

Figure 21: An apparently simple top orbit resulting in a surprisingly complex “knotted helix”(z1 = −0.4152, z2 = 0.2800, z3 = 1.026, S = −).

33

N

Figure 22: A torus knot with mZ = 5, mΦ = 3 and mζ = 1 (z1 = −0.4152, z2 = 0.3446, z3 = 1.026,S = −).

N

Figure 23: A torus knot with mZ = 7, mΦ = 4 and mζ = 1 (z1 = −0.4997, z2 = 0.4013, z3 = 1.037,S = −).

N

Figure 24: A supercoiled ring with mZ = 20, mΦ = 1 and mζ = 6 (z1 = −0.4269, z2 = 0.4171,z3 = 9.084, S = −).

34

IV Conclusions

In this paper we have shown how to classify the shapes of Kirchhoff filaments based on the geometryof the the spinning top solutions. To do so, we have pushed the Kirchhoff analogy to its extreme andsystematically obtained interesting properties of filaments based on the corresponding soutions ofthe Euler equations. We showed that the solutions of Kirchhoff equations can be extremely variedand that many interesting cases can be distinguished. In particular, we found explicit conditionson the boundary values for filaments to have points of self-tangency and multiple self-intersection.We also studied the case where filaments have points of vanishing curvature and show that theycorrespond to orbits of spinning tops with turn-back points. We gave a complete description oflocalizing solutions, that is solutions wich are homoclinic in the curvature-torsion space, that isfilaments in space which are asymptotic to a straight line. In the same way, we found conditionsto obtain filaments which have the topology of torus knots, that is bounded and periodic filaments(in the physical space). Finally we studied the behavior of generic filaments and show that on longlength scales they always behave like helical filaments.

Some of the particular solutions presented here have been obtained in various places by differentauthors. In this paper we have stressed on the geometry of these solutions and presented them ina unifying way based on the familiar framework of the spinning top.

The solutions of the Kirchhoff equations for rods with circular cross-sections are often usedas a first guess to study numerically physical filaments with different proeprties (e.g. non-circularcross-sections, intrinsic curvature or torsion, ...). We hope that the explicit solutions given in thispaper together with their geometric classification will be useful in this context.

Acknowledgements: M. N. is acknowledges support from the Fonds National de la RechercheScientifique (F.N.R.S) M.N. is an aspirant at the FNRS. This work is supported by NATO-CRG97/037.

A Elliptic Functions

Remark: In the following, the constant k is called the modulus of the elliptic functions and rangesbetween 0 and 1, while the constant n is called the characteristic and is a real number less than 1.

A.1 Elliptic Integrals

A.1.1 Incomplete Elliptic Integrals in Standard Form

The incomplete elliptic integrals of the first, second and third kind in standard form are, respec-tively, defined by:

FS(Φ | k) =∫ Φ

0

dΦ′√1− k2 sin2 Φ′

, (103.a)

ES(Φ | k) =∫ Φ

0dΦ′

√1− k2 sin2 Φ′, (103.b)

ΠS(Φ | n, k) =∫ Φ

0

dΦ′(1− n sin2 Φ′

)√1− k2 sin2 Φ′

. (103.c)

35

A.1.2 Incomplete Elliptic Integrals in Algebraic Form

The algebraic forms are obtained by carrying out the following change of variable into the standardforms:

u = sin Φ. (104)

This yields:

FA(u | k) =∫ u

0

du′√(1− u′2)(1− k2u′2)

, (105.a)

EA(u | k) =∫ u

0du′√

1− k2u′2

1− u′2 , (105.b)

ΠA(u | n, k) =∫ u

0

du′

(1− nu′2)√

(1− u′2)(1− k2u′2). (105.c)

The algebraic forms are those which are implemented in the symbolic calculus software Maple.Notice that for the change of variable (104) to be bijective, one must restrict Φ to the interval[−π

2 ,π2

].

A.1.3 Incomplete Elliptic Integrals in Practical Form

We call the following forms of the elliptic integrals of the second and third kind “practical” becausethese are the forms under which they appear the most naturally in the problems in Sections III.These forms are obtained by carrying out the following change of variable into the standard forms:

s = FS(Φ | k). (106)

This yields:

E(s | k) =∫ s

0ds′dn2(s′ | k), (107.a)

Π(s | n, k) =∫ s

0ds′

ds′

1− nsn2(s′ | k), (107.b)

where the functions sn and dn are defined in Section A.2.

A.1.4 Complete Elliptic Integrals

The complete elliptic integrals are defined by the expressions for the corresponding incompleteelliptic integrals evaluated at u = 1 in algebraic form. They are denoted in the following way:

K(k) = FA (1 | k) , (108.a)E(k) = EA (1 | k) , (108.b)Π(n, k) = ΠA (1 | n, k) . (108.c)

A.2 Jacobi’s Elliptic Functions

A.2.1 Definitions

The incomplete elliptic integrals, having positive integrands, define monotonous, hence invertible,functions. The function am is defined as the inverse of the standard form FS of the incompleteelliptic integral of the first kind:

am(s | k) = (FS)−1(s | k). (109)

36

We then define Jacobi’s elliptic functions sn, cn and dn as:

sn(s | k) = sin am(s | k), (110.a)cn(s | k) = cos am(s | k), (110.b)

dn(s | k) =√

1− k2sn2(s | k). (110.c)

Notice that the function sn itself, restricted to the interval [−K(k),K(k)], is the inverse of thealgebraic form FA of the incomplete elliptic integral of the first kind. The four functions sn, cn, dnand am are represented on Figures 25 to 26 for various values of k.

A.2.2 Elementary properties of am, sn, cn and dn

The function am obeys the relations:

am(2nK(k)± s | k) = nπ ± am(s | k), (111)

for any real s and integer n. Hence, the knowledge of am over the interval [0,K(k)], togetherwith (111), is sufficient to reconstruct the function over the whole real line. This holds too for theperiodic functions sn, cn and dn, which obey the following relations:

sn(2nK(k)± s | k) = ±(−1)nsn(s | k), (112.a)cn(2nK(k)± s | k) = (−1)ncn(s | k), (112.b)dn(2nK(k)± s | k) = dn(s | k). (112.c)

A.2.3 Limits of am, sn, cn and dn for k = 0 and k = 1

These limits are obtained by considering, for k = 0 and k = 1, the expression (103.a) defining thefunction FS, and then taking the limit of am to be the inverse function. For k = 0, one has:

am(s | 0) = s, (113.a)sn(s | 0) = sin s, (113.b)cn(s | 0) = cos s, (113.c)dn(s | 0) = 1, (113.d)

whereas k = 1 leads to:

am(s | 1) = arcsin tanh s, (114.a)sn(s | 1) = tanh s, (114.b)cn(s | 1) = sechs, (114.c)dn(s | 1) = sechs. (114.d)

37

A.2.4 Derivatives of am, sn, cn and dn

The derivatives of am, sn, cn and dn are obtained by differentiating (103.a) and the definitionequations (110.a) to (110.c). This leads to the following differential relations:

dds

am(s | k) = dn(s | k), (115.a)

dds

sn(s | k) = cn(s | k)dn(s | k), (115.b)

dds

cn(s | k) = −sn(s | k)dn(s | k), (115.c)

dds

dn(s | k) = −k2sn(s | k)cn(s | k). (115.d)

References

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[12] T. Schlick & W. K. Olson, Trefoil Knotting Revealed by Molecular Dynamics Simulations ofSupercoiled DNA, Science 257, 1110-1114 (1992).

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[14] T. R. Strick, J.-F Allemand, D. Bensimon, & V. Croquette, The Elasticity of a Single Super-coiled DNA Molecule, Science 271, 1835-1837 (1996).

38

[15] Y. Yang, I. Tobias& , W. K. Olson, Finite Element Analysis of DNA Supercoiling, J. Chem.Phys. 98, 1673-1686 (1993).

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39

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40

-1

-0.5

0

0.5

1

-6 -4 -2 2 4 6

sn(s | k)

s

k = 1

k = 63/64

k = 15/16k = 3/4k = 0

-1

-0.5

0

0.5

1

-6 -4 -2 2 4 6

cn(s | k)

s

k = 0 k = 3/4

k = 15/16

k = 63/64

k = 1

Figure 25: The function sn and cn for various values of k.

-6

-4

-2

0

2

4

6

-6 -4 -2 2 4 6

am(s | k)

s

k = 0

k = 3/4

k = 15/16

k = 63/64

k = 1

0

0.2

0.4

0.6

0.8

1

-6 -4 -2 2 4 6

dn(s | k)

s

k = 3/4

k = 15/16

k = 63/64

k = 0

k = 1

Figure 26: The function dn and am for various values of k.

41