Quaternions, Clifford Algebras and Relativistic Physics

Patrick R. Girard

Quaternions, Clifford Algebras and Relativistic Physics

Birkhäuser Basel . Boston . Berlin

Author: Patrick R. Girard INSA de Lyon Département Premier Cycle 20, avenue Albert Einstein F-69621 Villeurbanne Cedex France e-mail: [email protected]

Igor Ya. Subbotin Department of Mathematics and Natural Sciences National University Los Angeles Campus 3DFL¿F&RQFRXUVH'ULYH Los Angeles, CA 90045 USA e-mail: [email protected]

2000 Mathematical Subject Classification: 15A66, 20G20, 30G35, 35Q75, 78A25, 83A05, 83C05, 83C10

Library of Congress Control Number: 2006939566 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 .

ISBN 978-3-7643-7790-8 Birkhäuser Verlag AG, Basel – Boston – Berlin This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, re-use of illustrations, recitation, broadcasting, reproduction on microfilms or in other ways, and storage in data banks. For any kind of use permission of the copyright owner must be obtained. Originally published in French under the title “Quaternions, algèbre de Clifford et physique relativiste”. © 2004 Presses polytechniques et universitaires romandes, Lausanne All rights reserved © 2007 Birkhäuser Verlag AG, P.O. Box 133, CH-4010 Basel, Switzerland Part of Springer Science+Business Media Printed on acid-free paper produced from chlorine-free pulp. TCF f f Printed in Germany ISBN-10: 3-7643-7790-8 e-ISBN-10: 3-7643-7791-7 ISBN-13: 978-3-7643-7790-8 e-ISBN-13: 978-3-7643-7791-5 987654321

www.birkhauser.ch

To Isabelle, my wife, and to our children: Claire, B´eatrice, Thomas and Benoˆıt

Foreword

The use of Clifford algebra in mathematical physics and engineering has grown rapidly in recent years. Clifford had shown in 1878 the equivalence of two approaches to Clifford algebras: a geometrical one based on the work of Grassmann and an algebraic one using tensor products of quaternion algebras H. Recent developments have favored the geometric approach (geometric algebra) leading to an algebra (space-time algebra) complexified by the algebra H ⊗ H presented below and thus distinct from it. The book proposes to use the algebraic approach and to define the Clifford algebra intrinsically, independently of any particular matrix representation, as a tensor product of quaternion algebras or as a subalgebra of such a product. The quaternion group thus appears as a fundamental structure of physics. One of the main objectives of the book is to provide a pedagogical introduction to this new calculus, starting from the quaternion group, with applications to physics. The volume is intended for professors, researchers and students in physics and engineering, interested in the use of this new quaternionic Clifford calculus. The book presents the main concepts in the domain of, in particular, the quaternion algebra H, complex quaternions H(C), the Clifford algebra H ⊗ H real and complex, the multivector calculus and the symmetry groups: SO(3), the Lorentz group, the unitary group SU(4) and the symplectic unitary group USp(2, H). Among the applications in physics, we examine in particular, special relativity, classical electromagnetism and general relativity. I want to thank G. Casanova for having confirmed the validity of the interior and exterior products used in this book, F. Sommen for a confirmation of the Clifford theorem and A. Solomon for having attracted my attention, many years ago, to the quaternion formulation of the symplectic unitary group. Further thanks go to Professor Bernard Balland for reading the manuscript, the Docinsa library, the computer center and my colleagues: M.-P. Noutary for advice concerning Mathematica, G. Travin and A. Valentin for their help in Latex. For having initiated the project of this book in a conversation, I want to thank the Presses Polytechniques et Universitaires Romandes, in particular, P.-F. Pittet and O. Babel.

viii

Foreword

Finally, for the publication of the english translation, I want to thank Thomas Hempfling at Birkh¨ auser. Lyon, June 2006 Patrick R. Girard

Contents

Introduction

1

1 Quaternions 1.1 Group structure . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Finite groups of order n ≤ 8 . . . . . . . . . . . . . . . . . 1.3 Quaternion group . . . . . . . . . . . . . . . . . . . . . . . 1.4 Quaternion algebra H . . . . . . . . . . . . . . . . . . . . 1.4.1 Definitions . . . . . . . . . . . . . . . . . . . . . . 1.4.2 Polar form . . . . . . . . . . . . . . . . . . . . . . 1.4.3 Square root and nth root . . . . . . . . . . . . . . 1.4.4 Other functions and representations of quaternions 1.5 Classical vector calculus . . . . . . . . . . . . . . . . . . . 1.5.1 Scalar product and vector product . . . . . . . . . 1.5.2 Triple scalar and double vector products . . . . . . 1.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . .

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2 Rotation groups SO(4) and SO(3) 2.1 Orthogonal groups O(4) and SO(4) . . . . . . . 2.2 Orthogonal groups O(3) and SO(3) . . . . . . . 2.3 Crystallographic groups . . . . . . . . . . . . . 2.3.1 Double cyclic groups Cn (order N = 2n) 2.3.2 Double dihedral groups Dn (N = 4n) . 2.3.3 Double tetrahedral group (N = 24) . . . 2.3.4 Double octahedral group (N = 48) . . . 2.3.5 Double icosahedral group (N = 120) . . 2.4 Infinitesimal transformations of SO(4) . . . . . 2.5 Symmetries and invariants: Kepler’s problem . 2.6 Exercises . . . . . . . . . . . . . . . . . . . . .

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3 Complex quaternions 3.1 Algebra of complex quaternions H(C) . . . . . . . . . . . . . . . . 3.2 Lorentz groups O(1, 3) and SO(1, 3) . . . . . . . . . . . . . . . . . 3.2.1 Metric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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4 Clifford algebra 4.1 Clifford algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.2 Clifford algebra H ⊗ H over R . . . . . . . . . . . . . . . . . 4.2 Multivector calculus within H ⊗ H . . . . . . . . . . . . . . . . . . 4.2.1 Exterior and interior products with a vector . . . . . . . . . 4.2.2 Products of two multivectors . . . . . . . . . . . . . . . . . 4.2.3 General formulas . . . . . . . . . . . . . . . . . . . . . . . . 4.2.4 Classical vector calculus . . . . . . . . . . . . . . . . . . . . 4.3 Multivector geometry . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.1 Analytic geometry . . . . . . . . . . . . . . . . . . . . . . . 4.3.2 Orthogonal projections . . . . . . . . . . . . . . . . . . . . . 4.4 Differential operators . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.2 Infinitesimal elements of curves, surfaces and hypersurfaces 4.4.3 General theorems . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

57 57 57 58 59 59 61 62 64 64 64 66 69 69 69 71 72

5 Symmetry groups 5.1 Pseudo-orthogonal groups O(1, 3) and SO(1, 3) . . . . 5.1.1 Metric . . . . . . . . . . . . . . . . . . . . . . . 5.1.2 Symmetry with respect to a hyperplane . . . . 5.1.3 Pseudo-orthogonal groups O(1, 3) and SO(1, 3) 5.2 Proper orthochronous Lorentz group . . . . . . . . . . 5.2.1 Rotation group SO(3) . . . . . . . . . . . . . . 5.2.2 Pure Lorentz transformation . . . . . . . . . . 5.2.3 General Lorentz transformation . . . . . . . . . 5.3 Group of conformal transformations . . . . . . . . . . 5.3.1 Definitions . . . . . . . . . . . . . . . . . . . . 5.3.2 Properties of conformal transformations . . . .

75 75 75 75 77 78 78 79 81 82 82 83

3.3

3.4 3.5

3.6 3.7 3.8

3.2.2 Plane symmetry . . . . . . . . . . . . . 3.2.3 Groups O(1, 3) and SO(1, 3) . . . . . . . Orthochronous, proper Lorentz group . . . . . 3.3.1 Properties . . . . . . . . . . . . . . . . . 3.3.2 Infinitesimal transformations of SO(1, 3) Four-vectors and multivectors in H(C) . . . . . Relativistic kinematics via H(C) . . . . . . . . 3.5.1 Special Lorentz transformation . . . . . 3.5.2 General pure Lorentz transformation . . 3.5.3 Composition of velocities . . . . . . . . Maxwell’s equations . . . . . . . . . . . . . . . Group of conformal transformations . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . .

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91 . 91 . 91 . 92 . 94 . 94 . 94 . 97 . 99 . 99 . 100 . 103

7 Classical electromagnetism 7.1 Electromagnetic quantities . . . . . . . . . . . . . . 7.1.1 Four-current density and four-potential . . 7.1.2 Electromagnetic field bivector . . . . . . . . 7.2 Maxwell’s equations . . . . . . . . . . . . . . . . . 7.2.1 Differential formulation . . . . . . . . . . . 7.2.2 Integral formulation . . . . . . . . . . . . . 7.2.3 Lorentz force . . . . . . . . . . . . . . . . . 7.3 Electromagnetic waves . . . . . . . . . . . . . . . . 7.3.1 Electromagnetic waves in vacuum . . . . . . 7.3.2 Electromagnetic waves in a conductor . . . 7.3.3 Electromagnetic waves in a perfect medium 7.4 Relativistic optics . . . . . . . . . . . . . . . . . . . 7.4.1 Fizeau experiment (1851) . . . . . . . . . . 7.4.2 Doppler effect . . . . . . . . . . . . . . . . . 7.4.3 Aberration of distant stars . . . . . . . . . 7.5 Exercises . . . . . . . . . . . . . . . . . . . . . . .

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105 105 105 107 110 110 115 116 118 118 119 120 121 121 123 124 125

8 General relativity 8.1 Riemannian space . . . . . . 8.2 Einstein’s equations . . . . . 8.3 Equation of motion . . . . . . 8.4 Applications . . . . . . . . . . 8.4.1 Schwarzschild metric . 8.4.2 Linear approximation

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5.4

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5.3.3 Transformation of multivectors Dirac algebra . . . . . . . . . . . . . . 5.4.1 Dirac equation . . . . . . . . . 5.4.2 Unitary and symplectic unitary Exercises . . . . . . . . . . . . . . . .

6 Special relativity 6.1 Lorentz transformation . . . . . . . . . 6.1.1 Special Lorentz transformation 6.1.2 Physical consequences . . . . . 6.1.3 General Lorentz transformation 6.2 Relativistic kinematics . . . . . . . . . 6.2.1 Four-vectors . . . . . . . . . . . 6.2.2 Addition of velocities . . . . . . 6.3 Relativistic dynamics of a point mass . 6.3.1 Four-momentum . . . . . . . . 6.3.2 Four-force . . . . . . . . . . . . 6.4 Exercises . . . . . . . . . . . . . . . .

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xii

Contents

Conclusion

135

A Solutions

137

B Formulary: multivector products within H(C)

153

C Formulary: multivector products within H ⊗ H (over R)

157

D Formulary: four-nabla operator ∇ within H ⊗ H (over R)

161

E Work-sheet: H(C) (Mathematica)

163

F Work-sheet H ⊗ H over R (Mathematica)

165

G Work-sheet: matrices M2 (H) (Mathematica)

167

H Clifford algebras: isomorphisms

169

I

171

Clifford algebras: synoptic table

Bibliography

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Index

177

Introduction If one examines the mathematical tools used in physics, one finds essentially three calculi: the classical vector calculus, the tensor calculus and the spinor calculus. The three-dimensional vector calculus is used in nonrelativistic physics and also in classical electromagnetism which is a relativistic theory. This calculus, however, cannot describe the unity of the electromagnetic field and its relativistic features. As an example, a phenomenon as simple as the creation of a magnetic induction by a wire with a current is in fact a purely relativistic effect. A satisfactory treatment of classical electromagnetism, special relativity and general relativity is given by the tensor calculus. Yet, the tensor calculus does not allow a double representation of the Lorentz group and thus seems incompatible with relativistic quantum mechanics. A third calculus is then introduced, the spinor calculus, to formulate relativistic quantum mechanics. The set of mathematical tools used in physics thus appears as a succession of more or less coherent formalisms. Is it possible to introduce more coherence and unity in this set? The answer seems to reside in the use of Clifford algebra. One of the major benefits of Clifford algebras is that they yield a simple representation of the main covariance groups of physics: the rotation group SO(3), the Lorentz group, the unitary and symplectic unitary groups. Concerning SO(3), this is well known, since the quaternion algebra H which is a Clifford algebra (with two generators) allows an excellent representation of that group . The Clifford algebra H ⊗ H, the elements of which are simply quaternions having quaternions as coefficients, allows us to do the same for the Lorentz group. One shall notice that H ⊗ H is defined intrinsically independently of any particular matrix representation. By taking H ⊗ H (over C), one obtains the Dirac algebra and a simple representation of SU(4) and USp(2, H). Computations within these algebras have become straightforward with software like Mathematica which allows us to perform extended algebraic computations and to simplify them. One will find as appendices, worksheets which allow easy programming of the algebraic (or numerical) calculi presented here. One of the main objectives of this book is to show the interest in the use of Clifford algebra H ⊗ H in relativistic physics with applications such as classical electromagnetism, special relativity and general relativity.

Chapter 1

Quaternions The abstract quaternion group, discovered by William Rowan Hamilton in 1843, is an illustration of group structure. After having defined this fundamental concept of physics, the chapter examines as examples the finite groups of order n ≤ 8 and in particular, the quaternion group. Then the quaternion algebra and the classical vector calculus are treated as an application.

1.1 Group structure A set G of elements is a group if there exists an internal composition law ∗ defined for all elements and satisfying the following properties: 1. the law is associative (a ∗ b) ∗ c = a ∗ (b ∗ c),

∀a, b, c ∈ G,

2. the law admits an identity element e a ∗ e = e ∗ a = a,

∀a ∈ G,

3. any element a of G has an inverse a
a
∗ a = a ∗ a
= e. Let F and G be two groups. A composition law on F × G is defined by (f1 , g1 )(f2 , g2 ) = (f1 f2 , g1 g2 ),

(fi ∈ F, gi ∈ G, i = 1, 2);

the group F × G is called the direct product of the groups F and G. Examples.

1. Cyclic group Cn of order n the elements of which are (b, b2 , b3 , . . . , bn = e)

and where b represents, for example, a rotation of 2π/n around an axis.

4

Chapter 1. Quaternions 2. Dihedral group Dn of order 2n generated by two elements a and b such that a2 = bn = (ab)2 = e. One has in particular b−h a = abh (h = 1 · · · n); indeed, since (ab)−1 = b−1 a−1 = b−1 a = ab, one has

b−1 (b−1 a)b = b−2 ab = b−1 ab2 = ab3

and thus b−2 a = ab2 ; by proceeding similarly by recurrence, one establishes the above formula.

1.2 Finite groups of order n ≤ 8 The finite groups of order n ≤ 8, except the quaternion group which will be treated separately, are the following. 1. n = 1, there exists only one group (1 = e). 2. n = 2, only one group exists, the cyclic group C2 consisting of the elements (b, b2 = e). Examples.

(a) the group constituted by the elements (−1, 1);

(b) the group having the elements (b : rotation of ±π around an axis, b2 = e). 3. n = 3, only one group is possible: the cyclic group C3 of elements (b, b2 , b3 = e) where b, b2 are elements of order 3. 4. n = 4, two groups exist: (a) the cyclic group C4 constituted by the elements (b, b2 , b3 , b4 = e) where the element b2 is of order 2, and where (b, b3 ) are elements of order 4; (b) the Klein four-group defined by I 2 = J 2 = (IJ)2 = 1 or, equivalently I 2 = J 2 = K 2 = IJK = 1 with K = IJ and the multiplication table

1 I J K

1 1 I J K

I I 1 K J

J J K 1 I

K K J I 1

.

1.2. Finite groups of order n ≤ 8

5

The Klein four-group is isomorphic to the direct product of two cyclic groups C2 , 2 (−1,
1) × (b,2b = e)  = 1 ≡ (1, b ), I ≡ (1, b), J ≡ (−1, b), K ≡ (−1, b2 ) .

Example. The group constituted by the elements (I: rotation of π around the axis Ox, J: rotation of π around the axis Oy, K = IJ: rotation of π around the axis Oz). 5. n = 5, there exists only one group, the cyclic group C5 having the elements (b, b2 , b3 , b4 , b5 = e). 6. n = 6, two groups are possible: (a) the cyclic group C6 (b, b2 , b3 , b4 , b5 , b6 = e); (b) the dihedral group D3 defined by the relations a2 = b3 = (ab)2 = e, leading to the multiplication table

b b2 b3 = e a ab ba

b b2 e b ab ba b

b2 e b b2 ba a ab

b3 = e b b2 e a ab ba

a ba ab a e b2 b

ab a ba ab b e b2

ba ab a ba b2 b e

with b−h a = abh (h = 1, 2, 3). This group is the first noncommutative group of the series. Example. The symmetry group of the equilateral triangle (see Fig. 1.1). 7. n = 7, there exists only one group, the cyclic group C7 of elements (b, b2 , b3 , b4 , b5 , b6 , b7 = e). 8. n = 8, there exist five groups, among them the quaternion group which will be treated separately. (a) The cyclic group C8 of elements (b, b2 , b3 , b4 , b5 , b6 , b7 , b8 = e).

6

Chapter 1. Quaternions

B b(M )

y

a(M )

D x

Oz

ba(M )

C

M

b2 (M )

ab(M )

A

Figure 1.1: Symmetry group of the equilateral triangle; b represents a rotation of the point M by 2π/3 around the axis Oz, a a symmetry of M in the plane ABC with respect to the mediatrice CD and M an arbitrary point of the triangle. (b) The group S2×2×2 , direct product of the Klein four-group with C2 , (1, I, J, K) × (1, −1) = (±1, ±I, ±J, ±K) with 1 = (1, 1), −1 = (1, −1), ±I = (I, ±1), ±J = (J, ±1), ±K = (K, ±1) ; the multiplication table is given by 1 −1 I −I J −J K −K

1 1 −1 I −I J −J K −K

−1 −1 1 −I I −J J −K K

I I −I 1 −1 K −K J −J

−I −I I −1 1 −K K −J J

J J −J K −K 1 −1 I −I

−J −J J −K K −1 1 −I I

K K −K J −J I −I 1 −1

−K −K K −J ; J −I I −1 1

the group is commutative. (c) the group S4×2 , direct product of C4 with C2 and constituted by the elements (b, b2 , b3 , b4 = e) × (1, −1) = (±b, ±b2 , ±b3 , ±1); it is a commutative group.

1.3. Quaternion group

7

(d) The group D4 (noncommutative) defined by a2 = b4 = (ab)2 = e with the multiplication table

b b2 b3 b4 = e a ab ba ab2

b b2 b3 e b ab ab2 a ab

b2 b3 e b b2 ab2 ba ab a

b3 e b b2 b3 ba a b2 ab

b4 = e b b2 b3 e a ab ba ab2

a ba ab2 ab a e b3 b b2

ab a ba ab2 ab b e b2 b3

ba ab2 ab a ba b3 b2 e b

ab2 ab a ba ab2 b2 b b3 e

and b−h a = abh (h = 1, 2, 3, 4). Example. The symmetry group of the square (see Fig. 1.2). C

y

ba(M )

b(M )

b2 (M )

a(M )

Oz

x

ab2 (M ) D

B

M

b3 (M )

ab(M )

A

Figure 1.2: Symmetry group of the square; b is a rotation of π/2 around the axis Oz, a a symmetry with respect to the axis Ox in the plane ABCD and M an arbitrary point of the square.

1.3 Quaternion group The quaternion group (denoted Q) was discovered by William Rowan Hamilton in 1843 and is constituted by the eight elements ±1, ±i, ±j, ±k satisfying the

8

Chapter 1. Quaternions

relations i2 = j 2 = k 2 = ijk = −1, ij = −ji = k, jk = −kj = i, ki = −ik = j, with the multiplication table

1 −1 i −i j −j k −k

1 1 −1 i −i j −j k −k

−1 −1 1 −i i −j j −k k

i i −i −1 1 −k k j −j

−i −i i 1 −1 k −k −j j

j j −j k −k −1 1 −i i

−j −j j −k k 1 −1 i −i

k k −k −j j i −i −1 1

−k −k k j −j −i i 1 −1

the element of the first column being the first element to be multiplied and 1 being the identity element. The element −1 is of order 2 (i.e., its square is equal to 1) and the elements (±i, ±j, ±k) are of order 4. The subgroups of Q are (1) (1, −1) (1, −1, i, −i) (1, −1, j, −j) (1, −1, k, −k).

1.4 Quaternion algebra H 1.4.1 Definitions Consider the vector space of numbers called quaternions a, b, . . . constituted by four real numbers a = a0 + a1 i + a2 j + a3 k = (a0 , a1 , a2 , a3 ) = (a0 , a) = (a0 , a) where S(a) = a0 is the scalar part and V (a) = a = a the vectorial part. This

1.4. Quaternion algebra H

9

vector space is transformed into the associative algebra of quaternions (denoted H) via the multiplication ab = (a0 b0 − a1 b1 − a2 b2 − a3 b3 ) + (a0 b1 + a1 b0 + a2 b3 − a3 b2 )i + (a0 b2 + a2 b0 + a3 b1 − a1 b3 )j + (a0 b3 + a3 b0 + a1 b2 − a2 b1 )k and in a more condensed form ab = (a0 b0 − a · b, a0 b + b0 a + a × b) where a · b = (a1 b1 + a2 b2 + a3 b3 ) and a × b = (a2 b3 − a3 b2 )i + (a3 b1 − a1 b3 )j + (a1 b2 − a2 b1 )k are respectively the usual scalar and vector products. Historically, these two products were obtained by W. J. Gibbs [17] by taking a0 = b0 = 0 and by separating the quaternion product in two parts. The quaternion algebra constitutes a noncommutative field (without divisors of zero) containing R and C as particular cases. Let a = a0 + a1 i + a2 j + a3 k be a quaternion, the conjugate of a, the square of its norm and its norm are respectively ac = a0 − a1 i − a2 j − a3 k, 2

|a| = aac = a20 + a21 + a22 + a23 ,  |a| = |a|2 , with the following properties (ab)c = bc ac , |ab|2 = |a|2 |b|2 , the last relation deriving from (ab)c ab = bc ac ab = (aac )(bbc ) ; furthermore, S(ab) = S(ba) thus S [a (bc)] = S [(bc) a] = S [b (ca)] = S [(ca) b] and therefore S (abc) = S (bca) = S (cab) , ac , aac    −1 2 a 1  a  = ac = 2, aac aac |a| a−1 =

(a1 a2 · · · an )−1 =

(a1 a2 · · · an )c |a1 a2 · · · an |2

−1 −1 = a−1 n an−1 · · · a1 .

10

Chapter 1. Quaternions

To divide a quaternion a by the quaternion b (= 0), one simply has to resolve the equation xb = a or by = a with the respective solutions x = ab−1 = a y = b−1 a = and the relation |x| = |y| =

bc |b|2

,

bc a 2

|b|

|a| |b| .

Examples. Consider the quaternions a = 2 + 4i − 3j + k and b = 5 − 2i + j − 3k; 1. the vectorial parts a, b and the conjugates ac , bc are a = 4i − 3j + k, ac = 2 − 4i + 3j − k, 2. the norms are given by

b = −2i + j − 3k, bc = 5 + 2i − j + 3k; √ √ aac = 30,  √ bbc = 39;

|a| = |b| = 3. the inverses are a−1

=

b−1

=

ac 2

|a| bc

2

|b|

=

2 − 4i + 3j − k , 30

=

5 + 2i − j + 3k ; 39

4. one can realize the following operations a+b a−b

= 7 + 2i − 2j − 2k, = −3 + 6i − 4j + 4k,

ab

= 24 + 24i − 3j − 3k,

ba

= 24 + 8i − 23j + k,

(ab)−1 =

(ab)c |ab|

S(x)

= b−1 a−1 =

4 4i j k − + + , 195 195 390 390

8i 9j 13k −2 − + − ), 15 15 10 30 3k −2 16i 7j − + − ), = b, y = a−1 b = ( 15 15 30 10 13 = S(y), |x| = |y| = . 10

xa = b, ay

2

√ 1 170, √ |ba| = 1 170,

|ab| =

x = ba−1 = (

1.4. Quaternion algebra H

11

1.4.2 Polar form Any nonzero quaternion can be written a = a0 + a1 i + a2 j + a3 k = r(cos θ + u sin θ),

0 ≤ θ ≤ 2π

 with r = |a| = a20 + a21 + a22 + a23 being the norm of a and  ± a21 + a22 + a23 a0 cos θ = , sin θ = , r r |a| a0 tan θ = ± ; cot θ = ± , |a| a0 the unit vector u (uuc = 1) is given by u=

±(a1 i + a2 j + a3 k)  a21 + a22 + a23

with a21 + a22 + a23 = 0. Since u2 = −1, one has via the De Moivre theorem an = rn (cos nθ + u sin nθ). Example. Consider the quaternion a; let us determine its polar form a

=

|a| = tan θ

=

Answer: a

=

3 + i + j + k, √ √ √ 12 = 2 3, |a| = 3, |a| 1 θ = 30◦ , = √ , a0 3 

 √ i+j+k √ 2 3 cos 30◦ + sin 30◦ . 3

1.4.3 Square root and nth root Square root The square root of a quaternion a = a0 + a1 i + a2 j + a3 k can be obtained algebraically as follows. The equation b2 = a with b = b0 + b1 i + b2 j + b3 k leads to the following equations b20 − b21 − b22 − b23 = a0

(1.1)

2b0 b1 = a1 , 2b0 b2 = a2 ,

sgn(b0 b1 ) = sgn(a1 ), sgn(b0 b2 ) = sgn(a2 ),

(1.2) (1.3)

2b0 b3 = a3 ,

sgn(b0 b3 ) = sgn(a3 ).

(1.4)

12

Chapter 1. Quaternions

Writing t = b20 the above equation (1.1) leads to t−

a21 + a22 + a23 = a0 4t

or t2 − a 0 t −

a21 + a22 + a23 = 0. 4

One obtains  a20 + a21 + a22 + a23 t= = ≥0 2 a0 − a20 + a21 + a22 + a23 −(b21 + b22 + b23 ) = a0 − b20 = , 2 b20

hence ε b0 = √ 2

a0 +

 a20 + a21 + a22 + a23 + a0

(ε = ±1).

The equations (1.2), (1.3), (1.4) lead to −a21 , 4 −(a22 + a23 ) b20 (−b22 − b23 ) = , 4 b20 (−b21 ) =

(1.5) (1.6)

with −b22 − b23 = b21 + a0 − b20  a0 − a20 + a21 + a22 + a23 2 . = b1 + 2 Equation (1.6) then becomes with t = b21 and using (1.5) a21 4t

 t+

a0 −

thus a2 t= 1 2



2  a + a23 a20 + a21 + a22 + a23 =− 2 2 4   a20 + a21 + a22 + a23 − a0 a21 + a22 + a23

,

1.4. Quaternion algebra H

13

hence b1 (one proceeds similarly for b2 , b3 ); finally, one obtains  ε a20 + a21 + a22 + a23 + a0 (ε = ±1), b0 = √ 2  a1 ε b1 = √  2 a20 + a21 + a22 + a23 − a0 , 2 a1 + a22 + a23  a2 ε b2 = √  2 a20 + a21 + a22 + a23 − a0 , 2 a1 + a22 + a23  ε a3 b2 = √  2 a20 + a21 + a22 + a23 − a0 . 2 a1 + a22 + a23 Example. Consider the quaternion a = 1 + i + j + k ; find its square root b b0

=

Answer: b =

ε √ ε √ 3, b1 = b2 = b3 = √ , 2 6   j k 1 √ i ±√ 3+ √ + √ + √ . 2 3 3 3

nth roots The nth root of a quaternion a = r(cos ϕ + u sin ϕ), where ϕ can always be chosen within the interval [0, π] with an appropriate choice of u, is obtained as follows [9]. 1. Supposing sin ϕ = 0, the equation bn = a with b = R(cos θ+e sin θ), θ ∈ [0, π], leads to Rn = r, cos nθ = cos ϕ, sin nθ = sin ϕ, e = u and thus to 1

R = rn,

θ=

(ϕ + 2kπ) n

(k = 0, 1, . . . , n − 1);

finally, one has

(ϕ + 2kπ) 1 (ϕ + 2kπ) n b = r cos + u sin n n

(k = 0, 1, . . . , n − 1).

2. When sin ϕ = 0, the vector e in b is arbitrary. If a > 0, one has ϕ = 0 and thus θ = 2πm n (m √= 0, 1, . . . , n − 1). For n = 2, one obtains θ = 0, π and thus the real roots ± a. With n > 2, certain values of θ (= 0 or π) give nonreal roots, the vector e being arbitrary. With a < 0, one has ϕ = π, θ = (2m+1)π n (m = 0, 1, . . . , n − 1), certain values of θ = π give nonreal roots b , the vector e being arbitrary.

14

Chapter 1. Quaternions

Example. Find the cubic root of a

= 3+i+j+k

√ (i + j + k) ◦ ◦ √ sin 30 ; = 2 3 cos 30 + 3

Answer: b = θ

=

 √  13 (i + j + k) √ sin θ , 2 3 cos θ + 3 ◦ ◦ ◦ 10 , 130 , 250 .

1.4.4 Other functions and representations of quaternions The exponential ea is defined by [30] a a2 a3 + + + ··· 1! 2! 3! where a is an arbitrary quaternion. Furthermore, one defines ea = 1 +

a2 a4 a2p ea + e−a =1+ + + ···+ , 2 2! 4! (2p)! ea − e−a a a3 a5 a2p+1 sinh a = = + + + ···+ , 2 1! 3! 5! (2p + 1))!

cosh a =

and thus ea = cosh a + sinh a. For an arbitrary quaternion a, let U (V (a)) = u (u2 = −1) be a unit vector, and therefore one has ua = au ; consequently, one can define the trigonometric functions a2 a4 eua + e−ua =1− + + ··· , 2 2! 4! eua − e−ua a a3 a5 sin a = = − + + ··· . 2 1! 3! 5! Example. Let a = uθ be a quaternion without a scalar part with u ∈ Vec H, u2 = −1 and θ real; one has cos a =

cosh uθ sinh uθ

= =

cos θ, u sin θ,

euθ

=

cos θ + u sin θ,

cos uθ sin uθ

= =

cosh θ, u sinh θ.

In particular, if a = iθ, cosh iθ = cos θ,

sinh iθ = i sin θ,

cos iθ = cosh θ,

sin iθ = i sinh θ.

1.5. Classical vector calculus

15

One can represent a quaternion a = a0 + a1 i + a2 j + a3 k by a 2 × 2 complex matrix (with i
being the usual complex imaginary)

a0 + i
a3 −i
a1 + a2 A= −i
a1 − a2 a0 − i
a3 or by a 4 × 4 real matrix ⎡

a0 ⎢ a1 A=⎢ ⎣ a2 a3

−a1 a0 a3 −a2

−a2 −a3 a0 a1

⎤ −a3 a2 ⎥ ⎥. −a1 ⎦ a0

The differential of a product of quaternions is given by d(ab) = (da)b + a(db),

a, b ∈ H,

the order of the factors having to be respected.

1.5 Classical vector calculus 1.5.1 Scalar product and vector product Let a, b, c ∈ Vec H, be three quaternions without a scalar part, a = a1 i + a2 j + a3 k, b = b1 i + b2 j + b3 k, c = c1 i + c2 j + c3 k. The norm of a is  √ |a| = aac = a21 + a22 + a23 and ab + ba ab − ba + 2 2 = (−a · b, a × b).

ab =

One defines respectively the scalar product and the vector product of two vectors a, b by (ab + ba) = a1 b 1 + a2 b 2 + a3 b 3 , 2 (ab − ba) a×b≡a×b= 2 = (a2 b3 − a3 b2 )i + (a3 b1 − a1 b3 )j + (a1 b2 − a2 b1 )k,

(a, b) ≡ a · b = −

with ab = −(a, b) + a × b. Geometrically, one has (a, b) = |a| |b| cos α, |a × b| = |a| |b| sin α,

16

Chapter 1. Quaternions

α being the angle between the two vectors a and b. Furthermore, 2

2

2

|ab| = |a| |b| = (ab)(ab)c = [−(a, b) + a × b] [−(a, b) − a × b] = (a, b)2 − (a × b)2 2

= (a, b)2 + |a × b|

which is coherent with the above geometrical expressions. One has the properties (a, b) = (b, a), (λa, b) = λ(a, b),

λ ∈ R,

(a, b + c) = (a, b) + (a, c), a × b = −b × a, a × λb = λ(a × b), a × (b + c) = a × b + a × c.

1.5.2 Triple scalar and double vector products The triple scalar product is defined by [a(b × c) + (b × c)a] 2 [a(bc − cb) + (bc − cb)a] =− 2

(a, b × c) = −

and satisfies the relations (a, b × c) = (b, c × a) = (c, a × b), (a, b × c) = −(c, b × a) which are established using the relations S(abc) = S(bca) = S(cab) and S(abc) = S(abc)c = −S(cba) = −S(acb), S(abc) = −S(bac). In particular (a × b, a) = (a × b, b) = 0 which shows that a × b is orthogonal to a and b.

1.6. Exercises

17

To derive the expression of the double vector product a × (b × c), one can start from abc = a [−(b, c) + b × c] = −a(b, c) − (a, b × c) + a × (b × c) hence V (abc) = −a(b, c) + a × (b × c); since abc − (abc)c 2 abc + cba = 2 [abc + bac − bac − bca + bca + cba] = 2 (ab + ba)c b(ac + ca) (bc + cb)a = − + 2 2 2 = −(a, b)c + b(a, c) − a(b, c)

V (abc) =

one obtains a × (b × c) = b(a, c) − c(a, b). Knowing that (a × b) × c = −c × (a × b) = −a(c, b) + b(c, a), one notices that the vector product is not associative. From the above relations, one then obtains (with a, b, c, d ∈ Vec H) (a × b, c × d) = (a, c)(b, d) − (a, d)(b, c), (a × b) × (c × d) = c(a × b, d) − d(a × b, c).

1.6 Exercises E1-1 From the formulas i2 = j 2 = k 2 = ijk = −1, deduce the multiplication table of the quaternion group knowing that the element −1 commutes with all elements of the group and that (−1)i = −i, (−1)j = −j, (−1)k = −k. E1-2 Consider the quaternions a = 1 + i, b = 4 − 3j. Compute , |b|, a−1 , b−1 , √ √ |a| 1/3 a + b, ab, ba. Give the polar form of a and b. Compute a, b, a . E1-3 Solve in x ∈ H, the equation ax + xb = c (a, b, c ∈ H, aac = bbc ). N.A. : a = 2i, b = j, c = k, determine x. E1-4 Solve in x ∈ H, the equation axb + cxd = e (a, b, c, d, e ∈ H). N.A. : a = 2i, b = j, c = k, d = i, e = 3k. Find x. E1-5 Solve in x ∈ H, the equation x2 = xa + bx (a, b ∈ H). N.A. : a = −2j, b = −k, determine x.

Chapter 2

Rotation groups SO(4) and SO(3) In this chapter, the formulas of the rotation groups SO(4) and SO(3) are established from orthogonal symmetries. The crystallographic groups and Kepler’s problem are then examined as applications of these groups.

2.1 Orthogonal groups O(4) and SO(4) Consider two elements of a four-dimensional vector space x = x0 + x1 i + x2 j + x3 k, y = y0 + y1 i + y2 j + y3 k ∈ H, and the scalar product (x, x) = xxc = x20 + x21 + x22 + x23 . One deduces from it (x + y, x + y) = (x, x) + (y, y) + (x, y) + (y, x) and postulating the relation (x, y) = (y, x), one obtains 1 [(x + y, x + y) − (x, x) − (y, y)] 2 1 = [(x + y)(x + y)c − xxc − yyc ] 2 1 = (xyc + yxc ) 2 = x0 y0 + x1 y1 + x2 y2 + x3 y3 .

(x, y) =

Two quaternions x, y are orthogonal if (x, y) = 0; a quaternion is unitary if (x, x) = 1. A hyperplane is defined by the relation (a, x) = 0 where a is a quaternion perpendicular to the hyperplane. The expression of a plane symmetry is obtained as follows. Definition 2.1.1. The symmetric x
of x with respect to a hyperplane is obtained by drawing from x the perpendicular down to the hyperplane and by extending this perpendicular line by an equal length [11].

20

Chapter 2. Rotation groups SO(4) and SO(3)

We shall assume that the hyperplanes go through the origin. The vector x
− x is perpendicular to the hyperplane (and thus parallel to a) and (x
+ x)/2 is perpendicular to a. Hence, the relations x
= x + λa,   x
+ x a, = 0; 2 one then deduces

  λa a, x + = 0, 2

λ ∈ R,

−2(a, x) , (a, a) 2(a, x)a x
= x − (a, a) (axc + xac )a =x− aac axc a =− . aac λ=

Theorem 2.1.2. Any rotation of O(n) is the product of an even number ≤ n of symmetries; any inversion is the product of an odd number ≤ n of symmetries [11]. A rotation is a proper transformation of a determinant equal to 1; an inversion is an improper transformation of a determinant equal to −1. In combining, in an even number, plane symmetries x
= −mxc m, with mmc = 1, one obtains the rotation group SO(4), x
= axb, with a, b ∈ H and aac = bbc = 1. By including the inversions (odd number of symmetries) the expression of which are x
= −axc b with a, b ∈ H and aac = bbc = 1, one obtains the orthogonal group O(4) with six parameters. This group, by definition, conserves the scalar product; indeed, for SO(4), 1

[x yc + y
x
c ] 2 1 = [axbbc yac + aybbc xac ] 2 = (x, y)

(x
, y
) =

2.1. Orthogonal groups O(4) and SO(4)

21

and for the improper rotations 1

[x yc + y
x
c ] 2 1 = [axc bbc yac + ayc bbc xac ] 2 = (x, y).

(x
, y
) =

Any rotation of SO(4) can be written as a combination of a rotation of SO(3), examined below, rrc = 1 x = rxrc , and a transformation x
= axa,

a ∈ H, aac = 1.

It is sufficient to resolve the equation x
= f xg = arxrc a

(or r
a
x
a
rc
)

with f = ar, g = rc a and rrc = aac = 1. Writing the relations ([15], [16]) 2a2 = 2f g, a2 f g = (f g)2 , a2 (f g)c = 1, and adding, one obtains a2 [2 + f g + (f g)c ] = (1 + f g)2 ; hence, the solution a=

±(1 + f g) . |(1 + f g|

The rotation is given by r = ac f (1 + gc fc ) f |(1 + f g| ±(f + gc ) = . |(1 + f g| =±

One verifies that one has indeed the relations aac = rrc = 1. One solves similarly the equations f = r
a
, g = a
rc
with the solutions ±(1 + gf ) , |(1 + gf | ±(f + gc ) r
= , |(1 + gf |

a
=

with |(1 + f g| = |(1 + gf | since S(gf ) = S(f g).

22

Chapter 2. Rotation groups SO(4) and SO(3)

2.2 Orthogonal groups O(3) and SO(3) Consider the vectors x = x1 i + x2 j + x3 k, x
= x
1 i + x
2 j + x
3 k ∈ H, constituting a subvectorspace of H. A plane symmetry, in this subspace, is a particular case of the preceeding one and is expressed by x
= −axc a = axc ac with a ∈ Vec H, aac = 1, ac = −a. The improper rotations (odd number ≤ 3 of plane symmetries) are given by x
= f xc fc with f ∈ H, f fc = 1 and one has x
x
c = xxc . The SO(3) group is the set of proper rotations (even number ≤ 3 of symmetries). In combining two symmetries, one obtains x

= −axc a, x
= −bx

c b, hence x
= (bac )x(ac b) = rxrc with r (= bac ) ∈ H, rc = abc = ac b, rrc = 1. The unitary quaternion r can be expressed in the form   θ θ r = cos + u sin 2 2 where the unit vector u (u2 = −1) is the axis of rotation (going through the origin) and θ the angle of rotation of the vector x around u (θ is taken algebraically given the direction of u and using the right-handed screw rule). The conservation of the norm of x results from x
x
c = rxrc rxc rc = xxc . Furthermore, if one considers the transformation q
= rqrc with q ∈ H, one has S(q
) = S(rqrc ) = S(rc rq) = S(q)

2.2. Orthogonal groups O(3) and SO(3)

23

which shows that the scalar part of the quaternion is not affected by the rotation. The set of proper and improper rotations constitute the group O(3). In developing the formula x
= rxrc with x = x ∈ Vec H, one obtains     θ θ θ θ

x = x = cos + u sin x cos − u sin 2 2 2 2 θ θ = cos2 x + sin2 [(u · x)u − (u × x)u] + u × x sin θ; 2 2 furthermore (u × x)u = −(u × x) · u + (u × x) × u = (u × x) × u = x − (u · x)u; hence, the classical formula [26, p. 165] x
= x cos θ + u(u · x)(1 − cos θ) + u × x sin θ. In matrix form, this equation can be written x
= Ax with ⎡ ⎤ ⎤ ⎡ 0 0 ⎢ x
1 ⎥ ⎢ x1 ⎥
⎢
⎥, ⎥, x = x=⎢ ⎣ x2 ⎦ ⎣ x2 ⎦ x3 x
3      ⎤ ⎡ 2 u1 u1 u2 (1 − cos θ) u1 u3 (1 − cos θ) ⎢ +(u22 + u23 ) cos θ ⎥ −u3 sin θ +u2 sin θ ⎢ ⎥ ⎢    2    ⎥ ⎢ ⎥ u u (1 − cos θ) u (1 − cos θ) u u 1 2 2 3 2 ⎥; A=⎢ 2 2 ⎢ ⎥ +u −u sin θ +(u + u ) cos θ sin θ 3 1 1 3 ⎢ ⎥ ⎢  ⎥      ⎣ ⎦ u u (1 − cos θ) u2 u u (1 − cos θ) 1 3

−u2 sin θ

2 3

+u1 sin θ

3

+(u21 + u22 ) cos θ

one verifies that the matrix A is orthogonal (t A = A−1 ). If one combines the rotation r1 and the rotation r2 ,    α β α β , r2 = cos + b sin r1 = cos + a sin 2 2 2 2  γ γ r = r2 r1 = cos + c sin 2 2

one obtains with

α β α β γ = cos cos − (a, b) sin sin , 2 2 2 2 2 α β α β α β γ c sin = a sin cos + b cos sin − (a × b) sin sin , 2 2 2 2 2 2 2 cos

24

Chapter 2. Rotation groups SO(4) and SO(3)

which yields the Rodriguez formula c tan

a tan α2 + b tan β2 − (a × b) tan α2 tan β2 γ = . 2 1 − (a, b) tan α2 tan β2

2.3 Crystallographic groups If r belongs to a finite subgroup of real quaternions, the transformations q
= rqrc will constitute a subgroup of SO(3), with r and −r generating the same rotation. The finite subgroups of real quaternions ([48], [44]) are of five types.

2.3.1 Double cyclic groups Cn (order N = 2n) The elements of these groups are given by  π  2h 2h π n r = u n = cos + u sin 2 2  π h π = cos + u sin = bh n n with b = cos πn + u sin πn , the axis being oriented according to the unit vector u, with h = 1, . . . , 2n. If the rotation axis is oriented along Oz, one simply has  π π  2h 2h n r = k n = cos + k sin 2 2  π h π = cos + k sin . n n Example. Double group C3 (N = 6, rotation axis along Oz); the elements of the group are  π h π r = bh = cos + k sin , h = 1, . . . , 6 3 3 or explicitly ⎧ ⎫ √ √ ⎨ b = 1 (1 + k 3), b2 = 1 (−1 + k 3), b3 = −1, ⎬ 2 2 . ⎩ ⎭ b4 = −b, b5 = −b2 , b6 = 1

2.3.2 Double dihedral groups Dn (N = 4n) These groups are constituted by the elements 2h

r = u n al = b h al with

 π π u = cos + u sin , 2 2

 π π a = cos + a sin , 2 2

2

b = un,

2.3. Crystallographic groups

25

where u, a are two perpendicular vectors and S(au) = 0, a2 = −1, h = 1, . . . , 2n, l = 1, . . . , 4. 1. Double group D3 (order 12)  √ π π 1 = (1 + k 3), b = cos + k sin 3 3 2 √ √ 1 1 a = i, ab = (i − j 3), ba = (i + j 3); 2 2 
the elements of the group are ±b, ±b2 , ±b3 , ±a, ±ab, ±ba .

Examples.

2. Double group D4 (order 16); writing  π π 1 , b2 = (k), b = cos + k sin b3 = √ (−1 + k), 4 4 2 1 1 a = (i), ab = √ (i − j), ba = √ (i + j), 2 2 ab2 = (−j), a2 = (−1), a3 = (−i), a4 = 1, the group is constituted by the elements
 ±b, ±b2 , ±b3 , ±b4 , ±a, ±ab, ±ba, ±ab2 .

2.3.3 Double tetrahedral group (N = 24) This group is constitued by the 24 elements ±1, ±i, ±j, ±k, 1 (±1 ± i ± j ± k). 2 More explicitly, indicating the axes of multiple rotations, the group is composed of the elements iα , j α , k α ,  β 1+i+j+k , 2  β 1+i−j−k , 2

 

1−i−j+k 2 1−i+j−k 2

β , β ,

with α = 1, 2, 3, 4, β = 1, 2, 3, 4, 5, 6 (N = 24). Example. Consider the tetraeder having as vertices i+j+k √ , 3

−i − j + k √ , 3

i−j−k √ , 3

−i + j − k √ , 3

26

Chapter 2. Rotation groups SO(4) and SO(3)

as face centers − (i + j + k) √ , 3

−

(−i − j + k) √ , 3

−

(i − j − k) √ , 3

−

(−i + j − k) √ , 3

and as side centers ±i,

±j,

±k;

by taking for r the elements of the above group, the transformation x
= rxrc transforms the tetraeder in itself.

2.3.4 Double octahedral group (N = 48) The group is composed by the 48 elements ± 1, ±i, ±j, ±k, 1 (±1 ± i ± j ± k), 2 1 1 √ (±1 ± i) , √ (±1 ± j) , 2 2 1 1 √ (±i ± j) , √ (±j ± k) , 2 2

1 √ (±1 ± k) , 2 1 √ (±i ± k) . 2

Making explicit the axes of multiple rotations, the elements of the group are 

α



1+j √ 2

α



α 1+k √ , , , 2   β β 1−i−j+k 1+i+j+k , , 2 2   β β 1−i+j−k 1+i−j−k , , 2 2    γ γ γ i+k i+j j+k √ √ √ , , , 2 2 2    γ γ γ −i + k i−j j−k √ √ √ , , , 2 2 2 1+i √ 2

with α = 1, . . . , 8, β = 1, . . . , 6, γ = 1, . . . , 4 (N = 48). Example. Consider the octaeder having, in an orthonormal frame, for its 6 vertices the coordinates ±i, ±j, ±k, and for the centers of the 8 faces ±

i+j+k √ , 3

±

−i − j + k √ , 3

±

i−j−k √ , 3

±

−i + j − k √ , 3

2.3. Crystallographic groups

27

for the middle points of the 12 sides j±k ± √ , 2

±

(±i + k) √ , 2

i±j ± √ . 2

The octaeder transforms into itself under the rotation x
= rxrc , r being taken in the double octahedral group. The same property applies to the cube (dual of the octaeder) the 8 vertices of which are the centers of the faces of the above octaeder.

2.3.5 Double icosahedral group (N = 120) The 120 elements of this group are iα , j α , k α ,  α i + m
j + mk , 2   α i − m
j + mk , 2  α i + m
j − mk , 2  α i − m
j − mk , 2  β 1+i+j+k , 2  β 1+i−j−k , 2  β 1 + mj + m
k , 2  β 1 + mj − m
k , 2  γ m + m
j + k , 2   γ m + m
j − k , 2




α α mi + j + m
k m i + mj + k , , 2 2     α α −mi + j − m
k m
i − mj + k , , 2 2   α α mi + j − m
k −m
i − mj + k , , 2 2   α α −mi + j + m
k −m
i + mj + k , , 2 2  β 1−i−j+k , 2  β 1−i+j−k , 2   β β 1 + m
i + mk 1 + mi + m
j , , 2 2   β β 1 − m
i + mk 1 + mi − m
j , , 2 2   γ γ m + i + m
k m + m
i + j , , 2 2     γ γ m − i + m
k m + m
i − j , , 2 2

with α = 1, . . . , 4, β = 1, . . . , 6, γ = 1, . . . , 10 and √ 1+ 5 = 2 cos 36◦ , m= 2√ 1− 5 = −2 cos 72◦ . m
= 2

28

Chapter 2. Rotation groups SO(4) and SO(3)

One obtains another group, distinct from the first, by inverting m and m
. Example. Consider the icosaeder having, in an orthonormal frame, for the coordinates of the 12 vertices ±

m
j ± k √ , m
5

±

(±i + m
k) √ , m
5

±

m
i ± j √ ; m
5

for the centers of the 20 faces ±

i+j+k −i − j + k i−j−k −i + j − k √ √ √ , ± , ± √ , ± , 3 3 3 3 mj ± m
k (±m
i + mk) mi ± m
j √ √ ± , ± , ± √ ; 3 3 3

for the middles of the 30 sides ±i, ±

±j,

±k,

(±mi + j ± m
k) , 2

i ± m
j ± mk , 2
(±m i ± mj + k) ± . 2 ±

The transformation x
= rxrc transforms the icosaeder into itself; the same is true for its dual, the dodecaeder (having 20 vertices, 12 faces and 30 sides) the vertices of which are the centers of the faces of the icosaeder. The five groups (cyclic, dihedral, tetrahedral, octahedral, icosahedral) above, combined with the rotations x
= rxrc and the parity operator x
= xc = −x, generate the set of the 32 crystallographic groups([44], [43]).

2.4 Infinitesimal transformations of SO(4) Among the subgroups of SO(4) one has, in particular, the transformations x
= rxrc ,

x
= axa,

x
= f x,

x
= xg

with r, a, f, g ∈ H, rrc = aac = f fc = ggc = 1, x = (x0 , x) and x
= (x
0 , x
) ∈ H. For an infinitesimal rotation of SO(3) of dθ around the unit vector u = u1 i + u2 j + u3 k, one has     dθ dθ dθ + u sin r = cos  1+u 2 2 2 and

    dθ dθ x = rxrc = 1 + u x 1−u 2 2 dθ (ux − xu) = x + dθ u × x; =x+ 2


2.4. Infinitesimal transformations of SO(4) hence

29

dx = x
− x = dθ u × x.

In matrix notation, one obtains i ∈ (1, 2, 3)

dx = dθui Mi x, ⎤ x0 ⎢ x1 ⎥ ⎥ x=⎢ ⎣ x2 ⎦ , x3 ⎡

with

and

⎡

0 ⎢0 M1 = ⎢ ⎣0 0

0 0 0 0

⎡

⎤ 0 0 0 0 ⎥ ⎥, 0 −1 ⎦ 1 0

⎡

⎤ x
0 ⎢ x
1 ⎥ ⎥ x
= ⎢ ⎣ x
2 ⎦ , x
3

0 0 ⎢0 0 M2 = ⎢ ⎣0 0 0 −1

0 0 0 0

⎤ 0 1⎥ ⎥, 0⎦ 0

⎡

0 ⎢0 M3 = ⎢ ⎣0 0

⎤ 0 0 0 0 −1 0 ⎥ ⎥. 1 0 0⎦ 0 0 0

Introducing the commutator of two matrices A and B, [A, B] = AB − BA, one verifies the relations [M1 , M2 ] = M3 , [M1 , M3 ] = −M2 , [M2 , M3 ] = M1 , or [Mi , Mj ] = ijk Mk .


(2.1)

Concerning the subgroup x = axa, with a ∈ H and aac = 1, one has for the infinitesimal transformation     dβ dβ dβ + v sin a = cos  1+v ; 2 2 2 hence x
= axa = = x+ and

 1+v

dβ 2

   dβ x 1+v 2

dβ (vx + xv) = x + dβ (−v · x + x0 v) 2

dx = x
− x = dβ (−v · x + x0 v) .

30

Chapter 2. Rotation groups SO(4) and SO(3)

In matrix notation, one obtains i ∈ (1, 2, 3)

dx = dβv i Ni x, with

⎡

⎡

⎡

0 −1 0 0 ⎢1 0 0 0 N1 = ⎢ ⎣0 0 0 0 0 0 0 0

⎤ x0 ⎢ x1 ⎥ ⎥ x=⎢ ⎣ x2 ⎦ , x3 ⎡ ⎤ 0 ⎢0 ⎥ ⎥ , N2 = ⎢ ⎣1 ⎦ 0

0 0 0 0

x
0 ⎢ x
1 x
= ⎢ ⎣ x
2 x
3 ⎤ −1 0 0 0⎥ ⎥, 0 0⎦ 0 0

⎤ ⎥ ⎥, ⎦ ⎡

0 ⎢0 N3 = ⎢ ⎣0 1

0 0 0 0

⎤ 0 −1 0 0 ⎥ ⎥; 0 0 ⎦ 0 0

the matrices Ni satisfy the relations [N1 , N2 ] = M3 , [N1 , N3 ] = −M2 , [N2 , N3 ] = M1 , or, more concisely [Ni , Nj ] = ijk Mk

(2.2)

where Mi are the matrices defined previously; furthermore [Ni , Mj ] = ijk Nk .

(2.3)

Concerning the subgroup x
= f x, one obtains with f  1+ dα 2 f , f = f1 i+f2 j +f3 k (unit vector) dα [−f · x + x0 f + f × x] . dx = x
− x = 2 In matrix notation, one has dx = dαf i Fi x with

⎡

0 1⎢ 1 F1 = ⎢ 2⎣ 0 0

−1 0 0 0

⎤ 0 0 0 0 ⎥ ⎥, 0 −1 ⎦ 1 0 ⎡ 0 1⎢ 0 F3 = ⎢ 2⎣ 0 1

⎡

0 0 1 0

⎤ 0 0 −1 0 1⎢ 0 0 0 1 ⎥ ⎥, F2 = ⎢ ⎣ 1 0 0 0 ⎦ 2 0 −1 0 0 ⎤ 0 −1 −1 0 ⎥ ⎥ 0 0 ⎦ 0 0

2.4. Infinitesimal transformations of SO(4)

31

and [F1 , F2 ] = F3 , [F1 , F3 ] = −F2 , [F2 , F3 ] = F1 , or [Fi , Fj ] = ijk Fk . As to the subgroup x
= xg, with g  1 + dβ 2 g, one proceeds similarly and obtains dx = x
− x =

dβ [−g · x + x0 g − g × x] . 2

In matrix notation, dx = dβg i Gi x with ⎡

0 1⎢ 1 G1 = ⎢ 2⎣ 0 0

−1 0 0 0 0 0 0 −1

⎤ 0 0 ⎥ ⎥, 1 ⎦ 0 ⎡

0 1⎢ 0 G3 = ⎢ 2⎣ 0 1

⎡

0 1⎢ 0 G2 = ⎢ 2⎣ 1 0 0 0 −1 0

0 0 0 1

⎤ 0 −1 1 0 ⎥ ⎥ 0 0 ⎦ 0 0

and [G1 , G2 ] = −G3 , [G1 , G3 ] = G2 , [G2 , G3 ] = −G1 , or [Gi , Gj ] = − ijk Gk . Furthermore, one has [Fi , Gj ] = 0,

i, j = 1, 2, 3

the matrices Mi , Ni , Fi , Gi satisfying the relations Mi = Fi − Gi ,

Ni = Fi + Gi .

⎤ −1 0 0 −1 ⎥ ⎥, 0 0 ⎦ 0 0

32

Chapter 2. Rotation groups SO(4) and SO(3)

2.5 Symmetries and invariants: Kepler’s problem Let us consider the motion of a particle of mass m, of momentum p and gravitating at the distance r of a mass M . The Hamiltonian is given by H=

k p2 − , 2m r

k = GM m

and the equations of motion are q˙i =

∂H , ∂pi

p˙i =

−∂H . ∂qi

Let F (qi , pi , t) be a physical quantity of the motion; one has dF ∂F ∂F ∂qi ∂F ∂pi = + + dt ∂t ∂qi ∂t ∂pi ∂t

∂F ∂Hi ∂F ∂F ∂H = + − . ∂t ∂qi ∂pi ∂pi ∂qi Introducing Poisson’s bracket of two functions u and v [u, v] = one obtains

∂u ∂v ∂u ∂v − ∂qi ∂pi ∂pi ∂qi

∂F dF = + [F, H] . dt ∂t

dF If [F, H] = 0 and ∂F ∂t = 0, one has dt = 0, F is then an invariant of the motion. The angular momentum L = r × p and the Laplace-Runge-Lenz vector

A= p×L−

kmr r

satisfy the equations [L, H] = 0, [A, H] = 0, and thus are invariants. Let us consider a bound motion (with a total negative energy E < 0) and introduce the vector A D= √ −2mE with E=

p2 k − 2m r

2.5. Symmetries and invariants: Kepler’s problem

33

verifying the Poisson bracket [D, H] = 0; one has the relations [Li , Lj ] = ijk Lk , [Di , Dj ] = ijk Lk ,

(2.4) (2.5)

[Di , Lj ] = ijk Dk .

(2.6)

The relations (2.4), (2.5), (2.6) are respectively the same as those (2.1), (2.2), (2.3) concerning the infinitesimal transformations of SO(3) and the transformation q
= aqa of SO(4), which indicates that the symmetry group of the problem is SO(4). To see it more explicitly, let us develop A in the form  

km 2 A= r p − − p (p · r) r with

A = rp0 − pr0 D= √ −2mE

and

p2 − km p·r r p0 = √ , r0 = √ . −2mE −2mE One verifies immediately that D · L = 0; furthermore, 2

2

2

(L) = (r × p) = r2 p2 − (r · p) , hence 2

2

(D) + (L) = or H=

k 2 m2 −2mE

−k 2 m2  . 2 2 2 (D) + (L)

Introducing two quaternions r = (r0 , r),

p = (p0 , p)

and the quaternion K =r∧p=

1 (rpc − prc ) = (0, D − L) 2

one has 2

KKc = |0, D − L| = D2 + L2 , and thus H=

−k 2 m 2

2 |r ∧ p| which shows explicitly the invariance of the Hamiltonian with respect to a transformation of SO(4) of the type K
= aKb (a, b ∈ H, aac = bbc = 1) leading to K
Kc
= KKc .

34

Chapter 2. Rotation groups SO(4) and SO(3)

2.6 Exercises E2-1 From the general formula


x = rxrc ,

  θ θ r = cos + u sin 2 2

x ∈ Vec H, give in an orthonormal direct frame the matrix representation X
= AX of a rotation of angle α around the Ox axis of a point M (x, y, z), of a rotation of angle β around the Oy axis, of a rotation of angle γ around the Oz axis. Show that the matrices are orthogonal A−1 = At , det A = 1 (At : transposed matrix of A). E2-2 Consider the relation A = rA
rc where A are the components of a vector with respect to an orthonormal frame at rest and A
its components with respect to a mobile orthonormal frame. Show that dA = rDA
rc where DA
is the covariant differential with DA
= dA
+ dΩ
× A
, dΩ
= 2rc dr (dA
is the differential with respect to the components only). What does dr dΩ
= 2rc dt dt represent? E2-3 Consider the relations A = rA
rc = gA

gc with dA


= rDA
rc = gDA

gc ,

DA

= dA
+ dΩ
× A


DA



= dA

+ dΩ × A





(dΩ
= 2rc dr),



(dΩ

= 2gc dg)

(A represents the components of a vector with respect to an orthonormal basis at rest, A
, A

the components with respect to mobile orthonormal bases.). Find the

relation between dΩ and dΩ
. E2-4 Using the covariant derivative, express the velocity and the acceleration in polar coordinates (ρ, θ) and in cylindrical coordinates (ρ, θ, z). E2-5 Express r in the relation X = rX
rc for spherical coordinates with X = xi + yj + zk and X
= ρi. Determine dΩ
= 2rc dr and dΩ = 2(dr)rc . Express the basis vectors ei in the basis at rest. Find the velocity and the acceleration in the mobile basis.

2.6. Exercises

35

E2-6 Consider the Euler basis (O, x
, y
, z
) obtained via the following successive rotations (Figure 2.1). A first rotation of angle α (precession angle) around k transforms the basis i, j, k into the basis i
, j
, k
. A second rotation of angle β (nutation angle) around the vector i
transforms i
, j
, k
into i

, j

, k

. A third rotation of angle γ (proper rotation angle) around the vector k

transforms the basis i

, j

, k

into the basis e1 , e2 , e3 = k

. Give the quaternion r of the rotation X = rX
rc . Determine ω
= 2rc

dr , dt

ω=2

dr rc . dt

Give the components of the basis vectors ei . z y


z
β k

k

= e3

e2

j

e1 i


i

γ

y x


α x

Figure 2.1: Euler’s angles: α is the angle of precession, β the angle of nutation and γ the angle of proper rotation.

Chapter 3

Complex quaternions From the very beginning of special relativity, complex quaternions have been used to formulate that theory [45]. This chapter establishes the expression of the Lorentz group using complex quaternions and gives a few applications. Complex quaternions constitute a natural transition towards the Clifford algebra H ⊗ H.

3.1 Algebra of complex quaternions H(C) A complex quaternion is a quaternion a = a0 + a1 i + a2 j + a3 k having complex coefficients. Such a quaternion can be represented by the matrix A=

a0 + i
a3 −i
a1 − a2

−i
a1 + a2 a0 − i
a3



with the basis 1=

1 0

0 1



,

i=

0 −i


−i
0



,

j=

0 1 −1 0



,

k=

i
0

0 −i




where i
is the usual complex imaginary and ai ∈ C. The algebra of complex quaternions H(C) is isomorphic to 2 × 2 matrices over C and has zero divisors; indeed, the relation (1 + i
k)(1 − i
k) = 1 − (i
)2 k 2 = 0 shows that the product of two complex quaternions can be equal to zero without one of the complex quaternions being equal to zero.

38

Chapter 3. Complex quaternions

3.2 Lorentz groups O(1, 3) and SO(1, 3) 3.2.1 Metric With the advent of the special theory of relativity, space and time have been united into a four-dimensional pseudoeuclidean spacetime with the relativistic invariant

2 2 2 c2 t2 − x1 − x2 − x3 . Minkowski had the idea of using complex quaternions x = (i
ct, x1 , x2 , x3 ) = (i
ct + x) with the invariant

2 2 2 xxc = −c2 t2 + x1 + x2 + x3 .

To get the signature (+ − −−), we shall use complex quaternions in the form x = (ct, i
x1 , i
x2 , i
x3 ) = (ct + i
x) which we shall call minkowskian quaternions or minquats with the invariant

2 2 2 xxc = c2 t2 − x1 − x2 − x3 . A minquat is a complex quaternion such that x∗c = x where ∗ is the complex conjugation.

3.2.2 Plane symmetry Let us consider two minquats x = (x0 + i
x), y = (y 0 + i
y) with the scalar product

2 2 2 2 xxc = x0 − x1 − x2 − x3 . One has (x + y, x + y) = (x, x) + (y, y) + (x, y) + (y, x); postulating the property (x, y) = (y, x) one obtains 1 [(x + y, x + y) − (x, x) − (y, y)] 2 1 = [(x + y)(x + y)c − xxc − yyc ] 2 1 = (xyc + yxc ) 2 = x0 y 0 − x1 y 1 − x2 y 2 − x3 y 3 .

(x, y) =

Two minquats x, y are said to be orthogonal if (x, y) = 0. A minquat x is timelike if xxc > 0 and unitary if xxc = 1; a minquat x is spacelike if xxc < 0 and unitary if xxc = −1. A hyperplane is defined by the relation (a, x) = 0 where a is a minquat perpendicular to the hyperplane.

3.2. Lorentz groups O(1, 3) and SO(1, 3)

39

Definition 3.2.1. The symmetric x
of x with respect to a hyperplane x is obtained by drawing the perpendicular to the hyperplane and by extending this perpendicular by an equal length [11]. The hyperplanes are supposed to go through the origin. The vector x
− x is perpendicular to the hyperplane and thus parallel to a; furthermore, the vector (x
+ x)/2 is perpendicular to a. One has the relations x
= x + λa,   x
+ x a, = 0; 2

λ ∈ R,

hence,   λa a, x + = 0, 2 −2(a, x) , λ= (a, a) (axc + xac )a 2(a, x)a =x− , x
= x − (a, a) aac axc a x
= − . aac One shall distinguish time symmetries (with aac = 1) x
= −axc a from space symmetries (aac = −1) x
= axc a.

3.2.3 Groups O(1, 3) and SO(1, 3) Definition 3.2.2. The pseudo-orthogonal group O(1, 3) is the group of linear operators which leave invariant the quadratic form (x, y) = x0 y 0 − x1 y 1 − x2 y 2 − x3 y 3 with x = (x0 + i
x), y = (y 0 + i
y). Theorem 3.2.3. Any rotation of O(1, 3) is the product of an even number ≤ 4 of symmetries; any inversion is the product of an odd number ≤ 4 of symmetries [11]. A rotation is a proper transformation of a determinant equal to 1; an inversion is an improper transformation of a determinant equal to −1.

40

Chapter 3. Complex quaternions

Proper orthochronous Lorentz transformation Consider the transformation obtained by combining an even number of time symmetries and an even number of space symmetries. These transformations of determinant equal to +1 constitute a subgroup of O(1, 3), the special orthogonal group SO(1, 3). Let us take for example two time symmetries (minquats f and g) followed by two space symmetries (minquats m and n). One has  x
= n m [g (f xc f )c g]c m c n = (nmc gfc )x(fc gmc n); writing a = nmc gfc , one obtains ac = f gc mnc , a∗c = f ∗ gc∗ m∗ n∗c

aac = 1

= fc gmc n; hence, the formula is valid in the general case, x
= axa∗c with aac = 1, a ∈ H(C). Other Lorentz transformations Call n the number of time symmetries and p the number of space symmetries. In combining these symmetries, one obtains the following Lorentz transformations L: 1. n even, p odd, orthochronous, improper Lorentz transformation (det L = −1) x
= axc a∗c

(aac = −1);

2. n odd, p odd, antichronous, proper Lorentz transformation(det L = 1) x
= −axa∗c

(aac = −1);

3. n odd, p even, antichronous, improper Lorentz transformation (det L = −1) x
= −axc a∗c

(aac = 1).

Group O(1, 3): summarizing table The whole set of Lorentz transformations is given in the table below where n is the number of time symmetries and p the number of space symmetries [49].

3.3. Orthochronous, proper Lorentz group

orthochronous

antichronous

Rotation (det L = 1) n even, p even x
= axa∗c (aac = 1) n odd, p odd x
= −axa∗c (aac = −1)

41

Inversion (det L = −1) n even, p odd x
= axc a∗c (aac = −1) n odd, p even x
= −axc a∗c (aac = 1)

3.3 Orthochronous, proper Lorentz group 3.3.1 Properties The proper orthochronous Lorentz transformation x
= axa∗c with aac = 1, x = (ct + i
x), x
= (ct
+ i
x
) conserves, by definition, the norm xx
c = (axa∗c )(a∗ xc ac ) = xxc and the minquat type of x ∗ ∗ ∗ ∗ x
∗ c = (axac )c = axc ac

= axa∗c = x
. The composition of two transformations satisfies the rule x
= a2 (a1 xa∗1c )a∗2c = a3 xa∗3c with a3 = a2 a1 , a3 a3c = 1, a3 ∈ H(C). A (three-dimensional) rotation is given by the formula x
= rxrc

 with r = cos θ2 + u sin θ2 , rrc = 1, r ∈ H. A pure Lorentz transformation (without rotation) corresponds to the transformation x
= bxb∗c 

where b = cosh ϕ2 + i
v sinh ϕ2 is a minquat such that bbc = 1 and i
v is a unitary space-vector (i
v, i
v) = −1. A general transformation (proper, orthochronous) is

42

Chapter 3. Complex quaternions

obtained by combining a rotation and a pure Lorentz transformation. This can be done in two ways, x
= b(rxrc )b∗c , x

= r(bxb∗c )rc , br being in general distinct from rb. Reciprocally, a general Lorentz transformation can be decomposed into a pure Lorentz transformation and a rotation. The problem simply consists to resolve the equation a = br (or r
b
) where b is a unitary minquat (bbc = 1) and r a real unitary quaternion (rrc = 1). The equation a = br is solved in the following way ([15], [16]). Since a∗c = rc∗ b∗c = rc b one obtains aa∗c = b2 ; let us write d = aa∗c , with ddc = aa∗c a∗ ac = 1, hence, the equations 2b2 = 2d, b2 d = d2 , b2 dc = 1 and their sum b2 (2 + d + dc ) = (1 + d)2 . The solution therefore is ±(1 + d) |1 + d| ±(1 + aa∗c ) = |1 + aa∗c |

b=

with |1 + d| =

 (1 + d)(1 + d)c . The rotation is given by r = bc a =

±(a + a∗ ) . |1 + aa∗c |

Finally, one verifies that this is indeed a solution. For the equation a = r
b
, one finds in a similar way the solution ±(1 + a∗c a) , |1 + a∗c a| ±(a + a∗ ) r
= |1 + a∗c a| b
=

3.3. Orthochronous, proper Lorentz group

43

with |1 + a∗c a| = |1 + aa∗c | since S(aa∗c ) = S(a∗c a). One observes that in both cases (a = br or r
b
) the rotation is the same. The problem of the decomposition of a Lorentz transformation into a pure Lorentz transformation and a rotation is thus solved in the most general case. As an immediate application, consider the combination of two pure Lorentz transformations b1 , b2 . The quaternion b = b2 b1 will in general be a complex quaternion (and not a minquat) and will be written a = br. The resulting Lorentz transformation will thus contain a rotation; this is the principle of the Thomas precession ([52], [51]).

3.3.2 Infinitesimal transformations of SO(1, 3) Consider the pure Lorentz transformation x
= bxb∗c 

with x = (ct + i
x), x
= (ct
+ i
x
), b = cosh ϕ2 + i
v sinh ϕ2 and i
v a spacelike unitary minquat. For an infinitesimal transformation,   dϕ dϕ
+ i v sinh b  cosh 2 2   dϕ = 1 + i
v , 2 hence x
= bxb∗c     dϕ dϕ = 1 + i
v x 1 + i
v 2 2 dϕ (vx + xv) ; = x + i
2 consequently, dx = x
− x = dϕ (v · x + i
vx0 ) . Using real matrices,

⎡

⎤ x0 = ct ⎢ ⎥ x1 ⎥, X=⎢ 2 ⎣ ⎦ x 3 x

⎡

⎤ x
0 = ct
⎢ ⎥ x
1 ⎥ X
= ⎢
2 ⎣ ⎦ x
3 x

one can write dX = X
− X = dϕv i Ki X

44 with

Chapter 3. Complex quaternions ⎡

0 ⎢1 K1 = ⎢ ⎣0 0

1 0 0 0

0 0 0 0

⎤ 0 0⎥ ⎥, 0⎦ 0

⎡

0 ⎢0 K2 = ⎢ ⎣1 0

0 0 0 0

1 0 0 0

⎤ 0 0⎥ ⎥, 0⎦ 0

⎡

0 ⎢0 K3 = ⎢ ⎣0 1

0 0 0 0

0 0 0 0

⎤ 1 0⎥ ⎥. 0⎦ 0

The matrices Ki satisfy the relations

where Mi are in Chapter 2, ⎡ 0 ⎢0 M1 = ⎢ ⎣0 0

[Ki , Kj ] = − ijk Mk ,

(3.1)

[Ki , Mj ] = ijk Kk ,

(3.2)

the matrices defined for the infinitesimal transformations of SO(3) 0 0 0 0

⎤ 0 0 0 0 ⎥ ⎥, 0 −1 ⎦ 1 0

⎡

0 0 ⎢0 0 M2 = ⎢ ⎣0 0 0 −1

0 0 0 0

⎤ 0 1⎥ ⎥, 0⎦ 0

⎡

0 ⎢0 M3 = ⎢ ⎣0 0

⎤ 0 0 0 0 −1 0 ⎥ ⎥. 1 0 0⎦ 0 0 0

One observes that the relations (3.1) and (3.2) are those of the unbound Kepler problem, which identifies the corresponding symmetry group as being SO(1, 3).

3.4 Four-vectors and multivectors in H(C) Let x = (x0 + i
x), y = (y0 + i
y) be two four-vectors and their conjugates xc and yc ; one can define the exterior product x∧y =

1 (xyc − yxc ) 2 ⎡

⎢ ⎢ =⎢ ⎣

0, (x2 y3 − x3 y2 ) + i
(x1 y0 − x0 y1 ) , (x3 y1 − x1 y3 ) + i
(x2 y0 − x0 y2 ) , (x1 y2 − x2 y1 ) + i
(x3 y0 − x0 y3 )

⎤ ⎥ ⎥ ⎥ ⎦

= [0, x × y + i
(y0 x − x0 y)] with x ∧ y = −y ∧ x. The resulting quaternion is of the type B = (a + i
b) and is called a bivector; its real part, in the above example, gives the ordinary vector product but its nature differs from that of a four-vector. Under a Lorentz transformation (proper, orthochronous), a bivector transforms as B
= x
∧ y
=

1

(x yc − y
x
c ) 2

1 [(axa∗c ) (a∗ yc ac ) − (aya∗c ) (a∗ xc ac )] 2 = aBac . =

3.4. Four-vectors and multivectors in H(C)

45

The Lorentz transformation conserves the bivector type characterized by B = −Bc , Bc
= aBc ac = −aBac = −B
.  2 2 Furthermore, B 2 = − (a) + (b) − 2i
a · b is a relativistic invariant B
2 = B
B
= aBac aBac = aB 2 ac = B 2 . A trivector is a complex quaternion defined by 1 (xB ∗ + Bx) 2

 = [−i
x · a + x0 a + x × b] = i
t0 + t

T =x∧B =

and the product B∧x by postulating B∧x = x∧B. Under a proper, orthochronous Lorentz transformation, one has x
B
∗ + B
x
2 = aT a∗c

T
=

that is, T transforms as a four-vector. Furthermore, the transformation conserves the trivector type Tc∗ = −T , Tc
∗ = aTc∗ a∗c = a(−T )a∗c = −T
; T yields the relativistic invariant T
Tc
= aT a∗c a∗ Tc ac = T Tc

2 2 2 2 = t0 − t1 − t2 − t3 . The exterior product, defined above, is associative (x ∧ y) ∧ z = x ∧ (y ∧ z) . Indeed, (x ∧ y) ∧ z = z ∧ (x ∧ y) 1 ∗ z (xyc − yxc ) + (xyc − yxc ) z = 2 1 = [z (xc y − yc x) + (xyc − yxc ) z] , 2 1 ∗ x (yzc − zyc) + (yzc − zyc ) x x ∧ (y ∧ z) = 2 1 = [x (yc z − zc y) + (yzc − zyc) z] . 2

(3.3) (3.4) (3.5) (3.6) (3.7)

46

Chapter 3. Complex quaternions

Since (z, x)y = y(zc , xc ), one deduces the equality of the two equations (3.5), (3.7) and the associativity of the exterior product. A pseudoscalar is defined by the relation 1 P = x ∧ T = − (xTc + T x∗ ) 2 1 = − (xTc + T xc ) 2 where T is a trivector; by definition, one postulates T ∧ x = −x ∧ T . The pseudoscalar type is a pure imaginary P = i
s characterized by Pc = P and is invariant under a Lorentz transformation 1 P
= − (x
Tc
+ T
x
c ) = aP ac = P. 2 Examples. Consider the basis vectors e0 = 1, e1 = i
i, e2 = i
j, e3 = i
k; one obtains the following table 1 i
= e0 ∧ e1 ∧ e2 ∧ e3 1 = e0 i
= e1 ∧ e3 ∧ e2

i = e2 ∧ e3 i
i = e1 ∧ e0 i = e0 ∧ e2 ∧ e3 i
i = e1

j = e3 ∧ e1 i
j = e2 ∧ e0 j = e0 ∧ e3 ∧ e1 i
j = e2

k = e1 ∧ e2 i
k = e3 ∧ e0 k = e0 ∧ e1 ∧ e2 i
k = e3

One observes that distinct quantities occupy identical places in the H(C) algebra, a situation which one encounters also in the tridimensional vector calculus; this problem will be solved in the Clifford algebra H ⊗ H. Besides the exterior products, one can define interior products x · y = (x, y) =

1 (xyc + yxc ) , 2

1 (xB ∗ − Bx) ∈ four-vector, 2 1 x · T = − (xTc − T x∗ ) ∈ bivector; 2

x·B =

by definition, one supposes B · x = −x · B, T · x = x · T . More generally, one shall define the interior product between two multivectors Ap and Bq by [12, p. 14] Ap · Bq = (v1 ∧ v2 · · · ∧ vp−1 ) · (vp · Bq ) . Hence, one sees that H(C) already allows us to develop a few notions of a multivector calculus. These notions will be developed later on in the more satisfying framework of the Clifford algebra.

3.5. Relativistic kinematics via H(C)

47

3.5 Relativistic kinematics via H(C) 3.5.1 Special Lorentz transformation Consider the reference frame at rest K(O, x, y, z) and the reference frame K
(O
, x
, y
, z
) moving along the Ox axis with the constant velocity u (Figure 3.1). y


y

K


K O

u

O


x

x


z


z

Figure 3.1: Special Lorentz transformation (pure): the axes remain parallel to themselves and the reference frame moves along the Ox axis. The Lorentz transformation is expressed by X = bX
b∗c

(3.8)

 cosh ϕ2 + i
i sinh ϕ2 , bbc = 1,

with X = (x0 + i
x) , X
= (x
0 + i
x
), b = tanh ϕ = uc ; write γ = cosh ϕ, hence γ 2 = 1 + sinh2 ϕ = 1 + 1 γ=  1−

u2 c2

u2 2 γ , c2

.

Explicitly, equation (3.8) reads  u ct = γ ct
+ x
, c  u , x = γ x
+ ct
c

y=y, z=z.

48

Chapter 3. Complex quaternions

The inverse transformation is given by X
= bc Xb∗ and yields  u , ct
= γ ct − x c  u , x
= γ x − ct c y
= y, z
= z.

3.5.2 General pure Lorentz transformation A general pure Lorentz transformation is expressed by (Figure 3.2) X = bX
b∗c

 with b = cosh ϕ2 + i
uu sinh ϕ2 , where u is the velocity (of norm u), X = (ct+ i
x), X
= (ct
+ i
x
), γ = q 1 u2 , tanh ϕ = uc . Explicitly, one obtains [26, p. 280] 1− c2

  x
· u ct = γ ct
+ , c u x = x
+ n (n · x
) (γ − 1) + ct
nγ . c y
K
y

u O

K

O




x


z
x

z

Figure 3.2: General pure Lorentz transformation: the axes remain parallel to themselves but the reference frame K
moves in an arbitrary direction.

3.5.3 Composition of velocities Consider two reference frames K and K
with the special Lorentz transformation (Figure 3.1) X = bX
b∗c ,

3.5. Relativistic kinematics via H(C)

 b = cosh ϕ2 + i
i sinh ϕ2 , tanh ϕ =

49

u c,

γ = cosh ϕ =

four-velocity V transforms as

q 1 2 1− u c2

(ϕ constant). The

dX
∗ dX =b b ds ds c  with the relativistic invariant ds = c2 dt2 − dx2 − dy 2 − dz 2 ; ds can also be expressed as !   cdt
v2 cdt 2 2 ds = c dt 1 − 2 = =
. c γ γ V ≡

The four-velocity satisfies the relation V Vc =

dX dXc =1 ds ds

and can be written in the form

  v V = cosh θ + i
sinh θ v 
v = γ+iγ c

with tanh θ = vc , γ = cosh θ =

q 1 2 1− vc2

and v the velocity (of norm v). Consider a

particle moving along the Ox
axis with a velocity v
in the reference frame K
,
thus V
= (cosh θ1 + i
i sinh θ1 ) with tanh θ1 = vc . In the reference frame K, one 

has with b = cosh θ22 + i
i sinh θ22 , tanh θ2 = uc , V = bV
b∗c = [cosh (θ1 + θ2 ) + i
i sinh (θ1 + θ2 )] = (cosh θ + i
i sinh θ) hence, θ = θ1 + θ2 and u v
+ tanh θ1 + tanh θ2 v = c
c ; tanh θ = = vu c 1 + tanh θ1 tanh θ2 1+ 2 c finally v=

v
+ u .
1 + vc2u

When v
, u c one obtains the usual Galilean transformation. Furthermore, if v
= c, one has c+u v= = c, 1 + uc the velocity of light thus appears as a limit speed.

50

Chapter 3. Complex quaternions

3.6 Maxwell’s equations Let A = ( Vc + i
A) be the four-potential, D the relativistic four-nabla operator 

∂ ∂ ∂ ∂ D= , −i
1 , −i
2 , −i
3 c∂t ∂x ∂x ∂x   ∂
−i∇ = c∂t



and the conjugate operator  Dc =

 ∂
+i∇ . c∂t

Under a Lorentz transformation, D transforms as x = (x0 + i
x), D
= aDa∗c . To verify this, it is sufficient to use the relation ∂ ∂xα ∂ = ,
µ ∂x ∂xα ∂x
µ to develop x
= axa∗c , x = ac xa∗ and to compare the coefficients. Adopting the Lorentz gauge 1 ∂V + div A = 0 (D, A) = 2 c ∂t one obtains the electromagnetic field bivector F = Dc A = (D, A) + (D ∧ A)   i
E = (D ∧ A) = −B + . c Under a special Lorentz transformation, the bivector F transforms as F
= bF bc 

with b = cosh ϕ2 − i
i sinh ϕ2 , tanh ϕ = vc , γ = cosh ϕ, which yields the standard equations Ex
= Ex , Bx
= Bx ,   Ey v
By = γ By + 2 , Ey
= γ (Ey − vBz ) , c   Ey v Bz
= γ Bz − 2 , Ez
= γ (Ez + vBy ) . c

3.6. Maxwell’s equations

51

Furthermore, one has the relativistic invariant  

E2 2
E ·B F Fc = B − 2 + 2i . c c The exterior product D∧F =D∧D∧A=0

∂B rot E
− = −i div B , − c∂t c gives two of the Maxwell’s equations div B = 0,

rot E = −

∂B . ∂t

Introducing the four-current density C = (ρc, i
j), the equation DF ∗ − F D 2  

div(E
1 ∂E , i rot B − 2 = c c ∂t

D·F =

= µ0 C gives the two other Maxwell’s equations (in vacuum) ρ , 0

div E =

rot H = j +

∂D ∂t

with 0 µ0 c2 = 1. Furthermore, DF ∗ = D · F + D ∧ F = D(DAc )∗ = DDc A =
A = µ0 C where
=

∂2 c2 ∂t2

−

∂2 ∂

2 (x1 )

−

∂2 ∂

2 (x2 )

is the d’Alembertian. Hence, the equations ∂2V ρ − V = , 2 2 c ∂t 0 ∂2A − A = µ0 j. c2 ∂t2

−

∂2 ∂ (x3 )

2

52

Chapter 3. Complex quaternions

The entire set of Maxwell’s equations (in vacuum) can therefore be written DF ∗ = µ0 C ⎡

div E − i
div B ⎢ c =⎢     ⎣ 1 ∂E ∂B rot E
+i rot B − 2 + − c ∂t c∂t c = µ0 (ρc + i
j).

⎤ ⎥ ⎥ ⎦

The interior product F · C = −C · F

j·E + i
ρ (E + v ∧ B) = f = c gives the volumic four-force (of Minkowski).

3.7 Group of conformal transformations The group of conformal transformations is the group of transformations x
= f (x) (x, x
being minquats) such that dxdxc = 0 entails dx
dx
c = 0. This group includes spacetime translations, Lorentz transformations, dilatations x
= x + d, x
= axa∗c , x
= λx (d ∈ minquat, a ∈ H(C), aac = 1, λ ∈ R) and the transformations −1

x
= (1 + xac )

= x (1 + ac x) =

x

−1

x + a(x, x) 1 + 2(a, x) + (a, a)(x, x)

(3.9) (3.10) (3.11)

with a respective number of parameters of 4, 6, 1, 4 for a total of 15 parameters. The interest of this group in physics comes from the fact that Maxwell’s equations (without sources) are covariant with respect to this transformation group. The transformations (3.11) can also be expressed in the form −1

= x−1 (1 + xac ) ,

−1

= x−1 + ac ;

(x
) (x
)

3.7. Group of conformal transformations

53

the inverse transformation results from −1

x−1 = (x
)

− ac


−1

[1 − x
ac ]

= (x ) and thus

−1

x = (1 − x
ac )

x
.

The composition of two transformations gives −1

= x−1 + ac ,

−1

= (x
)

(x
) (x

)

−1

−1

=x

−1

=x

+ bc

+ ac + b c + cc

with c = a + b and thus belongs indeed to the group; if one permutes the two transformations, one obtains the same resulting transformation. As properties, one has xac x
= x
ac x, |dx
| = 2

|x
| = 2

|x
|2 |x|

2

(3.12) 2

(3.13)

2,

(3.14)

|dx| ,

|x|

2

|1 + xac |

−1

dx
= (1 + xac )

dx (1 + ac x)

−1

.

Equation (3.12) results from x
= (1 + xac )−1 x = x(1 + ac x)−1 , (1 + xac ) x
= x = x
(1 + ac x) which entails the relation. Furthermore, x = (1 + xac )x
, xxc = (1 + xac ) x
x
c (1 + axc ) , and thus |x
| = 2

2

|x|

|1 + xac |2

.

(3.15)

54

Chapter 3. Complex quaternions

Equation (3.13) which shows that the transformation is indeed a conformal transformation can be established as follows. Differentiating the relation qq −1 = qq −1 = 1, where q is a complex quaternion, one obtains

 d q −1 q + q −1 dq = 0,

 d q −1 = −q −1 dqq −1 ; hence







d x
−1 = − x
−1 dx
x
−1



 or d x
−1 = d x−1 , consequently


−1 

−1  −1  x dx x = x dxx−1 , x
c dx
x
c (x
x
c )2

xc dxxc

=

(xxc )2

.

By multiplying with the conjugate equation x
dx
c x
2 (x
x
c )

xdxc x

=

2

(xxc )

one obtains equation (3.13) dx
dx
c 2 (x
x
c )

=

dxdxc

2.

(xxc )

To obtain equation (3.15), one differentiates the equation x
= x (1 + ac x)

−1

,

−1

dx = dx (1 + ac x) − x (1 + ac x)−1 [d (1 + ac x)] (1 + ac x)−1 ,

x (1 + xc a) ac dx −1 = dx − (1 + ac x) , (1 + ac x) (1 + xc a)


−1

=

(1 + axc) dx (1 + ac x) (1 + ac x) (1 + xc a)

= (1 + xac )

−1

,

dx(1 + ac x)−1 .

3.8 Exercises E3-1 Express the matrices e1 =

1 0

0 0



,

e2 =

0 0

0 1



3.8. Exercises

55

using complex quaternions. Verify the relations e2i = ei , e1 + e2 = 1, e1 e2 = e2 e1 = 0. Take ∀a ∈ H(C), give the expression of the modules u = ae1 , v = e1 a. E3-2 Consider the complex quaternions x = 1 + i
i + (2 + i
)j + k y = 2 + 3i
k; compute x + y, xy, yx, x2 , y 2 , x−1 , y −1 , y −1 x−1 , (xy)−1 . E3-3 Consider the general Lorentz transformation X
= aXa∗c , √ √ √  1 a= 3 + i
i 5 − i
j 15 + k3 3 4 with aac = 1. Determine the rotation ri and the pure Lorentz transformation bi such that a = b1 r1 = r2 b2 with ∗

X
= b1 r1 X (b1 r1 )c = b1 (r1 X
r1c ) b∗1c or

X
= r2 (b2 Xb∗2c ) r2c .

E3-4 Consider the minquats x = 1 + i
(i + j) ,

y = 2 + i
k,

z = 3 + i
j,

w = 3i
i + i
j + i
k.

Compute x · y, B = x ∧ y, B
= z ∧ w, T = x ∧ (y ∧ z), T
= (x ∧ y) ∧ z, B · z, w · T , B ∧ B
, B · B
, B · T .


E3-5 Let K(O, t, x, y, z) be a reference frame at rest and K
(O
, t
, x ,y ,z ) a √ 3 5 reference frame moving along the Ox axis with a constant velocity v = 7 c. Write the Lorentz transformation X = bX
b∗c and give b. A particle is located at the point
X
= (0, i
, i
, 0) of K
and has a velocity vc = √17 in the direction O
y
; find its

 four-position and its four-velocity in the reference frame K. Let E
Ex
, Ey
, Ez
be the electric field in the reference frame K
; determine the electromagnetic field in K. E3-6 Consider a square ABCD of center 0 in the plane z = 0 and having its vertices located at the points A(0, 1, −1, 0), B(0, 1, 1, 0), C(0, −1, 1, 0), D(0, −1, −1, 0). Determine the transform of this square under the conformal transformation

x
= (1 + xac )−1 x

with a = i
k.

Chapter 4

Clifford algebra Clifford having demonstrated that the Clifford algebra is isomorphic to a tensor product of quaternion algebras or to a subalgebra thereof, this chapter develops within H ⊗ H the multivector calculus, multivectorial geometry and differential operators.

4.1 Clifford algebra 4.1.1 Definitions The Clifford algebra Cn over R is an associative algebra having n generators e0 , e1 , . . . , en−1 such that e2i = ±1,

ei ej = −ej ei

(i = j).

C + is the subalgebra constituted by products of an even number of ei . Clifford algebras are directly related to the quaternion algebra via the following theorem ([13],[14]). Theorem 4.1.1 (Clifford, 1878). If n = 2m (m integer), the Clifford algebra C2m is the tensor product of m quaternion algebras. If n = 2m − 1, the Clifford algebra C2m−1 is the tensor product of m − 1 quaternion algebras and the algebra (1, ω) where ω is the product of the 2m generators (ω = e0 e1 · · · e2m−1 ) of the algebra C2m . The tensor product of the algebras A and B is defined as follows [8, p. 57]. Consider two algebras A and B with x, y ∈ A and u, v ∈ B; the tensor product A ⊗ B is defined by the relation (x ⊗ u) (y ⊗ v) = (xy) ⊗ (uv) . Example (C ⊗ C). A general element of C ⊗ C is given by A = =

(a + i
b) ⊗ (f + i
g) (af ) 1 ⊗ 1 + (bf ) i
⊗ 1 + (ag) 1 ⊗ i
+ (bg) i
⊗ i
;

58

Chapter 4. Clifford algebra

write 1 = 1 ⊗ 1, i = i
⊗ 1, I = 1 ⊗ i
, iI = Ii = i
⊗ i
, one then has with I 2 = −1, A = =

(af + bf i) + I(ag + bgi) (α1 + Iα2 ).

The general element of C ⊗ C can thus be expressed as a complex number having complex coefficients, the imaginary I commuting with i.

4.1.2 Clifford algebra H ⊗ H over R The general element of the tensor product of two quaternion algebras H is expressed by A = (a0 + ia1 + ja2 + ka3 ) ⊗ (b0 + ib1 + jb2 + kb3 ) ⎡ a0 b0 1 ⊗ 1 + a0 b1 1 ⊗ i + a0 b2 1 ⊗ j + a0 b3 1 ⊗ k, ⎢ a1 b0 i ⊗ 1 + a1 b1 i ⊗ i + a1 b2 i ⊗ j + a1 b3 i ⊗ k, =⎢ ⎣ a2 b0 j ⊗ 1 + a2 b1 j ⊗ i + a2 b2 j ⊗ j + a2 b3 j ⊗ k, a3 b 0 k ⊗ 1 + a3 b 1 k ⊗ i + a3 b 2 k ⊗ j + a3 b 3 k ⊗ k

⎤ ⎥ ⎥; ⎦

let us write i = i ⊗ 1, j = j ⊗ 1, k = k ⊗ 1, I = 1 ⊗ i, J = 1 ⊗ j, K = 1 ⊗ k with i2 = j 2 = k 2 = ijk = −1, I 2 = J 2 = K 2 = IJK = −1. The general element A can thus be written A = α0 + α1 I + α2 J + α3 K where the coefficients αi = di + iai + jbi + kci are quaternions and where the lowercase i, j, k commute with the capital I, J, K (iJ = Ji, etc.). The element A is called a Clifford number and constitutes simply a quaternion having quaternions as coeffficients. Concisely, one can write A = (α0 ; α) with α = α1 I + α2 J + α3 K, αi ∈ H. The product of two Clifford numbers is given by ⎤ ⎡ α0 β0 − α1 β1 − α2 β2 − α3 β3 ; ⎢ α0 β1 + α1 β0 + α2 β3 − α3 β2 , ⎥ ⎥ (α0 ; α)(β0 ; β) = ⎢ ⎣ α0 β2 + α2 β0 + α3 β1 − α1 β3 , ⎦ α0 β3 + α3 β0 + α1 β2 − α2 β1 and in a more compact notation (α0 ; α)(β0 ; β) = [α0 β0 − α · β; α0 β + αβ0 + α × β]

4.2. Multivector calculus within H ⊗ H

59

where α · β, α × β are the ordinary scalar and vector products; the order of the terms has to be respected, the product of two quaternions being noncommutative. The generators of the Clifford algebra are e0 ≡ j,

e1 ≡ kI,

e2 ≡ kJ,

e3 ≡ kK

with e20 = −1, e21 = e22 = e23 = 1 and ei ej = −ej ei (i = j). A complete basis of the algebra is given in the following table. 1 i = e0 e1 e2 e3 j = e0 k = e1 e2 e3

I = e3 e2 iI = e0 e1 jI = e0 e3 e2 kI = e1

J = e1 e3 iJ = e0 e2 jJ = e0 e1 e3 kJ = e2

K = e2 e1 iK = e0 e3 jK = e0 e2 e1 kK = e3

A Clifford number can be written in the form [54] A = (a + ib + jc + kd; m + in + jr + ks) with m = m1 I + m2 J + m3 K and similarly for n, r and s. The Clifford algebra contains scalars a, pseudoscalars ib, vectors (four-dimensional) jc + ks, bivectors m + in and trivectors kd + jr. The conjugate of A is defined by transforming I, J, K into −I, −J, −K and j into −j, hence Ac = (a + ib − jc + kd; −m − in + jr − ks) with (AB)c = Bc Ac . The dual of A is defined by A∗ = iA and the commutator of two Clifford numbers by [A, B] =

1 (AB − BA) . 2

4.2 Multivector calculus within H ⊗ H 4.2.1 Exterior and interior products with a vector The product having been defined in the Clifford algebra, the interior and exterior products of two vectors x = jx0 + kx (x = x1 I + x2 J + x3 K), y = jy 0 + ky can be defined by the general formula [37] xy = λx · y + µx ∧ y

60

Chapter 4. Clifford algebra

where λ, µ are two nonzero coefficients. Adopting the choice λ = µ = −1, one has xy = −(x · y + x ∧ y), yx = −(y · x + y ∧ x); postulating a priori the relations x · y = y · x and x ∧ y = −y ∧ x one obtains 2x · y = −(xy + yx), 2x ∧ y = −(xy − yx). Explicitly, the formulas read x · y = x0 y 0 − x1 y 1 − x2 y 2 − x3 y 3 ∈ S (scalar), # " 



 x2 y 3 − x3 y 2 I + x3 y 1 − x1 y 3 J + x1 y 2 − x2 y 1 K 





x∧y = + x1 y 0 − x0 y 1 iI + x2 y 0 − x0 y 2 iJ + x3 y 0 − x0 y 3 iK 

 = x × y + i xy 0 − x0 y ∈ B with xc = −x, Bc = −B for a bivector B. The products of a vector with a multivector Ap = v1 ∧ v2 ∧ · · · ∧ vp are then defined by 2x · Ap = (−1)p [xAp − (−1)p Ap x] , 2x ∧ Ap = (−1)p [xAp + (−1)p Ap x] and Ap · x ≡ (−1)p x · Ap , Ap ∧ x ≡ (−1)p x ∧ Ap . The above formulas yield for a trivector (with x = jx0 + kx, B = a + ib) 1 T = x ∧ B = (xB + Bx) 2 

 = (−a · x) k + j x0 a + x × b with B ∧ x = x ∧ B and Tc = T . One verifies that the exterior product is indeed associative (x ∧ y) ∧ z = x ∧ (y ∧ z) ; one has, (x ∧ y) ∧ z = z ∧ (x ∧ y) 1 = − [z(xy − yx) + (xy − yx)z] , 2 1 x ∧ (y ∧ z) = − [x(yz − zy) + (yz − zy)x] . 2

(4.1) (4.2)

4.2. Multivector calculus within H ⊗ H

61

Since (z, x)y = y(z, x), i.e., (zx + xz)y = y(zx + xz), one verifies the equality of the equations (4.1), (4.2) and thus the associativity. A pseudoscalar is defined by (with T = kt0 + jt, x = jx0 + kx) 1 P = x ∧ T = − (xT − T x) 2 

= − x0 t0 + x · t i with T ∧ x ≡ −x ∧ T and Pc = P . As examples, let us express the standard basis in terms of the exterior product: 1 i = e0 ∧ e1 ∧ e2 ∧ e3 j = e0 k = e1 ∧ e3 ∧ e2

I = e2 ∧ e3 iI = e1 ∧ e0 jI = e0 ∧ e2 ∧ e3 kI = e1

J = e3 ∧ e1 iJ = e2 ∧ e0 jJ = e0 ∧ e3 ∧ e1 kJ = e2

K = e1 ∧ e2 iK = e3 ∧ e0 jK = e0 ∧ e1 ∧ e2 kK = e3

The interior products between a vector and a multivector are given by the formulas (with x = jx0 + kx, B = a + ib, T = kt0 + jt, P = is) 1 (xB − Bx) 2 

 = (−b · x) j + k −x0 b + x × a ∈ V, 1 x · T = − (xT + T x)  2 0  = x t + t0 x + i (x × t) ∈ B, 1 x · P = (xP − P x) 2 = −ksx0 + j (sx) ∈ T

x·B =

with B · x ≡ −x · B, T · x ≡ x · T, P · x ≡ −x · P.

4.2.2 Products of two multivectors The products of two multivectors Ap = v1 ∧v2 ∧· · ·∧vp and Bq = w1 ∧w2 ∧· · ·∧wp are defined [12] for p ≤ q, by Ap · Bq ≡ (v1 ∧ v2 ∧ · · · ∧ vp−1 ) · (vp · Bq ), Ap ∧ Bq ≡ v1 ∧ (v2 ∧ · · · ∧ vp ) ∧ Bq

62

Chapter 4. Clifford algebra

with Ap · Bq = (−1)p(q+1) Bq · Ap which defines Bq · Ap for q ≥ p. For products of multivectors one obtains, with S, P designating respectively the scalar and pseudoscalar parts of the multivector with B = a + ib, B
= a
+ ib
, T = ks0 + js, T
= ks
0 + js
, P = iw, P
= iw
,

1


B · B = S (BB + B B) 2 = [−a · a
+ b · b
] ∈ S,

1

B∧B =P (BB + B B) 2


= i [−b · a
− a · b
] ∈ P, 1 T · T
= − (T T
+ T
T ) 2 = s0 s
0 − s1 s
1 − s2 s
2 − s3 s
3 ∈ S, 1 (BP + P B) 2 = (−sb + isa) ∈ B,

B·P =

1 (T P − P T )

2  = jws0 − kws ∈ V,

T ·P =

1 (P P
+ P
P ) 2 = −ww
∈ S.

P · P
=

4.2.3 General formulas Among general formulas, one has x · (y ∧ z) = (x · y)z − (x · z)y, x · (y ∧ z) + y · (z ∧ x) + z · (x ∧ y) = 0, (x ∧ y) · B = x · (y · B) = −y · (x · B), (x ∧ y) · (z ∧ w) = (x · w)(y · z) − (x · z)(y · w), (B · T ) · V = (B ∧ V ) · T,

(4.3) (4.4) (4.5) (4.6) (4.7)

4.2. Multivector calculus within H ⊗ H

63

the Jacobi identity 0 = [F, [G, H]] + [G, [H, F ]] + [H, [F, G]] , (x · B1 ) · B2 = B1 · (B2 · x) + x · [B1 , B2 ] ,

(4.8) (4.9)

with x, y, z, w being (four-)vectors, B, B1 , B2 bivectors, T a trivector and F, G, H any Clifford numbers; the equations (4.4), (4.6), (4.9) are respectively consequences of equations (4.3), (4.5), (4.8). To establish equation (4.3), one just needs to write for any bivector B, 2x · B = (xB − Bx), hence 4x · (y ∧ z) = [−x(yz − zy) + (yz − zy)x] ; furthermore, 4(x · y)z = − [(xy + yx)z + z(xy + yx)] , −4(x · z)y = [(xz + zx)y + y(xz + zx)] ; adding the two last equations, one verifies equation (4.3). Equation (4.5) simply results from (x ∧ y) · B ≡ x · (y · B) = −(y ∧ x) · B = −y · (x · B); in particular, (x ∧ y) · (x ∧ y) = (x · y)2 − (x · x)(y · y). To prove equation (4.7), one writes 2B · T = (BT + T B), 4(B · T ) · V = − [(BT + T B)V + V (BT + T B)] , 2B ∧ V = 2V ∧ B = V B + BV, 4(B ∧ V ) · T = − [(V B + BV )T + T (V B + BV )] ; or B(T V − V T ) = (T V − V T )B because T V − V T is a pseudoscalar which commutes with B, hence the equation. The Jacobi identity results from 1 (ABC − BCA) 2 1 = [(AB − BA) C + (BAC − BCA)] 2 = [A, B] C + B [A, C] , [A, CB] = [A, C] B + C [A, B] , 1 [A, [B, C]] = {[A, B] C + B [A, C] − [A, C] B − C [A, B]} 2 = [[A, B] , C] + [B, [A, C]] [A, BC] =

64

Chapter 4. Clifford algebra

hence, the equation [A, [B, C]] + [B, [C, A]] + [C, [A, B]] = 0.

4.2.4 Classical vector calculus The link with the classical vector calculus is obtained as follows. Let x = kx, y = ky, z = kz, w = kw be four vectors without a temporal component, one obtains the relations x · (y ∧ z) = k [x × (y × z)] ∈ V, x ∧ (y ∧ z) = −k [x · (y × z)] ∈ T, (x ∧ y) · (z ∧ w) = −(x × y) · (z × w), 2

(x ∧ y) · (x ∧ y) = (x · y) − (x · x)(y · y), 2

= (x · y) − (x · x) (y · y) , = −(x × y) · (x × y) ∈ S, [x ∧ y, z ∧ w] = (x × y) × (z × w) ∈ B. The entire classical vector calculus thus constitutes a particular case of the multivector calculus.

4.3 Multivector geometry 4.3.1 Analytic geometry Straight line In the space of four dimensions, the equation of a straight line parallel to the vector u = ju0 + ku and going through the point a = ja0 + ka is given by (x − a) ∧ u = 0 with the immediate solution x − a = λu, x = λu + a

(λ ∈ R)

constituting the parametric equation of the straight line. Plane Let u, v be two linearly independent vectors and B = u ∧ v the corresponding plane; the equation of a plane parallel to B and going through the point a is expressed by (x − a) ∧ (u ∧ v) = 0

4.3. Multivector geometry

65

with the solution x − a = λu + µv, x = λu + µv + a

(λ, µ ∈ R)

giving the parametric equation of the plane parallel to B. A vector n is perpendicular to the plane B = u ∧ v if n is perpendicular to u and v (n · u = n · v = 0); for any vector x one has x · (u ∧ v) = (x · u)v − (x · v)u hence n · B = 0. Furthermore, one remarks that the vector x · (u ∧ v) is perpendicular to x, (x · B) · x = (x · u)(v · x) − (x · v)(u · x) = 0. A plane B1 = x ∧ y is perpendicular to a plane B2 = u ∧ v if the vectors x, y are perpendicular to the vectors u, v (x · u = y · u = x · v = y · v = 0); from the general formula (x ∧ y) · (u ∧ v) = (x · v)(y · u) − (x · u)(y · v) one obtains the orthogonality condition of two planes B1 · B2 = 0. In particular, the dual of a plane B ∗ = iB is perpendicular to that plane B ∗ · B = 0. Hyperplane A hyperplane is a subvector space of dimensions p = n − 1 = 3 (for n = 4). Let T = u ∧ v ∧ w be a trivector (hyperplane), the equation of a hyperplane parallel to T and going through the point a is given by (x − a) ∧ T = 0

(4.10)

with the solution x − a = λu + µv + γw, x = λu + µv + γw + a

(λ, µ, γ ∈ R)

giving the parametric equation of the hyperplane. Explicitly, equation (4.10) reads (with T = kt0 + jt, x = jx0 + kx, a = ja0 + ka) t0 x0 + t1 x1 + t2 x2 + t3 x3 − (a0 t0 + a · t) = 0

66

Chapter 4. Clifford algebra

which is indeed the equation of a hyperplane. A vector n is perpendicular to the hyperplane T = u ∧ v ∧ w if n is perpendicular to u, v, w (n · u = n · v = n · w = 0). From the general formula n · (u ∧ v ∧ w) = (n · u)(v ∧ w) + (n · v)(w ∧ u) + (n · w)(u ∧ w) one deduces n · T = 0.

(4.11)

If n is the dual of T , n = T ∗ = iT , one finds (with T = kt0 + jt, T ∗ = −jt0 + kt) 1 T ∗ · T = − (T ∗ T + T T ∗) = 0; 2 the dual of a hyperplane is thus perpendicular to that hyperplane. A plane B = x ∧ y is perpendicular to the hyperplane T = u ∧ v ∧ w if x, y are perpendicular to the hyperplane, hence T · B = B · T = (x ∧ y) · T ≡ x ∧ (y · T ) = −y ∧ (x · T ) = 0.

(4.12) (4.13)

4.3.2 Orthogonal projections Orthogonal projection of a vector on a vector Let u = u +u⊥ be the vector to project on the vector a with u⊥ ·a = 0, u ∧a = 0; since ua = −u · a − u ∧ a one obtains u a = −u · a = −u · a, u⊥ a = −u⊥ ∧ a = −u ∧ a, u = −(u · a)a−1 , u⊥ = −(u ∧ a)a−1 . Orthogonal projection of a vector on a plane Let u = u +u⊥ be the vector and let us represent the plane by a bivector B = a∧b (with u⊥ · B = 0, u ∧ B = 0); since uB = u · B + u ∧ B, Bu = B · u + u ∧ B, one obtains u = (u · B)B −1 = B −1 (B · u), u⊥ = (u ∧ B)B −1 = B −1 (u ∧ B).

4.3. Multivector geometry

67

Orthogonal projection of a vector on a hyperplane Let T be the hyperplane u = u + u⊥ with u⊥ · T = 0, u ∧ T = 0; one has uT = −u · T − u ∧ T, T u = −u · T + u ∧ T, hence u T = −u · T, u⊥ T = −u ∧ T,

T u = −u · T, T u⊥ = u ∧ T,

and finally u = −(u · T )T −1 = −T −1 (u · T ), u⊥ = −(u ∧ T )T

−1

=T

−1

(u ∧ T ).

(4.14) (4.15)

Orthogonal projection of a bivector on a plane Let B1 = B1 + B1⊥ be the bivector and B2 = a ∧ b the plane with B1⊥ · B2 = 0,  B1 ∧ B2 = 0 and B1 , B2 = 0; using the formula B1 B2 = B1 · B2 + B1 ∧ B2 + [B1 , B2 ] , one obtains B1 = (B1 · B2 )B2−1 , B1⊥ = {(B1 ∧ B2 ) + [B1 , B2 ]} B2−1 . Example. Take B1 = x ∧ y (with x = jx0 + kx, y = jy 0 + ky) and let us project on the plane x1 x2 , i.e., the plane B2 = K (B2−1 = −K); one finds B1

=

(B1 · B2 )B2−1

=

(−x2 y 1 + x1 y 2 )K = x ∧ y

(with x = x1 kI + x2 kJ, y = y 1 kI + y 2 kJ), B1⊥

= {(B1 ∧ B2 ) + [B1 , B2 ]} B −1 = (−x3 y2 + x2 y3 )I + (x3 y1 − x1 y3 )J +(x1 y0 − x0 y1 )iI + (x2 y0 − x0 y2 )iJ + (x3 y0 − x0 y3 )iK = B1 − B1 .

68

Chapter 4. Clifford algebra

Orthogonal projection of a bivector on a vector Let B = B + B⊥ be the bivector and a the vector (with B1 ∧ a = 0, B⊥ · a = 0); since aB = a · B + a ∧ B, Ba = B · a + a ∧ B, one obtains B = a−1 (a · B) = (B · a)a−1 , B⊥ = a−1 (a ∧ B) = (a ∧ B)a−1 . Orthogonal projection of a bivector on a hyperplane Let B = B + B⊥ be the bivector and T the hyperplane with B⊥ · T = 0 and [B , T ] = 0 (B ∈ T ). From the equation BT = B · T + [B, T ] one obtains B = (B · T )T −1 , B⊥ = [B, T ] T −1 . Example. Let F = −B + i Ec be the electromagnetic bivector, its orthogonal projection on the hyperplane T = k = e1 e2 e3 (T −1 = −k) yields F F⊥

= −(F · k)k = −B, E = − [F, k] k = . c

Orthogonal projection of a hyperplane T1 on the hyperplane T2 Let us write T1 = T1 + T1⊥ with T1 = x ∧ y ∧ z, T2 = u ∧ v ∧ w; from the relation T1 · T2 ≡ (x ∧ y) · [z · (u ∧ v ∧ w)] and the equation (4.11), one deduces 

T1⊥ · T2 = 0;

furthermore T1 , T2 = 0 (T1 ∈ T2 ). Since T1 T2 = −T1 · T2 + [T1 , T2 ] it follows that T1 = −(T1 · T2 )T2−1 , T1⊥ = [T1 , T2 ] T2−1 .

4.4. Differential operators

69

4.4 Differential operators 4.4.1 Definitions Consider the relativistic four-nabla operator, ∇   ∂ ∇ = j 0 − k∇ ∂x = ∂α eα = ∂ α eα with ∇=I ∂α =

∂ ∂xα

∂ ∂ ∂ + J 2 + K 3, ∂x1 ∂x ∂x

and eα the reciprocal basis defined by eα · eβ = δβα

(e0 = e0 , e1 = −e1 , e2 = −e2 , e3 = −e3 ). One can define the operators fourgradient ∂ ∇ϕ = j 0 − k grad ϕ (ϕ ∈ S) ∂x the four-divergence of a vector A = jA0 + kA ∇·A =

∂A1 ∂A2 ∂A3 ∂A0 + + + ∈S ∂x0 ∂x1 ∂x2 ∂x3 

the four-curl ∇ ∧ A = − rot A − i

∂A + grad A0 ∂x0

 ∈ B;

acting on a bivector B = a + ib, one can define the operators   ∂b + rot a ∈ V ∇ · B = j div b − k ∂x0   ∂a ∇ ∧ B = k div a + j − rot b ∈ T. ∂x0

4.4.2 Infinitesimal elements of curves, surfaces and hypersurfaces A curve in the four-dimensional space is defined by the parametric equations OM (α) = jx0 (α) + kx(α) defining the tangent vector at the point M , dOM =

∂OM dα ∂α

70

Chapter 4. Clifford algebra

where α is the parameter. For a surface, the parametric equations are OM (α, β) = jx0 (α, β) + kx(α, β) and the tangent plane to the surface at the point M is defined by   ∂OM ∂OM ∧ dS = dαdβ ∂α ∂β ⎧  2 3   3 1  ∂x ∂x ∂x3 ∂x2 ∂x ∂x ∂x1 ∂x3 ⎪ I + J − − ⎪ ∂α ∂β ∂α ∂β ∂α ∂β ⎪ ⎨ ∂α  ∂β   1 2 2 1 1 0 0 ∂x ∂x ∂x ∂x1 + ∂x K + ∂x iI − ∂x − ∂x = ∂α ∂β ∂α ∂β ∂α ∂β ∂α ∂β ⎪  2 0   3 0  ⎪ ⎪ ∂x ∂x ∂x0 ∂x2 ∂x ∂x ∂x0 ∂x3 ⎩ + ∂α ∂β − ∂α ∂β iJ + ∂α ∂β − ∂α ∂β iK

the bivector

⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭

dαdβ.

In abridged notation, one can write dS = dx2 dx3 I + dx3 dx1 J + dx1 dx2 K + dx1 dx0 iI + dx2 dx0 iJ + dx3 dx0 iK with D(x2 , x3 ) dαdβ D(α, β)    ∂x2 ∂x2     ∂α ∂β   dαdβ, etc. = 3 ∂x3   ∂x   ∂α ∂β

dx2 dx3 =

where dx2 dx3 is an undissociable symbol [3, p. 446]. For a hypersurface defined by OM (α, β, γ) = jx0 (α, β, γ) + kx(α, β, γ) the tangent hyperplane to the surface at the point M is given by the trivector   ∂OM ∂OM ∂OM ∧ ∧ dT = dαdβdγ ∂α ∂β ∂γ ⎫ ⎧  2 3 ∂x3 ∂x2 ∂x1 ∂x3 ∂x1 ∂x1 ∂x2 ⎪ ⎪ − ∂x − ∂x ⎪ ⎪ ∂α ∂β ∂γ ∂α ∂β ∂γ ∂α ∂β ∂γ ⎪ ⎪ ⎪ ⎪ k 1 3 2 2 1 3 1 2 3 ⎪ ⎪ ∂x ∂x ∂x ∂x ∂x ∂x ∂x ∂x ∂x ⎪ ⎪ + + − ⎪ ⎪ ⎪ ⎪ ∂α ∂β ∂γ ∂α ∂β ∂γ ∂α ∂β ∂γ  ⎪ ⎪ 3 2 0 2 3 0 3 0 2 ⎪ ⎪ ∂x ∂x ∂x ∂x ∂x ∂x ∂x ∂x ∂x ⎪ ⎪ + + − ⎪ ⎪ ⎪ ⎪ ∂α ∂β ∂γ ∂α ∂β ∂γ ∂α ∂β ∂γ ⎪ ⎪ jI + ⎪ ⎪ ⎬ ⎨ ∂x0 ∂x3 ∂x2 ∂x2 ∂x0 ∂x3 ∂x0 ∂x2 ∂x3 − ∂α ∂β ∂γ − ∂α ∂β ∂γ + ∂α ∂β ∂γ  dαdβdγ. = 3 1 0 1 3 0 3 0 1 ∂x ∂x ∂x ∂x ∂x ∂x ∂x ∂x ∂x ⎪ ⎪ ⎪ ⎪ ∂α ∂β ∂γ − ∂α ∂β ∂γ − ∂α ∂β ∂γ ⎪ ⎪ ⎪ + jJ ⎪ 0 1 0 ⎪ ⎪ ∂x3 ∂x1 ∂x0 ∂x3 ∂x1 ∂x3 ⎪ ⎪ + ∂x + ∂x − ∂x ⎪ ⎪ ⎪ ⎪ ∂α ∂β ∂γ ∂α ∂β ∂γ ∂α ∂β ∂γ  ⎪ ⎪ 2 1 0 1 2 0 2 0 1 ⎪ ⎪ ∂x ∂x ∂x ∂x ∂x ∂x ∂x ∂x ∂x ⎪ ⎪ + + − ⎪ ⎪ ⎪ ⎪ ∂α ∂β ∂γ ∂α ∂β ∂γ ∂α ∂β ∂γ ⎪ ⎪ jK + ⎪ ⎪ 0 2 1 1 0 2 0 1 2 ⎭ ⎩ ∂x ∂x ∂x ∂x ∂x ∂x ∂x ∂x ∂x − − + ∂α ∂β ∂γ

∂α ∂β ∂γ

∂α ∂β ∂γ

4.4. Differential operators

71

In short, one can write dT = kdx1 dx3 dx2 2

3

(4.16) 0

3

1

0

1

2

0

+ jIdx dx dx + jJdx dx dx + jKdx dx dx

(4.17)

with the symbol D(x1 , x3 , x2 ) dαdβdγ D(α, β, γ)    ∂x1 ∂x1 ∂x1     ∂α ∂β ∂γ      ∂x3 ∂x3 ∂x3  =  dαdβdγ, etc.  ∂α ∂β ∂γ     ∂x2 ∂x2 ∂x2     ∂α ∂β ∂γ 

dx1 dx3 dx2 =

4.4.3 General theorems Generalized Stokes theorem This theorem can be written % && A · dl = − (∇ ∧ A) · dS

(4.18)

with dl = jdx0 + kdx, A = jA0 + kA, dS = dS1 + idS2 and dS1 = dx2 dx3 I + dx3 dx1 J + dx1 dx2 K, dS2 = dx1 dx0 I + dx2 dx0 J + dx3 dx0 K. Explicitly, equation (4.18) is expressed in classical vector notation as  

% && ∂A A0 dx0 − A · dx = − rot A · dS1 + ∇A0 + 0 · dS2 . ∂x If dx0 = 0, one has dS2 = 0 and the formula reduces to the standard Stokes theorem. As to the orientation of the curve C, it results from that of the surface S [10, p. 39]. One takes two linearly independent four-vectors a, b ∈ S and one chooses the order (a, b) as being the positive orientation (dS = αa ∧ b, α > 0). On the curve C one chooses a vector f exterior to S and a vector g tangent to the curve such that (f, g) is ordered positively; the curve is then oriented along g. Generalized Gauss theorem The theorem is expressed as &&&

&&&& A ∧ dT =

(∇ · A) dτ

(4.19)

72

Chapter 4. Clifford algebra

with dτ = idx0 dx1 dx2 dx3 , dT = kdx1 dx3 dx2 + jt and t = dx2 dx3 dx0 I + dx3 dx1 dx0 J + dx1 dx2 dx0 K, the hypersurface being closed. Explicitly, equation (4.19) reads  &&& &&&&  0

 ∂A 0 1 3 2 −A dx dx dx + A · t i = + div A idx0 dx1 dx2 dx3 . ∂x0 The orientation of the hypersurface results from that of the four-volume. Let a, b, c, d be four linearly independent vectors of the four-volume a ∧ b ∧ c ∧ d = αi; if α > 0, the orientation of the four-volume is positive. On a point of the hypersurface, one chooses a vector m outside of the four-volume and three vectors p, q, r on the hypersurface; if m ∧ p ∧ q ∧ r has a positive sign, the orientation of the hypersurface is positive. Other formulas On a closed surface, one has && &&& F · dS = (∇ ∧ F ) · dT, && &&& F ∧ dS = − (∇ · F ) ∧ dT,

(4.20) (4.21)

with the bivector F = f + ig, dS = dS1 + idS2 , dT = kdx1 dx3 dx2 + jt; explicitly, formula (4.20) reads  

&&& && ∂f div f dx1 dx3 dx2 − − rot g ·t ; (−f · dS1 + g · dS2 ) = ∂x0 if dx0 = 0, dS2 = 0, t = 0, one obtains the standard Gauss theorem. Relation (4.21) gives  

&&& && ∂g i div gdx1 dx3 dx2 − + rot f ·t . i (−g · dS1 − f · dS2 ) = ∂x0 The orientation of dS proceeds from that of dT . One chooses three linearly independent vectors a, b, c of T and one defines the orientation a ∧ b ∧ c as being positive. On a point of the surface, one considers a vector m exterior to the trivolume T , and two vectors p, q on the surface; if m ∧ p ∧ q has the orientation of a ∧ b ∧ c, the orientation of the surface is positive.

4.5 Exercises E4-1 Consider the Clifford numbers A and B: A = I + 2J − iK, B = j + kI + 2kK.

4.5. Exercises

73

Determine Ac , Bc , A + B, A − B, AAc , BBc , A−1 , B −1 , AB, BA, (AB)c , (BA)c , −1 −1 (AB) (AB)c , (BA) (BA)c , (AB) , (BA) , [A, B]. E4-2 Consider the four-vectors x = j + kI + kJ, z = 3j + kJ,

y = 2j + kK, w = 3kI + kJ + kK.

Determine x · y, B = x ∧ y, B
= z ∧ w, T = x ∧ (y ∧ z), T
= (x ∧ y) ∧ z, B · z, w · T , B · B
, B ∧ B
, B · T . E4-3 Take an orthonormal reference frame with the components of the fourvectors x = (0, 1, 2, −1), y = (0, 3, 1, 1), z = (0, 1, 2, 1), w = (0, 2, 1, 5). Determine within the Clifford algebra H ⊗ H the surfaces S1 = x ∧ y, S2 = z ∧ w, the trivector x ∧ y ∧ z and the four-volume x ∧ y ∧ z ∧ w. Give the orthogonal projection of the vector w on S1 and the orthogonal projection of S1 on S2 .

Chapter 5

Symmetry groups This chapter formulates the Lorentz group and the group of conformal transformations within the Clifford algebra H ⊗ H over R. In complexifying this algebra, one obtains the Dirac algebra H ⊗ H over C, isomorphic to the subalgebra C + of H ⊗ H ⊗ H over R. Dirac’s equation, the unitary group SU(4) and the symplectic unitary group USp(2, H) are treated as applications of H ⊗ H over C.

5.1 Pseudo-orthogonal groups O(1, 3) and SO(1, 3) 5.1.1 Metric Consider two vectors x = jx0 + kx, y = jy 0 + ky with x = x1 I + x2 J + x3 K and the interior product, with xc = −x; x · x = xxc = −x2 , (x + y) · (x + y) = (x + y)(x + y)c = xxc + yyc + 2x · y, (xy + yx) (xyc + yxc ) =− 2 2 = x0 y 0 − x1 y 1 − x2 y 2 − x3 y 3 .

x·y =

A vector x is isotropic if xxc = 0, timelike if xxc > 0 and spacelike if xxc < 0.

5.1.2 Symmetry with respect to a hyperplane Let a = ja0 + ka be a vector, the hyperplane perpendicular to a is given by the dual of a, T = a∗ = ia = ka0 − ja

76

Chapter 5. Symmetry groups

+ and anticommutes with with T −1 = −ia aac (i commutes with all elements of C − those of C ). Let us suppose that T goes through the origin and let x = jx0 + kx be a vector. The orthogonal projections of x on T are given by the relations (4.14), (4.15)

x = −T −1 (x · T ), x⊥ = T −1 (x ∧ T ), hence x iaxia axa (xT − T x) x = + = − , 2 2 2aac 2 2aac axa x x = + 2 2aac

x⊥ = −T −1

with x = x + x⊥ . Definition 5.1.1. The symmetric of x with respect to a hyperplane is obtained by drawing the perpendicular to the hyperplane and by extending this perpendicular by an equal length [11]. Let x
be the symmetric of x with respect to the hyperplane T ; one has x − x
= 2x⊥ hence

x
=

axa . aac

More simply, one can write that x
− x is perpendicular to the hyperplane T (and
thus parallel to a) and x+x is parallel to the hyperplane. One obtains 2  a·

x
= x + λa,  x +x = 0, 2


hence   λa 2(a · x) , a· x+ = 0 =⇒ λ = − 2 a·a 2(a · x)a x
= x − a·a (ax + xa)a axa axc a =x+ = =− ; aac aac aac finally (with a2 = −aac ), one finds again the above expression of the symmetric of x with respect to the hyperplane. One shall distinguish the time symmetries

5.1. Pseudo-orthogonal groups O(1, 3) and SO(1, 3)

77

(with aac = 1) x
= axa = −axc a, from the space symmetries (with aac = −1) x
= −axa = axc a.

5.1.3 Pseudo-orthogonal groups O(1, 3) and SO(1, 3) Definition 5.1.2. The pseudo-orthogonal group O(1, 3) is the group of linear operators which leave invariant the quadratic form x · y = x0 y 0 − x1 y 1 − x2 y 2 − x3 y 3 with x = jx0 + kx. Theorem 5.1.3. Every rotation of O(1, 3) is the product of an even number ≤ 4 of symmetries, any inversion is the product of an odd number ≤ 4 of symmetries [11]. A rotation is a proper transformation of a determinant equal to 1; an inversion is an improper transformation of a determinant equal to −1. Proper orthochronous Lorentz transformation Consider an even number of time symmetries (s, r, . . .) and of space symmetries (t, u, . . .); one obtains x
= (utsr)x(rstu) = axac with a = utsr ∈ C , ac = rc sc tc uc = (−r)(−s)(−t)(−u) = rstu and aac = 1. +

Other transformations Let n be the number of time symmetries and p the number of space symmetries; their combinations give the following Lorentz transformations : 1. n odd, p odd: proper antichronous rotation x
= −axac ,

aac = −1,

a ∈ C+;

2. n even, p odd: improper orthochronous inversion x
= axac ,

aac = −1,

a ∈ C−;

3. n odd, p even: improper antichronous inversion x
= −axac ,

aac = 1,

a ∈ C−.

78

Chapter 5. Symmetry groups

Group O(1, 3): recapitulative table The entire set of Lorentz transformations L is given in the table below where n is the number of time symmetries and p the number of space symmetries.

orthochronous

antichronous

Inversion (det L = −1) n even, p odd x
= axac = −axc ac (aac = −1, a ∈ C − ) n odd, p even x
= −axc ac = axc ac (aac = 1, a ∈ C − )

Rotation (det L = 1) n even, p even x
= axac (aac = 1, a ∈ C + ) n odd, p odd x
= −axac (aac = −1, a ∈ C + )

5.2 Proper orthochronous Lorentz group 5.2.1 Rotation group SO(3) A subgroup of the proper orthochronous Lorentz group is the rotation group SO(3) x
= rxrc

(5.1)

with r = cos θ2 + u sin θ2 , u = u1 I + u2 J + u3 K, (u1 )2 + (u2 )2 + (u3 )2 = 1, x = jx0 + kx, x
= jx
0 + kx
, rrc = 1. One verifies that x
belongs indeed to the vector space of vectors x
c = rxc rc = −x
. In matrix form, equation (5.1) can be written ⎡ 0 ⎡
0 ⎤ x x ⎢ x1 ⎢ x
1 ⎥
⎥ X=⎢ X =⎢ ⎣ x2 ⎣ x
2 ⎦ ,
3 x x3

with ⎤ ⎥ ⎥, ⎦

xi , x
i ∈ R,

X
= AX, ⎡

1 0 ⎢ 0 a ⎢ A=⎣ 0 m 0 n with

0 f b p

⎤ 0 g ⎥ ⎥, h ⎦ c

2  2 2  cos θ, a = u1 + u2 + u3

2 2  1 2 3 2  b= u cos θ, + u + u  

2

2 2 c = u3 + u1 + u2 cos θ,

5.2. Proper orthochronous Lorentz group

79

f = u1 u2 (1 − cos θ) − u3 sin θ,

m = u1 u2 (1 − cos θ) + u3 sin θ,

g = u1 u3 (1 − cos θ) + u2 sin θ,

n = u1 u3 (1 − cos θ) − u2 sin θ,

h = u2 u3 (1 − cos θ) − u1 sin θ,

p = u2 u3 (1 − cos θ) + u1 sin θ.

One obtains the same expression as with quaternions, despite the distinct nature of r and x (r ∈ C + , x ∈ C − ). All considerations of Chapter 2 on the subgroups of SO(3) that is of r, apply here; it suffices to replace i, j, k by I, J, K respectively. For an infinitesimal rotation, one has with r  1 + u dθ 2 , x
= rxrc     dθ dθ = 1+u x 1−u 2 2 dθ (ux − xu) = x+ 2 dθ = x + k (ux − xu) 2 = x + dθku × x, dx = x
− x = dθku × x. In matrix notation dX = dθui Mi X, with

⎡

0 ⎢0 M1 = ⎢ ⎣0 0

0 0 0 0

⎤ 0 0 0 0 ⎥ ⎥, 0 −1 ⎦ 1 0

⎡

0 0 ⎢0 0 M2 = ⎢ ⎣0 0 0 −1

i ∈ (1, 2, 3) 0 0 0 0

⎤ 0 1 ⎥ ⎥, 0 ⎦ 0

⎡

0 ⎢ 0 M3 = ⎢ ⎣ 0 0

⎤ 0 0 0 0 −1 0 ⎥ ⎥ 1 0 0 ⎦ 0 0 0

and the relations [Mi , Mj ] = ijk Mk .

5.2.2 Pure Lorentz transformation A pure Lorentz transformation is given by x
= bxbc , b = cosh

ϕ ϕ + iv sinh 2 2

with v = v 1 I + v 2 J + v 3 K, (v 1 )2 + (v 2 )2 + (v 3 )2 = 1, x = jx0 + kx, x
= jx
0 + kx
, bbc = 1. One verifies that one has indeed x
c = bxc bc = −x
. In matrix formulation, one has with ⎡
0 ⎤ ⎡ 0 ⎤ x x ⎢ x
1 ⎥ ⎢ x1 ⎥
⎥ ⎥ X =⎢ X=⎢ xi , x
i ∈ R, ⎣ x
2 ⎦ , ⎣ x2 ⎦ , x
3 x3

80

Chapter 5. Symmetry groups X
= BX, ⎡

cosh ϕ ⎢ v 1 sinh ϕ B=⎢ ⎣ v 2 sinh ϕ v 3 sinh ϕ with

v 1 sinh ϕ v 2 sinh ϕ a f f b g h

2 a = 1 + v 1 (cosh ϕ − 1) ,

2 b = 1 + v 2 (cosh ϕ − 1) ,

2 c = 1 + v 3 (cosh ϕ − 1) ,

⎤ v 3 sinh ϕ ⎥ g ⎥, ⎦ h c

f = v 1 v 2 (cosh ϕ − 1) , g = v 1 v 3 (cosh ϕ − 1) , h = v 2 v 3 (cosh ϕ − 1) .

One will notice that the matrix is real and symmetric. For a pure infinitesimal Lorentz transformation, one obtains with b  1 + iv dϕ 2 and i anticommuting with x, x
= bxbc     dϕ dϕ = 1 + iv x 1 − iv 2 2 dϕ (vx + xv) =x+i 2   dϕ  0 v jx + kx + jx0 + kx v =x+i 2 

= x + dϕ kvx0 − j v 1 x1 + v 2 x2 + v 3 x3 ,  

dx = x
− x = dϕ kvx0 − j v 1 x1 + v 2 x2 + v 3 x3 . In matrix form, one has i ∈ (1, 2, 3)

dX = dϕv i Ki X, with ⎡

0 ⎢ 1 K1 = ⎢ ⎣ 0 0

1 0 0 0

0 0 0 0

⎤ 0 0 ⎥ ⎥, 0 ⎦ 0

⎡

0 ⎢ 0 K2 = ⎢ ⎣ 1 0

0 0 0 0

1 0 0 0

⎤ 0 0 ⎥ ⎥, 0 ⎦ 0

and the relations [Ki , Kj ] = − ijk Mk , [Ki , Mj ] = ijk Kk .

⎡

0 ⎢ 0 K3 = ⎢ ⎣ 0 1

0 0 0 0

0 0 0 0

⎤ 1 0 ⎥ ⎥ 0 ⎦ 0

5.2. Proper orthochronous Lorentz group

81

5.2.3 General Lorentz transformation Transformation of multivectors and Clifford numbers A general Lorentz transformation for vectors is expressed by x
= axac with a ∈ C + , aac = 1. A bivector of the type B = x ∧ y = − (xy−yx) transforms as 2 (x
y
− y
x
) 2 = a (x ∧ y) ac

B
= −

= aBac . Generally, for a product of two vectors xy one has the transformation x
y
= axac ayac = axyac ; consequently, any multivector A (and any Clifford number) which is a linear combination of such products transforms under a proper orthochronous Lorentz transformation according to the same formula A
= aAac

(a ∈ C + , aac = 1).

Decomposition into a rotation and a pure Lorentz transformation The decomposition a = br of a general Lorentz transformation into a rotation and a pure Lorentz transformation is obtained as follows ([15],[16]). Consider the element aac = 1; a = a0 + a + i (b0 + b) ∈ C + , let us introduce a conjugation (transforming i into −i), a = a0 + a − i (b0 + b) = br = bc r, ac = rc b. Let us write d = aac = brrc b = b2 (ddc = 1), hence 2b2 = 2d, b2 d = d2 , b2 dc = 1

(dc = aac );

82

Chapter 5. Symmetry groups

adding, one obtains 2

b2 (2 + d + dc ) = (1 + d) , (1 + d) b = ±√ 2 + d + dc r = bc a

(with d = aac ),

(1 + aac ) a = ±√ 2 + d + dc (a + a) = ±√ , 2 + d + dc with bbc = rrc = 1. For the decomposition a = r
b
, one obtains similarly (1 + d
) b
= ±  2 + d
+ d
c

(with d
= ac a, d
c = ac a),

r
= ab
c (a + a) = ± . 2 + d
+ d
c √ Example. Let a = 10K − 3iJ be an even Clifford number (a ∈ C + ) and such that aac = 1. Let us decompose a into a product (a = br = r
b
). One has √ √ a = 10 K + 3iJ, d = aac = 19 + 6 10 iI  √ 10 + 3iI , r = ±K; Answer: b = ± similarly √ d
= ac a = 19 − 6 10 iI √  10 − 3iI , r
= ±K. Answer: b
= ±

5.3 Group of conformal transformations 5.3.1 Definitions The treatment of conformal transformations within H(C) extends easily to H ⊗ H. The group of conformal transformations is constituted by the transformations such that if dx · dx = 0, then dx
· dx
= 0 with dx = jx0 + kx. The group contains the spacetime translations (x
= x + d), Lorentz transformations (x
= axac ), dilatations (x
= λx) and the transformations −1

(x
)

= x−1 + ac

(5.2)

5.3. Group of conformal transformations

83

xc . The above equation can also where a is a constant (four-)vector with x−1 = xx c be written  

−1 −1 −1 = x−1 + ac x
= (x
)   −1 −1 = x−1 (1 + xac ) = (1 + ac x) x−1

and using relation (AB) number

−1

= B −1 A−1 which is true for any invertible Clifford x
= (1 + xac )

−1

= x (1 + ac x)

x

−1

(5.3) .

(5.4)

Equivalently, one has 

 x +a xc xxc ,  x
= + ac = xc x xxc + ac +a xxc xxc x + a (x · x) . x
= 1 + 2(x · a) + (a · a)(x · x) 

−1

The inverse transformation is given by −1

x−1 = (x
)

− ac

hence −1

x = (1 − x
ac )

x



−1




= x (1 − ac x )

(5.5) .

(5.6)

5.3.2 Properties of conformal transformations For any invertible Clifford number A (AA−1 = A−1 A = 1), one has dA−1 A + A−1 dA = 0, dA−1 = −A−1 dAA−1 . Differentiating equation (5.2), one obtains  

 −1 = d x−1 , d (x
) −1

(x
)

−1

dx
(x
)

x
c dx
x
c 2 (x
x
c )

= x−1 dxx−1 , =

xc dxxc 2

(xxc )

;

(5.7) (5.8) (5.9)

84

Chapter 5. Symmetry groups

multiplying by the conjugate equation, one has x
c dx
x
c x
dx
c x


xc dxxc xdxc x

=

4 (x
x
c )

,

(5.10)

dxdxc

(5.11)

4

(xxc ) (x
.x
)

2

dx
dx
c =

(x.x)

2

which shows that the transformation is indeed a conformal transformation. Equation (5.8) gives (x
)

−1

dx
(x
)

−1

= x−1 dxx−1 ,

dx
= x
x−1 dxx−1 x
, and using equations (5.3), (5.4), one obtains the relation dx
= (1 + xac )−1 dx (1 + ac x)−1 .

(5.12)

Finally, from equations (5.3), (5.4) one has −1

x
= (1 + xac ) x = x (1 + ac x) (1 + xac ) x
= x = x
(1 + ac x) ,

−1

,

hence xac x
= x
ac x, xxc = (1 + x
x
c

(5.13)

xac ) x
x
c

(1 + axc ) , xxc . = 1 + 2(x · a) + (a · a)(x · x)

(5.14) (5.15)

5.3.3 Transformation of multivectors

 A conformal transformation is a relation of the type x
i = f i x0 , x1 , x2 , x3 with ∂x
i k dx . ∂xk

dx
i =

A contravariant (four-)vector A, by definition, transforms according to the formula ∂x
i k A . ∂xk

A
i = The equation (5.12)

−1

dx
= (1 + xac )

−1

dx (1 + ac x)

5.4. Dirac algebra

85

then gives the transformation of a vector A = jx0 + kA, A
= (1 + xac )

−1

A (1 + ac x)

−1

.

(5.16)

For a product of two vectors A, B one obtains A
B
= (1 + xac )−1 A (1 + ac x)−1 (1 + xac )−1 B (1 + ac x)−1 = K (1 + xac )

−1

−1

AB (1 + ac x)

with K=

1 ; 1 + 2(x · a) + (a · a)(x · x)

for a bivector A ∧ B = − (AB−BA) one then has 2 −1

A
∧ B
= K (1 + xac )

(A ∧ B) (1 + ac x)

−1

;

one verifies that A
∧ B
is indeed a bivector. Similarly, one obtains for a trivector and a pseudoscalar A
∧ B
∧ C
= K 2 (1 + xac )











−1

(A ∧ B ∧ C) (1 + ac x)

−1

,

A ∧ B ∧ C ∧ D = K (A ∧ B ∧ C ∧ D) . 4

Furthermore, A
A
c = K 2 AAc , A
· B
= K 2 (A · B) which shows that the conformal transformation conserves the angles.

5.4 Dirac algebra 5.4.1 Dirac equation The Dirac algebra is isomorphic to the Clifford algebra H ⊗ H over C and can be represented by 2 × 2 matrices over complex quaternions with the following generators:




i 0 i
i 0 e0 (≡ j) = (≡ kI) = , , e 1 0 −i
i
i 0



0 i
j 0 i
k (≡ kK) = , e , e2 (≡ kJ) = 3 i
j 0 i
k 0

86

Chapter 5. Symmetry groups

where i
is the ordinary complex imaginary (i
2 = −1) and e20 = −1, e21 = e22 = e23 = 1. The other matrices are given by





1 0 i 0 j 0 1= , I= , J= , 0 1 0 i 0 j K= iJ = jJ =

k 0

0 k



0 −j j 0 i
j 0

,

i=



,

0 −i
j

iK =

,

−1 0

0 1

jK =

0 k i
k 0



,

−k 0

iI =

,

0 −i
k

jI =

,

k=

The Dirac spinor can be expressed as a left ideal

f q1 f , E= Ψ = AE = 0 q2 f

0 0

0 i

−i 0

i
i 0 0 i


,

0 −i
i i
0

,

.



where E is a primitive idempotent (E 2 = E), f = (1 + i
k)/2 ; q1 , q2 being real quaternions and A a general element of the algebra. The hermitian norm is expressed by ΨΨ† = (q1 q1c + q2 q2c ) f where Ψ† is the transposed, quaternionic conjugated and complex conjugated matrix of Ψ. The Dirac equation in presence of an electromagnetic field is (ih∇ − eA − i
mc) Ψ = 0 ∂ − k∇, the vector potential A = j Vc + kA with the four-nabla operator ∇ = j c∂t and e the electric charge of the particle.

5.4.2 Unitary and symplectic unitary groups Having defined the Clifford algebra H ⊗ H over C, let A be a 2 × 2 matrix having as elements complex quaternions qi and its adjoint A† , A=

q1 q3

q2 q4

,

A† =



∗ q1c ∗ q2c

∗ q3c ∗ q4c

,

where the sympbols ∗ and c represent respectively the complex conjugation and the quaternionic conjugation. The adjunction transforms i
and i, j, k, I, J, K into their opposites, as one can verify directly on the basis matrices. A selfadjoint Clifford number (H = H † ) is consequently of the type H = (a + i
ib + i
jc + i
kd; i
p + iq + jr + ks)

5.4. Dirac algebra

87

1 2 3 with  p = p I + p J + p K (etc.) and real coefficients. The unitary group SU 2, H(C) , isomorphic to SU(4), is the set of matrices A such that

AA† = A† A = 1. If one restricts the matrices to real quaternion matrices, one obtains as a subgroup the symplectic unitary group USp(2, H) [39, p. 232]. The elements of the unitary
group being of the type ei H (with H = H † ), one can choose for the 15 generators of SU(4), " 



















eiθ , eIθ , eJθ , eKθ , ei jIθ , ei jJθ , ei jKθ , ei kIθ , ei kJθ , ei kKθ





 #




ejθ , ekθ , ei iIθ , ei iJθ , ei iKθ .

,

(5.17)

where θ is a real generic parameter with the usual series development eIθ = cos θ + I sin θ,


ei iJθ = cos θ + i
iJ sin θ

(etc.).

The 10 matrices within parentheses constitute the symplectic unitary group USp(2; H). Explicitly, one has eiθ = eJθ = eKθ =




ei jIθ = e e

i
jJθ

i
jKθ

e

i
kIθ




= = =

ei kKθ =



cos θ sin θ

− sin θ cos θ

cos θ + j sin θ 0



,

eIθ =

cos θ + i sin θ 0

0 cos θ + i sin θ

0 cos θ + j sin θ

,

cos θ + k sin θ 0 , 0 cos θ + k sin θ

cos θ − i sin θ 0 , 0 cos θ + i sin θ

cos θ − j sin θ 0 , 0 cos θ + j sin θ

cos θ − k sin θ 0 , 0 cos θ + k sin θ

cos θ −i sin θ cos θ i
kJθ = , e −i sin θ cos θ −j sin θ

cos θ −k sin θ . −k sin θ cos θ

−j sin θ cos θ

,

,

88

Chapter 5. Symmetry groups

One verifies that these matrices have indeed real quaternions as coefficients. The other elements of the SU(4) group are given by

e

i
iIθ




ei ikθ

cos θ + i
sin θ 0

0 cos θ − i
sin θ

cos θ −i
sin θ = , i
sin θ cos θ

cos θ −i
sin θ = . −i
sin θ cos θ

ejθ =



ekθ =

, e

i
iJθ

=



cos θ i
sin θ

i
sin θ cos θ

cos θ i
j sin θ

−i
j sin θ cos θ

,

,

To obtain a representation in terms of 4 × 4 complex matrices, one can choose for the generators of the Clifford algebra over C ⎤ ⎡
⎡ ⎤ 0 0 i 0 0 0 0 i
⎢ 0 i
0 ⎢ 0 0 ⎥ 0 i
0 ⎥ ⎥ ⎥ e0 (≡ j) = ⎢ e1 (≡ kI) = ⎢ ⎣ 0 0 −i
0 ⎦ , ⎣ 0 −i
0 0 ⎦ , 0 0 0 −i
−i
0 0 0 ⎡ ⎡ ⎤ ⎤
0 0 0 0 1 0 0 i ⎢ 0 0 −1 0 ⎥ ⎢ 0 0 0 −i
⎥ ⎥, ⎥ e2 (≡ kJ) = ⎢ e3 (≡ kK) = ⎢ ⎣ 0 −1 0 0 ⎦ , ⎣ −i
0 0 0 ⎦ 0 i
0 0 1 0 0 0 with e20 = −1, e21 = e22 = e23 = 1.

5.5 Exercises E5-1 Consider a special pure Lorentz transformation (b) along the Ox axis (velocity v = 3c ) followed by a rotation (r) of π4 around the same axis. Express the resulting Lorentz transformation X = aX
ac . E5-2 Consider a special Lorentz transformation (b1 ) along the Ox axis (velocity v = 3c ) followed by a pure Lorentz transformation (b2 ) along the Oy axis (with v = 3c ). Express the resulting Lorentz transformation X = aX
ac with a = b2 b1 . Decompose the resulting Lorentz transformation into a rotation followed by a pure Lorentz transformation (a = br). Give the direction of the Lorentz transformation, the velocity and the angle of rotation. E5-3 Consider an orthonormal system of axes (O, x, y, z) and a cube with vertices A(0, 0, 0), B(1, 0, 0), C(1, 1, 0), D(0, 1, 0), E(0, 1, 1), F (0, 0, 1), G(1, 0, 1), H(1, 1, 1). Determine the transform of the vertices of this cube under a conformal transformation x
= (1 + xac )−1 x with a = 2kI + kK , x = ctj + kx, x
= ct
j + kx
.

5.5. Exercises E5-4 Consider the matrices

1 j A= , k i

89

B=

Determine AB, BA, A−1 , B −1 , (AB)

1 k i j

−1



, −1

, (BA)

C=

1 k

i j

.

; does the matrix C −1 exist?

Chapter 6

Special relativity This chapter develops the special theory of relativity within the Clifford algebra H ⊗ H over R. The relativistic kinematics and relativistic dynamics of a point mass are examined.

6.1 Lorentz transformation 6.1.1 Special Lorentz transformation Consider a reference frame K(O, x, y, z) at rest and a reference frame K
(O
, x
, y
, z
) moving along the Ox axis with a constant velocity v (Figure 6.1). With X = jx0 + kx, X
= jx
0 + kx
(x0 = ct, x = xI + yJ + zK), b = cosh ϕ2 − iI sinh ϕ2 , the Lorentz transformation is expressed by X
= bXbc. y


y

K


K O

z

O


x

v x


z


Figure 6.1: Special Lorentz transformation (pure).

92

Chapter 6. Special relativity

Explicitly, one obtains x
0 = x0 cosh ϕ − x
sinh ϕ, x
= −x0 sinh ϕ + x
cosh ϕ, y
= y, z
= z. Writing tanh ϕ =

v c

= β, γ = cosh ϕ with γ 2 = 1 + sinh2 ϕ = 1 + γ 2 β 2 , 1 , γ=  1 − β2

the transformation becomes ct
= γ (ct − xβ) , x
= γ (x − βct) , y
= y, Taking β =

v c

z
= z.

1 (γ  1), one obtains the Galilean transformation t
= t, x
= x − vt, y
= y,

z
= z.

The inverse transformation is expressed by X = bc X
b with bc = cosh ϕ2 + iI sinh ϕ2 , hence ct = γ (ct
+ x
β) , x = γ (x
+ βct
) , y = y
, z = z
.

6.1.2 Physical consequences Contraction of length Consider a rod at rest in K
, parallel to the O
x
axis and of length l0
= x
2 − x
1 . In K, the length is l = x2 − x1 , where the abscisses x2 , x1 are determined at the same time t: x
2 = γ (x2 − βct) ,

x
1 = γ (x1 − βct) ,

6.1. Lorentz transformation

93

hence x
− x
1 l = x2 − x1 = 2 γ 
2 = l0 1 − β ≤ l0
; the observer in K concludes to a contraction. Reciprocally, let l0 = x2 − x1 be the length of a rod parallel to the Ox axis in K. The observer in K
measures at the same time t
, x2 = γ (x
2 + βct
) , x1 = γ (x
1 + βct
) ,

hence x2 − x1 l
= x
2 − x
1 = γ  = l0 1 − β 2 ≤ l0 ; the observer in K
concludes also to a contraction. Example. Take v = 300 km/s, β = 10−3 , β 2 = 10−6 , relative variation of l is 5 · 10−7 .

 1 − β 2 = 1 − 5 · 10−7 , the

Time dilatation A time interval t
= t
2 − t
1 measured at the same point x
= x
2 − x
1 = 0 of K
corresponds in K to a time interval t = t2 − t1 with ct2 = γ (ct
2 + x
2 β) ,

ct1 = γ (ct
1 + x
1 β) , hence t = t2 − t1 = γ  t
t
=  ≥ t
. 2 1−β The observer in K concludes to a slowing down of physical phenomena. Reciprocally, a time interval t = t2 − t1 measured at the same point x2 = x1 in K, gives in K
, ct
2 = γ (ct2 − x2 β) ,

ct
1 = γ (ct1 − x1 β) , t
= t
2 − t
1 t =  ≥ t. 1 − β2 Hence, the conclusion is the same.

94

Chapter 6. Special relativity

6.1.3 General Lorentz transformation For a general Lorentz transformation, one has a = br (or a
= rb) with r = cos θ2 + u sin θ2 , b = cosh ϕ2 + iv sinh ϕ2 (u · u = v · v = 1). Explicitly, one obtains ⎡

⎤ cos θ2 cosh ϕ2 + u cosh ϕ2 sin θ2 ⎦ a = br = ⎣ −iu · v sin θ2 sinh ϕ2  , +i v cos θ2 − u × v sin θ2 sinh ϕ2 ⎤ ⎡ cos θ2 cosh ϕ2 + u cosh ϕ2 sin θ2 ⎦ a
= rb = ⎣ −iu · v sin θ2 sinh ϕ2  +i v cos θ2 + u × v sin θ2 sinh ϕ2 with aac = a
a
c = 1. The general Lorentz transformation is simply expressed by X
= aXac

(or a
Xa
c )

with X = jx0 + kx, X
= jx
0 + kx
.

6.2 Relativistic kinematics 6.2.1 Four-vectors Transformation of a four-vector An arbitrary four-vector A = ja0 + ka transforms under a special Lorentz transformation as A
= bAbc with b = cosh ϕ2 − iI sinh ϕ2 ; explicitly, writing 1 γ = cosh ϕ =  1−

v2 c2

,

tanh ϕ =

one has  v , a
0 = γ a0 − a1 c  v a
1 = γ a1 − a0 , c a
2 = a2 , a
3 = a3 ; reciprocally, one has A = bc A
b

v , c

6.2. Relativistic kinematics

95

which yields  v a0 = γ a
0 + a
1 , c  v a1 = γ a
1 + a
0 , c a2 = a
2 , a3 = a
3 . If one considers a general Lorentz transformation, one obtains the following formulas ([40, p. 134], [53, p.123]) with b = cosh ϕ2 − i vv sinh ϕ2 where v is the velocity (of norm v) of the reference frame K
with respect to the reference frame K A
= bAbc or explicitly   v v a
0 = γ a0 − a · , v c  v v v a· (γ − 1) − a0 γ ; a
= a + v v c the reciprocal formulas are

A = bc A
b,

  v v , a0 = γ a
0 + a
· v c  v v v a
· (γ − 1) + a
0 γ . a = a
+ v v c Four-velocity Let X = jct + kx be the spacetime four-vector of a particle with dX = jcdt + kdx and the relativistic invariant

v2 2 2 2 2 2 dXdXc = c dt − (dx) = c dt 1 − 2 c 2 2 c dt = c2 dτ 2 = γ2 which defines the proper time dt = dt dτ = γ with γ = by

q 1 2 1− vc2

 1−

v2 c2

, v being the velocity of the particle. The four-velocity V is defined

V =

dX = jγc + kγv dτ

96 with v =

Chapter 6. Special relativity dx dt

and V Vc = c2 ; V can also be written in the form V = c (j cosh θ + km sinh θ)

with v = vm (m · m = 1) and tanh θ = vc , γ = cosh θ, sinh θ = γ vc . Four-acceleration Let V = jγc + kγv be the four-velocity and dτ = the four-acceleration is defined by A=

dt γ

the proper time differential;

dV = γ [jcγ˙ + k (γv ˙ + γa)] dτ

with v v˙ dγ = 2 γ3 dt c γ3 = (v · a) 2 c

γ˙ =

·

(6.1) (6.2)

where the relation v · v = v v˙ deduced from (v)2 = v 2 has been used. Furthermore, V · V = c2 from which one obtains by differentiating with respect to τ , V · A = 0 which gives again equation (6.2) (v · a) = c2

γ˙ . γ3

The four-acceleration can be written A = jγ 4 with γ =

q 1 2 1− vc2



(v · a) (v · a) v 2 + k γ4 + γ a c c2

; one has the relativistic invariant 2

(v · a) 2 AAc = −γ 6 − γ 4 (a) 2 c

v×a = γ 6 −a2 + c2 with (v × a)2 = v 2 a2 − (v · a)2 . The bivector V ∧ A is given by V ∧ A = (v × a) γ 3 − iacγ 3 ; since V A = − (V · A) − (V ∧ A) = − (V ∧ A), one has (V ∧ A) (V ∧ A)c = V AAc Vc = c2 AAc   2 = γ 6 −a2 c2 + (v × a) .

6.2. Relativistic kinematics

97

The four-acceleration in the proper frame (v = 0) is simply A = ka with 2

AAc = − (ap ) = −a2p . When a is parallel to v, one has AAc = −γ 6 a2 = −a2p , ap = γ 3 a. If a is perpendicular to v, one has AAc = −γ a

6 2

v2 1− 2 c



= −γ 4 a2 = −a2p , ap = γ 2 a; when γ  1, one remarks that the proper acceleration can be very much larger than the acceleration in the laboratory frame [42, p. 101]. Wave four-vector The wave four-vector is defined with k = K=j

2π λ n

by

ω + kk; c

writing X = jct + kr one has the relativistic invariants KKc =

ω2 2 − (k) , c2

K · X = ωt − k · r.

6.2.2 Addition of velocities Special Lorentz transformation Consider the special Lorentz transformation V
= bV bc

(6.3)

with b = cosh

ϕ ϕ − iI sinh , 2 2

tanh ϕ =

w , c

1 cosh ϕ = γ =  1−

w2 c2

the four-velocities V = c (j cosh θ + km sinh θ) ,

V
= c (j cosh θ
+ km
sinh θ
) ,

,

98

Chapter 6. Special relativity

and tanh θ = v = vm,

v , c

tanh θ
= v
= v
m

v
, c


(m · m = m
· m
= 1).

Equations (6.3) read cosh θ
= cosh θ cosh ϕ − m1 sinh θ sinh ϕ,

(6.4)


1




1

(6.5)


2




2

(6.6)


3




3

(6.7)

m sinh θ = m cosh ϕ sinh θ − cosh θ sinh ϕ, m sinh θ = m sinh θ, m sinh θ = m sinh θ. Dividing equation (6.5) by the equation (6.4), one obtains m
1 tanh θ
=

m1 tanh θ − tanh ϕ 1 − m1 tanh θ tanh ϕ

and thus v
1 =

v1 − w . v1 w 1− 2 c

(6.8)

Similarly, one obtains v2 ,  v1 w cosh ϕ 1 − 2 c 3 v   v
3 = v1 w cosh ϕ 1 − 2 c

v
2 =

with cosh ϕ = γ =

q 1 2 1− w c2

. The inverse formulas are (with β = v
1 + w , v
1 w 1+ 2 c
2 v 1 − β2 v2 = , v
1 w 1+ 2  c
3 1 − β2 v v3 = . v
1 w 1+ 2 c v1 =

(6.9)

(6.10)

w c)

(6.11)

(6.12)

(6.13)

6.3. Relativistic dynamics of a point mass

99

General pure Lorentz transformation It is expressed for four-velocities by V
= bV bc

(6.14) 

ϕ




with b = cosh ϕ2 − i w w sinh 2 , v = vm, v = v m (m · m = m · m = 1), w v v

tanh ϕ = c , tanh θ = c , tanh θ = c . Explicitly, equation (6.14) reads    w tanh θ tanh ϕ , cosh θ
= cosh θ cosh ϕ 1 − m · w w w

m sinh θ = m sinh θ − m· sinh θ w w  w w w m· sinh θ cosh ϕ − cosh θ sinh ϕ. + w w w Dividing equation (6.17) by equation (6.15), one obtains with β = [40, p. 75]   v · w    2 −1 1 − v 1 − β2 + w 1 − β w2 v
= v·w 1− 2 c and the inverse formula


   v ·w 
2 2 1− 1−β +1 v 1−β +w w2 . v= v
· w 1+ c2

w c

(6.15) (6.16) (6.17)

the formulas

6.3 Relativistic dynamics of a point mass 6.3.1 Four-momentum Let V = jγc + kγv be the four-velocity of a particle and m0 its mass, the fourmomentum is defined by P = m0 V = jm0 γc + kγm0 v E = j + kp c where E = γm0 c2 is the energy of the particle and p = γm0 v its momentum (of norm p). Furthermore, one has the relativistic invariant E2 2 − (p) c2 = m20 V Vc = m20 c2 ,

P Pc =

100

Chapter 6. Special relativity

hence E 2 = p2 c2 + m20 c4 . In the proper frame (v = 0), one has E0 = m0 c2 ; the kinetic energy is defined by T = m0 c2 (γ − 1) . Under a standard transformation, the four-momentum transforms as P
=

Lorentz  ϕ ϕ bP bc with b = cosh 2 − iI sinh 2 and tanh ϕ = wc = β, γ = q 1 w2 1−

c2

 E
= γ E − p1 w ,   E
1 1 p = γ p − 2w , c p
2 = p2 ,

p
3 = p3 .

The inverse transformation is P
= bc P b or 

E = γ E
+ p
1 w ,   E
p1 = γ p
1 + 2 w , c p2 = p
2 ,

p3 = p
3 .

Under a general pure Lorentz transformation, one has P
= bP bc with b = cosh ϕ2 − i vv sinh ϕ2 .

6.3.2 Four-force Let P = j Ec + kp be the four-momentum vector of the particle; the four-force vector is defined by d (m0 V ) dP = dτ dτ dm0 V = m0 A + dτ

F =

with V = jγc+ kγv [42, pp. 123-124]. If the four-force conserves m0 , then and d (jm0 γc + kγm0 v) F = m0 A = γ  dt dE = γ (v) j + kf cdt

dm0 dτ

=0

6.3. Relativistic dynamics of a point mass with, by definition f =

dp dt

=

d(γm0 v) . dt

101

Furthermore,

  dm0 2 dm0 V ·V = c m0 A + dτ dτ   dE −f ·v . = γ 2 (v) dt

F ·V =

If

dm0 dτ

= 0 = F · V , then

dE dt

= f · v and

(f · v) + kf . F = γ (v) j c Since d (γm0 v) d (γm0 ) = γm0 a + v dt dt dE (f · v) = γm0 a + 2 v = γm0 a + v, c dt c2

f=

one infers that a is coplanar to f and v but is not in general parallel to f . If dm0 dτ = 0, the force f satisfies the relation [42, p. 125] γ (v) f = m0

d2 x dt2

(6.18)

which can be established as d2 X F = m0 A = m0 2 dτ   dE + kf = γ (v) j cdt 2

d (jct) d2 x = m0 + k , dτ 2 dt2

(6.19) (6.20) (6.21)

hence the relation (6.18) by comparing equations (6.20) and (6.21).  Under a speQ + kf with Q = dE cial Lorentz transformation, the four-force F = γ (v) j c dt , 

ϕ ϕ transforms as F
= bF bc with b = cosh 2 − iI sinh 2 , tanh ϕ = wc et cosh ϕ = γ = q 1 w2 or explicitly 1−

c2

 w , F
0 = γ (w) F 0 − F 1 c  w , F
1 = γ (w) F 1 − F 0 c
2 2
3 3 F =F , F =F ,

102

Chapter 6. Special relativity

and equivalently  γ (v
) Q
= γ (w) γ (v) Q − f 1 w ,

Qw γ (v
) f
1 = γ (w) γ (v) f 1 − 2 , c γ (v
) f
2 = γ (w) f 2 ,


γ (v ) f


3

(6.22) (6.23) (6.24)

3

= γ (w) f .

(6.25)

Furthermore, one has the relation

γ (v
) v1 w = γ (w) 1 − 2 , γ (v) c which can be established as follows ([42, p. 69]); form the invariant

 
ds2 = dt2 c2 − v 2 = dt 2 c2 − v
2 and the relation  w cdt
= γ (w) cdt − dx c   v1 w = γ (w) cdt 1 − 2 . c Then one obtains

2

  v1 w 2 dt2 c2 − v 2 = dt2 γ 2 (w) 1 − 2 c − v
2 , c   

2 2 w c − v 2 1 − c2 

2 , c − v
2 =

2 1 1 − vc2w 1 1 1 1 = 2 , γ 2 (v
) γ (v) γ 2 (w) 1 − v12w 2 c

hence, the relation (6.26). Reciprocally, one has

v
1 w γ (v) = γ (w) 1 + . γ (v
) c2

(6.26)

6.4. Exercises

103

Finally, equations (6.22), (6.23), (6.24), (6.25) can be written [42, p. 124] Q
= f
1 =

Q − f 1w , 1 1 − vc2w f1 − 1−

wQ c2 , v1 w c2 2

f
2 =

f

γ (v) 1 −

v1 w c2

f
3 =

f3

γ (v) 1 −

v1 w c2

, .

6.4 Exercises E6-1 Two particles A and B move towards the origin O in opposite directions (on the Ox axis) with a uniform velocity 0, 8 c. Determine the relative velocity of B with respect to A for an observer at rest relatively to A. E6-2 Consider an isolated set of particles without interaction in a reference frame K at rest. Determine the total relativistic angular momentum ' L= Xi ∧ Pi , Xi = jct + kIxi + kJyi + kKzi , Ei + kIpxi + kJpyi + kKpzi . Pi = j c Define the center of energy of the set. Show that it moves with a constant velocity. E6-3 Consider a hyperbolic rectilinear motion of a particle whose acceleration is constant in the proper reference frame, at any instant. The particle being at rest at the origin of the axes and of the time, determine the four-vector X = jct + kx as a function of the parameter ϕ = gτ c where τ is the proper time of the particle. Show that one obtains the classical results of a uniformly accelerated motion for small velocities ϕ 1.

Chapter 7

Classical electromagnetism Classical electromagnetism is treated within the Clifford algebra H ⊗ H over R. This chapter develops Maxwell’s equations, electromagnetic waves and relativistic optics.

7.1 Electromagnetic quantities 7.1.1 Four-current density and four-potential Four-current density Let ρ0 be the charge density in the proper frame and V = jγc + kγv the fourvelocity; the four-current density is defined by C = ρ0 V = jγρ0 c + kγρ0 v = jρc + kj with ρ = γρ0 and j = ρv. One obtains the relativistic invariant   v2 2 2 2 2 2 2 CCc = ρ c − ρ (v) = ρ c 1 − 2 c 2 ρ = c2 2 = c2 ρ20 γ with γ =

q 1 2 1− vc2

. Under a pure special Lorentz transformation, the four-current

density transforms as C
= bCbc with b = cosh ϕ2 − iI sinh ϕ2 , tanh ϕ = cosh ϕ = γ = √ 1 2 ; explicitly, one has 1−β

 ρ
c = γ ρc − j 1 β ,

 j
1 = γ j 1 − ρcβ , j
2 = j 2 ,

j
3 = j 3 .

w c

= β and

106

Chapter 7. Classical electromagnetism

Consider a trivolume OM (α, β, γ) = jx0 (α, β, γ)+kx (α, β, γ) depending on three parameters α, β, γ. The infinitesimal hyperplane dT is expressed by (4.17)   ∂OM ∂OM ∂OM ∧ ∧ dT = dαdβdγ ∂α ∂β ∂γ  kdx1 dx3 dx2 + jIdx2 dx3 dx0 . = +jJdx3 dx1 dx0 + jKdx1 dx2 dx0 The dual dT ∗ of dT is given by dT ∗ = idT  jdx1 dx2 dx3 + kIdx2 dx3 dx0 = +kJdx3 dx1 dx0 + kKdx1 dx2 dx0 and is a four-vector orthogonal to dT . The relativistic invariant C · dT ∗ is given by  ρcdx1 dx2 dx3 − ρv 1 dx2 dx3 dx0 ∗ C · dT = ; −ρv 2 dx3 dx1 dx0 − ρv 3 dx1 dx2 dx0 in the proper frame C · dT ∗ = ρ0 cdx1 dx2 dx3 = cdq where dq is the electric charge contained in dT . The electric charge Q contained in the hyperplane T is given by &&& 1 Q= C · dT ∗ c T

for any Galilean frame. Furthermore, if one integrates over a closed four-volume τ , one obtains % % ∗ C · dT = (∇ · C) dx0 dx1 dx2 dx3 τ

T

1

2

3

∂ρc ∂ρv ∂ρv ∂ρv with the relation ∇ · C = ∂x = 0 expressing the con0 + ∂x1 + ∂x2 + ∂x3 servation of electric charge. If the four-volume τ is limited by the hypersurfaces H1 (t1 = const.), H2 (t2 = const.) and the lateral hypersurface H3 at infinity, one has (Figure 7.1) & & &

C · dT ∗ +

H1

C · dT ∗ +

H2

C · dT ∗ = 0;

H3

furthermore, the integral over H3 is nil in the absence of electric charges at infinity, hence & & ∗ Q2 = C · dT = − C · dT ∗ = Q1 H2

H1

which expresses the conservation of electric charge.

7.1. Electromagnetic quantities

107 t

H2

t2 H3

τ

t1

H1

Figure 7.1: Conservation of electric charge: H1 and H2 are hyperplanes (at t constant) and H3 is a lateral hypersurface at infinity.

Four-potential vector Let V be the scalar potential and A the potential vector, the four-potential vector 2 2 is defined by A = j Vc + kA yielding the relativistic invariant AAc = Vc2 − (A) . ϕ Under a special Lorentz transformation, one has A
= bAbc with b = cosh 2 − iI sinh ϕ2 , tanh ϕ = wc = β and cosh ϕ = γ = √ 1 2 , hence 1−β



 V 1 −A β , c   V
1 1 A =γ A − β , c V
=γ c

A
2 = A2 ,

A
3 = A3 .

7.1.2 Electromagnetic field bivector Four-nabla operator The four-nabla operator ∇=j

∂ ∂ ∂ ∂ − kI 1 − kJ 2 − kK 3 c∂t ∂x ∂x ∂x

transforms under a general Lorentz transformation as a four-vector, i.e., ∇
= a∇ac . Let us demonstrate it explicitly in the case of a pure special Lorentz trans-

108

Chapter 7. Classical electromagnetism

formation. The transformation formulas are ct = γ (ct + x
β) , x = γ (x
+ βct
) , y = y
, with β =

v c

and γ =

q 1 2 1− vc2

z = z
,

. One then obtains

∂ 1 ∂ ∂t ∂ ∂x ∂ ∂y ∂ ∂z = + + + c∂t
c ∂t ∂t
∂x ∂t
∂y ∂t
∂z ∂t
 

∂ ∂ − − =γ β c∂t ∂x with

∂t ∂t


= γ,

∂x ∂t


= γβc,

∂y ∂t


−

with

∂x ∂x


= γ,

∂t ∂x


=

∂z ∂t


= 0. Furthermore,

∂ ∂x ∂ ∂t ∂ = − + ∂x
∂x ∂x
∂t ∂x


∂ ∂ =γ − −β ∂x c∂t

= γ βc . Finally, one has  

∂ ∂ ∂ − − =γ β , c∂t
c∂t ∂x

∂ ∂ ∂ −β , −
=γ − ∂x ∂x c∂t ∂ ∂ ∂ ∂ −
=− , −
=− , ∂y ∂y ∂z ∂z

which shows that the four-nabla operator transforms indeed as a four-vector, the demonstration being similar in the general case.

Electromagnetic field bivector Let A = j Vc + kA be the four-potential vector, let us define the Clifford number F = ∇c A = −∇A = (∇ · A) + (∇ ∧ A)

7.1. Electromagnetic quantities

109

and adopt the Lorentz gauge (∇ · A) = the electromagnetic field bivector

∂ ( Vc ) c∂t

1

2

∂A + ∂A ∂x1 + ∂x2 +

∂A3 ∂x3

= 0, which gives

F = ∇∧A

  V ∂A = − rot A + i − grad − c c∂t E = −B + i c with the usual definitions of the magnetic induction B and the electric field E, E = − grad V −

B = rot A,

∂A . ∂t

(7.1)

Under a gauge transformation A → A
= A + ∇f where f is an arbitrary scalar function, one obtains the same electromagnetic field bivector F
= ∇ ∧ A
= ∇ ∧ A + (∇ ∧ ∇) f = ∇ ∧ A = F. The electromagnetic field bivector yields the relativistic invariant F2 = F · F + F ∧ F # " 2 (E) E·B 2 . = − (B) + 2 + 2i c c Under a pure special Lorentz transformation , one has F
= bF bc with b = cosh ϕ2 − iI sinh ϕ2 , tanh ϕ =

v c

= β and cosh ϕ = γ =

  E3 2 ⎢ −B I − B cosh ϕ + c sinh ϕ J ⎢   2 ⎢ ⎢ + −B 3 cosh ϕ + E sinh ϕ K ⎢ c ⎢  2  F
= ⎢ 1 ⎢ E E 3 ⎢ + iI + cosh ϕ − B sinh ϕ iJ ⎢ c c ⎢   ⎣ E3 cosh ϕ + B 2 sinh ϕ iK + c ⎡

1

= −B
+ i

E
c

q 1 2 1− vc2

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

, hence

110

Chapter 7. Classical electromagnetism

or B


1

 E3 β , B =γ B + c

 E
2 = γ E 2 − B 3 βc ,
2

1

=B ,

E
1 = E 1 ,

  E2 3 B =γ B − β , c

 E
3 = γ E 3 + B 2 βc .




3

2

Under a general pure Lorentz transformation, one has F
= bF bc with b = cosh ϕ2 − i vv sinh ϕ2 , tanh ϕ = vc = β and cosh ϕ = γ = q 1 v2 ; hence the for1− c2

mulas [40, p. 191]

 

1 1 v − 1 − 2 (v × E) , B
= γ B + 2 (v · B) v γ c 

 1 v E
= γ E + 2 (v · E) − 1 + (v × B) . v γ

7.2 Maxwell’s equations 7.2.1 Differential formulation In vacuum With the Lorentz gauge ∇ · A = 0, one has F = ∇ ∧ A = −B + i

E , c

∇ ∧ F = ∇ ∧ ∇ ∧ A = 0,   ∂B E = k (− div B) + j − − rot ; c∂t c hence, one obtains two of the Maxwell’s equations div B = 0,

rot E = −

∂B . c∂t

The two other equations are given by ∇ · F = µ0 C = j div

  1 ∂E E + k rot B − 2 c c ∂t

with the four-current density C = jρc + kρv; one obtains with ε0 µ0 c2 = 1, div E = µ0 ρc2 = rot B =

ρ , ε0

1 ∂E + µ0 j. c2 ∂t

7.2. Maxwell’s equations

111

The complete set of Maxwell’s equations reads ∇F = ∇∇c A = ∇ · F + ∇ ∧ F   ⎡ E 1 ∂E ⎢ j div c + k rot B − c2 ∂t   =⎢ ⎣ E ∂B − rot +k (− div B) + j − c∂t c

⎤ ⎥ ⎥ ⎦

=
A = µ0 C with the d’Alembertian operator
= ∇∇c = four-potential vector. Hence, the equations

∂2 c2 ∂t2

−  and A = j Vc + kA the

V ρ = ρµ0 c2 = , c ε0
A = µ0 j,




or equivalently div B = 0, rot E = −

div E = ∂B , ∂t

rot B = µ0 j +

ρ , ε0

∂ε0 E . ∂t

If one does not adopt the Lorentz gauge, one obtains F = ∇c A = −∇A  

∂V E = div A + −B+i c∂t c which is not a bivector but an element of C + . One obtains the same Maxwell’s equations ∇F = ∇∇c A =
A = µ0 C. Since ∇ · C = 0 expresses the conservation of electric charge, one has ∇ · (
A) = µ0 ∇ · C =
(∇ · A) = 0, a condition which is less restrictive than the Lorentz gauge ∇ · A = 0. The covariance of Maxwell’s equations is manifest since under a general Lorentz transformation any Clifford number X transforms as X
= aXac (aac = 1) and thus ∇
F
= µ0 C
= a∇F ac = aµ0 Cac ;

112

Chapter 7. Classical electromagnetism

hence, ∇F = µ0 C. Perfect dielectric or magnetic medium Consider the bivectors F = −B + i Ec , G = −H + icD. Maxwell’s equations are expressed by ∇ ∧ F = 0, ∇ · G = C = jρc + kρv   ∂D = j div (cD) + k rot H − ∂t or div B = 0,

div D = ρ,

∂D ∂B , rot H = j + . ∂t ∂t The bivector G yields the relativistic invariant rot E = −

G2 = G · G + G ∧ G = −H2 + c2 D2 + 2icD · H. Under a pure special Lorentz transformation, G transforms as G
= bGbc with b = cosh ϕ2 − iI sinh ϕ2 , tanh ϕ = vc = β and cosh ϕ = γ = q 1 v2 , hence 1− c2

Hx
= Hx , Hy
= γ (Hy + vDz ) , Hz
= γ (Hz − vDy ) , Dx
= Dx ,   Hz
Dy = γ Dy − v 2 , c   Hy
Dz = γ Dz + v 2 . c Under a general pure Lorentz transformation, one has G
= bGbc with b = cosh ϕ2 − i vv sinh ϕ2 , tanh ϕ = vc = β and cosh ϕ = γ = q 1 v2 ; hence 1− c2



1 v − 1 − (v × D) , H = γ H + 2 (v · H) v γ    

1 H v
−1 + v× 2 D = γ D + 2 (v · D) , v γ c






(7.2) (7.3)

7.2. Maxwell’s equations

113

and the inverse formulas  

1 v


H = γ H + 2 (v · H ) − 1 + (v × D ) , v γ    

1 H
v

D = γ D + 2 (v · D ) −1 − v× 2 . v γ c As an application, consider a perfect medium in the proper reference frame K
with B
= µ0 µr H
, D
= ε0 εr E
, moving with respect to a Galilean reference frame (at rest) K with a velocity v. Using the transformation formulas (7.2), (7.3), one obtains  

1 1 v − 1 − 2 (v × E) (7.4) B + 2 (v · B) v γ c 

 1 v = µ0 µr H + 2 (v · H) − 1 − (v × D) , (7.5) v γ    

1 v H D + 2 (v · D) −1 + v× 2 v γ c 

 1 v − 1 + (v × B) . = ε0 εr E + 2 (v · E) v γ

(7.6) (7.7)

Furthermore, from equations (7.5), (7.7) one deduces the relations v · B = µ0 µr v · H, v · D = ε0 εr v · E. Equations (7.5), (7.7) then become 1 (v × E) = µ0 µr [H − (v × D)] , c2  H D + v × 2 = ε0 εr [E + (v × B)] ; c

B−

(7.8) (7.9)

taking H from equation (7.8) and replacing it in equation (7.9) one finds [7, p. 239]     ε0 εr − µ1r E v× B−v× 2 . D = ε0 εr E + 1 − β2 c Similarly, taking D from equation (7.9) and replacing it in equation (7.8) one obtains   ε0 εr − µ1r B H= + v × (E + v × B) . µ0 µr 1 − β2 One observes that D and H depend on E and B as well as on the velocity of the material.

114

Chapter 7. Classical electromagnetism

Arbitrary dielectric or magnetic medium In an arbitrary dielectric or magnetic medium, with a polarization density P and a magnetization density M, one has the relations D = ε0 E + P, B = µ0 (H + M) which one can write in the form F = µ0 (G + N ) with the bivectors F = −B + i Ec , G = −H + icD and N = −M − icP. Maxwell’s equation ∇ · G = C (with C = jρc + kρv) then becomes

F ∇·G= ∇· −N =C µ0 or ∇ · F = µ0 (C + ∇ · N ) with   ∂P ∇ · N = j div (−cP) + k rot M + , ∂t     1 ∂E E + k rot B − 2 , ∇ · F = j div c c ∂t hence, the relations 1 (ρ − div P) , c   ∂E ∂P rot B = µ0 j + + rot M + ε0 µ0 . ∂t ∂t div E =

The entire set of Maxwell’s equations is expressed by (with ∇ ∧ F = 0) ∇F = ∇ · F + ∇ ∧ F =
A = µ0 C + µ0 ∇ · N or ρ − div P , ε  0 ∂P
A = µ0 j + + rot M . ∂t
V =

7.2. Maxwell’s equations

115

Consequently, one can replace the medium by a distribution, in vacuum, of charge density ρ
= div P and a current density j
= ∂P ∂t + rot M. Under a general pure Lorentz transformation, the bivector magnetization-polarization N transforms as N
= bN bc with b = cosh ϕ2 − i vv sinh ϕ2 , tanh ϕ = vc = β and cosh ϕ = γ = q 1 v2 , hence

1− c2



 1 v − 1 + (v × P) , M
= γ M + 2 (v · M) v γ    

1 M v P
= γ P + 2 (v · P) −1 − v× 2 v γ c

and the inverse formulas  

1 v


M = γ M + 2 (v · M ) − 1 − (v × P ) , v γ    

1 M
v

−1 + v× 2 P = γ P + 2 (v · P ) . v γ c For low velocities (γ  1), one obtains M = M
− v × P
, P = P
+ v ×

M
. c2

If, in the proper reference frame K
, one has M
= 0 and P
= 0, then in the reference frame at rest K one has M = −v × P
; if in K
, P
= 0, M
= 0, then one obtains in K a polarization P=v×

M
. c2

7.2.2 Integral formulation The relativistic integral formulation of Maxwell’s equations in vacuum results from the general formulas valid for any bivector F (4.20), (4.21) % & F · dS = (∇ ∧ F ) · dT, % & F ∧ dS = − (∇ · F ) ∧ dT, with dS = dS + idS
, dS = dydzI + dzdxJ + dxdyK, dS
= cdxdtI + cdydtJ + cdzdtK, dT = −kdxdydz + j (dydzcdtI + dzdxcdtJ + dxdycdtK) ,

116

Chapter 7. Classical electromagnetism

and where the integration is taken on a closed surface . Considering F = −B + i E c with ∇ ∧ F = 0 and ∇ · F = µ0 C, one obtains [4, p. 227] % 

% F · dS = %

= 0, i F ∧ dS = c

Bx dydz + By dzdx + Bz dxdy +Ex dxdt + Ey dydt + Ez dzdt

&  &

= iµ0



Ex dydz + Ey dzdx + Ez dxdy −Bx dxcdt − By dycdt − Bz dzcdt



(ρcdxdydz − jx dydzdt − jy dzdxcdt − jz dxdycdt) .

If one operates at t constant, one finds again the standard equations (in classical three-vector formulation) % % q B · dS = 0, E · dS = . ε0 Furthermore, using the general formula (4.18) % & A · dl = − (∇ ∧ A) · dS one obtains for the four-potential vector A (with the Lorentz gauge ∇ · A = 0) % & A · dl = − F · dS or [4, p. 231] & 

% A · dx − V dt =

Bx dydz + By dzdx + Bz dxdy +Ex dxdt + Ey dydt + Ez dzdt

 .

At t constant, one obtains the classical vector formulation of Stokes’ theorem % & & A · dl = rot A · dS = B · dS.

7.2.3 Lorentz force The four-force density (per unit volume of the laboratory frame) is expressed with F = −B + i Ec and C = jρc + kρv by f = F ·C = j (j · E) + k (ρE + j × B) ;

7.2. Maxwell’s equations

117

f can also be expressed in the form (with ∇F = µ0 C and ∇ = ∂α eα ) 1 F C − CF = [F (∇F ) − (∇F ) F ] 2 2µ0 1  = F (∂α eα F ) − (∂α eα F ) F 2µ0 1 = [∂α (F eα F ) − (∂α F ) eα F − (∂α eα F ) F ] . 2µ0

f=

Since ∇F + F ∇ 2 1 = [∂α (eα F ) + (∂α F ) eα ] = 0 2

∇∧F =

it follows that f=

1 ∂α (F eα F ) = −∂α tα 2µ0

with tα =

−F eα F = T αβ eβ 2µ0

where T αβ is the energy-momentum tensor; tα is a four-vector since (tα )c = −tα (with Fc = −F ). One obtains for tα , E×H (B · H + E · D) +k , 2 c ⎤ ⎡  2 E2  B + c2 E12 2 I − B − (E × H) k 2 1 c  1 ⎣   ⎦, 2 + t1 = j c µ0 − B1 B2 + E1c2E2 J − B1 B3 + E1c2E3 K  ⎤ ⎡  E1 E2 I − B B + 1 2 2 c (E × H)2 k ⎣  B2 + E2    ⎦, t2 = j + E22 E2 E3 c µ0 2 c2 J − B K − B − B + 2 2 3 2 2 2 c c   ⎤ ⎡ − B1 B3 + E1c2E3 I (E × H) k 3 ⎣   B2 + E2  ⎦.  + t3 = j E2 ´ c µ0 ce − B B + E2 E3 J + − B2 − 3 K

t0 = j

2

3

c2

2

3

c2

The four-vectors tα constitute the energy-current density (α = 0) and the momentum-current density along the three axes (x, y, z); the α component of the four-momentum of the electromagnetic field of a trivolume T is given by & P α = tα · (dT ∗ ) .

118

Chapter 7. Classical electromagnetism

The symmetry of the energy-momentum tensor T αβ of the electromagnetic field can be verified directly on the components of tα tα · e β = tβ · e α = T αγ eγ · eβ = T αβ = T βα ; it is sufficient to verify the equality

 (F eα F ) · eβ = eα · F eβ F ,

(7.10)

F a = aF

(7.11)

with a = eα F eβ − eβ F eα ∈ C + ; since a = ac , a is constituted by a scalar and a pseudoscalar, and thus commutes with F , which demonstrates the equality (7.10).

7.3 Electromagnetic waves 7.3.1 Electromagnetic waves in vacuum Starting from Maxwell’s equations ∇F = µ0 C with F = −B + i Ec , C = jρc + kj, one obtains in the absence of electric charges and currents ∇c ∇F =
F = 0 where
= ∇∇c =

∂2 c2 ∂t2

−  is the d’Alembertian operator; hence
B = 0,


E = 0.

A sinusoidal plane wave rectilinearly polarized is, in complex notation, of the type


F = Fm ei (ωt−k·r) = −B + i

E c

where Fm = −Bm + i Ecm is a constant bivector (Bm , Em real). One has, in a vacuum,  ω  ∇F = i
j + kk F ⎧ c  ⎨ j −k·E + k (k · B) c  = i
⎩ +j (k × E − Bω) + k −k × B − = 0;

ωE c2



⎫ ⎬ ⎭

7.3. Electromagnetic waves

119

hence, the relations characterizing the plane wave k · E = 0, k · B = 0, k×E , B= ω − (k × B) c2 E= . ω The relation ∇c ∇F =

(7.12) (7.13) (7.14) (7.15)

  2 ω2 j + kk F = − 2 + k · k F = 0 c c

ω

ω = c. gives the norm of k, |k| = ωc and the phase velocity v = |k| For a plane wave polarized elliptically, one just needs to take Fm complex.

7.3.2 Electromagnetic waves in a conductor Consider a conductor having a permittivity = ε0 , a permeability µ = µ0 and a conductivity γ (j = γE). Maxwell’s equations are expressed by ∇F = µ0 C, hence ∇c ∇F =
F = µ0 ∇c C

(7.16)

= µ0 (∇ · C + ∇ ∧ C) = µ0 (∇ ∧ C) 

 ∂j = µ0 − rot j + i − − (c) grad ρ c∂t

(7.17) (7.18)

with the conservation of electric charge (∇ · C = 0). Using the relations j = γE and ∇ ∧ F = 0, equation (7.18) becomes
F + γµ0

∂F = −i(µ0 c) grad ρ. ∂t

In the absence of an electric charge density (ρ = 0), one has
F + γµ0

∂F = 0. ∂t

(7.19)


Consider a sinusoidal plane wave rectilinearly polarized F = Fm ei t(ωt−k·r) with the constant bivector Fm = −Bm + i Ecm . Equation (7.19) gives with  

ω2 ∂F 2 = i
ωF ,
F = − 2 + |k| F ∂t c a complex expression of k ω2 − i
ωγµ0 . c2 The structure of the plane wave results from equation ∇F = µ0 C with C =
Cm ei (ωt−k·r) and yields the relations (7.12), (7.13), (7.14), (7.15) with k complex. |k|2 =

120

Chapter 7. Classical electromagnetism

7.3.3 Electromagnetic waves in a perfect medium Wave equation In a perfect medium, one has the relations ∇ ∧ F = 0,

∇·G= C

with F = −B + i Ec , G = −H + icD, C = jρc + kj. Hence ∇ ∧ (∇ · G) = ∇ ∧ C

(7.20)

with ( 

rot ∂D − rot (rot H)  ∂t  ∇ ∧ (∇ · G) = ∂2 D H +i c∂t2 − (c) grad (div D) − ∂ rot c∂t   ∂j − (c) grad ρ . ∇ ∧ C = − rot j + i − c∂t

) ,

Equation (7.20) can be written in the form G − εµ

∂2G ∂G + grad (div D) . = γµ 2 ∂t ∂t

In a nonconducting medium (γ = 0) and in the absence of an electric charge density, one obtains the wave equation G − εµ

∂ 2G = 0. ∂t2

(7.21)

∂
− k∇, G
= −H + vD, F
= −B + i E Introducing the operator ∇
= j v∂t v = µG , 1
C = jρv + kj and the constant v = √εµ , Maxwell’s equations are expressed by [5]

∇
∧ G
= 0, ∇
· G
= C
or ∇
G
= C
. Hence, the wave equation ∇
c ∇
G
= ∇
c C
=
G
. In a nonconducting medium (γ = 0), one finds again the equation (7.21).

7.4. Relativistic optics

121

Sinusoidal plane wave rectilineraly polarized


Consider G
= G
m ei (ωt−k·r) with the constant bivector G
m = −Hm + ivDm and v = √1εµ . In the absence of a four-current density C
, one has  ω  ∇
G
= i
j + kk G
( v ) j (−vk · D) + k (k · H)


 =i +j vk × D − Hω + k (−k × H − Dω) v = 0; hence, the relations characterizing the plane wave k · D = 0, k · H = 0, k×D , ω −k × H . D= ω

H = v2

7.4 Relativistic optics 7.4.1 Fizeau experiment (1851) In the Fizeau experiment (Figure 7.2), one observes at O a system of interference fringes when the water is at rest. When the water circulates, the fringes move and the experiment has determined the velocity of light in the moving water to be   1 c v = ± 1− 2 w n n whereas the classical theory predicts v=

c ± w. n

Classical theory The variation of the optical path δ between the two experiments is " # L L  − c  ∆δ = c c n −w n +w =

2Lcw 2Ln2 w .  c − w2

c2 n2

122

Chapter 7. Classical electromagnetism source

water inflow T1 w

w T2 L1

L2 L

f

Figure 7.2: Fizeau experiment (1851): when the water is at rest, one observes interference fringes; when the water circulates, the fringes move which allows one to determine the velocity of light in the moving water. The variation of the order of interference is ∆δ 2Ln2 w = . ∆pc = λ0 cλ0 Relativistic theory Let K
be the proper frame of the water in motion with nc the speed of light with respect to K
. With respect to the frame at rest K, the velocities of light with respect to the tubes T1 , T2 are respectively using the formula (6.11),   c 1 c n −w − 1 − 2 w, v1 = w  1 − nc n n   c 1 c n +w + 1 − 2 w, v2 = w  1 + nc n n hence, a variation of the optical path δ,   L L − ∆δ = c v1 v2

2   2Lcw n − 1 2Lw 2 = n −1 ;  2 2 2 c −n w c the variation of the order of interference is  ∆δ 2Lw 2 n −1 . = ∆pr = λ0 cλ0

7.4. Relativistic optics

123

Example. Take w = 10 m/s, L = 5 m, λ0 = 0, 6 µm, n = 4/3, one obtains ∆pc = 0, 99;

∆pr = 0, 43.

7.4.2 Doppler effect Consider an optical source at rest within the proper frame R
moving with a velocity w with respect to the frame at rest R (Figure 7.3). y


y

R


R O

x

O


w x


z


z

Figure 7.3: Doppler effect: an optical source (at rest in R
), of frequency f
, located at O
moves with a velocity w along the Ox axis. In the reference frame R, the measured frequency is f .


Let K
= j ωc + kk
be the wave four-vector in R
and K = j ωc + kk the wave four-vector in R. One has with kx
= k
cos θ and f, f
designating the frequencies   w w = γω
1 + cos θ , ω = γ ω
+ kx
c c   w . f = γf
1 + cos θ c Longitudinal Doppler effect 1. θ = 0 (the source moves towards the observer located at O, Figure 7.4) !  1 + wc w

f = γf 1 + =f ≥ f
; c 1 − wc 2. θ = π (the source recedes from the observer, Figure 7.4) !  1 − wc w

f = γf 1 − ≤ f
. =f c 1 + wc

124

Chapter 7. Classical electromagnetism y

y


O


w k




y

x

O

y


θ

O k




θ=0

O




w

x


θ=π

Figure 7.4: Longitudinal Doppler effect: in the case θ = 0, the wave is emitted in the direction of w towards the observer located at O in the frame R; for θ = π, the direction of propagation of the wave is opposed to that of the velocity w of the frame R
with respect to R. Transversal Doppler effect (θ = ± π2 ) In this case, one has

f
f = γf
=  1−

w2 c2

.

7.4.3 Aberration of distant stars With respect to the reference frame R of the sun, let v (vx = 0, vy = −c, vz = 0) be the velocity of light coming from a distant star located on the Oy axis (Figure 7.5). y

y
vx


R v

vy


v


x

O Solar

R


x


O
Earth

Figure 7.5: Aberration of distant stars: in the reference frame R of the sun, the velocity of light coming from a star located on the Oy axis is v; in the reference frame R
of the Earth having a velocity w with respect to R, the velocity of light is v
.

7.5. Exercises

125

The velocity v
in the mobile reference frame of the Earth is given by equation (6.8) vx − w = −w, 1 − vcx2w   2 vy 1 − wc2 w2
= −c 1 − 2 , vy = vx w 1 − c2 c  2 vz 1 − wc2
= 0. vz = 1 − vcx2w

vx
=

The angle θ under which one sees the star from the Earth is given by 
 v  w w =γ . tan θ =  x
 =  vy c w2 c 1− c2

7.5 Exercises E7-1 Consider a charge q located at the origin of a reference frame K
moving with a constant velocity v along the Ox axis of a reference frame K at rest. Determine the four-potential and the electromagnetic field in the frame K at rest. E7-2 Consider an infinite linear distribution of charges with a linear density λ0 (λ0 > 0) along the O
x
axis of a reference frame K
moving with a velocity v along the Ox axis of a reference frame K at rest. Along the Ox axis of the reference frame K, one has an infinite linear distribution with a linear density −λ (λ > 0) of charges at rest. Knowing that the total linear density of charges in the reference frame K at rest is nil, determine the electromagnetic field created in K by these charges at the point M
(x
, r
, 0). A particle of charge q moves with a velocity v in K, parallel to the Ox axis (at the distance r). Determine the force exerted by the two sets of charges in the reference frame K and in the proper frame of the particle. E7-3 Consider the electromagnetic field F = −I + 2

iJ c

(Bx = 1T, Ey = 2V /m)

in the reference frame of the laboratory K at rest . Determine the electromagnetic field in the reference frame moving with respect to K with a velocity 2c in the direction (1, 1, 0) of K. E7-4 Show that Maxwell’s equations in vacuum without sources are invariant under the transformation F → F ∗ = iF

(i.e., B →

E E , → −B). c c

Chapter 8

General relativity The general theory of relativity is developed within the Clifford algebra H⊗H over R. Einstein’s equations and the equation of motion are given as well as applications such as the Schwarzschild metric and the linear approximation.

8.1 Riemannian space Consider a four-dimensional space with the elementary displacement DM = ω i ei and the affine connection Dei = ωij ej . The covariant differentiation of the vector A is defined by DA = dA + dω · A with 2dω = ω ij ei ∧ ej et ω ij = ωki (ek · ej ). The reciprocal basis eα is defined by eµ · eυ = δυµ (e0 = e0 , e1 = −e1 , e2 = −e2 , e3 = −e3 ) where e0 , e1 , e2 , e3 are unitary orthogonal vectors. Under a Lorentz transformation A
= f Afc , one has DA
= f DAfc , dω
= f dωfc − 2df fc . A Riemannian space is a space without torsion but with a curvature. The absence of torsion is expressed by D2 (D1 M ) − D1 (D2 M ) = 0

(8.1)

where D1 , D2 are two linearly independent directions. Writing dω = Ik dxk , DM = σm dxm , condition (8.1) leads to the relations Ik · σm − Im · σk =

∂σk ∂σm − ∂xm ∂xk

which determine Ik when DM is given. The existence of a curvature is expressed by the relations (D2 D1 − D1 D2 )A = Ω · A

128

Chapter 8. General relativity

where Ω is a bivector defined by Ω = (d2 d1 − d1 d2 )ω + [d2 ω, d1 ω] = Ωkm dS km /2 * + ∂Ik ∂Im = − − [I , I ] d1 xk d2 xm k m ∂xm ∂xk ij with 2Ωkm = −Rkm ei ∧ ej and dS = D1 M ∧ D2 M = dS km ek ∧ em /2. Bianchi’s first identity is given by

Ωij · ek + Ωjk · ei + Ωki · ej = 0, Ωij · (ek ∧ em ) = Ωkm · (ei ∧ ej ),

(8.2) (8.3)

where relation (8.3) results from the preceeding equation. Bianchi’s second identity is expressed by Ωij;k + Ωjk;i + Ωki;j = 0 with D3 Ω(d2 , d1 ) = d3 Ω + [d3 ω, Ω] = Ωij;k ω i (d2 )ω j (d1 )ω k (d3 ). The Ricci tensor h Rik = Rihk and the curvature R = Rkk are obtained by the relations Rk = Ωik · ei = Rik ei , R = (Ωik · ei ) · ek = Ωik · (ei ∧ ek ).

8.2 Einstein’s equations To deduce Einstein’s equations from a variational principle, we shall use the method of the exterior calculus but within the framework of a Clifford algebra and without using the exterior product of differential forms. Adopting, for simplicity, an orthogonal curvilinear coordinate system, we can write & √ L = R −gdx0 dx1 dx2 dx3 & = Ωik · (ei ∧ ek )ω 0 ω 1 ω 2 ω 3 with

√ −gdx0 dx1 dx2 dx3 = ω 0 ω 1 ω 2 ω 3 . Taking the dual, we obtain & ∗

∗ L = Ωik ∧ ei ∧ ek ω 0 ω 1 ω 2 ω 3 .

The variation gives ∗

δL =

&

  ∗ Ωik ω I ω K ∧ δ ei ∧ ek ω G ω H &  ∗

+ δ Ωik ω I ω K ∧ ei ∧ ek ω G ω H

8.3. Equation of motion

129

with i = I, k = K and where the second integral vanishes. Furthermore, one has   ∗  ∗ 

= δ (eγ ) ∧ eγ · ei ∧ ek δ ei ∧ ek  ∗  = −δ (eγ ) ∧ ei ∧ ek ∧ eγ  ∗  . = −δ (eγ ) ∧ ei ∧ ek ∧ eγ Hence &



∗  Ωik ω I ω K ∧ δ (ω g eγ ) ∧ ei ∧ ek ∧ eγ ω h &   ∗

= − δ (ω g eγ ) ∧ Ωik ∧ ei ∧ ek ∧ eγ ω I ω K ω h = 0

δL∗ =

with g = γ, i = I, k = K (without summation). Finally, one obtains Einstein’s equations in a vacuum ∗

Ωij ∧ ei ∧ ej ∧ ek = 0. In the presence of an energy-momentum distribution, one has ∗

∗ 1 Ωij ∧ ei ∧ ej ∧ ek = κ T k 2

(8.4)

with T k = T ik ei , κ = 8πG/c4 where T ik is the energy-momentum tensor ([50], [23], [34]). The standard expression of Einstein’s equations is obtained as follows. Taking the dual of equation (8.4), one obtains 



1 − Ωij · ei ∧ ej ∧ ek = κ T k . 2 Using the formula (B · T ) · V = (B ∧ V ) · T where B, T, V are respectively a bivector, a trivector and a vector, one has

 

1 1 αβ − Ωij · ei ∧ ej ∧ ek · eµ = Rij (eα ∧ eβ ∧ eµ ) · ei ∧ ej ∧ ek 2 2 1 αβ ijk 1 βk = − Rij δαβµ = Rβµ − δµk R 2 2 = κT k · eµ = κT ik ei · eµ = κTµk .

8.3 Equation of motion To obtain the equation of motion, one writes & δS = −mc δ (ds) = 0

130

Chapter 8. General relativity 2

with δ (ds) = δ (ds) /2ds and ds2 = DM · DM = (σi · σk ) dxi dxk . One obtains & &  (σl · σk ) uk dδxl δS = −mc (δσi · σk ) ui uk ds − mc = A+B = 0 with ui = dxi /ds. The second term B can be rewritten in the form & &   k l B = −mc (σl · σk ) u δx + mc δxl d (σl · σk ) uk . Since, δσi = σi,l δxl + δω · σi with δω = Ik dxk , one obtains



∂σi d (σl · σk ) k duk i k − =0 + I · σ u u + + (σ · σ ) · σ u l i k l k ∂xl ds ds or, equivalently,

⎧ ∂σi ∂σl k i ⎪ ⎪ ⎨ − ∂xl + ∂xi − Il · σi + Ii · σl · σk u u   ⎪ ∂σk duk ⎪ i k ⎩ +σl · + I · σ i k u u + (σl · σk ) i ∂x ds

⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭

=0

where the expression in brackets vanishes in the absence of torsion. One thus obtains   du i σl · + Ii u · u = 0 ds with the four-velocity u = DM/ds = σi ui and the ordinary derivative du/ds. Finally, the equation of motion is expressed by du dω Du = + · u = 0. ds ds ds

8.4 Applications 8.4.1 Schwarzschild metric Consider a metric having a spherical symmetry and the elementary displacement DM , ds2 = e2a c2 dt2 − e2b dr2 − r2 dθ2 − r2 (sin θ)2 dϕ2 , DM = ea cdte0 + eb dre1 + rdθe2 + r sin θdϕe3 where a, b are functions of t and r. The bivector dω is given by  b˙ −a+b −b −b
a−b dω = cos θdϕI − e sin θdϕJ + e dθK + a e + e iI c

8.4. Applications

131

where the prime and the point designate respectively a partial derivative with respect to r and t. The bivectors Ωik are given by *

−2a +   2 e


2

−2b ¨ ˙ ˙ Ω01 = a + (a ) − a b e + −b − b + ba˙ iI , c2 Ω02 =

˙ −a−b a
e−2b be K+ iJ, rc r

b˙ b
e−2b K + e−a−b iJ, r rc 

1 − e−2b Ω23 = I. r2 Einstein’s equations, in a vacuum, are Ω12 =

Ω03 = −

˙ −a−b a
e−2b be J+ iK, rc r

Ω13 = −

b˙ b
e−2b J + e−a−b iK, r rc

Ω23 ∧ e1 − Ω13 ∧ e2 + Ω12 ∧ e3 = 0, −Ω23 ∧ e0 + Ω03 ∧ e2 − Ω02 ∧ e3 = 0, Ω13 ∧ e0 − Ω03 ∧ e1 + Ω01 ∧ e3 = 0, −Ω12 ∧ e0 + Ω02 ∧ e1 − Ω01 ∧ e2 = 0. Developing these four equations, one obtains respectively   1 − e−2b 2b
e−2b 2b˙ − + (8.5) k − e−a−b jI = 0, 2 r r cr   1 − e−2b 2a
e−2b 2b˙ −a−b − e − k = 0, (8.6) jI + r2 r cr ⎫ ⎧
b − a
−2b ⎪ ⎪ ⎪ ⎪ e ⎬ ⎨ r jJ = 0, (8.7) − ⎪ ⎪  e−2a ⎪


2

−2b ⎪ ⎭ ⎩ + (¨b + b˙ 2 − b˙ a) ˙ − (a + a − a b )e c2 ⎧
⎫
b − a −2b ⎪ ⎪ ⎪ ⎪ e ⎨ ⎬ r − jK = 0. (8.8) ⎪ ⎪  e−2a ⎪


2

−2b ⎪ ⎩ + (¨b + b˙ 2 − b˙ a) ⎭ ˙ − (a + a − a b )e c2 From equations (8.5), (8.6), one obtains b˙ = 0 and a
+ b
= 0 or equivalently log e2(a+b) = C where C = 0 because e2a = e2b = 1 at infinity; hence a + b = 0. Equation (8.5) then gives e−2b (2b
r − 1) + 1 = 0,

−2b 
= 0, re e−2b = 1 +

C1 , r

132

Chapter 8. General relativity

and taking C1 = −2m (with m = GM/c2 ), one has e2b = e−2a = 1 − which determines completely the Schwarzschild metric. Finally, one has    −1 2m 2 2 2m 2 ds = 1 − dr2 − r2 dθ2 − r2 sin2 θdϕ2 , c dt − 1 − r r  1/2 mcdt 2m (− sin θdϕJ + dθK) + 2 iI, dω = cos θdϕI + 1 − r r 2m m m Ω01 = − 3 iI, Ω02 = 3 iJ, Ω03 = 3 iK, r r r m m 2m Ω12 = − 3 K, Ω13 = 3 J, Ω23 = 3 I. r r r

 2m −1 r

The equation of motion is expressed simply in the form Du du dω = + · u = 0. ds ds ds Developing this expression along the axes e0 , e1 , e2 , e3 one obtains  

d 2m cdt 1− = 0, ds r ds ⎡ m cdt 

dθ 2 2 " #  − 12 12 +r − 2m 2m dr d ⎢ r2 ds ds 1− = 1− ⎣ 

ds r ds r dϕ 2 +r sin2 θ ds  2   dϕ d 2 dθ 2 , r = r sin θ ds ds ds   d dϕ r2 sin2 θ = 0. ds ds

respectively (8.9) ⎤ ⎥ ⎦,

Adopting θ = π/2, it follows from the equations (8.9), (8.12) that   dϕ 2m cdt = k, r2 = h, 1− r ds ds where k and h are constants. The relation uc u = 1 is expressed by  2   2  −1  2 dr dϕ cdt 2m 2m − 1− − r2 = 1, 1− r ds r ds ds or with w = 1/r, 2 dw d2 r dr 2 2d w = −h , = −h w . ds dϕ ds2 dϕ2 The projection of the equation on the axis e1 then leads to the relation

d2 w m + w = 2 + 3mw2 dϕ2 h

(8.10)

(8.11) (8.12)

8.4. Applications

133

which is the relativistic form of Binet’s equation and from which one can deduce the precession of the perihelion of Mercury in the usual way ([34]), ([46]).

8.4.2 Linear approximation Consider the elementary displacement DM = ω i ei with ω i = dxi + hij dxj where hij = hji = ηik hkj is small as compared to 1 [50, p. 186]. The bivectors dω and Ωkm are given by 2dω = (hik,j − hjk,i )ei ∧ ej dxk , 2Ωkm = (hik,jm − him,jk )ei ∧ ej . Using the relation (ei ∧ ej ) · eJ = ei (if J = j = i) and nil otherwise, one obtains, in a vacuum, with harmonic coordinates (2hmi,i = hii,m ), Rk = Ωkm · ei = −(
hm k )em = 0, the last equation being a gravitational wave equation (with
= ∂2 ∂2 ∂y 2 − ∂z 2 ). Writing 1 γkm = hkm − ηkm h 2 with h = hm m one obtains Einstein’s equations

∂2 c2 ∂t2

−

∂2 ∂x2

−

−
γ k = κT k (with γ k = γ mk em ). If one considers a homogeneous sphere (of mass M and radius r0 ) rotating slowly with an angular velocity ω around the axis Oz , one integrates the above equations and obtains the metric (with m = GM/c2 ) for r ≥ r0 ,      4GIω 2m 2 2 2m 2 2 ds = 1 − c dt − 1 + dx + dy 2 + dz 2 − 3 3 (ydx − xdy) cdt r r c r with I = 2M r02 /5. The elementary displacement is DM = ω i ei = (dxi + hij dxj )ei with h00 = h11 = h22 = h33 = −m/r, h01 = −GIωy/c3 r3 ,

h02 = −GIωx/c3 r3 ,

the other hij being equal to zero. The bivector dω = 12 (hik,j − hjk,i )ei ∧ ej dxk gives the equation of motion ⎡ ⎤ −j (u1 h00,1 + u2 h00,2 + u3 h00,3 ) Du ⎢ +kI (−h00,1 + u3 Ω2 − u2 Ω3 ) ⎥ ⎥ =⎢ ⎣ +kJ (−h00,2 + u1 Ω3 − u3 Ω1 ) ⎦ ds +kK (−h00,3 + u2 Ω1 − u1 Ω2 )

134

Chapter 8. General relativity

with h00,1 = mx/r3 , h00,2 = my/r3 , h00,3 = mz/r3 and Ω1 = h30,2 − h20,3 = 3GIωxz/c3r5 , Ω2 = h10,3 − h30,1 = 3GIωyz/c3r5 , 

Ω3 = h20,1 − h10,2 = GIω 2z 2 − x2 − y 2 /c3 r5 , which puts in evidence the Thirring precession with respect to the local inertial frame.

Conclusion From the abstract quaternion group, we have defined the quaternion algebra H, then the complex quaternion algebra H(C) and the Clifford algebra H ⊗ H. The quaternion algebra gives a representation of the rotation group SO(3) well-known for its simplicity and its immediate physical significance. The Clifford algebra H⊗H yields a double representation of the Lorentz group containing the SO(3) group as a particular case and having also an immediate physical meaning. Furthermore, the algebra H ⊗ H constitutes the framework of a relativistic multivector calculus, equipped with an associative exterior product and interior products generalizing the classical vector and scalar products. This calculus remains relatively close to the classical vector calculus which it contains as a particular case. The Clifford algebra H ⊗ H allows us to easily formulate special relativity, classical electromagnetism and general relativity. In complexifying H ⊗ H, one obtains the Dirac algebra, Dirac’s equation, relativistic quantum mechanics and a simple formulation of the unitary group SU(4) and the symplectic unitary group USp(2, H). Algebraic or numerical computations within the Clifford algebra H ⊗ H have become straightforward with software such as M athematica. We hope to have shown that the Clifford algebras H ⊗ H over R and C constitute a coherent, unified, framework of mathematical tools for special relativity, classical electromagnetism, general relativity and relativistic quantum mechanics. The quaternion group consequently appears via the Clifford algebra as a fundamental structure of physics revealing its deep harmony.

Appendix A

Solutions Chapter 1 S1-1 i2 = j 2 = k 2 = ijk = −1, (−i) ijk = jk = i, k = −ji, ijk (−k) = ij = k, j = −ik, i = −kj, etc. S1-2

√ 2,

√ 25 = 5,

1 1 (1 − i), b−1 = (4 + 3j), 2 5 a + b = 5 + i − 3j, ab = 4 + 4i − 3j − 3k, ba = 4 + 4i − 3j + 3k, √ 1+i √ a = 2 √ = 2 (cos 45◦ + i sin 45◦ ) , 2   4 − 3j b=5 = 5 (cos 36, 87◦ − sin 36, 87◦ ) , 5     √ √ 45 36, 87 45 36, 87 + i sin − i sin a = 21/2 cos b = 51/2 cos , , 2 2 2 2   45 45 + i sin a1/3 = 21/6 cos = 21/6 (cos 15◦ + i sin 15◦ ) . 3 3 |a| =

|b| =

a−1 =

S1-3 ac ax + ac xb = ac c, 

axb + xb2 = cb,

x aac + b2 + 2S(a)b = ac c + cbc ;

138

Appendix A. Solutions

the latter is a equation of the type xα = β, α, β ∈ H which one solves in x. 1 Answer: 2ix + xj = k, x = (2j − i). 3 S1-4 axbb−1 + cxdb−1 = eb−1 , c−1 ax + xdb−1 = c−1 eb−1 ; one finds an equation of the type of the previous exercice and solves similarly; Answer: x = 2 + i. S1-5 x−1 x2 = x−1 xa + x−1 bx, x = a + x−1 bx, xx−1 = ax−1 + x−1 b = 1; Answer: x =

1 (−6j + 3k.). 5

Chapter 2 S2-1

⎡

cos α Aα = ⎣ sin α 0

⎡ ⎤ − sin α 0 cos β cos α 0 ⎦ , 0 Aβ = ⎣ 0 1 − sin β ⎡ ⎤ 1 0 0 Aγ = ⎣ 0 cos γ − sin γ ⎦ . 0 sin γ cos γ

⎤ 0 sin β 1 0 ⎦, 0 cos β

S2-2 dA = rdA
rc + drA
rc + rA
drc = r (dA
+ rc drA
+ A
drc r) rc ; since rc dr = −drc r (which is obtained by differentiating rrc = 1), one obtains with dΩ
= 2rc dr, dA = r (dA
+ rc drA
− A
rc dr) rc  
dΩ



dΩ A −A = r dA + rc = r (dA
+ dΩ
× A
) rc = rDA
rc ; 2 2 DA
= dA
+ dΩ
× A
; dΩ
dt

is the angular velocity with respect to the moving frame.

139 S2-3 A

= f A
fc (f = gc r, f fc = 1), DA

= f DA
fc , dA

+ dΩ

× A

= f (dA
+ dΩ
× A
) fc , (df A fc + f dA fc + f A dfc ) + dΩ

× A

= f dA
fc + f (dΩ
× A
) fc ,








dΩ

× A

= f (dΩ
× A
) fc − df fc (f A
fc ) − (f A
fc ) f dfc = f dΩ
fc × A

− df fc A

+ A

df fc = (f dΩ
fc − 2df fc ) × A

; hence dΩ

= f dΩ
fc − 2df fc , dΩ
= fc dΩ

f − 2fc df. S2-4 X = rX
rc , r = cos

θ θ dΩ
dθ dr + k sin , = 2rc =k . 2 2 dt dt dt

Polar coordinates: X = xi + yj,

X
= ρi,

dX
dΩ
dθ DX
dρ = + × X
= i + ρ j; dt dt dt dt dt dV
dΩ
DV


γ = = + ×V dt dt " dt #  2

d2 θ d2 ρ dθ dρ dθ +ρ 2 . = −ρ i + j 2ρ dt2 dt dt dt dt

V
=

Cylindrical coordinates: X = xi + yj + zk,

X
= ρi + zk,

dX
dΩ
dθ dz DX
dρ = + × X
= i+ρ j+k ; dt dt dt dt dt dt dV
dΩ
DV


= + ×V γ = dt dt " dt #  2  

2

dθ 1 d d z d2 ρ 2 dθ = − ρ i + k. ρ j + dt2 dt ρ dt dt dt2

V
=

140

Appendix A. Solutions

S2-5 The moving frame (ρ, θ, ϕ) is obtained with the following three rotations: π π − j sin (rotation of − π2 around Oy), 4 4 θ θ r2 = cos − i sin (rotation of −θ around Ox), 2 2 2ϕ − π 2ϕ − π r3 = cos + k sin (rotation of −( π2 − ϕ) around Oz); 4 4 r1 = cos

r = r3 r2 r1

θ+ϕ 1 θ+ϕ + sin = cos 2 2 2

2θ − 2ϕ + π 2ϕ + 2θ − π 1 2ϕ − 2θ − π − j cos + k sin + √ i sin ; 4 4 4 2 dΩ
= 2rc dr = idϕ cos θ − jdϕ sin θ + kdθ, dΩ = 2drrc = −idθ sin ϕ + jdθ cos ϕ + kdϕ; e1 = rirc = i sin θ cos ϕ + j sin θ sin ϕ + k cos θ, e2 = rjrc = i cos θ cos ϕ + j cos θ sin ϕ − k sin θ, e3 = rkrc = −i sin ϕ + j cos ϕ; dX
dΩ
dθ dϕ DX
dρ = + × X
= i i + jρ + kρ sin θ, dt dt dt dt dt dt dV
dΩ
DV
γ
= = + ×V
dt dt dt ( "  #)  2 2 dθ dϕ d2 ρ 2 =i −ρ + sin θ dt2 dt dt (    2 ) dϕ 1 d 2 dθ +j ρ − ρ sin θ cos θ ρ dt dt dt *  + 1 d dϕ +k ρ2 sin2 θ . ρ sin θ dt dt

V
=

S2-6 α α + k sin (1st rotation), 2 2   β β β β r2 = cos + sin r1 ir1c = r1 cos + i sin r1c (2nd rotation), 2 2 2 2 γ γ r3 = cos + sin (r2 kr2c ) (3rd rotation); 2 2 r1 = cos

141 r = r3 r2 r1  γ γ r2c r2 r1 = r2 cos + k sin 2 2   γ β γ β r1 = r1 cos + i sin r1c cos + k sin 2 2 2 2    α β  γ γ α β cos + k sin = cos + k sin cos + i sin 2 2 2 2 2 2 because r1c

  γ γ γ γ cos + k sin r1 = cos + k sin ; 2 2 2 2

finally r = cos

α+γ β α−γ β α−γ β α+γ β cos + i sin cos + j sin sin + cos sin , 2 2 2 2 2 2 2 2  dβ dα cos γ + sin β sin γ dt dt   dα dβ sin γ + sin β cos γ +j − dt dt   dγ dα +k + cos β ; dt dt

dr =i ω = 2rc dt




 dβ dγ cos α + sin α sin β dt dt   dβ dγ +j sin α − cos α sin β dt dt   dα dγ +k + cos β . dt dt

dr ω = 2 rc = i dt



Chapter 3 S3-1

hence, a0 = 12 , a3 =

a0 + i
a3 = 1, −i
a1 + a2 = 0, −i
2 ,

a0 − i
a3 = 0, −i
a1 − a2 = 0,

a2 = 0 = a1 , or e1 =

1 (1 − i
k) ; 2

e21 = e1 ;

e2 =

1 (1 + i
k) ; 2

e22 = e2 .

similarly

142

Appendix A. Solutions a = (a0 + i
a
0 , a1 + i
a
1 , a2 + i
a
2 , a3 + i
a
3 ) , ⎤ ⎡ a0 + i
a3 + i
a
0 − a
3 , 1 ⎢ a1 − i
a2 + i
a
1 + a
2 , ⎥ ⎥, u = ae1 = ⎢ 2 ⎣ a1 − i
a2 + i
a
1 + a
2 , ⎦ −i
a0 + a3 + a
0 + i
a
3 ⎤ ⎡ a0 + i
a3 + i
a
0 − a
3 , 1 ⎢ a1 + i
a2 + i
a
1 − a
2 , ⎥ ⎥. v = e1 a = ⎢ 2 ⎣ −i
a1 + a2 + a
1 + i
a
2 , ⎦ −i
a0 + a3 + a
0 + i
a
3

S3-2 x + y = 3 + i
i + (2 + i
) j + (1 + 3i
) k, xy = (2 − 3i
) + (−3 + 8i
) i + (7 + 2i
) j + (2 + 3i
) k, yx = (2 − 3i
) + (3 − 4i
) i + (1 + 2i
) j + (2 + 3i
) k, x2 = (−2 − 4i
) + 2i
i + (4 + 2i
) j + 2k, xxc = 4 + 4i
,

y 2 = 13 + 12 i
k,

yyc = −5,

xc 1 = (1 − i
, −1 − i
, −3 + i
, −1 + i
) , xxc 8 1 = (−2 + 3i
k) , 5 1 (1 + 5i
, 5 + 11i
, 9 − 5i
, 5 + i
) = (xy)−1 . = 40

x−1 = y −1 y −1 x−1 S3-3

√  1 a + a∗ = ± 1 + 3k , |1 + d| 2 √  1 1+d = ± 3 + i
i 5 , b1 = ± |1 + d| 2  √  1 d = aa∗c = 7 + 3i
i 5 , 2

r1 = ±

√  1 a + a∗ = ± 3k , 1 + |1 + d
| 2 √  √ 1 1 + d


= ± 6 − i b2 = ± 5 − i j 15 , |1 + d
| 4 √ √  1 d
= a∗c a = 14 − 3i
i 5i − 3i
j 15 . 4

r2 = ±

143 S3-4 x·y = B =x∧y =

1 (xyc + yxc ) = 2, 2

1 (xyc − yxc ) = (1 + 2i
) i + (−1 + 2i
) j − i
k, 2

B
= z ∧ w = (1 − 9i
) i − 3i
j + (−3 − 3i
) k, F = y ∧ z = −i − 2i
j + 3i
k, 1 (xF ∗ + F x) = i
+ 2i − 3j − 2k, 2 T
= (x ∧ y) ∧ z = z ∧ (x ∧ y) = i
+ 2i − 3j − 2k, T = x ∧ (y ∧ z) =

1 B · z = −z · B = − (zB ∗ − Bz) = 2 + 6i
i + 6i
j − 2i
k ∈ V, 2 1 w · T = (wT ∗ + T wc ) = (3 + i
) i + (1 + 8i
) j + (1 − 11i
) k ∈ B, 2 

 BB
+ B
B = −22, 2   BB
+ B
B B ∧ B
= P = i
, 2 BT + T B ∗ = −5 − 6i
i + i
j − 10 i
k ∈ V. B·T = 2 B · B
= S

S3-5 X = ct + xi
i + yi
j + zi
k, X
= ct
+ x
i
i + y
i
j + z
i
k,   ϕ ϕ b = cosh , i
sinh , 0, 0 , 2 2 1 7 γ=  = cosh ϕ = , 2 v2 1− c2

cosh2 cosh

cosh ϕ + 1 ϕ = , 2 2

3 ϕ = , 2 2

cosh ϕ − 1 ϕ = , 2 2  √ √ ϕ 3
5 5 sinh = , b= ,i , 0, 0 , 2 2 2 2

sinh2

144

Appendix A. Solutions 

1




γ = 1−




v
2 c2

= cosh ϕ =

7 , 6




sinh ϕ =

1 , 6

X
= (0, i
, i
, 0),   7 1

, 0, i ,0 , V = 6 6 X = bX
b∗c ,

V = bV
b∗c ,

    7 7
1 105
1 ,i ,i ,0 , V = 2 6 2 2 6   Ey

Ez
E
E
= 0, i
x , i
,i F
= −B
+ i
, c c c c F = bF
bc Ex = Ex
, Bx = 0,

7
7 Ey , Ez = Ez
, 2 √ 2√ 3 5
3 5
E , E , By = − Bz = 2c z 2c y  
E F = 0, −B + i . c Ey =

S3-6 A (0, i
, −i
, 0) , B (0, i
, i
, 0) , 
 i
2i
i A
0, , − , − , 3 3 3   2i
i
i
C
0, − , , − , 3 3 3

C (0, −i
, −i
, 0) , D (0, −i
, −i
, 0) , 

 2i
i i B
0, , , − , 3 3 3   i
2i
i

D 0, − , − , − . 3 3 3

Chapter 4 S4-1 Ac = −I − 2J + iK,

Bc = −j − kI − 2kK,

A + B = j + I + 2J − iK + kI + 2kK, A − B = I + 2J − iK − j − kI − 2kK, AAc = 4,

BBc = −4,

145 A−1 =

Ac 1 = (−I − 2J + iK) , , AAc 4

B −1 =

1 (j + kI + 2kK) , 4

AB = −2j − k + jI + 3jJ + 4kI − 2kJ − 3kK, BA = −2j − k + jI + 3jJ − 4kI + 2kJ + 3kK, (AB)c = 2j − k + jI + 3jJ − 4kI + 2kJ + 3kK, (BA)c = −2j − k + jI + 3jJ + 4kI − 2kJ − 3kK, (AB) (AB)c = (AB)c (AB) = −16, (AB)

(BA)

−1

−1

(AB)c (AB) (AB)c 1 (−2j + k − jI + 4kI − 3jJ − 2kJ − 3kK) = B −1 A−1 = 16 1 (2j + k − jI − 4kJ − 3jJ + 2kJ + 3kK) , = 16 =

[A, B] = S4-2

(BA) (BA)c = −16,

1 (AB − BA) = −2j + 4kI − 2kJ − 3kK. 2 1 x · y = − (xy + yx) = 2, 2

1 B = x ∧ y = − (xy − yx) = I − J + 2iI + 2iJ − iK, 2 B
= z ∧ w = I − 3K − 9iI − 3iJ − 3iK, F = y ∧ z = −I − 2iJ + 3iK, T = x ∧ (y ∧ z) = x ∧ F =

1 (xF + F x) 2

= k + 2jI − 3jJ − 2jK, T
= (x ∧ y) ∧ z = z ∧ (x ∧ y) =

1 (zB + Bz) = T 2

= k + 2jI − 3jJ − 2jK, 1 B · z = −z · B = − (zB − Bz) = 2j + 6kI + −6kJ − 2kK, 2 1 w · T = − (wT + T w) = 3I + J + K + iI + 8iJ − 11 iK, 2

146

Appendix A. Solutions

1

(BB + B B) = −22, B·B = S 2

1


B ∧ B = PS (BB + B B) = i, 2 1 B · T = (BT + T B) = −5j − 6kI + kJ − 10 kK. 2


S4-3 1 S1 = − (xy − yx) = 3I − J − 5K, 2 1 S2 = − (zw − wz) = 9I − 3J − 3K, 2 1 (zS1 + S1 z) = 10 k, 2 S1 ∧ S2 = 0,

V = S1 ∧ z = z ∧ S1 =

w = w + w⊥ , 21 27 169 kI − kJ + kK, 50 25 10 46 23 69 −1 = S1 (S1 ∧ w) = − kI + kJ + kK, 50 25 10

w = (w · S1 ) S1−1 = S1−1 (S1 · w) = w⊥ = (w ∧ S1 ) S1−1

S1 = S1 + S1⊥ , 18 18 54 S1 = (S1 .S2 ) S2−1 = − I − J − K, 11 11 11 , 26 37 21 −1 S1⊥ = (S1 ∧ S2 ) + [S1 , S2 ] S2 = − I − J − K. 11 11 11

Chapter 5 S5-1

1 γ= 1−

v2 c2

3 = cosh ϕ = √ , 2 2

cosh ϕ + 1 cosh ϕ − 1 ϕ ϕ = , sinh2 = , 2 2 2 2 ! ! ϕ 1 1 ϕ 3 3 √ + 2 = α, √ − 2 = β, cosh = sinh = 2 2 2 2 2 2 ! ! 3 3 1 √ + 2 + iI √ − 2 , b= 2 2 2

cosh2

147 θ θ π π + I sin = cos + I sin , 2 2 8 8 π π π π π a = rb = α cos − β sin i + α sin + α sin I + β cos iI, 8 8 8 8 8 r = cos

X = aX
ac , X = ctj + kIx + kJy + kKz, X
= ct
j + kIx
+ kJy
+ kKz
. S5-2 !

!

3 √ −2 , 2 ! ! 3 3 1 √ + 2 + iJ √ − 2 , b2 = 2 2 2

1 b1 = 2

1 a = b2 b1 = 4

3 √ + 2 + iI 2

  

 1 3 1 3 + √ iI + √ iJ + −2 + √ K , 2+ √ 2 2 2 2

a = br, 1 3 9 1+d , d = aac = + √ iI + iJ, b = ±√ 8 2 2 8 2 + d + dc "√ # 1 17 3 + √ iI + √ iJ , b=± 4 34 4 17 a+a r = ±√ =± 2 + d + dc

 

  1 1 3 3 √ + −2 + √ √ K , 2+ √ 2 17 2 17

  3 1 √ , 2+ √ θ = 3, 37◦ , 2 17 √ ϕ 1 17 ϕ , sinh = , ϕ = 0, 494, cosh = 2 4" 2 4 # √ 1 3 17 + √ 4iI + √ iJ , b=± 4 4 34 17 θ cos = 2

 direction u

1 3 √ 4, √ , 0 34 17

 ,

tanh ϕ = 0, 4581 =

v . c

148

Appendix A. Solutions

S5-3   1 1 = − , 0, − , 2  2 1 1 1 x
D = − , , − ,  3 6 6 1 x
F = − , 0, 0 ,  2  1 1 1
xH = − , , − . 2 10 5

x
A

x
B

= (0, 0, 0) ,   3 1 2
, xC = − , , −  7 7 7 4 1 1 , x
E = − , , −  9 9 9 1 3 x
G = − , 0, − , 5 5

S5-4

1−k −1 + k

AB = A−1 =

1 2

−1

(AB)



−k −i

1 −j

1 = 4



−1 + k −1 + k



,



1+k −1 − k

B −1 =

, −1 − k −1 − k

0 i

BA =

,

(BA)

1 2

−1



j 0



1 −k

1 = 2



There is no inverse of C.

Chapter 6 S6-1 Velocity of B in the reference frame at rest vx = −v,

vy = 0,

vz = 0

velocity of B with respect to A vx
=

vx − w = −0, 975 c, 1 − vcx2w

vy
= 0,

vz
= 0;

The relative velocity remains smaller than c.

w = v = 0, 8 c,

, −i −j 0 −j

, −i 0

.

149 S6-2 L= hence

'

Xi ∧ Pi =

'

 '  Ei − ctpi = const. (ri × pi ) + i ri c

 '  Ei − ctpi = C2 , (ri × pi ) = C1 , ri c . Ei ri , R= . Ei . . Ei ri pi 2 . − tc . = C3 , Ei Ei . pi R = C3 + vt, v = c2 . = C4 . Ei

S6-3 In the proper frame (instantaneous, Galilean), the four-acceleration is A
= kIg with the invariant AAc = −g 2 ; In the frame at rest K, the four-velocity and the four-acceleration are respectively u = c (j cos hϕ + kI sinh ϕ) , A=c

tanh ϕ =

dϕ (j sinh ϕ + kI cosh ϕ) , dτ

v , c

τ : proper time

with  AAc = −g = c 2

 = −c2 hence

2

dϕ dτ

dϕ dτ

2

 sinh2 ϕ − cosh2 ϕ

2

g dϕ = , dτ c

ϕ=

gτ c

and  gτ  dX gτ + kI sinh = , u = c j cos h c c dτ   gτ gτ c2 j sin h + kI cosh = jct + kIx; X= g c c

150

Appendix A. Solutions

finally, one has t= and

c sinh ϕ, g

x=

c2 (cosh ϕ − 1) g

 2 c2 c4 x+ − (ct)2 = 2 . g g

For ϕ 1 t

c ϕ = τ, g

x=

c2 ϕ2 1 1 = gτ 2 = gt2 . g 2 2 2

Chapter 7 S7-1 A = bA
bc

 b = cosh

ϕ ϕ v + iI sinh , tanh ϕ = , 2 2 c  q , A
= 0, V
= 4πε0 r


V
V
+ kA
= j c c V A = j + kA, c 1/2

r
= x
2 + y
2 + z
2 , x
= γ (x − vt) , A
= j

V = γV
=

1 q  4πε0 2 (x − vt) +

y 2 +z γ2

1/2 , 2

y
= y, A=γ

z
= z,

v
v V = 2 V, 2 c c

(potentials of Lienard and Wiechert of an electric charge in rectilinear motion). E
, c

F
= −B
+ B
= 0,

F = −B + E
=

q r
, 4πε0 r
3

E , c

F = bF
bc , Bx = 0, v v q γvz
= −γEz
= −Ez , 4πε0 cr
3 c c v v q γvy
Bz = = γEy
= Ey , 4πε0 cr
3 c c

By = −

Ex = Ex
=

q x
, 4πε0 r
3

Ey = γEy
,

Ez = γEz
,

151 x
= γ (x − vt) , q (x − vt) , 4πε0 γ 2 r∗3/2 y q , Ey = 4πε0 γ 2 r∗3/2 q z Ez = . 4πε0 γ 2 r∗3/2

Ex =

y
= y,

z
= z,



1/2 y2 + z 2 2 r∗ = (x − vt) + , γ2

S7-2 F
= −B
+

E
, c

B
= 0,

E
=

E2
=

λ0 , 2πε0 r


λ0 n , 2πε0 r


E1
= E3
= 0,

F = bF
bc , ϕ v ϕ tanh ϕ = , b = cosh + iI sinh , 2 2 c µ0 i λ0 γv µ0 λv B3 = = , = 2πε0 r c2 2πr 2πr E3 =

λ λ0 γ = , 2πε0 r 2πε0 r

in K; the linear density of mobile charges is λK = γλ0 = λ (by hypothesis, λK − λ = 0) and i = λv (r = r
). At the point M in K, the total electric field is nil µ0 i ET = 0, ez . BT = 2πr In K, f = qv × B, f (M ) = −qvBey ; in the proper frame, the electromagnetic field is F

= −B

+

E

= bc FT b c

FT = −B,

B=

with

µ0 i ez , 2πr

152

Appendix A. Solutions

hence µ0 i γez , 2πr µ0 i γvey ; E

= − 2πr B

=

the particle being at rest in the proper frame, f

= qE

= − S7-3 F
= bc F b, tanh ϕ =

b=

qµ0 i γve = γf . 2πr

  iI ϕ iJ ϕ ϕ cosh + √ sinh + √ sinh , 2 2 2 2 2

1 v = , c 2

1 cosh ϕ = γ =  1−

cosh ϕ + 1 ϕ = = cosh 2 2 ! √2 + 1 ϕ 3 cosh = , 2 2 2

√2 3

+1 2

,

1 1 +√ , 2 3 1 1 B2
= − √ , 2 3  1 2 B3
= − , c 3

B1
=

v2 c2

2 = √ , 3

cosh ϕ − 1 ϕ sinh = = 2 2 ! √2 − 1 ϕ 3 sinh = , 2 2 2

2 E1
= 1 − √ , 3 2 E2
= 1 + √ , 3 c E1
= − √ . 6

S7-4 ∇F = 0, i∇F = −∇ (iF ) = 0, ∇F ∗ = 0.

√2 3

−1 2

,

Appendix B

Formulary: multivector products within H(C) Let x, y be four-vectors, B, B
bivectors, T, T
trivectors and P, P
pseudoscalars x = x0 + i
x1 + i
x2 + i
x3 , y = y 0 + i
y 1 + i
y 2 + i
y 3 , B = a + i
b,

B
= a
+ i
b
,

(a = a1 i + a2 j + a3 k, b = b1 i + b2 j + b3 k), T = i
t0 + t, P = i
s0 ,

T
= i
t
0 + t
(t = t1 i + t2 j + t3 k), P
= i
s
0 .

S [A], P [A] designate respectively the scalar and pseudoscalar parts of the complex quaternion A. For any two arbitrary boldface quantities (x, y, a, b, t) the following abridged notation is used: x · y = x1 y 1 + x2 y 2 + x3 y 3 , x × y = (x2 y 3 − x3 y 2 )i + (x3 y 1 − x1 y 3 )j + (x1 y 2 − x2 y 1 )k.

Products with four-vectors

x·y =

1 (xyc + yxc ) ∈ S 2

= x0 y 0 − x1 y 1 − x2 y 2 − x3 y 3 = y · x,

154

Appendix B. Formulary: multivector products within H(C) 1 (xyc − yxc ) ∈ B 2

 = x × y + i
y 0 x − x0 y = −y ∧ x.

x∧y =

Products with bivectors 1 (xB ∗ − Bx) ∈ V 2

 = −x · b + i
−x0 b + x × a

x·B =

≡ −B · x, 1 (xB ∗ + Bx) ∈ T 2

 = −i
x · a+ x0 a + x × b ≡ B ∧ x.

x∧B =

B · B
= S



1 (BB
+ B
B) 2



= −a · a
+ b · b
≡ B
· B,

1


(BB + B B) B∧B =P 2 = −i
(b · a
+ a · b
) ≡ B
∧ B.

Products with trivectors 1 (xT ∗ + T xc ) ∈ B 2 = x0 t + t0 x + i
(x × t)

x·T =

≡T ·x 1 (xT ∗ − T xc ) ∈ P 2  = i
−x0 t0 − x · t

x∧T =

≡ −T ∧ x,

B.0. Formulaire: produits multivectoriels dans H(C) 1 B · T = − (BT + T B ∗ ) ∈ V 2

 = (a · t) − i
t0 a + b × t ≡ T · B, 1 (T T
∗ + T
T ∗ ) ∈ S 2 = t0 t
0 − t · t


T · T
=

≡ T
· T.

Products with pseudoscalars 1 (xP ∗ − P x) ∈ T 2 = −i
s0 x0 + s0 x

x·P =

≡ −P · x, 1 (BP + P B) ∈ B 2 = −s0 b + i
s0 a

B·P =

≡ P · B, 1 T · P = − (T P ∗ − P T ) ∈ V 2 = −s0 t0 + i
s0 t ≡ −P · T,

1

(P P + P P ) ∈ S P.P = S 2




= −s0 s
0 ≡ P
· P.

155

Appendix C

Formulary: multivector products within H ⊗ H (over R) Let x, y be four-vectors, B, B
bivectors, T, T
trivectors and P, P
pseudoscalars

x = jx0 + kx

(x = x1 I + x2 J + x3 K),

y = jy 0 + ky, B = a + ib,

B
= a
+ ib
,

(a = a1 I + a2 J + a3 K, b = b1 I + b2 J + b3 K), T = kt0 + jt,

T
= kt
0 + jt
(t = t1 I + t2 J + t3 K),

P = is0 ,

P
= is
0 .

S [A], P [A] designate respectively the scalar and pseudoscalar parts of the Clifford number A. For any two arbitrary boldface quantities (x, y, a, b, t) the following abridged notation is used:

x · y = x1 y 1 + x2 y 2 + x3 y 3 , x × y = (x2 y 3 − x3 y 2 )I + (x3 y 1 − x1 y 3 )J + (x1 y 2 − x2 y 1 )K.

158

Appendix C. Formulary: multivector products within H ⊗ H (over R)

Products with four-vectors

1 x · y = − (xy + yx) ∈ S 2 0 0 = x y − x1 y 1 − x2 y 2 − x3 y 3 = y · x, 1 x ∧ y = − (xy − yx) ∈ B 2

 = x × y + i y 0 x − x0 y = −y ∧ x.

Products with bivectors

1 (xB − Bx) ∈ V 2

 = −jx · b + k −x0 b + x × a = −B · x,

x·B =

1 (xB + Bx) ∈ T 2

 = −kx · a+j x0 a + x × b ≡ B ∧ x.

x∧B =



1 B·B =S (BB
+ B
B) 2




= −a · a
+ b · b
= B
· B,

1 (BB
+ B
B) B ∧ B
= P 2 = −i (b · a
+ a · b
) ≡ B
∧ B.

159

Products with trivectors 1 x · T = − (xT + T x) ∈ B 2 = x0 t + t0 x + i (x × t) ≡ T · x, 1 x ∧ T = − (xT − T x) ∈ P

2 0 0  = i −x t − x · t ≡ −T ∧ x, 1 (BT + T B) ∈ V 2

 = j (a · t) − k t0 a + b × t

B·T =

≡ T · B, 1 T · T
= − (T T
+ T
T ) ∈ S 2 = t0 t
0 − t · t
≡ T
· T.

Products with pseudoscalars 1 (xP − P x) ∈ T 2 = −ks0 x0 + js0 x

x·P =

≡ −P · x, 1 (BP + P B) ∈ B 2 = −s0 b + is0 a

B·P =

≡ P · B, 1 (T P − P T ) ∈ V 2 = −js0 t0 + ks0 t

T ·P =

≡ −P · T,

160

Appendix C. Formulary: multivector products within H ⊗ H (over R)

1

(P P + P P ) ∈ S P ·P =S 2




= −s0 s
0 ≡ P
· P.

Appendix D

Formulary: four-nabla operator ∇ within H ⊗ H (over R) Four-nabla operator: ∇=j

∂ ∂ ∂ ∂ − kI − kJ − kK , c∂t ∂x ∂y ∂z

D’Alembertian operator:
=

∂2 c2 ∂t2

−

∂2 ∂2 ∂2 − 2− 2 ∂x ∂y ∂z

(four-vectors A = jA0 + kIA1 + kJA2 + kKA3 ,

B = jB 0 + kIB 1 + kJB 2 + kKB 3 ;

scalars : p, q). Then: ∇ ∧ (∇ ∧ A) = 0, ∇ ∧ (∇p) = (∇ ∧ ∇) p = 0, ∇ (∇p) = (∇ · ∇) p =
p, ∇ · (∇ ∧ A) =
A − ∇ (∇ · A) ,
(∇p) = ∇ (
p) ,
(∇ · A) = ∇ · (
A) ,
(∇ ∧ A) = ∇ ∧ (
A) , ∇ (pq) = p∇q + q∇p,
(pq) = p
q + q
p + 2 (∇p) · (∇q) , ∇ · (pA) = p (∇ · A) + A · (∇p) ,

162

Appendix D. Formulary: four-nabla operator ∇ within H ⊗ H (over R) ∇ · (∇ ∧ pA) =
pA − ∇ (∇ · pA) , ∇ ∧ pA = p (∇ ∧ A) + (∇p) ∧ A,
(A + B) =
A +
B, ∇ · (A + B) = ∇ · A + ∇ · B, ∇ ∧ (A + B) = ∇ ∧ A + ∇ ∧ B,
(p + q) =
p +
q, ∇ (p + q) = ∇p + ∇q, ∇ ∧ (A ∧ B) = B ∧ (∇ ∧ A) − A ∧ (∇ ∧ B) , ∇ · (A ∧ B) = (∇ · A) B + (A · ∇) B − (∇ · B) A − (B · ∇) A, ∇ (A · B) = −A · (∇ ∧ B) + (A · ∇) B − B · (∇ ∧ A) + (B · ∇) A.

Appendix E

Work-sheet: H(C) (Mathematica)