Electromagnetic An Introductory Course Waves
M.D. Verweij, P.M. van den Berg, H. Blok
Electromagnetic Waves — An Introductory Course
About the authors Martin Verweij (1961) graduated in 1992 from Delft University of Technology, Delft, Netherlands. From 19931997, he was a research fellow of the Royal Netherlands Academy of Arts and Sciences at the Laboratory of Electromagnetic Research of the Delft University of Technology, and since 1998 he is associate professor in the same laboratory. He has written a range of papers on integral transformation methods for electromagnetic, acoustic and elastic waves. He has been teaching various classes on basic and advanced electromagnetic wave theory since 1994 and was chosen Best Teacher of Electrical Engineering in 2003. Peter van den Berg (1943) was a member of the scientific staff of the Laboratory of Electromagnetic Research, Delft University of Technology, Delft, Netherlands. He graduated in 1971 from the Delft University of Technology, and during the academic year 19731974 he was Visiting Lecturer in the Department of Mathematics, University of Dundee, Scotland. He was appointed as a full professor at the Laboratory of Electromagnetic Research in 1981. Since 2003 he is a research professor in the Faculty of Applied Sciences of the Delft University of Technology. During his career, he has written an impressive amount of papers on the numerical analysis of forward and inverse electromagnetic wavefield problems, and he has been teaching various classes on electromagnetic, acoustic and elastic waves. Hans Blok (1935) was a member of the scientific staff of the Laboratory of Electromagnetic Research, Delft University of Technology, Delft, Netherlands. He graduated in 1970 from the Delft University of Technology, and he was appointed full professor at the Laboratory of Electromagnetic Research in 1980. He was dean of the Faculty of Electrical Engineering, Delft University of Technology in the period 19801982. During the academic year 19831984 he was a visiting scientist at SchlumbergerDoll Research, Ridgefield, Connecticut, U.S.A. During his career, he has written a number of papers on resonators and optical waveguides, and he has been teaching various classes on electromagnetic waves and signal theory. He is emeritus professor since 2000.
Electromagnetic Waves — An Introductory Course
M.D. Verweij P.M. van den Berg H. Blok
VSSD
© VSSD First edition 1999 Second edition 20012006 Published by: VSSD Leeghwaterstraat 42, 2628 CA Delft, The Netherlands tel. +31 15 278 2124, telefax +31 15 278 7585, email:
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EAN 9789040718366
Keywords: electromagnetic waves
Preface The course ”Electromagnetic Waves” oﬀers an introduction in the theoretical concepts of electromagnetic waves. This course book contains the basic material on timevarying waveﬁelds and their applications in electrical engineering, e.g., electromagnetic compatibility, communication and remote sensing. A prerequisite to this course is a standard course ”Electricity and Magnetism” where, from experimental laws, the Maxwell equations for timevarying electromagnetic ﬁelds are formulated as a system of partial diﬀerential equations. Chapter 1 reviews the necessary mathematical background, while Chapter 2 introduces the fundamental mathematical equations: the Maxwell equations, the constitutive relations and boundary conditions. The main line of the course is the construction of solutions to these equations in some simple conﬁgurations. The concept of an electromagnetic wave is introduced in Chapter 3, where onedimensional waves are discussed. A wave phenomenon can only be understood in connection with an electromagnetic source that generates a wave. For the excitation of onedimensional waves, the planarelectriccurrent sheet is chosen. As a simple example of onedimensional wave propagation, the parallelplate waveguide is discussed shortly. In Chapter 4, the twodimensional waves are studied, in particular speciﬁc properties as interference, Fresnel reﬂection/transmission factors, Brewster’s angle and total reﬂection are treated. In Chapter 5, the consequences of a weakly inhomogeneous medium are discussed and the theory of electromagnetic rays is introduced. Further, in Chapters 6 and 7, the theory of transmission lines and electromagnetic waveguides is treated. In view of communication applications, the closed parallelplate waveguide and the open dielectricslab waveguide are described in full detail. Finally, Chapter 8 deals with the excitation of twodimensional waves and the concept of the farﬁeld approximation is introduced. The student who has successfully completed the present introductory course on electromagnetic waves, has learned the basic concepts of electromagnetic wave propagation. By simplifying the problems in such a way
vi
preface
that a description in terms of onedimensional and twodimensional waves suﬃces, more attention can be given to the physical understanding of the propagation phenomena. However, it is stressed that in more realistic conﬁgurations of presentday technology, a full threedimensional description of electromagnetic waveﬁelds is needed. In this context, it is noted that the methodology of handling the radiation and scattering of electromagnetic waves in threedimensional conﬁgurations will be treated in more advanced courses of the electrical engineering curriculum. The authors acknowledge Dr. E. C. Slob for compiling the original set of exercises, problems and answers; Dr. M. D. Verweij for contributing to the material of Chapters 4 and 5, revising the exercises, problems and answers, and preparing the ﬁnal print version; and Mr. K. F. I. Haak for rechecking the answers.
Delft, January 1999
H. Blok P.M. van den Berg
Preface to the second edition This edition is identical to the ﬁrst edition, except that a number of errors have been corrected. The authors acknowledge their collegues of the Laboratory of Electromagnetic Research and in particular Dr. D. Quak and Mr. P. Jorna for reporting most of these errors.
Delft, September 2001
M.D. Verweij P.M. van den Berg H. Blok
Contents Preface
v
1 Introduction 1.1. Cartesian vectors and their properties . . . . . . . . . . . . .
1 4
1.1.1. Addition, subtraction and multiplication of vectors . .
4
1.1.2. Diﬀerentiation with respect to a parameter . . . . . .
6
1.1.3. Diﬀerentiation with respect to the spatial coordinates
6
1.2. Exercises and problems . . . . . . . . . . . . . . . . . . . . . 2 The Electromagnetic Field Equations
17 21
2.1. Force exerted on an electric point charge . . . . . . . . . . . .
21
2.2. Maxwell’s equations in vacuum . . . . . . . . . . . . . . . . .
24
2.3. Maxwell’s equations in matter . . . . . . . . . . . . . . . . . .
25
2.4. The constitutive relations . . . . . . . . . . . . . . . . . . . . 2.5. The system of ﬁeld equations . . . . . . . . . . . . . . . . . .
28 31
2.6. The boundary conditions . . . . . . . . . . . . . . . . . . . .
32
2.7. Frequencydomain representations . . . . . . . . . . . . . . .
36
2.7.1. The frequencydomain ﬁeld equations . . . . . . . . .
37
2.8. Polarization state . . . . . . . . . . . . . . . . . . . . . . . . . 2.9. Poynting’s theorem . . . . . . . . . . . . . . . . . . . . . . . .
39 41
2.10. Exercises and problems . . . . . . . . . . . . . . . . . . . . .
45
3 Onedimensional Electromagnetic Waves 3.1. The planar electriccurrent sheet as emitter . . . . . . . . . .
49 50
viii
contents
3.2. Steadystate analysis . . . . . . . . . . . . . . . . . . . . . . .
54
3.2.1. Lossless medium . . . . . . . . . . . . . . . . . . . . . 3.2.2. Lossy medium . . . . . . . . . . . . . . . . . . . . . .
55 57
3.3. Transient emission into a lossless medium . . . . . . . . . . . 3.4. Reﬂection and transmission problem . . . . . . . . . . . . . .
59 62
3.4.1. Electric ﬁeld analysis . . . . . . . . . . . . . . . . . . .
64
3.4.2. Magnetic ﬁeld analysis . . . . . . . . . . . . . . . . . .
65
3.5. Shielding problem . . . . . . . . . . . . . . . . . . . . . . . .
66
3.5.1. Electric ﬁeld analysis . . . . . . . . . . . . . . . . . . .
69
3.6. Parallelplate waveguide . . . . . . . . . . . . . . . . . . . . .
72
3.7. Exercises and problems . . . . . . . . . . . . . . . . . . . . .
77
4 Twodimensional Electromagnetic Waves
81
4.1. Plane waves in a homogeneous medium . . . . . . . . . . . .
83
4.1.1. Uniform plane waves . . . . . . . . . . . . . . . . . . .
87
4.2. Interference of two plane waves . . . . . . . . . . . . . . . . .
90
4.2.1. Steadystate analysis: lossless case . . . . . . . . . . .
92
4.3. Reﬂection of a plane wave by an electrically impenetrable halfspace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
95
4.4. Reﬂection and transmission of a plane wave incident upon a plane interface . . . . . . . . . . . . . . . . . . . . . . . . . . 101 4.4.1. Uniform plane waves in the frequency domain . . . . . 108 4.5. Exercises and problems . . . . . . . . . . . . . . . . . . . . . 116 5 Electromagnetic Rays in a Twodimensional Medium
121
5.1. Homogeneous, lossless medium . . . . . . . . . . . . . . . . . 122 5.2. Parallel polarization . . . . . . . . . . . . . . . . . . . . . . . 124 5.3. Perpendicular polarization . . . . . . . . . . . . . . . . . . . . 128 5.4. Ray trajectories . . . . . . . . . . . . . . . . . . . . . . . . . . 131 5.4.1. Ray trajectories in a horizontally layered medium . . . 133 5.4.2. Ray trajectories in a radially layered medium . . . . . 138 5.5. Exercises and problems . . . . . . . . . . . . . . . . . . . . . 143
contents
ix
6 Transmission Lines 145 6.1. TEMwaves . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146 6.2. Parallelplate waveguide . . . . . . . . . . . . . . . . . . . . . 151 6.3. Coaxial line . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152 6.4. Propagation properties . . . . . . . . . . . . . . . . . . . . . . 155 6.4.1. Twoconductor transmission line . . . . . . . . . . . . 156 6.4.2. Lossless transmission line: steadystate analysis . . . . 158 6.4.3. Transients on lossless transmission lines . . . . . . . . 160 6.5. Exercises and problems . . . . . . . . . . . . . . . . . . . . . 164 7 Electromagnetic Waveguides
167
7.1. Parallelplate waveguide . . . . . . . . . . . . . . . . . . . . . 172 7.2. Propagation properties of modes in a parallelplate waveguide 178 7.3. Dielectric slab waveguides . . . . . . . . . . . . . . . . . . . . 182 7.4. Propagation properties of guided modes in a dielectric slab waveguide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193 7.5. Exercises and problems . . . . . . . . . . . . . . . . . . . . . 198 8 Excitation of Twodimensional Electromagnetic Waves
203
8.1. The sheet emitter with a parallel electric current . . . . . . . 204 8.1.1. The farﬁeld approximation . . . . . . . . . . . . . . . 207 8.2. The sheet emitter with a perpendicular electric current . . . . 213 8.2.1. The farﬁeld approximation . . . . . . . . . . . . . . . 216 8.3. Exercises and problems . . . . . . . . . . . . . . . . . . . . . 221 Answers to Exercises
223
Bibliography
241
Index
243
Chapter 1
Introduction Electromagnetic waveﬁelds have a wide range of applications: from communication to medical treatment, from environmental sensing to energy radiation. When following an electromagnetic wave on its course, we start with its excitation by an electromagnetic source. Some of them are natural sources such as the sun and stars, others are artiﬁcial ones (a transmitting antenna, a laser). Once it has been generated, the wave propagates along a certain path from the source to the receiver. Depending on the properties of the medium through which the wave passes, this propagation can lead to continuous refraction by spatial changes in the medium parameters (for example, the atmosphere), or to discontinuous refraction by an abrupt change in the medium (for example, an interface between two diﬀerent media). Finally, the wave motion is picked up by an electromagnetic receiver (a receiving antenna, an optical detector). Each of these aspects is the subject of theoretical and experimental investigation. Usually, when the attention is focussed on a particular detail, the remaining circumstances are chosen as simple as possible. For example, when one wants to investigate the directional characteristics of a transmitting antenna, the surrounding medium will be taken of the utmost simplicity, as far as its electromagnetic properties are concerned, and of inﬁnite extent. When studying refraction phenomena during the propagation of an electromagnetic wave, the source will be taken a simple one, while the inﬂuence
2
introduction
of the receiver will be neglected at all. These simpliﬁcations are dictated by the impossibility to take into account the inﬂuence of all parameters simultaneously. The basic laws of macroscopic electromagnetic theory were formulated by James Clerk Maxwell and can be found in his famous book (Maxwell 1873). For a survey of the history of the subject the reader is referred to Whittaker (Whittaker 1953). From the theory it follows that there exist electromagnetic waves that travel with a ﬁnite speed which in vacuo seems to be a universal constant, independent of the state of motion in which the observer carries out his or her experiments. (The latter is not the case for waves in matter.) Since through a wave motion with constant speed the changes in position in space and the changes in time are interrelated in a rigid manner, electromagnetic waves in vacuum can serve to interconnect the spacetime observations for two observers in relative motion. This concept has led Einstein (Einstein 1956) to the theory of relativity. We shall conﬁne our analysis of electromagnetic waves to the case where the sources that generate the waveﬁeld, and the observer are at rest with respect to the material media in the conﬁguration. As in any type of wave motion, the physical quantities that describe the electromagnetic waves, depend on position and time. Their time dependence in the domain where the source is acting is impressed by the excitation mechanism of the source. The subsequent dependence on position and time elsewhere is governed by propagation laws. The physical laws that underly the properties of waves are induced from a series of basic standard experiments. To carry out these experiments, an observer must be able to register the position and the instant at which an observation is made. To register the position the existence of an isotropic background space is preassumed. In this space, distance can be measured along three mutually perpendicular directions with one and the same position and orientationindependent standard measuring rod. To register instants, the existence of a positionand orientationindependent standard clock is preassumed. The standard measuring rod is used to deﬁne, at a certain position which is denoted as the origin O, an orthogonal Cartesian reference frame consisting of three base vectors {i1 , i2 , i3 } that are of unit length each. The orientations of these three base vectors form a mutually perpendicular, righthanded triad (Fig. 1.1). (The property that each base vector speciﬁes geometrically a
3
introduction
length and an orientation, makes it a vectorial quantity, or a vector; notationally, vectors will be represented by bold face symbols.) Let {x1 , x2 , x3 } denote the three numbers that are needed to specify the position of an observer, then the vectorial position of the observer x is the linear combination (Fig. 1.2) (1.1) x = x1 i1 + x2 i2 + x3 i3 .
i1
...... ....... .... ... ... ... ... ... ... .... . . . .... ............ . . . . . . ....... .... . . . ....... . . . .... ....... . . . . . . . ....... ....... ......... ............. . ........... . . . . .
O
i2
i3
Figure 1.1. Standard measuring rod and Cartesian reference frame {O, i1 , i2 , i3 } in threedimensional space. .... ... .. ... ... ... .. ................ . . . . .. ....... . .. .. ....... . . .... ... ....... ....... ... ....... ... ....... ....... ... ...... ....... ... .......... . ....... ... ....... . . . ...... . ....... .. . .... ....... .. .......... . ....... ....... . ... ... .......... ... . . . .... . . .... ....... . ....... .. .... . ....... ...... ..... . .... ....... ....... ........ ... . .......................... . ... ... ..... . . . . . . . . . . . ... ............... .... .... .. . . . . . . . . . . . .. .... .......... . . . . ... . . ......... .......... . ............. .... ... ........... . . . . . . . . . . . . ... ........ ... .. ....... ... ......... . ....... ....... . .... ...... ....... ................ ....... . ....... . . . ... .... . . .... ... . . . . . . . . . . . . . . ...... ....... ... ....... . .... ....... ....... .......... . ....... ................ . . ....... ....... .... ....... ....... ....... ....... ...... .... .... ....... ...... . ....... ...... . .......... ...... .
i1
O
i2
x3
x
i3
x1
x2
Figure 1.2. Cartesian coordinates {x1 , x2 , x3 }.
4
introduction
The numbers {x1 , x2 , x3 } are denoted as the orthogonal Cartesian coordinates of the point of observation. The time coordinate is denoted by t. One of the purposes of the basic standard experiments is to deﬁne the units in terms of which the measured physical quantities are expressed. In accordance with international convention, we employ the International System of Units (Syst`eme International d’Unit´es), abbreviated to SI, for expressing the measured physical quantities. The mathematical framework by which the results from the standard basic experiments are cast into the macroscopic physical laws that govern the wave motion is furnished by vector calculus. For this reason, the next section summarizes those properties of Cartesian vectors that are needed in our further analysis.
1.1.
Cartesian vectors and their properties
The mathematical framework of the theory of electromagnetic waves is furnished by vector calculus. For this reason we summarize those properties of Cartesian vectors that are needed in our further analysis.
1.1.1.
Addition, subtraction and multiplication of vectors
Vectors can be subjected to the algebraic operations of addition, subtraction and multiplication. Let the components of v be given by v1 , v2 and v3 , and those of w by w1 , w2 and w3 , then the components of the sum (diﬀerence) of v and w is given by v ± w = (v1 ± w1 )i1 + (v2 ± w2 )i2 + (v3 ± w3 )i3 .
(1.2)
The product of the scalar ϕ and the vector v is given by ϕ v = ϕ v1 i1 + ϕ v2 i2 + ϕ v3 i3 .
(1.3)
The scalar (dot) product of the vectors v and w is given by v · w = v1 · w1 + v2 · w2 + v3 · w3 = w · v ,
(1.4)
5
cartesian vectors and their properties
The length of a vector v is denoted as 1
1
v = (v · v) 2 = (v12 + v22 + v32 ) 2 .
(1.5)
The vector (cross) product of the vectors v and w is given by v × w = (v2 w3−v3 w2 )i1 + (v3 w1−v1 w3 )i2 + (v1 w2−v2 w1 )i3 = −w × v , (1.6) or in matrix notation i 1 v × w = v1 w1
i2 i3 v2 v3 w2 w3
.
(1.7)
The scalar triple product of three vectors u, v and w is given by u · (v × w) = u1 (v2 w3 −v3 w2 ) + u2 (v3 w1 −v1 w3 ) + u3 (v1 w2 −v2 w1 ) , (1.8) or in matrix notation u 1 u · (v × w) = v1 w1
u2 u3 v2 v3 w2 w3
.
(1.9)
The scalar triple product has the property u · (v × w) = v · (w × u) = w · (u × v)
(1.10)
= −w · (v × u) = −v · (u × w) = −u · (w × v) . The vectorial triple product can be written as u × (v × w) = (u · w)v − (u · v)w .
(1.11)
As regards the diﬀerentiation of a vector, two cases have to be distinguished: diﬀerentiation with respect to a parameter, and diﬀerentiation with respect to the spatial (Cartesian) coordinates of the space in which the vector function is deﬁned.
6
1.1.2.
introduction
Diﬀerentiation with respect to a parameter
Let ϕ = ϕ(t) a scalar function and assume that ϕ is a diﬀerentiable function of the parameter t (in electromagnetics often the time coordinate). Then, the derivative ∂t ϕ = ∂ϕ/∂t is also a scalar function. Let v = v(t) be a vector function and assume that v is a diﬀerentiable function of the parameter t. Let v1 = v1 (t), v2 = v2 (t), and v3 = v3 (t) denote the components of v, then the derivative ∂t v of v is the vector ∂t v = (∂t v1 )i1 + (∂t v2 )i2 + (∂t v3 )i3 .
(1.12)
Let ϕ = ϕ(t) be a diﬀerentiable scalar function of the parameter t and let v = v(t) and w = w(t) be diﬀerentiable vector functions of the parameter t, then we have the following diﬀerentiation rules:
1.1.3.
∂t (ϕ v) = (∂t ϕ)v + ϕ∂t v ,
(1.13)
∂t (v × w) = (∂t v) × w + v × ∂t w .
(1.14)
Diﬀerentiation with respect to the spatial coordinates
Let ϕ be a scalar function and assume that ϕ = ϕ(x) = ϕ(x1 , x2 , x3 ) is a diﬀerentiable function of the spatial (Cartesian) coordinates x1 , x2 and x3 . Then, the derivatives ∂1 ϕ = ∂ϕ/∂x1 , ∂2 ϕ = ∂ϕ/∂x2 and ∂3 ϕ = ∂ϕ/∂x3 are also scalar functions. In this context, the gradient of ϕ = ϕ(x) is introduced as (1.15) grad ϕ = ∇ϕ = (∂1 ϕ)i1 + (∂2 ϕ)i2 + (∂3 ϕ)i3 , where ∇ = i1 ∂1 + i2 ∂2 + i3 ∂3
(1.16)
is the operator of Hamilton, the socalled nabla operator or del operator. This operator is a vector and acts as a spatial diﬀerentiation with respect to the three spatial coordinates. Let v be a vector function and assume that v = v(x) = v(x1 , x2 , x3 ) is a diﬀerentiable function of the spatial (Cartesian) coordinates x1 , x2 and x3 . The derivative ∂1 v is the vector ∂1 v = (∂1 v1 )i1 + (∂1 v2 )i2 + (∂1 v3 )i3 .
(1.17)
7
cartesian vectors and their properties
Similarly, we have ∂2 v = (∂2 v1 )i1 + (∂2 v2 )i2 + (∂2 v3 )i3 ,
(1.18)
∂3 v = (∂3 v1 )i1 + (∂3 v2 )i2 + (∂3 v3 )i3 .
(1.19)
These three derivatives operating on the vector function v can be combined in the divergence operator, deﬁned as div v = ∇ · v = ∂1 v1 + ∂2 v2 + ∂3 v3 ,
(1.20)
and in the curl operator, deﬁned as curl v = ∇ × v = (∂2 v3 −∂3 v2 )i1 + (∂3 v1 −∂1 v3 )i2 + (∂1 v2 −∂2 v1 )i3 . (1.21) We note that ∇ is a vector operator satisfying two sets of rules: • vector rules; • partial diﬀerentiation rules, including diﬀerentiation of a product. We now summarize the rules for the diﬀerentiation with respect to the spatial coordinates of the scalar functions ϕ = ϕ(x) and ψ = ψ(x), and of the vector functions v = v(x) and w = w(x). ∇(ϕ + ψ) = ∇ϕ + ∇ψ ,
(1.22)
∇ · (v + w) = ∇ · v + ∇ · w ,
(1.23)
∇ × (v + w) = ∇ × v + ∇ × w ,
(1.24)
∇(ϕ ψ) = (∇ϕ) ψ + ϕ ∇ψ ,
(1.25)
∇ · (ϕ v) = (∇ϕ) · v + ϕ ∇ · v ,
(1.26)
∇ × (ϕ v) = (∇ϕ) × v + ϕ ∇ × v ,
(1.27)
∇ · (v × w) = (∇ × v) · w − v · (∇ × w) , ∇ × (v × w) = (w · ∇)v − w ∇ · v − (v · ∇)w + v ∇ · w ,
(1.28) (1.29)
∇(v · w) = w × (∇ × v) + (w · ∇)v + v × (∇ × w) + (v · ∇)w . (1.30) We note that we have assumed that the functions ϕ, ψ, v and w are diﬀerentiable functions of the spatial coordinates. When we assume that ϕ = ϕ(x)
8
introduction
is also a twice diﬀerentiable function of x, we have the rules: ∇ · (∇ϕ) = (∇ · ∇)ϕ = (∂12 + ∂22 + ∂32 )ϕ ,
(1.31)
∇ × (∇ϕ) = 0 ,
(1.32)
∇ · (∇ × v) = 0 ,
(1.33)
∇ × (∇ × v) = ∇(∇ · v) − (∇ · ∇)v .
(1.34)
Subsequently, we present the rules for the spatial diﬀerentiation of a spatially dependent function f = f (x): x , x ∇xn = nxn−2 x , x ∇f (x) = ∂f (x) , x ∇x =
(1.35) (1.36) (1.37)
where ∂f is the derivative of f with respect to its argument. Further, we have: ∇·x = 3,
(1.38)
∇×x = 0, (∇ · ∇)x
n
= n(n + 1)x
(1.39) n−2
,
(1.40)
and when a is a constant vector: ∇(a · x) = a ,
(1.41)
(a · ∇)x = a ,
(1.42)
(a × ∇) × x = −2a .
(1.43)
Interpretation of grad ϕ We consider a continuously diﬀerentiable scalar function ϕ = ϕ(x) and we take the dot product of its gradient ∇ϕ (del ϕ) and an inﬁnitesimal increment of length dx = dx1 i1 + dx2 i2 + dx3 i3 .
(1.44)
Thus we obtain (∇ϕ) · dx = (∂1 ϕ)dx1 + (∂2 ϕ)dx2 + (∂3 ϕ)dx3 = dϕ ,
(1.45)
9
cartesian vectors and their properties
∇ϕ
. .... ......... .............. .. . . . .. ... ... .. ... ............... . . . . . . . . . . . . . . . ....... . .... ...... .......... ... ...... ........ .. ....... ...... ... ...... ....... ... . ...... . . . . . . . . . . . . ...... ....... ...... .... .... ....... ...... ... ................... ....... ....... ............. . .. ........ ............ .. .............. ........... ........ . . . . . . . . . . . . . . . . . . . . ......... . .. ........ ........... ........... ............................................. ... .............................................................. ... .. ... . . . ... .. ................................... ... ........... ...... ... . ........ . . . . . . . . . . . ...... . ...... ....... ... ...... ...... .. ...... ....... ... ...... ....... .............................. ...... ..... ....... ........................ . . . . . . . . ......... . ............... .. . ....... ............ ........ ........... .................... ............................... . . . . . . . . . . . . . . . . . . . . . . .................. . .. ....... ................... ... ........... ..... .. ............. ............................ ............. ..... . ......................................... ..... ..... ..... . . . . . ..... ..... ..... ..... ..... . . . . .. ..... ..... ..... ..... .. ..... . . . . ......... .... ... ..... ... .......... ... ....... ...... . .... .... ....... ............ .. ........... .......... ... ...........
ϕ(x) = C + ∆C
ϕ(x) = C
s
P
xP
dx
s
Q
i1
i2
O
i3
Figure 1.3. The gradient vector.
the change in the scalar function ϕ corresponding to a change in position dx. Now consider P and Q to be two points on a surface ϕ(x1 , x2 , x3 ) = constant. These points are chosen so that Q is a distance dx from P. Then, moving from P to Q, the change in ϕ(x1 , x2 , x3 ) = constant is given by dϕ = (∇ϕ) · dx = 0 ,
(1.46)
since we stay on the surface ϕ(x1 , x2 , x3 ) = constant. This shows that ∇ϕ is perpendicular to dx. The vectorial distance dx may have any direction from P as long as it stays in the surface ϕ = constant, point Q being restricted to this surface. For vanishing dx, we observe that ∇ϕ is oriented in a direction of the normal to the surface ϕ = constant (see Fig. 1.3). If we now permit dx to take us from one surface ϕ = C, C being a constant, to an adjacent surface ϕ = C + dC, then, dϕ = dC = (∇ϕ) · dx .
(1.47)
10
introduction
For a given dϕ, dx is a minimum when it is chosen parallel to ∇ϕ, or, for a given dx, the change in the scalar function ϕ is maximized by choosing dx parallel to ∇ϕ. This identiﬁes ∇ϕ as a vector having the direction of the maximum space rate of change of ϕ. Very often the notion of directional derivative occurs. When τ is a unit vector, the quantity τ · ∇ϕ is called the directional derivative of ϕ in the direction of τ , and equals the rate of change of ϕ in the direction of τ , viz., τ · ∇ϕ = ∂τ ϕ = τ1 ∂1 ϕ + τ2 ∂2 ϕ + τ3 ∂3 ϕ .
(1.48)
When τ is the tangent along a surface ϕ = constant, we obtain τ · ∇ϕ = ∂τ ϕ = 0 ,
(1.49)
which is consistent with Eq. (1.46). Interpretation of div v We consider a continuously diﬀerentiable vector function v = v(x). The divergence operator ∇ · v (del dot v) results in a scalar quantity indicating the outﬂow of a vector ﬁeld. It can be obtained from the limiting behavior of the net outﬂow integral for a vanishing small enclosed volume. To show this we ﬁrst compute the net outﬂow of a vector ﬁeld v over the elementary domain with volume dV = dx1 dx2 dx3 at the center of the elementary domain (see Fig. 1.4). This latter point is given by xP = { 12 dx1 , 12 dx2 , 12 dx3 }. By Taylor’s theorem, the ﬁeld component v1 is v1 (x) = v1 (xP ) + (∂1 v1 )(x1 − 12 dx1 ) + (∂2 v1 )(x2 − 12 dx2 ) + (∂3 v1 )(x3 − 12 dx3 ) + higher order terms .
(1.50)
The surface integral of the normal component of v (in the direction of the outward normal) over the top surface {x1 = dx1 , 0 < x2 < dx2 , 0 < x3 < dx3 } of the volume element, shown in Fig. 1.4, is dx2 dx3 x2 =0 x3 =0
v1 (dx1 , x2 , x3 ) dA =[v1 (xP ) + 12 (∂1 v1 ) dx1 ]dx2 dx3 + higher order terms .
(1.51)
11
cartesian vectors and their properties
i1
...... ......... ... .... .. ... .... .. ... ... ... ... ... ... . ......... . . . ... .... ................................................... . . . . .......................... .... ... . .......................... . . . ............ . ... . . . . . . . . .... ...... ..... .... ..... ... ...... ..... . . . ..... ....... . . . . ... ..... .............. . . . . . . . . . . ..... .... .... .... ..... ..... .. ............... ... ..... ..... .... ........................................................ ... ..... ........................... . . . ... ... ... . ........................... ..... ......... ... ... ... . ... ... ... . ... ... ... .... . ... ... ... . .. ... ... ... . .. .. ... ... . . ... ... .... .... . . .. ... . . .. .. ... .. . . ........................................ ... .... .... . . ... ... .... . . ... ... .... .... ... ... ... ... . . ... ... . . . ............................ . . . . ... ... . . ......................... . ... . . . . . . . . . . . . . ... . . ... . . . .......................... .... . . . . . . . . . . . . . . . ... . . . . . ............................ . ... . ......................... . . . . . ... . . . ................................... . ... ... . . . . . . . ... . . . ........ .... .... ..... ... .... ..... ..... . . . . ... ......... . . ... . .. . . .... ....... . . .... .. .... ..... ................................................... ........................... ..... ... ......... ........................... ... ..... .................................. ................. . . . . .....
dx1
∂V
s
xP
O
r
ν
dx3
i3
dx2
i2
Figure 1.4. Elementary domain in threedimensional space.
The surface integral of the normal component of v (in the direction of the outward normal) over the bottom surface {x1 = 0, 0 < x2 < dx2 , 0 < x3 < dx3 } of the volume element, shown in Fig. 1.4, is −
dx2 dx3
x2 =0 x3 =0
v1 (0, x2 , x3 ) dA = − [v1 (xP ) − 12 (∂1 v1 ) dx1 ]dx2 dx3 + higher order terms ,
(1.52)
the negative sign in front of the integral coming in because the outward pointing component of v is integrated, and the outward pointing component of v for the bottom surface is −v1 . The sum of the surface integrals over these two faces is therefore simply ∂1 v1 dx1 dx2 dx3 , to the order of approximation considered here. The contributions to the other faces depend on v2 and v3 and can be computed in a similar way. The net outﬂow integral from the volume element is therefore ⊂⊃
x∈∂V
ν · v dA = (∂1 v1 + ∂2 v2 + ∂3 v3 )dx1 dx2 dx3 = ∇ · v dV ,
(1.53)
12
introduction
in which ∂V denotes the boundary surface of the elementary domain, and ν denotes the normal to ∂V and is oriented away from the elementary domain V. The analogue of the divergence is the net outﬂow integral per unit volume at the point x and is written as
⊂⊃
div v = lim
V →0
ν · v dA x∈∂V = ∂1 v1 + ∂2 v2 + ∂3 v3 = ∇ · v , V
in which
(1.54)
V =
x∈V
dV
(1.55)
is the volume of the domain V. The integral x∈∂V ν · v dA is called the ﬂux of the vector ﬁeld v through the closed surface ∂V. Gauss’ integral theorem Let v be a continuously diﬀerentiable vector function of position deﬁned in a bounded domain D. Let, further, ∂D denote the boundary of D (Fig. 1.5). Because of the additive property, the net outﬂow integral for the domain D must equal the sum of the outﬂow integrals for all elementary domains included in the domain D. Then, we arrive at Gauss’ integral theorem, viz.,
∂D ν
D
Figure 1.5. Conﬁguration for the application of Gauss’ integral theorem.
13
cartesian vectors and their properties
⊂⊃
x∈∂D
ν · v dA =
x∈D
∇ · v dV =
x∈D
div v dV ,
(1.56)
in which ν is the unit vector normal to ∂D and oriented away from D. Two modiﬁcations of Gauss’ integral theorem are obtained as follows. With v = a ϕ, in which a is an arbitrary constant vector and ϕ is a continuous diﬀerentiable scalar function, we obtain
⊂⊃
x∈∂D
ν ϕ dA =
x∈D
∇ϕ dV =
x∈D
grad ϕ dV ,
(1.57)
With v = a × w, in which a is an arbitrary constant vector and w is a continuous diﬀerentiable vector function, we obtain
⊂⊃
x∈∂D
ν × w dA =
x∈D
∇ × w dV =
x∈D
curl w dV .
(1.58)
Interpretation of curl v We consider a continuously diﬀerentiable vector function v = v(x). The curl operator ∇ × v (del cross v) results in a vectorial quantity indicating the rate of circulation of a vector ﬁeld. It can be obtained from the limiting behavior of the net circulation integral around a vanishing small surface area. To show this we ﬁrst consider the case that the elementary area is perpendicular to the i1 axis. The center point of the area is given by xP = { 12 dx1 , 12 dx2 , 12 dx3 }. The circulation integral for the path shown in Fig. 1.6 is
x∈∂A
τ · vdl =
dx2
−
x2 =0
v2 ( 12 dx1 , l, 0) dl +
dx2 x2 =0
dx3 x3 =0
v2 ( 12 dx1 , l, dx3 ) dl −
v3 ( 12 dx1 , dx2 , l) dl
dx3 x3 =0
v3 ( 12 dx1 , 0, l) dl , (1.59)
14
introduction
i3
..... ........ .......... ... .... .. ... ... ... ... ... ... .. ..... ............................ .... ..... .. .. ..... ..... . . ... . .. ... ... ... ... ... ... ... ... .... .. ........ ....... .. .. ...... ....... ... .. ... .. ..... .. ... ... ... ... ... ... ... ... .. .. ............................................................................................................................................................................................................................. .. ..
dx3
τ
∂A
r xP
O
dx2
i2
Figure 1.6. Net circulation integral.
in which τ is the unit vector along the path and l is the arclength along the path. Substituting the Taylor’s series expansion (see Eq. (1.50)) for v2 and v3 we ﬁnd
x∈∂A
τ · vdl = v2 (xP ) dx2 − 12 (∂3 v2 )dx2 dx3 + v3 (xP ) dx3 + 12 (∂2 v3 )dx2 dx3 −v2 (xP ) dx2 − 12 (∂3 v2 )dx2 dx3 − v3 (xP ) dx3 + 12 (∂2 v3 )dx2 dx3 = (∂2 v3 − ∂3 v2 )dx2 dx3
(1.60)
plus higher order terms. Subsequently, the net circulation integral for the elementary surface area perpendicular to the i2 axis is (∂3 v1 − ∂1 v3 )dx3 dx1 and the net circulation integral for the elementary surface area perpendicular to the i3 axis is (∂1 v2 − ∂2 v1 )dx1 dx2 . Hence, the analogue of the curl is the net circulation integral per unit of surface area at the point x and is written as τ · v dl = ν · (∇ × v) , (1.61) ν · curl v = lim x∈∂A A→0 A
15
cartesian vectors and their properties
in which
dA (1.62) x∈A is the area of the surface A and ∂A is the closed boundary of the surface area A. In Eq. (1.61), ν is the unit vector normal to the surface area A and oriented toward the side of advance of a righthand screw as it is turned in the direction of τ around ∂A. A=
Stokes’ integral theorem Let v be a continuously diﬀerentiable vector function of position deﬁned on a bounded surface S. Let, further, ∂S denote the closed boundary of S (Fig. 1.7). We divide the surface S into its elements of area dA and add all the net circulation integrals for the elementary areas. Then, we arrive at Stokes’ integral theorem, viz.,
x∈∂S
τ · v dl =
x∈S
(ν × ∇) · v dA =
x∈S
ν · curl v dA ,
(1.63)
ν
S
∂S τ
Figure 1.7. Conﬁguration for the application of Stokes’ integral theorem.
16
introduction
in which ν is the unit vector normal to the surface area S and is oriented toward the side of advance of a righthand screw as it is turned in the direction of τ around ∂S. Two modiﬁcations of Stokes’ integral theorem are obtained as follows. With v = a ϕ, in which a is an arbitrary constant vector and ϕ is a continuous diﬀerentiable scalar function, we obtain
x∈∂S
τ ϕ dl =
x∈S
(ν × ∇)ϕ dA =
x∈S
ν × grad ϕ dA ,
(1.64)
With v = a × w, in which a is an arbitrary constant vector and w is a continuous diﬀerentiable vector function, we obtain
x∈∂S
τ × w dl =
x∈S
(ν × ∇) × w dA .
(1.65)
introduction
1.2.
17
Exercises and problems
Exercise 1.1 Calculate the following expressions and eliminate as many vector products as possible (a) a × a (b) (a × b) · c − a · (b × c) ( c ) (a × b) × c (d) (a × b) · (c × d) ( e ) (a × b) × (c × d). Exercise 1.2 What is the direction of the vector c = a × b with respect to the vectors a and b? Calculate (a × b) · a and (a × b) · b. Exercise 1.3 A twodimensional rectangle in the plane x3 = 0, with dimensions d1 , d2 , can said to be spanned by two vectors along the two base vectors of the Cartesian reference frame, viz., d1 = d1 i1 , d2 = d2 i2 . Let the unit normal ν point in the positive x3 direction. Give a geometrical interpretation of A = ν · (d1 × d2 ). Exercise 1.4 A threedimensional brick in space, with dimensions d1 , d2 , d3 , can said to be spanned by three vectors along the Cartesian reference frame base vectors, viz., d1 = d1 i1 , d2 = d2 i2 , d3 = d3 i3 . What is the geometrical meaning of the scalar triple product V = d1 · (d2 × d3 )? Exercise 1.5 Sketch the following vector ﬁelds by selecting several points and by drawing at each of these points an arrow with a direction that corresponds to the direction of the vector ﬁeld and with a length that is proportional to the magnitude of the vector ﬁeld (a) v(x) = −x1 i1 (b) v(x) = x2 i1
18
introduction
( c ) v(x) = −x1 i1 − x3 i3 (d) v(x) = x1 i3 − x3 i1 . Exercise 1.6 Determine the divergence and the rotation of the vector ﬁelds in Exercise 1.5. Exercise 1.7 The operator ∇ · ∇ in Eq. (1.31) is a scalar operator that occurs often in physics. It has therefore been given a name; it is called the Laplacian (operator) and is often denoted as ∇2 . Given the at least twice diﬀerentiable vector function A(x), write ∇2 A out in components. Exercise 1.8 Show that if an arbitrary vector function E(x) can be written as E = −∇V , where V (x) is a diﬀerentiable scalar function, the vector ﬁeld E is curl free, i.e. irrotational, and hence ∇ × E = 0. Exercise 1.9 Show that if an arbitrary vector function D(x) can be written as D = ∇ × C, where C(x) is a diﬀerentiable vector function, the vector ﬁeld D(x) is divergence free, i.e. solenoidal, and hence ∇ · D = 0. Exercise 1.10 The distance between two points with position vectors x and x is given by the length of the vector from x to x , d(x, x ) = x − x. When ∇ and ∇ denote partial diﬀerentiation with respect to x and x , respectively, determine (a) ∇d ,
∇ d
(b) ∇d−1 ,
∇ d−1
( c ) ∇ · ∇d , (d) ∇ · ∇d−1 ,
∇ · ∇ d ∇ · ∇ d−1 .
Exercise 1.11 Given the smooth surface S with unit normal ν, and a vector S = E × H. Show that the normal component ν · S of S can also be written as ν · S = ν · [(ν × E) × (ν × H)].
exercises and problems
19
Exercise 1.12 Given the smooth surface S with unit normal ν. Show that an arbitrary vector H can be decomposed into H = (ν · H)ν + (ν × H) × ν. Give a geometrical interpretation of the two terms.
Problem 1.1 Calculate the volume of a threedimensional parallepiped spanned by the arbitrarily oriented vectors a, b and c. Problem 1.2 Given the smooth surface S with unit normal ν. Decompose the operator ∇ in two parts, one part tangential and the other normal to the surface S. Problem 1.3 Use Gauss’ theorem to calculate x∈∂D ν dA, where ν is the outward unit normal to the closed boundary ∂D of the domain D (see Fig. 1.5). Problem 1.4 Use Stokes’ theorem to calculate x∈∂S τ dl, where τ denotes the unit tangent along the closed boundary contour ∂S of the surface S. The integration runs in the direction of circulation that forms a righthanded system with the unit normal ν to S (see Fig. 1.7).
Chapter 2
The Electromagnetic Field Equations The electromagnetic ﬁeld equations take on their simplest form in a vacuum domain. In such a domain, they follow, in principle, from a series of basic experiments in which an electric point charge is employed as measuring device, and the force exerted on it is measured. Through this force, the electric and the magnetic ﬁeld strengths, as experienced by an observer located in a vacuum domain, are introduced. The electromagnetic ﬁeld equations in matter are introduced through an axiomatic procedure that leaves in tact their structure in a vacuum domain in case the point of observation is located in a vacuum domain. Since external sources are essentially composed of matter, they are encompassed in the volume densities of electric and magnetic currents that describe the electromagnetic action of matter. The induced parts of the latter quantities then describe the passive reaction of a piece of matter to an electromagnetic ﬁeld.
2.1.
Force exerted on an electric point charge
From experience it is known that electrically charged particles exert forces on each other. These forces depend on the relative position and the relative
22
the electromagnetic field equations
state of motion of the charged particles. We take a single electrically charged particle and use this as a measuring device for deﬁning, in a vacuum domain, the electric and the magnetic ﬁeld strengths of an electromagnetic ﬁeld. It is assumed that an observer who is handling this measuring device can measure the force exerted on it (for example, by counterbalancing the force by an adjustable mechanical one). The experiments show the following: • the force F is proportional to the strength q of the point charge; • the force contains a term that is independent of the velocity v of the point charge with respect to the observer; • the force contains a term that is proportional to the velocity of the point charge with respect to the observer and has a direction perpendicular to it. In accordance with this, and with the conventions of the International System of Units, we postulate the expression (see Fig. 2.1) F = qE + qµ0 v × H .
(2.1)
In this expression F q v µ0 E H
= = = = = =
force (N), electric charge (C), velocity of point charge with respect to the observer (m/s), permeability in vacuum (H/m), electric ﬁeld strength (V/m), magnetic ﬁeld strength (A/m).
The value of µ0 is ﬁxed by SI as µ0 = 4π × 10−7 H/m. Assuming that the value of q is so small that the reaction of the test point charge on the sources that generate the electromagnetic ﬁeld can be neglected, Eq. (2.1) deﬁnes the value of the electric ﬁeld strength E = E(x, t) and the value of the magnetic ﬁeld strength H = H(x, t) at the position of observation x with respect to a given, ﬁxed, orthogonal Cartesian reference frame and the time of observation t. Since the observer must be free to position and to move his or her measuring device, the operational deﬁnition of Eq. (2.1) can only be applied in a vacuum domain and cannot be used in matter.
23
force exerted on an electric point charge
i1 .... ......... ... .. ....... ... ...... ... ...... ...... ... .... . . . . ... . ...... ... ....... ... ....... ....... ... ..... . . ... . . ..... ... ...... .. ....... ......... ....... ............. . . . . . ....... ...... ....... ....... ....... ....... ....... ....... ....... ....... . . ....... . . . . ......... ..... . . . . . . . . ............ ...........
........ ........ ...... ...... ...... . . . . . . ...... ...... ...... ...... ...... . . . . . . ...... ...... ...... ...... ...... . . . . . ... ...... ...... ..... ...... .......... ........................................................... . . . . . . ....... .......
x u q
F
v
O
i2
i3
Figure 2.1. Force F on an electric point charge of strength q and moving with velocity v with respect to an observer at rest in the reference frame {0, i1 , i2 , i3 }.
By ﬁrst choosing, in Eq. (2.1), v = 0, one determines by measuring F the value of E = E(x, t). By giving, subsequently, v three linearly independent directions, one can determine H = H(x, t) by measuring F and using the already known values of E. The quantity F = qE
(2.2)
is sometimes denoted as the electric force; the quantity F = qµ0 v × H
(2.3)
is sometimes denoted as the magnetic force or Lorentz force (after H.A. Lorentz). It is emphasized that the thus determined values of E and H have the meaning of electric and magnetic ﬁeld strengths, respectively, only for the observer who carries out the measurements and interprets them according to Eq. (2.1). A second observer who is in relative motion with respect to the ﬁrst one, measures diﬀerent values of the electric and the magnetic ﬁeld strengths. In particular, if the second observer moves with the point charge
24
the electromagnetic field equations
of the ﬁrst, we have, denoting the latter’s quantities by primed symbols, v = 0, and hence, F = qE , i.e., the second observer interprets the present electromagnetic ﬁeld as a pure electric one. (In accordance with the theory of relativity, we have put q = q.) The interrelations that exist between the values of the electric and the magnetic ﬁeld strengths of two observers who are in relative motion are investigated in the theory of relativity.
2.2.
Maxwell’s equations in vacuum
Using the electric point charge as a measuring device one can investigate the properties of an electromagnetic ﬁeld in a vacuum domain. In particular, it proves to be fruitful to investigate the relationships between the changes of the ﬁeld in space and the changes of the ﬁeld in time. Quantitatively, this is most simply done by mutually comparing the ﬁrstorder partial derivatives with respect to the spatial coordinates with the ﬁrstorder partial derivatives with respect to time. Carrying out the analysis, it is found that −∇ × H + ε0 ∂t E = 0 ,
(2.4)
∇ × E + µ0 ∂t H = 0 .
(2.5)
In these equations E H ε0 µ0
= = = =
electric ﬁeld strength (V/m), magnetic ﬁeld strength (A/m), permittivity in vacuum (F/m), permeability in vacuum (H/m).
The value of ε0 follows from ε0 =
1 , µ0 c20
(2.6)
where c0 = 299792458 m/s is the electromagnetic wave speed in vacuum. Substituting the values of µ0 and c0 , one obtains ε0 = 8.8541878 × 10−12 .
25
Maxwell’s equations in matter
Equations (2.4) and (2.5) are known as Maxwell’s equations (in vacuum). Further, it is found that ∇·E = 0,
(2.7)
∇·H = 0.
(2.8)
The latter equations are not independent of Maxwell’s equations and are a kind of compatibility relations. By noting that ∇ · (∇ × H) = 0, it follows from Eq. (2.4) that ∂t ∇ · E = 0. Hence, if at some instant ∇ · E = 0, this would be so at any instant. In particular, this applies when we study the causal ﬁelds that are generated by sources that are switched on at a certain instant, starting from an initially vanishing ﬁeld. The same reasoning applies to Eq. (2.5), from which, since ∇ · (∇ × E) = 0, it follows that ∂t ∇ · H = 0.
2.3.
Maxwell’s equations in matter
Methodologically the simplest way to account for the presence of matter is to retain the vectorial nature of the electromagnetic ﬁeld equations and introduce in the righthand sides of Eqs. (2.4) and (2.5) vectorial quantities that diﬀer from zero only in a domain occupied by matter (and necessarily reduce to zero in a vacuum domain). Let us write (Fig. 2.2) −∇ × H + ε0 ∂t E = −J mat , ∇ × E + µ0 ∂t H = −K
mat
(2.9) .
(2.10)
Here, J mat = K mat =
volume density of material electric current (A/m2 ), volume density of material magnetic current (V/m2 ).
As we have said, J mat = K mat = 0 in a vacuum domain .
(2.11)
26
the electromagnetic field equations
................................. .............. ....... ......... ..... ........ ..... ....... . . ... . . . .... ... . . . . . ... ... . . . . . ... .... . . . . ... .... . . . .. . ... . . . ... . .... . . . . .. ... . . . . . . ... . . . . . .. . . . . ... ... ... ... ... ... ... .. ... . . ... ... ... ... ... ... ... .. . . . .. ... ... ... ... ... ... .... . . ..... . ... .. ..... .... ..... ..... ... ..... . . ... . . ... ...... ..... ... ...... ... ...... ... ...... . . ..... . . . . ...... ..... ........ ........ ..................................................
J mat
vacuum
3 s @ @ R K mat @
matter
Figure 2.2. Piece of matter embedded in vacuum, with volume density of electric current J mat and volume density of magnetic current K mat .
First of all, we shall distinguish in J mat and K mat between an active part and a passive part. The active parts (also denoted as the source parts, or external parts) describe the action of sources that generate the ﬁeld and whose physical behavior is either irrelevant to or outside the scope of our analysis; they will be denoted as J ext and K ext , respectively. The active parts are taken to be ﬁeld independent. The passive parts (also denoted as the induced parts) describe the reaction of matter to the presence of an electromagnetic ﬁeld and are typically ﬁeld dependent; they will be denoted by J ind and K ind , respectively. Hence, we have J mat = J ind + J ext , K
mat
= K
ind
+K
ext
(2.12) .
(2.13)
The term with K ext has been introduced for symmetry and convenience reasons. It is noted that K ext has no physical meaning. Traditionally, the induced parts are written in a somewhat diﬀerent manner, viz. J ind = J + ∂t P , K
ind
= µ0 ∂t M ,
(2.14) (2.15)
27
Maxwell’s equations in matter
where J P M
= = =
volume density of electric current (A/m2 ), electric polarization (C/m2 ), magnetization (A/m).
The latter quantities are particularly useful for describing the electromagnetic properties of matter in case only ﬁelds varying slowly in time are present. Substitution of Eqs. (2.12)  (2.15) in Eqs. (2.9) and (2.10) leads to −∇ × H + ε0 ∂t E + J + ∂t P ∇ × E + µ0 ∂t H + µ0 ∂t M
= −J ext , = −K
ext
(2.16) .
(2.17)
Further, it is customary to introduce the quantities D = ε0 E + P ,
(2.18)
B = µ0 (H + M ) .
(2.19)
where D B
= =
electric ﬂux density (C/m2 ), magnetic ﬂux density (T).
With the aid of Eqs. (2.18) and (2.19), Eqs. (2.16) and (2.17) can be rewritten as −∇ × H + J + ∂t D = −J ext , ∇ × E + ∂t B = −K
ext
(2.20) .
(2.21)
Equations (2.20) and (2.21) are known as Maxwell’s equations (in matter). These equations constitute an incomplete system of equations since the number of equations is less than the number of unknown quantities, assuming that the righthand sides, which are representative of the action of external sources, are known. The supplementing equations are furnished by the
28
the electromagnetic field equations
constitutive relations that describe the reaction of passive matter to the penetration of an electromagnetic ﬁeld. As far as the compatibility relations are concerned, we observe that application of the divergence operator to Eqs. (2.20) and (2.21) yields ∇ · J + ∂t ∇ · D = −∇ · J ext , ∂t ∇ · B = −∇ · K
ext
(2.22) .
(2.23)
Historically, the volume density of electric charge (in C/m3 ) is introduced as ρ=∇·D. (2.24)
2.4.
The constitutive relations
The electromagnetic constitutive relations are representative for the macroscopic electromagnetic properties of matter. In their general form, they constitute an equivalent of three vectorial relations between the ﬁve vectorial quantities {J , D, B, E, H}. For reasons that are connected with the transfer of energy by electromagnetic waveﬁelds, the standard form expresses {J , D, B} in terms of {E, H}. Several terminological aspects of this relationship are enumerated below. In view of the assumed passivity of the medium, we require that for any type of matter we have {J , D, B} → 0 as {E, H} → 0 When the values of {J , D, B} are linearly related to the values of {E, H}, we denote the medium as linear. If this is not the case, the medium is denoted as nonlinear. When the operators that express the values of {J , D, B} in terms of the values of {E, H} are time invariant, the medium is denoted as time invariant. Otherwise, the medium is time variant or parametrically aﬀected. When the constitutive relations express the values of {J , D, B} at some instant in terms of the values of {E, H} at the same instant only, the medium is denoted as instantaneously reacting. When, on the other hand, the values of {J , D, B} are expressed in terms of the values of {E, H} at all previous
the constitutive relations
29
instants, the medium is said to show relaxation; the property that only the past is involved in relaxation phenomena, is known as the principle of causality. When the values of {J , D, B} at some position are related to the values of {E, H} at the same position only, the medium is denoted as locally reacting. Almost all media are locally reacting in their electromagnetic behavior. If at a point in space the constitutive operators are orientation invariant, the medium is denoted as isotropic at that point. If this property does not apply, the medium is denoted as anisotropic. In a domain in space where the constitutive operators are shift invariant, the medium is denoted as homogeneous; in a domain in space where the shift invariance does not apply, the medium is denoted as inhomogeneous or heterogeneous. For a wide class of materials, the quantities J and D only depend on E (and not on H), while the quantity B only depends on H (and not on E). For a medium that is, in addition, linear, time invariant, instantaneously reacting, locally reacting and isotropic in its electromagnetic behavior, we then have J (x, t) = σ(x)E(x, t) ,
(2.25)
D(x, t) = ε(x)E(x, t) ,
(2.26)
B(x, t) = µ(x)H(x, t) ,
(2.27)
where σ ε µ
= = =
conductivity (S/m), permittivity (F/m), permeability (H/m).
For electromagnetic ﬁelds that vary relatively slowly in time (frequency of operation less than a few MHz), the assumption of an instantaneously reacting material is realistic. In a (sub)domain where these constitutive coeﬃcients σ = σ(x), ε = ε(x) and µ = µ(x) change indeed with position,
30
the electromagnetic field equations
the medium is inhomogeneous; in a domain where they are constant, the medium is homogeneous. In vacuum we have σ = 0, ε = ε0 = µ = µ0 .
1 , µ0 c20
In the tables of physical constants (see Table 2.1) one customarily speciﬁes the dielectric properties of a material through εr = ε/ε0 = relative permittivity and its magnetic properties through µr = µ/µ0 = relative permeability; εr and µr are dimensionless.
Table 2.1. Relative permittivities and conductivities for electromagnetic ﬁelds that vary slowly in time (at 20 ◦ C)
air glass quartz bakelite dry earth wet earth fresh water seawater salt water (20%) aluminium copper silver
εr = ε/ε0
σ (S/m)
1.0006 6 4.6 4.8 5 10 81 81 81 1 1 1
1.0 × 10−12 1.0 × 10−12 1.0 × 10−9 10−4 10−2 10−3 4 20 3.5 × 107 5.8 × 107 6.17 × 107
31
the system of field equations
2.5.
The system of ﬁeld equations
Substitution of the constitutive relations in Maxwell’s equations (2.20) and (2.21) yields a system of two vectorial equations in two vectorial unknowns, viz., the electric ﬁeld strength and the magnetic ﬁeld strength. These electromagnetic ﬁeld equations are
−∇ × H + σE + ε∂t E = −J ext , ∇ × E + µ∂t H = −K
ext
(2.28) .
(2.29)
Let the subscripts {1, 2, 3} denote the components of a vector along the {i1 , i2 , i3 }axes of the chosen reference frame, respectively. Then the electromagnetic ﬁeld is governed by the six scalar equations for the three components of the electric ﬁeld strength and the three components of the magnetic ﬁeld strength. This system of electromagnetic ﬁeld equations is −(∂2 H3 − ∂3 H2 ) + σE1 + ε∂t E1 = −J1ext ,
(2.30)
−(∂3 H1 − ∂1 H3 ) + σE2 + ε∂t E2 =
,
(2.31)
,
(2.32)
−(∂1 H2 − ∂2 H1 ) + σE3 + ε∂t E3 =
−J2ext −J3ext
∂2 E3 − ∂3 E2 + µ∂t H1 = −K1ext , ∂3 E1 − ∂1 E3 + µ∂t H2 = ∂1 E2 − ∂2 E1 + µ∂t H3 =
−K2ext −K3ext
(2.33)
,
(2.34)
,
(2.35)
and is amenable to a mathematical solution procedure, either analytically or numerically. The transfer of electromagnetic power is governed by the Poynting vector S = E × H (in W/m2 ) ,
(2.36)
where the three components are deﬁned as S1 = E2 H3 − E3 H2 ,
(2.37)
S2 = E3 H1 − E1 H3 ,
(2.38)
S3 = E1 H2 − E2 H1 .
(2.39)
32
the electromagnetic field equations
The Poynting vector quantiﬁes the amount of electromagnetic power ﬂow per unit area.
2.6.
The boundary conditions
In those domains in a medium where the constitutive parameters change continuously with position, the electromagnetic ﬁeld vectors are continuously diﬀerentiable functions of position and satisfy the diﬀerential equations (2.28) and (2.29). Across certain boundary surfaces in a conﬁguration, the electromagnetic ﬁeld quantities may exhibit a discontinuous behavior. Since at those positions they are no longer continuously diﬀerentiable, the electromagnetic ﬁeld equations cease to hold. To interrelate the electromagnetic waveﬁeld quantities at either side of the interface, a certain set of boundary conditions is needed. Upon crossing the interface of two adjacent media that diﬀer in their electromagnetic properties, the electric and the magnetic ﬁeld strengths are, in general, no longer continuously diﬀerentiable in a domain that contains (part of) an interface, and Eqs. (2.28) and (2.29) cease to hold. Assuming that the properties of the media under consideration and the position of the interface are time invariant, the nondiﬀerentiability is restricted to the dependence on the spatial variables. To solve electromagnetic waveﬁeld problems in domains that contain abrupt boundaries, the electromagnetic ﬁeld equations must be supplemented by conditions that interrelate the ﬁeld values at either side of the interface, the socalled boundary conditions. The interface of two media Let ∂D denote the interface and assume that ∂D has everywhere a unique tangent plane. Let, further, ν denote the unit vector along the normal to ∂D such that upon traversing ∂D in the direction of ν, we pass from the domain D(2) to the domain D(1) , D(1) and D(2) being located at either side of ∂D (Fig. 2.3). Suppose, now, that some (or all) electromagnetic wave quantities jump across ∂D. In the direction parallel to ∂D, all electromagnetic wave quantities still vary in a continuously diﬀerentiable manner, and hence the
33
the boundary conditions
partial derivatives parallel to ∂D give no problem in Eqs. (2.28) and (2.29). The partial derivatives perpendicular to ∂D, on the contrary, meet functions that show a jump discontinuity across ∂D; these give rise to surface Dirac delta distributions (surface impulses) located on ∂D. Distributions of this kind would, however, physically be representative of surface sources located on ∂D. In the absence of such surface sources, the absence of surface impulses in the partial derivatives perpendicular to ∂D should be enforced. The latter is done by requiring that these normal derivatives only meet functions that are continuous across ∂D. To investigate the consequences of this reasoning, we write ∇ = (ν · ∇)ν + [∇ − (ν · ∇)ν] = (ν · ∇)ν + (ν × ∇) × ν ,
(2.40)
in which (ν · ∇)ν is the normal part of the derivatives in the ∇ operator, while (ν × ∇) × ν represents the tangential part. Considering Eqs. (2.28) and (2.29), and assuming that there are no external sources present at the interface ∂D, the quantities ∇ × H and ∇ × E may not contain any Dirac distribution at ∂D, when crossing the interface in the normal direction. This means that in the expressions (ν · ∇)ν × H = (ν · ∇)(ν × H) ,
(2.41)
(ν · ∇)ν × E = (ν · ∇)(ν × E) ,
(2.42)
... ......... ........ ....... ....... . . . . . . . ...... ...... ...... ...... ..... . . . . ..... ..... ..... ..... ..... . . . . . ..... ..... ..... ..... .... . . .... .... ... ... . . ... ... ... ... ... . . ... ... ... .. . ... ... ... .... . .. ... ... ..
∂D
D(1)
ν
@ I @
D(2)
Figure 2.3. Interface between two media with diﬀerent electromagnetic properties.
34
the electromagnetic field equations
the quantities ν × H and ν × E may not jump, when crossing ∂D in the normal direction. Hence, this reasoning leads to the requirements ν × H is continuous across ∂D ,
(2.43)
ν × E is continuous across ∂D .
(2.44)
and
Equations (2.43) and (2.44) are the boundary conditions at a sourcefree interface between two diﬀerent media. In these equations only the components of the electric and magnetic ﬁeld strengths tangential to ∂D occur. In view of the continuity of the tangential components of E and H across the interface ∂D, the normal component of the Poynting vector S, ν · S = ν · (E × H) = ν · [(ν × E) × (ν × H)], is continuous across this interface, as it should be on physical grounds. Electrically impenetrable object A material body, occupying the domain D in space, is denoted as electrically impenetrable if in it the electric ﬁeld strength is negligibly small, while the boundary condition of the continuity of the electric ﬁeld strength across the boundary surface ∂D of the object is maintained. Consequently, the boundary condition upon approaching the boundary surface ∂D of such a body via its exterior is given by (Fig. 2.4) lim ν × E(x + δν, t) = 0 for any x ∈ ∂D , δ↓0
(2.45)
where ν is the unit vector along the normal to ∂D pointing away from D. It is not allowed to prescribe boundary conditions for the tangential part of the magnetic ﬁeld strength in this case. In fact, the tangential part of the magnetic ﬁeld strength will, in general, exhibit a discontinuity upon crossing ∂D. Electrically impenetrable materials arise as limiting cases of materials whose conductivity and/or permittivity go to inﬁnity. In view of the vanishing tangential components of E at the boundary ∂D, the normal component of the Poynting vector S, ν · S = H · (ν × E), vanishes at this boundary as well.
35
the boundary conditions
ν
............................................. ...... ..... ... ... . . . .... ... . . . . . ... ... . . . . . ... .... . . . . .. .... . .. . . . ... .. . . . . .... .. . . . .. ... . . . . . ... .. . . . . .. . . . . ... ... ... ... ... .. ... .. . ... . . . ... ... ... ... ... ... ... .. . . . .. ... ... ... ... ... ... .... . . ..... . .... .... .... ..... ... ..... ... ..... . . . ... . ...... ... ...... ... ...... ... ...... .... ...... . . . . ..... . . ....... ...... ......... .......... .......................................
@ I @ * x + δν @ 1 x
O
∂D.................................. D
impenetrable object
Figure 2.4. Limiting procedure approaching the boundary of an impenetrable object.
Magnetically impenetrable object A material body, occupying the domain D in space, is denoted as magnetically impenetrable if in it the magnetic ﬁeld strength is negligibly small, while the boundary condition of the continuity of the magnetic ﬁeld strength across the boundary surface ∂D of the object is maintained. Consequently, the boundary condition upon approaching the boundary surface ∂D of such a body via its exterior is given by (Fig. 2.4) lim ν × H(x + δν, t) = 0 for any x ∈ ∂D , δ↓0
(2.46)
where ν is the unit vector along the normal to ∂D pointing away from D. It is not allowed to prescribe boundary conditions for the tangential part of the electric ﬁeld strength in this case. In fact, the tangential part of the electric ﬁeld strength will, in general, exhibit a discontinuity upon crossing ∂D. Magnetic impenetrable materials arise as limiting cases of materials whose permeability goes to inﬁnity. In view of the vanishing tangential components of H at the boundary ∂D, the normal component of the Poynting vector S, ν · S = −E · (ν × H), vanishes at this boundary as well.
36
2.7.
the electromagnetic field equations
Frequencydomain representations
In reality, any electromagnetic ﬁeld is transient in nature: it is generated by a source that has been switched on at some instant t = t0 in the ﬁnite past and decays in magnitude as t → ∞. An essential feature is further the physical condition of causality that requires that an eﬀect (the presence of an electromagnetic ﬁeld) is causally related to its cause (the action of the source). This implies that the ﬁeld that is causally related to the action of a source that is switched on at the instant t = t0 , necessarily vanishes prior to this instant, i.e., for t < t0 , at any point in space. In timeinvariant conﬁgurations the behavior of such transient ﬁelds can advantageously be described via their (complex) frequencydomain representations. Further, t0 may be chosen arbitrarily, hence, we simply take t0 = 0. Through these representations the principle of causality can automatically be taken into account, while the dependence on t (the fourth coordinate in spacetime) is replaced by the dependence on a (transform) parameter; the resulting reduction in dimensionality of the problem is often advantageous. The Laplace transformation combines the two features. Let f = f (x, t) denote an electromagnetic ﬁeld quantity that is causally related to the action of a source that has been switched on at the instant t = 0, then its Laplace transform (with respect to time) is deﬁned as fˆ(x, s) =
∞ t=0
exp(−st)f (x, t)dt for Re(s) > s0 ,
(2.47)
where s is the complex Laplace transform parameter and fˆ is analytic (i.e., diﬀerentiable with respect to s) in the right half of the complex splane Re(s) > s0 . An important property of the Laplace transformation is that the Laplace transform of ∂t f (x, t) is sfˆ(x, s) if f (x, t) is causal. Once fˆ = fˆ(x, s) has been evaluated, f = f (x, t) can be recovered by the application of the Bromwich inversion integral f (x, t) =
1 2πj
s0 +j∞ s=s0 −j∞
ˆ s)ds for all t , exp(st)f(x,
(2.48)
that automatically yields the value zero for t < 0 and for which numerical (Fast Fourier Transform) routines are available. In Eq. (2.48), j is the imaginary unit (j 2 = −1).
37
frequencydomain representations
In a number of cases one can take s0 = 0 and consider fˆ = fˆ(x, s) for imaginary values s = jω, where ω is the (real) angular frequency. Then, Eq. (2.47) can be rewritten as fˆ(x, jω) =
∞
exp(−jωt)f (x,t)dt for all real ω ,
(2.49)
t=0
and Eq. (2.48) as 1 f (x, t) = 2π
∞
exp(jωt)fˆ(x, jω)dω
for all t.
(2.50)
ω=−∞
Equations (2.49) and (2.50) express the Fourier transformation for causal ﬁelds. ˆ with components vˆ1 , vˆ2 and Note that the length of a complex vector v vˆ3 is deﬁned as ˆ ∗ ) 2 = (ˆ v1 vˆ1∗ + vˆ2 vˆ2∗ + vˆ3 vˆ3∗ ) 2 , ˆ v  = (ˆ v·v 1
1
(2.51)
which is in accordance with the deﬁnition of Eq. (1.5) for real vectors. The asterisk denotes complex conjugate.
2.7.1.
The frequencydomain ﬁeld equations
We subject the electromagnetic ﬁeld equations (2.28) and (2.29) to a Laplace transformation over the interval {t ∈ IR; t > 0}. We are only interested in the causal ﬁeld generated by sources that are switched on at the instant t = 0, we arrive at ext
ˆ + (σ+sε)E ˆ = −J ˆ , −∇ × H ˆ + sµH ˆ = −K ˆ ext , ∇×E
(2.52) (2.53)
or written out in their components as ˆ1 = −Jˆext , ˆ 3 − ∂3 H ˆ 2 ) + (σ+sε)E −(∂2 H 1 ˆ ˆ ˆ ˆ −(∂3 H1 − ∂1 H3 ) + (σ+sε)E2 = −J2ext , ˆ3 = −Jˆext , ˆ 2 − ∂2 H ˆ 1 ) + (σ+sε)E −(∂1 H 3
(2.54) (2.55) (2.56)
38
the electromagnetic field equations
ˆ 1 = −K ˆ ext , ˆ 3 − ∂3 E ˆ2 + sµH ∂2 E 1 ˆ ˆ ˆ ˆ ∂3 E1 − ∂1 E3 + sµH2 = −K2ext , ˆ 3 = −K ˆ ext . ˆ 2 − ∂2 E ˆ1 + sµH ∂1 E
(2.57) (2.58) (2.59)
3
The steadystate analysis The behavior of electromagnetic waves is often characterized by the results of a steadystate analysis. In such an analysis, all electromagnetic ﬁeld quantities are taken to depend sinusoidally on time with a common angular frequency ω. Each purely real quantity f (x, t) can then be associated with a complex counterpart fˆ(x, jω) and a common time factor exp(jωt). In doing so, the original quantity is found from its complex counterpart as
f (x, t) = Re fˆ(x, jω)exp(jωt) .
(2.60)
Substitution of these representations for the ﬁeld quantities occurring in the timedomain ﬁeld equations (2.28) and (2.29) yields, except for the common time factor exp(jωt), the set of ﬁeld equations ˆ + (σ+jωε)E ˆ = −J ˆ ext , −∇ × H ˆ + jωµH ˆ = −K ˆ ext . ∇×E
(2.61) (2.62)
or written out in their components as ˆ1 = −Jˆext , ˆ 3 − ∂3 H ˆ 2 ) + (σ+jωε)E −(∂2 H 1 ˆ ˆ ˆ ˆ −(∂3 H1 − ∂1 H3 ) + (σ+jωε)E2 = −J2ext , ˆ3 = −Jˆext , ˆ 2 − ∂2 H ˆ 1 ) + (σ+jωε)E −(∂1 H
(2.63)
ˆ 1 = −K ˆ ext , ˆ3 − ∂ 3 E ˆ2 + jωµH ∂2 E 1 ˆ 2 = −K ˆ ext , ˆ1 − ∂ 1 E ˆ3 + jωµH ∂3 E 2 ˆ 3 = −K ˆ ext , ˆ2 − ∂ 2 E ˆ1 + jωµH ∂1 E
(2.66)
3
3
(2.64) (2.65)
(2.67) (2.68)
which are similar to Eqs. (2.52)  (2.59), if we take s = jω. Hence, we interpret the steadystate analysis as a limiting case of the Laplace transform analysis, in which s → jω .
39
polarization state
2.8.
Polarization state
An important aspect of a timeharmonic electromagnetic ﬁeld deﬁned in the previous section is its polarization state (not to confuse with the electric polarization P ). The polarization state of a ﬁeld vector (here we will choose the electric ﬁeld strength of the electromagnetic ﬁeld) is described by the locus of the endpoint of the ﬁeld vector E(x, t) as time progresses. From (see Eq. (2.60))
ˆ (x, jω) sin(ωt) , ˆ (x, jω) cos(ωt) + E E(x, t) = E
(2.69)
where
ˆ ˆ (x, jω) = Re E(x, jω) , E
ˆ (x, jω) = −Im E(x, ˆ E jω) ,
(2.70) (2.71)
we observe that the ﬁeld vector E(x, t) for any instant is located in the plane ˆ (x, jω) while its endpoint describes an ellipse in ˆ (x, jω) and E through E that plane. The electric ﬁeld strength (and also) the electromagnetic ﬁeld is then said to be elliptically polarized at that point (Fig. 2.5). ˆ (x, jω) and E ˆ (x, jω) have the same direction, i.e., When E ˆ ˆ ∗ (x, jω) = 0 , E(x, jω) × E
(2.72)
the ellipse degenerates into a straight line. The electric ﬁeld strength is then ˆ (x, jω) said to be linearly polarized at that point (Fig. 2.5). Finally, when E ˆ (x, jω) have equal magnitudes and are perpendicular to each other, and E i.e., ˆ ˆ E(x, jω) · E(x, jω) = 0 ,
(2.73)
the ellipse changes into a circle. The electric ﬁeld strength is then said to be circularly polarized at that point (Fig. 2.6).
40
the electromagnetic field equations
ωt = 12 π
..................................................... .......... ...... ....... ........ ..... ............ ..... ....... .... ....... . . .. . . . ... .. ......... . . . ... . . ...... . . . ... . . . ... . . ... . . . . . . ... ... . . . . . . ... . ... . . . . . ... . ... . . . . ... . .. . . ... . . . .. . . . . . .. .. ... . . .. . . . . . . . . .. .. ... . . . ... ... ... ... ... ... .. .. .. .......... .... .. .. . . . . ..... ....... ....... ....... ....... ....... ............................................................................................... .. .. ... .. ... .. ... .. ... .. ... .... .. . . .... . . ... ... .. .. .... .. ... ... ... ... . . . ... .. ... .. ... .. ... .... ... ..... . ..... ... . .. . . . . ... ..... ... ..... .. ... ...... ... ..... ...... ..... ....... . . . . . . . . ...... .. ....... ....... ......... ........... .....................................
ωt = 12 π........ ...
.
........ ...........
t
ˆ ................. E .. .......... . ........ ............... ωt = 0 .. .
ˆ E
ˆ E
s
ωt = π
ωt =
ˆ........ E ...
. ... ... ... ... . . ... ... ... ...
s
ωt = 0
.
. ...
. ...
. ...
. ...
ωt = π............... . ...
3 2π
......... . . ...
. ...
.
ωt = 32 π
Figure 2.5. Elliptical polarization (left) and linear polarization (right) of the electric ﬁeld strength.
ωt = 12 π
........................................................... ......... ....... ....... ......... ...... ...... ...... . .. ... .. . . . . ..... .......... .... ..... ..... . . .... . . . . ... ... . . . . ... .. ... ... . . . . ... . . . . ... .. ... . . ... .... .... ... ... ... ... ... ... ... . ... ... .......... ....... ....... ....... ....... .......................................................................................... ... ... ........ .... . . ... . . ... ... .... ... . ... ... ... ... .... .. ... . . ... ... ... .... ... ... ... . .... ... . . ..... . . . .. ..... ... ..... ...... ..... ...... ...... ....... .. ....... ......... .................................................... ...
t
ωt = π
..
ˆ E
s
ˆ E
ωt = 0
ωt = 32 π
Figure 2.6. Circular polarization of the electric ﬁeld strength.
41
poynting’s theorem
At a ﬁxed position x, the polarization is said to be right or lefthanded ˆ , and the direction of propagation, form a rightˆ , E according to whether E or lefthanded triad, respectively.
2.9.
Poynting’s theorem
To calculate instantaneous power, we shall derive the instantaneous Poynting theorem from the ﬁeld equations (2.28) and (2.29). We take the dot product of Eq. (2.28) with E and the dot product of Eq. (2.29) with H, and add the results. Using the vector identity −E · (∇ × H) + H · (∇ × E) = ∇ · (E × H), we ﬁnd ∇ · (E × H) + σE · E + εE · ∂t E + µH · ∂t H = −E · J ext − H · K ext . (2.74) Noticing that E · ∂t E = ∂t ( 12 E · E) and H · ∂t H = ∂t ( 12 H · H), we can write ∇·(E×H)+σE·E+∂t ( 12 εE·E)+∂t ( 12 µH·H) = −E·J ext −H·K ext . (2.75) In order to arrive at the physical interpretation of the diﬀerent terms in this relation, we integrate Eq. (2.75) over an elementary domain D that is bounded by the surface ∂D with outward normal ν (Fig. 2.7). We obtain the power relation ˙ h + ∂t (W e + W m ) = P ext . P out + W The term
P
out
= =
x∈D
x∈∂D
(2.76)
∇ · (E × H)dV ν · (E × H)dA =
x∈∂D
ν · SdA ,
(2.77)
where Gauss’ integral theorem has been used, is written as an integral over the surface ∂D through which D is in contact with its surroundings. Therefore, it seems natural to interpret this term as the instantaneous power that is, across ∂D, transferred from D toward its surroundings.
42
the electromagnetic field equations
............................................. ...... ..... ... ... . . . . ... ..... . . . . ... ... . . . . . ... .... . . . .. . ... .. . . . . ... ... . . . . ... . .. . . . .. ... . . . . . . .... . . . . . .. . . . . .. ... ... ... ... ... ... .. ... . . ... ... ... ... ... ... ... .. . . . .. ... ... .. ... .. ... ... .... . . . ... . ..... .... .... ... ..... ... ..... ..... . ... . . . . ... ...... ... ..... ... ...... ...... .... ....... . ..... . . . . . ... ...... ......... ......... ...........................................
∂D...................................
ν
@ I @
D
σ, ε, µ
Figure 2.7. Representative elementary domain D.
The term ˙ h= W
x∈D
σE · EdV
(2.78)
is a loss term and represents the electromagnetic power that is irreversibly dissipated into heat.
The two terms We =
x∈D
1 2 εE
and Wm =
x∈D
1 2 µH
· EdV
(2.79)
· HdV
(2.80)
occur only as ∂t (W e + W m ) and are interpreted as the amount of energy that is reversibly stored in the electric ﬁeld and the magnetic ﬁeld in D, respectively. Finally, the term P
ext
=−
x∈D
(E · J ext + H · K ext )dV
(2.81)
43
poynting’s theorem
represents the electromagnetic power that is generated by the electromagnetic sources in D whose action has been accounted for by the volume densities of exterior electric current and magnetic current. The minus sign in this term is typical for the fact that for the sources to deliver a positive power to the system, the currents must be oriented opposite to the ﬁelds. Notice, that in this physical interpretation the law of conservation of energy holds. Now that Eq. (2.76) has been physically interpreted, we return to the local equation (2.75). In view of Eqs. (2.78)  (2.80) we introduce the volume density of electromagnetic power that is irreversibly dissipated into heat w˙ h = σE · E ,
(2.82)
the volume density of reversibly stored electric ﬁeld energy we = 12 εE · E ,
(2.83)
the volume density of reversibly stored magnetic ﬁeld energy wm = 12 µH · H .
(2.84)
Then, Poynting’s theorem is written as ∇ · S + w˙ h + ∂t (we + wm ) = −E · J ext − H · K ext ;
(2.85)
it expresses the local conservation of energy. Steadystate analysis In the steadystate analysis we deﬁne the time average Poynting’s vector ST as the average of the time domain quantity S(x, t) = E(x, t)×H(x, t). Let T = 2π/ω be the period in time of the ﬁelds, then we deﬁne the time average 1 t +T E(x, t) × H(x, t)dt . (2.86) ST = T t=t Using the complex representations of the type of Eq. (2.60), it follows that ST =
1 2
ˆ ×H ˆ∗ , Re E
(2.87)
44
the electromagnetic field equations
ˆ = E(x, ˆ ˆ = H(x, ˆ in which E jω) and H jω) are the complex electric and ˆ ˆ ∗ is known as the complex Poynting magnetic ﬁeld vectors. The vector E × H vector. To calculate the timeaverage power, we start with Eqs. (2.61) and (2.62), and subsequently we take the dot product of the complex conjugate of ˆ and the dot product of Eq. (2.62) with H ˆ ∗ , and add Eq. (2.61) with E the results. We then ﬁnd ∗
ˆ ·E ˆ ∗ +jωµH ˆ ·H ˆ ∗ = −E ˆ ·J ˆ ext − H ˆ ∗ ·K ˆ ext . (2.88) ˆ ×H ˆ ∗ )+(σ −jωε)E ∇·(E This equation is known as the complex Poynting’s theorem. The conservation of energy follows from this equation by taking half of its real part in accordance to Eq. (2.87), viz.
1 2 Re
∗
ˆ ×H ˆ ∗) + 1 σE ˆ ·E ˆ ∗ = − 1 Re E ˆ∗·K ˆ ext . ˆ ·J ˆ ext + H ∇ · (E 2 2 (2.89)
The term
ˆ ·E ˆ∗ w˙ h T = 12 σ E
(2.90)
represents the time average of the volume density of heat dissipated by electromagnetic ﬁeld, while the righthand side of Eq. (2.89) can be interpreted as the time average of the volume density of electromagnetic power that is generated by the speciﬁed source in D whose action has been accounted by the volume densities of exterior electric current and magnetic current.
the electromagnetic field equations
2.10.
45
Exercises and problems
Exercise 2.1 How many unknown quantities occur in Maxwell’s equations in matter? Exercise 2.2 Substitution of Eq. (2.24) in Eq. (2.22), with J = 0, yields the local form of the conservation law of electric charge, ∇ · J ext = −∂t ρ. If we integrate both sides of this equation over the source domain Dext we obtain the global ext form, x∈Dext ∇ · J dV = −∂t x∈Dext ρ dV . Use Gauss’ law to give a physical interpretation of the global form of the law of conservation of charge. Exercise 2.3 Let the plane x3 = 0 with unit normal ν = i3 be the sourcefree interface between two media with diﬀerent electromagnetic properties. Apply the reasoning, as it is used to derive the boundary conditions given in Section 2.6, to Eqs. (2.30)  (2.35) and derive the boundary conditions for E1 , E2 , H1 and H2 . Exercise 2.4 Apply the reasoning, as it is used to derive the boundary conditions given in Section 2.6, to Eqs. (2.22) and (2.23), and derive the boundary conditions for J , D and B at a sourcefree interface. Exercise 2.5 Find the polarization state (linear, circular or elliptical) of the following ﬁelds given at a ﬁxed position in space ˆ = ji1 + i2 (a) E ˆ = (1 + j)i1 + (1 − j)i2 (b) E ˆ = (1 + j)i2 − ji3 (c) E ˆ = (2 + j)i1 + (3 − j)i3 (d) E ˆ = ji1 + j2i2 . (e) E
46
the electromagnetic field equations
Exercise 2.6 Give the dimensions of the following quantities in SIunits (a) D · E ,
B·H
(b) J · E ,
E×H
(c) S ,
ST .
Exercise 2.7 ˆ = ( ε0 ) 21 (−ji1 + i2 ) at a ﬁxed ˆ = (i1 + ji2 ) and H Given the two ﬁelds E µ0 point in space, calculate ST . Then calculate E(t), H(t) and S(t).
Problem 2.1 Substitution of Eq. (2.24) in Eq. (2.22) for a conducting medium, with J ext = 0, yields ∇·J +∂t ρ = 0. In an isotropic homogeneous domain we can express ∇ · J in terms of ρ as ∇ · J = σε ρ. Substitution of this relation leads to a ﬁrst order ordinary diﬀerential equation for the electric charge density. The factor τ = σε is called the charge relaxation time of the medium and is a measure for the time it takes the medium to restore its equilibrium state once it has been disturbed by an electromagnetic wave. (a) Given a charge distribution ρ0 at t = 0, solve the ordinary diﬀerential equation for the electric charge density in the time domain. (b) Use Table 2.1 to ﬁnd the relaxation times for copper, sea water, fresh water and glass. In the frequency domain we can write the equation for the charge density ρ = ρ0 , where the righthand side comes from the initial charge as (jω + σε )ˆ distribution considered as a source input. If σ ωε the medium can be σ 1, considered a conductor for that frequency; this is equivalent to ωε σ where the factor ωε is known as the loss tangent also denoted as tan(δ). ( c ) Find the frequency range for which ‘wet earth’ can be regarded a conductor. Use Table 2.1 and the condition tan(δ) > 10. Problem 2.2 Let the plane x3 = 0 be the interface S between two media with diﬀerent
exercises and problems
47
electromagnetic properties, D(1) with x3 < 0 and D(2) with x3 > 0, respectively. In D(1) we have the constitutive constants ε(1) and µ(1) (while σ (1) = 0) and in D(2) we have the constitutive constants ε(2) , µ(2) and σ (2) . Let further in D(1) , in the limit of x3 ↑ 0, the electric ﬂux density and the ˆ (1) = 1 and Jˆ(1) = 0. Give the values electric current density be given by D 1 1 ˆ (2) and Jˆ(2) in the limit of x3 ↓ 0. of D 1 1 Problem 2.3 Show that an elliptically polarized ﬁeld can be decomposed in two circularly polarized ﬁelds, one lefthanded and the other righthanded. ˆ = (ai1 + bi2 ) with two arbitrary complex numbers a and Hint: let E ˆ = b, then let the superposition of two circularly polarized waves be E (a i1 + ja i2 ) + (b i1 − jb i2 ) and solve for a , b in terms of a, b. Problem 2.4 Let the timeharmonic varying electric and magnetic ﬁeld strengths at a ˆ = ( ε0 ) 21 i2 . Determine ˆ = i1 and H ﬁxed point in space be given by E µ0 (a) E(t) (b) H(t) ( c ) S(t). Use the timedomain expressions for the electric and magnetic ﬁelds and the Poynting vector to calculate (d) ST ( e ) ∂t (we + wm )T . Problem 2.5 Assume that the electromagnetic sources are switched on during a ﬁnite interval (0 < t < T ). Integrate both sides of Eq. (2.76) over all positive times and give a physical interpretation of the resulting terms in analogy of Eqs. (2.77)  (2.81).
Chapter 3
Onedimensional Electromagnetic Waves In Chapter 2 we observed that the space variations of the electric and magnetic ﬁeld components are related to the timevariations of the magnetic and electric ﬁeld components through a system of ﬁeld equations. This interdependence gives rise to the phenomenon of electromagnetic wave propagation. In the general case, electromagnetic wave propagation involves electric and magnetic ﬁelds having more than one component, each dependent on all three coordinates, in addition to time. However, a simple and very useful type of wave that serves as a building block in the study of the electromagnetic ﬁeld consists of electric and magnetic ﬁelds that are perpendicular to each other and to the direction of propagation. These waves are known as plane waves. By orienting the coordinate axes such that the electric ﬁeld strength is in the i1 direction and the magnetic ﬁeld strength is in the i2 direction, the propagation direction is in the i3 direction. Plane waves do not exist in practice because they cannot be produced by ﬁnitesized antennas. At large distances from the sources, however, the waves can be approximated as plane waves. Furthermore, the ﬁrst principles of guiding of electromagnetic waves along transmission lines and the ones of many other wave phenomena can be studied in terms of plane waves. In fact, any electromagnetic ﬁeld can be composed through the superposition of an inﬁnite number of diﬀerent plane waves. To illustrate the phenomenon of interaction of electric and magnetic ﬁelds
50
onedimensional electromagnetic waves
giving rise to plane electromagnetic wave propagation, and the principle of radiation of electromagnetic waves from an antenna, we shall consider a simple, idealized, electromagnetic source. This hypothetical source is the planar electriccurrent sheet with a uniform (but timevarying) impressed electric current distribution. Such a current sheet acts as an active device and emits in a homogeneous medium onedimensional electromagnetic waves.
3.1.
The planar electriccurrent sheet as emitter
A reference frame is introduced such that the sheet coincides with the plane x3 = 0 (Fig. 3.1). Let the impressed electric current be uniformly in the x2 direction and ﬂowing in the direction of decreasing x1 , then Jˆ1ext = −Iˆ∆ (s)δ(x3 ) , Jˆ2ext = 0 ,
Jˆ3ext = 0 ,
(3.1)
where δ(x3 ) denotes the onedimensional unit impulse (Dirac distribution), and Iˆ∆ (in A/m) is the electric current per unit length along the x2 direction. Since the electriccurrent sheet carries no magnetic current, we have ˆ ext = 0 , ˆ ext = 0 , K K 1 2
ˆ ext = 0 . K 3
(3.2)
Because the exciting electric current is independent of x1 and x2 , also the generated electromagnetic ﬁeld will be independent of x1 and x2 and, hence, ∂1 = 0 and ∂2 = 0. With this, Eqs. (2.54)  (2.56) and (2.57)  (2.59) are satisﬁed when ˆ3 = 0 . ˆ3 = 0 , H E
(3.3)
ˆ 1 are not related to the source, we can take ˆ2 and H However, since E ˆ1 = 0 . ˆ2 = 0 , H E ˆ 2 are ˆ1 and H The remaining equations for E
(3.4)
51
the planar electriccurrent sheet as emitter
σ, ε, µ E. ... .......... ... .. ... ... ....
S ..........................................×f H
x3 < 0
... ... ... ... ... ... ... ... ... ... ... ..... . ....... ... ............................ ... .... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ....... .. ........ ...... .
σ, ε, µ E .
i1
O r
J ext
i3
... .......... ... .. ... ... ....
r............................................ S f
H
x3 > 0
Figure 3.1. Electriccurrent sheet with impressed current as an emitter of onedimensional electromagnetic waves.
ˆ1 = Iˆ∆ (s)δ(x3 ) , ˆ 2 + (σ+sε)E ∂3 H ˆ2 = 0 , ˆ1 + sµH ∂3 E
(3.5) (3.6)
where σ, ε and µ are constants, since a homogeneous medium is assumed. ˆ 2 from these two diﬀerential equations of the ﬁrst order we Eliminating H end up with the second order diﬀerential equation, ˆ1 − γ 2 E ˆ1 = −sµIˆ∆ (s)δ(x3 ) , ∂3 ∂3 E
(3.7)
ˆ1 , where for the electric ﬁeld strength E 1
γ = [(σ + sε)sµ] 2
with Re(γ) ≥ 0.
(3.8)
In the domain outside the source distribution, i.e., for x3 = 0, the modiﬁed Helmholtz equation (3.7) has a zero righthand side and admits exponential functions of the type exp(±γx3 ) as their solutions. From Eq. (3.6) it follows that the magnetic ﬁeld also admits these exponential functions as their solutions. The quantity of γ is the (sdomain) propagation coeﬃcient. With
52
onedimensional electromagnetic waves
the chosen value of the square root in the righthand side, the propagation factor exp(−γx3 ) is bounded as x3 → ∞ and unbounded as x3 → −∞, while exp(γx3 ) is bounded as x3 → −∞ and unbounded as x3 → ∞. Since, due to causality, the electromagnetic ﬁeld in the halfspace x3 < 0 must remain bounded as x3 → −∞ and the electromagnetic ﬁeld in the halfspace x3 > 0 must remain bounded as x3 → ∞, we write ˆ1 = eˆ− (s) exp(γx3 ) for x3 < 0 , E ˆ − (s) exp(γx3 ) for x3 < 0 , ˆ2 = h H
(3.9) (3.10)
and ˆ1 = eˆ+ (s) exp(−γx3 ) for x3 > 0 , E ˆ + (s) exp(−γx3 ) for x3 > 0 . ˆ2 = h H
(3.11) (3.12)
As we will discuss later, Eqs. (3.9)  (3.10) represent an electromagnetic wave propagating in the negative x3 direction, while Eqs. (3.11)  (3.12) represent an electromagnetic wave propagating in the positive x3 direction. Substitution of Eqs. (3.9)  (3.12) in Eq. (3.6) and use of Eq. (3.8) lead to ˆ − = −Y eˆ− , h ˆ + = Y eˆ+ , h
(3.13) (3.14)
where Y = Y (s) is given by
Y =
σ + sε sµ
1 2
.
(3.15)
Y has the dimension of an admittance and it is therefore denoted as the wave admittance. We further observe that through Eqs. (3.13) and (3.14) the magnetic ﬁeld is expressed in terms of the electric ﬁeld. The analysis where the electric ﬁeld is considered as the fundamental unknown is called
the planar electriccurrent sheet as emitter
53
the electric ﬁeld analysis. One can also write ˆ− , eˆ− = −Z h ˆ+ , eˆ+ = Z h
(3.16) (3.17)
where Z = Z(s) is given by
Z=
sµ σ + sε
1 2
.
(3.18)
Z has the dimension of an impedance and it is therefore denoted as the wave impedance. We further observe that through Eqs. (3.16) and (3.17) the electric ﬁeld is expressed in terms of the magnetic ﬁeld. The analysis where the magnetic ﬁeld is considered as the fundamental unknown is called the magnetic ﬁeld analysis. Obviously, we have YZ =1.
(3.19)
The values of the as yet undetermined coeﬃcients in Eqs. (3.9)  (3.12) follow from the application of the excitation conditions ˆ 2 − lim H ˆ 2 = Iˆ∆ , lim H
x3 ↓0
x3 ↑0
(3.20)
which is a consequence of Eq. (3.5), and ˆ1 − lim E ˆ1 = 0 , lim E
x3 ↓0
x3 ↑0
(3.21)
which is a consequence of Eq. (3.6). Substitution of Eqs. (3.9)  (3.12) in Eqs. (3.20) and (3.21) leads to
and
ˆ − = Iˆ∆ ˆ+ − h h
(3.22)
eˆ+ − eˆ− = 0 .
(3.23)
Combining Eqs. (3.22) and (3.23), with Eqs. (3.16) and (3.17), we arrive at
54
THEORIES OF LIGHT: PARTICLES VERSUS WAVES
cation" theories by a series of experiments reported in Book I. Among Newton's other objections to the wave theory, perhaps the most famous is that if light "consisted in Pression or Motion either in an instant or in time, it would bend into the Shadow,"51 as with water waves and sound waves. But no such bending (no diffraction) had been observed. The wave theory, various forms of which were supported by Newton's critics Hooke and Huygens, was the rival to Newton's corpuscular theory. That there are objections to such theories seems to be at least part of Newton's reason for favoring a particle theory. Putting Queries 28 and 29 together, then, we might say that Newton is proposing a weak inference to a particle theory from two sets of considerations: the explanatory success of the particle theory (Query 29) and the objections to wave theories (Query 28). But how exactly is this inference supposed to proceed? What form does it take? If Newton is making such an inference—and I think it is plausible to suppose that he is — he does not spell it out. On the basis of the discussions in Queries 28 and 29 it is, I think, reasonable to say that Newton is arguing in some such way as this: The hypothesis that light is corpuscular explains a range of observed optical phenomena. The rival wave hypothesis has such and such difficulties. Therefore (probably) Light is corpuscular. The inference is intended to be weak, not strong. But even so, exactly how it is supposed to go and whether it is reasonable even as a weak inference is not at all clear. In what follows I will construct an idealized version of this argument, which, although I cannot claim it to be Newton's, may nevertheless reflect some features of his thought. (See Essay 3.) The argument will have two essential components: objections to the wave theory, and an appeal to the explanatory power of the particle theory. Moreover, it will be an argument whose conclusion is drawn with probability. This probability will be reasonably high but not high enough to achieve the certainty, or virtual certainty, of a "deduction from phenomena." In constructing the argument I will assume that the usual axioms of the probability calculus are satisfied. Probability can be construed here as representing rational degrees of belief. Let us assume to begin with that light is either a particle phenomenon or a wave phenomenon. Newton himself offers no explicit argument for such an assumption, although here is one that he was in a position to offer (and that is very similar to one in fact offered by his wavetheoretic opponents in the nineteenth century; see Essay 3): 51. Ibid.
NEWTON'S CORPUSCULAR QUERY AND EXPERIMENTAL PHILOSOPHY
55
Light is observed to travel in straight lines with uniform speed. In other cases, when something travels with uniform speed in a straight line this motion is always observed to be caused by a series of bodies or by a series of wave pulses produced in a medium (e.g., sound waves, water waves). Therefore Light is either a particle or a wave phenomenon. Let us assume that the premises are true, that they report "phenomena," and that this is a "deduction from phenomena," so that the conclusion has the highest certainty possible in experimental philosophy. Such an inference, a type of causal simplification, would be permitted by Newton's Rule 2 requiring that like causes be assigned to like effects, as far as possible. If in other cases motion is caused by particles or waves, then, unless there is evidence to the contrary, we should infer like causes in the case of the motion of light. Let T1 be the hypothesis that light consists of particles, T2 the hypothesis that light consists of waves, O the observed fact that light travels in straight lines with uniform speed, and b the accepted background information, which includes the information in the second premise above. We might express the results of this argument probabilistically, as follows:
That is, the probability that either T1 or T2 is true, given O and b, is equal to 1. (With minor alterations an argument similar to that which follows can be made if equation (1) is changed to say that the probability in question is close to 1; see Essay 3.) Let us suppose that by appeal to certain other observed facts about light — call them O'—we can show that the probability of the wave theory is low, say less than 1/2. Which facts these are, and how this is to be shown, will be taken up in a moment. For the present let us simply write
Now if p(T1 or T2/O&b) = 1, then p(T1 or T2/O&O'&b) = 1. Therefore from (1) and (2), since T1 and T2 are incompatible, we can infer
Let O1, . . . ,On be various observed facts about light (e.g., reflection, refraction, variety of colors, fits of easy reflection and transmission, etc.) other than those in O and O'. We would like to know how probable T1 is when these facts are considered in addition to O&O'&b. Is p(T 1 /O 1 , . . . ,On&O'&b) also high? Suppose that explanations of O1, . . . ,On by theory Tl can be con
56
THEORIES OF LIGHT: PARTICLES VERSUS WAVES
structed in such a way that O1, On follow deductively (in the ordinary sense) from T1 together possibly with the background information b. But if the particle theory Tl (together possibly with b) deductively entails O1, On, then it follows from the probability calculus that p(T1:/ O1, . . . ,On&O&O'&b) p(T1/O&O'&b). So from (3), given the existence of such explanations, we derive
which can be construed as the conclusion of the argument. It says that the particle theory is probable (more probable than not) given not just a few chosen optical phenomena but a range of them, including ones explained by that theory. The explanations of O1, . . . ,On provided by the particle theory do not create the high probability for that theory, but they do sustain it. (In Essay 4 I argue that, in general, derivational explanations, even those subject to certain further conditions, will not by themselves suffice to guarantee high probability, although they can increase it.) These explanations permit an inference from (3) to (4). In an attempt to establish high probability for the particle theory on the basis of a range of optical phenomena, this is an essential role played by such explanations. To create high probability in the first place, a type of eliminative argument is used in which the wave and particle theories exhaust the probability but the probability of the wave theory is low. How is the latter to be established? Newton offers arguments of two types, one direct, the other indirect. An argument appealing to diffraction is an example of the former. In the case of other wave motions such as sound waves and water waves, diffraction into the shadow of an obstacle is observed, but no such diffraction into the shadow had been observed by Newton or others in the case of light (though Newton had observed diffraction away from the shadow).52 So if we include in the background information the fact of observed diffraction with other wave phenomena and the absence of such observations in the case of light, then the probability of the wave theory, given b, is low. Second, Newton offers a more indirect type of argument. He points out that to explain certain observed optical phenomena, the wave theory introduces auxiliary assumptions that are either refuted by, or made very improbable by, observations. For example, to explain differences in refrangibility of rays emerging from a prism, wave theories introduce the auxiliary hypothesis that the prism modifies rather than separates the rays. Newton argues that this modification assumption is refuted or at least made extremely unlikely by further refraction experiments. (See Experiment 5, Book I, pp. 34ff.) Can we 52. See Roger H. Stuewer, "A Critical Analysis of Newton's Work on Diffraction," Isis 61 (1970), pp. 188205.
NEWTON'S CORPUSCULAR QUERY AND EXPERIMENTAL PHILOSOPHY
57
infer from this that the wave theory is improbable? We can if we suppose that the probability of the modification assumption, given the wave theory and the observations, is close to 1. Letting M be the modification assumption, T2 the wave theory, and O' the results of various of Newton's refraction experiments, if p(M/O&O'&b) is close to zero and p(M/T2&O&O'&b) is close to 1, then p(T2/O&O'&b) is close to zero. Even more generally, we get the same result if we suppose simply that p(M/T2&O&O'&b) is much, much larger than p(M/O&O'&b), without needing to suppose that the former probability is close to 1. (See Essay 3.) Accordingly, there are two sorts of arguments to show that the probability of the wave theory is low, that is, (2). Once this is shown, we can infer the high probability of the particle theory (3), and its continued high probability in light of the various observed facts that it explains, that is, (4). How "Newtonian" is the previous argument? In one respect, quite unNewtonian, since it explicitly invokes numerical probabilities and the probability calculus, neither of which, of course, Newton does. However, in certain other respects it reflects what Newton seems to be doing in Queries 28 and 29. Assuming, as I have been, that Newton intends to provide some reasons for believing the particle theory, albeit "weak" ones, it gives a basis for inferring that theory with probability rather than certainty. Moreover, in doing so it takes into account Newton's criticisms of the wave theory and the explanatory virtues of the particle theory (each of which Newton himself emphasizes), showing how both contribute to the probability of the particle theory. One objection that might be offered is that this argument is a type of eliminative one, whereas Newton at one point rejects (a certain form of) eliminative reasoning. In a letter to Oldenburg of July 1672 he writes: I cannot think it effectual for determining truth to examine the several ways by which phenomena may be explained, unless there can be a perfect enumeration of those ways. You know, the proper method for inquiring after the properties of things is to deduce them from experiments. And I told you that the theory, which I propounded [the theory of the heterogeneity of light rays] was evinced to me, not by inferring 'tis thus because not otherwise, that is, not by deducing it only from a confutation of contrary suppositions, but by deriving it from experiments concluding positively and directly.53
Newton here seems to be rejecting eliminative arguments of this form: E: Each of the hypotheses h1, . . . ,hn, if true, will correctly explain phenomenon p.
But hypotheses h2,. . . ,hn are false.
Therefore, hypothesis h1 is true.
53. I. B. Cohen, ed., Newton's Papers and Letters (Cambridge, Mass., 1978), p. 93.
58
THEORIES OF LIGHT: PARTICLES VERSUS WAVES
Such arguments, which infer the truth of an hypothesis from the falsity of competitors, are fallacious, unless a complete enumeration can be made of all the competitors. Newton appears to be thinking of deductive interpretations of E in which if the premises are true the conclusion must also be true. And he is correct in saying that arguments of form E, thus construed, are fallacious unless a complete enumeration of hypotheses is given. They are also fallacious if construed nondeductively, that is, as being such that the premises make the conclusion probable without entailing it. Unless some suitable assumption is made about the probability of the disjunction of hypotheses that are mentioned in the first premise, the conclusion that the probability of hl is high cannot be drawn. However, the particular eliminative argument I have constructed is not of form E. The first step in the argument, which leads to (1) above, is not an explanatory step, but one involving causal simplification. The claim is not that the particle and wave theories will both explain the finite rectilinear motion of light, but that in other observed cases when something travels with uniform speed in a straight line this motion is caused by a series of bodies or a series of wave pulses in a medium. Also, the claim in the first step is indeed exhaustive, since it assigns a probability of 1 to the disjunction of hypotheses. But even if it were not exhaustive in this sense, even if (1) were changed to read "p(T 1 or T2/O&b) is close to but not equal to 1," a fallacy would not emerge (although other changes would need to be made in the argument). I conclude that reconstructing what Newton does in Queries 28 and 29 in the form of a probabilistic argument that takes us from (1) to (4) above is in conformity with certain important aspects of Newton's methodology. It combines his explanatory reasoning in Query 29 with his criticisms of the wave theory in Query 28 to provide some reason, though not the highest possible in experimental philosophy, to believe the corpuscular hypothesis. Although it is an eliminative argument it is not one of the type Newton rejects. Because the hypothesis in question is inferred with a probability not sufficiently high to be a virtual certainty, Newton could not construe the argument as a "deduction from phenomena." While we should search for phenomena that will sanction such a "deduction," we should acknowledge that, assuming its premises are true, the present argument does provide a legitimate "weak" reason for believing the corpuscular hypothesis.
6. STRONG VERSUS WEAK INFERENCES: AN ASSESSMENT OF ONE TENET OF NEWTONIAN METHODOLOGY
A "strong" inference furnishes the proposition inferred with the highest evidence possible in experimental philosophy; a "weak" inference furnishes some evidence but not the highest possible. I shall suppose that this difference can
NEWTON'S CORPUSCULAR QUERY AND EXPERIMENTAL PHILOSOPHY
59
be interpreted as a difference over probabilities (construed as representing degrees of rational belief). In a strong inference from A to B, the probability of B given A is close to or equal to 1. In a weak inference the probability is high (say greater than 1/2), but is not close to or equal to 1. For Newton, both strong and weak inferences are based on "phenomena." Now I take it to be a tenet of Newtonian methodology that in experimental philosophy "deductions from phenomena," and only these, are strong inferences. Accordingly, I want to raise two questions of assessment: (a) Are Newton's "deductions from phenomena" guaranteed to be strong inferences? (b) Must other kinds of inferences fail to be strong? To answer the first question we must return to the definition of "deduction from phenomena" offered in section 2. Deductions from phenomena, we recall, include ordinary deductions, inductions, and causal simplification. Inductions are inferences from all observed members of a class to some members of the class that have not been observed, or to all members of the class. Let me simplify the discussion by considering deductive and inductive inferences but omitting causal simplification, which does not lend itself so readily to a general probabilistic treatment. Also, I shall discuss cases involving only deductions (in the ordinary sense) and those involving only inductions. Deductive cases: Let O1, . . . ,On be descriptions of phenomena the conjunction of which, together with background information b, deductively implies h. Then p(h/O 1 , . . . ,On&b) = 1. So here we have a "strong" inference from the O's and b to h. Inductive cases: To discuss these I shall first introduce some probability considerations and afterward apply them to the sorts of cases particularly relevant to Newtonian induction. Let h be a proposition that together with background information b deductively entails some observational statements O1,O2, . . . . The following claims are provable.54
(a) tells us that if the prior probability of h is not zero, then as the number n of observed consequences of h and b gets larger and larger, the probability that the n + 1 observational consequence of h and b will be true gets higher and higher, approaching 1 as a limit. (b) tells us that if the prior probability of h is not zero, then as the numbers m and n get larger, the probability that the
54. See John Earman, "Concepts of Projectibility and Problems of Induction," Nous 19 (1985), pp. 521535.
60
THEORIES OF LIGHT: PARTICLES VERSUS WAVES
next m observational consequences are true, given that n observational consequences obtain, gets higher and higher and approaches 1 as a limit. To introduce the third probability result some restrictions will need to be made on h and Oi. Let h be some universal generalization of the form (x)(Fx Gx). Let the O's be "instances" of h of the form Fai Gai. The following is provable:
(c) gives a set of sufficient conditions for the probability of (x)(Fx Gx), given observed instances of the form Fa, Gai, to approach 1 as a limit. Now let us apply these three probability results to Newtonian inductions. In all three cases let us consider h's of the form (x)(Fx Gx), and O's instances of the form Fai Gai. (a) tells us that if the prior probability of (x)(Fx Gx) is not zero, then as the number of observed instances of (x)(Fx Gx) increases, the probability that the next instance will obtain gets higher and higher, approaching 1 as a limit. A similar claim can be made for (b). (a) and (b) —so construed — correspond to Newton's inductions from some observed members of a class to some other member(s) of that class. (c) corresponds to Newton's inductions from some observed members of a class to all members. In all three cases the probability in question approaches 1 as a limit, under certain very weak assumptions. Intuitively, the probability that the next instance will satisfy (x)(Fx Gx), that the next m instances will, and that all instances will, gets higher and higher as more and more instances are observed. We get more and more certainty in these cases with more and more observed instances of (x)(Fx Gx). However, it is not the case that for every number n of observed instances, the probability that the next instance will satisfy (x)(Fx Gx), that the next m instances will (for any m), and that all instances will, is close to 1. Consider just the latter, and suppose that the prior probability of (x)(Fx Gx) is low, and the O's are such that, with sufficiently small n, the prior probability of the conjunction of O's is high. If (x)(Fx Gx) and b entails the O's, then by Bayes' theorem,
Now if p((x)(Fx
Gx)/b) is low and the O's are such that, with sufficiently
55. For proof see Earman, op. cit., p. 529.
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small n, p(O 1 , . . . ,On/b) is high, then the probability on the left will be small, despite the fact that all the observed O's satisfy (x)(Fx Gx). One case of this sort involves Goodmanesque properties such as "grue," where the prior probability of the proposition "All emeralds are grue" is very low but where, given appropriate background information, the probability that observed items are grue if they are emeralds is very high. Strange Goodmanesque properties or classes can prevent the probability on the left from being high for a given n. But as n increases without bound, the probability on the left will approach 1 as a limit, strange properties notwithstanding. However, Goodmanesque properties are not the only things that can prevent the probability on the left from being high for a given n. Recall the proof of Proposition 1 of the Opticks. Here an induction is made from observations of differences in refrangibility of blue and red rays in an experiment with the sorts of prisms used by Newton to differences in refrangibility of any differently colored rays in any sort of refraction, whether or not the latter is produced by a prism. A critic of Newton might argue as follows: (i) The number n of observed instances of Proposition 1 (that lights that differ in color differ in degrees of refrangibility) is quite low. (If we count as instances here the results of types of experiments, rather than specific trials, then the critic has some justification, since Newton cites only two experiments.) (ii) The critic might agree that the probability of getting the results Newton obtains with these types of experiments with prisms is high, while supposing that obtaining analogous refraction results with other sorts of prisms or without prisms is improbable.56 (iii) The critic might argue, on the basis of background information b, that the prior probability of Newton's Proposition 1 is very low. At least, the critic might complain, Newton does nothing to dispel doubts expressed in points (i) through (iii). But unless such doubts are removed, the probability of Newton's Proposition 1, given the results of the experiments Newton mentions, cannot be assumed to be high. The most we can say is that this probability will increase toward 1 as the number of observed instances of the proposition increases. Confining our attention to ordinary deductions and inductions, we can now answer the question "Are Newton's 'deductions from phenomena' guaranteed to be strong inferences?" in the following way. If they are deductions (in the ordinary sense), they are so guaranteed. Any deductive inference from O1, ,On and b to h—no matter what number n is — guarantees that the probability of h given O1, . . . ,On and b is maximal. By contrast, it is not the case that every particular inductive inference is guaranteed to be strong, no matter how many instances are involved and no matter what the character of
56. See Simon Schaffer, "Glass Works: Newton's Prisms and the Uses of Experiment," in David Gooding, Trevor Pinch, and Simon Schaffer, eds., The Uses of Experiment (Cambridge, England, 1989), pp. 67104.
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the properties or classes in question. If Newton's methodology requires a claim to the contrary, then it is mistaken. However, the previous probability results show that (under certain weak assumptions), as more and more instances are observed, then no matter what the character of properties or classes in question, the strength of the inference is guaranteed to increase and to approach the highest strength in the limit. Accordingly, there are several ways to interpret Newton's methodology (or to modify that methodology) so as to avoid the problems above. First, instead of saying that every inductive inference from phenomena is a strong one, Newton could say that some are, namely, those based on sufficiently many instances (provided that the prior probability of (x)(Fx Gx) is not zero). Second, Newton could restrict those inductions he will allow in the category of "deductions from phenomena" to ones based on sufficiently many instances. On both of these proposals, however, no particular number can be chosen that will count as "sufficiently many." In each case this will depend on the prior probability of (x)(Fx Gx) and on the prior probability of the conjunction of instances. Third, Newton could attempt to impose conditions on the character of the properties or classes that are subject to induction so that inductions involving such properties or classes will guarantee maximal probability no matter how many instances have been observed. Newton does not formulate any such conditions. Whether it would be possible to do so seems very dubious to me, though I shall not pursue this here. Finally, Newton could abandon entirely an "absolute" claim about the strength of inductions in favor of a "comparative" one. He could say simply that, under minimal assumptions, the strength of an induction increases as more and more instances are observed. Now turning to the other side of the coin we need to ask whether in experimental philosophy there are strong inferences that are not "deductions from phenomena." Is Newton correct in implying that only "deductions from phenomena" can have this feature? Let us return to result (a) above. (What I say here will be applicable to (b) as well, mutatis mutandis.) Although (a) allows h and Oi to be of forms (x)(Fx Gx) and Fa Ga, respectively, it does not require this. All that is necessary is that h and b deductively imply Oi. Accordingly, h might be some proposition that Newton would classify as an hypothesis, for example, that light consists of particles. This hypothesis is not "deduced from phenomena." Let the background information b include Newton's first law of motion that in the absence of forces particles travel with uniform speeds in straight lines. Hypothesis h + b deductively implies (O1) that in the absence of forces light travels in straight lines, and (O2) that in the absence of forces light travels with uniform speed. Now result (a) allows us to conclude that the probability that some consequence of an "hypothesis" (in the Newtonian sense) obtains gets higher and higher, approaching 1 as a limit, as more and more consequences of that hypothesis are observed. The only assumption needed is that the prior
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probability of this hypothesis is not zero. This, of course, does not imply that the probability of the "hypothesis" itself approaches certainty, but only that the probability of its deductive consequences does. Newton does not appear to be thinking of cases in which we make inferences to deductive consequences of "hypotheses." But such inferences can be strong ones, or at least they can get stronger and stronger as more and more consequences are observed to hold. To be sure, Newton could claim that he is classifying as "inductive" an inference from some observed consequences of h to other not yet observed consequences. But his inductions appear to be simply inductive generalizations from observed F's that are G's to other or all F's being G's. Let us turn, then, to result (c) involving the probability of h itself. And let us consider the more general case in which h is any proposition that, together with b, deductively implies O1, O2, Here we cannot obtain the result that lim p(h/O 1 , . . . ,On&b) = 1 because we cannot in general assume that n lim p(O 1 , . . . ,On/b) = p(h/b). Indeed, the following are provable: n— (d) Let h (together with b) entail O1, O2, . . . . If h has at least one incompatible competitor h' that together with b also entails O1, O2, . . . , and whose probability on b is greater than zero, then lim p(h/O 1 , . . . ,On&b) 1. (e) Let h together with b entail O1, O2, . . . . If h has at least one incompatible competitor h' that together with b also entails O1, O2, . . . , and is such that p(h'/b) p(h/b), then for any n, no matter how large, p(hlO 1 , . . . ,On&b) .5.57 So if h has competitors that, like h, deductively imply all the observable phenomena, then h's probability will not approach 1 as a limit; and if the prior probability of one of the competitors is at least as great as h's prior probability, then h's probability will not increase beyond .5, no matter how many deductive consequences of h are observed to be true. However, the quest for strong inferences to Newtonian "hypotheses" is not necessarily doomed. We need not insist that the limit of the probability of h be 1, but only that the probability of h given the observations be "very high" and remain so with more and more observations. To this end, I shall employ the concept of a partition of propositions on b, which is a set of mutually exclusive propositions, the probability of whose disjunction on b is 1, and the probability of each of which on b is not zero. The following is provable (see Essay 4):
57. See Earman, op. cit., pp. 528529.
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(f)
If h, h1, . . . , hk form a partition on b, then for any O, and for each hi ( h) in the partition, and for any number r greater than or equal to 0 and less than 1, p(h/O&b) > r if and only if p(h i /O&b) < 1  r. i=1
Now suppose that we have some observed phenomena O, O1, ,On and background information b, and we want to make a strong inference to h by showing that the probability of h given the observed phenomena and background information is greater than some threshold value r for "very high" probability. Using theorem (f), the following strategy is possible: Strategy for showing that h has a very high probability (greater than some threshold value r for very high probability), given observed phenomena O, O1, . . . , On and background information b:
1. Find some partition on b—h, h1, . . . ,hk—that includes h. 2. Show that phenomenon O is such that for each proposition hi, h in the partition,
p(ht /O&b) < 1  r. i 1
3. Show that O1, . . . ,On are derivable from h (together with b). If we complete steps 1 and 2 in this strategy, then, in accordance with theorem (f), we will have shown that p(h/O&b) > r. By completing step 3, we show that p(h/O& O1 , . . . ,On&b) > r, since O1, . . . ,On are derivable from h together with b. The question of interest is whether this strategy is applicable to propositions Newton would regard as hypotheses. In fact, it seems applicable to the hypothesis Newton considers in Query 29 of the Opticks—that light consists of particles. Indeed, the probabilistic argument I constructed in the last section can be suitably modified and shown to be a legitimate variant of this form. Let me recall the basic steps. We began with the observation O that light travels in straight lines with uniform speed, which, together with the background information b, yields a probability of 1 that light is corpuscular or undulatory, that is,
Accordingly, T1 and T2 form a partition on O&b, since T1 and T2 are incompatible. Second, we found some other observed facts O' that cast doubt on the wave theory T2. We noted this by writing p(T2 /O&O'&b) < 1/2. But this is
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too modest, even for Newton, since the actual facts cited, Newton thought, cast much more doubt on T2 than this. The two mentioned were diffraction and refraction. If light is a wave phenomenon, then, like water waves and sound waves, it ought to be diffracted into the shadow; but Newton observed no such diffraction. Second, Newton (as well as defenders of the wave theory) believed that, given the wave theory and the observations of differences in degrees of refraction, the probability that light is modified by the refracting prism is very high, say close to 1. But on the basis of his own refraction experiments, Newton pretty clearly thought he had refuted the modification assumption, that is, the probability of this assumption, given his experimental results, is close to zero. So, where M = the modification assumption, and O' includes the results of Newton's refraction experiments, we have the result ihat p(M/T2&O'&O&b) is close to 1, whereas p(M/O'&O&b) is close to zero. It follows that p(T2 /O'&O&b) is close to zero. Letting O' also contain the observed absence of diffraction into the shadow and b also contain observed diffraction in the case of sound and water, we write
This completes the second step in the strategy. From (1), since the probability of T1 or T2 is 1, it remains 1 if we add O'. So we have
Since T1 and T2 are incompatible, from (2) and (3) we infer
Since T1 + b deductively implies other optical phenomena O1, . . . ,On, from (4) we derive
which completes the third and final step of the strategy. Again, I must stress that it is not my claim that this is Newton's actual argument in Queries 28 and 29. Besides the attribution of the probability calculus, the main stumbling block lies in the use of the first step in the strategy, leading to (1) above (even if (1) were weakened by replacing "equals" with "is close to"). Although Newton considers only the wave and particle theories, he does not explicitly claim that the probability of their disjunction on the evidence is maximal (or even very high). Still in the previous section I
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indicated what type of argument for this claim Newton could have given that would be compatible with his general methodology. If the strategy is launched by this assumption in step (1), then Newton's own arguments against the wave theory can be used to justify the steps leading to the final (5). Let us suppose that (1) is justified by inference from observed phenomena. And let us assume that the remaining steps are also valid, so that the argument does establish the very high probability of a proposition given certain observations and background information. If so it provides the basis for a "strong" inference to that proposition from those observations and background information. Is the argument a "deduction from phenomena"? In certain ways it seems quite different from the sorts of arguments Newton has in mind when he uses this expression. First, unlike the "deductions" that Newton gives, it contains an inference to a disjunction of propositions in the first step. Second, the argument is eliminative, whereas the "deductions" Newton offers are not. Indeed, he rejects (certain types of) eliminative arguments. Third, and most important, it makes use of the probability calculus, which Newton never does. The inferences to (3), (4), and (5) are justified by principles of probability. Whether Newton would have been willing to classify such inferences as "deductive" is unclear. Yet reasons might be offered for classifying the argument as a "deduction from phenomena." First, the previous characterization of causal simplification (as well as that of induction) does not preclude an inference to a disjunction of propositions. Second, although it is eliminative, it is not an eliminative argument of the type that Newton rejects. Indeed, if the previous point is accepted, it is an eliminative argument that uses causal simplification to establish a disjunction of propositions and then to argue against one of the disjuncts. Third, the probability principles generating steps (3), (4), and (5) might be thought of as, or as akin to, mathematical principles, which for Newton can serve as a basis for "deductions." Accordingly, assuming the argument in question is valid, the following possibilities emerge: 1. In a broad sense the argument is a "deduction from phenomena." If so it does not refute the Newtonian claim that only "deductions from phenomena" guarantee strong inferences. However, if we construe it as a "deduction from phenomena," then with this argument we must deny that the Newtonian corpuscular hypothesis is an hypothesis. With this argument we will have "deduced" the corpuscular hypothesis from the phenomena and thus rendered it no longer hypothetical. 2. In a narrower sense (one that excludes probability arguments) the argument is not a "deduction from phenomena." Yet it provides the basis for a "strong" inference to the corpuscular hypothesis. So if this narrower sense is Newton's, then we need to reject his idea that only "deductions from phenomena" can provide the highest certainty in experimental philosophy.
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7. CONCLUSIONS
1. Although neither Newton's professed methodology, nor his actual practice, form consistent sets, my suggestion is that interpretation (3) in section 3 reflects a good deal of both. On that interpretation, the most certainty possible in experimental philosophy is achieved when, and only when, propositions are "deduced from phenomena." The latter involves deduction or induction or causal simplification from generally accepted facts established by observation. 2. However, on this interpretation one is allowed not only to consider propositions not "deduced from phenomena," that is, hypotheses, but to make weak inferences to them in cases in which "deductions" have not been achieved. But we must recognize that such inferences are weak, and we must continue to search for phenomena from which the propositions in question can be "deduced." 3. One sort of non"deductive" inference to hypotheses is illustrated in Queries 28 and 29 in the Opticks in the discussion of the particle and wave theories of light. Here Newton seems to be making a (weak) inference to the particle theory on the grounds that it explains a range of optical phenomena. In section 5 above this argument is reconstructed probabilistically in such a way as to reflect, at least in part, Newton's discussion in Queries 28 and 29, as well as his general methodology. 4. We cannot suppose, as Newton seems to, that "deductions from phenomena" will always yield the maximal certainty possible in experimental philosophy. In the case of induction, for example, such certainty is not guaranteed simply by observing positive instances of an inductive generalization and no negative ones. What we can say is that, granted certain minimal assumptions, an increase in the number of positive instances will increase the strength of such inferences toward maximality. Finally, assuming that probabilistic explanatory reasoning of the type constructed in section 6 can be valid, we may say this: If probabilistic arguments are not construed as "deductive," then we cannot suppose, as Newton seems to, that only "deductions from phenomena" can generate the highest certainty possible in experimental philosophy. *
*For very helpful suggestions I am indebted to Robert Rynasiewicz, Doren Recker, Robert Kargon, and Alan Shapiro.
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ESSAY 3
Light Hypotheses
1. INTRODUCTION
At the beginning of the nineteenth century Thomas Young published papers that defended the wave theory of light against the Newtonian particle theory. Following this there occurred a lengthy and sometimes heated dispute between particle theorists and wave theorists which, it is alleged, stemmed from deep divisions over scientific methodology. Particle theorists, it is said, particularly British ones, used the method of induction whereas wave theorists employed the antithetical method of hypothesis. Thus Geoffrey Cantor writes: Although in the eighteenth century almost every British natural philosopher accepted without question the corpuscular interpretation of Newton's writings on optics, by the 1830s most British natural philosophers had rejected Newton's corpuscular theory in favor of the wave theory of light. Intimately bound up with this scientific "revolution" in optical theory was a change in scientific methodology: the replacement of the method of induction by the method of hypothesis.1
According to Cantor, nineteenthcentury particle theorists "followed the [eighteenthcentury] commonsense philosophers in considering induction to be the proper scientific method" (p. 111), and in rejecting or limiting a reliance on hypotheses; nineteenthcentury "supporters of the wave theory, unlike its objectors, championed the method of hypothesis" (p. 114). Larry Laudan has also emphasized a change in methodology between the late eighteenth and early nineteenth centuries. The wave theory of light re1. Geoffrey Cantor, "The Reception of the Wave Theory of Light in Britain: A Case Study Illustrating the Role of Methodology in Scientific Debate," Historical Studies in the Physical Sciences 6 (1975), p. 109. Emphasis mine. See also his book Optics after Newton (Manchester, 1983), pp. 177186. 69
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quired an imperceptible luminiferous ether, which eighteenthcentury inductivists rejected as an untestable hypothesis. Speaking of the reception of ether theories in Scotland, Laudan writes: The primary reason for opposition to ether theories was the widespread acceptance among Scottish philosophers and scientists of a trenchant inductivism and empiricism, according to which speculative hypotheses and imperceptible entities were inconsistent with the search for reliable science.2
Laudan goes on to argue that nineteenthcentury defenders of the wave theory used a form of the method of hypothesis (which I shall consider in this essay). In 1803 Henry Brougham, a defender of the particle theory of light, wrote a scathing review of Thomas Young's "Bakerian Lecture on the Theory of Light and Colors." Brougham begins by saying: As this paper contains nothing which deserves the names either of experiment or discovery, and as it is in fact destitute of every species of merit, we should have allowed it to pass among the multitude of those articles which must always find admittance into the collections of a Society which is pledged to publish two or three volumes every year.3
Brougham's principal objection to Young's paper is that it is not based on inductions from experiments but involves simply the formulation of hypotheses to explain various facts. And Brougham writes: A discovery in mathematics, or a successful induction of facts, when once completed, cannot be too soon given to the world. But . . . an hypothesis is a work of fancy, useless in science, and fit only for the amusement of a vacant hour. . . . (p. 451). It is scarcely possible to conceive a wider difference than that which subsists between the philosophy of Newton and the philosophy of Dr. Young. While the former utterly rejects hypotheses, and asserts that our stock of facts upon the subject of the ether is insufficient; the latter says that we have enow [sic] of experiments, and that we only require to have a stock of hypotheses, (p. 455)
In this review Brougham defends the Newtonian particle theory on the grounds that it is inductively supported by experiments, while he rejects 2. Larry Laudan, "The Medium and its Message," in G. N. Cantor and M. J. Hodge, eds., Conceptions of the Ether (Cambridge, England, 1981), p. 170. Unlike Cantor, Laudan's principal aim in this paper is to compare the methodology of those who defended the wave theory in the nineteenth century with eighteenth (rather than nineteenth) century inductivist critiques of that theory (and of ether theories generally). Since my main interest in what follows is (like that of Cantor) in the debate between wave and particle theorists in the nineteenth century, Laudan's claim that is of special concern to me is that nineteenthcentury wave theorists utilized a method of hypothesis. 3. Edinburgh Review 1 (1803), p. 450.
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Young's defense of the wave theory on the grounds that it employs an unacceptable method of hypothesis. In what follows I propose the following: 1. To give an account of the method of hypothesis (or of various such methods). 2. To argue, contrary to Cantor and Laudan, that in their actual practice, as well as in their reflections on this practice, nineteenthcentury wave theorists such as Young, Fresnel, Lloyd, and Herschel typically employed a method that is significantly different from the method of hypothesis. 3. To argue that this method contains not only an explanatory component present in the method of hypothesis, but an "independent warrant" component that is not. For wave theorists the strategy for supplying independent warrant is an eliminative one that can be justified by appeal to probabilistic and inductive considerations. Of particular interest in this justification will be probability considerations introduced in section 5 that pertain to the introduction of auxiliary hypotheses. 4. To give an account of what nineteenthcentury British particle theorists such as Brougham meant by "induction" and how they utilized this method in developing the particle theory. In doing this some attention will need to be given to Newton's ideas, which exerted considerable influence on later particle theorists. 5. To argue that there are strong similarities, if not identities, between the inductivism of British particle theorists and the methodology of wave theorists, and that the important debate is over particles versus waves, not methodologies. Accordingly, the present case will not support a form of relativism that states that fundamentally different theories employ fundamentally different methodologies.
2. THE METHOD OF HYPOTHESIS
On this method one proposes an hypothesis to explain observed phenomena. If the hypothesis, if true, would correctly explain those phenomena, one can claim support for it, even if the hypothesis postulates unobserved or unobservable entities and processes. Here is a simple use of this method by David Hartley, in defending the hypothesis that an ether exists: Let us suppose the existence of the aether, with these its properties, to be destitute of all direct evidence, still, if it serves to explain a great variety of phenomena, it will have an indirect evidence in its favour by this means.4
4. David Hartley, Observations on Man, His Frame, His Duty, and His Expectations, 2nd ed., 2 vols. (London, 1791), vol. 1, p. 15. Quoted by Laudan, op. cit., p. 161.
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More generally, the method of hypothesis contains the following idea: Basic method of hypothesis: The fact that hypothesis h if true would correctly explain observed phenomena O1, . . . ,On constitutes at least some reason to think that h is true.
Frequently, especially in the mathematical physics developed by both wave and particle theorists, the explanations consist of deductive derivations of C\, . . . ,On from h. If so, we get the modern hypotheticodeductive viewpoint. However, the method of hypothesis is also close to the idea of retroduction introduced by Peirce at the end of the nineteenth century and developed in the middle of the present century by N. R. Hanson. According to Hanson, it involves an inference of the following form: Some surprising phenomenon P is observed. P would be explicable as a matter of course if h were true. Hence, there is reason to think that h is true.5 Now, according to Laudan, by the 1830s an important change had occurred in the method of hypothesis. Prior to this, hypothesists were willing to conclude that there is some reason to think an hypothesis true if it explains phenomena that have already been observed and for which the explanation was proposed in the first place. (This would make early versions of the method akin to Hanson's form of retroduction.) By the 1830s a requirement was instituted that the hypothesis has to explain states of affairs significantly different from those it was invented to explain. This can be accomplished if the hypothesis can predict (and explain) some new and as yet unobserved phenomenon or some known phenomenon that did not prompt the hypothesis in the first place.6 Such a view is expounded by William Whewell, who speaks in this connection of a consilience of inductions. We might formulate it like this: Method of hypothesis with consilience: Let h be some hypothesis proposed initially to explain O1 (and nothing else). The fact that h if true would correctly explain O1,
5. N. R. Hanson, Patterns of Discovery (Cambridge, England, 1958), p. 72. 6. See Laudan, op. cit., p. 175. At least one exception to Laudan's thesis is Huygens, who in 1690, in the Preface to his Treatise on Light, defends a form of the method of hypothesis that involves deriving or explaining not only known facts but new ones as well. He writes: . . . here [in his Treatise] the principles are verified by the conclusions to be drawn from them . . . especially when there are a great number of them, and further, principally, when one can imagine and foresee new phenomena which ought to follow from the hypotheses which one employs, and when one finds that therein the fact corresponds to our prevision. (Christian Huygens, Treatise on Light, reprinted in Robert Maynard Hutchins, ed., Newton, Huygens (Chicago, 1952), p. 551.)
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. . . ,On (where these are significantly different) provides some reason to think that h is true. In point of fact, Whewell's version of the method seems even more complex than this. Whewell notes that formulations like those above presuppose that the hypothesis with which we compare our fact [is] framed all at once, each of its parts being included in the original scheme. In reality, however, it often happens that the various suppositions which our system contains are added upon occasion of different researches.7 In modifications toward a true theory, Whewell notes: all the additional suppositions tend to simplicity and harmony; the new suppositions resolve themselves into the old ones, or at least require only some easy modification of the hypothesis first assumed: the system becomes more coherent as it is further extended.8 Perhaps, then, Whewell would have espoused the following: "Dynamical" method of hypothesis with consilience: Let h1 be some hypothesis proposed initially to explain O1 but not O2, . . . ,On, where the latter are different in kind from each other and from O1; let h2, . . . ,hk be hypotheses added to h1 to explain O2, . . . ,On. The fact that h1, . . . ,hk if true would correctly explain O1, . . . ,On, and in addition would correctly explain On+1, . . . ,On+p (different facts that did not prompt h1t . . . ,hk), provides some reason to believe h1t . . . ,hk, provided that h2, . . . ,hk are "natural" extensions of h1, so that h1, . . . ,hk has "coherence." These three formulations of the method, although by no means identical, have in common the basic idea that the fact that an hypothesis if true would correctly explain phenomena counts as some reason for believing that hypothesis. There may be restrictions on the kinds of phenomena explained (e.g., they should be different in kind from ones that prompted the hypothesis in the first place). And there may be restrictions on the additions to the hypothesis required for subsequent explanations. But there is no requirement that the hypothesis in question, or any subsequent one, be inductively inferable from any observations. More generally, there is no requirement that there be any independent warrant for the hypotheses introduced, that is, any reason for 7. William Whewell, The Philosophy of the Inductive Sciences (New York, 1967), vol. 2, p. 68. Emphasis in original.
8. Ibid., p. 68. Emphasis in original.
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believing such hypotheses other than the explanatory ones hypothesists mention.
3. WHAT METHOD DID NINETEENTHCENTURY WAVE THEORISTS EMPLOY IN PRACTICE? In publications setting forth arguments for their theory, a strategy wave theorists typically used is this: 1. Start with the assumption that light is either a wave phenomenon or a stream of particles. According to the wave theory, light consists of a wave motion or pulse transmitted through some medium; the medium itself may be composed of particles that vibrate rather than exhibit translational motion. According to the particle theory, light consists of discrete particles emanating from luminous bodies; these particles are subject to forces obeying Newton's laws of motion; if no such forces are acting the particles move in straight lines with constant finite velocity. The assumption that light is either a wave or particle phenomenon is made on the grounds that these are the two main theories that have been proposed by the physics community, or on the grounds of some empirical consideration regarding motion, or both. Taking the former line Young writes: It is allowed on all sides, that light either consists in the emission of very minute particles from luminous substances, which are actually projected, and continue to move with the velocity commonly attributed to light, or in the excitation of an undulatory motion, analogous to that which constitutes sound, in a highly light and elastic medium pervading the universe; but the judgments of philosophers of all ages have been much divided with respect to the preference of one or the other of these opinions.9
Humphrey Lloyd, after a few preliminaries, begins the body of his 1834 report on the present state of physical optics as follows: The first property of light which claims our notice is its progressive movement. Light we know, travels from one point of space to another in time, with a velocity 9. Thomas Young, A Course of Lectures on Natural Philosophy and the Mechanical Arts (London, 1845), p. 359. Also taking the former line, Fresnel, like Young, begins by noting that the particle and wave theories represent "the two systems which have up till now divided scientists with respect to the nature of light." A. Fresnel, "Memoir on the Diffraction of Light" (1816), reprinted (in part) in Henry Crew, ed., The Wave Theory of Light (New York, 1900). For parts of this material not in Crew, I have used a translation provided by Laurence Selim.
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of about 195,000 miles a second. The inquiry concerning the mode of this propagation involves that respecting the nature of light itself. There are two distinct and intelligible ways of conceiving such a motion. Either it is the selfsame body which is found at different times in distant points of space; or there are a multitude of moving bodies, occupying the entire interval, each of which vibrates continually with certain limits, while the vibratory motion is communicated from one to another, and so advances uniformly. Nature affords numerous examples of each of these modes of propagated movement; and in adopting one or the other to account for the phenomena of light, we fall upon one or other of the two rival systems, — the theories of Newton [particle theory] and of Huygens [wave theory].10 Lloyd's assumption that light is either a wave phenomenon or a particle phenomenon is based on the observation that light travels in space from one point to another with a finite velocity, that both particle and wave theories can account for this movement, and that in nature one observes motion from one point to another occurring by the motion of a body and by the motion of vibrations through a set of bodies. Herschel, another defender of the wave theory, begins his account of physical optics as follows: Among the theories which philosophers have imagined to account for the phenomena of light, two principally have commanded attention; the one conceived by Newton ... in which light is conceived to consist of excessively minute molecules of matter projected from luminous bodies. . . . The other hypothesis is that of Huygens . . . , which supposes light to consist, like sound, in undulations or pulses, propagated through an elastic medium.11 Although Herschel recognizes that other theories have been proposed, he notes that "these are the only mechanical theories which have been advanced." He seems to suppose that these are the most plausible theories. 2. Show how each theory explains various observed optical phenomena (e.g., rectilinear propagation, reflection, refraction, diffraction, dispersion, etc.). Young, for example, begins with diffraction ("when a portion of light is admitted through an aperture and spreads itself in a slight degree in every direction"): In this case it is maintained by Newton that the margin of the aperture possesses an attractive force, which is capable of inflecting the rays. . . . In the Huygensian 10. Humphrey Lloyd, "Report on the Progress and Present State of Physical Optics," Reports of the British Association for the Advancement of Science (1834), pp. 297298. 11. J. F. W. Herschel, "Light," Encyclopedia Metropolitana (1845), vol. 4, p. 439.
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system of undulation, this divergence or diffraction is illustrated by a comparison with the motions of waves of water and of sound, both of which diverge when they are admitted into a wide space through an aperture. . . . 12 Young continues by noting other observed optical phenomena and indicating whether or how the two leading theories account for them. Exactly similar strategies are followed by Fresnel, Lloyd, and Herschel. 3. Show that the particle theory, in explaining one or more of the observed optical phenomena, introduces improbable hypotheses, while the wave theory does not. Immediately after introducing the particle explanation of diffraction at an aperture as caused by an attractive force exerted at the margin, Young writes: But there is some improbability in supposing that bodies of different forms and of various refractive powers should possess an equal force of inflection, as they appear to do in the production of these effects; and there is reason to conclude from experiments, that such a force, if it existed, must extend to a very considerable distance from the surfaces concerned, at least a quarter of an inch, and perhaps more, which is a condition not easily reconciled with other phenomena.13 Fresnel offers a similar argument against the forces introduced by the particle theory to explain diffraction. He presents experiments on the basis of which he concludes that The phenomena of diffraction do not at all depend upon the nature, the mass, or the shape of the body which intercepts the light, but only upon the size of the intercepting body or upon the size of the aperture through which it passes. We must, therefore, reject any hypothesis which assigns these phenomena to attractive and repulsive forces whose action extends to a distance from the body as great as that at which rays are inflected.14
12. Young, op. cit., pp. 359360. Young's wave explanation of diffraction, which is rather sketchy, is perhaps this: waves (e.g., those of sound and water) diverge when admitted into a wide space through an aperture. Light is a wave motion. That is why it diverges similarly. It remained for Fresnel to give a quantitative wavetheoretic explanation of this and other diffraction phenomena. In one such phenomenon light bends around obstacles into the shadow as well as away from it. Newton (to whom Young refers in the quotation above) observed the external fringes but not the internal ones. This was one of his reasons for rejecting the wave theory (see section 7). 13. Ibid. Young proceeds to show how in explaining numbers of other phenomena the particle theory, by contrast to the wave theory, employs dubious hypotheses. This strategy is also evident in one of his earliest papers ("Outlines of Experiments and Inquiries Respecting Sound and Light," Philosophical Transactions of the Royal Society, 1800), in which he also argues in favor of the wave theory, though with somewhat less assurance (see pp. 613616). 14. Fresnel, op. cit., p. 99.
LIGHT HYPOTHESES
77
Earlier in this prize essay on diffraction Fresnel presents an argument against Newton's introduction of "fits of easy reflection and transmission" of light particles as an auxiliary hypothesis to explain Newton's rings. He concludes: Not only is the hypothesis of fits improbable because of its complexity, and difficult to reconcile with the facts in its consequences, but it does not even suffice in explaining the phenomenon of the colored rings, for which it was imagined.15
Fresnel goes on to argue that Newton's rings, as well as diffraction, can be explained as natural consequences of the wave theory, without introducing improbable hypotheses. Lloyd notes that discoveries of Bradley and Roemer lead to the conclusion that the velocity of light is the same whatever be the luminous origin: the light of the sun, the fixed stars, the planets and their satellites, being all propagated with the same swiftness. This conclusion must be allowed to present a formidable difficulty in the theory of emission.16
The difficulty is that if light consists of particles (with mass), then a massive object such as a fixed star should exert a force on them that "would be sufficient to destroy the whole momentum of the emitted molecules, and the star would be invisible at great distances" (p. 300). Lloyd asserts that the only way to explain the fact that the velocity we observe is the same for all luminous bodies is to adopt an hypothesis of Arago "that the molecules of light are originally projected with very different velocities: but that among these velocities is but one which is adapted to our organs of vision, and which produces the sensation of light" (pp. 300301). Lloyd notes that such a supposition has some support from discoveries of invisible rays of the spectrum. But he concludes that the supposition is not "easily reconciled with hypotheses which we are able to frame respecting the nature of vision" (p. 301; for additional discussion of this point, see my "Light Problems: Reply to Chen," Studies in History and Philosophy of Science 21 (1990)). By contrast, Lloyd asserts, the fact that the observed velocity of light is the same for all luminous bodies is explained without difficulty by the wave theory: This uniformity of velocity, on the other hand, is a necessary consequence of the principles of the wavetheory. The velocity with which vibratory movement is propagated in an elastic medium depends solely on the elasticity of that medium and on its density; and if these be uniform in the vast spaces which intervene between the material bodies of the universe, (and it is not easy to suppose it otherwise) the velocity must be the same, whatever be the originating source. (p. 301)
15. Ibid., sec. 7. 16. Lloyd, op. cit., p. 300.
78
THEORIES OF LIGHT: PARTICLES VERSUS WAVES
4. Conclude that the wave theory is (very probably) true, while the particle theory is (very probably) false. At the beginning of his paper Lloyd makes explicit the strategy he will follow: To take, in the first instance, a rapid survey of the several leading classes of optical phenomena which the labours of experimental philosophers have wrought out in such rich profusion, and afterward to examine how far they are reducible to one or other of the two rival theories which have alone advanced any claim to our consideration. This is, in fact, the only way in which the truth of a physical theory can be established; and the argument in its favor is essentially cumulative.17 Having seen how the two theories explain (or fail to explain) various optical phenomena, and having argued that the particle theorists introduce hypotheses in these explanations that are improbable, whereas wave theorists do not—or do not to such a great extent—it is concluded that the wave theory is probably true.
4. ANALYSIS OF THIS METHOD The method described in the previous section is a type of "eliminative" one that consists of four parts. Schematically: 1. Assume that either theory T1 or theory T2 is correct, and give grounds for such an assumption. 2. Show how T1 and T2 explain various observed phenomena. 3. Show that T2 in explaining one or more of these phenomena introduces improbable hypotheses, whereas T1, does not. 4. Conclude that T1 is probably true. Let us examine the three steps leading to the conclusion, beginning with the second. Step 2 introduces the idea of explanation. To conclude that T1 is probably true, a theorist using this strategy must show that T1 is capable of explaining certain phenomena (by producing the explanations). Following the method of hypothesis, we might say that such a theorist seeks a theory T that will satisfy a Basic explanatory condition: T if true would correctly explain observed phenomena O1t . . . ,On. 17. Ibid., p. 295.
LIGHT HYPOTHESES
79
In the light of the discussion in section 2 this explanatory condition, which is associated with the "basic" method of hypothesis, could in principle be broadened to include more sophisticated features of the "dynamical method of hypothesis with consilience." For example, it might be required that theory T explain phenomena different in kind from those that first prompted the theory, and that hypotheses added to T for this purpose be "natural" ones that result in a "coherence" among the theoretical assumptions.18 Despite the fact that wave theorists satisfied one or the other explanatory conditions of the method of hypothesis, there is a fundamental difference between the method they employed and the method(s) of hypothesis in section 2. In the latter case, when the explanatory condition is satisfied, it is concluded that there is some reason to think the theory is true. But with the method described in the previous section this is not a sufficient condition, only a necessary one. For each of the rival theories may satisfy the explanatory condition. The particle theory (no less than the wave theory), together with auxiliary assumptions needed, would, if true, correctly explain certain observed phenomena. Indeed, it might be the case that these auxiliary assumptions are "natural" extensions of the particle theory resulting in a "coherent" set of assumptions, and that if the enlarged theory is true it will correctly explain phenomena different from those prompting the theory in the first place. If so, more sophisticated "explanatory conditions" will be satisfied. For example, in explaining diffraction a particle theorist introduces the auxiliary hypothesis that the margins of the aperture exert an attractive force on the particles of light that is capable of causing the bending of their path. This is a "natural" extension of the particle theory, since the latter seeks to explain optical phenomena in terms of particles subject to forces. The objection to this explanation offered by nineteenthcentury wave theorists is not that it introduces ideas foreign to the particle theory, ones that render the set of theoretical assumptions "incoherent," but that it introduces hypotheses that are improbable, given observations made in this case and others. (Fresnel's objection to the Newtonian attractive force at the margins of the aperture in diffraction is based on experiments and observations that show that diffraction does "not at all depend upon the nature, the mass, or the shape of the body which intercepts the light.") If the satisfaction of an explanatory condition is not sufficient, what else is necessary? Here we must examine steps 1 and 3. In step 1 wave theorists begin with the assumption that light is either a particle or a wave phenomenon, that is, that one or the other of these two theories is true. This is not simply assumed for the sake of argument to see 18. These additional features of an explanation are emphasized by Lloyd, op. cit., pp. 349350.
80
THEORIES OF LIGHT: PARTICLES VERSUS WAVES
what follows. Rather, arguing in the manner of the previous section, wave theorists are committed to the idea that it is likely that one theory or the other is true, and reasons are given for thinking that this is so. What sorts of reasons are these? Lloyd, we recall, argues "there are two distinct and intelligible ways of conceiving" the observed motion of light, and that "in adopting one or the other [modes of propagated motion] to account for the phenomena of light" we obtain the particle or the wave theory. Accordingly, a part of Lloyd's reasoning is explanatory: both theories will explain the fact that light exhibits motion. But to this claim Lloyd adds importantly that "nature affords numerous examples of each of these modes of propagated movement." Had others been observed, presumably further ways of conceiving the propagation of light would have been noted. Accordingly, if, as seems plausible, Lloyd means that these are the only modes of communication known (or perhaps even just the most prevalent ones), then, in addition to the explanatory reason, there is an inductive one: light is observed to be communicated from one point to another in a finite time; the modes of communication observed in nature consist in the motion of bodies from one point to another and in the vibratory motion of a medium; so it is reasonable to assume that light is either a particle or a wave phenomenon. Sometimes (as with Young) the claim that light is either a particle or a wave phenomenon is defended by saying that these are the two leading theories proposed by physicists. If, as I am assuming, this is to be understood as providing some reason not simply to examine these theories but to think that one or the other is correct, then, even in this case, the reasoning involves some inductive steps: The fact that these are the leading theories proposed by physicists, together perhaps with an implicit appeal to the reputations of those supporting each theory (e.g., Newton vs. Huygens) and to their other successes, provides some reason to think that one of these theories is true. (Certainly a particle theorist such as Brougham had no qualms about defending his theory, at least in part, by appeal to the success of his authority Newton. And, as noted in Essay 1, Young in an 1802 paper also appealed to the authority of Newton in defense of some of the assumptions of the wave theory.) Since the claim that light is either a particle or a wave phenomenon is not simply assumed, but grounds are given for it, and since it is recognized that the resulting accounts are not the only possible ones,19 just the most likely, it
19. An alternative theory, which had some defenders until the early years of the nineteenth century, was that light is produced by the rectilinear motion, rather than the vibrations, of a fluid. According to most of these theorists this fluid consists of particles. However, unlike the particle theorists, fluid theorists refused to offer mechanical explanations of the interactions of light and matter, i.e., explanations in terms of attractive and repulsive forces obeying Newtonian
Chapter 4
Twodimensional Electromagnetic Waves In Chapter 3 we discussed the properties of onedimensional electromagnetic waves. These waves depend only on one spatial coordinate, viz. the x3 coordinate. We now consider the more general case of wave motion that depends on the two spatial coordinates x1 and x3 , but that has no variation in the x2 direction. In this case, we assume that the permittivity, the conductivity and the permeability are independent of the x2 coordinate. Hence, ε = ε(x1 , x3 ), σ = σ(x1 , x3 ), µ = µ(x1 , x3 ) .
(4.1)
We further assume that the sources of the electromagnetic wave ﬁeld are also independent of x2 , hence ˆ = K(x ˆ 1 , x3 , s) . ˆ =J ˆ (x1 , x3 , s), K J
(4.2)
Then, the generated electromagnetic ﬁeld will be independent of x2 , viz., ˆ = H(x ˆ 1 , x3 , s) , ˆ = E(x ˆ 1 , x3 , s), H E
(4.3)
and, hence, ∂2 = 0. With this, Eqs. (2.54)  (2.59) separate into ˆ1 = −Jˆext , ˆ 2 + (σ+sε)E ∂3 H 1 ˆ ˆ ˆ −∂1 H2 + (σ+sε)E3 = −J3ext , ˆ 2 = −K ˆ ext , ˆ1 − ∂ 1 E ˆ3 + sµH ∂3 E 2
(4.4) (4.5) (4.6)
82
twodimensional electromagnetic waves
and
ˆ 1 = −K ˆ ext , ˆ2 + sµH −∂3 E 1 ˆ ˆ ˆ ∂1 E2 + sµH3 = −K3ext . ˆ2 = −Jˆext . ˆ 1 − ∂1 H ˆ 3 ) + (σ+sε)E −(∂3 H 2
(4.7) (4.8) (4.9)
ˆ3 and H ˆ2 ˆ1 , E In the ﬁrst set of equations (4.4)  (4.6) the ﬁeld components E occur, while in the second set of equations (4.7)  (4.9) the ﬁeld components ˆ 1, H ˆ 3 and E ˆ2 occur. H Parallel polarization ˆ ext and Jˆext are equal to zero in all space, the only nonˆ ext , K When K 1 3 2 ˆ3 and H ˆ 2 . Since the electric ﬁeld vector ˆ1 , E zero ﬁeld components are E is parallel to the (x1 , x3 )plane of observation, this electromagnetic ﬁeld is called to be parallelly polarized. Since the magnetic ﬁeld vector is parallel to the x2 direction, this electromagnetic ﬁeld is also denoted as the case of Hpolarization. The system of equations (4.4)  (4.6) are the governing electromagnetic ﬁeld equations. In a sourcefree domain, we observe that ˆ1 , E ˆ3 follow from the magnetic ﬁeld component electric ﬁeld components E ˆ H2 as ˆ2 , ˆ1 = −(σ+sε)−1 ∂3 H E ˆ3 = (σ+sε)−1 ∂1 H ˆ2 . E
(4.10) (4.11)
In those parts of the medium where σ, ε and µ are constants, substituting these expressions for the electric ﬁeld components into the sourcefree counterpart of Eq. (4.6) results in the twodimensional modiﬁed Helmholtz ˆ 2 in a homogeneous domain, equation for the magnetic ﬁeld component H viz.,
ˆ2 = 0 . ˆ 2 + ∂3 ∂3 H ˆ 2 − (σ+sε)sµH ∂1 ∂1 H
(4.12)
plane waves in a homogeneous medium
83
Perpendicular polarization ˆ ext are equal to zero in all space, the only nonWhen Jˆ1ext , Jˆ3ext and K 2 ˆ 3 and E ˆ2 . Since the electric ﬁeld vector is ˆ 1, H zero ﬁeld components are H perpendicular to the (x1 , x3 )plane of observation, this electromagnetic ﬁeld is called to be perpendicularly polarized. Since the electric ﬁeld vector is parallel to the x2 direction, this electromagnetic ﬁeld is also denoted as the case of Epolarization. The system of equations (4.7)  (4.9) are the governing electromagnetic ﬁeld equations. In a sourcefree domain, we observe that ˆ 3 follow from the electric ﬁeld component ˆ 1, H magnetic ﬁeld components H ˆ2 as E ˆ 1 = (sµ)−1 ∂3 E ˆ2 , H ˆ2 . ˆ 3 = −(sµ)−1 ∂1 E H
(4.13) (4.14)
In those parts of the medium where σ, ε and µ are constants, substituting these expressions for the magnetic ﬁeld components into the sourcefree counterpart of Eq. (4.9) results in the twodimensional modiﬁed Helmholtz ˆ2 in a homogeneous domain, viz., equation for the electric ﬁeld component E
ˆ2 = 0 . ˆ 2 + ∂3 ∂3 E ˆ2 − (σ+sε)sµE ∂1 ∂1 E
4.1.
(4.15)
Plane waves in a homogeneous medium
In this section we discuss the important properties of plane electromagnetic waves. These waves exist in a homogeneous medium. The properties of such a medium are characterized by the spatially independent constants σ, ε and µ. Without loss of generality, we may assume that the propagation direction of a plane wave is in the (x1 , x3 )plane, and as such it is a particular example of twodimensional wave motion. It can be veriﬁed that, in a sourcefree domain, Eqs. (4.10)  (4.15) admit solutions of such a form that
84
twodimensional electromagnetic waves
the electromagnetic ﬁeld components depend on the spatial coordinates x1 and x3 through the factor exp(−γ1 x1 − γ3 x3 ), in which γ1 and γ3 are complex constants. We call an electromagnetic wave of this type a plane wave. In case γ1 = 0, the wave only depends on the x3 coordinate and the theory of Chapter 3 of onedimensional waves applies. To study the properties of plane waves, we write ˆ = e ˆ(s) exp(−γ1 x1 − γ3 x3 ) , E ˆ ˆ = h(s) H exp(−γ1 x1 − γ3 x3 ) ,
(4.16) (4.17)
ˆ = h(s) ˆ ˆ=e ˆ(s) and h in which e are spatially independent vectors. Substiˆ2 component ˆ tuting the H2 component of Eq. (4.17) into Eq. (4.12) and the E of Eq. (4.16) into Eq. (4.15), we observe that nonzero solutions exist if the complex propagation vector γ = γ1 i1 + γ3 i3 satisﬁes γ · γ = γ12 + γ32 = (σ + sε)sµ .
(4.18)
For the onedimensional waves with a propagation vector in the i3 direction only, Eq. (4.18) is equivalent to Eq. (3.8). Steadystate propagation properties For a single frequency component of angular frequency ω = 2πf , it is customary to separate the complex propagation vector into a real vector and and an imaginary vector according to
γ(jω) = α(ω) + jβ(ω) , where α β
= =
α1 i1 + α3 i3 β1 i1 + β3 i3
= =
attenuation vector (Np/m), phase vector (rad/m).
(4.19)
plane waves in a homogeneous medium
.. ........................................................................................... ....... .. ....... .... ... . ..........................
β
α
85
. . . . . . . . . . ... ... ... ... ... .... .... .... .... .... ... ... .. ... ... ... . . . . .. ... ... .. ... ... ... ... ... ... . . . . ... . . . . . ... ... .... .... .... .... .... .... .... .... ... .... .... .... .... ... .. .. .. .. ..... ..... ..... ..... ..... . . . . . . . . . . . . . . . . . ... . . . . . ... .... ... .... ... .... ... .... ... .... . .. . .. . .. .. ... .. ... ... .... ... .... ... ..... ... ... ... ... .. ... ... ... ... .. .. .. .. .. ... ... ... ... ... ... ... ... ... ...
nonuniform plane wave in a lossy medium
.. ........................................................................................ .. ... ..... ... ... ... .. ......... ...... .
β
α
... ... ... ... ... ....... .......... ....... ....... .......... ....... ....... .......... ....... ....... .......... ....... ....... ........... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... . . . ... . ... . . . . . . . . . . . . . ....... ........ ....... ....... ........ ....... ....... ........ ....... ....... ........ ....... ....... ........... ... .. .. ... .. .. .. ... ... ... ... ... ... ... ... ... . . . ... . ... . . ... . .. ... ... . . . ....... ........ ....... ....... ........ ....... ....... ........ ....... ....... ......... ....... ....... ........... ... .... .... .... .... ....
nonuniform plane wave in a lossless medium
.. .. .................................................................................................... .. ..
α
β
.... ... .. ... ... ... ... ... ... ... ... ... ... . ...
.. .. ... . ... .. ... .. .. .. ... ..
.... ... .. ... ... ... ... ... ... ... ... ... ... . ...
... . ..... .. .. ... .. .. .. ... ..
.... ... .. ... ... ... ... ... ... ... ... ... ... . ...
.. .. ... . ... .. ... .. .. .. ... ..
.... ... .. ... ... ... ... ... ... ... ... ... ... . ...
.. .. ... . ... .. .... .. .. ... ..
.... ... .. ... ... ... ... ... ... ... ... ... ... . ...
uniform plane wave in a lossy medium
... ..................................................................................... ..
β
... ... .. ... ... ... .... .. ... ... ... ... ... .. ..
... ... .. ... ... ... .... .. ... ... ... ... ... .. ..
... ... .. ... ... ... .... .. ... ... ... ... ... .. ..
... ... .. ... ... ... .... .. ... ... ... ... ... .. ..
... ... .. ... ... ... .... .. ... ... ... ... ... .. ..
uniform plane wave in a lossless medium
Figure 4.1. Planes of equal phase (−−−−−) and planes of equal amplitude (− − −).
86
twodimensional electromagnetic waves
Substitution of Eq. (4.19) in the propagation factor exp(−γ1 x1 − γ3 x3 ) leads to exp(−γ1 x1 − γ3 x3 ) = exp(−α1 x1 − α3 x3 ) exp(−jβ1 x1 − jβ3 x3 ) .
(4.20)
Since  exp(−γ1 x1 − γ3 x3 ) = exp(−α1 x1 − α3 x3 ) and arg[exp(−γ1 x1 − γ3 x3 )] = −β1 x1 −β3 x3 , the ﬁrst factor on the righthand side determines the amplitude of the propagation factor and is denoted as the attenuation factor, while the second factor on the righthand side determines the phase of the propagation factor and is denoted as the phase factor. The planes α1 x1 + α3 x3 = constant are the surfaces of equal amplitude, while the planes β1 x1 + β3 x3 = constant are the planes of equal phase. Substitution of Eq. (4.19) into Eq. (4.18) leads to α · α + 2jα · β − β · β = (σ + jωε)jωµ ,
(4.21)
α · α − β · β = −ω 2 εµ ,
(4.22)
or
2α · β = ωσµ .
(4.23)
Since ωσµ ≥ 0 we observe that the angle between the vectors α and β is less than or equal to 12 π. In a lossless medium we have σ = 0 and then either the case of α = 0 and β · β = ω 2 εµ occurs, or the case of α = 0 and β = 0 with α · β = 0 occurs. In the latter case, the planes of equal amplitudes are perpendicular to the planes of equal phase. A plane wave is called uniform when the vectors α and β have the same direction. A plane wave is called nonuniform when the vectors α and β have diﬀerent directions (see Fig. 4.1) The time average Poynting’s vector of a plane wave in the frequency domain is given by
ˆ ×H ˆ ∗ = S 0 exp[−(γ1 +γ ∗ )x1 − (γ3 +γ ∗ )x3 ] , ST = 12 Re E 1 3 in which
(4.24)
ˆ∗ . ˆ×h S 0 = 12 Re e
(4.25)
87
plane waves in a homogeneous medium
4.1.1.
Uniform plane waves
When the plane wave is uniform, the vectors α and β have the same direction. Let s be the unit vector in the direction of α and β. Then, we may write α = αs, β = βs, γ = γs , (4.26) in which (cf. Eq. (4.19)) γ = α + jβ
(4.27)
is the propagation coeﬃcient, α is the attenuation coeﬃcient and β is the phase coeﬃcient. Substitution of γ = γs into Eq. (4.18) yields, with s·s = 1, γ 2 = (σ + jωε)jωµ ,
(4.28)
which is identical to Eq. (3.8). The uniform plane wave is a onedimensional wave in the direction of s; the vector s = {s1 , 0, s3 } can have an arbitrary orientation with respect to the chosen coordinate system (in Chapter 2 we have either s = i3 or s = −i3 ). The phase factor exp(−jβ1 x1 − jβ3 x3 ) is periodic in the direction of s = s1 i1 + s3 i3 with period λ=
2π . β
(4.29)
This spatial period λ is denoted as the wavelength of the plane wave. Parallelly polarized plane wave ˆ3 and H ˆ 2 are the nonzero ˆ1 , E In the case of parallel polarization, E electromagnetic ﬁeld components. Substituting Eqs. (4.16)  (4.17) into Eqs. (4.10)  (4.11), it follows that the spatially independent components ˆ 2 and γ satisfy eˆ1 , eˆ3 , h γ3 ˆ h2 (4.30) eˆ1 = σ + sε and −γ1 ˆ h2 . (4.31) eˆ3 = σ + sε A particular linear combination of Eqs. (4.30)  (4.31) yields ˆ =0. γ1 eˆ1 + γ3 eˆ3 = γ · e
(4.32)
88
twodimensional electromagnetic waves
ˆ.... e ..
............ ....... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... .... .. ........... . . . . . ..... . . . . . .... ....... ....... ...... ..... . .... .. . . . . ..... ....... ....... .......
.... .............. ....... ..... ..... . . . .... ..... ..... .....
s
ˆ h
i.1
O
..... ........ .... ... .... .. ...... ..... . ... .... .. ... ... ... .. .... . . . . . ... ....... ... ....... .................................................................
r
r f
x
i3
Figure 4.2. The ﬁeld components of a parallelly polarized, uniform plane wave propagating in the (x1 , x3 )plane.
When the plane wave is uniform, we may use γ = γs = γs1 i1 + γs3 i3 . Then, Eqs. (4.30)  (4.32) can be written as ˆ 2 s3 , Zh ˆ 2 s1 , = −Z h
eˆ1 =
(4.33)
eˆ3
(4.34)
and ˆ =0, s1 eˆ1 + s3 eˆ3 = s · e
(4.35)
where we have used the expression for the wave impedance γ Z(s) = = σ + sε
sµ σ + sε
1 2
.
(4.36)
For a parallelly polarized, uniform, plane wave, we observe that the electric ˆ = {0, h ˆ 2 , 0} and ˆ = {ˆ ﬁeld strength e e1 , 0, eˆ3 }, the magnetic ﬁeld strength h the direction of propagation s = {s1 , 0, s3 } form a mutually perpendicular, righthanded triad (see Fig. 4.2). Further, in the frequency domain, the Poynting vector S 0 deﬁned in Eq. (4.25) becomes S0 =
1 2 Re
ˆ ∗ = 1 Re −ˆ ˆ ∗ i1 + eˆ1 h ˆ ∗ i3 ˆ×h e3 h e 2 2 2
89
plane waves in a homogeneous medium
=
1 ˆ ˆ∗ 2 Re [Z(jω)] h2 h2 s
.
(4.37)
Hence, the power ﬂow of the parallelly polarized, uniform, plane wave is in the direction of propagation. Perpendicularly polarized plane wave ˆ 1 and H ˆ 3 are the nonˆ2 , H In the case of perpendicular polarization, E zero electromagnetic ﬁeld components. Substituting Eqs. (4.16)  (4.17) into Eqs. (4.13)  (4.14), it follows that the spatially independent components eˆ2 , ˆ 3 and γ satisfy ˆ 1, h h ˆ 1 = −γ3 eˆ2 h (4.38) sµ and
ˆ 3 = γ1 eˆ2 . h sµ
(4.39)
A particular linear combination of Eqs. (4.38)  (4.39) yields ˆ =0. ˆ 1 + γ3 h ˆ3 = γ · h γ1 h
(4.40)
When the plane wave is uniform, we may use γ = γs = γs1 i1 + γs3 i3 . Then, Eqs. (4.38)  (4.40) can be written as ˆ 1 = −Y eˆ2 s3 , h ˆ 3 = Y eˆ2 s1 , h
(4.41)
ˆ =0, ˆ 1 + s3 h ˆ3 = s · h s1 h
(4.43)
(4.42)
and where we have used the expression for the wave admittance γ = Y (s) = sµ
σ + sε sµ
1 2
.
(4.44)
For a perpendicularly polarized, uniform, plane wave, we see that the electric ˆ = {h ˆ 1 , 0, h ˆ 3 } and ˆ = {0, eˆ2 , 0}, the magnetic ﬁeld strength h ﬁeld strength e the direction of propagation s = {s1 , 0, s3 } form a mutually perpendicular, righthanded triad (see Fig. 4.3).
90
twodimensional electromagnetic waves
... ................... .. ..... ..... . . . . . ..... ..... .... .....
i.1
O
...... ..... . .... .... .. ....... ... ... .. .... . . . . . .... ..... .... .. ... .... .. .. ... ... .. .... .... .. .... . . . . .. . ....... ... ...... ....... .... ....... .. ...... .................................................................
r
s
x rf ˆ .....e .....
....... ....... .... ....... ......
..... ..... ..... ..... ..... ..... ..... ..... ..... ..... . ................ ..
ˆ h
i3
Figure 4.3. The ﬁeld components of a perpendicularly polarized, uniform plane wave propagating in the (x1 , x3 )plane.
Further, in the frequency domain, the Poynting vector S 0 deﬁned in Eq. (4.25) becomes S0 = =
1 ˆ ∗ = 1 Re e ˆ×h ˆ∗ 2 Re e 2 1 ˆ2 eˆ∗2 s . 2 Re [Y (jω)] e
ˆ = 1 Re eˆ∗ h ˆ ˆ 1 i3 ×h ˆ∗2 h 2 3 i1 − e 2
(4.45)
Hence, the power ﬂow of the perpendicularly polarized, uniform, plane wave is in the direction of propagation.
4.2.
Interference of two plane waves
When in a linear medium two or more waves are present simultaneously, the total value of a ﬁeld strength in that domain is equal to the contributions of the constituent waveﬁelds (the principle of superposition). Even when the constituent waveﬁelds have a simple structure, the resulting waveﬁeld may depend on space and time in a complicated way. This phenomenon is called interference. In this section we discuss the interference of two plane waves in a homogeneous medium with constants σ, ε and µ. The ﬁrst wave is denoted as wave (1) with ﬁeld strengths
91
interference of two plane waves
(1) (1) ˆ (1) = e ˆ(1) exp(−γ1 x1 − γ3 x3 ) , E
ˆ H
(1)
(1)
ˆ = h
(1)
(1)
exp(−γ1 x1 − γ3 x3 ) .
(4.46) (4.47)
The second wave is denoted as wave (2) with ﬁeld strengths (2) (2) ˆ (2) = e ˆ(2) exp(−γ1 x1 − γ3 x3 ) , E
(4.48)
ˆ (2) exp(−γ (2) x1 − γ (2) x3 ) . ˆ (2) = h H 1 3
(4.49)
Since both waves are present in the same medium, we have (1)
(1)
(2)
(2)
[γ1 ]2 + [γ3 ]2 = [γ1 ]2 + [γ3 ]2 = (σ + sε)sµ .
(4.50)
In the domain where interference occurs, the total ﬁeld strengths are given by ˆ = E ˆ (1) + E ˆ (2) , E ˆ = H ˆ H
(1)
ˆ +H
(2)
(4.51) .
(4.52)
We now assume that both waves are uniform plane waves. Then (1)
(1)
γ3 = γs3 ,
(2)
(2)
γ3 = γs3 ,
γ1 = γs1 ,
(1)
(1)
(4.53)
(2)
(2)
(4.54)
and γ1 = γs1 , where 1
γ = [(σ + sε)sµ] 2 .
(4.55)
92
twodimensional electromagnetic waves
i.1
s(2)
... .......... ... ... ... ... ... ..... .... ............... ... ................... ....... . ... ..... ..... . . . . ... ..... . . ..... ..... .... ..... ..... ... .. ...... ... ....... .. ....... .... ..... .................................................................................................. ....
s(1)
θ
r
O
θ
i3
Figure 4.4. The propagation directions of the two uniform plane waves.
Without loss of generality we always can choose such a reference frame that the two waves both propagate in the positive x1 direction, while they have opposite propagation directions in the x3 direction (see Fig. 4.4). Let θ be the angle between s(1) and i3 and π − θ be the angle between s(2) and i3 . Then, (1) (1) (4.56) s(1) = s1 i1 + s3 i3 = sin(θ)i1 + cos(θ)i3 , and (2)
(2)
s(2) = s1 i1 + s3 i3 = sin(θ)i1 − cos(θ)i3 .
(4.57)
Substituting Eqs. (4.46)  (4.49) in Eqs. (4.51)  (4.52), we obtain
ˆ = exp[−γ sin(θ)x1 ] e ˆ(2) exp[γ cos(θ)x3 ] , (4.58) ˆ(1) exp[−γ cos(θ)x3 ] + e E ˆ (2) exp[γ cos(θ)x3 ] . (4.59) ˆ (1) exp[−γ cos(θ)x3 ] + h ˆ = exp[−γ sin(θ)x1 ] h H In order to discuss the total electromagnetic power ﬂow, we consider the lossless case and restrict ourselves to the steadystate analysis.
4.2.1.
Steadystate analysis: lossless case
In the lossless case with s = jω, we have 1 ω γ = j , with c = (εµ)− 2 . c The time average Poynting vector is given by
(4.60)
ˆ ×H ˆ∗ , ST = 12 Re E
(4.61)
93
interference of two plane waves
in which we have to substitute the expressions for the electric and magnetic ﬁeld strengths of Eqs. (4.58) and (4.59). We arrive at
ST
=
1 2 Re
ˆ e
(1)
+ 12 Re
∗
ˆ (1) + e ˆ (2) ˆ(2) × h ×h
∗
∗
ˆ (2) exp[−2jω cos(θ)x3 /c] ˆ(1) × h e ˆ +e
(2)
∗
ˆ (1) exp[2jω cos(θ)x3 /c] . ×h
(4.62)
The ﬁrst two terms of the righthand side of Eq. (4.62) represent the power ﬂows of the individual plane waves. In view of the obvious presence of the third and fourth term we conclude that the superposition principle only applies to the electromagnetic ﬁeld strengths, but not to the power ﬂow. In order to quantify the electromagnetic power ﬂow of two interfering waves we consider the cases of parallel and perpendicular polarization separately. Parallelly polarized plane waves Substituting Eq. (4.33) and (4.34) into Eq. (4.62), while observing that in the lossless case, 1 µ 2 (4.63) Z= ε is real valued, we obtain ST
=
1 2Z
∗
∗
ˆ (2) h ˆ (1) h ˆ (1) s(1) + h ˆ (2) s(2) h 2 2 2 2
∗
ˆ (1) h ˆ (2) sin(θ) exp[−2jω cos(θ)x3 /c] i1 . +Z Re h 2 2
(4.64)
Perpendicularly polarized plane waves Substituting Eq. (4.41) and (4.42) into Eq. (4.62), while observing that in the lossless case, 1 ε 2 (4.65) Y = µ is real valued, we obtain
94
twodimensional electromagnetic waves
ST
=
1 2Y
(1) (1)∗ (1)
eˆ2 eˆ2
s
(1) (2)∗
+Y Re eˆ2 eˆ2
(2) (2)∗ (2)
+ eˆ2 eˆ2
s
sin(θ) exp[−2jω cos(θ)x3 /c] i1 .
(4.66)
From the results of these cases of polarization, it follows that the power ﬂow in the i3 direction is position invariant, but the power ﬂow in the i1 direction is periodic in x3 with period λinterference =
λ , 2 cos(θ)
(4.67)
where λ is the wavelength of the uniform plane waves, see Eq. (4.29).
S1 T .... ........ .. ... ... ... ... ..
.... .... .... .... .... .... .... .... .. ..... ..... ... ... .... ... ... ... .... ... .... ... .... ... .... ... .... ... .. ... ... .. .. .. .. .. ... .. .. .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. .. .. .. ... .. .. .. .. .. .. .. ... .. ... .. ... ... ... ... .... ... ..... ... .... .. .... .. .... .. .... .. .... .. .... .. .... . . . . . . . . . . . . . . . . . . . . . . .. .. .. .. .. .. .. . . .. . . . .. . .. . . ... ... ... ... .. ... .. ... .. ... .. .. .. ... .. ... .. .. ... ... ... ... ... ... ... .. ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. ... ... ... ... ... .. .. .. .. ... .. ... .. .. .. ... .. ... ... ... ... .. .. .. ... .. ... ... .. .. .. .. .. . . ... . . . . . . . . . . . . . . . . . . . . . . . . . . ... ... ... ... ... ... ... ... ... ... .. .. .. .. ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. .. ... .. .. ... ... ... ... ... ... ... ... ... ... ... ... ... .... ... .... ... ... ... .... ... .... ... .... ... .... ... .... ... .... ... .. ... .... ... .. ... ... ... ... . . . . . . . . . . . . . ... .. . ... ... . . . . . . . . . . . ... .. ... .. ... .. ... .. ... . ... .. ... .. ... .. .. ... .. .. ... .. ... .. ... .. ... .. ... ... .. .. .. .. ... ... ... .. ... .. ....... ....... ...... ....... ...... ...... ...... .... .... ....
......................
.....................
............................................
x3
λ 2 cos(θ)
Figure 4.5. The power ﬂow in the i1 direction as a function of x3 in case of interference of two uniform plane waves propagating in directions given in Fig. 4.4.
reflection by an electrically impenetrable halfspace
4.3.
95
Reﬂection of a plane wave by an electrically impenetrable halfspace
When a wave is incident upon the boundary surface of an electrically impenetrable object, the wave is reﬂected by this boundary surface. In the case that the incident wave is a plane wave and the reﬂecting boundary surface is a plane boundary of an electrically impenetrable, semiinﬁnite medium, the reﬂected wave is a plane wave as well. Then, the pertaining reﬂection problem can be solved with the theory of plane waves. We take the Cartesian coordinate system in such a way that the plane x3 = 0 coincides with the boundary plane of the electrically impenetrable medium. The halfspace −∞ < x3 < 0 consists of a homogeneous medium with constitutive constants σ, ε and µ (see Fig. 4.6). The incident (plane) wave is given by (cf. Eqs. (4.16)  (4.17)) ˆi = e ˆi exp(−γ1i x1 − γ3i x3 ) , E ˆ H
i
i
ˆ exp(−γ i x1 − γ i x3 ) , = h 1 3
(4.68) (4.69)
while the reﬂected wave is given by ˆr = e ˆr exp(−γ1r x1 − γ3r x3 ) , E ˆ r exp(−γ r x1 − γ r x3 ) , ˆr = h H 1 3
(4.71)
(γ1i )2 + (γ3i )2 = (γ1r )2 + (γ3r )2 = (σ + sε)sµ .
(4.72)
(4.70)
in which
From this equation, it is observed that for given values of γ1i two solutions of γ3i exist. In view of causality with respect to the incident wave we take
γ3i = (σ + sε)sµ − (γ1i )2
1 2
,
(4.73)
where Re[γ3i ] ≥ 0. Similarly, in view of causality with respect to the reﬂected wave we take
96
twodimensional electromagnetic waves
σ, ε, µ
electrically impenetrable medium ................. ............ .... ........ ....... ........ ....... ....... ........ .
reﬂected wave
............................
i3
......................... .. . ....... ... ........ ....... . . . . . . ..... . . . . . . ......
incident wave
x3 < 0
x3 = 0
x3 > 0
Figure 4.6. Reﬂection of a plane wave by a plane reﬂector.
γ3r = − (σ + sε)sµ − (γ1r )2
1 2
,
(4.74)
where Re[γ3r ] ≤ 0. The total ﬁeld in the halfspace −∞ < x3 < 0 is obtained as ˆ r, H ˆ =H ˆi+H ˆr. ˆ =E ˆi +E E
(4.75)
At x3 = 0 the electromagnetic ﬁeld has to satisfy the boundary condition that the tangential component of the electric ﬁeld strength vanishes, i.e, ˆi +E ˆ r) = 0 , lim i3 × (E
(4.76)
ˆi + E ˆr ) = 0 , lim (E 1 1
(4.77)
ˆi + E ˆr ) = 0 . lim (E 2 2
(4.78)
x3 ↑0
or x3 ↑0 x3 ↑0
reflection by an electrically impenetrable halfspace
97
We now discuss the cases of parallel and perpendicular polarization separately. Parallel polarization ˆ3 and H ˆ 2 are the nonzero elecˆ1 , E In the case of parallel polarization, E tromagnetic ﬁeld components. We only have to meet the boundary condition ˆ1 . Using Eq. (4.10) and (4.75), the boundary condition at of Eq. (4.77) for E x3 = 0 becomes ˆ 2i + H ˆ 2r ) = 0 . (4.79) lim ∂3 (H x3 ↑0
Substituting the planewave representations of Eqs. (4.69) and (4.71) into Eq. (4.79), we obtain r ˆ i exp(−γ i x1 ) + γ r h ˆr γ3i h 2 1 3 2 exp(−γ1 x1 ) = 0 .
(4.80)
This equation can only be satisﬁed for all x1 , if γ1r = γ1i ,
(4.81)
while from Eqs. (4.73)  (4.74) it follows that
γ3r = −γ3i .
(4.82)
But, in that case, the boundary condition of Eq. (4.80) also requires
ˆi . ˆr = h h 2 2
(4.83)
The two other nonzero vectors follow with the help of Eqs. (4.10)  (4.11) as ei1 , eˆr1 = −ˆ
(4.84)
eˆr3
(4.85)
=
eˆi3
.
98
twodimensional electromagnetic waves
At this moment, the parallelly polarized reﬂected wave is determined completely. Perpendicular polarization ˆ 1 and H ˆ 3 are the nonˆ2 , H In the case of perpendicular polarization, E zero electromagnetic ﬁeld components. We only have to meet the boundary ˆ2 . Substituting the planewave representations condition of Eq. (4.78) for E of Eqs. (4.68) and (4.70) into Eq. (4.78), we obtain eˆi2 exp(−γ1i x1 ) + eˆr2 exp(−γ1r x1 ) = 0 .
(4.86)
This equation can only be satisﬁed for all x1 , if
γ1r = γ1i ,
(4.87)
while from Eqs. (4.73)  (4.74) it follows that
γ3r = −γ3i .
(4.88)
The boundary condition of Eq. (4.86) also requires
eˆr2 = −ˆ ei2 .
(4.89)
The two other nonzero vectors follow with the help of Eqs. (4.13)  (4.14) as ˆi , ˆr = h h 1 1 r ˆi . ˆ h3 = −h 3
(4.90) (4.91)
At this moment, the perpendicularly polarized reﬂected wave is determined completely.
99
reflection by an electrically impenetrable halfspace
σ, ε, µ
electrically impenetrable medium
sr
si
....... ....... .................. ......................... ............. ... . ... ....... ....... ... ........ ........ . . . . . ............... . ....... ............. ... ......................................................................... . ... .. ......... ... ...... . . . . . . .... . . ..... ....... ....... .......
θr θi
i3
Figure 4.7. Reﬂection of a uniform plane wave.
Uniform plane waves When the incident wave is a uniform plane wave, we may write 1
γ i = γ1i i1 + γ3i i3 = [(σ + sε)sµ] 2 si ,
(4.92)
where si is the unit vector in the direction of propagation of the incident wave. On account of Eqs. (4.81)  (4.82), or Eqs. (4.87)  (4.88), we may also write 1 (4.93) γ r = γ1r i1 + γ3r i3 = [(σ + sε)sµ] 2 sr , where sr is the unit vector in the direction of propagation of the reﬂected wave. The reﬂected wave is also a uniform plane wave. The angle θi between the vectors si and i3 is called the angle of incidence, while the angle θr between the vectors sr and −i3 is called the angle of reﬂection (see Fig. 4.7). From Eqs. (4.81)  (4.82), or Eqs. (4.87)  (4.88), we simply obtain θr = θi .
(4.94)
100
twodimensional electromagnetic waves
In the case of steadystate (s = jω), the total wave ﬁeld, being the superposition of the incident wave and the reﬂected wave, has the character of a travelling wave in the i1 direction and a standing wave in the i3 direction. This phenomenon is clearly observable when the medium in the halfspace −∞ < x3 < 0 is lossless. Then γ1i = jωsi1 /c ,
γ3i = jωsi3 /c ,
γ1r
γ3r
=
jωsi1 /c
,
=
(4.95)
−jωsi3 /c
,
(4.96)
where c = (εµ)− 2 . Combining all the previous results, we conclude that the total electromagnetic ﬁeld is given by 1
ˆ1 = −2jˆ E ei1 exp(−jωsi1 x1 /c) sin(ωsi3 x3 /c) , ˆ2 = 0 , E ˆ3 = 2ˆ ei exp(−jωsi x1 /c) cos(ωsi x3 /c) , E
(4.97)
ˆ1 = 0 , H ˆ i exp(−jωsi x1 /c) cos(ωsi x3 /c) , ˆ 2 = 2h H 2 1 3 ˆ H3 = 0 ,
(4.100)
3
1
3
(4.98) (4.99)
(4.101) (4.102)
for the case of parallel polarization, and ˆ1 = 0 , E ˆ2 = −2jˆ ei2 exp(−jωsi1 x1 /c) sin(ωsi3 x3 /c) , E ˆ3 = 0 , E
ˆ i exp(−jωsi x1 /c) cos(ωsi x3 /c) , ˆ 1 = 2h H 1 1 3 ˆ H2 = 0 , ˆ i exp(−jωsi x1 /c) sin(ωsi x3 /c) , ˆ 3 = −2j h H 3
1
for the case of perpendicular polarization.
3
(4.103) (4.104) (4.105)
(4.106) (4.107) (4.108)
101
reflection and transmission of a plane wave
4.4.
Reﬂection and transmission of a plane wave incident upon a plane interface
When a wave is incident upon the interface between two diﬀerent media, the wave is partly reﬂected by this interface and partly transmitted through this interface. In the case that the incident wave is a plane wave and the interface is a plane boundary between two semiinﬁnite media, the reﬂected and transmitted waves are plane waves as well. Then, the pertaining reﬂection problem can be solved with the theory of plane waves. We take the Cartesian coordinate system in such a way that the plane x3 = 0 coincides with the interface between the two media. The halfspace −∞ < x3 < 0 consists of a homogeneous medium with constitutive constants σ (1) , ε(1) and µ(1) and the halfspace 0 < x3 < ∞ consists of a homogeneous medium with constitutive constants σ (2) , ε(2) and µ(2) (see Fig. 4.8). The incident (plane) wave is given by (cf. Eqs. (4.16)  (4.17)) ˆi = e ˆi exp(−γ1i x1 − γ3i x3 ) , E ˆ H
(4.109)
i
i
ˆ exp(−γ i x1 − γ i x3 ) , = h 1 3
(4.110)
while the reﬂected wave is given by ˆr = e ˆr exp(−γ1r x1 − γ3r x3 ) , E r ˆ r exp(−γ r x1 − γ r x3 ) , ˆ = h H 1 3
(4.111) (4.112)
in which (γ1i )2 + (γ3i )2 = (γ1r )2 + (γ3r )2 = (σ (1) + sε(1) )sµ(1) .
(4.113)
In view of causality with respect to the incident wave we take
γ3i = (σ (1) + sε(1) )sµ(1) − (γ1i )2
1 2
,
(4.114)
102
twodimensional electromagnetic waves
σ (1) , ε(1) , µ(1)
σ (2) , ε(2) , µ(2)
transmitted wave
................. ............ ... ........ ....... ........ ....... ....... ........ .
............... .................. ............. .... ............. ............. . . . . . . . . . . . . ...
reﬂected wave
............................
i3
......................... .. . ....... ... ....... ........ . . . . . . ..... . . . . . . . ........ ....... ....... ....... ........ . . . . . . .
incident wave
x3 < 0
x3 = 0
x3 > 0
Figure 4.8. Reﬂection and transmission of a plane wave by a plane interface between two diﬀerent media.
where Re[γ3i ] ≥ 0. Similarly, in view of causality with respect to the reﬂected wave we take 1 γ3r = − (σ (1) + sε(1) )sµ(1) − (γ1r )2
2
,
(4.115)
where Re[γ3r ] ≤ 0. The total ﬁeld in the halfspace −∞ < x3 < 0 is obtained as ˆ r, H ˆ =H ˆi+H ˆr. ˆ =E ˆi +E E
(4.116)
The transmitted wave is given by ˆt = e ˆt exp(−γ1t x1 − γ3t x3 ) , E ˆ t exp(−γ t x1 − γ t x3 ) , ˆt = h H 1 3
(4.117) (4.118)
103
reflection and transmission of a plane wave
in which (γ1t )2 + (γ3t )2 = (σ (2) + sε(2) )sµ(2) .
(4.119)
In view of causality with respect to the transmitted wave we take
γ3t = (σ (2) + sε(2) )sµ(2) − (γ1t )2
1 2
,
(4.120)
where Re[γ3t ] ≥ 0. At x3 = 0 the electromagnetic ﬁeld has to satisfy the boundary conditions that the tangential components of the electric and magnetic ﬁeld strengths have to be continuous, i.e., ˆi +E ˆ r) = lim i3 × (E
x3 ↑0
ˆi+H ˆ r) = lim i3 × (H
x3 ↑0
ˆt , lim i3 × E
(4.121)
ˆt, lim i3 × H
(4.122)
ˆt , lim E 1
(4.123)
ˆt , lim E 2
(4.124)
ˆt , lim H 1
(4.125)
ˆt . lim H 2
(4.126)
x3 ↓0 x3 ↓0
or ˆi + E ˆr) = lim (E 1 1
x3 ↑0
ˆi + E ˆr) = lim (E 2 2
x3 ↑0
ˆi + H ˆ r) = lim (H 1 1
x3 ↑0
ˆi + H ˆ r) = lim (H 2 2
x3 ↑0
x3 ↓0 x3 ↓0 x3 ↓0 x3 ↓0
We now discuss the cases of parallel and perpendicular polarization separately. Parallel polarization ˆ3 and H ˆ 2 are the nonzero ˆ1 , E In the case of parallel polarization, E electromagnetic ﬁeld components. We only have to meet the boundary conˆ 2 . Using Eqs. (4.10) and ˆ1 and Eq. (4.126) for H ditions of Eq. (4.123) for E (4.116) the boundary conditions at x3 = 0 become lim
x3 ↑0 σ (1)
1 ˆi + H ˆ r) = ∂ 3 (H 2 2 + sε(1) ˆi + H ˆ r) = lim (H 2 2 x3 ↑0
lim
x3 ↓0 σ (2)
ˆt . lim H 2
x3 ↓0
1 ˆt , ∂3 H 2 + sε(2)
(4.127) (4.128)
104
twodimensional electromagnetic waves
Substituting the planewave representations of Eqs. (4.110), (4.112) and (4.118) into Eqs. (4.127)  (4.128), we obtain γ3i ˆ i exp(−γ i x1 ) + h 1 σ (1) + sε(1) 2
γ3r ˆ r exp(−γ r x1 ) h 1 σ (1) + sε(1) 2 t γ3 ˆ t exp(−γ t x1 ) , (4.129) h = 1 (2) σ + sε(2) 2 ˆ i exp(−γ i x1 ) + h ˆ r exp(−γ r x1 ) = h ˆ t exp(−γ t x1 ) . h (4.130) 2 1 2 1 2 1 Both equations can only be satisﬁed for all x1 , if γ1t = γ1r = γ1i
(4.131)
while from Eqs. (4.114), (4.115) and (4.120) it follows that
γ3r = −γ3i ,
γ3t = (σ (2) + sε(2) )sµ(2) − (γ1i )2
(4.132)
1 2
.
(4.133)
But, in that case, the boundary conditions of Eqs. (4.129)  (4.130) also require γ3r γ3t γ3i ˆi + ˆr = ˆt , h h h 2 2 σ (1) + sε(1) σ (1) + sε(1) σ (2) + sε(2) 2 ˆi + h ˆr = h ˆt . h 2 2 2
(4.134) (4.135)
Let us now introduce the reﬂection coeﬃcient R = R (s) and the transmission coeﬃcient T = T (s) via the relations ˆi , ˆ r = R h h 2 2 t i ˆ ˆ h2 = T h2 .
(4.136) (4.137)
105
reflection and transmission of a plane wave
Substituting the latter two expressions into Eqs. (4.134)  (4.135), dividˆ i , and using ing the resulting two equations by the nonzero quantity h 2 Eq. (4.132), we arrive at the system of two algebraic equations, γ3i γ3t γ3i − R = T , σ (1) + sε(1) σ (1) + sε(1) σ (2) + sε(2) 1 + R = T ,
(4.138) (4.139)
from which R and T are obtained as
γ3i γ3t − (1) (1) σ (2) + sε(2) , R = σ +i sε γ3t γ3 + σ (1) + sε(1) σ (2) + sε(2)
2 T =
γ3i σ (1) + sε(1)
γ3i γ3t + σ (1) + sε(1) σ (2) + sε(2)
(4.140)
.
(4.141)
At this moment, the parallelly polarized reﬂected and transmitted wave are determined completely. In the special case of normal incidence, we have γ1i = 0. Then, we have arrived at the reﬂection and transmission coeﬃcients of the onedimensional waves (cf. Eqs. (3.95) derived from a magnetic ﬁeld analysis), viz., R = T =
Z (1) − Z (2) , Z (1) + Z (2) 2Z (1) , Z (1) + Z (2)
in which Z
(1)
γ (1) = (1) = σ + sε(1)
sµ(1) σ (1) + sε(1)
(4.142) (4.143) 1 2
(4.144)
106
twodimensional electromagnetic waves
and Z
(2)
γ (2) = (2) = σ + sε(2)
sµ(2) σ (2) + sε(2)
1 2
(4.145)
are the wave impedances in medium (1) and (2), respectively. Perpendicular polarization ˆ 1 and H ˆ 3 are the nonˆ2 , H In the case of perpendicular polarization, E zero electromagnetic ﬁeld components. We only have to meet the boundary ˆ 1 . Using Eqs. (4.13) ˆ2 and Eq. (4.125) for H conditions of Eq. (4.124) for E and (4.116) the boundary conditions at x3 = 0 become ˆi + E ˆr) = lim (E 2 2
x3 ↑0
lim
x3 ↑0
1 ˆi + E ˆr) = ∂ 3 (E 2 2 sµ(1)
ˆt , lim E 2
(4.146)
x3 ↓0
lim
x3 ↓0
1 ˆt . ∂3 E 2 sµ(2)
(4.147)
Substituting the planewave representations of Eqs. (4.109), (4.111) and (4.117) into Eqs. (4.146)  (4.147), we obtain eˆi2 exp(−γ1i x1 ) + eˆr2 exp(−γ1r x1 ) = eˆt2 exp(−γ1t x1 ) , (4.148) i r t γ3 r γ3 t γ3 i eˆ2 exp(−γ1i x1 ) + (1) eˆ2 exp(−γ1r x1 ) = eˆ exp(−γ1t x1 ) . (4.149) (1) sµ sµ sµ(2) 2 Both equations can only be satisﬁed for all x1 , if γ1t = γ1r = γ1i ,
(4.150)
while from Eqs. (4.114), (4.115) and (4.120) it follows that γ3r = −γ3i ,
γ3t = (σ (2) + sε(2) )sµ(2) − (γ1i )2
(4.151)
1 2
.
(4.152)
reflection and transmission of a plane wave
107
But, in that case, the boundary conditions of Eqs. (4.148)  (4.149) also require eˆi2 + eˆr2 = eˆt2 , γ3r r γ3t t γ3i i e ˆ + e ˆ = eˆ . 2 2 sµ(1) sµ(1) sµ(2) 2
(4.153) (4.154)
Let us now introduce the reﬂection coeﬃcient R⊥ = R⊥ (s) and the transmission coeﬃcient T⊥ = T⊥ (s) via the relations eˆr2 = R⊥ eˆi2 ,
(4.155)
eˆt2
(4.156)
T⊥ eˆi2
=
.
Substituting the latter two expressions into Eqs. (4.153)  (4.154), dividing the resulting two equations by the nonzero quantity eˆi2 , and using Eq. (4.151) we arrive at the system of two algebraic equations, γ3i sµ(1)
1 + R⊥ = T⊥ , γ3i γ3t − (1) R⊥ = T⊥ , sµ sµ(2)
(4.157) (4.158)
from which R⊥ and T⊥ are obtained as γ3i γ3t − µ(1) µ(2) R⊥ = i , γ3t γ3 + µ(1) µ(2)
2 T⊥ =
γ3i µ(1)
γ3t γ3i + µ(1) µ(2)
.
(4.159)
(4.160)
At this moment, the perpendicularly polarized reﬂected and transmitted wave are determined completely.
108
twodimensional electromagnetic waves
In the special case of normal incidence, we have γ1i = 0. Then, we have arrived at the reﬂection and transmission coeﬃcients of the onedimensional waves (cf. Eqs. (3.85) derived from an electric ﬁeld analysis), viz., R⊥ = T⊥ =
Y (1) − Y (2) , Y (1) + Y (2) 2Y (1) , Y (1) + Y (2)
in which Y (1)
σ (1) + sε(1) = = γ (1)
(2)
σ (2) + sε(2) = = γ (2)
and Y
σ (1) + sε(1) sµ(1) σ (2) + sε(2) sµ(2)
(4.161) (4.162) 1 2
(4.163) 1 2
(4.164)
are the wave admittances in medium (1) and (2), respectively.
4.4.1.
Uniform plane waves in the frequency domain
We restrict ourselves to the analysis in the freqency domain with s = jω. When the incident wave is a uniform plane wave, we may write 1
γ i = γ1i i1 + γ3i i3 = [(σ (1) + jωε(1) )jωµ(1) ] 2 si ,
(4.165)
where si is the unit vector in the direction of propagation of the incident wave. On account of Eqs. (4.131)  (4.132), or Eqs. (4.150)  (4.151), we may also write 1 (4.166) γ r = γ1r i1 + γ3r i3 = [(σ (1) + jωε(1) )jωµ(1) ] 2 sr , where sr is the unit vector in the direction of propagation of the reﬂected wave. The reﬂected wave is also a uniform plane wave. The angle θi between the vectors si and i3 is called the angle of incidence, while the angle θr between the vectors sr and −i3 is called the angle of reﬂection (see Fig. 4.9). From Eqs. (4.131)  (4.132), or Eqs. (4.150)  (4.151), we simply obtain θr = θi .
(4.167)
109
reflection and transmission of a plane wave
This relation is identical to Eq. (4.94) of Section 4.3 dealing with the reﬂection by an electrically impenetrable halfspace. This relation is called Snell’s law of reﬂection. Subsequently, we investigate the conditions under which the transmitted wave is a uniform plane wave. Let us assume that 1
γ t = γ1t i1 + γ3t i3 = [(σ (2) + jωε(2) )jωµ(2) ] 2 st ,
(4.168)
where st is the unit vector in the direction of propagation of the transmitted wave. From Eq. (4.131), or Eq. (4.150), it follows that 1
1
[(σ (1) + jωε(1) )jωµ(1) ] 2 si1 = [(σ (2) + jωε(2) )jωµ(2) ] 2 st1 .
(4.169)
In the general case, real values of si1 and st1 do not meet this equation, and in general the transmitted wave is a nonuniform plane wave. However, for lossless media, Eq. (4.169) can be satisﬁed for real values of si1 and st1 . Lossless media In the case of lossless media, we introduce the (realvalued) index of refraction n(1) of medium (1) as 1
n(1) = c0 [ε(1) µ(1) ] 2 ,
(4.170)
and the index of refraction n(2) of medium (2) as 1
n(2) = c0 [ε(2) µ(2) ] 2 .
(4.171)
Then, we have (cf Eq. (4.169)) n(1) si1 = n(2) st1 .
(4.172)
The angle θt between the vectors st and i3 is called the angle of transmission (see Fig. 4.9). From Eq. (4.172) we obtain Snell’s law of transmission (also called Snell’s law of refraction),
n(1) sin(θi ) = n(2) sin(θt ) .
(4.173)
110
twodimensional electromagnetic waves
ε(1) , µ(1)
ε(2) , µ(2) n(2) .. n(1) ...... . . ......
....... .. ..... ...... .
1 .......
.......
..... ..
si
... ..
...... . ... ... . .................... ............................. .. ...... . .. ... . ...... .... .................... . . .. . . . ... . ........ ............... . . . . . . . ... . .. .... .. ..... .... . .............................................................................. .. ... .... .. . . . ... ... ... . .. .. .. . ...... . . . . ... .... .. .. ... .. ... ...... .... .. ... . ..... ... ..... . . . . . . . ... . . ...... ....... ..... .. ...... .... . ...... . ...... . . . . .. ....... .......
sr....................... . ...
θr θi
ε(1) , µ(1)
. ...
n(2) t s n(1)
ε(2) , µ(2)
.. ....
...... . . ......
1 .......
n(2) n
sr........................
....... .......
....... .... ... (1)
...... .
n(2) t ...... . n(1) s ... ................. si ..............
......... ... ........... . ... ................. ..... .. ......... . ..... .. ............. ... ...... ......... ..... ..... ........ ......... .. ... ......... .......................... .. ...... ... ... .... . . . ........ . ... ....... ....................................................................... ... ..... . ... . ... ....... . .. . . .. .... ... ... ....... ... . .. .. ...... . . . . . ... . ....... .. ...... .. ... .. ..... ....... .... .. . . . . . . ...... . ...... ....... ....... ....... ...... .......
θr
θt
i3
θt i3
θi
(a) n(1) < n(2)
(b) n(1) > n(2)
Figure 4.9. Reﬂection and transmission of a uniform plane wave.
Further, see Eq. (4.133) or (4.152),
γ3t = jω ε(2) µ(2)
⎧ 1 ⎨ 2
⎫1
2⎬ 2 n(1) i 1 − (2) sin(θ ) , ⎩ ⎭ n
(4.174)
has to be imaginary valued. This means that 0 ≤ sin(θi ) ≤
n(2) n(1)
for n(1) > n(2) ,
(4.175)
which yields a restriction of admissible values of the angle of incidence θi . When we are dealing with a uniform transmitted wave, the expressions for
111
reflection and transmission of a plane wave
the reﬂection and transmission coeﬃcients become
1
µ(1) ε(1)
R =
2
1
µ(1) ε(1)
2
cos(θi ) − cos(θi )
2 T =
µ(1) ε(1)
1 2
µ(1) ε(1)
2
cos(θi ) +
2
1
µ(2) ε(2)
+
1
1
µ(2) ε(2)
2
cos(θt ) ,
(4.176)
,
(4.177)
,
(4.178)
,
(4.179)
cos(θt )
cos(θi )
µ(2) ε(2)
1 2
cos(θt )
for parallel polarization, and
R⊥ =
ε(1) µ(1) ε(1) µ(1)
1 2
1 2
cos(θi ) − cos(θi ) +
2 T⊥ =
ε(1) µ(1)
1 2
ε(1) µ(1)
1 2
cos(θi ) +
ε(2) µ(2) ε(2) µ(2)
1 2
1 2
cos(θt ) cos(θt )
cos(θi )
ε(2) µ(2)
1 2
cos(θt )
for perpendicular polarization. These reﬂection and transmission coeﬃcients are known as the Fresnel reﬂection and transmission coeﬃcients. For lossless, dielectric media, the reﬂection coeﬃcients R and R⊥ as a function of the angle of incidence θi are presented in Fig. 4.10.
112
twodimensional electromagnetic waves
1.0 R  6
0.75
0.5
.. ... ... ..... ... ..... ... .......... .. ....... ...... ........ .... ...... ...... ........ ..... ..... ....... ε(2) ....... . ... . ε(1) ..... .... . . . . . . ..... . . . . . . . . . . ..... . ... .. . . . ... . . .. .. . . . ...... ...... ...... .... ... .. . . .. ...... ... . . . ...... ...... ... ... . . ...... .. .. . . . . ...... ... .. . . ...... ... .. . ..... ... ... . . .. . . .............................. . . . ..... . .. . . ........... ........ .... .... ... . ....... . . . .. ...... ... .... . ... . ...... ...... . . ..... ... . ..... . . .. . .... ..... ...... ... .. . . ....... .... .
1.0 R⊥  6
0.75
=8
0.5
0.25
2
0
1.0
0
30
. . . .... ... . .. . .. . . . . ε .... ... . ... ε(2) . .... .... . . . . ... ... . ... . . ... ... . ... .... . .. . ... .... . ... . .... ... . .. . ... ... . .. ... . .. . . . . ... .. . ... .. . .... . . .. . . . . . . ... ... . . ... . .. . .. . .. . . . . . . .. ... ...... ...... . . .. ... ...... . . .. ... ...... . ... .. .... .. . .... . ... ... . .... . .. .. . . .. ... . . ..................... . . .. . ........ . .. ... ... .............. . .. .. ............. .. . . . . . . .. . ... ...... ... .. .... .... ... ..... ... ... . ... . (1)
R  6
0.75
0.5
0.25
0
0
= 8 4
30
60
 θi
0
90
1.0 6
0.75
0.5
0.25
60

θi
90
2
0
30
. . . ... ... . . ... .. .. . . . ε . . ... .. ... . ε(2) . ... . ... .. . . . ... ... . ... . ... .... . . . . . ... ... . . ... ... .. . . .. . . . . . . ... . . ... ... . . ... . .. ... . . . ... ... . .. . ... ... . . .. . . . ... ... . ... ... .. ..... .. . . . . ..... ... ...... ..... ... .. ... . . . .... ..... ..... ...... ....... . . . . . . . . . . . ............ (1)
R⊥ 
2
=8 4
4
0.25
..... ....... . ...... . . . .. . . . ... . .... ... . ... .... . . .. . .. ... . ... .... . . . . .. . . . ... ... . . .. . ... .. . . . . . ... ... . . ... ... .. . . . .. ... . . (2) . . ε ..... ... . . .... ... .. ε(1) . . . . .. ..... .. . . . . . . . . . . . ... ... ..... ... .... .. .. ... . ... ... . . . . . . . . . ...... .... .. ...... .... ...... ...... ...... .... .... .... .... . . . . ..... ...... ...... ....... ......... . . . . . . . . . . ..............................
0
0
= 8 4
30
60
 θi
90
2
60
 θi
90
Figure 4.10. The reﬂection coeﬃcients as a function of the angle of incidence θ i . The media are lossless (σ (1) = σ (2) = 0) and pure dielectric (µ(1) = µ(2) = µ0 ).
reflection and transmission of a plane wave
113
Brewster angle It is possible that the Fresnel reﬂection coeﬃcient vanishes for a particular value of the angle of incidence. The pertaining angle of incidence is i called the Brewster angle. In the case of parallel polarization, this angle θB follows from
µ(1) ε(1)
1 2
cos(θ ) − i
µ(2) ε(2)
1 2
cos(θt ) = 0
(4.180)
and Snell’s law of refraction, Eq. (4.173). The result is ⎛ i )= tan(θB
µ(2) (1) ⎝µ (1) ε ε(2)
ε(2) ε(1) µ(2) µ(1)
− −
⎞12 ⎠ ,
for parallel polarization .
(4.181)
For dielectric media, where µ(1) = µ(2) = µ0 , this Brewster angle follows from i tan(θB )
=
ε(2) ε(1)
1 2
,
for parallel polarization .
(4.182)
i . In the case In this last case, we also have that θi + θt = 12 π, with θi = θB i of perpendicular polarization the Brewster angle θB follows from
ε(1) µ(1)
1 2
cos(θ ) − i
ε(2) µ(2)
1 2
cos(θt ) = 0
(4.183)
and Snell’s law of refraction, Eq. (4.173). The result is ⎛ i )= tan(θB
ε(2) (1) ⎝ε µ(1) µ(2)
− −
µ(2) µ(1) ε(2) ε(1)
⎞12 ⎠ ,
for perpendicular polarization .
(4.184)
For dielectric media, where µ(1) = µ(2) = µ0 , this Brewster angle does not exist; the righthand side of Eq. (4.184) is not real valued.
114
twodimensional electromagnetic waves
Total reﬂection If n(1) > n(2) and the angle of incidence is larger than the critical angle,
i
θ >
θci
n(2) = arcsin n(1)
,
(4.185)
Eq. (4.174) has to be written as ⎧ ⎫1 2 ⎬2 1 ⎨ n(1) i γ3t = ω ε(2) µ(2) 2 sin(θ ) − 1 . ⎩ n(2) ⎭
(4.186)
In this case γ3t is real valued, while on account of Eq. (4.131), or (4.150), γ1t is imaginary valued; hence, the transmitted wave is a nonuniform plane wave (Fig. 4.12). The planes of equal phase are perpendicular to the interface between the two diﬀerent media, while the planes of equal amplitude are parallel to the interface. Since we are dealing with lossless media, the planes of equal amplitude are perpendicular to the planes of equal phase. Moreover, since σ (1) = σ (2) = 0, γ3i is imaginary valued and γ3t is real valued, we observe from Eq. (4.140) that R  = 1, while from Eq. (4.159) we observe that R⊥  = 1. The reﬂected wave has an amplitude identical to the one of the incident wave; only the phase is diﬀerent. This phenomenon is called total reﬂection (see Fig. 4.11). There exists an electromagnetic ﬁeld in the halfspace 0 < x3 < ∞, but there is no time average power transported in the i3 direction. This is easily observed from (cf. Eq. (4.25)) ˆ t∗ ) · i3 ] exp(−2γ t x3 ) ˆt ×H ˆ t∗ ) · i3 ] = Re[(ˆ Re[(E et × h 3
(4.187)
and (cf. Eqs. (4.37) and (4.45)) ⎧ ⎪ γ3t ˆ t ˆ t∗ ⎪ ⎪ h h = 0 , for parallelly polarized waves, Re ⎪ ⎪ ⎪ jωε 2 2 ⎨
ˆ t∗ )·i3 ] = Re[(ˆ et × h
⎪ ⎪ t ⎪ γ ⎪ 3 t t∗ ⎪ ⎪ eˆ2 eˆ2 = 0 , for perpendicular polarized waves. ⎩ Re
jωµ
(4.188)
115
reflection and transmission of a plane wave
ε(1) , µ(1)
ε(2) , µ(2) ...... ....... ....... ....... ....... ...... .
.... t s.r.................... n(2) (1) s .................. . . . . .... n ....... ... . .....
si
ε(2) , µ(2) s........r. ....... ...... ....... ....... ....... .......... si
.. ............ ........... ....... ..... ....... .. ..... .. ... ... ..... .. ... .. ... ...... . . ... ... ... ... ... .. ... ... ... . .... . .. ... .. . ... . ... .. . .. . ... . . ... ... ... ..... . . ... ... . . . ... . .. . ............................................................................... .. . ... ... . .. . . . .. .. . .. ... ... . .. .. ... ... . . . . . . ... ... .... . (2) .. . ... .. ....... .... ... ..... ..... .. .. (1) . ........ ...... . . . . . . ...... .. . .... ....... ....... ... .. ...
..... ... . ....... ................. .... . . . . . . . . ........ . . . ..... . . . ... . . . . . ..... .... . . .. ..... .... ..... .... .. ..... ... ..... .. ..... ... .. ..... .... ......... ... ..... .. ..... . . . ..... .. ..... . . . . . . . . . ............................................................................ ... . . . . . . . .. . . . .... .. . ... . . . . . . . . . .. .. .. .. ..... ... . .. . .. ... ..... . . . . ... .... ... .... . (2) .. ... ... .. ....... .... ... ..... ..... ...... (1) . . ...... ...... . . ..... . . . . . ...... .. .. .... ....... ....... .....
. ...
i3
.. .....
i3
n n
n n
1
(a) si1 =
ε(1) , µ(1)
1
n(2) n(1)
(b) si1 >
n(2) n(1)
Figure 4.11. Total reﬂection of a uniform plane wave, when n(1) > n(2) .
γ1t
....... ............ . ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ... ............... ....... . . . . ... . . . . .... ..... ..... ..... ..... . . . . .... ..... ..... ..... ..... . . . . .... ..... ..... ..... ..... . . . . . ..... .....
θi = θci
... ...... ........ ... ... .... .. ... .... .. ... .... .. ... .... ..
.............................
i3
..... ............ ..... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ...... .............. . . . .. ... ... ... ... . . .. ... ... ... ... . . .. ... ... ... .. . . .. ... ... ... ... . . .. ... ... ...
θi > θci
γ1t
... ...... ........ ... ... .... .. ... .... .. ... .... .. ... .... .............................................................................. .....
γ3t
.............................
i3
Figure 4.12. Planes of equal phase ( ) and planes of equal amplitude ( ) for critical incidence (θ i = θci , γ1t is imaginary valued, γ3t = 0) and for above critical incidence (θ i > θci , γ1t is imaginary valued, γ3t is real valued).
116
4.5.
twodimensional electromagnetic waves
Exercises and problems
Exercise 4.1 A parallelly polarized plane wave in free space has an electric ﬁeld strength E = (3i1 − 4i3 ) sin[3π × 109 t − 2π(4x1 + 3x3 )] . (a) Find the frequency and the angle of propagation of this wave. (b) Find the frequency domain electric ﬁeld strength. ( c ) Find the frequency domain magnetic ﬁeld strength. (d) Find the time domain magnetic ﬁeld strength. Exercise 4.2 Given the electric ﬁelds of two parallelly polarized uniform plane waves, the total ﬁeld of which is given by Eq. (4.58) with γ = jω(ε0 µ0 )1/2 and e(2) . Show that this total electric ﬁeld corresponds to a wave ˆ e(1)  = ˆ standing in the x3 direction and propagating in the x1 direction. Hint: use ˆ = 0. the fact that s · e Exercise 4.3 A uniform perpendicularly polarized, steady state wave is incident from a lossless halfspace upon the plane boundary surface x3 = 0 of a perfect conductor. The incoming wave is uniform, and its electric ﬁeld is expressed as ˆ i = E0 exp[−(γ i x1 + γ i x3 )]i2 , for x3 < 0. Draw the amplitude of the total E 1 3 i ˆ r , as a function of x3 . ˆ ˆ electric ﬁeld, E = E + E Exercise 4.4 A uniform plane wave in free space, having an electric ﬁeld strength √ √ E i = ( 12 i1 − 12 3i3 ) E0 cos[6π × 109 t − 10π( 3x1 + x3 )] , is incident on the interface x3 = 0 between free space and a lossless dielectric medium, σ (2) = 0, ε(2) = 1.5ε0 and µ = µ0 . (a) Does this electric ﬁeld correspond to a parallelly or a perpendicularly polarized wave, why? (b) Find the angle of incidence θi , the angle of transmission θt , and the vectors si and st along the directions of propagation of the incident and the transmitted wave, respectively.
117
exercises and problems
( c ) Find the expressions for the reﬂected and transmitted electric ﬁeld strength. (d) Draw the electric ﬁeld vector and the direction of propagation of the incident, reﬂected and transmitted electric ﬁeld in the conﬁguration. Exercise 4.5 A perpendicularly polarized, plane electromagnetic wave is incident from medium (1) with ε(1) = 4ε0 , σ (1) = 0 and µ(1) = µ0 , upon the plane boundary of medium (2) with ε(2) = ε0 , σ (2) = 0 and µ(2) = µ0 , at an angle of incidence θi = π3 with the i3 axis, i3 being the normal of the plane interface. (a) What is the critical angle? 1
(b) Find γ1i and γ3i in terms of γ0 = jω(ε0 µ0 ) 2 . ( c ) Find γ3t in terms of γ0 . (d) Find in medium (2) the distance d at which the electric ﬁeld strength decays a factor of exp(−1). ( e ) Find the reﬂection coeﬃcient R⊥ and write this coeﬃcient as R⊥ = R⊥  exp(j arg[R⊥ ]). Exercise 4.6 If you are under water looking towards the water surface, only a small circle in the water surface is light while the rest of the surface is dark. Explain why there is only a cone of light connected with the circle. Calculate the angle θ of the cone (see Fig. 4.13). For optical frequencies the constitutive parameters of water are ε = 1.77ε0 , σ = 0 and µ = µ0 .
air ..... ... ..... ..... ..... ..... ..... ..... . ..... . . . ..... ... ..... ..... ..... ................. ..... ..... .......... .
θ
Figure 4.13. The cone of light.
water
118
twodimensional electromagnetic waves
Exercise 4.7 i of an air/glass Calculate the critical angle θci and the Brewster angle θB interface. At optical frequencies, glass is a lossless dielectric with ε = 2.25ε0 . Exercise 4.8 Show that for perfectly dielectric media and parallelly polarized waves the i ) = ( ε(2) ) 12 (see expression of the Brewster angle is indeed given by tan(θB ε(1) Eq. (4.182)).
Problem 4.1 A uniform plane wave is incident on a magnetically impenetrable wall at an angle of incidence of θi = π6 . Give for both the parallelly and perpendicularly polarized waves (a) the boundary conditions at the interface (b) the expressions for the total electric and magnetic ﬁeld strengths. Problem 4.2 A parallel polarized plane wave is incident from medium (1) on the plane boundary between medium (1) and medium (2). Both media are perfect dielectrics. You know that for an angle of incidence larger than the critical angle no time averaged power is transmitted into medium (2); also the reﬂected power in medium (1) is zero for an angle of incidence equal to the Brewster angle. Now imagine a situation where the Brewster angle is larger than the critical angle. Then the incident wave will not be reﬂected, because of the Brewster angle of incidence, and will not be transmitted, because of postcritical incidence. Solve this paradox. Problem 4.3 A sinusoidally varying uniform plane wave is incident on a plane interface between two lossless dielectric media at an angle larger than the critical angle, θi > θci . (a) Find the Fresnel reﬂection and transmission coeﬃcients. (b) Find the phase shift of the reﬂected wave at the interface. ( c ) Sketch the amplitude of the transmitted wave as a function of the distance to the interface.
119
exercises and problems
Problem 4.4 Indicate in the four graphs of Fig. 4.10 the Brewster angles and the critical angles. Give the expressions for the Brewster angles, both for parallel and perpendicular polarization, in case ε(1) = ε(2) while µ(1) = µ(2) . Problem 4.5 A parallelly polarized plane uniform wave in free space (n(1) = 1) is incident √ at an angle θi on a slab of lossless dielectric with index of refraction n(2) = 3 and thickness d = 3 m, while on the other side there is another lossless dielectric medium that extends to inﬁnity with index of refraction n(3) = 3 (see Fig. 4.14). Determine the angle of incidence in case the wave is totally transmitted through both interfaces.
n(1)
n(2)
n(3)
d Figure 4.14. A threemedia conﬁguration.
Problem 4.6 A lossless dielectric slab, with thickness d = λ4 m and ε = 2.25ε0 , is on one side coated with an electrically perfectly conducting material. A uniform plane wave in free space (n(1) = 1) impinges on the dielectric slab at normal incidence (see Fig. 4.15).
n(1)
n(2)
d Figure 4.15. The slab coated with a perfect conductor.
120
twodimensional electromagnetic waves
Problem 4.6 (continued) (a) Give the general expressions for the plane waves in the two regions. (b) Give the boundary conditions at the two interfaces. ( c ) Find the electric and magnetic ﬁeld strengths in the two regions by substituting the general expressions of the ﬁeld strengths into the boundary conditions at both interfaces. Problem 4.7 A parallelly polarized plane uniform wave in a lossless dielectric medium (1) with ε(1) = 12ε0 is incident at an angle θi = π6 on a lossless dielectric slab with index of refraction n(2) = 2 and thickness d = 1 m, while on the other side there is another lossless dielectric medium that extends to inﬁnity with index of refraction n(3) (see Fig. 4.14). Determine the maximum index of refraction n(3) , for which the wave is totally reﬂected back into medium (1).
Chapter 5
Electromagnetic Rays in a Twodimensional Medium In Chapter 3 we have seen that a planar source with a uniform current distribution, when it is placed in a homogeneous medium, will only generate plane electromagnetic waves. Further we have discussed in Chapter 4 that at a plane boundary between two diﬀerent homogeneous regions, a plane incident electromagnetic wave causes a plane reﬂected and eventually a plane transmitted electromagnetic wave. If we are dealing with either a nonplanar source, a nonuniform current distribution, an inhomogeneous medium, or a curved boundary – or combinations of these –, the electromagnetic waves that occur are in general not plane. In this chapter, we will ﬁrst introduce the concepts of a wavefront, a ray trajectory, and an electromagnetic ray by means of the sdomain analysis of electromagnetic waves in a homogeneous medium. Subsequently, we will derive an approximate solution for the electromagnetic ﬁeld that propagates in a weakly inhomogeneous medium. The applied approximation is valid in the limit s → ∞ and is known as the ray approximation. The resulting approximate solution in a sourcefree domain is called the electromagnetic ray. The wave propagation is very similar to the one of rays deﬁned in geometrical optics. In this chapter we will restrict the analysis to the twodimensional case, i.e., we assume that the permittivity, the conductivity, and the permeability
122
electromagnetic rays in a twodimensional medium
are independent of the x2 coordinate (cf. Eq. (4.1)), and that the sources of the electromagnetic wave ﬁeld are also independent of x2 (cf. Eq. (4.2)). Then, the generated electromagnetic ﬁeld will be independent of x2 (cf. Eq. (4.3)). From Chapter 4 we know that in this case the resulting total electromagnetic ﬁeld may be decomposed in two constituents, viz. the parallelly and perpendicularly polarized ﬁelds. Therefore, we separately consider the two cases of parallel and perpendicular polarization.
5.1.
Homogeneous, lossless medium
For the introduction of several notions we ﬁrst consider the case of a uniform plane electromagnetic wave in a homogeneous, lossless medium. According to Eqs. (3.38), (4.16), (4.17), and (4.18), the electric and the magnetic ﬁeld strengths in this case satisfy s ˆ = e ˆ(s) exp[− (s1 x1 + s3 x3 )] , E c s ˆ ˆ H = h(s) exp[− (s1 x1 + s3 x3 )] . c
(5.1) (5.2)
If we introduce the socalled eikonal L as L(x1 , x3 ) =
c0 (s1 x1 + s3 x3 ) , c
(5.3)
these equations may be written in the form s ˆ = e ˆ(s) exp[− L(x1 , x3 )] , E c0 ˆ exp[− s L(x1 , x3 )] . ˆ = h(s) H c0
(5.4) (5.5)
In the time domain these functions correspond to L(x1 , x3 ) L(x1 , x3 ) ] H[t − ], c0 c0 L(x1 , x3 ) L(x1 , x3 ) ] H[t − ]. H = h[t − c0 c0 E = e[t −
(5.6) (5.7)
123
homogeneous, lossless medium
For a given time instant t, the electric and magnetic ﬁeld strengths in all points of the planes given by (s1 x1 + s3 x3 ) L(x1 , x3 ) = constant = c0 c
(5.8)
have an equal value. The plane (s1 x1 + s3 x3 ) L(x1 , x3 ) =t = c0 c
(5.9)
is in a certain respect special: here the values of the electric and the magnetic ﬁeld strengths are given by E = e(0) and H = h(0), respectively. This implies that at this plane the disturbance of the source has just arrived. For this reason the plane given in Eq. (5.9) is called the wavefront. In Fig. 5.1 the wavefront is depicted for some time instants. Obviously, plane waves have plane wavefronts. These move away from the source.
L(x1 , x3 ) = t(1) c0
L(x1 , x3 ) = t(2) c0
(1) s
(2) s
L(x1 , x3 ) = t(3) c0
(3) s
Ray trajectory
Wavefronts
Figure 5.1. A sequence of wavefronts and a ray trajectory for a uniform plane electromagnetic wave in a homogeneous medium (t(1) < t(2) < t(3) ).
124
electromagnetic rays in a twodimensional medium
If we take a certain point on the wavefront at t = 0 and keep track of this point during the propagation of the wavefront through space, we obtain a line that is known as a ray trajectory. By deﬁnition, the tangent in every point of a ray trajectory coincides with the vector s in the propagation direction. In Fig. 5.1 an example of a ray trajectory has been drawn. The ray trajectories belonging to plane waves are straight lines, and (provided that the medium is isotropic) these lines are perpendicular to the wavefronts. The solutions of the form of Eqs. (5.4) and (5.5) are called electromagnetic rays. Knowing Eqs. (5.4) and (5.5), there exists an alternative way to ﬁnd the value of the electric and the magnetic ﬁeld strenghts at the wavefront. This method is based on Tauber’s theorem for the Laplace transformation, which states that lim sfˆ(s) = f (0) ,
(5.10)
lim sfˆ(s) exp(st1 ) = f (t1 ) ,
(5.11)
s→∞
and, more general s→∞
provided that the waves arrive at t = 0 and t = t1 , respectively, and the functions on the righthand side exist. In view of this theorem, we may write ˆ exp[ s L(x1 , x3 )] = lim sˆ e(s) = e(0) , (5.12) lim sE s→∞ s→∞ c0 ˆ ˆ exp[ s L(x1 , x3 )] = lim sh(s) lim sH = h(0) . (5.13) s→∞ s→∞ c0 Note that the results are equal to the results obtained above. However, from the latter two equations it may be seen that the time domain waveﬁeld at the wavefront follows from the behavior of its complex frequency domain counterpart for s → ∞.
5.2.
Parallel polarization
In this section we consider electromagnetic waves in an inhomogeneous medium with ε = ε(x1 , x3 ), σ = σ(x1 , x3 ), and µ = µ(x1 , x3 ). Assuming
parallel polarization
125
that the sources of the electromagnetic ﬁeld are also independent of x2 , the parallelly polarized ﬁeld forms an independent waveﬁeld constituent, for ˆ1 , E ˆ3 , and H ˆ 2 , are the nonzero electromagnetic ﬁeld components. which E In a sourcefree domain they satisfy the ﬁeld equations (cf. Eqs. (4.4)  (4.6)) ˆ1 = 0 , ˆ 2 + (σ+sε)E ∂3 H ˆ3 = 0 , ˆ 2 + (σ+sε)E −∂1 H ˆ 1 − ∂1 E ˆ3 + sµH ˆ2 = 0 . ∂3 E
(5.14) (5.15) (5.16)
In analogy with Eqs. (5.4) and (5.5), we now investigate a solution of the form ˆ1 = eˆ1 exp[− s L(x1 , x3 )] , E c0 ˆ3 = eˆ3 exp[− s L(x1 , x3 )] , E c0 s ˆ ˆ H2 = h2 exp[− L(x1 , x3 )] . c0
(5.17) (5.18) (5.19)
In the inhomogeneous situation, the eikonal L = L(x1 , x3 ) is a more intricate function of position than in the homogeneous case. Moreover, the amplitudes ˆ2 = h ˆ 2 (x1 , x3 , s), will now also eˆ1 = eˆ1 (x1 , x3 , s), eˆ3 = eˆ3 (x1 , x3 , s), and h depend on the position. In order to investigate the spatial behavior of eˆ1 , ˆ 2 , and L, we substitute Eqs. (5.17)  (5.19) into Eqs. (5.14)  (5.16) and eˆ3 , h divide by the nonzero factor exp(− cs0 L). We arrive at ˆ 2 + (σ+sε)ˆ ˆ 2 − s (∂3 L)h e1 = 0 , ∂3 h c0 ˆ 2 + (σ+sε)ˆ ˆ 2 + s (∂1 L)h −∂1 h e3 = 0 , c0 s s ˆ2 = 0 . e1 − ∂1 eˆ3 + (∂1 L)ˆ e3 + sµh ∂3 eˆ1 − (∂3 L)ˆ c0 c0
(5.20) (5.21) (5.22)
For most inhomogeneous media it is impossible to solve these equations exactly, and only approximate solutions may be obtained. Since the wavefront is the most interesting part of a propagating ﬁeld, it makes sense to derive an approximation that focusses on the behaviour at the wavefront. It follows from Tauber’s theorem that the behavior at the wavefront is found by letting s → ∞. In that case we may neglect in Eqs. (5.20)  (5.22) the terms
126
electromagnetic rays in a twodimensional medium
without the factor s and retain the terms with the factor s. In the resulting equations we divide by s/c0 . Then we obtain ˆ 2 + c0 εˆ e1 = 0 , −(∂3 L)h ˆ 2 + c0 εˆ e3 = 0 , (∂1 L)h ˆ2 = 0 . e1 + (∂1 L)ˆ e3 + c0 µ h −(∂3 L)ˆ
(5.23) (5.24) (5.25)
Away from the wavefront the approximation remains valid as long as eˆ1 , eˆ3 , ˆ 2 , and L, are slowly varying functions in space. This requires that ε, σ, and h µ, are slowly varying functions of position. The resulting approximation is called the ray approximation, and the corresponding approximate solutions of the form of Eqs. (5.17)  (5.19) are called electromagnetic rays. In each point of a domain where the ray approximation is valid, Eqs. (5.23)  (5.25) ˆ 2 , in which ∂1 L and ∂3 L have to are linear algebraic equations for eˆ1 , eˆ3 , and h be determined in such a way that nonzero solutions exist. In order to derive the condition to be satisﬁed by the partial derivatives of L, we eliminate eˆ1 and eˆ3 . Substituting Eqs. (5.23) and (5.24) into (5.25) yields ˆ2 = 0 . [(∂1 L)2 + (∂3 L)2 − c20 εµ]h
(5.26)
ˆ 2 = 0, L has to satisfy the nonlinear partial For a nonzero solution, h diﬀerential equation (∂1 L)2 + (∂3 L)2 = c20 εµ ,
(5.27)
which is known as the eikonal equation. Uniform electromagnetic rays For the interpretation of the electromagnetic rays of the form of Eqs. ˆ 2 , and L, are (5.17)  (5.19), we recall that we have assumed that eˆ1 , eˆ3 , h slowly varying functions in space. Then, the electromagnetic ﬁeld of the form of Eqs. (5.17)  (5.19) can locally be considered as a plane wave, of which the amplitude and propagation direction are slowly varying in space. As in the case of uniform plane waves, for general uniform rays the equation L(x1 , x3 ) =t c0
(5.28)
127
parallel polarization
deﬁnes the wavefront at time instant t, and connection of corresponding points at consecutive wavefronts yields a ray trajectory. In inhomogeneous media, the wavefronts and ray trajectories are in general curved (see Fig. 5.2). As long as the medium is isotropic, the wavefronts and the ray trajectories are perpendicular to each other. In each point the tangent to a ray trajectory coincides with the unit vector s in the propagation direction. Since ∇ L = ∂1 Li1 + ∂3 Li3 is a vector orthogonal to the surface L = constant, the unit vector s = s1 i1 + s3 i3 along the ray and in the direction of propagation is directed along the vector ∇ L. Therefore, we write ∂1 L = ns1 ∂3 L = ns3
s1 = n−1 ∂1 L ,
or
−1
or
L(x1 , x3 ) =t c0
s3 = n
(5.29)
∂3 L ,
(5.30)
L(x1 , x3 ) = t(3) c 0 (2)
L(x1 , x3 ) = t(1) c0 s(1) 1
(2) s
s(3) q
Ray trajectory
Wavefronts
Figure 5.2. A sequence of wavefronts and a ray trajectory for a uniform electromagnetic wave in an inhomogeneous medium (t(1) < t(2) < t(3) ).
128
electromagnetic rays in a twodimensional medium
in which n = n(x1 , x3 ) is the index of refraction. Using Eqs. (5.29)  (5.30), we rewrite Eqs. (5.23)  (5.24) as ˆ 2 + c0 εˆ −ns3 h e1 = 0 , ˆ e3 = 0 , ns1 h2 + c0 εˆ
(5.31)
ˆ =0, s1 eˆ1 + s3 eˆ3 = s · e
(5.33)
(5.32)
so that while from Eq. (5.27) it follows that 1
n = c0 (εµ) 2 .
(5.34)
The index of refraction n is independent of the propagation direction s of the particular ray. This is a consequence of the isotropic character of the medium. Equations (5.31)  (5.33) have the same form as Eqs. (4.33)  (4.35). ˆ 2 , s and n This leads to the conclusion that the relations between eˆ1 , eˆ3 , h are identical to the ones pertaining to the plane wave. This fact is recognized by stating that an electromagnetic ray behaves locally as a plane wave.
5.3.
Perpendicular polarization
In this section we consider electromagnetic waves in an inhomogeneous medium with ε = ε(x1 , x3 ), σ = σ(x1 , x3 ), and µ = µ(x1 , x3 ). Assuming that the sources of the electromagnetic ﬁeld are also independent of x2 , the perpendicularly polarized ﬁeld forms an independent waveﬁeld constituent, ˆ2 , H ˆ 1 and H ˆ 3 are the nonzero electromagnetic ﬁeld components. for which E In a sourcefree domain they satisfy the ﬁeld equations (cf. Eqs. (4.7)  (4.9)) ˆ1 = 0 , ˆ2 + sµH −∂3 E ˆ3 = 0 , ˆ2 + sµH ∂1 E ˆ2 = 0 . ˆ 1 − ∂1 H ˆ 3 ) + (σ+sε)E −(∂3 H
(5.35) (5.36) (5.37)
In analogy with Eqs. (5.4) and (5.5), we now investigate a solution of the form
perpendicular polarization
ˆ2 = eˆ2 exp[− s L(x1 , x3 )] , E c0 s ˆ ˆ H1 = h1 exp[− L(x1 , x3 )] , c0 ˆ 3 exp[− s L(x1 , x3 )] . ˆ3 = h H c0
129
(5.38) (5.39) (5.40)
In the inhomogeneous situation, the eikonal L = L(x1 , x3 ) is a more intricate function of position than in the homogeneous case. Moreover, the amplitudes ˆ1 = h ˆ 1 (x1 , x3 , s), and h ˆ3 = h ˆ 3 (x1 , x3 , s), will now also eˆ2 = eˆ2 (x1 , x3 , s), h depend on the position. In order to investigate the spatial behavior of eˆ2 , ˆ 1, h ˆ 3 , and L, we substitute Eqs. (5.38)  (5.40) into Eqs. (5.35)  (5.37) and h divide by the nonzero factor exp(− cs0 L). We arrive at s ˆ1 = 0 , (∂3 L)ˆ e2 + sµh c0 s ˆ3 = 0 , e2 + sµh ∂1 eˆ2 − (∂1 L)ˆ c0 ˆ 1 + ∂1 h ˆ 3 + (σ+sε)ˆ ˆ 1 + s (∂3 L)h ˆ 3 − s (∂1 L)h −∂3 h e2 = 0 . c0 c0 −∂3 eˆ2 +
(5.41) (5.42) (5.43)
For most inhomogeneous media it is impossible to solve these equations exactly, and only approximate solutions may be obtained. Since the wavefront is the most interesting part of a propagating ﬁeld, it makes sense to derive an approximation that focusses on the bahaviour at the wavefront. It follows from Tauber’s theorem that the behavior at the wavefront is found by letting s → ∞. In that case we may neglect in Eqs. (5.41)  (5.43) the terms without the factor s and retain the terms with the factor s. In the resulting equations we divide by s/c0 . Then we obtain ˆ1 = 0 , e2 + c0 µ h (∂3 L)ˆ ˆ3 = 0 , −(∂1 L)ˆ e2 + c0 µ h ˆ 1 − (∂1 L)h ˆ 3 + c0 εˆ e2 = 0 . (∂3 L)h
(5.44) (5.45) (5.46)
ˆ 1, Away from the wavefront the approximation remains valid as long as eˆ2 , h ˆ 3 , and L, are slowly varying functions in space. This requires that ε, σ, and h µ, are slowly varying functions of position. The resulting approximation is called the ray approximation, and the corresponding approximate solutions of the form of Eqs. (5.38)  (5.40) are called electromagnetic rays. In each
130
electromagnetic rays in a twodimensional medium
point of a domain where the ray approximation is valid, Eqs. (5.44)  (5.46) ˆ 1 , and h ˆ 3 , in which ∂1 L and ∂3 L are linear algebraic equations for eˆ2 , h have to be determined in such a way that nonzero solutions exist. In order to derive the condition to be satisﬁed by the partial derivatives of L, we ˆ 3 . Substituting Eqs. (5.44) and (5.45) into (5.46) yields ˆ 1 and h eliminate h e2 = 0 . [(∂1 L)2 + (∂3 L)2 − c20 εµ]ˆ
(5.47)
For a nonzero solution, eˆ2 = 0, L has to satisfy the nonlinear partial diﬀerential equation
(∂1 L)2 + (∂3 L)2 = c20 εµ ,
(5.48)
which is known as the eikonal equation. Uniform electromagnetic rays For the interpretation of the electromagnetic rays of the form of Eqs. ˆ 1, h ˆ 3 , and L, are (5.38)  (5.40), we recall that we have assumed that eˆ2 , h slowly varying functions in space. Then, the electromagnetic ﬁeld of the form of Eqs. (5.38)  (5.40) can locally be considered as a plane wave, of which the amplitude and propagation direction are slowly varying in space. As in the case of uniform plane waves, for general uniform rays the equation L(x1 , x3 ) =t c0
(5.49)
deﬁnes the wavefront at time instant t, and connection of corresponding points at consecutive wavefronts yields a ray trajectory. In inhomogeneous media, the wavefronts and ray trajectories are in general curved (see Fig. 5.2). As long as the medium is isotropic, the wavefronts and the ray trajectories are perpendicular to each other. In each point the tangent to a ray trajectory coincides with the unit vector s in the propagation direction. Since ∇ L = ∂1 Li1 + ∂3 Li3 is a vector orthogonal to the surface L = constant, the unit vector s = s1 i1 + s3 i3 along the ray and in the direction of propagation is directed along the vector ∇ L. Therefore, we write
131
ray trajectories
∂1 L = ns1 ∂3 L = ns3
or or
s1 = n−1 ∂1 L , −1
s3 = n
∂3 L ,
(5.50) (5.51)
in which n = n(x1 , x3 ) is the index of refraction. Using Eqs. (5.50)  (5.51), we rewrite Eqs. (5.44)  (5.45) as ˆ1 = 0 , ns3 eˆ2 + c0 µh ˆ3 = 0 , −ns1 eˆ2 + c0 µh
(5.52)
ˆ =0, ˆ 1 + s3 h ˆ3 = s · h s1 h
(5.54)
(5.53)
so that
while from Eq. (5.48) it follows that
1
n = c0 (εµ) 2 .
(5.55)
The index of refraction n is independent of the propagation direction s of the particular ray. This is a consequence of the isotropic character of the medium. Equations (5.52)  (5.54) have the same form as Eqs. (4.41)  (4.43). ˆ 1, h ˆ 3 , s and n This leads to the conclusion that the relations between eˆ2 , h are identical to the ones pertaining to the plane wave. This fact is recognized by stating that an electromagnetic ray behaves locally as a plane wave.
5.4.
Ray trajectories
By deﬁnition, a ray trajectory is a curve for which its tangent coincides with the unit vector s in the propagation direction of the uniform electromagnetic ray (see Fig. 5.3). The trajectory comes into being by following the ray in its propagation direction. Let x = x(l) = x1 (l)i1 + x3 (l)i3
(5.56)
132
electromagnetic rays in a twodimensional medium
the parametric representation of an electromagnetic ray, in which the para1 meter l is the arclength along the ray (dl = [(dx1 )2 + (dx3 )2 ] 2 > 0). Then, the unit vector τ along the tangent of the curve of Eq. (5.56) is given by τ = ∂l x = ∂l x1 i1 + ∂l x3 i3 .
(5.57)
Since Eq. (5.56) represents the ray trajectory, we simply have s = τ , hence, s1 = ∂l x1 ,
s3 = ∂l x3 .
(5.58)
With either Eqs. (5.29)  (5.30) or Eqs. (5.50)  (5.51) we ﬁnd n ∂l x1 (l) = ∂1 L ,
(5.59)
n ∂l x3 (l) = ∂3 L .
(5.60)
Diﬀerentiation of these equations with respect to l, while using the relation, ∂l L = (∂l x1 )∂1 L + (∂l x3 )∂3 L = s1 ∂1 L + s3 ∂3 L = n(s1 s1 + s3 s3 ) = n ,
(5.61)
τ =s
........ ...................... ................................................................ ................. .......... ........... .......... . ........ . . . . . . . . ....... .. ........ ...... .. ....... . ...... . . . . . . . ..... . ..... . . . . . ..... . . . .... . ..... . . . . . . ... .... . . . . . . . .... . .... . . .... . . . . . .... ... . .... . . . . . .... ..... . ... . . . . .... ...... . . . ... . . . ...... ... . . . . ... .......... ... . . . . . . ... ..... ..... ... ....... .... ...... ..... ... .... ... .. ...... .... ... .. ...... .... . . . . . . . . . ... . .. .. ... .... ...... .. .... ... ..... . . . . . . ... . .. . . . ... . . . ... ... . . . . ... ... .. . . . ... . . . . . ... . . ...... . ... ... . ...... . . . ... ...... .. ... . ... . ...... . ...... .... . ... . ....... . .. . . . . . .. .......... . . . . ...... .. ... ...... . . . . ...... . . . . .... .. ... ... ... ... ... ... ... . . .. ... ... .. ... . . .
t
L = L(xQ )
x
Q t
L = L(xP )
tP
t
O
Figure 5.3. The ray trajectory of a uniform electromagnetic ray.
133
ray trajectories
which follows from either Eqs. (5.29)  (5.30) or Eqs. (5.50)  (5.51), yields
∂l [n∂l x1 (l)] = ∂1 n ,
(5.62)
∂l [n∂l x3 (l)] = ∂3 n .
(5.63)
This is the diﬀerential equation of the electromagnetic ray trajectory. The coeﬃcients in this equation depend only on the spatially dependent index of refraction. By speciﬁcation of the starting position and the starting direction, this diﬀerential equation of the second order prescribes the ray trajectory completely. The ray trajectory can be determined analytically, numerically or graphically. Finally, we remark that in a homogeneous domain the index of refraction is constant, and, hence, ∂l ∂l x1 (l) = 0 ,
(5.64)
∂l ∂l x3 (l) = 0 ,
(5.65)
x1 (l) = x1 (0) + s1 (0)l ,
(5.66)
x3 (l) = x3 (0) + s3 (0)l ,
(5.67)
with the solution
in which x0 = x1 (0)i1 + x3 (0)i3 is the starting position of the ray and s0 = s1 (0)i1 + s3 (0)i3 is the starting direction of the ray. Equations (5.66)  (5.67) represent a straight line. Hence, in a homogeneous domain, the ray trajectories of uniform, electromagnetic rays are straight lines.
5.4.1.
Ray trajectories in a horizontally layered medium
In a horizontally layered medium, the medium parameters vary only in the vertical direction i3 , hence, ε = ε(x3 ), µ = µ(x3 ), n = n(x3 ) .
(5.68)
For a horizontally layered medium we have ∂1 n = 0 ,
(5.69)
134
electromagnetic rays in a twodimensional medium
and Eq. (5.62) becomes ∂l [n∂l x1 (l)] = 0 ,
(5.70)
n ∂l x1 (l) = constant
(5.71)
or equivalently along the ray trajectory. Since ∂l x1 = s1 , we may write n s1 = constant
(5.72)
along the ray trajectory. Let us now introduce θ as the angle between i3 and s, and θ0 as the angle between i3 and s0 (see Fig. 5.4), then s1 = sin(θ), and we obtain n(x3 ) sin(θ) = C0 ,
(5.73)
i.3
.... ....... .... ... 1 .... ....... ........ 2 .. ... ...... . . ... ................................................................... . . . . . . . . . . . . . . . . . . . . . . . . . . .... . . . . ............. ............ . . . . . . . . .......... . . .......... . . . . . . . . ............. ........ ... ................................. ... ....... . .......... . . . ........ . . . . . ... ..... . . . . . . . ...... . . . . .... . . .... ... ................. ........... ................ ... .......... . . . . . . .. ...... ...... ...... ...... ..... . . . . ...... ...... ..... ...... ..... . . . . ..... ..... ..... .....
i.3
i3
θ0
u
θ
s
θ= π shor
u
s0
u
x3
x3;hor
x3;0 i.3 O
.... ....... ..... ... .. ... .......................................... ....
r
i1
Figure 5.4. The ray trajectory of a uniform electromagnetic ray in a horizontally layered medium.
135
ray trajectories
in which C0 = n(x3;0 ) sin(θ0 ) ,
(5.74)
denotes the trajectory constant of the ray. Equation (5.73) is known as Snell’s law for a horizontally layered medium. A second result follows from Eq. (5.63) as ∂l [n cos(θ)] = ∂3 n ,
(5.75)
where we have used that ∂l x3 (l) = s3 = cos(θ). In Figs. 5.5 – 5.7, a number of trajectories of uniform electromagnetic rays have been plotted. These trajectories have been computed, by a numerical solution of Eqs. (5.73) and (5.75). For a chosen value of the starting position and the starting direction, the trajectory can have a point where θ = 12 π; in such a point the trajectory has a horizontal tangent. Let x3 = x3;hor the value of the vertical coordinate, where the tangent is in the horizontal direction (see Fig. 5.4), then substitution of θ = 12 π and x3 = x3;hor into Eq. (5.73), with use of Eq. (5.74), yields n(x3;hor ) = n(x3;0 ) sin(θ0 ) = C0 .
(5.76)
Since 0 ≤ sin(θ0 ) ≤ 1, Eq. (5.76) is only valid if n(x3;hor ) ≤ n(x3;0 ). After passing the point with horizontal tangent, the ray bends in the direction of increasing n. According to Eq. (5.73) the angle θ can keep the value of 12 π and the vertical coordinate x3 can keep the value x3;hor , but Eq. (5.75) shows that it is only possible if the level x3 = x3;hor coincides with the vertical level where ∂3 n = 0. Such a level can only be reached asymptotically, unless the trajectory starts in a horizontal direction at this level. In Fig. 5.7, this last situation is illustrated for a Gaussian proﬁle of the refraction index. The theory of electromagnetic rays in a horizontally layered medium ﬁnds its application in, e.g., the analysis of propagation of radio waves in the atmosphere over such a small distance that the curvature of the atmosphere can be neglected.
136
electromagnetic rays in a twodimensional medium
x..3 4 3 2 1
.. ......... .... .. ... .. ... ... ... ... ... ... ... ... ... ... .. ... .. ... .. .. ... .. .. ... .. ... .. . ... .. ... .. .. ... . .. ... . .. ... .. ... ... ... ... . ... . . ... ... ... .. .................................................................................... ...
0
1
x..3
4 θ = 15 3 2 1
n
2
4 3 2 1
0
1
4 3 2 1
n
3 2 1
0
1
2
θ = 60 ◦
2
4
6
8
x1
10
... ........ ... ◦ .... 0 . ◦ .. .. ... ... ..... 0 .. ... ..... . .... . .. . ... . . . .... .... ... ..... .. ... ....... ... ... ..... .............................. . ........ .. ....... ................. ........... ... . . ....... ...... ... ... .... ...... ...... ...... 0 ................................................................ ............................................... ..... ... . ........ .......... . ........ . . . . . . . .................... . . . . . . ... . . . . ....... ....... ..... ... ....... ..... ....... ....... ... ......... ......... ....... ..... . . . . . . . . . . ... . . . ......................................... . ...... ..... ... ....... ...... 0 ... ......... ....... ....................................... ... ... ... ... ... ... .......................................................................................................................................................................................................................................................................................................... ..
θ = 15
θ = 30
θ = 45 ◦
s
0
x..3 4
θ = 45
x..3
2
.... ........ .... .. ... .. ... ... ... ... ... ... ... ... ... ... .. ... .. ... .. .. ... .. .. ... .. ... .. . ... .. ... .. .. ... . .. ... . .. ... .. ... ... ... ... . ... . . ... ... ... .. .......................................................................................
θ = 30
s
0
x..3
... ........ ... .... .. .. ... ... ... ... ... ... .... ... .. ... .. ... .. ... .. .. ... .. .. ... .. ... .. ... .. ... .. .. ... . .. ... . .. ... .. ... ... ... ... . ... . . ... ... ... .. ..................................................................................... ..
. .......... .... .. ◦ ◦ ... 0 . .0 ◦ ....... ... .... ... ...... 0 . .. ...... . . ... . . . .. ... ....... ... .... .. ...... ............ ... . . ... ... ..... .......... ... ................. .......................... ...................... 0 ......... ... ........................... .......... ....... ....... ... .......... ...... ...... . ..... . ... . . ..... .... ... ..... .... ..... ... ..... ..... ..... ... ...... ..... . . . ...... ... . . ... ....... ... ....... .......... ................................... ... ... ... ... ... ... ......................................................................................................................................................................................................................................................................................................... ...
θ = 60 ◦
2
4
6
8
x1
10
x..3 4 3 2 1
n
.... ........ .... ◦ ◦ .. 0 0 ... . .. ◦ ....... ... .... ... ....... 0 . . .... . .. . . ... . . . . . ....... .... ... ... ...... ... . . . .. . . . ... . . . . . ..... ... ... ..... ....... ... ... ..... ......... ................................ 0 ......... ... ... ... ..... ....... ....... .... .. ... ........ ............ ... ...... . ....... ..... .. ... ..... ...... ...... ... ..... ..... . ... ... ..... ........ . ... . . . . ..... .... .. .. ... .... ... ..... ..... .. .... .... ..... ... ..... ................... ...... ...... ... ....... ........... ...... . . ........ . . ... . . . ................ .................. ... ...... ... ... ... ... ... ...........................................................................................................................................................................................................................................................................................................
0
θ = 15
θ = 30
θ = 45
θ = 60 ◦
s
2
4
6
8
x1
10
Figure 5.5. Ray trajectories for a piecewise linear proﬁle of the refraction index; n = 1 + 12 (x3 − 1) when 1 ≤ x3 < 2, n = 32 − 12 (x3 − 2) when 2 ≤ x3 ≤ 3.
137
ray trajectories
x..3 4 3 2 1
.. ......... .... .. ... .. ... ... ... ... ... ... ... ... ... .. .. ... . .. ... . .. ... . .. ... .. ... .. ... ... .. ... .. .. ... .. .. ... .. ... .. .. ... . ... ... ... ... ... ... ... .. .................................................................................... ...
0
1
x..3 4 3 2 1
n
2
4 3 2 1
0
1
4 3 2 1
n
3 2 1
0
1
2
s
2
4
6
8
x1
10
... ........ ... ◦ .... 0 .. . .. ◦ ... .. ... 0 .. ... ... . .. ... ... .. ... . .... . ... . .. ... .. .. ... ... .. ... ... .... ... ... ... ......... ... .. ... .. ....... ... ...... ... ... ... ... ... ... ... ... ... ... .......................................................................................................................................................................................................................................................................................................... ..
θ = 15
θ = 60
s
0
x..3 4
θ = 60
x..3
2
.... ........ .... .. ... .. ... ... ... ... ... ... ... ... ... .. .. ... . .. ... . .. ... . . ... .. .. ... ... ... .. .. ... .. ... .. .. ... .. ... .. .. ... . ... ... ... ... ... ... ... .. ......................................................................................
θ = 15
0
x..3
... ........ ... .... .. .. ... ... ... ... ... ... .... ... .. .. .. ... . .. ... . .. ... . . ... .. .. ... ... ... .. .. ... .. ... .. .. ... .. ... .. .. ... . ... ... ... ... ... ... ... .. .................................................................................... ..
. .......... .... ◦ .. 0 ... . ... ◦ ... ..... 0 ... ... ... ..... .... ... .. ..... . .. . . ... . . .. . ... .. ..... .. ..... ... .. .... ... .. ...... ... ........ ... ... ... ... ... ... ... ... ... ... ... ... ......................................................................................................................................................................................................................................................................................................... ...
2
4
6
8
x1
10
x..3
.... ........ .... ◦ ◦ .. 0. ... 0 . ◦ ... ... .. ... 0 ... . .. ... ... . . . . ... ... ... .. ... .. . . ... . . . .. .. .. ... ... .. .... ... .. .... .... .. ... ... .... . ... ... . . ... . . .. ... ... .... ....... ... ... .... ......... ... ... ... ...... ............................................................. ... . ....... . ... ...... ............. ...... ... ..... ... ..... ..... ... ..... ... ..... ... ..... ..... ... ..... ◦ ... . ....................................................................................................................................................0 ...........................................................................................................................................................
4 θ = 15 3 2 1
n
θ = 30
θ = 45
s
θ = 60
0
2
4
6
8
x1
10
Figure 5.6. Ray trajectories for a piecewise linear proﬁle of the refraction index; n = 32 − 12 (x3 − 1) when 1 ≤ x3 < 2, n = 1 + 12 (x3 − 2) when 2 ≤ x3 ≤ 3.
138
electromagnetic rays in a twodimensional medium
x..3 4 3 2 1
.. ......... .... .. ... .. ... .. .. ... .. ... ... ... .. .. ... .. ... .. .. ... . . ... .. .. ... .. ... .. ... ... .. ... .. .. ... .. ... .. .. ... .. .. ... .. ... .. .. ... .. ... .. .. ... ................................................................................... ...
0
1
2
x..3 4 3 2 1
n
. .......... .... .. ◦ ... .. 0 .... ... .... . . ... . ... ... ..... ... ..... ..... ... .... . . . ... . . ..... ... ...... ... ....... ........ ... .......... . . . . . . . . . ... . . . . . ..... ......................................................................................................................................................................... 0 ... ............................................................................................. ... ........... .................... ........ ............. ... .......... ....... . . . . . ... . . ...... ..... ... ..... ..... ..... ... ..... ..... ... ..... ..... ... ..... ... ..... ◦ ... .. .........................................................................................................................................................................................................................0 ........................................................................................
θ = 49
θ = 49.45 ◦
s
θ = 50
0
2
4
6
8
x1
10
Figure 5.7. Ray trajectories for a Gaussian proﬁle of the refraction index; n = 32 − 12 exp(−x3 − 22 ).
5.4.2.
Ray trajectories in a radially layered medium
In a radially layered medium, the medium parameters are only functions of the radial direction 1 r = (x21 + x23 ) 2 (5.77) in the (x1 , x3 )plane, hence, ε = ε(r), µ = µ(r), n = n(r) .
(5.78)
For a radially layered medium we have x1 ∂r n(r) , r x3 ∂3 n = ∂r n(r) , r
∂1 n =
(5.79) (5.80)
and Eqs. (5.62)  (5.63) become ∂l [n∂l x1 (l)] = ∂l [n∂l x3 (l)] =
x1 ∂r n(r) , r x3 ∂r n(r) . r
(5.81) (5.82)
139
ray trajectories
Multiplying Eq. (5.81) with x3 and Eq. (5.82) with −x1 , and adding the results, we arrive at x3 ∂l (n∂l x1 ) − x1 ∂l (n∂l x3 ) = 0 ,
(5.83)
which can be rewritten as ∂l (x3 n∂l x1 − x1 n∂l x3 ) = 0 ,
(5.84)
n(x3 ∂l x1 − x1 ∂l x3 ) = constant
(5.85)
or equivalently along the ray trajectory. Since ∂l x1 = s1 and ∂l x3 = s3 , we may write n(x3 s1 − x1 s3 ) = constant
(5.86)
along the ray trajectory. Let us now introduce φray as the angle between i3 and s, and φ as the angle between i3 and x, then s1 = sin(φray ) ,
s3 = cos(φray ) ,
(5.87)
x1 = r sin(φ) ,
x3 = r cos(φ) ,
(5.88)
and x3 s1 − x1 s3 = r sin(φray − φ) = r sin(θ) ,
(5.89)
where θ as the angle between x and s. Let us further introduce θ0 as the angle between x0 and s0 (see Fig. 5.8), we obtain r n(r) sin(θ) = C0 ,
(5.90)
C0 = r0 n(r0 ) sin(θ0 ) ,
(5.91)
in which denotes the trajectory constant of the ray. Equation (5.90) is known as Snell’s law for a radially layered medium. A second result follows from a multiplication of Eq. (5.81) with x1 and Eq. (5.82) with x3 , and adding the results, we arrive at x1 ∂l (n∂l x1 ) + x3 ∂l (n∂l x3 ) = r∂r n(r) .
(5.92)
140
electromagnetic rays in a twodimensional medium
x r.....
..... .. ... .... ........ .. .... ........ .. ... ................. .................................................................... ........... ....... ............... . . . . . . ........ . . ......... . . ....... ...... . . ..... . . . . . . . ...... .... . ..... . . . . . . . . . . . . . . ...... . ..... ..... ....... .. ..... .............. ...... ... ....... ...... .. . ...... ..... .. .... ......... .... ..... ......... . . . . . . . . ...... ... . .... .. ..... ... ................ ..... ..... . ... . . . . .... ... .... .. .. ....... . ............................ .... ........... . ... ...... .. ....... . .. . ... ..... . . . . ... ....... ... .... . . . . . . . . . ... .... . . . . . . . ... .... . . . . . ... .... .. ... .. .... .... .... ... .. .. .... .. .... ... .. . . . . .. ... . .... ..... .. ... ... . ... ... ..... ... .. . .. . . . . . . . . ... .. .... .. ... .. .. .... ... ... .. .. . . . . ... ..... .. . .. ..... ........ .....
t
x0 r0
θ
s
xang rang
t θ= 1π 2
s0
θ0
t
r
sang
rang
r0
t
O
Figure 5.8. The ray trajectory of a uniform electromagnetic ray in a radially layered medium.
The lefthand side of Eq. (5.92) can be rewritten as ∂l (x1 n∂l x1 + x3 n∂l x3 ) − n[(∂l x1 )2 + (∂l x3 )2 ] = ∂l (x1 ns1 + x3 ns3 ) − n(s21 + s23 ) = ∂l [r n cos(θ)] − n .
(5.93)
The righthand side of Eq. (5.92) can be rewritten as r∂r n(r) = ∂r [rn(r)] − n(r) .
(5.94)
Combining the results of the last three equations, we obtain
∂l [r n(r) cos(θ)] = ∂r [rn(r)] .
(5.95)
141
ray trajectories
For a chosen value of the starting position and the starting direction, the trajectory can have a point where θ = 12 π; in such a point the trajectory has a tangent in the angular direction. Let r = rang the value of the radial coordinate, where the tangent is in the angular direction (see Fig. 5.8), then substitution of θ = 12 π and r = rang into Eq. (5.90), with use of Eq. (5.91), yields (5.96) rang n(rang ) = r0 n(r0 ) sin(θ0 ) = C0 . Since 0 ≤ sin(θ0 ) ≤ 1, Eq. (5.96) is only valid if rang n(rang ) ≤ r0 n(r0 ). This means that in the starting position the trajectory has to start in the direction of decreasing r n. After passing the point with horizontal tangent, the ray bends in the direction of increasing r n. According to Eq. (5.90) the angle θ can remain at the value of 12 π and r can remain at the value rang , but Eq. (5.95) shows that it is only possible if the value r = rang coincides with the surface where ∂r (r n) = 0. Such a surface can only be reached asymptotically, unless the trajectory starts in a horizontal direction at this surface.
1 r2 2 n(r) = n0 2 − 2 a ..... ........
n0
..... ....... .................................................... ........ .............. ........... ...... ...................................................................................................................................................................................................................................................................................... ............ ................. .... . . . . . . . . . . . ......... ...... ..... ........ ..... ..... ....... ..... .............................................................................................................................................................................................................................................................. ...... ..... ............... . ...... .... ........... ...... .... . ... . . . . .......... ...... .. ... . . . . . . . . ......... ..... ... ..................................................................................................................................................................................................................................................... . . . . . . . ........ ... .................. ..... .. . . . . . . . . ... . . . . .............. ....... ..... ............ .. . ... . . . . . ....... ..... ........... . .. . . . . . . . . . . . . . . . . ................................................................................. ................................................................................................................................................................... .......... . ......................... ......... ........... ...... ..... .. .................... ......... ...... ... .. ... ................. . ........ ...... .. .. . . . . . . . . . . . . . . . . . . . . . ............. ....... ..... .. ... .................................................................................................................................................................................................................................................................................................... .................................................................................................... . ..................................... .... ................................ ... ..................................................................................................... .................................................................................................................................................................................................................................................................................................... ....... .................. .... ... . . . . . . . . ... . . . . . . .... ....... ... ... .......... . . . . . . . . . ... . . . . . . . . . . . . . . . . ............. .... ... . ...... .. ........................ .......... ........... ...... .... ............................................................................................................................................................................................................................................................ .......... . . .... ... ...... ........... . . . . . . . . . . . . . . . . ... ..... . .. ... ....... .... .............. ... ... .... ........ ................ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ................................................................................................................................................................................................................................. .. ...... ..... ... ......... ... ..... .......... ... ... ...... ........... .... ... ...... .............. ...... .... . . . . . .............................................................................................................................................................................................................................................................. . . . . . ..... .... ....... ...... ..... ........ ...... ......... ...... ... ............ ....... ...................................................................................................................................................................................................................................................................................... ....... . . . . . . ........... . ...................................................
q
Or
Figure 5.9. Luneberg’s lens of radius a.
142
electromagnetic rays in a twodimensional medium
The theory of electromagnetic rays in a radially layered medium ﬁnds its application in, e.g., the analysis of propagation of radio waves in the atmosphere over such a distance that the curvature of the atmosphere can be assumed to be constant. An interesting application is the lens of Luneberg. In our twodimensional case, the lens is a dielectric cylinder that is radially layered. The index of refraction increases from n0 , the index of refraction of the embedding, to √ 2 n0 . In Fig. 5.9, some ray trajectories have been plotted. These trajectories have been computed by a numerical solution of Eqs. (5.90) and (5.95). The lens of Luneberg focusses a beam of parallel rays in one point at the surface of the lens. Conversely, this lens is used to make a beam of parallel rays from a concentrated source located in a point at the surface of the lens. The location of the source is important. This is illustrated in Fig. 5.10, where the source is located at a distance 14 a from the surface of the lens. We then observe that the beam of parallel outgoing rays is lost.
1 r2 2 n(r) = n0 2 − 2 a ...... .......
n0
..... ....... ..................................................... ........... ........ .............. ........ .................. .......................................................... . . . . . . . . ............ ...... .................. ......... ..... ........ ......... ....... ......... ........... ....... .... ........ ..... ................................................. ............................................. .............. ...... .... . . . . . . . ........... ...... ... ................ .... .......... ..... ... ............. ......... ..... .. ... .................................................................... .................. ........ ........ ................. ....................................................................... . . . . . . . . . . ....... . . . .... ............. . ........................................... . . . . . . . . . . . . . . . . . . . . . . . . . . ...... ......... ........... ........................................... ...................................................................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ......................... ......... ......... ......... ................. ......................................................... . . . . . . . . . . . . . . . . . ... .. ... . . . . . . . . . . . . . . . . . . . . . . . . . . ................. . ................ ..................... .............. .............. ................... ... ................................................................................................................................ ............ ......... ..... .... ........................................................................................................... ........... ................ ................................................. ........................................................................................................ ............................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ....................... ...................................... . ..................... .......... ..... ............... ................... ... ... ........ .................................................................................................................................................................................................................................................................................. .......................................................................................................................................................... ........ . . . . . .......... .......... . . . . ... ................................................................................. ........................................................................ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. ......................................................................................................... .................... ................................................. ........... ...... .... .... ............................................................................................... ............ .................................. . . ................................................................................................................................. . . . . . . . . . . . . . ... ... . ....................... ........ ....... ..... .................. ..................... ........................................................................... ......... ........... .......... ................................ ................................................... ..................................................................... . . . . . . . . . . . . . . . .. ........................................ ... ........... ............. .................... ....... ......... ... ............ ............. ............................ ...... . . ............... ............. ....... ....... ....... ..................... .......................................................................................................... ........ ........ ............. ........ . . . ..... ... . . . . . ..................... ..... . . .... .... ....... . . . . . ...................... . . . . . . . . . . . . . . .. ........ .............................. ................................... ...... .... .......... ........ .... ....... ......... ........... .......... ........ ..... ................. .......... ...... ...... ............................................................. . . . . . ....... .. ......... ....... ............ ......... ................................................
q
Or
Figure 5.10. Source in front of the Luneberg’s lens.
s source
exercises and problems
5.5.
143
Exercises and problems
Exercise 5.1 Given the proﬁle for the index of refraction of a piecewise linear medium, n = 2 − 12 (x3 − 1) when 1 ≤ x3 ≤ 3. Sketch in Fig. 5.11, where the ‘bullets’ denote the starting positions, the ray trajectories for the starting angles θ0 = 45◦ , 60◦ , 90◦ and 120◦ . Whenever relevant, calculate the value of x3;hor where the ray has a horizontal tangent and the angle θexit with which a ray enters one of the homogeneous halfspaces. Exercise 5.2 √ Given the starting position x3;0 = 2 2 − 1 in a medium with the proﬁle indicated in Fig. 5.5, what is the range of starting angles for which the ray trajectories describe undulating paths that are conﬁned to the region 1 < x3 < 3? Exercise 5.3 Given a medium with the proﬁle indicated in Fig. 5.5. Does each starting position within the region 1 < x3 < 3 possess an associated range of starting angles such that the ray trajectories are conﬁned to the range 1 < x3 < 3? If so, give the range of starting angles for all starting positions, if not, explain your answer. Exercise 5.4 For a medium with a proﬁle as given in Fig. 5.6, ﬁnd the starting positions within the region 1 < x3 < 3 for which there exist rays with a horizontal tangent at some value of x3 = x3;hor . Exercise 5.5 For a medium with the proﬁle indicated in Fig. 5.6, ﬁnd the starting angle belonging to each starting position within the region 1 < x3;0 < 3 for which there exists a ray that reaches a horizontal tangent asymptotically at x3 = 2.
Problem 5.1 Given the proﬁle for the index of refraction of a piecewise radially layered medium, n(r) = 3 − r when 1 ≤ r ≤ 2. Consider a ray that comes from the region r < 1 and enters the region 1 ≤ r ≤ 2 with a starting angle θ0 . Give the range of starting angles for which the ray remains trapped.
144
electromagnetic rays in a twodimensional medium
x..3 4 3 2 1
.. ......... .... .. ... .. ... ... ... ... ... ... ... ... ... .... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .... ... ... ... ... ... .. ..................................................................................... ...
0
1
x..3 4 3 2 1
n
2
4 3 2 1
0
1
4 3 2 1
n
3 2 1
0
1
2
4
6
8
x1
10
... ........ ... .... .. ... ... ... .... .. ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .......................................................................................................................................................................................................................................................................................................... ..
s
0
x..3 4
2
x..3
2
.... ........ .... .. ... .. ... ... ... ... ... ... ... ... ... .... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .... ... ... ... ... ... .. ......................................................................................
s
0
x..3
... ........ ... .... .. .. ... ... ... ... ... ... .... ... .. .... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .... ... ... ... ... ... .. .................................................................................... ..
. .......... .... .. ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ......................................................................................................................................................................................................................................................................................................... ...
2
4
6
8
x1
10
x..3 4 3 2 1
n
.... ........ .... .. ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ...........................................................................................................................................................................................................................................................................................................
0
s
2
4
6
8
x1
10
Figure 5.11. Ray trajectories for a piecewise linear proﬁle of the refraction index; n = 2 − 12 (x3 − 1) when 1 ≤ x3 ≤ 3.
Chapter 6
Transmission Lines In Chapter 3 we have discussed the propagation of a transverse electromagnetic (TEM) wave along a parallelplate waveguide and we have shown that it can be seen as a transmission line. Transmission lines serve to guide the propagation of energy from one point to another, viz., from the source to the load. Some more common examples of transmission lines are a pair of parallel wires and coaxial cables. Transmission lines consisting of more than two conductors are called multiconductor transmission lines. In electronic systems they are present in the form of bundles of wires, while in power transmission systems they are present in the form of highvoltage threephase transmission lines above the earth or cables in the ground. The transmission lines, to be discussed in the present chapter, are uniform lines in the sense that the crosssection of the line does not change along the line. We shall assume that the lines are in a homogeneous medium, i.e., the medium in the domain D surrounding the conductors is homogeneous. We assume that we have a transmission line consisting of N inner conductors with outer boundaries Cn , n = 1, 2, · · · , N , and one outer conductor with inner boundary C0 , see Fig. 6.1. This is the most general representation of a multiconductor transmission line. In a particular case the outer conductor may not be present. We assume the conductors to be perfect or in other words electrically impenetrable. In this chapter we will investigate the propagation properties of transverse electromagnetic waves (TEM) waves along this transmission line.
146
transmission lines
..... ..... ..... ..... ..... . . . . .... ..... ..... ..... ..... . . . . ..... ..... ..... ..... .... . . . . ..... ..... ..... ..... .... . . . . .... ..... ..... ..... ..... . . . . ..... ........................................... ..... ..... ........... ............. ..... ................ ........ ..... ....... ..... ...... . ..... . . ...... ..... . ... . . . . ...... . . .... .. ... .. . ...... . . . . ..... . ................. .... ... . ..... ... . . . . ..... ... .. .................. . . . . .. . . . . . . . . . . . ..... .... .... .... ........ ... ....... ............................ ........... ..... ..... ........ ..... .... ... ..... ..... ..... ...... .... ........ ..... ..... ... ... ..... .... ..... ...... ..... . . . .... . . . . ... . . . . . . . ... ............. ... .... ... .... ... ... ... .... ......... ..... ..... ... .. ... ..... ..... .. ...... ... ... ... ..... ..... . . .... ... ... ........... ..... ..... . . . . . ..... . . . ... . . ... ... ...... .... ..... ... ... ......... ........... .... ..... ... .......... ... ..... ..... . ... . ..... ..... ............... ..... ....... ... ......................... . . . . . . . . . . . . . . ...... ..... .. ... .. .. ... ........ ... ..... .... ... ..... ..... .... .. ... ..... ... ......... .... ..... ..... ... .. ............. ......... ... ..... . ... . . . . . . . . . . . ... ... ... . ...... ..... ... ... ... .. ..... .... ... .......... ... ..... ..... .... ...... ... ..... ...... .... .. ......... .......... .......... . .... ....... ..... .. ... ..... .. ...... ..... ........ ..... .......... ...... . . . . . . ...... .... ...... ....... ..... ........ ..... ......... ...... ........... ....... . . . . . . . ..............................................
C2
i3
C0
i3 τ
τ
C1 i 3
D
τ
Figure 6.1. The multiconductor transmission line.
6.1.
TEMwaves
We start our analysis of the transmission lines with the frequencydomain ﬁeld equations in the sourcefree domain D between the conductors, viz. ˆ1 = 0 , ˆ 3 − ∂3 H ˆ 2 ) + (σ+sε)E −(∂2 H ˆ2 = 0 , ˆ 1 − ∂1 H ˆ 3 ) + (σ+sε)E −(∂3 H ˆ3 = 0 , ˆ 2 − ∂2 H ˆ 1 ) + (σ+sε)E −(∂1 H
(6.1)
ˆ1 = 0 , ˆ 3 − ∂3 E ˆ2 + sµH ∂2 E ˆ2 = 0 , ˆ 1 − ∂1 E ˆ3 + sµH ∂3 E ˆ3 = 0 . ˆ 2 − ∂2 E ˆ1 + sµH ∂1 E
(6.4)
(6.2) (6.3)
(6.5) (6.6)
147
temwaves
At the boundaries Cn , n = 0, 1, 2, · · · , N , of the conductors these equations have to be supplemented with the boundary conditions that the tangential component of the electric ﬁeld strength vanishes, viz., ˆ1 + τ2 E ˆ2 = 0 on Cn . τ1 E
(6.7)
Here, τ = {τ1 , τ2 , 0} is the unit vector tangent along the boundaries Cn . We now assume that a transverse electromagnetic (TEM) wave propagates along the transmission line in the longitudinal i3 direction. A TEM ﬁeld structure is one in which the electric and magnetic ﬁeld vectors at each point in space have no components in the direction of propagation. The electric and magnetic ﬁeld vectors at each point in space lie in a plane transverse ˆ 3 = 0. In Section 3.6, ˆ3 = 0 and H to the direction of propagation. Hence, E we have considered a special case of a TEMwave, where the ﬁeld vectors are independent of the position in the transversal plane, the (x1 , x2 )plane. In a general TEMwave, the ﬁeld vectors are not necessarily independent of position in this transversal plane. Therefore, we generalize the results of Eqs. (3.128)  (3.127) as ˆ1 = e1 (x1 , x2 ) Vˆ (x3 , s) , E ˆ2 = e2 (x1 , x2 ) Vˆ (x3 , s) , E ˆ3 = 0 , E
ˆ 3 , s) , ˆ 1 = h1 (x1 , x2 ) I(x H ˆ 3 , s) , ˆ 2 = h2 (x1 , x2 ) I(x H ˆ3 = 0 , H
(6.8) (6.9) (6.10)
(6.11) (6.12) (6.13)
in which e1 = e1 (x1 , x2 ), e2 = e2 (x1 , x2 ), h1 = h1 (x1 , x2 ) and h2 = h2 (x1 , x2 ) are real functions of the transversal coordinates x1 and x2 . Each has the dimension m−1 . Hence, the function Vˆ = Vˆ (x3 , s) is expressed in Volt ˆ 3 , s) is expressed in Amp`ere. The transversal vector functions and Iˆ = I(x e = {e1 , e2 , 0} and h = {h1 , h2 , 0} are normalized such that
(x1 ,x2 )∈D
(e × h) · i3 dA =
(x1 ,x2 )∈D
(e1 h2 − e2 h1 ) dA = 1 .
(6.14)
148
transmission lines
The TEMwave of the type represented by Eqs. (6.8)  (6.14) satisﬁes the partial diﬀerential equations (6.1)  (6.6) and the boundary conditions of Eq. (6.7). To investigate this, we substitute the representations of Eqs. (6.8)  (6.13) into Eqs. (6.1)  (6.7). This yields h2 ∂3 Iˆ + (σ+sε)e1 Vˆ −h1 ∂3 Iˆ + (σ+sε)e2 Vˆ
= 0,
(6.15)
= 0, ˆ −(∂1 h2 − ∂2 h1 ) I = 0 ,
(6.16) (6.17)
−e2 ∂3 Vˆ + sµh1 Iˆ = 0 , e1 ∂3 Vˆ + sµh2 Iˆ = 0 , (∂1 e2 − ∂2 e1 ) Vˆ = 0 .
(6.18)
(τ1 e1 + τ2 e2 )Vˆ = 0 on Cn .
(6.21)
(6.19) (6.20)
Subsequently, we apply the method of separation of variables. As Eqs. (6.15)  (6.20) have to hold for all x1 , x2 ∈ D and for all x3 , it follows that we arrive ˆ 3 , s), at a system of diﬀerential equations for Vˆ = Vˆ (x3 , s) and Iˆ = I(x ∂3 Iˆ + γY0 Vˆ = 0 , ∂3 Vˆ + γZ0 Iˆ = 0 ,
(6.22) (6.23)
with Z0 =
1 , Y0
(6.24)
while the transversal functions have to satisfy the equations −γY0 h2 + (σ+sε)e1 = 0 ,
(6.25)
γY0 h1 + (σ+sε)e2 = 0 ,
(6.26)
∂ 1 h2 − ∂ 2 h1 = 0 ,
(6.27)
γZ0 e2 + sµh1 = 0 ,
(6.28)
−γZ0 e1 + sµh2 = 0 ,
(6.29)
∂ 1 e2 − ∂ 2 e1 = 0 ,
(6.30)
149
temwaves
and the boundary conditions τ1 e1 + τ2 e2 = 0 on Cn .
(6.31)
Note that γY0 and γZ0 play the role of the constants of separation. In order that e1 and h2 satisfy Eqs. (6.25) and (6.29), and that e2 and h1 satisfy Eqs. (6.26) and (6.28), we require that 1
γ = [(σ + sε)sµ] 2
with Re(γ) ≥ 0 .
(6.32)
Then, Eqs. (6.25)  (6.26) and (6.28)  (6.29) may be replaced by
h1
σ + sε = − sµ
h2 =
σ + sε sµ
1 2
Z0 e2 ,
(6.33)
1 2
Z0 e1 .
(6.34)
Substitution of these expressions for h1 and h2 into Eq. (6.27) yields ∂ 1 e1 + ∂ 2 e2 = 0 .
(6.35)
This relation expressing that the vector ﬁeld e is divergencefree has to be satisﬁed together with Eqs. (6.30) and (6.31). Eq. (6.30) is satisﬁed identically if a twodimensional scalar potential function ϕ = ϕ(x1 , x2 ) is introduced such that
e1 = −∂1 ϕ ,
e2 = −∂2 ϕ .
(6.36)
On account of Eq. (6.35) the function ϕ has to satisfy the twodimensional Laplace equation ∂1 ∂1 ϕ + ∂2 ∂2 ϕ = 0 .
(6.37)
150
transmission lines
........................................................................... ............. .......... .......... ........ ........ ....... . . . . . . ....... ....... ...... ...... ...... . . . . . ..... .... . . ..... . . ..... .... . . . .... ... . .... . .. ... . . ... .. . . ... ... ... . ... .. . ... . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ....... ....... .... ... . ..... . . . . . . . . . . . . . . . . .... .... ..... ... ... ..... ..... ... . . ... .. . ... ... . ... . ..... .... . .. .. . .... ... ... .. .. ... . . . ... ... . . . . ... ... ..... . ..... . ... ... . . . . . . .... ..... ..... .. ... . .. . . . . . . ... . . . . . . . . . . . . . . . . . . ....... ....... .... ... ................................. ....... ................................. ....... ........ . . ... .. . ... .. . . ... .. ... ... ... ... ... ... ..... .... . ..... . . ..... ..... ..... ..... ...... ...... ...... ...... ....... ...... . . . ........ . . . ..... ......... ........... ......... ........... ............... .................................................................
D
C2
C1
C0
τ
τ
τ
Figure 6.2. Multiplyconnected crosssectional domain D of a multiconductor transmission line. The boundary of D is C0 + C1 + C2 .
The boundary conditions of Eq. (6.31) require ∂τ ϕ = 0 (see Eqs. (1.48) (1.49)), or (6.38) ϕ = constantn on Cn . The TEMwave can only exist if the twodimensional Laplace equation (6.37) supplemented with the boundary conditions of Eq. (6.38) has a nonzero solution. It can be proved that this is indeed the case if the crosssectional domain D is multiplyconnected. The boundary curve of D then consists of a number of nonintersecting curves C0 + C1 + · · · + CN , where N ≥ 1 (Fig. 6.2). Once the solution of this twodimensional potential problem has been solved, we can calculate e1 and e2 using Eq. (6.36). Subsequently, the expressions for e1 and e2 , and the expressions for h1 and h2 that follow from Eqs. (6.33)  (6.34), are substituted in the normalization integral of Eq. (6.14). This ﬁnally determines the socalled characteristic impedance Z0 of the transmission line. Once the characteristic impedance Z0 and its inverse, the characteristic admittance Y0 , have been determined, we can study the propagation properties of the TEMwave by investigating the transmissionline equations (6.22)  (6.23).
151
parallelplate waveguide
6.2.
Parallelplate waveguide
As a ﬁrst example, we solve the potential problem in the twodimensional transversal plane of a parallelplate waveguide (see Fig. 6.3). In view of the invariance of the conﬁguration and the boundary conditions in the x2 direction, we may assume that the potential function is independent of x2 . Hence, ϕ is a function of x1 only and ∂2 ϕ = 0 .
(6.39)
Then, the twodimensional Laplace equation is replaced by a onedimensional Laplace equation (6.40) ∂1 ∂1 ϕ = 0 , 0 < x1 < a , where a is the width of the parallelplate waveguide in the x1 direction. The boundary conditions become lim ϕ = 0 ,
(6.41)
lim ϕ = 1 .
(6.42)
x1 ↑a x1 ↓0
..............................................................................................
w
................................................................................................
x1 = a E
..... ........ ... .. ... .... .. .
×f............................................ H S
D i1
... .......... ... .... .............................
r
i2
O
Figure 6.3. Crosssection of a parallelplate waveguide.
x1 = 0
152
transmission lines
By convention the constants of Eq. (6.38) are simply chosen as 0 and 1, respectively. In order to satisfy Eq. (6.40), ϕ must be a linear function of x1 . With the boundary conditions of Eqs. (6.41)  (6.42) we obtain as solution x1 . (6.43) ϕ=1− a Hence, from Eq. (6.36) it follows that e1 =
1 , e2 = 0 , a
(6.44)
and from Eq. (6.33)  (6.34) we obtain
h1 = 0 ,
h2 =
σ + sε sµ
1 2
1 Z0 . a
(6.45)
Substituting Eqs. (6.44) and (6.45) in the normalization integral of Eq. (6.14) and integrating over the domain {(x1 , x2 ) ∈ D; 0 < x1 < a, 0 < x2 < w}, where w is a characteristic length in the x2 direction (see Fig. 6.3), we obtain the characteristic impedance as
Z0 =
sµ σ + sε
1
a , w
2
(6.46)
which is identical to the one of Eq. (3.140).
6.3.
Coaxial line
As a second example, we solve the potential problem in the twodimensional transversal plane of a coaxial line (see Fig. 6.4). Let the x3 axis of the Cartesian coordinate system be the axis of the coaxial line. In view of the rotational symmetry of the conﬁguration, we may assume that the potential function is a function of the radial coordinate 1
r = (x21 + x22 ) 2
(6.47)
153
coaxial line
.............................................................. ........... ......... ......... ........ ....... . ....... . . . . . . ...... ...... ...... ...... . ..... . . . ..... ... . . . . ..... .... . .... . . .... ... . . ... .. . ... . ... ... . . ... . . . . . . . . ......... ... ....... ....... ....... . ... . . . . . . . ... . ...... ....... .. . ... . . . . .. .... .. ... . ... .. . . . . .... ..... .......... ... . . . . . . . . . . . . . . . . . . . . . . . . . . ... . . . ...... ....... ... . ...... ........ . . . . . . . ... . . . . . . . .... ... ... ... .......... ..... .. . . ... . . ... ... ... . .... 1..... .. ... . .. ... ... ... .. . ... .. . ..... . . . .. . ......... ... . . . . . . .. . ... ... ............................ ... . ... . . . .. . . . . . ... .. . .... ... .. .. . . . . . . . . . . . . . . ... . ........... .. . . . ... . 2 . ... . . . . . .. .......... .... ... ... ... ..... ...... ................... ... ... ...... ......... ....... .. ... ..................... ........ .. ... .. ....... .. ... . . . . . . . . . ..... . ... ...... . ..... ... ... ....... ....... ... ... ... ....... ....... ....... ...... ... .. ... ............... ... .......... . . ... . ... ... ... ... .... ... .... .... . . ..... . ..... .... ..... ..... ...... ...... ...... ...... ....... ...... . . . . ........ . . ......... ....... ............ ......... .........................................................
D r
E
i
×f S
rO
H
a i
b
Figure 6.4. Crosssection of a coaxial line.
only. The general solution, ϕ(x1 , x2 ) = ϕ(r), of the twodimensional Laplace equation is obtained as ϕ = A + B ln(r) ,
a 0, hence  exp(−2sL/c) < 1. Then, it is allowed to expand vˆ+ (s) as Z0 VˆS (s) vˆ (s) = Z0 + ZS +
∞
2nL ) . (ΓS ΓL ) exp(−s c n=0 n
(6.91)
Applying the shift rule of the Laplace transformation, the corresponding time domain result is Z0 v (t) = Z0 + ZS +
∞
2nL ) . (ΓS ΓL ) VS (t − c n=0 n
(6.92)
Together with Eqs. (6.85) and (6.86) the transient time behavior of the propagation of waves on the transmission line is described completely. In order to interpret the diﬀerent terms of Eq. (6.85) and (6.92) we rearrange the diﬀerent terms as
V (x3 , t) =
Z0 [ Z0 + ZS
+ VS (t − +ΓL VS (t − +ΓS ΓL VS (t − +ΓL ΓS ΓL VS (t − +ΓS ΓL ΓS ΓL VS (t − +ΓL ΓS ΓL ΓS ΓL VS (t − + ···
x3 ) c 2L−x3 ) c x3 +2L ) c 4L−x3 ) c x3 +4L ) c 6L−x3 ) c
(6.93)
].
Suppose that at t = 0 the source generates a pulse VS (t). At the beginning, the wave does not know whether or not the transmission line is matched at the other end. The apparent input impedance at x3 = 0 is then equal to Z0 , the characteristic impedance of the transmission line. Hence, the voltage of the wave at the beginning of the transmission line is equal to
163
propagation properties
Z0 VS (t)/(Z0 + ZS ). This pulse travels along the transmission line with wave speed c and at t = L/c it arrives at the load. A reﬂection occurs with a reﬂection coeﬃcient ΓL . The reﬂected pulse travels with wave speed c toward the source and at t = 2L/c it arrives at the source, where a second reﬂection takes place with a reﬂection coeﬃcient ΓS . The reﬂected pulse travels again to the load with a voltage ΓS ΓL Z0 VS (t)/(Z0 + ZS ). This process is continued. In Fig. 6.8 we illustrate these phenomena by plotting the voltage V (x3 , t) at the middle, x3 = 12 L, of the transmission line as a function of time.
VS (t) .. 1 ....... ........ .. ... .. ... ... ..
0
Z0 = 50 Ω L = 0.6 m ZL = 10 Ω ZS = 150 Ω c = 3 × 108 m/s
... ....... ... ... .... .... .. .. ... ... ... .... .... .... .. .. ... .... ... ... .... .... .. .. ... .... ... .... ...................................................................... .. .. .. ... ... . ...
2 ................................................
t in (ns)
Z0 +ZS 1 Z0 ..V ( 2 L, t) ...... ......... ... .... .. ... ...
1
0
. ...... ... .... ... ... ..... .... . .. ... .... ... ... .... .... .. ... ... ... ... ... .. .... .... .... ..... .. .. .. .. .. .. .. ... .. .. .. .. ....... ...... .................................................... .... ......................................................................... .... ......................................................................... .... ............................................................................. ........................................................................................ ......................................... . .. .. ... .. .. ... .. .. .. . ... ... ... .. . . . . . ... .. ..... .. .. ... .. ... ... ... .... ................................................................................... ... .. ... ... .. .. ... ... ......
2
4
6
8
10
t (in ns)
Figure 6.8. The source voltage VS (t) and the voltage V ( 21 L, t).
164
6.5.
transmission lines
Exercises and problems
Exercise 6.1 (a) Calculate the inner radius of the outer conductor b of a lossless coaxial transmission line with Z0 = 50 Ω, ε = 1.5ε0 and an inner conductor with outer radius a = 4.11 mm. (b) Find the timeaveraged power ﬂow in the coaxial transmission line in case of a matched load. ( c ) If the coaxial transmission line has a prescribed maximum voltage of 100 V, then what is the maximum E in the cable? Exercise 6.2 For the circuit shown in Fig. 6.7, Z0 = 50Ω, ZS = 30Ω, ZL = 20Ω, L = 2 m, c = 1 × 108 m/s and VS = 1 V for 0 ≤ t ≤ 1 ns, otherwise VS = 0 V. (a) Calculate v + (t). (b) Give the expressions for V ( 12 L, t) and I( 12 L, t) and sketch them as a function of t. ( c ) Calculate the total energy WS delivered by the source. (d) Calculate the energy W + that is at the beginning transmission line. ( e ) Calculate the energy WL absorbed by the load. Exercise 6.3 For a twowire transmission line as depicted in Fig. 6.7, the following data are available: VS (t) = V0 H(t) V, H(t) denotes the unit step function starting at t = 0; Z0 = 100 Ω. The voltages at the beginning and at the end of the line are measured for the ﬁrst 5 µs, they are !
V (0, t) =
100 V, 0 < t < 4 µs, 90 V, 4 < t < 5 µs,
!
V (L, t) =
0 V, 0 < t < 2 µs, 75 V, 2 < t < 5 µs.
Find the values of V0 , ZS , ZL and the travel time T = L/c.
exercises and problems
165
Exercise 6.4 In a steadystate situation, for what value(s) of L will the transmission line terminated by a short circuit load behave like a capacitor? Conversely, for what value(s) of L will the open circuit load behave as an inductor? Exercise 6.5 The Voltage Standing Wave Ratio (VSWR) is a measure for the eﬀectiveness of the matching load at the end of a transmission line. In the steadystate situation it is found as VSWR = V max /V min = (1 + Γ)/(1 − Γ). A lossless, twowire line is terminated with a television set A giving a VSWR=3. When A is replaced by the set B, it is meaured that VSWR=1.5. Which of the two sets is better matched to the transmission line? Explain your answer. What is the amplitude of the reﬂection coeﬃcient of the set B?
Problem 6.1 Given a parallelplate transmission line, made up of perfect conductors, of width a = 1 mm and characteristic length w = 10 cm, with a perfect dielectric medium between the plates, µ = µ0 . The voltage along the line, for a uniform plane wave propagating along the line, is given by V (x3 , t) = 10 cos(3π × 108 t − 3πx3 ). Find (a) the electric ﬁeld strength E1 (x3 , t) of the wave (b) the magnetic ﬁeld strength H2 (x3 , t) of the wave ( c ) the electric current I(x3 , t) along the line (d) the power ﬂow P out (x3 , t) along the line. Problem 6.2 Consider the twowire transmission line of Fig. 6.7 but now the source is connected with the transmission line via a switch at x3 = 0. VS = 100 V, ZS = 40 Ω, Z0 = 60 Ω and T = L/c = 1 µs. If the switch is closed at t = 0 s, ﬁnd the value of ZL for the following situations (a) V ( 12 L, 1.6µs) = 75 V (b) V ( 13 L, 2.5µs) = 44 V ( c ) V ( 34 L, t → ∞) = 80 V.
166
transmission lines
Problem 6.3 A lossless transmission line is short circuited at the far end. A variable frequency voltage generator is connected as its input and the current drawn is measured. It is found that the current reaches a maximum value for f = 500 MHz while it reaches a minimum for f = 505 MHz. Determine whether the current drawn would be a maximum, minimum or neither for the following frequencies (a) f = 450 Mhz (b) f = 394 Mhz ( c ) f = 335 Mhz.
Chapter 7
Electromagnetic Waveguides At microwave and optical frequencies of the electromagnetic spectrum, the transmission of signals takes place through electromagnetic waves that carry the signal from one point to another. The simplest kind of waves in this category are the onedimensional and twodimensional plane waves. As we have seen in Chapters 3 and 4, each of these electromagnetic plane waves has a unique propagation coeﬃcient, so we can say that each of them provides a single wave channel for transmission. A similar observation can be made for a transverse electromagnetic (TEM) wave that propagates along a transmission line (see Chapter 6). Apart from the latter case, the electromagnetic wave that constitutes the transmission channel propagates in an unbounded, homogeneous isotropic medium along a straight line, while its intensity (or amplitude distribution) is uniform on the plane transverse to this line. In this chapter we discuss the problem of transmitting electromagnetic power from one point to another in such a way that the intensity of the wave is limited to a ﬁnite crosssection and that the intensity of the wave is not uniform on the plane transverse to the direction of propagation. Similar to the transmission lines, one can say that waveguides are conﬁgurations that are designed to carry energy or information, along a speciﬁc trajectory (not necessarily straight), from one point in space to another. As the shortest path joining two diﬀerent points in space is the straight line, the simplest type of waveguide is a cylindrical one, i.e., a structure that is uniform in a ﬁxed direction in space. When a waveguide is used in practice, for example
168
electromagnetic waveguides
as a transmission channel in a communication system, it will of course many times be necessary to deviate from the assumed uniformity and guide the wave along bends and corners. In this chapter we will restrict the discussion to cylindrical waveguides. The cylindrical waveguide has one direction in space in which it is uniform; this direction is called the axial or longitudinal direction. Any plane perpendicular to the axial direction will be called a transverse plane. In each transverse plane the conﬁguration is the same. As regards their properties in the transverse plane, the cylindrical waveguide can be classiﬁed into two categories: (a) closed waveguides and (b) open waveguides. A closed waveguide contains a closed cylindrical wall across which no transfer of electromagnetic energy can take place (Fig. 7.1). Energy or information is in this case carried by an electromagnetic waveﬁeld that is conﬁned to the interior region. In view of the manufacturing process, there is a lower bound on its crosssectional dimensions and this puts an upper limit on the frequencies of operation of the electromagnetic waves that can be sent along the waveguides of the closed type. In practice, waveguides of the closed type are used up to frequencies in the gigahertz region and the walls are made of highly conducting metal. Some examples of closed waveguides are shown in Fig. 7.2.
axial direction exterior domain .
electromagnetically impenetrable wall interior domain .
transverse plane
Figure 7.1. Closed waveguide.
169
electromagnetic waveguides
metallic wall . .
metallic wall (a)
(b)
.
metallic outer conductor metallic inner conductor
(c)
Figure 7.2. Closed waveguides: (a) waveguide with rectangular crosssection; (b) waveguide with circular crosssection; (c) coaxial waveguide.
In an optical communication system, the frequencies of operation are an order of magnitude higher and hence the use of closed waveguides is out of the question. For this reason, open waveguides should be used as waveguiding structures. Since the electromagnetic ﬁeld in the crosssection extends to inﬁnity, it should be operated in such a mode that the electromagnetic ﬁeld in the transverse plane shows an exponential decay away from the crosssectional structure. There are no fundamental lower limits due to the manufacturing process on the crosssectional dimensional sizes and many types are made of readily available material (glass). Fig. 7.3 shows the basic conﬁguration of an open waveguide. To transmit signals in the range of optical frequencies over long distances from one point to another, one usually employs a uniformly cylindrical, open waveguide of bounded crosssection. Such a waveguide is commonly designated as an optical ﬁber. The free space surrounding the ﬁber forms an unbounded part of its crosssection. If the electromagnetic properties of the bounded part of its crosssection vary continuously with position in the crosssection, we speak of a gradedindex ﬁber,
170
electromagnetic waveguides
axial direction
inhomogeneously distributed penetrable material
.
.
transverse plane
Figure 7.3. Open waveguide.
axis
axis mantle core
ﬁber
.
. .
(a)
(b)
Figure 7.4. Open waveguides: (a) Circularly cylindrical gradedindex ﬁber; (b) circularly cylindrical multistepindex ﬁber.
171
electromagnetic waveguides
free space
free space
layer
layers
.
. .
.
.
substrate
substrate
(a)
(b)
Figure 7.5. Open waveguides: (a) planar gradedindex ﬁber; (b) planar multistepindex ﬁber.
while, if those properties are piecewise constant, we speak of a (multi)stepindex ﬁber (Fig. 7.4). In almost all cases one aims at a conﬁguration that is rotationally symmetric and deviations from the rotational symmetry (in practice often unavoidable) are considered as imperfections. For local processing of signals in an integratedoptics system, one often employs a planar waveguide conﬁguration. This consists of a (semiinﬁnite) substrate on which a ﬁlm is deposited. A semiinﬁnite superstrate of free space completes now the open waveguide. Again, we distinguish the planar gradedindex waveguide and the planar (multi)stepindex waveguide (Fig. 7.5). In this chapter we will discuss the simplest closedwaveguiding structure, the parallelplate waveguide (see also Section 3.6), and the simplest open waveguiding structure, the dielectric slab waveguide.
172
electromagnetic waveguides
7.1.
Parallelplate waveguide
As we have seen in Section 3.6, the parallelplate waveguide consists of two parallel perfectly conducting plates a distance a apart (see Fig. 7.6). Between these plates a homogeneous and lossless medium is assumed with constitutive constants ε and µ. In contrast to the TEMwave that we discussed in Section 3.6, we will now investigate whether an electromagnetic wave of more general character can propagate in the parallelplate waveguide. We now assume that the waveﬁeld varies in the x1 direction and is independent of x2 , viz., ˆ = H(x ˆ 1 , x3 , s) . ˆ = E(x ˆ 1 , x3 , s) , H E
(7.1)
The problem is then twodimensional and in Chapter 4 we have seen that the frequencydomain ﬁeld equations (2.54)  (2.59) separate into two indepenˆ3 and ˆ1 , E dent sets of equations: one in which only the ﬁeld components E ˆ 2 occur (parallel polarization) and one in which only the ﬁeld components H ˆ 1, H ˆ 3 and E ˆ2 occur (perpendicular polarization). H In a sourcefree domain, we ﬁnd the following sets of equations (s = jω, ω ≥ 0, and σ = 0, cf. Eqs. (4.10)  (4.15)) • Parallel polarization ˆ2 = 0 , ˆ 2 + ∂3 ∂3 H ˆ 2 + ω 2 εµH ∂1 ∂1 H
(7.2)
ˆ2 , ˆ1 = −(jωε)−1 ∂3 H E ˆ2 . ˆ3 = (jωε)−1 ∂1 H E
(7.3) (7.4)
In literature we often speak in this case of Hpolarized waves or transverse magnetic waves or TMwaves. • Perpendicular polarization ˆ2 = 0 , ˆ 2 + ∂3 ∂3 E ˆ2 + ω 2 εµE ∂1 ∂1 E
(7.5)
ˆ2 , ˆ 1 = (jωµ)−1 ∂3 E H ˆ2 . ˆ 3 = −(jωµ)−1 ∂1 E H
(7.6) (7.7)
Here, we often speak in this case of Epolarized waves or transverse electric waves or TEwaves.
173
parallelplate waveguide
x1 = a ε, µ propagation direction .... ...................................................................................... ...
i1
..... ........ .... ........................... ....
r
i3
x1 = 0
O
Figure 7.6. A parallelplate waveguide.
TEwaves We will now ﬁrst discuss in detail the case of Epolarized waves or TEwaves. Since we are interested in the propagation of an electromagnetic wave in the positive x3 direction we will assume a solution of Eq. (7.5) of the type
ˆ 1 , jω) exp(−jk3 x3 ) , ˆ2 (x1 , x3 , jω) = E(x E
0 < x1 < a .
(7.8)
Substitution of Eq. (7.8) into the twodimensional Helmholtz equation (7.5) yields ˆ + k2E ˆ =0, (7.9) ∂1 ∂1 E 1 in which k12 = ω 2 εµ − k32 .
(7.10)
Equation (7.9) constitutes a secondorder diﬀerential equation with general solution ˆ exp(−jk1 x1 ) , ˆ = Aˆ exp(jk1 x1 ) + B E
(7.11)
174
electromagnetic waveguides
ˆ follow from the boundin which the frequencydependent constants Aˆ and B ary conditions at the perfectly conducting plates. At these plates we have the boundary conditions ˆ2 = 0 lim E
=⇒
ˆ2 = 0 lim E
=⇒
x1 ↑a
x1 ↓0
ˆ =0, lim E
(7.12)
ˆ =0. lim E
(7.13)
x1 ↑a x1 ↓0
From Eqs. (7.12)  (7.13) we see that ˆ exp(−jk1 a) = 0 , Aˆ exp(jk1 a) + B ˆ = 0, Aˆ + B
(7.14) (7.15)
ˆ = −Aˆ and which leads to B 2j Aˆ sin(k1 a) = 0 .
(7.16)
ˆ the latter equation can only be satisﬁed if For a nonzero value of A,
k1 a = mπ ,
for m = 0, 1, 2, · · · .
(7.17)
Substitution of this result into Eq. (7.10) leads to the dispersion relation k32 = ω 2 εµ − (
mπ 2 ) , a
m = 0, 1, 2, · · · .
(7.18)
Introducing the wavenumber 1
k = ω(εµ) 2 ,
(7.19)
we ﬁnd from Eq. (7.18) the propagation constants
k3 ≡ k3;m =
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨
mπ 2 ) k −( a 2
1 2
⎪ 1 ⎪ ⎪ 2 ⎪ mπ 2 ⎪ 2 ⎩ −j ( ) −k
a
when k ≥
mπ , a (7.20)
when k
0). The frequencydependent constants Aˆm are determined by the electromagnetic source strengths. It is noted that in contrast to the TEMwave discussed in Section 3.6, the ﬁeld distribution in the transverse plane is not uniform. For ka/π < m < ∞, we ﬁnd an inﬁnite number of nonpropagating or evanescent waves. The x3 dependence of the nonpropagating modes is given by exp(−jk3;m x3 ) = exp{−[(
1 mπ 2 ) − k 2 ] 2 x3 } . a
(7.24)
These modes decay in the positive x3 direction. From Eqs. (7.21)  (7.22) we ﬁnd for these modes the timeaveraged power ﬂow density,
1 2 Re
ˆ ×H ˆ ∗ ) · i3 (E
m
=
1 2 Re
ˆ2 H ˆ∗ −E 1
m
∗ k3;m mπ [sin( x1 )]2 exp(−2jk3;m x3 ) = 2 Re Aˆm 2 ωµ a
= 0,
(7.25)
since k3;m is imaginary. This result is typical for an evanescent waveﬁeld. TMwaves Secondly, we will discuss in detail the case of Hpolarized or TMwaves. Since we are interested in the propagation of an electromagnetic wave in the
176
electromagnetic waveguides
positive x3 direction we will assume a solution of Eq. (7.2) of the type ˆ 1 , jω) exp(−jk3 x3 ) , 0 < x1 < a . ˆ 2 (x1 , x3 , jω) = H(x H
(7.26)
Substitution of Eq. (7.26) into the twodimensional Helmholtz equation (7.2) yields ˆ + k2H ˆ =0, (7.27) ∂1 ∂1 H 1 in which k12 = ω 2 εµ − k32 .
(7.28)
Equation (7.27) constitutes a secondorder diﬀerential equation with general solution ˆ exp(−jk1 x1 ) , ˆ = Aˆ exp(jk1 x1 ) + B H
(7.29)
ˆ follow from the boundin which the frequencydependent constants Aˆ and B ary conditions at the perfectly conducting plates. At these plates we have the boundary conditions ˆ3 = 0 lim E
=⇒
ˆ3 = 0 lim E
=⇒
x1 ↑a x1 ↓0
ˆ =0, lim ∂1 H
(7.30)
ˆ =0, lim ∂1 H
(7.31)
x1 ↑a x1 ↓0
where Eq. (7.4) has been used. From Eqs. (7.30)  (7.31) we see that ˆ exp(−jk1 a) = 0 , Aˆ exp(jk1 a) − B ˆ = 0, Aˆ − B
(7.32) (7.33)
ˆ = Aˆ and which leads to B 2j Aˆ sin(k1 a) = 0 .
(7.34)
ˆ the latter equation can only be satisﬁed if For a nonzero value of A,
k1 a = mπ ,
for m = 0, 1, 2, · · · .
(7.35)
177
parallelplate waveguide
Substitution of this result into Eq. (7.28) leads to the dispersion relation mπ 2 ) , m = 0, 1, 2, · · · . (7.36) k32 = ω 2 εµ − ( a Introducing the wavenumber 1
k = ω(εµ) 2 ,
(7.37)
we ﬁnd from Eq. (7.36) the propagation constants
k3 ≡ k3;m =
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨
mπ 2 ) k −( a
1 2
2
⎪ 1 ⎪ ⎪ 2 ⎪ mπ 2 ⎪ 2 ⎩ −j ( ) −k
a
when k ≥
mπ , a (7.38)
when k
0 , βm = 0 , a
(7.50)
when ka/π < m < ∞, referring to nonpropagating modes. The expressions of Eqs. (7.49)  (7.50) can be cast into a diﬀerent form by introducing the cutoﬀ angular frequency ωc,m of the mth mode through ωc,m =
1 mπ (εµ)− 2 . a
(7.51)
179
propagation of modes in a parallelplate waveguide
Introducing of the normalized quantities ω
ωm =
ωc,m
, αm =
αm a βm a , βm = , mπ mπ
(7.52)
we rewrite Eq. (7.49) and Eq. (7.50) as 1
αm = 0 , β m = (ω 2m − 1) 2 ,
(7.53)
when 1 ≤ ω m < ∞, and 1
αm = (1 − ω 2m ) 2 ,
βm = 0 ,
(7.54)
when 0 ≤ ω m < 1. These results lead to the dispersion curve of the mth mode in a lossless parallelplate waveguide shown in Fig. 7.7.
αm βm 6
2.0
.
1.0
0
.
.
.
.
.
.
. . . .... ........................ . ... .......... .... . ........ . ....... ...... . ...... ... ..... . . ... ..... ... ...... . ..... ... . .. ... . ... .. . ... ... . ... . ... . ... .. . ... ... . ... .. . ... . .... . ... . .. . ..
0 "
1.0
#$ % ........ evanescent region
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.. ..... ... ....
.. .... ... ....
.. .... ... ....
.. .... ... ....
.. .... ... ...
. .... ... ....
αm
.... ... ....
βm
2.0
 ωm
propagation region
Figure 7.7. Dispersion diagram for the normalized attenuation and phase coeﬃcients.
180
electromagnetic waveguides
Planewave decomposition Each mode can be written as a superposition of two plane waves. For a TEm mode the electric ﬁeld strength may be written as (cf. Eq. (7.21)) ˆ2;m = Aˆm [exp(jk1;m x1 − jk3;m x3 ) − exp(−jk1;m x1 − jk3;m x3 )] , E
(7.55)
and for a TMm mode the magnetic ﬁeld strength may be written as (cf. Eq. (7.39) ˆ 2;m = Aˆm [exp(jk1;m x1 − jk3;m x3 ) + exp(−jk1;m x1 − jk3;m x3 )] , H where
(7.56)
1 mπ 2 , k3;m = [k 2 − k1;m ]2 . (7.57) a For propagating modes k3;m is realvalued and each planewave constituent of Eqs. (7.55)  (7.56) is a uniform one. The direction of propagation of such a uniform plane wave is determined by the vector {k1 , k3 } ≡ {k1;m , k3;m } for a wave propagating in the positive x1 direction and the positive x3 direction and by the vector {k1 , k3 } ≡ {−k1;m , k3;m } for a wave propagating in the negative x1 direction and the positive x3 direction. These directions of propagation can be constructed in the (k1 , k3 )plane (see Fig. 7.8). Each propagating mode becomes nonpropagating or evanescent in the x3 direction, when the operating frequency becomes below the cutoﬀ frequency for that mode. In Fig. 7.8, this phenomenon is also illustrated in the (k1 , k3 )plane. Finally, in Fig. 7.9, we present the amplitude distributions of three guided TEm modes, m = 1, 2, 3.
k1;m =
181
propagation of modes in a parallelplate waveguide
k1
..... ......... .......... ... .... .. ... ... ... ... .. ... 4π ...... ... ... a .... ... ... ... ......... .... ... . .............................................................. . . . . . . . . . . . . .... . . ........... ........ . . . . ......... . . . .... ....... . ........ ...... 3π . . . . . ... ... . . ...... ....... ... ... ........... ............ a ...... . . . ...... . . . . ... . ...... . .. ... . . . . . . . ... . . ..... . .. .... . . ..... . . . . ... . . ..... . ... . . . . . . . ... ..... . . . . ............. . ..... . . . . ... . . . .... . ... ......... . . . . . ... . . .... ..... . .. . . . . . ... . . . ... . . 2π .. ..... . .. . . . . . . . ............ ..... . . .. . . . ....... ..... . . a . . . . . ..... . ... . .. . . ... . . . . . . . . . . . . ..... . .. ... ..... .... ... ...... ... ... ..... ..... .. ... ... ... ..... ...... . .. . . ... . . . . . . . . . . ..... ... ... . . . . ... . . . . . . . . . . . . ... ..... ... ... . . .. .. .. ..... . . ... . . . . ... π . . ..... .... . . .. . . ... . . . . . . ..... ................... . . .. .... . . . . . . . . . ... . ..... . . . . . . .... . a ... ... ... . ........ . . ..... . . . ... . . . . . . . . . . . . . ... ..... ... . ...... . . . . ... . . . .. . . . . . . . . . . . . . . . ... ..... .. . ... .. ........... ..... ...... ... .... .... ... .......... .. . ..... . ..... . . . . . . . . . ... . . . . . . . . . ..... ... . ... ....... . ....... . . . . . .. . . . . . . . . . . . ..... .. .. ..... ........ .. . ...... . . . . . . . . . . . . . . . . . . . ..... .... ..... .......... .. ... .. .. ...................... . ........................................................................................................................................................................................................................................................................................................................................................................................................................ ..
evanescent modes
k3 = {k1;3 , k3;3 }
k2 = {k1;2 , k3;2 }
propagating modes
k
k1 = {k1;1 , k3;1 }
O
k3
Figure 7.8. The (k1 , k3 )plane for the lossless parallelplate waveguide.
x1 6 TE1 mode ........... .......... ........ ...... .... .. ...... ....... . . . . . . . . .... ...........
ˆ 1 , jω) E(x
x1 6 TE2 mode .............. ................ ............ ................ .................... .................. ........... . ..... .............. ...................
ˆ 1 , jω) E(x
x1 6 TE3 mode ........................... ......... .............. ............................. . . . . . . . . . . . . . . . . . . . . . . . .................. ............................. ......................... ..... ......... ...........................
ˆ 1 , jω) E(x
Figure 7.9. The amplitude distributions of three guided TEmodes.
182
electromagnetic waveguides
7.3.
Dielectric slab waveguides
In this section we will discuss the simplest openwaveguiding structure: the planarly layered, dielectric waveguide or dielectric slab waveguide. This class of waveguides plays an important role in integrated optics. The dielectric slab waveguide or thinﬁlm waveguide consists of a homogeneous dielectric slab or thin ﬁlm sandwiched between the semiinﬁnite substrate D(1) and the semiinﬁnite superstrate D(3) . The numbering of the domains is illustrated in Fig. 7.10. The nomenclature and material properties of the various domains are listed in Table 7.1. When the slab is homogeneous (the case we discuss here), we refer to the waveguide as a stepindex slab waveguide, otherwise we are dealing with a gradedindex slab waveguide. When sub and superstrate have the same electromagnetic properties, i.e., n(1) = n(3) , we have a symmetrical slab waveguide. Like before we will now investigate whether a timeharmonic electromagnetic wave can propagate in the slab waveguide. We assume that the waveﬁeld varies only in x1 direction and is independent of x2 , viz., ˆ = H(x ˆ 1 , x3 , jω) . ˆ = E(x ˆ 1 , x3 , jω) , H E
(7.58)
In Section 7.1 we have seen that two types of waves can exist independently: Epolarized waves or TEwaves (perpendicular polarization) with ﬁeld comˆ 1, H ˆ 3 and E ˆ2 that satisfy Eqs. (7.5)  (7.7), and Hpolarized waves ponents H ˆ3 and ˆ1 , E or TMwaves (parallel polarization) with the ﬁeld components E ˆ H2 that satisfy Eqs. (7.2)  (7.4).
Table 7.1. The various subdomains and their electromagnetic properties. subdomain
x1 coordinate
permittivity
permeability
refractive index
D(3) D(2) D(1)
a < x1 < ∞ 0 < x1 < a −∞ < x1 < 0
ε(3) ε(2) ε(1)
µ0 µ0 µ0
n(3) n(2) n(1)
183
dielectric slab waveguides
ε(3) , µ0
superstrate
slab
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Figure 7.10. A dielectric slab waveguide.
TEwaves We will now ﬁrst discuss in detail the case of Epolarized waves or TEwaves. We are looking for a solution of the equations, cf. Eqs. (7.5)  (7.7), ˆ 2 + ∂3 ∂3 E ˆ2 + ω 2 ε(i) µ0 E ˆ2 = 0 , in D(i) , i = 1, 2, 3 , ∂1 ∂1 E ˆ2 , ˆ 1 = (jωµ0 )−1 ∂3 E H ˆ2 , ˆ 3 = −(jωµ0 )−1 ∂1 E H
(7.59)
(7.60) (7.61)
together with the boundary conditions for the tangential ﬁeld components ˆ 3 at x1 = a and x1 = 0. We will further assume that the wave ˆ2 and H E we are looking for will propagate in the positive x3 direction and will have a nonuniform, transverse intensity distribution that is possibly restricted to the slab region. That means we will require that the waveﬁeld vanishes for
184
electromagnetic waveguides
x1  → ∞. In this respect it seems reasonable to assume a solution of the type, see Eq. (7.8), ˆ 1 , jω) exp(−jk3 x3 ) , −∞ < x1 < ∞ . ˆ2 (x1 , x3 , jω) = E(x E
(7.62)
Substitution of this representation into the twodimensional Helmholtz equation (7.59) yields ˆ + (k (i) )2 E ˆ =0, (7.63) ∂1 ∂1 E 1 in which (i)
(k1 )2 = ω 2 ε(i) µ0 − k32 , i = 1, 2, 3 .
(7.64)
Equation (7.63) constitutes a secondorder diﬀerential equation. In each subdomain D(i) , with i = 1, 2, 3, the general solution can be written as ˆ (i) exp(−jk (i) x1 ) . ˆ = Aˆ(i) exp(jk (i) x1 ) + B E 1 1
(7.65) 1
(i)
In order to satisfy the condition at x1  → ∞, k1 = (ω 2 ε(i) µ0 −k32 ) 2 is chosen (i) ˆ (1) = 0 and Aˆ(3) = 0. The such that Im(k1 ) ≤ 0, and in addition we take B (i) (i) ˆ follow from the boundary other frequencydependent constants Aˆ and B equations at x1 = a and x1 = 0, where ˆ2 = lim E ˆ2 lim E
=⇒
ˆ 3 = lim H ˆ3 lim H
=⇒
ˆ2 = lim E ˆ2 lim E
=⇒
ˆ 3 = lim H ˆ3 lim H
=⇒
x1 ↑a x1 ↑a
x1 ↓a
x1 ↓a
ˆ = lim E ˆ, lim E
(7.66)
ˆ = lim ∂1 E ˆ, lim ∂1 E
(7.67)
ˆ = lim E ˆ, lim E
(7.68)
ˆ = lim ∂1 E ˆ. lim ∂1 E
(7.69)
x1 ↑a x1 ↑a
x1 ↓a
x1 ↓a
and x1 ↓0 x1 ↓0
x1 ↑0
x1 ↑0
x1 ↓0 x1 ↓0
x1 ↑0
x1 ↑0
From Eqs. (7.66)  (7.69) we see that (2) ˆ (2) exp(−jk (2) a) = B ˆ (3) exp(−jk (3) a) , Aˆ(2) exp(jk1 a) + B 1 1 (2) (2) (2) ˆ (2) (2) (3) ˆ (3) (3) exp(−jk1 a) = −k1 B exp(−jk1 a) , k1 Aˆ(2) exp(jk1 a) − k1 B ˆ (2) = Aˆ(1) , Aˆ(2) + B (2) ˆ(2) (2) ˆ (2) (1) = k Aˆ(1) . (7.70) k A −k B 1
1
1
185
dielectric slab waveguides
ˆ (2) , B ˆ (3) and Aˆ(1) These four equations containing the four unknowns Aˆ(2) ,B have a nonzero solution only if the determinant of the matrix of their coefﬁcients is zero. In this manner we ﬁnd an equation which is known as the eigenvalue equation or characteristic equation, because it leads to speciﬁc values or eigenvalues for the still unknown propagation coeﬃcients k3 . In the lossless guiding structure, we expect those solutions of Eq. (7.70) for (1) (3) (2) which k1 and k1 are both negative imaginary and k1 is positive real. Therefore, it is advantageous to introduce the quantities (1)
κ(1) = jk1
(2)
κ(2) = k1
(3)
κ(3) = jk1
1
= [k32 − ω 2 ε(1) µ0 ] 2 > 0 , 1
= [ω 2 ε(2) µ0 − k32 ] 2 > 0 , 1
= [k32 − ω 2 ε(3) µ0 ] 2 > 0 .
(7.71)
Then, the characteristic equation which must be satisﬁed if a solution of Eq. (7.70) is to be obtained, is found as exp(jκ(2) a) (2) κ exp(jκ(2) a) 1 κ(2)
exp(−jκ(2) a) − exp(−κ(3) a) 0 (2) (2) −κ exp(−jκ a) −jκ(3) exp(−κ(3) a) 0 1 0 −1 (2) −κ 0 jκ(1)
=0.
(7.72) By straightforward algebra, Eq. (7.72) reduces to the form κ(1) κ(3) + (2) (2) (7.73) tan(κ(2) a) = κ (1)κ (3) . κ κ 1 − (2) (2) κ κ To obtain the relevant values of k3 , it is advantageous to use the addition formula tan(u) + tan(v) tan(u + v) = 1 − tan(u) tan(v) and to rewrite Eq. (7.73) as
κ(1) κ(2) a = arctan κ(2)
κ(3) + arctan κ(2)
+ mπ, m = 0, 1, 2, · · · , ∞ . (7.74)
186
electromagnetic waveguides
This equation, in which the quantities κ(i) are related to the propagation coeﬃcient k3 through Eq. (7.71), is called the dispersion equation. Its solutions k3 ≡ k3;m (ω) , m = 0, 1, 2, · · · , ∞ ,
form an innumerable set of real and complex numbers. The real values k3;m that satisfy the inequalities in Eq. (7.71) form a ﬁnite set with max[k (1) , k (3) ] < k3;m < k (2) ,
(7.75)
where 1
k (i) = ω(ε(i) µ0 ) 2 , i = 1, 2, 3 ,
(7.76)
with m = 0, 1, 2, · · · , M −1. Each value k3;m in Eq. (7.75) can be considered as the propagation constant of the mth guided mode of the thinﬁlm waveguide. The electromagnetic ﬁeld distribution corresponding to each k3;m is denoted as the TEm mode (m = 0, 1, 2, · · · , M−1) of the slab waveguide. The (i) quantities κ(i) corresponding to k3;m are denoted as κm . From Eq. (7.65) it then follows that
ˆ2;m = E
⎧ (2) (3) ⎪ cos(κm a−ψm ) exp[−κm (x1 −a)] exp(−jk3;m x3 ) ⎪ ⎪ ⎪ ⎪ when a < x1 < ∞ , ⎪ ⎪ ⎪ ⎨ (2)
(1) Aˆm cos(κm x1 −ψm ) exp(−jk3;m x3 ) cos(ψm ) ⎪ when 0 < x1 < a , ⎪ ⎪ ⎪ (1) ⎪ ⎪ cos(ψm ) exp(κm x1 ) exp(−jk3;m x3 ) ⎪ ⎪ ⎩ when − ∞ < x1 < 0 , (7.77)
in which (1)
ψm = arctan
κm
(2)
κm
.
(7.78)
dielectric slab waveguides
187
The magnetic ﬁeld strength then follows from Eqs. (7.60)  (7.61) as
ˆ 1;m = H
⎧ −k3;m (2) (3) ⎪ ⎪ cos(κm a−ψm ) exp[−κm (x1 −a)] exp(−jk3;m x3 ) ⎪ ⎪ ⎪ ωµ 0 ⎪ ⎪ ⎪ ⎪ when a < x1 < ∞ , ⎪ ⎪ ⎪ ⎪ −k 3;m (2) ⎨
(1) Aˆm cos(κm x1 −ψm ) exp(−jk3;m x3 ) ωµ0 ⎪ cos(ψm ) ⎪ ⎪ when 0 < x1 < a , ⎪ ⎪ ⎪ −k ⎪ 3;m (1) ⎪ ⎪ cos(ψm ) exp(κm x1 ) exp(−jk3;m x3 ) ⎪ ⎪ ωµ0 ⎪ ⎪ ⎩ when − ∞ < x1 < 0 , (7.79)
and
⎧ (3) ⎪ κm ⎪ (2) (3) ⎪ ⎪ cos(κm a−ψm ) exp[−κm (x1 −a)] exp(−jk3;m x3 ) ⎪ ⎪ jωµ ⎪ 0 ⎪ ⎪ ⎪ when a < x1 < ∞ , ⎪ ⎪ ⎪ (2) ⎪ ⎪ ⎨ κm (2)
(1) Aˆm sin(κm x1 −ψm ) exp(−jk3;m x3 ) jωµ0 cos(ψm ) ⎪ ⎪ ⎪ when 0 < x1 < a , ⎪ ⎪ ⎪ ⎪ (1) ⎪ ⎪ ⎪ −κm cos(ψ ) exp(κ(1) x ) exp(−jk x ) ⎪ m 1 ⎪ m 3;m 3 ⎪ jωµ0 ⎪ ⎪ ⎩ when − ∞ < x1 < 0 . (7.80) From Eqs. (7.77) and (7.80) it follows that the timeaveraged power ﬂow density in the x1 direction,
ˆ 3;m = H
1 ˆ 2 Re[(E m
∗ ˆ ∗ ) · i1 ] = 1 Re[E ˆ2;m H ˆ 3;m ×H ]=0, m 2
(7.81)
in each of the three domains. The timeaveraged power ﬂow density in the x3 direction follows from Eqs. (7.77) and (7.79) as
188
electromagnetic waveguides
1 ˆ 2 Re[(E m
ˆ ∗ ) · i3 ] = 1 Re[−E ˆ2;m H ˆ∗ ] ×H m 1;m 2
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 2 ⎪ (1) ⎨ A ˆ m = cos(ψm ) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
k3;m (2) (3) [cos(κm a−ψm )]2 exp[−2κm (x1 −a)] , ωµ0
a < x1 < ∞ ,
k3;m (2) [cos(κm x1 −ψm )]2 , ωµ0
0 < x1 < a ,
k3;m (1) [cos(ψm )]2 exp(2κm x1 ) , ωµ0
−∞ < x1 < 0 .
(7.82) From Eqs. (7.81) and (7.82) we observe that the timeaveraged power ﬂow of a guided mode is directed in the longitudinal direction and is conﬁned to (1) the slab region. The frequencydependent constant Aˆm is determined by the electromagnetic source strength. TMwaves We will now ﬁrst discuss in detail the case of Hpolarized waves or TMwaves. We are looking for a solution of the equations, cf. Eqs. (7.2)  (7.4), ˆ 2 = 0 , in D(i) , i = 1, 2, 3 , ˆ 2 + ∂3 ∂3 H ˆ 2 + ω 2 ε(i) µH ∂1 ∂1 H ˆ1 = −(jωε(i) )−1 ∂3 H ˆ 2 , i = 1, 2, 3 , E ˆ 2 , i = 1, 2, 3 , ˆ3 = (jωε(i) )−1 ∂1 H E
(7.83) (7.84) (7.85)
together with the boundary conditions for the tangential ﬁeld components ˆ 2 and E ˆ3 at x1 = a and x1 = 0. We will further assume that the wave H we are looking for will propagate in the positive x3 direction and will have a nonuniform, transverse intensity distribution that is possibly restricted to the slab region. That means we will require that the waveﬁeld vanishes for x1  → ∞. In this respect it seems reasonable to assume a solution of the type, see Eq. (7.26), ˆ 1 , jω) exp(−jk3 x3 ) , −∞ < x1 < ∞ . ˆ 2 (x1 , x3 , jω) = H(x H
(7.86)
Substitution of this representation into the twodimensional Helmholtz equation (7.83) yields ˆ + (k (i) )2 H ˆ =0, (7.87) ∂1 ∂1 H 1
189
dielectric slab waveguides
in which (i)
(k1 )2 = ω 2 ε(i) µ0 − k32 , i = 1, 2, 3 .
(7.88)
Equation (7.87) constitutes a secondorder diﬀerential equation. In each subdomain D(i) , with i = 1, 2, 3, the general solution can be written as ˆ (i) exp(−jk (i) x1 ) . ˆ = Aˆ(i) exp(jk (i) x1 ) + B H 1 1
(7.89)
1
(i)
In order to satisfy the condition at x1  → ∞, k1 = (ω 2 ε(i) µ0 −k32 ) 2 is chosen (i) ˆ (1) = 0 and Aˆ(3) = 0. The such that Im(k1 ) ≤ 0, and in addition we take B ˆ (i) follow from the boundary other frequencydependent constants Aˆ(i) and B conditions at x1 = a and x3 = 0, where ˆ 2 = lim H ˆ2 lim H
=⇒
ˆ3 = lim E ˆ3 lim E
=⇒
ˆ 2 = lim H ˆ2 lim H
=⇒
ˆ3 = lim E ˆ3 lim E
=⇒
x1 ↑a
x1 ↑a
x1 ↓a
x1 ↓a
ˆ = lim H ˆ, lim H
x1 ↑a
x1 ↓a
1
ˆ = lim 1 ∂1 H ˆ, x1 ↓a ε(3)
(7.90)
∂1 H x1 ↑a ε(2)
(7.91)
ˆ = lim H ˆ, lim H
(7.92)
lim
and x1 ↓0
x1 ↓0
x1 ↑0
x1 ↑0
x1 ↓0
lim
x1 ↓0
x1 ↑0
1 ˆ = lim 1 ∂1 H ˆ. ∂1 H x1 ↑0 ε(1) ε(2)
(7.93)
From Eqs. (7.90)  (7.93) we see that (2) ˆ (2) exp(−jk (2) a) = B ˆ (3) exp(−jk (3) a) , Aˆ(2) exp(jk1 a) + B 1 1 (2)
(2)
(3)
k1 ˆ(2) k1 ˆ (2) k1 ˆ (3) (2) (2) (3) A exp(jk1 a) − (2) B exp(−jk1 a) = − (3) B exp(−jk1 a) , (2) ε ε ε ˆ (2) = Aˆ(1) , Aˆ(2) + B (2)
(2)
(1)
k1 ˆ(1) k1 ˆ(2) k1 ˆ (2) = (1) A − (2) B A . (2) ε ε ε
(7.94)
ˆ (2) , B ˆ (3) and These four equations containing the four unknowns Aˆ(2) , B Aˆ(1) have a nonzero solution only if the determinant of the matrix of their coeﬃcients is zero. In this manner we ﬁnd an equation which is known as the eigenvalue equation or characteristic equation, because it leads to speciﬁc
190
electromagnetic waveguides
values or eigenvalues for the still unknown propagation coeﬃcients k3 . In the lossless guiding structure, we expect those solutions of Eq. (7.94) for (1) (3) (2) which k1 and k1 are both negative imaginary and k1 is positive real. Therefore, it is advantageous to introduce the quantities (1)
κ(1) = jk1
(2)
κ(2) = k1
(3)
κ(3) = jk1
1
= [k32 − ω 2 ε(1) µ0 ] 2 > 0 , 1
= [ω 2 ε(2) µ0 − k32 ] 2 > 0 , 1
= [k32 − ω 2 ε(3) µ0 ] 2 > 0 .
(7.95)
Then, the characteristic equation which must be satisﬁed if a solution of Eq. (7.94) is to be obtained, is found as exp(jκ(2) a) (2) κ (2) ε(2) exp(jκ a) 1 κ(2) (2)
ε
exp(−jκ(2) a) − exp(−κ(3) a) (2) κ jκ(3) − (2) exp(−jκ(2) a) − (3) exp(−κ(3) a) ε ε 1 0 (2) κ − (2) 0 ε
0 =0. −1 jκ(1) (1)
0
ε
(7.96)
By straightforward algebra, Eq. (7.96) reduces to the form ε(2) κ(1) ε(2) κ(3) + (3) (2) (1) (2) tan(κ(2) a) = ε κ(2) (1)ε (2)κ (3) . ε κ ε κ 1 − (1) (2) (3) (2) ε κ ε κ
(7.97)
To obtain the relevant values of k3 , it is advantageous to use the addition formula tan(u + v) =
tan(u) + tan(v) 1 − tan(u) tan(v)
and to rewrite Eq. (7.97) as
ε(2) κ(1) ε(2) κ(3) κ(2) a = arctan (1) (2) + arctan (3) (2) + mπ, m = 0, 1, 2, · · · , ∞ . ε κ ε κ (7.98)
191
dielectric slab waveguides
This equation, in which the quantities κ(i) are related to the propagation coeﬃcient k3 through Eq. (7.95), is called the dispersion equation. Its solutions k3 ≡ k3;m (ω) , m = 0, 1, 2, · · · , ∞ , form an innumerable set of real and complex numbers. The real values k3;m that satisfy the inequalities in Eq. (7.95) form a ﬁnite set with max[k (1) , k (3) ] < k3;m < k (2) , where
1
k (i) = ω(ε(i) µ0 ) 2 , i = 1, 2, 3 ,
(7.99) (7.100)
with m = 0, 1, 2, · · · , M−1. Each value k3;m in Eq. (7.99) can be considered as the propagation constant of the mth guided mode of the thinﬁlm waveguide. The electromagnetic ﬁeld distribution corresponding to each k3;m is denoted as the TMm mode (m = 0, 1, 2, · · · , M − 1) of the slab waveguide. The (i) quantities κ(i) corresponding to k3;m are denoted as κm . From Eq. (7.89) it then follows that
ˆ 2;m = H
⎧ (2) (3) ⎪ cos(κm a−ψm ) exp[−κm (x1 −a)] exp(−jk3;m x3 ) ⎪ ⎪ ⎪ ⎪ when a < x1 < ∞ , ⎪ ⎪ ⎪ ⎨ (2)
(1) Aˆm cos(κm x1 −ψm ) exp(−jk3;m x3 ) cos(ψm ) ⎪ when 0 < x1 < a , ⎪ ⎪ ⎪ (1) ⎪ ⎪ cos(ψm ) exp(κm x1 ) exp(−jk3;m x3 ) ⎪ ⎪ ⎩ when − ∞ < x1 < 0 , (7.101)
in which
ψm
(1)
ε(2) κm = arctan (1) (2) ε κm
.
(7.102)
The electric ﬁeld strength then follows from Eqs. (7.84)  (7.85) as
ˆ1;m = E
⎧ k3;m (2) (3) ⎪ ⎪ cos(κm a−ψm ) exp[−κm (x1 −a)] exp(−jk3;m x3 ) ⎪ ⎪ (3) ⎪ ωε ⎪ ⎪ ⎪ when a < x1 < ∞ , ⎪ ⎪ ⎪ ⎪ k ⎨ 3;m (2)
(1) Aˆm cos(κm x1 −ψm ) exp(−jk3;m x3 ) ωε(2) ⎪ cos(ψm ) ⎪ when 0 < x1 < a , ⎪ ⎪ ⎪ ⎪ k3;m ⎪ (1) ⎪ ⎪ cos(ψm ) exp(κm x1 ) exp(−jk3;m x3 ) ⎪ (1) ⎪ ⎪ ⎩ ωε when − ∞ < x1 < 0 , (7.103)
192
electromagnetic waveguides
and ⎧ (3) ⎪ −κm ⎪ (2) (3) ⎪ ⎪ cos(κm a−ψm ) exp[−κm (x1 −a)] exp(−jk3;m x3 ) ⎪ (3) ⎪ jωε ⎪ ⎪ ⎪ ⎪ when a < x1 < ∞ , ⎪ ⎪ ⎪ (2) ⎪ ⎪ ⎨ −κm (2)
(1) Aˆm sin(κm x1 −ψm ) exp(−jk3;m x3 ) jωε(2) cos(ψm ) ⎪ ⎪ ⎪ when 0 < x1 < a , ⎪ ⎪ ⎪ ⎪ (1) ⎪ ⎪ κm (1) ⎪ ⎪ cos(ψm ) exp(κm x1 ) exp(−jk3;m x3 ) ⎪ ⎪ (1) ⎪ jωε ⎪ ⎩ when − ∞ < x1 < 0 . (7.104) From Eqs. (7.101) and (7.104) it follows that the timeaveraged power ﬂow density in the x1 direction,
ˆ3;m = E
1 ˆ 2 Re[(E m
ˆ ∗ ) · i1 ] = 1 Re[−E ˆ3;m H ˆ∗ ] = 0 , ×H m 2;m 2
(7.105)
in each of the three domains. The timeaveraged power ﬂow density in the x3 direction follows from Eqs. (7.101) and (7.103) as 1 ˆ 2 Re[(E m
ˆ ∗ ) · i3 ] = 1 Re[E ˆ∗ ] ˆ1;m H ×H m 2;m 2
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 2⎪ (1) ⎨ A ⎪ ˆ m = cos(ψm ) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
k3;m (2) (3) [cos(κm a−ψm )]2 exp[−2κm (x1 −a)] , (3) ωε
a < x1 < ∞ ,
k3;m (2) [cos(κm x1 −ψm )]2 , ωε(2)
0 < x1 < a ,
k3;m (1) [cos(ψm )]2 exp(2κm x1 ) , ωε(1)
−∞ < x1 < 0 .
(7.106) From Eqs. (7.105) and (7.106) we observe that the timeaveraged power ﬂow of a guided mode is directed in the the longitudinal direction and is conﬁned (1) to the slab region. The frequencydependent constant Aˆm is determined by the electromagnetic source strength.
193
propagation properties of a dielectric slab waveguide
7.4.
Propagation properties of guided modes in a dielectric slab waveguide
It is noted that the propagation constant k3;m = k3;m (ω) is dependent on the angular frequency ω for both TEm and TMm modes, i.e., the guided modes in a slab waveguide like the propagating modes in a parallelplate waveguide show waveguiding dispersion or geometrical dispersion. This dispersion is illustrated in Fig. 7.11, where the eﬀective index of refraction (mode index) neﬀ ,m =
k3;m k0
(7.107)
of the mth guided mode is presented as a function of a/λ0 , in which λ0 =
2π ω , k0 = , k0 c0
(7.108)
where k0 is the freespace wavenumber. Note that the eﬀective index of a mode is located in max[n(1) , n(3) ] < neﬀ ,m < n(2) .
(7.109)
In terms of the phase velocity vφ,m of the mth guided mode vφ,m = we have neﬀ ,m =
ω k3;m c0 vφ,m
(7.110)
.
(7.111)
When the slab waveguide is used as a transmission channel, we often introduce the group velocity as vg,m =
1 . ∂ω k3;m
(7.112)
From Fig. 7.11 we observe that in general the guided (or conﬁned) modes have a cutoﬀ frequency. Assuming that n(1) > n(3) , the cutoﬀ frequencies of the guided modes follow from the condition κ(1) (ω) = 0. Substitution of k32 = ω 2 ε(1) µ0 = k02 (n(1) )2 in Eqs. (7.71) and (7.74) yields an equation from which the cutoﬀ frequencies of the TEmode can be derived as
194
electromagnetic waveguides
3.5
............................ .............................................................. .................................................... ............................................... ............................................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .............................. ...................... ........................................... ....................... ...................................... .................... ............................................ . . ..................... . . . . . . . . . . . . . . . . . . . . ... .... ....... ...... ........................ ...... ...... ....................... ...... ...... .......... ......... ...... ...... ......... ........ ......... ........................ ..... ......... . ..................... . . . . . . . . . . ..................... .. . .... .... ....... ........ .......... ......... ..... ....... .................. ........................... . . . . . . . . . . . . . . . . . . .. . .. .. ...... ...... ......... ......... ... ... ...... ...... ........ ....... ... ... ...... ..... ....... ....... ... ... ...... ...... ....... ....... .................. ..... ......... 0....... ....... . . . . . . . . . .. . ........ ...... ....... .. .. 1............ ............ ........ .... ...... ..... ... ... . . ....... ....... 2............................. ....... ....... .... ..... ... ... .................. .. ...... ... ....... . . . . . ... ... . 3 . . . . . . . . ..... ..... ...... ..... ... ... ... ... ... .... ..... ...... ..... ..... 0 ... ... ..... ....... ... .... ..... ...... ... ... ..... .... ...... ..... ... ... 1 ..... .......... ..... ........ . ... ..... . . .. ... . . . . . . . . . . . . . . ..... ..... 2 ... ... ..... ..... ... .... ..... ...... .... .... ... ... ..... ..... ... ... 3 .... ... ..... ..... ... ... ... ... ... ..... ..... ......... .. .... . . . . . . . . .... .... . . ..... ..... ... ... .. .. ... ... ..... ......... ... ..... . .. .... . . . .... .... . . . .. .. .... .... . . ... ... ... ... .... .... .. .. ... ... ... ... ... .... .. .. .. . ... ... ... .... ... ... .. .. ... ..... .... ...... .. ... . . ... .... . . . . .. .. .. .. .. .. ... .. ... ... ... ... .. .. .. ... ... ... ... .. .. .. ... .. .. ... ... .... ... .. ... ..... ... ...... .. .... . . ... .... . . . .. ... .. .. ... ..... .. .. .. .. ... ... .... .... .. .. ... ... ... .... .... .. .. ... .... .... .... ... ... .... ..... .....
n(3) = 1.0 n(2) = 3.5 n(1) = 3.2
neﬀ ,m 6
3.4
TE
TE
TM
3.3
3.2
0
TE
TM
0.5
TE
TM
1.0
TM
1.5
2.0

a/λ0
Figure 7.11. Dispersion curves for the guided modes of a GaAs ﬁlm located on a AlGaAs substrate (a = ﬁlm thickness, λ0 = 2π/k0 = wavelength in free space).
ωc,m =
c0
!
arctan
1
a[(n(2) )2 − (n(1) )2 ] 2
1
[(n(1) )2 − (n(3) )2 ] 2
&
1
[(n(2) )2 − (n(1) )2 ] 2
+ mπ .
(7.113) Substitution of k32 = ω 2 ε(1) µ0 = k02 (n(1) )2 in Eqs. (7.98) and (7.95) yields an equation from which the cutoﬀ frequencies of the TMmode can be derived as ωc,m =
c0 1
a[(n(2) )2 − (n(1) )2 ] 2
!
1
[(n(1) )2 − (n(3) )2 ] 2 ε(2) arctan 1 [(n(2) )2 − (n(1) )2 ] 2 ε(3)
&
+ mπ .
(7.114) The mode with the lowest cutoﬀ frequency (the TE0 mode) is often referred as the fundamental mode. Like for the parallelplate waveguide in Section 6.2 we show in Fig. 7.12, (2) the (k1 , k3 )plane for three diﬀerent TEmodes in a lossless, asymmetric, dielectric slab waveguide. Further it is interesting to investigate the inﬂuence of the operating frequency on the amplitude distribution of the guided mode.
195
propagation properties of a dielectric slab waveguide
(2)
k1
.. ....... ....... .. .... .. ... .... ... ... ... ... ... ... ... ... ... ... ... ... . ........................................................ . . . . . . . . . . . . . . ... . . ........... ........ . . . . ..... ......... . . . .... . . . ... ........ ..... . . . . . . ... . ....... ... .. ....... ....... ... .. ...... ...... . . . . . . ... ...... . ... . . ...... . . . . ... . . . .... . . . ............. .......... . . ... . . ... . ..... . . . . . ... . . . ..... . .. ............ . . . . ..... ... . . . .. .... ..... ........... ... .... ... ..... .... . .... . . . .... . . . ..... ... . .. .. . . . . . ... . . . ..... . . .. . ... . . . . . . . . ..... . . ... .. . . . . . . . . . ..... .. .. ........... ... . ..... . . . ......... .... . .. . ..... . . . . .. . . . . . . ... ..... . . .... . . . . . . . . . ... . . . . . . ..... ... . .... ... ..... ... .... ... ....... ... ..... ... ... ....... ... ... ... ..... ... ....... ... . .. . . . . . . ... . . . . . ..... ... .. ..... . .. . . ... . ..... . . . .... . ... . .... .. . ..... . . ... . . . ... ... . . . ..... . .... .. . . ... . . . . . . . . . . . . . . ..... . . ...................... .... . . . . . ... . . . . . . .. . . . . . . . . . . . . . . . . . . ..... .. . . .......... .... ... . . . . . . . . . . . ... . . . . . . . . ... . . . . . . . . . . ..... . ....... .. . .. ..... .... ..... ............. ................ .. .......... .. ..... .. ... ....... ................ ... ..... .. .......... .............................. .... .. .. ... ......................... ......................................................................................................................................................................................................................................................................................................................................................................................................................... ..
(2)
{k1;2 , k3;2 }
(2)
propagating modes
{k1;1 , k3;1 }
k (2)
(2)
{k1;0 , k3;0 }
O
k3
(2)
Figure 7.12. The (k1 , k3 )plane for the lossless dielectric slab waveguide.
In Fig. 7.13, we have shown the dispersion curve of a TE0 mode and its amplitude distribution for three diﬀerent values of the operating frequency. Close to cutoﬀ, a mode stretches far out in substrate and superstrate. Its eﬀective index of refraction neﬀ approaches the value n(1) . Far from cutoﬀ, the mode is entirely concentrated in the thin ﬁlm. As can be expected, now its eﬀective index neﬀ approaches the value n(2) . In Fig. 7.14, we have shown the dispersion curves of the TEm modes, m = 0, 1, 2, and their amplitude distributions for a particular frequency of operation. It is ﬁnally noted that from Eqs. (7.77) and (7.101), it can be seen that in the substrate and the superstrate a guided mode locally behaves as a nonuniform plane wave. The planes of equal phase and equal amplitude are perpendicular to each other.
196
electromagnetic waveguides
n(2)
.............................. ............................................................ ....................................... ................................. .................. ....................... . . . . . . . . . . . . . .. ........... ......... ........ ........ ....... . . . . . ...... ...... ..... ..... .... . . ... ... ... ... 0...... . ... ... ... . ... ... ... .. . ... ... .. .. ... .. ... ... .... . .. .. ... .... . ... .. .. .. . .. ...
n(3) = 1.0 n(2) = 3.5 n(1) = 3.2
neﬀ ,m 6
TE
n(1) 0
ω1
ω2
ω3

ωc,0
x1 6
x1 6
4πc0 /a
ω4
x1 6
ω
x1 6
... .... ..... .... ... .. ... ... ... .. ... ... ... ... . ... . ... ... . ... . . ... ... ..... . . . . . . . ...... ... ....... ... ....... ........... ....... ....... ..... ......... ..... . ................. . . . . . ......... . . . . ...... . . . . . . . . . . . . . . . . . . . ... . . . . ...... . ............. . . . . . . . .... . . . . . . ..... . . . . . . . . . . . . . . .... . . ..... . . . . . . ..... . . . . .... . . . . . . .... . . . . . . . . . . . . . . . . ........ . . . . ...... . . . . . ..... . . . . ..... . . . . . . . . . . . . . . . . .......... . . . . . .............. . . . . . ........ . . . . .... . . . . . . . . . . . . . . . . . ......... ... .... ......... ........ ...... ... ..... ..... ..... .. . . . . . . .. .. .. ... . . . . ... ... .... ... ... ... ... ... .. .. .... ... .. .. . . . . . .... . .. .. ..

ˆ 1 , jω1 ) E(x

ˆ 1 , jω2 ) E(x

ˆ 1 , jω3 ) E(x

n(3) .
n(2) .
n(1)
ˆ 1 , jω4 ) E(x
Figure 7.13. The inﬂuence of the operating frequency on the amplitude distribution of the TE0 mode.
.
.
197
propagation properties of a dielectric slab waveguide
n(2) neﬀ ,m 6
.............................. ............................................................ ....................................... ................................. .................. ....................... . . . . . . . . . . ....................... . . . ........ ..................... .......... .................. ................ ......... .............. ....... . . . . . . . . . . . . . . . . . .. .... ...... ........... ..... .......... ... ...... ......... ............ ........ ..... ........... ....... . ..... . . . .......... . . . . . ......... . ..... . . . .... . . . . . . . . . . . .. ..... ... ......... ...... ... ........ ...... ....... ... ...... ....... ... ..... ....... . . . . . . . 0...... . . ... ..... . ...... 1............ ...... ... ... 2.............. ... .... . ... . . . . . . .. .. ..... ... .... ..... ... ... ..... ... ... ..... ... .... .. . . . . . . .. .... ... ... ..... ... ... .... .. ... ... .. .. ... . . . . ... . .. .. ... ... ... ... ... ... ... ... ... .. . .... . . .. . ... ... .. .. ... ... ... ... ... .. . .. . . .... .. . .. ... .. ... ... .. .. .. .. .. .. ... .. . . . .. ... .. ... ... .....
n(3) = 1.0 n(2) = 3.5 n(1) = 3.2 TE
n(1) 0 ωc,0
TE
ωc,1
ωc,2
TE
4πc0 /a
ω 
x1 6
x1 6
... .... ... .... .. .. ... ... .... .. .. .. 0 1 . . . ...... . . . . . . .......... ............ . . . . . . . . . . . . . . . . . . . . . . . ........ . . . ........... . . . . . . . . . . . . . . . . . ................. . . . .......... . . . . . . . . . ................. . . . . . . . . . ... . . . . . . . . . ................ . . . . . . . . . . . . . . . . . . . . ......... . . . . . . ...... . . . . . . . . . . . . . ......... . . . . . . . . . ............ . . . . . . . . . . . . . .... ......... ............. ..... ....... ... .... .... .... .. .. ... ... .... ... .. .... . .
TE mode

ˆ 1 , jω ) E(x
TE mode

ˆ 1 , jω ) E(x
ω
x1 6
.... ... .. ... .... 2 .......... ........................ ... . . . . . . . . . . ................. . . . . . ......................... . . . . . . . . . ........................ . . . . . . . . ................. . . . . . . . . . . . . . . . . . . . . . ....................... . ...................................... . . . . . . . . . ........ ......... . . . . . . ..... . . . . ... ... ... .... .
TE mode

ˆ 1 , jω ) E(x
Figure 7.14. The amplitude distribution of three guided TEmodes at ω = ω = 2πc0 /a.
n(3) .
n
.
(2) .
n(1)
.
198
7.5.
electromagnetic waveguides
Exercises and problems
Exercise 7.1 A parallelplate waveguide is used to guide TEwaves. The conﬁguration is depicted in Fig. 7.6 with ε = 4ε0 , µ = µ0 and a = 10 cm while the frequency of operation is f = 1 GHz. (a) Find the propagating and the evanescent modes. (b) Give the expressions for the electric and magnetic ﬁeld strengths of the propagating mode(s). Now consider a perfectly conducting wall positioned at x3 = 0 inside the waveguide. ( c ) Find the total electromagnetic ﬁeld in the waveguide. Exercise 7.2 Find the maximum power that can be propagated in an airﬁlled parallelplate waveguide, with width a = 1 cm and characteristic length w = 5 cm, without causing breakdown. In air breakdown occurs at approximately E = 2 × 106 V/m. Use a safety factor of 10. Exercise 7.3 For frequencies above the cutoﬀ frequencies, the propagation constants of a parallelplate waveguide are associated with the socalled ‘guided wavefor ω > ωc,m . Find the guided wavelengths of lengths’, λg,m , as k3;m = λ2π g,m the TEm modes, m = 0, 1, 2, · · ·, with f = 300 MHz, c = 3 × 108 m/s and a = 0.75 m. Exercise 7.4 EHF electromagnetic waves, see Table 3.1, (also called millimeter waves) may be guided by dielectric slab waveguides. Consider a slab, with ε = 10ε0 , in free space. By selecting a proper thickness a it may be achieved that for any frequency of the entire EHF frequency band only the TE0 mode can propagate in the slab. What is the range of possible values of a? Exercise 7.5 Consider a symmetric dielectric slab waveguide in the following conﬁguration: Superstrate : a < x1 < ∞, ε(1) , ε(2) , Slab : −a < x1 < a, Substrate : −∞ < x1 < −a, ε(1) .
199
exercises and problems
Solve the following problems in case of TEwaves. ˆ (2) , B ˆ (3) that arise (a) Give the equations in the four unknowns Aˆ(1) , Aˆ(2) , B from application of the boundary conditions at the two interfaces. Use 1 (1) (2) the substitutions κ(1) = jk1 = [k32 − ω 2 ε(1) µ0 ] 2 > 0 and κ(2) = k1 = 1 [ω 2 ε(2) µ0 − k32 ] 2 > 0. ˆ (2) in terms of Aˆ(1) and also in terms (b) Solve the equations for Aˆ(2) and B (3) ˆ of B . ˆ (2) and ﬁnd ( c ) Use the results obtained in (b) to eliminate Aˆ(2) and B the two solutions with for each solution the corresponding relation (2) for κm a. Hint: One solution is found for m = 0, 2, 4, · · · (the even solution), while the other is found for m = 1, 3, 5, · · · (the odd solution). (d) Use the results in (b) and (c) to show that the electric ﬁeld strength is given by ⎧ (2) (1) ⎪ cos(κm a) exp[−κm (x1 − a)] exp(−jk3;m x3 ), ⎪ ⎪ ⎪ ⎪ ⎪ a < x1 < ∞, ⎪ ⎨ (2) (even) ˆ2;m = Cˆ −a < x1 < a, cos(κm x1 ) exp(−jk3;m x3 ), E m ⎪ ⎪ ⎪ (2) (1) ⎪ ⎪ cos(κm a) exp[κm (x1 + a)] exp(−jk3;m x3 ), ⎪ ⎪ ⎩
−∞ < x1 < −a,
with m = 0, 2, 4, · · ·, and
ˆ2;m = Cˆ (odd) E m
⎧ (2) (1) ⎪ sin(κm a) exp[−κm (x1 − a)] exp(−jk3;m x3 ), ⎪ ⎪ ⎪ ⎪ ⎪ a < x1 < ∞, ⎪ ⎨ (2)
sin(κm x1 ) exp(−jk3;m x3 ),
−a < x1 < a,
⎪ ⎪ ⎪ (2) (1) ⎪ ⎪ −sin(κm a) exp[κm (x1 + a)] exp(−jk3;m x3 ), ⎪ ⎪ ⎩
−∞ < x1 < −a,
(even) (odd) and Cˆm in with m = 1, 3, 5, · · ·, and express the coeﬃcients Cˆm (1) terms of Aˆm .
Problem 7.1 For a parallelplate waveguide ﬁnd the relations for the phase velocity and the group velocity, for propagating modes, and express them in terms of the actual wave speed, c, the angular radial frequency of operation, ω, and the modal cutoﬀ angular frequency, ωc,m .
200
electromagnetic waveguides
Problem 7.2 A dielectric slab is at x1 = 0 coated with a perfectly conducting foil, see Fig. 7.15. Find the solution for the electric ﬁeld strength in the conﬁguration in case of guided TEwaves in the slab. Hint: you may use your results in Exercise 7.5 to your advantage. ε(1) , µ0 x1 = a ε(2) , µ0
propagation direction ....... ................................................................ ....
i1
... .......... ... ... ....................
p
i3
x1 = 0
Figure 7.15. A coatedslab waveguide.
Problem 7.3 In this chapter we have considered one open waveguide structure, viz., the slab waveguide, and have found solutions for guided electromagnetic waves. One can also investigate whether guided wave solutions exist in a twomedia conﬁguration (see Fig. 7.16). In that case the guided wave is a surface wave that propagates along the interface between two diﬀerent dielectric media and is evanescent in both directions away from the interface. We simply assume a solution for TMwaves of the type ˆ 1 , jω) exp(−jk3 x3 ), −∞ < x1 < ∞ , ˆ 2 (x1 , x3 , jω) = H(x H and ˆ = B ˆ (1) exp(−κ(1) x1 ), 0 < x1 < ∞ , H (2) (2) ˆ ˆ H = A exp(κ x1 ), −∞ < x1 < 0 , ε(1) , µ0 ε(2) , µ
0
i1
.. .......... .... .. .....................
p
i3
x1 = 0
propagation direction ....... ................................................................ ....
Figure 7.16. A twomedia conﬁguration.
201
exercises and problems
(1)
1
(2)
1
with κ(1) = jk1 = [k32 − ω 2 ε(1) µ0 ] 2 > 0 and κ(2) = jk1 = [k32 − ω 2 ε(2) µ0 ] 2 > 0, ˆ 2 → 0 if x1  → ∞ is satisﬁed. where the condition H (a) Under what condition can there be a solution of this form? (b) For an electron plasma the permittivity has a negative sign, when the frequency is below the plasma frequency. When one of the two media ˆ 2 and draw the amplitude of is such a plasma, give the solution for H the magnetic ﬁeld as a function of x1 . ( c ) Investigate if there is a solution if ε(1) = ε(2) . If there is one, give the solution.
Chapter 8
Excitation of Twodimensional Electromagnetic Waves In Chapter 4 we have discussed the properties of twodimensional electromagnetic plane waves in a homogeneous medium. In Chapter 3 we have seen that an inﬁnite, planar, electriccurrent sheet emits a onedimensional plane wave, which is a special case of a twodimensional wave. We will discuss the generation of twodimensional waves by a planar electriccurrent sheet that has a ﬁnite dimension in the i1 direction and an inﬁnite dimension in the i2 direction. This antenna emits a twodimensional electromagnetic ﬁeld that can be composed through superposition of an inﬁnite number of diﬀerent plane waves, both of uniform and nonuniform character. We consider the cases that the electric current in the emitter ﬂows either in the i1 direction or in the i2 direction.
204
excitation of twodimensional electromagnetic waves
8.1.
The sheet emitter with a parallel electric current
A reference frame is introduced such that the sheet coincides with the plane x3 = 0 (Fig. 8.1). Let the impressed electric current be uniform as a function of x2 and ﬂow from x1 = 12 a to x1 = − 12 a, then !
Jˆ1ext =
−Iˆ∆ (x1 , s)δ(x3 ) , x1  < 12 a , x1  > 12 a ,
0,
Jˆ2ext = 0 , Jˆ3ext = 0 ,
(8.1)
where δ(x3 ) denotes the onedimensional unit impulse (Dirac distribution), and Iˆ∆ (in A/m) is the electric current per unit length (of the x2 direction). Since the electriccurrent sheet carries no magnetic current, we have ˆ ext = 0 , K ˆ ext = 0 . ˆ ext = 0 , K (8.2) K 1
2
3
The medium is assumed to be homogeneous and lossless with constitutive constants ε and µ. Since the conﬁguration is independent of x2 , the electromagnetic ﬁeld is twodimensional and in Chapter 4 we have seen that the frequencydomain ﬁeld equations (2.54)  (2.59) separate into two independent set of equations, cf. Eqs. (4.4)  (4.6) and Eqs. (4.7)  (4.9). ˆ ext and Jˆext are equal to ˆ ext , K From Eqs. (8.1)  (8.2) we conclude that K 1 3 2 zero, hence only a parallelly polarized electromagnetic ﬁeld occurs. The nonˆ1 , E ˆ3 and H ˆ 2 satisfy the following set of equations zero ﬁeld components E (s = jω, σ = 0, cf. Eqs. (4.4)  (4.6)) ! Iˆ∆ (x1 , jω)δ(x3 ) , x1  < 12 a , ˆ1 = ˆ 2 + jωεE (8.3) ∂3 H 0, x1  > 12 a , ˆ 2 + jωεE ˆ3 = 0 , −∂1 H (8.4) ˆ2 = 0 . ˆ 1 − ∂1 E ˆ3 + jωµH ∂3 E
(8.5)
In Chapter 4 we have seen that in a domain outside the source distribution, solutions of Eqs. (8.3)  (8.5) in the form of plane waves exist. We therefore assume that the general solution consists of an inﬁnite superposition of planewave constituents ⎧ 1 ∞ ⎪ ⎪ eˆ+ ⎪ 1 (k1 , jω) exp(−jk1 x1 − jk3 x3 )dk1 , x3 > 0 , ⎨ 2π k =−∞ 1 ˆ E1 (x1 , x3 , jω) = ⎪ 1 ∞ ⎪ ⎪ eˆ− (k1 , jω) exp(−jk1 x1 + jk3 x3 )dk1 , x3 < 0 , ⎩ 2π k1 =−∞ 1 (8.6)
205
sheet emitter with a parallel electric current
ε, µ
'
(................... x = 12 a
H
... 1 ... . ..... ........ .... ............................ ... ... ... ... ... .......... ....... .
O r ' ..... ....... ....... ............. . . . . ....
ε, µ
.............. ........
........... .......... ....... ....... ......
J ext
H
i1
i3
x1 = − 12 a
H
(
H
....... ....... ....... ......... ..........
x3 < 0
x3 > 0
Figure 8.1. Electriccurrent sheet with impressed current as an emitter of parallelly polarized electromagnetic waves.
⎧ 1 ∞ ⎪ ⎪ ⎪ ⎨
eˆ+ 3 (k1 , jω) exp(−jk1 x1 − jk3 x3 )dk1 , x3 > 0 , 2π k =−∞ 1 ˆ E3 (x1 , x3 , jω) = ⎪ 1 ∞ ⎪ ⎪ eˆ− (k1 , jω) exp(−jk1 x1 + jk3 x3 )dk1 , x3 < 0 , ⎩ 2π k1 =−∞ 3 (8.7) and ⎧ 1 ∞ ⎪ ⎪ ⎪ ⎨
ˆ + (k1 , jω) exp(−jk1 x1 − jk3 x3 )dk1 , x3 > 0, h 2 2π k =−∞ 1 ˆ 2 (x1 , x3 , jω) = H ∞ ⎪ 1 ⎪ ˆ − (k1 , jω) exp(−jk1 x1 + jk3 x3 )dk1 , x3 < 0, ⎪ h ⎩ 2π k1 =−∞ 2 (8.8) ⎧ ⎨
where k3 =
1
1
(ω 2 εµ − k12 ) 2 ,
k1  ≤ ω(εµ) 2 ,
⎩ −j(k 2 − ω 2 εµ) 21 , 1
k1  > ω(εµ) 2 .
1
(8.9)
Note that the deﬁnition of the square root of k3 takes care of the boundedness of the integrals as x3  → ∞. This guarantees the boundedness of the electromagnetic ﬁeld components as x3  → ∞. From Eqs. (8.3) and (8.4) it
206
excitation of twodimensional electromagnetic waves
ˆ+ ˆ− ˆ+ ˆ− ˆ− follows that the quantities eˆ+ 1, e 3, e 1 and e 3 are related to h2 and h2 as eˆ+ 1 = eˆ− 1 =
k3 ˆ + h , ωε 2 −k3 ˆ − h , ωε 2
eˆ+ 3 =
−k1 ˆ + h , ωε 2
(8.10)
−k1 ˆ − h . ωε 2
eˆ− 3 =
(8.11)
1
When x3 > 0 and k1  < ω(εµ) 2 we observe that the electromagnetic ﬁeld consists of a superposition of uniform plane waves propagating in the 1 {k1 , 0, k3 }direction, while when k1  > ω(εµ) 2 the constituents are (nonuniform) evanescent waves decaying exponentially in the x3 direction. The presence of the source distribution is accounted for by the application of the excitation conditions !
Iˆ∆ (x1 , jω) , x1  < 12 a ,
ˆ 2 (x1 , x3 , jω) − lim H ˆ 2 (x1 , x3 , jω) = lim H
x3 ↓0
x3 ↑0
x1  > 12 a , (8.12)
0,
which is a consequence of Eq. (8.3), and ˆ1 (x1 , x3 , jω) − lim E ˆ1 (x1 , x3 , jω) = 0 , lim E
x3 ↓0
x3 ↑0
−∞ < x1 < ∞ ,
(8.13)
which is a consequence of Eq. (8.5). From Eqs. (8.13) and (8.6) it directly follows that ˆ+ (8.14) eˆ− 1 =e 1 , and from Eqs. (8.10) and (8.11) it follows that ˆ+ . ˆ − = −h h 2 2
(8.15)
ˆ 2 and E ˆ3 are oddsymmetric funcWe observe that the ﬁeld components H ˆ tions of x3 , while E1 is an evensymmetric function of x3 . Application of Eqs. (8.12) and (8.15) to Eq. (8.8) leads to 1 2π
∞
ˆ + (k1 , jω) exp(−jk1 x1 )dk1 = h 2
! 1 ˆ 2 I∆ (x1 , jω) ,
x1  < 12 a ,
x1  > 12 a . (8.16) The lefthand side of this equation is a Fourier integral with respect to the spatial variable x1 and with the transform parameter k1 . Fourier’s inversion k1 =−∞
0,
207
sheet emitter with a parallel electric current
theorem yields
ˆ + (k1 , jω) = h 2
1 a 2
x1 =− 21 a
1ˆ 2 I∆ (x1 , jω) exp(jk1 x1 ) dx1
.
(8.17)
With Eqs. (8.17), (8.15), (8.10)  (8.11) and (8.6)  (8.8) the electromagnetic ﬁeld is determined completely. Once Iˆ∆ is prescribed, we can calculate the ˆ + is deintegral of Eq. (8.17) either analytically or numerically, and once h 2 termined we can calculate the Fourier integral of Eq. (8.8) either analytically or numerically. The other ﬁeld components can be computed in a similar way.
8.1.1.
The farﬁeld approximation
When the point of observation is far enough away from the emitter, i.e., 1
r = (x21 + x23 ) 2
(8.18)
is large enough, we can approximate the representations of the ﬁeld components. Let us start with the expression for the magnetic ﬁeld component ˆ2 = 1 H 2π
∞ k1 =−∞
ˆ + (k1 , jω) exp(−jk1 x1 − jk3 x3 )dk1 , h 2
(8.19)
when x3 > 0. In the case that r is large, the exponential function in the integrand oscillates violently along the integration path, except in a point where k1 x1 + k3 x3 is stationary. In such a stationary point the exponential function does not vary rapidly in a small interval along the integration path. In each part outside the stationary points the contribution to the integral is negligibly small. Around the stationary point the contribution to the integral is more signiﬁcant. This fact is employed in Kelvin’s principle of stationary phase. The stationary point k1 = k1s is the root of the equation
1
∂k1 [k1 x1 + k3 x3 ] = ∂k1 k1 x1 + (k 2 − k12 ) 2 x3 = 0 ,
(8.20)
208
excitation of twodimensional electromagnetic waves
where
1
(8.21)
k = ω(εµ) 2
has been used. Hence, this point k1 = k1s follows from x1 −
k1 1
(k 2 − k12 ) 2
as k1s = k
x3 = 0
x1 (x21 + x23 )
Note that
=k
1 2
(8.22)
x1 . r
(8.23)
x3 . (8.24) r Only around this stationary point the integrand of Eq. (8.19) will contribute ˆ + varies much slower than the signiﬁcantly. Assuming that the function h 2 exponential function, we obtain the approximation 1
k3s = [k 2 − (k1s )2 ] 2 = k
ˆ + (k x1 , jω) ˆ2 ≈ 1 h H 2π 2 r
∞ k1 =−∞
exp(−jk1 x1 − jk3 x3 )dk1 .
(8.25)
The argument of the exponential function has to be approximated with more care. We expand the argument of the exponential function around the stationary point in a Taylor series
k1 x1 + k3 x3 = k1s x1 + k3s x3 + 12 (k1 −k1s )2 ∂k x1 − 1
k1 1
(k 2 −k12 ) 2
x3 k1 = k1s
+··· ,
(8.26) where we omit higher order terms. Note that the term being linear in k1 −k1s has vanished in the stationary point. Some algebraic manipulation yields k1 x1 + k3 x3 = kr − 12 (k1 −k1s )2
r3 + ··· . kx23
(8.27)
Hence, ∞ k1 =−∞
exp(−jk1 x1 − jk3 x3 )dk1 ≈ exp(−jkr)
∞
exp k1 =−∞
2k ≈ exp(−jkr) r
1 2
x3 r
3 s 2 r 1 j(k −k ) 1 1 2 kx23
∞ κ=−∞
dk1
exp(jκ2 )dκ .
(8.28)
sheet emitter with a parallel electric current
209
The last integral can be calculated as follows ∞
exp(jκ2 )dκ
2
∞
=
−∞
κ=−∞
exp(jx2 )dx
=
2
(x,y)∈IR ∞
−∞
exp(jy 2 )dy
exp[j(x2 + y 2 )]dxdy
= 2π lim δ↓0
∞
ρ=0
exp[(j −δ)ρ2 ]ρdρ
−π = jπ . = lim δ↓0 j −δ
(8.29)
Combining Eqs. (8.25), (8.28) and (8.29) we arrive at
ˆ2 ≈ H
x3 ˆ + x1 k h2 (k , jω) r r 2πr
1 2
exp(−jkr + j 14 π) 1
when r = (x21 +x23 ) 2 → ∞ and x3 > 0 .
(8.30)
This is the farﬁeld representation of the magnetic ﬁeld component. The amplitude is directly related to the spatial Fourier transform of the electric current through the emitter, by taking in Eq. (8.17) the quantity k1 = kx1 /r as the transform parameter. The electricﬁeld components in the far ﬁeld are obtained from Eqs. (8.10) and (8.23)  (8.24) as
ˆ1 ≈ E
x3 + x1 k eˆ (k , jω) r 1 r 2πr
≈
x3 r
2 1
µ ε
2
1 2
exp(−jkr + j 14 π)
ˆ + (k x1 , jω) k h 2 r 2πr
1 2
exp(−jkr + j 14 π)
1
when r = (x21 +x23 ) 2 → ∞ and x3 > 0 ,
(8.31)
and
ˆ3 ≈ E
x3 + x1 k eˆ3 (k , jω) r r 2πr
x1 x3 ≈ − 2 r
1
µ ε
2
1 2
exp(−jkr + j 14 π)
ˆ + (k x1 , jω) k h 2 r 2πr 1
1 2
exp(−jkr + j 14 π)
when r = (x21 +x23 ) 2 → ∞ and x3 > 0 .
(8.32)
210
excitation of twodimensional electromagnetic waves
We observe that in the far ﬁeld the relation ˆ1 + x3 E ˆ3 = 0 x1 E
(8.33)
ˆ = {E ˆ3 } is perpendicular to the ˆ1 , 0, E holds, hence the electric ﬁeld vector E direction of observation x = {x1 , 0, x3 }, see Fig. 8.2. All the results pertaining to the electromagnetic ﬁeld quantities in the domain x3 < 0 can be derived in a similar way. The timeaveraged power ﬂow density in the far ﬁeld is given by 1 ˆ 2 Re[E
ˆ ∗] = ×H ≈
1 ˆ ˆ∗ 2 Re[−E3 H2 i1 2 1 x3 µ 2
≈
r x3 r
ε
2 1
µ ε
2
ˆ1 H ˆ ∗ i3 ] +E 2 x1 x3 k ˆ + x1 h2 (k , jω)2 [ i1 + i3 ] , 4πr r r r k ˆ + x1 x h (k , jω)2 , 4πr 2 r r 1
when r = (x21 +x23 ) 2 → ∞ ,
(8.34)
E .
...... ........ ... ... ... ... ... .
............. ........
(...................
......... .......... ....... ....... . . . . . . ...... ....... ....... ....... ....... ....... ...... . . . . . . . . ....... ...... .... . ....... ................. ....... .. .. ............ .............. ....... ....... .... ....... .......... ....... ... ............ ............. ..... ......................
x
S
H
θ
i1
O r
i3
Figure 8.2. The orientation of the electromagnetic ﬁeld vectors in the far ﬁeld of an emitter with parallel electric current.
211
sheet emitter with a parallel electric current
where x/r = x/x is the unit vector in the direction x of observation. The electromagnetic power ﬂow is in this direction and decays with 1/r as a function of r. In conclusion we observe that the electric and magnetic ﬁeld strengths have, in the farﬁeld region, the structure of a cylindrical wave that expands radially from the origin of the coordinate system. In the angular direction (see Fig. 8.2), the ﬁeld amplitudes depend on θ via cos(θ) =
x3 x1 , sin(θ) = . r r
(8.35)
The electric and magnetic ﬁeld strengths in the farﬁeld region are transverse to the local direction of propagation and behave locally as if the wave were a uniform plane wave, with exp(−jkr) = exp(−jk1 x1 −jk3 x3 ), traveling in the direction {k1 = k cos(θ), 0, k3 = k sin(θ)} away from the source. The timeaveraged power ﬂow vanishes in the i1 direction (θ = 0). All ﬁeld amplitudes ˆ + (k cos(θ), jω). Consequently, in the farﬁeld region are proportional to h 2 only the spatial Fourier transform of the electric current through the emitter (see Eq. (8.17)) at the subset of the transform parameter k1 = k cos(θ) is ”seen”. To study the angular dependence of the farﬁeld characteristic, the directive gain is introduced. This directive gain D(θ) is deﬁned as the power ﬂow in the observation direction, normalized to the angularaveraged power ﬂow in the far ﬁeld, viz., 1 2 Re
D(θ) = 1 2 Re
1 2π
ˆ ×H ˆ ∗) · x (E r
2π
ˆ ×H ˆ ∗ ) · x dθ (E r θ=0
.
(8.36)
Using Eqs. (8.34) and (8.35) we ﬁnd
D(θ) =
ˆ + (k cos(θ), jω)2  sin(θ) h 2
1 2π
2π
θ=0
ˆ + (k cos(θ), jω)2 dθ  sin(θ) h 2
.
(8.37)
In Fig. 8.3, the directive gain is presented for the case that Iˆ∆ is a constant function, i.e., (8.38) Iˆ∆ (x1 , jω) = ˆi∆ (jω) , x1  < 12 a ,
212
excitation of twodimensional electromagnetic waves
D(θ)
...... ......... ... .... .. ... .... .. . . .... . . . . . . . . . ... . . . . ... . . . .. ..... . . . . ... . . . . . . . ........... . . . . . . . . . . . . . . . ... . . . . . . . . .. .. ..... . . . . .. .. . . . . ......................... ............................ ... . . . . . . ....... ..... ..... . . ....... . . . . . . . . . . . . . .... . . .... .. .... . . ...... . . . . . . .... .. ... . . .... . .. .. . . . . . . . . . . . . . . . . . . . . . . . . ... . . . ... . . . . . .... .... .... . . .. . . . . . . . . . . . . . .. ....... .. . .. .. . . . . . . . . . ...... . . . . ... . . . . . . . . . . . . . . . . . . ..... . . ........... . ... . . . . . . . .... . . . . . . ... . ... . . ....... . . . . . . . .... . . . .. .. . . . . . .. . . . . . . . .. ... . . . . . . . . . . . .. . .. . . .. .. . .. . . ... .. . . . . . . ... .... . . . . ... .. . .. .. .. . . . .... . .... . . . . . .. ..... ..... .. .. . ....... . . . . . . . . . . .... .... . . .... . . .. . . . . . . . . . . . . . . . . ...... . . ..... .. . .. ........................... ........................... .. . . . . . . . . . .. . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . .. . . . . . . . . . . . . . . .
a/λ = 0.1
10
1
θ
0.1
D(θ)
.. ...... ....... .. .... .. .... .... . . . ...... . . . . . . . . . .... . . . . .... . ..... . . .. . . .... . . . . ... . .......... . . . . . . . . . . . . ... . . . . . . .. . .. . .... . . . ... . .. . . . ... . . . .. . . . . . . . . . . . . . . . . . . . . . ................ ........ . . . ... . . . .................................. . .... . . ...... ....... . . . ....... .. ....... . . . . . . . . . . . . . ..... . ........ . .. . ..... . . ... . . . . . . . . ... . . . ..... .. ..... . . . . . . . . .. . .. ... ... . . ... . . . ...... ... ...... . . . ... . . . . . . ... ... ... . . . . . . .... . . . . ... ......... . . ... . . . . . . . ........ . . . . . . . . .. ... . . . ... . . . . . . . . . ... . . . . . . ..... ...... . . . .. ... . .... . ... ... . . . . ... . . . . . . . . . .. . . .... . . ... . ... . . . . ........... ..... . .... . . ........ ....... . . . . ...... .. .. .... . .... ....... . . . . . . . ..................................... . . . . . .................... . . . . . . . . . . . . . . . .. . . . .. . . . . . . .. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . .
a/λ = 1
10
1
0.1
θ
D(θ)
...... ......... ... .... .. ... .... .. . . .... . . . . . . . . . ... . . . . ... . . . .. ..... . . . . ... . . . . . . . ........... . . . . . . . . . . . . . . . ... . . . . . . . . .. .. ..... . . . . . . .. . . . . ..................... ..................... ... . . . . . . .... ........ ..... ........ . . . . . . . . . . . . . . . .... .. .... . .... . . . ....... . . . . . . . . . . .... .. .... . . .... . ... .. . . . . . . . . . . . . . .. . . . . .. . . . . . . .... ... . . . . . . .... .... .... . . .. . . . . . . . . . . . . .. .. ...... . .. .. . . . . . . . . . . ...... . . . . ... . . . . . . . . . . . . . . . . . . ..... . . ............ . ... ... . . . . . . .... . . . . . ... . ... . . ........ . . . . . . . . .... . . . ... . . . . . . . . .. . . . . .. ... . . . . . . . . . . .. . .. . . . ... ... . . . . . . . . . ... . . . .. ... . . . . . .. . .. .. .. . . . ...... ... . . . . . .. . .... . ... . .... ..... . . . ...... . . ....... . . ..... . .......... . ..... . .. . . . . . . . . . . . . . . ....... ....... .. . .. ..................... .................... .. . . . . . . . . . .. . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . .. . . . . . . . . . . . . . . .
a/λ = 0.3
10 1
θ
0.1
D(θ)
.. ...... ....... .. .... .. .... .... . . . ...... . . . . . . . . . .... . . . . .... . ..... . . .. . . .... . . . . ... . .......... . . . . . . . . . . . . ... . . . . . . .. . .... .. . . . . . . . ... . . . ... . . . .. . . . . . . . . . . . . . . . . . . . . . .. . . .. . . .. . . . . ... . . . .. . . . . . . . . . ............ . .. . .............. . . . . . . . ..... . ...... ..... .. ...... ....... . .... . . . . .. . . . . . . ............................................................................................ .... ......................... ................................................................ . ..... ................... .............. .. .. ........... .................... ......... . .... . .......................................................... . . .. . . . . . . . . . . . . . . . . . . . . . . . ... . . . . . . . . . . . .. ..................... .. ...... . . ................................... .................................. . .. . ...... ................................................... .................................................................. . ............ . . . . . . . . . . . . . . . . . . . . . . . ....... ....... . ....... ................................ ............................... ...... .. ...... .. .... . . . . . . . . . . . . . . . . . . . . . .. ... . ... . . . . ..................... . .. . . . . .................. .. . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . .. . . . . . . .. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . .
a/λ = 3
10 1
θ
0.1
Figure 8.3. The directive gain of the sheet emitter with parallel electric current.
sheet emitter with a perpendicular electric current
213
so that Eq. (8.17) yields 1 ˆ + = sin( 2 k1 a) ˆi∆ . h 2 k1
(8.39)
The directive gain is presented for various values of a/λ, where λ = 2π/k ˆ + ≈ 1 a ˆi∆ is the wavelength. For values of a/λ → 0, we observe that h 2 2 2 and D(θ) ≈ 2 [sin(θ)] . From Fig. 8.3, we observe that this approximation ˆ+ ≈ holds up to a/λ = 0.3. For values of a/λ → ∞, we observe that h 2 1 πˆi∆ δ[k cos(θ)] and D(θ) vanishes everywhere, except at θ = ± 2 π, where it becomes inﬁnite. Hence, for large values of a/λ the emitter radiates the electromagnetic waveﬁeld mainly in forward and backward directions.
8.2.
The sheet emitter with a perpendicular electric current
A reference frame is introduced such that the sheet coincides with the plane x3 = 0 (Fig. 8.4). Let the impressed electric current ﬂow uniformly in the x2 direction. It is present at x3 = 0 from x1 = − 12 a to x1 = 12 a. Hence, !
Jˆ1ext = 0 , Jˆ2ext =
Iˆ∆ (x1 , s)δ(x3 ) , x1  < 12 a , 0,
x1  > 12 a ,
Jˆ3ext = 0 ,
(8.40)
where δ(x3 ) denotes the onedimensional unit impulse (Dirac distribution), and Iˆ∆ (in A/m) is the electric current per unit length (of the x1 direction). Since the electriccurrent sheet carries no magnetic current, we have ˆ ext = 0 , ˆ ext = 0 , K K 1 2
ˆ ext = 0 . K 3
(8.41)
The medium is assumed to be homogeneous and lossless with constitutive constants ε and µ. Since the conﬁguration is independent of x2 , the electromagnetic ﬁeld is twodimensional and in Chapter 4 we have seen that the frequencydomain ﬁeld equations (2.54)  (2.59) separate into two independent set of equations, cf. Eqs. (4.4)  (4.6) and Eqs. (4.7)  (4.9). ˆ ext are equal From Eqs. (8.40)  (8.41) we conclude that Jˆ1ext , Jˆ3ext and K 2 to zero, hence only a perpendicularly polarized electromagnetic ﬁeld occurs.
214
excitation of twodimensional electromagnetic waves
ˆ 3 and E ˆ2 satisfy the following set of ˆ 1, H The nonzero ﬁeld components H equations (s = jω, σ = 0, cf. Eqs. (4.7)  (4.9)) ˆ2 + jωµH ˆ1 = 0 , −∂3 E ˆ3 = 0 , ˆ2 + jωµH ∂1 E !
ˆ2 = ˆ 1 + ∂1 H ˆ 3 + jωεE −∂3 H
(8.42) (8.43) −Iˆ∆ (x1 , jω)δ(x3 ) , x1  < 12 a . x1  > 12 a ,
0,
(8.44)
In Chapter 4 we have seen that in a domain outside the source distribution, solutions of Eqs. (8.42)  (8.44) in the form of plane waves exist. We therefore assume that the general solution consists of an inﬁnite superposition of planewave constituents ⎧ 1 ∞ ⎪ ⎪ ⎪ ⎨
ˆ + (k1 , jω) exp(−jk1 x1 − jk3 x3 )dk1 , x3 > 0, h 1 2π k =−∞ 1 ˆ 1 (x1 , x3 , jω) = H ∞ ⎪ 1 ⎪ ˆ − (k1 , jω) exp(−jk1 x1 + jk3 x3 )dk1 , x3 < 0, ⎪ h ⎩ 2π k1 =−∞ 1 (8.45)
ε, µ
.......... ......... ....... ....... .......
'
E
' ....... ....... ....... ............ . . . . . .
x3 < 0
E
.............. ........
'...................
ε, µ
(x = 1 a ( 1 2 E . .... ( ........ i1 ( .... ...r........................... i3 O( ( ( J ext x1 = − 12 a '
E
....... ....... ....... .......... ..........
x3 > 0
Figure 8.4. Electriccurrent sheet with impressed current as an emitter of perpendicularly polarized electromagnetic waves.
215
sheet emitter with a perpendicular electric current ⎧ 1 ∞ ⎪ ⎪ ⎪ ⎨
ˆ + (k1 , jω) exp(−jk1 x1 − jk3 x3 )dk1 , x3 > 0 , h 3 2π k1 =−∞ ˆ 3 (x1 , x3 , jω) = H ∞ ⎪ ⎪ 1 ˆ − (k1 , jω) exp(−jk1 x1 + jk3 x3 )dk1 , x3 < 0 , ⎪ h ⎩ 2π k1 =−∞ 3 (8.46) and ⎧ 1 ∞ ⎪ ⎪ ⎪ ⎨
eˆ+ 2 (k1 , jω) exp(−jk1 x1 − jk3 x3 )dk1 , x3 > 0 , 2π k1 =−∞ ˆ2 (x1 , x3 , jω) = E ∞ ⎪ ⎪ 1 ⎪ eˆ− (k1 , jω) exp(−jk1 x1 + jk3 x3 )dk1 , x3 < 0 , ⎩ 2π k1 =−∞ 2 (8.47) where ⎧ 1 1 ⎨ (ω 2 εµ − k12 ) 2 , k1  ≤ ω(εµ) 2 , (8.48) k3 = ⎩ −j(k 2 − ω 2 εµ) 21 , k  > ω(εµ) 21 . 1 1 Note that the deﬁnition of the square root of k3 takes care of the boundedness of the integrals as x3  → ∞. This guarantees the boundedness of the electromagnetic ﬁeld components as x3  → ∞. From Eqs. (8.42) and (8.43) ˆ +, h ˆ − and h ˆ − are related to eˆ+ and eˆ− as ˆ +, h it follows that the quantities h 1 3 1 3 2 2 ˆ + = −k3 eˆ+ , h 1 ωµ 2
ˆ + = k1 eˆ+ , h 3 ωµ 2
(8.49)
ˆ − = k3 eˆ− , h 1 ωµ 2
ˆ − = k1 eˆ− . h 3 ωµ 2
(8.50)
1
When x3 > 0 and k1  < ω(εµ) 2 we observe that the electromagnetic ﬁeld consists of a superposition of uniform plane waves propagating in the 1 {k1 , 0, k3 }direction, while when k1  > ω(εµ) 2 the constituents are (nonuniform) evanescent waves decaying exponentially in the x3 direction. The presence of the source distribution is accounted for by the application of the excitation conditions !
ˆ 1 (x1 , x3 , jω) − lim H ˆ 1 (x1 , x3 , jω) = lim H
x3 ↓0
x3 ↑0
Iˆ∆ (x1 , jω) , x1  < 12 a , 0,
x1  > 12 a , (8.51)
which is a consequence of Eq. (8.44), and ˆ2 (x1 , x3 , jω) − lim E ˆ2 (x1 , x3 , jω) = 0 , lim E
x3 ↓0
x3 ↑0
−∞ < x1 < ∞ ,
(8.52)
216
excitation of twodimensional electromagnetic waves
which is a consequence of Eq. (8.42). From Eqs. (8.52) and (8.47) it directly follows that ˆ+ (8.53) eˆ− 2 =e 2 , and from Eqs. (8.49) and (8.50) it follows that ˆ+ . ˆ − = −h h 1 1
(8.54)
ˆ2 are evensymmetric funcˆ 3 and E We observe that the ﬁeld components H ˆ 1 is an oddsymmetric function of x3 . Application of tions of x3 , while H Eqs. (8.51) and (8.54) to Eq. (8.45) leads to 1 2π
∞
ˆ + (k1 , jω) exp(−jk1 x1 )dk1 = h 1
! 1 ˆ 2 I∆ (x1 , jω) ,
x1  < 12 a ,
x1  > 12 a . (8.55) The lefthand side of this equation is a Fourier integral with respect to the spatial variable x1 and with the transform parameter k1 . Fourier’s inversion theorem yields k1 =−∞
ˆ + (k1 , jω) = h 1
1 a 2
x1 =− 21 a
0,
1ˆ 2 I∆ (x1 , jω) exp(jk1 x1 ) dx1
.
(8.56)
With Eqs. (8.56), (8.54), (8.49)  (8.50) and (8.45)  (8.47) the electromagnetic ﬁeld is determined completely. Once Iˆ∆ is prescribed, we can calculate ˆ + is the integral of Eq. (8.56) either analytically or numerically, and once h 1 determined we can calculate the Fourier integral of Eq. (8.45) either analytically or numerically. The other ﬁeld components can be computed in a similar way.
8.2.1.
The farﬁeld approximation
When the point of observation is far enough away from the emitter, i.e., 1
r = (x21 + x23 ) 2
(8.57)
the sheet emitter with a perpendicular electric current
217
is large enough, we can approximate the representations of the ﬁeld components. Using the method of stationary phase of Section 8.1.1, we obtain the farﬁeld representation
1
x3 ˆ + x1 k h1 (k , jω) r r 2πr
ˆ1 ≈ H
2
exp(−jkr + j 14 π) 1
when r = (x21 +x23 ) 2 → ∞ and x3 > 0 .
(8.58)
The amplitude is directly related to the spatial Fourier transform of the electric current through the emitter, by taking in Eq. (8.56) the quantity k1 = kx1 /r as the transform parameter. The other ﬁeld components in the far ﬁeld are obtained from Eqs. (8.49) and (8.23)  (8.24) as
ˆ3 ≈ H
k x3 ˆ + x1 h3 (k , jω) r r 2πr
1 2
k x1 ˆ + x1 , jω) ≈ − h 1 (k r r 2πr
exp(−jkr + j 14 π)
1 2
exp(−jkr + j 14 π) 1
when r = (x21 +x23 ) 2 → ∞ and x3 > 0 , (8.59) and
ˆ2 ≈ E
x3 + x1 k eˆ2 (k , jω) r r 2πr 1
≈ −
µ ε
2
1 2
exp(−jkr + j 14 π)
ˆ + (k x1 , jω) k h 1 r 2πr
1 2
exp(−jkr + j 14 π) 1
when r = (x21 +x23 ) 2 → ∞ and x3 > 0 . (8.60) We observe that in the far ﬁeld the relation ˆ 1 + x3 H ˆ3 = 0 x1 H
(8.61)
ˆ 3 } is perpendicular to ˆ = {H ˆ 1 , 0, H holds, hence the magnetic ﬁeld vector H the direction of observation x = {x1 , 0, x3 }, see Fig. 8.5. All the results pertaining to the electromagnetic ﬁeld quantities in the domain x3 < 0 can be derived in a similar way.
218
excitation of twodimensional electromagnetic waves
H .
...... ........ ... ... ... ... ... .
............... ........
'...................
............... ........ ....... ....... . . . . . . ...... ....... ....... ....... ....... ...... ....... . . . . . . . . .... ...... .... . ....... ................. ....... .. .. ............ .............. .... ....... .... ....... .......... ....... ...... .... ............ .................................. ..
x
S
E
θ
i1
O r
i3
Figure 8.5. The orientation of the electromagnetic ﬁeld vectors in the far ﬁeld of an emitter with perpendicular electric current.
The timeaveraged power ﬂow density in the far ﬁeld is given by 1 ˆ 2 Re[E
ˆ ∗] = ×H ≈
1 ˆ ˆ∗ 2 Re[E2 H3 i1
1
µ ε
2
µ ε
2
1
≈
ˆ2 H ˆ ∗ i3 ] −E 1
x1 x3 k ˆ + x1 h1 (k , jω)2 [ i1 + i3 ] , 4πr r r r k ˆ + x1 x h1 (k , jω)2 , 4πr r r 1
when r = (x21 +x23 ) 2 → ∞ ,
(8.62)
where x/r = x/x is the unit vector in the direction x of observation. We observe that the power ﬂows in this direction and decays with 1/r as a function of r. In conclusion we observe that the electric and magnetic ﬁeld strengths have, in the farﬁeld region, the structure of a cylindrical wave that expands radially from the origin of the coordinate system. In the angular direction (see Fig. 8.5), the ﬁeld amplitudes depend on θ via cos(θ) =
x3 x1 , sin(θ) = . r r
(8.63)
The electric and magnetic ﬁeld strengths in the farﬁeld region are transverse to the local direction of propagation and behave locally as if the wave were
the sheet emitter with a perpendicular electric current
219
a uniform plane wave, with exp(−jkr) = exp(−jk1 x1 − jk3 x3 ), traveling in the direction {k1 = k cos(θ), 0, k3 = k sin(θ)} away from the source. All ˆ + (k cos(θ), jω). Consequently, in the ﬁeld amplitudes are proportional to h 1 farﬁeld region only the spatial Fourier transform of the electric current through the emitter (see Eq. (8.56)) at the subset of the transform parameter k1 = k cos(θ) is ”seen”. To study the angular dependence of the farﬁeld characteristic, the directive gain is introduced. This directive gain D(θ) is deﬁned as the power ﬂow in the observation direction, normalized to the angularaveraged power ﬂow in the far ﬁeld, viz., 1 2 Re
D(θ) = 1 2 Re
1 2π
ˆ ×H ˆ ∗) · x (E r
2π
ˆ ×H ˆ ∗ ) · x dθ (E r θ=0
.
(8.64)
Using Eqs. (8.62) and (8.63) we ﬁnd
D(θ) =
ˆ + (k cos(θ), jω)2 h
1 . 1 2π ˆ + h1 (k cos(θ), jω)2 dθ
2π
(8.65)
θ=0
In Fig. 8.6, the directive gain is presented for the case that Iˆ∆ is a constant function, i.e., (8.66) Iˆ∆ (x1 , jω) = ˆi∆ (jω) , x1  < 12 a , so that Eq. (8.56) yields 1 ˆ + = sin( 2 k1 a) ˆi∆ . h 1 k1
(8.67)
The directive gain is presented for various values of a/λ, where λ = 2π/k ˆ + ≈ 1 a ˆi∆ and is the wavelength. For values of a/λ → 0, we observe that h 1 2 D(θ) ≈ 1. The emitter is then an isotropic radiator in the (x1 , x3 )plane. From Fig. 8.6, we observe that this approximation holds up to a/λ = 0.3. ˆ + ≈ πˆi∆ δ[k cos(θ)] and D(θ) For values of a/λ → ∞, we observe that h 1 vanishes everywhere, except at θ = ± 12 π, where it becomes inﬁnite. Hence,
220
excitation of twodimensional electromagnetic waves
for large values of a/λ the emitter radiates the electromagnetic waveﬁeld mainly in forward and backward directions.
D(θ)
... ....... ....... .. ... .. .. .... .. . . . . ..... . . . . . . . . . .... . . . . ... . ..... . . ... . .. ... . . . ... . ........... . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ........ . . . ........ . ........ .... . ...... . . ........ ... ..... . . ..... ..... . . ... .. .... . . . .... . . . . . . . . . . . . .... . . . . . . . . . . . ...... . . .. . . . . . . . . .... . . . . ..... .. . . . . . . . . . ... . . . . . . . . . . . . . . .. . . .. . . .. .. .. . . .. . .. .. . . . . ... . . . . . ... .. . . ... . . . . . . . ... . ...... . .. . ... . . . . . . . . . . . . . . . . .. . . . . . . . . . . .. . . . . . .. . . . . .. .. . .. . . .. . . . . . . . .. . .. . . . . . ... . . . . .... .. . . . ... . . . . . . . . ... . . . . . ....... .. . . . . . . .... . . . . .. . . .... .... . .. . . ..... .. . .... ..... . . ......... . . ....... . ..... ....... . . . . . . . ............ . . . . . ................................ . . . . . . . . . . . . . . .. . .. . . . . . . . . . . . . . . .
a/λ = 0.1
10
1
θ
0.1
D(θ)
.. ...... ....... ... .... .. .. .... .. . . . . . ..... . . . . . . . . . . .... . . ... . ..... . . . .. ... .... . . . . . .......... . . . . .. . . . . . . . . ... . . . . . . .. . .. . .... . . . .. . .. . . . . . ... . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ....... ....... . . ... . ...... . . ... . . ............ ......... . . .. ....... ... .... . .... . . . . . . . . . . . . ..... . ... . . . . .. ...... . . . . . . . . . . . ... .... .. ... . . . . . .. .. . . . . . ... . . . . . . ..... ... .... . . . . . ... ... . . ........... . . . . ... . . . . .... . .............. . . . .... . . . . . .... . . . . ... . . . . . . .. . . .... ..... . ... . . . . . . . . . .... .... . . . ... . . ... ... .... . . . . . . . . . . . . ... . . . . . . ... . .... . . ... . ... . . . . .... .... .. .... . . . ..... . ...... . . . . ....... ..... . . ..... . . . . . ............. . . . . ......................................... . ................................. . . . . . . . . . . . . .. . . .. . . . . . . . . . . .. . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . .. . . . . . . . . . . . . . . .
a/λ = 1
10
1
0.1
θ
D(θ)
... ....... ....... .. ... .. .. .... .. . . . . ..... . . . . . . . . . .... . . . . ... . ..... . . ... . .. ... . . . ... . ........... . . . . . . ......... . . . . . . . . . . . . . . . . . . . . . .. .. ............... . . . . .............. .......... .. . ........ . ........... ... ..... . . ..... ..... . . ... .. ..... . . . .... . . . . . . . . . .... .. . . . . . . . . . . ....... . . . ... . . . . . . . .... . . . . ...... ... . . . . . . . . . . . . . .. . .... . . . . . . . . . . .. . . ..... ... .. . . .. . .. . . . . ... . . . . . .... .... .. . . ... . . . . .... . . . .. . ...... . . . .... . . . ... . . . . . . . . . . . . . . ... . . . . . . . . ... . . . . . . . .. . . . . ..... . ... . . . . . . . . .. . . ... . .... . . . . . . . . . ..... .. . .... . . . . . . . . . . . ... . . . . . ...... ... . . . . . . . . .... . . . . . . ..... .... . .. . . .. . ..... ..... ...... . . ......... . . ......... . ......... . ........... . . . . . . . . . . ...................................... . . . . . . . . . . . . . . . . . . . .. . .. . . . . . . . . . . . . . . .
a/λ = 0.3
10 1
θ
0.1
D(θ)
.. ...... ....... ... .... .. .. .... .. . . . . . ..... . . . . . . . . . . .... . . ... . ..... . . . .. ... .... . . . . . .......... . . . . . .. . . . . . . . ... . . . . . . .. . .. . .... . . . .. . . . . . . . . ... . .. . . . . . . . . . . . . . . . . ..... . . . . . . .. . . . . .. ................ . .... . ............... . . . . . . . . . . . . . . . . . . . . . . . . . . .... . ............ .... . ..... .. .... . .... ............ . .... . . ..... . ........ .... . .. .. .. . .. ........ . ..... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ............... . ................................... ................ ... .. .. . ............. . ................................. . ............... . ................... .......... ..... ... .......... .................... ..... ........ ............................ ....... .......... . . . ... . ........................................ .. ..... . . . . . . . . .. ................... .. ... ..... . . ............................................................................. . ... . ............ ..... . . .......................................... ........................ .... ..................... ............................................ . . . . . . . . . . . . . . . . . . . ........................ . ..... ................. . ... .... . ............... .......... ...................... . . . . . . ... . .... ....... . .... .. .. . . . . . .............. . .. .. .... ............ ...... ....... . . . .... . ............ . . . ................ .. . . . . . . .. . . . . . . . . . . . . . .. . . .. . . . . . . . . . . .. . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . .. . . . . . . . . . . . . . . .
a/λ = 3
10 1
θ
0.1
Figure 8.6. The directive gain of the sheet emitter with perpendicular electric current.
excitation of twodimensional electromagnetic waves
8.3.
221
Exercises and problems
Exercise 8.1 In the stationary phase method the argument of the exponential function is expanded in a Taylor series around the stationary point k1 = k1s . Why does the term linear in (k1 − k1s ) vanish? Exercise 8.2 In the case of a constant impressed current Iˆ∆ (x1 , jω) = ˆı∆ (jω), x1  ≤ 12 a, for what angles θ is the directive gain function D(θ) equal to zero? Exercise 8.3 Assume that the impressed current ﬂows uniformly in the x2 direction and is ˆ ext = ˆı∆ (jω)δ(x1 , x3 )i2 , where δ(x1 , x3 ) present at x1 = 0, x3 = 0. Hence, J is the twodimensional unit impulse operative at x1 = 0, x3 = 0. This implies that the electric current sheet is degenerated to a line source. ˆ +. (a) Give the expression for h 1 (b) Give in the farﬁeld approximation the expressions for the electric and magnetic ﬁeld strengths in the frequency domain. ( c ) Draw the farﬁeld radiation characteristic for the electromagnetic ﬁeld conform Fig. 8.5, for the angles of observation θ = nπ 6 , n ∈ IN . Indicate the magnitude of the ﬁeld strengths by the lengths of the vectors for ˆ and the radii of the circles for E. ˆ H (d) Calculate and draw the directive gain as a function of the angle of observation θ.
Problem 8.1 Let for a current sheet the impressed perpendicular electric current be given by Iˆ∆ (jω) = a−1ˆı∆ (jω), for x1  ≤ 12 a. Show that in case we let a → 0 ˆ + is obtained as in Exercise 8.3(a). This is most easily the same value for h 1 ˆ + = sin(k1 a/2) ˆı(jω). ˆ + with h obtained by taking lim h 1 1 k1 a a→0
Problem 8.2 Assume the impressed current ﬂows uniformly in the x1 direction; it is ˆ ext = −(1− 2x1  )ˆı∆ (jω)δ(x3 )i1 . present at x3 = 0 for x1  ≤ 12 a and is given by J a
222
excitation of twodimensional electromagnetic waves
ˆ +. (a) Give the expression for h 2 (b) Give in the farﬁeld approximation the expressions for the electric and magnetic ﬁeld strengths. ( c ) Draw the farﬁeld radiation characteristic for the electromagnetic ﬁeld conform Fig. 8.5, for the angles of observation θ = nπ 6 , n ∈ IN . Indicate the magnitude of the ﬁeld strengths by the lengths √ of the vectors for ˆ ˆ E and the radii of the circles for H. Take a = 4 2λ.
Answers to Exercises Exercise 1.1 (a) 0
(b) 0 ( c ) (a · c)b − (b · c)a (d) Replace one of the vector products by a new vector e Rewrite the expression Substitute the original vector product for e Again rewrite the expression to ﬁnally obtain (a · c)(b · d) − (a · d)(b · c) ( e ) Use the same procedure as in (d) to arrive at the result [d · (a × b)]c − [c · (a × b)]d.
Exercise 1.2 Since c · a = (a × b) · a = 0 and c · b = (a × b) · b = 0, the vector c is normal to both vectors a and b. Exercise 1.3 A = d1 d2 is the area of the rectangle. Exercise 1.4 V = d1 d2 d3 is the volume of the brick.
224
answers to exercises
Exercise 1.5 (a)
(b)
x2
x2
x1
x1
x3 (c)
x3 (d)
x1
x1
x3 x2
x3 x2
The vector ﬁelds are invariant in the direction of the coordinate axis that points forward. Exercise 1.6 (a) ∇ · v = −1 ,
∇×v =0
(b) ∇ · v = 0 ,
∇ × v = −i3
( c ) ∇ · v = −2 , (d) ∇ · v = 0 ,
∇×v =0 ∇ × v = −2i2 .
Exercise 1.7 ∇2 A = (∂12 + ∂22 + ∂32 )A1 i1 + (∂12 + ∂22 + ∂32 )A2 i2 + (∂12 + ∂22 + ∂32 )A3 i3 . Exercise 1.8 ∇ × E = −∇ × ∇V = 0, for all V (x) (see Eq. (1.32)). Exercise 1.9 ∇ · D = ∇ · (∇ × C) = 0, for all C(x) (see Eq. (1.33)).
answers to exercises
225
Exercise 1.10 Application of the chain rule of diﬀerentiation gives (a) ∇d = −
x − x , ∇d = −∇d x − x
(b) ∇d−1 =
x − x , ∇ d−1 = −∇d−1 x − x3
( c ) ∇ · ∇d =
x
2 , ∇ · ∇ d = ∇ · ∇d − x
(d) ∇ · ∇d−1 = 0 , ∇ · ∇ d−1 = 0. Exercise 1.11 Use Eq. (1.11) to obtain w × (ν × H) = (w · H)ν − (w · ν)H Substitute w = ν × E (ν × E) × (ν × H) = [(ν × E) · H] · ν, since (ν × E) · ν = 0 Take the scalar product of the result with ν ν · [(ν × E) × ν × H)] = (ν × E) · H = ν · (E × H) Exercise 1.12 (ν · H)ν − ν × (ν × H) = (ν · H)ν − (ν · H)ν + (ν · ν)H = H The term (ν · H)ν is the part of H that is normal to S; the term −ν × (ν × H) = (ν × H) × ν is the part of H that is tangential to S.
Exercise 2.1 There are ﬁve unknown vectorial quantities in Maxwell’s equations in matter: E, H, J , D and B, and hence ﬁfteen unknown scalar quantities. Exercise 2.2 Application of Gauss’ law yields x∈∂ D ext ν · J dA = −∂t x∈Dext ρ dV . This equation states that the amount of electric current that is ﬂowing out of the domain Dext through its boundary ∂Dext equals the decrease per unit time of the electric charge present in Dext . Exercise 2.3 Following the reasoning of Section 2.6 the continuity of those components that are diﬀerentiated in the i3 direction must be enforced. Thus we must enforce the continuity of H1 , H2 , E1 and E2 .
226
answers to exercises
Exercise 2.4 Following the reasoning of Section 2.6 the continuity of those components that are diﬀerentiated in the direction normal to the interface must be enforced. Thus ν · J + ∂t ν · D is continuous across S and ν · B is continuous across S. Exercise 2.5 (a) circular polarization (b) circular polarization ( c ) elliptical polarization (d) elliptical polarization ( e ) linear polarization. Exercise 2.6 (a) J/m3 , J/m3 (b) W/m3 ,
W/m2
( c ) W/m2 , W/m2 . An easy way to arrive at these results is to employ Poynting’s theorem and to realize that in physics all terms of an equation have equal dimensions. Exercise 2.7 ˆ ×H ˆ ∗] = ST = 12 Re[E
1 2
ˆ1 H ˆ2 H ˆ∗ − E ˆ ∗ ]i3 = ( ε0 ) 12 i3 . Re[E 2 1 µ0
E(t) = cos(ωt)i1 − sin(ωt)i2 , 1
1
H(t) = ( µε00 ) 2 [sin(ωt)i1 + cos(ωt)i2 ], 1
S = E(t) × H(t) = ( µε00 ) 2 [cos2 (ωt) + sin2 (ωt)]i3 = ( µε00 ) 2 i3 .
Exercise 3.1 The polarization of the wave is independent of the value of x3 (a) ˆ = −i1 ˆ = i2 and E Take x3 = 0 to simplify the problem E the wave is circularly polarized (cf. Exercise 2.2 (a)). After inversion the expression for the phase shows that wave propagates in the positive x3 direction. ˆ , E ˆ and the direction of propagation form a righthanded Since E triad, the polarization is righthanded (also called counter clockwise).
227
answers to exercises
(b) Use the approach under (a)
Circular, righthanded polarization
( c ) Use the approach under (a)
Elliptical, righthanded polarization
(d) Use the approach under (a)
Elliptical, lefthanded polarization
( e ) Use step 1 under (a)
Linear polarization No rotation.
Exercise 3.2 f = 1 MHz, µ = µ0 and γ = α + jβ = 0.04 + j0.1. (a) The wave is attenuated by a factor of exp{−Re[γx3 ]} tion of exp(−π) occurs over a distance x3 = 25π m. (b) The phase is determined by Im[γx3 ] over a distance x3 = 10π m.
An attenua
A phase change of π rad occurs
( c ) A 1 MHz wave travels the distance of one wavelength in 1µs The wavelength and hence the distance is equal to λ = 2π/Im[γ] = 20π m. ˆ H ˆ = Z and Z =  jωµ  = 73.3 Ω. (d) E/ γ ( e ) Z = Z exp(jφ)
φ = arctan( αβ ) = 0.381 rad (21.8◦).
Exercise 3.3 The velocity of electromagnetic waves in vacuum is c0 3 × 108 m/s f = cλ0 The range is 3.75 × 1014 Hz (red light)  1015 Hz (violet light), or 375 THz (terahertz)  1 PHz (petahertz). Exercise 3.4 1 1 2 2 ) 2 δCu = 9.3 mm and δAl = δCu ( 5.8 δ = ( ωµσ 3.5 ) = 12 mm. Only in the ﬁrst skin depth there is a signiﬁcant penetration of the electric ﬁeld that runs along the surface of the wire Only the ﬁrst skin depth carries a signiﬁcant part of the electric current Using a radius larger than the skin depth (δAl = 12 mm) would be a waste of (expensive) material since the inner part will hardly contribute to the power transport. Exercise 3.5 1 1 2 ) 2 = (πf µσ)− 2 The skin depth is proportional to the inverse δ = ( ωµσ of the square root of frequency The lower frequency (f = 100 kHz) determines the minimal thickness of the shield dAl = 5δAl = 1.3 mm and dCu = 5δCu = 1.0 mm.
228
answers to exercises
Exercise 3.6 I∆ (t) = Re[Iˆ∆ (jω) exp(jωt)] = sin(ωt) Iˆ∆ (jω) = −j A/m E1 (x3 , t) = 1 1 1 1 2 2 2 Z sin{ω[t − (εµ) x3 ]} and H2 (x3 , t) = 2 sin{ω[t − (εµ) x3 ]}. Exercise 3.7 ˆ ˆ H(0, t) = 1.5 cos(3π × 108 t)i2 = Re[H(0, jω) exp(jωt)] H(0, jω) = 8 H0 i2 with H0 = 1.5 A/m and ω = 3π × 10 rad/s. Assume that the ﬁeld propagates in the positive x3 direction ˆ ˆ 3 , jω) = H0 γ exp(−γx3 )i1 H(x3 , jω) = H0 exp(−γx3 )i2 E(x σ+jω ε α+jβ β E(x3 , t) = H0 exp(−αx3 ) Re{ σ+jω ε exp[jω(t − ω x3 )]}. Substitution of the actual parameter values gives 1 (a) γ = jβ = jω(εµ) 2 E(x3 , t) = 188 cos[3π × 108 (t − 10−8 x3 )]i1 V/m 1
(b) ωε = 7.5 × 10−2 σ γ = α + jβ = 0 + jω(εµ) 2 the same as in case (a)
The answer is
1
2 E(x3 , t) = ( c ) ωε = 7.5 × 10−2 σ γ = α + jβ = ( ωµσ 2 ) (1 + j) 77 77 8 8 11.5 exp(−77x3 ){cos[3π×10 (t− 3π×108 x3 )]−sin[3π×10 (t− 3π×10 8 x3 )]}× i1 V/m.
Exercise 3.8 Assume a unit amplitude of the incident electric wave, and apply the electric ﬁeld analysis ˆ i = exp(−γx3 )i1 E ˆ r = R exp[γ(x3 − 2L)]i1 E
ˆ i = ( ε0 ) 21 exp(−γx3 )i2 , H µ0 ˆ r = −( ε0 ) 21 R exp[γ(x3 − 2L)]i2 . H µ0
ˆ1 at x3 = L R = −1. Apply the boundary condition for E In free space and for steady state it follows that γ = jβ = j cω0 ˆ r = {exp[−j ω x3 ] − exp[j ω (x3 − 2L)]}i1 ˆ = E ˆi +E E c0
c0
= exp(−j cω0 L) {exp[−j cω0 (x3 − L)] − exp[j cω0 (x3 − L)]}i1 = −2j exp(−j cω0 L) sin[ cω0 (x3 − L)]i1 . Apply inversion E = 2 sin[ω(t − cL0 )] sin[ cω0 (x3 − L)]i1 Since the spatial behavior and the temporal behavior of E are decoupled, this is a standing wave. 1 In the same way, the total magnetic ﬁeld follows as H = 2( µε00 ) 2 × cos[ω(t − cL0 )] cos[ cω0 (x3 − L)]i2 the phase diﬀerence with the electric ﬁeld is π2 rad.
229
answers to exercises
Exercise 3.9 ˆ = 0 (a) Apply Eqs. (2.54)  (2.56) with ∂1 = ∂2 = 0 and J 1 ˆ 0 exp(−γ (1) x3 )i1 ( µε00 ) 2 H ˆ 0 exp(γ (1) x3 )i2 ˆ r = R H (b) H t ˆ = T H ˆ 0 exp(−γ (2) x3 )i2 H
r
Eˆ
i
=
1
Eˆ = −( µε00 ) 2 R Hˆ 0 exp(γ (1) x3 )i1 , 1 t Eˆ = 2( εµ00 ) 2 T Hˆ 0 exp(−γ (2) x3 )i1
( c ) From the modiﬁed Helmholtz equation and the principle of causality 1 1 it follows that γ (1) = s(ε0 µ0 ) 2 and γ (2) = 2s(ε0 µ0 ) 2 , i.e., in both cases the root of γ 2 has to be chosen in such a way that Re(γ) ≥ 0 (d) Apply the boundary conditions R = − 13 and T = 23
1 + R = T and 1 − R = 2T
ˆ0 H ˆ ∗× ˆt ×H ˆ ∗t ] = 1 Re[2( µ0 ) 12 T T ∗ H = 12 Re[E 0 2 ε0 1 1 µ0 2 4 µ0 2 ˆ 2 (2) ∗(2) 2 2 ˆ exp(−γ x3 − γ x3 )]i3 = ( ε0 ) T  H0  i3 = 9 ( ε0 ) H0  i3 .
( e ) s = jω
ST
Exercise 3.10 ˆ t = T⊥ exp(−γ (sea) x3 )i1 and Employ the electric ﬁeld analysis E ˆ t = Y (sea) T⊥ exp(−γ (sea) x3 )i2 . H The medium properties of seawater are ε = 81ε0 , µ = µ0 and σ = 4 S/m for f = 100 Hz it is found that in the sea ωσε = 8.9 × 106 1 1 σ γ (sea) = α + jβ = ( σµ20 ω ) 2 (1 + j) = 4 × 10−2 (1 + j) m−1 Y (sea) = γ (sea) = 50(1 − j) S. 1 Y (air) = Y0 = ( µε00 ) 2 = 2.6 mS. 2Y0 −5 (1 + j). T⊥ = Y +Y (sea) = 5.3 × 10 0 For the transmitted ﬁeld it follows that ˆt ×H ˆ ∗t ] ST = 1 Re[E = =
2 ∗ ∗(sea) 1 2 Re[T⊥ T⊥ Y 2 ∗(sea) 1 ] 2 T⊥  Re[Y −7
= 1.4 × 10
x3 = 0 m gives ST × 10−11 i
exp(−γ (sea) x3 ) exp(−γ ∗(sea) x3 )]i3 exp{−2 Re[γ (sea) x3 ]}i3
exp(−8 × 10−2 x3 )i3 W/m2
= 1.4 × 10−7 i3 W/m2 ; x3 = 100 m gives ST =
W/m2 .
4.7 3 ˆ = σE ˆ t = 4T⊥ exp(−γ (sea) x3 )i1 J (1 + j)i1 A/m2 .
x3
ˆ = 2.1 × 10−4 × = 0 m gives J
230
answers to exercises
Exercise 3.11 It could be included as a loss term in the inductance L line equations, Eqs. (3.112) and (3.113), would then be
The transmission
∂3 Iˆ + (G + sC)Vˆ = Iˆ∆ (s)wδ(x3 ) , ∂3 Vˆ + (R + sL)Iˆ = 0 . These are also known as the telegraph equations. The circuit parameter R is of course the series resistance, while the parameter G is the shunt conductance.
Exercise 4.1 (a)
ω = 3π × 109 rad/s f = 1.5 GHz. tan(θ) = ss13 = 43 θ = 0.927 rad (53.1◦ ).
ˆ exp(jωt)] (b) E = (3i1 − 4i3 ) sin[3π × 109 t − 2π(4x1 + 3x3 )] = Re[E ˆ E = −j(3i1 − 4i3 ) exp[−j2π(4x1 + 3x3 )]. (c)
ˆ =H ˆ 2 i2 . The wave is parallelly polarized H ext ˆ = ˆ Use Eq. (4.6) with K2 = 0 and s = jω H exp[−j2π(4x1 + 3x3 )]i2 .
ˆ exp(jωt)] = (d) H = Re[H
5 120π
50π jωµ0 ×
sin[3π × 109 t − 2π(4x1 + 3x3 )]i2 .
Exercise 4.2 ˆ (1) × ˆ(1) = E ˆ(1) = 0 and s(2) · e ˆ(2) = 0 e Parallel polarization s(1) · e (2) (2) ˆ ˆ = E [cos(θ)i1 + sin(θ)i3 ]. [cos(θ)i1 − sin(θ)i3 ] and e (1) (2) (1) ˆ ˆ (2) = E ˆ0 (assuming their phases to be equal). ˆ e  = ˆ e  E =E Subsitute in Eq. (4.58) and apply Euler’s formula ˆ = 2E ˆ0 exp[−j ω sin(θ)x1 ] E c0 ω ω × cos(θ) cos[ cos(θ)x3 ]i1 + j sin(θ) sin[ cos(θ)x3 ]i3 c0 c0
Propagating wave in the x1 direction, standing wave in the x3 direction.
231
answers to exercises
Exercise 4.3 ˆ i = E0 × Assume for x3 < 0 the presence of the incident ﬁeld E ˆ r=E0 R⊥ exp[−(γ r x1+γ r x3 )]i2 . exp[−(γ1i x1+γ3i x3 )]i2 and the reﬂected ﬁeld E 1 3 ˆ2 = 0 for x3 ↑ 0 must hold for all x1 γ r = γ i , The boundary condition E 1 1 γ3r = −γ3i and R⊥ = −1. In case of a lossless medium and steady state it is allowed to write γ1i = jβ1i and γ3i = jβ3i . For x3 < 0 the total ﬁeld is ˆ = E0 [exp(−jβ i x1 − jβ i x3 ) − exp(−jβ i x1 + jβ i x3 )]i2 E = =
1 3 1 3 i i i E0 exp(−jβ1 x1 ) [exp(−jβ3 x3 ) − exp(jβ3 x3 )]i2 −E0 exp(−jβ1i x1 ) 2j sin(β3i x3 )i2 .
Propagating wave in the x1 direction, standing wave in the x3 direction ˆ = 2E0   sin(β i x3 ) For x3 < 0 the amplitude of the total ﬁeld is E 3
Graph of the amplitude vs. x3 looks like a rectiﬁed sine with height 2E0  (n ∈ IN ). and with zeros at x3 = − nπ βi 3
Exercise 4.4 (a) Parallel polarization since the electric ﬁeld is entirely in the plane of propagation. √ √ si = 12 3i1 + 12 i3 sin(θi ) = si1 =√12 3 θi = π3 . (b) π i t t t According √ √ to Snell’s law, sin(θ ) = 1.5 sin(θ ) θ = 4 s = 1 1 2 2i1 + 2 2i3 . (c)
Parallel polarization Apply magnetic ﬁeld analysis. √ ˆ1 exp(jωt)] E1 = 12 E0 cos[6π × 109 t − 10π( 3x1 + x3 )] = Re[E √ √ ˆ i = 1 E0 exp[−j10π( 3x1 +x3 )] H ˆ i = H0 exp[−j10π( 3x1 +x3 )] E 1 2 2 1 ˆ i ). with H0 = ( µε00 ) 2 E0 (the same result may be deduced starting with E 3 √ ˆ r = R H0 exp[−j10π( 3x1 − x3 )] The reﬂected magnetic ﬁeld is H √ 2 √ Eˆ1r = −√12 R E0 exp[−j10π( 3x1 − x3 )] and Eˆ3r = − 12 3R E0 × exp[−j10π( 3x1 − x3 )] Apply, e.g., Eq. (4.176) R = −7.2 × 10−2 . √ Perform inversion E r = −( 12 i1 + 12 3i3 ) E0 R cos[6π × 109 t − √ 10π( 3x1 − x3 )]. √ In a similar way it follows that E t = 13 3(i1 −i3 ) E0 T cos[6π×109 t− √ 10π 3(x1 + x3 )], in which T = 0.928.
232
answers to exercises
(d)
... ... . ... ..... ......... . ..... ...... ............ ..... .... ....... ........... ............ ... ............ .. ... π ........... .............. π . 3 ...... ................ ..... 4 . .. π ............... 3 ...... ... .......... .......... ........... ........... ... . . ...
er et
ei
x3 = 0 Exercise 4.5 (2)
(2)
1
(a) θci = arcsin( nn(1) ) = arcsin[( εε(1) ) 2 ] = arcsin( 12 ) =
π 6
rad.
√ (b) γ1i = γ i sin(θi ) = γ0 3 and γ3i = γ i cos(θi ) = γ0 . 1
[−ω 2 ε(2) µ(2) − (γ i )2 ] 2 with Re[γ3t ] ≥ 0 ( c ) γ3t = √ −jγ0 2.
γ3t
1
= (γ02 − 3γ02 ) 2 =
An attenua(d) The wave is attenuated by a factor of exp{−Re[γ3t d]} √ tion of exp(−1) occurs over a distance x3 = 1/Re[γ3t ] = 12 2γ0 −1 . ( e ) Substitute the relevant quantities in Eq. (4.159) R⊥  = 1 and arg(R⊥ ) = 1.91 rad (109◦ ).
R⊥
=
√ 1+j √2 1−j 2
Exercise 4.6 All the light that comes in from the air enters the water at angles smaller than or equal to the critical angle; this is the reciprocal use of total reﬂection at angles of incidence larger than the critical angle for plane waves from water to air. The critical angle is θc = 0.85 rad (48.7◦ ). Exercise 4.7 nglass = 32 The critical angle exists for a wave incident on the glass/air nair interface and does not exist for a wave incident on the air/glass interface; this is independent of the type of polarization. glass εglass = 2.25 and µµair = 1 The Brewster angle does not exist in case εair of perpendicular polarization, whereas it exists for both the air/glass and the glass/air interfaces in case of parallel polarization.
233
answers to exercises
Interface
Polarization
Critical angle θci
i Brewster angle θB
Glass/air
⊥
arcsin( 23 )
—
arcsin( 23 )
arctan( 23 )
⊥
—
—
—
arctan( 32 )
Air/glass
Exercise 4.8 Take Eq. (4.181) and substitute ⎛ i tan(θB )
=⎝
1−
ε(2) ε(1)
ε(1) ε(2)
−1
⎞1 2
⎠ =
µ(2) µ(1)
=1
ε(2) (ε(1) − ε(2) ) ε(1) (ε(1) − ε(2) )
1
2
=
ε(2) ε(1)
1 2
Exercise 5.1 x3;0
θ0
Ray trajectory
2.50
45◦
Ray enters the halfspace x3 > 3 with θexit = 62◦
60◦
Ray turns at x3;hor = 2.84 and enters the halfspace x3 < 1 with θexit = 147◦
90◦
Ray starts at x3;hor = 2.50 and enters the halfspace x3 < 1 with θexit = 141◦
120◦
Ray enters the halfspace x3 < 1 with θexit = 147◦
45◦
Ray turns at x3;hor = 2.88 and enters the halfspace x3 < 1 with θexit = 148◦
60◦
Ray turns at x3;hor = 2.40 and enters the halfspace x3 < 1 with θexit = 139◦
90◦
Ray starts at x3;hor = 2.00 and enters the halfspace x3 < 1 with θexit = 131◦
120◦
Ray enters the halfspace x3 < 1 with θexit = 139◦
2.00
234
answers to exercises
(Continued) x3;0
θ0
Ray trajectory
1.50
45◦
Ray turns at x3;hor = 2.53 and enters the halfspace x3 < 1 with θexit = 142◦
60◦
Ray turns at x3;hor = 1.97 and enters the halfspace x3 < 1 with θexit = 131◦
90◦
Ray starts at x3;hor = 1.50 and enters the halfspace x3 < 1 with θexit = 119◦
120◦
Ray enters the halfspace x3 < 1 with θexit = 131◦
Exercise 5.2 The extreme positions where a ray may have a horizontal tangent are x3;hor = 1 and x3;hor = 3 For a horizontal tangent in between these extreme positions it is found that n(x3;hor ) > 1 For√an undulating ray, C0 = n(x3;hor ) sin( π2 ) > 1 C0 = n(x3;0 ) sin(θ0 ) = 2 sin(θ0 ) > 1 π 3π 4 < θ0 < 4 . Exercise 5.3 (1) (2) Yes: For any ray with 1 < x3;0 < 3 there exist a x3;hor and a x3;hor with (1)
(2)
(1)
(2)
1 < x3;hor < x3;0 < x3;hor < 3 and n(x3;0 ) > n(x3;hor ) = n(x3;hor ) > 1 For any 1 < x3;0 < 3 a range of θ0 may be found for which n(x3;0 ) sin(θ0 ) = (1,2) n(x3;hor ) sin( π2 ) > 1 arcsin( n(x13;0 ) ) < θ0 < π − arcsin( n(x13;0 ) ). Exercise 5.4 For any 1 < x3;0 < 2 there exists a x3;hor in the range x3;0 < x3;hor < 2 such that n(x3;0 ) sin(θ0 ) = n(x3;hor ) sin( π2 ) may be solved for θ0 Any starting position 1 < x3;0 < 2 has a ray with a horizontal tangent. For any 2 < x3;0 < 3 there exists a x3;hor in the range 2 < x3;hor < x3;0 such that n(x3;0 ) sin(θ0 ) = n(x3;hor ) sin( π2 ) may be solved for θ0 Any starting position 2 < x3;0 < 3 has a ray with a horizontal tangent. Since ∂3 n = 0 there is no ray with a horizontal tangent for x3;0 = 2. Exercise 5.5 For rays with a horizontal asymptote it must be possible to solve n(x3;0 ) sin(θ0 ) = n(2) sin( π2 ) for θ0 For all 1 < x3;0 < 2 a solution θ0 = arcsin[ n(x13;0 ) ] may be found, and for all 2 < x3;0 < 3 a solution
235
answers to exercises
θ0 = π − arcsin[ n(x13;0 ) ] exists. Since ∂3 n = 0 there is no ray with a horizontal tangent for x3;0 = 2.
Exercise 6.1 (a) b = 11.4 mm. (b) P = 12 Re[Vˆ Iˆ∗ ] = 12 Z0 ˆi+ 2 . 100 ( c ) Use Eq. (6.55) for φ = 100 E = (e21 + e22 )1/2 = r ln(b/a) 100 E has a maximum when r = a Emax = a ln(b/a) = 2.38 × 104 V/m.
Exercise 6.2 0 (a) v + (t) = Z0Z+Z S
)∞
n=0 (ΓS ΓL )
n V (t−2nL ) S c
=
5 8
)∞
3 n −8 n=0 ( 28 ) VS (t−4×10 n).
L ) + ΓL v + (t − 2L−L/2 ) = 58 [VS (t − 10−8 )− (b) V ( 12 L, t) = v + (t − 2c c 3 3 9 −8 −8 −8 7 VS (t − 3 × 10 ) + 28 VS (t − 5 × 10 ) − 196 VS (t − 7 × 10 ) + · · ·] and L 1 ) − ΓL v + (t − 2L−L/2 ) = 80 [VS (t − 10−8 )+ I( 12 L, t) = Z10 [v + (t − 2c c 3 3 9 −8 −8 −8 7 VS (t − 3 × 10 ) + 28 VS (t − 5 × 10 ) + 196 VS (t − 7 × 10 ) + · · ·].
( c ) WS = (d) W + = (e)
∞ 0
VS (t)I(0, t) dt =
10−9 0
1 Z0 +ZS
V (0, t)I(0, t) dt =
10−9 0
Z0 (Z0 +ZS )2
VS2 (t) dt = 12.5 pJ. 10−9 0
VS2 (t) dt = 7.8 pJ.
WL = 0∞ V (L, t)I(L, t) dt = 0∞ (1 + ΓL )v + (t − v + (t − Lc ) dt = Z10 (1 − Γ2L ) 0∞ [v + (t − Lc )]2 dt.
L 1 c ) Z0 (1
− ΓL )×
)∞
(2n+1)L 2 n 0 )] = ( Z0Z+Z )2 × n=0 (ΓS ΓL ) VS (t − c S )∞ (2n+1)L 2n 2 ) since VS (t− (2n+1)L )VS (t− (2m+1)L )=0 n=0 (ΓS ΓL ) VS (t− c c c 0 [v + (t − Lc )]2 = [ Z0Z+Z S
for all n = m. ∞ + Z0 L 2 2 )∞ (Γ Γ )2n ∞ V 2 (t− (2n+1)L ) dt = n=0 S L S 0 [v (t− c )] dt = ( Z0 +ZS ) 0 c 0 )2 ( Z0Z+Z S
1 1−(ΓS ΓL )2
× 10−9
WL
=
1−Γ2L Z0 (Z0 +ZS )2 1−(ΓS ΓL )2
× 10−9 =
6.5 pJ . Exercise 6.3 The 100 V signal travels in 2 µs from the beginning to the end
T
= 2 µs.
236
answers to exercises
At the end of the line the sum of the incoming 100 V signal and the once reﬂected signal gives a 75 V signal over the load A once reﬂected signal 1 L −Z0 ZL = 60 Ω. of −25 V signal originates at the end ΓL = Z ZL +Z0 = − 4 At the beginning of the line the reﬂected −25 V signal and the twice reﬂected signal gives a −10 V signal over the source A twice reﬂected 3 S −Z0 ZS = 25 Ω. signal of 15 V originates at the source ΓS = Z ZS +Z0 = − 5 The voltage divider existing of ZS = 25 Ω and Z0 = 100 Ω causes a 100 V signal right after t = 0 s V0 = 125 V. Exercise 6.4 To mimic a capacitor, the impedance at the beginning of the line must have −j the approximate form Zin ≈ ωC for some range of ω, say ω0 < ω < ω0 + ∆ω. The imput impedance of a shortcircuited transmission line may be writjZ0 . ten as Zin = cot(ωL/c) The approximation cot[ (1+2n)π + x] ≈ −x holds for x 1 and n ∈ ZZ 2 0 The transmission line behaves like a capacitor for L = (1+2m)πc = λ40 + mλ 2ω0 2 , ∆ω 2 with m ∈ IN , provided that ω0 (2m+1)π . Using an equivalent argumentation, it may be found that an open trans0 = λ40 + mλ mission line behaves like an inductor for L = (1+2m)πc 2ω0 2 , with 2 m ∈ IN , provided that ∆ω ω0 (2m+1)π . Exercise 6.5 −1 1 For a lossless transmission line, Γ = ΓL  ΓL  = VSWR VSWR+1 ΓL  = 2 1 for set A and ΓL  = 5 for set B Set B is better matched to the transmission line.
Exercise 7.1 1
(a) k = ω(εµ) 2 = 41.9 m−1 and πa = 31.4 m−1 k ≥ mπ a (propagating modes) for m = 1, k < mπ (evanescent modes) for m = 2, 3, 4, · · · . a (b) The ﬁeld strengths of the propagating TE1 mode follows from Eqs. (7.21)  (7.23) as ˆ2 = 2j Aˆ1 sin(31.4x1 ) exp(−27.7jx3 ), E ˆ 1 = −7.02 × 10−3 j Aˆ1 sin(31.4x1 ) exp(−27.7jx3 ), H ˆ 3 = −7.96 × 10−3 Aˆ1 cos(31.4x1 ) exp(−27.7jx3 ). H
237
answers to exercises
( c ) At x3 = 0 the reﬂection coeﬃcient of the electric ﬁeld is R⊥ = −1 ˆ2 = 4Aˆ1 sin(31.4x1 ) sin(27.7x3 ) (cf. ExThe total electric ﬁeld is E ˆ 1 = −1.40 × 10−2 j Aˆ1 × ercise 4.3) Use Eqs. (7.6) and (7.7) H −2 ˆ ˆ sin(31.4x1 ) cos(27.7x3 ), H3 = 1.59 × 10 j A1 cos(31.4x1 ) sin(27.7x3 ). Exercise 7.2 ˆ ×H ˆ ∗ ) · i3 ]mode dx1 For a given E0  the P = S · ν dA = 12 w 0a Re[(E 1 power is largest for the TM0  or TEMmode and equals P = 12 wa( µε ) 2 E0 2 . For safety take E0  = 2 × 105 V/m P = 26.5 kW. Exercise 7.3 2π k = 2πf c = λ
λg,m = k2π g,m
=√
1 1−4m2 /9
λg,0 = 1 m
and λg,1 = 1.34 m
while for m = 2, 3, 4, · · · the modes are evanescent and we can no longer speak of a guided wavelength. Exercise 7.4 The EHF band ranges from 30 GHz to 300 GHz The cutoﬀ frequency must be chosen in such a way that fc,0 ≤ 30 × 109 Hz < f < 300 × 109 Hz ≤ fc,1 . c0 Equation (7.113) with n3 = n1 and m = 0 gives fc,0 = 0 Hz and fc,1 = 6a fc,0 ≤ 30×109 Hz is satisﬁed for any value of a while fc,1 ≤ 300×109 Hz requires that a < 0.167 mm. Exercise 7.5 In each of the regions the general solution is ˆ (i) exp(−jk (i) x1 ). ˆ = Aˆ(i) exp(jk (i) x1 ) + B E 1 1 ˆ (1) = 0. Further it is known that Aˆ(3) = 0 and B (a) From the boundary conditions at x1 = a it follows that ˆ (2) exp(−jκ(2) a) = B ˆ (3) exp(−κ(1) a), Aˆ(2) exp(jκ(2) a) + B ˆ (2) exp(−jκ(2) a)] = jκ(1) B ˆ (3) exp(−κ(1) a). κ(2) [Aˆ(2) exp(jκ(2) a) − B From the boundary condition at x1 = −a it follows that ˆ (2) exp(jκ(2) a) = Aˆ(1) exp(−κ(1) a), Aˆ(2) exp(−jκ(2) a) + B ˆ (2) exp(−jκ(2) a)] = −jκ(1) Aˆ(1) exp(−κ(1) a). κ(2) [Aˆ(2) exp(jκ(2) a) − B
238
answers to exercises
(b) The ﬁrst two equations of (a) yield Aˆ(2) exp(jκ(2) a) = ˆ (2) exp(−jκ(2) a) = B
κ(2) + jκ(1) ˆ (3) B exp(−κ(1) a), 2κ(2) κ(2) − jκ(1) ˆ (3) B exp(−κ(1) a). 2κ(2)
The last two equations of (a) yield Aˆ(2) exp(−jκ(2) a) = ˆ (2) exp(jκ(2) a) = B
κ(2) − jκ(1) ˆ(1) A exp(−κ(1) a), 2κ(2) κ(2) + jκ(1) ˆ(1) A exp(−κ(1) a). (2) 2κ
( c ) From the ﬁrst and third equation of (b) and from the second and fourth equation of (b) it is found that κ(2) − jκ(1) ˆ(1) A exp(jκ(2) a) = 2κ(2) κ(2) + jκ(1) ˆ(1) A exp(−jκ(2) a) = 2κ(2)
κ(2) + jκ(1) ˆ (3) B exp(−jκ(2) a), 2κ(2) κ(2) − jκ(1) ˆ (3) B exp(jκ(2) a). 2κ(2)
ˆ (3) and Addition of these equations yields Aˆ(1) = B (1)
κ(2) m a = arctan(
κm
(2) κm
1 ) + mπ, m = 0, 2, 4, · · · . 2
ˆ (3) and Subtraction of these equations yields Aˆ(1) = −B (1)
κ(2) m a = arctan( (d)
κm
(2) κm
1 ) + mπ, m = 1, 3, 5, · · · . 2
ˆ (3) Substitute For m = 0, 2, 4, · · ·, part (c) shows that Aˆ(1) = B (2) ˆ (2) and Aˆ(2) cos(κm a) = 1 Aˆ(1) exp(−κ(1) in (b) Aˆ(2) = B m a) 2 ˆ2;m for even m, Substitute in the general solution This results in E (1) (even) (2) (1) = 2Aˆm = Am exp(−κ(2)m a) . in which Cˆm cos(κm a)
ˆ (3) Substitute For m = 1, 3, 5, · · ·, part (c) shows that Aˆ(1) = −B (2) ˆ (2) and Aˆ(2) sin(κm a) = 1 j Aˆ(1) exp(−κ(1) in (b) Aˆ(2) = −B m a) 2 ˆ Substitute in the general solution This results in E2;m for odd m, (1) (odd) (2) (1) in which Cˆm = 2j Aˆm = −Am exp(−κ(2)m a) . sin(κm a)
239
answers to exercises
Exercise 8.1 In the Taylor series expansion around the stationary point k1s the linear term is proportional to the ﬁrst derivative at k1s , which by deﬁnition is equal to zero. Exercise 8.2 Parallel electric current: D(θ) = 0 if sin(θ) = 0 θ = 0 and θ = π. ˆ + [k cos(θ), jω] = sin[ka cos(θ)/2] ˆı(jω) = 0 D(θ) = 0 if h 2 k cos(θ)
ka cos(θ) 2
= nπ
with n ∈ ZZ \{0} θ = Perpendicular electric current: ˆ + [k cos(θ), jω] = D(θ) = 0 if h 1
ka cos(θ) 2
= nπ
arccos( nλ a )
with n ∈ ZZ \{0}
θ=
with nλ ≤ a.
sin[ka cos(θ)/2] ˆı(jω) k cos(θ)
arccos( nλ a )
= 0
with nλ ≤ a.
Exercise 8.3 (a) Use Eq. (8.56) with a delta function in the integrand 1 ı(jω). 2ˆ (b)
(c)
hˆ +1 (k1 , jω) =
ˆ+ H ˆ ≈ Take together Eqs. (8.58)  (8.59) and substitute h 1 1 1 k 2 ı∆ (jω) ( 2πr ) exp(−jkr + j π4 ) [sin(θ)i1 − cos(θ)i3 ]. 2ˆ ˆ+ E ˆ ≈ − 1 ( µ ) 12 ˆı∆ (jω) ( k ) 21 × Take Eq. (8.60) and substitute h 1 2 ε 2πr exp(−jkr + j π4 )i2 . Note that the electric ﬁeld strength and the magnitude of the magnetic ﬁeld strength are not dependent on the angle of observation. θ=0 ⊗
......................... ........... .. ......... .... . .. . . ........ . . . . . . . . ... ....... ... . . ... ..... ... . ........ .. .. ... ... .. .. ... . .. .. ... . . . . . . . ... . ... ...... . .. . . . ....... . ........ . . . . . . ..... .... ....... ... ... .. ....... .. ....... .. . ... ....... ... ........... .. ....... . . . ......... . .. ....... .. .... .. ....... .... . . . . . . . ...................................... . ....... ....... ....... ....... ....... ....... ....... ....... ... .................. .... ... . . ...... ... ...... . . ... ....... ......... ....... . ... .... ....... ... ....... . ....... .. ... ...... ... ....... . ........... ... ... .. ..... . . . ....... ..... . . . ....... . ... . . . . . .. ... . .. ... ... . ........... .. .... ... ... . ... . ....... .. ... ... .......... . ....... ....... . .......... .. ....... .........................
⊗
⊗
⊗
⊗ E
i1
O p
θ = 32 π ⊗
H
⊗ θ = 12 π
i3
E⊗
⊗
H ⊗
⊗ θ=π
⊗
240
answers to exercises
* +1
ˆ ×H ˆ ∗ ] = ˆı∆ (jω)2 µ 2 k × (d) The timeaveraged power ﬂow is 12 Re[E ε 16πr [cos(θ)i1 + sin(θ)i3 ] D(θ) = 1 The directive gain is the unit circle.
Bibliography A. Einstein (1956), The Meaning of Relativity, Princeton University Press, Princeton. D.K. Cheng (1993), Fundamentals of Engineering Electromagnetics, AddisonWesley Publishing Company, Reading, Massachusetts. R.L. Coren (1989), Basic Engineering Electromagnetics, PrenticeHall, Englewood Cliﬀs, New Jersey. C.T.A. Jonk (1988), Engineering Electromagnetic Fields and Waves, John Wiley & Sons, New York. J.D. Kraus (1992), Electromagnetics, fourth edition, McGrawHill, New York. S.V. Marshall and G.G. Skitek (1990), Basic Engineering Electromagnetics, third edition, PrenticeHall, Englewood Cliﬀs, New Jersey. J.C. Maxwell (1873), A Treatise on Electricity and Magnetism, Clarendon Press, Oxford. N.N. Rao (1987), Elements of Engineering Electromagnetics, second edition, PrenticeHall, Englewood Cliﬀs, New Jersey. L.C. Shen and J.A. Kong (1983), Applied Electromagnetism, Brooks/Cole Engineering Division, Moneterey, California. K.F. Sanders and G.A.L. Reed (1986), Transmission and Propagation of Electromagnetic Waves, second edition, Cambridge University Press, Cambridge. E.T. Whittaker (1953), A history of the Theories of Aether and Electricity, vol. I: Classical theories; vol. II: Modern theories (19001926), Nelson, London.
Index Admittance, 52 Angle of incidence, 99, 108, 111, 113, 114 of reﬂection, 99, 108 of transmission, 109 Angular direction, 211, 218 Angular frequency, 37, 54, 84 Antenna, 49, 203 Attenuation coeﬃcient, 57, 84, 87, 178 Attenuation factor, 58, 86 Axial direction, 168
Base vector, 2 Boundary curved, 121 plane, 95, 101, 121 Boundary condition, 32, 65, 69, 73, 96–98, 103, 106, 147, 149, 151, 153, 157, 174, 176, 183, 184, 188, 189 Boundary surface, 95 Brewster angle, 113 Bromwich inversion integral, 36
Capacitance per unit length, 74 Cartesian coordinates, 4 Cartesian reference frame, 2 Cartesian vector, 4 Causal ﬁeld, 25 Causality, 29, 36, 52, 95, 101–103, 162 Characteristic admittance, 150 Characteristic equation, 185, 189
Characteristic impedance, 74, 150, 152, 154, 158, 159, 162 Circuit equivalent, 74, 79, 155 Circuit parameter, 73 Coaxial cable, 145 Coaxial line, 152, 156, 158 Compatibility relation, 25, 28 Conductance per unit length, 73 Conductivity, 29, 34, 81, 121 Conductor, 46 Conservation of electric charge, 45 of energy, 43, 44 Constitutive relation, 28, 31 Continuity of normal component of S, 34 of tangential components of E, 34, 64, 68, 103 of tangential components of H, 34, 64, 68, 103 Critical angle, 114 Curl, 7, 13 Current distribution nonuniform, 121 uniform, 121 Cutoﬀ angular frequency, 178 Cutoﬀ frequency, 180, 193 of TEmode, 193 of TMmode, 194 Cylindrical wave, 211, 218
Dielectric property, 30 Diﬀerentiation with respect to parameter, 6 with respect to spatial coordinates, 6
244
Dirac distribution, 33, 50, 204, 213 Direction of propagation, 54, 56, 58, 60, 61, 83, 88–90, 92, 99, 108, 109, 124, 126, 127, 130, 131, 180 Directional derivative, 10 Directive gain, 211, 213, 219 Dispersion geometrical, 193 waveguiding, 193 Dispersion curve, 179, 195 Dispersion equation, 186, 191 Dispersion relation, 174, 177 Distributed circuit, 75 Divergence, 7, 10
Epolarization, 83 Epolarized wave, 172, 173, 182, 183 Eigenvalue, 185, 190 Eigenvalue equation, 185, 189 Eikonal, 122, 125, 129 Eikonal equation, 126, 130 Electric charge, 22 volume density of, 28 Electric current, 73 volume density of, 25, 27 Electriccurrent sheet, 50, 203, 204, 213 Electric ﬁeld analysis, 53, 64, 69, 108 Electric ﬁeld strength, 22, 24, 88, 89 Electric ﬂux density, 27 Electric force, 23 Electric polarization, 27 Electric potential, 73 Electrically impenetrable halfspace, 95, 109 Electrically impenetrable object, 34 Electromagnetic ray, 121, 124, 126, 129 parametric representation of, 132 uniform, 126, 130, 131
index
Electromagnetic theory, 2 Electromagnetic wave, 1 Electromagnetic wave propagation, 49 Electromagnetic wave speed, 24 Energy stored in electric ﬁeld, 42 stored in magnetic ﬁeld, 42 volume density of electric ﬁeld, 43 volume density of magnetic ﬁeld, 43 Excitation, 1, 203 Excitation condition, 53, 206, 215
Farﬁeld approximation, 207, 216 Fast Fourier Transform, 36 Fiber gradedindex, 169 stepindex, 171 Field equations, 21, 31 in frequency domain, 37, 146 Flux, 12 Force, 21, 22 Fourier integral, 206, 207, 216 Fourier transform, 209, 211, 217, 219 Fourier transformation, 37 Fourier’s inversion theorem, 207, 216 Frequency, 54, 56 Frequency domain, 36, 108 Fresnel coeﬃcient for reﬂection, 111, 113 for transmission, 111
Gauss’ integral theorem, 12, 41 Geometrical optics, 121 Gradient, 6, 8 Group velocity, 193 Guided wavelength, 198
Hpolarization, 82 Hpolarized wave, 172, 175, 182, 188
index
Heat, 42 volume density of, 44 Helmholtz equation, 173, 176, 184, 188 modiﬁed, 51, 74, 82, 83
Imaginary unit, 36 Impedance, 53 Incident ﬁeld, 62, 67 Incident wave, 95, 101, 108, 121 Index of refraction, 109, 128, 131, 133, 135, 142 eﬀective, 193, 195 Inductance per unit length, 74 Infrared, 56 Input impedance, 157, 158, 161 capacitive, 159 inductive, 159 Integratedoptics system, 171 Interface, 32, 62, 101 Interference, 90 International System of Units, 4 Isotropic radiator, 219
Laplace equation, 149, 151, 153 Laplace transform, 161 Laplace transform parameter, 36 Laplace transformation, 36 shift rule of, 60, 161, 162 Tauber’s theorem for, 124, 125, 129 Lens of Luneberg, 142 Line source, 221 Load, 156 matched, 158, 159 opencircuit, 158, 159 shortcircuit, 158, 159 Load impedance, 156 Longitudinal direction, 168 Lorentz force, 23 Loss tangent, 80 Lumped circuit, 75
Magnetic current volume density of, 25
245
Magnetic ﬁeld analysis, 53, 65, 105 Magnetic ﬁeld strength, 22, 24, 88, 89 Magnetic ﬂux density, 27 Magnetic force, 23 Magnetic property, 30 Magnetically impenetrable object, 35 Magnetization, 27 Matter properties of, 28 Maxwell’s equations, 31 in matter, 25 in vacuum, 24 Maxwell, James Clerk, 2 Medium anisotropic, 29 dielectric, 111, 113 heterogeneous, 29 highly conducting, 59 homogeneous, 29, 121, 122 inhomogeneous, 29, 121, 124, 127, 128, 130 instantaneously reacting, 28 isotropic, 29, 124, 127, 128, 130, 131 linear, 28 locally reacting, 29 lossless, 55, 59, 86, 92, 100, 109, 111, 122, 160 lossy, 57 nonlinear, 28 parametrically aﬀected, 28 passive, 28 time invariant, 28 time variant, 28 Medium parameter, 133, 138 Mode evanescent, 180 fundamental, 194 guided, 186, 191, 193 nonpropagating, 175, 178, 180 propagating, 178, 180, 193 propagation properties of, 178, 193
246
TEm , 175, 180, 186, 193 TEM, 178 TMm , 177, 180, 191, 193 Mode index, 193
Normal incidence, 105, 108 Onedimensional wave, 49, 56, 58, 60, 61, 66, 87, 105, 108, 167, 203 Operator curl, 7, 13 del, 6 div, 7, 10 grad, 6, 8 Laplacian, 18 nabla, 6 of Hamilton, 6 Optical communication system, 169 Optical ﬁber, 169 Orientation of electric ﬁeld, 54 of magnetic ﬁeld, 54 Origin, 2
Parallel wires, 145 Penetration depht, 80 Permeability, 22, 24, 29, 35, 81, 121 relative, 30 Permittivity, 24, 29, 34, 81, 121, 201 relative, 30 Phase, 55 Phase coeﬃcient, 57, 87, 178 Phase factor, 58, 86, 87 Phase vector, 84 Phase velocity, 193 Physical laws, 4 Physical quantity, 2, 4 Plane of equal amplitude, 58, 86, 114, 195 of equal phase, 58, 86, 114, 195 of observation, 82, 83
index
Plane wave, 49, 83, 84, 90, 95, 101, 121, 126, 128, 130, 131, 167, 180, 204, 214 evanescent, 206, 215 nonuniform, 86, 109, 114, 195, 203, 206, 215 parallelly polarized, 87, 93 perpendicularly polarized, 89, 93 uniform, 86–89, 91, 99, 108, 109, 122, 126, 130, 203, 206, 211, 215, 219 Planewave decomposition, 180 Plasma, 201 Plasma frequency, 201 Point charge, 21 Polarization circular, 39 elliptical, 39 lefthanded, 41 linear, 39 parallel, 82, 87, 93, 97, 100, 103, 111, 113, 114, 122, 124, 172, 182, 204 perpendicular, 83, 89, 93, 98, 100, 106, 111, 113, 114, 122, 128, 172, 182, 213 righthanded, 41 Polarization state, 39 Position, 2 Power, 114, 159 dissipated into heat, 42 generated by sources, 43 instantaneous, 41 volume density of, 43, 44 Power ﬂow, 32, 89, 90, 92–94, 159 Power ﬂow density, 175, 187, 192, 210, 218 Power relation, 41 Poynting theorem, 41 complex, 44 instantaneous, 41 Poynting vector, 31, 43, 58, 61, 86, 88, 90, 92 complex, 44
247
index
Product cross, 5 dot, 4 scalar, 4 vector, 5 Propagation, 1 Propagation coeﬃcient, 51, 55, 57, 59, 62, 63, 67, 68, 87, 160, 167, 185, 186, 190, 191 Propagation constant, 174, 177, 178 Propagation factor, 58, 86 Propagation laws, 2 Propagation vector, 84
Radiation, 50 Ray approximation, 121, 126, 129 Ray trajectory, 121, 124, 127, 130, 131 in horizontally layered medium, 133 in radially layered medium, 138 Receiver, 1 Reﬂected ﬁeld, 62, 67 Reﬂected wave, 95, 101, 108, 121 Reﬂection, 62, 95, 101 Reﬂection coeﬃcient, 64, 66, 69, 104, 105, 107, 108, 111, 156, 157, 161, 163 reﬂection coeﬃcient, 158 Refraction, 1 Relativity, 2, 24 Relaxation, 29 Relaxation time, 46 Resistance, 79 Righthanded triad, 2
Scalar function, 6 derivative of, 6 Scalar potential, 149 Separation of variables, 148 Sheet emitter parallel electric current, 204 perpendicular electric current, 213 Shielding, 66
Shielding eﬀectiveness, 70 SI, 4 Skin depth, 59, 80 Snell’s law for horizontally layered medium, 135 for radially layered medium, 139 of reﬂection, 109 of refraction, 109, 113 of transmission, 109 Source, 1, 26, 50, 81, 122, 156 nonplanar, 121 planar, 121 Sourcefree interface, 34 Spatial period, 56, 87 Speciﬁc value, 185, 190 Standard clock, 2 Standard experiments, 2, 4 Standard measuring rod, 2 Stationary phase, 207, 217 Stationary point, 207 Steadystate, 38, 43, 54, 84, 92, 100, 158 Stokes’ integral theorem, 15 Substrate, 171, 182 Superposition, 49, 90, 93, 100, 175, 177, 180, 203, 204, 206, 214, 215 Superstrate, 171, 182 Surface impulse, 33 Surface source, 33
Taylor series, 208 TEwave, 172, 173, 182, 183 TEMwave, 72, 145, 146, 167, 172, 175 Time, 2 Timeinvariant conﬁguration, 36 TMwave, 172, 175, 182, 188 Total reﬂection, 114 Trajectory constant, 135, 139 Transient, 160 Transient emission, 59 Transient ﬁeld, 36 Transmission, 62, 101
248
Transmission coeﬃcient, 65, 69, 104, 105, 107, 108, 111 Transmission line, 49, 145, 167 lossless, 158, 160 multiconductor, 145 twoconductor, 156 Transmission line equations, 73, 79, 150 Transmission line equivalent, 73 Transmitted ﬁeld, 63, 68 Transmitted wave, 102, 109, 110, 114, 121 Transverse electric wave, 172 Transverse electromagnetic wave, 72, 145, 167 Transverse magnetic wave, 172 Transverse plane, 168 Triple product scalar, 5 vectorial, 5 Twodimensional medium, 121 Twodimensional wave, 81, 121, 167, 203
Uniform line, 145 Unit, 4 Unit impulse onedimensional, 50, 204, 213 twodimensional, 221
Vacuum, 30 Vector, 3 addition of, 4 components of, 4 length of, 5, 37 multiplication of, 4 properties of, 4 subtraction of, 4 Vector calculus, 4 Vector function, 6 derivative of, 6 Vectorial position, 3 Velocity of observer, 55 of point charge, 22
index
Voltage standing wave ratio, 165 VSWR, 165
Wave evanescent, 175, 177 nonpropagating, 175, 177 not plane, 121 plane, see Plane wave propagating, 175, 177 standing, 100 travelling, 100 Wave admittance, 52, 55, 62, 63, 68, 89, 108 Wave impedance, 53, 55, 64–66, 69, 88, 106 Wave speed, 56, 60, 61, 161, 163 Wavefront, 121, 123, 125, 127, 129, 130 Waveguide, 167 asymmetric dielectric slab, 194 closed, 168 cylindrical, 167, 168 dielectric slab, 171, 182, 193 gradedindex slab, 182 open, 168, 169 parallelplate, 72, 145, 151, 156, 158, 171, 172, 178 planar, 171 planar gradedindex, 171 planar stepindex, 171 stepindex slab, 182 symmetrical slab, 182 thinﬁlm, 182 uniformly cylindrical, 169 Wavelength, 56, 87, 94 Wavenumber, 174, 177, 193
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Mass Transfer in Multicomponent Mixtures J.A. Wesselingh and R. Krishna 2006 / 329 p. / ISBN 9071301583 http://www.vssd.nl/hlf/d004.htm Transport phenomena data companion L.P.B.M. Janssen and M.M.C.G. Warmoeskerken 2006 / 168 p. / ISBN 9071301591 http://www.vssd.nl/hlf/c017.htm
Mathematical Systems Theory G.J. Olsder and J.W. van der Woude 2008 / x+208 p. / ISBN 9071301400 http://www.vssd.nl/hlf/a003.htm Numerical Methods in Scientific Computing J. van Kan, A. Segal and F. Vermolen 2005 / xii+277 pp. / ISBN 9071301508 http://www.vssd.nl/hlf/a002.htm Order in Space JA.K. van der Vegt 2006 / 93 pp. / ISBN 9071301613 http://www.vssd.nl/hlf/a017.htm
Electromagnetic Waves – An Introductory Course Electromagnetic waves appear in many forms and their applications are extremely widespread. Without exaggeration it may be said that our ability to employ and manipulate electromagnetic waves forms one of the reasons that communication plays such an important role in society. The macroscopic theory of electromagnetic waves has been formulated by Maxwell in 1864. But the mathematicalphysical nature of the subject makes it difficult for students to master even today. The continuous stream of new college textbooks shows that many teachers encounter this problem and attempt to resolve it by presenting the theory in some suitable form. In the Electrical Engineering curriculum of the Delft University of Technology, the teaching of electromagnetic waves has been divided into three stages: 1) a basic course on Electricity and Magnetism, 2) an introductory course on Electromagnetic Waves, and 3) advanced courses on the application and computation of electromagnetic waves. The current book is written to facilitate the introductory course on Electromagnetic Waves. It is assumed that students are already acquainted with the basic phenomena and notions of the electric and the magnetic field, and that they know in which way Maxwell's equations describe the electromagnetic field. Starting from Maxwell's equations, this book deals with the derivation of plane wave propagation, plane wave reflection and transmission, electromagnetic rays, waves in twowire transmission lines, waves in planar waveguides, and the excitation of electromagnetic waves. As such, the aim of the book is to provide a solid understanding of how the basic ingredients that make up the more sophisticated applications follow from Maxwell's equations. The aim of the introductory course on Electromagnetic Waves is to teach students to manipulate the fundamental formulas in order to solve a problem at hand. To focus on this skill and to overcome the problem of having to learn many formulas by heart, an outline of this book is available from the website as a printable Repetitive Guide. Published by VSSD. ISBN 9040718369 EAN 9789040718366
URL on this book: http://www.vssd.nl/hlf/e016.htm
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