Girard P. R. - Quaternions, Clifford Algebras and Relativistic Physics (2007) PDF

Patrick R.
Girard
Quaternions,
Clifford Algebras and
Relativistic Physics
Birkhuser
Basel . Boston . Berlin
Author:
Patrick R. Girard
INSA de Lyon
Dpartement Premier Cycle
20, avenue Albert Einstein
F-69621 Villeurbanne Cedex
France
e-mail: patrick.girard@insa-lyon.fr
Igor Ya. Subbotin

Department of Mathematics and Natural Sciences
National University
Los Angeles Campus
3DFLF&RQFRXUVH'ULYH
Los Angeles, CA 90045
USA
e-mail: isubboti@nu.edu
2000 Mathematical Subject Classication: 15A66, 20G20, 30G35, 35Q75, 78A25, 83A05, 83C05, 83C10
Library of Congress Control Number: 2006939566

Bibliographic information published by Die Deutsche Bibliothek
Die Deutsche Bibliothek lists this publication in the Deutsche Nationalbibliograe;
detailed bibliographic data is available in the Internet at <http://dnb.ddb.de>.
ISBN 978-3-7643-7790-8 Birkhuser Verlag AG, Basel Boston Berlin

This work is subject to copyright. All rights are reserved, whether the whole or part of the material is
concerned, specically the rights of translation, reprinting, re-use of illustrations, recitation, broadcasting, reproduction on microlms or in other ways, and storage in data banks. For any kind of use
permission of the copyright owner must be obtained.
Originally published in French under the title Quaternions, algbre de Clifford et physique relativiste.
2004 Presses polytechniques et universitaires romandes, Lausanne
All rights reserved
2007 Birkhuser Verlag AG, P.O. Box 133, CH-4010 Basel, Switzerland
Part of Springer Science+Business Media
Printed on acid-free paper produced from chlorine-free pulp. TCF f
f
Printed in Germany
ISBN-10: 3-7643-7790-8
e-ISBN-10: 3-7643-7791-7
ISBN-13: 978-3-7643-7790-8
e-ISBN-13: 978-3-7643-7791-5
987654321
www.birkhauser.ch
To Isabelle, my wife, and to our children: Claire, Beatrice, Thomas and Benot
Foreword
The use of Cliord algebra in mathematical physics and engineering has grown
rapidly in recent years. Cliord had shown in 1878 the equivalence of two approaches to Cliord algebras: a geometrical one based on the work of Grassmann
and an algebraic one using tensor products of quaternion algebras H. Recent developments have favored the geometric approach (geometric algebra) leading to an
algebra (space-time algebra) complexied by the algebra H H presented below
and thus distinct from it. The book proposes to use the algebraic approach and
to dene the Cliord algebra intrinsically, independently of any particular matrix
representation, as a tensor product of quaternion algebras or as a subalgebra of
such a product. The quaternion group thus appears as a fundamental structure of
physics.
One of the main objectives of the book is to provide a pedagogical introduction to this new calculus, starting from the quaternion group, with applications to
physics. The volume is intended for professors, researchers and students in physics
and engineering, interested in the use of this new quaternionic Cliord calculus.
The book presents the main concepts in the domain of, in particular, the
quaternion algebra H, complex quaternions H(C), the Cliord algebra H H
real and complex, the multivector calculus and the symmetry groups: SO(3),
the Lorentz group, the unitary group SU(4) and the symplectic unitary group
USp(2, H). Among the applications in physics, we examine in particular, special
relativity, classical electromagnetism and general relativity.
I want to thank G. Casanova for having conrmed the validity of the interior
and exterior products used in this book, F. Sommen for a conrmation of the
Cliord theorem and A. Solomon for having attracted my attention, many years
ago, to the quaternion formulation of the symplectic unitary group.
Further thanks go to Professor Bernard Balland for reading the manuscript,
the Docinsa library, the computer center and my colleagues: M.-P. Noutary for
advice concerning Mathematica, G. Travin and A. Valentin for their help in Latex.
For having initiated the project of this book in a conversation, I want to
thank the Presses Polytechniques et Universitaires Romandes, in particular, P.-F.
Pittet and O. Babel.
viii
Foreword
Finally, for the publication of the english translation, I want to thank Thomas
Hemping at Birkh
auser.
Lyon, June 2006
Patrick R. Girard
Contents
Introduction
1 Quaternions
1.1 Group structure . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Finite groups of order n 8 . . . . . . . . . . . . . . . . .
1.3 Quaternion group . . . . . . . . . . . . . . . . . . . . . . .
1.4 Quaternion algebra H . . . . . . . . . . . . . . . . . . . .
1.4.1 Denitions . . . . . . . . . . . . . . . . . . . . . .
1.4.2 Polar form . . . . . . . . . . . . . . . . . . . . . .
1.4.3 Square root and nth root . . . . . . . . . . . . . .
1.4.4 Other functions and representations of quaternions
1.5 Classical vector calculus . . . . . . . . . . . . . . . . . . .
1.5.1 Scalar product and vector product . . . . . . . . .
1.5.2 Triple scalar and double vector products . . . . . .
1.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
3
3
4
7
8
8
11
11
14
15
15
16
17
2 Rotation groups SO(4) and SO(3)

2.1 Orthogonal groups O(4) and SO(4) . . . . . . .
2.2 Orthogonal groups O(3) and SO(3) . . . . . . .
2.3 Crystallographic groups . . . . . . . . . . . . .
2.3.1 Double cyclic groups Cn (order N = 2n)
2.3.2 Double dihedral groups Dn (N = 4n) .
2.3.3 Double tetrahedral group (N = 24) . . .
2.3.4 Double octahedral group (N = 48) . . .
2.3.5 Double icosahedral group (N = 120) . .
2.4 Innitesimal transformations of SO(4) . . . . .
2.5 Symmetries and invariants: Keplers problem .
2.6 Exercises . . . . . . . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
19
19
22
24
24
24
25
26
27
28
32
34
3 Complex quaternions
3.1 Algebra of complex quaternions H(C) . . . . . . . . . . . . . . . .
3.2 Lorentz groups O(1, 3) and SO(1, 3) . . . . . . . . . . . . . . . . .
3.2.1 Metric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
37
37
38
38
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Contents
.
.
.
.
.
.
.
.
.
.
.
.
.
38
39
41
41
43
44
47
47
48
48
50
52
54
4 Cliord algebra
4.1 Cliord algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.1.1 Denitions . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.1.2 Cliord algebra H H over R . . . . . . . . . . . . . . . . .
4.2 Multivector calculus within H H . . . . . . . . . . . . . . . . . .
4.2.1 Exterior and interior products with a vector . . . . . . . . .
4.2.2 Products of two multivectors . . . . . . . . . . . . . . . . .
4.2.3 General formulas . . . . . . . . . . . . . . . . . . . . . . . .
4.2.4 Classical vector calculus . . . . . . . . . . . . . . . . . . . .
4.3 Multivector geometry . . . . . . . . . . . . . . . . . . . . . . . . .
4.3.1 Analytic geometry . . . . . . . . . . . . . . . . . . . . . . .
4.3.2 Orthogonal projections . . . . . . . . . . . . . . . . . . . . .
4.4 Dierential operators . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4.1 Denitions . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4.2 Innitesimal elements of curves, surfaces and hypersurfaces
4.4.3 General theorems . . . . . . . . . . . . . . . . . . . . . . . .
4.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
57
57
57
58
59
59
61
62
64
64
64
66
69
69
69
71
72
5 Symmetry groups
5.1 Pseudo-orthogonal groups O(1, 3) and SO(1, 3) . . . .
5.1.1 Metric . . . . . . . . . . . . . . . . . . . . . . .
5.1.2 Symmetry with respect to a hyperplane . . . .
5.1.3 Pseudo-orthogonal groups O(1, 3) and SO(1, 3)
5.2 Proper orthochronous Lorentz group . . . . . . . . . .
5.2.1 Rotation group SO(3) . . . . . . . . . . . . . .
5.2.2 Pure Lorentz transformation . . . . . . . . . .
5.2.3 General Lorentz transformation . . . . . . . . .
5.3 Group of conformal transformations . . . . . . . . . .
5.3.1 Denitions . . . . . . . . . . . . . . . . . . . .
5.3.2 Properties of conformal transformations . . . .
75
75
75
75
77
78
78
79
81
82
82
83
3.3
3.4
3.5
3.6
3.7
3.8
3.2.2 Plane symmetry . . . . . . . . . . . . .

3.2.3 Groups O(1, 3) and SO(1, 3) . . . . . . .
Orthochronous, proper Lorentz group . . . . .
3.3.1 Properties . . . . . . . . . . . . . . . . .
3.3.2 Innitesimal transformations of SO(1, 3)
Four-vectors and multivectors in H(C) . . . . .
Relativistic kinematics via H(C) . . . . . . . .
3.5.1 Special Lorentz transformation . . . . .
3.5.2 General pure Lorentz transformation . .
3.5.3 Composition of velocities . . . . . . . .
Maxwells equations . . . . . . . . . . . . . . .
Group of conformal transformations . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Contents
xi
. . . .
. . . .
. . . .
groups
. . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
91
. 91
. 91
. 92
. 94
. 94
. 94
. 97
. 99
. 99
. 100
. 103
7 Classical electromagnetism
7.1 Electromagnetic quantities . . . . . . . . . . . . . .
7.1.1 Four-current density and four-potential . .
7.1.2 Electromagnetic eld bivector . . . . . . . .
7.2 Maxwells equations . . . . . . . . . . . . . . . . .
7.2.1 Dierential formulation . . . . . . . . . . .
7.2.2 Integral formulation . . . . . . . . . . . . .
7.2.3 Lorentz force . . . . . . . . . . . . . . . . .
7.3 Electromagnetic waves . . . . . . . . . . . . . . . .
7.3.1 Electromagnetic waves in vacuum . . . . . .
7.3.2 Electromagnetic waves in a conductor . . .
7.3.3 Electromagnetic waves in a perfect medium
7.4 Relativistic optics . . . . . . . . . . . . . . . . . . .
7.4.1 Fizeau experiment (1851) . . . . . . . . . .
7.4.2 Doppler eect . . . . . . . . . . . . . . . . .
7.4.3 Aberration of distant stars . . . . . . . . .
7.5 Exercises . . . . . . . . . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
105
105
105
107
110
110
115
116
118
118
119
120
121
121
123
124
125
8 General relativity
8.1 Riemannian space . . . . . .
8.2 Einsteins equations . . . . .
8.3 Equation of motion . . . . . .
8.4 Applications . . . . . . . . . .
8.4.1 Schwarzschild metric .
8.4.2 Linear approximation
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
127
127
128
129
130
130
133
5.4
5.5
5.3.3 Transformation of multivectors

Dirac algebra . . . . . . . . . . . . . .
5.4.1 Dirac equation . . . . . . . . .
5.4.2 Unitary and symplectic unitary
Exercises . . . . . . . . . . . . . . . .
6 Special relativity
6.1 Lorentz transformation . . . . . . . . .
6.1.1 Special Lorentz transformation
6.1.2 Physical consequences . . . . .
6.1.3 General Lorentz transformation
6.2 Relativistic kinematics . . . . . . . . .
6.2.1 Four-vectors . . . . . . . . . . .
6.2.2 Addition of velocities . . . . . .
6.3 Relativistic dynamics of a point mass .
6.3.1 Four-momentum . . . . . . . .
6.3.2 Four-force . . . . . . . . . . . .
6.4 Exercises . . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
84
85
85
86
88
xii
Contents
Conclusion
135
A Solutions
137
B Formulary: multivector products within H(C)
153
C Formulary: multivector products within H H (over R)
157
D Formulary: four-nabla operator within H H (over R)
161
E Work-sheet: H(C) (Mathematica)
163
F Work-sheet H H over R (Mathematica)
165
G Work-sheet: matrices M2 (H) (Mathematica)
167
H Cliord algebras: isomorphisms
169
171
Cliord algebras: synoptic table
Bibliography
173
Index
177
Introduction
If one examines the mathematical tools used in physics, one nds essentially three
calculi: the classical vector calculus, the tensor calculus and the spinor calculus.
The three-dimensional vector calculus is used in nonrelativistic physics and also
in classical electromagnetism which is a relativistic theory. This calculus, however,
cannot describe the unity of the electromagnetic eld and its relativistic features.
As an example, a phenomenon as simple as the creation of a magnetic induction by a wire with a current is in fact a purely relativistic eect. A satisfactory
treatment of classical electromagnetism, special relativity and general relativity
is given by the tensor calculus. Yet, the tensor calculus does not allow a double
representation of the Lorentz group and thus seems incompatible with relativistic
quantum mechanics. A third calculus is then introduced, the spinor calculus, to
formulate relativistic quantum mechanics. The set of mathematical tools used in
physics thus appears as a succession of more or less coherent formalisms. Is it
possible to introduce more coherence and unity in this set? The answer seems to
reside in the use of Cliord algebra. One of the major benets of Cliord algebras is that they yield a simple representation of the main covariance groups of
physics: the rotation group SO(3), the Lorentz group, the unitary and symplectic
unitary groups. Concerning SO(3), this is well known, since the quaternion algebra
H which is a Cliord algebra (with two generators) allows an excellent representation of that group . The Cliord algebra H H, the elements of which are simply
quaternions having quaternions as coecients, allows us to do the same for the
Lorentz group. One shall notice that H H is dened intrinsically independently
of any particular matrix representation. By taking H H (over C), one obtains the
Dirac algebra and a simple representation of SU(4) and USp(2, H). Computations
within these algebras have become straightforward with software like Mathematica which allows us to perform extended algebraic computations and to simplify
them. One will nd 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 Cliord algebra H H in relativistic
physics with applications such as classical electromagnetism, special relativity and
general relativity.
Chapter 1
Quaternions
The abstract quaternion group, discovered by William Rowan Hamilton in 1843, is
an illustration of group structure. After having dened this fundamental concept
of physics, the chapter examines as examples the nite groups of order n 8 and
in particular, the quaternion group. Then the quaternion algebra and the classical
vector calculus are treated as an application.
1.1 Group structure

A set G of elements is a group if there exists an internal composition law dened
for all elements and satisfying the following properties:
1. the law is associative
(a b) c = a (b c),
a, b, c G,
2. the law admits an identity element e

a e = e a = a,
a G,
3. any element a of G has an inverse a

a a = a a = e.
Let F and G be two groups. A composition law on F G is dened by
(f1 , g1 )(f2 , g2 ) = (f1 f2 , g1 g2 ),
(fi F, gi G, i = 1, 2);
the group F G is called the direct product of the groups F and G.

Examples.
1. Cyclic group Cn of order n the elements of which are

(b, b2 , b3 , . . . , bn = e)
and where b represents, for example, a rotation of 2/n around an axis.
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 bh a = abh (h = 1 n); indeed, since
(ab)1 = b1 a1 = b1 a = ab,
one has
b1 (b1 a)b = b2 ab = b1 ab2 = ab3
and thus b2 a = ab2 ; by proceeding similarly by recurrence, one establishes

the above formula.
1.2 Finite groups of order n 8

The nite groups of order n 8, except the quaternion group which will be treated
separately, are the following.
1. n = 1, there exists only one group (1 = e).
2. n = 2, only one group exists, the cyclic group C2 consisting of the elements
(b, b2 = e).
Examples.
(a) the group constituted by the elements (1, 1);
(b) the group having the elements (b : rotation of around an axis,

b2 = e).
3. n = 3, only one group is possible: the cyclic group C3 of elements (b, b2 ,
b3 = e) where b, b2 are elements of order 3.
4. n = 4, two groups exist:
(a) the cyclic group C4 constituted by the elements (b, b2 , b3 , b4 = e) where
the element b2 is of order 2, and where (b, b3 ) are elements of order 4;
(b) the Klein four-group dened by
I 2 = J 2 = (IJ)2 = 1
or, equivalently I 2 = J 2 = K 2 = IJK = 1 with K = IJ and the
multiplication table
1
I
J
K
1
1
I
J
K
I
I
1
K
J
J
J
K
1
I
K
K
J
I
1
1.2. Finite groups of order n 8
The Klein four-group is isomorphic to the direct product of two cyclic

groups C2 ,
2
(1,
1) (b,2b = e)

= 1 (1, b ), I (1, b), J (1, b), K (1, b2 ) .
Example. The group constituted by the elements (I: rotation of

around the axis Ox, J: rotation of around the axis Oy, K = IJ:
rotation of around the axis Oz).
5. n = 5, there exists only one group, the cyclic group C5 having the elements
(b, b2 , b3 , b4 , b5 = e).
6. n = 6, two groups are possible:
(a) the cyclic group C6 (b, b2 , b3 , b4 , b5 , b6 = e);
(b) the dihedral group D3 dened by the relations
a2 = b3 = (ab)2 = e,
leading to the multiplication table
b
b2
b3 = e
a
ab
ba
b
b2
e
b
ab
ba
b
b2
e
b
b2
ba
a
ab
b3 = e
b
b2
e
a
ab
ba
a
ba
ab
a
e
b2
b
ab
a
ba
ab
b
e
b2
ba
ab
a
ba
b2
b
e
with bh a = abh (h = 1, 2, 3). This group is the rst noncommutative

group of the series.
Example. The symmetry group of the equilateral triangle (see Fig. 1.1).
7. n = 7, there exists only one group, the cyclic group C7 of elements (b, b2 , b3 ,
b4 , b5 , b6 , b7 = e).
8. n = 8, there exist ve groups, among them the quaternion group which will
be treated separately.
(a) The cyclic group C8 of elements (b, b2 , b3 , b4 , b5 , b6 , b7 , b8 = e).
B
b(M )
a(M )
D
x
Oz
ba(M )
b2 (M )
ab(M )
Figure 1.1: Symmetry group of the equilateral triangle; b represents a rotation of

the point M by 2/3 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 S222 , direct product of the Klein four-group with C2 ,
(1, I, J, K) (1, 1) = (1, I, J, K)
with 1 = (1, 1), 1 = (1, 1), I = (I, 1), J = (J, 1), K =
(K, 1) ; the multiplication table is given by
1
1
I
I
J
J
K
K
1
1
1
I
I
J
J
K
K
1
1
1
I
I
J
J
K
K
I
I
I
1
1
K
K
J
J
I
I
I
1
1
K
K
J
J
J
J
J
K
K
1
1
I
I
J
J
J
K
K
1
1
I
I
K
K
K
J
J
I
I
1
1
K
K
K
J
;
J
I
I
1
1
the group is commutative.

(c) the group S42 , direct product of C4 with C2 and constituted by the
elements
(b, b2 , b3 , b4 = e) (1, 1) = (b, b2 , b3 , 1);
it is a commutative group.
1.3. Quaternion group
(d) The group D4 (noncommutative) dened by

a2 = b4 = (ab)2 = e
with the multiplication table
b
b2
b3
b4 = e
a
ab
ba
ab2
b
b2
b3
e
b
ab
ab2
a
ab
b2
b3
e
b
b2
ab2
ba
ab
a
b3
e
b
b2
b3
ba
a
b2
ab
b4 = e
b
b2
b3
e
a
ab
ba
ab2
a
ba
ab2
ab
a
e
b3
b
b2
ab
a
ba
ab2
ab
b
e
b2
b3
ba
ab2
ab
a
ba
b3
b2
e
b
ab2
ab
a
ba
ab2
b2
b
b3
e
and bh a = abh (h = 1, 2, 3, 4).

Example. The symmetry group of the square (see Fig. 1.2).
C
ba(M )
b(M )
b2 (M )
a(M )
Oz
ab2 (M )
D
b3 (M )
ab(M )
Figure 1.2: Symmetry group of the square; b is a rotation of /2 around the axis
Oz, a a symmetry with respect to the axis Ox in the plane ABCD and M an
arbitrary point of the square.
1.3 Quaternion group

The quaternion group (denoted Q) was discovered by William Rowan Hamilton
in 1843 and is constituted by the eight elements 1, i, j, k satisfying the
relations
i2 = j 2 = k 2 = ijk = 1,
ij = ji = k,
jk = kj = i,
ki = ik = j,
with the multiplication table
1
1
i
i
j
j
k
k
1
1
1
i
i
j
j
k
k
1
1
1
i
i
j
j
k
k
i
i
i
1
1
k
k
j
j
i
i
i
1
1
k
k
j
j
j
j
j
k
k
1
1
i
i
j
j
j
k
k
1
1
i
i
k
k
k
j
j
i
i
1
1
k
k
k
j
j
i
i
1
1
the element of the rst column being the rst element to be multiplied and 1 being
the identity element. The element 1 is of order 2 (i.e., its square is equal to 1)
and the elements (i, j, k) are of order 4. The subgroups of Q are
(1)
(1, 1)
(1, 1, i, i)
(1, 1, j, j)
(1, 1, k, k).
1.4 Quaternion algebra H

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

H) via the multiplication
ab = (a0 b0 a1 b1 a2 b2 a3 b3 )
+ (a0 b1 + a1 b0 + a2 b3 a3 b2 )i
+ (a0 b2 + a2 b0 + a3 b1 a1 b3 )j
+ (a0 b3 + a3 b0 + a1 b2 a2 b1 )k
and in a more condensed form
ab = (a0 b0 a b, a0 b + b0 a + a b)
where a b = (a1 b1 + a2 b2 + a3 b3 ) and a b = (a2 b3 a3 b2 )i + (a3 b1 a1 b3 )j +
(a1 b2 a2 b1 )k are respectively the usual scalar and vector products. Historically,
these two products were obtained by W. J. Gibbs [17] by taking a0 = b0 = 0 and
by separating the quaternion product in two parts.
The quaternion algebra constitutes a noncommutative eld (without divisors
of zero) containing R and C as particular cases. Let a = a0 + a1 i + a2 j + a3 k be a
quaternion, the conjugate of a, the square of its norm and its norm are respectively
ac = a0 a1 i a2 j a3 k,
2
|a| = aac = a20 + a21 + a22 + a23 ,

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

1 2
a
1
a = ac
= 2,
aac aac
|a|
a1 =
(a1 a2 an )1 =
(a1 a2 an )c
|a1 a2 an |2
1
1
= a1
n an1 a1 .
10
To divide a quaternion a by the quaternion b (= 0), one simply has to resolve the
equation
xb = a
or
by = a
with the respective solutions
x = ab1 = a
y = b1 a =
and the relation |x| = |y| =
bc
|b|2
bc a
2
|b|
|a|
|b| .
Examples. Consider the quaternions a = 2 + 4i 3j + k and b = 5 2i + j 3k;

1. the vectorial parts a, b and the conjugates ac , bc are
a = 4i 3j + k,
ac = 2 4i + 3j k,
2. the norms are given by
b = 2i + j 3k,
bc = 5 + 2i j + 3k;
aac = 30,

bbc = 39;
|a| =
|b| =
3. the inverses are
a1
b1
ac
2
|a|
bc
|b|
2 4i + 3j k
,
30
5 + 2i j + 3k
;
39
4. one can realize the following operations

a+b
ab
= 7 + 2i 2j 2k,
= 3 + 6i 4j + 4k,
ab
= 24 + 24i 3j 3k,
ba
= 24 + 8i 23j + k,
(ab)1 =
(ab)c
|ab|
S(x)
= b1 a1 =
4
4i
j
k
+
+
,
195 195 390 390
8i
9j
13k
2
),
15
15 10
30
3k
2 16i 7j
),
= b,
y = a1 b = (
15 15
30 10
13
= S(y),
|x| = |y| =
.
10
xa = b,
ay
1 170,
|ba| = 1 170,
|ab| =
x = ba1 = (
11
1.4.2 Polar form

Any nonzero quaternion can be written
a = a0 + a1 i + a2 j + a3 k
= r(cos + u sin ),
0 2

with r = |a| = a20 + a21 + a22 + a23 being the norm of a and

a21 + a22 + a23
a0
cos = ,
sin =
,
r
r
|a|
a0
tan = ;
cot = ,
|a|
a0
the unit vector u (uuc = 1) is given by
u=
(a1 i + a2 j + a3 k)

a21 + a22 + a23
with a21 + a22 + a23 = 0. Since u2 = 1, one has via the De Moivre theorem
an = rn (cos n + u sin n).
Example. Consider the quaternion a; let us determine its polar form
a
|a| =
tan
Answer: a
3 + i + j + k,
12 = 2 3,
|a| = 3,
|a|
1
= 30 ,
= ,
a0
3

i+j+k
2 3 cos 30 +
sin 30 .
3
1.4.3 Square root and nth root

Square root
The square root of a quaternion a = a0 + a1 i + a2 j + a3 k can be obtained algebraically as follows. The equation b2 = a with b = b0 + b1 i + b2 j + b3 k leads to the
following equations
b20 b21 b22 b23 = a0
(1.1)
2b0 b1 = a1 ,
2b0 b2 = a2 ,
sgn(b0 b1 ) = sgn(a1 ),
sgn(b0 b2 ) = sgn(a2 ),
(1.2)
(1.3)
2b0 b3 = a3 ,
sgn(b0 b3 ) = sgn(a3 ).
(1.4)
12
Writing t = b20 the above equation (1.1) leads to

t
a21 + a22 + a23

= a0
4t
or
t2 a 0 t
a21 + a22 + a23

= 0.
4
One obtains

a20 + a21 + a22 + a23
t= =
0
2
a0 a20 + a21 + a22 + a23
(b21 + b22 + b23 ) = a0 b20 =
,
2
b20
hence
b0 =
2
a0 +

a20 + a21 + a22 + a23 + a0
( = 1).
The equations (1.2), (1.3), (1.4) lead to

a21
,
4
(a22 + a23 )
b20 (b22 b23 ) =
,
4
b20 (b21 ) =
(1.5)
(1.6)
with
b22 b23 = b21 + a0 b20

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

t+
a0
thus
a2
t= 1
2

2

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

a20 + a21 + a22 + a23 a0
a21 + a22 + a23
13
hence b1 (one proceeds similarly for b2 , b3 ); nally, one obtains

a20 + a21 + a22 + a23 + a0

( = 1),
b0 =
2

a1
b1 = 2
a20 + a21 + a22 + a23 a0 ,
2 a1 + a22 + a23

a2
b2 = 2
a20 + a21 + a22 + a23 a0 ,
2 a1 + a22 + a23

a3
b2 = 2
a20 + a21 + a22 + a23 a0 .
2 a1 + a22 + a23
Example. Consider the quaternion a = 1 + i + j + k ; nd its square root b
b0
Answer: b =
3,
b1 = b2 = b3 = ,
2
6

j
k
1
i
3+ + +
.
2
3
3
3
nth roots
The nth root of a quaternion a = r(cos + u sin ), where can always be chosen
within the interval [0, ] with an appropriate choice of u, is obtained as follows [9].
1. Supposing sin = 0, the equation bn = a with b = R(cos +e sin ), [0, ],
leads to
Rn = r, cos n = cos , sin n = sin , e = u
and thus to
1
R = rn,
( + 2k)
n
(k = 0, 1, . . . , n 1);
nally, one has

( + 2k)
1
( + 2k)
n
b = r cos
+ u sin
n
n
(k = 0, 1, . . . , n 1).
2. When sin = 0, the vector e in b is arbitrary. If a > 0, one has = 0 and

thus = 2m
n (m
= 0, 1, . . . , n 1). For n = 2, one obtains = 0, and thus
the real roots a. With n > 2, certain values of (= 0 or ) give nonreal
roots, the vector e being arbitrary. With a < 0, one has = , = (2m+1)
n
(m = 0, 1, . . . , n 1), certain values of = give nonreal roots b , the vector
e being arbitrary.
14
Example. Find the cubic root of

a
= 3+i+j+k

(i + j + k)
sin 30 ;
= 2 3 cos 30 +
3
Answer: b =
13
(i + j + k)
sin ,
2 3
cos +
3
10 , 130 , 250 .
1.4.4 Other functions and representations of quaternions

The exponential ea is dened by [30]
a
a2
a3
+
+
+
1!
2!
3!
where a is an arbitrary quaternion. Furthermore, one denes
ea = 1 +
a2
a4
a2p
ea + ea
=1+
+
+ +
,
2
2!
4!
(2p)!
ea ea
a
a3
a5
a2p+1
sinh a =
=
+
+
+ +
,
2
1!
3!
5!
(2p + 1))!
cosh a =
and thus ea = cosh a + sinh a. For an arbitrary quaternion a, let U (V (a)) = u

(u2 = 1) be a unit vector, and therefore one has ua = au ; consequently, one
can dene the trigonometric functions
a2
a4
eua + eua
=1
+
+ ,
2
2!
4!
eua eua
a
a3
a5
sin a =
=
+
+ .
2
1!
3!
5!
Example. Let a = u be a quaternion without a scalar part with u Vec H,
u2 = 1 and real; one has
cos a =
cosh u
sinh u
=
=
cos ,
u sin ,
eu
cos + u sin ,
cos u
sin u
=
=
cosh ,
u sinh .
In particular, if a = i,
cosh i = cos ,
sinh i = i sin ,
cos i = cosh ,
sin i = i sinh .
1.5. Classical vector calculus
15
One can represent a quaternion a = a0 + a1 i + a2 j + a3 k by a 2 2 complex

matrix (with i being the usual complex imaginary)

a0 + i a3 i a1 + a2
A=
i a1 a2 a0 i a3
or by a 4 4 real matrix
a0
a1
A=
a2
a3
a1
a0
a3
a2
a2
a3
a0
a1
a3
a2
.
a1
a0
The dierential of a product of quaternions is given by

d(ab) = (da)b + a(db),
a, b H,
the order of the factors having to be respected.
1.5 Classical vector calculus

1.5.1 Scalar product and vector product
Let a, b, c Vec H, be three quaternions without a scalar part, a = a1 i + a2 j + a3 k,
b = b1 i + b2 j + b3 k, c = c1 i + c2 j + c3 k. The norm of a is
|a| = aac = a21 + a22 + a23

and
ab + ba ab ba
+
2
2
= (a b, a b).
ab =
One denes respectively the scalar product and the vector product of two vectors
a, b by
(ab + ba)
= a1 b 1 + a2 b 2 + a3 b 3 ,
2
(ab ba)
abab=
2
= (a2 b3 a3 b2 )i + (a3 b1 a1 b3 )j + (a1 b2 a2 b1 )k,
(a, b) a b =
with ab = (a, b) + a b. Geometrically, one has

(a, b) = |a| |b| cos ,
|a b| = |a| |b| sin ,
16
being the angle between the two vectors a and b. Furthermore,

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

= [(a, b) + a b] [(a, b) a b]
= (a, b)2 (a b)2
2
= (a, b)2 + |a b|
which is coherent with the above geometrical expressions. One has the properties
(a, b) = (b, a),
(a, b) = (a, b),
R,
(a, b + c) = (a, b) + (a, c),

a b = b a,
a b = (a b),
a (b + c) = a b + a c.
1.5.2 Triple scalar and double vector products

The triple scalar product is dened by
[a(b c) + (b c)a]
2
[a(bc cb) + (bc cb)a]
=
2
(a, b c) =
and satises the relations

(a, b c) = (b, c a) = (c, a b),
(a, b c) = (c, b a)
which are established using the relations
S(abc) = S(bca) = S(cab)
and
S(abc) = S(abc)c = S(cba) = S(acb),
S(abc) = S(bac).
In particular
(a b, a) = (a b, b) = 0
which shows that a b is orthogonal to a and b.
1.6. Exercises
17
To derive the expression of the double vector product a (b c), one can
start from
abc = a [(b, c) + b c]
= a(b, c) (a, b c) + a (b c)
hence
V (abc) = a(b, c) + a (b c);
since
abc (abc)c
2
abc + cba
=
2
[abc + bac bac bca + bca + cba]
=
2
(ab + ba)c b(ac + ca) (bc + cb)a
=
+
2
2
2
= (a, b)c + b(a, c) a(b, c)
V (abc) =
one obtains
a (b c) = b(a, c) c(a, b).
Knowing that
(a b) c = c (a b)
= a(c, b) + b(c, a),
one notices that the vector product is not associative. From the above relations,
one then obtains (with a, b, c, d Vec H)
(a b, c d) = (a, c)(b, d) (a, d)(b, c),
(a b) (c d) = c(a b, d) d(a b, c).
1.6 Exercises
E1-1 From the formulas i2 = j 2 = k 2 = ijk = 1, deduce the multiplication
table of the quaternion group knowing that the element 1 commutes with all
elements of the group and that (1)i = i, (1)j = j, (1)k = k.
E1-2 Consider the quaternions a = 1 + i, b = 4 3j. Compute
, |b|, a1 , b1 ,
|a|
1/3
a + b, ab, ba. Give the polar form of a and b. Compute a, b, a .
E1-3 Solve in x H, the equation ax + xb = c (a, b, c H, aac = bbc ).
N.A. : a = 2i, b = j, c = k, determine x.
E1-4 Solve in x H, the equation axb + cxd = e (a, b, c, d, e H).
N.A. : a = 2i, b = j, c = k, d = i, e = 3k. Find x.
E1-5 Solve in x H, the equation x2 = xa + bx (a, b H).
N.A. : a = 2j, b = k, determine x.
Chapter 2
Rotation groups SO(4) and SO(3)

In this chapter, the formulas of the rotation groups SO(4) and SO(3) are established from orthogonal symmetries. The crystallographic groups and Keplers
problem are then examined as applications of these groups.
2.1 Orthogonal groups O(4) and SO(4)

Consider two elements of a four-dimensional vector space x = x0 + x1 i + x2 j + x3 k,
y = y0 + y1 i + y2 j + y3 k H, and the scalar product
(x, x) = xxc = x20 + x21 + x22 + x23 .
One deduces from it
(x + y, x + y) = (x, x) + (y, y) + (x, y) + (y, x)
and postulating the relation (x, y) = (y, x), one obtains
1
[(x + y, x + y) (x, x) (y, y)]
2
1
= [(x + y)(x + y)c xxc yyc ]
2
1
= (xyc + yxc )
2
= x0 y0 + x1 y1 + x2 y2 + x3 y3 .
(x, y) =
Two quaternions x, y are orthogonal if (x, y) = 0; a quaternion is unitary if (x, x) =

1. A hyperplane is dened by the relation (a, x) = 0 where a is a quaternion
perpendicular to the hyperplane. The expression of a plane symmetry is obtained
as follows.
Denition 2.1.1. The symmetric x of x with respect to a hyperplane is obtained
by drawing from x the perpendicular down to the hyperplane and by extending
this perpendicular line by an equal length [11].
20
Chapter 2. Rotation groups SO(4) and SO(3)
We shall assume that the hyperplanes go through the origin. The vector
x x is perpendicular to the hyperplane (and thus parallel to a) and (x + x)/2
is perpendicular to a. Hence, the relations
x = x + a,

x + x
a,
= 0;
2
one then deduces

a
a, x +
= 0,
2
R,
2(a, x)
,
(a, a)
2(a, x)a
x = x
(a, a)
(axc + xac )a
=x
aac
axc a
=
.
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 inversion
is an improper transformation of a determinant equal to 1. In combining, in an
even number, plane symmetries
x = mxc m,
with mmc = 1, one obtains the rotation group SO(4),
x = axb,
with a, b H and aac = bbc = 1. By including the inversions (odd number of
symmetries) the expression of which are
x = axc b
with a, b H and aac = bbc = 1, one obtains the orthogonal group O(4) with six
parameters. This group, by denition, conserves the scalar product; indeed, for
SO(4),
1
[x yc + y xc ]
2
1
= [axbbc yac + aybbc xac ]
2
= (x, y)
(x , y ) =
2.1. Orthogonal groups O(4) and SO(4)
21
and for the improper rotations

1
[x yc + y xc ]
2
1
= [axc bbc yac + ayc bbc xac ]
2
= (x, y).
(x , y ) =
Any rotation of SO(4) can be written as a combination of a rotation of SO(3),

examined below,
rrc = 1
x = rxrc ,
and a transformation
x = axa,
a H, aac = 1.
It is sucient to resolve the equation

x = f xg = arxrc a
(or r a x a rc )
with f = ar, g = rc a and rrc = aac = 1. Writing the relations ([15], [16])
2a2 = 2f g,
a2 f g = (f g)2 ,
a2 (f g)c = 1,
and adding, one obtains
a2 [2 + f g + (f g)c ] = (1 + f g)2 ;
hence, the solution
a=
(1 + f g)
.
|(1 + f g|
The rotation is given by

r = ac f
(1 + gc fc ) f
|(1 + f g|
(f + gc )
=
.
|(1 + f g|
=
One veries that one has indeed the relations aac = rrc = 1. One solves similarly
the equations f = r a , g = a rc with the solutions
(1 + gf )
,
|(1 + gf |
(f + gc )
r =
,
|(1 + gf |
a =
with |(1 + f g| = |(1 + gf | since S(gf ) = S(f g).
22
2.2 Orthogonal groups O(3) and SO(3)

Consider the vectors x = x1 i + x2 j + x3 k, x = x1 i + x2 j + x3 k H, constituting
a subvectorspace of H. A plane symmetry, in this subspace, is a particular case of
the preceeding one and is expressed by
x = axc a
= axc ac
with a Vec H, aac = 1, ac = a. The improper rotations (odd number 3 of
plane symmetries) are given by
x = f xc fc
with f H, f fc = 1 and one has x xc = xxc . The SO(3) group is the set of proper
rotations (even number 3 of symmetries). In combining two symmetries, one
obtains
x = axc a,
x = bxc b,
hence
x = (bac )x(ac b)
= rxrc
with r (= bac ) H, rc = abc = ac b, rrc = 1. The unitary quaternion r can be
expressed in the form

r = cos + u sin
2
2
where the unit vector u (u2 = 1) is the axis of rotation (going through the origin)
and the angle of rotation of the vector x around u ( is taken algebraically given
the direction of u and using the right-handed screw rule). The conservation of the
norm of x results from
x xc = rxrc rxc rc = xxc .
Furthermore, if one considers the transformation
q = rqrc
with q H, one has
S(q ) = S(rqrc ) = S(rc rq) = S(q)
2.2. Orthogonal groups O(3) and SO(3)
23
which shows that the scalar part of the quaternion is not aected by the rotation.
The set of proper and improper rotations constitute the group O(3). In developing
the formula x = rxrc with x = x Vec H, one obtains

x = x = cos + u sin
x cos u sin
2
2
2
2
= cos2 x + sin2 [(u x)u (u x)u] + u x sin ;

2
2
furthermore
(u x)u = (u x) u + (u x) u
= (u x) u
= x (u x)u;
hence, the classical formula [26, p. 165]
x = x cos + u(u x)(1 cos ) + u x sin .
In matrix form, this equation can be written x = Ax with
0
0
x1
x1

,
,
x
=
x=
x2
x2
x3
x3

2
u1
u1 u2 (1 cos )
u1 u3 (1 cos )
+(u22 + u23 ) cos
u3 sin
+u2 sin

2

u
u
(1
cos
)
u
(1
cos
)
u
u
1 2
2 3
2
;
A=
2
2
+u
u
sin
+(u
+
u
)
cos
sin
3
1
1
3

u u (1 cos )
u2
u u (1 cos )
1 3
u2 sin
2 3
+u1 sin
+(u21 + u22 ) cos
one veries that the matrix A is orthogonal (t A = A1 ). If one combines the

rotation r1 and the rotation r2 ,

,
r2 = cos + b sin
r1 = cos + a sin
2
2
2
2

r = r2 r1 = cos + c sin
2
2
one obtains
with
= cos cos (a, b) sin sin ,

2
2
2
2
2
c sin = a sin cos + b cos sin (a b) sin sin ,

2
2
2
2
2
2
2
cos
24
which yields the Rodriguez formula

c tan
a tan 2 + b tan 2 (a b) tan 2 tan 2
=
.
2
1 (a, b) tan 2 tan 2
2.3 Crystallographic groups

If r belongs to a nite subgroup of real quaternions, the transformations q = rqrc
will constitute a subgroup of SO(3), with r and r generating the same rotation.
The nite subgroups of real quaternions ([48], [44]) are of ve types.
2.3.1 Double cyclic groups Cn (order N = 2n)

The elements of these groups are given by

2h
2h
n
r = u n = cos + u sin
2
2

h
= cos + u sin
= bh
n
n
with b = cos n + u sin n , 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

2h
2h
n
r = k n = cos + k sin
2
2

h
= cos + k sin
.
n
n
Example. Double group C3 (N = 6, rotation axis along Oz); the elements of the
group are

h
r = bh = cos + k sin
,
h = 1, . . . , 6
3
3
or explicitly
b = 1 (1 + k 3), b2 = 1 (1 + k 3), b3 = 1,
2
2
.
b4 = b, b5 = b2 , b6 = 1
2.3.2 Double dihedral groups Dn (N = 4n)

These groups are constituted by the elements
2h
r = u n al = b h al
with

u = cos + u sin
,
2
2

a = cos + a sin
,
2
2
b = un,
2.3. Crystallographic groups
25
where u, a are two perpendicular vectors and S(au) = 0, a2 = 1, h = 1, . . . , 2n,

l = 1, . . . , 4.
1. Double group D3 (order 12)

1
= (1 + k 3),
b = cos + k sin
3
3
2
1
1
a = i,
ab = (i j 3),
ba = (i + j 3);
2
2

the elements of the group are b, b2 , b3 , a, ab, ba .
Examples.
2. Double group D4 (order 16); writing

1
,
b2 = (k),
b = cos + k sin
b3 = (1 + k),
4
4
2
1
1
a = (i),
ab = (i j),
ba = (i + j),
2
2
ab2 = (j),
a2 = (1),
a3 = (i),
a4 = 1,
the group is constituted by the elements

b, b2 , b3 , b4 , a, ab, ba, ab2 .
2.3.3 Double tetrahedral group (N = 24)

This group is constitued by the 24 elements
1, i, j, k,
1
(1 i j k).
2
More explicitly, indicating the axes of multiple rotations, the group is composed of the elements
i , j , k ,

1+i+j+k
,
2

1+ijk
,
2

1ij+k
2
1i+jk
2

,

,
with = 1, 2, 3, 4, = 1, 2, 3, 4, 5, 6 (N = 24).
Example. Consider the tetraeder having as vertices
i+j+k
,
3
i j + k
,
3
ijk
,
3
i + j k
,
3
26
as face centers
(i + j + k)
,
3
(i j + k)
,
3
(i j k)
,
3
(i + j k)
,
3
and as side centers

i,
j,
k;
by taking for r the elements of the above group, the transformation x = rxrc
transforms the tetraeder in itself.
2.3.4 Double octahedral group (N = 48)

The group is composed by the 48 elements
1, i, j, k,
1
(1 i j k),
2
1
1
(1 i) , (1 j) ,
2
2
1
1
(i j) , (j k) ,
2
2
1
(1 k) ,
2
1
(i k) .
2
Making explicit the axes of multiple rotations, the elements of the group are

1+j

1+k
,
,
,
2

1ij+k
1+i+j+k
,
,
2
2

1i+jk
1+ijk
,
,
2
2

i+k
i+j
j+k
,
,
,
2
2
2

i + k
ij
jk
,
,
,
2
2
2
1+i
with = 1, . . . , 8, = 1, . . . , 6, = 1, . . . , 4 (N = 48).
Example. Consider the octaeder having, in an orthonormal frame, for its 6 vertices
the coordinates i, j, k, and for the centers of the 8 faces
i+j+k
,
3
i j + k
,
3
ijk
,
3
i + j k
,
3
2.3. Crystallographic groups
27
for the middle points of the 12 sides

jk
,
2
(i + k)
,
2
ij
.
2
The octaeder transforms into itself under the rotation x = rxrc , r being taken in
the double octahedral group. The same property applies to the cube (dual of the
octaeder) the 8 vertices of which are the centers of the faces of the above octaeder.
2.3.5 Double icosahedral group (N = 120)

The 120 elements of this group are
i , j , k ,

i + m j + mk
,
2

i m j + mk
,
2

i + m j mk
,
2

i m j mk
,
2

1+i+j+k
,
2

1+ijk
,
2

1 + mj + m k
,
2

1 + mj m k
,
2

m + m j + k
,
2

m + m j k
,
2

mi + j + m k
m i + mj + k
,
,
2
2

mi + j m k
m i mj + k
,
,
2
2

mi + j m k
m i mj + k
,
,
2
2

mi + j + m k
m i + mj + k
,
,
2
2

1ij+k
,
2

1i+jk
,
2

1 + m i + mk
1 + mi + m j
,
,
2
2

1 m i + mk
1 + mi m j
,
,
2
2

m + i + m k
m + m i + j
,
,
2
2

m i + m k
m + m i j
,
,
2
2
with = 1, . . . , 4, = 1, . . . , 6, = 1, . . . , 10 and
1+ 5
= 2 cos 36 ,
m=
2
1 5
= 2 cos 72 .
m =
2
28
One obtains another group, distinct from the rst, by inverting m and m .
Example. Consider the icosaeder having, in an orthonormal frame, for the coordinates of the 12 vertices
m j k
,
m 5
(i + m k)
,
m 5
m i j
;
m 5
for the centers of the 20 faces
i+j+k
i j + k
ijk
i + j k
,

,
,
3
3
3
3
mj m k
(m i + mk)
mi m j
,

;
3
3
3
for the middles of the 30 sides

i,
j,
k,
(mi + j m k)
,
2
i m j mk
,
2

(m i mj + k)
.
2
The transformation x = rxrc transforms the icosaeder into itself; the same is true
for its dual, the dodecaeder (having 20 vertices, 12 faces and 30 sides) the vertices
of which are the centers of the faces of the icosaeder.
The ve 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 Innitesimal transformations of SO(4)

Among the subgroups of SO(4) one has, in particular, the transformations
x = rxrc ,
x = axa,
x = f x,
x = xg
with r, a, f, g H, rrc = aac = f fc = ggc = 1, x = (x0 , x) and x = (x0 , x )

H. For an innitesimal rotation of SO(3) of d around the unit vector u =
u1 i + u2 j + u3 k, one has

d
d
d
+ u sin
r = cos
1+u
2
2
2
and

d
d
x = rxrc = 1 + u
x 1u
2
2
d
(ux xu) = x + d u x;
=x+
2
2.4. Innitesimal transformations of SO(4)

hence
29
dx = x x = d u x.
In matrix notation, one obtains

i (1, 2, 3)
dx = dui Mi x,
x0
x1
x=
x2 ,
x3
with
and
0
0
M1 =
0
0
0
0
0
0
0 0
0 0
,
0 1
1 0
x0
x1
x =
x2 ,
x3
0 0
0 0
M2 =
0 0
0 1
0
0
0
0
0
1
,
0
0
0
0
M3 =
0
0
0 0 0
0 1 0
.
1 0 0
0 0 0
Introducing the commutator of two matrices A and B,

[A, B] = AB BA,
one veries the relations
[M1 , M2 ] = M3 ,
[M1 , M3 ] = M2 ,
[M2 , M3 ] = M1 ,
or
[Mi , Mj ] = ijk Mk .
(2.1)
Concerning the subgroup x = axa, with a H and aac = 1, one has for the
innitesimal transformation

d
d
d
+ v sin
a = cos
1+v
;
2
2
2
hence
x = axa =
= x+
and

1+v
d
2

d
x 1+v
2
d
(vx + xv) = x + d (v x + x0 v)
2
dx = x x = d (v x + x0 v) .
30
In matrix notation, one obtains

i (1, 2, 3)
dx = dv i Ni x,
with
0 1 0 0
1 0 0 0
N1 =
0 0 0 0
0 0 0 0
x0
x1
x=
x2 ,
x3
0
0
, N2 =
1
0
0
0
0
x0
x1
x =
x2
x3
1 0
0 0
,
0 0
0 0
0
0
N3 =
0
1
0
0
0
0
0 1
0 0
;
0 0
0 0
the matrices Ni satisfy the relations

[N1 , N2 ] = M3 ,
[N1 , N3 ] = M2 ,
[N2 , N3 ] = M1 ,
or, more concisely
[Ni , Nj ] = ijk Mk
(2.2)
where Mi are the matrices dened previously; furthermore

[Ni , Mj ] = ijk Nk .
(2.3)
Concerning the subgroup x = f x, one obtains with f 1+ d

2 f , f = f1 i+f2 j +f3 k
(unit vector)
d
[f x + x0 f + f x] .
dx = x x =
2
In matrix notation, one has
dx = df i Fi x
with
0
1
1
F1 =
2 0
0
1
0
0
0
0 0
0 0
,
0 1
1 0
0
1
0
F3 =
2 0
1
0
0
1
0
0 0 1 0
1 0 0
0 1
,
F2 =
1
0
0 0
2
0 1 0 0
0 1
1 0
0
0
0
0
2.4. Innitesimal transformations of SO(4)
31
and
[F1 , F2 ] = F3 ,
[F1 , F3 ] = F2 ,
[F2 , F3 ] = F1 ,
or
[Fi , Fj ] = ijk Fk .
As to the subgroup x = xg, with g 1 + d
2 g, one proceeds similarly and obtains
dx = x x =
d
[g x + x0 g g x] .
2
In matrix notation,
dx = dg i Gi x
with
0
1
1
G1 =
2 0
0
1 0
0
0
0
0
0 1
0
0
,
1
0
0
1
0
G3 =
2 0
1
0
1
0
G2 =
2 1
0
0
0
1
0
0
0
0
1
0 1
1 0
0 0
0 0
and
[G1 , G2 ] = G3 ,
[G1 , G3 ] = G2 ,
[G2 , G3 ] = G1 ,
or
[Gi , Gj ] = ijk Gk .
Furthermore, one has
[Fi , Gj ] = 0,
i, j = 1, 2, 3
the matrices Mi , Ni , Fi , Gi satisfying the relations

Mi = Fi Gi ,
Ni = Fi + Gi .
1 0
0 1
,
0
0
0
0
32
2.5 Symmetries and invariants: Keplers problem

Let us consider the motion of a particle of mass m, of momentum p and gravitating
at the distance r of a mass M . The Hamiltonian is given by
H=
k
p2
,
2m r
k = GM m
and the equations of motion are

qi =
H
,
pi
pi =
H
.
qi
Let F (qi , pi , t) be a physical quantity of the motion; one has

dF
F
F qi
F pi
=
+
+
dt
t
qi t
pi t

F Hi
F
F H
=
+
.
t
qi pi
pi qi
Introducing Poissons bracket of two functions u and v
[u, v] =
one obtains
u v
u v
qi pi pi qi
F
dF
=
+ [F, H] .
dt
t
dF
If [F, H] = 0 and F
t = 0, one has dt = 0, F is then an invariant of the motion.
The angular momentum L = r p and the Laplace-Runge-Lenz vector
A= pL
kmr
r
satisfy the equations

[L, H] = 0,
[A, H] = 0,
and thus are invariants. Let us consider a bound motion (with a total negative
energy E < 0) and introduce the vector
A
D=
2mE
with
E=
p2
k
2m r
2.5. Symmetries and invariants: Keplers problem
33
verifying the Poisson bracket [D, H] = 0; one has the relations

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

km
2
A= r p
p (p r)
r
with
A
= rp0 pr0
D=
2mE
and
p2 km
pr
r
p0 =
,
r0 =
.
2mE
2mE
One veries immediately that D L = 0; furthermore,
2
(L) = (r p) = r2 p2 (r p) ,
hence
2
(D) + (L) =
or
H=
k 2 m2
2mE
k 2 m2

.
2
2
2 (D) + (L)
Introducing two quaternions

r = (r0 , r),
p = (p0 , p)
and the quaternion

K =rp=
1
(rpc prc ) = (0, D L)
2
one has
2
KKc = |0, D L| = D2 + L2 ,
and thus
H=
k 2 m
2
2 |r p|
which shows explicitly the invariance of the Hamiltonian with respect to a transformation of SO(4) of the type K = aKb (a, b H, aac = bbc = 1) leading to
K Kc = KKc .
34
2.6 Exercises
E2-1 From the general formula
x = rxrc ,

r = cos + u sin
2
2
x Vec H, give in an orthonormal direct frame the matrix representation X =

AX of a rotation of angle around the Ox axis of a point M (x, y, z), of a rotation
of angle around the Oy axis, of a rotation of angle around the Oz axis. Show
that the matrices are orthogonal A1 = At , det A = 1 (At : transposed matrix of
A).
E2-2 Consider the relation A = rA rc where A are the components of a vector
with respect to an orthonormal frame at rest and A its components with respect to
a mobile orthonormal frame. Show that dA = rDA rc where DA is the covariant
dierential with
DA = dA + d A ,
d = 2rc dr
(dA is the dierential with respect to the components only). What does
dr
d
= 2rc
dt
dt
represent?
E2-3 Consider the relations
A = rA rc = gA gc
with
dA
= rDA rc = gDA gc ,
DA
= dA + d A
DA
= dA + d A
(d = 2rc dr),

(d = 2gc dg)
(A represents the components of a vector with respect to an orthonormal basis at

rest, A , A the components with respect to mobile orthonormal bases.). Find the

relation between d and d .
E2-4 Using the covariant derivative, express the velocity and the acceleration in
polar coordinates (, ) and in cylindrical coordinates (, , z).
E2-5 Express r in the relation X = rX rc for spherical coordinates with X =
xi + yj + zk and X = i. Determine d = 2rc dr and d = 2(dr)rc . Express the
basis vectors ei in the basis at rest. Find the velocity and the acceleration in the
mobile basis.
2.6. Exercises
35
E2-6 Consider the Euler basis (O, x , y , z ) obtained via the following successive
rotations (Figure 2.1). A rst rotation of angle (precession angle) around k
transforms the basis i, j, k into the basis i , j , k . A second rotation of angle
(nutation angle) around the vector i transforms i , j , k into i , j , k . A third
rotation of angle (proper rotation angle) around the vector k transforms the
basis i , j , k into the basis e1 , e2 , e3 = k . Give the quaternion r of the rotation
X = rX rc . Determine
= 2rc
dr
,
dt
=2
dr
rc .
dt
Give the components of the basis vectors ei .

z
y
z
k = e3
e2
e1
i
y
x
Figure 2.1: Eulers angles: is the angle of precession, the angle of nutation and
the angle of proper rotation.
Chapter 3
Complex quaternions
From the very beginning of special relativity, complex quaternions have been used
to formulate that theory [45]. This chapter establishes the expression of the Lorentz
group using complex quaternions and gives a few applications. Complex quaternions constitute a natural transition towards the Cliord algebra H H.
3.1 Algebra of complex quaternions H(C)

A complex quaternion is a quaternion a = a0 + a1 i + a2 j + a3 k having complex
coecients. Such a quaternion can be represented by the matrix

A=
a0 + i a3
i a1 a2
i a1 + a2
a0 i a3
with the basis

1=
1
0
0
1
i=
0
i
i
0
j=
0 1
1 0
k=
i
0
0
i
where i is the usual complex imaginary and ai C. The algebra of complex

quaternions H(C) is isomorphic to 2 2 matrices over C and has zero divisors;
indeed, the relation
(1 + i k)(1 i k) = 1 (i )2 k 2 = 0
shows that the product of two complex quaternions can be equal to zero without
one of the complex quaternions being equal to zero.
38
Chapter 3. Complex quaternions
3.2 Lorentz groups O(1, 3) and SO(1, 3)

3.2.1 Metric
With the advent of the special theory of relativity, space and time have been united
into a four-dimensional pseudoeuclidean spacetime with the relativistic invariant
2 2 2
c2 t2 x1 x2 x3 .
Minkowski had the idea of using complex quaternions
x = (i ct, x1 , x2 , x3 ) = (i ct + x)
with the invariant
2 2 2
xxc = c2 t2 + x1 + x2 + x3 .
To get the signature (+ ), we shall use complex quaternions in the form

x = (ct, i x1 , i x2 , i x3 ) = (ct + i x)
which we shall call minkowskian quaternions or minquats with the invariant
2 2 2
xxc = c2 t2 x1 x2 x3 .
A minquat is a complex quaternion such that xc = x where is the complex
conjugation.
3.2.2 Plane symmetry

Let us consider two minquats x = (x0 + i x), y = (y 0 + i y) with the scalar product
2 2 2 2
xxc = x0 x1 x2 x3 .
One has
(x + y, x + y) = (x, x) + (y, y) + (x, y) + (y, x);
postulating the property (x, y) = (y, x) one obtains
1
[(x + y, x + y) (x, x) (y, y)]
2
1
= [(x + y)(x + y)c xxc yyc ]
2
1
= (xyc + yxc )
2
= x0 y 0 x1 y 1 x2 y 2 x3 y 3 .
(x, y) =
Two minquats x, y are said to be orthogonal if (x, y) = 0. A minquat x is timelike

if xxc > 0 and unitary if xxc = 1; a minquat x is spacelike if xxc < 0 and unitary if
xxc = 1. A hyperplane is dened 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
Denition 3.2.1. The symmetric x of x with respect to a hyperplane x is obtained

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

x + x
a,
= 0;
2
R,
hence,

a
a, x +
= 0,
2
2(a, x)
,
=
(a, a)
(axc + xac )a
2(a, x)a
=x
,
x = x
(a, a)
aac
axc a
x =
.
aac
One shall distinguish time symmetries (with aac = 1)
x = axc a
from space symmetries (aac = 1)
x = axc a.
3.2.3 Groups O(1, 3) and SO(1, 3)

Denition 3.2.2. The pseudo-orthogonal group O(1, 3) is the group of linear operators which leave invariant the quadratic form
(x, y) = x0 y 0 x1 y 1 x2 y 2 x3 y 3
with x = (x0 + i x), y = (y 0 + i y).
Theorem 3.2.3. Any rotation of O(1, 3) is the product of an even number 4 of
symmetries; any inversion is the product of an odd number 4 of symmetries
[11].
is an improper transformation of a determinant equal to 1.
40
Proper orthochronous Lorentz transformation

Consider the transformation obtained by combining an even number of time symmetries and an even number of space symmetries. These transformations of determinant equal to +1 constitute a subgroup of O(1, 3), the special orthogonal
group SO(1, 3). Let us take for example two time symmetries (minquats f and g)
followed by two space symmetries (minquats m and n). One has

x = n m [g (f xc f )c g]c m c n
= (nmc gfc )x(fc gmc n);
writing a = nmc gfc , one obtains
ac = f gc mnc ,
ac = f gc m nc
aac = 1
= fc gmc n;
hence, the formula is valid in the general case,
x = axac
with aac = 1, a H(C).
Other Lorentz transformations
Call n the number of time symmetries and p the number of space symmetries. In
combining these symmetries, one obtains the following Lorentz transformations L:
1. n even, p odd, orthochronous, improper Lorentz transformation (det L = 1)
x = axc ac
(aac = 1);
2. n odd, p odd, antichronous, proper Lorentz transformation(det L = 1)

x = axac
(aac = 1);
3. n odd, p even, antichronous, improper Lorentz transformation (det L = 1)

x = axc ac
(aac = 1).
Group O(1, 3): summarizing table

The whole set of Lorentz transformations is given in the table below where n is
the number of time symmetries and p the number of space symmetries [49].
3.3. Orthochronous, proper Lorentz group
orthochronous
antichronous
Rotation
(det L = 1)
n even, p even
x = axac
(aac = 1)
n odd, p odd
x = axac
(aac = 1)
41
Inversion
(det L = 1)
n even, p odd
x = axc ac
(aac = 1)
n odd, p even
x = axc ac
(aac = 1)
3.3 Orthochronous, proper Lorentz group

3.3.1 Properties
The proper orthochronous Lorentz transformation
x = axac
with aac = 1, x = (ct + i x), x = (ct + i x ) conserves, by denition, the norm
xxc = (axac )(a xc ac )
= xxc
and the minquat type of x

x
c = (axac )c = axc ac
= axac = x .
The composition of two transformations satises the rule
x = a2 (a1 xa1c )a2c
= a3 xa3c
with a3 = a2 a1 , a3 a3c = 1, a3 H(C). A (three-dimensional) rotation is given by
the formula
x = rxrc

with r = cos 2 + u sin 2 , rrc = 1, r H. A pure Lorentz transformation (without
rotation) corresponds to the transformation
x = bxbc

where b = cosh 2 + i v sinh 2 is a minquat such that bbc = 1 and i v is a unitary
space-vector (i v, i v) = 1. A general transformation (proper, orthochronous) is
42
obtained by combining a rotation and a pure Lorentz transformation. This can be

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

(1 + d)(1 + d)c . The rotation is given by
r = bc a
=
(a + a )
.
|1 + aac |
Finally, one veries that this is indeed a solution. For the equation a = r b , one
nds in a similar way the solution
(1 + ac a)
,
|1 + ac a|
(a + a )
r =
|1 + ac a|
b =
3.3. Orthochronous, proper Lorentz group
43
with |1 + ac a| = |1 + aac | since S(aac ) = S(ac a). One observes that in both cases
(a = br or r b ) the rotation is the same. The problem of the decomposition of
a Lorentz transformation into a pure Lorentz transformation and a rotation is
thus solved in the most general case. As an immediate application, consider the
combination of two pure Lorentz transformations b1 , b2 . The quaternion b = b2 b1
will in general be a complex quaternion (and not a minquat) and will be written
a = br. The resulting Lorentz transformation will thus contain a rotation; this is
the principle of the Thomas precession ([52], [51]).
3.3.2 Innitesimal transformations of SO(1, 3)

Consider the pure Lorentz transformation
x = bxbc

with x = (ct + i x), x = (ct + i x ), b = cosh 2 + i v sinh 2 and i v a spacelike
unitary minquat. For an innitesimal transformation,

d
d

+ i v sinh
b cosh
2
2

d
= 1 + i v
,
2
hence
x = bxbc

d
d
= 1 + i v
x 1 + i v
2
2
d
(vx + xv) ;
= x + i
2
consequently,
dx = x x
= d (v x + i vx0 ) .
Using real matrices,
x0 = ct
x1
,
X=
2
x
3
x
x0 = ct
x1
X =
2
x
3
x
one can write

dX = X X
= dv i Ki X
44
with
0
1
K1 =
0
0
1
0
0
0
0
0
0
0
0
0
,
0
0
0
0
K2 =
1
0
0
0
0
0
1
0
0
0
0
0
,
0
0
0
0
K3 =
0
1
0
0
0
0
0
0
0
0
1
0
.
0
0
The matrices Ki satisfy the relations
where Mi are
in Chapter 2,
0
0
M1 =
0
0
[Ki , Kj ] = ijk Mk ,
(3.1)
[Ki , Mj ] = ijk Kk ,
(3.2)
the matrices dened for the innitesimal transformations of SO(3)

0
0
0
0
0 0
0 0
,
0 1
1 0
0 0
0 0
M2 =
0 0
0 1
0
0
0
0
0
1
,
0
0
0
0
M3 =
0
0
0 0 0
0 1 0
.
1 0 0
0 0 0
One observes that the relations (3.1) and (3.2) are those of the unbound Kepler
problem, which identies the corresponding symmetry group as being SO(1, 3).
3.4 Four-vectors and multivectors in H(C)

Let x = (x0 + i x), y = (y0 + i y) be two four-vectors and their conjugates xc and
yc ; one can dene the exterior product
xy =
1
(xyc yxc )
2
0,
(x2 y3 x3 y2 ) + i (x1 y0 x0 y1 ) ,
(x3 y1 x1 y3 ) + i (x2 y0 x0 y2 ) ,
(x1 y2 x2 y1 ) + i (x3 y0 x0 y3 )
= [0, x y + i (y0 x x0 y)]

with x y = y x. The resulting quaternion is of the type B = (a + i b)
and is called a bivector; its real part, in the above example, gives the ordinary
vector product but its nature diers from that of a four-vector. Under a Lorentz
transformation (proper, orthochronous), a bivector transforms as
B = x y =
1
(x yc y xc )
2
1
[(axac ) (a yc ac ) (ayac ) (a xc ac )]
2
= aBac .
=
3.4. Four-vectors and multivectors in H(C)
45
The Lorentz transformation conserves the bivector type characterized by B =

Bc ,
Bc = aBc ac = aBac = B .

2
2
Furthermore, B 2 = (a) + (b) 2i a b is a relativistic invariant
B 2 = B B = aBac aBac = aB 2 ac = B 2 .
A trivector is a complex quaternion dened by
1
(xB + Bx)
2

= [i x a + x0 a + x b] = i t0 + t
T =xB =
and the product Bx by postulating Bx = xB. Under a proper, orthochronous

Lorentz transformation, one has
x B + B x
2
= aT ac
T =
that is, T transforms as a four-vector. Furthermore, the transformation conserves

the trivector type Tc = T ,
Tc = aTc ac = a(T )ac
= T ;
T yields the relativistic invariant
T Tc = aT ac a Tc ac = T Tc
2 2 2 2
= t0 t1 t2 t3 .
The exterior product, dened above, is associative
(x y) z = x (y z) .
Indeed,
(x y) z = z (x y)
1
z (xyc yxc ) + (xyc yxc ) z

=
2
1
= [z (xc y yc x) + (xyc yxc ) z] ,
2
1
x (yzc zyc) + (yzc zyc ) x

x (y z) =
2
1
= [x (yc z zc y) + (yzc zyc) z] .
2
(3.3)
(3.4)
(3.5)
(3.6)
(3.7)
46
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 dened by the
relation
1
P = x T = (xTc + T x )
2
1
= (xTc + T xc )
2
where T is a trivector; by denition, one postulates T x = x T . The pseudoscalar type is a pure imaginary P = i s characterized by Pc = P and is invariant
under a Lorentz transformation
1
P = (x Tc + T xc ) = aP ac = P.
2
Examples. Consider the basis vectors e0 = 1, e1 = i i, e2 = i j, e3 = i k; one
obtains the following table
1
i = e0 e1 e2 e3
1 = e0
i = e1 e3 e2
i = e2 e3
i i = e1 e0
i = e0 e2 e3
i i = e1
j = e3 e1
i j = e2 e0
j = e0 e3 e1
i j = e2
k = e1 e2
i k = e3 e0
k = e0 e1 e2
i k = e3
One observes that distinct quantities occupy identical places in the H(C)
algebra, a situation which one encounters also in the tridimensional vector calculus;
this problem will be solved in the Cliord algebra H H. Besides the exterior
products, one can dene interior products
x y = (x, y) =
1
(xyc + yxc ) ,
2
1
(xB Bx) four-vector,
2
1
x T = (xTc T x ) bivector;
2
xB =
by denition, one supposes B x = x B, T x = x T . More generally, one shall

dene the interior product between two multivectors Ap and Bq by [12, p. 14]
Ap Bq = (v1 v2 vp1 ) (vp Bq ) .
Hence, one sees that H(C) already allows us to develop a few notions of a multivector calculus. These notions will be developed later on in the more satisfying
framework of the Cliord algebra.
3.5. Relativistic kinematics via H(C)
47
3.5 Relativistic kinematics via H(C)

Consider the reference frame at rest K(O, x, y, z) and the reference frame
K (O , x , y , z ) moving along the Ox axis with the constant velocity u (Figure
3.1).
y
K
K
O
O
x
z
Figure 3.1: Special Lorentz transformation (pure): the axes remain parallel to
themselves and the reference frame moves along the Ox axis.
The Lorentz transformation is expressed by
X = bX bc
(3.8)

cosh 2 + i i sinh 2 , bbc = 1,
with X = (x0 + i x) , X = (x0 + i x ), b =

tanh = uc ; write = cosh , hence
2 = 1 + sinh2 = 1 +
1
=
1
u2
c2
u2 2
,
c2
Explicitly, equation (3.8) reads

u
ct = ct + x
,
c

u
,
x = x + ct
c

y=y,
z=z.
48
The inverse transformation is given by X = bc Xb and yields

u
,
ct = ct x
c

u
,
x = x ct
c
y = y,
z = z.
3.5.2 General pure Lorentz transformation

A general pure Lorentz transformation is expressed by (Figure 3.2)
X = bX bc

with b = cosh 2 + i uu sinh 2 , where u is the velocity (of norm u), X = (ct+ i x),
X = (ct + i x ), = q 1 u2 , tanh = uc . Explicitly, one obtains [26, p. 280]
1 c2

x u
ct = ct +
,
c
u
x = x + n (n x ) ( 1) + ct n .
c
y
K
y
u
O
x
z
x
Figure 3.2: General pure Lorentz transformation: the axes remain parallel to themselves but the reference frame K moves in an arbitrary direction.
3.5.3 Composition of velocities

Consider two reference frames K and K with the special Lorentz transformation
(Figure 3.1)
X = bX bc ,
3.5. Relativistic kinematics via H(C)

b = cosh 2 + i i sinh 2 , tanh =
49
u
c,
= cosh =
four-velocity V transforms as
q 1
2
1 u
c2
( constant). The
dX
dX
=b
b
ds
ds c

with the relativistic invariant ds = c2 dt2 dx2 dy 2 dz 2 ; ds can also be
expressed as
!

cdt
v2
cdt
2
2
ds = c dt 1 2 =
= .
c
The four-velocity satises the relation

V Vc =
dX dXc
=1
ds ds
and can be written in the form

v
V = cosh + i sinh
v

v
= +i
c
with tanh = vc , = cosh =
q 1
2
1 vc2
and v the velocity (of norm v). Consider a
particle moving along the Ox axis with a velocity v in the reference frame K ,

thus V = (cosh 1 + i i sinh 1 ) with tanh 1 = vc . In the reference frame K, one

has with b = cosh 22 + i i sinh 22 , tanh 2 = uc ,
V = bV bc
= [cosh (1 + 2 ) + i i sinh (1 + 2 )]
= (cosh + i i sinh )
hence, = 1 + 2 and
u
v
+
tanh 1 + tanh 2
v
= c c ;
tanh = =
vu
c
1 + tanh 1 tanh 2
1+ 2
c
nally
v=
v + u
.

1 + vc2u
When v , u c one obtains the usual Galilean transformation. Furthermore, if

v = c, one has
c+u
v=
= c,
1 + uc
the velocity of light thus appears as a limit speed.
50
3.6 Maxwells equations

Let A = ( Vc + i A) be the four-potential, D the relativistic four-nabla operator

D=
, i 1 , i 2 , i 3
ct
x
x
x

i
=
ct
and the conjugate operator

Dc =

+i .
ct
Under a Lorentz transformation, D transforms as x = (x0 + i x),

D = aDac .
To verify this, it is sucient to use the relation
x
=
,

x
x x
to develop x = axac , x = ac xa and to compare the coecients. Adopting the
Lorentz gauge
1 V
+ div A = 0
(D, A) = 2
c t
one obtains the electromagnetic eld bivector
F = Dc A = (D, A) + (D A)

i E
= (D A) = B +
.
c
Under a special Lorentz transformation, the bivector F transforms as
F = bF bc

with b = cosh 2 i i sinh 2 , tanh = vc , = cosh , which yields the standard
equations
Ex = Ex ,
Bx = Bx ,

Ey v

By = By + 2
,
Ey = (Ey vBz ) ,
c

Ey v
Bz = Bz 2
,
Ez = (Ez + vBy ) .
c
3.6. Maxwells equations
51
Furthermore, one has the relativistic invariant

E2
2
E B
F Fc =
B 2 + 2i
.
c
c
The exterior product
DF =DDA=0

B rot E
= i div B ,
ct
c
gives two of the Maxwells equations
div B = 0,
rot E =
B
.
t
Introducing the four-current density C = (c, i j), the equation

DF F D
2

div(E
1 E
, i rot B 2
=
c
c t
DF =
= 0 C
gives the two other Maxwells equations (in vacuum)
,
0
div E =
rot H = j +
D
t
with 0 0 c2 = 1. Furthermore,
DF = D F + D F
= D(DAc ) = DDc A
= A = 0 C
where
=
2
c2 t2
2
(x1 )
2
(x2 )
is the dAlembertian. Hence, the equations

2V
V = ,
2
2
c t
0
2A
A = 0 j.
c2 t2
2
(x3 )
52
The entire set of Maxwells equations (in vacuum) can therefore be written
DF = 0 C
div E
i div B
c
=

1 E
B rot E

+i rot B 2
+
c t
ct
c
= 0 (c + i j).
The interior product

F C = C F

jE
+ i (E + v B) = f
=
c
gives the volumic four-force (of Minkowski).
3.7 Group of conformal transformations

The group of conformal transformations is the group of transformations x = f (x)
(x, x being minquats) such that dxdxc = 0 entails dx dxc = 0. This group includes
spacetime translations, Lorentz transformations, dilatations
x = x + d,
x = axac ,
x = x
(d minquat, a H(C), aac = 1, R) and the transformations
1
x = (1 + xac )
= x (1 + ac x)
=
x + a(x, x)
1 + 2(a, x) + (a, a)(x, x)
(3.9)
(3.10)
(3.11)
with a respective number of parameters of 4, 6, 1, 4 for a total of 15 parameters.

The interest of this group in physics comes from the fact that Maxwells equations
(without sources) are covariant with respect to this transformation group. The
transformations (3.11) can also be expressed in the form
1
= x1 (1 + xac ) ,
= x1 + ac ;
(x )
(x )
3.7. Group of conformal transformations
53
the inverse transformation results from

1
x1 = (x )
ac
1
[1 x ac ]
= (x )
and thus
x = (1 x ac )
x .
The composition of two transformations gives

1
= x1 + ac ,
= (x )
(x )
(x )
=x
=x
+ bc
+ ac + b c
+ cc
with c = a + b and thus belongs indeed to the group; if one permutes the two
transformations, one obtains the same resulting transformation. As properties,
one has
xac x = x ac x,
|dx | =
2
|x | =
2
|x |2
|x|
(3.12)
2
(3.13)
2,
(3.14)
|dx| ,
|x|
|1 + xac |
dx = (1 + xac )
dx (1 + ac x)
Equation (3.12) results from

x = (1 + xac )1 x = x(1 + ac x)1 ,
(1 + xac ) x = x = x (1 + ac x)
which entails the relation. Furthermore,
x = (1 + xac )x ,
xxc = (1 + xac ) x xc (1 + axc ) ,
and thus
|x | =
2
|x|
|1 + xac |2
(3.15)
54
Equation (3.13) which shows that the transformation is indeed a conformal transformation can be established as follows. Dierentiating the relation qq 1 = qq 1 =
1, where q is a complex quaternion, one obtains

d q 1 q + q 1 dq = 0,

d q 1 = q 1 dqq 1 ;
hence

d x1 = x1 dx x1

or d x1 = d x1 , consequently
1 1 1
x
dx x
= x
dxx1 ,
xc dx xc
(x xc )2
xc dxxc
(xxc )2
By multiplying with the conjugate equation

x dxc x
2
(x xc )
xdxc x
(xxc )
one obtains equation (3.13)

dx dxc
2
(x xc )
dxdxc
2.
(xxc )
To obtain equation (3.15), one dierentiates the equation

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

x (1 + xc a) ac dx
1
= dx
(1 + ac x) ,
(1 + ac x) (1 + xc a)
(1 + axc) dx (1 + ac x)
(1 + ac x) (1 + xc a)
= (1 + xac )
dx(1 + ac x)1 .
3.8 Exercises
E3-1 Express the matrices

e1 =
1
0
0
0
e2 =
0
0
0
1
3.8. Exercises
55
using complex quaternions. Verify the relations e2i = ei , e1 + e2 = 1, e1 e2 = e2 e1 =

0. Take a H(C), give the expression of the modules u = ae1 , v = e1 a.
E3-2 Consider the complex quaternions
x = 1 + i i + (2 + i )j + k
y = 2 + 3i k;
compute x + y, xy, yx, x2 , y 2 , x1 , y 1 , y 1 x1 , (xy)1 .
E3-3 Consider the general Lorentz transformation
X = aXac ,

1
a=
3 + i i 5 i j 15 + k3 3
4
with aac = 1. Determine the rotation ri and the pure Lorentz transformation bi
such that a = b1 r1 = r2 b2 with
X = b1 r1 X (b1 r1 )c = b1 (r1 X r1c ) b1c

or
X = r2 (b2 Xb2c ) r2c .
E3-4 Consider the minquats

x = 1 + i (i + j) ,
y = 2 + i k,
z = 3 + i j,
w = 3i i + i j + i k.
Compute x y, B = x y, B = z w, T = x (y z), T = (x y) z, B z,
w T , B B, B B, B T .

E3-5 Let K(O, t, x, y, z) be a reference frame at rest and K (O , t , x
,y ,z ) a
3 5
reference frame moving along the Ox axis with a constant velocity v = 7 c. Write
the Lorentz transformation X = bX bc and give b. A particle is located at the point

X = (0, i , i , 0) of K and has a velocity vc = 17 in the direction O y ; nd its

four-position and its four-velocity in the reference frame K. Let E Ex , Ey , Ez
be the electric eld in the reference frame K ; determine the electromagnetic eld
in K.
E3-6 Consider a square ABCD of center 0 in the plane z = 0 and having its vertices located at the points A(0, 1, 1, 0), B(0, 1, 1, 0), C(0, 1, 1, 0), D(0, 1, 1, 0).
Determine the transform of this square under the conformal transformation
x = (1 + xac )1 x
with a = i k.
Chapter 4
Cliord algebra
Cliord having demonstrated that the Cliord algebra is isomorphic to a tensor
product of quaternion algebras or to a subalgebra thereof, this chapter develops
within H H the multivector calculus, multivectorial geometry and dierential
operators.
4.1 Cliord algebra

4.1.1 Denitions
The Cliord algebra Cn over R is an associative algebra having n generators
e0 , e1 , . . . , en1 such that
e2i = 1,
ei ej = ej ei
(i = j).
C + is the subalgebra constituted by products of an even number of ei . Cliord

algebras are directly related to the quaternion algebra via the following theorem
([13],[14]).
Theorem 4.1.1 (Cliord, 1878). If n = 2m (m integer), the Cliord algebra C2m
is the tensor product of m quaternion algebras. If n = 2m 1, the Cliord algebra
C2m1 is the tensor product of m 1 quaternion algebras and the algebra (1, )
where is the product of the 2m generators ( = e0 e1 e2m1 ) of the algebra
C2m .
The tensor product of the algebras A and B is dened as follows [8, p. 57].
Consider two algebras A and B with x, y A and u, v B; the tensor product
A B is dened by the relation
(x u) (y v) = (xy) (uv) .
Example (C C). A general element of C C is given by
A =
=
(a + i b) (f + i g)
(af ) 1 1 + (bf ) i 1 + (ag) 1 i + (bg) i i ;
58
Chapter 4. Cliord algebra
write 1 = 1 1, i = i 1, I = 1 i , iI = Ii = i i , one then has with I 2 = 1,

A =
=
(af + bf i) + I(ag + bgi)

(1 + I2 ).
The general element of C C can thus be expressed as a complex number having

complex coecients, the imaginary I commuting with i.
4.1.2 Cliord algebra H H over R

The general element of the tensor product of two quaternion algebras H is expressed by
A = (a0 + ia1 + ja2 + ka3 ) (b0 + ib1 + jb2 + kb3 )
a0 b0 1 1 + a0 b1 1 i + a0 b2 1 j + a0 b3 1 k,
a1 b0 i 1 + a1 b1 i i + a1 b2 i j + a1 b3 i k,
=
a2 b0 j 1 + a2 b1 j i + a2 b2 j j + a2 b3 j k,
a3 b 0 k 1 + a3 b 1 k i + a3 b 2 k j + a3 b 3 k k
let us write i = i 1, j = j 1, k = k 1, I = 1 i, J = 1 j, K = 1 k with

i2 = j 2 = k 2 = ijk = 1,
I 2 = J 2 = K 2 = IJK = 1.
The general element A can thus be written
A = 0 + 1 I + 2 J + 3 K
where the coecients i = di + iai + jbi + kci are quaternions and where the
lowercase i, j, k commute with the capital I, J, K (iJ = Ji, etc.). The element A
is called a Cliord number and constitutes simply a quaternion having quaternions
as coecients. Concisely, one can write
A = (0 ; )
with = 1 I + 2 J + 3 K, i H. The product of two Cliord numbers is given
by
0 0 1 1 2 2 3 3 ;
0 1 + 1 0 + 2 3 3 2 ,
(0 ; )(0 ; ) =
0 2 + 2 0 + 3 1 1 3 ,
0 3 + 3 0 + 1 2 2 1
and in a more compact notation
(0 ; )(0 ; ) = [0 0 ; 0 + 0 + ]
4.2. Multivector calculus within H H
59
where , are the ordinary scalar and vector products; the order of the
terms has to be respected, the product of two quaternions being noncommutative.
The generators of the Cliord algebra are
e0 j,
e1 kI,
e2 kJ,
e3 kK
with e20 = 1, e21 = e22 = e23 = 1 and ei ej = ej ei (i = j). A complete basis of the
algebra is given in the following table.
1
i = e0 e1 e2 e3
j = e0
k = e1 e2 e3
I = e3 e2
iI = e0 e1
jI = e0 e3 e2
kI = e1
J = e1 e3
iJ = e0 e2
jJ = e0 e1 e3
kJ = e2
K = e2 e1
iK = e0 e3
jK = e0 e2 e1
kK = e3
A Cliord number can be written in the form [54]

A = (a + ib + jc + kd; m + in + jr + ks)
with m = m1 I + m2 J + m3 K and similarly for n, r and s. The Cliord algebra
contains scalars a, pseudoscalars ib, vectors (four-dimensional) jc + ks, bivectors
m + in and trivectors kd + jr. The conjugate of A is dened by transforming
I, J, K into I, J, K and j into j, hence
Ac = (a + ib jc + kd; m in + jr ks)
with
(AB)c = Bc Ac .
The dual of A is dened by
A = iA
and the commutator of two Cliord numbers by
[A, B] =
1
(AB BA) .
2
4.2 Multivector calculus within H H

4.2.1 Exterior and interior products with a vector
The product having been dened in the Cliord algebra, the interior and exterior
products of two vectors x = jx0 + kx (x = x1 I + x2 J + x3 K), y = jy 0 + ky can
be dened by the general formula [37]
xy = x y + x y
60
where , are two nonzero coecients. Adopting the choice = = 1, one has
xy = (x y + x y),
yx = (y x + y x);
postulating a priori the relations x y = y x and x y = y x one obtains
2x y = (xy + yx),
2x y = (xy yx).
Explicitly, the formulas read
x y = x0 y 0 x1 y 1 x2 y 2 x3 y 3 S (scalar),
#
"

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

xy =
+ x1 y 0 x0 y 1 iI + x2 y 0 x0 y 2 iJ + x3 y 0 x0 y 3 iK

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

= (a x) k + j x0 a + x b
with B x = x B and Tc = T . One veries that the exterior product is indeed
associative
(x y) z = x (y z) ;
one has,
(x y) z = z (x y)
1
= [z(xy yx) + (xy yx)z] ,
2
1
x (y z) = [x(yz zy) + (yz zy)x] .
2
(4.1)
(4.2)
61
Since (z, x)y = y(z, x), i.e.,

(zx + xz)y = y(zx + xz),
one veries the equality of the equations (4.1), (4.2) and thus the associativity. A
pseudoscalar is dened by (with T = kt0 + jt, x = jx0 + kx)
1
P = x T = (xT T x)
2

= x0 t0 + x t i
with T x x T and Pc = P . As examples, let us express the standard basis
in terms of the exterior product:
1
i = e0 e1 e2 e3
j = e0
k = e1 e3 e2
I = e2 e3
iI = e1 e0
jI = e0 e2 e3
kI = e1
J = e3 e1
iJ = e2 e0
jJ = e0 e3 e1
kJ = e2
K = e1 e2
iK = e3 e0
jK = e0 e1 e2
kK = e3
The interior products between a vector and a multivector are given by the formulas
(with x = jx0 + kx, B = a + ib, T = kt0 + jt, P = is)
1
(xB Bx)
2

= (b x) j + k x0 b + x a V,
1
x T = (xT + T x)
2 0

= x t + t0 x + i (x t) B,
1
x P = (xP P x)
2
= ksx0 + j (sx) T
xB =
with
B x x B,
T x x T,
P x x P.
4.2.2 Products of two multivectors

The products of two multivectors Ap = v1 v2 vp and Bq = w1 w2 wp
are dened [12] for p q, by
Ap Bq (v1 v2 vp1 ) (vp Bq ),
Ap Bq v1 (v2 vp ) Bq
62
with
Ap Bq = (1)p(q+1) Bq Ap
which denes Bq Ap for q p. For products of multivectors one obtains, with
S, P designating respectively the scalar and pseudoscalar parts of the multivector
with B = a + ib, B = a + ib , T = ks0 + js, T = ks0 + js , P = iw, P = iw ,

1

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

1

BB =P
(BB + B B)
2
= i [b a a b ] P,
1
T T = (T T + T T )
2
= s0 s0 s1 s1 s2 s2 s3 s3 S,
1
(BP + P B)
2
= (sb + isa) B,
BP =
1
(T P P T )
2

= jws0 kws V,
T P =
1
(P P + P P )
2
= ww S.
P P =
4.2.3 General formulas

Among general formulas, one has
x (y z) = (x y)z (x z)y,
x (y z) + y (z x) + z (x y) = 0,
(x y) B = x (y B) = y (x B),
(x y) (z w) = (x w)(y z) (x z)(y w),
(B T ) V = (B V ) T,
(4.3)
(4.4)
(4.5)
(4.6)
(4.7)
63
the Jacobi identity

0 = [F, [G, H]] + [G, [H, F ]] + [H, [F, G]] ,
(x B1 ) B2 = B1 (B2 x) + x [B1 , B2 ] ,
(4.8)
(4.9)
with x, y, z, w being (four-)vectors, B, B1 , B2 bivectors, T a trivector and F, G, H

any Cliord numbers; the equations (4.4), (4.6), (4.9) are respectively consequences
of equations (4.3), (4.5), (4.8). To establish equation (4.3), one just needs to write
for any bivector B, 2x B = (xB Bx), hence
4x (y z) = [x(yz zy) + (yz zy)x] ;
furthermore,
4(x y)z = [(xy + yx)z + z(xy + yx)] ,
4(x z)y = [(xz + zx)y + y(xz + zx)] ;
adding the two last equations, one veries equation (4.3). Equation (4.5) simply
results from
(x y) B x (y B)
= (y x) B
= y (x B);
in particular,
(x y) (x y) = (x y)2 (x x)(y y).
To prove equation (4.7), one writes
2B T = (BT + T B),
4(B T ) V = [(BT + T B)V + V (BT + T B)] ,
2B V = 2V B = V B + BV,
4(B V ) T = [(V B + BV )T + T (V B + BV )] ;
or B(T V V T ) = (T V V T )B because T V V T is a pseudoscalar which
commutes with B, hence the equation. The Jacobi identity results from
1
(ABC BCA)
2
1
= [(AB BA) C + (BAC BCA)]
2
= [A, B] C + B [A, C] ,
[A, CB] = [A, C] B + C [A, B] ,
1
[A, [B, C]] = {[A, B] C + B [A, C] [A, C] B C [A, B]}
2
= [[A, B] , C] + [B, [A, C]]
[A, BC] =
64
hence, the equation

[A, [B, C]] + [B, [C, A]] + [C, [A, B]] = 0.
4.2.4 Classical vector calculus

The link with the classical vector calculus is obtained as follows. Let x = kx,
y = ky, z = kz, w = kw be four vectors without a temporal component, one
obtains the relations
x (y z) = k [x (y z)] V,
x (y z) = k [x (y z)] T,
(x y) (z w) = (x y) (z w),
2
(x y) (x y) = (x y) (x x)(y y),
2
= (x y) (x x) (y y) ,
= (x y) (x y) S,
[x y, z w] = (x y) (z w) B.
The entire classical vector calculus thus constitutes a particular case of the multivector calculus.
4.3 Multivector geometry

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

Plane
Let u, v be two linearly independent vectors and B = u v the corresponding
plane; the equation of a plane parallel to B and going through the point a is
expressed by
(x a) (u v) = 0
4.3. Multivector geometry
65
with the solution

x a = u + v,
x = u + v + a
(, R)
giving the parametric equation of the plane parallel to B. A vector n is perpendicular to the plane B = u v if n is perpendicular to u and v (n u = n v = 0);
for any vector x one has
x (u v) = (x u)v (x v)u
hence
n B = 0.
Furthermore, one remarks that the vector x (u v) is perpendicular to x,
(x B) x = (x u)(v x) (x v)(u x)
= 0.
A plane B1 = x y is perpendicular to a plane B2 = u v if the vectors x, y are
perpendicular to the vectors u, v (x u = y u = x v = y v = 0); from the general
formula
(x y) (u v) = (x v)(y u) (x u)(y v)
one obtains the orthogonality condition of two planes
B1 B2 = 0.
In particular, the dual of a plane B = iB is perpendicular to that plane
B B = 0.
Hyperplane
A hyperplane is a subvector space of dimensions p = n 1 = 3 (for n = 4). Let
T = u v w be a trivector (hyperplane), the equation of a hyperplane parallel
to T and going through the point a is given by
(x a) T = 0
(4.10)
with the solution

x a = u + v + w,
x = u + v + w + a
(, , R)
giving the parametric equation of the hyperplane. Explicitly, equation (4.10) reads
(with T = kt0 + jt, x = jx0 + kx, a = ja0 + ka)
t0 x0 + t1 x1 + t2 x2 + t3 x3 (a0 t0 + a t) = 0
66
which is indeed the equation of a hyperplane. A vector n is perpendicular to the

hyperplane T = u v w if n is perpendicular to u, v, w (n u = n v = n w = 0).
From the general formula
n (u v w) = (n u)(v w) + (n v)(w u) + (n w)(u w)
one deduces
n T = 0.
(4.11)
If n is the dual of T , n = T = iT , one nds (with T = kt0 + jt, T = jt0 + kt)

1
T T = (T T + T T ) = 0;
2
the dual of a hyperplane is thus perpendicular to that hyperplane. A plane B =
x y is perpendicular to the hyperplane T = u v w if x, y are perpendicular to
the hyperplane, hence
T B = B T = (x y) T
x (y T ) = y (x T ) = 0.
(4.12)
(4.13)
4.3.2 Orthogonal projections

Orthogonal projection of a vector on a vector
Let u = u +u be the vector to project on the vector a with u a = 0, u a = 0;
since
ua = u a u a
one obtains
u a = u a = u a,
u a = u a = u a,
u = (u a)a1 ,
u = (u a)a1 .
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 = ab
(with u B = 0, u B = 0); since
uB = u B + u B,
Bu = B u + u B,
one obtains
u = (u B)B 1 = B 1 (B u),
u = (u B)B 1 = B 1 (u B).
4.3. Multivector geometry
67
Orthogonal projection of a vector on a hyperplane

Let T be the hyperplane u = u + u with u T = 0, u T = 0; one has
uT = u T u T,
T u = u T + u T,
hence
u T = u T,
u T = u T,
T u = u T,
T u = u T,
and nally
u = (u T )T 1 = T 1 (u T ),
u = (u T )T
=T
(u T ).
(4.14)
(4.15)
Orthogonal projection of a bivector on a plane

Let B1 = B1 + B1 be the bivector and B2 = a b the plane with B1 B2 = 0,

B1 B2 = 0 and B1 , B2 = 0; using the formula
B1 B2 = B1 B2 + B1 B2 + [B1 , B2 ] ,
one obtains
B1 = (B1 B2 )B21 ,
B1 = {(B1 B2 ) + [B1 , B2 ]} B21 .
Example. Take B1 = x y (with x = jx0 + kx, y = jy 0 + ky) and let us project
on the plane x1 x2 , i.e., the plane B2 = K (B21 = K); one nds
B1
(B1 B2 )B21
(x2 y 1 + x1 y 2 )K = x y
(with x = x1 kI + x2 kJ, y = y 1 kI + y 2 kJ),

B1
= {(B1 B2 ) + [B1 , B2 ]} B 1
= (x3 y2 + x2 y3 )I + (x3 y1 x1 y3 )J
+(x1 y0 x0 y1 )iI + (x2 y0 x0 y2 )iJ + (x3 y0 x0 y3 )iK
= B1 B1 .
68
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 = a1 (a B) = (B a)a1 ,
B = a1 (a B) = (a B)a1 .
Orthogonal projection of a bivector on a hyperplane
Let B = B + B be the bivector and T the hyperplane with B T = 0 and
[B , T ] = 0 (B T ). From the equation
BT = B T + [B, T ]
one obtains
B = (B T )T 1 ,
B = [B, T ] T 1 .
Example. Let F = B + i Ec be the electromagnetic bivector, its orthogonal projection on the hyperplane T = k = e1 e2 e3 (T 1 = k) yields
F
F
= (F k)k = B,
E
= [F, k] k = .
c
Orthogonal projection of a hyperplane T1 on the hyperplane T2

Let us write T1 = T1 + T1 with T1 = x y z, T2 = u v w; from the relation
T1 T2 (x y) [z (u v w)]
and the equation (4.11), one deduces

T1 T2 = 0;
furthermore T1 , T2 = 0 (T1 T2 ). Since

T1 T2 = T1 T2 + [T1 , T2 ]
it follows that
T1 = (T1 T2 )T21 ,
T1 = [T1 , T2 ] T21 .
4.4. Dierential operators
69
4.4 Dierential operators

4.4.1 Denitions
Consider the relativistic four-nabla operator,

= j 0 k
x
= e = e
with
=I
=
+ J 2 + K 3,
x1
x
x
and e the reciprocal basis dened by

e e =
(e0 = e0 , e1 = e1 , e2 = e2 , e3 = e3 ). One can dene the operators fourgradient
= j 0 k grad
( S)
x
the four-divergence of a vector A = jA0 + kA
A =
A1
A2
A3
A0
+
+
+
S
x0
x1
x2
x3

the four-curl
A = rot A i
A
+ grad A0
x0

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

b
+ rot a V
B = j div b k
x0

a
B = k div a + j
rot b T.
x0
4.4.2 Innitesimal elements of curves, surfaces and hypersurfaces

A curve in the four-dimensional space is dened by the parametric equations
OM () = jx0 () + kx()
dening the tangent vector at the point M ,
dOM =
OM
d
70
where is the parameter. For a surface, the parametric equations are

OM (, ) = jx0 (, ) + kx(, )
and the tangent plane to the surface at the point M is dened by

OM OM
dS =
dd
2 3

3 1

x x
x3 x2
x x
x1 x3
I
+
J

1
2
2
1
1
0
0
x
x
x
x1
+ x
K + x
iI
x
x
=
2 0

3 0

x x
x0 x2
x x
x0 x3
+ iJ + iK
the bivector
dd.
In abridged notation, one can write

dS = dx2 dx3 I + dx3 dx1 J + dx1 dx2 K
+ dx1 dx0 iI + dx2 dx0 iJ + dx3 dx0 iK
with
D(x2 , x3 )
dd
D(, )

x2 x2

dd, etc.
=
3
x3
x

dx2 dx3 =
where dx2 dx3 is an undissociable symbol [3, p. 446]. For a hypersurface dened by
OM (, , ) = jx0 (, , ) + kx(, , )
the tangent hyperplane to the surface at the point M is given by the trivector

OM OM OM
dT =
ddd

2
3
x3 x2 x1
x3 x1
x1 x2
x
x
k
1
3
2
2
1
3
1
2
3
x x x
x x x
x x x
+
+
3
2
0
2
3
0
3
0
2
x
x
x
x
x
x
x
x
x
+
+
jI
+
x0 x3 x2
x2 x0 x3
x0 x2 x3
+

ddd.
=
3
1
0
1
3
0
3
0
1
x x x
x x x
x x x
+
jJ
0
1
0
x3 x1
x0 x3
x1 x3
+ x
+ x
x
2
1
0
1
2
0
2
0
1
x x x
x x x
x x x
+
+
jK
+
0
2
1
1
0
2
0
1
2
x x x
x x x
x x x
4.4. Dierential operators
71
In short, one can write

dT = kdx1 dx3 dx2
2
(4.16)
0
+ jIdx dx dx + jJdx dx dx + jKdx dx dx
(4.17)
with the symbol

D(x1 , x3 , x2 )
ddd
D(, , )

x1 x1 x1

x3 x3 x3
=
ddd, etc.

x2 x2 x2

dx1 dx3 dx2 =
4.4.3 General theorems

Generalized Stokes theorem
This theorem can be written
%
&&
A dl =
( A) dS
(4.18)
with dl = jdx0 + kdx, A = jA0 + kA, dS = dS1 + idS2 and

dS1 = dx2 dx3 I + dx3 dx1 J + dx1 dx2 K,
dS2 = dx1 dx0 I + dx2 dx0 J + dx3 dx0 K.
Explicitly, equation (4.18) is expressed in classical vector notation as

%
&&
A
A0 dx0 A dx =
rot A dS1 + A0 + 0 dS2 .
x
If dx0 = 0, one has dS2 = 0 and the formula reduces to the standard Stokes
theorem. As to the orientation of the curve C, it results from that of the surface
S [10, p. 39]. One takes two linearly independent four-vectors a, b S and one
chooses the order (a, b) as being the positive orientation (dS = a b, > 0). On
the curve C one chooses a vector f exterior to S and a vector g tangent to the
curve such that (f, g) is ordered positively; the curve is then oriented along g.
Generalized Gauss theorem
The theorem is expressed as
&&&
&&&&
A dT =
( A) d
(4.19)
72
with d = idx0 dx1 dx2 dx3 , dT = kdx1 dx3 dx2 + jt and

t = dx2 dx3 dx0 I + dx3 dx1 dx0 J + dx1 dx2 dx0 K,
the hypersurface being closed. Explicitly, equation (4.19) reads

&&&
&&&& 0

A
0
1
3
2
A dx dx dx + A t i =
+ div A idx0 dx1 dx2 dx3 .
x0
The orientation of the hypersurface results from that of the four-volume. Let
a, b, c, d be four linearly independent vectors of the four-volume a b c d =
i; if > 0, the orientation of the four-volume is positive. On a point of the
hypersurface, one chooses a vector m outside of the four-volume and three vectors
p, q, r on the hypersurface; if m p q r has a positive sign, the orientation of
the hypersurface is positive.
Other formulas
On a closed surface, one has
&&
&&&
F dS =
( F ) dT,
&&
&&&
F dS =
( F ) dT,
(4.20)
(4.21)
with the bivector F = f + ig, dS = dS1 + idS2 , dT = kdx1 dx3 dx2 + jt; explicitly,
formula (4.20) reads

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

&&&
&&
g
i div gdx1 dx3 dx2
+
rot
f
t .
i (g dS1 f dS2 ) =
x0
The orientation of dS proceeds from that of dT . One chooses three linearly independent vectors a, b, c of T and one denes the orientation a b c as being
positive. On a point of the surface, one considers a vector m exterior to the trivolume T , and two vectors p, q on the surface; if m p q has the orientation of
a b c, the orientation of the surface is positive.
4.5 Exercises
E4-1 Consider the Cliord 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 , A1 , B 1 , AB, BA, (AB)c , (BA)c ,

1
1
(AB) (AB)c , (BA) (BA)c , (AB) , (BA) , [A, B].
E4-2 Consider the four-vectors
x = j + kI + kJ,
z = 3j + kJ,
y = 2j + kK,
w = 3kI + kJ + kK.
Determine x y, B = x y, B = z w, T = x (y z), T = (x y) z, B z,
w T , B B, B B, B T .
E4-3 Take an orthonormal reference frame with the components of the fourvectors
x = (0, 1, 2, 1),
y = (0, 3, 1, 1),
z = (0, 1, 2, 1),
w = (0, 2, 1, 5).
Determine within the Cliord algebra H H the surfaces S1 = x y, S2 = z w,
the trivector x y z and the four-volume x y z w. Give the orthogonal
projection of the vector w on S1 and the orthogonal projection of S1 on S2 .
Chapter 5
Symmetry groups
This chapter formulates the Lorentz group and the group of conformal transformations within the Cliord 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. Diracs equation, the unitary group SU(4) and the symplectic
unitary group USp(2, H) are treated as applications of H H over C.
5.1 Pseudo-orthogonal groups O(1, 3) and SO(1, 3)

5.1.1 Metric
Consider two vectors x = jx0 + kx, y = jy 0 + ky with x = x1 I + x2 J + x3 K and
the interior product, with xc = x;
x x = xxc = x2 ,
(x + y) (x + y) = (x + y)(x + y)c
= xxc + yyc + 2x y,
(xy + yx)
(xyc + yxc )
=
2
2
= x0 y 0 x1 y 1 x2 y 2 x3 y 3 .
xy =
A vector x is isotropic if xxc = 0, timelike if xxc > 0 and spacelike if xxc < 0.
5.1.2 Symmetry with respect to a hyperplane

Let a = ja0 + ka be a vector, the hyperplane perpendicular to a is given by the
dual of a,
T = a = ia
= ka0 ja
76
Chapter 5. Symmetry groups
+
and anticommutes with
with T 1 = ia
aac (i commutes with all elements of C
those of C ). Let us suppose that T goes through the origin and let x = jx0 + kx
be a vector. The orthogonal projections of x on T are given by the relations (4.14),
(4.15)
x = T 1 (x T ),
x = T 1 (x T ),
hence
x iaxia
axa
(xT T x)
x
= +
=
,
2
2
2aac
2 2aac
axa
x
x = +
2 2aac
x = T 1
with x = x + x .
Denition 5.1.1. The symmetric of x with respect to a hyperplane is obtained by
drawing the perpendicular to the hyperplane and by extending this perpendicular
by an equal length [11].
Let x be the symmetric of x with respect to the hyperplane T ; one has
x x = 2x
hence
x =
axa
.
aac
More simply, one can write that x x is perpendicular to the hyperplane T (and

thus parallel to a) and x+x
is parallel to the hyperplane. One obtains
2

a
x = x + a,

x +x
= 0,
2
hence

a
2(a x)
,
a x+
= 0 = =
2
aa
2(a x)a
x = x
aa
(ax + xa)a
axa
axc a
=x+
=
=
;
aac
aac
aac
nally (with a2 = aac ), one nds again the above expression of the symmetric
of x with respect to the hyperplane. One shall distinguish the time symmetries
5.1. Pseudo-orthogonal groups O(1, 3) and SO(1, 3)
77
(with aac = 1)
x = axa
= axc a,
from the space symmetries (with aac = 1)
x = axa
= axc a.
5.1.3 Pseudo-orthogonal groups O(1, 3) and SO(1, 3)

Denition 5.1.2. The pseudo-orthogonal group O(1, 3) is the group of linear operators which leave invariant the quadratic form
x y = x0 y 0 x1 y 1 x2 y 2 x3 y 3
with x = jx0 + kx.
Theorem 5.1.3. Every rotation of O(1, 3) is the product of an even number 4
of symmetries, any inversion is the product of an odd number 4 of symmetries
[11].
is an improper transformation of a determinant equal to 1.
Proper orthochronous Lorentz transformation
Consider an even number of time symmetries (s, r, . . .) and of space symmetries
(t, u, . . .); one obtains
x = (utsr)x(rstu)
= axac
with a = utsr C , ac = rc sc tc uc = (r)(s)(t)(u) = rstu and aac = 1.
+
Other transformations
Let n be the number of time symmetries and p the number of space symmetries;
their combinations give the following Lorentz transformations :
1. n odd, p odd: proper antichronous rotation
x = axac ,
aac = 1,
a C+;
2. n even, p odd: improper orthochronous inversion

x = axac ,
aac = 1,
a C;
3. n odd, p even: improper antichronous inversion

x = axac ,
aac = 1,
a C.
78
Group O(1, 3): recapitulative table

The entire set of Lorentz transformations L is given in the table below where n is
the number of time symmetries and p the number of space symmetries.
orthochronous
antichronous
Inversion
(det L = 1)
n even, p odd
x = axac = axc ac
(aac = 1, a C )
n odd, p even
x = axc ac = axc ac
(aac = 1, a C )
Rotation
(det L = 1)
n even, p even
x = axac
(aac = 1, a C + )
n odd, p odd
x = axac
(aac = 1, a C + )
5.2 Proper orthochronous Lorentz group

5.2.1 Rotation group SO(3)
A subgroup of the proper orthochronous Lorentz group is the rotation group SO(3)
x = rxrc
(5.1)
with r = cos 2 + u sin 2 , u = u1 I + u2 J + u3 K, (u1 )2 + (u2 )2 + (u3 )2 = 1,

x = jx0 + kx, x = jx0 + kx , rrc = 1. One veries that x belongs indeed to the
vector space of vectors
xc = rxc rc = x .
In matrix form, equation (5.1) can be written
0
0
x
x
x1
x1
X=
X =
x2
x2 ,
3
x
x3
with
xi , xi R,
X = AX,
1 0
0 a
A=
0 m
0 n
with
0
f
b
p
0
g
,
h
c
2 2 2
cos ,
a = u1 + u2 + u3
2 2 1 2 3 2
b= u
cos ,
+ u
+ u

2
2 2
c = u3 + u1 + u2
cos ,
5.2. Proper orthochronous Lorentz group
79
f = u1 u2 (1 cos ) u3 sin ,
m = u1 u2 (1 cos ) + u3 sin ,
g = u1 u3 (1 cos ) + u2 sin ,
n = u1 u3 (1 cos ) u2 sin ,
h = u2 u3 (1 cos ) u1 sin ,
p = u2 u3 (1 cos ) + u1 sin .
One obtains the same expression as with quaternions, despite the distinct nature
of r and x (r C + , x C ). All considerations of Chapter 2 on the subgroups of
SO(3) that is of r, apply here; it suces to replace i, j, k by I, J, K respectively.
For an innitesimal rotation, one has with r 1 + u d
2 ,
x = rxrc

d
d
= 1+u
x 1u
2
2
d
(ux xu)
= x+
2
d
= x + k (ux xu)
2
= x + dku x,
dx = x x = dku x.
In matrix notation
dX = dui Mi X,
with
0
0
M1 =
0
0
0
0
0
0
0 0
0 0
,
0 1
1 0
0 0
0 0
M2 =
0 0
0 1
i (1, 2, 3)
0
0
0
0
0
1
,
0
0
0
0
M3 =
0
0
0 0 0
0 1 0
1 0 0
0 0 0
and the relations

[Mi , Mj ] = ijk Mk .
5.2.2 Pure Lorentz transformation

A pure Lorentz transformation is given by
x = bxbc ,
b = cosh
+ iv sinh
2
2
with v = v 1 I + v 2 J + v 3 K, (v 1 )2 + (v 2 )2 + (v 3 )2 = 1, x = jx0 + kx, x = jx0 + kx ,

bbc = 1. One veries that one has indeed xc = bxc bc = x . In matrix formulation,
one has with
0
0
x
x
x1
x1
X =
X=
xi , xi R,
x2 ,
x2 ,
x3
x3
80

X = BX,
cosh
v 1 sinh
B=
v 2 sinh
v 3 sinh
with
v 1 sinh v 2 sinh
a
f
f
b
g
h
2
a = 1 + v 1 (cosh 1) ,
2
b = 1 + v 2 (cosh 1) ,
2
c = 1 + v 3 (cosh 1) ,
v 3 sinh
g
,
h
c
f = v 1 v 2 (cosh 1) ,
g = v 1 v 3 (cosh 1) ,
h = v 2 v 3 (cosh 1) .
One will notice that the matrix is real and symmetric. For a pure innitesimal
Lorentz transformation, one obtains with b 1 + iv d
2 and i anticommuting with
x,
x = bxbc

d
d
= 1 + iv
x 1 iv
2
2
d
(vx + xv)
=x+i
2

d 0
v jx + kx + jx0 + kx v
=x+i
2

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

dx = x x = d kvx0 j v 1 x1 + v 2 x2 + v 3 x3 .
In matrix form, one has
i (1, 2, 3)
dX = dv i Ki X,
with
0
1
K1 =
0
0
1
0
0
0
0
0
0
0
0
0
,
0
0
0
0
K2 =
1
0
0
0
0
0
1
0
0
0
0
0
,
0
0
and the relations

[Ki , Kj ] = ijk Mk ,
[Ki , Mj ] = ijk Kk .
0
0
K3 =
0
1
0
0
0
0
0
0
0
0
1
0
0
0
5.2. Proper orthochronous Lorentz group
81

Transformation of multivectors and Cliord 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 = (xyyx)
transforms as
2
(x y y x )
2
= a (x y) ac
B =
= aBac .
Generally, for a product of two vectors xy one has the transformation
x y = axac ayac
= axyac ;
consequently, any multivector A (and any Cliord number) which is a linear combination of such products transforms under a proper orthochronous Lorentz transformation according to the same formula
A = aAac
(a C + , aac = 1).
Decomposition into a rotation and a pure Lorentz transformation

The decomposition a = br of a general Lorentz transformation into a rotation
and a pure Lorentz transformation is obtained as follows ([15],[16]). Consider the
element
aac = 1;
a = a0 + a + i (b0 + b) C + ,
let us introduce a conjugation (transforming i into i),
a = a0 + a i (b0 + b)
= br
= bc r,
ac = rc b.
Let us write d = aac = brrc b = b2 (ddc = 1), hence
2b2 = 2d,
b2 d = d2 ,
b2 dc = 1
(dc = aac );
82
adding, one obtains

2
b2 (2 + d + dc ) = (1 + d) ,
(1 + d)
b =
2 + d + dc
r = bc a
(with d = aac ),
(1 + aac ) a
=
2 + d + dc
(a + a)
=
,
2 + d + dc
with bbc = rrc = 1. For the decomposition a = r b , one obtains similarly
(1 + d )
b =
2 + d + dc
(with d = ac a, dc = ac a),
r = abc
(a + a)
=
.
2 + d + dc
Example. Let a = 10K 3iJ be an even Cliord number (a C + ) and such

that aac = 1. Let us decompose a into a product (a = br = r b ). One has
a = 10 K + 3iJ,
d = aac = 19 + 6 10 iI

10 + 3iI ,
r = K;
Answer: b =
similarly
d = ac a = 19 6 10 iI

10 3iI ,
r = K.
Answer: b =
5.3 Group of conformal transformations

5.3.1 Denitions
The treatment of conformal transformations within H(C) extends easily to H H.
The group of conformal transformations is constituted by the transformations such
that if dx dx = 0, then dx dx = 0 with dx = jx0 + kx. The group contains
the spacetime translations (x = x + d), Lorentz transformations (x = axac ),
dilatations (x = x) and the transformations
1
(x )
= x1 + ac
(5.2)
5.3. Group of conformal transformations
83
xc
. The above equation can also
where a is a constant (four-)vector with x1 = xx
c
be written

1
1 1
= x1 + ac
x = (x )

1
1
= x1 (1 + xac )
= (1 + ac x) x1
and using relation (AB)

number
= B 1 A1 which is true for any invertible Cliord

x = (1 + xac )
= x (1 + ac x)
(5.3)
.
(5.4)
Equivalently, one has

x
+a
xc
xxc
,

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

1
The inverse transformation is given by

1
x1 = (x )
ac
hence
1
x = (1 x ac )
x
1
= x (1 ac x )
(5.5)
.
(5.6)
5.3.2 Properties of conformal transformations

For any invertible Cliord number A (AA1 = A1 A = 1), one has
dA1 A + A1 dA = 0,
dA1 = A1 dAA1 .
Dierentiating equation (5.2), one obtains

1
= d x1 ,
d (x )
1
(x )
dx (x )
xc dx xc

2
(x xc )
= x1 dxx1 ,
=
xc dxxc
2
(xxc )
(5.7)
(5.8)
(5.9)
84
multiplying by the conjugate equation, one has

xc dx xc x dxc x
xc dxxc xdxc x
4
(x xc )
(5.10)
dxdxc
(5.11)
(xxc )
(x .x )
dx dxc =
(x.x)
which shows that the transformation is indeed a conformal transformation. Equation (5.8) gives
(x )
dx (x )
= x1 dxx1 ,
dx = x x1 dxx1 x ,
and using equations (5.3), (5.4), one obtains the relation
dx = (1 + xac )1 dx (1 + ac x)1 .
(5.12)
Finally, from equations (5.3), (5.4) one has

1
x = (1 + xac ) x = x (1 + ac x)
(1 + xac ) x = x = x (1 + ac x) ,
hence
xac x = x ac x,
xxc = (1 +
x xc
(5.13)
xac ) x xc
(1 + axc ) ,
xxc
.
=
1 + 2(x a) + (a a)(x x)
(5.14)
(5.15)
5.3.3 Transformation of multivectors

A conformal transformation is a relation of the type xi = f i x0 , x1 , x2 , x3 with
xi k
dx .
xk
dxi =
A contravariant (four-)vector A, by denition, transforms according to the formula

xi k
A .
xk
Ai =
The equation (5.12)
dx = (1 + xac )
dx (1 + ac x)
5.4. Dirac algebra
85
then gives the transformation of a vector A = jx0 + kA,

A = (1 + xac )
A (1 + ac x)
(5.16)
For a product of two vectors A, B one obtains

A B = (1 + xac )1 A (1 + ac x)1 (1 + xac )1 B (1 + ac x)1
= K (1 + xac )
AB (1 + ac x)
with
K=
1
;
1 + 2(x a) + (a a)(x x)
for a bivector A B = (ABBA)

one then has
2
1
A B = K (1 + xac )
(A B) (1 + ac x)
one veries that A B is indeed a bivector. Similarly, one obtains for a trivector
and a pseudoscalar
A B C = K 2 (1 + xac )
(A B C) (1 + ac x)
A B C D = K (A B C D) .
4
Furthermore,
A Ac = K 2 AAc ,
A B = K 2 (A B)
which shows that the conformal transformation conserves the angles.
5.4 Dirac algebra

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

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

0 i j
0 i k
(
kK)
=
,
e
,
e2 ( kJ) =
3
i j 0
i k 0
86
where i is the ordinary complex imaginary (i2 = 1) and e20 = 1, e21 = e22 =
e23 = 1. The other matrices are given by

1 0
i 0
j 0
1=
,
I=
,
J=
,
0 1
0 i
0 j

K=

iJ =

jJ =
k
0
0
k
0 j
j 0
i j
0
i=
0
i j
iK =
1
0
0
1
jK =
0
k
i k
0
k
0
iI =
0
i k
jI =
k=
The Dirac spinor can be expressed as a left ideal

f
q1 f
,
E=
= AE =
0
q2 f
0
0
0
i
i
0
i i
0
0
i
0
i i
i
0
where E is a primitive idempotent (E 2 = E), f = (1 + i k)/2 ; q1 , q2 being

real quaternions and A a general element of the algebra. The hermitian norm is
expressed by
= (q1 q1c + q2 q2c ) f
where is the transposed, quaternionic conjugated and complex conjugated matrix of . The Dirac equation in presence of an electromagnetic eld is
(ih eA i mc) = 0
k, the vector potential A = j Vc + kA

with the four-nabla operator = j ct
and e the electric charge of the particle.
5.4.2 Unitary and symplectic unitary groups

Having dened the Cliord algebra H H over C, let A be a 2 2 matrix having
as elements complex quaternions qi and its adjoint A ,

A=
q1
q3
q2
q4
A =
q1c
q2c
q3c
q4c
where the sympbols and c represent respectively the complex conjugation and
the quaternionic conjugation. The adjunction transforms i and i, j, k, I, J, K into
their opposites, as one can verify directly on the basis matrices. A selfadjoint
Cliord number (H = H ) is consequently of the type
H = (a + i ib + i jc + i kd; i p + iq + jr + ks)
5.4. Dirac algebra
87
1
2
3
with
p = p I + p J + p K (etc.) and real coecients. The unitary group
SU 2, H(C) , isomorphic to SU(4), is the set of matrices A such that
AA = A A = 1.
If one restricts the matrices to real quaternion matrices, one obtains as a subgroup
the symplectic unitary group USp(2, H) [39, p. 232]. The elements of the unitary

group being of the type ei H (with H = H ), one can choose for the 15 generators
of SU(4),
"
ei , eI , eJ , eK , ei jI , ei jJ , ei jK , ei kI , ei kJ , ei kK
#
ej , ek , ei iI , ei iJ , ei iK .
(5.17)
where is a real generic parameter with the usual series development

eI = cos + I sin ,
ei iJ = cos + i iJ sin
(etc.).
The 10 matrices within parentheses constitute the symplectic unitary group

USp(2; H). Explicitly, one has

ei =

eJ =

eK =

ei jI =
e
e
i jJ
i jK
i kI

=

=

=
ei kK =
cos
sin
sin
cos
cos + j sin
0
eI =
cos + i sin
0
0
cos + i sin
0
cos + j sin
cos + k sin
0
,
0
cos + k sin
cos i sin
0
,
0
cos + i sin
cos j sin
0
,
0
cos + j sin
cos k sin
0
,
0
cos + k sin

cos
i sin
cos
i kJ
=
,
e
i sin
cos
j sin
cos
k sin
.
k sin
cos
j sin
cos
88
One veries that these matrices have indeed real quaternions as coecients. The
other elements of the SU(4) group are given by

i iI
ei ik
cos + i sin
0
0
cos i sin

cos i sin
=
,
i sin
cos

cos
i sin
=
.
i sin
cos
ej =
ek =
,
e
i iJ
cos
i sin
i sin
cos
cos
i j sin
i j sin
cos
To obtain a representation in terms of 4 4 complex matrices, one can choose for

the generators of the Cliord algebra over C
0
0
i 0
0
0
0 i
0 i 0
0
0
0 i 0
e0 ( j) =
e1 ( kI) =
0 0 i 0 ,
0 i 0 0 ,
0 0
0 i
i 0
0 0

0
0 0
0 1
0
0 i
0 0 1 0
0
0 0 i
,
e2 ( kJ) =
e3 ( kK) =
0 1 0 0 ,
i 0 0
0
0 i 0
0
1 0
0 0
with e20 = 1, e21 = e22 = e23 = 1.
5.5 Exercises
E5-1 Consider a special pure Lorentz transformation (b) along the Ox axis (velocity v = 3c ) followed by a rotation (r) of 4 around the same axis. Express the
resulting Lorentz transformation X = aX ac .
E5-2 Consider a special Lorentz transformation (b1 ) along the Ox axis (velocity
v = 3c ) followed by a pure Lorentz transformation (b2 ) along the Oy axis (with
v = 3c ). Express the resulting Lorentz transformation X = aX ac with a = b2 b1 .
Decompose the resulting Lorentz transformation into a rotation followed by a pure
Lorentz transformation (a = br). Give the direction of the Lorentz transformation,
the velocity and the angle of rotation.
E5-3 Consider an orthonormal system of axes (O, x, y, z) and a cube with vertices
A(0, 0, 0), B(1, 0, 0), C(1, 1, 0), D(0, 1, 0), E(0, 1, 1), F (0, 0, 1), G(1, 0, 1),
H(1, 1, 1). Determine the transform of the vertices of this cube under a conformal
transformation
x = (1 + xac )1 x
with a = 2kI + kK , x = ctj + kx, x = ct j + kx .
5.5. Exercises
E5-4 Consider the matrices

1 j
A=
,
k i
89

B=
Determine AB, BA, A1 , B 1 , (AB)
1 k
i j

,
1
, (BA)
C=
1
k
i
j
; does the matrix C 1 exist?
Chapter 6
Special relativity
This chapter develops the special theory of relativity within the Cliord algebra
H H over R. The relativistic kinematics and relativistic dynamics of a point mass
are examined.
6.1 Lorentz transformation

Consider a reference frame K(O, x, y, z) at rest and a reference frame
K (O , x , y , z ) moving along the Ox axis with a constant velocity v (Figure
6.1). With X = jx0 + kx, X = jx0 + kx (x0 = ct, x = xI + yJ + zK),
b = cosh 2 iI sinh 2 , the Lorentz transformation is expressed by
X = bXbc.
y
K
K
O
O
v
x
z
Figure 6.1: Special Lorentz transformation (pure).
92
Chapter 6. Special relativity
Explicitly, one obtains

x0 = x0 cosh x sinh ,
x = x0 sinh + x cosh ,
y = y, z = z.
Writing tanh =
v
c
= , = cosh with
2 = 1 + sinh2 = 1 + 2 2 ,
1
,
=
1 2
the transformation becomes

ct = (ct x) ,
x = (x ct) ,
y = y,
Taking =
v
c
z = z.
1 ( 1), one obtains the Galilean transformation

t = t,
x = x vt,
y = y,
z = z.
The inverse transformation is expressed by

X = bc X b
with bc = cosh 2 + iI sinh 2 , hence
ct = (ct + x ) ,
x = (x + ct ) ,
y = y, z = z .
6.1.2 Physical consequences

Contraction of length
Consider a rod at rest in K , parallel to the O x axis and of length l0 = x2 x1 .
In K, the length is l = x2 x1 , where the abscisses x2 , x1 are determined at the
same time t:
x2 = (x2 ct) ,
x1 = (x1 ct) ,
6.1. Lorentz transformation
93
hence
x x1
l = x2 x1 = 2

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

= l0 1 2 l0 ;
the observer in K concludes also to a contraction.
Example. Take v = 300 km/s, = 103 , 2 = 106 ,
relative variation of l is 5 107 .

1 2 = 1 5 107 , the
Time dilatation
A time interval t = t2 t1 measured at the same point x = x2 x1 = 0 of
K corresponds in K to a time interval t = t2 t1 with
ct2 = (ct2 + x2 ) ,
ct1 = (ct1 + x1 ) ,

hence
t = t2 t1 = t
t
=
t .
2
1
The observer in K concludes to a slowing down of physical phenomena. Reciprocally, a time interval t = t2 t1 measured at the same point x2 = x1 in K, gives
in K ,
ct2 = (ct2 x2 ) ,
ct1 = (ct1 x1 ) ,
t = t2 t1
t
=
t.
1 2
Hence, the conclusion is the same.
94

For a general Lorentz transformation, one has a = br (or a = rb) with r =
cos 2 + u sin 2 , b = cosh 2 + iv sinh 2 (u u = v v = 1). Explicitly, one obtains
cos 2 cosh 2 + u cosh 2 sin 2
a = br = iu v sin 2 sinh 2
,
+i v cos 2 u v sin 2 sinh 2
cos 2 cosh 2 + u cosh 2 sin 2
a = rb = iu v sin 2 sinh 2

+i v cos 2 + u v sin 2 sinh 2
with aac = a ac = 1. The general Lorentz transformation is simply expressed by
X = aXac
(or a Xac )
with X = jx0 + kx, X = jx0 + kx .
6.2 Relativistic kinematics

6.2.1 Four-vectors
Transformation of a four-vector
An arbitrary four-vector A = ja0 + ka transforms under a special Lorentz transformation as
A = bAbc
with b = cosh 2 iI sinh 2 ; explicitly, writing
1
= cosh =
1
v2
c2
tanh =
one has

v
,
a0 = a0 a1
c

v
a1 = a1 a0
,
c
a2 = a2 , a3 = a3 ;
reciprocally, one has
A = bc A b
v
,
c
6.2. Relativistic kinematics
95
which yields

v
a0 = a0 + a1
,
c

v
a1 = a1 + a0
,
c
a2 = a2 , a3 = a3 .
If one considers a general Lorentz transformation, one obtains the following formulas ([40, p. 134], [53, p.123]) with b = cosh 2 i vv sinh 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

v v
a0 = a0 a
,
v c

v
v
v
a
( 1) a0 ;
a = a +
v
v
c
the reciprocal formulas are
A = bc A b,

v v
,
a0 = a0 + a
v c

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

v2
2
2 2
2 2
dXdXc = c dt (dx) = c dt 1 2
c
2 2
c dt
= c2 d 2
=
2
which denes the proper time
dt
= dt
d =
with =
by
q 1
2
1 vc2

1
v2
c2
, v being the velocity of the particle. The four-velocity V is dened
V =
dX
= jc + kv
d
96
with v =

dx
dt
and V Vc = c2 ; V can also be written in the form

V = c (j cosh + km sinh )
with v = vm (m m = 1) and tanh = vc , = cosh , sinh = vc .

Four-acceleration
Let V = jc + kv be the four-velocity and d =
the four-acceleration is dened by
A=
dt
the proper time dierential;
dV
= [jc + k (v
+ a)]
d
with
v v
d
= 2 3
dt
c
3
= (v a) 2
c
(6.1)
(6.2)
where the relation v v = v v deduced from (v)2 = v 2 has been used. Furthermore,
V V = c2 from which one obtains by dierentiating with respect to , V A = 0
which gives again equation (6.2)
(v a) = c2
.
3
The four-acceleration can be written

A = j 4
with =
q 1
2
1 vc2
(v a)
(v a) v
2
+ k 4
+
a
c
c2
; one has the relativistic invariant

2
(v a)
2
AAc = 6
4 (a)
2
c

va
= 6 a2 +
c2
with (v a)2 = v 2 a2 (v a)2 . The bivector V A is given by
V A = (v a) 3 iac 3 ;
since V A = (V A) (V A) = (V A), one has
(V A) (V A)c = V AAc Vc = c2 AAc

2
= 6 a2 c2 + (v a) .
6.2. Relativistic kinematics
97
The four-acceleration in the proper frame (v = 0) is simply A = ka with

2
AAc = (ap ) = a2p .

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

AAc = a
6 2
v2
1 2
c
= 4 a2 = a2p ,
ap = 2 a;
when 1, one remarks that the proper acceleration can be very much larger
than the acceleration in the laboratory frame [42, p. 101].
Wave four-vector
The wave four-vector is dened with k =
K=j
2
n
by
+ kk;
c
writing X = jct + kr one has the relativistic invariants

KKc =
2
2
(k) ,
c2
K X = t k r.
6.2.2 Addition of velocities

Special Lorentz transformation
Consider the special Lorentz transformation
V = bV bc
(6.3)
with
b = cosh
iI sinh ,
2
2
tanh =
w
,
c
1
cosh = =
1
w2
c2
the four-velocities
V = c (j cosh + km sinh ) ,
V = c (j cosh + km sinh ) ,
98
and
tanh =
v = vm,
v
,
c
tanh =
v = v m
v
,
c
(m m = m m = 1).
Equations (6.3) read

cosh = cosh cosh m1 sinh sinh ,
(6.4)
1
(6.5)
2
(6.6)
3
(6.7)
m sinh = m cosh sinh cosh sinh ,

m sinh = m sinh ,
m sinh = m sinh .
Dividing equation (6.5) by the equation (6.4), one obtains
m1 tanh =
m1 tanh tanh
1 m1 tanh tanh
and thus
v 1 =
v1 w
.
v1 w
1 2
c
(6.8)
Similarly, one obtains

v2
,

v1 w
cosh 1 2
c
3
v

v 3 =
v1 w
cosh 1 2
c
v 2 =
with cosh = =
q 1
2
1 w
c2
. The inverse formulas are (with =

v 1 + w
,
v 1 w
1+ 2
c
2
v
1 2
v2 =
,
v 1 w
1+ 2
c
3
1 2
v
v3 =
.
v 1 w
1+ 2
c
v1 =
(6.9)
(6.10)
w
c)
(6.11)
(6.12)
(6.13)
6.3. Relativistic dynamics of a point mass
99
General pure Lorentz transformation

It is expressed for four-velocities by
V = bV bc
(6.14)

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

tanh = c , tanh = c , tanh = c . Explicitly, equation (6.14) reads

w
tanh tanh ,
cosh = cosh cosh 1 m
w
w
w

m sinh = m sinh
m
sinh
w
w

w
w
w
m
sinh cosh cosh sinh .
+
w
w
w
Dividing equation (6.17) by equation (6.15), one obtains with =
[40, p. 75]

v w

2 1
1
v 1 2 + w
1
w2
v =
vw
1 2
c
and the inverse formula

v w

2
2
1 1 +1
v 1 +w
w2
.
v=
v w
1+
c2
w
c
(6.15)
(6.16)
(6.17)
the formulas
6.3 Relativistic dynamics of a point mass

6.3.1 Four-momentum
Let V = jc + kv be the four-velocity of a particle and m0 its mass, the fourmomentum is dened by
P = m0 V = jm0 c + km0 v
E
= j + kp
c
where E = m0 c2 is the energy of the particle and p = m0 v its momentum (of
norm p). Furthermore, one has the relativistic invariant
E2
2
(p)
c2
= m20 V Vc = m20 c2 ,
P Pc =
100
hence
E 2 = p2 c2 + m20 c4 .
In the proper frame (v = 0), one has E0 = m0 c2 ; the kinetic energy is dened by
T = m0 c2 ( 1) .
Under a standard
transformation,
the four-momentum transforms as P =
Lorentz

bP bc with b = cosh 2 iI sinh 2 and tanh = wc = , = q 1 w2

1
c2

E = E p1 w ,

E
1
1
p = p 2w ,
c
p2 = p2 ,
p3 = p3 .
The inverse transformation is P = bc P b or

E = E + p1 w ,

E
p1 = p1 + 2 w ,
c
p2 = p2 ,
p3 = p3 .
Under a general pure Lorentz transformation, one has P = bP bc with b = cosh 2

i vv sinh 2 .
6.3.2 Four-force
Let P = j Ec + kp be the four-momentum vector of the particle; the four-force
vector is dened by
d (m0 V )
dP
=
d
d
dm0
V
= m0 A +
d
F =
with V = jc+ kv [42, pp. 123-124]. If the four-force conserves m0 , then

and
d (jm0 c + km0 v)
F = m0 A =

dt
dE
= (v) j
+ kf
cdt
dm0
d
=0
6.3. Relativistic dynamics of a point mass

with, by denition f =
dp
dt
d(m0 v)
.
dt
101
Furthermore,

dm0 2
dm0
V V =
c
m0 A +
d
d

dE
f v .
= 2 (v)
dt
F V =
If
dm0
d
= 0 = F V , then
dE
dt
= f v and

(f v)
+ kf .
F = (v) j
c
Since
d (m0 v)
d (m0 )
= m0 a +
v
dt
dt
dE
(f v)
= m0 a + 2 v = m0 a +
v,
c dt
c2
f=
one infers that a is coplanar to f and v but is not in general parallel to f . If

dm0
d = 0, the force f satises the relation [42, p. 125]
(v) f = m0
d2 x
dt2
(6.18)
which can be established as

d2 X
F = m0 A = m0 2
d

dE
+ kf
= (v) j
cdt
2
d (jct)
d2 x
= m0
+
k
,
d 2
dt2
(6.19)
(6.20)
(6.21)
hence the relation (6.18) by comparing equations (6.20) and (6.21).

Under a speQ
+
kf
with Q = dE
cial Lorentz transformation, the four-force
F
=
(v)
j
c
dt ,

transforms as F = bF bc with b = cosh 2 iI sinh 2 , tanh = wc et cosh =

= q 1 w2 or explicitly
1
c2

w
,
F 0 = (w) F 0 F 1
c

w
,
F 1 = (w) F 1 F 0
c
2
2
3
3
F =F , F =F ,
102
and equivalently

(v ) Q = (w) (v) Q f 1 w ,

Qw
(v ) f 1 = (w) (v) f 1 2 ,
c
(v ) f 2 = (w) f 2 ,
(v ) f
3
(6.22)
(6.23)
(6.24)
= (w) f .
(6.25)
Furthermore, one has the relation

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

ds2 = dt2 c2 v 2 = dt 2 c2 v 2
and the relation

w
cdt = (w) cdt dx
c

v1 w
= (w) cdt 1 2 .
c
Then one obtains

2

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

2
2
w
c v 2 1 c2

2
,
c v 2 =

2
1
1 vc2w
1
1
1
1
= 2
,
2 (v )
(v) 2 (w) 1 v12w 2
c
hence, the relation (6.26). Reciprocally, one has

v 1 w
(v)
=
(w)
1
+
.
(v )
c2
(6.26)
6.4. Exercises
103
Finally, equations (6.22), (6.23), (6.24), (6.25) can be written [42, p. 124]
Q =
f 1 =
Q f 1w
,
1
1 vc2w
f1
1
wQ
c2
,
v1 w
c2
2
f 2 =
f

(v) 1
v1 w
c2
f 3 =
f3

(v) 1
v1 w
c2
,
.
6.4 Exercises
E6-1 Two particles A and B move towards the origin O in opposite directions
(on the Ox axis) with a uniform velocity 0, 8 c. Determine the relative velocity of
B with respect to A for an observer at rest relatively to A.
E6-2 Consider an isolated set of particles without interaction in a reference frame
K at rest. Determine the total relativistic angular momentum
'
L=
Xi Pi ,
Xi = jct + kIxi + kJyi + kKzi ,
Ei
+ kIpxi + kJpyi + kKpzi .
Pi = j
c
Dene the center of energy of the set. Show that it moves with a constant velocity.
E6-3 Consider a hyperbolic rectilinear motion of a particle whose acceleration is
constant in the proper reference frame, at any instant. The particle being at rest
at the origin of the axes and of the time, determine the four-vector X = jct + kx
as a function of the parameter = g
c where is the proper time of the particle.
Show that one obtains the classical results of a uniformly accelerated motion for
small velocities 1.
Chapter 7
Classical electromagnetism
Classical electromagnetism is treated within the Cliord algebra H H over R.
This chapter develops Maxwells equations, electromagnetic waves and relativistic
optics.
7.1 Electromagnetic quantities

7.1.1 Four-current density and four-potential
Four-current density
Let 0 be the charge density in the proper frame and V = jc + kv the fourvelocity; the four-current density is dened by
C = 0 V = j0 c + k0 v
= jc + kj
with = 0 and j = v. One obtains the relativistic invariant

v2
2
2 2
2
2 2
CCc = c (v) = c 1 2
c
2
= c2 2 = c2 20
with =
q 1
2
1 vc2
. Under a pure special Lorentz transformation, the four-current
density transforms as C = bCbc with b = cosh 2 iI sinh 2 , tanh =

cosh = = 1 2 ; explicitly, one has
1

c = c j 1 ,

j 1 = j 1 c ,
j 2 = j 2 ,
j 3 = j 3 .
w
c
= and
106
Chapter 7. Classical electromagnetism
Consider a trivolume OM (, , ) = jx0 (, , )+kx (, , ) depending on three

parameters , , . The innitesimal hyperplane dT is expressed by (4.17)

OM OM OM
dT =
ddd

kdx1 dx3 dx2 + jIdx2 dx3 dx0
.
=
+jJdx3 dx1 dx0 + jKdx1 dx2 dx0
The dual dT of dT is given by
dT = idT

jdx1 dx2 dx3 + kIdx2 dx3 dx0
=
+kJdx3 dx1 dx0 + kKdx1 dx2 dx0
and is a four-vector orthogonal to dT . The relativistic invariant C dT is given
by

cdx1 dx2 dx3 v 1 dx2 dx3 dx0
C dT =
;
v 2 dx3 dx1 dx0 v 3 dx1 dx2 dx0
in the proper frame C dT = 0 cdx1 dx2 dx3 = cdq where dq is the electric charge
contained in dT . The electric charge Q contained in the hyperplane T is given by
&&&
1
Q=
C dT
c
T
for any Galilean frame. Furthermore, if one integrates over a closed four-volume
, one obtains
%
%
C dT = ( C) dx0 dx1 dx2 dx3
c
v
v
v
with the relation C = x
= 0 expressing the con0 + x1 + x2 + x3
servation of electric charge. If the four-volume is limited by the hypersurfaces
H1 (t1 = const.), H2 (t2 = const.) and the lateral hypersurface H3 at innity, one
has (Figure 7.1)
&
&
&
C dT +
H1
C dT +
H2
C dT = 0;
H3
furthermore, the integral over H3 is nil in the absence of electric charges at innity,
hence
&
&
Q2 = C dT = C dT = Q1
H2
H1
which expresses the conservation of electric charge.
7.1. Electromagnetic quantities
107
t
H2
t2
H3
t1
H1
Figure 7.1: Conservation of electric charge: H1 and H2 are hyperplanes (at t constant) and H3 is a lateral hypersurface at innity.
Four-potential vector
Let V be the scalar potential and A the potential vector, the four-potential vector
2
2
is dened by A = j Vc + kA yielding the relativistic invariant AAc = Vc2 (A) .
Under a special Lorentz transformation, one has A = bAbc with b = cosh 2

iI sinh 2 , tanh = wc = and cosh = = 1 2 , hence
1

V
1
A ,
c

V
1
1
A = A ,
c
V
=
c
A2 = A2 ,
A3 = A3 .
7.1.2 Electromagnetic eld bivector

Four-nabla operator
The four-nabla operator
=j
kI 1 kJ 2 kK 3
ct
x
x
x
transforms under a general Lorentz transformation as a four-vector, i.e., =

aac . Let us demonstrate it explicitly in the case of a pure special Lorentz trans-
108
formation. The transformation formulas are

ct = (ct + x ) ,
x = (x + ct ) ,
y = y,
with =
v
c
and =
q 1
2
1 vc2
z = z,
. One then obtains
1 t
x
y
z
=
+
+
+
ct
c t t
x t
y t
z t

ct
x
with
t
t
= ,
x
t
= c,
y
t
with
x
x
= ,
t
x
z
t
= 0. Furthermore,

x
t
+
x
x x
t x

x
ct
= c . Finally, one has

=
,
ct
ct
x

,
=
x
x
ct
= ,
= ,
y
y
z
z
which shows that the four-nabla operator transforms indeed as a four-vector, the
demonstration being similar in the general case.
Electromagnetic eld bivector

Let A = j Vc + kA be the four-potential vector, let us dene the Cliord number
F = c A = A
= ( A) + ( A)
7.1. Electromagnetic quantities
109
and adopt the Lorentz gauge ( A) =

the electromagnetic eld bivector
( Vc )
ct
A
+ A
x1 + x2 +
A3
x3
= 0, which gives
F = A

V
A
= rot A + i grad
c
ct
E
= B + i
c
with the usual denitions of the magnetic induction B and the electric eld E,
E = grad V
B = rot A,
A
.
t
(7.1)
Under a gauge transformation

A A = A + f
where f is an arbitrary scalar function, one obtains the same electromagnetic eld
bivector
F = A = A + ( ) f
= A = F.
The electromagnetic eld bivector yields the relativistic invariant
F2 = F F + F F
#
"
2
(E)
EB
2
.
= (B) + 2 + 2i
c
c
Under a pure special Lorentz transformation , one has
F = bF bc
with b = cosh 2 iI sinh 2 , tanh =
v
c
= and cosh = =

E3
2
B I B cosh + c sinh J

2
+ B 3 cosh + E sinh K
2

F =
1
E
E
3
+ iI +
cosh
B
sinh
iJ
c
c

E3
cosh + B 2 sinh iK
+
c
= B + i
E
c
q 1
2
1 vc2
, hence
110
or
B
1

E3
,
B = B +
c

E 2 = E 2 B 3 c ,
2
=B ,
E 1 = E 1 ,

E2
3
B = B
,
c

E 3 = E 3 + B 2 c .
3
Under a general pure Lorentz transformation, one has F = bF bc with b =

cosh 2 i vv sinh 2 , tanh = vc = and cosh = = q 1 v2 ; hence the for1 c2
mulas [40, p. 191]

1
1
v
1 2 (v E) ,
B = B + 2 (v B)
v
c

1
v
E = E + 2 (v E)
1 + (v B) .
v
7.2 Maxwells equations

7.2.1 Dierential formulation
In vacuum
With the Lorentz gauge A = 0, one has
F = A = B + i
E
,
c
F = A = 0,

B
E
= k ( div B) + j
rot
;
ct
c
hence, one obtains two of the Maxwells equations
div B = 0,
rot E =
B
.
ct
The two other equations are given by

F = 0 C
= j div

1 E
E
+ k rot B 2
c
c t
with the four-current density C = jc + kv; one obtains with 0 0 c2 = 1,

div E = 0 c2 =
rot B =
,
0
1 E
+ 0 j.
c2 t
111
The complete set of Maxwells equations reads

F = c A = F + F

E
1 E
j div c + k rot B c2 t

=
E
B
rot
+k ( div B) + j
ct
c
= A = 0 C
with the dAlembertian operator = c =
four-potential vector. Hence, the equations
2
c2 t2
and A = j Vc + kA the
= 0 c2 = ,
c
0
A = 0 j,
or equivalently
div B = 0,
rot E =
div E =
B
,
t

rot B = 0 j +
,
0
0 E
.
t
If one does not adopt the Lorentz gauge, one obtains

F = c A = A

V
E
=
div A +
B+i
ct
c
which is not a bivector but an element of C + . One obtains the same Maxwells
equations
F = c A
= A = 0 C.
Since C = 0 expresses the conservation of electric charge, one has
(A) = 0 C
= ( A) = 0,
a condition which is less restrictive than the Lorentz gauge A = 0. The covariance of Maxwells equations is manifest since under a general Lorentz transformation any Cliord number X transforms as X = aXac (aac = 1) and thus
F = 0 C = aF ac
= a0 Cac ;
112
hence, F = 0 C.
Perfect dielectric or magnetic medium
Consider the bivectors F = B + i Ec , G = H + icD. Maxwells equations are
expressed by
F = 0,
G = C = jc + kv

D
= j div (cD) + k rot H
t
or
div B = 0,
div D = ,
D
B
,
rot H = j +
.
t
t
The bivector G yields the relativistic invariant
rot E =
G2 = G G + G G
= H2 + c2 D2 + 2icD H.
Under a pure special Lorentz transformation, G transforms as G = bGbc with
b = cosh 2 iI sinh 2 , tanh = vc = and cosh = = q 1 v2 , hence
1 c2
Hx = Hx ,
Hy = (Hy + vDz ) ,
Hz = (Hz vDy ) ,
Dx = Dx ,

Hz

Dy = Dy v 2 ,
c

Hy

Dz = Dz + v 2 .
c
Under a general pure Lorentz transformation, one has G = bGbc with b = cosh 2
i vv sinh 2 , tanh = vc = and cosh = = q 1 v2 ; hence
1 c2
1
v
1 (v D) ,
H = H + 2 (v H)
v

1
H
v

1 + v 2
D = D + 2 (v D)
,
v
c
(7.2)
(7.3)
113
and the inverse formulas

1
v

H = H + 2 (v H )
1 + (v D ) ,
v

1
H
v

D = D + 2 (v D )
1 v 2
.
v
c
As an application, consider a perfect medium in the proper reference frame K
with B = 0 r H , D = 0 r E , moving with respect to a Galilean reference frame
(at rest) K with a velocity v. Using the transformation formulas (7.2), (7.3), one
obtains

1
1
v
1 2 (v E)
(7.4)
B + 2 (v B)
v
c

1
v
= 0 r H + 2 (v H)
1 (v D) ,
(7.5)
v

1
v
H
D + 2 (v D)
1 + v 2
v
c

1
v
1 + (v B) .
= 0 r E + 2 (v E)
v
(7.6)
(7.7)
Furthermore, from equations (7.5), (7.7) one deduces the relations

v B = 0 r v H,
v D = 0 r v E.
Equations (7.5), (7.7) then become
1
(v E) = 0 r [H (v D)] ,
c2

H
D + v 2 = 0 r [E + (v B)] ;
c
(7.8)
(7.9)
taking H from equation (7.8) and replacing it in equation (7.9) one nds [7, p.
239]

0 r 1r
E
v Bv 2 .
D = 0 r E +
1 2
c
Similarly, taking D from equation (7.9) and replacing it in equation (7.8) one
obtains

0 r 1r
B
H=
+
v (E + v B) .
0 r
1 2
One observes that D and H depend on E and B as well as on the velocity of the
material.
114
Arbitrary dielectric or magnetic medium

In an arbitrary dielectric or magnetic medium, with a polarization density P and
a magnetization density M, one has the relations
D = 0 E + P,
B = 0 (H + M)
which one can write in the form
F = 0 (G + N )
with the bivectors F = B + i Ec , G = H + icD and N = M icP. Maxwells
equation G = C (with C = jc + kv) then becomes

F
G=
N =C
0
or
F = 0 (C + N )
with

P
N = j div (cP) + k rot M +
,
t

1 E
E
+ k rot B 2
,
F = j div
c
c t
hence, the relations
1
( div P) ,
c

E
P
rot B = 0 j +
+ rot M + 0 0
.
t
t
div E =
The entire set of Maxwells equations is expressed by (with F = 0)

F = F + F = A
= 0 C + 0 N
or
div P
,

0
P
A = 0 j +
+ rot M .
t
V =
115
Consequently, one can replace the medium by a distribution, in vacuum, of charge

density = div P and a current density j = P
t + rot M. Under a general pure
Lorentz transformation, the bivector magnetization-polarization N transforms as
N = bN bc with b = cosh 2 i vv sinh 2 , tanh = vc = and cosh = = q 1 v2 ,
hence
1 c2

1
v
1 + (v P) ,
M = M + 2 (v M)
v

1
M
v
P = P + 2 (v P)
1 v 2
v
and the inverse formulas

1
v

M = M + 2 (v M )
1 (v P ) ,
v

1
M
v

1 + v 2
P = P + 2 (v P )
.
v
c
For low velocities ( 1), one obtains
M = M v P ,
P = P + v
M
.
c2
If, in the proper reference frame K , one has M = 0 and P = 0, then in the
reference frame at rest K one has M = v P ; if in K , P = 0, M = 0, then
one obtains in K a polarization
P=v
M
.
c2
7.2.2 Integral formulation

The relativistic integral formulation of Maxwells equations in vacuum results from
the general formulas valid for any bivector F (4.20), (4.21)
%
&
F dS = ( F ) dT,
%
&
F dS = ( F ) dT,
with dS = dS + idS ,
dS = dydzI + dzdxJ + dxdyK,
dS = cdxdtI + cdydtJ + cdzdtK,
dT = kdxdydz + j (dydzcdtI + dzdxcdtJ + dxdycdtK) ,
116
and where the integration is taken on a closed surface . Considering F = B + i E

c
with F = 0 and F = 0 C, one obtains [4, p. 227]
%
%
F dS =
%
= 0,
i
F dS =
c
Bx dydz + By dzdx + Bz dxdy

+Ex dxdt + Ey dydt + Ez dzdt
&
&
= i0
Ex dydz + Ey dzdx + Ez dxdy

Bx dxcdt By dycdt Bz dzcdt
(cdxdydz jx dydzdt jy dzdxcdt jz dxdycdt) .
If one operates at t constant, one nds again the standard equations (in classical
three-vector formulation)
%
%
q
B dS = 0, E dS = .
0
Furthermore, using the general formula (4.18)
%
&
A dl = ( A) dS
one obtains for the four-potential vector A (with the Lorentz gauge A = 0)
%
&
A dl = F dS
or [4, p. 231]
&
%
A dx V dt =
Bx dydz + By dzdx + Bz dxdy

+Ex dxdt + Ey dydt + Ez dzdt

.
At t constant, one obtains the classical vector formulation of Stokes theorem

%
&
&
A dl = rot A dS = B dS.
7.2.3 Lorentz force

The four-force density (per unit volume of the laboratory frame) is expressed with
F = B + i Ec and C = jc + kv by
f = F C
= j (j E) + k (E + j B) ;
117
f can also be expressed in the form (with F = 0 C and = e )

1
F C CF
=
[F (F ) (F ) F ]
2
20
1
=
F ( e F ) ( e F ) F
20
1
=
[ (F e F ) ( F ) e F ( e F ) F ] .
20
f=
Since
F + F
2
1
= [ (e F ) + ( F ) e ] = 0
2
F =
it follows that
f=
1
(F e F ) = t
20
with
t =
F e F
= T e
20
where T is the energy-momentum tensor; t is a four-vector since (t )c = t

(with Fc = F ). One obtains for t ,
EH
(B H + E D)
+k
,
2
c
2 E2

B + c2
E12
2
I
(E
H)
k
2
1
c
1

,
2
+
t1 = j
c
0
B1 B2 + E1c2E2 J B1 B3 + E1c2E3 K

E1 E2
I
B
B
+
1
2
2
c
(E H)2
k
B2 + E2

,
t2 = j
+
E22
E2 E3
c
0
2
c2
J
B
K
B
+
2
2
3
2
2
2
c
c

B1 B3 + E1c2E3 I
(E
H)
k
3

B2 + E2
.

+
t3 = j
E2
c
0
ce
B B + E2 E3 J +
B2 3 K
t0 = j
c2
c2
The four-vectors t constitute the energy-current density ( = 0) and the momentum-current density along the three axes (x, y, z); the component of the
four-momentum of the electromagnetic eld of a trivolume T is given by
&
P = t (dT ) .
118
The symmetry of the energy-momentum tensor T of the electromagnetic eld

can be veried directly on the components of t
t e = t e
= T e e = T = T ;
it is sucient to verify the equality

(F e F ) e = e F e F ,
(7.10)
F a = aF
(7.11)
with a = e F e e F e 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 Maxwells equations F = 0 C with F = B + i Ec , C = jc + kj,
one obtains in the absence of electric charges and currents
c F = F = 0
where = c =
2
c2 t2
is the dAlembertian operator; hence

B = 0,
E = 0.
A sinusoidal plane wave rectilinearly polarized is, in complex notation, of the type
F = Fm ei (tkr) = B + i
E
c
where Fm = Bm + i Ecm is a constant bivector (Bm , Em real). One has, in a

vacuum,

F = i j + kk F
c

j kE + k (k B)
c

= i
+j (k E B) + k k B
= 0;
E
c2
7.3. Electromagnetic waves
119
hence, the relations characterizing the plane wave

k E = 0,
k B = 0,
kE
,
B=
(k B) c2
E=
.
The relation
c F =
(7.12)
(7.13)
(7.14)
(7.15)

2
2
j + kk F = 2 + k k F = 0
c
c
= c.
gives the norm of k, |k| = c and the phase velocity v = |k|
For a plane wave polarized elliptically, one just needs to take Fm complex.
7.3.2 Electromagnetic waves in a conductor

Consider a conductor having a permittivity = 0 , a permeability = 0 and a
conductivity (j = E). Maxwells equations are expressed by F = 0 C, hence
c F = F = 0 c C
(7.16)
= 0 ( C + C) = 0 ( C)

j
= 0 rot j + i
(c) grad
ct
(7.17)
(7.18)
with the conservation of electric charge ( C = 0). Using the relations j = E

and F = 0, equation (7.18) becomes
F + 0
F
= i(0 c) grad .
t
In the absence of an electric charge density ( = 0), one has

F + 0
F
= 0.
t
(7.19)
Consider a sinusoidal plane wave rectilinearly polarized F = Fm ei t(tkr) with

the constant bivector Fm = Bm + i Ecm . Equation (7.19) gives with

2
F
2
= i F ,
F =
2 + |k| F
t
c
a complex expression of k
2
i 0 .
c2
The structure of the plane wave results from equation F = 0 C with C =

Cm ei (tkr) and yields the relations (7.12), (7.13), (7.14), (7.15) with k complex.
|k|2 =
120
7.3.3 Electromagnetic waves in a perfect medium

Wave equation
In a perfect medium, one has the relations
F = 0,
G= C
with F = B + i Ec , G = H + icD, C = jc + kj. Hence

( G) = C
(7.20)
with
(
rot D
rot (rot H)
t

( G) =
2 D
H
+i ct2 (c) grad (div D) rot
ct

j
(c) grad .
C = rot j + i
ct
)
,
Equation (7.20) can be written in the form

G
2G
G
+ grad (div D) .
=
2
t
t
In a nonconducting medium ( = 0) and in the absence of an electric charge

density, one obtains the wave equation
G
2G
= 0.
t2
(7.21)

k, G = H + vD, F = B + i E
Introducing the operator = j vt
v = G ,
1

C = jv + kj and the constant v = , Maxwells equations are expressed by [5]
G = 0,
G = C
or G = C . Hence, the wave equation
c G = c C
= G .
In a nonconducting medium ( = 0), one nds again the equation (7.21).
7.4. Relativistic optics
121
Sinusoidal plane wave rectilineraly polarized

Consider G = Gm ei (tkr) with the constant bivector Gm = Hm + ivDm and
v = 1 . In the absence of a four-current density C , one has

G = i j + kk G
( v
)
j (vk D) + k (k H)

=i
+j vk D H
+ k (k H D)
v
= 0;
hence, the relations characterizing the plane wave
k D = 0,
k H = 0,
kD
,
k H
.
D=
H = v2
7.4 Relativistic optics

7.4.1 Fizeau experiment (1851)
In the Fizeau experiment (Figure 7.2), one observes at O a system of interference
fringes when the water is at rest. When the water circulates, the fringes move and
the experiment has determined the velocity of light in the moving water to be

1
c
v = 1 2 w
n
n
whereas the classical theory predicts
v=
c
w.
n
Classical theory
The variation of the optical path between the two experiments is
"
#
L
L
c

= c c
n w
n +w
=
2Lcw
2Ln2 w
.

c
w2
c2
n2
122

source
water inow
T1
w
w
T2
L1
L2
L
Figure 7.2: Fizeau experiment (1851): when the water is at rest, one observes
interference fringes; when the water circulates, the fringes move which allows one
to determine the velocity of light in the moving water.
The variation of the order of interference is
2Ln2 w
=
.
pc =
0
c0
Relativistic theory
Let K be the proper frame of the water in motion with nc the speed of light with
respect to K . With respect to the frame at rest K, the velocities of light with
respect to the tubes T1 , T2 are respectively using the formula (6.11),

c
1
c
n w
1 2 w,
v1 =
w
1 nc
n
n

c
1
c
n +w
+ 1 2 w,
v2 =
w
1 + nc
n
n
hence, a variation of the optical path ,

L
L
= c
v1
v2
2

2Lcw n 1
2Lw 2
=
n 1 ;

2
2
2
c n w
c
the variation of the order of interference is

2Lw 2
n 1 .
=
pr =
0
c0
7.4. Relativistic optics
123
Example. Take w = 10 m/s, L = 5 m, 0 = 0, 6 m, n = 4/3, one obtains

pc = 0, 99;
pr = 0, 43.
7.4.2 Doppler eect

Consider an optical source at rest within the proper frame R moving with a
velocity w with respect to the frame at rest R (Figure 7.3).
y
R
R
O
O
w
x
z
Figure 7.3: Doppler eect: an optical source (at rest in R ), of frequency f , located
at O moves with a velocity w along the Ox axis. In the reference frame R, the
measured frequency is f .
Let K = j c + kk be the wave four-vector in R and K = j c + kk the wave

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

w
w
= 1 + cos
,
= + kx
c
c

w
.
f = f 1 + cos
c
Longitudinal Doppler eect
1. = 0 (the source moves towards the observer located at O, Figure 7.4)
!

1 + wc
w

f = f 1 +
=f
f ;
c
1 wc
2. = (the source recedes from the observer, Figure 7.4)
!

1 wc
w

f = f 1
f .
=f
c
1 + wc
124

y
y
O
w
k
y
O
k
=0
x
Figure 7.4: Longitudinal Doppler eect: in the case = 0, the wave is emitted in
the direction of w towards the observer located at O in the frame R; for = ,
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 eect ( = 2 )
In this case, one has
f
f = 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 (Figure
7.5).
y
y
vx
R
v
vy
v
O
Solar
R
x
O
Earth
Figure 7.5: Aberration of distant stars: in the reference frame R of the sun, the
velocity of light coming from a star located on the Oy axis is v; in the reference
frame R of the Earth having a velocity w with respect to R, the velocity of light
is v .
7.5. Exercises
125
The velocity v in the mobile reference frame of the Earth is given by equation
(6.8)
vx w
= w,
1 vcx2w

2
vy 1 wc2
w2

= c 1 2 ,
vy =
vx w
1 c2
c

2
vz 1 wc2

= 0.
vz =
1 vcx2w
vx =
The angle under which one sees the star from the Earth is given by

v
w
w
= .
tan = x =
vy
c
w2
c 1
c2
7.5 Exercises
E7-1 Consider a charge q located at the origin of a reference frame K moving with
a constant velocity v along the Ox axis of a reference frame K at rest. Determine
the four-potential and the electromagnetic eld in the frame K at rest.
E7-2 Consider an innite linear distribution of charges with a linear density 0
(0 > 0) along the O x axis of a reference frame K moving with a velocity v along
the Ox axis of a reference frame K at rest. Along the Ox axis of the reference
frame K, one has an innite linear distribution with a linear density ( > 0)
of charges at rest. Knowing that the total linear density of charges in the reference
frame K at rest is nil, determine the electromagnetic eld created in K by these
charges at the point M (x , r , 0). A particle of charge q moves with a velocity v
in K, parallel to the Ox axis (at the distance r). Determine the force exerted by
the two sets of charges in the reference frame K and in the proper frame of the
particle.
E7-3 Consider the electromagnetic eld
F = I + 2
iJ
c
(Bx = 1T, Ey = 2V /m)
in the reference frame of the laboratory K at rest . Determine the electromagnetic

eld in the reference frame moving with respect to K with a velocity 2c in the
direction (1, 1, 0) of K.
E7-4 Show that Maxwells equations in vacuum without sources are invariant
under the transformation
F F = iF
(i.e., B
E E
,
B).
c c
Chapter 8
General relativity
The general theory of relativity is developed within the Cliord algebra HH over
R. Einsteins equations and the equation of motion are given as well as applications
such as the Schwarzschild metric and the linear approximation.
8.1 Riemannian space

Consider a four-dimensional space with the elementary displacement DM = i ei
and the ane connection Dei = ij ej . The covariant dierentiation of the vector
A is dened by
DA = dA + d A
with 2d = ij ei ej et ij = ki (ek ej ). The reciprocal basis e is dened by
e e = (e0 = e0 , e1 = e1 , e2 = e2 , e3 = e3 ) where e0 , e1 , e2 , e3 are
unitary orthogonal vectors. Under a Lorentz transformation A = f Afc , one has
DA = f DAfc ,
d = f dfc 2df fc .
A Riemannian space is a space without torsion but with a curvature. The absence
of torsion is expressed by
D2 (D1 M ) D1 (D2 M ) = 0
(8.1)
where D1 , D2 are two linearly independent directions. Writing d = Ik dxk , DM =

m dxm , condition (8.1) leads to the relations
Ik m Im k =
k
m
xm
xk
which determine Ik when DM is given. The existence of a curvature is expressed

by the relations
(D2 D1 D1 D2 )A = A
128
Chapter 8. General relativity
where is a bivector dened by

= (d2 d1 d1 d2 ) + [d2 , d1 ]
= km dS km /2
*
+
Ik
Im
=
[I
,
I
]
d1 xk d2 xm
k m
xm
xk
ij
with 2km = Rkm
ei ej and dS = D1 M D2 M = dS km ek em /2. Bianchis
rst identity is given by
ij ek + jk ei + ki ej = 0,
ij (ek em ) = km (ei ej ),
(8.2)
(8.3)
where relation (8.3) results from the preceeding equation. Bianchis second identity
is expressed by
ij;k + jk;i + ki;j = 0
with D3 (d2 , d1 ) = d3 + [d3 , ] = ij;k i (d2 ) j (d1 ) k (d3 ). The Ricci tensor
h
Rik = Rihk
and the curvature R = Rkk are obtained by the relations
Rk = ik ei = Rik ei ,
R = (ik ei ) ek = ik (ei ek ).
8.2 Einsteins equations

To deduce Einsteins equations from a variational principle, we shall use the
method of the exterior calculus but within the framework of a Cliord algebra
and without using the exterior product of dierential forms. Adopting, for simplicity, an orthogonal curvilinear coordinate system, we can write
&
L = R gdx0 dx1 dx2 dx3

&
= ik (ei ek ) 0 1 2 3
with
gdx0 dx1 dx2 dx3 = 0 1 2 3 . Taking the dual, we obtain

&

L = ik ei ek 0 1 2 3 .
The variation gives
L =
&

ik I K ei ek G H
&

+ ik I K ei ek G H
8.3. Equation of motion
129
with i = I, k = K and where the second integral vanishes. Furthermore, one has

= (e ) e ei ek
ei ek

= (e ) ei ek e

.
= (e ) ei ek e
Hence
&

ik I K ( g e ) ei ek e h
&

= ( g e ) ik ei ek e I K h = 0
L =
with g = , i = I, k = K (without summation). Finally, one obtains Einsteins

equations in a vacuum

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

1
ij ei ej ek = T k
2
(8.4)
with T k = T ik ei , = 8G/c4 where T ik is the energy-momentum tensor ([50],

[23], [34]). The standard expression of Einsteins equations is obtained as follows.
Taking the dual of equation (8.4), one obtains

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

1
1
ij ei ej ek e = Rij
(e e e ) ei ej ek
2
2
1 ijk
1
k
= Rij
= R
k R
2
2
= T k e = T ik ei e = Tk .
8.3 Equation of motion

To obtain the equation of motion, one writes
&
S = mc (ds) = 0
130

2
with (ds) = (ds) /2ds and ds2 = DM DM = (i k ) dxi dxk . One obtains
&
&

(l k ) uk dxl
S = mc (i k ) ui uk ds mc
= A+B = 0
with ui = dxi /ds. The second term B can be rewritten in the form
&
&

k
l
B = mc
(l k ) u x + mc xl d (l k ) uk .
Since, i = i,l xl + i with = Ik dxk , one obtains

i
d (l k ) k
duk
i k
=0
+
I
u
u
+
+
(
u
l
i
k
l
k
xl
ds
ds
or, equivalently,

i
l
k i
xl + xi Il i + Ii l k u u

k
duk
i k
+l
+
I
i
k u u + (l k )
i
x
ds
=0
where the expression in brackets vanishes in the absence of torsion. One thus
obtains

du
i
l
+ Ii u u = 0
ds
with the four-velocity u = DM/ds = i ui and the ordinary derivative du/ds.
Finally, the equation of motion is expressed by
du d
Du
=
+
u = 0.
ds
ds
ds
8.4 Applications
8.4.1 Schwarzschild metric
Consider a metric having a spherical symmetry and the elementary displacement
DM ,
ds2 = e2a c2 dt2 e2b dr2 r2 d2 r2 (sin )2 d2 ,
DM = ea cdte0 + eb dre1 + rde2 + r sin de3
where a, b are functions of t and r. The bivector d is given by

b a+b
b
b
ab
d = cos dI e sin dJ + e dK + a e
+ e
iI
c
8.4. Applications
131
where the prime and the point designate respectively a partial derivative with
respect to r and t. The bivectors ik are given by
*

2a +

2
e

2

2b
01 = a + (a ) a b e
+ b b + ba
iI
,
c2
02 =
ab
a e2b
be
K+
iJ,
rc
r
b
b e2b
K + eab iJ,
r
rc

1 e2b
23 =
I.
r2
Einsteins equations, in a vacuum, are
12 =
03 =
ab
a e2b
be
J+
iK,
rc
r
13 =
b
b e2b
J + eab iK,
r
rc
23 e1 13 e2 + 12 e3 = 0,
23 e0 + 03 e2 02 e3 = 0,
13 e0 03 e1 + 01 e3 = 0,
12 e0 + 02 e1 01 e2 = 0.
Developing these four equations, one obtains respectively

1 e2b
2b e2b
2b
+
(8.5)
k eab jI = 0,
2
r
r
cr

1 e2b
2a e2b
2b ab
k = 0,
(8.6)
jI
+
r2
r
cr

b a 2b
r
jJ = 0,
(8.7)

e2a

2
2b
+ (b + b 2 b a)
(a
+
a
a
b
)e
c2

b a 2b
jK = 0.
(8.8)

e2a

2
2b
+ (b + b 2 b a)
(a
+
a
a
b
)e
c2
From equations (8.5), (8.6), one obtains b = 0 and a + b = 0 or equivalently
log e2(a+b) = C where C = 0 because e2a = e2b = 1 at innity; hence a + b = 0.
Equation (8.5) then gives
e2b (2b r 1) + 1 = 0,
2b
= 0,
re
e2b = 1 +
C1
,
r
132

and taking C1 = 2m (with m = GM/c2 ), one has e2b = e2a = 1
which determines completely the Schwarzschild metric. Finally, one has

1
2m 2 2
2m
2
ds = 1
dr2 r2 d2 r2 sin2 d2 ,
c dt 1
r
r

1/2
mcdt
2m
( sin dJ + dK) + 2 iI,
d = cos dI + 1
r
r
2m
m
m
01 = 3 iI,
02 = 3 iJ,
03 = 3 iK,
r
r
r
m
m
2m
12 = 3 K,
13 = 3 J,
23 = 3 I.
r
r
r

2m 1
r
The equation of motion is expressed simply in the form

Du
du d
=
+
u = 0.
ds
ds
ds
Developing this expression along the axes e0 , e1 , e2 , e3 one obtains

d
2m cdt
1
= 0,
ds
r
ds
m cdt
d 2
2
"
#
12
12
+r
2m
2m
dr
d
r2 ds
ds
1
= 1

ds
r
ds
r
d
2
+r sin2
ds
2

d
d
2 d
2
,
r
= r sin
ds
ds
ds

d
d
r2 sin2
= 0.
ds
ds
respectively
(8.9)
Adopting = /2, it follows from the equations (8.9), (8.12) that

d
2m cdt
= k,
r2
= h,
1
r
ds
ds
where k and h are constants. The relation uc u = 1 is expressed by
2

2
1 2
dr
d
cdt
2m
2m
1
r2
= 1,
1
r
ds
r
ds
ds
or with w = 1/r,
2
dw
d2 r
dr
2 2d w
= h ,
=
h
w
.
ds
d
ds2
d2
The projection of the equation on the axis e1 then leads to the relation
d2 w
m
+ w = 2 + 3mw2
d2
h
(8.10)
(8.11)
(8.12)
8.4. Applications
133
which is the relativistic form of Binets equation and from which one can deduce
the precession of the perihelion of Mercury in the usual way ([34]), ([46]).
8.4.2 Linear approximation

Consider the elementary displacement DM = i ei with i = dxi + hij dxj where
hij = hji = ik hkj is small as compared to 1 [50, p. 186]. The bivectors d and
km are given by
2d = (hik,j hjk,i )ei ej dxk ,
2km = (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 = km ei = (hm
k )em = 0,
the last equation being a gravitational wave equation (with =
2
2
y 2 z 2 ). Writing
1
km = hkm km h
2
with h = hm
m one obtains Einsteins equations
2
c2 t2
2
x2
k = T k
(with k = mk em ).
If one considers a homogeneous sphere (of mass M and radius r0 ) rotating
slowly with an angular velocity around the axis Oz , one integrates the above
equations and obtains the metric (with m = GM/c2 ) for r r0 ,

4GI
2m 2 2
2m 2
2
ds = 1
c dt 1 +
dx + dy 2 + dz 2 3 3 (ydx xdy) cdt
r
r
c r
with I = 2M r02 /5. The elementary displacement is
DM = i ei = (dxi + hij dxj )ei
with
h00 = h11 = h22 = h33 = m/r,
h01 = GIy/c3 r3 ,
h02 = GIx/c3 r3 ,
the other hij being equal to zero. The bivector d = 12 (hik,j hjk,i )ei ej dxk
gives the equation of motion
j (u1 h00,1 + u2 h00,2 + u3 h00,3 )

Du
+kI (h00,1 + u3 2 u2 3 )
+kJ (h00,2 + u1 3 u3 1 )
ds
+kK (h00,3 + u2 1 u1 2 )
134
with h00,1 = mx/r3 , h00,2 = my/r3 , h00,3 = mz/r3 and

1 = h30,2 h20,3 = 3GIxz/c3r5 ,
2 = h10,3 h30,1 = 3GIyz/c3r5 ,

3 = h20,1 h10,2 = GI 2z 2 x2 y 2 /c3 r5 ,
which puts in evidence the Thirring precession with respect to the local inertial
frame.
Conclusion
From the abstract quaternion group, we have dened the quaternion algebra H,
then the complex quaternion algebra H(C) and the Cliord algebra H H. The
quaternion algebra gives a representation of the rotation group SO(3) well-known
for its simplicity and its immediate physical signicance. The Cliord algebra HH
yields a double representation of the Lorentz group containing the SO(3) group as
a particular case and having also an immediate physical meaning. Furthermore,
the algebra H H constitutes the framework of a relativistic multivector calculus,
equipped with an associative exterior product and interior products generalizing
the classical vector and scalar products. This calculus remains relatively close to
the classical vector calculus which it contains as a particular case. The Cliord
algebra H H allows us to easily formulate special relativity, classical electromagnetism and general relativity. In complexifying H H, one obtains the Dirac algebra, Diracs equation, relativistic quantum mechanics and a simple formulation of
the unitary group SU(4) and the symplectic unitary group USp(2, H). Algebraic or
numerical computations within the Cliord algebra H H have become straightforward with software such as M athematica. We hope to have shown that the
Cliord algebras H H over R and C constitute a coherent, unied, framework of
mathematical tools for special relativity, classical electromagnetism, general relativity and relativistic quantum mechanics. The quaternion group consequently
appears via the Cliord algebra as a fundamental structure of physics revealing
its deep harmony.
Appendix A
Solutions
Chapter 1
S1-1
i2 = j 2 = k 2 = ijk = 1,
(i) ijk = jk = i,
k = ji,
ijk (k) = ij = k,
j = ik,
i = kj, etc.
S1-2
2,
25 = 5,
1
1
(1 i),
b1 = (4 + 3j),
2
5
a + b = 5 + i 3j,
ab = 4 + 4i 3j 3k,
ba = 4 + 4i 3j + 3k,
1+i
a = 2 = 2 (cos 45 + i sin 45 ) ,
2

4 3j
b=5
= 5 (cos 36, 87 sin 36, 87 ) ,
5

45
36, 87
45
36, 87
+ i sin
i sin
a = 21/2 cos
b = 51/2 cos
,
,
2
2
2
2

45
45
+ i sin
a1/3 = 21/6 cos
= 21/6 (cos 15 + i sin 15 ) .
3
3
|a| =
|b| =
a1 =
S1-3
ac ax + ac xb = ac c,

axb + xb2 = cb,
x aac + b2 + 2S(a)b = ac c + cbc ;
138
Appendix A. Solutions
the latter is a equation of the type x = , , H which one solves in x.

1
Answer: 2ix + xj = k,
x = (2j i).
3
S1-4
axbb1 + cxdb1 = eb1 ,
c1 ax + xdb1 = c1 eb1 ;
one nds an equation of the type of the previous exercice and solves similarly;
Answer: x = 2 + i.
S1-5
x1 x2 = x1 xa + x1 bx,
x = a + x1 bx,
xx1 = ax1 + x1 b = 1;
Answer: x =
1
(6j + 3k.).
5
Chapter 2
S2-1
cos
A = sin
0
sin 0
cos
cos 0 ,
0
A =
0
1
sin
1
0
0
A = 0 cos sin .
0 sin
cos
0 sin
1
0 ,
0 cos
S2-2
dA = rdA rc + drA rc + rA drc
= r (dA + rc drA + A drc r) rc ;
since rc dr = drc r (which is obtained by dierentiating rrc = 1), one obtains
with d = 2rc dr,
dA = r (dA + rc drA A rc dr) rc

d

d
A A
= r dA +
rc = r (dA + d A ) rc = rDA rc ;
2
2
DA = dA + d A ;
d
dt
is the angular velocity with respect to the moving frame.
139
S2-3
A = f A fc (f = gc r, f fc = 1),
DA = f DA fc ,
dA + d A = f (dA + d A ) fc ,
(df A fc + f dA fc + f A dfc ) + d A = f dA fc + f (d A ) fc ,
d A = f (d A ) fc df fc (f A fc ) (f A fc ) f dfc

= f d fc A df fc A + A df fc
= (f d fc 2df fc ) A ;
hence
d = f d fc 2df fc ,
d = fc d f 2fc df.
S2-4
X = rX rc ,
r = cos
d
d
dr
+ k sin ,
= 2rc
=k .
2
2 dt
dt
dt
Polar coordinates:
X = xi + yj,
X = i,
dX
d
d
DX
d
=
+
X =
i + j;
dt
dt
dt
dt
dt
dV
d
DV

=
=
+
V
dt
dt
" dt
#
2

d2
d2
d
d d
+ 2 .
=
i + j 2
dt2
dt
dt dt
dt
V =
Cylindrical coordinates:
X = xi + yj + zk,
X = i + zk,
dX
d
d
dz
DX
d
=
+
X =
i+ j+k ;
dt
dt
dt
dt
dt
dt
dV
d
DV

=
+
V
=
dt
dt
" dt
#
2

d
1 d
d z
d2
2 d
=
i
+
k.
j
+
dt2
dt
dt
dt
dt2
V =
140
S2-5 The moving frame (, , ) is obtained with the following three rotations:
j sin (rotation of 2 around Oy),

4
4
r2 = cos i sin (rotation of around Ox),

2
2
2
2
r3 = cos
+ k sin
(rotation of ( 2 ) around Oz);
4
4
r1 = cos
r = r3 r2 r1

+
1
+
+ sin
=
cos
2
2
2

2 2 +
2 + 2
1
2 2
j cos
+ k sin
+ i sin
;
4
4
4
2
d = 2rc dr = id cos jd sin + kd,
d = 2drrc = id sin + jd cos + kd;
e1 = rirc = i sin cos + j sin sin + k cos ,
e2 = rjrc = i cos cos + j cos sin k sin ,
e3 = rkrc = i sin + j cos ;
dX
d
d
d
DX
d
=
+
X = i i + j + k
sin ,
dt
dt
dt
dt
dt
dt
dV
d
DV
=
=
+
V
dt
dt
dt
(
"
#)
2
2
d
d
d2
2
=i
+
sin
dt2
dt
dt
(

2 )
d
1 d
2 d
+j
sin cos
dt
dt
dt
*

+
1 d
d
+k
2 sin2
.
sin dt
dt
V =
S2-6
+ k sin (1st rotation),

2
2

r2 = cos + sin r1 ir1c = r1 cos + i sin

r1c (2nd rotation),
2
2
2
2
r3 = cos + sin (r2 kr2c ) (3rd rotation);

2
2
r1 = cos
141
r = r3 r2 r1

r2c r2 r1
= r2 cos + k sin
2
2

r1
= r1 cos + i sin
r1c cos + k sin
2
2
2
2

cos + k sin
= cos + k sin
cos + i sin
2
2
2
2
2
2
because
r1c

cos + k sin
r1 = cos + k sin
;
2
2
2
2
nally
r = cos
cos
+ i sin cos
+ j sin sin
+ cos sin
,
2
2
2
2
2
2
2
2

d
d
cos +
sin sin
dt
dt

d
d
sin +
sin cos
+j
dt
dt

d
d
+k
+
cos ;
dt
dt
dr
=i
= 2rc
dt

d
d
cos +
sin sin
dt
dt

d
d
+j
sin
cos sin
dt
dt

d d
+k
+
cos .
dt
dt
dr
= 2 rc = i
dt
Chapter 3
S3-1
hence, a0 = 12 , a3 =
a0 + i a3 = 1,
i a1 + a2 = 0,
i
2 ,
a0 i a3 = 0,
i a1 a2 = 0,
a2 = 0 = a1 , or
e1 =
1
(1 i k) ;
2
e21 = e1 ;
e2 =
1
(1 + i k) ;
2
e22 = e2 .
similarly
142
a = (a0 + i a0 , a1 + i a1 , a2 + i a2 , a3 + i a3 ) ,
a0 + i a3 + i a0 a3 ,
1 a1 i a2 + i a1 + a2 ,
,
u = ae1 =
2 a1 i a2 + i a1 + a2 ,
i a0 + a3 + a0 + i a3
a0 + i a3 + i a0 a3 ,
1 a1 + i a2 + i a1 a2 ,
.
v = e1 a =
2 i a1 + a2 + a1 + i a2 ,
i a0 + a3 + a0 + i a3
S3-2
x + y = 3 + i i + (2 + i ) j + (1 + 3i ) k,
xy = (2 3i ) + (3 + 8i ) i + (7 + 2i ) j + (2 + 3i ) k,
yx = (2 3i ) + (3 4i ) i + (1 + 2i ) j + (2 + 3i ) k,
x2 = (2 4i ) + 2i i + (4 + 2i ) j + 2k,
xxc = 4 + 4i ,
y 2 = 13 + 12 i k,
yyc = 5,
xc
1
= (1 i , 1 i , 3 + i , 1 + i ) ,
xxc
8
1
= (2 + 3i k) ,
5
1
(1 + 5i , 5 + 11i, 9 5i , 5 + i ) = (xy)1 .
=
40
x1 =
y 1
y 1 x1
S3-3

1
a + a
= 1 + 3k ,
|1 + d|
2

1
1+d
= 3 + i i 5 ,
b1 =
|1 + d|
2

1
d = aac =
7 + 3i i 5 ,
2
r1 =

1
a + a
=
3k ,
1
+
|1 + d |
2

1
1 + d

=
i
b2 =
5
i
j
15 ,
|1 + d |
4

1
d = ac a =
14 3i i 5i 3i j 15 .
4
r2 =
143
S3-4
xy =
B =xy =
1
(xyc + yxc ) = 2,
2
1
(xyc yxc ) = (1 + 2i ) i + (1 + 2i ) j i k,
2
B = z w = (1 9i ) i 3i j + (3 3i ) k,

F = y z = i 2i j + 3i k,
1
(xF + F x) = i + 2i 3j 2k,
2
T = (x y) z = z (x y) = i + 2i 3j 2k,
T = x (y z) =
1
B z = z B = (zB Bz) = 2 + 6i i + 6i j 2i k V,
2
1
w T = (wT + T wc ) = (3 + i ) i + (1 + 8i ) j + (1 11i) k B,
2

BB + B B
= 22,
2

BB + B B
B B = P
= i ,
2
BT + T B
= 5 6i i + i j 10 i k V.
BT =
2
B B = S
S3-5
X = ct + xi i + yi j + zi k,
X = ct + x i i + y i j + z i k,

b = cosh , i sinh , 0, 0 ,
2
2
1
7
=
= cosh = ,
2
v2
1
c2
cosh2
cosh
cosh + 1
=
,
2
2
= ,
2
2
cosh 1
=
,
2
2

3 5
5
sinh =
,
b=
,i
, 0, 0 ,
2
2
2
2
sinh2
144

=
1
v 2
c2
= cosh =
7
,
6

sinh =
1
,
6
X = (0, i , i , 0),

7
1

, 0, i
,0 ,
V =
6
6
X = bX bc ,
V = bV bc ,

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

Ey Ez
E
E
= 0, i x , i
,i
F = B + i
,
c
c
c
c
F = bF bc
Ex = Ex ,
Bx = 0,
7
7
Ey ,
Ez = Ez ,
2
2
3 5
3 5
E ,
E ,
By =
Bz =
2c z
2c y

E
F = 0, B + i
.
c
Ey =
S3-6
A (0, i , i , 0) , B (0, i , i , 0) ,

i
2i
i
A 0, , ,
,
3
3
3

2i
i i
C 0, , ,
,
3 3
3
C (0, i , i , 0) , D (0, i, i , 0) ,

2i
i i
B 0, , ,
,
3 3
3

i
2i
i

D 0, , ,
.
3
3
3
Chapter 4
S4-1
Ac = I 2J + iK,
Bc = j kI 2kK,
A + B = j + I + 2J iK + kI + 2kK,
A B = I + 2J iK j kI 2kK,
AAc = 4,
BBc = 4,
145
A1 =
Ac
1
= (I 2J + iK) , ,
AAc
4
B 1 =
1
(j + kI + 2kK) ,
4
AB = 2j k + jI + 3jJ + 4kI 2kJ 3kK,

BA = 2j k + jI + 3jJ 4kI + 2kJ + 3kK,
(AB)c = 2j k + jI + 3jJ 4kI + 2kJ + 3kK,
(BA)c = 2j k + jI + 3jJ + 4kI 2kJ 3kK,
(AB) (AB)c = (AB)c (AB) = 16,
(AB)
(BA)
(AB)c
(AB) (AB)c
1
(2j + k jI + 4kI 3jJ 2kJ 3kK) = B 1 A1
=
16
1
(2j + k jI 4kJ 3jJ + 2kJ + 3kK) ,
=
16
=
[A, B] =
S4-2
(BA) (BA)c = 16,
1
(AB BA) = 2j + 4kI 2kJ 3kK.
2
1
x y = (xy + yx) = 2,
2
1
B = x y = (xy yx) = I J + 2iI + 2iJ iK,
2
B = z w = I 3K 9iI 3iJ 3iK,
F = y z = I 2iJ + 3iK,
T = x (y z) = x F =
1
(xF + F x)
2
= k + 2jI 3jJ 2jK,

T = (x y) z = z (x y) =
1
(zB + Bz) = T
2
= k + 2jI 3jJ 2jK,

1
B z = z B = (zB Bz) = 2j + 6kI + 6kJ 2kK,
2
1
w T = (wT + T w) = 3I + J + K + iI + 8iJ 11 iK,
2
146

1

(BB + B B) = 22,
BB = S
2

1

B B = PS
(BB + B B) = i,
2
1
B T = (BT + T B) = 5j 6kI + kJ 10 kK.
2
S4-3
1
S1 = (xy yx) = 3I J 5K,
2
1
S2 = (zw wz) = 9I 3J 3K,
2
1
(zS1 + S1 z) = 10 k,
2
S1 S2 = 0,
V = S1 z = z S1 =
w = w + w ,
21
27
169
kI kJ + kK,
50
25
10
46
23
69
1
= S1 (S1 w) = kI + kJ + kK,
50
25
10
w = (w S1 ) S11 = S11 (S1 w) =

w = (w S1 ) S11
S1 = S1 + S1 ,
18
18
54
S1 = (S1 .S2 ) S21 = I J K,
11
11
11
,
26
37
21
1
S1 = (S1 S2 ) + [S1 , S2 ] S2 = I J K.
11
11
11
Chapter 5
S5-1
1
=
1
v2
c2
3
= cosh = ,
2 2
cosh + 1
cosh 1
=
,
sinh2 =
,
2
2
2
2
!
!
1
1
3
3
+ 2 = ,
2 = ,
cosh =
sinh =
2
2
2
2
2
2
!
!

3
3
1
+ 2 + iI 2 ,
b=
2
2
2
cosh2
147
+ I sin = cos + I sin ,

2
2
8
8
a = rb = cos sin i + sin + sin I + cos iI,

8
8
8
8
8
r = cos
X = aX ac ,
X = ctj + kIx + kJy + kKz,
X = ct j + kIx + kJy + kKz .
S5-2
!

3
2 ,
2
!
!

3
3
1
+ 2 + iJ 2 ,
b2 =
2
2
2
1
b1 =
2
1
a = b2 b1 =
4
3
+ 2 + iI
2

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

1
1
3
3
+ 2 +
K ,
2+
2
17
2
17

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

direction u
1
3
4, , 0
34
17

,
tanh = 0, 4581 =
v
.
c
148
S5-3

1
1
= , 0,
,
2
2
1 1 1
xD = , ,
,
3 6 6
1
xF = , 0, 0 ,
2

1
1 1

xH = , ,
.
2 10 5
xA
xB
= (0, 0, 0) ,

3 1 2

,
xC = , ,
7 7 7
4 1 1
,
xE = , ,
9 9 9
1
3
xG = , 0,
,
5
5
S5-4

1k
1 + k
AB =
A1 =
1
2
(AB)
k
i
1
j
1
=
4
1 + k
1 + k
1+k
1 k
B 1 =
,
1 k
1 k
0
i
BA =
(BA)
1
2
j
0
1
k
1
=
2
There is no inverse of C.
Chapter 6
S6-1 Velocity of B in the reference frame at rest
vx = v,
vy = 0,
vz = 0
velocity of B with respect to A

vx =
vx w
= 0, 975 c,
1 vcx2w
vy = 0,
vz = 0;
The relative velocity remains smaller than c.
w = v = 0, 8 c,
,
i
j
0
j
,
i
0
149
S6-2
L=
hence
'
Xi Pi =
'

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

' Ei
ctpi = C2 ,
(ri pi ) = C1 ,
ri
c
.
Ei ri
,
R= .
Ei
.
.
Ei ri
pi
2
.
tc .
= C3 ,
Ei
Ei
.
pi
R = C3 + vt,
v = c2 .
= C4 .
Ei
S6-3 In the proper frame (instantaneous, Galilean), the four-acceleration is

A = kIg
with the invariant
AAc = g 2 ;
In the frame at rest K, the four-velocity and the four-acceleration are respectively
u = c (j cos h + kI sinh ) ,
A=c
tanh =
d
(j sinh + kI cosh ) ,
d
v
,
c
: proper time
with

AAc = g = c
2

= c2
hence
d
d
d
d
2

sinh2 cosh2
2
g
d
= ,
d
c
g
c
and

g dX
g
+ kI sinh
=
,
u = c j cos h
c
c
d

g
g
c2
j sin h
+ kI cosh
= jct + kIx;
X=
g
c
c
150
nally, one has

t=
and
c
sinh ,
g
x=
c2
(cosh 1)
g

2
c2
c4
x+
(ct)2 = 2 .
g
g
For 1
t
c
= ,
g
x=
c2 2
1
1
= g 2 = gt2 .
g 2
2
2
Chapter 7
S7-1
A = bA bc

b = cosh
v
+ iI sinh , tanh =
,
2
2
c

q
,
A = 0, V =
40 r
V
V
+ kA = j
c
c
V
A = j + kA,
c
1/2

r = x2 + y 2 + z 2
,
x = (x vt) ,
A = j
V = V =
1
q

40
2
(x vt) +
y 2 +z
2
1/2 ,
2
y = y,
A=
z = z,
v
v
V = 2 V,
2
c
c
(potentials of Lienard and Wiechert of an electric charge in rectilinear motion).

E
,
c
F = B +
B = 0,
F = B +
E =
q r
,
40 r3
E
,
c
F = bF bc ,
Bx = 0,
v
v
q vz
= Ez = Ez ,
40 cr3
c
c
v
v
q vy
Bz =
= Ey = Ey ,
40 cr3
c
c
By =
Ex = Ex =
q x
,
40 r3
Ey = Ey ,
Ez = Ez ,
151
x = (x vt) ,
q (x vt)
,
40 2 r3/2
y
q
,
Ey =
40 2 r3/2
q
z
Ez =
.
40 2 r3/2
Ex =
y = y,
z = z,
1/2
y2 + z 2
2
r = (x vt) +
,
2
S7-2
F = B +
E
,
c
B = 0,
E =
E2 =
0
,
20 r
0 n
,
20 r
E1 = E3 = 0,
F = bF bc ,
tanh = ,
b = cosh + iI sinh ,
2
2
c
0 i
0 v
0 v
B3 =
=
,
=
20 r c2
2r
2r
E3 =
0
=
,
20 r
20 r
in K; the linear density of mobile charges is K = 0 = (by hypothesis,

K = 0) and i = v (r = r ). At the point M in K, the total electric eld is
nil
0 i
ET = 0,
ez .
BT =
2r
In K,
f = qv B,
f (M ) = qvBey ;
in the proper frame, the electromagnetic eld is
F = B +
E
= bc FT b
c
FT = B,
B=
with
0 i
ez ,
2r
152
hence
0 i
ez ,
2r
0 i
vey ;
E =
2r
B =
the particle being at rest in the proper frame,

f = qE =
S7-3
F = bc F b,
tanh =
b=
q0 i
ve = f .
2r

iI
iJ
cosh + sinh + sinh

,
2
2
2
2
2
1
v
= ,
c
2
1
cosh = =
1
cosh + 1
=
=
cosh
2
2
!
2 + 1
3
cosh =
,
2
2
2
2
3
+1
2
1
1
+ ,
2
3
1
1
B2 = ,
2
3

1 2
B3 =
,
c 3
B1 =
v2
c2
2
= ,
3
cosh 1
sinh
=
=
2
2
!
2 1
3
sinh =
,
2
2
2
2
E1 = 1 ,
3
2
E2 = 1 + ,
3
c
E1 = .
6
S7-4
F = 0,
iF = (iF ) = 0,
F = 0.
2
3
1
2
Appendix B
Formulary: multivector products

within H(C)
Let x, y be four-vectors, B, B bivectors, T, T trivectors and P, P pseudoscalars
x = x0 + i x1 + i x2 + i x3 ,
y = y 0 + i y 1 + i y 2 + i y 3 ,
B = a + i b,
B = a + i b ,
(a = a1 i + a2 j + a3 k, b = b1 i + b2 j + b3 k),
T = i t0 + t,
P = i s0 ,
T = i t0 + t (t = t1 i + t2 j + t3 k),
P = i s0 .
S [A], P [A] designate respectively the scalar and pseudoscalar parts of the
complex quaternion A. For any two arbitrary boldface quantities (x, y, a, b, t) the
following abridged notation is used:
x y = x1 y 1 + x2 y 2 + x3 y 3 ,
x y = (x2 y 3 x3 y 2 )i + (x3 y 1 x1 y 3 )j + (x1 y 2 x2 y 1 )k.
Products with four-vectors
xy =
1
(xyc + yxc ) S
2
= x0 y 0 x1 y 1 x2 y 2 x3 y 3
= y x,
154
Appendix B. Formulary: multivector products within H(C)

1
(xyc yxc ) B
2

= x y + i y 0 x x0 y
= y x.
xy =
Products with bivectors

1
(xB Bx) V
2

= x b + i x0 b + x a
xB =
B x,
1
(xB + Bx) T
2

= i x a+ x0 a + x b
B x.
xB =
B B = S
1
(BB + B B)
2
= a a + b b
B B,

1

(BB + B B)
BB =P
2
= i (b a + a b )
B B.
Products with trivectors

1
(xT + T xc ) B
2
= x0 t + t0 x + i (x t)
xT =
T x
1
(xT T xc ) P
2

= i x0 t0 x t
xT =
T x,
B.0. Formulaire: produits multivectoriels dans H(C)

1
B T = (BT + T B ) V
2

= (a t) i t0 a + b t
T B,
1
(T T + T T ) S
2
= t0 t0 t t
T T =
T T.
Products with pseudoscalars

1
(xP P x) T
2
= i s0 x0 + s0 x
xP =
P x,
1
(BP + P B) B
2
= s0 b + i s0 a
BP =
P B,
1
T P = (T P P T ) V
2
= s0 t0 + i s0 t
P T,
1

(P P + P P ) S
P.P = S
2
= s0 s0
P P.
155
Appendix C
Formulary: multivector products

within H H (over R)
Let x, y be four-vectors, B, B bivectors, T, T trivectors and P, P pseudoscalars
x = jx0 + kx
(x = x1 I + x2 J + x3 K),
y = jy 0 + ky,
B = a + ib,
B = a + ib ,
(a = a1 I + a2 J + a3 K, b = b1 I + b2 J + b3 K),
T = kt0 + jt,
T = kt0 + jt (t = t1 I + t2 J + t3 K),
P = is0 ,
P = is0 .
S [A], P [A] designate respectively the scalar and pseudoscalar parts of the
Cliord number A. For any two arbitrary boldface quantities (x, y, a, b, t) the
following abridged notation is used:
x y = x1 y 1 + x2 y 2 + x3 y 3 ,
x y = (x2 y 3 x3 y 2 )I + (x3 y 1 x1 y 3 )J + (x1 y 2 x2 y 1 )K.
158
Appendix C. Formulary: multivector products within H H (over R)
Products with four-vectors
1
x y = (xy + yx) S
2
0 0
= x y x1 y 1 x2 y 2 x3 y 3
= y x,
1
x y = (xy yx) B
2

= x y + i y 0 x x0 y
= y x.
Products with bivectors
1
(xB Bx) V
2

= jx b + k x0 b + x a
= B x,
xB =
1
(xB + Bx) T
2

= kx a+j x0 a + x b
B x.
xB =
1
BB =S
(BB + B B)
2
= a a + b b
= B B,

1
(BB + B B)
B B = P
2
= i (b a + a b )
B B.
159
Products with trivectors

1
x T = (xT + T x) B
2
= x0 t + t0 x + i (x t)
T x,
1
x T = (xT T x) P
2 0 0

= i x t x t
T x,
1
(BT + T B) V
2

= j (a t) k t0 a + b t
BT =
T B,
1
T T = (T T + T T ) S
2
= t0 t0 t t
T T.
Products with pseudoscalars

1
(xP P x) T
2
= ks0 x0 + js0 x
xP =
P x,
1
(BP + P B) B
2
= s0 b + is0 a
BP =
P B,
1
(T P P T ) V
2
= js0 t0 + ks0 t
T P =
P T,
160
Appendix C. Formulary: multivector products within H H (over R)
1

(P P + P P ) S
P P =S
2
= s0 s0
P P.
Appendix D
Formulary: four-nabla operator

within H H (over R)
Four-nabla operator:
=j
kI
kJ
kK ,
ct
x
y
z
DAlembertian operator:
=
2
c2 t2
2
2
2
2
2
x
y
z
(four-vectors
A = jA0 + kIA1 + kJA2 + kKA3 ,
B = jB 0 + kIB 1 + kJB 2 + kKB 3 ;
scalars : p, q). Then:

( A) = 0,
(p) = ( ) p = 0,
(p) = ( ) p = p,
( A) = A ( A) ,
(p) = ( p) ,
( A) = ( A) ,
( A) = ( A) ,
(pq) = pq + qp,
(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 +x1 i+x2 j +x3 k = [x0 , x1 , x2 , x3 ], xi C, I: usual complex
imaginary*)
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 end
of 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 Cliord numbers, a = a1 + a2 I + a3 J + a4 K, b = b1 + b2 I +
b3 J + b4 K, 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 dierence*)
Appendix F. Work-sheet H H over R (Mathematica)
166
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+ba abba
2 ,
2 *)
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]]]}
abba
(*products ab+ba
2 , 2 *)
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 = 2j k + jI + 4kI + 3jJ 2kJ 3kK; the function
Simplify simplies 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 M2(H)

(Mathematica)
(*product of two matrices 2 2 over H ,

a=
a1
a3
a2
a4
b=
b1
b3
b2
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
k
j
i
B=
1
i
k
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 M2 (H) (Mathematica)

(*result

AB =
1k
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
Cliord algebras: isomorphisms

General formulas ([18], p. 48)
Mn (R), Mn (C), Mn (H): square matrices of order n over R, C, H;
Mn (R) Mp (R) Mnp (R),
Mp (C) Mp (R) C,
Mp (H) Mp (R) H.
Cliord algebra H
H = M1 (H) = M1 (R) H,
C + = C.
Cliord algebra H H
H H M4 (R),
C + = H C M2 (C)
Cliord algebra H H H
H H H M4 (R) H M4 (H),
C + H H C M4 (R) C M4 (C),
H M2 (C) M2 [H (C)] (Dirac algebra).
Cliord algebra H H H H
H H H H M4 (R) M4 (R) M16 (R),
C + H H H C M4 (R) H C M4 [H (C)],
M4 (R) M2 (C) M8 (C).
Appendix I
Cliord algebras: synoptic table

H (R)
real quaternions
classical vector calculus
rotation group SO(3)
classical mechanics
H (C)
complex quaternions
Lorentz group
special relativity
classical electromagnetism
H H(R)
real Cliord numbers
relativistic multivector calculus
Lorentz group
classical electromagnetism
special relativity
general relativity
H H(C)
complex Cliord numbers
Dirac algebra
unitary group SU(4) and symplectic unitary group USp(2, H)
relativistic quantum mechanics
Bibliography
[1] S. L. Adler, Quaternionic Quantum Mechanics and Quantum Fields, Oxford University Press, New York, Oxford, 1995.
[2] S. L. Altmann, Rotations, Quaternions, and Double Groups, Clarendon
Press, Oxford, 1986.
[3] J. Bass, Cours de Mathematiques, Vol. I, Masson et Cie , Editeurs, Paris,
1964.
[4] J. Bateman, The Transformations of the Electrodynamical Equations, Proceedings of the London Mathematical Society, Second Series, 8 (1910), 223
264.
[5] W. Baylis, ed., Cliord (Geometric) Algebras with applications to physics,
mathematics, and engineering, Birkhauser, Boston, 1996.
[6] W. E. Baylis, Electrodynamics: A Modern Geometric Approach, Birkh
auser,
Boston, 1999.
[7] R. Becker, F. Sauter, Theorie der Elektrizit
at, Vol. 1, B. G. Teubner,
Stuttgart, 1973.
[8] A. Blanchard, Les corps non commutatifs, Presses Universitaires de France,
Collection SUP, 1972.
[9] L. Brand, Vector and Tensor Analysis, Wiley, New York, 1947, p. 412.
[10] E. Cartan, Les Systèmes dierentiels exterieurs et leurs applications
geometriques, Hermann, Paris, 1971.
[11] E. Cartan, Lecons sur la Theorie des Spineurs, Actualites Scientiques et
Industrielles, n 643, Hermann, Paris, 1937; I, pp. 1213.
[12] G. Casanova, LAlgèbre vectorielle, PUF, Paris, 1976.
[13] W. K. Clifford, Applications of Grassmanns extensive algebra, Amer. J.
Math., 1 (1878), 350358.
[14] W. K. Clifford, Mathematical Papers, ed. R.Tucker (1882; reprinted 1968,
Chelsea, New York), pp. 266276.
174
Bibliography
[15] A. W. Conway, Quaternions and Quantum Mechanics, Pont. Acad. Sci.

Acta, 12 (1948), 259.
[16] A. W. Conway, Selected Papers, ed. J. McConnell, Dublin Institute for
Advanced Studies, Dublin, 1953, pp. 206208.
[17] M. J. Crowe, A History of Vector Analysis: The Evolution of the Idea of a
Vectorial System, University of Notre Dame, Notre Dame, London, 1967.
[18] A. Crumeyrolle, Orthogonal and Symplectic Cliord Algebras: Spinor
Structures, Kluwer Academic Publishers, Dordrecht, Boston, London 1990.
[19] C. Doran, A. Lasenby, Geometric Algebra for Physicists, Cambridge University Press, Cambridge, 2003.
[20] L. Dorst, C. Doran and J. Lasenby, Editors, Applications of Geometric
Algebra in Computer Science and Engineering, Birkh
auser, Boston, 2002.
[21] P. R. Girard, Quaternions, algèbre de Cliord et physique relativiste,
Presses Polytechniques et Universitaires Romandes, Lausanne, 2004.
[22] P. R. Girard, Quaternions, Cliord Algebra and Symmetry Groups,
published in Applications of Geometric Algebra in Computer Science and
Engineering, L. Dorst, C. Doran and J. Lasenby, editors, Birkh
auser,
Boston, 2002, pp. 307315.
[23] P. R. Girard, Einsteins equations and Cliord algebra, Advances in
Applied Cliord Algebras, 9, 2 (1999), 225230.
[24] P. R. Girard, The quaternion group and modern physics, Eur. J. Phys., 5
(1984), pp. 2532.
[25] P. R. Girard, The Conceptual Development of Einsteins General Theory
of Relativity, Ph. D. Thesis, University of Wisconsin-Madison, published by
Microlms International, USA, order n 8105530, 1981.
[26] H. Goldstein, Classical Mechanics, second edition, Addison-Wesley, Reading, Massachusetts, 1980.
rlebeck, W. Spro
ssig, Quaternionic and Cliord Calculus for
[27] K. Gu
Physicists and Engineers, John Wiley & Sons, Chichester, New York, 1997.
rsey and H. C. Tze, Complex and quaternionic analyticity in chiral
[28] F. Gu
and gauge theories, Annals of Physics, 128 (1980), 29130.
[29] W. R. Hamilton, The Mathematical Papers, Cambridge University Press,
Cambridge, 193167, ed. A. W. Conway for the Royal Irish Academy.
[30] W. R. Hamilton, Elements of Quaternions, 2 vols (18991901); reprinted
Chelsea, New York, 1969.
[31] T. L. Hankins, Sir William Rowan Hamilton, Johns Hopkins University
Press, Baltimore and London, 1980.
Bibliography
175
[32] D. Hestenes, New Foundations for Classical Mechanics, 2nd edition,

Kluwer Academic Publishers, Dordrecht, Boston, 1999.
[33] D. Hestenes, Space-Time Algebra, Gordon and Breach, New York, 1966.
[34] R. dInverno, Introducing Einsteins Relativity, Clarendon Press, Oxford,
1992.
[35] B. Jancewicz, Multivectors and Cliord Algebra in Electrodynamics, World
Scientic, Singapore, New Jersey, 1988.
[36] J. B. Kuipers, Quaternions and Rotation Sequences: A Primer with
Applications to Orbits, Aerospace, and Virtual Reality, Princeton University
Press, Princeton, New Jersey, 1999.
[37] M. Lagally, Vorlesungen u
ber Vektorrechnung, Akademische Verlagsgesellschaft, Leipzig, 1956, p. 362.
[38] P. Lounesto, Cliord Algebras and Spinors, second edition, Cambridge University Press, Cambridge, 2001.
[39] R. Meinmn
e, F. Testard, Introduction a
` la theorie des groupes classiques,
Hermann, Paris, 1986.
ller, Relativit
[40] C. Mo
atstheorie, Bibliographisches Institut, B.I.-Wissenschaftsverlag, Wien, Z
urich, 1976.
[41] I. R. Porteous, Cliord Algebras and the Classical Groups, Cambridge University Press, Cambridge, 1995.
[42] W. Rindler, Relativity: Special, General, and Cosmological, Oxford
University Press, Oxford, 2001.
[43] J. B. Shaw, Vector Calculus with Applications to Physics, Van Nostrand,
New York, 1922.
[44] J. B. Shaw, Synopsis of Linear Associative Algebra: A Report on its Natural
Development and Results up to the Present Time, Carnegie Institution of
Washington, D.C., 1907, pp. 7374.
[45] L. Silberstein, The Theory of Relativity, Macmillan, London, 1914.
[46] J. Snygg, Cliord Algebra, A Computational Tool for Physicists, Oxford
University Press, New York, Oxford, 1995.
[47] F. Sommen, Radon and X-ray transforms in Cliord analysis, Complex variables, 11 (1989), 4970.
[48] W. I. Stringham, Determination of the Finite Quaternion Groups,
American Journal of Mathematics, 4 (1881), 345357.
[49] J. L. Synge, Quaternions, Lorentz Transformations and the Conway-DiracEddington Matrices, Dublin Institute for Advanced Studies, Dublin, 1972.
176
Bibliography
[50] W. Thirring, A Course in Mathematical Physics, Springer-Verlag, New

York, Vol. 2, 1992, p. 180.
[51] L. H. Thomas, Phil. Mag., 3 (1927), 1.
[52] L. H. Thomas, Nature, 117 (1926), 514.
[53] M.-A. Tonnelat, Les Principes de la theorie electromagnetique et de la
relativite, Masson et Cie , Editeurs, Paris, 1959.
[54] J. Vrbik, Dirac equation and Cliord algebra, J. Math. Phys. 35, 5 (1994),
23092314.
Index
aberration, 124
adjunction, 86
ane connection, 127
Bianchi, 128
Binet relativistic equation, 133
bivector, 44, 59, 60, 85
electromagnetic eld, 50
charge density, 105
Cliord algebra, 57
Cliord number, 58, 59
conjugate, 59
dual, 59
selfadjoint, 86
Cliord theorem, 57
commutator, 29, 59
complex imaginary, 86
conductor, 119
conformal transformation, 52, 82
conjugate, 9, 59
complex, 38, 86
quaternionic, 86
conservation, electric charge, 111
contraction, 93
covariance, 111
covariant dierentiation, 127
curvature, 127, 128
dAlembertian operator, 51, 111, 118
De Moivre theorem, 11
dielectric medium, 114
dilatation, 82
Dirac algebra, 85
Dirac spinor, 86
direct product, 3
Doppler eect, 123

dual, 28, 59
Einstein equations, 128
electric charge, conservation, 106
electromagnetic eld, 109
bivector, 109
energy, 99
energy-current density, 117
energy-momentum tensor, 117
equation of motion, 129
exterior product, 59, 61
Fizeau experiment, 121
four-acceleration, 96
proper, 97
four-curl, 69
four-current density, 51, 105, 110
four-divergence, 69
four-force , 100
density, 116
volumic, 52
four-gradient, 69
four-momentum, 99
four-nabla operator, 107
four-potential, 50
potential vector, 107, 111
four-vector, 44, 59
four-velocity, 48, 95
Galilean transformation, 92
gauge transformation, 109
Gauss theorem, generalized, 71
generators, 59, 85, 88
Gibbs, 9
gravitational wave, 133
178
group, 3
cyclic, 3
dihedral, 4
nite, 4
O(3), 22
of rotations SO(4), 20
orthogonal [Or](4), 20
pseudo-orthogonal Or(1,3), 39
SO(3), 22
special orthogonal SO(1,3), 40
crystallographic, 28
Hamilton, 3, 174
Hamiltonian, 32
harmonic coordinates, 133
hermitian norm, 86
hyperplane, 19, 38, 65
dual, 66
tangent, 70
hypersurface, 106
innitesimal transformation, 28
interference fringes, 121
interior product, 59
inversion, 39
improper antichronous, 77
improper orthochronous, 77
Jacobi identity, 63
Kepler, 32
kinetic energy, 100
Klein four-group, 4
Laplace-Runge-Lenz vector, 32
left ideal, 86
longitudinal Doppler eect, 123
Lorentz gauge, 50, 110, 111
Lorentz transformation,
antichronous, 40
general, 94
innitesimal, 80
orthochronous, 40
pure, 41, 79
special, 94
Index
Lorentz, special transformation, 47
magnetic induction, 109
magnetic medium, 114
magnetization density, 114
Maxwells equations, 52, 110
covariance, 111
Mercury perihelion precession, 133
metric, 130
Minkowski, 38
volumic four-force, 52
minquat, 38
spacelike, 38
timelike, 38
unitary, 38, 42
momentum, 99
momentum-current density, 117
multivector, 46, 61, 81
calculus, 46
nonconducting medium, 120
norm, 9
octaeder, 26
operator, conjugate, 50
operator, four-nabla, 50, 69
optical path, 121, 122
order of interference, 122
orthogonal projection, 66
permeability, 119
permittivity, 119
plane, 64
symmetry, 19
wave, 118
tangent, 70
Poisson bracket, 32
polarization density, 114, 115
primitive idempotent, 86
product, exterior, 44
product, interior, 46
proper frame, 97, 122
proper time, 95
pseudoeuclidean, 38
Index
pseudoscalar, 46, 59, 61
quadratic form, 39
quaternion, 3
algebra, 9
complex, 37
exponential, 14
group, 7
root nth , 13
square root, 11
unitary, 19, 42
reciprocal basis, 69, 127
Ricci tensor, 128
Riemannian space, 127
Rodriguez, 24
rotation, 39
group [SO](3), 78
proper antichronous, 77
scalar, 59
scalar product, 15
Schwarzschild metric, 132
space symmetry, 77
spacetime, pseudoeuclidean, 38
spherical symmetry, 130
Stokes theorem, 116
generalized, 71
straight line, 64
subalgebra C + , 57
subgroup, 8
symmetric, 39
symmetry group, of the equilateral
triangle, 5
symmetry group, of the square, 7
symmetry, space, 39
symmetry, time, 39
symplectic unitary group, 87
tensor product, 57
Thirring precession, 134
Thomas precession, 43
time dilatation, 93
time symmetry, 76
179
torsion, 127
transformation, improper, 39
transformation, proper, 39
transversal Doppler eect, 124
triple product, scalar, 16
trivector, 45, 59, 60
unitary group, 87
variational principle, 128
vector, 85
isotropic, 75
spacelike, 75
tangent, 69
timelike, 75
vector calculus, classical, 64
vector product, 15
vector product, double, 17
wave four-vector, 97
zero divisor, 37

Girard P. R. - Quaternions, Clifford Algebras and Relativistic Physics (2007) PDF

Uploaded by

Copyright:

Available Formats

Girard P. R. - Quaternions, Clifford Algebras and Relativistic Physics (2007) PDF

Uploaded by

Document Information

Original Description:

Original Title

Copyright

Available Formats

Share this document

Share or Embed Document

Sharing Options

Did you find this document useful?

Is this content inappropriate?

Copyright:

Available Formats

Girard P. R. - Quaternions, Clifford Algebras and Relativistic Physics (2007) PDF

Uploaded by

Copyright:

Available Formats

Patrick R.

Igor Ya. Subbotin

Library of Congress Control Number: 2006939566

ISBN 978-3-7643-7790-8 Birkhuser Verlag AG, Basel Boston Berlin

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

3.2.2 Plane symmetry . . . . . . . . . . . . .

5.3.3 Transformation of multivectors

B Formulary: multivector products within H(C)

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

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

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

F Work-sheet H H over R (Mathematica)

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

H Cliord algebras: isomorphisms

Cliord algebras: synoptic table

1.1 Group structure

2. the law admits an identity element e

3. any element a of G has an inverse a

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

1. Cyclic group Cn of order n the elements of which are

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

b1 (b1 a)b = b2 ab = b1 ab2 = ab3

and thus b2 a = ab2 ; by proceeding similarly by recurrence, one establishes

1.2 Finite groups of order n 8

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

(b) the group having the elements (b : rotation of around an axis,

1.2. Finite groups of order n 8

The Klein four-group is isomorphic to the direct product of two cyclic

Example. The group constituted by the elements (I: rotation of

with bh a = abh (h = 1, 2, 3). This group is the rst noncommutative

Figure 1.1: Symmetry group of the equilateral triangle; b represents a rotation of

the group is commutative.

1.3. Quaternion group

(d) The group D4 (noncommutative) dened by

and bh a = abh (h = 1, 2, 3, 4).

1.3 Quaternion group

1.4 Quaternion algebra H

1.4. Quaternion algebra H

vector space is transformed into the associative algebra of quaternions (denoted

|a| = aac = a20 + a21 + a22 + a23 ,

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

4. one can realize the following operations

1.4. Quaternion algebra H

1.4.2 Polar form

1.4.3 Square root and nth root

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

a21 + a22 + a23

a21 + a22 + a23

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

1.4. Quaternion algebra H

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

a20 + a21 + a22 + a23 + a0

nally, one has

2. When sin = 0, the vector e in b is arbitrary. If a > 0, one has = 0 and

Example. Find the cubic root of

1.4.4 Other functions and representations of quaternions

and thus ea = cosh a + sinh a. For an arbitrary quaternion a, let U (V (a)) = u

1.5. Classical vector calculus