Multiforms, Dyadics, and Electromagnetic Media IEEE Press 445 Hoes Lane Piscataway, NJ 08854

IEEE Press Editorial Board Tariq Samad, Editor in Chief

George W. Arnold Vladimir Lumelsky Linda Shafer Dmitry Goldgof Pui-In Mak Zidong Wang Ekram Hossain Jeffrey Nanzer MengChu Zhou Mary Lanzerotti Ray Perez George Zobrist

Kenneth Moore, Director of IEEE Book and Information Services (BIS) Multiforms, Dyadics, and Electromagnetic Media

Ismo V. Lindell

IEEE Antennas and Propagation Society, Sponsor

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Library of Congress Cataloging-in-Publication Data:

Lindell, Ismo V. Multiforms, dyadics, and electromagnetic media / Ismo V. Lindell. pages cm ISBN 978-1-118-98933-3 (cloth) 1. Electromagnetism–Mathematics. 2. Electromagnetism–Mathematical models. I. Title. QC760.4.M37L57 2015 537.01′515–dc23 2014039321

Printed in the United States of America

10987654321 Contents

Preface xi

1 Multivectors and Multiforms 1 1.1 Vectors and One-Forms, 1 1.1.1 Bar Product | 1 1.1.2 Basis Expansions 2 1.2 Bivectors and Two-Forms, 3 1.2.1 Wedge Product ∧ 3 1.2.2 Basis Expansions 4 1.2.3 Bar Product 5 1.2.4 Contraction Products ⌋ and ⌊ 6 1.2.5 Decomposition of Vectors and One-Forms 8 1.3 Multivectors and Multiforms, 8 1.3.1 Basis of Multivectors 9 1.3.2 Bar Product of Multivectors and Multiforms 10 1.3.3 Contraction of Trivectors and Three-Forms 11 1.3.4 Contraction of Quadrivectors and Four-Forms 12 1.3.5 Construction of Reciprocal Basis 13 1.3.6 Contraction of Quintivector 14 1.3.7 Generalized Bac-Cab Rules 14 1.4 Some Properties of Bivectors and Two-Forms, 16 1.4.1 Bivector Invariant 16 1.4.2 Natural Dot Product 17 1.4.3 Bivector as Mapping 17 Problems, 18

2Dyadics 21 2.1 Mapping Vectors and One-Forms, 21 2.1.1 Dyadics 21 2.1.2 Double-Bar Product || 23 2.1.3 Metric Dyadics 24 v vi Contents

2.2 Mapping Multivectors and Multiforms, 25 2.2.1 Bidyadics 25 ∧ 2.2.2 Double-Wedge Product ∧ 25 2.2.3 Double-Wedge Powers 28 2.2.4 Double Contractions ⌊⌊ and ⌋⌋ 30 2.2.5 Natural Dot Product for Bidyadics 31 2.3 Dyadic Identities, 32 2.3.1 Contraction Identities 32 2.3.2 Special Cases 33 2.3.3 More General Rules 35 2.3.4 Cayley–Hamilton Equation 36 2.3.5 Inverse Dyadics 36 2.4 Rank of Dyadics, 39 2.5 Eigenproblems, 41 2.5.1 Eigenvectors and Eigen One-Forms 41 2.5.2 Reduced Cayley–Hamilton Equations 42 2.5.3 Construction of Eigenvectors 43 2.6 Metric Dyadics, 45 2.6.1 Symmetric Dyadics 46 2.6.2 Antisymmetric Dyadics 47 2.6.3 Inverse Rules for Metric Dyadics 48 Problems, 49

3Bidyadics 53 3.1 Cayley–Hamilton Equation, 54 3.1.1 Coefficient Functions 55 3.1.2 Determinant of a Bidyadic 57 3.1.3 Antisymmetric Bidyadic 57 3.2 Bidyadic Eigenproblem, 58 3.2.1 Eigenbidyadic C− 60 3.2.2 Eigenbidyadic C+ 60 3.3 Hehl–Obukhov Decomposition, 61 3.4 Example: Simple Antisymmetric Bidyadic, 64 3.5 Inverse Rules for Bidyadics, 66 3.5.1 Skewon Bidyadic 67 3.5.2 Extended Bidyadics 70 3.5.3 3D Expansions 73 Problems, 74 Contents vii

4 Special Dyadics and Bidyadics 79 4.1 Orthogonality Conditions, 79 4.1.1 Orthogonality of Dyadics 79 4.1.2 Orthogonality of Bidyadics 81 4.2 Nilpotent Dyadics and Bidyadics, 81 4.3 Projection Dyadics and Bidyadics, 83 4.4 Unipotent Dyadics and Bidyadics, 85 4.5 Almost-Complex Dyadics, 87 4.5.1 Two-Dimensional AC Dyadics 89 4.5.2 Four-Dimensional AC Dyadics 89 4.6 Almost-Complex Bidyadics, 91 4.7 Modified Closure Relation, 93 4.7.1 Equivalent Conditions 94 4.7.2 Solutions 94 4.7.3 Testing the Two Solutions 96 Problems, 98

5 Electromagnetic Fields 101 5.1 Field Equations, 101 5.1.1 Differentiation Operator 101 5.1.2 Maxwell Equations 103 5.1.3 Potential One-Form 105 5.2 Medium Equations, 106 5.2.1 Medium Bidyadics 106 5.2.2 Potential Equation 107 5.2.3 Expansions of Medium Bidyadics 107 5.2.4 Gibbsian Representation 109 5.3 Basic Classes of Media, 110 5.3.1 Hehl–Obukhov Decomposition 110 5.3.2 3D Expansions 112 5.3.3 Simple Principal Medium 114 5.4 Interfaces and Boundaries, 117 5.4.1 Interface Conditions 117 5.4.2 Boundary Conditions 119 5.5 Power and Energy, 123 5.5.1 Bilinear Invariants 123 5.5.2 The Stress–Energy Dyadic 125 5.5.3 Differentiation Rule 127 viii Contents

5.6 Plane Waves, 128 5.6.1 Basic Equations 128 5.6.2 Dispersion Equation 130 5.6.3 Special Cases 132 5.6.4 Plane-Wave Fields 132 5.6.5 Simple Principal Medium 134 5.6.6 Handedness of Plane Wave 135 Problems, 136

6 Transformation of Fields and Media 141 6.1 Affine Transformation, 141 6.1.1 Transformation of Fields 141 6.1.2 Transformation of Media 142 6.1.3 Dispersion Equation 144 6.1.4 Simple Principal Medium 145 6.2 Duality Transformation, 145 6.2.1 Transformation of Fields 146 6.2.2 Involutionary Duality Transformation 147 6.2.3 Transformation of Media 149 6.3 Transformation of Boundary Conditions, 150 6.3.1 Simple Principal Medium 152 6.3.2 Plane Wave 152 6.4 Reciprocity Transformation, 153 6.4.1 Medium Transformation 153 6.4.2 Reciprocity Conditions 155 6.4.3 Field Relations 157 6.4.4 Time-Harmonic Fields 158 6.5 Conformal Transformation, 159 6.5.1 Properties of the Conformal Transformation 160 6.5.2 Field Transformation 164 6.5.3 Medium Transformation 165 Problems, 166

7 Basic Classes of Electromagnetic Media 169 7.1 Gibbsian Isotropy, 169 7.1.1 Gibbsian Isotropic Medium 169 7.1.2 Gibbsian Bi-isotropic Medium 170 7.1.3 Decomposition of GBI Medium 171 7.1.4 Affine Transformation 173 Contents ix

7.1.5 Eigenfields in GBI Medium 174 7.1.6 Plane Wave in GBI Medium 176 7.2 The Axion Medium, 178 7.2.1 Perfect Electromagnetic Conductor 179 7.2.2 PEMC as Limiting Case of GBI Medium 180 7.2.3 PEMC Boundary Problems 181 7.3 Skewon–Axion Media, 182 7.3.1 Plane Wave in Skewon–Axion Medium 184 7.3.2 Gibbsian Representation 185 7.3.3 Boundary Conditions 187 7.4 Extended Skewon–Axion Media, 192 Problems, 194

8 Quadratic Media 197 8.1 P Media and Q Media, 197 8.2 Transformations, 200 8.3 Spatial Expansions, 201 8.3.1 Spatial Expansion of Q Media 201 8.3.2 Spatial Expansion of P Media 203 8.3.3 Relation Between P Media and Q Media 204 8.4 Plane Waves, 205 8.4.1 Plane Waves in Q Media 205 8.4.2 Plane Waves in P Media 207 8.4.3 P Medium as Boundary Material 208 8.5 P-Axion and Q-Axion Media, 209 8.6 Extended Q Media, 211 8.6.1 Gibbsian Representation 211 8.6.2 Field Decomposition 214 8.6.3 Transformations 215 8.6.4 Plane Waves in Extended Q Media 215 8.7 Extended P Media, 218 8.7.1 Medium Conditions 218 8.7.2 Plane Waves in Extended P Media 219 8.7.3 Field Conditions 220 Problems, 221

9 Media Defined by Bidyadic Equations 225 9.1 Quadratic Equation, 226 9.1.1 SD Media 227 x Contents

9.1.2 Eigenexpansions 228 9.1.3 Duality Transformation 229 9.1.4 3D Representations 231 9.1.5 SDN Media 234 9.2 Cubic Equation, 235 9.2.1 CU Media 235 9.2.2 Eigenexpansions 236 9.2.3 Examples of CU Media 238 9.3 Bi-Quadratic Equation, 240 9.3.1 BQ Media 241 9.3.2 Eigenexpansions 242 9.3.3 3D Representation 244 9.3.4 Special Case 245 Problems, 246

10 Media Defined by Plane-Wave Properties 249 10.1 Media with No Dispersion Equation (NDE Media), 249 10.1.1 Two Cases of Solutions 250 10.1.2 Plane-Wave Fields in NDE Media 255 10.1.3 Other Possible NDE Media 257 10.2 Decomposable Media, 259 10.2.1 Special Cases 259 10.2.2 DC-Medium Subclasses 263 10.2.3 Plane-Wave Properties 267 Problems, 269

Appendix A Solutions to Problems 273 Appendix B Transformation to Gibbsian Formalism 369 Appendix C Multivector and Dyadic Identities 375 References 389 Index 395 Preface

This book is a continuation of a previous one by the same author (Dif- ferential Forms in Electromagnetics, [1]) on the application of multivectors, multiforms, and dyadics to electromagnetic problems. Main attention is focused on applying the formalism to the analysis of electromagnetic media, as inspired by the ongoing engineering interest in constructing novel metamaterials and metaboundaries. In this respect the present exposition can also be seen as an enlargement of a chapter in a recent book on metamaterials [2] by including substance from more recent studies by this author and collaborators. The present four- dimensional (4D) formalism has proved of advantage in simplifying expressions in the analysis of general media in comparison to the classical three-dimensional (3D) Gibbsian formalism. However, the step from electromagnetic media, defined by medium parameters, to actual metamaterials and metaboundaries, defined by physical structures, is beyond the scope of this book. The first four chapters are devoted to the algebra of multiforms and dyadics in order to introduce the formalism and useful analytic tools. Similar material presented in [1] has been extended. Chapter 5 summarizes basic electromagnetic concepts in the light of the present formalism. Chapter 6 discusses transformations useful for simplifying problems. In the final Chapters 7–10 different classes of electromagnetic media are defined on the basis of their various properties. Because the most general linear electromagnetic medium requires 36 parameters for its definition, it is not easy to understand the effect of all these parameters. This is why it becomes necessary to define medium classes with reduced numbers of parameters. In Chapter 7 the classes are defined in terms of a natural decomposition of the medium bidyadic in three components, independent of any basis representation. Chapter 8 considers media whose medium bidyadic can be expressed in terms of quadratic functions of dyadics defined by 16 parameters. In Chapter 9 medium classes are defined by the degree of the algebraic equation satisfied by

xi xii Preface the medium bidyadic. Finally, in Chapter 10 media are defined by certain restrictions imposed on plane waves propagating in the media. Main emphasis lies on the application of the present formalism in the definition and analysis of media. It turns out that certain concepts cannot be easily defined through the 3D Gibbsian vector and dyadic representation. For example, the perfect electromagnetic conductor (PEMC) medium generalizing both perfect electric conductor (PEC) and perfect magnetic conductor (PMC) media appears as the simplest possible medium in the present formalism while in terms of conventional engineering representation with Gibbsian medium dyadics it requires parameters of infinite magnitude. As another example, decomposable bi-anisotropic media, defined to generalize uniaxially anisotropic media in which fields can be decomposed in transverse electric (TE) and transverse magnetic (TM) components, can be represented in a compact 4D form while the original analysis applying 3D Gibbsian formulation produced extensive expressions. In addition to the economy in expression, the present analysis is able to reveal novel additional solutions. A number of details in the analysis has been skipped in the text and left as problems for the reader. Solutions to the problems can be found at the end of the book, which allows the book to be used for self-study. Because of the background of the author, the book is mainly directed to electrical engineers, although physicists and applied mathematicians may find the contents of interest as well. It has been attempted tomake the transition from 3D Gibbsian vector and dyadic formalism, famil- iar to most electrical engineers, to the 4D exterior calculus involving multivectors, multiforms, and dyadics, as small as possible by showing connections to the corresponding Gibbsian quantities in an appendix. The main idea for adopting the 4D formalism is not to emphasize time- domain analysis of electromagnetic fields but to obtain compactness in expression and analysis. In fact, harmonic time dependence exp(j𝜔t)is often tacitly assumed by allowing complex magnitudes for the medium parameters. Compared to the previous book [1], the present approach shows some changes in the terminology followed by an effort to make the pre- sentation more accessible. For example, to emphasize the most important dyadics defining electromagnetic media, they have been called bidyadics because they represent mappings between two-forms and/or bivectors. Preface xiii

The author thanks students of the postgraduate courses based on the material of this book for their comments and responses. Special thanks are due to professors Ari Sihvola and Friedrich Hehl and for doctors Alberto Favaro and Luzi Bergamin for their long-lasting interest and help in treating questions during the years behind this book.

Ismo V. Lindell

CHAPTER 1

Multivectors and Multiforms

1.1 VECTORS AND ONE-FORMS

Let us consider two four-dimensional (4D) linear spaces, that of vectors, E F E 1 and that of one-forms 1. The elements of 1 are most generally denoted by boldface lowercase Latin letters, E a, b, c, …∈ 1, (1.1) F while the elements of 1 are most generally denoted by boldface lowercase Greek letters 𝜶 𝜷 𝜸 F . , , , …∈ 1 (1.2) E F The space of scalars is denoted by 0 or 0 and its elements are in general represented by nonboldface Latin or Greek letters a, b, c, …, 𝛼, 𝛽, 𝛾, …. Exceptions are made for quantities with established conventional notation. For example, the electric and magnetic fields are one-forms which are respectively denoted by the boldface uppercase Latin letters E and H.

1 2 CHAPTER 1 Multivectors and Multiforms

Because of the sign, it will be called as the bar product. In the past it has also been known as the duality product or the inner product. The bar product should not be confused with the dot product. The dot product can be defined for two vectors as a ⋅ b or two one-forms as 𝜶 ⋅ 𝜷 and it depends on a particular metric dyadic as will be discussed later.

1.1.2 Basis Expansions

A set of four vectors, e1, e2, e3, e4, is called a basis if any vector a can be expressed as

a = a1e1 + a2e2 + a3e3 + a4e4, (1.4) in terms of some scalars ai. Similarly, any one-form can be expanded in 𝜺 𝜺 𝜺 𝜺 a basis of one-forms, 1, 2, 3, 4 as 𝜶 𝛼 𝜺 𝛼 𝜺 𝛼 𝜺 𝛼 𝜺 . = 1 1 + 2 2 + 3 3 + 4 4 (1.5) The expansion of the bar product yields ∑4 ∑4 |𝜶 𝛼 |𝜺 . a = ai jei j (1.6) i=1 j=1 The vector and one-form bases are called reciprocal to one another if they satisfy |𝜺 𝛿 ei j = i,j, (1.7) with 𝛿 𝛿 ≠ . i,i = 1, i,j = 0 i j (1.8) In this case the scalar coeffients in (1.4) and (1.5) satisfy 𝜺 | 𝛼 |𝜶 ai = i a, i = ei , (1.9) and the bar product can be expanded as ∑4 |𝜶 𝛼 𝛼 𝛼 𝛼 𝛼 . a = ai i = a1 1 + a2 2 + a3 3 + a4 4 (1.10) i=1 From here onwards we always assume that when the two bases are 𝜺 denoted by ei and j, they are reciprocal. Vectors can be visualized as yardsticks in the 4D spacetime, and they can be used for measuring one-forms. For example, measuring 1.2 Bivectors and Two-Forms 3

F the electric field one-form E ∈ 1 by a vector a yields the voltage U between the endpoints of the vector a|E = U, (1.11) provided E is constant in space or a is small in terms of wavelength. The bar product a|𝜶 is a bilinear function of a and 𝜶. Thus, a|𝜶 can be conceived as a linear scalar-valued function of 𝜶 for a given vector a. Conversely, any linear scalar-valued function f (𝜶) can be expressed as a bar product a|𝜶 in terms of some vector a. To prove this, we express 𝜶 𝜺 in a basis { i} and apply linearity, whence we have (∑ ) ∑ ∑ |𝜶 𝜶 𝛼 𝜺 𝛼 𝜺 𝜺 |𝜶 a = f ( ) = f i i = if ( i) = f ( i)ei , (1.12) in terms of the reciprocal vector basis {ei}. Thus, the vector a can be defined as ∑ 𝜺 . a = f ( i)ei (1.13)

1.2 BIVECTORS AND TWO-FORMS

1.2.1 Wedge Product ∧ The antisymmetric wedge product ∧ between two vectors a and b yields E a bivector, an element of the space 2 of bivectors, a ∧ b =−b ∧ a. (1.14) This implies a ∧ a = 0, (1.15) for any vector a. In general, bivectors are denoted by boldface uppercase Latin letters, E A, B, C, …∈ 2, (1.16) and they can be represented by a sum of wedge products of vectors, A = a ∧ b + c ∧ d + ⋯ . (1.17) Similarly, the wedge product of two one-forms 𝜶 and 𝜷 produces a two-form 𝜶 ∧ 𝜷 =−𝜷 ∧ 𝜶. (1.18) 4 CHAPTER 1 Multivectors and Multiforms

Two-forms are denoted by boldface uppercase Greek letters whenever it appears possible, 𝚪 𝚽 𝚿 F , , , …∈ 2, (1.19) and they are linear combinations of wedge products of one-forms, 𝚪 = 𝜶 ∧ 𝜷 + 𝜸 ∧ 𝜹 + ⋯ . (1.20) A bivector which can be expressed as a wedge product of two vectors, in the form A = a ∧ b, (1.21) is called a simple bivector. Similarly, two-forms of the special form 𝚪 = 𝜶 ∧ 𝜷, (1.22) are called simple two-forms. For the 4D vector space as considered here, the bivectors form a space of six dimensions as will be seen below. It is not possible to express the general bivector in the form of a simple bivector.

1.2.2 Basis Expansions

Expanding vectors in a vector basis {ei} induces a basis expansion of bivectors where the basis bivectors can be denoted by eij = ei ∧ ej. Because eii = 0 and six of the remaining twelve bivectors are linearly dependent of the other six, . e12 = e1 ∧ e2 =−e21, e23 = e2 ∧ e3 =−e32,etc, (1.23) the space of bivectors is six dimensional. Actually, the bivector basis need not be based on any vector basis. Any set of six linearly independent bivectors could do. A bivector can be expanded in the bivector basis as ∑ A = AJeJ J . = A12e12 + A23e23 + A31e31 + A14e14 + A24e24 + A34e34 (1.24) Here, J = ij is a bi-index containing two indices i, j taken in a suitable order. In the following we will apply the order J = 12, 23, 31, 14, 24, 34. (1.25) 1.2 Bivectors and Two-Forms 5

Similarly, a basis of two-forms can be built upon the basis of one-forms 𝜺 𝜺 𝜺 𝜺 as J = ij = i ∧ j. It helps in memorizing if we assume that the index 4 corresponds to the temporal basis element and 1, 2, 3 to the three spatial elements. In this case the spatial indices appear in cyclical order 1 → 2 → 3 → 1in J while the index 4 occupies the last position. It is useful to define an operation K(J) yielding the complementary bi-index of a given bi-index J as K(12) = 34, K(23) = 14, K(31) = 24, (1.26) K(14) = 23, K(24) = 31, K(34) = 12. (1.27) Obviously, the complementary index operation satisfies K(K(J)) = J. (1.28) The basis expansion (1.24) can be used to show that any bivector can be expressed as a sum of two simple bivectors, in the form A = a ∧ b + c ∧ d. (1.29) ≠ Such a representation is not unique. As an example, assuming A23 0 in (1.24), we can write ( ) ∑3 1 . A = (A31e1 − A23e2) ∧ (A12e1 − A23e3) + Ai4ei ∧ e4 (1.30) A23 i=1 Thus, any bivector can be expressed in the form

A = a1 ∧ a2 + a3 ∧ e4, (1.31) |𝜺 where the vectors ai are spatial, that is, they satisfy ai 4 = 0. a1 ∧ a2 is called the spatial part of A and a3 ∧ e4 its temporal part. Similar rules are valid for two-forms. In particular, any two-form can be expanded in terms of spatial and temporal one-forms as 𝚪 𝜶 𝜶 𝜶 𝜺 |𝜶 . = 1 ∧ 2 + 3 ∧ 4, e4 i = 0 (1.32)

1.2.3 Bar Product We can extend the definition of the bar product of a vector and aone- form to that of a bivector and a two-form, A|𝚽 = 𝚽|A. Starting from a simple bivector a ∧ b and a simple one-form 𝜶 ∧ 𝜷 the bar product is a 6 CHAPTER 1 Multivectors and Multiforms quadrilinear scalar function of the two vectors and two one-forms and it can be expressed in terms of the four possible bar products of vectors and one-forms as ( ) a|𝜶 a|𝜷 (a ∧ b)|(𝜶 ∧ 𝜷) = (a|𝜶)(b|𝜷) − (a|𝜷)(b|𝜶) = det . b|𝜶 b|𝜷 (1.33)

Such an expansion follows directly from the antisymmetry of the wedge product and assuming orthogonality of the basis bivectors and two-forms as |𝜺 훿 훿 . eij k𝓁 = i,k j,𝓁 (1.34) by assuming ordered indices. Equation (1.33) can be memorized from the corresponding rule for three-dimensional (3D) Gibbsian vectors E denoted by ag, bg, cg, dg ∈ 1, ⋅ ⋅ ⋅ ⋅ ⋅ . (ag × bg) (cg × dg) = (ag cg)(bg dg) − (ag dg)(bg cg) (1.35) Relations of multivectors and multiforms to Gibbsian vectors are summarized in Appendix B. As examples of spatial two-forms we may consider the electric and magnetic flux densities, for which we use the established symbols D and B. Bivectors can be visualized as surface regions with orientation (sense of rotation). They can be used to measure the flux of a two-form through the surface region. For example, the magnetic flux Φ (a scalar) of the magnetic spatial two-form B through the bivector a ∧ b is obtained as

Φ=(a ∧ b)|B. (1.36)

For more details on geometric interpretation of multiforms see, for example, [3, 4].

1.2.4 Contraction Products ⌋ and ⌊ Considering a bivector a ∧ b and a two-form 𝚽, the bar product (a ∧ b)|𝚽 can be conceived as a linear scalar-valued function of the vector a. Thus, there must exist a one-form 𝜶 in terms of which we can express

a|𝜶 = (a ∧ b)|𝚽 = 𝚽|(a ∧ b) =−𝚽|(b ∧ a) = 𝜶|a. (1.37) 1.2 Bivectors and Two-Forms 7

Obviously, the one-form 𝜶 is a linear function of both b and 𝚽 so that we can express it as a product of the vector b and the two-form 𝚽 and denote it either 𝜶 = b⌋𝚽, (1.38) or 𝜶 =−𝚽⌊b. (1.39) The operation denoted by the multiplication sign ⌋ or ⌊ will be called contraction, because the two-form 𝚽 is contracted (“shortened”) by the vector b from the left or from the right to yield a one-form. Thus, the contraction product obeys the simple rules a|(b⌋𝚽) = (a ∧ b)|𝚽, (1.40) (𝚽⌊b)|a = 𝚽|(b ∧ a). (1.41) Contraction of a bivector A by a one-form 𝜶 can be defined similarly. Applied to (1.33), with slightly changed symbols, yields (d ∧ a)|(𝜷 ∧ 𝜸) − ((d|𝜷)(a|𝜸) − (d|𝜸)(a|𝜷)) = d|[a⌋(𝜷 ∧ 𝜸)) − ((𝜷(a|𝜸) − 𝜸(a|𝜷)))] = 0, (1.42) which is valid for any vector d. Choosing d = ei for i = 1, … ,4,all components of the one-form expression in square brackets vanish. Thus, we immediately obtain the “bac-cab rule” valid for any vector a and one- forms 𝜷, 𝜸, a⌋(𝜷 ∧ 𝜸) = 𝜷(a|𝜸) − 𝜸(a|𝜷) = (𝜸 ∧ 𝜷)⌊a. (1.43) Equation (1.43) corresponds to the well-known bac-cab rule of 3D Gibbsian vectors, Appendix B, ⋅ ⋅ ag × (bg × cg) = bg(ag cg) − cg(ag bg) = (cg × bg) × ag, (1.44) which helps in memorizing the 4D rule (1.43). Useful contraction rules for basis vectors and one-forms can be obtained as special cases of (1.43) as ⌋ 𝜺 𝜺 ⌋𝜺 𝜺 ej ( i ∧ j) = ej ij = i, (1.45) ⌊𝜺 ⌊𝜺 . (ei ∧ ej) i = eij i = ej (1.46) They can be easily memorized as a way of canceling basis vectors and one-forms with the same index from the contraction operation. 8 CHAPTER 1 Multivectors and Multiforms

1.2.5 Decomposition of Vectors and One-Forms Two vectors a, b ≠ 0 are called parallel if they satisfy the relation a ∧ b = 0. (1.47) Applying the bac-cab rule (1.43) for parallel vectors a, b, 𝜶⌋(a ∧ b) = a(𝜶|b) − b(𝜶|a) = 0, (1.48) implies that parallel vectors are linearly dependent, that is, one is a multiple of the other one. Assuming a ∧ b ≠ 0and𝜶|a ≠ 0, we can write the following decomposition for a given vector b:

b = b∥ + b⟂, (1.49) 𝜶|b 𝜶⌋(b ∧ a) b = a, b⟂ = . (1.50) ∥ 𝜶|a 𝜶|a

Here, b∥ can be interpreted as the component parallel to a given vector a, while b⟂ can be called as the component perpendicular to a given one-form 𝜶, because it satisfies

𝜶|b⟂ = 𝜶|(𝜶⌋(a ∧ b)) = (𝜶 ∧ 𝜶)|(a ∧ b) = 0. (1.51) Similarly, we can decompose a one-form 𝜷 as 𝜷 𝜷 𝜷 = ∥ + ⟂, (1.52) a|𝜷 a⌋(𝜶 ∧ 𝜷) 𝜷 = 𝜶, 𝜷⟂ = , (1.53) ∥ a|𝜶 a|𝜶 in terms of a given one-form 𝜶 and a given vector a satisfying 𝜶|a ≠ 0.

1.3 MULTIVECTORS AND MULTIFORMS

Higher-order multivectors and multiforms are produced through wedge multiplication. The wedge product is associative so that we have a ∧ (b ∧ c) = (a ∧ b) ∧ c = a ∧ b ∧ c, (1.54) and the brackets can be omitted. Thus, trivectors and three-forms are obtained as ⋯ E k = a1 ∧ a2 ∧ a3 + b1 ∧ b2 ∧ b3 + ∈ 3, (1.55) 𝝅 𝜶 𝜶 𝜶 𝜷 𝜷 𝜷 ⋯ F . = 1 ∧ 2 ∧ 3 + 1 ∧ 2 ∧ 3 + ∈ 3 (1.56) 1.3 Multivectors and Multiforms 9

They will be denoted by lowercase Latin and Greek characters taken, if possible, from the end of the alphabets. Quadrivectors and four-forms can be constructed as ⋯ E qN = a1 ∧ a2 ∧ a3 ∧ a4 + b1 ∧ b2 ∧ b3 ∧ b4 + ∈ 4, (1.57) 𝜿 𝜶 𝜶 𝜶 𝜶 𝜷 𝜷 𝜷 𝜷 ⋯ F . N = 1 ∧ 2 ∧ 3 ∧ 4 + 1 ∧ 2 ∧ 3 ∧ 4 + ∈ 4 (1.58) The subscript N denoting the quadri-index N = 1234 is used to mark a quadrivector or a four-form. Because the space of vectors is 4D, there are no multivectors of higher order than four. In fact, because any vector a5 can be expressed as ≠ a linear combination of a basis a1 … a4 satisfying a1 ∧ a2 ∧ a3 ∧ a4 0, as will be shown below, we have a1 ∧ a2 ∧ a3 ∧ a4 ∧ a5 = 0. The spaces of trivectors and three-forms are 4D and, those of quadrivectors and four-forms, one dimensional.

1.3.1 Basis of Multivectors

The vector basis {ei} induces the trivector basis

e123 = e1 ∧ e2 ∧ e3, e234 = e23 ∧ e4 = e2 ∧ e3 ∧ e4, e314 = e31 ∧ e4 = e3 ∧ e1 ∧ e4, e124 = e12 ∧ e4 = e1 ∧ e2 ∧ e4, (1.59) whence the space of trivectors is 4D. There is only a single basis quadrivector denoted by

eN = e1234 = e1 ∧ e2 ∧ e3 ∧ e4, (1.60) based upon the vector basis. Similar definitions apply for the basis three- 𝜺 𝜺 𝜺 𝜺 𝜺 𝜺 𝜺 𝜺 forms ijk = i ∧ j ∧ k and the basis four-form N = 1234 = 1 ∧ 2 ∧ 𝜺 𝜺 3 ∧ 4. Recalling the definition of the complementary bi-index (1.26), (1.27), and applying the antisymmetry of the wedge product we obtain . e23 ∧ eK(23) = e23 ∧ e14 = e2 ∧ e3 ∧ e1 ∧ e4 = e1234 = eN (1.61) More generally, we can write . eJ ∧ eK(J) = eK(J) ∧ eJ = eN (1.62) Defining the bi-index Kronecker delta by 훿 ≠ 훿 I,J = 0, I J, I,J = 1, I = J, (1.63) 10 CHAPTER 1 Multivectors and Multiforms we can write even more generally, 훿 . eI ∧ eJ = I,K(J)eN (1.64) This means that, unless I equals K(J), that is, J equals K(I), the wedge product yields zero.

1.3.2 Bar Product of Multivectors and Multiforms Extending the bar product to multivectors and multiforms, we can define the orthogonality relations for the reciprocal basis multivector and multiforms as |𝜺 훿 훿 훿 eijk rst = i,r j,s k,t, (1.65) |𝜺 훿 훿 훿 훿 eijk𝓁 rstu = i,r j,s k,t 𝓁,u, (1.66) when the indices are ordered. From the antisymmetry of the wedge product we obtain the expansion rules ⎛a|𝜶 a|𝜷 a|𝜸⎞ | 𝜶 𝜷 𝜸 ⎜ |𝜶 |𝜷 |𝜸⎟ (a ∧ b ∧ c) ( ∧ ∧ ) = det ⎜b b b ⎟ , (1.67) ⎝c|𝜶 c|𝜷 c|𝜸⎠

⎛a|𝜶 a|𝜷 a|𝜸 a|𝜹⎞ ⎜ |𝜶 |𝜷 |𝜸 |𝜹⎟ | 𝜶 𝜷 𝜸 𝜹 b b b b . (a ∧ b ∧ c ∧ d) ( ∧ ∧ ∧ ) = det ⎜ |𝜶 |𝜷 |𝜸 |𝜹⎟ (1.68) ⎜c c c c ⎟ ⎝d|𝜶 d|𝜷 d|𝜸 d|𝜹⎠

All quadrivectors are multiples of a given basis quadrivector eN and 𝜺 all four-forms are multiples of a given basis four-form N: 𝜿 휅𝜺 qN = qeN, N = N, (1.69) with |𝜺 휅 |𝜿 . q = qN N, = eN N (1.70) Applying the expansion rule for the determinant in (1.67), we can expand the bar product ( ) ( ) b|𝜷 b|𝜸 b|𝜶 b|𝜸 (a ∧ b ∧ c)|(𝜶 ∧ 𝜷 ∧ 𝜸) = a|𝜶 det − a|𝜷 det c|𝜷 c|𝜸 c|𝜶 c|𝜸 ( ) b|𝜶 b|𝜷 + a|𝜸 det , (1.71) c|𝜶 c|𝜷 1.3 Multivectors and Multiforms 11 whence from (1.33) we obtain the rule

(a ∧ b ∧ c)|(𝜶 ∧ 𝜷 ∧ 𝜸) = (a|𝜶)(b ∧ c)|(𝜷 ∧ 𝜸) + (a|𝜷)(b ∧ c)|(𝜸 ∧ 𝜶) + (a|𝜸)(b ∧ c)|(𝜶 ∧ 𝜷)]. (1.72)

1.3.3 Contraction of Trivectors and Three-Forms Defining the contraction of a three-form by a bivector as arising from

(a ∧ b ∧ c)|(𝜶 ∧ 𝜷 ∧ 𝜸) = a|((b ∧ c)⌋(𝜶 ∧ 𝜷 ∧ 𝜸)) = ((𝜶 ∧ 𝜷 ∧ 𝜸)⌊(b ∧ c))|a, (1.73) from (1.72) we obtain the expansion rule

(b ∧ c)⌋(𝜶 ∧ 𝜷 ∧ 𝜸) = (𝜶 ∧ 𝜷 ∧ 𝜸)⌊(b ∧ c) = ((b ∧ c)|(𝜷 ∧ 𝜸))𝜶 + ((b ∧ c)|(𝜸 ∧ 𝜶))𝜷 + ((b ∧ c)|(𝜶 ∧ 𝜷))𝜸. (1.74)

From this it follows that if the three one-forms satisfy 𝜶 ∧ 𝜷 ∧ 𝜸 = 0, they must be linearly dependent. Rewriting (1.72) in the form

If 𝜶 ∧ 𝜷 ∧ 𝜸 = 0, the three two-forms 𝜷 ∧ 𝜸, 𝜸 ∧ 𝜶,and𝜶 ∧ 𝜷 must be linearly dependent, which also follows from the linear dependence of the three one-forms. The contraction rules (1.74) and (1.76) are similar to the bac-cab rule (1.43) and they can be easily memorized because of the cyclic symmetry. Other similar forms are obtained by replacing vectors by one-forms and one-forms by vectors in (1.74) and (1.76). Commutation 12 CHAPTER 1 Multivectors and Multiforms rules for the contraction product can be summarized as 𝜶⌋k = k⌊𝜶, 𝚽⌋k = k⌊𝚽, (1.77) a⌋𝝅 = 𝝅⌊a, A⌋𝝅 = 𝝅⌊A. (1.78) Here, a is a vector, A is a bivector and k is a trivector while 𝜶 is a one-form, 𝚽 is a two-form and 𝝅 is a three-form. Useful rules for the contraction operations involving basis trivectors and three-forms can be formed as ⌋𝜺 ⌋𝜺 𝜺 ejk ijk = ej ij = i, (1.79) ⌋𝜺 𝜺 ek ijk = ij, (1.80) showing how similar indices are canceled in contraction operations.

1.3.4 Contraction of Quadrivectors and Four-Forms Following the same path of reasoning, starting from (1.68) we can expand the contraction of a four-form by a trivector (b ∧ c ∧ d)⌋(𝜶 ∧ 𝜷 ∧ 𝜸 ∧ 𝜹) =−(𝜶 ∧ 𝜷 ∧ 𝜸 ∧ 𝜹)⌊(b ∧ c ∧ d) =−(b ∧ c ∧ d)|(𝜶 ∧ 𝜷 ∧ 𝜸)𝜹 + (b ∧ c ∧ d)|(𝜷 ∧ 𝜸 ∧ 𝜹)𝜶 + (b ∧ c ∧ d)|(𝜸 ∧ 𝜶 ∧ 𝜹)𝜷 + (b ∧ c ∧ d)|(𝜶 ∧ 𝜷 ∧ 𝜹)𝜸, (1.81) the contraction of a four-form by a bivector, (c ∧ d)⌋(𝜶 ∧ 𝜷 ∧ 𝜸 ∧ 𝜹) = (𝜶 ∧ 𝜷 ∧ 𝜸 ∧ 𝜹)⌊(c ∧ d) = (c ∧ d)|(𝜷 ∧ 𝜸)(𝜶 ∧ 𝜹) + (c ∧ d)|(𝜷 ∧ 𝜶)(𝜸 ∧ 𝜹) + (c ∧ d)|(𝜸 ∧ 𝜶)(𝜷 ∧ 𝜹) + (c ∧ d)|(𝜸 ∧ 𝜹)(𝜶 ∧ 𝜷) + (c ∧ d)|(𝜶 ∧ 𝜹)(𝜷 ∧ 𝜸) + (c ∧ d)|(𝜷 ∧ 𝜹)(𝜸 ∧ 𝜶), (1.82) and the contraction of a four-form by a vector, d⌋(𝜶 ∧ 𝜷 ∧ 𝜸 ∧ 𝜹) =−(𝜶 ∧ 𝜷 ∧ 𝜸 ∧ 𝜹)⌊d = (d|𝜶)(𝜷 ∧ 𝜸 ∧ 𝜹) + (d|𝜷)(𝜸 ∧ 𝜶 ∧ 𝜹) + (d|𝜸)(𝜶 ∧ 𝜷 ∧ 𝜹) − (d|𝜹)(𝜶 ∧ 𝜷 ∧ 𝜸). (1.83) The above expressions appear invariant to cyclic permutation of the one-forms 𝜶, 𝜷, 𝜸, which may help in memorizing and checking the formulas. If 𝜶 ∧ 𝜷 ∧ 𝜸 ∧ 𝜹 = 0, from (1.81) it follows that the four one-forms are linearly dependent, from (1.82) it further follows that also the six 1.3 Multivectors and Multiforms 13 two-forms are linearly dependent and from (1.83) it follows that the four 𝜿 three-forms are linearly dependent. The contraction of a four-form N or quadrivector qN obeys the commutation rules ⌋𝜿 𝜿 ⌊ 𝜶⌋ ⌊𝜶 a N =− N a, qN =−qN , (1.84) ⌋𝜿 𝜿 ⌊ 𝚽⌋ ⌊𝚽 A N = N A, qN = qN , (1.85) ⌋𝜿 𝜿 ⌊ 𝝅⌋ ⌊𝝅 k N =− N k, qN =−qN , (1.86) Equations (1.81)–(1.83) imply the following contraction rules for the basis multivectors and multiforms: ⌋𝜺 𝜺 ⌊𝜺 e𝓁 ijk𝓁 = ijk, eijk𝓁 i = ejk𝓁 (1.87) ⌋𝜺 𝜺 ⌊𝜺 ek𝓁 ijk𝓁 = ij, eijk𝓁 ij = ek𝓁 (1.88) ⌋𝜺 𝜺 ⌊𝜺 ejk𝓁 ijk𝓁 = i, eijk𝓁 ijk = e𝓁, (1.89) which can be applied for canceling indices in expressions involving contraction of basis multivectors and multiforms. From (1.81) to (1.83) we can see that contraction of a four-form can be applied to transform vectors to three-forms, bivectors to two-forms and trivectors to one- forms and conversely. The converse cases can be obtained by applying the rules ⌊ 𝜺 ⌊ ⌋𝜺 ⌋ eN ( N a) = (a N) eN =−a, (1.90) ⌊ 𝜺 ⌊ ⌋𝜺 ⌋ eN ( N A) = (A N) eN = A, (1.91) ⌊ 𝜺 ⌊ ⌋𝜺 ⌋ . eN ( N k) = (k N) eN =−k (1.92)

1.3.5 Construction of Reciprocal Basis 𝜿 Given a set of basis vectors ai, i = 1, … , 4, and a four-form N, we can form the reciprocal one-form basis as a ⌋𝜿 𝜶 K(i) N i = |𝜿 , (1.93) aN N where the aK(i) are four three-forms defined by . aK(1) = a234, aK(2) = a314, aK(3) = a124, aK(4) =−a123 (1.94) satisfying 훿 . aj ∧ aK(i) =−aK(i) ∧ aj = aN i,j (1.95) 14 CHAPTER 1 Multivectors and Multiforms

The rule (1.93) is easily checked: (a ∧ a )|𝜿 |𝜶 j K(i) N 훿 . aj i = |𝜿 = i,j (1.96) aN N When the indices are ordered in the above sense, we can define the two-form basis reciprocal to {aij}as a ⌋𝜿 𝜶 K(ij) N . ij = |𝜿 (1.97) aN N In fact, from (1.64) we have (a ∧ a )|𝜿 |𝜶 k𝓁 K(ij) N 훿 훿 . ak𝓁 ij = |𝜿 = i,k j,𝓁 (1.98) aN N Because all four-forms are equal except for a scalar factor, the definitions 𝜿 are independent of the chosen four-form N.

1.3.6 Contraction of Quintivector Because any quintivector Q = a ∧ b ∧ c ∧ d ∧ e based on the 4D vector space vanishes, continuing the previous pattern by expanding the con- 𝜿 ⌋ traction N Q we can obtain the following relation between the five vectors: |𝜿 |𝜿 |𝜿 a(b ∧ c ∧ d ∧ e) N − b(a ∧ c ∧ d ∧ e) N + c(a ∧ b ∧ d ∧ e) N |𝜿 |𝜿 . − d(a ∧ b ∧ c ∧ e) N + e(a ∧ b ∧ c ∧ d) N = 0 (1.99) Assuming that the four vectors a ⋯ d are linearly independent, we have |𝜿 ≠ (a ∧ b ∧ c ∧ d) N 0, whence from (1.99) we obtain a rule how a given vector e can be expressed as a linear combination of the other four vectors.

1.3.7 Generalized Bac-Cab Rules From the expansions of the previous sections we can derive useful operational rules similar to the bac-cab rule (1.43) involving more general bivectors or two-forms. Let us start from (1.74) which can be written as (b ∧ c)⌋(𝜶 ∧ 𝜷 ∧ 𝜸) = 𝜷(b ∧ c)|(𝜸 ∧ 𝜶) + 𝜶(f|𝜸) − 𝜸(f|𝜶), (1.100) 1.3 Multivectors and Multiforms 15 where we denote for brevity

f = (b ∧ c)⌊𝜷. (1.101)

Applying the bac-cab rule (1.43) as

𝜶(f|𝜸) − 𝜸(f|𝜶) = f⌋(𝜶 ∧ 𝜸), (1.102)

(1.100) can be rewritten in the form

(b ∧ c)⌋(𝜶 ∧ 𝜷 ∧ 𝜸) = 𝜷(b ∧ c)|(𝜸 ∧ 𝜶) + (𝜸 ∧ 𝜶)⌊((b ∧ c)⌊𝜷). (1.103)

Since this identity, valid for any vectors b, c and one-forms 𝜶, 𝜷, 𝜸,is linear in the simple bivector b∑∧ c it remains valid if we replace b ∧ c by a sum of bivector products i bi ∧ ci. Thus, it is actually valid if we replace b ∧ c by the general bivector A. Similarly, since the expression is linear in the two-form 𝜸 ∧ 𝜶, it can be replaced by the general two-form 𝚪. Thus, we have arrived at the more general operational rule

A⌋(𝜷 ∧ 𝚪) = 𝜷(A|𝚪) + 𝚪⌊(A⌊𝜷), (1.104) which is another bac-cab rule valid for any bivector A, one-form 𝜷 and two-form 𝚪. A sister formula is obtained immediately by swapping vectors and forms as

𝚪⌋(b ∧ A) = b(𝚪|A) + A⌊(𝚪⌊b). (1.105)

Through similar considerations we can derive the following bac-cab rules valid for any bivectors B, C, vector b and one-form 𝜶:

𝜶⌋(b ∧ C) = b ∧ (𝜶⌋C) + C(𝜶|b), (1.106) 𝜶⌋(B ∧ C) = B ∧ (𝜶⌋C) + C ∧ (𝜶⌋B). (1.107)

Equation (1.106) can be used to define a decomposition for a given bivector B in two components defined by a given vector a and one-form 𝜶 satisfying a|𝜶 ≠ 0as 𝜶⌋ . B = B∥ + B⟂, a ∧ B∥ = 0, B⟂ = 0 (1.108) The components can be identified from the rule (1.106) as a ∧ (𝜶⌋B) 𝜶⌋(a ∧ B) B =− , B⟂ = . (1.109) ∥ a|𝜶 a|𝜶 16 CHAPTER 1 Multivectors and Multiforms

1.4 SOME PROPERTIES OF BIVECTORS AND TWO-FORMS

Bivectors and two-forms play very central roles in electromagnetics. Let us consider some of their properties.

1.4.1 Bivector Invariant Two bivectors commute in the wedge product, A ∧ B = B ∧ A, (1.110) and the result is a quadrivector. Any bivector A has a quadrivector-valued quadratic invariant which can be defined as 1 A∧ = A ∧ A. (1.111) 2 For a vector representation in terms of four vectors A = a ∧ b + c ∧ d, (1.112) the invariant has the form A∧ = a ∧ b ∧ c ∧ d. (1.113)

Inserting an expansion in a vector basis {ei} we obtain 1 A∧ = (A e + A e + ⋯) ∧ (A e + A e + ⋯) 2 12 12 23 23 12 12 23 23 . = (A12A34 + A23A14 + A31A24)eN (1.114) Alternatively, we can write ∑ ∑ ∑ 1 1 A∧ = A e ∧ A e = A A e , (1.115) 2 I I J J 2 I K(I) N where summing goes over the ordered bi-indices. Obviously, the quadrivector invariant of a simple bivector vanishes, (a ∧ b)∧ = 0. (1.116) The converse is also true: if a bivector satisfies A∧ = 0, it must be simple. A corresponding invariant can be defined for two-forms with similar properties. 1.4 Some Properties of Bivectors and Two-Forms 17

1.4.2 Natural Dot Product 𝜺 A bivector A is mapped to a two-form by a four-form N and a two-form 𝚪 is mapped by a quadrivector eN to a bivector as 𝚪 𝜺 ⌊ ⌊𝚪. = N A, A = eN (1.117) A natural dot product for two bivectors A, B can be defined by ⋅ | 𝜺 ⌊ 𝜺 | ⋅ A B = A ( N B) = N (A ∧ B) = B A, (1.118) which yields a scalar. Similarly, a dot product for two two-forms is defined by 𝚪 ⋅ 𝚲 𝚪| ⌊𝚲 | 𝚪 𝚲 𝚲 ⋅ 𝚪. = (eN ) = eN ( ∧ ) = (1.119) Here one must note that the magnitude and sign of the scalars depend 𝜺 on the choice of the quadrivector eN and the reciprocal four-form N.A bivector A is simple exactly when A ⋅ A = 0 and a two-form 𝚪 is simple when 𝚪 ⋅ 𝚪 = 0.

1.4.3 Bivector as Mapping A given bivector G maps one-forms 𝜶 to vectors a as G⌊𝜶 = a. (1.120) The mapping may have an inverse of the form 𝚪⌊a = 𝜶. (1.121) To find the two-form 𝚪 for a given bivector G we can first apply the bac-cab rule (1.107) which for C = B = G can be rewritten as 1 G ∧ (𝜶⌋G) = 𝜶⌋(G ∧ G) = 𝜶⌋G∧ = 𝜶⌋e (𝜺 |G∧) 2 N N ⌊𝜶 𝜺 | ∧ . =−eN ( N G ) (1.122) On the other hand from (1.120) we have G ∧ (𝜶⌋G) =−G ∧ (G⌊𝜶) =−G ∧ a, (1.123) whence the two mappings are related by ⌊𝜶 𝜺 | ∧ . eN ( N G ) = G ∧ a (1.124) 18 CHAPTER 1 Multivectors and Multiforms

𝜺 ⌊ Operating this by N and applying (1.90) we obtain 𝜶 𝜺 | ∧ 𝜺 ⌊ 𝜺 ⌊ ⌊ . ( N G ) =− N (G ∧ a) =−( N G) a (1.125) Comparing with (1.121) we can now find the two-form 𝚪 as 𝜺 ⌊G 𝚪 =− N . (1.126) 𝜺 | ∧ N G Obviously, this is valid only when the bivector G is not simple, G ∧ G ≠ 0.

PROBLEMS

1.1 Show that for 훼훽 ≠ 0 the linear combination 훼A + 훽B of two simple bivectors A and B is simple exactly when A ∧ B = 0is valid. 1.2 Assuming a nonsimple bivector C, show that, for a given simple bivector A satisfying A ∧ C ≠ 0, we can find another simple bivector B and a scalar 훼 so that we can express C = 훼A + B. This proves that any bivector is either simple or it can be expressed as a sum of two simple bivectors.

1.3 Expressing∑ a bivector in terms of a given vector basis ai as B = < Bijai ∧ aj, i j, find the relation between the coefficients Bij so that B is simple. 1.4 Assuming A, a simple bivector, and B a nonsimple bivector, using basis expansions find the condition for B satisfying A ∧ B = 0. 1.5 Show that for a bivector C and one-form 𝜷 the condition 𝜷⌋C = 0 implies that either 𝜷 = 0orC is simple. 1.6 Assuming two simple bivectors A and B, show that if they satisfy A ∧ B = 0, there exist vectors a, b, c so that we can write A = a ∧ c and B = b ∧ c. 1.7 Show that a linear combination of two nonsimple bivectors can be so defined that it is a simple bivector. 𝚽 𝜺 ⌊ 1.8 Prove that if A is a simple bivector, = N A is a simple two- form. Problems 19

1.9 Find the one-form 𝜶 solution for the equation A⌊𝜶 = a, when A is a given nonsimple bivector and a is a given vector. Hint: write∑A = e1 ∧ e∑2 + e3 ∧ e4 where ei form a basis and expand 𝜶 훼 𝜺 𝜺 a = aiei, = j j when j is the reciprocal basis. 1.10 Derive the identity (A ∧ A)⌊𝖨T = 2A ∧ (A⌊𝖨T ) for a nonsimple bivector A and apply it for the solution of the equation of the previous problem. Hint: define A = e1 ∧ e2 + e3 ∧ e4 in terms of basis vectors ei. 1.11 Show that if a bivector A satisfies A⌊𝜶 = 0, A⌊𝜷 = 0 for two one-forms satisfying 𝜶 ∧ 𝜷 ≠ 0, it can be represented as ⌊ 𝜶 𝜷 A = kN ( ∧ )

in terms of some quadrivector kN. 1.12 Defining six nonsimple bivectors in terms of simple basis bivectors eij as

E1 = e12 + e34, E2 = e23 + e14, E3 = e31 + e24,

E4 = e12 − e34, E5 = e23 − e14, E6 = e31 − e24, show that they form a basis, that is, any bivector can be expanded as ∑ A = AiEi, ⋅ and that they are orthogonal in the dot product Ei Ej = 𝜺 | N (Ai ∧ Aj). 1.13 Derive (1.106) and (1.107). 1.14 Find the most general bivector B satisfying a ∧ B = 0 for a given vector a. 20 CHAPTER 1 Multivectors and Multiforms

1.15 Show that a bivector satisfying the equation 𝜶⌋B = 0 for a given one-form 𝜶 must be of the form B = 𝜶⌋k, where k may be any trivector. Show also that B must be simple bivector. 1.16 Derive the identity 𝜶⌋(k ∧ a) = a ∧ (𝜶⌋k) + (𝜶|a)k, where 𝜶 is a one-form, a is a vector and k is a trivector. Show that it is possible to decompose any trivector as k = k훼 + ka in 𝜶⌋ terms of two trivectors satisfying k훼 = 0anda ∧ ka = 0when a and 𝜶 are two given quantities satisfying a|𝜶 ≠ 0. 1.17 Derive the identity ⌊ 𝜶 𝚿 𝜶⌋ ⌊𝚿 eN ( ∧ ) =− (eN ) valid for any one-form 𝜶 and two-form 𝚿. 1.18 Derive the identity 𝜺 ⌊ ⌋ 𝜺 ⌊ ⌋ ( N k) A = ( N A) k, E E valid for any bivector A ∈ 2 and trivector k ∈ 3. Check the identity for k = e234, A = e12. 1.19 Find solutions X to the bivector equation A ∧ X = 0whenA satisfies A ∧ A ≠ 0. 1.20 Show that any simple bivector a ∧ b can be expressed in the form cs ∧ d where c is a spatial vector and d another vector. Assume 𝜺 | ≠ 𝜺 | ≠ 4 a 0and 4 b 0. CHAPTER 2

Dyadics

Dyadics are algebraic objects performing linear mappings between mul- E → E F → F E → F F → E tivectors and/or multiforms, p q, p q, p q, p q. Mapping between spaces of the same dimension, p = q or p = 4 − q, appear to be the most interesting ones because they may be inverted.

2.1 MAPPING VECTORS AND ONE-FORMS

2.1.1 Dyadics Any linear mapping between two vectors x → y can be expressed in the form ( ) ∑4 ∑4 𝜶 | 𝜶 | 𝖠| y = ai( i x) = ai i x = x, (2.1) i=1 i=1 in terms of suitably chosen vectors a and one-forms 𝜶 . Here we denote ∑i i 𝖠 𝜶 = ai i (2.2) which is called a dyadic polynomial or dyadic for short. Here we apply the original Gibbsian notation [5] denoting the dyadic product by “no sign” instead of the sign ⊗ encountered in mathematical literature. The representation (2.2) is not unique. By choosing a vector basis ei or a 𝜺 one-form basis j the same dyadic can be expressed in various forms, ∑ ∑ ∑∑ 𝖠 𝜶 𝜺 𝜺 . = ei i = ai i = Aijei j (2.3)

21 22 CHAPTER 2 Dyadics

The number of free parameters is 16 which can be defined through 𝜶 four one-forms i, four vectors ai, or 16 scalars Aij. Dyadics mapping E F vectors to vectors form a linear space which is denoted by 1 1. Similarly, dyadics mapping one-forms to one-forms form a linear F E space which is denoted by 1 1. They have the general form ∑ 𝖠 𝜶 . = iai (2.4) As a special example, the dyadic 𝖨 𝜺 𝜺 𝜺 𝜺 = e1 1 + e2 2 + e3 3 + e4 4, (2.5) called the unit dyadic, maps any vector to itself for any choice of recip- 𝜺 rocal vector and one-form bases ei, j, ∑ ∑ 𝖨| 𝜺 | . a = ei( i a) = aiei = a (2.6) The corresponding unit dyadic mapping any one-form to itself is denoted by 𝖨T 𝜺 𝜺 𝜺 𝜺 = 1e1 + 2e2 + 3e3 + 4e4, (2.7) where ()T denotes the transpose operation,

(a𝜶)T = 𝜶a. (2.8) E F For dyadics of the space 1 1 one can define the bar product of two dyadics which yields a dyadic of the same space, ∑ ∑ ∑∑ 𝖠|𝖡 𝜶 | 𝜷 𝜶 | 𝜷 . = ai i bj j = ( i bj)ai j (2.9) Dyadic powers are defined by

𝖠2 = 𝖠|𝖠, 𝖠3 = 𝖠|𝖠|𝖠, ⋯ (2.10)

The inverse dyadic 𝖠−1, when it exists, satisfies

𝖠|𝖠−1 = 𝖠−1|𝖠 = 𝖨, (2.11) and negative powers can be understood as

𝖠−p = (𝖠−1)p = (𝖠p)−1. (2.12) F E Similar properties are valid for dyadics belonging to the space 1 1. 2.1 Mapping Vectors and One-Forms 23

2.1.2 Double-Bar Product || ∑ ∑ 𝖠 𝜶 𝖡 𝜷 The double-bar product of a dyadic = ai i and a dyadic = bj j transposed is defined by ∑ ∑ ∑∑ 𝖠||𝖡T 𝜶 || 𝜷 |𝜷 𝜶 | . = ai i jbj = (ai j)( i bj) (2.13)

The result is a scalar which is a linear function of both dyadics. The double-bar product satisfies

𝖠||𝖡T = 𝖡T ||𝖠 = 𝖡||𝖠T = 𝖠T ||𝖡. (2.14)

E F F E Dyadics of the∑ spaces 1 1 and 1 1 have a scalar invariant called 𝖠 𝜶 the trace. For = ai i it is defined by ∑ ∑ 𝖠 𝖠||𝖨T |𝖨T |𝜶 |𝜶 𝖠T tr = = ai i = ai i = tr , (2.15) or ∑∑ ∑∑ ∑ 𝖠 𝜺 |𝜺 . tr = tr Ai,jei j = Ai,jei j = Ai,i (2.16)

Because 𝖨 is independent of the choice of the reciprocal bases, so is the trace of the dyadic. In particular, we have ∑ 𝖨 |𝜺 . tr = ei i = 4 (2.17) E F Trace of a product of two dyadics of the space 1 1 satisfies

tr(𝖠|𝖡) = 𝖠||𝖡T . (2.18)

The trace of a single dyadic can be expressed as

tr𝖠 = tr(𝖠|𝖨) = 𝖠||𝖨T = 𝖠T ||𝖨 = 𝖨||𝖠T = 𝖨T ||𝖠. (2.19)

𝖠 E F Any dyadic ∈ 1 1 can be uniquely decomposed as a sum of a multiple of the unit dyadic and a trace-free part as

1 𝖠 = tr𝖠 𝖨 + 𝖠 ,tr𝖠 = 0. (2.20) 4 o o F E Similar properties apply for dyadics of the space 1 1. 24 CHAPTER 2 Dyadics

2.1.3 Metric Dyadics F F E E Dyadics of the spaces 1 1 and 1 1 mapping respectively vectors to one-forms and one-forms to vectors are called metric dyadics. They can be expanded as ∑  𝜺 𝜺 F F = Γij i j ∈ 1 1 (2.21) and ∑ 𝖦 E E . = Gijeiej ∈ 1 1 (2.22) Because the transpose operation of a metric dyadic yields a dyadic in the same space, the dyadic eigenproblems 𝖦T = 𝜆𝖦, T = 𝜆 (2.23) make sense and have two eigenvalues 𝜆 = 1, −1 corresponding to symmetric and antisymmetric dyadics, respectively. Symmetric dyadics can be used to define a dot product for vectors and one-forms as ∑ a ⋅ b = b ⋅ a = a||b = Γ (a|𝜺 )(b|𝜺 ), (2.24) ∑ ii i i 𝜶 ⋅ 𝜷 𝜷 ⋅ 𝜶 𝜶|𝖦|𝜷 𝜶| 𝜷| . = = = Gii( ei)( bi) (2.25) The dot products for vectors and one-forms are compatible when the | F | F dot product for the mapped vectors a ∈ 1, b ∈ 1 equals the dot product of the vectors, (|a) ⋅ (|b) = a||𝖦||b = a||b = a ⋅ b (2.26) for all vectors a, b. This requires that 𝖦| = 𝖨, that is, that the two symmetric dyadics are inverses of each other,  = 𝖦−1, 𝖦 = −1. (2.27) The double-bar product 𝖠||𝖡 can be defined for two metric dyadics 𝖠 E E 𝖡 F F when the dyadics are taken from different spaces, ∈ 1 1, ∈ 1 1. Because of the property 𝖠||𝖡 = 𝖠T ||𝖡T , (2.28) the double-bar product vanishes if one of the metric dyadics is symmetric and the other one is antisymmetric. 2.2 Mapping Multivectors and Multiforms 25

2.2 MAPPING MULTIVECTORS AND MULTIFORMS

The previous definitions of dyadics can be extended to mappings between multivectors and/or multiforms.

2.2.1 Bidyadics A dyadic defined by ∑ 𝖣 𝜺 E F = DI,JeI J =∈ 2 2 (2.29) where the bi-indices I, J go from 12 to 34, maps bivectors to bivectors, ∑ ∑ ∑ 𝖣| 𝜺 | E B = A = DI,JeI J AKeK = DI,JAJeI ∈ 2, (2.30) 𝖣 E E and it can be called a bidyadic. Similarly, a dyadic ∈ 2 2 mapping two-forms to bivectors ∑ ∑ ∑ 𝖣|𝚽 | 𝜺 E B = = DI,JeIeJ KΦK = DI,JΦJeI ∈ 2 (2.31) K can be called a metric bidyadic. Because the electromagnetic fields are represented by two-forms, bidyadics play a central role in electromagnetics. In fact, electromagnetic media can be defined in terms of bidyadics. Bivectors are six-dimensional objects, whence a bidyadic has 6 × 6 = 36 parameters in the most general case. Another form for the 𝖣 E F bidyadic ∈ 2 2 is the expansion in bivectors and two-forms as ∑ 𝖣 𝚽 . = AI I (2.32)

Choosing any linearly independent set of six bivectors AI for a basis of bivectors, the bidyadic 𝖣 is uniquely determined by the set of six 𝚽 two-forms I and vice versa. The trace of a bidyadic is defined by ∑ ∑ 𝖣 |𝚽 . tr = AI I = DI,I (2.33)

∧ 2.2.2 Double-Wedge Product ∧ Multivectors and multiforms can be constructed from vectors and one- forms by applying the wedge product. Similarly, multidyadics can be formed from dyadics by applying the double-wedge product. 26 CHAPTER 2 Dyadics

The double-wedge product between two dyadic products of vectors and one-forms can be introduced by 𝜶 ∧ 𝜷 𝜶 𝜷 (a )∧(b ) = (a ∧ b)( ∧ ), (2.34) E F and the result lies in the space of bidyadics denoted by 2 2.More generally, we can define the double-wedge product of two dyadics as (∑ ) (∑ ) ∑∑ 𝖠∧𝖡 𝜶 ∧ 𝜷 𝜶 𝜷 . ∧ = ai i ∧ bj j = (ai ∧ bj)( i ∧ j) (2.35)

Since the wedge product is anticommutative and associative, the double- wedge product is commutative, 𝖠∧𝖡 𝖡∧𝖠 ∧ = ∧ , (2.36) and associative, 𝖠∧𝖡 ∧𝖢 𝖠∧ 𝖡∧𝖢 . ( ∧ )∧ = ∧( ∧ ) (2.37) Thus, dyadics in a chain of double-wedge products can be set in any order and there is no need for brackets to show the order of multiplication. Of course, a product chain of five or more dyadics yields zero. × The corresponding double-cross product × was introduced by Gibbs already in 1886 [6], Figure 2.1.

× ∧ Figure 2.1. The double-cross product × corresponding to the double-wedge product ∧ was originally introduced by Gibbs in 1886 [6]. Here the bar corresponds to the dyadic product (later replaced by “no sign” by Gibbs [5]). The expression on the last line stands for the nth double-cross power of the dyadic 𝚽. 2.2 Mapping Multivectors and Multiforms 27

The basic rule for the double-wedge product of two dyadics can be formed by starting from the expansion

𝜶∧ 𝜷 | 𝜶 𝜷 | (a ∧b ) (c ∧ d) = ((a ∧ b)( ∧ )) (c ∧ d) = (a ∧ b)((𝜶 ∧ 𝜷)|(c ∧ d)) = (a ∧ b)((𝜶|c)(𝜷|d) − (𝜶|d)(𝜷|c)) = (a𝜶|c) ∧ (b𝜷|d) − (a𝜶|d) ∧ (b𝜷|c). (2.38)

Since the first and last expressions are linear in the dyadic products a𝜶 𝜷 𝖠 E F and b , we can replace them by the respective general dyadics ∈ 1 1 𝖡 E F and ∈ 1 1, whence we have arrived at the rule 𝖠∧𝖡 | 𝖠| 𝖡| 𝖡| 𝖠| . ( ∧ ) (a ∧ b) = ( a) ∧ ( b) + ( a) ∧ ( b) (2.39) The right-hand side can be easily memorized by noticing that it is invariant in the interchange of 𝖠 and 𝖡 and changes sign in the interchange of a and b. Multiplying (2.39) dyadically by (𝜶 ∧ 𝜷) from the right we obtain

𝖠∧𝖡 | 𝜶 𝜷 𝖠∧𝖡 | 𝜶 ∧ 𝜷 ( ∧ ) (a ∧ b)( ∧ ) = ( ∧ ) (a )∧(b ) 𝖠| 𝜶 ∧ 𝖡| 𝜷 𝖡| 𝜶 ∧ 𝖠| 𝜷 . = ( a )∧( b ) + ( a )∧( b ) (2.40) Replacing again the dyadic products (a𝜶)and(b𝜷) by the respective general dyadics 𝖢 and 𝖣 we obtain an identity valid for any four dyadics E F of the space 1 1, 𝖠∧𝖡 | 𝖢∧𝖣 𝖠|𝖢 ∧ 𝖡|𝖣 𝖠|𝖣 ∧ 𝖡|𝖢 . ( ∧ ) ( ∧ ) = ( )∧( ) + ( )∧( ) (2.41) The form of (2.41) remains valid when the four dyadics are taken from F E the space 1 1. This can be shown by transposing both sides of the dyadic identity, applying

𝖠∧𝖡 T 𝖠T ∧𝖡T ( ∧ ) = ∧ , (2.42) and changing the symbols. Finally, (2.41) also remains to be valid 𝖠 𝖡 E E 𝖢 𝖣 F F for dyadics taken from metric spaces , ∈ 1 1 and , ∈ 1 1 or conversely. Applying the double-wedge product one can produce multidyadics 𝖠∧𝖡∧𝖢 E F 𝖠∧𝖡∧𝖢∧𝖣 E F of higher order, ∧ ∧ ∈ 3 3,and ∧ ∧ ∧ ∈ 4 4. Going through 28 CHAPTER 2 Dyadics algebraic steps similar to the previous ones, the corresponding basic multiplication rule for three dyadics and three vectors can be expressed as

𝖠∧𝖡∧𝖢 | 𝖠| 𝖡| 𝖢| ( ∧ ∧ ) (a ∧ b ∧ c) = ( a) ∧ ( b) ∧ ( c) + (𝖠|b) ∧ (𝖡|c) ∧ (𝖢|a) + (𝖠|c) ∧ (𝖡|a) ∧ (𝖢|b) − (𝖠|a) ∧ (𝖡|c) ∧ (𝖢|b) − (𝖠|c) ∧ (𝖡|b) ∧ (𝖢|a) − (𝖠|b) ∧ (𝖡|a) ∧ (𝖢|c), (2.43) which gives rise to the general rule for two sets of three dyadics becomes

𝖠∧𝖡∧𝖢 | 𝖣∧𝖤∧𝖥 𝖠|𝖣 ∧ 𝖡|𝖤 ∧ 𝖢|𝖥 ( ∧ ∧ ) ( ∧ ∧ ) = ( )∧( )∧( ) 𝖠|𝖤 ∧ 𝖡|𝖥 ∧ 𝖢|𝖣 𝖠|𝖥 ∧ 𝖡|𝖣 ∧ 𝖢|𝖤 + ( )∧( )∧( ) + ( )∧( )∧( ) 𝖠|𝖣 ∧ 𝖡|𝖥 ∧ 𝖢|𝖤 𝖠|𝖥 ∧ 𝖡|𝖤 ∧ 𝖢|𝖣 + ( )∧( )∧( ) + ( )∧( )∧( ) 𝖠|𝖤 ∧ 𝖡|𝖣 ∧ 𝖢|𝖥 . + ( )∧( )∧( ) (2.44) E F The double-wedge product of four dyadics of the space 1 1 is always 𝜺 a multiple of eN N: 𝖠∧𝖡∧𝖢∧𝖣 𝜺 | 𝖠∧𝖡∧𝖢∧𝖣 | 𝜺 . ∧ ∧ ∧ = ( N ( ∧ ∧ ∧ ) eN)eN N (2.45)

2.2.3 Double-Wedge Powers The double-wedge square of a dyadic can be defined as 1 𝖠(2) = 𝖠∧𝖠, (2.46) 2 ∧ and, because of (2.39), it obeys the rule

𝖠(2)|(a ∧ b) = (𝖠|a) ∧ (𝖠|b). (2.47)

The sign ()(p) is due to Reichardt [7], see Figure 2.2. From

𝖡(2)|𝖠(2)|(a ∧ b) = 𝖡(2)|((𝖠|a) ∧ (𝖠|b)) = (𝖡|𝖠|a) ∧ (𝖡|𝖠|b) (2.48) we come to the rule

(𝖡|𝖠)(2) = 𝖡(2)|𝖠(2). (2.49)

This follows also from (2.41) when choosing 𝖢 = 𝖠 and 𝖣 = 𝖡. 2.2 Mapping Multivectors and Multiforms 29

Figure 2.2. The notation (bracketed number in superscript) for the double-wedge power was introduced by Reichardt [7]. There is no sign denoting the bar product.

The double-wedge square of the unit dyadic 𝖨 ∑∑ ∑ 𝖨(2) 1 𝜺 𝜺 𝜺 = (ei ∧ ej)( i ∧ j) = eij ij, 2 i 4. From (2.55) we obtain (𝖠q)(p) = (𝖠(p))q, (2.56) 30 CHAPTER 2 Dyadics which remains valid for negative q values if the inverse 𝖠−1 exists. Let us denote for simplicity 𝖠(−p) = (𝖠−1)(p) = (𝖠(p))−1. (2.57) The dyadics ∑ 𝖨(3) 𝜺 = eijk ijk, (2.58) imatrix in any chosen basis.

2.2.4 Double Contractions ⌊⌊ and ⌋⌋ We can introduce double contractions ⌊⌊ and ⌋⌋ between a bidyadic A𝚪 and a dyadic 𝜶a as (A𝚪)⌊⌊(𝜶a) = (A⌊𝜶)(𝚪⌊a), (2.65) (𝜶a)⌋⌋(A𝚪) = (𝜶⌋A)(a⌋𝚪). (2.66) 2.2 Mapping Multivectors and Multiforms 31

E F The result belongs to the dyadic space 1 1. More generally, double 𝖠 E F 𝖡 E F contractions between a bidyadic ∈ 2 2 and a dyadic ∈ 1 1 can be defined as ∑ ∑ ∑ 𝖠⌊⌊𝖡T 𝚪 ⌊⌊ 𝜷 ⌊𝜷 𝚪 ⌊ = Aij ij kbk = (Aij k)( ij bk), (2.67) < < ∑i j ∑k i∑j,k 𝖡T ⌋⌋𝖠 𝜷 ⌋⌋ 𝚪 𝜷 ⌋ ⌋𝚪 . = kbk Aij ij = ( k Aij)(bk ij) (2.68) k i

2.2.5 Natural Dot Product for Bidyadics The natural dot product for two bivectors (1.118) ⋅ | 𝜺 ⌊ 𝜺 | A B = A ( N B) = N (A ∧ B) (2.70) and two two-forms (1.119) 𝚽 ⋅ 𝚿 𝚽| ⌊𝚿 | 𝚽 𝚿 = (eN ) = eN ( ∧ ) (2.71) can be directly applied to define a dot product for two bidyadics. The dot products are defined by the two metric bidyadics 𝖤 and 𝖤−1 defined by 𝖤 ⌊𝖨(2)T E E 𝖤−1 𝜺 ⌊𝖨(2) F F = eN ∈ 2 2, = N ∈ 2 2, (2.72) which are symmetric, ∑ ∑ 𝖤 ⌊ 𝜺 𝜺 𝜺 = eN eij ij = K(ij) ij

= e12e34 + e23e14 + e31e24 + e14e23 + e24e31 + e34e12 (2.73) ∑ ∑ 𝖤−1 𝜺 ⌊ 𝜺 𝜺 𝜺 = N eij ij = K(ij) ij 𝜺 𝜺 𝜺 𝜺 𝜺 𝜺 𝜺 𝜺 𝜺 𝜺 𝜺 𝜺 . = 12 34 + 23 14 + 31 24 + 14 23 + 24 31 + 34 12 (2.74) 𝖢 𝖣 E F For two bidyadics , ∈ 2 2 there exist four possible dot products 𝖢 ⋅ 𝖣T 𝖢| ⌊𝖣T 𝖢|𝖤|𝖣T E E = (eN ) = ∈ 2 2, (2.75) 𝖣 ⋅ 𝖢T 𝖣| ⌊𝖢T 𝖣|𝖤|𝖢T E E = (eN ) = ∈ 2 2, (2.76) 32 CHAPTER 2 Dyadics

𝖢T ⋅ 𝖣 𝖢T | 𝜺 ⌊𝖣 𝖢T |𝖤−1|𝖣 F F = ( N ) = ∈ 2 2 (2.77) 𝖣T ⋅ 𝖢 𝖣T | 𝜺 ⌊𝖢 𝖣T |𝖤−1|𝖢 F F . = ( N ) = ∈ 2 2 (2.78) The dot product of bidyadics has some obvious properties. The symmetric dyadics 𝖤 and 𝖤−1 act as unit mappings in the dot product. 𝖦 E E For example, for the metric bidyadic ∈ 2 2 we have 𝖤 ⋅ 𝖦 = 𝖤|𝖤−1|𝖦 = 𝖦. (2.79) 𝖦 𝖦 Also, we have for the metric bidyadics 1, 2 𝖦−1 ⋅ 𝖦−1 𝖦−1|𝖤|𝖦−1 𝖦 |𝖤−1|𝖦 −1 𝖦 ⋅ 𝖦 −1. 1 2 = 1 2 = ( 2 1) = ( 2 1) (2.80) Finally, we can easily prove 𝜺 𝜺 ⌊⌊𝖦 ⋅ 𝜺 𝜺 ⌊⌊𝖦 𝜺 𝜺 ⌊⌊ 𝖦 ⋅ 𝖦 . ( N N 1) ( N N 2) = N N ( 1 2) (2.81) Similar rules can be written for other bidyadics in slightly different form.

2.3 DYADIC IDENTITIES

Dyadic identities, equations valid for any dyadics, act as operational rules which can be applied in the analysis of equations encountered in electromagnetics. Use of identities replaces working with basis expansions of quantities and applying their relations. Let us now consider some dyadic identities which can be straightforwardly derived through basis expansions. Proofs of some of them are given here while the rest are left as exercises.

2.3.1 Contraction Identities Three identities involving contractions of the double-wedge product of two, three and four dyadics form a basis from which special identities 𝖠 𝖡 𝖢 𝖣 E F can be tailored. All dyadics , , , belong to the space 1 1.The F E identities are also valid for dyadics taken from the space 1 1 but in this case the unit dyadic 𝖨, wherever present, must be replaced by its transpose 𝖨T . The identity 𝖠∧𝖡 ⌊⌊𝖢T 𝖠||𝖢T 𝖡 𝖡||𝖢T 𝖠 𝖠|𝖢|𝖡 𝖡|𝖢|𝖠 ( ∧ ) = ( ) + ( ) − − , (2.82) 2.3 Dyadic Identities 33 can be derived by replacing the dyadics by three dyadic products of vectors and one-forms and applying the bac-cab rule twice, 𝜶∧ 𝜷 ⌊⌊ 𝜸 T ⌊𝜸 𝜶 𝜷 ⌊ (a ∧b ) (c ) = ((a ∧ b) )(( ∧ ) c), = (b(a|𝜸) − a(b|𝜸))(𝜷(𝜶|c) − 𝜶(𝜷|c)), = b𝜷(a𝜶||𝜸c) + a𝜶(b𝜷||𝜸c) − a𝜶|c𝜸|b𝜷 − b𝜷|c𝜸|a𝜶. (2.83) At the last stage the four terms were assembled to produce the original dyadic products. Since the resulting identity is linear in its three dyadic products of vectors and one-forms, replacing the products by general dyadics 𝖠, 𝖡, 𝖢 yields the dyadic identity (2.82). The following two identities can be derived similarly with somewhat more effort. 𝖠∧𝖡∧𝖢 ⌊⌊𝖣T 𝖠∧𝖡 𝖢||𝖣T 𝖡∧𝖢 𝖠||𝖣T 𝖢∧𝖠 𝖡||𝖣T ( ∧ ∧ ) = ( ∧ )( ) + ( ∧ )( ) + ( ∧ )( ) 𝖠∧ 𝖡|𝖣|𝖢 𝖢|𝖣|𝖡 𝖡∧ 𝖢|𝖣|𝖠 𝖠|𝖣|𝖢 − ∧( + ) − ∧( + ) 𝖢∧ 𝖠|𝖣|𝖡 𝖡|𝖣|𝖠 − ∧( + ) (2.84) 𝖠∧𝖡∧𝖢∧𝖣 ⌊⌊𝖤T 𝖠∧𝖡∧𝖢∧𝖣 𝖨(4)⌊⌊𝖤T ( ∧ ∧ ∧ ) = tr( ∧ ∧ ∧ )( ) 𝖠∧𝖡∧𝖢 𝖣||𝖤T 𝖠∧𝖡∧𝖣 𝖢||𝖤T 𝖠∧𝖢∧𝖣 𝖡||𝖤T = ( ∧ ∧ )( ) + ( ∧ ∧ )( ) + ( ∧ ∧ )( ) 𝖡∧𝖢∧𝖣 𝖠||𝖤T 𝖠∧𝖡 ∧ 𝖢|𝖤|𝖣 𝖣|𝖤|𝖢 + ( ∧ ∧ )( ) − ( ∧ )∧( + ) 𝖠∧𝖢 ∧ 𝖡|𝖤|𝖣 𝖣|𝖤|𝖡 𝖠∧𝖣 ∧ 𝖡|𝖤|𝖢 𝖢|𝖤|𝖡 − ( ∧ )∧( + ) − ( ∧ )∧( + ) 𝖡∧𝖢 ∧ 𝖠|𝖤|𝖣 𝖣|𝖤|𝖠 𝖡∧𝖣 ∧ 𝖠|𝖤|𝖢 𝖢|𝖤|𝖠 − ( ∧ )∧( + ) − ( ∧ )∧( + ) 𝖢∧𝖣 ∧ 𝖠|𝖤|𝖡 𝖡|𝖤|𝖠 − ( ∧ )∧( + ) (2.85)

2.3.2 Special Cases It is now easy to derive special cases of the above identities by replacing some of the dyadics by other dyadics. From (2.82) we obtain 𝖠(2)⌊⌊𝖡T = (𝖠||𝖡T )𝖠 − 𝖠|𝖡|𝖠, (2.86) 𝖨(2)⌊⌊𝖠T = (tr𝖠)𝖨 − 𝖠, (2.87) 𝖠(2)⌊⌊𝖨T = (tr𝖠)𝖠 − 𝖠2 = (𝖨(2)⌊⌊𝖠T )|𝖠, (2.88) Similarly, from (2.84) we obtain 𝖠(3)⌊⌊𝖡T 𝖠|𝖡 𝖠(2) 𝖠∧ 𝖠|𝖡|𝖠 = tr( ) − ∧( ), (2.89) 𝖨(3)⌊⌊𝖠T 𝖠 𝖨(2) 𝖨∧𝖠 = (tr ) − ∧ , (2.90) 𝖠(3)⌊⌊𝖨T 𝖠 𝖠(2) 𝖠∧𝖠2 𝖨(3)⌊⌊𝖠T |𝖠(2) = (tr ) − ∧ = ( ) , (2.91) 34 CHAPTER 2 Dyadics and, from (2.85), 𝖠(4)⌊⌊𝖡T 𝖠||𝖡T 𝖠(3) 𝖠(2)∧ 𝖠|𝖡|𝖠 = ( ) − ∧( ), (2.92) 𝖨(4)⌊⌊𝖠T 𝖠 𝖨(3) 𝖨(2)∧𝖠 = (tr ) − ∧ , (2.93) 𝖠(4)⌊⌊𝖨T 𝖠 𝖠(3) 𝖠(2)∧𝖠2 𝖨(4)⌊⌊𝖠T |𝖠(3). = (tr ) − ∧ = ( ) (2.94) The unit dyadic satisfies the relations 𝖨(4)⌊⌊𝖨(3)T = 𝖨, 𝖨(4)⌊⌊𝖨(2)T = 𝖨(2), 𝖨(4)⌊⌊𝖨 = 𝖨(3), (2.95) 𝖨(3)⌊⌊𝖨(2)T = 𝖨(2)⌊⌊𝖨T = 3𝖨. (2.96) Further identities are obtained by contracting previous identities by yet another dyadic and using other identities to expand the result. 𝖠∧𝖡 𝖨(2)|| 𝖠∧𝖡 T 𝖨(2)⌊⌊𝖠T ||𝖡T tr( ∧ ) = ( ∧ ) = ( ) , = (tr𝖠 𝖨 − 𝖠)||𝖡T = tr𝖠 tr𝖡 − tr(𝖠|𝖡), (2.97) 𝖨(3)⌊⌊ 𝖠∧𝖡 T 𝖨(3)⌊⌊𝖠T ⌊⌊𝖡T 𝖠 𝖨(2) 𝖨∧𝖠 ⌊⌊𝖡T ( ∧ ) = ( ) = (tr − ∧ ) , = (tr𝖠 tr𝖡 − tr(𝖠|𝖡))𝖨 − (tr𝖠)𝖡 − (tr𝖡)𝖠 + 𝖠|𝖡 + 𝖡|𝖠, 𝖠∧𝖡 𝖨 𝖠∧𝖡 ⌊⌊𝖨T . = tr( ∧ ) − ( ∧ ) (2.98) 𝖠(3)⌊⌊𝖡(2)T = tr(𝖠(2)|𝖡(2))𝖠 − tr(𝖠|𝖡)𝖠|𝖡|𝖠 + 𝖠|𝖡|𝖠|𝖡|𝖠 = tr(𝖠(2)|𝖡(2))𝖠 − 𝖠|(𝖡(2)⌊⌊𝖠T )|𝖠 (2.99) 𝖠(3)⌊⌊𝖨(2)T = tr(𝖠(2))𝖠 − (tr𝖠)𝖠2 + 𝖠3, (2.100) 𝖨(3)⌊⌊𝖠(2)T = (tr𝖠(2))𝖨 − (tr𝖠)𝖠 + 𝖠2, (2.101) 𝖨(4)⌊⌊ 𝖠∧𝖡 T 𝖠∧𝖡 𝖨(2) 𝖠∧𝖡 ⌊⌊𝖨T ∧𝖨 𝖠∧𝖡 ( ∧ ) = tr( ∧ ) − (( ∧ ) )∧ + ∧ , (2.102) 𝖨(4)⌊⌊ 𝖠∧𝖡∧𝖢 T 𝖠∧𝖡∧𝖢 𝖨 𝖠∧𝖡∧𝖢 ⌊⌊𝖨(2)T . ( ∧ ∧ ) = tr( ∧ ∧ ) − ( ∧ ∧ ) (2.103) These have the following simple interrelations 𝖠(4)⌊⌊𝖨(3)T = (𝖨(4)⌊⌊𝖠(3)T )|𝖠 = (tr𝖠(4))𝖨, (2.104) 𝖨(4)⌊⌊𝖠(3)T = (tr𝖠(3))𝖨 − 𝖠(3)⌊⌊𝖨(2)T , (2.105) 𝖠(3)⌊⌊𝖨(2)T = (𝖨(3)⌊⌊𝖠(2)T )|𝖠, (2.106) 𝖨(3)⌊⌊𝖠(2)T = (tr𝖠(2))𝖨 − 𝖠(2)⌊⌊𝖨T , (2.107) 𝖠(2)⌊⌊𝖨T = (𝖨(2)⌊⌊𝖠T )|𝖠, (2.108) 𝖨(2)⌊⌊𝖠T = (tr𝖠)𝖨 − 𝖠, (2.109) 𝖠(4)⌊⌊𝖨(2)T = (𝖨(4)⌊⌊𝖠(2)T )|𝖠(2). (2.110) 2.3 Dyadic Identities 35

Traces of different double-wedge powers of dyadics can be expanded as 1 tr𝖠(2) = ((tr𝖠)2 − tr𝖠2), (2.111) 2 1 tr𝖠(3) = ((tr𝖠)3 − 3tr𝖠tr𝖠2 + 2tr𝖠3), (2.112) 6 and 1 tr𝖠(4) = ((tr𝖠)4 − 6(tr𝖠)2tr𝖠2 + 8tr𝖠tr𝖠3 + 3(tr𝖠2)2 − 6tr𝖠4). 24 (2.113)

2.3.3 More General Rules E F The above identities are valid for dyadics in the space 1, 1.Itis possible to generalize some of them to involve bidyadics. For example, 𝖠∧𝖡 since the identity (2.102) is linear in the bidyadic ∧ we can replace it by an arbitrary bidyadic 𝖢 and arrive at the more general rule 𝖨(4)⌊⌊𝖢T 𝖢 𝖨(2) 𝖢⌊⌊𝖨T ∧𝖨 𝖢 𝖢 E F = tr − ( )∧ + , ∈ 2 2 (2.114) Similarly, from (2.98) we obtain 𝖨(3)⌊⌊𝖢T 𝖢 𝖨 𝖢⌊⌊𝖨T 𝖢 E F = tr − , ∈ 2 2 (2.115) and (2.99) has the more general form 𝖠(3)⌊⌊𝖢T 𝖠(2)|𝖢 𝖠 𝖠| 𝖢⌊⌊𝖠T |𝖠 𝖢 E F . = tr( ) − ( ) , ∈ 2 2 (2.116) This last result is based on the fact that an identity of the form f (𝖡(2)) = 0 𝖡 E F valid for any dyadic ∈ 1 1 where f () is a linear function of a bidyadic, 𝖢 𝖢 E F can be generalized to the form f ( ) = 0 for any bidyadic ∈ 2 2.The proof is left as an exercise. Finally, we can show that the identity (2.103) can be generalized to any tridyadic 𝖳 as 𝖨(4)⌊⌊𝖳T 𝖳 𝖨 𝖳⌊⌊𝖨(2)T 𝖳 E F . = tr − , ∈ 3 3 (2.117) 𝖠 E F However, since tridyadics can be reduced to dyadics ∈ 1 1 as 𝖳 = 𝖨(4)⌊⌊𝖠T , 𝖠 = 𝖨(4)⌊⌊𝖳T , (2.118) (2.117) follows from (2.87). 36 CHAPTER 2 Dyadics

2.3.4 Cayley–Hamilton Equation Eliminating the contraction terms stepwise from (2.104)–(2.109) we obtain an important identity: (tr𝖠(4))𝖨 = (𝖨(4)⌊⌊𝖠(3)T )𝖠 = (tr𝖠(3))𝖠 − (𝖠(3)⌊⌊𝖨(2)T )|𝖠 = (tr𝖠(3))𝖠 − (𝖨(3)⌊⌊𝖠(2)T )|𝖠2 = (tr𝖠(3))𝖠 − (tr𝖠(2))𝖠2 + (𝖠(2)⌊⌊𝖨T )|𝖠2 = (tr𝖠(3))𝖠 − (tr𝖠(2))𝖠2 + (𝖨(2)⌊⌊𝖠T )𝖠3 = (tr𝖠(3))𝖠 − (tr𝖠(2))𝖠2 + (tr𝖠)𝖠3 − 𝖠4, which can be written as 𝖠4 − (tr𝖠)𝖠3 + (tr𝖠(2))𝖠2 − (tr𝖠(3))𝖠 + (tr𝖠(4))𝖨 = 0. (2.119) This is called the Cayley–Hamilton equation and it is satisfied by any 𝖠 E F dyadic ∈ 1 1 operating in the 4D vector space. The Cayley–Hamilton equation was shown to be valid for 2 × 2 matrices by Cayley in 1858 in his “Memoir on the theory of matrices.” Healsomentionedhavingproveditfor3× 3 matrices and assumed that it would be valid for n × n matrices as well. Hamilton had given a version of the same equation for quaternions in 1853 in his “Lectures on quaternions.” The proof for a more general form in terms of n × n matrices was given by Frobenius in 1878.

2.3.5 Inverse Dyadics 𝖠 E F 𝖠(4) 𝖠 ≠ For dyadics ∈ 1 1 satisfying tr = det 0, the following com- 𝖠−1 E F pact analytic rule for the inverse ∈ 1 1 is obtained from the identity (2.104): 𝖨(4)⌊⌊𝖠(3)T 𝖠−1 = . (2.120) tr𝖠(4) F E 𝖨(4) Similar expression is valid for dyadics of the space 1 1 when is replaced by 𝖨(4)T . Multiplying (2.119) by 𝖠−1| we obtain an expansion for the inverse dyadic (2.120), −1 𝖠−1 = [𝖠3 − (tr𝖠)𝖠2 + (tr𝖠(2))𝖠 − (tr𝖠(3))𝖨]. (2.121) tr𝖠(4) 2.3 Dyadic Identities 37

Applying 𝖠(4) = (tr𝖠(4))𝖨(4), the identity (2.110) can be expressed as (tr𝖠(4))𝖨(2) = (𝖨(4)⌊⌊𝖠(2)T )|𝖠(2), (2.122) 𝖠 E F 𝖠(4) ≠ which is valid for any dyadic ∈ 1 1.Fortr 0, we obtain the second inverse formula 𝖨(4)⌊⌊𝖠(2)T 𝖠(−2) = (𝖠(2))−1 = (𝖠−1)(2) = , (2.123) tr𝖠(4) where we have applied the convention (2.57). This can be inverted as 𝖠(2) = 𝖨(4)⌊⌊(𝖨(4)T ⌊⌊𝖠(2)) = (tr𝖠(4))𝖨(4)⌊⌊𝖠(−2)T , (2.124) which also follows from (2.123) when replacing 𝖠 by 𝖠−1 and applying 1 tr𝖠(−4) = . (2.125) tr𝖠(4) Finally, (2.94) can be written as (tr𝖠(4))𝖨(3) = (𝖨(4)⌊⌊𝖠T )|𝖠(3), (2.126) which yields the third inverse formula 𝖨(4)⌊⌊𝖠T 𝖠(−3) = . (2.127) tr𝖠(4) This can also be obtained from (2.120) when replacing 𝖠 by 𝖠−1. The previous inverse rules can be given an easily memorizable form in terms of 1 𝖠(−4) = tr𝖠(−4) 𝖨(4) = 𝖨(4), (2.128) tr𝖠(4) as (4−p)T 𝖠(−p) = 𝖠(−4)⌊⌊𝖠 , (2.129) valid for 0 < p < 4. Similarly, we can write

(p−4)T 𝖠(p) = 𝖠(4)⌊⌊𝖠 . (2.130) Since a tridyadic 𝖳 can be expressed in terms of a dyadic 𝖠 as in (2.118), one can show that the inverses satisfy the relation 𝖳−1 = 𝖨(4)⌊⌊𝖠−1T , (2.131) 38 CHAPTER 2 Dyadics so that the inverse of the general tridyadic can be expressed in a compact analytic form as

𝖨(4)⌊⌊𝖠(3)T (𝖨(4)⌊⌊𝖳T )(3) 𝖳−1 = 𝖨(4)⌊⌊ = . (2.132) tr𝖠(4) tr(𝖨(4)⌊⌊𝖳T )(4) There does not seem to exist a simple analytic inverse rule for the general bidyadic. Special cases will be discussed in Chapter 3. 𝖠 E F Since any dyadic ∈ 1 1 can be expanded in terms of spatial 𝖠 𝜶 quantities s, as, s as 𝖠 𝖠 𝜺 𝜶 𝜺 = s + as 4 + e4 s + ae4 4, (2.133) the following expression for the inverse can be easily derived by solving a set of spatial linear equations,

𝖠−1| 𝜶 |𝖠−1 𝜺 ( as − e4)( s − 4) 𝖠−1 = 𝖠−1 + s s . (2.134) s 𝜶 |𝖠−1| a − s s as 𝖠 This expansion requires that the spatial inverse of s exists and that 𝜶 |𝖠−1| 𝖠−1 a − s s as does not vanish. The spatial inverse s is defined as a dyadic satisfying 𝖠 |𝖠−1 𝖠−1|𝖠 𝖨 𝖨 𝜺 . s s = s s = s = − e4 4 (2.135) 𝖠(3) One can show that the spatial inverse exists whenever tr s does not vanish and it obeys the rule 1 𝖠−1 = 𝖨(3)⌊⌊𝖠(2)T = 𝖠(−3)⌊⌊𝖠(2)T . (2.136) s 𝖠(3) s s s s tr s 𝖢 E F Similarly, the inverse of a spatial bidyadic s ∈ 2 2 can be expressed in the form 1 𝖢−1 = (𝖨(3)⌊⌊𝖢T )(2), (2.137) s Δ s s Cs with Δ = tr(𝖨(3)⌊⌊𝖢T )(3). (2.138) Cs s s Derivation of (2.136) and (2.137) are left as exercises. 2.4 Rank of Dyadics 39

2.4 RANK OF DYADICS

𝖠 E F Any dyadic ∈ 1 1 operating in the 4D vector space can be represented in the form 𝖠 𝜶 𝜶 𝜶 𝜶 = a1 1 + a2 2 + a3 3 + a4 4, (2.139) 𝜶 in terms of some vectors ai and one-forms i. Choosing either one of 𝜶 the sets {ai}or{ i} as a basis, the other set defines the dyadic uniquely. Dyadics can be classified in terms of their rank which equals the dimension of the space of the vectors b obtained through the mapping b = 𝖠|a, (2.140) when a goes through all possible vectors.

Dyadics of Rank 4 A dyadic 𝖠 is of full rank 4 when it cannot be represented with less than four terms as (2.139). This requires that the four vectors and one-forms 𝜶 {ai}and{ i} in (2.139) each form a basis, or, what is equivalent, 𝖠(4) 𝜶 𝜶 𝜶 𝜶 ≠ . = (a1 ∧ a2 ∧ a3 ∧ a4)( 1 ∧ 2 ∧ 3 ∧ 4) 0 (2.141) A dyadic of rank 4 has an inverse as expressed by (2.120). From 𝖠(4) 𝜺 | | 𝜶 𝜶 𝜶 𝜶 tr = N (a1 ∧ a2 ∧ a3 ∧ a4)eN ( 1 ∧ 2 ∧ 3 ∧ 4) (2.142) we see that the scalar condition tr𝖠(4) ≠ 0 is sufficient for the dyadic 𝖠 being of rank 4. The equations 𝖠|a = 0and𝜶|𝖠 = 0 only have the solutions a = 0, 𝜶 = 0.

Dyadics of Rank 3 The rank of a dyadic 𝖠 is 3 when it can be represented in the form 𝖠 𝜶 𝜶 𝜶 = a1 1 + a2 2 + a3 3, (2.143) but not in less than three terms. Thus, the dyadic must satisfy 𝖠(4) = 0, 𝖠(3) ≠ 0, (2.144) 𝜶 since the latter requires that the vectors {ai} and one-forms, { j}betwo linearly independent triplets. For a dyadic of rank 3 there exists a vector a ≠ 0 satisfying 𝖠|a = 0. (2.145) 40 CHAPTER 2 Dyadics

In fact, in terms of a quadrivector eN the vector a can be expressed as ⌊ 𝜶 𝜶 𝜶 ≠ a = eN ( 1 ∧ 2 ∧ 3) 0, (2.146) because it satisfies ∑3 𝖠| 𝜶 | ⌊ 𝜶 𝜶 𝜶 a = ai i (eN ( 1 ∧ 2 ∧ 3)) i=1 ∑3 𝜺 | 𝜶 𝜶 𝜶 𝜶 . = ai( N ( 1 ∧ 2 ∧ 3 ∧ i)) = 0 (2.147) i=1 ≠ Alternatively, choosing a vector b satisfying a1 ∧ a2 ∧ a3 ∧ b 0, we can write 𝖨(4)⌊⌊𝖠(3)T | ⌊ 𝜶 𝜶 𝜶 𝜺 ⌊ | a = ( ) b = eN ( 1 ∧ 2 ∧ 3)( N (a1 ∧ a2 ∧ a3)) b ⌊ 𝜶 𝜶 𝜶 𝜺 | ≠ . = eN ( 1 ∧ 2 ∧ 3) N (a1 ∧ a2 ∧ a3 ∧ b) 0 (2.148)

Dyadics of Rank 2 The rank of a dyadic 𝖠 is 2 when it can be expressed in the form 𝖠 𝜶 𝜶 = a1 1 + a2 2, (2.149) but not as a single dyadic term. Thus, the dyadic must satisfy 𝖠(3) = 0, 𝖠(2) ≠ 0. (2.150) 𝜶 𝜶 Because in this case the vectors a1, a2 and the one-forms 1, 2 must 𝜶 𝜶 ≠ be linearly independent, (a1 ∧ a2)( 1 ∧ 2) 0, there must exist two linearly independent vectors b1, b2 so that 𝖠| 𝖠| b1 = 0, b2 = 0 (2.151) are satisfied. Another way to express the same is to require that 𝖠⌋B = 0 (2.152) be valid for some simple bivector B. In fact, writing B = b1 ∧ b2 we can expand 𝖠⌋ 𝖠| 𝖠| . (b1 ∧ b2) = ( b2)b1 − ( b1)b2 = 0 (2.153) The two vectors can be defined as ⌊ 𝜶 𝜶 𝜶 ⌊ 𝜶 𝜶 𝜶 b1 = eN ( 1 ∧ 2 ∧ 3), b2 = eN ( 1 ∧ 2 ∧ 4), (2.154) 𝜶 𝜶 𝜶 𝜶 in terms of two one-forms 3, 4 which added to 1, 2 complete the one-form basis. 2.5 Eigenproblems 41

Dyadics of Rank 1 The rank of 𝖠 is 1 when it satisfies

𝖠(2) = 0, 𝖠 ≠ 0. (2.155)

Such a dyadic can be presented as a single dyad, 𝖠 𝜶 . = a1 1 (2.156)

2.5 EIGENPROBLEMS

2.5.1 Eigenvectors and Eigen One-Forms 𝖠 E F Any dyadic ∈ 1 1 defines an eigenvalue problem as 𝖠|x = 휆x, (2.157) where 휆 is an eigenvalue and x is an eigenvector. Because the dyadic 𝖠 − 휆𝖨 maps the corresponding eigenvector x to zero, its rank is less than 4 whence it satisfies

tr(𝖠 − 휆𝖨)(4) = 0, (2.158) or

휆4 − 휆3tr𝖠 + 휆2tr𝖠(2) − 휆tr𝖠(3) + tr𝖠(4) = 0. (2.159) 휆 In the general case there are four eigenvalues i satisfying 휆 휆 휆 휆 휆 휆 휆 휆 . ( − 1)( − 2)( − 3)( − 4) = 0 (2.160) Comparing this and (2.159) we obtain the relations 𝖠 휆 휆 휆 휆 tr = 1 + 2 + 3 + 4, (2.161) 𝖠(2) 휆 휆 휆 휆 휆 휆 휆 휆 휆 휆 휆 휆 tr = 1 2 + 2 3 + 3 1 + 1 4 + 2 4 + 3 4, (2.162) 𝖠(3) 휆 휆 휆 휆 휆 휆 휆 휆 휆 휆 휆 휆 tr = 1 2 3 + 2 3 4 + 3 1 4 + 1 2 4, (2.163) 𝖠(4) 휆 휆 휆 휆 𝖠. tr = 1 2 3 4 = det (2.164) The corresponding eigenproblem for the transposed dyadic is 𝖠T |𝝃 𝝃 |𝖠 휇𝝃 i = i = i, (2.165) 42 CHAPTER 2 Dyadics where 휇 is the (left) eigenvalue and 𝝃 the eigen one-form of the dyadic 𝖠. Since the equation for the eigenvalue 휇 equals (2.159), the set of 휇 휆 solutions is the same, i = i.From 𝝃 |𝖠| 휆 𝝃 | 휆 𝝃 | i xj = i i xj = j i xj (2.166) we obtain the orthogonality rule 휆 ≠ 휆 ⇒ 𝝃 | . i j i xj = 0 (2.167) Solutions for the eigenproblem of the dyadic 𝖠(2) can be found in terms of the eigensolutions of the dyadic 𝖠 as 𝖠(2)| 𝖠| 𝖠| 휆 휆 (xi ∧ xj) = ( xi) ∧ ( xj) = i jxi ∧ xj, (2.168) 𝝃 𝝃 |𝖠(2) 휆 휆 𝝃 𝝃 . ( i ∧ j) = i j i ∧ j (2.169)

2.5.2 Reduced Cayley–Hamilton Equations For certain dyadics the Cayley–Hamilton equation (2.119) can be reduced to dyadic equations of lower order than four. Actually, there 휆 are 11 basic forms each defining a class of dyadics. Assuming that i’s are not equal, the different cases can be represented as

𝖠 휆 𝖨 | 𝖠 휆 𝖨 | 𝖠 휆 𝖨 | 𝖠 휆 𝖨 1. ( − 1 ) ( − 2 ) ( − 3 ) ( − 4 ) = 0 𝖠 휆 𝖨 | 𝖠 휆 𝖨 | 𝖠 휆 𝖨 2 2. ( − 1 ) ( − 2 ) ( − 3 ) = 0 𝖠 휆 𝖨 | 𝖠 휆 𝖨 | 𝖠 휆 𝖨 3. ( − 1 ) ( − 2 ) ( − 3 ) = 0 𝖠 휆 𝖨 | 𝖠 휆 𝖨 3 4. ( − 1 ) ( − 2 ) = 0 𝖠 휆 𝖨 2| 𝖠 휆 𝖨 2 5. ( − 1 ) ( − 2 ) = 0 𝖠 휆 𝖨 | 𝖠 휆 𝖨 2 6. ( − 1 ) ( − 2 ) = 0 𝖠 휆 𝖨 | 𝖠 휆 𝖨 7. ( − 1 ) ( − 2 ) = 0 𝖠 휆 𝖨 4 8. ( − 1 ) = 0 𝖠 휆 𝖨 3 9. ( − 1 ) = 0 𝖠 휆 𝖨 2 10. ( − 1 ) = 0 𝖠 휆 𝖨 11. − 1 = 0

It is assumed that the equations for each case cannot be reduced to simpler equations on the lines below. 2.5 Eigenproblems 43

Let us just briefly consider some cases. The last class 11 defines dyadics of the form 𝖠 = A𝖨. Also, the classes 8–10 require that 𝖠 = A𝖨 + 𝖭,where𝖭 is any nilpotent dyadic of respective order p = 4 ⋯ 1, satisfying 𝖭p = 0. For example, case 10 can be shown to correspond to dyadics of the general form 𝖠 𝖨 𝜶 𝜶 |𝜶 . = A + a1 1 + a2 2, ai j = 0, i, j = 1, 2 (2.170) The class 7 condition can be rewritten as ( ) 휆 + 휆 2 (휆 − 휆 )2 𝖠 − 1 2 𝖨 = 1 2 𝖨, (2.171) 2 4 whence the dyadic in brackets must be a multiple of a unipotent dyadic. Such dyadics will be considered in Chapter 4. Any class 1 dyadic with four distinct eigenvalues can be expanded in terms of its eigenvectors and eigen one-forms as ∑ 𝖠 휆 𝝃 |𝝃 훿 . = ixi i, xi j = i,j (2.172) i

2.5.3 Construction of Eigenvectors 𝝃 The eigenvector xi and the eigen one-form i corresponding to a given 휆 eigenvalue i can be constructed in a straightforward manner as follows. Applying the inverse formula (2.120) in the form 𝖠|(𝖨(4)⌊⌊𝖠(3)T ) = (𝖨(4)⌊⌊𝖠(3)T )|𝖠 = tr𝖠(4) 𝖨, (2.173) 𝖠 𝖠 휆 𝖨 and replacing by the dyadic − i we have 𝖠 휆 𝖨 | 𝖨(4)⌊⌊ 𝖠 휆 𝖨 (3)T 𝖨(4)⌊⌊ 𝖠 휆 𝖨 (3)T | 𝖠 휆 𝖨 ( − i ) ( ( − i ) ) = ( ( − i ) ) ( − i ) 𝖠 휆 𝖨 (4)𝖨. = tr( − i ) (2.174) Because of (2.158) the last expression vanishes. Assuming 𝖠 휆 𝖨 (3) ≠ ( − i ) 0, (2.175) the first expression yields 𝖠| 𝖨(4)⌊⌊ 𝖠 휆 𝖨 (3)T | 휆 𝖨(4)⌊⌊ 𝖠 휆 𝖨 (3)T | ( ( − i ) ) a = i( ( − i ) ) a, (2.176) which must be valid for any vector a. Thus, the eigenvector correspond- 휆 ing to the eigenvalue i can be represented as 𝖨(4)⌊⌊ 𝖠 휆 𝖨 (3)T | xi = ( ( − i ) ) a, (2.177) 44 CHAPTER 2 Dyadics in terms of any vector a which yields a nonzero result. The left eigen one-form is obtained similarly as 𝝃 𝜶| 𝖨(4)⌊⌊ 𝖠 휆 𝖨 (3)T i = ( ( − i ) ) (2.178) in terms of any one-form 𝜶 yielding a nonzero result. Actually, from 𝖠 휆 𝖨 (2.175) and (2.158) we conclude that the dyadic − i must be of rank 3, whence we can write 𝖨(4)⌊⌊ 𝖠 휆 𝖨 (3)T 𝝃 ( − i ) = xi i (2.179) not caring about magnitudes of the eigenvectors and eigen one-forms. 휆 Obviously, this procedure is only valid when the eigenvalue i does not correspond to a multiple root of (2.159), because in the converse case 𝖠 휆 𝖨 (3) we would have ( − i ) = 0. 휆 휆 If 1 = 2 is a double but not a triple eigenvalue, we have two linearly independent eigenvectors x1, x2 corresponding to the same eigenvalue and satisfying 𝖠 휆 𝖨 ⌋ 𝖠 휆 𝖨 | . ( − 1 ) (x1 ∧ x2) = ( − 1 ) (x2x1 − x1x2) = 0 (2.180) 𝖠 휆 𝖨 (3) 𝖠 휆 𝖨 (2) ≠ 𝖠 In this case, we have ( − 1 ) = 0and( − 1 ) 0 whence − 휆 𝖨 1 must be of rank 2. It then follows that there must exist vectors a, b and one-forms 𝜶, 𝜷 such that we can write 𝖠 휆 𝖨 𝜶 𝜷 − 1 = a + b , (2.181) and 𝖠 휆 𝖨 (2) 𝜶 𝜷 ≠ . ( − 1 ) = (a ∧ b)( ∧ ) 0 (2.182) Comparing (2.180) to 𝖠 휆 𝖨 ⌋ 𝖨(4)⌊⌊ 𝖠 휆 𝖨 (2)T 𝜶 𝜷 ⌋ ⌊ 𝜶 𝜷 𝜺 ⌊ ( − 1 ) ( ( − 1 ) ) = (a + b ) (eN ( ∧ )) ( N (a ∧ b)) ⌊ 𝜶 𝜷 𝜶 ⌊ 𝜶 𝜷 𝜷 𝜺 ⌊ =−(aeN ( ∧ ∧ ) + beN ( ∧ ∧ )) ( N (a ∧ b)) = 0, (2.183) ⌊ 𝜶 𝜷 showsusthateN ( ∧ ) must be a multiple of x1 ∧ x2. Similarly, 𝜺 ⌊ 𝝃 𝝃 N (a ∧ b) must be a multiple of 1 ∧ 2, whence ignoring normalization we may write 𝖨(4)⌊⌊ 𝖠 휆 𝖨 (2)T 𝝃 𝝃 . ( − 1 ) = (x1 ∧ x2)( 1 ∧ 2) (2.184) This defines the two-dimensional (2D) subspaces of eigenvectors and 휆 eigen one-forms corresponding to the double eigenvalue 1. Another 2.6 Metric Dyadics 45 way to express the relation is 𝖠 휆 𝖨 (2) 𝖨(4)⌊⌊ 𝝃 𝝃 ⌊ 𝝃 𝝃 𝜺 ⌊ . ( − 1 ) = ( 1 ∧ 2)(x1 ∧ x2) = eN ( 1 ∧ 2) N (x1 ∧ x2) (2.185) 휆 휆 휆 ≠ 휆 This process can be continued so that, for 1 = 2 = 3 4 the three linearly independent eigenvectors x1, x2, x3 and eigen one-forms 𝝃 𝝃 𝝃 1, 2, 3 can be obtained as 𝖨(4)⌊⌊ 𝖠 휆 𝖨 T 𝝃 𝝃 𝝃 ( − 1 ) = (x1 ∧ x2 ∧ x3)( 1 ∧ 2 ∧ 3), (2.186) or 𝖠 휆 𝖨 𝖨(4)⌊⌊ 𝝃 𝝃 𝝃 − 1 = ( 1 ∧ 2 ∧ 3)(x1 ∧ x2 ∧ x3) ⌊ 𝝃 𝝃 𝝃 𝜺 ⌊ . = eN ( 1 ∧ 2 ∧ 3) N (x1 ∧ x2 ∧ x3) (2.187) 𝝃 The last dyadic term is actually a multiple of x4 4, the product of the 휆 eigenvector and eigen one-form corresponding to the eigenvalue 4.

2.6 METRIC DYADICS

Metric dyadics are dyadics mapping vectors to one-forms or one-forms F F E E to vectors, that is, members of the spaces 1 1 or 1 1. Because the transpose operation maps dyadics to the same space, symmetric and antisymmetric dyadics form subspaces in the general space of metric dyadics. 𝖡 E E Any metric dyadic ∈ 1 1 has a bivector invariant which can be defined in terms of its vector expansion as ∑ ∑ 𝖡∧ ∧ E = (aibi) = ai ∧ bi ∈ 2 i i . = a1 ∧ b1 + a2 ∧ b2 + a3 ∧ b3 + a4 ∧ b4 (2.188)  F F Similarly, any metric dyadic ∈ 1 1 has the two-form invariant ∑ ∑ ∧ 𝜶 𝜷 ∧ 𝜶 𝜷 F . = ( i i) = i ∧ i ∈ 2 (2.189) i i A metric dyadic 𝖡 is of full rank 4 if it satisfies 𝖡(4) ≠ 0. In the converse case, the rank is 3 for 𝖡(3) ≠ 0, 2 for 𝖡(3) = 0, 𝖡(2) ≠ 0and1 for 𝖡(2) = 0, 𝖡 ≠ 0. 46 CHAPTER 2 Dyadics

𝖡 E E 𝖡−1 The metric dyadic ∈ 1 1 of full rank has an inverse which F F is in the space 1 1. Transposing the equation 𝖡|𝖡−1 = 𝖨⇒𝖡−1T |𝖡T = 𝖨T , (2.190) and assuming symmetric or antisymmetric dyadic, 𝖡T =±𝖡, we obtain ± 𝖡−1T |𝖡 = 𝖨T ⇒ 𝖡−1T =±𝖡−1, (2.191) whence also the respective inverse is symmetric or antisymmetric. E E F F The same is also valid for metric dyadics of the spaces p p, p p for p = 2, 3, 4.

2.6.1 Symmetric Dyadics 𝖲 E E Symmetric dyadics ∈ 1 1 can be expanded as ∑ ∑ ∑ 𝖲 𝖲T 1 = aibi = = biai = (aibi + biai), (2.192) i i 2 i whence the bivector invariant of any symmetric dyadic vanishes, ∑ 𝖲∧ 1 . = (ai ∧ bi + bi ∧ ai) = 0 (2.193) 2 i Conversely, if the bivector invariant of a the metric dyadic 𝖡 vanishes, 𝖡∧ = 0, the dyadic is symmetric: ∑ ∑ ∑ 𝖡 1 1 = (aibi) = (aibi + biai) + (aibi − biai) i 2 2 ∑ ∑ ∑ 1 1 1 = (a b + b a ) − (a ∧ b )⌊𝖨T = (a b + b a ). 2 i i i i 2 i i 2 i i i i (2.194) Symmetric dyadics satisfying 𝖲(4) ≠ 0 can be used to define the dot product for one-forms as 𝜶 ⋅ 𝜷 = 𝜶|𝖲|𝜷 = 𝜷 ⋅ 𝜶. (2.195) Similarly, a dot product for vectors can be obtained in terms of a sym-  F F (4) ≠ metric dyadic ∈ 1 1 satisfying 0as a ⋅ b = a||b = b ⋅ a. (2.196) 2.6 Metric Dyadics 47

In Minkowskian metric, two complementary symmetric dyadics can be expressed in terms of reciprocal basis expansions as  𝜺 𝜺 𝜺 𝜺 𝜺 𝜺 𝜺 𝜺 (4) 𝜺 𝜺 = 1 1 + 2 2 + 3 3 − 4 4, =− N N, (2.197) 𝖦 𝖦(4) . = e1e1 + e2e2 + e3e3 − e4e4, =−eNeN (2.198) They satisfy the relations 𝖦| = 𝖨, 𝖦 = −1, (2.199)  𝜺 𝜺 ⌊⌊𝖦(3) 𝖦 ⌊⌊(3). =− N N , =−eNeN (2.200)

2.6.2 Antisymmetric Dyadics 𝖠 E E Expanding the general antisymmetric dyadic ∈ 1 1 as ∑ ∑ 𝖠 1 = aibi = (aibi − biai) i 2 i ∑ 1 ⌊𝖨T 1 𝖠∧ ⌊𝖨T =− (ai ∧ bi) =− ( ) , (2.201) 2 i 2 it is seen that an antisymmetric dyadic is determined by its bivector invariant (2.188). Thus, any dyadic of the form 𝖠 = A⌊𝖨T , (2.202) 𝖠 E E is antisymmetric and, conversely, any antisymmetric dyadic ∈ 1 1 can be expressed in terms of a bivector, 1 A =− 𝖠∧. (2.203) 2 Considering 3D spatial antisymmetric dyadics the bivector A is spatial and it can be expressed in terms of a spatial one-form 𝜶 as can be seen from the expansion 𝖠 = A12(e1e2 − e2e1) + A23(e2e3 − e3e2) + A31(e3e1 − e1e3) ⌊𝖨T =−(A12e1 ∧ e2 + A23e2 ∧ e3 + A31e3 ∧ e1) s ⌊ 𝜺 𝜺 𝜺 ⌊𝖨T ⌊𝜶 ⌊𝖨T . =−(e123 (A12 3 + A23 1 + A31 2)) = (e123 ) s (2.204) 𝖠 E E The most general 3D spatial antisymmetric bidyadic ∈ 2 2 can be expanded in terms of a spatial vector a, as can be seen from 48 CHAPTER 2 Dyadics the expansion 𝖠 = A1(e31e12 − e12e31) + A2(e12e23 − e23e12) + A3(e23e31 − e31e23) ⌊𝖨T ⌊𝖨T . =−e123 s ∧ (A1e1 + A2e2 + A3e3) = e123 ∧ a (2.205)

2.6.3 Inverse Rules for Metric Dyadics 𝖠 E F The expression (2.120) for the inverse of a dyadic ∈ 1 1 can be 𝖡 E E rewritten for the metric dyadic ∈ 1 1 in the form 𝜺 𝜺 ⌊⌊𝖡(3)T 𝖡−1 N N 𝜺 𝜺 ||𝖡(4) ≠ . = , N N 0 (2.206) 𝜺 𝜺 ||𝖡(4) N N Other inverse rules based on the same metric dyadic are

𝜺 𝜺 ⌊⌊𝖡(2)T 𝖡(−2) = N N , (2.207) 𝜺 𝜺 ||𝖡(4) N N 𝜺 𝜺 ⌊⌊𝖡T 𝖡(−3) = N N . (2.208) 𝜺 𝜺 ||𝖡(4) N N 𝖡 E E F F Since the inverse of a metric dyadic ∈ 1 1 is in the space 1 1 and since the inverse of an antisymmetric dyadic (2.202) is antisymmetric, there must exist a rule of the form

(A⌊𝖨T )−1 = 𝚽⌊𝖨, (2.209) where 𝚽 is a two-form depending on the bivector A. Leaving the details as an exercise, the result can be expressed in the form 𝜺 ⌊A 𝚽 =− N , (2.210) 𝜺 | ∧ N A which is valid if the bivector invariant A =−𝖠∕2 of the antisymmetric dyadic is not simple, A∧ = 1 A ∧ A ≠ 0. Because every spatial bivector 2 is simple, spatial antisymmetric dyadics have no inverse. 𝖡 E E Spatial inverse for a spatial metric dyadic s ∈ 1 1 can be expressed following the pattern of (2.136) as 1 𝖡−1 = 𝜺 𝜺 ⌊⌊𝖡(2)T , (2.211) s 𝜺 𝜺 ||𝖡(3) 123 123 s 123 123 s Problems 49

𝖡 𝖡(3) ≠ and it is valid for dyadics s satisfying s 0. Because antisymmetric 𝖡(3) spatial dyadics satisfy s = 0, they do not have a spatial inverse.

PROBLEMS

𝖨∧𝖠 𝖠 𝖠 E F 2.1 Provetheruletr(∑∧ ) = 3tr , ∑∈ 1 1 (1) by starting from the 𝖨 𝜺 𝖠 𝜶 expansions = ei i and = ej j and (2) by applying the rule 𝖡(2)⌊⌊𝖠T = (𝖡||𝖠T )𝖡 − 𝖡|𝖠|𝖡. 𝖠 E F 2.2 Assuming a given dyadic ∈ 1 1, solve the equation 𝖨(2)⌊⌊𝖡T = 𝖠. 𝖡 E F for the dyadic ∈ 1 1 by applying the rule of the previous problem. 2.3 Applying the multivector identity ⌊𝜹 |𝜹 (a1 ∧ a2 ∧ a3) = ((a1 ∧ a2)a3 + (a2 ∧ a3)a1 + (a3 ∧ a1)a2) 𝖠 ∧𝖠 ∧𝖠 ⌊⌊𝖡T develop an expansion rule for ( 1∧ 2∧ 3) involving four 𝖠 ⋯ 𝖡 E F 𝖠 𝖠 dyadics 1 ∈ 1 1. Form also the special cases i = and 𝖡 = 𝖠−1. Check the last case by setting 𝖠 = 𝖨 and taking the trace of both sides. 2.4 Applying the dyadic rule 𝖠∧𝖡 ⌊⌊𝖢T 𝖠||𝖢T 𝖡 𝖡||𝖢T 𝖠 𝖠|𝖢|𝖡 𝖡|𝖢|𝖠 ( ∧ ) = ( ) + ( ) − − 𝖠 𝖡 𝖢 E F 𝖡 valid for , , ∈ 1 1, find solution for in the dyadic equation 𝖠∧𝖡 𝖣 E F ∧ = ∈ 2 2, assuming that 𝖠−1 exists. Find a condition between 𝖠 and 𝖣 for the solution 𝖡 to exist. Hint: try 𝖢 = 𝖠−1. Check the result by setting 𝖣 = 𝖨(2) and 𝖠 = 𝖨. 2.5 Prove the identity 𝖠T |(𝚽⌊𝖠) = (𝚽|𝖠(2))⌊𝖨T , 𝖠 E E 𝚽 F where ∈ 1 1 is a metric dyadic∑ and ∈ 2 is a two-form, in 𝖠 𝚽 two ways: (1) expanding = aibi and (2) assuming that is a simple two-form. 50 CHAPTER 2 Dyadics

2.6 Solve the eigendyadic problem 𝖨(2)⌊⌊𝖷T = 휆𝖷. 휆 𝖷 E F for and ∈ 1 1. 𝖠 E E 2.7 Find the inverse of the general antisymmetric dyadic ∈ 1 1 which can be expressed in the form 𝖠 = A⌊𝖨T = 𝖨⌋A, E where A ∈ 2 is a bivector. What is the condition for the existence of the inverse dyadic? Hint: apply the bac-cab rule 𝜶⌋(B ∧ C) = B ∧ (𝜶⌋C) + C ∧ (𝜶⌋B). 2.8 Prove the identity satisfied by any bivector A, 1 (𝖨⌋A)(2) = AA − (A ∧ A)⌊𝖨(2)T . (2.212) 2 Hint: apply the expansion A = a ∧ b + c ∧ d. 𝖠 E F 2.9 Expanding two dyadics 1,2 ∈ 1 1 in 3D components as 𝖠 𝖡 𝜺 𝜷 𝜺 . i = i + bi 4 + e4 i + Biei i, i = 1, 2 ≠ 𝖡−1 assuming B2 0 and that the 3D inverse 1 exists, show that 𝖠 |𝖠 E 𝜸 F 1 2 = 0 implies the existence of c ∈ 1 and ∈ 1 such that 𝖠 𝜸 2 = c . 𝖣⌊⌊𝖨T 𝖣 𝖣 E F 2.10 Prove that = 0 implies = 0 for a dyadic ∈ 3 3. Hint: 𝖣 E F 𝖣 𝖨(4)⌊⌊𝖠T any dyadic ∈ 3 3 can be expressed as = in terms 𝖠 E F of some dyadic ∈ 1 1. 2.11 Derive the identities 𝖨(3)⌊⌊𝖠T 𝖠 𝖨(2) 𝖨∧𝖠 = (tr ) − ∧ , 𝖨(4)⌊⌊𝖠T 𝖠 𝖨(3) 𝖨(2)∧𝖠 = (tr ) − ∧ , 𝖠 E F valid for any dyadic ∈ 1 1, by applying the identity 𝖠(p)⌊a = (𝖠|a) ∧ 𝖠(p−1), where a is a vector and p is 2, 3 or 4. Hint: start by setting 𝖠 = a𝜶. Problems 51

2.12 Derive the identity 𝖨(2)∧𝖠 ⌊⌊𝖡T 𝖠|𝖡 𝖨(2) 𝖨∧𝖠 𝖡 𝖨∧ 𝖠|𝖡 𝖡|𝖠 𝖠∧𝖡 ( ∧ ) = (tr( )) + ∧ tr − ∧( + ) − ∧ , 𝖠 𝖡 E F . valid for any dyadics , ∈ 1 1 2.13 Derive the identity 𝖨(4)⌊⌊𝖢T 𝖢 𝖨(2) 𝖢⌊⌊𝖨T ∧𝖨 𝖢 = (tr ) − ( )∧ + , 𝖢 E F 𝖢 𝖠∧𝖡 valid for any bidyadic ∈ 2 2. Hint: start by setting = ∧ and use the identities of the two previous problems. 2.14 Show that the identity of the previous problem remains unchanged when both sides of the transposed identity are operated by 𝖨(4)⌊⌊ or when the bidyadic 𝖢 in the original is replaced by 𝖨(4)⌊⌊𝖢T . 𝖠(2) 𝖠 E F 2.15 Show that if an identity is of the form f ( ) = 0, ∈ 1 1 𝖣 E F where f is a linear function, it is valid for any bidyadic ∈ 2 2 as f (𝖣) = 0. 𝖲 E E 𝖠 2.16 A symmetric dyadic ∈ 1 1 and an antisymmetric dyadic = ⌊𝖨T E E 𝖲∧𝖠 A ∈ 1 1 defined by a bivector A form a bidyadic ∧ ∈ E E 2 2 which is obviously antisymmetric. Find a trace-free dyadic 𝖡 E F o ∈ 1 1 in terms of which we can express 𝖲∧ ⌊𝖨T 𝖡 ∧𝖨 ⌋ . ∧(A ) = ( o∧ ) eN Hint: because the expression is linear in both 𝖲 and A,sois 𝖡 𝖲→ → 𝖡 o. Replace ss and A e12,find o(ss, e12) and express 𝖡 𝖲 o( , A). Hint: the multivector rule a⌋(𝜶 ∧ 𝜷 ∧ 𝜸) = (a|𝜶)(𝜷 ∧ 𝜸) + 𝜶 ∧ (a⌋(𝜷 ∧ 𝜸)) may be useful in the derivation. 2.17 Derive the expression (2.134) for the dyadic inverse. 2.18 Find an expression for the inverse of a metric dyadic of the form 𝖠 + ab + ba where 𝖠 is an antisymmetric dyadic possessing an inverse. Check the special case b = a∕2. 2.19 Derive the spatial inverse rule (2.136). Hint: start from the Cayley–Hamilton equation for the spatial dyadic, 𝖠3 𝖠 𝖠2 𝖠(2) 𝖠 𝖠(3) 𝖨 . s − (tr s) s + (tr s ) s − (tr s ) s = 0 52 CHAPTER 2 Dyadics

𝖠 E F 𝖠| 2.20 Show that if a dyadic ∈ 1 1 satisfies (1) a = 0 for a vector a ≠ 0, it must satisfy 𝖠(4) = 0 and (2) if it satisfies 𝖠|a = 0and 𝖠|b = 0 for two vectors with a ∧ b ≠ 0, it must satisfy 𝖠(3) = 0. 𝖢 E F 2.21 Expressing the spatial bidyadic s ∈ 2 2 in terms of a spa- 𝖠 E F 𝖢 𝖨(3)⌊⌊𝖠T tial dyadic s ∈ 1 1 as s = s s ,derivetheinverserule 𝖢 𝖠 (2.137) for s starting from that of s (2.136). 2.22 Prove the identity 𝖠∧𝖲(2) ⌊⌊ 𝚽⌊𝖨 ∧ = eNeN ( ), 𝖠 ⌊𝖨T E E 𝖲 where = A ∈ 1 1 is an antisymmetric dyadic and ∈ E E 𝚽 1 1 is a symmetric dyadic. Show that the two-form depends on the bivector A as 𝚽 | 𝜺 𝜺 ⌊⌊𝖲(2) . =−A ( N N ) Hint: apply the result of Problem 2.5. CHAPTER 3

Bidyadics

E F F E E E F F Bidyadics, elements of the spaces 2 2, 2 2, 2 2 or 2 2, perform mappings between bivectors and/or two-forms. Elements of the latter two spaces may be called metric bidyadics because they define scalar 𝚽|𝖢 |𝚿 products between two bivectors or two two-forms like m for 𝖢 E E m ∈ 2 2. There are simple transformations between bidyadics and 𝖢 E F metric bidyadics. In fact, for every bidyadic ∈ 1 1 there exist metric 𝖢⌋ E E 𝜺 ⌊𝖢 F F bidyadics eN ∈ 2 2 and N ∈ 2 2. Bidyadics play a central role in the analysis of electromagnetic media. 𝖢 𝖣 There is a natural dot product between metric bidyadics m, m ∈ E E 2 2 induced by the natural dot product of bivectors, defined by 𝖢 ⋅ 𝖣 𝖢 |𝖤−1|𝖣 . m m = m m (3.1) 𝖤 E E The symmetric bidyadic ∈ 2 2 serves as the unit bidyadic in the dot product: 𝖤 ⋅ 𝖢 𝖢 ⋅ 𝖤 𝖢 . m = m = m (3.2) Powers of metric bidyadics must be understood in terms of the dot product as 𝖢2 𝖢 ⋅ 𝖢 𝖢p ⋅ 𝖢q 𝖢p+q m = m m, m m = m , (3.3) 𝖢0 𝖤 whence m = . The inverse of the metric bidyadic is defined with respect to the bar product as 𝖢 |𝖢−1 𝖨(2) 𝖢−1|𝖢 𝖨(2)T . m m = , m m = (3.4)

53 54 CHAPTER 3 Bidyadics

The inverse with respect to the dot product can be represented as 𝖢 ⋅ ⌊⌊𝖢−1 ⌊⌊𝖢−1 ⋅ 𝖢 𝖤. m (eNeN m ) = (eNeN m ) m = (3.5) Like dyadics in the previous chapter, bidyadics can be classified in terms of their rank, which is defined as the dimension of the bivector or two-form subspace formed by the image of the bidyadic. A bidyadic has an inverse only when the rank is 6. Bidyadics form a most important topic because, in the 4D representation of electromagnetic fields, linear media are defined in terms of bidyadics.

3.1 CAYLEY–HAMILTON EQUATION

𝖢 E F Any bidyadic ∈ 2 2 defines an eigenvalue problem 𝖢|A = 𝜆A. (3.6) 𝜆 In the general case, there are six eigenvalues i corresponding to six eigenbivectors Ai. The trace of the bidyadic equals the sum of the six eigenvalues, ∑ 𝖢 𝜆 tr = i, (3.7) i and the determinant (to be defined below) equals the product of the eigenvalues, 𝖢 𝜆 𝜆 𝜆 𝜆 𝜆 𝜆 . det = 1 2 3 4 5 6 (3.8) Similarly to the case handled in Chapter 2, the eigenvalues of the bidyadic satisfy a scalar characteristic equation of the sixth order, which can be written as 𝜆6 𝜆5 𝜆4 𝜆3 𝜆2 𝜆 . a0 + a1 + a2 + a3 + a4 + a5 + a6 = 0 (3.9) The corresponding sixth-order Cayley–Hamilton equation for the 𝖢 E F bidyadic ∈ 2 2 has the same form as the characteristic equation, 𝖢6 𝖢5 𝖢4 𝖢3 𝖢2 𝖢 𝖨(2) . a0 + a1 + a2 + a3 + a4 + a5 + a6 = 0 (3.10) 𝖢 𝖢⌋ E E For the metric bidyadic m = eN ∈ 2 2 (3.10) has a similar form in terms of powers defined by (3.3) 𝖢6 𝖢5 𝖢4 𝖢3 𝖢2 𝖢 𝖤 . a0 m + a1 m + a2 m + a3 m + a4 m + a5 m + a6 = 0 (3.11) 3.1 Cayley–Hamilton Equation 55

𝜆 𝜆 The scalar equation (3.9) for = i is obtained when multiplying (3.10) | by the eigenbivector as () Ai. When the bidyadic is transformed as 𝖢→𝖢 𝖣|𝖢|𝖣−1 1 = (3.12) 𝖣 E F in terms of some bidyadic ∈ 2 2 possessing an inverse, the Cayley–Hamilton equation (3.10) and the eigenvalues 𝜆 are the same 𝖢 for the bidyadic 1. Equation (3.12) is called the similarity transformation. All bidyadics satisfying the same Cayley–Hamilton equation are obtained from one known solution through different similarity transformations [8].

3.1.1 Coefficient Functions Derivation of the Cayley–Hamilton equation (3.10) is more complicated than that of the corresponding fourth-order equation (2.119). Let us just briefly summarize the result given in the literature. The coefficients 𝖢 ai are functions of the bidyadic and they can be found through the following recursion process [7]

a0 = 1 (3.13) 𝖢 a1 =−a0tr (3.14) 𝖢 𝖢2 2a2 =−(a1tr + a0tr ) (3.15) 𝖢 𝖢2 𝖢3 3a3 =−(a2tr + a1tr + a0tr ) (3.16) 𝖢 𝖢2 𝖢3 𝖢4 4a4 =−(a3tr + a2tr + a1tr + a0tr ) (3.17) 𝖢 𝖢2 𝖢3 𝖢4 𝖢5 5a5 =−(a4tr + a3tr + a2tr + a1tr + a0tr ) (3.18) 𝖢 𝖢2 𝖢3 𝖢4 𝖢5 𝖢6 . 6a6 =−(a5tr + a4tr + a3tr + a2tr + a1tr + a0tr ) (3.19) For example, in the special case 𝖢 = 𝖨(2) these yield a0 = 1, a1 =−6, a2 = 15, a3 =−20, a4 = 15, a5 =−6, a6 = 1, (3.20) whence (3.10) is reduced to (𝖢 − 𝖨(2))6 = 0. (3.21) For another special case 𝖢 = 𝖳(2),where𝖳 is the time reversion dyadic 𝖳 𝖨 𝜺 𝖳(2) 𝖨(2) 𝖨 ∧ 𝜺 = s − e4 4, = s − s∧e4 4, (3.22) 56 CHAPTER 3 Bidyadics we obtain

a0 = 1, a1 = 0, a2 =−3, a3 = 0, a4 = 3, a5 = 0, a6 =−1, (3.23) whence (3.10) is reduced to (𝖢2 − 𝖨(2))3 = (𝖢 − 𝖨(2))3|(𝖢 + 𝖨(2))3 = 0. (3.24) The recursion equations (3.13)–(3.19) can be solved for the coefficients as 𝖢 a1 =−tr (3.25) 𝖢 2 𝖢2 2a2 = (tr ) − tr (3.26) 𝖢 3 𝖢 𝖢2 𝖢3 3!a3 =−((tr ) − 3tr tr + 2tr ) (3.27) 𝖢 4 𝖢 2 𝖢2 𝖢2 2 𝖢 𝖢3 𝖢4 4!a4 = (tr ) − 6(tr ) tr + 3(tr ) + 8tr tr − 6tr (3.28) 𝖢 5 𝖢 3 𝖢2 𝖢2 2 𝖢 𝖢 2 𝖢3 5!a5 =−((tr ) − 10(tr ) tr + 15(tr ) tr + 20(tr ) tr − 30tr𝖢4 tr𝖢 − 20tr𝖢2 tr𝖢3 + 24tr𝖢5) (3.29) 𝖢 6 𝖢 4 𝖢2 𝖢 3 𝖢3 𝖢 2 𝖢2 2 6!a6 = (tr ) − 15(tr ) tr + 40(tr ) tr + 45(tr ) (tr ) − 120tr𝖢 tr𝖢2 tr𝖢3 − 90(tr𝖢)2tr𝖢4 + 144tr𝖢 tr𝖢5 + 40(tr𝖢3)2 − 15(tr𝖢2)3 + 90tr𝖢2tr𝖢4 − 120tr𝖢6. (3.30) It is easy to verify that these expressions reduce to (3.20) and (3.23) for the respective two special bidyadics 𝖢 = 𝖨(2) and 𝖢 = 𝖳(2). There is an interesting set of relations between the expressions of 𝖢 𝖢p the coefficients ai. If we consider x = tr to be independent of tr for p > 1, we can show that there exist the following relations 𝜕 xai(x) =−ai−1(x), (3.31) for all i from 1 to 6. Thus, starting from the expression of the coefficient < a6(x), we can obtain the expressions of ai, i 6 through successive differentiations as 𝜕 6−i . ai(x) = (− x) a6(x) (3.32) Thus, the Cayley–Hamilton equation can be expressed in terms of a bidyadic differential operator as ∑6 𝖫 𝜕 𝖫 𝜕 𝖢i 𝜕 i. ( x)a6(x) = 0, ( x) = (− x) (3.33) i=0 Details of this are left as an exercise. 3.1 Cayley–Hamilton Equation 57

3.1.2 Determinant of a Bidyadic The inverse of a bidyadic 𝖢 can be given the following analytic form,

𝖢−1 1 𝖢5 𝖢4 𝖢3 𝖢2 𝖢 𝖨(2) =− (a0 + a1 + a2 + a3 + a4 + a5 ), (3.34) a6 ≠ 𝖢 which is valid for a6 0. The determinant of a bidyadic can now be defined as a multiple of a6. The choice 𝖢 det = a6, (3.35) 𝖨 𝜺 (2) appears best because it yields det( s ± e4 4) =±1 which coincides with the results for the corresponding 6 × 6 matrices consisting of three 1’s and three ±1’s on its diagonal. The determinant function of a bidyadic 𝖢 has quite an extensive analytic form, 1 1 1 det𝖢 = (tr𝖢)6 − ((tr𝖢)4 + (tr𝖢2)2)tr𝖢2 + (tr𝖢 tr𝖢2)2 720 48 16 1 1 + ((tr𝖢)3 + tr𝖢3)tr𝖢3 − (tr𝖢6 + tr𝖢 tr𝖢2 tr𝖢3) 18 6 1 1 + (tr𝖢2 − (tr𝖢)2)tr𝖢4 + tr𝖢 tr𝖢5. (3.36) 8 5 For some special cases this expression can be simplified. For example, for a trace-free bidyadic 𝖢 satisfying tr𝖢 = 0, (3.36) is reduced to 1 1 1 1 det𝖢 =− (tr𝖢2)3 + (tr𝖢3)2 + tr𝖢2tr𝖢4 − tr𝖢6. (3.37) 48 18 8 6 Let us check this for the time reversal bidyadic (3.22) which satisfies tr𝖢 = 0. Expanding

𝖢i = 𝖨(2) ⇒ tr𝖢i = 6, i = 2, 4, 6, 𝖢i = 𝖢⇒tr𝖢i = 0, i = 1, 3, 5, (3.38) and substituting in (3.37) yields the known result det𝖢 =−1.

3.1.3 Antisymmetric Bidyadic 𝖢 𝖢⌋ 𝖢T Considering an antisymmetric bidyadic m = eN =− m,the bidyadic Cayley–Hamilton equation (3.11) can be split in its symmetric 58 CHAPTER 3 Bidyadics and antisymmetric parts as 𝖢6 𝖢4 𝖢2 𝖤 a0 m + a2 m + a4 m + a6 = 0, (3.39) 𝖢5 𝖢3 𝖢 a1 m + a3 m + a5 m = 0, (3.40) 𝖤 ⌊𝖨(2)T 𝖢 ⌋𝜺 p with = eN . Because we have tr( m N) = 0 for odd values of p, the coefficients a1, a3 and a5 actually vanish whence (3.40) is identically valid and the Cayley–Hamilton equation for an antisymmetric bidyadic 𝖢 m has the simpler form (3.39). Defining the bidyadic 𝖣 𝖢2 𝖢2 ⌋𝜺 𝖢 ⋅ 𝖢 ⌋𝜺 = = m N = m m N, (3.41) it satisfies 𝖣3 𝖣2 𝖣 𝖨(2) a0 + a2 + a4 + a6 = 0, (3.42) with the coefficients defined by

a0 = 1 (3.43) 1 a =− tr𝖣 (3.44) 2 2 1 1 a = (tr𝖣)2 − tr𝖣2 4 8 4 1 1 1 a =− (tr𝖣)3 + tr𝖣 tr𝖣2 − tr𝖣3. (3.45) 6 48 8 6 Thus, any bidyadic 𝖣 which can be expressed as a square of a bidyadic 𝖢 𝖢 𝖢 ⌋𝜺 𝖢 of the form = m N,where m is an antisymmetric bidyadic, satisfies a bidyadic equation of the third order. The form (3.42) of Cayley–Hamilton equation can also be expressed in operational form if we consider the coefficients as functions of x = tr𝖣 = tr𝖢2, ∑3 𝖫 휕 𝖢 𝖫 휕 𝖣 j 휕 j. ( x)det (x) = 0, ( x) = (− x∕2) (3.46) j=0

3.2 BIDYADIC EIGENPROBLEM

𝖨(4)⌊⌊ 𝜺 ⌊⌊ The double contraction = eN N can be applied to map between 𝖢 E F 𝖣 E F bidyadics ∈ 2 2 and ∈ 2 2 as 𝖣 = 𝖨(4)⌊⌊𝖢T , 𝖢 = 𝖨(4)⌊⌊𝖣T . (3.47) 3.2 Bidyadic Eigenproblem 59

Because of tr𝖣p = tr𝖢p, both 𝖢 and 𝖣 satisfy the same Cayley–Hamilton equation, an equation with the same coefficients ai. Being independent of any basis, such a mapping creates a natural eigenproblem for bidyadics,

𝖨(4)⌊⌊𝖢T = 휆𝖢. (3.48)

From

𝖢 = 𝖨(4)⌊⌊(𝖨(4)T ⌊⌊𝖢) = 휆𝖨(4)⌊⌊𝖢T = 휆2𝖢, (3.49) valid for any eigenbidyadic 𝖢, we arrive at two possible eigenvalues, 휆 . ± =±1 (3.50) The trace of (3.48) yields

tr(𝖨(4)⌊⌊𝖢T ) = tr𝖢 = 휆tr𝖢, (3.51) 𝖢 whence the eigenbidyadics ± corresponding to the eigenvalues ±1 satisfy 𝖢 𝖢 . tr ± =±tr ± (3.52) 𝖢 The eigenbidyadic − is trace free: 𝖢 . tr − = 0 (3.53) Applying the bidyadic identity (2.114) as 𝖨(4)⌊⌊𝖢T 𝖢 𝖨(2) 𝖢⌊⌊𝖨T ∧𝖨 𝖢 = (tr ) − ( )∧ + (3.54) to the two eigenbidyadics, we obtain the conditions 𝖢 𝖨(2) 𝖢 ⌊⌊𝖨T ∧𝖨 𝖢 𝖢 . (tr ±) − ( ± )∧ + ± =± ± (3.55) Further, we have 𝖨(4)⌊⌊𝖢Tp 𝖨(4)⌊⌊𝖢T p p𝖢p ± = ( ±) = (±1) ±, (3.56) 𝖢p p 𝖢p tr ± = (±1) tr ±, (3.57) 𝖢p whence tr − = 0 for odd values of p. Let us consider the two eigendyadics separately. 60 CHAPTER 3 Bidyadics

3.2.1 Eigenbidyadic C− Inserting (3.53) in (3.55) we obtain the following condition for the 𝖢 trace-free eigenbidyadic −: 1 𝖢 = (𝖢 ⌊⌊𝖨T )∧𝖨. (3.58) − 2 − ∧ 𝖡 E F This means that there must exist a dyadic o ∈ 1 1 in terms of which the eigenbidyadic can be expressed in the form 1 𝖢 = 𝖡 ∧𝖨, 𝖡 = (𝖢 ⌊⌊𝖨T ). (3.59) − o∧ o 2 − Applying again (3.53), we have 𝖢 𝖡 ∧𝖨 𝖡 || 𝖨⌋⌋𝖨(2)T 𝖡 tr − = tr( o∧ ) = o ( ) = 3tr o = 0, (3.60) 𝖡 whence the dyadic o must be trace free. 𝖢 The converse is also true: any bidyadic − expressed in the form 𝖡 E F (3.59) in terms of some trace-free dyadic o ∈ 1 1 is a solution of the 휆 𝖢 𝖡 ∧𝖨 eigenproblem (3.48) for =−1. In fact, inserting = o∧ in (2.114) and applying the expansion 𝖡 ∧𝖨 ⌊⌊𝖨T 𝖡 𝖨 𝖨 𝖡 𝖡 |𝖨|𝖨 𝖨|𝖨|𝖡 𝖡 ( o∧ ) = (tr o) + (tr ) o − o − o = 2 o, (3.61) we obtain 𝖨(4)⌊⌊ 𝖡 ∧𝖨 T 𝖡 ∧𝖨 ⌊⌊𝖨T ∧𝖨 𝖡 ∧𝖨 𝖡 ∧𝖨 ( o∧ ) =−(( o∧ ) )∧ + o∧ =− o∧ , (3.62) 𝖢 𝖡 ∧𝖨 𝖡 whence − = o∧ is an eigenbidyadic for any trace-free dyadic o corresponding to 휆 =−1.

3.2.2 Eigenbidyadic C+ 𝖢 For an eigenbidyadic + the identity (3.55) is reduced to 𝖢 ⌊⌊𝖨T ∧𝖨 𝖢 𝖨(2). ( + )∧ = (tr +) (3.63) Writing 1 1 𝖨(2) = 𝖨∧𝖨 = (𝖨(2)⌊⌊𝖨T )∧𝖨, (3.64) 2 ∧ 6 ∧ (3.63) can be expressed as 𝖢 ⌊⌊𝖨T ∧𝖨 ( +o )∧ = 0, (3.65) 3.3 Hehl–Obukhov Decomposition 61 where 1 𝖢 = 𝖢 − tr𝖢 𝖨(2), (3.66) +o + 6 + 𝖢 𝖢 is the trace-free part of +. Thus, the eigenbidyadic + can be decomposed in two components as 1 𝖢 = 𝖢 + A𝖨(2), A = tr𝖢 ,tr𝖢 = 0. (3.67) + +o 6 + +o In fact, inserting (3.67) in (3.55) we obtain (3.65) as the condition for 𝖢 the trace-free part of the eigenbidyadic +. Actually, (3.65) is equivalent with the condition 𝖢 ⌊⌊𝖨T . +o = 0 (3.68) 𝖡 E F 𝖡∧𝖨 To see this, let us assume that a dyadic ∈ 1 1 satisfies ∧ = 0. From the rule (2.82) we have 𝖡∧𝖨 ⌊⌊𝖨T 𝖡 𝖨 𝖡 . ( ∧ ) = (tr ) + 2 = 0 (3.69) 𝖡 𝖡∧𝖨 𝖡 The trace of this yields 6tr = 0, whence ∧ = 0 implies = 0. 𝖢 To conclude, the eigenbidyadic + can be decomposed in two parts 휆 as (3.67) both of which satisfy the eigenequation (3.48) for + =+1. 𝖢 The bidyadic +o is trace free and satisfies the condition (3.68). Con- 𝖢 𝖢 𝖢 ⌊⌊𝖨T versely, starting from a bidyadic o satisfying tr o = 0and o = 0, 𝖢 𝖢 inserting = o in (3.54) we obtain 𝖨(4)⌊⌊𝖢 T 𝖢 o = o, (3.70) 𝖢 휆 whence o is a trace-free eigenbidyadic corresponding to =+1. 𝖨(4)⌊⌊𝖨(2)T 𝖨(2) 𝖢 𝖢 𝖨(2) Because of = , + = o + A is a more general form of an eigenbidyadic corresponding to 휆 =+1.

3.3 HEHL–OBUKHOV DECOMPOSITION

Because of its significance in representing electromagnetic media, let 𝖬 F E us consider the bidyadic ∈ 2 2 mapping two-forms to two-forms. The eigenproblem (3.48) takes now the form 𝖨(4)⌊⌊𝖬 = 휆𝖬T . (3.71) 62 CHAPTER 3 Bidyadics

Any given bidyadic 𝖬 can be uniquely expanded as a sum of eigenbidyadics as 𝖬 𝖬 𝖬 . = + + − (3.72) In fact, we can write 1 𝖬 = (𝖬 ± 𝖨(4)T ⌊⌊𝖬T ). (3.73) ± 2 𝖬 Separating the trace-free and multiple of the unit bidyadic terms of +, a more complete decomposition can be expressed as 𝖬 𝖬 𝖬 𝖬 . = 1 + 2 + 3 (3.74) The terminology given here follows that introduced by Hehl and Obukhov in [9] and the decomposition will be called the Hehl–Obukhov decomposition for short. Here we denote 𝖬 𝖬 2 = −, (3.75) which is called the skewon part of 𝖬.Further, 𝖬 𝖬 𝖬 1 + 3 = +, (3.76) 𝖬 𝖬 in which 3 is called the axion part of , a multiple of the unit bidyadic,

1 𝖬 = tr𝖬 𝖨(2)T . (3.77) 3 6 𝖬 𝖬 Finally, 1 is called the principal part of and it equals the trace- 𝖬 free part of the eigenbidyadic +. Because of the condition (3.68), the principal part satisfies 𝖬 ⌊⌊𝖨 1 = 0, (3.78) 𝖬 implying tr 1 = 0, while the skewon part satisfies 𝖬 ⌊⌊𝖨 ∧𝖨T 𝖬 ( 2 )∧ = 2 2, (3.79) 𝖬 implying tr 2 = 0 and, the axion part satisfies 𝖬 ⌊⌊𝖨 ∧𝖨T 𝖬 . ( 3 )∧ = 6 3 (3.80) The form (3.79) shows us that the skewon part must be of the form 𝖬 𝖡 ∧𝖨 T 2 = ( o∧ ) , (3.81) 3.3 Hehl–Obukhov Decomposition 63

𝖡 E F where o ∈ 1 1 is a trace-free dyadic. The condition (3.78) can be used as a definition of a principal bidyadic, a bidyadic with noskewon and axion parts. In fact, from 𝖬⌊⌊𝖨 = 0 it follows that tr𝖬 = 0 whence 𝖬 𝖬 𝖬 𝖬 𝖬⌊⌊𝖨 ∧𝖨T = 1 + 2. From (3.79) we then obtain 2 2 = ( )∧ = 0. 𝖬 E E For the metric bidyadic m ∈ 2 2 mapping two-forms to bivectors and related to the bidyadic 𝖬 by

𝖬 ⌊𝖬 𝖬 𝜺 ⌊𝖬 m = eN , = N m, (3.82) the eigenproblem (3.71) corresponds to

𝖨(4)⌊⌊𝖬 𝜺 ⌊⌊ 𝜺 ⌊𝖬 𝖬 ⌋𝜺 휆𝖬T ⌋𝜺 = eN N ( N m) = m N = m N, (3.83) or,

𝖬T 휆𝖬 휆 . m = m, =±1 (3.84)

𝖬 𝖬 𝖬 𝖬 Thus, m+ and m− corresponding to the eigenbidyadics + and − are respectively symmetric and antisymmetric bidyadics. The corresponding Hehl–Obukhov decomposition for the metric bidyadic is, then,

𝖬 𝖬 𝖬 𝖬 m = m1 + m2 + m3, (3.85) 𝖬 𝖬 𝖬 where m2 equals the antisymmetric part of m while m3 isamul- ⌊𝖨(2)T 𝖬 tiple of the symmetric dyadic eN and m1 makes the rest of the 𝖬 𝜺 ⌊𝖬 symmetric part of m satisfying tr( N m1) = 0. 𝖬 E E From (3.59) it follows that any antisymmetric bidyadic m2 ∈ 2 2 𝖡 E F can be expressed in terms of a trace-free dyadic o ∈ 1 1 in the form

𝖬 ⌊ 𝖡 ∧𝖨 T 𝖡 . m2 = eN ( o∧ ) ,tro = 0 (3.86)

Let us check the number of parameters of the Hehl–Obukhov decomposition. Since an antisymmetric bidyadic corresponds to an antisymmetric 6 × 6 matrix, it has (36 − 6)∕2 = 15 free parameters. On the other hand, 𝖡 o corresponds to a trace-free 4 × 4 matrix whose number of parameters amounts to 16 − 1 = 15. Since the axion part has one free parameter, the number of parameters of the principal part equals 36 − 15 − 1 = 20. The Hehl–Obukhov decomposition plays an essential role in the definition of various classes of electromagnetic media. 64 CHAPTER 3 Bidyadics

3.4 EXAMPLE: SIMPLE ANTISYMMETRIC BIDYADIC

𝖴 F F Let us consider a simple antisymmetric bidyadic ∈ 2 2, mapping 횽 횿 F bivectors to two-forms, defined in terms of two two-forms , ∈ 2 as 1 𝖴 = (횽횿 − 횿횽). (3.87) 2 Such a bidyadic is of importance when considering quantities like electromagnetic stress, energy, and momentum. The two-forms are assumed to be linearly independent, because otherwise the bidyadic 𝖴 would vanish. Let us study some of its properties. 횽 ⋅ 횿 | 횽 Applying the natural dot product of two-forms, = eN ( ∧ 횿), we can write 𝖴2 ⋅ 횽 =Λ2횽, 𝖴2 ⋅ 횿 =Λ2횿, (3.88) with √ 1 Λ= (횿 ⋅ 횽)2 − (횿 ⋅ 횿)(횽 ⋅ 횽). (3.89) 2 Thus, for Λ ≠ 0 the bidyadic (𝖴∕Λ)2 maps any two-form of the form A횽 + B횿 to itself in the dot product. Considering the eigenproblem 𝖴 ⋅ 횵 = 휆횵, (3.90) from 𝖴2 ⋅ 횵 = 휆𝖴 ⋅ 횵 = 휆2횵 (3.91) we conclude that the solutions fall in two sets, either 휆 = 0, 횿 ⋅ 횵 = 횽 ⋅ 횵 = 0 (3.92) or 휆 횵 횽 횿 =±Λ, ± = A± + B± (3.93) for some A±, B±. Considering the latter case, from (𝖴2 −Λ2𝖤−1)|횷 = (𝖴 −Λ𝖤−1) ⋅ (𝖴 +Λ𝖤−1) ⋅ 횷 = (𝖴 +Λ𝖤−1) ⋅ (𝖴 −Λ𝖤−1) ⋅ 횷 = 0, (3.94) 3.4 Example: Simple Antisymmetric Bidyadic 65 valid for any two-form 횷 which is a linear combination of 횽 and 횿,we may express the eigen two-forms as 횵 𝖴 𝖤−1 ⋅ 횷 ± = ( ∓Λ ) , (3.95) 횷 횵 ≠ 횵 for any such yielding ± 0. The eigen two-forms ± satisfy the properties 횵 ⋅ 횵 횷 ⋅ 𝖴 𝖤−1 ⋅ 𝖴 𝖤−1 ⋅ 횷 ± ± = (− ∓Λ ) ( ∓Λ ) =−횷 ⋅ (𝖴2 −Λ2𝖤−1) ⋅ 횷 = 0, (3.96) 횵 ⋅ 횵 횷 ⋅ 𝖴 𝖤−1 ⋅ 𝖴 𝖤−1 ⋅ 횷 ± ∓ = (− ∓Λ ) ( ±Λ ) =−횷 ⋅ (𝖴2 +Λ2𝖤−1) ⋅ 횷 =−2Λ2횷 ⋅ 횷. (3.97) Here we have applied antisymmetry of 𝖴 and (3.88) for 횷,whichisa linear combination of 횽 and 횿. Assuming that the two-form 횷 is not ≠ 횵 ⋅ 횵 ≠ 𝖴 simple and Λ 0, we have + − 0. In this case can be expressed in terms of the eigen two-forms as 𝖴 Λ 횵 횵 횵 횵 . = 횵 ⋅ 횵 ( + − − − +) (3.98) + − This can be verified in terms of (3.90). Also, one can show thatthe bidyadic  1 ⌊ 횵 횵 횵 횵 E F = 횵 ⋅ 횵 eN ( + − + − +) ∈ 2 2 (3.99) + − satisfies 2 = , whence it can be interpreted as a projection bidyadic. From (3.96) it follows that each of the eigen two-forms must be simple, whence they can be expressed in terms of four one-forms 𝜶 𝜷 ±, ± as 횵 𝜶 𝜷 ± = ± ∧ ± (3.100) 횵 ⋅ 횵 ≠ Because of + − 0, the four one-forms make a basis. Let us rewrite 𝜺 the basis one-forms as multiples of i for convenience, 횵 휉 𝜺 횵 휉 𝜺 횵 ⋅ 횵 휉 휉 . + = + 12, − = − 34, + − = + − (3.101) The bidyadic (3.90) can now be expressed in the simple form 𝖴 Λ 횵 횵 횵 횵 = 휉 휉 ( + − − − +) + − 𝜺 𝜺 𝜺 𝜺 . =Λ( 12 34 − 34 12) (3.102) 66 CHAPTER 3 Bidyadics

It is easy to verify that (3.102) satisfies the eigenequation (3.90) for (3.101). Since 𝖴 is an antisymmetric bidyadic, it can be represented in terms 𝖡 E F of a trace-free dyadic o ∈ 1 1, in a form similar to (3.86), as 𝖴 𝜺 ⌊ 𝖡 ∧𝖨 𝖡 . = N ( o∧ ), tr o = 0 (3.103) 𝖡 𝖴 The dyadic o can be recovered from as 1 𝖡 = (e ⌊𝖴)⌊⌊𝖨T . (3.104) o 2 N Inserting (3.102) yields the representation Λ 𝖡 = (e 𝜺 − e 𝜺 )⌊⌊𝖨T o 2 34 34 12 12 Λ = (e 𝜺 + e 𝜺 − e 𝜺 − e 𝜺 ), (3.105) 2 3 3 4 4 1 1 2 2 𝖡 whence o satisfies the following remarkable property (Minkowski identity) Λ2 1 𝖡2 = 𝖨 = tr𝖡2 𝖨, (3.106) o 4 4 o 𝖡 independent of any basis. Thus, o is a multiple of what can be called a unipotent dyadic.

3.5 INVERSE RULES FOR BIDYADICS

To form an analytic expression for the inverse of a dyadic belonging to E F the space 1 1 we can apply the convenient rule (2.120). Alternatively, one can apply the rule (2.121) obtained from the Cayley–Hamilton E E equation. For the metric dyadic of the space 1 1, the inverse follows the rule (2.206). There does not seem to exist a simple universal ana- 𝖢 E F lytic inverse formula for the general bidyadic ∈ 2 2. However, one can always use the rule (3.34) due to the bidyadic Cayley–Hamilton equation, rewritten here as

𝖢−1 1 𝖢5 𝖢4 𝖢3 𝖢2 𝖢 𝖨(2) =− (a0 + a1 + a2 + a3 + a4 + a5 ), (3.107) a6 in which the extensive expressions (3.13)–(3.19) for the coefficients ai as functions of the bidyadic 𝖢 must be substituted. 3.5 Inverse Rules for Bidyadics 67

For some special bidyadics it is possible to find the inverse in ana- 𝖢 E F lytic form. For a bidyadic ∈ 2 2 which can be expressed in terms of a dyadic 𝖠 as 𝖢 = 𝖠(2) we can apply the rule (2.123) in the form 𝖨(4)T ⌊⌊𝖢T 𝖢−1 = 3 , 𝖢 = 𝖠(2). (3.108) tr𝖢(2) 𝖥 𝖡(2) E E Similarly, for special metric bidyadics of the form = ∈ 2 2 we can apply the rule (2.207) as 𝜺 𝜺 ⌊⌊𝖥T 𝖥−1 N N F F 𝖥 𝖡(2). = 3 ∈ 2 2, = (3.109) 𝜺 𝜺 ||𝖥(2) N N Let us consider other special cases of bidyadics for which we can form inverses in simple analytic form.

3.5.1 Skewon Bidyadic It turns out to be possible to find a simple rule for the inverse of the general skewon (or antisymmetric) bidyadic. The route to this result can be made through dyadic identities as was shown in [10]. A somewhat shorter route can be found based on the fact that the inverse of an antisymmetric metric bidyadic is antisymmetric, from which it follows that the inverse of a skewon bidyadic is also a skewon bidyadic. This is based on the property (2.191) which is valid for bidyadics as well. Thus, the inverse can be expressed in the form 𝖡 ∧𝖨 −1 𝖠 ∧𝖨. ( o∧ ) = o∧ (3.110) 𝖠 E F where o ∈ 1 1 is another trace-free dyadic. The problem is to find an 𝖠 𝖡 analytic expression for the dyadic o in terms of the given dyadic o. 𝖠 Expanding the basic equation for o through (2.41), 𝖡 ∧𝖨 | 𝖠 ∧𝖨 𝖡 |𝖠 ∧𝖨 𝖡 ∧𝖠 𝖨(2) ( o∧ ) ( o∧ ) = ( o o)∧ + o∧ o = (3.111) and contracting this by ⌊⌊𝖨T through (2.82) yields 𝖡 |𝖠 𝖨 𝖡 |𝖠 𝖡 |𝖠 𝖠 |𝖡 𝖨. tr( o o) + 2 o o − o o − o o = 3 (3.112) Solving the trace as 𝖡 |𝖠 𝖠 |𝖡 tr( o o) = tr( o o) = 3, (3.113) and inserting it in the previous equation yields 𝖡 |𝖠 𝖠 |𝖡 . o o = o o (3.114) 68 CHAPTER 3 Bidyadics

𝖠 𝖡 This means that o and o must have the same sets of eigenvectors and eigen one-forms. Assuming the eigenexpansion ∑4 𝖡 𝜺 o = Biei i, (3.115) i=1 𝖠 the corresponding expansion for o must be of the form ∑4 𝖠 𝜺 . o = Aiei i (3.116) i=1 After some simple steps, the two skewon bidyadics can be expanded as ∑ 𝖡 ∧𝖨 𝜺 o∧ = (Bi + Bj)eij ij, (3.117) < ∑i j 𝖠 ∧𝖨 𝜺 . o∧ = (Ai + Aj)eij ij (3.118) i

B4 =−(B1 + B2 + B3), A4 =−(A1 + A2 + A3), (3.120) and solve (3.119) for A1, A2 and A3 in the form 2 Bi Ai =− + A, i = 1, 2, 3, (3.121) (B1 + B2)(B2 + B3)(B3 + B1) with 2 2 2 B + B + B + B1B2 + B2B3 + B3B1 A = 1 2 3 . (3.122) 2(B1 + B2)(B2 + B3)(B3 + B1)

Applying these, from (3.120) we obtain the fourth coefficient A4 in a similar form, B2 4 . A4 =− + A (3.123) (B1 + B2)(B2 + B3)(B3 + B1) 3.5 Inverse Rules for Bidyadics 69

To interpret the result in coordinate-free form, let us expand 𝖡(3) tr o = B1B2B3 + B1B2B4 + B1B3B4 + B2B3B4 =−(B1 + B2)(B2 + B3)(B3 + B1), (3.124) 𝖡(2) tr o = B1B2 + B2B3 + B3B1 + B1B4 + B2B4 + B3B4 2 2 2 =−(B1 + B2 + B3 + B1B2 + B2B3 + B3B1), (3.125) in terms of which we can write B2 tr𝖡(2) A = i + o , i = 1, 2, 3, 4. (3.126) i 𝖡(3) 𝖡(3) tr o 2tr o 𝖠 Thus, finally, the dyadic o can be expressed as ( ) 1 1 𝖠 = 𝖡2 + tr𝖡(2)𝖨 , (3.127) o 𝖡(3) o 2 o tr o whence the inverse of the skewon bidyadic has the compact analytic form 1 (𝖡 ∧𝖨)−1 = 𝖠 ∧𝖨 = (𝖡2∧𝖨 + tr𝖡(2)𝖨(2)). (3.128) o∧ o∧ 𝖡(3) o∧ o tr o 𝖡(3) There is no inverse in the case of tr o = 0. (3.128) can be checked by starting from the converse relation ( ) 1 1 𝖡 = 𝖠2 + tr𝖠(2)𝖨 , (3.129) o 𝖠(3) o 2 o tr o 𝖡 𝖡 and substituting (3.127) to obtain the result o = o. Details are left as an exercise. Another form for the inverse of a skewon bidyadic can be found by 𝖢 𝖡(2)T starting from the rule (3.54) for = o and expanding 𝖨(4)⌊⌊𝖡(2)T 𝖡(2) 𝖨(2) 𝖡(2)⌊⌊𝖨T ∧𝖨 𝖡(2) o = tr o − ( o )∧ + o 𝖡(2) 𝖨(2) 𝖡2∧𝖨 𝖡(2) = tr o + o∧ + o 𝖡(3) 𝖡 ∧𝖨 −1 𝖡(2). = tr o ( o∧ ) + o (3.130) At the last stage we have applied (3.128). From this we obtain the expression 1 (𝖡 ∧𝖨)−1 = (𝖨(4)⌊⌊𝖡(2)T − 𝖡(2)). (3.131) o∧ 𝖡(3) o o tr o 70 CHAPTER 3 Bidyadics

As a check we can verify that this solution satisfies 𝖨(4)⌊⌊ 𝖡 ∧𝖨 −1 𝖡 ∧𝖨 −1T ( o∧ ) =−( o∧ ) , (3.132) which is the condition for a skewon bidyadic. A third expression for the inverse of a skewon bidyadic can be found through the special form (3.42) of the Cayley–Hamilton equation where 𝖣 𝖡 ∧𝖨 2 = ( o∧ ) . In this case we have 1 (𝖡 ∧𝖨)−1 = (𝖡 ∧𝖨)|𝖣−1 =− (𝖡 ∧𝖨)|(𝖣2 + a 𝖣 + a 𝖨(2)) o∧ o∧ a o∧ 2 4 ( 6 1 | 𝖡 ∧𝖨 5 𝖡2 𝖡 ∧𝖨 3 =− ( o∧ ) − (tr )( o∧ ) det(𝖡 ∧𝖨) o ( o∧ ) ) 1 − (tr𝖡2)2 − tr𝖡4 (𝖡 ∧𝖨) . (3.133) 4 o o o∧ Applying (3.37), the determinant of the skewon bidyadic can be shown to reduce to 𝖡 ∧𝖨 𝖡(3) 2 det( o∧ ) =−(tr o ) , (3.134) details of which are left as an exercise. 𝖣 E E The inverse of an antisymmetric bidyadic ∈ 2 2 can be found 𝖣 𝖡 ∧𝖨 ⌋ by expressing it in terms of a skewon bidyadic as = ( o∧ ) eN and applying the inverse rule (3.128) to 𝖣−1 𝖡 ∧𝖨 | 𝖨(2)⌋ −1 𝖨(2)⌋ −1| 𝖡 ∧𝖨 |−1 𝜺 ⌊ 𝖡 ∧𝖨 −1. = (( o∧ ) ( eN)) = ( eN) ( o∧ ) = N ( o∧ ) (3.135) Since it is known that the inverse of a symmetric and antisymmetric bidyadic is respectively symmetric and antisymmetric, it follows that the inverse of a skewon bidyadic is a skewon bidyadic and the inverse of a principal–axion bidyadic is a principal–axion bidyadic. However, the inverse of a principal bidyadic may contain an axion term and the inverse of a skewon–axion bidyadic may contain a principal term.

3.5.2 Extended Bidyadics 𝖢 E F If the inverse of a bidyadic ∈ 2 2 is known, the inverse of the extended bidyadic 𝖣 = 𝖢 + A횽 (3.136) 3.5 Inverse Rules for Bidyadics 71 can be formed for any bivector A and two-form 횽 satisfying 횽|𝖢−1|A ≠ −1. In fact, the expression 𝖢−1|A횽|𝖢−1 𝖣−1 = 𝖢−1 − (3.137) 1 + 횽|𝖢−1|A can be straightforwardly verified when multiplying by 𝖣|. Similarly, for 𝖣 𝖢 E E metric bidyadics , ∈ 2 2 related by 𝖣 = 𝖢 + AB (3.138) the inverse becomes 𝖢−1|AB|𝖢−1 𝖣−1 = 𝖢−1 − . (3.139) 1 + B|𝖢−1|A The previous rule (3.137) can be generalized. To simplify the notation without essential loss of generality, let us consider the rule for the bidyadic 𝖣 = 𝖨(2) + 𝖥, 𝖥 = A횽. (3.140) In this case the inverse can be expressed in the form 𝖥 𝖨(2) 𝖥 −1 𝖨(2) 𝖥. ( + ) = − , a1 =−tr (3.141) 1 − a1 The expression can be verified by applying the equation 𝖥2 𝖥 + a1 = 0 (3.142) satisfied by the bidyadic 𝖥. In fact, we obtain ( ) 𝖥 𝖥 + 𝖥2 𝖨(2) − |(𝖨(2) + 𝖥) = 𝖨(2) + 𝖥 − 1 − a1 1 − a1 𝖥 − a 𝖥 = 𝖨(2) + 𝖥 − 1 = 𝖨(2). (3.143) 1 − a1 Actually, the inverse is valid for any bidyadic 𝖥 satisfying (3.142), where a1 may be any scalar. Applying the rule (3.137) twice we can find the inverse rule for the bidyadic 𝖣 = 𝖨(2) + 𝖥, 𝖥 = A횽 + B횿. (3.144) 72 CHAPTER 3 Bidyadics

After some algebraic steps, details of which are left as an exercise, the inverse formula can be expressed as 𝖭 𝖣−1 = 𝖨(2) − , (3.145) D with 𝖭 = A횽 + B횿 + (A|횽)B횿 + (B|횿)A횽 − A횽|B횿 − B횿|A횽 (3.146) and D = 1 + A|횽 + B|횿 + (A|횽)(B|횿) − (A|횿)(B|횽). (3.147) However, the rule (3.145) can be written more compactly as (1 − a )𝖥 − 𝖥2 (𝖨(2) + 𝖥)−1 = 𝖨(2) − 1 , (3.148) 1 − a1 + a2 with a1 as above and 1 a = ((tr𝖥)2 − tr𝖥2). (3.149) 2 2 The expression (3.148) can be verified by first showing that, in this case, 𝖥 satisfies the equation 𝖥3 𝖥2 𝖥 . + a1 + a2 = 0 (3.150) The process can be continued by considering a bidyadic 𝖥 satisfying 𝖥4 𝖥3 𝖥2 𝖥 + a1 + a2 + a3 = 0, (3.151) with a1 and a2 as above and 1 a =− ((tr𝖥)3 − 3(tr𝖥)(tr𝖥2) + 2tr𝖥3). (3.152) 3 3! In this case the inverse becomes (1 − a + a )𝖥 − (1 − a )𝖥2 + 𝖥3 (𝖨 + 𝖥)−1 = 𝖨(2) − 1 2 1 . (3.153) 1 − a1 + a2 − a3 Again, (3.153) is valid for a bidyadic 𝖥 satisfying (3.151) for any three scalars a1, a2, a3. 3.5 Inverse Rules for Bidyadics 73

Finally, because any bidyadic 𝖥 satisfies the Cayley–Hamilton equation 𝖥6 𝖥5 𝖥4 𝖥2 𝖥 𝖨(2) + a1 + a2 + a3 + a5 + a6 = 0, (3.154) the expression for the inverse of 𝖨(2) + 𝖥 can be formed by continuing the previous pattern as ∑5 𝖨(2) 𝖥 −1 𝖨(2) 1 𝖥 6−i ( + ) = + Ai(− ) , (3.155) A6 j=0 with ∑i j ⋯ i . Ai = (−1) aj = 1 − a1 + + (−1) ai (3.156) j=0 As an example let us form the inverse of a skewon–axion bidyadic 𝖣 𝖢 𝖨 𝖢 𝖡 ∧𝖨. = o + , o = o∧ (3.157) The skewon bidyadic satisfies the reduced Cayley–Hamilton equation 𝖢6 𝖢4 𝖢2 𝖨(2) o + a2 o + a4 o + a6 = 0, (3.158) 𝖢 𝖢 with coefficients ai obtained from (3.13) to (3.19) with = o. Apply- ing the inverse rule (3.155) we obtain 𝖢 − 𝖢2 𝖨(2) 𝖢 −1 𝖨(2) o o | 𝖨(2) ( + o) = − ((1 + a2 + a4) 1 + a2 + a4 + a6 𝖢2 𝖢4 . + (1 + a2) o + o) (3.159) Verifying this is left as an exercise.

3.5.3 3D Expansions 𝖣 E F Expanding a bidyadic ∈ 2 2 in its spatial and temporal components as 𝖣 𝖠 𝖡 𝖢 𝜺 𝖣 𝜺 = s + e4 ∧ s + s ∧ 4 + e4 ∧ s ∧ 4, (3.160) it is possible to find its inverse in analytic form. Applying the corresponding expansion for the inverse bidyadic 𝖣−1 𝖠′ 𝖡′ 𝖢′ 𝜺 𝖣′ 𝜺 = s + e4 ∧ s + s ∧ 4 + e4 ∧ s ∧ 4, (3.161) 74 CHAPTER 3 Bidyadics we obtain the relations 𝖠 |𝖠′ 𝖢 |𝖡′ 𝖨(2) s s − s s = s , (3.162) 𝖠 |𝖢′ 𝖢 |𝖣′ s s − s s = 0, (3.163) 𝖡 |𝖠′ 𝖣 |𝖡′ s s − s s = 0, (3.164) 𝖡 |𝖢′ 𝖣 |𝖣′ 𝖨 . s s − s s = s (3.165) 𝖠′..𝖣′ To solve for the spatial dyadics s s requires certain assumptions on 𝖠 ..𝖣 the spatial inverses of some of the dyadics s s. For example, assuming 𝖠−1 𝖡−1 the existence of s and s , we obtain the expressions 𝖠′ 𝖠−1 𝖠−1|𝖢 | 𝖢 𝖠 |𝖡−1|𝖣 −1 s = s − s s ( s − s s s) , (3.166) 𝖡′ 𝖢 𝖠 |𝖡−1|𝖣 −1 s =−( s − s s s) , (3.167) 𝖢′ 𝖠−1|𝖢 | 𝖣 𝖡 |𝖠−1|𝖢 −1 s =− s s ( s − s s s) , (3.168) 𝖣′ 𝖣 𝖡 |𝖠−1|𝖢 −1. s =−( s − s s s) (3.169) Also, the two bracketed spatial dyadic expressions are assumed to have inverses. Similar expressions are obtained by starting from other possibilities. If all of the spatial dyadics have inverses, all expressions are simultaneously valid.

PROBLEMS

𝖢 E F 𝖢⌊ 3.1 Prove that if a bidyadic ∈ 2 2 satisfies a = 0 for any vector a,itmustvanish,𝖢 = 0. 3.2 Solve the bidyadic equation 𝖨∧𝖷 𝖸 ∧ = 𝖷 E F 𝖸 E F for the dyadic ∈ 1 1 and bidyadic ∈ 2 2, assuming that such a solution exists. Show that a necessary and sufficient condition for the solution to exist is that 𝖸 satisfy the bidyadic equation 𝖸 𝖨(2) 𝖸 𝖸⌊⌊𝖨T ∧𝖨 . 6 + 2 tr − 3( )∧ = 0 Hint: contract both sides of the equation by ()⌊⌊𝖨T . 𝖢 E E 3.3 An antisymmetric bidyadic ∈ 2 2 is called simple if it can be expressed in terms of two bivectors as AB − BA. Show that the most general antisymmetric bidyadic can be expressed as a sum Problems 75

of three simple antisymmetric bidyadics, that is, in terms of six linearly independent bivectors Ai in the form 𝖢 . = (A1A2 − A2A1) + (A3A4 − A4A3) + (A5A6 − A6A5) Hint: show that a sum of four simple antisymmetric bidyadics can always be reduced to a sum of three simple antisymmetric bidyadics. 3.4 Show that if one of the six bivectors in the expansion of the bidyadic in the previous problem is linearly dependent on the other five, the bidyadic can be expressed as a sum of two simple antisymmetric bidyadics.

3.5 Assuming that the six bivectors Ai of Problem 3.3 form a basis and 횷 the two-forms i make the reciprocal two-form basis satisfying |횷 훿 Ai j = ij, find an expression for the inverse of the general antisymmetric bidyadic 𝖢 of Problem 3.3. 3.6 Derive the inverse expression for the skewon–axion metric bidyadic 𝖣 = 𝖢 + 훼𝖤, 𝖢 = AB − BA, where A, B are two bivectors. Hint: start by expanding 𝖢3 to find a useful condition. 3.7 Derive the inverse of the bidyadic 𝖣 = 𝖨(2) + A횽 + B횿 by applying the rule (3.137) twice. Verify the result by forming the dyadic 𝖣|𝖣−1. 3.8 Check the expression for the inverse of a skewon bidyadic (3.128) by starting from the relation ( ) 1 1 𝖡 = 𝖠2 + tr𝖠(2)𝖨 , o 𝖠(3) o 2 o tr o 𝖠 𝖡 where the dyadic o is related to o by (3.127), by substituting it in the above expression. 3.9 Find the inverse of the bidyadic 𝖣 훼𝖨(2) 𝖭∧𝖨 E F = + ∧ ∈ 2 2, 76 CHAPTER 3 Bidyadics

𝖭 E F 𝖭2 where ∈ 1 1 is a nilpotent dyadic satisfying = 0. Hint: 𝖭∧𝖨 3 expand first ( ∧ ) . 𝖢 훼𝖨(2) 𝜶∧𝖨 훼 ≠ 3.10 Find the inverse of the bidyadic = + a ∧ , 0. 𝖢 E F 𝖢 3.11 Prove that if a bidyadic ∈ 2 2 satisfies (1) a ∧ = 0or(2) 𝖢⌊a = 0 for any vector a,wemusthave𝖢 = 0. 𝖬 F E 𝜶 𝖬⌊𝜶 3.12 Prove that if a bidyadic ∈ 2 2 satisfies ∧ = 0forany one-form 𝜶, 𝖬 must be a multiple of the unit bidyadic 𝖨(2)T .Show that the same problem can be expressed for a metric bidyadic 𝖬 E E 𝖬 ⌊⌊𝜶𝜶 𝜶 𝖬 m ∈ 2 2 in the form m = 0forall implies m = 훼 ⌊𝖨(2)T eN . 𝖰 E E 3.13 Find the solutions ∈ 1 1 to the bidyadic equation 𝖰(2) ⌊𝖨(2)T . = QeN Hint: operate both sides by ⌋⌋𝜶𝜶. 3.14 (1) Derive from (3.10) the Cayley–Hamilton equation for the 𝖢−1 𝖢 ≠ bidyadic , assuming det = a6 0. (2) Show that if the bidyadic 𝖢 satisfies the Cayley–Hamilton equation (3.10), the bidyadic 𝖣 = 𝖨(4)⌊⌊𝖢T satisfies the same equation. 3.15 Check that the bidyadic 𝖣 = 𝖢2 defined by 1 𝖢 = 𝖡 ∧𝖨, 𝖡 = (𝖨 − a𝜶), o∧ o 2 with a|𝜶 = 4 satisfies the equation (3.42) 𝖣3 𝖣2 𝖣 𝖨(2) a0 + a2 + A4 + a6 = 0 𝖢⌋ with coefficients defined by (3.43)–(3.45). Check also that eN is antisymmetric. 3.16 Find the inverse of a bidyadic of the form 𝖬 횷 횲 훼𝖨(2) F E = C + D + ∈ 2 2 defined by two two-forms 횷, 횲, two bivectors C, D and a scalar 훼, by assuming that the inverse bidyadic has the form 𝖬−1 = 횷C′ + 횲D′ + 훼′𝖨(2) Problems 77

where C, D and 훼 are replaced by the unknown quantities C′, D′, 훼′. 3.17 As a verification of the equation (3.42) show that the trace ofits left-hand side of vanishes. 3.18 Check the relations (3.31). 3.19 Verify that another form for the inverse of the skewon bidyadic 𝖡 ∧𝖨 𝖡 o∧ with tr o = 0, is 1 (𝖡 ∧𝖨)−1 = (𝖨(4)⌊⌊𝖡(2)T − 𝖡(2)). o∧ 𝖡(3) o o tr o 𝖬 F E 3.20 Show that if a bidyadic ∈ 2 2 is of full rank and satisfies 𝜶 ∧ 𝖬⌊𝜶 = 0 for all one-forms 𝜶,itmustbeoftheform𝖬 = 훼𝖨(2)T 𝖬 ⌊⌊𝜶𝜶 . As a conclusion show that the condition m = 0for 𝜶 𝖬 𝖤 𝖬 E E all yields m = A when m ∈ 2 2 is a metric bidyadic. 3.21 Find the form of the Cayley–Hamilton equation for the skewon 𝖢 𝖡 ∧𝖨 bidyadic = o∧ . Hint: apply the identity 𝖠∧𝖡 | 𝖢∧𝖣 𝖠|𝖢 ∧ 𝖡∧𝖣 𝖠|𝖣 ∧ 𝖡|𝖢 ( ∧ ) ( ∧ ) = ( )∧( ∧ ) + ( )∧( ) to expanding various powers 𝖢p. 3.22 Applying the results of the previous problem, find the determinant 𝖡 ∧𝖨 of the skewon bidyadic o∧ . 𝖢 E F 𝖨(3)⌊⌊𝖢T 3.23 Show that when a bidyadic ∈ 2 2 satisfies = 0, it is a principal bidyadic, possessing no axion or skewon part. Hint: 𝖨(3)⌊⌊𝜶 |𝜶 𝖨 𝖨∧ 𝜶 start from the identity a = (a ) − ∧a , valid for any vector a and one-form 𝜶. 3.24 Show by applying the result of Problem 3.6 for the metric bidyadic 𝖣 = 𝖤 + AB − BA that the inverse of a skewon–axion bidyadic may contain a principal part. 3.25 Verifythe rule (3.159) for the inverse of a skewon–axion bidyadic.

CHAPTER 4

Special Dyadics and Bidyadics

4.1 ORTHOGONALITY CONDITIONS

Let us consider conditions between dyadics and bidyadics which may loosely be called orthogonality conditions.

4.1.1 Orthogonality of Dyadics 𝖠 𝖡 E F Let us assume that the two dyadics , ∈ 1 1 satisfy the condition 𝖠|𝖡 = 0. (4.1) It is obvious that, if either one of the dyadics has an inverse, the other one must be zero. Let us ignore that case and assume that the rank of 𝖠 is less than 4 and find the conditions for 𝖡. From symmetry, similar conditions are valid for 𝖠 when starting from the properties of 𝖡.

AofRank3 Assuming that the dyadic 𝖠 is of rank 3 satisfying 𝖠(4) = 0, 𝖠(3) ≠ 0, (4.2) it can be expressed as 𝖠 𝜺 𝜺 𝜺 ≠ = a1 i + a2 2 + a3 3, a1 ∧ a2 ∧ a3 0, (4.3)

79 80 CHAPTER 4 Special Dyadics and Bidyadics

𝜺 where i, i = 1 − 4 is some basis of one-forms. Expanding the dyadic 𝖡 as ∑ 𝖡 𝜷 = ei i, (4.4) where the vectors ej make the reciprocal basis, from (4.1) we obtain 𝜷 𝜷 𝜷 a1 1 + a2 2 + a3 3 = 0, (4.5) 𝜷 𝜷 𝜷 𝖡 which implies 1 = 2 = 3 = 0. Thus, the dyadic must be of the form 𝖡 𝜷 . = e4 4 (4.6) This means that if 𝖠 is of rank 3, 𝖡 must be of rank 1, of the form 𝖡 = b𝜷 with b satisfying 𝖠|b = 0. Similarly, if 𝖡 is of rank 3, 𝖠 must be of rank 1.

AofRank2 Assuming that the dyadic 𝖠 is of rank 2 satisfying 𝖠(3) = 0, 𝖠(2) ≠ 0, (4.7) it can be expanded as 𝖠 𝜺 𝜺 𝖠(2) 𝜺 ≠ . = a1 1 + a2 2, = a1 ∧ a2 12 0 (4.8) Expressing the dyadic 𝖡 again as (4.4), we obtain 𝜷 𝜷 a1 1 + a2 2 = 0, (4.9) 𝜷 𝜷 𝖡 which from (4.8) leads to 1 = 2 = 0. Thus, must be of the form 𝖡 𝜷 𝜷 = e3 3 + e4 4, (4.10) whence it is at most of rank 2. Another way to express the relation is 𝖡 𝖨(4)⌊⌊ 𝖠(2)∧𝖢 T = ( ∧ ) , (4.11) 𝖢 E F where ∈ 1 1 is an arbitrary dyadic. Because of symmetry, if 𝖠 is of rank 1, 𝖡 maybeatmostofrank3. As a conclusion, the sum of the ranks of the two dyadics satisfying the orthogonality condition (4.1) must be ≤4. 4.2 Nilpotent Dyadics and Bidyadics 81

4.1.2 Orthogonality of Bidyadics 𝖢 𝖣 E F Assuming that , ∈ 2 2 are two bidyadics satisfying 𝖢|𝖣 = 0, (4.12) after reasoning similar to that above, the orthogonality condition (4.12) can be shown to require that the sum of the ranks of the two bidyadics may not exceed 6. Skipping the other possibilities, let us consider as a representative example the case when the bidyadic 𝖢 is of rank 3, that is, of the form 𝖢 𝚽 𝚽 𝚽 = A1 1 + A2 2 + A3 3, (4.13) 𝚽 where both the bivectors Ai and the two-forms i are linearly indepen- 𝚽 dent. If the two-forms i are completed to a basis by adding three two- 𝚽 𝚽 𝚽 forms 4, 5 and 6 and assuming a basis of bivectors Bi, i = 1, … ,6 𝚽 | 𝛿 𝖣 satisfying i Bj = i,j, expressing the bidyadic in the form ∑6 𝖣 𝚿 = Bj j, (4.14) j=1 (4.1) leads to 𝚿 𝚿 𝚿 . A1 1 + A2 2 + A3 3 = 0 (4.15) 𝚿 𝚿 From linear independence of the bivectors Aj we now obtain 1 = 2 = 𝚿 𝖣 3 = 0. Thus, is reduced to 𝖣 𝚿 𝚿 𝚿 . = B4 4 + B5 5 + B6 6 (4.16) In conclusion, the case when the two bidyadics satisfy the orthogonality condition (4.12) and 𝖢 is of rank 3, the bidyadic 𝖣 must be of rank 3 or less. The expansion (4.16) is only limited by requiring that the 𝖢| bivectors B4, B5 and B6 satisfy Bi = 0, that is, they are eigenbivectors of 𝖢 corresponding to the eigenvalue zero.

4.2 NILPOTENT DYADICS AND BIDYADICS

𝖠 E F Dyadics ∈ 1 1 are called nilpotent of grade p when they satisfy 𝖠p = 0, p > 1. (4.17) 82 CHAPTER 4 Special Dyadics and Bidyadics

Let us consider solutions to the grade 2 equation

𝖠2 = 0. (4.18)

Obviously, any solution to (4.18) satisfies (4.17) for any p > 2 as well, but the converse is not true in general. Replacing 𝖡 by 𝖠 in (4.1) we may conclude that the 𝖠 maybeatmostofrank2satisfying 𝖠(3) ⇒ 𝖠 𝜶 𝜶 . = 0, = a1 1 + a2 2 (4.19) Inserting this in (4.18) requires that the four conditions 𝜶 | i aj = 0, i, j = 1, 2 (4.20) must be satisfied by the two vectors and one-forms. The dyadic

𝖠 = a𝜶,tr𝖠 = a|𝜶 = 0 (4.21) is a simple special case. 𝖢 E F Considering the case of nilpotent bidyadics ∈ 2 2 of grade 2 satisfying

𝖢2 = 0, (4.22) the orthogonality condition (4.12) requires that it must be at most of rank 3, whence it can be represented in the form 𝖢 𝚽 𝚽 𝚽 . = A1 1 + A2 2 + A3 3 (4.23) Assuming that the rank is 3, the bivector and two-form triples must be linearly independent. Inserted in (4.18), on the left side we obtain a sum of nine terms each of which must vanish individually, |𝚽 . Ai j = 0, i, j = 1, 2, 3 (4.24) Thus, each term in (4.23) is a nilpotent bidyadic of rank 1 and each two terms are orthogonal bidyadics. As a simple example of a nilpotent bidyadic of grade 2 and rank 3 we may write 𝖢 𝜺 𝜺 𝜺 . = Ae12 34 + Be23 14 + Ce31 24 (4.25) Lower-rank nilpotent bidyadics can be obtained by omitting one or two terms of the expansion. 4.3 Projection Dyadics and Bidyadics 83

4.3 PROJECTION DYADICS AND BIDYADICS

 E F Projection dyadics ∈ 1 1 are defined by the condition 2 = , (4.26) which imply p = , p ≥ 1, (4.27) and trp = tr, p ≥ 1. (4.28) The dyadic defined by ′ = 𝖨 −  (4.29) can be called the complementary projection dyadic because it satisfies similar conditions, ′2 = 𝖨 − 2 + 2 = 𝖨 −  = ′, (4.30) ′p = ′,tr′p = tr′ = 4 − tr, p ≥ 1, (4.31) and the orthogonality |′ = ′| = 0. (4.32) Applying the identities 1 tr𝖠(2) = ((tr𝖠)2 − tr𝖠2), (4.33) 2 1 tr𝖠(3) = ((tr𝖠)3 − 3(tr𝖠)(tr𝖠2) + 2tr𝖠3), (4.34) 6 1 tr𝖠(4) = ((tr𝖠)4 − 6(tr𝖠)2(tr𝖠2) + 8(tr𝖠)(tr𝖠3) 24 + 3(tr𝖠2)2 − 6tr𝖠4), (4.35) 𝖠 E F valid for any dyadic ∈ 1 1, we obtain 1 tr(2) = (tr)(tr − 1), (4.36) 2 1 tr(3) = (tr)(tr − 1)(tr − 2), (4.37) 6 1 tr(4) = (tr)(tr − 1)(tr − 2)(tr − 3). (4.38) 24 84 CHAPTER 4 Special Dyadics and Bidyadics

Inserting these and (4.27) in the Cayley–Hamilton equation

4 − (tr)3 + (tr(2))2 − (tr(3)) + (tr(4))𝖨 = 0, (4.39) yields the following condition for the projection dyadic, ( ) 1 (tr − 1)(tr − 2)(tr − 3)  − (tr)𝖨 = 0. (4.40) 4 Vanishing of the last bracketed dyadic of (4.40) corresponds to  equaling zero or the unit dyadic. The other three solutions denoted by    1, 2, 3 are restricted by  . tr i = i, i = 1, 2, 3 (4.41) From (4.36)–(4.38), we now obtain  ⇒ (2) (3) (4) tr 1 = 1, tr 1 = tr 1 = tr 1 = 0, (4.42)  ⇒ (2) (3) (4) tr 2 = 2, tr 2 = 1, tr 2 = tr 2 = 0, (4.43)  ⇒ (2) (3) (4) . tr 3 = 3, tr 3 = 3, tr 3 = 1, tr 3 = 0 (4.44)

Inserting the projection dyadic 𝖠 =  in the dyadic identities 𝖠(4)⌊⌊𝖨T 𝖠 𝖠(3) 𝖠(2)∧𝖠2 𝖠(4) 𝖨(3) = (tr ) − ∧ = (tr ) , (4.45) 𝖠(3)⌊⌊𝖨T 𝖠 𝖠(2) 𝖠∧𝖠2 = (tr ) − ∧ , (4.46) 𝖠(2)⌊⌊𝖨T = (tr𝖠)𝖠 − 𝖠2, (4.47) they are respectively reduced to

(tr(4))𝖨(3) = (tr − 3)(3), (4.48) (3)⌊⌊𝖨T = (tr − 2)(2), (4.49) (2)⌊⌊𝖨T = (tr − 1). (4.50)

Applying (4.42)–(4.44) they are further reduced to (3) (2)  1 = 0, 1 = 0, tr 1 = 1, (4.51) (3) (2) 2 = 0, tr 2 = 1, (4.52) (3) . tr 3 = 1 (4.53) 4.4 Unipotent Dyadics and Bidyadics 85

From these conditions it follows that the different classes of projection dyadics can be represented by dyadic polynomials of the form  0 = 0, (4.54)  𝝅 1 = p1 1, (4.55)  𝝅 𝝅 2 = p1 1 + p2 2, (4.56)  𝝅 𝝅 𝝅 3 = p1 1 + p2 2 + p3 3, (4.57)  𝝅 𝝅 𝝅 𝝅 𝖨 4 = p1 1 + p2 2 + p3 3 + p4 4 = , (4.58) 𝝅 where the vectors pi and one-forms j are taken from any reciprocal system of basis vectors and one-forms. Actually, the different projection E F dyadics act as unit dyadics in certain subspaces of 1 1. Projection bidyadics can be studied along similar lines of analysis.  E F Adopting the same symbol, the projection bidyadic ∈ 2 2 and its complement ′ = 𝖨(2) −  satisfy 2 = , ′2 = ′, |′ = ′| = 0, (4.59) Without further proof we can extend the previous result to mappings of the six-dimensional bivector space as  0 = 0, (4.60)  𝚷 1 = P1 1, (4.61)  𝚷 𝚷 2 = P1 1 + P2 2, (4.62)  𝚷 𝚷 𝚷 3 = P1 1 + P2 2 + P3 3, (4.63)  𝚷 𝚷 𝚷 𝚷 4 = P1 1 + P2 2 + P3 3 + P4 4, (4.64)  𝚷 𝚷 𝚷 𝚷 𝚷 5 = P1 1 + P2 2 + P3 3 + P4 4 + P5 5, (4.65)  𝚷 𝚷 𝚷 𝚷 𝚷 𝚷 6 = P1 1 + P2 2 + P3 3 + P4 4 + P5 5 + P6 6, (4.66) |𝚷 훿 where the bivectors and two-forms satisfy Pi j = i,j. The last projec-  𝖨(2) tion bidyadic equals the unit bidyadic, 6 = .

4.4 UNIPOTENT DYADICS AND BIDYADICS

𝖪 E F The class of unipotent dyadics ∈ 1 1 is defined by the condition 𝖪2 = 𝖨. (4.67) 86 CHAPTER 4 Special Dyadics and Bidyadics

𝖪 must be of rank 4 because (𝖪2)(4) = (𝖪(4))2 = (tr𝖪(4))2𝖨(4) = 𝖨(4) (4.68) implies either tr𝖪(4) = 1or=−1. Considering solutions in terms of a dyadic  in the form 𝖪 =±(𝖨 − 2), (4.69) (4.67) is reduced to 𝖪2 = 𝖨 − 4 + 42 = 𝖨⇒2 = . (4.70) Because the dyadic  satisfies (4.26), it must be a projection dyadic. Thus, any unipotent dyadic can be defined in terms of some projection dyadic through the representation (4.69) and, conversely, any projection dyadic can be represented in terms of some unipotent dyadic. The unipotent dyadic 𝖪 can be expressed in a more symmetric form in terms of the complementary projection dyadic ′ defined by (4.29) as 𝖪 =±(′ − ). (4.71) Because of (4.54)–(4.58), the projection dyadics satisfy ′ 𝖨   i = − i = 4−i, (4.72) whence the possible unipotent dyadics have the following three basic forms, 𝖪 =±𝖨, (4.73) 𝖪 𝝅 𝝅 𝝅 𝝅 =±(p1 1 − p2 2 − p3 3 − p4 4), (4.74) 𝖪 𝝅 𝝅 𝝅 𝝅 . =±(p1 1 + p2 2 − p3 3 − p4 4) (4.75) 𝖪 E F The class of unipotent bidyadics ∈ 2 2 can be defined by the condition 𝖪2 = 𝖨(2). (4.76) Their classification can be made along the previous pattern in termsof  E F projection bidyadics i ∈ 2 2 as 𝖪 𝖨(2)  ′  . i =±( − 2 i) =±( i − i) (4.77) 4.5 Almost-Complex Dyadics 87

There are four different types of unipotent bidyadics which can be listed as follows: 𝖪 =±𝖨(2), (4.78) 𝖪 𝚷 𝚷 𝚷 𝚷 𝚷 𝚷 =±(P1 1 − P2 2 − P3 3 − P4 4 − P5 5 − P6 6), (4.79) 𝖪 𝚷 𝚷 𝚷 𝚷 𝚷 𝚷 =±(P1 1 + P2 2 − P3 3 − P4 4 − P5 5 − P6 6), (4.80) 𝖪 𝚷 𝚷 𝚷 𝚷 𝚷 𝚷 . =±(P1 1 + P2 2 + P3 3 − P4 4 − P5 5 − P6 6) (4.81)

4.5 ALMOST-COMPLEX DYADICS

An n-dimensional real-valued vector space is said to possess an almost complex structure [9] (or complex structure [8]) if there exists a real- 𝖩 E F valued dyadic ∈ 1 1 satisfying 𝖩2 =−𝖨. (4.82) Let us call such a dyadic 𝖩 by the name almost-complex (AC) dyadic. Inserting expansions of 𝖩 and 𝖨 in (4.82) as ( ) ( ) ∑ 2 ∑ ∑ ∑ 𝜺 𝜺 훿 𝜺 Ji,jei j = ei j Ji,kJk,j = (− i,j)ei j, (4.83) i,j i,j k i,j and requiring that the determinant of the coefficient matrices on each side must be the same: 2 훿 n (det[Ji,j]) = det[− i,j] = (−1) , (4.84) we can conclude that, for 𝖩 to be a real-valued dyadic, n must be an even number. For n ≤ 4, we may concentrate to vector spaces of dimensions n = 2andn = 4. For n = 4, the Cayley–Hamilton equation (2.119) for an AC dyadic 𝖩 is reduced to 𝖨 + tr𝖩 𝖩 − tr𝖩(2) 𝖨 − tr𝖩(3) 𝖩 + tr𝖩(4) 𝖨 = 0. (4.85) Since a real AC dyadic 𝖩 obviously cannot be a multiple of 𝖨,the coefficients of 𝖩 and 𝖨 must individually vanish, which leads to the conditions tr𝖩(3) − tr𝖩 = 0, (4.86) tr𝖩(4) − tr𝖩(2) + 1 = 0. (4.87) 88 CHAPTER 4 Special Dyadics and Bidyadics

Applying the identities (4.33)–(4.35) for 𝖠 replaced by 𝖩 these conditions take the form tr𝖩((tr𝖩)2 + 4) = 0, (4.88) (tr𝖩)2((tr𝖩)2 + 4) = 0. (4.89) For real 𝖩 we must have tr𝖩 = 0 which leads to a set of other conditions, tr𝖩 = tr𝖩3 = tr𝖩(3) = 0, (4.90) tr𝖩2 =−4, tr𝖩(2) = 2, tr𝖩4 = 4, tr𝖩(4) = 1. (4.91) This means that all AC dyadics 𝖩 satisfy the same Cayley–Hamilton 휆 equation, that is, they have the same set of four eigenvalues i. Actually, 휆 the eigenvalues i can be solved by writing (4.82) as (𝖩 − j𝖨)|(𝖩 + j𝖨) = 0, (4.92) whence the eigenproblem 𝖩| 휆 𝝃 |𝖩 휆 𝝃 xi = ixi i = i , (4.93) has four eigenvectors of the form 𝖩 𝖨 | 휆 xi = ( ± j ) a, i =±j, (4.94) ≠ 𝖩 for any vector a yielding xi 0. Because is trace free, each of +j and −j is a double eigenvalue. Denoting them by 휆 휆 휆 휆 1 = j, 2 =−j, 3 = j, 4 =−j, (4.95) they can be checked through the expressions (4.90), (4.91) as 𝖩 휆 휆 휆 휆 tr = 1 + 2 + 3 + 4 = 0, 𝖩(2) 휆 휆 휆 휆 휆 휆 휆 휆 휆 휆 tr = 1 2 + 1 3 + 1 4 + 2 3 + 3 4 = 2, 𝖩(3) 휆 휆 휆 휆 휆 휆 휆 휆 휆 휆 휆 휆 tr = 1 2 3 + 1 2 4 + 1 3 4 + 2 3 4 = 0, 𝖩(4) 휆 휆 휆 휆 . tr = 1 2 3 4 = 1 Applying the similarity transformation (3.12) for dyadics satisfying the same Cayley–Hamilton equation, two AC dyadics are similar in the sense that they can be transformed to one another as 𝖩 𝖣|𝖩 |𝖣−1 2 = 1 (4.96) 𝖣 E F in terms of some dyadic ∈ 1 1 possessing an inverse. Thus, knowing 𝖩 one AC dyadic 1, all other AC dyadics can be obtained through a suitable dyadic 𝖣 from (4.96). 4.5 Almost-Complex Dyadics 89

4.5.1 Two-Dimensional AC Dyadics To find explicit expressions for AC dyadics let us find a basis expansion of the 2D AC dyadic satisfying 𝖩2 𝖨 𝜺 𝜺 . =− 12 =−(e1 1 + e2 2) (4.97) Since in this case 𝖩(3) = 0and𝖩(4) = 0, (4.86) and (4.87) are reduced to tr𝖩 = 0, tr𝖩(2) = 1, (4.98) the latter of which can be written in the dyadic form as 𝖩(2) 𝖩(2) 𝖨(2) 𝖨(2) 𝜺 . = tr 12 = 12 = e12 12 (4.99) Actually, the conditions (4.98) are equivalent to (4.97) in the 2D case, as can be easily verified. One can show that the most general solution to (4.97) can be expressed in the form 𝖩 𝜺 𝜺 = e1 2 − e2 1, (4.100) 𝜺 𝜺 in terms of any two 2D vectors e1, e2 and one-forms 1, 2 satisfying |𝜺 훿 ei j = i,j. The result of the form (4.100) can be obtained by starting from the expansion 𝖩 𝜺 𝜺 𝜺 𝜺 . = J1,1e1 1 + J1,2e1 2 + J2,1e2 1 + J2,2e2 2 (4.101) Details of the proof are left as an exercise. The 2D AC dyadic (4.100) performs the operation 𝖩| 𝖩| x = (x1e1 + x2e2) =−x1e2 + x2e1, (4.102) which equals 90◦ rotation of vectors in a plane. A more general rotation by an angle 휃 can be represented by the dyadic 𝖱 휃 휃 𝖨 휃 𝖩 휃𝖩. ( ) = cos 12 + sin = e (4.103)

4.5.2 Four-Dimensional AC Dyadics To find a 4D solution of (4.82) let us consider the eigenproblem (4.93) whose eigenvalues are defined by (4.95). The corresponding eigenvec- 𝝃 tors and eigen one-forms xi, i can be assumed to satisfy the orthogonality conditions |𝝃 xi j = 0 (4.104) 90 CHAPTER 4 Special Dyadics and Bidyadics for all i ≠ j because of linear independence of the eigenvectors. Further, the eigenvectors and eigen one-forms can be assumed normalized as |𝝃 xi i = 1, (4.105) whence they form reciprocal sets of vector and one-form bases. In this case the unit dyadic can be represented by 𝖨 𝝃 𝝃 𝝃 𝝃 = x1 1 + x2 2 + x3 3 + x4 4, (4.106) and the AC dyadic has the eigenexpansion 𝖩 𝝃 𝝃 𝝃 𝝃 . = j(x1 1 − x2 2 + x3 3 − x4 4) (4.107) It can be easily verified that (4.107) satisfies (4.82) and the conditions (4.90), (4.91). For 𝖩 to be real, all eigenvectors and/or eigen one-forms cannot be real valued. The form of the 4D AC dyadic (4.107) suggests that it can be split in two 2D AC dyadics by defining 𝖩 𝖩 𝖩 = 12 + 34, (4.108) 𝖩 𝝃 𝝃 12 = j(x1 1 − x2 2), (4.109) 𝖩 𝝃 𝝃 34 = j(x3 3 − x4 4), (4.110) while 𝖨 can be split in two complementary projection dyadics as 𝖨 𝖨 𝖨 = 12 + 34, (4.111) 𝖨 𝝃 𝝃 12 = x1 1 + x2 2, (4.112) 𝖨 𝝃 𝝃 . 34 = x3 3 + x4 4 (4.113) 𝖩 𝖩 Because of the orthogonality conditions (4.104) 12 and 34 satisfy 𝖩2 𝖨 𝖩2 𝖨 𝖩 |𝖩 𝖩 |𝖩 . 12 =− 12 34 =− 34, 12 34 = 34 12 = 0 (4.114) 𝖩 𝖩 To express 12 and 34 in terms of real-valued components in the form of (4.100), let us require that the following vectors and one-forms are real, 1 1 e1 = √ (x1 − jx2) e2 = √ (x2 − jx1), (4.115) 2 2 1 1 e3 = √ (x3 − jx4) e4 = √ (x4 − jx3), (4.116) 2 2 4.6 Almost-Complex Bidyadics 91

𝜺 1 𝝃 𝝃 𝜺 1 𝝃 𝝃 1 = √ ( 1 + j 2) 2 = √ ( 2 + j 1), (4.117) 2 2 𝜺 1 𝝃 𝝃 𝜺 1 𝝃 𝝃 . 3 = √ ( 3 + j 4) 4 = √ ( 4 + j 3) (4.118) 2 2 Actually, these quantities form a reciprocal set of basis vectors and one-forms because they satisfy |𝜺 훿 . ei j = i,j (4.119) The two AC dyadics have the simple appearance 𝖩 𝜺 𝜺 𝖩 𝜺 𝜺 12 = e1 2 − e2 1 34 = e3 4 − e4 3, (4.120) It is left as an exercise to show that the eigenvectors and eigen one- forms must be alternatively real and imaginary valued to obtain the representation (4.108) in terms of real-valued dyadics (4.120). Applying the similarity transformation (4.96), any AC dyadic can be expressed in the form 𝖩 𝖣| 𝖩 𝖩 |𝖣−1 = ( 12 + 34) (4.121) 𝖣 E F in terms of a suitable dyadic ∈ 1 1 possessing an inverse [8], p. 156.

4.6 ALMOST-COMPLEX BIDYADICS

𝖩 E F AC bidyadics ∈ 2 2, are defined as solutions of the bidyadic equation 𝖩2 =−𝖨(2). (4.122) Since the underlying bivector space is even-dimensional, n = 6, one may expect to find real-valued bidyadic solutions for (4.122). The traces of different powers of the bidyadic 𝖩 satisfy tr𝖩6 =−tr𝖩4 = tr𝖩2 =−6, (4.123) tr𝖩5 =−tr𝖩3 = tr𝖩. (4.124) The equation (𝖩2 + 𝖨(2))3 = 𝖩6 + 3𝖩4 + 3𝖩2 + 𝖨(2) = 0, (4.125) 92 CHAPTER 4 Special Dyadics and Bidyadics implies that the AC bidyadic 𝖩 has triple eigenvalues of both +j and −j, whence its trace vanishes, ∑ 𝖩 휆 . tr = i = 0 (4.126) Taking this into account, (4.125) can be shown to equal the Cayley– Hamilton equation (3.10). Because any AC bidyadic can be obtained from a given AC bidyadic through a similarity transformation, let us construct a solution through 𝖯 E F 𝖯−1 E F a spatial dyadic s ∈ 2 1 possessing a spatial inverse s ∈ 1 2 satisfying 𝖯 |𝖯−1 𝖨(2) 𝖯−1|𝖯 𝖨 . s s = s , s s = s (4.127) 𝖭 𝖭 E F To simplify the notation, let us define two bidyadics a, b ∈ 2 2 by 𝖭 𝖯 𝜺 𝖭 𝖯−1. a = s ∧ 4, b = e4 ∧ s (4.128) They are actually nilpotent, 𝖭2 𝖭2 a = b = 0, (4.129) and give rise to two complementary projection bidyadics as  𝖭 |𝖭 𝖯 𝜺 | 𝖯−1 ab =− a b =−( s ∧ 4) (e4 ∧ s ) 𝖯 |𝖯−1 𝖨(2) = s s = s , (4.130)  𝖭 |𝖭 𝖯−1 | 𝖯 𝜺 ba =− b a =−(e4 ∧ s ) ( s ∧ 4) 𝖨 𝜺 𝖨 ∧ 𝜺 . =−e4 ∧ s ∧ 4 = s∧e4 4 (4.131) because they satisfy 2  2    𝖨(2). ab = ab, ba = ba, ab + ba = (4.132) Now one can show that an AC bidyadic can be expressed as 𝖩 𝖭 𝖭 . = a + b (4.133) In fact, applying the above properties we have 𝖩2 𝖭 |𝖭 𝖭 |𝖭   𝖨(2). = a b + b a =− ab − ba =− (4.134) As an example, let us consider the symmetric spatial dyadics 𝖦 s = e1e1 + e2e2 + e3e3, (4.135)  𝜺 𝜺 𝜺 𝜺 𝜺 𝜺 s = 1 1 + 2 2 + 3 3, (4.136) 4.7 Modified Closure Relation 93 in terms of which the two nilpotent bidyadics can be represented as 𝖭 ⌊ 𝜺 𝜺 𝜺 𝜺 a = e123 s ∧ 4 = e23 14 + e31 24 + e12 34, (4.137) 𝖭 𝖦 ⌋𝜺 𝜺 𝜺 𝜺 . b = e4 ∧ s 123 =−e14 23 − e24 31 − e34 12 (4.138) In this case the AC bidyadic becomes 𝖩 𝜺 𝜺 𝜺 𝜺 𝜺 𝜺 . = e23 14 + e31 24 + e12 34 − e14 23 − e24 31 − e34 12 (4.139)

4.7 MODIFIED CLOSURE RELATION

A condition of the form 𝖬|𝖬 = 훼𝖨(2)T , (4.140) 𝖬 F E is called the closure relation for bidyadics ∈ 2 2 [9, 11]. Actually, this requires that 𝖬 is a multiple of a unipotent bidyadic (4.76). A somewhat similar condition either for the bidyadic 𝖬 or for the metric 𝖬 ⌊𝖬 E E bidyadic m = eN ∈ 2 2 𝖬T ⋅ 𝖬 𝖬T | ⌊𝖬 훼𝖤 = (eN ) = , (4.141) 𝖬T ⋅ 𝖬 𝖬T | 𝜺 ⌊𝖬 훼𝖤. m m = m ( N m) = (4.142) can be called the modified closure relation [12] for 훼 ≠ 0. Equation (4.141) implies that 𝖬|𝚽 ⋅ 𝖬|𝚽 훼𝚽 ⋅ 𝚽 ( 1) ( 2) = 1 2, (4.143) 𝚽 𝚽 훼 𝖬 is valid for any two-forms 1, 2.When = 1, the bidyadic resembles a rotation operator in its property of preserving the dot product of two-forms. One can show through 3D expansions that, to satisfy (4.141), the bidyadic 𝖬 can be expressed in the form 𝖬 𝖯(2)T 𝖬 ⌊𝖯(2)T = , m = eN , (4.144) or in the form 𝖬 𝜺 ⌊𝖰(2) 𝖬 𝖰(2) = N , m = , (4.145) 𝖯 E F 𝖰 E E for some dyadic ∈ 1 1 or metric dyadic ∈ 1 1 [12]. Thus, the modified closure condition restricts the number of parameters defining the bidyadic 𝖬 from 6 × 6 = 36 to 4 × 4 = 16. The two cases (4.144) 94 CHAPTER 4 Special Dyadics and Bidyadics and (4.145) are called P-solutions and Q-solutions, respectively. Let us consider the problem without 3D expansions.

4.7.1 Equivalent Conditions Let us start from the result (see Problem 3.20) that if a metric bidyadic 𝖷 E E ∈ 2 2 satisfies 𝖷⌊⌊𝝂𝝂 = 0 (4.146) for all one-forms 𝝂,itmustbeoftheform 𝖷 = 훼𝖤, (4.147) for some scalar 훼. Thus, we may state that if the bidyadic 𝖬 satisfies (𝖬T ⋅ 𝖬)⌊⌊𝝂𝝂 = 0 (4.148) for all 𝝂, it must satisfy a modified closure relation of the form (4.141) for some 𝜶. Since the converse is also true, (4.141) and (4.148) are equivalent. 𝖠 E E Now it is also known that if a metric dyadic ∈ 1 1 satisfies 𝖠||𝝁𝝁 = 0 (4.149) 𝝁 F 𝖠 for all ∈ 1, the symmetric part of must vanish. Thus, if the symmet- 𝖠 𝖬T ⋅ 𝖬 ⌊⌊𝝂𝝂 E E ric dyadic = ( ) ∈ 1 1 satisfies (4.149), it must vanish. Combining these two conditions, we may state that if 𝖬 satisfies ((𝖬T ⋅ 𝖬)⌊⌊𝝂𝝂)||𝝁𝝁 = (𝖬|(𝝂 ∧ 𝝁)) ⋅ (𝖬|(𝝂 ∧ 𝝁)) = 0 (4.150) for all one-forms 𝝂 and 𝝁, it must satisfy the modified closure relation (4.141) for some scalar 𝜶. Since the converse is also true, (4.141) and (4.150) are equivalent. The same condition for the modified bidyadic is 𝖬T ⋅ 𝖬 ⌊⌊𝝂𝝂 ||𝝁𝝁 𝖬 | 𝝂 𝝁 ⋅ 𝖬 | 𝝂 𝝁 . (( m m) ) = ( m ( ∧ )) ( m ( ∧ )) = 0 (4.151)

4.7.2 Solutions The condition (4.150) can be interpreted so that a bidyadic 𝖬 satisfying the modified closure condition (4.141) maps any simple two-form 𝝂 ∧ 𝝁 to another simple two-form. A simple two-form can be expressed either in terms of two one-forms 𝜶, 𝜷 as 𝜶 ∧ 𝜷, or in terms of two vectors a, b, 4.7 Modified Closure Relation 95

𝜺 ⌊ as N (a ∧ b). These lead to two representations, 𝖬|(𝝂 ∧ 𝝁) = 𝜶 ∧ 𝜷, (4.152) 𝖬| 𝝂 𝝁 𝜺 ⌊ ( ∧ ) = N (a ∧ b), (4.153) the latter of which can be more conveniently expressed for the metric 𝖬 bidyadic m as 𝖬 | 𝝂 𝝁 . m ( ∧ ) = a ∧ b (4.154) Let us consider these two cases separately.

P-Solution Since the left side of (4.152) is linearly dependent on 𝝁, either of the one-forms 𝜶 or 𝜷 must be a linear function of 𝝁. There is no loss of generality if we express 𝜷 = 𝖯T |𝝁, (4.155) 𝖯 E F in terms of some dyadic ∈ 1 1.From ((𝖬⌊𝝂) − 𝜶 ∧ 𝖯T )|𝝁 = 0, (4.156) which must be valid for any one-form 𝝁, we obtain the condition 𝖬⌊𝝂 = 𝜶 ∧ 𝖯T . (4.157) Because this implies 𝜶 ∧ 𝖯T |𝝂 = 0, (4.158) the one-form 𝜶 must be of the form 𝜶 = M𝖯T |𝝂, (4.159) for some scalar M. The condition (4.157) then becomes 𝖬⌊𝝂 = M(𝖯T |𝝂) ∧ 𝖯T = M𝖯(2)T ⌊𝝂. (4.160) Since this must be valid for all 𝝂, we arrive at one solution to the modified closure relation, 𝖬 = M𝖯(2)T . (4.161) When the scalar M is absorbed in the dyadic 𝖯, this equals (4.144), the P-solution. 96 CHAPTER 4 Special Dyadics and Bidyadics

Q-Solution Starting from (4.154), we can proceed similarly. Assuming that the vector b depends linearly on 𝝁 as b = 𝖰|𝝁, (4.162) 𝖰 E E where ∈ 1 1 is some metric dyadic, (4.154) yields 𝖬 ⌊𝝂 𝖰. m = a ∧ (4.163) This implies a ∧ 𝖰|𝝂 = 0, (4.164) whence a must be of the form a = M𝖰|𝝂. (4.165) (4.163) now becomes 𝖬 ⌊𝝂 𝖰|𝝂 𝖰 𝖰(2)⌊𝝂. m = M( ) ∧ = M (4.166) Because this must be valid for all 𝝂, the modified medium bidyadic must be of the form 𝖬 𝖰(2). m = M (4.167) Again, if the scalar M is absorbed by the dyadic 𝖰, this yields (4.145), the Q-solution.

4.7.3 Testing the Two Solutions Since the modified closure relation (4.141) or (4.142) has two setsof solutions, one may ask how to distinguish whether a given solution 𝖬 𝖬 or m is a P-solution or a Q-solution? 𝖬 E E Considering the metric bidyadic m ∈ 2 2 satisfying the modified closure equation (4.142) let us first study the Q-solution by double contracting by one-forms 𝜶𝜶 and finding its double-wedge powers as 𝖬 ⌊⌊𝜶𝜶 𝖰(2)⌊⌊𝜶𝜶 𝖰||𝜶𝜶 𝖰 𝖰|𝜶 𝜶|𝖰 m = = ( ) − ( )( ), (4.168) 𝖬 ⌊⌊𝜶𝜶 (2) 𝖰||𝜶𝜶 𝖰||𝜶𝜶 𝖰(2) 𝖰∧ 𝖰|𝜶 𝜶|𝖰 ( m ) = ( )(( ) − ∧(( )( ))), (4.169) 𝖬 ⌊⌊𝜶𝜶 (3) 𝖰||𝜶𝜶 2 𝖰||𝜶𝜶 𝖰(3) 𝖰(2)∧ 𝖰|𝜶 𝜶|𝖰 ( m ) = ( ) (( ) − ∧(( )( ))) 𝖰||𝜶𝜶 2 𝜺 𝜺 ||𝖰(4) ⌊⌊𝜶𝜶 . = ( ) ( N N )(eNeN ) (4.170) 4.7 Modified Closure Relation 97

Let us now do the same operations with the P-solution. 𝖬 ⌊⌊𝜶𝜶 ⌊𝖯(2)T ⌊⌊𝜶𝜶 m = (eN ) ⌊ 𝜶 𝖯T |𝜶 𝖯T . = eN ( ∧ ( ) ∧ ) (4.171) Denoting the simple bivector by ⌊ 𝜶 𝖯T |𝜶 eN ( ∧ ( )) = a ∧ b, (4.172) we have 𝖬 ⌊⌊𝜶𝜶 ⌊𝖯T |𝖯T |𝖯T m = (a ∧ b) = b(a ) − a(b ), (4.173) 𝖬 ⌊⌊𝜶𝜶 (2) |𝖯(2)T ( m ) = (a ∧ b)(a ∧ b) , (4.174) 𝖬 ⌊⌊𝜶𝜶 (3) . ( m ) = 0 (4.175) The above conditions, valid for any one-form 𝜶, allow one to test the 𝖬 character of a metric bidyadic m satisfying the modified closure relation (4.142) in the case when it is of full rank, whence both 𝖰 and 𝖯 must be of rank 4, satisfying 𝖰(4) ≠ 0and𝖯(4) ≠ 0. The different possibilities 𝖬 ⌊⌊𝜶𝜶 can be listed as based on the rank of the dyadic m as follows. r 𝖬 𝖬 ⌊⌊𝜶𝜶 (3) ≠ 𝜶 m satisfies ( m ) 0 for some one-form . Because of (4.175), it cannot correspond to a P-solution, whence it must be a Q-solution. r 𝖬 𝖬 ⌊⌊𝜶𝜶 (3) 𝜶 𝖬 ⌊⌊𝜶𝜶 (2) ≠ m satisfies ( m ) = 0forall and ( m ) 0for some 𝜶. From (4.170) a full-rank Q-solution must then satisfy 𝖰||𝜶𝜶 𝖬 ⌊⌊𝜶𝜶 (2) ≠ = 0, which from (4.169) is contrary to ( m ) 0. Thus, this case must correspond to a P-solution. r 𝖬 𝖬 ⌊⌊𝜶𝜶 (2) 𝜶 𝖬 ⌊⌊𝜶𝜶 ≠ m satisfies ( m ) = 0forall and m 0forsome 𝜶. This is satisfied by a Q-solution with antisymmetric 𝖰 while 𝖬 ⌊⌊𝜶𝜶 ≠ from (4.174) we have a ∧ b = 0 whence m 0 cannot be satisfied by any full-rank P-solution. r 𝖬 𝖬 ⌊⌊𝜶𝜶 𝜶 𝖬 ≠ m satisfies m = 0forall and m 0. This requires 𝖬 ⌊𝖨(2)T that m = MeN which is again a P-solution. Thus, the Q- and P-solutions are distinct when 𝖬 is of full rank in which case 𝖯 or 𝖰 must also be of full rank. Actually, one can show that there are no full-rank solutions for 𝖰 and 𝖯 satisfying an equation of the form 𝖰(2) ⌊𝖯(2)T . = eN (4.176) In the case when 𝖬 is not of full rank, the P- and Q-solutions may both be possible. Further considerations are left as exercises. 98 CHAPTER 4 Special Dyadics and Bidyadics

PROBLEMS

4.1 Show that the condition 𝖠 |𝖡 s s = 0 𝖠 𝖡 E F 𝖠(2) for two spatial dyadics s, s ∈ 1 1 requires that either s = 0 𝖡(2) or s = 0 must be satisfied, whence one of the two dyadics 𝜶 must be of the form as s for some spatial vector as and spatial 𝜶 one-form s. 4.2 Show that the dyadic 𝖪 𝖨 𝜺 𝜺 𝜺 𝜺 = + e3 4 + e4 3 − e3 3 − e4 4 𝜺 is unipotent when ei and i form reciprocal vector and one-form bases. Find its expansion either in the form (4.74) or (4.75). 𝖠 E F 4.3 Show that a dyadic ∈ 1 1 satisfying 𝖠3 = 0, 𝖠2 ≠ 0 also satisfies tr𝖠i = 0fori = 1, … , 4. Hint: use the Cayley– Hamilton equation. 4.4 Show that the conditions (4.98) for the 2D AC dyadics are equiv- 𝖩(2)⌊⌊𝖨T alent to (4.97). Hint: start from the expression 12. 4.5 Show that the most general solution to (4.97) can be expressed in a form similar to (4.100)∑ by starting from the most general 𝖩 𝜺 2D dyadic expansion = Ji,jei j. Hint: It may help to denote 휃 휃 without loss of generality J1,1 = sinh , J1,2 = A cosh and J2,1 = −(1∕A)cosh휃. 4.6 Starting from (4.82) and the definition of the AC dyadic, show 𝖩 E F that the AC dyadic ∈ 1 1 satisfies the following properties: 𝖩(2)⌊⌊𝖨T = 𝖨, 𝖩(3)⌊⌊𝖨T 𝖩∧𝖨 = ∧ , 𝖩(3)⌊⌊𝖨(2)T = 𝖩, 𝖨(3)⌊⌊𝖩T 𝖨∧𝖩 =− ∧ , 𝖨(4)⌊⌊𝖩(3)T =−𝖩, 𝖨(3)⌊⌊𝖩(2)T = 𝖨. Problems 99

𝖬 F E 4.7 Show that a bidyadic ∈ 2 2 consisting of a real-valued AC dyadic 𝖩 defined by 𝖬 = M𝖩(2)T does not contain a skewon component. Show also that such a bidyadic contains a nonzero principal part. 𝖩 E F 𝖨(4)⌊⌊𝖩T 4.8 Show that if ∈ 2 2 is an AC bidyadic, is also an AC bidyadic. 4.9 Derive a general expression for the AC bidyadic 𝖩 by expressing 𝖩 in its spatial and temporal components as 𝖩 𝖠 𝖡 𝖢 𝜺 𝖣 𝜺 . = s + e4 ∧ s + s ∧ 4 + e4 ∧ s ∧ 4 𝖡 Assuming that s has an inverse, show that only two 3D equations are needed to define the problem. 𝝃 4.10 Show that the eigenvectors xi and eigen one-forms i of the AC dyadic 𝖩 must be alternatively real and imaginary valued to obtain the representation (4.108) in terms of real-valued dyadics (4.120). 4.11 Given two AC dyadics defined by 𝖩 𝜺 𝜺 𝜺 𝜺 = e1 2 − e2 1 + e3 4 − e4 3, 𝖩′ 𝜺 𝜺 𝜺 𝜺 = e1 3 + e2 4 − e3 1 − e4 2, (4.177) 𝖣 E F 𝖩′ 𝖣|𝖩|𝖣−1 find a dyadic ∈ 1 1 yielding the connection = between them. 훽 F E 4.12 Find the most general unipotent spatial dyadics s ∈ 1 1 and 훼 F E spatial bidyadics s ∈ 2 2. 4.13 Show that the equation 𝖰(2) ⌊𝖨(2)T = eN 𝖰 E E has no solutions ∈ 1 1. 4.14 Study possible solutions 𝖰 to the equation 𝖰(2) ⌊𝖯(2)T = eN 𝖯 E F for different values of the rank of the dyadic ∈ 1 1. 4.15 Show that, by applying the transformation bidyadic 𝖣 𝖨(2) 𝖦′| ∧ 𝜺 = s + ( s s)∧e4 4, 100 CHAPTER 4 Special Dyadics and Bidyadics

𝖦 an AC bidyadic defined by a spatial metric dyadic s,whichis not symmetric, can be transformed to an AC bidyadic defined by 𝖦′ a symmetric spatial dyadic s. 4.16 Consider the Cayley–Hamilton equation (2.119) for the unipotent dyadic 𝖪 and find the possible values oftr𝖪. 4.17 Show that for an AC bidyadic satisfying (4.122), the equation (4.125) actually equals the Cayley–Hamilton equation (3.10). CHAPTER 5

Electromagnetic Fields

The electromagnetic field quantities and their sources are represented by differential forms of different grades. Applying the notation introduced by Deschamps [1, 13], the electromagnetic fields are represented by 𝚿 𝚽 F 𝜸 F two-forms , ∈ 2 and their source by a three-form ∈ 3.The fields and the source can be decomposed in spatial and temporal parts as 𝚿 𝜺 = D − H ∧ 4, (5.1) 𝚽 𝜺 = B + E ∧ 4, (5.2) 𝜸 𝝔 𝜺 . = − J ∧ 4 (5.3) F Applying conventional notation, D, B ∈ 2 represent spatial electro- F magnetic two-forms while H, E ∈ 1 are two spatial electromagnetic 𝝔 F F one-forms. ∈ 3 denotes the spatial charge three-form and J ∈ 2 is 𝜺 𝜺 𝜺 the spatial current two-form. In the following, 1, 2, 3 form the spatial 𝜺 basis of one-forms while 4 denotes the temporal one-form.

5.1 FIELD EQUATIONS

5.1.1 Differentiation Operator E The spacetime vector x ∈ 1 is defined by 𝜏 𝜏 x = r + e4 , = ct, (5.4)

101 102 CHAPTER 5 Electromagnetic Fields

where r is the spatial part of x and e4 denotes the temporal basis vector. 𝜺 | ct = 4 x is the time variable multiplied by the velocity of light c.The F differential operator d ∈ 1 can be defined by the dyadic operation dx = 𝖨T . (5.5) Expanding x in Cartesian coordinates as ∑ x = e1x1 + e2x2 + e3x3 + e4x4 = eixi, (5.6) i where the basis vectors ei are assumed independent of x, (5.5) can be written as ∑ ∑ 𝖨T 𝜺 . dx = dxiei = = iei (5.7) i i 𝜺 The one-forms i defined by 𝜺 dxi = i (5.8) form a basis reciprocal to the vector basis {ei}. The differential operator can be expanded as ∑ d = 𝜺 𝜕 = d + 𝜺 𝜕𝜏 , (5.9) i xi s 4 i where ds denotes differentiation in the spatial coordinates x1, x2, x3.It is often of interest to consider time-harmonic electromagnetic sources and fields in terms of complex time dependence ej𝜔t = ejk𝜏 , k = 𝜔∕c. (5.10) In such a case the differential operator can be replaced by 𝜺 . d = ds + jk 4 (5.11) The differential operator d allows many operations on multivector and multiform functions of x: r Differentiation of a scalar 𝜙(x) yields a one-form, 𝜙 F . d (x) ∈ 1 (5.12) r Exterior differentiation of a p-form yields a p + 1-form, 𝝓 F 𝝓 F (x) ∈ 1, d ∧ (x) ∈ 2, (5.13) 𝚽 F 𝚽 F . (x) ∈ 2, d ∧ (x) ∈ 3 (5.14) 5.1 Field Equations 103

r Interior differentiation of vector yields a scalar, | E d f(x) ∈ 0, (5.15) r Contraction differentiation of a p-vector yields a p − 1-vector, E ⌋ E F(x) ∈ 2, d F(x) ∈ 1, (5.16) E ⌋ E . k(x) ∈ 3, d k(x) ∈ 2 (5.17) r “No-sign” differentiation yields a dyadic, 𝝓 F 𝝓 F F (x) ∈ 1, d (x) ∈ 1 1, (5.18) E F E . f(x) ∈ 1, df(x) ∈ 1 1 (5.19)

Various algebraic rules can be transformed to differentiation rules when replacing a one-form by the differential operator d and taking proper care of the differentiation. For example, the bac-cab rule (1.43) for two vector functions of x yields the differentiation rule ⌋ | | | | d (f ∧ g) = fc(d g) − gc(d f) + gc (df) − fc (dg), (5.20) where subscript ()c denotes that the quantity is considered constant, that is, independent of x in differentiation, while df and dg are dyadic quantities. The property d ∧ d ∧ 𝝃(x) = 0 is valid for any p-form function 𝝃 F (x) ∈ p. The inverse of this is known as de Rham’s theorem [14]. If d ∧ 𝝃 = 0forsomep-form, there exists a p − 1-form 𝜻 such that we can express 𝝃 = d ∧ 𝜻(x). However, 𝜻(x) may be multi-valued if it is defined in a multiply connected region such as the doughnut.

5.1.2 Maxwell Equations In differential-form formalism the Maxwell equations can be given a simple form [1, 4, 9, 13, 15–20] as d ∧ 𝚽 = 0, (5.21) d ∧ 𝚿 = 𝜸. (5.22) In some cases magnetic charges and currents may be included as equivalent sources. To distinguish the sources, (5.22) and (5.21) will 104 CHAPTER 5 Electromagnetic Fields then be replaced by 𝚽 𝜸 d ∧ = m, (5.23) 𝚿 𝜸 d ∧ = e, (5.24) with subscript ()e added to the electric sources, 𝜸 𝝔 𝜺 e = e − Je ∧ 4, (5.25) 𝜸 𝝔 𝜺 . m = m − Jm ∧ 4 (5.26)

When there are no magnetic sources present, the subscript ()e will be omitted. Inserting (5.2) in (5.21) and expanding, 𝚽 𝜺 d ∧ = d ∧ B + d ∧ E ∧ 4 (5.27) 𝜕 𝜺 = ds ∧ B + (ds ∧ E + 𝜏 B) ∧ 4 = 0, (5.28) after separating the spatial and temporal parts, the corresponding 3D equations for spatial field quantities are obtained,

ds ∧ B = 0, (5.29) 𝜕 . ds ∧ E + 𝜏 B = 0 (5.30) Similarly, inserting (5.1) and (5.3) in (5.22) yields 𝝔 ds ∧ D = , (5.31) 𝜕 . ds ∧ H − 𝜏 D = J (5.32) The spatial differential forms D, B, E, H can be transformed to Gibbsian vector fields Bg, Eg, Hg, Dg, Jg according to the rules given in Appendix B by ⌊ E Dg = e123 D ∈ 1, (5.33) ⌊ E Bg = e123 B ∈ 1, (5.34) 𝖦 | E Eg = s E ∈ 1, (5.35) 𝖦 | E Hg = s H ∈ 1, (5.36) where 𝖦 s = e1e1 + e2e2 + e3e3 (5.37) is a chosen spatial metric dyadic defining the dot product. The Gibbsian sources 𝜚 |𝜚 E g = e123 ∈ 0, (5.38) ⌊ E Jg = e123 J ∈ 1 (5.39) 5.1 Field Equations 105 are respectively scalar and vector valued. Applying these rules and 𝖦 | defining the vector operator ∇= s ds, the Maxwell equations can be written for the Gibbsian vector fields and sources as (details are leftas an exercise) ⋅ ∇ Bg = 0, (5.40) 𝜕 ∇×Eg + tBg = 0, (5.41) ⋅ 𝜚 ∇ Dg = g, (5.42) 𝜕 . ∇×Hg − tDg = Jg (5.43) In their original form, the Maxwell equations consisted of 20 scalar equations [21], which were later expressed by Heaviside [22] and Hertz [23] in a more compact form equivalent to (5.40)–(5.43).

5.1.3 Potential One-Form 𝜸 With no magnetic sources, m = 0, de Rham’s theorem and (5.21) imply that the two-form 𝚽 can be expressed in terms of a one-form, the potential 𝝓,as 𝚽 = d ∧ 𝝓. (5.44) Expanding 𝝓 in its spatial and temporal parts as 𝝓 𝜑𝜺 = A − 4, (5.45) F 𝝋 where A ∈ 1 is the spatial one-form and is the scalar potential, we obtain 𝚽 𝜕 𝜑 𝜺 = ds ∧ A + ( 𝜏 A − ds ) ∧ 4, (5.46) which can be split as

B = ds ∧ A, (5.47) 𝜕 𝜑. E = 𝜏 A − ds (5.48) The corresponding Gibbsian equations for the scalar potential 𝝋 and the 𝖦 | vector potential Ag = s A are ⋅ Bg =∇ Ag, (5.49) 𝜕 𝜑. Eg = 𝜏 Ag −∇ (5.50) 106 CHAPTER 5 Electromagnetic Fields

5.2 MEDIUM EQUATIONS

In macroscopic consideration, physical materials can be represented by a set of medium parameters which define relations between the electromagnetic field quantities.

5.2.1 Medium Bidyadics Assuming a simple homogeneous, time-invariant and linear medium, the electromagnetic two-forms are related by an algebraic equation of the form 𝚿 = 𝖬|𝚽, (5.51) 𝖬 F E where ∈ 2 2 is the medium bidyadic mapping two-forms to two- forms. Since 𝖬 corresponds to a 6 × 6 matrix, it can be represented by 36 parameters when expanded in some basis. Another way to define the medium equation is 𝚽 = 𝖭|𝚿, (5.52) and there is basically no reason to prefer one to the other one, although (5.51) will mostly be applied in the sequel. When the bidyadic 𝖬 has an inverse, we have the relations 𝖭 = 𝖬−1, 𝖬 = 𝖭−1. (5.53) However, if there is no inverse for 𝖬 or for 𝖭, only one of the equations (5.51), (5.52) makes sense. Another useful form for the medium equation, equivalent to (5.51), is ⌊𝚿 𝖬 |𝚽 E eN = m ∈ 2, (5.54) where 𝖬 ⌊𝖬 E E m = eN ∈ 2 2 (5.55) is a metric bidyadic called the modified medium bidyadic (in [1] the 𝖬 modified medium bidyadic was denoted by g). Similarly, the modified version of the medium bidyadic 𝖭 is defined by ⌊𝚽 𝖭 |𝚿 𝖭 ⌊𝖭. eN = m , m = eN (5.56) 5.2 Medium Equations 107

While the two medium bidyadics are related by (5.53), the two modified bidyadics satisfy the more complicated relations 𝖭 ⌊⌊𝖬−1 𝖬 ⌊⌊𝖭−1. m = eNeN m , m = eNeN m (5.57)

5.2.2 Potential Equation Knowledge of the medium bidyadic 𝖬 allows one to form an equation for the potential one-form 𝝓. For a homogeneous and time-invariant 𝖬 𝖬 medium and m have no functional dependence on x, whence inserting (5.51) and (5.44) in (5.22) yields d ∧ 𝚿 = d ∧ 𝖬|𝚽 = d ∧ 𝖬|(d ∧ 𝝓) = 𝜸. (5.58) Written in the form (d ∧ 𝖬⌊d)|𝝓 = 𝜸, (5.59) it represents a mapping from a one-form 𝝓 to a three-form 𝜸. Contracting ⌊ by eN and applying the rule ⌊ 𝜷 𝜸 𝜷⌋ ⌊𝜸 eN ( ∧ ) =− (eN ), (5.60) valid for any one-form 𝜷 and two-form 𝜸, yields ⌊ 𝖬⌊ |𝜶 ⌋𝖬 ⌊ |𝜶 ⌊𝜸. eN (d ∧ d) =−(d m d) = eN (5.61) This is of the form 𝖣(d)|𝝓 = g, (5.62) where the second-order dyadic operator is defined by 𝖣 ⌋𝖬 ⌊ 𝖬 ⌊⌊ E F (d) =−d m d = m dd ∈ 1 1, (5.63) and the source vector by ⌊𝜸 E . g = eN ∈ 1 (5.64)

5.2.3 Expansions of Medium Bidyadics The medium bidyadic can be expanded in its spatial and temporal parts as 𝖬 훼 휖′ 𝜺 휇−1 𝜺 훽 = + ∧ e4 + 4 ∧ + 4 ∧ ∧ e4, (5.65) 108 CHAPTER 5 Electromagnetic Fields where the four spatial dyadics belong to different spaces as 훼 F E 휖′ F E 휇−1 F E 훽 F E . ∈ 2 2, ∈ 2 1, ∈ 1 2, ∈ 1 1 (5.66) Because the dyadics 훼, 휖′, 휇−1, 훽 map spatial one-forms or two-forms to spatial one-forms or two-forms, each of them can be represented by 9 scalar parameters in a given basis. Inserting (5.65) in (5.51) we have 𝚿 𝜺 훼 𝜺 휇−1 | 휖′ 𝜺 훽 | = D − H ∧ 4 = ( + 4 ∧ ) B + ( + 4 ∧ ) E, (5.67) which corresponds to the spatial medium equations D = 훼|B + 휖′|E, (5.68) H = 휇−1|B + 훽|E. (5.69) In electrical engineering it is customary to express the Gibbsian field relations between other field quantities [24, 25]. In differential-form formalism the equations take the form D = 휖|E + 휉|H, (5.70) B = 휁|E + 휇|H. (5.71) In this notation all four dyadics are members of the same space, 휖 휉 휁 휇 F E 휖 휖′ , , , ∈ 2 1. One must note that and are different dyadics in general. They are the same when 훼 = 0and훽 = 0 or, equivalently, when 휉 = 휁 = 0. However, 휇 is the same dyadic in both representations. One can define media which cannot be represented in theform (5.68), (5.69) or in the form (5.70), (5.71). For example, if the dyadic 휇 does not have an inverse, (5.69) is obviously not a good choice. On the other hand, one can find examples of media for which 휇−1 exists but 휇 does not, in which case (5.70) and (5.71) cannot be applied. When (5.68), (5.69) and (5.70), (5.71) are both valid representations of the medium conditions, relations between the two sets of medium dyadics can be expressed as 휖′ = 휖 − 휉|휇−1|휁, 훼 = 휉|휇−1, 훽 =−휇−1|휁, (5.72) and 휖 = 휖′ − 훼|휇|훽, 휉 = 훼|휇, 휁 =−휇|훽. (5.73) Inserting (5.72) in (5.67) as 𝜺 휉|휇−1 𝜺 휇−1 | 휖 휉|휇−1|휁 𝜺 휇−1|휁 | D − H ∧ 4 = ( + 4 ∧ ) B + ( − − 4 ∧ ) E, (5.74) 5.2 Medium Equations 109 leads to another expansion for the medium bidyadic, 𝖬 휖 𝜺 𝖨T 휉 |휇−1| 𝖨(2)T 휁 . = ∧ e4 + ( 4 ∧ s + ) ( s − ∧ e4) (5.75) On the other hand, starting from the representation 𝜺 휁|휖−1 𝜺 휖−1 | 휇 𝜺 휖−1|휉 휁|휖−1|휉 | B + E ∧ 4 = ( − 4 ∧ ) D + ( + 4 ∧ − ) H, (5.76) yields the following expansion for the medium bidyadic 𝖭, 𝖭 𝖬−1 휇 𝜺 𝖨T 휁 |휖−1| 𝖨(2)T 휉 . = =− ∧ e4 − ( 4 ∧ s − ) ( s + ∧ e4) (5.77) One can verify that (5.75) and (5.77) satisfy

𝖬|𝖭 = 𝖭|𝖬 = 𝖨(2)T , (5.78) details of which are left as an exercise. Actually, there exist six possibilities to present the medium equations between two pairs of the four spatial fields E, H, B, D. Equations (5.68) and (5.69) could be inverted and presented as B, E in terms of D, H. Similarly, (5.70) and (5.71) could be written as E, H in terms of D, B. Yet another, equally valid, set of equations is

D = 𝖠|B + 𝖡|H, (5.79) E = 𝖢|B + 𝖣|H, (5.80) and, conversely, we can express B, H in terms of D, E. These last two possibilities have been ignored in the literature. However, there may well exist media whose simplest representation requires the form (5.79), (5.80) or its inverse.

5.2.4 Gibbsian Representation Let us now consider the connection of the medium dyadics and the corresponding Gibbsian dyadics. The Gibbsian vector fields Dg, Bg, Eg, Hg defined by (5.33)–(5.36) are related by Gibbsian medium dyadics 휖 휉 휁 휇 E E g, g, g, g, which all belong to the same space 1 1, 휖 ⋅ 휉 ⋅ 휖 | 휉 | Dg = g Eg + g Hg = g E + g H, (5.81) 휁 ⋅ 휇 ⋅ 휁 | 휇 | . Bg = g Eg + g Hg = g E + g H (5.82) 110 CHAPTER 5 Electromagnetic Fields

Here we apply rules given in Appendix B. Relations between the two sets of medium dyadics can be represented as 휖′ 𝜺 ⌊ 휖 휉 |휇−1|휁 = 123 ( g − g g g), (5.83) 휇 𝜺 ⌊휇 = 123 g, (5.84) 훼 𝜺 ⌊ 휉 |휇−1 ⌋ = 123 ( g g ) e123, (5.85) 훽 휇−1|휁 . =− g g (5.86) Inserting the latter set in (5.65) produces an expansion of the medium bidyadic in terms of the Gibbsian dyadics, 𝖬 𝜺 ⌊휖 𝜺 ⌊휉 𝜺 𝖨T |휇−1| 𝖨⌋ 휁 = 123 g ∧ e4 + ( 123 g + 4 ∧ ) g ( e123 − g ∧ e4), (5.87) and the corresponding expansion for the modified medium bidyadic is 𝖬 ⌊𝖬 obtained from m = eN as 𝖬 휖 ∧ ⌊𝖨T 휉 |휇−1| 𝖨⌋ 휁 . m = g∧e4e4 − (e123 + e4 ∧ g) g ( e123 − g ∧ e4) (5.88) Its inverse can be expressed following the form of (5.77) as 𝖬−1 𝜺 𝜺 ⌊⌊휇 𝜺 𝖨T 𝜺 ⌊휁 |휖−1| 𝖨 𝜺 휉 ⌋𝜺 . m =− 123 123 g − ( 4 ∧ − 123 g) g ( ∧ 4 + g 123) (5.89) One can verify that the two dyadics satisfy the relations 𝖬 |𝖬−1 𝖨(2) 𝖬−1|𝖬 𝖨(2)T m m = , m m = , (5.90) details of which are left as an exercise.

5.3 BASIC CLASSES OF MEDIA

5.3.1 Hehl–Obukhov Decomposition Based on the medium bidyadic 𝖬, the Hehl–Obukhov decomposition (3.74) 𝖬 𝖬 𝖬 𝖬 = 1 + 2 + 3, (5.91) and the corresponding modified medium bidyadic decomposition (3.85), 𝖬 𝖬 𝖬 𝖬 m = m1 + m2 + m3, (5.92) 5.3 Basic Classes of Media 111 allows one to define three basic classes of electromagnetic media. Properties of the different components in the decomposition as defined in Section 3.3 can be summarized as follows.

r 𝖬 The axion part 3 consists of the trace of the medium bidyadic 𝖬, 1 𝖬 = tr𝖬 𝖨(2)T . (5.93) 3 6 r 𝖬 The skewon part 2 satisfies 𝖨(4)⌊⌊𝖬 𝖬T 2 =− 2 , (5.94) 𝖬 whence it is trace free, tr 2 = 0. The modified medium bidyadic 𝖬 𝖬 m2 consists of the antisymmetric part of m, 1 𝖬 = (𝖬 − 𝖬T ). (5.95) m2 2 m m As was shown by (3.86), the skewon part of 𝖬 can be expressed 𝖡 E F in terms of a trace-free dyadic o ∈ 1 1 in the form 𝖬 𝜺 ⌊𝖬 𝖡 ∧𝖨 T 2 = N m2 = ( o∧ ) (5.96) 𝖡 𝖬 and the dyadic o is obtained from 2 through the operation 1 𝖡 = (𝖬 ⌊⌊𝖨)T . (5.97) o 2 2 r 𝖬 𝖬 𝖬 𝖬 The principal part is defined by 1 = − 2 − 3. It satisfies 𝖨(4)⌊⌊𝖬 𝖬T 𝖬 1 = 1 ,tr1 = 0, (5.98) 𝖬 whence the modified medium bidyadic m1 is symmetric. It 𝖬 equals the symmetric part of m without the axion part, 1 1 𝖬 = (𝖬 + 𝖬T ) − tr𝖬 e ⌊𝖨(2)T . (5.99) m1 2 m m 6 N In (3.78) it was shown that that if the medium bidyadic satisfies the condition 𝖬⌊⌊𝖨 = 0, (5.100) 𝖬 𝖬 it equals its principal part, = 1. Thus, (5.100) represents the definition of a medium with no skewon and axion components. 112 CHAPTER 5 Electromagnetic Fields

The Hehl–Obukhov decomposition of the medium bidyadic can be applied to define basic classes of media, independent of any basis systems by requiring that the medium bidyadic 𝖬 does not contain all 𝖬 of the three Hehl–Obukhov parts i. There are six different possibilities given in the following table.

𝖬 Medium class

𝖬 1 Principal media 𝖬 2 Skewon media 𝖬 3 Axion media 𝖬 𝖬 1 + 2 Axion-free media 𝖬 𝖬 1 + 3 Principal–axion media 𝖬 𝖬 2 + 3 Skewon–axion media

The classification could also be made through the medium bidyadic 𝖭 defined in (5.52). However, one must note that the medium classes do not coincide with those defined through 𝖬 in all cases. Because the inverse of a symmetric and an antisymmetric bidyadic is respectively symmetric and antisymmetric, the classes of skewon media and principal–axion media are the same in both definitions and so is the class of axion media. However, in the other cases the two definitions do not lead to the same classes in all cases. As an example, a principal medium defined through the bidyadic 𝖬 may contain an axion component when defined in terms of the bidyadic 𝖭. In fact, one can show that 𝖬⌊⌊𝖨 = 0 does not imply the condition 𝖬−1⌊⌊𝖨 = 0. However, there does exist media satisfying both conditions. 𝖬 𝜺 ⌊𝖰(2) As an example we may consider = N with a symmetric dyadic 𝖰 E E 𝖬 𝖬−1 ∈ 1 1. In this case one can show that tr = 0 implies tr = 0, whence both 𝖬 and 𝖭 = 𝖬−1 are principal bidyadics. Similarly, for the 𝖬 𝖡∧𝖨 T class of skewon–axion media, from = ( ∧ ) it does not follow that 𝖬−1 𝖠∧𝖨 T 𝖡 𝖡 the inverse bidyadic has a similar form = ( ∧ ) , unless = o happens to be trace free. In the following we will define the medium classes through the medium bidyadic 𝖬 whenever it exists.

5.3.2 3D Expansions The two solutions of the bidyadic eigenproblem (3.48) rewritten here as 𝖨(4)⌊⌊𝖬 𝖬T ± =± ±, (5.101) 5.3 Basic Classes of Media 113

𝖬 𝖬 𝖬 coincide with the principal–axion bidyadic + = 1 + 3 and the ske- 𝖬 𝖬 won bidyadic − = 2. Let us study the properties of the corresponding spatial medium dyadics by inserting the expansion (5.65) in (5.101) and equating similar terms on both sides. This leads to the set of double relations

훼 𝜺 ⌊⌊훽T ± =∓ 123e123 ±, (5.102) 휖′ 𝜺 ⌊⌊휖′T ± =± 123e123 ±, (5.103) 휇−1 𝜺 ⌊⌊휇−1T ± =± 123e123 ± , (5.104) 훽 𝜺 ⌊⌊훼T ± =∓ 123e123 ±, (5.105) for the principal–axion (+) and skewon (–) dyadics. They can be written more compactly as

( ) ( )T 휖′ 훼 휖′ −훼 𝖨(3)⌊⌊ =± . (5.106) s 훽 휇−1 훽 휇−1 ± − ± The relations (5.102) and (5.105) are actually equivalent. Expressing (5.103) and (5.105) as

⌊휖′ ⌊휖′ T E E e123 ± =±(e123 ±) ∈ 1 1, (5.107) ⌊휇−1 ⌊휇−1 T E E e123 ± =±(e123 ± ) ∈ 2 2, (5.108)

⌊휖′ ⌊휇−1 shows us that the dyadics e123 + and e123 + must be symmetric and ⌊휖′ ⌊휇−1 the dyadics e123 − and e123 − must be antisymmetric. Thus, for a skewon medium [26], neither 휖′ nor 휇−1 possesses a spatial inverse. In 휇 particular, since a dyadic − does not exist, there is no Gibbsian representation of the form (5.81), (5.82) for the skewon medium. In contrast, a Gibbsian representation exists for the principal–axion medium in which case the medium dyadic matrix is symmetric, ( ) ( ) ( ) 휖 휉 T 휖T 휁 T 휖 휉 g g g g g g . 휁 휇 = 휉T 휇T = 휁 휇 (5.109) g g g g + g g + + 휖 휇 It is known that symmetric medium dyadics g and g can be realized by dielectric and magnetic crystals while the symmetric part of the 휉 휁 T dyadic g + g exists in a Tellegen medium [27] and the antisymmetric part effectively arises when a medium is in uniform motion [28]. 114 CHAPTER 5 Electromagnetic Fields

𝖬 𝖬 𝖬 Considering the pure principal medium, = + with tr = 0, from (5.65) the latter condition becomes tr훼 = tr훽. (5.110) Expressing 훼 and 훽 in terms of the Gibbsian medium dyadics as 훼 𝜺 ⌊⌊ 휉 |휇−1 훽 휇−1|휁 = 123e123 ( g g ), =− g g, (5.111) (5.110) can be written as 휉 |휇−1 휇−1|휁 . tr( g g ) + tr( g g) = 0 (5.112) 휁 휉T 휇−1 휇−1T Since from (5.109) we obtain g = g and g = g , the axion-free condition (5.110) for a principal medium is further reduced to 휉 ||휇−1 . g g = 0 (5.113) 휇−1 Because g is symmetric, this does not restrict the choice of the anti- 휉 symmetric part of g. Equations (5.109) and (5.113) define the principal medium in terms of Gibbsian dyadics. The number of free parameters of a principal medium bidyadic equals 36 − 15 − 1 = 20.

5.3.3 Simple Principal Medium As an example of a simple electromagnetic medium let us consider one 𝖢 E F depending on a spatial dyadic s ∈ 2 1 possessing a spatial inverse 𝖢−1 E F s ∈ 1 2,definedby 훼 휖′ 휖𝖢 휇−1 휇𝖢 −1 훽 . = 0, = s, = ( s) , = 0 (5.114) The corresponding medium bidyadic has the form 𝖬 휖𝖢 𝜺 휇𝖢 −1. = s ∧ e4 + 4 ∧ ( s) (5.115) Because of the form (5.115), 𝖬 is trace free, whence the medium has no axion component. The modified medium bidyadic corresponding to (5.115) can be expanded as 𝖬 ⌊ 휖𝖢 𝜺 휇𝖢 −1 m = eN ( s ∧ e4 + 4 ∧ ( s) ) (5.116) 1 =−휖e ∧ (e ⌊𝖢 ) ∧ e − e ⌊𝖢−1 (5.117) 4 123 s 4 휇 123 s 1 휇휖 ⌊𝖢 ∧ ⌊⌊ ⌊𝖢 −1 . = 휇 ( (e123 s)∧e4e4 − e123e123 (e123 s) ) (5.118) 5.3 Basic Classes of Media 115

휖 휇 𝖢 Because of the scalar factors and , the dyadic s can be normalized 𝖦 without losing the generality. Let us define a spatial metric dyadic s ∈ E E 1 1 with the normalization, 𝖦 ⌊𝖢 𝜺 𝜺 ||𝖦(3) . s = e123 s, 123 123 s = 1 (5.119) The spatial inverse is obtained through (2.211) as 𝜺 𝜺 ⌊⌊𝖦(2)T 𝖦−1 = 123 123 s = 𝜺 𝜺 ⌊⌊𝖦(2)T . (5.120) s 𝜺 𝜺 ||𝖦(3) 123 123 s 123 123 s Thus, we obtain the representation, 1 𝖬 = (휇휖𝖦 ∧e e − 𝖦(2)T ). (5.121) m 휇 s∧ 4 4 s 𝖬 𝖦 To be a principal medium m must be symmetric, whence s must 𝖦 T 𝖦 be a symmetric dyadic, s = s. In this case the modified medium bidyadic takes the simple form

𝖬 1 𝖦 (2) 휇휖𝖦 ∧ 1 𝖦 휇휖 (2). m =−휇 ( s − s∧e4e4) =−휇 ( s − e4e4) (5.122) It was pointed out in the previous section that a medium defined by a 𝖬 𝜺 ⌊𝖰(2) 𝖰 bidyadic of a form = N for any symmetric belongs to the class of principal media. A medium defined by (5.122) with symmetric 𝖦 spatial dyadic s will be called by the name simple principal medium. Comparison with the expansions (5.75), (5.87), (5.88) leads to the following sets of 3D medium dyadics describing the simple principal medium 훼 휖′ 휖𝖢 휇−1 휇𝖢 −1 훽 = 0, = s, = ( s) , = 0, (5.123) 휖 휖𝖢 휇 휇𝖢 휉 휁 = s, = s, = = 0, (5.124) 휖 휖𝖦 휇 휇𝖦 휉 휁 . g = s, g = s, g = g = 0 (5.125) Choosing a vector basis properly, we can express the symmetric metric dyadic by 𝖦 s = e1e1 + e2e2 + e3e3, (5.126) 𝖦 𝖨 whence, identifying s with the Gibbsian unit dyadic g, the simple principal medium actually corresponds to an isotropic medium in Gibbsian formalism. In terms of (5.126), the modified medium bidyadic of the 116 CHAPTER 5 Electromagnetic Fields simple principal medium has the expansion √ 휖 𝖬 . m = 휇 (e14e14 + e24e24 + e34e34 − e12e12 − e23e23 − e31e31) (5.127) 𝖢 Because the dyadic s and its inverse satisfy the relations 𝖢 |𝖢−1 𝖨(2)T 𝖢−1|𝖢 𝖨T s s = s , s s = s , (5.128) we obtain 𝖬2 휖𝖢 | 𝜺 휇𝖢 −1 𝜺 휇𝖢 −1 | 휖𝖢 = ( s ∧ e4) ( 4 ∧ ( s) ) + ( 4 ∧ ( s) ) ( s ∧ e4) 휖 휖 휖 =− 𝖨(2)T + 𝜺 ∧ 𝖨T ∧ e =− 𝖨(2)T . (5.129) 휇 s 휇 4 s 4 휇 From this we conclude that the medium bidyadic is a multiple of an AC bidyadic or a unipotent bidyadic depending on the sign of 휖∕휇.For complex medium parameters these two cases cannot be distinguished. Let us consider the equation (5.62) for the potential one-form 𝝓 in the simple principal medium defined by (5.122). Applying the rule 𝖠(2)⌊⌊𝖡T = (𝖠||𝖡)𝖠 − 𝖠|𝖡|𝖠, (5.130) 𝖠 E E 𝖡 F F valid for ∈ 1 1 and ∈ 1 1, the dyadic differential operator can be expanded as

𝖣 1 𝖦 휇휖 (2)⌊⌊ (d) =−휇 ( s − e4e4) dd 1 𝖦 || 휇휖휕2 𝖦 휇휖 =−휇 ( s dd − 휏 )( s − e4e4) 1 𝖦 | 휇휖 휕 𝖦 | 휇휖 휕 . + 휇 ( s d − e4 휏 )( s d − e4 휏 ) (5.131) The potential is not unique because 𝝓 can be replaced by 𝝓 + d휆 for any scalar function 휆 without changing the field two-form 𝚽 = d ∧ 𝝓. For uniqueness we can restrict the potential by requiring that it satisfy an additional scalar-valued condition. Choosing the condition 𝖦 | 휇휖 휕 |𝝓 ( s d − e4 휏 ) = 0, (5.132) (the Lorenz condition), which can be expressed for the spatial potential one-form A and scalar 휑 as |𝖦 | 휇휖휕 휑 d s A + 휏 = 0, (5.133) 5.4 Interfaces and Boundaries 117

(5.62) is reduced to an equation with a scalar operator, 𝖦 || 휇휖휕2 𝖦 휇휖 |𝝓 휇 ⌊𝜸. ( s dd − 휏 )( s − e4e4) =− eN (5.134) This is equivalent with ( ) 1 (𝖦 ||dd − 휇휖휕2)𝝓 =− 휇𝖦−1 − e e |(e ⌊𝜸). (5.135) s 휏 s 휖 4 4 N where we can write 𝖦 ||dd = 휕2 + 휕2 + 휕2 . This is a second-order s x1 x2 x3 hyperbolic equation (wave equation) for 휇휖 > 0 and and elliptic equation for 휇휖 < 0.

5.4 INTERFACES AND BOUNDARIES

An interface is a surface which separates two media while a boundary is a mathematical concept, a surface where the region of interest ends. At an interface the fields on one side of the surface depend on thefields on the other side of the surface through continuity conditions. At a boundary the fields satisfy certain conditions and the question whatare the fields behind the boundary is irrelevant. As is known from amedium called perfect electric conductor (PEC), a medium interface may act as a boundary. To show some examples of media with a similar property, let us consider fields at a planar interface of two media.

5.4.1 Interface Conditions 𝖬 𝖬 Let us assume that two media a and b are separated by a planar interface S defined by x3 = 0 in terms of a Cartesian coordinate function 𝚽 𝚿 x3. Because of the discontinuity in the medium, the fields , may have a discontinuity at the interface. Denoting the fields on each side of the interface by subscripts a and b, the total field two-forms can be expressed as 𝚽 𝚽 𝚽 𝚿 𝚿 𝚿 = a + b, = a + b, (5.136) with ( ) ( ) 𝚽 (x) 𝚽′ (x) a = a P (r), (5.137) 𝚿 (x) 𝚿′ (x) a ( a ) ( a ) 𝚽 𝚽′ b(x) b(x) . 𝚿 = 𝚿′ Pb(r) (5.138) b(x) b(x) 118 CHAPTER 5 Electromagnetic Fields

The primed fields are assumed to be analytical continuations of thefields without primes, extending over the surface S with no discontinuity in medium, while the characteristic functions Pa and Pb defined by 휃 휃 Pa(r) = (x3), Pb(r) = (−x3) = 1 − Pa(x3), (5.139) extract the unphysical parts of the primed fields. Here, 휃(x) denotes the Heaviside unit step function satisfying 휃(x) = 1, x > 0, 휃(x) = 0, x < 0, (5.140) 휕 휃 훿 . x (x) = (x) (5.141) Assuming no sources at the interface, the Maxwell equations around the surface S can be expanded as ( ) ( ) ( ) 𝚽(x) 𝚽 (x) 𝚽 (x) d ∧ = d ∧ a + d ∧ b 𝚿(x) 𝚿 (x) 𝚿 (x) a ( ) b ( ) 𝚽′ (x) 𝚽′ (x) = dP (r) ∧ a + dP (r) ∧ b a 𝚿′ (x) b 𝚿′ (x) ( a ) ( )b 𝚽 𝚽 a(x) − b(x) 0 . = dPa(r) ∧ 𝚿 𝚿 = (5.142) a(x) − b(x) 0 Inserting dP (r) = d휃(x ) = dx 휕 휃(x ) = 𝜺 훿(x ), (5.143) a 3 3 x3 3 3 3 we obtain conditions for the fields at the interface x3 = 0: 𝜺 𝚽 𝜺 𝚽 𝜺 𝚿 𝜺 𝚿 . 3 ∧ a = 3 ∧ b, 3 ∧ a = 3 ∧ b (5.144) Expanded in spatial field components each of the interface conditions can be split in two conditions, 𝜺 𝜺 𝜺 𝜺 3 ∧ Ba = 3 ∧ Bb, 3 ∧ Ea = 3 ∧ Eb, (5.145) 𝜺 𝜺 𝜺 𝜺 3 ∧ Da = 3 ∧ Db, 3 ∧ Ha = 3 ∧ Hb, (5.146) which correspond to the well-known conditions for the Gibbsian fields, ⋅ ⋅ e3 Bga = e3 Bgb, e3 × Ega = e3 × Egb, (5.147) ⋅ ⋅ e3 Dga = e3 Dgb, e3 × Hga = e3 × Hgb, (5.148) where e3 denotes the unit vector normal to the interface. Being local conditions, (5.144) are actually valid for any smooth surface making the interface. 5.4 Interfaces and Boundaries 119

5.4.2 Boundary Conditions Since the interface conditions (5.144) were derived from the Maxwell equations, they are valid for any two media defined by the medium 𝖬 𝖬 bidyadics a and b or through other representations like (5.52) or (5.70) and (5.71) in each of the two media. For certain media b the fields in medium a can be shown to satisfy a set of conditions at the interface 𝖬 𝖭 depending only on the medium bidyadic b or its inverse bidyadic b no matter what the fields are in the medium b. Let us consider some examples of boundary conditions defined by media b.

PMC Boundary The perfect magnetic conductor (PMC) medium in the region b corresponds to the medium bidyadic 𝖬 . b = 0 (5.149) Since this requires that the field in medium b satisfy 𝚿 b(x) = 0, (5.150) 𝚿 from (5.144) we obtain for the field a at x3 = 0 the boundary condition 𝜺 𝚿 . 3 ∧ a = 0 (5.151) This corresponds to the conditions 𝜺 𝜺 3 ∧ Da = 0, 3 ∧ Ha = 0, (5.152) and the Gibbsian conditions ⋅ . e3 Dga = 0, e3 × Hga = 0 (5.153) The same PMC boundary conditions (5.151) can be obtained for the less restrictive medium condition 𝜺 𝖬 3 ∧ b = 0, (5.154) 𝖬 which allows the bidyadic b to have some nonzero components. In fact, (5.154) requires that the fields in medium b satisfy 𝜺 𝚿 ⇒ 𝜺 𝜺 3 ∧ b = 0, 3 ∧ Db = 0, 3 ∧ Hb = 0, (5.155) whence the PMC boundary conditions (5.151), (5.152) are also obtained for this more general medium. One should notice that if the one-form 120 CHAPTER 5 Electromagnetic Fields

𝜺 3 is not independent of x (nonplanar boundary), (5.154) requires that the medium b must be inhomogeneous in a certain way. PMC boundary has obtained an important role in antenna synthesis because, unlike the PEC boundary, it does not cancel the radiation of a current element parallel to it but, rather, enhances it. Various methods to realize the PMC condition by metasurfaces under the name “High-impedance electromagnetic surface” have been suggested since 1999 [29].

PEC Boundary The PEC medium is defined by 𝖭 b = 0 (5.156) in the representation (5.52). In this case the field condition is 𝚽 . b(x) = 0 (5.157) and we obtain at x3 = 0 the boundary condition 𝜺 𝚽 . 3 ∧ a = 0 (5.158) For the spatial field components this can be expressed as 𝜺 𝜺 3 ∧ Ba = 0, 3 ∧ Ea = 0, (5.159) and, for the Gibbsian vector fields, ⋅ . e3 Bga = 0, e3 × Ega = 0 (5.160) As in the PMC case, the medium condition can be actually relaxed to 𝜺 𝖭 3 ∧ b = 0 (5.161) for the same PEC boundary conditions.

PEMC Boundary The previous two boundary conditions can be generalized by considering the axion medium defined by 𝖬 𝖨(2)T b = M , (5.162) which corresponds to the field condition 𝚿 𝚽 . b(x) − M b(x) = 0 (5.163) 5.4 Interfaces and Boundaries 121

From (5.144) we then arrive at the boundary condition 𝜺 𝚿 𝚽 3 ∧ ( a − M a) = 0, (5.164) which can be split into 𝜺 𝜺 . 3 ∧ (Da − MBa) = 0, 3 ∧ (Ha + MEa) = 0 (5.165) Since this generalizes the PMC (M = 0) and PEC (1∕M = 0) conditions, the medium has been dubbed as the perfect electromagnetic conductor (PEMC) [30]. The Gibbsian form for the PEMC boundary conditions is ⋅ . e3 (Dga − MBga) = 0, e3 × (Hga + MEga) = 0 (5.166) Following the pattern of both of the previous examples, the PEMC conditions can also be obtained by a medium satisfying the less restricting medium condition 𝜺 𝖬 𝖨(2)T . 3 ∧ ( b − M ) = 0 (5.167) The PEMC boundary has the useful property of changing the polarization of a plane wave in reflection, whence some effort has been given to realize such a boundary by a metasurface [31, 32].

DB Boundary As another restriction to the medium b we may require a set of two scalar conditions 𝜺 𝜺 𝖬 𝜺 𝜺 𝖭 3 ∧ 4 ∧ b = 0, 3 ∧ 4 ∧ b = 0, (5.168) which imply the field conditions 𝜺 𝜺 𝚿 𝜺 𝜺 ⇒ 𝜺 3 ∧ 4 ∧ b = 3 ∧ 4 ∧ Db = 0, 3 ∧ Db = 0 (5.169) 𝜺 𝜺 𝚽 𝜺 𝜺 ⇒ 𝜺 . 3 ∧ 4 ∧ b = 3 ∧ 4 ∧ Bb = 0, 3 ∧ Bb = 0 (5.170) These give rise to the boundary conditions 𝜺 𝜺 . 3 ∧ Da = 0, 3 ∧ Ba = 0 (5.171) Written in Gibbsian form, ⋅ ⋅ . e3 Dga = 0, e3 Bga = 0 (5.172) 122 CHAPTER 5 Electromagnetic Fields they are known as DB boundary conditions [33, 34]. Inserting the expansions (5.75) and (5.77) in (5.168) we obtain 𝜺 𝜺 𝖬 𝜺 𝜺 휖 𝜺 𝜺 휉 |휇−1| 3 ∧ 4 ∧ b = 3 ∧ 4 ∧ b ∧ e4 + 3 ∧ 4 ∧ b b 𝖨(2)T 휁 × ( s − b ∧ e4) = 0, (5.173) 𝜺 𝜺 𝖭 𝜺 𝜺 휇 𝜺 𝜺 휁 |휖−1| 3 ∧ 4 ∧ b =− 3 ∧ 4 ∧ b ∧ e4 + 3 ∧ 4 ∧ b b 𝖨(2)T 휉 × ( s + b ∧ e4) = 0, (5.174) which can be split in four conditions for the spatial medium dyadics as 𝜺 휉 |휇−1 𝜺 휖 휉 |휇−1|휁 3 ∧ b b = 0, 3 ∧ ( b − b b b) = 0, (5.175) 𝜺 휁 |휖−1 𝜺 휇 휁 |휖−1|휉 . 3 ∧ b b = 0, 3 ∧ ( b − b b b) = 0 (5.176) 휉 휁 Considering a medium with no magnetoelectric dyadics, b = b = 0, these reduce to 𝜺 휖 𝜺 휇 3 ∧ b = 0, 3 ∧ b = 0, (5.177) which corresponds to a medium restricted by Gibbsian dyadics satisfying ⋅ 휖 휉 휁 ⋅ 휇 . e3 gb = 0, gb = 0, gb = 0, e3 gb = 0 (5.178) The DB boundary conditions were introduced already in 1959 [35]. Novel interest followed when they were discovered to have application in electromagnetic invisibility cloaking problems in the 2000s [36, 37]. Realization of the DB boundary by a metasurface was subsequently suggested in [38].

SH Boundary A case similar to the previous one is obtained by starting from medium b restricted by conditions of the form 𝜺 𝜺 𝖬 𝜺 𝜺 𝖭 . 2 ∧ 3 ∧ b = 0, 2 ∧ 3 ∧ b = 0 (5.179) These imply conditions for the fields 𝜺 𝜺 𝚿 𝜺 𝜺 𝜺 ⇒ 𝜺 𝜺 2 ∧ 3 ∧ b = 2 ∧ 3 ∧ 4 ∧ Hb = 0, 2 ∧ 3 ∧ Hb = 0 (5.180) 𝜺 𝜺 𝚽 𝜺 𝜺 𝜺 ⇒ 𝜺 𝜺 2 ∧ 3 ∧ b =− 2 ∧ 3 ∧ 4 ∧ Eb = 0, 2 ∧ 3 ∧ Eb = 0, (5.181) 5.5 Power and Energy 123 and, consequently, yield the boundary conditions 𝜺 𝜺 𝜺 𝜺 . 2 ∧ 3 ∧ Ea = 0, 2 ∧ 3 ∧ Ha = 0 (5.182) In Gibbsian form the conditions are ⋅ ⋅ ⋅ ⋅ e2 × e3 Ega = e1 Ega = 0, e2 × e3 Hga = e1 Hga = 0, (5.183) where the unit vector e1 is parallel to the boundary surface. These conditions have been labeled as the soft-and-hard boundary conditions or SH conditions in the past [39]. SH boundary conditions have been realized by corrugated conducting surfaces since the 1940s, for example, to obtain rotationally symmetric radiation patterns in horn antennas. In all of the previous examples the boundary conditions for fields in medium a were obtained directly from the medium conditions of medium b without having to consider further restrictions to the fields imposed by the Maxwell equations. Examples of other cases will be considered in subsequent chapters.

5.5 POWER AND ENERGY

5.5.1 Bilinear Invariants Let us consider quantities which are bilinear in the field two-forms 𝚽 and 𝚿 or quadratic in either of them. The four-forms 𝚽 ∧ 𝚽, 𝚿 ∧ 𝚿, 𝚽 ∧ 𝚿 (5.184) are examples of such quantities. Expanded in terms of 3D fields they become 𝚽 𝚽 𝜺 𝜺 ∧ = (B + E ∧ 4) ∧ (B + E ∧ 4) 𝜺 = 2B ∧ E ∧ 4, (5.185) 𝚿 𝚿 𝜺 𝜺 ∧ = (D − H ∧ 4) ∧ (D − H ∧ 4) 𝜺 =−2D ∧ H ∧ 4, (5.186) 𝚽 𝚿 𝜺 𝜺 ∧ = (B + E ∧ 4) ∧ (D − H ∧ 4) 𝜺 . = (D ∧ E − B ∧ H) ∧ 4 (5.187) These quantities are invariants, that is, independent of any basis systems. 124 CHAPTER 5 Electromagnetic Fields

The dyadic product of field two-forms 𝚽𝚿 is another example of a bilinear quantity. Its antisymmetric part denoted by 1 𝖴 = (𝚽𝚿 − 𝚿𝚽) =−𝖴T ∈ F F (5.188) 2 2 2 is of special interest. Properties of such a simple bidyadic (3.87) were already studied in Chapter 3. It can be expanded as 1 𝖴 = (BD − DB − 𝜺 ∧ (ED + HB) 2 4 𝜺 𝜺 𝜺 . − (DE + BH) ∧ 4 + 4 ∧ (EH − HE) ∧ 4) (5.189) Being antisymmetric, the bidyadic 𝖴 can be represented in terms of a 𝖡 E F certain trace-free dyadic o ∈ 1 1 as 𝖴 𝜺 ⌊ 𝖡 ∧𝖨 . = N ( o∧ ) (5.190) 𝖡 o serves as another electromagnetic bilinear invariant. Applying (5.97) ⌊𝖴 to the the skewon bidyadic eN , it can be extracted as 1 𝖡 = (e ⌊𝖴)⌊⌊𝖨T . (5.191) o 2 N ∑ 1 ⌊𝚿 ⌊𝜺 𝚽⌊ ⌊𝚽 ⌊𝜺 𝚿⌊ = ((eN ) i( ei) − (eN ) i( ei)) 4 i 1 =− e ⌊(𝚿 ∧ 𝖨T ⌋𝚽 − 𝚽 ∧ 𝖨T ⌋𝚿) 4 N 1 =− e ⌊𝖳. (5.192) 2 N 𝖳 𝜺 ⌊𝖡 F F The dyadic = 2 N o ∈ 3 1 is yet another bilinear invariant, known as the stress–energy dyadic [1, 40]. Applying (3.106), we can write 1 1 𝖡2 = (e ⌊𝖳)2 =− (e e ⌊⌊𝖳)|𝖳 o 4 N 4 N N 1 = tr((e ⌊𝖳)2) 𝖨, (5.193) 16 N 𝖡 ⌊𝖳 which shows us that o and, hence eN , is a multiple of an unipotent dyadic. Thus, the inverse of the stress–energy dyadic 𝖳 can be expressed as e e ⌊⌊𝖳 𝖳−1 =−4 N N , (5.194) ⌊𝖳 2 tr((eN ) ) ⌊𝖳 2 when tr((eN ) ) does not vanish. 5.5 Power and Energy 125

5.5.2 The Stress–Energy Dyadic The stress–energy density dyadic 𝖳 was obtained above as arising from bilinear invariants. It can be given a physical meaning by starting from the Gibbsian expression for energy density, represented by the scalar quantity 1 w = (D ⋅ E + B ⋅ H ). (5.195) g 2 g g g g The corresponding three-form energy-density is obtained through the rules of Appendix B as 1 w = (D ∧ E + B ∧ H). (5.196) 2 To express this in terms of the 4D electromagnetic two-forms we start by substituting 𝚽 𝜺 B = − E ∧ 4 (5.197) 𝚿 𝜺 D = + H ∧ 4 (5.198) 𝜺 ⌊ 𝚽⌊ E =−(E ∧ 4) e4 =− e4, (5.199) 𝜺 ⌊ 𝚿⌊ H =−(H ∧ 4) e4 = e4, (5.200) whence we arrive at the expression 1 w = (−𝚿 ∧ (𝚽⌊e ) + 𝚽 ∧ (𝚿⌊e ) + 2E ∧ H ∧ 𝜺 ). (5.201) 2 4 4 4 This suggests that we should rather concentrate on the 4D three-form 1 W = (−𝚿 ∧ (𝚽⌊e ) + 𝚽 ∧ (𝚿⌊e )) 2 4 4 𝜺 . = w − E ∧ H ∧ 4 (5.202) The term E ∧ H is the Poynting two-form which corresponds to the Poynting vector Eg × Hg in the Gibbsian sense. Thus, W can be called as the energy–power density three-form. Its expression still depends on the basis through the vector e4. To find an expression independent of 𝜺 any basis, let us first multiply W dyadically by 4 from the right as 1 W𝜺 = (𝚽 ∧ (𝚿⌊e 𝜺 ) − 𝚿 ∧ (𝚽⌊e 𝜺 )), (5.203) 4 2 4 4 4 4 126 CHAPTER 5 Electromagnetic Fields

∑ 𝜺 𝖨 𝜺 and generalize it by replacing e4 4 by = ei i. In this way we arrive 𝖳 F F at what is called the Maxwell stress–energy dyadic ∈ 3 1, 1 𝖳 = (𝚽 ∧ (𝚿⌊𝖨) − 𝚿 ∧ (𝚽⌊𝖨)). (5.204) 2 Applying the rule 𝚽⌊𝖨 =−𝖨T ⌋𝚽 we obtain the form 1 𝖳 = (𝚿 ∧ (𝖨T ⌋𝚽) − 𝚽 ∧ (𝖨T ⌋𝚿)). (5.205) 2 This derivation gives a meaning to the dyadic which was obtained through algebraic manipulation of quadratic invariants, in (5.192). The converse relation between the Maxwell stress–energy dyadic 𝖳 and the dyadic invariant 𝖴 can be found as ∑ 𝖳 1 𝚿 𝜺 ⌋𝚽 𝚽 𝜺 ⌋𝚿 = ( ∧ iei − ∧ iei ) 2 i ∑ ∑ 1 𝜺 𝚿𝚽 𝚽𝚿 ⌊ 𝜺 𝖴⌊ . =− i ∧ ( − ) ei = i ∧ ei (5.206) 2 i i The stress–energy dyadic can be written in yet another form as 𝖳 𝚿 𝖨T ⌋𝚽 𝜺 ⌊𝖨 = ∧ ( ) − T N 𝚽 𝖨T ⌋𝚿 𝜺 ⌊𝖨 =− ∧ ( )) + T N , (5.207) with the scalar 1 T = (𝚿 ∧ 𝚽)|e . (5.208) 2 N Let us study the components of the stress–energy dyadic (5.205). The spatial energy density three-form can be obtained as its spatial- temporal part, which yields the scalar |𝖳| | . e123 e4 = e123 w (5.209) On the other hand, the temporal–temporal part yields the Poynting two- form as ⌋𝖳| . e4 e4 = E ∧ H (5.210) The remaining components are the spatial–spatial part which can be expressed as the momentum two-form |𝖳⌋ | 𝖨T ⌋ ⌋ e123 e123 = e123 (D ∧ B) e123, (5.211) 5.5 Power and Energy 127 and the temporal–spatial part which corresponds to the stress dyadic 1 e ⌋𝖳⌋e = (DE + BH + H ∧ 𝖨T ⌋B + E ∧ 𝖨T ⌋D)⌋e . (5.212) 4 123 2 123

5.5.3 Differentiation Rule Differentiating the stress–energy dyadic leads to the expression 𝖳 𝚿 𝖨T ⌋𝚽 𝚿 𝖨T ⌋𝚽 2d ∧ = d ∧ ∧ c + d ∧ c ∧ 𝚽 𝖨T ⌋𝚿 𝚽 𝖨T ⌋𝚿 − d ∧ ∧ c − d ∧ c ∧ 𝚿 𝖨T ⌋𝚽 𝚽 𝖨T ⌋𝚿 = 2d ∧ ∧ c − 2d ∧ ∧ c 𝚿 𝚽 𝚽 𝚿 ⌊𝖨 + d ∧ (( c ∧ − c ∧ ) ), (5.213) where the subscript c marks a quantity which is kept constant in differentiation. Here we have applied the identity (𝚿 ∧ 𝚽)⌊𝖨 = (𝚽 ∧ 𝚿)⌊𝖨 = 𝚿 ∧ 𝖨T ⌋𝚽 + 𝚽 ∧ 𝖨T ⌋𝚿. (5.214) Invoking the Maxwell equations we obtain the relation 1 d ∧ 𝖳 = 𝜸 ∧ 𝖨T ⌋𝚽 − 𝜸 ∧ 𝖨T ⌋𝚿 + d ∧ ((𝚿 ∧ 𝚽 − 𝚽 ∧ 𝚿)⌊𝖨), e m 2 c c (5.215) 𝜸 𝜸 where e and m denote the respective electric and magnetic source three-forms. Let us consider the case when the last term of (5.215) vanishes for all possible fields, which is equivalent to 𝚽 ⋅ 𝚿 𝚿 ⋅ 𝚽 𝚽| 𝖬 𝖬T |𝚽 . d( c − c) = d ( m − m) c = 0 (5.216) This requires that the quantity behind d must be constant for any 𝚽(x), 𝖬 which is possible only when the modified medium bidyadic m is symmetric, 𝖬T 𝖬 m = m, (5.217) that is, there is no skewon component in the medium. Thus, the differentiation rule 𝖳 𝜸 𝖨T ⌋𝚽 𝜸 𝖨T ⌋𝚿 d ∧ = e ∧ ( ) − m ∧ ( ), (5.218) is valid for the principal–axion medium. 128 CHAPTER 5 Electromagnetic Fields

To interpret the right side of (5.218), we can write 𝜸 𝖨T ⌋𝚽 𝜺 휚 𝖨T ⌋ 𝜺 . e ∧ ( ) =− 4 ∧ ( eE + Je ∧ s B − (Je ∧ E) 4) (5.219) The Gibbsian counterparts of the consecutive terms in the brackets are 휚 the electric force on electric charge, egEg, the magnetic force on the ⋅ electric current, Jeg × Bg and the Joule loss Jeg Eg. In (5.218) they F F appear as dyadic density quantities of the space 4 1 which can be integrated over a 4D volume quadrivector to yield a one-form force quantity. The last term in (5.218) can be interpreted likewise as forces on magnetic sources.

5.6 PLANE WAVES

Assuming a medium which is space and time invariant, as defined by a 𝖬 𝖬 constant bidyadic or modified bidyadic m, the simplest solutions to the Maxwell equations appear to be plane waves whose sources may be ignored since they can be assumed to be outside the region under interest.

5.6.1 Basic Equations In time-harmonic Gibbsian analysis of complex fields the plane wave is described by an exponential function −jk⋅r Eg(r) = Ege , (5.220) where k is the (Gibbsian) wave vector. The corresponding 4D representation is 𝚽(x) = 𝚽 exp(𝝂|x), (5.221) where 𝚽 is an amplitude two-form. The wave one-form 𝝂 can be expanded as 𝝂 𝜷 𝜺 = − k 4, (5.222) where the spatial wave one-form 𝜷 corresponds to the Gibbsian k vector and the scalar k corresponds to the variable 휔∕c in the time-harmonic case. Equation (5.221) represents a plane wave because the field is constant on spatial planes r(휏) satisfying 𝜷|r = k휏. (5.223) Applying the medium equation (5.51), the field two-form 𝚿(x)has the same exponential dependence on x, 𝚿(x) = 𝖬|𝚽(x) = 𝚿 exp(𝝂|x). (5.224) 5.6 Plane Waves 129

The Maxwell equations become algebraic equations for the field two- forms 𝝂 ∧ 𝚽 = 0, (5.225) 𝝂 ∧ 𝚿 = 0. (5.226) From the form of these equations it follows that there must exist potential one-forms 𝝓 and 𝝍 in terms of which the field two-forms can be expressed as 𝚽 = 𝝂 ∧ 𝝓, (5.227) 𝚿 = 𝝂 ∧ 𝝍. (5.228) In fact, applying the rule h⌋(휈 ∧ 𝚽) = 휈 ∧ (h⌋𝚽) + (h|𝝂)𝚽 = 0 (5.229) we can write h⌋𝚽 𝝓 =− , (5.230) h|𝝂 for any vector h satisfying h|𝝂 ≠ 0. From (5.230) it can be seen that the potential 𝝓 is not unique, since it depends on the choice of the vector h. Once h is chosen, the potential satisfies the additional condition h|𝝓 = 0. (5.231) Because of (5.227) and (5.228), the field two-forms 𝚽 and 𝚿 of a plane wave must be simple and they satisfy three orthogonality conditions 𝚽 ∧ 𝚽 = 0, 𝚽 ∧ 𝚿 = 0, 𝚿 ∧ 𝚿 = 0, (5.232) or, equivalently, 𝚽 ⋅ 𝚽 = 0, 𝚽 ⋅ 𝚿 = 0, 𝚿 ⋅ 𝚿 = 0. (5.233) From (5.226) we obtain a three-form equation for the potential one- form 𝝓, 𝝂 ∧ 𝚿 = 𝝂 ∧ 𝖬|𝚽 = 𝝂 ∧ 𝖬|(𝝂 ∧ 𝝓) = 0. (5.234) ⌊ Contracted by eN this becomes a vector equation, ⌊ 𝝂 𝖬⌊𝝂 |𝝓 𝝂⌋ 𝖬 ⌊𝝂 |𝝓 eN ( ∧ ) =− ( m ) = 0, (5.235) which can be written as 𝖣(𝝂)|𝝓 = 0. (5.236) 130 CHAPTER 5 Electromagnetic Fields

𝖣 𝝂 E E The dyadic ( ) ∈ 1 1 defined by 𝖣 𝝂 𝝂⌋𝖬 ⌊𝝂 𝖬 ⌊⌊𝝂𝝂 ( ) =− m = m (5.237) is called the dispersion dyadic. One may note that (5.237) coincides with the dyadic differential operator (5.63) when d is replaced by 𝝂. Because of ⌊𝖨(2)T ⌊⌊𝝂𝝂 𝝂⌋ ⌊𝖨(2) ⌊𝝂 𝝂⌋ ⌊ 𝝂 𝖨T (eN ) =− (eN ) =− (eN ( ∧ )) ⌊ 𝝂 𝝂 𝖨T = eN ( ∧ ∧ ) = 0, (5.238) the axion part of the medium bidyadic 𝖬 can be ignored when forming the dispersion dyadic and it does not have any effect on a plane wave propagating in a homogeneous medium.

5.6.2 Dispersion Equation Since the dispersion dyadic satisfies (5.236) it has no inverse whence its rank can be at most 3. However, because it also satisfies 𝖣 𝝂 |𝝂 𝝂⌋𝖬 | 𝝂 𝝂 ( ) =− m ( ∧ ) = 0, (5.239) while for 𝚽 = 𝝂 ∧ 𝝓 ≠ 0, 𝝂 and 𝝓 are linearly independent, the rank of the dispersion dyadic 𝖣(𝝂) can be at most 2. This means that the dispersion dyadic satisfies 𝖣(3)(𝝂) = 0, (5.240) which is a dyadic equation for the wave one-form 𝝂. Actually, (5.240) is equivalent to a scalar equation called the dispersion equation. To find the scalar equation let us form the double-wedge square and cube of the dispersion dyadic, 1 𝖣(2)(𝝂) = 𝖣(𝝂)∧𝖣(𝝂) 2 ∧ 1 = 𝝂𝝂⌋⌋(𝖬 ∧𝖣(𝝂)) 2 m∧ 1 = 𝝂𝝂⌋⌋(𝖬 ∧(𝝂𝝂⌋⌋𝖬 )), (5.241) 2 m∧ m 1 𝖣(3)(𝝂) = 𝖣(𝝂)∧𝖣(2)(𝝂) 3 ∧ 1 = 𝝂𝝂⌋⌋(𝖬 ∧𝖣(2)(𝝂)). (5.242) 3 m∧ 5.6 Plane Waves 131

E E Because the last dyadic in brackets belongs to the space 4 4,itmust be a multiple of eNeN. Thus, we can express 𝖣(3) 𝝂 𝝂𝝂⌋⌋ 𝝂 ( ) = ( eNeN)D( ), (5.243) where the bracketed dyadic does not vanish for 𝝂 ≠ 0. The scalar dispersion function D(𝝂) depends on the medium through the modified 𝖬 medium bidyadic m. It has the following analytic form: 1 D(𝝂) = 𝜺 𝜺 ||(𝖬 ∧𝖣(2)(𝝂)) 3 N N m∧ 1 = 𝜺 𝜺 ||(𝖬 ∧(𝝂𝝂⌋⌋(𝖬 ∧(𝝂𝝂⌋⌋𝖬 )))). (5.244) 6 N N m∧ m∧ m Thus, for 𝝂 ≠ 0, we conclude that the dyadic dispersion equation (5.240) is equivalent to the scalar dispersion equation D(𝝂) = 0. (5.245) Substituting (5.222), the dispersion equation has the form f (k, 𝜷) = 0. Assuming that the scalar k is given and 𝜷 = 훽𝜺 where 𝜺 is a given spatial one-form, the dispersion equation becomes an algebraic equation for the scalar magnitude 훽 = 훽(𝜺, k) of the one-form 𝜷. Because the dispersion function D(𝝂) is a fourth-order polynomial in 𝝂, (5.245) represents an algebraic equation of the fourth order for 𝝂.For agivenvalueofk, it can be pictured as defining a surface of the fourth order in the 3D space of 𝜷. The dispersion equation could be alternatively obtained by starting from the medium equation (5.52) instead of (5.51). In fact, expressing the field two-form 𝚿 as (5.228) and following similar steps, the equation has the form 𝖣′ 𝝂 |𝝍 𝖣′ 𝝂 𝖭 ⌊⌊𝝂𝝂 ( ) = 0, ( ) = m , (5.246) and the dyadic dispersion equation becomes 𝖣′(3) 𝝂 ⌊⌊𝝂𝝂 ′ 𝝂 . ( ) = (eNeN )D ( ) = 0 (5.247) Now it is obvious that D′(𝝂) = 0 represents the same dispersion equation as (5.245) because they both define the wave one-form 𝝂 of the same plane wave in the same medium. Of course, when one of the bidyadics 𝖬 𝖭 m and m does not exist, the corresponding dispersion equation does not exist. 132 CHAPTER 5 Electromagnetic Fields

5.6.3 Special Cases There are various special cases of the dispersion equation depending on the medium in question.

1. D(𝝂) = 0 is a quartic equation for the general medium. 2. For some media the dispersion function can be factorized as 𝝂 𝝂 𝝂 a product of two second-order functions D( ) = D1( )D2( ), whence the quartic equation can be decomposed in two quadratic 𝝂 𝝂 equations D1( ) = 0andD2( ) = 0. The corresponding media can be called decomposable media (DC media). 3. For some DC media the two dispersion functions coincide, 𝝂 𝝂 2 D( ) = (D1( )) , whence there is only one quadratic equation 𝝂 D1( ) = 0. The corresponding media can be called nonbirefringent (NB media). 4. For some media the dispersion equation D(𝝂) = 0 is satisfied identically for any 𝝂, whence the choice of the one-form 𝝂 is not restricted by the medium. Such media can be called media with no dispersion equation (NDE media).

Examples of media with different dispersion equation properties will come up in subsequent chapters.

5.6.4 Plane-Wave Fields Let us assume that 𝝂 is a known solution for the dispersion equation of the form (5.245) or (5.240). Since the potential 𝝓 and the wave one-form 𝝂 satisfy 𝖣(𝝂)|𝝓 = 0and𝖣(𝝂)|𝝂 = 0, the field two-form 𝚽 = 𝝂 ∧ 𝝓 satisfies the dyadic equation 𝖣(𝝂)⌋𝚽 = (𝖣(𝝂)|𝝓)𝝂 − (𝖣(𝝂)|𝝂)𝝓 = 0. (5.248) To find the field two-form, we can start from the condition 𝖣⌋ 𝜺 ⌊𝖣(2)T ( N ) = 0, (5.249) 𝖣 E E which can be shown to be valid for any metric dyadic ∈ 1 1 satisfying 𝖣(3) = 0. Choosing 𝖣 equal to the dispersion dyadic 𝖣(𝝂)ofrank 2, satisfying 𝖣(2)(𝝂) ≠ 0, we can write a solution to (5.248) as 𝚽 𝜺 ⌊𝖣(2)T |𝚵 𝜺 ⌊ 𝚵|𝖣(2) = N = N ( ), (5.250) 5.6 Plane Waves 133 where 𝚵 may be any two-form yielding a nonzero result. As a check we can expand ⌊ 𝚽 𝝂 ⌊ 𝜺 ⌊ 𝚵|𝖣(2) 𝝂 eN ( ∧ ) = eN (( N ( )) ∧ ) ⌊ 𝜺 ⌊ 𝚵|𝖣(2) ⌊𝝂 = (eN ( N ( ))) = (𝚵|𝖣(2))⌊𝝂 = 𝚵|((𝖣|𝝂) ∧ 𝖣) = 0, (5.251) which implies 𝚽 ∧ 𝝂 = 0 because of (5.239). Thus, (5.250) satisfies (5.225). The corresponding potential solution is obtained from (5.230). Assuming that the chosen vector h satisfies h|𝝂 = 1, we obtain 𝝓 ⌋ 𝜺 ⌊ 𝚵|𝖣(2) 𝜺 ⌊ 𝚵|𝖣(2) . =−h ( N ( )) = N (h ∧ ( )) (5.252) The potential is not unique because of the vector h. The polarization of the field two-form is unique because the dispersion dyadic is ofrank 2 and can be expressed in the form 𝖣(𝝂) = ac + bd for some vectors 𝚽 𝜺 ⌊ 𝚵 a − d. Thus, is a multiple of N (c ∧ d) for any two-form . The rule (5.250) obviously fails for media with wave one-form 𝝂 satisfying 𝖣(2)(𝝂) = 0, in which case the dispersion dyadic 𝖣(𝝂)is of rank 1. Because in this case we can write 𝖣 = ac, any two-form satisfying c⌋𝚽 = 0 may represent a valid electromagnetic field. The two-form can be expressed in the form 𝚽 𝜺 ⌊ = N (c ∧ d), (5.253) where d may be any vector yielding a nonzero result. An equivalent form can be written as 𝚽 𝜺 ⌊ 𝜶|𝖣 = N (( ) ∧ d), (5.254) which depends on a one-form 𝜶 and a vector d. To check the validity of (5.225), we expand ⌊ 𝚽 𝝂 ⌊ 𝜺 ⌊ 𝜶|𝖣 𝝂 eN ( ∧ ) = eN ( N (( ) ∧ d) ∧ ) ⌊ 𝜺 ⌊ 𝜶|𝖣 ⌊𝝂 = (eN ( N (( ) ∧ d))) = ((𝜶|𝖣) ∧ d)⌊𝝂 = d(𝜶|𝖣|𝝂) − (d|𝝂)(𝜶|𝖣) =−(d|𝝂)(𝜶|𝖣), 134 CHAPTER 5 Electromagnetic Fields which requires that the vector d must satisfy d|𝝂 = 0forthegiven one-form 𝝂. With this, (5.254) satisfies (5.225) and represents a valid polarization of the plane wave field for any one-form 𝜶.

5.6.5 Simple Principal Medium As an example let us consider plane-wave propagation in the simple principal medium defined by the modified medium bidyadic (5.122),

𝖬 1 𝖦(2) m =−휇 , (5.255) 𝖦 E E with the symmetric dyadic ∈ 1 1 defined by 𝖦 𝖦 휇휖 𝖦 = s − e4e4, s = e1e1 + e2e2 + e3e3, (5.256) and satisfying 𝖦(3) 𝖦(4) 휇휖 . s = e123e123, =− eNeN (5.257) The dispersion dyadic and its double-wedge powers can now be expanded as 휇𝖣(𝝂) =−𝖦(2)⌊⌊𝝂𝝂 =−(𝖦||𝝂𝝂)𝖦 + (𝖦|𝝂)(𝖦|𝝂), (5.258) 휇2𝖣(2) 𝝂 𝖦||𝝂𝝂 𝖦(2) 𝖦||𝝂𝝂 𝖦∧ 𝖦|𝝂 𝖦|𝝂 ( ) = ( )( ( ) − ∧( )( )) (5.259) = (𝖦||𝝂𝝂)𝖦(3)⌊⌊𝝂𝝂, (5.260) 휇3𝖣(3) 𝝂 𝖦||𝝂𝝂 2 𝖦||𝝂𝝂 𝖦(3) 𝖦(2)∧ 𝖦|𝝂 𝖦|𝝂 ( ) =−( ) (( ) − ∧( )( )) =−(𝖦||𝝂𝝂)2(𝖦(4)⌊⌊𝝂𝝂) 휇휖 𝖦||𝝂𝝂 2 ⌊⌊𝝂𝝂 . = ( ) (eNeN ) (5.261) Here we have applied the following identity, valid for any metric dyadics 𝖦 E E  F F ∈ 1 1 and ∈ 1 1, 𝖦(p)⌊⌊T 𝖦||T 𝖦(p−1) 𝖦(p−2)∧ 𝖦||𝖦 . = ( ) − ∧( ), p = 4, 3 (5.262) From (5.261) the dispersion equation (5.240) is seen to reduce to 𝖦||𝝂𝝂 = 0, (5.263) or, in terms of spatial quantities, to 𝜷|𝖦 |𝜷 휇휖 2 . s − k = 0 (5.264) 5.6 Plane Waves 135

This corresponds to a special case when the dispersion surface in the spatial 𝜷 space becomes a single quadric. This means that the simple principal medium is a nonbirefringent medium. When 𝝂 is a solution to (5.263), from (5.259) it follows that the dyadic dispersion equation is of the more restricted form 𝖣(2)(𝝂) = 0 and the dispersion dyadic (5.258) is of rank 1 as 𝖣 𝝂 1 𝖦|𝝂 𝖦|𝝂 . ( ) = 휇 ( )( ) (5.265) The equation (5.236) for the corresponding potential one-form 𝝓 has now the simple scalar form 𝝂|𝖦|𝝓 = 0, (5.266) whence 𝝓 can be expressed in the form 𝝓 = 𝝂|𝖦⌋𝚵 (5.267) for any two-form 𝚵 yielding a nonzero result. The corresponding field two-form becomes 𝚽 = 𝝂 ∧ (𝝂|𝖦⌋𝚵). (5.268) It is left as an exercise to show that (5.268) corresponds to the previously obtained expression (5.254).

5.6.6 Handedness of Plane Wave 𝜿 F Spatial 3D trivectors s ∈ 3 are similar to scalars, because they are 𝜺 multiples of a given spatial trivector, say 123, 𝜿 휅𝜺 휅 𝜿 | . s = 123, = s e123 (5.269) 휅 𝜿 𝜺 If is positive, s and 123 have the same handedness, if negative, they have opposite handedness. A plane wave has a special spatial three- form, 𝜷 ∧ E ∧ H,where𝜷 is the spatial part of 𝝂. Let us assume that e123 is a right-handed trivector. It turns out that the handedness of the spatial trivector 𝜷 ∧ E ∧ H depends on the medium. For plane-wave fields, the energy–power density three-form (5.202) satisfies 1 𝝂 ∧ W = 𝝂 ∧ (−𝚿 ∧ (𝚽⌊e ) + 𝚽 ∧ (𝚿⌊e )) 2 4 4 1 = 𝝂 ∧ (−𝝂 ∧ 𝝍 ∧ (𝚽⌊e ) + 𝝂 ∧ 𝝓 ∧ (𝚿⌊e )) = 0. (5.270) 2 4 4 136 CHAPTER 5 Electromagnetic Fields

On the other hand, since it consists of the energy density three-form w and the Poynting two-form E ∧ H as 𝜺 W = w − E ∧ H ∧ 4, (5.271) we have | 𝝂 | 𝝂 𝜺 . eN ( ∧ w) = eN ( ∧ E ∧ H ∧ 4) (5.272) Expanding both sides of (5.272) as | 𝝂 | 𝜺 eN ( ∧ w) = eN (−k 4 ∧ (D ∧ E + B ∧ H)) | = ke123 (D ∧ E + B ∧ E) ⋅ ⋅ = k(Dg Eg + Bg Hg) (5.273) | 𝝂 𝜺 | 𝜷 𝜺 eN ( ∧ E ∧ H ∧ 4) = eN ( ∧ E ∧ H ∧ 4) | 𝜷 = e123 ( ∧ E ∧ H), (5.274) the condition (5.272) can be written for Gibbsian fields as ( ) ( ) 휖 휉 g g E e |(𝜷 ∧ E ∧ H) = k(E H ) ⋅ ⋅ g . (5.275) 123 g g 휁 휇 H g g g Because this is valid for any plane-wave fields, it links together the handedness of the three-form 𝜷 ∧ E ∧ H when compared with the trivector e123 and the definiteness of the matrix of Gibbsian medium dyadics. As an example, let us consider the simple principal medium with 휖 휖𝖦 휉 휁 휇 휇𝖦 g = s, g = 0, g = 0, g = s, (5.276) in which case we have | 𝜷 휖 ⋅ 휇 ⋅ e123 ( ∧ E ∧ H) = k( Eg Eg + Hg Hg), (5.277) |𝝂 휔 where k = e4 equals ∕c for a time-harmonic field. Thus, for positive 휖 and 휇 the three-form 𝜷 ∧ E ∧ H is right handed while for negative 휖 and 휇 it is left handed. This is the reason for the concept of “left-handed medium” launched by Veselago [41] for media with negative 휖 and 휇.

PROBLEMS

5.1 Show that the Gibbsian form of Maxwell equations (5.40)–(5.42) can be obtained from (5.21) and (5.22) through the rules (5.33)– (5.36). Problems 137

5.2 Find the equation for the potential representation of the field 𝚿 outside the source region, corresponding to (5.62), (5.63). 5.3 Derive the expansion (5.87) for the medium bidyadic in terms of Gibbsian medium dyadics, starting from (5.83)–(5.86). 5.4 Prove (5.97). 5.5 Verify that the bidyadic 𝖭 defined by the 3D expansion (5.77) is the inverse of the bidyadic 𝖬 defined by (5.75) and that (5.89) is the inverse of the modified medium bidyadic (5.88). 5.6 Show that a medium defined by the modified medium bidyadic 𝖬 𝖰(2) 𝖰 E E m = with symmetric dyadic ∈ 1 1 is a principal medium. 5.7 Find the Hehl–Obukhov decomposition of the medium bidyadic 𝖬 𝜺 ⌊𝖰(2) 𝖰 = N where the dyadic is antisymmetric, of the form 𝖰 ⌊𝖨T E = Q where Q ∈ 2 is some bidyadic. 5.8 Derive conditions for the spatial medium dyadics for the SH conditions starting from (5.179) and proceeding like for the DB conditions, (5.173) and (5.174). 5.9 Verify by basis expansion that the simple principal medium bidyadic

𝖬 1 𝖦 휇휖 (2) m =−휇 ( s − e4e4)

does not have an axion component. 5.10 Starting from medium conditions slightly different from those of (5.179),

𝜺 𝜺 𝜺 𝜺 2 ∧ 3 ∧ Eb = 0, 1 ∧ 3 ∧ Hb = 0,

study the boundary conditions at the interface x3 = 0definedby such a medium. Interpret the conditions in terms of Gibbsian vector fields. 𝖡 5.11 Verifythat the trace-free dyadic o defined by the simple antisym- 𝖴 𝖡2 훼𝖨 훼 metric bidyadic by (5.191) satisfies o = for some scalar . 𝖡2| Hint: show that, for any vector a, o a is a scalar multiple of a. 138 CHAPTER 5 Electromagnetic Fields

𝖬 𝖡∧𝖨 T 5.12 Show that for a skewon–axion medium defined by = ( ∧ ) 𝖡 E F where ∈ 1 1 is any dyadic, the dyadic dispersion equation (5.243) is satisfied identically. 5.13 Derive the expression (5.219) in detail. 5.14 Prove that the Maxwell energy–stress dyadic vanishes for any fields if and only if the medium bidyadic has only the axion term, that is, if it is PEMC. 5.15 Show that when the field two-form 𝚽 is simple, that is, satisfies 𝚽 ∧ 𝚽 = 0, there must exist a spatial one-form 𝜶 such that B = 𝜶 ∧ E. 5.16 Show that the condition for the modified medium bidyadic 𝖬 𝖬−1T ⌋⌋ A m + B m eNeN = 0 for some scalars A, B implies the following conditions for the Gibbsian dyadics: 휖T 휇T 휉T 휉 휁 T 휁 A g = B g , g =− g, g =− g, 휖 휇 provided the 3D inverses of the Gibbsian dyadics g, g exist. 5.17 Derive the dispersion equation for a plane wave in a medium defined by the modified medium bidyadic 𝖬 𝖰(2) m = + AB, 𝖰 E E where ∈ 1 1 is a metric dyadic and A, B are two bivectors. 5.18 Find the dispersion equation for a medium defined by the modified medium bidyadic 𝖬 𝖤 𝖤 ⌊𝖨(2)T m = M + A1C1 + A2C2, = eN

where Ai, Ci are two bivectors. 5.19 Find the dispersion equation for a medium defined by the modified medium bidyadic 𝖬 𝖤 𝖤 ⌊𝖨(2)T m = M + A1C1 + A2C2 + A3C3, = eN

where Ai, Ci are six bivectors. Show that the dispersion equation is split in two quadratic equations. Problems 139

5.20 Show that the expressions (5.268) and (5.254) correspond to the same field two-form of a plane wave in a simple principal medium. 𝖢 E E 5.21 Prove that (5.249) is satisfied for any metric dyadic ∈ 1 1 of rank 2.

CHAPTER 6

Transformation of Fields and Media

Applying various transformations of fields, sources, media and boundary conditions allows one to produce solutions for new problems from those of old problems without having to go through the solution process.

6.1 AFFINE TRANSFORMATION

Affine transformation is a linear transformation of spacetime which affects fields, sources, and media.

6.1.1 Transformation of Fields 𝖠 E F 𝖠(4) ≠ Assuming a dyadic ∈ 1 1 of rank 4, satisfying tr 0, vectors can be mapped to other vectors as → 𝖠| x xa = x, (6.1) which is called affine transformation [1, 25]. In its more general form affine transformation also involves translation by a vector [42], whichis omitted here. The differential operator d is simultaneously transformed as 𝖠−1T | da = d, (6.2) because it satisfies 𝖠−1T | |𝖠T 𝖨T daxa = dx = , (6.3)

141 142 CHAPTER 6 Transformation of Fields and Media when applying the property (5.5) dx = 𝖨T . (6.4) The wedge product of two transformed vectors, 𝖠| 𝖠| 𝖠(2)| xa ∧ ya = ( x) ∧ ( y) = (x ∧ y), (6.5) yields a transformation rule for any bivector A, 𝖠(2)| . Aa = A (6.6) E E Similarly, trivectors k ∈ 3 and quadrivectors qN ∈ 4 are transformed as 𝖠(3)| ka = k, (6.7) 𝖠(4) . qNa = tr qN (6.8) Transformation rule of a one-form 𝝓 follows that of the differential operator d, while that of a two-form 𝚽, a three-form 𝜸 and a four-form 𝜿 N can be expressed as 𝝓 𝖠−1T |𝝓 a = , (6.9) 𝚽 𝖠(−2)T |𝚽 a = , (6.10) 𝜸 𝖠(−3)T |𝜸 a = , (6.11) 𝜿 𝖠(−1) 𝜿 . Na = tr N (6.12) The Maxwell equations can be shown to be invariant in form in the affine transformation: 𝚿 𝜸 𝖠−1T | 𝖠(−2)T |𝚿 𝖠(−3)T |𝜸 da ∧ a − ea = ( d) ∧ ( ) − e 𝖠(−3)T | 𝚿 𝜸 = (d ∧ − e) = 0, (6.13) 𝚿 𝜸 𝖠−1T | 𝖠(−2)T |𝚿 𝖠(−3)T |𝜸 da ∧ a − ea = ( d) ∧ ( ) − e 𝖠(−3)T | 𝚽 𝜸 . = (d ∧ − m) = 0 (6.14)

6.1.2 Transformation of Media Expanding 𝚿 𝖠(−2)T |𝚿 𝖠(−2)T |𝖬|𝚽 a = = = 𝖠(−2)T |𝖬|𝖠(2)T |𝖠(−2)T |𝚽, 𝖠(−2)T |𝖬|𝖠(2)T |𝚽 = a 𝖬 |𝚽 = a a, (6.15) 6.1 Affine Transformation 143 we obtain the transformation rule for the medium bidyadic 𝖬, 𝖬 𝖠(−2)T |𝖬|𝖠(2)T . a = (6.16) From (6.16) it follows that the trace of the medium bidyadic is invariant in the affine transformation, 𝖬 𝖬. tr a = tr (6.17) The transformed modified medium bidyadic is defined as themod- ified transformed medium bidyadic, 𝖬 𝖬 ⌊𝖬 ma = ( a)m = eNa a, (6.18) with 𝖠(4) . eNa = tr eN (6.19) The transformation rule can be found from 𝖬 𝖠(4) ⌊𝖠(−2)T |𝖬|𝖠(2)T ma = tr eN 𝖠(4) ⌊𝖠(−2)T ⌋𝜺 |𝖬 |𝖠(2)T = tr eN N m 𝖠(2)|𝖬 |𝖠(2)T = m , (6.20) where we have applied the inverse rule (2.123) as 𝖠(2) 𝖠(4)𝖨(4)⌊⌊𝖠(−2)T 𝖠(4) ⌊𝖠(−2)T ⌋𝜺 . = tr = tr eN N (6.21) 𝖬 From (6.20) it follows that symmetric and antisymmetric bidyadics m 𝖬 are respectively mapped to symmetric and antisymmetric bidyadics ma in an affine transformation. Thus, in the Hehl–Obukhov decomposition (3.85) of the modified medium 𝖬 𝖬 𝖬 𝖬 m = m1 + m2 + m3, (6.22) 𝖬 𝖬 the skewon part m2 is mapped to the skewon part of ma. Since from (6.17) it follows that the axion part is invariant in the transformation, also the principal part is invariant. Thus, in conclusion, the principal, skewon, and axion parts of 𝖬 are transformed individually to the principal, 𝖬 skewon, and axion parts of a. To find the transformation rule of a skewon medium bidyadic we apply the representation (3.86) 𝖬 𝖡 ∧𝖨 T = ( o∧ ) (6.23) 144 CHAPTER 6 Transformation of Fields and Media

𝖡 E F in terms of a trace-free dyadic o ∈ 1 1. Invoking the rule (2.41) in the form 𝖠(2)| 𝖡 ∧𝖨 𝖠|𝖡 ∧𝖠 ( o∧ ) = ( o)∧ , (6.24) we can form the transformed medium bidyadic as 𝖬 𝖠(2)| 𝖡 ∧𝖨 |𝖠(−2) T a = ( ( o∧ ) ) 𝖠|𝖡 |𝖠−1 ∧ 𝖠|𝖠−1 T = (( o )∧( )) 𝖠|𝖡 |𝖠−1 ∧𝖨 T = (( o )∧ ) , (6.25) which has the skewon form, 𝖬 𝖡 ∧𝖨 T 𝖡 𝖠|𝖡 |𝖠−1. a = ( oa∧ ) , oa = o (6.26)

6.1.3 Dispersion Equation Considering a plane wave in the transformed medium, the dispersion dyadic (5.237) can be expanded as 𝖣 𝝂 𝖬 ⌊⌊𝝂𝝂 a( ) = ma 𝝂⌋ 𝖠(2)|𝖬 |𝖠(2)T ⌊𝝂 =− ( m ) 𝖠 𝝂|𝖠 |𝖬 | 𝖠T |𝝂 𝖠T =−( ∧ ( )) m (( ) ∧ ) 𝖠| 𝝂|𝖠 ⌋𝖬 ⌊ 𝖠T |𝝂 |𝖠T =− (( ) m ( )) 𝖠| 𝖬 ⌊⌊𝝂 𝝂 |𝖠T = ( m a a) 𝖠|𝖣 𝝂 |𝖠T = ( a) , (6.27) where we denote 𝝂 𝖠T |𝝂. a = (6.28) The dyadic dispersion equation (5.240) for the transformed medium can be expressed as 𝖣(3) 𝝂 𝖠(3)|𝖣(3) 𝝂 |𝖠(3)T a ( ) = ( a) = 0, (6.29) which implies 𝖣(3) 𝝂 . ( a) = 0 (6.30) Thus, the scalar dispersion equation (5.245) for the transformed medium becomes 𝝂 𝖠T |𝝂 D( a) = D( ) = 0, (6.31) where the dispersion function D involves the original medium bidyadic 𝖬. 6.2 Duality Transformation 145

One may conclude that the affine transformation does not change the character of the dispersion equation. In particular, if the original dispersion equation can be reduced to two quadratic equations, so can the transformed equation. If the medium is nonbirefringent, so is the transformed medium. If there is no dispersion equation for the original medium, there is no dispersion equation for the transformed medium.

6.1.4 Simple Principal Medium As an example, let us consider affine transformation of the simple principal medium defined by (5.115). Applying the spatial medium dyadics 𝛼 𝜖′ 𝜖𝜺 ⌊𝖦 𝜇 𝜇𝜺 ⌊𝖦 𝛽 = 0, = 123 s, = 123 s, = 0, (6.32) the modified medium bidyadic has the form (5.122)

𝖬 1 𝖦 𝜇𝜖 (2). m =−𝜇 ( s − e4e4) (6.33) The affine-transformed bidyadic now becomes 𝖬 𝖠(2)|𝖬 |𝖠(2)T ma = m 1 𝖠|𝖦 |𝖠T 𝜇𝜖 𝖠| 𝖠| (2) =−𝜇 ( s − ( e4)( e4)) 1 𝖦 𝜇𝜖 (2) =−𝜇 ( sa − e4ae4a) , (6.34) with ∑3 𝖠| 𝖦 . eia = ei, sa = eiaeia (6.35) 1 Thus, the transformed simple principal medium is another simple principal medium. In terms of Gibbsian dyadics, the simple principal medium defined 𝜖 𝜇T 𝜉 𝜁 by g = A g and g = g = 0 has been called affine-isotropic medium because one can find an affine transformation in terms of which itcan be transformed to an isotropic medium [25].

6.2 DUALITY TRANSFORMATION

Duality transformation is based on the symmetry of electric and magnetic quantities in the Maxwell equations. By changing symbols of fields 146 CHAPTER 6 Transformation of Fields and Media and sources the Maxwell equations remain the same in form which leads to corresponding similarity in different electromagnetic problems.

6.2.1 Transformation of Fields 𝚿 𝚽 𝚿 𝚽 Let us consider transformation of fields , to dual fields d, d as defined by the linear relation [1] ( ) ( )( ) 𝚿 𝜃 𝜃 𝚿 Zd d cos sin Z . 𝚽 = 𝜃 𝜃 𝚽 (6.36) d −sin cos More explicitly, the transformation is represented by 𝚿 1 𝜃 𝚿 𝜃 𝚽 d = (cos Z + sin ), (6.37) Zd 𝚽 𝜃 𝚿 𝜃 𝚽 d =−sin Z + cos , (6.38) 𝜃 as defined by three scalar parameters Z, Zd,and . The trivial case 𝜃 𝚿 𝚿 𝚽 Zd = Z,cos = 1 leads to the identity transformation, d = , d = 𝚽, which will be omitted in the sequel. The inverse of the duality transformation is ( ) ( )( ) 𝚿 𝜃 𝜃 𝚿 Z cos −sin Zd d . 𝚽 = 𝜃 𝜃 𝚽 (6.39) sin cos d The corresponding transformation of the electric and magnetic source three-forms follows a similar rule, ( ) ( )( ) 𝜸 𝜃 𝜃 𝜸 Zd ed cos sin Z e . 𝜸 = 𝜃 𝜃 𝜸 (6.40) md −sin cos m Considering the antisymmetric bidyadic function of fields (5.188), 1 𝖴 = (𝚽𝚿 − 𝚿𝚽) ∈ F F , (6.41) 2 2 2 its transformation can be written after substitution of (6.37) and (6.38) as 1 𝖴 = (𝚽 𝚿 − 𝚿 𝚽 ) d 2 d d d d Z = (𝚽𝚿 − 𝚿𝚽) 2Zd Z = 𝖴. (6.42) Zd 6.2 Duality Transformation 147

Because 𝖴 has a linear relation to the stress–energy dyadic 𝖳 (5.205) as (5.206), its dual is similarly obtained as

𝖳 Z 𝖳. d = (6.43) Zd

The same is also valid for all components of the stress–energy dyadic. For example, the dual of the Poynting two-form S = E ∧ H is then

Z . Sd = S (6.44) Zd

For the special choice of transformation parameters Zd = Z, these quantities are actually invariant in the duality transformation. The duality transformation rules (6.37), (6.38) can be decomposed in their spatial and temporal parts as ( ) ( )( ) Z D cos 𝜃 sin 𝜃 ZD d d = , (6.45) B −sin 𝜃 cos 𝜃 B ( d ) ( )( ) 𝜃 𝜃 E cos sin E . = 𝜃 𝜃 (6.46) ZdHd −sin cos ZH

6.2.2 Involutionary Duality Transformation Let us require that the transformation (6.36) be an involution, that is, that it for all 𝚿, 𝚽 the conditions

𝚿 1 𝜃 𝚿 𝜃𝚽 𝚿 ( d)d = (cos Z d + sin d) = , (6.47) Zd 𝚽 𝜃 𝚿 𝜃 𝚽 𝚽 ( d)d =−sin Z d + cos d = (6.48) are valid. Substituting from (6.37) and (6.38) we obtain two conditions for the three transformation parameters,

Z + Z d [(Z sin2 𝜃 + (Z − Z )cos2 𝜃)𝚿 + sin 𝜃 cos 𝜃𝚽] = 0, (6.49) 2 d d Zd Z + Z d (sin 𝜃 cos 𝜃Z𝚿 + sin2 𝜃𝚽) = 0. (6.50) Zd 148 CHAPTER 6 Transformation of Fields and Media

𝜃 Leaving as a free parameter, these conditions require Zd =−Z, whence the two-parameter involutionary transformation is defined by ( ) ( )( ) 𝚿 𝜃 𝜃 𝚿 Z d cos sin Z . 𝚽 =− 𝜃 𝜃 𝚽 (6.51) d sin − cos

Two simple special cases of (6.51) are obtained by choosing 𝜃 = 𝜋∕2 which reduces the transformation to ( ) ( )( ) 𝚿 𝚿 d 01∕Z 𝚽 =− 𝚽 , (6.52) d Z 0 or 𝜃 =−𝜋∕2, which changes the sign in (6.52). One should note that (6.51) does not contain the identity transformation as a special case 𝜃 𝚿 𝚿 𝚽 𝚽 𝜃 𝜋 since = 0 yields d =− , d = .For = ∕2 the spatial fields transform as

Bd =−ZD, Dd =−B∕Z, Ed = ZH, Hd = E∕Z, (6.53) which show the essence of the electric–magnetic duality. 𝖳 𝖳 The stress–energy dyadic is transformed as d =− in the involutionary transformation. The Poynting two-form is also reversed in the ⇄ transformation, Sd =−S, which corresponds to changing Eg Hg in the Gibbsian Poynting vector Eg × Hg. Any field can be split in two parts with respect to a given involutionary duality transformation: ( ) ( ) ( ) 𝚿 𝚿 𝚿 𝚽 = 𝚽 + 𝚽 , (6.54) sd ad 𝚿 𝚽 so that the field sd, sd is invariant (“self-dual”), while the field 𝚿 𝚽 ad, ad changes sign (“anti-self-dual”) in this duality transformation. The two partial fields are ( ) ( )( ) 𝚿 1 1 −1∕Z 𝚿 = , (6.55) 𝚽 −Z 1 𝚽 sd 2 ( ) ( )( ) 𝚿 1 11∕Z 𝚿 = . (6.56) 𝚽 Z 1 𝚽 ad 2 6.2 Duality Transformation 149

6.2.3 Transformation of Media Inserting the dual fields from (6.36) in the transformed medium equation, 𝚿 𝖬 |𝚽 d = d d (6.57) yields the relation 𝜃 𝖬|𝚽 𝜃 𝚽 𝖬 | 𝜃 𝖬|𝚽 𝜃 𝚽 cos Z + sin = Zd d (−sin Z + cos ), (6.58) which must be valid for any 𝚽. The transformation rule for the medium bidyadic can be extracted in the form 𝖬 𝜃 𝖨(2)T 𝜃 𝖬 | 𝜃 𝖨(2)T 𝜃 𝖬 −1 Zd d = (sin + cos Z ) (cos − sin Z ) , (6.59) or, more symmetrically, as 𝜃𝖨(2)T 𝜃 𝖬 | 𝜃𝖨(2)T 𝜃 𝖬 𝖨(2)T . (cos + sin Zd d) (cos − sin Z ) = (6.60) From this one can show that, in the general case, the medium bidyadics 𝖬 𝖬 𝖬|𝖬 𝖬 |𝖬 and d commute, d = d , whence they have the same set of eigen two-forms. Details are left as an exercise. A medium is called self-dual if one can find a nontrivial duality 𝖬 𝖬 transformation in which the medium bidyadic is invariant, d = .In a self-dual medium electromagnetic fields and sources can be combined with transformed fields and sources because the medium for both sets of fields and sources is the same. The transformation rule (6.59) can be expanded in terms of spatial medium dyadics as [1] 𝜖 𝜇 𝜖 𝜇 dZd + d∕Zd = Z + ∕Z, (6.61) 𝜉 𝜁 𝜉 𝜁 d − d = − , (6.62) 𝜖 𝜇 𝜖 𝜇 𝜃 𝜉 𝜁 𝜃 dZd − d∕Zd = ( Z − ∕Z)cos2 + ( + )sin2 , (6.63) 𝜉 𝜁 𝜖 𝜇 𝜃 𝜉 𝜁 𝜃. d + d =−( Z − ∕Z)sin2 + ( + )cos2 (6.64) It is worth noting that the dyadic 𝜉 − 𝜁 is invariant in any duality transformation. For the special involutionary duality transformation defined by 𝜃 = 𝜋 ∕2andZd =−Z, the medium bidyadic obeys the simple transformation rule 1 𝖬 = (Z2𝖬)−1 = 𝖭, (6.65) d Z2 150 CHAPTER 6 Transformation of Fields and Media with 𝖭 defined as in (5.52). This corresponds to 휖 휇 2 휇 휖 2 d =− ∕Z , d =− Z , (6.66) 휉 휁 휁 휉. d =− , d =− (6.67) The modified medium bidyadic is transformed as 1 1 1 𝖬 = e ⌊(𝜺 ⌊𝖬 )−1 = e e ⌊⌊𝖬−1 = 𝖭 , (6.68) md Z2 N N m Z2 N N m Z2 m 𝖭 ⌊𝖭 with m = eN . Equations (6.65) and (6.68) show us that inverse medium bidyadics have a close relation to duality-transformed medium bidyadics. The transformation rule (6.65) can actually be applied to finding the analytic expression of the inverse of a given medium bidyadic. In fact, inserting the spatial expansion (5.75) for the dual medium on the right side of (6.65) and substituting from (6.66), (6.67) yields 𝖬−1 2𝖬 = Z d (6.69) 2휖 2 𝜺 𝖨T 휉 |휇−1| 𝖨(2)T 휁 = Z d ∧ e4 + Z ( 4 ∧ + d) d ( − d ∧ e4) 휇 𝜺 𝖨T 휁 |휖−1| 𝖨(2)T 휉 =− ∧ e4 − ( 4 ∧ − ) ( + ∧ e4), (6.70) which coincides with (5.77).

6.3 TRANSFORMATION OF BOUNDARY CONDITIONS

Let us consider the set of boundary conditions of Section 5.4.2 in the duality transformation. r PMC Boundary Starting from the condition of the PMC boundary, 𝜺 𝚿 3 ∧ = 0, (6.71) the three-parameter inverse duality transformation (6.39) yields a condition of the form 𝜺 𝚿 𝚽 3 ∧ (A d + B d) = 0, (6.72) with 휃 휃. A = (Zd∕Z)cos , B =−(1∕Z)sin (6.73) 6.3 Transformation of Boundary conditions 151

In general case, the PMC boundary is transformed to a PEMC boundary. In particular, for 휃 =±휋∕2 the PMC boundary is transformed to the PEC boundary. r PEC Boundary Applying similar equations, the PEC boundary satisfying 𝜺 𝚽 3 ∧ = 0 (6.74) can also be transformed to a PEMC boundary. r PEMC Boundary Inverting the previous transformations, the perfect electromagnetic boundary with fields satisfying 𝜺 𝚿 𝚽 3 ∧ ( − M ) = 0 (6.75) can be transformed to the PMC or the PEC boundary. More generally, it can be transformed to another PEMC boundary satisfying a condition of the form (6.72) with Z 1 A = d (cos 휃 − MZ sin 휃), B =− (sin 휃 + MZ cos 휃). (6.76) Z Z Any PEMC boundary can be self-dual for a certain choice of transformation parameters. Details are left as an exercise. r DB Boundary The DB boundary conditions expressed as ( ) ( ) 𝚿 0 𝜺 ∧ 𝜺 ∧ = (6.77) 3 4 𝚽 0 are obviously self-dual in any duality transformations, as can be seen from (6.36). r SH Boundary Similarly, the SH boundary conditions ( ) ( ) 𝚿 0 𝜺 ∧ 𝜺 ∧ = (6.78) 2 3 𝚽 0 are also self-dual for any transformation parameters.

It has been shown that an object with a self-dual boundary and certain symmetry in its geometry has no back scattering for an incident plane wave [43]. Thus, such an object appears invisible for the radar. 152 CHAPTER 6 Transformation of Fields and Media

6.3.1 Simple Principal Medium As an example, let us consider the simple principal medium defined by the medium bidyadic 𝖬 휖𝖢 𝜺 휇𝖢 −1 = s ∧ e4 + 4 ∧ ( s) , (6.79) 𝖢 F E 𝖢 𝜺 ⌊𝖦 where the spatial dyadic s ∈ 2 1 can be expressed as s = 123 s 𝖦 in terms of a spatial symmetric dyadic s. Because (6.79) satisfies (5.129) as 𝖬2 =−M2𝖨(2)T , M2 = 휖∕휇, (6.80) it also satisfies (cos 휃𝖨(2)T + sin 휃 Z𝖬)|(cos 휃𝖨(2)T − sin 휃 Z𝖬) = (cos2 휃 + sin2 휃M2Z2)𝖨(2)T , (6.81) as can be verified after expanding the left side. Now one can show that the simple principal medium is actually self-dual in terms of a suitable transformation. In fact, choosing √ 휇 휖 Z = Zd = 1∕M = ∕ , (6.82) and comparing (6.81) to (6.60), we can identify from the two equations the condition 𝖬 𝖬. d = (6.83)

6.3.2 Plane Wave A plane wave in a medium defined by a bidyadic 𝖬 is governed by a wave one-form 𝝂 satisfying the dispersion equation (5.245). Making the duality transformation changes the medium and the fields but, unlike for the affine transformation, it does not change the wave one-form 𝝂. In fact, assuming that 𝝂 satisfies both 𝝂 ∧ 𝚿 = 0and𝝂 ∧ 𝚽 = 0, from (6.39) we have ( ) ( ) ( ) ( ) 𝚿 휃 휃 𝚿 𝝂 Z cos −sin 𝝂 Zd d 0 ∧ 𝚽 = 휃 휃 ∧ 𝚽 = , (6.84) sin cos d 0 𝝂 𝝂 𝚿 𝝂 𝚽 whence the same also satisfies ∧ d = 0and ∧ d = 0. Thus, we can express the transformed field two-forms in terms of potential 6.4 Reciprocity Transformation 153 one-forms as ( ) ( ) 𝚿 𝝍 d 𝝂 d 𝚽 = ∧ 𝝓 , (6.85) d d which satisfy the transformation rules ( ) ( )( ) 𝝍 휃 휃 𝝍 Zd d cos sin Z . 𝝓 = 휃 휃 𝝓 (6.86) d −sin cos In particular, considering the special involutory duality transfor- 휃 휋 mation defined by the parameters = ∕2andZd =−Z implying the 𝖬 𝖭 relation (6.68) between the modified bidyadics md and m, indicates that the dispersion equation (5.245) must have the same solutions as the dispersion equation (5.246).

6.4 RECIPROCITY TRANSFORMATION

Reciprocity (also called as Lorentz reciprocity [44]) in electromagnetics is related to the question whether it is possible to swap the positions of a transmitter and a receiver without a change in the received signal [45, 46]. This is only possible if the medium satisfies a certain property which is called the reciprocity condition.

6.4.1 Medium Transformation One can define a reciprocal medium by requiring that the medium bidyadic be invariant in a transformation called the reciprocity transformation, which can be defined as 𝖬→𝖬 𝖨(4)T ⌊⌊𝖬∗T . r =− (6.87) Here ()∗ denotes a special affine transformation called time reversion. It 𝖳 E F is defined for vectors and one-forms by the unipotent dyadic ∈ 1 1 𝖳 𝖨 𝜺 = s − e4 4, (6.88) 𝖳2 = 𝖨, (6.89) as a∗ = 𝖳|a, 𝜶∗ = 𝜶|𝖳. (6.90) 154 CHAPTER 6 Transformation of Fields and Media

For bivectors and two-forms the time reversion bidyadic becomes 𝖳(2) 𝖨(2) 𝖨 ∧ 𝜺 = s − s∧e4 4 𝜺 𝜺 𝜺 = e12 12 + e23 23 + e31 31 𝜺 𝜺 𝜺 − e14 14 − e24 24 − e34 34, (6.91) (𝖳(2))2 = 𝖨(2), (6.92) and it operates as A∗ = 𝖳(2)|A, 𝚽∗ = 𝚽|𝖳(2). (6.93) For medium bidyadics, the time reversion operation is defined by 𝖬∗ 𝖳(2)T |𝖬|𝖳(2)T 𝖬∗ 𝖳(2)|𝖬 |𝖳(2)T . = , m = m (6.94) Inserting the 3D expansion of the medium bidyadic we obtain 𝖬∗ 훼 휖′ 𝜺 휇−1 𝜺 훽 ∗ = ( + ∧ e4 + 4 ∧ + 4 ∧ ∧ e4) 훼 휖′ 𝜺 휇−1 𝜺 훽 . = − ∧ e4 − 4 ∧ + 4 ∧ ∧ e4 (6.95) Thus, the effect of time reversion to a medium is changing the sign of the dyadics 휖′ and 휇−1. The reciprocity-transformed medium bidyadic (6.87) can be expanded as 𝖬 훼 휖′ 𝜺 휇−1 𝜺 훽 r = r + r ∧ e4 + 4 ∧ r + 4 ∧ r ∧ e4 𝜺 ⌊𝖬∗T ⌋ =− N eN 𝜺 ⌊ 훼T 휖′T 휇−1T 𝜺 훽T 𝜺 ⌋ =− N ( + e4 ∧ + ∧ 4 + e4 ∧ ∧ 4) eN 𝜺 𝜺 ⌊⌊훼T 𝜺 ⌊⌊휖′T = 4 ∧ ( 123e123 ) ∧ e4 + ( 123e123 ) ∧ e4 𝜺 𝜺 ⌊⌊휇−1T 𝜺 ⌊⌊훽T + 4 ∧ ( 123e123 ) + 123e123 , (6.96) where we have applied the rules ⌊𝜶 ⌊𝜶 𝜶 ⌋ 𝜶 ⌋ eN s = e4 ∧ (e123 s), s eN =−( s e123) ∧ e4, (6.97) ⌊𝚽 ⌊𝚽 𝚽 ⌋ 𝚽 ⌋ eN s =−e4 ∧ (e123 s), s eN = ( s e123) ∧ e4, (6.98) 𝜶 𝚽 valid for spatial one-forms s and two-forms s. In terms of these rules, reciprocity transformations of the different spatial medium dyadics can be compactly expressed as ( ) ( ) 훼 휖′ 훽T 휖′T r r 𝜺 e ⌊⌊ . 휇−1 훽 = 123 123 휇−1T 훼T (6.99) r r 6.4 Reciprocity Transformation 155

From 훼 𝜺 ⌊⌊훽T 훽 tr r = tr( 123e123 ) = tr , (6.100) 훽 𝜺 ⌊⌊훼T 훼 tr r = tr( 123e123 ) = tr , (6.101) we obtain the relation 𝖬 훼 훽 훽 훼 𝖬. tr r = tr r − tr r = tr − tr =−tr (6.102) This means that the axion part of the medium bidyadic changes sign in the reciprocity transformation. For the modified medium bidyadic the reciprocity transformation is obtained from (6.87) as 𝖬 𝖬 ⌊𝖬 𝖬∗T ⌋ ∗⌊𝖬 ∗T 𝖬∗T mr = rm = eN r =− eN =−(eN ) = m , (6.103) ∗ with eN =−eN. For the Gibbsian medium parameters the transformation becomes ( ) ( ) 휖 휉 휖T −휁 T g g = g g . (6.104) 휁 휇 −휉T 휇T g g r g g

6.4.2 Reciprocity Conditions A reciprocal medium is defined by requiring invariance of the medium bidyadic in the the reciprocity transformation, 𝖬 𝖬 𝖨(4)T ⌊⌊𝖬T∗ 𝖨(4)⌊⌊𝖬 T∗ = r =− =−( ) , (6.105) or, 𝖬 𝖬 𝖬∗T . m = mr = m (6.106) A medium which fails to satisfy these conditions is called nonreciprocal. Media satisfying 𝖬 𝖬 𝖨(4)⌊⌊𝖬 T∗ =− r = ( ) , (6.107) or 𝖬 𝖬 𝖬∗T m =− mr =− m , (6.108) may be called anti-reciprocal. Because the reciprocity transformation is involutory, 𝖬 𝖨(4)⌊⌊ 𝖨(4)⌊⌊𝖬 T∗ T∗ 𝖨(4)T ⌊⌊ 𝖨(4)⌊⌊𝖬 𝖬 ( r)r =−( (− ) ) = ( )) = , (6.109) 𝖬 𝖬∗T 𝖬∗T ∗T 𝖬 ( mr)r = ( m )r = ( m ) = m, (6.110) 156 CHAPTER 6 Transformation of Fields and Media any medium bidyadic can be decomposed in a sum of reciprocal and anti-reciprocal parts as 1 1 𝖬 = (𝖬 − (𝖨(4)⌊⌊𝖬)∗T ) + (𝖬 + (𝖨(4)T ⌊⌊𝖬)∗T ), (6.111) 2 2 m 1 1 𝖬 = (𝖬 + 𝖬∗T ) + (𝖬 − 𝖬∗T ). (6.112) m 2 m m 2 m m Because of (6.102), any reciprocal medium satisfies tr𝖬 = 0. Thus, it does not contain an axion part and the axion medium is manifestly 𝖬 anti-reciprocal. The skewon part 2 in the Hehl–Obukhov decomposi- 𝖨(4)⌊⌊𝖬 𝖬T tion (5.91), satisfying 2 =− 2 , has the respective reciprocal and anti-reciprocal parts 1 1 𝖬 = (𝖬 + 𝖬∗) + (𝖬 − 𝖬∗), (6.113) 2 2 2 2 2 2 2 𝖬 while the principal part 1 has the corresponding decomposition 1 1 𝖬 = (𝖬 − 𝖬∗) + (𝖬 + 𝖬∗). (6.114) 1 2 1 1 2 1 1 𝖬 𝖬∗ Thus, if the skewon part is reciprocal, it must satisfy 2 = 2 , whence 휖′ 휇−1 the dyadics 2 and 2 must vanish. Correspondingly, if the principal 𝖬 𝖬∗ 훼 part is reciprocal, it must satisfy 1 =− 1 , whence the dyadics 1 and 훽 1 must vanish. The reciprocity conditions for the spatial medium dyadics are obtained from (6.99) as 훼 𝜺 ⌊⌊훽T = 123e123 , (6.115) 휖′ 𝜺 ⌊⌊휖′T = 123e123 , (6.116) 휇−1 𝜺 ⌊⌊휇−1T = 123e123 , (6.117) 훽 𝜺 ⌊⌊훼T = 123e123 , (6.118) or ⌊훼 ⌊훽 T ⌊휖′ ⌊휖′ T ⌊휇−1 ⌊휇−1 T . e123 = (e123 ) , e123 = (e123 ) , e123 = (e123 ) (6.119) For the Gibbsian medium dyadics, the reciprocity conditions can be expressed as [24] 휖 휖T 휉 휁 T 휁 휉T 휇 휇T . g = g , g =− g , g =− g , g = g (6.120) 휖 휇 Thus, g and g must be symmetric dyadics for a reciprocal medium. 6.4 Reciprocity Transformation 157

As an example, the modified bidyadic of the simple principal medium (5.122) is transformed as

𝖬 1 𝖦 휇휖 ∗T mr =−휇 ( s − e4e4) 1 𝖦 휇휖 =−휇 ( s − e4e4), (6.121) which means that the simple principal medium is invariant in the reciprocity transformation, that is, it is reciprocal. This can also be seen from 휖 휇 the Gibbsian representation (6.120) because g and g are symmetric 휉 휁 dyadics and g = g = 0. It is left as an exercise to prove that axion, skewon, and principal parts of a medium bidyadic are transformed individually to axion, skewon, and principal parts of the transformed medium bidyadic.

6.4.3 Field Relations The effect of medium reciprocity to electromagnetic fields can be elu- cidated by considering two media which are reciprocal to one another, 𝖬 𝖬 that is, whose medium bidyadics 1 and 2 are related through the reciprocity transformation. Assuming that the modified bidyadics satisfy 𝖬 𝖬 𝖬T∗ 1m = 2mr = 2m, (6.122) the fields 1 and 2 are related by | 𝚽∗ 𝚿 𝚽∗| ⌊𝚿 𝚽∗| 𝖬 |𝚿 eN ( 2 ∧ 1) = 2 (eN 1) = 2 ( 1m 1) 𝚽∗| 𝖬T∗|𝚽 𝖬 |𝚽 ∗|𝚽 = 2 ( 2m 1) = ( 2m 2) 1 ⌊𝚿 ∗|𝚽 ∗| 𝚿∗ 𝚽 = (eN 2) 1 = eN ( 2 ∧ 1) | 𝚿∗ 𝚽 =−eN ( 2 ∧ 1), (6.123) which equals the condition 𝚽∗ 𝚿 𝚽 𝚿∗ . 2 ∧ 1 + 1 ∧ 2 = 0 (6.124) In terms of spatial fields this becomes 𝜺 𝜺 𝜺 𝜺 − B2 ∧ H1 ∧ 4 − E2 ∧ D1 ∧ 4 + B1 ∧ H2 ∧ 4 + E1 ∧ D2 ∧ 4 = 0, (6.125) 158 CHAPTER 6 Transformation of Fields and Media which equals the condition . B2 ∧ H1 + E2 ∧ D1 = B1 ∧ H2 + E1 ∧ D2 (6.126) Working backwards, we can state that if (6.126) is satisfied for any fields 1 in medium 1 and any fields 2 in medium 2, the two medium bidyadics must be related by (6.122). If the two fields are in the same medium, this requires that the medium bidyadic satisfy the reciprocity condition (6.105) or (6.106).

6.4.4 Time-Harmonic Fields Let us finally interpret (6.126) in terms of time-harmonic sources and fields with time dependence exp(j휔t) = exp(jk휏). Assuming electric and 𝜸 𝜸 magnetic sources e1, m1 in medium 1, and a second set of sources 𝜸 𝜸 𝚽 𝚿 e2, m2 in medium 2, which produce the respective fields 1, 1 and 𝚽 𝚽 2, 2, the condition (6.126) can be expanded as

0 = jk(B2 ∧ H1 + E2 ∧ D1 − B1 ∧ H2 − E1 ∧ D2) 휕 휕 휕 휕 = ( 휏 B2) ∧ H1 + E2 ∧ ( 휏 D1) − ( 휏 B1) ∧ H2 − E1 ∧ ( 휏 D2)

=−(d ∧ E2 + Jm2) ∧ H1 + E2 ∧ (d ∧ H1 − Je1)

+ (d ∧ E1 + Jm1) ∧ H2 − E1 ∧ (d ∧ H2 − Je2)

=−(d ∧ E2) ∧ H1 + (d ∧ H1) ∧ E2 + (d ∧ E1) ∧ H2 − (d ∧ H2) ∧ E1 . − Jm2 ∧ H1 − E2 ∧ Je1 + Jm1 ∧ H2 + E1 ∧ Je2 (6.127) The resulting relation between the sources and fields can be rewritten in the form

(Je1 ∧ E2 − Jm1 ∧ H2) − (Je2 ∧ E1 − Jm2 ∧ H1) . = d ∧ (E1 ∧ H2 − E2 ∧ H1) (6.128) Integrating over the whole 3D space and assuming vanishing of the term on the right side expressed as a surface integral over a surface in infinity (the sources are assumed to be concentrated in a finite region), the left side yields

. ∫ (Je1 ∧ E2 − Jm1 ∧ H2)dV = ∫ (Je2 ∧ E1 − Jm2 ∧ H1)dV (6.129) In classical terminology, this can be interpreted as the reaction of field 2 to source 1 being equal to the reaction of field 1 to source 2, which forms the essence of reciprocity in electromagnetics [46]. 6.5 Conformal Transformation 159

6.5 CONFORMAL TRANSFORMATION

After Lorentz and Einstein demonstrated that the Maxwell equations remain form-invariant in the Lorentz transformation, another transformation defined by Bateman and Cunningham in 1909 [47, 48],was also shown to leave the Maxwell equations invariant. Labeled as the conformal transformation it was based on the inversion transformation, introduced in 1845 to 3D electrostatics by William Thomson (Kelvin). The inversion transformation for a four-vector x is defined by x x → I(x) = , (6.130) x ⋅ x where the dot product x ⋅ x = x||x is associated with the Minkowskian metric dyadic

 𝜺 𝜺 𝜺 𝜺 𝜺 𝜺 𝜺 𝜺 𝖦−1. = 1 1 + 2 2 + 3 3 − 4 4 = (6.131) The conformal transformation x → x′ is defined in terms of the inversion transformation by

x′ = C(x, a) = I(a + I(x)) (6.132) a + x ⋅ ( ) (x⋅x ) x + a(x x) = = 휎 , (6.133) a + x ⋅ a + x (x, a) x⋅x x⋅x with

휎(x, a) = 1 + 2a ⋅ x + (a ⋅ a)(x ⋅ x). (6.134)

The transformation has a vector parameter a (“the conformal accel- eration”) and depends on the metric dyadic  = 𝖦−1 defining the dot product. Cunningham’s paper from 1909 [48] contained a demonstra- tion that the Maxwell equations retain their form under the conformal transformation. Thus, any solution to a 4D field problem could be transformed to produce a solution to the transformed problem. If we express the a and x vectors in their respective spatial and temporal components as 휏 ⋅ a = as + ae4, x = r + e4, e4 e4 =−1, (6.135) 160 CHAPTER 6 Transformation of Fields and Media the transformation can be expressed as 1 r′ = (r + a (r2 − 휏2)), (6.136) 휎(x, a) s 1 휏′ = (휏 + a(r2 − 휏2)), (6.137) 휎(x, a) 휎 ⋅ 휏 2 휏2 ⋅ 2 . (x, a) = 1 + 2(r as − a ) + (r − )(as as − a ) (6.138) Because of its nonlinearity, the transformation of a time-harmonic field is no longer time-harmonic in general but obeys a more complicated time dependence. Also, a plane wave is transformed to a field of more general space dependence. Simple boundary surfaces constant in time are transformed to surfaces moving and deforming in time. After its introduction, the conformal transformation was applied to different branches of physics [49]. In electrical engineering the transformation appears lesser known, an exception being its use in studying certain beam-like forms of electromagnetic radiation (focus-wave modes) [50]. In addition to changing fields and their sources, the transformation has also the property of changing the electromagnetic media and boundaries. In the following just a few basic properties will be considered to open up the topic [51].

6.5.1 Properties of the Conformal Transformation Transformation of Vectors Let us start by considering the basic properties of the conformal transformation x → x′. Equation (6.133) can be interpreted by the procedure “invert x,adda and invert the result.” For a = 0 the transformation reduces to two inversions which returns the original vector,

x′ = C(x,0)= I(I(x)) = x. (6.139)

One can readily show that consecutive transformations with vectors a1, a2, … can be reduced to one transformation with the vector a1 + ⋯ a2 + . In fact, a transformation with a1 followed by a transformation with a2 results in

C(C(x, a1), a2) = I(a2 + I(I(a1 + I(x)))) = I(a2 + a1 + I(x)) . = C(x, a1 + a2) (6.140) 6.5 Conformal Transformation 161

From this it follows that the inverse of a transformation with the vector a equals the transformation with the vector −a: C(x, a) = y, ⇒ x = C(y, −a). (6.141) The transformed vector satisfies ⋅ ′ ⋅ ′ x x ( )1( ) x x = 휎 = , (6.142) (x, a) a + x ⋅ a + x x⋅x x⋅x whence x ⋅ x = 0 implies x′ ⋅ x′ = 0. For 휎(x, a) < 0 spatial vectors satisfying x ⋅ x > 0 are transformed to temporal vectors satisfying x′ ⋅ x′ < 0 and vice versa, which may violate causality. Vectors on the light cone (x ⋅ x = 0) are transformed to vectors x x′ = , (6.143) 1 + 2x ⋅ a which also lie on the light cone. The null vector is mapped to the null vector: x′ = C(0, a) = 0. (6.144) Substituting from (6.133) we can prove the relation (z − y) ⋅ (z − y) (z′ − y′) ⋅ (z′ − y′) = , (6.145) 휎(z, a)휎(y, a) for the transformation of any two vectors z, y. Setting now z = x + dx, y = x + dy, (6.146) where dx and dy are small and denoting the transformed vectors by z′ = x′ + dx′, y′ = x′ + dy′, (6.147) the relation (6.145) becomes (dx − dy) ⋅ (dx − dy) (dx′ − dy′) ⋅ (dx′ − dy′) = . (6.148) 휎2(x, a) In the two special cases with dy = 0ordx = 0, we have, respectively, dx ⋅ dx dy ⋅ dy dx′ ⋅ dx′ = , dy′ ⋅ dy′ = , (6.149) 휎2(x, a) 휎2(x, a) which inserted to (6.148) yield dx ⋅ dy dx′ ⋅ dy′ = . (6.150) 휎2(x, a) 162 CHAPTER 6 Transformation of Fields and Media

In analogy to 3D vectors we can define the angles 휑′, 휑 as dx′ ⋅ dy′ dx ⋅ dy cos 휑′ = √ √ = √ √ = cos 휑. dx′ ⋅ dx′ dy′ ⋅ dy′ dx ⋅ dx dy ⋅ dy (6.151) This relation serves as the reason why the transformation is called conformal: “the angle between dx and dy” is invariant in the transformation.

Differential Operations From the basic differentiation rule we can write d′x′ = 𝖨T . (6.152) To find the relation between the operators d′ and d, let us first express the relation between x and x′ as x′ x = a + . (6.153) x′ ⋅ x′ x ⋅ x This can be differentiated as x′ x d = d . (6.154) x′ ⋅ x′ x ⋅ x Applying the rules d(x ⋅ x) = 2(dx)||x = 2|x, (6.155) d(x′ ⋅ x′) = 2(dx′) ⋅ x′ = 2(dx′)||x′, (6.156) (6.154) is transformed to ( ) ( ) (|x′)x′ (|x)x (x ⋅ x)dx′| 𝖨T − 2 = (x′ ⋅ x′) 𝖨T − 2 . (6.157) x′ ⋅ x′ x ⋅ x The dyadics in brackets can be shown to be unipotent, whence they are equal to their inverses. Writing 휎(x, a) = 휎 for brevity and applying (6.142), (6.157) can be expressed as ( ) ( ) (|x′)x′ (|x)x 휎 𝖨T − 2 | 𝖨T − 2 |dx′ = 𝖨T . (6.158) x′ ⋅ x′ x ⋅ x Comparing with (6.152), we can now identify the transformation law for the differential operator d as d → d′ = 𝖢|d, (6.159) 6.5 Conformal Transformation 163 where the dyadic 𝖢 is a function of x defined by ( ) ( ) (|x′)x′ (|x)x 𝖢 = 휎 𝖨T − 2 | 𝖨T − 2 . (6.160) x′ ⋅ x′ x ⋅ x We can also define 𝖢 = 𝖢|(dx) = (𝖢|d)x = d′x. (6.161) while from (6.159) we have dx′ = 𝖢−1|d′x′ = 𝖢−1. (6.162)

Properties of the Dyadic C The dyadic 𝖢 satisfies the differentiation rules ′ 𝖢 ′ ′ ′ 𝖢(2) ′ 𝖢 ∧𝖢 d ∧ = d ∧ d x = 0, d ∧ = (d ∧ )∧ c = 0, (6.163) and 𝖢−1 ′ 𝖢(−2) 𝖢−1 ∧𝖢−1 d ∧ = d ∧ dx = 0, d ∧ = (d ∧ )∧ c = 0, (6.164) where the subscript c again marks a quantity considered constant in differentiation. The inverse of the dyadic 𝖢 can be expressed as ( ) ( ) 1 (|x)x (|x′)x′ 1 𝖢−1 = 𝖨T − 2 | 𝖨T − 2 = |𝖢T |𝖦. (6.165) 휎 x ⋅ x x′ ⋅ x′ 휎2

Thus, the 𝖢 dyadic satisfies the quadratic equations 𝖢||𝖢T = 휎2, 𝖢T |𝖦|𝖢 = 휎2𝖦. (6.166) Obviously, we have ( ) ( ) (|x′)x′ (p) (|x)x (p) 𝖢(p) = 휎p 𝖨T − 2 | 𝖨T − 2 , −4 ≤ p ≤ 4. x′ ⋅ x′ x ⋅ x (6.167) Applying 𝖨 𝜶 (4) 𝖨(4) 𝖨(3)∧ 𝜶 𝖨(4) 𝖨(3)∧ 𝜶 |𝜶 𝖨(4) ( + a ) = + ∧a = (1 + tr( ∧a )) = (1 + a ) , (6.168) 164 CHAPTER 6 Transformation of Fields and Media we obtain ( ) ( ) (|x)x (4) (|x′)x′ (4) 𝖨T − 2 = 𝖨T − 2 =−𝖨(4)T , (6.169) x ⋅ x x′ ⋅ x′ and 𝖢(4) = 휎4𝖨(4)T , 휎4 = tr𝖢(4). (6.170) Applying the rule of the inverse dyadic gives us

𝖨(4)T ⌊⌊𝖢(2)T 1 𝖢(−2) = = 𝖨(4)T ⌊⌊𝖢(2)T , (6.171) tr𝖢(4) 휎4 which combined with (6.165) yields the relation 𝖨(4)⌊⌊𝖢(2) = 𝖦(2)|𝖢(2)|(2). (6.172) Further, we have (x ⋅ x′)2 tr𝖢 = 4휎 = 4(1 + a ⋅ x)2 = 휎2tr𝖢−1, (6.173) (x ⋅ x)(x′ ⋅ x′)

6.5.2 Field Transformation Because the one-forms are transformed in the same manner as the operator d, 𝜶 → 𝜶′(x′) = 𝖢|𝜶(x), (6.174) the electromagnetic field two-forms and the source three-forms are transformed as 𝚽 → 𝚽′(x′) = 𝖢(2)|𝚽(x), (6.175) 𝚿 → 𝚿′(x′) = 𝖢(2)|𝚿(x), (6.176) 𝜸 → 𝜸′(x′) = 𝖢(3)|𝜸(x). (6.177) Since the transformation is not a linear one, the invariance of the Maxwell equations in the transformation is not obvious. However, taking into account the relation (6.163), we can expand d′ ∧ 𝚽′ = d′ ∧ (𝖢(2)|𝚽) ′ 𝖢(2) |𝚽 𝖢 | 𝖢(2)|𝚽 = (d ∧ ) c + ( c d) ∧ c 𝖢(3)| 𝚽 = c (d ∧ ) = 0, (6.178) 6.5 Conformal Transformation 165 and, similarly, the second one becomes ′ 𝚿′ 𝖢(3)| 𝚿 𝖢(3)|𝜸 𝜸′. d ∧ = c (d ∧ ) = = (6.179) This proves that the Maxwell equations appear form-invariant in the conformal transformation. As a consequence, solutions to the Maxwell equations (electromagnetic fields) remain solutions to the Maxwell equations after the conformal transformation (transformed electromagnetic fields).

6.5.3 Medium Transformation From the medium equation manipulated as ′ ′ 𝚿′ = 𝖢(2)|𝚿 = 𝖢(2)|𝖬|𝚽 = 𝖬 |𝚽′ = 𝖬 |𝖢(2)|𝚽, (6.180) we obtain the transformation rule for the medium bidyadic, ′ 𝖬 = 𝖢(2)|𝖬|𝖢(−2). (6.181) This resembles the rule for the affine transformation (6.16), but one must remember that 𝖢 is not a constant dyadic whence a medium loses the homogeneous and time-invariant character in the transformation in the general case. Because (6.181) implies ′ 𝖬 p = 𝖢(2)|𝖬p|𝖢(−2), (6.182) for all integer powers p, a medium whose medium bidyadic 𝖬 satisfies an algebraic equation with constant coefficients is transformed toa medium satisfying the same equation. The modified medium bidyadic is transformed as 𝖬′ ⌊𝖬′ ⌊𝖢(2)| 𝜺 ⌊⌊𝖬 |𝖢(−2) 𝖨(4)⌊⌊𝖢(2) |𝖬 |𝖢(−2). m = eN = eN ( N m) = ( ) m (6.183) Applying (6.171) we can write ′ 𝖬 = 휎4𝖢(−2)T |𝖬 |𝖢(−2) m ( m ) ( ) ( ) x′(|x′) (2) x(|x) (2) (|x)x (2) = 𝖨 − 2 | 𝖨 − 2 |𝖬 | 𝖨T − 2 | x′ ⋅ x′ x ⋅ x m x ⋅ x ( ) (|x′)x′ (2) × 𝖨T − 2 . (6.184) x′ ⋅ x′ 166 CHAPTER 6 Transformation of Fields and Media

From (6.181) we have ′ tr𝖬 = tr𝖬. (6.185)

PROBLEMS

𝖨(4)⌊⌊𝖬 𝖨(4)⌊⌊𝖬 6.1 Find the relation between the bidyadics and a, 𝖬 when a is the affine transformation of the medium bidyadic 𝖬 in terms of a given transformation dyadic 𝖠. 6.2 Find the affine-transformation rules for the 3D medium dyadics 훼 휖′ 휇−1 훽 휖 휉 휁 휇 , , , and g, g, g, g corresponding to the time reversion transformation defined by the dyadic 𝖳 𝖨 𝜺 . = s − e4 4 What media are invariant in time reversion? What about the Gibb- sian representation? 6.3 Defining the dyadic quotient function by [52] 𝖰(𝖷) = (𝖨 − 𝖷)|(𝖨 + 𝖷)−1 = (𝖨 + 𝖷)−1|(𝖨 − 𝖷), prove the following properties 𝖰(𝖰(𝖷)) = 𝖷, 𝖰(0) = 𝖨, 𝖰(𝖨) = 0 𝖰(−𝖷) = (𝖰(𝖷))−1, 𝖰(𝖷−1) =−𝖰(𝖷) 𝖰(𝖦|𝖷T |) = 𝖦|𝖰T (𝖷)| 𝖦 E E  where ∈ 1 1 is a symmetric dyadic and is its inverse. 6.4 Defining the dot product of two vectors a ⋅ b = a||b in terms of a  𝖦−1 F F symmetric dyadic = ∈ 1 1, show that an affine transfor- 𝖱 E F mation (rotation transformation) defined by the dyadic ∈ 1 1 satisfying the property (𝖱|a) ⋅ (𝖱|a) = a ⋅ a, for any vector a, can be represented by 𝖱 = 𝖰(𝖦⌋𝚷), where 𝖰(𝖷) is the dyadic quotient function of Problem 6.3 while 𝚷 F ∈ 2 is some two-form. Problems 167

6.5 Derive the duality transformation conditions for the medium dyadics (6.61)–(6.64). 6.6 Show that the Hehl–Obukhov decomposed parts of the general medium bidyadic are transformed individually in the reciprocity transformation. 6.7 Derive the transformation rule ′ 𝖳 = 𝖢(3)|𝖳|𝖢T for the stress–energy dyadic 𝖳 in the conformal transformation. 6.8 Show that the Hehl–Obukhov decomposed parts of the general medium bidyadic are transformed individually in the conformal transformation. 6.9 Expressing the relation between the four medium dyadics and their dual medium dyadics as 휖 휖 ⎛ dZd ⎞ ⎛ Z ⎞ ⎜휇 ⎟ ⎜휇 ⎟ d∕Zd  휃 ∕Z ⎜ 휉 ⎟ = ( ) ⎜ 휉 ⎟ , ⎜ d ⎟ ⎜ ⎟ ⎝ 휁 ⎠ ⎝ 휁 ⎠ d show that the 4 × 4matrix(휃) is a rotation matrix, that is, it satisfies (휃)T (휃) = ,where is the unit 4 × 4matrix. ′ 6.10 A transformation of medium bidyadics 𝖬→𝖬 can be defined as ′ 𝖬 = Y2𝖨(4)T ⌊⌊𝖬−1T , where Y is a scalar parameter and 𝖬 is a full-rank bidyadic. Find the corresponding transformation for the Gibbsian medium 휖 휉 휁 휇 dyadics g, g, g, g. 6.11 Show that a modified medium bidyadic of the form 𝖬 훼𝖤 𝖰(2) 𝖰 E E m = + , ∈ 1 1 can be transformed to a bidyadic of similar form 𝖬 훼 𝖤 𝖰(2) md = d + d in some duality transformation. 168 CHAPTER 6 Transformation of Fields and Media

𝖬 𝖬 𝖬|𝖬 6.12 Show that from (6.60) it follows that and d satisfy d = 𝖬 |𝖬 d . 6.13 Show that the parameters of the duality transformation can be chosen so that the PEMC boundary condition (6.75) appears self-dual. CHAPTER 7

Basic Classes of Electromagnetic Media

7.1 GIBBSIAN ISOTROPY

The simplest electromagnetic media in terms of Gibbsian representation are isotropic media, defined by equations of the form 𝜖 𝜇 . Dg = Eg, Bg = Hg (7.1) In fact, they do not depend on any special spatial direction and they involve just two scalar medium parameters 𝜖 and 𝜇. From a more general point of view such media can be called spatially isotropic, or invariant in spatial rotations.

7.1.1 Gibbsian Isotropic Medium In 4D electromagnetics, E and H are one-forms and D and B are two- forms whence the scalars 𝜖 and 𝜇 in (7.1) must be replaced by dyadics 𝜖 and 𝜇 mapping one-forms to two-forms. Thus, in the present formalism, (7.1) should be expressed as (see Appendix B) 𝜖| 𝜖𝖢 | 𝜇| 𝜇𝖢 | D = E = s E, B = H = s H, (7.2) with 𝖢 𝜺 ⌊𝖦 𝜺 𝜺 𝜺 F E s = 123 s = 12e3 + 23e1 + 31e2 ∈ 2 1, (7.3)

169 170 CHAPTER 7 Basic Classes of Electromagnetic Media

𝖦 in terms of a symmetric spatial metric dyadic s = e1e1 + e2e2 + e3e3. 𝖢 The dyadic s has the spatial inverse 𝖢−1 𝖦−1⌋ 𝜺 𝜺 𝜺 F E . s = s e123 = 1e23 + 2e31 + 3e12 ∈ 1 2 (7.4) The medium bidyadic corresponding to (7.1) has the form 𝖬 𝜖 𝜺 𝜇−1 = ∧ e4 + 4 ∧ 𝜖𝖢 𝜺 𝜇𝖢 −1 = s ∧ e4 + 4 ∧ ( s) , (7.5) and the medium can be called Gibbsian isotropic medium. Actually, the Gibbsian isotropic medium equals the simple principal medium defined by (5.115). The modified medium bidyadic has the compact form (5.122)

𝖬 1 𝖦 𝜇𝜖 (2) 1 𝖦(2). m =−𝜇 ( s − e4e4) =−𝜇 (7.6) Making an affine transformation with the dyadic 1 𝖠 = 𝖨 + √ e 𝜺 , (7.7) s 𝜇𝜖 4 4 the medium bidyadic (7.5) expressed as √ √ √ 𝖬 𝜖 𝜇 𝜇𝜖 𝖢 𝜺 𝜇𝜖 𝖢 −1 = ∕ (( s) ∧ e4 + 4 ∧ ( s) ), (7.8) is transformed to √ 𝜖 𝖬 = 𝖠(−2)T |𝖬|𝖠(2)T = (𝖢 ∧ e + 𝜺 ∧ 𝖢−1), (7.9) a 𝜇 s 4 4 s and the modified medium bidyadic is transformed to √ 𝜖 𝖬 𝖦 𝜺 (2) ma =− 𝜇 ( s − e4 4) , (7.10) with transformed parameters satisfying 𝜖𝜇 = 1.

7.1.2 Gibbsian Bi-isotropic Medium In the Gibbsian formalism, the bi-isotropic medium is defined by conditions of the form [28] 𝜖 𝜉 𝜁 𝜇 . Dg = Eg + Hg, Bg = Eg + Hg (7.11) 7.1 Gibbsian Isotropy 171

They correspond to the conditions ( ) ( ) ( ) 𝜖𝖢 𝜉𝖢 D s s | E = 𝜁𝖢 𝜇𝖢 , (7.12) B s s H defining a class of what can be called Gibbsian bi-isotropic (GBI) media. Writing these relations in the form ( ) ( ) ( ) D (𝜉∕𝜇)𝖨(2)T (𝜖 − 𝜉𝜁∕𝜇)𝖢 B = s s | , (7.13) H 𝜇𝖢 −1 𝜁 𝜇 𝖨T E ( s) −( ∕ ) s the medium bidyadic of the general GBI medium can be expressed as 𝖬 𝛼 𝜖′ 𝜺 𝜇−1 𝜺 𝛽 = + ∧ e4 + 4 ∧ + 4 ∧ ∧ e4 𝛼𝖨(2)T 𝜖′𝖢 𝜺 𝜇𝖢 −1 𝛽𝜺 𝖨T . = s + s ∧ e4 + 4 ∧ ( s) + 4 ∧ s ∧ e4 (7.14) The two sets of medium parameters are related as ( ) ( ) 𝛼𝜖′ 𝜉∕𝜇𝜖− 𝜉𝜁∕𝜇 = . (7.15) 1∕𝜇𝛽 1∕𝜇 −𝜁∕𝜇 For 𝜉 = 𝜁 = 0, or 𝛼 = 𝛽 = 0, the GBI medium (7.14) reduces to the Gibbsian isotropic medium (7.5).

7.1.3 Decomposition of GBI Medium Let us find the Hehl–Obukhov decomposition (5.91) of the general GBI medium. Denoting 𝛼 = A + B, 𝛽 = B − A, (7.16) (7.14) can be rewritten as 𝖬 𝖨(2)T 𝖨 𝜺 (2)T 𝜺 ⌊𝜖′𝖦 𝜺 𝜇𝖦 −1⌋ . = A + B( s − e4 4) + 123 s ∧ e4 + 4 ∧ ( s) e123 (7.17) Because the trace of each of the last three terms vanishes, the first term equals the axion part of the medium bidyadic. The coefficient 𝛼 − 𝛽 𝜉 + 𝜁 A = = (7.18) 2 2𝜇 is proportional to what is known as the Tellegen parameter of the medium, while the coefficient 𝛼 + 𝛽 𝜉 − 𝜁 B = = (7.19) 2 2𝜇 172 CHAPTER 7 Basic Classes of Electromagnetic Media is proportional to the chirality parameter of the medium [28]. After some steps of algebra, the medium bidyadic (7.17) can be expressed in the form 𝖬 𝖨(2)T 𝖳(2)T 1 𝜺 ⌊𝖦(2) = A + B − 𝜇 N , (7.20) 𝖳 𝖨 𝜺 where = s − e4 4 is the time reversion dyadic and the symmetric dyadic 𝖦 is here defined by 𝖦 𝖦 𝜇𝜖′ 𝖦 = s − e4e4, s = e1e1 + e2e2 + e3e3, (7.21) 𝖦(4) 𝜇𝜖′ satisfying =− eNeN. Since the metric bidyadic ⌊𝖳(2)T eN = e34e12 + e14e23 + e24e31 − e12e34 − e23e14 − e31e24 ⌊𝖳(2)T T =−(eN ) , (7.22) appears antisymmetric, the middle term in (7.20) can be identified as a 𝖦(2) 𝜺 ⌊𝖦(2) skewon bidyadic. Finally, since is symmetric and tr( N ) = 0, as is seen from (7.17), the last term in (7.20) makes the principal part of the medium bidyadic. Thus, the expansion (7.20) actually represents the Hehl–Obukhov decomposition of the modified GBI medium bidyadic, in terms of its respective axion, skewon, and principal parts. One can show that the GBI medium is reciprocal exactly when the axion term vanishes, that is, for tr𝖬 = 0. In fact, applying the time reversion (6.94) to the GBI medium bidyadic, 𝖬∗ = 𝖳(2)T |𝖬|𝖳(2)T 𝖨(2)T ∗ 𝖳(2)T ∗ 1 𝜺 ⌊𝖦(2) ∗ = A( ) + B( ) − 𝜇 ( N ) 𝖨(2)T 𝖳(2)T 1 𝜺 ⌊𝖦(2) = A + B + 𝜇 N , (7.23) the reciprocity transformation (6.96) yields 𝖬 𝖨(4)T ⌊⌊𝖬T∗ r =− 𝖨(4)T ⌊⌊𝖨(2) 𝖨(4)T ⌊⌊𝖳(2) 1 𝖨(4)T ⌊⌊ 𝜺 ⌊𝖦(2)T =−A − B − 𝜇 ( N ) 𝖨(2)T 𝖳(2)T 1 𝜺 ⌊𝖦(2). =−A + B − 𝜇 N (7.24) Comparing with (7.20), the transformation is seen to reverse the sign of the axion term. Thus, the GBI medium is reciprocal whenever A = 0, that is, when the Tellegen parameter vanishes [28]. 7.1 Gibbsian Isotropy 173

7.1.4 Affine Transformation Let us consider the affine-transformed GBI medium bidyadic

𝖬 𝖠(−2)T |𝖬|𝖠(2)T a = 𝖠(−2)T | 𝖨(2)T 𝖳(2)T 1 𝜺 ⌊𝖦(2) |𝖠(2)T . = (A + B − 휇 N ) (7.25)

The first (axion) term is seen to be invariant. The second (skewon) term becomes

𝖠(−2)T |𝖳(2)T |𝖠(2)T 𝖠|𝖳|𝖠−1 (2)T 𝖳(2)T B = B( ) = B a , (7.26) or it is another skewon term with 𝖳 replaced by

∑3 𝖳 𝜺 𝜺 a = eai ai − ea4 a4, (7.27) 1 as defined by the transformed basis vectors and one-forms

𝖠| 𝜺 𝖠−1T |𝜺 . eai = ei, ai = i (7.28)

Finally, the third (principal) term is transformed as

1 𝖠(−2)T | 𝜺 ⌊𝖦(2) |𝖠(2)T 1 𝜺 ⌊𝖠(2)|𝖦(2)|𝖠(2)T − ( N ) =− N 휇 휇tr𝖠(4) 1 =− 𝜺 ⌊𝖦(2), (7.29) 휇 aN a with

∑3 𝖦 휇휖′ . a = eaieai − ea4ea4 (7.30) 1 Thus, the GBI medium is transformed to another GBI medium with axion, skewon, and principal parts transformed individually to respective axion, skewon, and principal parts. This means, for example, that there is no affine transformation which would map the general GBI medium to a simpler Gibbsian isotropic medium with no axion and skewon components. 174 CHAPTER 7 Basic Classes of Electromagnetic Media

7.1.5 Eigenfields in GBI Medium The square of the GBI medium bidyadic can be expanded as

𝖬2 𝖨(2)T 𝖳(2)T 1 𝜺 ⌊𝖦(2) 2 = (A + B − 휇 N ) ( ) 휖′ 2 2 𝖨(2)T 𝖳(2)T 2A𝜺 ⌊𝖦(2)T = A + B − 휇 + 2AB − 휇 N ( ) 휖′ 𝖬 2 2 𝖨(2)T = 2A + −A + B − 휇 (7.31) by applying the following properties, which can be easily checked, (𝖳(2)T )2 = 𝖨(2)T (7.32) 𝜺 ⌊𝖦(2) 2 𝜺 𝜺 ⌊⌊𝖦(2) |𝖦(2) 𝜺 𝜺 ||𝖦(4) 𝖨(2)T ( N ) = ( N N ) = ( N N ) =−휇휖′𝖨(2)T (7.33) 𝖳(2)T | 𝜺 ⌊𝖦(2) 𝜺 ⌊𝖦(2) |𝖳(2)T . ( N ) =−( N ) (7.34) From (7.31) it follows that 𝖬 satisfies an algebraic equation of the second order. It can be written in the form 𝖬 𝖨(2)T | 𝖬 𝖨(2)T ( − M+ ) ( − M− ) = 0, (7.35) with √ 휖′ 휇 2 M± = A ± jS, S = ( ∕ ) − B , (7.36) where A and B are defined by (7.18) and (7.19). M+ and M− equal the two eigenvalues in the problem 𝖬|𝚽 𝚽 . ± = M± ± (7.37) 𝚽 From (7.35) it follows that the two eigen two-forms ± can be expressed as 𝚽 𝖬 𝖨(2)T |𝚯 ± = ( − M∓ ) , (7.38) 𝚯 𝚽 ≠ where may be any two-form yielding ± 0. Excluding the case ≠ S = 0, that is, assuming M+ M−, we can express 1 𝚯 =± 𝚽, (7.39) M+ − M− 7.1 Gibbsian Isotropy 175 in terms of some two-form 𝚽, whence from (7.38) we obtain 𝚽 ±1 𝖬 𝖨(2)T |𝚽. ± = ( − M∓ ) (7.40) M+ − M− Because the eigenfields satisfy 𝚽 𝚽 𝚽 + + − = , (7.41) (7.40) actually represents a rule how the two eigen two-forms components are obtained for a given two-form 𝚽 after solving for the eigenvalues M . The sum rule (7.41) is more obvious through the representation ± ( ) 1 1 1 𝚽 = 𝚽 ± B𝖳(2)T − 𝜺 ⌊𝖦(2) |𝚽. (7.42) ± 2 2jS 휇 N The excluded case S = 0 with coinciding eigenvalues arises when the medium bidyadic is of the special form 𝖬 = A𝖨(2)T + 𝖭 where 𝖭 is any nilpotent bidyadic satisfying 𝖭2 = 0. This can be directly seen from (7.31) written in the form (𝖬 − A𝖨(2)T )2 =−S2𝖨(2). (7.43) For any homogeneous and time-invariant GBI medium possessing ≠ constant eigenvalues M+ M−, the two Maxwell equations 𝚽 𝚽 𝚽 𝜸 d ∧ = d ∧ + + d ∧ − = m, (7.44) 𝚿 𝚽 𝚽 𝜸 d ∧ = M+d ∧ + + M−d ∧ − = e, (7.45) can be decomposed in two noninteracting parts. In fact, defining two source three-forms as 𝜸 1 𝜸 𝜸 ± = ( e − M∓ m), (7.46) M+ − M− the two eigenfields satisfy individually the Maxwell equations 𝚽 𝜸 𝚿 𝜸 d ∧ + = +, d ∧ + = M+ +, (7.47) 𝚽 𝜸 𝚿 𝜸 d ∧ − = −, d ∧ − = M− −, (7.48) whence the problem can be split in two separate parts. 𝚽 𝚽 One can show that, for each of the two eigenfields + and − the original GBI medium can be replaced by an equivalent Gibbsian isotropic (simple principal) medium. The medium equations for the eigenfields can be written as 𝚿 𝖬 |𝚽 ± = ± ±, (7.49) 176 CHAPTER 7 Basic Classes of Electromagnetic Media where the equivalent medium bidyadics have a simpler form corresponding to Gibbsian isotropic media, 𝖬 휖 𝖢 𝜺 휇 𝖢 −1. ± = ± s ∧ e4 + 4 ∧ ( ± s) (7.50) The medium parameters of (7.50) depend on the original medium parameters as √ M A ± B2 − 휖′∕휇 휖 휖′ ± 휖′ ± = = √ (7.51) M± − A − B −B ± B2 − 휖′∕휇 √ M − A + B B ± B2 − 휖′∕휇 휇 휇 ± 휇 . ± = = √ (7.52) M± A ± B2 − 휖′∕휇 Derivation of these expressions is left as an exercise. From the expression (5.205) one can notice that, for the eigenfields 𝚿 𝚽 ±, ± the stress–energy dyadic vanishes, 1 1 𝖳 = 𝚿 ∧ 𝖨T ⌋𝚽 − 𝚽 ∧ 𝖨T ⌋𝚿 ± 2 ± ± 2 ± ± 1 1 = M 𝚽 ∧ 𝖨T ⌋𝚽 − M 𝚽 ∧ 𝖨T ⌋𝚽 = 0. (7.53) 2 ± ± ± 2 ± ± ± Thus, to carry energy, the field must contain components of both eigenfields.

7.1.6 Plane Wave in GBI Medium The dispersion dyadic of a plane wave in a GBI medium, defined by the medium bidyadic (7.20), or the modified medium bidyadic

𝖬 ⌊ 𝖨(2)T 𝖳(2)T 1 𝖦(2) m = eN (A + B ) − 휇 , (7.54) can be expanded as 휇𝖣 𝝂 휇𝖬 ⌊⌊𝝂𝝂 ( ) = m 휇 ⌊𝖳(2)T ⌊⌊𝝂𝝂 𝖦(2)⌊⌊𝝂𝝂 = B(eN ) − 휇 ⌊ 𝖨T ∧𝜺 ⌊⌊𝝂𝝂 𝖦(2)⌊⌊𝝂𝝂. =−2 B(eN ( ∧ 4e4)) − (7.55) The axion term has no effect on plane-wave propagation due to (5.238). Applying the identity 𝖠∧𝖡 T ⌊𝜶 𝖠T |𝜶 𝖡T 𝖡T |𝜶 𝖠T ( ∧ ) = ( ) ∧ + ( ) ∧ , (7.56) 7.1 Gibbsian Isotropy 177 we have 휇 ⌊ 𝖨T ∧𝜺 ⌊⌊𝝂𝝂 휇 𝝂⌋ ⌊ 𝖨T ∧𝜺 ⌊𝝂 −2 B(eN ( ∧ 4e4)) = 2 B (eN ( ∧ 4e4)) 휇 ⌊ 𝝂 𝝂 𝜺 𝜺 |𝝂 𝖨T =−2 BeN ( ∧ ( ∧ 4e4 + ( 4e4 ) ∧ )) 휇 |𝝂 ⌊ 𝝂 𝜺 𝖨T =−2 B(e4 )eN ( ∧ 4 ∧ ) ⌊𝖨T . = As (7.57) The resulting dyadic is of antisymmetric form depending on the spatial bivector As defined by 훼 ⌊ 𝝂 𝜺 훼 ⌊𝝂 훼 휇 |𝝂 As = eN ( ∧ 4) = e123 s, =−2 B(e4 ), (7.58) and satisfying ⌊𝜺 ⌊휈 As ∧ As = 0, As 4 = 0, As = 0, (7.59) ⌊𝖨T (2) ⌊𝖨T (3) . (As ) = AsAs,(As ) = 0 (7.60) The dyadic dispersion equation (5.240) of the GBI medium can now be expanded as 휇3𝖣(3) 𝝂 ⌊𝖨T 𝖦(2)⌊⌊𝝂𝝂 (3) ( ) = (As − ) ⌊𝖨T (3) ⌊𝖨T (2)∧ 𝖦(2)⌊⌊𝝂𝝂 = (As ) − (As ) ∧( ) ⌊𝖨T ∧ 𝖦(2)⌊⌊𝝂𝝂 (2) 𝖦(2)⌊⌊𝝂𝝂 (3) + (As )∧( ) − ( ) ∧ 𝖦(2)⌊⌊𝝂𝝂 ⌊⌊𝖨T ∧ 𝖦(2)⌊⌊𝝂𝝂 (2) =−AsAs∧( ) + (As )∧( ) − (𝖦(2)⌊⌊𝝂𝝂)(3) = 0. (7.61) Since this is a dyadic equation, its symmetric and antisymmetric parts must vanish individually. Now one can show that the antisymmetric equation is identically valid. Writing the symmetric dyadic equation in the form 𝖣(3) 𝝂 𝝂 ⌊⌊𝝂𝝂 ( ) = D( )(eNeN ) = 0, (7.62) it is reduced to a scalar equation. Leaving the details of the analysis as an exercise, the resulting equation can be reduced to 휇3 𝝂 휇휖′ 𝝂 𝝂 . D( ) = D+( )D−( ) = 0 (7.63) The GBI medium is an example of a decomposable (DC) medium since its quartic dispersion equation can be decomposed in two quadratic dispersion equations, 𝝂 𝝂 D+( ) = 0, D−( ) = 0, (7.64) 178 CHAPTER 7 Basic Classes of Electromagnetic Media defined by two dispersion functions which can be given the form ofthe Gibbsian isotropic medium, 𝝂 𝖦 ||𝝂𝝂 D±( ) = ± (7.65) 𝖦 = 𝖦 − 휇 휖 e e , (7.66) ± s ± √± 4 4 휇 휖 휇 휇 2 휇휖′ 2. ± ± =−( B ± ( B) − ) (7.67) 휇 휖 휇휖′ As a check, setting B = 0 for vanishing chirality, we obtain ± ± = 𝖦 𝖦 and, hence, ± = . This corresponds to the Gibbsian isotropic (simple principal) medium case (5.263), for which the two dispersion equations coincide to 𝖦||𝝂𝝂 = 0. Actually, the previous results can be more directly obtained through the eigenfield expansions. Because for each eigenfield the GBI medium can be replaced by an equivalent Gibbsian isotropic medium with 𝖬 medium bidyadics ± defined by (7.50), the dispersion equations can be directly found from (7.65) and the relation (7.67) results from combining (7.51) and (7.52). It is interesting to note that the axion parameter 휖 휇 A, which has an effect on the parameters ± and ±, has no effect on the 휖 휇 product ± ± which governs the wave propagation.

7.2 THE AXION MEDIUM

The GBI medium (7.20) consisting of its axion part only, 𝖬 = M𝖨(2)T , (7.68) can be conceived as the simplest electromagnetic medium since it involves just a single parameter M [53]. Because the axion medium bidyadic is invariant in the affine transformation 𝖬 𝖠(−2)T |𝖬|𝖠(2)T 𝖬 a = = (7.69) 𝖠 E F for any dyadic ∈ 1 1 possessing an inverse (including all 4D rotations) the axion media could also be called 4D isotropic media. In contrast, the special GBI media with only the skewon term, 𝖬 𝖳(2)T 𝖨(2)T 𝖨T ∧𝜺 = B = B( − 2 ∧ 4e4), (7.70) also labeled as simple-skewon (SS) media [54], depend on the choice of 𝜺 the temporal basis vector e4 and one-form 4 as well as the coefficient B. 7.2 The Axion Medium 179

The medium equation of axion media, 𝚿 = 𝖬|𝚽 = M𝚽, (7.71) written for spatial fields as D = MB, H =−ME, (7.72) corresponds to spatial medium dyadics of the form 훼 𝖨(2)T 휖′ 휇−1 훽 𝖨T . = M s , = 0, = 0, =−M s (7.73) Because the dyadic 휇 does not exist, the axion medium cannot be rep- 휖 휇 휉 휁 resented in terms of the four Gibbsian medium dyadics g, g, g, g,at least not in a straightforward manner. In comparison, the medium equations for the SS medium (7.70) differ from (7.72) just for a sign [54]. D = BB, H = BE. (7.74)

7.2.1 Perfect Electromagnetic Conductor Because for M → 0 the medium conditions (7.71) and (7.72) of the axion medium reduce to those of the PMC 𝚿 = 0, ⇒ D = 0, H = 0, (7.75) and for 1∕M → 0 to those of the PEC, 𝚽 = 0, ⇒ B = 0, E = 0, (7.76) being a generalization of these, the axion medium has been given the name PEMC [30, 55, 56]. The coefficient M can then be called the PEMC admittance. 𝜸 𝜸 Assuming both electric and magnetic sources e, m in a PEMC medium with constant and finite M, the Maxwell equations written as 𝚽 𝜸 𝚽 𝜸 d ∧ = e∕M, d ∧ = m, (7.77) impose a condition for the sources, 𝜸 𝜸 e = M m, (7.78) restricting the choice of possible sources in a PEMC medium. Such a restriction is not too extraordinary. In fact, for the same reason, the sources in the PEC and PMC media must satisfy the respective condi- 𝜸 𝜸 tions m = 0and e = 0. 180 CHAPTER 7 Basic Classes of Electromagnetic Media

PEMC is an example of a medium for which the dispersion equation of a plane wave D(𝝂) = 0 is identically satisfied for any one-form 𝝂 due to (5.238). In fact, since the plane-wave field 𝚽 satisfies the equations

𝝂 ∧ 𝚽 = 0, M(𝝂 ∧ 𝚽) = 0, (7.79) which for finite M is the same equation, a solution can be found for any choice of 𝝂. Of course, at the interface of the PEMC medium, the interface conditions impose further restrictions for the one-form 𝝂 and the two-form 𝚽 of a plane wave transmitted into the medium. As another property, the antisymmetric bidyadic 𝚿𝚽 − 𝚽𝚿 vanishes identically for the PEMC medium whence, for example, the energy density and Poynting two-form in the medium are zero. Since PEMC has appeared most interesting when defining electromagnetic boundary conditions at its interface, in which case the exterior fields can be computed without knowing the interior fields, the question of interior fields can be leftto various realizations of the PEMC medium in terms of more conventional media.

7.2.2 PEMC as Limiting Case of GBI Medium Although it is not possible to define the PEMC medium in terms of 휖 휇 휉 휁 Gibbsian dyadics g, g, g, g in a simple manner, we may consider PEMC as a limiting case of skewon-free GBI medium by letting the principal part tend to zero. For that purpose, let us start from (7.20) by setting B = 0 and inserting an extra parameter q as

𝖬 𝖨(2)T 1 𝜺 ⌊𝖦(2) = M − 휇 ( N ) ( q ) 1 = M 𝖨(2)T + (휖𝖢 ∧ e + 𝜺 ∧ (휇𝖢 )−1) + 𝖨T ∧𝜺 e . (7.80) s q s 4 4 s s ∧ 4 4

Obviously, for q → ∞, the expression (7.80) approaches the medium bidyadic of a PEMC medium. The corresponding Gibbsian medium dyadics can be found after comparing (7.80) term wise with the expansion (5.87), written here as

𝖬 𝜺 ⌊휖 𝜺 ⌊휉 𝜺 𝖨T |휇−1| 𝖨⌋ 휁 . = 123 g ∧ e4 + ( 123 g + 4 ∧ s ) g ( e123 − g ∧ e4) (7.81) 7.2 The Axion Medium 181

This leads to the following Gibbsian representation for the PEMC medium ( ) ( ) 휖 휉 휇휖 2 g g M(1 + ( ∕q )) 1 𝖦 = lim q s, (7.82) 휁 휇 q→∞ 11∕M g g It is noteworthy that in this representation all four Gibbsian medium dyadics obtain infinite magnitudes in the limit. However, in spite of this, 𝖦 the determinant of the matrix multiplying the dyadic s has the finite value 휇휖∕q and approaches zero in the limit [57]. Also, the dyadic prod- 휖 ⋅ 휇−1 2𝖨 uct g g = M s remains finite. Applying the representation (7.82), computer codes designed for media expressed in terms of Gibbsian dyadics can be used for problems involving PEMC structures by letting q assume large values.

7.2.3 PEMC Boundary Problems PEC and PMC have been useful concepts because they define ideal boundary conditions at the interface. The same can be anticipated to be the case for the more general PEMC medium. Basic problems involving PEC boundaries concern closed regions like waveguides and cavity resonators and open problems like scattering from PEC objects. As was shown in Chapter 6, there exist duality transformations capable of transforming a PEMC boundary to a PEC boundary. Actually, it is possible to choose the transformation so that PEMC boundaries can be transformed to similar PEC boundaries without changing the adjacent medium when it is simple enough [55]. To show this, let us invoke the three-parameter duality transforma- 𝖬 𝖬 tion rule (6.59) which transforms a medium bidyadic to d as 𝖬 휃 𝖨(2)T 휃 𝖬 | 휃 𝖨(2)T 휃 𝖬 −1. Zd d = (sin + cos Z ) (cos − sin Z ) (7.83) Inserting the PEMC medium bidyadic 𝖬 = M𝖨(2)T in (7.83) we obtain 𝖬 𝖨(2)T another PEMC medium bidyadic d = Md obeying the following 휃 dependence on M and the transformation parameters Z, Zd, , sin 휃 + cos 휃 ZM Z M = . (7.84) d d cos 휃 − sin 휃 ZM Thus, if we wish to transform PEMC to PEC, that is, require that the 𝖬 휃 magnitude of d be infinitely large, we must choose to satisfy cot 휃 = ZM. (7.85) 182 CHAPTER 7 Basic Classes of Electromagnetic Media

The transformation rule (7.83) then becomes 𝖬 𝖨(2)T 2 𝖬 | 𝖨(2)T 𝖬 −1 Zd d = ( + Z M ) (ZM − Z ) , (7.86) where the parameters Z and Zd may still be arbitrary at this stage. To define these two parameters we may require that the medium outside the PEMC boundary be the same after the duality transformation. This is not possible for just any given medium, actually the medium must belong to the class of self-dual media. In Section 6.3.1, it was shown that the simple principal (Gibbsian isotropic) medium represented by the medium bidyadic

𝖬 1 𝜺 ⌊𝖦(2) 𝖦 𝖦 휇휖 =−휇 N , = s − e4e4 (7.87) is invariant if the parameters are chosen as √ 휇 휖. Z = Zd = ∕ (7.88) In conclusion, the three-parameter duality transformation defined√ by 휃 휃 휇 휖 parameters Z, Zd satisfying (7.88) and satisfying cot = M ∕ performs two tasks: changes PEMC boundaries to PEC boundaries and keeps the Gibbsian isotropic medium unchanged. After solving the transformed problem with the PEC boundary, the solution can be transformed back through the inverse duality transformation to yield the solution for the corresponding problem associated with the PEMC boundary.

7.3 SKEWON–AXION MEDIA

As was shown by (5.96), a medium defined by a bidyadic of the form 𝖬 𝖡 ∧𝖨 T 𝖡 = ( o∧ ) ,tro = 0, (7.89) 𝖡 E F where o ∈ 1 1 is a trace-free dyadic, has only a skewon part. More generally, media with no principal part can be represented by bidyadics of the form 𝖬 𝖡∧𝖨 T 𝖨∧𝖡 T = ( ∧ ) = ( ∧ ) , (7.90) in terms of a dyadic 𝖡 which may have nonzero trace. Such media are called skewon–axion media [9] (or “IB-media” [58]). Because the dyadic 𝖡 has 4 × 4 = 16 free parameters, the bidyadic 𝖬 is defined 7.3 Skewon–Axion Media 183 by the same number of parameters instead of 6 × 6 = 36 parameters corresponding to the most general medium bidyadic. Here we must (again) emphasize that the inverse bidyadic 𝖭 = 𝖬−1 is not generally 𝖠∧𝖨 T 𝖬 of the form ( ∧ ) . However, the inverse of a pure skewon is of the 𝖭 𝖠 ∧𝖨 T skewon form = ( o∧ ) as shown by (3.110). To be of the form (7.90), the medium bidyadic 𝖬 must satisfy a certain condition. It can be found by contracting (7.90) as

𝖬⌊⌊𝖨 𝖡∧𝖨 T ⌊⌊𝖨 𝖨 𝖡T 𝖡 𝖨T 𝖡T = ( ∧ ) = (tr ) + (tr ) − 2 (7.91) = 2𝖡T + (tr𝖡)𝖨T . (7.92)

Thetraceoftheleftsideistr(𝖬⌊⌊𝖨) = 2tr𝖬 while the right side yields 6tr𝖡, whence

tr𝖬 = 3tr𝖡, (7.93) which allows us to solve 𝖡 from (7.92) as 1 1 𝖡 = (𝖬⌊⌊𝖨)T − 𝖨(tr𝖬). (7.94) 2 6 Substituting (7.94) in (7.90) we obtain a condition for the medium bidyadic 𝖬 to be of the skewon–axion form (7.90),

𝖬 𝖬 𝖨(2)T 𝖬⌊⌊𝖨 ∧𝖨T . 6 + 2(tr ) − 3( )∧ = 0 (7.95) Because (7.95) follows from (7.90), it is a necessary condition. It is also sufficient because the form of (7.95) proves that 𝖬 can be expressed as 𝖨∧𝖡 𝖡 ∧ with defined by (7.94). The PEMC or axion medium bidyadic is a special case of (7.90) with M 1 𝖬 = M𝖨(2)T = (𝖡∧𝖨)T , 𝖡 = 𝖨, M = tr𝖡. (7.96) ∧ 2 2 The axion and skewon parts of the skewon–axion medium can be separated as 1 𝖬 = M𝖨(2)T + 𝖬 , M = tr𝖬, (7.97) o 6 𝖬 with tr o = 0. 184 CHAPTER 7 Basic Classes of Electromagnetic Media

7.3.1 Plane Wave in Skewon–Axion Medium Let us consider a plane wave in the skewon–axion medium defined by (7.90). Applying the identity

𝖠∧𝖡 T | 𝜶 𝜷 𝖠T |𝜶 𝖡T |𝜷 𝖡T |𝜶 𝖠T |𝜷 ( ∧ ) ( ∧ ) = ( ) ∧ ( ) + ( ) ∧ ( ), (7.98)

𝖠 𝖡 E F 𝜶 𝜷 valid for any dyadics , ∈ 1 1 and one-forms , , the dispersion equation for the potential one-form 𝝓 can be derived as

Assuming that 𝝂 and 𝝂|𝖡 are linearly independent: 𝝂 ∧ (𝝂|𝖡) ≠ 0, that is, 𝝂 is not a left eigen one-form of the dyadic 𝖡, 𝝓 must be linearly dependent on the two one-forms 𝝂 and 𝝂|𝖡 as

𝝓 𝝂 𝝂|𝖡. = c1 + c2 (7.100)

In this case the field two-forms can be expressed as

𝚽 𝝂 𝝓 𝝂 𝝂|𝖡 = ∧ = c2 ∧ ( ), (7.101) 𝚿 𝖡∧𝖨 T | 𝝂 𝝓 𝝂 𝝂|𝖡2 . = ( ∧ ) ( ∧ ) = c2 ∧ ( ) (7.102)

There is no restriction for the wave one-form 𝝂 except 𝝂 ∧ (𝝂|𝖡) ≠ 0 which excludes the pure axion medium. Actually, the wave one-form 𝝂 in the skewon–axion medium can be freely chosen, after which the potential one-form can be determined by (7.99). On the other hand, if 𝝂 is chosen to be an eigen one-form of 𝖡, satisfying

𝝂|𝖡 = 휆𝝂, 휆 ≠ 0, (7.103)

(7.99) is valid for any one-form 𝝓. In this case (7.101) and (7.102) are no longer valid. In such a case there is a lot of freedom for 𝚽 to choose from, because it is only required to satisfy 𝝂 ∧ 𝚽 = 0. The previous analysis shows that skewon–axion media fall in the class of NDE media, media with no dispersion equation. This can be 7.3 Skewon–Axion Media 185 shown more directly by expanding the dispersion dyadic (5.237) as 𝖣 𝝂 𝖬 ⌊⌊𝝂𝝂 𝝂⌋ ⌊ 𝖡∧𝖨 T ⌊𝝂 ( ) = m =− (eN ( ∧ ) ) 𝝂⌋ ⌊ 𝖡T |𝝂 𝖨T 𝝂 𝖡T =− (eN (( ) ∧ + ∧ )) ⌊ 𝝂 𝝂|𝖡 𝖨T = eN ( ∧ ( ) ∧ ) = F⌊𝖨T , (7.104) where the bivector F is defined by ⌊ 𝝂 𝝂|𝖡 . F = eN ( ∧ ( )) (7.105) Because of the form (7.104), the dispersion dyadic of a skewon–axion medium is antisymmetric. From 𝜺 | 𝝂 𝝂|𝖡 𝝂 𝝂|𝖡 F ∧ F = eN N ( ∧ ( ) ∧ ∧ ( )) = 0, (7.106) it follows that F is a simple bivector. Writing F = a ∧ b, the dyadic dispersion equation can be shown to be satisfied identically for any 𝝂,

𝖣(3)(𝝂) = (F⌊𝖨T )(3) = (ba − ab)(3) = 0. (7.107)

7.3.2 Gibbsian Representation To define a skewon–axion medium (7.90) in terms of Gibbsian dyadics, let us start by expanding the dyadic 𝖡 as 𝖡 𝖡 𝜺 𝜷 𝜺 = s + bs 4 + e4 s + be4 4, (7.108) whence the medium bidyadic has the form 𝖬 𝖨T ∧𝖡T 𝜺 𝖨T 𝜷 𝖨T 𝜺 𝖡T 𝖨T . = s ∧ s − 4 ∧ s ∧ bs − s ∧ s ∧ e4 − 4 ∧ ( s + b s ) ∧ e4 (7.109)

Comparing with (5.65), we obtain expressions for the spatial medium dyadics 훼 𝖨T ∧𝖡T 𝖡 𝖨(2)T 𝜺 ⌊⌊𝖡 = s ∧ s = tr s s − 123e123 s, (7.110) 휖′ 𝜷 𝖨T 𝖨(2)⌊𝜷 =− s ∧ s =− s, (7.111) 휇−1 ⌋𝖨(2)T 𝖨T =−bs s =− s ∧ bs, (7.112) 훽 𝖡T 𝖨T . =−( s + b s ) (7.113) 186 CHAPTER 7 Basic Classes of Electromagnetic Media

휖′|𝜷 |휇−1 휖′ 휇−1 Because of s = 0andbs = 0, and do not possess spatial inverses. The medium equations can be written as 𝖡 𝜺 ⌊𝖡 | ⌊ 𝜷 D = tr s B − 123 s (e123 B) − s ∧ E, (7.114) ⌋ 𝖡T | . H =−bs B − s E − bE (7.115) In terms of Gibbsian field vectors, (7.114) and (7.115) take the form (see Appendix B) 𝖡 𝖡 ⋅ 𝜷 Dg = (tr s)Bg − g Bg − g × Eg, (7.116) 𝖡T ⋅ Hg =−bs × Bg − g Eg − bEg, (7.117) where we define 𝖡 𝖡 |𝖦 𝜷 𝖦 |𝜷 . g = s s, g = s s (7.118) It appears instructive to check the previously derived plane-wave properties also through the Gibbsian representation of the skewon–axion medium, (7.116), (7.117). Assuming a time-harmonic plane wave with exp(−jk ⋅ r) dependence and denoting p = k∕휔, the Maxwell equations become

p × Eg = Bg, (7.119) . p × Hg =−Dg (7.120)

Substituting Bg in terms of Eg from (7.119) in (7.116) and (7.117) and substituting Dg and Hg in (7.120) leaves us with the following equation, 𝖡 ⋅ 𝜷 𝖡T 𝖡 ⋅ . (tr s + bs p − b)p × Eg − g × Eg − (p × g + g × p) Eg = 0 (7.121) The dyadic in brackets is actually antisymmetric and can be expressed in the form 𝖡T 𝖡 𝖡 ⋅ 𝖡 𝖨 p × g + g × p = (tr s p − p g) × g, (7.122) 𝖨 𝖦 where the Gibbsian unit dyadic equals the spatial metric dyadic, g = s. Derivation of (7.122) is left as an exercise. Inserting (7.122) in (7.121) yields an equation of the form 𝖣 ⋅ (p) Eg = q(p) × Eg = 0, (7.123) where the vector function of p is defined by ⋅ 𝜷 ⋅ 𝖡 . q(p) = (bs p − b)p − g + p g (7.124) 7.3 Skewon–Axion Media 187

In an analysis of more conventional media we proceed from here by requiring det𝖣(p) = 0, which is called the dispersion equation, yielding the possible values for the normalized propagation vector p as 𝖣 𝖨 its solutions. In the present case (p) = q(p) × g is an antisymmetric dyadic, whence det𝖣(p) = 0 is identically satisfied for any p,which means that a plane wave is possible for any chosen vector p.From (7.123) we see that the corresponding electric field vector must be parallel to the vector q(p). Also, the field vector Bg must be paral- 𝜷 ⋅ 𝖡 lel to p × q(p) = g × p − p g × p.However,ifp is chosen so that q(p) = 0, there is no restriction to the field vector Eg due to (7.123). It is also interesting to study the converse, what media would make the dispersion dyadic 𝖣(p) antisymmetric for any Gibbsian vector p. Inserting the general medium equations in (7.120) and applying (7.119), we obtain

𝖣 휇−1 훽 훼 휖′ . (p) = p × g × p + p × g + g × p + g = 0 (7.125)

Requiring 𝖣(p) to be antisymmetric for any vector p, terms with different orders of p must be individually antisymmetric. This implies that the 휇−1 휖′ 훼T 훽 dyadics g and g must be antisymmetric and g = g, which requires that the principal part of the medium bidyadic must vanish.

7.3.3 Boundary Conditions Like the pure axion medium, the skewon–axion media may also appear somewhat strange as electromagnetic media and one has to wait for engineering applications to change this attitude. However, the skewon– axion media do have importance in creating useful boundary conditions at the medium interface. This was already demonstrated by PEC, PMC and PEMC media, which are all special cases of the axion medium. In Chapter 5, it was shown that boundary conditions can be defined at the interface of media defined by certain medium bidyadics without having to know about the fields in the medium. In the present case wemust know some properties of the fields in the medium to be able to define boundary conditions at the interface. Let us study conditions for the plane-wave fields at the boundary 𝜺 | 𝜺 | < 3 x = 0 of certain skewon–axion media occupying the region 3 x 0. The medium is defined by the choice of the dyadic 𝖡. The pure axion 188 CHAPTER 7 Basic Classes of Electromagnetic Media medium is known to produce the PEMC boundary condition. Let us consider a few cases with simple extensions to the axion medium.

Case 1 The axion medium can be extended by adding a simple dyadic term a𝜶 to the defining dyadic 𝖡 as 𝖡 = B𝖨 + a𝜶, (7.126) 𝖡2 = B2𝖨 + (2B + a|𝜶)a𝜶, (7.127) where 𝜶 is a one-form and a is a vector. In this case the field two-forms of a plane wave in the medium (7.101) and (7.102) can be expressed as 𝚽 = 𝝂 ∧ 𝜶(a|𝝂), (7.128) 𝚿 = (2B + a|𝜶)𝝂 ∧ 𝜶(a|𝝂), (7.129) whence they satisfy the conditions 𝜶 ∧ 𝚽 = 0, 𝜶 ∧ 𝚿 = 0 (7.130) for any chosen one-form 𝝂. Let us first assume that 𝜶 is a temporal 𝜶 𝜺 one-form = 4. In this case the conditions (7.130) yield 𝜺 𝚽 𝜺 ⇒ 4 ∧ = 4 ∧ B = 0, B = 0, (7.131) 𝜺 𝚿 𝜺 ⇒ . 4 ∧ = 4 ∧ D = 0, D = 0 (7.132) Since the fields B and D of the plane wave must vanish in the medium, 𝜺 | from continuity at an interface defined by 3 x = 0, the fields must satisfy the boundary conditions 𝜺 𝜺 3 ∧ D = 0, 3 ∧ B = 0, (7.133) known as the DB conditions. The corresponding Gibbsian conditions are (5.172) ⋅ ⋅ e3 Dg = 0, e3 Bg = 0, (7.134) for the Gibbsian vector fields. Because (7.133) are linear conditions for the fields and independent of the one-form 𝝂, they are valid for any sum or integral of plane waves and, consequently, for fields created by any sources. Actually, they could have been derived for general fields without the plane-wave assumption [59]. 7.3 Skewon–Axion Media 189

Case 2 As another special case of (7.126), (7.127), let us assume that 𝜶 is a 𝜶 𝜺 spatial one-form, = 1. In this case the fields in the medium satisfy 𝜺 𝚽 ⇒ 𝜺 𝜺 1 ∧ = 0, 1 ∧ B = 0, 1 ∧ E = 0 (7.135) 𝜺 𝚿 ⇒ 𝜺 𝜺 . 1 ∧ = 0, 1 ∧ D = 0, 1 ∧ H = 0 (7.136) In this case the conditions for B and D are not continuous through the interface while those for E and H yield the boundary conditions 𝜺 𝜺 3 ∧ 1 ∧ E = 0, (7.137) 𝜺 𝜺 . 3 ∧ 1 ∧ H = 0 (7.138) For the Gibbsian fields they correspond to (5.183) ⋅ ⋅ (e3 × e1) Eg = e2 Eg = 0, (7.139) ⋅ ⋅ (e3 × e1) Hg = e2 Hg = 0, (7.140) and can be identified as the SH boundary conditions [39]. Here, e2 is a ⋅ unit vector tangential to the interface e3 r = 0.

Case 3 As the next possibility let us extend the previous dyadic 𝖡 by adding a second term b𝜷 as 𝖡 = B𝖨 + a𝜶 + b𝜷, (7.141) 𝖡2 = B2𝖨 + a′𝜶 + b′𝜷, (7.142) with a′ = 2Ba + a|(𝜶a + 𝜷b), (7.143) b′ = 2Bb + b|(𝜶a + 𝜷b). (7.144) Because the field two-forms in such a medium (7.101) and (7.102) have the form 𝚽 = 𝝂 ∧ (𝜶a + 𝜷b)|𝝂, (7.145) 𝚿 = 𝝂 ∧ (𝜶a′ + 𝜷b′)|𝝂, (7.146) they must satisfy the conditions 𝜶 ∧ 𝜷 ∧ 𝚽 = 0, (7.147) 𝜶 ∧ 𝜷 ∧ 𝚿 = 0. (7.148) 190 CHAPTER 7 Basic Classes of Electromagnetic Media

Let us first assume that one of the two one-forms is temporal andthe 𝜶 𝜺 𝜷 𝜺 other one spatial, = 3 and = 4. In this case we obtain the DB conditions 𝜺 𝜺 . 3 ∧ B = 0, 3 ∧ D = 0 (7.149)

As a second possibility we assume that both one-forms are spatial, 𝜶 𝜺 𝜷 𝜺 𝜺 | = 3 and = 1. In this case the fields at the interface 3 x = 0 satisfy 𝜺 𝜺 𝜺 𝜺 3 ∧ 1 ∧ E = 0, 3 ∧ 1 ∧ H = 0, (7.150) which yield the SH conditions (7.139), (7.140).

Case 4 𝜶 𝜺 𝜷 As a further generalization of Case 3, let us choose = 3 and = 𝜺 𝜺 1 + A 4,whereA is a scalar. The conditions (7.147), (7.148), for the fields in the skewon–axion medium now become

𝜺 𝜺 𝜺 𝜺 3 ∧ (A 4 ∧ B + 1 ∧ E ∧ 4) = 0, (7.151) 𝜺 𝜺 𝜺 𝜺 3 ∧ (A 4 ∧ D − 1 ∧ H ∧ 4) = 0, (7.152) and they are equivalent to the spatial conditions

𝜺 𝜺 3 ∧ (AB + 1 ∧ E) = 0, (7.153) 𝜺 𝜺 3 ∧ (AD − 1 ∧ H) = 0, (7.154) which, because of continuity, again serve as boundary conditions at the 𝜺 | interface 3 x = 0. The corresponding Gibbsian conditions are of the form ⋅ ⋅ Ae3 Bg + e2 Eg = 0, (7.155) ⋅ ⋅ . Ae3 Dg − e2 Hg = 0 (7.156)

Since these conditions generalize the SH conditions (obtained for A = 0) and the DB conditions (obtained for |A| → ∞), they have been dubbed as SHDB conditions [60]. It is left as an exercise to show that SHDB conditions can be also obtained through other definitions of skewon– axion media. 7.3 Skewon–Axion Media 191

Case 5 The electromagnetic medium can also be defined through the bidyadic 𝖭 in the alternative way of expressing the medium equation (5.52), 𝚽 = 𝖭|𝚿. (7.157) When 𝖬 has an inverse, we have 𝖭 = 𝖬−1. The previous analysis could 𝖭 𝖠∧𝖨 T have made by defining = ( ∧ ) and all of the results corresponding to the cases 1–5 could have arrived at through this route as well. However in the case when 𝖭−1 does not exist, there is no bidyadic 𝖬. As an example of such a case let us start by considering a skewon–axion medium defined by the bidyadic 𝖬 with a unipotent dyadic 𝖡 satisfying 𝖡2 = 𝖨. (7.158) From (7.102) we then obtain Ψ=𝝂 ∧ (𝝂|𝖡2) = 𝝂 ∧ 𝝂 = 0, (7.159) whence the plane wave satisfies the PMC conditions D = 0, H = 0, (7.160) in such a medium. This means that at any interface the Gibbsian PMC conditions ⋅ n Dg = 0, n × Hg = 0 (7.161) are valid. The corresponding medium yielding the PEC condition in the medium, 𝚽 = 0, and at the interface, ⋅ n Bg = 0, n × Eg = 0 (7.162) are obtained by defining a skewon–axion medium through a bidyadic 𝖭 in terms of a unipotent dyadic 𝖠 as 𝖭 𝖠∧𝖨 T 𝖠2 𝖨. = ( ∧ ) , = (7.163) As a conclusion, it was seen that same boundary conditions can be obtained at the interface of many different media. A more recent addition is the SHDB boundary generalizing the SH and DB boundary conditions which have been known since the 1950s. One can show that the SHDB conditions share the property of both SH and DB conditions as being 192 CHAPTER 7 Basic Classes of Electromagnetic Media self-dual, that is, invariant in any three-parameter duality transformation of the form (6.36) or ( ) ( )( ) 𝚿 휃 휃 𝚿 Zd d cos sin Z . 𝚽 = 휃 휃 𝚽 (7.164) d −sin cos In fact, expressing (7.153) and (7.154) as 𝜺 𝜺 f(E, B) = 3 ∧ (AB + 1 ∧ E) = 0, (7.165) 𝜺 𝜺 f(−H, D) = 3 ∧ (AD − 1 ∧ H) = 0, (7.166) we can write the relation ( ) ( )( ) 휃 휃 f(Ed, Bd) cos −Z sin f(E, B) = 휃 휃 , (7.167) f(−Hd, Dd) (1∕Zd)sin (Z∕Zd)cos f(−H, D) from which we obtain . f(Ed, Bd) = 0, f(−Hd, Dd) = 0 (7.168) Thus, an object with an SHDB boundary and suitable symmetry has zero backscattering, that is, it is invisible for the radar [43].

7.4 EXTENDED SKEWON–AXION MEDIA

The skewon–axion medium can be extended by adding a simple term to the medium bidyadic as 𝖬 𝖡∧𝖨 T 𝚫 = ( ∧ ) + B, (7.169) where 𝚫 is some two-form and B some bivector. The corresponding modified medium bidyadic has the form 𝖬 ⌊ 𝖡∧𝖨 T m = eN ( ∧ ) + AB, (7.170) ⌊𝚫 where A = eN is a bivector. Let us consider the effect of the added term to plane-wave properties. The dispersion dyadic (7.104) is extended correspondingly as 𝖣(𝝂) = F⌊𝖨T + ab, (7.171) with ⌊𝝂 ⌊𝝂 ⌊ 𝝂 𝝂|𝖡 . a = A , b = B , F = eN ( ∧ ( )) (7.172) 7.4 Extended Skewon–Axion Media 193

Because the bivector F is simple, it satisfies (F⌊𝖨T )(2) = FF,(F⌊𝖨T )(3) = 0, (7.173) whence the dyadic dispersion equation (5.240) becomes 𝖣(3) 𝝂 ⌊𝖨T (3) ⌊𝖨T (2)∧ ( ) = (F ) + (F ) ∧ab ∧ . = FF∧ab = 0 (7.174) Expanding 𝜺 𝜺 ⌊⌊ ∧ 𝜺 𝜺 ⌊⌊ ⌊⌊ N N (FF∧ab) = ( N N FF) ab = ((𝝂 ∧ (𝝂|𝖡))⌊a)((𝝂 ∧ (𝝂|𝖡))⌊b) = 𝝂𝝂(𝝂|𝖡|a)(𝝂|𝖡|b), (7.175) wherewehaveapplied𝝂|a = 𝝂|b = 0 in the bac-cab rule, the quartic dispersion equation is decomposed in two scalar dispersion equations of the second order, 𝝂|𝖡|a = 𝝂|(𝖡⌋A)|𝝂 = 0, (7.176) 𝝂|𝖡|b = 𝝂|(𝖡⌋B)|𝝂 = 0, (7.177) whose solutions define two characteristic waves which may respectively be called by the names A wave and B wave. Thus, the extended skewon– axion medium is an example of a DC medium. The previous analysis shows us that, extending the skewon–axion medium bidyadic by a term as in (7.170), the one-form 𝝂 can no longer be freely chosen but, instead, must satisfy either of the two conditions (7.176), (7.177). This is because the added term contains a nonzero principal component. In the general case, the A wave and the B wave obey different dispersion equations, whence such a medium is birefringent. The two equations become the same when the symmetric parts of the two dyadics 𝖡⌋A and 𝖡⌋B are multiples of each other. Inserting (7.171) in (5.236), the equation for the potential one-form 𝝓 of the plane wave in the extended skewon–axion medium becomes 𝖣(𝝂)|𝝓 = (F⌊𝖨T + ab)|𝝓 ⌊ 𝝂 𝝂|𝖡 𝝓 |𝝓 . = eN ( ∧ ( ) ∧ ) + a(b ) = 0 (7.178) Multiplying this by 𝝓| we obtain 𝝓|𝖣|𝝓 = F|(𝝓 ∧ 𝝓) + (a|𝝓)(b|𝝓) = (a|𝝓)(b|𝝓) = 0, (7.179) 194 CHAPTER 7 Basic Classes of Electromagnetic Media whence the potential one-form of the plane wave must satisfy one of the two polarization conditions

a|𝝓 = A|(𝝂 ∧ 𝝓) = A|𝚽 = 0, (7.180) b|𝝓 = B|(𝝂 ∧ 𝝓) = B|𝚽 = 0. (7.181)

To find the connection between the two dispersion equations (7.176), (7.177) and the two polarization conditions (7.180), (7.181), we multiply (7.178) by (𝝂|𝖡)| which yields

(𝝂|𝖡|a)(b|𝝓) = 0. (7.182)

Assuming b|𝝓 ≠ 0, whence a|𝝓 = 0, we must have 𝝂|𝖡|a = 0, whence (7.180) corresponds to (7.176) of the A-wave. Thus, we may expect that (7.181) must correspond to (7.177) of the B-wave. Details of the proof are left as an exercise. This reasoning fails when (7.176) and (7.177) are both simultaneously valid, in which case the medium is nonbirefringent and the polarization is not restricted by either of the conditions.

PROBLEMS

7.1 Show that if the Gibbsian isotropic medium bidyadic (7.5) has only a principal component, by requiring the condition 𝖬⌊⌊𝖨 = 0. ⌊𝖢 𝖦 E E The metric dyadic e123 s = s ∈ 1 1 must be symmetric. 7.2 Starting from the medium bidyadic (7.17) of the GBI medium, derive the expression (7.20). 7.3 Do the details in the derivation of (7.33) and (7.34) to arrive at (7.31). 𝖳(2) 𝖨 𝜺 (2) 7.4 Show that the bidyadic = ( s − e4 4) is a skewon bidyadic 𝖡 ∧𝖨 and that it can be expressed in the form o∧ ,intermsofa 𝖡 trace-free dyadic o. 7.5 Derive the expressions for equivalent medium parameters (7.51) and (7.52) of two Gibbsian isotropic medium bidyadics

1 𝖬 = 휖 𝖢 ∧ e + 𝜺 ∧ 𝖢−1 ± ± s 4 휇 4 s ± Problems 195

replacing a given GBI medium bidyadic 1 𝖬 = A𝖨(2)T + B𝖳(2)T + 휖′𝖢 ∧ e + 𝜺 ∧ 𝖢−1 s 4 휇 4 s 𝖬 |𝚽 𝚽 for eigenfields as ± ± = M± ±, by applying 3D decomposi- 𝚽 𝜺 tions = B + E ∧ 4. 7.6 Show that the antisymmetric part of the dyadic dispersion equation (7.61) vanishes identically for all 𝝂. The identity of Problem 2.22 may prove useful in the analysis. 7.7 Expand the symmetric part of the dyadic dispersion equation (7.61) in detail and verify that the dispersion functions are of the form (7.65)– (7.67). 7.8 Show that the general GBI-medium bidyadic (7.20) can be expressed as 𝖬 = A𝖨(2)T + B𝖴 + C𝖩, where 𝖴 is a unipotent bidyadic and 𝖩 is an AC bidyadic satisfying 𝖴2 =−𝖩2 = 𝖨(2)T . Show also that the three terms are mapped to similar terms in the similarity transformation 𝖣|𝖬|𝖣−1. 7.9 Show that a duality transformation can be defined so that a given GBI medium is self-dual, that is, invariant in the transformation. 7.10 Study the fields in the simple skewon–axion medium defined by the medium bidyadic 𝖬 𝖨(2)T 𝖳(2)T 훼𝖨(2)T 훽𝖨T ∧𝜺 = A + B = s − s ∧ 4e4, with B ≠ 0. Consider solutions to the Maxwell equations in a homogeneous medium and show that such a medium produces DB boundary conditions (5.171) at its interface. 7.11 Derive the Gibbsian dyadic rule (7.122). 7.12 Derive the Gibbsian medium equations (7.116) and (7.117) for the skewon–axion medium. 7.13 Show that if the dispersion dyadic corresponding to the medium bidyadic 𝖬 can be expressed in the form 𝖣 𝝂 ⌊ 𝜶 𝝂 𝜷 𝝂 𝖨T ( ) = eN ( ( ) ∧ ( ) ∧ ), 196 CHAPTER 7 Basic Classes of Electromagnetic Media

there is no dispersion equation, that is, D(𝝂) = 0 is satisfied for any 𝝂. Here the one-forms 𝜶 and 𝜷 are functions of 𝝂.Show also that in such a case the medium cannot have a principal 𝖢⌊⌊𝜸𝜸 𝜸 F component. Hint: apply the rule = 0forall ∈ 1 implies 𝖢 E E 훼𝖤 that ∈ 2 2 is of the form . 7.14 Study the possibility of defining boundary conditions at an 𝖬 𝖡∧𝖨 T interface of a skewon–axion medium = ( ∧ ) by defining 𝜶 𝜺 𝜺 = A 4 + 1 in (7.126) and (7.127). 𝖬 7.15 Derive the condition (7.95) by requiring that the trace-free part o 𝖬 𝖨(4)T ⌊⌊𝖬T 휆𝖬 of the medium bidyadic is an eigendyadic of o = o corresponding to the eigenvalue 휆 =−1. 7.16 Find the inverse of the extended skewon-medium bidyadic (7.169). 7.17 Find the conditions for the Gibbsian medium dyadics corresponding to a skewon medium by starting from the relations (5.102)– (5.105). 7.18 Show that a principal–axion medium satisfying 휖′ = 0and휇−1 = 0 is defined by medium equations of the form 𝜺 ⌊훽T ⌋ | 훽| . D =−( 123 e123) B, H = E Find the dispersion equation for a plane wave in such a medium. 7.19 Consider the special principal–axion medium of the previous problem in terms of Gibbsian symbols and find the dispersion equation of a plane wave. 7.20 Show in detail that the dispersion equation for a plane wave in an extended skewon–axion medium defined by (7.170) corresponding to the polarization condition b|𝝓 = 0is𝝂|(𝖡⌋B)|𝝂 = 0. CHAPTER 8

Quadratic Media

In this chapter we consider media whose medium bidyadic or modified medium bidyadic can be expressed as a double-wedge square of some dyadic. In both of these cases the the number of free parameters is 16 instead of 36. In addition, simple extensions of these media will be covered.

8.1 P MEDIA AND Q MEDIA

There are two possibilities in defining media in terms of bidyadics 𝖯 E F obeying a double-wedge square law. If ∈ 1 1 is a dyadic mapping vectors to vectors, a medium bidyadic can be defined as [61] 𝖬 𝖯(2)T F E . = ∈ 2 2 (8.1) 𝖰 E E On the other hand, if ∈ 1 1 is a dyadic mapping one-forms to vectors, a modified medium bidyadic can be defined as [1,62] 𝖬 𝖰(2) E E . m = ∈ 2 2 (8.2) Both (8.1) and (8.2) define certain classes of electromagnetic media which have been called by the respective names P media and Q media in the past. Alternatively, it is possible to define

𝖬 = M𝖯(2)T , (8.3) 𝖬 𝖰(2) m = M , (8.4)

197 198 CHAPTER 8 Quadratic Media by adding a coefficient M in which case we can impose a normalizing condition on the dyadics 𝖯 and 𝖰. For example, if the dyadics are of full rank, we could choose

𝖯(4) ΔP = tr = 1, (8.5) 𝜺 𝜺 ||𝖰(4) ΔQ = N N = 1, (8.6) or −1 instead of 1. However, since (8.1) and (8.2) appear simpler and are not limited by the full-rank requirement, they will be applied in the sequel even if the dyadics 𝖰 and 𝖯 may involve complex components. As an example, the simple principal medium (5.122) (Gibbsian isotropic medium), defined by the modified medium bidyadic ofthe form

𝖬 1 𝖦(2) 𝖦 𝖦 𝜇𝜖 m =−𝜇 , = s − e4e4, (8.7) belongs to a subclass√ of Q media with a symmetric bidyadic 𝖰(2) because we can set 𝖰 =±𝖦∕ −𝜇 with either sign. One can show that any spatial bidyadic or a modified spatial bidyadic possessing a spatial inverse can be uniquely expressed in terms of a spatial dyadic 𝖯 or a metric dyadic 𝖰 in the form (8.1) or (8.2). The proof of this property, which is known for Gibbsian 3D dyadics [25], is left as an exercise. However, since this is not valid in the general 4D case, bidyadics and modified bidyadics satisfying (8.1) or (8.2) form two definite classes of media. One can also show that P media and Q media do not have common elements when 𝖯 and 𝖰 are full-rank dyadics. To prove this, we may consider the equation 𝖰(2) ⌊𝖯(2)T = eN (8.8) 𝖰 𝖯 ≠ and show that there is no solution for any given satisfying ΔP 0. An equivalent equation is 𝖷(2) ⌊𝖨(2)T = eN , (8.9) for the metric dyadic 𝖷 = 𝖰|𝖯−1T . Details of the proof are left as an exercise. However, when the bidyadics are not restricted by the full- rank requirement, there exist solutions to (8.8). For example, choosing 𝖯 𝜺 𝜺 𝖰 = e1 3 + e2 4 we have a solution = e1e1 + e2e2. 8.1 P Media and Q Media 199

For full-rank dyadics 𝖯 and 𝖰, the corresponding medium bidyadics have inverses in both cases and their respective expressions can be obtained from the rules (2.123) and (2.207) as 1 1 𝖬−1 = 𝖯(−2)T = 𝖨(4)T ⌊⌊𝖯(2) = 𝖨(4)T ⌊⌊𝖬T , (8.10) ΔP ΔP 𝖬−1 𝖰(−2) 1 𝜺 𝜺 ⌊⌊𝖰(2)T 1 𝜺 𝜺 ⌊⌊𝖬T . m = = N N = N N m (8.11) ΔQ ΔQ 𝖭 𝖭 Considering the representation (5.52), the and m bidyadics can be expressed for the P medium as 𝖭 = 𝖬−1 = (𝖯−1T )(2), (8.12) and for the Q medium as 𝖭 ⌊𝖭 ⌊𝖬−1 m = eN = eN = e ⌊(𝜺 ⌊𝖬 )−1 = e ⌊𝖬−1⌋e N N m √N m N 1 𝖰(2)T 𝖰T (2). = = ( ∕ ΔQ) (8.13) ΔQ Consequently, the classes of P media and Q media can equally well defined through the 𝖭 bidyadic representation (5.52) in the case when the dyadics are of full rank. Because of the expansions 𝖯(2) ⋅ 𝖯(2)T 𝖯(2)| ⌊𝖯(2)T 𝖨(2)⌋ = eN =ΔP eN, (8.14) and 𝖰(2)T ⋅ 𝖰(2) 𝖰(2)T ⌋𝜺 |𝖰(2) ⌊𝖨(2)T = N =ΔQeN , (8.15) the medium bidyadics of P media and modified medium bidyadics of Q media satisfy the modified closure conditions (4.141) and (4.142) as 𝖬T ⋅ 𝖬 𝛼 ⌊𝖨(2)T = eN , (8.16) 𝖬T ⋅ 𝖬 𝛼 ⌊𝖨(2)T m m = eN , (8.17) 𝛼 where equals respectively ΔP or ΔQ. The converse is also true. In fact, solutions of the equation 𝖬T ⋅ 𝖬 𝛼 ⌊𝖨(2)T = eN (8.18) for 𝛼 ≠ 0 have been shown to be either of the P medium form (4.144) or the Q medium form (4.145). 200 CHAPTER 8 Quadratic Media

8.2 TRANSFORMATIONS

𝖠 E F An affine transformation (6.16) or (6.20) in terms of adyadic ∈ 1 1 yields for the P medium 𝖬 𝖠(−2)|𝖬|𝖠(2)T 𝖠(−2)|𝖯(2)T |𝖠(2)T a = = 𝖠−1|𝖯T |𝖠T (2) 𝖯(2)T = ( ) = a , (8.19) and, for the Q medium, 𝖬 𝖠(2)|𝖬 |𝖠(2)T 𝖠(2)|𝖰(2)|𝖠(2)T ma = m = 𝖠|𝖰|𝖠T (2) 𝖰(2). = ( ) = a (8.20) Thus, a P medium is transformed to another P medium and a Q medium is transformed to another Q medium as defined by the respective dyadics 𝖯 𝖠|𝖯|𝖠−1T 𝖰 𝖠|𝖰|𝖠T . a = , a = (8.21) 𝖰 𝖰 𝖰 In particular, if is symmetric or antisymmetric, so is a. Also, if is 𝖰(2) 𝖰 not symmetric but is symmetric, a has the same property. The three-parameter involutionary duality transformation of the medium bidyadic can be studied through the rule (6.65) as 1 1 𝖬 = 𝖬−1 = 𝖭, (8.22) d Z2 Z2 whence the P medium and the Q medium bidyadics are respectively transformed to 1 𝖬 = (𝖯(2)T )−1 = ((Z𝖯T )−1)(2)T , (8.23) d Z2 1 𝖬 = (𝜺 ⌊𝖰(2))−1 = (Z𝖰)(−2)⌋e . (8.24) d Z2 N N Thus, the P medium is transformed to another P medium with 𝖯 𝖯 −1 d = (Z ) , while the Q medium is transformed to another Q medium defined by 𝖬 ⌊𝖬 ⌊⌊ 𝖰 (−2) md = eN d = eNeN (Z ) 1 = 𝖰(2)T = 𝖰(2), (8.25) 2 d Z ΔQ 𝖰 1 𝖰T . d = √ (8.26) Z ΔQ 8.3 Spatial Expansions 201

For symmetric dyadics 𝖰 the Q medium is self-dual, that is,√ there exists an involutionary duality transformation defined by Z = 1∕ ΔQ which leaves the Q medium invariant.

8.3 SPATIAL EXPANSIONS

8.3.1 Spatial Expansion of Q Media Let us find the spatial medium dyadics of a Q medium in the expansion

𝖬 훼 휖′ 𝜺 휇−1 𝜺 훽 = + ∧ e4 + 4 ∧ + 4 ∧ ∧ e4, (8.27) by first expanding the metric dyadic 𝖰 as

𝖰 𝖰 = s + e4as + bse4 + ce4e4, (8.28) 𝖰 E E E where s ∈ 1 1 is a spatial dyadic, as, bs ∈ 1 are two spatial vectors and c is a scalar. The corresponding expansion for the double-wedge square is

𝖰(2) 𝖰(2) 𝖰 𝖰 𝜺 𝖰 = s − e4 ∧ s ∧ as − bs ∧ s ∧ e4 − 4 ∧ (c s − bsas) ∧ e4, (8.29) which yields

𝖬 𝜺 ⌊𝖰(2) = N 𝜺 𝜺 ⌊𝖰(2) 𝜺 ⌊𝖰 =− 4 ∧ ( 123 s ) + 123 s ∧ as 𝜺 𝜺 ⌊ 𝖰 𝜺 𝜺 ⌊ 𝖰 . + 4 ∧ ( 123 (bs ∧ s)) ∧ 4 + 123 (c s − bsas) ∧ e4 (8.30)

Comparing with (8.27) and equating the respective terms, leads to the following set of relations,

훼 𝜺 ⌊𝖰 = 123 s ∧ as, (8.31) 휖′ 𝜺 ⌊ 𝖰 = 123 (c s − bsas), (8.32) 휇−1 𝜺 ⌊𝖰(2) =− 123 s , (8.33) 훽 𝜺 ⌊ 𝖰 . = 123 (bs ∧ s) (8.34) 202 CHAPTER 8 Quadratic Media

One should note that the dyadics 훼 and 훽 of a Q medium do not have spatial inverses because they satisfy 훼 𝜺 ⌊𝖰 ∧ as = 123 s ∧ as ∧ as = 0, (8.35) |훽 𝜺 | 𝖰 . bs = 123 (bs ∧ bs ∧ s) = 0 (8.36) 𝖰 E E 𝖰(3)||𝜺 𝜺 ≠ Assuming that s ∈ 1 1 is of full rank 3 with ΔQs = s 123 123 0, it has the spatial inverse

𝖰−1 1 𝜺 𝜺 ⌊⌊𝖰(2)T . s = 123 123 s (8.37) ΔQs Applying the expansion (5.87), the Gibbsian medium dyadics of the Q medium can be identified as ( ) ( ) 휖 휉 휖𝖰 𝖨 ⌋X g g = s s s , (8.38) 휁 휇 𝖨 ⌋ 휇𝖰T g g s Zs s with the scalars 휖 |𝖰−1| 휇 = c − as s bs, =−1∕ΔQs (8.39) and the spatial bivectors ⌊𝖰−1T | ⌊𝖰−1| . Xs =−e123 s as, Zs = e123 s bs (8.40) 휖 휇 From (8.38) we conclude that the Gibbsian medium dyadics g and g of a Q medium must be related through a condition of the form 휖 휇T g = a g , (8.41) 휉 휁 while the Gibbsian medium dyadics g and g may be any antisymmetric dyadics. Let us compare the number of free parameters in the different definitions of the Q medium. In the Gibbsian representation, the permittivity and permeability are defined by 9 + 1 = 10 parameters because a dyadic 휇 휉 g and a scalar a can be chosen freely. The antisymmetric dyadics g and 휁 g involve two spatial bivectors Xs, Zs which correspond to 3 + 3 = 6 parameters. The total of 10 + 6 = 16 parameters equals 4 × 4 of the 4D dyadic 𝖰. 8.3 Spatial Expansions 203

8.3.2 Spatial Expansion of P Media To find the expansion of the medium bidyadic (8.27) for the Pmedium, let us expand 𝖯 𝖯 𝝅 𝜺 𝜺 = s + e4 s + ps 4 + pe4 4, (8.42) 𝖯 𝝅 where the dyadic s, vector ps and one-form s are spatial quantities, while p is a scalar. Inserting (8.42) in (8.1) we obtain 𝖬 𝖯(2)T 𝖯T ∧ 𝜺 𝝅 𝖯T 𝝅 ∧𝜺 . = s + s ∧( 4ps + se4) + (p s − sps)∧ 4e4 (8.43) Comparing termwise with (8.27), the spatial medium dyadics can be identified as 훼 𝖯(2)T = s , (8.44) 휖′ 𝝅 𝖯T =− s ∧ s , (8.45) 휇−1 𝖯T =− s ∧ ps, (8.46) 훽 𝝅 𝖯T . = sps − p s (8.47) The expansion parameters of a P medium can be shown to depend on each other through the relation 훽(3) 2 훼∧훽 휖′∧휇−1 . 3 = p ( ∧ + ∧ ) (8.48) Actually, this is equivalent to a scalar relation obtained by taking the trace operation of both sides of (8.48). From (8.44)–(8.47) it follows that the dyadic 훽 is essential for the existence of a P medium. In fact, assuming 훽 = 0andp ≠ 0, we have 𝖯 𝝅 훼 휖′ 휇−1 s = ps s∕p, and, consequently, = 0, = 0and = 0, which leads to vanishing of the whole medium bidyadic, 𝖬 = 0. 휇−1 휇−1 Because (8.46) implies ∧ ps = 0, the dyadic does not have a spatial inverse. Thus, unlike in the case of Q media, there is no simple way to express the medium equations in the form (5.70), (5.71) with D, B in terms of E, H. However, there exists another, less conventional, representation in terms of Gibbsian field vectors as ( ) ( ) ( ) 𝖠 𝖡 Bg g g ⋅ Dg . = 𝖢 𝖣 (8.49) Hg g g Eg 204 CHAPTER 8 Quadratic Media

In fact, the four Gibbsian medium dyadics turn out to take the quite simple appearance 𝖠 𝜺 ⌊⌊훼−1 |𝖦 𝖯 |𝖦 𝖯(3) g = (e123 123 ) s = s s∕tr s , (8.50) 𝖡 ⌊훼−1|휖′ 𝖨 g =−e123 = ag × g, (8.51) 𝖢 𝖦 | 휇−1|훼−1 ⌋𝜺 |𝖦 𝖨 g = ( s ( ) 123) s = g × bg, (8.52) 𝖣 𝖦 | 훽 휇−1|훼−1|휖′ 𝖦 |𝖯T . g = s ( − ) = a s s (8.53) 𝖨 𝖦 Here∑ we have applied the Gibbsian quantities of Appendix B, g = s = eiei and 𝝅 |𝖯−1|𝖦 𝖯−1| 𝝅 |𝖯−1| . ag = s s s, bg =− s ps, a = s s ps − p (8.54) Details of the derivation are left as an exercise. From (8.50) and (8.53) 𝖠 𝖣 it follows that, for the P medium, the dyadics g and g must satisfy the relation 𝖠 휆𝖣T g = g , (8.55) 𝖡 𝖢 while g and g may be any antisymmetric Gibbsian dyadics. This representation of the P medium bears close similarity to the Gibbsian representation (8.38) of the Q medium. In fact, the relation (8.55) appears similar to (8.41) while the off-diagonal dyadics may be arbitrary antisymmetric dyadics in both cases. As a check, the number of free param- 𝖣 휆 eters in the representation (8.49) is 9( g) + 1( ) + 3(ag) + 3(bg) = 16 equals that of the dyadic 𝖯.

8.3.3 Relation Between P Media and Q Media Comparison of the spatial expressions of Q media and P media gives an impression of certain complementarity of the two medium classes. In fact, while for Q media the dyadics 휖′, 휇−1 can be of full rank and 훼, 훽 do not have inverses, the opposite is true for P media. Actually, a relation between the spatial dyadic expressions (8.44)–(8.47) and (8.31)–(8.34) 𝖬 can be obtained through a connection between a P medium bidyadic P 𝖬 and a modified Q medium bidyadic mQ in the form

𝖬 𝖦(2)|𝖬 𝖬 (2)|𝖬 mQ = P, P = mQ, (8.56) 8.4 Plane Waves 205

𝖦 −1 E E by applying a symmetric metric dyadic of full rank, = ∈ 1 1. In fact, (8.56) is equivalent with the relations 𝖰 = 𝖦|𝖯T , 𝖯 = 𝖰T |. (8.57) The spatial dyadic expressions (8.31)–(8.34) for the Q medium are related to those of the P medium, (8.44)–(8.47), as 훼 𝜺 ⌊𝖦 |휇−1 Q =− 123 s P , (8.58) 휖′ 𝜺 ⌊𝖦 |훽 Q =− 123 s P, (8.59) 휇−1 𝜺 ⌊𝖦(2)|훼 Q =− 123 s P, (8.60) 훽 𝜺 ⌊𝖦(2)|휖′ Q =− 123 s P, (8.61) 𝖦 𝖦 휇−1 훼 where s is the spatial part of . Thus, it appears that and on one hand, and 휖′ and 훽 on the other hand, are related through this approach.

8.4 PLANE WAVES

8.4.1 Plane Waves in Q Media The condition (5.236) for the potential one-form 𝝓 of a plane wave in a Qmediumis 𝖣(𝝂)|𝝓 = (𝖰(2)⌊⌊𝝂𝝂)|𝝓 = 0. (8.62) The dispersion dyadic and its double-wedge powers can be expanded as 𝖣(𝝂) = (𝖰||𝝂𝝂)𝖰 − (𝖰|𝝂)(𝝂|𝖰), (8.63) 𝖣(2) 𝝂 𝖰||𝝂𝝂 𝖰||𝝂𝝂 𝖰(2) 𝖰∧ 𝖰|𝝂 𝝂|𝖰 ( ) = ( )(( ) − ∧( )( )), (8.64) 𝖣(3) 𝝂 𝖰||𝝂𝝂 2 𝖰||𝝂𝝂 𝖰(3) 𝖰(2)∧ 𝖰|𝝂 𝝂|𝖰 . ( ) = ( ) (( ) − ∧( )( )) (8.65) Applying the identities 𝖰(3)⌊⌊𝝂𝝂 𝖰||𝝂𝝂 𝖰(2) 𝖰∧ 𝖰|𝝂 𝝂|𝖰 = ( ) − ∧( )( ), (8.66) 𝖰(4)⌊⌊𝝂𝝂 𝖰||𝝂𝝂 𝖰(3) 𝖰(2)∧ 𝖰|𝝂 𝝂|𝖰 = ( ) − ∧( )( ), (8.67) we obtain the alternative representations 𝖣(2)(𝝂) = (𝖰||𝝂𝝂)(𝖰(3)⌊⌊𝝂𝝂), (8.68) 𝖣(3)(𝝂) = (𝖰||𝝂𝝂)2(𝖰(4)⌊⌊𝝂𝝂) (8.69) 𝖰||𝝂𝝂 2 ⌊⌊𝝂𝝂 . = ( ) ΔQ(eNeN ) (8.70) 206 CHAPTER 8 Quadratic Media

The dyadic dispersion equation (5.240),

𝖣(3)(𝝂) = 0, (8.71) can now be reduced for Q media to the scalar equation

𝖰||𝝂𝝂 2 . ( ) ΔQ = 0 (8.72)

Here we can separate two cases. For Q media with less than full-rank dyadics 𝖰, satisfying

𝜺 𝜺 ||𝖰(4) ΔQ = N N = 0, (8.73) the dispersion equation (8.71) is satisfied identically for any one-form 𝝂. Thus, a Q medium defined by a dyadic 𝖰 of rank 3 or lower, has no dispersion equation. This means that it belongs to the class of NDE media for which the one-form 𝝂 of the plane wave can be arbitrarily chosen. In the second case, assuming 𝖰 is a full-rank dyadic satisfying ≠ ΔQ 0, the dispersion equation has the form

𝖰||𝝂𝝂 = 0. (8.74)

Since (8.74) is a second-order equation, such a Q medium is nonbirefringent. Here one should note that the antisymmetric part of the dyadic 𝖰 does not affect the solution of (8.74). In particular, if 𝖰 is antisymmetric, the dispersion equation (8.71) is again satisfied identically. ≠ For ΔQ 0 we obtain a condition for the potential one-form by inserting (8.74) in (8.63), which yields

𝖣(𝝂)|𝝓 =−(𝖰|𝝂)(𝝂|𝖰)|𝝓 = 0, ⇒ 𝝂|𝖰|𝝓 = 0. (8.75)

Actually, any potential satisfying this condition is a valid solution. The general form for the potential of the plane wave can be expressed in the form

𝝓 = (𝝂|𝖰)⌋𝚪, (8.76)

𝚪 F where ∈ 2 may be any two-form yielding a nonzero result. It is left as an exercise to verify that (8.76) satisfies (8.62). 8.4 Plane Waves 207

8.4.2 Plane Waves in P Media The equation for the potential one-form of a plane wave in a P medium is obtained directly from the Maxwell equation as 𝝂 ∧ 𝚿 = 𝝂 ∧ 𝖯(2)T |(𝝂 ∧ 𝝓) = 𝝂 ∧ (𝖯T |𝝂) ∧ (𝖯T |𝝓) = 0. (8.77) Assuming that 𝝂 is not an eigen one-form of the dyadic 𝖯T ,thatis, 𝝂 ∧ (𝖯T |𝝂) ≠ 0, from (8.77) it follows that the one-form 𝖯T |𝝓 must be a linear combination of 𝝂 and 𝖯T |𝝂. Thus, for some parameters A, B we can write 𝝓 = A𝖯−1T |𝝂 + B𝝂 (8.78) The field two-forms then become 𝚽 = 𝝂 ∧ 𝝓 = A𝝂 ∧ (𝖯−1T |𝝂), (8.79) 𝚿 = 𝖯(2)T |𝚽 = A(𝖯T |𝝂) ∧ 𝝂. (8.80) No condition restricting the choice of the wave one-form 𝝂 appears here, whence any P medium belongs to the class of NDE media. Thus, a plane wave in the P medium can be defined for any chosen 𝝂 and the corresponding field two-forms are obtained from (8.79) and (8.80). A similar property was shown to be valid for skewon–axion media. To verify that the dispersion equation is identically valid, let us form the dispersion dyadic by 𝖣 𝝂 𝖬 ⌊⌊𝝂𝝂 ⌊ 𝝂 𝖯(2)T ⌊𝝂 ( ) = m =−eN ( ∧ ) ⌊ 𝝂 𝖯T |𝝂 𝖯T ⌊𝖨T |𝖯T =−eN ( ∧ ( ) ∧ ) = (F ) , (8.81) where the bivector F is defined by 𝝂 ⌊ 𝝂 𝖯T |𝝂 . F( ) = eN ( ∧ ( )) (8.82) F is a simple bivector because it satisfies ⋅ | 𝜺 ⌊ | 𝝂 𝖯T |𝝂 F F = F ( N F) = F ( ∧ ( )) | 𝝂 𝖯T |𝝂 𝝂 𝖯T |𝝂 . = eN ( ∧ ( ) ∧ ∧ ( )) = 0 (8.83) Applying the rule (see Problem 2.8) 1 (A⌊𝖨T )2 = AA − (A ∧ A)⌊𝖨(2)T (8.84) 2 we have (F⌊𝖨T )(2) = FF,(F⌊𝖨T )(3) = 0. (8.85) 208 CHAPTER 8 Quadratic Media

Because of this, the dyadic dispersion equation is satisfied identically, 𝖣(3)(𝝂) = (F(𝝂)⌊𝖨T )(3)|𝖯(3)T = 0, (8.86) for any one-form 𝝂 of a plane wave in a P medium.

8.4.3 P Medium as Boundary Material Properties of plane waves in P media and skewon–axion media appear somewhat similar because for both of them the dispersion equation is identically satisfied for any wave one-form 𝝂. Also, one can show that an interface of a P medium can produce boundary conditions similar to those of a skewon–axion medium. As an example, consider a P medium defined by 𝖯 = 𝖨 + a𝜶, a|𝜶 ≠ −1, (8.87) with a𝜶 𝖯−1 = 𝖨 − . (8.88) 1 + a|𝜶 From (8.79) and (8.80) the field two-forms of a plane wave in such aP medium can be expressed as A(a|𝜶) 𝚽 =− 𝝂 ∧ 𝜶, (8.89) 1 + a|𝜶 𝚿 =−A(a|𝜶)𝝂 ∧ 𝜶, (8.90) whence they satisfy the conditions 𝜶 ∧ 𝚽 = 0, 𝜶 ∧ 𝚿 = 0 (8.91) in this medium. Now these conditions coincide with (7.130) obtained for a special skewon–axion medium, whence after similar conclusions, DB boundary conditions (7.133) or SH boundary conditions (7.137) 𝜺 | and (7.138) can be obtained at the interface 3 x = 0. Actually, this is no wonder since for the choice (8.87) the P medium bidyadic can be written as 1 𝖬 = 𝖯(2)T = (𝖨∧𝖡)T , 𝖡 = 𝖨 + a𝜶, (8.92) ∧ 2 which is of the skewon–axion form (7.90). Because a nonzero principal part would distinguish a P medium from a skewon–axion medium, let us study the Hehl–Obukhov expansion of 8.5 P-Axion and Q-Axion Media 209 the general P medium bidyadic. Applying (3.73), the skewon part can be expressed as 1 𝖬 = 𝖬 = (𝖯(2)T − 𝖨(4)T ⌊⌊𝖯(2)). (8.93) 2 − 2 Because the axion part has the form (3.77), 1 𝖬 = (tr𝖯(2))𝖨(2)T , (8.94) 3 6 the principal part of the P-medium bidyadic can be written as 𝖬 𝖬 𝖬 𝖬 1 = − 2 − 3 1 1 = 𝖯(2)T + (tr𝖯(2)T )𝖨(2)T − (𝖯(2)T ⌊⌊𝖨)∧𝖨T . (8.95) 3 2 ∧ Since this does not vanish identically, the class of P media does not coincide with the class of skewon–axion media. Actually, one can show that it is possible to define a P-medium bidyadic consisting just of its principal part, that is, 𝖬 = 𝖯(3)T satisfying 𝖬⌊⌊𝖨 = 0. However, in such a case the medium bidyadic turns out to be of rank 1. Details are left as an exercise.

8.5 P-AXION AND Q-AXION MEDIA

A straightforward generalization of the class of P media and Q media can be obtained by adding an axion term to the definitions (8.1) and (8.2), 𝖬 = 𝖯(2)T + 훼𝖨(2)T , (8.96) 𝖬 𝖰(2) 훼𝖤 m = + , (8.97) 𝖤 ⌊𝖨(2)T with = eN . The number of free parameters is 17 in both cases. Since the axion term does not have an effect on the dispersion equation of the plane wave, there is no dispersion equation for the P-axion medium while (8.74) remains valid for the Q-axion medium. The effect of the axion term can be detected at an interface where the medium parameters change, as a change in reflection and transmission properties of plane waves. Certain special P-axion and Q-axion media have an indirect relation to skewon–axion media. The relation can be detected through the modified closure conditions (8.16) and (8.17). Let us consider the 210 CHAPTER 8 Quadratic Media condition (8.17) for the modified Q-axion medium bidyadic (8.97) written as 𝖬 훼𝖤 T ⋅ 𝖬 훼𝖤 𝖰(2)T ⋅ 𝖰(2) 𝖤 ( m − ) ( m − ) = =ΔQ , (8.98) or 𝖬T ⋅ 𝖬 훼 𝖬T 𝖬 훼2 𝖤 . m m − ( m + m) + ( −ΔQ) = 0 (8.99) Considering now a special Q-axion medium satisfying √ 훼 = ΔQ, (8.100) with either branch of the square root, whence the number of free parameters is reduced to 16. The closure condition becomes 𝖬T ⋅ 𝖬 훼 𝖬T 𝖬 . m m − ( m + m) = 0 (8.101) 𝖬−1 Assuming that the inverse bidyadic m exists, multiplying (8.101) by 𝖬−1T | |𝖬−1 m from the left and by m from the right yields the condition 1 𝖬−1T + 𝖬−1 − 𝖤 = 0, (8.102) m m 훼 which can be rewritten as ( ) 1 1 (𝖬−1 − 𝖤)T + 𝖬−1 − 𝖤 = 0. (8.103) m 2훼 m 2훼 This condition requires that the bidyadic in brackets must be antisym- 𝖷 F F metric, that is, there must exist an antisymmetric bidyadic ∈ 2 2 in terms of which we can write 1 𝖬−1 = 𝖷 + 𝖤. (8.104) m 2훼 Expressing this in terms of the medium bidyadic 𝖭 as defined by (5.52), we have 1 𝖭 = e e ⌊⌊𝖬−1 = 𝖸 + 𝖤, (8.105) m N N m 2훼 𝖸 ⌊⌊𝖷 E E where = eNeN ∈ 2 2 is another antisymmetric bidyadic. Obvi- ously, the right side of (8.105) is of the skewon–axion form. Since the skewon–axion medium has been defined through the bidyadic 𝖬 and not 𝖭, it is not, however, proper to talk about skewon–axion medium in this case. Actually this is a special Q-axion medium. A similar relation can be obtained by starting from a special P-axion medium bidyadic. 8.6 Extended Q Media 211

𝖬 Here one must note that a relation of the form (8.104) requires that m 𝖬 must have a nonzero antisymmetric part. In fact, assuming that m is 𝖬−1 symmetric, so must be its inverse m , in which case (8.104) would 𝖷 𝖬−1 𝖤 require = 0, whence m would be a multiple of , which would lead 𝖬 to a contradiction. Actually, it can be shown that if m defined by (8.97) with (8.100) is symmetric, it does not have an inverse. Details are left as an exercise.

8.6 EXTENDED Q MEDIA

The class of Q media can be extended in the same way as the class of skewon–axion media, by adding a simple dyadic term as 𝖬 𝜺 ⌊𝖰(2) 𝚫 = N + B, (8.106) 𝖬 𝖰(2) m = + AB, (8.107) 𝚫 𝜺 ⌊ where A and B are two bivectors and = N A. The corresponding class of media can be called that of extended Q media (or generalized Q media [1, 63]). When 𝖰 is of full rank and 1 + B|𝖰(−2)|A does not 𝖬 vanish, the bidyadic m has an inverse of the the same form as (8.107), (𝖰(−2)|A)(B|𝖰(−2)) 𝖬−1 = 𝖰(−2) − , (8.108) m 1 + B|𝖰(−2)|A as can be easily verified by multiplication. Applying (5.57) we obtain 𝖭 ⌊𝖬−1⌋ m = eN m eN 1 (e ⌊𝖰(−2)|A)(B|𝖰(−2)⌋e ) = 𝖰(2)T − N N , (8.109) ΔQ 1 + B|𝖰(−2)|A which is, again, of the similar form. Thus, the class of extended Q media can be defined through either the bidyadic 𝖬 or 𝖭.

8.6.1 Gibbsian Representation Expanding the two bivectors as ⌊𝜶 ⌊𝜷 A = e123 1 + a2 ∧ e4, B = e123 1 + b2 ∧ e4, (8.110) 212 CHAPTER 8 Quadratic Media

𝜶 𝜷 where 1, 1 are spatial one-forms and a2, b2 spatial vectors, we can write ⌊𝜶 ⌊𝜷 𝜷 ⌋ AB = (e123 1)(e123 1) − e4 ∧ a2( 1 e123) ⌊𝜶 + (e123 1)b2 ∧ e4 − e4 ∧ (a2b2) ∧ e4, (8.111) and 𝜺 ⌊ 𝜺 𝜶 𝜷 ⌋ 𝜺 ⌊ 𝜷 ⌋ N AB =− 4 ∧ 1( 1 e123) + ( 123 a2)( 1 e123) 𝜺 𝜶 𝜺 ⌊ . − 4 ∧ 1b2 ∧ e4 + ( 123 a2)b2 ∧ e4 (8.112) Combined with the expansion (8.28) we obtain expressions of the spatial medium dyadics, extended from those of the Q medium (8.31)–(8.34), as 훼 𝜺 ⌊𝖰 𝜺 ⌊ 𝜷 ⌋ = 123 s ∧ as + ( 123 a2)( 1 e123), (8.113) 휖′ 𝜺 ⌊ 𝖰 𝜺 ⌊ = 123 (c s − bsas) + ( 123 a2)b2, (8.114) 휇−1 𝜺 ⌊𝖰(2) 𝜶 𝜷 ⌋ =− 123 s − 1( 1 e123), (8.115) 훽 𝜺 ⌊ 𝖰 𝜶 . = 123 (bs ∧ s) − 1b2 (8.116) To find the corresponding Gibbsian medium dyadics we can start from 휇−1 휇−1⌋𝜺 g = 123 𝜺 𝜺 ⌊⌊𝖰(2) 𝜶 𝜷 =− 123 123 s − 1 1 𝖰−1T 𝜶 𝜷 . =−ΔQs s − 1 1 (8.117) Assuming Δ = 𝜺 𝜺 ||휇(−3) ≠ 0andΔ ≠ −𝜶 |𝖰 |𝜷 , its inverse Qs 123 123 Qs 1 s 1 can be formed as ( ) 1 (𝜶 |𝖰 )(𝖰 |𝜷 ) 휇 =− 𝖰T − 1 s s 1 . (8.118) g Δ s 𝜶 |𝖰 |𝜷 Qs ΔQs + 1 s 1 The other Gibbsian dyadics can now be found from 휉 𝜺 ⌊⌊훼 |휇 g = (e123 123 ) g, (8.119) 휁 휇 |훽 g =− g , (8.120) 휖 ⌊휖′ 휉 |휇−1|휁 . g = e123 + g g g (8.121) 8.6 Extended Q Media 213

Leaving the details as exercises, the results can be shown to take the form 휖 휖𝖰 g = s + p2q1, (8.122) 휉 𝖨⌋ g = Xs + p2q2, (8.123) 휁 𝖨⌋ g = Zs + p1q1, (8.124) 휇 휇𝖰T g = s + p1q2, (8.125) 휖 휇 where the scalars , and the spatial bivectors Xs, Zs are the same as in (8.38). Actually, the Gibbsian medium dyadics of the extended Q medium equal those of the Q medium (8.123)–(8.125) extended by terms involving four spatial vectors p1, p2, q1, q2 in the dyadic products 𝜶 |𝖰 𝜷 ⌋ ΔQs( 1 s)( 1 Zs + b2) p q =− , (8.126) 1 1 𝜶 |𝖰 |𝜷 ΔQs + 1 s 1 (𝜶 |𝖰 )(𝖰 |𝜷 ) p q = 1 s s 1 , (8.127) 1 2 𝜶 |𝖰 |𝜷 ΔQs + 1 s 1 ⌊𝜶 𝜷 ⌋ ΔQs(Xs 1 + a2)( 1 Zs + b2) p q = , (8.128) 2 1 휖 𝜶 |𝖰 |𝜷 (ΔQs + 1 s 1) (X ⌊𝜶 + a )(𝖰 |𝜷 ) p q =− s 1 2 s 1 . (8.129) 2 2 휖 𝜶 |𝖰 |𝜷 (ΔQs + 1 s 1) The expressions for the four vectors can be extracted as 𝜶 |𝖰 p1 =− 1 s, (8.130) 1 ⌊𝜶 p2 = 휖 (Xs 1 + a2), (8.131) 𝜷 ⌋ ΔQs( 1 Zs + b2) q = , (8.132) 1 𝜶 |𝖰 |𝜷 ΔQs + 1 s 1 𝖰 |𝜷 q =− s 1 . (8.133) 2 𝜶 |𝖰 |𝜷 ΔQs + 1 s 1 The economy in 4D notation (8.107) over (8.122)–(8.125), with (8.126)– (8.129) substituted, appears striking. 214 CHAPTER 8 Quadratic Media

8.6.2 Field Decomposition It will turn out that plane-wave fields in an extended Q medium can be decomposed in two sets each obeying a polarization condition of its own. To see this, let us apply the conditions (5.233) satisfied by the fields of any plane wave in any medium, 𝚽 ⋅ 𝚽 = 0, 𝚽 ⋅ 𝚿 = 0, 𝚿 ⋅ 𝚿 = 0. (8.134) Starting from the condition 𝖬 T ⋅ 𝖬 𝖰(2)T ⋅ 𝖰(2) 𝖤 ( m − AB) ( m − AB) = =ΔQ , (8.135) and multiplying this from both sides as 𝚽|()|𝚽, we can expand 𝚽|𝖤|𝚽 𝚽 ⋅ 𝚽 ΔQ =ΔQ = 0 𝖬 |𝚽 |𝚽 ⋅ 𝖬 |𝚽 |𝚽 = ( m − (B )A) ( m − A(B )) ⌊𝚿 |𝚽 ⋅ ⌊𝚿 |𝚽 = (eN − (B )A) (eN − A(B )) = 𝚿 ⋅ 𝚿 − 2(A|𝚿)(B|𝚽) + (A ⋅ A)(B|𝚽)2 =−(2A|𝚿 − A ⋅ A(B|𝚽))(B|𝚽) = 0. This condition is valid for any plane wave in the extended Q medium. It follows that any given plane wave must satisfy one of the two polarization conditions B|𝚽 = 0, (8.136) 2A|𝚿 − A ⋅ A(B|𝚽) = 0. (8.137) If A is a simple bivector, the latter condition reduces to A|𝚿 = 0. In the general case it can be rewritten in the alternative form A ⋅ (2𝖰(2) + AB) ⋅ 𝚽 = 0. (8.138) To conclude, all plane waves in an extended Q medium can be decomposed in two sets, satisfying either the first one or the second one of the two orthogonality conditions (8.136), (8.138). Since the conditions are independent of the wave one-form 𝝂 of the plane wave and linear in the field two-form 𝚽, actually any field which can be expressed as a sum or integral of plane waves can be decomposed in two fields satisfying these conditions. This is why such a medium has been called by the name decomposable (DC) medium in the past [64]. Because an axion term does not affect a plane wave in a homogeneous 8.6 Extended Q Media 215 medium, the same field decomposition is valid in more general media defined by 𝖬 훼 ⌊𝖨(2)T 𝖰(2) m = eN + + AB, (8.139) 𝖬 훼𝖨(2)T 𝜺 ⌊𝖰(2) 𝚫 . = + N + B (8.140)

8.6.3 Transformations The extended Q medium shares the property of the Q medium of being form-invariant in the affine transformation: 𝖬 𝖠(2)|𝖬 |𝖠(2)T 𝖰(2) ma = m = a + AaBa, (8.141) when the transformed quantities are defined by 𝖰 𝖠|𝖰|𝖠T 𝖠(2)| 𝖡 𝖠(2)| . a = , Aa = A, a = B (8.142) Considering the special involutionary duality transformation (6.65), 휃 휋 defined by the parameters Zd =−Z and = ∕2, the transformed extended Q medium bidyadic becomes 1 1 𝖬 = 𝖬−1 = (𝜺 ⌊(𝖰(2) + AB))−1. (8.143) d Z2 Z2 N Invoking (8.108), the result can be written in the form 𝖬 𝖰(2) md = d + AdBd, (8.144) with 𝖰 1 𝖰T d = √ , (8.145) Z ΔQ and ⋅ 𝖰(2) 𝖰(2) ⋅ (A )( B) . AdBd =− (8.146) 2 ⋅ 𝖰(2) ⋅ Z ΔQ(ΔQ + A B) Thus, the extended Q medium is form-invariant also in the involutionary duality transformation.

8.6.4 Plane Waves in Extended Q Media The dispersion dyadic corresponding to a plane wave in the extended Q medium can be expressed as 𝖣 𝝂 𝝂𝝂⌋⌋𝖬 𝝂𝝂⌋⌋𝖰(2) 𝝂⌋ 𝝂⌋ . ( ) = m = + ( A)( B) (8.147) 216 CHAPTER 8 Quadratic Media

Applying the identities (8.66) and (8.67), the double-wedge powers can be expanded as 𝖣(2) 𝝂 𝝂𝝂||𝖰 𝝂𝝂⌋⌋𝖰(3) 𝝂𝝂⌋⌋𝖰(2) ∧ 𝝂⌋ 𝝂⌋ ( ) = ( )( ) + ( )∧( A)( B), (8.148) 𝖣(3)(𝝂) = (𝝂𝝂||𝖰)2(𝝂𝝂⌋⌋𝖰(4)) 𝝂𝝂||𝖰 𝝂𝝂⌋⌋𝖰(3) ∧ 𝝂⌋ 𝝂⌋ . + ( )( )∧( A)( B) (8.149) Assuming that 𝖰 is of full rank, we can apply the rules 𝖰(3) ⌊⌊𝖰−1T 𝖰(4)⌊⌊𝖰−1T = eNeN ΔQ = (8.150) and 𝜶⌋ ⌊𝜷 ⌊ 𝜶 𝜷 ⌊ ( (eN )) ∧ a = eN (( ∧ ) a), (8.151) valid for any one-forms 𝜶, 𝜷 and a vector a. With these, we can further expand 𝝂𝝂⌋⌋𝖰(3) ∧ 𝝂⌋ 𝝂⌋ 𝝂𝝂⌋⌋ ⌊⌊𝖰−1T ∧ 𝝂⌋ 𝝂⌋ ( )∧( A)( B) =ΔQ( (eNeN ))∧( A)( B) ⌊⌊ 𝝂𝝂∧𝖰−1T ∧ 𝝂⌋ 𝝂⌋ =ΔQeNeN (( ∧ )∧( A)( B)) = (𝝂𝝂⌋⌋𝖰(4))(𝖰−1T ||(𝝂⌋A)(𝝂⌋B)) 𝝂𝝂⌋⌋ 𝝂𝝂|| 𝖰−1T ⌋⌋ . =ΔQ( eNeN) ( AB) (8.152) Inserting this in (8.149) we obtain 𝖣(3)(𝝂) = (𝝂𝝂⌋⌋𝖰(4))(𝝂𝝂||𝖰)(𝝂𝝂||(𝖰 + 𝖰−1T ⌋⌋AB)) 𝝂𝝂⌋⌋ 𝝂 . =ΔQ( eNeN)D( ) (8.153) The dispersion function defined by 𝝂 𝝂𝝂||𝖰 𝝂𝝂|| 𝖰 𝖰−1T ⌋⌋ 𝝂 𝝂 D( ) = ( ) ( + AB) = D1( )D2( ), (8.154) has a factorized form. Thus, for a full-rank dyadic 𝖰, the dispersion equation D(𝝂) = 0 can be split in two second-order equations as 𝝂 𝝂𝝂||𝖰 D1( ) = = 0, (8.155) 𝝂 𝝂𝝂|| 𝖰 𝖰−1T ⌋⌋ . D2( ) = ( + AB) = 0 (8.156) The former of these coincides with the dispersion equation of the pure Q 𝝓 medium (8.74), and the corresponding potential one-form 1 satisfies 𝖣 𝝂 |𝝓 𝖰|𝝂 𝝂 |𝖰|𝝓 𝝂 ⌋ 𝝂 ⌋ |𝝓 ( 1) 1 =−( 1)( 1 1) − ( 1 A)( 1 B) 1 = 0, (8.157) 8.6 Extended Q Media 217

𝝂 when 1 is a solution of (8.155). Choosing 𝝂 |𝖰|𝝓 1 1 = 0, (8.158) as the gauge condition for the potential, we obtain 𝝂 ⌋ |𝝓 | 𝝂 𝝓 |𝚽 − ( 1 B) 1 = B ( 1 ∧ 1) = B 1 = 0, (8.159) which represents the polarization condition for the field two-form. The medium condition for the fields of the plane wave 1 𝚿 𝖬|𝚽 𝜺 ⌊ 𝖰(2) |𝚽 𝜺 ⌊𝖰(2)|𝚽 1 = 1 = N ( + AB) 1 = N 1, (8.160) shows us that the the extended Q medium can be replaced by a simpler Q medium without changing the wave 1. 𝝂 𝝂 The second solution = 2 satisfies the dispersion equation (8.156), or 𝝂 𝝂 ||𝖰 𝝂 𝝂 ⌋⌋ ||𝖰−1T ( 2 2 ) =−( 2 2 AB) , (8.161) whence the corresponding potential one-form satisfies 𝖣 𝝂 |𝝓 𝝂 𝝂 ⌋⌋𝖰(2) 𝝂 𝝂 ⌋⌋ |𝝓 ( 2) 2 = ( 2 2 + ( 2 2 AB)) 2 𝝂 𝝂 ||𝖰 𝖰 𝖰|𝝂 𝝂 |𝖰 |𝝓 𝝂 𝝂 ⌋⌋ |𝝓 = (( 2 2 ) − 2 2 ) 2 + ( 2 2 AB) 2 𝝂 𝝂 ⌋⌋ ||𝖰−1T 𝖰|𝝓 𝖰|𝝂 𝝂 |𝖰|𝝓 =−(( 2 2 AB) ) 2 − ( 2)( 2 2) 𝝂 𝝂 ⌋⌋ |𝝓 . + ( 2 2 AB) 2 = 0 (8.162) Assuming now the same gauge condition (8.158) in the form 𝝂 |𝖰|𝝓 2 2 = 0, (8.163) the potential equation (8.156) is reduced to 𝝂 𝝂 ⌋⌋ ||𝖰−1T 𝖰|𝝓 𝝂 ⌋ 𝝂 ⌋ |𝝓 . (( 2 2 AB) ) 2 = ( 2 A)( 2 B) 2 (8.164) The first bracketed scalar factor can be assumed nonzero because otherwise (8.156) would reduce to (8.155). Thus, the vector 𝖰|𝝓 must be a 𝝂 ⌋ 𝝓 multiple of 2 A. This implies that the potential one-form 2 must be of the form 𝝓 휆𝖰−1| 𝝂 ⌋ 휆 𝖰−1 𝝂 | 2 = ( 2 A) = ( ∧ 2) A, (8.165) 휆 𝚽 for some scalar factor and, the two-form 2 must be of the form 𝚽 𝝂 𝝓 휆 𝝂 𝝂 ∧𝖰−1 | . 2 = 2 ∧ 2 =− ( 2 2∧ ) A (8.166) 218 CHAPTER 8 Quadratic Media

𝚽 The polarization condition for 2 is somewhat more tricky than (8.159). After some analysis one can show that the condition ( ) 1 𝜺 |(A ∧ 𝖰(2)) + (A ⋅ A)B |𝚽 = 0 (8.167) N 2 2 𝝂 is satisfied by the two-form defined by (8.166) when 2 satisfies (8.156). Writing (8.167) in the form 1 A|𝚿 − (A ⋅ A)B|𝚽 = 0, (8.168) 2 2 2 one can note that for the special case when A is a simple bivector, that is, for A ⋅ A = 0, the second orthogonality condition becomes similar in form to (8.159), |𝚿 . A 2 = 0 (8.169)

8.7 EXTENDED P MEDIA

8.7.1 Medium Conditions The class of P media can be extended following the pattern of the previous section by adding a term involving a two-form 𝚫 and a bivector C as

𝖬 = 𝖯(2)T + 𝚫C, (8.170) 𝖬 ⌊𝖯(2)T m = eN + DC, (8.171) ⌊𝚫 with D = eN . Media defined in this way are called extended P media (or generalized P media [61]). Considering the relation (8.56) connecting P media and Q media, one can show that extended P media and extended Q media obey a similar relation. To find the spatial expansion of the medium bidyadic, let us startby expressing the two bivectors as ⌊𝜹 ⌊𝜸 D = e123 s + ds ∧ e4, C = e123 s + cs ∧ e4, (8.172) 𝜸 𝜹 in terms of spatial one-forms s, s and spatial vectors cs, ds. Expanding 𝜺 ⌊ 𝜺 𝜹 𝜺 ⌊ ⌊𝜸 N DC = (− 4 ∧ s + 123 ds)(e123 s + c3 ∧ e4), (8.173) 8.7 Extended P Media 219 the spatial P medium parameter dyadics can be extended from those of the P medium (8.44)–(8.47) to 훼 𝖯(2)T 𝜺 ⌊⌊ 𝜸 = s + 123e123 ds s, (8.174) 휖′ 𝝅 𝖯T 𝜺 ⌊ =− s ∧ s + 123 dscs, (8.175) 휇−1 𝖯T 𝜹 𝜸 ⌋ =− s ∧ ps − s s e123, (8.176) 훽 𝝅 𝖯T 𝜹 . = ( sps − p s ) − scs (8.177) The extension makes it possible for the spatial dyadics 휖′ and 휇−1 to have spatial inverses. Following steps similar to those leading to the polarization conditions (8.136) and (8.138) for a plane wave in an extended Q medium, the conditions for the extended P medium become C|𝚽 = 0, (8.178) 2𝚫 ⋅ 𝚿 − (𝚫 ⋅ 𝚫)(C|𝚽) = 0, (8.179) where the latter can be replaced by 𝚫 ⋅ (2𝖯(2)T + 𝚫C)|𝚽 = 0. (8.180) Again, being linear in 𝚽 and independent of 𝝂, these conditions are valid for any fields which can be composed of plane waves.

8.7.2 Plane Waves in Extended P Media Considering a plane wave in the extended P medium, the dispersion dyadic (8.81) is extended to 𝖣 𝝂 ⌊ 𝝂 𝖯T |𝝂 𝖯T ⌊𝝂 ⌊𝝂 ( ) =−eN ( ∧ ( ) ∧ ) + (D )(C ) = (F(𝝂)⌊𝖨T )⌊𝖯T + (D⌊𝝂)(C⌊𝝂), (8.181) where the bivector function F(𝝂) equals that introduced in (8.82). Apply- ing (8.85), we can write 𝖣(2) 𝝂 |𝖯(2)T ⌊𝖨T |𝖯T ∧ ⌊𝝂 ⌊𝝂 ( ) = FF + ((F ) )∧(D )(C ), (8.182) and 𝖣(3) 𝝂 |𝖯(2)T ∧ ⌊𝝂 ⌊𝝂 ( ) = (FF )∧(D )(C ) = (F ∧ (D⌊𝝂))(F|𝖯(2)T ∧ (C⌊𝝂)). (8.183) 220 CHAPTER 8 Quadratic Media

Since the last dyadic appears factorized, the dyadic dispersion equation 𝖣(3)(𝝂) = 0 can be split in two trivector equations as ⌊𝝂 F1 ∧ (D 1) = 0 (8.184) 𝖯(2)| ⌊𝝂 ( F2) ∧ (C 2) = 0, (8.185) 𝝂 𝝂 wherewedenoteF1 = F( 1)andF2 = F( 2). The first equation (8.184) can be expanded as 𝜺 ⌊ ⌊𝝂 𝜺 ⌊ ⌊ ⌊𝝂 0 = N (F1 ∧ (D 1)) = ( N F1) (D 1) 𝝂 𝖯T |𝝂 ⌊ ⌊𝝂 = ( 1 ∧ ( 1)) (D 1) 𝖯T |𝝂 ⌊𝝂 |𝝂 𝝂 ⌊𝝂 | 𝖯T |𝝂 = ( 1)(D 1) 1 − 1(D 1) ( 1) 𝝂 ⌊𝖯T |𝝂 |𝝂 = 1(D 1) 1, (8.186) which yields the scalar dispersion equation 𝝂 𝝂 | ⌊𝖯T |𝝂 . D1( 1) = 1 (D ) 1 = 0 (8.187) The second equation (8.185) can be expanded as 𝜺 ⌊ 𝖯(−3)| 𝖯(2)| ⌊𝝂 0 = N ( (( F2) ∧ (C 2))) 𝜺 ⌊ 𝖯−1| ⌊𝝂 = N (F2 ∧ ( (C 2))) 𝜺 ⌊ ⌊ 𝖯−1| ⌊𝝂 = ( N F2) ( (C 2))) 𝝂 𝖯T |𝝂 ⌊ 𝖯−1| ⌊𝝂 = ( 2 ∧ ( 2)) ( (C 2))) 𝖯T |𝝂 𝝂 |𝖯−1| ⌊𝝂 𝝂 ⌊𝝂 |𝝂 = ( 2)( 2 (C 2)) − 2(C 2) 2 𝖯T |𝝂 𝝂 | ⌊𝖯−1T |𝝂 =−( 2) 2 (C 2), (8.188) whence the second scalar dispersion equation becomes 𝝂 𝝂 | ⌊𝖯−1T |𝝂 . D2( 2) = 2 (C ) 2 = 0 (8.189) Here we have assumed the existence of 𝖯−1,thatis,tr𝖯(4) ≠ 0. in the 𝖯−1T 𝜺 ⌊⌊𝖯(3) 𝖯(3) converse case we must replace in (8.189) by NeN .For = 0 (8.189) is an identity. As a check, in the case of the plain P medium with C = D = 0, both of the dispersion equations (8.187), (8.189) become identities, valid for any 𝝂.

8.7.3 Field Conditions Because the fields of any plane wave satisfy the conditions 𝚽 ⋅ 𝚽 = 0 𝚿 ⋅ 𝚿 𝖯(2)T ⋅ 𝖯(2) 𝖯(4) ⌊𝖨(2) and = 0, applying the rule = tr eN , we obtain Problems 221 the following condition for fields in a extended P medium 0 = 𝚿 ⋅ 𝚿 − tr𝖯(4)𝚽 ⋅ 𝚽 𝚽| 𝖬T ⋅ 𝖬 𝖯(4) ⌊𝖨(2)T |𝚽 = ( − tr eN ) = 𝚽|(𝖯(2) ⋅ 𝖯(2)T + C(D|𝖯(2)T ) + (D|𝖯(2)T )C ⋅ 𝖯(4) ⌊𝖨(2)T |𝚽 +(D D)CC − tr eN ) = (𝚽|C)(2D|𝖯(2)T + (D ⋅ D)C)|𝚽. (8.190) From this we conclude that the field must satisfy either of the two orthogonality conditions: |𝚽 ( )C = 0, (8.191) 1 D|𝖯(2) + (D ⋅ D)C |𝚽 = 0. (8.192) 2 In the special case when the bivector D is simple, (8.192) written as ( ) 1 D|𝖬 − (D ⋅ D)C |𝚽 = 0, (8.193) 2 can be seen to reduce to D|𝚿 = 0. (8.194) Obviously, the two orthogonality conditions (8.191) and (8.192) are associated with the two dispersion equations (8.187), (8.189). The question remains which one corresponds to which one. Let us consider the wave associated to (8.191). Because for a field satisfying this condition the extended P medium can be replaced by that of the P medium (8.1), the field two-form can be expressed as (8.79). Requiring that thefield satisfy (8.191), C|𝚽 = AC|(𝝂 ∧ 𝖯−1T |𝝂) =−A𝝂|(C⌊𝖯−1T)|𝝂 = 0, (8.195) we arrive at the dispersion equation (8.189). Thus, (8.189) corresponds to (8.191) and, consequently, (8.187) must correspond to (8.192).

PROBLEMS

𝖣 E E 8.1 Prove that any spatial bidyadic s ∈ 2 2 possessing a spatial 𝖣 𝖠(2) inverse can be uniquely expressed as s = s in terms of some 𝖠 E E spatial dyadic s ∈ 1 1. 222 CHAPTER 8 Quadratic Media

8.2 Derive the expressions in (8.38) for the Gibbsian medium dyadics of a Q medium. 8.3 Show that one can construct the Q medium bidyadic 𝖬 by start- 휉 휁 휖 휇 ing from given antisymmetric Gibbsian dyadics g, g and g, g satisfying (8.41). Hint: use the expansion (5.88). 8.4 Derive the dispersion equation for the Q medium starting from 1 D(𝝂) = 𝜺 𝜺 ||(𝖬 ∧(𝝂𝝂⌋⌋(𝖬 ∧(𝝂𝝂⌋⌋𝖬 )) = 0. 6 N N m∧ m∧ m 8.5 Verify that the potential 𝝓 defined by (8.76) satisfies the dispersion equation (8.62). The bac-cab rule A⌊(𝚪⌊a) = (A ∧ a)⌊𝚪 − (A|𝚪)a, where A is a bivector, 𝚪 is a two-form and a is a vector, may appear useful here. 𝖬 8.6 Show that if the modified medium bidyadic m of a special Q- axion medium, defined by (8.97) with (8.100), is symmetric, it does not have an inverse.

8.7 Find the conditions for the Gibbsian field vectors Eg, Hg corresponding to the conditions (8.136) and (8.138) for the extended Qmedium. 8.8 Find the most general P medium consisting of only the principal part, by requiring 𝖬⌊⌊𝖨 = 𝖯(2)T ⌊⌊𝖨 = 0and𝖬 ≠ 0. 8.9 Consider a Q-axion medium defined by 𝖬 𝖨(2)T 𝜺 ⌊𝖰(2) = A + B N , 𝖰(2) E E where ∈ 2 2 is a symmetric bidyadic normalized as 𝜺 𝜺 ||𝖰(4) 𝖩T 𝜺 ⌊𝖰(2) F E ΔQ = N N =−1. Show that = N ∈ 2 2 is an AC bidyadic and the medium bidyadic can be expressed in the form

𝖩T 𝖬 = Me휓 . 8.10 Find the eigenvalues and eigen two-forms corresponding to the equation 𝖬|𝚽 𝚽 i = Mi i Problems 223

for the symmetric Q-axion medium bidyadic of the previous prob- 𝚽 lem. Denoting the eigen two-forms by ±, show that that any 𝚽 𝚽 𝚽 given two-form can be decomposed as = + + − and that 𝚽 ⋅ 𝚽 they satisfy the orthogonality conditions + − = 0. 8.11 Show that the decomposition of the electromagnetic two-form 𝚽 for the symmetric Q-axion medium of the previous problem gives rise to decoupled Maxwell equations. Find the eigenfield expansion of the bidyadic 𝖴(𝚿, 𝚽) defined by (5.188), assuming ≠ M+ M−. 8.12 Derive an expression for the dispersion function D(𝝂)forthe extended Q medium starting from the general form (5.244). 8.13 Verify the orthogonality condition (8.167) by substituting the 𝝂 field two-form (8.166) when the wave one-form 2 is assumed to satisfy the dispersion equation (8.156) in an extended Q medium. 8.14 Derive the Gibbsian medium dyadic expressions (8.50)–(8.53) for the P medium. 8.15 Express the orthogonality conditions (8.191) and (8.192) for the plane-wave fields in the extended P medium in terms of the spatial field components D, B, E, H. 8.16 Derive the duality transformation rules for the extended Q medium (8.145), (8.146), and consider the possibility of defining a transformation in which the medium is self-dual. 8.17 Consider a P medium defined by the dyadic 𝖯 𝖨 𝖯 𝖯 𝜺 𝜺 𝜺 𝜺 . = P + 34, 34 = P3,3e3 3 + P3.4e3 4 + P4,3e4 3 + P4,4e4 4 𝜺 | Show that plane-wave fields at an interface 3 x = 0ofthe medium satisfy the DB boundary conditions.

CHAPTER 9

Media Defined by Bidyadic Equations

𝖬 𝔽 𝔼 It is known that each medium bidyadic ∈ 2 2 satisfies an algebraic equation of the sixth order, the Cayley–Hamilton equation (3.9). Equa- tions of lower order, 𝖬p 𝖬p−1 ⋯ 𝖨(2)T < + A1 + + Ap = 0, p 6, (9.1) restrict the generality of the medium bidyadic and, thus, define certain classes of media. As the simplest example, medium bidyadics satisfying the first-order equation, 𝖬 − M𝖨(2)T = 0, (9.2) form the class of axion (PEMC) media. ≠ There are some properties arising from (9.1). If Ap 0, the inverse bidyadic can be expressed as

𝖬−1 1 𝖬p−1 𝖬p−2 ⋯ 𝖨(2)T =− ( + A1 + + Ap−1 ), (9.3) Ap and it is a member of the same class. This is seen if (9.1) is multiplied 𝖬−p| 𝖬 by .IfAp = 0, either there is no inverse or belongs to a class defined by an order lower than p. 𝖬 If 1 is a solution of (9.1), so is 𝖬 𝖣|𝖬 |𝖣−1 = 1 (9.4)

225 226 CHAPTER 9 Media Defined by Bidyadic Equations

𝖣 𝔽 𝔼 for any bidyadic ∈ 2 2 possessing an inverse. Thus, one solution generates a family of solutions through this kind of similarity transformation. From 𝖣|𝖬 |𝖣−1 𝖣−1|𝖣|𝖬 𝖬 tr( 1 ) = tr( 1) = tr 1 (9.5) it follows that the axion part is invariant in this transformation. If 𝖬 belongs to a class defined by p, the bidyadic 𝖬 + 𝛼𝖨(2)T belongs to the same class. Thus, the axion part of 𝖬 does not affect the class defined by the order of the equation. It is the purpose of this chapter to study the some properties of the simplest cases, medium classes defined by quadratic, cubic, and biquadratic equations, with no claim of completeness on these topics.

9.1 QUADRATIC EQUATION

The class of media defined by medium bidyadics satisfying an equation of the second order, 𝖬2 − 2A𝖬 + B𝖨(2)T = 0, (9.6) will be called the class of SD media [59]. The equation involves two parameters A and B. As we have already seen, GBI media, defined by (7.17) as

𝖬 𝖨(2)T 𝖳(2)T 1 𝜺 ⌊ 𝖦 𝜖′𝜇 (2) = A + B − 𝜇 N ( s − e4e4) , (9.7) 𝖳 𝖨 𝜺 where = s − e4 4 is the time reversion dyadic, satisfy the second- order equation ( ) 𝜖′ 𝖬2 𝖬 2 2 𝖨(2)T . − 2A + A − B + 𝜇 = 0 (9.8)

Also, Q-axion media with a symmetric bidyadic 𝖰(2),definedby 𝖬 𝖨(2)T 𝜺 ⌊𝖰(2) = A + B N , (9.9) satisfy 𝖬2 𝖬 2 2 𝖨(2)T − 2A + (A − B ΔQ) = 0, (9.10) 9.1 QUADRATIC EQUATION 227 because of ⌊𝖰(2) 2 ⌊⌊𝖰(2) |𝖰(2) 𝖰(−2)T |𝖰(2) 𝖨(2)T . (eN ) = (eNeN ) =ΔQ =ΔQ (9.11) Thus, symmetric Q-axion media belong to the class of SD media.

9.1.1 SD Media Rewriting (9.6) as (𝖬 − A𝖨(2)T )2 = (A2 − B)𝖨(2)T , (9.12) we can consider three cases depending on the factor A2 − B. r For A2 − B > 0, the bidyadic 𝖬 − A𝖨(2)T is a multiple of a unipotent bidyadic 𝖪T , 𝖬 = A𝖨(2)T + C𝖪T , 𝖪2 = 𝖨(2), (9.13) √ with real C = A2 − B. r For A2 − B < 0, the bidyadic 𝖬 − A𝖨(2)T is a multiple of an AC bidyadic 𝖩T , 𝖬 = A𝖨(2)T + D𝖩T , 𝖩2 =−𝖨(2), (9.14) √ with real D = B − A2. r For A2 − B = 0, the bidyadic 𝖬 − A𝖨(2)T is a nilpotent bidyadic 𝖭T , 𝖬 = A𝖨(2)T + 𝖭T , 𝖭2 = 0. (9.15)

We could call bidyadics of the form (9.13), (9.14), and (9.15) respectively as SDK, SDJ, and SDN bidyadics. The pure axion medium 𝖬 = M𝖨(2)T is ruled out of the SDK and SDJ classes because of the assumption A2 ≠ B. One can verify that the inverse of a bidyadic belonging to each of these classes, if it exists, is a member of the same class. In fact, we have 1 (A𝖨(2)T + C𝖪T )−1 = (A𝖨(2)T − C𝖪T ), (9.16) B 1 (A𝖨(2)T + D𝖩T )−1 = (A𝖨(2)T − D𝖩T ), (9.17) B 1 (A𝖨(2)T + 𝖭T )−1 = (A𝖨(2)T − 𝖭T ). (9.18) A2 228 CHAPTER 9 Media Defined by Bidyadic Equations

Also, multiplication of two bidyadics of a subclass corresponding to a certain bidyadic 𝖪, 𝖩,or𝖭 leaves the bidyadic in the same subclass and so does adding an axion term. Finally, it is easy to verify that the similarity transformation (9.4) does not map media out of these three subclasses for any bidyadic 𝖣 possessing an inverse, although the bidyadics 𝖪, 𝖩, or 𝖭 may change in the transformation. As an example, the GBI medium defined by the bidyadic (9.7) belongs to the SDK class for B2 >휖′∕휇 and to the SDJ class for B2 < 휖′∕휇 while for B2 = 휖′∕휇 it belongs to the class SCN. Assuming time- harmonic fields with complex-valued medium parameters, SDK and SDJ classes cannot, however, be distinguished.

9.1.2 Eigenexpansions To simplify the analysis, let us consider only two subclasses, that of the (proper) SD media and that of SDN media by adopting the definition of SDJ media for SD media, in which case (9.13) will be written as (9.14) with imaginary-valued coefficient D. Defining √ M = A2 + D2,tan휓 = D∕A, (9.19)

(9.14) can be written in the compact form

𝖩T 𝖬 = M(cos 𝝍𝖨(2)T + sin 휓𝖩T ) = Me휓 . (9.20)

When the medium bidyadic equation (9.12) is factorized as 𝖬 𝖨(2)T | 𝖬 𝖨(2)T ( − M+ ) ( − M− ) = 0, (9.21) the eigenvalues M± can be identified as ±j휓 . M± = A ± jD = Me (9.22) From the assumption B − A2 = D2 ≠ 0 for the SD case it follows that the ≠ two eigenvalues are distinct, M+ M−. In the SDN case the eigenvalues coincide. For the special case A = 0wehaveM+ + M− = 0, whence tr𝖬 = 0. Thus, the cases when 𝖬 is a multiple of either 𝖩T or 𝖭T ,there is no axion component. The eigen two-forms satisfying 𝖬|𝚽 𝚽 ± = M± ± (9.23) 9.1 QUADRATIC EQUATION 229 can be expressed from (9.21) as 𝚽 훼 𝖬 𝖨(2)T |𝚽 ± = ±( − M∓ ) , 훼 휓𝖩T ∓j휓 𝖨(2)T |𝚽 = ±(Me − Me ) 훼 휓 𝖩T 휓𝖨(2)T |𝚽 = ±M(sin ± j sin ) 훼 휓 𝖨(2)T 𝖩T |𝚽. =±j ±M sin ( ∓ j ) Here 𝚽 is any two-form yielding nonzero results. Choosing the coefficients as 훼 1 1 ± =± 휓 =± , (9.24) 2jM sin M+ − M− the eigen two-forms are obtained by 1 𝚽 = (𝖨(2)T ∓ j𝖩T )|𝚽. (9.25) ± 2 Because of the relation 𝚽 𝚽 𝚽 = + + −, (9.26) any given field two-form 𝚽 can be decomposed as a sum of eigen two- 𝚽 𝚽 forms +, − which are obtained from (9.25) or, more compactly, from 𝚽 T |𝚽 ± = ± , (9.27)   where +, − are two complementary projection bidyadics defined by 1  = (𝖨(2) ∓ j𝖩), (9.28) ± 2 and obeying the properties 2    𝖨(2)  |  | . ± = ±, + + − = , + − = − + = 0 (9.29)

9.1.3 Duality Transformation 휃 The three-parameter (Zd, Z, ) duality transformation rule (6.59) for the medium bidyadic, 𝖬 휃 𝖨(2)T 휃 𝖬 | 휃 𝖨(2)T 휃 𝖬 −1 Zd d = (sin + cos Z ) (cos − sin Z ) , (9.30) can be shown to map SDK, SDJ, and SDN bidyadics to respective SDK, SDJ, and SDN bidyadics because (9.30) involves additions of axion terms, the inverse operation and multiplication of bidyadics. 230 CHAPTER 9 Media Defined by Bidyadic Equations

An SD medium is self-dual if there exists a duality transformation (excluding the trivial identity transformation), for which the medium 𝖬 𝖬 bidyadic is invariant, d = . From (9.30) we obtain an equation for the medium bidyadic defining a self-dual medium: Z − Z 1 𝖬2 − d cot 휃 𝖬 + 𝖨(2)T = 0, (9.31) ZdZ ZdZ 휃 which must be valid for some nontrivial parameters Zd, Z, . This shows us that any self-dual medium belongs to the class of SD media. On the other hand, one can show that any SD medium is self- 휃 dual. For this we must find the parameters Z, Zd, of the duality transformation corresponding to a given SD medium bidyadic. Since the SD medium equation involves two parameters and the transformation has three parameters, this leaves us with some freedom. Actually, we can consider the two-parameter involutory transformation (6.51) with Zd =−Z. In this case (9.31) is simplified to 2 1 0 = 𝖬2 − cot 휃 𝖬 − 𝖨(2)T , ( Z Z2 ) ( ) 1 1 = 𝖬 + tan(휃∕2)𝖨(2)T | 𝖬 − cot(휃∕2)𝖨(2)T , (9.32) Z Z from which we can choose 1 1 M =− tan(휃∕2), M = cot(휃∕2). (9.33) + Z − Z Comparing with (9.22), we now have a relation between the parameters of the transformation and the SD medium. Thus, “SD” in the name of the medium class may refer as well to “second” order equation as to “self-dual” medium. Applying the duality transformation just defined to the eigenfields (9.25) of the SD medium yields ( ) ( )( ) 𝚿 (Z∕Z )cos휃 (1∕Z )sin휃 𝚿 ±d = d d ± 𝚽 −Z sin 휃 cos 휃 𝚽 ±d ( )( )± − cos 휃 −(1∕Z)sin휃 M = ± 𝚽 −Z sin 휃 cos 휃 1 ± ( ) ( ) M 𝚿 ± 𝚽 ± =± ± =± 𝚽 , (9.34) 1 ± 9.1 QUADRATIC EQUATION 231 details of which are left as an exercise. This can be interpreted so that 𝚿 𝚽 𝚿 𝚽 the eigenfields +, + are self-dual and the eigenfields −, − are anti-self-dual with respect to the transformation whose parameters Z, 휃 are related to the parameters A, B of the original medium equation (9.6) as [59] 1 1 A = cot 휃, B =− . (9.35) Z Z2

9.1.4 3D Representations To define SD media in terms of spatial dyadics, let us start fromthe axion-free case, restricted by tr𝖬 = 0, and add the axion part at a later stage for more complete solutions. The axion-free medium bidyadic 𝖬 = M𝖩T satisfies 𝖬2 =−M2𝖨(2)T . (9.36) Thus, 𝖬 can be considered as a multiple of an AC bidyadic [12]. Sub- stituting the expansion (5.65) 𝖬 훼 휖′ 𝜺 휇−1 𝜺 훽 = + ∧ e4 + 4 ∧ + 4 ∧ ∧ e4, (9.37) in (9.36) yields four 3D equations, 훼2 휖′|휇−1 2𝖨(2)T − =−M s , (9.38) 훼|휖′ − 휖′|훽 = 0, (9.39) 휇−1|훼 − 훽|휇−1 = 0, (9.40) 훽2 휇−1|휖′ 2𝖨T . − + = M s (9.41) The trace-free condition requires tr𝖬 = tr훼 − tr훽 = 0. (9.42) To be able to proceed with the equations (9.38)–(9.41), we must make some assumption. Let us assume that the spatial dyadic 휇−1 has a spatial inverse 휇. A similar result will be obtained by assuming that the spatial dyadic 휖′ is invertible. Let us apply the transformation (9.4) to represent the medium 𝖬 bidyadic in terms of a simpler bidyadic 1 by considering the transformation bidyadic 𝖣 𝖨(2)T 훼|휇 = − ∧ e4, (9.43) 232 CHAPTER 9 Media Defined by Bidyadic Equations whose inverse is 𝖣−1 𝖨(2)T 훼|휇 = + ∧ e4, (9.44) as can be easily verified. Inserting (9.37) in the inverse of the transformation (9.4) and applying (9.38) and (9.40) we now obtain 𝖬 𝖣−1|𝖬|𝖣 2휇 𝜺 휇−1 1 = = M ∧ e4 + 4 ∧ , (9.45) which is of the form 𝖬 휖′ 𝜺 휇−1 1 = 1 ∧ e4 + 4 ∧ 1 , (9.46) 휖′ 2휇 휇 휇. 1 = M , 1 = (9.47) This is a simpler SD medium and can be shown to equal the Gibbsian ⌊휇 𝔼 𝔼 isotropic medium (7.5) when e123 ∈ 1 1 is a symmetric dyadic. One can also note that it corresponds to the representation (4.133) of an AC bidyadic. Thus, to define a SD medium we may take any spatial bidyadic 𝖬 𝔽 𝔼 휖′ 2휇 1 ∈ 2 2 of the form (9.46) with spatial dyadics related as 1 = M 1 and insert it with (9.47) in (9.4), from which we obtain a more general form for a SD medium bidyadic, 𝖬 𝖣|𝖬 |𝖣−1 = 1 𝖨(2)T 훼|휇 | 2휇 𝜺 휇−1 | 𝖨(2)T 훼|휇 = ( − ∧ e4) (M ∧ e4 + 4 ∧ ) ( + ∧ e4) 훼 2휇 𝜺 휇−1 𝜺 휇−1|훼|휇 훼2|휇 . = + M ∧ e4 + 4 ∧ + 4 ∧ ∧ e4 + ∧ e4 (9.48) Thus, under the assumption that 휇 exists, the axion-free SD media can be defined by conditions of the form D = 훼|B + (훼2|휇 + M2휇)|E, (9.49) H = 휇−1|B + (휇−1|훼|휇)|E. (9.50) The two spatial dyadics 훼 and 휇 are arbitrary except that 휇 must be invertible. The other two spatial dyadics are obtained from 휖′ = 훼2|휇 + M2휇, 훽 = 휇−1|훼|휇. (9.51) In the other representation of medium equations, D = 휖|E + 휉|H = M2휇|E + (훼|휇)|H, (9.52) B = 휁|E + 휇|H =−(훼|휇)|E + 휇|H, (9.53) 9.1 QUADRATIC EQUATION 233 the spatial medium dyadics are seen to obey the relations 휖 = M2휇, 휉 =−휁 (9.54) for the axion-free SD media. The bidyadic (9.48) can now be generalized by adding an axion 𝖨(2)T 훼 훼 𝖨(2)T term A . Because this corresponds to replacing by + A s and 훽 훽 𝖨T by − A s , the medium conditions (9.49) and (9.50) become 훼 𝖨(2)T | 훼2|휇 2휇 | D = ( + A s ) B + ( + M ) E, (9.55) 휇−1| 휇−1|훼|휇 𝖨(2)T | . H = B + ( − A s ) E (9.56) In another representation the medium conditions can be written as D = (A2M2휇)|E + (A휇 + 훼|휇)|H, (9.57) B = (A휇 − 훼|휇)|E + 휇|H, (9.58) which correspond to the relations 휖 = (A2 + M2)휇, 휉 + 휁 = 2A휇, 휉 − 휁 = 2훼|휇. (9.59) Thus, for general SD media, the spatial dyadics 휖, 휇,and휉 + 휁 must be multiples of the same dyadic while 휉 − 휁 may be any other dyadic. This result coincides with that for self-dual media in terms of Gibbsian vectors and dyadics [60]. The axion part is unchanged in the similarity transformation (9.4). The previous medium bidyadic solutions for the quadratic equation were based on the assumption of invertible spatial dyadic 휇−1 or 휖′. There are also other possibilities. For example, 𝖬 defined as a multiple of the time reversion bidyadic, 𝖬 𝖳(2)T 𝖨(2)T 𝖨T ∧𝜺 = jM = jM( s − s ∧ 4e4), (9.60) with the spatial medium dyadics 훼 𝖨(2)T 휖′ 휇−1 훽 𝖨T . = jM s , = 0, = 0, =−jM s (9.61) obviously satisfies (9.36), whence it serves as another candidate foran SD medium bidyadic. Actually, the corresponding medium conditions D = jMB, H = jME (9.62) coincide with those of the SS medium (7.74). Other solutions can be found through any bidyadic 𝖣. 234 CHAPTER 9 Media Defined by Bidyadic Equations

To demonstrate the property that any AC bidyadic can be obtained from any other AC bidyadic through the similarity transformation (9.4) involving a suitable bidyadic 𝖣 ([77], p.156), let us find a bidyadic 𝖣 𝖬 𝖬 mapping the bidyadic 1 of (9.46) to of (9.60). Leaving the details as 𝖬 𝖬 an exercise, one can show that, for the bidyadics and 1 so defined, 𝖣 𝖬 𝖬 the bidyadic = ( + 1)∕M satisfies 𝖣|𝖬 |𝖣−1 𝖬. 1 = (9.63) 𝖬 This actually demonstrates that an SD-medium bidyadic 1 corresponding to a medium with 훼 = 0and훽 = 0 can be transformed to another SD-medium bidyadic 𝖬 with 휖′ = 0, 휇−1 = 0.

9.1.5 SDN Media The SDN medium bidyadics are solutions for (9.6) with A2 − B = 0. Writing the quadratic equation in the form 𝖬2 − 2M𝖬 + M2𝖨(2)T = (𝖬 − M𝖨(2)T )2 = 0, (9.64) its solution can be written as 𝖬 = M𝖨(2)T + 𝖭, (9.65) 𝖭 𝔽 𝔼 𝖭 where, ∈ 2 2 (not to be mistaken as in (5.52)) may be any nilpotent bidyadic of grade 2. Let us expand the general nilpotent bidyadic 𝖭 in its spatiotemporal components (9.37) 𝖭 훼 휖′ 𝜺 휇−1 𝜺 훽 . = + ∧ e4 + 4 ∧ + 4 ∧ ∧ e4 (9.66) After squaring and requiring that all of its four components vanish, the following set of relations is obtained: 훼|휖′ = 휖′|훽, (9.67) 휇−1|훼 = 훽|휇−1, (9.68) 훼2 = 휖′|휇−1, (9.69) 훽2 = 휇−1|휖′. (9.70) Assuming again that 휇−1 has an inverse, from (9.68) and (9.69) we can express 훽 and 휖′ as 훽 = 휇−1|훼|휇, (9.71) 휖′ = 훼2|휇. (9.72) 9.2 CUBIC EQUATION 235

Inserting these in (9.67) and (9.70) one can show that they are identically valid, which means that two of the medium dyadics can be expressed in terms of the other two which can be freely chosen. The nilpotent bidyadic 𝖭 can be expressed as 𝖭 훼 훼2|휇 𝜺 휇−1 𝜺 휇−1|훼|휇 = + ∧ e4 + 4 ∧ + 4 ∧ ∧ e4 훼|휇 𝜺 𝖨T |휇−1| 𝖨(2)T 훼|휇 = ( + 4 ∧ s ) ( s + ∧ e4), (9.73) and its square can be easily shown to vanish. The medium equations corresponding to the nilpotent medium bidyadic can now be expressed as D = 훼|(B + (훼|휇)|E) (9.74) H = 휇−1|(B + (훼|휇)|E), (9.75) or D = (훼|휇)|H (9.76) B =−(훼|휇)E + 휇|H. (9.77) It is noteworthy that, while the dyadic 휖 vanishes, the dyadic 휖′ = 훼2|휇 does not vanish in general for this representation of the nilpotent bidyadic. For the more general SDN medium bidyadic (9.65) an axion term must be added. Again, other possible 𝖭-dyadics are obtained through the similarity transformation.

9.2 CUBIC EQUATION

In this section we consider solutions to the bidyadic cubic equation 𝖬3 + A𝖬2 + B𝖬 + C𝖨(2)T = 0, (9.78) which defines the class of media which can be called CU media for brevity.

9.2.1 CU Media As a first try one can study a medium bidyadic of theform 𝖬 휖 𝜺 휇−1 = ∧ e4 + 4 ∧ (9.79) 236 CHAPTER 9 Media Defined by Bidyadic Equations as a candidate of a CU medium. However, as one can show (details left as an exercise), assuming full-rank spatial dyadics 휖, 휇−1, any such 𝖬 satisfying the cubic equation (9.78), must actually satisfy a quadratic equation. Thus, a proper CU-medium bidyadic must have terms with nonzero 훼 and/or 훽 dyadics added to (9.79). As an example of a true CU medium, let us consider a bidyadic of the form

𝖬 𝖨(2)T 𝖢 = A + o, (9.80) 𝖢 where o is a simple trace-free bidyadic defined by 𝖢 𝜺 ⌊ 𝖢 o = N (AB − BA), tr o = 0, (9.81) A and B being two bivectors. Thus, (9.80) corresponds to a certain 𝖢 skewon–axion medium. It is not difficult to show that the bidyadic o satisfies the cubic equation

𝖢 3 ⋅ 2 ⋅ ⋅ 𝖢 o = ((A B) − (A A)(B B)) o 1 = (tr𝖢 2)𝖢 . (9.82) 2 o o

𝖢 𝖬 𝖨(2)T When substituting o = − A in (9.82), we find that the medium bidyadic 𝖬 satisfies the cubic equation 1 (𝖬 − A𝖨(2)T )3 − tr(𝖬 − A𝖨(2)T )(𝖬 − A𝖨(2)T ) = 0. (9.83) 2

9.2.2 Eigenexpansions Expressing (9.78) in factorized form as

𝖬 𝖨(2)T | 𝖬 𝖨(2)T | 𝖬 𝖨(2)T ( − Ma ) ( − Mb ) ( − Mc ) = 0, (9.84) where the bracketed bidyadics commute. The scalars Ma, Mb,andMc are related to the coefficients of (9.78) as

A =−(Ma + Mb + Mc), (9.85)

B = MaMb + MbMc + McMa, (9.86) . C =−MaMbMc (9.87) 9.2 CUBIC EQUATION 237

Following the pattern of the previous section for SD media, let us define ≠ ≠ ≠ the following bidyadics in the case Ma Mb Mc Ma: (𝖬 − M 𝖨(2)T )|(𝖬 − M 𝖨(2)T )  b c a = , (9.88) (Ma − Mb)(Ma − Mc) (𝖬 − M 𝖨(2)T )|(𝖬 − M 𝖨(2)T )  c a b = , (9.89) (Mb − Mc)(Mb − Ma) (𝖬 − M 𝖨(2)T )|(𝖬 − M 𝖨(2)T )  a b . c = (9.90) (Mc − Ma)(Mc − Mb) Because of (9.84) they satisfy the eigenequation 𝖬 𝖨(2)T | ( − Mi ) i = 0, (9.91) for i = a, b, c. Thus, the eigenvalue equation for the CU-medium bidyadic 𝖬|𝚽 𝚽 i = Mi i (9.92) has three solutions and the eigen two-forms can be represented in the form 𝚽  |𝚽 i = i , (9.93) 𝚽 𝚽 ≠  for any two-form producing i 0. The three bidyadics i can be shown to satisfy    𝖨(2)T a + b + c = , (9.94) which allows one to expand any two-form in its eigen two-forms as 𝚽    |𝚽 𝚽 𝚽 𝚽 . = ( a + b + c) = a + b + c (9.95) From (9.84) we obtain the orthogonality conditions  | ≠ i j = 0, i j, (9.96) while 2  i = i, (9.97)  follows from (9.94) and orthogonality. This means that the bidyadics i are actually projection bidyadics in the space of two-forms. Because of the property (9.91) we can write 𝖬 𝖬|    = ( a + b + c)    . = Ma a + Mb b + Mc c (9.98) 238 CHAPTER 9 Media Defined by Bidyadic Equations

This can be generalized to 𝖬m m m m = Ma a + Mb b + Mc c, (9.99) where m can be any integer, positive, or negative. As a check we can substitute the expansions of 𝖬3, 𝖬2, 𝖬,and𝖨(2)T  in (9.84) and find that the coefficient expressions of each bidyadic i vanish identically. For the special CU medium defined by (9.80) the cubic equation (9.83) can be written in the form [ ] 1 (𝖬 − A𝖨(2)T )| (𝖬 − A𝖨(2)T )2 − (tr𝖢 2)𝖨(2)T = 0, (9.100) 2 o from which the eigenvalues of the bidyadic 𝖬 are obtained as √ 1 M = A, M = A ± tr𝖢 2. (9.101) a b,c 2 o Being a skewon–axion medium, there is no dispersion equation for a plane wave in such a medium.

9.2.3 Examples of CU Media Let us consider some further examples of medium bidyadics satisfying a cubic equation. As a symmetric example let us consider projection bidyadics defined by  𝜺 𝜺 a = 12e12 + 34e34, (9.102)  𝜺 𝜺 b = 23e23 + 14e14, (9.103)  𝜺 𝜺 c = 31e31 + 24e24, (9.104) in terms of which we can construct the medium bidyadic as (9.98) by introducing any three coefficients Ma, Mb, Mc. From (9.99) we obtain 𝖬2 2 2 2 = Ma a + Mb b + Mc c, (9.105) 𝖬3 3 3 3 . = Ma a + Mb b + Mc c (9.106) Eliminating the projection dyadics from the expressions (9.98), (9.105), and (9.106) we find that the medium bidyadic 𝖬 really satisfies a cubic equation of the form (9.78) with coefficients defined by (9.85)–(9.87). In this case the three eigenvalues are double eigenvalues. 9.2 CUBIC EQUATION 239

As another, less symmetric, example let us consider the projection bidyadics  𝜺 a = 12e12, (9.107)  𝜺 𝜺 b = 13e13 + 23e23, (9.108)  𝜺 𝜺 𝜺 c = 14e14 + 24e24 + 34e34, (9.109) which satisfy the same orthogonality conditions (9.96). The medium bidyadic defined by (9.80) satisfies also (9.105) and (9.106), whence the medium bidyadic obeys the same cubic equation (9.78) with the same coefficients as in the previous example. However, in this case Ma is a single eigenvalue, Mb is a double eigenvalue, and Mc is a triple eigenvalue. Of course, there are many possibilities for choosing the projection bidyadics. As a second example, let us consider a generalization of the medium defined by (9.80), a medium bidyadic of theform

𝖬 = 훼𝖨(2)T + 𝖢, (9.110) where 𝖢 is the bidyadic 𝖢 𝜺 ⌊ = N (AC + BD), (9.111) defined by four bivectors A, B, C, D. One can show that the bidyadic 𝖢 defined by (9.111) satisfies a cubic equation of theform 1 𝖢3 − (tr𝖢)𝖢2 + ((tr𝖢)2 − tr𝖢2)𝖢 = 0. (9.112) 2 The proof is left as an exercise. From this it follows that the medium bidyadic (9.110) satisfies a cubic equation and, thus, defines aCU medium. A third example is based on the fact that an antisymmetric bidyadic satisfies a Cayley–Hamilton equation containing only even-powered terms (3.39). Thus, the Cayley–Hamilton equation of a skewon bidyadic can be written in the bicubic form 𝖡 ∧𝖨 6 𝖡 ∧𝖨 4 𝖡 ∧𝖨 2 𝖨(2) 𝖡 . ( o∧ ) + a2( o∧ ) + a4( o∧ ) + a6 = 0, tr o = 0 (9.113) Defining a medium bidyadic as 𝖬 𝖡 ∧𝖨 2T = ( o∧ ) , (9.114) 240 CHAPTER 9 Media Defined by Bidyadic Equations

𝖡 it satisfies the cubic equation (9.78) for any trace-free dyadic o. It turns out that such a medium bidyadic has no skewon component. In fact, we can expand 𝖨(4)⌊⌊𝖬 ⌊ 𝖡 ∧𝖨 T | 𝖨∧𝖡 T ⌋𝜺 = (eN (( o∧ ) ) (( ∧ o) ) N) ⌊ 𝖡 ∧𝖨 T ⌋𝜺 | ⌊ 𝖡 ∧𝖨 T ⌋𝜺 = (eN ( o∧ ) N) (eN ( o∧ ) N) (9.115) 𝖨(4)⌊⌊ 𝖡 ∧𝖨 T 2 𝖡 ∧𝖨 2 𝖬T = ( ( o∧ ) ) = (− o∧ ) = by applying the property (3.62), 𝖨(4)⌊⌊ 𝖨∧𝖡 T 𝖨∧𝖡 . ( ∧ o) =− ∧ o (9.116) The condition (9.115) means that 𝖬 is an eigenbidyadic of the eigenproblem (3.48) corresponding to the eigenvalue +1, which excludes the skewon component. From 𝖬⌊⌊𝖨 𝖡2T ∧𝖨T ⌊⌊𝖨 𝖡(2)T ⌊⌊𝖨 = ( o ∧ ) + 2 o 𝖡2 𝖨 𝖡2 𝖡 𝖡 𝖡2 = (tr o) + 2 o − 2(tr o o − o) (9.117) 𝖡2 𝖨 = tr o , we may conclude that 𝖬 defines a pure principal CU medium when- 𝖡 𝖡2 ever the trace-free dyadic o satisfies the additional condition tr o = 0. Finding the dispersion equation of a plane wave in such a medium is left as an exercise.

9.3 BI-QUADRATIC EQUATION

Instead of the general quartic equation, let us consider media defined by the bi-quadratic equation 𝖬4 − 2A𝖬2 + B𝖨(2)T = 0, (9.118) dubbed as BQ media. Comparing (9.6) and (9.118) shows us that the square of the medium bidyadic of any BQ medium serves as a possible 𝖬 𝖬2 medium bidyadic of a SD medium, SD = SQ. On the other hand, 𝖬 forming the square of the left-hand side of (9.6) shows us that SD is a solution of a more general quartic equation than the biquadratic equation considered here. 9.3 BI-QUADRATIC EQUATION 241

9.3.1 BQ Media As an attempt to define a simple example of a BQ medium let us consider a medium bidyadic of the restricted form 𝖬 휖 𝜺 휇−1. = ∧ e4 + 4 ∧ (9.119) Expanding 𝖬2 휖|휇−1 𝜺 휇−1|휖 =− + 4 ∧ ∧ e4, (9.120) 𝖬4 휖|휇−1 2 𝜺 휇−1|휖 2 = ( ) − 4 ∧ ( ) ∧ e4, (9.121) (9.118) is reduced to two spatial dyadic equations 휖|휇−1 2 휖|휇−1 𝖨(2)T ( ) + 2A + B s = 0, (9.122) 휇−1|휖 2 휇−1|휖 𝖨T . ( ) + 2A + B s = 0 (9.123) Multiplying (9.122) as 휇−1|()|휇 we obtain (9.123), whence they are actually the same equation when 휇 (or 휖−1) exists, which is assumed here. Expressing (9.123) as 휇−1|휖 𝖨T 2 2 𝖨T ( + A s ) = (A − B) s , (9.124) the dyadic in brackets can be expressed as a multiple of a spatial unipo- 𝖴T 𝔽 𝔼 tent dyadic s ∈ 1 1 satisfying 𝖴2 𝖨 𝖴 𝖨 | 𝖴 𝖨 . s = s,(s − s) ( s + s) = 0 (9.125) 𝖴 𝖨 𝖴 𝖨 Because one of the dyadics s − s and s + s must be of rank 1 (see Problem 4.1), we can set 𝖴 𝖨T 𝜶 |𝜶 s =±( s − sas), as s = 1, (9.126) 𝜶 where as and s are spatial vector and one-form, with either sign. From (9.124) it follows that the medium dyadics 휖 and 휇 must obey a relation of the form 휖 휇 𝚫 = c + sas, (9.127) 𝚫 where c is a scalar and s is a spatial two-form. One can readily show that (9.119) with (9.127) inserted satisfies the bi-quadratic equation (9.118) with coefficients 1 A =−c − a |휇−1|𝚫 , (9.128) 2 s s 2 |휇−1|𝚫 B = c + cas s, (9.129) 242 CHAPTER 9 Media Defined by Bidyadic Equations whence the bidyadic (9.119), with spatial dyadics related by (9.127), defines a CU medium. One can verify that a medium defined by (9.119) cannot satisfy a second-order equation. Of course, a medium bidyadic 𝖬 𝖬2 1 = as defined by (9.120) satisfies the quadratic equation (9.6).

9.3.2 Eigenexpansions 𝖬 The eigenvalues Mi of the BQ medium bidyadic satisfying 4 2 Mi − 2AMi + B = 0, (9.130) have the analytic form √ √ 2 Mi =± A ± A − B, (9.131) where the two double signs are independent and responsible for four eigenvalues. The biquadratic equation (9.118) can be written in factorized form as 𝖬 𝖨(2)T | 𝖬 𝖨(2)T | 𝖬 𝖨(2)T | 𝖬 𝖨(2)T ( − Ma ) ( + Ma ) ( − Mb ) ( + Mb ) = 0, (9.132) in which all of the bracketed bidyadics commute. The four eigenvalues denoted by ±Ma, ±Mb are related to the coefficients of (9.118) as 2 2 2 2. 2A = Ma + Mb, B = MaMb (9.133) Since the general medium bidyadic has six eigenvalues, some of the four eigenvalues ±Mi must be multiple ones and the eigen two-forms 𝚽 i corresponding to those eigenvalues must form subspaces in the six- 𝔽 dimensional space 2 of two-forms. Let us consider the case when the four eigenvalues are different, which also excludes the possibility of zero eigenvalues. When the eigenvalues Mi have been solved from (9.130), the corresponding eigen two-forms can be found in terms of the following four bidyadics: 𝖬2 2𝖨(2)T | 𝖬 𝖨(2)T ( − M ) ( ± Ma )  =± b , (9.134) a± 2 2 2Ma(Ma − Mb) 𝖬2 2𝖨(2)T | 𝖬 𝖨(2)T ( − M ) ( ± Mb )  =± a , (9.135) b± 2 2 2Mb(Mb − Ma) 9.3 BI-QUADRATIC EQUATION 243 which commute with the medium bidyadic 𝖬. Because of (9.132) the bidyadics satisfy 𝖬 𝖨(2)T | ⇒ 𝖬|  ( ∓ Ma ) a± = 0 a± =±Ma a±, (9.136) 𝖬 𝖨(2)T | ⇒ 𝖬|  ( ∓ Mb ) b± = 0 b± =±Mb b±, (9.137) whence the eigen two-forms can be represented as 𝚽  |𝚽 i = i (9.138) 𝚽 𝚽 ≠ in terms of any two-form producing i 0.  The four bidyadics i satisfy 𝖬2 − M2𝖨(2)T  +  = b , (9.139) a+ a− 2 2 Ma − Mb 𝖬2 − M2𝖨(2)T  +  = a , (9.140) b+ b− 2 2 Mb − Ma and     𝖨(2)T . a+ + a− + b+ + b− = (9.141) From (9.141) and (9.138) any given two-form 𝚽 can be expanded in 𝚽 terms of the eigen two-forms i as 𝚽     |𝚽 = ( a+ + a− + b+ + b−) 𝚽 𝚽 𝚽 𝚽 . = a+ + a− + b+ + b− (9.142)  Because the bidyadics i satisfy the conditions  | ≠ i j = 0, i j, (9.143) 2  i = i, (9.144) they are projection bidyadics. Applying the expansions 𝖬     = Ma a+ − Ma a− + Mb b+ − Mb b−, (9.145) 𝖬m m m m m = Ma a+ + (−Ma) a− + Mb b+ + (−Mb) b−, (9.146) for m = 2, 4 and (9.141) on the left-hand side of (9.132), the coefficients  of each bidyadic i can be shown to vanish identically. The expansion (9.145) allows one to construct medium bidyadics satisfying the bi-quadratic equation (9.118) for any choice of two  scalars Ma, Mb and four projection bidyadics i satisfying (9.143) 244 CHAPTER 9 Media Defined by Bidyadic Equations and (9.141). Applying the similarity transformation (9.4), all possible medium dyadics 𝖬 satisfying (9.118) for given A and B can be obtained 𝖬 𝖣 from a given solution 1 by applying different bidyadics .

9.3.3 3D Representation Substituting the spatiotemporal expansion (9.37) of the medium bidyadic 𝖬 in 𝖬2 𝖠 𝖡 𝜺 𝖢 𝜺 𝖣 = s + s ∧ e4 + 4 ∧ s + 4 ∧ s ∧ e4, (9.147) we can identify the four spatial dyadics as 𝖠 훼2 휖′|휇−1 s = − , (9.148) 𝖡 훼|휖′ 휖′|훽 s = − , (9.149) 𝖢 휇−1|훼 훽|휇−1 s = − , (9.150) 𝖣 휇−1|휖′ 훽2 s = − , (9.151) Inserting (9.147) in (9.132), or the equivalent equation 𝖬2 2𝖨(2)T | 𝖬2 2𝖨(2)T ( − Ma ) ( − Mb ) = 0, (9.152) and separating spatial and temporal components, the BQ medium conditions take the form of four 3D equations as 𝖠 2𝖨(2)T | 𝖠 2𝖨(2)T 𝖡 |𝖢 ( s − Ma s ) ( s − Mb s ) = s s, (9.153) 𝖠 2𝖨(2)T |𝖡 𝖡 | 𝖣 2𝖨T ( s − Ma s ) s = s ( s + Mb s ), (9.154) 𝖢 | 𝖠 2𝖨(2)T 𝖣 2𝖨T |𝖢 s ( s − Mb s ) = ( s + Ma s ) s, (9.155) 𝖣 2𝖨T | 𝖣 2𝖨T 𝖢 |𝖡 . ( s + Ma s ) ( s + Mb s ) = s s (9.156) These equations are not independent in the general case. For example, 𝖡 −1 if the inverse dyadic s exists, one can show that (9.153) and (9.154) imply (9.155) and (9.156), whence the latter ones can be omitted. At this instant it appears simpler to shift to the 휖, 휉, 휁, 휇 representation (which presumes the existence of the dyadic 휇) by applying the relations 휖′ = 휖 − 휉|휇−1|휁, 훼 = 휉|휇−1, 훽 =−휇−1|휁, (9.157) in terms of which we can write 𝖠 휉|휇−1| 휉 휁 |휇−1 휖|휇−1 s = ( + ) − , (9.158) 𝖡 휉|휇−1|휖 휉|휇−1| 휉 휁 |휇−1|휁 휖|휇−1|휁 s = − ( + ) + , (9.159) 9.3 BI-QUADRATIC EQUATION 245

𝖢 휇−1| 휉 휁 |휇−1 s = ( + ) , (9.160) 𝖣 휇−1|휖 휇−1| 휉 휁 |휇−1|휁. s = − ( + ) (9.161) When substituted in (9.153)–(9.156), the BQ medium conditions can be finally obtained for the medium dyadics 휖, 휉, 휁, 휇, which appears quite an involved task. The result

휇−1|휖|휇−1| 휉 휁 |휇−1|휖 2 2휇−1| 휉 휁 ( + ) = MaMb ( + ), (9.162) 휇−1|휖| 휇−1| 휉 휁 2 휇−1|휖 2𝖨T | 휇−1|휖 2𝖨T ( ( + )) = ( + Ma s ) ( + Mb s ) (9.163) has been derived in [44]. Assuming that 휇 and 휖−1 exist, the conditions can be given a more symmetric form as

휇−1| 휉 휁 |휇−1 2 2휖−1| 휉 휁 |휖−1 ( + ) = MaMb ( + ) (9.164) 휉 휁 |휇−1| 휉 휁 휖 2휇 |휖−1| 휖 2휇 . ( + ) ( + ) = ( + Ma ) ( + Mb ) (9.165) There is a double redundancy in the conditions (9.153)–(9.156). Actu- ally, two of the dyadic conditions are sufficient to define a BQ medium, provided the dyadic 휇−1|휖 has an inverse which is assumed here. It is worth noting that, in the conditions (9.162) and (9.163), the four medium dyadics appear through just two dyadics, 휇−1|휖 and 휖−1|(휉 + 휁). This means that the dyadic 휉 − 휁 is not restricted by the BQ medium conditions.

9.3.4 Special Case The problem is simplified by considering the special case of medium parameters satisfying the restriction

휉 + 휁 = 0. (9.166)

Since 휉 − 휁 may be any dyadic, the BQ conditions should actually be independent of the dyadic 휉 =−휁. Because (9.158)–(9.161) are now reduced to 𝖠 휖|휇−1 s =− , (9.167) 𝖡 휉|휇−1|휖 휖|휇−1|휉 s = − , (9.168) 𝖢 s = 0, (9.169) 𝖣 휇−1|휖 s = , (9.170) 246 CHAPTER 9 Media Defined by Bidyadic Equations from (9.147) we can expand 𝖬2 휖|휇−1 | 𝖨(2)T 휉 휉 𝜺 𝖨T | 휇−1|휖 =−( ) ( s + ∧ e4) + ( + 4 ∧ s ) ( ) ∧ e4, (9.171) 𝖬4 휖|휇−1 2| 𝖨(2)T 휉 휉 𝜺 𝖨T | 휇−1|휖 2 . = ( ) ( s + ∧ e4) − ( + 4 ∧ s ) ( ) ∧ e4 (9.172) 𝜺 ⌊⌊ Inserting these expressions in (9.118) and operating by e4 4 the equation is reduced to 휇−1|휖 2 휇−1|휖 𝖨T . ( ) + 2A( ) + B s = 0 (9.173) Similarly, multiplying both sides of (9.118), with (9.171) and (9.172) 𝖨(2)T | inserted, by s , yields 휖|휇−1 2 휖|휇−1 𝖨(2)T . ( ) + 2A( ) + B s = 0 (9.174) Actually, (9.173) and (9.174) represent the same equation when either 휇 or 휖−1 exists. Also, (9.173) coincides with (9.163) for this special case. Finally, inserting (9.171) and (9.172) in the original biquadratic equation (9.118), with (9.174) taken into account, can be shown to lead to the identity 0 = 0. This means that, for 휉 + 휁 = 0, there are no other restrictions to the dyadics 휉 and 휁 by a BQ medium. Comparing (9.173) with (9.123) and (9.174) with (9.122) we conclude that the special BQ medium bidyadic expressed by (9.119) can actually be generalized by introducing the dyadic 휉 as 𝖬 휖 휉 𝜺 𝖨T |휇−1| 𝖨(2)T 휉 = ∧ e4 + ( + 4 ∧ s ) ( s + ∧ e4), (9.175) and 휖 and 휇 are still related by (9.127).

PROBLEMS

9.1 Show that the extended Q medium defined by a medium bidyadic of the form 𝖬 𝖨(2)T 𝜺 ⌊ = M + N AB, where A and B are two bivectors, is an SD medium. 9.2 Do the details in proving that the SD-medium bidyadic 𝖬 (9.37) can be transformed to the form (9.45) when it satisfies (9.36). 9.3 Show that, by assuming that the spatial dyadic 휖′ has an inverse, the SD-medium bidyadic (9.37) satisfying (9.36) can be transformed to the form (9.46). Problems 247

9.4 Check that the representation (9.32) with (9.33) corresponds to (9.31) for Zd =−Z and find the relations of the transformation parameters Z, 휃 to those of the SD medium, (9.22). 9.5 Starting from the assumption 휖′ = 0and휇−1 = 0, find solutions to SD medium bidyadics 𝖬 satisfying (9.36). 9.6 Show that the similarity transformation rule implies the rule ′ ′ 𝖬 = 𝖣|𝖬|𝖣−1 ⇒ 𝖨(4)⌊⌊𝖬 = (𝖨(4)⌊⌊𝖣)|(𝖨(4)⌊⌊𝖬)|(𝖨(4)⌊⌊𝖣)−1. 9.7 Show that a similarity transformation (9.4) with the bidyadic 𝖣 = 𝖬 𝖬 ( + 1)∕M can be applied to map the SD-medium bidyadic 𝖬 𝖬 1 of (9.46) to the medium bidyadic corresponding to time reversion, 𝖬 𝖳(2)T 𝖨(2)T 𝖨T ∧𝜺 = jM = jM( s − s ∧ 4e4), and satisfying the condition (9.36) of an SD medium. 9.8 Study the possibility of defining a CU medium with medium bidyadic of the form 𝖬 휖 𝜺 휇−1 = ∧ e4 + 4 ∧ , where the spatial dyadics 휖, 휇−1 are assumed to be of full rank. 9.9 Find the eigenvalues and eigen two-forms corresponding to the CU-medium bidyadic 𝖬 𝖨(2)T 𝜺 ⌊ = A + N (AB − BA), where A and B are assumed to be linearly independent bivectors. 9.10 Show that the skewon–axion medium bidyadic defined by 𝖬 𝖨(2)T 𝖢 𝖢 𝜺 ⌊ = A + o, o = N (AB − BA) satisfies a cubic equation. 9.11 Verify the property (9.94). 9.12 Derive the cubic equation (9.112). 9.13 Verify that the medium defined by a medium bidyadic ofthe form (9.119) with the relation (9.127) satisfies the bi-quadratic equation (9.118) when the parameters are chosen as (9.128) and (9.129). 248 CHAPTER 9 Media Defined by Bidyadic Equations

𝚽 9.14 Show that the eigenfields i satisfy the orthogonality condition 𝚽 ⋅ 𝚽 ≠ i j = 0fori j in the case of a principal BQ medium. 9.15 Starting from the representation (9.148)–(9.151) find the 3D conditions for the BQ medium corresponding to (9.162) and (9.163). 9.16 Derive the expressions (9.171) and (9.172) and show that when they are inserted in (9.118), the biquadratic equation does not give any restriction to the dyadic 휉 when 휉 + 휁 = 0 is valid. 𝖣 𝔽 𝔼 9.17 Find a bidyadic ∈ 2 2 for a similarity transformation mapping a medium satisfying the BQ medium equation (9.118) to 휉 휁 𝖣 𝖨(2)T 𝖥 a medium with + = 0. Hint: start from = − s ∧ e4, 𝖣−1 𝖨(2)T 𝖥 𝖥 𝔽 𝔼 = + s ∧ e4,where s ∈ 2 1 is a spatial dyadic. 9.18 Find the dispersion equation for a plane wave in a CU medium defined by the medium bidyadic (9.114). CHAPTER 10

Media Defined by Plane-Wave Properties

Classes of media can also be defined by requiring that plane waves in such media are restricted by certain conditions. Here we only consider two examples: first, media in which the wave vector k or the corresponding wave one-form 𝝂 of a plane wave can be freely chosen without being restricted by a dispersion equation and, second, media in which plane waves are required to be decomposable in two sets in terms of their polarization properties. The media thus defined are respectively called “media with no dispersion equation” (NDE media) and “decomposable media” (DC media). Various subclasses of NDE media and DC media are presented in this chapter.

10.1 MEDIA WITH NO DISPERSION EQUATION (NDE MEDIA)

In the previous chapters, several cases of media have emerged with the property that a plane wave is not restricted by a dispersion equation, (3) either in dyadic form 𝖣 (𝝂) = 0 or in scalar form D(𝝂) = 0, because the equations are valid for any wave one-form 𝝂. For example, in Section 7.3 it was shown that a plane wave in a skewon medium is not restricted by a dispersion equation. The same was observed in Section 8.1 for Q media when the dyadic 𝖰 is not of full rank and in Section 8.3 for P media. Since an axion term can be added to the medium bidyadic

249 250 CHAPTER 10 Media Defined by Plane-Wave Properties without changing propagation properties of a plane-wave, the previous media can be extended to skewon–axion media, Q-axion media with low-rank dyadic 𝖰, and P-axion media. Thus, these serve as examples of NDE media. Here one must note that being an NDE medium does not mean that the medium is not dispersive, quite the opposite. Considering a time- harmonic plane wave propagating in the direction of a unit vector u with k = uk means that the magnitude of k is not limited by a dispersion equation and can be chosen. For a nondispersive NDE medium the group velocity equals the phase velocity 𝜔∕k and would exceed the velocity of light for k <𝜔∕c. To avoid this, the medium must be dispersive with function k = k(𝜔) satisfying 𝜕𝜔∕𝜕k < c.

10.1.1 Two Cases of Solutions (3) Because the dyadic dispersion equation requires vanishing of 𝖣 (𝝂), the class of NDE media is defined by requiring that the dispersion dyadic be of rank less than three for all possible one-forms 𝝂.Fromthis it follows that there must exist four vectors in terms of which we can write 𝖣 𝝂 𝖬 ⌊⌊𝝂𝝂 . ( ) = m = ac + bd (10.1) Different cases can be considered by recognizing that since the dispersion dyadic is a quadratic function of 𝝂, so must be the right side of (10.1). Let us first assume that the vectors a − d may be linear or quadratic functions of 𝝂, or independent of 𝝂 [48]. This gives us a few possibilities.

1. Each of the four vectors is a linear function of 𝝂 2. a and b are quadratic functions while c and d do not depend on 𝝂 3. a and d are quadratic functions while b and c do not depend on 𝝂 4. a is a quadratic function while b and d are linear functions and c does not depend on 𝝂

Other similar possibilities do not seem to bring any new solutions because the dispersion equation (5.245) is invariant to replacing the 𝖬 𝖬T modified medium bidyadic m by its transpose m which implies T (3) replacing the dispersion dyadic 𝖣(𝝂) by its transpose 𝖣 (𝝂)and𝖣 (𝝂) 10.1 Media With No Dispersion Equation (NDE Media) 251

(3)T by 𝖣 (𝝂). Let us consider here only Cases 1 and 2 because 3 and 4 can be reduced to them. Verifying this is left as an exercise. However, these cases do not contain all solutions. For example, some of the vectors a − d may be cubic or higher-order functions of 𝝂 divided by linear or higher-order scalar functions of 𝝂 as long as the right side of (10.1) is a quadratic function. These other possibilities will be only covered by an example at the end of this chapter.

Case 1 Because the dispersion dyadic satisfies 𝖣(𝝂)|𝝂 = 𝝂|𝖣(𝝂) = 0forall𝝂, the four vectors must satisfy a(c|𝝂) + b(d|𝝂) = 0, (𝝂|a)c + (𝝂|b)d = 0. (10.2) Assuming 𝖣(2) = (a ∧ b)(c ∧ d) ≠ 0, (10.3) for all 𝝂 ≠ 0, that is, that the dispersion dyadic be of rank 2, the vectors a, b must be linearly independent and so must c, d, whence we obtain a|𝝂 = b|𝝂 = c|𝝂 = d|𝝂 = 0, (10.4) for all 𝝂. This requires that there must exist bivectors A, B, C, D such that we can express a = A⌊𝝂, b = B⌊𝝂, c = C⌊𝝂, d = D⌊𝝂, (10.5) and 𝖣 𝝂 𝖬 ⌊⌊𝝂𝝂 ⌊⌊𝝂𝝂. ( ) = m = (AC + BD) (10.6) Because 𝖬 ⌊⌊𝝂𝝂 ( m − (AC + BD)) = 0 (10.7) must be valid for all 𝝂, applying the result of Problem 3.20, the modified 𝖬 medium bidyadic m must be of the form 𝖬 𝛼 ⌊𝖨(2)T m = AC + BD + eN , (10.8) for some scalar 𝛼. Equivalently, the medium bidyadic 𝖬 must be of the form (2)T 𝖬 = 𝚷C + 𝚲D + 𝛼𝖨 , (10.9) 252 CHAPTER 10 Media Defined by Plane-Wave Properties

𝚷 𝜺 ⌊ 𝚲 𝜺 ⌊ with two-forms = N A and = N B. The expression (10.9) defines in terms of two bivectors, two two-forms and a scalar one possible class of media, for which the dispersion equation is satisfied identically. When the medium bidyadic 𝖬 has an inverse, the dispersion equation can also be defined through the bidyadic 𝖭 = 𝖬−1 of (5.52). One can, however, show that 𝖭 = 𝖬−1 has a form similar to that of (10.9). (2) In the special case of 𝖣 (𝝂) = 0forall𝝂, (10.8) is reduced to the form 𝖬 𝛼 ⌊𝖨(2)T m = AC + eN , (10.10) (2)T while requiring 𝖣(𝝂) = 0forall𝝂 leads to 𝖬 = M𝖨 , that is, the axion medium. One may note that the medium defined by (10.10) includes Q-axion media defined by 𝖬 𝖰(2) 𝛽 ⌊𝖨(2)T m = M + eN , (10.11) 𝖰 𝔼 𝔼 in the special case when the dyadic ∈ 1 1 is antisymmetric. In fact, because an antisymmetric dyadic can be expressed in terms of some T bivector A as 𝖰 = A⌊𝖨 , expanding

T (2)T 𝖬 = M(A⌊𝖨 )(2) + 𝛽e ⌊𝖨 m ( N ) 1 (2)T (2)T = M AA − 𝜺 |(A ∧ A)𝖨 + 𝛽e ⌊𝖨 , (10.12) 2 N N we arrive at a special case of (10.10). It has been here tacitly assumed that none of the four bivectors in (10.8) is simple because, otherwise, the condition (10.3) cannot be valid for all 𝝂. Actually, in the converse case it turns out that 𝖣 has variable rank depending on 𝝂. Details of this are left as an exercise. Since media defined by (10.8) have no dispersion equation regardless of one ormore of the bivectors being simple, dispersion dyadics 𝖣(𝝂)ofvariablerank are included in addition to those of rank 2 or rank 1 for all 𝝂.

Case 2 As a second case let us now assume that in (10.1) the vectors c and d are independent of 𝝂, whence a and b must be quadratic functions of 𝝂. In this case it appears preferable to start from the representation 𝝂 𝖬⌊𝝂 𝜺 ⌊𝖣 𝝂 𝚪 𝚺 ∧ =− N ( ) = c + d, (10.13) 10.1 Media With No Dispersion Equation (NDE Media) 253 where the two three-forms 𝚪 𝜺 ⌊ 𝚺 𝜺 ⌊ 𝔽 =− N a, =− N b, ∈ 3 (10.14) are now quadratic functions of 𝝂. Since any two linearly independent three-forms can be interpreted as belonging to a basis of three-forms, which can be constructed in terms of a basis of one-forms, say 𝜶, 𝜷, 𝜸, 𝜹, we can express 𝚪 = 𝜶 ∧ 𝜷 ∧ 𝜸, 𝚺 = 𝜶 ∧ 𝜷 ∧ 𝜹. (10.15) Thus, (10.13) can be written as 𝝂 ∧ 𝖬⌊𝝂 = 𝜶 ∧ 𝜷 ∧ (𝜸c + 𝜹d), (10.16) or, more generally, as T 𝝂 ∧ 𝖬⌊𝝂 = 𝜶 ∧ 𝜷 ∧ 𝖯 . (10.17) Since terms of the form 𝜶a′ + 𝜷b′ can be added to the dyadic in brackets 𝖯 𝔼 𝔽 in (10.16), we may well assume that the dyadic ∈ 1 1 in (10.17) is of (2) full rank. The special case corresponding to 𝖣 (𝝂) = 0, that is, when the bracketed dyadic is reduced by assuming 𝜹d = 0, will be considered separately. It is left as an exercise to verify that when the medium bidyadic satisfies (10.17) for any 𝝂, the dispersion equation is satisfied identically. Considering (10.17), two conditions for the quantities on its right- hand side are obtained by noting that the left-hand side vanishes when multiplying by |𝝂 from the right or by 𝝂∧ from the left. These lead to the respective conditions T 𝜶 ∧ 𝜷 ∧ (𝖯 |𝝂) = 0, (10.18) T 𝝂 ∧ 𝜶 ∧ 𝜷 ∧ 𝖯 = 0. (10.19) T Equation (10.18) requires that the three one-forms 𝜶, 𝜷,and𝖯 |𝝂 be linearly dependent, whence they must satisfy a relation of the form T A𝜶 + B𝜷 + C(𝖯 |𝝂) = 0. (10.20) Considering different possibilities for the coefficients A, B,andC, one can show that (10.17) can be reduced to the form T T 𝝂 ∧ 𝖬⌊𝝂 = 𝜶 ∧ (𝖯 |𝝂) ∧ 𝖯 , (10.21) 254 CHAPTER 10 Media Defined by Plane-Wave Properties details of which are left as an exercise. Since the right-hand side of (10.21) must be a quadratic function of 𝝂, it appears that 𝖯 must be independent of 𝝂. Thus, 𝜶 must be a linear function of 𝝂 whence it can be expressed as 𝜶 = 𝖡T |𝝂 (10.22) 𝖡 𝔼 𝔽 in terms of some dyadic ∈ 1 1. Inserting this in (10.21) yields the representation T T T 𝝂 ∧ 𝖬⌊𝝂 = (𝖡 |𝝂) ∧ (𝖯 |𝝂) ∧ 𝖯 , (10.23) which leads to T T T 𝝂 ∧ (𝖡 |𝝂) ∧ (𝖯 |𝝂) ∧ 𝖯 = 0, (10.24) and it must be valid for any one-form 𝝂. Recalling that 𝖯 was assumed −1T to be a dyadic of full rank, we can multiply (10.24) by |𝖯 from the right. The condition is then reduced to T T T T T 𝝂 ∧ (𝖡 |𝝂) ∧ (𝖯 |𝝂) ∧ 𝖨 = 0 ⇒ 𝝂 ∧ (𝖡 |𝝂) ∧ (𝖯 |𝝂) = 0. (10.25) Since the three one-forms are linearly dependent for all 𝝂, the corresponding dyadics must be related as A𝖨 + B𝖡 + C𝖯 = 0, (10.26) for some scalars A, B, C. This leaves us the following three possibilities:

1. A = 0 which implies either 𝖯 = 0or𝖡 is a multiple of 𝖯. In both cases from (10.23) we have 𝝂 ∧ 𝖬⌊𝝂 = 0forall𝝂, whence (2)T 𝖬 = 𝛼𝖨 . (10.27) 2. A ≠ 0andB = 0 implies 𝖯 =−(A∕C)𝖨. In this case we have from (10.23) 𝝂 𝖬⌊𝝂 2 𝖡T |𝝂 𝝂 𝖨T 2𝝂 𝖡∧𝖨 T ⌊𝝂 ∧ = (A∕C) ( ) ∧ ∧ =−(A∕C) ∧ ( ∧ ) , (10.28) which further implies 𝖬 𝖡∧𝖨 T 𝛼𝖨(2)T . = ( ∧ ) + (10.29) 10.1 Media With No Dispersion Equation (NDE Media) 255

3. A ≠ 0andB ≠ 0 implies 𝖡 =−(A∕B)𝖨 − (C∕B)𝖯. In this case (10.23) yields T T (2)T 𝝂 ∧ 𝖬⌊𝝂 =−(A∕B)𝝂 ∧ (𝖯 |𝝂) ∧ 𝖯 =−(C∕B)𝝂 ∧ 𝖯 ⌊𝝂, (10.30) whence (2)T (2)T 𝖬 = 𝖯 + 훼𝖨 . (10.31)

In (10.29) and (10.31) the coefficients A, B, C have been absorbed in the dyadics 𝖡 or 𝖯. Thus, in Case 2 the media with no dispersion equation fall into two classes, those of skewon–axion media and P-axion media while axion media are special cases of both of the two. (2) The previous Case 2 analysis was based on the assumption 𝖣 (𝝂) ≠ 0forall𝝂. The converse case can be handled by replacing (10.1) by 𝖣(𝝂) = ac, (10.32) where a is a quadratic function of 𝝂 and c is independent of 𝝂.From 𝖣(𝝂)|𝝂 = a(c|𝝂) = 0, (10.33) satisfied for all 𝝂, we must have either a = 0orc = 0, both of which lead to 𝖣(𝝂) = 0 which corresponds to the axion medium case. One can show that both Case 1 and Case 2 classes of NDE media are closed in affine transformations, details of which are left as an exercise.

10.1.2 Plane-Wave Fields in NDE Media Let us consider the polarization of plane-wave fields in NDE media defined above under Case 1 and Case 2 for a given one-form 𝝂.

Case 1 Media Assuming that the dispersion dyadic 𝖣(𝝂) corresponding to the Case 1 medium (10.8) is of rank 2 for any 𝝂 ≠ 0, the vectors a = A⌊𝝂 and b = B⌊𝝂 must be linearly independent, in which case the potential must satisfy the two conditions c|𝝓 = (C⌊𝝂)|𝝓 = C|(𝝂 ∧ 𝝓) = 0, (10.34) d|𝝓 = (D⌊𝝂)|𝝓 = D|(𝝂 ∧ 𝝓) = 0, (10.35) 256 CHAPTER 10 Media Defined by Plane-Wave Properties which for the field two-form become

C|𝚽 = 0, D|𝚽 = 0. (10.36)

Because the potential one-form can be expressed as 𝝓 𝜺 ⌊ = N (p ∧ c ∧ d), (10.37) for any vector p yielding a nonzero result, this corresponds to choosing an additional condition

p|𝝓 = 0 (10.38) to make 𝝓 unique. Thus, the field two-form has the form 𝚽 𝝂 𝝓 𝝂 𝜺 ⌊ . = ∧ = ∧ N (p ∧ c ∧ d) (10.39) Both equations (10.36) follow from (10.39), which is seen from | 𝝂 𝜺 ⌊ | 𝜺 ⌊ 𝜺 | . C ( ∧ N (p ∧ c ∧ d)) = c ( N (p ∧ c ∧ d)) = N (p ∧ c ∧ d ∧ c) = 0 (10.40)

Since C and D are independent of 𝝂, the orthogonality conditions (10.36) are valid for all choices of the one-form 𝝂. Thus, because of linearity, the conditions are valid not just for a single plane-wave field but for the field of any linear combination of plane waves in such a medium.

Case 2 Media Considering the fields corresponding to the two main classes of Case2 NDE media, that of skewon–axion media and P-axion media, we can refer to results obtained in Chapters 7 and 8. In fact, fields in a skewon– axion medium were obtained in the form (7.101) and (7.102) rewritten here as

𝚽 = A𝝂 ∧ 𝖡T |𝝂, (10.41) 𝚿 = A𝝂 ∧ 𝖡2T |𝝂. (10.42)

In this case the polarization of the field depends on the chosen one-form 𝝂.If𝝂 is chosen as an eigen one-form of 𝖡, the above expressions are not valid. Actually, in such a case 𝚽 can be any two-form satisfying 𝝂 ∧ 𝚽 = 0. 10.1 Media With No Dispersion Equation (NDE Media) 257

For the P-axion medium defined by (10.31) the potential one-form 𝝓 obeys the same equation as for the P medium whence the field two-form 𝚽 is obtained from (8.79) rewritten here as −1T 𝚽 = A𝝂 ∧ 𝖯 |𝝂. (10.43) However, since the medium bidyadic has an extra axion term, the other field two-form 𝚿 is different from (8.80) and can be expressed as (2)T (2)T 𝚿 = (𝖯 + 훼𝖨 )|𝚽 T −1T = A((𝖯 |𝝂) ∧ 𝝂 + 훼𝝂 ∧ (𝖯 |𝝂)). (10.44) Also in this case the polarization of the field two-forms are uniquely determined by the choice of the wave one-form 𝝂.

10.1.3 Other Possible NDE Media Let us try to define other media satisfying the NDE condition outside the Case 1 and Case 2 definitions. Recalling that the medium equation could alternatively be defined in terms of a bidyadic 𝖭 in the form 𝚽 = 𝖭|𝚿, one may wonder whether other NDE media could be found through this 𝖣′ 𝝂 𝖭 ⌊⌊𝝂𝝂 route. Starting from the alternate dispersion dyadic ( ) = m we can work similarly and end up in Case 1 and Case 2 solutions based on the bidyadic 𝖭. To determine whether these differ from those obtained above we must find out what they correspond to in terms of the medium bidyadic 𝖬. Assuming full-rank bidyadics, we have the relation 𝖭 = 𝖬−1,in which case the problem is reduced to whether the inverse of Case 1 and Case 2 medium bidyadics 𝖭 falls under Case 1 and Case 2 form of bidyadics 𝖬. The answer turns out to be affirmative whence novel medium classes cannot be obtained through this way. Actually, the inverse of a Case 1 bidyadic 𝖭 can be shown to be a Case 1 bidyadic 𝖬. For Case 2, the situation is a bit more complicated. One can show that the inverse of the general P-axion bidyadic 𝖭 similar to (10.31), involving 17 scalar parameters, is a P-axion bidyadic 𝖬 but the inverse of a skewon–axion bidyadic 𝖭, which involves 16 scalar parameters, turns out to be a special P-axion bidyadic 𝖬 with coefficient 훼 satisfying (4) tr𝖯 = 훼2. Also, the inverse of this kind of special P-axion bidyadic 𝖭 turns out to be a skewon–axion bidyadic 𝖬. This was shown in Section 8.4. Applying another result from Section 4.7, one can conclude 258 CHAPTER 10 Media Defined by Plane-Wave Properties

Table 10.1. Relations Between Various Case 2 Medium Bidyadics

𝖬 𝖭

Axion Axion Skewon Skewon P medium P medium Skewon–axion Special P-axion Special P-axion Skewon–axion General P-axion General P-axion

𝖬 that if a medium bidyadic m is of full rank and satisfies the NDE 𝖬 ⌊⌊𝝂𝝂 (3) 𝖣(3) 𝝂 𝝂 condition ( m ) = ( ) = 0forall , it cannot be a Q medium. Because an axion term does not change this conclusion, the inverse of a skewon–axion medium bidyadic must be a P-axion bidyadic and not a Q-axion bidyadic which is not associated to an NDE medium anyway for full-rank dyadic 𝖰. Details are left as an exercise. The relations between different types of Case 2 NDE-medium bidyadics are shown in Table 10.1. Even if the consideration of Case 1 and Case 2 solutions through the bidyadic 𝖭 does not reveal new NDE media, the above media are not exhaustive as can be shown by an example. In fact, it was already observed by considering the expression (8.70) that a Q medium with dyadic 𝖰 of less than full rank satisfies the dyadic dispersion equation (5.240) for any one-form 𝝂. Obviously, such a Q medium can be completed to a Q-axion medium with the same property. The reason why it was not obtained from the previous analysis comes clear when assuming a dyadic 𝖰 of rank 3 for which we can write

𝖰 = e1q1 + e2q2 + e3q3, (10.45) 𝖰(2) . = e12q12 + e23q23 + e31q31 (10.46)

Assuming now

𝖬 𝖰(2) 훼 ⌊𝖨(2)T m = + eN , (10.47) 10.2 Decomposable Media 259

|𝝂 ≠ and e3 0, after some steps (left as an exercise), the dispersion dyadic can be expressed in the form (10.1) with 1 ⌊𝝂 𝝂|𝖰 ⌊𝝂 ac = |𝝂 ((e3 ∧ e1) )(( ∧ q1) ), (10.48) e3 1 ⌊𝝂 𝝂|𝖰 ⌊𝝂 . bd = |𝝂 ((e3 ∧ e2) )(( ∧ q2) ) (10.49) e3 These can be further combined to 𝖣 𝝂 1 𝝂|𝖰 ∧𝖰 ⌊⌊𝝂𝝂. ( ) = |𝝂 ((e3( ))∧ ) (10.50) e3 Obviously, the expansion of the dispersion dyadic (10.1) with (10.48) and (10.49) inserted corresponds to neither Case 1 nor Case 2 because |𝝂 𝝂 of the divisor e3 and cubic dependence on of both ac and bd.

10.2 DECOMPOSABLE MEDIA

The class of DC media is defined by the property that any field ina source-free region of the medium can be split in two independent partial fields each obeying a certain restriction on its polarization. Theidea can be understood by considering first TE/TM decomposition of plane- wave fields in a uniaxially anisotropic medium in terms of Gibbsian formalism.

10.2.1 Special Cases TE/TM Decomposition Let us consider time-harmonic fields with time dependence ej휔t in a uniaxially anisotropic medium defined by the Gibbsian medium dyadics 휖 휖 𝖨 휖 g = t t + zuzuz, (10.51) 휇 휇 𝖨 휇 g = t t + zuzuz, (10.52) 𝖨 where t = uxux + uyuy. Now, one can show that any plane wave in such a medium can be decomposed in two parts, called transverse electric (TE) and transverse magnetic (TM), as

Eg = ETE + ETM, (10.53)

Hg = HTE + HTM, (10.54) 260 CHAPTER 10 Media Defined by Plane-Wave Properties and restricted by the conditions ⋅ uz ETE = 0, (10.55) ⋅ . uz HTM = 0 (10.56)

Such a property was first published by Clemmow in 1963 [69]. The decomposition is unique when the medium parameters satisfy 휖 휇 휇 휖 ≠ t z − t z 0, (10.57) while in the converse case the decomposition is still valid but not unique. Actually, in the latter case the medium can be transformed to a Gibbsian isotropic medium through a suitable affine transformation [5]. To demonstrate the decomposition, we recall that the fields of a plane wave with space dependence exp(−jk ⋅ r) in any linear medium satisfy 휔 ⋅ ⋅ Eg Bg = Eg (k × Eg) = 0, (10.58) 휔 ⋅ ⋅ Hg Dg =−Hg (k × Hg) = 0, (10.59) whence in the uniaxial medium they must satisfy 휖 ⋅ 휇 ⋅ 휖 휇 휇 휖 ⋅ ⋅ . tEg Bg − tHg Dg = ( t z − t z)(uz Eg)(uz Hg) = 0 (10.60)

Thus, assuming (10.57), a plane wave in the uniaxial medium must either be a TE wave or a TM wave with respect to the axial direction defined by the unit vector uz. From this it follows that any electromagnetic field which can be decomposed in a spectrum of plane waves [37], can be uniquely decomposed in TE and TM parts in the uniaxial medium.

More General Decompositions The TE/TM decomposition theory can be generalized to certain media where the fields are decomposable to two parts satisfying either ⋅ ⋅ a Eg = 0orb Hg = 0wherea and b are two given (possibly complex) Gibbsian vectors. Even more generally, a decomposition by two polarization conditions ⋅ ⋅ ⋅ ⋅ a1 Eg + a2 Hg = 0, b1 Eg + b2 Hg = 0, (10.61) 10.2 Decomposable Media 261

in terms of four Gibbsian vectors a1 − b2 can be made in bi-anisotropic media with Gibbsian medium dyadics defined by [54] ( ) 1 T 휖 = −𝖡 + a b + b a , (10.62) g 2휏 g 2 1 2 1 1 휉 = x × 𝖨 + (a b + b a ), (10.63) g g 2휏 2 2 2 2 1 휁 = z × 𝖨 + (a b + b a ), (10.64) g g 2휂 1 1 1 1 1 휇 = (𝖡 + a b + b a ), (10.65) g 2휂 g 1 2 1 2 휏 휂 𝖨 where x, z are arbitrary Gibbsian vectors, , are arbitrary scalars, g is 𝖡 the Gibbsian unit dyadic and g is an arbitrary Gibbsian dyadic. Media defined by medium dyadics of the form (10.62)–(10.65) were called decomposable (DC) media. As a continuation to the theory it was shown that yet another scalar parameter 훼 can be added to the definition of the medium without changing the decomposition of the fields. In this case the definitions (10.62)–(10.65) are generalized (after some manipulations) to the form [21, 19] ( ) 휖 훼 𝖨 휂 𝖡T g = (z × g + a1b1 + b1a1) + − g + a2b1 + b2a1 , (10.66) 휉 = 휂(x × 𝖨 + a b + b a ) + 훼(𝖡 + a b + b a ), (10.67) g g 2 2 2 2 ( g 1 2 1 2 ) 휁 휏 𝖨 훼 𝖡T g = (z × g + a1b1 + b1a1) − − g + a2b1 + b2a1 , (10.68) 휇 훼 𝖨 휏 𝖡 . g =− (x × g + a2b2 + b2a2) + ( g + a1b2 + b1a2) (10.69)

Four-Dimensional Representation It turns out that the rather complicated looking conditions (10.66)– (10.69) of the DC medium can be represented in a compact way in terms of the 4D formalism [48]. Actually, applying an approach similar to that of [54, 21] to two-form representation of fields, additional solutions to media satisfying the DC-medium definition can be obtained. Let us recall that field two-forms 𝚽, 𝚿 with the plane-wave dependence exp(𝝂|x) satisfy the orthogonality conditions (5.232) 𝚽 ∧ 𝚽 = 0, 𝚽 ∧ 𝚿 = 0, 𝚿 ∧ 𝚿 = 0 (10.70) 262 CHAPTER 10 Media Defined by Plane-Wave Properties in any medium. Applying the natural dot product between two two-forms the four-form conditions (10.70) can be expressed as the respective scalar conditions (5.233) as

𝚽 ⋅ 𝚽 = 0, 𝚽 ⋅ 𝚿 = 𝚽 ⋅ 𝖬|𝚽 = 0, 𝚿 ⋅ 𝚿 = 𝚽|𝖬T ⋅ 𝖬|𝚽 = 0. (10.71)

Thus, for any medium bidyadic 𝖬, the two-form 𝚽 of any plane wave satisfies a condition of the form

𝚽| 훼 ⌊𝖨(2)T 훽 ⌊𝖬 훾𝖬T ⋅ 𝖬 |𝚽 ( eN + 2 eN + ) = 0, (10.72) for all choices of the scalars 훼, 훽, 훾. For the modified medium bidyadic this condition can be expressed as

𝚽| 훼𝖤 훽 𝖬 𝖬T 훾𝖬T ⋅ 𝖬 |𝚽 . ( + ( m + m) + m m) = 0 (10.73) Let us now assume that the medium requires that the field two-form 𝚽 of a given plane wave satisfies either of the two polarization conditions

A|𝚽 = 0, (10.74) B|𝚽 = 0, (10.75) involving two bivectors A and B. Expanding the bivectors in their spatial and temporal parts as

A = a1 ∧ a2 + e4 ∧ a3, B = b1 ∧ b2 + e4 ∧ b3, (10.76) the polarization conditions become |𝚽 | | A = (a1 ∧ a2) B + a3 E = 0, (10.77) |𝚽 | | B = (b1 ∧ b2) B + b3 E = 0, (10.78) which have a certain similarity to the Gibbsian conditions (10.61). Waves with fields satisfying (10.74) or (10.75) can be respectively called A-waves and B-waves and the medium, in analogy with the medium defined by (10.62)–(10.65), can be called a DC medium. Thus, any plane wave in such a medium is required to satisfy

(𝚽|A)(B|𝚽) = 𝚽|(AB)|𝚽 = 0, (10.79) 10.2 Decomposable Media 263 for some bivectors A, B. Following the pattern of [54], let us now define the class of DC media by combining (10.73) and (10.79) and requiring that the condition 𝚽| 훼𝖤 훽 𝖬 𝖬T 훾𝖬T ⋅ 𝖬 |𝚽 𝚽| |𝚽 ( + ( m + m) + m m) = (AB) (10.80) be satisfied for all two-forms 𝚽, which implies that the symmetric parts of the dyadics in brackets on both sides of (10.80) must be the same. Thus, a condition defining possible DC media can be expressed as 훼𝖤 훽 𝖬 𝖬T 훾𝖬T ⋅ 𝖬 + ( m + m) + m m = AB + BA, (10.81) where a disturbing factor 1∕2 has been absorbed by the dyadic AB.To 𝖬 find solutions m for possible DC media, we must separate two cases: either 훾 ≠ 0or훾 = 0. r The case 𝜸 ≠ 0 requires solving a symmetric quadratic bidyadic equation and the corresponding class of media can be called that of (proper) DC media. r In the case 훾 = 0 the bidyadic equation (10.81) is reduced to an equation of the first order. This requires separate consideration and it defines the class of what will be called special decomposable (SDC) media.

10.2.2 DC-Medium Subclasses Assuming 훾 ≠ 0 in (10.81) we can set 훾 = 1 without losing generality. In this case the DC-medium condition (10.81) can be expressed in compact form as 𝖬′T ⋅ 𝖬′ 훼′𝖤 m m = , (10.82) by defining 𝖬′ 𝖬 훽𝖤 m = m + − DC, (10.83) 훼′ = 훽2 − 훼. (10.84) Here, the bidyadic C equals C = A, (10.85) and the bidyadic D is related to A and B by 1 D ⋅ 𝖬 + 훽D − (D ⋅ D)C = B. (10.86) m 2 264 CHAPTER 10 Media Defined by Plane-Wave Properties

The condition (10.82) has the form of modified closure relation (4.142), whose solutions can be split in P- and Q-solutions, as were shown by (4.144) and (4.145). Accordingly, the class of DC media can be split in two subclasses. The first subclass assumes that there exist a dyadic 𝖯 𝔼 𝔽 ∈ 1 1 so that 𝖬′ ⌊𝖯(2)T . m = eN (10.87) 𝖰 𝔼 𝔼 The second subclass assumes that there exists a dyadic ∈ 1 1 so that 𝖬′ 𝖰(2). m = (10.88) Let us consider these two cases separately and respectively call them by the names PDC media and QDC media.

QDC Media Redefining the coefficient 훼, the QDC medium solution of (10.82) as obtained from (10.83) to (10.84) must be of the form 𝖬 훼𝖤 𝖰(2) m = + + DC, (10.89) or 𝖬 훼𝖨(2)T 𝜺 ⌊𝖰(2) 𝜺 ⌊ = + N + N DC, (10.90) 𝖰 𝔼 𝔼 depending on a metric dyadic ∈ 1 1, two bivectors C, D and a scalar 훼.For훼 = 0 the medium bidyadic has the form of (8.107), which corresponds to that of the extended Q medium. Thus, a QDC medium can also be called an extended Q-axion medium. The definition (10.86) can in this case be expressed more explicitly as (2) 1 B = D|𝖰 + (D ⋅ D)C. (10.91) 2 Assuming that the bivectors C and D are given, the bidyadics in the polarization conditions (10.74) and (10.75) satisfied by the field two- form in the QDC medium can be written from (10.85) and (10.91) as ( ) (2) 1 C|𝚽 = 0, D|𝖰 + (D ⋅ D)C |𝚽 = 0. (10.92) 2 10.2 Decomposable Media 265

One can show that the 3D Gibbsian medium dyadics corresponding to the general QDC medium (10.90) can be expressed in the general form (10.66)–(10.69).

PDC Media The second possibility in (10.82), that of the PDC medium, yields the following expansions for the medium bidyadics, 𝖬 훼𝖤 ⌊𝖯(2)T m = + eN + DC, (10.93) or 𝖬 훼𝖨(2)T 𝖯(2)T 𝜺 ⌊ = + + N DC, (10.94) 𝖯 𝔼 𝔽 훼 in terms of a dyadic ∈ 1 1, two bivectors C, D and a scalar .The definition (10.86) in this case is (2)T 1 B = D|𝖯 + (D ⋅ D)C. (10.95) 2 Thus, for given bidyadics C, D, the polarization conditions (10.74) and (10.75) satisfied by the field two-form in the PDC medium can bewritten from (10.85) and (10.95) as ( ) (2)T 1 C|𝚽 = 0, D|𝖯 + (D ⋅ D)C |𝚽 = 0. (10.96) 2 For 훼 = 0 the medium bidyadic has the form of (8.171), which corresponds to that of the extended P medium. Thus, a PDC medium can also be called an extended P-axion medium. Expansion of (10.94) in terms of 3D medium dyadics is left as an exercise.

SDC Media The SDC medium condition defined by 훾 = 0and 훽 = 1 in (10.81) yields a bidyadic equation of the first order, 훼𝖤 𝖬 𝖬T . + m + m = AB + BA (10.97) This equation can be interpreted so that the symmetric part of the 𝖬 𝖤 훼 bidyadic m − AB must be a multiple of . Thus, redefining ,the modified medium bidyadic must thus be of the form 𝖬 훼𝖤 𝖠 m = + AB + , (10.98) 266 CHAPTER 10 Media Defined by Plane-Wave Properties

𝖠 𝔼 𝔼 where ∈ 2 2 is an arbitrary antisymmetric bidyadic. From (3.86) it is known that such an antisymmetric bidyadic can be expressed in terms 𝖡 𝔼 𝔽 of a trace-free dyadic o ∈ 1 1 as

𝖠 ⌊ 𝖡 ∧𝖨 T 𝖡 . = eN ( o∧ ) ,tro = 0 (10.99)

It is easy to verify that a medium defined by (10.98) satisfies the decomposition condition (10.79). In fact, any plane wave in such a medium satisfies

𝚽 ⋅ 𝚿 훼𝚽 ⋅ 𝚽 𝚽| ⌊𝖬 훼𝖤 |𝚽 − = (eN − ) = 𝚽|(AB + 𝖠)|𝚽 = (A|𝚽)(B|𝚽) = 0. (10.100)

Redefining again 𝖠 and 훼, (10.98) can be expressed in the form

𝖬 훼𝖤 𝖠 m = + + AB + BA, (10.101) which corresponds to

𝖬 훼𝖨(2)T 𝜺 ⌊𝖠 𝜺 ⌊ . = + N + N (AB + BA) (10.102)

Without sacrificing generality, the bivectors in (10.102) can be assumed to satisfy A ⋅ B = 0, whence the third term is trace free. In this case the three terms correspond to the respective axion, skewon, and principal parts of the medium bidyadic 𝖬. While the axion and skewon parts may be arbitrary, the principal part is restricted to be of the simple form as defined by two bivectors A and B. Following the pattern of the QDC and PDC media, an SDC medium defined by (10.98) can be called an extended skewon–axion medium. In conclusion, the class of DC media as defined by the condition (10.81) consists of extended Q-axion (QDC) media, extended P-axion (PDC) media, and extended skewon–axion (SDC) media. This analysis has shown us that the present compact 4D formalism [48] has actually brought about two extra subclasses of solutions in addition to that of QDC media, first introduced through the more cumbersome 3D Gibbsian formalism in [54, 21]. 10.2 Decomposable Media 267

10.2.3 Plane-Wave Properties Let us finally consider properties of plane waves for the three medium subclasses defined above. Because the axion part plays no role in plane- wave propagation, it can again be ignored in all of the cases. Actually, the plane-wave problem has already been discussed for the extended-Q, extended-P, and extended-skewon media in the previous chapters. Thus, we may just adopt the results with some changes in symbols. In all of these cases the quartic dispersion equation D(𝝂) = 0 can be split in two quadratic equations corresponding to the decomposed waves. The converse, starting from the splitting of the quartic equation and deriving the polarization decomposition, can also be done [89, 53].

QDC Media In the case of QDC medium (extended Q-axion medium) the dispersion dyadic equals 𝖣 𝝂 𝝂𝝂⌋⌋𝖬 𝝂𝝂⌋⌋ 𝖰(2) . ( ) = m = ( + DC) (10.103) Comparing with (8.147), the two dispersion equations become 𝝂|𝖰|𝝂 = 0, (10.104) 𝝂|(𝖰 + CD⌊⌊𝖰−1)|𝝂 = 0, (10.105) (10.104) corresponds to a Q medium without the generalizing term, that is, with DC = 0. This is valid for fields satisfying the field condition C|𝚽 = A|𝚽 = 0, (10.106) whence (10.104) corresponds to the A-wave. To verify that (10.105) corresponds to the B-wave, that is, the condition (10.75), which from (10.91) equals ( ) 1 D ⋅ 𝖰(2) + (D ⋅ D)C |𝚽 = 0, (10.107) 2 is left as an exercise. (10.105) presumes that 𝖰 be of full rank. In the opposite case it must be replaced by 𝜺 𝜺 ⌊⌊𝖰(3) || 𝝂𝝂⌋⌋ 𝝂𝝂|| 𝜺 𝜺 ⌊⌊𝖰(3) ⌋⌋ . ( N N ) ( AB) = (( N N ) AB) = 0 (10.108) When 𝖰(3) = 0, there is no dispersion equation for the B-wave in a QDC medium. 268 CHAPTER 10 Media Defined by Plane-Wave Properties

PDC Media In the case of PDC medium (extended P-axion medium) the dispersion equation is decomposed as (8.184) and (8.185). In terms of the present symbols they have the form T 𝝂|(D⌊𝖯 )|𝝂 = 0, (10.109) −1T 𝝂|(C⌊𝖯 )|𝝂 = 0. (10.110) Omitting some details, in (8.191) it was shown that the field corresponding to the dispersion equation (10.110) satisfies the condition C|𝚽 = A|𝚽 = 0, (10.111) that is, it represents the A-wave. Similarly, the solutions of the dispersion equation (10.109) correspond to the field condition (8.192), which in the present form reads ( ) (2)T 1 D|𝖯 + (D ⋅ D)C |𝚽 = B|𝚽 = 0, (10.112) 2 due to the definition (10.95). Again, (10.110) assumes that 𝖯 has full rank. In the converse case it must be replaced by (4) (3) 𝝂|(C⌊(𝖨 ⌊⌊𝖯 ))|𝝂 = 0, (10.113) (3) and this is identically valid for 𝖯 = 0. For the P-axion medium without the extension, DC = 0, the PDC medium is an NDE medium because both dispersion equations vanish.

SDC Media Because the SDC medium equals the extended skewon–axion medium, properties of a plane wave in an SDC medium equal those discussed in Section 7.4. Expressing the dispersion dyadic as 𝖣 𝝂 𝝂𝝂⌋⌋ ⌊ 𝖡 ∧𝖨 T ( ) = (eN ( o∧ ) + AB + BA), (10.114) 𝖡 𝔼 𝔽 where o ∈ 1 1 is a trace-free dyadic, and denoting for simplicity 𝝂′ 𝖡T |𝝂 𝝂⌋ 𝝂⌋ = o , a = A, b = B, (10.115) after some steps left as an exercise we end up at the dispersion equation 𝖣(3) 𝝂 ⌊⌊𝝂𝝂 𝝂′| 𝝂′| . ( ) = 2(eNeN )( a)( b) = 0 (10.116) Thus, 𝝂′ must satisfy either 𝝂′|a = 0or𝝂′|b = 0. Problems 269

In conclusion, in the SDC medium case, the dispersion equation is decomposed to two equations as 𝝂′| 𝝂| 𝖡 ⌋ |𝝂 a = ( o A) = 0, (10.117) 𝝂′| 𝝂| 𝖡 ⌋ |𝝂 b = ( o B) = 0, (10.118) whence the plane-wave propagation depends on the two metric dyadics 𝖡 ⌋ 𝖡 ⌋ 𝔼 𝔼 o A and o B ∈ 1 1. The medium has no birefringence if the symmetric parts of these two dyadics are multiples of one another. To find the field conditions, the equation for the potential one-form can be expanded as 𝖣(𝝂)|𝝓 = (𝝂𝝂⌋⌋𝖠 + ab + ba)|𝝓 ⌊ 𝝂 𝝂′ 𝝓 |𝝓 |𝝓 . = eN ( ∧ ∧ ) + a(b ) + b(a ) = 0 (10.119) Multiplying by 𝝓| yields (a|𝝓)(b|𝝓) = 0. Multiplying by 𝝂′| leads to the condition (𝝂′|a)(b|𝝓) + (𝝂′|b)(a|𝝓) = 0, (10.120) from which we obtain the two possibilities 𝝂′| 𝝂| 𝖡 ⌋ |𝝂 ⇒ |𝝓 |𝚽 a = ( o A) = 0 a =−A = 0, (10.121) 𝝂′| 𝝂| 𝖡 ⌋ |𝝂 ⇒ |𝝓 |𝚽 . b = ( o B) = 0 b =−B = 0 (10.122) This verifies that the plane waves corresponding to the two dispersion equations (10.117) and (10.118) are respectively A-waves and B-waves. For vanishing extension, AB → 0, the SDC medium becomes an NDE medium since the dispersion equation is satisfied by any 𝝂.

PROBLEMS

10.1 Show that NDE media corresponding to Case 3 (a and d are quadratic functions while b and c do not depend on 𝝂 in (10.1)) are actually included in the Case 2 of NDE media. 10.2 Show that NDE media corresponding to Case 4 (a is a quadratic function while b and d are linear functions and c does not depend on 𝝂 in (10.1)) are actually included in the Case 1 of NDE media. 10.3 Assuming that the bidyadic A in (10.8) is simple, whence it can be expressed in terms of two vectors as A = a1 ∧ a2, show that the dispersion dyadic 𝖣(𝝂) is of variable rank for different 𝝂. 270 CHAPTER 10 Media Defined by Plane-Wave Properties

10.4 Show that a medium bidyadic 𝖬 restricted by a relation of the form (10.17) satisfies the dispersion equation for any 𝝂. 10.5 Show by considering different possibilities for the coefficients A, B,andC that the condition (10.17) can be reduced to the form (10.21). 10.6 Show that for the Case 2 of NDE media the medium bidyadic can be represented in a form combining the skewon–axion medium and P-axion medium in one expression as

𝖬 𝖡(2)T 𝖡∧𝖨 T 𝖨(2)T = A + B( ∧ ) + C

for some dyadic 𝖡 and scalars A, B, C. Verify that the dispersion equation (5.240) is satisfied for this medium bidyadic identically for any 𝝂. 10.7 Show that both Case 1 and Case 2 NDE media are closed in an affine transformation (6.16). 10.8 Show that when the medium is defined through the medium bidyadic 𝖭 as (5.52) in the form

(2)T (2)T (4) 𝖭 = 𝖯 − 훼 𝖨 ,tr𝖯 =Δ ≠ −훼2 1 1 1 P1 1

the medium is P-axion medium, that is, 𝖬 = 𝖭−1 is of the similar form. 10.9 Show that the orthogonality conditions (10.36) follow from the form (10.39). 10.10 Derive the expressions (10.48) and (10.49) to construct the dispersion dyadic (10.50) of a Q-axion medium with less than full-rank dyadic 𝖰 by assuming the expression (10.45). 10.11 Verify that the Q-axion medium bidyadic with 𝖰 defined by the rank 3 dyadic (10.45) satisfies the dyadic dispersion equation (5.240). 10.12 Expanding the expression (10.50) for the dispersion dyadic of a Q-axion medium with dyadic 𝖰 of the form (10.45), show that (2) it equals 𝖣(𝝂) = 𝖰 ⌊⌊𝝂𝝂. Problems 271

10.13 Define a condition for a Gibbsian anisotropic medium in which plane-wave fields are decomposed to TEa and TMb components ⋅ ⋅ satisfying the conditions a Ea = 0, b Hb = 0 for given Gibb- sain vectors a and b, by starting from a condition of the form 훼 ⋅ 훽 ⋅ Eg Bg + Hg Dg = 0. 10.14 Find the 3D representation for the SDC-medium bidyadic (10.94) by substituting the expansions 𝖯 𝖯 𝝅 𝜺 𝜺 = s + e4 + 4p + p 4e4,

DC = (d1 ∧ d2 + d3 ∧ e4)(c1 ∧ c2 + c3 ∧ e4), in (5.65). 10.15 Find the 3D representation for the SDC-medium bidyadic (10.98) by substituting the expansions 𝖡 𝖢 𝜸 𝜺 𝜺 𝖢 o = s + e4 s + cs 4 − e4 4(tr s), ⌊𝜶 ⌊𝜷 . A = e123 s + as ∧ e4, B = e123 s + bs ∧ e4 10.16 Verify that the QDC-medium dispersion equation (10.104) corresponds to the A-wave polarization condition. 10.17 Verify that the QDC-medium dispersion equation (10.105) corresponds to the B-wave polarization condition. 10.18 Derive the dyadic dispersion equation (10.116) for the SDC- medium case starting from (10.114).

APPENDIX A

Solutions to Problems

CHAPTER 1

1.1 Expand (𝛼A + 𝛽B) ∧ (𝛼A + 𝛽B) = 2𝛼𝛽A ∧ B. 1.2 Starting from

1.4 Defining∑ a vector basis so that A = e1 ∧ e2 = e12 and expanding B = i

From the expansion we see that B must be of the form B = e1 ∧ ≠ a + e2 ∧ b for some vectors a, b.FromB ∧ B = e1 ∧ a ∧ e2 ∧ b 0 it follows that the four vectors must be linearly independent.

273 274 APPENDIX A Solutions to Problems

1.5 Expressing C = a1 ∧ a2 + a3 ∧ a4 and applying the bac-cab rule twice we have 𝜷⌋ 𝜷| 𝜷| 𝜷| 𝜷| . C = a1( a2) − a2( a1) + a3( a4) − a4( a3) = 0 ≠ (1) For nonsimple C we have C ∧ C = 2a1 ∧ a2 ∧ a3 ∧ a4 0, 𝜶 whence the vectors ai make a basis with dual basis∑ i.From 𝜶 | 𝜷⌋ 𝜷| 𝜷 𝜶 |𝜷 i ( C) = 0, we obtain ai = 0foralli and = iai = 𝜷 ≠ 0. (2) For 0, the vectors ai cannot form a basis, whence C ∧ C = 0. 1.6 Writing A = x ∧ y, B = z ∧ w in terms of four vectors, from A ∧ B = x ∧ y ∧ z ∧ w = 0 it follows that they must be linearly dependent. Assuming x ∧ y ∧ z ≠ 0wemusthavew = 𝛼x + 𝛽y + 𝛾z for some scalars 𝛼, 𝛽, 𝛾. We can write A = x ∧ y = a ∧ c, B = z ∧ (𝛼x + 𝛽y) = b ∧ c by choosing a = x∕𝛽, b = z and c = 𝛼x + 𝛽y. The case x ∧ y ∧ z = 0 can be handled similarly. 1.7 Assuming A ∧ A ≠ 0andB ∧ B ≠ 0 and requiring 𝜆 𝜆 𝜺 | 𝜆 𝜆 (A + B) ∧ (A + B) = eN N ((A + B) ∧ (A + B)) = 0 for some scalar 𝜆 leads to the scalar equation 𝜺 | 𝜆𝜺 | 𝜆2𝜺 | N (A ∧ A) + 2 N (A ∧ B) + N (B ∧ B) = 0, from which 𝜆 can be solved. 𝜺 | 1.8 Because A ∧ A = 0and N (A ∧ A) = 0 are equivalent, expanding 𝜺 | | 𝜺 ⌊ |𝚽 ⌊ 𝜺 ⌊ |𝚽 N (A ∧ A) = A ( N A) = A = (eN ( N A)) | 𝚽 𝚽 = eN ( ∧ ) = 0 we obtain 𝚽 ∧ 𝚽 = 0. 1.9 Applying the bac-cab formula we can write ∑ ⌊𝜶 𝛼 𝛼 𝛼 𝛼 A = 1e2 − 2e1 + 3e4 − 4e3 = aiei 𝛼 Finding the coefficients i in terms of the coefficients ai the solution becomes 𝜶 𝜺 𝜺 𝜺 𝜺 𝜺 𝜺 𝜺 𝜺 ⌊ = a2 1 − a1 2 + a4 3 − a3 4 =−( 1 ∧ 2 + 3 ∧ 4) a 𝜺 ⌊ ⌊ 𝜺 ⌊ ⌊ =−( N (e1 ∧ e2 + e3 ∧ e4)) a =−( N A) a APPENDIX A Solutions to Problems 275

Because of A ∧ A = 2eN the solution can be written as 2𝜺 ⌊A 𝜶 N ⌊ . =−𝜺 | a N A ∧ A

1.10 From A ∧ A = 2e1234 = 2eN and expanding ∑ ⌊𝖨T ⌊𝜺 (A ∧ A) = 2 eN iei = 2(e234e1 + e314e2 + e124e3 − e123e4) i

= 2(e34 ∧ (e2e1 − e1e2) + e12 ∧ (e4e3 − e3e4)) ⌊𝖨T ⌊𝖨T = 2(e34 ∧ (e12 ) + e12 ∧ (e34 )) ⌊𝖨T = 2((e12 + e34) ∧ ((e12 + e34) ) = 2A ∧ (A⌊𝖨T ), we obtain ⌊𝜶 ⌊𝜶 ⇒ 𝜺 |𝜶 𝜺 ⌊ 𝜶 (A ∧ A) = 2A ∧ (A ) N = N (A ∧ A) 𝜺 ⌊ ⌊ ⌊𝜶 = 2( N A) (A ), from which 𝜶 can be solved. 𝜶 𝜺 𝜷 𝜺 1.11 Choosing a basis∑ of one-vectors we can assume = 1, = 2, and expand A = i

we obtain Aij = 0forallij except ij = 34. Thus, we can write ⌊ 𝜺 𝜺 ⌊ 𝜶 𝜷 . A = A34e34 = A34eN ( 1 ∧ 2) = kN ( ∧ ), kN = A34eN ∑ ∑ 1.12 Comparing the two expressions A = i

2A1 = A12 + A34,2A2 = A23 + A14,2A3 = A31 + A24, . 2A4 = A12 − A34,2A5 = A23 − A14,2A6 = A31 − A24

Thus, the bivectors Ei form a basis. Considering different i and j we can find the orthogonality rules ⋅ ≠ ⋅ Ei Ej = 0, i j, Ei Ei = 2, i = 1, 2, 3, ⋅ . Ei Ei =−2, i = 4, 5, 6 276 APPENDIX A Solutions to Problems

1.13 To derive (1.106) we can apply (1.76), written in another form, and proceed as 𝜶⌋(a ∧ b ∧ c) = (𝜶|a)(b ∧ c) + (𝜶|b)(c ∧ a) + (𝜶|c)(a ∧ b) = (𝜶|a)(b ∧ c) + 𝜶|(bc − cb) ∧ a = (𝜶|a)(b ∧ c) − (𝜶⌋(b ∧ c)) ∧ a, applying the bac-cab formula. Because this is a linear identity in the simple bivector b ∧ c we can replace it by the general bivector C whence the rule becomes 𝜶⌋(a ∧ C) = (𝜶|a)C − (𝜶⌋C) ∧ a which is equivalent to (1.106). To derive (1.107) we can start from (1.83) written as 𝜶⌋(a ∧ b ∧ c ∧ d) =−(𝜶|a)(b ∧ c ∧ d) + (𝜶|b)(c ∧ d ∧ a) − (𝜶|c)(d ∧ a ∧ b) + (𝜶|d)(a ∧ b ∧ c) =−𝜶|(ab − ba)∧(c∧d) − 𝜶|(cd − dc)∧(a∧b) = (𝜶⌋(a ∧ b)) ∧ (c ∧ d) + (𝜶⌋(c ∧ d)) ∧ (a ∧ b). Replacing a ∧ b and c ∧ d by the respective general bivectors B and C we arrive at the rule 𝜶⌋(B ∧ C) = (𝜶⌋B) ∧ C + (𝜶⌋C) ∧ B, which is equivalent to (1.107). 1.14 From (1.108) we see that if the bivector B is decomposed as B = B∥ + B⟂,wemusthaveB⟂ = 0, whence the solution can be expressed as a ∧ (𝜶⌋B) B = B =− , ∥ a|𝜶 for any one-form 𝜶 satisfying a|𝜶 ≠ 0. Thus any bivector of the form B = a ∧ b is a solution. ∑ 𝜶 𝜺 1.15 Choose a basis so that = 1 and expand B = i

we conclude that B12 = B31 = B14 = 0, whence 𝖡 = B23e23 + B24e24 + B34e34 APPENDIX A Solutions to Problems 277

satisfying B ∧ B = 0. Writing 𝜺 ⌋ B = 1 (B23e123 + B24e124 + B34e134) the bivector has the form 𝜶⌋k. 1.16 Starting from (1.83) and expanding 𝜶⌋(a ∧ b ∧ c ∧ d) =−(𝜶|a)(b ∧ c ∧ d) + (𝜶|b)(c ∧ d ∧ a) − (𝜶|c)(d ∧ a ∧ b) + (𝜶|d)(a ∧ b ∧ c) =−((𝜶|a)(b ∧ c) + (𝜶|b)(c ∧ a) + (𝜶|c)(a ∧ b)) ∧ d + (𝜶|d)(a ∧ b ∧ c) =−(𝜶⌋(a ∧ b ∧ c)) ∧ d + (𝜶|d)(a ∧ b ∧ c), we may replace a ∧ b ∧ c by the general trivector k and arrive at the rule 𝜶⌋(k ∧ d) =−(𝜶⌋k) ∧ d + (𝜶|d)k. Replacing d by a and assuming a|𝜶 ≠ 0 we can expand the trivector as 𝜶⌋(k ∧ a) a ∧ (𝜶⌋k) k = k훼 + k , k훼 = , k = . a a|𝜶 a a|𝜶 1.17 The identity can be more easily derived for the special case: ⌊ 𝜶 𝜺 ⌊𝜶 𝜶⌋ 𝜶⌋ ⌊𝜺 . eN ( ∧ 12) = e34 =− e34 =− (eN 12) 𝜺 Since choice of the basis is arbitrary, we can replace 12 by any simple two-form 𝚿 = 𝜷 ∧ 𝜸. Because of linearity, it remains valid for a sum of any two simple two-forms and, thus, for any two-form 𝚿. 𝜶 ⌊𝜶 1.18 Express k in terms of a one-form as eN and expand 𝜺 ⌊ ⌋ ⌊ 𝜺 ⌊ ( N A) k = k ( N A) ⌊𝜶 ⌊ 𝜺 ⌊ ⌊ 𝜶 𝜺 ⌊ = (eN ) ( N A) = eN ( ∧ ( N A)) ⌊ 𝜺 ⌊ 𝜶 ⌊ 𝜺 ⌊ ⌊𝜶 = eN (( N A) ∧ ) = eN ( N (A )) ⌊𝜶 𝜶⌋ 𝜺 ⌊ ⌋ . =−A = A = ( N k) A

To check: k = e234, A = e12, 𝜺 ⌊ ⌋ 𝜺 ⌋ ( N e12) e234 = 34 e234 = e2 𝜺 ⌊ ⌋ 𝜺 ⌋ . ( N e234 e12 =− 1 e12 = e2 278 APPENDIX A Solutions to Problems

1.19 We can choose∑ a vector basis {ei}sothatA = e12 + e34. Expand- ing X = Xijeij we obtain X12 + X34 = 0, which leaves 5 free ≠ parameters Xij, ij 34 for the solution. 1.20 Expanding

a = as + a4e4, b = bs + b4e4, we can write

a ∧ b = as ∧ bs + (b4as −(a4bs) ∧ e4) 1 . = (b4as − a4bs) ∧ bs + e4 b4 ≠ Assuming b4as − a4bs 0 we can identify . cs = b4as − a4bs, d = b∕b4

In the case b4as − a4bs = 0wehavea ∧ b = as ∧ bs.

CHAPTER 2 ∑ ∑ 𝖨 𝜺 𝖠 𝜶 2.1 (1) For = ei i and = ej j expand ∑ ∑ 𝖨∧𝖠 𝜺 𝜶 ||𝖨(2)T | 𝜺 𝜶 tr( ∧ ) = (ei ∧ ej)( i ∧ j) = (ei ∧ ej) ( i ∧ j) ∑i,j ∑ ∑ i,j |𝜺 |𝜶 |𝜶 |𝜺 = (ei i) (ej j) − (ei j)(ej i) i j i,j = tr𝖨 tr𝖠 − 𝖨T ||𝖠 = 3tr𝖠. (2) Applying the dyadic rule we can expand 𝖨∧𝖠 𝖨(2)|| 𝖨∧𝖠 T 𝖨(2)⌊⌊𝖨T ||𝖠T tr( ∧ ) = ( ∧ ) = ( ) = ((tr𝖨)𝖨 − 𝖨)||𝖠T = (4 − 1)𝖨||𝖠T = 3tr𝖠. 2.2 Applying the identity 𝖨(2)⌊⌊𝖡T = (tr𝖡)𝖨 − 𝖨|𝖡|𝖨 = (tr𝖡)𝖨 − 𝖡, we obtain 𝖡 = (tr𝖡)𝖨 − 𝖠. APPENDIX A Solutions to Problems 279

Taking the trace, tr𝖡 can be solved as 1 tr𝖡 = 4tr𝖡 − tr𝖠⇒tr𝖡 = tr𝖠, 3 whence the solution reads 1 𝖡 = tr𝖠 𝖨 − 𝖠. 3 2.3 Applying the multivector expansion twice on the right side of 𝜶 ∧ 𝜶 ∧ 𝜶 ⌊⌊𝜹 ⌊𝜹 𝜶 𝜶 𝜶 ⌊ (a1 1∧a2 2∧a3 3) d = ((a1 ∧ a2 ∧ a3) )(( 1 ∧ 2 ∧ 3) d) produces 3 × 3 = 9 terms invariant in cyclic permutations of the 𝜶 three dyadics ai i. The result can be expressed as 𝜶 ∧ 𝜶 ∧ 𝜶 ⌊⌊𝜹 𝜶 𝜶 𝜶 ||𝜹 (a1 1∧a2 2∧a3 3) d = (a1 1 ∧ a2 2)a3 3 d 𝜶 ∧ 𝜶 | 𝜹| 𝜶 𝜶 | 𝜹| 𝜶 ⋯ − a1 1∧(a2 2 d a3 3 + a3 3 d a2 2) + , where ⋯ corresponds to two permutations of the previous three 𝜶 𝖠 terms. Replacing ai i by i, the expansion rule becomes 𝖠 ∧𝖠 ∧𝖠 ⌊⌊𝖣T 𝖠 ∧𝖠 𝖠 ||𝖣T 𝖠 ∧𝖠 𝖠 ||𝖣T ( 1∧ 2∧ 3) = ( 1∧ 2)( 3 ) + ( 2∧ 3)( 1 ) 𝖠 ∧𝖠 𝖠 ||𝖣T 𝖠 ∧ 𝖠 |𝖣|𝖠 𝖠 |𝖣|𝖠 + ( 3∧ 1)( 2 ) − 1∧( 2 3 + 3 2) 𝖠 ∧ 𝖠 |𝖣|𝖠 𝖠 |𝖣|𝖠 𝖠 ∧ 𝖠 |𝖣|𝖠 𝖠 |𝖣|𝖠 . − 2∧( 3 1 + 1 3) − 3∧( 1 2 + 2 1) 𝖠 𝖠 𝖠 𝖠 For 1 = 2 = 3 = this reduces to 𝖠(3)⌊⌊𝖣T 𝖠(2) 𝖠||𝖣T 𝖠∧ 𝖠|𝖣|𝖠 = ( ) − ∧( ) and for 𝖣 = 𝖠−1 𝖠(3)⌊⌊𝖠−1T = 2𝖠(2). For 𝖠 = 𝖨 we have tr(𝖨(3)⌊⌊𝖨−1T ) = (𝖨(3)⌊⌊𝖨−1T )||𝖨(2)T = 3tr𝖨(3) = 3 ⋅ 4 = 12, tr(2𝖨(2)) = 2 ⋅ 6 = 12. 𝖢 𝖠T 𝖠∧𝖡 𝖣 2.4 Inserting = and ∧ = in the dyadic rule and applying 𝖠||𝖠−1T = tr(𝖠|𝖠−1) = tr𝖨 = 4 yields 𝖣⌊⌊𝖠−1T = 2𝖡 + (𝖡||𝖠−1T )𝖠. Operating this by ||𝖠−1T , the left side can be written as (𝖣⌊⌊𝖠−1T )||𝖠−1T = 2𝖣||(𝖠−1T )(2) 280 APPENDIX A Solutions to Problems

while the right side becomes 2𝖡||𝖠−1T + 4𝖡||𝖠−1T = 6𝖡||𝖠−1T . Combining these we obtain 𝖡||𝖠−1T = (1∕3)𝖣||𝖠(−2)T which inserted in the dyadic equation leads to the solution 1 1 𝖡 = 𝖣⌊⌊𝖠−1T − (𝖣||(𝖠(−2)T )𝖠. 2 6 A condition between 𝖠 and 𝖣 is obtained by substituting the solution 𝖡 in the original equation, 1 1 𝖣 = (𝖣⌊⌊𝖠−1T )∧𝖠 − (𝖣||(𝖠(−2)T )𝖠(2). 2 ∧ 3 As a simple check, setting 𝖣 = 𝖨(2) and 𝖠 = 𝖨 leads to 1 1 1 1 1 𝖡 = 𝖨(2)⌊⌊𝖨−1T − (tr(𝖨(2))𝖨 = 3𝖨 − 6𝖨 = 𝖨, 2 6 2 6 2 which obviously is the correct solution. ∑ 𝖠 2.5 (1) Expanding = aibi we have ∑ ∑ ∑ | 𝚽⌊ 𝚽| (bjaj) ( aibi) = bj( (ai ∧ aj))bi j i i,j ∑ 𝚽| = (ai ∧ aj)(bjbi) ∑i,j 𝚽| = (ai ∧ aj)(bjbi − bibj) < ∑i j 𝚽| ⌊𝖨T = (ai ∧ aj)(bi ∧ bj) i

2.6 Applying the rule of Problem 2.1 we have (tr𝖷)𝖨 − 𝖷 = 휆𝖷 taking the trace yields 4tr𝖷 − tr𝖷 − 휆tr𝖷 = tr𝖷(휆 − 3) = 0. One solution is 휆 =−1andany𝖷 satisfying tr𝖷 = 0. The other solution is 휆 = 3and𝖷 = 훼𝖨 for any scalar 훼. 2.7 Setting B = C = A in 𝜶⌋(B ∧ C) = B ∧ (𝜶⌋C) + C ∧ (𝜶⌋B), we obtain 1 A𝜶⌋e = A ∧ (𝜶⌋A), A = 𝜺 |(A ∧ A). N 2 N 𝜺 ⌊ 𝜶⌋ 𝜶 Applying N ( eN) = , this becomes 𝜶 𝜺 ⌊ 𝜶⌋ 𝜺 ⌊ ⌊𝖨T |𝜶. A = N (A ∧ ( A)) =− N (A ∧ (A )) Since this is valid for any 𝜶, we obtain 𝖨T 𝜺 ⌊ ⌊𝖨T 𝜺 ⌊ ⌊𝖨 | ⌊𝖨T . A =− N (A ∧ (A )) =−(( N A) ) (A ) The condition for the existence of the inverse is now A ≠ 0, or A must be a nonsimple bivector. In this case the inverse can be expressed in the form 𝜺 ⌊A ⌊𝖨T −1 𝚽⌊𝖨 𝚽 N . (A ) = , =−𝜺 | N (A ∧ A)∕2 This result proves that the inverse of an antisymmetric dyadic is antisymmetric. 2.8 Applying A = a ∧ b + c ∧ d and A ∧ A = 2a ∧ b ∧ c ∧ d, we can make use of the bac-cab formula as 𝖨⌋A = ba − ab + dc − cd. Expanding (𝖨⌋A)(2) = (a ∧ b)(a ∧ b) + (c ∧ d)(c ∧ d) − (b ∧ d)(c ∧ a) −(b ∧ c)(a ∧ d) − (a ∧ d)(b ∧ c) − (c ∧ a)(b ∧ d) this can be compiled to the form (𝖨⌋A)(2) = AA − 𝖣 − 𝖣T , 282 APPENDIX A Solutions to Problems

when denoting 𝖣 = (a ∧ b)(c ∧ d) + (b × c)(a ∧ d) + (c ∧ a)(b ∧ d). Actually we can write 1 𝖣 + 𝖣T = (a ∧ b ∧ c ∧ d)⌊𝖨(2)T = (A ∧ A)⌊𝖨(2)T , 2 whence the rule finally emerges in the form 1 (𝖨⌋A)(2) = AA − (A ∧ A)⌊𝖨(2)T . 2 𝖠 |𝖠 2.9 The condition 1 2 = 0 can be decomposed in four 3D equations 𝖡 |𝖡 𝜷 𝜷 | 1 2 + b1 2 = 0, 1 b2 + B1B2 = 0, 𝖡 | 𝜷 |𝖡 1 b2 + b1B2 = 0, 1 2 + B1b2 = 0, 𝖡 𝜷 which yield the condition 2 = b2 2∕B2 whence 𝖠 1 𝜷 𝜺 . 2 = (b2 + B2e4)( 2 + B2 4) B2 2.10 First we have 𝖣⌊⌊𝖨T 𝖣⌊⌊𝖨T ||𝖨(2)T || 𝖨∧𝖨(2) T 𝖣 tr( ) = ( ) = D ( ∧ ) = 3tr = 0, whence tr𝖣 = tr(𝖨(4)⌊⌊𝖠T ) = (𝖨(4)⌊⌊𝖠T )||𝖨(3)T = (𝖨(4)⌊⌊𝖨(3)T )||𝖠T ) = tr𝖠 = 0. Expanding 𝖣⌊⌊𝖨T 𝖨(4)⌊⌊𝖠T ⌊⌊𝖨T 𝖨(4)⌊⌊ 𝖠∧𝖨 T = ( ) = ( ∧ ) = (𝖨(4)⌊⌊𝖨T )⌊⌊𝖠T = 𝖨(3)T ⌊⌊𝖠T 𝖠 𝖨(2) 𝖨∧𝖠 𝖨∧𝖠 = (tr ) − ∧ = ∧ , and 𝖨∧𝖠 ⌊⌊𝖨T 𝖠 𝖨 𝖠 𝖠 ( ∧ ) = (tr ) + 2 = 2 lead to the following chain of conclusions: 𝖣⌊⌊𝖨T = 0 requires 𝖠∧𝖨 𝖠 𝖣 𝖨(4)⌊⌊𝖠T ∧ = 0and = 0, whence = = 0. APPENDIX A Solutions to Problems 283

2.11 Applying the bac-cab rule (1.106), we can expand 𝖨(3)⌊⌊𝜶a = 𝜶⌋𝖨(3)⌊a = 𝜶⌋(a ∧ 𝖨(2)) = a ∧ (𝜶⌋𝖨(2)) + (a|𝜶)𝖨(2) = a ∧ (𝖨 ∧ 𝜶) + (a|𝜶)𝖨(2) 𝖨∧ 𝜶 |𝜶 𝖨(2). =− ∧a + (a ) Replacing a𝜶 by 𝖠 we arrive at the identity 𝖨(3)⌊⌊𝖠T 𝖠 𝖨(2) 𝖨∧𝖠. = (tr ) − ∧ The other identity can be derived along similar lines. 𝖠 𝜶 𝖡 𝜷 𝖨(2) 𝜺 2.12 Setting = a and = b and = eij ij, we can write 𝖨(2)∧𝖠 ⌊⌊𝖡T 𝖨(2)∧ 𝜶 ⌊⌊𝜷 𝜷⌋ 𝜺 𝜶 ⌊ . ( ∧ ) = ( ∧a ) b = (a ∧ eij)( ij ∧ ) b Applying the bac-cab rule the two terms become 𝜷⌋ 𝜷⌋ |𝜷 (a ∧ eij) = a ∧ ( eij) + (a )eij, 𝜺 𝜶 ⌊ 𝜺 ⌊ 𝜶 |𝜶 𝜺 . ( ij ∧ ) b = ( ij b) ∧ + (b ) ij Combining these, we obtain a sum of four terms 𝖨(2)∧𝖠 ⌊⌊𝖡T 𝜷⌋𝖨(2)⌊ 𝜶 |𝜷 𝖨(2)⌊ 𝜶 ( ∧ ) = a ∧ ( b) ∧ + (a )( b) ∧ + (b|𝜶)a ∧ (𝜷⌋𝖨(2)) + (a|𝜷)(b|𝜶)𝖨(2), which can be expanded separately as 𝜷⌋𝖨(2)⌊ 𝜶 𝖨(2)⌊⌊𝜷 ∧ 𝜶 𝖨(2)⌊⌊𝖡T ∧𝖠 a ∧ ( b) ∧ = ( b)∧a = ( )∧ 𝖡 𝖨 𝖡 ∧𝖠 = (tr − )∧ |𝜷 𝖨(2)⌊ 𝜶 |𝜷 𝖨 𝜶 𝖨∧ 𝜷| 𝜶 (a )( b) ∧ = (a )(b ∧ ) ∧ =− ∧(b a ) 𝖨∧ 𝖡|𝖠 =− ∧( ) |𝜶 𝜷⌋𝖨(2) |𝜶 𝖨 𝜷 𝖨∧ 𝜶| 𝜷 (b )a ∧ ( ) = (b )a ∧ ( ∧ ) =− ∧(a b ) 𝖨∧ 𝖠|𝖡 =− ∧( ) (a|𝜷)(b|𝜶)𝖨(2) = tr(a𝜶|b𝜷)𝖨(2) = tr(𝖠|𝖡)𝖨(2). Combining these, we obtain the required identity, 𝖨(2)∧𝖠 ⌊⌊𝖡T 𝖠|𝖡 𝖨(2) 𝖡 𝖨∧𝖠 𝖨∧ 𝖠|𝖡 𝖡|𝖠 𝖠∧𝖡 ( ∧ ) = tr( ) + (tr ) ∧ − ∧( + ) − ∧ , 284 APPENDIX A Solutions to Problems

2.13 Applying the identities of the previous problems, we can expand 𝖨(4)⌊⌊ 𝖠∧𝖡 T 𝖨(4)⌊⌊𝖠T ⌊⌊𝖡T 𝖠 𝖨(3)⌊⌊𝖡T 𝖨(2)∧𝖠 ⌊⌊𝖡T ( ∧ ) = ( ) = (tr ) − ( ∧ )) = (tr𝖠 tr𝖡 − tr(𝖠|𝖡))𝖨(2) 𝖨∧ 𝖠|𝖡 𝖡|𝖠 𝖠 𝖡 𝖡 𝖠 𝖠∧𝖡. + ∧( + − (tr ) − (tr ) ) + ∧ This identity can be made more compact through the rules 𝖠∧𝖡 ⌊⌊𝖨T 𝖠 𝖡 𝖡 𝖠 𝖠|𝖡 𝖡|𝖠 ( ∧ ) = (tr ) + (tr ) − − , 1 tr(𝖠∧𝖡) = ((𝖠∧𝖡)⌊⌊𝖨T )||𝖨T = tr𝖠 tr𝖡 − tr(𝖠|𝖡), ∧ 2 ∧ which inserted gives us 𝖨(4)⌊⌊ 𝖠∧𝖡 T 𝖠∧𝖡 𝖨(2) 𝖠∧𝖡 ⌊⌊𝖨T ∧𝖨 𝖠∧𝖡. ( ∧ ) = tr( ∧ ) − (( ∧ ) )∧ + ∧ 𝖠∧𝖡 Because of linear dependence, the bidyadic ∧ can be replaced 𝖢 E F by an arbitrary bidyadic ∈ 2 2. 2.14 Operating the different terms of the identity as 𝖨(4)⌊⌊()T and applying the identities given of the previous problems we have 𝖨(4)⌊⌊(𝖨(4)T ⌊⌊𝖢) = 𝖢, 𝖨(4)⌊⌊(tr𝖢 𝖨(2)T ) = tr𝖢 𝖨(2) 𝖨(4)⌊⌊ 𝖢T ⌊⌊𝖨 ∧𝖨T 𝖨(4)⌊⌊𝖨T ⌊⌊ 𝖢T ⌊⌊𝖨 𝖨(3)⌊⌊ 𝖢T ⌊⌊𝖨 (( )∧ ) = ( ) ( ) = ( ) 𝖢T ⌊⌊𝖨 𝖨(2) 𝖨∧ 𝖢⌊⌊𝖨T = tr( ) − ∧( ) 𝖢 𝖨(2) 𝖢⌊⌊𝖨T ∧𝖨 = 2tr − ( )∧ 𝖨(4)⌊⌊𝖢T = 𝖨(4)⌊⌊𝖢T . Combining these we arrive at the original identity. The second case follows similar lines. 2.15 Inserting 𝖠 = 𝖡 + 𝖢 and applying linearity we obtain 𝖡 𝖢 (2) 𝖡(2) 𝖡∧𝖢 𝖢(2) 𝖡∧𝖢 . f (( + ) ) = f ( ) + 2f ( ∧ ) + f ( ) = 2f ( ∧ ) = 0 𝖡 𝜺 𝖢 𝜺 𝜺 Assuming = ei k and = ej 𝓁 we∑ have f (eij k𝓁) = 0, whence 𝖣 𝜺 expanding the general bidyadic = ij,k𝓁 Dij,k𝓁eij k𝓁 we obtain ( ) ∑ ∑ 𝖣 𝜺 𝜺 . f ( ) = f Dij,k𝓁eij k𝓁 = Dij,k𝓁f (eij k𝓁) = 0 ij,k𝓁 ij,k𝓁 APPENDIX A Solutions to Problems 285

𝖡 ∧𝖨 𝖲∧ ⌊𝖨T ⌋𝜺 E F 2.16 To find the representation o∧ = ( ∧(A )) N ∈ 2 2,we 𝖲 replace by ss and A by e12 and expand ∧ ⌊𝖨T ⌋𝜺 ∧ ⌋𝜺 (ss∧(e12 )) N = (ss∧(e2e1 − e1e2)) N ⌋𝜺 ⌋𝜺 = (s ∧ e2)(s ∧ e1) N − (s ∧ e1)(s ∧ e2) N ⌋𝜺 ⌋𝜺 =−(s ∧ e2)s 234 − (s ∧ e1)s 134 Applying the multivector rule we can further expand ∧ ⌊𝖨T ⌋𝜺 |𝜺 𝜺 𝜺 ⌋𝜺 (ss∧(e12 )) N =−(s ∧ e2)((s 2) 34 + 2 ∧ (s 34)) |𝜺 𝜺 𝜺 ⌋𝜺 − (s ∧ e1)((s 1) 34 + 1 ∧ (s 34)) 𝜺 𝜺 | 𝜺 =−s ∧ ((e2 2 + e1 1) s) 34 𝜺 𝜺 ⌋𝜺 . − s ∧ (e2 2 + e1 1) ∧ (s 34) Working on the latter term as 𝜺 𝜺 ⌋𝜺 𝖨 ⌋𝜺 s ∧ (e2 2 + e1 1) ∧ (s 34) = s ∧ ∧ (s 34) 𝜺 𝜺 ⌋𝜺 − s ∧ (e3 3 + e4 4) ∧ (s 34) ⌋𝜺 ∧𝖨 𝜺 | =−(ss 34)∧ + s ∧ (e3 3 s 𝜺 | 𝜺 + e4 4 s) 34, and inserting this in the previous expansion we obtain ∧ ⌊𝖨T ⌋𝜺 𝖨| 𝜺 ⌋𝜺 ∧𝖨 ⌋ ⌋𝜺 ∧𝖨. (ss∧(e12 )) N =−s ∧ ( s) 34 + (ss 34)∧ = (ss (e12 N))∧ 𝖲 Replacing ss by and e12 by A this can be written as the identity 𝖲∧ ⌊𝖨T ⌋𝜺 𝖲⌋ ⌋𝜺 ∧𝖨 𝖲 ⌋𝜺 ∧𝖨 ( ∧(A )) N = ( (A N))∧ = ((( ∧ A) N)∧ ), valid for any symmetric dyadic 𝖲 and a bivector A. The dyadic 𝖡 o can now be identified as 𝖡 𝖲⌋ ⌋𝜺 𝖲 ⌋𝜺 . o = (A N) = ( ∧ A) N 2.17 Starting from the equation 𝖠| (as + e4a4) = bs + e4b4 and splitting it in spatial and temporal parts as 𝖠 | s as + asa4 = bs 𝜶 | s bs + aa4 = b4 286 APPENDIX A Solutions to Problems

we can solve for as, a4 as 1 a = 𝖠−1|b + ((𝖠−1|a )(𝜶 |𝖠−1)|b − (𝖠−1|a )b ) s s s D s s s s s s s 4 1 a = (−(𝜶 |𝖠−1)|b + b ) 4 D s s s 4 𝜶 |𝖠−1| with D = a − s s as. Combining these as 𝖠−1| as + e4a4 = (bs + e4b4) we can identify the inverse dyadic, 1 𝖠−1 = 𝖠−1 + (𝖠−1|a 𝜺 − e 𝜶 |𝖠−1 − e 𝜺 ), s D s s 4 4 s s 4 4 which can be further written in the form of (2.134). 2.18 The inverse of the dyadic 𝖡 = 𝖠 + ab + ba can be found by solving the equation 𝖡|𝜸 = c for the one-form 𝜸 and expressing the solution in the form 𝜸 = 𝖡−1|c. Representing the solution as 𝜸 = 𝖠−1|c − 𝖠−1|(ab + ba)|𝜸, let us multiply this by a| and b| and express the equations in the form ( )( ) ( ) 1 + a|𝖠−1|ba|𝖠−1|a a|𝜸 a|𝖠−1|c = . b|𝖠−1|b 1 + b|𝖠−1|a b|𝜸 b|𝖠−1|c Assuming that the determinant Δ=(1 + a|𝖠−1|b)(1 + b|𝖠−1|a) − (a|𝖠−1|a)(b|𝖠−1|b) 𝖠−1|| 𝖠(−2)|| ∧ = 1 + (ab + ba) + (ab∧ba) does not vanish, we can solve for a|𝜸 and b|𝜸 and substitute them in the expression ( ) a|𝜸 𝜸 = 𝖠−1|c − (𝖠−1|b 𝖠−1|a) , b|𝜸 from which the inverse dyadic can be identified as 1 𝖡−1 = 𝖠−1 − 𝖠−1|((1 + b|𝖠−1|a)ba − (a|𝖠−1|a)bb Δ − (b|𝖠−1|b)aa + (1 + a|𝖠−1|b)ab)|𝖠−1 1 = 𝖠−1 − 𝖠−1|(ab + ba + 𝖠−1T ⌋⌋(ab∧ba))|𝖠−1. Δ ∧ APPENDIX A Solutions to Problems 287

For the special case b = a∕2 we obtain

𝖠−1|aa|𝖠−1 (𝖠 + aa)−1 = 𝖠−1 − , 1 + 𝖠−1||aa which can be easily verified by direct multiplication. 𝖠(3) ≠ 2.19 Assuming Δ=tr s 0, from the Cayley–Hamilton equation 𝖠3 𝖠 𝖠2 𝖠(2) 𝖠 𝖠(3) 𝖨 s − (tr s) s + (tr s ) s − (tr s ) s = 0 we obtain the inverse rule 1 𝖠−1 = (𝖠2 − (tr𝖠 )𝖠 + (tr𝖠(2))𝖨 ). s Δ s s s s s 𝖠 Applying the rule (2.101), written for s as 𝖨(3)⌊⌊𝖠(2)T 𝖠(2) 𝖨 𝖠 𝖠 𝖠2 s = (tr s ) − (tr s) s + s we can write 1 𝖠−1 = (𝖨(3)⌊⌊𝖠(2)T − (tr𝖠(2))𝖨 + (tr𝖠(2))𝖨 ) s Δ s s s s 1 = (𝖨(3)⌊⌊𝖠(2)T + (𝖨(2)∧e 𝜺 )⌊⌊𝖠(2)T − (tr𝖠(2))e 𝜺 ). Δ s s s ∧ 4 4 s s 4 4 𝖨(2)∧ 𝜺 ⌊⌊𝖠(2)T E F Obviously, the term ( s ∧e4 4) s ∈ 1 1 must be of the 훼 𝜺 훼 form e4 4. The coefficient can be expanded as 훼 𝖨(2)∧ 𝜺 ⌊⌊𝖠(2)T ||𝜺 = (( s ∧e4 4) s ) 4e4 𝖨(2)∧ 𝜺 ⌊⌊𝜺 ||𝖠(2)T 𝖠(2). = ( s ∧e4 4) 4e4) s = tr s Inserting this in the above expression, the inverse of the spatial dyadic is reduced to 1 𝖠−1 = 𝖨(3)⌊⌊𝖠(2)T . s Δ s s

2.20 (1) Assuming a = a1 is a vector in a basis {ai}, we have 𝖠(4)| 𝖠| 𝖠| | 𝖠| | 𝖠| (a1 ∧ a2 ∧ a3 ∧ a4) = ( a1) ∧ ( a2) ( a3) ( a4) = 0 𝖠(4)| 𝖠 which implies eN = 0. In other words, the dyadic must be of rank 3 or lower. 288 APPENDIX A Solutions to Problems

(2) Assuming a = a1 and b = a2 are vectors in a basis {ai}, we have 𝖠(3)| 𝖠| 𝖠| | 𝖠| (ai ∧ aj ∧ ak) = ( ai) ∧ ( aj) ( ak) = 0 for all four index combinations i, j, k since at least one of the indices must be either 1 or 2. Thus, 𝖠(3)|k = 0 for any trivector k and, hence, 𝖠(3) = 0. In other words, the dyadic 𝖠 must be of rank 2 or lower. 𝖢 𝖨(3)⌊⌊𝖠T 𝜺 ⌊⌊𝖠T 2.21 The representation s = s s = e123 123 s can be inverted 𝖠 𝜺 ⌊⌊𝖢T 𝖨(3)⌊⌊𝖢T as s = e123 123 s = s s . The corresponding relation between the inverses is obtained from 𝖢 |𝖢−1 ⌊𝖠T ⌋𝜺 |𝖢−1 𝖨(2) s s = (e123 s 123) s = s , which implies

𝜺 ⌊𝖢−1 𝖠−1T ⌋𝜺 ⇒ 𝖢−1 𝖨(3)⌊⌊𝖠−1T . 123 s = s 123 s = s s Invoking the inverse rule (2.136), we can write tr𝖠(3) = tr(𝖨(3)⌊⌊𝖢T )(3) =Δ , s s s Cs 1 1 𝖢−1 = 𝖨(3)⌊⌊𝖠−1T = 𝖨(3)⌊⌊(𝖨(3)T ⌊⌊𝖠(2)) = 𝖠(2) s s s Δ s s s Δ s Cs Cs 1 = (𝖨(3)⌊⌊𝖢T )(2), Δ s s Cs which coincides with (2.137). 2.22 Expand 𝜺 𝜺 ⌊⌊ 𝖠∧𝖲(2) 𝜺 𝜺 ⌊⌊𝖲(2) ⌊⌊𝖠 𝖲(−2)⌊⌊𝖠 N N ( ∧ ) = ( N N ) =ΔS 𝖲−1||𝖠 𝖲−1 𝖲−1|𝖠|𝖲−1 =ΔS(( ) − ) 𝖲−1| ⌊𝖨T |𝖲−1 𝖲−1| ⌊𝖲−1 . =−ΔS (A ) =−ΔS (A ) Applying now the identity of Problem 2.5, we can proceed as 𝜺 𝜺 ⌊⌊ 𝖠∧𝖲(2) |𝖲−2 ⌊𝖨 N N ( ∧ ) =−ΔS(A ) ⌊ ⌊⌊𝖲2 ⌊𝖨. =−(A (eNeN )) One should note that the resulting identity does not require 𝖲 to have an inverse. APPENDIX A Solutions to Problems 289

CHAPTER 3

𝖢⌊ | 𝖢| 3.1 Assume a = ei, whence ( ei) ej = eij = 0 for any basis bivector eij. Thus, ∑ 𝖢 𝖢|𝖨(2) 𝖢| 𝜺 . = = eij ij = 0 j

3.3 Assuming linear independence, the six bivectors Ai make a basis. Denoting the fourth∑ simple antisymmetric bidyadic by BC − CB 6 훽 and expanding B = iAi, we obtain i=1 ∑ 훽 . BC − CB = i(AiC − CAi) Because the sum of the four simple antisymmetric bivectors can be expressed in the form 𝖢 훽 훽 훽 훽 ⋯ = (A1 − 2C)(A2 + 1C) − (A2 + 1C)(A1 − 2C) + , 290 APPENDIX A Solutions to Problems

the resulting dyadic actually consists of three simple antisymmetric terms, only. ∑ 훼 ⋯ 3.4 Assuming A6 = iAi for i = 1 5, the antisymmetric bidyadic can be written in the form 𝖢 훼 훼 훼 훼 = (A1 − 1A5)(A2 + 2A5) − (A2 + 2A5)(A1 − 1A5) 훼 훼 훼 훼 + (A3 − 3A5)(A4 + 4A5) − (A4 + 4A5)(A3 − 3A5), as a sum of two simple antisymmetric dyadics. (2) 3.5 Because∑ the bivector unit dyadic can be expanded∑ as 𝖨 = 𝚷 𝖣 𝖢|𝖣 𝖨(2) 𝚷 Ai i, the dyadic satisfying = = Ai i can be written in the antisymmetric form 𝖣 𝚷 𝚷 𝚷 𝚷 𝚷 𝚷 𝚷 𝚷 𝚷 𝚷 𝚷 𝚷 = ( 2 1 − 1 2) + ( 4 3 − 3 4) + ( 6 5 − 5 6), as can most easily be verified. 3.6 To find the inverse for the special skewon–axion bidyadic 𝖣 we first derive a condition for the simple antisymmetric bidyadic 𝖢 by expanding 𝖢3 = (AB − BA) ⋅ (AB − BA) ⋅ (AB − BA) = ((A ⋅ B)(AB + BA) − (A ⋅ A)BB − (B ⋅ B)AA) ⋅ (AB − BA) = 휆2(AB − BA) = 휆2𝖢 휆2 = (A ⋅ B)2 − (A ⋅ A)(B ⋅ B). Thus, a simple antisymmetric bidyadic satisfies the simple condition 𝖢3 − 휆2𝖢 = 0. From this a condition for the bidyadic 𝖣 = 𝖢 + 훼𝖤 can be formed as 𝖢3 = (𝖣 − 훼𝖤) ⋅ (𝖣 − 훼𝖤) ⋅ (𝖣 − 훼𝖤) = 𝖣 ⋅ 𝖣 ⋅ 𝖣 − 3훼𝖣 ⋅ 𝖣 + 3훼2𝖣 − 훼3𝖤 = 휆2(𝖣 − 훼𝖤), or 𝖣 ⋅ (𝖣 ⋅ 𝖣 − 3훼𝖣 + (3훼2 − 휆2)𝖤) + (휆2훼 − 훼3)𝖤 = 0. APPENDIX A Solutions to Problems 291

Written in the form 𝖣 ⋅ 𝖷 = 훼(훼2 − 휆2)𝖤, 𝖷 = 𝖣 ⋅ 𝖣 − 3훼𝖣 + (3훼2 − 휆2)𝖤 or 𝖣| 𝜺 𝜺 ⌊⌊𝖷 훼 훼2 휆2 𝖨(2). ( N N ) = ( − ) the inverse of the bidyadic 𝖣 can be identified as 1 𝖣−1 = 𝜺 𝜺 ⌊⌊𝖷 훼(훼2 − 휆2) N N 1 = 𝜺 𝜺 ⌊⌊(𝖣 ⋅ 𝖣 − 3훼𝖣 + (3훼2 − 휆2)𝖤). 훼(훼2 − 휆2) N N 3.7 Denoting (𝖨(2) + A𝚽 + B𝚿)−1 = 𝖨(2) − 𝖷, and applying the inverse rule (3.137) twice we obtain ( ) ( ) 𝚽 𝚽 𝖨(2) − A |B𝚿| 𝖨(2) − A 𝖭 A𝚽 1+A|𝚽 1+A|𝚽 𝖷 = = + ( ) . |𝚽 𝚽 D 1 + A 1 + 𝚿| 𝖨(2) − A |B 1+A|𝚽 Combining the two terms, the numerator and the denominator can be written as 𝖭 = A𝚽 + B𝚿 + (A|𝚽)B𝚿 + (B|𝚿)A𝚽 − A𝚽|B𝚿 − B𝚿|A𝚽 D = 1 + A|𝚽 + B|𝚿 + (A|𝚽)(B|𝚿) − (A|𝚿)(B|𝚽). The bidyadic 𝖷 = 𝖭∕D can be shown to satisfy (A𝚽 + B𝚿)|𝖷 = A𝚽 + B𝚿 − 𝖷, after which we can verify the result through multiplication, (𝖨(2) + A𝚽 + B𝚿)|(𝖨(2) − 𝖷) = 𝖨(2) − 𝖷 + A𝚽 + B𝚿 − (A𝚽 + B𝚿 − 𝖷) = 𝖨(2). 3.8 The task is to substitute the dyadic (3.127) ( ) 1 1 𝖠 = 𝖥(𝖡 ) = 𝖡2 + tr𝖡(2)𝖨 o o 𝖡(3) o 2 o tr o 292 APPENDIX A Solutions to Problems

in the expression (3.129) ( ) 1 1 𝖡 = 𝖥(𝖠 ) = 𝖠2 + tr𝖠(2)𝖨 , o o 𝖠(3) o 2 o tr o as 𝖡 = 𝖥(𝖠 ) = 𝖥(𝖥(𝖡 )) o o ( o ) (𝖡2 − 1 tr𝖡2𝖨)2 − 1 tr(𝖡2 − 1 tr𝖡2𝖨)2𝖨 𝖡(3) o 4 o 4 o 4 o = tr o , tr(𝖡2 − 1 tr𝖡2𝖨)(3) o 4 o 𝖡 to see whether this is satisfied for any trace-free dyadic o. Invok- ing the Cayley–Hamilton equation (2.119) we obtain

𝖡2 1 𝖡2𝖨 2 𝖡4 1 𝖡2 𝖡2 1 𝖡2 2𝖨 ( o − tr o ) = o − tr o o + (tr o) 4 2 ( 16 ) 1 = tr𝖡(3)𝖡 − tr𝖡(4) − (tr𝖡2)2 𝖨 o o 16 o ( ) 1 2 1 tr 𝖡2 − tr𝖡2𝖨 =−4tr𝖡(4) + (tr𝖡2)2, o 4 o o 4 o whence the numerator of the dyadic inside the brackets melts 𝖡(3) 𝖡 to tr o. Applying twice the Cayley–Hamilton equation, and skipping the algebraic steps, the denominator can be expanded as ( ) ( ) 1 (3) 1 1 3 tr 𝖡2 − tr𝖡2𝖨 = tr 𝖡2 − tr𝖡2𝖨 = (tr𝖡(3))2, o 4 o 3 o 4 o o 𝖡 𝖡 which inserted reduces the expression to o = o. 3.9 Applying the identity 𝖠∧𝖡 | 𝖢∧𝖣 𝖠|𝖢 ∧ 𝖡|𝖣 𝖠|𝖣 ∧ 𝖡|𝖢 ( ∧ ) ( ∧ ) = ( )∧( ) + ( )∧( ) we can expand 𝖭∧𝖨 2 𝖭2∧𝖨 𝖭∧𝖭 𝖭∧𝖭 ( ∧ ) = ∧ + ∧ = ∧ , 𝖭∧𝖨 3 𝖭∧𝖭 | 𝖭∧𝖨 𝖭2∧𝖭 . ( ∧ ) = ( ∧ ) ( ∧ ) = 2 ∧ = 0 In terms of these we have 𝖣 훼𝖨(2) 3 𝖭∧𝖨 3 ( − ) = ( ∧ ) = 0, which corresponds to the equation 𝖣3 − 3훼𝖣2 + 3훼2𝖣 − 훼3𝖨(2) = 0. APPENDIX A Solutions to Problems 293

For 훼 ≠ 0 we obtain the inverse rule 1 𝖣−1 = (𝖣2 − 3훼𝖣 + 3훼2𝖨(2)) 훼3 1 = (훼2𝖨(2) − 훼𝖭∧𝖨 + 𝖭∧𝖭). 훼3 ∧ ∧ 3.10 Assuming that the vector a and one-form 𝜶 satisfy a|𝜶 ≠ 0, we 𝜶 |𝜶 𝜺 can temporarily denote a = (a )e4 4, whence 𝖢 훼𝖨(2) |𝜶 𝖨∧ 𝜺 = + (a ) ∧e4 4 훼𝖨(2) 훼 |𝜶 𝖨 ∧ 𝜺 . = s + ( + a ) s∧e4 4 Obviously, the inverse is 1 1 𝖢−1 = 𝖨(2) + 𝖨 ∧e 𝜺 훼 s 훼 + a|𝜶 s∧ 4 4 1 a|𝜶 = 𝖨(2) − 𝖨∧e 𝜺 훼 훼(훼 + a|𝜶) ∧ 4 4 which can be expressed with original quantities as 1 1 𝖢−1 = 𝖨(2) − 𝖨∧a𝜶. 훼 훼(훼 + a|𝜶) ∧ This result can be checked to yield 𝖢|𝖢−1 = 𝖨(2) also in the case a|𝜶 ≠ 0, whence the assumption can be released. ∑ 𝖢 𝚽 3.11 (1) Choose a = e1 and expand = eij ij in terms of six two- 𝚽 forms ij.From 𝖢 𝚽 𝚽 𝚽 e1 ∧ = e123 23 + e124 24 − e314 34 = 0 𝚽 𝚽 𝚽 we obtain 23 = 24 = 34 = 0. Choosing a = e2, e3, e4,we 𝚽 𝚽 𝚽 𝖢 similarly obtain 12 = 13 = 14 = 0. Thus, = 0. (2) Expand ∑ ∑ 𝖢 𝖢|𝖨(2) 𝖢| 𝜺 𝖢⌊ ⌊ 𝜺 . = = eij ij = ( ei) ej ij = 0 i

that is, 𝜶 is an eigen one-form of the dyadic 𝖠 for some eigenvalue A. Since this must be valid for any one-form 𝜶, A must be independent of 𝜶 and we have ∑ ∑ 𝖠 𝖠|𝖨T 𝖠|𝜺 𝜺 𝖨T . = = iei = A iei = A i i Thus, the bidyadic 𝖬 must satisfy 𝖬⌊𝜶 = A𝜶 ∧ 𝖨T = A𝖨(2)T ⌊𝜶 for any one-form 𝜶. Applying the result of the previous problem to the bidyadic 𝖬 − A𝖨(2)T , we obtain 𝖬 − A𝖨(2)T = 0, whence 𝖬 must be a multiple of the unit bidyadic 𝖨(2)T . Operating as ⌊ 𝜶 𝖬⌊𝜶 𝜶⌋ ⌊𝖬 ⌊𝜶 𝖬 ⌊⌊𝜶𝜶 eN ( ∧ ) =− (eN ) = m , 𝖬 ⌊𝖬 E E where m = eN ∈ 2 2, we can prove the second property. 3.13 Because we have ⌊𝖨(2)T ⌊⌊𝜶𝜶 ⌊ 𝜶 𝖨(2)T ⌊𝜶 ⌊ 𝜶 𝜶 𝖨T (eN ) = eN ( ∧ ( )) = eN ( ∧ ∧ ) = 0 identically for any one-form 𝜶, the solution 𝖰 must satisfy 𝖰(2)⌊⌊𝜶𝜶 = 0forany𝜶.From 𝖰(2)⌊⌊𝜶𝜶 = (𝖰||𝜶𝜶)𝖰 − (𝖰|𝜶)(𝜶|𝖰) = 0 we obtain two possibilities. (1) Assuming 𝖰||𝜶𝜶 ≠ 0 is valid for some 𝜶, we can express

(𝖰|𝜶)(𝜶|𝖰) 𝖰 = , 𝖰||𝜶𝜶

which corresponds to the solution 𝖰(2) = 0, Q = 0, that is, 𝖰 is of the form 𝖰 = ab. (2) For the converse case 𝖰||𝜶𝜶 = 0forall𝜶 we must have 𝖰 antisymmetric, that is, it can be expressed in terms of a bivector Q as 𝖰 = Q⌊𝖨T . Since we must have 𝖰|𝜶 = 0forany𝜶,this leads to Q = 0, whence the second solution would be 𝖰 = 0and Q = 0. Because Q = 0 in both cases, in particular, the equation 𝖰(2) ⌊𝖨(2) = eN does not have any solution. APPENDIX A Solutions to Problems 295

𝖢−6| 3.14 (1) Multiplying (3.10) by and dividing by a6 leads to the equation 𝖢−6 𝖢−5 𝖢−4 𝖢−3 + (a5∕a6) + (a4∕a6) + (a3∕a6) 𝖢−2 𝖢−1 𝖨(2) . + (a2∕a6) + (a1∕a6) + (1∕a6) = 0 (2) Because we have 𝖣2 ⌊𝖢T ⌋𝜺 | ⌊𝖢T ⌋𝜺 ⌊𝖢2T ⌋𝜺 𝖨(4)⌊⌊𝖢2T = (eN N) (eN N) = eN N = , and similarly for other powers of 𝖣, operating 3.10 by 𝖨(4)⌊⌊()T the Cayley–Hamilton equation for 𝖣 is obtained when replacing 𝖢 by 𝖣. 𝜺 𝜶 |𝜺 3.15 Let us define e1 = a and 1 = ∕4tosatisfye1 1 = 1. Thus, we have 1 1 𝖡 = (𝖨 − a𝜶) = (𝖨 − 4e 𝜺 ), o 2 2 1 1 𝖢 𝖡 ∧𝖨 𝖨(2) 𝜺 ∧𝖨 = o∧ = − 2e1 1∧ 𝜺 𝜺 𝜺 𝜺 𝜺 𝜺 =−e12 12 + e23 23 − e31 31 − e14 14 + e24 24 + e34 34, whence 𝖢⌋ eN =−e12e34 + e23e14 − e31e24 − e14e23 + e24e31 + e34e12 𝜺 ∧𝖨 2 𝜺 ∧𝖨 is antisymmetric. Because of (e1 1∧ ) = e1 1∧ ,wehave 𝖣 𝖢2 𝖡 ∧𝖨 2 𝖨(2) 𝜺 ∧𝖨 2 𝖨(2) = = ( o∧ ) = ( − 2e1 1∧ ) = and tr𝖣p = 4forp ≥ 0. The coefficients (3.43)–(3.45) become

a0 = 1, a2 =−2, a4 = 1, a6 = 0, and (3.42) is reduced to 𝖣3 − 2𝖣2 + 𝖣 = 𝖣|(𝖣 − 𝖨(2))2 = 0, which is satisfied for 𝖣 = 𝖨(2). 3.16 Inserting the expressions of 𝖬 and the assumed 𝖬−1 in 𝖬|𝖬−1 = 𝖨(2) and assuming that 𝚷 and 𝚲 are linearly independent and 훼′ = 1∕훼, we obtain a relation between the unknown and known bivectors as ( )( ) ( ) 𝚷|C + 훼 𝚲|C C′ C =−훼′ . 𝚷|D 𝚲|D + 훼 D′ D 296 APPENDIX A Solutions to Problems

This can be solved as ( ) ( )( ) C′ −1 𝚲|D + 훼 −𝚲|C C = , D′ 훼D −𝚷|D 𝚷|C + 훼 D D = (𝚷|C + 훼)(𝚲|D + 훼) − (𝚲|C)(𝚷|D). It is easy to verify the result by direct substitution. In the case D = 0, there is no inverse. For the special case D = 0, the bivector C′ is reduced to −1 C′ = C. 훼(𝚷|C + 훼) 3.17 Expanding from (3.43)–(3.45) 𝖣3 𝖣3 a0tr = tr 1 a tr𝖣2 =− tr𝖣 tr𝖣2 2 2 1 1 a tr𝖣 = (tr𝖣)3 − tr𝖣2tr𝖣 4 8 4 1 6 a tr𝖨(2) =− (tr𝖣)3 + tr𝖣 tr𝖣2 − tr𝖣3, 6 8 8 we can show that 𝖣3 𝖣2 𝖣 𝖨(2) a0tr + a2tr + a4tr + a6tr = 0, that is, the trace of (3.42) is an identity. 3.18 Denoting x = tr𝖢 we can write 6 4 𝖢2 3 𝖢3 2 𝖢2 2 6!a6(x) = x − 15x tr + 40x tr + 45x (tr ) − 120x tr𝖢2 tr𝖢3 − 90x2tr𝖢4 + 144x tr𝖢5 + 40(tr𝖢3)2 − 15(tr𝖢2)3 + 90tr𝖢2tr𝖢4 − 120tr𝖢6, which when differentiated yields 휕 5 3 𝖢2 2 𝖢3 𝖢2 2 6! xa6(x) = 6x − 60x tr + 120x tr + 90x(tr ) − 120tr𝖢2 tr𝖢3 − 180xtr𝖢4 + 144tr𝖢5 = 6(x5 − 10x3tr𝖢2 + 20x2tr𝖢3 + 15x(tr𝖢2)2 − 20tr𝖢2 tr𝖢3 − 30xtr𝖢4 + 24tr𝖢5).

The expression in brackets equals −5!a5(x). We can continue 휕 < similarly for i! xai, i 6. APPENDIX A Solutions to Problems 297

𝖢 𝖡(2) 3.19 Applying the rule (2.114) to = o 𝖨(4)⌊⌊𝖡(2)T 𝖡(2) 𝖨(2) 𝖡(2)⌊⌊𝖨T ∧𝖨 𝖡(2) o = tr o − ( o )∧ + o 𝖡(2) 𝖨(2) 𝖡2∧𝖨 𝖡(2) = tr o + o∧ + o and comparing with the rule (3.128) we have 𝖨(4)⌊⌊𝖡(2)T 𝖡(3) 𝖡 ∧𝖨 −1 𝖡(2). o = tr o ( o∧ ) + o 𝖡(3) ≠ Assuming tr o 0, we obtain another form for the inverse bidyadic, 1 (𝖡 ∧𝖨)−1 = (𝖨(4)⌊⌊𝖡(2)T − 𝖡(2)). o∧ 𝖡(3) o o tr o 3.20 Applying the bac-cab rule, valid for any one-form 𝜶, two-form 𝚷 and vector a, a⌋(𝜶 ∧ 𝚷) = 𝜶 ∧ (a⌋𝚷) + (a|𝜶)𝚷, and choosing a so that a|𝜶 = 1, we can write from a⌋(𝜶 ∧ 𝖬⌊𝜶) = 𝜶 ∧ (a⌋𝖬⌊𝜶) + 𝖬⌊𝜶 = 0, the representation 𝖬⌊𝜶 = 𝜶 ∧ 𝖠T , with 𝖠T =−a⌋𝖬⌊𝜶. Choos- 𝜶 𝜺 𝖠T 𝖬⌊⌊𝖨 ing a = ei, = i and summing over i we obtain = . Since 𝖬 is of full rank, so is 𝖠. The dyadic 𝖠 satisfies for any 𝜶 𝜶 ∧ 𝖠T |𝜶 = 𝖬|(𝜶 ∧ 𝜶) = 0, ∑ 𝖠 𝜺 𝜶 𝜺 whence expanding = Ai,jei j and taking = k, we obtain ∑ ∑ 𝜺 𝖠|𝜺 𝜺 𝜺 훿 𝜺 . k ∧ k = Ai,j( k ∧ j) ik = Ai,j ij = 0 ∑ ≠ 𝖠 𝜺 This implies Ai,j = 0fori j and, thus, = Ai,iei i. Continu- 𝜶 𝜺 𝜺 ≠ 𝓁 ing with = k + 𝓁, k ,wehave ∑ 𝜶 𝖠T |𝜶 𝜺 |𝜺 𝜺 |𝜺 𝜺 𝜺 ∧ = Ai,i( ki(ei 𝓁) + 𝓁i(ei k) = A𝓁,𝓁 k𝓁 + Ak,k 𝓁k 𝜺 ⇒ = (A𝓁,𝓁 − Ak,k) k𝓁 = 0, A𝓁,𝓁 = Ak,k from which we obtain 𝖠 = A𝖨. Thus, 𝖬⌊⌊𝖨 = A𝖨T and 𝖬⌊𝜶 = A𝜶 ∧ 𝖨T = A𝖨(2)T ⌊𝜶 ⇒ (𝖬 − A𝖨(2)T )⌊𝜶 = 0. 298 APPENDIX A Solutions to Problems

Choosing basis one-forms 𝜶 = 𝜺 , we finally obtain i ∑ 𝖬 − A𝖨(2)T = (𝖬 − A𝖨(2)T )| 𝜺 e ∑ ij ij 𝖬 𝖨(2)T ⌊𝜺 |𝜺 = (( − A ) i) jeij = 0, whence 𝖬 must be a multiple of the unit bidyadic 𝖨(2)T .As a consequence we also obtain the following rule for the met- 𝖬 ⌊𝖬 𝖬 ⌊⌊𝝂𝝂 ⌊ 𝜶 𝖬⌊𝜶 ric bidyadic m = eN :if m = eN ( ∧ ) = 0is satisfied for all 𝜶, the metric bidyadic must be of the form 𝖬 ⌊𝖬 ⌊𝖨(2)T 𝖤. m = eN = AeN = A 𝖢 𝖡 ∧𝖨 3.21 Applying the identity, the powers of = o∧ can be expressed in the following symmetric pattern, 𝖢 𝖡 ∧𝖨 𝖨∧𝖡 2 = o∧ + ∧ o 𝖢2 𝖡2∧𝖨 𝖡 ∧𝖡 𝖨∧𝖡2 2 = o∧ + 2 o∧ o + ∧ o 𝖢3 𝖡3∧𝖨 𝖡2∧𝖡 𝖡 ∧𝖡2 𝖨∧𝖡3 2 = o∧ + 3 o∧ o + 3 o∧ o + ∧ o 𝖢4 𝖡4∧𝖨 𝖡3∧𝖡 𝖡2∧𝖡2 𝖡 ∧𝖡3 𝖨∧𝖡4 2 = o∧ + 4 o∧ o + 6 o∧ o + 4 o∧ o + ∧ o 𝖢5 𝖡5∧𝖨 𝖡4∧𝖡 𝖡3∧𝖡2 𝖡2∧𝖡3 𝖡 ∧𝖡4 𝖨∧𝖡5 2 = o∧ + 5 o∧ o + 10 o∧ o + 10 o∧ o + 5 o∧ o + ∧ o 𝖢6 𝖡6∧𝖨 𝖡5∧𝖡 𝖡4∧𝖡2 𝖡3∧𝖡3 𝖡2∧𝖡4 2 = o∧ + 6 o∧ o + 15 o∧ o + 20 o∧ o + 15 o∧ o 𝖡 ∧𝖡5 𝖨∧𝖡6. + 6 o∧ o + ∧ o The traces can be expanded as 𝖢 𝖡 tr = 3tr o = 0 𝖢2 𝖡2∧𝖨 𝖡(2) 𝖡2 𝖡 2 𝖡2 tr = tr( o∧ ) + 2tr o = 3tr o + (tr o) − tr o 𝖡2 = 2tr o, 𝖢3 𝖡3 𝖡2∧𝖡 𝖡3 𝖡2 𝖡 𝖡3 tr = 3tr o + 3tr( o∧ o) = 3tr o + 3(tr o)(tr o) − 3tr o = 0, 𝖢4 𝖡4 𝖡3∧𝖡 𝖡2 (2) 𝖡4 𝖡2 2 𝖡4 tr = 3tr o + 4tr( o∧ o) + 6tr( o) =−tr o + 3(tr o) − 3tr o 𝖡4 𝖡2 2 =−4tr o + 3(tr o) 𝖢5 𝖡5 𝖡5 𝖡3 𝖡2 𝖡5 tr = 3tr o − 5tr o + 10(tr o)(tr o) − 10tr o 𝖡5 𝖡3 𝖡2 =−12tr o + 10(tr o)(tr o) 𝖢6 𝖡6 𝖡6 𝖡4 𝖡2 𝖡6 𝖡3 (2) tr = 3tr o − 6tr o + 15(tr o)(tr o) − 15tr o + 20tr(( o) ) 𝖡6 𝖡4 𝖡2 𝖡3 2 𝖡6 =−18tr o + 15(tr o)(tr o) + 10(tr o) − 10tr o 𝖡6 𝖡4 𝖡2 𝖡3 2. =−28tr o + 15(tr o)(tr o) + 10(tr o) APPENDIX A Solutions to Problems 299

𝖡 The trace-free dyadic o satisfies the Cayley–Hamilton equation 𝖡4 𝖡(2) 𝖡2 𝖡(3) 𝖡 𝖡(4) 𝖨 . o + (tr o ) o − (tr o ) o + (tr o ) = 0

𝖡(2) 𝖡2 Inserting tr o =−(1∕2)tr o and taking the trace, we can solve 1 tr𝖡(4) =− ((tr𝖡2)2 − 2tr𝖡4), o 8 o o whence 1 1 𝖡4 − (tr𝖡2)𝖡2 − (tr𝖡(3))𝖡 + ((tr𝖡2)2 − 2tr𝖡4)𝖨 = 0. o 2 o o o o 8 o o 𝖡 | Multiplying this by o and taking the trace yields 1 tr𝖡5 − (tr𝖡2)(tr𝖡3) − (tr𝖡(3))(tr𝖡2) = 0. o 2 o o o o

𝖡3 𝖡(3) Applying tr o = 3tr o we obtain the expansion 5 tr𝖡5 = (tr𝖡2)(tr𝖡3), o 6 o o which yields the simple result

𝖢5 𝖡2 𝖡3 𝖡3 𝖡2 . tr =−10(tr o)(tr o) + 10(tr o)(tr o) = 0

𝖡2| Further, multiplying the Cayley–Hamilton equation by o and taking the trace yields ( ) 1 1 1 1 tr𝖡6 = (tr𝖡2)(tr𝖡4) + (tr𝖡3)2 − (tr𝖡2)2 − (tr𝖡4)tr𝖡2 o 2 o o 3 o 8 o 4 o o 3 1 1 = (tr𝖡2)(tr𝖡4) + (tr𝖡3)2 − (tr𝖡2)3, 4 o o 3 o 8 o which allows us to expand 28 7 tr𝖢6 =−21(tr𝖡2)(tr𝖡4) − (tr𝖡3)2 + (tr𝖡2)3 o o 3 o 2 o 𝖡4 𝖡2 𝖡3 2 +15tr( o)(tr o) + 10tr( o) 2 7 =−6(tr𝖡2)(tr𝖡4) + (tr𝖡3)2 + (tr𝖡2)3. o o 3 o 2 o 300 APPENDIX A Solutions to Problems

In terms of these expressions for tr𝖢p, the Cayley–Hamilton equation coefficients for the skewon bidyadic are simplified as

a0 = 1 𝖢 a1 =−a0tr = 0 1 1 a =− (a tr𝖢 + a tr𝖢2) =− tr𝖢2 2 2 1 0 2 1 a =− (a tr𝖢 + a tr𝖢2 + a tr𝖢3) = 0 3 3 2 1 0 1 1 1 a =− (a tr𝖢 + a tr𝖢2 + a tr𝖢3 + a tr𝖢4) = (tr𝖢2)2 − tr𝖢4 4 4 3 2 1 0 8 4 1 a =− (a tr𝖢 + a tr𝖢2 + a tr𝖢3 + a tr𝖢4 + a tr𝖢5) = 0 5 5 4 3 2 1 0 1 a =− (a tr𝖢 + a tr𝖢2 + a tr𝖢3 + a tr𝖢4 + a tr𝖢5 + a tr𝖢6) 6 6 5 4 3 2 1 0 1 1 1 =− (tr𝖢2)3 + tr𝖢2tr𝖢4 − tr𝖢6. 48 8 6 Thus, the Cayley–Hamilton equation (3.10) for any skewon bidyadic 𝖢 can be expressed as ( ( ) ) 1 1 2 1 𝖢6 − (tr𝖢2)𝖢4 + tr𝖢2 − tr𝖢4 𝖢2 2 8 4 ( ) 1 1 1 + − (tr𝖢2)3 + tr𝖢2tr𝖢4 − tr𝖢6 𝖨(2) = 0. 48 8 6 As a simple check, the trace of the left side can be shown to vanish. 𝖢 𝖡 ∧𝖨 3.22 For the skewon bidyadic = o∧ , the determinant has the form 1 1 1 det𝖢 = a =− (tr𝖢2)3 + tr𝖢2tr𝖢4 − tr𝖢6, 6 48 8 6 in which we can insert

𝖢2 𝖡2 tr = 2tr o, 𝖢4 𝖡4 𝖡2 2 tr =−4tr o + 3(tr o) , ( )( ) ( ) ( ) 3 2 2 7 3 tr𝖢6 = tr𝖡2 tr𝖡4 + tr𝖡3 + tr𝖡2 . 4 o o 3 o 2 o APPENDIX A Solutions to Problems 301

The expression can be expanded as ( ) 1 1 3 det𝖢 =− (tr𝖢2)3 − tr𝖢2tr𝖢4 + tr𝖢6 . 6 8 4 ( ( ) ( )( ( ) ) 1 1 3 3 2 =− 2tr𝖡2 − 2tr𝖡2 − 4tr𝖡4 + 3 tr𝖡2 6 8 o 4 o o o ( )( ) ( ) ( ) ) 2 2 7 3 − 6 tr𝖡2 tr𝖡4 + tr𝖡3 + tr𝖡2 o o 3 o 2 o ( ) ( ) 1 2 2 =− tr𝖡3 =− tr𝖡(3) . 9 o o 3.23 The identity can be extended as 𝖨(3)⌊⌊ 𝜶 ∧ 𝜷 𝖨(3)⌊⌊ 𝜶 ⌊⌊ 𝜷 ( a∧b ) = (( ( a)) (b ) |𝜶 𝖨(2)⌊⌊ 𝜷 𝖨∧ 𝜶 ⌊⌊𝜷 = (a ) b − ( ∧a ) b = ((a|𝜶)(b|𝜷) − a𝜶||𝜷b)𝖨 − (a|𝜶)b𝜷 − (b|𝜷)a𝜶 + a𝜶|b𝜷 + b𝜷|a𝜶 𝜶∧ 𝜷 𝖨 𝜶∧ 𝜷 ⌊⌊𝖨T . = tr(a ∧b ) − (a ∧b ) 𝜶∧ 𝜷 Since this is linear and valid for any bidyad a ∧b , it is valid more generally for any bidyadic 𝖢 as 𝖨(3)⌊⌊𝖢T = tr𝖢 𝖨 − 𝖢⌊⌊𝖨T . Vanishing left side requires that trace of the right side vanish, or 4tr𝖢 = tr(𝖢⌊⌊𝖨T ) = (𝖢⌊⌊𝖨T )||𝖨T = 2tr𝖢, whence tr𝖢 = 0and𝖢⌊⌊𝖨T = 0. Thus, 𝖨(3)⌊⌊𝖢T = 0 is equivalent to 𝖢⌊⌊𝖨T = 0 as a definition of the principal bidyadic. 3.24 The inverse of the metric bidyadic 𝖣 involving only axion and skewon parts can be expressed as 1 𝖣−1 = 𝜺 𝜺 ⌊⌊(𝖣 ⋅ 𝖣 − 3훼𝖣 + (3훼2 − 휆2)𝖤). 훼(훼2 − 휆2) N N Since −3훼𝖣 + (3훼2 − 휆2)𝖤 contains no principal component, let us consider the term 𝖣 ⋅ 𝖣 훼2 ⌊𝖨(2)T 훼 = eN + 2 (AB − BA) + (A ⋅ B)(AB + BA) − (B ⋅ B)AA − (A ⋅ A)BB, which is symmetric. Now it is obvious that, in the general case, it ⌊𝖨(2)T cannot be a multiple of eN because it is not of full rank. In 302 APPENDIX A Solutions to Problems

conclusion, the inverse of such a skewon–axion bidyadic contains a principal part. Similarly, the inverse of a principal bidyadic may contain an axion part. 𝖣 𝖢 훼𝖨 𝖢 𝖡 ∧𝖨 𝖡 3.25 For the skewon–axion bidyadic = o + , o = o∧ ,tr o = 0 we multiply the rule (3.159) as 𝖷 𝖨(2) 𝖢 −1| 𝖨(2) 𝖢 𝖨(2) 𝖢 ( + o) ( + o) = + o − A6

with A6 = 1 + a2 + a4 + a6 and 𝖷 𝖢 𝖢2 | 𝖨(2) 𝖢2 𝖢4 | 𝖨(2) 𝖢 = ( o − o) ((1 + a2 + a4) + (1 + a2) o + o) ( + o) 𝖢 𝖢3 𝖢3 𝖢5 𝖢5 𝖢7 = (1 + a2 + a4)( o − o) + (1 + a2)( o − o) + o − o 𝖢 𝖢3 𝖢5 𝖢7 = (1 + a2 + a4) o − a4 o − a2 o − o 𝖢 𝖢 𝖢3 𝖢5 𝖢7 . = A6 o − (a6 + a4 o + a2 o + o) Applying the reduced Cayley–Hamilton equation 𝖢6 𝖢4 𝖢2 𝖨(2) o + a2 o + a4 o + a6 = 0 we have 𝖷 𝖢 𝖢 = (1 + a2 + a4 + a6) o = A6 o, whence we finally obtain 𝖨(2) 𝖢 −1| 𝖨(2) 𝖢 𝖨(2). ( + o) ( + o) =

CHAPTER 4

𝖠 ≠ 𝖡 ≠ 𝖡 𝖠 4.1 If det s 0ordet s 0, we respectively have s = 0or s = 0, 𝖠 𝖡 whence we must study the case det s = det s = 0. Expanding the dyadics in terms of four spatial vectors and one-forms as 𝖠 𝜺 𝜺 𝖡 𝜷 𝜷 s = a1 1 + a2 2, s = b1 1 + b2 2, 𝜺 𝜺 𝜺 where 1, 2, 3 form a spatial basis. Assuming that a1 ∧ a2, 𝜷 𝜷 𝖠(2) ≠ b1 ∧ b2 and 1 ∧ 2 do not vanish, we would have both s 0 𝖡(2) ≠ 𝖠 |𝖡 𝜺 | and s 0. However, the condition s s = 0 requires i b1 = 𝜺 | 𝜺 i b2 = 0fori = 1, 2, whence we must have b1 = b1 3 and b2 = 𝜺 ≠ 𝖡(2) ≠ b2 3 which is contrary to the assumption b1 ∧ b2 0and s 𝖠(2) ≠ 𝖡(2) ≠ 0. Thus, the case of both s 0and s 0 is not possible. APPENDIX A Solutions to Problems 303

4.2 Expressing 𝖪 = 𝖨 − 2, the dyadic  can be expanded as ( )( ) e − e 𝜺 − 𝜺  1 𝜺 𝜺 𝜺 𝜺 3 4 3 4 =− (e3 4 + e4 3 − e3 3 − e4 4) = √ √ , 2 2 2 which satisfies 2 =  and, hence, 𝖪2 = 𝖨. To express 𝖪 in the ′ form  − ,wedefine ′  = 𝖨 −  𝜺 𝜺 1 𝜺 𝜺 𝜺 𝜺 = e1 1 + e2 2 + (e3 4 + e4 3 + e3 3 + e4 4) (2 )( ) e + e 𝜺 + 𝜺 𝜺 𝜺 3 4 3 4 . = e1 1 + e2 2 + √ √ 2 2 The unipotent dyadic can now be given the standard form 𝖪 𝝅 𝝅 𝝅 𝝅 = p1 1 + p2 2 + p3 3 − p4 4 when the vectors and one-forms are defined by

e3 + e4 e3 − e4 p1 = e1, p2 = e2, p3 = √ , p4 = √ , 2 2 𝜺 + 𝜺 𝜺 − 𝜺 𝝅 𝜺 𝝅 𝜺 𝝅 3 4 𝝅 3 4 . 1 = 1, 2 = 2, 3 = √ , 4 = √ 2 2 4.3 From 𝖠3 = 0wehave𝖠4 = 0, whence tr𝖠3 = 0andtr𝖠4 = 0. From (𝖠3)(4) = (𝖠(4))3 = 𝖨(4)(tr𝖠(4))3 = 0 we have tr𝖠(4) = 0. The Cayley–Hamilton equation is then reduced to (tr𝖠(2))𝖠2 − (tr𝖠(3))𝖠 = 0. Multiplying this by 𝖠| leads to tr𝖠(3) = 0 and, further, to tr𝖠(2) = 0. From their expansions we obtain 1 tr𝖠(2) = ((tr𝖠)2 − tr𝖠2) = 0 ⇒ tr𝖠2 = (tr𝖠)2 2 1 1 tr𝖠(3) = ((tr𝖠)3 − 3tr𝖠 tr𝖠2 + 2tr𝖠3) =− (tr𝖠)3 = 0, 6 3 whence we also have tr𝖠 = 0andtr𝖠2 = 0. 304 APPENDIX A Solutions to Problems

4.4 The equivalence is seen by expanding 𝖩(2)⌊⌊𝖨T 𝖩 𝖩 𝖩2 12 = tr − , 1 tr𝖩(2) = ((tr𝖩)2 − tr𝖩2), 2 𝖩(2) 𝖨(2) and applying = 12 whence the two conditions become 𝖨(2)⌊⌊𝖨T 𝖨 𝖨 𝖨 𝖩 𝖩 𝖩2 𝖩 𝖩 𝖨 12 12 = 2 12 − 12 = 12 = tr − = tr + 12, 1 1 tr𝖩(2) = ((tr𝖩)2 + 2) = 1 + (tr𝖩)2. 2 2 𝖩(2) 𝖨(2) 𝖩 𝖩(2) Thus, = 12 implies tr = 0andtr = 1 and conversely. 4.5 From tr𝖩 = 0andtr𝖩(2) = 1 the conditions for the coefficients become . J11 + J22 = 0, J11J22 − J12J21 = 1 Defining 1 J =−J = sinh 휃, J = A cosh 휃, J =− cosh 휃, 11 22 12 21 A √ 휃 −1 2 with = sinh J11, A = J12∕ 1 + J11, the above conditions are obviously satisfied for any choice of 휃 and A. Inserted in the expansion we obtain 1 𝖩 = sinh 휃 e 𝜺 + A cosh 휃 e 𝜺 − cosh 휃 e 𝜺 − sinh 휃 e 𝜺 1 1 1 2 A 2 1 2 2 ( ) 1 휃 휃 = e (Ae − e )(A𝜺 + 𝜺 ) + e− (Ae + e )(A𝜺 − 𝜺 ) , 2A 1 2 2 1 1 2 2 1 when the vectors and one-forms are defined as 1 e−휃 a = (Ae − e ), a = (Ae + e ), 1 A 1 2 2 A 1 2 1 e휃 𝜶 =− (A𝜺 − 𝜺 ), 𝜶 = (A휖 + 𝜺 ). 1 2 2 1 2 2 2 1 Because they also satisfy the orthogonality conditions |𝜶 훿 ai j = ij, APPENDIX A Solutions to Problems 305

𝜶 𝜶 a1, a2 and 1, 2 form reciprocal bases. Thus, the most general 2D AC-dyadic can be written in the standard form of (4.100), 𝖩 𝜶 𝜶 = a1 2 − a2 1, satisfying 𝖩2 𝜶 𝜶 𝖨 . =−(a1 1 + a2 2) =− 12 4.6 Applying various dyadic rules we can expand 𝖩(2)⌊⌊𝖨T = (tr𝖩)𝖩 − 𝖩2 = 𝖨, 𝖩(3)⌊⌊𝖨T 𝖩 𝖩(2) 𝖩∧𝖩2 𝖩∧𝖨 = (tr ) − ∧ = ∧ , 1 1 𝖩(3)⌊⌊𝖨(2)T = (𝖩(3)⌊⌊𝖨T )⌊⌊𝖨T = (𝖩∧𝖨)⌊⌊𝖨T = 𝖩 2 2 ∧ 𝖨(3)⌊⌊𝖩T 𝖩 𝖨(2) 𝖨∧𝖩 𝖨∧𝖩 = (tr ) − ∧ =− ∧ , 1 1 𝖨(4)⌊⌊𝖩(3)T = (𝖨(4)⌊⌊𝖩(2)T )⌊⌊𝖩T = 𝖩(2)⌊⌊𝖩T =−𝖩, 3 3 1 1 𝖨(3)⌊⌊𝖩(2)T = (𝖨(3)⌊⌊𝖩T )⌊⌊𝖩T =− (𝖨∧𝖩)⌊⌊𝖩T = 𝖨 2 2 ∧ 4.7 Inserting 𝖬 = M𝖩(2)T in the rule (2.114) 𝖨(4)⌊⌊𝖬 𝖬 𝖨(2) 𝖬T ⌊⌊𝖨T ∧𝖨 𝖬T = (tr ) − ( )∧ + , and applying tr𝖩 = 0, whence M M tr𝖬 = Mtr𝖩(2) = ((tr𝖩)2 − tr𝖩2) = tr𝖨 = 2M, 2 2 𝖬T ⌊⌊𝖨T ∧𝖨 𝖩(2)⌊⌊𝖨T ∧𝖨 𝖩 𝖩 𝖩2 ∧𝖨 𝖨∧𝖨 𝖨(2) ( )∧ = M(( )∧ = M((tr ) − )∧ = M ∧ = 2 , we obtain 𝖨(4)⌊⌊𝖬 = M(2𝖨(2) − 2𝖨(2) + 𝖩(2)) = M𝖩(2) = 𝖬T . Thus, 𝖬 consists of its axion and principal parts, only: M 𝖬 = 𝖨(2)T + 𝖬 ,tr𝖬 = 0. 3 1 1 𝖬 Assuming vanishing of the principal part, 1 = 0, leads to 1 1 𝖩(2) = 𝖨(2), ⇒ (𝖩(2))2 = (𝖩2)(2) = 𝖨(2) = 𝖨(2), 3 9 a contradiction. Thus, for real 𝖩 such a bidyadic 𝖬 contains a nonzero principal part. 306 APPENDIX A Solutions to Problems

4.8 We can expand 𝖨(4)⌊⌊𝖩T 2 ⌊𝖩T ⌋𝜺 | ⌊𝖩T ⌋𝜺 ⌊𝖩2T ⌋𝜺 ( ) = (eN N) (eN N) = eN N =−𝖨(4)⌊⌊𝖨(2)T =−𝖨(2). 4.9 Expanding 𝖩2 𝖨(2) 𝖨 ∧ 𝜺 =− s − s∧e4 4 we obtain four equations 𝖠 |𝖠 𝖢 |𝖡 𝖨(2) s s − s s =− s 𝖡 |𝖠 𝖣 |𝖡 s s − s s = 0 𝖠 |𝖢 𝖢 |𝖣 s s − s s = 0 𝖡 |𝖢 𝖣 |𝖣 𝖨 . s s − s s = s 𝖡−1 𝖠 Assuming that s exists, the second equation yields s = 𝖡−1|𝖣 |𝖢 s s s, which substituted in the first equation yields the fourth equation. Similarly, from the second equation we have 𝖣 𝖡 |𝖠 |𝖡−1 s = s s s which inserted in the third equation yields 𝖠 |𝖢 |𝖡 𝖢 |𝖡 |𝖠 s s s − s s s = 0, which is valid because of the first equa- 𝖠 𝖡 tion. Thus, two of the dyadics can be chosen at will, say s and s 𝖣 𝖢 whence s was obtained from the second equation and s from the first equation as 𝖢 𝖡−1 𝖠2|𝖡 −1. s = s + s s The medium bidyadic can now be expressed in the form 𝖬 𝖡−1 𝖠 𝜺 𝖡 | 𝖨(2) 𝖠 |𝖡−1 . = s ∧ e4 + ( s + 4 ∧ s) ( s + s s ∧ e4) 4.10 One can find the following representations: 1 j x1 = √ (e1 + e2) x2 = √ (e1 − e2), 2 2 1 j x3 = √ (e3 + e4) x4 = √ (e3 − e4), 2 2 𝝃 j 𝜺 𝜺 𝝃 1 𝜺 𝜺 1 = √ ( 1 − 2) 2 = √ ( 1 + 2), 2 2 𝝃 j 𝜺 𝜺 𝝃 1 𝜺 𝜺 . 3 = √ ( 3 − 4) 4 = √ ( 3 + 4) 2 2 APPENDIX A Solutions to Problems 307

4.11 The following dyadic can be shown to yield the connection 𝖣 𝜺 𝜺 𝜺 𝜺 𝖣−1. = e1 1 + e2 3 + e3 2 + e4 4 = This can be verified as 𝖣|𝖩|𝖣−1 𝜺 𝜺 𝜺 𝜺 |𝖣−1 = (e1 2 − e3 1 + e2 4 − e4 3) 𝜺 𝜺 𝜺 𝜺 𝖣′. = e1 3 − e3 1 + e2 4 − e4 2 = 훽 F E 4.12 The unipotent spatial dyadic s ∈ 1 1 satisfies 훽2 𝖨T ⇒ 훽 𝖨T | 훽 𝖨T . s = s , ( s − s ) ( + s ) = 0 훽 𝖨T Excluding s =± s , neither of the bracketed terms can have an inverse, whence they must be of rank lower than 3. A product of two such dyadics can be zero only if the rank of one of them is lower than 2. Let us assume 훽 𝖨T 𝜷 𝜷 | s ± s = B 1b1, 1 b1 = 1 with either sign, whence 훽2 𝖨T 𝜷 | 𝜷 𝖨T s = s + (∓2 + B 1 b1) 1b1 = s implies B =±2. Thus, the dyadic has the form 훽 𝖨T 𝜷 𝜷 | s =±( s − 2 1b1), 1 b1 = 1, is the general solution for the unipotent 3D dyadic when excluding multiples of the unit dyadic. Similarly we obtain for the unipotent 훼 F E spatial bidyadic s ∈ 2 2 the representation 훼 𝖨(2)T 𝜶 𝜶 | . s =±( s − 2 12a12), 12 a12 = 1 4.13 Operating both sides of the equation by ⌊⌊𝜶𝜶, the left side can be expanded as 𝖰(2)⌊⌊𝜶𝜶 = (𝖰||𝜶𝜶)𝖰 − (𝖰|𝜶)(𝜶|𝖰), while the right side yields ⌊𝖨(2)T ⌊⌊𝜶 𝜶⌋ ⌊ 𝜶 𝖨T ⌊ 𝜶 𝜶 𝖨T . (eN ) = (eN ( ∧ )) = eN ( ∧ ∧ ) = 0 From these we have 𝖰|𝜶 𝜶|𝖰 (2) 𝖰||𝜶𝜶 2𝖰(2) 𝖰||𝜶𝜶 2 ⌊𝖨(2)T 0 = (( )( )) = ( ) = ( ) eN = 0, 308 APPENDIX A Solutions to Problems

which must be valid for any one-form 𝜶. This requires that 𝖰 be an antisymmetric dyadic, of the form 𝖰 = A⌊𝖨T . Defining 𝜺 | ΔA = N (A ∧ A)∕2, the original equation for the bivector A then becomes ⌊𝖨T (2) ⌊𝖨(2)T ⌊𝖨(2)T (A ) = AA −ΔAeN = eN or 𝜺 ⌊ 𝖨(2)T 𝖨(2)T . N AA −ΔA =

Taking the trace leads to 2ΔA − 6ΔA = 6, whence ΔA =−3∕2 𝜺 ⌊ 𝖨(2)T and N AA = (1∕2) which obviously cannot be satisfied by any bivector A. Thus, the original equation has no solution 𝖰. 4.14 Assuming that 𝖯 has full rank, multiplying the equation by |𝖯(−2)T we obtain 𝖰|𝖯−1 (2) ⌊𝖨(2)T ( ) = eN , whence from the result of the previous problem we conclude that there is no solution 𝖰. When 𝖯 has rank 1 or lower, we have 𝖯(2) = 0, whence 𝖰 may be any metric dyadic of rank 1 or lower. 𝖯 𝖯 𝜺 𝜺 When is of rank 2, we can write = p1 1 + p2 2, whence the equation becomes 𝖰(2) ⌊𝜺 . = eN 12p12 = e34p12 𝜺 Here we assume that {ei}and{ j} are reciprocal bases. This 𝖰 equation has an infinity of solutions, for example, = e3p1 + 𝖰 e4p2 and = e3p2 − e4p1 are two possibilities. Finally, for 𝖯 of rank 3, we may expand ∑4 𝖯 𝜺 𝜺 𝜺 𝖰 = p1 1 + p2 2 + p3 3, = eiqi, i=1 with p ∧ p ∧ p ≠ 0. From the equation 1 2 3∑ 𝖰(2) = eijqij = e34p12 + e14p23 + e24p31, i

we obtain conditions for the unknown vectors qi, . q12 = q23 = q31 = 0 q14 = p23, q24 = p31, q34 = p12 APPENDIX A Solutions to Problems 309

From the first three conditions it follows that the three vectors q1, q2, q3 must be multiples of the same vector. But this is contra- dictory to the last three conditions because it implies that q14, q24, and q34 would be multiples of the same bivector. Thus, we may conclude that the equation has solutions only when the rank of 𝖯 is at most 2, in which case the solution 𝖰 is not unique. 4.15 The inverse of the transformation dyadic 𝖣 is 𝖣−1 𝖨(2) 𝖦 |′ ∧ 𝜺 = s + ( s s)∧e4 4, as can be verified by direct multiplication because 𝖦 |′ | 𝖦′| 𝖦 | 𝖨 . ( s s) ( s s) = s s = s To prove the statement, we expand 𝖣|𝖩|𝖣−1 𝖣| ⌊ 𝜺 𝖦 ⌋𝜺 |𝖣−1 = (e123 s ∧ 4 + e4 ∧ s 123) ⌊ 𝜺 | 𝖦 |′ ∧ 𝜺 = (e123 s ∧ 4) (( s s)∧e4 4) 𝖦′| ∧ 𝜺 | 𝖦 ⌋𝜺 + (( s s)∧e4 4) (e4 ∧ s 123) ⌊ | 𝖦 |′ 𝜺 𝖦′| | 𝖦 ⌋𝜺 = (e123 s) ( s s ∧ 4) + (e4 ∧ s s) ( s 123) ⌊′ 𝜺 𝖦′⌋𝜺 . = e123 s ∧ 4 + e4 ∧ s 123 Thus, we have arrived at the AC bidyadic 𝖩′ ⌊′ 𝜺 𝖦′⌋𝜺 = e123 s ∧ 4 + e4 ∧ s 123 𝖦′ ′ with symmetric spatial metric dyadics s and s. 4.16 The Cayley–Hamilton equation (2.119) for the unipotent dyadic 𝖠 = 𝖪 is simplified to 𝖨 − (tr𝖪)𝖪 + (tr𝖪(2))𝖨 − (tr𝖪(3))𝖪 + (tr𝖪(4))𝖨 = 0. Obvious solutions are 𝖪 =±𝖨 with tr𝖪 =±4. For other solutions the equations are split to 1 + tr𝖪(2) + tr𝖪(4) = 0, tr𝖪 + tr𝖪(3) = 0. Applying the expansions (2.33)–(2.35) for 𝖠 = 𝖪, these equations can respectively be reduced to (tr𝖪)2((tr𝖪)2 − 4) = 0, tr𝖪((tr𝖪)2 − 4) = 0, 310 APPENDIX A Solutions to Problems

which can be compared with those of the AC dyadic, (4.88) and (4.89). Thus, the possible values of tr𝖪 for the unipotent dyadic are tr𝖪 = 0, tr𝖪 =±2andtr𝖪 =±4. 4.17 Since it was shown that (4.122) and (4.125) imply tr𝖩5 = tr𝖩3 = tr𝖩 = 0,

applying (4.123) we can expand the coefficients ai from (3.13) to (3.19) for 𝖢 = 𝖩 as

a0 = 1 𝖢2 2a2 =−tr = 6 𝖢2 𝖢4 4a4 =−(a2tr + tr ) = 12 𝖢2 𝖢4 𝖢6 . 6a6 =−(a4tr + a2tr + tr ) = 6

while a1 = a3 = a5 = 0. Inserting these in (3.10) yields 𝖩6 + 3𝖩4 + 3𝖩2 + 𝖨(2) = (𝖩2 + 𝖨(2))3 = 0.

CHAPTER 5

5.1 Since (5.29) and (5.31) are three-form equations, applying the bar-product rule from Appendix B, 𝜶 | 𝜶 | | 𝜶 ⋅ s as = g s as = g as the corresponding Gibbsian scalar equations can be written as | | ⌊ ⋅ e123 (ds ∧ B) = ds (e123 B) =∇ Bg = 0, | 𝝔 | ⌊ |𝝔 ⋅ 휚 . e123 (ds ∧ D − ) = ds (e123 D) − e123 =∇ Dg − g = 0 Applying the wedge-product rule ⌊ 𝜶 𝜷 𝜶 𝜷 e123 ( s ∧ s) = g × g the two-form equations (5.30) and (5.32) become vector equations as ⌊ 휕 휕 e123 (ds ∧ E + 휏 B) =∇×Eg + 휏 Bg = 0, ⌊ 휕 휕 . e123 (ds ∧ H − 휏 D − J) =∇×Hg − 휏 Dg − Jg = 0 APPENDIX A Solutions to Problems 311

5.2 From d ∧ 𝚿 = 0 we have the potential representation 𝚿 = d ∧ 𝝍, whence the second Maxwell equation yields d ∧ 𝚽 = d ∧ 𝖭|𝚿 = (d ∧ 𝖭⌊d)|𝝍 = 0. The equation corresponding to (5.62), (5.63) can be written as 𝖭 ⌊⌊ |𝝍 . ( m dd) = 0 5.3 Inserting (5.83)–(5.86) in (5.65) we can expand 𝖬 훼 휖′ 𝜺 휇−1 𝜺 훽 = + ∧ e4 + 4 ∧ + 4 ∧ ∧ ee 𝜺 ⌊휉 |휇−1⌋ 𝜺 ⌊ 휖 휉 |휇−1|휁 = 123 g g e123 + 123 ( g − g g g) ∧ e4 𝜺 𝜺 ⌊휇 −1 𝜺 휇−1|휁 + 4 ∧ ( 123 g) − 4 ∧ g g ∧ e4 𝜺 ⌊𝜺 ⌊휖 𝜺 ⌊휉 |휇−1| 𝖨⌋ 휁 = 123 123 g ∧ e4 + 123 g g ( e123 − g ∧ e4) 𝜺 휇−1| 𝖨⌋ 휁 + 4 ∧ g ( e123 − g ∧ e4) 𝜺 ⌊휖 𝜺 ⌊휉 𝜺 𝖨T |휇−1| 𝖨⌋ 휁 . = 123 g ∧ e4 + ( 123 g + 4 ∧ ) g ( e123 − g ∧ e4) 휇−1 휇 −1 F F Here the inverse g = ( g) is assumed to be in space 1 1. 𝖬 𝖡 ∧𝖨 T 5.4 The skewon bidyadic 2 = ( o∧ ) obeys the rule 𝖨(4)T ⌊⌊𝖬 𝖬 𝖨(2)T 𝖬 ⌊⌊𝖨 ∧𝖨T 𝖬 𝖬 . 2 = (tr 2) − ( 2 )∧ + 2 =− 2 Equating 𝖨(4)⌊⌊𝖬 𝖨(4)⌊⌊𝖨(2)T ||𝖬 𝖨(2)||𝖬 𝖬 tr( 2) = ( ) 2 = 2 = tr 2 𝖬 𝖬 to −tr 2 yields tr 2 = 0, whence we obtain 1 1 𝖬 = (𝖬 ⌊⌊𝖨)∧𝖨T , ⇒ 𝖡 = (𝖬 ⌊⌊𝖨)T . 2 2 2 ∧ o 2 2 5.5 Applying the orthogonality of the dyadic expressions 𝖨(2)T 휉 | 𝜺 𝖨 T 휉 휉 휉 | 𝜺 𝖨T ( s + ∧ e4) ( 4 ∧ s + ) = + ( ∧ e4) ( 4 ∧ s ) = 휉 − 휉 = 0, 𝖨(2)T 휁 | 𝜺 𝖨T 휁 휁 휁 | 𝜺 𝖨T −( s − ∧ e4) ( 4 ∧ − ) = + ( ∧ e4) ( 4 ∧ s ) = 휁 − 휁 = 0, one can directly verify that the bidyadic 𝖭 defined by the 3D expansion (5.77) is the inverse of the bidyadic 𝖬 defined by 312 APPENDIX A Solutions to Problems

(5.75). Similarly, we can apply the orthogonality properties 𝖨⌋ 휁 | 𝜺 𝖨T 𝜺 ⌊휁 휁 휁 ( e123 − g ∧ e4) ( 4 ∧ − 123 g) =− g + g = 0, 𝖨 𝜺 휉 ⌋𝜺 | ⌊𝖨T 휉 휉 휉 ( ∧ 4 + g 123) (e123 + e4 ∧ g) =− g + g = 0, to verify (5.90). 𝖬 5.6 Since m is symmetric, it defines a principal–axion∑ medium. To 𝖰 show that the axion part vanishes, let us write = qiqi.Any 𝖲 E E symmetric dyadic ∈ 1 1 can be written as a sum of symmetric dyadic products of vectors because of ∑ ∑ 𝖲 1 = aibi = (aibi + biai) ∑ 2 1 = ((a + b )(a + b ) − (a − b )(a − b )) 4 i i i i i i i i

(the minus sign makes some of the vectors qi imaginary). Forming the trace of the medium bidyadic we obtain ∑ ∑ 𝖬 𝜺 ⌊𝖰(2) 𝜺 ⌊ | 𝜺 | . tr = tr( N ) = ( N qij) qij = N (qij ∧ qij) = 0 i

5.7 When the bivector Q is not simple, we can write Q = e1 ∧ e2 + ≠ e3 ∧ e4, with e1 ∧ e2 ∧ e3 ∧ e4 = eN 0, whence 𝖰 ⌊𝖨T = (e12 + e34) = e2e1 − e1e2 + e4e3 − e3e4 and 𝖬 𝖰(2) m = = e12e12 + e34e34 − e24e31 − e23e14 − e14e23 − e31e24. Since this is symmetric, we have a principal–axion medium, 𝖬 = 𝖬 𝖬 𝖬 𝖨(2)T 𝖬 1 + 3. The axion component can be found as 3 = tr ∕6. Let us form 𝖬 𝜺 ⌊𝖬 𝜺 𝜺 𝜺 𝜺 = N m = 34e12 + 12e34 − 31e31 − 14e14

− e23e23 − e24e24, the trace of which equals tr𝖬 =−4. For a simple bivector Q 𝖬 𝖬 we can drop the term e3 ∧ e4.From m = e12e12 we have tr = 𝜺 tr( 34e12) = 0, whence in this case there is no axion component. APPENDIX A Solutions to Problems 313

5.8 Inserting the expansions (5.75) and (5.77) in (5.179) we obtain 𝜺 𝜺 𝜺 휇−1| 𝖨(2)T 휁 𝜺 2 ∧ 3 ∧ 4 ∧ b ( s − b ∧ 4) = 0, 𝜺 𝜺 𝜺 휖−1| 𝖨(2)T 휉 𝜺 − 2 ∧ 3 ∧ 4 ∧ b ( s + b ∧ 4) = 0, which reduce to 𝜺 𝜺 휇−1 𝜺 𝜺 휖−1 . 2 ∧ 3 ∧ b = 0, 2 ∧ 3 ∧ b = 0 These correspond to the Gibbsian conditions ⋅ 휇−1 ⋅ 휖−1 e1 gb = 0, e1 gb = 0 which require the field conditions (5.183) at the interface. 5.9 Expanding 휇𝖬 𝜺 ⌊ 𝖦 휇휖 (2) 𝜺 ⌊ 𝖦(2) 휇휖𝖦 ∧ − = N ( s − e4e4) = N ( s − s∧e4e4) 𝜺 ⌊ = N (e12e12 + e23e23 + e31e31) 휇휖𝜺 ⌊ − N (e14e14 + e24e24 + e34e34) 𝜺 𝜺 𝜺 휇휖 𝜺 𝜺 𝜺 = 34e12 + 14e23 + 24e31) − ( 23e14 + 31e24 + 12e34) The trace of each term vanishes, whence tr𝖬 = 0. 5.10 The boundary conditions are 𝜺 𝜺 𝜺 𝜺 2 ∧ 3 ∧ Ea = 0, 1 ∧ 3 ∧ Ha = 0, which are more general than the SH boundary conditions. For Gibbsian field vectors they take the form ⋅ ⋅ . e1 Ega = 0, e2 Hga = 0 ⌊𝚽 ⌊𝚿 5.11 Denoting∑ A =∑eN and B = eN , omitting the summation signs i and j, we can expand ⌊𝖴 ⌊⌊𝖨T 2| 𝚿 𝚽 ⌊⌊𝜺 | 𝚿 𝚽 ⌊⌊𝜺 | ((eN ) ) a = ((A − B ) iei) ((A − B ) jej) a = f(𝚽, 𝚿) − g(𝚽, 𝚿) − g(𝚿, 𝚽) + f(𝚿, 𝚽), with 𝚽 𝚿 ⌊𝜺 𝚿⌊ | ⌊𝜺 𝚿⌊ | f( , ) = (A i)( ei) (A j)( ej) a ⌊ 𝚽 𝚿⌊ | ⌊ 𝚽 𝜺 𝚿| = eN ( ∧ ei)( ei) ((eN ( ∧ j))( (ej ∧ a)), 𝚽 𝚿 ⌊𝜺 𝚿⌊ | ⌊𝜺 𝚽⌊ | g( , ) = (A i)( ei) (B i)( ei) a ⌊ 𝚽 𝜺 𝚿⌊ | ⌊ 𝚿 𝜺 𝚽| . = eN ( ∧ i)( ei) (eN ( ∧ i))( (ei ∧ a)) 314 APPENDIX A Solutions to Problems

We can do the summation with respect to j as ⌊ 𝚽 𝜺 𝚿| ⌊ 𝚽 𝜺 | ⌋𝚿 (eN ( ∧ j)( (ej ∧ a)) = (eN ( ∧ j)(ej (a )) ⌊ 𝚽 ⌋𝚿 = eN ( ∧ (a )), whence the functions become 𝚽 𝚿 ⌊ 𝚽 𝚿⌊ | ⌊ 𝚽 ⌋𝚿 f( , ) = eN ( ∧ ei)( ei) (eN ( ∧ (a ))) ⌊ 𝚽 | 𝚽 𝚿⌊ 𝚿⌊ =−eN ( ∧ ei)eN ( ∧ ( a) ∧ ( ei)), 𝚽 𝚿 ⌊ 𝚽 𝜺 𝚿⌊ | ⌊ 𝚿 ⌋𝚽 g( , ) = eN ( ∧ i)( ei) (eN ( ∧ (a ))) ⌊ 𝚽 𝜺 | 𝚿 𝚽⌊ 𝚿⌊ . =−eN ( ∧ i)eN ( ∧ ( a) ∧ ( ei)) At this point we can apply identities of the form 1 𝚿 ∧ (𝚿⌊b) = (𝚿 ⋅ 𝚿)𝜺 ⌊b 2 N (𝚿⌊b) ∧ (𝚿⌊c) = (𝚿⌊𝖨)(2)⌊(b ∧ c) = 𝚿𝚿|(b ∧ c) 1 − (𝚿 ⋅ 𝚿)𝜺 ⌊(b ∧ c), 2 N to the two functions and do the second summation as 𝚽 𝚿 ⌊ 𝚽 | 𝚽 𝚿⌊ 𝚿⌊ f( , ) =−eN ( ∧ ei)eN ( ∧ ( a) ∧ ( ei)) ⌊ 𝚽 | 𝚽 𝚿 𝚿| =−eN ( ∧ ei)eN ( ∧ )( (a ∧ ei)) 1 + (𝚿 ⋅ 𝚿)e ⌊(𝚽 ∧ e )e |(𝚽 ∧ 𝜺 ⌊(a ∧ e )), 2 N i N N i 1 =−(𝚽 ⋅ 𝚿)e ⌊(𝚽 ∧ (𝚿⌊a)) + (𝚿 ⋅ 𝚿)e ⌊(𝚽 ∧ (𝚽⌊a)), N 2 N 1 =−(𝚽 ⋅ 𝚿)e ⌊(𝚽 ∧ (𝚿⌊a)) + (𝚿 ⋅ 𝚿)(𝚽 ⋅ 𝚽)a, N 4 1 g(𝚽, 𝚿) =− (𝚿 ⋅ 𝚿)e |(𝚽 ∧ 𝜺 )e ⌊((𝚽⌊a) ∧ (𝜺 ⌊e )). 2 N i N N i 1 = (𝚿 ⋅ 𝚿)e ⌊(𝚽 ∧ 𝜺 )(𝚽⌊a)|e 2 N i i 1 = (𝚿 ⋅ 𝚿)e ⌊(𝚽 ∧ (𝚽⌊a)) 2 N 1 = (𝚿 ⋅ 𝚿)(𝚽 ⋅ 𝚽)a. 4 Summing the four terms and applying the identity 𝚿 𝚽⌊ 𝚽 𝚿⌊ 𝚽 ⋅ 𝚿 𝜺 ⌊ ∧ ( a) + ∧ ( a) = ( ) N a APPENDIX A Solutions to Problems 315

we finally obtain ⌊𝖴 ⌊⌊𝖨T 2| 𝚿 ⋅ 𝚿 𝚽 ⋅ 𝚽 ((eN ) ) a = ( )( )a 𝚽 ⋅ 𝚿 ⌊ 𝚽 𝚿⌊ 𝚿 ⋅ 𝚽 ⌊ 𝚿 𝚽⌊ − ( )eN ( ∧ ( a)) − ( )eN ( ∧ ( a)) = ((𝚿 ⋅ 𝚿)(𝚽 ⋅ 𝚽) − (𝚽 ⋅ 𝚿)2)a, which is valid for any vector a. Thus, we can write ⌊𝖴 ⌊⌊𝖨T 2 𝚿 ⋅ 𝚿 𝚽 ⋅ 𝚽 𝚽 ⋅ 𝚿 2 𝖨. ((eN ) ) = (( )( ) − ( ) ) 𝖬 𝖡∧𝖨 T 𝖡 5.12 A skewon–axion medium is defined by = ( ∧ ) where ∈ E F 1 1 may be any dyadic. The dispersion dyadic can be expanded as 𝖣 𝝂 𝖬 ⌊⌊𝝂𝝂 𝝂⌋ ⌊ 𝖡∧𝖨 T ⌊𝝂 ⌊ 𝝂 𝖡∧𝖨 ⌊𝝂 ( ) = m =− (eN ( ∧ ) ) = eN ( ∧ ( ∧ ) ) ⌊ 𝝂 𝖡T |𝝂 𝖨T = eN ( ∧ ( ) ∧ ) Denoting 𝝂′ = 𝖡T |𝝂,wehave 𝖣 𝝂 ⌊ 𝝂 𝝂′ ⌊𝖨T ⌊𝖨T ( ) = eN ( ∧ ) = A , which is an antisymmetric dyadic with simple bivector A because ⋅ ⌊ 𝝂 𝝂′ |𝜺 ⌊ ⌊ 𝝂 𝝂′ ⌊ 𝝂 𝝂′ | 𝝂 𝝂′ A A = (eN ( ∧ )) N (eN ( ∧ )) = (eN ( ∧ )) ( ∧ ) ⌊ 𝝂 𝝂′ 𝝂 𝝂′ . = eN ( ∧ ∧ ∧ ) = 0 Expressing it in the form A = a ∧ b, we can write 𝖣(𝝂) = (a ∧ b)⌊𝖨T = ba − ab. Because we have 𝖣(2)(𝝂) = (a ∧ b)(a ∧ b), 𝖣(3)(𝝂) = 0, the dyadic dispersion equation is satisfied identically for any 𝝂. 𝖷 𝜸 𝖨T ⌋𝚽 F F 5.13 To expand the dyadic = e ∧ ( ) ∈ 4 1 in 3D source and field quantities we insert 𝜸 𝝔 𝜺 𝚽 𝜺 e = e − Je ∧ 4, = B + E ∧ 4 and proceed 𝖷 𝝔 𝜺 𝖨T ⌋ 𝜺 = ( e − Je ∧ 4) ∧ ( (B + E ∧ 4) 𝝔 𝜺 ⌋ 𝜺 𝜺 𝖨T ⌋ 𝜺 = e ∧ ( 4e4) (E ∧ 4) − (Je ∧ 4) ∧ ( (B − E ∧ 4)) 𝝔 𝜺 𝜺 𝖨T ⌋ 𝜺 𝜺 𝜺 = e ∧ 4E − 4 ∧ (Je ∧ B) − ( 4 ∧ Je) ∧ ( 4E − E 4) 𝜺 휚 𝖨T ⌋ 𝜺 . =− 4 ∧ ( eE + Je ∧ s B − (Je ∧ E) 4) 316 APPENDIX A Solutions to Problems

5.14 The stress–energy dyadic 1 𝖳 = (𝚿 ∧ 𝖨T ⌋𝚽 − 𝚽 ∧ 𝖨T ⌋𝚿) 2 vanishes for any fields satisfying 𝚿 = M𝚽. To prove the converse, we expand 𝖳| 𝚿 𝖨T |𝚽 𝚽 𝖨T |𝚿 2 e4 =−( ∧ ∧ e4) + ( ∧ ∧ e4) =−𝚿 ∧ E − 𝚽 ∧ H. Requiring this to be zero leads to −H ∧ E + E ∧ H = 2E ∧ H = 0, D ∧ E + B ∧ H = 0. Assuming these to be valid for any nonzero E, B,theremustexist a scalar M such that H = ME, whence from (D − MB) ∧ E = 0 for any E we conclude that D − MB = 0. 𝚽 𝚽 𝜺 5.15 Expanding ∧ = 2B ∧ E ∧ 4 = 0 yields B ∧ E = 0. Apply- ing the bac-cab rule a⌋(E ∧ B) = E ∧ (a⌋B) + (a|E)B = 0, for a spatial vector a satisfying a|E ≠ 0 leads to the representation a⌋B B = 𝜶 ∧ E, 𝜶 = . a|E 5.16 Expanded in 3D medium dyadics the condition takes the form ⌊⌊휇−1 휉 |휇−1⌋ ⌊휇−1|휁 −Ae123e123 g − Ae4 ∧ g g e123 + Ae123 g g ∧ e4 휖 휉 |휇−1|휁 ∧ ⌊⌊휖−1T + A( g − g g g)∧e4e4) + Be123e123 g 휁 |휖−1⌋ T ⌊휖−1|휉 T − B(e4 ∧ g g e123) + B(e123 g g ∧ e4) 휇T 휉T |휖−1T |휁 T ∧ . − B( g − g g g )∧e4e4 = 0 Equating the four spatiotemporal terms separately leads to 휇−1 휖−1T ⇒ 휖 휇T −A g + B g = 0, A g = B g 휉 |휇−1 휉T |휖−1T ⇒ 휉T 휉 −A g g − B g g = 0, g =− g 휇−1|휁 휖−1T |휁 T ⇒ 휁 T 휁 . A g g + B g g = 0, g =− g APPENDIX A Solutions to Problems 317

The fourth condition 휖 휉 |휇−1|휁 휇T 휉T |휖−1T |휁 T A( g − g g g) − B( g − g g g ) = 0 is valid whenever the three other conditions are valid. 5.17 Starting from the Maxwell equation we can proceed as ⌊ 𝝂 𝚿 휈⌋ 𝖰(2) | 𝝂 𝝓 0 = eN ( ∧ ) =− ( + AB) ( ∧ ) =−𝝂⌋((𝖰|𝝂) ∧ (𝖰|𝝓) + AB|(𝝂 ∧ 𝝓)) =−(𝝂|𝖰|𝝓)𝖰|𝝂 + (𝝂|𝖰|𝝂)𝖰|𝝓 − (𝝂⌋A)(B⌊𝝂)|𝝓, which yields the equation for 𝝓. 5.18 The dispersion dyadic has the form 𝖣 𝝂 𝖤⌊⌊𝝂𝝂 ( ) = M + a1c1 + a2c2 ⌊𝝂 ⌊𝝂 with ai = Ai , ci = Ci . Because 𝖤⌊⌊𝝂𝝂 𝝂⌋ ⌊𝖨(2)T ⌊𝝂 𝝂⌋ ⌊ 𝝂 𝖨T =− (eN ) =− (eN ( ∧ )) ⌊ 𝝂 𝝂 𝖨T = eN ( ∧ ∧ ) = 0, the axion term falls out. Because we have 𝖣(2) 𝝂 ( ) = (a1 ∧ a2)(c1 ∧ c2) 𝖣(3)(𝝂) = 0, such a medium does not have a dispersion equation and the wave one-form 𝝂 of a plane wave can be freely chosen. 5.19 Since the first term does not affect the dispersion dyadic, we have 𝖣 𝝂 ( ) = a1c1 + a2c2 + a3c3, ⌊𝝂 ⌊𝝂 wherewedenoteai = Ai , ci = Ci . These vectors satisfy |𝝂 |𝝂 . ai = 0, ci = 0 The double-wedge powers are 𝖣(2) 𝝂 ( ) = a12c12 + a23c23 + a31c31, 𝖣(3) 𝝂 . ( ) = a123c123 The dyadic dispersion equation 𝖣(3) 𝝂 ( ) = a123c123 = 0 318 APPENDIX A Solutions to Problems

is split in two equations as ⌊𝝂 ⌊𝝂 ⌊𝝂 (A1 ) ∧ (A2 ) ∧ (A3 ) = 0, ⌊𝝂 ⌊𝝂 ⌊𝝂 . (C1 ) ∧ (C2 ) ∧ (C3 ) = 0 which appear cubic in 𝝂. Applying the bac-cab identity (B⌊𝜶) ∧ c = 𝜶⌋(B ∧ c) − (c|𝜶)B we can expand ⌊𝝂 𝝂⌋ |𝝂 a2 ∧ a3 = (A2 ) ∧ a3 = (A2 ∧ a3) − A2(a3 ) 𝝂⌋ . = (a3 ∧ A2) ⌊𝝂 . In particular we have (a2 ∧ a3) = 0 Applying another bac-cab rule ⌊𝜶 ⌊𝜶 ⌊𝜶 ⋅ ⌊𝜶 B ∧ (C ) + C ∧ (B ) = (B ∧ C) = (B C)eN , we can write 𝝂 a1 ∧ a2 ∧ a3 = (A1 ∧ ) ∧ a2 ∧ a3 ⌊𝝂 ⋅ ⌊𝝂. =−A1 ∧ ((a2 ∧ a3) ) + A1 (a2 ∧ a3)eN ⋅ ⌊𝝂. = A1 (a2 ∧ a3)eN The dispersion equations are actually only of the second order in 𝝂, and they have the scalar form ⋅ ⋅ ⌊𝝂 ⌊𝝂 A1 (a2 ∧ a3) = A1 ((A2 ) ∧ (A3 )) = 0, ⋅ ⋅ ⌊𝝂 ⌊𝝂 . C1 (c2 ∧ c3) = C1 ((C2 ) ∧ (C3 )) = 0 5.20 Inserting 𝖣(𝝂) = (𝖦|𝝂)(𝖦|𝝂)∕휇 in (5.254) and denoting a = 𝖦|𝝂 yields 𝚽 휆𝜺 ⌊ 휆 |𝜶 휇. = N (a ∧ d), = a ∕ Because d satisfies d|𝝂 = 0, we can express it as d = 𝝂⌋A in terms of some bivector A. Applying a bac-cab rule we can expand a ∧ d = a ∧ (𝝂⌋A) = 𝝂⌋(a ∧ A) − (a|𝝂)A. Taking a|𝝂 = 𝖦||𝝂𝝂 = 0 into account we can further expand 𝚽 휆𝜺 ⌊ 𝝂⌋ 휆𝝂 𝜺 ⌊ = N ( (a ∧ A)) = ∧ ( N (A ∧ a)) 휆𝝂 𝜺 ⌊ ⌊ 𝝂 ⌋𝚵 = ∧ (( N A) a) = ∧ (a ), 𝚵 휆𝜺 ⌊ . with =− N A This coincides with the expression (5.268). APPENDIX A Solutions to Problems 319

5.21 Writing 𝖢 = ac + bd in terms of four vectors we can expand 𝖢⌋ 𝜺 ⌊ 𝖢(2)T ⌋ 𝜺 ⌊ ( N ( )) = (ac + bd) ( N (c ∧ d)(a ∧ b)) 𝜺 ⌊ 𝜺 ⌊ =−(a( N (c ∧ d ∧ c) + b( N (c ∧ d ∧ d))(a ∧ b) = 0.

CHAPTER 6

𝖬 |𝖬 𝖬 |𝖬 6.1 Because tr( 1 2) = tr( 2 1), we have 𝖬 𝖠(−2)T |𝖬|𝖠(2)T) tr a = tr( ) = (𝖬|𝖠(2)T |𝖠(−2)T ) = tr𝖬. 𝖨(4)⌊⌊𝖬 ⌊ 𝖠(−2)T |𝖬|𝖠(2)T ⌋𝜺 a = eN ( N ⌊ 𝖠(−2)T ⌋𝜺 | ⌊𝖬⌋𝜺 | ⌊|𝖠(2)T ⌋𝜺 = (eN ( N) (eN N) (eN N) 𝖠(2) = |(𝖨(4)⌊⌊𝖬)|(𝖠(−2)tr𝖠(4)) tr𝖠(4) = 𝖠(2)|(𝖨(4)⌊⌊𝖬)|𝖠(−2). 6.2 The transformed medium bidyadic has the expansion 𝖬 𝖳(−2)T |𝖬|𝖳(2)T 𝖳(2)T |𝖬|𝖳(2)T a = = 𝖨(2)T 𝜺 ∧𝖨T |𝖬| 𝖨(2)T 𝜺 ∧𝖨T = ( s − 4e4∧ s ) ( s − 4e4∧ s ) 훼 휖′ 𝜺 휇−1 𝜺 훽 = − ∧ e4 − 4 ∧ + 4 ∧ ∧ e4, which yields the transformation rules 훼 훼 휖′ 휖′ 휇−1 휇−1 훽 훽. a = , a =− , a =− , a = Thus, a medium dyadic of the form 𝖬 훼 𝜺 훽 = + 4 ∧ ∧ e4 is invariant in time inversion. Since 휇−1 = 0, there is no Gibbsian representation of such a medium.

𝖰(𝖦|𝖷T |) = (𝖨 − 𝖦|𝖷T |)|(𝖨 + 𝖦|𝖷T |)−1 = (𝖦|(𝖨 − 𝖷)T |)|(𝖦|(𝖨 + 𝖷)T |)−1 = 𝖦|(𝖨 − 𝖷)T ||𝖦|(𝖨 + 𝖷)−1T | = 𝖦|𝖰T (𝖷)|. 6.4 Replacing a by a + b in the definition, the rotation dyadic 𝖱 satisfies (𝖱|a) ⋅ (𝖱|b) = a ⋅ b ⇒ a|(𝖱T ||𝖱 − )|b = 0 for any vectors a, b, whence the rotation dyadic satisfies 𝖱T ||𝖱 = ⇒𝖦|𝖱T | = 𝖱−1. Operating both sides of the latter equation by 𝖰() 𝖰(𝖦|𝖱T |) = 𝖰(𝖱−1), and applying properties of the quotient function in Problem 6.3 we obtain 𝖦|𝖰T (𝖱)| =−𝖰(𝖱), which equals 𝖰T (𝖱)| = (|𝖰(𝖱))T =−|𝖰(𝖱). Thus, |𝖰(𝖱) is an antisymmetric dyadic and it can be expressed in terms of some two-form 𝚷 as |𝖰(𝖱) = 𝖨T ⌋𝚷.Thisleadsto the representation 𝖰(𝖱) = 𝖦⌋𝚷, ⇒ 𝖱 = 𝖰(𝖦⌋𝚷) or 𝖱 = (𝖨 − 𝖦⌋𝚷)|(𝖨 + 𝖦⌋𝚷)−1. 6.5 Denoting s = sin 휃 and c = cos 휃 for brevity, we can expand both sides of (6.45) as ( ) ( ) ( ) ( ) ( ) Z D Z 휖 Z 휉 E Z 휖 휉 E d d = d d d d | d = d d d | d B 휁 휇 H 휁 휇 ∕Z Z H d ( d )d ( ) d ( d ) (d d ) (d d) Zc s D Zc s 휖 휉 E = | = | | . −Zs c B −Zs c 휁 휇 H APPENDIX A Solutions to Problems 321

On the second line we can insert from (6.46) ( ) ( )( ) ( )( ) E cs E cZs E d = = . ZdHd −sc ZH −sZc H Because (EH) is arbitrary, we can eliminate it and the result becomes ( )( ) ( ) ( ) 휖 휉 휖 휉 Zd d d cZs Zc s | 휁 휇 = 휁 휇 , d d∕Zd −sZc −Zs c or ( ) ( ) ( )( ) 휖 휉 휖 휉 Zd d d 1 Zc s | Zc −Zs . 휁 휇 = 휁 휇 d d∕Zd Z −Zs c sc This can be written componentwise as 휖 휖 2 휉 휁 휇 2 dZd = c Z + ( + )sc + s ∕Z, 휉 휖 휉 2 휁 2 휇 d =− scZ + c − s + sc∕Z, 휁 휖 휉 2 휁 2 휇 d =− scZ − s + c + sc∕Z, 휇 휖 2 휉 휁 휇 2 d∕Zd = s Z − ( + )sc + c ∕Z, and, finally, in the form 휖 휇 휖 휇 dZd + d∕Zd = Z + ∕Z, 휉 휁 휉 휁 d − d = − , 휖 휇 휖 휇 2 2 휉 휁 dZd − d∕Zd = ( Z − ∕Z)(c − s ) + ( + )2sc, 휉 휁 휖 휇 휉 휁 2 2 d + d =−( Z − ∕Z)2sc + ( + )(c − s ), which coincide with (6.61)–(6.64). 6.6 Inserting the Hehl–Obukhov decomposition in the reciprocity- transformation rule for the modified medium bidyadic (6.103) we obtain 𝖬 𝖬∗T 𝖬∗T 𝖬∗T 𝖬∗T . mr = m = m1 + m2 + m3 𝖬 The axion part 3 is transformed to 𝖬∗T ⌊𝖨(2) ∗ ∗⌊𝖨(2)∗ ⌊𝖨(2) m3 = M3(eN ) = M3eN =−M3eN 𝖬 =− m3, whence the axion part is transformed to an axion bidyadic which equals the original bidyadic with a change in sign. The skewon 322 APPENDIX A Solutions to Problems

𝖬 𝖡 ∧𝖨 T part 2 = ( o∧ ) is transformed to 𝖬∗T ⌊ 𝖡 ∧𝖨 T ∗T ⌊ 𝖡 ∧𝖨 T ∗ ∗⌊ 𝖡∗∧𝖨∗ T m2 = (eN ( o∧ ) ) =−(eN ( o∧ ) ) =−eN ( o ∧ ) ⌊ 𝖡∗∧𝖨 T = eN ( o ∧ ) , 𝖡 which yields an antisymmetric bidyadic because of tr or = 𝖡∗ 𝖡 ∗ 𝖬 tr o = (tr o) = 0. The principal part 1 is transformed to 𝖬∗T 𝖬∗ m1 = m1, whence the transformed principal bidyadic is symmetric. To test whether this is devoid of an axion component, let us consider the trace: 𝜺 ⌊𝖬 𝜺 ⌊𝖬∗T 𝖨(2)|| 𝜺 ⌊𝖬∗T 𝖨(2)|| 𝜺∗⌊𝖬T ∗ tr( N m1r) = tr( N m1) = ( N m1) = ( ( N m1) 𝖨(2)|| 𝜺 ⌊𝖬T ∗ 𝜺 ⌊𝖬 ∗. =−( ( N m1) =−tr( N m1) Since the last term is zero, so is the first one, whence there isno axion component in the transformed principal bidyadic. In conclusion, axion, skewon, and principal parts of a medium bidyadic are transformed individually to axion, skewon, and principal parts of the transformed medium bidyadic. 6.7 Because of the identity (𝖢(2)|(𝜶 ∧ 𝜷)) ∧ (𝖢(2)|(𝜸 ∧ 𝜹)) = (𝖢|𝜶) ∧ (𝖢|𝜷) ∧ (𝖢|𝜸) ∧ (𝖢|𝜹) = 𝖢(4)|(𝜶 ∧ 𝜷 ∧ 𝜸 ∧ 𝜹), the electromagnetic four-form 𝚿 ∧ 𝚽 is transformed as 𝚿′ ∧ 𝚽′ = (𝖢(2)|𝚿) ∧ (𝖢(2)|𝚽) = 𝖢(4)|(𝚿 ∧ 𝚽) = 휎4𝚿 ∧ 𝚽. To find the conformal transformation of the stress–energy dyadic we consider the transformed dyadic 𝚿′ ∧ 𝖨T ⌋⌋𝚽′ = (𝖢(2)|𝚿) ∧ (𝖢|𝖢−1)⌋⌋(𝖢(2)|𝚽) = 𝖢(3)|(𝚿 ∧ (𝖢−1⌋⌋𝖢(2))|𝚽). To proceed we can expand a|(𝖢−1⌋𝖢(2))|𝚽 = 𝖢 ∧ (a|𝖢−1|𝖢)|𝚽 = (𝖢 ∧ a)|𝚽 =−𝚽|(a ∧ ⌊⌊𝖢T ) = a|(𝚽⌊𝖢T ) = a|(𝚽⌊𝖨)|𝖢T = a|(𝖨T ⌋𝚽)|𝖢T , APPENDIX A Solutions to Problems 323

which is valid for any vector a and, thus, yields the rule (𝖢−1⌋𝖢(2))|𝚽 = (𝖨T ⌋𝚽)|𝖢T . In terms of this, we obtain 𝚿′ ∧ 𝖨T ⌋𝚽′ = 𝖢(3)|(𝚿 ∧ 𝖨T ⌋𝚽)|𝖢T , which finally leads to the following transformation rule for the Maxwell stress–energy dyadic, ′ 1 𝖳 = (𝚿′ ∧ 𝖨T ⌋𝚽′ − 𝚽′ ∧ 𝖨T ⌋𝚿′) = 𝖢(3)|𝖳|𝖢T . 2 6.8 Considering the Hehl–Obukhov decomposition of a medium 𝖬 𝖬 𝖬 𝖬 bidyadic as = 1 + 2 + 3, from (6.181) it follows that 𝖬 𝖨(2)T the axion part 3 = M remains invariant in the conformal transformation: 𝖬′ 𝖢(2)|𝖬 |𝖢(−2) 𝖢(2)|𝖨(2)T |𝖢(−2) 𝖬 . 3 = 3 = M = 3 Because of the symmetry involved in the transformation (6.184), each of the three parts is transformed individually in the conformal transformation. Thus a medium with only the principal part 𝖬 𝖬 = 1 is transformed to a medium whose medium bidyadic has only the principal part. The same applies to the skewon part. To check the last statement, we consider the transformation 𝖬 𝖡 ∧𝖨 T of a medium bidyadic of the form 2 = ( o∧ ) with a trace-free 𝖡 dyadic o. Applying (6.181) we can expand 𝖬′ 𝖢(2)|𝖬 |𝖢(−2) 𝖢(2)| 𝖡T ∧𝖨T |𝖢(−2) 2 = 2 = ( o ∧ ) 𝖢|𝖡T |𝖢−1 ∧ 𝖢|𝖨T |𝖢−1 𝖡′ ∧𝖨 T = ( o )∧( ) = ( o∧ ) , with the transformed dyadic defined by 𝖡′ 𝖢|𝖡T |𝖢−1 T 𝖢−1T |𝖡 |𝖢T . o = ( o ) = o 𝖡 𝖡′ 𝖬′ Because tr o = 0 implies tr o = 0, the transformed bidyadic 2 corresponds to a skewon medium. 6.9 From the relations between the medium dyadics and their duals 휖 휖 2 휃 휇 2 휃 휉 휁 휃 휃 dZd = Z cos + ( ∕Z)sin + ( + )sin cos 휇 휖 2 휃 휇 2 휃 휉 휁 휃 휃 d∕Zd = Z sin + ( ∕Z)cos − ( + )sin cos 휉 휖 휇 휃 휃 휉 2 휃 휁 2 휃 d =−( Z − ∕Z)sin cos + cos − sin 휁 휖 휇 휃 휃 휉 2 휃 휁 2 휃 d =−( Z − ∕Z)sin cos − sin + cos , 324 APPENDIX A Solutions to Problems

the 4 × 4 matrix can be identified as ( ) AB (휃) = CD

in terms of four 2 × 2 matrices ( ) ( ) ( ) c2 s2 11 c2 −s2 A = , B = sc =−CT , D = , s2 c2 −1 −1 −s2 c2 where c = cos 휃 and s = sin 휃. Forming the matrix product ( ) AAT + BBT ACT + BDT (휃)T (휃) = CAT + DBT CCT + DDT we have ( ) ( ) c4 + s4 + 2s2c2 2s2c2 − 2s2c2 10 AAT + BBT = = 2s2c2 − 2s2c2 c4 + s4 + 2s2c2 01 ( ) 00 ACT + BDT = CAT + DBT = 00 ( ) ( ) c4 + s4 + 2s2c2 −2s2c2 + 2s2c2 10 DDT + CCT = = . −2s2c2 + 2s2c2 c4 + s4 + 2s2c2 01

Thus, (휃)T (휃) becomes a 4 × 4 unit matrix. 6.10 The transformation rule for the corresponding modified medium bidyadics becomes 𝖬′ ⌊⌊𝖬′ 2𝖬−1T ⌋ 2 𝜺 ⌊𝖬 −1T ⌋ m = eN = Y eN = Y ( N m) eN 2 ⌊⌊𝖬−1T . = Y eNeN m Making the transformation twice returns the original medium dyadic 𝖬′′ 2 ⌊⌊𝖬′−1T ⌊⌊ ⌊⌊𝖬−1T −1T 𝖬−1T −1T m = Y eNeN m = eNeN (eNeN m ) = ( m ) 𝖬 . = m From (5.88) we have the expansion 𝖬′ 휖′ ∧ ⌊𝖨T 휉′ |휇′−1| 𝖨⌋ 휁 ′ m = g∧e4e4 − (e123 + e4 ∧ g) g ( e123 − g ∧ e4), APPENDIX A Solutions to Problems 325

and, from (5.89), 𝖬−1 𝜺 𝜺 ⌊⌊휇 m =− 123 123 g 𝜺 𝖨T 𝜺 ⌊휁 |휖−1| 𝖨 𝜺 휉 ⌋𝜺 . − ( 4 ∧ − 123 g) g ( ∧ 4 + g 123) Operating the latter as ⌊⌊𝖬−1T ⌊⌊휖−1T 휉T |휖−1T ⌋ eNeN m = e123e123 g − e4 ∧ g g e123 ⌊휖−1T |휁 T 휇T 휉T |휖−1T |휁 T ∧ + e123 g g ∧ e4 − ( g − g g g )∧e4e4, we obtain the relations 휇′−1 2휖−1T g =−Y g , 휉′ |휇′−1 2휉T |휖−1T g g = Y g g , 휇′−1|휁 ′ 2휖−1T |휁 T g g = Y g g , 휖′ 휉′ |휇′−1|휁 ′ 2 휇T 휉T |휖−1T |휁 T g − g g g =−Y ( g − g g g ), which lead to the following transformation rules: ( ) ( ) 휖′ 휉′ −Y2휇T −휉T g g = g g . 휁 ′ 휇′ 휁 T 휖T 2 g g − g − g ∕Y 6.11 The transformation rule (6.59) for the modified medium bidyadic is 𝖬 𝖤 휃 𝖬 | 휃 𝖤 𝖬 −1⌋ . Zd md = ( + cot Z m) (cot − Z m) eN To be able to find an analytic expression for the inverse bidyadic, let us choose the transformation parameters so that cot 휃 𝖤 − Z(훼𝖤 + 𝖰(2)) =−Z𝖰(2), that is, cot 휃 = Z훼, whence we obtain

𝖬 −1 𝖤 2훼 훼𝖤 𝖰(2) |𝖰(−2) md = ( + Z ( + )) ZdZ −1 2훼2 𝖤 2훼𝖰(2) |𝖰(−2)⌋ = ((1 + Z ) + Z ) eN ZdZ −1 2훼2 ⌊⌊𝖰(−2) 2훼𝖤 = ((1 + Z )eNeN + Z ) ZdZ −1 2훼𝖤 2훼2 𝖰(2T) . = (Z + (1 + Z ) ∕ΔQ) ZdZ 326 APPENDIX A Solutions to Problems

From this we can identify √ 2훼2 훼 Z 훼 𝖰 −(1 + Z )𝖰T . d =− , d = Zd ZdZΔQ √ 훼2 For the special choice Zd =−Z =±1∕ ΔQ − we obtain

𝖬 훼𝖤 𝖰(2)T . d = +

6.12 Equation (6.60) is of the general form

𝖬 |𝖬 𝖬 𝖬 𝖨(2)T . A d + B d + C + D = 0

Multiplying this equation as 𝖬|()|𝖬−1 and subtracting the original equation leaves us with a condition which can be written as

𝖬|𝖬 𝖬 𝖬 | 𝖬 𝖨(2)T . ( d − d ) (A + B ) = 0 𝖬 𝖬 The bidyadics , d commute in the general case, that is, when the bidyadic

𝖬 𝖨(2)T 휃 휃𝖬 휃𝖨(2)T A + B = Zd sin (−Z sin + cos )

has an inverse. 6.13 Starting from the PEMC condition (6.75) and applying (6.39) yields a condition of the form

𝜺 𝚿 𝚽 3 ∧ (A d − B d) = 0,

with Z 1 A = d (cos 휃 − MZ sin 휃), B = (sin 휃 + MZ cos 휃). Z Z To be self-dual, the duality transformation must depend on the PEMC parameter M as B = MA. This gives us a lot of freedom to 휃 choose. For the case sin = 0, Zd =−Z we obtain A = MZ and B = 1∕Z, whence we must have Z = 1∕M. APPENDIX A Solutions to Problems 327

CHAPTER 7

𝖬 휖𝖢 1 𝜺 𝖢 −1 𝖢 7.1 Starting from = s ∧ e4 + 휇 4 ∧ s and inserting s = ⌊𝖡 𝖢−1 𝖡−1⌋ e123 s, = e123, we obtain s ∑ s 𝖬⌊⌊𝖨 ⌋𝖬⌊𝜺 =− ei i i 휖 ⌋ 𝜺 ⌊𝖡 ⌊𝜺 1 ⌋ 𝜺 𝖡 −1⌋ ⌊𝜺 =− ei ( 123 s ∧ e4) i − 휇 ei ( 4 ∧ s e123) i 휖𝜺 ⌊ 𝖡 ⌊𝜺 1 ⌋ 𝜺 𝖡 −1 𝜺 ⌋ = 123 (ei ∧ s ∧ e4) i + 휇 ei ( 4 ∧ s ∧ i) e123, when omitting the summing sign. Applying the bac-cab rules 𝖡 ⌊𝜺 𝖡 |𝜺 𝖡 |𝜺 ( s ∧ e4) i = ( s i)e4 − s(e4 i) ⌋ 𝜺 𝖡 −1 𝜺 |𝖡−1 |𝜺 𝖡−1 ei ( 4 ∧ s ) = 4(ei s ) − (ei 4) s we can proceed as 𝖬⌊⌊𝖨 휖𝜺 ⌊ 𝖡 |𝜺 𝖡 = 123 ((ei ∧ s i)e4 − e4 ∧ s) 1 + (𝜺 (e |𝖡−1 ∧ 𝜺 ) − 𝖡−1 ∧ 𝜺 )⌋e 휇 4 i s i s 4 123 1 = 휖𝜺 ⌊((e ∧ 𝖡 |𝜺 )e ) + (𝜺 (e |𝖡−1 ∧ 𝜺 ))⌋e . 123 i s i 4 휇 4 i s i 123

Requiring 𝖬⌊⌊𝖨∑= 0, the two terms∑ must vanish separately. 𝖡 𝖡−1 훽 𝜺 𝜺 Expanding s = j,k Bj,kejek, s = j,k j,k j k we have ∑ ∑ ∑ ∑ 𝖡 |𝜺 |𝖡−1 𝜺 훽 𝜺 ei ∧ s i = Bj,ieij = 0, ei s ∧ i = i,k ki = 0, i i,j i i,k ≠ 훽 ≠ from which it follows that Bj,i = 0fori j and i,k = 0fori k. 𝖡 𝖡−1 Thus, both s and s must be symmetric spatial dyadics. 7.2 Inserting the expression for the inverse of the symmetric spatial 𝖦 metric dyadic s = e1e1 + e2e2 + e3e3, 𝜺 𝜺 ⌊⌊𝖦(2)T 𝖦−1 123 123 𝜺 𝜺 ⌊⌊𝖦(2) = = 123 123 s 𝜺 𝜺 ||𝖦(3) 123 123 the last term of (7.17) becomes 1 1 𝜺 ∧ (휇𝖦 )−1⌋e = 𝜺 ∧ (𝜺 𝜺 ⌊⌊𝖦(2))⌋e =− 𝜺 ⌊𝖦(2). 4 s 123 휇 4 123 123 123 휇 N s 328 APPENDIX A Solutions to Problems

The last two terms of (7.17) can be combined as 휖′𝜺 ⌊𝖦 𝜺 휇𝖦 −1⌋ 123 s ∧ e4 + 4 ∧ ( s) e123 1 = 휖′𝜺 ⌊𝖦 ∧ e − 𝜺 ⌊𝖦(2) 123 s 4 휇 N s 1 =− 𝜺 ⌊(휇휖′e ∧ 𝖦 ∧ e + 𝖦(2)) 휇 N 4 s 4 s 1 =− 𝜺 ⌊(−휇휖′𝖦 ∧e e + 𝖦(2)) 휇 N s∧ 4 4 s 1 𝜺 ⌊ 휇휖′ 𝖦 (2). =−휇 N (− e4e4 + s)

𝖦 𝖦 휇휖′ 𝖳 𝖨 𝜺 Defining = s − e4e4 and = s − e4 4, (7.17) reduces to 𝖬 𝖨(2)T 𝖳(2)T 1 𝜺 ⌊𝖦(2). = A + B − 휇 N 7.3 The condition (7.33) is clear from Problem 7.2. To derive (7.34), let us expand 𝜺 ⌊𝖦(2) 𝜺 ⌊ 𝖦(2) 휇휖′ 𝖦 N = N ( s + e4 ∧ s ∧ e4) 𝜺 𝜺 ⌊𝖦(2) 휇휖′𝜺 ⌊𝖦 =− 4 ∧ ( 123 s ) − 123 s ∧ e4, and substitute it in 𝖳(2)T | 𝜺 ⌊𝖦(2) 𝖨(2)T 𝜺 𝖨T | 𝜺 ⌊𝖦(2) ( N ) = ( s + 4 ∧ s ∧ e4) ( N ) 휇휖′𝜺 ⌊𝖦 𝜺 𝜺 ⌊𝖦(2) =− 123 s ∧ e4 + 4 ∧ ( 123 s ), 𝜺 ⌊𝖦(2) |𝖳(2)T 𝜺 ⌊𝖦(2) | 𝖨(2)T 𝜺 𝖨T ( N ) = ( N ) ( s + 4 ∧ s ∧ e4) 𝜺 𝜺 ⌊𝖦(2) 휇휖′𝜺 ⌊𝖦 . =− 4 ∧ ( 123 s ) + 123 s ∧ e4 Comparing these two expansions we obtain 𝖳(2)T | 𝜺 ⌊𝖦(2) 𝜺 ⌊𝖦(2) |𝖳(2)T ( N ) =−( N ) , whence the condition (7.31) is verified.

7.4 𝖳(2) 𝖨 𝜺 (2) 𝖨 𝜺 (2) = ( s − e4 4) = ( − 2e4 4) 1 = 𝖨(2) − 2e 𝜺 ∧𝖨 = (𝖨 − 4e 𝜺 )∧𝖨 4 4∧ 2 4 4 ∧ 𝖡 ∧𝖨. = o∧ The dyadic 𝖡 = 1 (𝖨 − 4e 𝜺 ) satisfies tr𝖡 = 0. o 2 4 4 o APPENDIX A Solutions to Problems 329

7.5 The eigenfields satisfy ( ) 1 𝖬|𝚽 = A𝖨(2)T + B𝖳(2)T + 휖′𝖢 ∧ e + 𝜺 ∧ 𝖢−1 ± s 4 휇 4 s | 𝜺 (B± + E± ∧ 4) 𝜺 휖′𝖢 | = (A + B)B± + (A − T)E± ∧ 4 + s E± 1 + 𝜺 ∧ 𝖢−1|B 휇 4 s ± 𝜺 . = M±(B± + E± ∧ 4) Separating the spatial and temporal parts we obtain the relations 휖′𝖢 | (A + B)B± + s E± = M±B±, 1 (A − B)E − 𝖢−1)|B = M E , ± 휇 s ± ± ± or 휖′ 𝖢 | B± = s E±, M± − A − B 1 E = 𝖢−1|B . ± 휇 s ± (M± − A + B) √ 2 휖′ 휇 These are the same equations if M± = A ± B − ∕ . Doing 𝖬 the same for the medium bidyadics ± we obtain, 휖 ± 𝖢 | B± = s E±, M± 1 E = 𝖢−1|B , ± 휇 s ± ±M± 2 휖 휇 which imply M± = ±∕ ±. Comparing the eigenfield relations of the two cases we can solve √ M A ± B2 − 휖′∕휇 휖 휖′ ± 휖′ ± = = √ , M± − A − B −B ± B2 − 휖′∕휇 √ M − A + B B ± B2 − 휖′∕휇 휇 휇 ± 휇 . ± = = √ M± A ± B2 − 휖′∕휇 휖 휖′ 휇 휇 As a simple check, for A = B = 0wehave ± = and ± = . 330 APPENDIX A Solutions to Problems

⌊⌊𝖨T E E 𝖦(2)⌊⌊𝝂𝝂 E E 7.6 Since As ∈ 1 1 is antisymmetric and ∈ 1 1 is symmetric, the antisymmetric part of the dispersion equation (7.61) is ⌊𝖨T ∧ 𝖦(2)⌊⌊𝝂𝝂 (2) . (As )∧( ) = 0 𝖲 𝖦(2)⌊⌊𝝂𝝂 E E Denoting the symmetric dyadic = ∈ 1 1, we can apply the identity of the Problem 2.22, 𝜺 𝜺 ⌊⌊ ⌊𝖨T ∧𝖲(2) 𝚽⌊𝖨 N N ((A )∧ ) = , 𝚽 | 𝜺 𝜺 ⌊⌊𝖲(2) F with =−A ( N N ) ∈ 2. Thus, equivalent to the above antisymmetric equation is vanishing of the two-form 𝚽 | 𝜺 𝜺 ⌊⌊𝖲(2) | 𝜺 ⌊𝖲(2) ⌋𝜺 =−A ( N N ) =−A ( N ) N 𝜺 ⌊ |𝖲(2)⌋𝜺 훼 𝝂 𝜺 |𝖲(2)⌋𝜺 . =−( N A) N =− ( ∧ 4) N Here we can apply the rule (𝜶 ∧ 𝜷)|𝖲(2) = (𝜶|𝖲) ∧ (𝜷|𝖲) and continue as 𝚽 훼 𝝂|𝖲 𝜺 |𝖲 ⌋𝜺 =− ( ) ∧ ( 4 ) N 훼 𝝂| 𝖦(2)⌊⌊𝝂𝝂 𝜺 | 𝖦(2)⌊⌊𝝂𝝂 ⌋𝜺 . =− ( ( )) ∧ ( 4 ( )) N But this vanishes identically, since 𝝂|(𝖦(2)⌊⌊𝝂𝝂) =−(𝝂 ∧ 𝝂)|𝖦(2)⌊𝝂 = 0forall𝝂. Thus, the antisymmetric part of the dispersion equation vanishes identically. 7.7 The symmetric part of the dispersion equation (7.61) is ∧ 𝖦(2)⌊⌊𝝂𝝂 𝖦(2)⌊⌊𝝂𝝂 (3) . AsAs∧( ) + ( ) = 0 𝖦(4) 휇휖′ Making use of (5.262) for p = 4and =− eNeN, we can expand (𝖦(2)⌊⌊𝝂𝝂)(3) = ((𝖦||𝝂𝝂)𝖦 − (𝖦|𝝂)(𝝂|𝖦))(3) 𝖦||𝝂𝝂 3𝖦(3) 𝖦||𝝂𝝂 2𝖦(2)∧ 𝖦|𝝂 𝝂|𝖦 = ( ) − ( ) ∧( )( ) 𝖦||𝝂𝝂 2 𝖦(4)⌊⌊𝝂𝝂 휇휖′ 𝖦||𝝂𝝂 2 ⌊⌊𝝂𝝂. = ( ) ( ) =− ( ) eNeN ∑ 𝖦(2) Writing = BiBi, we can expand i ∑ ∧ 𝖦(2)⌊⌊𝝂𝝂 ⌊𝝂 ⌊𝝂 . AsAs∧( ) = (As ∧ (Bi ))(As ∧ (Bi )) i APPENDIX A Solutions to Problems 331

Applying the identity ⌊𝜶 ⌊𝜶 ⌊𝜶 ⋅ ⌊𝜶 A ∧ (B ) + B ∧ (A ) = (A ∧ B) = (A B)eN valid for any bivectors A, B and one-form 𝜶,wehave ⌊𝝂 ⋅ 𝜺 ⌊𝝂 As ∧ (Bi ) = (As Bi) N , ⌊𝝂 because of As = 0. Thus, we can write ∧ 𝖦(2)⌊⌊𝝂𝝂 ⋅ 𝖦(2) ⋅ ⌊⌊𝝂𝝂 AsAs∧( ) = As As(eNeN ) and the dyadic dispersion equation is equivalent to the scalar equation 휇3 𝝂 ⋅ 𝖦(2) ⋅ 휇휖′ 𝖦||𝝂𝝂 2 . D( ) = As As − ( ) = 0 Expanding the first term as ⋅ 𝖦(2) ⋅ 𝜺 𝜺 || ∧𝖦(2) 𝜺 ⌊ 𝜺 ⌊ ||𝖦(2) As As = N N (AsAs)∧ ) = ( N As)( N As) 훼2 𝝂𝝂∧𝜺 𝜺 ||𝖦(2) = ( ∧ 4 4) 훼2 𝝂𝝂∧𝜺 𝜺 || 𝖦(2) 휇휖′𝖦 ∧ = ( ∧ 4 4) ( s − s∧e4e4) 휇휖′ 휇 휈 2 𝖦 ||𝝂𝝂 =− (2 B 4) ( s ), 휈 |𝝂 𝖦||𝝂𝝂 𝖦 ||𝝂𝝂 휇휖′휈2 where we denote 4 = e4 . Inserting = s − 4 , the dispersion equation becomes 휇3 𝝂 휇휖′ 휇 휈 2 𝖦 ||𝝂𝝂 휇휖′ 𝖦||𝝂𝝂 2 D( ) =− (2 B 4) ( s ) − ( ) 휇휖′ 휇 휈 2 𝖦 ||𝝂𝝂 𝖦 ||𝝂𝝂 2 =− ((2 B 4) ( s ) + ( s ) ) 휇휖′ 𝖦 ||𝝂𝝂 휈2 휇휖′ 2휈4 −2 ( s ) 4 + ( ) 4 )) 휇휖′ 𝖦 ||𝝂𝝂 2 휇 2 휇휖′ 𝖦 ||𝝂𝝂 휈2 =− (( s ) + ((2 B) − 2 )( s ) 4 휇휖′ 2휈4 + ( ) 4 ) 휇휖′ 𝖦 ||𝝂𝝂 2 휇휖′ 휇 2 𝖦 ||𝝂𝝂 휈2 =− (( s ) − 2( − 2( B) )( s ) 4 휇휖′ 2휈4 +( ) 4 ) = 0.

The last expression can be given the factorized form 휇3 𝝂 휇휖′ 𝝂 𝝂 𝝂 𝖦 ||𝝂𝝂 휇 휖′ 휈2 D( ) = D+( )D−( ), D±( ) = s − ± ± 4 332 APPENDIX A Solutions to Problems

𝖦 ||𝝂𝝂 휈2 by solving the second-order equation for x = ( s )∕ 4 = 휇 휖′ ± ±, x2 − 2(휇휖′ − 2(휇B)2)x + (휇휖′)2 = 0. The solutions can be expressed as √ 휇휖′ 휇 2 휇 휇 2 휇휖′ x = − 2(√B) ± 2 B ( B) − =−(휇B ± (휇B)2 − 휇휖′)2. 7.8 From the expression (7.20) we obtain (𝖳(2)T )2 = 𝖨(2)T , whence 𝖳(2)T = 𝖴. Also, from 𝜺 ⌊𝖦(2) 2 𝜺 𝜺 ⌊⌊𝖦(2) |𝖦(2) 𝖦(−2)T |𝖦(2) 휇휖′𝖨(2)T ( N ) = ( N N ) =ΔG =− √ 𝜺 ⌊𝖦(2) 휇 𝖩 휖′ 휇 we have − N ∕ = C with C = ∕ . Studying the trans- 𝖬 𝖣|𝖬|𝖣−1 formed medium bidyadic D = termwise, we have (𝖣|𝖴|𝖣−1)2 = 𝖣|𝖴2|𝖣−1 = 𝖨(2)T , (𝖣|𝖩|𝖣−1)2 = 𝖣|𝖩2|𝖣−1 =−𝖨(2)T , whence 𝖬 𝖨(2)T 𝖴 𝖩 D = A + B D + C D, 𝖴 𝖩 where D is a unipotent bidyadic and D is an AC bidyadic. How- ever, the medium so defined is not necessarily a GBI medium. In the special case when 𝖣 is of the form 𝖠(2) the similarity transformation is an affine transformation and a GDI medium is transformed to a GBI medium. 𝖬 𝖬 7.9 Substituting d = in the duality transformation rule (6.59) requires that the self-dual medium bidyadic must satisfy the second-order equation Z − Z 1 𝖬2 + cot 휃 d 𝖬 + 𝖨(2) = 0. ZZd ZZd 휃 If parameters Z − Zd and can be defined so that this equation equals (7.31) satisfied by the general GBI medium bidyadic, the medium is self-dual. Thus, the following conditions for the transformation parameters are obtained: Z − Z 휖′ 휃 d 1 2 2 . cot =−2A, = A − B + 휇 ZZd ZZd APPENDIX A Solutions to Problems 333

Since there are two equations for three parameters, it is obvious that a solution can be found by adding a third condition. Consid- ering the axion medium, 𝖬 = A𝖨(2)T , we can assume the involu- 휃 휋 tory transformation = ∕2, Zd = Z in which case the condition becomes 𝖬2 = 𝖨(2)T ∕Z2, which is satisfied by choosing Z = 1∕A or Z =−1∕A. 7.10 For the simple skewon–axion medium the conditions are D = (A + B)B = 훼B, H = (−A + B)E =−훽E. The Maxwell equations outside sources have the form 휕 ds ∧ B = 0, ds ∧ E + 휏 B = 0, 휕 ds ∧ D = 0, ds ∧ H − 휏 D = 0, the latter of which in the skewon–axion medium become 훼 훽 훼휕 . ds ∧ B = 0, − ds ∧ E − 휏 B = 0 For B ≠ 0wehave훼 ≠ 훽 whence we must have d ∧ E = 0and 휕휏 B = 0. Assuming B = 0 an initial value, we have B = 0and 휏 훽 휑 휑 D = 0forall while H =− E = ds where may be any 𝜺 | scalar potential function. Because at an interface 3 x = 0we 𝜺 𝜺 have 3 ∧ D = 0and 3 ∧ B = 0, the DB conditions are satisfied at the interface of an SS medium. ∑ 𝖡 7.11 The dyadic rule (7.122) can be obtained by writing g = risi and expanding

p × siri + risi × p = (p × si)ri − ri(p × si) 𝖨 = (ri × (p × si)) × g ⋅ ⋅ 𝖨 = (p(ri si) − (p ri)si) × g, 𝖡 ⋅ 𝖡 𝖨 . = (ptr g − p g) × g 7.12 Inserting the expressions (7.110)–(7.113), the medium equations can be written as D = 훼|B + 휖′|E 𝖡 𝜺 ⌊𝖡 ⌋ | 𝜷 = tr sB − ( 123 s e123) B − s ∧ E, H = 휇−1|B + 𝜷|E ⌋ 𝖡 𝖨T | . =−bs B − ( s + b s ) E 334 APPENDIX A Solutions to Problems

Applying the rules of Appendix B, we obtain ⌊ 𝖡 𝖡 | ⌊ 𝜷 Dg = e123 D = tr sBg − s Bg − e123 ( s ∧ E), 𝖡 𝖡 ⋅ 𝜷 = tr sBg − g Bg − g × Eg, 𝖦 | 𝖦 | ⌋ 𝖡T 𝖦 ⋅ . Hg = s H =− s (bs B) − ( g + b s) Eg 𝖡T ⋅ =−bs × Bg − g Eg − bEg 𝖡 𝖡 |𝖦 with g = s s. Here we have applied 𝖦 | ⌋ 𝖦 | ⌋ 𝜺 ⌊ 𝖦 | 𝜺 ⌊ s (bs B) = s (bs ( 123 Bg)) =− s ( 123 (Bg ∧ bs)) 훽 . =− g × bs 7.13 The dispersion dyadic can be written as 𝖣 𝝂 𝝂 ⌊𝖨T ⌊ 𝜶 𝜷 . ( ) = F( ) , F = eN ( ∧ ) F is a simple bivector, which can be expressed as F = a ∧ b,and 𝖣(𝝂) is an antisymmetric dyadic. Thus, the dyadic dispersion equation 𝖣(3)(𝝂) = ((a ∧ b)⌊𝖨T )(3) = (ba − ab)(3) = 0 is valid for any 𝝂. From the antisymmetry of 𝖣 we have 𝖣T 𝝂 𝖬 ⌊⌊𝝂𝝂 T 𝖬T ⌊⌊𝝂𝝂 𝖬 ⌊⌊𝝂𝝂 ( ) = ( m ) = m =− m , or 𝖬T 𝖬 ⌊⌊𝝂𝝂 ( m + m) = 0, which must be valid for any 𝝂. From this it follows that the bidyadic in brackets must be of the form 훼𝖤 and, thus, the sym- 𝖬 𝖤 metric part of m must be a multiple of , which rules out the principal part of 𝖬. 𝜶 𝜺 𝜺 7.14 Defining = A 4 + 1 in (7.126) and (7.127), the conditions (7.130) for the fields in the skewon–axion medium become 𝜺 𝜺 𝜺 A 4 ∧ B + 1 ∧ E ∧ 4 = 0 𝜺 𝜺 𝜺 A 4 ∧ D − 1 ∧ H ∧ 4 = 0, which equal 𝜺 AB + 1 ∧ E = 0, 𝜺 . AD − 1 ∧ H = 0 APPENDIX A Solutions to Problems 335

𝜺 Multiplying these by 3∧ we arrive at the SHDB conditions (7.153) and (7.154) and, in Gibbsian form, at (7.155) and (7.156). 7.15 Inserting 𝖬 = 𝖬 − 1 tr𝖬 𝖨(2)T in the eigenrule we obtain o 6 ( ) 𝖬⌊⌊𝖨 ∧𝖨T 1 𝖬 𝖨(2)T ⌊⌊𝖨 ∧𝖨T 𝖬 1 𝖬 𝖨(2)T − ( )∧ − tr ( )∧ + − tr ( 6 ) 6 1 =− 𝖬 − tr𝖬 𝖨(2)T 6 Combining terms we arrive at the condition (7.95). 7.16 The inverse bidyadic has the form (𝖠 ∧𝖨)T |𝚫B|(𝖠 ∧𝖨)T 𝖬−1 = (𝖠 ∧𝖨)T − o∧ o∧ , o∧ | 𝖠 ∧𝖨 |𝚫 1 + B ( o∧ ) 𝖠 where o is the dyadic defined by (3.127). The result can be verified by multiplication. 7.17 From (5.102)–(5.105) the conditions for the skewon-medium dyadics are obtained, 훼 𝜺 ⌊⌊훽T = 123e123 , 휖′ 𝜺 ⌊⌊휖′T =− 123e123 , 휇−1 𝜺 ⌊⌊휇−1T =− 123e123 , 훽 𝜺 ⌊⌊훼T . = 123e123 Applying the relations to the Gibbsian dyadics (5.83)–(5.86) as 훼 𝜺 ⌊ 휉 |휇−1 ⌋ = 123 ( g g ) e123, 휖′ 𝜺 ⌊ 휖 휉 |휇−1|휁 = 123 ( g − g g g), 휇−1 𝜺 ⌊휇−1 = 123 g , 훽 휇−1|휁 =− g g, the skewon-medium conditions become 휉 |휇−1 휇−1|휁 T g g =−( g g) , 휖 휉 |휇−1|휁 휖 휉 |휇−1|휁 T g − g g g =−( g − g g g) , 휇−1 휇−1T g =− g , 휇−1|휁 휉 |휇−1 T . g g =−( g g ) 336 APPENDIX A Solutions to Problems

From these we finally obtain the simple conditions 휇−1T 휇−1 휉T 휁 휖T 휖T . g =− g , g = g, g =− g 휇−1 휖 Since both g and g of a skewon medium are antisymmetric Gibbsian dyadics, they do not have inverses. 7.18 Applying the relations (5.102)–(5.105), the medium bidyadic of the special principal–axion medium can be written as 𝖬 훼 𝜺 훽 𝜺 ⌊훽T ⌋ 𝜺 훽 = + 4 ∧ ∧ e4 =− 123 e123 + 4 ∧ ∧ e4, and the corresponding modified medium bidyadic as 𝖬 ⌊ 𝜺 ⌊훽T ⌋ ⌊ 𝜺 훽 m =−eN ( 123 e123) + eN ( 4 ∧ ∧ e4) 훽T ⌋ ⌊훽 = e4 ∧ e123 − e123 ∧ e4 ⌊훽 T ⌊훽 =−(e123 ∧ e4) − e123 ∧ e4, which is symmetric. The dispersion dyadic becomes 𝖣 𝝂 𝖬 ⌊⌊𝝂𝝂 훽T ⌋ ⌊⌊𝝂𝝂 ⌊훽 ⌊⌊𝝂𝝂 ( ) = m = (e4 ∧ e123) − (e123 ∧ e4) 𝝂⌋ 훽T ⌋ ⌊𝝂 𝝂⌋ ⌊훽 ⌊𝝂 =− (e4 ∧ e123) + (e123 ∧ e4) 𝝂|훽T 𝝂| 훽T ⌋ ⌊𝝂 =−((e4( ) − ( e4) ) e123) 𝝂⌋ ⌊ 훽|𝝂 𝝂| 훽 + (e123 (( )e4 − ( e4) )), 𝖠 𝖠T = s + s + ase4 + e4as, 𝖠 E E where the spatial dyadic s ∈ 1 1 and the spatial vector as are 𝖠 |𝝂 훽T 𝝂 ⌋ ⌊ 𝝂 훽|𝝂 . defined by s =−(e4 )( ∧ ) e123 and as = e123 ( ∧ ) The dyadic dispersion equation becomes 𝖣(3) 𝝂 𝖠 𝖠T (3) 𝖠 𝖠T (2)∧ ( ) = ( a + s ) + ( a + s ) ∧(ase4 + e4as) 𝖠 𝖠T ∧ ∧ . −( a + s )∧(asas∧e4e4) = 0 This is equivalent to four spatial equations as 𝖠 𝖠T (3) ( s + s ) = 0, 𝖠 𝖠T (2) ( a + s ) ∧ as = 0, 𝖠 𝖠T (2) as ∧ ( s + s ) = 0, 𝖠 𝖠T ∧ . ( s + s )∧asas = 0 APPENDIX A Solutions to Problems 337

𝖠 From the last one we conclude that the dyadic s must satisfy 𝖠 𝖠T . s + s = asbs + bsas

for some vector bs. It turns out that the other three equations are then satisfied automatically. Thus, the last equation carries allthe |𝝂 휈 ≠ information of the dispersion equation. For e4 = 4 0 it can be expressed as [ ] ⌊ 𝝂 훽 훽T 𝝂 ⌋ ∧ ⌊ 𝝂 훽|𝝂 e123 ( s ∧ ) − ( ∧ s) e123 ∧(e123 ( s ∧ s)) ⌊ 𝝂 훽|𝝂 (e123 ( s ∧ s) = 0, 𝝂 𝝂 where s is the spatial part of . After a few applications of the bac-cab rule the dyadic dispersion equation can be reduced to ⌊⌊𝝂 𝝂 ⌊ 𝝂 훽|𝝂 훽2|𝝂 (e123e123 s s)e123 ( s ∧ ( s) ∧ ( s)) = 0, 𝝂 ≠ or, for s 0to 𝝂 훽|𝝂 훽2|𝝂 . s ∧ ( s) ∧ ( s) = 0 𝜺 It appears that, for a given spatial one-form s, there is no solution 𝝂 휈𝜺 𝜺 훽|𝜺 훽2|𝜺 s = s because s, s,and s are not linearly dependent in general. Because, for such a 𝝂, 𝖣−1(𝝂) exists, the fields of the plane wave must vanish. However, when 𝝂 is an eigen one- form of 훽, or a linear combination of two eigen one-forms, the dispersion equation is satisfied. 7.19 The medium conditions for the Gibbsian fields become ⌊ ⌊훼| 𝜺 ⌊ Dg = e123 D = e123 ( 123 Bg) 훽T | 훽T ⋅ =− Bg =− g Bg, 𝖦 | 𝖦 |훽| 훽 ⋅ . Hg = s H = s E = g Eg Here we define 훽 𝖦 |훽 𝖦 . g = s , s = e1e1 + e2e2 + e3e3 The equations for a time-harmonic plane wave are now 휔. p × Eg = Bg, p × Hg =−Dg, p = k∕ Applying the medium conditions we can expand 훽 ⋅ 훽T ⋅ 훽T ⋅ p × Hg = p × g Eg =−Dg = g Bg = ( g × p) Eg, 338 APPENDIX A Solutions to Problems

whence the field satisfies 𝖡 ⋅ 𝖡 훽 훽T 훽 훽 T . g(p) Eg = 0, g = p × g − g × p = p × g + (p × g) Applying 훽 (2) 𝖨 (2) ⋅ 훽(2) ⋅ 훽(2) 훽 훽−1T (p × g) = (p × ) g = pp g = det g g , 훽 where we assume that g is of full rank, we can expand 1 det𝖡 = 훽 ×훽 : 훽 g 6 g× g g 훽 (2) 훽T 훽 훽T (2) =−(p × g) :( g × p) + (p × g):( g × p) 훽 (2) 훽T ⋅ 훽(2) 훽T =−2(p × g) :( g × p) =−2(pp g ):( g × p) 훽 ⋅ 훽−1T ⋅ 훽 ⋅ ==2det g p g (p × g p) 훽 훽−1 ⋅ ⋅ 훽 ⋅ . = 2det g ( g p) × p ( g p) 훽 ⋅ The determinant vanishes only when the three vectors p, g p, 훽−1 ⋅ and g p are linearly dependent. Since for p = pu where u is 𝖡 a unit vector, the determinant of the dyadic g does not vanish in general, we must have Eg = 0 which implies vanishing of all field components. 7.20 Applying b|𝝓 = 0 in the dispersion equation (7.178) we obtain 𝝂 ∧ (𝝂|𝖡) ∧ 𝝓 = 0, whence the potential must be of the form 𝝓 = 훼𝝂 + 훽𝝂|𝖡. Thus, we must have b|𝝓 = 훼b|𝝂 + 훽𝝂|𝖡|b = 훽𝝂|(𝖡⌊B)|𝝂 = 0 because of b|𝝂 = 0.

CHAPTER 8

8.1 Applying the 3D inverse rule (2.137) for a spatial metric bidyadic 𝖢 F F s ∈ 2 2, ⌊⌊𝖢T (2) (e123e123 ) 𝖢−1 = s , s 𝜺 𝜺 || ⌊⌊𝖢 (3) 123 123 (e123e123 s) APPENDIX A Solutions to Problems 339

𝖡 𝖢−1 E E 𝖢 𝖡−1 and substituting s = s ∈ 2 2 and s = s , yields the rule 𝖡 for a full-rank spatial metric bidyadic s expressed as a double- wedge square of some spatial dyadic. This is not the case for the general bidyadic in the 4D space. 8.2 Starting from the expressions (8.31)–(8.34), we have

휇−1 =−𝜺 ⌊𝖰(2) =−Δ 𝖰−1T ⌋e , 123 s Qs 123 whence the Gibbsian dyadic becomes 1 휇 = e ⌊휇 =− e ⌊(𝜺 ⌊𝖰T ) = 휇𝖰T . g 123 Δ 123 123 s s Qs The other medium dyadics are obtained as 1 휉 = e ⌊휉 = e ⌊훼|휇 =− (𝖰 ∧ a )|(𝜺 ⌊𝖰T ) g 123 123 Δ s s 123 s Qs 𝖰 |𝖰(−2)⌋ 𝖨 |𝖰−1 ⌋ =−( s ∧ as) s e123 =−( s ∧ (as s )) e123 𝖨 ⌋ |𝖰−1 ⌋ 𝖨 ⌋ =− s (as s ) e123 = s Xs 1 휁 = e ⌊휁 =−e ⌊휇|훽 = 𝖰T |(𝜺 ⌊(b ∧ 𝖰 ) g 123 123 Δ s 123 s s Qs ⌊𝖰(−2)| 𝖰 ⌊ 𝖰−1| 𝖨T = e123 (bs ∧ s) = e123 (( bs) ∧ s ) ⌊ 𝖰−1| ⌊𝖨T ⌊𝖨T 𝖨 ⌋ = e123 ( bs) s = Zs s = s Zs 휖 ⌊ 휖′ 훼|휇|훽 𝖰 휉 |훽 g = e123 ( − ) = c s − bsas − g 1 = c𝖰 − b a + (𝖰 ∧ a )|(𝜺 ⌊𝖰T )|(e ⌊(b ∧ 𝖰 )). s s s Δ s s 123 s 123 s s Qs The last expression can be expanded as 1 (𝖰 ∧ a )|(𝜺 ⌊𝖰T )|(e ⌊(b ∧ 𝖰 )) Δ s s 123 s 123 s s Qs 𝖰 |𝖰(−2)| 𝖰 = ( s ∧ as) s (bs ∧ s) 𝖰 | 𝖰−1| 𝖨T |𝖰−1| 𝖰 = ( s ∧ as) (( s bs) ∧ s ) = bsas − (as s bs) s, whence 휖 𝖰 |𝖰−1| 𝖰 휖𝖰 . g = c s − (as s bs) s = s 340 APPENDIX A Solutions to Problems

8.3 Let us assume Gibbsian medium dyadics of the form 휖 휖𝖶 휇 휇𝖶T 휉 𝖨 ⌋ ⌊𝖨T 휁 𝖨 ⌋ ⌊𝖨T g = , g = , g = s X = X s , g = s Z = Z s , 𝖶 E E where ∈ 1 1 is a spatial metric dyadic and X, Z are two spatial bivectors. The task is to build the corresponding modified 𝖬 F E bidyadic m ∈ 2 2 by applying the expansion (5.88) 𝖬 휖 ∧ ⌊𝖨T 휉 |휇−1| 𝖨 ⌋ 휁 . m = g∧e4e4 − (e123 s + e4 ∧ g) g ( s e123 − g ∧ e4) Expanding 휖 ∧ 휖𝖶∧ g∧e4e4 = ∧e4e4 1 1 휇−1 = 𝖶−1T = 𝜺 𝜺 ⌊⌊𝖶(2) g 휇 휇 123 123 ΔW

⌊𝖨T 휉 ⌊𝖨T ⌊𝖨T e123 s + e4 ∧ g = e123 s + e4 ∧ (X s ) 𝖨 ⌋ 휁 𝖨 ⌋ 𝖨 ⌋ s e123 − g ∧ e4 = s e123 − ( s Z) ∧ e4 and, inserting the in the above expression, when defining 휇 = −1∕ΔW for simplicity, we have 𝖬 휖𝖶∧ m = ∧e4e4 ⌊𝖨T ⌊𝖨T |𝜺 𝜺 ⌊⌊𝖶(2)| 𝖨 ⌋ 𝖨 ⌋ +(e123 s + e4 ∧ (X s )) 123 123 ( s e123 − ( s Z) ∧ e4) 휖𝖶∧ 𝖶(2) 𝖶(2)⌋𝜺 ⌋ = ∧e4e4 + − (( 123) Z) ∧ e4 ⌊ 𝜺 ⌊𝖶(2) ⌊ 𝜺 ⌊𝖶(2)⌋𝜺 ⌋ . +e4 ∧ (X ( 123 )) − e4 ∧ (X ( 123 123) Z) ∧ e4 𝜶 𝜺 ⌊ ⌋𝜺 𝜷 𝜺 ⌊ ⌋𝜺 Defining = 123 X = X 123 and = 123 Z = Z 123,we can expand further ⌊ 𝜺 ⌊𝖶(2)⌋𝜺 ⌋ ⌋𝜺 ⌋𝖶(2)⌊ 𝜺 ⌊ X ( 123 123) Z = (−X 123) (− 123 Z) = 𝜶⌋𝖶(2)⌊𝜷 =−𝖶(2)⌊⌊𝜶𝜷 =−(𝖶||𝜶𝜷)𝖶 + (𝖶|𝜷)(𝜶|𝖶) ⌊ 𝜺 ⌊𝖶(2) 𝜶⌋𝖶(2) 𝖶 𝜶|𝖶 X ( 123 ) =− =− ∧ ( ) 𝖶(2)⌋𝜺 ⌋ 𝖶(2)⌊𝜷 𝖶|𝜷 𝖶. ( 123) Z =− =−( ) ∧ APPENDIX A Solutions to Problems 341

Denoting a = 𝜶|𝖶, b = 𝖶|𝜷 and combining these we have

𝖬 휖𝖶∧ 𝖶(2) 𝖶 𝖶 m = ∧e4e4 + − b ∧ ∧ e4 + e4 ∧ ∧ a 𝖶||𝜶𝜷 𝖶 +e4 ∧ (( ) − ba) ∧ e4 휆𝖶∧ 𝖶(2) 𝖶∧ 𝖶∧ ∧ = ∧e4e4 + + ∧be4 − ∧e4a − e4a∧be4 𝖶 휆 (2) 휆 휖 𝖶||𝜶𝜷. = ( + be4 − e4a + e4e4) , = − 𝖬 Thus, m has the form of a Q-medium bidyadic. 8.4

𝝂 𝜺 𝜺 || 𝖰(2)∧ 𝖰(2)⌊⌊𝝂𝝂 (2) 3D( ) = N N ( ∧( ) ) 𝜺 𝜺 || 𝖰(2)∧ 𝖰||𝝂𝝂 𝖰 𝖰|𝝂𝝂|𝖰 (2) = N N ( ∧(( ) − ) ) 𝖰||𝝂𝝂 𝜺 𝜺 || 𝖰(2)∧ 𝖰||𝝂𝝂 𝖰(2) 𝖰∧ 𝖰|𝝂𝝂|𝖰 = ( ) N N ( ∧(( ) − ∧( ))) 𝖰||𝝂𝝂 𝖰||𝝂𝝂 𝜺 𝜺 ||𝖰(4) 𝜺 𝜺 ⌊⌊𝖰(3) = ( )(( )6 N N − 3( N N ) ||(𝖰|𝝂𝝂|𝖰)) 𝖰||𝝂𝝂 𝜺 𝜺 ||𝖰(4) 𝖰||𝝂𝝂 𝖰−1T || 𝖰|𝝂𝝂|𝖰 = ( )( N N )(6( ) − 3 ( )) 𝖰||𝝂𝝂 2 𝜺 𝜺 ||𝖰(4) . = 3( ) ( N N ) = 0 8.5 Let us expand

𝖣(𝝂)|𝝓 =−(휈⌋𝖰(2)⌊𝝂)|(𝝂|𝖰⌋𝚪) = 휈⌋(𝖰(2)⌊(𝝂|𝖰⌋𝚪))|𝝂 =−휈⌋(𝖰(2)⌊(𝚪⌊(𝝂|𝖰))|𝝂,

and invoke the bac-cab rule

A⌊(𝚪⌊a) = (A ∧ a)⌊𝚪 − (A|𝚪)a

in the form

𝖰(2)⌊(𝚪⌊(𝝂|𝖰)) = (𝖰(2) ∧ (𝝂|𝖰))⌊𝚪 − (𝖰(2)|𝚪)𝝂|𝖰.

Inserting this, we can proceed as

𝖣(𝝂)|𝝓 =−휈⌋((𝖰(2) ∧ (𝝂|𝖰))⌊𝚪 − (𝖰(2)|𝚪)𝝂|𝖰)|𝝂 =−(휈 ∧ 𝝂)⌋𝖰(3)⌊𝚪 + (𝝂⌋𝖰(2)|𝚪)(𝝂|𝖰|𝝂).

Both terms of the last expression can be seen to vanish. 342 APPENDIX A Solutions to Problems

√ 𝖬 𝖰(2) 𝖤 8.6 Let us assume that a bidyadic m = + ΔQ is symmetric. Expanding √ √ 𝖰(2) 𝖤 T ⋅ 𝖰(2) 𝖤 ( + ΔQ ) ( − ΔQ ) √ 𝖰(2)T ⋅ 𝖰(2) 𝖰(2) 𝖰(2)T 𝖤 = + ΔQ( − ) −ΔQ ⌊𝖰(−2)|𝖰(2) ⌊𝖨(2)T =ΔQeN −ΔQeN = 0, √ 𝖰(2) 𝖤 from this we can conclude that the bidyadic ± ΔQ cannot have an inverse for either sign. In fact, assuming that one of these has an inverse, by multiplying this expression by the inverse, vanishing of the other one follows which would require that 𝖰(2) ⌊𝖨(2)T be a multiple of eN which is not possible (see Problem 4.14). 8.7 The Gibbsian relations (8.122)–(8.125) can be expressed more compactly in dyadic-matrix form as ( ) ( )[ ( ) ] 휖 휉 휖 휇 g g 0  −p1∕ 휁 휇 = 휇 + 휖 (q1 q2) , g g − 0 p2∕ where  is the antisymmetric dyadic matrix ( ) 𝖨⌋ 휇 𝖰T  − Zs∕ − s . = 𝖰 𝖨⌋ 휖 s Xs∕ Since fields of a plane wave in any medium satisfy the orthogo- ⋅ ⋅ nality conditions Eg Bg = 0andHg Dg = 0, we can form the following chain of relations in Gibbsian symbols: 1 ⋅ 1 ⋅ 0 =−휇 Eg Bg + 휖 Hg Dg 1 ⋅ 휁 ⋅ 휇 ⋅ 1 ⋅ 휖 ⋅ 휉 ⋅ =−휇 Eg ( g Eg + g Hg) + 휖 Hg ( g Eg + g Hg) ( )( ) ( ) 0 −1∕휇 휖 휉 E = (E H ) ⋅ g g ⋅ g g g 1∕휖 0 휁 휇 H [ ( ) g ]g ( )g −p ∕휇 E = (E H ) ⋅  + 1 (q q ) ⋅ g g g p ∕휖 1 2 H ( )2 ( ) g −p ∕휇 E = (E H ) ⋅ 1 (q q ) ⋅ g g g p ∕휖 1 2 H ( 2 ) g 1 ⋅ 1 ⋅ ⋅ ⋅ . = −휇 p1 Eg + 휖 p2 Hg (q1 Eg + q2 Hg) APPENDIX A Solutions to Problems 343

Thus, the Gibbsian field vectors must satisfy either of the two conditions 휖 ⋅ 휇 ⋅ ⋅ ⋅ p1 Eg − p2 Hg = 0, q1 Eg + q2 Hg = 0, which for the field one-forms take the corresponding form of 휖 | 휇 | | | . p1 E − p2 H = 0, q1 E + q2 H = 0 8.8 Expanding 𝖬⌊⌊𝖨 = 𝖯(2)⌊⌊𝖨 = (tr𝖯)𝖯 − 𝖯2 = 0, we may distinguish the two possibilities, tr𝖯 = 0andtr𝖯 ≠ 0. Assuming first tr𝖯 ≠ 0, we can define  = 𝖯∕tr𝖯 satisfying 2 =  and tr = 1. Thus,  is an example of a projection dyadic. From (4.42) and (4.45) it follows that this one must be of the form  = p𝝅 with p|𝝅 = 1. Thus, we have 𝖯 = tr𝖯(p𝝅)which implies 𝖬 = 𝖯(2)T = 0. This leaves us the second choice, tr𝖯 = 0. In this case 𝖯 satisfies 𝖯2 = 0, whence it must be nilpotent dyadic. From (4.19) we know that such a dyadic can be expressed as 𝖯 𝝅 𝝅 𝝅 | = p1 1 + p2 2, i pj = 0, i, j = 1, 2 Thus, any principal P-medium dyadic must be of the form 𝖬 𝚫 𝚫 𝝅 𝝅 . = A, = 1 ∧ 2, A = p1 ∧ p2 This is a valid solution, since it satisfies 𝖬⌊⌊𝖨 𝝅 | 𝝅 𝝅 | 𝝅 𝝅 |𝝅 𝝅 |𝝅 = ( 1 p1) 2p2 + ( 2 p2) 1p1 − 1p1 2p2 − 2p2 1p1 = 0. 8.9 For symmetric 𝖰(2) we can expand 𝖩T2 𝜺 ⌊𝖰(2) | 𝜺 ⌊𝖰(2) 𝜺 𝜺 ⌊⌊𝖰(2) |𝖰(2) = ( N ) ( N ) = ( N N ) 𝖰(−2)T |𝖰(2) 𝖨(2)T =ΔQ =ΔQ 𝖩T2 𝖨(2)T whence√ for ΔQ =−1wehave =− . Defining M = A2 + B2 and tan 휓 = B∕A, we can write

𝖩T 𝖬 = M(cos 휓 𝖨(2)T + sin 휓 𝖩T ) = Me휓 344 APPENDIX A Solutions to Problems

8.10 Expanding

𝖩T 𝖬2 = M2e2휓 = M2(cos 2휓 𝖨(2)T + sin 2휓 𝖩T ) = M2(2 cos2 휓 𝖨(2)T + 2cos휓 sin 휓 𝖩T − 𝖨(2)T ) = 2M cos 휓 𝖬 − M2𝖨(2)T ,

the equation satisfied by the bidyadic 𝖬 can be written as

(𝖬 − Mej휓 𝖨(2))|(𝖬 − Me−j휓 𝖨(2)) = 0.

⌊ 𝖨(2)T Here we have applied the symmetry of the bidyadic eN ( + 𝖩T ⌊𝖨(2)T 𝖰(2) j ) = eN + j . 8.11 Substituting the decomposition of the previous problem in terms 𝚽 of the eigenfields ± in 𝖬|𝚽 𝜸 𝚽 𝜸 d ∧ ( ) = e, d ∧ = m, APPENDIX A Solutions to Problems 345

the resulting equations 𝚽 𝚽 𝜸 M+d ∧ + + M−d ∧ − = e, 𝚽 𝚽 𝜸 d ∧ + + d ∧ − = m, can be split in two uncoupled equations as 𝚽 𝜸 𝚽 𝜸 d ∧ + = +, d ∧ − = − with source three-forms 𝜸 𝜸 𝜸 . ∓ =±(M± m − e)∕(M+ − M−)

The bidyadic 𝖴(𝚿, 𝚽) =−𝖴(𝚽, 𝚿) vanishes for each eigenfield, 𝖴 𝚿 𝚽 𝖴 𝚽 𝚽 ( ±, ±) = M± ( ±, ±) = 0, whence the eigenfields alone do not carry any energy. For total fields, energy transportation is possible through the interaction of both eigenfields: 𝖴 𝚿 𝚽 𝖴 𝚿 𝚽 𝖴 𝚿 𝚽 ( , ) = ( +, −) + ( −, +) 𝖴 𝚽 𝚽 𝖴 𝚽 𝚽 = M+ ( +, −) + M− ( −, +) 𝖴 𝚽 𝚽 . = (M+ − M−) ( +, −) 8.12 The dispersion function corresponding to the extended Q-medium has the form 𝝂 𝜺 𝜺 || 𝖰(2) ∧ 𝝂𝝂⌋⌋ 𝖰(2) (2) . D( ) = N N (( + AB)∧( ( + AB)) ) Applying the general rule 𝝂𝝂⌋⌋𝖰(p) 𝝂𝝂||𝖰 𝖰(p−1) 𝖰(p−2)∧ 𝖰|𝝂𝝂|𝖰 = ( ) − ∧( ), 𝖰(p−2)∧ where ∧ can be omitted for p = 2, and expanding 𝝂𝝂⌋⌋ 𝖰(2) (2) 𝝂𝝂⌋⌋𝖰(2) (2) 𝝂𝝂⌋⌋𝖰(2) ∧ 𝝂𝝂⌋⌋ ( ( + AB)) = ( ) + ( )∧( AB), we obtain the four-term expression ( ∑ 𝝂 𝜺 𝜺 || 𝖰(2)∧ 𝝂𝝂⌋⌋𝖰(2) (2) ∧ 𝝂𝝂⌋⌋𝖰(2) (2) D( ) = Di = N N ( ∧( ) + (AB)∧( ) ) 𝖰(2)∧ 𝝂𝝂⌋⌋𝖰(2) ∧ 𝝂𝝂⌋⌋ ∧ 𝝂𝝂⌋⌋𝖰(2) ∧ 𝝂𝝂⌋⌋ . + ( ∧( )∧( AB) + (AB)∧( )∧( AB) 346 APPENDIX A Solutions to Problems

The fourth term vanishes identically:

𝜺 𝜺 || ∧ 𝝂𝝂⌋⌋𝖰(2) ∧ 𝝂𝝂⌋⌋ D4 = N N ((AB)∧( )∧( AB)) 𝝂𝝂⌋⌋𝖰(2) ∧ 𝝂⌋ 𝝂⌋ = ( )∧(A ∧ ( A)(B ∧ ( B)) 1 = 𝜺 𝜺 ||((𝝂𝝂⌋⌋𝖰(2))∧(𝝂𝝂⌋⌋(AB∧AB))) 4 N N ∧ ∧ 1 = (𝜺 𝜺 ||(AB∧AB))(𝝂𝝂⌋⌋𝖰(2))∧(𝝂𝝂⌋⌋e e ) 4 N N ∧ ∧ N N 1 = (𝜺 𝜺 ||(AB∧AB))(𝝂𝝂⌋⌋𝖰(2))∧(𝝂𝝂⌋⌋e e ) = 0, 4 N N ∧ ∧ N N 𝝂⌋ 𝝂⌋ because ( C) ∧ ( eN) = 0 for any bivector C (check by setting 𝝂 𝜺 = 1). Applying the expansion

𝝂𝝂⌋⌋𝖰(2) (2) 𝝂𝝂||𝖰 2𝖰(2) 𝝂𝝂||𝖰 𝖰∧ 𝖰|𝝂𝝂|𝖰 ( ) = ( ) − ( ) ∧( ) = (𝝂𝝂||𝖰)(𝝂𝝂⌋⌋𝖰(3)) 𝝂𝝂||𝖰 𝝂𝝂⌋⌋ ⌊⌊𝖰−1T 𝜺 𝜺 ||𝖰(4) = ( )( (eNeN ))( N N ),

the first term becomes

𝜺 𝜺 || 𝖰(2)∧ 𝝂𝝂⌋⌋𝖰(2) (2) D1 = N N ( ∧( ) ) 𝝂𝝂||𝖰 𝜺 𝜺 ||𝖰(4) 𝜺 𝜺 || 𝖰(2)∧ 𝝂𝝂⌋⌋ ⌊⌊𝖰−1T = ( )( N N ) N N ( ∧( (eNeN ))) 𝝂𝝂||𝖰 𝜺 𝜺 ||𝖰(4) 𝖰(2)|| 𝜺 𝜺 ⌊⌊ ⌊⌊ 𝖰−1T ∧𝝂𝝂 = ( )( N N )( ( N N (eNeN ( ∧ ))) 𝝂𝝂||𝖰 𝜺 𝜺 ||𝖰(4) 𝖰(2)⌊⌊ 𝖰−1T ||𝝂𝝂 = ( )( N N )( ( ) 𝝂𝝂||𝖰 𝜺 𝜺 ||𝖰(4) 𝖰 𝖰|𝖰−1|𝖰 ||𝝂𝝂 = ( )( N N )(4 − ) 𝝂𝝂||𝖰 2 𝜺 𝜺 ||𝖰(4) . = ( ) ( N N )

This yields D(𝝂) for the Q-medium case, that is, for AB = 0. For the generalized Q-medium the second and third terms must be added. The second term can be expanded as

𝜺 𝜺 || ∧ 𝝂𝝂⌋⌋𝖰(2) (2) D2 = N N ((AB)∧( ) ) 𝝂𝝂||𝖰 𝜺 𝜺 ||𝖰(4) 𝜺 𝜺 || ∧ 𝝂𝝂⌋⌋ ⌊⌊𝖰−1T = ( )( N N ) N N ((AB)∧( (eNeN ))) 𝝂𝝂||𝖰 𝜺 𝜺 ||𝖰(4) 𝜺 𝜺 ⌊⌊ || ⌊⌊ 𝖰−1T ∧𝝂𝝂 = ( )( N N )(( N N AB) (eNeN ( ∧ )) 𝝂𝝂||𝖰 𝜺 𝜺 ||𝖰(4) ⌊⌊𝖰−1T ||𝝂𝝂 = ( )( N N )(AB ) , APPENDIX A Solutions to Problems 347

and the third term as 𝜺 𝜺 || 𝖰(2)∧ 𝝂𝝂⌋⌋𝖰(2) ∧ 𝝂𝝂⌋⌋ D3 = N N ( ∧( )∧( AB)) 𝜺 𝜺 || 𝖰||𝝂𝝂 𝖰(3) 𝖰(2)∧ 𝖰|𝝂𝝂|𝖰 ∧ 𝝂𝝂⌋⌋ = N N (3( ) − ∧( ))∧( AB)) 𝜺 𝜺 ⌊⌊ 𝖰||𝝂𝝂 𝖰(3) 𝝂𝝂⌋⌋𝖰(4) || 𝝂𝝂⌋⌋ = ( N N (2( ) + ) ( AB) 𝖰||𝝂𝝂 𝜺 𝜺 ⌊⌊𝖰(3) || 𝝂𝝂⌋⌋ = 2( )( N N ) ( AB) 𝖰||𝝂𝝂 𝜺 𝜺 ||𝖰(4) 𝖰−1T ⌋⌋ ||𝝂𝝂. = 2( )( N N )( AB) Summing up, we finally obtain 𝝂 𝜺 𝜺 ||𝖰(4) 𝝂𝝂||𝖰 𝖰 ⌊⌊𝖰−1T ||𝝂𝝂 D( ) = ( N N )( )( + AB ) , as the dispersion equation for the generalized Q-medium. 8.13 Expanding the first one of the two terms in (8.167), with (8.166) inserted, yields 𝜺 | 𝖰(2) | 𝝂 𝝂 ∧𝖰−1 | 𝜺 | 𝖰|𝝂 𝝂 ∧ 𝖰|𝖰−1 | N (A ∧ ) ( 2 2∧ ) A = N (A ∧ (( 2 2)∧( )) A 𝜺 | 𝖰|𝝂 𝖨 𝝂 | 𝜺 | 𝝂 ⌋ 𝖰|𝝂 =− N ((A ∧ ( 2) ∧ ) ∧ 2) A = N (A ∧ ( 2 A) ∧ ( 2)) 1 1 = (A ⋅ A)𝜺 |((𝝂 ⌋e ) ∧ (𝖰|𝝂 )) = (A ⋅ A)(𝝂 |𝖰|𝝂 ) 2 N 2 N 2 2 2 2 here we have applied the rule 1 1 A ∧ (𝝂 ⌋A) = 𝝂 ⌋(A ∧ A) = 𝝂 ⌋e (A ⋅ A). 2 2 2 2 2 N The second term can be expanded as 1 1 (A ⋅ A)B|(𝝂 𝝂 ∧𝖰−1)|A = (A ⋅ A)(𝝂 𝝂 ∧𝖰−1)||BA 2 2 2∧ 2 2 2∧ 1 1 = (A ⋅ A)𝝂 𝝂 ||(𝖰−1⌋⌋BA) =− (A ⋅ A)𝖰||𝝂 𝝂 , 2 2 2 2 2 2 the last step of which is due to the dispersion equation (8.156). Summing the two results yields zero, whence the condition (8.167) is verified. 8.14 Assuming that 훼 has a spatial inverse 훼−1, the medium conditions for the P-medium can be written in the form ( ) ( ) ( ) ( ) ( ) B 𝖠 𝖡 D 훼−1 −훼−1|휖′ D = | = | H 𝖢 𝖣 E 휇−1|훼−1 훽 − 휇−1|훼−1|휖′ E 348 APPENDIX A Solutions to Problems

whence, inserting (8.44)–(8.47), the medium dyadics can be expanded as 𝖠 = 𝖯(−2)T 𝖡 𝖯(−2)T | 𝝅 𝖯T 𝖯−1T |𝝅 𝖨T =− ( s ∧ s ) = ( s s) ∧ s 𝖢 𝖯T |𝖯(−2)T 𝖨T |𝖯−1T =−( s ∧ ps) s =− s ∧ (ps s ) 𝖣 𝝅 𝖯T 𝖯T |𝖯(−2)T | 𝝅 𝖯T = sps − p s − ( s ∧ ps) s ( s ∧ s ) 𝝅 |𝖯−1| 𝖯T . = ( s s ps − p) s 𝖠 𝖣 The corresponding expressions for the Gibbsian dyadics g − g are obtained (see Appendix B) through ( ) ( ) 𝖠 𝖡 ⌊𝖠⌋𝜺 |𝖦 ⌊𝖡 g g (e123 123) s e123 | 𝖢 𝖣 = 𝖦 |𝖢⌋𝜺 |𝖦 𝖦 |𝖣 g g ( s 123) s s after which (8.50)–(8.53) are straightforwardly obtained. 8.15 In terms of the spatial expansions (8.172) for the bivectors C and D the orthogonality condition (8.191) can be expressed as | 𝜸 | e123 ( s ∧ B) + cs E = 0, while the one corresponding to (8.192) has a more complicated form. Let us first expand ⋅ ⌊𝜹 |𝜺 ⌊ ⌊𝜹 D D = (e123 s + ds ∧ es) N (e123 s + ds ∧ es) ⌊𝜹 | 𝜺 𝜹 𝜺 ⌊ 𝜹 | = (e123 s + ds ∧ es) (− 4 ∧ s + 123 ds) = 2 s ds, and insert it, together with the spatial expansion (8.42) in (8.192). The result can be written as | 𝜹 𝖯(2)T |𝖯T 𝜹 | ⌊𝜸 | (e123 ( s ∧ s + ds s ∧ ps − ( s ds)e123 s) B | 𝜹 𝝅 𝖯T |𝖯T |𝝅 𝜹 | | . +(−(e123 ( s ∧ s s ) + pds s − (ds s)ps − ( s ds)cs) E = 0 8.16 From (8.143) we obtain 1 𝖬 = (𝖰(2) + AB)−1⌋e d Z2 N 𝖰(−2)| |𝖰(−2) 1 𝖰(−2)⌋ 1 ( A)(B )⌋ = eN − eN Z2 Z2 1 − B|𝖰(−2)|A APPENDIX A Solutions to Problems 349

and (𝖰(2)T ⌋𝜺 |A)(B|𝜺 ⌊𝖰(2)T ) 𝖬 1 𝖰(2)T 1 N N md = − Z2Δ Z2Δ |𝜺 ⌊𝖰(2)T |⌋𝜺 Q Q ΔQ − B N NA 1 1 (𝖰(2)T ⌋𝜺 |A)(B|𝜺 ⌊𝖰(2)T ) = 𝖰(2)T − N N Z2Δ Z2Δ |𝜺 ⌊𝖰(2)T |⌋𝜺 Q Q ΔQ − B N NA 1 1 (A ⋅ 𝖰(2))(𝖰(2) ⋅ B) = 𝖰(2)T − . Z2Δ Z2Δ ⋅ 𝖰(2) ⋅ Q Q ΔQ − A B The self-dual property requires 𝖰(2) 𝖰(2)T 2 ⇒ 𝖰(2)T 𝖰(2) 2 = ∕Z ΔQ =± , Z =±1∕ΔQ whence 𝖰(2) must be a symmetric or an antisymmetric bidyadic. Further, from ⋅ 𝖰(2) 𝖰(2) ⋅ (A )( B). AB = AdBd =∓ ⋅ 𝖰(2) ⋅ ΔQ − A B It follows that A and B must be eigenbivectors of the bidyadic 𝖰(2) ⋅ 𝖰(2) 휆 𝖰(2) ⋅ 휆 . A = aA, B = bB Because of ⋅ 𝖰(2) ⋅ 휆 ⋅ 휆 ⋅ A B = aA B = bA B, ⋅ 휆 휆 휆 either A B = 0or a = b = . In the latter case the eigenvalue must satisfy the equation 휆2 ⋅ 𝖰(2) ⋅ 휆 ⋅ . =∓(ΔQ − A B) =∓ΔQ ± (A B) 8.17 Plane-wave fields in a P-medium satisfy 𝝂 ∧ 𝚿 = 𝝂 ∧ (𝖯T |𝝂) ∧ (𝖯T |𝝓) = 0, whence there is no restriction to 𝝂 and, for some coefficients A, B, the potential and fields satisfy 𝝓 = 𝖯−1T |(A𝝂 + B(𝖯T |𝝂)) 𝚽 = 𝝂 ∧ 𝖯−1T |(A𝝂 + B(𝖯T |𝝂)) = A𝝂 ∧ (𝖯−1T |𝝂) 𝚿 = 𝖯(2)T |𝚽 = A(𝖯T |𝝂) ∧ 𝝂. 350 APPENDIX A Solutions to Problems

Inserting the special dyadic 𝖯 whose inverse can be expressed as 1 𝖯−1 = (e 𝜺 + e 𝜺 ) + 𝖯′ P 1 1 2 2 34 𝖯′ 𝜺 𝜺 𝖯 −1 34 = (Pe3 3 + Pe4 4 + 34) 1 = ((P + P )e 𝜺 − P e 𝜺 − P e 𝜺 + (P + P )e 𝜺 ) Δ 4,4 3 3 3,4 3 4 4,3 4 3 3,3 4 4 Δ=(P + P3,3)(P + P4,4) − P3,4P4.3, we obtain A 𝚽 = (𝝂 ∧ (−𝜺 e − 𝜺 e )|𝝂 + A𝝂 ∧ (𝖯′ T |𝝂) P 3 3 4 4 34 𝚿 𝖯 T |𝝂 𝝂. = A( 34 ) ∧ It follows that the field two-forms satisfy in the medium 𝜺 𝚽 𝜺 ⇒ 𝜺 34 ∧ = 34 ∧ B = 0, 3 ∧ B = 0, 𝜺 𝚿 𝜺 ⇒ 𝜺 34 ∧ = 34 ∧ D = 0, 3 ∧ D = 0, 𝜺 | . producing DB-boundary conditions at an interface 3 x = 0

CHAPTER 9

9.1 Expanding 𝖬2 2𝖨(2)T ⋅ 𝜺 ⌊ = M + (2M + A B) N AB = (2M + A ⋅ B)𝖬 − M(M + A ⋅ B)𝖨(2)T

the bidyadic 𝖬 is seen to satisfy the second-order equation. 9.2 Inserting the bidyadic 𝖣 and its inverse in (9.45) we can proceed as 𝖬 𝖣−1|𝖬|𝖣 1 = 훼 휖′ 𝜺 휇−1 𝜺 훽 = ( + ∧ e4 + 4 ∧ + 4 ∧ ∧ e4 훼|휇| 휇−1 훽 |𝖣 − ( + ∧ e4)) 휖′ 𝜺 휇−1 𝜺 훽 훼|휇|훽 |𝖣 = ( ∧ e4 + 4 ∧ + 4 ∧ ∧ e4 − ∧ e4) 휖′ 훼|휇|훽 𝜺 휇−1 𝜺 훽 휇−1|훼|휇 . = ( − ) ∧ e4 + 4 ∧ + 4 ∧ ( − ) ∧ e4 APPENDIX A Solutions to Problems 351

This is valid for any 𝖬. Applying 휇−1|훼 = 훽|휇−1 from (9.40) we have 𝖣−1|𝖬|𝖣 휖′ 훼|휇|훽 𝜺 휇−1. = ( ∧ e4 − ) ∧ e4 + 4 ∧ From (9.40) and (9.41) we obtain 훼|휇|훽 휇|훽2 휇| 휇−1|휖′ 2𝖨T 휖′ 2휇 = = ( − M s ) = − M , whence, finally, we arrive at the result 𝖬 2휇 𝜺 휇−1. 1 = M ∧ e4 + 4 ∧ 9.3 In analogy to the previous problem, we can define the transformation bidyadics 𝖣 𝖨(2)T 𝜺 훽|휖′−1 𝖣−1 𝖨(2)T 𝜺 훽|휖′−1 = + 4 ∧ , = − 4 ∧ and proceed similarly, 𝖬 𝖣−1|𝖬|𝖣 1 = 훼 휖′ 𝜺 휇−1 𝜺 훽 = ( + ∧ e4 + 4 ∧ + 4 ∧ ∧ e4 𝜺 훽|휖′−1| 훼 휖′ |𝖣 − 4 ∧ ( + ∧ e4)) 훼 휖′ 𝜺 휇−1 훽|휖′−1|훼 휖′|훽|휖′−1. = + ∧ e4 + 4 ∧ ( − ) − 휖′ 𝜺 휇−1 훽|휖′−1|훼 = ∧ e4 + 4 ∧ ( − ), wherewehaveapplied휖′|훽|휖′−1 = 훼 due to (9.39). Invoking (9.38) we can expand

훽|휖′−1|훼 = 휖′−1|훼2 = 휇−1 − M2휖′−1,

whence the result becomes 𝖬 휖′ 2𝜺 휖′−1. 1 = ∧ e4 + M 4 ∧ 9.4 Applying the trigonometric rules

tan(휃∕2) = sin 휃∕(1 + cos 휃) = (1 − cos 휃)∕ sin 휃 = 1∕ cot(휃∕2)

(9.33) can be written as ( ) 1 tan(휃∕2) sin 휃 M =∓ =∓ ± Z cot(휃∕2) Z(1 ± cos 휃) 352 APPENDIX A Solutions to Problems

whence (9.34) can be verified by ( ) ( )( ) 휃 휃 휃 휃 ± sin cos − 1 sin 휃 − cos −(1∕Z)sin M± Z(1±cos 휃) Z = 2 휃 −Z sin 휃 cos 휃 1 ± sin + cos 휃 1±cos 휃 ( ) 휃 ( ) − sin M = Z(1±cos 휃) =± ± . ±1 1

9.5 Assuming 휖′ = 0and휇−1 = 0, (9.38) and (9.41) yield 훼2 2𝖨(2)T 훼 𝖨(2)T | 훼 𝖨(2)T + M s = ( − jM ) ( + jM ) = 0, 훽2 2𝖨T 훽 𝖨T | 훽 𝖨T . + M s = ( − jM ) ( + jM ) = 0 Since we know (see Problem 4.12) that one of the dyadics in each product must be of rank 1 or 0, without losing generality we can set 훼 𝖨(2)T 𝚯 = jM( s − 2 A), 훽 𝖨T 𝜷 =±jM( s − 2 b), for some spatial quantities: two-form 𝚯, one-form 𝜷, bivector A, and vector b. The above conditions require 𝚯A = 0or𝚯|A = 1, 𝜷b = 0or𝜷|b = 1. The four basic cases can be treated separately as (1) 𝚯A = 0and𝜷b = 0, whence 𝖬 𝖨(2)T 𝖨 ∧𝜺 𝖨 𝜺 (2)T . = jM( s ± s∧ 4e4) = jM( s ± e4 4) In this case 𝖬 is either a multiple of 𝖨(2)T or 𝖳(2)T where 𝖳 = 𝖨 𝜺 s − e4 4 is the time reversion dyadic. (2) 𝚯A = 0, 𝜷|b = 1, whence 𝖬 𝖨 𝜺 (2)T 𝜷 ∧𝜺 . = jM(( s ± e4 4) ∓ 2 b∧ 4e4) (3) 𝚯|A = 1, 𝜷b = 0, whence 𝖬 𝖨 𝜺 (2)T 𝚯 . = jM(( s ± e4 4) − 2 A) APPENDIX A Solutions to Problems 353

(4) 𝚯|A = 𝜷|b = 1, whence 𝖬 𝖨 𝜺 (2)T 𝚯 𝜷 ∧𝜺 . = jM(( s ± e4 4) − 2( A ± b∧ 4e4)) The results can be easily verified to satisfy (9.36). 9.6 𝖨(4)⌊⌊𝖬′ ⌊ 𝖣|𝖬|𝖣−1 ⌋𝜺 = eN ( ) N ⌊𝖣⌋𝜺 | ⌊𝖬⌋𝜺 | ⌊𝖣−1⌋𝜺 = (eN N) (eN N) (eN N) = (𝖨(4)⌊⌊𝖣)|(𝖨(4)⌊⌊𝖬)|(𝖨(4)⌊⌊𝖣)−1. Here we have applied the rule ⌊𝖣−1⌋𝜺 ⌊𝖨(2)T |𝖣−1| 𝖨(2)T ⌋𝜺 eN N = (eN ) ( N) 𝜺 ⌊𝖨(2) −1|𝖣−1| 𝖨(2)⌋ −1 = ( N ) ( eN) ⌊𝖣⌋𝜺 −1 𝖨(4)⌊⌊𝖣 −1. = (eN N) = ( ) 9.7 The two bidyadics satisfy the properties 𝖬2 2𝖨(2)T 𝖬2 2𝖨(2)T 𝖬|𝖬 𝖬 |𝖬. − M , 1 − M , 1 =− 1 𝖣 𝖬 𝖬 Defining = ( + 1)∕M we have 1 1 𝖣2 = (𝖬 + 𝖬 )2 = (𝖬2 + 𝖬2) =−2𝖨(2)T , M2 1 M2 1 whence 𝖣−1 =−𝖣∕2. For this choice of 𝖣 we obtain 1 𝖣|𝖬 |𝖣−1 =− (𝖬 + 𝖬 )|𝖬 |(𝖬 + 𝖬 ) 1 2M2 1 1 1 1 =− (𝖬|𝖬 |𝖬 + 𝖬|𝖬2 + 𝖬2|𝖬 + 𝖬3) 2M2 1 1 1 1 1 =− (−𝖬2|𝖬 − M2𝖬 − M2𝖬 − M2𝖬 ) 2M2 1 1 1 =− (−2M2𝖬) = 𝖬. 2M2 9.8 Let us expand 𝖬 휖 𝜺 휇−1 = ∧ e4 + 4 ∧ , 𝖬2 휖|휇−1 𝜺 휇−1|휖 =− + 4 ∧ ∧ e4 𝖬3 휖|휇−1|휖 𝜺 휇−1|휖|휇−1. =− ∧ e4 − 4 ∧ 354 APPENDIX A Solutions to Problems

Requiring that (9.78) be satisfied leads to four conditions for the spatial medium dyadics 휖 and 휇−1, 휖|휇−1 𝖨(2)T A = C s , 휖|휇−1|휖 = B휖, 휇−1|휖|휇−1 = B휇−1, 휇−1|휖 𝖨T . A = C s Because 휇−1 is of full rank, the dyadic 휇 exists, whence the four conditions reduce to two as 휖 = B휇, A휖 = C휇. The coefficients must satisfy C = AB. Because we now have 𝖬2 =−B𝖨(2)T , the medium bidyadic actually satisfies the quadratic equation. 9.9 The eigenproblem 𝖨(2)T 𝜺 ⌊ 𝜺 ⌊ |𝚽 𝚽 (A + N AB − N BA) i = Mi i has a solution of the form 휇 |𝚽 |𝚽 c = A, A c = B c = 0 𝚽 for any c satisfying the two conditions. The other solutions must have the form 𝚽 훼 𝜺 ⌊ 훽 𝜺 ⌊ i = i N A + i N B 훼 훽 and Mi, i, i satisfy the equation ( )( ) ( ) ⋅ ⋅ 훼 A − Mi + A BBB i 0 . ⋅ ⋅ 훽 = −A A A − Mi − A B i 0 Requiring that the determinant of the matrix vanish, we obtain for i =± √ 휆 휆 ⋅ 2 ⋅ ⋅ M± = A ± , = (A B − (A A)(B B, while the corresponding field two-forms are defined by 𝚽 𝜺 ⌊ ⋅ 𝜺 ⌊ ⋅ ± = N A(B B) − N B(A − M± + A B) 𝜺 ⌊ ⋅ 휆𝜺 ⌊ . = N (AB − BA) B ± N B APPENDIX A Solutions to Problems 355

𝖢 9.10 Let us expand powers of the skewon bidyadic o 𝖢2 𝜺 ⌊ |𝜺 ⌊ o = N (AB − BA) N (AB − BA) 𝜺 ⌊ ⋅ ⋅ ⋅ ⋅ = N (A(B AB − B BA) − B(A AB − A BA)) ⋅ 𝜺 ⌊ ⋅ 𝜺 ⌊ ⋅ 𝜺 ⌊ = (A B) N (AB + BA) − (A A) N BB − (B B) N AA 𝖢3 𝜺 ⌊ ⋅ 𝖢2 o = N (AB − BA) o ⋅ 𝜺 ⌊ ⋅ ⋅ ⋅ = (A B) N ((B A)(AB − BA) + (B B)AA − (A A)BB) ⋅ 𝜺 ⌊ ⋅ ⋅ −(A A) N ((B B)AB − (A B)BB) ⋅ 𝜺 ⌊ ⋅ ⋅ −(B B) N ((B A)AA − (A A)BA) ⋅ 2 ⋅ ⋅ 𝜺 ⌊ . = ((A B) − (A A)(B B)) N (AB − BA) 1 = tr𝖢2 𝖢 . 2 o o The last step is obtained from 𝖢2 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ tr o = (A B)(A B + B A) − (A A)(B B) − (B B)(A A) = 2((A ⋅ B)2 − (A ⋅ A)(B ⋅ B)). 𝖬 𝖨(2)T 𝖢 Thus, the medium bidyadic = A + o satisfies the cubic equation 1 (𝖬 − A𝖨(2)T )3 − tr(𝖬 − A𝖨(2))2 (𝖬 − A𝖨(2)T ) = 0, 2 where A = tr𝖬∕6. 9.11 Expanding    (Ma − Mb)(Mb − Mc)(Mc − Ma)( a + b + c) 𝖬 𝖨(2)T | 𝖬 𝖨(2)T =−(Mb − Mc)( − Mb ) ( − Mc ) 𝖬 𝖨(2)T | 𝖬 𝖨(2)T −(Mc − Ma)( − Mc ) ( − Ma ) 𝖬 𝖨(2)T | 𝖬 𝖨(2)T −(Ma − Mb)( − Ma ) ( − Mb ) 𝖬2 =−((Mb − Mc) + (Mc − Ma) + (Ma − Mb)) 2 2 2 2 2 2 𝖬 +((Mb − Mc ) + (Mc − Ma) + (Ma − Mb)) −(((Mb − Mc)MbMc + (Mc − Ma)McMa 𝖨(2)T + (Ma − Mb)MaMb) 𝖨(2)T . = (Ma − Mb)(Mb − Mc)(Mc − Ma) The last step can be verified backwards by expanding the last expression. 356 APPENDIX A Solutions to Problems

9.12 Generalizing the procedure of Problem 9.10, we can expand different quantities as 𝖢 𝜺 ⌊ = N (AC + BD) tr𝖢 = A ⋅ C + B ⋅ D, ⌊𝖢2 ⋅ ⋅ ⋅ ⋅ eN = AC(A C) + AD(B C) + BC(A D) + BD(B D), tr𝖢2 = (A ⋅ C)2 + 2(A ⋅ D)(B ⋅ C) + (B ⋅ D)2, ⌊𝖢3 ⋅ ⋅ ⋅ ⋅ eN = (AD(B C) + BC(A D))(A C + B D) + (AC + BD)(A ⋅ D)(B ⋅ C) + AC(A ⋅ C)2 + BD(B ⋅ D)2 ⌊𝖢2 𝖢 ⋅ ⋅ ⋅ ⋅ = eN tr − (AC(A C) + BD(B D))(A C + B D) + (AC + BD)(A ⋅ D)(B ⋅ C) + AC(A ⋅ C)2 + BD(B ⋅ D)2, ⌊𝖢2 𝖢 ⌊𝖢 ⋅ ⋅ ⋅ ⋅ . = eN tr + eN ((A D)(B C) − (A C)(B D) Interpreting the last factor of 𝖢 from (tr𝖢)2 − tr𝖢2 = 2(A ⋅ C)(B ⋅ D) − 2(A ⋅ D)(B ⋅ C), the bidyadic 𝖢 can finally be shown to satisfy the cubic equation (9.110) 1 𝖢3 − (tr𝖢)𝖢2 + ((tr𝖢)2 − tr𝖢2)𝖢 = 0. 2 9.13 Inserting (9.119) with (9.127) in (9.120) and (9.121) we obtain 𝖬2 𝖨(2)T 𝚫 |휇−1 𝜺 𝖨T 휇−1|𝚫 =−(c s + s(as )) + 4 ∧ (c s + ( s)as) ∧ e4 𝖬4 𝖨(2)T 𝚫 |휇−1 2 𝜺 𝖨T 휇−1|𝚫 2 = (c s + s(as )) − 4 ∧ (c s + ( s)as) ∧ e4 2𝖨(2)T 휆 |휇−1 𝜺 2𝖨T 휆 휇−1|𝚫 = c s + Δs(as ) − 4 ∧ (c s + ( s)as) ∧ e4, 휆 |휇−1|𝚫 with = 2c + as s. Equation (9.118) can be split in two components as 2 𝖨(2)T 휆 𝚫 |휇−1 (c + 2Ac + B) + ( + 2A) s(as ) = 0 𝜺 2 𝖨T 휆 휇−1|𝚫 4 ∧ ((c + 2Ac + B) s + ( + 2A)( s)as) ∧ e4 = 0 These are valid for 1 휆 + 2A = 0 ⇒ A =−(c + a |휇−1|𝚫 ) 2 s s 2 ⇒ |휇−1|𝚫 c + 2Ac + B = 0 B = c(c + as s), which yield (9.128) and (9.129). APPENDIX A Solutions to Problems 357

9.14 In the case of the principal BQ medium the modified medium 𝖬 ⌊𝖬 bidyadic m = eN is symmetric. From this it follows that also ⌊𝖬n the bidyadics eN must be symmetric. In fact, we can expand ⌊𝖬2 ⌊𝖬 | 𝜺 ⌊ ⌊𝖬 ⌊𝖬 | 𝜺 ⌊𝖨(2) | ⌊𝖬 eN = (eN ) ( N (eN )) = (eN ) ( N ) (eN ), which is obviously symmetric. We can proceed similarly for higher powers of 𝖬. As a conclusion, for principal BQ media, the ⌊ bidyadics eN i are symmetric. This implies that, the “natural” 𝚽 ⋅ 𝚽 𝚽 | ⌊𝚽 𝚽 ⋅ 𝚽 dot product of two two-forms 1 2 = 1 (eN 2) = 2 1 of two eigen two-forms associated with different eigenvalues is zero. This is seen from the expansion 𝚽 ⋅ 𝚽 𝚽 | ⌊ |𝚽 ⌊𝚽 | |𝚽 ⌊ |𝚽 | |𝚽 i j = i (eN j) = (eN i) j = (eN i ) j 𝚽| ⌊ T | |𝚽 𝚽| ⌊  | |𝚽 ≠ = (eN i) j = (eN ( i j)) = 0, i j, and orthogonality of the projection dyadics. Thus, the eigenfields are orthogonal in the principal BQ medium. 9.15 Starting from the representation (9.148)–(9.151) find the 3D conditions for the BQ medium corresponding to (9.162) and (9.163). 𝖠 훼2 휖′|휇−1 s = − , 𝖡 훼|휖′ 휖′|훽 s = − , 𝖢 휇−1|훼 훽|휇−1 s = − , 𝖣 휇−1|휖′ 훽2 s = − ,

휇−1|휖|휇−1| 휉 휁 |휇−1|휖 2 2휇−1| 휉 휁 ( + ) = MaMb ( + ), 휇−1|휖| 휇−1| 휉 휁 2 휇−1|휖 2𝖨T | 휇−1|휖 2𝖨T ( ( + )) = ( + Ma s ) ( + Mb s ) 9.16 Inserting (9.167)–(9.170) in (9.147) we obtain 𝖬2 휖|휇−1 휉|휇−1|휖 휖|휇−1|휉 𝜺 휇−1|휖 =− + ∧ e4 − ∧ e4 + 4 ∧ ∧ e4 휖|휇−1 | 𝖨(2)T 휉 휉 𝜺 𝖨T | 휇−1|휖 . =−( ) ( s + ∧ e4) + ( + 4 ∧ s ) ( ) ∧ e4 𝖨(2)T 휉 | 휉 𝜺 𝖨T 휉 휉 Because ( s + ∧ e4) ( + 4 ∧ s ) = − = 0, we obtain 𝖬4 휖|휇−1 휉|휇−1|휖 휖|휇−1|휉 𝜺 휇−1|휖 2 = (− + ∧ e4 − ∧ e4 + 4 ∧ ∧ e4) 휖|휇−1 2| 𝖨(2)T 휉 휉 𝜺 𝖨T | 휇−1|휖 2 . = ( ) ( s + ∧ e4) − ( + 4 ∧ s ) ( ) ∧ e4 358 APPENDIX A Solutions to Problems

Inserting these in (9.118) the equation can be written as 휖|휇−1 2 휖|휇−1 𝖨(2)T 0 = ( ) + 2A( ) + B s 𝜺 휇−1|휖 2 휇−1|휖 𝖨T − 4 ∧ (( ) + 2A( ) + B s ) ∧ e4 휖|휇−1 2 휖|휇−1 |휉 휉| 휇−1|휖 2 휇−1|휖 . +(( ) + 2A( )) ∧ e4 − (( ) + 2A( )) ∧ e4 The three lines must equal zero separately. The second line yields 휇−1|휖 𝖨T 2 2 𝖨T ( + A s ) − (A − B) s = 0, whence √ 휖 휇 2 휇|𝖴T =−A ± A − B ( s ), 𝖴T F E where s ∈ 1 1 is a spatial unipotent dyadic. Applying this we have √ 휖|휇−1 𝖨(2)T 2 휇|𝖴T |휇−1 . =−A ± A − B ( s ) 휇|𝖴T |휇−1 F E The first line also vanishes because s ∈ 2 2 is another spatial unipotent dyadic. Applying these to the third line, it becomes 𝖨(2)T |휉 휉|𝖨T 휉 휉 . s ∧ e4 − s ∧ e4 = ∧ e4 − ∧ e4 = 0 Since this is valid for any dyadic 휉, the medium bidyadic (5.75) which in the present case reads 𝖬 휖 𝜺 𝖨T 휉 |휇−1| 𝖨(2)T 휉 = ∧ e4 + ( 4 ∧ s + ) ( s + ∧ e4), with 휖 and 휇 related in the above manner, represents a BQ medium for any 휉. 𝖣 𝖨(2)T 𝖥 𝖣−1 𝖨(2)T 𝖥 9.17 Starting from = − s ∧ e4 and = + s ∧ e4,and applying the expansion (5.75), we can expand 𝖣|𝖬|𝖣−1 𝖨(2)T 𝖥 |𝖬| 𝖨(2)T 𝖥 = ( − s ∧ e4) ( + s ∧ e4) 𝖬 𝖥 |휇−1 𝖥 |휇−1|휁 휉|휇−1|𝖥 = + s − s ∧ e4 + s ∧ e4 𝜺 휇−1|𝖥 𝖥 |휇−1|𝖥 + 4 ∧ s ∧ e4 + s s ∧ e4) 휉 𝖥 |휇−1 휖 휉 𝖥 |휇−1| 휁 𝖥 = ( + s) + ( − ( + s) ( − s)) ∧ e4 𝜺 휇−1 𝜺 휇−1| 휁 𝖥 + 4 ∧ − 4 ∧ ( − s) ∧ e4 𝖬 . = 1 APPENDIX A Solutions to Problems 359

The parameter dyadics are transformed as 휖 휖 휉 휉 𝖥 휁 휁 𝖥 휇 휇. 1 = , 1 = + s, 1 = − s, 1 = 𝖥 휁 휉 휉 휁 휉 휁 Choosing now s = ( − )∕2, we obtain 1 = 1 = ( + )∕2. Thus, we can represent the general medium dyadic 𝖬 as 𝖬 𝖣−1|𝖬 |𝖣 = 1 𝖬 휉 휁 in terms of a medium dyadic 1 satisfying 1 = 1. 9.18 The potential one-form 𝝓 of a plane wave in a CU-medium defined by the medium bidyadic (9.114) satisfies (𝝂 ∧ 𝖬⌊𝝂)|𝝓 = 0. Let us expand 𝝂 𝖬⌊𝝂 |𝝓 𝝂 𝖡2∧𝖨 T ⌊𝝂 𝝂 𝖡(2)T ⌊𝝂 |𝝓 ( ∧ ) = ( ∧ ( o∧ ) + 2 ∧ o ) 𝝂 𝖡2T |𝝂 𝖨T 𝝂 𝖡T |𝝂 𝖡T 𝝓 = ( ∧ ( o ) ∧ + 2 ∧ ( o ) ∧ o ) 𝝂 𝝂 𝝓 𝝂 𝝂 𝖡T |𝝓 = ( ∧ 2 ∧ + 2 ∧ 1 ∧ o ) = 0, 𝝂 𝖡iT |𝝂 denoting i = o .From 𝝂 𝝂 𝖬⌊𝝂 |𝝓 𝝂 𝝂 𝝂 𝝓 − 1 ∧ ( ∧ ) = ∧ 1 ∧ 2 ∧ = 0, the potential one-form can be represented in the form 𝝓 𝝂 𝝂 𝝂 . = A + B 1 + C 2 Substituting this in the potential equation we obtain 𝝂 𝝂 𝝂 𝝂 . ∧ 1 ∧ (B 2 + 2C 3) = 0 𝝂 𝝂 Multiplying by 3∧ and 2∧ we obtain the respective conditions 𝝂 𝝂 𝝂 𝝂 𝝂 𝝂 𝝂 𝝂 . B ∧ 1 ∧ 2 ∧ 3 = 0, C ∧ 1 ∧ 2 ∧ 3 = 0 𝚽 𝝂 𝝂 𝝂 𝝂 ≠ For = B ∧ 1 + C ∧ 2 0 at least one of the coefficients B and C must be nonzero, whence we must have 𝝂 𝝂 𝝂 𝝂 ∧ 1 ∧ 2 ∧ 3 = 0, | which, when multiplied by eN , can be considered as the scalar dispersion equation for 𝝂. This requires that for any solution 𝝂 the 𝝂 𝖡T |𝝂 𝖡2T |𝝂 𝖡3T |𝝂 one-forms , o , o ,and o must be linearly dependent. Obviously, this cannot be true for the general one-form 𝝂 when 𝖡 o is of full rank. Thus, the dispersion equation has no solution except, for example, when 𝝂 is an eigen one-form of the dyadic 360 APPENDIX A Solutions to Problems

𝖡T o , or, more generally, a linear combination of three eigen one- forms.

CHAPTER 10

10.1 For Case 3 the dispersion dyadic representation (10.1) is of the form 𝖬 ⌊⌊𝝂𝝂 𝖣 𝝂 𝝂 𝝂 m = ( ) = a( )c + bd( ) where the vectors c and b are independent of 𝝂.From b ∧ 𝖣(𝝂)|𝝂 = (b ∧ a)(c|𝝂) = 0, valid for all 𝝂, we conclude that either c = 0orb ∧ a = 0. In the latter case, we must have a(𝝂) = a(𝝂)b for some scalar a(𝝂). Thus, the dispersion dyadic must be of the form 𝖣(𝝂) = bd′(𝝂) with either d′(𝝂) = d(𝝂)ord′(𝝂) = a(𝝂)c + d(𝝂). From 𝝂|𝖣(𝝂) = 0 we obtain (𝝂|b)d′(𝝂) = 0forall𝝂, whence either b = 0ord′(𝝂) = 0. Since both lead to 𝖣(𝝂) = 0forall𝝂, Case 3 yields the axion medium which is already included in Case 1 or Case 2. 10.2 For Case 4, the dispersion dyadic representation (10.1) is of the form 𝖬 ⌊⌊𝝂𝝂 𝖣 𝝂 𝝂 𝝂 𝝂 m = ( ) = a( )c + b( )d( ), where c is independent of 𝝂. Requiring 𝖣(𝝂)|𝝂 = 0 and assuming a(𝝂) ∧ b(𝝂) ≠ 0 we obtain c|𝝂 = 0andd|𝝂 = 0forall𝝂. This requires c = 0andd = D⌊𝝂,whereD is a bidyadic, whence the dispersion dyadic must be of the form 𝖣(𝝂) = b(𝝂)D⌊𝝂. From 𝝂|𝖣(𝝂) = 0 we obtain either D = 0orb(𝝂) = B⌊𝝂.Since we now have 𝖣(𝝂) = BD⌊⌊𝝂𝝂, APPENDIX A Solutions to Problems 361

this belongs to Case 1 (10.8). Alternatively, from a(𝝂) ∧ b(𝝂) = 0 we obtain the relation a(𝝂) = a(𝝂)b(𝝂), where a(𝝂) is a linear function of 𝝂. In this case the dispersion dyadic becomes 𝖣(𝝂) = b(𝝂)(a(𝝂)c + d(𝝂)). From 𝝂|𝖣(𝝂) = 0wehave𝝂|b(𝝂) = 0 which leads to b(𝝂) = B⌊𝝂 and the previous case, or to 𝖣(𝝂) = 0 which corresponds to the axion medium case which is also included in Case 1 solution (10.10). 𝝂 |𝝂 10.3 Considering a 2D subspace of one-forms defined by a1 = |𝝂 𝝂 a2 = 0, for taken from that subspace we have ⌊𝝂 |𝝂 |𝝂 . a = A = (a2 )a1 − (a1 )a2 = 0 Thus, the dispersion dyadic 𝖣(𝝂) is of rank 1 for such a subspace of 𝝂 and of rank 2 for 𝝂 taken from the complementary subspace if the other three bivectors are not simple. ⌊ 𝜶 𝜷 10.4 Defining a bivector D = eN ( ∧ ), the dispersion dyadic can be expressed as 𝖣 𝝂 ⌊ 𝝂 𝖬⌊𝝂 ⌊𝖯T ⌊𝖨T |𝖯T ( ) = eN ( ∧ ) = D = (D ) , ⌊𝖨T E E where D ∈ 1 1 is an antisymmetric dyadic. The bivector D is simple because it satisfies 𝜺 | 𝜺 | ⌊ 𝜶 𝜷 ⌊ 𝜶 𝜷 N (D ∧ D) = N ((eN ( ∧ )) ∧ (eN ( ∧ ))) | 𝜶 𝜷 𝜶 𝜷 . = eN ( ∧ ∧ ∧ ) = 0 Since a simple bivector can be expressed in terms of two vectors as D = d1 ∧ d2, the antisymmetric dyadic satisfies ⌊𝖨T (3) ⌊𝖨T (3) (3) (D ) = ((d1 ∧ d2) ) = (d2d1 − d1d2) 1 = (d d − d d )∧(d d ∧d d ) = 0. 3 2 1 1 2 ∧ 1 2∧ 2 1 From this and the rule (𝖠|𝖡)(p) = 𝖠(p)|𝖡(p) it follows that the dispersion equation 𝖣(3)(𝝂) = (D⌊𝖨T )(3)|𝖯(3)T = 0 is satisfied for any 𝝂 and any medium bidyadic 𝖬 of the form (10.17). 362 APPENDIX A Solutions to Problems

10.5 Considering the relation (10.20) A𝜶 + B𝜷 + C(𝖯T |𝝂) = 0. For C = 0 the one-forms 𝜶 and 𝜷 satisfy 𝜶 ∧ 𝜷 = 0, whence from (10.17) we see that 𝝂 ∧ 𝖬⌊𝝂 = 0, and, hence, 𝖣(𝝂) = 0 which corresponds to the axion medium. For C ≠ 0 we can choose C =−1, whence the relation can be rewritten as 𝖯T |𝝂 = A𝜶 + B𝜷. Without sacrifying generality (𝜶 and 𝜷 appear interchangeable at this stage), we may additionally set B = 1, whence for non- axion solutions we must have 𝜷 = 𝖯T |𝝂 − A𝜶. Inserting this in (10.17), the condition (10.21) is obtained. 10.6 Combined Case 2 medium bidyadic of the form 𝖬 𝖡(2)T 𝖡∧𝖨 T 𝖨(2)T = A + B( ∧ ) + C corresponds to the skewon–axion medium only when A = 0, because for A ≠ 0evenifB ≠ 0 it yields the P-axion medium dyadic (10.31). This is seen by defining new symbols as B B2 𝖯 = 𝖡 + 𝖨, M = A, 훼 = C − . A A The dispersion dyadic for the combined bidyadic can be expanded as 𝖣 𝝂 ⌊ 𝝂 𝖬⌊𝝂 ( ) =−eN ( ∧ ) ⌊ 𝝂 𝖡(2)T ⌊𝝂 𝝂 𝖡∧𝖨 T ⌊𝝂 =−eN (A ∧ + B ∧ ( ∧ ) ) ⌊ 𝝂 𝖡T |𝝂 𝖡T 𝝂 𝖡T |𝝂 𝖨T =−eN (A ∧ ( ) ∧ + B ∧ ( ) ∧ ) ⌊ 𝝂 𝖡T |𝝂 ⌊ 𝖡 𝖨 T = (−eN ( ∧ ( ))) (A + B ) = (A⌊𝖨T )|(A𝖡 + B𝖨)T . ⌊ 𝝂 𝖡T |𝝂 Because the bivector A =−eN ( ∧ ( )) is simple and thus satisfiesA ( ⌊𝖨T )(3) = 0, the dispersion equation (5.240) 𝖣(3)(𝝂) = (A⌊𝖨T )(3)|(A𝖡 + B𝖨)(3)T = 0 is satisfied for any 𝝂. APPENDIX A Solutions to Problems 363

10.7 Applying (6.16), Case 1 medium bidyadics are mapped as 𝖬 𝖠(−2)T | 𝚷 𝚫 훼𝖨(2)T |𝖠(2)T a = ( C + D + ) 𝚷 𝚫 훼𝖨(2) = aCa + aDa + , with 𝚷 𝖠(−2)T |𝚷 𝚫 𝖠(−2)T |𝚫 a = , a = , 𝖠(2)| 𝖠(2)| Ca = C, Da = D, and Case 2 medium bidyadics (see the previous problem) are mapped as 𝖬 𝖠(−2)T | 𝖡(2)T 𝖡∧𝖨 T 𝖨(2)T |𝖠(2)T a = (A + B( ∧ ) + C ) 𝖡(2)T 𝖡 ∧𝖨 T 𝖨(2)T = A a + B( a∧ ) + C , with 𝖡 𝖠|𝖡|𝖠−1. a = Thus, in both cases, the form of the medium bidyadic is retained in the transformation. 10.8 Starting from the property (𝖭 − 훼 𝖨(2)T )T ⋅ (𝖭 − 훼 𝖨(2)T ) = 𝖯(2)|(e ⌊𝖯(2)T ) =Δ 𝖤. 1 1 1 N 1 P1 Multiplying by 𝖬T | from the left and by |𝖬 from the right we obtain (𝖨(2)T − 훼 𝖬)T ⋅ (𝖨(2)T − 훼 𝖬) =Δ 𝖬T ⋅ 𝖬 1 1 P1 which can be rewritten in the form (𝖬 − 훼𝖨(2)T )T ⋅ (𝖬 − 훼𝖨(2)T ) = 훽𝖤 with

훼 ΔP 훼 = 1 , 훽 = 1 훼2 −Δ (훼2 −Δ )2 1 P1 1 P1 Thus, the bidyadic 𝖬 must correspond to either a P-axion or a Q-axion medium. Since there is no dispersion equation for the medium defined through the bidyadic 𝖭, there is no dispersion equation when defined through the corresponding bidyadic 𝖬 = 𝖭−1. Now it is known that a full-rank Q-axion medium 364 APPENDIX A Solutions to Problems

has a dispersion equation, whence 𝖬 must represent a P-axion medium. 10.9 Applying the bac-cab rule c ∧ (D⌊𝝂) = (c|𝝂)D − (c ∧ D)⌊𝝂 to c = C⌊𝝂 we obtain (C⌊𝝂) ∧ (D⌊𝝂) =−((C⌊𝝂) ∧ D)⌊𝝂. Thus, from (10.39) we have 𝚽 𝝂| 𝜺 ⌊ ⌊𝝂 ⌊𝝂 𝝂| 𝜺 ⌊ ⌊𝝂 ⌊𝝂 = ( p) N ((C ) ∧ (D )) =−( p) N (((C ) ∧ D) ) 𝝂| 𝜺 ⌊ ⌊𝝂 𝝂 = ( p)( N ((C ) ∧ D)) ∧ , and |𝚽 𝝂| 𝜺 ⌊ ⌊𝝂 𝝂| C = ( p)( N ((C ) ∧ D)) ∧ C 𝝂| 𝜺 ⌊ ⌊𝝂 | ⌊𝝂 =−( p)( N ((C ) ∧ D)) (C ) 𝝂| 𝜺 | ⌊𝝂 ⌊𝝂 . =−( p) N ((C ) ∧ D ∧ (C )) = 0 Since 𝚽 is a symmetric function of C and D we also have D|𝚽 = 0. 10.10 The dispersion dyadic can be expanded as 𝖣 𝝂 𝖬 ⌊⌊𝝂𝝂 𝖰(2)⌊⌊𝝂𝝂 ( ) = m = ⌊𝝂 ⌊𝝂 ⌊𝝂 ⌊𝝂 ⌊𝝂 ⌊𝝂 = (e12 )(q12 ) + (e23 )(q23 ) + (e31 )(q31 )

= e1d1 + e2d2 + e3d3, where we denote 휈 ⌊𝝂 휈 ⌊𝝂 d1 =− 2(q12 ) + 3(q31 ), 휈 ⌊𝝂 휈 ⌊𝝂 d2 =− 3(q23 ) + 1(q12 ), 휈 ⌊𝝂 휈 ⌊𝝂 d3 =− 1(q31 ) + 2(q23 ), 휈 |𝝂 𝝂|𝖣 𝝂 and i = ei . Because of ( ) = 0, we have 휈 휈 휈 . 1d1 + 2d2 + 3d3 = 0 휈 ≠ Assuming 3 0 we can write 휈 휈 휈 d3 =−( 1d1 + 2d2)∕ 3, APPENDIX A Solutions to Problems 365

insert it in the expression of 𝖣(𝝂) above, and expand 휈 𝖣 𝝂 휈 휈 휈 휈 3 ( ) = ( 3e1 − 1e3)d1 + ( 3e2 − 2e3)d2 𝝂⌋ =− (e3 ∧ e1d1 + e3 ∧ e2d2) 𝝂⌋ 휈 휈 휈 휈 ⌊𝝂 =− (e3 ∧ e1( 3q31 − 2q12) + e3 ∧ e2( 1q12 − 3q23)) 𝝂⌋ 𝝂| =− (e3 ∧ e1( (e3q3 + e2q2) ∧ q1) 𝝂| ⌊𝝂 +e3 ∧ e2( (e1q1 + e3q3) ∧ q2)) 𝝂⌋ 𝝂|𝖰 𝝂|𝖰 ⌊𝝂. =− (e3 ∧ e1( ∧ q1) + e3 ∧ e2( ∧ q2)) This is of the form 𝖣 = ac + bd where the two dyads equal those of (10.48) and (10.49). Combining the two terms as 휈 𝖣 𝝂 𝝂⌋ 𝝂|𝖰 ∧ ⌊𝝂 3 ( ) =− (e3( )∧(e1q1 + e2q2)) 𝝂|𝖰 ∧𝖰 ⌊⌊𝝂𝝂 = (e3( )∧ ) , the expression (10.50) is obtained. 10.11 The double-wedge cube of the dispersion dyadic can be written as ( ) ∑ (3) 𝖣(3) 𝝂 ( ) = eidi = e123, 𝝂 휈 휈 휈 ⌊𝝂 ⌊𝝂 ⌊𝝂 d123( ) =− 2 3 1(q12 ) ∧ (q23 ) ∧ (q31 ) 휈 휈 휈 ⌊𝝂 ⌊𝝂 ⌊𝝂 + 3 1 2(q31 ) ∧ (q12 ) ∧ (q23 ), 𝝂 whence d123( ) = 0 identically, as expected. This implies that 𝝂 the three vector functions di( ) must be linearly dependent. 10.12 The expression (10.50) can be verified by applying the rule 𝖠∧𝖡 ⌊⌊𝖢 𝖠||𝖢T 𝖡 𝖡||𝖢T 𝖠 𝖠|𝖢T |𝖡 𝖡|𝖢T |𝖠 ( ∧ ) = ( ) + ( ) − − , 𝖠 𝖡 E E 𝖢 F F valid for dyadics , ∈ 1 1 and ∈ 1 1, by expanding 𝝂|𝖰 ∧𝖰 ⌊⌊𝝂𝝂 |𝝂 𝝂|𝖰|𝝂 𝖰 𝖰||𝝂𝝂 𝝂|𝖰 ((e3( ))∧ ) = (e3 )( ) + ( )e3( ) 𝝂|𝖰|𝝂 𝝂|𝖰 𝖰|𝝂 𝝂| 𝝂|𝖰 −e3( )( ) − ( )( e3)( ) |𝝂 𝝂|𝖰|𝝂 𝖰 𝖰|𝝂 𝝂|𝖰 = (e3 )(( ) − ( )( )) |𝝂 𝖰(2)⌊⌊𝝂𝝂 = (e3 ) |𝝂 𝖣 𝝂 . = (e3 ) ( ) 366 APPENDIX A Solutions to Problems

10.13 Requiring that an orthogonality condition of the form 훼 ⋅ 훽 ⋅ Eg Bg + Hg Dg = 0, satisfies by the fields of a plane wave in any medium, implies ⋅ ⋅ the polarization condition (a Eg)(b Hg) = 0inananisotropic medium, we set 훼 ⋅ 훽 ⋅ ⋅ ⋅ Eg Bg + Hg Dg − Eg ab Hg ⋅ 훼휇 훽휖T ⋅ . = Eg ( g + g − ab) Hg = 0

To be satisfied for any field vectors Eg, Hg requires that the medium dyadics satisfy the condition 훼휇 훽휖T . g + g − ab = 0 10.14 Substituting the 3D expansions in (5.65) for the PDC medium yields 훼 𝖯(2) 훼𝖨(2)T 𝜺 ⌊ = M s + s + ( 123 d3)(c1 ∧ c2), 휖′ 𝝅 𝖯 𝜺 ⌊ =− ∧ s + ( 123 d3)c3, 휇−1 𝖯 𝜺 ⌊ =− s ∧ p − 123 (d1 ∧ d2)(c1 ∧ c2), 훽 𝝅 𝖯 훼𝖨T 𝜺 ⌊ . = p − p s − s − 123 (d1 ∧ d2)c3 The dyadics 휖′ and 휇−1 have a restricted form. In the case DC = 0 they do not have an inverse, which is in contrast to the QDC medium. 𝖨 𝖨 𝜺 𝖨(2) 𝖨(2) 𝖨 ∧ 𝜺 10.15 Applying = s + e4 4 and = s + s∧e4 4,wehave 𝜺 ⌊𝖠 𝖡 ∧𝖨 T N = ( o∧ ) 𝖢 ∧𝖨 T 𝜺 𝖨T 𝜸 𝖨T = ( s∧ s) − 4 ∧ s ∧ cs − s ∧ s ∧ e4 𝜺 𝖢 𝖢 𝖨T . − 4 ∧ ( s − tr s s ) ∧ e4 Further, we can expand 𝜺 ⌊ 𝜺 𝜶 𝜺 ⌊ ⌊𝜷 N (AB + BA) = (− 4 ∧ s + 123 a3)(e123 s + bs ∧ e4) 𝜺 𝜷 𝜺 ⌊ ⌊𝜶 . +(− 4 ∧ s + 123 bs)(e123 s + as ∧ e4) Inserting the expansions in (10.102) we can identify one set of 3D medium dyadics as 훼 훼𝖨(2)T 𝖢 ∧𝖨 T 𝜺 ⌊⌊ 𝜷 𝜶 = s + ( s∧ s) + 123e123 (as s + bs s), 휖′ 𝜸 𝖨T 𝜺 ⌊ =− s ∧ s + 123 (asbs + bsas), APPENDIX A Solutions to Problems 367

휇−1 𝖨T 𝜶 𝜷 𝜷 𝜶 ⌋ =− s ∧ cs − ( s s + s s) e123, 훽 훼𝖨T 𝖢 𝖢 𝖨 T 𝜶 𝜷 . =− s − ( s − tr s s) − ( sbs + sas) ⌊휖′ 휇−1⌋𝜺 It can be seen that the dyadics e123 and 123 may have arbitrary antisymmetric parts while their symmetric parts are incomplete consisting of only two dyads. The decomposition can be written for the 3D fields E, B as | ⌊𝜶 | | ⌊𝜷 | . (B (e123 s) + E as)(B (e123 s) + E bs) = 0 10.16 To verify that (10.104) corresponds to the A-wave, applying (10.89) we can write for the QDC medium 𝖣 𝝂 |𝝓 𝝂𝝂⌋⌋ 𝖰(2) 훽 ⌊𝖨(2)T |𝝓 ( ) = (M + DC − eN ) = M(𝝂𝝂||𝖰)𝖰|𝝓 − M(𝖰|𝝂)𝝂|𝖰|𝝓 + dc|𝝓 = 0, where we denote d = 𝝂⌋D and c = 𝝂⌋C. Choosing the Lorenz condition 𝝂|𝖰|𝝓 = 0, the potential satisfies M(𝝂𝝂||𝖰)𝖰|𝝓 + d(c|𝝓) = 0. The solution corresponding to the dispersion condition (10.104) leads to c|𝝓 =−C|𝚽 =−A|𝚽 = 0, because of the definition (10.58). Thus, (10.104) corresponds to the A-wave. 10.17 To verify that the field corresponding to the dispersion equation (10.105) satisfies the B-wave condition B|𝚽 = 0 we can start from c|𝝓 ≠ 0and𝝂𝝂||𝖰 ≠ 0 with d and c as in the previous solution. Assuming the Lorenz condition 𝝂|𝖰|𝝓 = 0in 𝖣(𝝂)|𝝓 = M(𝝂𝝂||𝖰)𝖰|𝝓 − M(𝖰|𝝂)(𝝂|𝖰)|𝝓 + dc|𝝓 = M(𝝂𝝂||𝖰)𝖰|𝝓 + dc|𝝓 = 0 we see that the potential must be of the form 𝝓 = 휆𝖰−1|d, where 휆 is some scalar coefficient. The Lorenz condition then requires that 𝝂 must satisfy 𝝂|d = 0. Invoking (10.91) and the rule 𝝂⌋ 𝝂⌋ 𝝂⌋ ⋅ 2D ∧ ( D) = (D ∧ D) = ( eN)(D D), 368 APPENDIX A Solutions to Problems

we finally obtain 휆 B|𝚽 = 휆B|(𝝂 ∧ 𝖰−1|d) = (D ⋅ D)(M(𝝂|𝖰|𝝂) + c|𝖰−1|d) = 0, 2 because of (10.105). Thus, the corresponding solution can be identified as the B-wave. 10.18 Let us expand 𝝂𝝂⌋⌋𝖠 𝝂⌋ ⌊ 𝖡 ∧𝖨 T ⌊𝝂 ⌊ 𝝂 𝖡 |𝝂 ⌊𝖨 T ⌊𝖨T =− (eN ( o∧ ) ) = eN ( ∧ ( o )) ) = F , ⌊ 𝝂 𝝂′ where the bivector defined by F = eN ( ∧ ) is simple since it satisfies F ⋅ F = 0. Because of this, the antisymmetric dyadic ⌊𝖨T E E F ∈ 1 1 satisfies (F⌊𝖨T )(2) = FF,(F⌊𝖨T )(3) = 0. Applying (10.115) we can further expand 𝖣(3)(𝝂) = (F⌊𝖨T + ab + ba))(3) ∧ ⌊𝖨T ∧ (2). = (FF)∧(ab + ba) + (F )∧(ab + ba) Applying the orthogonality 𝝂|a = 𝝂|b = 0, we obtain ⌊ 𝝂 𝝂′ ⌊𝝂 𝝂′| F ∧ a = (eN ( ∧ )) ∧ a =−(eN )( a), and similarly for F ∧ b, whence we have ∧ ⌊⌊𝝂𝝂 𝝂′| 𝝂′| . (FF)∧(ab + ba) = 2(eNeN )( a)( b) The last term of the above expression of 𝖣(3)(𝝂) vanishes due to ⌊𝖨T ∧ ∧ ⌊𝝂 𝝂′| (F )∧(ab∧ba) = (eN ) (ab − ba) ∧ (a ∧ b) = 0, which leads to the dispersion equation (10.116). APPENDIX B

Transformation to Gibbsian Formalism

In this appendix we consider the connection between spatial multivectors and multiforms on one hand and Gibbsian vectors on the other hand. Explicit formulas allow one to interpret results obtained through the 4D analysis in terms of the Gibbsian formalism. The converse, interpreting a given Gibbsian vector in terms of the spatial part of a 4D quantity, is also possible if the nature of the quantity is given. In the following the subscript ()g denotes Gibbsian vectors and dyadics while the spatial part of a 4D vector or one-form is marked by the subscript ()s. The connection depends on a chosen vector basis ei which defines the spatial metric dyadic 𝖦 s = e1e1 + e2e2 + e3e3,(B.1) with its spatial inverse  𝖦−1 𝜺 𝜺 𝜺 𝜺 𝜺 𝜺 𝜺 𝜺 ⌊⌊𝖦(2). s = s = 1 1 + 2 2 + 3 3 = 123 123 (B.2) The dyadic satisfies 𝖦(2) ⌊⌊ . s = e12e12 + e23e23 + e31e31 = e123e123 s (B.3) 𝖦 𝖨 𝖦 𝔼 𝔼 s serves as the Gibbsian unit dyadic g = s ∈ 1 1 satisfying 𝖨 ⋅ ⋅ 𝖨 g xg = xg g = xg (B.4)

369 370 APPENDIX B Transformation to Gibbsian Formalism

for any Gibbsian vector xg. The following set of transformation rules can be easily verified applying the basis expansions 𝜶 𝛼 𝜺 𝛼 𝜺 𝛼 𝜺 s = 1 1 + 2 2 + 3 3,(B.5)

as = a1e1 + a2e2 + a3e3,(B.6) 𝜽 𝜃 𝜺 𝜃 𝜺 𝜃 𝜺 s = 12 12 + 23 23 + 31 31,(B.7) . As = A12e12 + A23e23 + A31e31 (B.8)

VECTOR AND ONE-FORM

Any spatial vector can be immediately understood in Gibbsian sense, 𝜶 ag = as. Transformation of a spatial one-form to a Gibbsian vector g is given through 𝜶 𝖦 |𝜶 𝛼 𝛼 𝛼 . g = s s = 1e1 + 2e2 + 3e3 (B.9) It can be inverted as 𝜶  |𝜶 . s = s g (B.10)

BIVECTOR AND TWO-FORM

𝜃 A spatial two-form s can be transformed to a Gibbsian vector as 𝜽 ⌊𝜽 𝜃 𝜃 𝜃 g = e123 s = 12e3 + 23e1 + 31e2, (B.11) and the inverse transformation is 𝜽 𝜺 ⌊𝜽 . s = 123 g (B.12)

The Gibbsian vector corresponding to a spatial bivector As can be 𝜺 ⌊ obtained by first transforming the bivector to the one-form 123 As, after which it can be transformed to the vector, 𝖦 | 𝜺 ⌊ . Ag = s ( 123 As) = e1A23 + e2A31 + e3A12 (B.13) The same result is obtained if the bivector is first transformed to a (2)| two-form as s As and ⌊ (2)| 𝖦 | 𝜺 ⌊ . Ag = e123 ( s As) = s ( 123 As) (B.14) APPENDIX B Transformation to Gibbsian Formalism 371

The inverse transformation is ⌊  | . As = e123 ( s Ag) (B.15)

TRIVECTOR AND THREE-FORM

𝜿 𝜺 𝜅 Spatial trivector ks = e123k and three-form s = 123 correspond to scalars in Gibbsian formalism: 𝜺 | k = 123 ks, (B.16) 𝜅 |𝜿 . = e123 s (B.17) Conversely, a Gibbsian scalar 𝛼 can be interpreted as a spatial trivector 𝛼 𝛼𝜺 e123 or a spatial three-form 123.

BAR PRODUCT

By definition, the bar product of a spatial vector as and a spatial one- 𝜶 form s equals the dot product of the corresponding Gibbsian vectors as ∑ |𝜶 | |𝜶 ⋅ 𝜶 𝛼 . as s = as s g = as g = ai i (B.18) 𝜽 The bar product of a spatial bivector As and a spatial two-form s can be shown to equal the dot product of the corresponding Gibbsian vectors as |𝜽 | 𝜺 ⌊ ⌊𝜽 𝜺 ⌊ | ⌊𝜽 As s = As ( 123 (e123 s)) = ( 123 As) (e123 s) 𝖦 | 𝜺 ⌊ | | ⌊𝜽 ⋅ 𝜽 . = ( s ( 123 As)) s (e123 s) = Ag g (B.19)

WEDGE PRODUCT

𝜶 𝜷 The wedge product of two spatial one-forms s, s has the following relation to the cross product of the corresponding Gibbsian vectors, 𝜶 𝜷 ⌊ 𝜶 𝜷 g × g = e123 ( s ∧ s) 𝛼 𝛽 𝛼 𝛽 𝛼 𝛽 𝛼 𝛽 𝛼 𝛽 𝛼 𝛽 . = e1( 2 3 − 3 2) + e2( 3 1 − 3 1) + e3( 1 2 − 2 3) (B.20) 372 APPENDIX B Transformation to Gibbsian Formalism

The wedge product of three one-forms transforms to a scalar as | 𝜶 𝜷 𝜸 ⌊ 𝜶 𝜷 |𝜸 𝜶 𝜷 |𝜸 e123 ( s ∧ s ∧ s) = (e123 ( s ∧ s)) s = ( g × g) s 𝛼 𝜷 ⋅ 𝜸 . = g × g g (B.21) The cross product of two spatial vectors can be interpreted in terms of their wedge product as ⌊  |  | ⌊ (2)| as × bs = e123 (( s as) ∧ ( s bs)) = e123 ( s (as ∧ bs))) 𝖦 | 𝜺 ⌊ . = s ( 123 (as ∧ bs)) (B.22)

CONTRACTION PRODUCTS

Contraction product of a spatial vector and a spatial two-form is related to the cross product of their Gibbsian counterparts as ⌋𝜽 𝜺 𝜃 𝜃 𝜺 𝜃 𝜃 𝜺 𝜃 𝜃 as s = 1(a2 12 − a3 31) + 2(a3 23 − a1 12) + 3(a1 31 − a2 23), ⌋ ⌊𝜽 𝜺 ⌊𝜃 ⌊ 𝜺 ⌊ 𝜽 = as (e123 g) =−( 123 g) as = 123 (as ∧ g)  | 𝜽 . = s (as × g) (B.23) 𝜶 For the spatial one-form s and bivector As we obtain 𝜶 ⌋ 𝜶 . s As = g × Ag (B.24) 𝜿 𝜅𝜺 Contraction with a spatial three-form s = 123 can be expressed as ⌋𝜿 𝜺 ⌊ 𝜅 as s = 123 ( as), (B.25) ⌋𝜿 𝜅𝜺 ⌊ ⌊  |  | 𝜅 . As s = 123 (e123 ( s Ag)) = s ( Ag) (B.26)

BAC-CAB RULE

As an example let us consider the bac-cab rule which for spatial vectors and one-forms reads ⌋ 𝜷 𝜸 𝜷 |𝜸 𝜸 |𝜷 . as ( s ∧ s) = s(as s) − s(as s) (B.27) Expanding first 𝜷 𝜸 ⌋𝜺 𝜷 𝜸 s ∧ s = xg 123, xg = g × g, (B.28) APPENDIX B Transformation to Gibbsian Formalism 373 we obtain ⌋ 𝜷 𝜸 ⌋ ⌋𝜺 ⌋𝜺  | . as ( s ∧ s) = as (xg 123) = (as ∧ xg) 123 = s (ag × xg) (B.29) Thus, the bac-cab rule becomes the corresponding Gibbsian rule, 𝜷 𝜸 𝜷 |𝜸 𝜸 |𝜷 ag × ( g × g) = ag × xg = g(as s) − g(as s) 𝜷 ⋅ 𝜸 𝜸 ⋅ 𝜷 . = g(ag g) − g(ag g) (B.30)

CROSS PRODUCT OF TWO GIBBSIAN TWO-FORMS

The cross product of two Gibbsian vectors has many counterparts depending how we interpret the vector quantities. Assuming that the Gibbsian vectors correspond to two spatial two-forms, their cross product has the form 𝜽 𝚷 ⌊휃 ⌊𝚷 g × g = (e123 s) × (e123 s) ⌊ ⌊𝜽 | ⌊𝚷 | . = e123 (((e123 s) s) ∧ ((e123 s) s)) (B.31) Now one can show through basis expansions that the following relation is valid 𝜽 𝚷 𝖦 | 𝜽 ⌊ ⌊𝚷 . g × g = s ( s (e123 s)) (B.32) As an example, the magnetic force law in Gibbsian form,

dFg = Jg × BgdV (B.33) where dFg is the force in the differential volume dV corresponds to  | ⌊ ⌊ ⌊ ⌋ ⌊ ⌋ . s (Jg × BgdV) = J (e123 B) =−(e123 B) JdV = (dVe123 J) B (B.34)

Here dVe123 can be interpreted as a trivector volume element which ⌊ makes dVe123 J the differential current moment vector.

APPENDIX C

Multivector and Dyadic Identities

In this appendix some useful identities are collected for convenience. Where not otherwise stated, the dimension n of the vector space equals 4. p and q are numbers in the range 0, … , n. Subscript ()s denotes spatial part, corresponding to coordinates 1, 2, 3, while the temporal coordinate is denoted by the index 4. Subindex ()g refers to Gibbsian vector or dyadic quantity. Where not indicated otherwise, dyadics are elements 𝔼 𝔽 𝔼 𝔽 𝔽 𝔼 𝔼 𝔼 𝔽 𝔽 of the space 1 1.Elementsof 2 2, 2 2, 2 2,and 2 2 are called bidyadics.

NOTATION

𝔼 𝜶 𝜷 𝜸 𝔽 vectors a, b, c, …∈ 1, one-forms , , , …∈ 1 𝔼 𝚽 𝚿 𝚪 𝔽 bivectors A, B, C, …∈ 2, one-forms , , , …∈ 2 𝔼 𝜿 𝝅 𝔽 trivectors k, p, …∈ 3, three-forms , , …∈ 3 𝔼 𝜿 𝝅 𝔽 quadrivectors aN, bN, …∈ 4, four-forms N, N, …∈ 4 p 𝔼 𝜶q 𝔽 p − vectors a ∈ p, q − forms ∈ q ⋯ 𝔼 p-index J = i1i2 ip, N = 1234, aN = a1234 = a1 ∧ a2 ∧ a3 ∧ a4 ∈ 4 𝜺 𝜺 𝜺 𝜺 vector basis e1, e2, e3, e4, reciprocal one-form basis 1, 2, 3, 4

375 376 APPENDIX C Multivector and Dyadic Identities

bi-index ordering i < j : ij = 12, 23, 31, 14, 24, 34 𝖠 𝖡 𝔼 𝔽 𝖠T 𝖡T 𝔽 𝔼 dyadics , , …∈ 1 1, , , …∈ 1 1 𝖠 𝖡 𝔼 𝔼   𝔽 𝔽 metric dyadics , , …∈ 1 1, , , …∈ 1 1 d = 𝜺 𝜕 + 𝜺 𝜕 + 𝜺 𝜕 + 𝜺 𝜕 = d + 𝜺 𝜕 1 x1 2 x2 3 x3 4 x4 s 4 x4

MULTIVECTORS AND MULTIFORMS

Multiplication p|𝜶p 𝜶p| p 𝔼 a = a ∈ 0 p q pq q p 𝔼 > a ∧ b = (−1) b ∧ a ∈ p+q (= 0forp + q 4) > p⌊𝜶q q(p−q)𝜶q⌋ p 𝔼 p q, a = (−1) a ∈ p−q p = q + r,(ap⌊𝜶q)|𝜷r = ap|(𝜶q ∧ 𝜷r) p > q + r,(ap⌊𝜶q)⌊𝜷r = ap⌊(𝜶q ∧ 𝜷r)

Reciprocal bases |𝜺 𝜺 | 𝛿 ei j = j ei = i,j ⋯ ⋯ 𝔼 eK(i) = e1 ∧ e2 ∧ ∧ ei−1 ∧ ei+1 ∧ ∧ en ∈ n−1 ⋯ ⋯ ⋯ 𝔼 eK(ij) = e1 ∧ e2 ∧ ∧ ei−1 ∧ ei+1 ∧ ∧ ej−1 ∧ ej+1 ∧ ∧ en ∈ n−2 i−1 4−i eN = e1234 = e1 ∧ e2 ∧ e3 ∧ e4 = (−1) ei ∧ eK(i) = (−1) eK(i) ∧ ei ⌊𝜺 i−1 ⌊𝜺 4−i eN i = (−1) eK(i), eN K(i) = (−1) ei ⌊𝜺 i+j−1 ⌊𝜺 i+j−1 eN ij = (−1) eK(ij), eN K(ij) = (−1) eij 𝜶⌋ 𝜶| 𝚽⌋ 𝚽| ( eN) ∧ a = ( a)eN,( eN) ∧ A = ( A)eN ⌊ p⌋𝜺 p ⌊ 𝜺 ⌊ p p p eN (a N) = a , eN ( N a ) = (−1) a ⌊ 𝜺 ⌊ p p < eN ( 123 a ) = a ∧ e4, p 3 ⌊ ⌋𝚽 ⌊𝚽 𝚽 𝔽 eN (a ) =−a ∧ (eN ), ∈ 2 ⌊ 𝜺 𝜶 i−1 ⌊𝜶. eN ( i ∧ ) = (−1) eK(i) APPENDIX C Multivector and Dyadic Identities 377

Bivector products (a ∧ b)|(𝜶 ∧ 𝜷) = (a|𝜶)(b|𝜷) − (a|𝜷)(b|𝜶) (a ∧ b)|(𝜶 ∧ 𝜷) = a|{b⌋(𝜶 ∧ 𝜷)} = {(a ∧ b)⌊𝜶}|𝜷 (a ∧ b)⌊𝜶 = b(a|𝜶) − a(b|𝜶) = (ba − ab)|𝜶 =−𝜶⌋(a ∧ b) 𝜶 ∧ (a⌋(𝜷 ∧ 𝜸)) =−(a⌋(𝜷 ∧ 𝜸)) ∧ 𝜶 = a⌋(𝜶 ∧ 𝜷 ∧ 𝜸) − (a|𝜶)(𝜷 ∧ 𝜸).

Bac–cab rules 𝜶⌋(b ∧ c) = b(𝜶|c) − c(𝜶|b) = (c ∧ b)⌊𝜶 𝜶⌋ 𝜶⌋ 𝜶| ⌊𝜶 𝔼 (b ∧ C) = b ∧ ( C) + C( b) = (C ∧ b) , C ∈ 2 𝜶⌋ 𝜶⌋ 𝜶⌋ ⌊𝜶 𝔼 (B ∧ C) = B ∧ ( C) + C ∧ ( B) =−(B ∧ C) , B, C ∈ 2 ⌋ 𝜷 𝚪 𝜷 |𝚪 𝚪⌊ ⌊𝜷 𝔼 𝚪 𝔽 A ( ∧ ) = (A ) + (A ), A ∈ 2, ∈ 2 𝜶⌋(bp ∧ cq) = bp ∧ (𝜶⌋cq) + (−1)pqcq ∧ (𝜶⌋bp) (𝜶 ∧ 𝜷)⌋(B ∧ C) = B(𝜶 ∧ 𝜷)|C + C(𝜶 ∧ 𝜷)|B + (𝜷⌋B) ∧ (𝜶⌋C) +(𝜷⌋C) ∧ (𝜶⌋B) (𝜶 ∧ 𝜷)⌋(bp ∧ cq) = (bp⌊(𝜶 ∧ 𝜷)) ∧ cq + bp ∧ ((𝜶 ∧ 𝜷)⌋cq) + (𝜶⌋bp) ∧ (cq⌊𝜷) −(𝜷⌋bp) ∧ (cq⌊𝜶).

Trivector products (a ∧ b ∧ c)|(𝜶 ∧ 𝜷 ∧ 𝜸) = ((a ∧ b ∧ c)⌊𝜶)|(𝜷 ∧ 𝜸) = 𝜶|((a ∧ b ∧ c)⌊(𝜷 ∧ 𝜸)) = 𝜶|(a(b ∧ c) + b(c ∧ a) + c(a ∧ b))|(𝜷 ∧ 𝜸) = (a|𝜶)(b ∧ c)|(𝜷 ∧ 𝜸) − (a⌋(𝜷 ∧ 𝜸))|(𝜶⌋(b ∧ c)) (a ∧ C)|(𝜶 ∧ 𝚪) = (a|𝜶)(C|𝚪) − (a⌋𝚪)|(𝜶⌋C) (a ∧ b ∧ c)⌊𝜶 = {(a ∧ b)c + (b ∧ c)a + (c ∧ a)b}|𝜶 (a ∧ b ∧ c)⌊𝚪 = {a(b ∧ c) + b(c ∧ a) + c(a ∧ b)}|𝚪 𝜶 𝜸⌊ 𝜶⌋ ⌋𝜸 𝜶 𝜸 ⌊ 𝜸 𝔽 𝔼 ∧ ( A) = ( A) + ( ∧ ) A, ∈ 3, A ∈ 2 𝜶 𝜺 ⌊ 𝜺 ⌊ 𝜶⌋ 𝔼 . ∧ ( N k) = N ( k), k ∈ 3 378 APPENDIX C Multivector and Dyadic Identities

Quadrivector products (a ∧ b ∧ c ∧ d)|(𝜶 ∧ 𝜷 ∧ 𝜸 ∧ 𝜹) = ((a ∧ b ∧ c ∧ d)⌊𝜶)|(𝜷 ∧ 𝜸 ∧ 𝜹) = 𝜶|(a(b ∧ c ∧ d) − b(c ∧ d ∧ a) + c(d ∧ a ∧ b) −d(a ∧ b ∧ c))|(𝜷 ∧ 𝜸 ∧ 𝜹) (a ∧ b ∧ c ∧ d)⌊𝜶 = d ∧ ((b ∧ c)a + (c ∧ a)b + (a ∧ b)c)|𝜶 −(a ∧ b ∧ c)(d|𝜶) (a ∧ b ∧ c ∧ d)⌊(𝜷 ∧ 𝜸 ∧ 𝜹) = (((a(b ∧ c + b(c ∧ a) + c(a ∧ b)) ∧ d − d(a ∧ b ∧ c))|(𝜷 ∧ 𝜸 ∧ 𝜹) (a ∧ b ∧ c ∧ d)|(𝜶 ∧ 𝜷 ∧ 𝜸 ∧ 𝜹) = (𝜸 ∧ 𝜹)|((a ∧ b ∧ c ∧ d)⌊(𝜶 ∧ 𝜷)) = (𝜸 ∧ 𝜹)|((a ∧ b)(c ∧ d) − (a ∧ c)(b ∧ d) + (a ∧ d)(b ∧ c) +(b ∧ c)(a ∧ d) − (b ∧ d)(a ∧ c) + (c ∧ d)(a ∧ b))|(𝜶 ∧ 𝜷) (a ∧ b ∧ c ∧ d)⌊(𝜶 ∧ 𝜷) = (((b ∧ c)a + (c ∧ a)b + (a ∧ b)c) ∧ d)|(𝜶 ∧ 𝜷) −(d ∧ (a(b ∧ c) + b(c ∧ a) + c(a ∧ b)))|(𝜶 ∧ 𝜷).

Special cases for n = 3 a(b ∧ c ∧ d) − b(c ∧ d ∧ a) + c(d ∧ a ∧ b) − d(a ∧ b ∧ c) = 0 (a(b ∧ c) + b(c ∧ a) + c(a ∧ b)) ∧ d = d(a ∧ b ∧ c) ((a ∧ b)c + (c ∧ a)b + (b ∧ c)a) ∧ d = d ∧ (a(b ∧ c) + b(c ∧ a) + c(a ∧ b)) ⌋𝜿 ⌋𝜿 ⌋𝜿 |𝜿 𝜿 𝔽 . ((a ∧ b) ) ∧ ((b ∧ c) ) = (b )((a ∧ b ∧ c) ), ∈ 3

Quintivector products 𝜶⌋(a ∧ b ∧ c ∧ d ∧ e) = 𝜶|(a(b ∧ c ∧ d ∧ e) − b(a ∧ c ∧ d ∧ e) +c(a ∧ b ∧ d ∧ e) − d(a ∧ b ∧ c ∧ e) + e(a ∧ b ∧ c ∧ d)) = (𝜶⌋(a ∧ b ∧ c)) ∧ (d ∧ e) + (a ∧ b ∧ c) ∧ (𝜶⌋(d ∧ e)) = 0 a(b ∧ c ∧ d ∧ e) − b(a ∧ c ∧ d ∧ e) + c(a ∧ b ∧ d ∧ e) −d(a ∧ b ∧ c ∧ e) + e(a ∧ b ∧ c ∧ d) = 0 (((a(b ∧ c) + b(c ∧ a) + c(a ∧ b)) ∧ d) − d(a ∧ b ∧ c)) ∧ e +e(a ∧ b ∧ c ∧ d) = 0 (𝜶⌋(a ∧ b ∧ c)) ∧ (d ∧ e) =−(a ∧ b ∧ c) ∧ (𝜶⌋(d ∧ e)). APPENDIX C Multivector and Dyadic Identities 379 p-vector rules ⋯ ⌊𝜶 p−1𝜶⌋ ⋯ (a1 ∧ a2 ∧ ∧ ap) = (−1) (a1 ∧ a2 ∧ ∧ ap) ∑p = 𝜶| (−1)i−1a a i Kp(i) i=1 = 𝜶|(a a − a a + ⋯ + (−1)p−1a a ), 2 ≤ p ≤ n 1 Kp(1) 2 Kp(2) p Kp(p) a = a ∧ ⋯ ∧ a ∧ a ∧ ⋯ ∧ a Kp(i) 1 i−1 i+1 p p⌋𝜺 | 𝜶p⌋ p|𝜶p. (a N) ( eN) = a

DYADICS

Basic rules a𝜶 ∈ 𝔼 𝔽 , ab ∈ 𝔼 𝔼 ,(a𝜶)T = 𝜶a ∈ 𝔽 𝔼 , 𝜶𝜷 ∈ 𝔽 𝔽 1 1 ∑ 1 1 1 1 1 1 𝖠 𝜶 𝜶 𝜶 𝜶 𝜶 = ai i = a1 1 + a2 2 + a3 3 + a4 4 ∑ ∑ ∑ 𝖠| |𝖠T 𝖠|𝖡 𝜶 | 𝜷 𝜶 | 𝜷 a = a , = ai i bj j = ( i bj)ai j i,j ∑ 𝖨 𝜺 𝖨|𝖠 𝖠|𝖨 𝖠 = ei i, = = (𝖠|𝖡)|𝖢 = 𝖠|(𝖡|𝖢), 𝖠p = 𝖠|𝖠p−1, 𝖠0 = 𝖨.

Double-bar products (a𝜶)||(b𝜷)T = (a𝜶)||(𝜷b) = (a|𝜷)(b|𝜶) ∑ ∑ 𝖠 𝜶 |𝜶 𝖠||𝖨T tr = tr ai i = ai i = 𝖠||𝖡T = 𝖡||𝖠T = tr(𝖠|𝖡) = (𝖠|𝖡)||𝖨T .

Double-wedge products 𝜶 ∧ 𝜷 𝜶 𝜷 ∧ (a )∧(b ) = (a ∧ b)( ∧ ), (ab)∧(cd) = (a ∧ c)(b ∧ d) 𝖠∧𝖡 𝖡∧𝖠 𝖠∧𝖡 T 𝖠T ∧𝖡T ∧ = ∧ ,(∧ ) = ∧ 𝖠∧𝖡 ∧𝖢 𝖠∧ 𝖡∧𝖢 𝖠∧𝖡∧𝖢 𝖡∧𝖠∧𝖢 ⋯ ( ∧ )∧ = ∧( ∧ ) = ∧ ∧ = ∧ ∧ = 𝖠∧𝖡 | 𝖠| 𝖡| 𝖡| 𝖠| ( ∧ ) (a ∧ b) = ( a) ∧ ( b) + ( a) ∧ ( b) 380 APPENDIX C Multivector and Dyadic Identities

𝖠∧𝖡 ⌊ 𝖠| 𝖡 𝖡| 𝖠 ( ∧ ) a = ( a) ∧ + ( a) ∧ ⌋ 𝖠∧𝖡 𝖠 |𝖡 𝖡 |𝖠 . a ( ∧ ) = ∧ (a ) + ∧ (a )

Double-wedge powers ∑ ∑ ∑ ∑ 𝖠(2) 1𝖠∧𝖠 1 𝜶 ∧ 𝜶 𝜶 𝜶 𝜶 = ∧ = ai i∧ aj j = (ai ∧ aj)( i ∧ j) = aij ij 2 2 i

𝖨(p)| ⋯ ⋯ (a1 ∧ a2 ∧ ∧ ap) = a1 ∧ a2 ∧ ap tr𝖨 = 4, tr(𝖨(2)) = 6, tr𝖨(3) = 4, tr𝖨(4) = 1 𝖨∧𝖠 𝖠 𝖨(2) 𝖨(3)⌊⌊𝖠T ∧ = (tr ) − 1 tr𝖠(2) = ((tr𝖠)2 − tr𝖠2) 2 1 tr𝖠(3) = ((tr𝖠)3 − 3tr𝖠 tr𝖠2 + 2tr𝖠3) 6 1 tr𝖠(4) = ((tr𝖠)4 − 6(tr𝖠)2 tr𝖠2 + 8tr𝖠 tr𝖠3 + 3(tr𝖠2)2 − 6tr𝖠4) 24 𝖨(2) 𝜺 𝜺 𝜺 𝖨(3) 𝜺 s = e12 12 + e23 23 + e31 31, s = e123 123 𝖨(2) 𝖨(2) 𝖨 ∧ 𝜺 = s + s∧e4 4 𝖨(3) 𝖨(3) 𝖨(2)∧ 𝜺 = s + s ∧e4 4 𝖤 ⌊𝖨(2)T 𝖨(2)⌋ ⌊𝖨(2)T T 𝖨(2)⌋ T = eN = eN = (eN ) = ( eN)

= e12e34 + e23e14 + e31e24 + e14e23 + e24e31 + e34e12 𝖨(2)⌊⌊𝖨T = 3𝖨, 𝖨(3)⌊⌊𝖨T = 2𝖨(2), 𝖨(3)⌊⌊𝖨(2)T = 3𝖨 𝖨(4)⌊⌊𝖨T = 𝖨(3), 𝖨(4)⌊⌊𝖨(2)T = 𝖨(2), 𝖨(4)⌊⌊𝖨(3)T = 𝖨.

Multivectors and unit dyadics (a ∧ b)⌊𝖨T = ba − ab =−𝖨⌋(a ∧ b) = (𝖨⌋(a ∧ b))T (a ∧ b ∧ c)⌊𝖨T = (a ∧ b)c + (b ∧ c)a + (c ∧ a)b = 𝖨(2)⌋(a ∧ b ∧ c) (a ∧ b ∧ c)⌊𝖨(2)T = a(b ∧ c) + b(c ∧ a) + c(a ∧ b) = 𝖨⌋(a ∧ b ∧ c) (a ∧ b ∧ c ∧ d)⌊𝖨T =−𝖨(3)⌋(a ∧ b ∧ c ∧ d) = d ∧ ((a ∧ b)c + (b ∧ c)a + (c ∧ a)b) − (a ∧ b ∧ c)d (a ∧ b ∧ c ∧ d)⌊𝖨(2)T = (a ∧ b)(c ∧ d) + (c ∧ a)(b ∧ d) + (b ∧ c)(a ∧ d) +(a ∧ d)(b ∧ c) + (b ∧ d)(c ∧ a) + (c ∧ d)(a ∧ b) = 𝖨(2)⌋(a ∧ b ∧ c ∧ d) = {(b ∧ c)a + (c ∧ a)b + (a ∧ b)c} ∧ d − d ∧ {a(b ∧ c) + b(c ∧ a) +c(a ∧ b)} (a ∧ b ∧ c ∧ d)⌊𝖨(3)T =−𝖨⌋(a ∧ b ∧ c ∧ d) 382 APPENDIX C Multivector and Dyadic Identities

= (a(b ∧ c) + b(c ∧ a) + c(a ∧ b)) ∧ d − d(a ∧ b ∧ c) ∑ ∑ 𝖨(2)⌋ 𝜺 ⌋ i+j−1 < eN = eij ij eN = (−1) eijeK(ij), i j ⌊𝖨T 𝖨⌋ 𝖨⌋ 𝖨(3)⌋ 𝔼 (A ∧ B) = A ∧ B + B ∧ A =− (A ∧ B), A, B ∈ 2 𝖨⌋(A ∧ B) = A⌊𝖨T ∧ B + B⌊𝖨T ∧ A =−(A ∧ B)⌋𝖨(3)T ⌊𝖨(2)T ⌊𝖨T ∧ ⌊𝖨T 𝖨(2)⌋ (A ∧ B) = AB + BA − (A )∧(B ) = (A ∧ B) 1 1 (A ∧ A)⌊𝖨(2)T = AA − (A⌊𝖨T )(2) = 𝖨(2)⌋(A ∧ A) 2 2 𝖠T | 횽⌊𝖠 횽|𝖠(2)⌊𝖨T 𝖠 𝔼 𝔽 횽 𝔽 ( ) = , ∈ 1 1, ∈ 2 ⌊𝖨T 𝖨⌋ 𝔼 a ∧ (kN ) = kNa,(kN) ∧ a = akN a ∈ 1 ⌊𝖨(2)T 𝖨(2)⌋ 𝔼 A ∧ (kN ) = kNA,( kN) ∧ A = AkN, A ∈ 2 (A⌊𝖨T ∧ B + B⌊𝖨T ∧ A) ∧ a = a(A ∧ B) 𝜺 ⌊ ⌊ ⌊𝖨T 𝜺 | 𝖨T ( N A) (A ) =− N (A ∧ A) ⌊𝖨T ∧ ⌊𝖨T ⌊𝖨(2)T (A )∧(B ) = AB + BA − (A ∧ B) 𝖨 ⌋ 𝖨 ⌋ ⌊𝖨T As ∧ s Bs + Bs ∧ s As = 0, As s ∧ A = 0 ⌊𝖨T ∧ ⌊𝖨T ⌊𝖨T (2) . (As s )∧(Bs s ) = AsBs + BsAs,(As s ) = AsAs 𝖲∧ ⌊𝖨T 𝖲 ⌋𝜺 ∧𝖨 ⌋ 𝖲 𝖲T 𝔼 𝔼 ∧(A ) = ((( ∧ A) N)∧ ) eN, = ∈ 1 1

Double multiplications 𝖠∧𝖡 𝖠 𝖡 𝖠|𝖡 tr( ∧ ) = (tr )(tr ) − tr( ) 𝖠∧𝖡 | 𝖢∧𝖣 𝖠|𝖢 ∧ 𝖡|𝖣 𝖠|𝖣 ∧ 𝖡|𝖢 ( ∧ ) ( ∧ ) = ( )∧( ) + ( )∧( ) 𝖠∧𝖡 | 𝖢∧𝖣 𝖠∧𝖡 || 𝖢∧𝖣 T tr(( ∧ ) ( ∧ )) = ( ∧ ) ( ∧ ) = (𝖠||𝖢T )(𝖡||𝖣T ) + (𝖠||𝖣T )(𝖡||𝖢T ) − (𝖠|𝖣)||(𝖡|𝖢)T − (𝖠|𝖢)||(𝖡|𝖣)T 𝖠∧𝖡 ⌊⌊𝖢T 𝖠||𝖢T 𝖡 𝖡||𝖢T 𝖠 𝖠|𝖢|𝖡 𝖡|𝖢|𝖠 ( ∧ ) = ( ) + ( ) − − 𝖠(2)⌊⌊𝖡T = (𝖠||𝖡T )𝖠 − 𝖠|𝖡|𝖠 𝖠(2)⌊⌊𝖨T = (tr𝖠)𝖠 − 𝖠2, 𝖨(2)⌊⌊𝖠T = (tr𝖠)𝖨 − 𝖠 𝖠∧𝖡∧𝖢 ⌊⌊𝖣T 𝖠∧𝖡 𝖢||𝖣T 𝖡∧𝖢 𝖠||𝖣T 𝖢∧𝖠 𝖡||𝖣T ( ∧ ∧ ) = ( ∧ )( ) + ( ∧ )( ) + ( ∧ )( ) 𝖠∧ 𝖡|𝖣|𝖢 𝖢|𝖣|𝖡 𝖡∧ 𝖢|𝖣|𝖠 𝖠|𝖣|𝖢 𝖢∧ 𝖠|𝖣|𝖡 𝖡|𝖣|𝖠 − ∧( + ) − ∧( + ) − ∧( + ) APPENDIX C Multivector and Dyadic Identities 383

𝖠(3)⌊⌊𝖡T 𝖠||𝖡T 𝖠(2) 𝖠∧ 𝖠|𝖡|𝖠 = ( ) − ∧( ) 𝖠(3)⌊⌊𝖨T 𝖠 𝖠(2) 𝖠∧𝖠2 𝖨(3)⌊⌊𝖠T 𝖠 𝖨(2) 𝖨∧𝖠 = (tr ) − ∧ , = (tr ) − ∧ 𝖨(3)⌊⌊ 𝖠∧𝖡 T 𝖠∧𝖡 𝖨 𝖠∧𝖡 ⌊⌊𝖨T ( ∧ ) = tr( ∧ ) − ( ∧ ) = (tr𝖠)(tr𝖡)𝖨 − tr(𝖠|𝖡)𝖨 − (tr𝖠)𝖡 − (tr𝖡)𝖠 + 𝖠|𝖡 + 𝖡|𝖠 𝖠(3)⌊⌊𝖢T 𝖠(2)|𝖢 𝖠 𝖠| 𝖢⌊⌊𝖠T |𝖠 𝖢 𝔼 𝔽 = tr( ) − (( ) , ∈ 2 2 𝖨(3)⌊⌊𝖢T 𝖢 𝖨 𝖢⌊⌊𝖨T 𝖢 𝔼 𝔽 = (tr ) − , ∈ 2 2 𝖠∧𝖡∧𝖢∧𝖣 ⌊⌊𝖤T 𝖠∧𝖡∧𝖢 𝖣||𝖤T 𝖠∧𝖡∧𝖣 𝖢||𝖤T ( ∧ ∧ ∧ ) = ( ∧ ∧ )( ) + ( ∧ ∧ )( ) 𝖠∧𝖢∧𝖣 𝖡||𝖤T 𝖡∧𝖢∧𝖣 𝖠||𝖤T 𝖠∧𝖡 ∧ 𝖢|𝖤|𝖣 𝖣|𝖤|𝖢 +( ∧ ∧ )( ) + ( ∧ ∧ )( ) − ( ∧ )∧( + ) 𝖠∧𝖢 ∧ 𝖡|𝖤|𝖣 𝖣|𝖤|𝖡 𝖠∧𝖣 ∧ 𝖡|𝖤|𝖢 𝖢|𝖤|𝖡 𝖡∧𝖢 ∧ 𝖠|𝖤|𝖣 −( ∧ )∧( + ) − ( ∧ )∧( + ) − ( ∧ )∧( 𝖣|𝖤|𝖠 𝖡∧𝖣 ∧ 𝖠|𝖤|𝖢 𝖢|𝖤|𝖠 𝖢∧𝖣 ∧ 𝖠|𝖤|𝖡 𝖡|𝖤|𝖠 + ) − ( ∧ )∧( + ) − ( ∧ )∧( + ) 𝖠(4)⌊⌊𝖡T 𝖠||𝖡T 𝖠(3) 𝖠(2)∧ 𝖠|𝖡|𝖠 = ( ) − ∧( ) 𝖠(4)⌊⌊𝖨T 𝖠 𝖠(3) 𝖠(2)∧𝖠2 𝖨(4)⌊⌊𝖠T 𝖠 𝖨(3) 𝖨(2)∧𝖠 = (tr ) − ∧ , = (tr ) − ∧ 𝖨(4)⌊⌊ 𝖠∧𝖡 T 𝖠∧𝖡 𝖨(2) 𝖠∧𝖡 ⌊⌊𝖨T ∧𝖨 𝖠∧𝖡 ( ∧ ) = tr( ∧ ) − (( ∧ ) )∧ + ∧ 𝖨(4)⌊⌊𝖢T 𝖢 𝖨(2) 𝖢⌊⌊𝖨T ∧𝖨 𝖢 𝖢 𝔼 𝔽 = (tr ) − ( )∧ + , ∈ 2 2 𝖨(4)⌊⌊ 𝖠∧𝖡∧𝖢 T 𝖠∧𝖡∧𝖢 𝖨 𝖠∧𝖡∧𝖢 ⌊⌊𝖨(2)T ( ∧ ∧ ) = tr( ∧ ∧ ) − ( ∧ ∧ ) 𝖨(4)⌊⌊𝖣T 𝖣 𝖨 𝖣⌊⌊𝖨(2)T 𝖣 𝔼 𝔽 = (tr ) − , ∈ 3 3 𝖠(p+1)⌊⌊𝖠−1T = (n − p)𝖠(p), 𝖠(p)∧𝖨(4−p) 𝖠(p) 𝖨(4) ∧ = (tr ) 𝖠(p)⌊⌊𝖨(p−1)T = (𝖨(p)⌊⌊𝖠(p−1)T )|𝖠, p > 1 𝖨(p+1)⌊⌊𝖠(p)T = (tr𝖠(p))𝖨 − (𝖨(p)⌊⌊𝖠(p−1)T )|𝖠, p > 1 = (tr𝖠(p))𝖨 + (tr𝖠(p−1))(−𝖠) + (tr𝖠(p−2))𝖠2 + ⋯ + (−𝖠)p 4 0 = tr𝖠(4) 𝖨 − tr𝖠(3) 𝖠 + tr𝖠(2) 𝖠2 − tr𝖠 𝖠3 + 𝖠 .

Inverse dyadics (𝖠|𝖡)−1 = 𝖡−1|𝖠−1,det𝖠 = tr𝖠(4) (𝖠(p))−1 = (𝖠−1)(p) = (def) 𝖠(−p),1< p < 4, 384 APPENDIX C Multivector and Dyadic Identities

𝖨(4)⌊⌊𝖠(3)T 𝖠−1 𝖠 𝔼 𝔽 = , ∈ 1 1 det𝖠 𝜺 𝜺 ⌊⌊𝖠(3)T 𝖠−1 N N 𝖠 𝔼 𝔼 = , ∈ 1 1 𝜺 𝜺 ||𝖠(4) N N 𝖨(4)⌊⌊𝖠(3)T 𝖠(−p) 𝖠 𝔼 𝔽 = , ∈ 1 1 det 𝖠 𝜺 𝜺 ⌊⌊𝖠(3)T 𝖠(−p) N N 𝖠 𝔼 𝔼 = , ∈ 1 1 𝜺 𝜺 ||𝖠(4) N N ⌊𝖨(p)T −1 p(4−p)𝜺 ⌊𝖨(4−p) (eN ) = (−1) N 𝖠−1| 𝜶|𝖠−1 𝖠 𝜶 −1 𝖠−1 a 𝖠 𝔼 𝔽 ( + a ) = − , ∈ 1 1 1 + 𝜶|𝖠−1|a 𝖢−1| |𝖢−1 𝖢 −1 𝖢−1 AB 𝖢 𝔼 𝔼 ( + AB) = − , ∈ 2 2 1 + B|𝖢−1|A ⌊𝖨T −1 𝜺 ⌊ ⌊𝖨 𝜺 | (A ) =−2( N A) ∕( N (A ∧ A)) 1 (𝖡 ∧𝖨)−1 = (𝖡2∧𝖨 + tr𝖡(2) 𝖨(2)), tr𝖡 = 0 o∧ 𝖡(3) o∧ o o tr o (𝖠−1|b − 𝜺 )(a |𝖠−1 − 𝜺 ) 𝖠 −1 𝖠−1 s s 4 s s 4 ( s + e4as + bse4 + ce4e4) = + s |𝖠−1| c − as s bs (𝖠−1|b − e )(𝜶 |𝖠−1 − 𝜺 ) 𝖠 𝜶 𝜺 𝜺 −1 𝖠−1 s s 4 s s 4 ( s + e4 s + bs 4 + ce4 4) = + s 𝜶 |𝖠−1| c − s s bs 𝖠 𝜶 𝜺 𝜺 | 𝖠(3)T 𝖠(2)T ∧𝜶 |𝜺 det( s + e4 s + bs 4 + ce4 4) = e123 (c s − s ∧ sbs) 123.

Inverse of spatial dyadics 1 𝖠−1 = 𝖨(3)⌊⌊𝖠(2)T , 𝖠 ∈ 𝔼 𝔽 s 𝖠(3) s s s 1 1 tr s

𝖢−1 1 𝖢T ⌋⌋𝖨(3) (2) 𝖢 𝔼 𝔽 . = ( ) , s ∈ 2 2 s 𝖨(3)⌊⌊𝖢T (3) s s tr( s s ) APPENDIX C Multivector and Dyadic Identities 385

METRIC DYADICS

Spatial metric dyadics 𝖦  𝜺 𝜺 𝜺 𝜺 𝜺 𝜺 s = e1e1 + e2e2 + e3e3, s = 1 1 + 2 2 + 3 3 𝖦(2) (2) 𝜺 𝜺 𝜺 𝜺 𝜺 𝜺 s = e12e12 + e23e23 + e31e31, s = 12 12 + 23 23 + 31 31 𝖦(3) (3) 𝜺 𝜺 s = e123e123, s = 123 123.

4D Minkowski space 𝖦 −1 = e1e1 + e2e2 + e3e3 − e4e4 =  𝜺 𝜺 𝜺 𝜺 𝜺 𝜺 𝜺 𝜺 𝖦−1 = 1 1 + 2 2 + 3 3 − 4 4 = 𝖦(2) 𝖦 ∧ (−2) = e12e12 + e23e23 + e31e31 − s∧e4e4 = (2) 𝜺 𝜺 𝜺 𝜺 𝜺 𝜺  ∧𝜺 𝜺 𝖦(−2) = 12 12 + 23 23 + 31 31 − s∧ 4 4 = 𝖦(3) 𝖦(2)∧ (−3) = e123e123 − s ∧e4e4 = (3) 𝜺 𝜺 (2)∧𝜺 𝜺 = 123 123 − s ∧ 4 4 𝖦(4) 𝖦(3)∧ (4) (3)∧𝜺 𝜺 𝜺 𝜺 . =− s ∧e4e4 =−eNeN, =− s ∧ 4 4 =− N N

ELECTROMAGNETICS

Maxwell equations 횽 𝜸 횿 𝜸 d ∧ = m, d ∧ = e 횽 𝜺 횿 𝜺 = B + E ∧ 4, = D − H ∧ 4 𝜸 𝝔 𝜺 𝜸 𝝔 𝜺 m = m − Jm ∧ 4, e = e − Je ∧ 4 휕 휕 ds ∧ E + 휏 B =−Jm, −ds ∧ H + 휏 D =−Je 𝝔 𝝔 ds ∧ B = m, ds ∧ D = e 휕 𝝔 휕 𝝔 ds ∧ Jm + 휏 m = 0, ds ∧ Je + 휏 e = 0 𝜸 ⇒ 횽 𝜶 𝜶 휙𝜺 . m = 0, = d ∧ , = A − 4 386 APPENDIX C Multivector and Dyadic Identities

Medium equations 횿 𝖬|횽 횽 𝖭|횿 𝖬 𝖭 𝔽 𝔼 = , = , , ∈ 2 2, ⌊횿 𝖬 |횽 ⌊횽 𝖭 |횿 ⌊⌊𝖬−1 |횿. eN = m , eN = m = (eNeN m ) 𝖬 ⌊⌊𝖬−1 𝔼 𝔼 𝖬−1 𝔽 𝔽 Modified medium dyadic m,(eNeN m ) ∈ 2 2, m ∈ 2 2 𝖬 ⌊𝖬 𝖬 𝜺 ⌊𝖬 m = eN , = N m 𝖬−1 𝖬−1⌋ 𝖬−1 𝖬−1⌋𝜺 . = m eN, m = N

Expansions ( ) ( ) ( ) D 훼 휖′ B = | , H 휇−1 훽 E

훼 𝔽 𝔼 휖′ 𝔽 𝔼 휇−1 𝔽 𝔼 훽 𝔽 𝔼 ∈ 2 2, ∈ 2 1, ∈ 1 2, ∈ 1 1 𝖬 훼 휖′ 𝜺 휇−1 𝜺 훽 = + ∧ e4 + 4 ∧ + 4 ∧ ∧ e4 ( ) ( ) ( ) D 휖 휉 E = | B 휁 휇 H

휖 휇 휉 휁 𝔽 𝔼 , , , ∈ 2 1 𝖬 휖 𝜺 𝖨T 휉 |휇−1| 𝖨(2)T 휁 = ∧ e4 + ( 4 ∧ + ) ( − ∧ e4) 𝖬−1 휇 𝜺 𝖨T 휁 |휖−1| 𝖨(2)T 휉 =− ∧ e4 − ( 4 ∧ − ) ( + ∧ e4) 휖 𝜺 휇−1 −1 휇 𝜺 휖−1 ( ∧ e4 + 4 ∧ ) =−( ∧ e4 + 4 ∧ ) ( ) ( ) ( ) ⌊ 휖 휉 e123 D g g | E . ⌊ = 휁 휇 e123 B g g H 휖 휇 휉 휁 𝔼 𝔼 Gibbsian dyadics g, g, g, g ∈ 1 1 𝖬 𝜺 ⌊휖 𝜺 ⌊휉 𝜺 𝖨T |휇−1| 𝖨⌋ 휁 = 123 g ∧ e4 + ( 123 g + 4 ∧ ) g ( e123 − g ∧ e4) 𝖬 휖 ∧ ⌊𝖨T 휉 |휇−1| 𝖨⌋ 휁 m = g∧e4e4 − (e123 + e4 ∧ g) g ( e123 − g ∧ e4) 𝖬−1 𝜺 𝜺 ⌊⌊휇 𝜺 𝖨T 𝜺 ⌊휁 |휖−1| 𝖨 𝜺 휉 ⌋𝜺 m =− 123 123 g − ( 4 ∧ − 123 g) g ( ∧ 4 + g 123) APPENDIX C Multivector and Dyadic Identities 387

Relations 휖′ = 휖 − 휉|휇−1|휁, 훼 = 휉|휇−1, 훽 =−휇−1|휁 휖 = 휖′ − 훼|휇|훽, 휉 = 훼|휇, 휁 =−휇|훽 휖′ 𝜺 ⌊ 휖 휉 |휇−1|휁 훼 𝜺 ⌊ 휉 |휇−1 ⌋ = 123 ( g − g g g), = 123 ( g g ) e123, 훽 휇−1|휁 휇 𝜺 ⌊휇 =− g g, = 123 g 휖 ⌊휖 휖 𝜺 ⌊휖 g = e123 , = 123 g,etc 휖−1 휖−1⌋ 휖−1 휖−1⌋𝜺 . = g e123, g = 123,etc

References

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affine transformation, 141, 173, 200, 215 DB boundary, 121, 151, 188, 208 almost-complex (AC) bidyadic, 91, 234 de Rahm’s theorem, 101 almost-complex (AC) dyadic, 87 decomposable (DC) media, 132, 177, antisymmetric bidyadic, 57 214, 259 antisymmetric dyadic, 47 decomposition rule, 8, 15 axion medium, 178, 225 determinant of bidyadic, 57 axion part, 62, 111 differentiation operator, 101 axion-free medium, 112 dispersion dyadic, 130, 250 dispersion equation, 130, 134, 177 bac-cab rule, 7, 14, 103, 372, 377 dot product, 2, 17, 24, 104 bar product, 1, 5, 10, 22, 376 double contraction, 30 basis, 2, 9, 15 double-bar product, 30, 379 bi-index, 4 double-cross product, 26 bidyadic, 25, 53 double-wedge power, 28, 380 bidyadic eigenproblem, 58 double-wedge product, 25, 379, 382 bilinear invariant, 123 duality transformation, 145, 181, 200, biquadratic (BQ) media, 241 229 bivector, 3, 17 dyadic, 21 bivector invariant, 16 dyadic identities, 32, 375 boundary conditions, 119, 150, 187 dyadic product, 21

Cayley-Hamilton equation, 36, 54 eigenbivector, 54 charge three-form, 101 eigenfields in GBI medium, 174 chirality parameter, 172 eigenproblem for bidyadics, 58 closure relation, 93 eigenvector, eigen one-form, 41 complementary index, 5 electromagnetic two-forms, 101 complementary projection dyadic, 83 energy-power three-form, 125 conformal transformation, 159 expansion of medium bidyadic, 107, 386 contraction identities, 11, 32, extended bidyadic, 70 contraction product, 6, 11, 376 extended P medium, 218, 265 corrugated surface, 123 extended Q medium, 211, 264 cross product, 372 extended skewon-axion medium, 192, CU media, 235 266

395 396 Index

Gibbsian bi-isotropic (GBI) medium, one-form, 1 170, 226 ordered indices, 4 Gibbsian formalism, 369 orthogonality of bidyadics, 81 Gibbsian isotropic medium, 169, 175 orthogonality of dyadics, 79 Gibbsian medium dyadics, 109 Gibbsian vector fields, 105 P medium, 197, 258 P-axion medium, 209, 258 handedness of plane wave, 135 P-solution, 95 Heaviside unit step, 118 PDC medium, 265 Hehl-Obukhov decomposition, 61, 110, PEC boundary, 120, 151 171 PEMC boundary, 120, 151, 181 PEMC (axion) medium, 179 IB medium, 182 plane wave, 128, 152, 176, 184, 267 interface conditions, 117 plane wave in extended P medium, 219 invariant of metric dyadic, 45 plane wave in extended Q medium, 215 inverse bidyadic, 66, 71, 384 plane wave in GBI medium, 176 inverse dyadic, 36, 48, 383 plane wave in P medium, 207 inverse of skewon bidyadic, 67 plane wave in Q medium, 205 inverse of spatial dyadic, 38, 48, 384 plane-wave fields, 132 inversion transformation, 159 PMC boundary, 119 involutionary duality transformation, potential equation, 107 147 potential one-form, 105 Power and energy, 123 left-handed medium, 136 Poynting two-form, 125, 147 Lorenz condition, 112 principal part, 62, 111 principal-axion medium, 112 magnetic sources, 104 projection dyadics and bidyadics, 83 Maxwell equations, 107, 385 Maxwell stress-energy dyadic, 126 Q medium, 197 media with no dispersion equation Q-axion medium, 209 (NDE), 132, 184, 207, 249 Q-solution, 96 medium bidyadic, 106 QDC medium, 264 medium equations, 106, 386 quadratic invariant, 123 metasurface, 121 quadratic media, 197 metric dyadic, 24, 45, 385 quadrivector, 9, 378 modified closure relation, 93, 199 quintivector, 14, 378 modified medium bidyadic, 106 multivector identities, 375 rank of bidyadic, 54 multivectors, multiforms, 8, 375 rank of dyadic, 39 reciprocal basis, 2, 13, 376 natural dot product, 17, 31, 53 reciprocal medium, 155 NDE media, 132, 184, 207, 249 reciprocity transformation, 153, 172 nilpotent dyadics and bidyadics, 81 non-birefringent medium, 132, 135 SDC medium, 265 no-sign product, 21 SDN medium, 234 Index 397 self-dual (SD) medium, 149, 227 symmetric dyadic, 46 SH boundary, 122, 151, 189, 208 symmetric Q-axion medium, 227 SHDB boundary, 190 similarity transformation, 55, 88, TE/TM decomposition, 259 234 Tellegen parameter, 113, 171 simple antisymmetric bidyadic, 64, 124, time reversion bidyadic, 55, 153, 172 146 time-harmonic fields, 158 simple bivector, 4 trace, 23, 29 simple principal medium, 114, 134, 145, trivector, three-form, 8, 377 152, 170, 198 two-form, 3 simple two-form, 4 skewon bidyadic, 67 uniaxially anisotropic medium, 259 skewon medium, 143 unipotent dyadic, 66, 85 skewon part, 62, 111 unipotent bidyadic, 85 skewon-axion medium, 112, 182, unit bidyadic, 29, 380 258 unit dyadic, 22, 380 soft-and-hard (SH) boundary, 122, 151, 189 vector, 1 source three-form, 101 spatial medium dyadics, 108 wave one-form, 128 stress-energy dyadic, 124, 147, 176 wedge product, 3 IEEE PRESS SERIES ON ELECTROMAGNETIC WAVE THEORY

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