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Birkhäuser

Basel.

Boston.

Berlin

Patrick R. Girard

Quaternions,Clifford Algebras and

Relativistic Physics



Author:

Patrick R. GirardINSA de LyonDépartement Premier Cycle

20, avenue Albert EinsteinF-69621 Villeurbanne CedexFrancee-mail: [email protected]

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

Igor Ya. Subbotin

Department of Mathematics and Natural Sciences

National UniversityLos Angeles Campus

Los Angeles, CA 90045

USA

e-mail: [email protected]

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 <http://dnb.ddb.de>.

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 isconcerned, specifically the rights of translation, reprinting, re-use of illustrations, recitation, broad-

casting, reproduction on microfilms or in other ways, and storage in data banks. For any kind of usepermission 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, SwitzerlandPart of Springer Science+Business MediaPrinted on acid-free paper produced from chlorine-free pulp. TCFPrinted in GermanyISBN-10: 3-7643-7790-8 e-ISBN-10: 3-7643-7791-7ISBN-13: 978-3-7643-7790-8 e-ISBN-13: 978-3-7643-7791-5

9 8 7 6 5 4 3 2 1 www.birkhauser.ch



To Isabelle, my wife, and to our children: Claire, Beatrice, Thomas and Benoıt



Foreword

The use of Clifford algebra in mathematical physics and engineering has grownrapidly in recent years. Clifford had shown in 1878 the equivalence of two ap-

proaches to Clifford algebras: a geometrical one based on the work of Grassmannand an algebraic one using tensor products of quaternion algebras H. Recent de-velopments have favored the geometric approach (geometric algebra ) leading to analgebra (space-time algebra ) complexified by the algebra H ⊗ H presented belowand thus distinct from it. The book proposes to use the algebraic approach andto define the Clifford algebra intrinsically, independently of any particular matrixrepresentation, 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 introduc-

tion to this new calculus, starting from the quaternion group, with applications tophysics. The volume is intended for professors, researchers and students in physicsand engineering, interested in the use of this new quaternionic Clifford calculus.

The book presents the main concepts in the domain of, in particular, thequaternion algebra H, complex quaternions H(C), the Clifford algebra H ⊗ Hreal and complex, the multivector calculus and the symmetry groups: SO(3),the Lorentz group, the unitary group SU(4) and the symplectic unitary groupUSp(2,H). Among the applications in physics, we examine in particular, specialrelativity, classical electromagnetism and general relativity.

I want to thank G. Casanova for having confirmed the validity of the interiorand exterior products used in this book, F. Sommen for a confirmation of theClifford theorem and A. Solomon for having attracted my attention, many yearsago, 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 foradvice 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 tothank 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 ThomasHempfling at Birkhauser.

Lyon, June 2006Patrick R. Girard



Contents

Introduction 1

1 Quaternions 31.1 Group structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Finite groups of order n ≤ 8 . . . . . . . . . . . . . . . . . . . . . . 41.3 Quaternion group . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.4 Quaternion algebra H . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.4.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.4.2 Polar form . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.4.3 Square root and nth root . . . . . . . . . . . . . . . . . . . 111.4.4 Other functions and representations of quaternions . . . . . 14

1.5 Classical vector calculus . . . . . . . . . . . . . . . . . . . . . . . . 15

1.5.1 Scalar product and vector product . . . . . . . . . . . . . . 151.5.2 Triple scalar and double vector products . . . . . . . . . . . 16

1.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2 Rotation groups SO(4) and SO(3) 192.1 Orthogonal groups O(4) and SO(4) . . . . . . . . . . . . . . . . . . 192.2 Orthogonal groups O(3) and SO(3) . . . . . . . . . . . . . . . . . . 222.3 Crystallographic groups . . . . . . . . . . . . . . . . . . . . . . . . 24

2.3.1 Double cyclic groups C n (order N = 2n) . . . . . . . . . . . 242.3.2 Double dihedral groups Dn (N = 4n) . . . . . . . . . . . . 24

2.3.3 Double tetrahedral group (N = 24) . . . . . . . . . . . . . . 252.3.4 Double octahedral group (N = 48) . . . . . . . . . . . . . . 262.3.5 Double icosahedral group (N = 120) . . . . . . . . . . . . . 27

2.4 Infinitesimal transformations of SO(4) . . . . . . . . . . . . . . . . 282.5 Symmetries and invariants: Kepler’s problem . . . . . . . . . . . . 322.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

3 Complex quaternions 373.1 Algebra of complex quaternions H(C) . . . . . . . . . . . . . . . . 373.2 Lorentz groups O(1, 3) and SO(1, 3) . . . . . . . . . . . . . . . . . 38

3.2.1 Metric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38



x Contents

3.2.2 Plane symmetry . . . . . . . . . . . . . . . . . . . . . . . . 383.2.3 Groups O(1, 3) and SO(1, 3) . . . . . . . . . . . . . . . . . . 39

3.3 Orthochronous, proper Lorentz group . . . . . . . . . . . . . . . . 41

3.3.1 Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413.3.2 Infinitesimal transformations of SO(1, 3) . . . . . . . . . . . 43

3.4 Four-vectors and multivectors in H(C) . . . . . . . . . . . . . . . . 443.5 Relativistic kinematics via H(C) . . . . . . . . . . . . . . . . . . . 47

3.5.1 Special Lorentz transformation . . . . . . . . . . . . . . . . 473.5.2 General pure Lorentz transformation . . . . . . . . . . . . . 483.5.3 Composition of velocities . . . . . . . . . . . . . . . . . . . 48

3.6 Maxwell’s equations . . . . . . . . . . . . . . . . . . . . . . . . . . 503.7 Group of conformal transformations . . . . . . . . . . . . . . . . . 523.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

4 Clifford algebra 574.1 Clifford algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

4.1.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . 574.1.2 Clifford algebra H⊗H over R . . . . . . . . . . . . . . . . . 58

4.2 Multivector calculus within H⊗H . . . . . . . . . . . . . . . . . . 594.2.1 Exterior and interior products with a vector . . . . . . . . . 594.2.2 Products of two multivectors . . . . . . . . . . . . . . . . . 614.2.3 General formulas . . . . . . . . . . . . . . . . . . . . . . . . 624.2.4 Classical vector calculus . . . . . . . . . . . . . . . . . . . . 64

4.3 Multivector geometry . . . . . . . . . . . . . . . . . . . . . . . . . 644.3.1 Analytic geometry . . . . . . . . . . . . . . . . . . . . . . . 644.3.2 Orthogonal projections . . . . . . . . . . . . . . . . . . . . . 66

4.4 Differential operators . . . . . . . . . . . . . . . . . . . . . . . . . . 694.4.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . 694.4.2 Infinitesimal elements of curves, surfaces and hypersurfaces 694.4.3 General theorems . . . . . . . . . . . . . . . . . . . . . . . . 71

4.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

5 Symmetry groups 755.1 Pseudo-orthogonal groups O(1, 3) and SO(1, 3) . . . . . . . . . . . 75

5.1.1 Metric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 755.1.2 Symmetry with respect to a hyperplane . . . . . . . . . . . 755.1.3 Pseudo-orthogonal groups O(1, 3) and SO(1, 3) . . . . . . . 77

5.2 Proper orthochronous Lorentz group . . . . . . . . . . . . . . . . . 785.2.1 Rotation group SO(3) . . . . . . . . . . . . . . . . . . . . . 785.2.2 Pure Lorentz transformation . . . . . . . . . . . . . . . . . 795.2.3 General Lorentz transformation . . . . . . . . . . . . . . . . 81

5.3 Group of conformal transformations . . . . . . . . . . . . . . . . . 825.3.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

5.3.2 Properties of conformal transformations . . . . . . . . . . . 83



Contents xi

5.3.3 Transformation of multivectors . . . . . . . . . . . . . . . . 845.4 Dirac algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

5.4.1 Dirac equation . . . . . . . . . . . . . . . . . . . . . . . . . 85

5.4.2 Unitary and symplectic unitary groups . . . . . . . . . . . . 865.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

6 Special relativity 916.1 Lorentz transformation . . . . . . . . . . . . . . . . . . . . . . . . . 91

6.1.1 Special Lorentz transformation . . . . . . . . . . . . . . . . 916.1.2 Physical consequences . . . . . . . . . . . . . . . . . . . . . 926.1.3 General Lorentz transformation . . . . . . . . . . . . . . . . 94

6.2 Relativistic kinematics . . . . . . . . . . . . . . . . . . . . . . . . . 946.2.1 Four-vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

6.2.2 Addition of velocities . . . . . . . . . . . . . . . . . . . . . . 976.3 Relativistic dynamics of a point mass . . . . . . . . . . . . . . . . . 996.3.1 Four-momentum . . . . . . . . . . . . . . . . . . . . . . . . 996.3.2 Four-force . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

6.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

7 Classical electromagnetism 1057.1 Electromagnetic quantities . . . . . . . . . . . . . . . . . . . . . . . 105

7.1.1 Four-current density and four-potential . . . . . . . . . . . 1057.1.2 Electromagnetic field bivector . . . . . . . . . . . . . . . . . 107

7.2 Maxwell’s equations . . . . . . . . . . . . . . . . . . . . . . . . . . 1107.2.1 Differential formulation . . . . . . . . . . . . . . . . . . . . 1107.2.2 Integral formulation . . . . . . . . . . . . . . . . . . . . . . 1157.2.3 Lorentz force . . . . . . . . . . . . . . . . . . . . . . . . . . 116

7.3 Electromagnetic waves . . . . . . . . . . . . . . . . . . . . . . . . . 1187.3.1 Electromagnetic waves in vacuum . . . . . . . . . . . . . . . 1187.3.2 Electromagnetic waves in a conductor . . . . . . . . . . . . 1197.3.3 Electromagnetic waves in a perfect medium . . . . . . . . . 120

7.4 Relativistic optics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1217.4.1 Fizeau experiment (1851) . . . . . . . . . . . . . . . . . . . 121

7.4.2 Doppler effect . . . . . . . . . . . . . . . . . . . . . . . . . . 1237.4.3 Aberration of distant stars . . . . . . . . . . . . . . . . . . 1247.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

8 General relativity 1278.1 Riemannian space . . . . . . . . . . . . . . . . . . . . . . . . . . . 1278.2 Einstein’s equations . . . . . . . . . . . . . . . . . . . . . . . . . . 1288.3 Equation of motion . . . . . . . . . . . . . . . . . . . . . . . . . . . 1298.4 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130

8.4.1 Schwarzschild metric . . . . . . . . . . . . . . . . . . . . . . 130

8.4.2 Linear approximation . . . . . . . . . . . . . . . . . . . . . 133



xii Contents

Conclusion 135

A Solutions 137

B Formulary: multivector products within H(C) 153

C Formulary: multivector products within H⊗H (overR) 157

D Formulary: four-nabla operator ∇ within H⊗H (overR) 161

E Work-sheet: H(C) (Mathematica) 163

F Work-sheet H⊗H over R (Mathematica) 165

G Work-sheet: matrices M 2

(H

) (Mathematica) 167H Clifford algebras: isomorphisms 169

I Clifford algebras: synoptic table 171

Bibliography 173

Index 177



Introduction

If one examines the mathematical tools used in physics, one finds essentially threecalculi: 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 induc-tion by a wire with a current is in fact a purely relativistic effect. A satisfactorytreatment of classical electromagnetism, special relativity and general relativityis given by the tensor calculus. Yet, the tensor calculus does not allow a doublerepresentation of the Lorentz group and thus seems incompatible with relativisticquantum mechanics. A third calculus is then introduced, the spinor calculus, toformulate relativistic quantum mechanics. The set of mathematical tools used inphysics 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 toreside in the use of Clifford algebra. One of the major benefits of Clifford alge-bras 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 symplecticunitary groups. Concerning SO(3), this is well known, since the quaternion algebraH which is a Clifford algebra (with two generators) allows an excellent representa-tion of that group . The Clifford algebra H⊗H, the elements of which are simplyquaternions having quaternions as coefficients, allows us to do the same for theLorentz group. One shall notice that H⊗H is defined intrinsically independentlyof 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). Computationswithin these algebras have become straightforward with software like Mathemat-

ica which allows us to perform extended algebraic computations and to simplifythem. 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 relativisticphysics with applications such as classical electromagnetism, special relativity andgeneral relativity.



Chapter 1

Quaternions

The abstract quaternion group, discovered by William Rowan Hamilton in 1843, isan illustration of group structure. After having defined this fundamental conceptof physics, the chapter examines as examples the finite groups of order n ≤ 8 andin particular, the quaternion group. Then the quaternion algebra and the classicalvector 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 ∗ definedfor 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

(f 1, g1)(f 2, g2) = (f 1f 2, g1g2), (f i ∈ 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 C n 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−ha = abh (h = 1 · · ·n); indeed, since

(ab)−1 = b−1a−1 = b−1a = ab,

one hasb−1(b−1a)b = b−2ab = b−1ab2 = ab3

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

1.2 Finite groups of order n ≤ 8

The finite groups of order n ≤ 8, except the quaternion group which will be treatedseparately, are the following.

1. n = 1, there exists only one group (1 = e).

2. n = 2, only one group exists, the cyclic group C 2 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 C 3 of elements (b, b2,b3 = e) where b, b2 are elements of order 3.

4. n = 4, two groups exist:

(a) the cyclic group C 4 constituted by the elements (b, b2, b3, b4 = e) wherethe 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 themultiplication 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 cyclicgroups C 2,

(−1, 1)× (b, b2 = e)=

1 ≡ (1, b2), I ≡ (1, b), J ≡ (−1, b), K ≡ (−1, b2)

.

Example. The group constituted by the elements (I : rotation of πaround 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 C 5 having the elements(b, b2, b3, b4, b5 = e).

6. n = 6, two groups are possible:

(a) the cyclic group C 6 (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 ba a abb2 e b b2 ab ba ab3 = e b b2 e a ab baa ab ba a e b b2

ab ba a ab b2 e bba b ab ba b b2 e

with b−h

a = abh

(h = 1, 2, 3). This group is the first noncommutativegroup 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 C 7 of elements (b, b2, b3,b4, b5, b6, b7 = e).

8. n = 8, there exist five groups, among them the quaternion group which willbe treated separately.

(a) The cyclic group C 8 of elements (b, b

2

, b

3

, b

4

, b

5

, b

6

, b

7

, b

8

= e).




ab(M )

M

a(M )b(M )

ba(M )

b2(M )

y

xOz

B

AC

D

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 S 2×2×2, direct product of the Klein four-group with C 2,

(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 S 4×2, direct product of C 4 with C 2 and constituted by theelements

(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 ba a ab2 abb2 b3 e b b2 ab2 ba ab ab3 e b b2 b3 ab ab2 a bab4 = e b b2 b3 e a ab ba ab2

a ab ab2 ba a e b b3 b2

ab ab2 ba a ab b3 e b2 b

ba a ab b

2

ba b b

2

e b

3

ab2 ab a ab ab2 b2 b3 b e

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

M

a(M )

b(M )

ab(M )

ba(M )

b2(M )

ab2(M )

b3(M )

y

xOz

A

BC

D

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




relations

i2 = j2 = k2 = ij k = −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 beingthe 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 DefinitionsConsider the vector space of numbers called quaternions a, b, . . . constituted byfour real numbers

a = a0 + a1i + a2 j + a3k

= (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 (denotedH) via the multiplication

ab = (a0b0 − a1b1 − a2b2 − a3b3)+ (a0b1 + a1b0 + a2b3 − a3b2)i

+ (a0b2 + a2b0 + a3b1 − a1b3) j

+ (a0b3 + a3b0 + a1b2 − a2b1)k

and in a more condensed form

ab = (a0b0 − a · b, a0b + b0a + a× b)

where a · b = (a1b1 + a2b2 + a3b3) and a× b = (a2b3 − a3b2)i + (a3b1 − a1b3) j +

(a1b2 − a2b1)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 andby separating the quaternion product in two parts.

The quaternion algebra constitutes a noncommutative field (without divisorsof zero) containing R and C as particular cases. Let a = a0 + a1i + a2 j + a3k be aquaternion, the conjugate of a, the square of its norm and its norm are respectively

ac = a0 − a1i − a2 j − a3k,

|a|2 = aac = a20 + a21 + a22 + a2

3,

|a| = |a|2,

with the following properties(ab)c = bcac,

|ab|2 = |a|2 |b|2 ,

the last relation deriving from (ab)cab = bcacab = (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) ,

a−1 = acaac

,

a−12 =

acaac

a

aac

=

1

|a|2 ,

(a1a2

· · ·an)

−1=

(a1a2 · · · an)c

|a1a2 · · · an|2 = a−1

n a−1n−1

· · ·a−11 .




To divide a quaternion a by the quaternion b (= 0), one simply has to resolve theequation

xb = a or by = a

with the respective solutions

x = ab−1 = a bc

|b|2 ,

y = b−1a = bca

|b|2

and the relation |x| = |y| = |a||b| .

Examples. Consider the quaternions a = 2 + 4i − 3 j + k and b = 5 − 2i + j − 3k;

1. the vectorial parts a, b and the conjugates ac, bc are

a = 4i − 3 j + k, b = −2i + j − 3k,

ac = 2 − 4i + 3 j − k, bc = 5 + 2i − j + 3k;

2. the norms are given by

|a| = √

aac =√

30,

|b| =

bbc =√

39;

3. the inverses are

a−1

= ac

|a|2 = 2−

4i + 3 j−

k

30 ,

b−1 = bc

|b|2 = 5 + 2i − j + 3k

39 ;

4. one can realize the following operations

a + b = 7 + 2i − 2 j − 2k,

a − b = −3 + 6i − 4 j + 4k,

ab = 24 + 24i− 3 j − 3k, |ab| =√

1170,

ba = 24 + 8i − 23 j + k, |ba| = √ 1170,

(ab)−1 = (ab)c

|ab|2 = b−1a−1 = 4

195 − 4i

195 +

j

390 +

k

390,

xa = b, x = ba−1 = (−2

15 − 8i

15 +

9 j

10 − 13k

30 ),

ay = b, y = a−1b = (−2

15 − 16i

15 +

7 j

30 − 3k

10),

S (x) = S (y), |x| = |y| = 13

10.




1.4.2 Polar form

Any nonzero quaternion can be written

a = a0 + a1i + a2 j + a3k= r(cos θ + u sin θ), 0 ≤ θ ≤ 2π

with r = |a| =

a20 + a21 + a22 + a23 being the norm of a and

cos θ = a0

r , sin θ =

±

a21 + a22 + a2

3

r ,

cot θ = ± a0|a| , tan θ = ±|a|

a0;

the unit vector u (uuc = 1) is given by

u = ±(a1i + a2 j + a3k)

a21 + a22 + a23

with a21 + a22 + a2

3 = 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 = 3 + i + j + k,

|a| =√

12 = 2√

3, |a| =√

3,

tan θ = |a|

a0=

1√ 3

, θ = 30,

Answer: a = 2√

3

cos30 +

i + j + k√

3

sin 30

.

1.4.3 Square root and nth root

Square root

The square root of a quaternion a = a0 + a1i + a2 j + a3k can be obtained alge-braically as follows. The equation b2 = a with b = b0 + b1i + b2 j + b3k leads to thefollowing equations

b20 − b21 − b22 − b23 = a0 (1.1)

2b0b1 = a1, sgn(b0b1) = sgn(a1), (1.2)

2b0b2 = a2, sgn(b0b2) = sgn(a2), (1.3)

2b0b3 = a3, sgn(b0b3) = sgn(a3). (1.4)




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

t−

a21 + a22 + a23

4t = a

0

or

t2 − a0t − a21 + a22 + a2

3

4 = 0.

One obtains

t = b20 = a0 +

a20 + a21 + a2

2 + a232

≥ 0

−(b21 + b22 + b23) = a0 − b20 = a0 − a20 + a21 + a22 + a23

2 ,

hence

b0 = ε√

2

a20 + a2

1 + a22 + a23 + a0 (ε = ±1).

The equations (1.2), (1.3), (1.4) lead to

b20(−b21) = −a2

14 , (1.5)

b20(−b22 − b23) = −(a22 + a23)

4 , (1.6)

with

−b22 − b23 = b21 + a0 − b20

= b21 + a0 −

a20 + a21 + a2

2 + a232

.

Equation (1.6) then becomes with t = b21 and using (1.5)

a214t

t +

a0 −

a20 + a21 + a22 + a23

2

= −

a22 + a23

4

thus

t = a2

1

2

a20 + a21 + a22 + a2

3 − a0

a2

1 + a2

2 + a2

3

,




hence b1 (one proceeds similarly for b2, b3); finally, one obtains

b0 = ε

√ 2 a20 + a2

1

+ a2

2

+ a2

3

+ a0 (ε =±

1),

b1 = ε√

2

a1 a21 + a2

2 + a23

a20 + a21 + a2

2 + a23 − a0,

b2 = ε√

2

a2 a21 + a2

2 + a23

a20 + a21 + a2

2 + a23 − a0,

b2 = ε√

2

a3 a21 + a2

2 + a23

a20 + a21 + a2

2 + a23 − a0.

Example. Consider the quaternion a = 1 + i + j + k ; find its square root b

b0 = ε√

2

√ 3, b1 = b2 = b3 =

ε√ 6

,

Answer: b = ± 1√ 2

√ 3 +

i√ 3

+ j√

3+

k√ 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

R = r1

n , θ = (ϕ + 2kπ)

n (k = 0, 1, . . . , n − 1);

finally, one has

b = r1

n

cos

(ϕ + 2kπ)

n + u sin

(ϕ + 2kπ)

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 andthus θ = 2πm

n (m = 0, 1, . . . , n− 1). For n = 2, one obtains θ = 0, π and thusthe 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)πn

(m = 0, 1, . . . , n− 1), certain values of θ = π give nonreal roots b , the vectore being arbitrary.




Example. Find the cubic root of

a = 3 + i + j + k

= 2√ 3 cos30 + (i + j + k)

√ 3 sin30

;

Answer: b =

2√

3 1

3

cos θ +

(i + j + k)√ 3

sin θ

,

θ = 10, 130, 250.

1.4.4 Other functions and representations of quaternions

The exponential ea is defined by [30]

ea = 1 + a

1! +

a2

2! +

a3

3! + · · ·

where a is an arbitrary quaternion. Furthermore, one defines

cosh a = ea + e−a

2 = 1 +

a2

2! +

a4

4! + · · · +

a2 p

(2 p)!,

sinh a = ea − e−a

2 =

a

1! +

a3

3! +

a5

5! + · · · +

a2 p+1

(2 p + 1))!,

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, onecan define the trigonometric functions

cos a = eua + e−ua

2 = 1 − a2

2! +

a4

4! + · · · ,

sin a = eua − e−ua

2 =

a

1! − a3

3! +

a5

5! + · · · .

Example. Let a = uθ be a quaternion without a scalar part with u ∈ Vec H,u2 = −1 and θ real; one has

coshu

θ = cos θ,sinhuθ = u sin θ,

euθ = cos θ + u sin θ,

cosuθ = cosh θ,

sinuθ = 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 + a1i + a2 j + a3k by a 2× 2 complexmatrix (with i being the usual complex imaginary)

A = a0 + ia3

−ia1 + a2

−ia1 − a2 a0 − ia3

or by a 4 × 4 real matrix

A =

a0 −a1 −a2 −a3a1 a0 −a3 a2a2 a3 a0 −a1a3 −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 = a1i +a2 j +a3k,b = b1i + b2 j + b3k, c = c1i + c2 j + c3k. The norm of a is

|a| =√

aac =

a21 + a22 + a23

and

ab = ab + ba

2 +

ab − ba

2= (−a · b,a× b).

One defines respectively the scalar product and the vector product of two vectorsa, b by

(a, b) ≡ a · b = − (ab + ba)

2 = a1b1 + a2b2 + a3b3,

a × b ≡ a× b = (ab − ba)

2= (a2b3 − a3b2)i + (a3b1 − a1b3) j + (a1b2 − a2b1)k,

with ab = −(a, b) + a × b. Geometrically, one has

(a, b) = |a| |b| cos α,

|a

×b

|=

|a

| |b

|sin α,




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

|ab|2 = |a|2 |b|2 = (ab)(ab)c

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

= (a, b)2 + |a × b|2

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) =

−

[a(b × c) + (b × c)a]

2= − [a(bc − cb) + (bc − cb)a]

2

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 canstart from

abc = a [

−(b, c) + b

×c]

= −a(b, c) − (a, b × c) + a × (b × c)

henceV (abc) = −a(b, c) + a × (b × c);

since

V (abc) = abc − (abc)c

2

= abc + cba

2

= [abc + bac

−bac

−bca + bca + cba]

2

= (ab + ba)c

2 − b(ac + ca)

2 +

(bc + cb)a

2= −(a, b)c + b(a, c) − a(b, c)

one obtainsa × (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 = j2 = k2 = ijk = −

1, deduce the multiplicationtable of the quaternion group knowing that the element −1 commutes with allelements of the group and that (−1)i = −i, (−1) j = − j, (−1)k = −k.E1-2 Consider the quaternions a = 1 + i, b = 4 − 3 j. Compute |a| , |b|, a−1, b−1,a + b, ab, ba. Give the polar form of a and b. Compute

√ a,√

b, a1/3.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 =

−2 j, 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 es-tablished from orthogonal symmetries. The crystallographic groups and Kepler’sproblem 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 + x1i + x2 j + x3k,y = y0 + y1i + y2 j + y3k ∈ H, and the scalar product

(x, x) = xxc = x2

0 + x2

1 + x2

2 + x2

3.

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

(x, y) = 1

2 [(x + y, x + y) − (x, x) − (y, y)]

= 1

2 [(x + y)(x + y)c − xxc − yyc]

= 1

2 (xyc + yxc)

= x0y0 + x1y1 + x2y2 + x3y3.

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 quaternionperpendicular to the hyperplane. The expression of a plane symmetry is obtainedas follows.

Definition 2.1.1. The symmetric x of x with respect to a hyperplane is obtainedby 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 vectorx − x is perpendicular to the hyperplane (and thus parallel to a) and (x + x)/2is perpendicular to a. Hence, the relations

x = x + λa, λ ∈ R,a,

x + x

2

= 0;

one then deduces a, x +

λa

2

= 0, λ =

−2(a, x)

(a, a) ,

x = x − 2(a, x)a

(a, a)

= x − (axc + xac)a

aac

= −axca

aac.

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 inversionis an improper transformation of a determinant equal to

−1. In combining, in an

even number, plane symmetries

x = −mxcm,

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

= −axcb

with a, b ∈ H and aac = bbc = 1, one obtains the orthogonal group O(4) with sixparameters. This group, by definition, conserves the scalar product; indeed, forSO(4),

(x, y) = 1

2 [xyc + yxc]

= 1

2 [axbbcyac + aybbcxac]

= (x, y)



2.1. Orthogonal groups O(4) and SO(4) 21

and for the improper rotations

(x, y) = 1

2 [xyc + yxc]

= 12

[axcbbcyac + aycbbcxac]

= (x, y).

Any rotation of SO(4) can be written as a combination of a rotation of SO(3),examined below,

x = rxrc, rrc = 1

and a transformation

x = axa, a ∈ H, aac = 1.

It is sufficient to resolve the equation

x = f xg = arxrca (or raxarc)

with f = ar, g = rca and rrc = aac = 1. Writing the relations ([15], [16])

2a2 = 2f g,

a2f 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 = acf

= ± (1 + gcf c) f

|(1 + f g|= ±(f + gc)|(1 + f g| .

One verifies that one has indeed the relations aac = rrc = 1. One solves similarlythe equations f = ra, g = arc with the solutions

a = ±(1 + gf )

|(1 + gf | ,

r = ±(f + gc)

|(1 + gf | ,

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




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

Consider the vectors x = x1i + x2 j + x3k, x = x1

i + x2 j + x

3k ∈ 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 = −axca

= axcac

with a ∈ Vec H, aac = 1, ac = −a. The improper rotations (odd number ≤ 3 of plane symmetries) are given by

x = f xcf c

with f ∈ H, f f c = 1 and one has xxc = xxc. The SO(3) group is the set of properrotations (even number ≤ 3 of symmetries). In combining two symmetries, oneobtains

x = −axca,

x = −bxc b,

hence

x

= (bac)x(acb)= rxrc

with r (= bac) ∈ H, rc = abc = acb, rrc = 1. The unitary quaternion r can beexpressed in the form

r =

cos

θ

2 + u sin

θ

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 thenorm of x results from

xxc = rxrcrxcrc = xxc.

Furthermore, if one considers the transformation

q = rqrc

with q ∈ H, one has

S (q

) = S (rqrc) = S (rcrq ) = 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 developingthe formula x = rxrc with x = x ∈ Vec H, one obtains

x = x =

cos θ2

+ u sin θ2

x

cos

θ2 − u sin

θ2

= cos2 θ

2x + sin2

θ

2 [(u · x)u − (u× x)u] + u× x sin θ;

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

x =

0x1

x2

x3

, x =

0x1

x2

x3

,

A =

u

2

1+(u2

2 + u2

3)cos θ

u1u2(1 − cos θ)−u3 sin θ

u1u3(1 − cos θ)+u2 sin θ


u22

+(u2

1 + u2

3)cos θ




u23

+(u2

1 + u2

2)cos θ

;

one verifies that the matrix A is orthogonal (tA = A−1). If one combines therotation r1 and the rotation r2,

r1 =

cos α2

+ a sin α2

, r2 =

cos β

2 + b sin β

2

one obtainsr = r2r1 =

cos

γ

2 + c sin

γ

2

with

cos γ

2 = cos

α

2 cos

β

2 − (a,b)sin

α

2 sin

β

2,

c sin γ

2

= a sin α

2

cos β

2

+ b cos α

2

sin β

2 −(a

×b)sin

α

2

sin β

2

,




which yields the Rodriguez formula

c tan γ

2

= a tan α

2 + b tan β

2 − (a× b)tan α

2 tan β

2

1 − (a,b)tan α

2 tan β

2

.

2.3 Crystallographic groups

If r belongs to a finite subgroup of real quaternions, the transformations q = rqrcwill 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 C n (order N = 2n)

The elements of these groups are given by

r = u2h

n =

cos π

2 + u sin

π

2

2h

n

=

cos π

n + u sin

π

n

h= bh

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

r = k2h

n = cos π

2 + k sin

π

2 2h

n

=

cos π

n + k sin

π

n

h.

Example. Double group C 3 (N = 6, rotation axis along Oz); the elements of thegroup are

r = bh =

cos π

3 + k sin

π

3

h, h = 1, . . . , 6

or explicitly

b =

1

2(1 + k

√ 3), b2 =

1

2(−1 + k

√ 3), b3 = −1,

b4 = −b, b5 = −b2, b6 = 1 .

2.3.2 Double dihedral groups Dn (N = 4n)

These groups are constituted by the elements

r = u2h

n al = bhal

with

u = cos π

2 + u sin

π

2 , a = cos

π

2 + a sin

π

2 , b = u

2

n ,



2.3. Crystallographic groups 25

where u, a are two perpendicular vectors and S (au) = 0, a2 = −1, h = 1, . . . , 2n,l = 1, . . . , 4.

Examples. 1. Double group D3 (order 12)

b =

cos π

3 + k sin

π

3

=

1

2(1 + k

√ 3),

a = i, ab = 1

2(i − j

√ 3), ba =

1

2(i + j

√ 3);

the elements of the group are±b,±b2,±b3,±a,±ab,±ba

.

2. Double group D4 (order 16); writing

b = cos π

4

+ k sin π

4 , b2 = (k), b3 =

1

√ 2(

−1 + k),

a = (i), ab = 1√

2(i − j), ba =

1√ 2

(i + j),

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

2 (±1 ± i ± j ± k).

More explicitly, indicating the axes of multiple rotations, the group is com-posed of the elements

iα, jα, kα,

1 + i + j + k2

β

,

1 − i − j + k2

β

,

1 + i − j − k

2

β

,

1 − i + j − k

2

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,




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 = rxrctransforms 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

2 (±1± i ± j ± k),

1√ 2

(±1 ± i) , 1√

2(±1 ± j) ,

1√ 2

(±1 ± k) ,

1√ 2

(±i ± j) , 1√

2(± j ± k) ,

1√ 2

(±i ± k) .

Making explicit the axes of multiple rotations, the elements of the group are1 + i√

2

α

,

1 + j√

2

α

,

1 + k√

2

α

,

1 + i + j + k

2

β

,

1 − i − j + k

2

β

,

1 + i − j − k

2

β

,

1 − i + j − k

2

β

,

j + k

√ 2 γ

, i + k

√ 2 γ

, i + j

√ 2 γ

,

j − k√

2

γ

,

−i + k√ 2

γ

,

i − j√

2

γ

,

with α = 1, . . . , 8, β = 1, . . . , 6, γ = 1, . . . , 4 (N = 48).

Example. Consider the octaeder having, in an orthonormal frame, for its 6 verticesthe 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 inthe double octahedral group. The same property applies to the cube (dual of theoctaeder) 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, jα, kα,

i + m

j + mk2

α

,

mi + j + m

k2

α

,

m

i + mj + k2

α

,

i − m j + mk

2

α

,

−mi + j − mk

2

α

,

mi − mj + k

2

α

,

i + m j − mk

2

α

,

mi + j − mk

2

α

,

−mi − mj + k

2

α

,

i − m j − mk

2

α

,

−mi + j + mk

2

α

,

−mi + mj + k

2

α

,

1 + i + j + k

2β

,1

−i−

j + k

2β

,

1 + i − j − k

2

β

,

1 − i + j − k

2

β

,

1 + mj + mk

2

β

,

1 + mi + mk

2

β

,

1 + mi + m j

2

β

,

1 + mj − mk

2

β

,

1 − mi + mk

2

β

,

1 + mi− m j

2

β

,

m + m j + k

2γ

,m + i + mk

2γ

,m + mi + j

2γ

,m + m j − k

2

γ

,

m − i + mk

2

γ

,

m + mi − j

2

γ

,

with α = 1, . . . , 4, β = 1, . . . , 6, γ = 1, . . . , 10 and

m = 1 +

√ 5

2 = 2 cos36,

m = 1 −√

5

2

=

−2cos72.




One obtains another group, distinct from the first, by inverting m and m.

Example. Consider the icosaeder having, in an orthonormal frame, for the coor-dinates of the 12 vertices

±m j ± k

m

√ 5

, ± (±i + mk)

m

√ 5

, ±mi ± j

m

√ 5

;

for the centers of the 20 faces

± i + j + k√ 3

, ±−i − j + k√ 3

, ± i − j − k√ 3

, ±−i + j − k√ 3

,

±mj ± mk√ 3

, ± (±mi + mk)√ 3

, ±mi ± m j√ 3

;

for the middles of the 30 sides

±i, ± j, ±k, ± i ± m j ± mk

2 ,

± (±mi + j ± mk)

2 , ± (±mi ± mj + k)

2 .

The transformation x = rxrc transforms the icosaeder into itself; the same is truefor its dual, the dodecaeder (having 20 vertices, 12 faces and 30 sides) the verticesof 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 f c = ggc = 1, x = (x0, x) and x = (x0

,x)

∈ H

. For an infinitesimal rotation of SO(3) of dθ around the unit vector u =u1i + u2 j + u3k, one has

r =

cos

dθ

2 + u sin

dθ

2

1 + u

dθ

2

and

x = rxrc =

1 + u

dθ

2

x

1 − u

dθ

2

= x + dθ

2

(ux

−xu) = x + dθu

×x;



2.4. Infinitesimal transformations of SO(4) 29

hencedx = x − x = dθ u× x.

In matrix notation, one obtains

dx = dθuiM ix, i ∈ (1, 2, 3)

with

x =

x0

x1

x2

x3

, x =

x0

x1

x2

x3

,

and

M 1 =

0 0 0 00 0 0 00 0 0 −10 0 1 0

, M 2 =

0 0 0 00 0 0 10 0 0 00 −1 0 0

, M 3 =

0 0 0 00 0 −1 00 1 0 00 0 0 0

.

Introducing the commutator of two matrices A and B ,

[A, B] = AB − BA,

one verifies the relations

[M 1, M 2] = M 3,[M 1, M 3] = −M 2,

[M 2, M 3] = M 1,

or[M i, M j ] = ijkM k. (2.1)

Concerning the subgroup x = axa, with a ∈ H and aac = 1, one has for theinfinitesimal transformation

a =

cos dβ

2 + v sin dβ

2

1 + vdβ

2

;

hence

x = axa =

1 + v

dβ

2

x

1 + v

dβ

2

= x + dβ

2 (vx + xv) = x + dβ (−v · x + x0v)

and

dx = x

− x = dβ (−v·x

+ x0v

) .




In matrix notation, one obtains

dx = dβviN ix, i ∈ (1, 2, 3)

with

x =

x0

x1

x2

x3

, x =

x0

x1

x2

x3

,

N 1 =

0 −1 0 01 0 0 00 0 0 00 0 0 0

, N 2 =

0 0 −1 00 0 0 01 0 0 00 0 0 0

, N 3 =

0 0 0 −10 0 0 00 0 0 01 0 0 0

;

the matrices N i satisfy the relations

[N 1, N 2] = M 3,

[N 1, N 3] = −M 2,

[N 2, N 3] = M 1,

or, more concisely

[N i, N j] = ijkM k (2.2)

where M i are the matrices defined previously; furthermore

[N i, M j ] = ijkN k. (2.3)

Concerning the subgroup x = f x, one obtains with f 1+ dα2 f , f = f 1i+f 2 j+f 3k

(unit vector)

dx = x − x = dα

2 [−f · x + x0f + f × x] .

In matrix notation, one has

dx = dαf iF ix

with

F 1 = 1

2

0 −1 0 01 0 0 00 0 0 −10 0 1 0

, F 2 =

1

2

0 0 −1 00 0 0 11 0 0 00 −1 0 0

,

F 3 = 1

2

0 0 0 −10 0 −1 00 1 0 0

1 0 0 0



2.4. Infinitesimal transformations of SO(4) 31

and

[F 1, F 2] = F 3,

[F 1, F 3] = −F 2,[F 2, F 3] = F 1,

or

[F i, F j ] = ijkF k.

As to the subgroup x = xg, with g 1 + dβ2 g, one proceeds similarly and obtains

dx = x − x = dβ

2 [−g · x + x0g − g× x] .

In matrix notation,

dx = dβgiGix

with

G1 = 1

2

0 −1 0 01 0 0 00 0 0 10 0 −1 0

, G2 =

1

2

0 0 −1 00 0 0 −11 0 0 00 1 0 0

,

G3 = 1

2

0 0 0 −10 0 1 00 −1 0 01 0 0 0

and

[G1, G2] = −G3,

[G1, G3] = G2,

[G2, G3] = −G1,

or

[Gi, Gj ] = −ijkGk.

Furthermore, one has

[F i, Gj ] = 0, i, j = 1, 2, 3

the matrices M i, N i, F i, Gi satisfying the relations

M i = F i−

Gi, N i = F i + Gi.




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 = p2

2m − k

r, k = GM m

and the equations of motion are

q i = ∂H

∂pi, ˙ pi =

−∂H

∂q i.

Let F (q i, pi, t) be a physical quantity of the motion; one has

dF dt

= ∂F ∂t

+ ∂F ∂q i

∂q i∂t

+ ∂F ∂pi

∂pi∂t

= ∂F

∂t +

∂F

∂q i

∂H i∂pi

− ∂F

∂pi

∂H

∂q i

.

Introducing Poisson’s bracket of two functions u and v

[u, v] = ∂u

∂q i

∂v

∂pi− ∂u

∂pi

∂v

∂q i

one obtainsdF dt

= ∂F ∂t

+ [F, H ] .

If [F, H ] = 0 and ∂F ∂t

= 0, one has dF 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 negativeenergy E < 0) and introduce the vector

D = A√ −2mE

with

E = p2

2m − k

r



2.5. Symmetries and invariants: Kepler’s problem 33

verifying the Poisson bracket [D, H ] = 0; one has the relations

[Li, Lj ] = ijkLk, (2.4)

[Di, Dj ] = ijkLk, (2.5)[Di, Lj ] = ijkDk. (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 transformationq = aqa of SO(4), which indicates that the symmetry group of the problem isSO(4). To see it more explicitly, let us develop A in the form

A =

r

p2 − km

r

− p (p · r)

with

D = A√ −2mE = r p0 − pr0

and

p0 = p2 − km

r√ −2mE , r0 =

p · r√ −2mE .

One verifies immediately that D · L = 0; furthermore,

(L)2

= (r× p)2

= r2 p2 − (r · p)2 ,

hence

(D)2

+ (L)2

= k

2

m

2

−2mE or

H = −k2m2

2

(D)2

+ (L)2

.

Introducing two quaternions

r = (r0, r), p = ( p0,p)

and the quaternion

K = r ∧ p = 12 (rpc − prc) = (0,D− L)

one hasKK c = |0,D− L|2 = D2 + L2,

and thus

H = −k2m

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

K

K

c = KK c.




2.6 Exercises

E2-1 From the general formula

x = rxrc, r =

cos

θ

2 + u sin

θ

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 rotationof angle β around the Oy axis, of a rotation of angle γ around the Oz axis. Showthat the matrices are orthogonal A−1 = At, det A = 1 (At : transposed matrix of A).

E2-2 Consider the relation A = rArc where A are the components of a vector

with respect to an orthonormal frame at rest and A

its components with respect toa mobile orthonormal frame. Show that dA = rDArc where DA is the covariantdifferential with

DA = dA + dΩ × A, dΩ = 2rcdr

(dA is the differential with respect to the components only). What does

dΩ

dt = 2rc

dr

dt

represent?

E2-3 Consider the relations

A = rArc = gAgc

with

dA = rDArc = gDAgc,

DA = dA + dΩ × A (dΩ = 2rcdr),

DA = dA + dΩ

×A

(dΩ = 2gcdg)

(A represents the components of a vector with respect to an orthonormal basis atrest, A, A the components with respect to mobile orthonormal bases.). Find therelation between dΩ

and dΩ.

E2-4 Using the covariant derivative, express the velocity and the acceleration inpolar 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 = 2rcdr and dΩ = 2(dr)rc. Express thebasis 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 successiverotations (Figure 2.1). A first rotation of angle α (precession angle) around ktransforms 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 thirdrotation of angle γ (proper rotation angle) around the vector k transforms thebasis i, j, k into the basis e1, e2, e3 = k. Give the quaternion r of the rotationX = rX rc. Determine

ω = 2rcdr

dt, ω = 2

dr

dtrc.

Give the components of the basis vectors ei.

x

y

z

x

y

z

α

β

γ i

j

k

ie1

e2k= e3

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



Chapter 3

Complex quaternions

From the very beginning of special relativity, complex quaternions have been usedto formulate that theory [45]. This chapter establishes the expression of the Lorentzgroup using complex quaternions and gives a few applications. Complex quater-nions 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 + a1i + a2 j + a3k having complexcoefficients. Such a quaternion can be represented by the matrix

A =

a0 + ia3 −ia1 + a2

−ia1 − a2 a0 − ia3

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 complexquaternions H(C) is isomorphic to 2 × 2 matrices over C and has zero divisors;indeed, the relation

(1 + ik)(1− ik) = 1− (i)2k2 = 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 unitedinto a four-dimensional pseudoeuclidean spacetime with the relativistic invariant

c2t2 − x12 −

x22 −

x32

.

Minkowski had the idea of using complex quaternions

x = (ict,x1, x2, x3) = (ict + x)

with the invariant

xxc =

−c2t2 + x12 + x22 + x32 .

To get the signature (+ −−−), we shall use complex quaternions in the form

x = (ct,ix1, ix2, ix3) = (ct + ix)

which we shall call minkowskian quaternions or minquats with the invariant

xxc = c2t2 − x12 −

x22 −

x32

.

A minquat is a complex quaternion such that x∗c = x where ∗ is the complexconjugation.

3.2.2 Plane symmetry

Let us consider two minquats x = (x0+ix), y = (y0+iy) with the scalar product

xxc =

x02 −

x12 −

x22 −

x32

.

One has(x + y, x + y) = (x, x) + (y, y) + (x, y) + (y, x);

postulating the property (x, y) = (y, x) one obtains

(x, y) = 1

2 [(x + y, x + y)

−(x, x)

−(y, y)]

= 1

2 [(x + y)(x + y)c − xxc − yyc]

= 1

2(xyc + yxc)

= x0y0 − x1y1 − x2y2 − x3y3.

Two minquats x, y are said to be orthogonal if (x, y) = 0. A minquat x is timelikeif 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 obtainedby drawing the perpendicular to the hyperplane and by extending this perpendic-ular by an equal length [11].

The hyperplanes are supposed to go through the origin. The vector x − xis 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, λ ∈ R,a,

x + x

2

= 0;

hence,

a, x + λa

2

= 0,

λ = −2(a, x)

(a, a) ,

x = x − 2(a, x)a

(a, a) = x − (axc + xac)a

aac,

x = −axca

aac.

One shall distinguish time symmetries (with aac = 1)

x = −axca

from space symmetries (aac = −1)

x = axca.

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 oper-

ators which leave invariant the quadratic form

(x, y) = x0y0 − x1y1 − x2y2 − x3y3

with x = (x0 + ix), y = (y0 + iy).

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.




Proper orthochronous Lorentz transformation

Consider the transformation obtained by combining an even number of time sym-metries and an even number of space symmetries. These transformations of de-terminant equal to +1 constitute a subgroup of O(1, 3), the special orthogonalgroup 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 xcf )c g]c mc n

= (nmcgf c)x(f cgmcn);

writing a = nmcgf c, one obtains

ac = f gcmnc, aac = 1

a∗c = f ∗g∗cm∗n∗c

= f cgmcn;

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. Incombining these symmetries, one obtains the following Lorentz transformations L:

1. n even, p odd, orthochronous, improper Lorentz transformation (det L = −1)

x = axca∗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 = −axca∗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 41

Rotation(det L = 1)

Inversion(det L = −1)

orthochronousn even, p even

x = axa∗c(aac = 1)

n even, p oddx = axca∗c(aac = −1)

antichronousn odd, p oddx = −axa∗c(aac = −1)

n odd, p evenx = −axca∗c

(aac = 1)

3.3 Orthochronous, proper Lorentz group

3.3.1 PropertiesThe proper orthochronous Lorentz transformation

x = axa∗c

with aac = 1, x = (ct + ix), x = (ct + ix) conserves, by definition, the norm

xxc = (axa∗c)(a∗xcac)

= xxc

and the minquat type of x

x∗c = (axa∗c)∗c = ax∗ca∗c

= axa∗c = x.

The composition of two transformations satisfies the rule

x = a2(a1xa∗1c)a∗2c= a3xa∗3c

with a3 = a2a1, a3a3c = 1, a3 ∈ H(C). A (three-dimensional) rotation is given bythe 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 + iv sinh ϕ

2 is a minquat such that bbc = 1 and iv is a unitary

space-vector (iv

, iv

) = −1. A general transformation (proper, orthochronous) is




obtained by combining a rotation and a pure Lorentz transformation. This can bedone in two ways,

x = b(rxrc)b∗c ,x = r(bxb∗c)rc,

br being in general distinct from rb. Reciprocally, a general Lorentz transformationcan be decomposed into a pure Lorentz transformation and a rotation. The prob-lem simply consists to resolve the equation a = br (or rb) where b is a unitaryminquat (bbc = 1) and r a real unitary quaternion (rrc = 1). The equation a = bris solved in the following way ([15], [16]). Since

a∗c = r∗c b∗c = rcb

one obtains aa∗c = b2; let us write d = aa∗c , with ddc = aa∗ca∗ac = 1, hence, theequations

2b2 = 2d,

b2d = d2,

b2dc = 1

and their sumb2(2 + d + dc) = (1 + d)2.

The solution therefore is

b = ±(1 + d)

|1 + d|= ±(1 + aa∗c)

|1 + aa∗c |

with |1 + d| =

(1 + d)(1 + d)c. The rotation is given by

r = bca

= ±(a + a∗)

|1 + aa∗c | .

Finally, one verifies that this is indeed a solution. For the equation a = rb, onefinds in a similar way the solution

b = ±(1 + a∗ca)

|1 + a∗ca| ,

r = ±(a + a∗)

|1 + a∗ca

|



3.3. Orthochronous, proper Lorentz group 43

with |1 + a∗ca| = |1 + aa∗c | since S (aa∗c) = S (a∗ca). One observes that in both cases(a = br or rb) 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 thecombination of two pure Lorentz transformations b1, b2. The quaternion b = b2b1will in general be a complex quaternion (and not a minquat) and will be writtena = br. The resulting Lorentz transformation will thus contain a rotation; this isthe 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 + ix), x = (ct + ix), b =

cosh ϕ2

+ iv sinh ϕ2

and iv a spacelike

unitary minquat. For an infinitesimal transformation,

b

cosh dϕ

2 + iv sinh

dϕ

2

=

1 + iv

dϕ

2

,

hence

x = bxb∗c

=

1 + iv

dϕ

2

x

1 + iv

dϕ

2

= x + idϕ

2 (vx + xv) ;

consequently,

dx = x − x

= dϕ (v · x + ivx0) .

Using real matrices,

X =

x0 = ctx1

x2

x3

, X =

x0 = ct

x1

x2

x3

one can write

dX = X − X

= dϕviK iX




with

K 1 =

0 1 0 01 0 0 0

0 0 0 00 0 0 0

, K 2 =

0 0 1 00 0 0 0

1 0 0 00 0 0 0

, K 3 =

0 0 0 10 0 0 0

0 0 0 01 0 0 0

.

The matrices K i satisfy the relations

[K i, K j] = −ijkM k, (3.1)

[K i, M j] = ijkK k, (3.2)

where M i are the matrices defined for the infinitesimal transformations of SO(3)in Chapter 2,

M 1 =

0 0 0 00 0 0 00 0 0 −10 0 1 0

, M 2 =

0 0 0 00 0 0 10 0 0 00 −1 0 0

, M 3 =

0 0 0 00 0 −1 00 1 0 00 0 0 0

.

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

3.4 Four-vectors and multivectors in H(C)

Let x = (x0 + ix), y = (y0 + iy) be two four-vectors and their conjugates xc andyc; one can define the exterior product

x ∧ y = 1

2 (xyc − yxc)

=

0,(x2y3 − x3y2) + i (x1y0 − x0y1) ,

(x3y1 − x1y3) + i (x2y0 − x0y2) ,

(x1y2 − x2y1) + i (x3y0 − x0y3)

= [0, x

×y + i (y0x

−x0y)]

with x ∧ y = −y ∧ x. The resulting quaternion is of the type B = (a + ib)and is called a bivector; its real part, in the above example, gives the ordinaryvector product but its nature differs from that of a four-vector. Under a Lorentztransformation (proper, orthochronous), a bivector transforms as

B = x ∧ y = 1

2 (xyc − yxc)

= 1

2 [(axa∗c ) (a∗ycac ) − (aya∗c ) (a∗xcac )]

= aBac.



3.4. Four-vectors and multivectors in H(C) 45

The Lorentz transformation conserves the bivector type characterized by B =−Bc,

Bc = aBcac = −aBac = −B.

Furthermore, B2 =− (a)2 + (b)2 − 2ia · b is a relativistic invariant

B2 = B B = aBacaBac = aB2ac = B2.

A trivector is a complex quaternion defined by

T = x ∧ B = 1

2 (xB∗ + Bx)

= [−ix · a + x0a + x× b] =

it0 + t

and the product B

∧x by postulating B

∧x = x

∧B. Under a proper, orthochronous

Lorentz transformation, one has

T = xB∗ + Bx

2= aT a∗c

that is, T transforms as a four-vector. Furthermore, the transformation conservesthe trivector type T ∗c = −T ,

T ∗c = aT ∗c a∗c = a(−T )a∗c

= −T ;T yields the relativistic invariant

T T c = aT a∗ca∗T cac = T T c

=

t02 −

t12 −

t22 −

t32

.

The exterior product, defined above, is associative

(x ∧ y) ∧ z = x ∧ (y ∧ z) .

Indeed,(x ∧ y) ∧ z = z ∧ (x ∧ y) (3.3)

= 1

2

z (xyc − yxc)

∗ + (xyc − yxc) z

(3.4)

= 1

2 [z (xcy − ycx) + (xyc − yxc) z] , (3.5)

x ∧ (y ∧ z) = 1

2

x (yzc − zyc)

∗ + (yzc − zyc) x

(3.6)

= 1

2

[x (ycz

−zcy) + (yzc

−zyc) z] . (3.7)




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 therelation

P = x ∧ T = −1

2 (xT c + T x∗)

= −1

2 (xT c + T xc)

where T is a trivector; by definition, one postulates T ∧ x = −x ∧ T . The pseu-doscalar type is a pure imaginary P = is characterized by P c = P and is invariantunder a Lorentz transformation

P = −1

2 (xT c + T xc) = aP ac = P.

Examples. Consider the basis vectors e0 = 1, e1 = ii, e2 = i j, e3 = ik; oneobtains the following table

1 i = e2 ∧ e3 j = e3 ∧ e1 k = e1 ∧ e2i = e0 ∧ e1 ∧ e2 ∧ e3 ii = e1 ∧ e0 i j = e2 ∧ e0 ik = e3 ∧ e0

1 = e0 i = e0 ∧ e2 ∧ e3 j = e0 ∧ e3 ∧ e1 k = e0 ∧ e1 ∧ e2i = e1 ∧ e3 ∧ e2 ii = e1 i j = e2 ik = 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 exteriorproducts, one can define interior products

x · y = (x, y) = 1

2 (xyc + yxc) ,

x · B = 1

2 (xB∗ − Bx) ∈ four-vector,

x · T = −1

2 (xT c − T x∗) ∈ bivector;

by definition, one supposes B · x = −x ·B, T · x = x · T . More generally, one shalldefine the interior product between two multivectors A p and Bq by [12, p. 14]

A p ·Bq = (v1 ∧ v2 · · · ∧ v p−1) · (v p · Bq) .

Hence, one sees that H(C) already allows us to develop a few notions of a mul-tivector 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 frameK (O, x, y, z) moving along the Ox axis with the constant velocity u (Figure3.1).

u

y

x

z

y

x

z

O O

K K

Figure 3.1: Special Lorentz transformation (pure): the axes remain parallel tothemselves and the reference frame moves along the Ox axis.

The Lorentz transformation is expressed by

X = bX b∗c (3.8)

with X = (x0 + ix) , X = (x0 + ix), b =

cosh ϕ2

+ ii sinh ϕ2

, bbc = 1,

tanh ϕ = uc

; write γ = cosh ϕ, hence

γ 2 = 1 + sinh2 ϕ = 1 + u2

c2 γ 2,

γ = 1 1− u2

c2

.

Explicitly, equation (3.8) reads

ct = γ

ct + xu

c

,

x = γ

x + ctu

c

,

y = y, z = z .




The inverse transformation is given by X = bcXb∗ and yields

ct = γ ct− xu

c ,

x = γ

x− ct uc

,

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+ix),

X = (ct + ix), γ = 1q 1−u

2

c2

, tanh ϕ = uc . Explicitly, one obtains [26, p. 280]

ct = γ

ct +

x · uc

,

x = x + n (n · x) (γ − 1) + ctnγ u

c.

uy

x

z

y

x

z

O

O

K

K

Figure 3.2: General pure Lorentz transformation: the axes remain parallel to them-selves 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) 49

b =

cosh ϕ2 + ii sinh ϕ

2

, tanh ϕ = u

c , γ = cosh ϕ = 1q 1−u

2

c2

(ϕ constant). The

four-velocity V transforms as

V ≡ dX ds

= b dX ds

b∗c

with the relativistic invariant ds =

c2dt2 − dx2 − dy2 − dz2 ; ds can also beexpressed as

ds =

c2dt2

1− v2

c2

=

cdt

γ =

cdt

γ .

The four-velocity satisfies the relation

V V c = dX

ds

dX c

ds

= 1

and can be written in the form

V =

cosh θ + iv

v sinh θ

=

γ + iγ v

c

with tanh θ = v

c, γ = cosh θ = 1q

1− v2

c2

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 + ii sinh θ1) with tanh θ1 = v

c . In the reference frame K , one

has with b =

cosh θ22 + ii sinh

θ22

, tanh θ2 = uc ,

V = bV b∗c= [cosh (θ1 + θ2) + ii sinh(θ1 + θ2)]

= (cosh θ + ii sinh θ)

hence, θ = θ1 + θ2 and

tanh θ = v

c =

tanh θ1 + tanh θ21 + tanh θ1 tanh θ2

=

v

c +

u

c

1 + vu

c2

;

finally

v = v + u

1 + vuc2

.

When v, u c one obtains the usual Galilean transformation. Furthermore, if v = c, one has

v = c + u

1 + uc

= c,

the velocity of light thus appears as a limit speed.




3.6 Maxwell’s equations

Let A = (V c + iA) be the four-potential, D the relativistic four-nabla operator

D =

∂

c∂t,−i

∂

∂x1,−i

∂

∂x2,−i

∂

∂x3

=

∂

c∂t − i∇

and the conjugate operator

Dc =

∂

c∂t + i∇

.

Under a Lorentz transformation, D transforms as x = (x0 + ix),

D = aDa∗c .

To verify this, it is sufficient to use the relation

∂

∂xµ = ∂

∂xα

∂xα

∂xµ ,

to develop x = axa∗c , x = acxa∗ and to compare the coefficients. Adopting the

Lorentz gauge(D, A) =

1

c2∂V

∂t + divA = 0

one obtains the electromagnetic field bivector

F = DcA = (D, A) + (D ∧ A)

= (D ∧A) =

−B +

iE

c

.

Under a special Lorentz transformation, the bivector F transforms as

F = bF bc

with b =

cosh ϕ2 − ii sinh ϕ

2

, tanh ϕ = v

c, γ = cosh ϕ, which yields the standard

equationsBx = Bx, E x = E x,

By = γ

By +

E yv

c2

, E y = γ (E y − vBz) ,

Bz = γ Bz

− E yv

c2 , E z = γ (E z + vBy) .



3.6. Maxwell’s equations 51

Furthermore, one has the relativistic invariant

F F c = B2

− E2

c2 + 2i

E ·Bc

.

The exterior product

D ∧ F = D ∧ D ∧ A = 0

=

−i divB , −∂ B

c∂t − rotE

c

gives two of the Maxwell’s equations

divB = 0, rotE = −∂ B

∂t .

Introducing the four-current density C = (ρc,i j), the equation

D · F = DF ∗ − F D

2

=

div(E

c , i

rotB − 1

c2∂ E

∂t

= µ0C

gives the two other Maxwell’s equations (in vacuum)

divE = ρ

0, rotH = j +

∂ D

∂t

with 0µ0c2 = 1. Furthermore,

DF ∗ = D · F + D ∧ F

= D(DAc)∗ = DDcA

= A = µ0C

where

= ∂ 2

c2∂t2 − ∂ 2

∂ (x1)2 − ∂ 2

∂ (x2)2 − ∂ 2

∂ (x3)2

is the d’Alembertian. Hence, the equations

∂ 2V

c2∂t2 − V =

ρ

0,

∂ 2A

c

2

∂t

2

− A = µ0 j.




The entire set of Maxwell’s equations (in vacuum) can therefore be written

DF ∗ = µ0C

=

divE

c − i divB

+i

rotB− 1

c2∂ E

∂t

−

∂ B

c∂t +

rotE

c

= µ0(ρc + i j).

The interior product

F · C = −C · F

= j ·E

c + iρ (E + v ∧B)

= f

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 dxdxc = 0. This group includesspacetime translations, Lorentz transformations, dilatations

x = x + d,

x = axa∗c ,

x = λx

(d ∈ minquat, a ∈ H(C), aac = 1, λ ∈ R) and the transformations

x = (1 + xac)−1 x (3.9)

= x (1 + acx)−1

(3.10)

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

(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. Thetransformations (3.11) can also be expressed in the form

(x)−1

= x−1 (1 + xac) ,

(x)−1

= x−1 + ac ;



3.7. Group of conformal transformations 53

the inverse transformation results from

x−1 = (x)−1 − ac

= (x)−1 [1 − xac]

and thus

x = (1− xac)−1

x.

The composition of two transformations gives

(x)−1

= x−1 + ac,

(x)−1

= (x)−1

+ bc

= x−1

+ ac + bc= x−1 + cc

with c = a + b and thus belongs indeed to the group; if one permutes the twotransformations, one obtains the same resulting transformation. As properties,one has

xacx = xacx, (3.12)

|dx

|2

= |x|2

|x|2 |dx|2 , (3.13)

|x|2 = |x|2|1 + xac|2

, (3.14)

dx = (1 + xac)−1 dx (1 + acx)

−1 . (3.15)

Equation (3.12) results from

x = (1 + xac)−1 x = x(1 + acx)−1,

(1 + xac) x = x = x(1 + acx)

which entails the relation. Furthermore,

x = (1 + xac)x,

xxc = (1 + xac) xxc (1 + axc) ,

and thus

|x|2 = |x|2|1 + xac|2

.




Equation (3.13) which shows that the transformation is indeed a conformal trans-formation 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 −1dq = 0,

d

q −1

= −q −1dqq −1;

hence

d

x−1

= − x−1

dx

x−1

or d

x−1

= d

x−1

, consequently

x−1

dx x−1

= x−1

dxx−1,

xcdxxc(xxc)2

= xcdxxc(xxc)2

.

By multiplying with the conjugate equation

xdxcx

(xxc)2 =

xdxcx

(xxc)2

one obtains equation (3.13)

dxdxc

(xxc)2 =

dxdxc

(xxc)2 .

To obtain equation (3.15), one differentiates the equation

x = x (1 + acx)−1 ,

dx = dx (1 + acx)−1 − x (1 + acx)

−1 [d (1 + acx)] (1 + acx)−1 ,

=

dx− x (1 + xca) acdx

(1 + acx) (1 + xca)

(1 + acx)

−1 ,

= (1 + axc)dx (1 + acx)

−1

(1 + acx) (1 + xca) ,

= (1 + xac)−1 dx(1 + acx)−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, e1e2 = e2e1 =0. Take ∀a ∈ H(C), give the expression of the modules u = ae1, v = e1a.

E3-2 Consider the complex quaternions

x = 1 + ii + (2 + i) j + k

y = 2 + 3ik;

compute x + y, xy, yx, x2, y2, x−1, y−1, y−1x−1, (xy)−1.

E3-3 Consider the general Lorentz transformation

X = aX a∗c ,

a = 1

4 3 + ii√

5− i j√

15 + k3√

3with aac = 1. Determine the rotation ri and the pure Lorentz transformation bisuch that a = b1r1 = r2b2 with

X = b1r1X (b1r1)∗c = b1 (r1X r1c) b∗1c

orX = r2 (b2Xb∗2c) r2c.

E3-4 Consider the minquats

x = 1 + i (i + j) , y = 2 + ik,

z = 3 + i j, w = 3ii + i j + ik.

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

reference frame moving along the Ox axis with a constant velocity v = 3√ 5

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 v

c = 1√

7 in the direction Oy; find its

four-position and its four-velocity in the reference frame K . Let E E x, E y, E z

be the electric field in the reference frame K ; determine the electromagnetic fieldin K .

E3-6 Consider a square ABCD of center 0 in the plane z = 0 and having its ver-tices 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 = ik.



Chapter 4

Clifford algebra

Clifford having demonstrated that the Clifford algebra is isomorphic to a tensorproduct of quaternion algebras or to a subalgebra thereof, this chapter developswithin H ⊗ H the multivector calculus, multivectorial geometry and differentialoperators.

4.1 Clifford algebra

4.1.1 Definitions

The Clifford algebra C n over R is an associative algebra having n generators

e0, e1, . . . , en−1 such that

e2i = ±1, eiej = −ejei (i = j).

C + is the subalgebra constituted by products of an even number of ei. Cliffordalgebras 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 C 2mis the tensor product of m quaternion algebras. If n = 2m− 1, the Clifford algebra C 2m−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 = e0e1 · · · e2m−1) of the algebra C 2m .

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 productA⊗ 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 + ib) ⊗ (f + ig)

= (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 = I i = 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 havingcomplex 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 ex-pressed by

A = (a0 + ia1 + ja2 + ka3)⊗ (b0 + ib1 + jb2 + kb3)

=

a0b01 ⊗ 1 + a0b11 ⊗ i + a0b21 ⊗ j + a0b31 ⊗ k,

a1b0i ⊗ 1 + a1b1i⊗ i + a1b2i ⊗ j + a1b3i ⊗ k,

a2b0 j ⊗ 1 + a2b1 j ⊗ i + a2b2 j ⊗ j + a2b3 j ⊗ k,a3b0k ⊗ 1 + a3b1k ⊗ i + a3b2k ⊗ j + a3b3k ⊗ k

;

let us write i = i ⊗ 1, j = j ⊗ 1, k = k ⊗ 1, I = 1 ⊗ i, J = 1 ⊗ j, K = 1 ⊗ k with

i2 = j2 = k2 = ij k = −1,

I 2 = J 2 = K 2 = I JK = −1.

The general element A can thus be written

A = α0 + α1I + α2J + α3K

where the coefficients αi = di + iai + j bi + kci are quaternions and where thelowercase i, j, k commute with the capital I , J, K (iJ = J i, etc.). The element A

is called a Clifford number and constitutes simply a quaternion having quaternionsas coeffficients. Concisely, one can write

A = (α0;α)

with α = α1I + α2J + α3K , αi ∈ H. The product of two Clifford numbers is givenby

(α0;α)(β 0;β) =

α0β 0 − α1β 1 − α2β 2 − α3β 3;α0β 1 + α1β 0 + α2β 3 − α3β 2,α0β 2 + α2β 0 + α3β 1 − α1β 3,

α0β 3 + α3β 0 + α1β 2 − α2β 1

and in a more compact notation

(α0;α

)(β 0;β

) = [α0

β 0 −

α·β

;α0β

+αβ

0 +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 theterms 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 eiej = −ejei (i = j ). A complete basis of thealgebra is given in the following table.

1 I = e3e2 J = e1e3 K = e2e1i = e0e1e2e3 iI = e0e1 iJ = e0e2 iK = e0e3

j = e0 jI = e0e3e2 jJ = e0e1e3 jK = e0e2e1k = e1e2e3 kI = e1 kJ = e2 kK = e3

A Clifford number can be written in the form [54]

A = (a + ib + jc + kd;m + in + jr + ks)

with m = m1I + m2J + m3K and similarly for n, r and s. The Clifford algebracontains scalars a, pseudoscalars ib, vectors (four-dimensional) jc + ks, bivectorsm + in and trivectors kd + jr. The conjugate of A is defined by transformingI, J, K into −I,−J,−K and j into − j, hence

Ac = (

a+

ib−

jc+

kd;−m−

in+

jr−

ks)

with

(AB)c = BcAc.

The dual of A is defined by

A∗ = iA

and the commutator of two Clifford numbers by

[A, B] = 1

2

(AB − BA) .

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 exteriorproducts of two vectors x = jx0 + kx (x = x1I + x2J + x3K ), y = j y0 + ky canbe defined by the general formula [37]

xy =

λx·

y +

µx∧

y




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 = x0y0 − x1y1 − x2y2 − x3y3 ∈ S (scalar),

x ∧ y =

x

2

y

3

− x

3

y

2I +

x

3

y

1

− x

1

y

3J +

x

1

y

2

− x

2

y

1K

+

x1y0 − x0y1

iI +

x2y0 − x0y2

iJ +

x3y0 − x0y3

iK

=x× y + i

xy0 − x0y

∈ B

with xc = −x, Bc = −B for a bivector B. The products of a vector with amultivector A p = v1 ∧ v2 ∧ · · · ∧ v p are then defined by

2x · A p = (−1) p [xA p − (−1) pA px] ,

2x ∧ A p = (−1) p [xA p + (−1) pA px]

and

A p · x ≡ (−1) px · A p,

A p ∧ x ≡ (−1) px ∧ A p.

The above formulas yield for a trivector (with x = jx0 + kx, B = a + ib)

T = x ∧ B = 1

2(xB + Bx)

=

(−a · x) k + j

x0a + x× b

with B ∧ x = x ∧ B and T c = T . One verifies that the exterior product is indeedassociative

(x ∧ y) ∧ z = x ∧ (y ∧ z) ;

one has,

(x ∧ y) ∧ z = z ∧ (x ∧ y)

= −1

2 [z(xy − yx) + (xy − yx)z] , (4.1)

x ∧ (y ∧ z) = −1

2

[x(yz − zy) + (yz − zy)x] . (4.2)




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. Apseudoscalar is defined by (with T = kt0 + jt, x = jx0 + kx)

P = x ∧ T = −1

2(xT − T x)

= −

x0t0 + x · t

i

with T ∧ x ≡ −x ∧ T and P c = P . As examples, let us express the standard basisin terms of the exterior product:

1 I = e2 ∧ e3 J = e3 ∧ e1 K = e1 ∧ e2

i = e0 ∧ e1 ∧ e2 ∧ e3 iI = e1 ∧ e0 iJ = e2 ∧ e0 iK = e3 ∧ e0 j = e0 jI = e0 ∧ e2 ∧ e3 jJ = e0 ∧ e3 ∧ e1 jK = e0 ∧ e1 ∧ e2

k = e1 ∧ e3 ∧ e2 kI = e1 kJ = 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)

x · B = 1

2(xB − Bx)

= (−b · x) j + k −x0b + x× a ∈ V,

x · T = −12

(xT + T x)

=

x0t + t0x

+ i (x× t)∈ B,

x · P = 1

2(xP − P x)

=−ksx0 + j (sx)

∈ T

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 A p = v1∧v2∧· · ·∧v p and Bq = w1∧w2∧· · ·∧w p

are defined [12] for p ≤ q , by

A p · Bq ≡ (v1 ∧ v2 ∧ · · · ∧ v p−1) · (v p · Bq),

A p ∧ Bq ≡ v1 ∧ (v2 ∧ · · · ∧ v p) ∧Bq




withA p · Bq = (−1) p(q+1)Bq · A p

which defines Bq · A p for q ≥ p. For products of multivectors one obtains, with

S, P designating respectively the scalar and pseudoscalar parts of the multivectorwith B = a + ib, B = a + ib, T = ks0 + js, T = ks0 + js, P = iw, P = iw,

B · B = S

1

2(BB + BB)

= [−a · a + b · b] ∈ S,

B ∧ B = P

1

2(BB + BB)

= i [−b · a − a · b] ∈ P,

T · T = −1

2(T T + T T )

= s0s0 − s1s1 − s2s2 − s3s3 ∈ S,

B · P = 1

2(BP + P B)

= (−sb + isa) ∈ B,

T · P = 1

2(T P − P T )

= jws0 − kws

∈ V,

P · P = 1

2(P P + P P )

= −ww ∈ S.

4.2.3 General formulas

Among general formulas, one has

x · (y ∧ z) = (x · y)z − (x · z)y, (4.3)

x · (y ∧ z) + y · (z ∧ x) + z · (x ∧ y) = 0, (4.4)

(x ∧ y) · B = x · (y · B) = −y · (x · B), (4.5)

(x ∧ y) · (z ∧ w) = (x · w)(y · z)− (x · z)(y · w), (4.6)

(B · T ) · V = (B ∧ V ) · T, (4.7)




the Jacobi identity

0 = [F, [G, H ]] + [G, [H, F ]] + [H, [F, G]] , (4.8)

(x · B1) · B2 = B1 · (B2 · x) + x · [B1, B2] , (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 consequencesof equations (4.3), (4.5), (4.8). To establish equation (4.3), one just needs to writefor 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) simplyresults 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 whichcommutes with B, hence the equation. The Jacobi identity results from

[A,BC ] = 1

2 (ABC − BC A)

= 1

2 [(AB − BA) C + (BAC − BC A)]

= [A, B] C + B [A, C ] ,

[A,CB] = [A, C ] B + C [A, B] ,

[A, [B, C ]] = 1

2[A, B] C + B [A, C ] − [A, C ] B − C [A, B]

= [[A, B] , C ] + [B, [A, C ]]




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, oneobtains 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),

(x ∧ y) · (x ∧ y) = (x · y)2

− (x · x)(y · y),= (x · y)

2 − (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 mul-tivector 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 thevector u = j u0 + ku and going through the point a = j a0 + 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 correspondingplane; the equation of a plane parallel to B and going through the point a isexpressed 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 perpen-dicular 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

hencen · 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 areperpendicular to the vectors u, v (x ·u = y ·u = x · v = y · v = 0); from the generalformula

(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). LetT = u ∧ v ∧ w be a trivector (hyperplane), the equation of a hyperplane parallelto 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 = j a0 + ka)

t0x0

+ t1x1

+ t2x2

+ t3x3

− (a0t0

+a·t

) = 0




which is indeed the equation of a hyperplane. A vector n is perpendicular to thehyperplane 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 deducesn · T = 0. (4.11)

If n is the dual of T , n = T ∗ = iT , one finds (with T = kt0 + jt, T ∗ = − jt0 + kt)

T ∗ · T = −1

2 (T ∗T + T T ∗) = 0;

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 (4.12)

≡ x ∧ (y · T ) = −y ∧ (x · T ) = 0. (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

ua = −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

uT = −u · T, T u = −u · T,

u⊥T = −u ∧ T, T u⊥ = u ∧ T,

and finally

u = −(u · T )T −1 = −T −1(u · T ), (4.14)

u⊥ = −(u ∧ T )T −1 = T −1(u ∧ T ). (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

B1B2 = B1 · B2 + B1 ∧ B2 + [B1, B2] ,

one obtains

B1 = (B1 · B2)B−12 ,

B1⊥ = (B1 ∧ B2) + [B1, B2]B−12 .

Example. Take B1 = x ∧ y (with x = j x0 + kx, y = j y0 + ky) and let us projecton the plane x1x2, i.e., the plane B2 = K (B−1

2 = −K ); one finds

B1 = (B1 · B2)B−12

= (−x2y1 + x1y2)K = x ∧ y

(with x = x1kI + x2kJ , y = y1kI + y2kJ ),

B1⊥ = (B1 ∧B2) + [B1, B2]B−1

= (−x3y2 + x2y3)I + (x3y1 − x1y3)J

+(x1y0 − x0y1)iI + (x2y0 − x0y2)iJ + (x3y0 − x0y3)iK

= B1 − B1.




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 + iEc

be the electromagnetic bivector, its orthogonal pro-

jection on the hyperplane T = k = e1e2e3 (T −1

= −k) yields

F = −(F · k)k = −B,

F ⊥ = − [F, k] k = E

c .

Orthogonal projection of a hyperplane T 1 on the hyperplane T 2

Let us write T 1 = T 1 + T 1⊥ with T 1 = x ∧ y ∧ z, T 2 = u ∧ v∧w; from the relation

T 1 · T 2 ≡ (x ∧ y) · [z · (u ∧ v ∧ w)]

and the equation (4.11), one deduces

T 1⊥ · T 2 = 0;

furthermore

T 1, T 2

= 0 (T 1 ∈ T 2). Since

T 1T 2 = −T 1 · T 2 + [T 1, T 2]

it follows that

T 1 = −(T 1 · T 2)T −12 ,

T 1⊥ = [T 1, T 2] T −12 .



4.4. Differential operators 69

4.4 Differential operators

4.4.1 Definitions

Consider the relativistic four-nabla operator, ∇

∇ =

j

∂

∂x0 − k∇

= ∂ αeα = ∂ αeα

with

∇ = I ∂

∂x1 + J

∂

∂x2 + K

∂

∂x3,

∂ α = ∂ ∂xα

and eα the reciprocal basis defined by

eα · eβ = δ αβ

(e0 = e0, e1 = −e1, e2 = −e2, e3 = −e3). One can define the operators four-gradient

∇ϕ = j ∂

∂x0 − k gradϕ (ϕ ∈ S )

the four-divergence of a vector A = j A0 + kA

∇ · A = ∂A0

∂x0

+ ∂A1

∂x1

+ ∂A2

∂x2

+ ∂A3

∂x3

∈ S

the four-curl

∇∧ A = − rotA− i

∂ A

∂x0 + gradA0

∈ B;

acting on a bivector B = a + ib, one can define the operators

∇ · B = j divb− k

∂ b

∂x0 + rota

∈ V

∇∧ B = k diva + j ∂ a

∂x0

− rotb ∈ T.

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

∂α dα




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 the bivector

dS =

∂OM

∂α ∧

∂OM

∂β

dαdβ

=

∂x2

∂α∂x3

∂β − ∂x3

∂α∂x2

∂β

I +

∂x3

∂α∂x1

∂β − ∂x1

∂α∂x3

∂β

J

+∂x1

∂α∂x2

∂β − ∂x2

∂α∂x1

∂β

K +

∂x1

∂α∂x0

∂β − ∂x0

∂α∂x1

∂β

iI

+∂x2

∂α∂x0

∂β − ∂x0

∂α∂x2

∂β

iJ +

∂x3

∂α∂x0

∂β − ∂x0

∂α∂x3

∂β

iK

dαdβ.

In abridged notation, one can write

dS = dx2dx3I + dx3dx1J + dx1dx2K

+ dx1dx0iI + dx2dx0iJ + dx3dx0iK

with

dx2dx3 = D(x2, x3)

D(α, β ) dαdβ

=

∂x2

∂α

∂x2

∂β

∂x3

∂α∂x

3

∂β

dαdβ, etc.

where dx2dx3 is an undissociable symbol [3, p. 446]. For a hypersurface defined by

OM (α, β, γ ) = j x0(α, β, γ ) + kx(α, β, γ )

the tangent hyperplane to the surface at the point M is given by the trivector

dT =

∂OM

∂α ∧

∂OM

∂β ∧

∂OM

∂γ

dαdβdγ

=

∂x3

∂α∂x2

∂β∂x1

∂γ − ∂x2

∂α∂x3

∂β∂x1

∂γ − ∂x3

∂α∂x1

∂β∂x2

∂γ

+∂x1

∂α∂x3

∂β∂x2

∂γ + ∂x2

∂α∂x1

∂β∂x3

∂γ − ∂x1

∂α∂x2

∂β∂x3

∂γ

k

+

− ∂x3

∂α∂x2

∂β∂x0

∂γ + ∂x2

∂α∂x3

∂β∂x0

∂γ + ∂x3

∂α∂x0

∂β∂x2

∂γ

−∂x0

∂α∂x3

∂β∂x2

∂γ − ∂x2

∂α∂x0

∂β∂x3

∂γ + ∂x0

∂α∂x2

∂β∂x3

∂γ

jI

+

∂x3

∂α∂x1

∂β∂x0

∂γ − ∂x1

∂α∂x3

∂β∂x0

∂γ − ∂x3

∂α∂x0

∂β∂x1

∂γ

+∂x0

∂α∂x3

∂β∂x1

∂γ + ∂x1

∂α∂x0

∂β∂x3

∂γ − ∂x0

∂α∂x1

∂β∂x3

∂γ

jJ

+

− ∂x2

∂α∂x1

∂β∂x0

∂γ + ∂x1

∂α∂x2

∂β∂x0

∂γ + ∂x2

∂α∂x0

∂β∂x1

∂γ

−∂x0

∂α∂x2

∂β∂x1

∂γ − ∂x1

∂α∂x0

∂β∂x2

∂γ + ∂x0

∂α∂x1

∂β∂x2

∂γ

jK

dαdβdγ.



4.4. Differential operators 71

In short, one can write

dT = kdx1dx3dx2 (4.16)

+ jIdx2

dx3

dx0

+ jJdx3

dx1

dx0

+ jKdx1

dx2

dx0

(4.17)with the symbol

dx1dx3dx2 = D(x1, x3, x2)

D(α, β, γ ) dαdβdγ

=

∂x1

∂α

∂x1

∂β

∂x1

∂γ

∂x3

∂α

∂x3

∂β

∂x3

∂γ

∂x2

∂α∂x

2

∂β ∂x

2

∂γ

dαdβdγ, etc.

4.4.3 General theorems

Generalized Stokes theorem

This theorem can be written A · dl = −

(∇∧A) · dS (4.18)

with dl = j dx0

+ kdx, A = jA0

+ kA, dS = dS1 + idS2 and

dS1 = dx2dx3I + dx3dx1J + dx1dx2K,

dS2 = dx1dx0I + dx2dx0J + dx3dx0K.

Explicitly, equation (4.18) is expressed in classical vector notation as A0dx0 −A · dx =

− rotA · dS1 +

∇A0 +

∂ A

∂x0

· dS2

.

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 surfaceS [10, p. 39]. One takes two linearly independent four-vectors a, b ∈ S and onechooses the order (a, b) as being the positive orientation (dS = αa ∧ b, α > 0). Onthe curve C one chooses a vector f exterior to S and a vector g tangent to thecurve 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)




with dτ = idx0dx1dx2dx3, dT = kdx1dx3dx2 + jt and

t = dx2dx3dx0I + dx3dx1dx0J + dx1dx2dx0K,

the hypersurface being closed. Explicitly, equation (4.19) reads −A0dx1dx3dx2 + A · t

i =

∂A0

∂x0 + divA

idx0dx1dx2dx3.

The orientation of the hypersurface results from that of the four-volume. Leta,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 thehypersurface, 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, (4.20)

F ∧ dS = −

(∇ · F ) ∧ dT, (4.21)

with the bivector F = f + ig, dS = dS1 + idS2, dT = kdx1dx3dx2 + jt; explicitly,formula (4.20) reads

(−f · dS1 + g · dS2) =

div f dx1dx3dx2 −

∂ f

∂x0 − rotg

· t

;

if dx0 = 0, dS2 = 0, t = 0, one obtains the standard Gauss theorem. Relation(4.21) gives

i (−g · dS1 − f · dS2) =

i

div gdx1dx3dx2 −

∂ g

∂x0 + rotf

· t

.

The orientation of dS proceeds from that of dT . One chooses three linearly in-dependent 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 trivol-ume 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,

(AB) (AB)c, (BA) (BA)c, (AB)−1

, (BA)−1

, [A, B].

E4-2 Consider the four-vectors

x = j + kI + kJ, y = 2 j + kK,

z = 3 j + kJ, 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 four-vectors

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 S 1 = x ∧ y, S 2 = z ∧w,the trivector x ∧ y ∧ z and the four-volume x ∧ y ∧ z ∧ w. Give the orthogonalprojection of the vector w on S 1 and the orthogonal projection of S 1 on S 2.



Chapter 5

Symmetry groups

This chapter formulates the Lorentz group and the group of conformal transfor-mations 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 symplecticunitary 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 = jy0 + ky with x = x1I + x2J + x3K andthe interior product, with xc = −x;

x · x = xxc = −x2,

(x + y) · (x + y) = (x + y)(x + y)c

= xxc + yyc + 2x · y,

x · y = (xyc + yxc)

2 = − (xy + yx)

2= x0y0

−x1y1

−x2y2

−x3y3.

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 thedual of a,

T = a∗ = ia

= ka0

− ja



76 Chapter 5. Symmetry groups

with T −1 = −iaaac

(i commutes with all elements of C + and anticommutes with

those of C −). Let us suppose that T goes through the origin and let x = jx0 + kxbe 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⊥ = −T −1 (xT − T x)

2 =

x

2 +

iaxia

2aac=

x

2 − axa

2aac,

x = x

2

+ axa

2aacwith x = x + x⊥.

Definition 5.1.1. The symmetric of x with respect to a hyperplane is obtained bydrawing the perpendicular to the hyperplane and by extending this perpendicularby an equal length [11].

Let x be the symmetric of x with respect to the hyperplane T ; one has

x− x = 2x⊥

hencex =

axaaac

.

More simply, one can write that x − x is perpendicular to the hyperplane T (and

thus parallel to a) and x+x

2 is parallel to the hyperplane. One obtains

x = x + λa,

a ·

x + x

2

= 0,

hence

a ·

x + λa

2

= 0 =⇒ λ = −2(a · x)

a · a ,

x = x − 2(a · x)a

a · a

= x + (ax + xa)a

aac=

axa

aac= −axca

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

=−

axca,

from the space symmetries (with aac = −1)

x = −axa

= axca.

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 oper-ators which leave invariant the quadratic form

x · y = x0

y0

− x1

y1

− x2

y2

− x3

y3

with x = jx0 + kx.

Theorem 5.1.3. Every rotation of O(1, 3) is the product of an even number ≤ 4of 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 inversionis 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 = rcsctcuc = (−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 −.




Group O(1, 3): recapitulative table

The entire set of Lorentz transformations L is given in the table below where n isthe number of time symmetries and p the number of space symmetries.

Rotation(det L = 1)

Inversion(det L = −1)

orthochronousn even, p even

x = axac(aac = 1, a ∈ C +)

n even, p oddx = axac = −axcac(aac = −1, a ∈ C −)

antichronousn odd, p oddx = −axac

(aac = −1, a ∈ C +)

n odd, p evenx = −axcac = axcac

(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 = u1I + u2J + u3K , (u1)2 + (u2)2 + (u3)2 = 1,

x = jx0 + kx, x = jx0 + kx, rrc = 1. One verifies that x belongs indeed to thevector space of vectors

xc = rxcrc = −x.In matrix form, equation (5.1) can be written with

X =

x0

x1

x2

x3

, X =

x0

x1

x2

x3

, xi, xi ∈ R,

X = AX,

A =

1 0 0 00 a f g0 m b h0 n p c

,

with

a =

u12

+

u22

+

u32

cos θ,

b =

u22

+

u12

+

u32

cos θ,

c = u32

+ u12

+ u22

cos θ,



5.2. Proper orthochronous Lorentz group 79

f = u1u2(1− cos θ)− u3 sin θ, m = u1u2(1− cos θ) + u3 sin θ,

g = u1u3(1− cos θ) + u2 sin θ, n = u1u3(1− cos θ)− u2 sin θ,

h = u2u3(1

−cos θ)

−u1 sin θ, p = u2u3(1

−cos θ) + u1 sin θ.

One obtains the same expression as with quaternions, despite the distinct natureof 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 + udθ

2 ,

x = rxrc

=

1 + u

dθ

2

x

1− u

dθ

2

= x + dθ

2

(ux

−xu)

= x + dθ

2 k (ux− xu)

= x + dθku× x,

dx = x − x = dθku× x.

In matrix notationdX = dθuiM iX, i ∈ (1, 2, 3)

with

M 1 =

0 0 0 0

0 0 0 00 0 0 −10 0 1 0

, M 2 =

0 0 0 0

0 0 0 10 0 0 00 −1 0 0

, M 3 =

0 0 0 0

0 0 −1 00 1 0 00 0 0 0

and the relations

[M i, M j ] = ijkM k.

5.2.2 Pure Lorentz transformation

A pure Lorentz transformation is given by

x = bxbc,

b = cosh ϕ

2 + iv sinh

ϕ

2

with v = v1I + v2J + v3K , (v1)2 + (v2)2 + (v3)2 = 1, x = jx0 + kx, x = jx0 + kx,bbc = 1. One verifies that one has indeed x

c = bxcbc = −x. In matrix formulation,one has with

X =

x0

x1

x2

x3

, X =

x0

x1

x2

x3

, xi, xi ∈ R,




X = BX,

B =

cosh ϕ v1

sinh ϕ v2

sinh ϕ v3

sinh ϕv1 sinh ϕ a f gv2 sinh ϕ f b hv3 sinh ϕ g h c

,

with

a = 1 +

v12

(cosh ϕ− 1) , f = v1v2 (cosh ϕ− 1) ,

b = 1 +

v22

(cosh ϕ− 1) , g = v1v3 (cosh ϕ− 1) ,

c = 1 +

v3

2

(cosh ϕ− 1) , h = v2v3 (cosh ϕ− 1) .

One will notice that the matrix is real and symmetric. For a pure infinitesimalLorentz transformation, one obtains with b 1 + ivdϕ

2 and i anticommuting with

x,

x = bxbc

=

1 + iv

dϕ

2

x

1− iv

dϕ

2

= x + idϕ

2 (vx + xv)

= x + i dϕ2v jx0 + kx

+ jx0 + kx

v

= x + dϕ

kvx0 − j

v1x1 + v2x2 + v3x3

,

dx = x − x = dϕ

kvx0 − j

v1x1 + v2x2 + v3x3

.

In matrix form, one has

dX = dϕviK iX, i ∈ (1, 2, 3)

with

K 1 =

0 1 0 01 0 0 00 0 0 00 0 0 0

, K 2 =

0 0 1 00 0 0 01 0 0 00 0 0 0

, K 3 =

0 0 0 10 0 0 00 0 0 01 0 0 0

and the relations

[K i, K j] = −ijkM k,

[K i, M j] = ijkK k.



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)2 transforms as

B = − (xy − yx)

2= a (x ∧ y) ac

= aBac.

Generally, for a product of two vectors xy one has the transformation

xy = axacayac

= axyac;

consequently, any multivector A (and any Clifford number) which is a linear com-bination of such products transforms under a proper orthochronous Lorentz trans-formation 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 rotationand a pure Lorentz transformation is obtained as follows ([15],[16]). Consider theelement

a = a0 + a + i (b0 + b) ∈ C +, aac = 1;

let us introduce a conjugation (transforming i into −i),

a = a0 + a

−i (b0 + b)

= br

= bcr,

ac = rcb.

Let us write d = aac = brrcb = b2 (ddc = 1), hence

2b2 = 2d,

b2d = d2,

b2dc = 1 (dc = aac);




adding, one obtains

b2 (2 + d + dc) = (1 + d)2 ,

b = ± (1 + d)√ 2 + d + dc

(with d = aac),

r = bca

= ± (1 + aac) a√ 2 + d + dc

= ± (a + a)√ 2 + d + dc

,

with bbc = rrc = 1. For the decomposition a = rb, one obtains similarly

b = ± (1 + d) 2 + d + dc

(with d = aca, dc = aca),

r = abc

= ± (a + a) 2 + d + dc

.

Example. Let a =√

10K − 3iJ be an even Clifford number (a ∈ C +) and suchthat aac = 1. Let us decompose a into a product (a = br = rb). One has

a = √ 10 K + 3iJ, d = aac = 19 + 6√ 10 iI

Answer: b = ±√

10 + 3iI

, r = ±K ;

similarly

d = aca = 19− 6√

10 iI

Answer: b = ±√

10− 3iI

, r = ±K.

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 suchthat if dx · dx = 0, then dx · dx = 0 with dx = jx0 + kx. The group containsthe spacetime translations (x = x + d), Lorentz transformations (x = axac),dilatations (x = λx) and the transformations

(x

)

−1

= x

−1

+ ac (5.2)



5.3. Group of conformal transformations 83

where a is a constant (four-)vector with x−1 = xcxxc

. The above equation can alsobe written

x

=

(x

)

−1−1

=

x−1

+ ac−1

=

x−1 (1 + xac)−1

=

(1 + acx) x−1−1

and using relation (AB)−1

= B−1A−1 which is true for any invertible Cliffordnumber

x = (1 + xac)−1 x (5.3)

= x (1 + acx)−1 . (5.4)

Equivalently, one has

x =

xcxxc

+ ac

−1

=

x

xxc+ a

xcxxc

+ ac

x

xxc+ a

,

x = x + a (x · x)

1 + 2(x · a) + (a · a)(x · x).

The inverse transformation is given by

x−1

= (x

)

−1

− ac

hence

x = (1− xac)−1

x (5.5)

= x (1 − acx)−1

. (5.6)

5.3.2 Properties of conformal transformations

For any invertible Clifford number A (AA−1 = A−1A = 1), one has

dA−1A + A−1dA = 0,

dA−1 = −A−1dAA−1.

Differentiating equation (5.2), one obtains

d

(x)−1

= d

x−1

, (5.7)

(x)−1

dx (x)−1

= x−1dxx−1, (5.8)

xcdxx

c

(xxc)2 =

xcdxxc

(xxc)2 ; (5.9)




multiplying by the conjugate equation, one has

xcdxx

cxdxcx

(x

xc)4 =

xcdxxcxdxcx

(xxc)4 , (5.10)

dxdxc =

(x.x)2

(x.x)2

dxdxc (5.11)

which shows that the transformation is indeed a conformal transformation. Equa-tion (5.8) gives

(x)−1

dx (x)−1

= x−1dxx−1,

dx = xx−1dxx−1x,

and using equations (5.3), (5.4), one obtains the relation

dx = (1 + xac)−1 dx (1 + acx)−1 . (5.12)

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

x = (1 + xac)−1 x = x (1 + acx)

−1 ,

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

hence

xacx = xacx, (5.13)

xxc = (1 + xac) xxc (1 + axc) , (5.14)

xxc =

xxc1 + 2(x · a) + (a · a)(x · x)

. (5.15)

5.3.3 Transformation of multivectors

A conformal transformation is a relation of the type xi = f i

x0, x1, x2, x3

with

dxi = ∂xi

∂xk dxk.

A contravariant (four-)vector A, by definition, transforms according to the formula

Ai = ∂xi

∂xk Ak.

The equation (5.12)

dx

= (1 + xac)

−1

dx (1 + acx)

−1



5.4. Dirac algebra 85

then gives the transformation of a vector A = jx0 + kA,

A = (1 + xac)−1

A (1 + acx)−1

. (5.16)

For a product of two vectors A, B one obtains

AB = (1 + xac)−1 A (1 + acx)−

1(1 + xac)−

1 B (1 + acx)−1

= K (1 + xac)−1 AB (1 + acx)

−1

with

K = 1

1 + 2(x · a) + (a · a)(x · x);

for a bivector A ∧B = −(AB−BA)

2 one then has

A ∧B = K (1 + xac)−1

(A ∧B) (1 + acx)−1

;

one verifies that A ∧B is indeed a bivector. Similarly, one obtains for a trivectorand a pseudoscalar

A ∧B ∧ C = K 2 (1 + xac)−1

(A ∧B ∧ C ) (1 + acx)−1 ,

A ∧B ∧ C ∧D = K 4 (A ∧B ∧ C ∧D) .

Furthermore,

AAc = K 2AAc,

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 canbe represented by 2 × 2 matrices over complex quaternions with the followinggenerators:

e0(≡ j) =

i 0

0 −i

, e1(≡ kI ) =

0 iiii 0

,

e2(

≡kJ ) = 0 i j

i

j 0 , e3(

≡kK ) = 0 ik

i

k 0 ,




where i is the ordinary complex imaginary (i2 = −1) and e20 = −1, e21 = e22 =e23 = 1. The other matrices are given by

1 = 1 0

0 1

, I = i 0

0 i

, J = j 0

0 j

,

K =

k 00 k

, i =

0 −11 0

, iI =

0 −i

i 0

,

iJ =

0 − j j 0

, iK =

0 −kk 0

, jI =

ii 0

0 −ii

,

jJ = i j 0

0 −

i j , jK = ik 0

0 −

ik , k = 0 i

i 0 .

The Dirac spinor can be expressed as a left ideal

Ψ = AE =

q 1f q 2f

, E =

f 00 0

where E is a primitive idempotent (E 2 = E ), f = (1 + ik)/2 ; q 1, q 2 beingreal quaternions and A a general element of the algebra. The hermitian norm isexpressed by

ΨΨ† = (q 1q 1c + q 2q 2c) f

where Ψ†

is the transposed, quaternionic conjugated and complex conjugated ma-trix of Ψ. The Dirac equation in presence of an electromagnetic field is

(ih∇− eA− imc) Ψ = 0

with the four-nabla operator ∇ = j ∂ c∂t − k∇, the vector potential A = j V

c + kA

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 q i and its adjoint A†,

A =

q 1 q 2q 3 q 4

, A† =

q ∗1c q ∗3cq ∗2c q ∗4c

,

where the sympbols ∗ and c represent respectively the complex conjugation andthe quaternionic conjugation. The adjunction transforms i and i, j, k, I,J,K intotheir opposites, as one can verify directly on the basis matrices. A selfadjointClifford number (H = H †) is consequently of the type

H = (a + i

ib + i

jc + i

kd; i

p+ i

q+ j

r+ k

s)



5.4. Dirac algebra 87

with p = p1I + p2J + p3K (etc.) and real coefficients. The unitary groupSU

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 subgroupthe symplectic unitary group USp(2,H) [39, p. 232]. The elements of the unitarygroup being of the type ei

H (with H = H †), one can choose for the 15 generatorsof SU(4),

eiθ, eIθ , eJθ , eKθ , ei

jIθ, eijJθ , ei

jKθ, eikIθ, ei

kJθ, eikKθ

ejθ

, ekθ

, eiiIθ

, eiiJθ

, eiiKθ

. , (5.17)

where θ is a real generic parameter with the usual series development

eIθ = cos θ + I sin θ,

eiiJθ = cos θ + iiJ sin θ (etc.).

The 10 matrices within parentheses constitute the symplectic unitary group

USp(2;H

). Explicitly, one has

eiθ =

cos θ − sin θ

sin θ cos θ

, eIθ =

cos θ + i sin θ 0

0 cos θ + i sin θ

,

eJθ =

cos θ + j sin θ 0

0 cos θ + j sin θ

,

eKθ =

cos θ + k sin θ 0

0 cos θ + k sin θ

,

eijIθ = cos θ

−i sin θ 0

0 cos θ + i sin θ ,

eijJθ =

cos θ − j sin θ 0

0 cos θ + j sin θ

,

eijKθ =

cos θ − k sin θ 0

0 cos θ + k sin θ

,

eikIθ =

cos θ −i sin θ−i sin θ cos θ

, ei

kJθ =

cos θ − j sin θ− j sin θ cos θ

,

eikKθ =

cos θ −k sin θ

−k sin θ cos θ .




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

ejθ =

cos θ + i

sin θ 00 cos θ − i sin θ

, ekθ =

cos θ i

sin θi sin θ cos θ

,

eiiIθ =

cos θ −i sin θi sin θ cos θ

, ei

iJθ =

cos θ −i j sin θi j sin θ cos θ

,

eiikθ =

cos θ −i sin θ−i sin θ cos θ

.

To obtain a representation in terms of 4 × 4 complex matrices, one can choose forthe generators of the Clifford algebra over C

e0(≡ j) =

i 0 0 00 i 0 00 0 −i 00 0 0 −i

, e1(≡ kI ) =

0 0 0 i0 0 i 00 −i 0 0−i 0 0 0

,

e2(≡ kJ ) =

0 0 0 10 0 −1 00 −1 0 01 0 0 0

, e3(≡ kK ) =

0 0 i 00 0 0 −i

−i 0 0 00 i 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 (ve-locity v = c

3) followed by a rotation (r) of π4 around the same axis. Express theresulting Lorentz transformation X = aX ac.

E5-2 Consider a special Lorentz transformation (b1) along the Ox axis (velocityv = c

3 ) followed by a pure Lorentz transformation (b2) along the Oy axis (withv = c

3). Express the resulting Lorentz transformation X = aX a

c with a = b

2b1

.Decompose the resulting Lorentz transformation into a rotation followed by a pureLorentz 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 verticesA(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 conformaltransformation

x = (1 + xac)−1x

with a = 2kI + kK , x = ctj + kx

, x

= ct

j + kx

.



5.5. Exercises 89

E5-4 Consider the matrices

A = 1 jk i , B =

1 ki j , C =

1 ik j .

Determine AB, BA, A−1, B−1, (AB) −1, (BA)−1

; does the matrix C −1 exist?



Chapter 6

Special relativity

This chapter develops the special theory of relativity within the Clifford algebraH⊗H over R. The relativistic kinematics and relativistic dynamics of a point massare 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 frameK (O, x, y, z) moving along the Ox axis with a constant velocity v (Figure6.1). With X = jx0 + kx, X = jx0 + kx (x0 = ct, x = xI + yJ + zK ),b = cosh ϕ

2 − iI sinh ϕ

2, the Lorentz transformation is expressed by

X = bX bc.

v

y

x

z

y

x

z

O O

K K

Figure 6.1: Special Lorentz transformation (pure).



92 Chapter 6. Special relativity

Explicitly, one obtains

x0 = x0 cosh ϕ− x sinh ϕ,

x

= −x0 sinh ϕ + x

cosh ϕ,y = y, z = z.

Writing tanh ϕ = vc = β , γ = cosh ϕ with

γ 2 = 1 + sinh2 ϕ = 1 + γ 2β 2,

γ = 1

1− β 2,

the transformation becomes

ct = γ (ct− xβ ) ,

x = γ (x− βct) ,

y = y, z = z.

Taking β = vc 1 (γ 1), one obtains the Galilean transformation

t = t,

x = x − vt,

y = y, z = z .

The inverse transformation is expressed by

X = bcX b

with bc = cosh ϕ2 + iI sinh ϕ

2 , hence

ct = γ (ct + xβ ) ,

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 Ox 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 thesame time t:

x

2 = γ (x2 − βct) ,

x

1 = γ (x1 − βct) ,



6.1. Lorentz transformation 93

hence

l = x2 − x1 = x

2 − x

1

γ

= l 0

1− β 2 ≤ l0;

the observer in K concludes to a contraction. Reciprocally, let l0 = x2−x1 be thelength of a rod parallel to the Ox axis in K . The observer in K measures at thesame time t,

x2 = γ (x

2 + βct) ,

x1 = γ (x

1 + βct) ,

hence

l

= x

2 − x

1 =

x2 − x1

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,

1− β 2 = 1− 5 · 10−7, therelative variation of l is 5 · 10−7.

Time dilatation

A time interval t = t2 − t1 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 (ct2 + x

2β ) ,

ct1 = γ (ct1 + x

1β ) ,

hence

t = t2 − t1 = γ t

= t

1− β 2≥ t.

The observer in K concludes to a slowing down of physical phenomena. Recipro-cally, a time interval t = t2− t1 measured at the same point x2 = x1 in K , givesin K ,

ct2 = γ (ct2 − x2β ) ,

ct1 = γ (ct1 − x1β ) ,

t = t2 − t1

= t

1− β 2≥ t.

Hence, the conclusion is the same.




6.1.3 General Lorentz transformation

For a general Lorentz transformation, one has a = br (or a = rb) with r =cos θ

2 + u sin θ

2, b = cosh ϕ

2 + iv sinh ϕ

2 (u · u = v · v = 1). Explicitly, one obtains

a = br =

cos θ2 cosh ϕ

2 + u cosh ϕ2 sin θ

2

−iu · v sin θ2 sinh ϕ

2

+iv cos θ2 − u× v sin θ

2 sinh ϕ2

,

a = rb =

cos θ2 cosh ϕ

2 + u cosh ϕ2 sin θ

2

−iu · v sin θ2 sinh ϕ

2

+iv cos θ2 + u× v sin θ

2 sinh ϕ2

with aac = aa

c = 1. The general Lorentz transformation is simply expressed by

X = aX ac (or aXa

c)

with X = j x0 + kx, X = jx0 + kx.

6.2 Relativistic kinematics

6.2.1 Four-vectors

Transformation of a four-vector

An arbitrary four-vector A = j a0 + ka transforms under a special Lorentz trans-formation as

A = bAbc

with b = cosh ϕ2 − iI sinh ϕ

2 ; explicitly, writing

γ = cosh ϕ = 1

1− v2

c2

, tanh ϕ = v

c,

one has

a0 = γ

a0 − a1 v

c

,

a1 = γ

a1 − a0 v

c

,

a2 = a2, a3 = a3;

reciprocally, one has

A =

bc

Ab



6.2. Relativistic kinematics 95

which yields

a0 = γ

a0 + a1 v

c,

a1 = γ

a1 + a0 vc

,

a2 = a2, a3 = a3.

If one considers a general Lorentz transformation, one obtains the following for-mulas ([40, p. 134], [53, p.123]) with b = cosh ϕ

2 − iv

v sinh ϕ

2 where v is the velocity(of norm v) of the reference frame K with respect to the reference frame K

A = bAbc

or explicitly

a0 = γ

a0 −

a · v

v

v

c

,

a = a + v

v

a · v

v

(γ − 1)− a0γ

v

c

;

the reciprocal formulas areA = bcAb,

a0 = γ

a0 +a ·

v

v

v

c

,

a = a + vv

a · v

v

(γ − 1) + a0γ v

c

.

Four-velocity

Let X = j ct + kx be the spacetime four-vector of a particle with dX = jcdt + kdx

and the relativistic invariant

dXdX c = c2dt2 − (dx)2

= c2dt2

1− v2

c2

=

c2dt2

γ 2 = c

2

dτ

2

which defines the proper time

dτ = dt

γ = dt

1−

v2

c2

with γ = 1q 1− v

2

c2

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

by

V = dX

dτ = j γc + kγ v




with v = dxdt and V V c = 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 = j γc + kγ v be the four-velocity and dτ = dtγ

the proper time differential;the four-acceleration is defined by

A = dV

dτ = γ [ jcγ + k (γ v + γ a)]

with

γ = dγ dt

= vvc2

γ 3 (6.1)

= (v · a) γ 3

c2 (6.2)

where the relation v · ·

v = vv deduced from (v)2 = v2 has been used. Furthermore,V · V = c2 from which one obtains by differentiating with respect to τ , V ·A = 0which gives again equation (6.2)

(v · a) = c2 γ

γ 3.

The four-acceleration can be written

A = jγ 4(v · a)

c + k

γ 4

(v · a)v

c2 + γ 2a

with γ = 1q 1− v

2

c2

; one has the relativistic invariant

AAc = −γ 6(v · a)

2

c2 − γ 4 (a)

2

= γ 6−a2 + v × a

c2

with (v × a)2 = v2a2 − (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 AAcV c = c2AAc

= γ 6 −a2c2 + (v × a)2

.



6.2. Relativistic kinematics 97

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

AAc = − (a p)2

= −a2 p.

When a is parallel to v, one has

AAc = −γ 6a2 = −a2 p,

a p = γ 3a.

If a is perpendicular to v, one has

AAc = −γ 6a2

1−

v2

c2

= −γ 4a2 = −a2 p,

a p = γ 2a;

when γ 1, one remarks that the proper acceleration can be very much largerthan the acceleration in the laboratory frame [42, p. 101].

Wave four-vector

The wave four-vector is defined with k = 2πλ n by

K = jω

c + kk;

writing X = j ct + kr one has the relativistic invariants

KK c = ω2

c2 − (k)

2,

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

2 − iI sinh

ϕ

2 , tanh ϕ =

w

c , cosh ϕ = γ =

1 1− w2

c2

,

the four-velocities

V =

c( j

coshθ

+ km

sinhθ

), V

= c

( j

coshθ

+ km

sinhθ

),




and

tanh θ = v

c, tanh θ =

v

c ,

v = vm, v = vm (m ·m = m ·m = 1).

Equations (6.3) read

cosh θ = cosh θ cosh ϕ−m1 sinh θ sinh ϕ, (6.4)

m1 sinh θ = m1 cosh ϕ sinh θ − cosh θ sinh ϕ, (6.5)

m2 sinh θ = m2 sinh θ, (6.6)

m3 sinh θ = m3 sinh θ. (6.7)

Dividing equation (6.5) by the equation (6.4), one obtains

m1 tanh θ = m1 tanh θ − tanh ϕ

1−m1 tanh θ tanh ϕ

and thus

v1 = v1 − w

1− v1w

c2

. (6.8)

Similarly, one obtains

v2 = v2

cosh ϕ

1−

v1w

c2

, (6.9)

v3 = v3

cosh ϕ

1−

v1w

c2

(6.10)

with cosh ϕ = γ = 1q 1−w

2

c2

. The inverse formulas are (with β = wc )

v1 = v1 + w

1 + v1w

c2

, (6.11)

v2 = v2

1− β 2

1 + v1w

c2

, (6.12)

v3 = v3

1− β 2

1 + v1w

c2

. (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)

with b =

cosh ϕ2 − iw

w sinh ϕ

2

, v = vm, v = vm (m · m = m · m = 1),

tanh ϕ = wc , tanh θ = v

c, tanh θ = v

c . Explicitly, equation (6.14) reads

cosh θ = cosh θ cosh ϕ

1−m ·

w

w

tanh θ tanh ϕ

, (6.15)

m sinh θ = m sinh θ − w

w

m ·

w

w

sinh θ (6.16)

+ w

wm ·

w

w sinh θ cosh ϕ−

w

w

cosh θ sinh ϕ. (6.17)

Dividing equation (6.17) by equation (6.15), one obtains with β = wc

the formulas[40, p. 75]

v =v

1− β 2 + w

v ·w

w2

1−

1− β 2

− 1

1− v ·w

c2

and the inverse formula

v =

v

1− β 2 + w v ·w

w2 1− 1− β 2 + 1

1 + v ·w

c2

.

6.3 Relativistic dynamics of a point mass

6.3.1 Four-momentum

Let V = jγ c + kγ v be the four-velocity of a particle and m0 its mass, the four-momentum is defined by

P = m0V = j m0γc + kγ m0v

= jE

c + kp

where E = γ m0c2 is the energy of the particle and p = γ m0v its momentum (of norm p). Furthermore, one has the relativistic invariant

P P c = E 2

c2 − (p)

2

= m20V V c = m2

0c2,




hence

E 2 = p2c2 + m20c4.

In the proper frame (v = 0), one has E 0 = m0c2

; the kinetic energy is defined by

T = m0c2 (γ − 1) .

Under a standard Lorentz transformation, the four-momentum transforms as P =bP bc with b =

cosh ϕ

2 − iI sinh ϕ2

and tanh ϕ = w

c = β , γ = 1q

1−w2

c2

E = γ

E − p1w

,

p1 = γ p1 − E

c2

w ,

p2 = p2, p3 = p3.

The inverse transformation is P = bcP b or

E = γ

E + p1w

,

p1 = γ

p1 +

E

c2 w

,

p2 = p2, p3 = p3.

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

ivv

sinh ϕ2 .

6.3.2 Four-force

Let P = j E c

+ kp be the four-momentum vector of the particle; the four-forcevector is defined by

F = dP

dτ

= d (m0V )

dτ = m0A +

dm0

dτ V

with V = j γc +kγ v [42, pp. 123-124]. If the four-force conserves m0, then dm0

dτ = 0

and

F = m0A = γ d ( jm0γc + kγm0v)

dt

= γ (v)

j

dE

cdt + kf



6.3. Relativistic dynamics of a point mass 101

with, by definition f = dpdt

= d(γm0v)dt

. Furthermore,

F · V = m0A + dm0

dτ

V · V = dm0

dτ

c2

= γ 2 (v)

dE

dt − f · v

.

If dm0

dτ = 0 = F · V , then dE

dt = f · v and

F = γ (v)

j

(f · v)

c + kf

.

Since

f = d (γm0v)

dt = γ m0a +

d (γm0)

dt v

= γm0a + dE

c2dtv = γ m0a +

(f · v)

c2 v,

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 = m0d2x

dt2 (6.18)

which can be established as

F = m0A = m0d2X

dτ 2 (6.19)

= γ (v)

j

dE

cdt + kf

(6.20)

= m0

d2 ( jct)

dτ 2 + k

d2x

dt2

, (6.21)

hence the relation (6.18) by comparing equations (6.20) and (6.21). Under a spe-cial Lorentz transformation, the four-force F = γ (v)

j Q

c + kf

with Q = dE

dt ,

transforms as F = bF bc with b =

cosh ϕ2 − iI sinh ϕ

2

, tanh ϕ = w

c et cosh ϕ =

γ = 1q 1−w

2

c2

or explicitly

F 0 = γ (w)

F 0 − F 1w

c

,

F 1 = γ (w)

F 1 − F 0w

c

,

F 2 = F 2, F 3 = F 3,




and equivalently

γ (v) Q = γ (w) γ (v) Q− f 1w , (6.22)

γ (v) f 1 = γ (w) γ (v)

f 1 −

Qw

c2

, (6.23)

γ (v) f 2 = γ (w) f 2, (6.24)

γ (v) f 3 = γ (w) f 3. (6.25)

Furthermore, one has the relation

γ (v)

γ (v) = γ (w) 1−

v1w

c2 , (6.26)

which can be established as follows ([42, p. 69]); form the invariant

ds2 = dt2

c2 − v2

= dt2

c2 − v2

and the relation

cdt

= γ (w)

cdt− dx

w

c

= γ (w) cdt

1−

v1w

c2

.

Then one obtains

dt2

c2 − v2

= dt2γ 2 (w)

1−

v1w

c2

2 c2 − v2

,

c2 − v2

=

c

2 −v

2 1− w2

c2

1− v1wc2

2 ,

1

γ 2 (v) =

1

γ 2 (v)

1

γ 2 (w)

11− v1w

c2

2 ,

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

γ (v)

γ (v

)

= γ (w) 1 + v1w

c2 .



6.4. Exercises 103

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

Q = Q− f 1w

1− v1wc2

,

f 1 = f 1 − wQ

c2

1− v1wc2

,

f 2 = f 2

γ (v)

1− v1wc2

,

f 3 = f 3

γ (v)

1− v1wc2

.

6.4 ExercisesE6-1 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 frameK at rest. Determine the total relativistic angular momentum

L =

X i ∧ P i,

X i = j ct + kI xi + kJ yi + kK zi,

P i = jE i

c + kI pxi + kJ pyi + kK pzi.

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 isconstant in the proper reference frame, at any instant. The particle being at restat 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 forsmall 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 relativisticoptics.

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 + kγ v the four-velocity; the four-current density is defined by

C = ρ0V = jγρ0c + kγρ0v

= jρc + k j

with ρ = γρ0 and j = ρv. One obtains the relativistic invariant

CC c = ρ2c2 − ρ2 (v)2

= ρ2c2

1− v2

c2

= c2 ρ2

γ 2 = c2

ρ2

0

with γ = 1q 1− v2

c2

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

density transforms as C = bC bc with b = cosh ϕ2 − iI sinh ϕ

2, tanh ϕ = w

c = β and

cosh ϕ = γ = 1√ 1−β2 ; explicitly, one has

ρc = γ

ρc− j1β

,

j1 = γ j1 − ρcβ

,

j2 = j2, j3 = j3.



106 Chapter 7. Classical electromagnetism

Consider a trivolume OM (α,β,γ ) = jx0 (α,β,γ )+kx (α,β,γ ) depending on threeparameters α, β,γ . The infinitesimal hyperplane dT is expressed by (4.17)

dT =

∂OM ∂α ∧ ∂OM ∂β ∧ ∂OM ∂γ

dαdβdγ

=

kdx1dx3dx2 + jI dx2dx3dx0

+ jJ dx3dx1dx0 + jK dx1dx2dx0

.

The dual dT ∗ of dT is given by

dT ∗ = idT

= jdx1dx2dx3 + kIdx2dx3dx0

+kJdx3

dx1

dx0

+ kKdx1

dx2

dx0

and is a four-vector orthogonal to dT . The relativistic invariant C · dT ∗ is givenby

C · dT ∗ =

ρcdx1dx2dx3 − ρv1dx2dx3dx0

−ρv2dx3dx1dx0 − ρv3dx1dx2dx0

;

in the proper frame C · dT ∗ = ρ0cdx1dx2dx3 = cdq where dq is the electric chargecontained in dT . The electric charge Q contained in the hyperplane T is given by

Q =

1

c T

C · dT ∗

for any Galilean frame. Furthermore, if one integrates over a closed four-volumeτ , one obtains

T

C · dT ∗ =

τ

(∇ · C ) dx0dx1dx2dx3

with the relation ∇ · C = ∂ρc∂x0

+ ∂ρv1

∂x1 + ∂ρv2

∂x2 + ∂ρv3

∂x3 = 0 expressing the con-

servation of electric charge. If the four-volume τ is limited by the hypersurfaces

H 1 (t1 = const.), H 2 (t2 = const.) and the lateral hypersurface H 3 at infinity, onehas (Figure 7.1)

H 1

C · dT ∗ +

H 2

C · dT ∗ +

H 3

C · dT ∗ = 0;

furthermore, the integral over H 3 is nil in the absence of electric charges at infinity,hence

Q2 =

H 2

C · dT ∗ = − H 1

C · dT ∗ = Q1

which expresses the conservation of electric charge.



7.1. Electromagnetic quantities 107

t

t2

t1

H 2

H 1

H 3 τ

Figure 7.1: Conservation of electric charge: H 1 and H 2 are hyperplanes (at t con-stant) and H 3 is a lateral hypersurface at infinity.

Four-potential vector

Let V be the scalar potential and A the potential vector, the four-potential vector

is defined by A = j V c

+ kA yielding the relativistic invariant AAc = V 2

c2 − (A)

2.

Under a special Lorentz transformation, one has A = bAbc with b = cosh ϕ2 −

iI sinh ϕ

2

, tanh ϕ = wc = β and cosh ϕ = γ = 1

√ 1−β2

, hence

V

c = γ

V

c −A1β

,

A1 = γ

A1 − V

c β

,

A2 = A2, A3 = A3.

7.1.2 Electromagnetic field bivectorFour-nabla operator

The four-nabla operator

∇ = j ∂

c∂t − kI

∂

∂x1 − kJ

∂

∂x2 − kK

∂

∂x3

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-




formation. The transformation formulas are

ct = γ (ct + xβ ) ,

x = γ (x + βct) ,

y = y , z = z ,

with β = vc

and γ = 1q 1− v2

c2

. One then obtains

∂

c∂t =

1

c

∂

∂t

∂t

∂t +

∂

∂x

∂x

∂t +

∂

∂y

∂y

∂t +

∂

∂z

∂z

∂t

= γ ∂

c∂t − − ∂

∂xβ

with ∂t∂t

= γ , ∂x∂t

= γβc, ∂y∂t

= ∂z∂t

= 0. Furthermore,

− ∂

∂x = −

∂

∂x

∂x

∂x + ∂

∂t

∂t

∂x

= γ

− ∂

∂x − β

∂

c∂t

with ∂x∂x = γ , ∂t∂x = γ βc . Finally, one has

∂

c∂t = γ

∂

c∂t −

− ∂

∂x

β

,

− ∂

∂x = γ

− ∂

∂x − β

∂

c∂t

,

− ∂

∂y = − ∂

∂y, − ∂

∂z = − ∂

∂z ,

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

Electromagnetic field bivector

Let A = j V c

+ kA be the four-potential vector, let us define the Clifford number

F = ∇cA = −∇A

= (

∇ ·A) + (

∇∧A)



7.1. Electromagnetic quantities 109

and adopt the Lorentz gauge (∇ · A) = ∂ (V

c )c∂t

+ ∂A1

∂x1 + ∂A2

∂x2 + ∂A3

∂x3 = 0, which gives

the electromagnetic field bivector

F = ∇∧A

= − rotA + i

− grad V

c − ∂ A

c∂t

= −B+ iE

c

with the usual definitions of the magnetic induction B and the electric field E,

B = rotA, E = −gradV − ∂ A

∂t . (7.1)

Under a gauge transformation

A → A = A +∇f

where f is an arbitrary scalar function, one obtains the same electromagnetic fieldbivector

F = ∇ ∧A = ∇∧A + (∇ ∧∇) f

= ∇ ∧A = F.

The electromagnetic field bivector yields the relativistic invariant

F 2 = F ·

F + F ∧

F

=

− (B)

2+

(E)2

c2 + 2i

E ·Bc

.

Under a pure special Lorentz transformation , one has

F = bF bc

with b = cosh ϕ2 − iI sinh ϕ

2 , tanh ϕ = vc

= β and cosh ϕ = γ = 1q 1− v2

c2

, hence

F =

−B1

I − B

2

cosh ϕ +

E 3

c sinh ϕ

J

+

−B3 cosh ϕ +

E 2

c sinh ϕ

K

+E 1

c iI +

E 2

c cosh ϕ−B3 sinh ϕ

iJ

+

E 3

c cosh ϕ + B2 sinh ϕ

iK

=

−B + i

E

c




or

B1 = B1, B2 = γ B2 + E 3

c β , B3 = γ B3 − E 2

c β ,

E 1 = E 1, E 2 = γ

E 2 −B3βc

, E 3 = γ

E 3 + B2βc

.

Under a general pure Lorentz transformation, one has F = bF bc with b =cosh ϕ

2 − ivv

sinh ϕ2 , tanh ϕ = v

c = β and cosh ϕ = γ = 1q

1− v2

c2

; hence the for-

mulas [40, p. 191]

B = γ

B +

v

v2 (v ·B)

1

γ − 1

− 1

c2 (v×E)

,

E = γ E + v

v2

(v

·E) 1

γ −1 + (v

×B) .

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,

= k (− divB) + j

−∂ B

c∂t − rot

E

c

;

hence, one obtains two of the Maxwell’s equations

divB = 0, rotE = −∂ B

c∂t.

The two other equations are given by

∇ · F = µ0C

= j div E

c + k

rotB− 1

c2∂ E

∂t

with the four-current density C = jρc + kρv; one obtains with ε0µ0c2 = 1,

divE = µ0ρc2 = ρ

ε0,

rotB = 1

c

2

∂ E

∂t

+ µ0 j.




The complete set of Maxwell’s equations reads

∇F = ∇∇cA = ∇ · F +∇ ∧ F

=

j div E

c + krotB− 1c2 ∂

E

∂t

+k (− divB) + j

−∂ B

c∂t − rot

E

c

= A = µ0C

with the d’Alembertian operator = ∇∇c = ∂ 2

c2∂t2 − and A = j V

c + kA the

four-potential vector. Hence, the equations

V

c = ρµ0c2 =

ρ

ε0,

A = µ0 j,

or equivalently

divB = 0, divE = ρ

ε0,

rotE = −∂ B

∂t ,

rotB = µ0

j +

∂ε0E

∂t

.

If one does not adopt the Lorentz gauge, one obtainsF = ∇cA = −∇A

=

divA +

∂V

c∂t

−B + i

E

c

which is not a bivector but an element of C +. One obtains the same Maxwell’sequations

∇F = ∇∇cA

= A = µ0C.

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 covari-ance of Maxwell’s equations is manifest since under a general Lorentz transforma-tion any Clifford number X transforms as X = aX ac (aac = 1) and thus

∇F = µ0C = a∇F ac

= aµ0Cac;




hence, ∇F = µ0C .

Perfect dielectric or magnetic medium

Consider the bivectors F = −B + iEc , G = −H + icD. Maxwell’s equations are

expressed by

∇∧ F = 0,

∇ · G = C = jρc + kρv

= j div(cD) + k

rotH− ∂ D

∂t

ordivB = 0, divD = ρ,

rotE = −∂ B

∂t , rotH = j +

∂ D

∂t .

The bivector G yields the relativistic invariant

G2 = G · G + G ∧G

= −H2 + c2D2 + 2icD ·H.

Under a pure special Lorentz transformation, G transforms as G = bGbc withb = cosh ϕ

2 −iI sinh ϕ

2, tanh ϕ = v

c = β and cosh ϕ = γ = 1

q 1− v

2

c2

, hence

H x = H x,

H y = γ (H y + vDz) ,

H z = γ (H z − vDy) ,

Dx = Dx,

Dy = γ

Dy − v

H zc2

,

Dz = γ

Dz + vH yc2

.

Under a general pure Lorentz transformation, one has G = bGbc with b = cosh ϕ2−

ivv

sinh ϕ2 , tanh ϕ = v

c = β and cosh ϕ = γ = 1q

1− v2

c2

; hence

H = γ

H +

v

v2 (v ·H)

1

γ − 1

− (v×D)

, (7.2)

D = γ D+ v

v2 (v ·D)

1

γ − 1 + v ×

H

c2 , (7.3)




and the inverse formulas

H = γ H +

v

v2 (v ·H)

1

γ − 1 + (v×D) ,

D = γ

D +

v

v2 (v ·D)

1

γ − 1

−v × H

c2

.

As an application, consider a perfect medium in the proper reference frame K

with B = µ0µrH, D = ε0εrE

, moving with respect to a Galilean reference frame(at rest) K with a velocity v. Using the transformation formulas (7.2), (7.3), oneobtains

B + v

v2 (v ·B)

1

γ − 1

− 1

c2 (v× E)

(7.4)

= µ0µrH+ v

v2 (v ·H)

1γ − 1

− (v ×D)

, (7.5)

D +

v

v2 (v ·D)

1

γ − 1

+

v× H

c2

(7.6)

= ε0εr

E +

v

v2 (v ·E)

1

γ − 1

+ (v ×B)

. (7.7)

Furthermore, from equations (7.5), (7.7) one deduces the relations

v ·B = µ0µrv ·H,v ·D = ε0εrv ·E.

Equations (7.5), (7.7) then become

B− 1

c2 (v× E) = µ0µr [H− (v ×D)] , (7.8)

D +

v × H

c2

= ε0εr [E + (v ×B)] ; (7.9)

taking H from equation (7.8) and replacing it in equation (7.9) one finds [7, p.

239]

D = ε0εrE +ε0

εr − 1

µr

1− β 2

v×B− v × E

c2

.

Similarly, taking D from equation (7.9) and replacing it in equation (7.8) oneobtains

H = B

µ0µr+

ε0

εr − 1

µr

1 − β 2

v × (E + v ×B) .

One observes that D and H depend on E and B as well as on the velocity of the

material.




Arbitrary dielectric or magnetic medium

In an arbitrary dielectric or magnetic medium, with a polarization density P anda magnetization density M, one has the relations

D = ε0E + P,

B = µ0 (H + M)

which one can write in the form

F = µ0 (G + N )

with the bivectors F = −B+ iEc , G = −H+ icD and N = −M− icP. Maxwell’s

equation

∇ ·G = C (with C = jρc + kρv) then becomes

∇ ·G = ∇ ·

F

µ0−N

= C

or∇ · F = µ0 (C + ∇ ·N )

with

∇ · N = j div (−cP) + k

rotM+

∂ P

∂t

,

∇ · F = j div

E c

+ k

rotB− 1

c2∂ E∂t

,

hence, the relations

divE = 1

c (ρ− divP) ,

rotB = µ0

j +

∂ P

∂t + rotM

+ ε0µ0

∂ E

∂t .

The entire set of Maxwell’s equations is expressed by (with ∇ ∧

F = 0)

∇F = ∇ · F +∇ ∧ F = A

= µ0C + µ0∇ ·N

or

V = ρ − div P

ε0,

A = µ0 j + ∂ P

∂t + rotM .




Consequently, one can replace the medium by a distribution, in vacuum, of chargedensity ρ = divP and a current density j = ∂ P

∂t + rotM. Under a general pure

Lorentz transformation, the bivector magnetization-polarization N transforms as

N = bN bc with b = cosh ϕ

2 − iv

v sinh ϕ

2 , tanh ϕ = vc = β and cosh ϕ = γ =

1q 1− v2

c2,

hence

M = γ

M +

v

v2 (v ·M)

1

γ − 1

+ (v ×P)

,

P = γ

P +

v

v2 (v ·P)

1

γ − 1

−v × M

c2

and the inverse formulas

M = γ M +

v

v2 (v ·M) 1

γ − 1− (v×P)

,

P = γ

P +

v

v2 (v ·P)

1

γ − 1

+

v × M

c2

.

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 thereference frame at rest K one has M = −v ×P; if in K , P = 0, M = 0, thenone 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 fromthe 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 ) ,




and where the integration is taken on a closed surface . Considering F = −B+ iEc

with ∇ ∧ F = 0 and ∇ · F = µ0C , one obtains [4, p. 227]

F · dS =

Bxdydz + Bydzdx + Bzdxdy+E xdxdt + E ydydt + E zdzdt

= 0,

F ∧ dS = i

c

E xdydz + E ydzdx + E zdxdy−Bxdxcdt −Bydycdt −Bzdzcdt

= iµ0

(ρcdxdydz − jxdydzdt− jydzdxcdt − jzdxdycdt) .

If one operates at t constant, one finds again the standard equations (in classicalthree-vector formulation)

B · dS = 0,

E · dS =

q

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 =

Bxdydz + Bydzdx + Bzdxdy

+E xdxdt + E ydydt + E zdzdt

.

At t constant, one obtains the classical vector formulation of Stokes’ theorem

A · dl =

rotA · dS =

B · dS.

7.2.3 Lorentz force

The four-force density (per unit volume of the laboratory frame) is expressed withF = −B + iE

c and C = jρc + kρv by

f = F · C

= j ( j

·E) + k (ρE + j

×B) ;




f can also be expressed in the form (with ∇F = µ0C and ∇ = ∂ αeα)

f = F C − CF

2

= 1

2µ0

[F (

∇F )

−(

∇F ) F ]

= 1

2µ0

F (∂ αeαF ) − (∂ αeαF ) F

=

1

2µ0[∂ α (F eαF )− (∂ αF ) eαF − (∂ αeαF ) F ] .

Since

∇ ∧ F = ∇F + F ∇

2

=

1

2 [∂ α (eα

F ) + (∂ αF ) eα

] = 0

it follows that

f = 1

2µ0∂ α (F eαF ) = −∂ αtα

with

tα = −F eαF

2µ0= T αβeβ

where T αβ is the energy-momentum tensor; tα is a four-vector since (tα)c = −tα

(with F c =−

F ). One obtains for tα,

t0 = j(B ·H + E ·D)

2 + k

E×H

c ,

t1 = j(E×H)1

c +

k

µ0

B2+E

2

c2

2 −B2

1 − E 21

c2

I

−

B1B2 + E 1E 2c2

J −

B1B3 + E 1E 3

c2

K

,

t2 = j(E×H)2

c +

k

µ0

−

B1B2 + E 1E 2c2

I

B2+E

2

c2

2

−B22

− E 2

2

c2 J

− B2B3 + E 2E 3c2 K

,

t3 = j(E×H)3

c +

k

µ0

−

B1B3 + E 1E 3

c2

I

−

B2B3 + E 2E 3c2

J +

B2+E

2

ce

2 − B23 − E 2

3

c2

K

.

The four-vectors tα constitute the energy-current density (α = 0) and the mo-mentum-current density along the three axes (x,y,z); the α component of thefour-momentum of the electromagnetic field of a trivolume T is given by

P α = tα

·(dT ∗) .




The symmetry of the energy-momentum tensor T αβ of the electromagnetic fieldcan be verified directly on the components of tα

tα

· eβ

= tβ

· eα

= T αγ eγ · eβ = T αβ = T βα;

it is sufficient to verify the equality

(F eαF ) · eβ = eα · F eβF

, (7.10)

F a = aF (7.11)

with a = eαF eβ − eβF 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 = µ0C with F = −B+ iEc , C = jρc + k j,

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 = F mei(ωt−k·r) = −B + i

E

c

where F m = −Bm + iEm

c is a constant bivector (Bm,Em real). One has, in a

vacuum,

∇F = i j

ω

c + kk

F

= i

j−k·Ec

+ k (k ·B)

+ j (k×E−Bω) + k−k×B− ωE

c2

= 0 ;



7.3. Electromagnetic waves 119

hence, the relations characterizing the plane wave

k ·E = 0, (7.12)

k ·B = 0, (7.13)B =

k× E

ω , (7.14)

E = − (k×B) c2

ω . (7.15)

The relation

∇c∇F =ω

c j + kk

2F =

−ω2

c2 + k · k

F = 0

gives the norm of k,

|k

|=ω

c and the phase velocity v = ω

|k

| = c.

For a plane wave polarized elliptically, one just needs to take F m complex.

7.3.2 Electromagnetic waves in a conductor

Consider a conductor having a permittivity = ε0, a permeability µ = µ0 and aconductivity γ ( j = γ E). Maxwell’s equations are expressed by ∇F = µ0C , hence

∇c∇F = F = µ0∇cC (7.16)

= µ0 (∇ · C + ∇ ∧C ) = µ0 (∇ ∧ C ) (7.17)

= µ0− rotj + i

− ∂ j

c∂t − (c)grad ρ

(7.18)

with the conservation of electric charge (∇ · C = 0). Using the relations j = γ Eand ∇ ∧ F = 0, equation (7.18) becomes

F + γµ0∂F

∂t = −i(µ0c)grad ρ.

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

F + γµ0∂F

∂t = 0. (7.19)

Consider a sinusoidal plane wave rectilinearly polarized F = F meit(ωt−k·r) withthe constant bivector F m = −Bm + iEm

c . Equation (7.19) gives with

∂F

∂t = iωF , F =

−ω2

c2 + |k|2

F

a complex expression of k

|k|2 = ω2

c2 − iωγµ0.

The structure of the plane wave results from equation ∇F = µ0C with C =

C me

i(ωt

−k

·r)

and yields the relations (7.12), (7.13), (7.14), (7.15) with k

complex.




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+ iEc , G = −H + icD, C = jρc + k j. Hence

∇∧ (∇ ·G) = ∇ ∧ C (7.20)

with

∇∧ (∇ ·G) =

rot ∂ D

∂t − rot (rotH)

+i

∂ 2Dc∂t2

− (c)grad (divD)− ∂ rotHc∂t

,

∇ ∧ C = − rot j + i

− ∂ j

c∂t − (c)grad ρ

.

Equation (7.20) can be written in the form

G

−εµ

∂ 2G

∂t2

= γµ∂G

∂t

+ grad (divD) .

In a nonconducting medium (γ = 0) and in the absence of an electric chargedensity, one obtains the wave equation

G− εµ∂ 2G

∂t2 = 0. (7.21)

Introducing the operator ∇ = j ∂ v∂t −k∇, G = −H + vD, F = −B+ iE

v = µG,

C = jρv + k j and the constant v = 1√ εµ

, Maxwell’s equations are expressed by [5]

∇ ∧G = 0,

∇ · G = C

or ∇G = C . Hence, the wave equation

∇c∇G = ∇

cC

= 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 = Gmei

(ωt−k·r) with the constant bivector Gm = −Hm + ivDm and

v = 1

√ εµ. In the absence of a four-current density C , one has

∇G = i

jω

v + kk

G

= i

j (−vk ·D) + k (k ·H)

+ j

vk×D− Hωv

+ k (−k×H−Dω)

= 0;

hence, the relations characterizing the plane wave

k

·D = 0,

k ·H = 0,

H = v2k×D

ω ,

D = −k×H

ω .

7.4 Relativistic optics

7.4.1 Fizeau experiment (1851)

In the Fizeau experiment (Figure 7.2), one observes at O a system of interferencefringes when the water is at rest. When the water circulates, the fringes move andthe experiment has determined the velocity of light in the moving water to be

v = c

n ±

1 − 1

n2

w

whereas the classical theory predicts

v = c

n ± w.

Classical theory

The variation of the optical path δ between the two experiments is

∆δ = c

Lcn − w

− Lcn + w

= 2Lcwc2

n2

−w2 2Ln2w

c .




L1 L2

T 1

T 2

L f

water inflowsource

w

w

Figure 7.2: Fizeau experiment (1851): when the water is at rest, one observesinterference fringes; when the water circulates, the fringes move which allows oneto determine the velocity of light in the moving water.

The variation of the order of interference is

∆ pc = ∆δ

λ0=

2Ln2w

cλ0.

Relativistic theory

Let K be the proper frame of the water in motion with cn

the speed of light withrespect to K . With respect to the frame at rest K , the velocities of light withrespect to the tubes T 1, T 2 are respectively using the formula (6.11),

v1 =cn − w

1− wnc

c

n −

1 − 1

n2

w,

v2 =cn

+ w

1 + w

nc

c

n

+ 1

−

1

n2w,

hence, a variation of the optical path δ ,

∆δ = c

L

v1− L

v2

= 2Lcw

n2 − 1

c2 − n2w2

2Lw

c

n2 − 1

;

the variation of the order of interference is

∆ pr = ∆δ

λ0=

2Lw

cλ0 n2 − 1 .



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 avelocity w with respect to the frame at rest R (Figure 7.3).

w

y

x

z

y

x

z

O O

R R

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

Let K = j ω

c + kk be the wave four-vector in R and K = j ω

c + kk the wave

four-vector in R. One has with kx = k cos θ and f, f designating the frequencies

ω = γ

ω + kx

w

c

= γω

1 + cos θ

w

c

,

f = γf

1 + cos θw

c

.

Longitudinal Doppler effect

1. θ = 0 (the source moves towards the observer located at O, Figure 7.4)

f = γf

1 + w

c

= f

1 + w

c

1− wc

≥ f ;

2. θ = π (the source recedes from the observer, Figure 7.4)

f = γf

1− w

c

= f

1− w

c

1 + wc

≤ f .




θ

yy y y y

x x

O O O

O

w

k

k w

θ = 0 θ = π

Figure 7.4: Longitudinal Doppler effect: in the case θ = 0, the wave is emitted inthe 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 = γf = f

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 (Figure7.5).

y y

x x

O O

v

v v

y

v

x

R R

EarthSolar

Figure 7.5: Aberration of distant stars: in the reference frame R of the sun, thevelocity of light coming from a star located on the Oy axis is v; in the referenceframe 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 =

vx

−w

1 − vxwc2 = −w,

vy =vy

1 − w2

c2

1− vxwc2

= −c

1− w2

c2 ,

vz =vz

1 − w2

c2

1− vxwc2

= 0.

The angle θ under which one sees the star from the Earth is given by

tan θ =vx

vy

= w

c

1 − w2

c2= γ

w

c .

7.5 Exercises

E7-1 Consider a charge q located at the origin of a reference frame K moving witha constant velocity v along the Ox axis of a reference frame K at rest. Determinethe 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 Ox axis of a reference frame K moving with a velocity v alongthe Ox axis of a reference frame K at rest. Along the Ox axis of the referenceframe 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 referenceframe K at rest is nil, determine the electromagnetic field created in K by thesecharges at the point M (x, r, 0). A particle of charge q moves with a velocity vin K , parallel to the Ox axis (at the distance r). Determine the force exerted bythe two sets of charges in the reference frame K and in the proper frame of theparticle.

E7-3 Consider the electromagnetic field

F = −I + 2 iJ c

(Bx = 1T, E y = 2V/m)

in the reference frame of the laboratory K at rest . Determine the electromagneticfield in the reference frame moving with respect to K with a velocity c

2 in thedirection (1, 1, 0) of K .

E7-4 Show that Maxwell’s equations in vacuum without sources are invariantunder the transformation

F → F ∗ = iF (i.e., B→ E

c , E

c → −B).



Chapter 8

General relativity

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

8.1 Riemannian space

Consider a four-dimensional space with the elementary displacement DM = ωieiand the affine connection Dei = ωj

i ej . The covariant differentiation of the vectorA is defined by

DA = dA + dω·

A

with 2dω = ωijei ∧ ej et ωij = ωik(ek · ej). The reciprocal basis eα is defined by

eµ · eυ = δ µυ (e0 = e0, e1 = −e1, e2 = −e2, e3 = −e3) where e0, e1, e2, e3 areunitary orthogonal vectors. Under a Lorentz transformation A = f Af c, one has

DA = f DAf c,

dω = fdωf c − 2df f c.

A Riemannian space is a space without torsion but with a curvature. The absenceof torsion is expressed by

D2(D1M ) − D1(D2M ) = 0 (8.1)

where D1, D2 are two linearly independent directions. Writing dω = I kdxk , DM =σmdxm, condition (8.1) leads to the relations

I k · σm − I m · σk = ∂σk

∂xm − ∂σm

∂xk

which determine I k when DM is given. The existence of a curvature is expressedby the relations

(D2D1 − D1D2)A = Ω · A



128 Chapter 8. General relativity

where Ω is a bivector defined by

Ω = (d2d1 − d1d2)ω + [d2ω, d1ω]

= ΩkmdS km

/2

=

∂ I k∂xm

− ∂I m∂xk

− [I k, I m]

d1xkd2xm

with 2Ωkm = −Rijkmei ∧ ej and dS = D1M ∧ D2M = dS kmek ∧ em/2. Bianchi’s

first identity is given by

Ωij · ek + Ωjk · ei + Ωki · ej = 0, (8.2)

Ωij · (ek ∧ em) = Ωkm · (ei ∧ ej), (8.3)

where relation (8.3) results from the preceeding equation. Bianchi’s second identityis expressed by

Ωij;k + Ωjk;i + Ωki;j = 0

with D3Ω(d2, d1) = d3Ω + [d3ω, Ω] = Ωij;kωi(d2)ωj(d1)ωk(d3). The Ricci tensorRik = Rh

ihk and the curvature R = Rkk are obtained by the relations

Rk = Ωik · ei = Rikei,

R = (Ωik · ei) · ek = Ωik · (ei ∧ ek).

8.2 Einstein’s equations

To deduce Einstein’s equations from a variational principle, we shall use themethod of the exterior calculus but within the framework of a Clifford algebraand without using the exterior product of differential forms. Adopting, for sim-plicity, an orthogonal curvilinear coordinate system, we can write

L =

R√ −gdx0dx1dx2dx3

=

Ωik · (ei ∧ ek)ω0ω1ω2ω3

with √ −gdx0dx1dx2dx3 = ω0ω1ω2ω3. Taking the dual, we obtain

L∗ =

Ωik ∧

ei ∧ ek

∗ω0ω1ω2ω3.

The variation gives

δL∗ =

ΩikωI ωK ∧ δ

ei ∧ ek

∗ωGωH

+ δ ΩikωI ωK

∧ ei ∧ ek∗

ωGωH



8.3. Equation of motion 129

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

δ ei ∧ ek∗

= δ (eγ ) ∧ eγ · ei ∧ ek∗

= −δ (eγ ) ∧

ei ∧ ek ∧ eγ

∗= −δ (eγ ) ∧

ei ∧ ek ∧ eγ

∗.

Hence

δL∗ =

ΩikωI ωK ∧

δ (ωgeγ ) ∧

ei ∧ ek ∧ eγ

∗ωh

= − δ (ωgeγ ) ∧ Ωik ∧ ei ∧ ek ∧ eγ ∗

ωI ωK ωh

= 0

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

Ωij ∧

ei ∧ ej ∧ ek∗

= 0.

In the presence of an energy-momentum distribution, one has

1

2Ωij ∧

ei ∧ ej ∧ ek

∗= κ

T k∗

(8.4)

with T k = T ikei,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

2Ωij ·

ei ∧ ej ∧ ek

= κ

T k

.

Using the formula (B · T ) · V = (B ∧ V ) · T where B, T , V are respectively abivector, a trivector and a vector, one has

− 1

2Ωij · ei ∧ ej ∧ ek · eµ =

1

2Rαβij (eα ∧ eβ ∧ eµ) · ei ∧ ej ∧ ek

= −12

Rαβij δ ijkαβµ = Rβk

βµ − 12

δ kµR

= κ T k · eµ = κ T ikei · eµ = κ T kµ .

8.3 Equation of motion

To obtain the equation of motion, one writes

δS =

−mc δ (ds) = 0




with δ (ds) = δ (ds)2 /2ds and ds2 = DM ·DM = (σi · σk) dxidxk. One obtains

δS = −mc (δσi · σk) uiukds − mc (σl · σk) uk dδxl

= A + B = 0

with ui = dxi/ds. The second term B can be rewritten in the form

B = −mc

(σl · σk) uk

δxl + mc

δxld

(σl · σk) uk

.

Since, δσi = σi,lδxl + δω · σi with δω = I kdxk, one obtains

− ∂σi

∂xl + I l · σi · σkuiuk +

d (σl · σk)

ds uk + (σl · σk) duk

ds = 0

or, equivalently,

−∂σi

∂xl +

∂σl

∂xi − I l · σi + I i · σl

· σkukui

+σl ·

∂σk∂xi

+ I i · σk

uiuk + (σl · σk)

duk

ds

= 0

where the expression in brackets vanishes in the absence of torsion. One thusobtains

σl · du

ds + I iui

· u

= 0

with the four-velocity u = DM/ds = σiui and the ordinary derivative du/ds.Finally, the equation of motion is expressed by

Du

ds =

du

ds +

dω

ds · u = 0.

8.4 Applications

8.4.1 Schwarzschild metric

Consider a metric having a spherical symmetry and the elementary displacementDM ,

ds2 = e2ac2dt2 − e2bdr2 − r2dθ2 − r2(sin θ)2dϕ2,

DM = eacdte0 + ebdre1 + rdθe2 + r sin θdϕe3

where a, b are functions of t and r. The bivector dω is given by

dω = cos θdϕI − e−b sin θdϕJ + e−bdθK + aea−b +b

ce−a+b iI



8.4. Applications 131

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

Ω01 =

a

+ (a

)2

− a

b

e−2b

+−b − b

2+ ba

e−2a

c2

iI ,

Ω02 =be−a−b

rc K +

ae−2b

r iJ, Ω03 = − be−a−b

rc J +

ae−2b

r iK,

Ω12 = be−2b

r K +

b

rce−a−biJ, Ω13 = −be−2b

r J +

b

rce−a−biK,

Ω23 =

1 − e−2b

r2

I.

Einstein’s equations, in a vacuum, are

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

r2 +

2be−2b

r k − 2b

cre−a−b jI = 0, (8.5)

−

1 − e−2b

r2 − 2ae−2b

r

jI +

2b

cre−a−bk = 0, (8.6)

−

b − a

r e−2b

+

(b + b2 − ba)e−2a

c2 − (a + a2 − ab)e−2b

jJ = 0, (8.7)

−

b − a

r e−2b

+

(b + b2 − ba) e−2a

c2 − (a + a2 − ab)e−2b

jK = 0. (8.8)

From equations (8.5), (8.6), one obtains b = 0 and a + b = 0 or equivalentlylog e2(a+b) = C where C = 0 because e2a = e2b = 1 at infinity; hence a + b = 0.Equation (8.5) then gives

e−2b (2br − 1) + 1 = 0,re−2b

= 0,

e−2b = 1 + C 1

r

,




and taking C 1 = −2m (with m = GM/c2), one has e2b = e−2a =

1 − 2mr

−1

which determines completely the Schwarzschild metric. Finally, one has

ds2 =

1 − 2m

r

c2dt2 −1 − 2m

r−1

dr2 − r2dθ2 − r2 sin2 θdϕ2,

dω = cos θdϕI +

1 − 2m

r

1/2

(− sin θdϕJ + dθK ) + mcdt

r2 iI,

Ω01 = −2m

r3 iI, Ω02 =

m

r3iJ, Ω03 =

m

r3iK,

Ω12 = −m

r3K, Ω13 =

m

r3J, Ω23 =

2m

r3 I.

The equation of motion is expressed simply in the form

Duds

= duds

+ dωds · u = 0.

Developing this expression along the axes e0, e1, e2, e3 one obtains respectively

d

ds

1 − 2m

r

cdt

ds

= 0, (8.9)

d

ds

1 − 2m

r

−

1

2 dr

ds

=

1− 2m

r

1

2

−m

r2cdt

ds

2+ rdθ

ds

2+r sin2 θ

dϕ

ds 2

, (8.10)

d

ds

r2

dθ

ds

= r2 sin θ

dϕ

ds

2, (8.11)

d

ds

r2 sin2 θ

dϕ

ds

= 0. (8.12)

Adopting θ = π/2, it follows from the equations (8.9), (8.12) that1 − 2m

r

cdt

ds = k, r2

dϕ

ds = h,

where k and h are constants. The relation ucu = 1 is expressed by1 − 2m

r

cdt

ds

2

−

1 − 2m

r

−1

dr

ds

2

− r2

dϕ

ds

2

= 1,

or with w = 1/r,dr

ds = −h

dw

dϕ,

d2r

ds2 = −h2w2 d2w

dϕ2 .

The projection of the equation on the axis e1 then leads to the relation

d2w

dϕ2

+ w = m

h2

+ 3mw2



8.4. Applications 133

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

8.4.2 Linear approximation

Consider the elementary displacement DM = ωiei with ωi = dxi + hijdxj where

hij = hji = ηikhkj is small as compared to 1 [50, p. 186]. The bivectors dω andΩkm are given by

2dω = (hik,j − hjk,i)ei ∧ ejdxk,

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 = −(hmk )em = 0,

the last equation being a gravitational wave equation (with = ∂ 2

c2∂t2 − ∂ 2

∂x2 −∂ 2

∂y2 − ∂ 2

∂z2 ). Writing

γ km = hkm − 1

2ηkmh

with h = hmm one obtains Einstein’s equations

−γ k = κ T k

(with γ k

= γ mk

em).If one considers a homogeneous sphere (of mass M and radius r0) rotatingslowly with an angular velocity ω around the axis Oz , one integrates the aboveequations and obtains the metric (with m = GM/c2) for r ≥ r0,

ds2 =

1 − 2m

r

c2dt2 −

1 +

2m

r

dx2 + dy2 + dz2

− 4GIω

c3r3 (ydx − xdy) cdt

with I = 2M r20/5. The elementary displacement is

DM = ωiei = (dxi + hijdxj)ei

withh00 = h11 = h22 = h33 = −m/r,

h01 = −GIωy/c3r3, h02 = −GIωx/c3r3,

the other hij being equal to zero. The bivector dω = 12

(hik,j − hjk,i)ei ∧ ejdxk

gives the equation of motion

Du

ds =

− j (u1h00,1 + u2h00,2 + u3h00,3)

+kI (−h00,1 + u3Ω2 − u2Ω3)+kJ (−h00,2 + u1Ω3 − u3Ω1)

+kK (−h00,3 + u2Ω1 − u1Ω2)




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 = GI ω

2z2 − x2 − y2

/c3r5,

which puts in evidence the Thirring precession with respect to the local inertialframe.



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. Thequaternion algebra gives a representation of the rotation group SO(3) well-known

for its simplicity and its immediate physical significance. The Clifford algebraH⊗Hyields a double representation of the Lorentz group containing the SO(3) group asa 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 generalizingthe classical vector and scalar products. This calculus remains relatively close tothe classical vector calculus which it contains as a particular case. The Cliffordalgebra H⊗H allows us to easily formulate special relativity, classical electromag-netism and general relativity. In complexifying H⊗H, one obtains the Dirac alge-bra, 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 ornumerical computations within the Clifford algebra H⊗H have become straight-forward with software such as Mathematica. We hope to have shown that theClifford algebras H⊗H over R and C constitute a coherent, unified, framework of mathematical tools for special relativity, classical electromagnetism, general rel-ativity and relativistic quantum mechanics. The quaternion group consequentlyappears via the Clifford algebra as a fundamental structure of physics revealingits deep harmony.



Appendix A

Solutions

Chapter 1S1-1

i2 = j2 = k2 = ij k = −1,

(−i) ijk = jk = i,

k = − ji,

ijk (−k) = ij = k,

j = −ik,

i = −kj, etc.S1-2

|a| =√

2, |b| =√

25 = 5, a−1 = 1

2(1 − i), b−1 =

1

5(4 + 3 j),

a + b = 5 + i − 3 j, ab = 4 + 4i − 3 j − 3k, ba = 4 + 4i − 3 j + 3k,

a =√

21 + i√

2=√

2(cos45 + i sin45) ,

b = 54− 3 j

5 = 5 (cos36, 87 − sin36, 87) ,

√ a = 21/2

cos

45

2 + i sin

45

2

,

√ b = 51/2

cos

36, 87

2 − i sin

36, 87

2

,

a1/3 = 21/6

cos 45

3 + i sin

45

3

= 21/6 (cos 15 + i sin15) .

S1-3

acax + acxb = acc,

axb + xb2 = cb,

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



138 Appendix A. Solutions

the latter is a equation of the type xα = β, α, β ∈ H which one solves in x.

Answer: 2ix + xj = k, x = 1

3(2 j − i).

S1-4axbb−1 + cxdb−1 = eb−1,

c−1ax + xdb−1 = c−1eb−1;

one finds an equation of the type of the previous exercice and solves similarly;Answer: x = 2 + i.

S1-5

x−1x2 = x−1xa + x−1bx,

x = a + x−1

bx,xx−1 = ax−1 + x−1b = 1;

Answer: x = 1

5 (−6 j + 3k.).

Chapter 2

S2-1

Aα = cos α

−sin α 0

sin α cos α 00 0 1

, Aβ =

cos β 0 sin β

0 1 0− sin β 0 cos β

,

Aγ =

1 0 0

0 cos γ − sin γ

0 sin γ cos γ

.

S2-2

dA = rdArc + drArc + rAdrc

= r (dA + rcdrA + Adrcr) rc ;

since rcdr = −drcr (which is obtained by differentiating rrc = 1), one obtainswith dΩ = 2rcdr,

dA = r (dA + rcdrA − Arcdr) rc

= r

dA +

dΩ

2 A − A dΩ

2

rc = r (dA + dΩ × A) rc = rDArc;

DA = dA + dΩ × A ;

dΩ

dt is the angular velocity with respect to the moving frame.



139

S2-3

A = f Af c (f = gcr,ff c = 1),

DA = f DAf c,dA + dΩ × A = f (dA + dΩ × A) f c,

(df Af c + f dAf c + f Adf c) + dΩ × A = f dAf c + f (dΩ × A) f c,

dΩ × A = f (dΩ × A) f c − df f c (f Af c) − (f Af c) f df c

= f dΩf c × A − df f cA + Adf f c

= (f dΩf c − 2df f c) × A;

hence

dΩ = f dΩf c − 2df f c,

dΩ = f cdΩf − 2f cdf.

S2-4

X = rX rc,

r = cos θ

2 + k sin

θ

2, dΩ

dt = 2rc

dr

dt = k

dθ

dt.

Polar coordinates:

X = xi + yj, X = ρi,

V = DX

dt =

dX

dt +

dΩ

dt × X =

dρ

dti + ρ

dθ

dt j;

γ = DV

dt =

dV

dt +

dΩ

dt × V

=

d2ρ

dt2 − ρ

dθ

dt

2

i + j

2ρ

dρ

dt

dθ

dt + ρ

d2θ

dt2

.

Cylindrical coordinates:

X = xi + yj + zk, X = ρi + zk,

V = DX

dt =

dX

dt +

dΩ

dt × X =

dρ

dti + ρ

dθ

dt j + k

dz

dt;

γ = DV

dt =

dV

dt +

dΩ

dt × V

=

d2ρ

dt2 − ρ

dθ

dt2

i +

1

ρ

d

dt ρ2

dθ

dt j +

d2z

dt2 k.




S2-5 The moving frame (ρ,θ,ϕ) is obtained with the following three rotations:

r1 = cos π

4 − j sin

π

4 (rotation of −π

2 around Oy),

r2 = cos θ2 − i sin θ

2 (rotation of −θ around Ox),

r3 = cos 2ϕ− π

4 + k sin

2ϕ− π

4 (rotation of −(π

2 − ϕ) around Oz);

r = r3r2r1

= 1

2

cos

θ + ϕ

2 + sin

θ + ϕ

2

+ 1

√ 2 i sin 2ϕ − 2θ − π

4 − j cos

2θ − 2ϕ + π

4 + k sin

2ϕ + 2θ − π

4 ;

dΩ = 2rcdr = 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;

V = DX

dt =

dX

dt +

dΩ

dt × X = i

dρ

dti + jρ

dθ

dt + kρ

dϕ

dt sin θ,

γ = DV

dt =

dV

dt +

dΩ

dt × V

= i

d2ρ

dt2 − ρ

dθ

dt

2

+

dϕ

dt

2

sin2 θ

+ j

1

ρ

d

dt

ρ2

dθ

dt

− ρ sin θ cos θ

dϕ

dt

2

+ k

1

ρ sin θ

d

dt

ρ2 sin2 θ

dϕ

dt

.

S2-6

r1 = cos α

2 + k sin

α

2 (1st rotation),

r2 = cos β

2 + sin

β

2r1ir1c = r1

cos

β

2 + i sin

β

2

r1c (2nd rotation),

r3 = cos γ

2

+ sin γ

2

(r2kr2c) (3rd rotation);



141

r = r3r2r1

= r2

cos

γ

2 + k sin

γ

2r2cr2r1

= r1

cos β 2

+ i sin β 2

r1c

cos γ

2 + k sin γ

2

r1

=

cos α

2 + k sin

α

2

cos

β

2 + i sin

β

2

cos

γ

2 + k sin

γ

2

because

r1c

cos

γ

2 + k sin

γ

2

r1 =

cos

γ

2 + k sin

γ

2

;

finally

r = cos

β

2 cos

α + γ

2 + i sin

β

2 cos

α

−γ

2 + j sin

β

2 sin

α

−γ

2 + cos

β

2 sin

α + γ

2 ,

ω = 2rcdr

dt = i

dβ

dt cos γ +

dα

dt sin β sin γ

+ j

−dβ

dt sin γ +

dα

dt sin β cos γ

+ k

dγ

dt +

dα

dt cos β

;

ω = 2dr

dtrc = i

dβ

dt cos α +

dγ

dt sin α sin β

+ j

dβ

dt sin α − dγ

dt cos α sin β

+ k

dα

dt +

dγ

dt cos β

.

Chapter 3

S3-1a0 + ia3 = 1, a0 − ia3 = 0,

−ia1 + a2 = 0, −ia1 − a2 = 0,

hence, a0 = 1

2, a3 = −i

2 , a2 = 0 = a1, or

e1 = 1

2 (1 − ik) ; e21 = e1;

similarly

e2 = 1

2 (1 + ik) ; e22 = e2.




a = (a0 + ia0, a1 + ia1, a2 + ia2, a3 + ia3) ,

u = ae1 =

1

2

a0 + ia3 + ia0 − a3,a1−

ia2 + ia1 + a2,

a1 − ia2 + ia1 + a2,−ia0 + a3 + a0 + ia3

,

v = e1a = 1

2

a0 + ia3 + ia0 − a3,a1 + ia2 + ia1 − a2,

−ia1 + a2 + a1 + ia2,−ia0 + a3 + a0 + ia3

.

S3-2

x + y = 3 + ii + (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) + 2ii + (4 + 2i) j + 2k, y2 = 13 + 12 ik,

xxc = 4 + 4i, yyc = −5,

x−1 = xc

xxc=

1

8 (1 − i,−1− i,−3 + i,−1 + i) ,

y−1 = 15

(−2 + 3ik) ,

y−1x−1 = 1

40 (1 + 5i, 5 + 11i, 9− 5i, 5 + i) = (xy)

−1.

S3-3

r1 = ±a + a∗

|1 + d| = ±1

2

1 +

√ 3k

,

b1 = ± 1 + d

|1 + d

| = ±1

2 3 + ii√

5 ,

d = aa∗c = 1

2

7 + 3ii

√ 5

,

r2 = ± a + a∗

|1 + d| = ±1

2

1 +

√ 3k

,

b2 = ± 1 + d

|1 + d| = ±1

4

6− i

√ 5 − i j

√ 15

,

d = a∗ca = 1

414

−3ii

√ 5i

−3i j

√ 15 .



143

S3-4

x · y = 1

2 (xyc + yxc) = 2,

B = x ∧ y = 1

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

B = z ∧ w = (1 − 9i) i − 3i j + (−3− 3i) k,

F = y ∧ z = −i − 2i j + 3ik,

T = x ∧ (y ∧ z) = 1

2 (xF ∗ + F x) = i + 2i− 3 j − 2k,

T = (x ∧ y) ∧ z = z ∧ (x ∧ y) = i + 2i − 3 j − 2k,

B · z = −z · B = −1

2 (zB∗ − Bz) = 2 + 6ii + 6i j − 2ik ∈ V,

w · T = 1

2 (wT ∗ + T wc) = (3 + i) i + (1 + 8i) j + (1− 11i) k ∈ B,

B · B = S

BB + BB

2

= −22,

B ∧ B = P

BB + BB2

= i,

B · T = BT + T B∗

2 = −5− 6ii + i j − 10 ik ∈ V.

S3-5

X = ct + xii + yi j + zik,

X = ct + xii + yi j + zik,

b =

cosh ϕ

2, i sinh

ϕ

2, 0, 0

,

γ = 1

1 − v2c2= cosh ϕ =

7

2,

cosh2 ϕ

2 =

cosh ϕ + 1

2 , sinh2 ϕ

2 =

cosh ϕ − 1

2 ,

cosh ϕ

2 =

3

2, sinh

ϕ

2 =

√ 5

2 , b =

3

2, i√

5

2 , 0, 0 ,




γ = 1

1− v2c2

= cosh ϕ =

7

6, sinh ϕ =

1

6,

X = (0, i, i, 0),

V =

7

6, 0, i

1

6, 0

,

X = bX b∗c , V = bV b∗c,

V =

7

2 7

6, i

1

2 105

2 , i

1

6, 0

,

F = −B + iE

c =

0, i

E xc

, iE y

c , i

E zc

,

F = bF bc

E x = E x, E y = 7

2E y, E z =

7

2E z ,

Bx = 0, By = −3√

5

2c E z, Bz =

3√

5

2c E y,

F =

0,−B+ iE

c

.

S3-6

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

A

0, i

3,− i

3,−2i

3

, B

0,

i

3, i

3,−2i

3

,

C

0,− i

3, i

3,−2i

3

, D

0,− i

3,− i

3,−2i

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

AAc=

1

4 (−I − 2J + iK ) , , B−1 =

1

4 ( j + kI + 2kK ) ,

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

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

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

(AB) (AB)c = (AB)c (AB) = −16, (BA) (BA)c = −16,

(AB)−1

= (AB)

c(AB) (AB)c

= 1

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

(BA)−1

= 1

16 (2 j + k − jI − 4kJ − 3 jJ + 2kJ + 3kK ) ,

[A, B] = 1

2 (AB − BA) = −2 j + 4kI − 2kJ − 3kK.

S4-2

x · y = −12

(xy + yx) = 2,

B = x ∧ y = −1

2 (xy − yx) = I − J + 2iI + 2iJ − iK,

B = z ∧ w = I − 3K − 9iI − 3iJ − 3iK,

F = y ∧ z = −I − 2iJ + 3iK,

T = x ∧ (y ∧ z) = x ∧ F =

1

2 (xF + F x)= k + 2 jI − 3 jJ − 2 jK,

T = (x ∧ y) ∧ z = z ∧ (x ∧ y) = 1

2 (zB + Bz) = T

= k + 2 jI − 3 jJ − 2 jK,

B · z = −z · B = −1

2 (zB − Bz ) = 2 j + 6kI + −6kJ − 2kK,

w·

T =−

1

2 (wT + T w) = 3I + J + K + iI + 8iJ

−11 iK,




B · B = S

1

2 (BB + BB)

= −22,

B∧

B = P S 1

2 (BB + BB) = i,

B · T = 1

2 (BT + T B) = −5 j − 6kI + kJ − 10 kK.

S4-3

S 1 = −1

2 (xy − yx) = 3I − J − 5K,

S 2 = −1

2 (zw − wz) = 9I − 3J − 3K,

V = S 1 ∧ z = z ∧ S 1 = 12

(zS 1 + S 1z) = 10 k,

S 1 ∧ S 2 = 0,

w = w + w⊥,

w = (w · S 1) S −11 = S −1

1 (S 1 · w) = 169

50 kI − 21

25kJ +

27

10kK,

w⊥ = (w ∧ S 1) S −11 = S −1

1 (S 1 ∧ w) = −69

50kI +

46

25kJ +

23

10kK,

S 1 = S 1 + S 1⊥,

S 1 = (S 1.S 2) S −12 = −54

11I − 18

11J − 18

11K,

S 1⊥ =

(S 1 ∧ S 2) + [S 1, S 2]

S −12 = −21

11I − 26

11J − 37

11K.

Chapter 5

S5-1

γ = 1 1 − v2c2

= cosh ϕ = 32√

2,

cosh2 ϕ

2 =

cosh ϕ + 1

2 , sinh2

ϕ

2 =

cosh ϕ − 1

2 ,

cosh ϕ

2 =

1

2

3√

2+ 2 = α, sinh

ϕ

2 =

1

2

3√

2− 2 = β,

b = 1

2 3√

2

+ 2 + iI 3√

2

− 2 ,



147

r = cos θ

2 + I sin

θ

2 = cos

π

8 + I sin

π

8,

a = rb = α cos π

8 −β sin

π

8i + α sin

π

8 + α sin

π

8I + β cos

π

8iI,

X = aX ac,X = ctj + kI x + kJ y + kKz,

X = ct j + kI x + kJ y + kK z.

S5-2

b1 = 1

2 3

√ 2+ 2 + iI 3

√ 2 −2 ,

b2 = 1

2

3√

2+ 2 + iJ

3√

2− 2

,

a = b2b1 = 1

4

2 +

3√ 2

+

1√ 2

iI + 1√

2iJ +

−2 +

3√ 2

K

,

a = br,

b = ± 1 + d√ 2 + d + dc

, d = aac = 9

8 +

1

2√

2iI +

3

8iJ,

b = ±√

17

4 +

1√ 34

iI + 3

4√

17iJ

,

r = ± a + a√ 2 + d + dc

= ±

2 + 3√

2

1√

17+

−2 +

3√ 2

1√

17K

,

cos θ

2 =

2 +

3√ 2

1√

17, θ = 3, 37,

cosh ϕ

2 =

√ 17

4 , sinh

ϕ

2 =

1

4, ϕ = 0, 494,

b = ±√

17

4 +

1

4√

344iI +

3√ 17

iJ

,

direction u 1

√ 34

4, 3

√ 17

, 0 , tanh ϕ = 0, 4581 = v

c

.




S5-3

xA = (0, 0, 0) , xB = −1

2

, 0,

−1

2 ,

xC =−3

7, 1

7,−2

7

, xD =

−1

3, 1

6,−1

6

,

xE =−4

9, 1

9,−1

9

, xF =

−1

2, 0, 0

,

xG =

−3

5, 0,−1

5

, xH =

−1

2,

1

10,−1

5

.

S5-4

AB =

1 − k −1 + k

−1 + k −1 + k

, BA =

0 j

i 0

,

A−1 = 1

2

1 −k

− j −i

, B−1 =

1

2

1 −i

−k − j

,

(AB)−1

= 1

4

1 + k −1− k

−1− k −1− k

, (BA)

−1=

1

2

0 −i

− j 0

.

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

1− vxwc2= −0, 975 c, w = v = 0, 8 c,

vy = 0, vz = 0;

The relative velocity remains smaller than c

.



149

S6-2

L =

X i ∧ P i =

(ri × pi) + i

ri

E i

c − ctpi

= const.

hence

(ri × pi) = C1 ,

riE i

c − ctpi

= C2,

R =

E iri

E i,

E iri

E i− tc2

piE i

= C3,

R = C3

+vt, v

= c2pi

E i = C4.

S6-3 In the proper frame (instantaneous, Galilean), the four-acceleration is

A = kIg

with the invariant

AAc = −g2;

In the frame at rest K , the four-velocity and the four-acceleration are respectively

u = c ( j cos hϕ + kI sinh ϕ) , tanh ϕ = vc

,

A = cdϕ

dτ ( j sinh ϕ + kI cosh ϕ) , τ : proper time

with

AAc = −g2 = c2

dϕ

dτ

2 sinh2 ϕ − cosh2 ϕ

=

−c2dϕ

dτ 2

hencedϕ

dτ =

g

c, ϕ =

gτ

c

and

u = c

j cos hgτ

c + kI sinh

gτ

c

=

dX

dτ ,

X = c2

g j sin hgτ

c + kI cosh

gτ

c = j ct + kI x;




finally, one has

t = c

g sinh ϕ, x =

c2

g (cosh ϕ− 1)

and x +

c2

g

2

− (ct)2

= c4

g2.

For ϕ 1

t c

gϕ = τ , x =

c2

g

ϕ2

2 =

1

2gτ 2 =

1

2gt2.

Chapter 7

S7-1

A = bAbc

b = cosh ϕ

2 + iI sinh

ϕ

2, tanh ϕ =

v

c

,

A = jV

c + kA = j

V

c

A = 0, V =

q

4πε0r

,

A = jV

c + kA,

r =

x2 + y2 + z2

1/2

, x = γ (x − vt) , y = y, z = z,

V = γ V = q 4πε0

1(x − vt)

2+ y

2+z2

γ 2

1/2 , A = γ vc2

V = vc2

V,

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

F = −B + E

c , F = −B +

E

c ,

B = 0, E = q

4πε0

r

r3,

F = bF bc,

Bx = 0,

By = − q

4πε0

γvz

cr3 = −γE z

v

c = −E z

v

c,

Bz = q

4πε0

γvy

cr3 = γ E y

v

c = E y

v

c,

E x = E x = q

4πε0

x

r3

, E y = γ E y, E z = γ E z,



151

x = γ (x − vt) , y = y, z = z,

E x = q

4πε0

(x−

vt)

γ 2r∗3/2 , r∗ =

(x− vt)2

+ y2 + z2

γ 21/2

,

E y = q

4πε0

y

γ 2r∗3/2,

E z = q

4πε0

z

γ 2r∗3/2.

S7-2

F = −B + E

c ,

B = 0, E = λ0

2πε0

n

r,

E 2 = λ0

2πε0r, E 1 = E 3 = 0,

F = bF bc,

b = cosh ϕ

2 + iI sinh

ϕ

2, tanh ϕ =

v

c,

B3 =

λ0

2πε0r

γv

c2 =

µ0λv

2πr =

µ0i

2πr ,

E 3 = λ0γ

2πε0r =

λ

2πε0r,

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 isnil

ET = 0, BT = µ0i

2πrez.

In K ,

f = q v ×B,

f (M ) = −qvBey;

in the proper frame, the electromagnetic field is

F = −B + E

c = bcF T b

with

F T =−B, B =

µ0i

2πrez,




hence

B= µ0i

2πrγ ez,

E= − µ0i

2πrγvey;

the particle being at rest in the proper frame,

f = qE = − qµ0i

2πr γ ve = γ f .

S7-3

F = bcF b, b = cosh ϕ

2 +

iI √ 2

sinh ϕ

2 +

iJ √ 2

sinh ϕ

2 ,

tanh ϕ = v

c =

1

2, cosh ϕ = γ =

1 1 − v2c2

= 2√

3,

cosh2 ϕ

2 =

cosh ϕ + 1

2 =

2√ 3

+ 1

2 , sinh2

ϕ

2 =

cosh ϕ − 1

2 =

2√ 3 − 1

2 ,

cosh ϕ

2 =

2√ 3

+ 1

2 , sinh

ϕ

2 =

2√ 3 − 1

2 ,

B1 = 1

2 + 1√

3, E 1 = 1 − 2√

3,

B2 =

1

2 − 1√

3, E 2 = 1 +

2√ 3

,

B3 = −1

c

2

3, E 1 = − c√

6.

S7-4

∇F = 0,

i∇F = −∇ (iF ) = 0,

∇F ∗ = 0.



Appendix B

Formulary: multivector productswithin H(C)

Let x, y be four-vectors, B, B bivectors, T, T trivectors and P, P pseudoscalars

x = x0 + ix1 + ix2 + ix3,

y = y0 + iy1 + iy2 + iy3,

B = a + ib, B = a + ib,

(a = a1i + a2 j + a3k, b = b1i + b2 j + b3k),

T = it0

+ t, T = it0

+ t(t = t1

i + t2

j + t3

k),P = is0, P = is0.

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

x · y = x1y1 + x2y2 + x3y3,

x× y = (x2y3 − x3y2)i + (x3y1 − x1y3) j + (x1y2 − x2y1)k.

Products with four-vectors

x · y = 1

2 (xyc + yxc) ∈ S

= x0y0 − x1y1 − x2y2 − x3y3

= y

·x,



154 Appendix B. Formulary: multivector products within H(C)

x ∧ y = 1

2 (xyc − yxc) ∈ B

= x × y + i y0x− x0y= −y ∧ x.

Products with bivectors

x · B = 1

2 (xB∗ − Bx) ∈ V

= −x · b + i

−x0b + x× a

≡ −B · x,

x ∧B = 1

2 (xB∗ + Bx) ∈ T

= −ix · a+

x0a + x× b

≡ B ∧ x.

B · B = S

1

2 (BB + BB)

=−a

·a + b

·b

≡ B · B,

B ∧ B = P

1

2 (BB + BB)

= −i (b · a + a · b)≡ B ∧ B.

Products with trivectors

x · T = 1

2 (xT ∗ + T xc) ∈ B

= x0t + t0x + i (x× t)

≡ T · x

x ∧ T = 1

2 (xT ∗ − T xc) ∈ P

= i

−x0t0 − x · t

≡ −T

∧x,



B.0. Formulaire: produits multivectoriels dans H(C) 155

B · T = −1

2 (BT + T B∗) ∈ V

= (a · t) − i t0a + b× t≡ T · B,

T · T = 1

2 (T T ∗ + T T ∗) ∈ S

= t0t0 − t · t≡ T · T.

Products with pseudoscalars

x · P = 12

(xP ∗ − P x) ∈ T

= −is0x0 + s0x

≡ −P · x,

B · P = 1

2 (BP + P B) ∈ B

= −s0b + is0a

≡P ·

B,

T · P = −1

2 (T P ∗ − P T ) ∈ V

= −s0t0 + is0t

≡ −P · T,

P.P = S 1

2 (P P + P P ) ∈ S

= −s0s0

≡ P · P.



Appendix C

Formulary: multivector products

within H⊗H (overR)

Let x, y be four-vectors, B, B bivectors, T, T trivectors and P, P pseudoscalars

x = j x0

+ kx (x = x1

I + x2

J + x3

K ),y = j y0 + ky,

B = a + ib, B = a + ib,

(a = a1I + a2J + a3K, b = b1I + b2J + b3K ),

T = kt0 + jt, T = kt0 + jt (t = t1I + t2J + t3K ),

P = is0, P = is0.

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

x · y = x1y1 + x2y2 + x3y3,

x

×y = (x2y3

−x3y2)I + (x3y1

−x1y3)J + (x1y2

−x2y1)K.



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

Products with four-vectors

x · y = −1

2 (xy + yx) ∈ S

= x0y0 − x1y1 − x2y2 − x3y3

= y · x,

x ∧ y = −1

2 (xy − yx) ∈ B

= x × y + i

y0x− x0y

=−

y∧

x.

Products with bivectors

x · B = 1

2 (xB − Bx) ∈ V

= − jx · b + k−x0b + x× a

= −B · x,

x ∧ B = 1

2 (xB + Bx) ∈ T

= −kx · a+ j

x0a + x× b

≡ B ∧ x.

B · B = S

1

2 (BB + BB)

= −a · a + b · b= B · B,

B ∧ B = P

1

2 (BB + BB)

= −i (b · a + a · b)

≡B

∧B.



159

Products with trivectors

x · T = −12

(xT + T x) ∈ B

= x0t + t0x + i (x× t)

≡ T · x,

x ∧ T = −1

2 (xT − T x) ∈ P

= i−x0t0 − x · t

≡ −T ∧ x,

B · T = 1

2 (BT + T B) ∈ V

= j (a · t) − k

t0a + b× t

≡ T · B,

T · T = −1

2 (T T + T T ) ∈ S

= t0t0 − t · t≡ T · T.

Products with pseudoscalars

x · P = 1

2 (xP − P x) ∈ T

= −ks0x0 + js0x

≡ −P · x,

B · P = 12

(BP + P B) ∈ B

= −s0b + is0a

≡ P · B,

T · P = 1

2 (T P − P T ) ∈ V

= − js0t0 + ks0t

≡ −P

·T,



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

P · P = S

1

2 (P P + P P )

∈ S

=

−s0s0

≡ P · P.



Appendix D

Formulary: four-nabla operator ∇within H⊗H (overR)

Four-nabla operator:

∇ = j ∂

c∂t − kI

∂

∂x − kJ

∂

∂y − kK

∂

∂z,

D’Alembertian operator:

= ∂ 2

c2∂t2 − ∂ 2

∂x2 − ∂ 2

∂y2 − ∂ 2

∂z

(four-vectors

A = j A0 + kI A1 + kJ A2 + kK A3, B = j B0 + kI B1 + kJ B2 + kK B3;

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)

<<Algebra Quaternions`

(*example x = x0+x1i+x2 j +x3k = [x0, x1, x2, x3], xi ∈ C, I : usual compleximaginary*)

x=Quaternion[1,I,2+I,1]

y=Quaternion[2,0,0,3I]

x**y

(*the two stars indicate a quaternion product; place the pointer at the endof the last program line and click “Enter” on the numerical pad; result*)

out=Quaternion[2-3I,-3+8I,7+2I,2+3I]



Appendix F

Work-sheet H⊗H over R(Mathematica)


(*product of two Clifford numbers, a = a1 + a2I + a3J + a4K, b = b1 + b2I +b3J + b4K, ai, bi ∈ H*)

CP[a_,b_]:=

(a[[1]]**b[[1]])-(a[[2]]**b[[2]])

-(a[[3]]**b[[3]])-(a[[4]]**b[[4]]),

(a[[2]]**b[[1]])+(a[[1]]**b[[2]])

-(a[[4]]**b[[3]])+(a[[3]]**b[[4]]),

(a[[3]]**b[[1]])+(a[[4]]**b[[2]])

+(a[[1]]**b[[3]])-(a[[2]]**b[[4]]),

(a[[4]]**b[[1]])-(a[[3]]**b[[2]])

+(a[[2]]**b[[3]])+(a[[1]]**b[[4]])

(*conjugate*)

K[a_]:=Quaternion[a[[1,1]],a[[1,2]],-a[[1,3]],a[[1,4]]],

Quaternion[-a[[2,1]],-a[[2,2]],a[[2,3]],-a[[2,4]]],

Quaternion[-a[[3,1]],-a[[3,2]],a[[3,3]],-a[[3,4]]],

Quaternion[-a[[4,1]],-a[[4,2]],a[[4,3]],-a[[4,4]]]

(*sum and difference*)



166 Appendix F. Work-sheet H⊗H over R ( Mathematica )

csum[a_,b_]:=a[[1]]+b[[1]],

a[[2]]+b[[2]],a[[3]]+b[[3]],a[[4]]+b[[4]]

cdif[a_,b_]:=a[[1]]-b[[1]],

a[[2]]-b[[2]],a[[3]]-b[[3]],a[[4]]-b[[4]]

(*multiplication by a scalar f *)

fclif[f_,a_]:=f*a[[1]],f*a[[2]],f*a[[3]],f*a[[4]]

(*products ab+ba2

, ab−ba2

*)

int[a_,b_]:=fclif[1/2,csum[CP[a,b],CP[b,a]]]

ext[a_,b_]:=fclif[1/2,cdif[CP[a,b],CP[b,a]]]

(*products −ab+ba2

, −ab−ba2

*)

mint[a_,b_]:=fclif[-1/2,csum[CP[a,b],CP[b,a]]]

mext[a_,b_]:=fclif[-1/2,cdif[CP[a,b],CP[b,a]]]

(*example A = I + 2J − iK , B = j + kI + 2kK , product w = AB*)

A:=Quaternion[0,0,0,0],Quaternion[1,0,0,0],

Quaternion[2,0,0,0],Quaternion[0,-1,0,0]

B:=Quaternion[0,0,1,0],Quaternion[0,0,0,1],

Quaternion[0,0,0,0],Quaternion[0,0,0,2]

w=CP[A,B]

Simplify[%]

(*result w = AB = −2 j − k + jI + 4kI + 3 jJ − 2kJ − 3kK ; the function“Simplify” simplifies numerically or algebraically the result of the line above*)

Out=Quaternion[0,0,-2,-1],Quaternion[0,0,1,4],

Quaternion[0,0,3,-2],Quaternion[0,0,0,-3]



Appendix G

Work-sheet: matrices M 2(H)(Mathematica)


(*product of two matrices 2 ∗ 2 over H ,

a =

a1 a2a3 a4

, b =

b1 b2b3 b4

, ai, bi ∈ H*)

CP[a_,b_]:=(a[[1]]**b[[1]])+(a[[2]]**b[[3]]),

(a[[1]]**b[[2]])+(a[[2]]**b[[4]]),

(a[[3]]**b[[1]])+(a[[4]]**b[[3]]),

(a[[3]]**b[[2]])+(a[[4]]**b[[4]])

(*example

A = 1 j

k i , B = 1 k

i j *)

A:=Quaternion[1,0,0,0],Quaternion[0,0,1,0],


B:=Quaternion[1,0,0,0],Quaternion[0,0,0,1],


CP[A,B]



168 Appendix G. Work-sheet: matrices M 2(H) ( Mathematica )

(*result

AB =

1− k −1 + k−1 + k −1 + k

*)

Out=Quaternion[1,0,0,-1],Quaternion[-1,0,0,1],

Quaternion[-1,0,0,1],Quaternion[-1,0,0,1]



Appendix H

Clifford algebras: isomorphisms

General formulas ([18], p. 48)M n(R), M n(C), M n(H): square matrices of order n over R, C, H;

M n(R) ⊗ M p(R) M np(R),

M p(C) M p(R) ⊗ C,

M p(H) M p(R) ⊗H.

Clifford algebra H

H = M 1(H) = M 1(R) ⊗H,C + = C.

Clifford algebra H⊗H

H⊗H M 4(R),C + = H⊗ C M 2(C)

Clifford algebra H⊗H⊗H

H⊗H⊗H M 4(R) ⊗H M 4(H),C + H⊗H⊗ C M 4(R) ⊗ C M 4(C), H⊗ M 2(C) M 2 [H (C)] (Dirac algebra).

Clifford algebra H⊗H⊗H⊗H

H⊗H⊗H⊗H M 4(R) ⊗ M 4(R) M 16(R),C + H⊗H⊗H⊗ C M 4(R) ⊗H⊗ C M 4 [H (C)],

M 4(R) ⊗ M 2(C) M 8 (C).



Appendix I

Clifford algebras: synoptic table

H (R)

real quaternionsclassical vector calculus

rotation group SO(3)classical mechanics

H (C)

complex quaternionsLorentz group

special relativityclassical electromagnetism

H⊗H(R)

real Clifford numbersrelativistic multivector calculus

Lorentz groupclassical electromagnetism

special relativitygeneral relativity

H⊗H(C)

complex Clifford numbersDirac algebra

unitary group SU(4) and symplectic unitary group USp(2,H)relativistic quantum mechanics



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Index

aberration, 124adjunction, 86affine connection, 127

Bianchi, 128

Binet relativistic equation, 133bivector, 44, 59, 60, 85

electromagnetic field, 50

charge density, 105Clifford algebra, 57Clifford number, 58, 59

conjugate, 59dual, 59selfadjoint, 86

Clifford theorem, 57commutator, 29, 59complex imaginary, 86conductor, 119conformal transformation, 52, 82conjugate, 9, 59

complex, 38, 86quaternionic, 86

conservation, electric charge, 111contraction, 93

covariance, 111covariant differentiation, 127curvature, 127, 128

d’Alembertian operator, 51, 111, 118De Moivre theorem, 11dielectric medium, 114dilatation, 82Dirac algebra, 85Dirac spinor, 86

direct product, 3

Doppler effect, 123dual, 28, 59

Einstein equations, 128electric charge, conservation, 106

electromagnetic field, 109bivector, 109

energy, 99energy-current density, 117energy-momentum tensor, 117equation of motion, 129exterior product, 59, 61

Fizeau experiment, 121four-acceleration, 96

proper, 97four-curl, 69four-current density, 51, 105, 110four-divergence, 69four-force , 100

density, 116volumic, 52

four-gradient, 69four-momentum, 99four-nabla operator, 107

four-potential, 50potential vector, 107, 111four-vector, 44, 59four-velocity, 48, 95

Galilean transformation, 92gauge transformation, 109Gauss theorem, generalized, 71generators, 59, 85, 88Gibbs, 9

gravitational wave, 133



178 Index

group, 3cyclic, 3dihedral, 4

finite, 4O(3), 22of rotations SO(4), 20orthogonal [Or](4), 20pseudo-orthogonal Or(1,3), 39SO(3), 22special orthogonal SO(1,3), 40crystallographic, 28

Hamilton, 3, 174

Hamiltonian, 32harmonic coordinates, 133hermitian norm, 86hyperplane, 19, 38, 65

dual, 66tangent, 70

hypersurface, 106

infinitesimal transformation, 28interference fringes, 121interior product, 59

inversion, 39improper antichronous, 77improper orthochronous, 77

Jacobi identity, 63

Kepler, 32kinetic energy, 100Klein four-group, 4

Laplace-Runge-Lenz vector, 32left ideal, 86longitudinal Doppler effect, 123Lorentz gauge, 50, 110, 111Lorentz transformation,

antichronous, 40general, 94infinitesimal, 80orthochronous, 40pure, 41, 79

special, 94

Lorentz, special transformation, 47

magnetic induction, 109

magnetic medium, 114magnetization density, 114Maxwell’s equations, 52, 110

covariance, 111Mercury perihelion precession, 133metric, 130Minkowski, 38

volumic four-force, 52minquat, 38

spacelike, 38

timelike, 38unitary, 38, 42momentum, 99momentum-current density, 117multivector, 46, 61, 81

calculus, 46

nonconducting medium, 120norm, 9

octaeder, 26operator, conjugate, 50operator, four-nabla, 50, 69optical path, 121, 122order of interference, 122orthogonal projection, 66

permeability, 119permittivity, 119plane, 64

symmetry, 19wave, 118tangent, 70

Poisson bracket, 32polarization density, 114, 115primitive idempotent, 86product, exterior, 44product, interior, 46proper frame, 97, 122proper time, 95

pseudoeuclidean, 38



Index 179

pseudoscalar, 46, 59, 61

quadratic form, 39

quaternion, 3algebra, 9complex, 37exponential, 14group, 7root nth, 13square root, 11unitary, 19, 42

reciprocal basis, 69, 127

Ricci tensor, 128Riemannian space, 127Rodriguez, 24rotation, 39

group [SO](3), 78proper antichronous, 77

torsion, 127transformation, improper, 39transformation, proper, 39

transversal Doppler effect, 124triple product, scalar, 16trivector, 45, 59, 60

unitary group, 87

variational principle, 128vector, 85

isotropic, 75spacelike, 75tangent, 69timelike, 75

vector calculus, classical, 64vector product, 15vector product, double, 17

wave four-vector, 97

girard p. r. - quaternions, clifford algebras and relativistic physics(2007)(179).pdf

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