Rigid Body Mechanics: Mathematics, Physics and Applications (Physics Textbook)

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William B. Heard Rigid Body Mechanics

William B. Heard

Rigid Body Mechanics Mathematics, Physics and Applications

WILEY-VCH Verlag GmbH & Co. KGaA

The Author William B. Heard Alexandria, VA USA

For a Solutions Manual, lecturers should contact the editorial department at [email protected], stating their affiliation and the course in which they wish to use the book

All books published by Wiley-VCH are carefully produced. Nevertheless, authors, editors, and publisher do not warrant the information contained in these books, including this book, to be free of errors. Readers are advised to keep in mind that statements, data, illustrations, procedural details or other items may inadvertently be inaccurate. Library of Congress Card No.: applied for British Library Cataloguing-inPublication Data A catalogue record for this book is available from the British Library. Bibliographic information published by Die Deutsche Bibliothek Die Deutsche Bibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data is available in the Internet at .

© 2006 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim All rights reserved (including those of translation into other languages). No part of this book may be reproduced in any form – by photoprinting, microfilm, or any other means – nor transmitted or translated into a machine language without written permission from the publishers. Registered names, trademarks, etc. used in this book, even when not specifically marked as such, are not to be considered unprotected by law. Printed in the Federal Republic of Germany Printed on acid-free paper Printing betz-druck GmbH, Darmstadt Binding Litges & Dopf Buchbinderei GmbH, Heppenheim

ISBN-13: 978-3-527-40620-3 ISBN-10: 3-527-40620-4

This book is dedicated to Peggy and the memory of my Mother and Father

VII

Contents

Preface

XI 1

1

Rotations

1.1 1.1.1 1.1.2 1.1.3 1.1.4 1.1.5 1.1.6 1.2 1.2.1 1.2.2 1.2.3 1.2.4 1.2.5 1.3 1.3.1 1.3.2 1.4 1.5

Rotations as Linear Operators 1 Vector Algebra 1 Rotation Operators on R3 7 Rotations Specified by Axis and Angle 8 The Cayley Transform 12 Reflections 13 Euler Angles 15 Quaternions 17 Quaternion Algebra 20 Quaternions as Scalar–Vector Pairs 21 Quaternions as Matrices 21 Rotations via Unit Quaternions 23 Composition of Rotations 25 Complex Numbers 29 Cayley–Klein Parameters 30 Rotations and the Complex Plane 31 Summary 35 Exercises 37 39

2

Kinematics, Energy, and Momentum

2.1 2.1.1 2.1.2

Rigid Body Transformation 40 A Rigid Body has 6 Degrees of Freedom 40 Any Rigid Body Transformation is Composed of Translation and Rotation 40 The Rotation is Independent of the Reference Point 41

2.1.3

Rigid Body Mechanics: Mathematics, Physics and Applications. William B. Heard Copyright © 2006 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim ISBN: 3-527-40620-4

VIII

Contents

2.1.4 2.1.5 2.2 2.2.1 2.2.2 2.2.3 2.3 2.4 2.5 2.6

Rigid Body Transformations Form the Group SE(3) Chasles’ Theorem 42 Angular Velocity 44 Angular Velocity in Euler Angles 47 Angular Velocity in Quaternions 49 Angular Velocity in Cayley–Klein Parameters 50 The Inertia Tensor 50 Angular Momentum 54 Kinetic Energy 54 Exercises 57 59

3

Dynamics

3.1 3.1.1 3.1.2 3.2 3.2.1 3.2.2 3.2.3 3.3 3.3.1 3.3.2 3.3.3 3.4

Vectorial Mechanics 61 Translational and Rotational Motion 61 Generalized Euler Equations 62 Lagrangian Mechanics 68 Variational Methods 68 Natural Systems and Connections 72 Poincaré’s Equations 75 Hamiltonian Mechanics 84 Momenta Conjugate to Euler Angles 86 Andoyer Variables 88 Brackets 92 Exercises 98 99

4

Constrained Systems

4.1 4.2 4.2.1 4.2.2 4.2.3 4.3 4.3.1 4.3.2 4.3.3 4.3.4 4.4 4.4.1 4.4.2 4.5 4.6

Constraints 99 Lagrange Multipliers 102 Using Projections to Eliminate the Multipliers 103 Using Reduction to Eliminate the Multipliers 104 Using Connections to Eliminate the Multipliers 107 Applications 108 Sphere Rolling on a Plane 108 Disc Rolling on a Plane 111 Two-Wheeled Robot 114 Free Rotation in Terms of Quaternions 116 Alternatives to Lagrange Multipliers 117 D’Alembert’s Principle 118 Equations of Udwadia and Kalaba 119 The Fiber Bundle Viewpoint 122 Exercises 127

41

Contents

5

5.1 5.1.1 5.1.2 5.1.3 5.1.4 5.2 5.2.1 5.2.2 5.3 5.3.1 5.3.2 5.4 5.5 5.5.1 5.5.2 5.6

Integrable Systems 129 Free Rotation 129 Integrals of Motion 129

Reduction to Quadrature 130 Free Rotation in Terms of Andoyer Variables 136 Poinsot Construction and Geometric Phase 140 Lagrange Top 145 Integrals of Motion and Reduction to Quadrature 145 Motion of the Top’s Axis 147 The Gyrostat 148 Bifurcation of the Phase Portrait 149 Reduction to Quadrature 150 Kowalevsky Top 153 Liouville Tori and Lax Equations 154 Liouville Tori 155 Lax Equations 156 Exercises 159 161

6

Numerical Methods

6.1 6.2 6.3 6.4 6.5 6.6

Classical ODE Integrators 161 Symplectic ODE Integrators 168 Lie Group Methods 170 Differential–Algebraic Systems 179 Wobblestone Case Study 182 Exercises 187 189

7

Applications

7.1 7.1.1 7.1.2 7.1.3 7.2 7.2.1 7.2.2 7.2.3 7.3 7.4

Precession and Nutation 189 Gravitational Attraction 190 Precession and Nutation via Lagrangian Mechanics 192 A Hamiltonian Formulation 195 Gravity Gradient Stabilization of Satellites 197 Kinematics in Terms of Yaw-Pitch-Roll 198 Rotation Equations for an Asymmetric Satellite 198 Linear Stability Analysis 200 Motion of a Multibody: A Robot Arm 203 Molecular Dynamics 210

IX

X

Contents

Appendix

213

A

Spherical Trigonometry

B

Elliptic Functions

B.1 B.2 B.3 B.3.1

Elliptic Functions Via the Simple Pendulum 218 Algebraic Relations Among Elliptic Functions 222 Differential Equations Satisfied by Elliptic Functions The Addition Formulas 224

C

Lie Groups and Lie Algebras

C.1 C.2 C.2.1 C.3 C.3.1 C.3.2 C.4 C.4.1 C.4.2

Infinitesimal Generators of Rotations 225 Lie Groups 227 Examples 230 Lie Algebras 231 Examples 232 Adjoint Operators 234 Lie Group–Lie Algebra Relations 237 The Exp Map 237 The Derivative of exp A(t) 238

D

Notation

241

References Index

247

243

218

225

224

XI

Preface This is a textbook on rigid body mechanics written for graduate and advanced undergraduate students of science and engineering. The primary reason for writing the book was to give an account of the subject which was firmly grounded in both the classical and geometrical foundations of the subject. The book is intended to be accessible to a student who is well prepared in linear algebra and advanced calculus, who has had an introductory course in mechanics and who has a certain degree of mathematical maturity. Any mathematics needed beyond this is included in the text. Chapter 1 deals with the rotations, the basic operation in rigid body theory. Rotations are presented in several parameterizations including axis angle, Euler angle, quaternion, and Cayley–Klein parameters. The rotations form a Lie group which underlies all of rigid body mechanics. Chapter 2 studies rigid body motions, angular velocity, and the physical concepts of angular momentum and kinetic energy. The fundamental idea of angular velocity is straight from the Lie algebra theory. These concepts are illustrated with several examples from physics and engineering. Chapter 3 studies rigid body dynamics in vector, Lagrangian, and Hamiltonian formulations. This chapter introduces many geometric concepts as dynamics occurs on differential manifolds and for rigid body mechanics the manifold is often a Lie group. The idea of the adjoint action is seen to be basic to the rigid body equations of motion. This chapter contains many examples from physics and engineering. Chapter 4 considers the dynamics of constrained systems. Here Lagrange multipliers are introduced and several ways of determining or eliminating them are considered. This topic is rich in geometrical interactions and there are several examples, some standard and some not. Chapter 5 considers the integrable problems of free rotation, Lagrange’s top, and the gyrostat. The Kowalevsky top and Lax equations are also considered. Geometrical topics include the Poinsot construction, the geometric phase, and


XII

Preface

Liouville tori. Verification and validation are of the utmost importance in the world of scientific and engineering computing and analytical solutions are treasured. In addition to the validation service they also play an important role in developing our intuitive understanding of the subject, not to mention their intrinsic and historical worth. The importance of numerical methods in today’s applications of mechanics cannot be overstated. Chapter 6 discusses classical numerical methods and more recent methods tailored specifically for Lie groups. This chapter includes a case study of a complicated rigid body motion, the wobblestone, which is naturally studied with numerical methods. The final chapter, Chapter 7, applies the previous material to phenomena ranging in scale from the astronomical to the molecular. The largest scale concerns precession and nutation of the Earth. The Earth is nonspherical – an oblate spheroid to first approximation – and the axis of the Earth’s rotation is observed to move on the celestial sphere. Most of this motion is attributed to the torque exerted on the nonspherical Earth by the Moon and the Sun and rigid body dynamics explains the effect. Next we study satellite gravity gradient stabilization. If Earth’s satellites are not stabilized by some mechanism, they will tumble – as do the asteroids – and will not be useful platforms. One stabilization mechanism uses the gradient in the Earth’s gravitational field and the basics of this mechanism are explained by rigid body theory. On the same scale, but down to the Earth, we consider the motion of a multibody, a mechanical system consisting of interconnected rigid bodies. Rigid body dynamics is used to study the motion of a robot arm. At the smallest scale we examine the techniques of molecular dynamics. Physicists and chemists study properties of matter by simulating the motions of a large number of interacting molecules. At the most basic level this is a problem in quantum mechanics. However, there are properties which can be calculated by idealizing the material to be a collection of interacting rigid bodies. Three appendices are provided on spherical trigonometry, elliptic functions and Lie groups and Lie algebras. Lie groups and Lie algebras unify the subject. Appendix C provides background on all the Lie group concepts used in the text. Some choices made in the course of writing the book should be mentioned. The application of mechanics often boils down to making a calculation and getting the useful number. I have tried to keep calculations foremost in mind. There are no theorem–proof structures in the book. Rather observations are made and substantiated by methods which are decidedly computational. Rarely are equations put in dimensionless form because one can rely on dimensional checks to avoid algebra mistakes and misconceived physics. Of course, when it comes time to compute one is well advised to pay attention to

Preface

the scalings. Various notations are used in the text ranging from “old tensor” to intrinsic operators independent of coordinates and the concise notations of [1, 23] which facilitate the computations. There is a glossary in Appendix D of the notations used. Examples are set aside in italic type and end with the symbol ♦. William B. Heard Alexandria, Virginia, U.S.A., 23 August 2005

XIII

Rigid Body Mechanics William B. Heard © 2006 WILEY-VCH Verlag GmbH & Co.

1

1

Rotations This chapter is devoted to rotations in three-dimensional space. Rotations are fundamental to rigid body dynamics because there is a one-one correspondence between orientations of a rigid body and rotations in three-dimensional space. The theory of rotations is a classical subject with a rich history and a variety of modes of expression. We shall begin by expressing the elements of the theory in terms of vectors and linear operators. Next, quaternions will be introduced and additional elements of the theory developed with them. This will lead to some elegant connections between rotations and spherical geometry. This leads, via the stereographic projection and Möbius transformations, to a description of rotations in terms of complex variables.

1.1 Rotations as Linear Operators

One way to approach rotations is to study their effect on spatial objects. The language of vectors and matrices provides a natural calculus. This section reviews some basic algebra of vector spaces and establishes our notation. Then the angle of rotation and axis of rotation as well as the Euler angles are studied as ways to parameterize rotation matrices. 1.1.1 Vector Algebra

Let us first establish some notation. Let V be a finite-dimensional vector space over the real numbers, R. The elements of V, the vectors, will be denoted by bold, lower case Latin letters, u. A basis of V is a set of vectors {ei } having the property that every vector has a unique representation as a linear combination of basis elements. The basis vectors will be indexed with subscripts. Let V∗ be the dual vector space of V – the space of all linear, real valued functions on V. The elements of V∗ , the covectors, will be denoted by bold, lower case Greek Rigid Body Mechanics: Mathematics, Physics and Applications. William B. Heard Copyright © 2006 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim ISBN: 3-527-40620-4

2

1 Rotations

letters, υ. For each basis {ei } of V there is a basis {i } of V∗ defined by i (e j ) = δij where δij is the Kronecker delta. If u = ui ei , 1 then e and u will denote 

e = ( e1

...

en )

 u1   u =  ...  un

and the representation of a vector u in the basis e is denoted by u = eu The basis e will also be referred to as a frame. Similarly, if υ = υi i , then and υ will denote  1   ..  υ = υ1 · · · υn = .  n and the representation of covector υ in the basis is denoted by υ = υ This notation is also found in [1] which is a good source of material on geometrical aspects of mechanics. The distinction between a vector u = eu and its components u relative to a basis e is of prime importance. The former is invariant under a change of basis and the latter, of course, is not. We shall always reserve the bold-face font for the invariant form and the regular typeface components relative to a basis. A linear operator A on V takes vector x ∈ V to vector A( x) ∈ V and preserves the operations of the vector space

A( ax + by) = aA( x) + bA(y), for all a, b ∈ R and x, y ∈ V Given a linear operator A and a basis {ei } define the matrix representation A = [ Aij ] (row superscript i and column subscript j) by

A(e j ) = Aij ei The action of A on any u = u j e j is then represented as

Au = Aij u j ei 1) The summation convention that repeated indices indicate a sum over their range is used here and throughout the text.


The matrix can be regarded as operating on a row of basis vectors from the right according to

A(u) = u j f j

f j = Aij ei

with

This can be expressed in a matrix form as

f1

f2

f3

=

e2

e1

e3



A11  A2 1 A31

A12 A22 A32

 A12 A23  A33

or in the shorthand notation f = eA Alternatively it can be regarded as operating on a column of components from the left according to

A(u) = vi ei

vi = Aij u j

with

This can be expressed as 

  1 v1 A1  v 2  =  A2 1 v3 A31

A12 A22 A32

 1  u A12 2   A3 u2  3 A3 u3

or v = Au Given a pair of vectors u, v, a scalar or inner product on V assigns a nonnegative, real number u, v which has the following properties: if u = 0 then u, u > 0

u, v = v, u a i ei , b j e j = a i b j ei , e j An inner product may be expressed in the following equivalent ways:

u, v = u · v = (ui ei ) · (v j e j ) = ui vi Gij where the real numbers Gij = ei · e j are components of a symmetric, positive definite matrix G called the metric tensor. The Euclidean inner product is distinguished by Gij = δij . In the Euclidean case u · v = ut v = v t u

3

4

1 Rotations

Given an inner product we can define the length or norm of a vector and the angle between two vectors. The norm of u is u = u, u The angle between u and v is

arccos

u, v uv

Tensors are important objects in rigid body mechanics and we now set down the basics of tensor algebra. We start with the algebraic definition of tensors of rank 2.2 A tensor of rank 2 assigns a real number to a pair of vectors or covectors and is linear in each argument. A covariant tensor T of rank 2 assigns a real number to pairs of vectors T (u, v) ∈ R. The metric tensor is an example of a covariant tensor. A contravariant tensor T of rank 2 assigns a real number to pairs of covectors T (υ, ν) ∈ R. A mixed tensor T of rank 2 assigns a real number to a vector–covector pair T (u, ν) ∈ R. The components of a tensor are its values on basis vectors. Thus, a covariant tensor A has components aij = A(ei , e j ) and a mixed tensor B has components bij = B(ei , j ). Tensors can be formed from tensor products of vectors and covectors. The tensor products are denoted with the symbol ⊗ and are defined by their action on their arguments. Thus we define a covariant tensor υ ⊗ ν by υ ⊗ ν(u, v) = υ(u)ν(v) and define the mixed tensor u ⊗ ν by u ⊗ ν(υ, v) = υ(u)ν(v). It is not the case that every tensor is a tensor product but every tensor is a linear combination of tensor products of basis vectors. For example, T ( µi i , v j e j ) = µi v j T ( i , e j ) Addition and scalar multiplication of the tensors can be defined by their action on vectors

( aT + bS)(u, v) = aT (u, v) + bS(u, v) In the case of mixed tensors, the result of this construction is a new vector space V ⊗ V ∗ . When V has dimension n, V ⊗ V ∗ has dimension n2 . The same construction can be carried through for pairs of covectors and the resulting vector space V ∗ ⊗ V ∗ again has dimension n2 . The subspace 2 V ∗ of V ∗ ⊗ V ∗ consists of skew-symmetric covariant tensors, that is, of covariant tensors Ω which satisfy Ω(u, v) = −Ω(v, u) 2) The development extends to any rank but we will need only rank 2.


Example 1.1 The tensor T = ν ⊗ υ − υ ⊗ ν is skew-symmetric because T (u, v) = ν(u)υ(v) − υ(v)ν(u) and T (v, u) = ν(v)υ(u) − υ(u)ν(v) = − T (u, v).

♦

The members of 2 V ∗ are called 2-forms over V. Covectors, members of V ∗ , are also called 1-forms. There is a product, the wedge product, which produces a 2-form ω ∧ µ from two 1-forms ω and µ. The wedge product is defined by its action on its arguments ω ∧ µ(u, v) = ω(u)µ(v) − ω(v)µ(u) The wedge product is basic to the geometric treatment of Hamiltonian mechanics. Now we follow [1] to establish the connection between rank 2 mixed tensors and linear operators. If A is a linear operator, let TA be the tensor defined by the action TA (v, ν) = ν(A(v)). The components of TA are given by TA ij = TA (e j , i ) = i (A(e j )) = Aij Thus the components of TA are the same as those of A. We will exploit this correspondence, borrowing from the theory of dyadics, to represent linear transformations by

A(uk ek ) = ei Aij j (uk ek ) = ei Aij uk δjk = ei Aij u j

(1.1)

Thus the action of the second-rank tensor depends on its object. If applied to a vector–covector pair it produces a scalar and if applied to a vector it returns another vector. This is reminiscent of the dot and double-dot products of dyadics [2] which are closely related to second-rank tensors. The combinations e and e ⊗ arise frequently in the manipulation of tensors. The product e is the identity matrix because  1  2  e1 e2 e3 = [i (e j )] = I The product e ⊗ is the identity operator because  1 e1 e2 e3 ⊗  2  = e i ⊗ i and for any vector v = v i ei

(ei ⊗ i )(vk ek ) = vk ei δik = v

5

6

1 Rotations

The above representation leads to a succinct representation of linear operators A = eA which is shorthand for A = Aij ei ⊗ i so that

A(v) = eA(ev) = eAv Now consider the effect of a change of basis. Suppose e and e¯ are bases of V linearly related by e = e¯ B Then a vector v has the representations v = ev = e¯ Bv = e¯ v¯ so that the components in the two bases are related by v¯ = Bv The covector relations I = e = e¯ B = e¯ ¯ give ¯ = B Thus a covector µ has the representations µ = µ = µB−1 ¯ = µ¯ ¯ so that the components in the two bases are related by ¯ µ = µB The effect of change of basis on a linear operator follows immediately

A = eA = e¯ BAB−1 ¯ = e¯ A¯ ¯ or

A¯ = BAB−1

Given the linear operators A, B with the matrix representations A, B, the matrix representation of the composite map B ◦ A, A followed by B , is the matrix product BA. The basis vectors transform as e = eBA and the components transform as v = BAv


For any n the vector space R n has the standard basis {ei } where ei is the column of length n having a single nonzero entry, 1 in row i. The inner product in the standard basis is u · v = ut = v u = ui v i where no distinction is made between ui and ui . In R3 there is also a vector product or cross product

( ai ei ) × (bj e j ) = ijk ai bj ek = ( a2 b3 − a3 b2 )e1 + ( a3 b1 − a1 b3 )e2 + ( a1 b2 − a2 b1 )e3 where kij is the permutation symbol  

1 −1  0

kij =

if i j k is an even permutation of 123 if i j k is an odd permutation of 123 otherwise

We will have no further need in this chapter to distinguish between subscripts and superscripts, so subscripts will be used. In this case matrix entries will be denoted by Aij with row index i and column index j. Superscripts will return with a vengeance when generalized coordinates are considered. 1.1.2 Rotation Operators on R 3

A rotation is a linear transformation, R, that fixes the origin, preserves the lengths of vectors, and preserves the orientation of bases. That is,

R : V → V : x → R( x) R(0) = 0 R( x) · R( x) = x · x R(e1 ) · (R(e2 ) × R(e3 )) = e1 · (e2 × e3 ) The length-preserving property becomes, in a matrix form,

R( x) · R( x) = ( x j Rij ei ) · ( xl Rkl ek ) = x j xl Rij Ril = xi xi for all { xi } ∈ R3 which implies that Rij Ril = δjl

or

Rt R = I

This defines an orthogonal matrix. The orientation preserving condition becomes Rk1 kij Ri2 R j3 = 1 or det R = +1 When R and S are orthogonal, then ( RS)t RS = St Rt RS = I. When R is orthogonal R−1 = Rt and ( R−1 )t R−1 = RRt = I. Therefore RS and R−1 are

7

8

1 Rotations

also orthogonal. Clearly I is orthogonal. The product of orthogonal matrices preserves the determinant, that is, if det R = det S = 1 then det RS = det R det S = 1. If det R = 1 then det R−1 = 1 because det RR−1 = det I = 1. It follows from these facts that the orthogonal matrices of determinant 1 form a group.3 The group is called SO(3) – the special orthogonal group of order 3. This group has the additional structure of a three-dimensional manifold and is therefore a Lie group. In the next section we begin to study parameterizations, or coordinates, of SO(3). The basics of the Lie group theory are outlined in Appendix C. In addition to preserving lengths, orthogonal matrices preserve angles because they preserve inner products

( Ru)t Rv = ut Rt Rv = ut v Members of SO(3) also preserve R3 vector products in the sense that A( x × y) = Ax × Ay. To prove this, it is enough to show it for e1 , e2 , e3 , the standard basis of R3 . Let A ∈ SO(3) be presented in terms of its orthonormal columns, A = [c1 c2 c3 ] so that Aei = ci . Then A(ei × e j ) = A(ijk ek ) = ijk Aek = ijk ck = ci × c j = Aei × Ae j Every member of SO(3) fixes not only the origin but actually an entire line. This follows from the structure of eigensystems of rotation matrices. The length preserving property of a rotation requires that eigenvalues have magnitude 1. To show this we must allow for complex eigenvalues and eigenvectors and use the norm x 2 = ( x ∗ )t x where x ∗ is the complex conjugate of x. Then Rx = λx implies that Rx 2 = λ∗ λ( x ∗ )t x = |λ|2 x 2 . In other words Rx = x implies |λ| = ±1. The characteristic polynomial of a rotation matrix is a cubic and one of its roots must be +1 because det R = 1. Thus, any eigenvector corresponding to λ = 1 is fixed and real. The set of all such eigenvectors forms the axis of rotation. 1.1.3 Rotations Specified by Axis and Angle

First consider plane rotations. Represent vectors as complex numbers ( x, y) ↔ x + ıy = ρ exp(ıθ ). Then a counterclockwise rotation by angle φ is simply multiplication by exp(ıφ): z → exp(ıφ)z = ρ exp[ı(θ + φ)]. In rectangular components x + ıy = z → z = eıθ ( x + ıy) = ( x cos θ − y sin θ ) + ı( x sin θ + y cos θ ) 3) A group G is a set equipped with a binary operation such that if a, b ∈ G then ab ∈ G. There is an identity element e such that ea = a for every a ∈ G and for every a ∈ G there is an inverse a−1 such that aa−1 = e.


This shows that rotations by angle θ about the x, y, z axes are represented by 

1  0 0

0 c s

 0 −s  c



c  0 −s

0 1 0

 s 0  c

 and

c  s 0

 −s 0 c 0  0 1

respectively, where c = cos θ, s = sin θ. These matrices have alternative expressions as operators which emphasize the role of the axis of rotation and the rotation angle. To state the alternative forms we first introduce the idea of duality between R3 and so(3), which is the set of all skew-symmetric 3 × 3 matrices, that is matrices which have the property that At = − A. In fact so(3) is the Lie algebra of the Lie group SO(3) (see Appendix C). To each vector v there corresponds a skew-symmetric matrix v, 

  v1 0    v= v2 ←→ v3 v3 −v2

−v3 0 v1

 v2 −v1  = v 0

and to each skew-symmetric matrix M there is a vector M, 

0 M =  u3 − u2

− u3 0 u1

   u2 u1 −u1  ←→  u2  = M 0 u3

In the language of Lie algebra v = ad v (see Section C.3.2). Using the canonical basis {e1 , e2 , e3 } = {i, j, k}, this is expressed in terms of tensor products as 

 u1  u2  ←→ u1i + u2j + u3 k u3 where i = k ⊗ j − j ⊗ k

j = i ⊗ k − k ⊗ i

= j⊗i−i⊗j k

are the invariant forms of the operator. Here we have identified R3 and (R3 )∗ via   u1  u2  ←→ u1 u2 u3 u3 or u ⊗ v ←→ uvt

9

10

1 Rotations

With these preliminaries we can write that the rotations about the coordinate axes are i ⊗ i + c( I − i ⊗ i) + si j ⊗ j + c( I − j ⊗ j) + sj k ⊗ k + c ( I − k ⊗ k ) + sk These expressions can be generalized to an arbitrary axis of rotation determined by the unit vector n. The form of the expressions suggests that one might construct a basis containing n and write immediately that

Rn (θ ) = n ⊗ n + cos θ ( I − n ⊗ n) + sin θ n and we now proceed to justify this. Let P represent the operator relative to the standard basis which projects a vector onto the axis of rotation (Fig. 1.1) P = nnt P has the property that, for any r, Pr is a multiple of n Pr = nnt r = λn

λ = nt r

and P is idempotent P2 = (nnt )(nnt ) = n(nt n)nt = nnt = P I − P then represents the operator which projects a vector onto the plane normal to n because for any r nt ( I − P)r = nt r − nt nnt r = 0 and

( I − P )2 = I − P − P + P2 = I − P . Any rotation of r through an angle θ leaves Pr invariant and Let N = n rotates ( I − P)r by an angle θ in the plane spanned by ( I − P)r, n × r. Thus the rotation is given by r = Rn (θ )r with Rn (θ ) = nnt + cos θ I − nnt + sin θ N It is easy to show that the operators P, I − P, N form a closed system defined by Table 1.1. Here we sketch the proof of one entry as an illustration N 2 r = n × ( n × r ) = (r · n ) n − r = ( P − I )r The relations in Table 1.1 can be used to recast the rotation operator entirely in terms of N, R = I + sin θN + (1 − cos θ ) N 2 (1.2)


Fig. 1.1 Rotation about the axis n by an angle θ. . Table 1.1 Composition of projection operators P = nn t , I − P, and N = n

◦ P I−P N

P P 0 0

I−P 0 I−P N

N 0 N −( I − P )

◦ P N N2

P P 0 0

N 0 N2 −N

N2 0 −N −N2

We close this section with a derivation of the general rotation matrix using general principles of linear algebra. A general maxim in linear algebra is that problems should be worked in the “right” basis, i.e., the basis in which the problem assumes its simplest, most revealing form. Since rotations are given by an axis, n, and an angle, α, the “right” basis in which to study rotations would naturally include n. Accordingly, let b1 = n; choose unit vector b2 ⊥n and b3 = b1 × b2 . In this basis we can write immediately 

1 0 Rn ( α ) =  0 c 0 s

 0 −s  c

where c = cos α and s = sin α and α is reckoned from b2 in the b2 –b3 plane. To work in another basis – the standard basis (e1 , e2 , e3 ), for example – we can use the change of basis relations

R = bRn (α) β = eARn (α) At = eR A is the change of basis matrix such that b = eA

Aij = b j · ei

11

12

1 Rotations

so column j of A contains the components b j of vector b j in the basis e (b j = eb j ). That is A = (b1 b2 b2 ) Thus 

1 R = (b1 b2 b3 ) 0 0

0 c s

 t b1 0 −s  b2t  = b1 b1t + c(b2 b2t + b3 b3t ) + s(b3 b2t − b2 b3t ) c b3t

Now we use the identities b1 = n, I = b1 b1t + b2 b2t + b3 b3t and

b1 = b3 b2t − b2 b3t

to recover R = nnt + c( I − nnt ) + sN

(1.3)

1.1.4 The Cayley Transform

Equation (1.2) can be written in terms of t = tan 12 θ by using the identities sin θ =

2t 1 + t2

Thus R=I+

1 − cos θ =

2t2 1 + t2

2t 2t2 N + N2 1 + t2 1 + t2

This form of R leads to the Cayley transform of R and yet another form of the rotation matrix which plays an important role in numerical calculations. We need the series expansion ∞ 1 = ∑ (−1)n t2n 2 1+t n =0

and the results obtained from Table 1.1, N 2n = (−1)n+1 N 2 N 2n+1 = (−1)n N


These provide R

=

I+2

∞

∑ tn N n

n =1

∞

= ( I + tN ) ∑ tn N n 0

= ( I + tN )( I − tN )−1 = ( I − tN )−1 ( I + tN ) and we have arrived at the Cayley transform. Given a skew-symmetric matrix S, its Cayley transform is cay(S) = ( I − S)−1 ( I + S) The importance of cay(S) is that it is always orthogonal cay(S)cay(S)t

= ( I − S)−1 ( I + S)( I + St )( I − St )−1 = ( I − S)−1 ( I + S)( I − S)( I + S)−1 =

I

The inverse of the Cayley transform is expressed as if St = −S and R = cayS then S = ( R − I )( R + I )−1 Orthogonality of R insures that S is skew-symmetric. 1.1.5 Reflections

Reflections are more basic than rotations in the sense that every rotation can be obtained by composing two reflections. First consider reflections in a plane through a line containing the complex number exp(ıφ). Then z = exp(2ıφ) x ∗

(1.4)

is the reflection of z through the line. If we apply two reflections, say through the lines containing exp(ıφ) and exp(ıψ), then z = exp(2ıψ)[exp(2ıφ)z∗ ]∗ = exp[2ı(ψ − φ)]z which is a rotation by an angle 2(ψ − φ). The representation of any rotation as a composition of reflections generalizes to SO(3). We first need the matrix representation for reflections in planes

13

14

1 Rotations

in R3 and the approach used to derive (1.3) also yields the matrix representation for a reflection. The appropriate basis includes a normal n to the plane of reflection b1 = n b2 ⊥n and b3 = b1 × b2 Then the reflection is represented by (M for mirror)  −1 0 0 M= 0 1 0  0 0 1 

and use of the change of basis matrix to the standard basis gives 

 t  −1 0 0 b1 Mn = (b1 b2 b3 )  0 1 0   b2t  = −b1 b1t + b2 b2t + b3 b3t 0 0 1 b3t and using b1 = n and I = b1 b1t + b2 b2t + b3 b3t we obtain Mn = I − 2nnt The invariant form of the reflection operator is

Mn = I − 2n ⊗ n Now one can show that every rotation is the product of two reflections. Let the rotation be Rn (α). Choose unit vectors q, p such that p·n = 0 q·n = 0

and

1 p · q = cos α 2

The product of reflections in q and p is 1 A = M p Mq = I − 2p ⊗ p − 2q ⊗ q + 4 cos αp ⊗ q 2 It is easily verified that

An = n n · Ap = n · Aq = 0, 1 p · Ap = −1 + 2 cos2 α = cos α 2 and q · Aq = cos α which shows that

M p M q = Rn ( α )


1.1.6 Euler Angles

The specification of a rotation by axis and angle is convenient and subject to direct geometrical interpretation. It is, however, inconvenient as a basis for rigid body dynamics because of redundancy. Any rotation matrix can be specified by three parameters (there are nine entries in the orthogonal matrix and six constraints imposed on the inner products of the columns). The redundancy arises because all three components of the axis are used as parameters even though only two are independent. The correspondence is also double-valued in that n, α yield the same rotation as −n, −α. We now consider the Euler angles which provide a three-parameter specification of a rotation.

Fig. 1.2 Definition of the Euler angles φ, θ, ψ.

The Euler angle parameterization of a rotation is the composite of three intermediate rotations. The rotations are defined relative to the standard basis e, ¯ and the effect of the intermediate rotations on this basis will be denoted by primes (Fig. 1.2). 1. Rotate about e¯3 by an angle φ taking e¯1 ,e¯2 to e1 ,e2 and leaving e3 = e¯3 . 2. Rotate about e1 by an angle θ taking e2 ,e3 to e2 ,e3 and leaving e1 = e1 . 3. Rotate about e3 by an angle ψ taking e1 ,e2 to e1 ,e2 and leaving e = e3 . The angles φ, θ, ψ are sometimes called the angles of precession, nutation, and spin, respectively. The intermediate rotations have the following matrix representations:   cos φ − sin φ 0 e = e¯  sin φ ¯ e3 ( φ ) cos φ 0  = eR 0 0 1

15

16

1 Rotations



1 0 e = e  0 cos θ 0 sin θ and



cos ψ e = e  sin ψ 0

 0 − sin θ  = e Re1 (θ ) cos θ

 − sin ψ 0 cos ψ 0  = e Re3 (ψ) 0 1

The composite rotation is e = e¯ Re3 (φ) Re1 (θ ) Re3 (ψ) = e¯ R(φ, θ, ψ) The operator is represented by the fixed basis e¯ by using

R = e¯ = e¯ R(φ, θ, ψ). ¯ The matrix products yield R(φ, θ, ψ ) =  cos φ cos ψ − sin φ cos θ sin ψ sin φ cos ψ + cos φ cos θ sin ψ sin θ sin ψ

− cos φ sin ψ − sin φ cos θ cos ψ − sin φ sin ψ + cos φ cos θ cos ψ sin θ cos ψ

(1.5)  sin φ sin θ − cos φ sin θ  cos θ

Given a rotation parameterized by the rotation angle α and rotation axis n, the Euler angle parameterization can be found as follows: cos θ = Rn (α)(e¯3 ) · e¯3 cos φ = (e¯3 × Rn (α)(e¯3 )) · e¯1 / sin θ cos ψ = (e¯3 × Rn (α)(e¯3 )) · Rn (α)(e¯2 )/ sin θ This conversion is not possible if sin θ = 0. In that case the Euler angle parameterizations is not unique – φ and ψ cannot be separated. There are many ways to represent a rotation as successive rotations about intermediate axes. For example, in aeronautics and astronautics one uses a parameterization dubbed yaw, pitch, and roll defined by the following: 1. Rotate about e¯3 by the yaw angle ψ. 2. Rotate about e2 by the pitch angle θ. 3. Rotate about e1 by the roll angle φ. In fact [3] there are 12 possible rotation sequences each defining a variant of the Euler angles. These are labeled by n1 − n2 − n3 corresponding to the rotation e = e¯ Re¯n1 Re n Re n 2

3

1.2 Quaternions

The Euler angle parameterization discussed here would be labeled 3 − 1 − 3 while yaw, pitch, and roll would be 3 − 2 − 1. There are 27 possible n1 − n2 − n3 combinations but it is necessary that n1 = n2 and n2 = n3 and 12 combinations survive 1−2−1 1−3−1 1−2−3 1−3−2

2−1−2 2−3−2 2−3−1 2−1−3

3−1−3 3−2−3 3−1−2 3−2−1

1.2 Quaternions

This section introduces a new number systems called quaternions which are intimately connected to three-dimensional rotations (we have already seen that complex numbers are intimately connected to plane rotations). As a prelude, the construction of more familiar number systems will be reviewed. Quaternions will then seem quite natural. A good reference for these matters is [4]. We begin with the integers, Z. In Z we can add, multiply, and subtract but there is no division. In other words, there is no multiplicative inverse in Z of any integers other than ±1. To obtain multiplicative inverses we need rational numbers, traditionally denoted by Q (for quotients). They are constructed as equivalence classes of ordered pairs of integers (n, m), m = 0, based on the equivalence relation

(n, m) ∼ ( p, q), if and only if nq = mp The equivalence classes are the sets

[(n, m)] = {( p, q)|( p, q) ∼ (n, m)} which is merely an abstract way to declare that rational numbers should be reduced to lowest terms. Having made the distinction between the pairs and the equivalence classes we will now dispense with the square brackets for notational simplicity. The operations of addition and multiplication of rational numbers are defined by

(n, m) + ( p, q) = (nq + mp, mq) (n, m)( p, q) = (np, mq) Now every pair (n, m) has an additive inverse (−n, m)

(n, m) + (−n, m) = (0, m) and, if n = 0 a multiplicative inverse (m, n)

(n, m)(m, n) = (1, 1)

17

18

1 Rotations

This is quite transparent when translated to grade school notation

(n, m) ↔

n m

The rational numbers are incomplete in the sense that there is no rational number which represents the hypotenuse of a right triangle having unit legs. The real numbers R are constructed from pairs of sets of rational numbers, called Dedekind cuts, ( D1 , D2 ), D1 and D2 ⊂ Q which have the following four properties: each x ∈ D1 is less than every y ∈ D2 , D1 has no maximum element, x ∈ D1 and y < x then y ∈ D1 and x ∈ D2 and y > x then y ∈ D2 For example, let D1 = { r ∈ Q | r 2 < 2 } and let D2 be the set complement of D1 in Q . The Dedekind cut ( D1, D2 ) √ represents the real number 2. The real numbers are incomplete in the sense that there is no solution in R to the polynomial equation x2 + 1 = 0 We obtain an algebraically complete number system, the complex numbers C, from ordered pairs of real numbers z = ( x, y)

x and y ∈ R

with addition and multiplication defined as

( x, y) + (u, v) = ( x + u, y + v) and

( x, y)(u, v) = ( xu − yv, xv + yu) There is also a new operation, multiplication of a complex number by a real number: a( x, y) = ( a, 0)( x, y) = ( ax, ay) This appears more familiar in high school notation

( x, y) ↔ x + ıy

1.2 Quaternions

The Irish mathematician Rowan Hamilton struggled in vain to extend complex numbers to three dimensions. Eventually he realized that it is necessary to go to four dimensions and he invented a new number system called the quaternions. We will introduce quaternions, H for Hamilton, as ordered pairs of complex numbers. Although Hamilton did not use the ordered pair construction for quaternions, he was the inventor of the pair construction for complex numbers. A quaternion, then, is an ordered pair of complex numbers q = ( z1 , z2 )

z1 and z2 ∈ C

with addition and multiplication defined by

( z1 , z2 ) + ( w1 , w2 ) = ( z1 + w1 , z2 + w2 ) (z1 , z2 )(w1 , w2 ) = (z1 w1 − z2 w2∗ , z1 w2 + z2 w1∗ ) and a(z, w) = ( a, 0)(z, w) = ( az, aw)

a∈R

It turns out that multiplication is not commutative. That is, in general rq = qr We shall denote quaternions with sans serif typeface as in the last equation. This construction by pairs ties in nicely with the constructions of the rational, real, and complex numbers but is not the traditional approach. If we single out three special pairs and attach Hamilton’s notation to them i = (ı, 0)

j = (0, 1)

k = (0, ı)

and identify

( a, 0) ↔ a

a∈R

then we find

(q0 + ıq1 , q2 + ıq3 ) = q0 + q1 i + q2 j + q3 k

qi ∈ R

which is the form Hamilton used to express quaternions. This form makes it quite clear that quaternions are a four-dimensional generalization of complex numbers. The quaternions i, j,k satisfy the following relations: i2 = j2 = k2 = ijk = −1

(1.6)

These are the rules Hamilton inscribed on Broome Bridge in Dublin on October 16, 1843 [5].

19

20

1 Rotations

In the language of abstract algebra, the quaternions form a noncommutative, normed division algebra over R. The eight-dimensional octonians O [5] can be constructed from pairs of quaternions but there the chain ends. The only normed division algebras over R are R, C, H, and O. 1.2.1 Quaternion Algebra

This section summarizes the essentials of the algebra of quaternions. First we express the basic operations in the i, j,k form. Addition becomes

( q0 + q1 i + q2 j + q3 k ) + ( p 0 + p 1 i + p 2 j + p 3 k ) = ( q0 + p 0 ) + ( q1 + p 1 ) i + ( q2 + p 2 ) j + ( q3 + p 3 ) k and multiplication becomes

(q0 + q1 i + q2 j + q3 k)( p0 + p1 i + p2 j + p3 k) = q0 p0 − q1 p1 − q2 p2 − q3 p3 + ( q0 p 1 + q1 p 0 + q2 p 3 − q3 p 2 ) i + ( q0 p 2 + q2 p 0 + q3 p 1 − q1 p 3 ) j + ( q0 p 3 + q3 p 0 + q1 p 2 − q2 p 1 ) k There are many patterns in this last expression and they will become clear by the end of the next section. Given q = q0 + q1 i + q2 j + q3 k, q0 is called the scalar part of q S ( q ) = q0 and q1 i + q2 j + q3 k is called the vector part of q V ( q ) = q1 i + q2 j + q3 k The conjugate of q is q¯ = q0 − q1 i − q2 j − q3 k The square of the norm of q is

|q|2 = q¯q = q20 + q21 + q22 + q23 ∈ R where we have taken the liberty of identifying a real number a and the quaternion q with S(q) = a, V (q) = 0. The inverse of q is, when |q| = 0, q−1 =

1 q¯ |q|2

Clearly qq−1 = 1 + 0i + 0j + 0k. Note also that qp = p¯ q¯ and (qp)−1 = p−1 q−1 .

1.2 Quaternions

1.2.2 Quaternions as Scalar–Vector Pairs

At this point we have two ways to express quaternions – the complex number pair form and the i, j, k form. There is a third way which is convenient for anyone accustomed to vector algebra. This third way represents quaternions as scalar–vector pairs q0 + q1 i + q2 j + q3 k ↔ ( q0 , v )

with

v = q1 e1 + q2 e2 + q3 e3

In this way, the quaternion algebra is expressed as

( q0 , u ) + ( p 0 , v ) = ( q0 + p 0 , u + v ) (q0 , u)( p0, v) = (q0 p0 − u · v, q0 v + p0 u + u × v) ( q0 , u ) = ( q0 , −u ) |(q0 , u)|2 = q20 + u · u 1 ( q0 , u ) − 1 = 2 ( q0 , −u ) q0 + u · u A perusal of the multiplication formula now reveals the patterns to be elements of the scalar and vector products. A pure quaternion (0, u) is one with zero scalar part and a unit quaternion has unit norm q¯q = 1. The product of two pure quaternions encapsulates the scalar and vector products

(0, u)(0, v) = (−u · v, u × v) 1.2.3 Quaternions as Matrices

Complex numbers can be represented as 2 × 2 matrices with real entries. Let

0 −1 and note that J 2 = − I. Then Z = aI + bJ and W = cI + dJ J = 1 0 satisfy the rules for addition and multiplication of complex numbers Z + W = ( a + c) I + (b + d) J ZW = ( aI + bJ )(cI + dJ ) = ( ac − bd) I + ( ad + bc) J Similarly, quaternions can be represented as 2 × 2 matrices of complex numbers. If z1 = q0 + ıq1 z2 = q2 + ıq3 w1 = r0 + ır1

w2 = r2 + ır3

21

22

1 Rotations

then the correspondences

z1 Q1 = −z2∗ and

Q2 =

w1 −w2∗

z2 z1∗

w2 w1∗

↔ q0 + q1 i + q2 j + q3 k

↔ r0 + r1 i + r2 j + r3 k

are preserved by the matrix product Q1 Q2 ↔ (q0 + q1 i + q2 j + q3 k)(r0 + r1 i + r2 j + r3 k) The quaternion product is linear in the components of each factor. This allows us to express the quaternion operations in terms of linear operations on R4 . If we associate   q0  q1   q = q0 + q1 i + q2 j + q3 k ↔ q =   q2  q3 and



 p0  p1   p = p0 + p1 i + p2 j + p3 k ↔ p =   p2  p3

then left multiplication by q is represented by  q0 − q1 − q2  q1 q0 − q3 L (q) =   q2 q3 q0 q3 − q2 q1

 − q3 q2   − q1  q0

(1.7)

that is, qp ↔ L(q) p This is expressed explicitly in a component form      q0 − q1 − q2 − q3 p0 q0 p 0 − q1 p 1 − q2 p 2 − q3 p 3  q1   p 1   q0 p 1 + q1 p 0 + q2 p 3 − q3 p 2 q − q q 0 3 2  . =  q2 q3 q0 − q1   p 2   q0 p 2 + q2 p 0 + q3 p 1 − q1 p 3 q3 − q2 q1 q0 p3 q0 p 3 + q3 p 0 + q1 p 2 − q2 p 1 Right multiplication by q is represented by  q0 − q1 − q2  q1 q0 q3 R (q) =   q2 − q3 q0 q3 q2 − q1

 − q3 − q2   q1  q0

   

(1.8)

1.2 Quaternions

and pq ↔ R(q) p These are the representations of the left and right translations Lq and Rq on the Lie group S3 ∼ = H1 (Appendix C). Since Lp Lp = Lpq , it follows that L(p) L(q) = L(pq) There is the analogous result for right multiplication R(p) R(q) = R(qp) L(p) and R(p) have the following properties: L (q) p

= − L(p)q if qt p = 0

R (q) p

= − R(p)q if qt p = 0

L(q¯ )

=

L (q) t

L(q¯ ) L(q)

= ( qt q) I

R(q¯ ) R(q)

= ( qt q) I

If q ∈ H1 then R(q) and L(q) are orthogonal matrices. These relations are discussed and applied in [6, 7]. This correspondence is a generalization of the − → v ↔ M correspondence of R3 and so(3). 1.2.4 Rotations via Unit Quaternions

Quaternions are useful in rigid body dynamics because they provide succinct expressions for quantities which can be rather unwieldy in matrix form. A good example is the composition of rotations which is a simple quaternion product. Rotations are effected in the quaternions by means of unit quaternions. We denote the set of all unit quaternions by H1 . Let r = (0, v) be a pure quaternion and q = (q0 , q) be a unit quaternion. Let r = qrq−1 = qrq¯ Now we compute r = (0, [q20 − q · q]v + 2v · qq + 2q0 q × v) and observe that if an angle θ and a unit vector n are defined by 1 q0 = cos θ 2

and

1 q ≡ (q1 , q2 , q3 ) = sin θ n 2

23

24

1 Rotations

then

r = (0, [1 − cos θ ]nn · v + cos θv + sin θn × v)

which reproduces the effect of Rn (θ ) on the vector part of r. The qi arise most naturally in the context of quaternions. They were, however, introduced much earlier by Euler and Rodrigues. For that reason they are sometimes called the Euler parameters or Euler–Rodrigues parameters. Now it is natural to ask how the rotation matrix and the quaternion are related. First, let us express the rotation matrix R = Rn (θ ) explicitly in terms of the components of n and the rotation angle via c = cos θ, s = sin θ, 

(1 − c)n21 + c  R= (1 − c)n2 n1 + sn3 (1 − c)n3 n1 − sn2

(1 − c)n1 n2 − sn3 (1 − c)n22 + c (1 − c)n3 n2 + sn1

 (1 − c)n1 n3 + sn2 (1 − c)n2 n3 − sn1  (1 − c)n23 + c

(1.9)

Then, to express R in terms of the quaternion components qi , we use the identities c = q20 − q21 − q22 − q23

1 − c = 2(1 − q20 )

(1 − c)ni n j = 2qi q j

and sn = 2q0 q. Thus 

q20 + q21 − q22 − q23 R =  2 ( q2 q1 + q0 q3 ) 2 ( q3 q1 − q0 q2 )

2 ( q1 q2 − q0 q3 ) 2 q0 − q21 + q22 − q23 2 ( q3 q2 + q0 q1 )

 2 ( q1 q3 + q0 q2 ) 2 ( q2 q3 − q0 q1 )  2 q0 − q21 − q22 + q23

(1.10)

Note that quaternions q and −q correspond the same rotation matrix. Apart from its intrinsic interest, this expression is of practical importance because it is a purely algebraic version of the rotation matrix. No transcendental functions are involved. This is relevant when computing requirements are an issue. The rotation defined by quaternion (q0 , q) can also be expressed in the axisangle form as 1 Rn (θ ) = 2q ⊗ q + (q20 − q · q) I + 2q0 q q = sin θ n 2 The process can be reversed. That is, given a rotation matrix, R, one can construct the components of the associated quaternion. Let τ = Trace( R) – the sum of the diagonal elements of R. From (1.9) we find τ = 1 + 2 cos θ

or

cos θ =

1 ( τ − 1) 2

(1.11)

1.2 Quaternions

and

1√ 1 τ + 1. (1.12) q0 = cos θ = 2 2 The vector part of the quaternion is found from the off-diagonal elements of R q1 =

1 ( R32 − R23 ) 4q0

(1.13)

q2 =

1 ( R − R31 ) 4q0 13

(1.14)

q3 =

1 ( R − R12 ) 4q0 21

(1.15)

This may be expressed in matrix notation as

(q) R − Rt = 4q0 V Quaternions also handle reflections concisely. To reflect through the plane with unit normal n we have

(0, n)(0, −r )(0, −n) = (0, r − 2nn · r ) Conjugation applied to a pure quaternion is the same as negation. Therefore the reflection operation is expressed in a slightly simpler form as

(0, n)(0, r )(0, n) = (0, r − 2nn · r ) Let two reflections be defined by the unit vectors n and m. The rotation determined by the composition of the two reflections is simply

(0, n)(0, m) = (−n · m, −n × m) = −(cos θ, sin θ

n×m ) n × m

where θ is the angle between n and m. Therefore the angle of rotation is twice the angle between the unit vectors and the axis of rotation is parallel to n × m (recall that rotations defined by quaternions q and −q are the same). 1.2.5 Composition of Rotations

The composition of two rotations is itself a rotation because the product of two elements of SO(3) is itself in SO(3). Now we wish to inquire about the rotation axis and rotation angle for the composite of two rotations. Consider two rotations with axes specified by unit vectors a, b and angles by α, β. We wish to find the axis and angle for the rotation composed of Ra (α) followed by Rb ( β). Construct the spherical triangle abc on the unit sphere

25

26

1 Rotations

Fig. 1.3 Defining figure for the Rodrigues spherical triangle.

Fig. 1.4 A sequence of finite rotations taking point a to a , then to a and then back to a illustrating the Rodrigues triangle.

shown in Fig. 1.3. The angles a and b are half the rotation angles and the orientation is such that rotation about a would bring arc ac toward arc ab. We will refer to this triangle as the Rodrigues triangle after Olinde Rodrigues who discovered the construction [8].

1.2 Quaternions

Fig. 1.5√ A sequence of finite rotations of a cube: 180◦ about n = [0 1/2 3/2] followed by 180◦ about k = [0 0 1]. The composite rotation is about i = [1 0 0] by 60◦ as the shown by the Rodrigues spherical triangle displayed in the last panel.

The rotation axis and angle for the composite rotation can be obtained by an elegant geometrical argument [9–11]. Rotate arc ac about a by angle α taking point c to c . Then rotate arc bc about axis b by angle β. This returns c to c thus showing that c is fixed under the composite rotation and must lie on the axis of rotation. In more formal terms, consider the reflections Ma , Mb , and Mc (using the polar triangle notation established in Appendix A). We know that reflections preserve lengths of vectors and therefore reflections map the unit sphere into itself. The relation between reflections and rotations provides

Ra ( α ) = M c M b Rb ( β ) = M a M c and therefore

Rb ( β)Ra (α) = Ma Mc Mc Mb = Ma Mb = R−c (2π − γ) = Rc ( γ ) It is instructive to use quaternions to calculate the composite rotation angle. Let the rotations be represented by the unit quaternions p = (cos 12 α, sin 12 αa) and q = (cos 12 β, sin 12 βb) and let a · b = cos φ . The effect of rotation 1 followed by rotation 2 is r = q(prp−1 )q−1 = (qp)r(qp)−1

27

28

1 Rotations

so that the composition is obtained from the product r = qp 1 1 1 1 qp = (1.16) cos α cos β − sin α sin β cos φ, 2 2 2 2 1 1 1 1 1 1 cos α sin β b + cos β sin α a + sin α sin β a × b 2 2 2 2 2 2 Thus

1 1 1 1 1 cos γ = S(qp) = cos α cos β − sin α sin β cos φ 2 2 2 2 2 Now, referring to Appendix A, Eq. (A.15), we find that 1 1 1 1 cos C = − cos α cos β + sin α sin β cos φ 2 2 2 2

Therefore or

1 cos γ = − cos C 2

1 C=π− γ 2 and the construction is validated. With γ in hand, the axis of the composite rotation is simply 1 n = V (qp)/ sin γ 2 The Rodrigues triangle construction may also be given by the following equivalent description. Refer to the spherical triangle constructed in Fig. 1.4. Given a spherical triangle with vertices p, q, and r and vertex angles α/2, β/2 and γ/2 successive rotations about p by angle α, q by β, and r by γ yield the identity transformation.

Rr (γ)Rq ( β)Rp (α) = Mq Mp Mp Mr Mr Mq =

I

Example 1.2 An illustration of the composition of rotations of a cube and the associated Rodrigues spherical triangle is shown in Fig. 1.5. The analytic expression of that composition is

Rk (π )Rn (π ) = R−i (2π − π/3) = Ri (π/3) Note the orientation of the Rodrigues triangle. ♦ We close this section with an application of the Rodrigues triangle. Let a, b, and b = Ra (α)b be points on the unit sphere. Construct the spherical triangles shown in Fig. 1.6. If each triangle is interpreted as a Rodrigues triangle for composition of rotations, we obtain the relation

Ra (α)Rb ( β) = Rb ( β)Ra (α) = Rc (γ)

1.3 Complex Numbers

Fig. 1.6 Transposition of rotations about a, b, and b = Ra ( α)b.

This is referred to as Rodrigues’ transposition of rotations [3, 12] or rotation reversal [13]. Example 1.3 This relation can be used to clarify the order of factors occurring in the operator and matrix versions of the Euler angle parameterization. In operator notation, R = Re (ψ)Re (θ )Re3 (φ) 3

1

which can be subjected to transpositions to obtain

R = Re (θ )Re (ψ)Re3 (φ) = Re (θ )Re3 (φ)Re3 (ψ) = Re3 (φ)Re1 (θ )Re3 (ψ) 1

3

1

This leads immediately, in the notation of Section 1.1.6, to the representation

R = eRe3 (φ) Re1 (θ ) Re3 (ψ). ♦ 1.3 Complex Numbers

In this section we describe rotations parameterized by Möbius transformations of the complex plane. The development begins by recasting the representation of quaternions as 2 × 2 matrices with complex entries. Section 1.2.3 examined a mapping from the quaternion components to the complex entries in the matrices. Here we explore a different mapping which will lead to interesting connections between rotations and complex functions.4 4) There are actually 24 such mappings which are isomorphisms.

29

30

1 Rotations

1.3.1 Cayley–Klein Parameters

Define the following matrices.5

J1 =

0 ı ı 0

J2 =

−1 0

0 1

J3 =

ı 0

0 −ı

These matrices obey the quaternion multiplication rules (1.6), namely J12 = J22 = J32 = J1 J2 J3 = − I, and the quaternions are mapped to the set of real, linear combinations of I, J1 , J2 , J3 ,

q0 + ıq3 −q2 + ıq1 q0 + q1 i + q2 j + q3 k ↔ q0 I + q1 J1 + q2 J2 + q3 J3 = q2 + ıq1 q0 − ıq3 It is easy to show that if p, q ∈ H are mapped to P, Q, then p + q and pq are mapped to P + Q and PQ. Conjugation in the quaternions corresponds to taking the adjoint of the associated matrix q¯ ↔ Q† = ( Qt )∗ In other words, H is isomorphic to the algebra over R of complex 2 × 2 matrices of the form

u v −v ∗ u∗ Now if

Q=

q0 + ıq3 q2 + ıq1

−q2 + ıq1 q0 − ıq3

,

q = q0 + q1 i + q2 j + q3 k

then det Q = q20 + q21 + q22 + q23 = |q|2 which shows that the subalgebra of unit quaternions, H1 , is isomorphic to the subgroup of GL(2, C ) in which each element satisfies QQ† = I and det Q = 1. This group is called SU (2), the group of special unitary matrices of order 2. Given the correspondence between SU (2) and H1 it follows that rotations can be represented in SU (2) as follows. Let X be the matrix corresponding to the pure quaternion r = (0, r ), r = ( x y x ) and let q ∈ H1 . Then

−y + ıx ız r↔ =X y + ıx −ız and qrq¯ ↔ QXQ† 5) These matrices are related to the Pauli matrices σi of quantum physics by J1 = ıσ1 , J2 = − ıσ2 , and J3 = ıσ3 .

1.3 Complex Numbers

In the context of rigid body mechanics the elements of the matrices in SU (2) are called the Cayley–Klein parameters. These parameters can be expressed in terms of the Euler angles. From Eqs. (1.5) and (1.12–1.15) we obtain 1 1 4q20 = τ + 1 = (1 + cos θ )(1 + cos(φ + ψ)) = 4 cos2 θ cos2 (φ + ψ) 2 2 and 4q0 q1

= sin θ (cos φ + cos ψ)

1 1 1 1 = sin θ cos (φ + ψ) + (φ − ψ) + cos (φ + ψ) − (φ − ψ) 2 2 2 2 1 1 1 1 = 4 cos θ sin θ cos (φ + ψ) cos (φ − ψ) 2 2 2 2

Thus q0 q1

1 1 = cos θ cos (φ + ψ) 2 2 1 1 = sin θ cos (φ − ψ) 2 2

(1.17) (1.18)

and similarly q2 q3

1 1 = sin θ sin (φ − ψ) 2 2 1 1 = cos θ sin (φ + ψ) 2 2

(1.19) (1.20)

With the identifications α = q0 + ıq3 , β = −q2 + ıq1 , γ = − β∗ , δ = α∗ the SU (2) representation of the rotation matrix becomes

1 1 α β cos 12 θe 2 ı(φ+ψ) ı sin 12 θe 2 ı(φ−ψ) = 1 1 γ δ ı sin 12 θe− 2 ı(φ−ψ) cos 12 θe− 2 ı(φ+ψ) 1.3.2 Rotations and the Complex Plane

We have considered rotations in the SO(3), H1 , and SU (2) settings. The last setting we wish to consider is the complex plane with point at infinity. This is the simplest of the compact Riemann surfaces. the so-called Riemann sphere, C, There is a one–one map, called the stereographic projection, from the unit sphere Imagine the complex plane intersecting a unit sphere in its equator. To to C. map the point r = (ξ η ζ ) on the sphere to the complex plane, one draws a

31

32

1 Rotations

Fig. 1.7 Stereographic projection from the unit sphere to the complex plane.

line from the north pole of the sphere, say k, through the point r to the place z where the line intersects the complex plane (Fig. 1.7) z = St(ξ, η, ζ ) =

ξ + ıη = x + ıy 1−ζ

(1.21)

Note that points on the equator (ζ = 0) are fixed points of the map, k (ζ = 1) is mapped to the point at infinity, and −k (ζ = −1) is mapped to the origin. The upper hemisphere is mapped outside the unit circle and the lower hemisphere is mapped inside the unit circle. The inverse map is

2x 2y x 2 + y2 − 1 r = St−1 ( x + ıy) = , , (1.22) x 2 + y2 + 1 x 2 + y2 + 1 x 2 + y2 + 1 A rotation of R3 induces a single-valued map of the unit sphere into itself and also induces, via the stereographic projection, a map of the complex plane into itself with the same properties. The images of the poles of the rotation will be fixed points of the map. The orbits of the rotation on the sphere are circles in planes normal to the axis of rotation. These circles are mapped to circles in the complex plane (Fig. 1.8). Note that if the circles are traversed clockwise (seen from the exterior of the sphere) their images will be traversed counterclockwise. We now sketch the argument that the transformation z → z induced by a rotation must take the form [14] z − n+ z − n+ =C z − n− z − n− where z = f (z) is the image of z induced by the rotation and we have denoted by n+ and n− the images of the poles of the rotation, i.e., the intersections of

1.3 Complex Numbers

Fig. 1.8 Rotation of the sphere about n induces, via stereographic projection of the orbits, a map of the complex plane with the fixed point n .

the axis of rotation n with the sphere. The details of the argument can be found in the beautifully rendered [11]. We begin by examining the transformation induced by a reflection since any rotation is composed of two reflections. The simplest case is reflection in the equatorial plane. If z = St( x), then St(Mk ( x)) = 1/z∗ This transformation is inversion in the unit circle. It can be shown that stereographic projection maps circles on the sphere to circles in the plane and that it preserves angles of intersection (angles between tangents at the point of intersection). Inversion in the unit circle generalizes to inversion in any circle, say radius R and center a, and the mapping is InvR,a , InvR,a (z) = a +

R2 z∗ − a∗

Inv also preserves circles and angles. Now consider a rotation composed of the reflections M a and Mb . The vectors a and b are poles of great circles, say Ca and Cb . St(Ca ) and St(Cb ) are circles in the complex plane and they intersect at points n+ and n− which are the stereographic images of the axis of rotation and are the fixed points of the induced transformation. Thus the induced transformation is the composition of two inversions and is clearly of the form az + b z = cz + d Functions of this form with ad − bc = 0 are called Möbius transformations. For each pair of complex triplets q, r, s and q , r , s there is a unique Möbius

33

34

1 Rotations

transformation taking one to the other z − q s − r z−q s−r = z −r s −q z−r s−q If we choose q = n+ , r = n− , and s = ∞ z − n+ z − n+ z − n+ = ∞ z − n− z∞ − n− z − n− By (1.21) n+ =

n1 + ın2 1 − n3

n− = −

n1 + ın2 1 + n3

The image of the point at infinity in the complex plane, z∞ , is the stereographic projection of the image of k under the rotation which is, in the standard basis, (1 − cos χ)nnt + cos χI + sin χ n k Rn (χ)(k) =   2(n1 n3 sin2 21 χ + n2 sin 12 χ cos 12 χ)   2 1 1 1  =   2(n2 n3 sin 2 χ − n1 sin 2 χ cos 2 χ)  2n23 sin2 12 χ − sin2 21 χ + cos2 21 χ   2 ( q1 q3 + q0 q2 ) =  2 ( q2 q3 − q0 q1 )  2(q20 + q23 ) − 1

Stereographic projection yields z∞

= ı

(n1 + ın2 )(cos 12 χ + ı sin 12 χ) (1 − n23 ) sin2 12 χ

from which it follows that C=

z∞ − n+ = e−ıχ z∞ − n−

Now it is just a matter of algebra to show that z = f (z) =

αz + β − β∗ z + α∗

where α = q0 + ıq3

β = −q2 + ıq1

(1.23)

are the Cayley–Klein parameters. The form of this function is not unique because the numerator and denominator can be multiplied by a common factor

1.4 Summary

without affecting its value. In other words, there is an equivalence class of functions which correspond to each rotation of the sphere. We have chosen, as is always possible if αδ − βγ = 0, the function satisfying z = f (z) =

αz + β − β∗ z + α∗

αα∗ + ββ∗ = 1

These equivalence classes of functions are called normalized Möbius transformations, Mo (C ), and there is a one–one correspondence between them and matrices in SU (2)

uz + v u v ↔ −v ∗ u∗ −v ∗ z + u∗ The correspondence yields an isomorphism. If A, B ∈ SU (2) and f , g ∈ Mo (C ) with A ↔ f , B ↔ g, then AB ↔ f ◦ g and A−1 ↔ f −1 .

1.4 Summary

Table 1.2 Summary of rotation parameterizations. Object H1

Typical element

( q0 , q ) − q2 + ıq1 q0 − ıq3

Group operation

Representation of rotation

Quaternion multiply

qrq¯

Matrix multiply

AXA †

SU (2)

q0 + ıq3 q2 + ıq1

Mo (C )

( q0 + ıq3 ) z − q2 + ıq1 ( q2 + ıq1 ) z + q0 − ıq3

Composition

f ( z)

SO (3)

2qq t + ( q20 − q · q ) I + 2q0 q

Matrix multiply

Rv

The aspects of the theory of rotations in R3 covered in this chapter are summarized in Table 1.2 . The first three objects are isomorphic but they each bear a 2–1 relationship with SO(3). This state of affairs is sometimes expressed as “SU (2) is a double cover for SO(3).” This has profound implications in particle physics where the duplicity represents spin. We close this chapter with comment on the topology of the systems used to represent rotations. H1 , and hence SU (2), is homeomorphic to the threedimensional sphere S3 = [ x ∈ R 4 | x · x = 1 ] On the other hand, SO(3) is homeomorphic to S3 with antipodal points identified. This is the projective space RP3 – the space of all lines through the origin in R4 . The former is simply connected (any loop can be contracted to a point)

35

36

1 Rotations

but the latter is not. In H1 or SU (2) it is possible, to “unwind” the loops which project to noncontractible loops in SO(3). For example, an arc from a point on S3 to its antipode projects to a closed loop in SO(3). This can actually be visualized with strings attached to a rotated body and is very nicely illustrated in [10]. It is also related to the “waiter’s trick” described in [15]. If one holds a tray on one’s outstretched, upturned hand one can rotate the tray by 360◦ using only motion of the wrist and the arm. The “trick” is that in doing so the elbow winds up pointed skyward. A further rotation by 360◦ using only motions of the wrist and the arm restores the elbow to its original position.

1.5 Exercises

1.5 Exercises

Exercise 1.1 Find the axis of rotation and rotation angle for the following matrices:     0 0 1 1 0 0 (a)  1 0 0  (b)  0 −1 0  0 1 0 0 0 −1 Exercise 1.2 Show that a × b = [a , b] where [ A, B] = AB − BA is the commutator of A and B. = A Exercise 1.3 Show that if A ∈ SO(3) and x ∈ R3 then Ax x A −1 . Exercise 1.4 Show that the reflection operator Mn has the following properties: • Mn fixes the plane having normal n and passing through the origin, • Mn is involutory, that is Mn ◦ Mn = Mn , • Mn preserves norms, that is Mn (r) = r. Exercise 1.5 Show that the quaternion multiplication rules in the form i2 = j2 = k2 = ijk = −1 imply ij = −ji = k

jk = −kj = i

ki = −ik = j

Exercise 1.6 Derive the rotation matrix parameterized by yaw, pitch, and roll. Exercise 1.7 Show that the fixed points of the Möbius transformation z = ¯ + u¯ ) are (uz + v)/(−vz

1 2 z∗ = u¯ − u ± (u¯ − u) − 4v v¯ 2v¯ and that for the u, v of (1.23) z∗ =

n1 + ın2 n + ın2 , − 1 1 − n3 1 + n3

which are the images of the axis of rotation under stereographic projection. Exercise 1.8 Show that if A, B ∈ SU (2) and f , g ∈ Mo (C ) with A ↔ f , B ↔ g, then AB ↔ f ◦ g.

37


39

2

Kinematics, Energy, and Momentum Definition A rigid body is a collection of particles moving in such a way that the relative distances between the particles do not change. This book is devoted to the dynamics of rigid bodies – the study of the motion of rigid bodies as determined by the shape and mass distribution of the body as well as the applied forces and torques. The subject is interdisciplinary and has engaged some of the most prominent mathematicians, physicists, astronomers, and engineers. The history spans the time from mid-eighteenth century, the work of d’Alembert and Euler, to the present day when it is still an area of active research. Rigid body dynamics occupies a unique niche in the realm of theoretical mechanics. The theory generalizes particle dynamics by endowing the objects with finite extent, but it does not entertain deformations that fluids and elastic bodies experience.1 Rigid body theory is an idealization which captures the dynamics of many objects from planets to molecules. This chapter considers general rigid body transformations and rigid body motions as well as basic concepts from physics which underlie rigid body mechanics. A rigid body transformation is any change of position which preserves the relative distances of all the points of the body. A rigid body motion is a time sequence of rigid body transformations. The rate of change of a rigid body motion leads to the concept of angular velocity. We consider how it is defined and how it is computed using the various representations of rotations. The angular velocity leads us to angular momentum and kinetic energy which involve the inertia tensor. We begin with rigid body transformations. Mathematically, the angular velocity is an element of so(3), the Lie algebra of the Lie group SO(3) (see Appendix C). 1) Euler derived the basic equations of rigid body dynamics, elastic bodies, and fluid dynamics [16, 17].


40

2 Kinematics, Energy, and Momentum

2.1 Rigid Body Transformation

The basic facts about the group of rigid body transformations are established here. Rotations are a subgroup and each element of the subgroup has a fixed point (fixed axis in fact). The group of rigid body transformations includes translations which do not have fixed points. 2.1.1 A Rigid Body has 6 Degrees of Freedom

Consider a rigid body composed of N + 1 > 3 particles, enumerated p0, . . . , p N , whose relative separation distances are known. We wish to determine the number of parameters that are needed to specify the position of all the particles. Clearly the 3N + 3 coordinates of the particles would suffice but the rigid body constraint can reduce the necessary number substantially. Three parameters will fix the spatial position of particle p0. Particle p1 (located at a fixed distance from particle p0) can be located with two additional parameters – say spherical coordinates relative to particle p0 . Particle p2 has a fixed distance from particles p0 and p1, so it may be located by a rotation – one degree of freedom – about the axis r1 − r0 . Once the position of three points are fixed the positions of the remaining N − 2 points are also fixed. Thus, six parameters will determine the coordinates of every particle in the body. 2.1.2 Any Rigid Body Transformation is Composed of Translation and Rotation

Any rigid body transformation can be composed of translations of a reference point r0 and a rotation about that reference point. A translation T a moves each point r of space to r + a. Consider a rigid body transformation such that ri −→ ri

i = 0, . . . , N

Apply Tr −r0 to each point in the body. This produces intermediate positions 0

r˜0 = r0

and

r˜i = Tr −r0 (ri ) = ri + (r0 − r0 ) 0

i = 1, . . . , N

Now consider a transformation which fixes r0 and moves each r˜ i to ri . This is a rigid body transformation with the fixed point r0 . Therefore it is a rotation, say R, and the action on ri is ri = r0 + (r0 − r0 ) + R(r˜ i − r0 ) = r0 + a + R(ri − r0 )

a = r0 − r0


2.1.3 The Rotation is Independent of the Reference Point

The rotational part of a rigid body transformation does not depend on the reference point. Suppose particle rk had been chosen as the reference point. Then ri = r0 + a + R [(ri − rk ) + (rk − r0 )] = rk + b + R(ri − rk ) where b = r0 − rk + a + R(rk − r0 ) Thus the transformation is again expressed in terms of a translation Tb applied to the new reference point rk and the rotation R which is unchanged by the altered reference point. 2.1.4 Rigid Body Transformations Form the Group SE (3)

It will now be convenient to take r0 = 0. Then the generic rigid body transformation is r = Rr + a If this is followed by a second rigid body transformation r

= S r + b = SRr + S a + b

the result is another rigid body transformation with rotation T = SR and translation c = S a + b. The inverse of rigid body transformation r = Rr + a is the rigid body transformation

R −1 r − R −1 a = r Therefore the rigid body transformations form a group, the special Euclidean group SE(3). This group consists of pairs (R, b), R a rotation and b a vector, with a binary operation ◦,

(R, a) ◦ (S , b) = (RS , Rb + a) and inverse

(R, a)−1 = (R−1 , −R−1 a)

41

42


If

  r1  r = r2  r3

we may represent points of R3 in homogeneous coordinates

r 1 in which case the rigid body transformation (R, a) is represented by the 4×4 matrix written in partitioned form as

R a 0 1 where R is the representation of R and it is understood that the zero in the lower-left-hand corner represents a column of three zeros. Thus

R a r Rr + a = (2.1) 1 0 1 1 2.1.5 Chasles’ Theorem

Fig. 2.1 (a) The geometry of the rigid body transformation ( Rn ( θ ), a) decomposed into a rotation about an axis containing n and a translation along that axis. (b) The two-dimensional case wherein the rigid body transformation is affected by rotation alone.

Chasles’ theorem states that every rigid body transformation can be composed of a rotation about an axis and a translation along that axis. This is


the transformation executed by a turning screw and is referred to as screw transformation. The simplest instance of screw transformation involves an axis λn through the origin where λ ∈ R and n ∈ R3 with nt n = 1. The transformation consists a rotation about n, say by angle θ, and a translation along n by distance pθ/2π, where p is called the screw pitch. This transformation is the action of the group element

θ Rn (θ ) p 2π n 0 1 To obtain the screw transformation about an axis λn + x through a point x, again with λ ∈ R and n ∈ R3 with nt n = 1, one translates by − x to the origin, performs the transformation at the origin, and then translates by x. In other words, the transformation is the action of the group element

θ I x I −x Rn (θ ) p 2π n 0 1 0 1 0 1 which is

Rn ( θ ) 0

b 1

(2.2)

with

θ n 2π To prove Chasles’ theorem we must find n, θ, x, and p such that the matrix (2.2) matches a given general rigid body transformation (2.1). In other words, we need to find n and θ such that Rn (θ ) = R and x and p such that b = a. The axis n and angle θ can be determined directly from R as in (1.12)–(1.15). From b = [ I − Rn (θ )] x + p

nt a = nt b = p we obtain p=

θ 2π

2π t na θ

Finally, to obtain x we have

( I − R) x = a − nnt a It is not possible to invert I − R (n is an eigenvector with eigenvalue 0). Let v1 ,v2 be two vectors orthogonal to n such that {v1 , v2 , n} form an orthonormal basis. In this basis let x = αv1 + βv2 + γn, a = a1 v1 + a2 v2 + a3 n. In this basis R = Re3 (θ ). Thus      1 − cos θ α a1 sin θ 0  − sin θ 1 − cos θ 0  β  =  a2  0

0

0

γ

0

43

44


We see that γ is arbitrary and take it to be zero. The remaining components are obtained from

sin θ 1 − cos θ a α = 1 − sin θ 1 − cos θ β a2 This linear system is nonsingular if θ = 0. The geometry of Chasles’ theorem is illustrated in Fig. 2.1(a). The two-dimensional version of Chasles’ theorem has particularly transparent algebraic and geometrical derivations. In two dimensions every rigid body transformation has a fixed point and the axis through this point and perpendicular to the plane is the axis of Chasles’ theorem. If we have in mind a motion of an actual rigid body the fixed point need not lie in the body but we can apply the transformation to all points of the plane and thereby discover the fixed point. Algebraically we find the fixed point p from the equation p = Rp + a which is satisfied by

p = ( I − R ) −1 a

This is the same problem to which the three-dimensional case reduced. In this case the rigid body transformation does not involve translation along the screw axis. The geometry of the two-dimensional case is illustrated in Fig. 2.1(b).

2.2 Angular Velocity

e3 e2 e¯ 3 6 I @ @ @ @ @ H HH HH j e¯ 1 e¯ 2 e1

Fig. 2.2 Fixed reference frame {e¯ i } and moving reference frame {ei }.


Consider rotational motion of a rigid body, i.e., a rigid body motion leaving one point fixed. Let r (0) be the position of a point in the body at t = 0, r (t) its position at time t, and R(t) the rotation operator which carries r (0) to r (t) r (t) = R(t)r (0)

R(t) thus traces the orientation of the rigid body in configuration space SO(3). The velocity of the point is ˙ −1 r r˙ = R˙ r (0) = RR or r˙ = Ωr

˙ −1 Ω = RR

(2.3)

The operator Ω is the angular velocity operator. It is skew-symmetric because 0=

d ˙ t + RR˙ t = Ω + Ωt (RRt ) = RR dt

where the fact that R−1 = Rt has been used.2 Now consider two orthonormal reference frames (Fig. 2.2). The first is an inertial frame {e¯ i } fixed in space and the second is a moving frame {ei } (perhaps fixed in a moving rigid body). The frames are related just as r(t) and r(0) are related, namely ei = Re¯ i Let R = [ Rij ] be the matrix representation of R in the fixed frame, R = eR. Then the frames satisfy e = e¯ R¯ (e¯ ) = e¯ R

and

e¯ = eRt

(2.4)

The components of R are explicitly given by the dot products Rij = e j · e¯ i Now ˙ t ¯ Ω = e¯ R˙ ¯ ¯ e Rt ¯ = e¯ RR ˙ t is the matrix representation of Ω in frame e. To obtain the Thus Ω = RR representation in the moving frame we have ˙ t ¯ = eRt R ˙ Ω = e¯ RR Thus Ω = Rt R˙ is the matrix representation of Ω in frame e. Note that ¯ = RΩRt Ω 2) Recall that if R is a linear operator on a vector space V over R3 , then a basis-free definition of R t is (a, Rb) = (R t a, b) for all a, b ∈ V.

45

46


By virtue of the correspondence so(3) ↔ R3 there is an angular velocity vector

− → − → ω = eΩ = eΩ We denote the components as

− → ω= Ω

and

− → ω= Ω

and we have ω¯ = Rω In frame e, Eq. (2.3) takes the form r˙ = ω × r

(2.5)

The following physical argument is often used to arrive at (2.5). Consider an infinitesimal rotation

[(1 − cos θ )nnt + cos θI + sin θ n]|θ =∆φ = I + ∆φ n + O(∆φ2 ) Then

r = ( I + ∆φ n)r = r + ∆φ nr

or ∆r = ∆φn × r In the limit ∆φ → 0 r˙ = ω × r

with

˙ ω = φn

The angular velocity can be obtained directly from the time derivatives of the basis vectors according to ω=

1 e × e˙ i 2 i

To obtain this result we note that e˙ i = ω × ei because e˙ i = e j Ω ji = e j jik ωk = ω × ei . Therefore ei × e˙ i = ei × (ω × ei ) = ωei · ei − ω · ei ei = 3ω − ω = 2ω The concept of angular velocity is basic to rigid body theory. Mathematically, the angular velocity belongs to the Lie algebra of the rotation group – the tangent space at the identity of the group – which is endowed with a special binary relation related to the vector cross product. One moves from velocity on the manifold (Lie group) to angular velocity at the identity by left or right translations – depending on the frame of reference used. In Lie group terminology, the angular velocity operator Ω = R−1 R˙ is the translation to the


˙ In other words, it selects the left invariant identity of the tangent vector R. ˙ When we are working vector field in the Lie algebra which corresponds to R. with vector language and the angular velocity we have transferred our problem from the highly nonlinear world of SO(3) to the linear world of so(3) (see Appendix C for more details). 2.2.1 Angular Velocity in Euler Angles

From Section 1.1.6, the Euler angle parameterization of the rotation matrix which takes the fixed frame to the rotating frame is 

cos φ  R(θ, φ, ψ) = sin φ 0

 − sin φ 0 1 0   cos φ 0 0 cos θ 0 1 0 sin θ

 0 cos ψ   sin ψ − sin θ 0 cos θ

 − sin ψ 0 cos ψ 0 0 1

= R1 ( φ ) R2 ( θ ) R3 ( ψ ) So its time derivative is ˙ 1 (φ) R (θ ) R3 (ψ) + ψR ˙ 1 (φ) R2 (θ ) R2 (ψ) + θR ˙ 1 (φ) R2 (θ ) R3 (ψ) R˙ = φR 2 The angular velocity matrix in the moving frame is Ω

= =

which yields

Rt R˙ ˙ t Rt R R3 + ψR ˙ 3t R2t R1t R1 R2 R3 + θR ˙ 3t R3 φR 3 2 2

˙ t e1 R3 + ψ˙ e3 ˙ 3t R2t e3 R2 R3 + θR = φR 3

(2.6)

t t ˙ t e1 ˙ 3 R2 + ψ˙ e3 + θR ω = φR 3

(2.7)

Lie group concepts are implicit in this result. Equation (2.6) can be expressed in Lie group language −1 −1 ˙ −1 ˙ ˙ Ω = φAd R3 AdR2 ade3 + θAdR3 ade1 + ψade3

Using the R3 representation of so(3) we obtain (2.7). Note that ultimately only one matrix–matrix multiply is required. This result can be expressed in two convenient forms. First, there is a matrix expression relating the components of ω to the derivatives of the Euler angles, 

sin θ sin ψ ω =  sin θ cos ψ cos θ

cos ψ − sin ψ 0

 0 0  1

 φ˙ θ˙  ψ˙

(2.8)

47

48


This is most often used in the inverse form, the so-called kinematic equation relating derivatives of the angles to the angular velocity,   sin ψ/ sin θ φ˙  θ˙  =  cos ψ − cos θ sin ψ/ sin θ ψ˙ 

cos ψ/ sin θ − sin ψ − cos θ cos ψ/ sin θ

  ω1 0 0   ω2  (2.9) 1 ω3

Second, there is an invariant, vector expression for the angular velocity ˙ + ψe ˙ e3 + θe ˙ 3 ω = φ¯ 1

(2.10)

The line containing the vector e1 is sometimes referred to as the line of the nodes. The angular velocity matrix in the fixed frame is ¯ Ω

= =

˙ t RR ˙ 1 R Rt Rt + ψR ˙ 1 R1t + θR ˙ 1 R2 R3 R3t R2t R1t φR 2 2 1

˙ 1 ˙ e3 + θR ˙ 1 R2 e1 R1t + ψR e3 R2t R1t = φ and the resulting components of angular velocity vector are 

0 ω¯ =  0 1

cos φ sin φ 0

 sin φ sin θ − cos φ sin θ   cos θ

 φ˙ θ˙  ψ˙

(2.11)

It is shown in [18] that Eq. (2.10) generalizes to the following sequence of rotations: e1 = e¯ R1 ( β 1 ) e2 = e1 R2 ( β 2 )

R1 ( β 1 ) = Re¯i ( β 1 ) 1

R2 ( β 2 ) = R e 1 ( β 2 )

···

i2

e N = e¯ R1 ( β 1 ) R2 ( β 2 ) · · · R N ( β N ), R N ( β N ) = Re N−1 ( β N ) iN

where e j are intermediate frames (analogs of our e and e ) and the subscripts 1 ≤ ik ≤ 3 designate a particular vector in frame k.The representation of the angular velocity operator in the body fixed frame is Ω = RtN · · · R2t R1t β˙ 1 R1 R2 · · · R N + β˙ 2 R1 R2 · · · R N + · · · + β˙ N R1 R2 · · · RN = β˙ 1 Rt · · · Rt Rt R R2 · · · R N + β˙ 2 Rt · · · Rt R · · · R N + · · · + β˙ N Rt R N

= =

2

1

1

N

2

2

N

ei1 R2 · · · R N + β˙ 2 RtN · · · ei2 · · · R N + · · · + β˙ N e β˙ 1 RtN · · · R2t iN ˙β 1 Ad−1 · · · Ad−1 ade + β˙ 2 Ad−1 · · · Ad−1 ade + · · · + β˙ N ade RN

R2

i1

RN

R3

i2

iN

from which it follows that the angular velocity vector components are ω = β˙ 1 RtN · · · R2t ei1 + β˙ 2 RtN · · · R3t ei2 + · · · + β˙ N ei N

N


Therefore the angular velocity vector is ω

= eω = β˙ 1 eRtN · · · R2t ei1 + β˙ 2 eRtN · · · R3t ei2 + · · · + β˙ N eei N = β˙ 1 e1 + β˙ 2 e2 + · · · + β˙ N e N i1

i2

iN

2.2.2 Angular Velocity in Quaternions

Recall the representation of rotations in the quaternions r = qrq¯ with q¯q = (q0 , q)(q0, −q) = (1, 0) and r = (0, r ) pure. We regard r as the rotated position and r as the constant, reference position. Then the velocity is the vector part of r˙ r˙

= qr ˙ q¯ + qrq¯˙ = q¯ ˙ qqrq¯ + qr(−q¯ q¯ ˙ q) = [w, r ]

where [, ] denotes the commutator and w = q¯ ˙ q. We calculate w = (0, q0 q˙ − q˙0 q + q × q˙ ) = (0, w) and

[w, r] = [0, 2w × r ] Thus w is a pure quaternion and 2V (w) = 2w is the angular velocity. Since the vector part of w is linear in q˙ it can be expressed in matrix terminology. We find

0 ω



q0  − q1 = 2  − q2 − q3

q1 q0 q3 − q2

q2 − q3 q0 q1

 q3 q˙0  q˙1 q2   −q1   q˙2 q˙3 q0

   

(2.12)

This expression follows directly from (1.8) ˙ q ↔ R(q¯ )q˙ = R(q)t q˙ w = q¯ From the properties of R(q) and the fact that qt q˙ = 0 we also obtain the alternative

0 = R(q)t q˙ = − R(q˙ )t q w

49

50


Since q is a unit quaternion, the matrix in (2.12) is orthogonal and it is easily inverted to obtain the kinematical differential equation for q˙ in terms of ω 

  q˙0  q˙1    = 1  q˙2  2 q˙3

− q1 q0 − q3 q2

q0 q1 q2 q3

− q2 q3 q0 − q1

 − q3

− q2  0  q1  ω q0

(2.13)

The Euler angle parameterization of q [(1.17–1.20)], when substituted into (2.12), yields ω = (θ˙ cos ψ + φ˙ sin θ sin ψ, −θ˙ sin ψ + φ˙ sin θ cos ψ, φ˙ cos θ + ψ˙ )t which recovers (2.8). 2.2.3 Angular Velocity in Cayley–Klein Parameters

The representation of a rotation in SU (2) is X = SX0 S† with S of the form

S=

where

γ = − β∗

α γ

β δ

δ = α∗

αα∗ + ββ∗ = 1

Thus ˙ † SX0 S† − SX0 S† SS ˙ † = [W, X ] X˙ = SS with ˙ †= W = SS

˙ ∗ ˙ ∗ + ββ αα α˙ ∗ β∗ − α∗ β˙ ∗

˙ + α β˙ −αβ ∗ α˙ α + β˙ ∗ β

=

˙ ˙ − βγ αδ ˙ −δγ + δγ˙

˙ + α β˙ −αβ ˙δα − γβ ˙

2.3 The Inertia Tensor

Consider a rigid body whose mass distribution is characterized by the differential mass element dm = ρ(r) dr where ρ (r) is the local density of the body. The zeroth, first, and second-order moments of dm capture important properties the body. The zeroth-order moment, a scalar, is the total mass m=

B

dm


The first-order moment, a vector, is the center of mass r¯ =

1 m

B

dm r

The second-order moment about a reference point r0 is the inertia tensor, I ,

I (r0 , B) =

B

dm [(r − r0 ) · (r − r0 ) I − (r − r0 ) ⊗ (r − r0 )]

When the context is clear we will abbreviate I (r¯, B) = I . It is sometimes convenient to regard a rigid body to be formed of discrete masses mi at positions ri in which case the integral formulation can be used with the differential mass element being dm = dr

∑ m i δ ( r − ri ) i

The integrals are thereby converted to sums m = ∑ mi i

r¯ =

1 m

∑ m i ri i

I (r0 , B) = ∑ mi [(ri − r0 ) · (ri − r0 ) I − (ri − r0 ) ⊗ (ri − r0 )] i

The inertia tensor depends only on the distribution of the mass in the body and not on the state of motion. The inertia tensor is positive definite because v · (r · rI − r ⊗ r) · v = (v · v)(r · r) − (v · r)2 > 0

for all v = 0 or r

Every symmetric, positive definite operator has positive eigenvalues and associated orthogonal eigenvectors. The orthogonal eigenvectors of the inertia tensor are known as the principal axes of the body. When the coordinates are chosen such that r0 = r¯ = 0 the inertial tensor relative to the principal axes assumes the diagonal form 

y 2 + z2  I= ρ 0 B 0

0 2 x + z2 0

  I1 0   dr = 0 0 x 2 + y2 0

0 I2 0

 0 0  I3

From this diagonal form we see that the principal moments of inertia I1 , I2 , I3 satisfy the triangle equalities I1 ≤ I2 + I3

I2 ≤ I1 + I3

I3 ≤ I1 + I2

(2.14)

51

52


If the reference point is translated relative to the center of mass, r0 = r¯ + c, then Steiner’s theorem relates the inertia tensor relative to the two reference points: I(r0 , B) = I(r¯, B) + M [c · cI − c ⊗ c] (2.15) This is true because

I (r¯ + c, B) =

B

dm [(r − r¯ − c) · (r − r¯ − c) I − (r − r¯ − c) ⊗ (r − r¯ − c)]

= I(r¯ , B) + M [c · cI − c ⊗ c] +

B

dm [−2(r − r¯ ) · cI + (r − r¯ ) ⊗ c + c ⊗ (r − r¯ )]

= I(r¯ , B) + M [c · cI − c ⊗ c] Suppose B is composed of n disjoint sub-bodies

B = B1 ∪ B2 ∪ · · · ∪ B n

Bi ∩ B j = ∅

for i = j

Then additivity of the Riemann integral gives

I (r0 , B) =

n

∑

i =1 B i

dm [(r − r0 )2 I − (r − r0 ) ⊗ (r − r0 )] =

n

∑ I (r0, Bi )

i =1

The inertia tensor of each sub-body is most naturally calculated relative to its center of mass. Steiner’s theorem (2.15) can be used to relate the constituent inertia tensors to a reference body, say B1 . Let m i ri = Then

Bi

r dm

mi =

Bi

dm

ci = r¯ i − r¯1

n I (r1 , B) = I (r1 , B1 ) + ∑ I (ri , Bi ) + mi [c2i I − ci ⊗ ci ] i =2

Finally, we must choose a basis for the representation of I (ri , B). Let us use a principal axis basis for B1 . Let e(i) be a principal axis basis for Bi and let e(i) = e(1) Ri . Then

I (ri , Bi ) = e(i) Ii (i) = e(1) Ri Ii Rti (1) no summation Therefore, the representation of the inertia tensor in the chosen reference frame is " ! n (1) t t t I (r1 , B) = e I1 + ∑ R i I i R i + m i ( c i c i I − c i c i ) ( 1 ) (2.16) i =2


Fig. 2.3 Scale definition for spacecraft with solar panels.

Example 2.1 Let us calculate the inertia tensor for the spacecraft with solar panels illustrated in Fig. 2.3. We will consider the body to be a cylinder and the two panels to be rectangular hexahedra, all of uniform density and of the dimensions indicated in Fig. 2.3. Let the x-axis be the axis of the panel passing through the center of the b − −c face and the y-axis the symmetry axis of the cylinder. Choose the x-axis to be a principal axis for the cylinder. Let the panels each be inclined at angles π/2 − θ to the generators of the cylinder. Let B1 be the cylinder and B2 , B3 the (identical) panels. Elementary calculations provide

and

 3R2 + H 2 1  0 I1 = m 1 12 0

0 2 3R + H 2 0

 0 0  6R2

 2 b + c2 1 I2 = I3 = m 2  0 12 0

0 a2 + c2 0

 0  0 a 2 + b2

The translations from the center of mass of B1 to the centers of mass of B2 and B3 are c2 = ( R + 12 a)e1 and c3 = −( R + 12 a)e1 and the rotation matrices are 

1 R1 = R2 =  0 0

0 cos θ sin θ

 0 − sin θ  cos θ

Now we can assemble the inertia matrix as given by (2.16) 

I11 I= 0 0

0 I22 I32

 0 I23  I33

53

54


with 1 1 m1 (3R2 + H 2 ) + m2 (b2 + c2 ), 6 6 1 1 1 2 2 = m1 (3R + H ) + m2 ( a2 + b2 sin2 θ + c2 cos2 θ ) + 2m2 ( R + a)2 , 6 6 2 1 1 2 2 2 2 2 2 2 = m1 R + m2 ( a + b cos θ + c sin θ ) + 2m2 ( R + a) , 6 2 1 = I32 = sin θ cos θm2 (c2 − b2 ). ♦ 6

I11 = I22 I33 I23

2.4 Angular Momentum

The angular momentum of a particle relative to the reference point r0 is h p = m(r p − r0 ) ×

d (r p − r0 ) dt

This is used in an integral form to obtain the total angular momentum of a rigid body relative to the reference point r0 as h

= =

B B

dm (r − r0 ) ×

d (r − r0 ) dt

dm (r − r0 ) × [ω × (r − r0 )]

= I (r0 , B)(ω) Steiner’s theorem can be used to express the angular momentum in two additional useful forms h

= I (r, B) + M (r − r0 ) × [ω × (r − r0 )] = I (r, B) + M (r − r0 ) × (r˙ − r˙ 0 )

2.5 Kinetic Energy

Consider a rigid body B and let r0 be the reference point chosen to mark the translational motion. As shown in Section 2.1 the motion of any point in the body can be expressed as r(t) = r0 (t) + R(t)(r(0) − r0 (0))

2.5 Kinetic Energy

Then the velocity of the point is r˙

˙ r(0) − r0 (0)) = r˙ 0 + R( ˙ t (r − r0 (t)) = r˙ 0 + RR

= r˙ 0 + ω × (r − r0 ) where ω is the angular velocity of the body. This can be written in the useful form as r˙ − r˙ 0 = ω × (r − r0 ) (2.17) The total kinetic energy of the body is T=

1 2

B

dm r˙ · r˙

where dm = ρ(r) dr with ρ (r) the mass density at point r in B . Thus T

= = =

1 dm [r˙ 0 + ω × (r − r0 )]2 2 B # $ 1 dm r˙ 20 + 2˙r0 · ω × (r − r0 ) + (ω × (r − r0 ))2 2 B 1 2 r˙ 0 dm + r˙ 0 · dm ω × (r − r0 ) 2 B B $ # 1 dm (r − r0 )2 I + (r − r0 ) ⊗ (r − r0 ) · ω + ω· 2 B

This result can be rewritten in terms of the total mass m, the center of mass r¯, and the moment of inertia tensor I (r0 , B) as T=

1 2 1 m˙r + m˙r0 · ω × (r¯ − r0 ) + ω · I(r0 , B)ω 2 0 2

(2.18)

Note that if one chooses r0 = r¯, then Eq. (2.18) simplifies to T=

1 2 1 mr¯˙ + + ω · I · ω 2 2

which expresses the kinetic energy as the sum of translational and rotational parts with the translational part concentrated at the center of mass. Example 2.2 The rotational kinetic energy will be expressed in the moving frame and in terms of the Euler angles. From Eq. (2.7) one obtains t  sin θ sin ψ cos ψ 0 2T = φ˙ θ˙ ψ˙  sin θ cos ψ − sin ψ 0  cos θ 0 1     φ˙ A1 0 sin θ sin ψ cos ψ 0 0  0 A2 0   sin θ cos ψ − sin ψ 0   θ˙  ψ˙ 0 0 A3 cos θ 0 1

55

56


or T

=

1# A1 (φ˙ sin θ sin ψ + θ˙ cos ψ)2 + A2 (φ˙ sin θ cos ψ − θ˙ sin ψ)2 + 2 $ A3 (φ˙ cos θ + ψ˙ )2 ♦ (2.19)

Example 2.3 A gyrostat is a rigid body B , called the platform, to which there are attached axi-symmetric rotors. The gyrostat is not a rigid body but rather a collection of rigid bodies linked in a specific way. We wish to calculate the rotational kinetic energy for a gyrostat which has three rotors Ri each aligned with a principal axis of B . Let the inertia tensors in the principal axes of the sub-bodies be 

A0 I(r0 , B) =  0 0

0 B0 0

 0 0  C0



Ai I(ri , Ri ) =  0 0

0 Bi 0

 0 0  Ci

Let the translations from the center of mass of B to the center of mass of Ri be ci = di ei . Here the rotation matrices are identities because the rotors are aligned with the principle axes. Let ω be the components of the angular velocity of B in the principal axis frame of B and let ω + si ei (no summation) be the components of the angular velocity of Ri in the same frame. Then we can express the kinetic energy T as 2T

= ω t I(r0 , B)ω + 3 ∑ (ω + si ei )t I(ri , Ri ) + mi (cti ci I − ci cti ) (ω + si ei ) i −1

=

Aω12 + Bω22 + Cω32 + A1 s21 + B2 s22 + C3 s23 + m1 d21 (ω22 + ω32 ) + m2 d22 (ω12 + ω32 ) + m3 d23 (ω12 + ω22 ) + 2( A1 ω1 s1 + B2 ω2 s2 + C3 ω3 s3 )

with A = A0 + A1 + A2 + A3 B = B0 + B1 + B2 + B3 C = C0 + C1 + C2 + C3 .

♦

2.6 Exercises

2.6 Exercises

Exercise 2.1 Represent r0 , r¯, and c = r0 − r¯ in an orthonormal basis. Show that I(r0 , B) = I(r¯, B) − mcˆ2 Exercise 2.2 Consider a close-packed, two-layer stack of cannon balls, four in the first layer and one in the second layer. Assume the cannon balls are homogeneous and identical having radius a and mass m. Find the inertia tensor of the assemblage relative to the center of the ball in the second layer. Exercise 2.3 Express the rotational kinetic energy of a triaxial rigid body (three unequal principal moments of inertia) in terms of the Cayley–Klein parameters. Exercise 2.4 The so-called (w, z) parameterization of the rotation matrix [19] is 3 1 R(z, w1, w2 ) = F1 (z) F2 (w1 , w2 ) 1 + w21 + w22 where  F1 (z)

F2 (w1 , w2 )

=

=

 cos z − sin z 0  sin z cos z 0 0 0 1  2 2 2w1 w2 1 + w1 − w2  2w1 w2 1 − w21 + w22 −2w2 2w1

 2w2 −2w1  1 − w21 − w22

Find the angular velocity in terms of the parameters z, w1 , w2 . Exercise 2.5 A cone of apex angle 2β rolls inside a fixed cone of apex angle 2α (β < α). Find the angular velocity of the rolling cone.

3) The matrix listed here is the inverse of that in [19] to conform with the notation in this text.

57


59

3

Dynamics This chapter lays the foundation for studying the dynamics of rigid bodies. The focus is on the derivation of the equations of motion. Solutions of the equations for specific cases of rigid body systems are considered in later chapters. Classical dynamics has three threads which originate with Newton, Lagrange, and Hamilton. The Newtonian thread is based on the celebrated Newton’s second law for motion of a particle, m¨r = F relating mass, m, force, F, and acceleration r¨ and is expressed in a fixed Cartesian frame. It was from this point that d’Alembert and Euler first developed equations of motion for a rigid body. The second thread is based on Lagrange’s generalization of Newton’s equations to generalized coordinates q1 , . . . , qn . The function we now call the Lagrangian was introduced L(q1 , . . . , qn , q˙1 , . . . , q˙n ) = T (q1 , . . . , qn , q˙1 , . . . , q˙n ) − U (q1 , . . . , qn ) where T is the kinetic energy and U is the potential energy. The equations of motion in this formulation are ∂L d ∂L = dt ∂q˙ i ∂qi q˙ 1 , . . . , q˙ n are called the generalized velocities. Poincaré generalized these equations so that the Lagrangian can assume the form L = L ( q1 , . . . , q n , s1 , . . . , s n ) where si are linear functions of q˙i with coefficients depending on qi . Hamilton introduced the momenta ∂L pi = ∂q˙i and what we now call the Hamiltonian H ( q1 , . . . , q n , p 1 , . . . , p n ) = T ( q1 , . . . , q n , p 1 , . . . , p n ) + U ( q1 , . . . , q n ) Rigid Body Mechanics: Mathematics, Physics and Applications. William B. Heard Copyright © 2006 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim ISBN: 3-527-40620-4

60

3 Dynamics

He found that this led to a very symmetric form of the equations of motion ∂H dqi = dt ∂pi

dpi ∂H =− dt ∂qi

This approach to mechanics is conducive to the use of coordinate transformations to simplify problems and in turn to the development of perturbation techniques. Modern treatments of mechanics make frequent use of the geometrical language. Underlying every mechanical system is a configuration space. This is the space described by the generalized coordinates of the problem. The configuration space is a differential manifold M which means intuitively that locally the space looks like a region of R n and that the local pieces are blended together smoothly to form the total manifold. An aspect of the differential manifold which is important to us is the existence of vector fields, the assignment of tangent vectors to each point of the manifold in a smooth way. The tangent vectors can be thought of as tangents to curves, or particle trajectories, on the manifold. The collection of all the tangent vectors to the manifold forms the tangent manifold T M. The tangent manifold T M, called the state space, is the setting for Lagrangian mechanics. There is also the cotangent manifold T ∗ M which is the collection of covectors, or cotangent vectors – duals of the tangent vectors (Section 1.1). The cotangent vectors form the cotangent bundle, called the phase space, and this is the setting for Hamiltonian mechanics. The calculus on cotangent manifolds is carried out with differential forms which are generalizations of line, area, and volume elements familiar from calculus. The generalization of calculus to this setting is a basic tool of mechanics. We will not make extensive use of this other than discussing the symplectic 2-form (generalization of the element of area) which is basic to Hamiltonian mechanics and the differential form version of Stoke’s theorem. The notation in the subject can be an obstacle upon first exposure but that is easily overcome. For example, a holonomic or coordinate basis for the tangent space is denoted as {∂qi } which is apparently at odds with the calculus notation for partial derivatives. It is, however, actually not unfamiliar because in calculus we learn that on a surface having coordinates u, v, the prototype of a two-dimensional manifold, the derivatives ∂r/∂u and ∂r/∂v are tangent to the surface. The basics of the differential manifold theory are summarized in Section C.2 and the basics of calculus on manifolds are included in Sections 3.2, 3.3, and Chapter 4. Frankel’s book [1] is a very accessible and systematic treatment of the subject. There are also more advanced [17] and physics oriented [20] sources. This chapter begins with the vector version of the equations of translational and rotational motion. Then the Euler equations for rotational motion are derived. Next, Euler–Lagrange equations for unconstrained motion are obtained

3.1 Vectorial Mechanics

by variational methods. This is followed by derivations of the Poincaré equations, which generalize the Euler–Lagrange equations beyond the L(q, q˙ ) setting on both manifolds and Lie groups. Finally the Hamiltonian formulation and Poisson brackets are explored.


3.1.1 Translational and Rotational Motion

The equation of translational motion of a rigid body follows from Newton’s second law for particle motion m p r¨ p = Fp For a rigid body, this is used in an integral form as B

dm r¨ =

B

dm f

where f is the force per unit mass acting on an infinitesimal element of mass.1 Now r˙ is expressed in terms the velocity of the reference point r0 and the angular velocity ω (2.17), r˙ = r˙ 0 + ω × (r − r0 ) r¨

= r¨0 + ω˙ × (r − r0 ) + ω × (r˙ − r˙0 ) = r¨0 + ω˙ × (r − r0 ) + ω × [ω × (r − r0 )]

and upon integration we obtain M {r¨0 + ω˙ × (r¯ − r0 ) + ω × [ω × (r¯ − r0 )]} = F where F= When r0 = r¯ this reduces to

(3.1)

B

dm f

Mr¨¯ = F

(3.2)

and the body behaves as if it were a point mass concentrated at the center of mass. 1) The use of f dm or dF to derive the equations of motion for rigid bodies is attributed to Euler [21].

61

62

3 Dynamics

The vector form of the equation of rotational motion of a rigid body is derived from the moment, relative to r0 , of Newton’s second law for particle motion m(r p − r0 ) × r¨ p = (r p − r0 ) × Fp which can be rewritten as d d m (r p − r0 ) × (r p − r0 ) + m(r p − r0 ) × r¨0 = (r p − r0 ) × Fp dt dt and this yields the equation for the evolution of the angular momentum of a particle d h p + m(r p − r0 ) × r¨0 = (r p − r0 ) × Fp dt This is used in an integral form to obtain the rate of change of angular momentum for a rigid body, % & d d dm (r − r0 ) × (r − r0 ) + m(r − r0 ) × r¨0 = dm (r − r0 ) × f dt dt B B or

d h + m(r − r0 ) × r¨0 = τ dt

where τ=

B

(3.3)

dm (r − r0 ) × f

is the total torque about r0 . Again f is the force per unit mass acting on an infinitesimal element of mass dm. As in the derivation of Eq. (3.2) the time derivative can be taken outside the integral. 2 Therefore the equation of rotational motion of a rigid body is d I(r0 , B)(ω) + m(r¯ − r0 ) × r¨0 = τ dt

(3.4)

and in the case r0 = r¯ or r¨0 = 0 then the equation of rotational motion becomes d I(r0 , B)(ω) = τ dt

(3.5)

3.1.2 Generalized Euler Equations

The Euler equations express the rigid body total angular momentum equation d h=τ dt 2) Alternatively, we can use the formula for the differentiation of an ' ' ' ' d dω + ∂B v ω. Here integral [1], dt B t ω = B t Lv ω = B t v t the boundary integral vanishes and the volume integral involves the time derivative via v ω.


in a principal axis, moving body frame. We have seen in the derivation of Eq. (3.3) that this equation is valid if r0 = r¯ or r¨0 = 0, that is, either the reference point is the center of mass or the reference point is not accelerating. In this section we derive a generalized version of the Euler equations valid in any rotating or fixed frame. We note that a a distinction is made between generalized and modified Euler equations in [22]. Here we lump both categories together as the generalized Euler equations. The basic idea is very simple. Express h in an arbitrary frame f and take the derivative h

=

( f I( ω

h˙

=

(˙ ( + f I(˙ ω ( + f I( ω f˙ I( ω

To evaluate the derivatives we need to know the relationship between f and the space fixed and body fixed frames e¯ and e. Let the frames be related by rotations R(t), S(t), and T (t) e(t) = e¯ R(t)

f (t) = e¯ S(t)

f ( t ) = e( t ) T ( t )

T is determined by R and S T = Rt S The derivative of the frame is (f f˙ = e¯ S˙ = f St S˙ = f Ω ( f = St S˙ is the operator for the angular velocity of the frame f relative where Ω to the fixed frame e¯ . We can also evaluate f˙ via the frame e ( r + T t ΩT ) = f (Ω (r + Ω () f˙ = eT ˙ + e T˙ = f (Ω ( r = T t T˙ is the operator for angular velocity of the frame f relative to where Ω the body fixed frame e. The second term T t ΩT is the operator for the angular velocity of the body expressed in the frame f . Each operator has its associated angular velocity vector ( =ω ˜ Ω

(f =ω ˜ f Ω

(r = ω ˜ r Ω

To evaluate the derivative of the inertia tensor we use the change of basis theorem to place express the time dependence of I( in terms of T (t),

I( = T t I T From this we find

(r − Ω ( r I( = [I(, Ω ( r] I(˙ = I( Ω

63

64

3 Dynamics

Assembling these results we obtain the generalized Euler equations ( r ]ω˜ = τ ( f I( ω˜ + [I(, Ω I( ω˜˙ + Ω

(3.6)

From this general form we derive several important special cases. The generalized Euler equations are usually used when I(˙ = 0 as is the case, for example, when one of the f axes lies on a line of symmetry. In this case the equations become ( f I( ω ( = τ( (˙ + Ω I( ω (3.7) If f = e

( = Ω = Ωf Ω

(=ω ω

and the equations become

I ω˙ + ΩI ω = τ If f = e and the coordinates are aligned with the principal axes, then

I( = I = diag[ A1 , A2 , A3 ] and we obtain the original Euler equations A1 ω˙ 1 + ( A3 − A2 )ω3 ω2 = τ1 A2 ω˙ 2 + ( A1 − A3 )ω1 ω3 = τ2

(3.8)

A3 ω˙ 3 + ( A2 − A1 )ω2 ω1 = τ3 This can be restated in terms of column vectors as h˙ + ω × h = τ h, T = τ as or in terms of skew-symmetric matrices H = H˙ + [Ω, H ] = T If f = e¯ , then

I( = I (t) (=ω ω

( =Ω Ω

Ωf = 0

Ωr = − Ω

and the equations of motion are

I ω˙ + [Ω, I]ω = τ This may be further simplified to

I ω˙ + Ω I ω = τ because Ω ω = 0.

(3.9)


The orientation of the body depends on the parameters used to parameterize the rotation matrix, Euler angles, or quaternions for example. To obtain the parameters once the angular velocity is known we use the kinematic equations which relate the derivatives of the parameters to the angular velocity. For Euler angles and quaternions the kinematic equations are (2.9) and (2.13), respectively.

Fig. 3.1 Definitions for the spinning top.

Example 3.1 The spinning top is a rigid body T rotating in a constant gravitational field with one point fixed (see Fig. 3.1). Assign the reference point r0 to be the fixed point. This simplifies the problem because it forces r0 = r˙ 0 = r¨0 = 0. There are two forces acting on the top: the constant gravity force acting at the center of mass and the reaction force acting at the fixed point. The total gravitational force acting on the top is f= dm(− ge3 ) = −mge3 T

The torque due to gravity is τ=f=

T

dm(− gr × e3 ) = mge3 × r

where r is the position of the center of mass. The reaction force exerts no torque about the fixed point. The angular momentum of the top is

I1 = I + m(r · rI − r ⊗ r) where I = I(r, T ). In a principal axis frame at the fixed point the equations of motion are therefore I1 ω˙ + ω × I1 ω = mge3 × r

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The coordinates of e¯ 3 in the body frame depend on the parameters used to parameterize the rotation matrix. We need to include the appropriate kinematic equations such as (2.9) or (2.13). The reaction force can be determined from the equation of motion of the reference point which here is fixed 0 = m¨r = −mge¯ 3 + fr ⇒ fr = mge¯ 3.

♦

Fig. 3.2 Reference frames for a gyrocompass on a rotating planet.

Example 3.2 A gyrocompass is a device consisting of an axially symmetric rotor mounted in one or more freely moving gimbals (see Fig. 3.2). Gyrocompasses are key components of nautical, air, and space vehicle navigation and control systems. Here we consider a single gimbal gyrocompass with the rotor axis free to move in the horizontal plane of a rotating spherical planet [13]. That is, the axis about which the gimbal rotates is vertical. We wish to write the equations of motion of the gyrocompass in the frame of the gimbal – not the rotor – thereby removing the rapid rotation rate of the rotor from the reference frame. The gimbal is assumed to be massless. There are four coordinate frames involved in the formulation of the problem. • Inertial frame: e¯ The angular velocity of the planet is collinear with e¯ 3 . • Local reference frame: e e3 is the local vertical at longitude φ and latitude θ, e2 points northward, e1 = e2 × e3 .


• Gimbal reference frame: f f3 = e3 . f2 = axis of the rotor, f1 = f2 × f3 • Rotor fixed frame : e e2 = f2 = axis of the rotor. The frames are related by 

 − sin φ − sin θ cos φ cos θ cos φ e = e¯  cos φ − sin θ sin φ cos θ sin φ  = e¯ R1 (φ) 0 cos θ sin θ   cos α − sin α 0 f = e  sin α cos α 0  = e R2 (α) 0 0 1 and 

cos β e= f 0 sin β

0 1 0

 − sin β  = f R3 ( β ) 0 cos β

No θ dependence is exhibited in R1 because θ is constant whereas φ increases at constant rate φ˙ because of the rotation of the planet. In calculating the angular velocity matrices we need   0 − sin θ cos θ t R1 R1 =  sin φ 0 0  − cos θ 0 0 and 

0 R2t R2 =  1 0

 −1 0 0 0  0 0

Since f = e¯ R1 R2 , the angular velocity matrix Ω f for the frame f is obtained from ˙ 2t R1t R1 R2 + αR ˙ 2t R2 ) = f Ω f f˙ = f (φR which yields 

0 Ω f = φ˙  sin θ − cos α cos θ

− sin θ 0 − sin α cos θ

  0 cos α cos θ sin α cos θ  + α˙  1 0 0

 −1 0 0 0  0 0

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3 Dynamics

The angular velocity operator of the rotor in the frame f is   0 0 1 ( = Ω f + β˙  0 0 0  Ω

−1 0 0

Let the principal moments of inertia be A, B and A. Then from (3.7) the equations of motion for the rotor in the frame f are Aω˜˙ 1 + ( A − B)ω˜ 2 ω˜ 3 − A β˙ ω˜ 3 = τ(1 Aω˜˙ 2 = τ(2 Aω˜˙ 3 + ( B − A)ω˜ 1 ω˜ 2 + A β˙ ω˜ 1 = 0. The torque τ(3 is set to zero so that the device can rotate freely about the vertical axis. The problem formulation is completed by expressing the angular velocity components in terms of the angles and their derivatives. We obtain these from the appropriate ( entries of Ω, ω˜ 1 = −φ˙ sin α cos θ ω˜ 2 = φ˙ cos α cos θ + β˙ ω˜ 3 = φ˙ sin θ − α˙ This system of six first order ordinary differential equations is reduced to a second order differential equation for α and two algebraic equations in τ1 and τ2 by eliminating ˙ the equation for α is ωi . In the limit β˙ φ, α¨ +

B ˙˙ βφ cos θ sin α = 0 A

This is the equation for the simple pendulum (Appendix B) and we see that the gyrocompass oscillates about the northerly direction. ♦

3.2 Lagrangian Mechanics

Here we review the derivation of the Euler–Lagrange equations by variational methods. We also introduce natural dynamical systems, the notion of an affine connection, d’Alembert’s principle and Hamilton’ principles. 3.2.1 Variational Methods

Consider a mechanical system governed by a Lagrangian L(q, q˙ ) = T (q, q˙ ) − U (q)


where q = (q1 (t), . . . , qn (t)) and q˙ = (q˙ 1 (t), . . . , q˙ n (t)) are state space coordinates for T M and M is the configuration space for the system. T is the kinetic energy and U is the potential energy. Consider a trajectory q(t) which starts at q a at time a and ends at qb at time b. A variation of q is, roughly speaking,3 an embedding of the trajectory in a two parameter family of curves q(t, u) such that4 q( a, u) = q a

q(b, u) = qb

q(t, 0) = q(t)

Let w be the tangent, evaluated at u = 0, to the curve q(t, u) with t constant, that is w = q u | u =0 Then

) ) ) d d i )) d d i )) d d i )) q˙ ) q) q) = = = wi du u=0 du dt u=0 dt du u=0 dt

(3.10)

Given a function f (q, q˙ ), let ) ) ) d f )) = f q˙ · q˙ u )u=0 + f q · qu )u=0 = f q˙ · wt + f q · w δf = du )u=0 The equations of motion are now derived from Hamilton’s principle that the actual trajectories yield a stationary point of the action functional S=

b a

L(q(t), q˙ (t)) dt

The variation of S is δS

= = =

b a

δL dt

b

Lq · w + Lq˙ · wt dt a )b b

) d d ( Lq˙ · w))) + Lq − Lq˙ · w dt dt dt a a

Integration by parts has been used to obtain the last line. The first term vanishes because w( a) = w(b) = 0. Since w is arbitrary, the part of the integrand in parentheses must also vanish. Therefore we obtain the Euler–Lagrange equations d Lq˙ − Lq = 0 dt 3) For more precise definitions of a variation see [1, 17, 23, 24] 4) We are using q in two different senses which is an abuse of the notation, but it will be used only when there is no ambiguity.

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Example 3.3 A symmetric rigid body rotates about a fixed point in a conservative force field. Let the coordinates be the Euler angles q1 = φ

q2 = θ

q3 = ψ

so that the potential function is U = U (φ, θ, ψ). The inertia tensor written in a principal axis frame is diag{ A, B, C } and the symmetry condition is A = B. Then from (2.19), the Lagrangian in the principal axis moving frame is $ 1 # ˙2 L(q, q˙ ) = A(φ sin2 θ + θ˙ 2 ) + C (φ˙ cos θ + ψ˙ )2 − U (φ, θ, ψ) 2 The Euler–Lagrange equations yield $ d #˙ φ( A sin2 θ + C cos2 θ ) + C cos θ ψ˙ = −Uφ dt Aθ¨ − A sin θ cos θ φ˙ 2 + C φ˙ sin θ (φ˙ cos θ + ψ˙ ) = −Uθ C

d ˙ (φ cos θ + ψ˙ ) = −Uψ dt

♦

The discussion of the Euler–Lagrange equations has made no special mention of rigid bodies. The derivation and the resulting equations apply equally well to particles and rigid bodies. When applied to rigid bodies, the generalized coordinates include those which completely specify the orientation of the body or bodies, such as the Euler angles. The gradient of the potential yields a generalized torque acting on the body or bodies. To demonstrate this last point we consider an infinitesimal displacement δr of an infinitesimal element of mass and the consequent variation in u, the potential energy per unit mass. To relate gradients of the potential energy and the torque one views the variation in u in two different ways. On one hand, we can view u as dependent on the position of the particle u = u ( r) On the other hand, being a rigid body, each particle position is uniquely dependent on the generalized coordinates qi which determine the position of the reference point and the orientation of the body or bodies. u = u(q1 , . . . , qn ) = u(r(q1 , . . . , qn )) For an infinitesimal displacement δr δU

= =

B B

= v·

dm ∇r u · δr dm f · (vdt + ω × rdt)

B

dm f dt + ω ·

= (v · F + ω · τ )dt

B

dm r × f dt


where f is the force per unit mass. On the other hand, δU =

B

dm ∇q u · δq = δq · ∇q U = q˙ · ∇U dt

Since dt is arbitrary it follows that q˙ · ∇q U = v · F + ω · τ Now let qr be the generalized coordinates which determine the orientation and assume we may find vectors ui such that the angular velocity has components q˙ ri , that is ω = q˙ ri ui Then q˙ ri

∂U = q˙ri (τ · ui ) ∂qri

Since the system is unconstrained q˙ ri are arbitrary and ∂U = τ · ui ∂qri which shows that the derivative of the potential function with respect to the orienting coordinates produces components of the torque. For example, in terms of Euler angles ˙ + ψe ˙ e3 + θe ˙ 3 ω = φ¯ 1 so that u1 = e¯ 3 and

∂U = τ · e¯ 3 ∂φ

u2 = e1

u3 = e 3

∂U = τ · e1 ∂θ

∂U = τ · e3 ∂ψ

If there are forces or torques not derived from a potential function they are included on the right-hand side as d Lq˙ − Lq = τ dt This follows from generalized Hamilton’s principle that the actual trajectory of a dynamical system is the one which satisfies b a

(δT − δW )dt = 0

where T (q, q˙ ) is the kinetic energy and δW is the virtual work. The virtual work is δW =

N

∑ Fi · δri i

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where Fi is the force or torque, expressed in a fixed Cartesian frame, acting on particle or body i out of a total of N particles and bodies. Using the same techniques as before we express these variations in terms of generalized coordinates q1 , . . . , qn as δT

=

δW

=

d d ( Tq˙ · δq) − ( Tq˙ − Tq ) · δq dt dt N n N ∂ri j ∑ Fi · ∑ ∂q j δq = ∑ Q j δq j i =1 j =1 j

where Q j = ∑iN=1 Fi · ∂ri /∂q j is called the generalized force. Therefore we obtain the Lagrange equations of the second kind d Tq˙ − Tq = Q dt If the forces are potential, Q = −∇q U (q), these equations reduce to the Euler– Lagrange equations just derived.

D’Alembert’s Principle

Hamilton’s principle is, in the case of unconstrained motion and potential imposed forces, equivalent to d’Alembert’s principle that n

d (3.11) ∑ dt Lq˙i − Lqi · δqi = 0 i for arbitrary virtual displacements δqi . A system with an n-dimensional configuration space M is unconstrained if any tangent vector in T M is a suitable initial condition. That is, there is no restriction imposed on the possible initial conditions. In the unconstrained case a virtual displacement is simply a tangent vector, δqi ∈ T M, and we can find n linearly dependent ones. The fact that δqi are arbitrary in (3.11) implies that each term must vanish which in turn implies the Euler–Lagrange equations. In the next chapter we will consider constrained systems for which virtual displacements acquire restrictions. 3.2.2 Natural Systems and Connections

An important special case is that of a natural system which has a Lagrangian of the form 1 L = T (q, q˙ ) − V (q) = q˙ t M (q)q˙ − V (q) (3.12) 2 The Euler-Lagrange equations for a natural system take the form ˙ q˙ − Tq + Vq = 0. M q¨ + M

(3.13)


˙ q˙ − Tq comprises the gyroscopic terms. In a component form The group c = M these terms are 1 1 ci = Mij,qk q˙ j q˙ k − M jk,qi q˙ j q˙ k = ( Mij,qk + Mik,q j − M jk,qi )q˙ j q˙ k = Γijk q˙ j q˙ k 2 2 and expressed in terms of the Christoffel symbols of the first kind Γijk for the metric Mij , ∂M jk ∂M ji 1 ∂Mik + − Γijk = (3.14) 2 ∂q j ∂qi ∂qk The Christoffel symbols of the second kind are Γijk = ( M −1 )il Γl jk

(3.15)

Thus the equations of motion for a natural system become qï + Γijk q˙j q˙k = −( M −1 )il Vql Example 3.4 Consider the toroidal coordinates defined by x

= ( a + ξ cos θ ) cos φ

y

= ( a + ξ cos θ ) sin φ

z

= ξ sin θ

which map 0 ≤ θ, φ < 2π, 0 ≤ ξ < a into the interior of the “apple” in R3 . We can find the Christoffel symbols in this coordinate system by writing down the Lagrangian for the motion of a free particle in these coordinates. The velocity components in the two-coordinate systems are related by ˙ sin θ cos φ + ξ˙ cos θ cos φ = −φ˙ ( a + ξ cos θ ) sin φ − θξ ˙ sin θ sin φ + ξ˙ cos θ sin φ y˙ = φ˙ ( a + ξ cos θ ) cos φ − θξ ˙ cos θ + ξ˙ sin θ z˙ = θξ

x˙

Now we write the Lagrangian L

= =

1 2 ( x˙ + y˙ 2 + z˙ 2 ) 2 $ 1 # ˙2 φ ( a + ξ cos θ )2 + θ˙ 2 ξ 2 + ξ˙2 2

The equations of motion are ξ¨ − φ˙ 2 ( a + ξ cos θ ) cos θ − ξ θ˙ 2 d 2˙ ξ θ + φ˙ 2 ( a + ξ cos θ )ξ sin θ dt $ d # ( a + ξ cos θ )2 φ˙ dt

= 0 = 0 = 0

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or ξ¨ − φ˙ 2 ( a + ξ cos θ ) cos θ − ξ θ˙ 2 2 a θ¨ + θ˙ ξ˙ + sin θ ( + cos θ )φ˙ 2 ξ ξ 2 ˙ sin θ )φ˙ (ξ˙ cos θ − θξ φ¨ + ( a + ξ cos θ )

= 0 = 0 = 0

Given the general form of the equations of motion for a free particle qï + Γijk q˙ j q˙ k = 0 and the ordering q1 = ξ, q2 = θ, q3 = φ we can read the nonzero Christoffel symbols directly Γ133 = −( a + ξ cos θ ) cos θ Γ212 = Γ221 = 1/ξ Γ313 = Γ331 = cos θ/( a + ξ cos θ )

Γ122 = −ξ Γ233 = sin θ ( a/ξ + cos θ ) Γ323 = Γ332 = −ξ sin θ/( a + ξ cos θ ) ♦

The modern differential geometric view [25] of this development is that the Christoffel symbols are components of an affine connection ∇. The coefficients are derived from the covariant derivative ∇ X , where X is a vector field. For vector field Y and function f , ∇ X Y, ∇ f X Y, and ∇ X f Y are vector fields which satisfy

∇ X Y is bilinear in X and Y ∇ f X Y = f ∇X Y ∇ X f Y = f ∇ X Y + X ( f )Y If e is a frame, that is a set of vector fields forming a basis for Tq M at each point q in a configuration manifold M, then the components of a connection ∇ are the coefficients Gijk which satisfy ∇ei e j = Gijk ek . Given X = X k ek and Y = Y k ek the component expression for ∇ X Y is

∇ X Y = X (Y i )ei + Y i X (ei ) = [ X (Y k ) + X i Y j Gijk ]ek The covariant derivative is extended to higher order tensors by applying the Leibniz rule. For example, let P = eP be a linear operator relative to a frame e and its dual . Since ei , j = δij we have 0 = ∇ X ei , j + ei, ∇ X j , j

which implies ∇ei j = − Gki k . Now we can calculate

∇X P

= (∇ X ei ) Pji j + ei (∇ X Pji ) j + ei Pji (∇ X j ) # $ i l j = ei X ( Pji ) + X k Pjl Gkl − X k Pli Gkj

3.2 Lagrangian Mechanics j

j

The covariant derivative based on the Christoffel symbols Gik = Γik (3.14) has the following expression in coordinates relative to a natural or holonomic basis j ∇∂ i Y = (∂qi Y j + Y k Γik )∂q j q

This connection is known as the Levi-Civita connection. The Christoffel symbols are symmetric in the lower indices but this is not a general feature of connections. The Levi-Civita connection is distinguished by its unique compatibility with the metric which means that is ∇ X G = ∇ X gij bi b j = 0 for all X where {bi } is the natural basis {∂qi } and bi is its dual. Using the procedure just followed we calculate

∇∂ i G = (∂qi g jk − glk Γlij − g jl Γlik )b j bk q

Since ∇∂ i G vanishes we have q

∂qi g jk − glk Γlij − g jl Γlik

= 0

∂q j gki −

gkl Γlji

= 0

∂qk gij − gl j Γlki − gil Γlkj

= 0

gli Γljk

−

where the last two equations follow from the first by cyclic permutation of the indices. From this we find glk Γlij =

1 (∂ i g + ∂q j gik − ∂qk gij ) 2 q jk

and

1 mk g (∂qi g jk + ∂q j gik − ∂qk gij ) 2 Thus, the Christoffel symbols are recovered. The Euler–Lagrange equations for a natural system can be expressed in the connection notation as ∇q˙ q˙ + M −1 Vq = 0 (3.16) Γm ij =

We will need to invoke a more general connection in the chapter on constrained motion. 3.2.3 Poincaré’s Equations

The Euler–Lagrange equations are tied to the variables q, q˙ although these variables are not always the most advantageous. In the case of rigid body

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motion, for example, it is natural to use angular velocity in place of the coordinate derivatives. A generalized set of equations due to Poincaré [26] makes this possible. The derivation of Poincaré’s equations uses the concepts of vector fields on manifolds, structure coefficients for Lie algebras, and variational equations. References for this section are [17, 20, 23, 24, 26]. The works [27, 28] include basic results for Lie groups.

On a Manifold

Given an n-dimensional manifold M with local coordinates {qi }, the natural, or holonomic, basis for the tangent space is {∂/∂qi } which we can abbreviate as {∂qi }. We wish to consider an alternative nonholonomic basis for vector fields on M. Let the alternative basis be denoted by Xi , i = 1, . . . , n with components

µ

Xi = Xi ∂ q µ acting on functions according to µ

Xi ( f ) = Xi ∂ q µ f The time derivative of a function f (q) may be expressed as either ∂f df = q˙ µ µ dt ∂q or

df = s µ Xµ f dt in the holonomic and nonholonomic bases, respectively. In an operator form d d = q˙ µ ∂qµ or = s µ Xµ dt dt

(3.17)

Therefore the components in the two bases are related by q˙ i = sµ Xµi Some works, the textbooks [20, 22] for example, refer to the nonholonomic basis as a quasi-velocity. Now consider a variation qi (t, u) of qi (t), described in Section 3.2, and let the variational derivative be expressed in the nonholonomic basis as d = w µ Xµ du

(3.18)


The equations of motion in terms of {q, s} are found, as before, by using the variation d/du|u=0 to find a stationary point of the action integral. If X ji = δji then si = q˙ i and the development reverts to the Euler–Lagrange case (Section 3.2). The first step in the derivation is to determine how to vary s. The analogous question in the Euler–Lagrange problem of how to vary q˙ was solved in (3.10) by using the condition d d d d (3.19) = dt du du dt Here we will use the same condition to relate dwi /dt and dsi /du thereby generalizing (3.10). From (3.17), (3.18), relation (3.19) is equivalent to s k Xk ( w j X j ) = w k Xk ( s j X j ) which yields

( s k Xk w j ) X j + s k w j Xk X j = ( w k Xk s j ) X j + w k s j Xk X j or

d j d s − wj du dt

Xj

= s k w j Xk X j − X j Xk

= s k w j [ Xk , X j ] i = sk w j Ckj Xi

= [s, w]i Xi i are the structure coefficients relative to the basis { X } for the Lie where Ckj i

algebra Tq M with the Lie derivative bracket. [s, w]i denotes the ith component i is of the Lie bracket. The expression for Ckj µ µ i Ckj = Yli Xk ∂qµ X jl − X j ∂qµ Xkl where the functions Yji are defined by the condition j

Yji Xk = δki We conclude that d d d d i s = wi + [s, w]i or s = w + [s, w] du dt du dt because Xi are linearly independent.

(3.20)

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Given a Lagrangian L(q, s) = T (q, s) − U (q) with the kinetic energy T and the potential energy U, the action integral is S=

b a

L[q(t, u), s(t, u)] dt

The variation of S is δS = with

=

δL

=

b a

δL dt

) d )) L) du u=0

) ) ( Lqi qiu + Lsi siu ))

u =0

Thus, using (3.20), we obtain

=

δS

b# a

i Lqi qiu + Lsi wit + Lsi sk w j Ckj )

$ u =0

dt

The first term of the integrand may be rearranged to Lqi qiu = Lqi w j X j qi = w j X ji Lqi = w j X j ( L) and the second Lsi wit = ( Lsi wi )t − ( Lsi )t wi leading to δS =

b % a

i Xk ( L) + Lsi s j Cjk −

& d Ls k wk + ( Ls i wi )t d t dt

After integration, the fact that w( a, 0) = w(b, 0) = 0 eliminates the last term. Since w is arbitrary the part of the integrand in brackets must vanish. Therefore we obtain Poincaré’s equations d i L j − Lsi sk Ckj = X j ( L) dt s Insight and ease of application are gained by connecting Poincaré’s equations to the Lie algebra operators ad s and ad∗s . Recall from Section C.3 that if s is a matrix Lie algebra, then for s ∈ s, ad s is the matrix operating on s according to i [ ads ]ij = sk Ckj Similarly, if s∗ is the Lie algebra dual to s, then for s ∈ s, ad ∗s is the matrix operating on s∗ according to

[ ad∗s ]ij = sk Cki j


We note again that

ad ∗s = ad ts

and that the relation between ad s and ad ∗s simply restates the matrix identity

x, Ay = At x, y With this we can rewrite the Poincaré equations in the intrinsic form d Ls − ad∗s Ls = X ( L) dt where X ( L ) = [ X1 ( l ) · · · X n ] t The complicated sum over the structure coefficients is replaced with a matrix multiplied by ad ∗s . Example 3.5 A point mass moves freely in space and we wish to obtain its equations of motion in spherical coordinates q and the “physical” components of velocity s q = {r, θ, φ},

˙ r sin θ φ˙ } ≡ {u, v, w} ˙ r θ, s = {r,

(3.21)

The general relation defining the nonholonomic basis is q˙ i ∂qi = s j X j For this example we see by inspection that 1 1 ∂φ } X = { ∂r , ∂ θ , r r sin θ The nonzero Lie brackets are 1 1 [ X1 , X2 ] = − X2 [ X1 , X3 ] = − X3 r r so the nonzero structure coefficients are 1 r Then, from the Lagrangian 2 2 C12 = −C21 =−

3 3 C13 = −C31 =−

L=T=

1 r

[ X2 , X3 ] = −

3 3 C23 = −C32 =−

$ 1# 2 u + v2 + w2 − U (r, θ, φ) 2

the Poincaré equations yield u˙

=

v˙

=

w˙

=

cot θ X3 r

v 2 + w2 − Ur r −uv + w2 cot θ 1 − Uθ r r uw + vw cot θ 1 Uφ − − r r sin θ

♦

cot θ r

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Example 3.6 Consider the rotational motion of a rigid body with fixed center of mass. Let the configuration space, SO(3), be parameterized by the Euler angles. The Lagrangian in body fixed, principal axis coordinates in terms of the angular velocity is 1 L = Ai ωi2 − U (φ, θ, ψ) 2 From Eq. (2.9) we have      ω1 sin ψ/ sin θ cos ψ/ sin θ 0 φ˙  θ˙  =  − sin ψ 0  ω2  (3.22) cos ψ − cos θ sin ψ/ sin θ − cos θ cos ψ/ sin θ 1 ω3 ψ˙ The defining relation for the nonholonomic basis is now φ˙ ∂φ + θ˙ ∂θ + ψ˙ ∂ψ = ωi Xi and from (3.22) we can write immediately X1 = (sin ψ/ sin θ )∂φ + cos ψ ∂θ − (cos θ sin ψ/ sin θ )∂ψ

(3.23)

X2 = (cos ψ/ sin θ )∂φ − sin ψ ∂θ − (cos θ cos ψ/ sin θ )∂ψ

(3.24)

X3 = ∂ ψ

(3.25)

Calculating the Lie brackets we find

[ Xi , X j ] = ijk Xk so the structure coefficients are Cijk = ijk From this we can write ad ∗ω immediately t = −ω ad ∗ω = ω and from that the Poincaré equations are dωi + ijk ω j Ak ωk = Xi (U ) (no sum on i ) dt The forced Euler equations are recovered. ♦ Ai

In Example 3.6 the fact that the structure coefficients are the constants ijk implies that the manifold is a Lie group and the tangent space span{ X1 , X2 , X3 } at each point on the manifold is isomorphic (as a Lie algebra) to so(3). The vector fields ∂φ , ∂θ , and ∂ψ are matrices in the tangent space TSO(3). We can express them (Section 1.1.6) in terms of the matrix factors Re3 (φ), Re1 (θ ), and Re3 (ψ) as ∂φ

=

∂θ

=

∂ψ

=

Re3 (φ) Re1 (θ ) Re3 (ψ) Re3 (φ) Re1 (θ ) Re3 (ψ) Re3 (φ) Re1 (θ ) Re3 (ψ)


Now we can exploit the Lie group structure to left translate these vector fields to the identity and to discover the true significance of the vector fields Xi . We find R −1 ∂ ψ

= e3 = ade3

R −1 ∂ θ

=

R3t e1 R3 = ad e

R −1 ∂ φ

=

R3t R2t e3 R2 R3 = ade¯3

1

Applying the coefficients from Eqs. (3.23–3.25) we find the result R−1 Xi = adei This shows that Xi are the infinitesimal generators of rotations about the body fixed axes left translated to the tangent space of the Lie group. This is a particularly nice example of insight gained from the Lie group point of view.

On a Lie Group

This section considers the specialization of the Poincaré equations to systems where the configuration space M is a Lie group. That is, M = G where G a matrix group with Lie algebra g. The functional form of the Lagrangian in this case is L = L( G, G˙ ), ( G, G˙ ) ∈ T G The Lie group structure allows us to work in the trivialization G × g ∼ = TG where the Lagrangian assumes the functional form L = L( G, v), ( G, v) ∈ G × g, v = G −1 G˙ Whereas the derivation of the Poincaré equations on a manifold relied heavily on the parameterization, or the coordinates, on the manifold our treatment of the Lie group case will emphasize operations with the group elements themselves which are, of course, matrices. The vector fields Xi played a key role in the manifold case and in the case of the Lie group they are replaced by the linear operator D defined by its action on L( G, v). For any w ∈ g ) ) d w L(e G, v))) DL, w = d =0 where a, b = − 12 Tr ab is an inner product on g [27, 28]. We can motivate this operation by recalling that for functions on R n the gradient can be defined as the vector ∇ f ( x ) which, for any v ∈ R n , satisfies ) ) d f ( x + v ))) ∇ f (x) · v = d =0

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This is the derivative along a curve obtained by translation. The analog of x + v in a Lie group is ev G. To determine how to vary elements of g we consider a variation G (t, u) of G (t), as described in Section 3.2 and define s, w ∈ g by s = G −1

dG dG and w = G −1 dt du

The derivatives ds/du and dw/dt are ds dG −1 dG d2 d2 G G = −ws + G −1 G = − G −1 + G −1 du du dt du dt du dt and

dw dG −1 dG d2 d2 = − G −1 + G −1 G G = −sw + G −1 G dt dt du dt du dt du From the difference of these two expressions and use of the equality of the second-order derivatives we obtain dw dw ds = + [s, w] = + ads w du dt dt

This is the analog for a matrix group of (3.20) for the manifold. The variation of the action is δS =

b a

δL dt

with5 δL

= = =

d ds L(ew G, s)|=0 + Ls , d du dw + ads w DL, w + Ls , dt )b ) d d DL − Ls + ad∗s Ls , w + Ls , w)) dt dt a

Therefore δS = 0 and w arbitrary imply d Ls − ad∗s Ls = DL dt 5) We are taking liberties with the notation here. Strictly speaking, L s should be the functional derivative δL/δs [17] which is defined as δL/δs i , δs = lim→0 ( L( s., . . . , s i + δs i , . . . , s n ) − L( s ))/. Operationally, however, nothing changes for our applications. For example δ( w2 ) /δw = ( w2 ) w = 2w.


Example 3.7 A Lagrange top is an axi-symmetric top (Example 3.1) whose center of mass lies on the axis of symmetry and which has one point of the axis of symmetry is fixed. Symmetry implies equality of two of the principal moments of inertia, say A = B. The Lagrangian for this system, expressed in terms of angular velocity and Euler coordinate θ, is L(θ, ω) =

1 1 Aω · ω + (C − A)(e3 · ω)2 − mg(e3 · e¯ 3 ) 2 2

where m is the mass of the top, g is the acceleration of gravity and is the distance from the tip of the top to the center of mass. e3 lies on the axis of symmetry. The goal of this example is to rewrite the Lagrangian on SU (2) and derive the equations of motion using the Poincaré equations for a Lie group. The Lagrangian will have the form [28] L = L(S, v )

S ∈ SU (2)

v ∈ su(2)

To translate the Lagrange top Lagrangian to this setting one replaces the vector dot product with the Lie algebra inner product. Following Sections 2.2.3 and 1.3.1 the vectors ω and e3 are replaced by the matrices

w=

˙ ˙ − βγ αδ ˙ + δγ˙ −δγ

˙ + α β˙ −αβ ˙ − γβ ˙ δα

k=

ı 0

0 −ı

and the vector e3 is replaced by

k = SkS†

S=

α γ

β δ

where α, β, γ, and δ are the Cayley–Klein parameters. The result is the Lagrangian L(S, w) =

1 1 Aw, w + (C − A)SkS† , w2 − mgSkS† , k 2 2

To derive the equations of motion we first recall that SU (2) has the same Lie algebra as SO(3) so that the structure constants are the Levi-Civita symbols ijk . We need ∂L/∂w which is ∂L = Aw + (C − A)k, wk ∂w

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Next we apply the operator D to L and calculate d v e SkS† e−v , k|=0 = (vev SkS† e−v , k)|=0 + (ev SkS† (−v)e−v , k)|=0 d = vSkS† − SkS† v, k

= [v, k], k = [k, k], v, d v e SkS† e−v , w2 |=0 = 2w, k[v, k], w d = 2w, k[k, w], v Therefore the equations of motion can be expressed as d p + [ p, w] = ( A − C )k, w[k, w] + mg[k, k] dt where p = Aw + (C − A)k, wk.

♦

3.3 Hamiltonian Mechanics

This section discusses the application of Hamiltonian mechanics to rigid body dynamics. Recall that from a Lagrangian L(q, q˙ ) one obtains the equations of motion d Lq˙ = Lq dt Generalized momenta are then defined as p = Lq˙ and the Hamiltonian is introduced by means of the Legendre transformation, H = p · q˙ − L In the new variables q, p the equations of motion are q˙ = H p

p˙ = − Hq

Mechanics in its modern geometric form sets Lagrangian mechanics in the tangent manifold T M of the configuration space M. Hamiltonian mechanics is set in the cotangent manifold T ∗ M. These two manifolds are related in local coordinates by the maps

(q, q˙ ) → (q, p) = (q, Lq˙ ) (q, p) → (q, q˙ ) = (q, H p )


We will use some basic theory of differential forms here and in Chapter 5 and now summarize the prerequisites. The book [1] can be consulted for more details. The operators ∂qi form a basis for T Mq , the tangent space at q ∈ M. The tangent vectors v ∈ T Mq are linear combinations of the basis vectors and can be expressed concisely as ui ∂qi . The dual basis for the covectors in T ∗ Mq consists of the differential 1-forms dqi defined by their action on the basis vectors dqi (∂q j ) = δij . The cotangent vectors ω ∈ T ∗ M are linear combinations of the basis 1-forms and are expressed as ω = p i dqi . The algebra of differential forms includes the wedge product or outer product . The wedge product of the basis 1-forms is defined by dqi ∧ dq j = dqi ⊗ dq j − dq j ⊗ dqi where ⊗ is the vector space tensor product which was used in Chapter 1. The 2-form dqi ∧ dq j acts on pairs of vectors according to dqi ∧ dq j (uk ∂qk , v l ∂ql ) = ui v j − u j vi Note that dqi ∧ dq j = −dq j ∧ dqi . As in Chapter 1 the wedge product is extended by bilinearity to all of T ∗ Mq ωij dqi ∧ dq j (uk ∂qk , v l ∂ql ) = ωij (ui v j − u j vi ) The defining property of the co-basis {dqi } can be written in terms of a vector contraction as ∂qi dq j = δij and this extends to any vector ui ∂qi

ω j dq j = ui ωi

This extends to the 2-form α where u ui ∂qi

α returns a 1-form according to

α jk dq j ∧ dqk = ui αij dq j

In general, if α is an n-form, then u α is an (n − 1)-form. If f = f (q) is a real-valued function on M, then the differential of f is the 1-form ω = f qi dqi . The differential operator can be applied to any form. The differential operator applied to n-forms results in an (n + 1)-form. If ω = ωi dqi , then dω = ωi,q j dq j ∧ dqi Note that d(dqi ) = 0. An n-form ω is exact if there is an (n − 1)-form σ such that ω = dσ. An example of an exact form is d f . An n-form ω is closed if dω = 0. Every exact form is closed but there are closed forms which are not exact. A 2-form ω = ωij dqi ∧ dq j is nondegenerate if det[ωij ] = 0. Example 3.8 The form d f is closed. By definition of the differential operator d(d f ) = d( f qi dqi ) = f qi ,q j dqi ∧ dq j

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and if this is applied to any vector pair (u, v ) = (ui ∂qi , v i ∂qi ), then

[d(d f )](u, v) = f qi ,q j ui v j − f q j ,qi ui v j = 0 ♦ The geometric view of Hamiltonian mechanics gives prominence to a symplectic form which is a closed, nondegenerate 2-form Ω on a differential manifold of an even dimension, Ω : T M × T M → R : X, Y → Ω( X, Y ) ∈ R The symplectic form defines the Hamiltonian vector field X H associated with function H = H (q, p ) (the Hamiltonian) implicitly by the requirement that for every vector field Y Ω( X H , Y ) = dH (Y ) The Darboux theorem [17, 29] asserts that for any symplectic form Ω there is a coordinate system in which it can be written as Ω = ∑ dqi ∧ dpi i

This may be abbreviated as Ω = dq ∧ dp. In these coordinates let X H = X1i ∂qi + X2i ∂ pi and Y = Y1i ∂qi + Y2i ∂ pi be an arbitrary vector field. Then Ω( X H , Y ) = dq ∧ dp X1 · ∂q + X2 · ∂ p , Y1 · ∂q + Y2 · ∂ p = X1 · Y2 − X2 · Y1 and dH (Y ) = Hq · Y1 + H p · Y2 . Since Y is arbitrary it follows that X1 = H p

X2 = − H q

which are the classical Hamiltonian equations. In symplectic coordinates they assume the matrix form

q 0 I z˙ = J Hz , z = J= p −I 0 J is the canonical symplectic matrix. The advantage of the abstract formulation is that it does not require that we work in symplectic coordinates. We have already seen that the analogous freedom in Lagrangian mechanics made possible by the Poincaré equations is quite useful. 3.3.1 Momenta Conjugate to Euler Angles

Consider a rigid body in body fixed, principal axis coordinates with rotations parameterized by the Euler angles. Here the configuration space is SO(3) with


Euler angle coordinates and the tangent manifold T M has coordinates (q, q˙ ) where the abbreviation {φ, θ, φ} = q is used. We wish to find the corresponding coordinates on the cotangent manifold T ∗ M. In other words, we want to find the momenta conjugate to the coordinates q. The Lagrangian for this system is (2.19) L

= =

1 ( A1 ω12 + A2 ω22 + A3 ω32 ) − U (φ, θ, ψ) 2 1 [ A1 (φ˙ sin θ sin ψ + θ˙ cos ψ)2 + A2 (φ˙ sin θ cos ψ − θ˙ sin ψ)2 2 + A3 (φ˙ cos θ + ψ˙ )2 ] − U (φ, θ, ψ)

From this we calculate the conjugate momenta pφ =

∂L = A1 ω1 sin ψ sin θ + A2 ω2 cos ψ sin θ + A3 ω3 cos θ ∂φ˙

∂L = A1 ω1 cos ψ − A2 ω2 sin ψ ∂θ˙ ∂L pψ = = A3 ω3 ∂ψ˙ pθ =

Note that e¯ 3 · h = pφ e 1 · h = p θ and e3 · h = p ψ so that the conjugate momenta are, geometrically, projections of the angular momentum onto the e3 , e¯ 3 , e 1 axes. In matrix notation      pφ A1 sin ψ sin θ A2 cos ψ sin θ A3 cos θ ω1      A1 cos ψ − A2 sin ψ 0  pθ  =    ω2  (3.26) pψ

0

0

ω3

A3

The inverse relation is needed to express the kinetic momenta and thus form the Hamiltonian. We obtain    cos ψ sin ψ − cos θ sin ψ ω1 A1 A1 sin θ   A1 sin θ     sin ψ − cos θ cos ψ  ω2  =  Acossinψ θ − A A2 sin θ 2   2  1 ω3 0 0 A3

energy in terms of the 

 pφ

    pθ  pψ

   

(3.27)

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Therefore the Hamiltonian for this system is6 *

2 1 1 sin ψ cos θ sin ψ pφ − pψ + K = (3.28) cos ψ p θ + 2 A1 sin θ sin θ +

2 cos ψ cos θ cos ψ 1 2 1 pφ − pψ + − sin ψ pθ + p + U (φ, θ, ψ) A2 sin θ sin θ A3 ψ This rather formidable Hamiltonian will not be used directly. Rather, it will be transformed to a new set of canonical variables, the Andoyer variables [12, 30], which gain considerable simplicity. 3.3.2 Andoyer Variables

Fig. 3.3 The arc containing angle g is on the great circle with pole h, the angular momentum (plane Π2 ). The arc containing angle l is on the great circle with pole e3 (plane Π3 ). The arc containing angle h is on the great circle with pole e¯ 3 (plane Π1 ). 6) K is used to denote the Hamiltonian here for reasons which will soon be clear.


We will present the Andoyer variables from two points of view. First, following [31], Andoyer variables are derived by specifying the momenta and evaluating a generating function which produces a symplectic transformation from (φ, θ, φ, p φ, p θ , p φ ) to the new variables which are denoted as (l, g, h, L, G, H ). Second, following [32], the defining relations are shown directly to produce symplectic transformation. The momenta specified for the Andoyer variables are L = pψ G = h H = pφ From (3.27) we see that G is related to the momenta p by # $1 2 G = p2θ + ( pφ − pψ cos θ )2 / sin2 θ + p2ψ The inverse relations are pψ = L

(3.29) #

pθ = ± G2 − L2 − ( H − L cos θ )2 / sin2 θ

$1 2

(3.30)

pφ = H

(3.31)

Let us abbreviate the Euler angles by q, the momenta conjugate to the Euler angles by p, the Andoyer momenta by P, and the conjugate coordinates by Q. Since the transformation from q, p to Q, P is to be symplectic the differential forms p dq and Q dP can differ only by a closed form. This introduces the generating function S(q, P) p dq + Q dP = dS(q, P) This relation insures that the transformation is symplectic, i.e., dq dp = dQ dP, and provides the generating equations p = Sq

Q = SP

We use the first of these equations to obtain the generating function itself S

= = =

p dq

p φ ( P) dφ + pθ ( P) dθ + pψ ( P) dψ

Lψ + Hφ +

θ# θ∗

G2 − L2 − ( H − L cos θ )2 / sin2 θ

$1 2

dθ

(3.32)

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The lower limit of the integral, θ∗ , is the larger root of G2 cos2 θ − 2LH cos θ + L2 + H 2 − G2 = 0 Let us define cos η = H/G

and

cos λ = L/G

Then θ∗ is the larger root of cos2 θ − 2 cos η cos λ cos θ + cos η 2 + cos λ2 − 1 = 0 which is θ∗ = η + λ Now we use the second of the generating function equations to obtain the conjugate coordinates l, g, h. For l: θ# $− 1 2 l−ψ = × G2 − L2 − ( H − L cos θ )2 / sin2 θ θ # ∗ $ − L + cos θ ( H − L cos θ )/ sin2 θ dθ.

= =

θ θ∗

cos η cos θ − cos λ

sin θ + 2 cos η cos λ cos θ − cos2 η − cos2 λ cos η − cos λ cos θ π − arcsin 2 sin η sin θ sin θ

2

dθ

from which it follows that cos η = cos λ cos θ + sin λ sin θ cos(ψ − l )

(3.33)

For g: g

=

θ θ∗

= arcsin

sin θ sin2 θ + 2 cos η cos λ cos θ − cos2 η − cos2 λ

dθ

π cos η cos λ − cos θ − sin λ sin η 2

from which it follows that cos θ = cos η cos λ + sin η sin λ cos g For h: h−φ

=

θ θ∗

sin θ

= arcsin

cos η − cos λ cos θ sin θ + 2 cos η cos λ cos θ − cos2 η − cos2 λ 2

π cos λ − cos η cos θ − sin η sin θ 2

(3.34)

dθ


from which it follows that cos λ = cos η cos θ + sin η sin θ cos(φ − h) Restoring L, G, H in Eqs. (3.33–3.35) we obtain , H = L cos θ + G2 − L2 sin θ cos(ψ − l ) , , G2 cos θ = HL + G2 − H 2 G2 − L2 cos g , L = H cos θ + G2 − H 2 sin θ cos(φ − h)

(3.35)

(3.36) (3.37) (3.38)

Now, we refer to the law of cosines for spherical triangles (A.13–A.15) to connect the analytical expressions (3.33–3.35) to the following geometrical description (Fig. 3.3). The spherical triangle formed on the unit sphere by the planes with normals e¯ 3 , e3 , and h has angles η, π − θ, λ and sides g, φ − h, ψ − l. Having derived a coordinate–conjugate momentum pair we follow [32] to verify directly, by a rather concise derivation, that this pair is the result of a symplectic transformation. This is, of course, redundant but enlightening nonetheless. We begin by referring to the spherical triangle in Appendix A and setting the correspondences { A, B, C, α, β, γ} and {η, π − θ, λ, ψ − l, g, φ − h}. It follows from (A.11) that cos g = cos(ψ − l ) cos(φ − h) − sin(ψ − l ) sin(φ − h) cos θ From the differential of this last equation, and using (A.8), (A.18), (A.19), we obtain dg = cos λ d(ψ − l ) + cos η d(φ − h) − sin λ sin(ψ − l )dθ From (3.30), (3.36) it follows that (with the right choice of sign) , pθ = − G2 − L2 sin(ψ − l ) Now multiply (3.39) by G and use (3.34) to obtain G dg = L d(ψ − l ) + H d (φ − h) + pθ dθ or L dl + G dg + H dh = p φ dφ + pθ dθ + pψ dψ which proves that the transformation is symplectic.

(3.39)

(3.40)

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Example 3.9 We want to derive the Hamiltonian for free rotation in terms of the Andoyer variables. Combining Eq. (3.27) and Eqs. (3.29), (3.31), (3.40) we find , h1 = G sin λ sin l = G2 − L2 sin l (3.41) , h2 = G sin λ cos l = G2 − L2 cos l (3.42) h3

= G cos λ

(3.43)

The Hamiltonian follows immediately K=

$ 1# 2 ( G − L2 ) a1 sin2 l + a2 cos2 l + a3 L2 2

with ai =

1 . Ai

(3.44)

♦

Example 3.10 Example 2.3 derived the rotational kinetic energy for a three-rotor gyrostat. Here we want to express this in Andoyer variables. The procedure is the same as in Example 3.9. From (2.20) and (3.41) ,

2 G2 − L2 sin2 l + µ1 2 , + (b G2 − L2 cos2 l + µ2

a 2K = (

+ (b( L + µ3 )2

(3.45) (3.46) (3.47)

where ( a = a + m2 d22 a2 + m3 d23 a2 ( b = b + m2 d22 b2 + m3 d23 b2 c( = c + m2 d22 c2 + m3 d23 c2 µ1 = aA1 s1 /( a

µ2 = bB2 s2 /( b

µ3 = cC3 s3 /( c.

The constant terms, which are irrelevant as far as the equations of motion are concerned, have been modified to obtain the Hamiltonian in a convenient form. ♦ 3.3.3 Brackets

This section reviews the Hamiltonian formulation in terms of Poisson brackets and its generalization to Lie–Poisson brackets [16, 17, 29]. Given conjugate coordinates and momenta q, p and functions f (q, p ) and g (q, p), the classical Poisson bracket { f , g} is defined by

{ f , g} = ∑ i

∂ f ∂g ∂ f ∂g − i i ∂qi ∂pi ∂p ∂q


or, in abbreviated notation

{ f , g } = f q · g p − f p · gq If H is the Hamiltonian, then

{ f , H } = f q · H p − f p · Hq = f q · q˙ + f p · p˙ = f˙ so { f , H } is the rate of change of f along trajectories of the system and the equations of motion can be expressed as q˙ = {q, H }

p˙ = { p, H }

Example 3.11 Let us use the classical Poisson bracket to derive the equations for rotational motion in terms of the components of angular momentum. The Hamiltonian in the principal axis coordinates can be written in terms of Euler angles and the components of angular momentum as h23 h22 1 h21 H= + + + U (φ, θ, φ) 2 A1 A2 A3 Note hi are not the conjugate momenta for the Euler angles and we are not deriving the equations of motion in the Hamiltonian form. Rather, we are using the bracket formalism to derive the equations of motion in hi and Euler angles. h i are expressed in terms of Euler angles and the conjugate momenta as in (3.27), h1

=

h2

=

h3

=

sin ψ cos θ sin ψ pφ + cos ψp θ − pψ sin θ sin θ cos ψ cos θ cos ψ pφ − sin ψp θ − pψ sin θ sin θ pψ

It is straightforward to calculate the brackets

{h2 , h1 } = h3 = −{h1 , h2 } {h1 , h3 } = h2 = −{h3 , h1 } {h3 , h2 } = h1 = −{h2 , h3 } Now h˙ 1

= {h1 , H } h21 h23 h22 1 {h1 , } + {h1 , } + {h1 , } + {h1 , U } = 2 A1 A2 A3

h1 h2 h3 {h , h } + {h , h2 } + {h , h3 } + {h1 , U } = A1 1 1 A2 1 A3 1

1 ∂U 1 sin ψ ∂U cos θ sin ψ ∂U − − − cos ψ + = h2 h3 A3 A2 sin θ ∂φ ∂θ sin θ ∂ψ

The Leibniz property has been used.

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Similarly h˙ 2

= {h2 , H } 1 1 cos ψ ∂U cos θ cos ψ ∂U ∂U − )− + sin ψ + = h1 h3 ( A1 A3 sin θ ∂φ ∂θ sin θ ∂ψ

h˙ 3

= {h3 , H } 1 1 ∂U . ♦ − )− = h2 h1 ( A2 A1 ∂ψ

Example 3.12 The bracket formalism can be used to derive the Euler–Lagrange equations for a given Hamiltonian. Care must be taken to precisely identify function arguments in this derivation. Let the Hamiltonian be H (q, p ) and let the Legendre transformation be expressed as H (q, p ) = p · q˙ (q, p) − ( L (q, p ) with

( L(q, p) = L(q, q˙ (q, p))

Then d Lq˙ dt

= { Lq˙ , H } = { p, p · q˙ (q, p) − (L(q, p)} = −q˙ q p + Lq + q˙ q Lq˙ =

Lq .

♦

Example 3.13 The bracket formalism can also be used to derive the Poincaré equations for a given Hamiltonian. Now it is paramount to precisely identify function arguments. In this instance the Lagrangian is a function of the configuration coordinates q and the generalized velocities s L = L(q, s) Let the Hamiltonian be H (q, p) and express the Legendre transformation as L (q, p ) H (q, p ) = p · q˙ (q, p) − ( with

( L (q, p) = L [q, s (q, q˙ (q, p ))]

The notation to be used for this derivation is to display all the indices, to denote derivatives by subscripts separated by commas and to use the summation convention. Recall from Section 3.2.3 that q˙ i = sµ Xµi The inverse of this is

si = q˙ µ Yµi


where

µ

µ

j

j

j

Xi Yµ = Yi Xµ = δi

(3.48)

The conjugate momenta are related to derivatives of L by µ

µ

pi = Lsµ sq˙ i = Yi L,sµ and the inverse relation is

µ

L,si = Xi pµ µ

j

j

We will need the fact that from Yi Xµ = δi it follows that ν l ν k Yλ,q µ = −Yλ Yk Xl,q µ

Now we compute d L i dt ,s

= { Xiν pν , pν q˙ ν − ( L} ν µ ν ν λ = Xi,q µ p ν ( q˙ + p ν q˙ ,p − L,s ν s λ q˙ ,p ) µ µ ,q˙ µ

ν ν ν λ − Xi ( pν q˙ ,q µ − L,q µ − L,s ν s,q µ − L,s ν s λ q˙ ,q ) µ ,q˙ µ

µ

µ

ν λ σ ν λ σ = Xi,q µ Xσ Yν L,s λ s − Xσ,q µ X Yν L,s λ s + X L,q µ i i λ σ = Cσi s L,sλ + Xi ( L)

In the last step the fact that µ

µ

λ ν ν Cσi = Yνλ ( Xi,q µ Xσ − Xσ,q µ X ) i λ is has been used. An alternative form for Cσi µ

λ λ λ Cσi = Xσν Xi (Yν,q µ − Yµ,q ν )

which is obtained by using (3.48). ♦ We now want to explore the brackets in a coordinate-free, intrinsic form. First we note that the classical Poisson bracket satisfies the conditions: skew-symmetry bilinearity Jacobi identity Leibniz identity

{ f , g} = −{ g, f } { a f + bg, h} = a{ f , h} + b{ g, h} {h, a f + bg} = a{h, f } + b{h, g} for a, b ∈ R { f , { g, h}} + { g, {h, f }} + {h, { f , g}} = 0 { f , gh} = h{ f , g} + g{ f , h}

Any binary operation on functions, denoted as before by { f , g}, which satisfies the conditions of skew-symmetry and bilinearity and the Jacobi and Leibniz identities, is called a Poisson bracket. Every Poisson bracket defines a vector field X H associated with a function H via XH ( f ) = { f , H }

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Working this out in symplectic coordinates for the bracket defined above X H ( f ) = X1 · f q + X2 · f p and

{ f , H } = f q · H p − f p · Hq Therefore, X1 = H p

X2 = − H q

as before. There is an intricate connection between Poisson brackets and Lie brackets. For any functions f , g [ X f , Xg ] = − X{ f ,g} The proof of this is a very good exercise in the use of the brackets. On one hand, for any function h

[ X f , Xg ](h) = X f ({h, g}) − Xg ({h, f }) = {{h, g}, f } − {{h, f }, g} On the other hand, X{ f ,g} (h)

= {h, { f , g}} = −{ f , { g, h}} − { g, {h, f }} = { f , {h, g}} − { g, {h, f }}

The Jacobi identity was used to obtain the second line and skew-symmetry was used to obtain the last. Now let us ask for the form of the vector field X H . That is, we want to determine the coefficients ξ i so that XH = ∑ ξ i ∂x i i

Applying X H to the coordinate function yields X H ( x i ) = { x i , H } = ξ i . Thus X H = ∑{ x i , H }∂ x i i

Now use this last identity replacing H by x i to obtain Xx i = ∑{ x j , x i }∂ x j j

and

{ x i , H } = −{ H, x i } = − Xx i ( H ) = ∑{ x i , x j }∂x i H j


Therefore

X H = ∑{ x i , x j }∂ x j H∂ x i i,j

This leads to the equations of motion being expressed in the vector form as x˙ = J Hx where J = [{ x i , x j }]. In the classical case matrix J reduces to

0 I J= −I 0 and the classical form of the equations is recovered. Finally, the payoff for rigid body mechanics. Let G be a Lie group with Lie i be the structure coefficients for g. Identify g and its dual algebra g and let Cjk g∗ . A Lie–Poisson bracket of two functions f , g on g is defined by

{ f , g} =

∑ Cjki x i ∂x j f ∂xk g

i,j,k

For a given Hamiltonian, H, this gives the equations of motion i i x˙ j = { x j , H } = ∑ Cjk x ∂x k H i,k

If G = SO(3) then g = R3 with the vector product and the equations of motion take the special vector form x˙ = x × Hx The intrinsic version [17] starts with

{ f , g} = x, [ f x , gx ] Then x˙ j = { x j , H } = x, [ x x , Hx ] = − x, [ Hx , x x ] = − x, ad Hx x x = − ad ∗Hx x, x x j

or

j

x˙ = − ad ∗Hx x

j

j

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98

3 Dynamics

3.4 Exercises

Exercise 3.1 Derive the Poincaré equations directly from the Euler–Lagrange equations by making the replacement q˙ i = su Xµi (q) Exercise 3.2 Derive the Poincaré equations for free motion of a particle in toroidal coordinates (Example 3.4) using the natural velocity components as a nonholonomic basis for the tangent space. Exercise 3.3 Use the Poincaré equations to derive the equations of motion for a top without a fixed point, that is, the point of contact is free to slide without friction on a horizontal surface using Cartesian velocity components and angular velocity components as a nonholonomic basis for the tangent space [28]. Exercise 3.4 Derive the equations of motion for a one-rotor gyrostat using Poisson brackets (see Example 3.11). Exercise 3.5 Show that { f n , g} = n f n−1 { f , g} and that { f n , gm } = nm f n−1 gm−1 { f , g}. Exercise 3.6 Verify that { f , g} = x, [ f x , gx ].


99

4

Constrained Systems This chapter considers rigid body motion that is subject to constraints. The need to include constraints in rigid body mechanics arises, for example, when a body is rolling without slipping or is interconnected to other rigid bodies endowed with various degrees of freedom. The chapter starts by examining the types of constraints that are encountered in mechanics, then describes the use of Lagrange multipliers to formulate the equations of motion for constrained systems and examines examples from the rigid body mechanics literature. We also consider alternatives to the Lagrange multiplier formalism. Finally, we describe the formulation of constrained mechanical systems in the language of fiber bundles. 4.1 Constraints

Constraints are conditions imposed on the state of a mechanical system, which restrict access to state space or phase space. For example, one can consider the free motion of a massive particle under the influence of gravity. If the particle is constrained to lie a given distance a from a fixed point, x i xi − a2 = 0, the system becomes a spherical pendulum. If, in addition, it is constrained to lie in a given vertical plane, x1 − mx2 = 0, it becomes a simple pendulum. Constraints are termed holonomic when they are expressed as an algebraic function of the coordinates, such as f ( q1 , . . . , q m ) = 0

(4.1)

The spherical pendulum and the simple pendulum illustrate holonomic constraints. A holonomic constraint restricts access to the configuration space and to velocity or momentum space. Constraints are termed nonholonomic when they cannot be expressed solely in terms of the coordinates. Examples include constraints defined by inequalities; a particle confined to a box, for example; or constraints imposed on the velocity components. We shall be concerned Rigid Body Mechanics: Mathematics, Physics and Applications. William B. Heard Copyright © 2006 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim ISBN: 3-527-40620-4

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4 Constrained Systems

primarily with constraints that are expressed in terms of differential forms. That is, there is a 1-form ω=

n

∑ ai (q1 , . . . , qm ) dqi

(4.2)

i =1

and the velocity is required to satisfy ω (v) = 0. If {∂qi } is the basis of the vector fields dual to the differential forms dqi and the velocity in this basis is v = q˙ i ∂qi , then ω (v) = q˙ i ai = 0 A nonholonomic constraint restricts access to velocity or momentum space. The differential form constraint is nonholonomic when there is no function f such that the constraint coefficients assume the form ai = Clearly, if

∂f ∂qi

∂a j ∂ai = ∂qi ∂q j

the constraint cannot be holonomic. The holonomic constraint f (q) = 0 can, however, be expressed in the differential form as n

∂f

∑ ∂qi dqi = 0

i =1

There are more general forms of constraints. In the holonomic case a constraint could acquire time dependence and assume the form φ(q1 , . . . , qn , t) = 0.

(4.3)

A spherical pendulum with variable length a(t) is an example. A nonholonomic constraint could also acquire time dependence and assume the form ω=

n

∑ ai (q1 , . . . , qm , t) dqi + a0 (q1 , . . . , qm , t) dt.

(4.4)

i =1

The more general forms of the constraints lead to additional classifications of constraints beyond holonomic and nonholonomic. A constraint is sclereonomic if there is no time dependence. Scleronomic constraints assume the forms (4.1) and (4.2) in the algebraic and differential cases, respectively. A constraint is rheonomic if it has form (4.3) or (4.4). A catastatic holonomic constraint is a constraint of the form (4.1). A nonholonomic constraint is catastatic if it has the form ω=

n

∑ ai (q1, . . . , qm , t) dqi.

i =1

(4.5)

4.1 Constraints

A constraint is acatastatic if it is not catastatic. Scleronomic constraints are a subset of the catastatic constraints but not vice versa. The basic geometrical structure underlying constraints is the distribution [1]. A k-dimensional distribution D on a configuration manifold M is a kdimensional subspace of Tq M assigned smoothly to each q ∈ M. Given a set of independent constraint forms ω α = aαi dqi + a0α dt

α = 1, . . . , r

the distribution D associated with these constraints is the set of vectors annihilated by the homogeneous part of the constraint forms. In other words

D = {v ∈ Tq M| aαi dqi (v) = 0

α = 1, . . . , r

A distribution is a submanifold of the tangent bundle T M. In the case of the spherical pendulum the distribution of the constraint is r t r = 2 and the constraint form is ω = r t dr = xdx + ydy + zdz. Given vector v = (ξ, η, ζ ), ω (v) = 0 if and only if v t r = 0. This describes the plane at r normal to r. Equivalently, the distribution is the space spanned by the vectors X1 = −y∂ x + x∂y ,

X2 = z∂y − y∂z

(4.6)

at the point r = ( x, y, z) ∈ R3 . Given a distribution D one can ask if there are coordinates for M such that the distribution assumes a holonomic form. That is, are there functions f i (q) such that, locally at least, the planes of the distribution are tangent to the surfaces f i = const. The distribution is said to be an integrable distribution if such coordinates exist. A distribution D is integrable if and only if X, Y ∈ D implies [ X, Y ] ∈ D [1]. In other words the distribution is integrable if and only if it is closed under Lie brackets. This is sometimes expressed [D , D] ⊆ D . We verify that the distribution defined by (4.6) is closed under Lie brackets. Indeed, ω ( X3 ) = 0

where X3 = [ X1 , X2 ] = − x∂z + z∂ x

Alternatively, X3 = −(z/y ) X1 + ( x/y ) X2 shows that X3 ⊆ D . Holonomic constraints reduce the number of degrees of freedom. Each independent equation of constraint reduces the number of degrees of freedom by one. For example, in the case of the simple pendulum the free motion in R3 is reduced to one-dimensional motion by the constraints x 2 + y2 + z2 − a2 = 0 and x = my. Nonholonomic constraints are completely different in this regard – they do not entail loss of degrees of freedom. The motion of an upright penny rolling on a plane is (as we shall see presently) an example of a nonholonomic system. Yet, any configuration is accessible to any other. A configuration is determined by the x, y coordinates of the center of the coin, the orientation angle φ of the plane of the coin about a vertical axis and the

101

102


angle θ of Abraham Lincoln’s line of sight. To change from ( x1 , y1 , φ1 , θ1 ) to ( x2 , y2 , φ2 , θ2 ) first roll the penny along the line joining ( x1 , y1 ) to ( x2 , y2 ). Record the difference between the resulting line of sight and the desired line of sight. Move along the extended line until one-half the difference in line of sight is attained, revolve the coin 180◦ about the vertical and roll back to ( x2 , y2 ). Finish by rotating about the vertical axis to the orientation angle φ2 . Rolling without slipping is a common source of nonholonomic constraints. The motion of a rigid body consists of the translational motion of a point of reference r0 in the body and rotation about that point. As a result, each point r of the body has a unique velocity, v = v0 + ω × (r − r0 ) If the body is to roll without slipping on, say, a plane then the velocity at the point of contact rc must vanish. That is, v0 + ω × (rc − r0 ) = 0

4.2 Lagrange Multipliers

A fundamental technique used in the analysis of constraints is the Lagrange multiplier. This is best seen in the context of the variational derivation of the Lagrange equations. Recall that by finding a stationary point of the action integral we were led the condition d ∂L ∂L − i δqi = 0 dt ∂q˙ i ∂q It was then possible to argue that each term in sum vanished because the δqi were independent. In the presence of constraints the coordinate variations are not independent and it cannot be concluded that the individual terms vanish. Let us consider constraints expressed in the form aki δx i = 0

i = 1, . . . , n

k = 1, . . . , s

Let λk , k = 1, . . . , s be arbitrary functions of q, the Lagrange multipliers. In view of the constraint equations we may write d ∂L ∂L − i + λk aki δqi = 0 dt ∂q˙ i ∂q Now fix the values of the multipliers so that ∂L d ∂L − i + λk aki = 0 i dt ∂q˙ ∂q

k = 1, . . . , s


Then the remaining n − s variations δqi are independent so we may conclude ∂L d ∂L − i + λk aki = 0 i dt ∂q˙ ∂q

k = s + 1, . . . , n

Thus, s unknowns λk have been added to the problem but there are exactly s constraint equations which allow their determination and we arrive at the following formulation of the problem: d Lq˙ − Lq + At λ dt Aq˙

= 0,

(4.7)

= 0

(4.8)

This is a system of n second-order differential equations and s algebraic equa˙ and the s Lagrange multipliers λ. An alternative tions for the 2n variables q, q, form based on the Euler–Lagrange equations of the second kind is d Tq˙ − Tq dt Aq˙

= Q − At λ, = 0

(4.9) (4.10)

where Q represents the generalized applied forces. In this form we see clearly the physical interpretation of the Lagrange multipliers. The term − At λ is the force exerted by the constraint. If an unconstrained particle or rigid body is to satisfy the constraints there must be a force in addition to the imposed force and the additional force is the force of constraint. Thus knowing the Lagrange multipliers is tantamount to knowing the force of constraint. 4.2.1 Using Projections to Eliminate the Multipliers

Let us consider a natural mechanical system having Lagrangian L=

1 t q˙ M (q)q˙ − V (q) = T (q, q˙ ) − V (q) 2

The Lagrange multiplier problem becomes M q¨

= − M˙ q˙ + Tq − Vq − At λ,

(4.11)

Aq˙

= 0.

(4.12)

˙ q˙ − Tq can be expressed in terms of the Christoffel The gyroscopic terms c = M symbols of the first kind for the metric Mij (3.14). Upon differentiating the constraint equation we obtain Aq¨ + A˙ q˙ = 0

103

104


which allows the elimination of q¨ in Eq. (4.11) and leads to an expression for λ which contains no derivatives of order higher than the first λ = ( AM −1 At )−1 [ A˙ q˙ − AM −1 (c + Vq )] The existence of the inverse ( AM −1 At )−1 is guaranteed if the rows of A are linearly independent. This condition will be satisfied for any valid set of constraints. Now substitute this expression for λ into (4.11) to obtain ˙ q˙ − Tq + Vq = 0 P M q¨ + M (4.13) where

P = I − At ( AM −1 At )−1 AM −1

(4.14)

It is convenient to introduce matrices K = ( AM −1 At )−1 and N = The matrix N is idempotent, N 2 = N, and P = I − N is also idempotent. Thus N and P are projection operators. Note that APt = 0 so that if v ∈ Tx M, then Pt v satisfies the constraints. In other words, v is projected into the distribution by Pt . We will refer to this method of deriving equations of motion for nonholonomic systems as the projection method. We now want to introduce terminology from [33], and the series of papers referred to therein, which leads to a revealing form of the equations. We will go into the results of [33], which are considerably more general than the results of this section, in more detail in a later section. First rewrite the equations of motion in the form At KAM −1 .

M q¨

= Q − At λ

Aq¨

= b

˙ q˙ + Tq − Vq and b = − A˙ q. ˙ Let a = M −1 Q be the generalwhere Q = − M ized acceleration the system would have it were unconstrained. The projected equations become M q¨ = Q − At K (b − Aa) This form of equations clearly displays the constraint forces, the second term on the right-hand side, and reveals that the constraint forces are driven by the residual in the constraint equation evaluated at the unconstrained acceleration, b − Aa. 4.2.2 Using Reduction to Eliminate the Multipliers

There is another path to the equations of motion when coordinates can be partitioned q = (s, r ) such that the constraint equations (p in number) assume the form s˙ a + A aj r˙ j = 0 a = 1, . . . , p j = 1, . . . , n − p


This is possible when the original matrix A has rank p. In this case there is another method for eliminating Lagrange multipliers [34]. We will begin with a natural mechanical system. Let the mass matrix be partitioned according to

M=

M1 Nt

N M2

with M1 p × p and M2 (n − p) × (n − p). Then the Lagrange multiplier problem is partitioned as M1 s¨ + N r¨ + M˙ 1 s˙ + N˙ r˙ − Ts + Vs + λ N t s¨ + M2 r¨ + N˙ t r˙ + M˙ 2 s˙ − Tr + Vr + At λ

= 0

(4.15)

= 0

(4.16)

− Ar˙

= s˙

(4.17)

The multipliers λ are obtained immediately λ

= −( M1 s¨ + N r¨ + M˙ 1 s˙ + N˙ r˙ − Ts + Vs ) = ( M1 A − N )r¨ + ( M˙ 1 A + M1 A˙ − N˙ )r˙ + Ts − Vs

Now substitute this equation for λ into (4.16) and use the constraint equation to obtain a in r, s, r˙ ˙ r˙ − ( Tr − Vr ) + At ( Ts − Vs ) = A˙ t Ls˙ Mr¨ + M

(4.18)

where

M = M2 + At M1 A − N t A − At N is the reduced mass matrix. It is important to note that in Eq. (4.18) s˙ has been replaced by − Ar˙ after derivatives Tr , Ts and Ls˙ have been formed. Now introduce the constrained Lagrangian R(r, s, r˙ ) =

1 t (−V r˙ Mr˙ − V (r, s) = T 2

Then we find that d (R ) − A T ( Tr − T (R ) Rr˙ − Rr + At Rs = A˙ t Ls˙ + ( Tr − T dt Rather than work out the right-hand side in detail for the special case of a natural system, it is advantageous to turn to the general case. The general Lagrangian is considered in [34] and the treatment below follows [34] – some notational changes have been made to accommodate the current notation. It will be useful when working with the partitioned variables to index the r variables in the center of the alphabet (i, j, k, . . .) and the s variables at the beginning of the alphabet ( a, b, c, . . .). In this way we can see at a

105

106


glance the range of the indices. When the variables are partitioned the general Lagrange multiplier equations become d Lr˙ − Lri + A ai λ a dt i d Ls˙ − Ls a + λ a dt a

= 0 = 0.

The elimination of the multipliers is immediate and we obtain d d Lr˙ − Lri = A ai ( Ls˙ a − Ls a ) dt i dt

(4.19)

The next step is to eliminate the s˙ variables entirely thus reducing the problem to one in r, s, r˙ alone. Introduce the constrained Lagrangian ˙ − Ar, ˙ r˙ ), R(r, r˙ ) = L(r, s, r,

A = A(r, s)

The equations of motion are derived using the relations Rr i

=

Lri − Ls˙ a A aj,ri r˙ j

Rs a

=

Ls a − Ls˙ b Abj,s a r˙ j

Rr˙ i

=

Lr˙ i − Ls˙ a A ai

Note that in these equations s˙ has been replaced by − Ar˙ after derivatives Tr , Ts , and Ls˙ have been formed. Then we find d Rr˙ − Rri + A ai Rs a dt i

= =

d d Lr˙ − Lri − A ai ( Ls˙ a − Ls a ) + Baji Ls˙ a r˙ j dt i dt Baji Ls˙ a r˙ j

by virtue of (4.19) and where Baij = A ai,r j − A aj,ri + Abi A aj,sb − Abj A ai,rb Baij is an object akin to the Riemann curvature of a Riemannian manifold. As is the case for Riemann curvature, Baji can be efficiently calculated using differential forms. To accomplish this one introduces the differential constraint 1-forms σa = ds a + A ai dri The differential of σa is the 2-form dσa = A ai,r j dr j ∧ dri − A ai,sb Abk drk ∧ dri Then ˙ ∂ri ) = Bajir˙ j dσa (q,


and the reduced equations may be written as d ˙ ∂r i ) Rr˙ − Rri + A ai Rs a = Ls˙ a dσa (q, dt i

(4.20)

The terms on the right-hand side are called the curvature terms. The geometrical content of these equations is discussed in [34]. We will touch on these matters in the last section of this chapter. 4.2.3 Using Connections to Eliminate the Multipliers

The multipliers can also be eliminated using the notion of a connection on the configuration space [25]. In Section 3.2 we introduced the abstract connection, the Levi-Civita connection ∇ and wrote the equations of motion in connection form (3.16). The Lagrange multiplier problem for a natural system can be expressed in terms of the connection ∇ and the projection P. Let P = I − Pt , the projection into the orthogonal complement of D . Then we have

∇q˙ q˙ + M −1 Vq + M −1 At λ = 0

P q˙

= 0

(4.21) (4.22)

Apply P to (4.21) and use the fact that P satisfies P M −1 At = M −1 At to obtain (4.23) P ∇q˙ q˙ + P M −1 Vq + M −1 At λ = 0 By applying ∇q˙ to (4.22) we obtain 0

= (∇q˙ P )q˙ + P ∇q˙ q˙ = (∇q˙ P )q˙ − P M −1 Vq − M −1 At λ

Now we can eliminate λ between this last equation and (4.21) to obtain

∇q˙ q˙ + (∇q˙ P )q˙ + ( I − P ) M −1 Vq = 0 or where

∇q˙ q˙ + ( I − P ) M −1Vq = 0 ∇X = ∇X + ∇X P

To show that ∇ X is a connection we note that ∇ X Y is bilinear in X and Y because ∇ X Y is and clearly ∇ f X Y = f ∇ X Y. From the component form of ∇

∇ X Y = [ X (Y i ) + X ( Pji )Y j + X k Y j (Γijk + Γijl Pkl − Γljk Pli )]ei

107

108


we also note that ∇ X ( f Y ) = f ∇ X Y + X ( f )Y and that the components of ∇ are i Γ jk = Γijk + ∂k Pji + Γijl Pkl − Γljk Pli

4.3 Applications

In this section we consider four examples. The first three are well-known in the rigid body mechanics literature. The equations of motion are formulated for a sphere rolling on a plane [35, 36], a disc rolling on a plane [35, 36] and a two-wheeled robot [37]. The projection method is illustrated for the first three systems and reduction is illustrated for the second and third. The fourth example derives the equations of free rotation in terms of the Euler parameters or quaternion components. 4.3.1 Sphere Rolling on a Plane

Consider a sphere, symmetric about its center, of mass m and radius a rolling on an inclined plane. We will work a frame fixed at the center of the sphere and parallel to a fixed frame. The inertia tensor is constant in this frame. Let the z-axis be normal to the plane and the y-axis be horizontal and let the angle of inclination be α. To obtain the constraints we note that in the chosen frame the point of contact of the sphere with the plane is rc = (0, 0, − a). Thus, the constraints which insure rolling without slipping are x˙ 1 − aω2

= 0,

x˙ 2 + aω1

= 0,

x˙ 3

= 0

The total force F and the total torque τ acting on the sphere are obtained from the integrals F=

B

dm f = m f

and

τ=

B

dm r × f = − f ×

B

dm r = 0

where f = g(cos α, 0, − sin α). The torque vanishes because the reference point is at the center of mass. The inertia tensor is I = AI where A = k2 ma2 and ka is the radius of gyration. In the case of a homogeneous sphere k2 = 2/5. The kinetic energy of the rolling sphere is 1 2 1 ( x˙ + x˙ 22 + x˙ 32 ) + A(ω12 + ω22 + ω32 ) 2 1 2

4.3 Applications

We will use Poincaré’s equations with Lagrange multipliers and generalized forces to derive the equations of motion with nonholonomic coordinates s = (ω, r˙ )t . The configuration space is the Lie group SE(3) which allows us to calculate the structure coefficients ckij on the Lie algebra se(3). From Section 2.1.4 we know SE(3) is represented by matrices of the form R a 0 1 where R is a rotation matrix and a is a 3-column. Therefore se(3) can be represented by matrices of the form R −1 − R −1 a R−1 R˙ R−1 a˙ R˙ a˙ = 0 1 0 0 0 0 Therefore the six matrices ei 0 , ri = 0 0

ti =

0 0

ei 0

i = 1, 2, 3

are a basis for se(3). The commutators of the basis vectors are

[ri , r j ] = ijk rk

[ri , t j ] = ijk tk

[ti , t j ] = 0

The dual adjoint operator is constructed from the structure coefficients v ω ad∗(ω,v) = − 0 ω Alternatively, one can calculate the commutator Σ w Ω v Σ ad(ω,v)(σ, w) = − 0 0 0 0 0 [Ω, Σ] Ωw − Σv = 0 0 [Ω, Σ] Ωw + vσ = 0 0

w 0

Ω 0

v 0

where the last equality follows from the fact that Σ = σ and v are skew6 symmetric. Translating this to the R setting we find 0 ω ad (ω,v) = v ω

109

110


and ad ∗(ω,v)

=−

ω 0

v ω

Given the dual adjoint operator we can write the equations of motion which are Aω˙ 1 + aλ2

= 0

Aω˙ 2 − aλ1

= 0

Aω˙ 3

= 0

m x¨ 1 − m x˙ 2 ω3 + λ1

= mg sin α

m x¨ 2 + m x˙ 1 ω3 + λ2

= 0

m x¨3 + m x˙ 1 ω2 − m x˙ 2 ω1 + λ3

= −mg cos α

Now eliminate ω˙ 1 and ω˙ 2 from the rotation equations and the derivative of the constraint equations to find the following expressions for the multipliers: λ1

=

λ2

=

λ3

=

mk2 ( g sin α + m x˙ 2 ω3 ) 1 + k2 mk2 − m x˙ 1 ω3 1 + k2 −mg cos α − m x˙ 1 ω2 + m x˙ 2 ω1

The multiplier λ3 is not needed for the dynamics problem, but it is needed to calculate the normal constraint force. From this we eliminate the multipliers from the equations of motion

( A + ma2 )ω˙ 1 + ma x˙ 1 ω3

= 0

( A + ma )ω˙ 2 + ma x˙ 2 ω3

= mga sin α,

2

Aω˙ 3

= 0

( A + ma ) x¨1 − m x˙ 2 ω3

= mga2 sin α

( A + ma2 ) x¨2 + m x˙ 1 ω3

= 0

x¨3

= 0

2

The spin ω3 remains constant at its initial value. If the sphere starts with ω3 = 0 its translational behavior is the same as a particle in a reduced gravitational field 1 ma2 = g g A + ma2 1 + k2

4.3 Applications

If there were no rolling constraint on a spin free sphere the force on the sphere would be mg sin α. The difference between this and the actual force mg/(1 + k2 ) is the Lagrange multiplier λ1 illustrating the role of multipliers in constraint forces. 4.3.2 Disc Rolling on a Plane

Fig. 4.1 Frame of reference used for the rolling disc problem.

Consider a disc, symmetric about the normal axis through its center, of negligible thickness and rolling without slipping on a horizontal plane (Fig. 4.1). A convenient frame f for studying this system consists of f3 normal to the disc oriented so that when the disc lies in the x, y plane f3 is parallel to e¯ 3 , f1 in the plane of the coin and parallel to the x, y plane and f2 = f3 × f1 . The inertia tensor is constant in the frame f because of symmetry about the f3 -axis. The orientation of the disc is described by the Euler angles φ, θ, ψ. Let the reference point be located at the center of the disk. The equations for unconstrained motion written in the moving frame are ( f I( ω (˙ + Ω ( = τ( I( ω ( × v() = f( m(v(˙ + ω We will drop the tilde to simplify the notation. In doing so we write ω = ωi f i ( i f i and ω f = ω ( f i f i . The frame f is related to and ω f = ω f i f i instead of ω = ω the space fixed and body fixed frames by f = eRe3 (φ) Re1 (θ ) = eRe3 (ψ)

111

112


From these relations it follows that the angular velocity of the reference frame is ω f = θ˙ f1 + φ˙ sin θ f2 + φ˙ cos θ f3 and the angular velocity of the disc is ω = θ˙ f1 + φ˙ sin θ f2 + (φ˙ cos θ + ψ˙ ) f3

(4.24)

The acceleration of gravity in the frame f is g = − g sin θ f2 − g cos θ f3 There is no applied torque because of the choice of reference point. To derive the constraints we first note that point of contact of the disc with the plane is r = − f2 . Therefore the constraint that the disc is rolling without slipping is r˙ − a f2 × ω = 0 In the moving frame f the coordinate version of the constraints is x˙ + aω3

= 0

y˙

= 0

z˙ − aω1

= 0

Now we can write the equations of motion in component form as A1 ω˙ 1 + A3 ω f 2 ω3 − A1 ω f 3 ω2

= aλ3

A1 ω˙ 2 + A1 ω f 3 ω1 − A3 ω f 1 ω3

= 0

A3 ω˙ 3 m( x¨ + ω2 z˙ ) m(y¨ + ω3 x˙ − ω1 z˙ ) m(z¨ − ω2 x˙ )

= − aλ1 = − λ1 = −mg sin θ = −mg cos θ − λ3

The Lagrange multipliers λ1 and λ3 are obtained by eliminating the secondorder derivatives between the translation equations and the derivatives of the constraint equations. The result is λ1

= ma(ω˙ 3 − ω1 ω2 )

λ2

= −ma(ω˙ 1 + ω2 ω3 ) − mg cos θ

To obtain the final form of the rotation equations we substitute the values just obtained for the multipliers and use the relations ωf3 =

sin θ sin θ ω = ω2 cos θ f 2 cos θ

4.3 Applications

The final form is

( A1 + ma2 )ω˙ 1 + [( A3 + ma2 )ω3 − A1 cot θω2 ]ω2

= −mga cos θ (4.25)

A1 ω˙ 2 + ( A1 cot θω2 − A3 ω3 )ω1

= 0

(4.26)

( A3 + ma )ω˙ 3 − ma2 ω1 ω2

= 0

(4.27)

2

We will now use reduction to derive the equations of motion for a rolling disc without using Lagrange multipliers. This is done in the context of Lagrangian mechanics and the Lagrangian is written in terms of Euler angles and rectangular coordinates relative to a fixed frame of reference L=

1 1 1 m( x˙ 2 + y˙ 2 + z˙ 2 ) + A1 (φ˙ 2 sin2 θ + θ˙ 2 ) + A3 (ψ˙ + φ˙ cos θ )2 − mga sin θ 2 2 2

˙ in the The constraints r˙ − af2 × ω are written in terms of the velocities x˙ and y, fixed frame, and the derivatives of the Euler angles x˙ + aφ˙ cos φ cos θ − aθ˙ sin φ sin θ + aψ˙ cos φ y˙ + aφ˙ sin φ cos θ + aθ˙ cos φ sin θ + aψ˙ sin φ

= 0

z˙ − aθ˙ cos θ

= 0

= 0

An obvious partition is r = (φ, θ, ψ) and s = ( x, y, z). This leads to the constrained Lagrangian R(r, r˙ ) =

1# ( A1 sin2 θ + A3 cos2 θ + ma2 cos2 θ )φ˙ 2 + ( A1 + ma2 )θ˙ 2 2 $ + ( A3 + ma2 )ψ˙ 2 + θ˙ ψ˙ ( A3 + ma2 ) cos θ − mga sin θ

From the differentials of the constraint 1-forms dσ1

= − a sin φ dφ ∧ dψ

dσ2

=

dσ3

= 0

a cos φ dφ ∧ dψ

we calculate the curvature terms ˙ ∂φ ) Ls a dσa (q,

= ma2 sin θ θ˙ ψ˙

˙ ∂θ ) Ls a dσa (q,

= 0

˙ ∂ψ ) Ls a dσa (q,

= −ma2 sin θ θ˙ φ˙

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Therefore, the reduced equations of motion are d ˙ [φ( A1 sin2 θ + A3 cos2 θ + ma2 cos2 θ ) + ψ˙ ( A3 + ma2 ) cos θ ] dt = −ma2 θ˙ ψ˙ sin θ θ¨( A1 + ma2 ) + φ˙ ( A3 + ma2 )(ψ˙ + φ˙ cos θ ) sin θ

= − A1 φ˙ 2 sin θ cos θ − mga cos θ d [( A3 + ma2 )ψ˙ + ( A3 + ma2 )θ˙ cos θ ] = ma2 θ˙ ψ˙ sin θ dt

(4.28) (4.29) (4.30)

One verifies that equations derived by the two methods are one in the same by substituting the components of (4.24) into (4.25)–(4.27). Before leaving the rolling disc we will examine the special case where θ is constant. This leads to a simple solution of the equations describing a disc rolling and turning while keeping its angle of inclination constant. In this case Eqs. (4.28)–(4.30) reduce to φ¨ = 0

(4.31)

φ˙ ( A3 + ma )(ψ˙ + φ˙ cos θ ) sin θ = − A1 φ sin θ cos θ − mga cos θ 2

( A3 + ma2 )

˙2

d ˙ (ψ + φ˙ cos θ ) = 0 dt

(4.32) (4.33)

From (4.31) we conclude that φ = νt + φ0 for constants ν and φ0 . From (4.33) we conclude ψ˙ + φ˙ cos θ = ψ˙ + ν cos θ = h and ψ = (h − ν cos θ )t + ψ0 for constants h and ψ0 . Using these results the translation equations become x˙ = − a(h + ν cos θ ) cos(νt) y˙ = − a(h + ν cos θ ) sin(νt) This system of ordinary differential equations describes motion on a circle. Therefore the rolling disc of constant inclination angle rolls on a circle. 4.3.3 Two-Wheeled Robot

Consider the idealized robot shown in Fig. 4.2. The robot consists of a body and two wheels and it is driven by torques τ1 and τ2 applied to the wheels. Let B0 be the body, Bi the ith wheel, B = B0 ∪ B1 ∪ B2 and mi , ri are the mass and the center of mass of Bi , respectively. The inertia tensor for the collection

4.3 Applications

Fig. 4.2 Two-wheeled robot.

of rigid bodies is

I(B , r0 ) = I(B0 , r0 ) + I(B1 , r1 ) + I(B2 , r2 ) 1 1 + a2 m1 ( I − i ⊗ i) + a2 m2 ( I − i ⊗ i) 2 2 = I(B0 , r0 ) + 2I(B1 , r1 ) + a2 m1 ( I − i ⊗ i) = I0 + 2 I1 + m 1 a 2 ( j ⊗ j + k ⊗ k ) Therefore the Lagrangian for the system is L=

1 1 1 m( x˙ 2 + y˙ 2 ) + I θ˙ 2 + I1 (φ˙ 12 + φ˙ 22 ) 2 2 2

where m = m0 + 2m1 , I = I0 + 2I1 + m1 a2 . The rolling constraints are x˙

=

y˙

=

aθ˙

=

1 ˙ b(φ1 + φ˙ 2 ) cos θ 2 1 ˙ b(φ1 + φ˙ 2 ) sin θ 2 b(φ˙1 − φ˙ 2 )

The equations of motions are

1 I1 φ¨ 1 − b cos θλ1 − 2 1 I1 φ¨ 2 − b cos θλ1 − 2

m x¨ + λ1

= 0

my¨ + λ2 I θ¨ + aλ3

= 0

1 b sin θλ2 − bλ3 2 1 b sin θλ2 + bλ3 2

= 0 = τ1 = τ2

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The Lagrange multipliers are λ1

= −m x¨

λ2

= −my¨

λ3

= −(I /a)θ¨

Derivatives of the constraint equations are then used to eliminate x, y, θ which gives the equations in φ1 , φ2

1 2 1 2 b2 ¨ mb − I1 + mb + 2 I φ1 + 4 4 a

1 2 b2 1 mb − 2 I φ¨ 1 + I1 + mb2 + 4 4 a

b2 I a2 b2 I a2

φ¨ 2

= τ1

φ¨ 2

= τ2

We can apply the reduction method with s = ( x, y, θ ) and r = (φ1 , φ2 ). The constraint differential forms are σ1 σ2 σ3

1 = dx − b cos θ (φ1 + φ2 ) 2 1 = dy − b sin θ (φ1 + φ2 ) 2 b = dθ − (φ1 − φ2 ) a

and their differentials are dσ1

=

dσ2

=

dσ3

=

b2 sin θdφ1 ∧ dφ2 a b2 − cos θdφ1 ∧ dφ2 a 0

The resulting curvature terms are all zero. The reduced Lagrangian is R=

1 2 1 1 2 2 mb (φ˙1 + φ˙2 )2 + I (φ˙1 − φ˙2 )2 + Iw (φ˙1 + φ˙2 ) 8 2 2

This Lagrangian leads to equations of motion identical to those derived by the projection method. 4.3.4 Free Rotation in Terms of Quaternions

This example expresses the equations of motion for free rotation in terms of quaternions. Consider the Euler parameters q = q0 + q1 i + q2 j + q3 k characterizing the state of a rigid body in SO(3) (Section 1.2.4). Let I be the inertia

4.4 Alternatives to Lagrange Multipliers

tensor in principal axes with origin at the center of mass of the body. Let W (q) = R(q)t . From (2.12) the angular velocity ω satisfies

0 = 2W (q)q˙ ω where q = (q0 q1 q2 q3 )t . Thus we may write the Lagrangian for free rotation as L(q, q˙ ) = 2q˙ t W (q)t I+ W (q)q˙ where

I+ =

0 I

1 0

This is valid only for q ∈ S3 . Therefore we have the holonomic constraint qt q = 1 Expressed in differential form the constraint is qt q˙ = 0 and the constraint matrix is A = qt . Since qt q˙ = 0 W has the property that W (q)q˙ = −W (q˙ )q. This allows us to write the Lagrangian in the alternative form L(q, q˙ ) = 2qt W (q˙ )t I+ W (q˙ )q Using the two forms of the Lagrangian we find Lq˙ = 4W (q)t I+ W (q)q˙

Lq = 4W (q˙ )t I+ W (q˙ )q

The equations of motion are 4W (q)t I+ W (q)q¨ + 4W (q)t I+ W (q˙ )q˙ − 8W (q˙ )t I+ W (q˙ )q + λq

= 0.

qt q˙

= 0

The property that dW (q)/dt = W (q˙ ) has been used. Upon elimination of the Lagrange multiplier λ we obtain

(1 − qqt )[W (q)t I+ W (q)q¨ + W (q)t I+ W (q˙ )q˙ − 2W (q˙ )t I+ W (q˙ )q] = 0 which projects the equations of motion from R4 into TS3 . See [6, 7] for related developments.


In this section we consider d’Alembert’s principle and the equations of Udwadia and Kalaba. The former is of great historical importance and the latter is a

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118


recent development, which generalizes Lagrange mechanics to systems where the d’Alembert principle does not apply. 4.4.1 D’Alembert’s Principle

Consider a dynamical system with configuration space M of dimension n. D’Alembert’s principle depends on the notion of a virtual displacement. In Section 3.2.1 we introduced virtual displacements for unconstrained systems where a virtual displacement is a tangent vector δq ∈ T M. In a constrained system a virtual displacement is a tangent vector which satisfies the homogeneous part of the constraints. If we express each of the constraints, including the holonomic ones, in differential form we obtain the system of constraints as Aq˙ − b = 0 and a virtual displacement δq is a tangent vector which satisfies Aδq = 0 If we write the equations of motion in the form M q¨ = Q − At λ where λi are the Lagrange multipliers, then the generalized forces of constraint are Qc = − At λ. The work done in a displacement by the forces of constraint is δWc = δqt Qc = −δqt At λ = −( Aδq)t λ and if the displacement is a virtual displacement the work vanishes. D’Alembert’s principle asserts that the work done by all the forces, including the inertial force M q¨ is zero δqt ( M q¨ − Q − Qc ) = 0 The work done by the constraint δqt Qc vanishes and d’Alembert’s principle ultimately does not involve the forces of constraint δqt ( M q¨ − Q) = 0 To apply d’Alembert’s principle to formulate equations of motion one finds a basis for the distribution D associated with the constraint forms. If the system has n degrees of freedom and r independent constraints then n − r independent virtual displacements δq1 , . . . , δqn−r span D and the equations of motion are (δqi )t ( M q¨ − Q) = 0 i = 1, . . . , n − r Aq¨ + A˙ q˙ − b˙ = 0


Example 4.1 The spherical pendulum is described in R3 by r¨ = − ge3

r t r = 2

There is one imposed force, − ge3 , and one constraint, the length of the pendulum is . The constraint form is ω = r t dr so that A = r t and a virtual displacement δr satisfies xδx + yδy + zδz = 0. The distribution is two-dimensional and the vectors δr1 = [−z, 0, x ], δr2 = [0, −z, x ] are a basis. Thus, d’Alembert’s principle provides the equations of motion −z x¨ + x z¨ + gx = 0

−zy¨ + yz¨ + gy = 0 There is the additional equation obtained from the second derivative of the constraint x x¨ + y y¨ + zz¨ = − x˙ 2 − y˙ 2 − z˙ 2 .

♦

Example 4.2 Consider the dynamical system on R3 r¨ = f subject to the constraints x˙ 1 = α x˙ 3 The constrain matrix is

A=

x˙ 2 = β x˙ 3 1 0 0 1

α β

and the constraint forms are ω 1 = dx1 − αdx3

ω 2 = dx2 − βdx3

The distribution is one-dimensional with basis vector (α, β, 1). D’Alembert’s principle and the constraints provide the equations of motion α( x¨1 − f 1 ) + β( x¨2 − f 2 ) + x¨3 − f 3 = 0 x¨1 − α x¨3 = 0 x¨2 − β x¨3 = 0.

♦

4.4.2 Equations of Udwadia and Kalaba

A constraint is said to be ideal if no work is done in a virtual displacement, otherwise it is nonideal. The equations of Lagrangian mechanics are based on ideal constraints and nonideal constraints are excluded. This is a significant omission because any constraint involving friction is nonideal.

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120


Udwadia and Kalaba have developed a general set of dynamical equations without invoking d’Alembert’s principle and these equations admit nonideal constraints. The starting point for their analysis [38, 39] is the system ˙ t) + Qc (q, q, ˙ t) M (q, t)q¨ = Q(q, q, where q is an n vector of generalized coordinates, M is a symmetric positive definite n × n matrix, Q is an n vector of imposed forces and gyroscopic terms and Qc is an n vector of forces of constraint. The system is subject to the constraint ˙ t)q¨ = b(q, q, ˙ t) A(q, q, which is general enough to accommodate any constraint we have discussed. The initial conditions q(0) and q˙ (0) are to satisfy the constraints. Introduce the unconstrained acceleration a = M −1 Q and write the equations in the form q¨ = a + M −1 Qc Now multiply the equations by the constraint matrix to obtain Aq¨ = b = Aa + AM −1 Qc or

AM −1 Qc = b − Aa

Introduce the matrix B = AM − 2 and rewrite the last equation as 1

B( M − 2 Qc ) = b − Aa 1

(4.34)

The solution of this linear equation for M − 2 Qc is obtained in terms of the Moore–Penrose (MP) inverse B+ as 1

M − 2 Qc = B+ (b − Aa) + ( I − B+ B)z 1

where z(t) is an arbitrary n vector. The equations of motion can then be written as 1 1 M q¨ = Q + M 2 B+ (b − Aa) + M 2 ( I − B+ B)z (4.35) The above argument is certainly technically correct but we have gotten ahead of ourselves. To make sense of the derivation we need to explain the MP inverse and its use in the derivation and we need to interpret these new equations. The former is a mathematical activity and latter belongs to the realm of physics. Given a m × n real matrix A of rank k there is a unique matrix A+ , the MP inverse of A, which has the following properties [40, 41]: AA+ A

=

A

+

+

=

A+

( AA+ )t

=

AA+

( A+ A)t

=

A+ A

A AA


If A is invertible it is easy to see that A−1 has the MP properties. If m ≥ n and rank( A) = n then A+ = ( At A)−1 At . On the other hand if n ≥ m and rank( A) = m then A+ = At ( AAt )−1 . The latter form was used earlier when we discussed the elimination of Lagrange multipliers by projection. In general, the MP inverse is defined in terms of the singular value decomposition. There are orthogonal matrices U (m × m) and V (n × n) such that A = UΣV t where   σ1   ..   D 0 . k k × n − k = Σ=   0m−k ×k 0m−k ×n−k σk   The σk , called the singular values, are positive and all off-diagonal elements of Σ are zero. The SVD amounts to finding bases for domain and range of A such that the matrix assumes its simplest form, namely Σ. The MP inverse is expressed in terms of the SVD as A+ = VΣ+ U t where   1/σ1   ..   Dk−1 0k ×m−k . +   Σ = = 0n−k ×k 0n−k ×m−k 1/σk  

Note that ΣΣ+ =

Ik

0k ×m−k

0m−k ×k

0m−k ×m−k

and Σ+ Σ =

Ik

0k ×n−k

0n−k ×k

0n−k ×n−k

Let us verify that this indeed satisfies the MP properties. AA+ A = UΣV t VΣ+ U t UΣV t = UΣV t = A A+ AA+ = VΣ+ U t UΣV t VΣ+ U t = VΣ+ U t = A

( AA+ )t = (UΣV t VΣ+ U t )t = AA+ because (ΣΣ+ )t = ΣΣ+ and similarly for A+ A. The MP properties imply that A+ A is idempotent

( A+ A)( A+ A) = A+ ( AA+ A) = A+ A Therefore A+ A and I − A+ A are projection operators. Now we can justify the use of the MP inverse in the derivation of the equations of motion. Let y = B+ c + ( I − B+ B)z, then B+ By = B+ BB+ c + [ B+ B − ( B+ B)( B+ B)]z = B+ c

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122


Therefore B+ c + ( I − B+ B)z is the solution of the linear system B+ By = B+ c for arbitrary z. Multiplying Eq. (4.34) by B+ we get B+ B( M − 2 Qc ) = B+ (b − Aa) 1

and

M − 2 Qc = B+ (b − Aa) + ( I − B+ B)z 1

for any vector z. The interpretation of the equations of motion (4.35) is that the second term on the right-hand side represents the force of constraint for the ideal constraints and the last term represents the force of constraint for the nonideal constraints. The fact that the arbitrary z appears is an indication that additional physics is needed to model the nonideal constraints. This is analogous to the constitutive relations which arise in continuum mechanics. The general theory dictates the form of the equations, but additional physics is required to specify certain terms. For a complete discussion of the interpretation of the equations see [38].

4.5 The Fiber Bundle Viewpoint

Fig. 4.3 Local trivialization of a fiber bundle.

In this section we will describe the development of the theory of nonholonomic mechanical systems in the language of fiber bundles. This is the setting for the modern treatments of reduction of mechanical systems. That is, given a mechanical system described by a Lagrangian which is invariant under a symmetry group, use the symmetry to eliminate some of the variables and thus reduce the equations to ones with fewer variables. We have chosen to include this material at this point in the development because much of the modern research on nonholonomic systems is expressed in this language.


Fig. 4.4 Transition functions of a fiber bundle.

Our first task is to answer the question: What is a fiber bundle? Before giving a formal answer to the question we consider the Möbius strip and the equations that describe it because a Möbius strip is the simplest example of a fiber bundle. Recall that a Möbius strip is constructed from a strip of paper by twisting the strip and pasting the two ends together. To obtain the equations, we imagine a line drawn down the middle of the strip of paper so that it becomes a circle after the gluing. Imagine, also, that the paper is inscribed with line segments normal to the central circle. The circle can be parameterized by the equations [ x, y, z] = [cos u, sin u, 0] 0 ≤ u ≤ 2π The line segments generate the Möbius strip and we need merely to write an equation for a line segment attached to each point on the circle in such a way that the line segment rotates through π radians as the circle is traversed. This rotation introduces the twist. Therefore, an appropriate set of equations for the Möbius strip is

1 1 1 [ x, y, z] = 1 + v sin u cos u, 1 + v sin u sin u, v cos u 2 2 2 0 ≤ u ≤ 2π − c ≤ v ≤ c or

−π ≤ u ≤ π

−c ≤ v ≤ c

where c is a positive real number sufficiently small to prevent intersections of any of the line segments. The relevance of this example is that the Möbius strip is built from a base circle with line segments and a twist and these are

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precisely the ingredients of a fiber bundle. Namely, a fiber bundle consists of the elements { E, M, F, π, G} where E is a manifold (the total space), M is a manifold (the base space), π : E → M is a projection of E onto M, F is a manifold (the fiber) and G is a group acting on F. The elements satisfy the following conditions: • There is an open covering {Ui } of M such that E is locally trivial, that is, π −1 (Ui ) ∼ = Ui × F. In other words, there are diffeomorphisms φi : π −1 (Ui ) → Ui × F : φi (u) → ( p, f ) such that π [φi−1 ( p, f )] = p (Fig. 4.3). • If Ui and Uj are not disjoint, then the transition functions cij = φj ◦ φi−1 are elements of the group G (Fig. 4.4). A notation sometimes used to describe a fiber bundle { E, M, F, π, G} is π

F → E → M The group is described separately. The Möbius strip is a fiber bundle in which E is the surface, M is the circle, F is a line segment and G is the two element group {1, a| a2 = 1}. There is an open cover of the circle with two open sets: say −π < u < 12 π and 0 < u < 32 π. These intervals intersect on two subsets (0, 12 π ) and (π, 32 π ). On the first of these intervals the transition function reflects the twist, element a of the group, and on the second subset the transition function is the identity. Example 4.3 The torus is an example of a trivial fiber bundle, T2 = S1 × S1 , where the trivialization and the transition functions are the identity mapping. Any circle of constant longitude serves as the base and the fibers are circles of constant latitude. ♦ The vertical vectors Vq at base point q ∈ E of a fiber bundle are tangent vectors aligned with the fibers, Vq = ker( Dπ ) = [v ∈ Tq E| Dπ (v) = 0].1 An Ehresmann connection A on a fiber bundle is a projection of the tangent vectors onto the vertical vectors • v ∈ Tq M =⇒ A(v) ∈ Vq , • v ∈ Vq =⇒ A(v) = v. The role of connections in differential geometry is to define curvature and parallel transport; the most widely known connection is the Levi-Civita connection specified by the Christoffel symbols Γkij = gkl Γijl (see Section 4.2). The application of these concepts to mechanics is to describe inertial forces – the simplest of which are the centrifugal and Coriolis forces in rotating mechanical systems. 1) See Appendix C for an explanation of the tangent map D.


The connection can be written as the tensor product of a vector and a differential form. Suppose that we use coordinates s a , a = 1, . . . , p for the fibers and ri , i = 1, . . . , n − p for the base space. Let A(s˙ a ∂s a + r˙ i ∂ri ) = v a ∂s a Then the tensor product form for A is A = ∂s a ⊗ ω a where ω a = An− p+ a i dri + An− p+ a b dsb and are known as the connection f orms and the connection can be written in the succinct notation [1] as A = eω where e is a row vector containing a frame for the vertical vectors and ω column containing the connection forms. This is the nonlinear version of the notation set out for vector spaces in Chapter 1. Example 4.4 To illustrate these notions we consider the concrete example of Hopf fibration π S1 → S3 → S2 with G = S1 , acting on itself. S3 is relevant to rigid body mechanics because we have used both the unit quaternions H1 and the Lie group SU2 as configuration spaces for rigid body problems and S2 is the natural object to use for specifying axes of rotation. The parameterizations of S3 and S2 in R4 and R3 , and the projection π, are expressed in the following: 1 1 1 1 cos θ cos ψ1 , cos θ sin ψ1 , sin θ cos ψ2 , sin θ sin ψ2 2 2 2 2 π

−→ [sin θ cos(ψ1 − ψ2 ), sin θ sin(ψ1 − ψ2 ), cos θ ] It is convenient to introduce φ = 12 (ψ1 − ψ2 ) and χ = 12 (ψ1 + ψ2 ). Then we have

1 1 1 1 1 1 1 1 cos θ cos (φ + χ), cos θ sin (φ + χ), sin θ cos (φ − χ), sin θ sin (φ − χ) 2 2 2 2 2 2 2 2

π

−→ [sin θ cos φ, sin θ sin cos φ, cos θ ]

It is then clear that the fibers are the circles {θ, φ = const.}, 0 ≤ χ ≤ 2π. Next consider the derivative of the projection Dπ. It is clear that ∂θ ∈ Tq S3 → ∂θ ∈ Tπ (q)S2 , ∂φ ∈ Tq S3 → ∂φ ∈ Tπ (q) S2 and ∂χ ∈ Tq S3 → 0. Therefore, the matrix representation of the derivative in the natural basis is

1 0 0 Dπ = 0 1 0

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126


Obviously Vq = ker( Dπ ) = [α∂χ |α = α(q) = α(r, s)] The most general projection onto Vq may be represented in the natural basis as 

0 A= 0 a ( q)

0 0 b( q)

 0 0  1

The tensor product form of the connection is therefore A = ∂χ (dχ + a dθ + b dφ).

♦

With these preliminaries, it is quite natural to describe mechanical systems with nonholonomic constraints in the fiber-bundle language. The fiber bundle is the configuration space Q with coordinates {s, r } in which the p constraints are expressed by p 1-forms ω a = ds a + A ai dri . The base space is the submanifold with coordinates {r } and the projection is π {s, r } = {r }. The vertical space contains the tangent vectors tangent to the si coordinate lines and horizontal vectors are those satisfying the constraints. The constraints define an Ehresmann connection on TQ and the associated curvature tensor. The curvature tensor is zero only if the constraints are holonomic [34, 42].

4.6 Exercises

4.6 Exercises

Exercise 4.1 Consider an idealized set of bar-bells, two thin disks of radius a and mass m connected by a massless rod of length b. Derive the equations of motion using Lagrange multipliers assuming the disks roll without slipping. Determine the Lagrange multipliers. Exercise 4.2 Find the distribution associated with the constraints for the twowheeled robot. Exercise 4.3 Consider a sphere of radius a whose center of mass is off-set from the center of the sphere by distance c. Write the equations of motion for the sphere rolling without slipping on an inclined plane. Exercise 4.4 Consider a right circular cone with base radius a and height b. Write the equations of motion for the cone rolling without slipping on an inclined plane assuming the line of contact between the cone and plane is always a generator of the cone. Exercise 4.5 Let A be an m × n matrix which can be written as the product of an m × r matrix P and an r × matrix Q A = PQ Show that if ( QQt )−1 and ( Pt P)−1 exist then the Moore–Penrose inverse of A is A+ = Qt ( QQt )−1 ( Pt P)−1 Pt

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5

Integrable Systems This chapter explores rigid body systems which are integrable. By this it is meant that the equations of motion for these systems admit enough integrals of motion to reduce the problem to quadrature. An integral of motion is a function of the state variables which is constant on the trajectories of the system. A problem is reduced to quadrature when it is expressed as an integral which is a known function. The integrals most often encountered in rigid body problems are evaluated in terms of elliptic or hyperelliptic functions.

5.1 Free Rotation

This problem concerns the rotational motion of a rigid body which has a fixed point and does not have a net torque exerted on it. This system is sometimes called the Euler top or the Euler–Poinsot top in honor of seminal contributions to its study. The equations of motion for this system in a set of principal axes are Ai ω˙ i + ( Ak − A j )ω j ωk = 0 with i, j, k are cyclic permutation of 1, 2, 3 and no summation on repeated subscripts (3.8). Without loss of generality we may assume A1 ≤ A2 ≤ A3 . 5.1.1 Integrals of Motion

There are two integrals of motion, total momentum and energy, which follow from the relations A21 ω1 ω˙ 1 + A22 ω2 ω˙ 2 + A23 ω3 ω˙ 3 = 0 and A1 ω1 ω˙ 1 + A2 ω2 ω˙ 2 + A3 ω3 ω˙ 3 = 0 Rigid Body Mechanics: Mathematics, Physics and Applications. William B. Heard Copyright © 2006 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim ISBN: 3-527-40620-4

130

5 Integrable Systems

The integrals are then expressed in terms of the components of angular momentum h21 + h22 + h23 = G2 (5.1) and a1 h21 + a2 h22 + a3 h23 = 2E

(5.2)

with ai = 1/Ai ( a3 ≤ a2 ≤ a1 ). The tip of the vector h traces a curve which is the intersection of a sphere (given by the momentum integral) and an ellipsoid (given by the energy integral), when these surfaces intersect. Let a I = 2E/G2 . If a1 < a I the ellipsoid lies outside the sphere and if a3 > a I the sphere lies outside the ellipsoid. Therefore for intersection to occur it is necessary that a3 ≤ a I ≤ a1 If G2 × (5.2) is subtracted from 2E × (5.1) one obtains

( a I − a1 )h21 + ( a I − a2 )h22 + ( a I − a3 )h23 = 0 which is the equation of a cone. The orientation of the cone is determined by a I . If a2 > a I (Case I) then

( a1 − a I )h21 + ( a2 − a I )h21 = ( a I − a3 )h23 wherein all coefficients are positive and the axis of the cone is the z-axis. If a2 < a I (Case II) then

( a I − a3 )h21 + ( a I − a2 )h21 = ( a1 − a I )h21 wherein all coefficients are positive and the axis of the cone is the x-axis. If a2 = 1, the cone reduces to the two planes

( a I − a3 )h21 = ( a1 − a I )h21 which contain the intersection of the sphere and the ellipsoid. 5.1.2 Reduction to Quadrature

The equations of motion in terms of the components of angular momentum are h˙ 1 = ( a3 − a2 )h2 h3 h˙ 2 = ( a1 − a3 )h1 h3 h˙ 3 = ( a2 − a1 )h2 h1

5.1 Free Rotation

The three equations of motion are reduced to a single one by use of the momentum and energy integrals. Eliminating h1 and h3 between the integrals yields the relations h22 ( a2 − a1 ) + h23 ( a3 − a1 ) = G2 ( a I − a1 )

(5.3)

h22 ( a2 − a3 ) + h21 ( a1 − a3 ) = G2 ( a I − a3 )

(5.4)

and which express h1 and h3 in terms of h2 . Substituting these into the second angular momentum equation leads to ˙h2 = G4 ( a1 − a I )( a I − a3 ) 1 − a1 − a2 h2 1 − a2 − a3 h2 (5.5) 2 G 2 ( a1 − a I ) 2 G 2 ( a I − a3 ) 2 which is a single equation in h2 that can be reduced to quadratures in the form of a standard elliptic integral. The reduction depends on the ratio of the coefficients of h22 in the brackets. In Case I (a3 < a I ≤ a2 < a1 ) let k21 =

a I − a3 a1 − a2 a2 − a3 a1 − a I

η2 =

a2 − a3 h2 2 G ( a I − a3 ) 2

and λ21 = G2 ( a1 − a I )( a2 − a3 ) Then η˙ 2 = λ21 (1 − η 2 )(1 − k21 η 2 )

0 ≤ k21 ≤ 1

In Case II (a3 < a2 < a I < a1 ) let k22 =

a2 − a3 a1 − a I a I − a3 a1 − a2

η2 =

a1 − a2 h2 G 2 ( a1 − a I ) 2

and

λ22 = G2 ( a1 − a2 )( a I − a3 ) Then, η˙ 2 = λ22 (1 − η 2 )(1 − k22 η 2 )

0 ≤ k22 ≤ 1

In both cases we have λ( t + τ ) =

η 0

,

du

(1 − u2 )(1 − k2 u2 )

=

sin−1 η o

,

dx 1 − k2 sin2 x

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132


where τ is an integration constant determined by the initial value of η. Upon inversion of the integral (Appendix B.1) we arrive at the Jacobi elliptic function η = sn(λ(t + τ ), k) We evaluate η at t = 0 obtaining sn−1 η0 and η = sn(λ + sn−1 η0 ) λ

τ=

Now, the addition formula for sn (B.3.1) supplies η=

snλt cn sn−1 η0 dnsn−1 η0 + η0 cnλt dnλt 1 − k2 η02 sn2 λt

which, together with the identities (B.1) dn sn−1 η0 = 1 − k2 η02 cn sn−1 η0 = 1 − η02 show that

η (t, k, λ, t, η0 ) =

1 − η02

1 − k2 η02 snλt + η0 cnλt dnλt 1 − k2 η02 sn2 λt

Therefore the solution for h2 is h2 = Bi η (t, k i , λi , t, η0 ) a − a3 B1 = G2 I , a I < a2 , a2 − a3 a − aI B2 = G2 1 , a2 < a I a1 − a2 The solutions for h1 and h3 are obtained by the same procedure. For h1 we obtain % A ξ (t, k1 , λ1 , t, ξ 0 ) a I < a2 h1 = A ζ (t, k2 , λ2 t, ξ 0 ) a2 < a I where A2 = G 2

a I − a3 a1 − a3

√

and ξ (t, k, λ, ξ 0 ) =

ξ 0 cnλt −

ζ (t, k, λt, ξ 0 ) =

1 − k2

1 − k2 + k2 ξ 02 snλt dnλt

1 − k2 (1 − ξ 02 )sn2 λt ξ 0 dnλt − 1 − ξ 02 k2 − 1 + ξ 02 snλt cnλt 1 − (1 − ξ 02 )sn2 λt

5.1 Free Rotation

for h3 we obtain

% h3 =

C ζ (t, k1 , λ1 , ζ 0 ) C ξ (t, k2 , λ2 , ζ 0 )

with C2 = G2

a I < a2 a2 < a I

a1 − a I a1 − a3

Note that if ξ 0 = ζ 0 = 1 and η0 = 0, then the solutions reduce to the textbook cases ξ = cn, η = sn and ζ = dn. In this limit, the second case for η follows from the first by the reciprocal modulus transformation of elliptic integrals of the first kind [43] φ 0

,

du 1 − k2 sin u 2

=

1 k

φ 0

du 1 − (1/k2 ) sin2 u

where sin φ = k sin φ. The trajectory of the angular momentum vector in the moving frame is periodic because the functions sn, cn, and dn are periodic. Functions sn and cn have period 4K, where K is the complete elliptic integral of the first kind K=

1 0

,

du (1 − u2 )(1 − k2 u2 )

The function dn has period 2K. Therefore the period for h is T = 4K/λ The solution is completed by finding expressions for the Eutherian angles which specify the orientation of the body. Two of these θ, ψ are obtained algebraically and the third, φ, requires a further quadrature. We can, without loss of generality, choose the fixed frame so that h = Ge¯ 3 Then hi ei = Ge¯ 3

hi = Ge¯ 3 · ei

From (1.5) we have e¯ 3 = sin θ sin ψe1 + sin θ cos ψe2 + cos θe3 Therefore h1 = G sin θ sin ψ h2 = G sin θ cos ψ h3 = G cos θ

133

134


The angles θ and ψ are spherical coordinates for h/G (co-latitude and longitude, respectively). This provides the expressions for cos θ and tan ψ % h23 ( a1 − a I ) dn(λt, k) a I < a2 cos θ = 2 = cn(λt, k) a2 < a I ( a1 − a3 ) G  ( a2 − a3 ) cn2 (λt, k)   a I < a2  2 h ( a1 − a3 ) sn2 (λt, k) tan2 ψ = 12 =  ( a I − a3 )( a1 − a2 ) dn2 (λt, k) h2   a2 < a I ( a1 − a I )( a1 − a3 ) sn2 (λt, k) 2

To obtain the equation for φ we begin with (2.8) ω1 = φ˙ sin ψ sin θ + θ˙ cos ψ ω2 = φ˙ cos ψ sin θ − θ˙ sin ψ which yield φ˙ sin θ = a1 h1 sin ψ + a2 h2 cos ψ a h + a2 h 2 φ˙ = G 1 12 h1 + h22 2

2

The right-hand side of this equation is known as a function of time and φ is obtained by a single quadrature. The integral involved is an elliptic integral of the third kind. The following is the result of substituting the known solutions for h1 , h2 and manipulating the equation into its canonical form  a1 − a2   a I < a2  Ga3 + 2 1 + ( a − a ) / 1 2 ( a2 − a3 )sn ( λt, k ) φ˙ =  a1 − a3  a2 < a I  Ga3 + 1 + ( a1 − a I )/( a I − a3 )sn2 (λt, k) and upon integration    Ga3 t + ( a1 − a2 )Π[t/λ, ( a1 − a2 )/( a2 − a3 ), k] φ(t) =   Ga3 t + ( a1 − a3 )Π[t/λ, ( a1 − a I )/( a I − a3 ), k]

a I < a2 a2 < a I

where Π(t, α, κ ) is the elliptic integral of the third kind [44]. To completely describe the solution to the free rotation problem, we must mention special cases which we have neglected so far. There are six stationary solutions

±1

0

0

0

±1

0

0

0

±1

5.1 Free Rotation

Fig. 5.1 Traces of the angular velocity vector ω projected onto a unit sphere in the body fixed frame. The curves were generated with a1 , a2 , a3 = 4, 2, 0.5 for a range of values of E ∈ [ 14 a2 G2 , a2 G2 ].

Fig. 5.2 Traces of the vector e3 ( t) in the fixed frame. The curves were generated with a1 : a I : a2 : a3 = 1.2 : 1.1 : 1 : 0.8.

135

136


We shall see below that the stationary solutions on the x- and z-axes are stable, whereas the stationary solutions on the y-axis are unstable. The unstable stationary solutions are connected by four heteroclinic orbits or separatrices. These occur when a I = a2 which implies that the elliptic modulus k is then unity. Equation (5.5) in this case is h˙ 22 = ( a1 − a2 )( a2 − a3 )( G2 − h22 )2 The solution is elementary and is h2 = G tanh λt

λ=G

( a1 − a2 )( a2 − a3 )

We now examine some numerical results which illustrate free rotation of a rigid body. Figure 5.1 shows the paths of the angular velocity vector, ω (t), projected onto a unit sphere. The curves are closed and the motion is periodic on them. Stable loops are seen around the ω1 - and ω3 -axes. The closer the initial conditions are to the ω1 or ω3 axes the smaller the loops. On, the other hand, the closer the initial conditions are to the ω2 -axis the wider the excursions will be. The different curves were generated by varying a I from a3 to a1 . As a I is increased (increasing energy for fixed angular momentum) the orbits grow from small loops around the ω3 -axis, merge at the separatrix with the companion orbit reflected through the plane ω3 = 0 then shrink to the ω1 -axis. The lowest energy orbit is at the ω3 -axis and the highest at the ω1 -axis. Figure 5.2 shows the path of spin axis e3 = sin φ sin θ e¯ 1 + cos φ sin θ e¯ 2 + cos θ e¯ 3 of the body. This illustrates the spatial orientation of the body, in the inertial frame. In general this is not a closed curve. In other words the motion of the body is not periodic in general. This occurs because there are two frequencies involved in free rotation , λ for θ and ψ and Ga3 for φ, and they are generally incommensurate, that is they are not linearly dependent over the integers. 5.1.3 Free Rotation in Terms of Andoyer Variables

This section considers the free rotation problem formulated in terms of the Andoyer variables (Section 3.3.2). This problem has been studied by [12, 31, 32]. The Hamiltonian for the free rotation problem is (3.44)

1 1 1 1 2 K= cos2 l + sin2 l ( G2 − L2 ) + L 2 A1 A2 A3 which can be rewritten as 1 1 K = [( a1 + a2 ) + ( a2 − a1 ) cos 2l ] ( G2 − L2 ) + a3 L2 4 2 with ai = 1/Ai .

(5.6)

5.1 Free Rotation

From this we obtain the equations of motion 1 1 l˙ = (2a3 − a1 − a2 ) L + ( a1 − a2 ) cos 2l 2 2 1 1 g˙ = ( a1 + a2 ) G + ( a2 − a1 ) cos 2l 2 2 h˙ = 0 1 L˙ = ( a2 − a1 )( G2 − L2 ) sin 2l 2 G˙ = 0 H˙ = 0

(5.7)

We see immediately that h, G, H are constants and that g can be obtained by quadrature once l and L are known. Thus the problem is reduced to one degree of freedom at the outset. Let us now examine the phase-portrait structure in l, L phase space. This phase space is a cylinder with angular coordinate l and axial coordinate L. The Hamiltonian vector field has singularities at

1 3 π, 0 , (π, 0), π, 0 (l0 , L0 ) = (0, 0), 2 2 The lines L2 = G2 can also be regarded as singularities because dL/dt vanishes there and l is undefined there because in this case λ vanishes and the planes Π2 and Π3 of Fig. 3.3 collapse on one another. The equations linearized at the singular points (l0 , L0 ) and are written in terms of ξ = l − l0 and η = L − L0 . At (0, 0), (π, 0): ξ˙ = ( a3 − a2 )η η˙ = ( a2 − a1 ) G2 ξ ξ¨ = ( a3 − a2 )( a2 − a1 ) G2 ξ and at ( 12 π, 0), ( 32 π, 0) ξ˙ = ( a3 − a1 )η η˙ = ( a1 − a2 ) G2 ξ ξ¨ = ( a3 − a1 )( a1 − a2 ) G2 ξ The first two are saddle points lying on a heteroclinic orbit or separatrix connecting them. The last two are centers. Thus the phase plane is as depicted in Fig. (5.3). A trajectory is either a circulation, libration about a center or the separatrix. These correspond to the x-axis cone, the z-axis cone and the two intersecting planes of Section 5.1.1.

137

138


Fig. 5.3 L − l phase-portraits showing libration, circulation, and a heteroclinic orbit. L0 and l0 are initial points on the general circulating and librating cases, respectively.

Circulation

The one degree of freedom system is reduced to quadrature by using the energy integral to eliminate l in the L equation. In the case of circulation we will evaluate the energy integral at the point l = 0, L = L0 . The circulating trajectories form a one parameter family parameterized by L0 . The energy integral is 1 1 1 1 [( a1 + a2 ) + ( a2 − a1 ) cos 2l ] ( G2 − L2 ) + a3 L2 = a2 ( G2 − L20 ) + a3 L20 4 2 2 2 which is used to eliminate l from the third of Eq. (5.7) to obtain

dL dt

2

# $ = ( a3 − a2 )( L2 − L20 ) L2 ( a1 − a3 ) + G2 ( a2 − a1 ) + L20 ( a3 − a2 )

The right-hand side is a biquadratic which implies that the solution is a Jacobi elliptic function. From Section B.3 it is seen that by defining the modulus to be ( a1 − a2 )( G2 − L20 ) k2 = 2 G ( a1 − a2 ) + L20 ( a2 − a3 ) and defining the frequency λ2 = L20 ( a2 − a3 )( a1 − a3 )/(1 − k2 )

5.1 Free Rotation

the solution becomes L = L0 nd(λt, k) Note that k2 ranges from 0 to 1 (the separatrix) as L0 ranges from G to 0. Libration

In the case of circulation we will evaluate the energy integral at the point l = l0 , L = 0. The librating trajectories form a one-parameter family parameterized by l0 . The energy integral becomes G2 ( a2 − a1 )(cos 2l − cos 2l0 ) − L2 [( a1 + a2 ) + ( a2 − a1 ) = 0 Now use the energy integral to eliminate l from the third of Eq. (5.7) to obtain

2 # $# $ dL = L2 ( a1 − a3 ) + G2 ( a2 − a1 ) cos2 l0 L2 ( a3 − a2 ) + G2 ( a2 − a1 ) sin2 l0 dt The right-hand side is again a biquadratic. Two of the Jacobi elliptic functions are candidates – sn and sd = sn/dn. The signs of the coefficients determine sd to be the correct choice. Let us introduce a scaling to put the equation in its canonical form L2 = y2 G2 ( a1 − a2 ) cos2 l0 sin2 l0 /λ2 where λ2 = ( a2 − a3 ) cos2 l0 + ( a1 − a3 ) sin2 l0 Then the equation becomes

dy dt

2

( a − a3 ) sin2 l0 y2 ( a2 − a3 ) cos2 l0 y2 = G 2 ( a 1 − a 2 ) λ2 1 − 1 − 1 λ2 λ2

If the modulus is defined to be k2 =

( a2 − a3 ) cos2 l0 λ2

the differential equation assumes the form

dy dt

2

= G2 ( a1 − a2 )λ2 (1 + k2 y2 )[1 − (1 − k2 )y2 ]

Referring to Section B.3 the solution is seen to be

√ y = sd ( G a1 − a2 λt, k) Now k2 ranges from 0 to 1 as l0 ranges from 12 π to π.

139

140


Fig. 5.4 Polhode curves on the ellipsoid of inertia.

(a)

(b)

Fig. 5.5 Traces of the angular velocity vector on invariable plane (the herpolhode curves). The curves were generated with a1 : a I : a2 : a3 = 1.4 : 1.1 : 1 : 0.5 (a) and a1 : a I : a2 : a3 = 1.4 : 0.5 : 1 : 0.5 (b).

5.1.4 Poinsot Construction and Geometric Phase

There are two elegant geometric results describing free rotation of a rigid body. The first, due to Poinsot [45], constructs the motion from an ellipsoid rolling on a plane and the second, of a much more recent origin [46], gives an elegant relation between the paths followed by the angular velocity vector and the axes of the body.

5.1 Free Rotation

Fig. 5.6 The polhode rolls without slipping along the herpolhode which lies in the invariable plane.

The Poinsot Construction

To derive the Poinsot construction, we begin by recalling the energy and momentum integrals expressed in invariant vector form as h · ω = 2E h · h = G2 The projection of the angular velocity on the angular momentum vector is constant in free rotation because

1 d d 1 h·ω = (2E) = 0 dt G G dt We will chose the fixed frame, without loss of generality, such that e¯ 3 is parallel to h. In this case ω¯ 3 is constant. The ellipsoid of inertia1 is the surface defined by f (r) = r · I r − 2E = 0 which is an ellipsoid because the inertia tensor is positive definite. Imagine the ellipsoid as a massless shell attached to the body and moving with it. The utility of the ellipsoid is that it affords an elegant description of free rotation. The energy integral is a statement of the fact that the instantaneous angular velocity vector ω(t) lies on the instantaneous ellipsoid of inertia. 1) This is not the standard definition. Most books evaluate the ellipsoid of inertia with 1 rather than 2E.

141

142


The path followed by the angular velocity on the moving ellipsoid of inertia is called a polhode. It is a closed curve and from the point of view of a body fixed observer the motion of ω on the polhode is periodic. The polhode is illustrated in Fig. 5.4. The path followed by the angular velocity vector in the fixed frame, ω¯ (t), is called a herpolhode.2 Since the angular velocity component ω¯ 3 is constant, the herpolhode is a plane curve. It is generally not a closed curve and can have a complicated shape (Fig. 5.5, for two examples). The curve does not close because, as we noted before, there are two frequencies involved in free rotation, and they are generally incommensurate. It is shown in [47] that the herpolhode has no points of inflection. At time t the herpolhode and the polhode intersect in ω(t). In fact they share the tangent vector ω. ˙ Therefore the polhode rolls on the herpolhode. A vector normal to the ellipsoid at r is parallel to the gradient ∇ f = 2I r. When evaluated at ω ∇ f (ω) = 2I ω = 2h Since the angular momentum h is constant the tangent plane has a fixed orientation. The distance from the center of the ellipsoid to the tangent plane is

1 h = 2E/G ω· G which is constant. Summing up: the tangent plane has fixed orientation and fixed distance from the center of the ellipse which is fixed. Therefore the tangent plane is fixed. It is called the invariable plane. The ellipsoid of inertia rolls on the invariable with point of contact at ω. Since the point of contact is the instantaneous axis of rotation, E rolls without slipping. This is all put rather poetically by the phrase “the polhode rolls without slipping on the herpolhode in the invariable plane.” These relations are illustrated in Fig. 5.6. When the body symmetric, A1 = A2 , both polhode and herpolhode become circles. The circular polhode rolls along the circular herpolhode. In this case the angular velocity vector generates two cones based on the two circles and the motion can be described by the rolling of the polhode cone on the herpolhode cone. 2) MacMillan [35] attributes the terminology to Andoyer and gives their meaning derived from the Greek for “axis path” and “serpentine path.”

5.1 Free Rotation

The Angle of Rotation and the Geometric Phase

The polhode is closed. Therefore, as the ellipsoid of inertia rolls on the invariable plane an observer fixed on the body sees the point of contact move along the polhode and return to its original position after period T. If we arrange the fixed frame such that h = he¯ 3 , with no loss of generality, and transport the fixed frame along the polhode the vectors e¯ 1 and e¯ 2 will appear to rotate. The question addressed in this section is: what is the rotation angle after one period T? An elegant geometrical answer to the question was presented in [46] and we follow [46] and [48] here. The qualitative description given in the previous paragraph can be made quantitative. The vectors h(t) and h¯ are related by h (t) = R(t)h¯ and h is periodic h(t0 ) = h(t0 + T ). Therefore, h ( t0 + T ) = R( t0 + T ) Rt ( t0 ) h ( t0 ) The matrix R(t0 + T ) Rt (t0 ) is a rotation matrix and its axis of rotation is h¯ because Rt (t0 )h¯ = Rt (t0 + T )h¯

or

R(t0 + T ) Rt (t0 )h¯ = h¯

and therefore R(t0 + T ) Rt (t0 ) = Rh¯ (∆φ) The question addressed in [46] is: what is the angle of rotation ∆φ ? The answer is obtained by integrating the 1-form α = p φ dφ + pθ dθ + pψ dψ around a specially constructed curve C. Let the curve C1 be the path followed by the body in phase space in one period. That is, C1 is given in parametric form by [φ(t), θ (t), ψ(t), p φ(t), p θ (t), p ψ (t)], 0 ≤ t ≤ T. Let C2 be the path followed in phase space by a rotation about h by angle ∆φ. The parametric form of C2 is [∆φt/T, θ (0), ψ(0), p φ(0), p θ (0), p ψ (0)]. The curves C1 and C2 intersect at t = 0 and t = T. Let the closed curve C consist of C1 followed by C2 in reverse. The value the 1-form α on C1 can be obtained from the Lagrangian. Denote the Euler angles by {qi } = {φ, θ, ψ} and express the Lagrangian in terms of the angular velocity components L=

1 1 A ω 2 = hi ωi 2 i i 2

By definition pi = Therefore, α = hj

∂ω j ∂L = hj ∂q˙ i ∂q˙ i

∂ω j ∂ω j q˙ dt = A j ω 2j dt = 2E dt dqi = h j ∂q˙ i ∂q˙ i i

143

144


because ω j is a linear function of the q˙ i . The value of α on curve C2 is α = G dφ. The integrals of α along the curves are simply

C1

α = 2ET

and

C2

α = G∆φ

At this point in the development Stokes’ theorem is used in its generalized form which states ω = dω ∂S

S

where ω is a differential form, S is a domain, and ∂S is its boundary. Recall the discussion of differential forms from Section 3.3. As an example, let A be a closed region of the plane, C the bounding curve of A, and ω the 1-form ω = f dx + gdy then application of Stokes’ theorem gives

C

( f dx + gdy) =

A

( gx − f y ) dxdy

which is a familiar form of Stokes’ theorem. In fact the curve need not be a plane curve and the general form of Stokes’ theorem for ω = f dx + gdy + hdz gives C

( f dx + gdy + hdz) =

A

( gx − f y ) dx ∧ dy + (hy − gz ) dy ∧ dz + ( f z − h x ) dz ∧ dx

The general form encompasses the divergence theorem as well. Let ω be the 2-form f dx ∧ dy + gdy ∧ dz + hdz ∧ dx and let S be a surface bounding volume V, then Stokes’ theorem provides S

ω=

V

dω =

V

( f x + gy + hz ) dx ∧ dy ∧ dz

The calculation of the rotation angle is completed by using Stokes’ theorem C

α=

C1

α−

C2

α=

S

dα

where S is the portion of the angular momentum sphere enclosed by the ∂S = C. To calculate dα we use ˙ + ψe ˙ e3 + θe ˙ 3 ) dt = G (φ˙ + cos θ ψ˙ ) dt = G (dφ + cos θdψ) α = h · ω dt = Ge¯ 3 · (φ¯ 1 Therefore, dα = G sin θ dψdθ Assembling the above results we obtain ∆φ =

2ET −Σ G

5.2 Lagrange Top

where Σ=

S

sin θ dψdθ

is the integral over the projection of the area enclosed by the polhode on the momentum sphere to the unit sphere. Σ is referred to as the geometric phase. A different derivation of this result which uses the Gauss–Bonnet theorem is presented in [49]. This calculation could be carried out in terms of elliptic functions, of course. In fact, [50] derives the equation of the herpolhode in terms of elliptic functions. It is, however the economy and conceptual clarity of the geometrical approach which is so appealing.

5.2 Lagrange Top

The Lagrange top refers to the rotational motion of a symmetric rigid body in a uniform gravitational field. The center of mass lies on the axis of symmetry and one point if the axis of symmetry is fixed. Being symmetrical means that two of the principal moments of inertia are equal A1 = A2 The Lagrangian for this system expressed in terms of the Euler angles is $ # ˙ ψ˙ ) = 1 A1 (φ˙ 2 sin2 θ + θ˙ 2 ) + A3 (φ˙ cos θ + ψ˙ )2 − U (θ ) ˙ θ, L(θ, φ, 2 and the potential for the uniform gravitational field is U (θ ) =

B

gz dm = Mg z¯ = Mg cos θ

where M is the mass of the body, g is the acceleration of gravity, and is the distance from the fixed point to the center of mass of the body. This Lagrangian yields the following equations of motion: $ d # ˙ A1 φ sin2 θ + A3 cos θ (φ˙ cos θ + ψ˙ ) = 0 dt A1 θ¨ − A1 φ˙ 2 sin θ cos θ + A3 φ˙ sin θ (φ˙ cos θ + ψ˙ ) − Mg sin θ = 0 d [ A3 (φ˙ cos θ + ψ˙ )] = 0 dt 5.2.1 Integrals of Motion and Reduction to Quadrature

The first and third equations of motion immediately give integrals of motion A1 φ˙ sin2 θ + A3 cos θ (φ˙ cos θ + ψ˙ ) = h1

145

146


Fig. 5.7 Root structure for the cubic associated with Lagrange top.

and A3 (φ˙ cos θ + ψ˙ ) = h2 These integrals are p ψ and pφ for symmetric rigid bodies. In addition there is the energy integral derived from the Hamiltonian H = q˙ · Lq − L $ 1# A1 (φ˙ 2 sin2 θ + θ˙ 2 ) + A3 (φ˙ cos θ + ψ˙ )2 + Mg cos θ = E 2 Substituting the first two integrals into the energy integral yields a single firstorder equation in θ A21 sin2 θ θ˙ 2 = −(h1 − h2 cos θ )2 − 2Mg A1 sin2 θ cos θ + 2EA1 sin2 θ This is simplified by substituting u = cos θ u˙ 2 = −

2E (h1 − h2 u)2 2Mg − ( 1 − u2 ) u + ( 1 − u2 ) = f ( u ) 2 A1 A1 A1

The right-hand side of this equation, f (u), is a cubic polynomial in u. f (−1) < 0 and f (1) < 0 while f → +∞ as u → +∞ and f → −∞ as u → −∞. In the case of real motion there must be two roots in the interval [−1, 1] and one root greater than 1 as shown in Fig. 5.7. Thus, u˙ 2 =

2Mg (u − e1 )(u − e2 )(u − e3 ) A1

(5.8)

with

−1 ≤ e1 ≤ e2 ≤ 1 ≤ e3 This is a classic equation with solution of the form u(t) = a + b sn2 (k, αt)

(5.9)

Substituting (5.9) into (5.8) we find the following relations: a = e1

b = e2 − e1

k2 =

e2 − e1 e3 − e1

α2 =

Mg ( e3 − e1 ) 2A1

5.2 Lagrange Top

An equation for φ˙ is obtained by eliminating φ˙ from the first integrals of motion φ˙

= = =

1 (h1 − h2 cos θ ) A1 sin2 θ h1 − h2 u A1 ( 1 − u2 ) h1 + h2 1 h1 − h2 + 2A1 (1 + a) 1 + b sn2 αt 2A1 (1 − a) 1 − 1+ a

1 b 2 1− a sn αt

which is the canonical form for the elliptic integral of the third kind. Therefore, b h1 + h2 h1 − h2 −b φ= Π k, , αt + Π k, , αt 2A1 (1 + a)α 1+a 2A1 (1 − a)α 1−a Starting from the second integral and performing a similar sequence of manipulations we obtain an equation for ψ˙ h h h1 + h2 ψ˙ = 3 + 3 − A3 A1 2A1 (1 + a) 1 +

1 b 2 1+ a sn αt

+

h2 − h1 2A1 (1 − a) 1 −

1 b 2 1− a sn αt

and a solution in terms of Π(k, α, t), the elliptic integral of the third kind, ψ

1 1 + )t + A1 A3 b h1 + h2 h2 − h1 −b Π k, , αt + Π k, , αt 2A1 (1 + a)α 1+a 2A1 (1 − a)α 1−a

= h2 (

5.2.2 Motion of the Top’s Axis

The motion of the axis of the top is given completely by θ (t) and φ(t) as we saw in the analysis of the free rotational motion of a rigid body (Fig. 5.2). The angle of nutation, θ, is bounded in the range e1 ≤ cos θ ≤ e2 reflecting of the confinement of u to the range e1 ≤ u ≤ e2 as shown in Fig. 5.7. If the motion of the angle of precession, φ, is monotonic then the trajectory in θ, φ-space (the unit sphere) is sinusoidal in form. To assess that possibility we examine the differential equation satisfied by φ h − h2 u φ˙ = 1 A1 ( 1 − u2 ) If h1 − h2 u = 0 for e1 ≤ u ≤ e2 then there can be no reversal and the motion is monotonic. On the other hand, if h1 − h2 u = 0 has a root in [e1 , e2 ] reversal will

147

148


Fig. 5.8 Trajectories of the axis of the top on the unit sphere.

occur and the motion is either looped (generically) or cusped (at the boundary between looped and monotonic motions). This is illustrated in Fig. 5.8.

5.3 The Gyrostat

In Section 3.3.2 we derived the Hamiltonian for a three-rotor gyrostat in Andoyer variables. Here we will consider a special case – that of a single rotor R on the z-axis of the platform B . The Hamiltonian for this problem is K=

1 [ a1 ( G2 − L2 ) sin2 l + a2 ( G2 − L2 ) cos2 l + a3 ( L + µ)2 ] 2

which is more conveniently written as K=

1 2 1 ( G − L2 )[( a2 + a1 ) + ( a2 − a1 ) cos 2l ] + a3 ( L + µ)2 4 2

5.3 The Gyrostat

where ai = 1/Ai , Ai being the principal moments of inertia assumed ordered by A1 ≤ A2 ≤ A3 . The equations of motion are then 1 d l = L[2a3 − ( a1 + a2 ) + ( a2 − a1 ) cos 2l ] + µa3 dt 2 d 1 g = G [( a2 + a1 ) + ( a2 − a1 ) cos 2l ] dt 2 d h=0 dt d 1 L = ( G2 − L2 )( a2 − a1 ) sin 2l dt 2 d G=0 dt d H=0 dt We see immediately that h, G, and H are integrals of motion and the problem is of one degree of freedom in l and L. The variable g is obtained by quadrature once l (t) and L(t) are available. 5.3.1 Bifurcation of the Phase Portrait

Since the angular momentum is constant we may represent the phase space as the unit sphere to be with coordinates l and λ = arccos( L/G ). The singular points in phase space are those points where l˙ = L˙ = 0. There are either 6, 4, or 2 singularities depending on the magnitude of µ. A bifurcation occurs at values of µ for which the number of singularities changes. For the single-rotor gyrostat problem bifurcations occur for µ = ( a2 − a3 )/a3 and ( a1 − a3 )/a3 (Fig. 5.9). If 0 ≤ |µ| < ( a2 − a3 )/a3 there are 6 singularities l

0

π/2

π

3π/2

–

–

L

µa3 /( a3 − a2 )

µa3 /( a3 − a1 )

µa3 /( a3 − a2 )

µa3 /( a3 − a1 )

G

G

These 6 singularities correspond to the four stable centers and two unstable saddle points in the free rotation problem. If ( a2 − a3 )/a3 ≤ |µ| < ( a1 − a3 )/a3 there are 4 singularities l

π/2

3π/2

–

–

L

µa3 /( a3 − a1 )

µa3 /( a3 − a1 )

G

G

The unstable saddle points have disappeared. If ( a2 − a1 )/a3 ≤ |µ| there are 2 singularities

149

150


Fig. 5.9 Bifurcations for the single-rotor gyrostat. The curves were calculated for ai =4, 2, 0.5 and the values of µ are shown above the spheres. The bifurcations occur for µ = 3, 7.

l

–

–

L

G

G

A complete analysis of the bifurcations will be found in [51]. 5.3.2 Reduction to Quadrature

Here one follows the strategy used in the Andoyer variable treatment of the free rotation problem. The energy integral is used to eliminate the l dependence of dL/dt. Let (π/2, L0 ) be a point on a trajectory. The energy integral becomes 1 1 2 ( G − L2 ) ( a2 + a1 ) + ( a2 − a1 ) cos 2l + a3 ( L + µ)2 4 2 1 1 = ( G2 − L20 ) a1 + a3 ( L0 + µ)2 2 2

5.3 The Gyrostat

This leads to the reduced form of the dL/dt equation

dL dt

2

=

( L2 − L20 )( a1 − a3 ) − 2a3 µ( L − L0 )

(5.10) 2 G ( a2 − a1 ) − L20 ( a3 − a1 ) + L2 ( a3 − a2 ) − 2a3 µ( L − L0 )

As in the free-rotation problem this is a biquadratic and the solution is given in terms of Jacobi elliptic functions although more special cases arise because of the bifurcations in the phase portrait. We will consider only one example case in detail. Complete analyses in terms of elliptic functions will be found in [52, 53]. Let 0 ≤ |µ| < ( a2 − a3 )/a3 and consider the case of circulation when L0 lies above the separatrix. The first step toward converting (5.10) into a canonical form is to move the coefficients of L2 outside the brackets and to factor the first term

2 dL 2a3 µ = ( a1 − a3 )( a2 − a3 )( L − L0 ) L + L0 − (5.11) dt a1 − a3 2a3 µ 2 2 ( a1 − a2 ) 2 ( a1 − a3 ) ( L − L0 ) + G − L0 L − ( a2 − a3 ) ( a2 − a3 ) ( a2 − a3 ) Now we are motivated by the fact [43] that the coordinate transformation z2 =

β−δ y−α α−δ y−γ

transforms the differential form dy 1 , 2 (y − α)(y − β)(y − γ)(y − δ) to the canonical form

dz (1 − z2 )(1 − k2 z2 )

with k2 =

β−γα−δ α−γ β−δ

In this way the solution to y˙ 2 = (y − α)(y − β)(y − γ)(y − δ) is found to be y=

β(α − δ)sn2 (t + c, k) + α( β − δ) (α − δ)sn2 (t + c, k) − ( β − δ)

151

152


Thus we posit a solution of the form L=A

sn2 (αt, k) + a sn2 (αt, k) + b

(5.12)

The constants A, a, b, k2 , and α will be determined by matching the coefficients in (5.11). From (5.12) we calculate L˙ 2

=

4α2 ( L − A)( Aa − bL) A2 ( b − a )2

[(1 + b) L − A( a + 1)][(1 + k2 b) L − A(k2 a + 1)] Now we make the following substitutions: A = L0

a 2µa3 = −1 + b L0 ( a 1 − a 3 )

and let f = L/l0 . Also introduce Ln = µa3 /( a1 − a3 ) and Ls = µa3 /( a2 − a3 ), the values of L at the node and stationary point on the separatrix, along with γ = G/L0 , λn = Ln /L0 , and λs = Ls /L0 . Then one obtains f˙2

4α2 b(1 + b)(1 + k2 b) = − (1 − f )( f + 1 − 2λn ) ( b − a )2

1 + a 1 + k2 a 1+a 1 + k2 a 2 + f+ f − 1 + b 1 + k2 b 1+b 1 + k2 b

Matching the coefficients in (5.11) provides equations for a, b, k2 λs − ∆1/2 − 1 λs + 2λn − ∆1/2 − 1 λs − ∆1/2 − 1 λs + 2λn − ∆1/2 − 1 λs + ∆1/2 − 1 a − b(λs + ∆1/2 )

a

= (1 + 2λn )

(5.13)

b

=

(5.14)

k2

=

(5.15)

where ∆ = λ2s − 2λs − γ2 ( a1 − a2 )/( a2 − a3 ) + ( a1 − a3 )( a2 − a3 ). There is one free parameter, α, remaining which is determined by matching the constant factor in (5.12). As a check we evaluate the modulus for L0 = G and L0 = L∗ where (l, L) = (π/2, L∗ ) lies on the separatrix. From the energy integral evaluated at (0, Ls ) we find a − a3 a − a2 − γ2∗ 1 =0 λ2∗ − 2λ∗ + 1 a2 − a3 a2 − a3 where λ∗ = Ls /L∗ and γ∗ = G/L∗ . Then for L0 = L∗ we find k2 = 1 and for L0 = G we find k2 = 0, which are the correct values.

5.4 Kowalevsky Top

5.4 Kowalevsky Top

The equations of motion for a general top (no restrictions on moments of inertia or position of the center of mass) are

I ω˙ γ˙

= I ω × ω − mg¯r × γ

(5.16)

= γ×ω

(5.17)

where I is the inertia tensor referred to the fixed point and in the principal axis frame, r¯ is the position of the center of mass (a constant) and e¯3 = γi ei . These equations have three integrals of motion 2E h¯ 3 1

= ω · I ω + 2mg¯r · γ

(5.18)

= γ · Iω

(5.19)

= γ·γ

(5.20)

The tops we have studied were defined by the relative position of the fixed point and the center of mass and symmetry conditions. The center of mass and the fixed point coincide for the Euler top. The Lagrange top satisfies the axial symmetry condition A1 = A2 and the center of mass and the fixed point lie on the axis of symmetry. In each case there was an additional, fourth, integral of motion, total angular momentum I ω · I ω in the Euler case and ω3 in the Lagrange case. These four integrals for the six degree of freedom problem enable it to be reduced to a two degree of freedom problem, which can be reduced to quadrature. The Kowalevsky top [54] is defined by the symmetry condition A1 = A2 = 2A3 , where Ai are principal moments of inertia relative the fixed point, and the condition that the center of mass lie in the e1 − −e2 plane where the origin of the e frame is at the fixed point. Kowalevsky discovered that there is a new fourth integral of the problem which can be expressed [55] as 2 2 ¯ ¯ κ = [ A(ω12 − ω22 ) − xmgγ 2) 1 ] + (2Aω1 ω2 − xmgγ

(5.21)

We will follow [35] to derive the fourth integral. The component form of the equations of motion in the Kowalevsky case is 2Aω˙ 1

=

2Aω˙ 2

¯ 3 = − Aω1 ω3 + mg xγ

(5.23)

¯ 2 = −mg xγ

(5.24)

γ˙ 1

= γ2 ω 3 − γ3 ω 2

(5.25)

γ˙ 2

= γ3 ω 1 − γ1 ω 3

(5.26)

γ˙ 3

= γ1 ω 2 − γ2 ω 1

(5.27)

Aω˙ 3

Aω2 ω3

(5.22)

153

154


The x-axis has been chosen to pass through the center of mass so that r¯ = ¯ 0, 0) and we have set A3 = A. Now form (5.22)+ı(5.23) and (5.25)+ı(5.26) ( x, 2A

d (ω1 + ıω2 ) dt d (γ + ıγ2 ) dt 1

¯ 3 = −ıAω3 (ω1 + ıω2 ) + ımg xγ

(5.28)

= −ıω3 (γ1 + ıγ2 ) + ıγ3 (ω1 + ıω2 )

(5.29)

Eliminate γ3 between (5.28) and (5.29) to obtain d σ = −ıω3 σ dt where σ = A(ω1 + ıω2 )2 − mg x¯ (γ1 + ıγ2 ) From this it is follows that d d d (σσ ∗ ) = σ ∗ σ + σ σ ∗ = 0 dt dt dt and we obtain the integral of motion σσ ∗ = κ which is the same as (5.21). This new integral is of historical importance as it was discovered by a novel method which analyzed singularities of the solution in the complex time plane. The equations are integrable in terms of hyperelliptic functions which are inverse functions of the hyperelliptic integrals

dx , P( x )

where P( x ) is a polynomial of degree 5. Reference [50] derives the fourth integral in terms of the Euler angles, [56] gives a modern mathematical account of the problem using geometroalgebraic methods and [55] discusses the analysis of singularities of ordinary differential equations.

5.5 Liouville Tori and Lax Equations

The introduction to this chapter gave a pragmatic description of integrability. There is, however, a vast literature on the subject of integrable systems (see [57,58] for example) and many technical definitions of types of integrability. The goal of those studies is to identify systems which are “regular,” “reducible to quadratures” or “separable” versus systems which are “chaotic,” “ergodic” or “not expressible in terms of classical functions” and to study the geometry and topology of the trajectories for integrable systems. In this section we will describe two notions from the theory of integrable systems. The


first, the Liouville tori, describes the geometry underlying a class of integrable Hamiltonian systems. Our discussion here will follow [59]. The second, the Lax equation, is a technique for finding integrals of motion and finding solutions using geometro-algebraic concepts. Our treatment of this subject follows [56] and [57]. 5.5.1 Liouville Tori

Consider the Lagrange top. It is a system with a six-dimensional phase space (perhaps parameterized by the Euler angles and the conjugate momenta). There are four functionally independent integrals (energy and three components of angular momentum). Thus the motion for a system with given values of the four integrals generically occurs on a two-dimensional sub-manifold M of phase space. There are nonsingular trajectories which fill M and tangent vectors to these trajectories provide a nonvanishing vector field on M. The only orientable, two-dimensional manifold which a nonvanishing vector field is the torus. Therefore the motion occurs on a torus. The Poisson brackets { H, pi } vanish because the pi are integrals and by the definition of the bracket the { pi , p j } vanish. This situation is but a special case of Liouville-integrable systems and in general the tori are called Liouville tori because of the following theorem: Liouville–Arnold Theorem: Consider a Hamiltonian system with n degrees of freedom q˙ = H p

p˙ = − Hq

q = ( q1 , . . . , q n )

having n integrals of motion F1 , . . . , Fn

{ H, Fi } = 0

p = ( p1 , . . . , p n )

in involution

{ Fi , Fj } = 0

Then if the equations Fi = Ii define a n-dimensional connected, compact manifold M and at each point of M the Fi are functionally independent then (1) M is an n-dimensional torus T n and neighborhood of M is the direct product T n × R n , (2) this neighborhood admits action-angle coordinates ( I, φ), I˙ = 0

φ˙ = H I = ω ( I )

The proof is beyond the scope of this book – see [59].

155

156


5.5.2 Lax Equations

A Lax equation is differential equation satisfied by a pair of matrices A(t) and B(t) A˙ = [ A, B] where [ A, B] = AB − BA, the commutator of A and B. The matrices A and are skew-symmetric a, Ω = − ω B are called a Lax pair. For example, if A = then A˙ = [ A, Ω] is the same as a˙ = ω × a, which we recognize as an important kinematic relation in rigid body mechanics. Our use of Lax equations will involve matrices which depend on indeterminate parameter, say λ, and this will be denoted by adding a subscript to the matrices A˙λ = [ Aλ , Bλ ] A λ = A0 + A1 λ + · · · + A n λ n Bλ = B0 + B1 λ + · · · + Bm λm If A and B are p × p this represents a system of up to p2 (n + 1) differential equations – an unknown for each power of λ in each matrix entry. An isospectral deformation of a matrix A0 is a time dependent similarity transformation A ( t ) = U ( t ) A0 U ( t ) −1 The term isospectral refers to the fact that similarity transformations preserve eigenvalues of a matrix. Thus, the time dependent matrix A(t) has constant eigenvalues and the coefficients of the characteristic polynomial of A(t) are also constants because they can be expressed in terms of the eigenvalues. Solutions of Lax equations are isospectral transformations of the initial conditions. If A(t) = U (t) A0U (t)−1 then ˙ −1 A − AUU ˙ −1 ] ˙ −1 = [ A, −UU A˙ = UU Given A˙ = [ A, B], obtain U (t) from U˙ (t) = − B(t)U (t)

U (0) = I

then A = U A0 U −1 satisfies ˙ −1 ] = [ A, B] A˙ = [ A, −UU

A (0 ) = A0

The importance of Lax equations to the study of integrable systems is that they come with an ample supply of integrals of motion. The Lagrange top problem can be formulated as a Lax equation. To that end let us write the Lagrange top equations in terms of skew-symmetric matrices. The basic rotational equation of motion is h˙ = τ


where τ is the torque. The torque for the Lagrange top is τ=

B

r × (− ge¯ 3 ) dm = ge¯ 3 ×

B

r dm = mge¯ 3 × e3

where is the distance from the fixed point to the center of mass of the top. Since e¯ 3 is constant, we also have e¯˙ 3 = 0 These equations can be expressed in a principal axis frame at the center of mass using h = eh, e¯ 3 = eγ and e˙ = eΩ h˙ + Ωh = mgγ × e3 γ˙ + Ωγ = 0 Let

H= h

Γ=γ

E3 = e3

then the equations can be written in terms of skew-symmetric matrices as H˙ = [ H, Ω] + mg[Γ, E3 ] Γ˙ = [Γ, Ω] Now we want to formulate the problem in terms of a Lax equation. To that end let 1 Aλ = Γ + ηHλ + E3 λ2 η= , mg A1 .

and Bλ = Ω + νE3 λ

ν=

mg A1

Then

[ Aλ , Bλ ] = [Γ, Ω] + (η [ H, Ω] + ν[Γ, E3 ]) λ The coefficients of λ2 and λ3 in [ Aλ , Bλ ] are both zero. Since Γ˙ = [Γ, Ω], H˙ = [ H, Ω] + mg[Γ, E3 ] and E˙ 3 = 0 we find A˙ λ = [ Aλ , Bλ ] and the problem is cast as a Lax equation. The characteristic polynomial for a 3 × 3 skew-symmetric matrix A is det( A − µI ) = µ(µ2 + A2 ) where

− → − → 1 1 A2 = Tr( AAt ) = − Tr( A2 ) = A · A 2 2

157

158


For the Lagrange top,

1 1 2 2 A = 1 + λ , h · γ + λ 2γ · e3 + h · h + λ3 h · e3 + λ4 mg A1 mg A1 mg 2

The isospectral property of Lax equations implies that the coefficients of the characteristic polynomial are constant. Consequently, we obtain the following integrals of motion: h · γ = h 1 γ1 + h 2 γ2 + h 3 γ3 = c 1 h · e3 = A 3 ω 3 = c 2 1 h · h + mgγ · e3 = c3 2A1 The third of these integrals is equivalent to the energy integral in the case of symmetry A1 = A2 because A2 1 h · h = h · ω − A3 ω32 + 3 ω32 = h · ω + c22 2A1 A1 Therefore, 1 h · ω + mgγ · e3 = c3 + c22 2

1 1 − A1 A3

1 1 − A1 A3

= c4

5.6 Exercises

5.6 Exercises

Exercise 5.1 Use the energy integral for free rotation evaluated in terms of l = π/2 and L0 to eliminate l from the equation for dL/dt and find the solution of the resulting equation in terms of Jacobian elliptic functions. Exercise 5.2 Derive the equation for the herpolhode curve for free rotation in the case of symmetry A1 = A2 . Exercise 5.3 Verify Eqs. (5.13–5.15). Exercise 5.4 Find a generalization the Poinsot polhode/herpolhode construction which is valid for the single-rotor gyrostat [35, 48]. Exercise 5.5 Consider the single-rotor gyrostat of Section 5.3. Use Poisson brackets to find differential equations for the components of angular momentum in the body fixed, principal axis frame. Exercise 5.6 Verify that (5.18–5.20) are integrals of motion for the general top equations (5.16–5.17). Exercise 5.7 Derive the Hamiltonian H for the Kowalevsky top and show that {K, H } = 0, where K is the Kowalevsky integral (5.21). Exercise 5.8 Show that the problem of the Lagrange top without a fixed point, the contact point being free to slide without friction on a horizontal surface, is integrable (as in the case of the Kowalevsky top the actual quadrature involves hyperelliptic functions [28]).

159


161

6

Numerical Methods This chapter considers classical and modern numerical techniques suitable for the numerical integration of rigid body equations of motion. These techniques are used when the problem at hand has no known analytical solution. Analytic solutions supply continuous functions of time which define the exact motion of the dynamical system insofar as the model used is itself accurate. In other words, since all mathematical models are merely approximations to physical reality, the virtue of the analytic solution is that it introduces no further approximations. Numerical techniques, on the other hand, introduce another level of approximation. In place of continuous functions there are approximations to the solution at a discrete sequence of time points. The primary requirement of a numerical method is that its approximation approaches the true solution as the spacing between the discrete time step approaches zero. In this chapter we first consider classical techniques for approximating solutions to ordinary differential equation (ODE) initial boundary problems (IVP). Then we consider the modern geometrical methods, symplectic and Lie group methods. Multibody rigid body problems are often cast as differential– algebraic systems (DAE) and we briefly consider them here. The chapter ends with a case study – the wobblestone – which exhibits fascinatingly complicated behavior. 6.1 Classical ODE Integrators

Let us start with a single differential equation dy = f (y, t) dt

y (0) = y0

We wish to find approximate values of y (t1 ), y(t2), . . . , y (t N ). For the time being let the time intervals all have the same length so that t1 = h, t2 = 2h, . . ., t N = Nh. Our objective is accomplished if we can provide a method to Rigid Body Mechanics: Mathematics, Physics and Applications. William B. Heard Copyright © 2006 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim ISBN: 3-527-40620-4

162

6 Numerical Methods

Fig. 6.1 Illustration of discrete solution with local and global truncation errors.

advance the solution from y (t) to y(t + h). Accordingly, we can use the Taylor series expansion y(t + h) = y(t) + h

d y ( t ) + O( h 2 ) = y ( t ) + h f ( y ( t ) , t ) + O( h 2 ) dt

By neglecting term of order h2 we arrive at the forward Euler method y n = y n −1 + h f ( y n −1 , t n −1 ) where y n ≈ y(tn ) is the approximation to the solution. There are three sources of error in the y n . First, there is round-off error inherent to any digital device. The round-off error depends on the number of bits representing the mantissa and the exponent of a floating point number in memory and the method of rounding (chopping or rounding up, for example). Second, there is local truncation error which in this case is of order h2 because these terms are neglected in the Taylor series expansion. Third, there is the global truncation error which is an accumulating error incurred as the starting point for approximating y n is not y(tn−1 ) but rather y n−1 . The global truncation error is generally an order of magnitude larger than the local error. Thus, the forward Euler method is regarded as a first-order method. If the time step is halved the global error will be halved. The truncation errors are illustrated in Fig. (6.1). We can obtain a companion method, the backward Euler method by evaluating the Taylor series at t + h y(t) = y(t + h) − h

d y ( t + h ) + O( h2 ) = y ( t + h ) − h f ( y ( t + h ), t + h ) + O( h2 ) dt

6.1 Classical ODE Integrators

Now the neglect of second-order terms gives y n − h f ( y n , t n ) = y n −1 Notice that y n is now given implicitly. One must solve the equation y n − h f (yn , tn ) − yn−1 = 0 for y n . This equation is usually nonlinear and its solution can add considerable overhead to the computing time. The forward and backward Euler methods are also called the explicit Euler and implicit Euler , respectively. A reason one might prefer the implicit method to the explicit one is stability. This is seen in its most elementary form in the population equation dy = ay dt

a 2/| a| the solution oscillates to infinity. Thus, the forward Euler method is conditionally stable. The implicit Euler method gives y n = yn−1 /(1 − ah ) For any a < 0, we have 1 − ah > 1 and the approximate solution approaches zero monotonically as it should. The backward Euler method is unconditionally stable, that is, it is stabile for any step size. The Euler methods generalize easily to systems of equations. One simply replaces scalar arguments for vector arguments. For example, consider the Euler equations for free rotation Aω˙ 1 + (C − B)ω2 ω3

= 0

Bω˙ 2 + ( A − C )ω3 ω1

= 0

C ω˙ 3 + ( B − A)ω1 ω2

= 0

163

164

6 Numerical Methods

The Euler methods for this system are

(C − B) ω2,n ω3,n A ( A − C) ω3,n ω1,n ω2,n+1 = ω2,n + h B ( B − A) ω1,n ω2,n ω3,n+1 = ω3,n + h C ω1,n+1 = ω1,n + h

and

(C − B) ω2,n+1ω3,n+1 = ω1,n A ( A − C) ω3,n+1ω1,n+1 = ω2,n ω2,n+1 − h B ( B − A) ω1,n+1ω2,n+1 = ω3,n ω3,n+1 − h C ω1,n+1 − h

The implicit method requires the solution of a system of nonlinear algebraic equations. We have seen an illustration of the stability characteristics of the Euler methods. Let us now examine their accuracy. The example of simple harmonic motion is instructive. The problem is d2 u = −u dt2

u (0) = 1

ddtu(0) = 0

This is easily cast as a system of two first-order equations u˙

=

v, u(0) = 1,

v˙

= −u, v(0) = 0

The explicit Euler method gives un

= un−1 + hvn−1

vn

= vn−1 − hun−1

while the implicit method gives un

= un−1 + hvn

vn

= vn−1 − hun

The energy u2 + v2 is an integral of motion for this system. The Euler solutions provide u2n + v2n = (u2n−1 + v2n−1 )(1 + h2 )


and u2n + v2n = (u2n−1 + v2n−1 )/(1 + h2 ) for the explicit and implicit cases, respectively. Thus the phase trajectory spirals outward in the explicit case, inward in the implicit case; and a major qualitative feature of the exact solution is not captured by the numerical solution. If these methods were applied to the free rotation of a rigid body the integrals of energy and momentum would not preserved. The simplest example of this is the rotational kinetic energy for the explicit Euler method where we find

+

2 2 2 2 2 2 Aω1,n +1 + Bω1,n+1 + Cω1,n+1 = Aω1,n + Bω2,n + Cω3,n+1 ( C − B )2 2 2 ( A − C )2 2 2 ( B − A )2 2 2 ω2,n ω3,n + ω3,n ω1,n + ω1,n+1ω2,n h2 A B C

Higher order methods could also be considered. For example, let us construct a second-order method using the Taylor series expansions 1 = y(t) + hy˙ (t) + h2 y¨(t) + O(h3 ), 2 y˙ (t) = f (y, t),

y(t + h)

y¨(t)

=

f y y˙ + f t = f y f + f t

Therefore, a second-order method is 1 yn+1 = yn + h f (yn , tn ) + h2 [ f y (yn , tn ) f (yn , tn ) + f t (yn , tn )] 2 For a system of equations this becomes 1 y n +1 = y n + h f ( y n , t n ) + h 2 [ J f ( y n , t n ) + f t ] 2 *

with y, f ∈ R

n

J=

∂ fi ∂y j

+

The second-order method applied to the Euler equations for free rotation yields the following method: ω1,n+1

(C − B) ω2,n ω3,n + A 1 2 (C − B)( A − C ) 2 h ω1,n ω3,n + 2 A 1 2 (C − B)( B − A) 2 h ω1,n ω2,n 2 A

= ω1,n + h

165

166

6 Numerical Methods

( A − C) ω3,n ω1,n + B 1 2 ( A − C )( B − A) 2 h ω2,n ω1,n + 2 B 1 2 ( A − C )(C − B) 2 h ω2,n ω3,n 2 B ( B − A) ω1,n ω2,n + = ω3,n + h C 1 2 ( B − A)(C − B) 2 h ω3,n ω2,n + 2 C 1 2 ( B − A)( A − C ) 2 h ω3,n ω1,n . 2 C = ω2,n + h

ω2,n+1

ω3,n+1

This can be expressed more concisely in vector notation as ω˙

= −I −1 [ω × I ω],

ω ¨

= −I −1 [ ω˙ × I ω + ω × I ω˙ ] , 1 = ωk + hω˙ (ωk ) + h2 ω¨ (ωk ) 2

ωk+1

To construct a third-order method one needs d3 y/dt3 which is d3 y = f f y2 + f y f t + f 2 f yy + 2 f f yt + f tt dt3 It becomes clear that this is not a viable route to higher order methods. The derivatives become increasingly complex and intractable for most systems. There is also the issue of computational complexity, that is, the number of floating point operations per time step. The higher order Taylor series method will be very inefficient. The drawbacks to the higher order Taylor series methods are the motivation for the Runge–Kutta methods. The idea is to avoid calculating higher order derivatives by using judiciously chosen function evaluations. This is best illustrated by the following example – the second-order explicit Runge–Kutta method for the equation y˙ = f (y, t).

k1

=

1 (k 1 + k 2 ) , 2 f ( y n , t n ),

k2

=

f (yn + hk1 , tn + h)

y n +1

= yn +

If the right-hand side of the first equation is expanded in Taylor series about y n , tn , we obtain 1 y n +1 = y n + h f ( y n , t n ) + h 2 ( f y f + f t ) + O( h 3 ) 2 which is precisely the second-order Taylor series method.


The Runge–Kutta methods have the general form s

y n +1

= y n + h ∑ bi k i i =1

ki

=

f

s

yn + h ∑ aij k j , tn + ci h j =1

and are expressed with the array (the so-called Butcher tableau) c1 .. .

a11 .. .

...

cs

as1

...

ass

b1

...

bs

a1s .. .

or in abbreviated form

c

A b

We can use this notation to denote the method just described as RK2: 0

0

0

1

1

0

1 2

1 2

A very widely used ODE method is the fourth-order Runge–Kutta RK4: 0

0

0

0

0

1 2

1 2

0

0

0

1 2

0

1 2

0

0

1

0

0

1

0

1 6

1 3

1 3

1 6

It is instructive to construct Taylor series expansions of the Runge–Kutta formulas about y n , tn . In that way we see how constraints on A, b, c arise. To completely construct a Runge–Kutta method of order n would require the nthorder Taylor series. Here we will examine only the first-order expansions of the k i which are ki

=

f (yn , t) + h f y (yn , t)(∑ aij k j ) + hci f t (yn , t) + O(h2 ) j

=

f (yn , t) + h(∑ aij ) f y (yn , t) f (yn , t) + hci f t (yn , t) + O(h2 ) j

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6 Numerical Methods

where f y (yn , t) is the matrix with ij entry ∂ f i /∂y j . Substituting this last result and the series expansion for y (tn + h) into the Runge–Kutta formula we obtain 1 yn + h y˙ + h2 ( f y f + f t ) + O(h3 ) 2

= yn + h(∑ bi ) f + h2 (∑ bi ∑ aij ) f y i

i

j

+ h ( ∑ bi c i ) f t + O ( h ) 2

3

i

Function arguments, suppressed in this equation are understood to be y n , t. Now we see that if ∑i bi = 1 the method is consistent, that is, the discretized equation converges to the differential equation as h → 0. Consistency is a necessary condition for the approximate solution to converge to the exact solution, yn → y(tn ). Furthermore, if ∑ j aij = ci the method is at least second order accurate. Runge–Kutta methods in which the matrix A is lower triangular (aij = 0 for j ≥ i) are explicit methods. Otherwise they are implicit methods.

6.2 Symplectic ODE Integrators

The classical numerical methods can introduce spurious effects when applied to Hamiltonian systems. The reason for this is that they do not respect the Hamiltonian structure. For example, to use the backward Euler method on a Hamiltonian system is to introduce dissipation which is antithetical to a Hamiltonian system. The past two or three decades have seen intense research activity aimed at the development of numerical methods which are amenable to Hamilton systems. These new methods are called symplectic integrators because they are built on symplectic maps. Consider a Hamiltonian system with phase-space T ∗ M and with coordinates q and conjugate momenta p. A map φ : T ∗ M → T ∗ M : z → φ(z), z = (q, p )t is symplectic if Φt JΦ = J where Φ is the Jacobian of the map and J is the canonical symplectic matrix   * + ∂φi  0 I  J= Φ= . ∂z j −I 0 If H (q, p ) is the Hamiltonian for the system the equations of motion can be written z˙ = J Hz z = (q, p)t

6.2 Symplectic ODE Integrators

The map induced by the flow of the Hamiltonian system, z(0) → z(t), has Jacobian Φ = z ( t ) z (0) and from the equations of motion we find that d Φ = J Hzz Φ dt

* Hzz =

∂2 H ∂zi ∂z j

+

where Hzz is the Hessian of H. To show that the flow map is symplectic we examine (d/dt)(Φt JΦ) d t Φ JΦ = Φt Hzz J t JΦ + Φt J J Hzz Φ = Φt Hzz Φ − Φt Hzz Φ = 0 dt because J t J = − J J = I. This, coupled with the fact that Φ(0) = I, proves that Φ is symplectic. The map of interest to developers of numerical integration methods is the discrete version of the flow map which takes zk to zk+1 , successive approximate solutions at tk and tk+1 = tk + h. In the case of the forward Euler method applied to a system with Hamiltonian H (z) we have zk+1 = zk + hJ Hz (zk ) Here φ(z) = z + hJ Hz (z)

and

Φ = I + hJ Hzz

Checking for symplecticity we find Φt JΦ = ( I − hHzz J ) J ( I + hJ Hzz ) = J and the map is not symplectic. In the case of the backward Euler method zk+1 = zk + hJ Hz (zk+1 ) The expression for this map is implicit and its Jacobian is obtained from Φ = I + hJ Hzz Φ

Φ = ( I − hJ Hzz )−1

It also fails to be symplectic. The simplest symplectic method is the mid-point method [60]

1 zk+1 = zk + hJ Hz ( z + z k +1 ) 2 k We obtain the Jacobian Φ from 1 Φ = I + hJ Hzz ( I + Φ) 2

1 1 Φ = ( I − hJ Hzz )−1 ( I + hJ Hzz ) 2 2

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6 Numerical Methods

Let K = 12 hJ Hzz . Then the symplecticity condition is

( I + K t )( I − K t )−1 J ( I − K )−1 ( I + K ) = J It is easy to verify directly that

( I + K) J ( I + Kt ) = ( I − K) J ( I − Kt )

(6.1)

because JK t = −K J If one takes the inverse of (6.1) and moves all factors involving K to the righthand side one verifies the symplecticity condition. If the mid-point method is applied to the harmonic oscillator with Hamiltonian 1 H (z) = zt z Hz = z 2 one obtains the following update scheme: −1 1 1 z k +1 = I − hJ I + hJ zk 2 2   1 2 − h h 1 1   4 =  zk 1 2  1 + 4h −h 1 − 14 h2 It is easily verified that z2k+1 = z2k , so that energy is conserved exactly by this method. Although this is not a general feature of the symplectic methods it does foreshadow a great improvement in performance that is found for a wide class of problems.

6.3 Lie Group Methods

It often happens that the differential equation at hand evolves on a matrix Lie group (a continuous subgroup of GL(n)). In the case of many rigid body problems the group is SO(3) × so(3). The classical methods employ vector space operations which typically do not respect a Lie group structure. One cannot form a linear combination of orthogonal matrices, for example, and expect to obtain an orthogonal matrix. The recently developed methods called Lie group methods are designed to preserve the Lie group structure [60, 61]. Consider the differential equation y˙ (t) = A(y, t)y(t) y(0) = y0

y(t) ∈ G

A(y, t) ∈ g

(6.2)

where G is a Lie group and g is its Lie algebra (Appendix C). The general approach taken with Lie group methods is to represent y as a transform of


an element in the Lie algebra, to advance the solution one step in g and to invert the transform to obtain the advanced solution in the Lie group. The Lie algebra is a vector space where the classical methods can be applied (Fig. 6.2). The transform is chosen so that the advanced value always lies in G .

yn

y˙ = A(y, t)y -yn+1 6

σ = φ −1 ( y )

G

y = φ(σ )

? σn

σ˙ = f (σ, t)

-

g

σn+1

Fig. 6.2 Lie group methods transform to the Lie algebra to insure the solution is constrained to lie in the Lie group.

A commonly used transform is y(t) = exp(σ (t))yn

t ∈ [ t n , t n +1 ]

(6.3)

To obtain the transformed differential equation σ˙ = f (σ, t) we need to calculate d exp(σ (t))/dt. This requires the dexpx function discussed in Section C.4 and the result is d exp(σ (t)) = dexpσ σ˙ exp σ dt where ∞ 1 dexpσ ( A) = ∑ ad kσ A ( k + 1 ) ! 0 and the iterated commutator adkσ is defined recursively by ad0σ = identity

adkσ+1 = [σ, adkσ ]

The matrix inverse of dexpx is (Section C.4) 1 dexp− x =

∞

∑

k =1

Bk k ad k! x

where Bk , k = 1, 2, 3, . . . is the kth Bernoulli number. Now if we take the derivative of (6.3) we obtain y˙ (t) = dexpσ σ˙ (t) exp(σ (t))yn

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6 Numerical Methods

Comparing this to (6.2) we see that A(y, t)y(t) = dexpσ σ˙ (t)y(t) and consequently the problem is transformed to 1 σ˙ (t) = dexp− σ A (exp( σ ( t )) y n , t )

(6.4)

Equation (6.4) is the differential equation on the Lie algebra to which the classical numerical methods can be applied. The initial condition σ (tn ) = 0 follows from y(tn ) = exp(σ (tn ))yn and the identification of y (tn ) and yn . In (6.2) A multiplies y on the left and we will refer to this as the left-handed version of the problem. There is a right-handed version which will be relevant to our rigid body problems y˙ (t) = y(t) A(y, t) y(0) = y0

y(t) ∈ G

A(y, t) ∈ g

Using the result (C.6) we obtain the transformed, Lie algebra equation for the right-handed version 1 σ˙ (t) = dexp− −σ A ( y n exp( σ ( t )), t )

As example of using classical numerical methods on the Lie algebra, the Runge–Kutta/Munthe–Kaas (RKMK3) method is based on the explicit third order Runge–Kutta method RK3: 0

0

0

0

1 2

1 2

0

0

1

−1

2

0

1 6

2 3

1 6.

When implemented on the Lie algebra it leads to the RK3MK method k1 k2

=

A(yk , tk )

1 1 1 = dexp−1 hk A e 2 hk1 yk , tk + h 2 2 1

1 1 1 ≈ 1 − ad 1 hk A e 2 hk1 yk , tk + h 2 2 1 2

1 −hk 1 +2hk 2 = dexp− yk , tk + h) −hk 1 +2hk 2 A ( e

1 ≈ 1 − ad −hk1 +2hk2 A(e−hk1 +2hk2 yk , tk + h) 2

1 2 1 ∆ = h k + k + k 6 1 3 2 6 3 yk+1 = exp (∆)

k3


The expressions for k2 and k3 incorporate the truncated dexp−1 function 1 1 1 dexp− x y ≈ y − 2 ad x y = y − 2 [ x, y ] Now we will focus on two particular Lie group methods, the second-order SE method (symplectic Euler) and the fourth-order RK4MK method (Runge– Kutta/Munthe–Kaas) [62]. We will define the methods and, following [63], see how to apply them to free rotation of a rigid body having a fixed point (Euler top) and the Lagrange top. In the case of the Euler top, the results from the two methods will be compared to the exact solution to verify the implementation and to assess their accuracy. Let us consider the differential equation y˙ = A(y, t)y with y ∈ G and A ∈ g (G , g a Lie group, Lie algebra pair). Let y(t) = exp(σ (t))y(tn) in the interval tn ≤ t ≤ tn+1 . As we have seen, σ˙ = dexpσ−1 A(exp(σ )y n ). The SE method for advancing the solution from σn−1 to σn is

1 1 A exp ( σ + σ ) y σn = σn−1 + h dexp− n n −1 n 1 2 2 ( σn + σn −1 ) Pseudo-code (recognizable to a programmer as a template for actual code in a computer language such as Fortran or C) expresses the algorithm as assign initial condition y for number_of_steps do update σ:

1 1 solve σ = hdexp− 1 A exp( 2 σ ) y 2σ

update y: y ← exp(σ )y end do After n = number_o f _steps steps the algorithm returns y which is the approximate solution at time t0 + nh. The variable σ in the pseudo-code represents σn . The method is based on the discretized derivative (σn − σn−1 )/h but the initial condition for each update is σn−1 = 0 hence the form of the update σ step. The factor dexpσ−1 is implemented by truncating the series expansion at an appropriate point. For a second-order method it can be truncated at the second term. The update sigma step involves the solution of a system of equations which is generally nonlinear. The fixed point iteration method can be effective for small problems.

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6 Numerical Methods

The algorithm for applying the RK4MK method to this problem is expressed, without any further ado, in pseudo-code as assign initial group variable y for number_of_steps do update σ:

=

k1

A(y)

1 1 = dexp−1 hk A exp( hk1 )y 1 2 2

1 1 = dexp−1 hk A exp( hk2 )y 2 2 2

k2 k3

−1 = dexphk A (exp(hk3 )y) 3

k4

h (k1 + 2k2 + 2k3 + k4 ) 6

=

σ update y: y ← exp(σ )y end do

Here dexp−1 can be truncated at the fourth term. The right-handed version of RK4MK is assign initial group variable y for number_of_steps do update σ: k1 k2 k3 k4 σ update y: y ← y exp(σ ) end do

=

A(y)

1 1 hk = dexp− A y exp 1 − 21 hk 1 2

1 1 hk = dexp− A y exp 2 − 21 hk 2 2 1 = dexp− −hk 3 A (y exp( hk 3 ))

=

h (k1 + 2k2 + 2k3 + k4 ) 6


The relevant Lie group for the Euler and the Lagrange tops is TSO(3). We will follow [63, 64] and work in the direct product trivialization SO(3) × so(3) which is isomorphic, as a Lie group, to TSO(3). Thus the equations will be written in terms of pairs ( R, ω) with R ∈ SO(3) and ω ∈ so(3) ∼ = R3 . The group operation is

( R1 , ω1 )( R2 , ω2 ) = ( R1 R2 , ω1 + ω2 ) The Lie algebra of TSO(3) is so(3) × so(3) ∼ = R3 × R3 . With x1 , x2 , y1 , y2 ∈ R3 and α ∈ R the Lie algebra operations are addition and multiplication by a scalar

( x1 , y1 ) + ( x2 , y2 ) = ( x1 + x2 , y1 + y2 ) α( x1 , y1 ) = (αx1 , αy1 ) and the Lie bracket

[( x1 , y1 ), ( x2 , y2 )] = ( x1 , y1 )( x2 , y2 ) − ( x2 , y2 )( x1 , y1 ) = ([ x1 , x2 ], 0) The function exp maps so(3) × so(3) to SO(3) × so(3) according to exp( x, y)( R, ω ) = (exp( x ) R, ω + y) and dexp−1 is defined on so(3) × so(3) via the Lie brackets 1 −1 dexp− ( x ,x ) ( y1 , y2 ) = (dexpx1 y1 , y2 ). 1

2

We note that in [65] the semi-direct product SO(3) so(3) was used and some accuracy benefits were noted. Mastering the direct product implementation is good preparation for implementing the semi-direct product version. To implement the SE method we need to express the update σ and update y steps in the direct product Lie algebra. Let the group variable be y = ( R, ω ) and let the algebra variable be σ = (σ1 , σ2 ). The action exp(σ )y becomes exp(σ )y = (exp σ1 , σ2 )( R, ω ) = ((exp σ1 ) R, ω + σ2 ) The equation to be solved in the σ update becomes

1 1 1 σ σ (σ1 , σ2 ) = h dexp− + A exp R, ω 1 1 2 , ( 21 σ1 , 21 σ2 ) 2 2

1 1 A2 exp σ1 R, ω + σ2 2 2

1 1 = h dexp−1 σ1 A1 exp σ1 R, ω + σ2 , 1 2 2 2

1 1 A2 exp σ1 R, ω + σ2 2 2

175

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6 Numerical Methods

The equations for our sample problems written in body fixed, principal axis frame are

I ω˙ = I ω × ω + τ R˙ = RΩ ˆ Thus we see that we are dealing with the right-handed version where Ω = ω. of the problem. The torque is τ = 0 for the Euler top and τ = mglRe¯3 for the Lagrange top. Let f ( R, ω ) = I −1 ((I ω ) × ω + τ ). Thus the functions Ai are A1 ( R, ω ) = ω

A2 ( R, ω ) = f ( R, ω )

Now we can assemble these results to obtain the following algorithm for applying the SE method to the top problems: assign initial group variable y = ( R, ω ) assign initial algebra variable σ = (0, ω ) for number_of_steps do update σ: solve: σ1

=

σ2

=

1 + σ2 ω 2

1 1 + σ2 ω + h f R exp σ1 , ω 2 2 hdexp−11 − 2 σ1

update ( R, ω ): R ← R exp(σ1 ) ω ← σ2 The details of applying the RK4MK method to the top problems are given in the following algorithm: assign initial condition ( R, ω ) for number_of_steps do update (σ1 , σ2 ):


ˆ f ( R, ω )) (k11 , k12 ) = (ω,

1 1 ˆ 1 −1 ˆ (k21 , k22 ) = dexp− 1 hk ωˆ + hk12 , f R exp hk11 , ω + hk12 11 2 2 2 2

1 ˆ 1 ˆ 1 −1 ˆ (k31 , k32 ) = dexp− 1 hk ω + hk22 , f R exp hk21 , ω + hk22 21 2 2 2 2

−1 ˆ ˆ ˆ (k41 , k42 ) = dexp− 1 hk (ω + hk32 ), f ( R exp(hk31 ), ω + hk32 ) 2

31

h (k + 2k21 + 2k31 + k41 ) 6 11 h σ2 = (k12 + 2k22 + 2k32 + k42 ) 6 σ1 =

update ( R, ω ): R ← R exp(σ1 ) ω ← ω + σ2

Fig. 6.3 Error in free body rotation as a function of step size for the SE and RK4MK numerical methods.

The exact solution of the Euler top problem is given in terms of Jacobi elliptic functions in Section 5.1.2. This offers the opportunity to verify the numerical methods under consideration. Verification is the process of determining that an implementation of a numerical model correctly approximates the solution to the underlying differential equation. A important test is to determine

177

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6 Numerical Methods

Fig. 6.4 Relative error in the (a) rotational kinetic energy | E − ESE |/G 2 | /G2 . and (b) angular momentum | G2 − GSE

Fig. 6.5 Relative error in the (a) rotational kinetic energy | E − 2 ERK4MK |/G and (b) angular momentum | G2 − GRK4MK |/G2 .

the accuracy of the approximate solution as a function of the level of refinement of the discretization. In our case this is the size of the time step. If accuracy improves systematically as the discretization is refined it is an indication that the approximate solutions do converge to the exact solution. Our verification is based of numerical experiments for the case { A1 , A2 , A3 } = {1, 2, 4},

6.4 Differential–Algebraic Systems

the initial angular velocity ω0 = {1, 1, 1} and the initial rotation matrix is Re¯1 (π/16). In addition to indicating convergence, the discretization refinement tests can determine the order of accuracy of the numerical methods. The analytical solution is compared to the numerical approximations using a range of step sizes. The results of our comparison are shown in Fig. 6.3. The error refers to the maximum difference between the actual and computed angular velocity components = max{|ωi,numerical(t) − ωi,analytical (t)|, i = 1, 2, 3} over a fixed number of time steps (100 in our case). The time steps used were h = 2−n , n = 1, . . . , 10. The results verify that SE is performing as a secondorder method (halving the time step reduces the error by a factor of four) and the RK4MK is behaving as a fourth-order method (halving the time step reduces the error by a factor of sixteen). The results also indicate convergence. The Euler top problem has two integrals: total angular momentum G and total rotational kinetic energy E (Section 5.1.1). The numerical methods can be judged on their ability to maintain these integrals. Figures 6.4 and 6.5 show the relative error (∆E/E and ∆G2 /G2 ) for the two methods over 1000 time steps with h = 0.01. The results indicate that the integrals are determined precisely by the numerical methods but not exactly. Exact preservation of the integrals would be indicated by accuracy of the order of machine precision (∼ 10−14). Secular drifts are also seen in the errors (more pronounced for SE) so that the errors will steadily increase as the number of time steps is increased.


Differential–algebraic equations (DAE) combine ordinary differential equations and algebraic equations. The formal definition of DAE is a set of equations dz F (t, z, z˙ ) = 0 z˙ = z∈R z = z(t) dt such that rankFz˙ < dim(z). The rank condition declares that F (t, z, z˙ ) = 0 cannot be solved for z˙ as a function of t and z. That is, the system cannot be reduced to a system of ordinary differential equations. A prototypical example is x˙ − f ( x, λ)

= 0

(6.5)

g( x, λ)

= 0

(6.6)

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180

6 Numerical Methods

with x ∈ R m , λ ∈ R n−m and z = ( x, λ)t . Then we have    x˙ − f ( x, λ) F (z) =   g( x, λ) 

and

I Fz˙ =  0

 0  0

which is rank deficient if m < n. Note that we have denoted the second arguments of f and g as λ to suggest the Lagrange multipliers of rigid body mechanics. The importance of DAE systems to rigid body mechanics is that the equations for constrained systems are DAE. We have seen (Section 4.2) that the equations for a natural mechanical system subject to nonholonomic constraints have the form ˙ q˙ − Tq + Vq + At λ M q¨ + M

= 0,

Aq˙

= 0

˙ Then we can write the with q ∈ R n , λ ∈ R m . Let z = (q, v, λ) with v = q. equations as first-order system F (t, z, z˙ ) = 0 with   ˙ q − v      t ˙ (6.7) F=  M v˙ + Mv − Tq + Vq + A λ    Av The Jacobian Fz˙ is the block matrix  I  0   0

0 M 0

 0  0   0

Here I and M are n × n matrices and the zero blocks in the last row have m rows. Clearly these are DAE. There is an index, more precisely the differential index, which indicates the difficulty of a given DAE system. Informally, the index is the number of times the equations must by differentiated to obtain an ODE system. Higher index systems are typically more difficult to solve. Two examples will make this clearer.


First, consider the DAE systems (6.5) and (6.6). If we differentiate the second equation we obtain g x x˙ + gλ λ˙ = 0 If gλ is nonsingular we obtain λ˙ = − gλ−1 gx x˙ which yields the ODE system x˙ − f ( x, λ) ˙λ + g−1 ( x, λ) gx ( x, λ) f ( x, λ) λ

= 0 = 0

This system is therefore has index 1. Now consider the DAE system x˙ − f ( x, λ)

= 0

g( x )

= 0

Differentiating the second equation once yields gx ( x ) x˙ = gx ( x ) f ( x, λ) = 0 Let φ( x, λ) = g x ( x ) f ( x, λ) and differentiate φ to obtain φx x˙ + φλ λ˙ = 0 If φλ is nonsingular then

λ˙ = −φλ−1 φx x˙

This leads to the DAE x˙ − f ˙λ + [ gx f λ ]−1 [ gxx f + gx f x ] f

= 0 = 0

Therefore this system has index 2. The equations for a natural system subject to a nonholonomic constraint (6.7) are of index two. The constraint equation is differentiated once to solve ˙ for λ and then λ is differentiated once to obtain the equation for λ. The numerical methods used for DAE are either backward difference methods (BDF) or implicit Runge–Kutta methods (IRK). The standard software for index 1 DAE systems is the DASSL program developed by Petzold [66] . DASSL is based on BDF methods. The standard software for index 2 or 3 DAE systems is RADAUII developed by Hairer [67]. RADAUII is based on IRK methods.

181

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6 Numerical Methods

The BDF discretization for the ODE system x˙ = f ( x, t) has the form xn =

k

∑ ai x n−i + f ( x n , tn )

i =0

The prototype BDF method is the backward Euler method. Let us work through its application to the index one DAE systems (6.5) and (6.6) The discretized version of the problem is x n − h f ( x n , λn ) g ( x n , λn )

= x n −1 = 0

This presents a system of equations, usually nonlinear, which can be solved by Newton’s method. Let x nk be the kth iterate for x n and let   x − x − h f ( x, λ ) n   F ( x, λ) =   g( x, λ) The Jacobian for the system is  I − h fx Fz =  gx



−h f λ   gλ

The Newton iteration can now be written as Fz ( xnk , λkn )δz = − F ( xnk , λkn )     k + 1 k  xn   xn  Update :   =   + δz λkn+1 λkn

Solve :

The bulk of the computational work comes from evaluating the functions and the Jacobian and solving the linear system. The art of developing good DAE software involves finding robust and efficient methods for these computations. 6.5 Wobblestone Case Study

The raison d’etre of numerical methods is to enable the study of systems which are so complicated that analytical solutions or perturbation theory approximations are unavailable. This section illustrates this activity by presenting a

6.5 Wobblestone Case Study

Fig. 6.6 Geometry of the wobblestone modeled as a semi-ellipsoid rolling without slipping on a horizontal plane Π. The vector r goes from the center of mass, c.m., to the point of contact.

case study of a complicated rigid body problem. The particular system is the wobblestone (also known as a celt or a rattleback). The behavior we wish to simulate is nicely described in the following quote by Jearl Walker [68] If you spin a certain type of stone in the “wrong” direction, it will quickly stop, rattle up and down for a few seconds and then spin in the opposite direction. Going in the “right” direction, however, it will usually spin stably. The stone is apparently biased toward one direction of spin. It will even develop a spin in that direction if you just tap one end downward. The rocking of the stone caused by the tap is quickly converted into a spin. The source of the quote also describes several of the stones many of which are shaped like a tri-axial ellipsoid sliced in half in the plane containing the two longest semi-axes. Our first step is to construct a mathematical model. The body will be the semi-ellipsoid described in the last paragraph with mass distributed such that the principal axes of inertia are skewed to the axes of the ellipsoid. We shall also assume that all the ellipsoid axes are unequal. The geometry is shown in Fig. 6.6.

183

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6 Numerical Methods

To derive the equations of motion we need to be able to relate the orientation of the body to the contact point. The normal to body surface at the contact point is −e¯ 3 . Let the normal be γ = {γ1 , γ2 , γ3 } in body fixed coordinates with origin at the center of mass and oriented along the ellipsoid axes (not the principal axes. Let the equation of the ellipsoid be

E : g( x, y, z) = x2 /a2 + y2 /b2 + (z + z¯)2 /c2 − 1 = 0 The distance from the origin to the tangent plane is Π : f ( x, y, z) = r t γ = xγ1 + yγ2 + zγ3 The contact point can be found by finding the minimum of f subject to the constraint E which is solved by Lagrange multipliers. Let L = f + λg Then ∇ L = 0 implies 2λx = − a2 γ1

2λy = −b2 γ2

2λ(z + z¯) = −c2 γ3

These, together with the constraint provide 2λ = a2 γ12 + b2 γ22 + c2 γ32 which in turn yields

{ x, y, z + z¯} =

−1 a2 γ12

+ b2 γ22

+ c2 γ32

{ a 2 γ1 , b 2 γ2 , c 2 γ3 } = φ ( γ )

(6.8)

We will model the motion wobblestone as the semi-ellipsoid rigid body rolling, without slipping, on a horizontal plane. The equations of motion are the rates of change of linear and angular momentum in the moving frame and the attendant kinematic relations m(v˙ + ω × v)

= −mgγ + F

I ω˙ + ω × I ω

= r×F

(6.9) (6.10)

where F is the constraint force and I = I(B , r¯ ). The body is subject to the nonholonomic constraint v+ω×r = 0 To determine the attitude we have the kinematic equation R˙ = RΩ

6.5 Wobblestone Case Study

and the contact point is obtained from (6.8) r = φ(γ) where γ is the third row of R (e¯ = eRt , e¯3 = ei R3i ). The constraint force is found by eliminating v from (6.9) and (6.10) F = m[r˙ × ω + r × ω˙ − ω × (ω × r ) + gγ] Therefore the torque is r × F = m[(rr t − r t rI )ω˙ + r × (r˙ × ω ) − (r t ω )r × ω + gr × γ] Substituting the expression for the torque into (6.10) gives

[I + m(r t rI − rr t )]ω˙ + ω × I ω = m[r × (r˙ × ω ) − (r t ω )(r × ω ) + gr × γ] Note that I + m(r t rI − rr t ) = I(B , r¯ 0 ) where r0 is the contact point. For the model to simulate the observed behavior it is essential to include off-diagonal terms in the inertia tensor I . The wobblestones seen in demonstrations are basically semi-ellipsoids with contrivances to produce nonzero products of inertia.

Fig. 6.7 Angular velocity component ω3 as a function of time for a wobblestone starting from rest in a nose-down position.

These model equations evolve on the Lie group SO(3) × so(3) and one may use the RK4MK method to obtain numerical solutions of the equations.

185

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6 Numerical Methods

The results of one such simulation is shown in Fig. (6.7). For this simulation the moments of inertia were calculated for a semi-ellipsoid with axes ( a, b, c) = (20, 3, 2) and density ρ = 1 gm/cm3 . All products of inertia were set to −100. The wobblestone was started at rest in a nose down position (γ = (− sin(π/30) 0 cos(π/30))t ). The body began to rock and then to acquire an angular velocity component parallel to e¯3 . Rotation reversal is also seen. The behavior was generally as described by the quote at the beginning of the section.

6.6 Exercises

6.6 Exercises

Exercise 6.1 Write a computer program to use the implicit and explicit Euler methods and the Runge–Kutta RK4 method to numerically integrate the equations of motion for a simple harmonic oscillator. Determine the order of accuracy of the methods by finding how the integration error depends on the step size. Exercise 6.2 Write out the right-handed versions of the RK3MK and SE methods. Exercise 6.3 Implement the SE method and the RK4MK methods and compare their performance when applied to the wobblestone problem. Since there is no analytical solution you will need to devise a suitable standard of reference. Exercise 6.4 Consider the following DAE [69]: x˙ + x − λ1 = 0

= 0

x − λ1 + λ2

= 0

λ1

= sin t

• Show that the system has index 1. • Find the exact solution. • Write a computer program to solve the system using the order backward difference formula (BDF) and compare the results to the exact solution. Exercise 6.5 Consult [65] for the implementation of RK4MK in the semi-direct product trivialization. Implement the method and compare the results to those obtain with RK4MK in the direct product implementation. (This exercise has a higher degree of difficulty than the others and might best be considered as a course project for advanced classes).

187


189

7

Applications This chapter shows how to apply the methods discussed in this book to a diverse collection of applied problems. The scales of the mechanical systems to be considered range from the astronomical (precession and nutation of the Earth) to the molecular (molecular dynamics). In between these extremes we will consider gravity gradient stabilization of satellites and motion of a multibody from robotics.

7.1 Precession and Nutation

This section is concerned with the rotational motion of the Earth. The axis of rotation of the Earth moves slowly on the celestial sphere as if it were on a cone centered on the pole of the ecliptic. The secular part of this motion is called precession and it has been known since the second century B.C. when the Greek astronomer Hipparchus compared his measured positions of stars with those reported a century or so earlier. One of the triumphs of Newton’s theory of gravitation was to identify, after nearly eighteen centuries, the physical mechanism which produced the precession of the equinoxes. He attributed precession to the torque produced by the gravitational attraction of the Sun and the Moon on the equatorial bulge of the Earth’s figure which he had inferred from the rotation of the Earth. His analysis of the phenomenon in the Principia claimed to deduce correct numerical value for the precession, about 50 /year. Newton’s claim was criticized by d’Alembert [70] and was considered to be fortuitous by Chandrasekar [71] and indeed rigid body mechanics did not exist at the time the Principia was written. It remained for d’Alembert to develop the beginnings of rigid body mechanics and to construct the first quantitatively correct theory [70]. In addition to the secular effect there are small oscillations, called nutation, about the cone of precession which were discovered by Bradley late in the 18th century.


190

7 Applications

This section uses Lagrangian mechanics with the Euler angle generalized coordinates to derive the dominant precession and nutation effects. A more precise description can be obtained by the use of perturbation theory based on Hamiltonian mechanics. Here we present a Hamiltonian formulation, a simplified version of the Kinoshita theory [72], which is based on Andoyer variables. The dominant precession effect is derived from the secular terms of the Hamiltonian. This provides an introduction to the full Kinoshita theory. A similar approach was taken in [31] and this section has benefitted from that material. Additional references for the material in this section are [13, 73–77]. 7.1.1 Gravitational Attraction

Fig. 7.1 Coordinates used to derive the potential for a rigid body.

The Lagrange equations and the Hamiltonian equations both require the potential function for gravitational attraction of a point mass on a rigid body of finite extent. This subsection derives the lowest order terms of the potential where the order of accuracy is measured by the ratio of the size of the Earth to the distance of the point mass. For the Earth–Moon system the small parameter is ≈ 6378/384000 ≈ 1/60. For the Earth–Sun system the small parameter is ≈ 6378/152000000 ≈ 1/23800. Let x, y, z be a principal axis coordinate system at the center of mass of the Earth with the zaxis normal to the equatorial plane. The body is acted upon by point mass, M , which moves in an orbit of radius r (t) in a plane inclined at angle to the equatorial plane (see Fig. 7.1). The coordinates of the point mass are ( x, y, z) = r (cos λ cos β, sin λ cos β, sin β). Let ρ = (ξ, η, ζ ) be the coordinates


of an infinitesimal mass dm in the body and let ∆ = r − ρ. The gravitational potential for dm is GM dm |∆| − 21

GM ξ 2 + η2 + ζ 2 xξ + yη + zζ + dm 1−2 r r2 r2 GM xξ + yη + zζ ξ 2 + η2 + ζ 2 − + 1+ r r2 2r2

=

dU

= =

4 ξ 3 2 2 2 2 2 2 ( x ξ + y η + z ζ + 2xyξη + 2xzξζ + 2yzηζ ) + O dm 4 2r r4 GM dm xξ + yη + zζ − 1−2 r r2 1 2 2 ξ (y + z2 − 2x2 ) + η 2 ( x2 + z2 − 2y2 ) + ζ 2 ( x2 + y2 − 2z2 ) 2r4

4 $ ξ +2xyξη + 2xzξζ + 2yzηζ + O 4 r

=

Integrating this expression over the body yields the potential energy U=

dU GM = M+ r GM = M+ r B

1 2 2 2 A + B + C − 3 ( Ax + By + Cz ) 2r4 1 A+B+C 2r2

− 3( A cos2 λ cos2 β + B sin2 λ cos2 β + C sin2 β)

where A, B, and C are the principal moments of inertia and β and λ are the latitude and longitude of the point mass relative to the equatorial plane, respectively. We have used the results that the chosen coordinate system imply

and

ξ dm =

η dm =

ζ dm =

ξη dm =

ξζ dm =

ηζ dm = 0

1 1 (ξ 2 + η 2 + ξ 2 + ζ 2 − η 2 − ζ 2 ) dm = ( B + C − A) 2 2 and the companion results obtained by cyclic permutation of the coordinates. The Earth is very nearly axisymmetric, ( A − C )/C ≈ 1/305. In that case the potential energy simplifies to GM C−A 2 U= (1 − 3 sin β) M+ r 2r2 ξ 2 dm =

191

192

7 Applications

As far as the rotational motion is concerned the only contributing term is the last one. The orbits of the Earth and Moon are very nearly circular. In the circular case, r (t) = a, we can use Kepler’s third law n2 a3 = G ( M + M ), where n is the mean motion 2π/period. Thus the potential function for the precession and nutation problem becomes U=

1 2 n µ(C − A) sin2 β 2

where µ = M /( M + M ). 7.1.2 Precession and Nutation via Lagrangian Mechanics

Fig. 7.2 Reference planes used in the study of precession and nutation. (Note: in the figure γ0 ≡ 0 and Ω ≡ )

The treatment here follows [75, 76]. We will assume that the Earth is a rigid body acted upon by the Moon and Sun which move in circular orbits. We want to apply the Euler–Lagrange equations and we need to express the torque acting on the Earth in terms of the Euler angles (which describe the orientation of the Earth) and the time (which specifies the positions of the Moon and the Sun). The fixed reference frame e¯ has e¯ 1 pointing to the vernal equinox of epoch, 0 , and e¯ 3 normal to the ecliptic of epoch (the modifiers of epoch mean that the position of the moving ecliptic is fixed at a reference date). The moving frame r has e1 in the Earth’s equatorial plane pointing to the Greenwich


meridian and e3 is normal to the equatorial plane. This is a principal axis frame because we have assumed symmetry of the Earth. In astronomical parlance the Euler angles are the obliquity of the ecliptic (θ), right ascension of the ascending node of the Earth’s orbit (φ − π) and Greenwich siderial time ψ. We also need a frame f associated with the orbit. We set f1 along the radius vector from the center of the Earth to the orbiter and f3 normal to the orbit plane (see Fig. 7.2). Let us abbreviate the notation for rotation matrices Rei (α) to Ri (α). Then the fixed frame is related to the body fixed frame by e = e¯ R3 (φ) R1 (θ ) R3 (ψ) where φ, θ, ψ are the Euler angles. The orbit frame is related to the fixed frame by f = e¯ R3 () R1 (i ) R3 (u) where is the longitude of the ascending node of the orbit, i is the inclination of the orbit on the ecliptic, and u the argument of longitude. To express the potential in terms of these variable we use sin β

= f1 · e3 = (fe1 ) · (ee3 ) = e1t R3t (u) R1t (i ) R3t () R3 (φ) R1 (θ ) R3 (ψ)e3 = sin(φ − ) cos u sin θ − cos(φ − ) sin u sin θ cos i + cos θ sin u sin i

The inclination of the Sun’s orbit is zero and it is small (5◦ 9 ) for the Moon. We will therefore retain only terms linear in sin i and that means we can set cos i to one. Thus we arrive at the following expression for sin β: sin β = − sin θ sin(u + − φ) + cos θ sin u sin i From this we find that the potential function, up to a constant, is U=

# 3 2 n µ( A − C ) sin2 θ [1 − cos 2(φ − u − )] 4

− sin 2θ sin i [cos( − φ) − cos(2u + − φ)] The Lagrangian for this system (see Example 3.3) is L

= =

1 ( Aω12 + Aω22 + Cω32 ) − U 2 1 1 A(θ˙ 2 + φ˙ 2 sin2 θ ) + C (ψ˙ + φ˙ cos θ )2 − U 2 2

$

193

194

7 Applications

The equations of motion are therefore d [ Aφ˙ sin2 θ + C cos θ (ψ˙ + φ˙ cos θ )] dt

= Uφ

d ( Aθ˙ ) − Aφ˙ 2 sin θ cos θ + C φ˙ sin θ (ψ˙ + φ˙ cos θ ) = Uθ dt d ˙ (φ cos θ + ψ˙ ) = 0 dt Let us denote σ = ψ˙ + φ˙ cos θ, then the last equation asserts σ is constant and since precession is a very slow phenomenon in comparison to the Earth’s daily rotation we have θ˙ φ˙ σ Thus, the first two equations are well approximated by Cσ sin θ θ˙ Cσ sin θ φ˙ Let N=

= − Uφ = Uθ

3 2 C−A n µ 2σ C

then φ˙

=

θ˙

=

1 U sin θ θ −1 Uφ sin θ

1 Uφ = N [− sin θ sin 2(u + − φ) sin θ − cos θ sin i (sin( − φ) − sin(2u + − φ))] # −1 Uθ = N − cos θ (1 − cos 2(u + − φ)) sin θ $ cos 2θ sin i − (sin( − φ) − sin(2u + − φ)) . cos θ Since θ and φ change very slowly we can treat them as constants where they appear on the right-hand sides of the equations. Furthermore, we may consider i to be a constant and u and to be linear functions of time u = nt + const.

= n t + const.

With these approximations the differential equations can be integrated immediately 1 θ = −N cos θ sin 2(u + − φ) 2 ( n + n )

1 1 cos( − φ) − sin(2u + − φ) − cos θ sin i − n 2n + n


φ

=

1 N − cos θ t − sin 2(u + − φ) 2 ( n + n )

cos 2θ sin i 1 1 − − sin( − φ) − sin(2u + − φ) sin θ n 2n + n

Numerical values for the separate effects of the Moon and the Sun can be obtained from the following physical parameters: n σ

n σ

C− A C

M M ⊕ + M

0

σ

1 27.32

1 365.25

1 304.5

1 81

23.◦ 5

rad 2π × 365.25 year

The results are

φ˙ = 35.37 year

φ˙ = 16.03

year

The combination of the two effects is the luni-solar precession φ˙ ls = 51.40 /year with a period of 25200 years. This approximate value differs from the accepted value [73] of 50.39 /year by about 2%. The periodic or nutation term having the largest amplitude, 9.15 , is the one associated with the small divisor n . This is the nodal regression rate for the Moon which has a period of about 18.6 years. Thus the largest nutation term has a period of about 18.6 years. It is interesting to note that Bradley, the first to observe this effect, followed one complete period before publishing his results. 7.1.3 A Hamiltonian Formulation

Here we formulate the problem of the Earth’s rotation as a Hamiltonian problem in terms of the Andoyer variables and express the torque in terms of these variables and time. We assume as before the point mass moves with constant velocity on a circular orbit and at inclination i to the ecliptic. The inclination will be assumed to be small enough that terms of order sin2 i can be neglected. As noted before, for the Moon i ∼ 5◦ . It happens that, for the Earth, the angle λ between the angular momentum vector and the pole of the equator is very small, ∼ 10−6 [72]. Thus we shall further assume that λ is negligible and this has the following consequences: η=θ

L=G

ψ= l+g

φ=h

The reference frames will be defined as in the previous section. Now, sin β

= f1 · e3 = e1t R3t (u) R1t (i ) R3t () R3 (φ) R3 (h) R1 (η ) R3 (l + g)e3 = sin(h − ) cos u sin η − cos(h − ) cos i sin u sin η + sin i sin u cos η ≈ sin η sin(h − u − ) + sin i sin u cos η

195

196

7 Applications

where the final approximation obtains when the inclination i is small. From this we find that the potential function, up to a constant and terms linear in i, is 1 3 2 M 1 n sin2 η − sin 2η sin i cos( − h) U = ( A − C) 4 M⊕ + M 2 4 1 1 2 + sin 2η sin i cos(2u + − h) − sin η cos(2u + 2 − 2h) 4 2 Therefore, we obtain the Hamiltonian accounting for a single orbiter as K = K0 + K1 with

G2 K1 = U 2C The approximation L = G has been used to obtain K0 . In [72] the Hori Lie series perturbation method [78] is applied to obtain a first order accurate approximate solution to this problem. The Hamiltonian, as formulated, depends explicitly on time t via the longitudes of the Moon and the Sun. Explicit time dependence would rule out the application of the Hori method but it is possible to remove the time dependence by working in extended phase space (l, g, h, k, L, G, H, K ) by using the modified Hamiltonian K0 =

( = K0 ( G ) + U ( G, H, h, k, u, ) + U ( G, H, h, k, u ) + K K The differential equations for k and K are ( dk dK ∂K =1 =− dt dt ∂k The perturbation theory provides a symplectic transformation q, p → q∗ , p ∗ such that the transformed equations eliminate the short period variables and are integrable. We will not pursue this in any more detail since we have not provided the necessary background on the perturbation theory. The complete details will be found in [72, 78]. We will be content to obtain the precession effect from the secular, or averaged, Hamiltonian. At order one, the short period terms are those containing u or u . Eliminating the periodic terms we obtain the secular Hamiltonian as

$ 3 H2 # 2 n µ + n2 K¯ 1 = ( A − C ) 1 − 2 8 G The precession of the equinoxes is obtained from the solution of the order ¯ H, ¯ G, ¯ l¯ are the constants and two system in which L, $ 3 H2 # g¯ = ( A − C ) 3 n2 µ + n2 t 4 G $ 3 H # h¯ = − ( A − C ) 2 n2 µ + n2 t 4 G

7.2 Gravity Gradient Stabilization of Satellites

Making the substitution G = Cσ we obtain g¯ =

1 cos2 η ( N + N ) t 2

1 h¯ = − cos η ( N + N ) t 2 7.2 Gravity Gradient Stabilization of Satellites

The success of an artificial Earth’s satellite mission often depends on controlling the attitude of the platform so that antennas or sensors are in proper positions. This section is concerned with one method that is used to stabilize artificial Earth’s satellite attitudes. The method is called the gravity gradient stabilization because it exploits the radial variation of the Earth’s gravity. This application involves the yaw-pitch-roll angles and a linear stability analysis. For an alternative treatment which emphasizes the technology of the problem see [79].

Fig. 7.3 Reference frames used to analyze gravity gradient stabilization of a satellite.

197

198

7 Applications

7.2.1 Kinematics in Terms of Yaw-Pitch-Roll

We assume the satellite orbit to be circular of radius a. There is a local reference frame f¯ such that f¯3 points vertically downward to the center of the Earth, f¯1 is tangent to the orbit and f¯2 is normal to the orbit plane (see Fig. 7.3). Following Section 1.1.6, rotations of the satellite can be specified by composing rotations about f¯3 by angle ψ, about f2 by angle θ and about f1 by angle φ. The angles ψ, θ, φ are the yaw, pitch, and roll angles, respectively. Thus the rotation matrix R is obtained from ¯ e (ψ) Re (θ ) Re (φ) = fR ¯ (7.1) f = fR 3 2 1 and we find R(ψ, θ, φ) = (7.2)   cos ψ cos θ − cos φ sin ψ + sin φ sin θ cos ψ sin φ sin ψ + cos φ sin θ cos ψ      sin ψ cos θ cos φ cos ψ + sin φ sin θ sin ψ − sin φ cos ψ + cos φ sin θ sin ψ    − sin θ cos θ sin φ cos θ cos φ

The angular velocity relative ω f to f is obtained in terms of yaw-pitch-roll from the derivative of (7.1) as in Section 2.2 ˙ te (ψ) ˙ te (ψ) Rte (θ ) ˙ e1 ] ee3 Re2 (θ ) Re1 (ψ) + θR ee2 Re1 (ψ) + φ f˙ = f[ψR 2 1 1 from which it follows that ˙ t (ψ)e2 + φe ˙ + φf ˙ 1t (ψ) R2t (θ )e3 + θR ˙ 1 ] = ψ˙ f¯3 + θf ˙ 1 ω f = fω f = f[ψR 2 1 The matrix relating the angular velocity components and the angle derivatives is    ˙ − sin θ 0 1   ψ      ωf =  cos φ 0  cos θ sin φ   θ˙     cos θ cos φ − sin φ 0 φ˙ 7.2.2 Rotation Equations for an Asymmetric Satellite

To write the Euler equations for satellite rotation we need to obtain the torque exerted by Earth’s gravity (assumed to be a point mass) and relate the angular velocity to derivatives of the yaw-pitch-roll variables. The force acting on an


infinitesimal element of mass dm is, to order ξ 4 /r4 ,

= ∇ρ dU GM = [ x − ξ, y − η, z − ζ ]dm r3 3GM + 5 [ x2 ξ + xyη + xzζ, y2 η + xyξ + yzζ, z2 ζ + xzξ + yzη ]dm r

dF

which produces the infinitesimal torque dτ

= ρ × dF GM dm 3GMdm = ρ×r+ [ xzηξ − xyζξ + yz(η 2 − ζ 2 ) + (z2 − y2 )ζη r3 r5 xyηζ − yzξη + xz(ζ 2 − ξ 2 ) + ( x2 − z2 )ζξ yzξζ − xzηζ + xy(ξ 2 − η 2 ) + (y2 − x2 )ξη ]

The total torque is found by integrating the infinitesimal torque over the body τ=

B

dτ =

3GM [(C − B)yz, ( A − C ) xz, ( B − A) xy] r5

(7.3)

When the orbit is circular, r (t) = a this simplifies to τ=

3n2 [(C − B)yz, ( A − C ) xz, ( B − A) xy]t a2

where n is the mean motion and Kepler’s third law n2 a3 = GM has been used. Using the rotation matrix (7.2) we express the coordinates in terms of the yaw-pitch-roll angles

( x, y, z)t = ae¯3 = − a(sin θ, cos θ sin φ, cos θ cos φ)t Therefore, the torque is expressed in terms of the yaw-pitch-roll angles τ=

3n2 [(C − B) cos2 θ sin 2φ, (C − A) sin 2θ cos φ, ( A − B) sin 2θ sin φ]t 2

Now let e¯ be a fixed frame such that e¯ 1 and e¯ 2 lie in the plane of the orbit and e¯ 3 is normal to it. Let Ro be the rotation matrix relating f¯ and e¯ , f¯ = e¯ Ro ,    − sin u 0 − cos u   Ro =  sin u  − cos u 0    0 1 0 To find the angular velocity of the frame f¯ we use f¯˙ = e¯ ( R˙ o R + Ro R˙ ) = f( Rt Rto R˙ o R + Rt R˙ ) = f(−nRt eˆ2 R + Ωr ) = f(Ω f + Ωr )

199

200

7 Applications

From this we obtain the angular velocity components of the body fixed, principal axis frame   ˙ ˙ −n sin ψ cos θ − ψ sin θ + φ      ˙ ˙ ω=  −n(cos ψ cos φ + sin ψ sin θ sin φ) + ψ cos θ sin φ + θ cos φ  (7.4)   −n(− cos ψ sin φ + sin ψ sin θ cos φ) + ψ˙ cos θ cos φ − θ˙ sin φ The Euler equations for this system are Aω˙ 1 + (C − B)ω2 ω3

=

Bω˙ 2 + ( A − C )ω3 ω3

=

C ω˙ 3 + ( B − A)ω1 ω2

=

3 2 n (C − B) cos2 θ sin 2φ 2 3 2 n (C − A) sin 2θ cos φ 2 3 2 n ( A − B) sin 2θ sin φ 2

Substituting (7.4) into these equation we obtain three second order ODE in the yaw-pitch-roll angles A

d ˙ [φ − ψ˙ sin θ − n sin ψ cos θ ] dt + (C − B)[θ˙ cos φ + ψ˙ cos θ sin φ − n(cos ψ cos φ + sin ψ sin θ sin φ)] × [−θ˙ sin φ + ψ˙ cos θ cos φ − n(− cos ψ sin φ + sin ψ sin θ cos φ)]

= (C − B) cos2 θ sin 2φ B

d ˙ [θ cos φ + ψ˙ cos θ sin φ − n(− cos ψ cos φ + sin ψ sin θ sin φ)] dt + ( A − C )[φ˙ − ψ˙ sin θ − n sin ψ cos θ ] × [−θ˙ sin φ + ψ˙ cos θ cos φ − n(− cos ψ sin φ + sin ψ sin θ cos φ)]

= (C − A) sin 2θ cos φ C

d [−θ˙ sin φ + ψ˙ cos θ cos φ − n(− cos ψ sin φ + sin ψ sin θ cos φ)] dt + ( B − A)[θ˙ cos φ + ψ˙ cos θ sin φ − n(cos ψ cos φ + sin ψ sin θ sin φ)] × [φ˙ − ψ˙ sin θ − n sin ψ cos θ ]

= ( A − B) sin 2θ sin φ 7.2.3 Linear Stability Analysis

We now wish to investigate the stability of the equilibrium state ψ = θ = φ = 0. The goal of the analysis is to find the region in ( A, B, C )-space for which the


equilibrium is stable. The equations are linearized about the equilibrium state using the linearized angular velocity components derived from (7.4)   ˙φ − nψ      ˙ (= ω  −n + θ    ψ˙ + nφ Substituting these into the Euler equation and retaining only the first-order terms we obtain A

d ˙ (φ − nψ) − n(C − B)(nφ + ψ˙ ) = 3n2 (C − B)φ dt Bθ¨ = 3n2 (C − A)θ

d ˙ (ψ + nφ) + n( B − A)(nψ − φ˙ ) = 0. dt Note that θ is decoupled from ψ, φ and that ψ, φ are coupled. In other words, pitch (in-plane) rotation is decoupled from yaw and roll (out-of-plane) rotation but yaw and roll are coupled. The characteristic equation is the polynomial in λ which must be satisfied if eλ is to be a solution of a linear system. The characteristic polynomial for the gravity gradient stabilization is   2 + 4n2 α λ 0 − n ( 1 − α ) λ     =0  2 2 det  0 λ + 3n β 0    2 2 n (1 − γ ) λ 0 λ +n γ C

where

B−C A−C B− A β= γ= A B C Expanding the determinant we obtain the sixth-order polynomial α=

(λ2 + 3n2 β)[λ4 + λ2 n2 (1 + 3α + αγ) + 4n4 αγ] = 0 For stability it is necessary that all roots of the characteristic equation are purely imaginary. That is, λ2 must be negative real. For λ2 to be negative we see from the first factor of the polynomial that necessarily β≥0 The roots of the second factor are 1 2 2 2 λ = n −(1 + 3α + αγ) ± (1 + 3α + αγ) − 16αγ 2

201

202

7 Applications

From this we see that λ2 can be real only if

(1 + 3α + αγ)2 ≥ 16αγ and can be negative only if 1 + 3α + αγ ≥ 0 αγ ≥ 0

Fig. 7.4 (a) Parametric curve √ for α. (b) Parametric curve for γ with vertical asymptotes at η = 2 ± 3, horizontal asymptote at γ = −3. (c) The curve of γ ( α), Eq. (7.5), with vertical asymptotes at α = 0, 1 and a horizontal asymptote at γ = −3. (d) The regions of stability √ √ (shaded). The curve exits the unit square at γ = 2 − 6 and α = −( 3/2 − 1)2 and is tangent to the α-axis at (−1/3, 0).

The first condition requires A ≥ C. To analyze the remaining three conditions we recall the triangle inequalities (2.14) A ≤ B + C, B ≤ A + C, C ≤ A + B These imply that

−1 ≤ α ≤ 1

−1 ≤ β ≤ 1

−1 ≤ γ ≤ 1

7.3 Motion of a Multibody: A Robot Arm

Then we have the following implications: • αγ ≥ 0 ⇒ α ≥ 0 and γ ≥ 0, or α ≤ 0 and γ ≤ 0 • A ≥ C and B ≤ A + C ⇒ α − γ =

( A −C)( A +C− B ) AC

≥ 0 ⇒ α ≥ γ.

Finally, we must analyze the curve

√ 1 + 3α + αγ = 4 αγ

(7.5)

which separates the stable and unstable regions. This curve can be simplified by the change of variables (α, γ) → (ξ, η )

√

ξ=α

η=

α=ξ

γ = η 2 /ξ

αγ

The curve assumes the form 1 + 3ξ + η 2 = 4η from which ξ is obtained as a function of η ξ=

−1 2 (η − 4η + 1) 3

which is the parabola in Fig. 7.4(a). This in turn allows the following parametric representation of the original curve: 1 α = − (η 2 − 4η + 1) 3

γ=

η2

−3η 2 − 4η + 1

The second of these curves is shown in Fig. 7.4(b). Given α(η ) and γ(η ) we can plot γ(α) as shown in Fig. 7.4(c). Focussing on the square region −1 ≤ α, γ ≤ 1 we find the region of stability as shown in Fig. 7.4(d). The analysis of the curve is illustrated in Fig. 7.4. The region of stability is shown in Fig. 7.4(d).


Technological applications of rigid body theory often involve multiple rigid bodies which are interconnected by various kinds of joints. This section considers the techniques which enable the application of rigid body mechanics to these systems. One can mention models of the human torso such as the much studied crash dummy, robots, automobiles, and space stations as a diverse set of examples. The general multibody problem can be quite complex because

203

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7 Applications

Fig. 7.5 A plane chain of two links specified by angles θ1 , θ2 and center of mass coordinates ( x1 , y1 ) and ( x2 , y2 ).

of the combinatorially large number of configurations which must be considered. In this section we will examine a simple mechanism to illustrate (1) a general approach and (2) a special technique which is much more efficient (and of course much less general). The special technique is nicely phrased in the language of Lie groups. The simple mechanism is a two-link chain of rods with one end anchored, the other end free, and one joint which we will take to be a planar, rotational joint. Let the length, mass, and inertia tensor relative to the center of mass of link i be Li , mi and Ii , respectively, with i = 1, 2 and let the links and joints be labeled serially from the anchor point. Let the distance from joint i to the center of mass of link i be denoted by ai and let bi = Li − ai . Clearly, these conventions could also describe a chain of an arbitrary number of links. The general technique we wish to explore involves deriving the Lagrangian in terms of orientation angles θi and center of mass coordinates ( x i , y i ) (Fig. 7.5). We will use the center of mass as the reference point for each link. Thus we obtain the Lagrangian L(θ1 , θ2 , x˙1 , y˙1 , θ˙1 , x˙2 , y˙2 , θ˙2 )

=

1 1 m ( x˙ 2 + y˙ 21 ) + I1 θ˙12 + m2 ( x˙ 22 + y˙ 22 ) + I2 θ˙22 2 1 1 2 −m1 ga1 sin θ1 − m2 ga2 sin(θ1 + θ2 )

To complete the problem specification we need to impose constraints which produce a chain. These constraints are holonomic and are four in number x1 − a1 cos θ1 = 0 y1 − a1 sin θ1 = 0 x1 + b1 cos θ1 − x2 + a2 cos(θ1 + θ2 ) = 0 y1 + b1 sin θ1 − y2 + a2 sin(θ1 + θ2 ) = 0


The first pair of constraints fixes node 1 at the origin and the second connects the links at the joint. In this case the constraints can be eliminated by first expressing x1 , y1 in terms of θ1 using the first two constraints, then x2 , y2 in terms of θ1 , and θ2 using the last two. The result is the reduced Lagrangian L ( θ1 , θ2 )

=

1 1 (I1 + m1 a21 + m2 L21 )θ˙12 + (I2 + m2 a22 )(θ˙ 1 + θ˙2 )2 2 2 +m2 a2 L1 cos θ2 θ˙1 (θ˙1 + θ˙2 ) (7.6)

− g(m1 a1 + m2 L1 ) sin θ1 − gm2 a2 sin(θ1 + θ2 ) Now the equations of motion are d L ˙ − Lθi = 0 dt θi

i = 1, 2

and have the form M θ¨ + f (θ ) = 0 where θ = (θ1 , θ2 )t and M is the mass matrix  

I1 + m1 a21 + m2 L21 + I2 + m2 a22 + m2 a2 L1 cos θ2 1 (I 2 2

+ m2 a22 + m2 a2 L1 cos θ2 )

1 (I 2 2

+ m2 a22 + m2 a2 L1 cos θ2 ) I2 + m2 a22

 

This procedure obviously generalizes to a planar chain having an arbitrary number of links. If the chain becomes three-dimensional the constraints become more complicated and that leads us to the geometrical, Lie group method. In the Lie group method, the configuration of each link is represented as an element of the special Euclidian group SE(3). Recall that elements of SE(3) can be thought of either as pairs ( A, a) with A ∈ SO(3) and a ∈ R3 , in which case the group operation is ( A, a) ◦ ( B, b) = ( AB, Ab + a) and inverse ( A, a)−1 = ( At , − At a), or as 4 × 4 matrices of the form   a A     A ∈ SO(3) a ∈ R3 0 1 In the matrix representation the group operations are the matrix multiply and matrix inverse. SE(3) acts on R3 according to ( A, a)v = Av + a, with v ∈ R3 or by the use of homogeneous coordinates      a A   v   Av + a    =   1 1 0 1

205

206

7 Applications

The Lie algebra of SE(3) is se(3) which is represented by 4 × 4 matrices of the form    u v    0 0

u, v ∈ R3

and se(3) is isomorphic as a Lie algebra to R6 under the correspondence     u v   u   ↔  v 0 0 where the Lie group operation in R6 is       a × u a u          ,   =  a×v+b×u b v If s = (ut , v t )t ∈ R6 then s will denote the 4 × 4 matrix    u v    0 0 We will restrict our attention to single degree of freedom joints, although the formalism can be developed without this restriction. For example, a joint might act as a hinge (called a cylindrical joint) or as a one-dimensional translation (as in a sliding radio antenna joint). Both types of motion are described by what is called a screw motion. The simplest instance is screw motion about an axis λn through the origin where λ ∈ R and n ∈ R3 with nt n = 1. The motion consists a rotation about n, say by angle θ, and a translation along n by distance pθ/2π, where p is called the pitch. This motion is the action of the group element    Rn ( θ )  0

θ p 2π n   1

To obtain the screw motion about an axis λn + x through a point x, again with λ ∈ R and n ∈ R3 with nt n = 1, one translates by − x to the origin, performs the motion at the origin, and then translates by x. In other words, the motion is the action of the group element     θ x ( θ ) p n − x R I I 2π  n        0 1 0 1 0 1




which is

 Rn ( θ )  0

[ I − Rn (θ )] x +

θ p 2π n

  

(7.7)

1

These matrices can be obtained as exponentials of elements of se(3). Recall from Section C.4 that for u, v ∈ R3     1 t  u v   exp(u) u [ I − exp(u)](n × v) + nn v  exp  =  0 0 0 1 Thus, to obtain the matrix (7.7) we need to choose u and v to satisfy the conditions 1. u = θn θ 2. nt v = p 2π

3.

1 θn×v

= x.

From the last two items it follows that p n+x×n v= 2π Now we can return to the description of the configuration of the chain. We start by specifying an initial configuration of the chain. For example, in the case of two links we may start from the links extended in a line. Let ξ 0 be a point in link i in the initial configuration. Let the joint parameter, hinge angle or translation distance for example, be denoted by qi . Then the point ξ 0 is moved to s1 ) M2 exp(q2 s2 ) · · · Mi exp(qi si ) ξ 0 ξ = M1 exp(q1 where Mi specifies the position of link i relative to link i − 1 in the initial configuration and s j ∈ R6 are the parameters of the screw motion unique to joint j. In the two link example consistent choices for the Mi are     I I 0 L e 1 1     M1 =   , M2 =   0 1 0 1 The joint parameter is the rotation angle θi and the screw parameters are   e 0   3 si =   i = 1, 2 0 0

207

208

7 Applications

Introduce the notation s1 ) M2 exp(q2 s2 ) · · · Mi exp(qi si ) Ti = M1 exp(q1

si ) Di = Mi exp(qi

Then we have the recursion relation Ti+1 = Ti Di+1 The velocity of link i in its body fixed frame is vi = Ti−1 T˙ i Note that this is a generalization of the angular velocity from so(3) to se(3). For T1 we have v1 = q˙ 1 s1 The velocity satisfies the recursion relation vi

= =

si ) + q˙ i Ti−1 Mi exp(qi si ) si ] Mi−1 exp(−qi si ) Ti−1 [ T˙ i−1 Mi exp(qi 1 i−1 + q˙ i si Ad− D v i

With this relation one can start with the velocity of link 1 and generate the velocity of each link in the chain recursively. This then enables the writing of the Lagrangian. This can be illustrated with the two-link example. Here the velocity of the stationary base is v0 = 0. For the first link    e3  s1 v1 = θ˙1 v1 = θ˙1 s1 = θ˙1   0 For the second link

1 ˙ v2 = Ad− D2 v1 + θ2 s2

1 and we need to calculate Ad − D2 . If W ∈ SE (3) and X ∈ se(3)

  R W= 0 then

 a   1

  ω X= 0



  AdW X = 

 v   0

t RωR 0

ta + − RωR 0

Rv  


t and To translate this to R6 we use the fact that 1 Rω = RωR a Rω −1 Rωa = −( Rω ) × a = a × ( Rω ) = Therefore, AdW acting on R6 is  AdW

 R = aR

and we calculate the inverse to be −1 AdW

 0   R



 0   = . a Rt − Rt Rt

Now we can evaluate the expression for v2      ˙ 0 e3   R3 (−θ2 )   θ 1 e3  v2 = θ˙2   +    e1 R3 (−θ2 ) 0 − L1 R3 (−θ2 ) 0     ˙ e θ e 1 3  3   = θ˙2   +   0 L1 R3 (−θ2 )e2   0       0      ˙   θ1 + θ˙2   =     L1 θ˙1 sin θ2       ˙   L1 θ1 cos θ2    0 Now we may calculate the kinetic energy using K= where

1 t 1 v J1 v1 + v2t J2 v2 2 1 2





2 e2 Ii − m i ai 1

Ji = 

e1 −m i ai

e1  mi ai  m1 I

i = 1, 2

209

210

7 Applications

The result is the same as (7.6) but the method used has greater generality. Any three-dimensional configuration can be analyzed once the screw parameters si have been specified. The general theory of Lie groups applied to robotics is the subject of [80]. The two-link case is presented in detail in [81]. A review of the use of Lie groups in the Lagrangian formulation is the subject of [82].

7.4 Molecular Dynamics

Fig. 7.6 (a) A small chain molecule illustrating the bond length s, bond angle θ, and dihedral angle φ. (b) Form of the Lennard–Jones potential.

Ensembles of molecules can be modeled as interacting (perhaps colliding) rigid bodies or multibodies. This section examines the use of rigid body mechanics in this context. The description of molecular dynamics given below follows [83, 84]. Molecular dynamics models the behavior of gases, liquids, and solids on spatial scales of 10–100 Å (10−9–10−8 m). The intermolecular spacing must be resolved. The time resolution must be able to resolve intermolecular collision rates and possible oscillations of bond lengths or angles and can attain trajectory lengths of about ns (10−9 s). Bench mark calculations have been carried out for systems as large as 1000 Å or for times as long as a microsecond. The molecules are modeled as a collection of rigid bodies which interact through electrostatic and molecular forces and collisions and the molecules may be multibodies which experience intramolecular forces. The bodies may be as simple as hard spheres (the only interaction being collisions) or complicated multibodies representing organic molecules. The equations of motion for translational and rotational motion (if nonspherical) are numerically integrated and statistical measures of the state of the material are calculated.

7.4 Molecular Dynamics

Typical time steps are 0.5–1 fs (10−15 s). Some of the measures of interest are equations of state, phase diagrams and internal structure. An example of a small molecule is shown on the left in Fig. 7.6 illustrating the bond length s, bond angle θ, and dihedral angle φ. The potentials for these bonds can be modeled as quadratics Vl = Kl (r − r0 )2 Vb = Kb (θ − θ0 )2 Vd = Kd (φ − φ0 )2 Water, for example, consists of one oxygen and two hydrogen atoms; the length of the H–O bond length is about 1 Å and the H–O–H bond angle is about 109◦ . In addition to these bond potentials one might include electrostatic interaction among the atoms or clusters of atoms making up the molecule. The intermolecular interactions are often modeled by the Lennard-Jones potential shown on the right in Fig. 7.6. The expression for this potential is σ 12 σ 6 − V (r ) = r r where is a parameter related to the depth of the well and σ is a parameter related to the molecule diameter or scattering parameter. Typical equations which are integrated are Newton’s equations for translational motion and Euler’s equations for rotational motion. d i x dt d p dt i d Iω dt i i

=

1 p mi i

= ∇ x i U ( x 1 , · · · , x N , q1 , · · · q N ) = − ω i × I i ω i + ∇ θ i U ( x 1 , · · · , x N , q1 , · · · q N )

where x i , p i , ωi are the position, linear momentum, and angular velocity of molecule i, respectively. The Euler angles of molecule i are represented by θi and the orientation is parameterized by quaternions so that an additional differential equation is needed which is obtained from (2.13) d q = Q (qi ) ω i dt i and θi is obtained from qi . The indices run over the number of particles N. At the dawn of molecular dynamics computer technology would support N = O(102 ) but now benchmark calculations have been performed for N = 109.

211

212

7 Applications

A popular method for integrating the linear momentum equation has been the so-called velocity Verlet method. In this method the position and velocity are advanced one time step h according to 1 p(t + h) 2

=

q( t + h )

=

p(t + h)

=

1 p(t) + h ∇U (q(t)) 2 h 1 q( t ) + p ( t + h ) m 2 1 1 p(t + h ) + h ∇U (q(t + h)) 2 2

This is a symplectic method [85]. If the bond constants are large there can be very high frequency oscillations in the bond length or bond angle. If the goal of the calculation is better served by a longer trajectory (better statistics) than resolution of the oscillations then the potential description of the bond can be replaced with a holonomic constraint. In that case, the velocity Verlet method is modified. The following modification is called the Rattle method [85]: 1 p(t + h) 2

=

q( t + h )

=

f (q(t + h))

=

p(t + h)

=

A(q(t + h))q˙ (t + h)

=

1 p(t) + h (∇U (q(t)) + At (q(t))λ(t)) 2 h 1 q( t ) + p ( t + h ) m 2 0 1 1 p(t + h) + h (∇U (q(t + h) + A(q(t + h))µ(t))) 2 2 0

where the holonomic constraints are f (q) = 0, A = ∇q f ≡ f q and λ and µ are Lagrange multipliers used to satisfy the holonomic constraint and the differential form of the constraint Aq˙ = 0. The first three lines of the method specify p(t + 12 h), q(t + h), and λ(t) and require an iterative solution. The last two lines specify p(t + h) and µ(t) and are also solved iteratively.


213

Appendix A

Spherical Trigonometry This brief account of spherical trigonometry is included to make the treatment of composition of rotations self-contained. Classical treatments of the subject, such as [86], are primarily analytical and can be a featureless collection of formulas. There is, however a geometric point of view [87, 88] which unifies and motivates the collection of formulas which define the subject. The account given here follows [87, 88]. We consider a spherical triangle T defined by unit vectors a, b, and c (Fig. A.1). The lengths of the sides of the triangle, α, β and γ, are the angles between these unit vectors (assumed ≤ π) and the vertex angles, A, B and C, are the dihedral angles between the planes formed by pairs of the unit vectors. Associated with this spherical triangle is its polar triangle T , whose vertices are the poles of the great circles completing the sides of T . This relation is fully reciprocal – T is the polar of T .

Fig. A.1 (a) A triad of unit vectors and defining T and the reciprocal set defining T . (b) Spherical triangle T and its polar T on the unit sphere.


214

A Spherical Trigonometry

The geometrical relations have the following vector expressions: a · b = cos γ a · b = − cos C a × b = sin γ c

a × b = sin C c

b · c = cos α

c · a = cos β

b · c = − cos A b × c = sin α a

b × c = sin A a

(A.1)

c · a = − cos B c × a = sin β b

c × a = sin B b

(A.2) (A.3) (A.4)

The minus signs appear in Eqs. (A.2) because the dihedral angle is the supplement of the angle between the normals to the associated planes. Note that here, and in the formulas below, once a relation is obtained others follow by permuting the symbols. The first series of formulas, the so-called law of sines for spherical triangles, follows from the vector identity a×b·c = b×c·a = c×a·b

(A.5)

so that (A.3) and (A.4) yield a × b · c = sin γ c · c = sin β b · b = sin α a · a

(A.6)

a × b · c = sin C c · c = sin B b · b = sin A a · a

(A.7)

from which the law of sines follows sin γ/ sin C = sin β/ sin B = sin α/ sin A.

(A.8)

The second series of formulas, the so-called law of cosines for spherical triangles, follows from the vector identity

( a × b) · (c × d) = ( a · c)(b · d) − ( a · d)(b · c)

(A.9)

so that (A.3) and (A.4) yield cos α = cos β cos γ + sin β sin γ cos A

(A.10)

cos β = cos γ cos α + sin γ sin α cos B

(A.11)

cos γ = cos α cos β + sin α sin β cos C

(A.12)

cos A = − cos B cos C + sin B sin C cos α

(A.13)

cos B = − cos C cos A + sin C sin A cos β

(A.14)

cos C = − cos A cos B + sin A sin B cos γ

(A.15)

The final series of formulas comes from the following pattern of substitution [86] starting from (A.10): sin β sin γ cos A

= cos α − cos β cos γ = cos α − cos β(cos α cos β + sin α sin β cos C ) = cos α sin2 β − cos β sin α sin β cos C


The complete series is sin γ cos A = cos α sin β − sin α cos β cos C

(A.16)

sin γ cos B = cos β sin α − sin β cos α cos C

(A.17)

sin β cos C = cos γ sin α − sin γ cos α cos B

(A.18)

sin β cos A = cos α sin γ − sin α cos γ cos B

(A.19)

sin α cos B = cos β sin γ − sin β cos γ cos A

(A.20)

sin α cos C = cos γ sin β − sin γ cos β cos A

(A.21)

sin C cos α = cos A sin B + sin A cos B cos γ

(A.22)

sin C cos β = cos B sin A + sin B cos A cos γ

(A.23)

sin B cos γ = cos C sin A + sin C cos A cos β

(A.24)

sin B cos α = cos A sin C + sin A cos C cos β

(A.25)

sin A cos β = cos B sin C + sin B cos C cos α

(A.26)

sin A cos γ = cos C sin B + sin C cos B cos α

(A.27)

The cosine laws (A.10–A.15) are tied together nicely in [88] using matrix notation. First, we obtain (A.10–A.12). Let

{v1 , v2 , v3 } = {a, b, c} Rescale the polar vectors to obtain the reciprocal basis as & % a sin α b sin β c sin γ , , {w1 , w2 , w3 } = a×b·c a×b·c a×b·c From (A.7) the reciprocal relations are vi · w j = δij Let matrices V and W consist of columns of these vectors represented in the standard basis V = [v1 v2 v3 ] W = [ w1 w2 w3 ] then the reciprocal relation can be expressed WtV = VtW = I Let

{θ12 , θ13 , θ23 } = {α, β, γ} {φ1 , φ2 , φ3 } = { A, B, C } and let c12 = cos θ12

s12 = sin θ12

etc.

215

216


Now consider the matrices V t V and W t W.   1 c12  VtV =   c12 1  c13 c23

 c13   c23    1

and W t W = (V t V )−1 because the reciprocal relations yield W t WV t V = V −1 (V t )−1 V t V = I Thus  1 WW= det(V t V ) t

s223

   c c −c  13 23 12  c12 c23 − c13

c13 c23 − c12 s213 c12 c13 − c23



c12 c23 − c13   c12 c13 − c23    s212

It follows that cos φ3 (A.12), for example, is cos φ3 = −

w1 · w2 c c − c12 (W t W )12 , = −, = − 13 23 w1 w2 s23 s13 (W t W )11 (W t W )22

or c12 = c13 c23 + s23 s13 cos φ3 That is cos γ = cos α cos β + sin α sin β cos C The relations for cos φ1 and cos φ2 follow in the same way from the 2 3 and 1 3 entries. To obtain (A.13–A.15) the roles of v1 and wi are reversed. Let

{w1 , w2 , w3 } = {a , b , c } Rescale the vertex vectors to obtain the reciprocal basis % & a sin α b sin β c sin γ {v1 , v2 , v3 } = , , a×b·c a×b·c a×b·c The reciprocal relations are again WtV = VtW = I from which again

W t W = ( V t V ) −1


Now

 1

  WW=  − c3  − c2

− c3

t

and

1

− c1

 1 VV= det(W t W ) t

s21

   c +c c 1 2  3  c2 + c1 c3

 − c2   − c1    1

c3 + c1 c2 s22 c1 + c2 c3

 c2 + c1 c3   c1 + c2 c3    2 s3

It follows that cos θ12 (A.15), for example, is cos θ12 =

(V t V )12 v1 · v2 c + c1 c2 , = , = 3 t t v1 v2 s1 s2 (V V )11 (V V )22

or c3 = −c1 c2 + s1 s2 cos θ12 That is cos C = − cos A cos B + sin A sin B cos γ

217


218

Appendix B

Elliptic Functions This appendix presents a miscellany of the elliptic function theory which is useful for the treatment of integrable cases of rigid body motion.

B.1 Elliptic Functions Via the Simple Pendulum

Elliptic functions can be introduced in many ways. Here we shall use the mechanical problem of the motion of the simple pendulum. Standard references are [89, 90]. See [91] for a current, concise and eclectic view of the subject. The Lagrangian for the pendulum is 1 2 ˙2 m θ − mg(1 − cos θ ) 2 where is the length of the pendulum, m its mass, g the acceleration of gravity, and θ is the angle of departure from the vertical. The equation of motion is L=

2 θ¨ = − g sin θ The energy is an integral of motion 1 ˙2 1 θ − ω 2 cos θ = θ˙02 − ω 2 cos θ0 2 2 2 where ω = g/. It is convenient to cast the energy integral in terms of the half-angle 1 ˙2 1 θ = c2 − ω 2 sin2 θ 4 2 1 ˙2 2 1 2 2 where c = 4 θ0 + ω sin 2 θ0

Libration

If c2 < ω 2 let k2 =

c2 ω2

0 ≤ k2 < 1



1 ˙2 2 2 2 1 θ = ω k − sin θ . 4 2

Then

To insure positivity of the right-hand side it is necessary that sin φ ≤ k. This shows that the oscillations are of limited amplitude or, in other words, that the pendulum librates. Introduce a new variable ψ by means of 1 sin θ = k sin ψ 2 then it follows that ψ˙ 2 = ω 2 (1 − k2 sin2 ψ) or ω (t + τ ) =

sin−1

sin 12 θ k

0

dψ

(B.1)

1 − k2 sin2 ψ

This brings us to the elliptic function theory. The incomplete elliptic integral of the first kind is F ( x, k) =

x 0

,

dx

k2 ≤ 1

1 − k2 sin2 x

where x and k are referred to as the argument and modulus, respectively. This is the direct integral, that is, given x and k, F ( x, k) is produced. The integral can be inverted to obtain a function am am u = F −1 (u, k) Inversion of integrals is familiar in the elementary case of the sin function sin−1 x =

x 0

√

dx 1 − x2

x=

sin x 0

√

dx 1 − x2

The analogous relations for am are am−1 u =

u 0

,

dx 1 − k2 sin2 x

u=

am u 0

One can now introduce the Jacobi elliptic functions sn(u, k) = sin[am(u, k)] cn(u, k) = cos[am(u, k)] dn(u, k) = 1 − k2 sn2 (u, k)

,

dx 1 − k2 sin2 x

(B.2)

219

220

B Elliptic Functions

The argument k will be omitted when there is no chance of ambiguity. There is the obvious identity sn2 + cn2 = 1 We also record the inverse relations sn

−1

−1

u = am

sin

−1

u

cn

−1

√

u = am

−1

cos

−1

u

dn

−1

u = sn

−1

1 − u2 k

The complete elliptic integral of the first kind is K (k) =

1 2π

0

,

dx 1 − k2 sin2 x

K plays a role for elliptic functions similar to that of π/2 for trigonometric functions. The functions sn and cn are periodic of period 4K and dn is periodic of period 2K. sn(u + 4K ) = sn(u)

cn(u + 4K ) = cn(u)

dn(u + 2K ) = dn(u)

Let k = 1 − k2 . k is called the complementary modulus. Table B.1 gives special values of elliptic functions. 2

Table B.1 Special values of elliptic functions. 0

K

2K

3K

4K

sn

0

1

0

−1

0

cn

1

0

−1

0

1

dn

1

k

1

k

1

To obtain derivatives we start by taking the derivative of (B.2b) to obtain

1 d √ amu (B.3) 1= du 1 − k2 sn2 u or

d amu = dnu du This makes it clear that amu increases without bound because dnu is never negative. It follows from (B.3) that d snu = cnu dnu du

d cnu = −snu dnu du

d dnu = −k2 snu cnu du

Now let us return to the pendulum. Equation (B.1) can now be expressed as + * 1 −1 −1 sin 2 θ ω (t + τ ) = am sin k


or ω (t + τ ) = sn−1

sin 12 θ k

from which it follows that 1 sin θ = k sn[ω (t + τ )] 2 It remains to determine the integration constant τ. This is done by requiring θ0 = θ (0 ) 1 sin θ0 = k sn(ωτ ) 2 1 −1 sin 12 θ0 τ = sn ω k It is easily verified that θ˙ (0) = θ˙0 Circulation

If ω 2 < c2 let k2 = Then

ω2 1 ˙2 θ = 2 4 k

ω2 c2

1 1 − k2 sin2 θ 2

and we obtain immediately ω (t + τ ) 1 = am−1 ( θ ) k 2

or θ = 2am with

ω (t + τ ) k

k τ = am−1 ω

1 θ0 2

221

222


B.2 Algebraic Relations Among Elliptic Functions

In addition to sn, cn, and dn, there are (in the Glaisher notation) ns = 1/sn

nc = 1/cn

sc = sn/cn

sd = sn/dn

cs = cn/sn

cd = cn/dn

nd = 1/dn

ds = dn/sn dc = dn/cn Table B.2 lists algebraic relations among these functions. For example, the second row shows how to express the functions in the first row in terms of sn alone. A full 12 × 12 table could be constructed using all 12 of Glaisher’s functions (just such a table appears in [92]).

cn2

k2 − 1 + dn2 k2 (k2 − 1)nd2 + 1 k2 nd2

ns2 − 1 ns2

1 nc2

1 + (k2 − 1)sd2 1 + k2 sd2

1 − cn2

1 − dn2 k2

nd2 − 1 k2 nd2

1 ns2

nc2 − 1 nc2

sd2 1 + k2 sd2

cn2

dn2

nd2

ns2

nc2

sd2

1 1 + k2 sd2

ns2 − k2 ns2 (1 − k2 )nc2 + k2 nc2

1 nd2

dn2

1 − k2 + k2 cn2

1 − k2 sn2

1 − sn2

sn2

dn2

cn2

sn2

sn2

Table B.2 Algebraic relations among elliptic functions.

1 + k2 sd2

nc2 2 (1 − k )nc2

ns2 − k2

ns2

nd2

+ k2

1 1 − cn2

1 1 − k2 + k2 cn2

1 + k2 sd2 sd2

nc2 nc2 − 1

ns2

k2 nd2 nd2 − 1

k2 1 − dn2

1 sn2

1 1 − k2 sn2

1 dn2

ns2

nd2

1 + k2 sd2 1 + (k2 − 1)sd2

nc2

ns2 ns2 − 1

k2 nd2 − 1)nd2 + 1

k2 − 1 + dn2

( k2

k2

1 cn2

1 1 − sn2

nc2

sd2

nc2 − 1 (1 − k2 )nc2 + k2

1

ns2 − k2

nd2 − 1 k2

1 − dn2 k2 dn2

1 − cn2 1 − k2 + k2 cn2

sn2 1 − k2 sn2

sd2

B.2 Algebraic Relations Among Elliptic Functions 223

224


B.3 Differential Equations Satisfied by Elliptic Functions

Each elliptic function f satisfies a differential equation of the form ( f )2 = F ( f ). Table B.3 lists the F for some of the elliptic functions. Table B.3 Differential equations for elliptic functions. f

F( f )

sn

cn dn

(1 − f 2 )(1 − k2 f 2 )

cn

−sn dn

(1 − f 2 )(1 − k2 + k2 f 2 )

dn

− k2 sn cn

(1 − f 2 )(k2 − 1 + f 2 )

nd

k2 nd2 sn cn

(1 − f 2 )[(1 − k2 ) f 2 − 1]

ns

ns2 cn dn

(1 − f 2 )(k2 − f 2 )

nc

nc2 sn dn

(1 − f 2 )[(1 − k2 ) f 2 + k2 ]

sd

nd2 cn

( k2 f 2 + 1)[(1 − k2 ) f 2 − 1]

f

B.3.1 The Addition Formulas

The functions sn, cn, and dn satisfy the addition formulas snu cnv dnv + cnu dnu snv 1 − k2 sn2 u sn2 v snu cnv dnv + cnu dnu snv dnv cn(u + v) = 1 − k2 sn2 u sn2 v dnu dnv − k2 snu cnu snu cnv dn(u + v) = 1 − k2 sn2 u sn2 v sn(u + v) =


225

Appendix C

Lie Groups and Lie Algebras This appendix summarizes aspects of the Lie group theory used in modern treatments of rigid body dynamics and numerical methods for solving rigid body problems. The approach is informal and more details can be found in [17, 29, 93]. We begin with some classical results which will then be translated to the abstract setting.

C.1 Infinitesimal Generators of Rotations

Let us begin with rotation about the x-axis 



0  1  Re1 ( θ ) =   0 cos θ  0 sin θ

0

  − sin θ    cos θ

Take the derivative of Re1 (θ ) and evaluate it at θ = 0 to obtain   0 0  d Re1 (0) =  M1 =  0 0 dθ  0 1

 0   −1    0

M1 = diag(0, − J ) 1 is skew-symmetric and satisfies M12 = diag(0, − I )  1) Here J = 

 0

1

−1

0

M13 = diag(0, J ) 

 and I = 

and

M14 = diag(0, I )

 1

0

0

1

.


226

C Lie Groups and Lie Algebras

We wish to show that the matrix exponential of the skew-symmetric M1 produces the rotation matrix Re1 . That is, Re1 (θ ) = exp{θM1 } Recall that exp A is defined by the power series exp A =

∞

1 k A k! r =0

∑

and that the series converges for any real or complex matrix A. Applying this to M1 we obtain exp{θM1 }

=

1 1 I + θM1 + θ 2 M12 + θ 3 M13 + 2 3! 1 1 I + θM1 − θ 2 M14 − θ 3 M1 + 2 3! 1 e1 ⊗ e1 + (θ − θ 3 + . . .) M1 3! 1 1 +(1 − θ 2 + θ 4 + . . .) M14 2 4! e1 ⊗ e1 + sin θM1 + cos θM14

=

Re1 ( θ )

= = =

1 4 4 θ M1 + . . . 4! 1 4 4 θ M1 + . . . 4!

There are the obvious companion results Re2 (θ ) = exp{θM2 }

and

Re3 (θ ) = exp{θM3 }

where Mi = eî . It is less obvious, but true, that for any unit vector n Rn (θ ) = exp{θ nˆ } To see this we use the results of Table 1.1 with N = nˆ N 3 = − N, N 4 = − N 2 , N 5 = N, and N 2 = P − I thus exp θN

=

=

1 3 1 θ + θ 5 + . . .) N 3! 5! 1 1 1 ( θ 2 − θ 4 + θ 6 + . . .) N 2 2 4! 6! P + sin θN − cos θN 2 I + (θ −

This also shows that Rn (θ ) = exp{θ (n1 M1 + n2 M2 + n3 M3 )}

C.2 Lie Groups

This ability of exponentials of skew-symmetric matrices to generate rotation matrices is a special case of a general property of Lie group–Lie algebra pairs, to which we now turn. C.2 Lie Groups

Fig. C.1 Differentiable manifold terminology.

A Lie group, G , is a set equipped with a smooth differential structure and a group structure (multiplication and inversion). An n-dimensional differential structure (Fig. C.1) on G consists of a covering of G by sets Ui

G = ∪ Ui and one–one coordinate functions φi mapping the Ui to open sets in R n , φi : U i → R n

φi ( p ) = ( q 1 , . . . , q n )

which are smoothly related on overlapping Ui . That is φj−1 φi : φj (Uj ) ∩ φi (Ui ) → φj (Uj ) ∩ φi (Ui ) is a smooth function on R n . The group operations are continuous in the topology inherited from R n in which the open sets are images under the φi−1 of open sets in R n .

227

228


For example, the sphere S2 can be embedded in R3 and covered with the six hemispheres r ∈ S2 , xi ≷ 0, i = 1, 2, 3. Coordinates can be assigned to points in each hemisphere by projection onto the plane normal to the pole of the hemisphere. Let φ1 and φ2 refer to the hemispheres x > 0 and z < 0, respectively. A point in the intersection of these hemispheres will have coordinates (y, z) under φ1 and ( x, y) under φ2 . The map φ2 φ1−1 , is (y, z) → ( 1 − y2 − z2 , y) which is smooth. The choice of the open cover and coordinate functions is not unique. The differential structure on the sphere could have been constructed from stereographic projections from the North and South poles. In that case the cover would contain two sets S2 \ n and S2 \ −n. A curve γ(t) in a Lie group, G , is a smooth map from a neighborhood of 0 in R to G . Tangents to a curve can be calculated using the coordinate functions. If the curve is γ(t) = (q1 (t), . . . , qn (t)) the tangent is γ (t) = (q1 (t), . . . , qn (t)). The tangent space Tg G at g ∈ G is the collection of all tangent vectors of curves passing through g.2 Tg G is a vector space of dimension n. Given a mapping F : G → G : g → F ( g), the tangent map is the linear map DF ( g) : Tg G → TF ( g) G : ( g, v ) → ( F ( g), DF ( g)v) In local coordinates the tangent map is    DF ( g) =   

∂F1 ( q) ∂q1

...

∂Fn ( q) ∂q1

...

.. .

∂F1 ( q) ∂qn

.. .

∂Fn ( q) ∂qn

     

A coordinate free definition of the tangent map is obtained by composition F (γ(t)) which gives a curve passing through F ( g) whose tangent is DF ( g)(γ ). Given a real valued function f : G → R defined in a neighborhood of a curve γ(t), one can differentiate f along the curve f =

d ∂f f (q1 (t), . . . , qn (t)) = ∑ q dt ∂qi i

Thus it is seen that tangent vectors are differential operators acting on functions. A tangent vector ∂ X = ∑ Xi ∂qi applied to a function f yields X f = ∑ Xi

∂f ∂qi

2) More precisely, the set of equivalence classes of curves.

C.2 Lie Groups

If tangent vectors are composed as, for example, * + ∂X j ∂ ∂ ∂ ∂2 YX = ∑ Yi Xj = ∑ ∑ Yi + Yi X j ∂qi ∑ ∂q j ∂qi ∂q j ∂qi ∂q j i j i j the result is not a tangent vector because of the second derivative terms. However, we can form the Lie bracket

[ X, Y ] ≡ XY − YX to eliminate the offending second-order derivatives and thereby obtain new tangent vector. Thus, the tangent space supports an anti-symmetric ([ x, y] = −[y, x ]) binary operation, the Lie bracket [ X, Y ] which satisfies the Jacobi identity [ X, [Y, Z ]] + [Y, [ Z, X ]] + [ Z, [ X, Y ]] = 0 We now consider vector fields on G . A vector field is a smooth assignment of a tangent vector to each point of G . One member is chosen from each Tg G , g ∈ G in such a way that the components of the vectors are smooth functions of position on G . In R n , one can create a vector field by choosing a vector at the origin and translating it without rotation to every point in R n . A similar construction is available in any Lie group. A translation in a Lie group G by an element g ∈ G is the mapping L g : G → G : h → gh (left translation) or R g : G → G : h → hg (right translation) The translation maps can be used to generate a vector field Xv on G from v ∈ Te G as follows Xv ( g) = DL g (e)v A left invariant vector field is one which satisfies X ( L g h) = DL g X (h) By construction the vector fields Xv are left invariant. It is instructive to calculate DL g for the case of matrix groups. In this case L A B = AB or ( L A B)ij = Aik Bkj and

   0

∂( L A B)ij =  ∂Bkl  Aik

if l = j if l = j

229

230


Thus

[( DL A )v] = Av The abundance of vector fields distinguishes Lie groups among the category of differentiable manifolds. Many differential manifolds will support no nonvanishing, smooth vector fields. For example, in the family of spheres, only S1 , S3 , and S7 will support them. S1 and S3 spheres are Lie groups, the set of units of the complex numbers and quaternions and S7 is the set of units of octonions. C.2.1 Examples

Let M (n, R ) denote the set of order n matrices with real entries. The general linear group is GL(n, R ) = { A ∈ M (n, R )| det A = 0} This is a group because a matrix A is invertible if det A = 0 and det( AB) = det A det B. GL(n, R ) is a manifold because it is the inverse image of the open set R \ 0 under the mapping det. GL(n, R ) is an n2 -dimensional manifold by virtue of the n2 real entries in each of the matrices. There is also the complex counterpart GL(n, C ) = { A ∈ M (n, C )| det A = 0} where M (n, C ) denotes the set of order n matrices with complex entries. Subgroups of the general linear groups are Lie groups and are generally referred to as matrix groups. Not every Lie group is a matrix group but all Lie groups we shall need are matrix groups. Two examples which figure prominently in the theory of rotations are the special orthogonal group SO(n) = { A ∈ M (n, R )| At A = I, det A = 1} and the special unitary group SU (n) = { A ∈ M (n, C )| AA† = I, det A = 1}. These groups are subgroups of the larger matrix groups – the orthogonal group O(n) = { A ∈ M (n, R )| At A = I } and the unitary group U (n) = { A ∈ M (n, C )| AA† = I }. The symplectic group is



 Sp(2n) = { A ∈ M (2n, R )| At J A = J }, where J = 

 0

−I

I   0

C.3 Lie Algebras

and I is the order n identity matrix. Sp (2n) is prominent in Hamiltonian mechanics because tangent maps of canonical transformations are symplectic matrices. Our last example is the affine group A(n) which consists of matrices of the form    M a    0 1 where M ∈ GL(n, R ) and a is a column of n real numbers. A(n) acts on R n by v → Av + a which is the result of the matrix multiplication     M a  v    . 0 1 1 An important special case of A(n) is the Euclidian group SE(n) which consists of matrices of the form    R a    0 1 where R ∈ SO(3) and a is a column of three real numbers. SE(3) acts on R3 by v → Rv + a and this is the most general displacement of a rigid body.

C.3 Lie Algebras

A Lie algebra, L, is a vector space closed under a binary operation [ x, y], the abstract bracket, which has the properties

[ x, y] = −[y, x ] [ x, [y, z]] + [y, [z, x ]] + [z, [ x, y]] = 0 For example,

R3

is a Lie algebra under the vector cross product

[ x, y] = x × y

231

232


and M (n, R ) is a Lie algebra with the matrix commutator [ A, B] = AB − BA serving as the abstract bracket. In general, any vector space of matrices which is closed under commutation is a Lie algebra. The Lie algebra of a Lie group is a linearization of the group. Its utility is that it is easier to calculate and to construct proofs in a Lie algebra setting and there are techniques for translating results from Lie algebras to Lie groups. We have seen an example of this in the infinitesimal generators of rotations. The set of skew-symmetric matrices Sk(n) = A ∈ M (n, R )| At = − A is a Lie algebra because, for A, B ∈ Sk(n)

[ A, B]t = ( AB − BA)t = Bt At − At Bt = BA − AB = −[ A, B] and consequently [ A, B] ∈ Sk(n) In every Lie group G the tangent space at the identity Te G is a Lie algebra with the abstract bracket being vector field Lie bracket. For x, y ∈ Te G

[ x, y] = [ Xx , Xy ](e) Thus every Lie group, G , has an associated Lie algebra, g,3 g = Te G Since every v ∈ g generates a left invariant field, Xv , g can be regarded as the set of left invariant fields on G The bracket operation for an n-dimensional Lie algebra g is completely encoded in a set of n3 parameters called the structure coefficients Cijk . Let { f i } be a basis for g. Since g is closed under the Lie bracket operation we may express the Lie bracket operation on the basis vectors by

[ f i , f j ] = Cijk f k Then the bracket of any two vectors is expressed in terms of the Cijk by

[ ai f i , b j f j ] = ai b j Cijk f k C.3.1 Examples

To find the Lie algebras associated with the matrix groups we have discussed, we consider curves passing through the identity in the Lie group and evaluate their tangents at the identity. 3) We shall use lower case Germanic letters to name Lie algebras associated with Lie groups.

C.3 Lie Algebras

so(n) Let A(t) be a curve in SO(n) for which A(0) = I. Then A(t)t A(t) = I implies A (t)t A(t) + A(t)t A (t) = 0 and evaluating this at A(0) = I gives x + xt = 0 where x = A (0). Thus

so(n) ⊂ Sk(n)

Now let A ∈ Sk(n)

and

M = exp( A)

then M t M = exp( At ) exp( A) = exp(− A) exp( A) = I and Sk(n) ⊂ so(n) This identifies so(n) with Sk(n). gl(n): Let M ∈ M (n, R ) be arbitrary and consider the curve A(t) = I + tM. A(0) = I and, for t sufficiently small, A(t) ∈ GL(n, R ). The fact that A (0) = M and the arbitrariness of M show that gl(n) = M (n, R ) su(n): Let A(t) be a curve in SU (n) for which A(0) = I. Then A(t)† A(t) = I implies A (t)† A(t) + A(t)† A (t) = 0 and evaluating this at A(0) = I shows that x is skew-Hermitian, that is x + x† = 0 where x = A (0). Thus su(n) ⊂ skew-Hermitian matrices Now let A be skew-Hermitian and let M = exp( A), then M † M = exp( A† ) exp( A) = exp(− A) exp( A) = I Therefore all skew-Hermitian matrices are in o(n).

233

234


Elements of SU (n) ⊂ U must satisfy det U = 1 and this imposes an additional condition on elements of so(n). To arrive at the additional condition we need the result det(exp( A)) = eTrA which follows from the fact that the determinant of a matrix is equal to the product of its eigenvalues, the trace is equal to the sum of the eigenvalues and if A has eigenvalues λi then exp A has eigenvalues {λi }. Now let A be skew-Hermitian so that exp A is unitary, then 1 = det(exp( A)) = eTrA ⇒ TrA = 0 This identifies so(n) as the set of skew-Hermitian matrices having zero trace. C.3.2 Adjoint Operators

Fig. C.2 The mappings Ad and ad on g. Artistic license has been exercised to display ad a at the tip of Ada b rather than the identity.

Let G be a matrix group with Lie algebra g. For A ∈ G the conjugation operator C A B = ABA−1 is a linear operator (C A [ a1 A1 + a2 A2 ] = a1 C A A1 + a2 C A A2 ) which leaves the identity fixed (C A ( I ) = AA−1 = I ). Thus C A maps a curve through the origin to another curve through the origin and an element of g, a tangent to the curve, into another element of g. Formally we define the adjoint operator Ad A as the derivative of C A at the identity C A B(t) = AB(t) A−1

d C B(t)|t=0 = AB (0) A−1 = Ad A x dt A

C.3 Lie Algebras

with x = B (0). This is referred to as the action of Ad on g. If A(t) is a curve through the identity at t = 0 then # $ d Ad A(t) x |0 = A yA−1 − AyA−1 A A−1 = xy − yx = adx y dt t =0 with y = A (0) provides a linear action of g on itself. A graphical depiction of the relation between Ad and ad is shown in Fig. C.2. For example, let us find the ad action for so(3). A basis for so(3) is provided by the matrices       0 0 0 0 0 1 0 − 1 0                  f f f1 =  = =  0 0 −1  2  0 0 0  3  1 0 0        0 0 0 0 1 0 −1 0 0 We calculate the commutators

[ f1 , f2 ] = f3

[ f2 , f3 ] = f1

[ f3, f1 ] = f2

Thus if v = a f 1 + b f 2 + c f 3 then

[ f 1 , v ] = −c f 2 + b f 3 [ f2 , v] = c f1 − a f3 [ f 3 , v ] = −b f 1 + a f 2 and we find that the matrix representations of ad f i are      0 0 0 0 0 1    0            ad f 1 =   0 0 −1  ad f 2 =  0 0 0  ad f 3 =  1      0 0 1 0 −1 0 0

 −1 0   0 0    0 0

The fact that f i = ad f i is peculiar to so(3) but the method is generally applicable. To calculate the representation for adx for an arbitrary Lie algebra we can use the structure coefficients. Let { f i } be a set of basis vectors. Since ad f i a j f j = [ f i , a j f j ] = a j Cijk f k we conclude that the representation of ad f i is the matrix whose entry in row k and column j is Cijk . When we consider Lie group numerical methods we will need the iterated adjoint operators adkx which are defined by ad0x = 1

adkx+1 = adx (adkx )

235

236


The first three terms in the series are ad1x y = xy − yx

ad2x y = x2 − 2xyx + yx 2 , ad3x y = x3 y − 3x2 yx + 3xyx 2 − yx3

In general adkx y

k k −r r x yx = ∑ (−1) r r =0 k

r

(C.1)

which is easily proved by induction. It is clearly true for k = 0. If it is true for k then

k k k −r +1 r k k −r r +1 yx − adkx+1 y = ∑ (−1)r x x yx r r r =0

k k k k +1 r = x y + ∑ (−1) + x k−r +1 yx r + yx k+1 r r−1 r =1

k +1 r k+1 = ∑ (−1) x k+1−r yx r r r =0 Therefore it is true for all k. The last line used the Pascal triangle identity. We can put this right to work to prove the identity Adexp x = exp adx The left-hand side of this equation yields the series Adexp x y

= (exp x )y(exp − x ) 1 1 = (1 + x + x2 + · · · )(y − yx + yx2 + · · · ) 2 2

The general term involving k + 1 factors is Tk+1 =

(−1)k k!

k

∑ (−1)r

r =0

k k −r r (−1)k k adx y x yx = r k!

which is precisely the corresponding term in exp adx y.

(C.2)

C.4 Lie Group–Lie Algebra Relations


We now restrict our attention to matrix groups, G , and their Lie algebras, g. C.4.1 The Exp Map

For each x ∈ g there is a map exp from g to a neighborhood of the identity in G exp : g → G : x → exp( x ) exp( x ) = I +

∞

1 n x n! n =1

∑

We have already seen this used in the section on infinitesimal rotations and exp

the section on adjoint operators. In that case of so(3) → SO(3) the map exp is surjective, i.e., all elements of SO(3) are images of a skew symmetric matrix under exp. This is not true in general, however. Consider A ∈ sl2   a b   A= ∆ = det A  c −a Then

and

 √ √   cos ∆I + √1 sin ∆A  ∆   , , exp( A) = cosh |∆| I + √1 sin |∆| A  | ∆ |     I+A  √   2 cos ∆    , Tr [exp( A)] = 2 cosh |∆|      2

if ∆ > 0 if ∆ < 0 if ∆ = 0

if ∆ > 0 if ∆ < 0 if ∆ = 0

Thus Tr [exp( A)] ≥ −2 and any M ∈ SL2 of the form   − a 0   M=  0 −1/a with a > 2 cannot be an image of an element of A ∈ sl2 under exp.

237

238


C.4.2 The Derivative of exp A(t )

The calculation of d exp( A(t))/dt is complicated by the fact that A and A˙ do not, in general, commute. Thus, from the series expansion d 1 ˙ 1 ˙ 2 ˙ + A2 A˙ ) + · · · exp( A(t)) = A˙ + ( AA + A A˙ ) + ( AA + A AA dt 2 3! it is clear that the function identity d x (t) e = e x (t) x˙ dt does not generalize to matrix groups. The correct result involves a new function dexp A which satisfies d exp( A(t)) = dexp A A˙ exp A dt and our next goal is to construct the function dexp A . We will do this by calculating the product of the infinite series * + k −1 1 1 d k − 1 − i i ˙ + A A˙ ) + · · · + ˙ +··· AA A A˙ + ( AA exp A exp(− A) = dt 2 k! i∑ =0 * + 1 2 (−1)k k A ··· I − A+ A +···+ 2 k! The generic term having k factors is Tk =

1 k!

where cr =

k

˙ r −1 ∑ cr Ak−r AA

r =1 k

∑ (−1)

k+i

i =r

k i

Using the Pascal triangle identity

k−1 k−1 k = + i i−1 i we find cr = (−1) Therefore,

∞ (−1)k d A exp(− A) = ∑ dt k! k =1

k −r

k −1

k−1 r−1

∑ (−1)r+1

r =0

k−1 ˙ k −r −1 Ar AA r

(C.3)


˙ We now wish to express the above series in a form that allows it to act on A. The key to this expression is the iterated ad A operator. Substitute (C.1) for the second sum in (C.3) to obtain d exp A = dexp A A˙ exp A dt where dexpx y =

∞

1 k −1 ad y k! x k =1

∑

(C.4)

The coefficients of the dexp series agree with those of (e x − 1)/x and the result is sometimes expressed symbolically as dexpx =

exp(adx ) − 1 adx

(C.5)

which should be interpreted as a power series in adx . Here dexpx has been developed from the right translation of the derivative to the identity, that is, multiplication on the right by the inverse of exp A. This is sometimes expressed as the right trivialization because T G can be represented as the trivial bundle (Section 4.5) g × G . Our treatment is in agreement with [61] but is at odds with [93] where left translation is used. The following argument from [61] shows how the two are related

d d exp A exp(− A) A˙ exp(− A) exp A = Adexp(− A) dt dt exp(ad A ) − 1 ˙ = exp ad− A A ad A 1 − exp(ad− A ) ˙ A = ad A where Eq. (C.2) has been used in the second line. Therefore, exp(− A)

d exp A = dexp− A A˙ dt

(C.6)

Equation (C.5) suggests the matrix inverse of dexp, namely 1 dexp− x =

∞ adx 1 = ∑ Bk adkx exp(adx ) − 1 k! k =0

where Bk is the kth Bernoulli number for which x/(e x − 1) is a generating function. The sequence of Bernoulli numbers begins % & 1 −1 1 1 , 0, , 0, · · · { B0 , B1, B2 , B3 , B4, B5 , B6 , B7 , · · · } = 1, , , 0, 2 6 30 42

239

240


If k is odd and k > 1 then Bk = 0. There is a simple recursion relation to calculate the Bernoulli numbers

1 k −1 k + 1 Bk = − Br k + 1 r∑ r =0 This is equivalent to the easily remembered relation k +1

∑

r =1

k+1 Bk + 1 − r = 0 r

For example, 5B4 + 10B3 + 10B2 + 5B1 + B0 = 0


241

Appendix D

Notation

R

1

C

29

the complex numbers

ˆ C

29

the Riemann sphere

V

1

finite-dimensional vector space over R

V∗

1

dual vector space over R

u

1

vector, element of a vector space

υ

1

co-vector, element of a dual space

X→Y

7

mapping of a set

x → y

7

mapping of elements of a set

ei

7

standard basis vector for R n

u×v

7

vector or cross product

u·v

3

scalar or dot product

u⊗v

4

tensor product

Gij

3

matrix element in row i, column j

Mt

3

transpose of matrix M

det M

7

determinant of matrix M

|z|

8

modulus of complex number z

u

4

norm of vector u

the real numbers


242

D Notation

uˆ

9

skew-symmetric matrix associated with vector u

M

9

vector associated with skew-symmetric matrix M

z∗

8

complex conjugate of complex number z

Sk n

9

skew-symmetric matrices of order n

Rn ( θ )

10

operator for rotation about axis n by angle θ

Rn ( θ )

10

matrix representation of Rn (θ )

q

19

quaternion q

q¯

20

conjugate of quaternion q

(R, b)

41

rotation–translation pair in SE(3)

Ω

45

angular velocity operator

ω

46

angular velocity vector

I (r0 , B)

51

the inertia tensor

h

54

angular momentum vector

85

vector contraction

D

101

distribution

Ad A

234

adjoint operator , the action of G on g

ad a

235

adjoint operator , the action of g on g

exp

237

the exponential map

dexp

238

the dexp function


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247

Index Mo (C ) 35 SO (3) 8, 230 SU (2) 30, 230 H 19 H1 23 so(3) 233 su(2) 233

a adjoint operators 234 affine connection 74 Andoyer variables 88, 136, 195 angular momentum 54 angular velocity operator 45 angular velocity vector 46 axis of rotation 8 b base space 124 basis holonomic 76 natural 76 vector space 1 bifurcation 149 bilinearity 95 bracket 92 classical Poisson 92 Lie bracket 229 Lie–Poisson 97 Poisson 95 c Cayley transform 13 Cayley–Klein parameters 31 change of basis 6 Chasles’ theorem 42 Christoffel symbols 73 closed form 85 commutator 232 conditionally stable 163 configuration space 60

conjugate 20 conjugation operator 234 constraint 99 acatastatic 101 catastatic 100 holonomic 99 ideal 119 nonholonomic 99 nonideal 119 rheonomic 100 sclereonomic 100 contraction 85 vector 85 covariant derivative 74 covector 1, 60, 85 curvature terms 107

d d’Alembert’s principle 72, 118 Darboux theorem 86 Dedekind cuts 18 differential manifold 60 differential structure 227 distribution 101, 118 dual vector space 1 e Ehresmann connection 124 ellipsoid of inertia 141 elliptic integral of the third kind 134 Euclidean 3 Euler equations 62, 64 generalized 64 Euler parameters 24 Euler top 129 Euler–Poinsot top 129 Euler–Rodrigues parameters 24 exact form 85 explicit Euler method 163 f fiber 124


248

Index fiber bundle 123 force of constraint frame 2

103

g generalized force 72 generalized velocities 59 geometric phase 143, 145 gyroscopic terms 73 gyrostat 148 h Hamilton’s equations 84 Hamilton’s principle 69 herpolhode 142 holonomic basis 76 Hopf fibration 125 i idempotent 10 implicit Euler method 163 inertia tensor 51 infinitesimal generators of rotations 225 inner product 3 integrable distribution 101 integral of motion 129 invariable plane 142 inverse 20 involutory 37 isospectral deformation 156 j Jacobi identity 95, 229 k kinematic equation 48 kinematic equations 65 Kowalevsky top 153 l Lagrange multiplier 102 Lagrange top 145 Lagrangian 59 Lax equation 156 Lax pair 156 left invariant vector field 229 Leibniz identity 95 Lennard-Jones potential 211 Levi-Civita connection 75 Lie algebra 231 Lie group 227 Lie group methods 170 line of the nodes 48 linear operator 2 Liouville tori 155 Liouville-integrable 155 locally trivial 124

m Möbius transformations 35 matrix group 230 n natural system 72 nonholonomic basis 76 norm 20 nutation 15 o orthogonal matrix 7 outer product 85 p phase space 60 pitch 16 Poinsot Construction 141 polhode 142 precession 15 principal axes 51 principal moments of inertia 51 projection method 104 pure quaternion 21 q quasi-velocity 76 quaternions 17 free rotation 116 r reduced equation 105 reduction 104 reduced equation 105 representation of a covector 2 representation of a vector 2 Riemann sphere 31 right trivialization 239 rigid body motion 39 rigid body transformation 39 RK4MK method 173 Rodrigues triangle 26 roll 16 rolling disc 111 rolling sphere 108 rotation reversal 29 Runge–Kutta methods 166 s scalar part 20 screw pitch 43 screw transformation 43 SE method 173 singular value decomposition 121 skew-symmetry 95 special orthogonal group 8 special unitary matrices 30 spin 15

Index standard basis 7 state space 60 Steiner’s theorem 52 stereographic projection 31 SVD 121 symplectic form 86

t tangent manifold 60 tangent map 228 tangent space 228 top 65 Euler 129 Kowalevsky 153 Lagrange 145 torque 62 total space 124 translation 40, 229 transposition of rotations 29 triangle equalities 51 two-wheeled robot 114

u unconditionally stable 163 unconstrained 72 unit quaternion 21 v variation 69 vector field 229 vector part 20 vector product 7 vector space 1 vertical vectors 124 virtual displacement 118 w wedge product 85 wobblestone 183 y yaw 16

249

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