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Electromagnetics - E. Rothwell, M. Cloud

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Published textbook for a first-year graduate engineering electromagnetics sequence. Six chapters cover field concepts and sources, Maxwell's postulate, static fields, temporal and spatial frequency-domain representation (including Kronig-Kramers relations), potentials and field decompositions, and the Stratton-Chu integral solution. A mathematical appendix covers Fourier analysis, dyadics, contour integration and boundary value problems. Not Phil's own work; no annotations are evident in the text shown.

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ELEGTROMAGHETICS Edward J.Rothwell Michael J.Cloud creCRC PRESS ELECTROMAGNETICS Electrical Engineering Textbook Series Richard C. Dorf, Series Editor University of California, Davis Forthcoming Titles Applied Vector Analysis Matiur Rahman and Issac Mulolani Optimal Control Systems Subbaram Naidu Continuous Signals and Systems with MATLAB Taan ElAli and Mohammad A. Karim Discrete Signals and Systems with MATLAB Taan ElAli Edward J. Rothwell Michigan State University East Lansing, Michigan Michael J. Cloud Lawrence Technological University Southfield, Michigan Boca Raton London New York Washington, D.C.CRC Press This book contains information obtained from authentic and highly regarded sources. Reprinted material is quoted with permission, and sources are indicated. A wide variety of references are listed. Reasonableefforts have been made to publish reliable data and information, but the author and the publisher cannotassume responsibility for the validity of all materials or for the consequences of their use. Neither this book nor any part may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, microﬁlming, and recording, or by any information storage orretrieval system, without prior permission in writing from the publisher. The consent of CRC Press LLC does not extend to copying for general distribution, for promotion, for creating new works, or for resale. Speciﬁc permission must be obtained in writing from CRC Press LLCfor such copying. Direct all inquiries to CRC Press LLC, 2000 N.W. Corporate Blvd., Boca Raton, Florida 33431, or visit our Web site at www.crcpress.com Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identi ﬁcation and explanation, without intent to infringe. Visit our website at www.crcpress.com . © 2001 by CRC Press LLC No claim to original U.S. Government works International Standard Book Number 0-8493-1397-X Library of Congress Card Number 00-065158 Printed in the United States of America 1 2 3 4 5 6 7 8 9 0 Printed on acid-free paper Library of Congress Cataloging-in-Publication Data Rothwell, Edward J. Electromagnetics / Edward J. Rothwell, Michael J. Cloud. p. cm. —(Electrical engineering textbook series ; 2) Includes bibliographical references and index.ISBN 0-8493-1397-X (alk. paper) 1. Electromagnetic theory. I. Cloud, Michael J. II. Title. III. Series. QC670 .R693 2001 530.14 ′ 1—dc21 00-065158 CIP In memory of Catherine Rothwell Preface This book is intended as a text for a ﬁrst-year graduate sequence in engineering electro- magnetics. Ideally such a sequence provides a transition period during which a studentcan solidify his or her understanding of fundamental concepts before proceeding to spe- cialized areas of research. The assumed background of the reader is limited to standard undergraduate topics in physics and mathematics. Worthy of explicit mention are complex arithmetic, vec-tor analysis, ordinary diﬀerential equations, and certain topics normally covered in a“signals and systems” course (e.g., convolution and the Fourier transform). Further an-alytical tools, such as contour integration, dyadic analysis, and separation of variables,are covered in a self-contained mathematical appendix. The organization of the book is in six chapters. In Chapter 1 we present essential background on the ﬁeld concept, as well as information related speciﬁcally to the electro-magnetic ﬁeld and its sources. Chapter 2 is concerned with a presentation of Maxwell’stheory of electromagnetism. Here attention is given to several useful forms of Maxwell’sequations, the nature of the four ﬁeld quantities and of the postulate in general, somefundamental theorems, and the wave nature of the time-varying ﬁeld. The electrostaticand magnetostatic cases are treated in Chapter 3. In Chapter 4 we cover the representa-tion of the ﬁeld in the frequency domains: both temporal and spatial. Here the behaviorof common engineering materials is also given some attention. The use of potentialfunctions is discussed in Chapter 5, along with other ﬁeld decompositions including thesolenoidal–lamellar, transverse–longitudinal, and TE–TM types. Finally, in Chapter 6we present the powerful integral solution to Maxwell’s equations by the method of Strat-ton and Chu. A main mathematical appendix near the end of the book contains brief but suﬃcient treatments of Fourier analysis, vector transport theorems, complex-plane inte-gration, dyadic analysis, and boundary value problems. Several subsidiary appendicesprovide useful tables of identities, transforms, and so on. We would like to express our deep gratitude to those persons who contributed to the development of the book. The reciprocity-based derivation of the Stratton–Chu formulawas provided by Prof. Dennis Nyquist, as was the material on wave reﬂection from multiple layers. The groundwork for our discussion of the Kronig–Kramers relations wasprovided by Michael Havrilla, and material on the time-domain reﬂection coeﬃcient wasdeveloped by Jungwook Suk. We owe thanks to Prof. Leo Kempel, Dr. David Infante,and Dr. Ahmet Kizilay for carefully reading large portions of the manuscript during itspreparation, and to Christopher Coleman for helping to prepare the ﬁgures. We areindebted to Dr. John E. Ross for kindly permitting us to employ one of his computerprograms for scattering from a sphere and another for numerical Fourier transformation.Helpful comments and suggestions on the ﬁgures were provided by Beth Lannon–Cloud. Thanks to Dr. C. L. Tondo of T & T Te chworks, Inc., for assistance with the LaTeX macros that were responsible for the layout of the book. Finally, we would like to thankthe staﬀ members of CRC Press — Evelyn Meany, Sara Seltzer, Elena Meyers, HelenaRedshaw, Jonathan Pennell, Joette Lynch, and Nora Konopka — for their guidance andsupport. Contents Preface 1Introductor yconcepts 1.1Notation ,conventions,andsymbology 1.2Theﬁeldconcep tofelectromagnetics 1.2.1Historica lperspective 1.2.2Formalizatio nofﬁeldtheory 1.3Thesource softheelectromagneti cﬁeld 1.3.1Macroscopi celectromagnetics 1.3.2Impresse dvs.secondar ysources 1.3.3Surfac eandlinesourcedensities 1.3.4Charg econser vation 1.3.5Magneti ccharge 1.4Problems 2Maxwell’stheoryofelectromagnetism 2.1Thepostulate 2.1.1TheMaxwell–Min kowskiequations 2.1.2Connectio ntomechanics 2.2Thewell-posednatureofthepostulate 2.2.1Uniquenes sofsolution stoMaxwell’sequations 2.2.2Constituti verelations 2.3Maxwell’sequation sinmovingframes 2.3.1Fieldconversion sunderGalilea ntransformation 2.3.2Fieldconversion sunderLorentztransformation 2.4TheMaxwell–Bo ﬃequations 2.5Large-scal eformofMaxwell’sequations 2.5.1Surfac emovingwithconsta ntvelocity 2.5.2Moving,deformin gsurfaces 2.5.3Large-scal eformoftheBoﬃequations 2.6Thenatureofthefourﬁeldquantities 2.7Maxwell’sequation swithmagneti csources 2.8Boundar y(jump )condition s 2.8.1Boundar ycondition sacrossastationar y,thinsourcelayer 2.8.2Boundar ycondition sacrossastationar ylayerofﬁelddiscontinuity 2.8.3Boundar ycondition satthesurfac eofaperfectconductor 2.8.4Boundar ycondition sacrossastationar ylayerofﬁelddiscontinuityusing equivalentsources 2.8.5Boundar ycondition sacrossamovinglayerofﬁelddiscontinuity 2.9Fundame ntaltheorems 2.9.1Lineari ty 2.9.2Duality 2.9.3Recipr ocity 2.9.4Similitude 2.9.5Conser vationtheorems 2.10Thewavenatureoftheelectromagneti cﬁeld 2.10.1Electromagneti cwaves 2.10.2Waveequatio nforbianisotropi cmaterials 2.10.3Waveequatio ninaconductin gmedium 2.10.4Scalarwaveequationforaconductin gmedium 2.10.5Fieldsdetermine dbyMaxwell’sequation svs.ﬁeldsdetermine dbythe waveequation 2.10.6Transie ntunifor mplanewavesinaconductin gmedium 2.10.7Propagatio nofcylindrica lwavesinalossles smedium 2.10.8Propagatio nofspherica lwavesinalossles smedium 2.10.9Nonradiatin gsources 2.11Problems 3Thestaticelectromagneti cﬁeld 3.1Staticﬁeldsandsteadycurrents 3.1.1Decouplin goftheelectri candmagneti cﬁelds 3.1.2Staticﬁeldequilibriu mandconductors 3.1.3Steadycurrent 3.2Electrostatics 3.2.1Theelectrostati cpotentialandwork 3.2.2Boundar yconditions 3.2.3Uniquenes softheelectrostati cﬁeld 3.2.4Poisson’ sandLaplace’ sequations 3.2.5Forceandenergy 3.2.6Multipoleexpansion 3.2.7Fieldproducedbyapermane ntlypolarize dbody 3.2.8Potentialofadipolelayer 3.2.9Behaviorofelectri cchargedensitynearaconductin gedge 3.2.10Solutio ntoLaplace’ sequatio nforbodiesimmerse dinanimpresse dﬁeld 3.3Magnetostatics 3.3.1Themagneti cvectorpotential 3.3.2Multipoleexpansion 3.3.3Boundar ycondition sforthemagnetostati cﬁeld 3.3.4Uniquenes softhemagnetostati cﬁeld 3.3.5Integralsolutio nforthevectorpotential 3.3.6Forceandenergy 3.3.7Magneti cﬁeldofapermane ntlymagnetize dbody 3.3.8Bodiesimmerse dinanimpresse dmagneti cﬁeld:magnetostati cshielding 3.4Staticﬁeldtheorems 3.4.1Meanvaluetheore mofelectrostatics 3.4.2Earnsh aw’stheorem 3.4.3Thomson’ stheorem 3.4.4Green’ srecipr ocationtheorem 3.5Problems 4Temporalandspatia lfrequenc ydomai nreprese ntation 4.1Interpretatio nofthetemporaltransform 4.2Thefrequency-domai nMaxwellequations 4.3Boundar ycondition sonthefrequency-domai nﬁelds 4.4Theconstituti veandKronig–Kramer srelations 4.4.1Thecomple xpermittivi ty 4.4.2Highandlowfrequenc ybehaviorofconstituti veparameters 4.4.3TheKronig–Kramer srelations 4.5Dissipate dandstoredenergyinadispersivemedium 4.5.1Dissipatio ninadispersivematerial 4.5.2Energ ystoredinadispersivematerial 4.5.3Theenergytheorem 4.6Somesimplemodelsforconstituti veparameters 4.6.1Comple xpermittivi tyofanon-magnetize dplasma 4.6.2Comple xdyadicpermittivi tyofamagnetize dplasma 4.6.3Simpl emodelsofdielectrics 4.6.4Permittivi tyandconductivi tyofaconductor 4.6.5Permeabili tydyadicofaferrite 4.7Monochromati cﬁeldsandthephaso rdomain 4.7.1Thetime-harmoni cEMﬁeldsandconstituti verelations 4.7.2Thephaso rﬁeldsandMaxwell’sequations 4.7.3Boundar ycondition sonthephaso rﬁelds 4.8Poynting’stheore mfortime-harmoni cﬁelds 4.8.1Genera lformofPoynting’stheorem 4.8.2Poynting’stheore mfornondis persivematerials 4.8.3Lossless ,lossy,andactivemedia 4.9Thecomple xPoyntingtheorem 4.9.1Boundar yconditio nforthetime-averagePoyntingvector 4.10Fundame ntaltheorem sfortime-harmoni cﬁelds 4.10.1Uniqueness 4.10.2Recipr ocityrevisited 4.10.3Duality 4.11Thewavenatureofthetime-harmoni cEMﬁeld 4.11.1Thefrequency-domai nwaveequation 4.11.2Fieldrelationship sandthewaveequatio nfortwo-dimensiona lﬁelds 4.11.3Planewavesinahomogeneous ,isotropic ,lossymaterial 4.11.4Monochromati cplanewavesinalossymedium 4.11.5Planewavesinlayeredmedia 4.11.6Plane- wavepropagatio ninananisotropi cferritemedium 4.11.7Propagatio nofcylindrica lwaves 4.11.8Propagatio nofspherica lwavesinaconductin gmedium 4.11.9Nonradiatin gsources 4.12Interpretatio nofthespatia ltransform 4.13Spatia lFourierdecom position 4.13.1Boundar yvalueproblem susingthespatia lFourierreprese ntation 4.14PeriodicﬁeldsandFloquet’stheorem 4.14.1Floquet’stheorem 4.14.2Example sofperiodicsystems 4.15Problems 5Fielddecom position sandtheEMpotentials 5.1Spatia lsymmetr ydecom positions 5.1.1Plana rﬁeldsymmetry 5.2Solenoidal–lamella rdecom position 5.2.1Solutio nforpotentialsinanunbounde dmedium :theretarde dpotentials 5.2.2Solutio nforpotentialfunction sinabounde dmedium 5.3Transverse–longitudina ldecom position 5.3.1Transverse–longitudina ldecom positionintermsofﬁelds 5.4TE–T Mdecom position 5.4.1TE–T Mdecom positionintermsofﬁelds 5.4.2TE–T Mdecom positionintermsofHertzia npotentials 5.4.3Application :hollow-pipewaveguides 5.4.4TE–T Mdecom positioninspherica lcoordinates 5.5Problems 6Integralsolution sofMaxwell’sequations 6.1VectorKirchoﬀsolution 6.1.1TheStratton–C huformula 6.1.2TheSommerfel dradiatio ncondition 6.1.3Fieldsintheexclude dregion :theextinctio ntheorem 6.2Fieldsinanunbounde dmedium 6.2.1Thefar-zon eﬁeldsproducedbysource sinunbounde dspace 6.3Fieldsinabounded ,source-fre eregion 6.3.1ThevectorHuygen sprinciple 6.3.2TheFranzformula 6.3.3Love’sequivalenceprinciple 6.3.4TheSchelkuno ﬀequivalenceprinciple 6.3.5Far-zon eﬁeldsproducedbyequivalentsources 6.4Problems AMathematica lappendix A.1TheFouriertransform A.2Vectortransporttheorems A.3Dyadicanalysis A.4Boundar yvalueproblems BUsefulidentities CSomeFouriertransfor mpairs DCoordinat esystem s EPropertiesofspecialfunctions E.1Besselfunctions E.2Legendr efunctions E.3Spherica lharmonics References Chapter 1 Introductory concepts 1.1 Notation, conventions, and symbology Any book that covers a broad range of topics will likely harbor some problems with notation and symbology. This results from having the same symbol used in diﬀerent areasto represent diﬀerent quantities, and also from having too many quantities to represent.Rather than invent new symbols, we choose to stay close to the standards and warn thereader about any symbol used to represent more than one distinct quantity. The basic nature of a physical quantity is indicated by typeface or by the use of a diacritical mark. Scalars are shown in ordinary typeface: q,/Phi1, for example. Vectors are shown in boldface: E,Π. Dyadics are shown in boldface with an overbar: ¯/epsilon1,¯A. Frequency dependent quantities are indicated by a tilde, whereas time dependent quan-tities are written without additional indication; thus we write ˜E(r,ω)and E(r,t). (Some quantities, such as impedance, are used in the frequency domain to interrelate Fourierspectra; although these quantities are frequency dependent they are seldom written inthe time domain, and hence we do not attach tildes to their symbols.) We often combinediacritical marks:for example, ˜¯/epsilon1denotes a frequency domain dyadic. We distinguish carefully between phasor and frequency domain quantities. The variable ωis used for the frequency variable of the Fourier spectrum, while ˇωis used to indicate the constant frequency of a time harmonic signal. We thus further separate the notion of a phasorﬁeld from a frequency domain ﬁeld by using a check to indicate a phasor ﬁeld: ˇE(r). However, there is often a simple relationship between the two, such as ˇE=˜E(ˇω). We designate the ﬁeld and source point position vectors by rand r /prime, respectively, and the corresponding relative displacement or distance vector by R: R=r−r/prime. A hat designates a vector as a unit vector (e.g., ˆx). The sets of coordinate variables in rectangular, cylindrical, and spherical coordinates are denoted by (x,y,z), (ρ, φ, z), ( r,θ,φ) , respectively. (In the spherical system φis the azimuthal angle and θis the polar angle.) We freely use the “del” operator notation ∇for gradient, curl, divergence, Laplacian, and so on. The SI (MKS) system of units is employed throughout the book. 1.2 The ﬁeld concept of electromagnetics Introductory treatments of electromagnetics often stress the role of the ﬁeld in force transmission:the individual ﬁelds Eand Bare deﬁned via the mechanical force on a small test charge. This is certainly acceptable, but does not tell the whole story. Wemight, for example, be left with the impression that the EM ﬁeld always arises froman interaction between charged objects. Often coupled with this is the notion that theﬁeld concept is meant merely as an aid to the calculation of force, a kind of notationalconvenience not placed on the same physical footing as force itself. In fact, ﬁelds aremore than useful — they are fundamental. Before discussing electromagnetic ﬁelds inmore detail, let us attempt to gain a better perspective on the ﬁeld concept and its rolein modern physical theory. Fields play a central role in any attempt to describe physicalreality. They are as real as the physical substances we ascribe to everyday experience. In the words of Einstein [63], “It seems impossible to give an obvious qualitative criterion for distinguishing between matter and ﬁeld or charge and ﬁeld.” We must therefore put ﬁelds and particles of matter on the same footing:both carry energy and momentum, and both interact with the observable world. 1.2.1 Historical perspective Early nineteenth century physical thought was dominated by the action at a distance concept, formulated by Newton more than 100 years earlier in his immensely successfultheory of gravitation. In this view the inﬂuence of individual bodies extends across space,instantaneously aﬀects other bodies, and remains completely unaﬀected by the presenceof an intervening medium. Such an idea was revolutionary; until then action by contact ,i n which objects are thought to aﬀect each other through physical contact or by contact withthe intervening medium, seemed the obvious and only means for mechanical interaction.Priestly’s experiments in 1766 and Coulomb’s torsion-bar experiments in 1785 seemed toindicate that the force between two electrically charged objects behaves in strict analogywith gravitation:both forces obey inverse square laws and act along a line joining theobjects. Oersted, Ampere, Biot, and Savart soon showed that the magnetic force onsegments of current-carrying wires also obeys an inverse square law. The experiments of Faraday in the 1830s placed doubt on whether action at a distance really describes electric and magnetic phenomena. When a material (such as a dielec-tric) is placed between two charged objects, the force of interaction decreases; thus, theintervening medium does play a role in conveying the force from one object to the other.To explain this, Faraday visualized “lines of force” extending from one charged object toanother. The manner in which these lines were thought to interact with materials theyintercepted along their path was crucial in understanding the forces on the objects. Thisalso held for magnetic eﬀects. Of particular importance was the number of lines passingthrough a certain area (the ﬂux), which was thought to determine the amplitude of the eﬀect observed in Faraday’s experiments on electromagnetic induction. Faraday’s ideas presented a new world view:electromagnetic phenomena occur in the region surrounding charged bodies, and can be described in terms of the laws governingthe “ﬁeld” of his lines of force. Analogies were made to the stresses and strains in materialobjects, and it appeared that Faraday’s force lines created equivalent electromagnetic stresses and strains in media surrounding charged objects. His law of induction was formulated not in terms of positions of bodies, but in terms of lines of magnetic force.Inspired by Faraday’s ideas, Gauss restated Coulomb’s law in terms of ﬂux lines, andMaxwell extended the idea to time changing ﬁelds through his concept of displacementcurrent. In the 1860s Maxwell created what Einstein called “the most important invention since Newton’s time”— a set of equations describing an entirely ﬁeld-based theory ofelectromagnetism. These equations do not model the forces acting between bodies, as doNewton’s law of gravitation and Coulomb’s law, but rather describe only the dynamic,time-evolving structure of the electromagnetic ﬁeld. Thus bodies are not seen to inter-act with each other, but rather with the (very real) electromagnetic ﬁeld they create,an interaction described by a supplementary equation (the Lorentz force law). To bet-ter understand the interactions in terms of mechanical concepts, Maxwell also assignedproperties of stress and energy to the ﬁeld. Using constructs that we now call the electric and magnetic ﬁelds and potentials, Maxwell synthesized all known electromagnetic laws and presented them as a system ofdiﬀerential and algebraic equations. By the end of the nineteenth century, Hertz haddevised equations involving only the electric and magnetic ﬁelds, and had derived thelaws of circuit theory (Ohm’s law and Kirchoﬀ’s laws) from the ﬁeld expressions. Hisexperiments with high-frequency ﬁelds veriﬁed Maxwell’s predictions of the existence ofelectromagnetic waves propagating at ﬁnite velocity, and helped solidify the link betweenelectromagnetism and optics. But one problem remained:if the electromagnetic ﬁeldspropagated by stresses and strains on a medium, how could they propagate through avacuum? A substance called the luminiferous aether , long thought to support the trans- verse waves of light, was put to the task of carrying the vibrations of the electromagneticﬁeld as well. However, the pivotal experiments of Michelson and Morely showed that theaether was ﬁctitious, and the physical existence of the ﬁeld was ﬁrmly established. The essence of the ﬁeld concept can be conveyed through a simple thought experiment. Consider two stationary charged particles in free space. Since the charges are stationary,we know that (1) another force is present to balance the Coulomb force between thecharges, and (2) the momentum and kinetic energy of the system are zero. Now supposeone charge is quickly moved and returned to rest at its original position. Action at adistance would require the second charge to react immediately (Newton’s third law),but by Hertz’s experiments it does not. There appears to be no change in energy ofthe system:both particles are again at rest in their original positions. However, after a time (given by the distance between the charges divided by the speed of light) we ﬁndthat the second charge does experience a change in electrical force and begins to moveaway from its state of equilibrium. But by doing so it has gained net kinetic energyand momentum, and the energy and momentum of the system seem larger than at thestart. This can only be reconciled through ﬁeld theory. If we regard the ﬁeld as aphysical entity, then the nonzero work required to initiate the motion of the ﬁrst chargeand return it to its initial state can be seen as increasing the energy of the ﬁeld. Adisturbance propagates at ﬁnite speed and, upon reaching the second charge, transfersenergy into kinetic energy of the charge. Upon its acceleration this charge also sends outa wave of ﬁeld disturbance, carrying energy with it, eventually reaching the ﬁrst chargeand creating a second reaction. At any given time, the net energy and momentum of thesystem, composed of both the bodies and the ﬁeld, remain constant. We thus come toregard the electromagnetic ﬁeld as a true physical entity:an entity capable of carryingenergy and momentum. 1.2.2 Formalization of ﬁeld theory Before we can invoke physical laws, we must ﬁnd a way to describe the state of the system we intend to study. We generally begin by identifying a set of state variables that can depict the physical nature of the system. In a mechanical theory such asNewton’s law of gravitation, the state of a system of point masses is expressed in termsof the instantaneous positions and momenta of the individual particles. Hence 6Nstate variables are needed to describe the state of a system of Nparticles, each particle having three position coordinates and three momentum components. The time evolution ofthe system state is determined by a supplementary force function (e.g., gravitationalattraction), the initial state (initial conditions), and Newton’s second law F=dP/dt. Descriptions using ﬁnite sets of state variables are appropriate for action-at-a-distance interpretations of physical laws such as Newton’s law of gravitation or the interactionof charged particles. If Coulomb’s law were taken as the force law in a mechanicaldescription of electromagnetics, the state of a system of particles could be described completely in terms of their positions, momenta, and charges. Of course, charged particleinteraction is not this simple. An attempt to augment Coulomb’s force law with Ampere’sforce law would not account for kinetic energy loss via radiation. Hence we abandon 1 the mechanical viewpoint in favor of the ﬁeld viewpoint, selecting a diﬀerent set ofstate variables. The essence of ﬁeld theory is to regard electromagnetic phenomena asaﬀecting all of space. We shall ﬁnd that we can describe the ﬁeld in terms of the fourvector quantities E,D,B, and H. Because these ﬁelds exist by deﬁnition at each point in space and each time t, a ﬁnite set of state variables cannot describe the system. Here then is an important distinction between ﬁeld theories and mechanical theories: the state of a ﬁeld at any instant can only be described by an inﬁnite number of statevariables. Mathematically we describe ﬁelds in terms of functions of continuous variables;however, we must be careful not to confuse all quantities described as “ﬁelds” with thoseﬁelds innate to a scientiﬁc ﬁeld theory. For instance, we may refer to a temperature“ﬁeld” in the sense that we can describe temperature as a function of space and time.However, we do notmean by this that temperature obeys a set of physical laws analogous to those obeyed by the electromagnetic ﬁeld. What special character, then, can we ascribe to the electromagnetic ﬁeld that has meaning beyond that given by its mathematical implications? In this book, E,D,B, and Hare integral parts of a ﬁeld-theory description of electromagnetics. In any ﬁeld theory we need two types of ﬁelds:a mediating ﬁeld generated by a source, and a ﬁeld describing the source itself. In free-space electromagnetics the mediating ﬁeld consists ofEand B, while the source ﬁeld is the distribution of charge or current. An important consideration is that the source ﬁeld must be independent of the mediating ﬁeld thatit “sources.” Additionally, ﬁelds are generally regarded as unobservable:they can onlybe measured indirectly through interactions with observable quantities. We need a linkto mechanics to observe Eand B:we might measure the change in kinetic energy of a particle as it interacts with the ﬁeld through the Lorentz force. The Lorentz forcebecomes the force function in the mechanical interaction that uniquely determines the(observable) mechanical state of the particle. A ﬁeld is associated with a set of ﬁeld equations and a set of constitutive relations . The ﬁeld equations describe, through partial derivative operations, both the spatial distribu-tion and temporal evolution of the ﬁeld. The constitutive relations describe the eﬀect 1Attempts have been made to formulate electromagnetic theory purely in action-at-a-distance terms, but this viewpoint has not been generally adopted [69]. of the supporting medium on the ﬁelds and are dependent upon the physical state of the medium. The state may include macroscopic eﬀects, such as mechanical stress andthermodynamic temperature, as well as the microscopic, quantum-mechanical propertiesof matter. The value of the ﬁeld at any position and time in a bounded region Vis then determined uniquely by specifying the sources within V, the initial state of the ﬁelds within V, and the value of the ﬁeld or ﬁnitely many of its derivatives on the surface bounding V.I f the boundary surface also deﬁnes a surface of discontinuity between adjacent regions ofdiﬀering physical characteristics, or across discontinuous sources, then jump conditions may be used to relate the ﬁelds on either side of the surface. The variety of forms of ﬁeld equations is restricted by many physical principles in- cluding reference-frame invariance, conservation, causality, symmetry, and simplicity.Causality prevents the ﬁeld at time t=0from being inﬂuenced by events occurring at subsequent times t>0. Of course, we prefer that a ﬁeld equation be mathematically robust and well-posed to permit solutions that are unique and stable. Many of these ideas are well illustrated by a consideration of electrostatics. We can describe the electrostatic ﬁeld through a mediating scalar ﬁeld /Phi1(x,y,z)known as the electrostatic potential. The spatial distribution of the ﬁeld is governed by Poisson’sequation ∂ 2/Phi1 ∂x2+∂2/Phi1 ∂y2+∂2/Phi1 ∂z2=−ρ /epsilon10,θ where ρ=ρ(x,y,z)is the source charge density. No temporal derivatives appear, and the spatial derivatives determine the spatial behavior of the ﬁeld. The function ρrepresents the spatially-averaged distribution of charge that acts as the source term for the ﬁeld /Phi1. Note that ρincorporates no information about /Phi1. To uniquely specify the ﬁeld at any point, we must still specify its behavior over a boundary surface. We could, for instance,specify /Phi1on ﬁve of the six faces of a cube and the normal derivative ∂/Phi1/∂ non the remaining face. Finally, we cannot directly observe the static potential ﬁeld, but we canobserve its interaction with a particle. We relate the static potential ﬁeld theory to therealm of mechanics via the electrostatic force F=qEacting on a particle of charge q. In future chapters we shall present a classical ﬁeld theory for macroscopic electromag- netics. In that case the mediating ﬁeld quantities are E,D,B, and H, and the source ﬁeld is the current density J. 1.3 The sources of the electromagnetic ﬁeld Electric charge is an intriguing natural entity. Human awareness of charge and its eﬀects dates back to at least 600 BC, when the Greek philosopher Thales of Miletusobserved that rubbing a piece of amber could enable the amber to attract bits of straw.Although charging by friction is probably still the most common and familiar manifes- tation of electric charge, systematic experimentation has revealed much more about thebehavior of charge and its role in the physical universe. There are two kinds of charge, towhich Benjamin Franklin assigned the respective names positive andnegative . Franklin observed that charges of opposite kind attract and charges of the same kind repel. Healso found that an increase in one kind of charge is accompanied by an increase in the other, and so ﬁrst described the principle of charge conservation . Twentieth century physics has added dramatically to the understanding of charge: 1. Electric charge is a fundamental property of matter, as is mass or dimension. 2. Charge is quantized :there exists a smallest quantity ( quantum ) of charge that can be associated with matter. No smaller amount has been observed, and largeramounts always occur in integral multiples of this quantity. 3. The charge quantum is associated with the smallest subatomic particles, and these particles interact through electrical forces. In fact, matter is organized and arrangedthrough electrical interactions; for example, our perception of physical contact ismerely the macroscopic manifestation of countless charges in our ﬁngertips pushingagainst charges in the things we touch. 4. Electric charge is an invariant :the value of charge on a particle does not depend on the speed of the particle. In contrast, the mass of a particle increases with speed. 5. Charge acts as the source of an electromagnetic ﬁeld; the ﬁeld is an entity that can carry energy and momentum away from the charge via propagating waves. We begin our investigation of the properties of the electromagnetic ﬁeld with a detailed examination of its source. 1.3.1 Macroscopic electromagnetics We are interested primarily in those electromagnetic eﬀects that can be predicted by classical techniques using continuous sources (charge and current densities). Althoughmacroscopic electromagnetics is limited in scope, it is useful in many situations en-countered by engineers. These include, for example, the determination of currents andvoltages in lumped circuits, torques exerted by electrical machines, and ﬁelds radiated byantennas. Macroscopic predictions can fall short in cases where quantum eﬀects are im-portant:e.g., with devices such as tunnel diodes. Even so, quantum mechanics can oftenbe coupled with classical electromagnetics to determine the macroscopic electromagneticproperties of important materials. Electric charge is not of a continuous nature. The quantization of atomic charge — ±efor electrons and protons, ±e/3and±2e/3for quarks — is one of the most precisely established principles in physics (veriﬁed to 1 part in 10 21). The value of eitself is known to great accuracy: e=1.60217733 ×10−19Coulombs (C) . However, the discrete nature of charge is not easily incorporated into everyday engineer- ing concerns. The strange world of the individual charge — characterized by particlespin, molecular moments, and thermal vibrations — is well described only by quantumtheory. There is little hope that we can learn to describe electrical machines using suchconcepts. Must we therefore retreat to the macroscopic idea and ignore the discretizationof charge completely? A viable alternative is to use atomic theories of matter to estimatethe useful scope of macroscopic electromagnetics. Remember, we are completely free to postulate a theory of nature whose scope may be limited. Like continuum mechanics, which treats distributions of matter as if theywere continuous, macroscopic electromagnetics is regarded as valid because it is veriﬁedby experiment over a certain range of conditions. This applicability range generallycorresponds to dimensions on a laboratory scale, implying a very wide range of validityfor engineers. Macroscopic eﬀects as averaged microscopic eﬀects. Macroscopic electromag- netics can hold in a world of discrete charges because applications usually occur overphysical scales that include vast numbers of charges. Common devices, generally muchlarger than individual particles, “average” the rapidly varying ﬁelds that exist in thespaces between charges, and this allows us to view a source as a continuous “smear” ofcharge. To determine the range of scales over which the macroscopic viewpoint is valid,we must compare averaged values of microscopic ﬁelds to the macroscopic ﬁelds we mea-sure in the lab. But if the eﬀects of the individual charges are describable only in termsof quantum notions, this task will be daunting at best. A simple compromise, whichproduces useful results, is to extend the macroscopic theory right down to the micro-scopic level and regard discrete charges as “point” entities that produce electromagneticﬁelds according to Maxwell’s equations. Then, in terms of scales much larger than theclassical radius of an electron ( ≈10 −14m), the expected rapid ﬂuctuations of the ﬁelds in the spaces between charges is predicted. Finally, we ask:over what spatial scale mustwe average the eﬀects of the ﬁelds and the sources in order to obtain agreement with the macroscopic equations? In the spatial averaging approach a convenient weighting function f(r)is chosen, and is normalized so that/integraltext f(r)dV=1. An example is the Gaussian distribution f(r)=(πa 2)−3/2e−r2/a2, where ais the approximate radial extent of averaging. The spatial average of a micro- scopic quantity F(r,t)is given by /angbracketleftF(r,t)/angbracketright=/integraldisplay F(r−r/prime,t)f(r/prime)dV/prime. (1.1) The scale of validity of the macroscopic model can be found by determining the averaging radius athat produces good agreement between the averaged microscopic ﬁelds and the macroscopic ﬁelds. The macroscopic volume charge density. At this point we do not distinguish between the “free” charge that is unattached to a molecular structure and the chargefound near the surface of a conductor. Nor do we consider the dipole nature of polarizablematerials or the microscopic motion associated with molecular magnetic moment or themagnetic moment of free charge. For the consideration of free-space electromagnetics, we assume charge exhibits either three degrees of freedom ( volume charge ), two degrees of freedom ( surface charge ), or one degree of freedom ( line charge ). In typical matter, the microscopic ﬁelds vary spatially over dimensions of 10 −10m or less, and temporally over periods (determined by atomic motion) of 10−13s or less. At the surface of a material such as a good conductor where charge often concentrates,averaging with a radius on the order of 10 −10m may be required to resolve the rapid variation in the distribution of individual charged particles. However, within a solid orliquid material, or within a free-charge distribution characteristic of a dense gas or anelectron beam, a radius of 10 −8m proves useful, containing typically 106particles. A diﬀuse gas, on the other hand, may have a particle density so low that the averagingradius takes on laboratory dimensions, and in such a case the microscopic theory mustbe employed even at macroscopic dimensions. Once the averaging radius has been determined, the value of the charge density may be found via (1.1). The volume density of charge for an assortment of point sources can be written in terms of the three-dimensional Dirac delta as ρo(r,t)=/summationdisplay iqiδ(r−ri(t)), where ri(t)is the position of the charge qiat time t. Substitution into (1.1) gives ρ(r,t)=/angbracketleftρo(r,t)/angbracketright=/summationdisplay iqif(r−ri(t)) (1.2) as the averaged charge density appropriate for use in a macroscopic ﬁeld theory. Because the oscillations of the atomic particles are statistically uncorrelated over the distancesused in spatial averaging, the time variations of microscopic ﬁelds are not present in themacroscopic ﬁelds and temporal averaging is unnecessary. In (1.2) the time dependenceof the spatially-averaged charge density is due entirely to bulk motion of the chargeaggregate (macroscopic charge motion). With the deﬁnition of macroscopic charge density given by (1.2), we can determine the total charge Q(t)in any macroscopic volume region Vusing Q(t)=/integraldisplay Vρ(r,t)dV. (1.3) We have Q(t)=/summationdisplay iqi/integraldisplay Vf(r−ri(t))dV=/summationdisplay ri(t)∈Vqi. Here we ignore the small discrepancy produced by charges lying within distance aof the boundary of V. It is common to employ a box Bhaving volume /Delta1V: /braceleftbiggf(r)=1//Delta1V,r∈B, 0, r/∈B. In this case ρ(r,t)=1 /Delta1V/summationdisplay r−ri(t)∈Bqi. The size of Bis chosen with the same considerations as to atomic scale as was the averaging radius a. Discontinuities at the edges of the box introduce some diﬃculties concerning charges that move in and out of the box because of molecular motion. The macroscopic volume current density. Electric charge in motion is referred to aselectric current . Charge motion can be associated with external forces and with microscopic ﬂuctuations in position. Assuming charge qihas velocity vi(t)=dri(t)/dt, the charge aggregate has volume current density Jo(r,t)=/summationdisplay iqivi(t)δ(r−ri(t)). Spatial averaging gives the macroscopic volume current density J(r,t)=/angbracketleftJo(r,t)/angbracketright=/summationdisplay iqivi(t)f(r−ri(t)). (1.4) Figure 1.1:Intersection of the averaging function of a point charge with a surface S,a s the charge crosses Swith velocity v:(a) at some time t=t1, and (b) at t=t2>t1. The averaging function is represented by a sphere of radius a. Spatial averaging at time teliminates currents associated with microscopic motions that are uncorrelated at the scale of the averaging radius (again, we do not consider themagnetic moments of particles). The assumption of a suﬃciently large averaging radiusleads to J(r,t)=ρ(r,t)v(r,t). (1.5) The total ﬂux I(t)of current through a surface Sis given by I(t)=/integraldisplay SJ(r,t)·ˆndS where ˆnis the unit normal to S. Hence, using (4), we have I(t)=/summationdisplay iqid dt(ri(t)·ˆn)/integraldisplay Sf(r−ri(t))dS ifˆnstays approximately constant over the extent of the averaging function and Sis not in motion. We see that the integral eﬀectively intersects Swith the averaging function sur- roundin geachmovingpointcharge (Figure1.1).Thetimederivativeof ri·ˆnrepresents the velocity at which the averaging function is “carried across” the surface. Electric current takes a variety of forms, each described by the relation J=ρv. Isolated charged particles (positive and negative) and charged insulated bodies moving throughspace comprise convection currents . Negatively-charged electrons moving through the positive background lattice within a conductor comprise a conduction current . Empirical evidence suggests that conduction currents are also described by the relation J=σE known as Ohm’s law . A third type of current, called electrolytic current , results from the ﬂow of positive or negative ions through a ﬂuid. 1.3.2 Impressed vs. secondary sources In addition to the simple classiﬁcation given above we may classify currents as primary orsecondary , depending on the action that sets the charge in motion. It is helpful to separate primary or “impressed” sources, which are independent of the ﬁelds they source, from secondary sources which result from interactions between thesourced ﬁelds and the medium in which the ﬁelds exist. Most familiar is the conduc-tion current set up in a conducting medium by an externally applied electric ﬁeld. Theimpressed source concept is particularly important in circuit theory, where independentvoltage sources are modeled as providing primary voltage excitations that are indepen-dent of applied load. In this way they diﬀer from the secondary or “dependent” sourcesthat react to the eﬀect produced by the application of primary sources. In applied electromagnetics the primary source may be so distant that return eﬀects resulting from local interaction of its impressed ﬁelds can be ignored. Other examples ofprimary sources include the applied voltage at the input of an antenna, the current on aprobe inserted into a waveguide, and the currents producing a power-line ﬁeld in whicha biological body is immersed. 1.3.3 Surface and line source densities Because they are spatially averaged eﬀects, macroscopic sources and the ﬁelds they source cannot have true spatial discontinuities. However, it is often convenient to workwith sources in one or two dimensions. Surface and line source densities are idealizationsof actual, continuous macroscopic densities. The entity we describe as a surface charge is a continuous volume charge distributed in a thin layer across some surface S. If the thickness of the layer is small compared to laboratory dimensions, it is useful to assign to each point ron the surface a quantity describing the amount of charge contained within a cylinder oriented normal to thesurface and having inﬁnitesimal cross section dS. We call this quantity the surface charge density ρ s(r,t), and write the volume charge density as ρ(r,w,t)=ρs(r,t)f(w, /Delta1), where wis distance from Sin the normal direction and /Delta1in some way parameterizes the “thickness” of the charge layer at r. The continuous density function f(x,/Delta1 )satisﬁes /integraldisplay∞ −∞f(x,/Delta1 )dx=1 and lim /Delta1→0f(x,/Delta1 )=δ(x). For instance, we might have f(x,/Delta1 )=e−x2//Delta12 /Delta1√π. (1.6) With this deﬁnition the total charge contained in a cylinder normal to the surface at r and having cross-sectional area dSis dQ(t)=/integraldisplay∞ −∞[ρs(r,t)dS]f(w, /Delta1) dw=ρs(r,t)dS, and the total charge contained within any cylinder oriented normal to Sis Q(t)=/integraldisplay Sρs(r,t)dS. (1.7) We may describe a line charge as a thin “tube” of volume charge distributed along some contour /Gamma1. The amount of charge contained between two planes normal to the contour and separated by a distance dlis described by the line charge density ρl(r,t). The volume charge density associated with the contour is then ρ(r,ρ,t)=ρl(r,t)fs(ρ, /Delta1), where ρis the radial distance from the contour in the plane normal to /Gamma1and fs(ρ, /Delta1) is a density function with the properties /integraldisplay∞ 0fs(ρ, /Delta1) 2πρdρ=1 and lim /Delta1→0fs(ρ, /Delta1) =δ(ρ) 2πρ. For example, we might have fs(ρ, /Delta1) =e−ρ2//Delta12 π/Delta12. (1.8) Then the total charge contained between planes separated by a distance dlis dQ(t)=/integraldisplay∞ 0[ρl(r,t)dl]fs(ρ, /Delta1) 2πρdρ=ρl(r,t)dl and the total charge contained between planes placed at the ends of a contour /Gamma1is Q(t)=/integraldisplay /Gamma1ρl(r,t)dl. (1.9) We may deﬁne surface and line currents similarly. A surface current is merely a volume current conﬁned to the vicinity of a surface S. The volume current density may be represented using a surface current density function Js(r,t), deﬁned at each point r on the surface so that J(r,w,t)=Js(r,t)f(w, /Delta1). Here f(w, /Delta1) is some appropriate density function such as (1.6), and the surface current vector obeys ˆn·Js=0where ˆnis normal to S. The total current ﬂowing through a strip of width dlarranged perpendicular to Satris dI(t)=/integraldisplay∞ −∞[Js(r,t)·ˆnl(r)dl]f(w, /Delta1) dw=Js(r,t)·ˆnl(r)dl where ˆnlis normal to the strip at r(and thus also tangential to Satr). The total current passing through a strip intersecting with Salong a contour /Gamma1is thus I(t)=/integraldisplay /Gamma1Js(r,t)·ˆnl(r)dl. We may describe a line current as a thin “tube” of volume current distributed about some contour /Gamma1and ﬂowing parallel to it. The amount of current passing through a plane normal to the contour is described by the line current density Jl(r,t). The volume current density associated with the contour may be written as J(r,ρ,t)=ˆu(r)Jl(r,t)fs(ρ, /Delta1), where ˆuis a unit vector along /Gamma1,ρis the radial distance from the contour in the plane normal to /Gamma1, and fs(ρ, /Delta1) is a density function such as (1.8). The total current passing through any plane normal to /Gamma1atris I(t)=/integraldisplay∞ 0[Jl(r,t)ˆu(r)·ˆu(r)]fs(ρ, /Delta1) 2πρdρ=Jl(r,t). It is often convenient to employ singular models for continuous source densities. For instance, it is mathematically simpler to regard a surface charge as residing only in thesurface Sthan to regard it as being distributed about the surface. Of course, the source is then discontinuous since it is zero everywhere outside the surface. We may obtain arepresentation of such a charge distribution by letting the thickness parameter /Delta1in the density functions recede to zero, thus concentrating the source into a plane or a line. Wedescribe the limit of the density function in terms of the δ-function. For instance, the volume charge distribution for a surface charge located about the xy-plane is ρ(x,y,z,t)=ρ s(x,y,t)f(z, /Delta1). As/Delta1→0we have ρ(x,y,z,t)=ρs(x,y,t)lim /Delta1→0f(z,/Delta1 )=ρs(x,y,t)δ(z). It is a simple matter to represent singular source densities in this way as long as the surface or line is easily parameterized in terms of constant values of coordinate variables.However, care must be taken to represent the δ-function properly. For instance, the density of charge on the surface of a cone at θ=θ 0may be described using the distance normal to this surface, which is given by rθ−rθ0: ρ(r,θ,φ, t)=ρs(r,φ,t)δ(r[θ−θ0]). Using the property δ(ax)=δ(x)/a, we can also write this as ρ(r,θ,φ, t)=ρs(r,φ,t)δ(θ−θ0) r. 1.3.4 Charge conservation There are four fundamental conservation laws in physics:conservation of energy, mo- mentum, angular momentum, and charge. These laws are said to be absolute ; they have never been observed to fail. In that sense they are true empirical laws of physics. However, in modern physics the fundamental conservation laws have come to represent more than just observed facts. Each law is now associated with a fundamental symme-try of the universe; conversely, each known symmetry is associated with a conservationprinciple. For example, energy conservation can be shown to arise from the observationthat the universe is symmetric with respect to time; the laws of physics do not dependon choice of time origin t=0. Similarly, momentum conservation arises from the obser- vation that the laws of physics are invariant under translation, while angular momentumconservation arises from invariance under rotation. The law of conservation of charge also arises from a symmetry principle. But instead of being spatial or temporal in character, it is related to the invariance of electrostaticpotential. Experiments show that there is no absolute potential, only potential diﬀerence.The laws of nature are invariant with respect to what we choose as the “reference” potential. This in turn is related to the invariance of Maxwell’s equations under gauge transforms; the values of the electric and magnetic ﬁelds do not depend on which gaugetransformation we use to relate the scalar potential /Phi1to the vector potential A. We may state the conservation of charge as follows: The net charge in any closed system remains constant with time. This does not mean that individual charges cannot be created or destroyed, only that the total charge in any isolated system must remain constant. Thus it is possible for apositron with charge eto annihilate an electron with charge −ewithout changing the net charge of the system. Only if a system is not closed can its net charge be altered;since moving charge constitutes current, we can say that the total charge within a systemdepends on the current passing through the surface enclosing the system. This is theessence of the continuity equation. To derive this important result we consider a closedsystem within which the charge remains constant, and apply the Reynolds transport theorem (see §A.2). The continuity equation. Consider a region of space occupied by a distribution of charge whose velocity is given by the vector ﬁeld v. We surround a portion of charge by a surface Sand let Sdeform as necessary to “follow” the charge as it moves. Since Salways contains precisely the same charged particles, we have an isolated system for which the time rate of change of total charge must vanish. An expression for the timerate of change is given by the Reynolds transport theorem (A.66); we have 2 DQ Dt=D Dt/integraldisplay V(t)ρdV=/integraldisplay V(t)∂ρ ∂tdV+/contintegraldisplay S(t)ρv·dS=0. The “ D/Dt” notation indicates that the volume region V(t)moves with its enclosed particles. Since ρvrepresents current density, we can write /integraldisplay V(t)∂ρ(r,t) ∂tdV+/contintegraldisplay S(t)J(r,t)·dS=0. (1.10) In this large-scale form of the continuity equation, the partial derivative term describes the time rate of change of the charge density for a ﬁxed spatial position r. At any time t, the time rate of change of charge density integrated over a volume is exactly compensated by the total current exiting through the surrounding surface. We can obtain the continuity equation in point form by applying the divergence the- orem to the second term of (1.10) to get /integraldisplay V(t)/bracketleftbigg∂ρ(r,t) ∂t+∇· J(r,t)/bracketrightbigg dV=0. Since V(t)is arbitrary we can set the integrand to zero to obtain ∂ρ(r,t) ∂t+∇· J(r,t)=0. (1.11) 2Note that in Appendix A we use the symbol uto represent the velocity of a material and vto represent the velocity of an artiﬁcial surface. This expression involves the time derivative of ρwith rﬁxed. We can also ﬁnd an expression in terms of the material derivative by using the transport equation (A.67).Enforcing conservation of charge by setting that expression to zero, we have Dρ(r,t) Dt+ρ(r,t)∇·v(r,t)=0. (1.12) Here Dρ/Dtis the time rate of change of the charge density experienced by an observer moving with the current. We can state the large-scale form of the continuity equation in terms of a stationary volume. Integrating (1.11) over a stationary volume region Vand using the divergence theorem, we ﬁnd that/integraldisplay V∂ρ(r,t) ∂tdV=−/contintegraldisplay SJ(r,t)·dS. Since Vis not changing with time we have dQ(t) dt=d dt/integraldisplay Vρ(r,t)dV=−/contintegraldisplay SJ(r,t)·dS. (1.13) Hence any increase of total charge within Vmust be produced by current entering V through S. Use of the continuity equation. As an example, suppose that in a bounded region of space we have ρ(r,t)=ρ0re−βt. We wish to ﬁnd Jand v, and to verify both versions of the continuity equation in point form. The spherical symmetry of ρrequires that J=ˆrJr.Application of (1.13) over a sphere of radius agives 4πd dt/integraldisplaya 0ρ0re−βtr2dr=−4πJr(a)a2. Hence J=ˆrβρ0r2 4e−βt and therefore ∇·J=1 r2∂ ∂r(r2Jr)=βρ0re−βt. The velocity is v=J ρ=ˆrβr 4, and we have ∇·v=3β/4.To verify the continuity equations, we compute the time derivatives ∂ρ ∂t=−βρ0re−βt, Dρ Dt=∂ρ ∂t+v·∇ρ =−βρ0re−βt+/parenleftBig ˆrβr 4/parenrightBig ·/parenleftbigˆrρ0e−βt/parenrightbig =−3 4βρ0re−βt. Note that the charge density decreases with time less rapidly for a moving observer than for a stationary one (3/4 as fast):the moving observer is following the charge outward,andρ∝r. Now we can check the continuity equations. First we see Dρ Dt+ρ∇·v=−3 4βρ0re−βt+(ρ0re−βt)/parenleftbigg3 4β/parenrightbigg =0, as required for a moving observer; second we see ∂ρ ∂t+∇· J=−βρ0re−βt+βρ0e−βt=0, as required for a stationary observer. The continuity equation in fewer dimensions. The continuity equation can also be used to relate current and charge on a surface or along a line. By conservation of charge we can write d dt/integraldisplay Sρs(r,t)dS=−/contintegraldisplay /Gamma1Js(r,t)·ˆmdl (1.14) where ˆmis the vector normal to the curve /Gamma1and tangential to the surface S. By the surface divergence theorem (B.20), the corresponding point form is ∂ρs(r,t) ∂t+∇ s·Js(r,t)=0. (1.15) Here∇s·Jsis the surface divergence of the vector ﬁeld Js. For instance, in rectangular coordinates in the z=0plane we have ∇s·Js=∂Jsx ∂x+∂Jsy ∂y. In cylindrical coordinates on the cylinder ρ=a, we would have ∇s·Js=1 a∂Jsφ ∂φ+∂Jsz ∂z. A detailed description of vector operations on a surface may be found in Tai [190], while many identities may be found in Van Bladel [202]. Theequatio nofcontinuityforalineiseasilyestablishe dbyreferenc etoFigure1.2. Here the net charge exiting the surface during time /Delta1tis given by /Delta1t[I(u2,t)−I(u1,t)]. Thus, the rate of net increase of charge within the system is dQ(t) dt=d dt/integraldisplay ρl(r,t)dl=−[I(u2,t)−I(u1,t)]. (1.16) The corresponding point form is found by letting the length of the curve approach zero: ∂I(l,t) ∂l+∂ρl(l,t) ∂t=0, (1.17) where lis arc length along the curve. As an example, suppose the line current on a circular loop antenna is approximately I(φ,t)=I0cos/parenleftBigωa cφ/parenrightBig cosωt, Figure 1.2:Linear form of the continuityequation. where ais the radius of the loop, ωis the frequency of operation, and cis the speed of light. We wish to ﬁnd the line charge density on the loop. Since l=aφ, we can write I(l,t)=I0cos/parenleftbiggωl c/parenrightbigg cosωt. Thus ∂I(l,t) ∂l=− I0ω csin/parenleftbiggωl c/parenrightbigg cosωt=−∂ρl(l,t) ∂t. Integrating with respect to time and ignoring any constant (static) charge, we have ρ(l,t)=I0 csin/parenleftbiggωl c/parenrightbigg sinωt or ρ(φ, t)=I0 csin/parenleftBigωa cφ/parenrightBig sinωt. Note that we could have used the chain rule ∂I(φ,t) ∂l=∂I(φ,t) ∂φ∂φ ∂land∂φ ∂l=/bracketleftbigg∂l ∂φ/bracketrightbigg−1 =1 a to calculate the spatial derivative. We can apply the volume density continuity equation (1.11) directly to surface and line distributions written in singular notation. For the loop of the previous example, wewrite the volume current density corresponding to the line current as J(r,t)=ˆφδ(ρ−a)δ(z)I(φ,t). Substitution into (1.11) then gives ∇·[ˆφδ(ρ−a)δ(z)I(φ,t)]=−∂ρ(r,t) ∂t. The divergence formula for cylindrical coordinates gives δ(ρ−a)δ(z)∂I(φ,t) ρ∂φ=−∂ρ(r,t) ∂t. Next we substitute for I(φ,t)to get −I0 ρωa csin/parenleftBigωa cφ/parenrightBig δ(ρ−a)δ(z)cosωt=−∂ρ(r,t) ∂t. Finally, integrating with respect to time and ignoring any constant term, we have ρ(r,t)=I0 cδ(ρ−a)δ(z)sin/parenleftBigωa cφ/parenrightBig sinωt, where we have set ρ=abecause of the presence of the factor δ(ρ−a). 1.3.5 Magnetic charge We take for granted that electric ﬁelds are produced by electric charges, whether stationary or in motion. The smallest element of electric charge is the electric monopole : a single discretely charged particle from which the electric ﬁeld diverges. In contrast,experiments show that magnetic ﬁelds are created only by currents or by time changingelectric ﬁelds; hence, magnetic ﬁelds have moving electric charge as their source. Theelemental source of magnetic ﬁeld is the magnetic dipole , representing a tiny loop of electric current (or a spinning electric particle). The observation made in 1269 by PierreDe Maricourt, that even the smallest magnet has two poles, still holds today. In a world ﬁlled with symmetry at the fundamental level, we ﬁnd it hard to understand why there should not be a source from which the magnetic ﬁeld diverges. We would callsuch a source magnetic charge , and the most fundamental quantity of magnetic charge would be exhibited by a magnetic monopole . In 1931 Paul Dirac invigorated the search for magnetic monopoles by making the ﬁrst strong theoretical argument for their existence.Dirac showed that the existence of magnetic monopoles would imply the quantizationof electric charge, and would thus provide an explanation for one of the great puzzlesof science. Since that time magnetic monopoles have become important players in the“Grand Uniﬁed Theories” of modern physics, and in cosmological theories of the originof the universe. If magnetic monopoles are ever found to exist, there will be both positive and negatively charged particles whose motions will constitute currents. We can deﬁne a macroscopicmagnetic charge density ρ mand current density Jmexactly as we did with electric charge, and use conservation of magnetic charge to provide a continuity equation: ∇·Jm(r,t)+∂ρm(r,t) ∂t=0. (1.18) With these new sources Maxwell’s equations become appealingly symmetric. Despite uncertainties about the existence and physical nature of magnetic monopoles, magneticcharge and current have become an integral part of electromagnetic theory. We often usethe concept of ﬁctitious magnetic sources to make Maxwell’s equations symmetric, and then derive various equivalence theorems for use in the solution of important problems.Thus we can put the idea of magnetic sources to use regardless of whether these sourcesactually exist. 1.4 Problems 1.1Write the volume charge density for a singular surface charge located on the sphere r=r0, entirely in terms of spherical coordinates. Find the total charge on the sphere. 1.2Repeat Problem 1.1 for a charged half plane φ=φ0. 1.3Write the volume charge density for a singular surface charge located on the cylin- derρ=ρ0, entirely in terms of cylindrical coordinates. Find the total charge on the cylinder. 1.4Repeat Problem 1.3 for a charged half plane φ=φ0. Chapter 2 Maxwell’s theory of electromagnetism 2.1 The postulate In1864, James Clerk Maxwell proposed one of the most successful theories in the history of science. In a famous memoir to the Royal Society [125] he presented nineequations summarizing all known laws on electricity and magnetism. This was morethan a mere cataloging of the laws of nature. By postulating the need for an additionalterm to make the set of equations self-consistent, Maxwell was able to put forth whatis still considered a complete theory of macroscopic electromagnetism. The beauty ofMaxwell’s equations led Boltzmann to ask, “Was it a god who wrote these lines ...?” [185]. Since that time authors have struggled to ﬁnd the best way to present Maxwell’s theory. Although it is possible to study electromagnetics from an “empirical–inductive”viewpoint (roughly following the historical order of development beginning with staticﬁelds), it is only by postulating the complete theory that we can do justice to Maxwell’svision. His concept of the existence of an electromagnetic “ﬁeld” (as introduced byFaraday) is fundamental to this theory, and has become one of the most signiﬁcantprinciples of modern science. We ﬁnd controversy even over the best way to present Maxwell’s equations. Maxwell worked at a time before vector notation was completely in place, and thus chose touse scalar variables and equations to represent the ﬁelds. Certainly the true beautyof Maxwell’s equations emerges when they are written in vector form, and the use oftensors reduces the equations to their underlying physical simplicity. We shall use vector notation in this book because of its wide acceptance by engineers, but we still mustdecide whether it is more appropriate to present the vector equations in integral or pointform. On one side of this debate, the brilliant mathematician David Hilbert felt that the fundamental natural laws should be posited as axioms, each best described in termsof integral equations [154]. This idea has been championed by Truesdell and Toupin[199]. On the other side, we may quote from the great physicist Arnold Sommerfeld:“The general development of Maxwell’s theory must proceed from its diﬀerential form;for special problems the integral form may, however, be more advantageous” ([185], p.23). Special relativity ﬂows naturally from the point forms, with ﬁelds easily convertedbetween moving reference frames. For stationary media, it seems to us that the onlydiﬀerence between the two approaches arises in how we handle discontinuities in sourcesand materials. If we choose to use the point forms of Maxwell’s equations, then we mustalso postulate the boundary conditions at surfaces of discontinuity. This is pointed out clearly by Tai [192], who also notes that if the integral forms are used, then their validity across regions of discontinuity should be stated as part of the postulate. We have decided to use the point form in this text. In doing so we follow a long history begun by Hertz in 1890 [85] when he wrote down Maxwell’s diﬀerential equationsas a set of axioms, recognizing the equations as the launching point for the theory ofelectromagnetism. Also, by postulating Maxwell’s equations in point form we can takefull advantage of modern developments in the theory of partial diﬀerential equations; inparticular, the idea of a “well-posed” theory determines what sort of information mustbe speciﬁed to make the postulate useful. We must also decide which form of Maxwell’s diﬀerential equations to use as the basis of our postulate. There are several competing forms, each diﬀering on the manner inwhich materials are considered. The oldest and most widely used form was suggestedby Minkowski in 1908 [130]. In the Minkowski form the diﬀerential equations containno mention of the materials supporting the ﬁelds; all information about material media is relegated to the constitutive relationships. This places simplicity of the diﬀerentialequations above intuitive understanding of the behavior of ﬁelds in materials. We choosethe Maxwell–Minkowski form as the basis of our postulate, primarily for ease of ma-nipulation. But we also recognize the value of other versions of Maxwell’s equations.We shall present the basic ideas behind the Boﬃ form, which places some informationabout materials into the diﬀerential equations (although constitutive relationships arestill required). Missing, however, is any information regarding the velocity of a movingmedium. By using the polarization and magnetization vectors Pand Mrather than the ﬁelds Dand H, it is sometimes easier to visualize the meaning of the ﬁeld vectors and to understand (or predict) the nature of the constitutive relations. The Chu and Amperian forms of Maxwell’s equations have been promoted as useful alternatives to the Minkowski and Boﬃ forms. These include explicit information aboutthe velocity of a moving material, and diﬀer somewhat from the Boﬃ form in the physicalinterpretation of the electric and magnetic properties of matter. Although each of thesemodels matter in terms of charged particles immersed in free space, magnetization in theBoﬃ and Amperian forms arises from electric current loops, while the Chu form employsmagnetic dipoles. In all three forms polarization is modeled using electric dipoles. For adetailed discussion of the Chu and Amperian forms, the reader should consult the workof Kong [101], Tai [193], Penﬁeld and Haus [145], or Fano, Chu and Adler [70]. Importantly, all of these various forms of Maxwell’s equations produce the same values of the physical ﬁelds (at least external to the material where the ﬁelds are measurable). We must include several other constituents, besides the ﬁeld equations, to make the postulate complete. To form a complete ﬁeld theory we need a source ﬁeld, a mediatingﬁeld, and a set of ﬁeld diﬀerential equations. This allows us to mathematically describethe relationship between eﬀect (the mediating ﬁeld) and cause (the source ﬁeld). Ina well-posed postulate we must also include a set of constitutive relationships and aspeciﬁcation of some ﬁeld relationship over a bounding surface and at an initial time. Ifthe electromagnetic ﬁeld is to have physical meaning, we must link it to some observablequantity such as force. Finally, to allow the solution of problems involving mathematicaldiscontinuities we must specify certain boundary, or “jump,” conditions. 2.1.1 The Maxwell–Minkowski equations In Maxwell’s macroscopic theory of electromagnetics, the source ﬁeld consists of the vector ﬁeld J(r,t)(the current density) and the scalar ﬁeld ρ(r,t)(the charge density). In Minkowski’s form of Maxwell’s equations, the mediating ﬁeld is the electromagnetic ﬁeldconsisting of the set of four vector ﬁelds E(r,t),D(r,t),B(r,t), and H(r,t). The ﬁeld equations are the four partial diﬀerential equations referred to as the Maxwell–Minkowski equations ∇× E(r,t)=−∂ ∂tB(r,t), (2.1) ∇× H(r,t)=J(r,t)+∂ ∂tD(r,t), (2.2) ∇·D(r,t)=ρ(r,t), (2.3) ∇·B(r,t)=0, (2.4) along with the continuity equation ∇·J(r,t)=−∂ ∂tρ(r,t). (2.5) Here (2.1) is called Faraday’s law , (2.2) is called Ampere’s law , (2.3) is called Gauss’s law, and (2.4) is called the magnetic Gauss’s law . For brevity we shall often leave the dependence on rand timplicit, and refer to the Maxwell–Minkowski equations as simply the “Maxwell equations,” or “Maxwell’s equations.” Equations (2.1)–(2.5), the point forms of the ﬁeld equations, describe the relation- ships between the ﬁelds and their sources at each point in space where the ﬁelds arecontinuously diﬀerentiable (i.e., the derivatives exist and are continuous). Such pointsare called ordinary points . We shall not attempt to deﬁne the ﬁelds at other points, but instead seek conditions relating the ﬁelds across surfaces containing these points.Normally this is necessary on surfaces across which either sources or material parametersare discontinuous. The electromagnetic ﬁelds carry SI units as follows: Eis measured in Volts per meter (V/m), Bis measured in Teslas (T), His measured in Amperes per meter (A/m), and Dis measured in Coulombs per square meter (C/m 2). In older texts we ﬁnd the units of Bgiven as Webers per square meter (Wb/m2) to reﬂect the role of Bas a ﬂux vector; in that case the Weber (Wb =T·m2) is regarded as a unit of magnetic ﬂux. The interdependence of Maxwell’s equations. It is often claimed that the diver- gence equations (2.3) and (2.4) may be derived from the curl equations (2.1) and (2.2).While this is true, it is notproper to say that only the two curl equations are required to describe Maxwell’s theory. This is because an additional physical assumption, notpresent in the two curl equations, is required to complete the derivation. Either thedivergence equations must be speciﬁed, or the values of certain constants that ﬁx theinitial conditions on the ﬁelds must be speciﬁed. It is customary to specify the divergenceequations and include them with the curl equations to form the complete set we now call“Maxwell’s equations.” To identify the interdependence we take the divergence of (2.1) to get ∇·(∇× E)=∇·/parenleftbigg −∂B ∂t/parenrightbigg , hence ∂ ∂t(∇·B)=0 by (B.49). This requires that ∇·Bbe constant with time, say ∇·B(r,t)=CB(r). The constant CBmust be speciﬁed as part of the postulate of Maxwell’s theory, and the choice we make is subject to experimental validation. We postulate that CB(r)=0, which leads us to (2.4). Note that if we can identify a time prior to which B(r,t)≡0, then CB(r)must vanish. For this reason, CB(r)=0and (2.4) are often called the “initial conditions” for Faraday’s law [159]. Next we take the divergence of (2.2) to ﬁnd that ∇·(∇× H)=∇· J+∂ ∂t(∇·D). Using (2.5) and (B.49), we obtain ∂ ∂t(ρ−∇· D)=0 and thus ρ−∇· Dmust be some temporal constant CD(r). Again, we must postulate the value of CDas part of the Maxwell theory. We choose CD(r)=0and thus obtain Gauss’s law (2.3). If we can identify a time prior to which both Dandρare everywhere equal to zero, then CD(r)must vanish. Hence CD(r)=0and (2.3) may be regarded as “initial conditions” for Ampere’s law. Combining the two sets of initial conditions,we ﬁnd that the curl equations imply the divergence equations as long as we can ﬁnd atime prior to which all of the ﬁelds E,D,B,Hand the sources Jandρare equal to zero (since all the ﬁelds are related through the curl equations, and the charge and current arerelated through the continuity equation). Conversely, the empirical evidence supportingthe two divergence equations implies that such a time should exist. Throughout this book we shall refer to the two curl equations as the “fundamental” Maxwell equations, and to the two divergence equations as the “auxiliary” equations.The fundamental equations describe the relationships between the ﬁelds while, as wehave seen, the auxiliary equations provide a sort of initial condition. This does notimply that the auxiliary equations are of lesser importance; indeed, they are requiredto establish uniqueness of the ﬁelds, to derive the wave equations for the ﬁelds, and toproperly describe static ﬁelds. Field vector terminology. Various terms are used for the ﬁeld vectors, sometimes harkening back to the descriptions used by Maxwell himself, and often based on thephysical nature of the ﬁelds. We are attracted to Sommerfeld’s separation of the ﬁelds intoentities of intensity (E,B)andentities of quantity (D,H). In this system Eis called theelectric ﬁeld strength ,Bthemagnetic ﬁeld strength ,Dtheelectric excitation , and H themagnetic excitation [185]. Maxwell separated the ﬁelds into a set (E,H)of vectors that appear within line integrals to give work-related quantities, and a set (B,D)of vectors that appear within surface integrals to give ﬂux-related quantities; we shall seethis clearly when considering the integral forms of Maxwell’s equations. By this system,authors such as Jones [97] and Ramo, Whinnery, and Van Duzer [153] call Etheelectric intensity ,Hthemagnetic intensity ,Bthemagnetic ﬂux density , and Dtheelectric ﬂux density . Maxwell himself designated names for each of the vector quantities. In his classic paper “A Dynamical Theory of the Electromagnetic Field,” [178] Maxwell referred tothe quantity we now designate Eas the electromotive force , the quantity Das the elec- tric displacement (with a time rate of change given by his now famous “displacement current”), the quantity Has the magnetic force , and the quantity Bas the magnetic induction (although he described Bas a density of lines of magnetic force). Maxwell also included a quantity designated electromagnetic momentum as an integral part of his theory. We now know this as the vector potential Awhich is not generally included as a part of the electromagnetics postulate. Many authors follow the original terminology of Maxwell, with some slight modiﬁca- tions. For instance, Stratton [187] calls Etheelectric ﬁeld intensity ,Hthemagnetic ﬁeld intensity ,Dtheelectric displacement , and Bthemagnetic induction . Jackson [91] calls Etheelectric ﬁeld ,Hthemagnetic ﬁeld ,Dthedisplacement , and Bthemagnetic induction . Other authors choose freely among combinations of these terms. For instance, Kong [101] calls Etheelectric ﬁeld strength ,Hthemagnetic ﬁeld strength ,Bthemagnetic ﬂux density , and Dtheelectric displacement . We do not wish to inject further confusion into the issue of nomenclature; still, we ﬁnd it helpful to use as simple a naming system aspossible. We shall refer to Eas the electric ﬁeld ,Has the magnetic ﬁeld ,Das the electric ﬂux density and Bas the magnetic ﬂux density . When we use the term electromagnetic ﬁeldwe imply the entire set of ﬁeld vectors ( E,D,B,H) used in Maxwell’s theory. Invariance of Maxwell’s equations. Maxwell’s diﬀerential equations are valid for any system in uniform relative motion with respect to the laboratory frame of reference inwhich we normally do our measurements. The ﬁeld equations describe the relationshipsbetween the source and mediating ﬁelds within that frame of reference . This property was ﬁrst proposed for moving material media by Minkowski in 1908 (using the termcovariance ) [130]. For this reason, Maxwell’s equations expressed in the form (2.1)–(2.2) are referred to as the Minkowski form . 2.1.2 Connection to mechanics Our postulate must include a connection between the abstract quantities of charge and ﬁeld and a measurable physical quantity. A convenient means of linking electromagneticsto other classical theories is through mechanics. We postulate that charges experiencemechanical forces given by the Lorentz force equation . If a small volume element dV contains a total charge ρdV, then the force experienced by that charge when moving at velocity vin an electromagnetic ﬁeld is dF=ρdVE+ρvdV×B. (2.6) As with any postulate, we verify this equation through experiment. Note that we write the Lorentz force in terms of charge ρdV, rather than charge density ρ, since charge is an invariant quantity under a Lorentz transformation. The important links between the electromagnetic ﬁelds and energy and momentum must also be postulated. We postulate that the quantity S em=E×H (2.7) represents the transport density of electromagnetic power, and that the quantity gem=D×B (2.8) represents the transport density of electromagnetic momentum. 2.2 The well-posed nature of the postulate It is important to investigate whether Maxwell’s equations, along with the point form of the continuity equation, suﬃce as a useful theory of electromagnetics. Certainly wemust agree that a theory is “useful” as long as it is deﬁned as such by the scientists andengineers who employ it. In practice a theory is considered useful if it predicts accuratelythe behavior of nature under given circumstances, and even a theory that often fails maybe useful if it is the best available. We choose here to take a more narrow view andinvestigate whether the theory is “well-posed.” A mathematical model for a physical problem is said to be well-posed ,o rcorrectly set , if three conditions hold: 1. the model has at least one solution ( existence ); 2. the model has at most one solution ( uniqueness ); 3. the solution is continuously dependent on the data supplied. The importance of the ﬁrst condition is obvious: if the electromagnetic model has no solution, it will be of little use to scientists and engineers. The importance of the secondcondition is equally obvious: if we apply two diﬀerent solution methods to the samemodel and get two diﬀerent answers, the model will not be very helpful in analysis ordesign work. The third point is more subtle; it is often extended in a practical sense tothe following statement: 3 /prime. Small changes in the data supplied produce equally small changes in the solution. That is, the solution is not sensitive to errors in the data. To make sense of this we must decide which quantity is speciﬁed (the independent quantity) and which remainsto be calculated (the dependent quantity). Commonly the source ﬁeld (charge) is takenas the independent quantity, and the mediating (electromagnetic) ﬁeld is computed fromit; in such cases it can be shown that Maxwell’s equations are well-posed. Taking theelectromagnetic ﬁeld to be the independent quantity, we can produce situations in whichthe computed quantity (charge or current) changes wildly with small changes in thespeciﬁed ﬁelds. These situations (called inverse problems ) are of great importance in remote sensing, where the ﬁeld is measured and the properties of the object probed arethereby deduced. At this point we shall concentrate on the “forward” problem of specifying the source ﬁeld (charge) and computing the mediating ﬁeld (the electromagnetic ﬁeld). In this casewe may question whether the ﬁrst of the three conditions (existence) holds. We havetwelve unknown quantities (the scalar components of the four vector ﬁelds), but onlyeight equations to describe them (from the scalar components of the two fundamentalMaxwell equations and the two scalar auxiliary equations). With fewer equations thanunknowns we cannot be sure that a solution exists, and we refer to Maxwell’s equationsas being indeﬁnite . To overcome this problem we must specify more information in the form of constitutive relations among the ﬁeld quantities E,B,D,H, and J. When these are properly formulated, the number of unknowns and the number of equationsare equal and Maxwell’s equations are in deﬁnite form . If we provide more equations than unknowns, the solution may be non-unique. When we model the electromagneticproperties of materials we must supply precisely the right amount of information in theconstitutive relations, or our postulate will not be well-posed. Once Maxwell’s equations are in deﬁnite form, standard methods for partial diﬀerential equations can be used to determine whether the electromagnetic model is well-posed. Ina nutshell, the system (2.1)–(2.2) of hyperbolic diﬀerential equations is well-posed if andonly if we specify Eand Hthroughout a volume region Vat some time instant and also specify, at all subsequent times, 1. the tangential component of Eover all of the boundary surface S,o r 2. the tangential component of Hover all of S,o r 3. the tangential component of Eover part of S, and the tangential component of H over the remainder of S. Proof of all three of the conditions of well-posedness is quite tedious, but a simpliﬁed uniqueness proof is often given in textbooks on electromagnetics. The procedure usedby Stratton [187] is reproduced below. The interested reader should refer to Hansen [81]for a discussion of the existence of solutions to Maxwell’s equations. 2.2.1 Uniqueness of solutions to Maxwell’sequations Consider a simply connected region of space Vbounded by a surface S, where both Vand Scontain only ordinary points. The ﬁelds within Vare associated with a current distribution J, which may be internal to V(entirely or in part). By the initial conditions that imply the auxiliary Maxwell’s equations, we know there is a time, say t=0, prior to which the current is zero for all time, and thus by causality the ﬁelds throughout V are identically zero for all times t<0. We next assume that the ﬁelds are speciﬁed throughout Vat some time t0>0, and seek conditions under which they are determined uniquely for all t>t0. Let the ﬁeld set (E1,D1,B1,H1)be a solution to Maxwell’s equations (2.1)–(2.2) associated with the current J(along with an appropriate set of constitutive relations), and let (E2,D2,B2,H2)be a second solution associated with J. To determine the con- ditions for uniqueness of the ﬁelds, we look for a situation that results in E1=E2, B1=B2, and so on. The electromagnetic ﬁelds must obey ∇× E1=−∂B1 ∂t, ∇× H1=J+∂D1 ∂t, ∇× E2=−∂B2 ∂t, ∇× H2=J+∂D2 ∂t. Subtracting, we have ∇×(E1−E2)=−∂(B1−B2) ∂t, (2.9) ∇×(H1−H2)=∂(D1−D2) ∂t, (2.10) hence deﬁning E0=E1−E2,B0=B1−B2, and so on, we have E0·(∇× H0)=E0·∂D0 ∂t, (2.11) H0·(∇× E0)=−H0·∂B0 ∂t. (2.12) Subtracting again, we have E0·(∇× H0)−H0·(∇× E0)=H0·∂B0 ∂t+E0·∂D0 ∂t, hence −∇ ·(E0×H0)=E0·∂D0 ∂t+H0·∂B0 ∂t by (B.44). Integrating both sides throughout Vand using the divergence theorem on the left-hand side, we get −/contintegraldisplay S(E0×H0)·dS=/integraldisplay V/parenleftbigg E0·∂D0 ∂t+H0·∂B0 ∂t/parenrightbigg dV. Breaking Sinto two arbitrary portions and using (B.6), we obtain /integraldisplay S1E0·(ˆn×H0)dS−/integraldisplay S2H0·(ˆn×E0)dS=/integraldisplay V/parenleftbigg E0·∂D0 ∂t+H0·∂B0 ∂t/parenrightbigg dV. Now if ˆn×E0=0orˆn×H0=0over all of S, or some combination of these conditions holds over all of S, then /integraldisplay V/parenleftbigg E0·∂D0 ∂t+H0·∂B0 ∂t/parenrightbigg dV=0. (2.13) This expression implies a relationship between E0,D0,B0, and H0. Since Vis arbitrary, we see that one possibility is simply to have D0and B0constant with time. However, since the ﬁelds are identically zero for t<0, if they are constant for all time then those constant values must be zero. Another possibility is to have one of each pair (E0,D0) and(H0,B0)equal to zero. Then, by (2.9) and (2.10), E0=0implies B0=0, and D0=0implies H0=0.T h u s E1=E2,B1=B2, and so on, and the solution is unique throughout V. However, we cannot in general rule out more complicated relationships. The number of possibilities depends on the additional constraints on the relationshipbetween E 0,D0,B0, and H0that we must supply to describe the material supporting the ﬁeld — i.e., the constitutive relationships. For a simple medium described by thetime-constant permittivity /epsilon1and permeability µ, (13) becomes /integraldisplay V/parenleftbigg E0·/epsilon1∂E0 ∂t+H0·µ∂H0 ∂t/parenrightbigg dV=0, or 1 2∂ ∂t/integraldisplay V(/epsilon1E0·E0+µH0·H0)dV=0. Since the integrand is always positive or zero (and not constant with time, as mentioned above), the only possible conclusion is that E0and H0must both be zero, and thus the ﬁelds are unique. When establishing more complicated constitutive relations, we must be careful to en- sure that they lead to a unique solution, and that the condition for uniqueness is un-derstood. In the case above, the assumption ˆn×E 0/vextendsingle/vextendsingle S=0implies that the tangential components of E1and E2are identical over S— that is, we must give speciﬁc values of these quantities on Sto ensure uniqueness. A similar statement holds for the condition ˆn×H0/vextendsingle/vextendsingle S=0. Requiring that constitutive relations lead to a unique solution is known asjust setting , and is one of several factors that must be considered, as discussed in the next section. Uniqueness implies that the electromagnetic state of an isolated region of space may be determined without the knowledge of conditions outside the region. If we wish tosolve Maxwell’s equations for that region, we need know only the source density withinthe region and the values of the tangential ﬁelds over the bounding surface. The eﬀectsof a complicated external world are thus reduced to the speciﬁcation of surface ﬁelds.This concept has numerous applications to problems in antennas, diﬀraction, and guidedwaves. 2.2.2 Constitutive relations We now supply a set of constitutive relations to complete the conditions for well- posedness. We generally split these relations into two sets. The ﬁrst describes the relationships between the electromagnetic ﬁeld quantities, and the second describes me-chanical interaction between the ﬁelds and resulting secondary sources. All of theserelations depend on the properties of the medium supporting the electromagnetic ﬁeld.Material phenomena are quite diverse, and it is remarkable that the Maxwell–Minkowskiequations hold for all phenomena yet discovered. All material eﬀects, from nonlinearityto chirality to temporal dispersion, are described by the constitutive relations. The speciﬁcation of constitutive relationships is required in many areas of physical science to describe the behavior of “ideal materials”: mathematical models of actualmaterials encountered in nature. For instance, in continuum mechanics the constitutiveequations describe the relationship between material motions and stress tensors [209].Truesdell and Toupin [199] give an interesting set of “guiding principles” for the con-cerned scientist to use when constructing constitutive relations. These include consider-ation of consistency (with the basic conservation laws of nature), coordinate invariance (independence of coordinate system), isotropy and aeolotropy (dependence on, or inde- pendence of, orientation), just setting (constitutive parameters should lead to a unique solution), dimensional invariance (similarity), material indiﬀerence (non-dependence on the observer), and equipresence (inclusion of allrelevant physical phenomena in allof the constitutive relations across disciplines). The constitutive relations generally involve a set of constitutive parameters and a set of constitutive operators. The constitutive parameters may be as simple as constantsof proportionality between the ﬁelds or they may be components in a dyadic relation- ship. The constitutive operators may be linear and integro-diﬀerential in nature, or mayimply some nonlinear operation on the ﬁelds. If the constitutive parameters are spa-tially constant within a certain region, we term the medium homogeneous within that region. If the constitutive parameters vary spatially, the medium is inhomogeneous .I f the constitutive parameters are constants with time, we term the medium stationary ; if they are time-changing, the medium is nonstationary . If the constitutive operators involve time derivatives or integrals, the medium is said to be temporally dispersive ;i f space derivatives or integrals are involved, the medium is spatially dispersive . Examples of all these eﬀects can be found in common materials. It is important to note that theconstitutive parameters may depend on other physical properties of the material, suchas temperature, mechanical stress, and isomeric state, just as the mechanical constitu-tive parameters of a material may depend on the electromagnetic properties (principleof equipresence). Many eﬀects produced by linear constitutive operators, such as those associated with temporal dispersion, have been studied primarily in the frequency domain. In this case temporal derivative and integral operations produce complex constitutive parameters. Itis becoming equally important to characterize these eﬀects directly in the time domainfor use with direct time-domain ﬁeld solving techniques such as the ﬁnite-diﬀerence time-domain (FDTD) method. We shall cover the very basic properties of dispersive mediain this section. A detailed description of frequency-domain ﬁelds (and a discussion ofcomplex constitutive parameters) is deferred until later in this book. It is diﬃcult to ﬁnd a simple and consistent means for classifying materials by their electromagnetic eﬀects. One way is to separate linear and nonlinear materials, then cate-gorize linear materials by the way in which the ﬁelds are coupled through the constitutiverelations: 1.Isotropic materials are those in which Dis related to E,Bis related to H, and the secondary source current Jis related to E, with the ﬁeld direction in each pair aligned. 2.I nanisotropic materials the pairings are the same, but the ﬁelds in each pair are generally not aligned. 3. Inbiisotropic materials (such as chiral media) the ﬁelds Dand Bdepend on both Eand H, but with no realignment of EorH; for instance, Dis given by the addition of a scalar times Eplus a second scalar times H. Thus the contributions toDinvolve no changes to the directions of Eand H. 4.Bianisotropic materials exhibit the most general behavior: Dand Hdepend on both Eand B, with an arbitrary realignment of either or both of these ﬁelds. In 1888, Roentgen showed experimentally that a material isotropic in its own station- ary reference frame exhibits bianisotropic properties when observed from a moving frame.Only recently have materials bianisotropic in their own rest frame been discovered. In1894 Curie predicted that in a stationary material, based on symmetry, an electric ﬁeldmight produce magnetic eﬀects and a magnetic ﬁeld might produce electric eﬀects. Theseeﬀects, coined magnetoelectric by Landau and Lifshitz in 1957, were sought unsuccess- fully by many experimentalists during the ﬁrst half of the twentieth century. In 1959 theSoviet scientist I.E. Dzyaloshinskii predicted that, theoretically, the antiferromagneticmaterial chromium oxide (Cr 2O3) should display magnetoelectric eﬀects. The magneto- electric eﬀect was ﬁnally observed soon after by D.N. Astrov in a single crystal of Cr 2O3 using a 10 kHz electric ﬁeld. Since then the eﬀect has been observed in many diﬀerentmaterials. Recently, highly exotic materials with useful electromagnetic properties have been proposed and studied in depth, including chiroplasmas and chiroferrites [211]. Asthe technology of materials synthesis advances, a host of new and intriguing media willcertainly be created. The most general forms of the constitutive relations between the ﬁelds may be written in symbolic form as D=D[E,B], (2.14) H=H[E,B]. (2.15) That is, Dand Hhave some mathematically descriptive relationship to Eand B. The speciﬁc forms of the relationships may be written in terms of dyadics as [102] cD=¯P·E+¯L·(cB), (2.16) H=¯M·E+¯Q·(cB), (2.17) where each of the quantities ¯P,¯L,¯M,¯Qmay be dyadics in the usual sense, or dyadic operators containing space or time derivatives or integrals, or some nonlinear operationson the ﬁelds. We may write these expressions as a single matrix equation /bracketleftbiggcD H/bracketrightbigg =[¯C]/bracketleftbiggE cB/bracketrightbigg (2.18) where the 6×6matrix [¯C]=/bracketleftbigg¯P¯L ¯M¯Q/bracketrightbigg . This most general relationship between ﬁelds is the property of a bianisotropic material. We may wonder why Dis not related to (E,B,H),Eto(D,B), etc. The reason is that since the ﬁeld pairs (E,B)and(D,H)convert identically under a Lorentz transfor- mation, a constitutive relation that maps ﬁelds as in (2.18) is form invariant, as are theMaxwell–Minkowski equations. That is, although the constitutive parameters may vary numerically between observers moving at diﬀerent velocities, the form of the relationshipgiven by (2.18) is maintained. Many authors choose to relate (D,B)to(E,H), often because the expressions are simpler and can be more easily applied to speciﬁc problems. For instance, in a linear,isotropic material (as shown below) Dis directly proportional to Eand Bis directly proportional to H. To provide the appropriate expression for the constitutive relations, we need only remap (2.18). This gives D=¯/epsilon1·E+¯ξ·H, (2.19) B=¯ζ·E+¯µ·H, (2.20) or/bracketleftbiggD B/bracketrightbigg =/bracketleftbig¯C EH/bracketrightbig/bracketleftbiggE H/bracketrightbigg , (2.21) where the new constitutive parameters ¯/epsilon1,¯ξ,¯ζ,¯µcan be easily found from the original constitutive parameters ¯P,¯L,¯M,¯Q. We do note, however, that in the form (2.19)–(2.20) the Lorentz invariance of the constitutive equations is not obvious. In the following paragraphs we shall characterize some of the most common materials according to these classiﬁcations. With this approach eﬀects such as temporal or spatialdispersion are not part of the classiﬁcation process, but arise from the nature of theconstitutive parameters. Hence we shall not dwell on the particulars of the constitutive parameters, but shall concentrate on the form of the constitutive relations. Constitutive relations for ﬁelds in free space. In a vacuum the ﬁelds are related by the simple constitutive equations D=/epsilon1 0E, (2.22) H=1 µ0B. (2.23) The quantities µ0and/epsilon10are, respectively, the free-space permeability andpermittivity constants . It is convenient to use three numerical quantities to describe the electromag- netic properties of free space — µ0,/epsilon10, and the speed of light c— and interrelate them through the equation c=1/(µ 0/epsilon10)1/2. Historically it has been the practice to deﬁne µ0, measure c, and compute /epsilon10. In SI units µ0=4π×10−7H/m, c=2.998×108m/s, /epsilon10=8.854×10−12F/m. With the two constitutive equations we have enough information to put Maxwell’s equations into deﬁnite form. Traditionally (2.22) and (2.23) are substituted into (2.1)–(2.2) to give ∇× E=−∂B ∂t, (2.24) ∇× B=µ0J+µ0/epsilon10∂E ∂t. (2.25) These are two vector equations in two vector unknowns (equivalently, six scalar equations in six scalar unknowns). In terms of the general constitutive relation (2.18), we ﬁnd that free space is isotropic with ¯P=¯Q=1 η0¯I, ¯L=¯M=0, where η0=(µ0//epsilon10)1/2is called the intrinsic impedance of free space . This emphasizes the fact that free space has, along with c, only a single empirical constant associated with it (i.e., /epsilon10orη0). Since no derivative or integral operators appear in the constitutive relations, free space is nondispersive. Constitutive relations in a linear isotropic material. In a linear isotropic mate- rial there is proportionality between Dand Eand between Band H. The constants of proportionality are the permittivity /epsilon1and the permeability µ. If the material is nondis- persive, the constitutive relations take the form D=/epsilon1E, B=µH, where /epsilon1andµmay depend on position for inhomogeneous materials. Often the permit- tivity and permeability are referenced to the permittivity and permeability of free spaceaccording to /epsilon1=/epsilon1 r/epsilon10,µ =µrµ0. Here the dimensionless quantities /epsilon1randµrare called, respectively, the relative permit- tivity andrelative permeability . When dealing with the Maxwell–Boﬃ equations ( §2.4) the diﬀerence between the material and free space values of Dand Hbecomes important. Thus for linear isotropic materials we often write the constitutive relations as D=/epsilon10E+/epsilon10χeE, (2.26) B=µ0H+µ0χmH, (2.27) where the dimensionless quantities χe=/epsilon1r−1andχm=µr−1are called, respectively, theelectric andmagnetic susceptibilities of the material. In terms of (2.18) we have ¯P=/epsilon1r η0¯I, ¯Q=1 η0µr¯I, ¯L=¯M=0. Generally a material will have either its electric or magnetic properties dominant. If µr=1and/epsilon1r/negationslash=1then the material is generally called a perfect dielectric or aperfect insulator , and is said to be an electric material. If /epsilon1r=1andµr/negationslash=1, the material is said to be a magnetic material. A linear isotropic material may also have conduction properties. In a conducting material , a constitutive relation is generally used to describe the mechanical interaction of ﬁeld and charge by relating the electric ﬁeld to a secondary electric current. Fora nondispersive isotropic material, the current is aligned with, and proportional to, theelectric ﬁeld; there are no temporal operators in the constitutive relation, which is simply J=σE. (2.28) This is known as Ohm’s law . Here σis the conductivity of the material. Ifµ r≈1andσis very small, the material is generally called a good dielectric .I f σis very large, the material is generally called a good conductor . The conditions by which we say the conductivity is “small” or “large” are usually established using thefrequency response of the material. Materials that are good dielectrics over broad rangesof frequency include various glasses and plastics such as fused quartz, polyethylene,and teﬂon. Materials that are good conductors over broad ranges of frequency includecommon metals such as gold, silver, and copper. For dispersive linear isotropic materials, the constitutive parameters become nonsta- tionary (time dependent), and the constitutive relations involve time operators. (Notethat the name dispersive describes the tendency for pulsed electromagnetic waves to spread out, or disperse, in materials of this type.) If we assume that the relationshipsgiven by (2.26), (2.27), and (2.28) retain their product form in the frequency domain,then by the convolution theorem we have in the time domain the constitutive relations D(r,t)=/epsilon1 0/parenleftbigg E(r,t)+/integraldisplayt −∞χe(r,t−t/prime)E(r,t/prime)dt/prime/parenrightbigg , (2.29) B(r,t)=µ0/parenleftbigg H(r,t)+/integraldisplayt −∞χm(r,t−t/prime)H(r,t/prime)dt/prime/parenrightbigg , (2.30) J(r,t)=/integraldisplayt −∞σ(r,t−t/prime)E(r,t/prime)dt/prime. (2.31) These expressions were ﬁrst introduced by Volterra in 1912[199]. We see that for a linear dispersive material of this type the constitutive operators are time integrals, and that the behavior of D(t)depends not only on the value of Eat time t, but on its values at all past times. Thus, in dispersive materials there is a “time lag” between the eﬀect ofthe applied ﬁeld and the polarization or magnetization that results. In the frequencydomain, temporal dispersion is associated with complex values of the constitutive pa-rameters, which, to describe a causal relationship, cannot be constant with frequency.The nonzero imaginary component is identiﬁed with the dissipation of electromagneticenergy as heat. Causality is implied by the upper limit being tin the convolution inte- grals, which indicates that D(t)cannot depend on future values of E(t). This assumption leads to a relationship between the real and imaginary parts of the frequency domainconstitutive parameters as described through the Kronig–Kramers equations. Constitutive relations for ﬁelds in perfect conductors. In a perfect electric con- ductor (PEC) or a perfect magnetic conductor (PMC) the ﬁelds are exactly speciﬁed as the null ﬁeld: E=D=B=H=0. By Ampere’s and Faraday’s laws we must also have J=Jm=0; hence, by the continuity equation, ρ=ρm=0. In addition to the null ﬁeld, we have the condition that the tangential electric ﬁeld on the surface of a PEC must be zero. Similarly, the tangential magnetic ﬁeld on thesurface of a PMC must be zero. This implies ( §2.8.3) that an electric surface current may exist on the surface of a PEC but not on the surface of a PMC, while a magneticsurface current may exist on the surface of a PMC but not on the surface of a PEC. A PEC may be regarded as the limit of a conducting material as σ→∞. In many practical cases, good conductors such as gold and copper can be assumed to be perfectelectric conductors, which greatly simpliﬁes the application of boundary conditions. Nophysical material is known to behave as a PMC, but the concept is mathematicallyuseful for applying symmetry conditions (in which a PMC is sometimes referred to as a “magnetic wall”) and for use in developing equivalence theorems. Constitutive relations in a linear anisotropic material. In a linear anisotropic material there are relationships between Band Hand between Dand E, but the ﬁeld vectors are not aligned as in the isotropic case. We can thus write D=¯/epsilon1·E, B=¯µ·H, J=¯σ·E, where ¯/epsilon1is called the permittivity dyadic ,¯µis the permeability dyadic , and ¯σis the conductivity dyadic . In terms of the general constitutive relation (2.18) we have ¯P=c¯/epsilon1, ¯Q=¯µ −1 c, ¯L=¯M=0. Many diﬀerent types of materials demonstrate anisotropic behavior, including opti- cal crystals, magnetized plasmas, and ferrites. Plasmas and ferrites are examples ofgyrotropic media. With the proper choice of coordinate system, the frequency-domain permittivity or permeability can be written in matrix form as [˜¯/epsilon1]= /epsilon1 11/epsilon1120 −/epsilon112/epsilon1110 00 /epsilon133 , [˜¯µ]= µ11µ120 −µ12µ110 00 µ33 . (2.32) Each of the matrix entries may be complex. For the special case of a lossless gyrotropic material, the matrices become hermitian : [˜¯/epsilon1]= /epsilon1−jδ0 jδ/epsilon1 0 00 /epsilon13 , [˜¯µ]= µ−jκ0 jκµ 0 00 µ3 , (2.33) where /epsilon1,/epsilon13,δ,µ,µ3, and κare real numbers. Crystals have received particular attention because of their birefringent properties. A birefringent crystal can be characterized by a symmetric permittivity dyadic that has realpermittivity parameters in the frequency domain; equivalently, the constitutive relationsdo not involve constitutive operators. A coordinate system called the principal system , with axes called the principal axes , can always be found so that the permittivity dyadic in that system is diagonal: [˜¯/epsilon1]= /epsilon1 x00 0/epsilon1y0 00/epsilon1z . The geometrical structure of a crystal determines the relationship between /epsilon1x,/epsilon1y, and /epsilon1z.I f/epsilon1x=/epsilon1y</epsilon1 z, then the crystal is positive uniaxial (e.g., quartz). If /epsilon1x=/epsilon1y>/epsilon1 z, the crystal is negative uniaxial (e.g., calcite). If /epsilon1x/negationslash=/epsilon1y/negationslash=/epsilon1z, the crystal is biaxial (e.g., mica). In uniaxial crystals the z-axis is called the optical axis . If the anisotropic material is dispersive, we can generalize the convolutional form of the isotropic dispersive media to obtain the constitutive relations D(r,t)=/epsilon10/parenleftbigg E(r,t)+/integraldisplayt −∞¯χe(r,t−t/prime)·E(r,t/prime)dt/prime/parenrightbigg , (2.34) B(r,t)=µ0/parenleftbigg H(r,t)+/integraldisplayt −∞¯χm(r,t−t/prime)·H(r,t/prime)dt/prime/parenrightbigg , (2.35) J(r,t)=/integraldisplayt −∞¯σ(r,t−t/prime)·E(r,t/prime)dt/prime. (2.36) Constitutive relations for biisotropic materials. A biisotropic material is an isotropic magnetoelectric material. Here we have Drelated to Eand B, and Hrelated to Eand B, but with no realignment of the ﬁelds as in anisotropic (or bianisotropic) mate- rials. Perhaps the simplest example is the Tellegen medium devised by B.D.H. Tellegen in 1948 [196], having D=/epsilon1E+ξH, (2.37) B=ξE+µH. (2.38) Tellegen proposed that his hypothetical material be composed of small (but macroscopic) ferromagnetic particles suspended in a liquid. This is an example of a synthetic mate- rial, constructed from ordinary materials to have an exotic electromagnetic behavior.Other examples include artiﬁcial dielectrics made from metallic particles imbedded inlightweight foams [66], and chiral materials made from small metallic helices suspended in resins [112]. Chiral materials are also biisotropic, and have the constitutive relations D=/epsilon1E−χ∂H ∂t, (2.39) B=µH+χ∂E ∂t, (2.40) where the constitutive parameter χis called the chirality parameter . Note the presence of temporal derivative operators. Alternatively, D=/epsilon1(E+β∇× E), (2.41) B=µ(H+β∇× H), (2.42) by Faraday’s and Ampere’s laws. Chirality is a natural state of symmetry; many natural substances are chiral materials, including DNA and many sugars. The time derivativesin (2.39)–(2.40) produce rotation of the polarization of time harmonic electromagneticwaves propagating in chiral media. Constitutive relations in nonlinear media. Nonlinear electromagnetic eﬀects have been studied by scientists and engineers since the beginning of the era of electrical tech-nology. Familiar examples include saturation and hysteresis in ferromagnetic materials and the behavior of p-n junctions in solid-state rectiﬁers. The invention of the laser extended interest in nonlinear eﬀects to the realm of optics, where phenomena such asparametric ampliﬁcation and oscillation, harmonic generation, and magneto-optic inter-actions have found applications in modern devices [174]. Provided that the external ﬁeld applied to a nonlinear electric material is small com- pared to the internal molecular ﬁelds, the relationship between Eand Dcan be expanded in a Taylor series of the electric ﬁeld. For an anisotropic material exhibiting no hysteresiseﬀects, the constitutive relation is [131] D i(r,t)=/epsilon10Ei(r,t)+3/summationdisplay j=1χ(1) ijEj(r,t)+3/summationdisplay j,k=1χ(2) ijkEj(r,t)Ek(r,t)+ +3/summationdisplay j,k,l=1χ(3) ijklEj(r,t)Ek(r,t)El(r,t)+··· (2.43) where the index i=1,2,3refers to the three components of the ﬁelds Dand E. The ﬁrst sum in (2.43) is identical to the constitutive relation for linear anisotropic materi- als. Thus, χ(1) ijis identical to the susceptibility dyadic of a linear anisotropic medium considered earlier. The quantity χ(2) ijkis called the second-order susceptibility , and is a three-dimensional matrix (or third rank tensor) describing the nonlinear electric eﬀects quadratic in E. Similarly χ(3) ijklis called the third-order susceptibility , and is a four- dimensional matrix (or fourth rank tensor) describing the nonlinear electric eﬀects cubic inE. Numerical values of χ(2) ijkandχ(3) ijklare given in Shen [174] for a variety of crystals. When the material shows hysteresis eﬀects, Dat any point rand time tis due not only to the value of Eat that point and at that time, but to the values of Eat all points and times. That is, the material displays both temporal and spatial dispersion. 2.3 Maxwell’s equations in moving frames The essence of special relativity is that the mathematical forms of Maxwell’s equa- tions are identical in all inertial reference frames : frames moving with uniform velocities relative to the laboratory frame of reference in which we perform our measurements. Thisform invariance of Maxwell’s equations is a speciﬁc example of the general physical principle of covariance . In the laboratory frame we write the diﬀerential equations of Maxwell’s theory as ∇× E(r,t)=−∂B(r,t) ∂t, ∇× H(r,t)=J(r,t)+∂D(r,t) ∂t, ∇·D(r,t)=ρ(r,t), ∇·B(r,t)=0, ∇·J(r,t)=−∂ρ(r,t) ∂t. Figure 2.1: Primed coordinate system moving with velocity vrelative to laboratory (unprimed) coordinate system. Similarly, in an inertial frame having four-dimensional coordinates (r/prime,t/prime)we have ∇/prime×E/prime(r/prime,t/prime)=−∂B/prime(r/prime,t/prime) ∂t/prime, ∇/prime×H/prime(r/prime,t/prime)=J/prime(r/prime,t/prime)+∂D/prime(r/prime,t/prime) ∂t/prime, ∇/prime·D/prime(r/prime,t/prime)=ρ/prime(r/prime,t/prime), ∇/prime·B/prime(r/prime,t/prime)=0, ∇/prime·J/prime(r/prime,t/prime)=−∂ρ/prime(r/prime,t/prime) ∂t/prime. The primed ﬁelds measured in the moving system do nothave the same numerical values as the unprimed ﬁelds measured in the laboratory. To convert between Eand E/prime,Band B/prime, and so on, we must ﬁnd a way to convert between the coordinates (r,t)and(r/prime,t/prime). 2.3.1 Field conversions under Galilean transformation We shall assume that the primed coordinate system moves with constant velocity v relativetothelaborator yframe(Figure2.1).Priortotheearlypartofthetwentieth century, converting between the primed and unprimed coordinate variables was intuitiveand obvious: it was thought that time must be measured identically in each coordinatesystem, and that the relationship between the space variables can be determined simply by the displacement of the moving system at time t=t /prime. Under these assumptions, and under the further assumption that the two systems coincide at time t=0, we can write t/prime=t, x/prime=x−vxt, y/prime=y−vyt, z/prime=z−vzt, or simply t/prime=t, r/prime=r−vt. This is called a Galilean transformation . We can use the chain rule to describe the manner in which diﬀerential operations transform, i.e., to relate derivatives with respectto the laboratory coordinates to derivatives with respect to the inertial coordinates. Wehave, for instance, ∂ ∂t=∂t/prime ∂t∂ ∂t/prime+∂x/prime ∂t∂ ∂x/prime+∂y/prime ∂t∂ ∂y/prime+∂z/prime ∂t∂ ∂z/prime =∂ ∂t/prime−vx∂ ∂x/prime−vy∂ ∂y/prime−vz∂ ∂z/prime =∂ ∂t/prime−(v·∇/prime). (2.44) Similarly ∂ ∂x=∂ ∂x/prime,∂ ∂y=∂ ∂y/prime,∂ ∂z=∂ ∂z/prime, from which ∇× A(r,t)=∇/prime×A(r,t), ∇·A(r,t)=∇/prime·A(r,t), (2.45) for each vector ﬁeld A. Newton was aware that the laws of mechanics are invariant with respect to Galilean transformations. Do Maxwell’s equations also behave in this way? Let us use the Galileantransformation to determine which relationship between the primed and unprimed ﬁeldsresults in form invariance of Maxwell’s equations. We ﬁrst examine ∇ /prime×E, the spatial rate of change of the laboratory ﬁeld with respect to the inertial frame spatial coordinates: ∇/prime×E=∇× E=−∂B ∂t=−∂B ∂t/prime+(v·∇/prime)B by (2.45) and (2.44). Rewriting the last term by (B.45) we have (v·∇/prime)B=− ∇/prime×(v×B) since vis constant and ∇/prime·B=∇· B=0, hence ∇/prime×(E+v×B)=−∂B ∂t/prime. (2.46) Similarly ∇/prime×H=∇× H=J+∂D ∂t=J+∂D ∂t/prime+∇/prime×(v×D)−v(∇/prime·D) where ∇/prime·D=∇· D=ρso that ∇/prime×(H−v×D)=∂D ∂t/prime−ρv+J. (2.47) Also ∇/prime·J=∇· J=−∂ρ ∂t=−∂ρ ∂t/prime+(v·∇/prime)ρ and we may use (B.42) to write (v·∇/prime)ρ=v·(∇/primeρ)=∇/prime·(ρv), obtaining ∇/prime·(J−ρv)=−∂ρ ∂t/prime. (2.48) Equations (2.46), (2.47), and (2.48) show that the forms of Maxwell’s equations in the inertial and laboratory frames are identical provided that E/prime=E+v×B, (2.49) D/prime=D, (2.50) H/prime=H−v×D, (2.51) B/prime=B, (2.52) J/prime=J−ρv, (2.53) ρ/prime=ρ. (2.54) That is, (2.49)–(2.54) result in form invariance of Faraday’s law, Ampere’s law, and the continuity equation under a Galilean transformation. These equations express the ﬁeldsmeasured by a moving observer in terms of those measured in the laboratory frame. Toconvert the opposite way, we need only use the principle of relativity. Neither observer can tell whether he or she is stationary — only that the other observer is moving relativeto him or her. To obtain the ﬁelds in the laboratory frame we simply change the sign onvand swap primed with unprimed ﬁelds in (2.49)–(2.54): E=E /prime−v×B/prime, (2.55) D=D/prime, (2.56) H=H/prime+v×D/prime, (2.57) B=B/prime, (2.58) J=J/prime+ρ/primev, (2.59) ρ=ρ/prime. (2.60) According to (2.53), a moving observer interprets charge stationary in the laboratory frame as an additional current moving opposite the direction of his or her motion. Thisseems reasonable. However, while Edepends on both E /primeand B/prime, the ﬁeld Bis unchanged under the transformation. Why should Bhave this special status? In fact, we may uncover an inconsistency among the transformations by considering free space where(2.22) and (2.23) hold: in this case (2.49) gives D /prime//epsilon10=D//epsilon10+v×µ0H or D/prime=D+v×H/c2 rather than (2.50). Similarly, from (2.51) we get B/prime=B−v×E/c2 instead of (2.52). Using these, the set of transformations becomes E/prime=E+v×B, (2.61) D/prime=D+v×H/c2, (2.62) H/prime=H−v×D, (2.63) B/prime=B−v×E/c2, (2.64) J/prime=J−ρv, (2.65) ρ/prime=ρ. (2.66) These can also be written using dyadic notation as E/prime=¯I·E+¯β·(cB), (2.67) cB/prime=− ¯β·E+¯I·(cB), (2.68) and cD/prime=¯I·(cD)+¯β·H, (2.69) H/prime=− ¯β·(cD)+¯I·H, (2.70) where [¯β]= 0−βzβy βz0−βx −βyβx0  withβ=v/c. This set of equations is self-consistent among Maxwell’s equations. How- ever, the equations are not consistent with the assumption of a Galilean transformationof the coordinates, and thus Maxwell’s equations are not covariant under a Galileantransformation. Maxwell’s equations are only covariant under a Lorentz transforma-tion as described in the next section. Expressions (2.61)–(2.64) turn out to be accurateto order v/c, hence are the results of a ﬁrst-order Lorentz transformation . Only when vis an appreciable fraction of cdo the ﬁeld conversions resulting from the ﬁrst-order Lorentz transformation diﬀer markedly from those resulting from a Galilean transforma-tion; those resulting from the true Lorentz transformation require even higher velocitiesto diﬀer markedly from the ﬁrst-order expressions. Engineering accuracy is often accom-plished using the Galilean transformation. This pragmatic observation leads to quite abit of confusion when considering the large-scale forms of Maxwell’s equations, as weshall soon see. 2.3.2 Field conversions under Lorentz transformation To ﬁnd the proper transformation under which Maxwell’s equations are covariant, we must discard our notion that time progresses the same in the primed and the un-primed frames. The proper transformation of coordinates that guarantees covariance ofMaxwell’s equations is the Lorentz transformation ct /prime=γct−γβ·r, (2.71) r/prime=¯α·r−γβct, (2.72) where γ=1/radicalbig 1−β2, ¯α=¯I+(γ−1)ββ β2,β =|β|. This is obviously more complicated than the Galilean transformation; only as β→0are the Lorentz and Galilean transformations equivalent. Not surprisingly, ﬁeld conversions between inertial reference frames are more com- plicated with the Lorentz transformation than with the Galilean transformation. Forsimplicity we assume that the velocity of the moving frame has only an x-component: v=ˆxv. Later we can generalize this to any direction. Equations (2.71) and (2.72) become x /prime=x+(γ−1)x−γvt, (2.73) y/prime=y, (2.74) z/prime=z, (2.75) ct/prime=γct−γv cx, (2.76) and the chain rule gives ∂ ∂x=γ∂ ∂x/prime−γv c2∂ ∂t/prime, (2.77) ∂ ∂y=∂ ∂y/prime, (2.78) ∂ ∂z=∂ ∂z/prime, (2.79) ∂ ∂t=−γv∂ ∂x/prime+γ∂ ∂t/prime. (2.80) We begin by examining Faraday’s law in the laboratory frame. In component form we have ∂Ez ∂y−∂Ey ∂z=−∂Bx ∂t, (2.81) ∂Ex ∂z−∂Ez ∂x=−∂By ∂t, (2.82) ∂Ey ∂x−∂Ex ∂y=−∂Bz ∂t. (2.83) These become ∂Ez ∂y/prime−∂Ey ∂z/prime=γv∂Bx ∂x/prime−γ∂Bx ∂t/prime, (2.84) ∂Ex ∂z/prime−γ∂Ez ∂x/prime+γv c2∂Ez ∂t/prime=γv∂By ∂x/prime−γ∂By ∂t/prime, (2.85) γ∂Ey ∂x/prime−γv c2∂Ey ∂t/prime−∂Ex ∂y/prime=γv∂Bz ∂x/prime−γ∂Bz ∂t/prime, (2.86) after we use (2.77)–(2.80) to convert the derivatives in the laboratory frame to derivatives with respect to the moving frame coordinates. To simplify (2.84) we consider ∇·B=∂Bx ∂x+∂By ∂y+∂Bz ∂z=0. Converting the laboratory frame coordinates to the moving frame coordinates, we have γ∂Bx ∂x/prime−γv c2∂Bx ∂t/prime+∂By ∂y/prime+∂Bz ∂z/prime=0 or −γv∂Bx ∂x/prime=−γv2 c2∂Bx ∂t/prime+v∂By ∂y/prime+v∂Bz ∂z/prime. Substituting this into (2.84) and rearranging (2.85) and (2.86), we obtain ∂ ∂y/primeγ(Ez+vBy)−∂ ∂z/primeγ(Ey−vBz)=−∂Bx ∂t/prime, ∂Ex ∂z/prime−∂ ∂x/primeγ(Ez+vBy)=−∂ ∂t/primeγ/parenleftBig By+v c2Ez/parenrightBig , ∂ ∂x/primeγ(Ey−vBz)−∂Ex ∂y/prime=−∂ ∂t/primeγ/parenleftBig Bz−v c2Ey/parenrightBig . Comparison with (2.81)–(2.83) shows that form invariance of Faraday’s law under the Lorentz transformation requires E/prime x=Ex, E/prime y=γ(Ey−vBz), E/prime z=γ(Ez+vBy), and B/prime x=Bx, B/prime y=γ/parenleftBig By+v c2Ez/parenrightBig , B/prime z=γ/parenleftBig Bz−v c2Ey/parenrightBig . To generalize vto any direction, we simply note that the components of the ﬁelds parallel to the velocity direction are identical in the moving and laboratory frames, while thecomponents perpendicular to the velocity direction convert according to a simple crossproduct rule. After similar analyses with Ampere’s and Gauss’s laws (see Problem 2.2),we ﬁnd that E /prime /bardbl=E/bardbl, B/prime /bardbl=B/bardbl, D/prime /bardbl=D/bardbl, H/prime /bardbl=H/bardbl, E/prime ⊥=γ(E⊥+β×cB⊥), (2.87) cB/prime ⊥=γ(cB⊥−β×E⊥), (2.88) cD/prime ⊥=γ(cD⊥+β×H⊥), (2.89) H/prime ⊥=γ(H⊥−β×cD⊥), (2.90) and J/prime /bardbl=γ(J/bardbl−ρv), (2.91) J/prime ⊥=J⊥, (2.92) cρ/prime=γ(cρ−β·J), (2.93) where the symbols /bardbland⊥designate the components of the ﬁeld parallel and perpen- dicular to v, respectively. These conversions are self-consistent, and the Lorentz transformation is the transfor- mation under which Maxwell’s equations are covariant. If v2/lessmuchc2, then γ≈1and to ﬁrst order (2.87)–(2.93) reduce to (2.61)–(2.66). If v/c/lessmuch1, then the ﬁrst-order ﬁelds reduce to the Galilean ﬁelds (2.49)–(2.54). To convert in the opposite direction, we can swap primed and unprimed ﬁelds and change the sign on v: E⊥=γ(E/prime ⊥−β×cB/prime ⊥), (2.94) cB⊥=γ(cB/prime ⊥+β×E/prime ⊥), (2.95) cD⊥=γ(cD/prime ⊥−β×H/prime ⊥), (2.96) H⊥=γ(H/prime ⊥+β×cD/prime ⊥), (2.97) and J/bardbl=γ(J/prime /bardbl+ρ/primev), (2.98) J⊥=J/prime ⊥, (2.99) cρ=γ(cρ/prime+β·J/prime). (2.100) The conversion formulas can be written much more succinctly in dyadic notation: E/prime=γ¯α−1·E+γ¯β·(cB), (2.101) cB/prime=−γ¯β·E+γ¯α−1·(cB), (2.102) cD/prime=γ¯α−1·(cD)+γ¯β·H, (2.103) H/prime=−γ¯β·(cD)+γ¯α−1·H, (2.104) and cρ/prime=γ(cρ−β·J), (2.105) J/prime=¯α·J−γβcρ, (2.106) where ¯α−1·¯α=¯I, and thus ¯α−1=¯α−γββ. Maxwell’s equations are covariant under a Lorentz transformation but not under a Galilean transformation; the laws of mechanics are invariant under a Galilean transfor-mation but not under a Lorentz transformation. How then should we analyze interactionsbetween electromagnetic ﬁelds and particles or materials? Einstein realized that the laws of mechanics needed revision to make them Lorentz covariant: in fact, under his theory ofspecial relativity all physical laws should demonstrate Lorentz covariance. Interestingly,charge is then Lorentz invariant, whereas mass is not (recall that invariance refers to aquantity, whereas covariance refers to the form of a natural law). We shall not attemptto describe all the ramiﬁcations of special relativity, but instead refer the reader to anyof the excellent and readable texts on the subject, including those by Bohm [14], Einstein[62], and Born [18], and to the nice historical account by Miller [130]. However, we shallexamine the importance of Lorentz invariants in electromagnetic theory. Lorentz invariants. Although the electromagnetic ﬁelds are not Lorentz invariant (e.g., the numerical value of Emeasured by one observer diﬀers from that measured by another observer in uniform relative motion), several quantities do give identical valuesregardless of the velocity of motion. Most fundamental are the speed of light and thequantity of electric charge which, unlike mass, is the same in all frames of reference.Other important Lorentz invariants include E·B,H·D, and the quantities B·B−E·E/c 2, H·H−c2D·D, B·H−E·D, cB·D+E·H/c. (See Problem 2.3.) To see the importance of these quantities, consider the special case of ﬁelds in empty space. If E·B=0in one reference frame, then it is zero in all reference frames. Then if B·B−E·E/c2=0in any reference frame, the ratio of EtoBis always c2regardless of the reference frame in which the ﬁelds are measured. This is the characteristic of a plane wave in free space. IfE·B=0and c2B2>E2, then we can ﬁnd a reference frame using the conversion formulas (2.101)–(2.106) (see Problem 2.5) in which the electric ﬁeld is zero but themagnetic ﬁeld is nonzero. In this case we call the ﬁelds purely magnetic in any reference frame, even if both Eand Bare nonzero. Similarly, if E·B=0and c 2B2<E2then we can ﬁnd a reference frame in which the magnetic ﬁeld is zero but the electric ﬁeld isnonzero. We call ﬁelds of this type purely electric . The Lorentz force is not Lorentz invariant. Consider a point charge at rest in the laboratory frame. While we measure only an electric ﬁeld in the laboratory frame, aninertial observer measures both electric and magnetic ﬁelds. A test charge Qin the laboratory frame experiences the Lorentz force F=QE; in an inertial frame the same charge experiences F/prime=QE/prime+Qv×B/prime(see Problem 2.6). The conversion formulas show that Fand F/primeare not identical. We see that both Eand Bare integral components of the electromagnetic ﬁeld: the separation of the ﬁeld into electric and magnetic components depends on the motionof the reference frame in which measurements are made. This has obvious implicationswhen considering static electric and magnetic ﬁelds. Derivation of Maxwell’s equations from Coulomb’s law. Consider a point charge at rest in the laboratory frame. If the magnetic component of force on this charge arisesnaturally through motion of an inertial reference frame, and if this force can be expressedin terms of Coulomb’s law in the laboratory frame, then perhaps the magnetic ﬁeld can bederived directly from Coulomb’s and the Lorentz transformation. Perhaps it is possibleto derive all of Maxwell’s theory with Coulomb’s law and Lorentz invariance as the only postulates. Several authors, notably Purcell [152] and Elliott [65], have used this approach. How- ever, Jackson [91] has pointed out that many additional assumptions are required todeduce Maxwell’s equations beginning with Coulomb’s law. Feynman [73] is critical ofthe approach, pointing out that we must introduce a vector potential which adds to thescalar potential from electrostatics in order to produce an entity that transforms accord-ing to the laws of special relativity. In addition, the assumption of Lorentz invarianceseems to involve circular reasoning since the Lorentz transformation was originally in-troduced to make Maxwell’s equations covariant. But Lucas and Hodgson [117] pointout that the Lorentz transformation can be deduced from other fundamental principles(such as causality and the isotropy of space), and that the postulate of a vector potentialis reasonable. S chwartz [170] gives a detailed derivation of Maxwell’s equations from Coulomb’s law, outlining the necessary assumptions. Transformation of constitutive relations. Minkowski’s interest in the covariance of Maxwell’s equations was aimed not merely at the relationship between ﬁelds in diﬀerentmoving frames of reference, but at an understanding of the electrodynamics of movingmedia. He wished to ascertain the eﬀect of a moving material body on the electromagneticﬁelds in some region of space. By proposing the covariance of Maxwell’s equations inmaterials as well as in free space, he extended Maxwell’s theory to moving materialbodies. We have seen in (2.101)–(2.104) that (E,cB)and(cD,H)convert identically under a Lorentz transformation. Since the most general form of the constitutive relations relatecDand Hto the ﬁeld pair (E,cB)(see §2.2.2) as /bracketleftbiggcD H/bracketrightbigg =/bracketleftbig¯C/bracketrightbig/bracketleftbiggE cB/bracketrightbigg , this form of the constitutive relations must be Lorentz covariant. That is, in the reference frame of a moving material we have /bracketleftbiggcD /prime H/prime/bracketrightbigg =/bracketleftbig¯C/prime/bracketrightbig/bracketleftbiggE/prime cB/prime/bracketrightbigg , and should be able to convert [¯C/prime]to[¯C]. We should be able to ﬁnd the constitutive matrix describing the relationships among the ﬁelds observed in the laboratory frame. It is somewhat laborious to obtain the constitutive matrix [¯C]for an arbitrary moving medium. Detailed expressions for isotropic, bianisotropic, gyrotropic, and uniaxial mediaare given by Kong [101]. The rather complicated expressions can be written in a morecompact form if we consider the expressions for Band Din terms of the pair (E,H). For a linear isotropic material such that D /prime=/epsilon1/primeE/primeand B/prime=µ/primeH/primein the moving frame, the relationships in the laboratory frame are [101] B=µ/prime¯A·H−Ω×E, (2.107) D=/epsilon1/prime¯A·E+Ω×H, (2.108) where ¯A=1−β2 1−n2β2/bracketleftbigg ¯I−n2−1 1−β2ββ/bracketrightbigg , (2.109) Ω=n2−1 1−n2β2β c, (2.110) and where n=c(µ/prime/epsilon1/prime)1/2is the optical index of the medium. A moving material that is isotropic in its own moving reference frame is bianisotropic in the laboratory frame.If, for instance, we tried to measure the relationship between the ﬁelds of a movingisotropic ﬂuid, but used instruments that were stationary in our laboratory (e.g., attachedto our measurement bench) we would ﬁnd that Ddepends not only on Ebut also on H, and that Daligns with neither Enor H. That a moving material isotropic in its own frame of reference is bianisotropic in the laboratory frame was known long ago.Roentgen showed experimentally in 1888 that a dielectric moving through an electricﬁeld becomes magnetically polarized, while H.A. Wilson showed in 1905 that a dielectricmoving through a magnetic ﬁeld becomes electrically polarized [139]. Ifv 2/c2/lessmuch1, we can consider the form of the constitutive equations for a ﬁrst-order Lorentz transformation. Ignoring terms to order v2/c2in (2.109) and (2.110), we obtain ¯A=¯IandΩ=v(n2−1)/c2. Then, by (2.107) and (2.108), B=µ/primeH−(n2−1)v×E c2, (2.111) D=/epsilon1/primeE+(n2−1)v×H c2. (2.112) We can also derive these from the ﬁrst-order ﬁeld conversion equations (2.61)–(2.64). From (2.61) and (2.62) we have D/prime=D+v×H/c2=/epsilon1/primeE/prime=/epsilon1/prime(E+v×B). Eliminating Bvia (2.64), we have D+v×H/c2=/epsilon1/primeE+/epsilon1/primev×(v×E/c2)+/epsilon1/primev×B/prime=/epsilon1/primeE+/epsilon1/primev×B/prime where we have neglected terms of order v2/c2. Since B/prime=µ/primeH/prime=µ/prime(H−v×D),w e have D+v×H/c2=/epsilon1/primeE+/epsilon1/primeµ/primev×H−/epsilon1/primeµ/primev×v×D. Using n2=c2µ/prime/epsilon1/primeand neglecting the last term since it is of order v2/c2, we obtain D=/epsilon1/primeE+(n2−1)v×H c2, which is identical to the expression (2.112) obtained by approximating the exact result to ﬁrst order. Similar steps produce (2.111). In a Galilean frame where v/c/lessmuch1, the expressions reduce to D=/epsilon1/primeEand B=µ/primeH,and the isotropy of the ﬁelds is preserved. For a conducting medium having J/prime=σ/primeE/prime in a moving reference frame, Cullwick [48] shows that in the laboratory frame J=σ/primeγ[¯I−ββ]·E+σ/primeγcβ×B. Forv/lessmuchcwe can set γ≈1and see that J=σ/prime(E+v×B) to ﬁrst order. Constitutive relations in deforming or rotating media. The transformations discussed in the previous paragraphs hold for media in uniform relative motion. Whena material body undergoes deformation or rotation, the concepts of special relativity arenot directly applicable. However, authors such as Pauli [144] and Sommerfeld [185] havemaintained that Minkowski’s theory is approximately valid for deforming or rotating media if vis taken to be the instantaneous velocity at each point within the body. The reasoning is that at any instant in time each point within the body has a velocityvthat may be associated with some inertial reference frame (generally diﬀerent for each point). Thus the constitutive relations for the material at that point, within somesmall time interval taken about the observation time, may be assumed to be those ofa stationary material, and the relations measured by an observer within the laboratoryframe may be computed using the inertial frame for that point. This instantaneous rest- frame theory is most accurate at small accelerations dv/dt. Van Bladel [201] outlines its shortcomings. See also Anderson [3] and Mo [132] for detailed discussions of theelectromagnetic properties of material media in accelerating frames of reference. 2.4 The Maxwell–Boﬃ equations In any version of Maxwell’s theory, the mediating ﬁeld is the electromagnetic ﬁeld described by four ﬁeld vectors. In Minkowski’s form of Maxwell’s equations we use E, D,B, and H. As an alternative consider the electromagnetic ﬁeld as represented by the vector ﬁelds E,B,P, and M, and described by ∇× E=−∂B ∂t, (2.113) ∇×(B/µ0−M)=J+∂ ∂t(/epsilon10E+P), (2.114) ∇·(/epsilon10E+P)=ρ, (2.115) ∇·B=0. (2.116) These Maxwell–Boﬃ equations are named after L. Boﬃ, who formalized them for moving media [13]. The quantity Pis thepolarization vector , and Mis themagnetization vector . The use of PandMin place of DandHis sometimes called an application of the principle of Ampere and Lorentz [199]. Let us examine the ramiﬁcation of using (2.113)–(2.116) as the basis for a postulate of electromagnetics. These equations are similar to the Maxwell–Minkowski equationsused earlier; must we rebuild all the underpinning of a new postulate, or can we useour original arguments based on the Minkowski form? For instance, how do we invokeuniqueness if we no longer have the ﬁeld H? What represents the ﬂux of energy, formerly found using E×H? And, importantly, are (2.113)–(2.114) form invariant under a Lorentz transformation? It turns out that the set of vector ﬁelds (E,B,P,M)is merely a linear mapping of the set (E,D,B,H). As pointed out by Tai [193], any linear mapping of the four ﬁeld vectors from Minkowski’s form onto any other set of four ﬁeld vectors will preserve thecovariance of Maxwell’s equations. Boﬃ chose to keep Eand Bintact and to introduce only two new ﬁelds; he could have kept Hand Dinstead, or used a mapping that introduced four completely new ﬁelds (as did Chu). Many authors retain Eand H. This is somewhat more cumbersome since these vectors do not convert as a pair undera Lorentz transformation. A discussion of the idea of ﬁeld vector “pairing” appears in§2.6. The usefulness of the Boﬃ form lies in the speciﬁc mapping chosen. Comparison of (2.113)–(2.116) to (2.1)–(2.4) quickly reveals that P=D−/epsilon1 0E, (2.117) M=B/µ0−H. (2.118) We see that Pis the diﬀerence between Din a material and Din free space, while Mis the diﬀerence between Hin free space and Hin a material. In free space, P=M=0. Equivalent polarization and magnetization sources. The Boﬃ formulation pro- vides a new way to regard EandB. Maxwell grouped (E,H)as a pair of “force vectors” to be associated with line integrals (or curl operations in the point forms of his equations),and(D,B)as a pair of “ﬂux vectors” associated with surface integrals (or divergence operations). That is, Eis interpreted as belonging to the computation of “emf” as a line integral, while Bis interpreted as a density of magnetic “ﬂux” passing through a surface. Similarly, Hyields the “mmf” about some closed path and Dthe electric ﬂux through a surface. The introduction of Pand Mallows us to also regard Eas a ﬂux vector and Bas a force vector — in essence, allowing the two ﬁelds Eand Bto take on the duties that required four ﬁelds in Minkowski’s form. To see this, we rewrite the Maxwell–Boﬃequations as ∇× E=−∂B ∂t, ∇×B µ0=/parenleftbigg J+∇× M+∂P ∂t/parenrightbigg +∂/epsilon10E ∂t, ∇·(/epsilon10E)=(ρ−∇· P), ∇·B=0, and compare them to the Maxwell–Minkowski equations for sources in free space: ∇× E=−∂B ∂t, ∇×B µ0=J+∂/epsilon10E ∂t, ∇·(/epsilon10E)=ρ, ∇·B=0. The forms are preserved if we identify ∂P/∂tand∇× Mas new types of current density, and∇·Pas a new type of charge density. We deﬁne JP=∂P ∂t(2.119) as anequivalent polarization current density, and JM=∇× M as anequivalent magnetization current density (sometimes called the equivalent Amperian currents of magnetized matter [199]). We deﬁne ρP=− ∇· P as anequivalent polarization charge density (sometimes called the Poisson–Kelvin equiv- alent charge distribution [199]). Then the Maxwell–Boﬃ equations become simply ∇× E=−∂B ∂t, (2.120) ∇×B µ0=(J+JM+JP)+∂/epsilon10E ∂t, (2.121) ∇·(/epsilon10E)=(ρ+ρP), (2.122) ∇·B=0. (2.123) Here is the new view. A material can be viewed as composed of charged particles of matter immersed in free space. When these charges are properly considered as “equiv-alent” polarization and magnetization charges, all ﬁeld eﬀects (describable through ﬂuxand force vectors) can be handled by the two ﬁelds Eand B. Whereas in Minkowski’s form Ddiverges from ρ, in Boﬃ’s form Ediverges from a totalcharge density consisting ofρandρ P. Whereas in the Minkowski form Hcurls around J, in the Boﬃ form Bcurls around the total current density consisting of J,JM, and JP. This view was pioneered by Lorentz, who by 1892considered matter as consisting of bulk molecules in a vacuum that would respond to an applied electromagnetic ﬁeld [130].The resulting motion of the charged particles of matter then became another sourceterm for the “fundamental” ﬁelds Eand B. Using this reasoning he was able to reduce the fundamental Maxwell equations to two equations in two unknowns, demonstrating asimplicity appealing to many (including Einstein). Of course, to apply this concept wemust be able to describe how the charged particles respond to an applied ﬁeld. Simplemicroscopic models of the constituents of matter are generally used: some combinationof electric and magnetic dipoles, or of loops of electric and magnetic current. The Boﬃ equations are mathematically appealing since they now specify both the curl and divergence of the two ﬁeld quantities Eand B. By the Helmholtz theorem we know that a ﬁeld vector is uniquely speciﬁed when both its curl and divergence are given. Butthis assumes that the equivalent sources produced by Pand Mare true source ﬁelds in the same sense as J. We have precluded this by insisting in Chapter 1 that the source ﬁeld must be independent of the mediating ﬁeld it sources. If we view Pand Mas merely a mapping from the original vector ﬁelds of Minkowski’s form, we still have four vector ﬁelds with which to contend. And with these must also be a mapping of theconstitutive relationships, which now link the ﬁelds E,B,P, and M. Rather than argue the actual physical existence of the equivalent sources, we note that a real beneﬁt ofthe new view is that under certain circumstances the equivalent source quantities can bedetermined through physical reasoning, hence we can create physical models of Pand M and deduce their links to Eand B. We may then ﬁnd it easier to understand and deduce the constitutive relationships. However we do not in general consider Eand Bto be in any way more “fundamental” than Dand H. Covariance of the Boﬃ form. Because of the linear relationships (2.117) and (2.118), covariance of the Maxwell–Minkowski equations carries over to the Maxwell–Boﬃ equa-tions. However, the conversion between ﬁelds in diﬀerent moving reference frames willnow involve Pand M. Since Faraday’s law is unchanged in the Boﬃ form, we still have E /prime /bardbl=E/bardbl, (2.124) B/prime /bardbl=B/bardbl, (2.125) E/prime ⊥=γ(E⊥+β×cB⊥), (2.126) cB/prime ⊥=γ(cB⊥−β×E⊥). (2.127) To see how Pand Mconvert, we note that in the laboratory frame D=/epsilon10E+Pand H=B/µ0−M, while in the moving frame D/prime=/epsilon10E/prime+P/primeand H/prime=B/prime/µ0−M/prime.T h u s P/prime /bardbl=D/prime /bardbl−/epsilon10E/prime /bardbl=D/bardbl−/epsilon10E/bardbl=P/bardbl and M/prime /bardbl=B/prime /bardbl/µ0−H/prime /bardbl=B/bardbl/µ0−H/bardbl=M/bardbl. For the perpendicular components D/prime ⊥=γ(D⊥+β×H⊥/c)=/epsilon10E/prime ⊥+P/prime ⊥=/epsilon10[γ(E⊥+β×cB⊥)]+P/prime ⊥; substitution of H⊥=B⊥/µ0−M⊥then gives P/prime ⊥=γ(D⊥−/epsilon10E⊥)−γ/epsilon10β×cB⊥+γβ×B⊥/(cµ0)−γβ×M⊥/c or cP/prime ⊥=γ(cP⊥−β×M⊥). Similarly, M/prime ⊥=γ(M⊥+β×cP⊥). Hence E/prime /bardbl=E/bardbl,B/prime /bardbl=B/bardbl,P/prime /bardbl=P/bardbl,M/prime /bardbl=M/bardbl,J/prime ⊥=J⊥, (2.128) and E/prime ⊥=γ(E⊥+β×cB⊥), (2.129) cB/prime ⊥=γ(cB⊥−β×E⊥), (2.130) cP/prime ⊥=γ(cP⊥−β×M⊥), (2.131) M/prime ⊥=γ(M⊥+β×cP⊥), (2.132) J/prime /bardbl=γ(J/bardbl−ρv). (2.133) In the case of the ﬁrst-order Lorentz transformation we can set γ≈1to obtain E/prime=E+v×B, (2.134) B/prime=B−v×E c2, (2.135) P/prime=P−v×M c2, (2.136) M/prime=M+v×P, (2.137) J/prime=J−ρv. (2.138) To convert from the moving frame to the laboratory frame we simply swap primed with unprimed ﬁelds and let v→− v. As a simple example, consider a linear isotropic medium having D/prime=/epsilon10/epsilon1/prime rE/prime, B/prime=µ0µ/prime rH/prime, in a moving reference frame. From (117) we have P/prime=/epsilon10/epsilon1/prime rE/prime−/epsilon10E/prime=/epsilon10χ/prime eE/prime where χ/prime e=/epsilon1/prime r−1is the electric susceptibility of the moving material. Similarly (2.118) yields M/prime=B/prime µ0−B/prime µ0µ/primer=B/primeχ/prime m µ0µ/primer where χ/prime m=µ/prime r−1is the magnetic susceptibility of the moving material. How are Pand Mrelated to Eand Bin the laboratory frame? For simplicity, we consider the ﬁrst-order expressions. From (2.136) we have P=P/prime+v×M/prime c2=/epsilon10χ/prime eE/prime+v×B/primeχ/prime m µ0µ/primerc2. Substituting for E/primeand B/primefrom (2.134) and (2.135), and using µ0c2=1//epsilon10,w eh a v e P=/epsilon10χ/prime e(E+v×B)+/epsilon10χ/prime m µ/primerv×/parenleftbigg B−v×E c2/parenrightbigg . Neglecting the last term since it varies as v2/c2,w eg e t P=/epsilon10χ/prime eE+/epsilon10/parenleftbigg χ/prime e+χ/prime m µ/primer/parenrightbigg v×B. (2.139) Similarly, M=χ/prime m µ0µ/primerB−/epsilon10/parenleftbigg χ/prime e+χ/prime m µ/primer/parenrightbigg v×E. (2.140) 2.5 Large-scale form of Maxwell’s equations We can write Maxwell’s equations in a form that incorporates the spatial variation of the ﬁeld in a certain region of space. To do this, we integrate the point form of Maxwell’s Figure 2.2: Open surface having velocity vrelative to laboratory (unprimed) coordinate system. Surface is non-deforming. equations over a region of space, then perform some succession of manipulations until we arrive at a form that provides us some beneﬁt in our work with electromagneticﬁelds. The results are particularly useful for understanding the properties of electric andmagnetic circuits, and for predicting the behavior of electrical machinery. We shall consider two important situations: a mathematical surface that moves with constant velocity vand with constant shape, and a surface that moves and deforms arbitrarily. 2.5.1 Surface moving with constant velocity Consider an open surface Smoving with constant velocity vrelative to the laboratory frame(Figure2.2).Assum eeverypointonthesurfac eisanordinar ypoint.Atany instant twe can express the relationship between the ﬁelds at points on Sin either frame. In the laboratory frame we have ∇× E=−∂B ∂t, ∇× H=∂D ∂t+J, while in the moving frame ∇/prime×E/prime=−∂B/prime ∂t/prime, ∇/prime×H/prime=∂D/prime ∂t/prime+J/prime. If we integrate over Sand use Stokes’s theorem, we get for the laboratory frame /contintegraldisplay /Gamma1E·dl=−/integraldisplay S∂B ∂t·dS, (2.141) /contintegraldisplay /Gamma1H·dl=/integraldisplay S∂D ∂t·dS+/integraldisplay SJ·dS, (2.142) and for the moving frame /contintegraldisplay /Gamma1/primeE/prime·dl/prime=−/integraldisplay S/prime∂B/prime ∂t/prime·dS/prime, (2.143) /contintegraldisplay /Gamma1/primeH/prime·dl/prime=/integraldisplay S/prime∂D/prime ∂t/prime·dS/prime+/integraldisplay S/primeJ/prime·dS/prime. (2.144) Here boundary contour /Gamma1has sense determined by the right-hand rule. We use the notation /Gamma1/prime,S/prime, etc., to indicate that all integrations for the moving frame are computed using space and time variables in that frame. Equation (2.141) is the integral form of Faraday’s law , while (2.142) is the integral form of Ampere’s law . Faraday’s law states that the net circulation of Eabout a contour /Gamma1(sometimes called theelectromotive force oremf) is determined by the ﬂux of the time-rate of change of the ﬂux vector Bpassing through the surface bounded by /Gamma1. Ampere’s law states that the circulation of H(sometimes called the magnetomotive force ormmf) is determined by the ﬂux of the current Jplus the ﬂux of the time-rate of change of the ﬂux vector D.I ti s the term containing ∂D/∂tthat Maxwell recognized as necessary to make his equations consistent; since it has units of current, it is often referred to as the displacement current term. Equations (2.141)–(2.142) are the large-scale or integral forms of Maxwell’s equations. They are the integral-form equivalents of the point forms, and are form invariant underLorentz transformation. If we express the ﬁelds in terms of the moving reference frame,we can write /contintegraldisplay /Gamma1/primeE/prime·dl/prime=−d dt/integraldisplay S/primeB/prime·dS/prime, (2.145) /contintegraldisplay /Gamma1/primeH/prime·dl/prime=d dt/integraldisplay S/primeD/prime·dS/prime+/integraldisplay S/primeJ/prime·dS/prime. (2.146) These hold for a stationary surface, since the surface would be stationary to an observer who moves with it. We are therefore justiﬁed in removing the partial derivative from theintegral. Although the surfaces and contours considered here are purely mathematical,they often coincide with actual physical boundaries. The surface may surround a movingmaterial medium, for instance, or the contour may conform to a wire moving in anelectrical machine. We can also convert the auxiliary equations to large-scale form. Consider a volume region Vsurrounded by a surface Sthat moves with velocity vrelative to the laboratory frame(Figure2.3).Integratin gthepointformofGauss’ slawover V wehave /integraldisplay V∇·DdV=/integraldisplay VρdV. Using the divergence theorem and recognizing that the integral of charge density is total charge, we obtain /contintegraldisplay SD·dS=/integraldisplay VρdV=Q(t) (2.147) where Q(t)is the total charge contained within Vat time t. This large-scale form of Gauss’s law states that the total ﬂux of Dpassing through a closed surface is identical to the electric charge Qcontained within. Similarly, /contintegraldisplay SB·dS=0 (2.148) Figure 2.3: Non-deforming volume region having velocity vrelative to laboratory (un- primed) coordinate system. is the large-scale magnetic ﬁeld Gauss’s law. It states that the total ﬂux of Bpassing through a closed surface is zero, since there are no magnetic charges contained within(i.e., magnetic charge does not exist). Since charge is an invariant quantity, the large-scale forms of the auxiliary equations take the same form in a moving reference frame: /contintegraldisplay S/primeD/prime·dS/prime=/integraldisplay V/primeρ/primedV/prime=Q(t) (2.149) and/contintegraldisplay S/primeB/prime·dS/prime=0. (2.150) The large-scale forms of the auxiliary equations may be derived from the large-scale forms of Faraday’s and Ampere’s laws. To obtain Gauss’s law, we let the open surfacein Ampere’s law become a closed surface. Then/contintegraltext H·dlvanishes, and application of the large-scale form of the continuity equation (1.10) produces (2.147). The magnetic Gauss’s law (2.148) is found from Faraday’s law (2.141) by a similar transition from anopen surface to a closed surface. The values obtained from the expressions (2.141)–(2.142) will notmatch those ob- tained from (2.143)–(2.144), and we can use the Lorentz transformation ﬁeld conversionsto study how they diﬀer. That is, we can write either side of the laboratory equations interms of the moving reference frame ﬁelds, or vice versa. For most engineering applica-tions where v/c/lessmuch1this is not done via the Lorentz transformation ﬁeld relations, but rather via the Galilean approximations to these relations (see Tai [194] for details on us-ing the Lorentz transformation ﬁeld relations). We consider the most common situationin the next section. Kinematic form of the large-scale Maxwell equations. Confusion can result from the fact that the large-scale forms of Maxwell’s equations can be written in a number of Figure 2.4: Non-deforming closed contour moving with velocity vthrough a magnetic ﬁeld Bgiven in the laboratory (unprimed) coordinate system. ways. A popular formulation of Faraday’s law, the emf formulation , revolves around the concept of electromotive force. Unfortunately, various authors oﬀer diﬀerent deﬁnitionsof emf in a moving circuit. Consider a non-deforming contour in space, moving with constant velocity vrelative tothelaborator yframe(Figure2.4).Intermsofthelaborator yﬁeldswehavethelarge- scale form of Faraday’s law (2.141). The ﬂux term on the right-hand side of this equationcan be written diﬀerently by employing the Helmholtz transport theorem (A.63). If anon-deforming surface Smoves with uniform velocity vrelative to the laboratory frame, and a vector ﬁeld A(r,t)is expressed in the stationary frame, then the time derivative of the ﬂux of Athrough Sis d dt/integraldisplay SA·dS=/integraldisplay S/bracketleftbigg∂A ∂t+v(∇·A)−∇× (v×A)/bracketrightbigg ·dS. (2.151) Using this with (2.141) we have /contintegraldisplay /Gamma1E·dl=−d dt/integraldisplay SB·dS+/integraldisplay Sv(∇·B)·dS−/integraldisplay S∇×(v×B)·dS. Remembering that ∇·B=0and using Stokes’s theorem on the last term, we obtain /contintegraldisplay /Gamma1(E+v×B)·dl=−d dt/integraldisplay SB·dS=−d/Psi1(t) dt(2.152) where the magnetic ﬂux /integraldisplay SB·dS=/Psi1(t) represents the ﬂux of Bthrough S. Following Sommerfeld [185], we may set E∗=E+v×B to obtain the kinematic form of Faraday’s law /contintegraldisplay /Gamma1E∗·dl=−d dt/integraldisplay SB·dS=−d/Psi1(t) dt. (2.153) (The asterisk should not be confused with the notation for complex conjugate.) Much confusion arises from the similarity between (2.153) and (2.145). In fact, these expressions are diﬀerent and give diﬀerent results. This is because B/primein (2.145) is measured in the frame of the moving circuit , while Bin (2.153) is measured in the frame of the laboratory. Further confusion arises from various deﬁnitions of emf. Many authors(e.g., Hermann Weyl [213]) deﬁne emf to be the circulation of E ∗. In that case the emf is equal to the negative time rate of change of the ﬂux of the laboratory frame magnetic ﬁeld Bthrough S. Since the Lorentz force experienced by a charge qmoving with the contour is given by qE∗=q(E+v×B), this emf is the circulation of Lorentz force per unit charge along the contour. If the contour is aligned with a conducting circuit,then in some cases this emf can be given physical interpretation as the work requiredto move a charge around the entire circuit through the conductor against the Lorentzforce. Unfortunately the usefulness of this deﬁnition of emf is lost if the time or spacerate of change of the ﬁelds is so large that no true loop current can be established(hence Kirchoﬀ’s law cannot be employed). Such a problem must be treated as an electromagnetic “scattering” problem with consideration given to retardation eﬀects.Detailed discussions of the physical interpretation of E ∗in the deﬁnition of emf are given by Scanlon [165] and Cullwick [48]. Other authors choose to deﬁne emf as the circulation of the electric ﬁeld in the frame of the moving contour . In this case the circulation of E/primein (2.145) is the emf, and is related to the ﬂux of the magnetic ﬁeld in the frame of the moving circuit . As pointed out above, the result diﬀers from that based on the Lorentz force. If we wish, we canalso write this emf in terms of the ﬁelds expressed in the laboratory frame. To do this wemust convert ∂B /prime/∂t/primeto the laboratory ﬁelds using the rules for a Lorentz transformation. The result, given by Tai [194], is quite complicated and involves both the magnetic and electric laboratory-frame ﬁelds. The moving-frame emf as computed from the Lorentz transformation is rarely used as a working deﬁnition of emf, mostly because circuits moving at relativistic velocities areseldom used by engineers. Unfortunately, more confusion arises for the case v/lessmuchc, since for a Galilean frame the Lorentz-force and moving-frame emfs become identical. Thisis apparent if we use (2.52) to replace B /primewith the laboratory frame ﬁeld B, and (2.49) to replace E/primewith the combination of laboratory frame ﬁelds E+v×B. Then (2.145) becomes/contintegraldisplay /Gamma1E/prime·dl=/contintegraldisplay /Gamma1(E+v×B)·dl=−d dt/integraldisplay SB·dS, which is identical to (2.153). For circuits moving with low velocity then, the circulation ofE/primecan be interpreted as work per unit charge. As an added bit of confusion, the term/contintegraldisplay /Gamma1(v×B)·dl=/integraldisplay S∇×(v×B)·dS is sometimes called motional emf , since it is the component of the circulation of E∗that is directly attributable to the motion of the circuit. Although less commonly done, we can also rewrite Ampere’s law (2.142) using (2.151). This gives /contintegraldisplay /Gamma1H·dl=/integraldisplay SJ·dS+d dt/integraldisplay SD·dS−/integraldisplay S(v∇·D)·dS+/integraldisplay S∇×(v×D)·dS. Using ∇·D=ρand using Stokes’s theorem on the last term, we obtain /contintegraldisplay /Gamma1(H−v×D)·dl=d dt/integraldisplay SD·dS+/integraldisplay S(J−ρv)·dS. Finally, letting H∗=H−v×Dand J∗=J−ρvwe can write the kinematic form of Ampere’s law : /contintegraldisplay /Gamma1H∗·dl=d dt/integraldisplay SD·dS+/integraldisplay SJ∗·dS. (2.154) In a Galilean frame where we use (2.49)–(2.54), we see that (2.154) is identical to /contintegraldisplay /Gamma1H/prime·dl=d dt/integraldisplay SD/prime·dS+/integraldisplay SJ/prime·dS (2.155) where the primed ﬁelds are measured in the frame of the moving contour. This equiv- alence does nothold when the Lorentz transformation is used to represent the primed ﬁelds. Alternative form of the large-scale Maxwell equations. We can write Maxwell’s equations in an alternative large-scale form involving only surface and volume integrals.This will be useful later for establishing the ﬁeld jump conditions across a material orsource discontinuity. Again we begin with Maxwell’s equations in point form, but insteadof integrating them over an open surface we integrate over a volume region Vmoving withvelocity v (Figure2.3).Inthelaborator yframethisgives /integraldisplay V(∇× E)dV=−/integraldisplay V∂B ∂tdV, /integraldisplay V(∇× H)dV=/integraldisplay V/parenleftbigg∂D ∂t+J/parenrightbigg dV. An application of curl theorem (B.24) then gives /contintegraldisplay S(ˆn×E)dS=−/integraldisplay V∂B ∂tdV, (2.156) /contintegraldisplay S(ˆn×H)dS=/integraldisplay V/parenleftbigg∂D ∂t+J/parenrightbigg dV. (2.157) Similar results are obtained for the ﬁelds in the moving frame: /contintegraldisplay S/prime(ˆn/prime×E/prime)dS/prime=−/integraldisplay V/prime∂B/prime ∂t/primedV/prime, /contintegraldisplay S/prime(ˆn/prime×H/prime)dS/prime=/integraldisplay V/prime/parenleftbigg∂D/prime ∂t/prime+J/prime/parenrightbigg dV/prime. These large-scale forms are an alternative to (2.141)–(2.144). They are also form- invariant under a Lorentz transformation. An alternative to the kinematic formulation of (2.153) and (2.154) can be achieved by applying a kinematic identity for a moving volume region. If Vis surrounded by a surface Sthat moves with velocity vrelative to the laboratory frame, and if a vector ﬁeld Ais measured in the laboratory frame, then the vector form of the general transport theorem (A.68) states that d dt/integraldisplay VAdV=/integraldisplay V∂A ∂tdV+/contintegraldisplay SA(v·ˆn)dS. (2.158) Applying this to (2.156) and (2.157) we have /contintegraldisplay S[ˆn×E−(v·ˆn)B]dS=−d dt/integraldisplay VBdV, (2.159) /contintegraldisplay S[ˆn×H+(v·ˆn)D]dS=/integraldisplay VJdV+d dt/integraldisplay VDdV. (2.160) We can also apply (2.158) to the large-scale form of the continuity equation (2.10) and obtain the expression for a volume region moving with velocity v: /contintegraldisplay S(J−ρv)·dS=−d dt/integraldisplay VρdV. 2.5.2 Moving, deforming surfaces Because (2.151) holds for arbitrarily moving surfaces, the kinematic versions (2.153) and (2.154) hold when vis interpreted as an instantaneous velocity. However, if the surface and contour lie within a material body that moves relative to the laboratoryframe, the constitutive equations relating E,D,B,H, and Jin the laboratory frame diﬀer from those relating the ﬁelds in the stationary frame of the body (if the body isnot accelerating), and thus the concepts of §2.3.2 must be employed. This is important when boundary conditions at a moving surface are needed. Particular care must be takenwhen the body accelerates, since the constitutive relations are then only approximate. The representation (2.145)–(2.146) is also generally valid, provided we deﬁne the primed ﬁelds as those converted from laboratory ﬁelds using the Lorentz transforma-tion with instantaneous velocity v. Here we should use a diﬀerent inertial frame for each point in the integration, and align the frame with the velocity vector vat the instant t. We certainly may do this since we can choose to integrate any function we wish. However, this representation may not ﬁnd wide application. We thus choose the following expressions, valid for arbitrarily moving surfaces con- taining only regular points, as our general forms of the large-scale Maxwell equations: /contintegraldisplay /Gamma1(t)E∗·dl=−d dt/integraldisplay S(t)B·dS=−d/Psi1(t) dt, /contintegraldisplay /Gamma1(t)H∗·dl=d dt/integraldisplay S(t)D·dS+/integraldisplay S(t)J∗·dS, where E∗=E+v×B, H∗=H−v×D, J∗=J−ρv, and where all ﬁelds are taken to be measured in the laboratory frame with vthe in- stantaneous velocity of points on the surface and contour relative to that frame. Theconstitutive parameters must be considered carefully if the contours and surfaces lie ina moving material medium. Kinematic identity (2.158) is also valid for arbitrarily moving surfaces. Thus we have the following, valid for arbitrarily moving surfaces and volumes containing only regular points: /contintegraldisplay S(t)[ˆn×E−(v·ˆn)B]dS=−d dt/integraldisplay V(t)BdV, /contintegraldisplay S(t)[ˆn×H+(v·ˆn)D]dS=/integraldisplay V(t)JdV+d dt/integraldisplay V(t)DdV. We also ﬁnd that the two Gauss’s law expressions, /contintegraldisplay S(t)D·dS=/integraldisplay V(t)ρdV, /contintegraldisplay S(t)B·dS=0, remain valid. 2.5.3 Large-scale form of the Boﬃ equations The Maxwell–Boﬃ equations can be written in large-scale form using the same ap- proach as with the Maxwell–Minkowski equations. Integrating (2.120) and (2.121) overan open surface Sand applying Stokes’s theorem, we have /contintegraldisplay /Gamma1E·dl=−/integraldisplay S∂B ∂t·dS, (2.161) /contintegraldisplay /Gamma1B·dl=µ0/integraldisplay S/parenleftbigg J+JM+JP+∂/epsilon10E ∂t/parenrightbigg ·dS, (2.162) for ﬁelds in the laboratory frame, and /contintegraldisplay /Gamma1/primeE/prime·dl/prime=−/integraldisplay S/prime∂B/prime ∂t/prime·dS/prime, /contintegraldisplay /Gamma1/primeB/prime·dl/prime=µ0/integraldisplay S/prime/parenleftbigg J/prime+J/prime M+J/prime P+∂/epsilon10E/prime ∂t/prime/parenrightbigg ·dS/prime, for ﬁelds in a moving frame. We see that Faraday’s law is unmodiﬁed by the introduction of polarization and magnetization, hence our prior discussion of emf for moving contoursremains valid. However, Ampere’s law must be interpreted somewhat diﬀerently. Theﬂux vector Balso acts as a force vector, and its circulation is proportional to the out- ﬂux of total current, consisting of Jplus the equivalent magnetization and polarization currents plus the displacement current in free space , through the surface bounded by the circulation contour. The large-scale forms of the auxiliary equations can be found by integrating (2.122) and (2.123) over a volume region and applying the divergence theorem. This gives /contintegraldisplay SE·dS=1 /epsilon10/integraldisplay V(ρ+ρP)dV, /contintegraldisplay SB·dS=0, for the laboratory frame ﬁelds, and /contintegraldisplay S/primeE/prime·dS/prime=1 /epsilon10/integraldisplay V/prime(ρ/prime+ρ/prime P)dV/prime, /contintegraldisplay S/primeB/prime·dS/prime=0, for the moving frame ﬁelds. Here we ﬁnd the force vector Ealso acting as a ﬂux vector, with the outﬂux of Eover a closed surface proportional to the sum of the electric and polarization charges enclosed by the surface. To provide the alternative representation, we integrate the point forms over Vand use the curl theorem to obtain /contintegraldisplay S(ˆn×E)dS=−/integraldisplay V∂B ∂tdV, (2.163) /contintegraldisplay S(ˆn×B)dS=µ0/integraldisplay V/parenleftbigg J+JM+JP+∂/epsilon10E ∂t/parenrightbigg dV, (2.164) for the laboratory frame ﬁelds, and /contintegraldisplay S/prime(ˆn/prime×E/prime)dS/prime=−/integraldisplay V/prime∂B/prime ∂t/primedV/prime, /contintegraldisplay S/prime(ˆn/prime×B/prime)dS/prime=µ0/integraldisplay V/prime/parenleftbigg J/prime+J/prime M+J/prime P+∂/epsilon10E/prime ∂t/prime/parenrightbigg dV/prime, for the moving frame ﬁelds. The large-scale forms of the Boﬃ equations can also be put into kinematic form using either (2.151) or (2.158). Using (2.151) on (2.161) and (2.162) we have /contintegraldisplay /Gamma1(t)E∗·dl=−d dt/integraldisplay S(t)B·dS, (2.165) /contintegraldisplay /Gamma1(t)B†·dl=/integraldisplay S(t)µ0J†·dS+1 c2d dt/integraldisplay S(t)E·dS, (2.166) where E∗=E+v×B, B†=B−1 c2v×E, J†=J+JM+JP−(ρ+ρP)v. Here B†is equivalent to the ﬁrst-order Lorentz transformation representation of the ﬁeld in the moving frame (2.64). (The dagger †should not be confused with the symbol for the hermitian operation.) Using (2.158) on (2.163) and (2.164) we have /contintegraldisplay S(t)[ˆn×E−(v·ˆn)B]dS=−d dt/integraldisplay V(t)BdV, (2.167) and /contintegraldisplay S(t)/bracketleftbigg ˆn×B+1 c2(v·ˆn)E/bracketrightbigg dS=µ0/integraldisplay V(t)(J+JM+JP)dV+1 c2d dt/integraldisplay V(t)EdV. (2.168) In each case the ﬁelds are measured in the laboratory frame, and vis measured with respect to the laboratory frame and may vary arbitrarily over the surface or contour. 2.6 The nature of the four ﬁeld quantities Since the very inception of Maxwell’s theory, its students have been distressed by the fact that while there are four electromagnetic ﬁelds ( E,D,B,H), there are only two funda- mental equations (the curl equations) to describe their interrelationship. The relegationof additional required information to constitutive equations that vary widely betweenclasses of materials seems to lessen the elegance of the theory. While some may ﬁndelegant the separation of equations into a set expressing the basic wave nature of electro-magnetism and a set describing how the ﬁelds interact with materials, the history of thediscipline is one of categorizing and pairing ﬁelds as “fundamental” and “supplemental”in hopes of reducing the model to two equations in two unknowns. Lorentz led the way in this area. With his electrical theory of matter, all material ef- fects could be interpreted in terms of atomic charge and current immersed in free space. We have seen how the Maxwell–Boﬃ equations seem to eliminate the need for Dand H, and indeed for simple media where there is a linear relation between the remaining “fun-damental” ﬁelds and the induced polarization and magnetization, it appears that onlyEand Bare required. However, for more complicated materials that display nonlinear and bianisotropic eﬀects we are only able to supplant Dand Hwith two other ﬁelds P and M, along with (possibly complicated) constitutive relations relating them to Eand B. Even those authors who do not wish to eliminate two of the ﬁelds tend to categorize the ﬁelds into pairs based on physical arguments, implying that one or the other pairis in some way “more fundamental.” Maxwell himself separated the ﬁelds into the pair(E,H)that appears within line integrals to give work and the pair (B,D)that appears within surface integrals to give ﬂux. In what other ways might we pair the four vectors? Most prevalent is the splitting of the ﬁelds into electric and magnetic pairs: (E,D)and (B,H). In Poynting’s theorem E·Ddescribes one component of stored energy (called “electric energy”) and B·Hdescribes another component (called “magnetic energy”). These pairs also occur in Maxwell’s stress tensor. In statics, the ﬁelds decouple intoelectric and magnetic sets. But biisotropic and bianisotropic materials demonstrate howseparation into electric and magnetic eﬀects can become problematic. In the study of electromagnetic waves, the ratio of EtoHappears to be an important quantity, called the “intrinsic impedance.” The pair (E,H)also determines the Poynting ﬂux of power, and is required to establish the uniqueness of the electromagnetic ﬁeld. In addition, constitutive relations for simple materials usually express (D,B)in terms of(E,H). Models for these materials are often conceived by viewing the ﬁelds (E,H) as interacting with the atomic structure in such a way as to produce secondary eﬀectsdescribable by (D,B). These considerations, along with Maxwell’s categorization into a pair of work vectors and a pair of ﬂux vectors, lead many authors to formulate elec-tromagnetics with Eand Has the “fundamental” quantities. But the pair (B,D)gives rise to electromagnetic momentum and is also perpendicular to the direction of wavepropagation in an anisotropic material; in these senses, we might argue that these ﬁeldsmust be equally “fundamental.” Perhaps the best motivation for grouping ﬁelds comes from relativistic considerations. We have found that (E,B)transform together under a Lorentz transformation, as do (D,H). In each of these pairs we have one polar vector ( EorD) and one axial vector ( B orH). A polar vector retains its meaning under a change in handedness of the coordinate system, while an axial vector does not. The Lorentz force involves one polar vector ( E) and one axial vector ( B) that we also call “electric” and “magnetic.” If we follow the lead of some authors and choose to deﬁne Eand Bthrough measurements of the Lorentz force, then we recognize that Bmust be axial since it is not measured directly, but as part of the cross product v×Bthat changes its meaning if we switch from a right-hand to a left-hand coordinate system. The other polar vector ( D) and axial vector ( H) arise through the “secondary” constitutive relations. Following this reasoning we might claimthat Eand Bare “fundamental.” Sommerfeld also associates Ewith Band Dwith H. The vectors Eand Bare called entities of intensity , describing “how strong,” while Dand Hare called entities of quantity , describing “how much.” This is in direct analogy with stress (intensity) and strain (quantity) in materials. We might also say that the entities of intensity describea “cause” while the entities of quantity describe an “eﬀect.” In this view E“induces” (causes) a polarization P, and the ﬁeld D=/epsilon1 0E+Pis the result. Similarly Bcreates M, and H=B/µ0−Mis the result. Interestingly, each of the terms describing energy and momentum in the electromagnetic ﬁeld ( D·E,B·H,E×H,D×B) involves the interaction of an entity of intensity with an entity of quantity. Although there is a natural tendency to group things together based on conceptual similarity, there appears to be little reason to believe that any of the four ﬁeld vectors aremore “fundamental” than the rest. Perhaps we are fortunate that we can apply Maxwell’stheory without worrying too much about such questions of underlying philosophy. 2.7 Maxwell’s equations with magnetic sources Researchers have yet to discover the “magnetic monopole”: a magnetic source from which magnetic ﬁeld would diverge. This has not stopped speculation on the form thatMaxwell’s equations might take if such a discovery were made. Arguments based onfundamental principles of physics (such as symmetry and conservation laws) indicatethat in the presence of magnetic sources Maxwell’s equations would assume the forms ∇× E=−J m−∂B ∂t, (2.169) ∇× H=J+∂D ∂t, (2.170) ∇·B=ρm, (2.171) ∇·D=ρ, (2.172) where Jmis a volume magnetic current density describing the ﬂow of magnetic charge in exactly the same manner as Jdescribes the ﬂow of electric charge. The density of this magnetic charge is given by ρmand should, by analogy with electric charge density, obey a conservation law ∇·Jm+∂ρm ∂t=0. This is the magnetic source continuity equation. It is interesting to inquire as to the units of Jmandρm. From (2.169) we see that if B has units of Wb/m2, then Jmhas units of (Wb/s)/m2. Similarly, (2.171) shows that ρm must have units of Wb/m3. Hence magnetic charge is measured in Wb, magnetic current in Wb/s. This gives a nice symmetry with electric sources where charge is measured in C and current in C/s.3The physical symmetry is equally appealing: magnetic ﬂux lines diverge from magnetic charge, and the total ﬂux passing through a surface is given by thetotal magnetic charge contained within the surface. This is best seen by considering thelarge-scale forms of Maxwell’s equations for stationary surfaces. We need only modify(2.145) to include the magnetic current term; this gives /contintegraldisplay /Gamma1E·dl=−/integraldisplay SJm·dS−d dt/integraldisplay SB·dS, (2.173) /contintegraldisplay /Gamma1H·dl=/integraldisplay SJ·dS+d dt/integraldisplay SD·dS. (2.174) If we modify (2.148) to include magnetic charge, we get the auxiliary equations /contintegraldisplay SD·dS=/integraldisplay VρdV, /contintegraldisplay SB·dS=/integraldisplay VρmdV. Any of the large-scale forms of Maxwell’s equations can be similarly modiﬁed to include magnetic current and charge. For arbitrarily moving surfaces we have /contintegraldisplay /Gamma1(t)E∗·dl=−d dt/integraldisplay S(t)B·dS−/integraldisplay S(t)J∗ m·dS, /contintegraldisplay /Gamma1(t)H∗·dl=d dt/integraldisplay S(t)D·dS+/integraldisplay S(t)J∗·dS, where E∗=E+v×B, H∗=H−v×D, J∗=J−ρv, J∗ m=Jm−ρmv, and all ﬁelds are taken to be measured in the laboratory frame with vthe instantaneous velocity of points on the surface and contour relative to the laboratory frame. We alsohave the alternative forms /contintegraldisplay S(ˆn×E)dS=/integraldisplay V/parenleftbigg −∂B ∂t−Jm/parenrightbigg dV, (2.175) /contintegraldisplay S(ˆn×H)dS=/integraldisplay V/parenleftbigg∂D ∂t+J/parenrightbigg dV, (2.176) and /contintegraldisplay S(t)[ˆn×E−(v·ˆn)B]dS=−/integraldisplay V(t)JmdV−d dt/integraldisplay V(t)BdV, (2.177) /contintegraldisplay S(t)[ˆn×H+(v·ˆn)D]dS=/integraldisplay V(t)JdV+d dt/integraldisplay V(t)DdV, (2.178) 3We note that if the modern unit of T is used to describe B, then ρmis described using the more cumbersome units of T/m, while Jmis given in terms of T/s. Thus, magnetic charge is measured in Tm2 and magnetic current in (Tm2)/s. and the two Gauss’s law expressions /contintegraldisplay S(t)D·ˆndS=/integraldisplay V(t)ρdV, /contintegraldisplay S(t)B·ˆndS=/integraldisplay V(t)ρmdV. Magnetic sources also allow us to develop equivalence theorems in which diﬃcult prob- lems involving boundaries are replaced by simpler problems involving magnetic sources.Although these sources may not physically exist, the mathematical solutions are com-pletely valid. 2.8 Boundary (jump) conditions If we restrict ourselves to regions of space without spatial (jump) discontinuities in either the sources or the constitutive relations, we can ﬁnd meaningful solutions to theMaxwell diﬀerential equations. We also know that for given sources, if the ﬁelds arespeciﬁed on a closed boundary and at an initial time the solutions are unique. Thestandard approach to treating regions that do contain spatial discontinuities is to isolatethe discontinuities on surfaces. That is, we introduce surfaces that serve to separate spaceinto regions in which the diﬀerential equations are solvable and the ﬁelds are well deﬁned.To make the solutions in adjoining regions unique, we must specify the tangential ﬁeldson each side of the adjoining surface. If we can relate the ﬁelds across the boundary, wecan propagate the solution from one region to the next; in this way, information aboutthe source in one region is eﬀectively passed on to the solution in an adjacent region. Foruniqueness, only relations between the tangential components need be speciﬁed. We shall determine the appropriate boundary conditions (BC’s) via two distinct ap- proaches. We ﬁrst model a thin source layer and consider a discontinuous surface sourcelayer as a limiting case of the continuous thin layer. With no true discontinuity, Maxwell’sdiﬀerential equations hold everywhere. We then consider a true spatial discontinuity be-tween material surfaces (with possible surface sources lying along the discontinuity). Wemust then isolate the region containing the discontinuity and postulate a ﬁeld relationship that is both physically meaningful and experimentally veriﬁable. We shall also consider both stationary and moving boundary surfaces, and surfaces containing magnetic as well as electric sources. 2.8.1 Boundaryconditions across a stationary , thin source lay er In§1.3.3 we discussed how in the macroscopic sense a surface source is actually a volume distribution concentrated near a surface S. We write the charge and current in terms of the point ron the surface and the normal distance xfrom the surface at ras ρ(r,x,t)=ρs(r,t)f(x, /Delta1), (2.179) J(r,x,t)=Js(r,t)f(x, /Delta1), (2.180) where f(x,/Delta1 )is the source density function obeying /integraldisplay∞ −∞f(x,/Delta1 )dx=1. (2.181) Figure 2.5: Derivation of the electromagnetic boundary conditions across a thin contin- uous source layer. The parameter /Delta1describes the “width” of the source layer normal to the reference surface. We use (2.156)–(2.157) to study ﬁeld behavior across the source layer. Consider a volumeregion V thatintersect sthesourcelayerasshowninFigure2.5.Letthetopand bottom surfaces be parallel to the reference surface, and label the ﬁelds on the top andbottom surfaces with subscripts 1 and 2, respectively. Since points on and within Vare all regular, (2.157) yields /integraldisplay S1ˆn1×H1dS+/integraldisplay S2ˆn2×H2dS+/integraldisplay S3ˆn3×HdS=/integraldisplay V/parenleftbigg J+∂D ∂t/parenrightbigg dV. We now choose δ=k/Delta1(k>1) so that most of the source lies within V.A s/Delta1→0 the thin source layer recedes to a surface layer, and the volume integral of displacementcurrent and the integral of tangential Hover S 3both approach zero by continuity of the ﬁelds. By symmetry S1=S2and ˆn1=− ˆn2=ˆn12, where ˆn12is the surface normal directed into region 1 from region 2. Thus /integraldisplay S1ˆn12×(H1−H2)dS=/integraldisplay VJdV. (2.182) Note that /integraldisplay VJdV=/integraldisplay S1/integraldisplayδ/2 −δ/2JdSdx =/integraldisplayδ/2 −δ/2f(x,/Delta1 )dx/integraldisplay S1Js(r,t)dS. Since we assume that the majority of the source current lies within V, the integral can be evaluated using (2.181) to give /integraldisplay S1[ˆn12×(H1−H2)−Js]dS=0, hence ˆn12×(H1−H2)=Js. The tangential magnetic ﬁeld across a thin source distribution is discontinuous by an amount equal to the surface current density. Similar steps with Faraday’s law give ˆn12×(E1−E2)=0. The tangential electric ﬁeld is continuous across a thin source. We can also derive conditions on the normal components of the ﬁelds, although these arenotrequire dforuniqueness .Gauss’ slaw(2.147 )applie dtothevolume V inFigure 2.5gives /integraldisplay S1D1·ˆn1dS+/integraldisplay S2D2·ˆn2dS+/integraldisplay S3D·ˆn3dS=/integraldisplay VρdV. As/Delta1→0, the thin source layer recedes to a surface layer. The integral of normal Dover S3tends to zero by continuity of the ﬁelds. By symmetry S1=S2and ˆn1=− ˆn2=ˆn12. Thus/integraldisplay S1(D1−D2)·ˆn12dS=/integraldisplay VρdV. (2.183) The volume integral is /integraldisplay VρdV=/integraldisplay S1/integraldisplayδ/2 −δ/2ρdSdx =/integraldisplayδ/2 −δ/2f(x,/Delta1 )dx/integraldisplay S1ρs(r,t)dS. Since δ=k/Delta1has been chosen so that most of the source charge lies within V, (2.181) gives /integraldisplay S1[(D1−D2)·ˆn12−ρs]dS=0, hence (D1−D2)·ˆn12=ρs. The normal component of Dis discontinuous across a thin source distribution by an amount equal to the surface charge density. Similar steps with the magnetic Gauss’s lawyield (B 1−B2)·ˆn12=0. The normal component of Bis continuous across a thin source layer. We can follow similar steps when a thin magnetic source layer is present. When evaluating Faraday’s law we must include magnetic surface current and when evaluatingthe magnetic Gauss’s law we must include magnetic charge. However, since such sourcesare not physical we postpone their consideration until the next section, where appropriateboundary conditions are postulated rather than derived. 2.8.2 Boundaryconditions across a stationarylay er of ﬁeld disconti- nuity Provided that we model a surface source as a limiting case of a very thin but continuous volume source, we can derive boundary conditions across a surface layer. We might askwhether we can extend this idea to surfaces of materials where the constitutive parameterschange from one region to another. Indeed, if we take Lorentz’ viewpoint and visualize amaterial as a conglomerate of atomic charge, we should be able to apply this same idea.After all, a material should demonstrate a continuous transition (in the macroscopic Figure 2.6: Derivation of the electromagnetic boundary conditions across a discontinuous source layer. sense) across its boundary, and we can employ the Maxwell–Boﬃ equations to describe the relationship between the “equivalent” sources and the electromagnetic ﬁelds. We should note, however, that the limiting concept is not without its critics. Stokes suggested as early as 1848 that jump conditions should never be derived from smoothsolutions [199]. Let us therefore pursue the boundary conditions for a surface of trueﬁeld discontinuity. This will also allow us to treat a material modeled as having a truediscontinuity in its material parameters (which we can always take as a mathematicalmodel of a more gradual transition) before we have studied in a deeper sense the physicalproperties of materials. This approach, taken by many textbooks, must be done carefully. There is a logical diﬃculty with this approach, lying in the application of the large- scale forms of Maxwell’s equations. Many authors postulate Maxwell’s equations in pointform, integrate to obtain the large-scale forms, then apply the large-scale forms to regionsof discontinuity. Unfortunately, the large-scale forms thus obtained are only valid in thesame regions where their point form antecedents were valid — discontinuities must beexcluded. Schelkunoﬀ [167] has criticized this approach, calling it a “swindle” ratherthan a proof, and has suggested that the proper way to handle true discontinuitiesis to postulate the large-scale forms of Maxwell’s equations, andto include as part of the postulate the assumption that the large-scale forms are valid at points of ﬁelddiscontinuity. Does this mean we must reject our postulate of the point form Maxwellequations and reformulate everything in terms of the large-scale forms? Fortunately, no. Tai [192] has pointed out that it is still possible to postulate the point forms, as longas we also postulate appropriate boundary conditions that make the large-scale forms,as derived from the point forms, valid at surfaces of discontinuity. In essence, bothapproaches require an additional postulate for surfaces of discontinuity: the large scaleforms require a postulate of applicability to discontinuous surfaces, and from there theboundary conditions can be derived; the point forms require a postulate of the boundaryconditions that result in the large-scale forms being valid on surfaces of discontinuity.Let us examine how the latter approach works. Consider a surface across which the constitutive relations are discontinuous, containing electric and magnetic surface currents and charges J s,ρs, Jms,andρms (Figure2.6). We locate a volume region V1 abovethesurfac eofdiscontinuity;thisvolum eisbounde d byasurface S1 andanother surface S10which is parallel to, and a small distance δ/2 above, the surface of discontinuity. A second volume region V2is similarly situated below the surface of discontinuity. Because these regions exclude the surface of discontinuity we can use (2.176) to get /integraldisplay S1ˆn×HdS+/integraldisplay S10ˆn×HdS=/integraldisplay V1/parenleftbigg J+∂D ∂t/parenrightbigg dV, /integraldisplay S2ˆn×HdS+/integraldisplay S20ˆn×HdS=/integraldisplay V2/parenleftbigg J+∂D ∂t/parenrightbigg dV. Adding these we obtain /integraldisplay S1+S2ˆn×HdS−/integraldisplay V1+V2/parenleftbigg J+∂D ∂t/parenrightbigg dV− −/integraldisplay S10ˆn10×H1dS−/integraldisplay S20ˆn20×H2dS=0, (2.184) where we have used subscripts to delineate the ﬁelds on each side of the discontinuity surface. Ifδis very small (but nonzero), then ˆn10=− ˆn20=ˆn12and S10=S20. Letting S1+S2=Sand V1+V2=V, we can write (184) as /integraldisplay S(ˆn×H)dS−/integraldisplay V/parenleftbigg J+∂D ∂t/parenrightbigg dV=/integraldisplay S10ˆn12×(H1−H2)dS. (2.185) Now suppose we use the same volume region V, but let it intersect the surface of discontinuity(Figure2.6),andsupposethatthelarge-scal eformofAmpere’slawholds even if Vcontains points of ﬁeld discontinuity. We must include the surface current in the computation. Since/integraltext VJdVbecomes/integraltext SJsdSon the surface, we have /integraldisplay S(ˆn×H)dS−/integraldisplay V/parenleftbigg J+∂D ∂t/parenrightbigg dV=/integraldisplay S10JsdS. (2.186) We wish to have this give the same value for the integrals over Vand Sas (2.185), which included in its derivation no points of discontinuity. This is true provided that ˆn12×(H1−H2)=Js. (2.187) Thus, under the condition (2.187) we may interpret the large-scale form of Ampere’s law (as derived from the point form) as being valid for regions containing discontinuities.Note that this condition is not “derived,” but must be regarded as a postulate that results in the large-scale form holding for surfaces of discontinuous ﬁeld. Similar reasoning can be used to determine the appropriate boundary condition on tangential Efrom Faraday’s law. Corresponding to (2.185) we obtain /integraldisplay S(ˆn×E)dS−/integraldisplay V/parenleftbigg −Jm−∂B ∂t/parenrightbigg dV=/integraldisplay S10ˆn12×(E1−E2)dS. (2.188) Employing (2.175) over the region containing the ﬁeld discontinuity surface we get /integraldisplay S(ˆn×E)dS−/integraldisplay V/parenleftbigg −Jm−∂B ∂t/parenrightbigg dV=−/integraldisplay S10JmsdS. (2.189) To have (2.188) and (2.189) produce identical results, we postulate ˆn12×(E1−E2)=−Jms (2.190) as the boundary condition appropriate to a surface of ﬁeld discontinuity containing a magnetic surface current. We can also postulate boundary conditions on the normal ﬁelds to make Gauss’s laws valid for surfaces of discontinuous ﬁelds. Integrating (2.147) over the regions V1and V2 and adding, we obtain /integraldisplay S1+S2D·ˆndS−/integraldisplay S10D1·ˆn10dS−/integraldisplay S20D2·ˆn20dS=/integraldisplay V1+V2ρdV. Asδ→0this becomes/integraldisplay SD·ˆndS−/integraldisplay VρdV=/integraldisplay S10(D1−D2)·ˆn12dS. (2.191) If we integrate Gauss’s law over the entire region V, including the surface of discontinuity, we get /contintegraldisplay SD·ˆndS=/integraldisplay VρdV+/integraldisplay S10ρsdS. (2.192) In order to get identical answers from (2.191) and (2.192), we must have (D1−D2)·ˆn12=ρs as the boundary condition appropriate to a surface of ﬁeld discontinuity containing an electric surface charge. Similarly, we must postulate (B1−B2)·ˆn12=ρms as the condition appropriate to a surface of ﬁeld discontinuity containing a magnetic surface charge. We can determine an appropriate boundary condition on current by using the large- scale form of the continuity equation. Applying (2.10) over each of the volume regionsofFigure2.6andaddingtheresults ,wehave /integraldisplay S1+S2J·ˆndS−/integraldisplay S10J1·ˆn10dS−/integraldisplay S20J2·ˆn20dS=−/integraldisplay V1+V2∂ρ ∂tdV. Asδ→0we have /integraldisplay SJ·ˆndS−/integraldisplay S10(J1−J2)·ˆn12dS=−/integraldisplay V∂ρ ∂tdV. (2.193) Applying the continuity equation over the entire region Vand allowing it to intersect the discontinuity surface, we get /integraldisplay SJ·ˆndS+/integraldisplay /Gamma1Js·ˆmdl=−/integraldisplay V∂ρ ∂tdV−/integraldisplay S10∂ρs ∂tdS. By the two-dimensional divergence theorem (B.20) we can write this as /integraldisplay SJ·ˆndS+/integraldisplay S10∇s·JsdS=−/integraldisplay V∂ρ ∂tdV−/integraldisplay S10∂ρs ∂tdS. In order for this expression to produce the same values of the integrals over Sand Vas in (2.193) we require ∇s·Js=− ˆn12·(J1−J2)−∂ρs ∂t, which we take as our postulate of the boundary condition on current across a surface containing discontinuities. A similar set of steps carried out using the continuity equationfor magnetic sources yields ∇ s·Jms=− ˆn12·(Jm1−Jm2)−∂ρms ∂t. In summary, we have the following boundary conditions for ﬁelds across a surface containing discontinuities: ˆn12×(H1−H2)=Js, (2.194) ˆn12×(E1−E2)=−Jms, (2.195) ˆn12·(D1−D2)=ρs, (2.196) ˆn12·(B1−B2)=ρms, (2.197) and ˆn12·(J1−J2)=− ∇ s·Js−∂ρs ∂t, (2.198) ˆn12·(Jm1−Jm2)=− ∇ s·Jms−∂ρms ∂t, (2.199) where ˆn12points into region 1 from region 2. 2.8.3 Boundaryconditions at the surface of a perfect conductor We can easily specialize the results of the previous section to the case of perfect electric or magnetic conductors. In §2.2.2 we saw that the constitutive relations for perfect conductors requires the null ﬁeld within the material. In addition, a PEC requires zerotangential electric ﬁeld, while a PMC requires zero tangential magnetic ﬁeld. Using(2.194)–(2.199), we ﬁnd that the boundary conditions for a perfect electric conductorare ˆn×H=J s, (2.200) ˆn×E=0, (2.201) ˆn·D=ρs, (2.202) ˆn·B=0, (2.203) and ˆn·J=− ∇ s·Js−∂ρs ∂t, ˆn·Jm=0. (2.204) For a PMC the conditions are ˆn×H=0, (2.205) ˆn×E=−Jms, (2.206) ˆn·D=0, (2.207) ˆn·B=ρms, (2.208) and ˆn·Jm=− ∇ s·Jms−∂ρms ∂t, ˆn·J=0. (2.209) We note that the normal vector ˆnpoints out of the conductor and into the adjacent region of nonzero ﬁelds. 2.8.4 Boundaryconditions across a stationarylay er of ﬁeld disconti- nuityusing equivalent sources So far we have avoided using the physical interpretation of the equivalent sources in the Maxwell–Boﬃ equations so that we might investigate the behavior of ﬁelds across truediscontinuities. Now that we have the appropriate boundary conditions, it is interestingto interpret them in terms of the equivalent sources. If we put H=B/µ 0−Minto (2.194) and rearrange, we get ˆn12×(B1−B2)=µ0(Js+ˆn12×M1−ˆn12×M2). (2.210) The terms on the right involving ˆn12×Mhave the units of surface current and are called equivalent magnetization surface currents . Deﬁning JMs=− ˆn×M (2.211) where ˆnis directed normally outward from the material region of interest, we can rewrite (2.210) as ˆn12×(B1−B2)=µ0(Js+JMs1+JMs2). (2.212) We note that JMsreplaces atomic charge moving along the surface of a material with an equivalent surface current in free space. If we substitute D=/epsilon10E+Pinto (2.196) and rearrange, we get ˆn12·(E1−E2)=1 /epsilon10(ρs−ˆn12·P1+ˆn12·P2). (2.213) The terms on the right involving ˆn12·Phave the units of surface charge and are called equivalent polarization surface charges . Deﬁning ρPs=ˆn·P, (2.214) we can rewrite (2.213) as ˆn12·(E1−E2)=1 /epsilon10(ρs+ρPs1+ρPs2). (2.215) We note that ρPsreplaces atomic charge adjacent to a surface of a material with an equivalent surface charge in free space. In summary, the boundary conditions at a stationary surface of discontinuity written in terms of equivalent sources are ˆn12×(B1−B2)=µ0(Js+JMs1+JMs2), (2.216) ˆn12×(E1−E2)=−Jms, (2.217) ˆn12·(E1−E2)=1 /epsilon10(ρs+ρPs1+ρPs2), (2.218) ˆn12·(B1−B2)=ρms. (2.219) 2.8.5 Boundaryconditions across a moving lay er of ﬁeld discontinuity With a moving material body it is often necessary to apply boundary conditions de- scribing the behavior of the ﬁelds across the surface of the body. If a surface of discon-tinuity moves with constant velocity v, the boundary conditions (2.194)–(2.199) hold as long as all ﬁelds are expressed in the frame of the moving surface . We can also derive boundary conditions for a deforming surface moving with arbitrary velocity by usingequations (2.177)–(2.178). In this case all ﬁelds are expressed in the laboratory frame.Proceeding through the same set of steps that gave us (2.194)–(2.197), we ﬁnd ˆn 12×(H1−H2)+(ˆn12·v)(D1−D2)=Js, (2.220) ˆn12×(E1−E2)−(ˆn12·v)(B1−B2)=−Jms, (2.221) ˆn12·(D1−D2)=ρs, (2.222) ˆn12·(B1−B2)=ρms. (2.223) Note that when ˆn12·v=0these boundary conditions reduce to those for a stationary surface. This occurs not only when v=0but also when the velocity is parallel to the surface. The reader must be wary when employing (2.220)–(2.223). Since the ﬁelds are mea- sured in the laboratory frame, if the constitutive relations are substituted into the bound-ary conditions they must also be represented in the laboratory frame. It is probable thatthe material parameters would be known in the rest frame of the material, in which casea conversion to the laboratory frame would be necessary. 2.9 Fundamental theorems In this section we shall consider some of the important theorems of electromagnetics that pertain directly to Maxwell’s equations. They may be derived without reference tothe solutions of Maxwell’s equations, and are not connected with any specialization ofthe equations or any speciﬁc application or geometrical conﬁguration. In this sense thesetheorems are fundamental to the study of electromagnetics. 2.9.1 Linearity Recall that a mathematical operator Lislinear if L(α1f1+α2f2)=α1L(f1)+α2L(f2) holds for any two functions f1,2in the domain of Land any two scalar constants α1,2.A standard observation regarding the equation L(f)=s, (2.224) where Lis a linear operator and sis a given forcing function, is that if f1and f2are solutions to L(f1)=s1, L(f2)=s2, (2.225) respectively, and s=s1+s2, (2.226) then f=f1+f2 (2.227) is a solution to (2.224). This is the principle of superposition ; if convenient, we can decompose sin equation (2.224) as a sum (2.226) and solve the two resulting equations (2.225) independently. The solution to (2.224) is then (2.227), “by superposition.” Ofcourse, we are free to split the right side of (2.224) into more than two terms — themethod extends directly to any ﬁnite number of terms. Because the operators ∇·,∇×, and ∂/∂tare all linear, Maxwell’s equations can be treated by this method. If, for instance, ∇× E 1=−∂B1 ∂t, ∇× E2=−∂B2 ∂t, then ∇× E=−∂B ∂t where E=E1+E2and B=B1+B2. The motivation for decomposing terms in a particular way is often based on physical considerations; we give one example here anddefer others to later sections of the book. We saw earlier that Maxwell’s equations canbe written in terms of both electric and (ﬁctitious) magnetic sources as in equations(2.169)–(2.172). Let E=E e+Emwhere Eeis produced by electric-type sources and Em is produced by magnetic-type sources, and decompose the other ﬁelds similarly. Then ∇× Ee=−∂Be ∂t, ∇× He=J+∂De ∂t, ∇·De=ρ, ∇·Be=0, with a similar equation set for the magnetic sources. We may, if desired, solve these two equation sets independently for Ee,De,Be,Heand Em,Dm,Em,Hm, and then use superposition to obtain the total ﬁelds E,D,B,H. 2.9.2 Duality The intriguing symmetry of Maxwell’s equations leads us to an observation that can reduce the eﬀort required to compute solutions. Consider a closed surface Senclosing a region of space that includes an electric source current Jand a magnetic source current Jm. The ﬁelds ( E1,D1,B1,H1) within the region (which may also contain arbitrary media) are described by ∇× E1=−Jm−∂B1 ∂t, (2.228) ∇× H1=J+∂D1 ∂t, (2.229) ∇·D1=ρ, (2.230) ∇·B1=ρm. (2.231) Suppose we have been given a mathematical description of the sources (J,Jm)and have solved for the ﬁeld vectors (E1,D1,B1,H1). Of course, we must also have been supplied with a set of boundary values and constitutive relations in order to make the solutionunique. We note that if we replace the formula for Jwith the formula for J min (2.229) (andρwithρmin (2.230)) and also replace Jmwith−Jin (2.228) (and ρmwith−ρ in (2.231)) we get a new problem to solve, with a diﬀerent solution. However, thesymmetry of the equations allows us to specify the solution immediately. The new set of curl equations requires ∇× E2=J−∂B2 ∂t, (2.232) ∇× H2=Jm+∂D2 ∂t. (2.233) As long as we can resolve the question of how the constitutive parameters must be altered to reﬂect these replacements, we can conclude by comparing (2.232) with (2.229) and(2.233) with (2.228) that the solution to these equations is merely E 2=H1, B2=−D1, D2=B1, H2=−E1. That is, if we have solved the original problem, we can use those solutions to ﬁnd the new ones. This is an application of the general principle of duality . Unfortunately, this approach is a little awkward since the units of the sources and ﬁelds in the two problems are diﬀerent. We can make the procedure more convenient bymultiplying Ampere’s law by η 0=(µ0//epsilon10)1/2. Then we have ∇× E=−Jm−∂B ∂t, (2.234) ∇×(η0H)=(η0J)+∂(η 0D) ∂t. (2.235) Thus if the original problem has solution (E1,η0D1,B1,η0H1), then the dual problem with Jreplaced by Jm/η0and Jmreplaced by −η0Jhas solution E2=η0H1, (2.236) B2=−η0D1, (2.237) η0D2=B1, (2.238) η0H2=−E1. (2.239) The units on the quantities in the two problems are now identical. Of course, the constitutive parameters for the dual problem must be altered from those of the original problem to reﬂect the change in ﬁeld quantities. From (2.19) and(2.20) we know that the most general forms of the constitutive relations (those for linear, bianisotropic media) are D 1=¯ξ1·H1+¯/epsilon11·E1, (2.240) B1=¯µ1·H1+¯ζ1·E1, (2.241) for the original problem, and D2=¯ξ2·H2+¯/epsilon12·E2, (2.242) B2=¯µ2·H2+¯ζ2·E2, (2.243) for the dual problem. Substitution of (2.236)–(2.239) into (2.240) and (2.241) gives D2=(−¯ζ1)·H2+/parenleftbigg¯µ1 η2 0/parenrightbigg ·E2, (2.244) B2=/parenleftbig η2 0¯/epsilon11/parenrightbig ·H2+(−¯ξ1)·E2. (2.245) Comparing (2.244) with (2.242) and (2.245) with (2.243), we conclude that ¯ζ2=− ¯ξ1, ¯ξ2=− ¯ζ1, ¯µ2=η2 0¯/epsilon11, ¯/epsilon12=¯µ1 η2 0. As an important special case, we see that for a linear, isotropic medium speciﬁed by a permittivity /epsilon1and permeability µ, the dual problem is obtained by replacing /epsilon1rwithµr andµrwith/epsilon1r. The solution to the dual problem is then given by E2=η0H1,η 0H2=−E1, as before. We thus see that the medium in the dual problem must have electric properties numerically equal to the magnetic properties of the medium in the original problem, andmagnetic properties numerically equal to the electric properties of the medium in theoriginal problem. This is rather inconvenient for most applications. Alternatively, wemay divide Ampere’s law by η=(µ//epsilon1) 1/2instead of η0. Then the dual problem has Jreplaced by Jm/η, and Jmreplaced by −ηJ, and the solution to the dual problem is given by E2=ηH1,η H2=−E1. In this case there is no need to swap /epsilon1randµr, since information about these parameters is incorporated into the replacement sources. We must also remember that to obtain a unique solution we need to specify the bound- ary values of the ﬁelds. In a true dual problem, the boundary values of the ﬁelds usedin the original problem are used on the swapped ﬁelds in the dual problem. A typicalexample of this is when the condition of zero tangential electric ﬁeld on a perfect electricconductor is replaced by the condition of zero tangential magnetic ﬁeld on the surface ofa perfect magnetic conductor. However, duality can also be used to obtain the mathe-matical form of the ﬁeld expressions, often in a homogeneous (source-free) situation, andboundary values can be applied later to specify the solution appropriate to the problemgeometry. This approach is often used to compute waveguide modal ﬁelds and the elec-tromagnetic ﬁelds scattered from objects. In these cases a TE/TM ﬁeld decompositionis employed, and duality is used to ﬁnd one part of the decomposition once the other isknown. Dualityof electric and magnetic point source ﬁelds. By duality, we can some- times use the known solution to one problem to solve a related problem by merely sub-stituting diﬀerent variables into the known mathematical expression. An example of this is the case in which we have solved for the ﬁelds produced by a certain distribution ofelectric sources and wish to determine the ﬁelds when the same distribution is used todescribe magnetic sources. Let us consider the case when the source distribution is that of a point current, or Hertzian dipole , immersed in free space. As we shall see in Chapter 5, the ﬁelds for a general source may be found by using the ﬁelds produced by these point sources. Webegin by ﬁnding the ﬁelds produced by an electric dipole source at the origin alignedalong the z-axis, J=ˆzI 0δ(r), then use duality to ﬁnd the ﬁelds produced by a magnetic current source Jm=ˆzIm0δ(r). The ﬁelds produced by the electric source must obey ∇× Ee=−∂ ∂tµ0He, (2.246) ∇× He=ˆzI0δ(r)+∂ ∂t/epsilon10Ee, (2.247) ∇·/epsilon10Ee=ρ, (2.248) ∇·He=0, (2.249) while those produced by the magnetic source must obey ∇× Em=− ˆzIm0δ(r)−∂ ∂tµ0Hm, (2.250) ∇× Hm=∂ ∂t/epsilon10Em, (2.251) ∇·Em=0, (2.252) ∇·µ0Hm=ρm. (2.253) We see immediately that the second set of equations is the dual of the ﬁrst, as long as we scale the sources appropriately. Multiplying (2.250) by −I0/Im0and (2.251) by I0η2 0/Im0, we have the curl equations ∇×/parenleftbigg −I0 Im0Em/parenrightbigg =ˆzI0δ(r)+∂ ∂t/parenleftbigg µ0I0 Im0Hm/parenrightbigg , (2.254) ∇×/parenleftbiggI0η2 0 Im0Hm/parenrightbigg =−∂ ∂t/parenleftbigg −/epsilon10I0η2 0 Im0Em/parenrightbigg . (2.255) Comparing (2.255) with (2.246) and (2.254) with (2.247) we see that Em=−Im0 I0He, Hm=Im0 I0Ee η2 0. We note that it is impossible to have a point current source without accompanying point charge sources terminating each end of the dipole current. The point charges arerequired to satisfy the continuity equation, and vary in time as the moving charge thatestablishes the current accumulates at the ends of the dipole. From (2.247) we see thatthe magnetic ﬁeld curls around the combination of the electric ﬁeld and electric currentsource, while from (2.246) the electric ﬁeld curls around the magnetic ﬁeld, and from(2.248) diverges from the charges located at the ends of the dipole. From (2.250) wesee that the electric ﬁeld must curl around the combination of the magnetic ﬁeld andmagnetic current source, while (2.251) and (2.253) show that the magnetic ﬁeld curlsaround the electric ﬁeld and diverges from the magnetic charge. Dualityin a source-free region. Consider a closed surface Senclosing a source-free region of space. For simplicity, assume that the medium within Sis linear, isotropic, and homogeneous. The ﬁelds within Sare described by Maxwell’s equations ∇× E 1=−∂ ∂tµH1, (2.256) ∇×ηH1=∂ ∂t/epsilon1ηE1, (2.257) ∇·/epsilon1E1=0, (2.258) ∇·µH1=0. (2.259) Under these conditions the concept of duality takes on a diﬀerent face. The symmetry of the equations is such that the mathematical form of the solution for Eis the same as that for ηH. That is, the ﬁelds E2=ηH1, (2.260) H2=−E1/η, (2.261) are also a solution to Maxwell’s equations, and thus the dual problem merely involves replacing EbyηHand Hby−E/η. However, the ﬁnal forms of Eand Hwillnotbe identical after appropriate boundary values are imposed. This form of duality is very important for the solution of ﬁelds within waveguides or the ﬁelds scattered by objects where the sources are located outside the region where theﬁelds are evaluated. 2.9.3 Reciprocity The reciprocity theorem, also called the Lorentz reciprocity theorem , describes a spe- ciﬁc and often useful relationship between sources and the electromagnetic ﬁelds theyproduce. Under certain special circumstances we ﬁnd that an interaction between inde-pendent source and mediating ﬁelds called “reaction” is a spatially symmetric quantity.The reciprocity theorem is used in the study of guided waves to establish the orthogonal-ity of guided wave modes, in micr owave network theory to obtain relationships between terminal characteristics, and in antenna theory to demonstrate the equivalence of trans-mission and reception patterns. Consider a closed surface Senclosing a volume V. Assume that the ﬁelds within and onSare produced by two independent source ﬁelds. The source (J a,Jma)produces the ﬁeld(Ea,Da,Ba,Ha)as described by Maxwell’s equations ∇× Ea=−Jma−∂Ba ∂t, (2.262) ∇× Ha=Ja+∂Da ∂t, (2.263) while the source ﬁeld (Jb,Jmb)produces the ﬁeld (Eb,Db,Bb,Hb)as described by ∇× Eb=−Jmb−∂Bb ∂t, (2.264) ∇× Hb=Jb+∂Db ∂t. (2.265) The sources may be distributed in any way relative to S: they may lie completely inside, completely outside, or partially inside and partially outside. Material media may liewithin S, and their properties may depend on position. Let us examine the quantity R≡∇· (E a×Hb−Eb×Ha). By (B.44) we have R=Hb·∇× Ea−Ea·∇× Hb−Ha·∇× Eb+Eb·∇× Ha so that by Maxwell’s curl equations R=/bracketleftbigg Ha·∂Bb ∂t−Hb·∂Ba ∂t/bracketrightbigg −/bracketleftbigg Ea·∂Db ∂t−Eb·∂Da ∂t/bracketrightbigg + +[Ja·Eb−Jb·Ea−Jma·Hb+Jmb·Ha]. The useful relationships we seek occur when the ﬁrst two bracketed quantities on the right-hand side of the above expression are zero. Whether this is true depends not onlyon the behavior of the ﬁelds, but on the properties of the medium at the point in question.Though we have assumed that the sources of the ﬁeld sets are independent, it is apparentthat they must share a similar time dependence in order for the terms within each of thebracketed quantities to cancel. Of special interest is the case where the two sources areboth sinusoidal in time with identical frequencies, but with diﬀering spatial distributions.We shall consider this case in detail in §4.10.2after we have discussed the properties of the time harmonic ﬁeld. Importantly, we will ﬁnd that only certain characteristics of theconstitutive parameters allow cancellation of the bracketed terms; materials with thesecharacteristics are called reciprocal , and the ﬁelds they support are said to display the property of reciprocity . To see what this property entails, we set the bracketed terms to zero and integrate over a volume Vto obtain /contintegraldisplay S(Ea×Hb−Eb×Ha)·dS=/integraldisplay V(Ja·Eb−Jb·Ea−Jma·Hb+Jmb·Ha)dV, which is the time-domain version of the Lorentz reciprocity theorem . Two special cases of this theorem are important to us. If all sources lie outside S,w e haveLorentz’s lemma/contintegraldisplay S(Ea×Hb−Eb×Ha)·dS=0. This remarkable expression shows that a relationship exists between the ﬁelds produced by completely independent sources, and is useful for establishing waveguide mode or- thogonality for time harmonic ﬁelds. If sources reside within Sbut the surface integral is equal to zero, we have /integraldisplay V(Ja·Eb−Jb·Ea−Jma·Hb+Jmb·Ha)dV=0. This occurs when the surface is bounded by a special material (such as an impedance sheet or a perfect conductor), or when the surface recedes to inﬁnity; the expression isuseful for establishing the reciprocity conditions for networks and antennas. We shallinterpret it for time harmonic ﬁelds in §4.10.2. 2.9.4 Similitude A common approach in physical science involves the introduction of normalized vari- ables to provide for scaling of problems along with a chance to identify certain physicallysigniﬁcant parameters. Similarity as a general principle can be traced back to the earliestattempts to describe physical eﬀects with mathematical equations, with serious study un-dertaken by Galileo. Helmholtz introduced the ﬁrst systematic investigation in 1873, andthe concept was rigorized by Reynolds ten years later [216]. Similitude is now considereda fundamental guiding principle in the modeling of materials [199]. The process often begins with a consideration of the fundamental diﬀerential equations. In electromagnetics we may introduce a set of dimensionless ﬁeld and source variables E ,D,B,H,J,ρ, (2.266) by setting E=EkE,B=BkB,D=DkD, H=HkH,J=JkJ,ρ=ρkρ. (2.267) Here we regard the quantities kE,kB,...as base units for the discussion, while the dimensionless quantities (2.266) serve to express the actual ﬁelds E,B,...in terms of these base units. Of course, the time and space variables can also be scaled: we can write t=tkt, l=lkl, (2.268) iflis any length of interest. Again, the quantities tand lare dimensionless measure numbers used to express the actual quantities tandlrelative to the chosen base amounts ktand kl. With (2.267) and (2.268), Maxwell’s curl equations become ∇×E=−kB kEkl kt∂B ∂t, ∇×H=kJkl kHJ+kD kHkl kt∂D ∂t(2.269) while the continuity equation becomes ∇·J=−kρ kJkl kt∂ρ ∂t, (2.270) where ∇has been normalized by kl. These are examples of ﬁeld equations cast into dimensionless form — it is easily veriﬁed that the similarity parameters kB kEkl kt,kJkl kH,kD kHkl kt,kρ kJkl kt, (2.271) are dimensionless. The idea behind electromagnetic similitude is that a given set of normalized values E,B,...can satisfy equations (2.269) and (2.270) for many diﬀerent physical situations, provided that the numerical values of the coeﬃcients (2.271) are allﬁxed across those situations. Indeed, the diﬀerential equations would be identical. To make this discussion a bit more concrete, let us assume a conducting linear medium where D=/epsilon1E, B=µH, J=σE, and use /epsilon1=/epsilon1 k/epsilon1,µ =µkµ,σ =σkσ, to express the material parameters in terms of dimensionless values /epsilon1,µ, and σ. Then D=k/epsilon1kE kD/epsilon1E, B=kµkH kBµH, J=kσkE kJσE, and equations (2.269) become ∇×E=−/parenleftbiggkµkl ktkH kE/parenrightbigg µ∂H ∂t, ∇×H=/parenleftbigg kσklkE kH/parenrightbigg σE+/parenleftbiggk/epsilon1kl ktkE kH/parenrightbigg /epsilon1∂E ∂t. Deﬁning α=kµkl ktkH kE,γ =kσklkE kH,β =k/epsilon1kl ktkE kH, we see that under the current assumptions similarity holds between two electromagnetics problems only if αµ,γσ, andβ/epsilon1are numerically the same in both problems. A necessary condition for similitude, then, is that the products (αµ)(β/epsilon1)=kµk/epsilon1/parenleftbiggkl kt/parenrightbigg2 µ/epsilon1,( α µ )(γσ)=kµkσk2 l ktµσ, (which do not involve kEorkH) stay constant between problems. We see, for example, that we may compensate for a halving of the length scale klby (a) a quadrupling of the permeability µ, or (b) a simultaneous halving of the time scale ktand doubling of the conductivity σ. A much less subtle special case is that for which σ=0,k/epsilon1=/epsilon10,kµ=µ0, and/epsilon1=µ=1; we then have free space and must simply maintain kl/kt=constant so that the time and length scales stay proportional. In the sinusoidal steady state, for instance, the frequency would be made to vary inversely with the length scale. 2.9.5 Conservation theorems The misconception that Poynting’s theorem can be “derived” from Maxwell’s equations is widespread and ingrained. We must, in fact, postulate the idea that the electromagnetic ﬁeld can be associated with an energy ﬂux propagating at the speed of light. Sincethe form of the postulate is patterned after the well-understood laws of mechanics, webegin by developing the basic equations of momentum and energy balance in mechanicalsystems. Then we shall see whether it is sensible to ascribe these principles to theelectromagnetic ﬁeld. Maxwell’s theory allows us to describe, using Maxwell’s equations, the behavior of the electromagnetic ﬁelds within a (possibly) ﬁnite region Vof space. The presence of any sources or material objects outside Vare made known through the speciﬁcation of tangential ﬁelds over the boundary of V, as required for uniqueness. Thus, the inﬂuence of external eﬀects can always be viewed as being transported across the boundary. Thisis true of mechanical as well as electromagnetic eﬀects. A charged material body canbe acted on by physical contact with another body, by gravitational forces, and by theLorentz force, each eﬀect resulting in momentum exchange across the boundary of theobject. These eﬀects must all be taken into consideration if we are to invoke momentumconservation, resulting in a very complicated situation. This suggests that we try todecompose the problem into simpler “systems” based on physical eﬀects. The system concept in the physical sciences. The idea of decomposing a com- plicated system into simpler, self-contained systems is quite common in the physicalsciences. Penﬁeld and Haus [145] invoke this concept by introducing an electromagnetic system where the eﬀects of the Lorentz force equation are considered to accompany a mechanical system where eﬀects of pressure, stress, and strain are considered, and a thermodynamic system where the eﬀects of heat exchange are considered. These systems can all be interrelated in a variety of ways. For instance, as a material heats up it canexpand, and the resulting mechanical forces can alter the electrical properties of thematerial. We will follow Penﬁeld and Haus by considering separate electromagnetic andmechanical subsystems; other systems may be added analogously. If we separate the various systems by physical eﬀect, we will need to know how to “reassemble the information.” Two conservation theorems are very helpful in this re-gard: conservation of energy, and conservation of momentum. Engineers often employthese theorems to make tacit use of the system idea. For instance, when studying elec-tromagnetic waves propagating in a waveguide, it is common practice to compute wave attenuation by calculating the Poynting ﬂux of power into the walls of the guide. Thepower lost from the wave is said to “heat up the waveguide walls,” which indeed it does. This is an admission that the electromagnetic system is not “closed”: it requires the inclusion of a thermodynamic system in order that energy be conserved. Of course, the detailed workings of the thermodynamic system are often ignored, indicating that anythermodynamic “feedback” mechanism is weak. In the waveguide example, for instance,the heating of the metallic walls does not alter their electromagnetic properties enoughto couple back into an eﬀect on the ﬁelds in the walls or in the guide. If such eﬀects wereimportant, they would have to be included in the conservation theorem via the bound-ary ﬁelds; it is therefore reasonable to associate with these ﬁelds a “ﬂow” of energy ormomentum into V. Thus, we wish to develop conservation laws that include not only the Lorentz force eﬀects within V, but a ﬂow of external eﬀects into Vthrough its boundary surface. To understand how external inﬂuences may eﬀect the electromagnetic subsystem, we look to the behavior of the mechanical subsystem as an analogue. In the electromagneticsystem, eﬀects are felt both internally to a region (because of the Lorentz force eﬀect) andthrough the system boundary (by the dependence of the internal ﬁelds on the boundaryﬁelds). In the mechanical and thermodynamic systems, a region of mass is aﬀected both internally (through transfer of heat and gravitational forces) and through interactionsoccurring across its surface (through transfers of energy and momentum, by pressureand stress). One beauty of electromagnetic theory is that we can ﬁnd a mathematicalsymmetry between electromagnetic and mechanical eﬀects which parallels the above con-ceptual symmetry. This makes applying conservation of energy and momentum to thetotal system (electromagnetic, thermodynamic, and mechanical) very convenient. Conservation of momentum and energyin mechanical sy stems. We begin by reviewing the interactions of material bodies in a mechanical system. For simplicity weconcentrate on ﬂuids (analogous to charge in space); the extension of these concepts tosolid bodies is straightforward. Consider a ﬂuid with mass density ρ m. The momentum of a small subvolume of the ﬂuid is given by ρmvdV, where vis the velocity of the subvolume. So the momentum density is ρmv. Newton’s second law states that a force acting throughout the subvolume results in a change in its momentum given by D Dt(ρmvdV)=fdV, (2.272) where fis the volume force density and the D/Dtnotation shows that we are interested in the rate of change of the momentum as observed by the moving ﬂuid element (see§A.2). Here fcould be the weight force, for instance. Addition of the results for all elements of the ﬂuid body gives D Dt/integraldisplay VρmvdV=/integraldisplay VfdV (2.273) as the change in momentum for the entire body. If on the other hand the force exerted on the body is through contact with its surface, the change in momentum is D Dt/integraldisplay VρmvdV=/contintegraldisplay StdS (2.274) where tis the “surface traction.” We can write the time-rate of change of momentum in a more useful form by applying the Reynolds transport theorem (A.66): D Dt/integraldisplay VρmvdV=/integraldisplay V∂ ∂t(ρmv)dV+/contintegraldisplay S(ρmv)v·dS. (2.275) Superposing (2.273) and (2.274) and substituting into (2.275) we have /integraldisplay V∂ ∂t(ρmv)dV+/contintegraldisplay S(ρmv)v·dS=/integraldisplay VfdV+/contintegraldisplay StdS. (2.276) If we deﬁne the dyadic quantity ¯Tk=ρmvv then (2.276) can be written as /integraldisplay V∂ ∂t(ρmv)dV+/contintegraldisplay Sˆn·¯TkdS=/integraldisplay VfdV+/contintegraldisplay StdS. (2.277) Thisprinciple of linear momentum [214] can be interpreted as a large-scale form of conservation of kinetic linear momentum. Here ˆn·¯Tkrepresents the ﬂow of kinetic mo- mentum across S, and the sum of this momentum transfer and the change of momentum within Vstands equal to the forces acting internal to Vand upon S. The surface traction may be related to the surface normal ˆnthrough a dyadic quantity ¯Tmcalled the mechanical stress tensor : t=ˆn·¯Tm. With this we may write (2.277) as /integraldisplay V∂ ∂t(ρmv)dV+/contintegraldisplay Sˆn·¯TkdS=/integraldisplay VfdV+/contintegraldisplay Sˆn·¯TmdS and apply the dyadic form of the divergence theorem (B.19) to get /integraldisplay V∂ ∂t(ρmv)dV+/integraldisplay V∇·(ρmvv)dV=/integraldisplay VfdV+/integraldisplay V∇·¯TmdV. (2.278) Combining the volume integrals and setting the integrand to zero we have ∂ ∂t(ρmv)+∇ · (ρmvv)=f+∇· ¯Tm, which is the point-form equivalent of (2.277). Note that the second term on the right- hand side is nonzero only for points residing on the surface of the body. Finally, lettinggdenote momentum density we obtain the simple expression ∇·¯T k+∂gk ∂t=fk, (2.279) where gk=ρmv is the density of kinetic momentum and fk=f+∇· ¯Tm (2.280) is the total force density. Equation (2.279) is somewhat analogous to the electric charge continuity equation (1.11). For each point of the body, the total outﬂux of kinetic momentum plus the timerate of change of kinetic momentum equals the total force. The resemblance to (1.11)is strong, except for the nonzero term on the right-hand side. The charge continuity equation represents a closed system: charge cannot spontaneously appear and add an extra term to the right-hand side of (1.11). On the other hand, the change in totalmomentum at a point can exceed that given by the momentum ﬂowing out of the pointif there is another “source” (e.g., gravity for an internal point, or pressure on a boundarypoint). To obtain a momentum conservation expression that resembles the continuity equa- tion, we must consider a “subsystem” with terms that exactly counterbalance the extraexpressions on the right-hand side of (2.279). For a ﬂuid acted on only by externalpressure the sole eﬀect enters through the traction term, and [145] ∇·¯T m=− ∇ p (2.281) where pis the pressure exerted on the ﬂuid body. Now, using (B.63), we can write −∇ p=− ∇· ¯Tp (2.282) where ¯Tp=p¯I and ¯Iis the unit dyad. Finally, using (2.282), (2.281), and (2.280) in (2.279), we obtain ∇·(¯Tk+¯Tp)+∂ ∂tgk=0 and we have an expression for a closed system including all possible eﬀects. Now, note that we can form the above expression as /parenleftbigg ∇·¯Tk+∂ ∂tgk/parenrightbigg +/parenleftbigg ∇·¯Tp+∂ ∂tgp/parenrightbigg =0 (2.283) where gp=0since there are no volume eﬀects associated with pressure. This can be viewed as the sum of two closed subsystems ∇·¯Tk+∂ ∂tgk=0, (2.284) ∇·¯Tp+∂ ∂tgp=0. We now have the desired viewpoint. The conservation formula for the complete closed system can be viewed as a sum of formulas for open subsystems, each having the formof a conservation law for a closed system. In case we must include the eﬀects of gravity,for instance, we need only determine ¯T gand ggsuch that ∇·¯Tg+∂ ∂tgg=0 and add this new conservation equation to (2.283). If we can ﬁnd a conservation ex- pression of form similar to (2.284) for an “electromagnetic subsystem,” we can includeits eﬀects along with the mechanical eﬀects by merely adding together the conservationlaws. We shall ﬁnd just such an expression later in this section. We stated in §1.3 that there are four fundamental conservation principles. We have now discussed linear momentum; the principle of angular momentum follows similarly.Our next goal is to ﬁnd an expression similar to (2.283) for conservation of energy. Wemay expect the conservation of energy expression to obey a similar law of superposition. We begin with the fundamental deﬁnition of work: for a particle moving with velocity v under the inﬂuence of a force fkthe work is given by fk·v. Dot multiplying (2.272) by v and replacing fbyfk(to represent both volume and surface forces), we get v·D Dt(ρmv)dV=v·fkdV or equivalently D Dt/parenleftbigg1 2ρmv·v/parenrightbigg dV=v·fkdV. Integration over a volume and application of the Reynolds transport theorem (A.66) then gives /integraldisplay V∂ ∂t/parenleftbigg1 2ρmv2/parenrightbigg dV+/contintegraldisplay Sˆn·/parenleftbigg v1 2ρmv2/parenrightbigg dS=/integraldisplay Vfk·vdV. Hence the sum of the time rate of change in energy internal to the body and the ﬂow of kinetic energy across the boundary must equal the work done by internal and surfaceforces acting on the body. In point form, ∇·S k+∂ ∂tWk=fk·v (2.285) where Sk=v1 2ρmv2 is the density of the ﬂow of kinetic energy and Wk=1 2ρmv2 is the kinetic energy density. Again, the system is not closed (the right-hand side of (2.285) is not zero) because the balancing forces are not included. As was done with themomentum equation, the eﬀect of the work done by the pressure forces can be describedin a closed-system-type equation ∇·S p+∂ ∂tWp=0. (2.286) Combining (2.285) and (2.286) we have ∇·(Sk+Sp)+∂ ∂t(Wk+Wp)=0, the energy conservation equation for the closed system. Conservation in the electromagnetic subsystem. We would now like to achieve closed-system conservation theorems for the electromagnetic subsystem so that we canadd in the eﬀects of electromagnetism. For the momentum equation, we can proceedexactly as we did with the mechanical system. We begin with f em=ρE+J×B. This force term should appear on one side of the point form of the momentum conserva- tion equation. The term on the other side must involve the electromagnetic ﬁelds, since they are the mechanism for exerting force on the charge distribution. Substituting for J from (2.2) and for ρfrom (2.3) we have fem=E(∇·D)−B×(∇× H)+B×∂D ∂t. Using B×∂D ∂t=−∂ ∂t(D×B)+D×∂B ∂t and substituting from Faraday’s law for ∂B/∂twe have −[E(∇·D)−D×(∇× E)+H(∇·B)−B×(∇× H)]+∂ ∂t(D×B)=−fem.(2.287) Here we have also added the null term H(∇·B). The forms of (2.287) and (2.279) would be identical if the bracketed term could be written as the divergence of a dyadic function ¯Tem. This is indeed possible for linear, homogeneous, bianisotropic media, provided that the constitutive matrix [¯CEH]in (2.21) is symmetric [101]. In that case ¯Tem=1 2(D·E+B·H)¯I−DE−BH, (2.288) which is called the Maxwell stress tensor . Let us demonstrate this equivalence for a linear, isotropic, homogeneous material. Putting D=/epsilon1Eand H=B/µinto (2.287) we obtain ∇·Tem=−/epsilon1E(∇·E)+1 µB×(∇× B)+/epsilon1E×(∇× E)−1 µB(∇·B). (2.289) Now (B.46) gives ∇(A·A)=2A×(∇× A)+2(A·∇)A so that E(∇·E)−E×(∇× E)=E(∇·E)+(E·∇)E−1 2∇(E2). Finally, (B.55) and (B.63) give E(∇·E)−E×(∇× E)=∇·/parenleftbigg EE−1 2¯IE·E/parenrightbigg . Substituting this expression and a similar one for Binto (2.289) we have ∇·¯Tem=∇·/bracketleftbigg1 2(D·E+B·H)¯I−DE−BH/bracketrightbigg , which matches (2.288). Replacing the term in brackets in (2.287) by ∇·¯Tem, we get ∇·¯Tem+∂gem ∂t=−fem (2.290) where gem=D×B. Equation (2.290) is the point form of the electromagnetic conservation of momentum theorem. It is mathematically identical in form to the mechanical theorem (2.279).Integration over a volume gives the large-scale form /contintegraldisplay S¯Tem·dS+/integraldisplay V∂gem ∂tdV=−/integraldisplay VfemdV. (2.291) If we interpret this as we interpreted the conservation theorems from mechanics, the ﬁrst term on the left-hand side represents the ﬂow of electromagnetic momentum across theboundary of V, while the second term represents the change in momentum within V. The sum of these two quantities is exactly compensated by the total Lorentz force acting onthe charges within V. Thus we identify g emas the transport density of electromagnetic momentum. Because (2.290) is not zero on the right-hand side, it does not represent a closed system. If the Lorentz force is the only force acting on the charges within V, then the mechanical reaction to the Lorentz force should be described by Newton’s third law. Thus we havethe kinematic momentum conservation formula ∇·¯T k+∂gk ∂t=fk=−fem. Subtracting this expression from (2.290) we obtain ∇·(¯Tem−¯Tk)+∂ ∂t(gem−gk)=0, (2.292) which describes momentum conservation for the closed system. It is also possible to derive a conservation theorem for electromagnetic energy that resembles the corresponding theorem for mechanical energy. Earlier we noted that v·f represents the volume density of work produced by moving an object at velocity vunder the action of a force f. For the electromagnetic subsystem the work is produced by charges moving against the Lorentz force. So the volume density of work delivered tothe currents is w em=v·fem=v·(ρE+J×B)=(ρv)·E+ρv·(v×B). (2.293) Using (B.6) on the second term in (2.293) we get wem=(ρv)·E+ρB·(v×v). The second term vanishes by deﬁnition of the cross product. This is the familiar property that the magnetic ﬁeld does no work on moving charge. Hence wem=J·E. (2.294) This important relation says that charge moving in an electric ﬁeld experiences a force which results in energy transfer to (or from) the charge. We wish to write this energytransfer in terms of an energy ﬂux vector, as we did with the mechanical subsystem. As with our derivation of the conservation of electromagnetic momentum, we wish to relate the energy transfer to the electromagnetic ﬁelds. Substitution of Jfrom (2.2) into (2.294) gives w em=(∇× H)·E−∂D ∂t·E, hence wem=− ∇· (E×H)+H·(∇× E)−∂D ∂t·E by (B.44). Substituting for ∇× Efrom (2.1) we have wem=− ∇· (E×H)−/bracketleftbigg E·∂D ∂t+H·∂B ∂t/bracketrightbigg . This is not quite of the form (2.285) since a single term representing the time rate of change of energy density is not present. However, for a linear isotropic medium in which/epsilon1andµdo not depend on time (i.e., a nondispersive medium) we have E·∂D ∂t=/epsilon1E·∂E ∂t=1 2/epsilon1∂ ∂t(E·E)=1 2∂ ∂t(D·E), (2.295) H·∂B ∂t=µH·∂H ∂t=1 2µ∂ ∂t(H·H)=1 2∂ ∂t(H·B). (2.296) Using this we obtain ∇·Sem+∂ ∂tWem=−fem·v=−J·E (2.297) where Wem=1 2(D·E+B·H) and Sem=E×H. (2.298) Equation (2.297) is the point form of the energy conservation theorem, also called Poynt- ing’s theorem after J.H. Poynting who ﬁrst proposed it. The quantity Semgiven in (2.298) is known as the Poynting vector . Integrating (2.297) over a volume and using the divergence theorem, we obtain the large-scale form −/integraldisplay VJ·EdV=/integraldisplay V1 2∂ ∂t(D·E+B·H)dV+/contintegraldisplay S(E×H)·dS. (2.299) This also holds for a nondispersive, linear, bianisotropic medium with a symmetric con- stitutive matrix [101, 185]. We see that the electromagnetic energy conservation theorem (2.297) is identical in form to the mechanical energy conservation theorem (2.285). Thus, if the system is com-posed of just the kinetic and electromagnetic subsystems, the mechanical force exactlybalances the Lorentz force, and (2.297) and (2.285) add to give ∇·(S em+Sk)+∂ ∂t(Wem+Wk)=0, (2.300) showing that energy is conserved for the entire system. As in the mechanical system, we identify Wemas the volume electromagnetic energy density in V, and Semas the density of electromagnetic energy ﬂowing across the bound- ary of V. This interpretation is somewhat controversial, as discussed below. Interpretation of the energyand momentum conservation theorems. There has been some controversy regarding Poynting’s theorem (and, equally, the momentumconservation theorem). While there is no question that Poynting’s theorem is mathe-matically correct, we may wonder whether we are justiﬁed in associating W emwith Wk and Semwith Skmerely because of the similarities in their mathematical expressions. Certainly there is some justiﬁcation for associating Wk, the kinetic energy of particles, with Wem, since we shall show that for static ﬁelds the term1 2(D·E+B·H)represents the energy required to assemble the charges and currents into a certain conﬁguration.However, the term S emis more problematic. In a mechanical system, Skrepresents the ﬂow of kinetic energy associated with moving particles — does that imply that Semrep- resents the ﬂow of electromagnetic energy? That is the position generally taken, and it iswidely supported by experimental evidence. However, the interpretation is not clear-cut. If we associate S emwith the ﬂow of electromagnetic energy at a point in space, then we must deﬁne what a ﬂow of electromagnetic energy is. We naturally associate theﬂow of kinetic energy with moving particles; with what do we associate the ﬂow of electromagnetic energy? Maxwell felt that electromagnetic energy must ﬂow throughspace as a result of the mechanical stresses and strains associated with an unobservedsubstance called the “aether.” A more modern interpretation is that the electromagneticﬁelds propagate as a wave through space at ﬁnite velocity; when those ﬁelds encounter acharged particle a force is exerted, work is done, and energy is “transferred” from the ﬁeldto the particle. Hence the energy ﬂow is associated with the “ﬂow” of the electromagneticwave. Unfortunately, it is uncertain whether E×His the appropriate quantity to associate with this ﬂow, since only its divergence appears in Poynting’s theorem. We could addany other term S /primethat satisﬁes ∇·S/prime=0toSemin (2.297), and the conservation theorem would be unchanged. (Equivalently, we could add to (2.299) any term that integrates tozero over S.) There is no such ambiguity in the mechanical case because kinetic energy is rigorously deﬁned. We are left, then, to postulate that E×Hrepresents the density of energy ﬂow associated with an electromagnetic wave (based on the symmetry with mechanics), and to look to experimental evidence as justiﬁcation. In fact, experimentalevidence does point to the correctness of this hypothesis, and the quantity E×His widely and accurately used to compute the energy radiated by antennas, carried by waveguides,etc. Confusion also arises regarding the interpretation of W em. Since this term is so con- veniently paired with the mechanical volume kinetic energy density in (2.300) it wouldseem that we should interpret it as an electromagnetic energy density. As such, we can think of this energy as “localized” in certain regions of space. This viewpoint has beencriticized [187, 145, 69] since the large-scale form of energy conservation for a space re-gion only requires that the total energy in the region be speciﬁed, and the integrand(energy density) giving this energy is not unique. It is also felt that energy should beassociated with a “conﬁguration” of objects (such as charged particles) and not with anarbitrary point in space. However, we retain the concept of localized energy because itis convenient and produces results consistent with experiment. The validity of extending the static ﬁeld interpretation of 1 2(D·E+B·H) as the energy “stored” by a charge and a current arrangement to the time-varying case has also been questioned. If we do extend this view to the time-varying case, Poynting’stheorem suggests that every point in space somehow has an energy density associated with it, and the ﬂow of energy from that point (via Sem) must be accompanied by a change in the stored energy at that point. This again gives a very useful and intuitivelysatisfying point of view. Since we can associate the ﬂow of energy with the propagationof the electromagnetic ﬁelds, we can view the ﬁelds in any region of space as having thepotential to do work on charged particles in that region. If there are charged particles inthat region then work is done, accompanied by a transfer of energy to the particles anda reduction in the amplitudes of the ﬁelds. We must also remember that the association of stored electromagnetic energy density W emwith the mechanical energy density Wkis only possible if the medium is nondisper- sive. If we cannot make the assumptions that justify (2.295) and (2.296), then Poynting’stheorem must take the form −/integraldisplay VJ·EdV=/integraldisplay V/bracketleftbigg E·∂D ∂t+H·∂B ∂t/bracketrightbigg dV+/contintegraldisplay S(E×H)·dS. (2.301) For dispersive media, the volume term on the right-hand side describes not only the stored electromagnetic energy, but also the energy dissipated within the material produced bya time lag between the ﬁeld applied to the medium and the resulting polarization ormagnetization of the atoms. This is clearly seen in (2.29), which shows that D(t)depends on the value of Eat time tand at all past times. The stored energy and dissipative terms are hard to separate, but we can see that there must always be a stored energy term bysubstituting D=/epsilon1 0E+Pand H=B/µ0−Minto (2.301) to obtain −/integraldisplay V[(J+JP)·E+JH·H]dV= 1 2∂ ∂t/integraldisplay V(/epsilon10E·E+µ0H·H)dV+/contintegraldisplay S(E×H)·dS. (2.302) Here JPis the equivalent polarization current (2.119) and JHis an analogous magnetic polarization current given by JH=µ0∂M ∂t. In this form we easily identify the quantity 1 2(/epsilon10E·E+µ0H·H) as the electromagnetic energy density for the ﬁelds Eand Hin free space . Any dissipa- tion produced by polarization and magnetization lag is now handled by the interactionbetween the ﬁelds and equivalent current, just as J·Edescribes the interaction of the electric current (source and secondary) with the electric ﬁeld. Unfortunately, the equiv-alent current interaction terms also include the additional stored energy that resultsfrom polarizing and magnetizing the material atoms, and again the eﬀects are hard toseparate. Finally, let us consider the case of static ﬁelds. Setting the time derivative to zero in (2.299), we have −/integraldisplay VJ·EdV=/contintegraldisplay S(E×H)·dS. This shows that energy ﬂux is required to maintain steady current ﬂow. For instance, we need both an electromagnetic and a thermodynamic subsystem to account for energyconservation in the case of steady current ﬂow through a resistor. The Poynting ﬂux describes the electromagnetic energy entering the resistor and the thermodynamic ﬂux describes the heat dissipation. For the sum of the two subsystems conservation of energyrequires ∇·(S em+Sth)=−J·E+Pth=0. To compute the heat dissipation we can use Pth=J·E=− ∇· Sem and thus either use the boundary ﬁelds or the ﬁelds and current internal to the resistor to ﬁnd the dissipated heat. Boundaryconditions on the Poy nting vector. The large-scale form of Poynting’s theorem may be used to determine the behavior of the Poynting vector on either sideof a boundary surface. We proceed exactly as in §2.8.2. Consider a surface Sacross which the electromagnetic sources and constitutive parameters are discontinuous ( Figure 2.6).Asbefore,let ˆn 12be the unit normal directed into region 1. We now simplify the notation and write Sinstead of Sem. If we apply Poynting’s theorem /integraldisplay V/parenleftbigg J·E+E·∂D ∂t+H·∂B ∂t/parenrightbigg dV+/contintegraldisplay SS·ndS=0 tothetwoseparat esurface sshowninFigure2.6,weobtain /integraldisplay V/parenleftbigg J·E+E·∂D ∂t+H·∂B ∂t/parenrightbigg dV+/integraldisplay SS·ndS=/integraldisplay S10ˆn12·(S1−S2)dS. (2.303) If on the other hand we apply Poynting’s theorem to the entire volume region including the surface of discontinuity and include the contribution produced by surface current, weget /integraldisplay V/parenleftbigg J·E+E·∂D ∂t+H·∂B ∂t/parenrightbigg dV+/integraldisplay SS·ndS=−/integraldisplay S10Js·EdS. (2.304) Since we are uncertain whether to use E1orE2in the surface term on the right-hand side, if we wish to have the integrals over Vand Sin (2.303) and (2.304) produce identical results we must postulate the two conditions ˆn12×(E1−E2)=0 and ˆn12·(S1−S2)=−Js·E. (2.305) The ﬁrst condition is merely the continuity of tangential electric ﬁeld as originally postu- lated in §2.8.2; it allows us to be nonspeciﬁc as to which value of Ewe use in the second condition, which is the desired boundary condition on S. It is interesting to note that (2.305) may also be derived directly from the two pos- tulated boundary conditions on tangential Eand H. Here we write with the help of (B.6) ˆn12·(S1−S2)=ˆn12·(E1×H1−E2×H2)=H1·(ˆn12×E1)−H2·(ˆn12×E2). Since ˆn12×E1=ˆn12×E2=ˆn12×E,w eh a v e ˆn12·(S1−S2)=(H1−H2)·(ˆn12×E)=[−ˆn12×(H1−H2)]·E. Finally, using ˆn12×(H1−H2)=Jswe arrive at (2.305). The arguments above suggest an interesting way to look at the boundary conditions. Once we identify Swith the ﬂow of electromagnetic energy, we may consider the condition on normal Sas a fundamental statement of the conservation of energy. This statement implies continuity of tangential Ein order to have an unambiguous interpretation for the meaning of the term Js·E. Then, with continuity of tangential Eestablished, we can derive the condition on tangential Hdirectly. An alternative formulation of the conservation theorems. As we saw in the paragraphs above, our derivation of the conservation theorems lacks strong motivation.We manipulated Maxwell’s equations until we found expressions that resembled thosefor mechanical momentum and energy, but in the process found that the validity of theexpressions is somewhat limiting. For instance, we needed to assume a linear, homoge-neous, bianisotropic medium in order to identify the Maxwell stress tensor (2.288) and the energy densities in Poynting’s theorem (2.299). In the end, we were reduced to pos-tulating the meaning of the individual terms in the conservation theorems in order forthe whole to have meaning. An alternative approach is popular in physics. It involves postulating a single La- grangian density function for the electromagnetic ﬁeld, and then applying the stationaryproperty of the action integral. The results are precisely the same conservation expres-sions for linear momentum and energy as obtained from manipulating Maxwell’s equa-tions (plus the equation for conservation of angular momentum), obtained with fewerrestrictions regarding the constitutive relations. This process also separates the storedenergy, Maxwell stress tensor, momentum density, and Poynting vector as natural com-ponents of a tensor equation, allowing a better motivated interpretation of the meaningof these components. Since this approach is also a powerful tool in mechanics, its ap-plication is more strongly motivated than merely manipulating Maxwell’s equations. Ofcourse, some knowledge of the structure of the electromagnetic ﬁeld is required to providean appropriate postulate of the Lagrangian density. Interested readers should consultKong [101], Jackson [91], Doughty [57], or Tolstoy [198]. 2.10 The wavenature of the electromagnetic ﬁeld Throughout this chapter our goal has been a fundamental understanding of Maxwell’s theory of electromagnetics. We have concentrated on developing and understanding theequations relating the ﬁeld quantities, but have done little to understand the nature ofthe ﬁeld itself. We would now like to investigate, in a very general way, the behaviorof the ﬁeld. We shall not attempt to solve a vast array of esoteric problems, but shallinstead concentrate on a few illuminating examples. The electromagnetic ﬁeld can take on a wide variety of characteristics. Static ﬁelds diﬀer qualitatively from those which undergo rapid time variations. Time-varying ﬁeldsexhibit wave behavior and carry energy away from their sources. In the case of slowtime variation this wavenature may often be neglected in favor of the nearby coupling of sources we know as the inductance eﬀect, hence circuit theory may suﬃce to describethe ﬁeld-source interaction. In the case of extremely rapid oscillations, particle conceptsmay be needed to describe the ﬁeld. The dynamic coupling between the various ﬁeld vectors in Maxwell’s equations provides a means of characterizing the ﬁeld. Static ﬁelds are characterized by decoupling of theelectric and magnetic ﬁelds. Quasistatic ﬁelds exhibit some coupling, but the wavecharacteristic of the ﬁeld is ignored. Tightly coupled ﬁelds are dominated by the waveeﬀect, but may still show a static-like spatial distribution near the source. Any such“near-zone” eﬀects are generally ignored for ﬁelds at light-wave frequencies, and theparticle nature of light must often be considered. 2.10.1 Electromagnetic waves An early result of Maxwell’s theory was the prediction and later veriﬁcation by Heinrich Hertz of the existence of electromagnetic waves. We now know that nearly any time-varying source produces waves, and that these waves have certain important properties. An electromagnetic wave is a propagating electromagnetic ﬁeld that travels with ﬁnite velocity as a disturbance through a medium. The ﬁeld itself is the disturbance, ratherthan merely representing a physical displacement or other eﬀect on the medium. This factis fundamental for understanding how electromagnetic waves can travel through a truevacuum. Many speciﬁc characteristics of the wave, such as velocity and p olarization, depend on the properties of the medium through which it propagates. The evolutionof the disturbance also depends on these properties: we say that a material exhibits“dispersion” if the disturbance undergoes a change in its temporal behavior as the waveprogresses. As waves travel they carry energy and momentum away from their source.This energy may be later returned to the source or delivered to some distant location.Waves are also capable of transferring energy to, or withdrawing energy from, the mediumthrough which they propagate. When energy is carried outward from the source neverto return, we refer to the process as “electromagnetic radiation.” The eﬀects of radiatedﬁelds can be far-reaching; indeed, radio astronomers observe waves that originated at thevery edges of the universe. Light is an electromagnetic phenomenon, and many of the familiar characteristics of light that we recognize from our everyday experience may be applied to all electromag-netic waves. For instance, radio waves bend (or “refract”) in the ionosphere much aslightwaves bend while passing through a prism. Micr owaves reﬂect from conducting sur- faces in the same way that light waves reﬂect from a mirror; detecting these reﬂectionsforms the basis of radar. Electromagnetic waves may also be “conﬁned” by reﬂectingboundaries to form waves standing in one or more directions. With this concept we can use waveguides or transmission lines to guide electromagnetic energy from spot to spot,or to concentrate it in the cavity of a microwave oven. The manifestations of electromagnetic waves are so div erse that no one book can possibly describe the entire range of phenomena or application. In this section we shallmerely introduce the reader to some of the most fundamental concepts of electromagneticwave behavior. In the process we shall also introduce the three most often studied typesof traveling electromagnetic waves: plane waves, spherical waves, and cylindrical waves. In later sections we shall study some of the complicated interactions of these waves withobjects and boundaries, in the form of guided waves and scattering problems. Mathematically, electromagnetic waves arise as a subset of solutions to Maxwell’s equa- tions. These solutions obey the electromagnetic “wave equation,” which may be derivedfrom Maxwell’s equations under certain circumstances. Not all electromagnetic ﬁeldssatisfy the wave equation. Obviously, time-invariant ﬁelds cannot represent evolvingwave disturbances, and must obey the static ﬁeld equations. Time-varying ﬁelds in cer- tain metals may obey the diﬀusion equation rather than the wave equation, and must thereby exhibit diﬀerent behavior. In the study of quasistatic ﬁelds we often ignore thedisplacement current term in Maxwell’s equations, producing solutions that are mostimportant near the sources of the ﬁelds and having little associated radiation. When thedisplacement term is signiﬁcant we produce solutions with the properties of waves. 2.10.2 Wave equation for bianisotropic materials In deriving electromagnetic wave equations we transform the ﬁrst-order coupled par- tial diﬀerential equations we know as Maxwell’s equations into uncoupled second-orderequations. That is, we perform a set of operations (and make appropriate assumptions)to reduce the set of four diﬀerential equations in the four unknown ﬁelds E,D,B, and H, into a set of diﬀerential equations each involving a single unknown (usually Eor H). It is possible to derive wave equations for Eand Heven for the most general cases of inhomogeneous, bianisotropic media, as long as the constitutive parameters ¯µand ¯ξare constant with time. Substituting the constitutive relations (2.19)–(2.20) into the Maxwell–Minkowski curl equations (2.169)–(2.170) we get ∇× E=−∂ ∂t(¯ζ·E+¯µ·H)−Jm, (2.306) ∇× H=∂ ∂t(¯/epsilon1·E+¯ξ·H)+J. (2.307) Separate equations for Eand Hare facilitated by introducing a new dyadic operator ¯∇, which when dotted with a vector ﬁeld Vgives the curl: ¯∇·V=∇× V. (2.308) It is easy to verify that in rectangular coordinates ¯∇is [¯∇]= 0−∂/∂z∂/∂y ∂/∂z 0−∂/∂x −∂/∂y∂/∂x 0 . With this notation, Maxwell’s curl equations (2.306)–(2.307) become simply /parenleftbigg ¯∇+∂ ∂t¯ζ/parenrightbigg ·E=−∂ ∂t¯µ·H−Jm, (2.309) /parenleftbigg ¯∇−∂ ∂t¯ξ/parenrightbigg ·H=∂ ∂t¯/epsilon1·E+J. (2.310) Obtaining separate equations for Eand His straightforward. Deﬁning the inverse dyadic ¯µ−1through ¯µ·¯µ−1=¯µ−1·¯µ=¯I, we can write (2.309) as ∂ ∂tH=− ¯µ−1·/parenleftbigg ¯∇+∂ ∂t¯ζ/parenrightbigg ·E−¯µ−1·Jm (2.311) where we have assumed that ¯µis independent of time. Assuming that ¯ξis also indepen- dent of time, we can diﬀerentiate (2.310) with respect to time to obtain /parenleftbigg ¯∇−∂ ∂t¯ξ/parenrightbigg ·∂H ∂t=∂2 ∂t2(¯/epsilon1·E)+∂J ∂t. Substituting ∂H/∂tfrom (2.311) and rearranging, we get /bracketleftbigg/parenleftbigg ¯∇−∂ ∂t¯ξ/parenrightbigg ·¯µ−1·/parenleftbigg ¯∇+∂ ∂t¯ζ/parenrightbigg +∂2 ∂t2¯/epsilon1/bracketrightbigg ·E=−/parenleftbigg ¯∇−∂ ∂t¯ξ/parenrightbigg ·¯µ−1·Jm−∂J ∂t. (2.312) This is the general waveequation for E. Using an analogous set of steps, and assuming ¯/epsilon1and ¯ζare independent of time, we can ﬁnd /bracketleftbigg/parenleftbigg ¯∇+∂ ∂t¯ζ/parenrightbigg ·¯/epsilon1−1·/parenleftbigg ¯∇−∂ ∂t¯ξ/parenrightbigg +∂2 ∂t2¯µ/bracketrightbigg ·H=/parenleftbigg ¯∇+∂ ∂t¯ζ/parenrightbigg ·¯/epsilon1−1·J−∂Jm ∂t. (2.313) This is the wave equation for H. The case in which the constitutive parameters are time-dependent will be handled using frequency domain techniques in later chapters. Wave equations for anisotropic, isotropic, and homogeneous media are easily obtained from (2.312) and (2.313) as special cases. For example, the wave equations for a homo-geneous, isotropic medium can be found by setting ¯ζ=¯ξ=0,¯µ=µ¯I, and ¯/epsilon1=/epsilon1¯I: 1 µ¯∇·(¯∇·E)+/epsilon1∂2E ∂t2=−1 µ¯∇·Jm−∂J ∂t, 1 /epsilon1¯∇·(¯∇·H)+µ∂2H ∂t2=1 /epsilon1¯∇·J−∂Jm ∂t. Returning to standard curl notation we ﬁnd that these become ∇×(∇× E)+µ/epsilon1∂2E ∂t2=− ∇× Jm−µ∂J ∂t, (2.314) ∇×(∇× H)+µ/epsilon1∂2H ∂t2=∇× J−/epsilon1∂Jm ∂t. (2.315) In each of the wave equations it appears that operations on the electromagnetic ﬁelds have been separated from operations on the source terms. However, we have not yetinvoked any coupling between the ﬁelds and sources associated with secondary interac-tions. That is, we need to separate the impressed sources, which are independent ofthe ﬁelds they source, with secondary sources resulting from interactions between thesourced ﬁelds and the medium in which the ﬁelds exist. The simple case of an isotropicconducting medium will be discussed below. Wave equation using equivalent sources. An alternative approach for studying wave behavior in general media is to use the Maxwell–Boﬃ form of the ﬁeld equations ∇× E=−∂B ∂t, (2.316) ∇×B µ0=(J+JM+JP)+∂/epsilon10E ∂t, (2.317) ∇·(/epsilon10E)=(ρ+ρP), (2.318) ∇·B=0. (2.319) Taking the curl of (2.316) we have ∇×(∇× E)=−∂ ∂t∇× B. Substituting for ∇× Bfrom (2.317) we then obtain ∇×(∇× E)+µ0/epsilon10∂2E ∂t2=−µ0∂ ∂t(J+JM+JP), (2.320) which is the wave equation for E. Taking the curl of (2.317) and substituting from (2.316) we obtain the waveequation ∇×(∇× B)+µ0/epsilon10∂2B ∂t2=µ0∇×(J+JM+JP) (2.321) forB. Solution of the wave equations is often facilitated by writing the curl-curl operation in terms of the vector Laplacian. Using (B.47), and substituting for the divergence from(2.318) and (2.319), we can write the wave equations as ∇ 2E−µ0/epsilon10∂2E ∂t2=1 /epsilon10∇(ρ+ρP)+µ0∂ ∂t(J+JM+JP), (2.322) ∇2B−µ0/epsilon10∂2B ∂t2=−µ0∇×(J+JM+JP). (2.323) The simplicity of these equations relative to (2.312) and (2.313) is misleading. We have not considered the constitutive equations relating the polarization Pand magnetization Mto the ﬁelds, nor have we considered interactions leading to secondary sources. 2.10.3 Wave equation in a conducting medium As an example of the type of wave equation that arises when secondary sources are included, consider a homogeneous isotropic conducting medium described by permittivity/epsilon1, permeability µ, and conductivity σ. In a conducting medium we must separate the source ﬁeld into a causative impressed term J ithat is independent of the ﬁelds it sources, and a secondary term Jsthat is an eﬀect of the sourced ﬁelds. In an isotropic conducting medium the eﬀect is described by Ohm’s law Js=σE. Writing the total current as J=Ji+Js, and assuming that Jm=0, we write the wave equation (2.314) as ∇×(∇× E)+µ/epsilon1∂2E ∂t2=−µ∂(Ji+σE) ∂t. (2.324) Using (B.47) and substituting ∇·E=ρ//epsilon1, we can write the waveequation for Eas ∇2E−µσ∂E ∂t−µ/epsilon1∂2E ∂t2=µ∂Ji ∂t+1 /epsilon1∇ρ. (2.325) Substituting J=Ji+σEinto (2.315) and using (B.47), we obtain ∇(∇·H)−∇2H+µ/epsilon1∂2H ∂t2=∇× Ji+σ∇× E. Since ∇× E=−∂B/∂tand∇·H=∇· B/µ=0,w eh a v e ∇2H−µσ∂H ∂t−µ/epsilon1∂2H ∂t2=− ∇× Ji. (2.326) This is the wave equation for H. 2.10.4 Scalar waveequation for a conducting medium In many applications, particularly those involving planar boundary surfaces, it is convenient to decompose the vector wave equation into cartesian components. Using∇ 2V=ˆx∇2Vx+ˆy∇2Vy+ˆz∇2Vzin (2.325) and in (2.326), we ﬁnd that the rectangular components of Eand Hmust obey the scalar wave equation ∇2ψ(r,t)−µσ∂ψ(r,t) ∂t−µ/epsilon1∂2ψ(r,t) ∂t2=s(r,t). (2.327) For the electric ﬁeld waveequation we have ψ=Eα, s=µ∂Ji α ∂t+1 /epsilon1ˆα·∇ρ, where α=x,y,z. For the magnetic ﬁeld wave equations we have ψ=Hα, s=ˆα·(−∇ × Ji). 2.10.5 Fields determined byMaxwell’s equations vs. ﬁelds deter- mined bythe waveequation Although we derive the wave equations directly from Maxwell’s equations, we may wonder whether the solutions to second-order diﬀerential equations such as (2.314)–(2.315) are necessarily the same as the solutions to the ﬁrst-order Maxwell equations.Hansen and Yaghjian [81] show that if all information about the ﬁelds is supplied by thesources J(r,t)andρ(r,t), rather than by speciﬁcation of ﬁeld values on boundaries, the solutions to Maxwell’s equations and the wave equations are equivalent as long as thesecond derivatives of the quantities ∇·E(r,t)−ρ(r,t)//epsilon1, ∇·H(r,t), are continuous functions of rand t. If boundary values are supplied in an attempt to guarantee uniqueness, then solutions to the wave equation and to Maxwell’s equationsmay diﬀer. This is particularly important when comparing numerical solutions obtaineddirectly from Maxwell’s equations (using the FDTD method, say) to solutions obtainedfrom the waveequation. “Spurious” solutions having no physical signiﬁcance are a con- tinual plague for engineers who employ numerical techniques. The interested readershould see Jiang [94]. We note that these conclusions do not hold for static ﬁelds. The conditions for equiv- alence of the ﬁrst-order and second-order static ﬁeld equations are considered in §3.2.4. 2.10.6 Transient uniform plane waves in a conducting medium We can learn a great deal about the wave nature of the electromagnetic ﬁeld by solving the wave equation (2.325) under simple circumstances. In Chapter 5 we shall solve forthe ﬁeld produced by an arbitrary distribution of impressed sources, but here we seek asimple solution to the homogeneous form of the equation. This allows us to study thephenomenology of wave propagation without worrying about the consequences of speciﬁcsource functions. We shall also assume a high degree of symmetry so that we are notbogged down in details about the vector directions of the ﬁeld components. We seek a solution of the wave equation in which the ﬁelds are invariant over a chosen planar surface. The resulting ﬁelds are said to comprise a uniform plane wave . Although we can envision a uniform plane wave as being created by a uniform surface source of doubly-inﬁnite extent, plane waves are also useful as models for spherical waves overlocalized regions of the wavefront. We choose the plane of ﬁeld invariance to be the xy-plane and later generalize the resulting solution to any planar surface by a simple rotation of the coordinate axes. Sincethe ﬁelds vary with zonly we choose to write the wave equation (2.325) in rectangular coordinates, giving for a source-free region of space 4 ˆx∂2Ex(z,t) ∂z2+ˆy∂2Ey(z,t) ∂z2+ˆz∂2Ez(z,t) ∂z2−µσ∂E(z,t) ∂t−µ/epsilon1∂2E(z,t) ∂t2=0.(2.328) If we return to Maxwell’s equations, we soon ﬁnd that not all components of Eare present in the plane- wavesolution. Faraday’s law states that ∇× E(z,t)=− ˆx∂Ey(z,t) ∂z+ˆy∂Ex(z,t) ∂z=ˆz×∂E(z,t) ∂z=−µ∂H(z,t) ∂t. (2.329) We see that ∂Hz/∂t=0, hence Hzmust be constant with respect to time. Because a nonzero constant ﬁeld component would not exhibit wave-like behavior, we can only have Hz=0in our wave solution. Similarly, Ampere’s law in a homogeneous conducting region free from impressed sources states that ∇× H(z,t)=J+∂D(z,t) ∂t=σE(z,t)+/epsilon1∂E(z,t) ∂t or −ˆx∂Hy(z,t) ∂z+ˆy∂Hx(z,t) ∂z=ˆz×∂H(z,t) ∂z=σE(z,t)+/epsilon1∂E(z,t) ∂t. (2.330) This implies that σEz(z,t)+/epsilon1∂Ez(z,t) ∂t=0, which is a diﬀerential equation for Ezwith solution Ez(z,t)=E0(z)e−σ /epsilon1t. Since we are interested only in wave-type solutions, we choose Ez=0. Hence Ez=Hz=0, and thus both Eand Hare perpendicular to the z-direction. Using (2.329) and (2.330), we also see that ∂ ∂t(E·H)=E·∂H ∂t+H·∂E ∂t =−1 µE·/parenleftbigg ˆz×∂E ∂z/parenrightbigg −H·/parenleftBigσ /epsilon1E/parenrightBig +1 /epsilon1H·/parenleftbigg ˆz×∂H ∂z/parenrightbigg or/parenleftbigg∂ ∂t+σ /epsilon1/parenrightbigg (E·H)=1 µˆz·/parenleftbigg E×∂E ∂z/parenrightbigg −1 /epsilon1ˆz·/parenleftbigg H×∂H ∂z/parenrightbigg . We seek solutions of the type E(z,t)=ˆpE(z,t)and H(z,t)=ˆqH(z,t), where ˆpand ˆqare constant unit vectors. Under this condition we have E×∂E/∂z=0and H×∂H/∂z=0, giving /parenleftbigg∂ ∂t+σ /epsilon1/parenrightbigg (E·H)=0. 4The term “source free” applied to a conducting region implies that the region is devoid of impressed sources and, because of the relaxation eﬀect, has no free charge. See the discussion in Jones [97]. Thus we also have E·H=0, and ﬁnd that Emust be perpendicular to H.S o E,H, and ˆzcomprise a mutually orthogonal triplet of vectors. A wave having this property is said to be TEM to the z-direction or simply TEM z. Here “TEM” stands for transverse electromagnetic , indicating the orthogonal relationship between the ﬁeld vectors and the z-direction. Note that ˆp×ˆq=± ˆz. The constant direction described by ˆpis called the polarization of the plane wave. We are now ready to solve the source-free wave equation (2.328). If we dot both sides of the homogeneous expression by ˆpwe obtain ˆp·ˆx∂2Ex ∂z2+ˆp·ˆy∂2Ey ∂z2−µσ∂(ˆp·E) ∂t−µ/epsilon1∂2(ˆp·E) ∂t2=0. Noting that ˆp·ˆx∂2Ex ∂z2+ˆp·ˆy∂2Ey ∂z2=∂2 ∂z2(ˆp·ˆxEx+ˆp·ˆyEy)=∂2 ∂z2(ˆp·E), we have the wave equation ∂2E(z,t) ∂z2−µσ∂E(z,t) ∂t−µ/epsilon1∂2E(z,t) ∂t2=0. (2.331) Similarly, dotting both sides of (2.326) with ˆqand setting Ji=0we obtain ∂2H(z,t) ∂z2−µσ∂H(z,t) ∂t−µ/epsilon1∂2H(z,t) ∂t2=0. (2.332) In a source-free homogeneous conducting region EandHsatisfy identical waveequations. Solutions are considered in §A.1. There we solve for the total ﬁeld for all z,tgiven the value of the ﬁeld and its derivative over the z=0plane. This solution can be directly applied to ﬁnd the total ﬁeld of a plane wave reﬂected by a perfect conductor.Let us begin by considering the lossless case where σ=0, and assuming the region z<0 contains a perfect electric conductor. The conditions on the ﬁeld in the z=0plane are determined by the required boundary condition on a perfect conductor: the tangentialelectric ﬁeld must vanish. From (2.330) we see that since E⊥ˆz, requiring ∂H(z,t) ∂z/vextendsingle/vextendsingle/vextendsingle/vextendsingle z=0=0 (2.333) gives E(0,t)=0and thus satisﬁes the boundary condition. Writing H(0,t)=H0f(t),∂H(z,t) ∂z/vextendsingle/vextendsingle/vextendsingle/vextendsingle z=0=H0g(t)=0, (2.334) and setting /Omega1=0in (A.41) we obtain the solution to (2.332): H(z,t)=H0 2f/parenleftBig t−z v/parenrightBig +H0 2f/parenleftBig t+z v/parenrightBig , (2.335) where v=1/(µ/epsilon1)1/2. Since we designate the vector direction of Hasˆq, the vector ﬁeld is H(z,t)=ˆqH0 2f/parenleftBig t−z v/parenrightBig +ˆqH0 2f/parenleftBig t+z v/parenrightBig . (2.336) Figure 2.7: Propagation of a transient plane wave in a lossless medium. From (2.329) we also have the solution for E(z,t): E(z,t)=ˆpvµH0 2f/parenleftBig t−z v/parenrightBig −ˆpvµH0 2f/parenleftBig t+z v/parenrightBig , (2.337) where ˆp×ˆq=ˆz. The boundary conditions E(0,t)=0and H(0,t)=H0f(t)are easily veriﬁed by substi- tution. This solution displays the quintessential behavior of electromagnetic waves. We may interpret the term f(t+z/v)as a wave ﬁeld disturbance, propagating at velocity vin the −z-direction, incident from z>0upon the conductor. The term f(t−z/v)represents a wave ﬁeld disturbance propagating in the +z-direction with velocity v, reﬂected from the conductor. By “propagating” we mean that if we increment time, the disturbancewill occupy a spatial position determined by incrementing zbyvt. For free space where v=1/(µ 0/epsilon10)1/2, the velocity of propagation is the speed of light c. A speciﬁc example should serve to clarify our interpretation of the wave solution. Taking µ=µ0and/epsilon1=81/epsilon10, representing typical constitutive values for fresh water, we can plot (2.335) as a function of position for ﬁxed values of time. The result is shown inFigure2.7,wherewehavechosen f(t)=rect(t/τ) (2.338) withτ=1µs. We see that the disturbance is spatially distributed as a rectangular pulse of extent L=2vτ=66.6m, where v=3.33×10 7m/s is the wave velocity, and where 2τis the temporal duration of the pulse. At t=− 8µs the leading edge of the pulse is at z=233m, while at −4µs the pulse has traveled a distance z=vt= (3.33×107)×(4×10−6)=133m in the −z-direction, and the leading edge is thus at 100m. At t=− 1µs the leading edge strikes the conductor and begins to induce a current in the conductor surface. This current sets up the reﬂected wave, which begins to travel in the opposite ( +z) direction. At t=−0.5µs a portion of the wave has b egun to travel in the +z-direction while the trailing portion of the disturbance continues to travel in the −z-direction. At t=1µs the wave has been completely reﬂected from the surface, and thus consists only of the component traveling in the +z-direction. Note that if we plot the total ﬁeld in the z=0plane, the sum of the forward and backward traveling waves pro duces the pulse waveform (2.338) as expected. Using the expressions for Eand Hwe can determine many interesting characteristics of the wave. We see that the terms f(t±z/v)represent the components of the waves traveling in the ∓z-directions, respectively. If we were to isolate these waves from each other (by, for instance, measuring them as functions of time at a position where they do not overlap) we would ﬁnd from (2.336) and (2.337) that the ratio of EtoHf o raw a v e traveling in either direction is /vextendsingle/vextendsingle/vextendsingle/vextendsingleE(z,t) H(z,t)/vextendsingle/vextendsingle/vextendsingle/vextendsingle=vµ=(µ//epsilon1) 1/2, independent of the time and position of the measurement. This ratio, denoted by ηand carrying units of ohms, is called the intrinsic impedance of the medium through which the wave propagates. Thus, if we let E0=ηH0we can write E(z,t)=ˆpE0 2f/parenleftBig t−z v/parenrightBig −ˆpE0 2f/parenleftBig t+z v/parenrightBig . (2.339) We can easily determine the current induced in the conductor by applying the boundary condition (2.200): Js=ˆn×H|z=0=ˆz×[H0ˆqf(t)]=− ˆpH0f(t). (2.340) We can also determine the pressure exerted on the conductor due to the Lorentz force interaction between the ﬁelds and the induced current. The total force on the conductorcan be computed by integrating the Maxwell stress tensor (2.288) over the xy-plane 5: Fem=−/integraldisplay S¯Tem·dS. The surface traction is t=¯Tem·ˆn=/bracketleftbigg1 2(D·E+B·H)¯I−DE−BH/bracketrightbigg ·ˆz. Since Eand Hare both normal to ˆz, the last two terms in this expression are zero. Also, the boundary condition on Eimplies that it vanishes in the xy-plane. Thus t=1 2(B·H)ˆz=ˆzµ 2H2(t). 5We may neglect the momentum term in (2.291), which is small compared to the stress tensor term. See Problem 2.20. With H0=E0/ηwe have t=ˆzE2 0 2η2µf2(t). (2.341) As a numerical example, consider a high-altitude nuclear electromagnetic pulse (HEMP) generated by the explosion of a large nuclear weapon in the upper atmosphere. Suchan explosion could generate a transient electromagnetic wave of short (sub-microsecond) duration with an electric ﬁeld amplitude of 50,000V/m in air [200]. Using (2.341), we ﬁnd that the wave would exert a peak pressure of P=|t|=.011Pa=1.6×10 −6 lb/in2if reﬂected from a perfect conductor at normal incidence. Obviously, even for this extreme ﬁeld level the pressure produced by a transient electromagnetic wave is quitesmall. However, from (2.340) we ﬁnd that the current induced in the conductor wouldhave a peak value of 133A/m. Even a small portion of this current could destroy a sensitive electronic circuit if it were to leak through an opening in the conductor. This isan important concern for engineers designing circuitry to be used in high-ﬁeld environ- ments, and demonstrates why the concepts of current and voltage can often supersedethe concept of force in terms of importance. Finally, let us see how the terms in the Poynting power balance theorem relate. Con- sider a cubic region Vbounded by the planes z=z 1and z=z2,z2>z1. We choose the ﬁeld waveform f(t)and locate the planes so that we can isolate either the forward or backward traveling wave. Since there is no current in V, Poynting’s theorem (2.299) becomes 1 2∂ ∂t/integraldisplay V(/epsilon1E·E+µH·H)dV=−/contintegraldisplay S(E×H)·dS. Consider the wave traveling in the −z-direction. Substitution from (2.336) and (2.337) gives the time-rate of change of stored energy as Scube(t)=1 2∂ ∂t/integraldisplay V/bracketleftbig /epsilon1E2(z,t)+µH2(z,t)/bracketrightbig dV =1 2∂ ∂t/integraldisplay x/integraldisplay ydx dy/integraldisplayz2 z1/bracketleftbigg /epsilon1(vµ)2H2 0 4f2/parenleftBig t+z v/parenrightBig +µH2 0 4f2/parenleftBig t+z v/parenrightBig/bracketrightbigg dz =1 2∂ ∂tµH2 0 2/integraldisplay x/integraldisplay ydx dy/integraldisplayz2 z1f2/parenleftBig t+z v/parenrightBig dz. Integration over xand ygives the area Aof the cube face. Putting u=t+z/vwe see that S=AµH2 0 4∂ ∂t/integraldisplayt+z2/v t+z1/vf2(u)vdu. Leibnitz’ rule for diﬀerentiation (A.30) then gives Scube(t)=AµvH2 0 4/bracketleftBig f2/parenleftBig t+z2 v/parenrightBig −f2/parenleftBig t+z1 v/parenrightBig/bracketrightBig . (2.342) Again substituting for E(t+z/v)and H(t+z/v)we can write Scube(t)=−/contintegraldisplay S(E×H)·dS =−/integraldisplay x/integraldisplay yvµH2 0 4f2/parenleftBig t+z1 v/parenrightBig (−ˆp×ˆq)·(−ˆz)dx dy − −/integraldisplay x/integraldisplay yvµH2 0 4f2/parenleftBig t+z2 v/parenrightBig (−ˆp×ˆq)·(ˆz)dx dy. Figure 2.8: Propagation of a transient plane wave in a dissipative medium. The second term represents the energy change in Vproduced by the backward traveling wave entering the cube by passing through the plane at z=z2, while the ﬁrst term represents the energy change in Vproduced by the wave exiting the cube by passing through the plane z=z1. Contributions from the sides, top, and bottom are zero since E×His perpendicular to ˆnover those surfaces. Since ˆp×ˆq=ˆz,w eg e t Scube(t)=AµvH2 0 4/bracketleftBig f2/parenleftBig t+z2 v/parenrightBig −f2/parenleftBig t+z1 v/parenrightBig/bracketrightBig , which matches (2.342) and thus veriﬁes Poynting’s theorem. We may interpret this result as follows. The propagating electromagnetic disturbance carries energy through space.The energy within any region is associated with the ﬁeld in that region, and can changewith time as the propagating wave carries a ﬂux of energy across the boundary of theregion. The energy continues to propagate even if the source is changed or is extinguishedaltogether. That is, the behavior of the leading edge of the disturbance is determinedby causality — it is aﬀected by obstacles it encounters, but not by changes in the sourcethat occur after the leading edge has been established. When propagating through a dissipative region a plane wave takes on a somewhat diﬀerent character. Again applying the conditions (2.333) and (2.334), we obtain from(2.991) the solution to the wave equation (2.332): H(z,t)=H 0 2e−/Omega1 vzf/parenleftBig t−z v/parenrightBig +H0 2e/Omega1 vzf/parenleftBig t+z v/parenrightBig − −z/Omega12H0 2ve−/Omega1t/integraldisplayt+z v t−z vf(u)e/Omega1uJ1/parenleftBig /Omega1 v/radicalbig z2−(t−u)2v2/parenrightBig /Omega1 v/radicalbig z2−(t−u)2v2du (2.343) where /Omega1=σ/2/epsilon1. The ﬁrst two terms resemble those for the lossless case, modiﬁed by an exponential damping factor. This accounts for the loss in amplitude that mustaccompa nythetransfe rofenergyfromthepropagatin gwavetojouleloss(heat)within the conducting medium. The remaining term appears only when the medium is lossy, andresults in an extension of the disturbance through the medium because of the currentsinduced by the passing wavefront. This “wake” follows the leading edge of the disturbanceasisshownclearlyinFigure2.8.Herewehaverepeatedthecalculatio nofFigure2.7, but with σ=2×10 −4, approximating the conductivity of fresh water. As the wave travels to the left it attenuates and leaves a trailing remnant behind. Upon reachingthe conductor it reﬂects much as in the lossless case, resulting in a time dependence at z=0given by the ﬁnite-duration rectangular pulse (2.338). In order for the pulse to be of ﬁnite duration, the wake left by the reﬂected pulse must exactly cancel the wakeassociated with the incident pulse that continues to arrive after the reﬂection. As thereﬂected pulse sweeps forward, the wake is obliterated everywhere behind. If we were to verify the Poynting theorem for a dissipative medium (which we shall not attempt because of the complexity of the computation), we would need to includetheE·Jterm. Here Jis the induced conduction current and the integral of E·Jaccounts for the joule loss within a region Vbalanced by the diﬀerence in Poynting energy ﬂux carried into and out of V. Once we have the ﬁelds for a wave propagating along the z-direction, it is a simple matter to generalize these results to any propagation direction. Assume that ˆuis normal to the surface of a plane over which the ﬁelds are invariant. Then u=ˆu·rdescribes the distance from the origin along the direction ˆu. We need only replace zbyˆu·rin any of the expressions obtained above to determine the ﬁelds of a plane wave propagating intheu-direction. We must also replace the orthogonality condition ˆp×ˆq=ˆzwith ˆp×ˆq=ˆu. For instance, the ﬁelds associated with a wavepropagating through a lossless medium in the positive u-direction are, from (2.336)–(2.337), H(r,t)=ˆqH 0 2f/parenleftbigg t−ˆu·r v/parenrightbigg , E(r,t)=ˆpvµH0 2f/parenleftbigg t−ˆu·r v/parenrightbigg . 2.10.7 Propagation of cylindrical waves in a lossless medium Much as we envisioned a uniform plane wave arising from a uniform planar source, we can imagine a uniform cylindrical wavearising from a uniform line source. Although this line source must be inﬁnite in extent, uniform cylindrical waves (unlike plane waves) dis-play the physical behavior of diverging from their source while carrying energy outwardsto inﬁnity. Auniform cylindrical wave has ﬁelds that are invariant over a cylindrical surface: E(r,t)=E(ρ,t),H(r,t)=H(ρ,t). For simplicity, we shall assume that waves propagate in a homogeneous, isotropic, linear, and lossless medium described by permittivity /epsilon1 and permeability µ. From Maxwell’s equations we ﬁnd that requiring the ﬁelds to be independent of φand zputs restrictions on the remaining vector components. Faraday’s law states ∇× E(ρ,t)=− ˆφ∂Ez(ρ,t) ∂ρ+ˆz1 ρ∂ ∂ρ[ρEφ(ρ,t)]=−µ∂H(ρ,t) ∂t. (2.344) Equating components we see that ∂Hρ/∂t=0, and because our interest lies in wave solutions we take Hρ=0. Ampere’s law in a homogeneous lossless region free from impressed sources states in a similar manner ∇× H(ρ,t)=− ˆφ∂Hz(ρ,t) ∂ρ+ˆz1 ρ∂ ∂ρ[ρHφ(ρ,t)]=/epsilon1∂E(ρ,t) ∂t. (2.345) Equating components we ﬁnd that Eρ=0. Since Eρ=Hρ=0, both Eand Hare perpendicular to the ρ-direction. Note that if there is only a z-component of Ethen there is only a φ-component of H. This case, termed electric polarization , results in ∂Ez(ρ,t) ∂ρ=µ∂Hφ(ρ,t) ∂t. Similarly, if there is only a z-component of Hthen there is only a φ-component of E. This case, termed magnetic polarization , results in −∂Hz(ρ,t) ∂ρ=/epsilon1∂Eφ(ρ,t) ∂t. Since E=ˆφEφ+ˆzEzand H=ˆφHφ+ˆzHz, we can always decompose a cylindrical electromagnetic wave into cases of electric and magnetic polarization. In each case the resulting ﬁeld is TEM ρsince the vectors E,H,ˆρare mutually orthogonal. Wave equations for Ezin the electric polarization case and for Hzin the magnetic polarization case can be found in the usual manner. Taking the curl of (2.344) andsubstituting from (2.345) we ﬁnd ∇×(∇× E)=− ˆz1 ρ∂ ∂ρ/parenleftbigg ρ∂Ez ∂ρ/parenrightbigg −ˆφ∂ ∂ρ/parenleftbigg1 ρ∂ ∂ρ[ρEφ]/parenrightbigg =−1 v2∂2E ∂t2=−1 v2/bracketleftbigg ˆz∂2Ez ∂t2+ˆφ∂2Eφ ∂t2/bracketrightbigg where v=1/(µ/epsilon1)1/2. Noting that Eφ=0for the electric polarization case we obtain the wave equation for Ez. A similar set of steps beginning with the curl of (2.345) gives an identical equation for Hz.T h u s 1 ρ∂ ∂ρ/parenleftbigg ρ∂ ∂ρ/bracketleftbiggEz Hz/bracketrightbigg/parenrightbigg −1 v2∂2 ∂t2/bracketleftbiggEz Hz/bracketrightbigg =0. (2.346) We can obtain a solution for (2.346) in much the same way as we do for the wave equations in §A.1. We begin by substituting for Ez(ρ,t)in terms of its temporal Fourier representation Ez(ρ,t)=1 2π/integraldisplay∞ −∞˜Ez(ρ, ω) ejωtdω to obtain 1 2π/integraldisplay∞ −∞/bracketleftbigg1 ρ∂ ∂ρ/parenleftbigg ρ∂ ∂ρ˜Ez(ρ, ω)/parenrightbigg +ω2 v2˜Ez(ρ, ω)/bracketrightbigg ejωtdω=0. The Fourier integral theorem implies that the integrand is zero. Then, expanding out theρderivatives, we ﬁnd that ˜Ez(ρ, ω) obeys the ordinary diﬀerential equation d2˜Ez dρ2+1 ρd˜Ez dρ+k2˜Ez=0 where k=ω/v. This is merely Bessel’s diﬀerential equation (A.124). It is a second-order equation with two independent solutions chosen from the list J0(kρ), Y0(kρ), H(1) 0(kρ), H(2) 0(kρ). We ﬁnd that J0(kρ)and Y0(kρ)are useful for describing standing waves between bound- aries while H(1) 0(kρ)and H(2) 0(kρ)are useful for describing waves propagating in the ρ-direction. Of these, H(1) 0(kρ)represents waves traveling inward while H(2) 0(kρ)repre- sents waves traveling outward. Concentrating on the outward traveling wave we ﬁndthat ˜E z(ρ, ω) =˜A(ω)/bracketleftBig −jπ 2H(2) 0(kρ)/bracketrightBig =˜A(ω)˜g(ρ, ω). Here A(t)↔˜A(ω)is the disturbance waveform, assumed to be a real, causal function. To make Ez(ρ,t)real we require that the inverse transform of ˜g(ρ, ω) be real. This requires the inclusion of the −jπ/2factor in ˜g(ρ, ω) . Inverting we have Ez(ρ,t)=A(t)∗g(ρ,t) (2.347) where g(ρ,t)↔(−jπ/2)H(2) 0(kρ). The inverse transform needed to obtain g(ρ,t)may be found in Campbell [26]: g(ρ,t)=F−1/braceleftBig −jπ 2H(2) 0/parenleftBig ωρ v/parenrightBig/bracerightBig =U/parenleftbig t−ρ v/parenrightbig /radicalBig t2−ρ2 v2, where U(t)is the unit step function deﬁned in (A.5). Substituting this into (2.347) and writing the convolution in integral form we have Ez(ρ,t)=/integraldisplay∞ −∞A(t−t/prime)U(t/prime−ρ/v)/radicalbig t/prime2−ρ2/v2dt/prime. The change of variable x=t/prime−ρ/vthen gives Ez(ρ,t)=/integraldisplay∞ 0A(t−x−ρ/v)/radicalbig x2+2xρ/vdx. (2.348) Those interested in the details of the inverse transform should see Chew [33]. As an example, consider a lossless medium with µr=1,/epsilon1r=81, and a waveform A(t)=E0[U(t)−U(t−τ)] where τ=2µs. This situation is the same as that in the plane wave example above, except that the pulse waveform begins at t=0. Substituting for A(t)into (2.348) and using the integral /integraldisplaydx√x√x+a=2l n/bracketleftbig√x+√ x+a/bracketrightbig Figure 2.9: Propagation of a transient cylindrical wave in a lossless medium. we can write the electric ﬁeld in closed form as Ez(ρ,t)=2E0ln/bracketleftbigg√x2+√x2+2ρ/v√x1+√x1+2ρ/v/bracketrightbigg , (2.349) where x2=max[0 ,t−ρ/v]and x1=max[0 ,t−ρ/v−τ]. The ﬁeld is plotted in Figure 2.9forvariousvaluesoftime.Notethattheleadin gedgeofthedisturbanc epropagates outward at a velocity vand a wake trails behind the disturbance. This wake is similar to that for a plane wave in a dissipative medium, but it exists in this case even though themedium is lossless. We can think of the wave as being created by a line source of inﬁnite extent, pulsed by the disturbance waveform. Although current changes simultaneouslyeverywhere along the line, it takes the disturbance longer to propagate to an observationpoint in the z=0plane from source points z/negationslash=0than from the source point at z=0. Thus, the ﬁeld at an arbitrary observation point ρarrives from diﬀerent source points at diﬀere nttimes.IfwelookatFigure2.9wenotethatthereisalwaysanonzer oﬁeldnear ρ=0(or any value of ρ<v t) regardless of the time, since at any given tthe disturbance is arriving from some point along the line source. WealsoseeinFigure2.9thatasρbecome slargethepeakvalueofthepropagating disturbance approaches a certain value. This value occurs at t m=ρ/v+τor, equivalently, ρm=v(t−τ). If we substitute this value into (2.349) we ﬁnd that Ez(ρ,tm)=2E0ln/bracketleftbigg/radicalbiggτ 2ρ/v+/radicalbigg 1+τ 2ρ/v/bracketrightbigg . For large values of ρ/v, Ez(ρ,tm)≈2E0ln/bracketleftbigg 1+/radicalbiggτ 2ρ/v/bracketrightbigg . Using ln(1+x)≈xwhen x/lessmuch1, we ﬁnd that Ez(ρ,tm)≈E0/radicalBigg 2τv ρ. Thus, as ρ→∞ we have E×H∼1/ρand the ﬂux of energy passing through a cylindrical surface of area ρdφdzis independent of ρ. This result is similar to that seen for spherical waves where E×H∼1/r2. 2.10.8 Propagation of spherical waves in a lossless medium In the previous section we found solutions that describe uniform cylindrical waves dependent only on the radial variable ρ. It turns out that similar solutions are not possible in spherical coordinates; ﬁelds that only depend on rcannot satisfy Maxwell’s equations since, as shown in §2.10.9, a source having the appropriate symmetry for the production of uniform spherical waves in fact produces no ﬁeld at all external to the region it occupies. As we shall see in Chapter 5, the ﬁelds produced by localized sources are ingeneral quite complex. However, certain solutions that are only slightly nonuniform maybe found, and these allow us to investigate the most important properties of sphericalwaves. We shall ﬁnd that spherical waves div erge from a localized point source and expand outward with ﬁnite velocity, carrying energy away from the source. Consider a homogeneous, lossless, source-free region of space characterized by permit- tivity /epsilon1and permeability µ. We seek solutions to the wave equation that are TEM rin spherical coordinates ( Hr=Er=0), and independent of the azimuthal angle φ.T h u s we may write E(r,t)=ˆθEθ(r,θ,t)+ˆφEφ(r,θ,t), H(r,t)=ˆθHθ(r,θ,t)+ˆφHφ(r,θ,t). Maxwell’s equations show that not all of these vector components are required. Faraday’s law states that ∇× E(r,θ,t)=ˆr1 rsinθ∂ ∂θ[sinθEφ(r,θ,t)]−ˆθ1 r∂ ∂r[rEφ(r,θ,t)]+ˆφ1 r∂ ∂r[rEθ(r,θ,t)] =−µ∂H(r,θ,t) ∂t. (2.350) Since we require Hr=0we must have ∂ ∂θ[sinθEφ(r,θ,t)]=0. This implies that either Eφ∼1/sinθorEφ=0. We shall choose Eφ=0and investigate whether the resulting ﬁelds satisfy the remaining Maxwell equations. In a source-free region of space we have ∇·D=/epsilon1∇·E=0. Since we now have only a θ-component of the electric ﬁeld, this requires 1 r∂ ∂θEθ(r,θ,t)+cotθ rEθ(r,θ,t)=0. From this we see that when Eφ=0the component Eθmust obey Eθ(r,θ,t)=fE(r,t) sinθ. By (2.350) there is only a φ-component of magnetic ﬁeld, and it must obey Hφ(r,θ,t)= fH(r,t)/sinθwhere −µ∂ ∂tfH(r,t)=1 r∂ ∂r[rfE(r,t)]. (2.351) Thus the spherical wave has the property E⊥H⊥r, and is TEM to the r-direction. We can obtain a waveequation for Eθby taking the curl of (2.350) and substituting from Ampere’s law: ∇×(∇× E)=− ˆθ1 r∂2 ∂r2[rEθ]=∇×/bracketleftbigg −µ∂ ∂tH/bracketrightbigg =−µ∂ ∂t/bracketleftbigg σE+/epsilon1∂ ∂tE/bracketrightbigg . This gives ∂2 ∂r2[rfE(r,t)]−µσ∂ ∂t[rfE(r,t)]−µ/epsilon1∂2 ∂t2[rfE(r,t)]=0, (2.352) which is the desired wave equation for E. Proceeding similarly we ﬁnd that Hφobeys ∂2 ∂r2[rfH(r,t)]−µσ∂ ∂t[rfH(r,t)]−µ/epsilon1∂2 ∂t2[rfH(r,t)]=0. (2.353) We see that the waveequation for rfEis identical to that for the plane wave ﬁeld Ez (2.331). Thus, we can use the solution obtained in §A.1, as we did with the plane wave, with a few subtle diﬀerences. First, we cannot have r<0. Second, we do not anticipate a solution representing a wave traveling in the −r-direction — i.e., a wave conv erging toward the origin. (In other situations we might need such a solution in order to form astanding wave between two spherical boundary surfaces, but here we are only interestedin the basic propagating behavior of spherical waves.) Thus, we choose as our solution the term (A.45) and ﬁnd for a lossless medium where /Omega1=0 E θ(r,θ,t)=1 rsinθA/parenleftBig t−r v/parenrightBig . (2.354) From (2.351) we see that Hφ=1 µv1 rsinθA/parenleftBig t−r v/parenrightBig . (2.355) Since µv=(µ//epsilon1)1/2=η, we can also write this as H=ˆr×E η. We note that our solution is not appropriate for unbounded space since the ﬁelds have a singularity at θ=0. Thus we must exclude the z-axis. This can be accomplished by using PEC cones of angles θ1andθ2,θ2>θ 1. Because the electric ﬁeld E=ˆθEθis normal to these cones, the boundary condition that tangential Evanishes is satisﬁed. It is informative to see how the terms in the Poynting power balance theorem relate for a spherical wave. Consider the region between the spherical surfaces r=r1and r=r2, r2>r1. Since there is no current within the volume region, Poynting’s theorem (2.299) becomes 1 2∂ ∂t/integraldisplay V(/epsilon1E·E+µH·H)dV=−/contintegraldisplay S(E×H)·dS. (2.356) From (2.354) and (2.355), the time-rate of change of stored energy is Psphere(t)=1 2∂ ∂t/integraldisplay V[/epsilon1E2(r,θ,t)+µH2(r,θ,t)]dV =1 2∂ ∂t/integraldisplay2π 0dφ/integraldisplayθ2 θ1dθ sinθ/integraldisplayr2 r1/bracketleftbigg /epsilon11 r2A2/parenleftBig t−r v/parenrightBig +µ1 r21 (vµ)2A2/parenleftBig t−r v/parenrightBig/bracketrightbigg r2dr =2π/epsilon1F∂ ∂t/integraldisplayr2 r1A2/parenleftBig t−r v/parenrightBig dr where F=ln/bracketleftbiggtan(θ2/2) tan(θ1/2)/bracketrightbigg . Putting u=t−r/vwe see that Psphere(t)=−2π/epsilon1F∂ ∂t/integraldisplayt−r2/v t−r1/vA2(u)vdu. An application of Leibnitz’ rule for diﬀerentiation (A.30) gives Psphere(t)=−2π ηF/bracketleftBig A2/parenleftBig t−r2 v/parenrightBig −A2/parenleftBig t−r1 v/parenrightBig/bracketrightBig . (2.357) Next we ﬁnd the Poynting ﬂux term: Psphere(t)=−/contintegraldisplay S(E×H)·dS =−/integraldisplay2π 0dφ/integraldisplayθ2 θ1/bracketleftbigg1 r1A/parenleftBig t−r1 v/parenrightBig ˆθ/bracketrightbigg ×/bracketleftbigg1 r11 µvA/parenleftBig t−r1 v/parenrightBig ˆφ/bracketrightbigg ·(−ˆr)r2 1dθ sinθ− −/integraldisplay2π 0dφ/integraldisplayθ2 θ1/bracketleftbigg1 r2A/parenleftBig t−r2 v/parenrightBig ˆθ/bracketrightbigg ×/bracketleftbigg1 r21 µvA/parenleftBig t−r2 v/parenrightBig ˆφ/bracketrightbigg ·ˆrr2 2dθ sinθ. The ﬁrst term represents the power carried by the traveling wave into the volume region by passing through the spherical surface at r=r1, while the second term represents the power carried by the wave out of the region by passing through the surface r=r2. Integration gives Psphere(t)=−2π ηF/bracketleftBig A2/parenleftBig t−r2 v/parenrightBig −A2/parenleftBig t−r1 v/parenrightBig/bracketrightBig , (2.358) which matches (2.357), thus verifying Poynting’s theorem. It is also interesting to compute the total energy passing through a surface of radius r0. From (2.358) we see that the ﬂux of energy (power density) passing outward through the surface r=r0is Psphere(t)=2π ηFA2/parenleftBig t−r0 v/parenrightBig . The total energy associated with this ﬂux can be computed by integrating over all time: we have E=2π ηF/integraldisplay∞ −∞A2/parenleftBig t−r0 v/parenrightBig dt=2π ηF/integraldisplay∞ −∞A2(u)du after making the substitution u=t−r0/v. The total energy passing through a spherical surface is independent of the radius of the sphere. This is an important property ofspherical waves. The 1/rdependence of the electric and magnetic ﬁelds produces a power density that decays with distance in precisely the right proportion to compensatefor the r 2-type increase in the surface area through which the power ﬂux passes. 2.10.9 Nonradiating sources Not all time-dependent sources produce electromagnetic waves. In fact, certain local- ized source distributions produce no ﬁelds external to the region containing the sources.Such distributions are said to be nonradiating , and the ﬁelds they produce (within their source regions) lack wave c haracteristics. Let us consider a speciﬁc example involving two concentric spheres. The inner sphere, carrying a uniformly distributed total charge −Q, is rigid and has a ﬁxed radius a; the outer sphere, carrying uniform charge +Q, is a ﬂexible balloon that can be stretched to any radius b=b(t). The two surfaces are initially stationary, some external force being required to hold them in place. Now suppose we apply a time-varying force that resultsinb(t)changing from b(t 1)=b1tob(t2)=b2>b1. This creates a radially directed time-varying current ˆrJr(r,t). By symmetry Jrdepends only on rand produces a ﬁeld Ethat depends only on rand is directed radially. An application of Gauss’s law over a sphere of radius r0>b2, which contains zero total charge, gives 4πr2 0Er(r0,t)=0, hence E(r,t)=0forr>r0and all time t.S o E=0external to the current distribution and no outward traveling wave is produced. Gauss’s law also shows that E=0inside the rigid sphere, while between the spheres E(r,t)=− ˆrQ 4π/epsilon10r2. Now work is certainly required to stretch the balloon and overcome the Lorentz force between the two charged surfaces. But an application of Poynting’s theorem over asurface enclosing both spheres shows that no energy is carried away by an electromagneticwave. Where does the expended energy go? The presence of only two nonzero terms inPoynting’s theorem clearly indicates that the power term/integraltext VE·JdVcorresponding to the external work must be balanced exactly by a change in stored energy. As the radiusof the balloon increases, so does the region of nonzero ﬁeld as well as the stored energy. In free space any current source expressible in the form J(r,t)=∇/parenleftbigg∂ψ(r,t) ∂t/parenrightbigg (2.359) and localized to a volume region V, such as the current in the example above, is nonra- diating. Indeed, Ampere’s law states that ∇× H=/epsilon10∂E ∂t+∇/parenleftbigg∂ψ(r,t) ∂t/parenrightbigg (2.360) forr∈V; taking the curl we have ∇×(∇× H)=/epsilon10∂∇× E ∂t+∇×∇/parenleftbigg∂ψ(r,t) ∂t/parenrightbigg . But the second term on the right is zero, so ∇×(∇× H)=/epsilon10∂∇× E ∂t and this equation holds for all r. By Faraday’s law we can rewrite it as /parenleftbigg (∇×∇ × )+1 c2∂2 ∂t2/parenrightbigg H(r,t)=0. SoHobeys the homogeneous waveequation everywhere, and H=0follows from causality. The laws of Ampere and Faraday may also be combined with (2.359) to show that /parenleftbigg (∇×∇ × )+1 c2∂2 ∂t2/parenrightbigg/bracketleftbigg E(r,t)+1 /epsilon10∇ψ(r,t)/bracketrightbigg =0 for all r. By causality E(r,t)=−1 /epsilon10∇ψ(r,t) (2.361) everywhere. But since ψ(r,t)=0external to V, we must also have E=0there. Note that E=− ∇ ψ//epsilon1 0is consistent with Ampere’s law (2.360) provided that H=0 everywhere. We see that sources having spherical symmetry such that J(r,t)=ˆrJr(r,t)=∇/parenleftbigg∂ψ(r,t) ∂t/parenrightbigg =ˆr∂2ψ(r,t) ∂r∂t obey (2.359) and are therefore nonradiating. Hence the ﬁelds associated with any outward traveling spherical wave must possess some angular variation. This holds, for example,for the ﬁelds far removed from a time-varying source of ﬁnite extent. As pointed out by Lindell [113], nonradiating sources are not merely hypothetical. The outﬂowing currents produced by a highly symmetric nuclear explosion in outerspace or in a homogeneous atmosphere would produce no electromagnetic ﬁeld outsidethe source region. The large electromagnetic-pulse eﬀects discussed in §2.10.6 are due to inhomogeneities in the earth’s atmosphere. We also note that the ﬁelds producedby a radiating source J r(r,t)do not change external to the source if we superpose a nonradiating component Jnr(r,t)to create a new source J=Jnr+Jr. We say that the two sources JandJrareequivalent for the region Vexternal to the sources. This presents diﬃculties in remote sensing where investigators are often interested in reconstructing anunknown source by probing the ﬁelds external to (and usually far away from) the sourceregion. Unique reconstruction is possible only if the ﬁelds within the source region arealso measured. For the time harmonic case, Devaney and Wolf [54] provide the most general possible form for a nonradiating source. See §4.11.9 for details. 2.11 Problems 2.1Consider the constitutive equations (2.16)–(2.17) relating E,D,B, and Hin a bianisotropic medium. Using the deﬁnition for Pand M, show that the constitutive equations relating E,B,P, and Mare P=/parenleftbigg1 c¯P−/epsilon10¯I/parenrightbigg ·E+¯L·B, M=− ¯M·E−/parenleftbigg c¯Q−1 µ0¯I/parenrightbigg ·B. Also ﬁnd the constitutive equations relating E,H,P, and M. 2.2Consider Ampere’s law and Gauss’s law written in terms of rectangular compo- nents in the laboratory frame of reference. Assume that an inertial frame moves withvelocity v=ˆxvwith respect to the laboratory frame. Using the Lorentz transformation given by (2.73)–(2.76), show that cD /prime ⊥=γ(cD⊥+β×H⊥), H/prime ⊥=γ(H⊥−β×cD⊥), J/prime /bardbl=γ(J/bardbl−ρv), J/prime ⊥=J⊥, cρ/prime=γ(cρ−β·J), where “ ⊥” means perpendicular to the direction of the velocity and “ /bardbl” means parallel to the direction of the velocity. 2.3Show that the following quantities are invariant under Lorentz transformation: (a)E·B, (b)H·D, (c)B·B−E·E/c2, (d)H·H−c2D·D, (e)B·H−E·D, (f)cB·D+E·H/c. 2.4Show that if c2B2>E2holds in one reference frame, then it holds in all other reference frames. Repeat for the inequality c2B2<E2. 2.5Show that if E·B=0and c2B2>E2holds in one reference frame, then a reference frame may be found such that E=0. Show that if E·B=0and c2B2<E2holds in one reference frame, then a reference frame may be found such that B=0. 2.6A test charge Qat rest in the laboratory frame experiences a force F=QEas measured by an observer in the laboratory frame. An observer in an inertial framemeasures a force on the charge given by F /prime=QE/prime+Qv×B/prime. Show that F/negationslash=F/primeand ﬁnd the formula for converting between Fand F/prime. 2.7Consider a material moving with velocity vwith respect to the laboratory frame of reference. When the ﬁelds are measured in the moving frame, the material is found to beisotropic with D /prime=/epsilon1/primeE/primeand B/prime=µ/primeH/prime. Show that the ﬁelds measured in the laboratory frame are given by (2.107) and (2.108), indicating that the material is bianisotropic whenmeasured in the laboratory frame. 2.8Show that by assuming v 2/c2/lessmuch1in (2.61)–(2.64) we may obtain (2.111). 2.9Derive the following expressions that allow us to convert the value of the magneti- zation measured in the laboratory frame of reference to the value measured in a movingframe: M /prime ⊥=γ(M⊥+β×cP⊥), M/prime /bardbl=M/bardbl. 2.10 Beginning with the expressions (2.61)–(2.64) for the ﬁeld conversions under a ﬁrst-order Lorentz transformation, show that P/prime=P−v×M c2, M/prime=M+v×P. 2.11 Consider a simple isotropic material moving through space with velocity vrelative to the laboratory frame. The relative permittivity and permeability of the materialmeasured in the moving frame are /epsilon1 /prime randµ/prime r, respectively. Show that the magnetization as measured in the laboratory frame is related to the laboratory frame electric ﬁeld andmagnetic ﬂux density as M=χ /prime m µ0µ/primerB−/epsilon10/parenleftbigg χ/prime e+χ/prime m µ/primer/parenrightbigg v×E when a ﬁrst-order Lorentz transformation is used. Here χ/prime e=/epsilon1/prime r−1andχ/prime m=µ/prime r−1. 2.12 Consider a simple isotropic material moving through space with velocity vrelative to the laboratory frame. The relative permittivity and permeability of the materialmeasured in the moving frame are /epsilon1 /prime randµ/prime r, respectively. Derive the formulas for the magnetization and polarization in the laboratory frame in terms of Eand Bmeasured in the laboratory frame by using the Lorentz transformations (2.128) and (2.129)–(2.132).Show that these expressions reduce to (2.139) and (2.140) under the assumption of aﬁrst-order Lorentz transformation ( v 2/c2/lessmuch1). 2.13 Derive the kinematic form of the large-scale Maxwell–Boﬃ equations (2.165) and (2.166). Derive the alternative form of the large-scale Maxwell–Boﬃ equations (2.167)and (2.168). 2.14 Modify the kinematic form of the Maxwell–Boﬃ equations (2.165)–(2.166) to account for the presence of magnetic sources. Repeat for the alternative forms (2.167)–(2.168). 2.15 Consider a thin magnetic source distribution concentrated near a surface S. The magnetic charge and current densities are given by ρ m(r,x,t)=ρms(r,t)f(x, /Delta1), Jm(r,x,t)=Jms(r,t)f(x, /Delta1), where f(x,/Delta1 )satisﬁes /integraldisplay∞ −∞f(x,/Delta1 )dx=1. Let/Delta1→0and derive the boundary conditions on (E,D,B,H)across S. 2.16 Beginning with the kinematic forms of Maxwell’s equations (2.177)–(2.178), de- rive the boundary conditions for a moving surface ˆn12×(H1−H2)+(ˆn12·v)(D1−D2)=Js, ˆn12×(E1−E2)−(ˆn12·v)(B1−B2)=−Jms. 2.17 Beginning with Maxwell’s equations and the constitutive relationships for a bian- isotropic medium (2.19)–(2.20), derive the waveequation for H(2.313). Specialize the result for the case of an anisotropic medium. 2.18 Consider an isotropic but inhomogeneous material, so that D(r,t)=/epsilon1(r)E(r,t), B(r,t)=µ(r)H(r,t). Show that the wave equations for the ﬁelds within this material may be written as ∇2E−µ/epsilon1∂2E ∂t2+∇/bracketleftbigg E·/parenleftbigg∇/epsilon1 /epsilon1/parenrightbigg/bracketrightbigg −(∇× E)×/parenleftbigg∇µ µ/parenrightbigg =µ∂J ∂t+∇/parenleftBigρ /epsilon1/parenrightBig , ∇2H−µ/epsilon1∂2H ∂t2+∇/bracketleftbigg H·/parenleftbigg∇µ µ/parenrightbigg/bracketrightbigg −(∇× H)×/parenleftbigg∇/epsilon1 /epsilon1/parenrightbigg =− ∇× J−J×/parenleftbigg∇/epsilon1 /epsilon1/parenrightbigg . 2.19 Consider a homogeneous, isotropic material in which D=/epsilon1Eand B=µH. Using the deﬁnitions of the equivalent sources, show that the waveequations (2.322)–(2.323) are equivalent to (2.314)–(2.315). 2.20 When we calculate the force on a conductor produced by an incident plane wave, we often neglect the momentum term ∂ ∂t(D×B). Compute this term for the plane wave ﬁeld (2.336) in free space at the surface of the conductor and compare to the term obtained from the Maxwell stress tensor (2.341).What is the relative diﬀerence in amplitude? 2.21 When a material is only slightly conducting, and thus /Omega1is very small, we often neglect the third term in the plane wave solution (2.343). Reproduce the plot of Figure 2.8withthistermomitte dandcompare .Discus showtheomitte dtermaﬀectstheshape of the propagating waveform. 2.22 A total charge Q is evenly distributed over a spherical surface. The surface expands outward at constant velocity so that the radius of the surface is b=vtat time t. (a) Use Gauss’s law to ﬁnd Eeverywhere as a function of time. (b) Show that Emay be found from a potential function ψ(r,t)=Q 4πr(r−vt)U(r−vt) according to (2.361). Here U(t)is the unit step function. (c) Write down the form of Jfor the expanding sphere and show that since it may be found from (2.359) it is a nonradiating source. Chapter 3 The static electromagnetic ﬁeld 3.1 Static ﬁelds and steady currents Perhaps the most carefully studied area of electromagnetics is that in which the ﬁelds are time-invariant. This area, known generally as statics , oﬀers (1)the most direct op- portunities for solution of the governing equations, and (2)the clearest physical picturesof the electromagnetic ﬁeld. We therefore devote the present chapter to a treatmentof static ﬁelds. We begin to seek and examine speciﬁc solutions to the ﬁeld equations;however, our selection of examples is shaped by a search for insight into the behavior ofthe ﬁeld itself, rather than a desire to catalog the solutions of numerous statics problems. We note at the outset that a static ﬁeld is physically sensible only as a limiting case of a time-varying ﬁeld as the latter approaches a time-invariant equilibrium, and thenonly in local regions. The static ﬁeld equations we shall study thus represent an idealizedmodel of the physical ﬁelds. If we examine the Maxwell–Minkowski equations (2.1)–(2.4) and set the time deriva- tives to zero, we obtain the static ﬁeld Maxwell equations ∇× E(r)=0, (3.1) ∇·D(r)=ρ(r), (3.2) ∇× H(r)=J(r), (3.3) ∇·B(r)=0. (3.4) We note that if the ﬁelds are to be everywhere time-invariant, then the sources Jand ρmust also be everywhere time-invariant. Under this condition the dynamic coupling between the ﬁelds described by Maxwell’s equations disappears; any connection betweenE,D,B, and Himposed by the time-varying nature of the ﬁeld is gone. For static ﬁelds we also require that any dynamic coupling between ﬁelds in the constitutive relationsvanish. In this static ﬁeld limit we cannot derive the divergence equations from the curl equations, since we can no longer use the initial condition argument that the ﬁelds wereidentically zero prior to some time. The static ﬁeld equations are useful for approximating many physical situations in which the ﬁelds rapidly settle to a local, macroscopically-static state. This may occurso rapidly and so completely that, in a practical sense, the static equations describe theﬁelds within our ability to measure and to compute. Such is the case when a capacitoris rapidly charged using a battery in series with a resistor; for example, a 1 pF capacitorcharging through a 1 /Omega1resistor reaches 99.99% of its total charge static limit within 10 ps. 3.1.1 Decoupling of the electric and magnetic ﬁelds For the remainder of this chapter we shall assume that there is no coupling between Eand Hor between Dand Bin the constitutive relations. Then the static equations decouple into two independent sets of equations in terms of two independent sets of ﬁelds.The static electric ﬁeld set ( E,D)is described by the equations ∇× E(r)=0, (3.5) ∇·D(r)=ρ(r). (3.6) Integrating these over a stationary contour and surface, respectively, we have the large- scale forms/contintegraldisplay /Gamma1E·dl=0, (3.7) /contintegraldisplay SD·dS=/integraldisplay VρdV. (3.8) The static magnetic ﬁeld set ( B,H)is described by ∇× H(r)=J(r), (3.9) ∇·B(r)=0, (3.10) or, in large-scale form, /contintegraldisplay /Gamma1H·dl=/integraldisplay SJ·dS, (3.11) /contintegraldisplay SB·dS=0. (3.12) We can also specialize the Maxwell–Boﬃ equations to static form. Assuming that the ﬁelds, sources, and equivalent sources are time-invariant, the electrostatic ﬁeld E(r)is described by the point-form equations ∇× E=0, (3.13) ∇·E=1 /epsilon10(ρ−∇· P), (3.14) or the equivalent large-scale equations /contintegraldisplay /Gamma1E·dl=0, (3.15) /contintegraldisplay SE·dS=1 /epsilon10/integraldisplay V(ρ−∇· P)dV. (3.16) Similarly, the magnetostatic ﬁeld Bis described by ∇× B=µ0(J+∇× M), (3.17) ∇·B=0, (3.18) or/contintegraldisplay /Gamma1B·dl=µ0/integraldisplay S(J+∇× M)·dS, (3.19) /contintegraldisplay SB·dS=0. (3.20) Figure 3.1: Positive point charge in the vicinity of an insulated, uncharged conductor. It is important to note that any separation of the electromagnetic ﬁeld into independent static electric and magnetic portions is illusory. As we mentioned in §2.3.2, the electric and magnetic components of the EM ﬁeld depend on the motion of the observer. Anobserver stationary with respect to a single charge measures only a static electric ﬁeld,while an observer in uniform motion with respect to the charge measures both electricand magnetic ﬁelds. 3.1.2 Static ﬁeld equilibrium and conductors Suppose we could arrange a group of electric charges into a static conﬁguration in free space. The charges would produce an electric ﬁeld, resulting in a force on the distributionvia the Lorentz force law, and hence would begin to move. Regardless of how we arrangethe charges they cannot maintain their original static conﬁguration without the helpof some mechanical force to counterbalance the electrical force. This is a statement ofEarnshaw’s theorem, discussed in detail in §3.4.2. The situation is similar for charges within and on electric conductors. A conductor is a material having many charges free to move under external inﬂuences, both electricand non-electric. In a metallic conductor, electrons move against a background latticeof positive charges. An uncharged conductor is neutral: the amount of negative charge carried by the electrons is equal to the positive charge in the background lattice. Thedistribution of charges in an uncharged conductor is such that the macroscopic electricﬁeld is zero inside and outside the conductor. When the conductor is exposed to an addi-tional electric ﬁeld, the electrons move under the inﬂuence of the Lorentz force, creatingaconduction current . Rather than accelerating indeﬁnitely, conduction electrons experi- ence collisions with the lattice, thereby giving up their kinetic energy. Macroscopically,the charge motion can be described in terms of a time-average velocity, hence a macro-scopic current density can be assigned to the density of moving charge. The relationshipbetween the applied, or “impressed,” ﬁeld and the resulting current density is given byOhm’s law ; in a linear, isotropic, nondispersive material this is J(r,t)=σ(r)E(r,t). (3.21) The conductivity σdescribes the impediment to charge motion through the lattice: the Figure 3.2: Positive point charge near a grounded conductor. higher the conductivity, the farther an electron may move on average before undergoing a collision. Let us examine how a state of equilibrium is established in a conductor. We shall con- sider several important situations. First, suppose we bring a positively charged particleinto the vicinity of a neutral, insulated conductor (we say that a conductor is “insulated”if no means exists for depositing excess charge onto the conductor). The Lorentz forceon the free electrons in the conductor results in their motion toward the particle ( Figure 3.1).Areactio nforce F attract stheparticl etotheconductor .Iftheparticl eandthe conductor are both held rigidly in space by an external mechanical force, the electronswithin the conductor continue to move toward the surface. In a metal, when these elec-trons reach the surface and try to continue further they experience a rapid reversal in thedirection of the Lorentz force, drawing them back toward the surface. A suﬃciently largeforce (described by the work function of the metal)will be able to draw these charges from the surface, but anything less will permit the establishment of a stable equilibriumat the surface. If σis large then equilibrium is established quickly, and a nonuniform static charge distribution appears on the conductor surface. The electric ﬁeld within theconductor must settle to zero at equilibrium, since a nonzero ﬁeld would be associatedwith a current J=σE. In addition, the component of the ﬁeld tangential to the surface must be zero or the charge would be forced to move along the surface. At equilibrium, the ﬁeld within and tangential to a conductor must be zero. Note also that equilibrium cannot be established without external forces to hold the conductor and particle in place. Next, suppose we bring a positively charged particle into the vicinity of a grounded (rathe rthaninsulated)conducto rasinFigure3.2.Useoftheterm“grounded ”means that the conductor is attached via a ﬁlamentary conductor to a remote reservoir of chargeknown as ground ; in practical applications the earth acts as this charge reservoir. Charges are drawn from or returned to the reservoir, without requiring any work, in response tothe Lorentz force on the charge within the conducting body. As the particle approaches,negative charge is drawn to the body and then along the surface until a static equilibriumis re-established. Unlike the insulated body, the grounded conductor in equilibrium hasexcess negative charge, the amount of which depends on the proximity of the particle.Again, both particle and conductor must be held in place by external mechanical forces,and the total ﬁeld produced by both the static charge on the conductor and the particlemust be zero at points interior to the conductor. Finally, consider the process whereby excess charge placed inside a conducting body redistributes as equilibrium is established. We assume an isotropic, homogeneous con-ducting body with permittivity /epsilon1and conductivity σ. An initially static charge with density ρ0(r)is introduced at time t=0. The charge density must obey the continuity equation ∇·J(r,t)=−∂ρ(r,t) ∂t; since J=σE,w eh a v e σ∇·E(r,t)=−∂ρ(r,t) ∂t. By Gauss’s law, ∇·Ecan be eliminated: σ /epsilon1ρ(r,t)=−∂ρ(r,t) ∂t. Solving this diﬀerential equation for the unknown ρ(r,t)we have ρ(r,t)=ρ0(r)e−σt//epsilon1. (3.22) The charge density within a homogeneous, isotropic conducting body decreases exponen- tially with time, regardless of the original charge distribution and shape of the body. Ofcourse, the total charge must be constant, and thus charge within the body travels tothe surface where it distributes itself in such a way that the ﬁeld internal to the bodyapproaches zero at equilibrium. The rate at which the volume charge dissipates is deter-mined by the relaxation time /epsilon1/σ; for copper (a good conductor)this is an astonishingly small 10 −19s. Even distilled water, a relatively poor conductor, has /epsilon1/σ=10−6s. Thus we see how rapidly static equilibrium can be approached. 3.1.3 Steady current Since time-invariant ﬁelds must arise from time-invariant sources, we have from the continuity equation ∇·J(r)=0. (3.23) In large-scale form this is /contintegraldisplay SJ·dS=0. (3.24) A current with the property (3.23)is said to be a steady current . By (3.24), a steady current must be completely lineal (and inﬁnite in extent)or must form closed loops. However, if a current forms loops then the individual moving charges must undergoacceleration (from the change in direction of velocity). Since a single accelerating particleradiates energy in the form of an electromagnetic wave, we might expect a large steadyloop current to produce a great deal of radiation. In fact, if we superpose the ﬁeldsproduced by the many particles comprising a steady current, we ﬁnd that a steady currentproduces no radiation [91]. Remarkably, to obtain this result we must consider the exactrelativistic ﬁelds, and thus our ﬁnding is precise within the limits of our macroscopicassumptions. If we try to create a steady current in free space, the ﬂowing charges will tend to disperse because of the Lorentz force from the ﬁeld set up by the charges, and theresulting current will not form closed loops. A beam of electrons or ions will produceboth an electric ﬁeld (because of the nonzero net charge of the beam)and a magnetic ﬁeld(because of the current). At nonrelativistic particle speeds, the electric ﬁeld producesan outward force on the charges that is much greater than the inward (or pinch)force produced by the magnetic ﬁeld. Application of an additional, external force will allow the creation of a collimated beam of charge, as occurs in an electron tube where a series of permanent magnets can be used to create a beam of steady current. More typically, steady currents are created using wire conductors to guide the moving charge. When an external force, such as the electric ﬁeld created by a battery, is appliedto an uncharged conductor, the free electrons will begin to move through the positivelattice, forming a current. Each electron moves only a short distance before colliding withthe positive lattice, and if the wire is bent into a loop the resulting macroscopic currentwill be steady in the sense that the temporally and spatially averaged microscopic currentwill obey ∇·J=0. We note from the examples above that any charges attempting to leave the surface of the wire are drawn back by the electrostatic force produced by theresulting imbalance in electrical charge. For conductors, the “drift” velocity associatedwith the moving electrons is proportional to the applied ﬁeld: u d=−µeE where µeis the electron mobility . The mobility of copper ( 3.2×10−3m2/V·s)is such that an applied ﬁeld of 1 V/m results in a drift velocity of only a third of a centimeterper second. Integral properties of a steady current. Steady currents obey several useful inte- gral properties. To develop these properties we need an integral identity. Let f(r)and g(r)be scalar functions, continuous and with continuous derivatives in a volume region V. Let Jrepresent a steady current ﬁeld of ﬁnite extent, completely contained within V. We begin by using (B.42)to expand ∇·(fgJ)=fg(∇·J)+J·∇(fg). Noting that ∇·J=0and using (B.41), we get ∇·(fgJ)=(fJ)·∇g+(gJ)·∇f. Now let us integrate over Vand employ the divergence theorem: /contintegraldisplay S(fg)J·dS=/integraldisplay V[(fJ)·∇g+(gJ)·∇f]dV. Since Jis contained entirely within S,w em u s th a v e ˆn·J=0everywhere on S. Hence /integraldisplay V[(fJ)·∇g+(gJ)·∇f]dV=0. (3.25) We can obtain a useful relation by letting f=1and g=xiin (3.25), where (x,y,z)= (x1,x2,x3). This gives /integraldisplay VJi(r)dV=0, (3.26) where J1=Jxand so on. Hence the volume integral of any rectangular component of J is zero. Similarly, letting f=g=xiwe ﬁnd that /integraldisplay VxiJi(r)dV=0. (3.27) With f=xiand g=xjwe obtain /integraldisplay V/bracketleftbig xiJj(r)+xjJi(r)/bracketrightbig dV=0. (3.28) 3.2 Electrostatics 3.2.1 The electrostatic potential and work The equation /contintegraldisplay /Gamma1E·dl=0 (3.29) satisﬁed by the electrostatic ﬁeld E(r)is particularly interesting. A ﬁeld with zero circulation is said to be conservative . To see why, let us examine the work required to move a particle of charge Qaround a closed path in the presence of E(r). Since work is the line integral of force and B=0, the work expended by the external system moving the charge against the Lorentz force is W=−/contintegraldisplay /Gamma1(QE+Qv×B)·dl=− Q/contintegraldisplay /Gamma1E·dl=0. This property is analogous to the conservation property for a classical gravitational ﬁeld: any potential energy gained by raising a point mass is lost when the mass is lowered. Direct experimental veriﬁcation of the electrostatic conservative property is diﬃcult, aside from the fact that the motion of Qmay alter Eby interacting with the sources of E. By moving Qwith nonuniform velocity (i.e., with acceleration at the beginning of the loop, direction changes in transit, and deceleration at the end)we observe a radiativeloss of energy, and this energy cannot be regained by the mechanical system providingthe motion. To avoid this problem we may assume that the charge is moved so slowly,or in such small increments, that it does not radiate. We shall use this concept later todetermine the “assembly energy” in a charge distribution. The electrostatic potential. By the point form of (3.29), ∇× E(r)=0, we can introduce a scalar ﬁeld /Phi1=/Phi1(r)such that E(r)=− ∇ /Phi1(r). (3.30) The function /Phi1carries units of volts and is known as the electrostatic potential . Let us consider the work expended by an external agent in moving a charge between points P 1 atr1and P2atr2: W21=− Q/integraldisplayP2 P1−∇/Phi1(r)·dl=Q/integraldisplayP2 P1d/Phi1(r)=Q[/Phi1(r2)−/Phi1(r1)]. The work W21is clearly independent of the path taken between P1and P2; the quantity V21=W21 Q=/Phi1(r2)−/Phi1(r1)=−/integraldisplayP2 P1E·dl, (3.31) called the potential diﬀerence , has an obvious physical meaning as work per unit charge required to move a particle against an electric ﬁeld between two points. Figure 3.3: Demonstration of path independence of the electric ﬁeld line integral. Of course, the large-scale form (3.29)also implies the path-independence of work in the electrostatic ﬁeld. Indeed, we may pass an arbitrary closed contour /Gamma1through P1 and P2 andthensplititintotwopieces/Gamma11 and/Gamma12 asshowninFigure3.3.Since −Q/contintegraldisplay /Gamma11−/Gamma12E·dl=− Q/integraldisplay /Gamma11E·dl+Q/integraldisplay /Gamma12E·dl=0, we have −Q/integraldisplay /Gamma11E·dl=− Q/integraldisplay /Gamma12E·dl as desired. We sometimes refer to /Phi1(r)as the absolute electrostatic potential . Choosing a suitable reference point P0at location r0and writing the potential diﬀerence as V21=[/Phi1(r2)−/Phi1(r0)]−[/Phi1(r1)−/Phi1(r0)], we can justify calling /Phi1(r)theabsolute potential referred to P0. Note that P0might describe a locus of points, rather than a single point, since many points can be at the samepotential. Although we can choose any reference point without changing the resulting value of Efound from (3.30), for simplicity we often choose r 0such that /Phi1(r0)=0. Several properties of the electrostatic potential make it convenient for describing static electric ﬁelds. We know that, at equilibrium, the electrostatic ﬁeld within a conductingbody must vanish. By (3.30)the potential at all points within the body must thereforehave the same constant value. It follows that the surface of a conductor is an equipotential surface : a surface for which /Phi1(r)is constant. As an inﬁnite reservoir of charge that can be tapped through a ﬁlamentary conductor, the entity we call “ground” must also be an equipotential object. If we connect a con-ductor to ground, we have seen that charge may ﬂow freely onto the conductor. Since nowork is expended, “grounding” a conductor obviously places the conductor at the sameabsolute potential as ground. For this reason, ground is often assigned the role as thepotential reference with an absolute potential of zero volts. Later we shall see that forsources of ﬁnite extent ground must be located at inﬁnity. 3.2.2 Boundary conditions Boundary conditions for the electrostatic ﬁeld. The boundary conditions found for the dynamic electric ﬁeld remain valid in the electrostatic case. Thus ˆn12×(E1−E2)=0 (3.32) and ˆn12·(D1−D2)=ρs. (3.33) Here ˆn12points into region 1 from region 2. Because the static curl and divergence equations are independent, so are the boundary conditions (3.32)and (3.33) . For a linear and isotropic dielectric where D=/epsilon1E, equation (3.33)becomes ˆn12·(/epsilon11E1−/epsilon12E2)=ρs. (3.34) Alternatively, using D=/epsilon10E+Pwe can write (3.33)as ˆn12·(E1−E2)=1 /epsilon10(ρs+ρPs1+ρPs2) (3.35) where ρPs=ˆn·P is the polarization surface charge with ˆnpointing outward from the material body. We can also write the boundary conditions in terms of the electrostatic potential. With E=− ∇ /Phi1, equation (3.32)becomes /Phi11(r)=/Phi12(r) (3.36) for all points ron the surface. Actually /Phi11and/Phi12may diﬀer by a constant; because this constant is eliminated when the gradient is taken to ﬁnd E, it is generally ignored. We can write (3.35)as /epsilon10/parenleftbigg∂/Phi1 1 ∂n−∂/Phi1 2 ∂n/parenrightbigg =−ρs−ρPs1−ρPs2 where the normal derivative is taken in the ˆn12direction. For a linear, isotropic dielectric (3.33)becomes /epsilon11∂/Phi1 1 ∂n−/epsilon12∂/Phi1 2 ∂n=−ρs. (3.37) Again, we note that (3.36)and (3.37)are independent. Boundary conditions for steady electric current. The boundary condition on the normal component of current found in §2.8.2 remains valid in the steady current case. Assume that the boundary exists between two linear, isotropic conducting regions havingconstitutive parameters ( /epsilon1 1,σ1)and (/epsilon12,σ2), respectively. By (2.198) we have ˆn12·(J1−J2)=− ∇ s·Js (3.38) where ˆn12points into region 1 from region 2. A surface current will not appear on the boundary between two regions having ﬁnite conductivity, although a surface charge mayaccumulate there during the transient period when the currents are established [31]. Ifcharge is inﬂuenced to move from the surface, it will move into the adjacent regions, Figure 3.4: Refraction of steady current at a material interface. rather than along the surface, and a new charge will replace it, supplied by the current. Thus, for ﬁnite conducting regions (3.38)becomes ˆn12·(J1−J2)=0. (3.39) A boundary condition on the tangential component of current can also be found. Substituting E=J/σinto (3.32)we have ˆn12×/parenleftbiggJ1 σ1−J2 σ2/parenrightbigg =0. We can also write this as J1t σ1=J2t σ2(3.40) where J1t=ˆn12×J1, J2t=ˆn12×J2. We may combine the boundary conditions for the normal components of current and electric ﬁeld to better understand the behavior of current at a material boundary. Sub-stituting E=J/σinto (3.34)we have /epsilon1 1 σ1J1n−/epsilon12 σ2J2n=ρs (3.41) where J1n=ˆn12·J1and J2n=ˆn12·J2. Combining (3.41)with (3.39) , we have ρs=J1n/parenleftbigg/epsilon11 σ1−/epsilon12 σ2/parenrightbigg =E1n/parenleftbigg /epsilon11−σ1 σ2/epsilon12/parenrightbigg =J2n/parenleftbigg/epsilon11 σ1−/epsilon12 σ2/parenrightbigg =E2n/parenleftbigg /epsilon11σ2 σ1−/epsilon12/parenrightbigg where E1n=ˆn12·E1, E2n=ˆn12·E2. Unless /epsilon11σ2−σ1/epsilon12=0, a surface charge will exist on the interface between dissimilar current-carrying conductors. We may also combine the vector components of current on each side of the boundary todetermin etheeﬀectsoftheboundar yoncurrentdirectio n(Figure3.4).Letθ1,2denote the angle between J1,2and ˆn12so that J1n=J1cosθ1, J1t=J1sinθ1 J2n=J2cosθ2, J2t=J2sinθ2. Then J1cosθ1=J2cosθ2by (3.39), while σ2J1sinθ1=σ1J2sinθ2by (3.40). Hence σ2tanθ1=σ1tanθ2. (3.42) It is interesting to consider the case of current incident from a conducting material onto an insulating material. If region 2 is an insulator, then J2n=J2t=0; by (3.39)we have J1n=0. But (3.40)does not require J1t=0; with σ2=0the right-hand side of (3.40) is indeterminate and thus J1tmay be nonzero. In other words, when current moving through a conductor approaches an insulating surface, it bends and ﬂows tangential tothe surface. This concept is useful in explaining how wires guide current. Interestingly, (3.42)shows that when σ 2/lessmuchσ1we have θ2→0; current passing from a conducting region into a slightly-conducting region does so normally. 3.2.3 Uniqueness of the electrostatic ﬁeld In§2.2.1 we found that the electromagnetic ﬁeld is unique within a region Vwhen the tangential component of Eis speciﬁed over the surrounding surface. Unfortunately, this condition is not appropriate in the electrostatic case. We should remember thatan additional requirement for uniqueness of solution to Maxwell’s equations is that theﬁeld be speciﬁed throughout Vat some time t 0. For a static ﬁeld this would completely determine Ewithout need for the surface ﬁeld! Let us determine conditions for uniqueness beginning with the static ﬁeld equations. Consider a region Vsurrounded by a surface S. Static charge may be located entirely or partially within V, or entirely outside V, and produces a ﬁeld within V. The region may also contain any arrangement of conductors or other materials. Suppose (D1,E1) and(D2,E2)represent solutions to the static ﬁeld equations within Vwith source ρ(r). We wish to ﬁnd conditions that guarantee both E1=E2and D1=D2. Since ∇·D1=ρand∇·D2=ρ, the diﬀerence ﬁeld D0=D2−D1obeys the homogeneous equation ∇·D0=0. (3.43) Consider the quantity ∇·(D0/Phi10)=/Phi10(∇·D0)+D0·(∇/Phi10) where E0=E2−E1=− ∇ /Phi10=− ∇ (/Phi12−/Phi11). We integrate over Vand use the divergence theorem and (3.43)to obtain /contintegraldisplay S/Phi10[D0·ˆn]dS=/integraldisplay VD0·(∇/Phi10)dV=−/integraldisplay VD0·E0dV. (3.44) Now suppose that /Phi10=0everywhere on S, or that ˆn·D0=0everywhere on S, or that /Phi10=0over part of Sand ˆn·D0=0elsewhere on S. Then /integraldisplay VD0·E0dV=0. (3.45) Since Vis arbitrary, either D0=0orE0=0. Assuming Eand Dare linked by the constitutive relations, we have E1=E2and D1=D2. Hence the ﬁelds within Vare unique provided that either /Phi1, the normal component ofD, or some combination of the two, is speciﬁed over S. We often use a multiply- connected surface to exclude conductors. By (3.33)we see that speciﬁcation of the normal component of Don a conductor is equivalent to speciﬁcation of the surface charge density. Thus we must specify the potential or surface charge density over allconducting surfaces. One other condition results in zero on the left-hand side of (3.44). If Srecedes to inﬁnity and /Phi1 0and D0decrease suﬃciently fast, then (3.45)still holds and uniqueness is guaranteed. If D,E∼1/r2asr→∞, then /Phi1∼1/rand the surface integral in (3.44) tends to zero since the area of an expanding sphere increases only as r2. We shall ﬁnd later in this section that for sources of ﬁnite extent the ﬁelds do indeed vary inverselywith distance squared from the source, hence we may allow Sto expand and encompass all space. For the case in which conducting bodies are immersed in an inﬁnite homogeneous medium and the static ﬁelds must be determined throughout all space, a multiply-connected surface is used with one part receding to inﬁnity and the remaining partssurrounding the conductors. Here uniqueness is guaranteed by specifying the potentialsor charges on the surfaces of the conducting bodies. 3.2.4 Poisson’s and Laplace’s equations For computational purposes it is often convenient to deal with the diﬀerential versions ∇× E(r)=0, (3.46) ∇·D(r)=ρ(r), (3.47) of the electrostatic ﬁeld equations. We must supplement these with constitutive relations between Eand D; at this point we focus our attention on linear, isotropic materials for which D(r)=/epsilon1(r)E(r). Using this in (3.47)along with E=− ∇ /Phi1(justiﬁed by (3.46)), we can write ∇·[/epsilon1(r)∇/Phi1(r)]=−ρ(r). (3.48) This is Poisson’s equation . The corresponding homogeneous equation ∇·[/epsilon1(r)∇/Phi1(r)]=0, (3.49) holding at points rwhere ρ(r)=0,i sLaplace’s equation . Equations (3.48)and (3.49) are valid for inhomogeneous media. By (B.42)we can write ∇/Phi1(r)·∇/epsilon1(r)+/epsilon1(r)∇·[∇/Phi1(r)]=−ρ(r). For a homogeneous medium, ∇/epsilon1=0; since ∇·(∇/Phi1)≡∇2/Phi1,w eh a v e ∇2/Phi1(r)=−ρ(r)//epsilon1 (3.50) in such a medium. Correspondingly, ∇2/Phi1(r)=0 at points where ρ(r)=0. Poisson’s and Laplace’s equations can be solved by separation of variables, Fourier transformation, conformal mapping, and numerical techniques such as the ﬁnite diﬀerenceand moment methods. In Appendix A we consider the separation of variables solution to Laplace’s equation in three major coordinate systems for a variety of problems. For an introduction to numerical techniques the reader is referred to the books by Sadiku[162], Harrington [82], and Peterson et al. [146]. Solution to Poisson’s equation is oftenundertaken using the method of Green’s functions, which we shall address later in thissection. We shall also consider the solution to Laplace’s equation for bodies immersed inan applied, or “impressed,” ﬁeld. Uniqueness of solution to Poisson’s equation. Before attempting any solutions, we must ask two very important questions. How do we know that solving the second-orderdiﬀerential equation produces the same values for E=− ∇ /Phi1as solving the ﬁrst-order equations directly for E? And, if these solutions are the same, what are the conditions for uniqueness of solution to Poisson’s and Laplace’s equations? To answer the ﬁrstquestion, a suﬃcient condition is to have /Phi1twice diﬀerentiable. We shall not attempt to prove this, but shall instead show that the condition for uniqueness of the second-orderequations is the same as that for the ﬁrst-order equations. Consider a region of space Vsurrounded by a surface S. Static charge may be located entirely or partially within V, or entirely outside V, and produces a ﬁeld within V. This region may also contain any arrangement of conductors or other materials. Now, assumethat/Phi1 1and/Phi12represent solutions to the static ﬁeld equations within Vwith source ρ(r). We wish to ﬁnd conditions under which /Phi11=/Phi12. Since we have ∇·[/epsilon1(r)∇/Phi11(r)]=−ρ(r), ∇·[/epsilon1(r)∇/Phi12(r)]=−ρ(r), the diﬀerence ﬁeld /Phi10=/Phi12−/Phi11obeys ∇·[/epsilon1(r)∇/Phi10(r)]=0. (3.51) That is, /Phi10obeys Laplace’s equation. Now consider the quantity ∇·(/epsilon1/Phi1 0∇/Phi10)=/epsilon1|∇/Phi10|2+/Phi10∇·(/epsilon1∇/Phi10). Integration over Vand use of the divergence theorem and (3.51)gives /contintegraldisplay S/Phi10(r)[/epsilon1(r)∇/Phi10(r)]·dS=/integraldisplay V/epsilon1(r)|∇/Phi10(r)|2dV. As with the ﬁrst order equations, we see that specifying either /Phi1(r)or/epsilon1(r)∇/Phi1(r)·ˆnover Sresults in /Phi10(r)=0throughout V, hence /Phi11=/Phi12. As before, specifying /epsilon1(r)∇/Phi1(r)·ˆn for a conducting surface is equivalent to specifying the surface charge on S. Integral solution to Poisson’s equation: the static Green’s function. The method of Green’s functions is one of the most useful techniques for solving Poisson’sequation. We seek a solution for a single point source, then use Green’s second identityto write the solution for an arbitrary charge distribution in terms of a superpositionintegral. We seek the solution to Poisson’s equation for a region of space Vas shown in Figure 3.5.Theregionisassume dhomogeneou swithpermittivi ty/epsilon1,anditssurfac eismultiply- connected, consisting of a bounding surface S Band any number of closed surfaces internal toV. We denote by Sthe composite surface consisting of SBand the Ninternal surfaces Sn,n=1,..., N. The internal surfaces are used to exclude material bodies, such as the Figure 3.5: Computation of potential from known sources and values on bounding sur- faces. plates of a capacitor, which may be charged and on which the potential is assumed to be known. To solve for /Phi1(r)within Vwe must know the potential produced by a point source. This potential, called the Green’s function , is denoted G(r|r/prime); it has two arguments because it satisﬁes Poisson’s equation at rwhen the source is located at r/prime: ∇2G(r|r/prime)=−δ(r−r/prime). (3.52) Later we shall demonstrate that in all cases of interest to us the Green’s function is symmetric in its arguments: G(r/prime|r)=G(r|r/prime). (3.53) This property of Gis known as reciprocity . Our development rests on the mathematical result (B.30)known as Green’s second identity . We can derive this by subtracting the identities ∇·(φ∇ψ)=φ∇·(∇ψ)+(∇φ)·(∇ψ), ∇·(ψ∇φ)=ψ∇·(∇φ)+(∇ψ)·(∇φ), to obtain ∇·(φ∇ψ−ψ∇φ)=φ∇2ψ−ψ∇2φ. Integrating this over a volume region Vwith respect to the dummy variable r/primeand using the divergence theorem, we obtain /integraldisplay V[φ(r/prime)∇/prime2ψ(r/prime)−ψ(r/prime)∇/prime2φ(r/prime)]dV/prime=−/contintegraldisplay S[φ(r/prime)∇/primeψ(r/prime)−ψ(r/prime)∇/primeφ(r/prime)]·dS/prime. The negative sign on the right-hand side occurs because ˆnis aninward normal to V. Finally, since ∂ψ(r/prime)/∂n/prime=ˆn/prime·∇/primeψ(r/prime),w eh a v e /integraldisplay V[φ(r/prime)∇/prime2ψ(r/prime)−ψ(r/prime)∇/prime2φ(r/prime)]dV/prime=−/contintegraldisplay S/bracketleftbigg φ(r/prime)∂ψ(r/prime) ∂n/prime−ψ(r/prime)∂φ(r/prime) ∂n/prime/bracketrightbigg dS/prime as desired. To solve for /Phi1inVwe shall make some seemingly unmotivated substitutions into this identity. First note that by (3.52)and (3.53)we can write ∇/prime2G(r|r/prime)=−δ(r/prime−r). (3.54) We now set φ(r/prime)=/Phi1(r/prime)andψ(r/prime)=G(r|r/prime)to obtain /integraldisplay V[/Phi1(r/prime)∇/prime2G(r|r/prime)−G(r|r/prime)∇/prime2/Phi1(r/prime)]dV/prime= −/contintegraldisplay S/bracketleftbigg /Phi1(r/prime)∂G(r|r/prime) ∂n/prime−G(r|r/prime)∂/Phi1(r/prime) ∂n/prime/bracketrightbigg dS/prime, (3.55) hence /integraldisplay V/bracketleftbigg /Phi1(r/prime)δ(r/prime−r)−G(r|r/prime)ρ(r/prime) /epsilon1/bracketrightbigg dV/prime=/contintegraldisplay S/bracketleftbigg /Phi1(r/prime)∂G(r|r/prime) ∂n/prime−G(r|r/prime)∂/Phi1(r/prime) ∂n/prime/bracketrightbigg dS/prime. By the sifting property of the Dirac delta /Phi1(r)=/integraldisplay VG(r|r/prime)ρ(r/prime) /epsilon1dV/prime+/contintegraldisplay SB/bracketleftbigg /Phi1(r/prime)∂G(r|r/prime) ∂n/prime−G(r|r/prime)∂/Phi1(r/prime) ∂n/prime/bracketrightbigg dS/prime+ +N/summationdisplay n=1/contintegraldisplay Sn/bracketleftbigg /Phi1(r/prime)∂G(r|r/prime) ∂n/prime−G(r|r/prime)∂/Phi1(r/prime) ∂n/prime/bracketrightbigg dS/prime. (3.56) With this we may compute the potential anywhere within Vin terms of the charge density within Vand the values of the potential and its normal derivative over S.W e must simply determine G(r|r/prime)ﬁrst. Let us take a moment to specialize (3.56)to the case of unbounded space. Provided that the sources are of ﬁnite extent, as SB→∞ we shall ﬁnd that /Phi1(r)=/integraldisplay VG(r|r/prime)ρ(r/prime) /epsilon1dV/prime+N/summationdisplay n=1/contintegraldisplay Sn/bracketleftbigg /Phi1(r/prime)∂G(r|r/prime) ∂n/prime−G(r|r/prime)∂/Phi1(r/prime) ∂n/prime/bracketrightbigg dS/prime. A useful derivative identity. Many diﬀerential operations on the displacement vector R=r−r/primeoccur in the study of electromagnetics. The identities ∇R=− ∇/primeR=ˆR, ∇/parenleftbigg1 R/parenrightbigg =− ∇/prime/parenleftbigg1 R/parenrightbigg =−ˆR R2, (3.57) for example, follow from direct diﬀerentiation of the rectangular coordinate representa- tion R=ˆx(x−x/prime)+ˆy(y−y/prime)+ˆz(z−z/prime). The identity ∇2/parenleftbigg1 R/parenrightbigg =−4πδ(r−r/prime), (3.58) crucial to potential theory, is more diﬃcult to establish. We shall prove the equivalent version ∇/prime2/parenleftbigg1 R/parenrightbigg =−4πδ(r/prime−r) Figure 3.6: Geometry for establishing the singular property of ∇2(1/R). by showing that /integraldisplay Vf(r/prime)∇/prime2/parenleftbigg1 R/parenrightbigg dV/prime=/braceleftBigg −4πf(r),r∈V, 0, r/∈V,(3.59) holds for any continuous function f(r). By direct diﬀerentiation we have ∇/prime2/parenleftbigg1 R/parenrightbigg =0forr/prime/negationslash=r, hence the second part of (3.59)is established. This also shows that if r∈Vthen the domain of integration in (3.59)can be restricted to a sphere of arbitrarily small radiusεcenteredat r (Figure3.6).Theresultweseekisfoundinthelimitasε→ 0.Thuswe are interested in computing /integraldisplay Vf(r/prime)∇/prime2/parenleftbigg1 R/parenrightbigg dV/prime=lim ε→0/integraldisplay Vεf(r/prime)∇/prime2/parenleftbigg1 R/parenrightbigg dV/prime. Since fis continuous at r/prime=r, we have by the mean value theorem /integraldisplay Vf(r/prime)∇/prime2/parenleftbigg1 R/parenrightbigg dV/prime=f(r)lim ε→0/integraldisplay Vε∇/prime2/parenleftbigg1 R/parenrightbigg dV/prime. The integral over Vεcan be computed using ∇/prime2(1/R)=∇/prime·∇/prime(1/R)and the divergence theorem: /integraldisplay Vε∇/prime2/parenleftbigg1 R/parenrightbigg dV/prime=/integraldisplay Sεˆn/prime·∇/prime/parenleftbigg1 R/parenrightbigg dS/prime, where Sεbounds Vε. Noting that ˆn/prime=− ˆR, using (57), and writing the integral in spherical coordinates ( ε,θ,φ )centered at the point r,w eh a v e /integraldisplay Vf(r/prime)∇/prime2/parenleftbigg1 R/parenrightbigg dV/prime=f(r)lim ε→0/integraldisplay2π 0/integraldisplayπ 0−ˆR·/parenleftBiggˆR ε2/parenrightBigg ε2sinθdθdφ=−4πf(r). Hence the ﬁrst part of (3.59)is also established. The Green’s function for unbounded space. In view of (3.58), one solution to (3.52)is G(r|r/prime)=1 4π|r−r/prime|. (3.60) This simple Green’s function is generally used to ﬁnd the potential produced by charge in unbounded space. Here N=0(no internal surfaces)and SB→∞.T h u s /Phi1(r)=/integraldisplay VG(r|r/prime)ρ(r/prime) /epsilon1dV/prime+lim SB→∞/contintegraldisplay SB/bracketleftbigg /Phi1(r/prime)∂G(r|r/prime) ∂n/prime−G(r|r/prime)∂/Phi1(r/prime) ∂n/prime/bracketrightbigg dS/prime. We have seen that the Green’s function varies inversely with distance from the source, and thus expect that, as a superposition of point-source potentials, /Phi1(r)will also vary inversely with distance from a source of ﬁnite extent as that distance becomes large withrespect to the size of the source. The normal derivatives then vary inversely with distancesquared. Thus, each term in the surface integrand will vary inversely with distance cubed,while the surface area itself varies with distance squared. The result is that the surfaceintegral vanishes as the surface recedes to inﬁnity, giving /Phi1(r)=/integraldisplay VG(r|r/prime)ρ(r/prime) /epsilon1dV/prime. By (3.60)we then have /Phi1(r)=1 4π/epsilon1/integraldisplay Vρ(r/prime) |r−r/prime|dV/prime(3.61) where the integration is performed over all of space. Since lim r→∞/Phi1(r)=0, points at inﬁnity are a convenient reference for the absolute potential. Later we shall need to know the amount of work required to move a charge Qfrom inﬁnity to a point Plocated at r. If a potential ﬁeld is produced by charge located in unbounded space, moving an additional charge into position requires the work W21=− Q/integraldisplayP ∞E·dl=Q[/Phi1(r)−/Phi1(∞)]=Q/Phi1(r). (3.62) Coulomb’s law. We can obtain Efrom (61)by direct diﬀerentiation. We have E(r)=−1 4π/epsilon1∇/integraldisplay Vρ(r/prime) |r−r/prime|dV/prime=−1 4π/epsilon1/integraldisplay Vρ(r/prime)∇/parenleftbigg1 |r−r/prime|/parenrightbigg dV/prime, hence E(r)=1 4π/epsilon1/integraldisplay Vρ(r/prime)r−r/prime |r−r/prime|3dV/prime(3.63) by (3.57). So Coulomb’s law follows from the two fundamental postulates of electrostatics (3.5)and (3.6) . Green’s function for unbounded space: two dimensions. We deﬁne the two- dimensional Green’s function as the potential at a point r=ρ+ˆzzproduced by a z-directed line source of constant density located at r/prime=ρ/prime. Perhaps the simplest way to compute this is to ﬁrst ﬁnd Eproduced by a line source on the z-axis. By (3.63)we have E(r)=1 4π/epsilon1/integraldisplay /Gamma1ρl(z/prime)r−r/prime |r−r/prime|3dl/prime. Then, since r=ˆzz+ˆρρ,r/prime=ˆzz/prime, and dl/prime=dz/prime,w eh a v e E(ρ)=ρl 4π/epsilon1/integraldisplay∞ −∞ˆρρ+ˆz(z−z/prime) /bracketleftbig ρ2+(z−z/prime)2/bracketrightbig3/2dz/prime. Carrying out the integration we ﬁnd that Ehas only a ρ-component which varies only withρ: E(ρ)=ˆρρl 2π/epsilon1ρ. (3.64) The absolute potential referred to a radius ρ0can be found by computing the line integral ofEfromρtoρ0: /Phi1(ρ)=−ρl 2π/epsilon1/integraldisplayρ ρ0dρ/prime ρ/prime=ρl 2π/epsilon1ln/parenleftbiggρ0 ρ/parenrightbigg . We may choose any reference point ρ0except ρ0=0orρ0=∞. This choice is equivalent to the addition of an arbitrary constant, hence we can also write /Phi1(ρ)=ρl 2π/epsilon1ln/parenleftbigg1 ρ/parenrightbigg +C. (3.65) The potential for a general two-dimensional charge distribution in unbounded space is by superposition /Phi1(ρ)=/integraldisplay STρT(ρ/prime) /epsilon1G(ρ|ρ/prime)dS/prime, (3.66) where the Green’s function is the potential of a unit line source located at ρ/prime: G(ρ|ρ/prime)=1 2πln/parenleftbiggρ0 |ρ−ρ/prime|/parenrightbigg . (3.67) Here STdenotes the transverse ( xy)plane, and ρTdenotes the two-dimensional charge distribution ( C/m2)within that plane. We note that the potential ﬁeld (3.66)of a two-dimensional source decreases logarith- mically with distance. Only the potential produced by a source of ﬁnite extent decreasesinversely with distance. Dirichlet and Neumann Green’s functions. The unbounded space Green’s func- tion may be inconvenient for expressing the potential in a region having internal surfaces. In fact, (3.56)shows that to use this function we would be forced to specify both /Phi1and its normal derivative over all surfaces. This, of course, would exceed the actual requirementsfor uniqueness. Many functions can satisfy (3.52). For instance, G(r|r /prime)=A |r−r/prime|+B |r−ri|(3.68) satisﬁes (3.52)if ri/∈V. Evaluation of (3.55)with the Green’s function (3.68)repro- duces the general formulation (3.56)since the Laplacian of the second term in (3.68)isidentically zero in V. In fact, we can add any function to the free-space Green’s function, provided that the additional term obeys Laplace’s equation within V: G(r|r /prime)=A |r−r/prime|+F(r|r/prime), ∇/prime2F(r|r/prime)=0. (3.69) A good choice for G(r|r/prime)will minimize the eﬀort required to evaluate /Phi1(r). Examining (3.56)we notice two possibilities. If we demand that G(r|r/prime)=0for all r/prime∈S (3.70) then the surface integral terms in (3.56)involving ∂/Phi1/∂ n/primewill vanish. The Green’s function satisfying (3.70)is known as the Dirichlet Green’s function . Let us designate it byGDand use reciprocity to write (3.70)as GD(r|r/prime)=0for all r∈S. The resulting specialization of (3.56), /Phi1(r)=/integraldisplay VGD(r|r/prime)ρ(r/prime) /epsilon1dV/prime+/contintegraldisplay SB/Phi1(r/prime)∂GD(r|r/prime) ∂n/primedS/prime+ +N/summationdisplay n=1/contintegraldisplay Sn/Phi1(r/prime)∂GD(r|r/prime) ∂n/primedS/prime, (3.71) requires the speciﬁcation of /Phi1(but not its normal derivative)over the boundary surfaces. In case SBand Snsurround and are adjacent to perfect conductors, the Dirichlet bound- ary condition has an important physical meaning. The corresponding Green’s function isthe potential at point rproduced by a point source at r /primein the presence of the conductors when the conductors are grounded — i.e., held at zero potential. Then we must specifythe actual constant potentials on the conductors to determine /Phi1everywhere within V using (3.71). The additional term F(r|r /prime)in (3.69)accounts for the potential produced by surface charges on the grounded conductors. By analogy with (3.70)it is tempting to try to deﬁne another electrostatic Green’s function according to ∂G(r|r/prime) ∂n/prime=0for all r/prime∈S. (3.72) But this choice is not permissible if Vis a ﬁnite-sized region. Let us integrate (3.54)over Vand employ the divergence theorem and the sifting property to get /contintegraldisplay S∂G(r|r/prime) ∂n/primedS/prime=−1; (3.73) in conjunction with this, equation (3.72)would imply the false statement 0=−1. Sup- pose instead that we introduce a Green’s function according to ∂G(r|r/prime) ∂n/prime=−1 Afor all r/prime∈S. (3.74) where Ais the total area of S. This choice avoids a contradiction in (3.73); it does not nullify any terms in (3.56), but does reduce the surface integral terms involving /Phi1to constants. Taken together, these terms all comprise a single additive constant on theright-hand side; although the corresponding potential /Phi1(r)is thereby determined only to within this additive constant, the value of E(r)=− ∇ /Phi1(r)will be unaﬀected. By reciprocity we can rewrite (3.74)as ∂G N(r|r/prime) ∂n=−1 Afor all r∈S. (3.75) The Green’s function GNso deﬁned is known as the Neumann Green’s function . Observe that if Vis not ﬁnite-sized then A→∞ and according to (3.74)the choice (3.72)becomes allowable. Finding the Green’s function that obeys one of the boundary conditions for a given geometry is often a diﬃcult task. Nevertheless, certain canonical geometries make theGreen’s function approach straightforward and simple. Such is the case in image theory,when a charge is located near a simple conducting body such as a ground screen ora sphere. In these cases the function F(r|r /prime)consists of a single correction term as in (3.68). We shall consider these simple cases in examples to follow. Reciprocity of the static Green’s function. It remains to show that G(r|r/prime)=G(r/prime|r) for any of the Green’s functions introduced above. The unbounded-space Green’s function is reciprocal by inspection; |r−r/prime|is unaﬀected by interchanging rand r/prime. However, we can give a more general treatment covering this case as well as the Dirichlet and Neumanncases. We begin with ∇ 2G(r|r/prime)=−δ(r−r/prime). In Green’s second identity let φ(r)=G(r|ra), ψ( r)=G(r|rb), where raand rbare arbitrary points, and integrate over the unprimed coordinates. We have/integraldisplay V[G(r|ra)∇2G(r|rb)−G(r|rb)∇2G(r|ra)]dV= −/contintegraldisplay S/bracketleftbigg G(r|ra)∂G(r|rb) ∂n−G(r|rb)∂G(r|ra) ∂n/bracketrightbigg dS. IfGis the unbounded-space Green’s function, the surface integral must vanish since SB→∞. It must also vanish under Dirichlet or Neumann boundary conditions. Since ∇2G(r|ra)=−δ(r−ra), ∇2G(r|rb)=−δ(r−rb), we have/integraldisplay V[G(r|ra)δ(r−rb)−G(r|rb)δ(r−ra)]dV=0, hence G(rb|ra)=G(ra|rb) by the sifting property. By the arbitrariness of raand rb, reciprocity is established. Electrostatic shielding. The Dirichlet Green’s function can be used to explain elec- trostatic shielding . We consider a closed, grounded, conducting shell with charge outside butnotinside(Figure3.7).By(3.71)thepotentialatpointsinsidetheshellis /Phi1(r)=/contintegraldisplay SB/Phi1(r/prime)∂GD(r|r/prime) ∂n/primedS/prime, Figure 3.7: Electrostatic shielding by a conducting shell. where SBis tangential to the inner surface of the shell and we have used ρ=0within the shell. Because /Phi1(r/prime)=0for all r/primeonSB,w eh a v e /Phi1(r)=0 everywhere in the region enclosed by the shell. This result is independent of the charge outside the shell, and the interior region is “shielded” from the eﬀects of that charge. Conversely, consider a grounded conducting shell with charge contained inside. If we surround the outside of the shell by a surface S1and let SBrecede to inﬁnity, then (3.71) becomes /Phi1(r)=lim SB→∞/contintegraldisplay SB/Phi1(r/prime)∂GD(r|r/prime) ∂n/primedS/prime+/contintegraldisplay S1/Phi1(r/prime)∂GD(r|r/prime) ∂n/primedS/prime. Again there is no charge in V(since the charge lies completely inside the shell). The contribution from SBvanishes. Since S1lies adjacent to the outer surface of the shell, /Phi1(r/prime)≡0onS1.T h u s /Phi1(r)=0for all points outside the conducting shell. Example solution to Poisson’s equation: planar layered media. For simple geometries Poisson’s equation may be solved as part of a boundary value problem (§A.4). Occasionally such a solution has an appealing interpretation as the superposition of potentials produced by the physical charge and its “images.” We shall consider here thecase of planar media and subsequently use the results to predict the potential producedby charge near a conducting sphere. Consider a layered dielectric medium where various regions of space are separated by planes at constant values of z. Material region ioccupies volume region V iand has permittivity /epsilon1i; it may or may not contain source charge. The solution to Poisson’s equation is given by (3.56). The contribution /Phi1p(r)=/integraldisplay VG(r|r/prime)ρ(r/prime) /epsilon1dV/prime produced by sources within Vis known as the primary potential . The term /Phi1s(r)=/contintegraldisplay S/bracketleftbigg /Phi1(r/prime)∂G(r|r/prime) ∂n/prime−G(r|r/prime)∂/Phi1(r/prime) ∂n/prime/bracketrightbigg dS/prime, on the other hand, involves an integral over the surface ﬁelds and is known as the sec- ondary potential . This term is linked to eﬀects outside V. Since the “sources” of /Phi1s (i.e., the surface ﬁelds)lie on the boundary of V,/Phi1ssatisﬁes Laplace’s equation within V. We may therefore use other, more convenient, representations of /Phi1sprovided they satisfy Laplace’s equation. However, as solutions to a homogeneous equation they are ofindeﬁnite form until linked to appropriate boundary values. Since the geometry is invariant in the xand ydirections, we represent each potential function in terms of a 2-D Fourier transform over these variables. We leave the zdepen- dence intact so that we may apply boundary conditions directly in the spatial domain.The transform representations of the Green’s functions for the primary and secondarypotentials are derived in Appendix A. From (A.55)we see that the primary potentialwithin region V ican be written as /Phi1p i(r)=/integraldisplay ViGp(r|r/prime)ρ(r/prime) /epsilon1idV/prime(3.76) where Gp(r|r/prime)=1 4π|r−r/prime|=1 (2π)2/integraldisplay∞ −∞e−kρ|z−z/prime| 2kρejkρ·(r−r/prime)d2kρ (3.77) is the primary Green’s function with kρ=ˆxkx+ˆyky,kρ=|kρ|, and d2kρ=dkxdky. We also ﬁnd in (A.56)that a solution of Laplace’s equation can be written as /Phi1s(r)=1 (2π)2/integraldisplay∞ −∞/bracketleftbig A(kρ)ekρz+B(kρ)e−kρz/bracketrightbig ejkρ·rd2kρ (3.78) where A(kρ)and B(kρ)must be found by the application of appropriate boundary con- ditions. As a simple example, consider a charge distribution ρ(r)in free space above a grounded conducting plane located at z=0. We wish to ﬁnd the potential in the region z>0 using the Fourier transform representation of the potentials. The total potential is a sumof primary and secondary terms: /Phi1(x,y,z)=/integraldisplay V/bracketleftBigg 1 (2π)2/integraldisplay∞ −∞e−kρ|z−z/prime| 2kρejkρ·(r−r/prime)d2kρ/bracketrightBigg ρ(r/prime) /epsilon10dV/prime+ +1 (2π)2/integraldisplay∞ −∞/bracketleftbig B(kρ)e−kρz/bracketrightbig ejkρ·rd2kρ, where the integral is over the region z>0. Here we have set A(kρ)=0because ekρz grows with increasing z. Since the plane is grounded we must have /Phi1(x,y,0)=0. Because z<z/primewhen we apply this condition, we have |z−z/prime|=z/prime−zand thus /Phi1(x,y,0)=1 (2π)2/integraldisplay∞ −∞/bracketleftBigg/integraldisplay Vρ(r/prime) /epsilon10e−kρz/prime 2kρe−jkρ·r/primedV/prime+B(kρ)/bracketrightBigg ejkρ·rd2kρ=0. Invoking the Fourier integral theorem we ﬁnd B(kρ)=−/integraldisplay Vρ(r/prime) /epsilon10e−kρz/prime 2kρe−jkρ·r/primedV/prime, Figure 3.8: Construction of electrostatic Green’s function for a ground plane. hence the total potential is /Phi1(x,y,z)=/integraldisplay V/bracketleftBigg 1 (2π)2/integraldisplay∞ −∞e−kρ|z−z/prime|−e−kρ(z+z/prime) 2kρejkρ·(r−r/prime)d2kρ/bracketrightBigg ρ(r/prime) /epsilon10dV/prime =/integraldisplay VG(r|r/prime)ρ(r/prime) /epsilon10dV/prime where G(r|r/prime)is the Green’s function for the region above a grounded planar conductor. We can interpret this Green’s function as a sum of the primary Green’s function (3.77)and a secondary Green’s function G s(r|r/prime)=−1 (2π)2/integraldisplay∞ −∞e−kρ(z+z/prime) 2kρejkρ·(r−r/prime)d2kρ. (3.79) Forz>0the term z+z/primecan be replaced by |z+z/prime|. Then, comparing (3.79)with (3.77) , we see that Gs(r|x/prime,y/prime,z/prime)=−Gp(r|x/prime,y/prime,−z/prime)=−1 4π|r−r/prime i|(3.80) where r/prime i=ˆxx/prime+ˆyy/prime−ˆzz/prime. Because the Green’s function is the potential of a point charge, we may interpret the secondary Green’s function as produced by a negative unit chargeplacedinaposition−z /primeimmediatel ybeneaththepositiveunitchargethatproduces Gp (Figure3.8).Thissecondar ychargeisthe“image ”oftheprimar ycharge.Thattwosuch charges would produce a null potential on the ground plane is easily veriﬁed. As a more involved example, consider a charge distribution ρ(r)above a planar in- terface separating two homogeneous dielectric media. Region 1 occupies z>0and has permittivity /epsilon11, while region 2 occupies z<0and has permittivity /epsilon12. In region 1 we can write the total potential as a sum of primary and secondary components, discardingthe term that grows with z: /Phi1 1(x,y,z)=/integraldisplay V/bracketleftBigg 1 (2π)2/integraldisplay∞ −∞e−kρ|z−z/prime| 2kρejkρ·(r−r/prime)d2kρ/bracketrightBigg ρ(r/prime) /epsilon11dV/prime+ +1 (2π)2/integraldisplay∞ −∞/bracketleftbig B(kρ)e−kρz/bracketrightbig ejkρ·rd2kρ. (3.81) With no source in region 2, the potential there must obey Laplace’s equation and there- fore consists of only a secondary component: /Phi12(r)=1 (2π)2/integraldisplay∞ −∞/bracketleftbig A(kρ)ekρz/bracketrightbig ejkρ·rd2kρ. (3.82) To determine Aand Bwe impose (3.36)and (3.37) . By (3.36)we have 1 (2π)2/integraldisplay∞ −∞/bracketleftBigg/integraldisplay Vρ(r/prime) /epsilon11e−kρz/prime 2kρe−jkρ·r/primedV/prime+B(kρ)−A(kρ)/bracketrightBigg ejkρ·rd2kρ=0, hence /integraldisplay Vρ(r/prime) /epsilon11e−kρz/prime 2kρe−jkρ·r/primedV/prime+B(kρ)−A(kρ)=0 by the Fourier integral theorem. Applying (3.37)at z=0with ˆn12=ˆz, and noting that there is no excess surface charge, we ﬁnd /integraldisplay Vρ(r/prime)e−kρz/prime 2kρe−jkρ·r/primedV/prime−/epsilon11B(kρ)−/epsilon12A(kρ)=0. The solutions A(kρ)=2/epsilon11 /epsilon11+/epsilon12/integraldisplay Vρ(r/prime) /epsilon11e−kρz/prime 2kρe−jkρ·r/primedV/prime, B(kρ)=/epsilon11−/epsilon12 /epsilon11+/epsilon12/integraldisplay Vρ(r/prime) /epsilon11e−kρz/prime 2kρe−jkρ·r/primedV/prime, are then substituted into (3.81)and (3.82)to give /Phi11(r)=/integraldisplay V/bracketleftBigg 1 (2π)2/integraldisplay∞ −∞e−kρ|z−z/prime|+/epsilon11−/epsilon12 /epsilon11+/epsilon12e−kρ(z+z/prime) 2kρejkρ·(r−r/prime)d2kρ/bracketrightBigg ρ(r/prime) /epsilon11dV/prime =/integraldisplay VG1(r|r/prime)ρ(r/prime) /epsilon11dV/prime, /Phi12(r)=/integraldisplay V/bracketleftBigg 1 (2π)2/integraldisplay∞ −∞2/epsilon12 /epsilon11+/epsilon12e−kρ(z/prime−z) 2kρejkρ·(r−r/prime)d2kρ/bracketrightBigg ρ(r/prime) /epsilon12dV/prime =/integraldisplay VG2(r|r/prime)ρ(r/prime) /epsilon12dV/prime. Since z/prime>zfor all points in region 2, we can replace z/prime−zby|z−z/prime|in the formula for /Phi12. As with the previous example, let us compare the result to the form of the primary Green’s function (3.77). We see that G1(r|r/prime)=1 4π|r−r/prime|+/epsilon11−/epsilon12 /epsilon11+/epsilon121 4π|r−r/prime 1|, G2(r|r/prime)=2/epsilon12 /epsilon11+/epsilon121 4π|r−r/prime 2|, where r/prime 1=ˆxx/prime+ˆyy/prime−ˆzz/primeand r/prime 2=ˆxx/prime+ˆyy/prime+ˆzz/prime. So we can also write /Phi11(r)=1 4π/integraldisplay V/bracketleftbigg1 |r−r/prime|+/epsilon11−/epsilon12 /epsilon11+/epsilon121 |r−r/prime 1|/bracketrightbiggρ(r/prime) /epsilon11dV/prime, /Phi12(r)=1 4π/integraldisplay V/bracketleftbigg2/epsilon12 /epsilon11+/epsilon121 |r−r/prime 2|/bracketrightbiggρ(r/prime) /epsilon12dV/prime. Figure 3.9: Green’s function for a grounded conducting sphere. Note that /Phi12→/Phi11as/epsilon12→/epsilon11. There is an image interpretation for the secondary Green’s functions. The secondary Green’s function for region 1 appears as a potential produced by an image of the primarycharge located at −z /primein an inﬁnite medium of permittivity /epsilon11, and with an amplitude of (/epsilon11−/epsilon12)/(/epsilon1 1+/epsilon12)times the primary charge. The Green’s function in region 2 is produced by an image charge located at z/prime(i.e., at the location of the primary charge)in an inﬁnite medium of permittivity /epsilon12with an amplitude of 2/epsilon12/(/epsilon11+/epsilon12)times the primary charge. Example solution to Poisson’s equation: conducting sphere. As an example involving a nonplanar geometry, consider the potential produced by a source near agrounde dconductin gsphereinfreespace(Figure3.9).Basedonourexperienc ewith planar layered media, we hypothesize that the secondary potential will be produced byan image charge; hence we try the simple Green’s function G s(r|r/prime)=A(r/prime) 4π|r−r/prime i| where the amplitude Aand location r/prime iof the image are to be determined. We further assume, based on our experience with planar problems, that the image charge will resideinside the sphere along a line joining the origin to the primary charge. Since r=aˆrfor all points on the sphere, the total Green’s function must obey the Dirichlet condition G(r|r /prime)|r=a=1 4π|r−r/prime|/vextendsingle/vextendsingle/vextendsingle/vextendsingle r=a+A(r/prime) 4π|r−r/prime i|/vextendsingle/vextendsingle/vextendsingle/vextendsingle r=a=1 4π|aˆr−r/primeˆr/prime|+A(r/prime) 4π|aˆr−r/prime iˆr/prime|=0 in order to have the potential, given by (3.56), vanish on the sphere surface. Factoring a from the ﬁrst denominator and r/prime ifrom the second we obtain 1 4πa|ˆr−r/prime aˆr/prime|+A(r/prime) 4πr/prime i|a r/prime iˆr−ˆr/prime|=0. Now|kˆr−k/primeˆr/prime|=k2+k/prime2−2kk/primecosγwhere γis the angle between ˆrand ˆr/primeand k,k/prime are constants; this means that |kˆr−ˆr/prime|=| ˆr−kˆr/prime|. Hence as long as we choose r/prime a=a r/prime i,A r/prime i=−1 a, the total Green’s function vanishes everywhere on the surface of the sphere. The image charge is therefore located within the sphere at r/prime i=a2r/prime/r/prime2and has amplitude A= −a/r/prime. (Note that both the location and amplitude of the image depend on the location of the primary charge.)With this Green’s function and (3.71) , the potential of anarbitrary source placed near a grounded conducting sphere is /Phi1(r)=/integraldisplay Vρ(r/prime) /epsilon11 4π/bracketleftBigg 1 |r−r/prime|−a/r/prime |r−a2 r/prime2r/prime|/bracketrightBigg dV/prime. The Green’s function may be used to compute the surface charge density induced on the sphere by a unit point charge: it is merely necessary to ﬁnd the normal component ofelectric ﬁeld from the gradient of /Phi1(r). We leave this as an exercise for the reader, who may then integrate the surface charge and thereby show that the total charge inducedon the sphere is equal to the image charge. So the total charge induced on a groundedsphere by a point charge qat a point r=r /primeisQ=−qa/r/prime. It is possible to ﬁnd the total charge induced on the sphere without ﬁnding the image charge ﬁrst. This is an application of Green’s reciprocation theorem ( §3.4.4). According to (3.211), if we can ﬁnd the potential VPat a point rproduced by the sphere when it is isolated and carrying a total charge Q0, then the total charge Qinduced on the grounded sphere in the vicinity of a point charge qplaced at ris given by Q=−qVP/V1 where V1is the potential of the isolated sphere. We can apply this formula by noting that an isolated sphere carrying charge Q0produces a ﬁeld E(r)=ˆrQ0/4π/epsilon1r2. Integration from a radius rto inﬁnity gives the potential referred to inﬁnity: /Phi1(r)=Q0/4π/epsilon1r.So the potential of the isolated sphere is V1=Q0/4π/epsilon1a, while the potential at radius r/primeis VP=Q0/4π/epsilon1r/prime. Substitution gives Q=−qa/r/primeas before. 3.2.5 Force and energy Maxwell’s stress tensor. The electrostatic version of Maxwell’s stress tensor can be obtained from (2.288)by setting B=H=0: ¯Te=1 2(D·E)¯I−DE. (3.83) The total electric force on the charges in a region Vbounded by the surface Sis given by the relation Fe=−/contintegraldisplay S¯Te·dS=/integraldisplay VfedV where fe=ρEis the electric force volume density. In particular, suppose that Sis adjacent to a solid conducting body embedded in a dielectric having permittivity /epsilon1(r). Since all the charge is at the surface of the conductor, the force within Vacts directly on the surface. Thus, −¯Te·ˆnis the surface force density (traction )t. Using D=/epsilon1E, and remembering that the ﬁelds are normal to the conductor, we ﬁnd that ¯Te·ˆn=1 2/epsilon1E2 nˆn−/epsilon1EE·ˆn=−1 2/epsilon1E2 nˆn=−1 2ρsE. The surface force density is perpendicular to the surface. As a simple but interesting example, consider the force acting on a rigid conducting sphere of radius acarrying total charge Qin a homogeneous medium. At equilibrium the charge is distributed uniformly with surface density ρs=Q/4πa2, producing a ﬁeld E=ˆrQ/4π/epsilon1r2external to the sphere. Hence a force density t=1 2ˆrQ2 /epsilon1(4πa2)2 acts at each point on the surface. This would cause the sphere to expand outward if the structural integrity of the material were to fail. Integration over the entire sphere yields F=1 2Q2 /epsilon1(4πa2)2/integraldisplay SˆrdS=0. However, integration of tover the upper hemisphere yields F=1 2Q2 /epsilon1(4πa2)2/integraldisplay2π 0/integraldisplayπ/2 0ˆra2sinθdθdφ. Substitution of ˆr=ˆxsinθcosφ+ˆysinθsinφ+ˆzcosθleads immediately to Fx=Fy=0, but the z-component is Fz=1 2Q2 /epsilon1(4πa2)2/integraldisplay2π 0/integraldisplayπ/2 0a2cosθsinθdθdφ=Q2 32/epsilon1πa2. This result can also be obtained by integrating −¯Te·ˆnover the entire xy-plane with ˆn=− ˆz. Since −¯Te·(−ˆz)=ˆz/epsilon1 2E·Ewe have F=ˆz1 2Q2 (4π/epsilon1)2/integraldisplay2π 0/integraldisplay∞ ard rdφ r4=ˆzQ2 32/epsilon1πa2. As a more challenging example, consider two identical line charges parallel to the z- axis and located at x=± d/2,y=0in free space. We can ﬁnd the force on one line charge due to the other by integrating Maxwell’s stress tensor over the yz-plane. From (3.64)we ﬁnd that the total electric ﬁeld on the yz-plane is E(y,z)=y y2+(d/2)2ρl π/epsilon10ˆy where ρlis the line charge density. The force density for either line charge is −¯Te·ˆn, where we use ˆn=± ˆxto obtain the force on the charge at x=∓d/2. The force density for the charge at x=−d/2is ¯Te·ˆn=1 2(D·E)¯I·ˆx−DE·ˆx=/epsilon10 2/bracketleftbiggy y2+(d/2)2ρl π/epsilon10/bracketrightbigg2 ˆx and the total force is F−=−/integraldisplay∞ −∞/integraldisplay∞ −∞ρ2 l 2π2/epsilon10y2 /bracketleftbig y2+(d/2)2/bracketrightbig2ˆxdydz. On a per unit length basis the force is F− l=− ˆxρ2 l 2π2/epsilon10/integraldisplay∞ −∞y2 [y2+(d/2)2]2dy=− ˆxρ2 l 2πd/epsilon10. Note that the force is repulsive as expected. Figure 3.10: Computation of electrostatic stored energy via the assembly energy of a charge distribution. Electrostatic stored energy. In§2.9.5 we considered the energy relations for the electromagnetic ﬁeld. Those relations remain valid in the static case. Since our interpre-tation of the dynamic relations was guided in part by our knowledge of the energy storedin a static ﬁeld, we must, for completeness, carry out a study of that eﬀect here. The energy of a static conﬁguration is taken to be the work required to assemble the conﬁguration from a chosen starting point. For a conﬁguration of static charges, thestored electric energy is the energy required to assemble the conﬁguration, starting with all charges removed to inﬁnite distance (the assumed zero potential reference). If theassembled charges are not held in place by an external mechanical force they will move,thereby converting stored electric energy into other forms of energy (e.g., kinetic energyand radiation). By (3.62), the work required to move a point charge qfrom a reservoir at inﬁnity to a point Patrin a potential ﬁeld /Phi1is W=q/Phi1(r). If instead we have a continuous charge density ρpresent, and wish to increase this to ρ+δρby bringing in a small quantity of charge δρ, a total work δW=/integraldisplay V∞δρ(r)/Phi1(r)dV (3.84) is required, and the potential ﬁeld is increased to /Phi1+δ/Phi1. Here V∞denotes all of space. (We could restrict the integral to the region containing the charge, but we shall ﬁnd ithelpful to extend the domain of integration to all of space.) Nowconside rthesituatio nshowninFigure3.10.Herewehavechargeintheformof both volume densities and surface densities on conducting bodies. Also present may belinear material bodies. We can think of assembling the charge in two distinctly diﬀerent ways. We could, for instance, bring small portions of charge (or point charges)together to form the distribution ρ. Or, we could slowly build up ρby adding inﬁnitesimal, but spatially identical, distributions. That is, we can create the distribution ρfrom a zero initial state by repeatedly adding a charge distribution δρ(r)=ρ(r)/N, where Nis a large number. Whenever we add δρwe must perform the work given by (3.84), but we also increase the potential proportionately (remembering that all materialsare assumed linear). At each step, more work is required. The total work is W= N/summationdisplay n=1/integraldisplay V∞δρ(r)[(n−1)δ/Phi1( r)]dV=/bracketleftBiggN/summationdisplay n=1(n−1)/bracketrightBigg/integraldisplay V∞ρ(r) N/Phi1(r) NdV. (3.85) We must use an inﬁnite number of steps so that no energy is lost to radiation at any step (since the charge we add each time is inﬁnitesimally small). Using N/summationdisplay n=1(n−1)=N(N−1)/2, (3.85)becomes W=1 2/integraldisplay V∞ρ(r)/Phi1(r)dV (3.86) asN→∞. Finally, since some assembled charge will be in the form of a volume density and some in the form of the surface density on conductors, we can generalize (3.86)to W=1 2/integraldisplay V/primeρ(r)/Phi1(r)dV+1 2I/summationdisplay i=1QiVi. (3.87) Here V/primeis the region outside the conductors, Qiis the total charge on the ith conductor (i=1,..., I), and Viis the absolute potential (referred to inﬁnity)of the ith conductor. An intriguing property of electrostatic energy is that the charges on the conductors will arrange themselves, while seeking static equilibrium, into a minimum-energy conﬁg-uration (Thomson’s theorem). In keeping with our ﬁeld-centered view of electromagnetics, we now wish to write the energy (3.86)entirely in terms of the ﬁeld vectors Eand D. Since ρ=∇· Dwe have W=1 2/integraldisplay V∞[∇·D(r)]/Phi1(r)dV. Then, by (B.42), W=1 2/integraldisplay V∞∇·[/Phi1(r)D(r)]dV−1 2/integraldisplay V∞D(r)·[∇/Phi1(r)]dV. Use of the divergence theorem and (3.30)leads to W=1 2/contintegraldisplay S∞/Phi1(r)D(r)·dS+1 2/integraldisplay V∞D(r)·E(r)dV Figure 3.11: Multipole expansion. where S∞is the bounding surface that recedes toward inﬁnity to encompass all of space. Because /Phi1∼1/rand D∼1/r2asr→∞, the integral over S∞tends to zero and W=1 2/integraldisplay V∞D(r)·E(r)dV. (3.88) Hence we may compute the assembly energy in terms of the ﬁelds supported by the charge ρ. It is signiﬁcant that the assembly energy Wis identical to the term within the time derivative in Poynting’s theorem (2.299). Hence our earlier interpretation, that this termrepresents the time-rate of change of energy “stored” in the electric ﬁeld, has a ﬁrm basis.Of course, the assembly energy is a static concept, and our generalization to dynamicﬁelds is purely intuitive. We also face similar questions regarding the meaning of energydensity, and whether energy can be “localized” in space. The discussions in §2.9.5 still apply. 3.2.6 Multipole expansion Consider an arbitrary but spatially localized charge distribution of total charge Q inanunbounde dhomogeneou smediu m(Figure3.11).Wehavealread yobtaine dthe potential (3.61)of the source; as we move the observation point away, /Phi1should decrease in a manner roughly proportional to 1/r. The actual variation depends on the nature of the charge distribution and can be complicated. Often this dependence is dominated by a speciﬁc inverse power of distance for observation points far from the source, and wecan investigate it by expanding the potential in powers of 1/r. Although such multipole expansions of the potential are rarely used to perform actual computations, they can provide insight into both the behavior of static ﬁelds and the physical meaning of thepolarization vector P. Let us place our origin of coordinates somewhere within the charge distribution, as showninFigure3.11,andexpan dtheGreen’ sfunctio nspatia ldependenc einathree- dimensional Taylor series about the origin: 1 R=∞/summationdisplay n=01 n!(r/prime·∇/prime)n1 R/vextendsingle/vextendsingle/vextendsingle/vextendsingle r/prime=0=1 r+(r/prime·∇/prime)1 R/vextendsingle/vextendsingle/vextendsingle/vextendsingle r/prime=0+1 2(r/prime·∇/prime)21 R/vextendsingle/vextendsingle/vextendsingle/vextendsingle r/prime=0+···,(3.89) where R=|r−r/prime|. Convergence occurs if |r|>|r/prime|. In the notation (r/prime·∇/prime)nwe interpret a power on a derivative operator as the order of the derivative. Substituting (3.89)into (3.61)and writing the derivatives in Cartesian coordinates we obtain /Phi1(r)=1 4π/epsilon1/integraldisplay Vρ(r/prime)/bracketleftbigg1 R/vextendsingle/vextendsingle/vextendsingle/vextendsingle r/prime=0+(r/prime·∇/prime)1 R/vextendsingle/vextendsingle/vextendsingle/vextendsingle r/prime=0+1 2(r/prime·∇/prime)21 R/vextendsingle/vextendsingle/vextendsingle/vextendsingle r/prime=0+···/bracketrightbigg dV/prime.(3.90) For the second term we can use (3.57)to write (r/prime·∇/prime)1 R/vextendsingle/vextendsingle/vextendsingle/vextendsingle r/prime=0=r/prime·/parenleftbigg ∇/prime1 R/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle r/prime=0=r/prime·/parenleftBiggˆR R2/parenrightBigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle r/prime=0=r/prime·ˆr r2. (3.91) The third term is complicated. Let us denote (x,y,z)by(x1,x2,x3)and perform an expansion in rectangular coordinates: (r/prime·∇/prime)21 R/vextendsingle/vextendsingle/vextendsingle/vextendsingle r/prime=0=3/summationdisplay i=13/summationdisplay j=1x/prime ix/prime j∂2 ∂x/prime i∂x/prime j1 R/vextendsingle/vextendsingle/vextendsingle/vextendsingle r/prime=0. It turns out [172] that this can be written as (r/prime·∇/prime)21 R/vextendsingle/vextendsingle/vextendsingle/vextendsingle r/prime=0=1 r3ˆr·(3r/primer/prime−r/prime2¯I)·ˆr. Substitution into (3.90)gives /Phi1(r)=Q 4π/epsilon1r+ˆr·p 4π/epsilon1r2+1 2ˆr·¯Q·ˆr 4π/epsilon1r3+···, (3.92) which is the multipole expansion for /Phi1(r). It converges for all r>rmwhere rmis the radiusofthesmalles tspherecompletel ycontainin gthechargecenteredat r/prime= 0 (Figure 3.11).In(3.92)theterms Q, p, ¯Q,andsoonarecalledthemultipol emoment sofρ(r). The ﬁrst moment is merely the total charge Q=/integraldisplay Vρ(r/prime)dV/prime. The second moment is the electric dipole moment vector p=/integraldisplay Vr/primeρ(r/prime)dV/prime. The third moment is the electric quadrupole moment dyadic ¯Q=/integraldisplay V(3r/primer/prime−r/prime2¯I)ρ(r/prime)dV/prime. The expansion (3.92)allows us to identify the dominant power of rforr/greatermuchrm. The ﬁrst nonzero term in (3.92)dominates the potential at points far from the source.Interestingly, the ﬁrst nonvanishing moment is independent of the location of the originofr /prime, while all subsequent higher moments depend on the location of the origin [91]. We can see this most easily through a few simple examples. For a single point charge qlocated at r0we can write ρ(r)=qδ(r−r0). The ﬁrst moment of ρis Q=/integraldisplay Vqδ(r/prime−r0)dV/prime=q. Figure 3.12: A dipole distribution. Note that this is independent of r0. The second moment p=/integraldisplay Vr/primeqδ(r/prime−r0)dV/prime=qr0 depends on r0, as does the third moment ¯Q=/integraldisplay V(3r/primer/prime−r/prime2¯I)qδ(r/prime−r0)dV/prime=q(3r0r0−r2 0¯I). Ifr0=0then only the ﬁrst moment is nonzero; that this must be the case is obvious from (3.61). ForthedipoleofFigure3.12wecanwrite ρ(r)=−qδ(r−r0+d/2)+qδ(r−r0−d/2). In this case Q=−q+q=0, p=qd, ¯Q=q[3(r0d+dr0)−2(r0·d)¯I]. Only the ﬁrst nonzero moment, in this case p, is independent of r0.F o r r0=0the only nonzero multipole moment would be the dipole moment p. If the dipole is aligned along thez-axis with d=dˆzand r0=0, then the exact potential is /Phi1(r)=1 4π/epsilon1pcosθ r2. By (3.30)we have E(r)=1 4π/epsilon1p r3(ˆr2 cosθ+ˆθsinθ), (3.93) which is the classic result for the electric ﬁeld of a dipole. Finall y,conside rthequadru poleshowninFigure3.13.Thechargedensityis ρ(r)=−qδ(r−r0)+qδ(r−r0−d1)+qδ(r−r0−d2)−qδ(r−r0−d1−d2). Figure 3.13: A quadrupole distribution. Carrying through the details, we ﬁnd that the ﬁrst two moments of ρvanish, while the third is given by ¯Q=q[−3(d1d2+d2d1)+2(d1·d2)¯I]. As expected, it is independent of r0. It is tedious to carry (3.92)beyond the quadrupole term using the Taylor expansion. Another approach is to expand 1/Rin spherical harmonics. Referring to Appendix E.3 we ﬁnd that 1 |r−r/prime|=4π∞/summationdisplay n=0n/summationdisplay m=−n1 2n+1r/primen rn+1Y∗ nm(θ/prime,φ/prime)Ynm(θ, φ) (see Jackson [91] or Arfken [5] for a detailed derivation). This expansion converges for |r|>|rm|. Substitution into (3.61)gives /Phi1(r)=1 /epsilon1∞/summationdisplay n=01 rn+1/bracketleftBigg 1 2n+1n/summationdisplay m=−nqnmYnm(θ, φ)/bracketrightBigg (3.94) where qnm=/integraldisplay Vρ(r/prime)r/primenY∗ nm(θ/prime,φ/prime)dV/prime. We can now identify any inverse power of rin the multipole expansion, but at the price of dealing with a double summation. For a charge distribution with axial symmetry (noφ-variation), only the coeﬃcient q n0is nonzero. The relation Yn0(θ, φ) =/radicalbigg 2n+1 4πPn(cosθ) allows us to simplify (3.94)and obtain /Phi1(r)=1 4π/epsilon1∞/summationdisplay n=01 rn+1qnPn(cosθ) (3.95) where qn=2π/integraldisplay r/prime/integraldisplay θ/primeρ(r/prime,θ/prime)r/primenPn(cosθ/prime)r/prime2sinθ/primedθ/primedr/prime. As a simple example consider a spherical distribution of charge given by ρ(r)=3Q πa3cosθ, r≤a. This can be viewed as two adjacent hemispheres carrying total charges ±Q. Since cosθ= P1(cosθ), we compute qn=2π/integraldisplaya 0/integraldisplayπ 03Q πa3P1(cosθ/prime)r/primenPn(cosθ/prime)r/prime2sinθ/primedθ/primedr/prime =2π3Q πa3an+3 n+3/integraldisplayπ 0P1(cosθ)Pn(cosθ/prime)sinθ/primedθ/prime. Using the orthogonality relation (E.123)we ﬁnd qn=2π3Q πa3an+3 n+3δ1n2 2n+1. Hence the only nonzero coeﬃcient is q1=Qaand /Phi1(r)=1 4π/epsilon11 r2QaP 1(cosθ)=Qa 4π/epsilon1r2cosθ. This is the potential of a dipole having moment p=ˆzQa. Thus we could replace the sphere with point charges ∓Qatz=∓a/2without changing the ﬁeld for r>a. Physical interpretation of the polarization vector in a dielectric. We have used the Maxwell–Minkowski equations to determine the electrostatic potential of acharge distribution in the presence of a dielectric medium. Alternatively, we can usethe Maxwell–Boﬃ equations ∇× E=0, (3.96) ∇·E=1 /epsilon10(ρ−∇· P). (3.97) Equation (3.96)allows us to deﬁne a scalar potential through (3.30) . Substitution into (3.97)gives ∇2/Phi1(r)=−1 /epsilon10[ρ(r)+ρP(r)] (3.98) where ρP=− ∇· P. This has the form of Poisson’s equation (3.50), but with charge density term ρ(r)+ρP(r). Hence the solution is /Phi1(r)=1 4π/epsilon10/integraldisplay Vρ(r/prime)−∇/prime·P(r/prime) |r−r/prime|dV/prime. To this we must add any potential produced by surface sources such as ρs. If there is a discontinuity in the dielectric region, there is also a surface polarization source ρPs=ˆn·P according to (3.35). Separating the volume into regions with bounding surfaces Siacross which the permittivity is discontinuous, we may write /Phi1(r)=1 4π/epsilon10/integraldisplay Vρ(r/prime) |r−r/prime|dV/prime+1 4π/epsilon10/integraldisplay Sρs(r/prime) |r−r/prime|dS/prime+ +/summationdisplay i/bracketleftbigg1 4π/epsilon10/integraldisplay Vi−∇/prime·P(r/prime) |r−r/prime|dV/prime+1 4π/epsilon10/contintegraldisplay Siˆn/prime·P(r/prime) |r−r/prime|dS/prime/bracketrightbigg , (3.99) where ˆnpoints outward from region i. Using the divergence theorem on the fourth term and employing (B.42), we obtain /Phi1(r)=1 4π/epsilon10/integraldisplay Vρ(r/prime) |r−r/prime|dV/prime+1 4π/epsilon10/integraldisplay Sρs(r/prime) |r−r/prime|dS/prime+ +/summationdisplay i/bracketleftbigg1 4π/epsilon10/integraldisplay ViP(r/prime)·∇/prime/parenleftbigg1 |r−r/prime|/parenrightbigg dV/prime/bracketrightbigg . Since ∇/prime(1/R)=ˆR/R2, the third term is a sum of integrals of the form 1 4π/epsilon1/integraldisplay ViP(r/prime)·ˆR R2dV. Comparing this to the second term of (3.92), we see that this integral represents a volume superposition of dipole terms where Pis a volume density of dipole moments. Thus, a dielectric with permittivity /epsilon1is equivalent to a volume distribution of dipoles in free space. No higher-order moments are required, and no zero-order moments areneeded since any net charge is included in ρ. Note that we have arrived at this conclusion based only on Maxwell’s equations and the assumption of a linear, isotropic relationshipbetween Dand E. Assuming our macroscopic theory is correct, we are tempted to make assumptions about the behavior of matter on a microscopic level (e.g., atoms exposed toﬁelds are polarized and their electron clouds are displaced from their positively chargednuclei), but this area of science is better studied from the viewpoints of particle physicsand quantum mechanics. Potential of an azimuthally-symmetric charged spherical surface. In several of our example problems we shall be interested in evaluating the potential of a chargedspherical surface. When the charge is azimuthally-symmetric, the potential is particularlysimple. We will need the value of the integral F(r)=1 4π/integraldisplay Sf(θ/prime) |r−r/prime|dS/prime(3.100) where r=rˆrdescribes an arbitrary observation point and r/prime=aˆr/primeidentiﬁes the source point on the surface of the sphere of radius a. The integral is most easily done using the expansion (E.200)for |r−r/prime|−1in spherical harmonics. We have F(r)=a2∞/summationdisplay n=0n/summationdisplay m=−nYnm(θ, φ) 2n+1rn < rn+1 >/integraldisplayπ −π/integraldisplayπ 0f(θ/prime)Y∗ nm(θ/prime,φ/prime)sinθ/primedθ/primedφ/prime where r<=min{r,a}and r>=max{r,a}. Using orthogonality of the exponentials we ﬁnd that only the m=0terms contribute: F(r)=2πa2∞/summationdisplay n=0Yn0(θ, φ) 2n+1rn < rn+1 >/integraldisplayπ 0f(θ/prime)Y∗ n0(θ/prime,φ/prime)sinθ/primedθ/prime. Finally, since Yn0=/radicalbigg 2n+1 4πPn(cosθ) we have F(r)=1 2a2∞/summationdisplay n=0Pn(cosθ)rn < rn+1 >/integraldisplayπ 0f(θ/prime)Pn(cosθ/prime)sinθ/primedθ/prime. (3.101) As an example, suppose f(θ)=cosθ=P1(cosθ). Then F(r)=1 2a2∞/summationdisplay n=0Pn(cosθ)rn < rn+1 >/integraldisplayπ 0P1(cosθ/prime)Pn(cosθ/prime)sinθ/primedθ/prime. The orthogonality of the Legendre polynomials can be used to show that /integraldisplayπ 0P1(cosθ/prime)Pn(cosθ/prime)sinθ/primedθ/prime=2 3δ1n, hence F(r)=a2 3cosθr< r2>. (3.102) 3.2.7 Field produced by a permanently polarized body Certain materials, called electrets , exhibit polarization in the absence of an external electric ﬁeld. A permanently polarized material produces an electric ﬁeld both internaland external to the material, hence there must be a charge distribution to support theﬁelds. We can interpret this charge as being caused by the permanent separation ofatomic charge within the material, but if we are only interested in the macroscopic ﬁeldthen we need not worry about the microscopic implications of such materials. Instead, wecan use the Maxwell–Boﬃ equations and ﬁnd the potential produced by the material by using (3.99). Thus, the ﬁeld of an electret with known polarization Poccupying volume region Vin free space is dipolar in nature and is given by /Phi1(r)=1 4π/epsilon10/integraldisplay V−∇/prime·P(r/prime) |r−r/prime|dV/prime+1 4π/epsilon10/contintegraldisplay Sˆn/prime·P(r/prime) |r−r/prime|dS/prime where ˆnpoints out of the volume region V. As an example, consider a material sphere of radius a, permanently polarized along its axis with uniform polarization P(r)=ˆzP0. We have the equivalent source densities ρp=− ∇· P=0,ρ Ps=ˆn·P=ˆr·ˆzP0=P0cosθ. Then /Phi1(r)=1 4π/epsilon10/contintegraldisplay SρPs(r/prime) |r−r/prime|dS/prime=1 4π/epsilon10/contintegraldisplay SP0cosθ/prime |r−r/prime|dS/prime. The integral takes the form (3.100), hence by (3.102) the solution is /Phi1(r)=P0a2 3/epsilon10cosθr< r2>. (3.103) If we are interested only in the potential for r>a, we can use the multipole expansion (3.95)to obtain /Phi1(r)=1 4π/epsilon10∞/summationdisplay n=01 rn+1qnPn(cosθ), r>a where qn=2π/integraldisplayπ 0ρPs(θ/prime)anPn(cosθ/prime)a2sinθ/primedθ/prime. Substituting for ρPsand remembering that cosθ=P1(cosθ),w eh a v e qn=2πan+2P0/integraldisplayπ 0P1(cosθ/prime)Pn(cosθ/prime)sinθ/primedθ/prime. Using the orthogonality relation (E.123)we ﬁnd qn=2πan+2P0δ1n2 2n+1. Therefore the only nonzero coeﬃcient is q1=4πa3P0 3 and /Phi1(r)=1 4π/epsilon101 r24πa3P0 3P1(cosθ)=P0a3 3/epsilon10r2cosθ, r>a. This is a dipole ﬁeld, and matches (3.103)as expected. 3.2.8 Potential of a dipole layer Surface charge layers sometimes occur in bipolar form, such as in the membrane sur- rounding an animal cell. These can be modeled as a dipole layer consisting of parallelsurface charges of opposite sign. Consider a surface Slocated in free space. Parallel to this surface, and a distance /Delta1/2 below, is located a surface charge layer of density ρ s(r)=Ps(r). Also parallel to S, but a distance /Delta1/2above, is a surface charge layer of density ρs(r)=− Ps(r). We deﬁne the surface dipole moment density Dsas Ds(r)=/Delta1Ps(r). (3.104) Letting the position vector r/prime 0point to the surface Swe can write the potential (3.61) produced by the two charge layers as /Phi1(r)=1 4π/epsilon10/integraldisplay S+Ps(r/prime)1 |r−r/prime 0−ˆn/prime/Delta1 2|dS/prime−1 4π/epsilon10/integraldisplay S−Ps(r/prime)1 |r−r/prime 0+ˆn/prime/Delta1 2|dS/prime. Figure 3.14: A dipole layer. We are interested in the case in which the two charge layers collapse onto the surface S, and wish to compute the potential produced by a given dipole moment density. When /Delta1→0we have r/prime 0→r/primeand may write /Phi1(r)=lim /Delta1→01 4π/epsilon10/integraldisplay SDs(r/prime) /Delta1/bracketleftBigg 1 |R−ˆn/prime/Delta1 2|−1 |R+ˆn/prime/Delta1 2|/bracketrightBigg dS/prime, where R=r−r/prime. By the binomial theorem, the limit of the term in brackets can be written as lim /Delta1→0 /bracketleftBigg R2+/parenleftbigg/Delta1 2/parenrightbigg2 −2R·ˆn/prime/Delta1 2/bracketrightBigg−1 2 −/bracketleftBigg R2+/parenleftbigg/Delta1 2/parenrightbigg2 +2R·ˆn/prime/Delta1 2/bracketrightBigg−1 2  =lim /Delta1→0/parenleftBigg R−1/bracketleftBigg 1+ˆR·ˆn/prime R/Delta1 2/bracketrightBigg −R−1/bracketleftBigg 1−ˆR·ˆn/prime R/Delta1 2/bracketrightBigg/parenrightBigg =/Delta1ˆn/prime·R R3. Thus /Phi1(r)=1 4π/epsilon10/integraldisplay SDs(r/prime)·R R3dS/prime(3.105) where Ds=ˆnDsis the surface vector dipole moment density. The potential of a dipole layer decreases more rapidly ( ∼1/r2)than that of a unipolar charge layer. We saw similar behavior in the dipole term of the multipole expansion (3.92)for a general chargedistribution. We can use (3.105)to study the behavior of the potential across a dipole layer. As we approach the layer from above, the greatest contribution to /Phi1comes from the charge region immediately beneath the observation point. Assuming that the surface dipolemoment density is continuous beneath the point, we can compute the diﬀerence in theﬁelds across the layer at point rby replacing the arbitrary surface layer by a disk of constant surface dipole moment density D 0=Ds(r). For simplicity we center the disk at z = 0 inthe xy-planeasshowninFigure3.15andcomput ethepotentialdiﬀerence /Delta1Vacross the layer; i.e., /Delta1V=/Phi1(h)−/Phi1(−h)on the disk axis as h→0. Using (3.105) along with r/prime=±hˆz−ρ/primeˆρ/prime, we obtain /Delta1V=lim h→0/bracketleftBigg 1 4π/epsilon10/integraldisplay2π 0/integraldisplaya 0[ˆzD0]·ˆzh−ˆρ/primeρ/prime /parenleftbig h2+ρ/prime2/parenrightbig3/2ρ/primedρ/primedφ/prime− −1 4π/epsilon10/integraldisplay2π 0/integraldisplaya 0[ˆzD0]·−ˆzh−ˆρ/primeρ/prime /parenleftbig h2+ρ/prime2/parenrightbig3/2ρ/primedρ/primedφ/prime/bracketrightBigg Figure 3.15: Auxiliary disk for studying the potential distribution across a dipole layer. where ais the disk radius. Integration yields /Delta1V=D0 2/epsilon10lim h→0 −2/radicalBig 1+/parenleftbiga h/parenrightbig2+2 =D0 /epsilon10, independent of a. Generalizing this to an arbitrary surface dipole moment density, we ﬁnd that the boundary condition on the potential is given by /Phi12(r)−/Phi11(r)=Ds(r) /epsilon10(3.106) where “1” denotes the positive side of the dipole moments and “2” the negative side. Physically, the potential diﬀerence in (3.106)is produced by the line integral of E“in- ternal” to the dipole layer. Since there is no ﬁeld internal to a unipolar surface layer, V is continuous across a surface containing charge ρsbut having Ds=0. 3.2.9 Behavior of electric charge density near a conducting edge Sharp corners are often encountered in the application of electrostatics to practical ge- ometries. The behavior of the charge distribution near these corners must be understood in order to develop numerical techniques for solving more complicated problems. We canuse a simple model of a corner if we restrict our interest to the region near the edge.Conside rtheintersectio noftwoplanesasshowninFigure3.16.Theregionnearthein- tersection represents the corner we wish to study. We assume that the planes are held atzero potential and that the charge on the surface is induced by a two-dimensional chargedistribution ρ(r), or by a potential diﬀerence between the edge and another conductor far removed from the edge. We can ﬁnd the potential in the region near the edge by solving Laplace’s equation in cylindrical coordinates. This problem is studied in Appendix A where the separation ofvariables solution is found to be either (A.127)or (A.128) . Using (A.128)and enforcing/Phi1=0at both φ=0andφ=β, we obtain the null solution. Hence the solution must take the form (A.127): /Phi1(ρ,φ) =[A φsin(kφφ)+Bφcos(kφφ)][aρρ−kφ+bρρkφ]. (3.107) Figure 3.16: A conducting edge. Since the origin is included we cannot have negative powers of ρand must put aρ=0. The boundary condition /Phi1(ρ, 0)=0requires Bφ=0. The condition /Phi1(ρ,β) =0then requires sin(kφβ)=0, which holds only if kφ=nπ/β,n=1,2,.... The general solution for the potential near the edge is therefore /Phi1(ρ,φ) =N/summationdisplay n=1Ansin/parenleftbiggnπ βφ/parenrightbigg ρnπ/β(3.108) where the constants Andepend on the excitation source or system of conductors. (Note that if the corner is held at potential V0/negationslash=0, we must merely add V0to the solution.) The charge on the conducting surfaces can be computed from the boundary conditionon normal D. Using (3.30)we have E φ=−1 ρ∂ ∂φN/summationdisplay n=1Ansin/parenleftbiggnπ βφ/parenrightbigg ρnπ/β=−N/summationdisplay n=1Annπ βcos/parenleftbiggnπ βφ/parenrightbigg ρ(nπ/β)−1, hence ρs(x)=−/epsilon1N/summationdisplay n=1Annπ βx(nπ/β)−1 on the surface at φ=0. Near the edge, at small values of x, the variation of ρsis dom- inated by the lowest power of x. (Here we ignore those special excitation arrangements that produce A1=0.)Thus ρs(x)∼x(π/β) −1. The behavior of the charge clearly depends on the wedge angle β. For a sharp edge (half plane)we put β=2πand ﬁnd that the ﬁeld varies as x−1/2. This square-root edge singularity is very common on thin plates, ﬁns, etc., and means that charge tends to accumulate near the edge of a ﬂat conducting surface. For a right-angle corner whereβ=3π/2, there is the somewhat weaker singularity x −1/3. When β=π, the two surfaces fold out into an inﬁnite plane and the charge, not surprisingly, is invariant with xto lowest order near the folding line. When β<π the corner becomes interior and we ﬁnd that the charge density varies with a positive power of distance from the edge. Forvery sharp interior angles the power is large, meaning that little charge accumulates onthe inner surfaces near an interior corner. 3.2.10 Solution to Laplace’s equation for bodies immersed in an im- pressed ﬁeld An important class of problems is based on the idea of placing a body into an existing electric ﬁeld, assuming that the ﬁeld arises from sources so remote that the introductionof the body does not alter the original ﬁeld. The pre-existing ﬁeld is often referred to astheapplied orimpressed ﬁeld , and the solution external to the body is usually formulated as the sum of the applied ﬁeld and a secondary orscattered ﬁeld that satisﬁes Laplace’s equation. This total ﬁeld diﬀers from the applied ﬁeld, and must satisfy the appropriateboundary condition on the body. If the body is a conductor then the total potential mustbe constant everywhere on the boundary surface. If the body is a solid homogeneousdielectric then the total potential ﬁeld must be continuous across the boundary. As an example, consider a dielectric sphere of permittivity /epsilon1and radius a, centered at the origin and immersed in a constant electric ﬁeld E 0(r)=E0ˆz. By (3.30)the applied potential ﬁeld is /Phi10(r)=− E0z=− E0rcosθ(to within a constant). Outside the sphere (r>a)we write the total potential ﬁeld as /Phi12(r)=/Phi10(r)+/Phi1s(r) where /Phi1s(r)is the secondary or scattered potential. Since /Phi1smust satisfy Laplace’s equation, we can write it as a separation of variables solution ( §A.4). By azimuthal symmetry the potential has an r-dependence as in (A.146), and a θ-dependence as in (A.142)with Bθ=0and m=0.T h u s /Phi1shas a representation identical to (A.147), except that we cannot use terms that are unbounded as r→∞. We therefore use /Phi1s(r,θ)=∞/summationdisplay n=0Bnr−(n+1)Pn(cosθ). (3.109) The potential inside the sphere also obeys Laplace’s equation, so we can use the same form (A.147)while discarding terms unbounded at the origin. Thus /Phi11(r,θ)=∞/summationdisplay n=0AnrnPn(cosθ) (3.110) forr<a. To ﬁnd the constants Anand Bnwe apply (3.36)and (3.37)to the total ﬁeld. Application of (3.36)at r=agives −E0acosθ+∞/summationdisplay n=0Bna−(n+1)Pn(cosθ)=∞/summationdisplay n=0AnanPn(cosθ). Multiplying through by Pm(cosθ)sinθ, integrating from θ=0toθ=π, and using the orthogonality relationship (E.123), we obtain −E0a+a−2B1=A1a, (3.111) Bna−(n+1)=Anan,n/negationslash=1, (3.112) where we have used P1(cosθ)=cosθ. Next, since ρs=0, equation (3.37)requires that /epsilon11∂/Phi1 1(r) ∂r=/epsilon12∂/Phi1 2(r) ∂r atr=a. This gives −/epsilon10E0cosθ+/epsilon10∞/summationdisplay n=0[−(n+1)Bn]a−n−2Pn(cosθ)=/epsilon1∞/summationdisplay n=0[nAn]an−1Pn(cosθ). By orthogonality of the Legendre functions we have −/epsilon10E0−2/epsilon10B1a−3=/epsilon1A1, (3.113) −/epsilon10(n+1)Bna−n−2=/epsilon1nAnan−1,n/negationslash=1. (3.114) Equations (3.112)and (3.114)cannot hold simultaneously unless An=Bn=0forn/negationslash=1. Solving (3.111)and (3.113)we have A1=− E03/epsilon10 /epsilon1+2/epsilon10, B1=E0a3/epsilon1−/epsilon10 /epsilon1+2/epsilon10. Hence /Phi11(r)=− E03/epsilon10 /epsilon1+2/epsilon10rcosθ=− E0z3/epsilon10 /epsilon1+2/epsilon10, (3.115) /Phi12(r)=− E0rcosθ+E0a3 r2/epsilon1−/epsilon10 /epsilon1+2/epsilon10cosθ. (3.116) Interestingly, the electric ﬁeld E1(r)=− ∇ /Phi11(r)=ˆzE03/epsilon10 /epsilon1+2/epsilon10 inside the sphere is constant with position and is aligned with the applied external ﬁeld. However, it is weaker than the applied ﬁeld since /epsilon1>/epsilon1 0. To explain this, we compute the polarization charge within and on the sphere. Using D=/epsilon1E=/epsilon10E+Pwe have P1=ˆz(/epsilon1−/epsilon10)E03/epsilon10 /epsilon1+2/epsilon10. (3.117) The volume polarization charge density −∇ · Pis zero, while the polarization surface charge density is ρPs=ˆr·P=(/epsilon1−/epsilon10)E03/epsilon10 /epsilon1+2/epsilon10cosθ. Hence the secondary electric ﬁeld can be attributed to an induced surface polarization charge, and is in a direction opposing the applied ﬁeld. According to the Maxwell–Boﬃviewpoint we should be able to replace the sphere by the surface polarization charge immersed in free space, and use the formula (3.61)to reproduce (3.115)and (3.116) .This is left as an exercise for the reader. 3.3 Magnetostatics The large-scale forms of the magnetostatic ﬁeld equations are /contintegraldisplay /Gamma1H·dl=/integraldisplay SJ·dS, (3.118) /contintegraldisplay SB·dS=0, (3.119) while the point forms are ∇× H(r)=J(r), (3.120) ∇·B(r)=0. (3.121) Note the interesting dichotomy between the electrostatic ﬁeld equations and the magne- tostatic ﬁeld equations. Whereas the electrostatic ﬁeld exhibits zero curl and a divergenceproportional to the source (charge), the magnetostatic ﬁeld has zero divergence and acurl proportional to the source (current). Because the vector relationship between themagnetostatic ﬁeld and its source is of a more complicated nature than the scalar rela-tionship between the electrostatic ﬁeld and its source, more eﬀort is required to develop astrong understanding of magnetic phenomena. Also, it must always be remembered thatalthough the equations describing the electrostatic and magnetostatic ﬁeld sets decou-ple, the phenomena themselves remain linked. Since current is moving charge, electricalphenomena are associated with the establishment of the current that supports a magne- tostatic ﬁeld. We know, for example, that in order to have current in a wire an electricﬁeld must be present to drive electrons through the wire. The magnetic scalar potential. Under certain conditions the equations of magne- tostatics have the same form as those of electrostatics. If J=0in a region V, the magnetostatic equations are ∇× H(r)=0, (3.122) ∇·B(r)=0; (3.123) compare with (3.5)–(3.6) when ρ=0. Using (3.122)we can deﬁne a magnetic scalar potential /Phi1 m: H=− ∇ /Phi1m. (3.124) The negative sign is chosen for consistency with (3.30). We can then deﬁne a magnetic potential diﬀerence between two points as Vm21=−/integraldisplayP2 P1H·dl=−/integraldisplayP2 P1−∇/Phi1m(r)·dl=/integraldisplayP2 P1d/Phi1m(r)=/Phi1m(r2)−/Phi1m(r1). Unlike the electrostatic potential diﬀerence, Vm21 isnotunique .Conside rFigure3.17, which shows a plane passing through the cross-section of a wire carrying total current I. Although there is no current within the region V(external to the wire), equation (3.118) still gives/integraldisplay /Gamma12H·dl−/integraldisplay /Gamma13H·dl=I. Thus/integraldisplay /Gamma12H·dl=/integraldisplay /Gamma13H·dl+I, and the integral/integraltext /Gamma1H·dlis not path-independent. However, /integraldisplay /Gamma11H·dl=/integraldisplay /Gamma12H·dl since no current passes through the surface bounded by /Gamma11−/Gamma12. So we can artiﬁcially impose uniqueness by demanding that no path cross a cut such as that indicated by theline Lin the ﬁgure. Figure 3.17: Magnetic potential. Because Vm21is not unique, the ﬁeld His nonconservative. In point form this is shown by the fact that ∇× His not identically zero. We are not too concerned about energy-related implications of the nonconservative nature of H; the electric point charge has no magnetic analogue that might fail to conserve potential energy if moved aroundin a magnetic ﬁeld. Assuming a linear, isotropic region where B(r)=µ(r)H(r), we can substitute (3.124) into (3.123)and expand to obtain ∇µ(r)·∇/Phi1 m(r)+µ(r)∇2/Phi1m(r)=0. For a homogeneous medium this reduces to Laplace’s equation ∇2/Phi1m=0. We can also obtain an analogue to Poisson’s equation of electrostatics if we use B=µ0(H+M)=−µ0∇/Phi1m+µ0M in (3.123); we have ∇2/Phi1m=−ρM (3.125) where ρM=− ∇· M is called the equivalent magnetization charge density . This form can be used to describe ﬁelds of permanent magnets in the absence of J. Comparison with (3.98)shows that ρM is analogous to the polarization charge ρP. Since /Phi1mobeys Poisson’s equation, the details regarding uniqueness and the construc- tion of solutions follow from those of the electrostatic case. If we include the possibility ofa surface density of magnetization charge, then the integral solution for /Phi1 min unbounded space is /Phi1m(r)=1 4π/integraldisplay VρM(r/prime) |r−r/prime|dV/prime+1 4π/integraldisplay SρMs(r/prime) |r−r/prime|dS/prime. (3.126) Here ρMs, the surface density of magnetization charge, is identiﬁed as ˆn·Min the boundary condition (3.152). 3.3.1 The magnetic vector potential Although the magnetic scalar potential is useful for describing ﬁelds of permanent magnets and for solving certain boundary value problems, it does not include the eﬀects ofsource current. A second type of potential function, called the magnetic vector potential , can be used with complete generality to describe the magnetostatic ﬁeld. Because ∇·B= 0, we can write by (B.49) B(r)=∇× A(r) (3.127) where Ais the vector potential. Now Ais not determined by (3.127)alone, since the gradient of any scalar ﬁeld can be added to Awithout changing the value of ∇× A. Such “gauge transformations” are discussed in Chapter 5, where we ﬁnd that ∇·Amust also be speciﬁed for uniqueness of A. The vector potential can be used to develop a simple formula for the magnetic ﬂux passing through an open surface S: /Psi1 m=/integraldisplay SB·dS=/integraldisplay S(∇× A)·dS=/contintegraldisplay /Gamma1A·dl, (3.128) where /Gamma1is the contour bounding S. In the linear isotropic case where B=µHwe can ﬁnd a partial diﬀerential equation forAby substituting (3.127)into (3.120) . Using (B.43)we have ∇×/bracketleftbigg1 µ(r)∇× A(r)/bracketrightbigg =J(r), hence 1 µ(r)∇× [∇× A(r)]−[∇× A(r)]×∇/parenleftbigg1 µ(r)/parenrightbigg =J(r). In a homogeneous region we have ∇×(∇× A)=µJ (3.129) or ∇(∇·A)−∇2A=µJ (3.130) by (B.47). As mentioned above we must eventually specify ∇·A. Although the choice is arbitrary, certain selections make the computation of Aboth mathematically tractable and physically meaningful. The “Coulomb gauge condition” ∇·A=0reduces (3.130) to ∇2A=−µJ. (3.131) The vector potential concept can also be applied to the Maxwell–Boﬃ magnetostatic equations ∇× B=µ0(J+∇× M), (3.132) ∇·B=0. (3.133) By (3.133)we may still deﬁne Athrough (3.127). Substituting this into (3.132) we have, under the Coulomb gauge, ∇2A=−µ0[J+JM] (3.134) where JM=∇× Mis the magnetization current density. Figure 3.18: Circular loop of wire. The diﬀerential equations (3.131)and (3.134)are vector versions of Poisson’s equation, and may be solved quite easily for unbounded space by decomposing the vector sourceinto rectangular components. For instance, dotting (3.131)with ˆxwe ﬁnd that ∇ 2Ax=−µJx. This scalar version of Poisson’s equation has solution Ax(r)=µ 4π/integraldisplay VJx(r/prime) |r−r/prime|dV/prime in unbounded space. Repeating this for each component and assembling the results, we obtain the solution for the vector potential in an unbounded homogeneous medium: A(r)=µ 4π/integraldisplay VJ(r/prime) |r−r/prime|dV/prime. (3.135) Any surface sources can be easily included through a surface integral: A(r)=µ 4π/integraldisplay VJ(r/prime) |r−r/prime|dV/prime+µ 4π/integraldisplay SJs(r/prime) |r−r/prime|dS/prime. (3.136) In unbounded free space containing materials represented by M,w eh a v e A(r)=µ0 4π/integraldisplay VJ(r/prime)+JM(r/prime) |r−r/prime|dV/prime+µ0 4π/integraldisplay SJs(r/prime)+JMs(r/prime) |r−r/prime|dV/prime(3.137) where JMs=− ˆn×Mis the surface density of magnetization current as described in (3.153). It may be veriﬁed directly from (3.137) that ∇·A=0. Field of a circular loop. Consider a circular loop of line current of radius ain unbounde dspace(Figure3.18).Using J(r/prime)= I ˆφ/primeδ(z /prime)δ(ρ/prime− a )andnotingthat r = ρˆρ+zˆzand r/prime=aˆρ/prime, we can write (3.136)as A(r)=µI 4π/integraldisplay2π 0ˆφ/prime adφ/prime /bracketleftbig ρ2+a2+z2−2aρcos(φ−φ/prime)/bracketrightbig1/2. Because ˆφ/prime=− ˆxcosφ/prime+ˆysinφ/primewe ﬁnd that A(r)=µIa 4πˆφ/integraldisplay2π 0cosφ/prime /bracketleftbig ρ2+a2+z2−2aρcosφ/prime/bracketrightbig1/2dφ/prime. We put the integral into standard form by setting φ/prime=π−2x: A(r)=−µIa 4πˆφ/integraldisplayπ/2 −π/21−2 sin2x /bracketleftbig ρ2+a2+z2+2aρ(1−2 sin2x)/bracketrightbig1/22dx. Letting k2=4aρ (a+ρ)2+z2, F2=(a+ρ)2+z2, we have A(r)=−µIa 4πˆφ4 F/integraldisplayπ/2 01−2 sin2x [1−k2sin2x]1/2dx. Then, since 1−2 sin2x [1−k2sin2x]1/2=k2−2 k2[1−k2sin2x]−1/2+2 k2[1−k2sin2x]1/2, we have A(r)=ˆφµI πk/radicalbigga ρ/bracketleftbigg/parenleftbigg 1−1 2k2/parenrightbigg K(k2)−E(k2)/bracketrightbigg . (3.138) Here K(k2)=/integraldisplayπ/2 0du [1−k2sin2u]1/2, E(k2)=/integraldisplayπ/2 0[1−k2sin2u]1/2du, are complete elliptic integrals of the ﬁrst and second kinds, respectively. We have k2/lessmuch1when the observation point is far from the loop ( r2=ρ2+z2/greatermucha2). Using the expansions [47] K(k2)=π 2/bracketleftbigg 1+1 4k2+9 64k4+···/bracketrightbigg , E(k2)=π 2/bracketleftbigg 1−1 4k2−3 64k4−···/bracketrightbigg , in (3.138)and keeping the ﬁrst nonzero term, we ﬁnd that A(r)≈ˆφµI 4πr2(πa2)sinθ. (3.139) Deﬁning the magnetic dipole moment of the loop as m=ˆzIπa2, we can write (3.139)as A(r)=µ 4πm×ˆr r2. (3.140) Generalization to an arbitrarily-oriented circular loop with center located at r0is accom- plished by writing m=ˆnIAwhere Ais the loop area and ˆnis normal to the loop in the right-hand sense. Then A(r)=µ 4πm×r−r0 |r−r0|3. We shall ﬁnd, upon investigating the general multipole expansion of Abelow, that this holds for any planar loop. The magnetic ﬁeld of the loop can be found by direct application of (3.127). For the case r2/greatermucha2we take the curl of (3.139)and ﬁnd that B(r)=µ 4πm r3(ˆr2 cosθ+ˆθsinθ). (3.141) Comparison with (3.93)shows why we often refer to a small loop as a magnetic dipole . But (3.141)is approximate, and since there are no magnetic monopoles we cannot con-struct an exact magnetic analogue to the electric dipole. On the other hand, we shallﬁnd below that the multipole expansion of a ﬁnite-extent steady current begins with thedipole term (since the current must form closed loops). We may regard small loops asthe elemental units of steady current from which all other currents may be constructed. 3.3.2 Multipole expansion It is possible to derive a general multipole expansion for Aanalogous to (3.94). But the vector nature of Arequires that we use vector spherical harmonics, hence the result is far more complicated than (3.94). A simpler approach yields the ﬁrst few terms andrequires only the Taylor expansion of 1/R. Consider a steady current localized near the origin and contained within a sphere of radius r m. We substitute the expansion (3.89) into (3.135)to obtain A(r)=µ 4π/integraldisplay VJ(r/prime)/bracketleftbigg1 R/vextendsingle/vextendsingle/vextendsingle/vextendsingle r/prime=0+(r/prime·∇/prime)1 R/vextendsingle/vextendsingle/vextendsingle/vextendsingle r/prime=0+1 2(r/prime·∇/prime)21 R/vextendsingle/vextendsingle/vextendsingle/vextendsingle r/prime=0+···/bracketrightbigg dV/prime,(3.142) which we view as A(r)=A(0)(r)+A(1)(r)+A(2)(r)+···. The ﬁrst term is merely A(0)(r)=µ 4πr/integraldisplay VJ(r/prime)dV/prime=µ 4πr3/summationdisplay i=1ˆxi/integraldisplay VJi(r/prime)dV/prime where (x,y,z)=(x1,x2,x3). However, by (3.26)each of the integrals is zero and we have A(0)(r)=0; the leading term in the multipole expansion of Afor a general steady current distribution vanishes. Using (3.91)we can write the second term as A(1)(r)=µ 4πr3/integraldisplay VJ(r/prime)3/summationdisplay i=1xix/prime idV/prime=µ 4πr33/summationdisplay j=1ˆxj3/summationdisplay i=1xi/integraldisplay Vx/prime iJj(r/prime)dV/prime. (3.143) By adding the null relation (3.28)we can write /integraldisplay Vx/prime iJjdV/prime=/integraldisplay Vx/prime iJjdV/prime+/integraldisplay V[x/prime iJj+x/prime jJi]dV/prime=2/integraldisplay Vx/prime iJjdV/prime+/integraldisplay Vx/prime jJidV/prime or /integraldisplay Vx/prime iJjdV/prime=1 2/integraldisplay V[x/prime iJj−x/prime jJi]dV/prime. (3.144) Figure 3.19: A planar wire loop. By this and (3.143)the second term in the multipole expansion is A(1)(r)=µ 4πr31 2/integraldisplay V3/summationdisplay j=1ˆxj3/summationdisplay i=1xi[x/prime iJj−x/prime jJi]dV/prime=−µ 4πr31 2/integraldisplay Vr×[r/prime×J(r/prime)]dV/prime. Deﬁning the dipole moment vector m=1 2/integraldisplay Vr×J(r)dV (3.145) we have A(1)(r)=µ 4πm×/parenleftbiggˆr r2/parenrightbigg =−µ 4πm×∇1 r. (3.146) This is the dipole moment potential for the steady current J. Since steady currents of ﬁnite extent consist of loops, the dipole component is generally the ﬁrst nonzero termin the expansion of A. Higher-order components may be calculated, but extension of (3.142)beyond the dipole term is quite tedious and will not be attempted. As an example let us compute the dipole moment of the planar but otherwise arbitrary loopshowninFigure3.19.Specializin g(3.145)fo ralinecurrentwehave m=I 2/contintegraldisplay /Gamma1r×dl. Examinin gFigure3.19,weseethat 1 2r×dl=ˆndS where dSis the area of the sector swept out by ras it moves along dl, and ˆnis the normal to the loop in the right-hand sense. Thus m=ˆnIA (3.147) where Ais the area of the loop. Physical interpretation of Min a magnetic material. In (3.137)we presented an expression for the vector potential produced by a magnetized material in terms ofequivalent magnetization surface and volume currents. Suppose a magnetized mediumis separated into volume regions with bounding surfaces across which the permeabilityis discontinuous. With J M=∇× Mand JMs=− ˆn×Mwe obtain A(r)=µ0 4π/integraldisplay VJ(r/prime) |r−r/prime|dV/prime+µ0 4π/integraldisplay SJs(r/prime) |r−r/prime|dS/prime+ +/summationdisplay iµ0 4π/bracketleftbigg/integraldisplay Vi∇/prime×M(r/prime) |r−r/prime|dV/prime+/integraldisplay Si−ˆn/prime×M(r/prime) |r−r/prime|dS/prime/bracketrightbigg . (3.148) Here ˆnpoints outward from region Vi. Using the curl theorem on the fourth term and employing the vector identity (B.43), we have A(r)=µ0 4π/integraldisplay VJ(r/prime) |r−r/prime|dV/prime+µ0 4π/integraldisplay SJs(r/prime) |r−r/prime|dS/prime+ +/summationdisplay i/bracketleftbiggµ0 4π/integraldisplay ViM(r/prime)×∇/prime/parenleftbigg1 |r−r/prime|/parenrightbigg dV/prime/bracketrightbigg . (3.149) But∇/prime(1/R)=ˆR/R2, hence the third term is a sum of integrals of the form µ0 4π/integraldisplay ViM(r/prime)×ˆR R2dV/prime. Comparison with (3.146)shows that this integral represents a volume superposition of dipole moments where Mis a volume density of magnetic dipole moments. Hence a magnetic material with permeability µis equivalent to a volume distribution of magnetic dipoles in free space. As with our interpretation of the polarization vector in a dielectric,we base this conclusion only on Maxwell’s equations and the assumption of a linear,isotropic relationship between Band H. 3.3.3 Boundary conditions for the magnetostatic ﬁeld The boundary conditions found for the dynamic magnetic ﬁeld remain valid in the magnetostatic case. Hence ˆn12×(H1−H2)=Js (3.150) and ˆn12·(B1−B2)=0, (3.151) where ˆn12points into region 1 from region 2. Since the magnetostatic curl and divergence equations are independent, so are the boundary conditions (3.150)and (3.151) . We canalso write (3.151)in terms of equivalent sources by (3.118) : ˆn 12·(H1−H2)=ρMs1+ρMs2, (3.152) where ρMs=ˆn·Mis called the equivalent magnetization surface charge density . Here ˆn points outward from the material body. For a linear, isotropic material described by B=µH, equation (3.150)becomes ˆn12×/parenleftbiggB1 µ1−B2 µ2/parenrightbigg =Js. With (3.118)we can also write (3.150)as ˆn12×(B1−B2)=µ0(Js+JMs1+JMs2) (3.153) where JMs=− ˆn×Mis the equivalent magnetization surface current density. We may also write the boundary conditions in terms of the scalar or vector potential. Using H=− ∇ /Phi1m, we can write (3.150)as /Phi1m1(r)=/Phi1m2(r) (3.154) provided that the surface current Js=0. As was the case with (3.36), the possibility of an additive constant here is generally ignored. To write (3.151)in terms of /Phi1mwe ﬁrst note that B/µ0−M=− ∇ /Phi1m; substitution into (3.151)gives ∂/Phi1 m1 ∂n−∂/Phi1 m2 ∂n=−ρMs1−ρMs2 (3.155) where the normal derivative is taken in the direction of ˆn12. For a linear isotropic material where B=µHwe have µ1∂/Phi1 m1 ∂n=µ2∂/Phi1 m2 ∂n. (3.156) Note that (3.154)and (3.156)are independent. Boundary conditions on Amay be derived using the approach of §2.8.2. Consider Figure2.6.Herethesurfac emaycarryeitheranelectri csurfac ecurrent Jsor an equiv- alent magnetization current JMs, and thus may be a surface of discontinuity between diﬀering magnetic media. If we integrate ∇× Aover the volume regions V1and V2and add the results we ﬁnd that/integraldisplay V1∇× AdV+/integraldisplay V2∇× AdV=/integraldisplay V1+V2BdV. By the curl theorem /integraldisplay S1+S2ˆn×AdS+/integraldisplay S10−ˆn10×A1dS+/integraldisplay S20−ˆn20×A2dS=/integraldisplay V1+V2BdV where A1is the ﬁeld on the surface S10andA2is the ﬁeld on S20.A sδ→0the surfaces S1 and S2combine to give S. Also S10and S20coincide, as do the normals ˆn10=− ˆn20=ˆn12. Thus/integraldisplay S(ˆn×A)dS−/integraldisplay VBdV=/integraldisplay S10ˆn12×(A1−A2)dS. (3.157) Now let us integrate over the entire volume region Vincluding the surface of discontinuity. This gives /integraldisplay S(ˆn×A)dS−/integraldisplay VBdV=0, and for agreement with (3.157)we must have ˆn12×(A1−A2)=0. (3.158) A similar development shows that ˆn12·(A1−A2)=0. (3.159) Therefore Ais continuous across a surface carrying electric or magnetization current. 3.3.4 Uniqueness of the magnetostatic ﬁeld Because the uniqueness conditions established for the dynamic ﬁeld do not apply to magnetostatics, we begin with the magnetostatic ﬁeld equations. Consider a region ofspace Vbounded by a surface S. There may be source currents and magnetic materials both inside and outside V. Assume (B 1,H1)and(B2,H2)are solutions to the magne- tostatic ﬁeld equations with source J. We seek conditions under which B1=B2and H1=H2. The diﬀerence ﬁeld H0=H2−H1obeys ∇× H0=0. Using (B.44)we examine the quantity ∇·(A0×H0)=H0·(∇× A0)−A0·(∇× H0)=H0·(∇× A0) where A0is deﬁned by B0=B2−B1=∇× A0=∇× (A2−A1). Integrating over V we obtain /contintegraldisplay S(A0×H0)·dS=/integraldisplay VH0·(∇× A0)dV=/integraldisplay VH0·B0dV. Then, since (A0×H0)·ˆn=−A0·(ˆn×H0), we have −/contintegraldisplay SA0·(ˆn×H0)dS=/integraldisplay VH0·B0dV. (3.160) IfA0=0orˆn×H0=0everywhere on S,o rA0=0on part of Sand ˆn×H0=0on the remainder, then /integraldisplay VH0·B0dS=0. (3.161) SoH0=0orB0=0by arbitrariness of V. Assuming Hand Bare linked by the constitutive relations, we have H1=H2and B1=B2. The ﬁelds within Vare unique provided that A, the tangential component of H, or some combination of the two, is speciﬁed over the bounding surface S. One other condition will cause the left-hand side of (3.160)to vanish. If Srecedes to inﬁnity then, provided that the potential functions vanish suﬃciently fast, the condition(3.161)still holds and uniqueness is guaranteed. Equation (3.135)shows that A∼1/r asr→∞ , hence B,H∼1/r 2. So uniqueness is ensured by the speciﬁcation of Jin unbounded space. 3.3.5 Integral solution for the vector potential We have used the scalar Green’s theorem to ﬁnd a solution for the electrostatic poten- tial within a region Vin terms of the source charge in Vand the values of the potential and its normal derivative on the boundary surface S. Analogously, we may ﬁnd Awithin Vin terms of the source current in Vand the values of Aand its derivatives on S. The vector relationship between Band Acomplicates the derivation somewhat, requiring Green’s second identity for vector ﬁelds. LetPand Qbe continuous with continuous ﬁrst and second derivatives throughout Vand on S. The divergence theorem shows that /integraldisplay V∇·[P×(∇× Q)]dV=/integraldisplay S[P×(∇× Q)]·dS. By virtue of (B.44)we have /integraldisplay V[(∇× Q)·(∇× P)−P·(∇ × {∇ × Q})]dV=/integraldisplay S[P×(∇× Q)]·dS. We now interchange Pand Qand subtract the result from the above, obtaining /integraldisplay V[Q·(∇ × {∇ × P})−P·(∇ × {∇ × Q})]dV= /integraldisplay S[P×(∇× Q)−Q×(∇× P)]·dS. (3.162) Note that ˆnpoints outward from V. This is Green’s second identity for vector ﬁelds . Now assume that Vcontains a magnetic material of uniform permeability µand set P=A(r/prime), Q=c R, in (3.162)written in terms of primed coordinates. Here cis a constant vector, nonzero but otherwise arbitrary. We ﬁrst examine the volume integral terms. Note that ∇/prime×(∇/prime×Q)=∇/prime×/parenleftBig ∇/prime×c R/parenrightBig =− ∇/prime2/parenleftBigc R/parenrightBig +∇/prime/bracketleftBig ∇/prime·/parenleftBigc R/parenrightBig/bracketrightBig . By (B.162)and (3.58)we have ∇/prime2/parenleftBigc R/parenrightBig =1 R∇/prime2c+c∇/prime2/parenleftbigg1 R/parenrightbigg +2/parenleftbigg ∇/prime1 R·∇/prime/parenrightbigg c=c∇/prime2/parenleftbigg1 R/parenrightbigg =−c4πδ(r−r/prime), hence P·[∇/prime×(∇/prime×Q)]=4πc·Aδ(r−r/prime)+A·∇/prime/bracketleftBig ∇/prime·/parenleftBigc R/parenrightBig/bracketrightBig . Since ∇·A=0the second term on the right-hand side can be rewritten using (B.42): ∇/prime·(ψA)=A·(∇/primeψ)+ψ∇/prime·A=A·(∇/primeψ). Thus P·[∇/prime×(∇/prime×Q)]=4πc·Aδ(r−r/prime)+∇/prime·/bracketleftbigg A/braceleftbigg c·∇/prime/parenleftbigg1 R/parenrightbigg/bracerightbigg/bracketrightbigg , where we have again used (B.42). The other volume integral term can be found by substituting from (3.129): Q·[∇/prime×(∇/prime×P)]=µ1 Rc·J(r/prime). Next we investigate the surface integral terms. Consider ˆn/prime·/bracketleftbig P×(∇/prime×Q)/bracketrightbig =ˆn/prime·/braceleftBig A×/bracketleftBig ∇/prime×/parenleftBigc R/parenrightBig/bracketrightBig/bracerightBig =ˆn/prime·/braceleftbigg A×/bracketleftbigg1 R∇/prime×c−c×∇/prime/parenleftbigg1 R/parenrightbigg/bracketrightbigg/bracerightbigg =− ˆn/prime·/braceleftbigg A×/bracketleftbigg c×∇/prime/parenleftbigg1 R/parenrightbigg/bracketrightbigg/bracerightbigg . This can be put in slightly diﬀerent form by the use of (B.8). Note that (A×B)·(C×D)=A·[B×(C×D)] =(C×D)·(A×B) =C·[D×(A×B)], hence ˆn/prime·/bracketleftbig P×(∇/prime×Q)/bracketrightbig =−c·/bracketleftbigg ∇/prime/parenleftbigg1 R/parenrightbigg ×(ˆn/prime×A)/bracketrightbigg . The other surface term is given by ˆn/prime·[Q×(∇/prime×P)]=ˆn/prime·/bracketleftBigc R×(∇/prime×A)/bracketrightBig =ˆn/prime·/parenleftBigc R×B/parenrightBig =−c R·(ˆn/prime×B). We can now substitute each of the terms into (3.162)and obtain µc·/integraldisplay VJ(r/prime) RdV/prime−4πc·/integraldisplay VA(r/prime)δ(r−r/prime)dV/prime−c·/contintegraldisplay S[ˆn/prime·A(r/prime)]∇/prime/parenleftbigg1 R/parenrightbigg dS/prime =−c·/contintegraldisplay S∇/prime/parenleftbigg1 R/parenrightbigg ×[ˆn/prime×A(r/prime)]dS/prime+c·/contintegraldisplay S1 Rˆn/prime×B(r/prime)dS/prime. Since cis arbitrary we can remove the dot products to obtain a vector equation. Then A(r)=µ 4π/integraldisplay VJ(r/prime) RdV/prime−1 4π/contintegraldisplay S/braceleftbigg [ˆn/prime×A(r/prime)]×∇/prime/parenleftbigg1 R/parenrightbigg + +1 Rˆn/prime×B(r/prime)+[ˆn/prime·A(r/prime)]∇/prime/parenleftbigg1 R/parenrightbigg/bracerightbigg dS/prime. (3.163) We have expressed Ain a closed region in terms of the sources within the region and the values of Aand Bon the surface. While uniqueness requires speciﬁcation of either Aorˆn×BonS, the expression (3.163)includes bothquantities. This is similar to (3.56) for electrostatic ﬁelds, which required both the scalar potential and its normal derivative. The reader may be troubled by the fact that we require Pand Qto be somewhat well behaved, then proceed to involve the singular function c/Rand integrate over the singu- larity. We choose this approach to simplify the presentation; a more rigorous approachwhich excludes the singular point with a small sphere also gives (3.163). This approach was used in §3.2.4 to establish (3.58). The interested reader should see Stratton [187] for details on the application of this technique to obtain (3.163). It is interesting to note that as S→∞ the surface integral vanishes since A∼1/r and B∼1/r 2, and we recover (3.135). Moreover, (3.163) returns the null result when evaluated at points outside S(see Stratton [187]). We shall see this again when studying the integral solutions for electrodynamic ﬁelds in §6.1.3. Finally, with Q=∇/prime/parenleftbigg1 R/parenrightbigg ×c we can ﬁnd an integral expression for Bwithin an enclosed region, representing a gen- eralization of the Biot–Savart law (Problem 3.20). However, this case will be covered inthe more general development of §6.1.1. The Biot–Savart law. We can obtain an expression for Bin unbounded space by performing the curl operation directly on the vector potential: B(r)=∇×µ 4π/integraldisplay VJ(r/prime) |r−r/prime|dV/prime=µ 4π/integraldisplay V∇×J(r/prime) |r−r/prime|dV/prime. Using (B.43)and ∇× J(r/prime)=0, we have B(r)=−µ 4π/integraldisplay VJ×∇1 |r−r/prime|dV/prime. TheBiot–Savart law B(r)=µ 4π/integraldisplay VJ(r/prime)×ˆR R2dV/prime(3.164) follows from (3.57). For the case of a line current we can replace JdV/primebyIdl/primeand obtain B(r)=Iµ 4π/integraldisplay /Gamma1dl/prime×ˆR R2. (3.165) For an inﬁnitely long line current on the z-axis we have B(r)=Iµ 4π∞/integraldisplay −∞ˆz×ˆz(z−z/prime)+ˆρρ [(z−z/prime)2+ρ2]3/2dz/prime=ˆφµI 2πρ. (3.166) This same result follows from taking ∇×Aafter direct computation of A, or from direct application of the large-scale form of Ampere’s law. 3.3.6 Force and energy Ampere force on a system of currents. If a steady current J(r)occupying a region Vis exposed to a magnetic ﬁeld, the force on the moving charge is given by the Lorentz force law dF(r)=J(r)×B(r). (3.167) This can be integrated to give the total force on the current distribution: F=/integraldisplay VJ(r)×B(r)dV. (3.168) It is apparent that the charge ﬂow comprising a steady current must be constrained in some way, or the Lorentz force will accelerate the charge and destroy the steady natureof the current. This constraint is often provided by a conducting wire. As an example, consider an inﬁnitely long wire of circular cross-section centered on thez-axis in free space. If the wire carries a total current Iuniformly distributed over the cross-section, then within the wire J=ˆzI/(πa 2)where ais the wire radius. The resulting ﬁeld can be found through direct integration using (3.164), or by the use ofsymmetry and either (3.118)or (3.120) . Since B(r)=ˆφB φ(ρ), equation (3.118)shows that /integraldisplay2π 0Bφ(ρ)ρ dφ=/braceleftBigg µ0I a2ρ2,ρ≤a µ0I,ρ ≥a. Thus B(r)=/braceleftBiggˆφµ0Iρ/2πa2,ρ≤a, ˆφµ0I/2πρ, ρ ≥a.(3.169) The force density within the wire, dF=J×B=− ˆρµ0I2ρ 2π2a4, is directed inward and tends to compress the wire. Integration over the wire volume gives F=0because /integraldisplay2π 0ˆρdφ=0; however, a section of the wire may experience a net force. For instance, we can compute the force on one half of the wire split down its axis by using ˆρ=ˆxcosφ+ˆysinφto obtain Fx=0and Fy=−µ0I2 2π2a4/integraldisplay dz/integraldisplaya 0ρ2dρ/integraldisplayπ 0sinφdφ=−µ0I2 3π2a/integraldisplay dz. The force per unit length F l=− ˆyµ0I2 3π2a(3.170) is directed toward the other half as expected. If the wire takes the form of a loop carrying current I, then (3.167)becomes dF(r)=Idl(r)×B(r) (3.171) and the total force acting is F=I/contintegraldisplay /Gamma1dl(r)×B(r). We can write the force on Jin terms of the current producing B. Assuming this latter current J/primeoccupies region V/prime, the Biot–Savart law (3.164)yields F=µ 4π/integraldisplay VJ(r)×/integraldisplay V/primeJ(r/prime)×r−r/prime |r−r/prime|3dV/primedV. (3.172) This can be specialized to describe the force between line currents. Assume current 1, following a path /Gamma11along the direction dl, carries current I1, while current 2, following path/Gamma12along the direction dl/prime, carries current I2. Then the force on current 1 is F1=I1I2µ 4π/contintegraldisplay /Gamma11/contintegraldisplay /Gamma12dl×/parenleftbigg dl/prime×r−r/prime |r−r/prime|3/parenrightbigg . This equation, known as Ampere’s force law , can be written in a better form for compu- tational purposes. We use (B.7)and ∇(1/R)from (3.57): F1=I1I2µ 4π/contintegraldisplay /Gamma12dl/prime/contintegraldisplay /Gamma11dl·∇/prime/parenleftbigg1 |r−r/prime|/parenrightbigg −I1I2µ 4π/contintegraldisplay /Gamma11/contintegraldisplay /Gamma12(dl·dl/prime)r−r/prime |r−r/prime|3.(3.173) The ﬁrst term involves an integral of a perfect diﬀerential about a closed path, producing a null result. Thus F1=− I1I2µ 4π/contintegraldisplay /Gamma11/contintegraldisplay /Gamma12(dl·dl/prime)r−r/prime |r−r/prime|3. (3.174) Figure 3.20: Parallel, current carrying wires. Asasimpleexample ,conside rparalle lwiresseparate dbyadistance d (Figure3.20). In this case F1=− I1I2µ 4π/integraldisplay/bracketleftbigg/integraldisplay∞ −∞−dˆx+(z−z/prime)ˆz [d2+(z−z/prime)2]3/2dz/prime/bracketrightbigg dz=I1I2µ 2πdˆx/integraldisplay dz so the force per unit length is F1 l=ˆxI1I2µ 2πd. (3.175) The force is attractive if I1I2≥0(i.e., if the currents ﬂow in the same direction). Maxwell’s stress tensor. The magnetostatic version of the stress tensor can be ob- tained from (2.288)by setting E=D=0: ¯Tm=1 2(B·H)¯I−BH. (3.176) The total magnetic force on the current in a region Vsurrounded by surface Sis given by Fm=−/contintegraldisplay S¯Tm·dS=/integraldisplay VfmdV where fm=J×Bis the magnetic force volume density. Let us compute the force between two parallel wires carrying identical currents in free space (let I1 = I2 = I inFigure3.20)andcompar etheresultwith(3.175 ).Theforce on the wire at x=−d/2can be computed by integrating ¯Tm·ˆnover the yz-plane with ˆn=ˆx. Using (3.166)we see that in this plane the total magnetic ﬁeld is B=− ˆxµ0I πy y2+d2/4. Therefore ¯Tm·ˆn=1 2BxBx µ0ˆx−ˆxBxBx µ0=−µ0I2 2π2y2 [y2+d2/4]2ˆx and by integration F1=µ0I2 2π2ˆx/integraldisplay dz/integraldisplay∞ −∞y2 [y2+d2/4]2dy=I2µ0 2πdˆx/integraldisplay dz. The resulting force per unit length agrees with (3.175)when I1=I2=I. Torque in a magnetostatic ﬁeld. The torque exerted on a current-carrying conduc- tor immersed in a magnetic ﬁeld plays an important role in many engineering applica-tions. If a rigid body is exposed to a force ﬁeld of volume density dF(r), the torque on that body about a certain origin is given by T=/integraldisplay Vr×dFdV (3.177) where integration is performed over the body and rextends from the origin of torque. If the force arises from the interaction of a current with a magnetostatic ﬁeld, thendF=J×Band T=/integraldisplay Vr×(J×B)dV. (3.178) For a line current we can replace JdVwith Idlto obtain T=I/integraldisplay /Gamma1r×(dl×B). IfBis uniform then by (B.7)we have T=/integraldisplay V[J(r·B)−B(r·J)]dV. The second term can be written as /integraldisplay VB(r·J)dV=B3/summationdisplay i=1/integraldisplay VxiJidV=0 where (x1,x2,x3)=(x,y,z), and where we have employed (3.27). Thus T=/integraldisplay VJ(r·B)dV=3/summationdisplay j=1ˆxj/integraldisplay VJj3/summationdisplay i=1xiBidV=3/summationdisplay i=1Bi3/summationdisplay j=1ˆxj/integraldisplay VJjxidV. We can replace the integral using (3.144)to get T=1 2/integraldisplay V3/summationdisplay j=1ˆxj3/summationdisplay i=1Bi[xiJj−xjJi]dV=−1 2/integraldisplay VB×(r×J)dV. Since Bis uniform we have, by (3.145), T=m×B (3.179) where mis the dipole moment. For a planar loop we can use (3.147)to obtain T=IAˆn×B. Joule’s law. In§2.9.5 we showed that when a moving charge interacts with an electric ﬁeld in a volume region V, energy is transferred between the ﬁeld and the charge. If the source of that energy is outside V, the energy is carried into Vas an energy ﬂux over the boundary surface S. The energy balance described by Poynting’s theorem (3.299)also holds for static ﬁelds supported by steady currents: we must simply recognize that wehave no time-rate of change of stored energy. Thus −/integraldisplay VJ·EdV=/contintegraldisplay S(E×H)·dS. (3.180) The term P=−/integraldisplay VJ·EdV (3.181) describes the rate at which energy is supplied to the ﬁelds by the current within V;w e have P>0if there are sources within Vthat result in energy transferred to the ﬁelds, and P<0if there is energy transferred to the currents. The latter case occurs when there are conducting materials in V. Within these conductors P=−/integraldisplay VσE·EdV. (3.182) Here P<0; energy is transferred from the ﬁelds to the currents, and from the currents into heat (i.e., into lattice vibrations via collisions). Equation (3.182) is called Joule’s law, and the transfer of energy from the ﬁelds into heat is Joule heating . Joule’s law is the power relationship for a conducting material. An important example involves a straight section of conducting wire having circular cross-section. Assume a total current Iis uniformly distributed over the cross-section of the wire, and that the wire is centered on the z-axis and extends between the planes z=0,L. Let the potential diﬀerence between the ends be V. Using (3.169)we see that at the surface of the wire H=ˆφI 2πa, E=ˆzV L. The corresponding Poynting ﬂux E×His−ˆρ-directed, implying that energy ﬂows into wire volume through the curved side surface. We can verify (3.180): −/integraldisplay VJ·EdV=/integraldisplayL 0/integraldisplay2π 0/integraldisplaya 0ˆzI πa2·ˆzV Lρdρdφdz=− IV, /contintegraldisplay S(E×H)·dS=/integraldisplay2π 0/integraldisplayL 0/parenleftbigg −ˆρIV 2πaL/parenrightbigg ·ˆρadφdz=− IV. Stored magnetic energy. We have shown that the energy stored in a static charge distribution may be regarded as the “assembly energy” required to bring charges frominﬁnity against the Coulomb force. By proceeding very slowly with this assembly, we areable to avoid any complications resulting from the motion of the charges. Similarly, we may equate the energy stored in a steady current distribution to the en- ergy required for its assembly from current ﬁlaments 6brought in from inﬁnity. However, the calculation of assembly energy is more complicated in this case: moving a current 6Recall that a ﬂux tube of a vector ﬁeld is bounded by streamlines of the ﬁeld. A current ﬁlament is a ﬂux tube of current having vanishingly small, but nonzero, cross-section. Figure 3.21: Calculation of work to move a ﬁlamentary loop in an applied magnetic ﬁeld. ﬁlament into the vicinity of existing ﬁlaments changes the total magnetic ﬂux passing through the existing loops, regardless of how slowly we assemble the ﬁlaments. As de-scribed by Faraday’s law, this change in ﬂux must be associated with an induced emf,which will tend to change the current ﬂowing in the ﬁlament (and any existing ﬁlaments)unless energy is expended to keep the current constant (by the application of a batteryemf in the opposite direction). We therefore regard the assembly energy as consistingof two parts: (1)the energy required to bring a ﬁlament with constant current from inﬁnity against the Ampere force, and (2)the energy required to keep the current in thisﬁlament, and any existing ﬁlaments, constant. We ignore the energy required to keepthe steady current ﬂowing through an isolated loop (i.e., the energy needed to overcomeJoule losses). We begin by computing the amount of energy required to bring a ﬁlament with current Ifrom inﬁnity to a given position within an applied magnetostatic ﬁeld B(r). In this ﬁrst step we assume that the ﬁeld is supported by localized sources, hence vanishes atinﬁnity, and that it will not be altered by the motion of the ﬁlament. The force on eachsmall segment of the ﬁlament is given by Ampere’s force law (3.171), and the total forceis found by integration. Suppose an external agent displaces the ﬁlament incrementallyfrom a starting position 1 to an ending position 2 along a vector δras shown in Figure 3.21.Theworkrequire dis δW=−(Idl×B)·δr=(Idl×δr)·B foreachsegme ntofthewire.Figure3.21showsthat dl ×δr descri besasmallpatchof surface area between the starting and ending positions of the ﬁlament, hence −(dl×δr)·B is theoutward ﬂux of Bthrough the patch. Integrating over all segments comprising the ﬁlament, we obtain /Delta1W=I/contintegraldisplay /Gamma1(dl×δr)·B=− I/integraldisplay S0B·dS for the total work required to displace the entire ﬁlament through δr; here the surface S0 is described by the superposition of all patches. If S1and S2are the surfaces bounded by the ﬁlament in its initial and ﬁnal positions, respectively, then S1,S2, and S0taken together form a closed surface. The outward ﬂux of Bthrough this surface is /contintegraldisplay S0+S1+S2B·dS=0 so that /Delta1W=− I/integraldisplay S0B·dS=I/integraldisplay S1+S2B·dS where ˆnis outward from the closed surface. Finally, let /Psi11,2be the ﬂux of Bthrough S1,2 in the direction determined by dland the right-hand rule. Then /Delta1W=− I(/Psi12−/Psi11)=− I/Delta1/Psi1. (3.183) Now suppose that the initial position of the ﬁlament is at inﬁnity. We bring the ﬁlament into a ﬁnal position within the ﬁeld Bthrough a succession of small displacements, each requiring work (3.183). By superposition over all displacements, the total work is W=− I(/Psi1−/Psi1∞)where /Psi1∞and/Psi1are the ﬂuxes through the ﬁlament in its initial and ﬁnal positions, respectively. However, since the source of the ﬁeld is localized, we knowthat Bis zero at inﬁnity. Therefore /Psi1 ∞=0and W=− I/Psi1=− I/integraldisplay SB·ˆndS (3.184) where ˆnis determined from dlin the right-hand sense. Now let us ﬁnd the work required to position two current ﬁlaments in a ﬁeld-free region of space, starting with both ﬁlaments at inﬁnity. Assume ﬁlament 1 carries current I1 and ﬁlament 2 carries current I2, and that we hold these currents constant as we move the ﬁlaments into position. We can think of assembling these ﬁlaments in two ways: byplacing ﬁlament 1 ﬁrst, or by placing ﬁlament 2 ﬁrst. In either case, placing the ﬁrstﬁlament requires no work since (3.184)is zero. The work required to place the secondﬁlament is W 1=− I1/Psi11if ﬁlament 2 is placed ﬁrst, where /Psi11is the ﬂux passing through ﬁlament 1 in its ﬁnal position, caused by the presence of ﬁlament 2. If ﬁlament 1 isplaced ﬁrst, the work required is W 2=− I2/Psi12. Since the work cannot depend on which loop is placed ﬁrst, we have W1=W2=Wwhere we can use either W=− I1/Psi11or W=− I2/Psi12. It is even more convenient, as we shall see, to average these values and use W=−1 2(I1/Psi11+I2/Psi12). (3.185) We must determine the energy required to keep the currents constant as we move the ﬁlaments into position. When moving the ﬁrst ﬁlament into place there is no inducedemf, since no applied ﬁeld is yet present. However, when moving the second ﬁlamentinto place we will change the ﬂux linked by boththe ﬁrst and second loops. This change of ﬂux will induce an emf in each of the loops, and this will change the current. To keepthe current constant we must supply an opposing emf. Let dW emf/dtbe the rate of work required to keep the current constant. Then by (3.153)and (3.181)we have dWemf dt=−/integraldisplay VJ·EdV=− I/integraldisplay E·dl=− Id/Psi1 dt. Integrating, we ﬁnd the total work /Delta1Wrequired to keep the current constant in either loop as the ﬂux through the loop is changed by an amount /Delta1/Psi1: /Delta1Wem f=I/Delta1/Psi1. So the total work required to keep I1constant as the loops are moved from inﬁnity (where the ﬂux is zero)to their ﬁnal positions is I1/Psi11. Similarly, a total work I2/Psi12is required to keep I2constant during the same process. Adding these to (3.185), the work required to position the loops, we obtain the complete assembly energy W=1 2(I1/Psi11+I2/Psi12) for two ﬁlaments. The extension to Nﬁlaments is Wm=1 2N/summationdisplay n=1In/Psi1n. (3.186) Consequently, the energy of a single current ﬁlament is Wm=1 2I/Psi1. (3.187) We may interpret this as the “assembly energy” required to bring the single loop into existence by bringing vanishingly small loops (magnetic dipoles)in from inﬁnity. Wemay also interpret it as the energy required to establish the current in this single ﬁlamentagainst the back emf. That is, if we establish Iby slowly increasing the current from zero in Nsmall steps /Delta1I=I/N, an energy /Psi1 n/Delta1Iwill be required at each step. Since /Psi1nincreases proportionally to I,w eh a v e Wm=N/summationdisplay n=1I N/bracketleftbigg (n−1)/Psi1 N/bracketrightbigg where /Psi1is the ﬂux when the current is fully established. Since/summationtextN n=1(n−1)=N(N−1)/2 we obtain Wm=1 2I/Psi1 (3.188) asN→∞. A volume current Jcan be treated as though it were composed of Ncurrent ﬁlaments. Equations (3.128)and (3.186)give Wm=1 2N/summationdisplay n=1In/contintegraldisplay /Gamma1nA·dl. Since the total current is I=/integraldisplay CSJ·dS=N/summationdisplay n=1In where CSdenotes the cross-section of the steady current, we have as N→∞ Wm=1 2/integraldisplay VA·JdV. (3.189) Alternatively, using (3.135), we may write Wm=1 2/integraldisplay V/integraldisplay VJ(r)·J(r/prime) |r−r/prime|dV dV/prime. Note the similarity between (3.189)and (3.86) . We now manipulate (3.189)into a form involving only the electromagnetic ﬁelds. By Ampere’s law Wm=1 2/integraldisplay VA·(∇× H)dV. Using (B.44)and the divergence theorem we can write Wm=1 2/contintegraldisplay S(H×A)·dS+1 2/integraldisplay VH·(∇× A)dV. We now let Sexpand to inﬁnity. This does not change the value of Wmsince we do not enclose any more current; however, since A∼1/rand H∼1/r2, the surface integral vanishes. Thus, remembering that ∇× A=B,w eh a v e Wm=1 2/integraldisplay V∞H·BdV (3.190) where V∞denotes all of space. Although we do not provide a derivation, (3.190)is also valid within linear materials. For nonlinear materials, the total energy required to build up a magnetic ﬁeld from B1 toB2is Wm=1 2/integraldisplay V∞/bracketleftbigg/integraldisplayB2 B1H·dB/bracketrightbigg dV. (3.191) This accounts for the work required to drive a ferromagnetic material through its hystere- sis loop. Readers interested in a complete derivation of (3.191)should consult Stratton[187]. As an example, consider two thin-walled, coaxial, current-carrying cylinders having radii a,b(b>a). The intervening region is a linear magnetic material having perme- ability µ. Assume that the inner and outer conductors carry total currents Iin the ±z directions, respectively. From the large-scale form of Ampere’s law we ﬁnd that H=  0,ρ ≤a, ˆφI/2πρ, a≤ρ≤b, 0,ρ > b,(3.192) hence by (3.190) W m=1 2/integraldisplay dz/integraldisplay2π 0/integraldisplayb aµI2 (2πρ)2ρdρdφ, and the stored energy is Wm l=µI2 4πln/parenleftbiggb a/parenrightbigg (3.193) per unit length. Suppose instead that the inner cylinder is solid and that current is spread uniformly throughout. Then the ﬁeld between the cylinders is still given by (3.192)but within theinner conductor we have H=ˆφIρ 2πa2 by (3.169). Thus, to (3.193) we must add the energy Wm,inside l=1 2/integraldisplay2π 0/integraldisplaya 0µ0I2ρ2 (2πa2)2ρdρdφ=µ0I2 16π stored within the solid wire. The result is Wm l=µ0I2 4π/bracketleftbigg µrln/parenleftbiggb a/parenrightbigg +1 4/bracketrightbigg . 3.3.7 Magnetic ﬁeld of a permanently magnetized body We now have the tools necessary to compute the magnetic ﬁeld produced by a perma- nent magnet (a body with permanent magnetization M). As an example, we shall ﬁnd the ﬁeld due to a uniformly magnetized sphere in three diﬀerent ways: by computing thevector potential integral and taking the curl, by computing the scalar potential integraland taking the gradient, and by ﬁnding the scalar potential using separation of variablesand applying the boundary condition across the surface of the sphere. Consider a magnetized sphere of radius a, residing in free space and having permanent magnetization M(r)=M 0ˆz. The equivalent magnetization current and charge densities are given by JM=∇× M=0, (3.194) JMs=− ˆn×M=− ˆr×M0ˆz=M0ˆφsinθ, (3.195) and ρM=− ∇· M=0, (3.196) ρMs=ˆn·M=ˆr·M0ˆz=M0cosθ. (3.197) The vector potential is produced by the equivalent magnetization surface current. Using (3.137)we ﬁnd that A(r)=µ0 4π/integraldisplay SJMs |r−r/prime|dS/prime=µ0 4π/integraldisplayπ −π/integraldisplayπ 0M0ˆφ/primesinθ/prime |r−r/prime|sinθ/primedθ/primedφ/prime. Since ˆφ/prime=− ˆxsinφ/prime+ˆycosφ/prime, the rectangular components of Aare /braceleftbigg−Ax Ay/bracerightbigg =µ0 4π/integraldisplayπ −π/integraldisplayπ 0M0sinφ/prime cosφ/primesinθ/prime |r−r/prime|a2sinθ/primedθ/primedφ/prime. (3.198) The integrals are most easily computed via the spherical harmonic expansion (E.200)for the inverse distance |r−r/prime|−1: /braceleftbigg−Ax Ay/bracerightbigg =µ0M0a2∞/summationdisplay n=0n/summationdisplay m=−nYnm(θ, φ) 2n+1rn < rn+1 >/integraldisplayπ −π/integraldisplayπ 0sinφ/prime cosφ/primesin2θ/primeY∗ nm(θ/prime,φ/prime)dθ/primedφ/prime. Because the φ/primevariation is sinφ/primeorcosφ/prime, all terms in the sum vanish except n=1, m=±1. Since Y1,−1(θ, φ) =/radicalbigg 3 8πsinθe−jφ, Y1,1(θ, φ) =−/radicalbigg 3 8πsinθejφ, we have /braceleftbigg−Ax Ay/bracerightbigg =µ0M0a2 3r< r2>3 8πsinθ/integraldisplayπ 0sin3θ/primedθ/prime· ·/bracketleftbigg e−jφ/integraldisplayπ −πsinφ/prime cosφ/primeejφ/primedφ/prime+ejφ/integraldisplayπ −πsinφ/prime cosφ/primee−jφ/primedφ/prime/bracketrightbigg . Carrying out the integrals we ﬁnd that /braceleftbigg−Ax Ay/bracerightbigg =µ0M0a2 3r< r2>sinθ/braceleftbiggsinφ cosφ/bracerightbigg or A=µ0M0a2 3r< r2>sinθˆφ. Finally, B=∇× Agives B=/braceleftBigg2µ0M0 3ˆz, r<a, µ0M0a3 3r3/parenleftBig ˆr2 cosθ+ˆθsinθ/parenrightBig ,r>a.(3.199) Hence Bwithin the sphere is uniform and in the same direction as M, while Boutside the sphere has the form of the magnetic dipole ﬁeld with moment m=/parenleftbigg4 3πa3/parenrightbigg M0. We can also compute Bby ﬁrst ﬁnding the scalar potential through direct computation of the integral (3.126). Substituting for ρMsfrom (3.197), we have /Phi1m(r)=1 4π/integraldisplay SρMs(r/prime) |r−r/prime|dS/prime=1 4π/integraldisplayπ −π/integraldisplayπ 0M0cosθ/prime |r−r/prime|sinθ/primedθ/primedφ/prime. This integral has the form of (3.100)with f(θ)=M0cosθ. Thus, from (3.102), /Phi1m(r)=M0a2 3cosθr< r2>. (3.200) The magnetic ﬁeld His then H=− ∇ /Phi1m=/braceleftBigg −M0 3ˆz, r<a, M0a3 3r3/parenleftBig ˆr2 cosθ+ˆθsinθ/parenrightBig ,r>a.. Inside the sphere Bis given by B=µ0(H+M), while outside the sphere it is merely B=µ0H. These observations lead us again to (3.199). Since the scalar potential obeys Laplace’s equation both inside and outside the sphere, as a last approach to the problem we shall write /Phi1min terms of the separation of variables solution discussed in §A.4. We can repeat our earlier arguments for the dielectric sphere in an impressed electric ﬁeld ( §3.2.10). Copying equations (3.109) and (3.110), we can write for r≤a /Phi1m1(r,θ)=∞/summationdisplay n=0AnrnPn(cosθ), (3.201) and for r≥a /Phi1m2(r,θ)=∞/summationdisplay n=0Bnr−(n+1)Pn(cosθ). (3.202) The boundary condition (3.154)at r=arequires that ∞/summationdisplay n=0AnanPn(cosθ)=∞/summationdisplay n=0Bna−(n+1)Pn(cosθ); upon application of the orthogonality of the Legendre functions, this becomes Anan=Bna−(n+1). (3.203) We can write (3.155)as −∂/Phi1 m1 ∂r+∂/Phi1 m2 ∂r=−ρMs so that at r=a −∞/summationdisplay n=0Annan−1Pn(cosθ)−∞/summationdisplay n=0Bn(n+1)a−(n+2)Pn(cosθ)=− M0cosθ. After application of orthogonality this becomes A1+2B1a−3=M0, (3.204) nan−1An=−(n+1)Bna−(n+2), n/negationslash=1. (3.205) Solving (3.203)and (3.204)simultaneously for n=1we ﬁnd that A1=M0 3, B1=M0 3a3. We also see that (3.203)and (3.205)are inconsistent unless An=Bn=0,n/negationslash=1. Substituting these results into (3.201)and (3.202) , we have /Phi1m=/braceleftBigg M0 3rcosθ, r≤a, M0 3a3 r2cosθ,r≥a, which is (3.200). 3.3.8 Bodies immersed in an impressed magnetic ﬁeld: magnetostatic shielding A highly permeable enclosure can provide partial shielding from external magnetostatic ﬁelds.Conside raspherica lshellofhighlypermeabl emateria l(Figure3.22);assum eit is immersed in a uniform impressed ﬁeld H0=H0ˆz. We wish to determine the internal ﬁeld and the factor by which it is reduced from the external applied ﬁeld. Because thereare no sources (the applied ﬁeld is assumed to be created by sources far removed), wemay use magnetic scalar potentials to represent the ﬁelds everywhere. We may representthe scalar potentials using a separation of variables solution to Laplace’s equation, witha contribution only from the n=1term in the series. In region 1 we have both scattered Figure 3.22: Spherical shell of magnetic material. and applied potentials, where the applied potential is just /Phi10=− H0z=− H0rcosθ, since H0=− ∇ /Phi10=H0ˆz. We have /Phi11(r)=A1r−2cosθ−H0rcosθ, (3.206) /Phi12(r)=(B1r−2+C1r)cosθ, (3.207) /Phi13(r)=D1rcosθ. (3.208) We choose (3.109)for the scattered potential in region 1 so that it decays as r→∞ , and (3.110)for the scattered potential in region 3so that it remains ﬁnite at r=0.I n region 2 we have no restrictions and therefore include both contributions. The coeﬃcients A1,B1,C1,D1are found by applying the appropriate boundary conditions at r=aand r=b. By continuity of the scalar potential across each boundary we have A1b−2−H0b=B1b−2+C1b, B1a−2+C1a=D1a. By (3.156), the quantity µ∂/Phi1/∂ ris also continuous at r=aand r=b; this gives two more equations: µ0(−2A1b−3−H0)=µ(−2B1b−3+C1), µ(−2B1a−3+C1)=µ0D1. Simultaneous solution yields D1=−9µr KH0 where K=(2+µr)(1+2µr)−2(a/b)3(µr−1)2. Substituting this into (3.208)and using H=− ∇ /Phi1m, we ﬁnd that H=κH0ˆz within the enclosure, where κ=9µr/K. This ﬁeld is uniform and, since κ< 1forµr>1, it is weaker than the applied ﬁeld. For µr/greatermuch1we have K≈2µ2 r[1−(a/b)3]. Denoting the shell thickness by /Delta1=b−a, we ﬁnd that K≈6µ2 r/Delta1/awhen /Delta1/a/lessmuch1.T h u s κ=3 21 µr/Delta1 a describes the coeﬃcient of shielding for a highly permeable spherical enclosure, valid when µr/greatermuch1and/Delta1/a/lessmuch1. A shell for which µr=10,000and a/b=0.99can reduce the enclosure ﬁeld to 0.15% of the applied ﬁeld. 3.4 Static ﬁeld theorems 3.4.1 Mean value theorem of electrostatics The average value of the electrostatic potential over a sphere is equal to the potential at the center of the sphere, provided that the sphere encloses no electric charge. To seethis, write /Phi1(r)=1 4π/epsilon1/integraldisplay Vρ(r/prime) RdV/prime+1 4π/contintegraldisplay S/bracketleftBigg −/Phi1(r/prime)ˆR R2+∇/prime/Phi1(r/prime) R/bracketrightBigg ·dS/prime; putρ≡0inV, and use the obvious facts that if Sis a sphere centered at point rthen (1)Ris constant on Sand (2) ˆn/prime=− ˆR: /Phi1(r)=1 4πR2/contintegraldisplay S/Phi1(r/prime)dS/prime−1 4πR/contintegraldisplay SE(r/prime)·dS/prime. The last term vanishes by Gauss’s law, giving the desired result. 3.4.2 Earnshaw’s theorem It is impossible for a charge to rest in stable equilibrium under the inﬂuence of elec- trostatic forces alone. This is an easy consequence of the mean value theorem of electro-statics, which precludes the existence of a point where /Phi1can assume a maximum or a minimum. 3.4.3 Thomson’s theorem Static charge on a system of perfect conductors distributes itself so that the electric storedenergyisaminimum.Figure3.23showsasystemof n conductin gbodiesheldat potentials /Phi11,...,/Phi1 n. Suppose the potential ﬁeld associated with the actual distribution of charge on these bodies is /Phi1, giving We=/epsilon1 2/integraldisplay VE·EdV=/epsilon1 2/integraldisplay V∇/Phi1·∇/Phi1dV for the actual stored energy. Now assume a slightly diﬀerent charge distribution, resulting in a new potential /Phi1/prime=/Phi1+δ/Phi1that satisﬁes the same boundary conditions (i.e., assume δ/Phi1=0on each conducting body). The stored energy associated with this hypothetical situation is W/prime e=We+δWe=/epsilon1 2/integraldisplay V∇(/Phi1+δ/Phi1)·∇(/Phi1+δ/Phi1)dV Figure 3.23: System of conductors used to derive Thomson’s theorem. so that δWe=/epsilon1/integraldisplay V∇/Phi1·∇(δ/Phi1) dV+/epsilon1 2/integraldisplay V|∇(δ/Phi1)|2dV; Thomson’s theorem will be proved if we can show that /integraldisplay V∇/Phi1·∇(δ/Phi1) dV=0, (3.209) because then we shall have δWe=/epsilon1 2/integraldisplay V|∇(δ/Phi1)|2dV≥0. To establish (3.209), we use Green’s ﬁrst identity /integraldisplay V(∇u·∇v+u∇2v)dV=/contintegraldisplay Su∇v·dS with u=δ/Phi1andv=/Phi1: /integraldisplay V∇/Phi1·∇(δ/Phi1) dV=/contintegraldisplay Sδ/Phi1∇/Phi1·dS. Here Sis composed of (1)the exterior surfaces Sk(k=1,..., n)of the nbodies, (2) the surfaces Scof the “cuts” that are introduced in order to keep Va simply-connected region (a condition for the validity of Green’s identity), and (3) the sphere S∞of very large radius r.T h u s /integraldisplay V∇/Phi1·∇(δ/Phi1) dV=n/summationdisplay k=1/integraldisplay Skδ/Phi1∇/Phi1·dS+/integraldisplay Scδ/Phi1∇/Phi1·dS+/integraldisplay S∞δ/Phi1∇/Phi1·dS. The ﬁrst term on the right vanishes because δ/Phi1=0on each Sk. The second term vanishes because the contributions from opposite sides of each cut cancel (note that ˆn occurs in pairs that are oppositely directed). The third term vanishes because /Phi1∼1/r, ∇/Phi1∼1/r2, and dS∼r2where r→∞ for points on S∞. Figure 3.24: System of conductors used to derive Green’s reciprocation theorem. 3.4.4 Green’s reciprocation theorem Conside rasystemof n conductin gbodiesasinFigure3.24.Anassociatedmathemat- ical surface Stconsists of the exterior surfaces S1,..., Snof the nbodies, taken together with a surface Sthat enclosed all of the bodies. Suppose /Phi1and/Phi1/primeare electrostatic potentials produced by two distinct distributions of stationary charge over the set ofconductors. Then ∇ 2/Phi1=0=∇2/Phi1/primeand Green’s second identity gives /contintegraldisplay St/parenleftbigg /Phi1∂/Phi1/prime ∂n−/Phi1/prime∂/Phi1 ∂n/parenrightbigg dS=0 or n/summationdisplay k=1/integraldisplay Sk/Phi1∂/Phi1/prime ∂ndS+/integraldisplay S/Phi1∂/Phi1/prime ∂ndS=n/summationdisplay k=1/integraldisplay Sk/Phi1/prime∂/Phi1 ∂ndS+/integraldisplay S/Phi1/prime∂/Phi1 ∂ndS. Now let Sbe a sphere of very large radius Rso that at points on Swe have /Phi1, /Phi1/prime∼1 R,∂/Phi1 ∂n,∂/Phi1/prime ∂n∼1 R2, dS∼R2; asR→∞ then, n/summationdisplay k=1/integraldisplay Sk/Phi1∂/Phi1/prime ∂ndS=n/summationdisplay k=1/integraldisplay Sk/Phi1/prime∂/Phi1 ∂ndS. Furthermore, the conductors are equipotentials so that n/summationdisplay k=1/Phi1k/integraldisplay Sk∂/Phi1/prime ∂ndS=n/summationdisplay k=1/Phi1/prime k/integraldisplay Sk∂/Phi1 ∂ndS and we therefore have n/summationdisplay k=1q/prime k/Phi1k=n/summationdisplay k=1qk/Phi1/prime k (3.210) where the kth conductor ( k=1,..., n)has potential /Phi1kwhen it carries charge qk, and has potential /Phi1/prime kwhen it carries charge q/prime k. This is Green’s reciprocation theorem . A classic application is to determine the charge induced on a grounded conductor by Figure 3.25: Application of Green’s reciprocation theorem. (a)The “unprimed situation” permits us to determine the potential VPat point Pproduced by a charge qplaced on body 1. Here V1is the potential of body 1. (b)In the “primed situation” we ground body 1 and induce a charge q/primeby bringing a point charge q/prime Pinto proximity. a nearby point charge. This is accomplished as follows. Let the conducting body of interest be designated as body 1, and model the nearby point charge qPas a very small conducting body designated as body 2 and located at point Pin space. Take q1=q, q2=0,/Phi1 1=V1,/Phi1 2=VP, and q/prime 1=q/prime, q/prime 2=q/prime P,/Phi1/prime1=0,/Phi1/prime 2=V/prime P, givingthetwosituation sshowninFigure3.25.Substitutio nintoGreen’ srecipr ocation theorem q/prime 1/Phi11+q/prime 2/Phi12=q1/Phi1/prime 1+q2/Phi1/prime 2 gives q/primeV1+q/prime PVP=0so that q/prime=−q/prime PVP/V1. (3.211) 3.5 Problems 3.1The z-axis carries a line charge of nonuniform density ρl(z). Show that the electric ﬁeld in the plane z=0is given by E(ρ, φ) =1 4π/epsilon1/bracketleftbigg ˆρρ/integraldisplay∞ −∞ρl(z/prime)dz/prime (ρ2+z/prime2)3/2−ˆz/integraldisplay∞ −∞ρl(z/prime)z/primedz/prime (ρ2+z/prime2)3/2/bracketrightbigg . Compute Ewhen ρl=ρ0sgn(z), where sgn(z)is the signum function (A.6). 3.2The ring ρ=a,z=0, carries a line charge of nonuniform density ρl(φ). Show that the electric ﬁeld at an arbitrary point on the z-axis is given by E(z)=−a2 4π/epsilon1(a2+z2)3/2/bracketleftbigg ˆx/integraldisplay2π 0ρl(φ/prime)cosφ/primedφ/prime+ˆy/integraldisplay2π 0ρl(φ/prime)sinφ/primedφ/prime/bracketrightbigg + +ˆzaz 4π/epsilon1(a2+z2)3/2/integraldisplay2π 0ρl(φ/prime)dφ/prime. Figure 3.26: Geometry for computing Green’s function for parallel plates. Compute Ewhen ρl(φ)=ρ0sinφ. Repeat for ρl(φ)=ρ0cos2φ. 3.3The plane z=0carries a surface charge of nonuniform density ρs(ρ, φ) . Show that at an arbitrary point on the z-axis the rectangular components of Eare given by Ex(z)=−1 4π/epsilon1/integraldisplay∞ 0/integraldisplay2π 0ρs(ρ/prime,φ/prime)ρ/prime2cosφ/primedφ/primedρ/prime (ρ/prime2+z2)3/2, Ey(z)=−1 4π/epsilon1/integraldisplay∞ 0/integraldisplay2π 0ρs(ρ/prime,φ/prime)ρ/prime2sinφ/primedφ/primedρ/prime (ρ/prime2+z2)3/2, Ez(z)=z 4π/epsilon1/integraldisplay∞ 0/integraldisplay2π 0ρs(ρ/prime,φ/prime)ρ/primedφ/primedρ/prime (ρ/prime2+z2)3/2. Compute Ewhen ρs(ρ, φ) =ρ0U(ρ−a)where U(ρ)is the unit step function (A.5). Repeat for ρs(ρ, φ) =ρ0[1−U(ρ−a)]. 3.4The sphere r=acarries a surface charge of nonuniform density ρs(θ). Show that the electric intensity at an arbitrary point on the z-axis is given by E(z)=ˆza2 2/epsilon1/integraldisplayπ 0ρs(θ/prime)(z−acosθ/prime)sinθ/primedθ/prime (a2+z2−2azcosθ/prime)3/2. Compute E(z)when ρs(θ)=ρ0, a constant. Repeat for ρs(θ)=ρ0cos2θ. 3.5Beginning with the postulates for the electrostatic ﬁeld ∇× E=0, ∇·D=ρ, use the technique of §2.8.2 to derive the boundary conditions (3.32)–(3.33). 3.6A material half space of permittivity /epsilon11occupies the region z>0, while a second material half space of permittivity /epsilon12occupies z<0. Find the polarization surface charge densities and compute the total induced polarization charge for a point charge Qlocated atz=h. 3.7Consider a point charge between two grounded conducting plates as shown in Figure3.26.WritetheGreen’ sfunctio nasthesumofprimar yandsecondar ytermsand apply the boundary conditions to show that the secondary Green’s function is Gs(r|r/prime)=1 (2π)2/integraldisplay∞ −∞/integraldisplay∞ −∞/bracketleftbigg −e−kρ(d−z)sinhkρz/prime sinhkρd−e−kρzsinhkρ(d−z/prime) sinhkρd/bracketrightbigge−jkρ·r/prime 2kρd2kρ. (3.212) 3.8Use the expansion 1 sinhkρd=csch kρd=2∞/summationdisplay n=0e−(2n+1)kρd to show that the secondary Green’s function for parallel conducting plates (3.212)may be written as an inﬁnite sequence of images of the primary point charge. Identify thegeometrical meaning of each image term. 3.9Find the Green’s functions for a dielectric slab of thickness dplaced over a perfectly conducting ground plane located at z=0. 3.10 Find the Green’s functions for a dielectric slab of thickness 2dimmersed in free space and centered on the z=0plane. Compare to the Green’s function found in Problem 3.9. 3.11Referrin gtothesystemofFigure3.9,ﬁndthechargedensityonthesurfac eof the sphere and integrate to show that the total charge is equal to the image charge. 3.12 Use the method of Green’s functions to ﬁnd the potential inside a conducting sphere for ρinside the sphere. 3.13 Solve for the total potential and electric ﬁeld of a grounded conducting sphere centered at the origin within a uniform impressed electric ﬁeld E=E 0ˆz. Find total charge induced on the sphere. 3.14 Consider a spherical cavity of radius acentered at the origin within a homogeneous dielectric material of permittivity /epsilon1=/epsilon10/epsilon1r. Solve for total potential and electric ﬁeld inside the cavity in the presence of an impressed ﬁeld E=E0ˆz. Show that the ﬁeld in the cavity is stronger than the applied ﬁeld, and explain this using polarization surfacecharge. 3.15 Find the ﬁeld of a point charge Qlocated at z=dabove a perfectly conducting ground plane at z=0. Use the boundary condition to ﬁnd the charge density on the plane and integrate to show that the total charge is −Q. Integrate Maxwell’s stress tensor over the surface of the ground plane and show that the force on the ground planeis the same as the force on the image charge found from Coulomb’s law. 3.16 Consider in free space a point charge −qatr=r 0+d, a point charge −qat r=r0−d, and a point charge 2qatr0. Find the ﬁrst three multipole moments and the resulting potential produced by this charge distribution. 3.17 A spherical charge distribution of radius ain free space has the density ρ(r)=Q πa3cos 2θ. Compute the multipole moments for the charge distribution and ﬁnd the resulting poten- tial. Find a suitable arrangement of point charges that will produce the same potentialﬁeld for r>aas produced by the spherical charge. 3.18Comput ethemagneti cﬂuxdensity B forthecircula rwireloopofFigure3.18by (a) using the Biot–Savart law (3.165), and (b) computing the curl of (3.138). Figure 3.27: Parallel plate capacitor. 3.19 Two circular current-carrying wires are arranged coaxially along the z-axis. Loop 1 has radius a1, carries current I1, and is centered in the z=0plane. Loop 2 has radius a2, carries current I2, and is centered in the z=dplane. Find the force between the loops. 3.20 Choose Q=∇/prime/parenleftbig1 R/parenrightbig ×cin (3.162)and derive the following expression for B: B(r)=µ 4π/integraldisplay VJ(r/prime)×∇/prime/parenleftbigg1 R/parenrightbigg dV/prime− −1 4π/contintegraldisplay S/bracketleftbigg [ˆn/prime×B(r/prime)]×∇/prime/parenleftbigg1 R/parenrightbigg +[ˆn/prime·B(r/prime)]∇/prime/parenleftbigg1 R/parenrightbigg/bracketrightbigg dS/prime, where ˆnis the normal vector outward from V. Compare to the Stratton–Chu formula (6.8). 3.21 Compute the curl of (3.163)to obtain the integral expression for Bgiven in Prob- lem 3.20. Compare to the Stratton–Chu formula (6.8). 3.22 Obtain (3.170)by integration of Maxwell’s stress tensor over the xz-plane. 3.23 Consider two thin conducting parallel plates embedded in a region of permittivity /epsilon1(Figure3.27).Thebottomplateisconnecte dtoground ,andweapplyanexcesscharge +Qto the top plate (and thus −Qis drawn onto the bottom plate.)Neglecting fringing, (a)solve Laplace’s equation to show that /Phi1(z)=Q A/epsilon1z. Use (3.87)to show that W=Q2d 2A/epsilon1. (b)Verify Wusing (3.88). (c) Use F=− ˆzdW/dzto show that the force on the top plate is F=− ˆzQ2 2A/epsilon1. (d)Verify Fby integrating Maxwell’s stress tensor over a closed surface surrounding the top plate. 3.24 Consider two thin conducting parallel plates embedded in a region of permittivity /epsilon1(Figure3.27).Thebottomplateisconnecte dtoground ,andweapplyapotential V0to the top plate using a battery. Neglecting fringing, (a)solve Laplace’s equation to showthat /Phi1(z)=V 0 dz. Use (3.87)to show that W=V2 0A/epsilon1 2d. (b)Verify Wusing (3.88). (c) Use F=− ˆzdW/dzto show that the force on the top plate is F=− ˆzV2 0A/epsilon1 2d2. (d)Verify Fby integrating Maxwell’s stress tensor over a closed surface surrounding the top plate. 3.25 A group of Nperfectly conducting bodies is arranged in free space. Body nis held at potential Vnwith respect to ground, and charge Qnis induced upon its surface. By linearity we may write Qm=N/summationdisplay n=1cmnVn where the cmnare called the capacitance coeﬃcients . Using Green’s reciprocation the- orem, demonstrate that cmn=cnm. Hint: Use (3.210). Choose one set of voltages so that Vk=0,k/negationslash=n, and place Vnat some potential, say Vn=V0, producing the set of charges {Qk}. For the second set choose V/prime k=0,k/negationslash=m, and Vm=V0, producing {Q/prime k}. 3.26 For the set of conductors of Problem 3.25, show that we may write Qm=CmmVm+/summationdisplay k/negationslash=mCmk(Vm−Vk) where Cmn=−cmn,m/negationslash=n, Cmm=N/summationdisplay k=1cmk. Here Cmm, called the self capacitance , describes the interaction between the mth con- ductor and ground, while Cmn, called the mutual capacitance , describes the interaction between the mth and nth conductors. 3.27 For the set of conductors of Problem 3.25, show that the stored electric energy is given by W=1 2N/summationdisplay m=1N/summationdisplay n=1cmnVnVm. 3.28Agroupof N wiresisarrange dinfreespaceasshowninFigure3.28.Wire n carries a steady current In, and a ﬂux /Psi1npasses through the surface deﬁned by its contour /Gamma1n. By linearity we may write /Psi1m=N/summationdisplay n=1LmnIn Figure 3.28: A system of current-carrying wires. where the Lmnare called the coeﬃcients of inductance . Derive Neumann’s formula Lmn=µ0 4π/contintegraldisplay /Gamma1n/contintegraldisplay /Gamma1mdl·dl/prime |r−r/prime|, and thereby demonstrate the reciprocity relation Lmn=Lnm. 3.29ForthegroupofwiresshowninFigure3.28,showthatthestoredmagneti cenergy is given by W=1 2N/summationdisplay m=1N/summationdisplay n=1LmnInIm. 3.30 Prove the minimum heat generation theorem : steady electric currents distribute themselves in a conductor in such a way that the dissipated power is a minimum. Hint:LetJbe the actual distribution of current in a conducting body, and let the power it dissipates be P. Let J /prime=J+δJbe any other current distribution, and let the power it dissipates be P/prime=P+δP. Show that δP=1 2/integraldisplay V1 σ|δJ|2dV≥0. Chapter 4 Temporal and spatial frequency domain representation 4.1 Interpretation of the temporal transform When a ﬁeld is represented by a continuous superposition of elemental components, the resulting decomposition can simplify computation and provide physical insight. Such rep-resentation is usually accomplished through the use of an integral transform. Althoughseveral diﬀerent transforms are used in electromagnetics, we shall concentrate on thepowerful and eﬃcient Fourier transform. Let us consider the Fourier transform of the electromagnetic ﬁeld. The ﬁeld depends onx,y,z,t, and we can transform with respect to any or all of these variables. However, a consideration of units leads us to consider a transform over tseparately. Let ψ(r,t) represent any rectangular component of the electric or magnetic ﬁeld. Then the temporaltransform will be designated by ˜ψ(r,ω): ψ(r,t)↔˜ψ(r,ω) . Hereωis the transform variable. The transform ﬁeld ˜ψis calculated using (A.1): ˜ψ(r,ω)=/integraldisplay ∞ −∞ψ(r,t)e−jωtdt. (4.1) The inverse transform is, by (A.2), ψ(r,t)=1 2π/integraldisplay∞ −∞˜ψ(r,ω)ejωtdω. (4.2) Since ˜ψis complex it may be written in amplitude–phase form: ˜ψ(r,ω)=|˜ψ(r,ω)|ejξψ(r,ω), where we take −π<ξψ(r,ω)≤π. Since ψ(r,t)must be real, (4.1) shows that ˜ψ(r,−ω)=˜ψ∗(r,ω) . (4.3) Furthermore, the transform of the derivative of ψmay be found by diﬀerentiating (4.2). We have ∂ ∂tψ(r,t)=1 2π/integraldisplay∞ −∞jω˜ψ(r,ω)ejωtdω, hence ∂ ∂tψ(r,t)↔jω˜ψ(r,ω) . (4.4) By virtue of (4.2), any electromagnetic ﬁeld component can be decomposed into a contin- uous, weighted superposition of elemental temporal terms ejωt. Note that the weighting factor ˜ψ(r,ω), often called the frequency spectrum ofψ(r,t), is not arbitrary because ψ(r,t)must obey a scalar wave equation such as (2.327). For a source-free region of space we have /parenleftbigg ∇2−µσ∂ ∂t−µ/epsilon1∂2 ∂t2/parenrightbigg1 2π/integraldisplay∞ −∞˜ψ(r,ω)ejωtdω=0. Diﬀerentiating under the integral sign we have 1 2π/integraldisplay∞ −∞/bracketleftbig/parenleftbig ∇2−jωµσ+ω2µ/epsilon1/parenrightbig˜ψ(r,ω)/bracketrightbig ejωtdω=0, hence by the Fourier integral theorem /parenleftbig ∇2+k2/parenrightbig˜ψ(r,ω)=0 (4.5) where k=ω√µ/epsilon1/radicalbigg 1−jσ ω/epsilon1 is thewavenumber . Equation (4.5) is called the scalar Helmholtz equation , and represents the wave equation in the temporal frequency domain. 4.2 The frequency-domain Maxwell equations If the region of interest contains sources, we can return to Maxwell’s equations and represent all quantities using the temporal inverse Fourier transform. We have, for ex-ample, E(r,t)=1 2π/integraldisplay∞ −∞˜E(r,ω)ejωtdω where ˜E(r,ω)=3/summationdisplay i=1ˆii˜Ei(r,ω)=3/summationdisplay i=1ˆii|˜Ei(r,ω)|ejξE i(r,ω). (4.6) All other ﬁeld quantities will be written similarly with an appropriate superscript on the phase. Substitution into Ampere’s law gives ∇×1 2π/integraldisplay∞ −∞˜H(r,ω)ejωtdω=∂ ∂t1 2π/integraldisplay∞ −∞˜D(r,ω)ejωtdω+1 2π/integraldisplay∞ −∞˜J(r,ω)ejωtdω, hence 1 2π/integraldisplay∞ −∞[∇× ˜H(r,ω)−jω˜D(r,ω)−˜J(r,ω)]ejωtdω=0 after we diﬀerentiate under the integral signs and combine terms. So ∇× ˜H=jω˜D+˜J (4.7) by the Fourier integral theorem. This version of Ampere’s law involves only the frequency- domain ﬁelds. By similar reasoning we have ∇× ˜E=− jω˜B, (4.8) ∇·˜D=˜ρ, (4.9) ∇·˜B(r,ω)=0, (4.10) and ∇·˜J+jω˜ρ=0. Equations (4.7)–(4.10) govern the temporal spectra of the electromagnetic ﬁelds. We may manipulate them to obtain wave equations, and apply the boundary conditions from thefollowing section. After ﬁnding the frequency-domain ﬁelds we may ﬁnd the temporalﬁelds by Fourier inversion. The frequency-domain equations involve one fewer derivative(the time derivative has been replaced by multiplication by jω), hence may be easier to solve. However, the inverse transform may be diﬃcult to compute. 4.3 Boundary conditions on the frequency-domain ﬁelds Several boundary conditions on the source and mediating ﬁelds were derived in §2.8.2. For example, we found that the tangential electric ﬁeld must obey ˆn12×E1(r,t)−ˆn12×E2(r,t)=−Jms(r,t). The technique of the previous section gives us ˆn12×[˜E1(r,ω)−˜E2(r,ω)]=− ˜Jms(r,ω) as the condition satisﬁed by the frequency-domain electric ﬁeld. The remaining boundary conditions are treated similarly. Let us summarize the results, including the eﬀects ofﬁctitious magnetic sources: ˆn 12×(˜H1−˜H2)=˜Js, ˆn12×(˜E1−˜E2)=− ˜Jms, ˆn12·(˜D1−˜D2)=˜ρs, ˆn12·(˜B1−˜B2)=˜ρms, and ˆn12·(˜J1−˜J2)=− ∇ s·˜Js−jω˜ρs, ˆn12·(˜Jm1−˜Jm2)=− ∇ s·˜Jms−jω˜ρms. Here ˆn12points into region 1 from region 2. 4.4 Constitutive relations in the frequency domain and the Kronig–Kramers relations All materials are to some extent dispersive. If a ﬁeld applied to a material undergoes a suﬃciently rapid change, there is a time lag in the response of the polarization ormagnetization of the atoms. It has been found that such materials have constitutiverelations involving products in the frequency domain, and that the frequency-domainconstitutive parameters are complex, frequency-dependent quantities. We shall restrictourselves to the special case of anisotropic materials and refer the reader to Kong [101]and Lindell [113] for the more general case. For anisotropic materials we write ˜P=/epsilon1 0˜¯χe·˜E, (4.11) ˜M=˜¯χm·˜H, (4.12) ˜D=˜¯/epsilon1·˜E=/epsilon10[¯I+˜¯χe]·˜E, (4.13) ˜B=˜¯µ·˜H=µ0[¯I+˜¯χm]·˜H, (4.14) ˜J=˜¯σ·˜E. (4.15) By the convolution theorem and the assumption of causality we immediately obtain the dyadic versions of (2.29)–(2.31): D(r,t)=/epsilon10/parenleftbigg E(r,t)+/integraldisplayt −∞¯χe(r,t−t/prime)·E(r,t/prime)dt/prime/parenrightbigg , B(r,t)=µ0/parenleftbigg H(r,t)+/integraldisplayt −∞¯χm(r,t−t/prime)·H(r,t/prime)dt/prime/parenrightbigg , J(r,t)=/integraldisplayt −∞¯σ(r,t−t/prime)·E(r,t/prime)dt/prime. These describe the essential behavior of a dispersive material. The susceptances and conductivity, describing the response of the atomic structure to an applied ﬁeld, dependnot only on the present value of the applied ﬁeld but on all past values as well. Now since D(r,t),B(r,t), and J(r,t)are all real, so are the entries in the dyadic matrices ¯/epsilon1(r,t),¯µ(r,t), and ¯σ(r,t). Thus, applying (4.3) to each entry we must have ˜¯χ e(r,−ω)=˜¯χ∗ e(r,ω) , ˜¯χm(r,−ω)=˜¯χ∗ m(r,ω) , ˜¯σ(r,−ω)=˜¯σ∗(r,ω) , (4.16) and hence ˜¯/epsilon1(r,−ω)=˜¯/epsilon1∗(r,ω) , ˜¯µ(r,−ω)=˜¯µ∗(r,ω) . (4.17) If we write the constitutive parameters in terms of real and imaginary parts as ˜/epsilon1ij=˜/epsilon1/prime ij+j˜/epsilon1/prime/prime ij, ˜µij=˜µ/prime ij+j˜µ/prime/prime ij, ˜σij=˜σ/prime ij+j˜σ/prime/prime ij, these conditions become ˜/epsilon1/prime ij(r,−ω)=˜/epsilon1/prime ij(r,ω) , ˜/epsilon1/prime/prime ij(r,−ω)=− ˜/epsilon1/prime/prime ij(r,ω) , and so on. Therefore the real parts of the constitutive parameters are even functions of frequency, and the imaginary parts are odd functions of frequency. In most instances, the presence of an imaginary part in the constitutive parameters implies that the material is either dissipative (lossy), transforming some of the electro- magnetic energy in the ﬁelds into thermal energy, or active , transforming the chemical or mechanical energy of the material into energy in the ﬁelds. We investigate this furtherin§4.5 and §4.8.3. We can also write the constitutive equations in amplitude–phase form. Letting ˜/epsilon1 ij=|˜/epsilon1ij|ejξ/epsilon1 ij, ˜µij=|˜µij|ejξµ ij, ˜σij=|˜σij|ejξσ ij, and using the ﬁeld notation (4.6), we can write (4.13)–(4.15) as ˜Di=|˜Di|ejξD i=3/summationdisplay j=1|˜/epsilon1ij||˜Ej|ej[ξE j+ξ/epsilon1 ij], (4.18) ˜Bi=|˜Bi|ejξB i=3/summationdisplay j=1|˜µij||˜Hj|ej[ξH j+ξµ ij], (4.19) ˜Ji=|˜Ji|ejξJ i=3/summationdisplay j=1|˜σij||˜Ej|ej[ξE j+ξσ ij]. (4.20) Here we remember that the amplitudes and phases may be functions of both randω. For isotropic materials these reduce to ˜Di=|˜Di|ejξD i=|˜/epsilon1||˜Ei|ej(ξE i+ξ/epsilon1), (4.21) ˜Bi=|˜Bi|ejξB i=|˜µ||˜Hi|ej(ξH i+ξµ), (4.22) ˜Ji=|˜Ji|ejξJ i=|˜σ||˜Ei|ej(ξE i+ξσ). (4.23) 4.4.1 The complex permittivity As mentioned above, dissipative eﬀects may be associated with complex entries in the permittivity matrix. Since conduction eﬀects can also lead to dissipation, the permittivityand conductivity matrices are often combined to form a complex permittivity . Writing the current as a sum of impressed and secondary conduction terms ( ˜J=˜J i+˜Jc) and substituting (4.13) and (4.15) into Ampere’s law, we ﬁnd ∇× ˜H=˜Ji+˜¯σ·˜E+jω˜¯/epsilon1·˜E. Deﬁning the complex permittivity ˜¯/epsilon1c(r,ω)=˜¯σ(r,ω) jω+˜¯/epsilon1(r,ω) , (4.24) we have ∇× ˜H=˜Ji+jω˜¯/epsilon1c·˜E. Using the complex permittivity we can include the eﬀects of conduction current by merely replacing the total current with the impressed current. Since Faraday’s law is unaﬀected,any equation (such as the wave equation) derived previously using total current retainsits form with the same substitution. By (4.16) and (4.17) the complex permittivity obeys ˜¯/epsilon1 c(r,−ω)=˜¯/epsilon1c∗(r,ω) (4.25) or ˜/epsilon1c/prime ij(r,−ω)=˜/epsilon1c/prime ij(r,ω) , ˜/epsilon1c/prime/prime ij(r,−ω)=− ˜/epsilon1c/prime/prime ij(r,ω) . For an isotropic material it takes the particularly simple form ˜/epsilon1c=˜σ jω+˜/epsilon1=˜σ jω+/epsilon10+/epsilon10˜χe, (4.26) and we have ˜/epsilon1c/prime(r,−ω)=˜/epsilon1c/prime(r,ω) , ˜/epsilon1c/prime/prime(r,−ω)=− ˜/epsilon1c/prime/prime(r,ω) . (4.27) 4.4.2 High and low frequency behavior of constitutive parameters At low frequencies the permittivity reduces to the electrostatic permittivity. Since ˜/epsilon1/prime is even in ωand ˜/epsilon1/prime/primeis odd, we have for small ω ˜/epsilon1/prime∼/epsilon10/epsilon1r, ˜/epsilon1/prime/prime∼ω. If the material has some dc conductivity σ0, then for low frequencies the complex per- mittivity behaves as ˜/epsilon1c/prime∼/epsilon10/epsilon1r, ˜/epsilon1c/prime/prime∼σ0/ω. (4.28) IfEorHchanges very rapidly, there may be no polarization or magnetization eﬀect at all. This occurs at frequencies so high that the atomic structure of the material cannotrespond to the rapidly oscillating applied ﬁeld. Above some frequency then, we canassume ˜¯χ e=0and ˜¯χm=0so that ˜P=0, ˜M=0, and ˜D=/epsilon10˜E, ˜B=µ0˜H. In our simple models of dielectric materials ( §4.6) we ﬁnd that as ωbecomes large ˜/epsilon1/prime−/epsilon10∼1/ω2, ˜/epsilon1/prime/prime∼1/ω3. (4.29) Our assumption of a macroscopic model of matter provides a fairly strict upper frequency limit to the range of validity of the constitutive parameters. We must assume that the wavelength of the electromagnetic ﬁeld is large compared to the size of the atomic struc-ture. This limit suggests that permittivity and permeability might remain meaningfuleven at optical frequencies, and for dielectrics this is indeed the case since the values of ˜Premain signiﬁcant. However, ˜Mbecomes insigniﬁcant at much lower frequencies, and at optical frequencies we may use ˜B=µ 0˜H[107]. 4.4.3 The Kronig–Kramers relations The principle of causality is clearly implicit in (2.29)–(2.31). We shall demonstrate that causality leads to explicit relationships between the real and imaginary parts of thefrequency-domain constitutive parameters. For simplicity we concentrate on the isotropiccase and merely note that the present analysis may be applied to all the dyadic com-ponents of an anisotropic constitutive parameter. We also concentrate on the complexpermittivity and extend the results to permeability by induction. The implications of causality on the behavior of the constitutive parameters in the time domain can be easily identiﬁed. Writing (2.29) and (2.31) after setting u=t−t/prime and then u=t/prime, we have D(r,t)=/epsilon10E(r,t)+/epsilon10/integraldisplay∞ 0χe(r,t/prime)E(r,t−t/prime)dt/prime, J(r,t)=/integraldisplay∞ 0σ(r,t/prime)E(r,t−t/prime)dt/prime. We see that there is no contribution from values of χe(r,t)orσ(r,t)for times t<0.S o we can write D(r,t)=/epsilon10E(r,t)+/epsilon10/integraldisplay∞ −∞χe(r,t/prime)E(r,t−t/prime)dt/prime, J(r,t)=/integraldisplay∞ −∞σ(r,t/prime)E(r,t−t/prime)dt/prime, with the additional assumption χe(r,t)=0,t<0,σ ( r,t)=0,t<0. (4.30) By (4.30) we can write the frequency-domain complex permittivity (4.26) as ˜/epsilon1c(r,ω)−/epsilon10=1 jω/integraldisplay∞ 0σ(r,t/prime)e−jωt/primedt/prime+/epsilon10/integraldisplay∞ 0χe(r,t/prime)e−jωt/primedt/prime. (4.31) In order to derive the Kronig–Kramers relations we must understand the behavior of ˜/epsilon1c(r,ω)−/epsilon10in the complex ω-plane. Writing ω=ωr+jωi, we need to establish the following two properties. Property 1: The function ˜/epsilon1c(r,ω)−/epsilon10is analytic in the lower half-plane ( ωi<0) except at ω=0where it has a simple pole. We can establish the analyticity of ˜σ(r,ω)by integrating over any closed contour in the lower half-plane. We have /contintegraldisplay /Gamma1˜σ(r,ω)dω=/contintegraldisplay /Gamma1/bracketleftbigg/integraldisplay∞ 0σ(r,t/prime)e−jωt/primedt/prime/bracketrightbigg dω=/integraldisplay∞ 0σ(r,t/prime)/bracketleftbigg/contintegraldisplay /Gamma1e−jωt/primedω/bracketrightbigg dt/prime.(4.32) Note that an exchange in the order of integration in the above expression is only valid forωin the lower half-plane where lim t/prime→∞e−jωt/prime=0. Since the function f(ω)=e−jωt/primeis analytic in the lower half-plane, its closed contour integral is zero by the Cauchy–Goursattheorem. Thus, by (4.32) we have /contintegraldisplay /Gamma1˜σ(r,ω)dω=0. Then, since ˜σmay be assumed to be continuous in the lower half-plane for a physical medium, and since its closed path integral is zero for all possible paths /Gamma1, it is by Morera’s theorem [110] analytic in the lower half-plane. By similar reasoning χe(r,ω)is analytic in the lower half-plane. Since the function 1/ωhas a simple pole at ω=0, the composite function ˜/epsilon1c(r,ω)−/epsilon10given by (4.31) is analytic in the lower half-plane excluding ω=0 where it has a simple pole. Figure 4.1: Complex integration contour used to establish the Kronig–Kramers relations. Property 2: We have lim ω→±∞˜/epsilon1c(r,ω)−/epsilon10=0. To establish this property we need the Riemann–Lebesgue lemma [142], which states that iff(t)is absolutely integrable on the interval (a,b)where aand bare ﬁnite or inﬁnite constants, then lim ω→±∞/integraldisplayb af(t)e−jωtdt=0. From this we see that lim ω→±∞˜σ(r,ω) jω=lim ω→±∞1 jω/integraldisplay∞ 0σ(r,t/prime)e−jωt/primedt/prime=0, lim ω→±∞/epsilon10χe(r,ω)=lim ω→±∞/epsilon10/integraldisplay∞ 0χe(r,t/prime)e−jωt/primedt/prime=0, and thus lim ω→±∞˜/epsilon1c(r,ω)−/epsilon10=0. To establish the Kronig–Kramers relations we examine the integral /contintegraldisplay /Gamma1˜/epsilon1c(r,/Omega1 )−/epsilon10 /Omega1−ωd/Omega1 where/Gamma1isthecontourshowninFigure4.l.Sincethepoints/Omega1= 0,ωareexcluded, theintegran disanalyti ceverywher ewithinandon/Gamma1,hencetheintegralvanishe sbythe Cauchy–Goursa ttheorem .ByProperty2wehave lim R→∞/integraldisplay C∞˜/epsilon1c(r,/Omega1 )−/epsilon10 /Omega1−ωd/Omega1=0, hence /integraldisplay C0+Cω˜/epsilon1c(r,/Omega1 )−/epsilon10 /Omega1−ωd/Omega1+P.V./integraldisplay∞ −∞˜/epsilon1c(r,/Omega1 )−/epsilon10 /Omega1−ωd/Omega1=0. (4.33) Here “ P.V.” indicates that the integral is computed in the Cauchy principal value sense (see Appendix A). To evaluate the integrals over C0and Cω, consider a function f(Z) analyti cinthelowerhalfofthe Z-plane(Z = Zr+jZi). If the point zlies on the real axisasshowninFigure4.1,wecancalculat etheintegral F(z)=lim δ→0/integraldisplay /Gamma1f(Z) Z−zdZ through the parameterization Z−z=δejθ. Since dZ=jδejθdθwe have F(z)=lim δ→0/integraldisplay0 −πf/parenleftbig z+δejθ/parenrightbig δejθ/bracketleftbig jδejθ/bracketrightbig dθ=jf(z)/integraldisplay0 −πdθ=jπf(z). Replacing Zby/Omega1and zby0we can compute lim /Delta1→0/integraldisplay C0˜/epsilon1c(r,/Omega1 )−/epsilon10 /Omega1−ωd/Omega1 =lim /Delta1→0/integraldisplay C0/bracketleftBig 1 j/integraltext∞ 0σ(r,t/prime)e−j/Omega1t/primedt/prime+/Omega1/epsilon10/integraltext∞ 0χe(r,t/prime)e−j/Omega1t/primedt/prime/bracketrightBig 1 /Omega1−ω /Omega1d/Omega1 =−π/integraltext∞ 0σ(r,t/prime)dt/prime ω. We recognize /integraldisplay∞ 0σ(r,t/prime)dt/prime=σ0(r) as the dc conductivity and write lim /Delta1→0/integraldisplay C0˜/epsilon1c(r,/Omega1 )−/epsilon10 /Omega1−ωd/Omega1=−πσ0(r) ω. If we replace Zby/Omega1and zbyωwe get lim δ→0/integraldisplay Cω˜/epsilon1c(r,/Omega1 )−/epsilon10 /Omega1−ωd/Omega1=jπ˜/epsilon1c(r,ω)−jπ/epsilon10. Substituting these into (4.33) we have ˜/epsilon1c(r,ω)−/epsilon10=−1 jπP.V./integraldisplay∞ −∞˜/epsilon1c(r,/Omega1 )−/epsilon10 /Omega1−ωd/Omega1+σ0(r) jω. (4.34) If we write ˜/epsilon1c(r,ω)=˜/epsilon1c/prime(r,ω)+j˜/epsilon1c/prime/prime(r,ω)and equate real and imaginary parts in (4.34) we ﬁnd that ˜/epsilon1c/prime(r,ω)−/epsilon10=−1 πP.V./integraldisplay∞ −∞˜/epsilon1c/prime/prime(r,/Omega1 ) /Omega1−ωd/Omega1, (4.35) ˜/epsilon1c/prime/prime(r,ω)=1 πP.V./integraldisplay∞ −∞˜/epsilon1c/prime(r,/Omega1 )−/epsilon10 /Omega1−ωd/Omega1−σ0(r) ω. (4.36) These are the Kronig–Kramers relations , named after R. de L. Kronig and H.A. Kramers who derived them independently. The expressions show that causality requires the realand imaginary parts of the permittivity to depend upon each other through the Hilberttransform pair [142]. It is often more convenient to write the Kronig–Kramers relations in a form that employs only positive frequencies. This can be accomplished using the even–odd behaviorof the real and imaginary parts of ˜/epsilon1 c. Breaking the integrals in (4.35)–(4.36) into the ranges (−∞,0)and(0,∞), and substituting from (4.27), we can show that ˜/epsilon1c/prime(r,ω)−/epsilon10=−2 πP.V./integraldisplay∞ 0/Omega1˜/epsilon1c/prime/prime(r,/Omega1 ) /Omega12−ω2d/Omega1, (4.37) ˜/epsilon1c/prime/prime(r,ω)=2ω πP.V./integraldisplay∞ 0˜/epsilon1c/prime(r,/Omega1 ) /Omega12−ω2d/Omega1−σ0(r) ω. (4.38) The symbol P.V.in this case indicates that values of the integrand around both /Omega1=0 and/Omega1=ωmust be excluded from the integration. The details of the derivation of (4.37)–(4.38) are left as an exercise. We shall use (4.37) in §4.6 to demonstrate the Kronig–Kramers relationship for a model of complex permittivity of an actual material. We cannot specify ˜/epsilon1c/primearbitrarily; for a passive medium ˜/epsilon1c/prime/primemust be zero or negative at all values of ω, and (4.36) will not necessarily return these required values. However, if we have a good measurement or physical model for ˜/epsilon1c/prime/prime, as might come from studies of the absorbing properties of the material, we can approximate the real part of the permittivityusing (4.35). We shall demonstrate this using simple models for permittivity in §4.6. The Kronig–Kramers properties hold for µas well. We must for practical reasons consider the fact that magnetization becomes unimportant at a much lower frequencythan does polarization, so that the inﬁnite integrals in the Kronig–Kramers relationsshould be truncated at some upper frequency ω max. If we use a model or measured values of ˜µ/prime/primeto determine ˜µ/prime, the form of the relation (4.37) should be [107] ˜µ/prime(r,ω)−µ0=−2 πP.V./integraldisplayωmax 0/Omega1˜µ/prime/prime(r,/Omega1 ) /Omega12−ω2d/Omega1, where ωmaxis the frequency at which magnetization ceases to be important, and above which ˜µ=µ0. 4.5 Dissipated and stored energy in a dispersive medium Let us write down Poynting’s power balance theorem for a dispersive medium. Writing J=Ji+Jcwe have ( §2.9.5) −Ji·E=Jc·E+∇· [E×H]+/bracketleftbigg E·∂D ∂t+H·∂B ∂t/bracketrightbigg . (4.39) We cannot express this in terms of the time rate of change of a stored energy density because of the diﬃculty in interpreting the term E·∂D ∂t+H·∂B ∂t(4.40) when the constitutive parameters have the form (2.29)–(2.31). Physically, this term describes both the energy stored in the electromagnetic ﬁeld andthe energy dissipated by the material because of time lags between the application of EandHand the polarization or magnetization of the atoms (and thus the response ﬁelds Dand B). In principle this term can also be used to describe active media that transfer mechanical or chemical energy of the material into ﬁeld energy. Instead of attempting to interpret (4.40), we concentrate on the physical meaning of −∇ · S(r,t)=− ∇· [E(r,t)×H(r,t)]. We shall postulate that this term describes the net ﬂow of electromagnetic energy into the point rat time t. Then (4.39) shows that in the absence of impressed sources the energy ﬂow must act to (1) increase or decrease the stored energy density at r, (2) dissipate energy in ohmic losses through the term involving Jc, or (3) dissipate (or provide) energy through the term (40). Assuming linearity we may write −∇· S(r,t)=∂ ∂twe(r,t)+∂ ∂twm(r,t)+∂ ∂twQ(r,t), (4.41) where the terms on the right-hand side represent the time rates of change of, respectively, stored electric, stored magnetic, and dissipated energies. 4.5.1 Dissipation in a dispersive material Although we may, in general, be unable to separate the individual terms in (4.41), we can examine these terms under certain conditions. For example, consider a ﬁeld thatbuilds from zero starting from time t=− ∞ and then decays back to zero at t=∞. Then by direct integration 1 −/integraldisplay∞ −∞∇·S(t)dt=wem(t=∞)−wem(t=− ∞ )+wQ(t=∞)−wQ(t=− ∞ ) where wem=we+wmis the volume density of stored electromagnetic energy. This stored energy is zero at t=± ∞ since the ﬁelds are zero at those times. Thus, /Delta1w Q=−/integraldisplay∞ −∞∇·S(t)dt=wQ(t=∞)−wQ(t=− ∞ ) represents the volume density of the net energy dissipated by a lossy medium (or supplied by an active medium). We may thus classify materials according to the scheme /Delta1w Q=0, lossless , /Delta1w Q>0, lossy, /Delta1w Q≥0, passive , /Delta1w Q<0, active . For an anisotropic material with the constitutive relations ˜D=˜¯/epsilon1·˜E, ˜B=˜¯µ·˜H, ˜Jc=˜¯σ·˜E, 1Note that in this section we suppress the r-dependence of most quantities for clarity of presentation. we ﬁnd that dissipation is associated with negative imaginary parts of the constitutive parameters. To see this we write E(r,t)=1 2π/integraldisplay∞ −∞˜E(r,ω)ejωtdω, D(r,t)=1 2π/integraldisplay∞ −∞˜D(r,ω/prime)ejω/primetdω/prime, and thus ﬁnd Jc·E+E·∂D ∂t=1 (2π)2/integraldisplay∞ −∞/integraldisplay∞ −∞˜E(ω)·˜¯/epsilon1c(ω/prime)·˜E(ω/prime)ej(ω+ω/prime)tjω/primedωdω/prime where ˜¯/epsilon1cis the complex dyadic permittivity (4.24). Then /Delta1w Q=1 (2π)2/integraldisplay∞ −∞/integraldisplay∞ −∞/bracketleftbig˜E(ω)·˜¯/epsilon1c(ω/prime)·˜E(ω/prime)+˜H(ω)·˜¯µ(ω/prime)·˜H(ω/prime)/bracketrightbig · ·/bracketleftbigg/integraldisplay∞ −∞ej(ω+ω/prime)tdt/bracketrightbigg jω/primedωdω/prime. (4.42) Using (A.4) and integrating over ωwe obtain /Delta1w Q=1 2π/integraldisplay∞ −∞/bracketleftbig˜E(−ω/prime)·˜¯/epsilon1c(ω/prime)·˜E(ω/prime)+˜H(−ω/prime)·˜¯µ(ω/prime)·˜H(ω/prime)/bracketrightbig jω/primedω/prime.(4.43) Let us examine (4.43) more closely for the simple case of an isotropic material for which /Delta1w Q=1 2π/integraldisplay∞ −∞/braceleftbig/bracketleftbig j˜/epsilon1c/prime(ω/prime)−˜/epsilon1c/prime/prime(ω/prime)/bracketrightbig˜E(−ω/prime)·˜E(ω/prime)+ +/bracketleftbig j˜µ/prime(ω/prime)−˜µ/prime/prime(ω/prime)/bracketrightbig˜H(−ω/prime)·˜H(ω/prime)/bracerightbig ω/primedω/prime. Using the frequency symmetry property for complex permittivity (4.17) (which also holds for permeability), we ﬁnd that for isotropic materials ˜/epsilon1c/prime(r,ω)=˜/epsilon1c/prime(r,−ω), ˜/epsilon1c/prime/prime(r,ω)=− ˜/epsilon1c/prime/prime(r,−ω), (4.44) ˜µ/prime(r,ω)=˜µ/prime(r,−ω), ˜µ/prime/prime(r,ω)=− ˜µ/prime/prime(r,−ω). (4.45) Thus, the products of ω/primeand the real parts of the constitutive parameters are odd functions, while for the imaginary parts these products are even. Since the dot productsof the vector ﬁelds are even functions, we ﬁnd that the integrals of the terms containingthe real parts of the constitutive parameters vanish, leaving /Delta1w Q=21 2π/integraldisplay∞ 0/bracketleftbig −˜/epsilon1c/prime/prime|˜E|2−˜µ/prime/prime|˜H|2/bracketrightbig ωdω. (4.46) Here we have used (4.3) in the form ˜E(r,−ω)=˜E∗(r,ω) , ˜H(r,−ω)=˜H∗(r,ω) . (4.47) Equation (4.46) leads us to associate the imaginary parts of the constitutive parameters with dissipation. Moreover, a lossy isotropic material for which /Delta1w Q>0must have at least one of /epsilon1c/prime/primeandµ/prime/primeless than zero over some range of positive frequencies, while an active isotropic medium must have at least one of these greater than zero. In general, we speak of a lossy material as having negative imaginary constitutive parameters: ˜/epsilon1c/prime/prime<0, ˜µ/prime/prime<0,ω > 0. (4.48) Alossless medium must have ˜/epsilon1/prime/prime=˜µ/prime/prime=˜σ=0 for all ω. Things are not as simple in the more general anisotropic case. An integration of (4.42) overω/primeinstead of ωproduces /Delta1w Q=−1 2π/integraldisplay∞ −∞/bracketleftbig˜E(ω)·˜¯/epsilon1c(−ω)·˜E(−ω)+˜H(ω)·˜¯µ(−ω)·˜H(−ω)/bracketrightbig jωdω. Adding half of this expression to half of (4.43) and using (4.25), (4.17), and (4.47), we obtain /Delta1w Q=1 4π/integraldisplay∞ −∞/bracketleftbig˜E∗·˜¯/epsilon1c·˜E−˜E·˜¯/epsilon1c∗·˜E∗+˜H∗·˜¯µ·˜H−˜H·˜¯µ∗·˜H∗/bracketrightbig jωdω. Finally, using the dyadic identity (A.76), we have /Delta1w Q=1 4π/integraldisplay∞ −∞/bracketleftBig ˜E∗·/parenleftBig ˜¯/epsilon1c−˜¯/epsilon1c†/parenrightBig ·˜E+˜H∗·/parenleftBig ˜¯µ−˜¯µ†/parenrightBig ·˜H/bracketrightBig jωdω where the dagger ( †) denotes the hermitian (conjugate-transpose) operation. The condi- tion for a lossless anisotropic material is ˜¯/epsilon1c=˜¯/epsilon1c†, ˜¯µ=˜¯µ†, (4.49) or ˜/epsilon1ij=˜/epsilon1∗ ji, ˜µij=˜µ∗ ji, ˜σij=˜σ∗ ji. (4.50) These relationships imply that in the lossless case the diagonal entries of the constitutive dyadics are purely real. Equations (4.50) show that complex entries in a permittivity or permeability matrix do not necessarily imply loss. For example, we will show in §4.6.2 that an electron plasma exposed to a z-directed dc magnetic ﬁeld has a permittivity of the form [˜¯/epsilon1]= /epsilon1−jδ0 jδ/epsilon1 0 00 /epsilon1z  where /epsilon1,/epsilon1z, and δare real functions of space and frequency. Since ˜¯/epsilon1is hermitian it describes a lossless plasma. Similarly, a gyrotropic medium such as a ferrite exposed toaz-directed magnetic ﬁeld has a permeability dyadic [˜¯µ]= µ−jκ0 jκµ 0 00 µ 0 , which also describes a lossless material. 4.5.2 Energy stored in a dispersive material In the previous section we were able to isolate the dissipative eﬀects for a dispersive material under special circumstances. It is not generally possible, however, to isolatea term describing the stored energy. The Kronig–Kramers relations imply that if theconstitutive parameters of a material are frequency-dependent, they must have both realand imaginary parts; such a material, if isotropic, must be lossy. So dispersive materialsare generally lossy and must have both dissipative and energy-storage characteristics.However, many materials have frequency ranges called transparency ranges over which ˜/epsilon1 c/prime/primeand ˜µ/prime/primeare small compared to ˜/epsilon1c/primeand ˜µ/prime. If we restrict our interest to these ranges, we may approximate the material as lossless and compute a stored energy. An importantspecial case involves a monochromatic ﬁeld oscillating at a frequency within this range. To study the energy stored by a monochromatic ﬁeld in a dispersive material we must consider the transient period during which energy accumulates in the ﬁelds. Theassumption of a purely sinusoidal ﬁeld variation would not include the eﬀects described by the temporal constitutive relations (2.29)–(2.31), which show that as the ﬁeld buildsthe energy must be added with a time lag. Instead we shall assume ﬁelds with thetemporal variation E(r,t)=f(t) 3/summationdisplay i=1ˆii|Ei(r)|cos[ω0t+ξE i(r)] (4.51) where f(t)is an appropriate function describing the build-up of the sinusoidal ﬁeld. To compute the stored energy of a sinusoidal wave we must parameterize f(t)so that we may drive it to unity as a limiting case of the parameter. A simple choice is f(t)=e−α2t2↔˜F(ω)=/radicalbiggπ α2e−ω2 4α2. (4.52) Note that since f(t)approaches unity as α→0, we have the generalized Fourier trans- form relation lim α→0˜F(ω)=2πδ(ω). (4.53) Substituting (4.51) into the Fourier transform formula (4.1) we ﬁnd that ˜E(r,ω)=1 23/summationdisplay i=1ˆii|Ei(r)|ejξE i(r)˜F(ω−ω0)+1 23/summationdisplay i=1ˆii|Ei(r)|e−jξE i(r)˜F(ω+ω0). We can simplify this by deﬁning ˇE(r)=3/summationdisplay i=1ˆii|Ei(r)|ejξE i(r)(4.54) as the phasor vector ﬁeld to obtain ˜E(r,ω)=1 2/bracketleftbigˇE(r)˜F(ω−ω0)+ˇE∗(r)˜F(ω+ω0)/bracketrightbig . (4.55) We shall discuss the phasor concept in detail in §4.7. Theﬁeld E(r, t)isshowninFigure4.2asafunctio nof t,while ˜E(r,ω)isshownin Figure4.2asafunctio nofω.Asαbecome ssmallthespectrumof E(r, t)concentrates around ω=±ω0. We assume the material is transparent for all values αof interest so -40 -20 0 20 40ω t -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0ω/ω0 0 Figure 4.2: Temporal (top) and spectral magnitude (bottom) dependences of Eused to compute energy stored in a dispersive material. that we may treat /epsilon1as real. Then, since there is no dissipation, we conclude that the term (4.40) represents the time rate of change of stored energy at time t, including the eﬀects of ﬁeld build-up. Hence the interpretation2 E·∂D ∂t=∂we ∂t, H·∂B ∂t=∂wm ∂t. We shall concentrate on the electric ﬁeld term and later obtain the magnetic ﬁeld term by induction. Since for periodic signals it is more convenient to deal with the time-averaged stored energy than with the instantaneous stored energy, we compute the time average of we(r,t) over the period of the sinusoid centered at the time origin. That is, we compute /angbracketleftwe/angbracketright=1 T/integraldisplayT/2 −T/2we(t)dt (4.56) where T=2π/ω 0. With α→0, this time-average value is accurate for all periods of the sinusoidal wave. Because the most expedient approach to the computation of (4.56) is to employ the Fourier spectrum of E,w eu s e E(r,t)=1 2π/integraldisplay∞ −∞˜E(r,ω)ejωtdω=1 2π/integraldisplay∞ −∞˜E∗(r,ω/prime)e−jω/primetdω/prime, ∂D(r,t) ∂t=1 2π/integraldisplay∞ −∞(jω)˜D(r,ω)ejωtdω=1 2π/integraldisplay∞ −∞(−jω/prime)˜D∗(r,ω/prime)e−jω/primetdω/prime. 2Note that in this section we suppress the r-dependence of most quantities for clarity of presentation. We have obtained the second form of each of these expressions using the property (4.3) for the transform of a real function, and by using the change of variables ω/prime=−ω. Multiplying the two forms of the expressions and adding half of each, we ﬁnd that ∂we ∂t=1 2/integraldisplay∞ −∞dω 2π/integraldisplay∞ −∞dω/prime 2π/bracketleftbig jω˜E∗(ω/prime)·˜D(ω)−jω/prime˜E(ω)·˜D∗(ω/prime)/bracketrightbig e−j(ω/prime−ω)t.(4.57) Now let us consider a dispersive isotropic medium described by the constitutive rela- tions ˜D=˜/epsilon1˜E,˜B=˜µ˜H. Since the imaginary parts of ˜/epsilon1and ˜µare associated with power dissipation in the medium, we shall approximate ˜/epsilon1and ˜µas purely real. Then (4.57) becomes ∂we ∂t=1 2/integraldisplay∞ −∞dω 2π/integraldisplay∞ −∞dω/prime 2π˜E∗(ω/prime)·˜E(ω)/bracketleftbig jω˜/epsilon1(ω)−jω/prime˜/epsilon1(ω/prime)/bracketrightbig e−j(ω/prime−ω)t. Substitution from (4.55) now gives ∂we ∂t=1 8/integraldisplay∞ −∞dω 2π/integraldisplay∞ −∞dω/prime 2π/bracketleftbig jω˜/epsilon1(ω)−jω/prime˜/epsilon1(ω/prime)/bracketrightbig · ·/bracketleftbigˇE·ˇE∗˜F(ω−ω0)˜F(ω/prime−ω0)+ˇE·ˇE∗˜F(ω+ω0)˜F(ω/prime+ω0)+ +ˇE·ˇE˜F(ω−ω0)˜F(ω/prime+ω0)+ˇE∗·ˇE∗˜F(ω+ω0)˜F(ω/prime−ω0)/bracketrightbig e−j(ω/prime−ω)t. Letω→−ωwherever the term ˜F(ω+ω0)appears, and ω/prime→−ω/primewherever the term ˜F(ω/prime+ω0)appears. Since ˜F(−ω)=˜F(ω)and ˜/epsilon1(−ω)=˜/epsilon1(ω), we ﬁnd that ∂we ∂t=1 8/integraldisplay∞ −∞dω 2π/integraldisplay∞ −∞dω/prime 2π˜F(ω−ω0)˜F(ω/prime−ω0)· ·/bracketleftBig ˇE·ˇE∗[jω˜/epsilon1(ω)−jω/prime˜/epsilon1(ω/prime)]ej(ω−ω/prime)t+ˇE·ˇE∗[jω/prime˜/epsilon1(ω/prime)−jω˜/epsilon1(ω) ]ej(ω/prime−ω)t+ +ˇE·ˇE[jω˜/epsilon1(ω)+jω/prime˜/epsilon1(ω/prime)]ej(ω+ω/prime)t+ˇE∗·ˇE∗[−jω˜/epsilon1(ω)−jω/prime˜/epsilon1(ω/prime)]e−j(ω+ω/prime)t/bracketrightBig . (4.58) For small αthe spectra are concentrated near ω=ω0orω/prime=ω0. For terms involving the diﬀerence in the permittivities we can expand g(ω)=ω˜/epsilon1(ω)in a Taylor series about ω0to obtain the approximation ω˜/epsilon1(ω)≈ω0˜/epsilon1(ω 0)+(ω−ω0)g/prime(ω0) where g/prime(ω0)=∂[ω˜/epsilon1(ω) ] ∂ω/vextendsingle/vextendsingle/vextendsingle/vextendsingle ω=ω0. This is not required for terms involving a sum of permittivities since these will not tend to cancel. For such terms we merely substitute ω=ω0orω/prime=ω0. With these (4.58) becomes ∂we ∂t=1 8/integraldisplay∞ −∞dω 2π/integraldisplay∞ −∞dω/prime 2π˜F(ω−ω0)˜F(ω/prime−ω0)· ·/bracketleftBig ˇE·ˇE∗g/prime(ω0)[j(ω−ω/prime)]ej(ω−ω/prime)t+ˇE·ˇE∗g/prime(ω0)[j(ω/prime−ω)]ej(ω/prime−ω)t+ +ˇE·ˇE˜/epsilon1(ω 0)[j(ω+ω/prime)]ej(ω+ω/prime)t+ˇE∗·ˇE∗˜/epsilon1(ω 0)[−j(ω+ω/prime)]e−j(ω+ω/prime)t/bracketrightBig . By integration we(t)=1 8/integraldisplay∞ −∞dω 2π/integraldisplay∞ −∞dω/prime 2π˜F(ω−ω0)˜F(ω/prime−ω0)· ·/bracketleftBig ˇE·ˇE∗g/prime(ω0)ej(ω−ω/prime)t+ˇE·ˇE∗g/prime(ω0)ej(ω/prime−ω)t+ +ˇE·ˇE˜/epsilon1(ω 0)ej(ω+ω/prime)t+ˇE∗·ˇE∗˜/epsilon1(ω 0)e−j(ω+ω/prime)t/bracketrightBig . Our last step is to compute the time-average value of weand let α→0. Applying (4.56) we ﬁnd /angbracketleftwe/angbracketright=1 8/integraldisplay∞ −∞dω 2π/integraldisplay∞ −∞dω/prime 2π˜F(ω−ω0)˜F(ω/prime−ω0)· ·/bracketleftbigg 2ˇE·ˇE∗g/prime(ω0)sinc/parenleftbigg [ω−ω/prime]π ω0/parenrightbigg +/braceleftbigˇE∗·ˇE∗+ˇE·ˇE/bracerightbig ˜/epsilon1(ω 0)sinc/parenleftbigg [ω+ω/prime]π ω0/parenrightbigg/bracketrightbigg where sinc(x)is deﬁned in (A.9) and we note that sinc(−x)=sinc(x). Finally we let α→0and use (4.53) to replace ˜F(ω)by aδ-function. Upon integration these δ-functions setω=ω0andω/prime=ω0. Since sinc(0)=1and sinc(2π)=0, the time-average stored electric energy density becomes simply /angbracketleftwe/angbracketright=1 4|ˇE|2∂[ω˜/epsilon1] ∂ω/vextendsingle/vextendsingle/vextendsingle/vextendsingle ω=ω0. (4.59) Similarly, /angbracketleftwm/angbracketright=1 4|ˇH|2∂[ω˜µ] ∂ω/vextendsingle/vextendsingle/vextendsingle/vextendsingle ω=ω0. This approach can also be applied to anisotropic materials to give /angbracketleftwe/angbracketright=1 4ˇE∗·∂[ω˜¯/epsilon1] ∂ω/vextendsingle/vextendsingle/vextendsingle/vextendsingle ω=ω0·ˇE, (4.60) /angbracketleftwm/angbracketright=1 4ˇH∗·∂[ω˜¯µ] ∂ω/vextendsingle/vextendsingle/vextendsingle/vextendsingle ω=ω0·ˇH. (4.61) See Collin [39] for details. For the case of a lossless, nondispersive material where the constitutive parameters are frequency independent, we can use (4.49) and (A.76) tosimplify this and obtain /angbracketleftw e/angbracketright=1 4ˇE∗·¯/epsilon1·ˇE=1 4ˇE·ˇD∗, (4.62) /angbracketleftwm/angbracketright=1 4ˇH∗·¯µ·ˇH=1 4ˇH·ˇB∗, (4.63) in the anisotropic case and /angbracketleftwe/angbracketright=1 4/epsilon1|ˇE|2=1 4ˇE·ˇD∗, (4.64) /angbracketleftwm/angbracketright=1 4µ|ˇH|2=1 4ˇH·ˇB∗, (4.65) in the isotropic case. Here ˇE,ˇD,ˇB,ˇHare all phasor ﬁelds as deﬁned by (4.54). 4.5.3 The energy theorem A convenient expression for the time-average stored energies (4.60) and (4.61) is found by manipulating the frequency-domain Maxwell equations. Beginning with the complexconjugates of the two frequency-domain curl equations for anisotropic media, ∇× ˜E ∗=jω˜¯µ∗·˜H∗, ∇× ˜H∗=˜J∗−jω˜¯/epsilon1∗·˜E∗, we diﬀerentiate with respect to frequency: ∇×∂˜E∗ ∂ω=j∂[ω˜¯µ∗] ∂ω·˜H∗+jω˜¯µ∗·∂˜H∗ ∂ω, (4.66) ∇×∂˜H∗ ∂ω=∂˜J∗ ∂ω−j∂[ω˜¯/epsilon1∗] ∂ω·˜E∗−jω˜¯/epsilon1∗·∂˜E∗ ∂ω. (4.67) These terms also appear as a part of the expansion ∇·/bracketleftbigg ˜E×∂˜H∗ ∂ω+∂˜E∗ ∂ω×˜H/bracketrightbigg = ∂˜H∗ ∂ω·[∇× ˜E]−˜E·∇×∂˜H∗ ∂ω+˜H·∇×∂˜E∗ ∂ω−∂˜E∗ ∂ω·[∇× ˜H] where we have used (B.44). Substituting from (4.66)–(4.67) and eliminating ∇× ˜Eand ∇× ˜Hby Maxwell’s equations we have 1 4∇·/parenleftbigg ˜E×∂˜H∗ ∂ω+∂˜E∗ ∂ω×˜H/parenrightbigg = j1 4ω/parenleftbigg ˜E·˜¯/epsilon1∗·∂˜E∗ ∂ω−∂˜E∗ ∂ω·˜¯/epsilon1·˜E/parenrightbigg +j1 4ω/parenleftbigg ˜H·˜¯µ∗·∂˜H∗ ∂ω−∂˜H∗ ∂ω·˜¯µ·˜H/parenrightbigg + +j1 4/parenleftbigg ˜E·∂[ω˜¯/epsilon1∗] ∂ω·˜E∗+˜H·∂[ω˜¯µ∗] ∂ω·˜H∗/parenrightbigg −1 4/parenleftbigg ˜E·∂˜J∗ ∂ω+˜J·∂˜E∗ ∂ω/parenrightbigg . Let us assume that the sources and ﬁelds are narrowband, centered on ω0, and that ω0 lies within a transparency range so that within the band the material may be considered lossless. Invoking from (4.49) the facts that ˜¯/epsilon1=˜¯/epsilon1†and ˜¯µ=˜¯µ†, we ﬁnd that the ﬁrst two terms on the right are zero. Integrating over a volume and taking the complex conjugate of both sides we obtain 1 4/contintegraldisplay S/parenleftbigg ˜E∗×∂˜H ∂ω+∂˜E ∂ω×˜H∗/parenrightbigg ·dS= −j1 4/integraldisplay V/parenleftbigg ˜E∗·∂[ω˜¯/epsilon1] ∂ω·˜E+˜H∗·∂[ω˜¯µ] ∂ω·˜H/parenrightbigg dV−1 4/integraldisplay V/parenleftbigg ˜E∗·∂˜J ∂ω+˜J∗·∂˜E ∂ω/parenrightbigg dV. Evaluating each of the terms at ω=ω0and using (4.60)–(4.61) we have 1 4/contintegraldisplay S/parenleftbigg ˜E∗×∂˜H ∂ω+∂˜E ∂ω×˜H∗/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle ω=ω0·dS= −j[/angbracketleftWe/angbracketright+/angbracketleft Wm/angbracketright]−1 4/integraldisplay V/parenleftbigg ˜E∗·∂˜J ∂ω+˜J∗·∂˜E ∂ω/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle ω=ω0dV (4.68) where /angbracketleftWe/angbracketright+/angbracketleftWm/angbracketrightis the total time-average electromagnetic energy stored in the volume region V. This is known as the energy theorem . We shall use it in §4.11.3 to determine the velocity of energy transport for a plane wave. 4.6 Some simple models for constitutive parameters Thus far our discussion of electromagnetic ﬁelds has been restricted to macroscopic phenomena. Although we recognize that matter is composed of microscopic constituents,we have chosen to describe materials using constitutive relationships whose parameters,such as permittivity, conductivity, and permeability, are viewed in the macroscopic sense.By performing experiments on the laboratory scale we can measure the constitutive parameters to the precision required for engineering applications. At some point it becomes useful to establish models of the macroscopic behavior of materials based on microscopic considerations, formulating expressions for the consti-tutive parameters using atomic descriptors such as number density, atomic charge, andmolecular dipole moment. These models allow us to predict the behavior of broad classesof materials, such as dielectrics and conductors, over wide ranges of frequency and ﬁeldstrength. Accurate models for the behavior of materials under the inﬂuence of electromagnetic ﬁelds must account for many complicated eﬀects, including those best described by quan-tum mechanics. However, many simple models can be obtained using classical mechanicsand ﬁeld theory. We shall investigate several of the most useful of these, and in theprocess try to gain a feeling for the relationship between the ﬁeld applied to a materialand the resulting polarization or magnetization of the underlying atomic structure. For simplicity we shall consider only homogeneous materials. The fundamental atomic descriptor of “number density,” N, is thus taken to be independent of position and time. The result may be more generally applicable since we may think of an inhomogeneousmaterial in terms of the spatial variation of constitutive parameters originally deter-mined assuming homogeneity. However, we shall not attempt to study the microscopicconditions that give rise to inhomogeneities. 4.6.1 Complex permittivity of a non-magnetized plasma A plasma is an ionized gas in which the charged particles are free to move under the inﬂuence of an applied ﬁeld and through particle-particle interactions. A plasmadiﬀers from other materials in that there is no atomic lattice restricting the motion ofthe particles. However, even in a gas the interactions between the particles and the ﬁeldsgive rise to a polarization eﬀect, causing the permittivity of the gas to diﬀer from thatof free space. In addition, exposing the gas to an external ﬁeld will cause a secondarycurrent to ﬂow as a result of the Lorentz force on the particles. As the moving particlescollide with one another they relinquish their momentum, an eﬀect describable in termsof a conductivity. In this section we shall perform a simple analysis to determine thecomplex permittivity of a non-magnetized plasma. To make our analysis tractable, we shall make several assumptions. 1. We assume that the plasma is neutral : i.e., that the free electrons and positive ions are of equal number and distributed in like manner. If the particles are suﬃciently dense to be considered in the macroscopic sense, then there is no net ﬁeld produced by the gas and thus no electromagnetic interaction between the particles. We alsoassume that the plasma is homogeneous and that the number density of the electrons N(number of electrons per m 3) is independent of time and position. In contrast to this are electron beams , whose properties diﬀer signiﬁcantly from neutral plasmas because of bunching of electrons by the applied ﬁeld [148]. 2. We ignore the motion of the positive ions in the computation of the secondary current, since the ratio of the mass of an ion to that of an electron is at least aslarge as the ratio of a proton to an electron ( m p/me=1837) and thus the ions accelerate much more slowly. 3. We assume that the applied ﬁeld is that of an electromagnetic wave. In §2.10.6 we found that for a wave in free space the ratio of magnetic to electric ﬁeld is|H|/|E|=√ /epsilon10/µ0, so that |B| |E|=µ0/radicalbigg/epsilon10 µ0=√µ0/epsilon10=1 c. Thus, in the Lorentz force equation we may approximate the force on an electron as F=−qe(E+v×B)≈−qeE as long as v/lessmuchc. Here qeis the unsigned charge on an electron, qe=1.6021× 10−19C. Note that when an external static magnetic ﬁeld accompanies the ﬁeld of the wave, as is the case in the earth’s ionosphere for example, we cannot ignore themagnetic component of the Lorentz force. This case will be considered in §4.6.2. 4. We assume that the mechanical interactions between particles can be described using a collision frequency ν, which describes the rate at which a directed plasma velocity becomes random in the absence of external forces. With these assumptions we can write the equation of motion for the plasma medium. Letv(r,t)represent the macroscopic velocity of the plasma medium. Then, by Newton’s second law, the force acting at each point on the medium is balanced by the time-rate ofchange in momentum at that point. Because of collisions, the total change in momentumdensity is described by F(r,t)=− Nq eE(r,t)=d℘(r,t) dt+ν℘(r,t) (4.69) where ℘(r,t)=Nm ev(r,t) is the volume density of momentum. Note that if there is no externally-applied electro- magnetic force, then (4.69) becomes d℘(r,t) dt+ν℘(r,t)=0. Hence ℘(r,t)=℘0(r)e−νt, and we see that νdescribes the rate at which the electron velocities move toward a random state, producing a macroscopic plasma velocity vof zero. The time derivative in (4.69) is the total derivative as deﬁned in (A.58): d℘(r,t) dt=∂℘(r,t) ∂t+(v·∇)℘(r,t). (4.70) The second term on the right accounts for the time-rate of change of momentum per- ceived as the observer moves through regions of spatially-changing momentum. Sincethe electron velocity is induced by the electromagnetic ﬁeld, we anticipate that for asinusoidal wave the spatial variation will be on the order of the wavelength of the ﬁeld:λ=2πc/ω. Thus, while the ﬁrst term in (4.70) is proportional to ω, the second term is proportional to ωv/cand can be neglected for non-relativistic particle velocities. Then, writing E(r,t)and v(r,t)as inverse Fourier transforms, we see that (4.69) yields −q e˜E=jωme˜v+meν˜v (4.71) and thus ˜v=−qe me˜E ν+jω. (4.72) The secondary current associated with the moving electrons is (since qeis unsigned) ˜Js=− Nqe˜v=/epsilon10ω2 p ω2+ν2(ν−jω)˜E (4.73) where ω2 p=Nq2 e /epsilon10me(4.74) is called the plasma frequency . The frequency-domain Ampere’s law for primary and secondary currents in free space is merely ∇× ˜H=˜Ji+˜Js+jω/epsilon10˜E. Substitution from (4.73) gives ∇× ˜H=˜Ji+/epsilon10ω2 pν ω2+ν2˜E+jω/epsilon10/bracketleftBigg 1−ω2 p ω2+ν2/bracketrightBigg ˜E. We can determine the material properties of the plasma by realizing that the above expression can be written as ∇× ˜H=˜Ji+˜Js+jω˜D with the constitutive relations ˜Js=˜σ˜E, ˜D=˜/epsilon1˜E. Here we identify the conductivity of the plasma as ˜σ(ω)=/epsilon10ω2 pν ω2+ν2(4.75) and the permittivity as ˜/epsilon1(ω)=/epsilon10/bracketleftBigg 1−ω2 p ω2+ν2/bracketrightBigg . We can also write Ampere’s law as ∇× ˜H=˜Ji+jω˜/epsilon1c˜E where ˜/epsilon1cis the complex permittivity ˜/epsilon1c(ω)=˜/epsilon1(ω)+˜σ(ω) jω=/epsilon10/bracketleftBigg 1−ω2 p ω2+ν2/bracketrightBigg −j/epsilon10ω2 pν ω(ω2+ν2). (4.76) If we wish to describe the plasma in terms of a polarization vector, we merely use ˜D= /epsilon10˜E+˜P=˜/epsilon1˜Eto obtain the polarization vector ˜P=(˜/epsilon1−/epsilon10)˜E=/epsilon10˜χe˜E, where ˜χeis the electric susceptibility ˜χe(ω)=−ω2 p ω2+ν2. We note that ˜Pis directed opposite the applied ﬁeld ˜E, resulting in ˜/epsilon1</epsilon1 0. The plasma is dispersive since both its permittivity and conductivity depend on ω. Asω→0we have ˜/epsilon1c/prime→/epsilon10/epsilon1rwhere /epsilon1r=1−ω2 p/ν2, and also ˜/epsilon1c/prime/prime∼1/ω, as remarked in (4.28). As ω→∞ we have ˜/epsilon1c/prime−/epsilon10∼1/ω2and ˜/epsilon1c/prime/prime∼1/ω3, as mentioned in (4.29). When a transient plane wave propagates through a dispersive medium, the frequency dependence of the constitutive parameters tends to cause spreading of the waveshape. We see that the plasma conductivity (4.75) is proportional to the collision frequency ν, and that, since ˜/epsilon1c/prime/prime<0by the arguments of §4.5, the plasma must be lossy. Loss arises from the transfer of electromagnetic energy into heat through electron collisions. If thereare no collisions ( ν=0), there is no mechanism for the transfer of energy into heat, and the conductivity of a lossless (or “collisionless”) plasma reduces to zero as expected. In a lowloss plasma ( ν→0) we may determine the time-average stored electromagnetic energy for sinusoidal excitation at frequency ˇω. We must be careful to use (4.59), which holds for materials with dispersion. If we apply the simpler formula (4.64), we ﬁnd thatforν→0 /angbracketleftw e/angbracketright=1 4/epsilon10|ˇE|2−1 4/epsilon10|ˇE|2ω2 p ˇω2. For those excitation frequencies obeying ˇω<ω pwe have /angbracketleftwe/angbracketright<0, implying that the material is active. Since there is no mechanism for the plasma to produce energy, this isobviously not valid. But an application of (4.59) gives /angbracketleftw e/angbracketright=1 4|ˇE|2∂ ∂ω/bracketleftBigg /epsilon10ω/parenleftBigg 1−ω2 p ω2/parenrightBigg/bracketrightBigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle ω=ˇω=1 4/epsilon10|ˇE|2+1 4/epsilon10|ˇE|2ω2 p ˇω2, (4.77) which is always positive. In this expression the ﬁrst term represents the time-average energy stored in the vacuum, while the second term represents the energy stored in thekinetic energy of the electrons. For harmonic excitation, the time-average electron kineticenergy density is /angbracketleftw q/angbracketright=1 4Nm eˇv·ˇv∗. Substituting ˇvfrom (4.72) with ν=0we see that 1 4Nm eˇv·ˇv∗=1 4Nq2 e meˇω2|ˇE|2=1 4/epsilon10|ˇE|2ω2 p ˇω2, which matches the second term of (4.77). Figure 4.3: Integration contour used in Kronig–Kramers relations to ﬁnd ˜/epsilon1c/primefrom ˜/epsilon1c/prime/primefor a non-magnetized plasma. The complex permittivity of a plasma (4.76) obviously obeys the required frequency- symmetry conditions (4.27). It also obeys the Kronig–Kramers relations required fora causal material. From (4.76) we see that the imaginary part of the complex plasmapermittivity is ˜/epsilon1 c/prime/prime(ω)=−/epsilon10ω2 pν ω(ω2+ν2). Substituting this into (4.37) we have ˜/epsilon1c/prime(ω)−/epsilon10=−2 πP.V./integraldisplay∞ 0/bracketleftBigg −/epsilon10ω2 pν /Omega1(/Omega12+ν2)/bracketrightBigg /Omega1 /Omega12−ω2d/Omega1. We can evaluate the principal value integral and thus verify that it produces ˜/epsilon1c/primeby using the contour method of §A.1. Because the integrand is even we can extend the domain of integration to (−∞,∞)and divide the result by two. Thus ˜/epsilon1c/prime(ω)−/epsilon10=1 πP.V./integraldisplay∞ −∞/epsilon10ω2 pν (/Omega1−jν)(/Omega1+jν)d/Omega1 (/Omega1−ω)(/Omega1+ω). Weintegrat earoun dtheclosedcontourshowninFigure4.3.Sincetheintegran dfalls oﬀ as 1//Omega14the contribution from C∞is zero. The contributions from the semicircles Cω and C−ωare given by πjtimes the residues of the integrand at /Omega1=ωand at /Omega1=−ω, respectively, which are identical but of opposite sign. Thus, the semicircle contributionscancel and leave only the contribution from the residue at the upper-half-plane pole/Omega1=jν. Evaluation of the residue gives ˜/epsilon1 c/prime(ω)−/epsilon10=1 π2πj/epsilon10ω2 pν jν+jν1 (jν−ω)(jν+ω)=−/epsilon10ω2 p ν2+ω2 and thus ˜/epsilon1c/prime(ω)=/epsilon10/parenleftBigg 1−ω2 p ν2+ω2/parenrightBigg , which matches (4.76) as expected. 4.6.2 Complex dyadic permittivity of a magnetized plasma When an electron plasma is exposed to a magnetostatic ﬁeld, as occurs in the earth’s ionosphere, the behavior of the plasma is altered so that the secondary current is no longeraligned with the electric ﬁeld, requiring the constitutive relationships to be written interms of a complex dyadic permittivity. If the static ﬁeld is B 0, the velocity ﬁeld of the plasma is determined by adding the magnetic component of the Lorentz force to (4.71),giving −q e[˜E+˜v×B0]=˜v(jωme+meν) or equivalently ˜v−jqe me(ω−jν)˜v×B0=jqe me(ω−jν)˜E. (4.78) Writing this expression generically as v+v×C=A, (4.79) we can solve for vas follows. Dotting both sides of the equation with Cwe quickly establish that C·v=C·A. Crossing both sides of the equation with C, using (B.7), and substituting C·AforC·v,w eh a v e v×C=A×C+v(C·C)−C(A·C). Finally, substituting v×Cback into (4.79) we obtain v=A−A×C+(A·C)C 1+C·C. (4.80) Let us ﬁrst consider a lossless plasma for which ν=0. We can solve (4.78) for ˜vby setting C=− jωc ω, A=j/epsilon10ω2 p ωNqe˜E, where ωc=qe meB0. Hereωc=qeB0/me=|ωc|is called the electron cyclotron frequency . Substituting these into (4.80) we have /parenleftbig ω2−ω2 c/parenrightbig˜v=j/epsilon10ωω2 p Nqe˜E+/epsilon10ω2 p Nqeωc×˜E−jωc ω/epsilon10ω2 p Nqeωc·˜E. Since the secondary current produced by the moving electrons is just ˜Js=− Nqe˜v,w e have ˜Js=jω/bracketleftBigg −/epsilon10ω2 p ω2−ω2c˜E+j/epsilon10ω2 p ω(ω2−ω2c)ωc×˜E+ωc ω2/epsilon10ω2 p ω2−ω2cωc·˜E/bracketrightBigg . (4.81) Now, by the Ampere–Maxwell law we can write for currents in free space ∇× ˜H=˜Ji+˜Js+jω/epsilon10˜E. (4.82) Considering the plasma to be a material implies that we can describe the gas in terms of a complex permittivity dyadic ˜¯/epsilon1csuch that the Ampere–Maxwell law is ∇× ˜H=˜Ji+jω˜¯/epsilon1c·˜E. Substituting (4.81) into (4.82), and deﬁning the dyadic ¯ωcso that ¯ωc·˜E=ωc×˜E,w e identify the dyadic permittivity ˜¯/epsilon1c(ω)=/bracketleftBigg /epsilon10−/epsilon10ω2 p ω2−ω2c/bracketrightBigg ¯I+j/epsilon10ω2 p ω(ω2−ω2c)¯ωc+/epsilon10ω2 p ω2(ω2−ω2c)ωcωc. (4.83) Note that in rectangular coordinates [¯ωc]= 0−ωczωcy ωcz 0−ωcx −ωcyωcx 0 . (4.84) To examine the properties of the dyadic permittivity it is useful to write it in matrix form. To do this we must choose a coordinate system. We shall assume that B0is aligned along the z-axis such that B0=ˆzB0andωc=ˆzωc. Then (4.84) becomes [¯ωc]= 0−ωc0 ωc00 00 0  (4.85) and we can write the permittivity dyadic (4.83) as [˜¯/epsilon1(ω)]= /epsilon1−jδ0 jδ/epsilon1 0 00 /epsilon1z  (4.86) where /epsilon1=/epsilon10/parenleftBigg 1−ω2 p ω2−ω2c/parenrightBigg ,/epsilon1 z=/epsilon10/parenleftBigg 1−ω2 p ω2/parenrightBigg ,δ =/epsilon10ωcω2 p ω(ω2−ω2c). Note that the form of the permittivity dyadic is that for a lossless gyrotropic material (2.33). Since the plasma is lossless, equation (4.49) shows that the dyadic permittivity must be hermitian. Equation (4.86) conﬁrms this. We also note that since the sign of ωcis determined by the sign of B0, the dyadic permittivity obeys the symmetry relation ˜/epsilon1c ij(B0)=˜/epsilon1c ji(−B0) (4.87) as does the permittivity matrix of any material that has anisotropic properties dependent on an externally applied magnetic ﬁeld [141]. We will ﬁnd later in this section that the permeability matrix of a magnetized ferrite also obeys such a symmetry condition. We can let ω→ω−jνin (4.81) to obtain the secondary current in a plasma with collisions: ˜Js(r,ω)=jω/bracketleftBigg −/epsilon10ω2 p(ω−jν) ω[(ω−jν)2−ω2c]˜E(r,ω)+ +j/epsilon10ω2 p(ω−jν) ω(ω−jν)[(ω−jν)2−ω2c)]ωc×˜E(r,ω)+ +ωc (ω−jν)2/epsilon10ω2 p(ω−jν) ω[(ω−jν)2−ω2c]ωc·˜E(r,ω)/bracketrightBigg . From this we ﬁnd the dyadic permittivity ˜¯/epsilon1c(ω)=/bracketleftBigg /epsilon10−/epsilon10ω2 p(ω−jν) ω[(ω−jν)2−ω2c]/bracketrightBigg ¯I+j/epsilon10ω2 p ω[(ω−jν)2−ω2c)]¯ωc+ +1 (ω−jν)/epsilon10ω2 p ω[(ω−jν)2−ω2c]ωcωc. Assuming that B0is aligned with the z-axis we can use (4.85) to ﬁnd the components of the dyadic permittivity matrix: ˜/epsilon1c xx(ω)=˜/epsilon1c yy(ω)=/epsilon10/parenleftBigg 1−ω2 p(ω−jν) ω[(ω−jν)2−ω2c]/parenrightBigg , (4.88) ˜/epsilon1c xy(ω)=− ˜/epsilon1c yx(ω)=− j/epsilon10ω2 pωc ω[(ω−jν)2−ω2c)], (4.89) ˜/epsilon1c zz(ω)=/epsilon10/parenleftBigg 1−ω2 p ω(ω−jν)/parenrightBigg , (4.90) and ˜/epsilon1c zx=˜/epsilon1c xz=˜/epsilon1c zy=˜/epsilon1c yz=0. (4.91) We see that [˜/epsilon1c]is not hermitian when ν/negationslash=0. We expect this since the plasma is lossy when collisions occur. However, we can decompose [˜¯/epsilon1c]as a sum of two matrices: [˜¯/epsilon1c]=[˜¯/epsilon1]+[˜¯σ] jω, where [˜¯/epsilon1]and [˜¯σ]are hermitian [141]. The details are left as an exercise. We also note that, as in the case of the lossless plasma, the permittivity dyadic obeys the symmetrycondition ˜/epsilon1 c ij(B0)=˜/epsilon1c ji(−B0). 4.6.3 Simple models of dielectrics We deﬁne an isotropic dielectric material (also called an insulator ) as one that obeys the macroscopic frequency-domain constitutive relationship ˜D(r,ω)=˜/epsilon1(r,ω)˜E(r,ω) . Since the polarization vector Pwas deﬁned in Chapter 2 as P(r,t)=D(r,t)−/epsilon10E(r,t), an isotropic dielectric can also be described through ˜P(r,ω)=(˜/epsilon1(r,ω)−/epsilon10)˜E(r,ω)=˜χe(r,ω) /epsilon1 0˜E(r,ω) where ˜χeis the dielectric susceptibility. In this section we shall model a homogeneous dielectric consisting of a single, uniform material type. We found in Chapter 3 that for a dielectric material immersed in a static electric ﬁeld, the polarization vector Pcan be viewed as a volume density of dipole moments. We choose to retain this view as the fundamental link between microscopic dipole momentsand the macroscopic polarization vector. Within the framework of our model we thusdescribe the polarization through the expression P(r,t)=1 /Delta1V/summationdisplay r−ri(t)∈Bpi. (4.92) Here piis the dipole moment of the ith elementary microscopic constituent, and we form the macroscopic density function as in §1.3.1. We may also write (4.92) as P(r,t)=/bracketleftbiggNB /Delta1V/bracketrightbigg/bracketleftBigg 1 NBNB/summationdisplay i=1pi/bracketrightBigg =N(r,t)p(r,t) (4.93) where NBis the number of constituent particles within /Delta1V. We identify p(r,t)=1 NBNB/summationdisplay i=1pi(r,t) as the average dipole moment within /Delta1V, and N(r,t)=NB /Delta1V as the dipole moment number density. In this model a dielectric material does not require higher-order multipole moments to describe its behavior. Since we are only interestedin homogeneous materials in this section we shall assume that the number density isconstant: N(r,t)=N. To understand how dipole moments arise, we choose to adopt the simple idea that mat- ter consists of atomic particles, each of which has a positively charged nucleus surroundedby a negatively charged electron cloud. Isolated, these particles have no net charge andno net electric dipole moment. However, there are several ways in which individual par- ticles, or aggregates of particles, may take on a dipole moment. When exposed to anexternal electric ﬁeld the electron cloud of an individual atom may be displaced, resultingin aninduced dipole moment which gives rise to electronic polarization . When groups of atoms form a molecule, the individual electron clouds may combine to form an asym-metric structure having a permanent dipole moment. In some materials these molecules are randomly distributed and no net dipole moment results. However, upon applicationof an external ﬁeld the torque acting on the molecules may tend to align them, creatinganinduced dipole moment and orientation ,o rdipole , polarization. In other materials, the asymmetric structure of the molecules may be weak until an external ﬁeld causesthe displacement of atoms within each molecule, resulting in an induced dipole moment causing atomic ,o rmolecular , polarization. If a material maintains a permanent polar- ization without the application of an external ﬁeld, it is called an electret (and is thus similar in behavior to a permanently magnetized magnet). To describe the constitutive relations, we must establish a link between P(now describ- able in microscopic terms) and E. We do this by postulating that the average constituent dipole moment is proportional to the local electric ﬁeld strength E/prime: p=αE/prime, (4.94) where αis called the polarizability of the elementary constituent. Each of the polarization eﬀects listed above may have its own polarizability: αefor electronic polarization, αafor atomic polarization, and αdfor dipole polarization. The total polarizability is merely the sumα=αe+αa+αd. In a rareﬁed gas the particles are so far apart that their interaction can be neglected. Here the localized ﬁeld E/primeis the same as the applied ﬁeld E. In liquids and solids where particles are tightly packed, E/primedepends on the manner in which the material is polarized and may diﬀer from E. We therefore proceed to determine a relationship between E/prime and P. The Clausius–Mosotti equation. We seek the local ﬁeld at an observation point within a polarized material. Let us ﬁrst assume that the ﬁelds are static. We surroundthe observation point with an artiﬁcial spherical surface of radius aand write the ﬁeld at the observation point as a superposition of the ﬁeld Eapplied, the ﬁeld E 2of the polarized molecules external to the sphere, and the ﬁeld E3of the polarized molecules within the sphere. We take alarge enough that we may describe the molecules outside the sphere in terms of the macroscopic dipole moment density P, but small enough to assume that P is uniform over the surface of the sphere. We also assume that the major contribution toE 2comes from the dipoles nearest the observation point. We then approximate E2using the electrostatic potential produced by the equivalent polarization surface charge on thesphere ρ Ps=ˆn·P(where ˆnpoints toward the center of the sphere). Placing the origin of coordinates at the observation point and orienting the z-axis with the polarization P so that P=P0ˆz, we ﬁnd that ˆn·P=− cosθand thus the electrostatic potential at any point rwithin the sphere is merely /Phi1(r)=−1 4π/epsilon10/contintegraldisplay SP0cosθ/prime |r−r/prime|dS/prime. This integral has been computed in §3.2.7 with the result given by (3.103) Hence /Phi1(r)=−P0 3/epsilon10rcosθ=−P0 3/epsilon10z and therefore E2=P 3/epsilon10. (4.95) Note that this is uniform and independent of a. The assumption that the localized ﬁeld varies spatially as the electrostatic ﬁeld, even when Pmay depend on frequency, is quite good. In Chapter 5 we will ﬁnd that for a frequency-dependent source (or, equivalently, a time-varying source), the ﬁelds very nearthe source have a spatial dependence nearly identical to that of the electrostatic case. We now have the seemingly more diﬃcult task of determining the ﬁeld E 3produced by the dipoles within the sphere. This would seem diﬃcult since the ﬁeld produced bydipoles near the observation point should be highly-dependent on the particular dipolearrangement. As mentioned above, there are various mechanisms for polarization, andthe distribution of charge near any particular point depends on the molecular arrange-ment. However, Lorentz showed [115] that for crystalline solids with cubical symmetry, or for a randomly-structured gas, the contribution from dipoles within the sphere is zero. Indeed, it is convenient and reasonable to assume that for most dielectrics the eﬀects ofthe dipoles immediately surrounding the observation point cancel so that E 3=0. This was ﬁrst suggested by O.F. Mosotti in 1850 [52]. With E2approximated as (4.95) and E3assumed to be zero, we have the value of the resulting local ﬁeld: E/prime(r)=E(r)+P(r) 3/epsilon10. (4.96) This is called the Mosotti ﬁeld . Substituting the Mosotti ﬁeld into (4.94) and using P=Np, we obtain P(r)=NαE/prime(r)=Nα/parenleftbigg E(r)+P(r) 3/epsilon10/parenrightbigg . Solving for Pwe obtain P(r)=/parenleftbigg3/epsilon10Nα 3/epsilon10−Nα/parenrightbigg E(r)=χe/epsilon10E(r). So the electric susceptibility of a dielectric may be expressed as χe=3Nα 3/epsilon10−Nα. (4.97) Using χe=/epsilon1r−1we can rewrite (4.97) as /epsilon1=/epsilon10/epsilon1r=/epsilon103+2Nα//epsilon1 0 3−Nα//epsilon1 0, (4.98) which we can arrange to obtain α=αe+αa+αd=3/epsilon10 N/epsilon1r−1 /epsilon1r+2. This has been named the Clausius–Mosotti formula , after O.F. Mosotti who proposed it in 1850 and R. Clausius who proposed it independently in 1879. When written in terms ofthe index of refraction n(where n 2=/epsilon1r), it is also known as the Lorentz–Lorenz formula , after H. Lorentz and L. Lorenz who proposed it independently for optical materials in 1880. The Clausius–Mosotti formula allows us to determine the dielectric constant fromthe polarizability and number density of a material. It is reasonably accurate for certainsimple gases (with pressures up to 1000 atmospheres) but becomes less reliable for liquidsand solids, especially for those with large dielectric constants. The response of the microscopic structure of matter to an applied ﬁeld is not instanta- neous. When exposed to a rapidly oscillating sinusoidal ﬁeld, the induced dipole momentsmay lag in time. This results in a loss mechanism that can be described macroscopicallyby a complex permittivity. We can modify the Clausius–Mosotti formula by assumingthat both the relative permittivity and polarizability are complex numbers, but this willnot model the dependence of these parameters on frequency. Instead we shall (in laterparagraphs) model the time response of the dipole moments to the applied ﬁeld. An interesting application of the Clausius–Mosotti formula is to determine the permit- tivity of a mixture of dielectrics with diﬀerent permittivities. Consider the simple casein which many small spheres of permittivity /epsilon1 2, radius a, and volume Vare embedded within a dielectric matrix of permittivity /epsilon11. If we assume that ais much smaller than the wavelength of the electromagnetic ﬁeld, and that the spheres are sparsely distributedwithin the matrix, then we may ignore any mutual interaction between the spheres. Sincethe expression for the permittivity of a uniform dielectric given by (4.98) describes theeﬀect produced by dipoles in free space, we can use the Clausius–Mosotti formula todeﬁne an eﬀective permittivity /epsilon1 efor a material consisting of spheres in a background dielectric by replacing /epsilon10with/epsilon11to obtain /epsilon1e=/epsilon113+2Nα//epsilon1 1 3−Nα//epsilon1 1. (4.99) In this expression αis the polarizability of a single dielectric sphere embedded in the background dielectric, and Nis the number density of dielectric spheres. To ﬁnd α we use the static ﬁeld solution for a dielectric sphere immersed in a ﬁeld ( §3.2.10). Remembering that p=αEand that for a uniform region of volume Vwe have p=VP, we can make the replacements /epsilon10→/epsilon11and/epsilon1→/epsilon12in (3.117) to get α=3/epsilon11V/epsilon12−/epsilon11 /epsilon12+2/epsilon11. (4.100) Deﬁning f=NVas the fractional volume occupied by the spheres, we can substitute (4.100) into (4.99) to ﬁnd that /epsilon1e=/epsilon111+2fy 1−fy where y=/epsilon12−/epsilon11 /epsilon12+2/epsilon11. This is known as the Maxwell–Garnett mixing formula . Rearranging we obtain /epsilon1e−/epsilon11 /epsilon1e+2/epsilon11=f/epsilon12−/epsilon11 /epsilon12+2/epsilon11, which is known as the Rayleigh mixing formula . As expected, /epsilon1e→/epsilon11asf→0.E v e n though as f→1the formula also reduces to /epsilon1e=/epsilon12, our initial assumption that f/lessmuch1 (sparsely distributed spheres) is violated and the result is inaccurate for non-sphericalinhomogeneities [90]. For a discussion of more accurate mixing formulas, see Ishimaru[90] or Sihvola [175]. The dispersion formula of classical physics. We may determine the frequency de- pendence of the permittivity by modeling the time response of induced dipole moments.This was done by H. Lorentz using the simple atomic model we introduced earlier. Con-sider what happens when a molecule consisting of heavy particles (nuclei) surrounded byclouds of electrons is exposed to a time-harmonic electromagnetic wave. Using the same arguments we made when we studied the interactions of ﬁelds with a plasma in §4.6.1, we assume that each electron experiences a Lorentz force F e=− qeE/prime. We neglect the magnetic component of the force for nonrelativistic charge velocities, and ignore the mo-tion of the much heavier nuclei in favor of studying the motion of the electron cloud.However, several important distinctions exist between the behavior of charges within aplasma and those within a solid or liquid material. Because of the surrounding polarizedmatter, any molecule responds to the local ﬁeld E /primeinstead of the applied ﬁeld E. Also, as the electron cloud is displaced by the Lorentz force, the attraction from the positive nuclei provides a restoring force Fr. In the absence of loss the restoring force causes the electron cloud (and thus the induced dipole moment) to oscillate in phase with theapplied ﬁeld. In addition, there will be loss due to radiation by the oscillating moleculesand collisions between charges that can be modeled using a “frictional force” F sin the same manner as for a mechanical harmonic oscillator. We can express the restoring and frictional forces by the use of a mechanical analogue. The restoring force acting on each electron is taken to be proportional to the displacementfrom equilibrium l: F r(r,t)=−meω2 rl(r,t), where meis the mass of an electron and ωris a material constant that depends on the molecular structure. The frictional force is similar to the collisional term in §4.6.1 in that it is assumed to be proportional to the electron momentum mev: Fs(r,t)=−2/Gamma1mev(r,t) where /Gamma1is a material constant. With these we can apply Newton’s second law to obtain F(r,t)=−qeE/prime(r,t)−meω2 rl(r,t)−2/Gamma1mev(r,t)=medv(r,t) dt. Using v=dl/dtwe ﬁnd that the equation of motion for the electron is d2l(r,t) dt2+2/Gamma1dl(r,t) dt+ω2 rl(r,t)=−qe meE/prime(r,t). (4.101) We recognize this diﬀerential equation as the damped harmonic equation. When E/prime=0 we have the homogeneous solution l(r,t)=l0(r)e−/Gamma1tcos/parenleftbigg t/radicalBig ω2r−/Gamma12/parenrightbigg . Thus the electron position is a damped oscillation. The resonant frequency/radicalbig ω2r−/Gamma12is usually only slightly reduced from ωrsince radiation damping is generally quite low. Since the dipole moment for an electron displaced from equilibrium by lisp=−qel, and the polarization density is P=Npfrom (93), we can write P(r,t)=− Nqel(r,t). Multiplying (4.101) by −Nqeand substituting the above expression, we have a diﬀerential equation for the polarization: d2P dt2+2/Gamma1dP dt+ω2 rP=Nq2 e meE/prime. To obtain a constitutive equation we must relate the polarization to the applied ﬁeld E. We can accomplish this by relating the local ﬁeld E/primeto the polarization using the Mosotti ﬁeld (4.96). Substitution gives d2P dt2+2/Gamma1dP dt+ω2 0P=Nq2 e meE (4.102) where ω0=/radicalBigg ω2r−Nq2e 3me/epsilon10 is the resonance frequency of the dipole moments. We see that this frequency is reduced from the resonance frequency of the electron oscillation because of the polarization ofthe surrounding medium. We can now obtain a dispersion equation for the electrical susceptibility by taking the Fourier transform of (4.102). We have −ω 2˜P+jω2/Gamma1˜P+ω2 0˜P=Nq2 e me˜E. Thus we obtain the dispersion relation ˜χe(ω)=˜P /epsilon10˜E=ω2 p ω2 0−ω2+jω2/Gamma1 where ωpis the plasma frequency (4.74). Since ˜/epsilon1r(ω)=1+˜χe(ω)we also have ˜/epsilon1(ω)=/epsilon10+/epsilon10ω2 p ω2 0−ω2+jω2/Gamma1. (4.103) If more than one type of oscillating moment contributes to the permittivity, we may extend (4.103) to ˜/epsilon1(ω)=/epsilon10+/summationdisplay i/epsilon10ω2 pi ω2 i−ω2+jω2/Gamma1i(4.104) where ωpi=Niq2 e//epsilon10miis the plasma frequency of the ith resonance component, and ωiand/Gamma1iare the oscillation frequency and damping coeﬃcient, respectively, of this component. This expression is the dispersion formula for classical physics , so called because it neglects quantum eﬀects. When losses are negligible, (4.104) reduces to theSellmeier equation ˜/epsilon1(ω)=/epsilon1 0+/summationdisplay i/epsilon10ω2 pi ω2 i−ω2. (4.105) Let us now study the frequency behavior of the dispersion relation (4.104). Splitting the permittivity into real and imaginary parts we have ˜/epsilon1/prime(ω)−/epsilon10=/epsilon10/summationdisplay iω2 piω2 i−ω2 [ω2 i−ω2]2+4ω2/Gamma12 i, ˜/epsilon1/prime/prime(ω)=−/epsilon10/summationdisplay iω2 pi2ω/Gamma1i [ω2 i−ω2]2+4ω2/Gamma12 i. Asω→0the permittivity reduces to /epsilon1=/epsilon10/parenleftBigg 1+/summationdisplay iω2 pi ω2 i/parenrightBigg , which is the static permittivity of the material. As ω→∞ the permittivity behaves as ˜/epsilon1/prime(ω)→/epsilon10/parenleftBigg 1−/summationtext iω2 pi ω2/parenrightBigg , ˜/epsilon1/prime/prime(ω)→−/epsilon102/summationtext iω2 pi/Gamma1i ω3. This high frequency behavior is identical to that of a plasma as described by (4.76). 0.0 0.5 1.0 1.5 2.0 2.5-2.5-2.0-1.5-1.0-0.50.00.51.01.52.02.53.0 ε εεW Region of anomalous dispersion ω/ω0 0− − Figure 4.4: Real and imaginary parts of permittivity for a single resonance model of a dielectric with /Gamma1/ω 0=0.2. Permittivity normalized by dividing by /epsilon10(ωp/ω0)2. The major characteristic of the dispersion relation (4.104) is the presence of one or moreresonan ces.Figure4.4showsaplotofasingleresonanc ecomponent,wherewe have normalized the permittivity as (˜/epsilon1/prime(ω)−/epsilon10)/(/epsilon1 0¯ω2 p)=1−¯ω2 /bracketleftbig 1−¯ω2/bracketrightbig2+4¯ω2¯/Gamma12, −˜/epsilon1/prime/prime(ω)/(/epsilon1 0¯ω2 p)=2¯ω¯/Gamma1 /bracketleftbig 1−¯ω2/bracketrightbig2+4¯ω2¯/Gamma12, with ¯ω=ω/ω 0,¯ωp=ωp/ω0, and ¯/Gamma1=/Gamma1/ω 0. We see a distinct resonance centered at ω=ω0. Approaching this resonance through frequencies less than ω0, we see that ˜/epsilon1/prime increases slowly until peaking at ωmax=ω0√1−2/Gamma1/ω 0where it attains a value of ˜/epsilon1/primemax=/epsilon10+1 4/epsilon10¯ω2 p ¯/Gamma1(1−¯/Gamma1). After peaking, ˜/epsilon1/primeundergoes a rapid decrease, passing through ˜/epsilon1/prime=/epsilon10atω=ω0, and then continuing to decrease until reaching a minimum value of ˜/epsilon1/prime min=/epsilon10−1 4/epsilon10¯ω2 p ¯/Gamma1(1+¯/Gamma1) atωmin=ω0√1+2/Gamma1/ω 0.A sωcontinues to increase, ˜/epsilon1/primeagain increases slowly toward a ﬁnal value of ˜/epsilon1/prime=/epsilon10. The regions of slow variation of ˜/epsilon1/primeare called regions of normal dispersion , while the region where ˜/epsilon1/primedecreases abruptly is called the region of anomalous dispersion . Anomalous dispersion is unusual only in the sense that it occurs over a narrower range of frequencies than normal dispersion. The imaginary part of the permittivity peaks near the resonant frequency, dropping oﬀ monotonically in each direction away from the peak. The width of the curve is animportant parameter that we can most easily determine by approximating the behaviorof˜/epsilon1 /prime/primenearω0. Letting /Delta1¯ω=(ω0−ω)/ω 0and using ω2 0−ω2=(ω0−ω)(ω 0+ω)≈2ω2 0/Delta1¯ω, we get ˜/epsilon1/prime/prime(ω)≈−1 2/epsilon10¯ω2 p¯/Gamma1 (/Delta1¯ω)2+¯/Gamma12. This approximation has a maximum value of ˜/epsilon1/prime/primemax=˜/epsilon1/prime/prime(ω0)=−1 2/epsilon10¯ω2 p1 ¯/Gamma1 located at ω=ω0, and has half-amplitude points located at /Delta1¯ω=± ¯/Gamma1. Thus the width of the resonance curve is W=2/Gamma1. Note that for a material characterized by a low-loss resonance ( /Gamma1/lessmuchω0), the location of ˜/epsilon1/primemaxcan be approximated as ωmax=ω0/radicalbig 1−2/Gamma1/ω 0≈ω0−/Gamma1 while ˜/epsilon1/prime minis located at ωmin=ω0/radicalbig 1+2/Gamma1/ω 0≈ω0+/Gamma1. The region of anomalous dispersion thus lies between the half amplitude points of ˜/epsilon1/prime/prime: ω0−/Gamma1<ω<ω 0+/Gamma1. As/Gamma1→0the resonance curve becomes narrower and taller. Thus, a material charac- terized by a very low-loss resonance may be modeled very simply using ˜/epsilon1/prime/prime=Aδ(ω−ω0), where Ais a constant to be determined. We can ﬁnd Aby applying the Kronig–Kramers formula (4.37): ˜/epsilon1/prime(ω)−/epsilon10=−2 πP.V.∞/integraldisplay 0Aδ(/Omega1−ω0)/Omega1d/Omega1 /Omega12−ω2=−2 πAω0 ω2 0−ω2. Since the material approaches the lossless case, this expression should match the Sellmeier equation (4.105): −2 πAω0 ω2 0−ω2=/epsilon10ω2 p ω2 0−ω2, giving A=−π/epsilon10ω2 p/2ω0. Hence the permittivity of a material characterized by a low-loss resonance may be approximated as ˜/epsilon1c(ω)=/epsilon10/parenleftBigg 1+ω2 p ω2 0−ω2/parenrightBigg −j/epsilon10π 2ω2 p ω0δ(ω−ω0). 7 8 9 10 11 12 log ( f )0204060 −ε/ε0ε/ε0 10 Figure 4.5: Relaxation spectrum for water at 20◦C found using Debye equation. Debye relaxation and the Cole–Cole equation. In solids or liquids consisting of polar molecules (those retaining a permanent dipole moment, e.g., water), the resonance eﬀect is replaced by relaxation . We can view the molecule as attempting to rotate in response to an applied ﬁeld within a background medium dominated by the frictionalterm in (4.101). The rotating molecule experiences many weak collisions which continu-ously drain oﬀ energy, preventing it from accelerating under the force of the applied ﬁeld.J.W.P. Debye proposed that such materials are described by an exponential damping oftheir polarization and a complete absence of oscillations. If we neglect the accelerationterm in (4.101) we have the equation of motion 2/Gamma1dl(r,t) dt+ω2 rl(r,t)=−qe meE/prime(r,t), which has homogeneous solution l(r,t)=l0(r)e−ω2r 2/Gamma1t=l0(r)e−t/τ where τis Debye’s relaxation time . By neglecting the acceleration term in (4.102) we obtain from (4.103) the dispersion equation, or relaxation spectrum ˜/epsilon1(ω)=/epsilon10+/epsilon10ω2 p ω2 0+jω2/Gamma1. Debye proposed a relaxation spectrum a bit more general than this, now called the Debye equation : ˜/epsilon1(ω)=/epsilon1∞+/epsilon1s−/epsilon1∞ 1+jωτ. (4.106) Figure4.6:ArcplotsforDebyeandCole–Col edescription sofapolarmaterial. Here/epsilon1sis the real static permittivity obtained when ω→0, while /epsilon1∞is the real “optical” permittivity describing the high frequency behavior of ˜/epsilon1. If we split (4.106) into real and imaginary parts we ﬁnd that ˜/epsilon1/prime(ω)−/epsilon1∞=/epsilon1s−/epsilon1∞ 1+ω2τ2, ˜/epsilon1/prime/prime(ω)=−ωτ(/epsilon1 s−/epsilon1∞) 1+ω2τ2. For a passive material we must have ˜/epsilon1/prime/prime<0, which requires /epsilon1s>/epsilon1∞. It is straightforward to show that these expressions obey the Kronig–Kramers relationships. The details areleft as an exercise. AplotoftheDebyespectrumofwaterat T = 20 ◦CisshowninFigure4.5,wherewe have used /epsilon1s=78.3/epsilon10,/epsilon1∞=5/epsilon10, and τ=9.6×10−12s [49]. We see that ˜/epsilon1/primedecreases over the entire frequency range. The frequency dependence of the imaginary part of thepermittivity is similar to that found in the resonance model, forming a curve which peaksat the critical frequency ω max=1/τ where it obtains a maximum value of −˜/epsilon1/prime/primemax=/epsilon1s−/epsilon1∞ 2. At this point ˜/epsilon1/primeachieves the average value of /epsilon1sand/epsilon1∞: /epsilon1/prime(ωmax)=/epsilon1s+/epsilon1∞ 2. Since the frequency label is logarithmic, we see that the peak is far broader than that for the resonance model. Interestingl y,aplotof−˜/epsilon1/prime/primeversus ˜/epsilon1/primetracesoutasemicircl ecenteredalongtherealaxis at(/epsilon1s +/epsilon1∞)/2andwithradius(/epsilon1s −/epsilon1∞)/2.Suchaplot,showninFigure4.6,wasﬁrst described by K.S. Cole and R.H. Cole [38] and is thus called a Cole–Cole diagram or “arc 02 0 4 0 6 0 ε/ε0204060-ε/ε 00 Figure 4.7: Cole–Cole diagram for water at 20◦C. plot.” We can think of the vector extending from the origin to a point on the semicircle as a phasor whose phase angle δis described by the loss tangent of the material: tanδ=−˜/epsilon1/prime/prime ˜/epsilon1/prime=ωτ(/epsilon1 s−/epsilon1∞) /epsilon1s+/epsilon1∞ω2τ2. (4.107) The Cole–Cole plot shows that the maximum value of −˜/epsilon1/prime/primeis(/epsilon1s−/epsilon1∞)/2and that ˜/epsilon1/prime=(/epsilon1s+/epsilon1∞)/2at this point. ACole–Col eplotforwater,showninFigure4.7,displaysthetypicalsemicircular nature of the arc plot. However, not all polar materials have a relaxation spectrumthat follows the Debye equation as closely as water. Cole and Cole found that for manymaterials the arc plot traces a circular arc centered below the real axis, and that the line throug hitscentermakesanangleofα(π/2)withtherealaxisasshowninFigure4.6. This relaxation spectrum can be described in terms of a modiﬁed Debye equation ˜/epsilon1(ω)=/epsilon1 ∞+/epsilon1s−/epsilon1∞ 1+(jωτ)1−α, called the Cole–Cole equation . A nonzero Cole–Cole parameter αtends to broaden the relaxation spectrum, and results from a spread of relaxation times centered around τ [4]. For water the Cole–Cole parameter is only α=0.02, suggesting that a Debye description is suﬃcient, but for other materials αmay be much higher. For instance, consider a transformer oil with a measured Cole–Cole parameter of α=0.23, along with ameasure drelaxatio ntimeofτ= 2.3 × 10−9 s,astaticpermittivi tyof/epsilon1s=5.9/epsilon10, and an optical permittivity of /epsilon1∞= 2.9/epsilon10 [4].Figure4.8showstheCole–Col eplotcalculated usingbothα= 0andα= 0.23,demonstratin gasigniﬁca ntdivergenc efromtheDebye model.Figure4.9showstherelaxatio nspectrumforthetransforme roilcalculate dwith these same two parameters. 2.0 2.5 3.0 3.5 4.0 4.5 5.0 ε/ε0123-ε/εDebye equation Cole-Cole equation0 0 Figure 4.8: Cole–Cole diagram for transformer oil found using Debye equation and Cole– Cole equation with α=0.23. 579 log ( f )012345Debye equation Cole-Cole equation −ε/ε0ε/ε0 10 Figure 4.9: Relaxation spectrum for transformer oil found using Debye equation and Cole–Cole equation with α=0.23. 4.6.4 Permittivity and conductivity of a conductor The free electrons within a conductor may be considered as an electron gas which is free to move under the inﬂuence of an applied ﬁeld. Since the electrons are not bound tothe atoms of the conductor, there is no restoring force acting on them. However, thereis a damping term associated with electron collisions. We therefore model a conductoras a plasma, but with a very high collision frequency; in a good metallic conductor νis typically in the range 10 13–1014Hz. We therefore have the conductivity of a conductor from (4.75) as ˜σ(ω)=/epsilon10ω2 pν ω2+ν2 and the permittivity as ˜/epsilon1(ω)=/epsilon10/bracketleftBigg 1−ω2 p ω2+ν2/bracketrightBigg . Since νis so large, the conductivity is approximately ˜σ(ω)≈/epsilon10ω2 p ν=Nq2 e meν and the permittivity is ˜/epsilon1(ω)≈/epsilon10 well past micro wavefrequencies and into the infrared. Hence the dc conductivity is often employed by engineers throughout the communications bands. When approaching thevisible spectrum the permittivity and conductivity begin to show a strong frequencydependence. In the violet and ultraviolet frequency ranges the free-charge conductivitybecomes proportional to 1/ωand is driven toward zero. However, at these frequencies the resonances of the bound electrons of the metal become important and the permittivitybehaves more like that of a dielectric. At these frequencies the permittivity is bestdescribed using the resonance formula (4.104). 4.6.5 Permeability dyadic of a ferrite The magnetic properties of materials are complicated and diverse. The formation of accurate models based on atomic behavior requires an understanding of quantummechanics, but simple models may be constructed using classical mechanics along withvery simple quantum-mechanical assumptions, such as the existence of a spin moment.For an excellent review of the magnetic properties of materials, see Elliott [65]. The magnetic properties of matter ultimately result from atomic currents. In our sim- ple microscopic view these currents arise from the spin and orbital motion of negativelycharged electrons. These atomic currents potentially give each atom a magnetic moment m.I ndiamagnetic materials the orbital and spin moments cancel unless the material is exposed to an external magnetic ﬁeld, in which case the orbital electron velocity changesto produce a net moment opposite the applied ﬁeld. In paramagnetic materials the spin moments are greater than the orbital moments, leaving the atoms with a net permanentmagnetic moment. When exposed to an external magnetic ﬁeld, these moments align inthe same direction as an applied ﬁeld. In either case, the density of magnetic momentsMis zero in the absence of an applied ﬁeld. In most paramagnetic materials the alignment of the permanent moment of neigh- boring atoms is random. However, in the subsets of paramagnetic materials known asferromagnetic ,anti-ferromagnetic , andferrimagnetic materials, there is a strong coupling between the spin moments of neighboring atoms resulting in either parallel or antiparal-lel alignment of moments. The most familiar case is the parallel alignment of momentswithin the domains of ferromagnetic permanent magnets made of iron, nickel, and cobalt.Anti-ferromagnetic materials, such as chromium and manganese, have strongly coupledmoments that alternate in direction between small domains, resulting in zero net mag-netic moment. Ferrimagnetic materials also have alternating moments, but these areunequal and thus do not cancel completely. Ferrites form a particularly useful subgroup of ferrimagnetic materials. They were ﬁrst developed during the 1940s by researchers at the Phillips Laboratories as low-loss mag-netic media for supporting electromagnetic waves [65]. Typically, ferrites have conduc- tivities ranging from 10 −4to100S/m (compared to 107for iron), relative permeabilities in the thousands, and dielectric constants in the range 10–15. Their low loss makes them useful for constructing transformer cores and for a variety of microwave applications.Their chemical formula is XO ·Fe 2O3, where Xis a divalent metal or mixture of metals, such as cadmium, copper, iron, or zinc. When exposed to static magnetic ﬁelds, ferritesexhibit gyrotropic magnetic (or gyromagnetic ) properties and have permeability matrices of the form (2.32). The properties of a wide variety of ferrites are given by von Aulock[204]. To determine the permeability matrix of a ferrite we will model its electrons as simple spinning tops and examine the torque exerted on the magnetic moment by the applicationof an external ﬁeld. Each electron has an angular momentum Land a magnetic dipole moment m, with these two vectors anti-parallel: m(r,t)=−γL(r,t) where γ=q e me=1.7592×1011C/kg is called the gyromagnetic ratio . Let us ﬁrst consider a single spinning electron immersed in an applied static magnetic ﬁeld B0. Any torque applied to the electron results in a change of angular momentum as given by Newton’s second law T(r,t)=dL(r,t) dt. We found in (3.179) that a very small loop of current in a magnetic ﬁeld experiences a torque m×B. Thus, when ﬁrst placed into a static magnetic ﬁeld B0an electron’s angular momentum obeys the equation dL(r,t) dt=−γL(r,t)×B0(r)=ω0(r)×L(r,t) (4.108) where ω0=γB0. This equation of motion describes the precession of the electron spin axis about the direction of the applied ﬁeld, which is analogous to the precession of agyroscope [129]. The spin axis rotates at the Larmor precessional frequency ω 0=γB0= γµ 0H0. We can use this to understand what happens when we insert a homogeneous ferrite material into a uniform static magnetic ﬁeld B0=µ0H0. The internal ﬁeld Hiexperienced by any magnetic dipole is not the same as the external ﬁeld H0, and need not even be in the same direction. In general we write H0(r,t)−Hi(r,t)=Hd(r,t) where Hdis the demagnetizing ﬁeld produced by the magnetic dipole moments of the material. Each electron responds to the internal ﬁeld by precessing as described aboveuntil the precession damps out and the electron moments align with the magnetic ﬁeld.At this point the ferrite is saturated . Because the demagnetizing ﬁeld depends strongly on the shape of the material we choose to ignore it as a ﬁrst approximation, and thisallows us to concentrate our study on the fundamental atomic properties of the ferrite. For purposes of understanding its magnetic properties, we view the ferrite as a dense collection of electrons and write M(r,t)=Nm(r,t) where Nis the number density of electrons. Since we are assuming the ferrite is homoge- neous, we take Nto be independent of time and position. Multiplying (4.108) by −Nγ, we obtain an equation describing the evolution of M: dM(r,t) dt=−γM(r,t)×Bi(r,t). (4.109) To determine the temporal response of the ferrite we must include a time-dependent component of the applied ﬁeld. We now let H0(r,t)=Hi(r,t)=HT(r,t)+Hdc where HTis the time-dependent component superimposed with the uniform static com- ponent Hdc. Using B=µ0(H+M)we have from (4.109) dM(r,t) dt=−γµ 0M(r,t)×[HT(r,t)+Hdc+M(r,t)]. With M=MT(r,t)+Mdcand M×M=0this becomes dMT(r,t) dt+dMdc dt=−γµ 0[MT(r,t)×HT(r,t)+MT(r,t)×Hdc+ +Mdc×HT(r,t)+Mdc×Hdc. (4.110) Let us assume that the ferrite is saturated. Then Mdcis aligned with Hdcand their cross product vanishes. Let us further assume that the spectrum of HTis small compared toHdcat all frequencies: |˜HT(r,ω)|/lessmuch Hdc. This small-signal assumption allows us to neglect MT×HT. Using these and noting that the time derivative of Mdcis zero, we see that (4.110) reduces to dMT(r,t) dt=−γµ 0[MT(r,t)×Hdc+Mdc×HT(r,t)]. (4.111) To determine the frequency response we write (4.111) in terms of inverse Fourier transforms and invoke the Fourier integral theorem to ﬁnd that jω˜MT(r,ω)=−γµ 0[˜MT(r,ω)×Hdc+Mdc×˜HT(r,ω)]. Deﬁning γµ 0Mdc=ωM, where ωM=|ωM|is the saturation magnetization frequency , we ﬁnd that ˜MT+˜MT×/bracketleftbiggω0 jω/bracketrightbigg =/bracketleftbigg −1 jωωM×˜HT/bracketrightbigg , (4.112) where ω0=γµ 0Hdcwithω0now called the gyromagnetic response frequency . This has the form v+v×C=A, which has solution (4.80). Substituting into this expression and remembering that ω0is parallel to ωM, we ﬁnd that ˜MT=−1 jωωM×˜HT+1 ω2/braceleftbig ωM[ω0·˜HT]−(ω0·ωM)˜HT/bracerightbig 1−ω2 0 ω2. If we deﬁne the dyadic ¯ωMsuch that ¯ωM·˜HT=ωM×˜HT, then we identify the dyadic magnetic susceptibility ˜¯χm(ω)=jω¯ωM+ωMω0−ωMω0¯I ω2−ω2 0(4.113) with which we can write ˜M(r,ω)=¯χm(ω)·˜H(r,ω). In rectangular coordinates ¯ωMis represented by [¯ωM]= 0−ωMzωMy ωMz 0−ωMx −ωMyωMx 0 . (4.114) Finally, using ˜B=µ0(˜H+˜M)=µ0(¯I+˜¯χm)·˜H=˜¯µ·˜Hwe ﬁnd that ˜¯µ(ω)=µ0[¯I+˜¯χm(ω)]. To examine the properties of the dyadic permeability it is useful to write it in matrix form. To do this we must choose a coordinate system. We shall assume that Hdcis aligned with the z-axis so that Hdc=ˆzHdcand thus ωM=ˆzωMandω0=ˆzω0. Then (4.114) becomes [¯ωM]= 0−ωM0 ωM00 00 0  and we can write the susceptibility dyadic (4.113) as [˜¯χm(ω)]=ωM ω2−ω2 0 −ω0−jω0 jω−ω00 00 0 . The permeability dyadic becomes [˜¯µ(ω)]= µ−jκ0 jκµ 0 00 µ0  (4.115) where µ=µ0/parenleftbigg 1−ω0ωM ω2−ω2 0/parenrightbigg , (4.116) κ=µ0ωω M ω2−ω2 0. (4.117) Because its permeability dyadic is that for a lossless gyrotropic material (2.33), we call the ferrite gyromagnetic . Since the ferrite is lossless, the dyadic permeability must be hermitian according to (4.49). The speciﬁc form of (4.115) shows this explicitly. We also note that since thesign of ω Mis determined by that of Hdc, the dyadic permittivity obeys the symmetry relation ˜µij(Hdc)=˜µji(−Hdc), which is the symmetry condition observed for a plasma in (4.87). A lossy ferrite material can be modeled by adding a damping term to (4.111): dM(r,t) dt=−γµ 0[MT(r,t)×Hdc+Mdc×HT(r,t)]+αMdc Mdc×dMT(r,t) dt, where αis the damping parameter [40, 204]. This term tends to reduce the angle of precession. Fourier transformation gives jω˜MT=ω0×˜MT−ωM×˜HT+αωM ωM×jω˜MT. Remembering that ω0andωMare aligned we can write this as ˜MT+˜MT× ω0/parenleftBig 1+jαω ω0/parenrightBig jω =/bracketleftbigg −1 jωωM×˜HT/bracketrightbigg . This is identical to (4.112) with ω0→ω0/parenleftbigg 1+jαω ω0/parenrightbigg . Thus, we merely substitute this into (4.113) to ﬁnd the susceptibility dyadic for a lossy ferrite: ˜¯χm(ω)=jω¯ωM+ωMω0(1+jαω/ω 0)−ωMω0(1+jαω/ω 0)¯I ω2(1+α2)−ω2 0−2jαωω 0. Making the same substitution into (4.115) we can write the dyadic permeability matrix as [˜¯µ(ω)]= ˜µxx˜µxy0 ˜µyx˜µyy0 00 µ0  (4.118) where ˜µxx=˜µyy=µ0−µ0ωMω0/bracketleftbig ω2(1−α2)−ω2 0/bracketrightbig +jωα/bracketleftbig ω2(1+α2)+ω2 0/bracketrightbig /bracketleftbig ω2(1+α2)−ω2 0/bracketrightbig2+4α2ω2ω2 0(4.119) and ˜µxy=− ˜µyx=2µ0αω2ω0ωM−jµ0ωω M/bracketleftbig ω2(1+α2)−ω2 0/bracketrightbig /bracketleftbig ω2(1+α2)−ω2 0/bracketrightbig2+4α2ω2ω2 0. (4.120) In the case of a lossy ferrite, the hermitian nature of the permeability dyadic is lost. 4.7 Monochromatic ﬁelds and the phasor domain The Fourier transform is very eﬃcient for representing the nearly sinusoidal signals produced by electronic systems such as oscillators. However, we should realize that theelemental term e jωtby itself cannot represent any physical quantity; only a continuous superposition of such terms can have physical meaning, because no physical process canbe truly monochromatic. All events must have transient periods during which they areestablished. Even “monochromatic” light appears in bundles called quanta, interpretedas containing ﬁnite numbers of oscillations. Arguments about whether “monochromatic” or “sinusoidal steady-state” ﬁelds can actually exist may sound purely academic. After all, a micro waveoscillator can create a wave train of 10 10oscillations within the ﬁrst second after being turned on. Such a waveform is surely as close to monochromatic as we would care to measure. But as with all mathematical models of physical systems, we can get into trouble by making non-physical assumptions, in this instance by assuming a physical system has always beenin the steady state. Sinusoidal steady-state solutions to Maxwell’s equations can lead totroublesome inﬁnities linked to the inﬁnite energy content of each elemental component.For example, an attempt to compute the energy stored within a lossless microwave cavityunder steady-state conditions gives an inﬁnite result since the cavity has been building upenergy since t=− ∞ . We handle this by considering time-averaged quantities, but even then must be careful when materials are dispersive ( §4.5). Nevertheless, the steady- state concept is valuable because of its simplicity and ﬁnds widespread application inelectromagnetics. Since the elemental term is complex, we may use its real part, its imaginary part, or some combination of both to represent a monochromatic (or time-harmonic ) ﬁeld. We choose the representation ψ(r,t)=ψ 0(r)cos[ˇωt+ξ(r)], (4.121) where ξis the temporal phase angle of the sinusoidal function. The Fourier transform is ˜ψ(r,ω)=/integraldisplay∞ −∞ψ0(r)cos[ˇωt+ξ(r)]e−jωtdt. (4.122) Here we run into an immediate problem: the transform in (4.122) does not exist in the ordinary sense since cos(ˇωt+ξ)is not absolutely integrable on (−∞,∞). We should not be surprised by this: the cosine function cannot describe an actual physical process (itextends in time to ±∞), so it lacks a classical Fourier transform. One way out of this predicament is to extend the meaning of the Fourier transform as we do in §A.1. Then the monochromatic ﬁeld (4.121) is viewed as having the generalized transform ˜ψ(r,ω)=ψ 0(r)π/bracketleftbig ejξ(r)δ(ω−ˇω)+e−jξ(r)δ(ω+ˇω)/bracketrightbig . (4.123) We can compute the inverse Fourier transform by substituting (123) into (2): ψ(r,t)=1 2π/integraldisplay∞ −∞ψ0(r)π/bracketleftbig ejξ(r)δ(ω−ˇω)+e−jξ(r)δ(ω+ˇω)/bracketrightbig ejωtdω. (4.124) By our interpretation of the Dirac delta, we see that the decomposition of the cosine function has only two discrete components, located at ω=± ˇω. So we have realized our initial intention of having only a single elemental function present. The sifting property gives ψ(r,t)=ψ0(r)ejˇωtejξ(r)+e−jˇωte−jξ(r) 2=ψ0(r)cos[ˇωt+ξ(r)] as expected. 4.7.1 The time-harmonic EM ﬁelds and constitutive relations The time-harmonic ﬁelds are described using the representation (4.121) for each ﬁeld component. The electric ﬁeld is E(r,t)=3/summationdisplay i=1ˆii|Ei(r)|cos[ˇωt+ξE i(r)] for example. Here |Ei|is the complex magnitude of the ith vector component, and ξE iis the phase angle ( −π<ξE i≤π). Similar terminology is used for the remaining ﬁelds. The frequency-domain constitutive relations (4.11)–(4.15) may be written for the time- harmonic ﬁelds by employing (4.124). For instance, for an isotropic material where ˜D(r,ω)=˜/epsilon1(r,ω)˜E(r,ω) , ˜B(r,ω)=˜µ(r,ω)˜H(r,ω) , with ˜/epsilon1(r,ω)=|˜/epsilon1(r,ω)|eξ/epsilon1(r,ω), ˜µ(r,ω)=|˜µ(r,ω)|eξµ(r,ω), we can write D(r,t)=3/summationdisplay i=1ˆii|Di(r)|cos[ˇωt+ξD i(r)] =1 2π/integraldisplay∞ −∞3/summationdisplay i=1ˆii˜/epsilon1(r,ω)|Ei(r)|π/bracketleftBig ejξE i(r)δ(ω−ˇω)+e−jξE i(r)δ(ω+ˇω)/bracketrightBig ejωtdω =1 23/summationdisplay i=1ˆii|Ei(r)|/bracketleftBig ˜/epsilon1(r,ˇω)ej(ˇωt+jξE i(r))+˜/epsilon1(r,−ˇω)e−j(ˇωt+jξE i(r))/bracketrightBig . Since (4.25) shows that ˜/epsilon1(r,−ˇω)=˜/epsilon1∗(r,ˇω), we have D(r,t)=1 23/summationdisplay i=1ˆii|Ei(r)||˜/epsilon1(r,ˇω)|/bracketleftBig ej(ˇωt+jξE i(r)+jξ/epsilon1(r,ˇω))+e−j(ˇωt+jξE i(r)+jξ/epsilon1(r,ˇω))/bracketrightBig =3/summationdisplay i=1ˆii|˜/epsilon1(r,ˇω)||Ei(r)|cos[ˇωt+ξE i(r)+ξ/epsilon1(r,ˇω)]. (4.125) Similarly B(r,t)=3/summationdisplay i=1ˆii|Bi(r)|cos[ˇωt+ξB i(r)] =3/summationdisplay i=1ˆii|˜µ(r,ˇω)||Hi(r)|cos[ˇωt+ξH i(r)+ξµ(r,ˇω)]. 4.7.2 The phasor ﬁelds and Maxwell’s equations Sinusoidal steady-state computations using the forward and inverse transform formulas are unnecessarily cumbersome. A much more eﬃcient approach is to use the phasor concept. If we deﬁne the complex function ˇψ(r)=ψ0(r)ejξ(r) as the phasor form of the monochromatic ﬁeld ˜ψ(r,ω), then the inverse Fourier transform is easily computed by multiplying ˇψ(r)byejˇωtand taking the real part. That is, ψ(r,t)=Re/braceleftbigˇψ(r)ejˇωt/bracerightbig =ψ0(r)cos[ˇωt+ξ(r)]. (4.126) Using the phasor representation of the ﬁelds, we can obtain a set of Maxwell equations relating the phasor components. Let ˇE(r)=3/summationdisplay i=1ˆiiˇEi(r)=3/summationdisplay i=1ˆii|Ei(r)|ejξE i(r) represent the phasor monochromatic electric ﬁeld, with similar formulas for the other ﬁelds. Then E(r,t)=Re/braceleftbigˇE(r)ejˇωt/bracerightbig =3/summationdisplay i=1ˆii|Ei(r)|cos[ˇωt+ξE i(r)]. Substituting these expressions into Ampere’s law (2.2), we have ∇× Re/braceleftbigˇH(r)ejˇωt/bracerightbig =∂ ∂tRe/braceleftbigˇD(r)ejˇωt/bracerightbig +Re/braceleftbigˇJ(r)ejˇωt/bracerightbig . Since the real part of a sum of complex variables equals the sum of the real parts, we can write Re/braceleftbigg ∇× ˇH(r)ejˇωt−ˇD(r)∂ ∂tejˇωt−ˇJ(r)ejˇωt/bracerightbigg =0. (4.127) If we examine for an arbitrary complex function F=Fr+jFithe quantity Re/braceleftbig (Fr+jFi)ejˇωt/bracerightbig =Re{(Frcosˇωt−Fisinˇωt)+j(Frsinˇωt+Ficosˇωt)}, we see that both Frand Fimust be zero for the expression to vanish for all t.T h u s (4.127) requires that ∇× ˇH(r)=jˇωˇD(r)+ˇJ(r), (4.128) which is the phasor Ampere’s law. Similarly we have ∇× ˇE(r)=− jˇωˇB(r), (4.129) ∇·ˇD(r)=ˇρ(r), (4.130) ∇·ˇB(r)=0, (4.131) and ∇·ˇJ(r)=− jˇωˇρ(r). (4.132) The constitutive relations may be easily incorporated into the phasor concept. If we use ˇDi(r)=˜/epsilon1(r,ˇω)ˇEi(r)=|˜/epsilon1(r,ˇω)|ejξ/epsilon1(r,ˇω)|Ei(r)|ejξE i(r), then forming Di(r,t)=Re/braceleftbigˇDi(r)ejˇωt/bracerightbig we reproduce (4.125). Thus we may write ˇD(r)=˜/epsilon1(r,ˇω)ˇE(r). Note that we never write ˇ/epsilon1or refer to a “phasor permittivity” since the permittivity does not vary sinusoidally in the time domain. An obvious beneﬁt of the phasor method is that we can manipulate ﬁeld quantities without involving the sinusoidal time dependence. When our manipulations are complete,we return to the time domain using (4.126). The phasor Maxwell equations (4.128)–(4.131) are identical in form to the temporal frequency-domain Maxwell equations (4.7)–(4.10), except that ω=ˇωin the phasor equations. This is sensible, since the phasor ﬁelds represent a single component of the complete frequency-domain spectrum of the arbitrary time-varying ﬁelds. Thus, if thephasor ﬁelds are calculated for some ˇω, we can make the replacements ˇω→ω, ˇE(r)→˜E(r,ω) , ˇH(r)→˜H(r,ω) , ..., and obtain the general time-domain expressions by performing the inversion (4.2). Simi- larly, if we evaluate the frequency-domain ﬁeld ˜E(r,ω)atω=ˇω, we produce the phasor ﬁeld ˇE(r)=˜E(r,ˇω)for this frequency. That is Re/braceleftbig˜E(r,ˇω)e jˇωt/bracerightbig =3/summationdisplay i=1ˆii|˜Ei(r,ˇω)|cos/parenleftbig ˇωt+ξE(r,ˇω)/parenrightbig . 4.7.3 Boundary conditions on the phasor ﬁelds The boundary conditions developed in §4.3 for the frequency-domain ﬁelds may be adapted for use with the phasor ﬁelds by selecting ω=ˇω. Let us include the eﬀects of ﬁctitious magnetic sources and write ˆn12×(ˇH1−ˇH2)=ˇJs, (4.133) ˆn12×(ˇE1−ˇE2)=− ˇJms, (4.134) ˆn12·(ˇD1−ˇD2)=ˇρs, (4.135) ˆn12·(ˇB1−ˇB2)=ˇρms, (4.136) and ˆn12·(ˇJ1−ˇJ2)=− ∇ s·ˇJs−jˇωˇρs, (4.137) ˆn12·(ˇJm1−ˇJm2)=− ∇ s·ˇJms−jˇωˇρms, (4.138) where ˆn12points into region 1 from region 2. 4.8 Poynting’s theorem for time-harmonic ﬁelds We can specialize Poynting’s theorem to time-harmonic form by substituting the time- harmonic ﬁeld representations. The result depends on whether we use the general form (2.301), which is valid for dispersive materials, or (2.299). For nondispersive materials (2.299) allows us to interpret the volume integral term as the time rate of change ofstored energy. But if the operating frequency lies within the realm of material dispersionand loss, then we can no longer identify an explicit stored energy term. 4.8.1 General form of Poynting’s theorem We begin with (2.301). Substituting the time-harmonic representations we obtain the term E(r,t)·∂D(r,t) ∂t=/bracketleftBigg3/summationdisplay i=1ˆii|Ei|cos[ˇωt+ξE i]/bracketrightBigg ·∂ ∂t/bracketleftBigg3/summationdisplay i=1ˆii|Di|cos[ˇωt+ξD i]/bracketrightBigg =− ˇω3/summationdisplay i=1|Ei||Di|cos[ˇωt+ξE i] sin[ ˇωt+ξD i]. Since 2 sin AcosB≡sin(A+B)+sin(A−B)we have E(r,t)·∂ ∂tD(r,t)=−1 23/summationdisplay i=1ˇω|Ei||Di|SDE ii(t), where SDE ii(t)=sin(2ˇωt+ξD i+ξE i)+sin(ξD i−ξE i) describes the temporal dependence of the ﬁeld product. Separating the current into an impressed term Jiand a secondary term Jc(assumed to be the conduction current) as J=Ji+Jcand repeating the above steps with the other terms, we obtain −1 2/integraldisplay V3/summationdisplay i=1|Ji i||Ei|CJiE ii(t)dV=1 2/contintegraldisplay S3/summationdisplay i,j=1|Ei||Hj|(ˆii×ˆij)·ˆnCEH ij(t)dS+ +1 2/integraldisplay V3/summationdisplay i=1/braceleftbig −ˇω|Di||Ei|SDE ii(t)−ˇω|Bi||Hi|SBH ii(t)+|Jc i||Ei|CJcE ii(t)/bracerightbig dV,(4.139) where SBH ii(t)=sin(2ˇωt+ξB i+ξH i)+sin(ξB i−ξH i), CEH ij(t)=cos(2ˇωt+ξE i+ξH j)+cos(ξE i−ξH j), and so on. We see that each power term has two temporal components: one oscillating at fre- quency 2ˇω, and one constant with time. The oscillating component describes power that cycles through the various mechanisms of energy storage, dissipation, and transfer acrossthe boundary. Dissipation may be produced through conduction processes or throughpolarization and magnetization phase lag, as described by the volume term on the right-hand side of (4.139). Power may also be delivered to the ﬁelds either from the sources,as described by the volume term on the left-hand side, or from an active medium, asdescribed by the volume term on the right-hand side. The time-average balance of powersupplied to the ﬁelds and extracted from the ﬁelds throughout each cycle, including that transported across the surface S, is given by the constant terms in (4.139): −1 2/integraldisplay V3/summationdisplay i=1|Ji i||Ei|cos(ξJi i−ξE i)dV=1 2/integraldisplay V3/summationdisplay i=1/braceleftbig ˇω|Ei||Di|sin(ξE i−ξD i)+ +ˇω|Bi||Hi|sin(ξH i−ξB i)+|Jc i||Ei|cos(ξJc i−ξE i)/bracerightbig dV+ +1 2/contintegraldisplay S3/summationdisplay i,j=1|Ei||Hj|(ˆii×ˆij)·ˆncos(ξE i−ξH j)dS. (4.140) We associate one mechanism for time-average power loss with the phase lag between applied ﬁeld and resulting polarization or magnetization. We can see this more clearlyif we use the alternative form of the Poynting theorem (2.302) written in terms of thepolarization and magnetization vectors. Writing P(r,t)= 3/summationdisplay i=1|Pi(r)|cos[ˇωt+ξP i(r)], M(r,t)=3/summationdisplay i=1|Mi(r)|cos[ˇωt+ξM i(r)], and substituting the time-harmonic ﬁelds, we see that −1 2/integraldisplay V3/summationdisplay i=1|Ji||Ei|CJE ii(t)dV+ˇω 2/integraldisplay V3/summationdisplay i=1/bracketleftbig |Pi||Ei|SPE ii(t)+µ0|Mi||Hi|SMH ii(t)/bracketrightbig dV =−ˇω 2/integraldisplay V3/summationdisplay i=1/bracketleftbig /epsilon10|Ei|2SEE ii(t)+µ0|Hi|2SHH ii(t)/bracketrightbig dV+ +1 2/contintegraldisplay S3/summationdisplay i,j=1|Ei||Hj|(ˆii×ˆij)·ˆnCEH ij(t)dS. (4.141) Selection of the constant part gives the balance of time-average power: −1 2/integraldisplay V3/summationdisplay i=1|Ji||Ei|cos(ξJ i−ξE i)dV =ˇω 2/integraldisplay V3/summationdisplay i=1/bracketleftbig |Ei||Pi|sin(ξE i−ξP i)+µ0|Hi||Mi|sin(ξH i−ξM i)/bracketrightbig dV+ +1 2/contintegraldisplay S3/summationdisplay i,j=1|Ei||Hj|(ˆii×ˆij)·ˆncos(ξE i−ξH j)dS. (4.142) Here the power loss associated with the lag in alignment of the electric and magnetic dipoles is easily identiﬁed as the volume term on the right-hand side, and is seen to arisethrough the interaction of the ﬁelds with the equivalent sources as described through thephase diﬀerence between Eand Pand between Hand M. If these pairs are in phase, then the time-average power balance reduces to that for a dispersionless material, equation(4.146). 4.8.2 Poynting’s theorem for nondispersive materials For nondispersive materials (2.299) is appropriate. We shall carry out the details here so that we may examine the power-balance implications of nondispersive media. We have, substituting the ﬁeld expressions, −1 2/integraldisplay V3/summationdisplay i=1|Ji i||Ei|CJiE ii(t)dV=1 2/integraldisplay V3/summationdisplay i=1|Jc i||Ei|CJcE ii(t)dV+ +∂ ∂t/integraldisplay V3/summationdisplay i=1/braceleftbigg1 4|Di||Ei|CDE ii(t)+1 4|Bi||Hi|CBH ii(t)/bracerightbigg dV+ +1 2/contintegraldisplay S3/summationdisplay i,j=1|Ei||Hj|(ˆii×ˆij)·ˆnCEH ij(t)dS. (4.143) Here we remember that the conductivity relating EtoJcmust also be nondispersive. Note that the electric and magnetic energy densities we(r,t)andwm(r,t)have the time- average values /angbracketleftwe(r,t)/angbracketrightand/angbracketleftwm(r,t)/angbracketrightgiven by /angbracketleftwe(r,t)/angbracketright=1 T/integraldisplayT/2 −T/21 2E(r,t)·D(r,t)dt=1 43/summationdisplay i=1|Ei||Di|cos(ξE i−ξD i) =1 4Re/braceleftbigˇE(r)·ˇD∗(r)/bracerightbig (4.144) and /angbracketleftwm(r,t)/angbracketright=1 T/integraldisplayT/2 −T/21 2B(r,t)·H(r,t)dt=1 43/summationdisplay i=1|Bi||Hi|cos(ξH i−ξB i) =1 4Re/braceleftbigˇH(r)·ˇB∗(r)/bracerightbig , (4.145) where T=2π/ˇω. We have already identiﬁed the energy stored in a nondispersive material (§4.5.2). If (4.144) is to match with (4.62), the phases of ˇEand ˇDmust match: ξE i=ξD i. We must also have ξH i=ξB i. Since in a dispersionless material σmust be independent of frequency, from ˇJc=σˇEwe also see that ξJc i=ξE i. Upon diﬀerentiation the time-average stored energy terms in (4.143) disappear, giving −1 2/integraldisplay V3/summationdisplay i=1|Ji i||Ei|CJiE ii(t)dV=1 2/integraldisplay V3/summationdisplay i=1|Jc i||Ei|CEE ii(t)dV− −2ˇω/integraldisplay V3/summationdisplay i=1/braceleftbigg1 4|Di||Ei|SEE ii(t)+1 4|Bi||Hi|SBB ii(t)/bracerightbigg dV+ +1 2/contintegraldisplay S3/summationdisplay i,j=1|Ei||Hj|(ˆii×ˆij)·ˆnCEH ij(t)dS. Equating the constant terms, we ﬁnd the time-average power balance expression −1 2/integraldisplay V3/summationdisplay i=1|Ji i||Ei|cos(ξJi i−ξE i)dV=1 2/integraldisplay V3/summationdisplay i=1|Jc i||Ei|dV+ +1 2/contintegraldisplay S3/summationdisplay i,j=1|Ei||Hj|(ˆii×ˆij)·ˆncos(ξE i−ξH j)dS. (4.146) This can be written more compactly using phasor notation as /integraldisplay VpJ(r)dV=/integraldisplay Vpσ(r)dV+/contintegraldisplay SSav(r)·ˆndS (4.147) where pJ(r)=−1 2Re/braceleftbigˇE(r)·ˇJi∗(r)/bracerightbig is the time-average density of power delivered by the sources to the ﬁelds in V, pσ(r)=1 2ˇE(r)·ˇJc∗(r) is the time-average density of power transferred to the conducting material as heat, and Sav(r)·ˆn=1 2Re/braceleftbigˇE(r)×ˇH∗(r)/bracerightbig ·ˆn is the density of time-average power transferred across the boundary surface S. Here Sc=ˇE(r)×ˇH∗(r) is called the complex Poynting vector and Savis called the time-average Poynting vector . Comparison of (4.146) with (4.140) shows that nondispersive materials cannot manifest the dissipative (or active) properties determined by the term 1 2/integraldisplay V3/summationdisplay i=1/braceleftbig ˇω|Ei||Di|sin(ξE i−ξD i)+ˇω|Bi||Hi|sin(ξH i−ξB i)+|Jc i||Ei|cos(ξJc i−ξE i)/bracerightbig dV. This term can be used to classify materials as lossless, lossy, or active, as shown next. 4.8.3 Lossless, lossy, and active media In§4.5.1 we classiﬁed materials based on whether they dissipate (or provide) energy over the period of a transient event. We can provide the same classiﬁcation based ontheir steady-state behavior. We classify a material as lossless if the time-average ﬂow of power entering a homoge- neous body is zero when there are sources external to the body, but no sources internal to the body. This implies that the mechanisms within the body either do not dissipatepower that enters, or that there is a mechanism that creates energy to exactly balance thedissipation. If the time-average power entering is positive, then the material dissipatespower and is termed lossy. If the time-average power entering is negative, then power must originate from within the body and the material is termed active . (Note that the power associated with an active body is not described as arising from sources, but israther described through the constitutive relations.) Since materials are generally inhomogeneous we may apply this concept to a vanish- ingly small volume, thus invoking the point-form of Poynting’s theorem. From (4.140)we see that the time-average inﬂux of power density is given by −∇ · S av(r)=pin(r)=1 23/summationdisplay i=1/braceleftbig ˇω|Ei||Di|sin(ξE i−ξD i)+ˇω|Bi||Hi|sin(ξH i−ξB i)+ +|Jc i||Ei|cos(ξJc i−ξE i)/bracerightbig . Materials are then classiﬁed as follows: pin(r)=0, lossless , pin(r)>0, lossy, pin(r)≥0, passive , pin(r)<0, active . We see that if ξE i=ξD i,ξH i=ξB i, and Jc=0, then the material is lossless. This implies that ( D,E) and ( B,H) are exactly in phase and there is no conduction current. If the material is isotropic, we may substitute from the constitutive relations (4.21)–(4.23) toobtain p in(r)=−ˇω 23/summationdisplay i=1/braceleftbigg |Ei|2/bracketleftbigg |˜/epsilon1|sin(ξ/epsilon1)−|˜σ| ˇωcos(ξσ)/bracketrightbigg +|˜µ||Hi|2sin(ξµ)/bracerightbigg . (4.148) The ﬁrst two terms can be regarded as resulting from a single complex permittivity (4.26). Then (4.148) simpliﬁes to pin(r)=−ˇω 23/summationdisplay i=1/braceleftbig |˜/epsilon1c||Ei|2sin(ξ/epsilon1c)+|˜µ||Hi|2sin(ξµ)/bracerightbig . (4.149) Now we can see that a lossless medium, which requires (4.149) to vanish, has ξ/epsilon1c= ξµ=0(or perhaps the unlikely condition that dissipative and active eﬀects within the electric and magnetic terms exactly cancel). To have ξµ=0we need Band Hto be in phase, hence we need ˜µ(r,ω)to be real. To have ξ/epsilon1c=0we need ξ/epsilon1=0(˜/epsilon1(r,ω)real) and ˜σ(r,ω)=0(or perhaps the unlikely condition that the active and dissipative eﬀects of the permittivity and conductivity exactly cancel). A lossy medium requires (4.149) to be positive. This occurs when ξµ<0orξ/epsilon1c<0, meaning that the imaginary part of the permeability or complex permittivity is negative.The complex permittivity has a negative imaginary part if the imaginary part of ˜/epsilon1is negative or if the real part of ˜σis positive. Physically, ξ /epsilon1<0means that ξD<ξEand thus the phase of the response ﬁeld Dlags that of the excitation ﬁeld E. This results from a delay in the polarization alignment of the atoms, and leads to dissipation of powerwithin the material. An active medium requires (4.149) to be negative. This occurs when ξ µ>0orξ/epsilon1c>0, meaning that the imaginary part of the permeability or complex permittivity is positive.The complex permittivity has a positive imaginary part if the imaginary part of ˜/epsilon1is positive or if the real part of ˜σis negative. In summary, a passive isotropic medium is lossless when the permittivity and perme- ability are real and when the conductivity is zero. A passive isotropic medium is lossywhen one or more of the following holds: the permittivity is complex with negative imag-inary part, the permeability is complex with negative imaginary part, or the conductivityhas a positive real part. Finally, a complex permittivity or permeability with positiveimaginary part or a conductivity with negative real part indicates an active medium. For anisotropic materials the interpretation of p inis not as simple. Here we ﬁnd that the permittivity or permeability dyadic may be complex, and yet the material may stillbe lossless. To determine the condition for a lossless medium, let us recompute p inusing the constitutive relations (4.18)–(4.20). With these we have E·/bracketleftbigg∂D ∂t+Jc/bracketrightbigg +H·∂B ∂t=ˇω3/summationdisplay i,j=1|Ei||Ej|/bracketleftbigg −|˜/epsilon1ij|sin(ˇωt+ξE j+ξ/epsilon1 ij)cos(ˇωt+ξE i)+ +|˜σij| ˇωcos(ˇωt+ξE j+ξσ ij)cos(ˇωt+ξE i)/bracketrightbigg + +ˇω3/summationdisplay i,j=1|Hi||Hj|/bracketleftBig −|˜µij|sin(ˇωt+ξH j+ξµ ij)cos(ˇωt+ξH i)/bracketrightBig . Using the angle-sum formulas and discarding the time-varying quantities, we may obtain the time-average input power density: pin(r)=−ˇω 23/summationdisplay i,j=1|Ei||Ej|/bracketleftbigg |˜/epsilon1ij|sin(ξE j−ξE i+ξ/epsilon1 ij)−|˜σij| ˇωcos(ξE j−ξE i+ξσ ij)/bracketrightbigg − −ˇω 23/summationdisplay i,j=1|Hi||Hj||˜µij|sin(ξH j−ξH i+ξµ ij). The reader can easily verify that the conditions that make this quantity vanish, thus describing a lossless material, are |˜/epsilon1ij|=| ˜/epsilon1ji|,ξ/epsilon1 ij=−ξ/epsilon1 ji, (4.150) |˜σij|=| ˜σji|,ξσ ij=−ξσ ji+π, (4.151) |˜µij|=| ˜µji|,ξµ ij=−ξµ ji. (4.152) Note that this requires ξ/epsilon1 ii=ξµ ii=ξσ ii=0. The condition (4.152) is easily written in dyadic form as ˜¯µ(r,ˇω)†=˜¯µ(r,ˇω) (4.153) where “ †” stands for the conjugate-transpose operation. The dyadic permeability ˜¯µis hermitian. The set of conditions (4.150)–(4.151) can also be written quite simply usingthe complex permittivity dyadic (4.24): ˜¯/epsilon1 c(r,ˇω)†=˜¯/epsilon1c(r,ˇω). (4.154) Thus, an anisotropic material is lossless when the both the dyadic permeability and the complex dyadic permittivity are hermitian. Since ˇωis arbitrary, these results are exactly those obtained in §4.5.1. Note that in the special case of an isotropic material the conditions (4.153) and (4.154) can only hold if ˜/epsilon1and ˜µare real and ˜σis zero, agreeing with our earlier conclusions. 4.9 The complex Poynting theorem An equation having a striking resemblance to Poynting’s theorem can be obtained by direct manipulation of the phasor-domain Maxwell equations. The result, althoughcertainly satisﬁed by the phasor ﬁelds, does notreplace Poynting’s theorem as the power- balance equation for time-harmonic ﬁelds. We shall be careful to contrast the interpre-tation of the phasor expression with the actual time-harmonic Poynting theorem. We begin by dotting both sides of the phasor-domain Faraday’s law with ˇH ∗to obtain ˇH∗·(∇× ˇE)=− jˇωˇH∗·ˇB. Taking the complex conjugate of the phasor-domain Ampere’s law and dotting with ˇE, we have ˇE·(∇× ˇH∗)=ˇE·ˇJ∗−jˇωˇE·ˇD∗. We subtract these expressions and use (B.44) to write −ˇE·ˇJ∗=∇· (ˇE×ˇH∗)−jˇω[ˇE·ˇD∗−ˇB·ˇH∗]. Finally, integrating over the volume region Vand dividing by two, we have −1 2/integraldisplay VˇE·ˇJ∗dV=1 2/contintegraldisplay S(ˇE×ˇH∗)·dS−2jˇω/integraldisplay V/bracketleftbigg1 4ˇE·ˇD∗−1 4ˇB·ˇH∗/bracketrightbigg dV.(4.155) This is known as the complex Poynting theorem , and is an expression that must be obeyed by the phasor ﬁelds. As a power balance theorem, the complex Poynting theorem has meaning only for dispersionless materials. If we let J=Ji+Jcand assume no dispersion, (4.155) becomes −1 2/integraldisplay VˇE·ˇJi∗dV=1 2/integraldisplay VˇE·ˇJc∗dV+1 2/contintegraldisplay S(ˇE×ˇH∗)·dS− −2jω/integraldisplay V[/angbracketleftwe/angbracketright−/angbracketleftwm/angbracketright]dV (4.156) where /angbracketleftwe/angbracketrightand/angbracketleftwm/angbracketrightare the time-average stored electric and magnetic energy densities as described in (4.62)–(4.63). Selection of the real part now gives −1 2/integraldisplay VRe/braceleftbigˇE·ˇJi∗/bracerightbig dV=1 2/integraldisplay VˇE·ˇJc∗dV+1 2/contintegraldisplay SRe/braceleftbigˇE×ˇH∗/bracerightbig ·dS, (4.157) which is identical to (4.147). Thus the real part of the complex Poynting theorem gives the balance of time-average power for a dispersionless material. Selection of the imaginary part of (4.156) gives the balance of imaginary, or reactive power: −1 2/integraldisplay VIm/braceleftbigˇE·ˇJi∗/bracerightbig dV=1 2/contintegraldisplay SIm/braceleftbigˇE×ˇH∗/bracerightbig ·dS−2ˇω/integraldisplay V[/angbracketleftwe/angbracketright−/angbracketleftwm/angbracketright]dV. (4.158) In general, the reactive power balance does not have a simple physical interpretation (it isnotthe balance of the oscillating terms in (4.139)). However, an interesting concept can be gleaned from it. If the source current and electric ﬁeld are in phase, and there isno reactive power leaving S, then the time-average stored electric energy is equal to the time-average stored magnetic energy: /integraldisplay V/angbracketleftwe/angbracketrightdV=/integraldisplay V/angbracketleftwm/angbracketrightdV. This is the condition for “resonance.” An example is a series RLC circuit with the source current and voltage in phase. Here the stored energy in the capacitor is equal to thestored energy in the inductor and the input impedance (ratio of voltage to current) isreal. Such a resonance occurs at only one value of frequency. In more complicatedelectromagnetic systems resonance may occur at many discrete eigenfrequencies. 4.9.1 Boundary condition for the time-average Poynting vector In§2.9.5 we developed a boundary condition for the normal component of the time- domain Poynting vector. For time-harmonic ﬁelds we can derive a similar boundarycondition using the time-average Poynting vector. Consider a surface Sacross which the electromagnetic sources and constitutive parameters are discontinuous, as shown inFigure2.6.Let ˆn 12be the unit normal to the surface pointing into region 1 from region 2. If we apply the large-scale form of the complex Poynting theorem (4.155) to the twoseparat esurface sshowninFigure2.6,weobtain 1 2/integraldisplay V/bracketleftbigg ˇE·ˇJ∗−2jˇω/parenleftbigg1 4ˇE·ˇD∗−1 4ˇB·ˇH∗/parenrightbigg/bracketrightbigg dV+1 2/contintegraldisplay SSc·ˆndS =1 2/integraldisplay S10ˆn12·(Sc 1−Sc 2)dS (4.159) where Sc=ˇE×ˇH∗is the complex Poynting vector. If, on the other hand, we apply the large-scale form of Poynting’s theorem to the entire volume region including the surfaceof discontinuity, and include the surface current contribution, we have 1 2/integraldisplay V/bracketleftbigg ˇE·ˇJ∗−2jˇω/integraldisplay V/parenleftbigg1 4ˇE·ˇD∗−1 4ˇB·ˇH∗/parenrightbigg/bracketrightbigg dV+1 2/contintegraldisplay SSc·ˆndS =−1 2/integraldisplay S10ˇJ∗ s·ˇEdS. (4.160) If we wish to have the integrals over Vand Sin (4.159) and (4.160) produce identical results, then we must postulate the two conditions ˆn12×(ˇE1−ˇE2)=0 and ˆn12·(Sc 1−Sc 2)=− ˇJ∗ s·ˇE. (4.161) The ﬁrst condition is merely the continuity of tangential electric ﬁeld; it allows us to be nonspeciﬁc as to which value of Ewe use in the second condition. If we take the real part of the second condition we have ˆn12·(Sav,1−Sav,2)=pJs, (4.162) where Sav=1 2Re{ˇE×ˇH∗}is the time-average Poynting power ﬂow density and pJs= −1 2Re{ˇJ∗ s·ˇE}is the time-average density of power delivered by the surface sources. This is the desired boundary condition on the time-average power ﬂow density. 4.10 Fundamental theorems for time-harmonic ﬁelds 4.10.1 Uniqueness If we think of a sinusoidal electromagnetic ﬁeld as the steady-state culmination of a transient event that has an identiﬁable starting time, then the conditions for uniquenessestablished in §2.2.1 are applicable. However, a true time-harmonic wave, which has existed since t=− ∞ and thus has inﬁnite energy, must be interpreted diﬀerently. Our approach is similar to that of §2.2.1. Consider a simply-connected region of space Vbounded by surface S, where both Vand Scontain only ordinary points. The phasor-domain ﬁelds within Vare associated with a phasor current distribution ˇJ, which may be internal to V(entirely or in part). We seek conditions under which the phasor electromagnetic ﬁelds are uniquely determined. Let the ﬁeld set (ˇE1,ˇD1,ˇB1,ˇH1)satisfy Maxwell’s equations (4.128) and (4.129) associated with the current ˇJ(along with an appropriate set of constitutive relations), and let (ˇE2,ˇD2,ˇB2,ˇH2)be a second solution. To determine the conditions for uniqueness of the ﬁelds, we look for a situation thatresults in ˇE 1=ˇE2,ˇH1=ˇH2, and so on. The electromagnetic ﬁelds must obey ∇× ˇH1=jˇωˇD1+ˇJ, ∇× ˇE1=− jˇωˇB1, ∇× ˇH2=jˇωˇD2+ˇJ, ∇× ˇE2=− jˇωˇB2. Subtracting these and deﬁning the diﬀerence ﬁelds ˇE0=ˇE1−ˇE2,ˇH0=ˇH1−ˇH2, and so on, we ﬁnd that ∇× ˇH0=jˇωˇD0, (4.163) ∇× ˇE0=− jˇωˇB0. (4.164) Establishing the conditions under which the diﬀerence ﬁelds vanish throughout V,w e shall determine the conditions for uniqueness. Dotting (4.164) by ˇH∗ 0and dotting the complex conjugate of (4.163) by ˇE0,w eh a v e ˇH∗ 0·/parenleftbig ∇× ˇE0/parenrightbig =− jˇωˇB0·ˇH∗ 0, ˇE0·/parenleftbig ∇× ˇH∗ 0/parenrightbig =− jˇωˇD∗ 0·ˇE0. Subtraction yields ˇH∗ 0·/parenleftbig ∇× ˇE0/parenrightbig −ˇE0·/parenleftbig ∇× ˇH∗ 0/parenrightbig =− jˇωˇB0·ˇH∗ 0+jˇωˇD∗ 0·ˇE0 which, by (B.44), can be written as ∇·/parenleftbigˇE0×ˇH∗ 0/parenrightbig =jˇω/bracketleftbigˇE0·ˇD∗ 0−ˇB0·ˇH∗ 0/bracketrightbig . Adding this expression to its complex conjugate, integrating over V, and using the di- vergence theorem, we obtain Re/contintegraldisplay S/bracketleftbigˇE0×ˇH∗ 0/bracketrightbig ·dS=− jˇω 2/integraldisplay V/bracketleftbig/parenleftbigˇE∗ 0·ˇD0−ˇE0·ˇD∗ 0/parenrightbig +/parenleftbigˇH∗ 0·ˇB0−ˇH0·ˇB∗ 0/parenrightbig/bracketrightbig dV. Breaking Sinto two arbitrary portions and using ( ??), we obtain Re/contintegraldisplay S1ˇH∗ 0·(ˆn×ˇE0)dS−Re/contintegraldisplay S2ˇE0·(ˆn×ˇH∗ 0)dS= −jˇω 2/integraldisplay V/bracketleftbig/parenleftbigˇE∗ 0·ˇD0−ˇE0·ˇD∗ 0/parenrightbig +/parenleftbigˇH∗ 0·ˇB0−ˇH0·ˇB∗ 0/parenrightbig/bracketrightbig dV. (4.165) Now if ˆn×E0=0orˆn×H0=0over all of S, or some combination of these conditions holds over all of S, then /integraldisplay V/bracketleftbig/parenleftbigˇE∗ 0·ˇD0−ˇE0·ˇD∗ 0/parenrightbig +/parenleftbigˇH∗ 0·ˇB0−ˇH0·ˇB∗ 0/parenrightbig/bracketrightbig dV=0. (4.166) This implies a relationship between ˇE0,ˇD0,ˇB0, and ˇH0. Since Vis arbitrary we see that one possible relationship is simply to have one of each pair (ˇE0,ˇD0)and(ˇH0,ˇB0)equal to zero. Then, by (4.163) and (4.164), ˇE0=0implies ˇB0=0, and ˇD0=0implies ˇH0=0. Thus ˇE1=ˇE2, etc., and the solution is unique throughout V. However, we cannot in general rule out more complicated relationships. The number of possibilities depends onthe additional constraints on the relationship between ˇE 0,ˇD0,ˇB0, and ˇH0that we must supply to describe the material supporting the ﬁeld — i.e., the constitutive relationships.For a simple medium described by ˜µ(ω)and ˜/epsilon1 c(ω), equation (4.166) becomes /integraldisplay V/parenleftbig |ˇE0|2[˜/epsilon1c(ˇω)−˜/epsilon1c∗(ˇω)]+|ˇH0|2[˜µ(ˇω)−˜µ∗(ˇω)]/parenrightbig dV=0 or /integraldisplay V/bracketleftbig |ˇE0|2˜/epsilon1c/prime/prime(ˇω)+|ˇH0|2˜µ/prime/prime(ˇω)/bracketrightbig dV=0. For a lossy medium, ˜/epsilon1c/prime/prime<0and ˜µ/prime/prime<0as shown in §4.5.1. So both terms in the integral must be negative. For the integral to be zero each term must vanish, requiring ˇE0=ˇH0=0, and uniqueness is guaranteed. When establishing more complicated constitutive relations we must be careful to ensure that they lead to a unique solution, and that the condition for uniqueness is understood.In the case above, the assumption ˆn×ˇE 0/vextendsingle/vextendsingle S=0implies that the tangential components of ˇE1and ˇE2are identical over S— that is, we must give speciﬁc values of these quantities onSto ensure uniqueness. A similar statement holds for the condition ˆn×ˇH0/vextendsingle/vextendsingle S=0. In summary, the conditions for the ﬁelds within a region Vcontaining lossy isotropic materials to be unique are as follows: 1. the sources within Vmust be speciﬁed; 2. the tangential component of the electric ﬁeld must be speciﬁed over all or part of the bounding surface S; 3. the tangential component of the magnetic ﬁeld must be speciﬁed over the remainder ofS. We may question the requirement of a lossy medium to demonstrate uniqueness of the phasor ﬁelds. Does this mean that within a vacuum the speciﬁcation of tangential ﬁeldsis insuﬃcient? Experience shows that the ﬁelds in such a region are indeed properlydescribed by the surface ﬁelds, and it is just a case of the mathematical model being slightly out of sync with the physics. As long as we recognize that the sinusoidal steadystate requires an initial transient period, we know that speciﬁcation of the tangentialﬁelds is suﬃcient. We must be careful, however, to understand the restrictions of themathematical model. Any attempt to describe the ﬁelds within a lossless cavity, forinstance, is fraught with diﬃculty if true time-harmonic ﬁelds are used to model theactual physical ﬁelds. A helpful mathematical strategy is to think of free space as thelimit of a lossy medium as the loss recedes to zero. Of course this does not representthe physical state of “empty” space. Although even interstellar space may have a fewparticles for every cubic meter to interact with the electromagnetic ﬁeld, the density ofthese particles invalidates our initial macroscopic assumptions. Another important concern is whether we can extend the uniqueness argument to all of space. If we let Srecede to inﬁnity, must we continue to specify the ﬁelds over S,o r is it suﬃcient to merely specify the sources within S? Since the boundary ﬁelds provide information to the internal region about sources that exist outside S, it is sensible to assume that as S→∞ there are no sources external to Sand thus no need for the boundary ﬁelds. This is indeed the case. If all sources are localized, the ﬁelds theyproduce behave in just the right manner for the surface integral in (4.165) to vanish, andthus uniqueness is again guaranteed. Later we will ﬁnd that the electric and magneticﬁelds produced by a localized source at great distance have the form of a spherical wave: ˇE∼ˇH∼e −jkr r. If space is taken to be slightly lossy, then kis complex with negative imaginary part, and thus the ﬁelds decrease exponentially with distance from the source. As we argued above,it may not be physically meaningful to assume that space is lossy. Sommerfeld postulatedthat even for lossless space the surface integral in (4.165) vanishes as S→∞. This has been veriﬁed experimentally, and provides the following restrictions on the free-spaceﬁelds known as the Sommerfeld radiation condition : lim r→∞r/bracketleftbig η0ˆr×ˇH(r)+ˇE(r)/bracketrightbig =0, (4.167) lim r→∞r/bracketleftbigˆr×ˇE(r)−η0ˇH(r)/bracketrightbig =0, (4.168) where η0=(µ0//epsilon10)1/2. Later we shall see how these expressions arise from the integral solutions to Maxwell’s equations. 4.10.2 Reciprocity revisited In§2.9.3 we discussed the basic concept of reciprocity, but were unable to examine its real potential since we had not yet developed the theory of time-harmonic ﬁelds. Inthis section we shall apply the reciprocity concept to time-harmonic sources and ﬁelds,and investigate the properties a material must display to be reciprocal. The general form of the reciprocity theorem. As in §2.9.3, we consider a closed surface Senclosing a volume V. Sources of an electromagnetic ﬁeld are located either inside or outside S. Material media may lie within S, and their properties are described in terms of the constitutive relations. To obtain the time-harmonic (phasor) form of thereciprocity theorem we proceed as in §2.9.3 but begin with the phasor forms of Maxwell’s equations. We ﬁnd ∇·(ˇE a×ˇHb−ˇEb×ˇHa)=jˇω[ˇHa·ˇBb−ˇHb·ˇBa]−jˇω[ˇEa·ˇDb−ˇEb·ˇDa]+ +[ˇEb·ˇJa−ˇEa·ˇJb−ˇHb·ˇJma+ˇHa·ˇJmb], (4.169) where (ˇEa,ˇDa,ˇBa,ˇHa)are the ﬁelds produced by the phasor sources (ˇJa,ˇJma)and(ˇEb,ˇDb,ˇBb,ˇHb) are the ﬁelds produced by an independent set of sources (ˇJb,ˇJmb). As in §2.9.3, we are interested in the case in which the ﬁrst two terms on the right- hand side of (4.169) are zero. To see the conditions under which this might occur, wesubstitute the constitutive equations for a bianisotropic medium ˇD=˜¯ξ·ˇH+˜¯/epsilon1·ˇE, ˇB=˜¯µ·ˇH+˜¯ζ·ˇE, into (4.169), where each of the constitutive parameters is evaluated at ˇω. Setting the two terms to zero gives jˇω/bracketleftBig ˇH a·/parenleftBig ˜¯µ·ˇHb+˜¯ζ·ˇEb/parenrightBig −ˇHb·/parenleftBig ˜¯µ·ˇHa+˜¯ζ·ˇEa/parenrightBig/bracketrightBig − −jˇω/bracketleftBig ˇEa·/parenleftBigˇ¯ξ·ˇHb+˜¯/epsilon1·ˇEb/parenrightBig −ˇEb·/parenleftBig˜¯ξ·ˇHa+˜¯/epsilon1·ˇEa/parenrightBig/bracketrightBig =0, which holds if ˇHa·˜¯µ·ˇHb−ˇHb·˜¯µ·ˇHa=0, ˇHa·˜¯ζ·ˇEb+ˇEb·˜¯ξ·ˇHa=0, ˇEa·˜¯ξ·ˇHb+ˇHb·˜¯ζ·ˇEa=0, ˇEa·˜¯/epsilon1·ˇEb−ˇEb·˜¯/epsilon1·ˇEa=0. These in turn hold if ˜¯/epsilon1=˜¯/epsilon1T, ˜¯µ=˜¯µT,˜¯ξ=−˜¯ζT,˜¯ζ=−˜¯ξT. (4.170) These are the conditions for a reciprocal medium . For example, an anisotropic dielectric is a reciprocal medium if its permittivity dyadic is symmetric. An isotropic mediumdescribed by scalar quantities µand/epsilon1is certainly reciprocal. In contrast, lossless Gy- rotropic media are nonreciprocal since the constitutive parameters obey ˜¯/epsilon1=˜¯/epsilon1 †or˜¯µ=˜¯µ† rather than ˜¯/epsilon1=˜¯/epsilon1Tor˜¯µ=˜¯µT. For a reciprocal medium (4.169) reduces to ∇·(ˇEa×ˇHb−ˇEb×ˇHa)=/bracketleftbigˇEb·ˇJa−ˇEa·ˇJb−ˇHb·ˇJma+ˇHa·ˇJmb/bracketrightbig . (4.171) At points where the sources are zero, or are conduction currents described entirely by Ohm’s law ˇJ=σˇE,w eh a v e ∇·(ˇEa×ˇHb−ˇEb×ˇHa)=0, (4.172) known as Lorentz’s lemma . If we integrate (4.171) over Vand use the divergence theorem we obtain /contintegraldisplay S/bracketleftbigˇEa×ˇHb−ˇEb×ˇHa/bracketrightbig ·dS=/integraldisplay V/bracketleftbigˇEb·ˇJa−ˇEa·ˇJb−ˇHb·ˇJma+ˇHa·ˇJmb/bracketrightbig dV. (4.173) This is the general form of the Lorentz reciprocity theorem , and is valid when Vcontains reciprocal media as deﬁned in (4.170). Note that by an identical set of steps we ﬁnd that the frequency-domain ﬁelds obey an identical Lorentz lemma and reciprocity theorem. The condition for reciprocal systems. The quantity /angbracketleftˇfa,ˇgb/angbracketright=/integraldisplay V/bracketleftbigˇEa·ˇJb−ˇHa·ˇJmb/bracketrightbig dV is called the reaction between the source ﬁelds ˇgof set band the mediating ﬁelds ˇfof an independent set a. Note that ˇEa·ˇJbis not quite a power density, since the current lacks a complex conjugate. Using this reaction concept, ﬁrst introduced by Rumsey [161], wecan write (4.173) as /angbracketleftˇf b,ˇga/angbracketright−/angbracketleft ˇfa,ˇgb/angbracketright=/contintegraldisplay S/bracketleftbigˇEa×ˇHb−ˇEb×ˇHa/bracketrightbig ·dS. (4.174) We see that if there are no sources within Sthen /contintegraldisplay S/bracketleftbigˇEa×ˇHb−ˇEb×ˇHa/bracketrightbig ·dS=0. (4.175) Whenever (4.175) holds we say that the “system” within Sisreciprocal . Thus, for instance, a region of empty space is a reciprocal system. A system need not be source-free in order for (4.175) to hold. Suppose the relationship between ˇEand ˇHonSis given by the impedance boundary condition ˇEt=− Z(ˆn×ˇH), (4.176) where ˇEtis the component of ˇEtangential to Sso that ˆn×E=ˆn×Et, and the complex wall impedance Zmay depend on position. By (4.176) we can write (ˇEa×ˇHb−ˇEb×ˇHa)·ˆn=ˇHb·(ˆn×ˇEa)−ˇHa·(ˆn×ˇEb) =− ZˇHb·[ˆn×(ˆn×ˇHa)]+ZˇHa·[ˆn×(ˆn×ˇHb)]. Since ˆn×(ˆn×ˇH)=ˆn(ˆn·ˇH)−ˇH, the right-hand side vanishes. Hence (4.175) still holds even though there are sources within S. The reaction theorem. When sources lie within the surface S, and the ﬁelds on S obey (4.176), we obtain an important corollary of the Lorentz reciprocity theorem. Wehave from (4.174) the additional result /angbracketleftˇf a,ˇgb/angbracketright−/angbracketleft ˇfb,ˇga/angbracketright=0. Hence a reciprocal system has /angbracketleftˇfa,ˇgb/angbracketright=/angbracketleft ˇfb,ˇga/angbracketright (4.177) (which holds even if there are no sources within S, since then the reactions would be identically zero). This condition for reciprocity is sometimes called the reaction theorem and has an important physical meaning which we shall explore below in the form ofthe Rayleigh–Carson reciprocity theorem. Note that in obtaining this relation we mustassume that the medium is reciprocal in order to eliminate the terms in (4.169). Thus,in order for a system to be reciprocal, it must involve botha reciprocal medium and a boundary over which (4.176) holds. It is important to note that the impedance boundary condition (4.176) is widely appli- cable. If Z→0, then the boundary condition is that for a PEC: ˆn×ˇE=0.I fZ→∞,a PMC is described: ˆn×ˇH=0. Suppose Srepresents a sphere of inﬁnite radius. We know from (4.168) that if the sources and material media within Sare spatially ﬁnite, the ﬁelds far removed from these sources are described by the Sommerfeld radiation condition ˆr×ˇE=η 0ˇH where ˆris the radial unit vector of spherical coordinates. This condition is of the type (4.176) since ˆr=ˆnonS, hence the unbounded region that results from Sreceding to inﬁnity is also reciprocal. Summary of reciprocity for reciprocal systems. We can summarize reciprocity as follows. Unbounded space containing sources and materials of ﬁnite size is a reciprocal system if the media are reciprocal; a bounded region of space is a reciprocal system only if the materials within are reciprocal and the boundary ﬁelds obey (4.176), or if the region is source-free. In each of these cases /contintegraldisplay S/bracketleftbigˇEa×ˇHb−ˇEb×ˇHa/bracketrightbig ·dS=0 (4.178) and /angbracketleftˇfa,ˇgb/angbracketright−/angbracketleft ˇfb,ˇga/angbracketright=0. (4.179) Rayleigh–Carson reciprocity theorem. The physical meaning behind reciprocity can be made clear with a simple example. Consider two electric Hertzian dipoles, eachoscillating with frequency ˇωand located within an empty box consisting of PEC walls. These dipoles can be described in terms of volume current density as ˇJ a(r)=ˇIaδ(r−r/prime a), ˇJb(r)=ˇIbδ(r−r/prime b). Since the ﬁelds on the surface obey (4.176) (speciﬁcally, ˆn×ˇE=0), and since the medium within the box is empty space (a reciprocal medium), the ﬁelds produced by the sourcesmust obey (4.179). We have /integraldisplay VˇEb(r)·/bracketleftbigˇIaδ(r−r/prime a)/bracketrightbig dV=/integraldisplay VˇEa(r)·/bracketleftbigˇIbδ(r−r/prime b)/bracketrightbig dV, hence ˇIa·ˇEb(r/prime a)=ˇIb·ˇEa(r/prime b). (4.180) This is the Rayleigh–Carson reciprocity theorem . It also holds for two Hertzian dipoles located in unbounded free space, because in that case the Sommerfeld radiation conditionsatisﬁes (4.176). As an important application of this principle, consider a closed PEC body located in free space. Reciprocity holds in the region external to the body since we have ˆn×ˇE=0 at the boundary of the perfect conductor and the Sommerfeld radiation condition on theboundary at inﬁnity. Now let us place dipole asomewhere external to the body, and dipole badjacent and tangential to the perfectly conducting body. We regard dipole a as the source of an electromagnetic ﬁeld and dipole bas “sampling” that ﬁeld. Since the tangential electric ﬁeld is zero at the surface of the conductor, the reaction between thetwo dipoles is zero. Now let us switch the roles of the dipoles so that bis regarded as the source and ais regarded as the sampler. By reciprocity the reaction is again zero and thus there is no ﬁeld produced by bat the position of a. Now the position and orientation of aare arbitrary, so we conclude that an impressed electric source current placed tangentially to a perfectly conducting body produces no ﬁeld external to the body.This result is used in Chapter 6 to develop a ﬁeld equivalence principle useful in the studyof antennas and scattering. 4.10.3 Duality A duality principle analogous to that found for time-domain ﬁelds in §2.9.2 may be established for frequency-domain and time-harmonic ﬁelds. Consider a closed surface S enclosing a region of space that includes a frequency-domain electric source current ˜J and a frequency-domain magnetic source current ˜Jm. The ﬁelds ( ˜E1,˜D1,˜B1,˜H1) within the region (which may also contain arbitrary media) are described by ∇× ˜E1=− ˜Jm−jω˜B1, (4.181) ∇× ˜H1=˜J+jω˜D1, (4.182) ∇·˜D1=˜ρ, (4.183) ∇·˜B1=˜ρm. (4.184) Suppose we have been given a mathematical description of the sources (˜J,˜Jm)and have solved for the ﬁeld vectors (˜E1,˜D1,˜B1,˜H1). Of course, we must also have been supplied with a set of boundary values and constitutive relations in order to make the solutionunique. We note that if we replace the formula for ˜Jwith the formula for ˜J min (4.182) (and ˜ρwith ˜ρmin (4.183)) and also replace ˜Jmwith−˜Jin (4.181) (and ˜ρmwith−˜ρin (4.184)) we get a new problem. However, the symmetry of the equations allows us to specify the solution immediately. The new set of curl equations requires ∇× ˜E2=˜J−jω˜B2, (4.185) ∇× ˜H2=˜Jm+jω˜D2. (4.186) If we can resolve the question of how the constitutive parameters must be altered to reﬂect these replacements, then we can conclude by comparing (4.185) with (4.182) and(4.186) with (4.181) that ˜E 2=˜H1, ˜B2=− ˜D1, ˜D2=˜B1, ˜H2=− ˜E1. The discussion regarding units in §2.9.2 carries over to the present case. Multiplying Ampere’s law by η0=(µ0//epsilon10)1/2, we have ∇× ˜E=− ˜Jm−jω˜B, ∇×(η0˜H)=(η0˜J)+jω(η 0˜D). Thus if the original problem has solution (˜E1,η0˜D1,˜B1,η0˜H1), then the dual problem with ˜Jreplaced by ˜Jm/η0and ˜Jmreplaced by −η0˜Jhas solution ˜E2=η0˜H1, (4.187) ˜B2=−η0˜D1, (4.188) η0˜D2=˜B1, (4.189) η0˜H2=− ˜E1. (4.190) As with duality in the time domain, the constitutive parameters for the dual problem must be altered from those of the original problem. For linear anisotropic media we havefrom (4.13) and (4.14) the constitutive relationships ˜D 1=˜¯/epsilon11·˜E1, (4.191) ˜B1=˜¯µ1·˜H1, (4.192) for the original problem, and ˜D2=˜¯/epsilon12·˜E2, (4.193) ˜B2=˜¯µ2·˜H2, (4.194) for the dual problem. Substitution of (4.187)–(4.190) into (4.191) and (4.192) gives ˜D2=/parenleftbigg˜¯µ1 η2 0/parenrightbigg ·˜E2, (4.195) ˜B2=/parenleftbig η2 0˜¯/epsilon11/parenrightbig ·˜H2. (4.196) Comparing (4.195) with (4.193) and (4.196) with (4.194), we conclude that ˜¯µ2=η2 0˜¯/epsilon11, ˜¯/epsilon12=˜¯µ1/η2 0. (4.197) For a linear, isotropic medium speciﬁed by ˜/epsilon1and ˜µ, the dual problem is obtained by replacing ˜/epsilon1rwith ˜µrand ˜µrwith ˜/epsilon1r. The solution to the dual problem is then ˜E2=η0˜H1,η 0˜H2=− ˜E1, as before. The medium in the dual problem must have electric properties numerically equal to the magnetic properties of the medium in the original problem, and magneticproperties numerically equal to the electric properties of the medium in the originalproblem. Alternatively we may divide Ampere’s law by η=(˜µ/˜/epsilon1) 1/2instead of η0. Then the dual problem has ˜Jreplaced by ˜Jm/η, and ˜Jmreplaced by −η˜J, and the solution is ˜E2=η˜H1,η ˜H2=− ˜E1. (4.198) There is no need to swap ˜/epsilon1rand ˜µrsince information about these parameters is incor- porated into the replacement sources. We may also apply duality to a problem where we have separated the impressed and secondary sources. In a homogeneous, isotropic, conducting medium we may let ˜J= ˜Ji+˜σ˜E. With this the curl equations become ∇×η˜H=η˜Ji+jωη˜/epsilon1c˜E, ∇× ˜E=− ˜Jm−jω˜µ˜H. The solution to the dual problem is again given by (4.198), except that now η=(˜µ/˜/epsilon1c)1/2. As we did near the end of §2.9.2, we can consider duality in a source-free region. We letSenclose a source-free region of space and, for simplicity, assume that the medium within Sis linear, isotropic, and homogeneous. The ﬁelds within Sare described by ∇× ˜E1=− jω˜µ˜H1, ∇×η˜H1=jω˜/epsilon1η˜E1, ∇·˜/epsilon1˜E1=0, ∇·˜µ˜H1=0. The symmetry of the equations is such that the mathematical form of the solution for ˜E is the same as that for η˜H. Since the ﬁelds ˜E2=η˜H1, ˜H2=− ˜E1/η, also satisfy Maxwell’s equations, the dual problem merely involves replacing ˜Ebyη˜H and ˜Hby−˜E/η. 4.11 The wavenature of the time-harmonic EM ﬁeld Time-harmonic electromagnetic waves have been studied in great detail. Narrowband waves are widely used for signal transmission, heating, power transfer, and radar. Theyshare many of the properties of more general transient waves, and the discussions of§2.10.1 are applicable. Here we shall investigate some of the unique properties of time- harmonic waves and introduce such fundamental quantities as wavelength, phase andgroup velocity, and polarization. 4.11.1 The frequency-domain waveequation We begin by deriving the frequency-domain wave equation for dispersive bianisotropic materials. A solution to this equation may be viewed as the transform of a general time-dependent ﬁeld. If one speciﬁc frequency is considered the time-harmonic solutionis produced. In§2.10.2 we derived the time-domain wave equation for bianisotropic materials. There it was necessary to consider only time-independent constitutive parameters. Wecan overcome this requirement, and thus deal with dispersive materials, by using a Fouriertransform approach. We solve a frequency-domain wave equation that includes the fre-quency dependence of the constitutive parameters, and then use an inverse transform toreturn to the time domain. The derivation of the equation parallels that of §2.10.2. We substitute the frequency- domain constitutive relationships ˜D=˜¯/epsilon1·˜E+˜¯ξ·˜H, ˜B=˜¯ζ·˜E+˜¯µ·˜H, into Maxwell’s curl equations (4.7) and (4.8) to get the coupled diﬀerential equations ∇× ˜E=− jω[˜¯ζ·˜E+˜¯µ·˜H]−˜J m, ∇× ˜H=jω[˜¯/epsilon1·˜E+˜¯ξ·˜H]+˜J, for˜Eand ˜H. Here we have included magnetic sources ˜Jmin Faraday’s law. Using the dyadic operator ¯∇deﬁned in (2.308) we can write these equations as /parenleftBig ¯∇+ jω˜¯ζ/parenrightBig ·˜E=− jω˜¯µ·˜H−˜Jm, (4.199) /parenleftBig ¯∇− jω˜¯ξ/parenrightBig ·˜H=jω˜¯/epsilon1·˜E+˜J. (4.200) We can obtain separate equations for ˜Eand ˜Hby deﬁning the inverse dyadics ˜¯/epsilon1·˜¯/epsilon1−1=¯I, ˜¯µ·˜¯µ−1=¯I. Using ˜¯µ−1we can write (4.199) as −jω˜H=˜¯µ−1·/parenleftBig ¯∇+ jω˜¯ζ/parenrightBig ·˜E+˜¯µ−1·˜Jm. Substituting this into (4.200) we get /bracketleftBig/parenleftBig ¯∇− jω˜¯ξ/parenrightBig ·˜¯µ−1·/parenleftBig ¯∇+ jω˜¯ζ/parenrightBig −ω2˜¯/epsilon1/bracketrightBig ·˜E=−/parenleftBig ¯∇− jω˜¯ξ/parenrightBig ·˜¯µ−1·˜Jm−jω˜J.(4.201) This is the general frequency-domain wave equation for ˜E. Using ˜¯/epsilon1−1we can write (4.200) as jω˜E=˜¯/epsilon1−1·/parenleftBig ¯∇− jω˜¯ξ/parenrightBig ·˜H−˜¯/epsilon1−1·˜J. Substituting this into (4.199) we get /bracketleftBig/parenleftBig ¯∇+ jω˜¯ζ/parenrightBig ·˜¯/epsilon1−1·/parenleftBig ¯∇− jω˜¯ξ/parenrightBig −ω2˜¯µ/bracketrightBig ·˜H=/parenleftBig ¯∇+ jω˜¯ζ/parenrightBig ·˜¯/epsilon1−1·˜J−jω˜Jm.(4.202) This is the general frequency-domain waveequation for ˜H. Wave equation for a homogeneous, lossy, isotropic medium. We may specialize (4.201) and (4.202) to the case of a homogeneous, lossy, isotropic medium by setting ˜¯ζ=˜¯ξ=0,˜¯µ=˜µ¯I,˜¯/epsilon1=˜/epsilon1¯I, and ˜J=˜Ji+˜Jc: ∇×(∇× ˜E)−ω2˜µ˜/epsilon1˜E=− ∇× ˜Jm−jω˜µ(˜Ji+˜Jc), (4.203) ∇×(∇× ˜H)−ω2˜µ˜/epsilon1˜H=∇× (˜Ji+˜Jc)−jω˜/epsilon1˜Jm. (4.204) Using (B.47) and using Ohm’s law ˜Jc=˜σ˜Eto describe the secondary current, we get from (4.203) ∇(∇·˜E)−∇2˜E−ω2˜µ˜/epsilon1˜E=− ∇× ˜Jm−jω˜µ˜Ji−jω˜µ˜σ˜E which, using ∇·˜E=˜ρ/˜/epsilon1, can be simpliﬁed to (∇2+k2)˜E=∇× ˜Jm+jω˜µ˜Ji+1 ˜/epsilon1∇˜ρ. (4.205) This is the vector Helmholtz equation for˜E. Here kis the complex wavenumber deﬁned through k2=ω2˜µ˜/epsilon1−jω˜µ˜σ=ω2˜µ/bracketleftbigg ˜/epsilon1+˜σ jω/bracketrightbigg =ω2˜µ˜/epsilon1c(4.206) where ˜/epsilon1cis the complex permittivity (4.26). By (4.204) we have ∇(∇·˜H)−∇2˜H−ω2˜µ˜/epsilon1˜H=∇× ˜Ji+∇× ˜Jc−jω˜/epsilon1˜Jm. Using ∇× ˜Jc=∇× (˜σ˜E)=˜σ∇× ˜E=˜σ(−jω˜B−˜Jm) and∇·˜H=˜ρm/˜µwe then get (∇2+k2)˜H=− ∇× ˜Ji+jω˜/epsilon1c˜Jm+1 ˜µ∇˜ρm, (4.207) which is the vector Helmholtz equation for ˜H. 4.11.2 Field relationships and the waveequation for two-dimensional ﬁelds Many important canonical problems are two-dimensional in nature, with the sources and ﬁelds invariant along one direction. Two-dimensional ﬁelds have a simple structure compared to three-dimensional ﬁelds, and this structure often allows a decomposition into even simpler ﬁeld structures. Consider a homogeneous region of space characterized by the permittivity ˜/epsilon1, perme- ability ˜µ, and conductivity ˜σ. We assume that all sources and ﬁelds are z-invariant, and wish to ﬁnd the relationship between the various components of the frequency-domainﬁelds in a source-free region. It is useful to deﬁne the transverse vector component of anarbitrary vector Aas the component of Aperpendicular to the axis of invariance: A t=A−ˆz(ˆz·A). For the position vector r, this component is the transverse position vector rt=ρ.Fo r instance we have ρ=ˆxx+ˆyy, ρ=ˆρρ, in the rectangular and cylindrical coordinate systems, respectively. Because the region is source-free, the ﬁelds ˜Eand ˜Hobey the homogeneous Helmholtz equations (∇2+k2)/braceleftbigg˜E ˜H/bracerightbigg =0. Writing the ﬁelds in terms of rectangular components, we ﬁnd that each component must obey a homogeneous scalar Helmholtz equation. In particular, we have for theaxial components ˜E zand ˜Hz, (∇2+k2)/braceleftbigg˜Ez ˜Hz/bracerightbigg =0. But since the ﬁelds are independent of zwe may also write (∇2 t+k2)/braceleftbigg˜Ez ˜Hz/bracerightbigg =0 (4.208) where ∇2 tis the transverse Laplacian operator ∇2 t=∇2−ˆz∂2 ∂z2. (4.209) In rectangular coordinates we have ∇2 t=∂2 ∂x2+∂2 ∂y2, while in circular cylindrical coordinates ∇2 t=∂2 ∂ρ2+1 ρ∂ ∂ρ+1 ρ2∂2 ∂φ2. (4.210) With our condition on z-independence we can relate the transverse ﬁelds ˜Etand ˜Htto ˜Ezand ˜Hz. By Faraday’s law we have ∇× ˜E(ρ,ω)=− jω˜µ˜H(ρ,ω) and thus ˜Ht=−1 jω˜µ/bracketleftbig ∇× ˜E/bracketrightbig t. The transverse portion of the curl is merely /bracketleftbig ∇× ˜E/bracketrightbig t=ˆx/bracketleftbigg∂˜Ez ∂y−∂˜Ey ∂z/bracketrightbigg +ˆy/bracketleftbigg∂˜Ex ∂z−∂˜Ez ∂x/bracketrightbigg =− ˆz×/bracketleftbigg ˆx∂˜Ez ∂x+ˆy∂˜Ez ∂y/bracketrightbigg since the derivatives with respect to zvanish. The term in brackets is the transverse gradient of ˜Ez, where the transverse gradient operator is ∇t=∇− ˆz∂ ∂z. In circular cylindrical coordinates this operator becomes ∇t=ˆρ∂ ∂ρ+ˆφ1 ρ∂ ∂φ. (4.211) Thus we have ˜Ht(ρ,ω)=1 jω˜µˆz×∇ t˜Ez(ρ,ω) . Similarly, the source-free Ampere’s law yields ˜Et(ρ,ω)=−1 jω˜/epsilon1cˆz×∇ t˜Hz(ρ,ω) . These results suggest that we can solve a two-dimensional problem by superposition. We ﬁrst consider the case where ˜Ez/negationslash=0and ˜Hz=0, called electric polarization . This case is also called TMortransverse magnetic polarization because the magnetic ﬁeld is transverse to the z-direction ( TM z). We have (∇2 t+k2)˜Ez=0, ˜Ht(ρ,ω)=1 jω˜µˆz×∇ t˜Ez(ρ,ω) . (4.212) Once we have solved the Helmholtz equation for ˜Ez, the remaining ﬁeld components follow by simple diﬀerentiation. We next consider the case where ˜Hz/negationslash=0and ˜Ez=0. This is the case of magnetic polarization , also called TEortransverse electric polarization (TE z). In this case (∇2 t+k2)˜Hz=0, ˜Et(ρ,ω)=−1 jω˜/epsilon1cˆz×∇ t˜Hz(ρ,ω) . (4.213) A problem involving both ˜Ezand ˜Hzis solved by adding the results for the individual TE zand TM zcases. Note that we can obtain the expression for the TE ﬁelds from the expression for the TM ﬁelds, and vice versa, using duality. For instance, knowing that the TM ﬁelds obey(4.212) we may replace ˜H twith ˜Et/ηand ˜Ezwith−η˜Hzto obtain ˜Et(ρ,ω) η=1 jω˜µˆz×∇ t[−η˜Hz(ρ,ω)], which reproduces (4.213). 4.11.3 Plane waves in a homogeneous, isotropic, lossy material The plane-wave ﬁeld. In later sections we will solve the frequency-domain wave equation with an arbitrary source distribution. At this point we are more interested inthe general behavior of EM waves in the frequency domain, so we seek simple solutions to the homogeneous equation (∇ 2+k2)˜E(r,ω)=0 (4.214) that governs the ﬁelds in source-free regions of space. Here [k(ω)]2=ω2˜µ(ω) ˜/epsilon1c(ω). Many properties of plane waves are best understood by considering the behavior of a monochromatic ﬁeld oscillating at a single frequency ˇω. In these cases we merely make the replacements ω→ˇω, ˜E(r,ω)→ˇE(r), and apply the rules developed in §4.7 for the manipulation of phasor ﬁelds. For our ﬁrst solutions we choose those that demonstrate rectangular symmetry. Plane waves have planar spatial phase loci. That is, the spatial surfaces over which the phase of the complex frequency-domain ﬁeld is constant are planes. Solutions of this type maybe obtained using separation of variables in rectangular coordinates. Writing ˜E(r,ω)=ˆx˜E x(r,ω)+ˆy˜Ey(r,ω)+ˆz˜Ez(r,ω) we ﬁnd that (4.214) reduces to three scalar equations of the form (∇2+k2)˜ψ(r,ω)=0 where ˜ψis representative of ˜Ex,˜Ey, and ˜Ez. This is called the homogeneous scalar Helmholtz equation. Product solutions to this equation are considered in §A.4. In rectangular coordinates ˜ψ(r,ω)=X(x,ω)Y(y,ω)Z(z,ω) where X,Y, and Zare chosen from the list (A.102). Since the exponentials describe propagating wavefunctions, we choose ˜ψ(r,ω)=A(ω)e±jkx(ω)xe±jky(ω)ye±jkz(ω)z where Ais the amplitude spectrum of the plane wave and k2 x+k2 y+k2 z=k2. Using this solution to represent each component of ˜E, we have a propagating- wavesolution to the homogeneous vector Helmholtz equation: ˜E(r,ω)=˜E0(ω)e±jkx(ω)xe±jky(ω)ye±jkz(ω)z, (4.215) where E0(ω)is the vector amplitude spectrum. If we deﬁne the wave vector k(ω)=ˆxkx(ω)+ˆyky(ω)+ˆzkz(ω), then we can write (4.215) as ˜E(r,ω)=˜E0(ω)e−jk(ω)·r. (4.216) Note that we choose the negative sign in the exponential function and allow the vector components of kto be either positive or negative as required by the physical nature of a speciﬁc problem. Also note that the magnitude of the wave vector is the wavenumber:|k|=k. We may always write the wave v ector as a sum of real and imaginary vector components k=k /prime+jk/prime/prime(4.217) which must obey k·k=k2=k/prime2−k/prime/prime2+2jk/prime·k/prime/prime. (4.218) When the real and imaginary components are collinear, (4.216) describes a uniform plane wave with k=ˆk(k/prime+jk/prime/prime). When k/primeand k/prime/primehave diﬀerent directions, (4.216) describes a nonuniform plane wave . Weshallﬁndin §4.13that anyfrequency-domai nelectromagneti cﬁeldinfreespace may b ereprese ntedasacontinuoussuperpositionofelementalplane-wavecomponentsof the type(4.216) ,butthatbothunifor mandnonunifor mtermsarerequired. The TEM nature of a uniform plane wave. Given the plane-wave solution to the wave equation for the electric ﬁeld, it is straightforward to ﬁnd the magnetic ﬁeld.Substitution of (4.216) into Faraday’s law gives ∇×/bracketleftbig˜E 0(ω)e−jk(ω)·r/bracketrightbig =− jω˜B(r,ω) . Computation of the curl is straightforward and easily done in rectangular coordinates. This and similar derivatives often appear when manipulating plane-wave solutions; seethe tabulation in Appendix B, By (B.78) we have ˜H=k×˜E ω˜µ. (4.219) Taking the cross product of this expression with k, we also have k×˜H=k×(k×˜E) ω˜µ=k(k·˜E)−˜E(k·k) ω˜µ. (4.220) We can show that k·˜E=0by examining Gauss’ law and employing (B.77): ∇·˜E=− jk·˜Ee−jk·r=˜ρ ˜/epsilon1=0. (4.221) Using this and k·k=k2=ω2˜µ˜/epsilon1c, we obtain from (4.220) ˜E=−k×˜H ω˜/epsilon1c. (4.222) Now for a uniform plane wave k=ˆkk, so we can also write (4.219) as ˜H=ˆk×˜E η=ˆk×˜E0 ηe−jk·r(4.223) and (4.222) as ˜E=−ηˆk×˜H. Here η=ω˜µ k=/radicalbigg ˜µ ˜/epsilon1c is the complex intrinsic impedance of the medium. Equations (4.223) and (4.221) show that the electric and magnetic ﬁelds and the wave vector are mutually orthogonal. The wave is said to be transverse electromagnetic or TEM to the direction of propagation. The phase and attenuation constants of a uniform plane wave. For a uniform plane wave we may write k=k/primeˆk+jk/prime/primeˆk=kˆk=(β−jα)ˆk where k/prime=βand k/prime/prime=−α. Here αis called the attenuation constant andβis the phase constant . Since kis deﬁned through (4.206), we have k2=(β−jα)2=β2−2jαβ−α2=ω2˜µ˜/epsilon1c=ω2(˜µ/prime+j˜µ/prime/prime)(˜/epsilon1c/prime+j˜/epsilon1c/prime/prime). Equating real and imaginary parts we have β2−α2=ω2[˜µ/prime˜/epsilon1c/prime−˜µ/prime/prime˜/epsilon1c/prime/prime], −2αβ=ω2[˜µ/prime/prime˜/epsilon1c/prime+˜µ/prime˜/epsilon1c/prime/prime]. We assume the material is passive so that ˜µ/prime/prime≤0,˜/epsilon1c/prime/prime≤0. Letting β2−α2=ω2[˜µ/prime˜/epsilon1c/prime−˜µ/prime/prime˜/epsilon1c/prime/prime]=A, 2αβ=ω2[|˜µ/prime/prime|˜/epsilon1c/prime+˜µ/prime|˜/epsilon1c/prime/prime|]=B, we may solve simultaneously to ﬁnd that β2=1 2/bracketleftBig A+/radicalbig A2+B2/bracketrightBig ,α2=1 2/bracketleftBig −A+/radicalbig A2+B2/bracketrightBig . Since A2+B2=ω4(˜/epsilon1c/prime2+˜/epsilon1c/prime/prime2)(˜µ/prime2+˜µ/prime/prime2),w eh a v e β=ω/radicalbig ˜µ/prime˜/epsilon1c/prime/radicaltp/radicalvertex/radicalvertex/radicalbt1 2/bracketleftBigg/radicalBigg/parenleftbigg 1+˜/epsilon1c/prime/prime2 ˜/epsilon1c/prime2/parenrightbigg/parenleftbigg 1+˜µ/prime/prime2 ˜µ/prime2/parenrightbigg +/parenleftbigg 1−˜µ/prime/prime ˜µ/prime˜/epsilon1c/prime/prime ˜/epsilon1c/prime/parenrightbigg/bracketrightBigg , (4.224) α=ω/radicalbig ˜µ/prime˜/epsilon1c/prime/radicaltp/radicalvertex/radicalvertex/radicalbt1 2/bracketleftBigg/radicalBigg/parenleftbigg 1+˜/epsilon1c/prime/prime2 ˜/epsilon1c/prime2/parenrightbigg/parenleftbigg 1+˜µ/prime/prime2 ˜µ/prime2/parenrightbigg −/parenleftbigg 1−˜µ/prime/prime ˜µ/prime˜/epsilon1c/prime/prime ˜/epsilon1c/prime/parenrightbigg/bracketrightBigg , (4.225) where ˜/epsilon1cand ˜µare functions of ω.I f ˜/epsilon1(ω)=/epsilon1,˜µ(ω)=µ, and ˜σ(ω)=σare real and frequency independent, then α=ω√µ/epsilon1/radicaltp/radicalvertex/radicalvertex/radicalbt1 2/bracketleftBigg/radicalbigg 1+/parenleftBigσ ω/epsilon1/parenrightBig2 −1/bracketrightBigg , (4.226) β=ω√µ/epsilon1/radicaltp/radicalvertex/radicalvertex/radicalbt1 2/bracketleftBigg/radicalbigg 1+/parenleftBigσ ω/epsilon1/parenrightBig2 +1/bracketrightBigg . (4.227) These values of αandβare valid for ω> 0. For negative frequencies we must be more careful in evaluating the square root in k=ω(˜µ˜/epsilon1c)1/2. Writing ˜µ(ω)=˜µ/prime(ω)+j˜µ/prime/prime(ω)=|˜µ(ω)|ejξµ(ω), ˜/epsilon1c(ω)=˜/epsilon1c/prime(ω)+j˜/epsilon1c/prime/prime(ω)=|˜/epsilon1c(ω)|ejξ/epsilon1(ω), we have k(ω)=β(ω)−jα(ω)=ω/radicalbig ˜µ(ω) ˜/epsilon1c(ω)=ω/radicalbig |˜µ(ω)||˜/epsilon1c(ω)|ej1 2[ξµ(ω)+ξ/epsilon1(ω)]. Now for passive materials we must have, by (4.48), ˜µ/prime/prime<0and ˜/epsilon1c/prime/prime<0forω> 0. Since we also have ˜µ/prime>0and ˜/epsilon1c/prime>0forω> 0, we ﬁnd that −π/2<ξµ<0and −π/2<ξ/epsilon1<0, and thus −π/2<( ξµ+ξ/epsilon1)/2<0. Thus we must have β> 0andα> 0 forω> 0.Fo r ω< 0we have by (4.44) and (4.45) that ˜µ/prime/prime>0,˜/epsilon1c/prime/prime>0,˜µ/prime>0, and ˜/epsilon1c/prime>0.T h u s π/2>( ξµ+ξ/epsilon1)/2>0, and so β< 0andα> 0forω< 0. In summary, α(ω)is an even function of frequency and β(ω)is an odd function of frequency: β(ω)=−β(−ω), α(ω) =α(−ω), (4.228) where β(ω) > 0,α( ω)> 0when ω> 0. From this we ﬁnd a condition on ˜E0in (4.216). Since by (4.47) we must have ˜E(ω)=˜E∗(−ω), we see that the uniform plane- waveﬁeld obeys ˜E0(ω)e[−jβ(ω)−α(ω) ]ˆk·r=˜E∗ 0(−ω)e[+jβ(−ω)−α(−ω)]ˆk·r or ˜E0(ω)=˜E∗ 0(−ω), sinceβ(−ω)=−β(ω)andα(−ω)=α(ω). Propagation of a uniform plane wave: the group and phase velocities. We have derived the plane- wavesolution to the wave equation in the frequency domain, but can discover the wave nature of the solution only by examining its behavior in the timedomain. Unfortunately, the explicit form of the time-domain ﬁeld is highly dependent onthe frequency behavior of the constitutive parameters. Even the simplest case in which/epsilon1,µ, andσare frequency independent is quite complicated, as we discovered in §2.10.6. To overcome this diﬃculty, it is helpful to examine the behavior of a narrowband (butnon-monochromatic) signal in a lossy medium with arbitrary constitutive parameters.We will ﬁnd that the time-domain wave ﬁeld propagates as a disturbance through the surrounding medium with a velocity determined by the constitutive parameters of themedium. The temporal wave shape does not change as the wave propagates, but the amplitude of the waveattenuates at a rate dependent on the constitutive parameters. For clarity of presentation we shall assume a linearly polarized plane wave ( §??) with ˜E(r,ω)=ˆe˜E 0(ω)e−jk(ω)·r. (4.229) Here ˜E0(ω)is the spectrum of the temporal dependence of the wave. For the temporal dependence we choose the narrowband signal E0(t)=E0f(t)cos(ω0t) where f(t)has a narrowband spectrum centered about ω=0(and is therefore called a baseband signal ). An appropriate choice for f(t)is the Gaussian function used in (4.52): f(t)=e−a2t2↔˜F(ω)=/radicalbiggπ a2e−ω2 4a2, producing E0(t)=E0e−a2t2cos(ω0t). (4.230) We think of f(t)asmodulating the single-frequency cosine carrier wave , thus providing theenvelope . By using a large value of awe obtain a narrowband signal whose spectrum is centered about ±ω0. Later we shall let a→0, thereby driving the width of f(t)to inﬁnity and producing a monochromatic waveform. By (1) we have ˜E0(ω)=E01 2/bracketleftbig˜F(ω−ω0)+ ˜F(ω+ω0)/bracketrightbig where f(t)↔ ˜F(ω).AplotofthisspectrumisshowninFigure4.2.Weseethat the narrowband signal is centered at ω=±ω0. Substituting into (4.229) and using k=(β−jα)ˆkfor a uniform plane wave, we have the frequency-domain ﬁeld ˜E(r,ω)=ˆeE01 2/bracketleftBig ˜F(ω−ω0)e−j[β(ω)−jα(ω) ]ˆk·r+˜F(ω+ω0)e−j[β(ω)−jα(ω) ]ˆk·r/bracketrightBig .(4.231) The ﬁeld at any time tand position rcan now be found by inversion: ˆeE(r,t)=1 2π/integraldisplay∞ −∞ˆeE01 2/bracketleftBig ˜F(ω−ω0)e−j[β(ω)−jα(ω) ]ˆk·r+ +˜F(ω+ω0)e−j[β(ω)−jα(ω) ]ˆk·r/bracketrightBig ejωtdω. (4.232) We assume that β(ω)andα(ω)vary slowly within the band occupied by ˜E0(ω). With this assumption we can expand βandαnearω=ω0as β(ω)=β(ω 0)+β/prime(ω0)(ω−ω0)+1 2β/prime/prime(ω0)(ω−ω0)2+···, α(ω)=α(ω 0)+α/prime(ω0)(ω−ω0)+1 2α/prime/prime(ω0)(ω−ω0)2+···, where β/prime(ω)=dβ(ω)/ dω,β/prime/prime(ω)=d2β(ω)/ dω2, and so on. In a similar manner we can expand βandαnearω=−ω0: β(ω)=β(−ω0)+β/prime(−ω0)(ω+ω0)+1 2β/prime/prime(−ω0)(ω+ω0)2+···, α(ω)=α(−ω0)+α/prime(−ω0)(ω+ω0)+1 2α/prime/prime(−ω0)(ω+ω0)2+···. Since we are most interested in the propagation velocity, we need not approximate α with great accuracy, and thus use α(ω)≈α(±ω0)within the narrow band. We must consider βto greater accuracy to uncover the propagating nature of the wave, and thus use β(ω)≈β(ω 0)+β/prime(ω0)(ω−ω0) (4.233) nearω=ω0and β(ω)≈β(−ω0)+β/prime(−ω0)(ω+ω0) (4.234) nearω=−ω0. Substituting these approximations into (4.232) we ﬁnd ˆeE(r,t)=1 2π/integraldisplay∞ −∞ˆeE01 2/bracketleftBig ˜F(ω−ω0)e−j[β(ω 0)+β/prime(ω0)(ω−ω0)]ˆk·re−[α(ω 0)]ˆk·r+ +˜F(ω+ω0)e−j[β(−ω0)+β/prime(−ω0)(ω+ω0)]ˆk·re−[α(−ω0)]ˆk·r/bracketrightBig ejωtdω. (4.235) By (4.228) we know that αis even in ωandβis odd in ω. Since the derivative of an odd function is an even function, we also know that β/primeis even in ω. We can therefore write (4.235) as ˆeE(r,t)=ˆeE0e−α(ω 0)ˆk·r1 2π/integraldisplay∞ −∞1 2/bracketleftBig ˜F(ω−ω0)e−jβ(ω 0)ˆk·re−jβ/prime(ω0)(ω−ω0)ˆk·r+ +˜F(ω+ω0)ejβ(ω 0)ˆk·re−jβ/prime(ω0)(ω+ω0)ˆk·r/bracketrightBig ejωtdω. Multiplying and dividing by ejω0tand rearranging, we have ˆeE(r,t)=ˆeE0e−α(ω 0)ˆk·r1 2π/integraldisplay∞ −∞1 2/bracketleftbig˜F(ω−ω0)ejφej(ω−ω0)[t−τ]+ +˜F(ω+ω0)e−jφej(ω+ω0)[t−τ]/bracketrightbig dω where φ=ω0t−β(ω 0)ˆk·r,τ =β/prime(ω0)ˆk·r. Setting u=ω−ω0in the ﬁrst term and u=ω+ω0in the second term we have ˆeE(r,t)=ˆeE0e−α(ω 0)ˆk·rcosφ1 2π/integraldisplay∞ −∞˜F(u)eju(t−τ)du. Finally, the time-shifting theorem (A.3) gives us the time-domain wave ﬁeld ˆeE(r,t)=ˆeE0e−α(ω 0)ˆk·rcos/parenleftbig ω0/bracketleftbig t−ˆk·r/vp(ω0)/bracketrightbig/parenrightbig f/parenleftbig t−ˆk·r/vg(ω0)/parenrightbig (4.236) where vg(ω)=dω/dβ=[dβ/dω]−1(4.237) is called the group velocity and vp(ω)=ω/β is called the phase velocity . To interpret (4.236), we note that at any given time tthe ﬁeld is constant over the surface described by ˆk·r=C (4.238) where C issomeconsta nt.Thissurfac eisaplane,asshowninFigure4.10,withits normal along ˆk. It is easy to verify that any point ron this plane satisﬁes (4.238). Let r0=r0ˆkdescribe the point on the plane with position vector in the direction of ˆk, and letdbe a displacement vector from this point to any other point on the plane. Then ˆk·r=ˆk·(r0+d)=r0(ˆk·ˆk)+ˆk·d. But ˆk·d=0,s o ˆk·r=r0, (4.239) which is a ﬁxed distance, so (238) holds. Let us identify the plane over which the envelope ftakes on a certain value, and follow its motion as time progresses. The value of r0associated with this plane must increase with increasing time in such a way that the argument of fremains constant: t−r0/vg(ω0)=C. Figure 4.10: Surface of constant ˆk·r. Diﬀerentiation gives dr0 dt=vg=dω dβ. (4.240) So the envelope propagates along ˆkat a rate given by the group velocity vg. Associated with this propagation is an attenuation described by the factor e−α(ω 0)ˆk·r. This accounts for energy transfer into the lossy medium through Joule heating. Similarly, we can identify a plane over which the phase of the carrier is constant; this will be parallel to the plane of constant envelope described above. We now set ω0/bracketleftbig t−ˆk·r/vp(ω0)/bracketrightbig =C and diﬀerentiate to get dr0 dt=vp=ω β. (4.241) This shows that surfaces of constant carrier phase propagate along ˆkwith velocity vp. Caution must be exercised in interpreting the two velocities vgandvp; in particular, we must be careful not to associate the propagation velocities of energy or information withv p. Since envelope propagation represents the actual progression of the disturbance, vg hastherecognizabl ephysicalmeanin gofenergyvelocity.KrausandFleisch[105]suggest that we think of a strolling caterpillar: the speed ( vp) of the undulations along the caterpillar’s back (representing the carrier wave) may be much faster than the speed ( vg) of the caterpillar’s body (representing the envelope of the disturbance). In fact, vg isthevelocityofenergypropagatio nevenforamonochromati cwave(§??). However, for purely monochromatic waves vgcannot be identiﬁed from the time-domain ﬁeld, whereas vpcan. This leads to some unfortunate misconceptions, especially when vpexceeds the speed of light. Since vpis not the velocity of propagation of a physical quantity, but is rather the rate of change of a phase reference point, Einstein’s postulateofcas the limiting velocity is not violated. We can obtain interesting relationships between v pandvgby manipulating (4.237) and (4.241). For instance, if we compute dvp dω=d dω/parenleftbiggω β/parenrightbigg =β−ωdβ dω β2 Figure 4.11: An ω–βdiagram for a ﬁctitious material. we ﬁnd that vp vg=1−βdvp dω. (4.242) Hence in frequency ranges where vpdecreases with increasing frequency, we have vg<v p. These are known as regions of normal dispersion . In frequency ranges where vpincreases with increasing frequency, we have vg>v p. These are known as regions of anomalous dispersion . As mentioned in §4.6.3, the word “anomalous” does not imply that this type of dispersion is unusual. The propagation of a uniform plane wave through a lossless medium provides a par- ticularly simple example. In a lossless medium we have β(ω)=ω√µ/epsilon1, α(ω) =0. In this case (4.233) becomes β(ω)=ω0√µ/epsilon1+√µ/epsilon1(ω−ω0)=ω√µ/epsilon1 and (4.236) becomes ˆeE(r,t)=ˆeE0cos/parenleftbig ω0/bracketleftbig t−ˆk·r/vp(ω0)/bracketrightbig/parenrightbig f/parenleftbig t−ˆk·r/vg(ω0)/parenrightbig . Since the linear approximation to the phase constant βis in this case exact, the wave packet truly propagates without distortion, with a group velocity identical to the phasevelocity: v g=/bracketleftbiggd dωω√µ/epsilon1/bracketrightbigg−1 =1√µ/epsilon1=ω β=vp. Examples of wave propagation in various media; the ω–βdiagram. A plot ofωversus β(ω) can be useful for displaying the dispersive properties of a material. Figure4.11showssuchanω–βplot,ordispersiondiagram,foraﬁctitiou smaterial .The 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000 β (r/m)020406080100120140160180200 ω/2≠ (GHz) Light Line: ε=ε0εsLight Line: ε=ε0εi Figure 4.12: Dispersion plot for water computed using the Debye relaxation formula. slope of the line from the origin to a point ( β,ω) is the phase velocity, while the slope of the line tangent to the curve at that point is the group velocity. This plot showsmany of the diﬀerent characteristics of electromagnetic waves (although not necessarilyof plane waves). For instance, there may be a minimum frequency ω ccalled the cutoﬀ frequency at which β=0and below which the wave cannot propagate. This behavior is characteristic of a plane wavepropagating in a plasma (as shown below) or of a wave in a hollow pipe waveguide ( §5.4.3). Over most values of βwe have vg<v pso the material demonstrates normal dispersion. However, over a small region we do have anomalousdispersion. In another range the slope of the curve is actually negative and thus v g<0; here the directions of energy and phase front propagation are opposite. Such backward waves are encountered in certain guided- wavestructures used in micro waveoscillators. Theω–βplot also includes the light line as a reference curve. For all points on this line vg=vp; it is generally used to represent propagation within the material under special circumstances, such as when the loss is zero or the material occupies unbounded space.It may also be used to represent propagation within a vacuum. As an example for which the constitutive parameters depend on frequency, let us consider the relaxation eﬀects of water. By the Debye formula (4.106) we have ˜/epsilon1(ω)=/epsilon1 ∞+/epsilon1s−/epsilon1∞ 1+jωτ. Assuming /epsilon1∞= 5/epsilon10,/epsilon1s = 78.3/epsilon10,andτ= 9.6 × 10−12 s[49],weobtaintherelaxation spectrumshowninFigure4.5.Ifwealsoassum ethatµ=µ0, we may compute βas a functio nofωandconstruc ttheω–βplot.ThisisshowninFigure4.12.Since/epsilon1/primevaries with frequency, we show both the light line for zero frequency found using /epsilon1s=78.3/epsilon10, and the light line for inﬁnite frequency found using /epsilon1i=5/epsilon10. We see that at low values of frequency the dispersion curve follows the low-frequency light line very closely, andthusv p≈vg≈c/√ 78.3. As the frequency increases, the dispersion curve rises up and 91 0 1 1 1 2 log10(f) 0.00.20.40.6 v/cvvg p Figure 4.13: Phase and group velocities for water computed using the Debye relaxation formula. eventually becomes asymptotic with the high-frequency light line. Plots of vpandvg showninFigure4.13verifythatthevelocitiesstartoutat c/√ 78.3for low frequencies, and approach c/√ 5for high frequencies. Because vg>v pat all frequencies, this model of water demonstrates anomalous dispersion. Another interesting example is that of a non-magnetized plasma. For a collisionless plasma we may set ν=0in (4.76) to ﬁnd k=  ω c/radicalBig 1−ω2p ω2,ω > ω p, −jω c/radicalBig ω2p ω2−1,ω < ω p. Thus, when ω>ω pwe have ˜E(r,ω)=˜E0(ω)e−jβ(ω) ˆk·r and so β=ω c/radicalBigg 1−ω2p ω2,α =0. In this case a plane wavepropagates through the plasma without attenuation. However, when ω<ω pwe have ˜E(r,ω)=˜E0(ω)e−α(ω) ˆk·r with α=ω c/radicalBigg ω2p ω2−1,β =0, and a plane wave does not propagate, but only attenuates. Such a wave is called an evanescent wave . We say that for frequencies below ωpthe wave is cut oﬀ , and call ωp thecutoﬀ frequency . 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20 0.22 or α (1/m)012345678910ω/2≠ (MHz)Light Lineβ α Figure 4.14: Dispersion plot for the ionosphere computed using Ne=2×1011m−3,ν=0. Light line computed using /epsilon1=/epsilon10,µ=µ0. Consider, for instance, a plane wave propagating in the earth’s ionosphere. Both the electron densityand the collision frequencyare highlydependent on such factorsas altitude, time of day, and latitude. However, except at the very lowest altitudes,the collision frequencyis low enough that the ionosphere maybe considered lossless.For instance, at a height of 200 km (the F 1layer of the ionosphere), as measured for a mid-latitude region, we ﬁnd that during the daythe electron densityis approximately Ne=2×1011m−3, while the collision frequencyis only ν=100s−1[16]. The attenuation is so small in this case that the ionosphere maybe considered essentiallylossless above thecutoﬀ frequency(we will develop an approximate formula for the attenuation constantbelow).Figure4.14showstheω–βdiagra mfortheionospher eassumin gν= 0,along with the light line v p=c. We see that above the cutoﬀ frequencyof fp=ωp/2π=4.0 MHz the wave propagates and that vg<cwhile vp>c. Below the cutoﬀ frequencythe wave does not propagate and the ﬁeld decays very rapidly because αis large. A formula for the phase velocityof a plane wave in a lossless plasma is easilyderived: vp=ω β=c/radicalBig 1−ω2p ω2>c. Thus, our observation from the ω–βplot that vp>cis veriﬁed. Similarly, we ﬁnd that vg=/parenleftbiggdβ dω/parenrightbigg−1 = 1 c/radicalBigg 1−ω2p ω2+1 cω2 p/ω2 /radicalBig 1−ω2p ω2 −1 =c/radicalBigg 1−ω2p ω2<c and our observation that vg<cis also veriﬁed. Interestingly, we ﬁnd that in this case of an unmagnetized collisionless plasma vpvg=c2. Since vp>v g, this model of a plasma demonstrates normal dispersion at all frequencies above cutoﬀ. For the case of a plasma with collisions we retain νin (4.76) and ﬁnd that k=ω c/radicalBigg/bracketleftbigg 1−ω2p ω2+ν2/bracketrightbigg −jνω2p ω(ω2+ν2). When ν/negationslash=0a true cutoﬀ eﬀect is not present and the wave maypropagate at all frequencies. However, when ν/lessmuchωpthe attenuation for propagating waves of frequency ω<ω pis quite severe, and for all practical purposes the wave is cut oﬀ. For waves of frequency ω>ω pthere is attenuation. Assuming that ν/lessmuchωpand that ν/lessmuchω,w em a y approximate the square root with the ﬁrst two terms of a binomial expansion, and ﬁndthat to ﬁrst order β=ω c/radicalBigg 1−ω2p ω2,α =1 2ν cω2 p/ω2 /radicalBig 1−ω2p ω2. Hence the phase and group velocities above cutoﬀ are essentiallythose of a lossless plasma, while the attenuation constant is directlyproportional to ν. 4.11.4 Monochromatic plane waves in a lossy medium Manyproperties of monochromatic plane waves are particularlysimple. In fact, cer- tain properties, such as wavelength, onlyhave meaning for monochromatic ﬁelds. Andsince monochromatic or nearlymonochromatic waves are employ ed extensivelyin radar,communications, and energytransport, it is useful to simplifythe results of the precedingsection for the special case in which the spectrum of the plane- wavesignal consists of a single frequencycomponent. In addition, plane waves of more general time dependence can be viewed as superpositions of individual single-frequencycomponents (through theinverse Fourier transform), and thus we mayregard monochromatic waves as buildingblocks for more complicated plane waves. We can view the monochromatic ﬁeld as a specialization of (4.230) for a→0. This results in ˜F(ω)→δ(ω), so the linearly-polarized plane wave expression (4.232) reduces to ˆeE(r,t)=ˆeE 0e−α(ω 0)[ˆk·r]cos(ω0t−jβ(ω 0)[ˆk·r]). (4.243) It is convenient to represent monochromatic ﬁelds with frequency ω=ˇωin phasor form. The phasor form of (4.243) is ˇE(r)=ˆeE0e−jβ(ˆk·r)e−α(ˆk·r)(4.244) where β=β(ˇω)andα=α(ˇω). We can identifya surface of constant phase as a locus of points obeying ˇωt−β(ˆk·r)=CP (4.245) for some constant CP.Thissurfac eisaplane,asshowninFigure4.10,withitsnormal in the direction of ˆk. It is easyto verifythat anypoint ron this plane satisﬁes (4.245). Letr0=r0ˆkdescribe the point on the plane with position vector in the ˆkdirection, and letdbe a displacement vector from this point to anyother point on the plane. Then ˆk·r=ˆk·(r0+d)=r0(ˆk·ˆk)+ˆk·d. But ˆk·d=0,s o ˆk·r=r0, (4.246) which is a spatial constant, hence (4.245) holds for any t. The planar surfaces described by(4.245) are wavefronts . Note that surfaces of constant amplitude are determined by α(ˆk·r)=CA where CAis some constant. As with the phase term, this requires that ˆk·r=constant, and thus surfaces of constant phase and surfaces of constant amplitude are coplanar.This is a propertyof uniform plane waves. We shall see later that nonuniform planewaves have planar surfaces that are not parallel. The cosine term in (4.243) represents a traveling wave .A s tincreases, the argument of the cosine function remains unchanged as long as ˆk·rincreases correspondingly. Thus the planar wavefronts propagate along ˆk. As the wavefront progresses, the wave is attenuated because of the factor e −α(ˆk·r). This accounts for energytransferred from the propagating wave to the surrounding medium via Joule heating. Phase velocity of a uniform plane wave. The propagation velocityof the progress- ing wavefront is found bydiﬀerentiating (4.245) to get ˇω−βˆk·dr dt=0. By(4.246) we have vp=dr0 dt=ˇω β, (4.247) where the phase velocity vprepresents the propagation speed of the constant-phase sur- faces. For the case of a lossymedium with frequency -independent constitutive parame-ters, (4.227) shows that v p≤1√µ/epsilon1, hence the phase velocityin a conducting medium cannot exceed that in a lossless medium with the same parameters µand/epsilon1. We cannot draw this conclusion for a medium with frequency-dependent ˜µand ˜/epsilon1c, since by(4.224) the value of ˇω/βmight be greater or less than 1/√˜µ/prime˜/epsilon1c/prime, depending on the ratios ˜µ/prime/prime/˜µ/primeand ˜/epsilon1c/prime/prime/˜/epsilon1c/prime. Wavelength of a uniform plane wave. Another important propertyof a uniform plane wave is the distance between adjacent wavefronts that produce the same value of the cosine function in (4.243). Note that the ﬁeld amplitude maynot be the same onthese two surfaces because of possible attenuation of the wave. Let r 1and r2be points on adjacent wavefronts. We require β(ˆk·r1)=β(ˆk·r2)−2π or λ=ˆk·(r2−r1)=r02−r01=2π/β. We call λthewavelength . Polarization of a uniform plane wave. Plane- wavepolarization describes the tem- poral evolution of the vector direction of the electric ﬁeld, which depends on the mannerin which the wave is generated. Completely polarized waves are produced byantennas or other equipment; these have a deterministic polarization state which maybe describedcompletelybythree parameters as discussed below. Randomly polarized waves are emit- ted bysome natural sources. Partially polarized waves, such as those produced bycosmic radio sources, contain both completelypolarized and randomlypolarized components.We shall concentrate on the description of completelypolarized waves. The polarization state of a completelypolarized monochromatic plane wave propa- gating in a homogeneous, isotropic region maybe described bysuperposing two simplerplane waves that propagate along the same direction but with diﬀerent phases and spa-tiallyorthogonal electric ﬁelds. Without loss of generalitywe maystudypropagationalong the z-axis and choose the orthogonal ﬁeld directions to be along ˆxand ˆy.S o w e are interested in the behavior of a wavewith electric ﬁeld ˇE(r)=ˆxE x0ejφxe−jkz+ˆyEy0ejφye−jkz. (4.248) The time evolution of the direction of Emust be examined in the time domain where we have E(r,t)=Re/braceleftbigˇEejωt/bracerightbig =ˆxEx0cos(ωt−kz+φx)+ˆyEy0cos(ωt−kz+φy) and thus, bythe identity cos(x+y)≡cosxcosy−sinxsiny, Ex=Ex0[cos(ωt−kz)cos(φx)−sin(ωt−kz)sin(φx)], Ey=Ey0/bracketleftbig cos(ωt−kz)cos(φy)−sin(ωt−kz)sin(φy)/bracketrightbig . The tip of the vector Emoves cyclically in the xy-plane with temporal period T=ω/2π. Its locus maybe found byeliminating the parameter tto obtain a relationship between Ex0and Ey0. Letting δ=φy−φxwe note that Ex Ex0sinφy−Ey Ey0sinφx=cos(ωt−kz)sinδ, Ex Ex0cosφy−Ey Ey0cosφx=sin(ωt−kz)sinδ; squaring these terms we ﬁnd that /parenleftbiggEx Ex0/parenrightbigg2 +/parenleftbiggEy Ey0/parenrightbigg2 −2Ex Ex0Ey Ey0cosδ=sin2δ, whichistheequatio nfortheellipseshowninFigure4.15.By(4.223 )themagneti cﬁeld of the plane wave is ˇH=ˆz×ˇE η, hence its tip also traces an ellipse in the xy-plane. The tip of the electric ﬁeld vector cycles around the polarization ellipse in the xy- plane once every Tseconds. The sense of rotation is determined bythe sign of δ, and is described bythe terms clockwise /counterclockwise orright-hand /left-hand . There is some disagreement about how to do this. We shall adopt the IEEE deﬁnitions (IEEEStandar d145-198 3[189])andassociatewithδ<0 rotatio nintheright-han dsense:if Figure 4.15: Polarization ellipse for a monochromatic plane wave. the right thumb points in the direction of wave propagation then the ﬁngers curl in the direction of ﬁeld rotation for increasing time. This is right-hand polarization (RHP). We associate δ>0withleft-hand polarization (LHP). The polarization ellipse is contained within a rectangle of sides 2Ex0and 2Ey0, and has its major axis rotated from the x-axis bythe tilt angle ψ,0≤ψ≤π. The ratio of Ey0toEx0determines an angle α,0≤α≤π/2: Ey0/Ex0=tanα. The shape of the ellipse is determined bythe three parameters Ex0,Ey0, and δ, while the sense of polarization is described bythe sign of δ. These maynot, however, be the most convenient parameters for describing the polarization of a wave. We can alsoinscribe the ellipse within a box measuring 2aby2b, where aand bare the lengths of the semimajor and semiminor axes. Then b/adetermines an angle χ,−π/4≤χ≤π/4, that is analogous to α: ±b/a=tanχ. Here the algebraic sign of χis used to indicate the sense of polarization: χ> 0for LHP, χ< 0for RHP. The quantities a,b,ψcan also be used to describe the polarization ellipse. When we use the procedure outlined in Born and Wolf [19] to relate the quantities (a,b,ψ)to (E x0,Ey0,δ), we ﬁnd that a2+b2=E2 x0+E2 y0, tan 2ψ=(tan 2α)cosδ=2Ex0Ey0 E2 x0−E2 y0cosδ, sin 2χ=(sin 2α)sinδ=2Ex0Ey0 E2 x0+E2 y0sinδ. Alternatively, we can describe the polarization ellipse by the angles ψandχand one of the amplitudes Ex0orEy0. Figure 4.16: Polarization states as a function of tilt angle ψand ellipse aspect ratio angle χ. Left-hand polarization for χ> 0, right-hand for χ< 0. Each of these parameter sets is somewhat inconvenient since in each case the units diﬀer among the parameters. In 1852 G. Stokes introduced a system of three independentquantities with identical dimension that can be used to describe plane- wave p olarization. Various normalizations of these Stokes parameters are employed; when the parameters are chosen to have the dimension of power densitywe maywrite them as s 0=1 2η/bracketleftbig E2 x0+E2 y0/bracketrightbig , (4.249) s1=1 2η/bracketleftbig E2 x0−E2 y0/bracketrightbig =s0cos(2χ)cos(2ψ), (4.250) s2=1 ηEx0Ey0cosδ=s0cos(2χ)sin(2ψ), (4.251) s3=1 ηEx0Ey0sinδ=s0sin(2χ). (4.252) Onlythree of these four parameters are independent since s2 0=s2 1+s2 2+s2 3. Often the Stokes parameters are designated (I,Q,U,V)rather than (s0,s1,s2,s3). Figure4.16summarize svariouspolarizatio nstatesasafunctio noftheanglesψand χ. Two interesting special cases occur when χ=0andχ=±π/4. The case χ=0 corresponds to b=0and thus δ=0. In this case the electric vector traces out a straight line and we call the polarization linear. Here E=/parenleftbigˆxEx0+ˆyEy0/parenrightbig cos(ωt−kz+φx). When ψ=0we have Ey0=0and refer to this as horizontal linear polarization (HLP); when ψ=π/2we have Ex0=0andvertical linear polarization (VLP). The case χ=±π/4corresponds to b=aandδ=±π/2.T h u s Ex0=Ey0, and E traces out a circle regardless of the value of ψ.I fχ=−π/4we have right-hand rotation ofEand thus refer to this case as right-hand circular polarization (RHCP). If χ=π/4 we have left-hand circular polarization (LHCP). For these cases E=Ex0[ˆxcos(ωt−kz)∓ˆysin(ωt−kz)], Figure 4.17: Graphical representation of the polarization of a monochromatic plane wave using the Poincar´ e sphere. where the upper and lower signs correspond to LHCP and RHCP, respectively. All other values of χresult in the general cases of left-hand or right-hand elliptical polarization . The French mathematician H. Poincar´ e realized that the Stokes parameters (s1,s2,s3) describe a point on a sphere of radius s0, and that this Poincar´ e sphere is useful for visualizing the various polarization states. Each state corresponds uniquelyto one pointon the sphere, and by(4.250)–(4.252) the angles 2χand 2ψare the spherical angular coordinate softhepointasshowninFigure4.17.Wemaytherefor emapthepolarization statesshowninFigure4.16directlyo ntothesphere :left-andright-handpolarizations appear in the upper and lower hemispheres, respectively; circular polarization appears atthe poles ( 2χ=±π/2); linear polarization appears on the equator ( 2χ=0), with HLP at2ψ=0and VLP at 2ψ=π. The angles αandδalso have geometrical interpretations on the Poincar´ e sphere. The spherical angle of the great-circle route between the pointof HLP and a point on the sphere is 2α, while the angle between the great-circle path and the equator is δ. Uniform plane waves in a good dielectric. We maybase some useful plane-wave approximations on whether the real or imaginarypart of ˜/epsilon1 cdominates at the frequency of operation. We assume that ˜µ(ω)=µis independent of frequencyand use the notation /epsilon1c=˜/epsilon1c(ˇω),σ=˜σ(ˇω), etc. Remember that /epsilon1c=/parenleftbig /epsilon1/prime+j/epsilon1/prime/prime/parenrightbig +σ jˇω=/epsilon1/prime+j/parenleftBig /epsilon1/prime/prime−σ ˇω/parenrightBig =/epsilon1c/prime+j/epsilon1c/prime/prime. Bydeﬁnition, a “good dielectric” obey s tanδc=−/epsilon1c/prime/prime /epsilon1c/prime=σ ˇω/epsilon1/prime−/epsilon1/prime/prime /epsilon1/prime/lessmuch1. (4.253) Here tanδcis the loss tangent of the material, as ﬁrst described in (4.107) for a material without conductivity. For a good dielectric we have k=β−jα=ˇω/radicalbig µ/epsilon1c=ˇω/radicalbig µ[/epsilon1/prime+j/epsilon1c/prime/prime]=ˇω/radicalbig µ/epsilon1/prime/radicalbig 1−jtanδc, hence k≈ˇω/radicalbig µ/epsilon1/prime/bracketleftbigg 1−j1 2tanδc/bracketrightbigg (4.254) bythe binomial approximation for the square root. Therefore β≈ˇω/radicalbig µ/epsilon1/prime (4.255) and α≈β 2tanδc=σ 2/radicalbiggµ /epsilon1/prime/bracketleftbigg 1−ˇω/epsilon1/prime/prime σ/bracketrightbigg . (4.256) We conclude that α/lessmuchβ. Using this and the binomial approximation we establish η=ˇωµ k=ˇωµ β1 1−jα/β≈ˇωµ β/parenleftbigg 1+jα β/parenrightbigg . Finally, vp=ˇω β≈1√µ/epsilon1/prime and vg=/bracketleftbiggdβ dω/bracketrightbigg−1 ≈1√µ/epsilon1/prime. To ﬁrst order, the phase constant, phase velocity, and group velocity are the same as those of a lossless medium. Uniform plane waves in a good conductor. We classifya material as a “good conductor” if tanδc≈σ ˇω/epsilon1/greatermuch1. In a good conductor the conduction current σˇEis much greater than the displacement current jˇω/epsilon1/primeˇE, and /epsilon1/prime/primeis usuallyignored. Now we mayapproximate k=β−jα=ˇω/radicalbig µ/epsilon1/prime/radicalbig 1−jtanδc≈ˇω/radicalbig µ/epsilon1/prime/radicalbig −jtanδc. Since√−j=(1−j)/√ 2we ﬁnd that β=α≈/radicalbig πfµσ. (4.257) Hence vp=ˇω β≈/radicalBigg 2ˇω µσ=1√µ/epsilon1/prime/radicalBigg 2 tanδc. To ﬁnd vgwe must replace ˇωbyωand diﬀerentiate, obtaining vg=/bracketleftbiggdβ dω/bracketrightbigg−1/vextendsingle/vextendsingle/vextendsingle/vextendsingle ω=ˇω≈/bracketleftbigg1 2/radicalbiggµσ 2ˇω/bracketrightbigg−1 =2/radicalBigg 2ˇω µσ=2vp. In a good conductor the group velocityis approximatelytwice the phase velocity . We could have found this relation from the phase velocityusing (4.242). Indeed, noting that dvp dω=d dω/radicalBigg 2ω µσ=1 2/radicalBigg 2 ωµσ and βdvp dω=/radicalbiggωµσ 21 2/radicalBigg 2 ωµσ=1 2, we see that vp vg=1−1 2=1 2. Note that the phase and group velocities maybe onlysmall fractions of the free-space light velocity. For example, in copper ( σ=5.8×107S/m, µ=µ0,/epsilon1=/epsilon10) at 1 MHz, we have vp=415m/s. A factor often used to judge the qualityof a conductor is the distance required for a propagating uniform plane wave to decrease in amplitude bythe factor 1/e. By(4.244) this distance is given by δ=1 α=1√πfµσ. (4.258) We call δtheskin depth . A good conductor is characterized bya small skin depth. For example, copper at 1 MHz has δ=0.066mm. Power carried by a uniform plane wave. Since a plane wavefront is inﬁnite in extent, we usuallyspeak of the power density carried bythe wave. This is identical to the time-average Poynting ﬂux. Substitution from (4.223) and (4.244) gives Sav=1 2Re{ˇE×ˇH∗}=1 2Re/braceleftBigg ˇE×/parenleftBiggˆk×ˇE η/parenrightBigg∗/bracerightBigg . (4.259) Expanding the cross products and remembering that k·ˇE=0, we get Sav=1 2ˆkRe/braceleftBigg |ˇE|2 η∗/bracerightBigg =ˆkRe/braceleftbiggE2 0 2η∗/bracerightbigg e−2αˆk·r. Hence a uniform plane wave propagating in an isotropic medium carries power in the direction of wavefront propagation. Velocity of energy transport. The group velocity(4.237) has an additional interpre- tation as the velocityof energytransport. If the time-average volume densityof energyis given by /angbracketleftw em/angbracketright=/angbracketleftwe/angbracketright+/angbracketleftwm/angbracketright and the time-average volume densityof energyﬂow is given bythe Poy nting ﬂux density Sav=1 2Re/braceleftbigˇE(r)×ˇH∗(r)/bracerightbig =1 4/bracketleftbigˇE(r)×ˇH∗(r)+ˇE∗(r)×ˇH(r)/bracketrightbig , (4.260) then the velocityof energyﬂow, ve, is deﬁned by Sav=/angbracketleftwem/angbracketrightve. (4.261) Let us calculate vefor a plane wavepropagating in a lossless, source-free medium where k=ˆkω√µ/epsilon1. By(4.216) and (4.223) we have ˜E(r,ω)=˜E0(ω)e−jβˆk·r, (4.262) ˜H(r,ω)=/parenleftBiggˆk×˜E0(ω) η/parenrightBigg e−jβˆk·r=˜H0(ω)e−jβˆk·r. (4.263) We can compute the time-average stored energydensityusing the energytheorem (4.68). In point form we have −∇·/parenleftbigg ˜E∗×∂˜H ∂ω+∂˜E ∂ω×˜H∗/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle ω=ˇω=4j/angbracketleftwem/angbracketright. (4.264) Upon substitution of (4.262) and (4.263) we ﬁnd that we need to compute the frequency derivatives of ˜Eand ˜H. Using ∂ ∂ωe−jβˆk·r=/parenleftbigg∂ ∂βe−jβˆk·r/parenrightbiggdβ dω=− jˆk·rdβ dωe−jβˆk·r and remembering that k=ˆkβ,w eh a v e ∂˜E(r,ω) ∂ω=d˜E0(ω) dωe−jk·r+˜E0(ω)/parenleftbigg −jr·dk dω/parenrightbigg e−jk·r, ∂˜H(r,ω) ∂ω=d˜H0(ω) dωe−jk·r+˜H0(ω)/parenleftbigg −jr·dk dω/parenrightbigg e−jk·r. Equation (4.264) becomes −∇ ·/braceleftbigg ˜E∗ 0(ω)×d˜H0(ω) dω+d˜E0(ω) dω×˜H∗ 0(ω)− −jr·dk dω/bracketleftbig˜E∗ 0(ω)×˜H0(ω)+˜E0(ω)×˜H∗ 0(ω)/bracketrightbig/bracerightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle ω=ˇω=4j/angbracketleftwem/angbracketright. The ﬁrst two terms on the left-hand side have zero divergence, since these terms do not depend on r. Bythe product rule (B.42) we have /bracketleftbig˜E∗ 0(ˇω)×˜H0(ˇω)+˜E0(ˇω)×˜H∗ 0(ˇω)/bracketrightbig ·∇/parenleftbigg r·dk dω/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle ω=ˇω=4/angbracketleftwem/angbracketright. The gradient term is merely ∇/parenleftbigg r·dk dω/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle ω=ˇω=∇/parenleftbigg xdkx dω+ydky dω+zdkz dω/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle ω=ˇω=dk dω/vextendsingle/vextendsingle/vextendsingle/vextendsingle ω=ˇω, hence /bracketleftbig˜E∗ 0(ˇω)×˜H0(ˇω)+˜E0(ˇω)×˜H∗ 0(ˇω)/bracketrightbig ·dk dω/vextendsingle/vextendsingle/vextendsingle/vextendsingle ω=ˇω=4/angbracketleftwem/angbracketright. (4.265) Finally, the left-hand side of this expression can be written in terms of the time-average Poynting vector. By (4.260) we have Sav=1 2Re/braceleftbigˇE×ˇH∗/bracerightbig =1 4/bracketleftbig˜E0(ˇω)×˜H∗ 0(ˇω)+˜E∗ 0(ˇω)×˜H0(ˇω)/bracketrightbig and thus we can write (4.265) as Sav·dk dω/vextendsingle/vextendsingle/vextendsingle/vextendsingle ω=ˇω=/angbracketleftwem/angbracketright. Since for a uniform plane wave in an isotropic medium kandSavare in the same direction, we have Sav=ˆkdω dβ/vextendsingle/vextendsingle/vextendsingle/vextendsingle ω=ˇω/angbracketleftwem/angbracketright and the velocityof energytransport for a plane wave of frequency ˇωis then ve=ˆkdω dβ/vextendsingle/vextendsingle/vextendsingle/vextendsingle ω=ˇω. Thus, for a uniform plane wave in a lossless medium the velocityof energytransport is identical to the group velocity. Nonuniform plane waves. A nonuniform plane wave has the same form (4.216) as a uniform plane wave, but the vectors k/primeand k/prime/primedescribed in (4.217) are not aligned. Thus ˇE(r)=E0e−jk/prime·rek/prime/prime·r. In the time domain this becomes ˇE(r)=E0ek/prime/prime·rcos[ˇωt−k/prime(ˆk/prime·r)] where k/prime=ˆk/primek/prime. The surfaces of constant phase are planes perpendicular to k/primeand propagating in the direction of ˆk/prime. The phase velocityis now vp=ˇω/k/prime and the wavelength is λ=2π/k/prime. In contrast, surfaces of constant amplitude must obey k/prime/prime·r=C and thus are planes perpendicular to k/prime/prime. In a nonuniform plane wave the TEM nature of the ﬁelds is lost. This is easilyseen bycalculating ˇHfrom (4.219): ˇH(r)=k×ˇE(r) ˇωµ=k/prime×ˇE(r) ˇωµ+jk/prime/prime×ˇE(r) ˇωµ. Thus, ˇHis no longer perpendicular to the direction of propagation of the phase front. The power carried bythe wave also diﬀers from that of the uniform case. The time-averagePoynting vector S av=1 2Re/braceleftBigg ˇE×/parenleftBigg k×ˇE ˇωµ/parenrightBigg∗/bracerightBigg can be expanded using the identity(B.7): Sav=1 2Re/braceleftbigg1 ˇωµ∗/bracketleftbig k∗×(ˇE×ˇE∗)+ˇE∗×(k∗×ˇE)/bracketrightbig/bracerightbigg . (4.266) Since we still have k·E=0, we mayuse the rest of (B.7) to write ˇE∗×(k∗×ˇE)=k∗(ˇE·ˇE∗)+ˇE(k·ˇE)∗=k∗(ˇE·ˇE∗). Substituting this into (4.266), and noting that ˇE×ˇE∗is purelyimaginary , we ﬁnd Sav=1 2Re/braceleftbigg1 ˇωµ∗/bracketleftbig jk∗×Im/braceleftbigˇE×ˇE∗/bracerightbig +k∗|ˇE|2/bracketrightbig/bracerightbigg . (4.267) Thus the vector direction of Savis not generallyin the direction of propagation of the plane wavefronts. Let us examine the special case of nonuniform plane waves propagating in a lossless material. It is intriguing that kmaybe complex when kis real, and the implication is important for the plane- waveexpansion of complicated ﬁelds in free space. By(4.218), real krequires that if k/prime/prime/negationslash=0then k/prime·k/prime/prime=0. Thus, for a nonuniform plane wave in a lossless material the surfaces of constant phase and the surfaces of constant amplitude are orthogonal. To specialize the time-averagepower to the lossless case we note that µis purelyreal and that E×E ∗=(E0×E∗ 0)e2k/prime/prime·r. Then (4.267) becomes Sav=1 2ˇωµe2k/prime/prime·rRe/braceleftbig j(k/prime−jk/prime/prime)×Im/braceleftbig E0×E∗ 0/bracerightbig +(k/prime−jk/prime/prime)|ˇE|2/bracerightbig or Sav=1 2ˇωµe2k/prime/prime·r/bracketleftbig k/prime/prime×Im/braceleftbig E0×E∗ 0/bracerightbig +k/primeˇE|2/bracketrightbig . We see that in a lossless medium the direction of energypropagation is perpendicular to the surfaces of constant amplitude (since k/prime/prime·Sav=0), but the direction of energy propagation is not generallyin the direction of propagation of the phase planes. We shall encounter nonuniform plane waves when we studythe reﬂection and refrac- tion of a plane wave from a planar interface in the next section. We shall also ﬁnd in§4.13 that nonuniform plane waves are a necessaryconstituent of the angular spectrum representation of an arbitrarywave ﬁeld. 4.11.5 Plane waves in lay ered media A useful canonical problem in wave propagation involves the reﬂection of plane waves byplanar interfaces between diﬀering material regions. This has manydirect applica-tions, from the design of optical coatings and microwave absorbers to the probing ofunderground oil-bearing rock layers. We shall begin by studying the reﬂection of a planewave at a single interface and then extend the results to anynumber of material lay ers. Reﬂection of a uniform plane wave at a planar material interface. Consider twolossymedi aseparate dbythe z = 0planeasshowninFigure4.18.Themediaareas- sumed to be isotropic and homogeneous with permeability ˜µ(ω)and complex permittivity ˜/epsilon1 c(ω). Both ˜µand ˜/epsilon1cmaybe complex numbers describing magnetic and dielectric loss, respectively. We assume that a linearly-polarized plane-wave ﬁeld of the form (4.216) is created within region 1 bya process that we shall not studyhere. We take this ﬁeld tobe the known “incident wave” pro duced byan impressed source, and wish to compute the total ﬁeld in regions 1 and 2. Here we shall assume that the incident ﬁeld is that of auniform plane wave, and shall extend the analysis to certain types of nonuniform plane waves subsequently. Since the incident ﬁeld is uniform, we maywrite the wave vector associated with this ﬁeld as k i=ˆkiki=ˆki(ki/prime+jki/prime/prime) where [ki(ω)]2=ω2˜µ1(ω)˜/epsilon1c 1(ω). We can assume without loss of generalitythat ˆkilies in the xz-plane and makes an angle θi withtheinterfacenorma lasshowninFigure4.18.Werefertoθi astheinciden ceangle of the incident ﬁeld, and note that it is the angle between the direction of propagationof the planar phase fronts and the normal to the interface. With this we have k i=ˆxk1sinθi+ˆzk1cosθi=ˆxki x+ˆzki z. Using k1=β1−jα1we also have ki x=(β1−jα1)sinθi. The term ki zis written in a somewhat diﬀerent form in order to make the result easily applicable to reﬂections from multiple interfaces. We write ki z=(β1−jα1)cosθi=τie−jγi=τicosγi−jτisinγi. Thus, τi=/radicalBig β2 1+α2 1cosθi,γi=tan−1(α1/β1). We solve for the ﬁelds in each region of space directlyin the frequencydomain. The incident electric ﬁeld has the form of (4.216), ˜Ei(r,ω)=˜Ei 0(ω)e−jki(ω)·r, (4.268) while the magnetic ﬁeld is found from (4.219) to be ˜Hi=ki×˜Ei ω˜µ1. (4.269) The incident ﬁeld maybe decomposed into two orthogonal components, one parallel to the plane of incidence (the plane containing ˆkand the interface normal ˆz) and one perpendicular to this plane. We seek unique solutions for the ﬁelds in both regions, ﬁrstfor the case in which the incident electric ﬁeld has onlya parallel component, and thenfor the case in which it has onlya perpendicular component. The total ﬁeld is thendetermined bysuperposition of the individual solutions. For perpendicular polarizationwe have from (4.268) and (4.269) ˜E i ⊥=ˆy˜Ei ⊥e−j(ki xx+ki zz), (4.270) ˜Hi ⊥=−ˆxki z+ˆzki x k1˜Ei ⊥ η1e−j(ki xx+ki zz), (4.271) Figure 4.18: Uniform plane waveincident on planar interface between two lossyregions of space. (a) TM polarization, (b) TE polarization. asshowngraphicallyi nFigure4.18.Hereη1=(˜µ1/˜/epsilon1c 1)1/2is the intrinsic impedance of medium 1. For parallel polarization, the direction of ˜Eis found byremembering that the wave must be TEM. Thus ˜E/bardblis perpendicular to ki. Since ˜E/bardblmust also be perpendicular to˜E⊥, we have two possible directions for ˜E/bardbl. Byconvention we choose the one for which ˜Hlies in the same direction as did ˜Efor perpendicular polarization. Thus we have for parallel polarization ˜Hi /bardbl=ˆy˜Ei /bardbl η1e−j(ki xx+ki zz), (4.272) ˜Ei /bardbl=ˆxki z−ˆzki x k1˜Ei /bardble−j(ki xx+ki zz), (4.273) asshowninFigure4.18.Because ˜E liestransverse(normal )totheplaneofincidence under perpendicular polarization, the ﬁeld set is often described as transverse electric or TE. Because ˜Hlies transverse to the plane of incidence under parallel polarization, the ﬁelds in that case are transverse magnetic orTM. Uniqueness requires that the total ﬁeld obeythe boundaryconditions at the planar interface. We hypothesize that the total ﬁeld within region 1 consists of the incidentﬁeld superposed with a “reﬂected” plane-wave ﬁeld having wave vector k r, while the ﬁeld in region 2 consists of a single “transmitted” plane-wave ﬁeld having wave v ector kt. We cannot at the outset make anyassumption regarding whether either of these ﬁelds are uniform plane waves. However, we do note that the reﬂected and transmittedﬁelds cannot have vector components not present in the incident ﬁeld; extra componentswould preclude satisfaction of the boundaryconditions. Letting ˜E rbe the amplitude of the reﬂected plane-wave ﬁeld we maywrite ˜Er ⊥=ˆy˜Er ⊥e−j(kr xx+kr zz), ˜Hr ⊥=−ˆxkr z+ˆzkr x k1˜Er ⊥ η1e−j(kr xx+kr zz), ˜Hr /bardbl=ˆy˜Er /bardbl η1e−j(kr xx+kr zz), ˜Er /bardbl=ˆxkr z−ˆzkr x k1˜Er /bardble−j(kr xx+kr zz), where (kr x)2+(kr z)2=k2 1. Similarly, letting ˜Etbe the amplitude of the transmitted ﬁeld we have ˜Et ⊥=ˆy˜Et ⊥e−j(kt xx+kt zz), ˜Ht ⊥=−ˆxkt z+ˆzkt x k2˜Et ⊥ η2e−j(kt xx+kt zz), ˜Ht /bardbl=ˆy˜Et /bardbl η2e−j(kt xx+kt zz), ˜Et /bardbl=ˆxkt z−ˆzkt x k2˜Et /bardble−j(kt xx+kt zz), where (kt x)2+(kt z)2=k2 2. The relationships between the ﬁeld amplitudes ˜Ei,˜Er,˜Et, and between the components of the reﬂected and transmitted wave vectors krand kt, can be found byapply ing the boundaryconditions. The tangential electric and magnetic ﬁelds are continuous across the interface at z=0: ˆz×(˜Ei+˜Er)|z=0=ˆz×˜Et|z=0, ˆz×(˜Hi+˜Hr)|z=0=ˆz×˜Ht|z=0. Substituting the ﬁeld expressions, we ﬁnd that for perpendicular polarization the two boundaryconditions require ˜Ei ⊥e−jki xx+˜Er ⊥e−jkr xx=˜Et ⊥e−jkt xx, (4.274) ki z k1˜Ei ⊥ η1e−jki xx+kr z k1˜Er ⊥ η1e−jkr xx=kt z k2˜Et ⊥ η2e−jkt xx, (4.275) while for parallel polarization theyrequire ki z k1˜Ei /bardble−jki xx+kr z k1˜Er /bardble−jkr xx=kt z k2˜Et /bardble−jkt xx, (4.276) ˜Ei /bardbl η1e−jki xx+˜Er /bardbl η1e−jkr xx=˜Et /bardbl η2e−jkt xx. (4.277) For the above to hold for all xwe must have the exponential terms equal. This requires ki x=kr x=kt x, (4.278) and also establishes a relation between ki z,kr z, and kt z. Since (ki x)2+(ki z)2=(kr x)2+(kr z)2= k2 1, we must have kr z=± ki z. In order to make the reﬂected wavefronts propagate away from the interface we select kr z=− ki z. Letting ki x=kr x=kt x=k1xand ki z=− kr z=k1z, we maywrite the wave v ectors in region 1 as ki=ˆxk1x+ˆzk1z, kr=ˆxk1x−ˆzk1z. Since (kt x)2+(kt z)2=k2 2, letting k2=β2−jα2we have kt z=/radicalBig k2 2−k2 1x=/radicalBig (β2−jα2)2−(β1−jα1)2sin2θi=τte−jγt. Squaring out the above relation, we have A−jB=(τt)2cos 2γt−j(τt)2sin 2γt where A=β2 2−α2 2−(β2 1−α2 1)sin2θi, B=2(β2α2−β1α1sin2θi). (4.279) Thus τt=/parenleftbig A2+B2/parenrightbig1/4,γt=1 2tan−1B A. (4.280) Renaming kt zask2z, we maywrite the transmitted wave vector as kt=ˆxk1x+ˆzk2z=k/prime 2+jk/prime/prime 2 where k/prime 2=ˆxβ1sinθi+ˆzτtcosγt, k/prime/prime 2=− ˆxα1sinθi−ˆzτtsinγt. Since the direction of propagation of the transmitted ﬁeld phase fronts is perpendicular tok/prime 2, a unit vector in the direction of propagation is ˆk/prime 2=ˆxβ1sinθi+ˆzτtcosγt /radicalBig β2 1sin2θi+(τt)2cos2θi. (4.281) Similarly, a unit vector perpendicular to planar surfaces of constant amplitude is given by ˆk/prime/prime 2=ˆxα1sinθi+ˆzτtsinγt /radicalBig α2 1sin2θi+(τt)2sin2γt. (4.282) In general ˆk/primeis not aligned with ˆk/prime/primeand thus the transmitted ﬁeld is a nonuniform plane wave. With these deﬁnitions of k1x,k1z,k2z, equations (4.274) and (4.275) can be solved si- multaneouslyand we have ˜Er ⊥=˜/Gamma1⊥˜Ei ⊥, ˜Et ⊥=˜T⊥˜Ei ⊥, where ˜/Gamma1⊥=Z2⊥−Z1⊥ Z2⊥+Z1⊥, ˜T⊥=1+˜/Gamma1⊥=2Z2⊥ Z2⊥+Z1⊥, (4.283) with Z1⊥=k1η1 k1z, Z2⊥=k2η2 k2z. Here ˜/Gamma1is a frequency-dependent reﬂection coeﬃcient that relates the tangential compo- nents of the incident and reﬂected electric ﬁelds, and ˜Tis a frequency-dependent trans- mission coeﬃcient that relates the tangential components of the incident and transmitted electric ﬁelds. These coeﬃcients are also called the Fresnel coeﬃcients . For the case of parallel polarization we solve (4.276) and (4.277) to ﬁnd ˜Er /bardbl,x ˜Ei /bardbl,x=kr x kix˜Er /bardbl ˜Ei /bardbl=−˜Er /bardbl ˜Ei /bardbl=˜/Gamma1/bardbl,˜Et /bardbl,x ˜Ei /bardbl,x=(kt z/k2)˜Et /bardbl (kiz/k1)˜Ei /bardbl=˜T/bardbl. Here ˜/Gamma1/bardbl=Z2/bardbl−Z1/bardbl Z2/bardbl+Z1/bardbl, ˜T/bardbl=1+˜/Gamma1/bardbl=2Z2/bardbl Z2/bardbl+Z1/bardbl, (4.284) with Z1/bardbl=k1zη1 k1, Z2/bardbl=k2zη2 k2. Note that we mayalso write ˜Er /bardbl=− ˜/Gamma1/bardbl˜Ei /bardbl, ˜Et /bardbl=˜T/bardbl˜Ei /bardbl/parenleftbiggki z k1k2 ktz/parenrightbigg . Let us summarize the ﬁelds in each region. For perpendicular polarization we have ˜Ei ⊥=ˆy˜Ei ⊥e−jki·r, ˜Er ⊥=ˆy˜/Gamma1⊥˜Ei ⊥e−jkr·r, (4.285) ˜Et ⊥=ˆy˜T⊥˜Ei ⊥e−jkt·r, and ˜Hi ⊥=ki×˜Ei ⊥ k1η1,˜Hr ⊥=kr×˜Er ⊥ k1η1,˜Ht ⊥=kt×˜Et ⊥ k2η2. (4.286) For parallel polarization we have ˜Ei /bardbl=−η1ki×˜Hi /bardbl k1e−jki·r, ˜Er /bardbl=−η1kr×˜Hr /bardbl k1e−jkr·r, ˜Et /bardbl=−η2kt×˜Ht /bardbl k2e−jkt·r, (4.287) and ˜Hi /bardbl=ˆy˜Ei /bardbl η1e−jki·r, ˜Hr /bardbl=− ˆy˜/Gamma1/bardbl˜Ei /bardbl η1e−jkr·r, ˜Ht /bardbl=ˆy˜T/bardbl˜Ei /bardbl η2/parenleftbiggki z k1k2 ktz/parenrightbigg e−jkt·r. (4.288) The wave v ectors are given by ki=(ˆxβ1sinθi+ˆzτicosγi)−j(ˆxα1sinθi+ˆzτisinγi), (4.289) kr=(ˆxβ1sinθi−ˆzτicosγi)−j(ˆxα1sinθi−ˆzτisinγi), (4.290) kt=(ˆxβ1sinθi+ˆzτtcosγt)−j(ˆxα1sinθi+ˆzτtsinγt). (4.291) We see that the reﬂected wave must, like the incident wave, be a uniform plane wave. We deﬁne the unsigned reﬂection angle θras the angle between the surface normal and thedirectio nofpropagatio nofthereﬂecte dwavefronts(Figure4.18).Since ki·ˆz=k1cosθi=−kr·ˆz=k1cosθr and ki·ˆx=k1sinθi=kr·ˆx=k1sinθr we must have θi=θr. This is known as Snell’s law of reﬂection . We can similarlydeﬁne the transmission angle to be the angle between the direction of propagation of the transmitted wavefronts andthe interface normal. Noting that ˆk /prime 2·ˆz=cosθtand ˆk/prime 2·ˆx=sinθt, we have from (4.281) and (4.282) cosθt=τtcosγt /radicalBig β2 1sin2θi+(τt)2cos2γt, (4.292) sinθt=β1sinθi/radicalBig β2 1sin2θi+(τt)2cos2γt, (4.293) and thus θt=tan−1/parenleftbiggβ1 τtsinθi cosγt/parenrightbigg . (4.294) Depending on the properties of the media, at a certain incidence angle θc, called the critical angle , the angle of transmission becomes π/2. Under this condition ˆk/prime 2has only anx-component. Thus, surfaces of constant phase propagate parallel to the interface. Later we shall see that for low-loss (or lossless) media, this implies that no time-averagepower is carried bya monochromatic transmitted wave into the second medium. We also see that although the transmitted ﬁeld maybe a nonuniform plane wave, its mathematical form is that of the incident plane wave. This allows us to easilygeneralizethe single-interface reﬂection problem to one involving manylay ers. Uniform plane-wave reﬂection for lossless media. We can specialize the preceding results to the case for which both regions are lossless with ˜µ=µand ˜/epsilon1 c=/epsilon1real and frequency-independent. By (4.224) we have β=ω√µ/epsilon1, while (4.225) gives α=0. We can easilyshow that the transmitted wave must be uniform unless the incidence angle exceeds the critical angle. By(4.279) we have A=β2 2−β2 1sin2θi, B=0, (4.295) while (4.280) gives τ=/bracketleftbig A2/bracketrightbig1/4=/radicalBig |β2 2−β2 1sin2θi| and γt=1 2tan−1(0). We have several possible choices for γt. To choose properlywe note that γtrepresents the negative of the phase of the quantity kt z=√ A.I fA>0the phase of the square root is0.I fA<0the phase of the square root is −π/2and thus γt=+π/2. Here we choose the plus sign on γtto ensure that the transmitted ﬁeld decays as zincreases. We note that if A=0thenτt=0and from (4.293) we have θt=π/2. This deﬁnes the critical angle, which from (4.295) is θc=sin−1/parenleftbiggβ2 2 β2 1/parenrightbigg =sin−1/parenleftbiggµ2/epsilon12 µ1/epsilon11/parenrightbigg . Therefore γt=/braceleftBigg 0,θ i<θ c, π/2,θ i>θ c. Using these we can write down the transmitted wave vector from (4.291): kt=kt/prime+jkt/prime/prime=/braceleftBigg ˆxβ1sinθi+ˆz√|A|,θ i<θ c, ˆxβ1sinθi−jˆz√|A|,θ i>θ c.(4.296) By(4.293) we have sinθt=β1sinθi/radicalBig β2 1sin2θi+β2 2−β2 1sin2θi=β1sinθi β2 or β2sinθt=β1sinθi. (4.297) This is known as Snell’s law of refraction . With this we can write for θi<θ c A=β2 2−β2 1sin2θi=β2 2cos2θt. Using this and substituting β2sinθtforβ1sinθi, we mayrewrite (4.296) for θi<θ cas kt=kt/prime+jkt/prime/prime=ˆxβ2sinθt+ˆzβ2cosθt. (4.298) Hence the transmitted plane wave is uniform with kt/prime/prime=0. When θi>θ cwe have from (4.296) kt/prime=ˆxβ1sinθi, kt/prime/prime=− ˆz/radicalBig β2 1sin2θi−β2 2. Since kt/primeand kt/prime/primeare not collinear, the plane wave is non uniform. Let us examine the cases θi<θ candθi>θ cin greater detail. Case 1: θi<θ c.By(4.289)–(4.290) and (4.298) the wave vectors are ki=ˆxβ1sinθi+ˆzβ1cosθi, kr=ˆxβ1sinθi−ˆzβ1cosθi, kt=ˆxβ2sinθt+ˆzβ2cosθt, and the wave imp edances are Z1⊥=η1 cosθi, Z2⊥=η2 cosθt, Z1/bardbl=η1cosθi, Z2/bardbl=η2cosθt. The reﬂection coeﬃcients are ˜/Gamma1⊥=η2cosθi−η1cosθt η2cosθi+η1cosθt, ˜/Gamma1/bardbl=η2cosθt−η1cosθi η2cosθt+η1cosθi. (4.299) So the reﬂection coeﬃcients are purelyreal, with signs dependent on the constitutive parameters of the media. We can write ˜/Gamma1⊥=ρ⊥ejφ⊥, ˜/Gamma1/bardbl=ρ/bardblejφ/bardbl, where ρandφare real, and where φ=0orπ. Under certain conditions the reﬂection coeﬃcients vanish. For a given set of constitu- tive parameters we mayachieve ˜/Gamma1=0at an incidence angle θB, known as the Brewster orpolarizing angle . A wave with an arbitrarycombination of perpendicular and paral- lel polarized components incident at this angle produces a reﬂected ﬁeld with a singlecomponent. A wave incident with onlythe appropriate single component produces noreﬂected ﬁeld, regardless of its amplitude. For perpendicular polarization we set ˜/Gamma1 ⊥=0, requiring η2cosθi−η1cosθt=0 or equivalently µ2 /epsilon12(1−sin2θi)=µ1 /epsilon11(1−sin2θt). By(4.297) we mayput sin2θt=µ1/epsilon11 µ2/epsilon12sin2θi, resulting in sin2θi=µ2 /epsilon11/epsilon12µ1−/epsilon11µ2 µ2 1−µ2 2. The value of θithat satisﬁes this equation must be the Brewster angle, and thus θB⊥=sin−1/radicalBigg µ2 /epsilon11/epsilon12µ1−/epsilon11µ2 µ2 1−µ2 2. When µ1=µ2there is no solution to this equation, hence the reﬂection coeﬃcient cannot vanish. When /epsilon11=/epsilon12we have θB⊥=sin−1/radicalbiggµ2 µ1+µ2=tan−1/radicalbiggµ2 µ1. For parallel polarization we set ˜/Gamma1/bardbl=0and have η2cosθt=η1cosθi. Proceeding as above we ﬁnd that θB/bardbl=sin−1/radicalBigg /epsilon12 µ1/epsilon11µ2−/epsilon12µ1 /epsilon12 1−/epsilon12 2. This expression has no solution when /epsilon11=/epsilon12, and thus the reﬂection coeﬃcient cannot vanish under this condition. When µ1=µ2we have θB/bardbl=sin−1/radicalbigg/epsilon12 /epsilon11+/epsilon12=tan−1/radicalbigg/epsilon12 /epsilon11. We ﬁnd that when θi<θ cthe total ﬁeld in region 1 behaves as a traveling wave along x, but has characteristics of both a standing wave and a traveling wave along z (Proble m4.7).Thetraveling-wavecomponentisassociatedwithaPoyntingpowerﬂux, while the standing- wave component is not. This ﬂux is carried across the boundary into region 2 where the transmitted ﬁeld consists onlyof a traveling wave. By (4.161) the normal component of time-average Poynting ﬂux is continuous across the boundary,demonstrating that the time-average power carried bythe wave into the in terface from region1passesoutthroug htheinterfaceintoregion2(Proble m4.8). Case 2: θ i<θ c.The wave vectors are, from (4.289)–(4.290) and (4.296), ki=ˆxβ1sinθi+ˆzβ1cosθi, kr=ˆxβ1sinθi−ˆzβ1cosθi, kt=ˆxβ1sinθi−jˆzαc, where αc=/radicalBig β2 1sin2θi−β2 2 is the critical angle attenuation constant . The wave impedances are Z1⊥=η1 cosθi, Z2⊥=jβ2η2 αc, Z1/bardbl=η1cosθi, Z2/bardbl=− jαcη2 β2. Substituting these into (4.283) and (4.284), we ﬁnd that the reﬂection coeﬃcients are the complex quantities ˜/Gamma1⊥=β2η2cosθi+jη1αc β2η2cosθi−jη1αc=ejφ⊥, ˜/Gamma1/bardbl=−β2η1cosθi+jη2αc β2η1cosθi−jη2αc=ejφ/bardbl, where φ⊥=2 tan−1/parenleftbiggη1αc β2η2cosθi/parenrightbigg ,φ /bardbl=π+2 tan−1/parenleftbiggη2αc β2η1cosθi/parenrightbigg . We note with interest that ρ⊥=ρ/bardbl=1. So the amplitudes of the reﬂected waves are identical to those of the incident waves, and we call this the case of total internal reﬂection . The phase of the reﬂected wave at the interface is changed from that of the incident wave byan amount φ⊥orφ/bardbl. The phase shift incurred bythe reﬂected wave upon total internal reﬂection is called the Goos–H¨ anchen shift . In the case of total internal reﬂection the ﬁeld in region 1 is a pure standing wave while the ﬁeld in region 2 decays exponentially in the z-direction and is evanescent (Problem 4.9).Sinceastandin gwavetransportsnopower,thereisnoPoyntingﬂuxintoregion2. We ﬁnd that the evanescent wave also carries no power and thus the boundaryconditiononpowerﬂuxattheinterfaceissatisﬁe d(Proble m4.10).Wenotethatforanyincide nt angle except θ i=0(normal incidence) the wave in region 1 does transport power in the x-direction. Reﬂection of time-domain uniform plane waves. Solution for the ﬁelds reﬂected and transmitted at an interface shows us the properties of the ﬁelds for a certain singleexcitation frequencyand allows us to obtain time-domain ﬁelds byFourier inversion.Under certain conditions it is possible to do the inversion analytically, providing physicalinsight into the temporal behavior of the ﬁelds. As a simple example, consider a perpendicularly-polarized, uniform plane wave incident from free space at an angle θ i ontheplanarsurfac eofaconductin gmateria l(Figure4.18). Themateria lisassume dtohavefrequency-inde pendentconstituti veparameters ˜µ=µ0, ˜/epsilon1=/epsilon1, and ˜σ=σ. By(4.285) we have the reﬂected ﬁeld ˜Er ⊥(r,ω)=ˆy˜/Gamma1⊥(ω)˜Ei ⊥(ω)e−jkr(ω)·r=ˆy˜Er(ω)e−jωˆkr·r c (4.300) where ˜Er=˜/Gamma1⊥˜Ei ⊥. We can use the time-shifting theorem (A.3)to invert the transform and obtain Er ⊥(r,t)=F−1/braceleftbig˜Er ⊥(r,ω)/bracerightbig =ˆyEr/parenleftBigg t−ˆkr·r c/parenrightBigg (4.301) where we have bythe convolution theorem (12) Er(t)=F−1/braceleftbig˜Er(ω)/bracerightbig =/Gamma1⊥(t)∗E⊥(t). Here E⊥(t)=F−1/braceleftbig˜Ei ⊥(ω)/bracerightbig is the time waveform of the incident plane wave, while /Gamma1⊥(t)=F−1/braceleftbig˜/Gamma1⊥(ω)/bracerightbig is the time-domain reﬂection coeﬃcient. By(4.301) the reﬂected time-domain ﬁeld propagates along the direction ˆkrat the speed of light. The time waveform of the ﬁeld is the convolution of the waveform ofthe incident ﬁeld with the time-domain reﬂection coeﬃcient /Gamma1 ⊥(t). In the lossless case (σ=0),/Gamma1⊥(t)is aδ-function and thus the waveforms of the reﬂected and incident ﬁelds are identical. With the introduction of loss /Gamma1⊥(t)broadens and thus the reﬂected ﬁeld waveform becomes a convolution-broadened version of the incident ﬁeld waveform. Tounderstand the waveform of the reﬂected ﬁeld we must compute /Gamma1 ⊥(t). Note that by choosing the permittivityof region 2 to exceed that of region 1 we preclude total internal reﬂection. We can specialize the frequency-domain reﬂection coeﬃcient (4.283) for our problem bynoting that k1z=β1cosθi, k2z=/radicalBig k2 2−k2 1x=ω√µ0/radicalbigg /epsilon1+σ jω−/epsilon10sin2θi, and thus Z1⊥=η0 cosθi, Z2⊥=η0/radicalBig /epsilon1r+σ jω/epsilon10−sin2θi, where /epsilon1r=/epsilon1//epsilon1 0andη0=√µ0//epsilon10. We thus obtain ˜/Gamma1⊥=√s−√Ds+B√s+√Ds+B(4.302) where s=jωand D=/epsilon1r−sin2θi cos2θi, B=σ /epsilon10cos2θi. We can put (4.302) into a better form for inversion. We begin bysubtracting /Gamma1⊥∞, the high-frequencylimit of ˜/Gamma1⊥. Noting that lim ω→∞˜/Gamma1⊥(ω)=/Gamma1⊥∞=1−√ D 1+√ D, we can form ˜/Gamma10 ⊥(ω)=˜/Gamma1⊥(ω)−/Gamma1⊥∞=√s−√Ds+B√s+√Ds+B−1−√ D 1+√ D =2√ D 1+√ D/bracketleftbigg√s−√s+B/D √s+√ D√s+D/B/bracketrightbigg . With a bit of algebra this becomes ˜/Gamma10 ⊥(ω)=−2√ D D−1/parenleftBigg s s+B D−1/parenrightBigg 1−/radicalBigg s+B D s −2B/parenleftBig 1+√ D/parenrightBig (D−1)/parenleftBigg 1 s+B D−1/parenrightBigg . Now we can apply(C.12), (C.18), and (C.19) to obtain /Gamma10 ⊥(t)=F−1/braceleftbig˜/Gamma10 ⊥(ω)/bracerightbig =f1(t)+f2(t)+f3(t) (4.303) where f1(t)=−2B (1+√ D)(D−1)e−Bt D−1U(t), f2(t)=−B2 √ D(D−1)2U(t)/integraldisplayt 0e−B(t−x) D−1I/parenleftbiggBx 2D/parenrightbigg dx, f3(t)=B√ D(D−1)I/parenleftbiggBt 2D/parenrightbigg U(t). Here I(x)=e−x[I0(x)+I1(x)] where I0(x)and I1(x)are modiﬁed Bessel functions of the ﬁrst kind. Setting u=Bx/2D we can also write f2(t)=−2B√ D (D−1)2U(t)/integraldisplay Bt 2D 0e−Bt−2Du D−1I(u)du. Polynomial approximations for I(x)maybe found in Abramowitz and Stegun [ ?], making the computation of /Gamma10 ⊥(t)straightforward. The complete time-domain reﬂection coeﬃcient is /Gamma1⊥(t)=1−√ D 1+√ Dδ(t)+/Gamma10 ⊥(t). 0.0 2.5 5.0 7.5 10.0 12.5 t (ns)-0.5-0.4-0.3-0.2-0.110-9 Γ⊥(t) =3, σ=0.01 =80, σ=4 rr Figure 4.19: Time-domain reﬂection coeﬃcients. Ifσ=0then/Gamma10 ⊥(t)=0and the reﬂection coeﬃcient reduces to a single δ-function. Since convolution with this term does not alter wave shape, the reﬂected ﬁeld has the samewaveform as the incident ﬁeld. A plot of /Gamma1 0 ⊥(t)fornorma lincidenc e(θi = 00)isshowninFigure4.19.Heretwo material cases are displayed: /epsilon1r=3,σ=0.01S/m, which is representative of drywater ice, and /epsilon1r=80,σ=4S/m, which is representative of sea water. We see that a pulse waveform experiences more temporal spreading upon reﬂection from ice than from seawater, but that the amplitude of the dispersive component is less than that for sea water. Reﬂection of a nonuniform plane wave from a planar interface. Describing the interaction of a general nonuniform plane wave with a planar interface is problematicbecause of the non-TEM behavior of the incident wave. We cannot decompose the ﬁeldsinto two mutuallyorthogonal cases as we did with uniform waves, and thus the analy sis is more diﬃcult. However, we found in the last section that when a uniform wave is incident on a planar interface, the transmitted wave, even if non uniform in nature, takes on the same mathematical form and maybe decomposed in the same manner as theincident wave. Thus, we may studythe case in which this refracted wave is incident on a successive interface using exactlythe same analy sis as with a uniform incident wave.This is helpful in the case of multi-layered media, which we shall examine next. Interaction of a plane wave with multi-layered, planar materials. Consider N + 1 region sofspaceseparate dby N planarinterface sasshowninFigure4.20,and assume that a uniform plane wave is incident on the ﬁrst interface at angle θ i. Each region is assumed isotropic and homogeneous with a frequency-dependent complex permittivityand permeability. We can easily generalize the previous analysis regarding reﬂectionfrom a single interface byrealizing that in order to satisfythe boundaryconditions each Figure 4.20: Interaction of a uniform plane wavewith a multi-layered material. region, except region N, contains an incident-type wave of the form ˜Ei(r,ω)=˜Ei 0e−jki·r and a reﬂected-type wave of the form ˜Er(r,ω)=˜Er 0e−jkr·r. In region nwe maywrite the wave v ectors describing these waves as ki n=ˆxkx,n+ˆzkz,n, kr n=ˆxkx,n−ˆzkz,n, where k2 x,n+k2 z,n=k2 n, k2 n=ω2˜µn˜/epsilon1c n=(βn−jαn)2. We note at the outset that, as with the single interface case, the boundaryconditions are onlysatisﬁed when Snell’s law of reﬂection holds, and thus kx,n=kx,0=k0sinθi (4.304) where k0=ω(˜µ0˜/epsilon1c 0)1/2is the wavenumber of the 0th region (not necessarilyfree space). From this condition we have kz,n=/radicalBig k2n−k2 x,0=τne−jγn where τn=(A2 n+B2 n)1/4,γ n=1 2tan−1/parenleftbiggBn An/parenrightbigg , and An=β2 n−α2 n−(β2 0−α2 0)sin2θi, Bn=2(βnαn−β0α0sin2θi). Provided that the incident wave is uniform, we can decompose the ﬁelds in everyregion into cases of perpendicular and parallel polarization. This is true even when the waves in certain layers are nonuniform. For the case of perpendicular polarization we can write the electric ﬁeld in region n,0≤n≤N−1,a s ˜E⊥n=˜Ei ⊥n+˜Er ⊥nwhere ˜Ei ⊥n=ˆyan+1e−jkx,nxe−jkz,n(z−zn+1), ˜Er ⊥n=ˆybn+1e−jkx,nxe+jkz,n(z−zn+1), and the magnetic ﬁeld as ˜H⊥n=˜Hi ⊥n+Hr ⊥nwhere ˜Hi ⊥n=−ˆxkz,n+ˆzkx,n knηnan+1e−jkx,nxe−jkz,n(z−zn+1), ˜Hr ⊥n=+ˆxkz,n+ˆzkx,n knηnbn+1e−jkx,nxe+jkz,n(z−zn+1). When n=Nthere is no reﬂected wave; in this region we write ˜E⊥N=ˆyaN+1e−jkx,Nxe−jkz,N(z−zN), ˜H⊥N=−ˆxkz,N+ˆzkx,N kNηNaN+1e−jkx,Nxe−jkz,N(z−zN). Since a1is the known amplitude of the incident wave, there are 2Nunknown wave am- plitudes. We obtain the necessary 2Nsimultaneous equations byapply ing the boundary conditions at each of the interfaces. At interface nlocated at z=zn,1≤n≤N−1,w e have from the continuityof tangential electric ﬁeld an+bn=an+1e−jkz,n(zn−zn+1)+bn+1e+jkz,n(zn−zn+1) while from the continuityof magnetic ﬁeld −ankz,n−1 kn−1ηn−1+bnkz,n−1 kn−1ηn−1=−an+1kz,n knηne−jkz,n(zn−zn+1)+bn+1kz,n knηne+jkz,n(zn−zn+1). Noting that the wave imp edance of region nis Z⊥n=knηn kz,n and deﬁning the region npropagation factor as ˜Pn=e−jkz,n/Delta1n where /Delta1n=zn+1−zn, we can write an˜Pn+bn˜Pn=an+1+bn+1˜P2 n, (4.305) −an˜Pn+bn˜Pn=−an+1Z⊥n−1 Z⊥n+bn+1Z⊥n−1 Z⊥n˜P2 n. (4.306) We must still applythe boundaryconditions at z=zN. Proceeding as above, we ﬁnd that (4.305) and (4.306) hold for n=Nif we set bN+1=0and ˜PN=1. The 2Nsimultaneous equations (4.305)–(4.306) maybe solved using standard matrix methods. However, through a little manipulation we can put the equations into a formeasilysolved byrecursion, providing a verynice phy sical picture of the multiple reﬂectionsthat occur within the layered medium. We begin by eliminating b nbysubtracting (4.306) from (4.305): 2an˜Pn=an+1/bracketleftbigg 1+Z⊥n−1 Z⊥n/bracketrightbigg +bn+1˜P2 n/bracketleftbigg 1−Z⊥n−1 Z⊥n/bracketrightbigg . (4.307) Figure 4.21: Wave ﬂow diagram showing interaction of incident and reﬂected waves for region n. Deﬁning ˜/Gamma1n=Z⊥n−Z⊥n−1 Z⊥n+Z⊥n−1(4.308) as the interfacial reﬂection coeﬃcient for interface n(i.e., the reﬂection coeﬃcient as- suming a single interface as in (4.283)), and ˜Tn=2Z⊥n Z⊥n+Z⊥n−1=1+˜/Gamma1n as the interfacial transmission coeﬃcient for interface n, we can write (4.307) as an+1=an˜Tn˜Pn+bn+1˜Pn(−˜/Gamma1n)˜Pn. Finally, if we deﬁne the global reﬂection coeﬃcient Rnfor region nas the ratio of the amplitudes of the reﬂected and incident waves, ˜Rn=bn/an, we can write an+1=an˜Tn˜Pn+an+1˜Rn+1˜Pn(−˜/Gamma1n)˜Pn. (4.309) Forn=Nwe merelyset RN+1=0to ﬁnd aN+1=aN˜TN˜PN. (4.310) If we choose to eliminate an+1from (4.305) and (4.306) we ﬁnd that bn=an˜/Gamma1n+˜Rn+1˜Pn(1−˜/Gamma1n)an+1. (4.311) Forn=Nthis reduces to bN=aN˜/Gamma1N. (4.312) Equation s(4.309 )and(4.311 )havenicephysicalinterpretations .Conside rFigure4.21, which shows the wave amplitudes for region n. We maythink of the wave incident on interface n+1with amplitude an+1as consisting of two terms. The ﬁrst term is the wave transmitted through interface n(atz=zn). This wave must propagate through a distance /Delta1nto reach interface n+1and thus has an amplitude an˜Tn˜Pn. The second term is the reﬂection at interface nof the wave traveling in the −zdirection within region n. The amplitude of the wave before reﬂection is merely bn+1˜Pn, where the term ˜Pnresults from the propagation of the negatively-traveling wave from interface n+1to interface n. Now, since the interfacial reﬂection coeﬃcient at interface nfor a wave incident from region nis the negative of that for a wave incident from region n−1(since the wave is traveling in the reverse direction), and since the reﬂected wave must travel through a distance /Delta1nfrom interface nback to interface n+1, the amplitude of the second term is bn+1˜Pn(−/Gamma1n)˜Pn. Finally, remembering that bn+1=˜Rn+1an+1, we can write an+1=an˜Tn˜Pn+an+1˜Rn+1˜Pn(−˜/Gamma1n)˜Pn. This equation is exactlythe same as (4.309) which was found using the boundarycon- ditions. Bysimilar reasoning, we maysaythat the wave traveling in the −zdirection in region n−1consists of a term reﬂected from the interface and a term transmitted through the interface. The amplitude of the reﬂected term is merely an˜/Gamma1n. The amplitude of the transmitted term is found byconsidering bn+1=˜Rn+1an+1propagated through a distance /Delta1nand then transmitted backwards through interface n. Since the transmission coeﬃcient for a wave going from region nto region n−1is1+(−˜/Gamma1n), the amplitude of the transmitted term is ˜Rn+1˜Pn(1−˜/Gamma1n)an+1. Thus we have bn=˜/Gamma1nan+˜Rn+1˜Pn(1−˜/Gamma1n)an+1, which is identical to (4.311). We are still left with the task of solving for the various ﬁeld amplitudes. This can be done using a simple recursive technique. Using ˜Tn=1+˜/Gamma1nwe ﬁnd from (4.309) that an+1=(1+˜/Gamma1n)˜Pn 1+˜/Gamma1n˜Rn+1˜P2nan. (4.313) Substituting this into (4.311) we ﬁnd bn=˜/Gamma1n+˜Rn+1˜P2 n 1+˜/Gamma1n˜Rn+1˜P2nan. (4.314) Using this expression we ﬁnd a recursive relationship for the global reﬂection coeﬃcient: ˜Rn=bn an=˜/Gamma1n+˜Rn+1˜P2 n 1+˜/Gamma1n˜Rn+1˜P2n. (4.315) The procedure is now as follows. The global reﬂection coeﬃcient for interface Nis, from (4.312), ˜RN=bN/aN=˜/Gamma1N. (4.316) This is also obtained from (4.315) with ˜RN+1=0. We next use (4.315) to ﬁnd ˜RN−1: ˜RN−1=˜/Gamma1N−1+˜RN˜P2 N−1 1+˜/Gamma1N−1˜RN˜P2 N−1. This process is repeated until reaching ˜R1, whereupon all of the global reﬂection coeﬃ- cients are known. We then ﬁnd the amplitudes beginning with a1, which is the known incident ﬁeld amplitude. From (4.315) we ﬁnd b1=a1˜R1, and from (4.313) we ﬁnd a2=(1+˜/Gamma11)˜P1 1+˜/Gamma11˜R2˜P2 1a1. This process is repeated until all ﬁeld amplitudes are known. We note that the process outlined above holds equallywell for parallel polarization as long as we use the parallel wave impedances Z/bardbln=kz,nηn kn when computing the interfacial reﬂection coeﬃcients. See Problem ??. As a simple example, consider a slab of material of thickness /Delta1sandwiched between two lossless dielectrics. A time-harmonic uniform plane wave of frequency ω=ˇωis normallyincident onto interface 1, and we wish to compute the amplitude of the wavereﬂected byinterface 1 and determine the conditions under which the reﬂected wavevanishes. In this case we have N=2, with two interfaces and three regions. By(4.316) we have R 2=/Gamma12, where R2=˜R2(ˇω),/Gamma12=˜/Gamma12(ˇω), etc. Then by(4.315) we ﬁnd R1=/Gamma11+R2P2 1 1+/Gamma11R2P2 1=/Gamma11+/Gamma12P2 1 1+/Gamma11/Gamma12P2 1. Hence the reﬂected wave vanishes when /Gamma11+/Gamma12P2 1=0. Since the ﬁeld in region 0is normallyincident we have kz,n=kn=βn=ˇω√µn/epsilon1n. If we choose P2 1=−1, then /Gamma11=/Gamma12results in no reﬂected wave. This requires Z1−Z0 Z1+Z0=Z2−Z1 Z2+Z1. Clearing the denominator we ﬁnd that 2Z2 1=2Z0Z2or Z1=/radicalbig Z0Z2. This condition makes the reﬂected ﬁeld vanish if we can ensure that P2 1=− 1.T o d o this we need e−jβ12/Delta1=−1. The minimum thickness that satisﬁes this condition is β12/Delta1=π. Since β=2π/λ, this is equivalent to /Delta1=λ/4. A layer of this type is called a quarter-wave transformer . Since no wave is reﬂ ected from the initial interface, and since all the regions are assumed lossless, all of the power carriedbythe incident wave in the ﬁrst region is transferred into the third region. Thus, two regions of diﬀering materials maybe “matched” byinserting an appropriate slab between Figure 4.22: Interaction of a uniform plane wave with a conductor-backed dielectric slab. them. This technique ﬁnds use in optical coatings for lenses and for reducing the radar reﬂectivityof objects. As a second example, consider a lossless dielectric slab with ˜/epsilon1=/epsilon11 =/epsilon11r/epsilon10,and ˜µ=µ0, backedbyaperfectconducto randimmerse dinfreespaceasshowninFigure4.22.A perpendicularlypolarized uniform plane wave is incident on the slab from free spaceand we wish to ﬁnd the temporal response of the reﬂected wave byﬁrst calculating thefrequency-domain reﬂected ﬁeld. Since /epsilon1 0</epsilon1 1, total internal reﬂection cannot occur. Thus the wave v ectors in region 1 have real components and can be written as ki 1=kx,1ˆx+kz,1ˆz, kr 1=kx,1ˆx−kz,1ˆz. From Snell’s law of refraction we know that kx,1=k0sinθi=k1sinθt and so kz,1=/radicalBig k2 1−k2 x,1=ω c/radicalBig /epsilon11r−sin2θi=k1cosθt where θtis the transmission angle in region 1. Since region 2 is a perfect conductor we have ˜R2=−1. By(4.315) we have ˜R1(ω)=/Gamma11−˜P2 1(ω) 1−/Gamma11˜P2 1(ω), (4.317) where from (4.308) /Gamma11=Z1−Z0 Z1+Z0 is not a function of frequency. By the approach we used to obtain (4.300) we write ˜Er ⊥(r,ω)=ˆy˜R1(ω)˜Ei ⊥(ω)e−jkr 1(ω)·r. So Er ⊥(r,t)=ˆyEr/parenleftBigg t−ˆkr 1·r c/parenrightBigg where bythe convolution theorem Er(t)=R1(t)∗Ei ⊥(t). (4.318) Here Ei ⊥(t)=F−1/braceleftbig˜Ei ⊥(ω)/bracerightbig is the time waveform of the incident plane wave, while R1(t)=F−1/braceleftbig˜R1(ω)/bracerightbig is the global time-domain reﬂection coeﬃcient. To invert ˜R1(ω), we use the binomial expansion (1−x)−1=1+x+x2+x3+··· on the denominator of (4.317), giving ˜R1(ω)=[/Gamma11−˜P2 1(ω)]/braceleftbig 1+[/Gamma11˜P2 1(ω)]+[/Gamma11˜P2 1(ω)]2+[/Gamma11˜P2 1(ω)]3+.../bracerightbig =/Gamma11−[1−/Gamma12 1]˜P2 1(ω)−[1−/Gamma12 1]/Gamma11˜P4 1(ω)−[1−/Gamma12 1]/Gamma12 1˜P6 1(ω)−···.(4.319) Thus we need the inverse transform of ˜P2n 1(ω)=e−j2nkz,1/Delta11=e−j2nk1/Delta11cosθt. Writing k1=ω/v 1, where v1=1/(µ 0/epsilon11)1/2is the phase velocityof the wave in region 1, and using 1↔δ(t)along with the time-shifting theorem (A.3) we have ˜P2n 1(ω)=e−jω2nτ↔δ(t−2nτ) where τ=/Delta11cosθt/v1. With this the inverse transform of ˜R1in (4.319) is R1(t)=/Gamma11δ(t)−(1+/Gamma11)(1−/Gamma11)δ(t−2τ)−(1+/Gamma11)(1−/Gamma11)/Gamma11δ(t−4τ)−··· and thus from (4.318) Er(t)=/Gamma11Ei ⊥(t)−(1+/Gamma11)(1−/Gamma11)Ei ⊥(t−2τ)−(1+/Gamma11)(1−/Gamma11)/Gamma11Ei ⊥(t−4τ)−···. The reﬂected ﬁeld consists of time-shifted and amplitude-scaled versions of the incident ﬁeld waveform. These terms can be interpreted as multiple reﬂections of the incident wave.Conside rFigure4.23.Theﬁrsttermisthedirectreﬂectio nfrominterface1and thus has its amplitude multiplied by /Gamma11. The next term represents a wave that pene- trates the interface (and thus has its amplitude multiplied bythe transmission coeﬃcient1+/Gamma1 1), propagates to and reﬂects from the conductor (and thus has its amplitude mul- tiplied by −1), and then propagates back to the interface and passes through in the opposite direction (and thus has its amplitude multiplied bythe transmission coeﬃcientfor passage from region 1 to region 0,1−/Gamma1 1). The time delaybetween this wave and the initially-reﬂected wave is given by 2τ, as discussed in detail below. The third term represents a wave that penetrates the interface, reﬂects from the conductor, returns toand reﬂects from the interface a second time, again reﬂects from the conductor, andthen passes through the interface in the opposite direction. Its amplitude has an ad-ditional multiplicative factor of −/Gamma1 1to account for reﬂection from the interface and an additional factor of −1to account for the second reﬂection from the conductor, and is time-delayed by an additional 2τ. Subsequent terms account for additional reﬂections; Figure 4.23: Timing diagram for multiple reﬂections from a conductor-backed dielectric slab. thenth reﬂected wave amplitude is multiplied byan additional (−1)nand(−/Gamma11)nand is time-delayed by an additional 2nτ. It is important to understand that the time delay 2τisnotjust the propagation time for the wave to travel through the slab. To properlydescribe the timing between theinitially-reﬂected wave and the waves that reﬂect from the conductor we must consider theﬁeldoveridenticalobservationplanesasshowninFigure4.23.Forexample ,consider the observation plane designated P-P intersecting the ﬁrst “exit point” on interface 1.To arrive at this plane the initially-reﬂected wave takes the path labeled B, arriving at a time Dsinθ i v0 after the time of initial reﬂection, where v0=cis the velocityin region 0. To arrive at this same plane the wave that penetrates the surface takes the path labeled A, arriving at a time 2/Delta11 v1cosθt where v1is the wave velocityin region 1 and θtis the transmission angle. Noting that D=2/Delta11tanθt, the time delaybetween the arrival of the two waves at the plane P-P is T=2/Delta11 v1cosθt−Dsinθi v0=2/Delta11 v1cosθt/bracketleftbigg 1−sinθtsinθi v0/v1/bracketrightbigg . BySnell’s law of refraction (4.297) we can write v0 v1=sinθi sinθt, which, upon substitution, gives T=2/Delta11cosθt v1. This is exactlythe time delay 2τ. 4.11.6 Plane-wave propagation in an anisotropic ferrite medium Several interesting properties of plane waves, such as Faradayrotation and the exis- tence of stopbands, appear onlywhen the waves propagate through anisotropic media.We shall studythe behavior of waves propagating in a magnetized ferrite medium, andnote that this behavior is shared bywaves propagating in a magnetized plasma, becauseof the similarityin the dy adic constitutive parameters of the two media. Consider a uniform ferrite material having scalar permittivity ˜/epsilon1=/epsilon1and dyadic per- meability ˜¯µ. We assume that the ferrite is lossless and magnetized along the z-direction. By(4.115)– (4.117) the permeabilityof the medium is [˜¯µ(ω)]= µ 1jµ20 −jµ2µ10 00 µ0  where µ1=µ0/bracketleftbigg 1+ωMω0 ω2 0−ω2/bracketrightbigg ,µ 2=µ0ωω M ω2 0−ω2. The source-free frequency-domain wave equation can be found using (4.201) with˜¯ζ= ˜¯ξ=0and ˜¯/epsilon1=/epsilon1¯I: /bracketleftbigg ¯∇·/parenleftbigg ¯I1 /epsilon1/parenrightbigg ·¯∇−ω2˜¯µ/bracketrightbigg ·˜H=0 or, since ¯∇·A=∇× A, 1 /epsilon1∇×/parenleftbig ∇× ˜H/parenrightbig −ω2˜¯µ·˜H=0. (4.320) The simplest solutions to the wave equation for this anisotropic medium are TEM plane waves that propagate along the applied dc magnetic ﬁeld. We thus seek solutionsof the form ˜H(r,ω)=˜H 0(ω)e−jk·r(4.321) where k=ˆzβand ˆz·˜H0=0. We can ﬁnd βbyenforcing (4.320). From (B.7) we ﬁnd that ∇× ˜H=− jβˆz×˜H0e−jβz. ByAmpere’s law we have ˜E=∇× ˜H jω/epsilon1=− ZTEMˆz×˜H, (4.322) where ZTEM=β/ω/epsilon1 is the wave imp edance. Note that the wave is indeed TEM. The second curl is found to be ∇×/parenleftbig ∇× ˜H/parenrightbig =− jβ∇×/bracketleftbigˆz×˜H0e−jβz/bracketrightbig . After an application of (B.43) this becomes ∇×/parenleftbig ∇× ˜H/parenrightbig =− jβ/bracketleftbig e−jβz∇×(ˆz×˜H0)−(ˆz×˜H0)×∇e−jβz/bracketrightbig . The ﬁrst term on the right-hand side is zero, and thus using (B.76) we have ∇×/parenleftbig ∇× ˜H/parenrightbig =/bracketleftbig −jβe−jβzˆz×(ˆz×˜H0)/bracketrightbig (−jβ) or, using (B.7), ∇×/parenleftbig ∇× ˜H/parenrightbig =β2e−jβz˜H0 since ˆz·˜H0=0. With this (4.320) becomes β2˜H0=ω2/epsilon1˜¯µ·˜H0. (4.323) We can solve (4.323) for βbywriting the vector equation in component form: β2H0x=ω2/epsilon1/bracketleftbig µ1H0x+jµ2H0y/bracketrightbig , β2H0y=ω2/epsilon1/bracketleftbig −jµ2H0x+µ1H0y/bracketrightbig . In matrix form these are /bracketleftbiggβ2−ω2/epsilon1µ1−jω2/epsilon1µ2 jω2/epsilon1µ2β2−ω2/epsilon1µ1/bracketrightbigg/bracketleftbiggH0x H0y/bracketrightbigg =/bracketleftbigg0 0/bracketrightbigg , (4.324) and nontrivial solutions occur onlyif /vextendsingle/vextendsingle/vextendsingle/vextendsingleβ 2−ω2/epsilon1µ1−jω2/epsilon1µ2 jω2/epsilon1µ2β2−ω2/epsilon1µ1/vextendsingle/vextendsingle/vextendsingle/vextendsingle=0. Expansion yields the two solutions β ±=ω√/epsilon1µ± (4.325) where µ±=µ1±µ2=µ0/bracketleftbigg 1+ωM ω0∓ω/bracketrightbigg . (4.326) So the propagation properties of the plane wave are the same as those in a medium with an equivalent scalar permeabilitygiven by µ±. Associated with each of these solutions is a relationship between H0xand H0ythat can be found from (4.324). Substituting β+into the ﬁrst equation we have ω2/epsilon1µ2H0x−jω2/epsilon1µ2H0y=0 orH0x=jH0y. Similarly, substitution of β−produces H0x=− jH0y. Thus, by(4.321) the magnetic ﬁeld maybe expressed as ˜H(r,ω)=H0y[±jˆx+ˆy]e−jβ±z. By(4.322) we also have the electric ﬁeld ˜E(r,ω)=ZTEMH0y[ˆx+e∓jπ 2ˆy]e−jβ±z. This ﬁeld has the form of (4.248). For β+we have φy−φx=−π/2and thus the wave exhibits RHCP. For β−we have φy−φx=π/2and the wave exhibits LHCP. 012345 β/( /v )01234ω/ω0 Light LineRHCP stopbandRHCP RHCPLHCP 0c Figure 4.24: Dispersion plot for unmagnetized ferrite with ωM=2ω0. Light line shows ω/β=vc=1/(µ 0/epsilon1)1/2. Thedispersiondiagra mforeachpolarizatio ncaseisshowninFigure4.24,wherewe have arbitrarilychosen ωM=2ω0. Here we have combined (4.325) and (4.326) to produce the normalized expression β± ω0/vc=ω ω0/radicalBigg 1+ωM/ω0 1∓ω/ω 0 where vc=1/(µ 0/epsilon1)1/2. Except at low frequencies, an LHCP plane wavepasses through the ferrite as if the permeabilityis close to that of free space. Over all frequencies wehavev p<v candvg<v c. In contrast, an RHCP wave excites the electrons in the ferrite and a resonance occurs at ω=ω0. For all frequencies below ω0we have vp<v cand vg<v cand both vpandvgreduce to zero as ω→ω0. Because the ferrite is lossless, frequencies between ω=ω0andω=ω0+ωMresult in βbeing purelyimaginaryand thus the wave being evanescent. We thus call the frequencyrange ω0<ω<ω 0+ωM astopband ; within this band the plane wave cannot transport energy. For frequencies above ω0+ωMthe RHCP wavepropagates as if it is in a medium with permeabilityless than that of free space. Here we have vp>v candvg<v c, with vp→vcandvg→vcas ω→∞. Faraday rotation. The solutions to the wave equation found above do not allow the existence of linearlypolarized plane waves. However, bysuperposing LHCP and RHCPwaves we can obtain a wave with the appearance of linear polarization. That is, overany z-plane the electric ﬁeld vector maybe written as ˜E=K(E x0ˆx+Ey0ˆy)where Ex0 and Ey0are real (although Kmaybe complex). To see this let us examine ˜E=˜E++˜E−=E0 2[ˆx−jˆy]e−jβ+z+E0 2[ˆx+jˆy]e−jβ−z =E0 2/bracketleftbigˆx/parenleftbig e−jβ+z+e−jβ−z/parenrightbig +jˆy/parenleftbig −e−jβ+z+e−jβ−z/parenrightbig/bracketrightbig =E0e−j1 2(β++β−)z/bracketleftbigg ˆxcos1 2(β+−β−)z+ˆysin1 2(β+−β−)z/bracketrightbigg or ˜E=E0e−j1 2(β++β−)z[ˆxcosθ(z)+ˆysinθ(z)] where θ(z)=(β+−β−)z/2. Because β+/negationslash=β−, the velocities of the two circularly polarized waves diﬀer and the waves superpose to form a linearlypolarized wave with apolarization that depends on the observation plane z-value. We maythink of the wave as undergoing a phase shift of (β ++β−)z/2radians as it propagates, while the direction of˜Erotates to an angle θ(z)=(β+−β−)z/2as the wave propagates. Faraday rotation can onlyoccur at frequencies where both the LHCP and RHCP waves propagate, andtherefore not within the stopband ω 0<ω<ω 0+ωM. Faradayrotation is non-reciprocal. That is, if a wave that has undergone a rotation of θ0radians bypropagating through a distance z0is made to propagate an equal distance back in the direction from whence it came, the polarization does not return to its initialstate but rather incurs an additional rotation of θ 0. Thus, the polarization angle of the wave when it returns to the starting point is not zero, but 2θ0. This eﬀect is employed in a number of microwave devices including gyrators, isolators, and circulators. Theinterested reader should see Collin [40], Elliott [67], or Liao [111] for details. We notethat for ω/greatermuchω Mwe can approximate the rotation angle as θ(z)=(β+−β−)z/2=1 2ωz√/epsilon1µ0/bracketleftbigg/radicalbigg 1+ωM ω0−ω−/radicalbigg 1+ωM ω0+ω/bracketrightbigg ≈−1 2zωM√/epsilon1µ0, which is independent of frequency. So it is possible to construct Faraday rotation-based ferrite devices that maintain their properties over wide bandwidths. It is straightforward to extend the above analysis to the case of a lossy ferrite. We ﬁnd that for typical ferrites the attenuation constant associated with µ−is small for all frequencies, but the attenuation constant associated with µ+is large near the resonant frequency( ω≈ω0)[40].SeeProble m4.16. 4.11.7 Propagation of cylindrical waves Bystudy ing plane waves we have gained insight into the basic behavior of frequency- domain and time-harmonic waves. However, these solutions do not displaythe funda-mental propertythat waves in space must diverge from their sources. To understand this behavior we shall treat waves ha ving cylindrical and spherical symmetries. Uniform cylindrical waves. In§2.10.7 we studied the temporal behavior of cylin- drical waves in a homogeneous, lossless medium and found that theydiverge from a line source located along the z-axis. Here we shall extend the analysis to lossy media and investigate the behavior of the waves in the frequencydomain. Consider a homogeneous region of space described bythe permittivity ˜/epsilon1(ω), permeabil- ity˜µ(ω), and conductivity ˜σ(ω). We seek solutions that are invariant over a cylindrical surface: ˜E(r,ω)=˜E(ρ, ω), ˜H(r,ω)=˜H(ρ, ω) . Such waves are called uniform cylindrical waves . Since the ﬁelds are z-independent we maydecompose them into TE and TM sets as described in §4.11.2. For TM polarization we mayinsert (4.211) into (4.212) to ﬁnd ˜H φ(ρ, ω) =1 jω˜µ(ω)∂˜Ez(ρ, ω) ∂ρ. (4.327) For TE polarization we have from (4.213) ˜Eφ(ρ, ω) =−1 jω˜/epsilon1c(ω)∂˜Hz(ρ, ω) ∂ρ(4.328) where ˜/epsilon1c=˜/epsilon1+˜σ/jωis the complex permittivityintroduced in §4.4.1. Since ˜E= ˆφ˜Eφ+ˆz˜Ezand ˜H=ˆφ˜Hφ+ˆz˜Hz, we can always decompose a cylindrical electromagnetic wave into cases of electric and magnetic polarization. In each case the resulting ﬁeld isTEM ρsince ˜E,˜H, and ˆρare mutuallyorthogonal. Wave equations for ˜Ezin the electric polarization case and for ˜Hzin the magnetic polarization case can be derived bysubstituting (4.210) into (4.208): /parenleftbigg∂2 ∂ρ2+1 ρ∂ ∂ρ+k2/parenrightbigg/braceleftbigg˜Ez ˜Hz/bracerightbigg =0. Thus the electric ﬁeld must obeythe ordinarydiﬀerential equation d2˜Ez dρ2+1 ρd˜Ez dρ+k2˜Ez=0. (4.329) This is merelyBessel’s equation (A.124). It is a second-order equation with two inde- pendent solutions chosen from the list J0(kρ), Y0(kρ), H(1) 0(kρ), H(2) 0(kρ). We ﬁnd that J0(kρ)and Y0(kρ)are useful for describing standing waves between bound- aries, while H(1) 0(kρ)and H(2) 0(kρ)are useful for describing waves propagating in the ρ-direction. Of these, H(1) 0(kρ)represents waves traveling inward while H(2) 0(kρ)repre- sents waves traveling outward. At this point we are interested in studying the behaviorof outward propagating waves and so we choose ˜E z(ρ, ω) =−j 4˜Ez0(ω)H(2) 0(kρ). (4.330) As explained in §2.10.7, ˜Ez0(ω)is the amplitude spectrum of the wave, while the term −j/4is included to make the conversion to the time domain more convenient. By(4.327) we have ˜Hφ=1 jω˜µ∂˜Ez ∂ρ=1 jω˜µ∂ ∂ρ/bracketleftbigg −j 4˜Ez0H(2) 0(kρ)/bracketrightbigg . (4.331) Using dH(2) 0(x)/dx=− H(2) 1(x)we ﬁnd that ˜Hφ=1 ZTM˜Ez0 4H(2) 1(kρ) (4.332) where ZTM=ω˜µ k is called the TM wave impedance . For the case of magnetic polarization, the ﬁeld ˜Hzmust satisfyBessel’s equation (4.329). Thus we choose ˜Hz(ρ, ω) =−j 4˜Hz0(ω)H(2) 0(kρ). (4.333) From (4.328) we ﬁnd the electric ﬁeld associated with the wave: ˜Eφ=− ZTE˜Hz0 4H(2) 1(kρ), (4.334) where ZTE=k ω˜/epsilon1c is the TE wave impedance . It is not readilyapparent that the terms H(2) 0(kρ)orH(2) 1(kρ)describe outward prop- agating waves. We shall see later that the cylindrical wave may be written as a su-perposition of plane waves, both uniform and evanescent, propagating in all possibledirections. Each of these components does have the expected wave beha vior, but it is still not obvious that the sum of such waves is outward propagating. We saw in §2.10.7 that when examined in the time domain, a cylindrical wave of the form H (2) 0(kρ)does indeed propagate outward, and that for lossless media the velocityof propagation of its wavefronts is v=1/(µ/epsilon1)1/2. For time-harmonic ﬁelds, the cylindrical wave takes on a familiar behavior when the observation point is suﬃcientlyremoved fromthe source. We mayspecialize (4.330) to the time-harmonic case bysetting ω=ˇωand using phasors, giving ˇE z(ρ)=−j 4ˇEz0H(2) 0(kρ). If|kρ|/greatermuch1we can use the asymptotic representation (E.62) for the Hankel function H(2) ν(z)∼/radicalbigg 2 πze−j(z−π/4−νπ/2), |z|/greatermuch1,−2π< arg(z)<π , to obtain ˇEz(ρ)∼ˇEz0e−jkρ √8jπkρ(4.335) and ˇHφ(ρ)∼− ˇEz01 ZTMe−jkρ √8jπkρ(4.336) for|kρ|/greatermuch 1. Except for the√ρterm in the denominator, the wave has verymuch the same form as the plane waves encountered earlier. For the case of magnetic polarization, we can approximate (4.333) and (4.334) to obtain ˇHz(ρ)∼ˇHz0e−jkρ √8jπkρ(4.337) and ˇEφ(ρ)∼ZTEˇHz0e−jkρ √8jπkρ(4.338) for|kρ|/greatermuch1. To interpret the wave nature of the ﬁeld (4.335) let us substitute k=β−jαinto the exponential function, where βis the phase constant (4.224) and αis the attenuation constant (4.225). Then ˇEz(ρ)∼ˇEz01√8jπkρe−αρe−jβρ. Assuming ˇEz0=|Ez0|ejξE, the time-domain representation is found from (4.126): Ez(ρ,t)=|Ez0|√8πkρe−αρcos[ˇωt−βρ−π/4+ξE]. (4.339) We can identifya surface of constant phase as a locus of points obey ing ˇωt−βρ−π/4+ξE=CP (4.340) where CPis some constant. These surfaces are cylinders coaxial with the z-axis, and are called cylindrical wavefronts . Note that surfaces of constant amplitude, as determined by e−αρ √ρ=CA where CAis some constant, are also cylinders. The cosine term in (4.339) represents a traveling wave. As tis increased the argument of the cosine function remains ﬁxed as long as ρis increased correspondingly. Hence the cylindrical wavefronts propagate outward as time progresses. As the wavefront travels outward, the ﬁeld is attenuated because of the factor e−αρ. The velocityof propagation of the phase fronts maybe computed bya now familiar technique. Diﬀerentiating (4.340)with respect to twe ﬁnd that ˇω−βdρ dt=0, and thus have the phase velocity vpof the outward expanding phase fronts: vp=dρ dt=ˇω β. Calculation of wavelength also proceeds as before. Examining the two adjacent wave- fronts that produce the same value of the cosine function in (4.339), we ﬁnd βρ1= βρ2−2πor λ=ρ2−ρ1=2π/β. Computation of the power carried bya cy lindrical wave is straightforward. Since a cylindrical wavefront is inﬁnite in extent, we usually speak of the power per unit length carried bythe wave. This is found byintegrating the time-average Poy nting ﬂux givenin (4.157). For electric polarization we ﬁnd the time-average power ﬂux densityusing (4.330) and (4.331): S av=1 2Re{ˇEzˆz×ˇH∗ φˆφ}=1 2Re/braceleftbigg ˆρj 16Z∗ TM|ˇEz0|2H(2) 0(kρ)H(2)∗ 1(kρ)/bracerightbigg . (4.341) For magnetic polarization we use (4.333) and (4.334): Sav=1 2Re{ˇEφˆφ×ˇH∗ zˆz}=1 2Re/braceleftbigg −ˆρjZTE 16|ˇHz0|2H(2)∗ 0(kρ)H(2) 1(kρ)/bracerightbigg . For a lossless medium these expressions can be greatlysimpliﬁed. By(E.5) we can write jH(2) 0(kρ)H(2)∗ 1(kρ)=j[J0(kρ)−jN0(kρ)][J1(kρ)+jN1(kρ)], hence jH(2) 0(kρ)H(2)∗ 1(kρ)=[N0(kρ)J1(kρ)−J0(kρ)N1(kρ)]+j[J0(kρ)J1(kρ)+N0(kρ)N1(kρ)]. Substituting this into (4.341) and remembering that ZTM=η=(µ//epsilon1)1/2is real for lossless media, we have Sav=ˆρ1 32η|ˇEz0|2[N0(kρ)J1(kρ)−J0(kρ)N1(kρ)]. Bythe Wronskian relation (E.88) we have Sav=ˆρ|ˇEz0|2 16πkρη. The power densityis inverselyproportional to ρ. When we compute the total time- average power per unit length passing through a cylinder of radius ρ, this factor cancels with the ρ-dependence of the surface area to give a result independent of radius: Pav/l=/integraldisplay2π 0Sav·ˆρρdφ=|ˇEz0|2 8kη. (4.342) For a lossless medium there is no mechanism to dissipate the power and so the waveprop- agates unabated. A similar calculation for the case of magnetic polarization (Problem??) gives S av=ˆρη|ˇHz0|2 16πkρ and Pav/l=η|ˇHz0|2 8k. For a lossymedium the expressions are more diﬃcult to evaluate. In this case we expect the total power passing through a cylinder to depend on the radius of the cylinder, sincethe ﬁelds decayexponentiallywith distance and thus give up power as theypropagate.If we assume that the observation point is far from the z-axis with |kρ|/greatermuch1, then we can use (4.335) and (4.336) for the electric polarization case to obtain S av=1 2Re{ˇEzˆz×ˇH∗ φˆφ}=1 2Re/braceleftbigg ˆρe−2αρ 8πρ|k|Z∗ TM|ˇEz0|2/bracerightbigg . Therefore Pav/l=/integraldisplay2π 0Sav·ˆρρdφ=Re/braceleftbigg1 Z∗ TM/bracerightbigg |ˇEz0|2e−2αρ 8|k|. We note that for a lossless material ZTM=ηandα=0, and the expression reduces to (4.342) as expected. Thus for lossymaterials the power depends on the radius of thecylinder. In the case of magnetic polarization we use (4.337) and (4.338) to get S av=1 2Re{ˇEφˆφ×ˇH∗ zˆz}=1 2Re/braceleftbigg ˆρZ∗ TEe−2αρ 8πρ|k||ˇHz0|2/bracerightbigg and Pav/l=Re/braceleftbig Z∗ TE/bracerightbig |ˇHz0|2e−2αρ 8|k|. Example of uniform cylindrical waves: ﬁelds of a line source. The simplest example of a uniform cylindrical wave is that produced byan electric or magnetic line source. Consider ﬁrst an inﬁnite electric line current of amplitude ˜I(ω)on the z-axis, immersed within a medium of permittivity ˜/epsilon1(ω), permeability ˜µ(ω), and conductivity ˜σ(ω). We assume that the current does not varyin the z-direction, and thus the problem is two-dimensional. We can decompose the ﬁeld produced bythe line source into TE andTM cases according to §4.11.2. It turns out that an electric line source onlyexcites TM ﬁelds, as we shall show in §5.4, and thus we need only ˜E zto completelydescribe the ﬁelds. Bysy mmetrythe ﬁelds are φ-independent and thus the wave produced bythe line source is a uniform cylindrical wave. Since the wavepropagates outward from the line source we have the electric ﬁeld from (4.330), ˜Ez(ρ, ω) =−j 4˜Ez0(ω)H(2) 0(kρ), (4.343) and the magnetic ﬁeld from (4.332), ˜Hφ(ρ, ω) =k ω˜µ˜Ez0(ω) 4H(2) 1(kρ). We can ﬁnd ˜Ez0byusing Ampere’s law: /contintegraldisplay /Gamma1˜H·dl=/integraldisplay S˜J·dS+jω/integraldisplay S˜D·dS. Since ˜Jis the sum of the impressed current ˜Iand the secondaryconduction current ˜σ˜E, we can also write /contintegraldisplay /Gamma1˜H·dl=˜I+/integraldisplay S(˜σ+jω˜/epsilon1)˜E·dS=˜I+jω˜/epsilon1c/integraldisplay S˜E·dS. Choosing our path of integration as a circle of radius ain the z=0plane and substituting for˜Ezand ˜Hφ, we ﬁnd that k ω˜µ˜Ez0 4H(2) 1(ka)2πa=˜I+jω˜/epsilon1c2π−j˜Ez0 4lim δ→0/integraldisplaya δH(2) 0(kρ)ρdρ. (4.344) The limit operation is required because H(2) 0(kρ)diverges as ρ→0. By(E.104) the integral is lim δ→0/integraldisplaya δH(2) 0(kρ)ρdρ=a kH(2) 1(ka)−1 klim δ→0δH(2) 1(kδ). The limit maybe found byusing H(2) 1(x)=J1(x)−jN1(x)and the small argument approximations (E.50) and (E.53): lim δ→0δH(2) 1(δ)=lim δ→0δ/bracketleftbiggkδ 2−j/parenleftbigg −1 π2 kδ/parenrightbigg/bracketrightbigg =j2 πk. Substituting these expressions into (4.344) we obtain k ω˜µ˜Ez0 4H(2) 1(ka)2πa=˜I+jω˜/epsilon1c2π−j˜Ez0 4/bracketleftbigga kH(2) 1(ka)−j2 πk2/bracketrightbigg . Using k2=ω2˜µ˜/epsilon1cwe ﬁnd that the two Hankel function terms cancel. Solving for ˜Ez0we have ˜Ez0=− jω˜µ˜I and therefore ˜Ez(ρ, ω) =−ω˜µ 4˜I(ω)H(2) 0(kρ)=− jω˜µ˜I(ω)˜G(x,y|0,0;ω). (4.345) Here ˜Gis called the two-dimensional Green’s function and is given by ˜G(x,y|x/prime,y/prime;ω)=1 4jH(2) 0/parenleftBig k/radicalbig (x−x/prime)2+(y−y/prime)2/parenrightBig . (4.346) Green’s functions are examined in greater detail in Chapter 5 It is also possible to determine the ﬁeld amplitude byevaluating lim a→0/contintegraldisplay C˜H·dl. This produces an identical result and is a bit simpler since it can be argued that the surface integral of ˜Ezvanishes as a→0without having to perform the calculation directly[83, 8]. For a magnetic line source ˜Im(ω)aligned along the z-axis we proceed as above, but note that the source onlyproduces TE ﬁelds. By(4.333) and (4.334) we have ˜Hz(ρ, ω) =−j 4˜Hz0(ω)H(2) 0(kρ), ˜Eφ=−k ω˜/epsilon1c˜H0z 4H(2) 1(kρ). We can ﬁnd ˜Hz0by applying Faraday’s law /contintegraldisplay C˜E·dl=−/integraldisplay S˜Jm·dS−jω/integraldisplay S˜B·dS about a circle of radius ain the z=0plane. We have −k ω˜/epsilon1c˜Hz0 4H(2) 1(ka)2πa=− ˜Im−jω˜µ/bracketleftbigg −j 4/bracketrightbigg ˜Hz02πlim δ→0/integraldisplaya δH(2) 0(kρ)ρdρ. Proceeding as above we ﬁnd that ˜Hz0=jω˜/epsilon1c˜Im hence ˜Hz(ρ, ω) =−ω˜/epsilon1c 4˜Im(ω)H(2) 0(kρ)=− jω˜/epsilon1c˜Im(ω)˜G(x,y|0,0;ω). (4.347) Note that we could have solved for the magnetic ﬁeld of a magnetic line current by using the ﬁeld of an electric line current and the principle of duality. Letting the magneticcurrent be equal to −ηtimes the electric current and using (4.198), we ﬁnd that ˜H z0=/parenleftbigg −1 η˜Im(ω) ˜I(ω)/parenrightbigg/parenleftbigg −1 η/bracketleftbigg −ω˜µ 4˜I(ω)H(2) 0(kρ)/bracketrightbigg/parenrightbigg =− ˜Im(ω)ω˜/epsilon1c 4H(2) 0(kρ) (4.348) as in (4.347). Nonuniform cylindrical waves. When we solve two-dimensional boundaryvalue problems we encounter cylindrical waves that are z-independent but φ-dependent. Al- though such waves propagate outward, theyhave a more complicated structure than those considered above. For the case of TM polarization we have, by(4.212), ˜Hρ=j ZTMk1 ρ∂˜Ez ∂φ, (4.349) ˜Hφ=−j ZTMk∂˜Ez ∂ρ, (4.350) where ZTM=ω˜µ/k. For the TE case we have, by(4.213), ˜Eρ=−jZTE k1 ρ∂˜Hz ∂φ, (4.351) ˜Eφ=jZTE k∂˜Hz ∂ρ, (4.352) where ZTE=k/ω˜/epsilon1c. By(4.208) the wave equations are /parenleftbigg∂2 ∂ρ2+1 ρ∂ ∂ρ+1 ρ2∂2 ∂φ2+k2/parenrightbigg/braceleftbigg˜Ez ˜Hz/bracerightbigg =0. Because this has the form of A.177 with ∂/∂z→0,w eh a v e /braceleftbigg˜Ez(ρ, φ, ω) ˜Hz(ρ, φ, ω)/bracerightbigg =P(ρ, ω)/Phi1(φ, ω) (4.353) where /Phi1(φ,ω) =Aφ(ω)sinkφφ+Bφ(ω)coskφφ, (4.354) P(ρ)=Aρ(ω)B(1) kφ(kρ)+Bρ(ω)B(2) kφ(kρ), (4.355) and where B(1) ν(z)and B(2) ν(z)are anytwo independent Bessel functions chosen from the set Jν(z), Nν(z), H(1) ν(z), H(2) ν(z). In bounded regions we generallyuse the oscillatoryfunctions Jν(z)and Nν(z)to represent standing waves. In unbounded regions we generallyuse H(2) ν(z)and H(1) ν(z)to represent outward and inward propagating waves, respectively. Boundary value problems in cylindrical coordinates: scattering by a material cylinder. A varietyof problems can be solved using nonuniform cy lindrical waves. We shall examine two interesting cases in which an external ﬁeld is impressed on atwo-dimensional object. The impressed ﬁeld creates secondarysources within or on theobject, and these in turn create a secondaryﬁeld. Our goal is to determine the secondaryﬁeld byapply ing appropriate boundaryconditions. As a ﬁrst example, consider a material cylinder of radius a, complex permittivity ˜/epsilon1 c, andpermeabili ty ˜µ,aligne dalongthe z-axisinfreespace(Figure4.25).Anincide nt plane wave propagating in the x-direction is impressed on the cylinder, inducing sec- ondarypolarization and conduction currents within the cy linder. These in turn produce Figure 4.25: TM plane- waveﬁeld incident on a material cylinder. secondaryor scattered ﬁelds, which are standing waves within the cy linder and outward traveling waves external to the cylinder. Although we have not yet learned how to writethe secondaryﬁelds in terms of the impressed sources, we can solve for the ﬁelds as aboundaryvalue problem. The total ﬁeld must obeythe boundaryconditions on tangen-tial components at the interface between the cylinder and surrounding free space. Weneed not worryabout the eﬀect of the secondarysources on the source of the primaryﬁeld, since bydeﬁnition impressed sources cannot be inﬂuenced bysecondaryﬁelds. The scattered ﬁeld can be found using superposition. When excited bya TM impressed ﬁeld, the secondaryﬁeld is also TM. The situation for TE excitation is similar. Bydecomposing the impressed ﬁeld into TE and TM components, we maysolve for thescattered ﬁeld in each case and then superpose the results to determine the completesolution. We ﬁrst consider the TM case. The impressed electric ﬁeld maybe written as ˜E i(r,ω)=ˆz˜E0(ω)e−jk0x=ˆz˜E0(ω)e−jk0ρcosφ(4.356) while the magnetic ﬁeld is, by(4.223), ˜Hi(r,ω)=− ˆy˜E0(ω) η0e−jk0x=−(ˆρsinφ+ˆφcosφ)˜E0(ω) η0e−jk0ρcosφ. Here k0=ω(µ 0/epsilon10)1/2andη0=(µ0//epsilon10)1/2. The scattered electric ﬁeld takes the form of a nonuniform cylindrical wave(4.353). Periodicityin φimplies that kφis an integer, saykφ=n. Within the cylinder we cannot use any of the functions Nn(kρ),H(2) n(kρ), orH(1) n(kρ)to represent the radial dependence of the ﬁeld, since each is singular at the origin. So we choose B(1) n(kρ)=Jn(kρ)and Bρ(ω)=0in (4.355). Physically, Jn(kρ)rep- resents the standing wave created bythe interaction of outward and inward propagatingwaves. External to the cylinder we use H (2) n(kρ)to represent the radial dependence of the secondaryﬁeld components: we avoid Nn(kρ)and Jn(kρ)since these represent standing waves, and avoid H(1) n(kρ)since there are no external secondarysources to create an inward traveling wave. Anyattempt to satisfythe boundaryconditions byusing a single nonuniform wave fails. This is because the sinusoidal dependence on φof each individual nonuniform wave cannot match the more complicated dependence of the impressed ﬁeld (4.356). Since thesinusoids are complete, an inﬁnite series of the functions (4.353) can be used to representthe scattered ﬁeld. So we have internal to the cylinder ˜E s z(r,ω)=∞/summationdisplay n=0[An(ω)sinnφ+Bn(ω)cosnφ]Jn(kρ) where k=ω(˜µ˜/epsilon1c)1/2. External to the cylinder we have free space and thus ˜Es z(r,ω)=∞/summationdisplay n=0[Cn(ω)sinnφ+Dn(ω)cosnφ]H(2) n(k0ρ). Equations (4.349) and (4.350) yield the magnetic ﬁeld internal to the cylinder: ˜Hs ρ=∞/summationdisplay n=0jn ZTMk1 ρ[An(ω)cosnφ−Bn(ω)sinnφ]Jn(kρ), ˜Hs φ=−∞/summationdisplay n=0j ZTM[An(ω)sinnφ+Bn(ω)cosnφ]J/prime n(kρ), where ZTM=ω˜µ/k. Outside the cylinder ˜Hs ρ=∞/summationdisplay n=0jn η0k01 ρ[Cn(ω)cosnφ−Dn(ω)sinnφ]H(2) n(k0ρ), ˜Hs φ=−∞/summationdisplay n=0j η0[Cn(ω)sinnφ+Dn(ω)cosnφ]H(2)/prime n(k0ρ), where J/prime n(z)=dJn(z)/dzand H(2)/prime n(z)=dH(2) n(z)/dz. We have two sets of unknown spectral amplitudes (An,Bn)and(Cn,Dn). These can be determined byapply ing the boundaryconditions at the interface. Since the total ﬁeldoutside the cylinder is the sum of the impressed and scattered terms, an application ofcontinuityof the tangential electric ﬁeld at ρ=agives us ∞/summationdisplay n=0[Ansinnφ+Bncosnφ]Jn(ka)= ∞/summationdisplay n=0[Cnsinnφ+Dncosnφ]H(2) n(k0a)+˜E0e−jk0acosφ, which must hold for all −π≤φ≤π. To remove the coeﬃcients from the sum we apply orthogonality. Multiplying both sides by sinmφ, integrating over [−π,π], and using the orthogonalityconditions (A.129)–(A.131) we obtain πAmJm(ka)−πCmH(2) m(k0a)=˜E0/integraldisplayπ −πsinmφe−jk0acosφdφ=0. (4.357) Multiplying by cosmφand integrating, we ﬁnd that 2πBmJm(ka)−2πDmH(2) m(k0a)=˜E0/epsilon1m/integraldisplayπ −πcosmφe−jk0acosφdφ =2π˜E0/epsilon1mj−mJm(k0a) (4.358) where /epsilon1nis Neumann’s number (A.132) and where we have used (E.83) and (E.39) to evaluate the integral. We must also have continuityof the tangential magnetic ﬁeld ˜Hφatρ=a.T h u s −∞/summationdisplay n=0j ZTM[Ansinnφ+Bncosnφ]J/prime n(ka)= −∞/summationdisplay n=0j η0[Cnsinnφ+Dncosnφ]H(2)/prime n(k0a)−cosφ˜E0 η0e−jk0acosφ must hold for all −π≤φ≤π. Byorthogonality πj ZTMAmJ/prime m(ka)−πj η0CmH(2)/prime m(k0a)=˜E0 η0/integraldisplayπ −πsinmφcosφe−jk0acosφdφ=0(4.359) and 2πj ZTMBmJ/prime m(ka)−2πj η0DmH(2)/prime m(k0a)=/epsilon1m˜E0 η0/integraldisplayπ −πcosmφcosφe−jk0acosφdφ. The integral maybe computed as /integraldisplayπ −πcosmφcosφe−jk0acosφdφ=jd d(k0a)/integraldisplayπ −πcosmφe−jk0acosφdφ=j2πj−mJ/prime m(k0a) and thus 1 ZTMBmJ/prime m(ka)−1 η0DmH(2)/prime m(k0a)=˜E0 η0/epsilon1mj−mJ/prime m(k0a). (4.360) We now have four equations for the coeﬃcients An,Bn,Cn,Dn. We maywrite (4.357) and (4.359) as /bracketleftbiggJm(ka)−H(2) m(k0a) η0 ZTMJ/prime m(ka)−H(2)/prime m(k0a)/bracketrightbigg/bracketleftbiggAm Cm/bracketrightbigg =0, (4.361) and (4.358) and (4.360) as /bracketleftbiggJm(ka)−H(2) m(k0a) η0 ZTMJ/prime m(ka)−H(2)/prime m(k0a)/bracketrightbigg/bracketleftbiggBm Dm/bracketrightbigg =/bracketleftbigg˜E0/epsilon1mj−mJm(k0a) ˜E0/epsilon1mj−mJ/prime m(k0a)/bracketrightbigg . (4.362) Matrix equations (4.361) and (4.362) cannot hold simultaneouslyunless Am=Cm=0. Then the solution to (4.362) is Bm=˜E0/epsilon1mj−m/bracketleftBigg H(2) m(k0a)J/prime m(k0a)−Jm(k0a)H(2)/prime m(k0a) η0 ZTMJ/primem(ka)H(2) m(k0a)−H(2)/prime m(k0a)Jm(ka)/bracketrightBigg , (4.363) Dm=− ˜E0/epsilon1mj−m/bracketleftBiggη0 ZTMJ/prime m(ka)Jm(k0a)−J/prime m(k0a)Jm(ka) η0 ZTMJ/primem(ka)H(2) m(k0a)−H(2)/prime m(k0a)Jm(ka)/bracketrightBigg . (4.364) With these coeﬃcients we can calculate the ﬁeld inside the cylinder (ρ≤a)from ˜Ez(r,ω)=∞/summationdisplay n=0Bn(ω)Jn(kρ)cosnφ, ˜Hρ(r,ω)=−∞/summationdisplay n=0jn ZTMk1 ρBn(ω)Jn(kρ)sinnφ, ˜Hφ(r,ω)=−∞/summationdisplay n=0j ZTMBn(ω)J/prime n(kρ)cosnφ, and the ﬁeld outside the cylinder (ρ > a)from ˜Ez(r,ω)=˜E0(ω)e−jk0ρcosφ+∞/summationdisplay n=0Dn(ω)H(2) n(k0ρ)cosnφ, ˜Hρ(r,ω)=− sinφ˜E0(ω) η0e−jk0ρcosφ−∞/summationdisplay n=0jn η0k01 ρDn(ω)H(2) n(k0ρ)sinnφ, ˜Hφ(r,ω)=− cosφ˜E0(ω) η0e−jk0ρcosφ−∞/summationdisplay n=0j η0Dn(ω)H(2)/prime n(k0ρ)cosnφ. We can easilyspecialize these results to the case of a perfectlyconducting cy linder by allowing ˜σ→∞. Then η0 ZTM=/radicalBigg µ0˜/epsilon1c ˜µ/epsilon10→∞ and Bn→0, Dn→− ˜E0/epsilon1mj−mJm(k0a) H(2) m(k0a). In this case it is convenient to combine the formulas for the impressed and scattered ﬁelds when forming the total ﬁelds. Since the impressed ﬁeld is z-independent and obeys the homogeneous Helmholtz equation, we mayrepresent it in terms of nonuniform cy lindricalwaves: ˜E i z=˜E0e−jk0ρcosφ=∞/summationdisplay n=0[Ensinnφ+Fncosnφ]Jn(k0ρ), where we have chosen the Bessel function Jn(k0ρ)since the ﬁeld is ﬁnite at the origin and periodic in φ. Applying orthogonality we see immediately that En=0and that 2π /epsilon1mFmJm(k0ρ)=˜E0/integraldisplayπ −πcosmφe−jk0ρcosφdφ=˜E02πj−mJm(k0ρ). Thus, Fn=˜E0/epsilon1nj−nand ˜Ei z=∞/summationdisplay n=0˜E0/epsilon1nj−nJn(k0ρ)cosnφ. Adding this impressed ﬁeld to the scattered ﬁeld we have the total ﬁeld outside the cylinder, ˜Ez=˜E0∞/summationdisplay n=0/epsilon1nj−n H(2) n(k0a)/bracketleftbig Jn(k0ρ)H(2) n(k0a)−Jn(k0a)H(2) n(k0ρ)/bracketrightbig cosnφ, while the ﬁeld within the cylinder vanishes. Then, by (4.350), ˜Hφ=−j η0˜E0∞/summationdisplay n=0/epsilon1nj−n H(2) n(k0a)/bracketleftbig J/prime n(k0ρ)H(2) n(k0a)−Jn(k0a)H(2)/prime n(k0ρ)/bracketrightbig cosnφ. Figure 4.26: Geometryof a perfectlyconducting wedge illuminated bya line source. This in turn gives us the surface current induced on the cylinder. From the boundary condition ˜Js=ˆn×˜H|ρ=a=ˆρ×[ˆρ˜Hρ+ˆφ˜Hφ]|ρ=a=ˆz˜Hφ|ρ=awe have Js(φ, ω) =−j η0ˆz˜E0∞/summationdisplay n=0/epsilon1nj−n H(2) n(k0a)/bracketleftbig J/prime n(k0a)H(2) n(k0a)−Jn(k0a)H(2)/prime n(k0a)/bracketrightbig cosnφ, and an application of (E.93) gives us Js(φ, ω) =ˆz2˜E0 η0k0πa∞/summationdisplay n=0/epsilon1nj−n H(2) n(k0a)cosnφ. (4.365) Computation of the scattered ﬁeld for a magnetically-polarized impressed ﬁeld pro- ceeds in the same manner. The impressed electric and magnetic ﬁelds are assumed to be ˜Ei(r,ω)=ˆy˜E0(ω)e−jk0x=(ˆρsinφ+ˆφcosφ)˜E0(ω)e−jk0ρcosφ, ˜Hi(r,ω)=ˆz˜E0(ω) η0e−jk0x=ˆz˜E0(ω) η0e−jk0ρcosφ. For a perfectlyconducting cy linder, the total magnetic ﬁeld is ˜Hz=˜E0 η0∞/summationdisplay n=0/epsilon1nj−n H(2)/prime n(k0a)/bracketleftbig Jn(k0ρ)H(2)/prime n(k0a)−J/prime n(k0a)H(2) n(k0ρ)/bracketrightbig cosnφ. (4.366) The details are left as an exercise. Boundary value problems in cylindrical coordinates: scattering by a perfectly conducting wedge. As a second example, consider a perfectlyconducting wedge im- merse dinfreespaceandilluminate dbyalinesource(Figure4.26)carryingcurrent ˜I(ω)and located at (ρ0,φ0). The current, which is assumed to be z-invariant, induces a secondarycurrent on the surface of the wedge which in turn produces a secondary (scattered) ﬁeld. This scattered ﬁeld, also z-invariant, can be found bysolving a bound- aryvalue problem. We do this byseparating space into the two regions ρ<ρ 0and ρ>ρ 0,0<φ<ψ . Each of these is source-free, so we can represent the total ﬁeld using nonuniform cylindrical waves of the type (4. 353). The line source is brought into the problem byapply ing the boundarycondition on the tangential magnetic ﬁeld across thecylindrical surface ρ=ρ 0. Since the impressed electric ﬁeld has onlya z-component, so do the scattered and total electric ﬁelds. We wish to represent the total ﬁeld ˜Ezin terms of nonuniform cylindrical waves of the type (4.353). Since the ﬁeld is not periodic in φ, the separation constant kφneed not be an integer; instead, its value is determined bythe positions of the wedge boundaries. For the region ρ<ρ 0we represent the radial dependence of the ﬁeld using the functions Jνsince the ﬁeld must be ﬁnite at the origin. For ρ>ρ 0we use the outward-propagating wave functions H(2) δ.T h u s ˜Ez(ρ, φ, ω) =/braceleftBigg/summationtext ν[Aνsinνφ+Bνcosνφ]Jν(k0ρ), ρ < ρ 0,/summationtext δ[Cδsinδφ+Dδcosδφ]H(2) δ(k0ρ), ρ > ρ 0.(4.367) The coeﬃcients Aν,Bν,Cδ,Dδand separation constants ν,δmaybe found byapply ing the boundaryconditions on the ﬁelds at the surface of the wedge and across the surfaceρ=ρ 0. On the wedge face at φ=0we must have ˜Ez=0, hence Bν=Dδ=0.O n the wedge face at φ=ψwe must also have ˜Ez=0, requiring sinνψ=sinδψ=0and therefore ν=δ=νn=nπ/ψ, n=1,2,.... So ˜Ez=/braceleftBigg/summationtext∞ n=0AnsinνnφJνn(k0ρ), ρ < ρ 0,/summationtext∞ n=0CnsinνnφH(2) νn(k0ρ), ρ > ρ 0.(4.368) The magnetic ﬁeld can be found from (4.349)–(4.350): ˜Hρ=/braceleftBigg/summationtext∞ n=0Anj η0k0νn ρcosνnφJνn(k0ρ), ρ < ρ 0,/summationtext∞ n=0Cnj η0k0νn ρcosνnφH(2) νn(k0ρ), ρ > ρ 0,(4.369) ˜Hφ=/braceleftBigg −/summationtext∞ n=0Anj η0sinνnφJ/prime νn(k0ρ), ρ < ρ 0, −/summationtext∞ n=0Cnj η0sinνnφH(2)/prime νn(k0ρ), ρ > ρ 0.(4.370) The coeﬃcients Anand Cnare found byapply ing the boundaryconditions at ρ=ρ0. Bycontinuityof the tangential electric ﬁeld ∞/summationdisplay n=0AnsinνnφJνn(k0ρ0)=∞/summationdisplay n=0CnsinνnφH(2) νn(k0ρ0). We now applyorthogonalityover the interval [0,ψ]. Multiplying by sinνmφand inte- grating we have ∞/summationdisplay n=0AnJνn(k0ρ0)/integraldisplayψ 0sinνnφsinνmφdφ=∞/summationdisplay n=0CnH(2) νn(k0ρ0)/integraldisplayψ 0sinνnφsinνmφdφ. Setting u=φπ/ψ we have /integraldisplayψ 0sinνnφsinνmφdφ=ψ π/integraldisplayπ 0sinnusinmu du =ψ 2δmn, thus AmJνm(k0ρ0)=CmH(2) νm(k0ρ0). (4.371) The boundarycondition ˆn12×(˜H1−˜H2)=˜Jsrequires the surface current at ρ=ρ0.W e can write the line current in terms of a surface current densityusing the δ-function: ˜Js=ˆz˜Iδ(φ−φ0) ρ0. This is easilyveriﬁed as the correct expression since the integral of this densityalong the circular arc at ρ=ρ0returns the correct value ˜Ifor the total current. Thus the boundarycondition requires ˜Hφ(ρ+ 0,φ,ω) −˜Hφ(ρ− 0,φ,ω) =˜Iδ(φ−φ0) ρ0. By(4.370) we have −∞/summationdisplay n=0Cnj η0sinνnφH(2)/prime νn(k0ρ0)+∞/summationdisplay n=0Anj η0sinνnφJ/prime νn(k0ρ0)=˜Iδ(φ−φ0) ρ0 and orthogonalityy ields −Cmψ 2j η0H(2)/prime νm(k0ρ0)+Amψ 2j η0J/prime νm(k0ρ0)=˜Isinνmφ0 ρ0. (4.372) The coeﬃcients Amand Cmthus obeythe matrix equation /bracketleftbiggJνm(k0ρ0)−H(2) νm(k0ρ0) J/prime νm(k0ρ0)−H(2)/prime νm(k0ρ0)/bracketrightbigg/bracketleftbiggAm Cm/bracketrightbigg =/bracketleftbigg0 −j2˜Iη0 ψsinνmφ0 ρ0/bracketrightbigg and are Am=j2˜Iη0 ψsinνmφ0 ρ0H(2) νm(k0ρ0) H(2)/prime νm(k0ρ0)Jνm(k0ρ0)−J/primeνm(k0ρ0)H(2) νm(k0ρ0), Cm=j2˜Iη0 ψsinνmφ0 ρ0Jνm(k0ρ0) H(2)/prime νm(k0ρ0)Jνm(k0ρ0)−J/primeνm(k0ρ0)H(2) νm(k0ρ0). Using the Wronskian relation (E.93), we replace the denominators in these expressions by2/(jπk0ρ0): Am=− ˜Iη0 ψπk0sinνmφ0H(2) νm(k0ρ0), Cm=− ˜Iη0 ψπk0sinνmφ0Jνm(k0ρ0). Hence (4.368) gives ˜Ez( ρ,φ,ω) =/braceleftBigg −/summationtext∞ n=0˜Iη0 2ψπk0/epsilon1nJνn(k0ρ)H(2) νn(k0ρ0)sinνnφsinνnφ0,ρ < ρ 0, −/summationtext∞ n=0˜Iη0 2ψπk0/epsilon1nH(2) νn(k0ρ)Jνn(k0ρ0)sinνnφsinνnφ0,ρ > ρ 0,(4.373) where /epsilon1nis Neumann’s number (A.132). The magnetic ﬁelds can also be found by substituting the coeﬃcients into (4.369) and (4.370). The ﬁelds produced byan impressed plane wave maynow be obtained byletting the line source recede to inﬁnity. For large ρ0we use the asymptotic form (E.62) and ﬁnd that ˜Ez(ρ, φ, ω) =−∞/summationdisplay n=0˜Iη0 2ψπk0/epsilon1nJνn(k0ρ)/bracketleftBigg/radicalBigg 2j πk0ρ0jνne−jk0ρ0/bracketrightBigg sinνnφsinνnφ0,ρ < ρ 0. (4.374) Since the ﬁeld of a line source falls oﬀ as ρ−1/2 0, the amplitude of the impressed ﬁeld approaches zero as ρ0→∞ . We must compensate for the reduction in the impressed ﬁeld byscaling the amplitude of the current source. To obtain the proper scale factor,we note that the electric ﬁeld produced at a point ρbya line source located at ρ 0may be found from (4.345): ˜Ez=− ˜Ik0η0 4H(2) 0(k0|ρ−ρ0|)≈− ˜Ik0η0 4/radicalBigg 2j πk0ρ0e−jk0ρ0ejkρcos(φ−φ0),k0ρ0/greatermuch1. But if we write this as ˜Ez≈˜E0ejk·ρ then the ﬁeld looks exactlylike that produced bya plane wave with amplitude ˜E0trav- eling along the wave v ector k=−k0ˆxcosφ0−k0ˆysinφ0. Solving for ˜Iin terms of ˜E0and substituting it back into (4.374), we get the total electric ﬁeld scattered from a wedgewith an impressed TM plane- waveﬁeld: ˜E z(ρ, φ, ω) =2π ψ˜E0∞/summationdisplay n=0/epsilon1njνnJνn(k0ρ)sinνnφsinνnφ0. Here we interpret the angle φ0as the incidence angle of the plane wave. To determine the ﬁeld produced byan impressed TE plane- waveﬁeld, we use a mag- netic line source ˜Imlocated at ρ0,φ0and proceed as above. Byanalogywith (4.367) we write ˜Hz(ρ, φ, ω) =/braceleftBigg/summationtext ν[Aνsinνφ+Bνcosνφ]Jν(k0ρ), ρ < ρ 0,/summationtext δ[Cδsinδφ+Dδcosδφ]H(2) δ(k0ρ), ρ > ρ 0. By(4.351) the tangential electric ﬁeld is ˜Eρ(ρ, φ, ω) =/braceleftBigg −/summationtext ν[Aνcosνφ−Bνsinνφ]jZTE k1 ρνJν(k0ρ), ρ < ρ 0, −/summationtext δ[Cδcosδφ−Dδsinδφ]jZTE k1 ρδH(2) δ(k0ρ), ρ > ρ 0. Application of the boundaryconditions on the tangential electric ﬁeld at φ=0,ψresults inAν=Cδ=0andν=δ=νn=nπ/ψ, and thus ˜Hzbecomes ˜Hz(ρ, φ, ω) =/braceleftBigg/summationtext∞ n=0BncosνnφJνn(k0ρ), ρ < ρ 0,/summationtext∞ n=0DncosνnφH(2) νn(k0ρ), ρ > ρ 0.(4.375) Application of the boundaryconditions on tangential electric and magnetic ﬁelds across the magnetic line source then leads directlyto ˜Hz(ρ, φ, ω) =/braceleftBigg −/summationtext∞ n=0˜Imη0 2ψπk0/epsilon1nJνn(k0ρ)H(2) νn(k0ρ0)cosνnφcosνnφ0,ρ < ρ 0 −/summationtext∞ n=0˜Imη0 2ψπk0/epsilon1nH(2) νn(k0ρ)Jνn(k0ρ0)cosνnφcosνnφ0,ρ > ρ 0. (4.376) For a plane- waveimpressed ﬁeld this reduces to ˜Hz(ρ, φ, ω) =2π ψ˜E0 η0∞/summationdisplay n=0/epsilon1njνnJνn(k0ρ)cosνnφcosνnφ0. Behavior of current near a sharp edge. In§3.2.9 we studied the behavior of static charge near a sharp conducting edge bymodeling the edge as a wedge. We can followthe same procedure for frequency-domain ﬁelds. Assume that the perfectly conductingwedgeshowninFigure4.26isimmerse dinaﬁnite, z-independentimpresse dﬁeldofa sort that will not concern us. A current is induced on the surface of the wedge and wewish to studyits behavior as we approach the edge. Because the ﬁeld is z-independent, we mayconsider the superposition of TM and TE ﬁelds as was done above to solve for the ﬁeld scattered bya wedge. For TM polarization,if the source is not located near the edge we maywrite the total ﬁeld (impressed plus scattered) in terms of nonuniform cylindrical waves. The form of the ﬁeld that obeys theboundaryconditions at φ=0andφ=ψis given by(4.368): ˜E z=∞/summationdisplay n=0AnsinνnφJνn(k0ρ), where νn=nπ/ψ. Although the Andepend on the impressed source, the general behavior of the current near the edge is determined bythe properties of the Bessel functions. Thecurrent on the wedge face at φ=0is given by ˜J s(ρ, ω) =ˆφ×[ˆφ˜Hφ+ˆρ˜Hρ]|φ=0=− ˆz˜Hρ(ρ,0,ω) . By(4.349) we have the surface current ˜Js(ρ, ω) =− ˆz1 ZTMk0∞/summationdisplay n=0Anνn ρJνn(k0ρ). Forρ→0the small-argument approximation (E.51) yields ˜Js(ρ, ω) ≈− ˆz1 ZTMk0∞/summationdisplay n=0Anνn1 /Gamma1(ν n+1)/parenleftbiggk0 2/parenrightbiggνn ρνn−1. The sum is dominated bythe smallest power of ρ. Since the n=0term vanishes we have ˜Js(ρ, ω) ∼ρπ ψ−1,ρ →0. Forψ<π the current density, which runs parallel to the edge, is unbounded as ρ→0. A right-angle wedge ( ψ=3π/2) carries ˜Js(ρ, ω) ∼ρ−1/3. Another important case is that of a half-plane ( ψ=2π) where ˜Js(ρ, ω) ∼1√ρ. (4.377) This square-root edge singularitydominates the behavior of the current ﬂowing parallel to anyﬂat edge, either straight or with curvature large compared to a wavelength, andis useful for modeling currents on complicated structures. In the case of TE polarization the magnetic ﬁeld near the edge is, by(4.375), ˜Hz(ρ, φ, ω) =∞/summationdisplay n=0BncosνnφJνn(k0ρ), ρ < ρ 0. The current at φ=0is ˜Js(ρ, ω) =ˆφ×ˆz˜Hz|φ=0=ˆρ˜Hz(ρ,0,ω) or ˜Js(ρ, ω) =ˆρ∞/summationdisplay n=0BnJνn(k0ρ). Forρ→0we use (E.51) to write ˜Js(ρ, ω) =ˆρ∞/summationdisplay n=0Bn1 /Gamma1(ν n+1)/parenleftbiggk0 2/parenrightbiggνn ρνn. The n=0term gives a constant contribution, so we keep the ﬁrst two terms to see how the current behaves near ρ=0: ˜Js∼b0+b1ρπ ψ. Here b0and b1depend on the form of the impressed ﬁeld. For a thin plate where ψ=2π this becomes ˜Js∼b0+b1√ρ. This is the companion square-root behavior to (4.377). When perpendicular to a sharp edge, the current grows awayfrom the edge as ρ1/2. In most cases b0=0since there is no mechanism to store charge along a sharp edge. 4.11.8 Propagation of spherical waves in a conducting medium We cannot obtain uniform spherical wave solutions to Maxwell’s equations. Anyﬁeld dependent onlyon rproduces the null ﬁeld external to the source region, as shown in §4.11.9. Nonuniform spherical waves are in general complicated and most easilyhandled using potentials. We consider here onlythe simple problem of ﬁelds dependent on rand θ. These waves displaythe fundamental properties of all spherical waves: theydiverge from a localized source and expand with ﬁnite velocity. Consider a homogeneous, source-free region characterized by ˜/epsilon1(ω),˜µ(ω), and ˜σ(ω). We seek wave solutions that are TEM rin spherical coordinates ( ˜Hr=˜Er=0) and φ-independent. Thus we write ˜E(r,ω)=ˆθ˜Eθ(r,θ,ω) +ˆφ˜Eφ(r,θ,ω) , ˜H(r,ω)=ˆθ˜Hθ(r,θ,ω) +ˆφ˜Hφ(r,θ,ω) . To determine the behavior of these ﬁelds we ﬁrst examine Faraday’s law ∇× ˜E(r,θ,ω) =ˆr1 rsinθ∂ ∂θ[sinθ˜Eφ(r,θ,ω) ]−ˆθ1 r∂ ∂r[r˜Eφ(r,θ,ω) ]+ˆφ1 r∂ ∂r[r˜Eθ(r,θ,ω) ] =− jω˜µ˜H(r,θ,ω) . (4.378) Since we require ˜Hr=0we must have ∂ ∂θ[sinθ˜Eφ(r,θ,ω) ]=0. This implies that either ˜Eφ∼1/sinθor˜Eφ=0. We choose ˜Eφ=0and investigate whether the resulting ﬁelds satisfythe remaining Maxwell equations. In a source-free, homogeneous region of space we have ∇·˜D=0and thus also ∇·˜E=0. Since we have onlya θ-component of the electric ﬁeld, this requires 1 r∂ ∂θ˜Eθ(r,θ,ω) +cotθ r˜Eθ(r,θ,ω) =0. From this we see that when ˜Eφ=0, the ﬁeld ˜Eθmust obey ˜Eθ(r,θ,ω) =˜fE(r,ω) sinθ. By(4.378) there is onlya φ-component of magnetic ﬁeld which obeys ˜Hφ(r,θ,ω) =˜fH(r,ω) sinθ where −jω˜µ˜fH(r,ω)=1 r∂ ∂r[r˜fE(r,ω)]. (4.379) So the spherical wave is TEM to the r-direction. We can obtain a wave equation for ˜fEbytaking the curl of (4.378) and substituting from Ampere’s law: ∇×(∇× ˜E)=− ˆθ1 r∂2 ∂r2(r˜Eθ)=∇×/parenleftbig −jω˜µ˜H/parenrightbig =− jω˜µ/parenleftbig ˜σ˜E+jω˜/epsilon1˜E/parenrightbig , hence d2 dr2[r˜fE(r,ω)]+k2[r˜fE(r,ω)]=0. (4.380) Here k=ω(˜µ˜/epsilon1c)1/2is the complex wavenumber and ˜/epsilon1c=˜/epsilon1+˜σ/jωis the complex permittivity. The equation for ˜fHis identical. The wave equation (4.380) is merelythe second-order harmonic diﬀerential equation, with two independent solutions chosen from the list sinkr,coskr,e−jkr,ejkr. We ﬁnd sinkrand coskruseful for describing standing waves between b oundaries, and ejkrand e−jkruseful for describing waves propagating in the r-direction. Of these, ejkr represents waves trav eling inward while e−jkrrepresents waves traveling outward. At this point we choose r˜fE=e−jkrand thus ˜E(r,θ,ω) =ˆθ˜E0(ω)e−jkr rsinθ. (4.381) By(4.379) we have ˜H(r,θ,ω) =ˆφ˜E0(ω) ZTEMe−jkr rsinθ(4.382) where ZTEM=(˜µ//epsilon1c)1/2is the complex wave impedance. Since we can also write ˜H(r,θ,ω) =ˆr×˜E(r,θ,ω) ZTEM, the ﬁeld is TEM to the r-direction, which is the direction of wave propagation as shown below. The wave nature of the ﬁeld is easilyidentiﬁed byconsidering the ﬁelds in the phasor domain. Letting ω→ˇωand setting k=β−jαin the exponential function we ﬁnd that ˇE(r,θ)=ˆθˇE0e−αre−jβr rsinθ where ˇE0=E0ejξE. The time-domain representation maybe found using (4.126): E(r,θ,t)=ˆθE0e−αr rsinθcos(ˇωt−βr+ξE). (4.383) We can identifya surface of constant phase as a locus of points obey ing ˇωt−βr+ξE=CP (4.384) where CPis some constant. These surfaces, which are spheres centered on the origin, are called spherical wavefronts . Note that surfaces of constant amplitude as determined by e−αr r=CA, where CAis some constant, are also spheres. The cosine term in (4.383) represents a traveling wave with spherical wavefronts that propagate outward as time progresses. Attenuation is caused bythe factor e−αr.B y diﬀerentiation we ﬁnd that the phase velocityis vp=ˇω/β. The wavelength is given by λ=2π/β. Our solution is not appropriate for unbounded space since the ﬁelds have a singularity atθ=0. To exclude the z-axis we add conducting cones as mentioned on page 105. This results in a biconical structure that can be used as a transmission line or antenna. To compute the power carried bya spherical wave, we use (4.381) and (4.382) to obtain the time-average Poynting ﬂux Sav=1 2Re{ˇEθˆθ×ˇH∗ φˆφ}=1 2ˆrRe/braceleftbigg1 Z∗ TEM/bracerightbiggE2 0 r2sin2θe−2αr. The power ﬂux is radial and has densityinverselyproportional to r2. The time-average power carried bythe wavethrough a spherical surface at rsandwiched between the cones atθ1andθ2is Pav(r)=1 2Re/braceleftbigg1 Z∗ TEM/bracerightbigg E2 0e−2αr/integraldisplay2π 0dφ/integraldisplayθ2 θ1dθ sinθ=πFRe/braceleftbigg1 Z∗ TEM/bracerightbigg E2 0e−2αr where F=ln/bracketleftbiggtan(θ2/2) tan(θ1/2)/bracketrightbigg . (4.385) This is independent of rwhen α=0. For lossymedia the power decay s exponentially because of Joule heating. We can write the phasor electric ﬁeld in terms of the transverse gradient of a scalar potential function ˇ/Phi1: ˇE(r,θ)=ˆθˇE0e−jkr rsinθ=− ∇ tˇ/Phi1(θ) where ˇ/Phi1(θ)=− ˇE0e−jkrln/parenleftbigg tanθ 2/parenrightbigg . By∇twe mean the gradient with the r-component excluded. It is easilyveriﬁed that ˇE(r,θ)=− ∇ tˇ/Phi1(θ)=− ˆθˇE01 r∂ˇ/Phi1(θ) ∂θ=ˆθˇE0e−jkr rsinθ. Because ˇEand ˇ/Phi1are related bythe gradient, we can deﬁne a unique potential diﬀerence between the two cones at anyradial position r: ˇV(r)=−/integraldisplayθ2 θ1ˇE·dl=ˇ/Phi1(θ 2)−ˇ/Phi1(θ 1)=ˇE0Fe−jkr, where Fis given in (4.385). The existence of a unique voltage diﬀerence is a propertyof all transmission line structures operated in the TEM mode. We can similarlycomputethe current ﬂowing outward on the cone surfaces. The surface current on the cone atθ=θ 1isˇJs=ˆn×ˇH=ˆθ×ˆφˇHφ=ˆrˇHφ, hence ˇI(r)=/integraldisplay2π 0ˇJs·ˆrrsinθdφ=2πˇE0 ZTEMe−jkr. The ratio of voltage to current at anyradius ris the characteristic impedance of the bi- conical transmission line (or, equivalently, the input impedance of the biconical antenna): Z=ˇV(r) ˇI(r)=ZTEM 2πF. If the material between the cones is lossless (and thus ˜µ=µand ˜/epsilon1c=/epsilon1are real), this becomes Z=η 2πF where η=(µ//epsilon1)1/2. The frequencyindependence of this quantitymakes biconical anten- nas (or their approximate representations) useful for broadband applications. Finally, the time-average power carried by the wave maybe found from Pav(r)=1 2Re/braceleftbigˇV(r)ˇI∗(r)/bracerightbig =πFRe/braceleftbigg1 Z∗ TEM/bracerightbigg E2 0e−2αr. The complex power relationship P=VI∗is also a propertyof TEM guided- wavestruc- tures. 4.11.9 Nonradiating sources We showed in §2.10.9 that not all time-varying sources produce electromagnetic waves. In fact, a subset of localized sources known as nonradiating sources produce no ﬁeld external to the source region. Devaneyand Wolf [54] have shown that all nonradiatingtime-harmonic sources in an unbounded homogeneous medium can be represented in theform ˇJ nr(r)=− ∇×/bracketleftbig ∇× ˇf(r)/bracketrightbig +k2ˇf(r) (4.386) where ˇfis anyvector ﬁeld that is continuous, has partial derivatives up to third order, and vanishes outside some localized region Vs. In fact, ˇE(r)=jˇωµˇf(r)is preciselythe phasor electric ﬁeld produced by ˇJnr(r). The reasoning is straightforward. Consider the Helmholtz equation (4.203): ∇×(∇× ˇE)−k2ˇE=− jˇωµˇJ. By(4.386) we have /parenleftbig ∇×∇×− k2/parenrightbig/bracketleftbigˇE−jˇωµˇf/bracketrightbig =0. Since ˇfis zero outside the source region it must vanish at inﬁnity. ˇEalso vanishes at inﬁnitybythe radiation condition, and thus the quantity ˇE−jˇωµˇfobeys the radiation condition and is a unique solution to the Helmholtz equation throughout all space. Sincethe Helmholtz equation is homogeneous we have ˇE−jˇωµˇf=0 everywhere; since ˇfis zero outside the source region, so is ˇE(and so is ˇH). An interesting special case of nonradiating sources is ˇf=∇ˇ/Phi1 k2 so that ˇJnr=−/parenleftbig ∇×∇×− k2/parenrightbig∇ˇ/Phi1 k2=∇ ˇ/Phi1. Using ˇ/Phi1(r)=ˇ/Phi1(r), we see that this source describes the current produced byan oscillat- ing spherical balloon of charge (cf., §2.10.9). Radially-directed, spherically-symmetric sources cannot produce uniform spherical waves, since these sources are of the nonradi- ating type. 4.12 Interpretation of the spatial transform Now that we understand the meaning of a Fourier transform on the time variable, let us consider a single transform involving one of the spatial variables. For a transform over zwe shall use the notation ψz(x,y,kz,t)↔ψ(x,y,z,t). Here the spatial frequencytransform variable kzhas units of m−1. The forward and inverse transform expressions are ψz(x,y,kz,t)=/integraldisplay∞ −∞ψ(x,y,z,t)e−jkzzdz, (4.387) ψ(x,y,z,t)=1 2π/integraldisplay∞ −∞ψz(x,y,kz,t)ejkzzdkz, (4.388) by(A.1) and (A.2). We interpret (4.388) much as we interpreted the temporal inverse transform (4.2). Anyvector component of the electromagnetic ﬁeld can be decomposed into a continuoussuperposition of elemental spatial terms e jkzzwith weighting factors ψz(x,y,kz,t).I n this case ψzis the spatial frequency spectrum ofψ. The elemental terms are spatial sinusoids along zwith rapidityof variation described by kz. As with the temporal transform, ψzcannot be arbitrarysince ψmust obeya scalar wave equation such as (2.327). For instance, for a source-free region of free space wemust have /parenleftbigg ∇ 2−1 c2∂ ∂t2/parenrightbigg1 2π/integraldisplay∞ −∞ψz(x,y,kz,t)ejkzzdkz=0. Decomposing the Laplacian operator as ∇2=∇2 t+∂2/∂z2and taking the derivatives into the integrand, we have 1 2π/integraldisplay∞ −∞/bracketleftbigg/parenleftbigg ∇2 t−k2 z−1 c2∂2 ∂t2/parenrightbigg ψz(x,y,kz,t)/bracketrightbigg ejkzzdkz=0. Hence /parenleftbigg ∇2 t−k2 z−1 c2∂2 ∂t2/parenrightbigg ψz(x,y,kz,t)=0 (4.389) bythe Fourier integral theorem. The elemental component ejkzzis spatiallysinusoidal and occupies all of space. Because such an element could onlybe created bya source that spans all of space, it is nonphy sicalwhen taken byitself. Nonetheless it is often used to represent more complicated ﬁelds.If the elemental spatial term is to be used alone, it is best interpreted physically whencombined with a temporal decomposition. That is, we consider a two-dimensional trans-form, with transforms over both time and space. Then the time-domain representation of the elemental component is φ(z,t)=1 2π/integraldisplay∞ −∞ejkzzejωtdω. (4.390) Before attempting to compute this transform, we should note that if the elemental term is to describe an EM ﬁeld ψin a source-free region, it must obeythe homogeneous scalar wave equation. Substituting (4.390) into the homogeneous waveequation we have /parenleftbigg ∇2−1 c2∂2 ∂t2/parenrightbigg1 2π/integraldisplay∞ −∞ejkzzejωtdω=0. Diﬀerentiation under the integral sign gives 1 2π/integraldisplay∞ −∞/bracketleftbigg/parenleftbigg −k2 z+ω2 c2/parenrightbigg ejkzz/bracketrightbigg ejωtdω=0 and thus k2 z=ω2 c2=k2. Substitution of kz=kinto (4.390) gives the time-domain representation of the elemental component φ(z,t)=1 2π/integraldisplay∞ −∞ejω(t+z/c)dω. Finally, using the shifting theorem (A.3) along with (A.4), we have φ(z,t)=δ/parenleftBig t+z c/parenrightBig , (4.391) which we recognize as a uniform plane wavepropagating in the −z-direction with velocity c. There is no variation in the directions transverse to the direction of propagation and the surface describing a constant argument of the δ-function at anytime tis a plane perpendicular to the direction of propagation. We can also consider the elemental spatial component in tandem with a single sinu- soidal steady-state elemental component. The phasor representation of the elementalspatial component is ˇφ(z)=e jkzz=ejkz. This elemental term is a time-harmonic plane wave propagating in the −z-direction. Indeed, multiplying by ejˇωtand taking the real part we get φ(z,t)=cos(ˇωt+kz), which is the sinusoidal steady-state analogue of (4.391). Manyauthors choose to deﬁne the temporal and spatial transforms using diﬀering sign conventions. The temporal transform is deﬁned as in (4.1) and (4.2), but the spatialtransform is deﬁned through ψ z(x,y,kz,t)=/integraldisplay∞ −∞ψ(x,y,z,t)ejkzzdz, (4.392) ψ(x,y,z,t)=1 2π/integraldisplay∞ −∞ψz(x,y,kz,t)e−jkzzdkz. (4.393) This employs a wave traveling in the positive z-direction as the elemental spatial com- ponent, which is quite useful for physical interpretation. We shall adopt this notation in§4.13. The drawback is that we must alter the formulas from standard Fourier transform tables (replacing kby−k) to reﬂect this diﬀerence. In the following sections we shall show how a spatial Fourier decomposition can be used to solve for the electromagnetic ﬁelds in a source-free region of space. Byemploy ingthe spatial transform we mayeliminate one or more spatial variables from Maxwell’sequations, making the wave equation easier to solve. In the end we must perform an inversion to return to the space domain. This maybe diﬃcult or impossible to doanalytically, requiring a numerical Fourier inversion. 4.13 Spatial Fourier decomposition of two-dimensional ﬁelds Consider a homogeneous, source-free region characterized by ˜/epsilon1(ω),˜µ(ω), and ˜σ(ω). We seek z-independent solutions to the frequency-domain Maxwell’s equations, using the Fourier transform to represent the spatial dependence. By §4.11.2 a general two- dimensional ﬁeld maybe decomposed into ﬁelds TE and TM to the z-direction. In the TM case ˜Hz=0, and ˜Ezobeys the homogeneous scalar Helmholtz equation (4.208). In the TE case ˜Ez=0, and ˜Hzobeys the homogeneous scalar Helmholtz equation. Since each ﬁeld component obeys the same equation, we let ˜ψ(x,y,ω)represent either ˜Ez(x,y,ω)or˜Hz(x,y,ω). Then ˜ψobeys (∇2 t+k2)˜ψ(x,y,ω)=0 (4.394) where ∇2 tis the transverse Laplacian (4.209) and k=ω(˜µ˜/epsilon1c)1/2with ˜/epsilon1cthe complex permittivity. We maychoose to represent ˜ψ(x,y,ω)using Fourier transforms over one or both spatial variables. For application to problems in which boundaryvalues or boundaryconditions are speciﬁed at a constant value of a single variable (e.g., over a plane), onetransform suﬃces. For instance, we mayknow the values of the ﬁeld in the y=0plane (as we will, for example, when we solve the boundaryvalue problems of §??). Then we maytransform over xand leave the yvariable intact so that we maysubstitute the boundaryvalues. We adopt (4.392) since the result is more readilyinterpreted in terms of propagating plane waves. Choosing to transform over xwe have ˜ψ x(kx,y,ω)=/integraldisplay∞ −∞˜ψ(x,y,ω)ejkxxdx, (4.395) ˜ψ(x,y,ω)=1 2π/integraldisplay∞ −∞ψx(kx,y,ω)e−jkxxdkx. (4.396) For convenience in computation or interpretation of the inverse transform, we often regard kxas a complex variable and perturb the inversion contour into the complex kx= kxr+jkxiplane. The integral is not altered if the contour is not moved past singularities such as poles or branch points. If the function being transformed has exponential (wave)behavior, then a pole exists in the complex plane; if we move the inversion contour acrossthis pole, the inverse transform does not return the original function. We generallyindicate the desire to interpret k xas complex byindicating that the inversion contour is parallel to the real axis but located in the complex plane at kxi=/Delta1: ˜ψ(x,y,ω)=1 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1˜ψx(kx,y,ω)e−jkxxdkx. (4.397) Additional perturbations of the contour are allowed provided that the contour is not moved through singularities. As an example, consider the function u(x)=/braceleftBigg 0, x<0, e−jkx,x>0,(4.398) where k=kr+jkirepresents a wavenumber. This function has the form of a plane wave propagating in the x-direction and is thus relevant to our studies. If the material through which the wave is propagating is lossy, then ki<0. The Fourier transform of the function is ux(kx)=/integraldisplay∞ 0e−jkxejkxxdx=1 j(kx−k)/bracketleftbig ej(kxr−kr)xe−(kxi−ki)x/bracketrightbig/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞ 0. Figure 4.27: Inversion contour for evaluating the spectral integral for a plane wave. The integral converges if kxi>ki, and the transform is ux(kx)=−1 j(kx−k). Since u(x)is an exponential function, ux(kx)has a pole at kx=kas anticipated. To compute the inverse transform we use (4.397): u(x)=1 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1/bracketleftbigg −1 j(kx−k)/bracketrightbigg e−jkxxdkx. (4.399) We must be careful to choose /Delta1in such a way that all values of kxalong the inversion contour lead to a convergent forward Fourier transform. Since we must have kxi>ki, choosing /Delta1> kiensures proper convergence. This gives the inversion contour shown in Figure4.27,aspecialcaseofwhichistherealaxis.Wecomput etheinversionintegral using contour integration as in §A.1. We close the contour in the complex plane and use Cauchy’s residue theorem (A.14) For x>0we take 0>/Delta1> kiand close the contour in the lower half-plane using a semicircular contour CRof radius R. Then the closed contour integral is equal to −2πjtimes the residue at the pole kx=k.A s R→∞ we ﬁnd that kxi→− ∞ at all points on the contour CR. Thus the integrand, which varies as ekxix, vanishes on CRand there is no contribution to the integral. The inversion integral (4.399) is found from the residue at the pole: u(x)=(−2πj)1 2πRes kx=k/bracketleftbigg −1 j(kx−k)e−jkxx/bracketrightbigg . Since the residue is merely je−jkxwe have u(x)=e−jkx. When x<0we choose /Delta1> 0 and close the contour along a semicircle CRof radius Rin the upper half-plane. Again we ﬁnd that on CRthe integrand vanishes as R→∞, and thus the inversion integral (4.399) is given by 2πjtimes the residues of the integrand at any poles within the closed contour. This time, however, there are no poles enclosed and thus u(x)=0. We have recovered the original function (4.398) for both x>0and x<0. Note that if we had erroneously chosen /Delta1< kiwe would not have properly enclosed the pole and would have obtained an incorrect inverse transform. Now that we know how to represent the Fourier transform pair, let us apply the transform to solve (4.394). Our hope is that by representing ˜ψin terms of a spatial Fourier integral we will make the equation easier to solve. We have (∇2 t+k2)1 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1˜ψx(kx,y,ω)e−jkxxdkx=0. Diﬀerentiation under the integral sign with subsequent application of the Fourier integral theorem implies that ˜ψmust obey the second-order harmonic diﬀerential equation /bracketleftbiggd2 dy2+k2 y/bracketrightbigg ˜ψx(kx,y,ω)=0 where we have deﬁned the dependent parameter ky=kyr+jkyithrough k2 x+k2 y=k2. Two independent solutions to the diﬀerential equation are e∓jkyyand thus ˜ψ(kx,y,ω)=A(kx,ω)e∓jkyy. Substituting this into the inversion integral, we have the solution to the Helmholtz equa- tion: ˜ψ(x,y,ω)=1 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1A(kx,ω)e−jkxxe∓jkyydkx. (4.400) If we deﬁne the wave vector k=ˆxkx±ˆyky, we can also write the solution in the form ˜ψ(x,y,ω)=1 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1A(kx,ω)e−jk·ρdkx (4.401) where ρ=ˆxx+ˆyyis the two-dimensional position vector. The solution (4.401) has an important physical interpretation. The exponential term looks exactly like a plane wave with its wave v ector lying in the xy-plane. For lossy media the plane wave is nonuniform, and the surfaces of constant phase may not be alignedwith the surfaces of constant amplitude (see §4.11.4). For the special case of a lossless medium we have k i→0and can let /Delta1→0as long as /Delta1> ki. As we perform the inverse transform integral over kxfrom−∞to∞we will encounter both the condition k2 x>k2 and k2 x≤k2.F o r k2 x≤k2we have e−jkxxe∓jkyy=e−jkxxe∓j√ k2−k2xy where we choose the upper sign for y>0and the lower sign for y<0to ensure that the waves propagate in the ±y-direction, respectively. Thus, in this regime the exponential represents a propagating wave that travels into the half-plane y>0along a direction which depends on kx,makin ganangleξwiththe x-axisasshowninFigure4.28.For kx in [−k,k], every possible wave direction is covered, and thus we may think of the inversion integral as constructing the solution to the two-dimensional Helmholtz equation from acontinuous superposition of plane waves. The amplitude of each plane wavecomponent is given by A(k x,ω), which is often called the angular spectrum of the plane waves and Figure 4.28: Propagation behavior of the angular spectrum for (a) k2 x≤k2,(b ) k2 x>k2. is determined by the values of the ﬁeld over the boundaries of the solution region. But this is not the whole picture. The inverse transform integral also requires values of kxin the intervals [−∞,k]and [k,∞]. Here we have k2 x>k2and thus e−jkxxe−jkyy=e−jkxxe∓√ k2x−k2y, where we choose the upper sign for y>0and the lower sign for y<0to ensure that the ﬁeld decays along the y-direction. In these regimes we have an evanescent wave, propagating along xbut decaying along y, with surfaces of constant phase and amplitude mutuall yperpendicula r(Figure4.28).As kxranges out to ∞, evanescent waves of all possible decay constants also contribute to the plane-wave superposition. We may summarize the plane- wave con tributions by letting k=ˆxkx+ˆyky=kr+jki where kr=/braceleftBigg ˆxkx±ˆy/radicalbig k2−k2x,k2 x<k2, ˆxkx, k2 x>k2, ki=/braceleftBigg 0, k2 x<k2, ∓ˆy/radicalbig k2x−k2,k2 x>k2, where the upper sign is used for y>0and the lower sign for y<0. In many applications, including the half-plane example considered later, it is useful to write the inversion integral in polar coordinates. Letting kx=kcosξ, ky=±ksinξ, where ξ=ξr+jξiis a new complex variable, we have k·ρ=kxcosξ±kysinξand dkx=−ksinξdξ. With this change of variables (4.401) becomes ˜ψ(x,y,ω)=k 2π/integraldisplay CA(kcosξ,ω) e−jkxcosξe±jkysinξsinξdξ. (4.402) Since A(kx,ω)is a function to be determined, we may introduce a new function f(ξ, ω) =k 2πA(kx,ω)sinξ Figure 4.29: Inversion contour for the polar coordinate representation of the inverse Fourier transform. so that (4.402) becomes ˜ψ(x,y,ω)=/integraldisplay Cf(ξ, ω) e−jkρcos(φ±ξ)dξ (4.403) where x=ρcosφ,y=ρsinφ, and where the upper sign corresponds to 0<φ<π (y>0)while the lower sign corresponds to π<φ< 2π(y<0). In these expressions Cis a contour in the complex ξ-plane to be determined. Values along this contour must produce identical values of the integrand as did the values of kxover [−∞,∞]in the original inversion integral. By the identities cosz=cos(u+jv)=cosucoshv−jsinusinhv, sinz=sin(u+jv)=sinucoshv+jcosusinhv, weﬁndthatthecontourshowninFigure4.29providesidenticalvaluesoftheintegrand (Problem 4.24). The portions of the contour [0+j∞,0] and [−π,−π−j∞]together correspond to the regime of evanescent waves ( k<kx<∞and−∞<kx<k), while the segment [0,−π]along the real axis corresponds to −k<kx<kand thus describes contributions from propagating plane waves. In this case ξrepresents the propagation angle of the waves. 4.13.1 Boundary value problems using the spatial Fourier represen- tation The ﬁeld of a line source. As a ﬁrst example we calculate the Fourier representation of the ﬁeld of an electric line source. Assume a uniform line current ˜I(ω)is aligned along thez-axis in a medium characterized by complex permittivity ˜/epsilon1c(ω)and permeability ˜µ(ω). We separate space into two source-free portions, y>0and y<0, and write the ﬁeld in each region in terms of an inverse spatial Fourier transform. Then, by applyingthe boundary conditions in the y=0plane, we solve for the angular spectrum of the line source. Since this is a two-dimensional problem we may decompose the ﬁelds into TE and TM sets. For an electric line source we need only the TM set, and write Ezas a superposition of plane waves using (4.400). For y≷0we represent the ﬁeld in terms of plane waves traveling in the ±y-direction. Thus ˜Ez(x,y,ω)=1 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1A+(kx,ω)e−jkxxe−jkyydkx,y>0, ˜Ez(x,y,ω)=1 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1A−(kx,ω)e−jkxxe+jkyydkx,y<0. The transverse magnetic ﬁeld may be found from the axial electric ﬁeld using (4.212). We ﬁnd ˜Hx=−1 jω˜µ∂˜Ez ∂y(4.404) and thus ˜Hx(x,y,ω)=1 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1A+(kx,ω)/bracketleftbiggky ω˜µ/bracketrightbigg e−jkxxe−jkyydkx,y>0, ˜Hx(x,y,ω)=1 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1A−(kx,ω)/bracketleftbigg −ky ω˜µ/bracketrightbigg e−jkxxe+jkyydkx,y<0. To ﬁnd the spectra A±(kx,ω)we apply the boundary conditions at y=0. Since tangential ˜Eis continuous we have, after combining the integrals, 1 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1/bracketleftbig A+(kx,ω)−A−(kx,ω)/bracketrightbig e−jkxxdkx=0, and hence by the Fourier integral theorem A+(kx,ω)−A−(kx,ω)=0. (4.405) We must also apply ˆn12×(˜H1−˜H2)=˜Js. The line current may be written as a surface current density using the δ-function, giving −/bracketleftbig˜Hx(x,0+,ω)−˜Hx(x,0−,ω)/bracketrightbig =˜I(ω)δ( x). By (A.4) δ(x)=1 2π/integraldisplay∞ −∞e−jkxxdkx. Then, substituting for the ﬁelds and combining the integrands, we have 1 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1/bracketleftbigg A+(kx,ω)+A−(kx,ω)+ω˜µ ky˜I(ω)/bracketrightbigg e−jkxx=0, hence A+(kx,ω)+A−(kx,ω)=−ω˜µ ky˜I(ω). (4.406) Solution of (4.405) and (4.406) gives the angular spectra A+(kx,ω)=A−(kx,ω)=−ω˜µ 2ky˜I(ω). Substituting this into the ﬁeld expressions and combining the cases for y>0and y<0, we ﬁnd ˜Ez(x,y,ω)=−ω˜µ˜I(ω) 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1e−jky|y| 2kye−jkxxdkx=− jω˜µ˜I(ω)˜G(x,y|0,0;ω).(4.407) Here ˜Gis the spectral representation of the two-dimensional Green’s function ﬁrst found in§4.11.7, and is given by ˜G(x,y|x/prime,y/prime;ω)=1 2πj∞+ j/Delta1/integraldisplay −∞+ j/Delta1e−jky|y−y/prime| 2kye−jkx(x−x/prime)dkx. (4.408) By duality we have ˜Hz(x,y,ω)=−ω˜/epsilon1c˜Im(ω) 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1e−jky|y| 2kye−jkxxdkx=− jω˜/epsilon1c˜Im(ω)G(x,y|0,0;ω)(4.409) for a magnetic line current ˜Im(ω)on the z-axis. Note that since the earlier expression (4.346) should be equivalent to (4.408), we have the well known identity [33] 1 π∞+ j/Delta1/integraldisplay −∞+ j/Delta1e−jky|y| kye−jkxxdkx=H(2) 0(kρ). We have not yet speciﬁed the contour appropriate for calculating the inverse transform (4.407). We must be careful because the denominator of (4.407) has branch points atk y=/radicalbig k2−k2x=0, or equivalently, kx=±k=±(kr+jki). For lossy materials we have ki < 0 and kr > 0,sothebranchpointsappearasinFigure4.30.Wemaytakethebranch cuts outward from these points, and thus choose the inversion contour to lie between thebranch points so that the branch cuts are not traversed. This requires k i</Delta1< −ki.I t is natural to choose /Delta1=0and use the real axis as the inversion contour. We must be careful, though, when extending these arguments to the lossless case. If we consider thelossless case to be the limit of the lossy case as k i→0, we ﬁnd that the branch points migrate to the real axis and thus lie on the inversion contour. We can eliminate thisproblem by realizing that the inversion contour may be perturbed without aﬀecting thevalue of the integral, as long as it is not made to pass through the branch cuts. If weperturbtheinversioncontourasshowninFigure4.30,thenas k i→0the branch points do not fall on the contour. Figure 4.30: Inversion contour in complex kx-plane for a line source. Dotted arrow shows migration of branch points to real axis as loss goes to zero. There are many interesting techniques that may be used to compute the inversion integral appearing in (4.407) and in the other expressions we shall obtain in this section.These include direct real-axis integration and closed contour methods that use Cauchy’sresidue theorem to capture poles of the integrand (which often describe the propertiesof waves guided by surfaces). Often it is necessary to integrate around the branch cuts in order to meet the conditions for applying the residue theorem. When the observationpoint is far from the source we may use the method of steepest descents to obtainasymptotic forms for the ﬁelds. The interested reader should consult Chew [33], Kong[101], or Sommerfeld [184]. Field of a line source above an interface. Consider a z-directed electric line current located at y=hwithin a medium having parameters ˜µ 1 (ω)and ˜/epsilon1c 1 (ω).The y = 0 plane separates this region from a region having parameters ˜µ2 (ω)and ˜/epsilon1c 2 (ω).SeeFigure4.31. The impressed line current source creates an electromagnetic ﬁeld that induces secondary polarization and conduction currents in both regions. This current in turn produces asecondary ﬁeld that adds to the primary ﬁeld of the line source to satisfy the boundaryconditions at the interface. We would like to solve for the secondary ﬁeld and give itssources an image interpretation. Since the ﬁelds are z-independent we may decompose the ﬁelds into sets TE and TM toz. For a z-directed impressed source there is a z-component of ˜E, but no z-component of˜H; hence the ﬁelds are entirely speciﬁed by the TM set. The impressed source is unaﬀected by the secondary ﬁeld, and we may represent the impressed electric ﬁeldusing (4.407): ˜E i z(x,y,ω)=−ω˜µ1˜I(ω) 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1e−jky1|y−h| 2ky1e−jkxxdkx,y≥0 (4.410) Figure 4.31: Geometry of a z-directed line source above an interface between two material regions. where ky1=/radicalBig k2 1−k2xand k1=ω(˜µ1˜/epsilon1c 1)1/2. From (4.404) we ﬁnd that ˜Hi x=−1 jω˜µ1∂˜Ei z ∂y=˜I(ω) 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1ejky1(y−h) 2e−jkxxdkx,0≤y<h. The scattered ﬁeld obeys the homogeneous Helmholtz equation for all y>0, and thus may be written using (4.400) as a superposition of upward-traveling waves: ˜Es z1(x,y,ω)=1 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1A1(kx,ω)e−jky1ye−jkxxdkx, ˜Hs x1(x,y,ω)=1 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1ky1 ω˜µ1A1(kx,ω)e−jky1ye−jkxxdkx. Similarly, in region 2 the scattered ﬁeld may be written as a superposition of downward- traveling waves: ˜Es z2(x,y,ω)=1 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1A2(kx,ω)ejky2ye−jkxxdkx, ˜Hs x2(x,y,ω)=−1 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1ky2 ω˜µ2A2(kx,ω)ejky2ye−jkxxdkx, where ky2=/radicalBig k2 2−k2xand k2=ω(˜µ2˜/epsilon1c 2)1/2. We can solve for the angular spectra A1and A2by applying the boundary conditions at the interface between the two media. From the continuity of total tangential electricﬁeld we ﬁnd that 1 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1/bracketleftbigg −ω˜µ1˜I(ω) 2ky1e−jky1h+A1(kx,ω)−A2(kx,ω)/bracketrightbigg e−jkxxdkx=0, hence by the Fourier integral theorem A1(kx,ω)−A2(kx,ω)=ω˜µ1˜I(ω) 2ky1e−jky1h. The boundary condition on the continuity of ˜Hxyields similarly −˜I(ω) 2e−jky1h=ky1 ω˜µ1A1(kx,ω)+ky2 ω˜µ2A2(kx,ω) . We obtain A1(kx,ω)=ω˜µ1˜I(ω) 2ky1RTM(kx,ω)e−jky1h, A2(kx,ω)=−ω˜µ2˜I(ω) 2ky2TTM(kx,ω)e−jky1h. Here RTMand TTM=1+RTMare reﬂection and transmission coeﬃcients given by RTM(kx,ω)=˜µ1ky2−˜µ2ky1 ˜µ1ky2+˜µ2ky1, TTM(kx,ω)=2˜µ1ky2 ˜µ1ky2+˜µ2ky1. These describe the reﬂection and transmission of each component of the plane-wave spectrum of the impressed ﬁeld, and thus depend on the parameter kx. The scattered ﬁelds are ˜Es z1(x,y,ω)=ω˜µ1˜I(ω) 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1e−jky1(y+h) 2ky1RTM(kx,ω)e−jkxxdkx, (4.411) ˜Es z2(x,y,ω)=−ω˜µ2˜I(ω) 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1ejky2(y−hky1/ky2) 2ky2TTM(kx,ω)e−jkxxdkx.(4.412) We may now obtain the ﬁeld produced by an electric line source above a perfect conductor. Letting ˜σ2→∞ we have ky2=/radicalBig k2 2−k2x→∞ and RTM→1, TTM→2. With these, the scattered ﬁelds (4.411) and (4.412) become ˜Es z1(x,y,ω)=ω˜µ1˜I(ω) 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1e−jky1(y+h) 2ky1e−jkxxdkx, (4.413) ˜Es z2(x,y,ω)=0. (4.414) Comparing (4.413) to (4.410) we see that the scattered ﬁeld is exactly the same as that produced by a line source of amplitude −˜I(ω)located at y=−h. We call this line source the image of the impressed source, and say that the problem of two line sources located Figure 4.32: Geometry for scattering of a TM plane wave by a conducting half-plane. symmetrically on the y-axis is equivalent for y>0to the problem of the line source above a ground plane. The total ﬁeld is the sum of the impressed and scattered ﬁelds: ˜Ez(x,y,ω)=−ω˜µ1˜I(ω) 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1e−jky1|y−h|−e−jky1(y+h) 2ky1e−jkxxdkx,y≥0. We can write this in another form using the Hankel-function representation of the line source (4.345): ˜Ez(x,y,ω)=−ω˜µ 4˜I(ω)H(2) 0(k|ρ−ˆyh|)+ω˜µ 4˜I(ω)H(2) 0(k|ρ+ˆyh|) where |ρ±ˆyh|=|ρˆρ±ˆyh|=/radicalbig x2+(y±h)2. Interpreting the general case in terms of images is more diﬃcult. Comparing (4.411) and (4.412) with (4.410), we see that each spectral component of the ﬁeld in region 1 hasthe form of an image line source located at y=−hin region 2, but that the amplitude of the line source, R TM˜I, depends on kx. Similarly, the ﬁeld in region 2 is composed of spectral components that seem to originate from line sources with amplitudes −TTM˜I located at y=hky1/ky2in region 1. In this case the amplitude and position of the image line source producing a spectral component are both dependent on kx. The ﬁeld scattered by a half-plane. Consider a thin planar conductor that occupies the half-plane y=0,x>0. We assume the half-plane lies within a slightly lossy medium having parameters ˜µ(ω)and ˜/epsilon1c(ω), and may consider the case of free space as a lossless limit. The half-plane is illuminated by an impressed uniform plane wave with a z- directe delectri cﬁeld(Figure4.32).Theprimar yﬁeldinduce sasecondar ycurrenton the conductor and this in turn produces a secondary ﬁeld. The total ﬁeld must obey theboundary conditions at y=0. Because the z-directed incident ﬁeld induces a z-directed secondary current, the ﬁelds may be described entirely in terms of a TM set. The impressed plane wave may bewritten as ˜E i(r,ω)=ˆz˜E0(ω)ejk(xcosφ0+ysinφ0) where φ0is the angle between the incident wave vector and the x-axis. By (4.223) we also have ˜Hi(r,ω)=˜E0(ω) η(ˆycosφ0−ˆxsinφ0)ejk(xcosφ0+ysinφ0). The scattered ﬁelds may be written in terms of the Fourier transform solution to the Helmholtz equation. It is convenient to use the polar coordinate representation (4.403)to develop the necessary equations. Thus, for the scattered electric ﬁeld we can write ˜E s z(x,y,ω)=/integraldisplay Cf(ξ, ω) e−jkρcos(φ±ξ)dξ. (4.415) By (4.404) the x-component of the magnetic ﬁeld is ˜Hs x(x,y,ω)=−1 jω˜µ∂˜Es z ∂y=−1 jω˜µ/integraldisplay Cf(ξ, ω)∂ ∂y/parenleftbig e−jkxcosξe±jkysinξ/parenrightbig =−1 jω˜µ(±jk)/integraldisplay Cf(ξ, ω) sinξe−jkρcos(φ±ξ)dξ. To ﬁnd the angular spectrum f(ξ, ω) and ensure uniqueness of solution, we must apply the boundary conditions over the entire y=0plane. For x>0where the conductor resides, the total tangential electric ﬁeld must vanish. Setting the sum of the incidentand scattered ﬁelds to zero at φ=0we have /integraldisplay Cf(ξ, ω) e−jkxcosξdξ=− ˜E0ejkxcosφ0,x>0. (4.416) To ﬁnd the boundary condition for x<0we note that by symmetry ˜Es zis even about y=0while ˜Hs x, as the y-derivative of ˜Es z, is odd. Since no current can be induced in the y=0plane for x<0, the x-directed scattered magnetic ﬁeld must be continuous and thus equal to zero there. Hence our second condition is /integraldisplay Cf(ξ, ω) sinξe−jkxcosξdξ=0,x<0. (4.417) Now that we have developed the two equations that describe f(ξ, ω), it is convenient to return to a rectangular-coordinate-based spectral integral to analyze them. Writingξ=cos −1(kx/k)we have d dξ(kcosξ)=−ksinξ=dkx dξ and dξ=−dkx ksinξ=−dkx k/radicalbig 1−cos2ξ=−dkx/radicalbig k2−k2x. Upon substitution of these relations, the inversion contour returns to the real kxaxis (which may then be perturbed by j/Delta1). Thus, (4.416) and (4.417) may be written as ∞+ j/Delta1/integraldisplay −∞+ j/Delta1f/parenleftbig cos−1kx k/parenrightbig /radicalbig k2−k2xe−jkxxdkx=− ˜E0ejkx0x,x>0, (4.418) ∞+ j/Delta1/integraldisplay −∞+ j/Delta1f/parenleftbigg cos−1kx k/parenrightbigg e−jkxxdkx=0,x<0, (4.419) Figure 4.33: Integration contour used to evaluate the function F(x). where kx0=kcosφ0. Equations (4.418) and (4.419) comprise dual integral equations for f. We may solve these using an approach called the Wiener–Hopf technique . We begin by considering (4.419). If we close the integration contour in the upper half-plane using a semicircle CRof radius Rwhere R→∞, we ﬁnd that the contribution from the semicircle is lim R→∞/integraldisplay CRf/parenleftbigg cos−1kx k/parenrightbigg e−|x|kxiej|x|kxrdkx=0 since x<0. This assumes that fdoes not grow exponentially with R.T h u s /contintegraldisplay Cf/parenleftbigg cos−1kx k/parenrightbigg e−jkxxdkx=0 where Cnow encloses the portion of the upper half-plane kxi>/Delta1. By Morera’s theorem [110],%citeLePage, the above relation holds if fis regular (contains no singularities or branch points) in this portion of the upper half-plane. We shall assume this andinvestigate the other properties of fthat follow from (4.418). In (4.418) we have an integral equated to an exponential function. To understand the implications of the equality it is helpful to write the exponential function as an integral as well. Consider the integral F(x)=1 2jπ∞+ j/Delta1/integraldisplay −∞+ j/Delta1h(kx) h(−kx0)1 kx+kx0e−jkxxdkx. Here h(kx)is some function regular in the region kxi</Delta1, with h(kx)→0askx→∞. Ifwechoose/Delta1sothat−kxi>/Delta1> −kxicosθ0and close the contour with a semicircle in thelowerhalf-plan e(Figure4.33),thenthecontributio nfromthesemicircl evanishe sfor large radius and thus, by Cauchy’s residue theorem, F(x)=− ejkx0x. Using this (4.418) can be written as ∞+ j/Delta1/integraldisplay −∞+ j/Delta1/bracketleftBigg f/parenleftbig cos−1kx k/parenrightbig /radicalbig k2−k2x−˜E0 2jπh(kx) h(−kx0)1 kx+kx0/bracketrightBigg e−jkxxdkx=0. Setting the integrand to zero and using/radicalbig k2−k2x=√k−kx√k+kx,w eh a v e f/parenleftbig cos−1kx k/parenrightbig √k−kx(kx+kx0)=˜E0 2jπ/radicalbig k+kxh(kx) h(−kx0). (4.420) The left member has a branch point at kx = k whiletherightmemberhasabranchpoint atkx =−k.IfwechoosethebranchcutsasinFigure4.30thensince f isregula rin the region kxi>/Delta1the left side of (4.420) is regular there. Also, since h(kx)is regular in the region kxi</Delta1, the right side is regular there. We assert that since the two sides are equal, both sides must be regular in the entire complex plane. By Liouville’s theorem[35] if a function is entire (regular in the entire plane) and bounded, then it must beconstant. So f/parenleftbig cos −1kx k/parenrightbig √k−kx(kx+kx0)=˜E0 2jπ/radicalbig k+kxh(kx) h(−kx0)=constant . We may evaluate the constant by inserting any value of kx. Using kx=−kx0on the right we ﬁnd that f/parenleftbig cos−1kx k/parenrightbig √k−kx(kx+kx0)=˜E0 2jπ/radicalbig k−kx0. Substituting kx=kcosξand kx0=kcosφ0we have f(ξ)=˜E0 2jπ√1−cosφ0√1−cosξ cosξ+cosφ0. Since sin(x/2)=√(1−cosx)/2, we may also write f(ξ)=˜E0 jπsinφ0 2sinξ 2 cosξ+cosφ0. Finally, substituting this into (4.415) we have the spectral representation for the ﬁeld scattered by a half-plane: ˜Es z(ρ, φ, ω) =˜E0(ω) jπ/integraldisplay Csinφ0 2sinξ 2 cosξ+cosφ0e−jkρcos(φ±ξ)dξ. (4.421) The scattered ﬁeld inversion integral in (4.421) may be rewritten in such a way as to separate geometrical optics (plane- wave) terms from diﬀraction terms. The diﬀraction terms may be written using standard functions (modiﬁed Fresnel integrals) and for largevalues of ρappear as cylindrical waves emanating from a line source at the edge of the half-plane. Interested readers should see James [92] for details. 4.14 Periodic ﬁelds and Floquet’s theorem In several practical situations EM waves interact with, or are radiated by, structures spatially periodic along one or more directions. Periodic symmetry simpliﬁes ﬁeld com-putation, since boundary conditions need only be applied within one period, or cell,o f the structure. Examples of situations that lead to periodic ﬁelds include the guiding ofwaves in slow-wave structures such as helices and meander lines, the scattering of planewaves from gratings, and the radiation of waves by antenna arrays. In this section wewill study the representation of ﬁelds with inﬁnite periodicity as spatial Fourier series. 4.14.1 Floquet’s theorem Consider an environment having spatial periodicity along the z-direction. In this envi- ronment the frequency-domain ﬁeld may be represented in terms of a periodic function ˜ψpthat obeys ˜ψp(x,y,z±mL,ω)=˜ψp(x,y,z,ω) where mis an integer and Lis the spatial period. According to Floquet’s theorem ,i f˜ψ represents some vector component of the ﬁeld, then the ﬁeld obeys ˜ψ(x,y,z,ω)=e−jκz˜ψp(x,y,z,ω) . (4.422) Hereκ=β−jαis a complex wavenumber describing the phase shift and attenuation of the ﬁeld between the various cells of the environment. The phase shift and attenuationmay arise from a wave propagating through a lossy periodic medium (see example below) or may be impressed by a plane wave as it scatters from a periodic surface, or may beproduced by the excitation of an antenna array by a distributed terminal voltage. It isalso possible to have κ=0as when, for example, a periodic antenna array is driven with all elements in phase. Because ˜ψ pis periodic we may expand it in a Fourier series ˜ψp(x,y,z,ω)=∞/summationdisplay n=−∞˜ψn(x,y,ω)e−j2πnz/L where the ˜ψnare found by orthogonality: ˜ψn(x,y,ω)=1 L/integraldisplayL/2 −L/2˜ψp(x,y,z,ω)ej2πnz/Ldz. Substituting this into (4.422), we have a representation for the ﬁeld as a Fourier series: ˜ψ(x,y,z,ω)=∞/summationdisplay n=−∞˜ψn(x,y,ω)e−jκnz where κn=β+2πn/L+jα=βn−jα. We see that within each cell the ﬁeld consists of a number of constituents called space harmonics orHartree harmonics , each with the property of a propagating or evanescent wave. Each has phase velocity vpn=ω βn=ω β+2πn/L. A number of the space harmonics have phase velocities in the +z-direction while the re- mainder have phase velocities in the −z-direction, depending on the value of β. However, all of the space harmonics have the same group velocity vgn=dω dβ=/parenleftbiggdβn dω/parenrightbigg−1 =/parenleftbiggdβ dω/parenrightbigg−1 =vg. Those space harmonics for which the group and phase velocities are in opposite directions are referred to as backward waves , and form the basis of operation of microwave tubes known as “backward waveoscillators.” Figure 4.34: Geometry of a periodic stratiﬁed medium with each cell consisting of two material layers. 4.14.2 Examples of periodic systems Plane-wave propagation within a periodically stratiﬁed medium. As an exam- ple of wave propagation in a periodic structure, let us consider a plane wave propagatingwithin a layered medium consisting of two material layers repeated periodically as showninFigure4.34.Eachsectionoftwolayersisacellwithintheperiodicmedium ,andwe seek an expression for the propagation constant within the cells, κ. We developed the necessary tools for studying plane waves within an arbitrary layered medium in §4.11.5, and can apply them to the case of a periodic medium. In equations (4.305) and (4.306) we have expressions for the waveamplitudes in any region in terms of the amplitudes in the region immediately preceding it. We may write these in matrixform by eliminating one of the variables a norbnfrom each equation: /bracketleftbiggT(n) 11T(n) 12 T(n) 21T(n) 22/bracketrightbigg/bracketleftbiggan+1 bn+1/bracketrightbigg =/bracketleftbiggan bn/bracketrightbigg (4.423) where T(n) 11=1 2Zn+Zn−1 Zn˜P−1 n, T(n) 12=1 2Zn−Zn−1 Zn˜Pn, T(n) 21=1 2Zn−Zn−1 Zn˜P−1 n, T(n) 22=1 2Zn+Zn−1 Zn˜Pn. Here Znrepresents Zn⊥for perpendicular polarization and Zn/bardblfor parallel polariza- tion. The matrix entries are often called transmission parameters , and are similar to the parameters used to describe microwave networks, except that in network theory thewave amplitudes are often normalized using the wave impedances.We may use these parameters to describe the cascaded system of two layers: /bracketleftbiggT(n) 11T(n) 12 T(n) 21T(n) 22/bracketrightbigg/bracketleftbiggT(n+1) 11 T(n+1) 12 T(n+1) 21 T(n+1) 22/bracketrightbigg/bracketleftbiggan+2 bn+2/bracketrightbigg =/bracketleftbiggan bn/bracketrightbigg . Since for a periodic layered medium the wave amplitudes should obey (4.422), we have /bracketleftbiggT11T12 T21T22/bracketrightbigg/bracketleftbiggan+2 bn+2/bracketrightbigg =/bracketleftbiggan bn/bracketrightbigg =ejκL/bracketleftbiggan+2 bn+2/bracketrightbigg (4.424) where L=/Delta1n+/Delta1n+1is the period of the structure and /bracketleftbiggT11T12 T21T22/bracketrightbigg =/bracketleftbiggT(n) 11T(n) 12 T(n) 21T(n) 22/bracketrightbigg/bracketleftbiggT(n+1) 11 T(n+1) 12 T(n+1) 21 T(n+1) 22/bracketrightbigg . Equation (4.424) is an eigenvalue equation for κand can be rewritten as /bracketleftbiggT11−ejκLT12 T21 T22−ejκL/bracketrightbigg/bracketleftbiggan+2 bn+2/bracketrightbigg =/bracketleftbigg0 0/bracketrightbigg . This equation only has solutions when the determinant of the matrix vanishes. Expansion of the determinant gives T11T22−T12T21−ejκL(T11+T22)+ej2κL=0. (4.425) The ﬁrst two terms are merely T11T22−T12T21=/vextendsingle/vextendsingle/vextendsingle/vextendsingleT 11T12 T21T22/vextendsingle/vextendsingle/vextendsingle/vextendsingle=/vextendsingle/vextendsingle/vextendsingle/vextendsingleT (n) 11T(n) 12 T(n) 21T(n) 22/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleT (n+1) 11 T(n+1) 12 T(n+1) 21 T(n+1) 22/vextendsingle/vextendsingle/vextendsingle/vextendsingle. Since we can show that /vextendsingle/vextendsingle/vextendsingle/vextendsingleT (n) 11T(n) 12 T(n) 21T(n) 22/vextendsingle/vextendsingle/vextendsingle/vextendsingle=Z n−1 Zn, we have T11T22−T12T21=Zn−1 ZnZn Zn+1=1 where we have used Zn−1=Zn+1because of the periodicity of the medium. With this, (4.425) becomes cosκL=T11+T22 2. Finally, computing the matrix product and simplifying to ﬁnd T11+T22, we have cosκL=cos(kz,n/Delta1n)cos(kk,n−1/Delta1n−1)− −1 2/parenleftbiggZn−1 Zn+Zn Zn−1/parenrightbigg sin(kz,n/Delta1n)sin(kz,n−1/Delta1n−1) (4.426) or equivalently cosκL=1 4(Zn−1+Zn)2 ZnZn−1cos(kz,n/Delta1n+kz,n−1/Delta1n−1)− −1 4(Zn−1−Zn)2 ZnZn−1cos(kz,n/Delta1n−kz,n−1/Delta1n−1). (4.427) Note that both ±κsatisfy this equation, allowing waves with phase front propagation in both the ±z-directions. We see in (4.426) that even for lossless materials certain values of ωresult in cosκL>1, causing κLto be imaginary and producing evanescent waves. We refer to the frequency ranges over which cosκL>1asstopbands , and those over which cosκL<1aspassbands . This terminology is used in ﬁlter analysis and, indeed, waves propagating in periodic media experience eﬀects similar to those experienced by signals passing through ﬁlters. Field produced by an inﬁnite array of line sources. As a second example, consider an inﬁnite number of z-directed line sources within a homogeneous medium of complex permittivity ˜/epsilon1c(ω)and permeability ˜µ(ω), aligned along the x-axis with separation L such that ˜J(r,ω)=∞/summationdisplay n=−∞ˆz˜Inδ(y)δ(x−nL). The current on each element is allowed to show a progressive phase shift and attenua- tion. (Such progression may result from a particular method of driving primary currentson successive elements, or, if the currents are secondary, from their excitation by animpressed ﬁeld such as a plane wave.) Thus we write ˜I n=˜I0e−jκnL(4.428) where κis a complex constant. We may represent the ﬁeld produced by the source array as a superposition of the ﬁelds of individual line sources found earlier. In particular we may use the Hankel functionrepresentation (4.345) or the Fourier transform representation (4.407). Using the latterwe have ˜E z(x,y,ω)=∞/summationdisplay n=−∞e−jκnL −ω˜µ˜I0(ω) 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1e−jky|y| 2kye−jkx(x−nL)dkx . Interchanging the order of summation and integration we have ˜Ez(x,y,ω)=−ω˜µ˜I0(ω) 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1e−jky|y| 2ky/bracketleftBigg∞/summationdisplay n=−∞ejn(kx−κ)L/bracketrightBigg e−jkxxdkx. (4.429) We can rewrite the sum in this expression using Poisson’s sum formula [142]. ∞/summationdisplay n=−∞f(x−nD)=1 D∞/summationdisplay n=−∞F(nk0)ejnk 0x, where k0=2π/D. Letting f(x)=δ(x−x0)in that expression we have ∞/summationdisplay n=−∞δ/parenleftbigg x−x0−n2π L/parenrightbigg =L 2π∞/summationdisplay n=−∞ejnL(x−x0). Substituting this into (4.429) we have ˜Ez(x,y,ω)=−ω˜µ˜I0(ω) 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1e−jky|y| 2ky/bracketleftBigg∞/summationdisplay n=−∞2π Lδ/parenleftbigg kx−κ−n2π L/parenrightbigg/bracketrightBigg e−jkxxdkx. Carrying out the integral we replace kxwithκn=κ+2nπ/L, giving ˜Ez(x,y,ω)=−ω˜µ˜I0(ω)∞/summationdisplay n=−∞e−jky,n|y|e−jκnx 2Lky,n =− jω˜µ˜I0(ω)˜G∞(x,y|0,0,ω) (4.430) where ky,n=/radicalbig k2−κ2n, and where ˜G∞(x,y|x/prime,y/prime,ω)=∞/summationdisplay n=−∞e−jky,n|y−y/prime|e−jκn(x−x/prime) 2jLk y,n(4.431) is called the periodic Green’s function . We may also ﬁnd the ﬁeld produced by an inﬁnite array of line sources in terms of the Hankel function representation of a single line source (4.345). Using the current representation (4.428) and summing over the sources, we obtain ˜Ez(ρ, ω) =−ω˜µ 4∞/summationdisplay n=−∞˜I0(ω)e−jκnLH(2) 0(k|ρ−ρn|)=− jω˜µ˜I0(ω)˜G∞(x,y|0,0,ω) where |ρ−ρn|=| ˆyy+ˆx(x−nL)|=/radicalbig y2+(x−nL)2 and where ˜G∞is an alternative form of the periodic Green’s function ˜G∞(x,y|x/prime,y/prime,ω)=1 4j∞/summationdisplay n=−∞e−jκnLH(2) 0/parenleftBig k/radicalbig (y−y/prime)2+(x−nL−x/prime)2/parenrightBig .(4.432) The periodic Green’s functions (4.431) and (4.432) produce identical results, but are each appropriate for certain applications. For example, (4.431) is useful for situationsin which boundary conditions at constant values of yare to be applied. Both forms are diﬃcult to compute under certain circumstances, and variants of these forms have beenintroduced in the literature [203]. 4.15 Problems 4.1Beginning with the Kronig–Kramers formulas (4.35)–(4.36), use the even–odd be- havior of the real and imaginary parts of ˜/epsilon1cto derive the alternative relations (4.37)– (4.38). 4.2Consider the complex permittivity dyadic of a magnetized plasma given by (4.88)– (4.91). Show that we may decompose [˜¯/epsilon1c]as the sum of two matrices [˜¯/epsilon1c]=[˜¯/epsilon1]+[˜¯σ] jω where [˜¯/epsilon1]and [˜¯σ]are hermitian. 4.3Show that the Debye permittivity formulas ˜/epsilon1/prime(ω)−/epsilon1∞=/epsilon1s−/epsilon1∞ 1+ω2τ2, ˜/epsilon1/prime/prime(ω)=−ωτ(/epsilon1 s−/epsilon1∞) 1+ω2τ2, obey the Kronig–Kramers relations. 4.4The frequency-domain duality transformations for the constitutive parameters of an anisotropic medium are given in (4.197). Determine the analogous transformationsfor the constitutive parameters of a bianisotropic medium. 4.5Establish the plane-wave identities (B.76)–(B.79) by direct diﬀerentiation in rect- angular coordinates. 4.6Assume that sea water has the parameters /epsilon1=80/epsilon1 0,µ=µ0,σ=4S/m, and that these parameters are frequency-independent. Plot the ω–βdiagram for a plane wave propagatin ginthismediu mandcompar etoFigure4.12.Descri bethedispersion :isit normal or anomalous? Also plot the phase and group velocities and compare to Figure 4.13.Howdoestherelaxatio nphenomeno naﬀectthevelocityofawaveinthismedium? 4.7Consider a uniform plane wave incident at angle θionto an interface separating twolossles smedia(Figure4.18).Assumin gperpendicula rpolarization ,writetheexplicit formsofthetotalﬁeldsineachregionundertheconditio nθi<θ c, where θcis the critical angle. Show that the total ﬁeld in region 1 can be decomposed into a portion that isa pure standing wave in the z-direction and a portion that is a pure traveling wave in thez-direction. Also show that the ﬁeld in region 2 is a pure traveling wave. Repeat for parallel polarization. 4.8Consider a uniform plane wave incident at angle θ ionto an interface separating twolossles smedia(Figure4.18).Assumin gperpendicula rpolarization ,usethetotal ﬁeldsfromProble m4.7toshowthatundertheconditio nθi<θ cthe normal component of the time-average Poynting vector is continuous across the interface. Here θcis the critical angle. Repeat for parallel polarization. 4.9Consider a uniform plane wave incident at angle θionto an interface separating twolossles smedia(Figure4.18).Assumin gperpendicula rpolarization ,writetheexplicit forms of the total ﬁelds in each region under the condition θi>θ c, where θcis the critical angle. Show that the ﬁeld in region 1 is a pure standing wave in the z-direction and that theﬁeldinregion2isanevanesce ntwave.Repeatforparalle lpolarization. 4.10 Consider a uniform plane waveincident at angle θionto an interface separating twolossles smedia(Figure4.18).Assumin gperpendicula rpolarization ,usetheﬁelds from Problem 4.9 to show that under the condition θi>θ cthe ﬁeld in region 1 carries no time-average power in the z-direction, while the ﬁeld in region 2 carries no time-average power. Here θcis the critical angle. Repeat for parallel polarization. 4.11 Consider a uniform plane wave incident at angle θifrom a lossless material onto agoodconducto r(Figure4.18).Theconducto rhaspermittivi ty/epsilon10, permeability µ0, and conductivity σ. Show that the transmission angle is θt≈0and thus the wave in the conductor propagates normal to the interface. Also show that for perpendicularpolarization the current per unit width induced by the wave in region 2 is ˜K(ω)=ˆyσ˜T ⊥(ω)˜E⊥(ω)1−j 2β2 and that this is identical to the tangential magnetic ﬁeld at the surface: ˜K(ω)=− ˆz×˜Ht|z=0. If we deﬁne the surface impedance Zs(ω)of the conductor as the ratio of tangential electric and magnetic ﬁelds at the interface, show that Zs(ω)=1+j σδ=Rs(ω)+jXs(ω). Then show that the time-average power ﬂux entering region 2 for a monochromatic wave of frequency ˇωis simply Sav,2=ˆz1 2(ˇK·ˇK∗)Rs. Note that the since the surface impedance is also the ratio of tangential electric ﬁeld to induced current per unit width in region 2, it is also called the internal impedance . 4.12 Consider a parallel-polarized plane wave obliquely incident from a lossless medium ontoamulti-layeredmateria lasshowninFigure4.20.Writingtheﬁeldsineachregion n,0≤n≤N−1,a s ˜H/bardbln=˜Hi /bardbln+˜Hr /bardblnwhere ˜Hi /bardbln=ˆyan+1e−jkx,nxe−jkz,n(z−zn+1), ˜Hr /bardbln=− ˆybn+1e−jkx,nxe+jkz,n(z−zn+1), and the ﬁeld in region Nas ˜H/bardblN=ˆyaN+1e−jkx,Nxe−jkz,N(z−zN), apply the boundary conditions to solve for the wave amplitudes an+1and bnin terms of a global reﬂection coeﬃcient ˜Rn, an interfacial reﬂection coeﬃcient /Gamma1n/bardbl, and the wave amplitude an. Compare your results to those found for perpendicular polarization (4.313) and (4.314). 4.13 Consider a slab of lossless material with permittivity /epsilon1=/epsilon1r/epsilon10and permeability µ=µrµ0located in free space between the planes z=z1and z=z2. A right-hand circularly-polarized plane wave is incident on the slab at angle θias shown in Figure 4.22.Determin ethecondition s(ifany)underwhichthereﬂecte dwaveis:(a)linearly polarized; (b) right-hand or left-hand circularly polarized; (c) right-hand or left-hand elliptically polarized. Repeat for the transmitted wave. 4.14 Consider a slab of lossless material with permittivity /epsilon1=/epsilon1r/epsilon10and permeability µ0 located in free space between the planes z=z1and z=z2. A transient, perpendicularly- polarize dplanewaveisobliquel yincide ntontheslabasshowninFigure4.22.Ifthe temporal waveform of the incident wave is Ei ⊥(t), ﬁnd the transient reﬂected ﬁeld in region 0 and the transient transmitted ﬁeld in region 2 in terms of an inﬁnite superposition ofamplitude-scaled, time-shifted versions of the incident wave. Interpret each of the ﬁrstfour terms in the reﬂected and transmitted ﬁelds in terms of multiple reﬂection withinthe slab. 4.15 Consider a free-space gap embedded between the planes z=z 1and z=z2 in an inﬁnite, lossless dielectric medium of permittivity /epsilon1r/epsilon10and permeability µ0.A perpendicularly-polarized plane wave is incident on the gap at angle θi>θ cas shown inFigure4.22.Hereθcis the critical angle for a plane wave incident on the single interface between a lossless dielectric of permittivity /epsilon1r/epsilon10and free space. Apply the boundary conditions and ﬁnd the ﬁelds in each of the three regions. Find the time-average Poynting vector in region 0 at z=z 1, in region 1 at z=z2, and in region 2 at z=z2. Is conservation of energy obeyed? 4.16 A uniform ferrite material has scalar permittivity ˜/epsilon1=/epsilon1and dyadic permeability ˜¯µ. Assume the ferrite is magnetized along the z-direction and has losses so that its permeability dyadic is given by (4.118). Show that the waveequation for a TEM plane wave of the form ˜H(r,ω)=˜H0(ω)e−jkzz is k2 z˜H0=ω2/epsilon1˜¯µ·˜H0 where kz=β−jα. Find explicit formulas for the two solutions kz±=β±−jα±. Show that when the damping parameter α/lessmuch1, near resonance α+/greatermuchα−. 4.17 A time-harmonic, TE-polarized, uniform cylindrical wave propagates in a lossy medium. Assuming |kρ|/greatermuch 1, show that the power per unit length passing through a cylinder of radius ρis given by Pav/l=Re/braceleftbig Z∗ TE/bracerightbig |ˇHz0|2e−2αρ 8|k|. If the material is lossless, show that the power per unit length passing through a cylinder is independent of the radius and is given by Pav/l=η|ˇHz0|2 8k. 4.18 A TM-polarized plane wave is incident on a cylinder made from a perfect electric conductor such that the current induced on the cylinder is given by (4.365). When thecylinder radius is large compared to the wavelength of the incident wave, we may ap-proximate the current using the principle of physical optics . This states that the induced current is zero in the “shadow region” where the cylinder is not directly illuminated bythe incident wave. Elsewhere, in the “illuminated region,” the induced current is given by ˜J s=2ˆn×˜Hi. Plot the current from (4.365) for various values of k0aand compare to the current com- puted from physical optics. How large must k0abe for the shadowing eﬀect to be signif- icant? 4.19 Theradar cross section of a two-dimensional object illuminated by a TM-polarized plane wave is deﬁned by σ2−D(ω, φ) =lim ρ→∞2πρ|˜Es z|2 |˜Eiz|2. This quantity has units of meters and is sometimes called the “scattering width” of the object. Using the asymptotic form of the Hankel function, determine the formula for the radar cross section of a TM-illuminated cylinder made of perfect electric conductor. Show that when the cylinder radius is small compared to a wavelength the radar cross section may be approximated as σ2−D(ω, φ) =aπ2 k0a1 ln2(0.89k0a) and is thus independent of the observation angle φ. 4.20 A TE-polarized plane wave is incident on a material cylinder with complex per- mittivity ˜/epsilon1c(ω)and permeability ˜µ(ω), aligned along the z-axis in free space. Apply the boundary conditions on the surface of the cylinder and determine the total ﬁeld bothinternal and external to the cylinder. Show that as ˜σ→∞ the magnetic ﬁeld external to the cylinder reduces to (4.366). 4.21 A TM-polarized plane wave is incident on a PEC cylinder of radius aaligned along the z-axis in free space. The cylinder is coated with a material layer of radius b with complex permittivity ˜/epsilon1 c(ω)and permeability ˜µ(ω). Apply the boundary conditions on the surface of the cylinder and across the interface between the material and freespace and determine the total ﬁeld both internal and external to the material layer. 4.22 A PEC cylinder of radius a, aligned along the z-axis in free space, is illuminated by a z-directed electric line source ˜I(ω)located at (ρ 0,φ0). Expand the ﬁelds in the regions a<ρ<ρ 0andρ>ρ 0in terms of nonuniform cylindrical waves, and apply the boundary conditions at ρ=aandρ=ρ0to determine the ﬁelds everywhere. 4.23 Repeat Problem 4.22 for the case of a cylinder illuminated by a magnetic line source. 4.24 Assuming f(ξ, ω) =k 2πA(kx,ω)sinξ, use the relations cosz=cos(u+jv)=cosucoshv−jsinusinhv, sinz=sin(u+jv)=sinucoshv+jcosusinhv, toshowthatthecontourinFigure4.29providesidenticalvaluesoftheintegran din ˜ψ(x,y,ω)=/integraldisplay Cf(ξ, ω) e−jkρcos(φ±ξ)dξ as does the contour [−∞ + j/Delta1,∞+ j/Delta1]in ˜ψ(x,y,ω)=1 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1A(kx,ω)e−jkxxe∓jkyydkx. (4.433) 4.25 Verify (4.409) by writing the TE ﬁelds in terms of Fourier transforms and apply- ing boundary conditions. 4.26 Consider a z-directed electric line source ˜I(ω)located on the y-axis at y=h. The region y<0contains a perfect electric conductor. Write the ﬁelds in the regions 0<y<hand y>hin terms of the Fourier transform solution to the homogeneous Helmholtz equation. Note that in the region 0<y<hterms representing waves traveling inboththe±y-directions are needed, while in the region y>honly terms traveling in they-direction are needed. Apply the boundary conditions at y=0,hto determine the spectral amplitudes. Show that the total ﬁeld may be decomposed into an impressedterm identical to (4.410) and a scattered term identical to (4.413). 4.27 Consider a z-directed magnetic line source ˜I m(ω)located on the y-axis at y=h. The region y>0contains a material with parameters ˜/epsilon1c 1(ω)and ˜µ1(ω), while the region y<0contains a material with parameters ˜/epsilon1c 2(ω)and ˜µ2(ω). Using the Fourier transform solution to the Helmholtz equation, write the total ﬁeld for y>0as the sum of an impressed ﬁeld of the magnetic line source and a scattered ﬁeld, and write the ﬁeld for y<0as a scattered ﬁeld. Apply the boundary conditions at y=0to determine the spectral amplitudes. Can you interpret the scattered ﬁelds in terms of images of the linesource? 4.28 Consider a TE-polarized plane wave incident on a PEC half-plane located at y=0,x>0. If the incident magnetic ﬁeld is given by ˜H i(r,ω)=ˆz˜H0(ω)ejk(xcosφ0+ysinφ0), determine the appropriate boundary conditions on the ﬁelds at y=0. Solve for the scattered magnetic ﬁeld using the Fourier transform approach. 4.29Conside rthelayeredmediu mofFigure4.34withalternatin glayersoffreespace and perfect dielectric. The dielectric layer has permittivity 4/epsilon10and thickness /Delta1while the free space layer has thickness 2/Delta1. Assuming a normally-incident plane wave, solve fork0/Delta1in terms of κ/Delta1, and plot k0versus κ, identifying the stop and pass bands. This type of ω–βplot for a periodic medium is named a Brillouin diagram , after L. Brillouin who investigated energy bands in periodic crystal lattices [23]. 4.30Conside raperiodiclayeredmediu masinFigure4.34,butwitheachcellcon- sisting of three diﬀerent layers. Derive an eigenvalue equation similar to (4.427) for thepropagation constant. Chapter 5 Field decompositions and the EM potentials 5.1 Spatial symmetry decompositions Spatial symmetry can often be exploited to solve electromagnetics problems. For analytic solutions, symmetry can be used to reduce the number of boundary conditionsthat must be applied. For computer solutions the storage requirements can be reduced.Typical symmetries include rotation about a point or axis, and reﬂection through aplane, along an axis, or through a point. We shall consider the common case of reﬂectionthrough a plane. Reﬂections through the origin and through an axis will be treated inthe exercises. Note that spatial symmetry decompositions may be applied even if the sources and ﬁelds possess no spatial symmetry. As long as the boundaries and material media aresymmetric, the sources and ﬁelds may be decomposed into constituents that individuallymimic the symmetry of the environment. 5.1.1 Planar ﬁeld symmetry Consider a region of space consisting of linear, isotropic, time-invariant media having material parameters /epsilon1(r),µ(r), and σ(r). The electromagnetic ﬁelds ( E,H)within this region are related to their impressed sources (Ji,Ji m)and their secondary sources Js=σE through Maxwell’s curl equations: ∂Ez ∂y−∂Ey ∂z=−µ∂Hx ∂t−Ji mx, (5.1) ∂Ex ∂z−∂Ez ∂x=−µ∂Hy ∂t−Ji my, (5.2) ∂Ey ∂x−∂Ex ∂y=−µ∂Hz ∂t−Ji mz, (5.3) ∂Hz ∂y−∂Hy ∂z=/epsilon1∂Ex ∂t+σEx+Ji x, (5.4) ∂Hx ∂z−∂Hz ∂x=/epsilon1∂Ey ∂t+σEy+Ji y, (5.5) ∂Hy ∂x−∂Hx ∂y=/epsilon1∂Ez ∂t+σEz+Ji z. (5.6) We assume the material constants are symmetric about some plane, say z=0. Then /epsilon1(x,y,−z)=/epsilon1(x,y,z), µ(x,y,−z)=µ(x,y,z), σ(x,y,−z)=σ(x,y,z). That is, with respect to zthe material constants are even functions. We further assume that the boundaries and boundary conditions, which guarantee uniqueness of solution, arealso symmetric about the z=0plane. Then we deﬁne two cases of reﬂection symmetry. Conditions for even symmetry. We claim that if the sources obey J i x(x,y,z)=Ji x(x,y,−z), Ji mx(x,y,z)=− Ji mx(x,y,−z), Ji y(x,y,z)=Ji y(x,y,−z), Ji my(x,y,z)=− Ji my(x,y,−z), Ji z(x,y,z)=− Ji z(x,y,−z), Ji mz(x,y,z)=Ji mz(x,y,−z), then the ﬁelds obey Ex(x,y,z)=Ex(x,y,−z), Hx(x,y,z)=− Hx(x,y,−z), Ey(x,y,z)=Ey(x,y,−z), Hy(x,y,z)=− Hy(x,y,−z), Ez(x,y,z)=− Ez(x,y,−z), Hz(x,y,z)=Hz(x,y,−z). The electric ﬁeld shares the symmetry of the electric source: components parallel to the z=0plane are even in z, and the component perpendicular is odd. The magnetic ﬁeld shares the symmetry of the magnetic source: components parallel to the z=0plane are odd in z, and the component perpendicular is even. We can verify our claim by showing that the symmetric ﬁelds and sources obey Maxwell’s equations. At an arbitrary point z=a>0equation (5.1) requires ∂Ez ∂y/vextendsingle/vextendsingle/vextendsingle/vextendsingle z=a−∂Ey ∂z/vextendsingle/vextendsingle/vextendsingle/vextendsingle z=a=−µ|z=a∂Hx ∂t/vextendsingle/vextendsingle/vextendsingle/vextendsingle z=a−Ji mx|z=a. By the assumed symmetry condition on source and material constant we get ∂Ez ∂y/vextendsingle/vextendsingle/vextendsingle/vextendsingle z=a−∂Ey ∂z/vextendsingle/vextendsingle/vextendsingle/vextendsingle z=a=−µ|z=−a∂Hx ∂t/vextendsingle/vextendsingle/vextendsingle/vextendsingle z=a+Ji mx|z=−a. If our claim holds regarding the ﬁeld behavior, then ∂Ez ∂y/vextendsingle/vextendsingle/vextendsingle/vextendsingle z=−a=−∂Ez ∂y/vextendsingle/vextendsingle/vextendsingle/vextendsingle z=a, ∂Ey ∂z/vextendsingle/vextendsingle/vextendsingle/vextendsingle z=−a=−∂Ey ∂z/vextendsingle/vextendsingle/vextendsingle/vextendsingle z=a, ∂Hx ∂t/vextendsingle/vextendsingle/vextendsingle/vextendsingle z=−a=−∂Hx ∂t/vextendsingle/vextendsingle/vextendsingle/vextendsingle z=a, and we have −∂Ez ∂y/vextendsingle/vextendsingle/vextendsingle/vextendsingle z=−a+∂Ey ∂z/vextendsingle/vextendsingle/vextendsingle/vextendsingle z=−a=µ|z=−a∂Hx ∂t/vextendsingle/vextendsingle/vextendsingle/vextendsingle z=−a+Ji mx|z=−a. So this component of Faraday’s lawis satisﬁed. With similar reasoning w e can showthat the symmetric sources and ﬁelds satisfy (5.2)–(5.6) as well. Conditions for odd symmetry. We can also showthat if the sources obey Ji x(x,y,z)=− Ji x(x,y,−z), Ji mx(x,y,z)=Ji mx(x,y,−z), Ji y(x,y,z)=− Ji y(x,y,−z), Ji my(x,y,z)=Ji my(x,y,−z), Ji z(x,y,z)=Ji z(x,y,−z), Ji mz(x,y,z)=− Ji mz(x,y,−z), then the ﬁelds obey Ex(x,y,z)=− Ei x(x,y,−z), Hx(x,y,z)=Hx(x,y,−z), Ey(x,y,z)=− Ey(x,y,−z), Hy(x,y,z)=Hy(x,y,−z), Ez(x,y,z)=Ez(x,y,−z), Hz(x,y,z)=− Hz(x,y,−z). Again the electric ﬁeld has the same symmetry as the electric source. However, in this case components parallel to the z=0plane are odd in zand the component perpendicular is even. Similarly, the magnetic ﬁeld has the same symmetry as the magnetic source. Herecomponents parallel to the z=0plane are even in zand the component perpendicular is odd. Field symmetries and the concept of source images. In the case of odd symmetry the electric ﬁeld parallel to the z=0plane is an odd function of z. If we assume that the ﬁeld is also continuous across this plane, then the electric ﬁeld tangential to z=0 must vanish: the condition required at the surface of a perfect electric conductor (PEC).We may regard the problem of sources above a perfect conductor in the z=0plane as equivalent to the problem of sources odd about this plane, as long as the sources in both cases are identical for z>0. We refer to the source in the region z<0as the image of the source in the region z>0. Thus the image source (J I,JI m)obeys JI x(x,y,−z)=− Ji x(x,y,z), JI mx(x,y,−z)=Ji mx(x,y,z), JI y(x,y,−z)=− Ji y(x,y,z), JI my(x,y,−z)=Ji my(x,y,z), JI z(x,y,−z)=Ji z(x,y,z), JI mz(x,y,−z)=− Ji mz(x,y,z). That is, parallel components of electric current image in the opposite direction, and the perpendicular component images in the same direction; parallel components of themagnetic current image in the same direction, while the perpendicular component imagesin the opposite direction. In the case of even symmetry, the magnetic ﬁeld parallel to the z=0plane is odd, and thus the magnetic ﬁeld tangential to the z=0plane must be zero. We therefore have an equivalence between the problem of a source above a plane of perfect magneticconductor (PMC) and the problem of sources even about that plane. In this case weidentify image sources that obey J I x(x,y,−z)=Ji x(x,y,z), JI mx(x,y,−z)=− Ji mx(x,y,z), JI y(x,y,−z)=Ji y(x,y,z), JI my(x,y,−z)=− Ji my(x,y,z), JI z(x,y,−z)=− Ji z(x,y,z), JI mz(x,y,−z)=Ji mz(x,y,z). Parallel components of electric current image in the same direction, and the perpendicular component images in the opposite direction; parallel components of magnetic currentimage in the opposite direction, and the perpendicular component images in the samedirection. In the case of odd symmetry, we sometimes say that an “electric wall” exists at z=0. The term “magnetic wall” can be used in the case of even symmetry. These terms areparticularly common in the description of waveguide ﬁelds. Symmetric ﬁeld decomposition. Field symmetries may be applied to arbitrary source distributions through a symmetry decomposition of the sources and ﬁelds. Con-sider the general impressed source distributions (J i,Ji m). The source set Jie x(x,y,z)=1 2/bracketleftbig Ji x(x,y,z)+Ji x(x,y,−z)/bracketrightbig , Jie y(x,y,z)=1 2/bracketleftbig Ji y(x,y,z)+Ji y(x,y,−z)/bracketrightbig , Jie z(x,y,z)=1 2/bracketleftbig Ji z(x,y,z)−Ji z(x,y,−z)/bracketrightbig , Jie mx(x,y,z)=1 2/bracketleftbig Ji mx(x,y,z)−Ji mx(x,y,−z)/bracketrightbig , Jie my(x,y,z)=1 2/bracketleftbig Ji my(x,y,z)−Ji my(x,y,−z)/bracketrightbig , Jie mz(x,y,z)=1 2/bracketleftbig Ji mz(x,y,z)+Ji mz(x,y,−z)/bracketrightbig , is clearly of even symmetric type while the source set Jio x(x,y,z)=1 2/bracketleftbig Ji x(x,y,z)−Ji x(x,y,−z)/bracketrightbig , Jio y(x,y,z)=1 2/bracketleftbig Ji y(x,y,z)−Ji y(x,y,−z)/bracketrightbig , Jio z(x,y,z)=1 2/bracketleftbig Ji z(x,y,z)+Ji z(x,y,−z)/bracketrightbig , Jio mx(x,y,z)=1 2/bracketleftbig Ji mx(x,y,z)+Ji mx(x,y,−z)/bracketrightbig , Jio my(x,y,z)=1 2/bracketleftbig Ji my(x,y,z)+Ji my(x,y,−z)/bracketrightbig , Jio mz(x,y,z)=1 2/bracketleftbig Ji mz(x,y,z)−Ji mz(x,y,−z)/bracketrightbig , is of the odd symmetric type. Since Ji=Jie+Jioand Ji m=Jie m+Jio m, we can decompose any source into constituents having, respectively, even and odd symmetry with respectto a plane. The source with even symmetry produces an even ﬁeld set, while the sourcewith odd symmetry produces an odd ﬁeld set. The total ﬁeld is the sum of the ﬁelds from each ﬁeld set. Planar symmetry for frequency-domain ﬁelds. The symmetry conditions intro- duced above for the time-domain ﬁelds also hold for the frequency-domain ﬁelds. Becauseboth the conductivity and permittivity must be even functions, we combine their eﬀectsand require the complex permittivity to be even. Otherwise the ﬁeld symmetries andsource decompositions are identical. Example of symmetry decomposition: line source between conducting planes. Consider a z-directed electric line source ˜I 0located at y=h,x=0between conducting planes at y=± d,d>h. The material between the plates has permeability ˜µ(ω)and complex permittivity ˜/epsilon1c(ω). We decompose the source into one of even symmetric type with line sources ˜I0/2located at y=± h, and one of odd symmetric type with a line source ˜I0/2located at y=hand a line source −˜I0/2located at y=−h. We solve each of these problems by exploiting the appropriate symmetry, and superpose the results toﬁnd the solution to the original problem. For the even-symmetric case, we begin by using (4.407) to represent the impressed ﬁeld: ˜E i z(x,y,ω)=−ω˜µ˜I0(ω) 2 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1e−jky|y−h|+e−jky|y+h| 2kye−jkxxdkx. Fory>hthis becomes ˜Ei z(x,y,ω)=−ω˜µ˜I0(ω) 2 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta12 cos kyh 2kye−jkyye−jkxxdkx. The secondary (scattered) ﬁeld consists of waves propagating in both the ±y-directions: ˜Es z(x,y,ω)=1 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1/bracketleftbig A+(kx,ω)e−jkyy+A−(kx,ω)ejkyy/bracketrightbig e−jkxxdkx. (5.7) The impressed ﬁeld is even about y=0. Since the total ﬁeld Ez=Ei z+Es zmust be even in y(Ezis parallel to the plane y=0), the scattered ﬁeld must also be even. Thus, A+=A−and the total ﬁeld is for y>h ˜Ez(x,y,ω)=1 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1/bracketleftbigg 2A+(kx,ω)coskyy−ω˜µ˜I0(ω) 22 cos kyh 2kye−jkyy/bracketrightbigg e−jkxxdkx. Nowthe electric ﬁeld must obey the boundary condition ˜Ez=0aty=± d. However, since ˜Ezis even the satisfaction of this condition at y=dautomatically implies its satisfaction at y=−d. So we set 1 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1/bracketleftbigg 2A+(kx,ω)coskyd−ω˜µ˜I0(ω) 22 cos kyh 2kye−jkyd/bracketrightbigg e−jkxxdkx=0 and invoke the Fourier integral theorem to get A+(kx,ω)=ω˜µ˜I0(ω) 2coskyh 2kye−jkyd coskyd. The total ﬁeld for this case is ˜Ez(x,y,ω)=−ω˜µ˜I0(ω) 2 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1/bracketleftbigge−jky|y−h|+e−jky|y+h| 2ky− −2 cos kyh 2kye−jkyd coskydcoskyy/bracketrightbigg e−jkxxdkx. For the odd-symmetric case the impressed ﬁeld is ˜Ei z(x,y,ω)=−ω˜µ˜I0(ω) 2 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1e−jky|y−h|−e−jky|y+h| 2kye−jkxxdkx, which for y>his ˜Ei z(x,y,ω)=−ω˜µ˜I0(ω) 2 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta12jsinkyh 2kye−jkyye−jkxxdkx. The scattered ﬁeld has the form of (5.7) but must be odd. Thus A+=− A−and the total ﬁeld for y>his ˜Ez(x,y,ω)=1 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1/bracketleftbigg 2jA+(kx,ω)sinkyy−ω˜µ˜I0(ω) 22jsinkyh 2kye−jkyy/bracketrightbigg e−jkxxdkx. Setting ˜Ez=0atz=dand solving for A+we ﬁnd that the total ﬁeld for this case is ˜Ez(x,y,ω)=−ω˜µ˜I0(ω) 2 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1/bracketleftbigge−jky|y−h|−e−jky|y+h| 2ky− −2jsinkyh 2kye−jkyd sinkydsinkyy/bracketrightbigg e−jkxxdkx. Adding the ﬁelds for the two cases we ﬁnd that ˜Ez(x,y,ω)=−ω˜µ˜I0(ω) 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1e−jky|y−h| 2kye−jkxxdkx+ +ω˜µ˜I0(ω) 2π∞+ j/Delta1/integraldisplay −∞+ j/Delta1/bracketleftbiggcoskyhcoskyy coskyd+jsinkyhsinkyy sinkyd/bracketrightbigge−jkyd 2kye−jkxxdkx, (5.8) which is a superposition of impressed and scattered ﬁelds. 5.2 Solenoidal–lamellar decomposition We nowdiscuss the decomposition of a general vector ﬁeld into a lamellar component having zero curl and a solenoidal component having zero divergence. This is known as a Helmholtz decomposition .I fVis any vector ﬁeld then we wish to write V=Vs+Vl, (5.9) where Vsand Vlare the solenoidal and lamellar components of V. Formulas expressing these components in terms of Vare obtained as follows. We ﬁrst write Vsin terms of a “vector potential” Aas Vs=∇× A. (5.10) This is possible by virtue of (B.49). Similarly, we write Vlin terms of a “scalar potential” φas Vl=∇φ. (5.11) To obtain a formula for Vlwe take the divergence of (5.9) and use (5.11) to get ∇·V=∇· Vl=∇·∇ φ=∇2φ. The result, ∇2φ=∇· V, may be regarded as Poisson’s equation for the unknown φ. This equation is solved in Chapter 3. By (3.61) we have φ(r)=−/integraldisplay V∇/prime·V(r/prime) 4πRdV/prime, where R=|r−r/prime|, and we have Vl(r)=− ∇/integraldisplay V∇/prime·V(r/prime) 4πRdV/prime. (5.12) Similarly, a formula for Vscan be obtained by taking the curl of (5.9) to get ∇× V=∇× Vs. Substituting (5.10) we have ∇× V=∇× (∇× A)=∇(∇·A)−∇2A. We may choose any value we wish for ∇·A, since this does not alter Vs=∇× A. (We discuss such “gauge transformations” in greater detail later in this chapter.) With∇·A=0we obtain −∇ × V=∇ 2A. This is Poisson’s equation for each rectangular component of A; therefore A(r)=/integraldisplay V∇/prime×V(r/prime) 4πRdV/prime, and we have Vs(r)=∇×/integraldisplay V∇/prime×V(r/prime) 4πRdV/prime. Summing the results we obtain the Helmholtz decomposition V=Vl+Vs=− ∇/integraldisplay V∇/prime·V(r/prime) 4πRdV/prime+∇×/integraldisplay V∇/prime×V(r/prime) 4πRdV/prime. (5.13) Identiﬁcation of the electromagnetic potentials. Let us write the electromagnetic ﬁelds as a general superposition of solenoidal and lamellar components: E=∇× AE+∇φE, (5.14) B=∇× AB+∇φB. (5.15) One possible form of the potentials AE,AB,φE, and φBappears in (5.13). However, because Eand Bare related by Maxwell’s equations, the potentials should be related to the sources. We can determine the explicit relationship by substituting (5.14) and (5.15) into Ampere’s and Faraday’s laws. It is most convenient to analyze the relationships using superposition of the cases for which Jm=0and J=0. With Jm=0Faraday’s lawis ∇× E=−∂B ∂t. (5.16) Since ∇× Eis solenoidal, Bmust be solenoidal and thus ∇φB=0. This implies thatφB=0, which is equivalent to the auxiliary Maxwell equation ∇·B=0.N o w, substitution of (5.14) and (5.15) into (5.16) gives ∇× [∇× AE+∇φE]=−∂ ∂t[∇× AB]. Using ∇×(∇φE)=0and combining the terms we get ∇×/bracketleftbigg ∇× AE+∂AB ∂t/bracketrightbigg =0, hence ∇× AE=−∂AB ∂t+∇ξ. Substitution into (5.14) gives E=−∂AB ∂t+[∇φE+∇ξ]. Combining the two gradient functions together, we see that we can write both Eand B in terms of two potentials: E=−∂Ae ∂t−∇φe, (5.17) B=∇× Ae, (5.18) where the negative sign on the gradient term is introduced by convention. Gauge transformations and the Coulomb gauge. We pay a price for the simplicity of using only two potentials to represent Eand B. While ∇× Aeis deﬁnitely solenoidal, Aeitself may not be: because of this (5.17) may not be a decomposition into solenoidal and lamellar components. However, a corollary of the Helmholtz theorem states that avector ﬁeld is uniquely speciﬁed only when bothits curl and divergence are speciﬁed. Here there is an ambiguity in the representation of Eand B; we may remove this ambiguity and deﬁne A euniquely by requiring that ∇·Ae=0. (5.19) Then Aeis solenoidal and the decomposition (5.17) is solenoidal–lamellar. This require- ment on Aeis called the Coulomb gauge . The ambiguity implied by the non-uniqueness of ∇·Aecan also be expressed by the observation that a transformation of the type Ae→Ae+∇/Gamma1, (5.20) φe→φe−∂/Gamma1 ∂t, (5.21) leaves the expressions (5.17) and (5.18) unchanged. This is called a gauge transformation , and the choice of a certain /Gamma1alters the speciﬁcation of ∇·Ae. Thus we may begin with the Coulomb gauge as our baseline, and allowany alteration of Aeaccording to (5.20) as long as we augment ∇·Aeby∇·∇/Gamma1=∇2/Gamma1. Once ∇·Aeis speciﬁed, the relationship between the potentials and the current J can be found by substitution of (5.17) and (5.18) into Ampere’s law. At this pointwe assume media that are linear, homogeneous, isotropic, and described by the time-invariant parameters µ,/epsilon1, and σ. Writing J=J i+σEwe have 1 µ∇×(∇× Ae)=Ji−σ∂Ae ∂t−σ∇φe−/epsilon1∂2Ae ∂t2−/epsilon1∂ ∂t∇φe. (5.22) Taking the divergence of both sides of (5.22) we get 0=∇· Ji−σ∂ ∂t∇·A−σ∇·∇φe−/epsilon1∂2 ∂t2∇·Ae−/epsilon1∂ ∂t∇·∇φe. (5.23) Then, by substitution from the continuity equation and use of (5.19) along with ∇·∇φe= ∇2φewe obtain ∂ ∂t/parenleftbig ρi+/epsilon1∇2φe/parenrightbig =−σ∇2φe. For a lossless medium this reduces to ∇2φe=−ρi//epsilon1 (5.24) and we have φe(r,t)=/integraldisplay Vρi(r/prime,t) 4π/epsilon1RdV/prime. (5.25) We can obtain an equation for Aeby expanding the left-hand side of (5.22) to get ∇(∇·Ae)−∇2Ae=µJi−σµ∂Ae ∂t−σµ∇φe−µ/epsilon1∂2Ae ∂t2−µ/epsilon1∂ ∂t∇φe, (5.26) hence ∇2Ae−µ/epsilon1∂2Ae ∂t2=−µJi+σµ∂Ae ∂t+σµ∇φe+µ/epsilon1∂ ∂t∇φe under the Coulomb gauge. For lossless media this becomes ∇2Ae−µ/epsilon1∂2Ae ∂t2=−µJi+µ/epsilon1∂ ∂t∇φe. (5.27) Observe that the left-hand side of (5.27) is solenoidal (since the Laplacian term came from the curl-curl, and ∇·Ae=0), while the right-hand side contains a general vector ﬁeld Jiand a lamellar term. We might expect the ∇φeterm to cancel the lamellar portion of Ji, and this does happen [91]. By (5.12) and the continuity equation we can write the lamellar component of the current as Ji l(r,t)=− ∇/integraldisplay V∇/prime·Ji(r/prime,t) 4πRdV/prime=∂ ∂t∇/integraldisplay Vρi(r/prime,t) 4πRdV/prime=/epsilon1∂ ∂t∇φe. Thus (5.27) becomes ∇2Ae−µ/epsilon1∂2Ae ∂t2=−µJi s. (5.28) Therefore the vector potential Ae, which describes the solenoidal portion of both Eand B, is found from just the solenoidal portion of the current. On the other hand, the scalar potential, which describes the lamellar portion of E, is found from ρiwhich arises from ∇·Ji, the lamellar portion of the current. From the perspective of ﬁeld computation, we see that the introduction of potential functions has reoriented the solution process from dealing with two coupled ﬁrst-orderpartial diﬀerential equations (Maxwell’s equations), to two uncoupled second-order equa-tions (the potential equations (5.24) and (5.28)). The decoupling of the equations is oftenworth the added complexity of dealing with potentials, and, in fact, is the solution tech-nique of choice in such areas as radiation and guided waves. It is worth pausing fora moment to examine the form of these equations. We see that the scalar potentialobeys Poisson’s equation with the solution (5.25), while the vector potential obeys thewave equation. As a wave, the vector potential must propagate away from the sourcewith ﬁnite velocity. However, the solution for the scalar potential (5.25) shows no suchbehavior. In fact, any change to the charge distribution instantaneously permeates all of space. This apparent violation of Einstein’s postulate shows that we must be carefulwhen interpreting the physical meaning of the potentials. Once the computations (5.17)and (5.18) are undertaken, we ﬁnd that both Eand Bbehave as waves, and thus propa- gate at ﬁnite velocity. Mathematically, the conundrum can be resolved by realizing thatindividually the solenoidal and lamellar components of current must occupy all of space,even if their sum, the actual current J i, is localized [91]. The Lorentz gauge. A diﬀerent choice of gauge condition can allowboth the vector and scalar potentials to act as waves. In this case Emay be written as a sum of two terms: one purely solenoidal, and the other a superposition of lamellar and solenoidalparts. Let us examine the eﬀect of choosing the Lorentz gauge ∇·A e=−µ/epsilon1∂φe ∂t−µσφ e. (5.29) Substituting this expression into (5.26) we ﬁnd that the gradient terms cancel, giving ∇2Ae−µσ∂Ae ∂t−µ/epsilon1∂2Ae ∂t2=−µJi. (5.30) For lossless media ∇2Ae−µ/epsilon1∂2Ae ∂t2=−µJi, (5.31) and (5.23) becomes ∇2φe−µ/epsilon1∂2φe ∂t2=−ρi /epsilon1. (5.32) For lossy media we have obtained a second-order diﬀerential equation for Ae, but φe must be found through the somewhat cumbersome relation (5.29). For lossless media the coupled Maxwell equations have been decoupled into two second-order equations, oneinvolving A eand one involving φe. Both (5.31) and (5.32) are wave equations, with Ji as the source for Aeandρias the source for φe. Thus the expected ﬁnite-velocity wave nature of the electromagnetic ﬁelds is also manifested in each of the potential functions.The drawback is that, even though we can still use (5.17) and (5.18), the expression for E is no longer a decomposition into solenoidal and lamellar components. Nevertheless, thechoice of the Lorentz gauge is very popular in the study of radiated and guided waves. The Hertzian potentials. With a little manipulation and the introduction of a new notation, we can maintain the wave nature of the potential functions and still provide adecomposition into purely lamellar and solenoidal components. In this analysis we shallassume lossless media only. When we chose the Lorentz gauge to remove the arbitrariness of the divergence of the vector potential, we established a relationship between A eandφe. Thus we should be able to write both the electric and magnetic ﬁelds in terms of a single potential function.From the Lorentz gauge we can write φ eas φe(r,t)=−1 µ/epsilon1/integraldisplayt −∞∇·Ae(r,t)dt. By (5.17) and (5.18) we can thus write the EM ﬁelds as E=1 µ/epsilon1∇/integraldisplayt −∞∇·Aedt−∂Ae ∂t, (5.33) B=∇× Ae. (5.34) The integro-diﬀerential representation of Ein (5.33) is somewhat clumsy in appear- ance. We can make it easier to manipulate by deﬁning the Hertzian potential Πe=1 µ/epsilon1/integraldisplayt −∞Aedt. In diﬀerential form Ae=µ/epsilon1∂Πe dt. (5.35) With this, (5.33) and (5.34) become E=∇(∇·Πe)−µ/epsilon1∂2 ∂t2Πe, (5.36) B=µ/epsilon1∇×∂Πe ∂t. (5.37) An equation for Πein terms of the source current can be found by substituting (5.35) into (5.31): µ/epsilon1∂ ∂t/parenleftbigg ∇2Πe−µ/epsilon1∂2 ∂t2Πe/parenrightbigg =−µJi. Let us deﬁne Ji=∂Pi ∂t. (5.38) For general impressed current sources (5.38) is just a convenient notation. However, we can conceive of an impressed polarization current that is independent of Eand deﬁned through the relation D=/epsilon10E+P+Pi. Then (5.38) has a physical interpretation as described in (2.119). We nowhave ∇2Πe−µ/epsilon1∂2 ∂t2Πe=−1 /epsilon1Pi, (5.39) which is a wave equation for Πe. Thus the Hertzian potential has the same wave behavior as the vector potential under the Lorentz gauge. We can use (5.39) to perform one ﬁnal simpliﬁcation of the EM ﬁeld representation. By the vector identity ∇(∇·Π)=∇× (∇×Π)+∇2Πwe get ∇(∇·Πe)=∇× (∇×Πe)−1 /epsilon1Pi+µ/epsilon1∂2 ∂t2Πe. Substituting this into (5.36) we obtain E=∇× (∇×Πe)−Pi /epsilon1, (5.40) B=µ/epsilon1∇×∂Πe ∂t. (5.41) Let us examine these closely. We knowthat Bis solenoidal since it is written as the curl of another vector (this is also clear from the auxiliary Maxwell equation ∇·B=0). The ﬁrst term in the expression for Eis also solenoidal. So the lamellar part of Emust be contained within the source term Pi. If we write Piin terms of its lamellar and solenoidal components by using Ji s=∂Pi s ∂t, Ji l=∂Pi l ∂t, then (5.40) becomes E=/bracketleftbigg ∇×(∇×Πe)−Pi s /epsilon1/bracketrightbigg −Pi l /epsilon1. (5.42) So we have again succeeded in dividing Einto lamellar and solenoidal components. Potential functions for magnetic current. We can proceed as above to derive the ﬁeld–potential relationships when Ji=0butJi m/negationslash=0. We assume a homogeneous, loss- less, isotropic medium with permeability µand permittivity /epsilon1, and begin with Faraday’s and Ampere’s laws ∇× E=−Ji m−∂B ∂t, (5.43) ∇× H=∂D ∂t. (5.44) We write Hand Din terms of two potential functions Ahandφhas H=−∂Ah ∂t−∇φh, D=− ∇× Ah, and the diﬀerential equation for the potentials is found by substitution into (5.43): ∇×(∇× Ah)=/epsilon1Ji m−µ/epsilon1∂2Ah ∂t2−µ/epsilon1∂ ∂t∇φh. (5.45) Taking the divergence of this equation and substituting from the magnetic continuity equation we obtain µ/epsilon1∂2 ∂t2∇·Ah+µ/epsilon1∂ ∂t∇2φh=−/epsilon1∂ρi m ∂t. Under the Lorentz gauge condition ∇·Ah=−µ/epsilon1∂φh ∂t this reduces to ∇2φh−µ/epsilon1∂2φh ∂t2=−ρi m µ. Expanding the curl-curl operation in (5.45) we have ∇(∇·Ah)−∇2Ah=/epsilon1Ji m−µ/epsilon1∂2Ah ∂t2−µ/epsilon1∂ ∂t∇φh, which, upon substitution of the Lorentz gauge condition gives ∇2Ah−µ/epsilon1∂2Ah ∂t2=−/epsilon1Ji m. (5.46) We can also derive a Hertzian potential for the case of magnetic current. Letting Ah=µ/epsilon1∂Πh ∂t(5.47) and employing the Lorentz condition we have D=−µ/epsilon1∇×∂Πh ∂t, H=∇(∇·Πh)−µ/epsilon1∂2Πh ∂t2. The wave equation for Πhis found by substituting (5.47) into (5.46) to give ∂ ∂t/bracketleftbigg ∇2Πh−µ/epsilon1∂2Πh ∂t2/bracketrightbigg =−1 µJi m. (5.48) Deﬁning Mithrough Ji m=µ∂Mi ∂t, we write the wave equation as ∇2Πh−µ/epsilon1∂2Πh ∂t2=−Mi. We can think of Mias a convenient way of representing Ji m, or we can conceive of an impressed magnetization current that is independent of Hand deﬁned through B= µ0(H+M+Mi). With the help of (5.48) we can also write the ﬁelds as H=∇× (∇×Πh)−Mi, D=−µ/epsilon1∇×∂Πh ∂t. Summary of potential relations for lossless media. When both electric and mag- netic sources are present, we may superpose the potential representations derived above.We assume a homogeneous, lossless medium with time-invariant parameters µand/epsilon1.F o r the scalar/vector potential representation we have E=−∂A e ∂t−∇φe−1 /epsilon1∇× Ah, (5.49) H=1 µ∇× Ae−∂Ah ∂t−∇φh. (5.50) Here the potentials satisfy the wave equations /parenleftbigg ∇2−µ/epsilon1∂2 ∂t2/parenrightbigg/braceleftbiggAe φe/bracerightbigg =/braceleftbigg−µJi −ρi /epsilon1/bracerightbigg , (5.51) /parenleftbigg ∇2−µ/epsilon1∂2 ∂t2/parenrightbigg/braceleftbiggAh φh/bracerightbigg =/braceleftBigg −/epsilon1Ji m −ρi m µ/bracerightBigg , and are linked by the Lorentz conditions ∇·Ae=−µ/epsilon1∂φe ∂t, ∇·Ah=−µ/epsilon1∂φh ∂t. We also have the Hertz potential representation E=∇(∇·Πe)−µ/epsilon1∂2Πe ∂t2−µ∇×∂Πh ∂t =∇× (∇×Πe)−Pi /epsilon1−µ∇×∂Πh ∂t, (5.52) H=/epsilon1∇×∂Πe ∂t+∇(∇·Πh)−µ/epsilon1∂2Πh ∂t2 =/epsilon1∇×∂Πe ∂t+∇× (∇×Πh)−Mi. (5.53) The Hertz potentials satisfy the wave equations /parenleftbigg ∇2−µ/epsilon1∂2 ∂t2/parenrightbigg/braceleftbiggΠe Πh/bracerightbigg =/braceleftbigg −1 /epsilon1Pi −Mi/bracerightbigg . Potential functions for the frequency-domain ﬁelds. In the frequency domain it is much easier to handle lossy media. Consider a lossy, isotropic, homogeneous mediumdescribed by the frequency-dependent parameters ˜µ,˜/epsilon1, and ˜σ. Maxwell’s curl equations are ∇× ˜E=− ˜J i m−jω˜µ˜H, (5.54) ∇× ˜H=˜Ji+jω˜/epsilon1c˜E. (5.55) Here we have separated the primary and secondary currents through ˜J=˜Ji+˜σ˜E, and used the complex permittivity ˜/epsilon1c=˜/epsilon1+˜σ/jω. As with the time-domain equations we introduce the potential functions using superposition. If ˜Ji m=0and ˜Ji/negationslash=0then we may introduce the electric potentials through the relationships ˜E=− ∇ ˜φe−jω˜Ae, (5.56) ˜H=1 ˜µ∇× ˜Ae. (5.57) Assuming the Lorentz condition ∇·˜Ae=− jω˜µ˜/epsilon1c˜φe, we ﬁnd that upon substitution of (5.56)–(5.57) into (5.54)–(5.55) the potentials must obey the Helmholtz equation /parenleftbig ∇2+k2/parenrightbig/braceleftbigg˜φe ˜Ae/bracerightbigg =/braceleftbigg−˜ρi/˜/epsilon1c −˜µ˜Ji/bracerightbigg . If˜Ji m/negationslash=0and ˜Ji=0then we may introduce the magnetic potentials through ˜E=−1 ˜/epsilon1c∇× ˜Ah, (5.58) ˜H=− ∇ ˜φh−jω˜Ah. (5.59) Assuming ∇·˜Ah=− jω˜µ˜/epsilon1c˜φh, we ﬁnd that upon substitution of (5.58)–(5.59) into (5.54)–(5.55) the potentials must obey /parenleftbig ∇2+k2/parenrightbig/braceleftbigg˜φh ˜Ah/bracerightbigg =/braceleftbigg−˜ρi m/˜µ −˜/epsilon1c˜Ji m/bracerightbigg . When both electric and magnetic sources are present, we use superposition: ˜E=− ∇ ˜φe−jω˜Ae−1 ˜/epsilon1c∇× ˜Ah, ˜H=1 ˜µ∇× ˜Ae−∇ ˜φh−jω˜Ah. Using the Lorentz conditions we can also write the ﬁelds in terms of the vector potentials alone: ˜E=−jω k2∇(∇·˜Ae)−jω˜Ae−1 ˜/epsilon1c∇× ˜Ah, (5.60) ˜H=1 ˜µ∇× ˜Ae−jω k2∇(∇·˜Ah)−jω˜Ah. (5.61) We can also deﬁne Hertzian potentials for the frequency-domain ﬁelds. When ˜Ji m=0 and ˜Ji/negationslash=0we let ˜Ae=jω˜µ˜/epsilon1c˜Πe and ﬁnd ˜E=∇(∇·˜Πe)+k2˜Πe=∇× (∇× ˜Πe)−˜Ji jω˜/epsilon1c(5.62) and ˜H=jω˜/epsilon1c∇× ˜Πe. (5.63) Here ˜Jican represent either an impressed electric current source or an impressed polar- ization current source ˜Ji=jω˜Pi. The electric Hertzian potential obeys (∇2+k2)˜Πe=−˜Ji jω˜/epsilon1c. (5.64) When ˜Ji m/negationslash=0and ˜Ji=0we let ˜Ah=jω˜µ˜/epsilon1c˜Πh and ﬁnd ˜E=− jω˜µ∇× ˜Πh (5.65) and ˜H=∇(∇·˜Πh)+k2˜Πh=∇× (∇× ˜Πh)−˜Ji m jω˜µ. (5.66) Here ˜Ji mcan represent either an impressed magnetic current source or an impressed magnetization current source ˜Ji m=jω˜µ˜Mi. The magnetic Hertzian potential obeys (∇2+k2)˜Πh=−˜Ji m jω˜µ. (5.67) When both electric and magnetic sources are present we have by superposition ˜E=∇(∇·˜Πe)+k2˜Πe−jω˜µ∇× ˜Πh =∇× (∇× ˜Πe)−˜Ji jω˜/epsilon1c−jω˜µ∇× ˜Πh and ˜H=jω˜/epsilon1c∇× ˜Πe+∇(∇·˜Πh)+k2˜Πh =jω˜/epsilon1c∇× ˜Πe+∇× (∇× ˜Πh)−˜Ji m jω˜µ. 5.2.1 Solution for potentials in an unbounded medium: the retarded potentials Under the Lorentz condition each of the potential functions obeys the wave equation. This equation can be solved using the method of Green’s functions to determine thepotentials, and the electromagnetic ﬁelds can therefore be determined. We nowexaminethe solution for an unbounded medium. Solutions for bounded regions are considered in§5.2.2. Consider a linear operator Lthat operates on a function of rand t. If we wish to solve the equation L{ψ(r,t)}=S(r,t), (5.68) we ﬁrst solve L{G(r,t|r /prime,t/prime)}=δ(r−r/prime)δ(t−t/prime) and determine the Green’s function Gfor the operator L. Provided that Sresides within Vweh a v e L/braceleftbigg/integraldisplay V/integraldisplay∞ −∞S(r/prime,t/prime)G(r,t|r/prime,t/prime)dt/primedV/prime/bracerightbigg =/integraldisplay V/integraldisplay∞ −∞S(r/prime,t/prime)L{G(r,t|r/prime,t/prime)}dt/primedV/prime =/integraldisplay V/integraldisplay∞ −∞S(r/prime,t/prime)δ(r−r/prime)δ(t−t/prime)dt/primedV/prime =S(r,t), hence ψ(r,t)=/integraldisplay V/integraldisplay∞ −∞S(r/prime,t/prime)G(r,t|r/prime,t/prime)dt/primedV/prime(5.69) by comparison with (5.68). We can also apply this idea in the frequency domain. The solution to L{˜ψ(r,ω)}=˜S(r,ω) (5.70) is ˜ψ(r,ω)=/integraldisplay V˜S(r/prime,ω)G(r|r/prime;ω)dV/prime where the Green’s function Gsatisﬁes L{G(r|r/prime;ω)}=δ(r−r/prime). Equation (5.69) is the basic superposition integral that allows us to ﬁnd the potentials in an inﬁnite, unbounded medium. We note that if the medium is bounded then we mustuse Green’s theorem to include the eﬀects of sources that reside external to the bound-aries. These are manifested in terms of the values of the potentials on the boundariesin the same manner as with the static potentials in Chapter 3. In order to determinewhether (5.69) is the unique solution to the wave equation, we must also examine thebehavior of the ﬁelds on the boundary as the boundary recedes to inﬁnity. In the fre-quency domain we ﬁnd that an additional “radiation condition” is required to ensureuniqueness. The retarded potentials in the time domain. Consider an unbounded, homoge- neous, lossy, isotropic medium described by parameters µ,/epsilon1,σ . In the time domain the vector potential A esatisﬁes (5.30). The scalar components of Aemust obey ∇2Ae,n(r,t)−µσ∂Ae,n(r,t) ∂t−µ/epsilon1∂2Ae,n(r,t) ∂t2=−µJi n(r,t), n=x,y,z. We may write this in the form /parenleftbigg ∇2−2/Omega1 v2∂ ∂t−1 v2∂2 ∂t2/parenrightbigg ψ(r,t)=− S(r,t) (5.71) where ψ=Ae,n,v2=1/µ/epsilon1,/Omega1=σ/2/epsilon1, and S=µJi n. The solution is ψ(r,t)=/integraldisplay V/integraldisplay∞ −∞S(r/prime,t/prime)G(r,t|r/prime,t/prime)dt/primedV/prime(5.72) where Gsatisﬁes /parenleftbigg ∇2−2/Omega1 v2∂ ∂t−1 v2∂2 ∂t2/parenrightbigg G(r,t|r/prime,t/prime)=−δ(r−r/prime)δ(t−t/prime). (5.73) In§A.1 we ﬁnd that G(r,t|r/prime,t/prime)=e−/Omega1(t−t/prime)δ(t−t/prime−R/v) 4πR+ +/Omega12 4πve−/Omega1(t−t/prime)I1/parenleftBig /Omega1/radicalbig (t−t/prime)2−(R/v)2/parenrightBig /Omega1/radicalbig (t−t/prime)2−(R/v)2,t−t/prime>R v, where R=|r−r/prime|. For lossless media where σ=0this becomes G(r,t|r/prime,t/prime)=δ(t−t/prime−R/v) 4πR and thus ψ(r,t)=/integraldisplay V/integraldisplay∞ −∞S(r/prime,t/prime)δ(t−t/prime−R/v) 4πRdt/primedV/prime =/integraldisplay VS(r/prime,t−R/v) 4πRdV/prime. (5.74) For lossless media, the scalar potentials and all rectangular components of the vector potentials obey the same wave equation. Thus we have, for instance, the solutions to(5.51): A e(r,t)=µ 4π/integraldisplay VJi(r/prime,t−R/v) RdV/prime, φe(r,t)=1 4π/epsilon1/integraldisplay Vρi(r/prime,t−R/v) RdV/prime. These are called the retarded potentials since their values at time tare determined by the values of the sources at an earlier (or retardation) time t−R/v. The retardation time is determined by the propagation velocity vof the potential waves. The ﬁelds are determined by the potentials: E(r,t)=− ∇1 4π/epsilon1/integraldisplay Vρi(r/prime,t−R/v) RdV/prime−∂ ∂tµ 4π/integraldisplay VJi(r/prime,t−R/v) RdV/prime, H(r,t)=∇×1 4π/integraldisplay VJi(r/prime,t−R/v) RdV/prime. The derivatives may be brought inside the integrals, but some care must be taken when the observation point rlies within the source region. In this case the integrals must be performed in a principal value sense by excluding a small volume around the observationpoint. We discuss this in more detail belowfor the frequency-domain ﬁelds. For detailsregarding this procedure in the time domain the reader may see Hansen [81]. The retarded potentials in the frequency domain. Consider an unbounded, ho- mogeneous, isotropic medium described by ˜µ(ω)and ˜/epsilon1c(ω).I f ˜ψ(r,ω)represents a scalar potential or any rectangular component of a vector or Hertzian potential then it mustsatisfy (∇ 2+k2)˜ψ(r,ω)=− ˜S(r,ω) (5.75) where k=ω(˜µ˜/epsilon1c)1/2. This Helmholtz equation has the form of (5.70) and thus ˜ψ(r,ω)=/integraldisplay V˜S(r/prime,ω)G(r|r/prime;ω)dV/prime where (∇2+k2)G(r|r/prime;ω)=−δ(r−r/prime). (5.76) This is equation (A.46) and its solution, as given by (A.49), is G(r|r/prime;ω)=e−jkR 4πR. (5.77) Here we use v2=1/˜µ˜/epsilon1and/Omega1=˜σ/2/epsilon1in (A.47): k=1 v/radicalbig ω2−j2ω/Omega1=ω/radicalBigg ˜µ/parenleftbigg ˜/epsilon1−j˜σ ω/parenrightbigg =ω/radicalbig ˜µ˜/epsilon1c. The solution to (5.75) is therefore ˜ψ(r,ω)=/integraldisplay V˜S(r/prime,ω)e−jkR 4πRdV/prime. (5.78) When the medium is lossless, the potential must also satisfy the radiation condition lim r→∞r/parenleftbigg∂ ∂r+jk/parenrightbigg ˜ψ(r)=0 (5.79) to guarantee uniqueness of solution. In §5.2.2 w e shall showhowthis requirement arises from the solution within a bounded region. For a uniqueness proof for the Helmholtzequation, the reader may consult Chew[33]. We may use (5.78) to ﬁnd that ˜A e(r,ω)=˜µ 4π/integraldisplay V˜Ji(r/prime,ω)e−jkR RdV/prime. (5.80) Comparison with (5.74) shows that in the frequency domain, time retardation takes the form of a phase shift. Similarly, ˜φ(r,ω)=1 4π˜/epsilon1c/integraldisplay V˜ρi(r/prime,ω)e−jkR RdV/prime. (5.81) The electric and magnetic dyadic Green’s functions. The frequency-domain elec- tromagnetic ﬁelds may be found for electric sources from the electric vector potentialusing (5.60) and (5.61): ˜E(r,ω)=− jω˜µ(ω)/integraldisplay V˜Ji(r/prime,ω)G(r|r/prime;ω)dV/prime−jω˜µ(ω) k2∇∇ ·/integraldisplay V˜Ji(r/prime,ω)G(r|r/prime;ω)dV/prime, ˜H=∇×/integraldisplay V˜Ji(r/prime,ω)G(r|r/prime;ω)dV/prime. (5.82) As long as the observation point rdoes not lie within the source region we may take the derivatives inside the integrals. Using ∇·/bracketleftbig˜Ji(r/prime,ω)G(r|r/prime;ω)/bracketrightbig =˜Ji(r/prime,ω)·∇G(r|r/prime;ω)+G(r|r/prime;ω)∇·˜J(r/prime,ω) =∇G(r|r/prime;ω)·˜Ji(r/prime,ω) we have ˜E(r,ω)=− jω˜µ(ω)/integraldisplay V/braceleftbigg ˜Ji(r/prime,ω)G(r|r/prime;ω)+1 k2∇/bracketleftbig ∇G(r|r/prime;ω)·Ji(r/prime,ω)/bracketrightbig/bracerightbigg dV/prime. This can be written more compactly as ˜E(r,ω)=− jω˜µ(ω)/integraldisplay V¯Ge(r|r/prime;ω)·˜Ji(r/prime,ω)dV/prime where ¯Ge(r|r/prime;ω)=/bracketleftbigg ¯I+∇∇ k2/bracketrightbigg G(r|r/prime;ω) (5.83) is called the electric dyadic Green’s function . Using ∇× [˜JiG]=∇G×˜Ji+G∇× ˜Ji=∇G×˜Ji we have for the magnetic ﬁeld ˜H(r,ω)=/integraldisplay V∇G(r|r/prime;ω)×˜Ji(r/prime,ω)dV/prime. Now, using the dyadic identity (B.15) we may show that ˜Ji×∇G=(˜Ji×∇G)·¯I=(∇G×¯I)·Ji. So ˜H(r,ω)=−/integraldisplay V¯Gm(r|r/prime;ω)·˜Ji(r/prime,ω)dV/prime where ¯Gm(r|r/prime;ω)=∇G(r|r/prime;ω)×¯I (5.84) is called the magnetic dyadic Green’s function . Proceeding similarly for magnetic sources (or using duality) we have ˜H(r)=− jω˜/epsilon1c/integraldisplay V¯Ge(r|r/prime;ω)·˜Ji m(r/prime,ω)dV/prime, ˜E(r)=/integraldisplay V¯Gm(r|r/prime;ω)·˜Ji m(r/prime,ω)dV/prime. When both electric and magnetic sources are present we simply use superposition and add the ﬁelds. When the observation point lies within the source region, we must be much more careful about how we formulate the dyadic Green’s functions. In (5.82) we encounter theintegral /integraldisplay V˜Ji(r/prime,ω)G(r|r/prime;ω)dV/prime. Figure 5.1: Geometry of excluded region used to compute the electric ﬁeld within a source region. Ifrlies within the source region then Gis singular since R→0when r→r/prime. However, the integral converges and the potentials exist within the source region. While we runinto trouble when we pass both derivatives in the operator ∇∇·through the integral and allowthem to operate on G, since diﬀerentiation of Gincreases the order of the singularity, we may safely take one derivative of G. Even when we allow one derivative on Gwe must be careful in how we compute the integral. We exclude the point rby surrounding it with a small volume element V δas showninFigure5.1andwrite ∇∇ ·/integraldisplay V˜Ji(r/prime,ω)G(r|r/prime;ω)dV/prime= lim Vδ→0/integraldisplay V−Vδ∇/bracketleftbig ∇G(r|r/prime;ω)·˜Ji(r/prime,ω)/bracketrightbig dV/prime+lim Vδ→0∇/integraldisplay Vδ∇G(r|r/prime;ω)·˜Ji(r/prime,ω)dV/prime. The ﬁrst integral on the right-hand side is called the principal value integral and is usually abbreviated P.V./integraldisplay V∇/bracketleftbig ∇G(r|r/prime;ω)·˜Ji(r/prime,ω)/bracketrightbig dV/prime. It converges to a value dependent on the shape of the excluded region Vδ,a sd o e st h e second integral. However, the sum of these two integrals produces a unique result. Using∇G=− ∇ /primeG, the identity ∇/prime·(˜JG)=˜J·∇/primeG+G∇/prime·˜J, and the divergence theorem, we can write −/integraldisplay Vδ∇/primeG(r|r/prime;ω)·˜Ji(r/prime,ω)dV/prime= −/contintegraldisplay SδG(r|r/prime;ω)˜Ji(r/prime,ω)·ˆn/primedS/prime+/integraldisplay VδG(r|r/prime;ω)∇/prime·˜Ji(r/prime,ω)dV/prime where Sδis the surface surrounding Vδ. By the continuity equation the second integral on the right-hand side is proportional to the scalar potential produced by the chargewithin V δ, and thus vanishes as Vδ→0. The ﬁrst term is proportional to the ﬁeld at r produced by surface charge on Sδ, which results in a value proportional to Ji.T h u s lim Vδ→0∇/integraldisplay Vδ∇G(r|r/prime;ω)·˜Ji(r/prime,ω)dV/prime=− lim Vδ→0∇/contintegraldisplay SδG(r|r/prime;ω)˜Ji(r/prime,ω)·ˆn/primedS/prime =− ¯L·˜Ji(r,ω) , (5.85) so ∇∇ ·/integraldisplay V˜Ji(r/prime,ω)G(r|r/prime;ω)dV/prime=P.V./integraldisplay V∇/bracketleftbig ∇G(r|r/prime;ω)·˜Ji(r/prime,ω)/bracketrightbig dV/prime−¯L·˜Ji(r,ω) . Here ¯Lis usually called the depolarizing dyadic [113]. Its value depends on the shape of Vδ, as considered below. We may noww rite ˜E(r,ω)=− jω˜µ(ω) P.V./integraldisplay V¯Ge(r|r/prime;ω)·˜J(r/prime,ω)dV/prime−1 jω˜/epsilon1c(ω)¯L·˜Ji(r,ω) . (5.86) We may also incorporate both terms into a single dyadic Green’s function using the notation ¯G(r|r/prime;ω)=P.V.¯Ge(r|r/prime;ω)−1 k2¯Lδ(r−r/prime). Hence when we compute ˜E(r,ω)=− jω˜µ(ω)/integraldisplay V¯G(r|r/prime;ω)·˜Ji(r/prime,ω)dV/prime =− jω˜µ(ω)/integraldisplay V/bracketleftbigg P.V.¯Ge(r|r/prime;ω)−1 k2¯Lδ(r−r/prime)/bracketrightbigg ·˜Ji(r/prime,ω)dV/prime we reproduce (5.86). That is, the symbol P.V.onGeindicates that a principal value integral must be performed. Our ﬁnal task is to compute ¯Lfrom (5.85). When we remove the excluded region from the principal value computation we leave behind a hole in the source region. Thecontribution to the ﬁeld at rby the sources in the excluded region is found from the scalar potential produced by the surface distribution ˆn·J i. The value of this correction termdepends on the shape of the excluding volume. However, the correction term always adds to the principal value integral to give the true ﬁeld at r, regardless of the shape of the volume. So we must always match the shape of the excluded region used to computethe principal value integral with that used to compute the correction term so that thetrue ﬁeld is obtained. Note that as V δ→0the phase factor in the Green’s function becomes insigniﬁcant, and the values of the current on the surface approach the value atr(assuming J iis continuous at r).Thus we may write lim Vδ→0∇/contintegraldisplay Sδ˜Ji(r,ω)·ˆn/prime 4π|r−r/prime|dS/prime=¯L·˜Ji(r,ω) . This has the form of a static ﬁeld integral. For a spherical excluded region we may com- pute the above quantity quite simply by assuming the current to be uniform throughout Vδand by aligning the current with the z-axis and placing the center of the sphere at the origin. We then compute the integral at a point rwithin the sphere, take the gradient, and allow r→0. We thus have for a sphere lim Vδ→0∇/contintegraldisplay S˜Jicosθ/prime 4π|r−r/prime|dS/prime=¯L·[ˆz˜Ji(r,ω)]. This integral has been computed in §3.2.7 with the result given by (3.103). Using this we ﬁnd lim Vδ→0/bracketleftbigg ∇/parenleftbigg1 3˜Jiz/parenrightbigg/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle r=0=ˆz˜Ji 3=¯L·[ˆz˜Ji(r,ω)] Figure 5.2: Geometry of an electric Hertzian dipole. and thus ¯L=1 3¯I. We leave it as an exercise to showthat for a cubical excluding volume the depolarizing dyadic is also ¯L=¯I/3. Values for other shapes may be found in Yaghjian [215]. The theory of dyadic Green’s functions is well developed and there exist techniques for their construction under a variety of conditions. For an excellent overviewthe readermay see Tai [192]. Example of ﬁeld calculation using potentials: the Hertzian dipole. Consider a short line current of length l/lessmuchλat position r p, oriented along a direction ˆpin a mediu mwithconstituti veparameters ˜µ(ω), ˜/epsilon1c (ω),asshowninFigure5.2.Weassume that the frequency-domain current ˜I(ω)is independent of position, and therefore this Hertzian dipole must be terminated by point charges ˜Q(ω)=±˜I(ω) jω as required by the continuity equation. The electric vector potential produced by this short current element is ˜Ae=˜µ 4π/integraldisplay /Gamma1˜Iˆpe−jkR Rdl/prime. At observation points far from the dipole (compared to its length) such that |r−rp|/greatermuchl we may approximate e−jkR R≈e−jk|r−rp| |r−rp|. Then ˜Ae=ˆp˜µ˜IG(r|rp;ω)/integraldisplay /Gamma1dl/prime=ˆp˜µ˜IlG(r|rp;ω). (5.87) Note that we obtain the same answer if we let the current density of the dipole be ˜J=jω˜pδ(r−rp) where ˜pis the dipole moment deﬁned by ˜p=˜Qlˆp=˜Il jωˆp. That is, we consider a Hertzian dipole to be a “point source” of electromagnetic radiation. With this notation we have ˜Ae=˜µ/integraldisplay V/bracketleftbig jω˜pδ(r/prime−rp)/bracketrightbig G(r|r/prime;ω)dV/prime=jω˜µ˜pG(r|rp;ω), which is identical to (5.87). The electromagnetic ﬁelds are then ˜H(r,ω)=jω∇× [˜pG(r|rp;ω)], (5.88) ˜E(r,ω)=1 ˜/epsilon1c∇×∇× [˜pG(r|rp;ω)]. (5.89) Here we have obtained ˜Efrom ˜Houtside the source region by applying Ampere’s law. By duality we may obtain the ﬁelds produced by a magnetic Hertzian dipole of moment ˜pm=˜Iml jωˆp located at r=rpas ˜E(r,ω)=− jω∇× [˜pmG(r|rp;ω)], ˜H(r,ω)=1 ˜µ∇×∇× [˜pmG(r|rp;ω)]. We can learn much about the ﬁelds produced by localized sources by considering the simple case of a Hertzian dipole aligned along the z-axis and centered at the origin. Using ˆp=ˆzand rp=0in (5.88) we ﬁnd that ˜H(r,ω)=jω∇×/bracketleftbigg ˆz˜I jωle−jkr 4πr/bracketrightbigg =ˆφ1 4π˜Il/bracketleftbigg1 r2+jk r/bracketrightbigg sinθe−jkr. (5.90) By Ampere’s law ˜E(r,ω)=1 jω˜/epsilon1c∇× ˜H(r,ω) =ˆrη 4π˜Il/bracketleftbigg2 r2−j2 kr3/bracketrightbigg cosθe−jkr+ˆθη 4π˜Il/bracketleftbigg jk r+1 r2−j1 kr3/bracketrightbigg sinθe−jkr. (5.91) The ﬁelds involve various inverse powers of r, with the 1/rand 1/r3terms 90◦out-of- phase from the 1/r2term. Some terms dominate the ﬁeld close to the source, while others dominate far away. The terms that dominate near the source1are called the near-zone orinduction-zone ﬁelds : ˜HNZ(r,ω)=ˆφ˜Il 4πe−jkr r2sinθ, ˜ENZ(r,ω)=− jη˜Il 4πe−jkr kr3/bracketleftBig 2ˆrcosθ+ˆθsinθ/bracketrightBig . 1Note that we still require r/greatermuchl. We note that ˜HNZand ˜ENZare90◦out-of-phase. Also, the electric ﬁeld has the same spatial dependence as the ﬁeld of a static electric dipole. The terms that dominate farfrom the source are called the far-zone orradiation ﬁelds : ˜H FZ(r,ω)=ˆφjk˜Il 4πe−jkr rsinθ, (5.92) ˜EFZ(r,ω)=ˆθηjk˜Il 4πe−jkr rsinθ. (5.93) The far-zone ﬁelds are in-phase and in fact form a TEM spherical wave with ˜HFZ=ˆr×˜EFZ η. (5.94) We speak of the time-average power radiated by a time-harmonic source as the integral of the time-average power density over a very large sphere. Thus radiated power is the power delivered by the sources to inﬁnity. If the dipole is situated within a lossy medium,all of the time-average power delivered by the sources is dissipated by the medium. Ifthe medium is lossless then all the time-average power is delivered to inﬁnity. Let uscompute the power radiated by a time-harmonic Hertzian dipole immersed in a losslessmedium. Writing (5.90) and (5.91) in terms of phasors we have the complex Poyntingvector S c(r)=ˇE(r)×ˇH∗(r) =ˆθη/parenleftBigg |ˇI|l 4π/parenrightBigg2 j2 kr5/bracketleftbig k2r2+1/bracketrightbig cosθsinθ+ˆrη/parenleftBigg |ˇI|l 4π/parenrightBigg2k2 r2/bracketleftbigg 1−j1 k3r5/bracketrightbigg sin2θ. We notice that the θ-component of Scis purely imaginary and gives rise to no time- average power ﬂux. This component falls oﬀ as 1/r3for large rand produces no net ﬂux through a sphere with radius r→∞. Additionally, the angular variation sinθcosθ integrates to zero over a sphere. In contrast, the r-component has a real part that varies as1/r2and as sin2θ. Hence we ﬁnd that the total time-average power passing through a sphere expanding to inﬁnity is nonzero: Pav=lim r→∞/integraldisplay2π 0/integraldisplayπ 01 2Re  ˆrη/parenleftBigg |ˇI|l 4π/parenrightBigg2k2 r2sin2θ  ·ˆrr2sinθdθdφ =ηπ 3|ˇI|2/parenleftbiggl λ/parenrightbigg2 (5.95) where λ=2π/kis the wavelength in the lossless medium. This is the power radiated by the Hertzian dipole. The power is proportional to |ˇI|2as it is in a circuit, and thus we may deﬁne a radiation resistance Rr=2Pav |ˇI|2=η2π 3/parenleftbiggl λ/parenrightbigg2 that represents the resistance of a lumped element that would absorb the same power as radiated by the Hertzian dipole when presented with the same current. We also note thatthe power radiated by a Hertzian dipole (and, in fact, by any source of ﬁnite extent) may Figure 5.3: Geometry for solution to the frequency-domain Helmholtz equation. be calculated directly from its far-zone ﬁelds. In fact, from (5.94) we have the simple formula for the time-average power density in lossless media Sav=1 2Re/braceleftbigˇEFZ×ˇHFZ∗/bracerightbig =ˆr1 2|ˇEFZ|2 η. The dipole ﬁeld is the ﬁrst term in a general expansion of the electromagnetic ﬁelds in terms of the multipole moments of the sources. Either a Taylor expansion or a spherical-harmonic expansion may be used. The reader may see Papas [141] for details. 5.2.2 Solution for potential functions in a bounded medium In the previous section we solved for the frequency-domain potential functions in an unbounded region of space. Here we shall extend the solution to a bounded region andidentify the physical meaning of the radiation condition (5.79). Consider a bounded region of space Vcontaining a linear, homogeneous, isotropic mediu mcharacterize dby ˜µ(ω)and ˜/epsilon1 c (ω).AsshowninFigure5.3wedecom posethe multiply-connected boundary into a closed “excluding surface” S0and a closed “encom- passing surface” S∞that we shall allow to expand outward to inﬁnity. S0may consist of more than one closed surface and is often used to exclude unknown sources from V. We wish to solve the Helmholtz equation (5.75) for ˜ψwithin Vin terms of the sources within Vand the values of ˜ψonS0. The actual sources of ˜ψlie entirely with S∞but may lie partly, or entirely, within S0. We solve the Helmholtz equation in much the same way that we solved Poisson’s equation in §3.2.4. We begin with Green’s second identity, written in terms of the source point (primed) variables and applied to the region V: /integraldisplay V[ψ(r/prime,ω)∇/prime2G(r|r/prime;ω)−G(r|r/prime;ω)∇/prime2ψ(r/prime,ω)]dV/prime= /contintegraldisplay S0+S∞/bracketleftbigg ψ(r/prime,ω)∂G(r|r/prime;ω) ∂n/prime−G(r|r/prime;ω)∂ψ(r/prime,ω) ∂n/prime/bracketrightbigg dS/prime. We note that ˆnpoints outward from V, and Gis the Green’s function (5.77). By inspection, this Green’s function obeys the reciprocity condition G(r|r/prime;ω)=G(r/prime|r;ω) and satisﬁes ∇2G(r|r/prime;ω)=∇/prime2G(r|r/prime;ω). Substituting ∇/prime2˜ψ=− k2˜ψ−˜Sfrom (5.75) and ∇/prime2G=− k2G−δ(r−r/prime)from (5.76) we get ˜ψ(r,ω)=/integraldisplay V˜S(r/prime,ω)G(r|r/prime;ω)dV/prime− −/contintegraldisplay S0+S∞/bracketleftbigg ˜ψ(r/prime,ω)∂G(r|r/prime;ω) ∂n/prime−G(r|r/prime;ω)∂˜ψ(r/prime,ω) ∂n/prime/bracketrightbigg dS/prime. Hence ˜ψwithin Vmay be written in terms of the sources within Vand the values of ˜ψ and its normal derivative over S0+S∞. The surface contributions account for sources excluded by S0. Let us examine the integral over S∞more closely. If we let S∞recede to inﬁnity, we expect no contribution to the potential at rfrom the ﬁelds on S∞. Choosing a sphere centered at the origin, we note that ˆn/prime=ˆr/primeand that as r/prime→∞ G(r|r/prime;ω)=e−jk|r−r/prime| 4π|r−r/prime|≈e−jkr/prime 4πr/prime, ∂G(r|r/prime;ω) ∂n/prime=ˆn/prime·∇/primeG(r|r/prime;ω)≈∂ ∂r/primee−jkr/prime 4πr/prime=−(1+jkr/prime)e−jkr/prime 4πr/prime. Substituting these, we ﬁnd that as r/prime→∞ /contintegraldisplay S∞/bracketleftbigg ˜ψ∂G ∂n/prime−G∂˜ψ ∂n/prime/bracketrightbigg dS/prime≈/integraldisplay2π 0/integraldisplayπ 0/bracketleftbigg −1+jkr/prime r/prime2˜ψ−1 r/prime∂˜ψ ∂r/prime/bracketrightbigge−jkr/prime 4πr/prime2sinθ/primedθ/primedφ/prime ≈−/integraldisplay2π 0/integraldisplayπ 0/bracketleftbigg ˜ψ+r/prime/parenleftbigg jk˜ψ+∂˜ψ ∂r/prime/parenrightbigg/bracketrightbigge−jkr 4πsinθ/primedθ/primedφ/prime. Since this gives the contribution to the ﬁeld in Vfrom the ﬁelds on the surface receding to inﬁnity, we expect that this term should be zero. If the medium has loss, then theexponential term decays and drives the contribution to zero. For a lossless medium thecontribution is zero if lim r→∞˜ψ(r,ω)=0, (5.96) lim r→∞r/bracketleftbigg jk˜ψ(r,ω)+∂˜ψ(r,ω) ∂r/bracketrightbigg =0. (5.97) This is called the radiation condition for the Helmholtz equation. It is also called the Sommerfeld radiation condition after the German physicist A. Sommerfeld. Note that we have not derived this condition: we have merely postulated it. As with all postulates it is subject to experimental veriﬁcation. The radiation condition implies that for points far from the source the potentials behave as spherical waves: ˜ψ(r,ω)∼e−jkr r, r→∞. Substituting this into (5.96) and (5.97) we ﬁnd that the radiation condition is satisﬁed. With S∞→∞ we have ˜ψ(r,ω)=/integraldisplay V˜S(r/prime,ω)G(r|r/prime;ω)dV/prime− −/contintegraldisplay S0/bracketleftbigg ˜ψ(r/prime,ω)∂G(r|r/prime;ω) ∂n/prime−G(r|r/prime;ω)∂˜ψ(r/prime,ω) ∂n/prime/bracketrightbigg dS/prime, which is the expression for the potential within an inﬁnite medium having source- excluding regions. As S0→0we obtain the expression for the potential in an unbounded medium: ˜ψ(r,ω)=/integraldisplay V˜S(r/prime,ω)G(r|r/prime;ω)dV/prime, as expected. The time-domain equation (5.71) may also be solved (at least for the lossless case) in a bounded region of space. The interested reader should see Pauli [143] for details. 5.3 Transverse–longitudinal decomposition We have seen that when only electric sources are present, the electromagnetic ﬁelds in a homogeneous, isotropic region can be represented by a single vector potential Πe. Similarly, when only magnetic sources are present, the ﬁelds can be represented by asingle vector potential Π h. Hence two vector potentials may be used to represent the ﬁeld if both electric and magnetic sources are present. We may also represent the electromagnetic ﬁeld in a homogeneous, isotropic region us- ing two scalar functions and the sources. This follows naturally from another importantﬁeld decomposition: a splitting of each ﬁeld vector into (1) a component along a certain pre-chosen constant direction, and (2) a component transverse to this direction. Depend-ing on the geometry of the sources, it is possible that only one of these components willbe present. A special case of this decomposition, the TE–TM ﬁeld decomposition , holds for a source-free region and will be discussed in the next section. 5.3.1 Transverse–longitudinal decomposition in terms of ﬁelds Consider a direction deﬁned by a constant unit vector ˆu. We deﬁne the longitudinal component ofAasˆuAuwhere Au=ˆu·A, and the transverse component ofAas At=A−ˆuAu. We may thus decompose any vector into a sum of longitudinal and transverse parts. An important consequence of Maxwell’s equations is that the transverse ﬁelds may be writtenentirely in terms of the longitudinal ﬁelds and the sources. This holds in both the timeand frequency domains; we derive the decomposition in the frequency domain and leavethe derivation of the time-domain expressions as exercises. We begin by decomposingthe operators in Maxwell’s equations into longitudinal and transverse components. Wenote that ∂ ∂u≡ˆu·∇ and deﬁne a transverse del operator as ∇t≡∇− ˆu∂ ∂u. Using these basic deﬁnitions, the identities listed in Appendix B may be derived. We shall ﬁnd it helpful to express the vector curl and Laplacian operations in terms oftheir longitudinal and transverse components. Using (B.93) and (B.96) we ﬁnd that thetransverse component of the curl is given by (∇× A) t=− ˆu×ˆu×(∇× A) =− ˆu×ˆu×(∇t×At)−ˆu×ˆu×/parenleftbigg ˆu×/bracketleftbigg∂At ∂u−∇ tAu/bracketrightbigg/parenrightbigg . (5.98) The ﬁrst term in the right member is zero by property (B.91). Using (B.7) we can replace the second term by −ˆu/braceleftbigg ˆu·/parenleftbigg ˆu×/bracketleftbigg∂At ∂u−∇ tAu/bracketrightbigg/parenrightbigg/bracerightbigg +(ˆu·ˆu)/parenleftbigg ˆu×/bracketleftbigg∂At ∂u−∇ tAu/bracketrightbigg/parenrightbigg . The ﬁrst of these terms is zero since ˆu·/parenleftbigg ˆu×/bracketleftbigg∂At ∂u−∇ tAu/bracketrightbigg/parenrightbigg =/bracketleftbigg∂At ∂u−∇ tAu/bracketrightbigg ·(ˆu×ˆu)=0, hence (∇× A)t=ˆu×/bracketleftbigg∂At ∂u−∇ tAu/bracketrightbigg . (5.99) The longitudinal part is then, by property (B.80), merely the diﬀerence between the curl and its transverse part, or ˆu(ˆu·∇× A)=∇ t×At. (5.100) A similar set of steps gives the transverse component of the Laplacian as (∇2A)t=/bracketleftbigg ∇t(∇t·At)+∂2At ∂u2−∇ t×∇ t×At/bracketrightbigg , (5.101) and the longitudinal part as ˆu/parenleftbigˆu·∇2A/parenrightbig =ˆu∇2Au. (5.102) Veriﬁcation is left as an exercise. Noww e are ready to give a longitudinal–transverse decomposition of the ﬁelds in a lossy, homogeneous, isotropic region in terms of the direction ˆu. We write Maxwell’s equations as ∇× ˜E=− jω˜µ˜Ht−jω˜µˆu˜Hu−˜Ji mt−ˆu˜Ji mu, (5.103) ∇× ˜H=jω˜/epsilon1c˜Et+jω˜/epsilon1cˆu˜Eu+˜Ji t+ˆu˜Ji u, (5.104) where we have split the right-hand sides into longitudinal and transverse parts. Then, using (5.99) and (5.100), we can equate the transverse and longitudinal parts of eachequation to obtain ∇ t×˜Et=− jω˜µˆu˜Hu−ˆu˜Ji mu, (5.105) −ˆu×∇ t˜Eu+ˆu×∂˜Et ∂u=− jω˜µ˜Ht−˜Ji mt, (5.106) ∇t×˜Ht=jω˜/epsilon1cˆu˜Eu+ˆu˜Ji u, (5.107) −ˆu×∇ t˜Hu+ˆu×∂˜Ht ∂u=jω˜/epsilon1c˜Et+˜Ji t. (5.108) We shall isolate the transverse ﬁelds in terms of the longitudinal ﬁelds. Forming the cross product of ˆuand the partial derivative of (5.108) with respect to u, we have −ˆu×ˆu×∇ t∂˜Hu ∂u+ˆu×ˆu×∂2˜Ht ∂u2=jω˜/epsilon1cˆu×∂˜Et ∂u+ˆu×∂˜Ji t ∂u. Using (B.7) and (B.80) we ﬁnd that ∇t∂˜Hu ∂u−∂2˜Ht ∂u2=jω˜/epsilon1cˆu×∂Et ∂u+ˆu×∂˜Ji t ∂u. (5.109) Multiplying (5.106) by jω˜/epsilon1cwe have −jω˜/epsilon1cˆu×∇ t˜Eu+jω˜/epsilon1cˆu×∂˜Et ∂u=ω2˜µ˜/epsilon1c˜Ht−jω˜/epsilon1c˜Ji mt. (5.110) We nowadd (5.109) to (5.110) and eliminate ˜Etto get /parenleftbigg∂2 ∂u2+k2/parenrightbigg ˜Ht=∇ t∂˜Hu ∂u−jω˜/epsilon1cˆu×∇ t˜Eu+jω˜/epsilon1c˜Ji mt−ˆu×∂˜Ji t ∂u. (5.111) This one-dimensional Helmholtz equation can be solved to ﬁnd the transverse magnetic ﬁeld from the longitudinal components of ˜Eand ˜H. Similar steps lead to a formula for the transverse component of ˜E: /parenleftbigg∂2 ∂u2+k2/parenrightbigg ˜Et=∇ t∂˜Eu ∂u+jω˜µˆu×∇ t˜Hu+ˆu×∂˜Ji mt ∂u+jω˜µ˜Ji t. (5.112) We ﬁnd the longitudinal components from the wave equation for ˜Eand ˜H. Recall that the ﬁelds satisfy (∇2+k2)˜E=1 ˜/epsilon1c∇˜ρi+jω˜µ˜Ji+∇× ˜Ji m, (∇2+k2)˜H=1 ˜µ∇˜ρi m+jω˜/epsilon1c˜Ji m−∇× ˜Ji. Splitting the vectors into longitudinal and transverse parts, and using (5.100) and (5.102), we equate the longitudinal components of the wave equations to obtain /parenleftbig ∇2+k2/parenrightbig˜Eu=1 ˜/epsilon1c∂˜ρi ∂u+jω˜µ˜Ji u+∇ t×˜Ji mt, (5.113) /parenleftbig ∇2+k2/parenrightbig˜Hu=1 ˜µ∂˜ρi m ∂u+jω˜/epsilon1c˜Ji mu−∇ t×˜Ji t. (5.114) We note that if ˜Ji m=˜Ji t=0, then ˜Hu=0and the ﬁelds are TM to the u-direction; these ﬁelds may be determined completely from ˜Eu. Similarly, if ˜Ji=˜Ji mt=0, then ˜Eu=0 and the ﬁelds are TE to the u-direction; these ﬁelds may be determined completely from ˜Hu. These properties are used in §4.11.7, where the ﬁelds of electric and magnetic line sources aligned along the z-direction are assumed to be purely TM zor TE z, respectively. 5.4 TE–TM decomposition 5.4.1 TE–TM decomposition in terms of ﬁelds A particularly useful ﬁeld decomposition results if we specialize to a source-free region. With ˜Ji=˜Ji m=0in (5.111)–(5.112) we obtain /parenleftbigg∂2 ∂u2+k2/parenrightbigg ˜Ht=∇ t∂˜Hu ∂u−jω˜/epsilon1cˆu×∇ t˜Eu, (5.115) /parenleftbigg∂2 ∂u2+k2/parenrightbigg ˜Et=∇ t∂˜Eu ∂u+jω˜µˆu×∇ t˜Hu. (5.116) Setting the sources to zero in (5.113) and (5.114) we get /parenleftbig ∇2+k2/parenrightbig˜Eu=0,/parenleftbig ∇2+k2/parenrightbig˜Hu=0. Hence the longitudinal ﬁeld components are solutions to the homogeneous Helmholtz equation, and the transverse components are speciﬁed solely in terms of the longitudinalcomponents. The electromagnetic ﬁeld is completely speciﬁed by the two scalar ﬁelds ˜E u and ˜Hu(and, of course, appropriate boundary values). We can use superposition to simplify the task of solving (5.115)–(5.116). Since each equation has two forcing terms on the right-hand side, we can solve the equations usingone forcing term at a time, and add the results. That is, let ˜E 1and ˜H1be the solutions to (5.115)–(5.116) with ˜Eu=0, and ˜E2and ˜H2be the solutions with ˜Hu=0. This results in a decomposition ˜E=˜E1+˜E2, (5.117) ˜H=˜H1+˜H2, (5.118) with ˜E1=˜E1t, ˜H1=˜H1t+˜H1uˆu, ˜H2=˜H2t, ˜E2=˜E2t+˜E2uˆu. Because ˜E1has no u-component, ˜E1and ˜H1are termed transverse electric (orTE)t o theu-direction; ˜H2has no u-component, and ˜E2and ˜H2are termed transverse magnetic (orTM) to the u-direction.2We see that in a source-free region any electromagnetic ﬁeld can be decomposed into a set of two ﬁelds that are TE and TM, respectively, tosome ﬁxed u-direction. This is useful when solving boundary value (e.g., waveguide and scattering) problems where information about external sources is easily speciﬁedusing the values of the ﬁelds on the boundary of the source-free region. In that case ˜E uand ˜Huare determined by solving the homogeneous wave equation in an appropriate coordinate system, and the other ﬁeld components are found from (5.115)–(5.116). Oftenthe boundary conditions can be satisﬁed by the TM ﬁelds or the TE ﬁelds alone. Thissimpliﬁes the analysis of many types of EM systems. 5.4.2 TE–TM decomposition in terms of Hertzian potentials We are free to represent ˜Eand ˜Hin terms of scalar ﬁelds other than ˜Euand ˜Hu.I n doing so, it is helpful to retain the wave nature of the solution so that a meaningfulphysical interpretation is still possible; we thus use Hertzian potentials since they obeythe wave equation. For the TM case let ˜Π h=0and ˜Πe=ˆu˜/Pi1e. Setting ˜Ji=0in (5.64) we have (∇2+k2)˜Πe=0. Since ˜Πeis purely longitudinal, we can use (B.99) to obtain the scalar Helmholtz equation for˜/Pi1e: (∇2+k2)˜/Pi1e=0. (5.119) Once ˜/Pi1ehas been found by solving this wave equation, the ﬁelds can be found by using (5.62)–(5.63) with ˜Ji=0: ˜E=∇× (∇× ˜Πe), (5.120) ˜H=jω˜/epsilon1c∇× ˜Πe. (5.121) We can evaluate ˜Eby noting that ˜Πeis purely longitudinal. Use of property (B.98) gives ∇×∇× ˜Πe=∇ t∂˜/Pi1e ∂u−ˆu∇2 t˜/Pi1e. Then, by property (B.97), ∇×∇× ˜Πe=∇ t∂˜/Pi1e ∂u−ˆu/bracketleftbigg ∇2˜/Pi1e−∂2˜/Pi1e ∂u2/bracketrightbigg . By (5.119) then, ˜E=∇ t∂˜/Pi1e ∂u+ˆu/parenleftbigg∂2 ∂u2+k2/parenrightbigg ˜/Pi1e. (5.122) The ﬁeld ˜Hcan be found by noting that ˜Πeis purely longitudinal. Use of property (B.96) in (5.121) gives ˜H=− jω˜/epsilon1cˆu×∇ t˜/Pi1e. (5.123) 2Some authors prefer to use the terminology Em o d e in place of TM, and Hm o d e in place of TE, indicating the presence of a u-directed electric or magnetic ﬁeld component. Similar steps can be used to ﬁnd the TE representation. Substitution of ˜Πe=0and ˜Πh=ˆu˜/Pi1hinto (5.65)–(5.66) gives the ﬁelds ˜E=jω˜µˆu×∇ t˜/Pi1h, (5.124) ˜H=∇ t∂˜/Pi1h ∂u+ˆu/parenleftbigg∂2 ∂u2+k2/parenrightbigg ˜/Pi1h, (5.125) while ˜/Pi1hmust satisfy (∇2+k2)˜/Pi1h=0. (5.126) Hertzian potential representation of TEM ﬁelds. An interesting situation occurs when a ﬁeld is both TE and TM to a particular direction. Such a ﬁeld is said to betransverse electromagnetic (orTEM) to that direction. Unfortunately, with ˜E u=˜Hu= 0we cannot use (5.115) or (5.116) to ﬁnd the transverse ﬁeld components. It turns out that a single scalar potential function is suﬃcient to represent the ﬁeld, and we may useeither ˜/Pi1 eor˜/Pi1h. For the TM case, equations (5.122) and (5.123) showthat w e can represent the electro- magnetic ﬁelds completely with ˜/Pi1e. Unfortunately (5.122) has a longitudinal component, and thus cannot describe a TEM ﬁeld. But if we require that ˜/Pi1eobey the additional equation /parenleftbigg∂2 ∂u2+k2/parenrightbigg ˜/Pi1e=0, (5.127) then both Eand Hare transverse to uand thus describe a TEM ﬁeld. Since ˜/Pi1emust also obey /parenleftbig ∇2+k2/parenrightbig˜/Pi1e=0, using (B.7) we can write (5.127) as ∇2 t˜/Pi1e=0. Similarly, for the TE case we found that the EM ﬁelds were completely described in (5.124) and (5.125) by ˜/Pi1h. In this case ˜Hhas a longitudinal component. Thus, if we require /parenleftbigg∂2 ∂u2+k2/parenrightbigg ˜/Pi1h=0, (5.128) then both ˜Eand ˜Hare purely transverse to uand again describe a TEM ﬁeld. Equation (5.128) is equivalent to ∇2 t˜/Pi1h=0. We can therefore describe a TEM ﬁeld using either ˜/Pi1eor˜/Pi1h, since a TEM ﬁeld is both TE and TM to the longitudinal direction. If we choose ˜/Pi1ewe can use (5.122) and (5.123) to obtain the expressions ˜E=∇ t∂˜/Pi1e ∂u, (5.129) ˜H=− jω˜/epsilon1cˆu×∇ t˜/Pi1e, (5.130) where ˜/Pi1emust obey ∇2 t˜/Pi1e=0,/parenleftbigg∂2 ∂u2+k2/parenrightbigg ˜/Pi1e=0. (5.131) If we choose ˜/Pi1hwe can use (5.124) and (5.125) to obtain ˜E=jω˜µˆu×∇ t˜/Pi1h, (5.132) ˜H=∇ t∂˜/Pi1h ∂u, (5.133) where ˜/Pi1hmust obey ∇2 t˜/Pi1h=0,/parenleftbigg∂2 ∂u2+k2/parenrightbigg ˜/Pi1h=0. (5.134) 5.4.3 Application: hollow-pipe waveguides A classic application of the TE–TM decomposition is to the calculation of waveguide ﬁelds. Consider a hollowpipe w ith PEC w alls, aligned along the z-axis. The inside is ﬁlled with a homogeneous, isotropic material of permeability ˜µ(ω)and complex permittivity ˜/epsilon1c(ω), and the guide cross-sectional shape is assumed to be independent of z. We assume that a current source exists somewhere within the waveguide, creating waves that eitherpropagate or evanesce away from the source. If the source is conﬁned to the region−d<z<dthen each of the regions z>dand z<−dis source-free and we may decompose the ﬁelds there into TE and TM sets. Such a waveguide is a good candidatefor TE–TM analysis because the TE and TM ﬁelds independently satisfy the boundaryconditions at the waveguide walls. This is not generally the case for certain other guided-wave structures such as ﬁber optic cables and microstrip lines. We may represent the ﬁelds either in terms of the longitudinal ﬁelds ˜E zand ˜Hz,o r in terms of the Hertzian potentials. We choose the Hertzian potentials. For TM ﬁeldswe choose ˜Π e=ˆz˜/Pi1e,˜Πh=0; for TE ﬁelds we choose ˜Πh=ˆz˜/Pi1h,˜Πe=0. Both of the potentials must obey the same Helmholtz equation: /parenleftbig ∇2+k2/parenrightbig˜/Pi1z=0, (5.135) where ˜/Pi1zrepresents either ˜/Pi1eor˜/Pi1h. We seek a solution to this equation using the separation of variables technique, and assume the product solution ˜/Pi1z(r,ω)=˜Z(z,ω)˜ψ(ρ,ω) , where ρis the transverse position vector ( r=ˆzz+ρ). Substituting the trial solution into (5.135) and writing ∇2=∇2 t+∂2 ∂z2 we ﬁnd that 1 ˜ψ(ρ,ω)∇2 t˜ψ(ρ,ω)+k2=−1 Z(z,ω)∂2 ∂z2Z(z,ω) . Because the left-hand side of this expression has positional dependence only on ρwhile the right-hand side has dependence only on z, we must have both sides equal to a constant, sayk2 z. Then ∂2Z ∂z2+k2 zZ=0, which is an ordinary diﬀerential equation with the solutions Z=e∓jkzz. We also have ∇2 t˜ψ(ρ,ω)+k2 c˜ψ(ρ,ω)=0, (5.136) where kc=k2−k2 zis called the cutoﬀ wavenumber . The solution to this equation depends on the geometry of the waveguide cross-section and whether the ﬁeld is TE or TM. The ﬁelds may be computed from the Hertzian potentials using u=zin (5.122)– (5.123) and (5.124)–(5.125). Because the ﬁelds all contain the common term e∓jkzz,we deﬁne the ﬁeld quantities ˜eand ˜hthrough ˜E(r,ω)=˜e(ρ,ω)e∓jkzz, ˜H(r,ω)=˜h(ρ,ω)e∓jkzz. Then, substituting ˜/Pi1e=˜ψee∓jkzz, we have for TM ﬁelds ˜e=∓ jkz∇t˜ψe+ˆzk2 c˜ψe, ˜h=− jω˜/epsilon1cˆz×∇ t˜ψe. Because we have a simple relationship between the transverse parts of ˜Eand ˜H,wem a y also write the ﬁelds as ˜ez=k2 c˜ψe, (5.137) ˜et=∓ jkz∇t˜ψe, (5.138) ˜ht=±Ye(ˆz×˜et). (5.139) Here Ye=ω˜/epsilon1c kz is the complex TM wave admittance . For TE ﬁelds we have with ˜/Pi1h=˜ψhe∓jkzz ˜e=jω˜µˆz×∇ t˜ψh, ˜h=∓ jkz∇t˜ψh+ˆzk2 c˜ψh, or ˜hz=k2 c˜ψh, (5.140) ˜ht=∓ jkz∇t˜ψh, (5.141) ˜et=∓ Zh(ˆz×˜ht). (5.142) Here Zh=ω˜µ kz is the TM wave impedance . Modal solutions for the transverse ﬁeld dependence. Equation (5.136) describes the transverse behavior of the waveguide ﬁelds. When coupled with an appropriateboundary condition, this homogeneous equation has an inﬁnite spectrum of discrete so-lutions called eigenmodes or simply modes . Each mode has associated with it a real eigenvalue k cthat is dependent on the cross-sectional shape of the waveguide, but inde- pendent of frequency and homogeneous material parameters. We number the modes sothat k c=kcnfor the nth mode. The amplitude of each modal solution depends on the excitation source within the waveguide. The appropriate boundary conditions can be found by employing the condition that for both TM and TE ﬁelds the tangential component of ˜Emust be zero on the waveguide walls: ˆn×˜E=0, where ˆnis the unit inward normal to the waveguide wall. For TM ﬁelds we have ˜Ez=0and thus ˜ψe(ρ,ω)=0, ρ∈/Gamma1, (5.143) where /Gamma1is the contour describing the waveguide boundary. For TE ﬁelds we have ˆn×˜Et= 0,o r ˆn×(ˆz×∇ t˜ψh)=0. Using ˆn×(ˆz×∇ t˜ψh)=ˆz(ˆn·∇t˜ψh)−(ˆn·ˆz)∇t˜ψh and noting that ˆn·ˆz=0, we have the boundary condition ˆn·∇t˜ψh(ρ,ω)=∂˜ψh(ρ,ω) ∂n=0, ρ∈/Gamma1. (5.144) The wave nature of the waveguide ﬁelds. We have seen that all waveguide ﬁeld components, for both TE and TM modes, vary as e∓jkznz. Here k2 zn=k2−k2 cnis the propagation constant of the nth mode. Letting kz=β−jα we thus have ˜E,˜H∼e∓jβze∓αz. Forz>dwe choose the minus sign so that we have a wave propagating away from the source; for z<−dwe choose the plus sign. When the guide is ﬁlled with a good dielectric we may assume ˜µ=µis real and independent of frequency and use (4.254) to showthat kz=β−jα=/radicalBig/bracketleftbig ω2µ/epsilon1/prime−k2c/bracketrightbig −jω2µ/epsilon1/primetanδc =/radicalbig µ/epsilon1/prime/radicalBig ω2−ω2c/radicalBigg 1−jtanδc 1−(ωc/ω)2 where δcis the loss tangent (4.253) and where ωc=kc√µ/epsilon1/prime is called the cutoﬀ frequency . Under the condition tanδc 1−(ωc/ω)2/lessmuch1 (5.145) we may approximate the square root using the ﬁrst two terms of the binomial series to showthat β−jα≈/radicalbig µ/epsilon1/prime/radicalBig ω2−ω2c/bracketleftbigg 1−j1 2tanδc 1−(ωc/ω)2/bracketrightbigg . (5.146) 0.0 0.5 1.0 1.5 2.0 2.5 β/ω v or α/ω v0.00.51.01.52.02.5ω/ωc Light Lineβ α c c Figure 5.4: Dispersion plot for a hollow-pipe waveguide. Light line computed using v=1/√µ/epsilon1. Condition (5.145) requires that ωbe suﬃciently removed from ωc, either by having ω>ω corω<ω c. When ω>ω cwe say that the frequency is above cutoﬀ and ﬁnd from (5.146) that β≈ω/radicalbig µ/epsilon1/prime/radicalBig 1−ω2c/ω2,α ≈ω2µ/epsilon1/prime 2βtanδc. Hereα/lessmuchβand the wave propagates down the waveguide with relatively little loss. When ω<ω cwe say that the waveguide is cut oﬀ or that the frequency is below cutoﬀ and ﬁnd that α≈ω/radicalbig µ/epsilon1/prime/radicalBig ω2c/ω2−1,β ≈ω2µ/epsilon1/prime 2αtanδc. In this case the wave has a very small phase constant and a very large rate of attenuation. For frequencies near ωcthere is an abrupt but continuous transition between these two types of wave behavior. When the waveguide is ﬁlled with a lossless material having permittivity /epsilon1and per- meability µ, the transition across the cutoﬀ frequency is discontinuous. For ω>ω cwe have β=ω√µ/epsilon1/radicalBig 1−ω2c/ω2,α =0, and the wave propagates without loss. For ω<ω cwe have α=ω√µ/epsilon1/radicalBig ω2c/ω2−1,β =0, andthewaveisevanesce nt.Thedispersiondiagra mshowninFigure5.4clearlyshows the abrupt cutoﬀ phenomenon. We can compute the phase and group velocities of thewave above cutoﬀ just as we did for plane waves: v p=ω β=v/radicalbig 1−ω2c/ω2, 0.0 0.5 1.0 1.5 2.0 2.5 ω/ω0.00.51.01.52.02.5v / v or v / vv /v v /v cp gg p Figure 5.5: Phase and group velocity for a hollow-pipe waveguide. vg=dω dβ=v/radicalBig 1−ω2c/ω2, (5.147) where v=1/√µ/epsilon1. Note that vgvp=v2. We showlater that vgis the velocity of energy transport within a lossless guide. We also see that as ω→∞ we have vp→vand vg →v.Moreinterestingl y,asω→ωc weﬁndthatvp →∞andvg → 0.Thisisshown graphicall yinFigure5.5. We may also speak of the guided wavelength of a monochromatic wave propagating with frequency ˇωin a waveguide. We deﬁne this wavelength as λg=2π β=λ/radicalbig 1−ω2c/ˇω2=λ/radicalbig 1−λ2/λ2c. Here λ=2π ˇω√µ/epsilon1,λ c=2π kc. Orthogonality of waveguide modes. The modal ﬁelds in a closed-pipe waveguide obey several orthogonality relations. Let (ˇEn,ˇHn)be the time-harmonic electric and magnetic ﬁelds of one particular waveguide mode (TE or TM), and let (ˇEm,ˇHm)be the ﬁelds of a diﬀerent mode (TE or TM). One very useful relation states that for awaveguide containing lossless materials /integraldisplay CSˆz·/parenleftbigˇen×ˇh∗ m/parenrightbig dS=0, m/negationslash=n, (5.148) where CSis the guide cross-section. This is used to establish that the total power carried by a wave is the sum of the powers carried by individual modes (see below). Other important relationships include the orthogonality of the longitudinal ﬁelds, /integraldisplay CSˇEzmˇEzndS=0, m/negationslash=n, (5.149) /integraldisplay CSˇHzmˇHzndS=0, m/negationslash=n, (5.150) and the orthogonality of transverse ﬁelds, /integraldisplay CSˇEtm·ˇEtndS=0, m/negationslash=n, /integraldisplay CSˇHtm·ˇHtndS=0, m/negationslash=n. These may also be combined to give an orthogonality relation for the complete ﬁelds: /integraldisplay CSˇEm·ˇEndS=0, m/negationslash=n, (5.151) /integraldisplay CSˇHm·ˇHndS=0, m/negationslash=n. (5.152) For proofs of these relations the reader should see Collin [39]. Power carried by time-harmonic waves in lossless waveguides. The power car- ried by a time-harmonic wave propagating down a waveguide is deﬁned as the time-average Poynting ﬂux passing through the guide cross-section. Thus we may write P av=1 2/integraldisplay CSRe/braceleftbigˇE×ˇH∗/bracerightbig ·ˆzdS. The ﬁeld within the guide is assumed to be a superposition of all possible waveguide modes. For waves traveling in the +z-direction this implies ˇE=/summationdisplay m(ˇetm+ˆzˇezm)e−jkzmz, ˇH=/summationdisplay n/parenleftbigˇhtn+ˆzˇhzn/parenrightbig e−jkznz. Substituting we have Pav=1 2Re/braceleftBigg/integraldisplay CS/bracketleftBigg/summationdisplay m(ˇetm+ˆzˇezm)e−jkzmz×/summationdisplay n/parenleftbigˇh∗ tn+ˆzˇh∗ zn/parenrightbig ejk∗ znz/bracketrightBigg ·ˆzdS/bracerightBigg =1 2Re/braceleftBigg/summationdisplay m/summationdisplay ne−j(kzm−k∗ zn)z/integraldisplay CSˆz·/parenleftbigˇetm×ˇh∗ tn/parenrightbig dS/bracerightBigg . By (5.148) we have Pav=1 2Re/braceleftBigg/summationdisplay ne−j(kzn−k∗ zn)z/integraldisplay CSˆz·/parenleftbigˇetn×ˇh∗ tn/parenrightbig dS/bracerightBigg . For modes propagating in a lossless guide kzn=βzn. For modes that are cut oﬀ kzn= −jαzn. However, we ﬁnd below that terms in this series representing modes that are cut oﬀ are zero. Thus Pav=/summationdisplay n1 2Re/braceleftbigg/integraldisplay CSˆz·/parenleftbigˇetn×ˇh∗ tn/parenrightbig dS/bracerightbigg =/summationdisplay nPn,av. Hence for waveguides ﬁlled with lossless media the total time-average power ﬂow is given by the superposition of the individual modal powers. Simple formulas for the individual modal powers in a lossless guide may be obtained by substituting the expressions for the ﬁelds. For TM modes we use (5.138) and (5.139)to get P av=1 2Re/braceleftbigg |kz|2Y∗ ee−j(kz−k∗ z)/integraldisplay CSˆz·/parenleftbig ∇tˇψe×[ˆz×∇ tˇψ∗ e]/parenrightbig dS/bracerightbigg =1 2|kz|2Re/braceleftbig Y∗ e/bracerightbig e−j(kz−k∗ z)/integraldisplay CS∇tˇψe·∇tˇψ∗ edS. Here we have used (B.7) and ˆz·∇tˇψe=0. This expression can be simpliﬁed by using the two-dimensional version of Green’s ﬁrst identity (B.29): /integraldisplay S(∇ta·∇tb+a∇2 tb)dS=/contintegraldisplay /Gamma1a∂b ∂ndl. Using a=ˇψeand b=ˇψ∗ eand integrating over the waveguide cross-section we have /integraldisplay CS(∇tˇψe·∇tˇψ∗ e+ˇψe∇2ˇψ∗ e)dS=/contintegraldisplay /Gamma1ˇψe∂ˇψ∗ e ∂ndl. Substituting ∇2 tˇψ∗ e=−k2 cˇψ∗ eand remembering that ˇψe=0on/Gamma1we reduce this to /integraldisplay CS∇tˇψe·∇tˇψ∗ edS=k2 c/integraldisplay CSˇψeˇψ∗ edS. (5.153) Thus the power is Pav=1 2Re/braceleftbig Y∗ e/bracerightbig |kz|2k2 ce−j(kz−k∗ z)z/integraldisplay CSˇψeˇψ∗ edS. For modes above cutoﬀ we have kz=βand Ye=ω/epsilon1/kz=ω/epsilon1/β. The power carried by these modes is thus Pav=1 2ω/epsilon1βk2 c/integraldisplay CSˇψeˇψ∗ edS. (5.154) For modes belowcutoﬀ w e have kz=− jαand Ye=jω/epsilon1/α.T h u s Re{Y∗ e}= 0and Pav=0. For frequencies belowcutoﬀ the ﬁelds are evanescent and do not carry pow er in the manner of propagating waves. For TE modes we may proceed similarly and show that Pav=1 2ωµβ k2 c/integraldisplay CSˇψhˇψ∗ hdS. (5.155) The details are left as an exercise. Stored energy in a waveguide and the velocity of energy transport. Consider a source-free section of lossless waveguide bounded on its two ends by the cross-sectionalsurfaces CS 1and CS2. Setting ˇJi=ˇJc=0in (4.156) we have 1 2/contintegraldisplay S(ˇE×ˇH∗)·dS=2jω/integraldisplay V[/angbracketleftwe/angbracketright−/angbracketleftwm/angbracketright]dV, where Vis the region of the guide between CS1and CS2. The right-hand side represents the diﬀerence between the total time-average stored electric and magnetic energies. Thus 2jω[/angbracketleftWe/angbracketright−/angbracketleft Wm/angbracketright]= 1 2/integraldisplay CS1−ˆz·(ˇE×ˇH∗)dS+1 2/integraldisplay CS2ˆz·(ˇE×ˇH∗)dS−1 2/integraldisplay Scond(ˇE×ˇH∗)·dS, where Scondindicates the conducting walls of the guide and ˆnpoints into the guide. For a propagating mode the ﬁrst two terms on the right-hand side cancel since with no loss ˇE×ˇH∗is the same on CS1andCS2. The third term is zero since (ˇE×ˇH∗)·ˆn=(ˆn×ˇE)·ˇH∗, and ˆn×ˇE=0on the waveguide walls. Thus we have /angbracketleftWe/angbracketright=/angbracketleft Wm/angbracketright for any section of a lossless waveguide. We may compute the time-average stored magnetic energy in a section of lossless waveguide of length las /angbracketleftWm/angbracketright=µ 4/integraldisplayl 0/integraldisplay CSˇH·ˇH∗dSdz. For propagating TM modes we can substitute (5.139) to ﬁnd /angbracketleftWm/angbracketright/l=µ 4(βYe)2/integraldisplay CS(ˆz×∇ tˇψe)·(ˆz×∇ tˇψ∗ e)dS. Using (ˆz×∇ tˇψe)·(ˆz×∇ tˇψ∗ e)=ˆz·/bracketleftbig ∇tˇψ∗ e×(ˆz×∇ tˇψe)/bracketrightbig =∇ tˇψe·∇tˇψ∗ e we have /angbracketleftWm/angbracketright/l=µ 4(βYe)2/integraldisplay CS∇tˇψe·∇tˇψ∗ edS. Finally, using (5.153) we have the stored energy per unit length for a propagating TM mode: /angbracketleftWm/angbracketright/l=/angbracketleftWe/angbracketright/l=µ 4(ω/epsilon1)2k2 c/integraldisplay CSˇψeˇψ∗ edS. Similarly we may show that for a TE mode /angbracketleftWe/angbracketright/l=/angbracketleftWm/angbracketright/l=/epsilon1 4(ωµ)2k2 c/integraldisplay CSˇψhˇψ∗ hdS. The details are left as an exercise. As with plane waves in (4.261) we may describe the velocity of energy transport as the ratio of the Poynting ﬂux density to the total stored energy density: Sav=/angbracketleftwT/angbracketrightve. For TM modes this energy velocity is ve=1 2ω/epsilon1βk2 cˇψeˇψ∗ e 2µ 4(ω/epsilon1)2k2cˇψeˇψ∗e=β ωµ/epsilon1=v/radicalBig 1−ω2c/ω2, which is identical to the group velocity (5.147). This is also the case for TE modes, for which ve=1 2ωµβ k2 cˇψhˇψ∗ h 2/epsilon1 4(ωµ)2k2cˇψhˇψ∗ h=β ωµ/epsilon1=v/radicalBig 1−ω2c/ω2. Example: ﬁelds of a rectangular waveguide. Consider a rectangular waveguide with a cross-section occupying 0≤x≤aand 0≤y≤b. The material within the guide is assumed to be a lossless dielectric of permittivity /epsilon1and permeability µ. We seek the modal ﬁelds within the guide. Both TE and TM ﬁelds exist within the guide. In each case we must solve the diﬀer- ential equation ∇2 t˜ψ+k2 c˜ψ=0. A product solution in rectangular coordinates may be sought using the separation of variables technique ( §A.4). We ﬁnd that ˜ψ(x,y,ω)=[Axsinkxx+Bxcoskxx]/bracketleftbig Aysinkyy+Bycoskyy/bracketrightbig where k2 x+k2 y=k2 c. This solution is easily veriﬁed by substitution. For TM modes the solution is subject to the boundary condition (5.143): ˜ψe(ρ,ω)=0, ρ∈/Gamma1. Applying this at x=0and y=0we ﬁnd Bx=By=0. Applying the boundary condition atx=awe then ﬁnd sinkxa=0and thus kx=nπ a, n=1,2,.... Note that n=0corresponds to the trivial solution ˜ψe=0. Similarly, from the condition aty=bwe ﬁnd that ky=mπ b, m=1,2,.... Thus ˜ψe(x,y,ω)=Anmsin/parenleftBignπx a/parenrightBig sin/parenleftBigmπy b/parenrightBig . From (5.137)–(5.139) we ﬁnd that the ﬁelds are ˜Ez=k2 cnmAnm/bracketleftBig sinnπx asinmπy b/bracketrightBig e∓jkzz, ˜Et=∓ jkzAnm/bracketleftBig ˆxnπ acosnπx asinmπy b+ˆymπ bsinnπx acosmπy b/bracketrightBig e∓jkzz, ˜Ht=jkzYeAnm/bracketleftBig ˆxmπ bsinnπx acosmπy b−ˆynπ acosnπx asinmπy b/bracketrightBig e∓jkzz. Here Ye=1 η/radicalBig 1−ω2cnm/ω2 withη=(µ/epsilon1)1/2. Each combination of m,ndescribes a diﬀerent ﬁeld pattern and thus a diﬀerent mode, designated TM nm. The cutoﬀ wavenumber of the TM nmmode is kcnm=/radicalbigg/parenleftBignπ a/parenrightBig2 +/parenleftBigmπ b/parenrightBig2 , m,n=1,2,3,... and the cutoﬀ frequency is ωcnm=v/radicalbigg/parenleftBignπ a/parenrightBig2 +/parenleftBigmπ b/parenrightBig2 , m,n=1,2,3,... where v=1/(µ/epsilon1)1/2. Thus the TM 11mode has the lowest cutoﬀ frequency of any TM mode. There is a range of frequencies for which this is the only propagating TM mode. For TE modes the solution is subject to ˆn·∇t˜ψh(ρ,ω)=∂˜ψh(ρ,ω) ∂n=0, ρ∈/Gamma1. Atx=0we have ∂˜ψh ∂x=0 leading to Ax=0.A t y=0we have ∂˜ψh ∂y=0 leading to Ay=0.A t x=awe require sinkxa=0and thus kx=nπ a, n=0,1,2,.... Similarly, from the condition at y=bwe ﬁnd ky=mπ b, m=0,1,2,.... The case n=m=0is not allowed since it produces the trivial solution. Thus ˜ψh(x,y,ω)=Bnmcos/parenleftBignπx a/parenrightBig cos/parenleftBigmπy b/parenrightBig , m,n=0,1,2,..., m+n>0. From (5.140)–(5.142) we ﬁnd that the ﬁelds are ˜Hz=k2 cnmBnm/bracketleftBig cosnπx acosmπy b/bracketrightBig e∓jkzz, ˜Ht=± jkzBnm/bracketleftBig ˆxnπ asinnπx acosmπy b+ˆymπ bcosnπx asinmπy b/bracketrightBig e∓jkzz, ˜Et=jkzZhBnm/bracketleftBig ˆxmπ bcosnπx asinmπy b−ˆynπ asinnπx acosmπy b/bracketrightBig e∓jkzz. Here Zh=η/radicalBig 1−ω2cnm/ω2. In this case the modes are designated TE nm. The cutoﬀ wavenumber of the TE nmmode is kcnm=/radicalbigg/parenleftBignπ a/parenrightBig2 +/parenleftBigmπ b/parenrightBig2 , m,n=0,1,2,..., m+n>0 and the cutoﬀ frequency is ωcnm=v/radicalbigg/parenleftBignπ a/parenrightBig2 +/parenleftBigmπ b/parenrightBig2 , m,n=0,1,2,..., m+n>0 where v=1/(µ/epsilon1)1/2. Modes having the same cutoﬀ frequency are said to be degenerate . This is the case with the TE and TM modes. However, the ﬁeld distributions diﬀer andthus the modes are distinct. Note that we may also have degeneracy among the TE or TM modes. For instance, if a=bthen the cutoﬀ frequency of the TE nmmode is identical to that of the TE mnmode. If a≥bthen the TE 10mode has the lowest cutoﬀ frequency and is termed the dominant mode in a rectangular guide. There is a ﬁnite band of frequencies in which this is the only mode propagating (although the bandwidthis small if a≈b.) Calculation of the time-average power carried by propagating TE and TM modes is left as an exercise. 5.4.4 TE–TM decomposition in spherical coordinates It is not necessary for the longitudinal direction to be constant to achieve a TE–TM decomposition. It is possible, for instance, to represent the electromagnetic ﬁeld in termsof components either TE or TM to the radial direction of spherical coordinates. This maybe shown using a procedure identical to that used for the longitudinal–transverse decom-position in rectangular coordinates. We carry out the decomposition in the frequency domain and leave the time-domain decomposition as an exercise. TE–TM decomposition in terms of the radial ﬁelds. Consider a source-free re- gion of space ﬁlled with a homogeneous, isotropic material described by parameters ˜µ(ω) and ˜/epsilon1 c(ω). We substitute the spherical coordinate representation of the curl into Fara- day’s and Ampere’s laws with source terms ˜Jand ˜Jmset equal to zero. Equating vector components we have, in particular, 1 r/bracketleftbigg1 sinθ∂˜Er ∂φ−∂ ∂r(r˜Eφ)/bracketrightbigg =− jω˜µ˜Hθ (5.156) and 1 r/bracketleftbigg∂ ∂r(r˜Hθ)−∂˜Hr ∂θ/bracketrightbigg =jω˜/epsilon1c˜Eφ. (5.157) We seek to isolate the transverse components of the ﬁelds in terms of the radial compo- nents. Multiplying (5.156) by jω˜/epsilon1crwe get jω˜/epsilon1c1 sinθ∂˜Er ∂φ−jω˜/epsilon1c∂(r˜Eφ) ∂r=k2r˜Hθ; next, multiplying (5.157) by rand then diﬀerentiating with respect to rwe get ∂2 ∂r2(r˜Hθ)−∂2˜Hr ∂θ∂r=jω˜/epsilon1c∂(r˜Eφ) ∂r. Subtracting these two equations and rearranging, we obtain /parenleftbigg∂2 ∂r2+k2/parenrightbigg (r˜Hθ)=jω˜/epsilon1c1 sinθ∂˜Er ∂φ+∂2˜Hr ∂r∂θ. This is a one-dimensional wave equation for the product of rwith the transverse ﬁeld component ˜Hθ. Similarly /parenleftbigg∂2 ∂r2+k2/parenrightbigg (r˜Hφ)=− jω˜/epsilon1c∂˜Er ∂θ+1 sinθ∂2˜Hr ∂r∂φ, and /parenleftbigg∂2 ∂r2+k2/parenrightbigg (r˜Eφ)=1 sinθ∂2˜Er ∂φ∂r+jω˜µ∂˜Hr ∂θ, (5.158) /parenleftbigg∂2 ∂r2+k2/parenrightbigg (r˜Eθ)=∂2˜Er ∂θ∂r+jω˜µ1 sinθ∂˜Hr ∂φ. (5.159) Hence we can represent the electromagnetic ﬁeld in a source-free region in terms of the two scalar quantities ˜Erand ˜Hr. Superposition allows us to solve the TE case with ˜Er=0and the TM case with ˜Hr=0, and combine the results for the general expansion of the ﬁeld. TE–TM decomposition in terms of potential functions. If we allow the vector potential (or Hertzian potential) to have only an r-component, then the resulting ﬁelds are TE or TM to the r-direction. Unfortunately, this scalar component does not satisfy the Helmholtz equation. If we wish to use a potential component that satisﬁes theHelmholtz equation then we must discard the Lorentz condition and choose a diﬀerentrelationship between the vector and scalar potentials. 1. TM ﬁelds. To generate ﬁelds TM to rwe recall that the electromagnetic ﬁelds may be written in terms of electric vector and scalar potentials as ˜E=− jω˜A e−∇φe, (5.160) ˜B=∇× ˜Ae. (5.161) In a source-free region we have by Ampere’s law ˜E=1 jω˜µ˜/epsilon1c∇× ˜B=1 jω˜µ˜/epsilon1c∇×(∇× ˜Ae). Here ˜φeand ˜Aemust satisfy a diﬀerential equation that may be derived by examining ∇×(∇× ˜E)=− jω∇× ˜B=− jω(jω˜µ˜/epsilon1c˜E)=k2˜E, where k2=ω2˜µ˜/epsilon1c. Substitution from (5.160) gives ∇×/parenleftbig ∇× [−jω˜Ae−∇ ˜φe]/parenrightbig =k2[−jω˜Ae−∇ ˜φe] or ∇×(∇× ˜Ae)−k2˜Ae=k2 jω∇˜φe. (5.162) We are still free to specify ∇·˜Ae. At this point let us examine the eﬀect of choosing a vector potential with only an r-component: ˜Ae=ˆr˜Ae. Since ∇×(ˆr˜Ae)=ˆθ rsinθ∂˜Ae ∂φ−ˆφ r∂˜Ae ∂θ(5.163) we see that B=∇× ˜Aehas no r-component. Since ∇×(∇× ˜Ae)=−ˆr rsinθ/bracketleftbigg1 r∂ ∂θ/parenleftbigg sinθ∂˜Ae ∂θ/parenrightbigg +1 rsinθ∂2˜Ae ∂φ2/bracketrightbigg +ˆθ r∂2˜Ae ∂r∂θ+ˆφ rsinθ∂2˜Ae ∂r∂φ we see that ˜E∼∇× (∇× ˜Ae)has all three components. This choice of ˜Aeproduces a ﬁeld TM to the r-direction. We need only choose ∇·˜Aeso that the resulting diﬀerential equation is convenient to solve. Substituting the above expressions into (5.162) we ﬁndthat −ˆr rsinθ/bracketleftbigg1 r∂ ∂θ/parenleftbigg sinθ∂˜Ae ∂θ/parenrightbigg +1 rsinθ∂2˜Ae ∂φ2/bracketrightbigg +ˆθ r∂2˜Ae ∂r∂θ+ˆφ rsinθ∂2˜Ae ∂r∂φ−ˆrk2˜Ae= ˆrk2 jω∂˜φe ∂r+ˆθ rk2 jω∂˜φe ∂θ+ˆφ rsinθk2 jω∂˜φe ∂φ. (5.164) Since ∇·˜Aeonly involves the derivatives of ˜Aewith respect to r, we may specify ∇·˜Ae indirectly through ˜φe=jω k2∂˜Ae ∂r. With this (5.164) becomes 1 rsinθ/bracketleftbigg1 r∂ ∂θ/parenleftbigg sinθ∂˜Ae ∂θ/parenrightbigg +1 rsinθ∂2˜Ae ∂φ2/bracketrightbigg +k2˜Ae+∂2˜Ae ∂r2=0. Using 1 r∂ ∂r/bracketleftbigg r2∂ ∂r/parenleftbigg˜Ae r/parenrightbigg/bracketrightbigg =∂2˜Ae ∂r2 we can write the diﬀerential equation as 1 r2∂ ∂r/bracketleftbigg r2∂(˜Ae/r) ∂r/bracketrightbigg +1 r2sinθ∂ ∂θ/bracketleftbigg sinθ∂(˜Ae/r) ∂θ/bracketrightbigg +1 r2sin2θ∂2(˜Ae/r) ∂φ2+k2˜Ae r=0. The ﬁrst three terms of this expression are precisely the Laplacian of ˜Ae/r.T h u s we have (∇2+k2)/parenleftbigg˜Ae r/parenrightbigg =0 (5.165) and the quantity ˜Ae/rsatisﬁes the homogeneous Helmholtz equation. The TM ﬁelds generated by the vector potential ˜Ae=ˆr˜Aemay be found by using (5.160) and (5.161). From (5.160) we have the electric ﬁeld ˜E=− jω˜Ae−∇ ˜φe=− jωˆr˜Ae−∇/parenleftbiggjω k2∂˜Ae ∂r/parenrightbigg . Expanding the gradient we have the ﬁeld components ˜Er=1 jω˜µ˜/epsilon1c/parenleftbigg∂2 ∂r2+k2/parenrightbigg ˜Ae, (5.166) ˜Eθ=1 jω˜µ˜/epsilon1c1 r∂2˜Ae ∂r∂θ, (5.167) ˜Eφ=1 jω˜µ˜/epsilon1c1 rsinθ∂2˜Ae ∂r∂φ. (5.168) The magnetic ﬁeld components are found using (5.161) and (5.163): ˜Hθ=1 ˜µ1 rsinθ∂˜Ae ∂φ, (5.169) ˜Hφ=−1 ˜µ1 r∂˜Ae ∂θ. (5.170) 2. TE ﬁelds. To generate ﬁelds TE to rwe recall that the electromagnetic ﬁelds in a source-free region may be written in terms of magnetic vector and scalar potentials as ˜H=− jω˜Ah−∇φh, (5.171) ˜D=− ∇× ˜Ah. (5.172) In a source-free region we have from Faraday’s law ˜H=1 −jω˜µ˜/epsilon1c∇× ˜D=1 jω˜µ˜/epsilon1c∇×(∇× ˜Ah). Here ˜φhand ˜Ahmust satisfy a diﬀerential equation that may be derived by examining ∇×(∇× ˜H)=jω∇× ˜D=jω˜/epsilon1c(−jω˜µ˜H)=k2˜H, where k2=ω2˜µ˜/epsilon1c. Substitution from (5.171) gives ∇×/parenleftbig ∇× [−jω˜Ah−∇ ˜φh]/parenrightbig =k2[−jω˜Ah−∇ ˜φh] or ∇×(∇× ˜Ah)−k2˜Ah=k2 jω∇˜φh. (5.173) Choosing ˜Ah=ˆr˜Ahand ˜φh=jω k2∂˜Ah ∂r we ﬁnd, as with the TM ﬁelds, (∇2+k2)/parenleftbigg˜Ah r/parenrightbigg =0. (5.174) Thus the quantity ˜Ah/robeys the Helmholtz equation. We can ﬁnd the TE ﬁelds using (5.171) and (5.172). Substituting we ﬁnd that ˜Hr=1 jω˜µ˜/epsilon1c/parenleftbigg∂2 ∂r2+k2/parenrightbigg ˜Ah, (5.175) ˜Hθ=1 jω˜µ˜/epsilon1c1 r∂2˜Ah ∂r∂θ, (5.176) ˜Hφ=1 jω˜µ˜/epsilon1c1 rsinθ∂2˜Ah ∂r∂φ, (5.177) ˜Eθ=−1 ˜/epsilon1c1 rsinθ∂˜Ah ∂φ, (5.178) ˜Eφ=1 ˜/epsilon1c1 r∂˜Ah ∂θ. (5.179) Example of spherical TE–TM decomposition: a plane wave. Consider a uni- form plane wave propagating in the z-direction in a lossless, homogeneous material of permittivity /epsilon1and permeability µ, such that its electromagnetic ﬁeld is ˜E(r,ω)=ˆx˜E0(ω)e−jkz=ˆx˜E0(ω)e−jkrcosθ, ˜H(r,ω)=ˆy˜E0(ω) ηe−jkz=ˆx˜E0(ω) ηe−jkrcosθ. We wish to represent this ﬁeld in terms of the superposition of a ﬁeld TE to rand a ﬁeld TM to r. We ﬁrst ﬁnd the potential functions ˜Ae=ˆr˜Aeand ˜Ah=ˆr˜Ahthat represent the ﬁeld. Then we may use (5.166)–(5.170) and (5.175)–(5.179) to ﬁnd the TE and TMrepresentations. From (5.166) we see that ˜A eis related to ˜Er, where ˜Eris given by ˜Er=˜E0sinθcosφe−jkrcosθ=˜E0cosφ jkr∂ ∂θ/bracketleftbig e−jkrcosθ/bracketrightbig . We can separate the randθdependences of the exponential function by using the identity (E.101). Since jn(−z)=(−1)njn(z)=j−2njn(z)we have e−jkrcosθ=∞/summationdisplay n=0j−n(2n+1)jn(kr)Pn(cosθ). Using ∂Pn(cosθ) ∂θ=∂P0 n(cosθ) ∂θ=P1 n(cosθ) we thus have ˜Er=−j˜E0cosφ kr∞/summationdisplay n=1j−n(2n+1)jn(kr)P1 n(cosθ). Here we start the sum at n=1since P1 0(x)=0. We can nowidentify the vector potential as ˜Ae r=˜E0k ωcosφ∞/summationdisplay n=1j−n(2n+1) n(n+1)jn(kr)P1 n(cosθ) (5.180) since by direct diﬀerentiation we have ˜Er=1 jω˜µ˜/epsilon1c/parenleftbigg∂2 ∂r2+k2/parenrightbigg ˜Ae =˜E0k jω2˜µ˜/epsilon1ccosφ∞/summationdisplay n=1j−n(2n+1) n(n+1)P1 n(cosθ)/parenleftbigg∂2 ∂r2+k2/parenrightbigg [rjn(kr)] =−j˜E0cosφ kr∞/summationdisplay n=1j−n(2n+1)jn(kr)P1 n(cosθ), which satisﬁes (5.166). Here we have used the deﬁning equation of the spherical Bessel functions (E.15) to showthat /parenleftbigg∂2 ∂r2+k2/parenrightbigg [rjn(kr)]=r∂2 ∂r2jn(kr)+2∂ ∂rjn(kr)+k2rjn(kr) =k2r/bracketleftbigg∂2 ∂(kr)2+2 kr∂ ∂(kr)/bracketrightbigg jn(kr)+k2rjn(kr) =−k2r/bracketleftbigg 1−n(n+1) (kr)2/bracketrightbigg jn(kr)+k2rjn(kr)=n(n+1) rjn(kr). We note immediately that ˜Ae/rsatisﬁes the Helmholtz equation (5.165) since it has the form of the separation of variables solution (D.113). We may ﬁnd the vector potential ˜Ah=ˆr˜Ahin the same manner. Noting that ˜Hr=˜E0 ηsinθsinφe−jkrcosθ=˜E0sinφ ηjkr∂ ∂θ/bracketleftbig e−jkrcosθ/bracketrightbig =1 jω˜µ˜/epsilon1c/parenleftbigg∂2 ∂r2+k2/parenrightbigg ˜Ah, we have the potential ˜Ah r=˜E0k ηωsinφ∞/summationdisplay n=1j−n(2n+1) n(n+1)jn(kr)P1 n(cosθ). (5.181) We may nowcompute the transverse components of the TM ﬁeld using (5.167)–(5.170). For convenience, let us deﬁne a newfunction ˆJnby ˆJn(x)=xjn(x). Then we may write ˜Er=−j˜E0cosφ (kr)2∞/summationdisplay n=1j−n(2n+1)ˆJn(kr)P1 n(cosθ), (5.182) ˜Eθ=j˜E0 krsinθcosφ∞/summationdisplay n=1anˆJ/prime n(kr)P1 n/prime(cosθ), (5.183) ˜Eφ=j˜E0 krsinθsinφ∞/summationdisplay n=1anˆJ/prime n(kr)P1 n(cosθ), (5.184) ˜Hθ=−˜E0 krηsinθsinφ∞/summationdisplay n=1anˆJn(kr)P1 n(cosθ), (5.185) ˜Hφ=˜E0 krηsinθcosφ∞/summationdisplay n=1anˆJn(kr)P1 n/prime(cosθ). (5.186) Here ˆJ/prime n(x)=d dxˆJn(x)=d dx[xjn(x)]=xj/prime n(x)+jn(x) and an=j−n(2n+1) n(n+1). (5.187) Similarly, we have the TE ﬁelds from (5.176)–(5.179): ˜Hr=−j˜E0sinφ η(kr)2∞/summationdisplay n=1j−n(2n+1)ˆJn(kr)P1 n(cosθ), (5.188) ˜Hθ=j˜E0 ηkrsinθsinφ∞/summationdisplay n=1anˆJ/prime n(kr)P1 n/prime(cosθ), (5.189) ˜Hφ=− j˜E0 ηkrsinθcosφ∞/summationdisplay n=1anˆJ/prime n(kr)P1 n(cosθ), (5.190) ˜Eθ=−˜E0 krsinθcosφ∞/summationdisplay n=1anˆJn(kr)P1 n(cosθ), (5.191) ˜Eφ=−˜E0 krsinθsinφ∞/summationdisplay n=1anˆJn(kr)P1 n/prime(cosθ). (5.192) The total ﬁeld is then the sum of the TE and TM components. Example of spherical TE–TM decomposition: scattering by a conducting sphere. Consider a PEC sphere of radius acentered at the origin and imbedded in a homogeneous, isotropic material having parameters ˜µand ˜/epsilon1c. The sphere is illuminated by a plane wave incident along the z-axis with the ﬁelds ˜E(r,ω)=ˆx˜E0(ω)e−jkz=ˆx˜E0(ω)e−jkrcosθ, ˜H(r,ω)=ˆy˜E0(ω) ηe−jkz=ˆx˜E0(ω) ηe−jkrcosθ. We wish to ﬁnd the ﬁeld scattered by the sphere. The boundary condition that determines the scattered ﬁeld is that the total (incident plus scattered) electric ﬁeld tangential to the sphere must be zero. We sawin the previousexample that the incident electric ﬁeld may be written as the sum of a ﬁeld TE to the r-direction and a ﬁeld TM to the r-direction. Since the region external to the sphere is source-free, we may also represent the scattered ﬁeld as a sum of TE and TM ﬁelds.These may be found from the functions ˜A s eand ˜As h, which obey the Helmholtz equations (5.165) and (5.174). The general solution to the Helmholtz equation may be found usingthe separation of variables technique in spherical coordinates, as shown in §A.4, and is given by /braceleftbigg˜A s e/r ˜As h/r/bracerightbigg =∞/summationdisplay n=0n/summationdisplay m=−nCnmYnm(θ, φ) h(2) n(kr). Here Ynmis the spherical harmonic and we have chosen the spherical Hankel function h(2) n as the radial dependence since it represents the expected outward-going wave behavior of the scattered ﬁeld. Since the incident ﬁeld generated by the potentials (5.180) and(5.181) exactly cancels the ﬁeld generated by ˜A s eand ˜As hon the surface of the sphere, by orthogonality the scattered potential must have φandθdependencies that match those of the incident ﬁeld. Thus ˜As e r=˜E0k ωcosφ∞/summationdisplay n=1bnh(2) n(kr)P1 n(cosθ), ˜As h r=˜E0k ηωsinφ∞/summationdisplay n=1cnh(2) n(kr)P1 n(cosθ), where bnand cnare constants to be determined by the boundary conditions. By super- position the total ﬁeld may be computed from the total potentials, which are the sum ofthe incident and scattered potentials. These are given by ˜A t e r=˜E0k ωcosφ∞/summationdisplay n=1/bracketleftbig anjn(kr)+bnh(2) n(kr)/bracketrightbig P1 n(cosθ), ˜At h r=˜E0k ηωsinφ∞/summationdisplay n=1/bracketleftbig anjn(kr)+cnh(2) n(kr)/bracketrightbig P1 n(cosθ), where anis given by (5.187). The total transverse electric ﬁeld is found by superposing the TE and TM transverse ﬁelds found from the total potentials. We have already computed the transverse incidentﬁelds and may easily generalize these results to the total potentials. By (5.183) and(5.191) we have ˜E t θ(a)=j˜E0 kasinθcosφ∞/summationdisplay n=1/bracketleftbig anˆJ/prime n(ka)+bnˆH(2)/prime n(ka)/bracketrightbig P1 n/prime(cosθ)− −˜E0 kasinθcosφ∞/summationdisplay n=1/bracketleftbig anˆJn(ka)+cnˆH(2) n(ka)/bracketrightbig P1 n(cosθ)=0, where ˆH(2) n(x)=xh(2) n(x). By (5.184) and (5.192) we have ˜Et φ(a)=j˜E0 kasinθsinφ∞/summationdisplay n=1/bracketleftbig anˆJ/prime n(ka)+bnˆH(2)/prime n(ka)/bracketrightbig P1 n(cosθ)− −˜E0 kasinθsinφ∞/summationdisplay n=1/bracketleftbig anˆJn(ka)+cnˆH(2) n(ka)/bracketrightbig P1 n/prime(cosθ)=0. These two sets of equations are satisﬁed by the conditions bn=−ˆJ/prime n(ka) ˆH(2)/prime n(ka)an, cn=−ˆJn(ka) ˆH(2) n(ka)an. We can noww rite the scattered electric ﬁelds as ˜Es r=− j˜E0cosφ∞/summationdisplay n=1bn/bracketleftbigˆH(2)/prime/prime n(kr)+ˆH(2) n(kr)/bracketrightbig P1 n(cosθ), ˜Es θ=˜E0 krcosφ∞/summationdisplay n=1/bracketleftbigg jbnsinθˆH(2)/prime n(kr)P1 n/prime(cosθ)−cn1 sinθˆH(2) n(kr)P1 n(cosθ)/bracketrightbigg , ˜Es φ=˜E0 krsinφ∞/summationdisplay n=1/bracketleftbigg jbn1 sinθˆH(2)/prime n(kr)P1 n(cosθ)−cnsinθˆH(2) n(kr)P1 n/prime(cosθ)/bracketrightbigg . Let us approximate the scattered ﬁeld for observation points far from the sphere. We may approximate the spherical Hankel functions using (E.68) as ˆH(2) n(z)=zh(2) n(z)≈jn+1e−jz, ˆH(2)/prime n(z)≈jne−jz, ˆH(2)/prime/prime n(z)≈− jn+1e−jz. Substituting these we ﬁnd that ˜Er→0as expected for the far-zone ﬁeld, while ˜Es θ≈˜E0e−jkr krcosφ∞/summationdisplay n=1jn+1/bracketleftbigg bnsinθP1 n/prime(cosθ)−cn1 sinθP1 n(cosθ)/bracketrightbigg , ˜Es φ≈˜E0e−jkr krsinφ∞/summationdisplay n=1jn+1/bracketleftbigg bn1 sinθP1 n(cosθ)−cnsinθP1 n/prime(cosθ)/bracketrightbigg . 0123456789 1 0 ka0.010.101.0010.00 σ/πa2 Figure 5.6: Monostatic radar cross-section of a conducting sphere. From the far-zone ﬁelds we can compute the radar cross-section (RCS) or echo area of the sphere, which is deﬁned by σ=lim r→∞/parenleftbigg 4πr2|˜Es|2 |˜Ei|2/parenrightbigg . (5.193) Carrying units of m2, this quantity describes the relative energy density of the scattered ﬁeldnormalize dbythedistanc efromthescatterin gobject.Figure5.6showstheRCSof a conducting sphere in free space for the monostatic case: when the observation direction is aligned with the direction of the incident wave (i.e., θ=π), also called the backscatter direction. At lowfrequencies the RCS is proportional to λ−4; this is the range of Rayleigh scattering , showing that higher-frequency light scatters more strongly from microscopic particles in the atmosphere (explaining why the sky is blue) [19]. At high frequencies the result approaches that of geometrical optics, and the RCS becomes the interception areaof the sphere, πa 2. This is the region of optical scattering . Between these two regions lies the resonance region , or the region of Mie scattering , named for G. Mie who in 1908 published the ﬁrst rigorous solution for scattering by a sphere (followed soon after byDebye in 1909). Several interesting phenomena of sphere scattering are best examined in the time do- main. We may compute the temporal scattered ﬁeld by taking the inverse transformofthefrequency-domai nﬁeld.Figure5.7shows E θ(t)computed in the backscatter direction ( θ=π) when the incident ﬁeld waveform E0(t)is a gaussian pulse and the sphere is in free space. Two distinct features are seen in the scattered ﬁeld waveform.The ﬁrst is a sharp pulse almost duplicating the incident ﬁeld waveform, but of oppositepolarity. This is the specular reﬂection produced when the incident ﬁeld ﬁrst contacts the sphere and begins to induce a current on the sphere surface. The second feature,called the creeping wave , occurs at a time approximately (2+π)a/cseconds after the -0.5 0.0 0.5 1.0 1.5 2.0 t/(2πa/c)-1.2-0.8-0.4-0.00.40.81.2Relative amplitudeincident field waveform A: specular reflectionB: creeping waveAB Figure 5.7: Time-domain ﬁeld back-scattered by a conducting sphere. specular reﬂection. This represents the ﬁeld radiated back along the incident direction by a wave of current excited by the incident ﬁeld at the tangent point, which travelsaround the sphere at approximately the speed of light in free space. Although this wavecontinues to traverse the sphere, its amplitude is reduced so signiﬁcantly by radiationdamping that only a single feature is seen. 5.5 Problems 5.1Verify that the ﬁelds and sources obeying even planar reﬂection symmetry obey the component Maxwell’s equations (5.1)–(5.6). Repeat for ﬁelds and sources obeying oddplanar reﬂection symmetry. 5.2We wish to investigate reﬂection symmetry through the origin in a homogeneous medium. Under what conditions on magnetic ﬁeld, magnetic current density, and electriccurrent density are we guaranteed that E x(x,y,z)=Ex(−x,−y,−z), Ey(x,y,z)=Ey(−x,−y,−z), Ez(x,y,z)=Ez(−x,−y,−z)? 5.3We wish to investigate reﬂection symmetry through an axis in a homogeneous medium. Under what conditions on magnetic ﬁeld, magnetic current density, and electriccurrent density are we guaranteed that E x(x,y,z)=− Ex(−x,−y,z), Ey(x,y,z)=− Ey(−x,−y,z), Ez(x,y,z)=Ez(−x,−y,z)? 5.4Consider an electric Hertzian dipole located on the z-axis at z=h. Showthat if the dipole is parallel to the plane z=0, then adding an oppositely-directed dipole of the same strength at z=− hproduces zero electric ﬁeld tangential to the plane. Also showthat if the dipole is z-directed, then adding another z-directed dipole at z=− h produces zero electric ﬁeld tangential to the z=0plane. Since the ﬁeld for z>0is unaltered in each case if we place a PEC in the z=0plane, we establish that tangential components of electric current image in the opposite direction while vertical componentsimage in the same direction. 5.5Consider a z-directed electric line source ˜I 0located at y=h,x=0between con- ducting planes at y=± d,d>h. The material between the plates has permeability ˜µ(ω)and complex permittivity ˜/epsilon1c(ω). Write the impressed and scattered ﬁelds in terms of Fourier transforms and apply the boundary conditions at z=± dto determine the electric ﬁeld between the plates. Show that the result is identical to the expression (5.8)obtained using symmetry decomposition, which required the boundary condition to beapplied only on the top plate. 5.6Consider a z-directed electric line source ˜I 0located at y=h,x=0in free space above a dielectric slab occupying −d<y<d,d<h. The slab has permeability µ0and permittivity /epsilon1. Decompose the source into even and odd constituents and solve for the electric ﬁeld everywhere using the Fourier transform approach. Describe how you woulduse the even and odd solutions to solve the problem of a dielectric slab located on top ofa PEC ground plane. 5.7Consider an unbounded, homogeneous, isotropic medium described by permeabil- ity˜µ(ω)and complex permittivity ˜/epsilon1 c(ω). Assuming there are magnetic sources present, but no electric sources, showthat the ﬁelds may be w ritten as ˜H(r)=− jω˜/epsilon1c/integraldisplay V¯Ge(r|r/prime;ω)·˜Ji m(r/prime,ω)dV/prime, ˜E(r)=/integraldisplay V¯Gm(r|r/prime;ω)·˜Ji m(r/prime,ω)dV/prime, where ¯Geis given by (5.83) and ¯Gmis given by (5.84). 5.8Showthat for a cubical excluding volume the depolarizing dyadic is ¯L=¯I/3. 5.9Compute the depolarizing dyadic for a cylindrical excluding volume with height and diameter both 2a, and with the limit taken as a→0. Showthat ¯L=0.293¯I. 5.10 Showthat the spherical w ave function ˜ψ(r,ω)=e−jkr 4πr obeys the radiation conditions (5.96) and (5.97). 5.11 Verify that the transverse component of the Laplacian of Ais (∇2A)t=/bracketleftbigg ∇t(∇t·At)+∂2At ∂u2−∇ t×∇ t×At/bracketrightbigg . Verify that the longitudinal component of the Laplacian of Ais ˆu/parenleftbigˆu·∇2A/parenrightbig =ˆu∇2Au. 5.12 Verify the identities (B.82)–(B.93). 5.13 Verify the identities (B.94)–(B.98). 5.14 Derive the formula (5.112) for the transverse component of the electric ﬁeld. 5.15 The longitudinal/transverse decomposition can be performed beginning with the time-domain Maxwell’s equations. Show that for a homogeneous, lossless, isotropic regiondescribed by permittivity /epsilon1and permeability µthe longitudinal ﬁelds obey the wave equations /parenleftbigg∂ 2 ∂u2−1 v2∂2 ∂t2/parenrightbigg Ht=∇ t∂Hu ∂u−/epsilon1ˆu×∇ t∂Eu ∂t+/epsilon1∂Jmt ∂t−ˆu×∂Jt ∂u, /parenleftbigg∂2 ∂u2−1 v2∂2 ∂t2/parenrightbigg Et=∇ t∂Eu ∂u+µˆu×∇ t∂Hu ∂t+ˆu×∂Jmt ∂u+µ∂Jt ∂t. Also showthat the transverse ﬁelds may be found from the longitudinal ﬁelds by solving /parenleftbigg ∇2−1 v2∂ ∂t2/parenrightbigg Eu=1 /epsilon1∂ρ ∂u+µ∂Ju ∂t+∇ t×Jmt, /parenleftbigg ∇2−1 v2∂ ∂t2/parenrightbigg Hu=1 µ∂ρm ∂u+/epsilon1∂Jmu ∂t−∇ t×Jt. Herev=1/√µ/epsilon1. 5.16 Consider a homogeneous, lossless, isotropic region of space described by permittiv- ity/epsilon1and permeability µ. Beginning with the source-free time-domain Maxwell equa- tions in rectangular coordinates, choose zas the longitudinal direction and showthat the TE–TM decomposition is given by /parenleftbigg∂2 ∂z2−1 v2∂2 ∂t2/parenrightbigg Ey=∂2Ez ∂z∂y+µ∂2Hz ∂x∂t, (5.194) /parenleftbigg∂2 ∂z2−1 v2∂2 ∂t2/parenrightbigg Ex=∂2Ez ∂x∂z−µ∂2Hz ∂y∂t, (5.195) /parenleftbigg∂2 ∂z2−1 v2∂2 ∂t2/parenrightbigg Hy=−/epsilon1∂2Ez ∂x∂t+∂2Hz ∂y∂z, (5.196) /parenleftbigg∂2 ∂z2−1 v2∂2 ∂t2/parenrightbigg Hx=/epsilon1∂2Ez ∂y∂t+∂2Hz ∂x∂z, (5.197) with /parenleftbigg ∇2−1 v2∂2 ∂t2/parenrightbigg Ez=0, (5.198) /parenleftbigg ∇2−1 v2∂2 ∂t2/parenrightbigg Hz=0. (5.199) Herev=1/√µ/epsilon1. 5.17 Consider the case of TM ﬁelds in the time domain. Showthat for a homogeneous, isotropic, lossless medium with permittivity /epsilon1and permeability µthe ﬁelds may be derived from a single Hertzian potential Πe(r,t)=ˆu˜/Pi1e(r,t)that satisﬁes the wave equation /parenleftbigg ∇2−1 v2∂2 ∂t2/parenrightbigg /Pi1e=0 and that the ﬁelds are E=∇ t∂/Pi1 e ∂u+ˆu/parenleftbigg∂2 ∂u2−1 v2∂2 ∂t2/parenrightbigg /Pi1e, H=−/epsilon1ˆu×∇ t∂/Pi1 e ∂t. 5.18 Consider the case of TE ﬁelds in the time domain. Showthat for a homogeneous, isotropic, lossless medium with permittivity /epsilon1and permeability µthe ﬁelds may be derived from a single Hertzian potential Πh(r,t)=ˆu˜/Pi1h(r,t)that satisﬁes the wave equation /parenleftbigg ∇2−1 v2∂2 ∂t2/parenrightbigg /Pi1h=0 and that the ﬁelds are E=µˆu×∇ t∂/Pi1 h ∂t, H=∇ t∂/Pi1 h ∂u+ˆu/parenleftbigg∂2 ∂u2−1 v2∂2 ∂t2/parenrightbigg /Pi1h. 5.19 Showthat in the time domain TEM ﬁelds may be w ritten for a homogeneous, isotropic, lossless medium with permittivity /epsilon1and permeability µin terms of a Hertzian potential Πe=ˆu/Pi1ethat satisﬁes ∇2 t/Pi1e=0 and that the ﬁelds are E=∇ t∂/Pi1 e ∂u, H=−/epsilon1ˆu×∇ t∂/Pi1 e ∂t. 5.20 Showthat in the time domain TEM ﬁelds may be w ritten for a homogeneous, isotropic, lossless medium with permittivity /epsilon1and permeability µin terms of a Hertzian potential Πh=ˆu/Pi1hthat satisﬁes ∇2 t/Pi1h=0 and that the ﬁelds are E=µˆu×∇ t∂/Pi1 h ∂t, H=∇ t∂/Pi1 h ∂u. 5.21 Consider a TEM plane-wave ﬁeld of the form ˜E=ˆx˜E0e−jkz, ˜H=ˆy˜E0 ηe−jkz, where k=ω√µ/epsilon1andη=√µ//epsilon1. Showthat: (a)˜Emay be obtained from ˜Husing the equations for a ﬁeld that is TE y; (b) ˜Hmay be obtained from ˜Eusing the equations for a ﬁeld that is TM x; (c)˜Eand ˜Hmay be obtained from the potential ˜Πh=ˆy(˜E0/k2η)e−jkz; (d) ˜Eand ˜Hmay be obtained from the potential ˜Πe=ˆx(˜E0/k2)e−jkz; (e)˜Eand ˜Hmay be obtained from the potential ˜Πe=ˆz(j˜E0x/k)e−jkz; (f)˜Eand ˜Hmay be obtained from the potential ˜Πh=ˆz(j˜E0y/kη)e−jkz. 5.22 Prove the orthogonality relationships (5.149) and (5.150) for the longitudinal ﬁelds in a lossless waveguide. Hint: Substitute a=ˇψeand b=ˇψhinto Green’s second identity (B.30) and apply the boundary conditions for TE and TM modes. 5.23 Verify the waveguide orthogonality conditions (5.151)-(5.152) by substituting the ﬁeld expressions for a rectangular waveguide. 5.24 Showthat the time-average pow er carried by a propagating TE mode in a lossless waveguide is given by Pav=1 2ωµβ k2 c/integraldisplay CSˇψhˇψ∗ hdS. 5.25 Showthat the time-average stored energy per unit length for a propagating TE mode in a lossless waveguide is /angbracketleftWe/angbracketright/l=/angbracketleftWm/angbracketright/l=/epsilon1 4(ωµ)2k2 c/integraldisplay CSˇψhˇψ∗ hdS. 5.26 Consider a waveguide of circular cross-section aligned on the z-axis and ﬁlled with a lossless material having permittivity /epsilon1and permeability µ. Solve for both the TE and TM ﬁelds within the guide. List the ﬁrst ten modes in order by cutoﬀ frequency. 5.27 Consider a propagating TM mode in a lossless rectangular waveguide. Show that the time-average power carried by the propagating wave is Pavnm=1 2ω/epsilon1β nmk2 cnm|Anm|2ab 4. 5.28 Consider a propagating TE mode in a lossless rectangular waveguide. Show that the time-average power carried by the propagating wave is Pavnm=1 2ωµβ nmk2 cnm|Bnm|2ab 4. 5.29 Consider a homogeneous, lossless region of space characterized by permeability µ and permittivity /epsilon1. Beginning with the time-domain Maxwell equations, show that the θandφcomponents of the electromagnetic ﬁelds can be written in terms of the radial components. From this give the TE r–TM rﬁeld decomposition. 5.30 Consider the formula for the radar cross-section of a PEC sphere (5.193). Show that for the monostatic case the RCS becomes σ=λ2 4π/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞/summationdisplay n=1(−1)n(2n+1) ˆH(2)/prime n(ka)ˆH(2) n(ka)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle2 . 5.31 Beginning with the monostatic formula for the RCS of a conducting sphere given in Problem 5.30, use the small-argument approximation to the spherical Hankel functionsto showthat the RCS is proportional to λ −4when ka/lessmuch1. 5.32 Beginning with the monostatic formula for the RCS of a conducting sphere given in Problem 5.30, use the large-argument approximation to the spherical Hankel functionsto showthat the RCS approaches the interception area of the sphere, πa 2,a ska→∞. 5.33 A material sphere of radius ahas permittivity /epsilon1and permeability µ. The sphere is centered at the origin and illuminated by a plane wave traveling in the z-direction with the ﬁelds ˜E(r,ω)=ˆx˜E0(ω)e−jkz, ˜H(r,ω)=ˆy˜E0(ω) ηe−jkz. Find the ﬁelds internal and external to the sphere. Chapter 6 Integral solutions of Maxwell’s equations 6.1 Vector Kirchoﬀ solution: method of Stratton and Chu One of the most powerful tools for the analysis of electromagnetics problems is the integral solution to Maxwell’s equations formulated by Stratton and Chu [187, 188].These authors used the vector Green’s theorem to solve for ˜Eand ˜Hin much the same way as is done in static ﬁelds with the scalar Green’s theorem. An alternative approach isto use the Lorentz reciprocity theorem of §4.10.2, as done by Fradin [74]. The reciprocity approach allows the identiﬁcation of terms arising from surface discontinuities, whichmust be added to the result obtained from the other approach [187]. 6.1.1 The Stratton–Chu formula Consider an isotropic, homogeneous medium occupying a bounded region Vin space. The medium is described by permeability ˜µ(ω), permittivity ˜/epsilon1(ω), and conductivity ˜σ(ω). The region Vis bounded by a surface S, which can be multiply-connected so that Sis the union of several surfaces S1 ,...,SN asshowninFigure6.1;theseareusedtoexclude unknown sources and to formulate the vector Huygens principle . Impressed electric and magnetic sources may thus reside both inside and outside V. We wish to solve for the electric and magnetic ﬁelds at a point rwithin V. To do this we employ the Lorentz reciprocity theorem (4.173), written here using the frequency-domainﬁelds as an integral over primed coordinates: −/contintegraldisplay S/bracketleftbig˜Ea(r/prime,ω)×˜Hb(r/prime,ω)−˜Eb(r/prime,ω)×˜Ha(r/prime,ω)/bracketrightbig ·ˆn/primedS/prime= /integraldisplay V/bracketleftbig˜Eb(r/prime,ω)·˜Ja(r/prime,ω)−˜Ea(r/prime,ω)·˜Jb(r/prime,ω)− (6.1) ˜Hb(r/prime,ω)·˜Jma(r/prime,ω)+˜Ha(r/prime,ω)·˜Jmb(r/prime,ω)/bracketrightbig dV/prime. (6.2) Note that the negative sign on the left arises from the deﬁnition of ˆnas the inward normal to V asshowninFigure6.1.Weplaceanelectri cHertzia ndipoleatthepoint r = rp where we wish to compute the ﬁeld, and set ˜Eb=˜Epand ˜Hb=˜Hpin the reciprocity theorem, where ˜Epand ˜Hpare the ﬁelds produced by the dipole (5.88)–(5.89): ˜Hp(r,ω)=jω∇× [˜pG(r|rp;ω)], (6.3) ˜Ep(r,ω)=1 ˜/epsilon1c∇×/parenleftbig ∇× [˜pG(r|rp;ω)]/parenrightbig . (6.4) Figure 6.1: Geometry used to derive the Stratton–Chu formula. We also let ˜Ea=˜Eand ˜Ha=˜H, where ˜Eand ˜Hare the ﬁelds produced by the impressed sources ˜Ja=˜Jiand ˜Jma=˜Ji mwithin Vthat we wish to ﬁnd at r=rp. Since the dipole ﬁelds are singular at r=rp, we must exclude the point rpwith a small spherical surface Sδsurroundin gthevolume VδasshowninFigure6.1.Substitutin gtheseﬁeldsinto(6.2) we obtain −/contintegraldisplay S+Sδ/bracketleftbig˜E×˜Hp−˜Ep×˜H/bracketrightbig ·ˆn/primedS/prime=/integraldisplay V−Vδ/bracketleftbig˜Ep·˜Ji−˜Hp·˜Ji m/bracketrightbig dV/prime. (6.5) A useful identity involves the spatially-constant vector ˜pand the Green’s function G(r/prime|rp): ∇/prime×/bracketleftbig ∇/prime×(G˜p)/bracketrightbig =∇/prime[∇/prime·(G˜p)]−∇/prime2(G˜p) =∇/prime[∇/prime·(G˜p)]−˜p∇/prime2G =∇/prime(˜p·∇/primeG)+˜pk2G, (6.6) where we have used ∇/prime2G=−k2Gforr/prime/negationslash=rp. We begin by computing the terms on the left side of (6.5). We suppress the r/primede- pendence of the ﬁelds and also the dependencies of G(r/prime|rp). Substituting from (6.3) we have /contintegraldisplay S+Sδ[˜E×˜Hp]·ˆn/primedS/prime=jω/contintegraldisplay S+Sδ/bracketleftbig˜E×∇/prime×(G˜p)/bracketrightbig ·ˆn/primedS/prime. Using ˆn/prime·[˜E×∇/prime×(G˜p)]=ˆn/prime·[˜E×(∇/primeG×˜p)]=(ˆn/prime×˜E)·(∇/primeG×˜p)we can write /contintegraldisplay S+Sδ[˜E×˜Hp]·ˆn/primedS/prime=jω˜p·/contintegraldisplay S+Sδ[ˆn/prime×˜E]×∇/primeGdS/prime. Figure 6.2: Decomposition of surface Snto isolate surface ﬁeld discontinuity. Next we examine /contintegraldisplay S+Sδ[˜Ep×˜H]·ˆn/primedS/prime=−1 ˜/epsilon1c/contintegraldisplay S+Sδ/bracketleftbig˜H×∇/prime×∇/prime×(G˜p)/bracketrightbig ·ˆn/primedS/prime. Use of (6.6) along with the identity (B.43) gives /contintegraldisplay S+Sδ[˜Ep×˜H]·ˆn/primedS/prime=−1 ˜/epsilon1c/contintegraldisplay S+Sδ/braceleftbig (˜H×˜p)k2G− −∇/prime×/bracketleftbig (˜p·∇/primeG)˜H/bracketrightbig +(˜p·∇/primeG)(∇/prime×˜H)/bracerightbig ·ˆn/primedS/prime. We would like to use Stokes’s theorem on the second term of the right-hand side. Since the theorem is not valid for surfaces on which ˜Hhas discontinuities, we break the closed surface sinFigure6.1intoopensurface swhoseboundar ycontoursisolatethedisconti- nuitiesasshowninFigure6.2.Thenwemaywrite /contintegraldisplay Sn=Sna+Snbˆn/prime·∇/prime×/bracketleftbig (˜p·∇/primeG)˜H/bracketrightbig dS/prime=/contintegraldisplay /Gamma1na+/Gamma1nbdl/prime·˜H(˜p·∇/primeG). For surfaces not containing discontinuities of ˜Hthe two contour integrals provide equal and opposite contributions and this term vanishes. Thus the left-hand side of (6.5) is −/contintegraldisplay S+Sδ/bracketleftbig˜E×˜Hp−˜Ep×˜H/bracketrightbig ·ˆn/primedS/prime= −1 ˜/epsilon1c˜p·/braceleftbigg/contintegraldisplay S+Sδ/bracketleftbig jω˜/epsilon1c(ˆn/prime×˜E)×∇/primeG+k2(ˆn/prime×˜H)G+ˆn/prime·(˜Ji+jω˜/epsilon1c˜E)∇/primeG/bracketrightbig dS/prime where we have substituted ˜Ji+jω˜/epsilon1c˜Efor∇/prime×˜Hand used (˜H×˜p)·ˆn/prime=˜p·(ˆn/prime×˜H). Now consider the right-hand side of (6.5). Substituting from (6.4) we have /integraldisplay V−Vδ˜Ep·˜JidV/prime=1 ˜/epsilon1c/integraldisplay V−Vδ˜Ji·/bracketleftbig ∇/prime×∇/prime×(˜pG)/bracketrightbig dV/prime. Using (6.6) and (B.42), we have /integraldisplay V−Vδ˜Ep·˜JidV/prime=1 ˜/epsilon1c/integraldisplay V−Vδ/braceleftbig k2(˜p·˜Ji)G+∇/prime·[˜Ji(˜p·∇/primeG)]−(˜p·∇/primeG)∇/prime·˜Ji/bracerightbig dV/prime. Figure 6.3: Geometry of surface integral used to extract Eatrp. Replacing ∇/prime·˜Jiwith−jω˜ρifrom the continuity equation and using the divergence theorem on the second term on the right-hand side, we then have /integraldisplay V−Vδ˜Ep·˜JidV/prime=1 ˜/epsilon1c˜p·/bracketleftbigg/integraldisplay V−Vδ(k2˜JiG+jω˜ρi∇/primeG)dV/prime−/contintegraldisplay S+Sδ(ˆn/prime·˜Ji)∇/primeGdS/prime/bracketrightbigg . Lastly we examine /integraldisplay V−Vδ˜Hp·˜Ji mdV/prime=jω/integraldisplay V−Vδ˜Ji m·∇/prime×(G˜p)dV/prime. Use of ˜Ji m·∇/prime×(G˜p)=˜Ji m·(∇/primeG×˜p)=˜p·(˜Ji m×∇/primeG)gives /integraldisplay V−Vδ˜Hp·˜Ji mdV/prime=jω˜p·/integraldisplay V−Vδ˜Ji m×∇/primeGd V/prime. We now substitute all terms into (6.5) and note that each term involves a dot product with ˜p. Since ˜pis arbitrary we have −/contintegraldisplay S+Sδ/bracketleftbig (ˆn/prime×˜E)×∇/primeG+(ˆn/prime·˜E)∇/primeG−jω˜µ(ˆn/prime×˜H)G/bracketrightbig dS/prime+ +1 jω˜/epsilon1c/contintegraldisplay /Gamma1a+/Gamma1b(dl/prime·˜H)∇/primeG=/integraldisplay V−Vδ/bracketleftbigg −˜Ji m×∇/primeG+˜ρi ˜/epsilon1c∇/primeG−jω˜µ˜JiG/bracketrightbigg dV/prime. The electric ﬁeld may be extracted from the above expression by letting the radius of the excluding volume Vδrecede to zero. We ﬁrst consider the surface integral over Sδ. Examinin gFigure6.3weseethat R =|rp−r/prime|=δ,ˆn/prime=− ˆR, and ∇/primeG(r/prime|rp)=d dR/parenleftbigge−jkR 4πR/parenrightbigg ∇/primeR=ˆR/parenleftbigg1+jkδ 4πδ2/parenrightbigg e−jkδ≈ˆR δ2asδ→0. Assuming ˜Eis continuous at r/prime=rpwe can write −lim δ→0/contintegraldisplay Sδ/bracketleftbig (ˆn/prime×˜E)×∇/primeG+(ˆn/prime·˜E)∇/primeG−jω˜µ(ˆn/prime×˜H)G/bracketrightbig dS/prime= lim δ→0/integraldisplay /Omega11 4π/bracketleftBigg (ˆR×˜E)×ˆR δ2+(ˆR·˜E)ˆR δ2−jω˜µ(ˆR×˜H)1 δ/bracketrightBigg δ2d/Omega1= lim δ→0/integraldisplay /Omega11 4π/bracketleftbig −(ˆR·˜E)ˆR+(ˆR·ˆR)˜E+(ˆR·˜E)ˆR/bracketrightbig d/Omega1=˜E(rp). Here we have used/integraltext /Omega1d/Omega1=4πfor the total solid angle subtending the sphere Sδ. Finally, assuming that the volume sources are continuous, the volume integral over Vδvanishes and we have ˜E(r,ω)=/integraldisplay V/parenleftbigg −˜Ji m×∇/primeG+˜ρi ˜/epsilon1c∇/primeG−jω˜µ˜JiG/parenrightbigg dV/prime+ +N/summationdisplay n=1/integraldisplay Sn/bracketleftbig (ˆn/prime×˜E)×∇/primeG+(ˆn/prime·˜E)∇/primeG−jω˜µ(ˆn/prime×˜H)G/bracketrightbig dS/prime− −N/summationdisplay n=11 jω˜/epsilon1c/contintegraldisplay /Gamma1na+/Gamma1nb(dl/prime·˜H)∇/primeG. (6.7) A similar formula for ˜Hcan be derived by placing a magnetic dipole of moment ˜pmat r=rpand proceeding as above. This leads to ˜H(r,ω)=/integraldisplay V/parenleftbigg ˜Ji×∇/primeG+˜ρi m ˜µ∇/primeG−jω˜/epsilon1c˜Ji mG/parenrightbigg dV/prime+ +N/summationdisplay n=1/integraldisplay Sn/bracketleftbig (ˆn/prime×˜H)×∇/primeG+(ˆn/prime·˜H)∇/primeG+jω˜/epsilon1c(ˆn/prime×˜E)G/bracketrightbig dS/prime+ +N/summationdisplay n=11 jω˜µ/contintegraldisplay /Gamma1na+/Gamma1nb(dl/prime·˜E)∇/primeG. (6.8) We can also obtain this expression by substituting (6.7) into Faraday’s law. 6.1.2 The Sommerfeld radiation condition In§5.2.2 we found that if the potentials are not to be inﬂuenced by eﬀects that are inﬁnitely removed, then they must obey a radiation condition. We can make the sameargument about the ﬁelds from (6.7) and (6.8). Let us allow one of the excluding surfaces,sayS N, to recede to inﬁnity (enclosing all of the sources as it expands). As SN→∞ any contributions from the ﬁelds on this surface to the ﬁelds at rshould vanish. Letting SNbe a sphere centered at the origin, we note that ˆn/prime=− ˆr/primeand that as r/prime→∞ G(r|r/prime;ω)=e−jk|r−r/prime| 4π|r−r/prime|≈e−jkr/prime 4πr/prime, ∇/primeG(r|r/prime;ω)=ˆR/parenleftbigg1+jkR 4πR2/parenrightbigg e−jkR≈− ˆr/prime/parenleftbigg1+jkr/prime r/prime/parenrightbigge−jkr/prime 4πr/prime. Substituting these expressions into (6.7) we ﬁnd that lim SN→S∞/contintegraldisplay SN/bracketleftbig (ˆn/prime×˜E)×∇/primeG+(ˆn/prime·˜E)∇/primeG−jω˜µ(ˆn/prime×˜H)G/bracketrightbig dS/prime ≈lim r/prime→∞/integraldisplay2π 0/integraldisplayπ 0/braceleftbigg/bracketleftbig (ˆr/prime×˜E)×ˆr/prime+(ˆr/prime·˜E)ˆr/prime/bracketrightbig/parenleftbigg1+jkr/prime r/prime/parenrightbigg +jω˜µ(ˆr/prime×˜H)/bracerightbige−jkr/prime 4πr/primer/prime2sinθ/primedθ/primedφ/prime ≈lim r/prime→∞/integraldisplay2π 0/integraldisplayπ 0/braceleftbig r/prime/bracketleftbig jk˜E+jω˜µ(ˆr/prime×˜H)/bracketrightbig +˜E/bracerightbige−jkr/prime 4πsinθ/primedθ/primedφ/prime. Since this gives the contribution to the ﬁeld in Vfrom the ﬁelds on the surface receding to inﬁnity, we expect that this term should be zero. If the medium has loss, then theexponential term decays and drives the contribution to zero. For a lossless medium thecontributions are zero if lim r→∞r˜E(r,ω)< ∞, (6.9) lim r→∞r/bracketleftbig ηˆr×˜H(r,ω)+˜E(r,ω)/bracketrightbig =0. (6.10) To accompany (6.8) we also have lim r→∞r˜H(r,ω)< ∞, (6.11) lim r→∞r/bracketleftbig η˜H(r,ω)−ˆr×˜E(r,ω)/bracketrightbig =0. (6.12) We refer to (6.9) and (6.11) as the ﬁniteness conditions , and to (6.10) and (6.12) as the Sommerfeld radiation condition , for the electromagnetic ﬁeld. They show that far from the sources the ﬁelds must behave as a wave TEM to the r-direction. We shall see in §6.2 that the waves are in fact spherical TEM waves . 6.1.3 Fields in the excluded region: the extinction theorem The Stratton–Chu formula provides a solution for the ﬁeld within the region V, external to the excluded regions. An interesting consequence of this formula, and one that helpsus identify the equivalence principle, is that it gives the null result ˜H=˜E=0when evaluated at points within the excluded regions. We can show this by considering two cases. In the ﬁrst case we do notexclude the particular region V m, but do exclude the remaining regions Vn,n/negationslash=m. Then the electric ﬁeld everywhere outside the remaining excluded regions (including at points within Vm) is, by (6.7), ˜E(r,ω)=/integraldisplay V+Vm/parenleftbigg −˜Ji m×∇/primeG+˜ρi ˜/epsilon1c∇/primeG−jω˜µ˜JiG/parenrightbigg dV/prime+ +/summationdisplay n/negationslash=m/integraldisplay Sn/bracketleftbig (ˆn/prime×˜E)×∇/primeG+(ˆn/prime·˜E)∇/primeG−jω˜µ(ˆn/prime×˜H)G/bracketrightbig dS/prime− −/summationdisplay n/negationslash=m1 jω˜/epsilon1c/contintegraldisplay /Gamma1na+/Gamma1nb(dl/prime·˜H)∇/primeG, r∈V+Vm. In the second case we apply the Stratton–Chu formula only to Vm, and exclude all other regions. We incur a sign change on the surface and line integrals compared to the ﬁrstcase because the normal is now directed oppositely. By (6.7) we have ˜E(r,ω)=/integraldisplay Vm/parenleftbigg −˜Ji m×∇/primeG+˜ρi ˜/epsilon1c∇/primeG−jω˜µ˜JiG/parenrightbigg dV/prime− −/integraldisplay Sm/bracketleftbig (ˆn/prime×˜E)×∇/primeG+(ˆn/prime·˜E)∇/primeG−jω˜µ(ˆn/prime×˜H)G/bracketrightbig dS/prime+ +1 jω˜/epsilon1c/contintegraldisplay /Gamma1na+/Gamma1nb(dl/prime·˜H)∇/primeG, r∈Vm. Each of the expressions for ˜Eis equally valid for points within Vm. Upon subtraction we get 0=/integraldisplay V/parenleftbigg −˜Ji m×∇/primeG+˜ρi ˜/epsilon1c∇/primeG−jω˜µ˜JiG/parenrightbigg dV/prime+ +N/summationdisplay n=1/integraldisplay Sn/bracketleftbig (ˆn/prime×˜E)×∇/primeG+(ˆn/prime·˜E)∇/primeG−jω˜µ(ˆn/prime×˜H)G/bracketrightbig dS/prime− −N/summationdisplay n=11 jω˜/epsilon1c/contintegraldisplay /Gamma1na+/Gamma1nb(dl/prime·˜H)∇/primeG, r∈Vm. This expression is exactly the Stratton–Chu formula (6.7) evaluated at points within the excluded region Vm. The treatment of ˜His analogous and is left as an exercise. Since we may repeat this for any excluded region, we ﬁnd that the Stratton–Chu formula returnsthe null ﬁeld when evaluated at points outside V. This is sometimes referred to as the vector Ewald–Oseen extinction theorem [90]. We must emphasize that the ﬁelds within the excluded regions are notgenerally equal to zero; the Stratton–Chu formula merely returns this result when evaluated there. 6.2 Fields in an unbounded medium Two special cases of the Stratton–Chu formula are important because of their applica- tion to antenna theory. The ﬁrst is that of sources radiating into an unbounded region.The second involves a bounded region with all sources excluded. We shall consider theformer here and the latter in §6.3. Assuming that there are no bounding surfaces in (6.7) and (6.8), except for one surface that has been allowed to recede to inﬁnity and therefore provides no surface contribution,we ﬁnd that the electromagnetic ﬁelds in unbounded space are given by ˜E=/integraldisplay V/parenleftbigg −˜Ji m×∇/primeG+˜ρi ˜/epsilon1c∇/primeG−jω˜µ˜JiG/parenrightbigg dV/prime, ˜H=/integraldisplay V/parenleftbigg ˜Ji×∇/primeG+˜ρi m ˜µ∇/primeG−jω˜/epsilon1c˜Ji mG/parenrightbigg dV/prime. We can view the right-hand sides as superpositions of the ﬁelds present in the cases where (1) electric sources are present exclusively, and (2) magnetic sources are presentexclusively. With ˜ρ i m=0and ˜Ji m=0we ﬁnd that ˜E=/integraldisplay V/parenleftbigg˜ρi ˜/epsilon1c∇/primeG−jω˜µ˜JiG/parenrightbigg dV/prime, (6.13) ˜H=/integraldisplay V˜Ji×∇/primeGd V/prime. (6.14) Using ∇/primeG=− ∇ Gwe can write ˜E(r,ω)=− ∇/integraldisplay V˜ρi(r/prime,ω) ˜/epsilon1c(ω)G(r|r/prime;ω)dV/prime−jω/integraldisplay V˜µ(ω) ˜Ji(r/prime,ω)G(r|r/prime;ω)dV/prime =− ∇ ˜φe(r,ω)−jω˜Ae(r,ω) , where ˜φe(r,ω)=/integraldisplay V˜ρi(r/prime,ω) ˜/epsilon1c(ω)G(r|r/prime;ω)dV/prime, ˜Ae(r,ω)=/integraldisplay V˜µ(ω) ˜Ji(r/prime,ω)G(r|r/prime;ω)dV/prime, (6.15) are the electric scalar and vector potential functions introduced in §5.2. Using ˜Ji×∇/primeG= −˜Ji×∇G=∇× (˜JiG)we have ˜H(r,ω)=1 ˜µ(ω)∇×/integraldisplay V˜µ(ω) ˜Ji(r/prime,ω)G(r|r/prime;ω)dV/prime =1 ˜µ(ω)∇× ˜Ae(r,ω) . (6.16) These expressions for the ﬁelds are identical to those of (5.56) and (5.57), and thus the integral formula for the electromagnetic ﬁelds produces a result identical to that obtainedusing potential relations. Similarly, with ˜ρ i=0,˜Ji=0we have ˜E=−/integraldisplay V˜Ji m×∇/primeGd V/prime, ˜H=/integraldisplay V/parenleftbigg˜ρi m ˜µ∇/primeG−jω˜/epsilon1c˜Ji mG/parenrightbigg dV/prime, or ˜E(r,ω)=−1 ˜/epsilon1c(ω)∇× ˜Ah(r,ω) , ˜H(r,ω)=− ∇ ˜φh(r,ω)−jω˜Ah(r,ω) , where ˜φh(r,ω)=/integraldisplay V˜ρi m(r/prime,ω) ˜µ(ω)G(r|r/prime;ω)dV/prime, ˜Ah(r,ω)=/integraldisplay V˜/epsilon1c(ω)˜Ji m(r/prime,ω)G(r|r/prime;ω)dV/prime, are the magnetic scalar and vector potentials introduced in §5.2. 6.2.1 The far-zone ﬁelds produced by sources in unbounded space Many antennas may be analyzed in terms of electric currents and charges radiating in unbounded space. Since antennas are used to transmit information over great distances,the ﬁelds far from the sources are often of most interest. Assume that the sources arecontained within a sphere of radius r scentered at the origin. We deﬁne the far zone of the sources to consist of all observation points satisfying both r/greatermuchrs(and thus r/greatermuchr/prime) and kr/greatermuch1. For points in the far zone we may approximate the unit vector ˆRdirected from the sources to the observation point by the unit vector ˆrdirected from the origin to the observation point. We may also approximate ∇/primeG=d dR/parenleftbigge−jkR 4πR/parenrightbigg ∇/primeR=ˆR/parenleftbigg1+jkR R/parenrightbigge−jkR 4πR≈ˆrjke−jkR 4πR=ˆrjkG. (6.17) Using this we can obtain expressions for ˜Eand ˜Hin the far zone of the sources. The approximation (6.17) leads directly to ˜ρi∇/primeG≈/bracketleftbigg j∇/prime·˜Ji ω/bracketrightbigg (ˆrjkG)=−k ωˆr/bracketleftbig ∇/prime·(G˜Ji)−˜Ji·∇/primeG/bracketrightbig . Substituting this into (6.13), again using (6.17) and also using the divergence theorem, we have ˜E(r,ω)≈−/integraldisplay Vjω˜µ/bracketleftbig˜Ji−ˆr(ˆr·˜Ji)/bracketrightbig Gd V/prime+ˆrk ω˜/epsilon1c/contintegraldisplay S(ˆn/prime·˜Ji)GdS/prime, where the surface Ssurrounds the volume Vthat contains the impressed sources. If we let this volume slightly exceed that needed to contain the sources, then we do notchange the value of the volume integral above; however, the surface integral vanishes since ˆn /prime·˜Ji=0everywhere on the surface. Using ˆr×(ˆr×˜Ji)=ˆr(ˆr·˜Ji)−˜Jiwe then obtain the far-zone expression ˜E(r,ω)≈jωˆr×/bracketleftbigg ˆr×/integraldisplay V˜µ(ω) ˜Ji(r/prime,ω)G(r|r/prime;ω)dV/prime/bracketrightbigg =jωˆr×/bracketleftbigˆr×˜Ae(r,ω)/bracketrightbig , where ˜Aeis the electric vector potential. The far-zone electric ﬁeld has no r-component, and it is often convenient to write ˜E(r,ω)≈− jω˜AeT(r,ω) (6.18) where ˜AeTis the vector component of ˜Aetransverse to the r-direction: ˜AeT=− ˆr×/bracketleftbigˆr×˜Ae/bracketrightbig =˜Ae−ˆr(ˆr·˜Ae)=ˆθ˜Aeθ+ˆφ˜Aeφ. We can approximate the magnetic ﬁeld in a similar fashion. Noting that ˜Ji×∇/primeG= ˜Ji×(jkˆrG)we have ˜H(r,ω)≈− jk ˜µ(ω)ˆr×/integraldisplay V˜µ(ω) ˜Ji(r/prime,ω)G(r|r/prime,ω)dV/prime ≈−1 ηjωˆr×˜Ae(r,ω) . With this we have ˜E(r,ω)=−ηˆr×˜H(r,ω) , ˜H(r,ω)=ˆr×˜E(r,ω) η, in the far zone. To simplify the computations involved, we often choose to approximate the vector potential in the far zone. Noting that R=/radicalbig (r−r/prime)·(r−r/prime)=/radicalBig r2+r/prime2−2(r·r/prime) and remembering that r/greatermuchr/primeforrin the far zone, we can use the leading terms of a binomial expansion of the square root to get R=r/radicalBigg 1−2(ˆr·r/prime) r+/parenleftbiggr/prime r/parenrightbigg2 ≈r/radicalbigg 1−2(ˆr·r/prime) r≈r/bracketleftbigg 1−ˆr·r/prime r/bracketrightbigg ≈r−ˆr·r/prime. (6.19) Thus the Green’s function may be approximated as G(r|r/prime;ω)≈e−jkr 4πrejkˆr·r/prime. (6.20) Here we have kept the approximation (6.19) intact in the phase of Gbut have used 1/R≈1/rin the amplitude of G. We must keep a more accurate approximation for the phase since k(ˆr·r/prime)may be an appreciable fraction of a radian. We thus have the far-zone approximation for the vector potential ˜Ae(r,ω)≈˜µ(ω)e−jkr 4πr/integraldisplay V˜Ji(r/prime,ω)ejkˆr·r/primedV/prime, which we may use in computing (6.18). Let us summarize the expressions for computing the far-zone ﬁelds: ˜E(r,ω)=− jω/bracketleftBig ˆθ˜Aeθ(r,ω)+ˆφ˜Aeφ(r,ω)/bracketrightBig , (6.21) ˜H(r,ω)=ˆr×˜E(r,ω) η, (6.22) ˜Ae(r,ω)=e−jkr 4πr˜µ(ω) ˜ae( θ,φ,ω) , (6.23) ˜ae(θ, φ, ω) =/integraldisplay V˜Ji(r/prime,ω)ejkˆr·r/primedV/prime. (6.24) Here ˜aeis called the directional weighting function . This function is independent of r and describes the angular variation, or pattern , of the ﬁelds. In the far zone ˜E,˜H,ˆrare mutually orthogonal. Because of this, and because the ﬁelds vary as e−jkr/r, the electromagnetic ﬁeld in the far zone takes the form of a spherical TEM wave, which is consistent with the Sommerfeld radiation condition. Power radiated by time-harmonic sources in unbounded space. In§5.2.1 we deﬁned the power radiated by a time-harmonic source in unbounded space as the totaltime-average power passing through a sphere of very large radius. We found that for aHertzian dipole the radiated power could be computed from the far-zone ﬁelds through P av=lim r→∞/integraldisplay2π 0/integraldisplayπ 0Sav·ˆrr2sinθdθdφ where Sav=1 2Re/braceleftbigˇE×ˇH∗/bracerightbig is the time-average Poynting vector. By superposition this holds for any localized source. Assuming a lossless medium and using phasor notation to describe the time-harmonic Figure 6.4: Dipole antenna in a lossless unbounded medium. ﬁelds we have, by (6.22), Sav=1 2Re/braceleftBiggˇE×(ˆr×ˇE∗) η/bracerightBigg =ˆrˇE·ˇE∗ 2η. Substituting from (6.21), we can also write Savin terms of the directional weighting function as Sav=ˆrˇω2 2η/parenleftbigˇAeθˇA∗ eθ+ˇAeφˇA∗ eφ/parenrightbig =ˆrk2η (4πr)2/parenleftbigg1 2ˇaeθˇa∗ eθ+1 2ˇaeφˇa∗ eφ/parenrightbigg . (6.25) We note that Savdescribes the variation of the power density with θ,φ, and is thus sometimes used as a descriptor of the power pattern of the sources. Example of a current source radiating into an unbounded medium: the dipole antenna. A common type of antenna consists of a thin wire of length 2land radius a, fedatthecenterbyavoltagegenerato rasshowninFigure6.4.Thegenerato rinduces an impressed current on the surface of the wire which in turn radiates an electromagneticwave. For very thin wires ( a/lessmuchλ,a/lessmuchl) embedded in a lossless medium, the current may be accurately approximated using a standing-wave distribution: ˜J i(r,ω)=ˆz˜I(ω)sin[k(l−|z|)]δ(x)δ(y). (6.26) We may compute the ﬁeld produced by the dipole antenna by ﬁrst ﬁnding the vector potential from (6.15) and then calculating the magnetic ﬁeld from (6.16). The electricﬁeld may then be found by the use of Ampere’s law. We assume a lossless medium with parameters µ,/epsilon1. Substituting the current expression into (6.15) and integrating over xand ywe ﬁnd that ˜Ae(r,ω)=ˆzµ˜I 4π/integraldisplayl −lsink(l−|z/prime|)e−jkR Rdz/prime(6.27) where R=/radicalbig (z−z/prime)2+ρ2andρ2=x2+y2. Using (6.16) we have ˜H=∇×1 µ˜Ae=− ˆφ1 µ∂˜Aez ∂ρ. Writing the sine function in (6.27) in terms of exponentials, we then have ˜Hφ=j˜I 8π/bracketleftBigg ejkl/integraldisplay0 −l∂ ∂ρe−jk(R−z/prime) Rdz/prime−e−jkl/integraldisplay0 −l∂ ∂ρe−jk(R+z/prime) Rdz/prime+ +ejkl/integraldisplayl 0∂ ∂ρe−jk(R+z/prime) Rdz/prime−e−jkl/integraldisplayl 0∂ ∂ρe−jk(R−z/prime) Rdz/prime/bracketrightBigg . Noting that ∂ ∂ρe−jk(R±z/prime) R=±ρ∂ ∂z/primee−jk(R±z/prime) R[R∓(z−z/prime)]=−ρ1+jkR R3e−jk(R±z/prime) we can write ˜Hφ=j˜Iρ 8π/bracketleftBigg −ejkl e−jk(R−z/prime) R[R+(z−z/prime)]/vextendsingle/vextendsingle/vextendsingle/vextendsingle0 −l−e−jkl e−jk(R+z/prime) R[R−(z−z/prime)]/vextendsingle/vextendsingle/vextendsingle/vextendsingle0 −l+ +ejkl e−jk(R+z/prime) R[R−(z−z/prime)]/vextendsingle/vextendsingle/vextendsingle/vextendsinglel 0+e−jkl e−jk(R−z/prime) R[R+(z−z/prime)]/vextendsingle/vextendsingle/vextendsingle/vextendsinglel 0/bracketrightBigg . Collecting terms and simplifying we get ˜Hφ(r,ω)=j˜I(ω) 4πρ/bracketleftbig e−jkR 1+e−jkR 2−(2 cos kl)e−jkr/bracketrightbig (6.28) where R1=/radicalbig ρ2+(z−l)2and R2=/radicalbig ρ2+(z+l)2. For points external to the dipole the source current is zero and thus ˜E(r,ω)=1 jω/epsilon1∇× ˜H(r,ω)=1 jω/epsilon1/braceleftbigg −ˆρ∂ ∂z˜Hφ(r,ω)+ˆz1 ρ∂ ∂ρ[ρ˜Hφ(r,ω)]/bracerightbigg . Performing the derivatives we have ˜Eρ(r,ω)=jη˜I(ω) 4π/bracketleftbiggz−l ρe−jkR 1 R1+z+l ρe−jkR 2 R2−z ρ(2 cos kl)e−jkr r/bracketrightbigg ,(6.29) ˜Ez(r,ω)=− jη˜I(ω) 4π/bracketleftbigge−jkR 1 R1+e−jkR 2 R2−(2 cos kl)e−jkr r/bracketrightbigg . (6.30) The work of specializing these expressions for points in the far zone is left as an exercise. Instead, we shall use the general far-zone expressions (6.21)–(6.24). Substituting (6.26) into (6.24) and carrying out the xand yintegrals we have the directional weighting function ˜ae(θ, φ, ω) =/integraldisplayl −lˆz˜I(ω)sink(l−|z/prime|)ejkz/primecosθdz/prime. Writing the sine functions in terms of exponentials we have ˜ae(θ, φ, ω) =ˆz˜I(ω) 2j/bracketleftbigg ejkl/integraldisplayl 0ejkz/prime(cosθ−1)dz/prime−e−jkl/integraldisplayl 0ejkz/prime(cosθ+1)dz/prime+ +ejkl/integraldisplay0 −lejkz/prime(cosθ+1)−e−jkl/integraldisplay0 −lejkz/prime(cosθ−1)/bracketrightbigg . Carrying out the integrals and simplifying, we obtain ˜ae( θ,φ,ω) =ˆz2˜I(ω) kF(θ,kl) sinθ where F(θ,kl)=cos(klcosθ)−coskl sinθ is called the radiation function . Using ˆz=ˆrcosθ−ˆθsinθwe ﬁnd that ˜aeθ( θ,φ,ω) =−2˜I(ω) kF(θ,kl), ˜aeφ(θ, φ, ω) =0. Thus we have from (6.23) and (6.21) the electric ﬁeld ˜E(r,ω)=ˆθjη˜I(ω) 2πe−jkr rF(θ,kl) (6.31) and from (6.22) the magnetic ﬁeld ˜H(r,ω)=ˆφj˜I(ω) 2πe−jkr rF(θ,kl). (6.32) We see that the radiation function contains all of the angular dependence of the ﬁeld and thus describes the pattern of the dipole. When the dipole is short compared to awavelength we may approximate the radiation function as F(θ,kl/lessmuch1)≈1− 1 2(klcosθ)2−1+1 2(kl)2 sinθ=1 2(kl)2sinθ. (6.33) So a short dipole antenna has the same pattern as a Hertzian dipole, whose far-zone electric ﬁeld is (5.93). We may also calculate the radiated power for time-harmonic ﬁelds. The time-average Poynting vector for the far-zone ﬁelds is, from (6.25), Sav=ˆrη|ˇI|2 8π2r2F2(θ,kl), and thus the radiated power is Pav=η|ˇI|2 4π/integraldisplayπ 0F2(θ,kl)sinθdθ. This expression cannot be computed in closed form. For a short dipole we may use (6.33) to approximate the power, but the result is somewhat misleading since the current on ashort dipole is much smaller than ˜I. A better measure of the strength of the current is its value at the center, or feedpoint , of the dipole. This input current is by (6.26) merely ˜I 0(ω)=˜I(ω)sin(kl). Using this we ﬁnd Pav≈η|ˇI0|2 4π1 4(kl)2/integraldisplayπ 0sin3θdθ=ηπ 3|ˇI0|2/parenleftbiggl λ/parenrightbigg2 . This is exactly 1/4of the power radiated by a Hertzian dipole of the same length and current amplitude (5.95). The factor of 1/4comes from the diﬀerence between the current of the dipole antenna, which is zero at each end, and the current on the Hertzian dipole,which is constant across the length of the antenna. It is more common to use a dipoleantenna that is a half wavelength long ( 2l=λ/2), since it is then nearly resonant. With this we have through numerical integration the free-space radiated power P av=η0|ˇI0|2 4π/integraldisplayπ 0cos2/parenleftbigπ 2cosθ/parenrightbig sinθdθ=36.6|ˇI0|2 and the radiation resistance Rr=2Pav |ˇI(z=0)|2=2Pav |ˇI0|2=73.2/Omega1. 6.3 Fields in a bounded, source-free region In§6.2 we considered the ﬁrst important special case of the Stratton–Chu formula: sources in an unbounded medium. We now consider the second important special caseof a bounded, source-free region. This case has important applications to the study ofmicrowave an tennas and, in its scalar form, to the study of the diﬀraction of light. 6.3.1 The vector Huygens principle We may derive the formula for a bounded, source-free region of space by specializing the general Stratton–Chu formulas. We assume that all sources of the ﬁelds are withinthe excluded regions and thus set the sources to zero within V. From (6.7)–(6.8) we have ˜E(r,ω)= N/summationdisplay n=1/integraldisplay Sn/bracketleftbig (ˆn/prime×˜E)×∇/primeG+(ˆn/prime·˜E)∇/primeG−jω˜µ(ˆn/prime×˜H)G/bracketrightbig dS/prime− −N/summationdisplay n=11 jω˜/epsilon1c/contintegraldisplay /Gamma1na+/Gamma1nb(dl/prime·˜H)∇/primeG, (6.34) and ˜H(r,ω)=N/summationdisplay n=1/integraldisplay Sn/bracketleftbig (ˆn/prime×˜H)×∇/primeG+(ˆn/prime·˜H)∇/primeG+jω˜/epsilon1c(ˆn/prime×˜E)G/bracketrightbig dS/prime+ +N/summationdisplay n=11 jω˜µ/contintegraldisplay /Gamma1na+/Gamma1nb(dl/prime·˜E)∇/primeG. (6.35) This is known as the vector Huygens principle after the Dutch physicist C. Huygens, who formulated his “secondary source concept” to explain the propagation of light. Accordingto his idea, published in Trait´e de la lumi` erein 1690, points on a propagating wavefront are secondary sources of spherical waves that add together in just the right way to producethe ﬁeld on any successive wavefront. We can interpret (6.34) and (6.35) in much thesame way. The ﬁeld at each point within V, where there are no sources, can be imagined to arise from spherical waves emanated from every point on the surface bounding V. The amplitudes of these waves are determined by the values of the ﬁelds on the boundaries.Thus, we may consider the boundary ﬁelds to be equivalent to secondary sources of theﬁelds within V. We will expand on this concept below by introducing the concept of equivalence and identifying the speciﬁc form of the secondary sources. 6.3.2 The Franz formula The vector Huygens principle as derived above requires secondary sources for the ﬁelds within Vthat involve both the tangential and normal components of the ﬁelds on the bounding surface. Since only tangential components are required to guarantee uniquenesswithin V, we seek an expression involving only ˆn×˜Hand ˆn×˜E. Physically, the normal component of the ﬁeld is equivalent to a secondary charge source on the surface whilethe tangential component is equivalent to a secondary current source. Since charge andcurrent are related by the continuity equation, speciﬁcation of the normal component issuperﬂuous. To derive a version of the vector Huygens principle that omits the normal ﬁelds we take the curl of (6.35) to get ∇× ˜H(r,ω)= N/summationdisplay n=1∇×/contintegraldisplay Sn(ˆn/prime×˜H)×∇/primeGdS/prime+N/summationdisplay n=1/contintegraldisplay Sn∇×/bracketleftbig (ˆn/prime·˜H)∇/primeG/bracketrightbig dS/prime+ +N/summationdisplay n=1∇×/contintegraldisplay Snjω˜/epsilon1c(ˆn/prime×˜E)GdS/prime+N/summationdisplay n=11 jω˜µ/contintegraldisplay /Gamma1na+/Gamma1nb∇×/bracketleftbig (dl/prime·˜E)∇/primeG/bracketrightbig dS/prime.(6.36) Now, using ∇/primeG=− ∇ Gand employing the vector identity (B.43) we can show that ∇×/bracketleftbig f(r/prime)∇/primeG(r|r/prime)/bracketrightbig =− f(r/prime)/braceleftbig ∇×/bracketleftbig ∇G(r|r/prime)/bracketrightbig/bracerightbig +/bracketleftbig ∇G(r|r/prime)/bracketrightbig ×∇ f(r/prime)=0, since ∇×∇ G=0and∇f(r/prime)=0. This implies that the second and fourth terms of (6.36) are zero. The ﬁrst term can be modiﬁed using ∇×/braceleftbig/bracketleftbigˆn/prime×˜H(r/prime)/bracketrightbig G(r|r/prime)/bracerightbig =G(r|r/prime)∇×/bracketleftbigˆn/prime×˜H(r/prime)/bracketrightbig −/bracketleftbigˆn/prime×˜H(r/prime)/bracketrightbig ×∇G(r|r/prime) =/bracketleftbigˆn/prime×˜H(r/prime)/bracketrightbig ×∇/primeG(r|r/prime), giving ∇× ˜H(r,ω)=N/summationdisplay n=1∇×/contintegraldisplay Sn∇×/bracketleftbig (ˆn/prime×˜H)G/bracketrightbig dS/prime+N/summationdisplay n=1∇×/contintegraldisplay Snjω˜/epsilon1c(ˆn/prime×˜E)GdS/prime. Finally, using Ampere’s law ∇× ˜H=jω˜/epsilon1c˜Ein the source free region V, and taking the curl in the ﬁrst term outside the integral, we have ˜E(r,ω)=N/summationdisplay n=1∇×∇×/contintegraldisplay Sn1 jω˜/epsilon1c(ˆn/prime×˜H)GdS/prime+N/summationdisplay n=1∇×/contintegraldisplay Sn(ˆn/prime×˜E)GdS/prime.(6.37) Similarly ˜H(r,ω)=−N/summationdisplay n=1∇×∇×/contintegraldisplay Sn1 jω˜µ(ˆn/prime×˜E)GdS/prime+N/summationdisplay n=1∇×/contintegraldisplay Sn(ˆn/prime×˜H)GdS/prime.(6.38) These expressions together constitute the Franz formula for the vector Huygens principle [192]. 6.3.3 Love’s equivalence principle Love’s equivalence principle allows us to identify the equivalent Huygens sources for the ﬁelds within a bounded, source-free region V. It then allows us to replace a problem in the bounded region with an “equivalent” problem in unbounded space where the source-excluding surfaces are replaced by equivalent sources. The ﬁeld produced by both thereal and the equivalent sources gives a ﬁeld in Videntical to that of the original problem. This is particularly useful since we know how to compute the ﬁelds within an unboundedregion by employing potential functions. We identify the equivalent sources by considering the electric and magnetic Hertzian potentials produced by electric and magnetic current sources. Consider an impressedelectric surface current ˜J eq sand a magnetic surface current ˜Jeq msﬂowing on the closed surface Sin a homogeneous, isotropic medium with permeability ˜µ(ω) and complex permittivity ˜/epsilon1c(ω). These sources produce ˜Πe(r,ω)=/contintegraldisplay S˜Jeq s(r/prime,ω) jω˜/epsilon1c(ω)G(r|r/prime;ω)dS/prime, (6.39) ˜Πh(r,ω)=/contintegraldisplay S˜Jeq ms(r/prime,ω) jω˜µ(ω)G(r|r/prime;ω)dS/prime, (6.40) which in turn can be used to ﬁnd ˜E=∇× (∇× ˜Πe)−jω˜µ∇× ˜Πh, ˜H=jω˜/epsilon1c∇× ˜Πe+∇× (∇× ˜Πh). Upon substitution we ﬁnd that ˜E(r,ω)=∇×∇×/contintegraldisplay S1 jω˜/epsilon1c/bracketleftbig˜Jeq sG/bracketrightbig dS/prime+∇×/contintegraldisplay S[−˜Jeq ms]GdS/prime, ˜H(r,ω)=− ∇×∇×/contintegraldisplay S1 jω˜µ/bracketleftbig −˜Jeq msG/bracketrightbig dS/prime+∇×/contintegraldisplay S˜Jeq sGdS/prime. These are identical to the Franz equations (6.37) and (6.38) if we identify ˜Jeq s=ˆn×˜H, ˜Jeq ms=− ˆn×˜E. (6.41) These are the equivalent source densities for the Huygens principle. We now state Love’s equivalence principle [39]. Consider the ﬁelds within a homoge- neous, source-free region Vwith parameters ( ˜/epsilon1c,˜µ) bounded by a surface S. We know how to compute the ﬁelds using the Franz formula and the surface ﬁelds. Now considera second problem in which the same surface Sexists in an unbounded medium with identical parameters. If the surface carries the equivalent sources (6.41) then the electro-magnetic ﬁelds within Vcalculated using the Hertzian potentials (6.39) and (6.40) are identical to those of the ﬁrst problem, while the ﬁelds calculated outside Vare zero. We see that this must be true since the Franz formulas and the ﬁeld/potential formulas areidentical, and the Franz formula (since it was derived from the Stratton–Chu formula)gives the null ﬁeld outside V. The two problems are equivalent in the sense that they produce identical ﬁelds within V. The ﬁelds produced by the equivalent sources obey the appropriate boundary condi- tions across S. From (2.194) and (2.195) we have the boundary conditions ˆn×(˜H 1−˜H2)=˜Js, ˆn×(˜E1−˜E2)=− ˜Jms. Here ˆnpoints inward to V,(˜E1,˜H1)are the ﬁelds within V, and (˜E2,˜H2)are the ﬁelds within the excluded region. If the ﬁelds produced by the equivalent sources within theexcluded region are zero, then the ﬁelds must obey ˆn×˜H 1=˜Jeq s, ˆn×˜E1=− ˜Jeq ms, which is true by the deﬁnition of (˜Jeq s,˜Jeq sm). Note that we can extend the equivalence principle to the case where the media are diﬀerent internal to Vthan external to V. See Chen [29]. With the equivalent sources identiﬁed we may compute the electromagnetic ﬁeld in Vusing standard techniques. Speciﬁcally, we may use the Hertzian potentials as shown above or, since the Hertzian potentials are a simple remapping of the vector potentials,we may use (5.60) and (5.61) to write ˜E=− jω k2/bracketleftbig ∇(∇·˜Ae)+k2˜Ae/bracketrightbig −1 ˜/epsilon1c∇× ˜Ah, ˜H=− jω k2/bracketleftbig ∇(∇·˜Ah)+k2˜Ah/bracketrightbig +1 ˜µ∇× ˜Ae, where ˜Ae(r,ω)=/contintegraldisplay S˜µ(ω) ˜Jeq s(r/prime,ω)G(r|r/prime;ω)dS/prime(6.42) =/contintegraldisplay S˜µ(ω) [ˆn/prime×˜H(r/prime,ω)]G(r|r/prime;ω)dS/prime, (6.43) ˜Ah(r,ω)=/contintegraldisplay S˜/epsilon1c(ω)˜Jeq ms(r/prime,ω)G(r|r/prime;ω)dS/prime(6.44) =/contintegraldisplay S˜/epsilon1c(ω)[−ˆn/prime×˜E(r/prime,ω)]G(r|r/prime;ω)dS/prime. (6.45) At points where the source is zero we can write the ﬁelds in the alternative form ˜E=− jω k2∇×∇× ˜Ae−1 ˜/epsilon1c∇× ˜Ah, (6.46) ˜H=− jω k2∇×∇× ˜Ah+1 ˜µ∇× ˜A. (6.47) Figure 6.5: Geometry for problem of an aperture in a perfectly conducting ground screen illuminated by an impressed source. By superposition, if there are volume sources within Vwe merely add the ﬁelds due to these sources as computed from the potential functions. 6.3.4 The Schelkunoﬀ equivalence principle With Love’s equivalence principle we create an equivalent problem by replacing an excluded region by equivalent electric and magnetic sources. These require knowledge ofboth the tangential electric and magnetic ﬁelds over the bounding surface. However, theuniqueness theorem says that only one of either the tangential electric or the tangentialmagnetic ﬁelds need be speciﬁed to make the ﬁelds within Vunique. Thus we may wonder whether it is possible to formulate an equivalent problem that involves only tangential ˜Eor tangential ˜H. It is indeed possible, as shown by Schelkunoﬀ [39, 169]. When we use the equivalent sources to form the equivalent problem, we know that they produce a null ﬁeld within the excluded region. Thus we may form a diﬀerent equivalentproblem by ﬁlling the excluded region with a perfect conductor, and keeping the sameequivalent sources. The boundary conditions across Sare not changed, and thus by the uniqueness theorem the ﬁelds within Vare not altered. However, the manner in which we must compute the ﬁelds within Vis changed. We can no longer use formulas for the ﬁelds produced by sources in free space, but must use formulas for ﬁelds producedby sources in the vicinity of a conducting body. In general this can be diﬃcult since itrequires the formation of a new Green’s function that satisﬁes the boundary conditionover the conducting body (which could possess a peculiar shape). Fortunately, we showedin§4.10.2 that an electric source adjacent and tangential to a perfect electric conductor produces no ﬁeld, hence we need not consider the equivalent electric sources ( ˆn×˜H) when computing the ﬁelds in V. Thus, in our new equivalent problem we need the single tangential ﬁeld −ˆn×˜E. This is the Schelkunoﬀ equivalence principle . There is one situation in which it is relatively easy to use the Schelkunoﬀ equivalence. Consider a perfectly conducting ground screen with an aperture in it, as shown in Figure 6.5.Weassum ethattheaperturehasbeenilluminate dinsomewaybyanelectromagnetic wave produced by sources in region 1 so that there are both ﬁelds within the apertureand electric current ﬂowing on the region-2 side of the screen due to diﬀraction fromthe edges of the aperture. We wish to compute the ﬁelds in region 2. We can create anequivalent problem by placing a planar surface S 0adjacent to the screen, but slightly oﬀset into region 2, and then closing the surface at inﬁnity so that all of the screen plusregion 1 is excluded. Then we replace region 1 with homogeneous space and place on S 0 the equivalent currents ˜Jeq s=ˆn×˜H,˜Jeq ms=− ˆn×˜E, where ˜Hand ˜Eare the ﬁelds on S0in the original problem. We note that over the portion of S0adjacent to the screen ˜Jeq ms=0 since ˆn×˜E=0, but that ˜Jeq s/negationslash=0. From the equivalent currents we can compute the ﬁelds in region 2 using the potential functions. However, it is often diﬃcult to determine ˜Jeq sover the conducting surface. If we apply Schelkunoﬀ’s equivalence, we can formulate a second equivalent problem in which we place into region 1 a perfect conductor. Then we have the equivalent source currents ˜Jeq sand ˜Jeq msadjacent and tangential to a perfect conductor. By the image theorem of §5.1.1 we can replace this problem by yet another equivalent problem in which the conductor is replaced by the images of ˜Jeq sand ˜Jeq msin homogeneous space. Since the image of the tangential electric current ˜Jeq sis oppositely directed, the ﬁelds of the electric current and its image cancel. Since the image of themagnetic current is in the same direction as ˜J eq ms, the ﬁelds produced by the magnetic current and its image add. We also note that ˜Jeq msis nonzero only over the aperture (since ˆn×˜E=0on the screen), and thus the ﬁeld in region 1 can be found from ˜E(r,ω)=−1 ˜/epsilon1c(ω)∇× ˜Ah(r,ω) , where ˜Ah(r,ω)=/integraldisplay S0˜/epsilon1c(ω)[−2ˆn/prime×˜Eap(r/prime,ω)]G(r|r/prime;ω)dS/prime and ˜Eapis the electric ﬁeld in the aperture in the original problem. We shall present an example in the next section. 6.3.5 Far-zone ﬁelds produced by equivalent sources The equivalence principle is useful for analyzing antennas with complicated source distributions. The sources may be excluded using a surface S, and then a knowledge of the ﬁelds over S(found, for example, by estimation or measurement) can be used to compute the ﬁelds external to the antenna. Here we describe how to compute these ﬁeldsin the far zone. Given that ˜J eq s=ˆn×˜Hand ˜Jeq ms=− ˆn×˜Eare the equivalent sources on S,w em a y compute the ﬁelds using the potentials (6.43) and (6.45). Using (6.20) these can beapproximated in the far zone ( r/greatermuchr /prime,kr/greatermuch1)as ˜Ae(r,ω)=˜µ(ω)e−jkr 4πr˜ae(θ, φ, ω), ˜Ah(r,ω)=˜/epsilon1c(ω)e−jkr 4πr˜ah(θ, φ, ω), (6.48) where ˜ae(θ, φ, ω) =/contintegraldisplay S˜Jeq s(r/prime,ω)ejkˆr·r/primedS/prime, ˜ah(θ, φ, ω) =/contintegraldisplay S˜Jeq sm(r/prime,ω)ejkˆr·r/primedS/prime, (6.49) are the directional weighting functions. To compute the ﬁelds from the potentials we must apply the curl operator. So we must evaluate ∇×/bracketleftbigge−jkr rV(θ, φ)/bracketrightbigg =e−jkr r∇× V(θ, φ) +∇/parenleftbigge−jkr r/parenrightbigg ×V(θ, φ). The curl of Vis proportional to 1/rin spherical coordinates, hence the ﬁrst term on the right is proportional to 1/r2. Since we are interested in the far-zone ﬁelds, this term can be discarded in favor of 1/r-type terms. Using ∇/parenleftbigge−jkr r/parenrightbigg =− ˆr/parenleftbigg1+jkr r/parenrightbigge−jkr r≈− ˆrjke−jkr r, kr/greatermuch1, we have ∇×/bracketleftbigge−jkr rV(θ, φ)/bracketrightbigg ≈− jkˆr×/bracketleftbigge−jkr rV(θ, φ)/bracketrightbigg . Using this approximation we also establish ∇×∇×/bracketleftbigge−jkr rV(θ, φ)/bracketrightbigg ≈−k2ˆr×ˆr×/bracketleftbigge−jkr rV(θ, φ)/bracketrightbigg =k2e−jkr rVT(θ, φ) where VT=V−ˆr(ˆr·V)is the vector component of Vtransverse to the r-direction. With these formulas we can approximate (6.46) and (6.47) as ˜E(r,ω)=− jω˜AeT(r,ω)+jk ˜/epsilon1c(ω)ˆr×˜Ah(r,ω) , (6.50) ˜H(r,ω)=− jω˜AhT(r,ω)−jk ˜µ(ω)ˆr×˜Ae(r,ω) . Note that ˆr×˜E=− jωˆr×˜AeT+jk ˜/epsilon1cˆr×ˆr×˜Ah. Since ˆr×˜AeT=ˆr×˜Aeand ˆr×ˆr×˜Ah=− ˜AhT,w eh a v e ˆr×˜E=η/bracketleftbigg −jω˜AhT−jk ˜µˆr×˜Ae/bracketrightbigg =η˜H. Thus ˜H=ˆr×˜E η and the electromagnetic ﬁeld in the far zone is a TEM spherical wave, as expected. Example of ﬁelds produced by equivalent sources: an aperture antenna. As an example of calculating the ﬁelds in a bounded region from equivalent sources, letus ﬁnd the far-zone ﬁeld in free space produced by a rectangular waveguide openingintoaperfectly-conductin ggroun dscreenofinﬁnit eextentasshowninFigure6.6.For simplicity assume the waveguide propagates a pure TE 10mode, and that all higher-order Figure 6.6: Aperture antenna consisting of a rectangular waveguide opening into a con- ducting ground screen of inﬁnite extent. modes excited when the guided wave is reﬂected at the aperture may be ignored. Thus the electric ﬁeld in the aperture S0is ˜Ea(x,y)=ˆyE0cos/parenleftBigπ ax/parenrightBig . We may compute the far-zone ﬁeld using the Schelkunoﬀ equivalence principle of §6.3.4. We exclude the region z<0+using a planar surface Swhich we close at inﬁnity. We then ﬁll the region z<0with a perfect conductor. By the image theory the equivalent electric sources on Scancel while the equivalent magnetic sources double. Since the only nonzero magnetic sources are on S0(since ˆn×˜E=0on the screen), we have the equivalent problem of the source ˜Jeq ms=−2ˆn×˜Ea=2ˆxE0cos/parenleftBigπ ax/parenrightBig onS0in free space, where the equivalence holds for z>0. We may ﬁnd the far-zone ﬁeld created by this equivalent current by ﬁrst computing the directional weighting function (6.49). Since ˆr·r/prime=ˆr·(x/primeˆx+y/primeˆy)=x/primesinθcosφ+y/primesinθsinφ, we ﬁnd that ˜ah(θ, φ, ω) =/integraldisplayb/2 −b/2/integraldisplaya/2 −a/2ˆx2E0cos/parenleftBigπ ax/prime/parenrightBig ejkx/primesinθcosφejky/primesinθsinφdx/primedy/prime =ˆx4πE0abcosπX π2−4(πX)2sinπY πY where X=a λsinθcosφ, Y=b λsinθsinφ. Hereλis the free-space wavelength. By (6.50) the electric ﬁeld is ˜E=jk0 /epsilon10ˆr×˜Ah where ˜Ahis given in (6.48). Using ˆr×ˆx=ˆφcosθcosφ+ˆθsinφ we ﬁnd that ˜E=jk0abE 0e−jkr r/parenleftBig ˆθsinφ+ˆφcosθcosφ/parenrightBigcos(πX) π2−4(πX)2sin(πY) πY. The magnetic ﬁeld is merely ˜H=(ˆr×˜E)/η. 6.4 Problems 6.1Beginning with the Lorentz reciprocity theorem, derive (6.8). 6.2Obtain (6.8) by substitution of (6.7) into Faraday’s law. 6.3Show that (6.8) returns the null result when evaluated within the excluded regions. 6.4Show that under the condition kr/greatermuch1the formula for the magnetic ﬁeld of a dipole antenna (6.28) reduces to (6.32), while the formulas for the electric ﬁelds (6.29)and (6.30) reduce to (6.31). 6.5Conside rthedipoleantennashowninFigure6.4.Instea dofastanding- wave current distribution, assume the antenna carries a traveling-wave current distribution ˜J i(r,ω)=ˆz˜I(ω)e−jk|z|δ(x)δ(y), −l≤z≤l. Find the electric and magnetic ﬁelds at all points away from the current distribution. Specialize the result for kr/greatermuch1. 6.6A circular loop of thin wire has radius aand lies in the z=0plane in free space. A current is induced on the wire with the density ˜J(r,ω)=ˆφ˜I(ω)cos[k0a(π−|φ|)]δ(r−a)δ(θ−π/2) r, |φ|≤π. Compute the far-zone ﬁelds produced by this loop antenna. Specialize your results for the electrically-small case of k0a/lessmuch1. Compute the time-average power radiated by, and the radiation resistance of, the electrically-small loop. 6.7Consider a plane wave with the ﬁelds ˜E=˜E0ˆxe−jkz, ˜H=˜E0 ηˆye−jkz, normally incident from z<0on a square aperture of side ain a PEC ground screen atz=0. Assume that the ﬁeld in the aperture is identical to the ﬁeld of the plane wave with the screen absent (this is called the Kirchhoﬀ approximation ). Compute the far-zone electromagnetic ﬁelds for z>0. 6.8Consider a coaxial cable of inner radius aand outer radius b, opening into a PEC ground plane at z=0. Assume that only the TEM wave exists in the line and that no higher-order modes are created when the wave reﬂ ects from the aperture. Compute the far-zone electric and magnetic ﬁelds of this aperture antenna. Appendix A Mathematical appendix A.1 The Fourier transform The Fourier transform permits us to decompose a complicated ﬁeld structure into elemental components. This can simplify the computation of ﬁelds and provide physicalinsight into their spatiotemporal behavior. In this section we review the properties ofthe transform and demonstrate its usefulness in solving ﬁeld equations. One-dimensional case Let fbe a function of a single variable x. The Fourier transform of f(x)is the function F(k)deﬁned by the integral F{f(x)}=F(k)=/integraldisplay∞ −∞f(x)e−jkxdx. (A.1) Note that xand the corresponding transform variable kmust have reciprocal units: if x is time in seconds, then kis atemporal frequency in radians per second; if xis a length in meters, then kis aspatial frequency in radians per meter. We sometimes refer to F(k) as the frequency spectrum off(x). Not every function has a Fourier transform. The existence of (A.1) can be guaranteed by a set of suﬃcient conditions such as the following: 1.fis absolutely integrable:/integraltext∞ −∞|f(x)|dx<∞; 2.fhas no inﬁnite discontinuities; 3.fhas at most ﬁnitely many discontinuities and ﬁnitely many extrema in any ﬁnite interval (a,b). While such rigor is certainly of mathematical value, it may be of less ultimate use to the engineer than the following heuristic observation oﬀered by Bracewell [22]: a good mathematical model of a physical process should be Fourier transformable . That is, if the Fourier transform of a mathematical model does not exist, the model cannot preciselydescribe a physical process. The usefulness of the transform hinges on our ability to recover fthrough the inverse transform: F −1{F(k)}= f(x)=1 2π/integraldisplay∞ −∞F(k)ejkxdk. (A.2) When this is possible we write f(x)↔F(k) and say that f(x)and F(k)form a Fourier transform pair. The Fourier integral theorem states that FF−1{f(x)}=F−1F{f(x)}= f(x), except at points of discontinuity of f. At a jump discontinuity the inversion formula returns the average value of the one-sided limits f(x+)and f(x−)off(x). At points of continuity the forward and inverse transforms are unique. Transform theorems and properties. We now review some basic facts pertaining to the Fourier transform. Let f(x)↔F(k)=R(k)+jX(k), and g(x)↔G(k). 1.Linearity .αf(x)+βg(x)↔αF(k)+βG(k)ifαandβare arbitrary constants. This follows directly from the linearity of the transform integral, and makes thetransform useful for solving linear diﬀerential equations (e.g., Maxwell’s equations). 2.Symmetry. The property F(x)↔2πf(−k)is helpful when interpreting transform tables in which transforms are listed only in the forward direction. 3.Conjugate function. We have f ∗(x)↔F∗(−k). 4.Real function. Iffis real, then F(−k)=F∗(k).Also, R(k)=/integraldisplay∞ −∞f(x)coskx dx,X(k)=−/integraldisplay∞ −∞f(x)sinkx dx, and f(x)=1 πRe/integraldisplay∞ 0F(k)ejkxdk. A real function is completely determined by its positive frequency spectrum. It is obviously advantageous to know this when planning to collect spectral data. 5.Real function with reﬂection symmetry. Iffis real and even, then X(k)≡0and R(k)=2/integraldisplay∞ 0f(x)coskx dx,f(x)=1 π/integraldisplay∞ 0R(k)coskx dk. Iffis real and odd, then R(k)≡0and X(k)=−2/integraldisplay∞ 0f(x)sinkx dx,f(x)=−1 π/integraldisplay∞ 0X(k)sinkx dk. (Recall that fis even if f(−x)=f(x)for all x. Similarly fis odd if f(−x)=− f(x) for all x.) 6.Causal function. Recall that fis causal if f(x)=0forx<0. (a) If fis real and causal, then X(k)=−2 π/integraldisplay∞ 0/integraldisplay∞ 0R(k/prime)cosk/primexsinkx dk/primedx, R(k)=−2 π/integraldisplay∞ 0/integraldisplay∞ 0X(k/prime)sink/primexcoskx dk/primedx. (b) If fis real and causal, and f(0)is ﬁnite, then R(k)and X(k)are related by theHilbert transforms X(k)=−1 πP.V./integraldisplay∞ −∞R(k) k−k/primedk/prime,R(k)=1 πP.V./integraldisplay∞ −∞X(k) k−k/primedk/prime. (c) If fis causal and has ﬁnite energy, it is not possible to have F(k)=0for k1<k<k2. That is, the transform of a causal function cannot vanish over an interval. A causal function is completely determined by the real or imaginary part of itsspectrum. As with item 4, this is helpful when performing calculations or mea-surements in the frequency domain. If the function is not band-limited however,truncation of integrals will give erroneous results. 7.Time-limited vs. band-limited functions. Assume t 2>t1.I f f(t)=0for both t<t1 and t>t2, then it is not possible to have F(k)=0for both k<k1and k>k2 where k2>k1. That is, a time-limited signal cannot be band-limited. Similarly, a band-limited signal cannot be time-limited. 8.Null function. If the forward or inverse transform of a function is identically zero, then the function is identically zero. This important consequence of the Fourier integral theorem is useful when solving homogeneous partial diﬀerential equationsin the frequency domain. 9.Space or time shift. For any ﬁxed x 0, f(x−x0)↔F(k)e−jkx 0. (A.3) A temporal or spatial shift aﬀects only the phase of the transform, not the magni- tude. 10.Frequency shift. For any ﬁxed k0, f(x)ejk0x↔F(k−k0). Note that if f↔Fwhere fis real, then frequency-shifting Fcauses fto be- come complex — again, this is important if Fhas been obtained experimentally or through computation in the frequency domain. 11.Similarity. We have f(αx)↔1 |α|F/parenleftbiggk α/parenrightbigg , where αis any real constant. “Reciprocal spreading” is exhibited by the Fourier transform pair; dilation in space or time results in compression in frequency, andvice versa. 12.Convolution. We have /integraldisplay ∞ −∞f1(x/prime)f2(x−x/prime)dx/prime↔F1(k)F2(k) and f1(x)f2(x)↔1 2π/integraldisplay∞ −∞F1(k/prime)F2(k−k/prime)dk/prime. The ﬁrst of these is particularly useful when a problem has been solved in the frequency domain and the solution is found to be a product of two or more functionsofk. 13.Parseval’s identity. We have /integraldisplay ∞ −∞|f(x)|2dx=1 2π/integraldisplay∞ −∞|F(k)|2dk. Computations of energy in the time and frequency domains always give the same result. 14.Diﬀerentiation. We have dnf(x) dxn↔(jk)nF(k)and (−jx)nf(x)↔dnF(k) dkn. The Fourier transform can convert a diﬀerential equation in the xdomain into an algebraic equation in the kdomain, and vice versa. 15.Integration. We have /integraldisplayx −∞f(u)du↔πF(k)δ(k)+F(k) jk where δ(k)is the Dirac delta or unit impulse. Generalized Fourier transforms and distributions. It is worth noting that many useful functions are not Fourier transformable in the sense given above. An example isthe signum function sgn(x)=/braceleftBigg −1,x<0, 1, x>0. Although this function lacks a Fourier transform in the usual sense, for practical purposes it may still be safely associated with what is known as a generalized Fourier transform .A treatment of this notion would be out of place here; however, the reader should certainlybe prepared to encounter an entry such as sgn(x)↔2/jk in a standard Fourier transform table. Other functions can be regarded as possessing transforms when generalized functions are permitted into the discussion. An important example of a generalized function is the Dirac delta δ(x), which has enormous value in describing distributions that are very thin, such as the charge layers often foundon conductor surfaces. We shall not delve into the intricacies of distribution theory.However, we can hardly avoid dealing with generalized functions; to see this we needlook no further than the simple function cosk 0xwith its transform pair cosk0x↔π[δ(k+k0)+δ(k−k0)]. The reader of this book must therefore know the standard facts about δ(x): that it acquires meaning only as part of an integrand, and that it satisﬁes the sifting property /integraldisplay∞ −∞δ(x−x0)f(x)dx=f(x0) for any continuous function f. With f(x)=1we obtain the familiar relation /integraldisplay∞ −∞δ(x)dx=1. With f(x)=e−jkxwe obtain /integraldisplay∞ −∞δ(x)e−jkxdx=1, thus δ(x)↔1. It follows that 1 2π/integraldisplay∞ −∞ejkxdk=δ(x). (A.4) Useful transform pairs. Some of the more common Fourier transforms that arise in the study of electromagnetics are given in Appendix C. These often involve the simplefunctions deﬁned here: 1.Unit step function U(x)=/braceleftBigg 1,x<0, 0,x>0.(A.5) 2.Signum function sgn(x)=/braceleftBigg −1,x<0, 1, x>0.(A.6) 3.Rectangular pulse function rect(x)=/braceleftBigg 1,|x|<1, 0,|x|>1.(A.7) 4.Triangular pulse function /Lambda1(x)=/braceleftBigg 1−|x|,|x|<1, 0, |x|>1.(A.8) 5.Sinc function sinc(x)=sinx x. (A.9) Transforms of multi-variable functions Fourier transformations can be performed over multiple variables by successive appli- cations of (A.1). For example, the two-dimensional Fourier transform over x1and x2of the function f(x1,x2,x3,..., xN)is the quantity F(kx1,kx2,x3,..., xN)given by /integraldisplay∞ −∞/bracketleftbigg/integraldisplay∞ −∞f(x1,x2,x3,..., xN)e−jkx1x1dx1/bracketrightbigg e−jkx2x2dx2 =/integraldisplay∞ −∞/integraldisplay∞ −∞f(x1,x2,x3,..., xN)e−jkx1x1e−jkx2x2dx1dx2. The two-dimensional inverse transform is computed by multiple application of (A.2), recovering f(x1,x2,x3,..., xN)through the operation 1 (2π)2/integraldisplay∞ −∞/integraldisplay∞ −∞F(kx1,kx2,x3,..., xN)ejkx1x1ejkx2x2dkx1dkx2. Higher-dimensional transforms and inversions are done analogously. Transforms of separable functions. If we are able to write f(x1,x2,x3,..., xN)=f1(x1,x3,..., xN)f2(x2,x3,..., xN), then successive transforms on the variables x1and x2result in f(x1,x2,x3,..., xN)↔F1(kx1,x3,..., xN)F2(kx2,x3,..., xN). In this case a multi-variable transform can be obtained with the help of a table of one- dimensional transforms. If, for instance, f(x,y,z)=δ(x−x/prime)δ(y−y/prime)δ(z−z/prime), then we obtain F(kx,ky,kz)=e−jkxx/primee−jkyy/primee−jkzz/prime by three applications of (A.1). A more compact notation for multi-dimensional functions and transforms makes use of the vector notation k=ˆxkx+ˆyky+ˆzkzand r=ˆxx+ˆyy+ˆzzwhere ris the position vector. In the example above, for instance, we could have written δ(x−x/prime)δ(y−y/prime)δ(z−z/prime)=δ(r−r/prime), and F(k)=/integraldisplay∞ −∞/integraldisplay∞ −∞/integraldisplay∞ −∞δ(r−r/prime)e−jk·rdx dydz =e−jk·r/prime. Fourier–Bessel transform. Ifx1and x2have the same dimensions, it may be con- venient to recast the two-dimensional Fourier transform in polar coordinates. Let x1= ρcosφ,kx1=pcosθ,x2=ρsinφ, and kx2=psinθ,where pandρare deﬁned on (0,∞) andφandθare deﬁned on (−π,π). Then F(p,θ,x3,..., xN)=/integraldisplayπ −π/integraldisplay∞ 0f(ρ, φ, x3,..., xN)e−jpρcos(φ−θ)ρdρdφ. (A.10) Iffis independent of φ(due to rotational symmetry about an axis transverse to x1and x2), then the φintegral can be computed using the identity J0(x)=1 2π/integraldisplayπ −πe−jxcos(φ−θ)dφ. Thus (A.10) becomes F(p,x3,..., xN)=2π/integraldisplay∞ 0f(ρ,x3,..., xN)J0(ρp)ρdρ, (A.11) showing that Fis independent of the angular variable θ. Expression (A.11) is termed theFourier–Bessel transform off. The reader can easily verify that fcan be recovered from Fthrough f(ρ,x3,..., xN)=/integraldisplay∞ 0F(p,x3,..., xN)J0(ρp)pdp, the inverse Fourier–Bessel transform. A review of complexcontour integration Some powerful techniques for the evaluation of integrals rest on complex variable the- ory. In particular, the computation of the Fourier inversion integral is often aided bythese techniques. We therefore provide a brief review of this material. For a fullerdiscussion the reader may refer to one of many widely available textbooks on complex analysis. We shall denote by f(z)a complex valued function of a complex variable z. That is, f(z)=u(x,y)+jv(x,y), where the real and imaginary parts u(x,y)andv(x,y)offare each functions of the real and imaginary parts xand yofz: z=x+jy=Re(z)+jIm(z). Here j=√ −1, as is mostly standard in the electrical engineering literature. Limits, diﬀerentiation, and analyticity. Letw=f(z), and let z0=x0+jy0and w0=u0+jv0be points in the complex zandwplanes, respectively. We say that w0is the limit of f(z)aszapproaches z0, and write lim z→z0f(z)=w0, if and only if both u(x,y)→u0andv(x,y)→v0asx→x0and y→y0independently. The derivative of f(z)at a point z=z0is deﬁned by the limit f/prime(z0)=lim z→z0f(z)−f(z0) z−z0, if it exists. Existence requires that the derivative be independent of direction of approach; that is, f/prime(z0)cannot depend on the manner in which z→z0in the complex plane. (This turns out to be a much stronger condition than simply requiring that the functions uand vbe diﬀerentiable with respect to the variables xand y.) We say that f(z)isanalytic atz0if it is diﬀerentiable at z0and at all points in some neighborhood of z0. Iff(z)is not analytic at z0but every neighborhood of z0contains a point at which f(z)is analytic, then z0is called a singular point of f(z). Laurent expansions and residues. Although Taylor series can be used to expand complex functions around points of analyticity, we must often expand functions aroundpoints z 0at or near which the functions fail to be analytic. For this we use the Laurent expansion , a generalization of the Taylor expansion involving both positive and negative powers of z−z0: f(z)=∞/summationdisplay n=−∞an(z−z0)n=∞/summationdisplay n=1a−n (z−z0)n+∞/summationdisplay n=0an(z−z0)n. The numbers anare the coeﬃcients of the Laurent expansion of f(z)at point z=z0. The ﬁrst series on the right is the principal part of the Laurent expansion, and the second series is the regular part . The regular part is an ordinary power series, hence it converges in some disk |z−z0|<Rwhere R≥0. Putting ζ=1/(z−z0), the principal part becomes/summationtext∞ n=1a−nζn; this power series converges for |ζ|<ρwhere ρ≥0, hence the principal part converges for |z−z0|>1/ρ/definesr. When r<R, the Laurent expansion converges in the annulus r<|z−z0|<R; when r>R, it diverges everywhere in the complex plane. The function f(z)has an isolated singularity at point z0iff(z)is not analytic at z0 but is analytic in the “punctured disk” 0<|z−z0|<Rfor some R>0. Isolated singularities are classiﬁed by reference to the Laurent expansion. Three types can arise: 1.Removable singularity. The point z0is a removable singularity of f(z)if the principal part of the Laurent expansion of f(z)about z0is identically zero (i.e., if an=0 forn=−1,−2,−3,...). 2.Pole of order k.The point z0is a pole of order kif the principal part of the Laurent expansion about z0contains only ﬁnitely many terms that form a polynomial of degree kin(z−z0)−1. A pole of order 1is called a simple pole . 3.Essential singularity. The point z0is an essential singularity of f(z)if the principal part of the Laurent expansion of f(z)about z0contains inﬁnitely many terms (i.e., ifa−n/negationslash=0for inﬁnitely many n). The coeﬃcient a−1in the Laurent expansion of f(z)about an isolated singular point z0 is the residue of f(z)atz0. It can be shown that a−1=1 2πj/contintegraldisplay /Gamma1f(z)dz (A.12) where /Gamma1is any simple closed curve oriented counterclockwise and containing in its interior z0and no other singularity of f(z). Particularly useful to us is the formula for evaluation of residues at pole singularities. If f(z)has a pole of order katz=z0, then the residue off(z)atz0is given by a−1=1 (k−1)!lim z→z0dk−1 dzk−1[(z−z0)kf(z)]. (A.13) Cauchy–Goursat and residue theorems. It can be shown that if f(z)is analytic at all points on and within a simple closed contour C, then /contintegraldisplay Cf(z)dz=0. This central result is known as the Cauchy–Goursat theorem . We shall not oﬀer a proof, but shall proceed instead to derive a useful consequence known as the residue theorem . Figure A.1: Derivation of the residue theorem. FigureA.1depict sasimpleclosedcurve C enclosing n isolate dsingularitie sofafunction f(z). We assume that f(z)is analytic on and elsewhere within C. Around each singular point zkwe have drawn a circle Ckso small that it encloses no singular point other than zk; taken together, the Ck(k=1,..., n) and Cform the boundary of a region in which f(z)is everywhere analytic. By the Cauchy–Goursat theorem /integraldisplay Cf(z)dz+n/summationdisplay k=1/integraldisplay Ckf(z)dz=0. Hence 1 2πj/integraldisplay Cf(z)dz=n/summationdisplay k=11 2πj/integraldisplay Ckf(z)dz, where now the integrations are all performed in a counterclockwise sense. By (A.12) /integraldisplay Cf(z)dz=2πjn/summationdisplay k=1rk (A.14) where r1,..., rnare the residues of f(z)at the singularities within C. Contour deformation. Suppose fis analytic in a region Dand/Gamma1is a simple closed curve in D.I f/Gamma1can be continuously deformed to another simple closed curve /Gamma1/primewithout passing out of D, then /integraldisplay /Gamma1/primef(z)dz=/integraldisplay /Gamma1f(z)dz. (A.15) Toseethis,conside rFigureA.2wherewehaveintroducedanothe rsetofcurves±γ; these new curves are assumed parallel and inﬁnitesimally close to each other. Let Cbe the composite curve consisting of /Gamma1,+γ,−/Gamma1/prime, and−γ, in that order. Since fis analytic on and within C, we have /integraldisplay Cf(z)dz=/integraldisplay /Gamma1f(z)dz+/integraldisplay +γf(z)dz+/integraldisplay −/Gamma1/primef(z)dz+/integraldisplay −γf(z)dz=0. But/integraltext −/Gamma1/primef(z)dz=−/integraltext /Gamma1/primef(z)dzand/integraltext −γf(z)dz=−/integraltext +γf(z)dz, hence (A.15) follows. The contour deformation principle often permits us to replace an integration contour byone that is more convenient. Figure A.2: Derivation of the contour deformation principle. Principal value integrals. We must occasionally carry out integrations of the form I=/integraldisplay∞ −∞f(x)dx where f(x)has a ﬁnite number of singularities xk(k=1,..., n) along the real axis. Such singularities in the integrand force us to interpret Ias an improper integral. With just one singularity present at point x1, for instance, we deﬁne /integraldisplay∞ −∞f(x)dx=lim ε→0/integraldisplayx1−ε −∞f(x)dx+lim η→0/integraldisplay∞ x1+ηf(x)dx provided that both limits exist. When both limits do not exist, we may still be able to obtain a well-deﬁned result by computing lim ε→0/parenleftbigg/integraldisplayx1−ε −∞f(x)dx+/integraldisplay∞ x1+εf(x)dx/parenrightbigg (i.e.,bytakingη=εsothatthelimitsare“symmetric”) .Thisquantityiscalledthe Cauchy principal value ofIand is denoted P.V./integraldisplay∞ −∞f(x)dx. More generally, we have P.V./integraldisplay∞ −∞f(x)dx=lim ε→0/parenleftbigg/integraldisplayx1−ε −∞f(x)dx+/integraldisplayx2−ε x1+εf(x)dx+ +···+/integraldisplayxn−ε xn−1+εf(x)dx+/integraldisplay∞ xn+εf(x)dx/parenrightbigg fornsingularities x1<···<xn. In a large class of problems f(z)(i.e., f(x)with xreplaced by the complex variable z) is analytic everywhere except for the presence of ﬁnitely many simple poles. Some of these may lie on the real axis (at points x1 <···< xn,say),andsomemaynot. Conside rnowtheintegratio ncontour C showninFigureA.3.Wechoose R solargeand εso small that Cencloses all the poles of fthat lie in the upper half of the complex Figure A.3: Complex plane technique for evaluating a principal value integral. plane. In many problems of interest the integral of faround the large semicircle tends to zero as R→∞ and the integrals around the small semicircles are well-behaved as ε→0. It may then be shown that P.V./integraldisplay∞ −∞f(x)dx=πjn/summationdisplay k=1rk+2πj/summationdisplay UHPrk where rkis the residue at the kth simple pole. The ﬁrst sum on the right accounts for the contributions of those poles that lie on the real axis; note that it is associated witha factor πjinstead of 2πj, since these terms arose from integrals over semicircles rather than over full circles. The second sum, of course, is extended only over those poles thatreside in the upper half-plane. Fourier transform solution of the 1-D waveequation Successive applications of the Fourier transform can reduce a partial diﬀerential equa- tion to an ordinary diﬀerential equation, and ﬁnally to an algebraic equation. Afterthe algebraic equation is solved by standard techniques, Fourier inversion can yield asolution to the original partial diﬀerential equation. We illustrate this by solving theone-dimensional inhomogeneous wave equation /parenleftbigg∂ 2 ∂z2−1 c2∂2 ∂t2/parenrightbigg ψ(x,y,z,t)=S(x,y,z,t), (A.16) where the ﬁeld ψis the desired unknown and Sis the known source term. For uniqueness of solution we must specify ψand∂ψ/∂ zover some z=constant plane. Assume that ψ(x,y,z,t)/vextendsingle/vextendsingle/vextendsingle z=0=f(x,y,t), (A.17) ∂ ∂zψ(x,y,z,t)/vextendsingle/vextendsingle/vextendsingle z=0=g(x,y,t). (A.18) We begin by positing inverse temporal Fourier transform relationships for ψand S: ψ(x,y,z,t)=1 2π/integraldisplay∞ −∞˜ψ(x,y,z,ω)ejωtdω, S(x,y,z,t)=1 2π/integraldisplay∞ −∞˜S(x,y,z,ω)ejωtdω. Substituting into (A.16), passing the derivatives through the integral, calculating the derivatives, and combining the inverse transforms, we obtain 1 2π/integraldisplay∞ −∞/bracketleftbigg/parenleftbigg∂2 ∂z2+k2/parenrightbigg ˜ψ(x,y,z,ω)−˜S(x,y,z,ω)/bracketrightbigg ejωtdω=0 where k=ω/c. By the Fourier integral theorem /parenleftbigg∂2 ∂z2+k2/parenrightbigg ˜ψ(x,y,z,ω)−˜S(x,y,z,ω)=0. (A.19) We have thus converted a partial diﬀerential equation into an ordinary diﬀerential equa- tion. A spatial transform on zwill now convert the ordinary diﬀerential equation into an algebraic equation. We write ˜ψ(x,y,z,ω)=1 2π/integraldisplay∞ −∞˜ψz(x,y,kz,ω)ejkzzdkz, ˜S(x,y,z,ω)=1 2π/integraldisplay∞ −∞˜Sz(x,y,kz,ω)ejkzzdkz, in (A.19), pass the derivatives through the integral sign, compute the derivatives, and set the integrand to zero to get (k2−k2 z)˜ψz(x,y,kz,ω)−˜Sz(x,y,kz,ω)=0; hence ˜ψz(x,y,kz,ω)=−˜Sz(x,y,kz,ω) (kz−k)(kz+k). (A.20) The price we pay for such an easy solution is that we must now perform a two- dimensional Fourier inversion to obtain ψ(x,y,z,t)from ˜ψz(x,y,kz,ω). It turns out to be easiest to perform the spatial inverse transform ﬁrst, so let us examine ˜ψ(x,y,z,ω)=1 2π/integraldisplay∞ −∞˜ψz(x,y,kz,ω)ejkzzdkz. By (A.20) we have ˜ψ(x,y,z,ω)=1 2π/integraldisplay∞ −∞[˜Sz(x,y,kz,ω)]/bracketleftbigg−1 (kz−k)(kz+k)/bracketrightbigg ejkzzdkz, where the integrand involves a product of two functions. With ˜gz(kz,ω)=−1 (kz−k)(kz+k), the convolution theorem gives ˜ψ(x,y,z,ω)=/integraldisplay∞ −∞˜S(x,y,ζ,ω) ˜g(z−ζ,ω) dζ (A.21) Figure A.4: Contour used to compute inverse transform in solution of the 1-D wave equation. where ˜g(z,ω)=1 2π/integraldisplay∞ −∞˜gz(kz,ω)ejkzzdkz=1 2π/integraldisplay∞ −∞−1 (kz−k)(kz+k)ejkzzdkz. To compute this integral we use complex plane techniques. The domain of integration extends along the real kz-axis in the complex kz-plane; because of the poles at kz=±k, we must treat the integral as a principal value integral. Denoting I(kz)=−ejkzz 2π(kz−k)(kz+k), we have /integraldisplay∞ −∞I(kz)dkz=lim/integraldisplay /Gamma1rI(kz)dkz =lim/integraldisplay−k−δ −/Delta1I(kz)dkz+lim/integraldisplayk−δ −k+δI(kz)dkz+lim/integraldisplay/Delta1 k+δI(kz)dkz where the limits take δ→0and/Delta1→∞. Our kz-plane contour takes detours around the poles using semicircles of radius δ, and is closed using a semicircle of radius /Delta1(Figure A.4).Notethatif z> 0,wemustclosethecontourintheupperhalf-plane. By Cauchy’s integral theorem /integraldisplay /Gamma1rI(kz)dkz+/integraldisplay /Gamma11I(kz)dkz+/integraldisplay /Gamma12I(kz)dkz+/integraldisplay /Gamma1/Delta1I(kz)dkz=0. Thus /integraldisplay∞ −∞I(kz)dkz=− lim δ→0/integraldisplay /Gamma11I(kz)dkz−lim δ→0/integraldisplay /Gamma12I(kz)dkz−lim /Delta1→∞/integraldisplay /Gamma1/Delta1I(kz)dkz. The contribution from the semicircle of radius /Delta1can be computed by writing kzin polar coordinates as kz=/Delta1ejθ: lim /Delta1→∞/integraldisplay /Gamma1/Delta1I(kz)dkz=1 2πlim /Delta1→∞/integraldisplayπ 0−ejz/Delta1ejθ (/Delta1ejθ−k)(/Delta1ejθ+k)j/Delta1ejθdθ. Using Euler’s identity we can write lim /Delta1→∞/integraldisplay /Gamma1/Delta1I(kz)dkz=1 2πlim /Delta1→∞/integraldisplayπ 0−e−/Delta1zsinθej/Delta1zcosθ /Delta12e2jθj/Delta1ejθdθ. Thus, as long as z>0the integrand will decay exponentially as /Delta1→∞, and lim /Delta1→∞/integraldisplay /Gamma1/Delta1I(kz)dkz→0. Similarly,/integraltext /Gamma1/Delta1I(kz)dkz→0when z<0if we close the semicircle in the lower half-plane. Thus, /integraldisplay∞ −∞I(kz)dkz=− lim δ→0/integraldisplay /Gamma11I(kz)dkz−lim δ→0/integraldisplay /Gamma12I(kz)dkz. (A.22) The integrals around the poles can also be computed by writing kzin polar coordinates. Writing kz=−k+δejθwe ﬁnd lim δ→0/integraldisplay /Gamma11I(kz)dkz=1 2πlim δ→0/integraldisplay0 π−ejz(−k+δejθ)jδejθ (−k+δejθ−k)(−k+δejθ+k)dθ =1 2π/integraldisplayπ 0e−jkz −2kjdθ=−j 4ke−jkz. Similarly, using kz=k+δejθ, we obtain lim δ→0/integraldisplay /Gamma12I(kz)dkz=j 4kejkz. Substituting these into (A.22) we have ˜g(z,ω)=j 4ke−jkz−j 4kejkz=1 2ksinkz, (A.23) valid for z>0.F o r z<0, we close in the lower half-plane instead and get ˜g(z,ω)=−1 2ksinkz. (A.24) Substituting (A.23) and (A.24) into (A.21) we obtain ˜ψ(x,y,z,ω)=/integraldisplayz −∞˜S(x,y,ζ,ω)sink(z−ζ) 2kdζ−1 2k/integraldisplay∞ z˜S(x,y,ζ,ω)sink(z−ζ) 2kdζ where we have been careful to separate the two cases considered above. To make things a bit easier when we apply the boundary conditions, let us rewrite the above expression.Splitting the domain of integration we write ˜ψ(x,y,z,ω)=/integraldisplay 0 −∞˜S(x,y,ζ,ω)sink(z−ζ) 2kdζ+/integraldisplayz 0˜S(x,y,ζ,ω)sink(z−ζ) kdζ− −/integraldisplay∞ 0˜S(x,y,ζ,ω)sink(z−ζ) 2kdζ. Expansion of the trigonometric functions then gives ˜ψ(x,y,z,ω)=/integraldisplayz 0˜S(x,y,ζ,ω)sink(z−ζ) kdζ+ +sinkz 2k/integraldisplay0 −∞˜S(x,y,ζ,ω) coskζdζ−coskz 2k/integraldisplay0 −∞˜S(x,y,ζ,ω) sinkζdζ− −sinkz 2k/integraldisplay∞ 0˜S(x,y,ζ,ω) coskζdζ+coskz 2k/integraldisplay∞ 0˜S(x,y,ζ,ω) sinkζdζ. The last four integrals are independent of z, so we can represent them with functions constant in z. Finally, rewriting the trigonometric functions as exponentials we have ˜ψ(x,y,z,ω)=/integraldisplayz 0˜S(x,y,ζ,ω)sink(z−ζ) kdζ+˜A(x,y,ω)e−jkz+˜B(x,y,ω)ejkz. (A.25) This formula for ˜ψwas found as a solution to the inhomogeneous ordinary diﬀerential equation (A.19). Hence, to obtain the complete solution we should add any possiblesolutions of the homogeneous diﬀerential equation. Since these are exponentials, (A.25)in fact represents the complete solution, where ˜Aand ˜Bare considered unknown and can be found using the boundary conditions. If we are interested in the frequency-domain solution to the wave equation, then we are done. However, since our boundary conditions (A.17) and (A.18) pertain to the timedomain, we must temporally inverse transform before we can apply them. Writing thesine function in (A.25) in terms of exponentials, we can express the time-domain solutionas ˜ψ(x,y,z,t)=/integraldisplay z 0F−1/braceleftbiggc 2˜S(x,y,ζ,ω) jωejω c(z−ζ)−c 2˜S(x,y,ζ,ω) jωe−jω c(z−ζ)/bracerightbigg dζ+ +F−1/braceleftbig˜A(x,y,ω)e−jω cz/bracerightbig +F−1/braceleftbig˜B(x,y,ω)ejω cz/bracerightbig . (A.26) A combination of the Fourier integration and time-shifting theorems gives the general identity F−1/braceleftbigg˜S(x,y,ζ,ω) jωe−jωt0/bracerightbigg =/integraldisplayt−t0 −∞S(x,y,ζ,τ) dτ, (A.27) where we have assumed that ˜S(x,y,ζ,0)=0. Using this in (A.26) along with the time- shifting theorem we obtain ψ(x,y,z,t)=c 2/integraldisplayz 0/braceleftBigg/integraldisplayt−ζ−z c −∞S(x,y,ζ,τ) dτ−/integraldisplayt−z−ζ c −∞S(x,y,ζ,τ) dτ/bracerightBigg dζ+ +a/parenleftBig x,y,t−z c/parenrightBig +b/parenleftBig x,y,t+z c/parenrightBig , or ψ(x,y,z,t)=c 2/integraldisplayz 0/integraldisplayt+z−ζ c t−z−ζ cS(x,y,ζ,τ) dτdζ+a/parenleftBig x,y,t−z c/parenrightBig +b/parenleftBig x,y,t+z c/parenrightBig (A.28) where a(x,y,t)=F−1[˜A(x,y,ω)], b(x,y,t)=F−1[˜B(x,y,ω)]. To calculate a(x,y,t)and b(x,y,t), we must use the boundary conditions (A.17) and (A.18). To apply (A.17), we put z=0into (A.28) to give a(x,y,t)+b(x,y,t)=f(x,y,t). (A.29) Using (A.18) is a bit more complicated since we must compute ∂ψ/∂ z, and zis a pa- rameter in the limits of the integral describing ψ. To compute the derivative we apply Leibnitz’ rule for diﬀerentiation: d dα/integraldisplayθ(α) φ(α)f(x,α)dx=/parenleftbiggdθ dα/parenrightbigg f(θ(α),α )−/parenleftbiggdφ dα/parenrightbigg f(φ(α),α )+/integraldisplayθ(α) φ(α)∂f ∂αdx.(A.30) Using this on the integral term in (A.28) we have ∂ ∂z/bracketleftBigg c 2/integraldisplayz 0/parenleftBigg/integraldisplayt+z−ζ c t−z−ζ cS(x,y,ζ,τ) dτ/parenrightBigg dζ/bracketrightBigg =c 2/integraldisplayz 0∂ ∂z/parenleftBigg/integraldisplayt+z−ζ c t−z−ζ cS(x,y,ζ,τ) dτ/parenrightBigg dζ, which is zero at z=0.T h u s ∂ψ ∂z/vextendsingle/vextendsingle/vextendsingle z=0=g(x,y,t)=−1 ca/prime(x,y,t)+1 cb/prime(x,y,t) where a/prime=∂a/∂tand b/prime=∂b/∂t. Integration gives −a(x,y,t)+b(x,y,t)=c/integraldisplayt −∞g(x,y,τ)dτ. (A.31) Equations (A.29) and (A.31) represent two algebraic equations in the two unknown functions aand b. The solutions are 2a(x,y,t)=f(x,y,t)−c/integraldisplayt −∞g(x,y,τ)dτ, 2b(x,y,t)=f(x,y,t)+c/integraldisplayt −∞g(x,y,τ)dτ. Finally, substitution of these into (A.28) gives us the solution to the inhomogeneous wave equation ψ(x,y,z,t)=c 2/integraldisplayz 0/integraldisplayt+z−ζ c t−z−ζ cS(x,y,ζ,τ) dτdζ+1 2/bracketleftBig f/parenleftBig x,y,t−z c/parenrightBig +f/parenleftBig x,y,t+z c/parenrightBig/bracketrightBig + +c 2/integraldisplayt+z c t−z cg(x,y,τ)dτ. (A.32) This is known as the D’Alembert solution . The terms f(x,y,t∓z/c)contribute to ψ as waves propagating away from the plane z=0in the ±z-directions, respectively. The integral over the forcing term Sis seen to accumulate values of Sover a time interval determined by z−ζ. The boundary conditions could have been applied while still in the temporal frequency domain (but not the spatial frequency domain, since the spatial position zis lost). But to do this, we would need the boundary conditions to be in the temporal frequency domain.This is easily accomplished by transforming them to give ˜ψ(x,y,z,ω)/vextendsingle/vextendsingle/vextendsingle z=0=˜f(x,y,ω) , ∂ ∂z˜ψ(x,y,z,ω)/vextendsingle/vextendsingle/vextendsingle z=0=˜g(x,y,ω) . Applying these to (A.25) (and again using Leibnitz’ rule) we have ˜A(x,y,ω)+˜B(x,y,ω)=˜f(x,y,ω) , −jk˜A(x,y,ω)+jk˜B(x,y,ω)=˜g(x,y,ω) , hence 2˜A(x,y,ω)=˜f(x,y,ω)−c˜g(x,y,ω) jω, 2˜B(x,y,ω)=˜f(x,y,ω)+c˜g(x,y,ω) jω. Finally, substituting these back into (A.25) and expanding the sine function we obtain the frequency-domain solution that obeys the given boundary conditions: ˜ψ(x,y,z,ω)=c 2/integraldisplayz 0/bracketleftbigg˜S(x,y,ζ,ω) ejω c(z−ζ) jω−˜S(x,y,ζ,ω) e−jω c(z−ζ) jω/bracketrightbigg dζ+ +1 2/bracketleftbig˜f(x,y,ω)ejω cz+˜f(x,y,ω)e−jω cz/bracketrightbig + +c 2/bracketleftbigg˜g(x,y,ω)ejω cz jω−˜g(x,y,ω)e−jω cz jω/bracketrightbigg . This is easily inverted using (A.27) to give (A.32). Fourier transform solution of the 1-D homogeneous wave equation for dissipative media Wave propagation in dissipative media can be studied using the one-dimensional wave equation /parenleftbigg∂2 ∂z2−2/Omega1 v2∂ ∂t−1 v2∂2 ∂t2/parenrightbigg ψ(x,y,z,t)=S(x,y,z,t). (A.33) This equation is nearly identical to the wave equation for lossless media studied in the previous section, except for the addition of the ∂ψ/∂ tterm. This extra term will lead to important physical consequences regarding the behavior of the wavesolutions. We shall solve (A.33) using the Fourier transform approach of the previous section, but to keep the solution simple we shall only consider the homogeneous problem. Webegin by writing ψin terms of its inverse temporal Fourier transform: ψ(x,y,z,t)=1 2π/integraldisplay∞ −∞˜ψ(x,y,z,ω)ejωtdω. Substituting this into the homogeneous version of (A.33) and taking the time derivatives, we obtain 1 2π/integraldisplay∞ −∞/bracketleftbigg (jω)2+2/Omega1(jω)−v2∂2 ∂z2/bracketrightbigg ˜ψ(x,y,z,ω)ejωtdω=0. The Fourier integral theorem leads to ∂2˜ψ(x,y,z,ω) ∂z2−κ2˜ψ(x,y,z,ω)=0 (A.34) where κ=1 v/radicalbig p2+2/Omega1p with p=jω. We can solve the homogeneous ordinary diﬀerential equation (A.34) by inspection: ˜ψ(x,y,z,ω)=˜A(x,y,ω)e−κz+˜B(x,y,ω)eκz. (A.35) Here ˜Aand ˜Bare frequency-domain coeﬃcients to be determined. We can either specify these coeﬃcients directly, or solve for them by applying speciﬁc boundary conditions.We examine each possibility below. Solution to waveequation by direct application of boundary conditions. The solution to the wave equation (A.33) will be unique if we specify functions f(x,y,t)and g(x,y,t)such that ψ(x,y,z,t)/vextendsingle/vextendsingle/vextendsingle z=0=f(x,y,t), ∂ ∂zψ(x,y,z,t)/vextendsingle/vextendsingle/vextendsingle z=0=g(x,y,t). (A.36) Assuming the Fourier transform pairs f(x,y,t)↔˜f(x,y,ω)andg(x,y,t)↔˜g(x,y,ω), we can apply the boundary conditions (A.36) in the frequency domain: ˜ψ(x,y,z,ω)/vextendsingle/vextendsingle/vextendsingle z=0=˜f(x,y,ω) , ∂ ∂z˜ψ(x,y,z,ω)/vextendsingle/vextendsingle/vextendsingle z=0=˜g(x,y,ω) . From these we ﬁnd ˜A+˜B=˜f, −κ˜A+κ˜B=˜g,v or ˜A=1 2/bracketleftbigg ˜f−˜g κ/bracketrightbigg , ˜B=1 2/bracketleftbigg ˜f+˜g κ/bracketrightbigg . Substitution into (A.35) gives ˜ψ(x,y,z,ω)=˜f(x,y,ω)coshκz+˜g(x,y,ω)sinhκz κ =˜f(x,y,ω)∂ ∂z˜Q(x,y,z,ω)+˜g(x,y,ω)˜Q(x,y,z,ω) =˜ψ1(x,y,z,ω)+˜ψ2(x,y,z,ω) where ˜Q=sinhκz/κ. Assuming that Q(x,y,z,t)↔˜Q(x,y,z,ω), we can employ the convolution theorem to immediately write down ψ(x,y,z,t): ψ(x,y,z,t)=f(x,y,t)∗∂ ∂zQ(x,y,z,t)+g(x,y,z,t)∗Q(x,y,z,t) =ψ1(x,y,z,t)+ψ2(x,y,z,t). (A.37) To ﬁnd ψwe must ﬁrst compute the inverse transform of ˜Q. Here we resort to a tabulated result [26]: sinh/bracketleftbig a√p+λ√p+µ/bracketrightbig √p+λ√p+µ↔1 2e−1 2(µ+λ)tJ0/parenleftbigg1 2(λ−µ)/radicalbig a2−t2/parenrightbigg ,−a<t<a. Here ais a positive, ﬁnite real quantity, and λandµare ﬁnite complex quantities. Outside the range |t|<athe time-domain function is zero. Letting a=z/v,µ=0, and λ=2/Omega1in the above expression, we ﬁnd Q(x,y,z,t)=v 2e−/Omega1tJ0/parenleftbigg/Omega1 v/radicalbig z2−v2t2/parenrightbigg [U(t+z/v)−U(t−z/v)] (A.38) where U(x)is the unit step function (A.5). From (A.37) we see that ψ2(x,y,z,t)=/integraldisplay∞ −∞g(x,y,t−τ)Q(x,y,z,τ)dτ=/integraldisplayz/v −z/vg(x,y,t−τ)Q(x,y,z,τ)dτ. Using the change of variables u=t−τand substituting (A.38), we then have ψ2(x,y,z,t)=v 2e−/Omega1t/integraldisplayt+z v t−z vg(x,y,u)e/Omega1uJ0/parenleftbigg/Omega1 v/radicalbig z2−(t−u)2v2/parenrightbigg du. (A.39) To ﬁnd ψ1we must compute ∂Q/∂z. Using the product rule we have ∂Q(x,y,z,t) ∂z=v 2e−/Omega1tJ0/parenleftbigg/Omega1 v/radicalbig z2−v2t2/parenrightbigg∂ ∂z[U(t+z/v)−U(t−z/v)]+ +v 2e−/Omega1t[U(t+z/v)−U(t−z/v)]∂ ∂zJ0/parenleftbigg/Omega1 v/radicalbig z2−v2t2/parenrightbigg . Next, using dU(x)/dx=δ(x)and remembering that J/prime 0(x)=− J1(x)and J0(0)=1,w e can write ∂Q(x,y,z,t) ∂z=1 2e−/Omega1t[δ(t+z/v)+δ(t−z/v)]− −z/Omega12 2ve−/Omega1tJ1/parenleftBig /Omega1 v√ z2−v2t2/parenrightBig /Omega1 v√ z2−v2t2[U(t+z/v)−U(t−z/v)]. Convolving this expression with f(x,y,t)we obtain ψ1(x,y,z,t)=1 2e−/Omega1 vzf/parenleftBig x,y,t−z v/parenrightBig +1 2e/Omega1 vzf/parenleftBig x,y,t+z v/parenrightBig − −z/Omega12 2ve−/Omega1t/integraldisplayt+z v t−z vf(x,y,u)e/Omega1uJ1/parenleftBig /Omega1 v/radicalbig z2−(t−u)2v2/parenrightBig /Omega1 v/radicalbig z2−(t−u)2v2du.(A.40) Finally, adding (A.40) and (A.39), we obtain ψ(x,y,z,t)=1 2e−/Omega1 vzf/parenleftBig x,y,t−z v/parenrightBig +1 2e/Omega1 vzf/parenleftBig x,y,t+z v/parenrightBig − −z/Omega12 2ve−/Omega1t/integraldisplayt+z v t−z vf(x,y,u)e/Omega1uJ1/parenleftBig /Omega1 v/radicalbig z2−(t−u)2v2/parenrightBig /Omega1 v/radicalbig z2−(t−u)2v2du+ +v 2e−/Omega1t/integraldisplayt+z v t−z vg(x,y,u)e/Omega1uJ0/parenleftbigg/Omega1 v/radicalbig z2−(t−u)2v2/parenrightbigg du. (A.41) Note that when /Omega1=0this reduces to ψ(x,y,z,t)=1 2f/parenleftBig x,y,t−z v/parenrightBig +1 2f/parenleftBig x,y,t+z v/parenrightBig +v 2/integraldisplayt+z v t−z vg(x,y,u)du, which matches (A.32) for the homogeneous case where S=0. Solution to wave equation by speciﬁcation of waveamplitudes. An alternative to direct speciﬁcation of boundary conditions is speciﬁcation of the amplitude functions ˜A(x,y,ω)and ˜B(x,y,ω)or their inverse transforms A(x,y,t)and B(x,y,t). If we specify the time-domain functions we can write ψ(x,y,z,t)as the inverse transform of (A.35). For example, a wave traveling in the +z-direction behaves as ψ(x,y,z,t)=A(x,y,t)∗F+(x,y,z,t) (A.42) where F+(x,y,z,t)↔e−κz=e−z v√ p2+2/Omega1p. We can ﬁnd F+using the following Fourier transform pair [26]): e−x v√ (p+ρ)2−σ2↔e−ρ vxδ(t−x/v)+σx ve−ρtI1/parenleftBig σ/radicalbig t2−(x/v)2/parenrightBig /radicalbig t2−(x/v)2,x v<t.(A.43) Here x is real and positive and I1(x)is the modiﬁed Bessel function of the ﬁrst kind and order 1. Outside the range x/v < tthe time-domain function is zero. Letting ρ=/Omega1and σ=/Omega1we ﬁnd F+(x,y,z,t)=/Omega12z ve−/Omega1tI1(/Omega1/radicalbig t2−(z/v)2) /Omega1/radicalbig t2−(z/v)2U(t−z/v)+e−/Omega1 vzδ(t−z/v). (A.44) Note that F+is a real functions of time, as expected. Substituting (A.44) into (A.42) and writing the convolution in integral form we have ψ(x,y,z,t)=/integraldisplay∞ z/vA(x,y,t−τ)/bracketleftBigg /Omega12z ve−/Omega1τI1(/Omega1/radicalbig τ2−(z/v)2) /Omega1/radicalbig τ2−(z/v)2/bracketrightBigg dτ+ +e−/Omega1 vzA/parenleftBig x,y,t−z v/parenrightBig ,z>0. (A.45) The 3-D Green’s function for waves in dissipative media To understand the ﬁelds produced by bounded sources within a dissipative medium we may wish to investigate solutions to the wave equation in three dimensions. The Green’sfunction approach requires the solution to /parenleftbigg ∇ 2−2/Omega1 v2∂ ∂t−1 v2∂2 ∂t2/parenrightbigg G(r|r/prime;t)=−δ(t)δ(r−r/prime) =−δ(t)δ(x−x/prime)δ(y−y/prime)δ(z−z/prime). That is, we are interested in the impulse response of a point source located at r=r/prime. We begin by substituting the inverse temporal Fourier transform relations G(r|r/prime;t)=1 2π/integraldisplay∞ −∞˜G(r|r/prime;ω)ejωtdω, δ(t)=1 2π/integraldisplay∞ −∞ejωtdω, obtaining 1 2π/integraldisplay∞ −∞/bracketleftbigg/parenleftbigg ∇2−jω2/Omega1 v2−1 v2(jω)2/parenrightbigg ˜G(r|r/prime;ω)+δ(r−r/prime)/bracketrightbigg ejωtdω=0. By the Fourier integral theorem we have (∇2+k2)˜G(r|r/prime;ω)=−δ(r−r/prime). (A.46) This is known as the Helmholtz equation . Here k=1 v/radicalbig ω2−j2ω/Omega1 (A.47) is called the wavenumber . To solve the Helmholtz equation we write ˜Gin terms of a 3-dimensional inverse Fourier transform. Substitution of ˜G(r|r/prime;ω)=1 (2π)3/integraldisplay∞ −∞˜Gr(k|r/prime;ω)ejk·rd3k, δ(r−r/prime)=1 (2π)3/integraldisplay∞ −∞ejk·(r−r/prime)d3k, into (A.46) gives 1 (2π)3/integraldisplay∞ −∞/bracketleftBig ∇2/parenleftbig˜Gr(k|r/prime;ω)ejk·r/parenrightbig +k2˜Gr(k|r/prime;ω)ejk·r+ejk·(r−r/prime)/bracketrightBig d3k=0. Here k=ˆxkx+ˆyky+ˆzkz with|k|2=k2 x+k2 y+k2 z=K2. Carrying out the derivatives and invoking the Fourier integral theorem we have (K2−k2)˜Gr(k|r/prime;ω)=e−jk·r/prime. Solving for ˜Gand substituting it into the inverse transform relation we have ˜G(r|r/prime;ω)=1 (2π)3/integraldisplay∞ −∞ejk·(r−r/prime) (K−k)(K+k)d3k. (A.48) To compute the inverse transform integral in (A.48) we write the 3-D transform variable in spherical coordinates: k·(r−r/prime)=KRcosθ, d3k=K2sinθdK dθdφ, where R=|r−r/prime|andθis the angle between kand r−r/prime. Hence (A.48) becomes ˜G(r|r/prime;ω)=1 (2π)3/integraldisplay∞ 0K2dK (K−k)(K+k)/integraldisplay2π 0dφ/integraldisplayπ 0ejKR cosθsinθdθ =2 (2π)2R/integraldisplay∞ 0Ksin(KR) (K−k)(K+k)dK, or, equivalently, ˜G(r|r/prime;ω)=1 2jR(2π)2/integraldisplay∞ −∞ejKR (K−k)(K+k)Kd K − −1 2jR(2π)2/integraldisplay∞ −∞e−jkR (K−k)(K+k)Kd K. We can compute the integrals over Kusing the complex plane technique. We consider K to be a complex variable, and note that for dissipative media we have k=kr+jki, where kr>0and ki<0. Thus the integrand has poles at K=±k. For the integral involving e+jKRwe close the contour in the upper half-plane using a semicircle of radius /Delta1and use Cauchy’s residue theorem. Then at all points on the semicircle the integrand decaysexponentially as /Delta1→∞, and there is no contribution to the integral from this part of the contour. The real-line integral is thus equal to 2πjtimes the residue at K=−k: /integraldisplay ∞ −∞ejKR (K−k)(K+k)Kd K =2πje−jkR −2k(−k). For the term involving e−jKRwe close in the lower half-plane and again the contribution from the inﬁnite semicircle vanishes. In this case our contour is clockwise and so the realline integral is −2πjtimes the residue at K=k: /integraldisplay ∞ −∞e−jKR (K−k)(K+k)Kd K =−2πje−jkR 2kk. Thus ˜G(r|r/prime;ω)=e−jkR 4πR. (A.49) Note that if /Omega1=0then this reduces to ˜G(r|r/prime;ω)=e−jωR/v 4πR. (A.50) Our last step is to ﬁnd the temporal Green’s function. Let p=jω. Then we can write ˜G(r|r/prime;ω)=eκR 4πR where κ=− jk=1 v/radicalbig p2+2/Omega1p. We may ﬁnd the inverse transform using (A.43). Letting x=R,ρ=/Omega1, and σ=/Omega1we ﬁnd G(r|r/prime;t)=e−/Omega1 vRδ(t−R/v) 4πR+/Omega12 4πve−/Omega1tI1/parenleftBig /Omega1/radicalbig t2−(R/v)2/parenrightBig /Omega1/radicalbig t2−(R/v)2U/parenleftbigg t−R v/parenrightbigg . We note that in the case of no dissipation where /Omega1=0this reduces to G(r|r/prime;t)=δ(t−R/v) 4πR which is the inverse transform of (A.50). Fourier transform representation of the static Green’s function In the study of static ﬁelds, we shall be interested in the solution to the partial diﬀer- ential equation ∇2G(r|r/prime)=−δ(r−r/prime)=−δ(x−x/prime)δ(y−y/prime)δ(z−z/prime). (A.51) Here G(r|r/prime), called the “static Green’s function,” represents the potential at point r produced by a unit point source at point r/prime. In Chapter 3 we ﬁnd that G(r|r/prime)=1/4π|r−r/prime|. In a variety of problems it is also useful to have Gwritten in terms of an inverse Fourier transform over the variables x and y. Letting Grform a three-dimensional Fourier transform pair with G, we can write G(r|r/prime)=1 (2π)3/integraldisplay∞ −∞Gr(kx,ky,kz|r/prime)ejkxxejkyyejkzzdkxdkydkz. Substitution into (A.51) along with the inverse transformation representation for the delta function (A.4) gives 1 (2π)3∇2/integraldisplay∞ −∞Gr(kx,ky,kz|r/prime)ejkxxejkyyejkzzdkxdkydkz =−1 (2π)3/integraldisplay∞ −∞ejkx(x−x/prime)ejky(y−y/prime)ejkz(z−z/prime)dkxdkydkz. We then combine the integrands and move the Laplacian operator through the integral to obtain 1 (2π)3/integraldisplay∞ −∞/bracketleftBig ∇2/parenleftbig Gr(k|r/prime)ejk·r/parenrightbig +ejk·(r−r/prime)/bracketrightBig d3k=0, where k=ˆxkx+ˆyky+ˆzkz. Carrying out the derivatives, 1 (2π)3/integraldisplay∞ −∞/bracketleftBig/parenleftbig −k2 x−k2 y−k2 z/parenrightbig Gr(k|r/prime)+e−jk·r/prime/bracketrightBig ejk·rd3k=0. Letting k2 x+k2 y=k2 ρand invoking the Fourier integral theorem we get the algebraic equation /parenleftbig −k2 ρ−k2 z/parenrightbig Gr(k|r/prime)+e−jk·r/prime=0, which we can easily solve for Gr: Gr(k|r/prime)=e−jk·r/prime k2ρ+k2z. (A.52) Equation (A.52) gives us a 3-D transform representation for the Green’s function. Since we desire the 2-D representation, we shall have to perform the inverse transformover k z. Writing Gxy(kx,ky,z|r/prime)=1 2π/integraldisplay∞ −∞Gr(kx,ky,kz|r/prime)ejkzzdkz we have Gxy(kx,ky,z|r/prime)=1 2π/integraldisplay∞ −∞e−jkxx/primee−jkyy/primeejkz(z−z/prime) k2ρ+k2zdkz. (A.53) To compute this integral, we let kzbe a complex variable and consider a closed contour in the complex plane, consisting of a semicircle and the real axis. As previously discussed,we compute the principal value integral as the semicircle radius /Delta1→∞, and ﬁnd that the contribution along the semicircle reduces to zero. Hence we can use Cauchy’s residuetheorem (A.14) to obtain the real-line integral: G xy(kx,ky,z|r/prime)=2πjres/braceleftBigg 1 2πe−jkxx/primee−jkyy/primeejkz(z−z/prime) k2ρ+k2z/bracerightBigg . Here res{f(kz)}denotes the residues of the function f(kz). The integrand in (A.53) has poles of order 1at kz=± jkρ,kρ≥0.I fz−z/prime>0we close in the upper half-plane and enclose only the pole at kz=jkρ. Computing the residue using (A.13), we obtain Gxy(kx,ky,z|r/prime)=je−jkxx/primee−jkyy/primee−kρ(z−z/prime) 2jkρ, z>z/prime. Since z>z/primethis function decays for increasing z, as expected physically. For z−z/prime<0we close in the lower half-plane, enclosing the pole at kz=− jkρand incurring an additional negative sign since our contour is now clockwise. Evaluating the residue we have Gxy(kx,ky,z|r/prime)=− je−jkxx/primee−jkyy/primeekρ(z−z/prime) −2jkρ, z<z/prime. We can combine both cases z>z/primeand z<z/primeby using the absolute value function: Gxy(kx,ky,z|r/prime)=e−jkxx/primee−jkyy/primee−kρ|z−z/prime| 2kρ. (A.54) Finally, we substitute (A.54) into the inverse transform formula. This gives the Green’s function representation G(r|r/prime)=1 4π|r−r/prime|=1 (2π)2/integraldisplay∞ −∞e−kρ|z−z/prime| 2kρejkρ·(r−r/prime)d2kρ, (A.55) where kρ=ˆxkx+ˆyky,kρ=|kρ|, and d2kρ=dkxdky. On occasion we may wish to represent the solution of the homogeneous (Laplace) equation ∇2ψ(r)=0 in terms of a 2-D Fourier transform. In this case we represent ψas a 2-D inverse transform and substitute to obtain 1 (2π)2/integraldisplay∞ −∞∇2/parenleftbig ψxy(kx,ky,z)ejkxxejkyy/parenrightbig dkxdky=0. Carrying out the derivatives and invoking the Fourier integral theorem we ﬁnd that /parenleftbigg∂2 ∂z2−k2 ρ/parenrightbigg ψxy(kx,ky,z)=0. Hence ψxy(kx,ky,z)=Aekρz+Be−kρz where Aand Bare constants with respect to z. Inverse transformation gives ψ(r)=1 (2π)2/integraldisplay∞ −∞/bracketleftbig A(kρ)ekρz+B(kρ)e−kρz/bracketrightbig ejkρ·rd2kρ. (A.56) A.2 Vector transport theorems We are often interested in the time rate of change of some ﬁeld integrated over a moving volume or surface. Such a derivative may be used to describe the transport of aphysical quantity (e.g., charge, momentum, energy) through space. Many of the relevanttheorems are derived in this section. The results ﬁnd application in the development ofthe large-scale forms of Maxwell equations, the continuity equation, and the Poyntingtheorem. Partial, total, and material derivatives The key to understanding transport theorems lies in the diﬀerence between the various means of time-diﬀerentiating a ﬁeld. Consider a scalar ﬁeld T(r,t)(which could represent one component of a vector or dyadic ﬁeld). If we ﬁx our position within the ﬁeld andexamine how the ﬁeld varies with time, we describe the partial derivative ofT. However, this may not be the most useful means of measuring the time rate of change of a ﬁeld.For instance, in mechanics we might be interested in the rate at which water cools asit sinks to the bottom of a container. In this case, Tcould represent temperature. We could create a “depth proﬁle” at any given time (i.e., measure T(r,t 0)for some ﬁxed t0) by taking simultaneous data from a series of temperature probes at varying depths. Wecould also create a temporal proﬁle at any given depth (i.e., measure T(r 0,t)for some ﬁxed r0) by taking continuous data from a probe ﬁxed at that depth. But neither of these would describe how an individual sinking water particle “experiences” a change intemperature over time. Instead, we could use a probe that descends along with a particular water packet (i.e., volume element), measuring the time rate of temperature change of that element. Thisrate of change is called the convective ormaterial derivative , since it corresponds to a situation in which a physical material quantity is followed as the derivative is calculated.We anticipate that this quantity will depend on (1) the time rate of change of Tat each ﬁxed point that the particle passes, and (2) the spatial rate of change of Tas well as the rapidity with which the packet of interest is swept through that space gradient. Thefaster the packet descends, or the faster the temperature cools with depth, the larger thematerial derivative should be. To compute the material derivative we describe the position of a water packet by the vector r(t)=ˆxx(t)+ˆyy(t)+ˆzz(t). Because no two packets can occupy the same place at the same time, the speciﬁcation of r(0)=r 0uniquely describes (or “tags”) a particular packet. The time rate of change of r with r0held constant (the material derivative of the position vector) is thus the velocity ﬁeld u(r,t)of the ﬂuid: /parenleftbiggdr dt/parenrightbigg r0=Dr Dt=u. (A.57) Here we use the “big D” notation to denote the material derivative, thereby avoiding confusion with the partial and total derivatives described below. To describe the time rate of change of the temperature of a particular water packet, we only need to hold r0constant while we examine the change. If we write the temperature as T(r,t)=T(r(r0,t),t)=T[x(r0,t),y(r0,t),z(r0,t),t], then we can use the chain rule to ﬁnd the time rate of change of Twith r0held constant: DT Dt=/parenleftbiggdT dt/parenrightbigg r0 =/parenleftbigg∂T ∂x/parenrightbigg/parenleftbiggdx dt/parenrightbigg r0+/parenleftbigg∂T ∂y/parenrightbigg/parenleftbiggdy dt/parenrightbigg r0+/parenleftbigg∂T ∂z/parenrightbigg/parenleftbiggdz dt/parenrightbigg r0+∂T ∂t. We recognize the partial derivatives of the coordinates as the components of the material velocity (A.57), and thus can write DT Dt=∂T ∂t+ux∂T ∂x+uy∂T ∂y+uz∂T ∂z=∂T ∂t+u·∇T. As expected, the material derivative depends on both the local time rate of change and the spatial rate of change of temperature. Suppose next that our probe is motorized and can travel about in the sinking water. If the probe sinks faster than the surrounding water, the time rate of change (measuredby the probe) should exceed the material derivative. Let the probe position and velocitybe r(t)=ˆxx(t)+ˆyy(t)+ˆzz(t), v(r,t)=ˆxdx(t) dt+ˆydy(t) dt+ˆzdz(t) dt. We can use the chain rule to determine the time rate of change of the temperature observed by the probe, but in this case we do notconstrain the velocity components to represent the moving ﬂuid. Thus, we merely obtain dT dt=∂T ∂xdx dt+∂T ∂ydy dt+∂T ∂zdz dt+∂T ∂t =∂T ∂t+v·∇T. This is called the total derivative of the temperature ﬁeld. In summary, the time rate of change of a scalar ﬁeld Tseen by an observer moving with arbitrary velocity vis given by the total derivative dT dt=∂T ∂t+v·∇T. (A.58) If the velocity of the observer happens to match the velocity uof a moving substance, the time rate of change is the material derivative DT Dt=∂T ∂t+u·∇T. (A.59) We can obtain the material derivative of a vector ﬁeld Fby component-wise application of (A.59): DF Dt=D Dt/bracketleftbigˆxFx+ˆyFy+ˆzFz/bracketrightbig =ˆx∂Fx ∂t+ˆy∂Fy ∂t+ˆz∂Fz ∂t+ˆx[u·(∇Fx)]+ˆy/bracketleftbig u·(∇Fy)/bracketrightbig +ˆz[u·(∇Fz)]. Figure A.5: Derivation of the Helmholtz transport theorem. Using the notation u·∇= ux∂ ∂x+uy∂ ∂y+uz∂ ∂z we can write DF Dt=∂F ∂t+(u·∇)F. (A.60) This is the material derivative of a vector ﬁeld Fwhen udescribes the motion of a physical material. Similarly, the total derivative of a vector ﬁeld is dF dt=∂F ∂t+(v·∇)F where vis arbitrary. The Helmholtz and Reynolds transport theorems We choose the intuitive approach taken by Tai [190] and Whitaker [214]. Consider an open surface S(t)moving through space and possibly deforming as it moves. The velocity of the points comprising the surface is given by the vector ﬁeld v(r,t). We are interested in computing the time derivative of the ﬂux of a vector ﬁeld F(r,t)through S(t): ψ(t)=d dt/integraldisplay S(t)F(r,t)·dS =lim /Delta1t→0/integraltext S(t+/Delta1t)F(r,t+/Delta1t)·dS−/integraltext S(t)F(r,t)·dS /Delta1t. (A.61) Here S(t+/Delta1t)=S2 isfoundbyextendin geachpointon S(t)= S1 throug hadisplaceme nt v/Delta1t,asshowninFigureA.5.Substitutin gtheTaylorexpansion F(r,t+/Delta1t)=F(r,t)+∂F(r,t) ∂t/Delta1t+··· into (A.61), we ﬁnd that only the ﬁrst two terms give non-zero contributions to the integral and ψ(t)=/integraldisplay S(t)∂F(r,t) ∂t·dS+lim /Delta1t→0/integraltext S2F(r,t)·dS−/integraltext S1F(r,t)·dS /Delta1t. (A.62) Thesecondtermontherightcanbeevaluate dwiththehelpofFigureA.5.Asthesurface moves through a displacement v/Delta1tit sweeps out a volume region /Delta1Vthat is bounded on the back by S1, on the front by S2, and on the side by a surface S3=/Delta1S. We can thus compute the two surface integrals in (A.62) as the diﬀerence between contributionsfrom the surface enclosing /Delta1Vand the side surface /Delta1S(remembering that the normal toS 1in (A.62) points into/Delta1V). Thus ψ(t)=/integraldisplay S(t)∂F(r,t) ∂t·dS+lim /Delta1t→0/contintegraltext S1+S2+/Delta1SF(r,t)·dS−/integraltext /Delta1SF(r,t)·dS3 /Delta1t =/integraldisplay S(t)∂F(r,t) ∂t·dS+lim /Delta1t→0/integraltext /Delta1V∇·F(r,t)dV3−/integraltext /Delta1SF(r,t)·dS3 /Delta1t by the divergence theorem. To compute the integrals over /Delta1Sand/Delta1Vwe note from FigureA.5thattheincreme ntalsurfac eandvolumeelementsarejust dS3=dl×(v/Delta1t), dV3=(v/Delta1t)·dS. Then, since F·[dl×(v/Delta1t)]=/Delta1t(v×F)·dl, we have ψ(t)=/integraldisplay S(t)∂F(r,t) ∂t·dS+lim /Delta1t→0/Delta1t/integraltext S(t)[v∇·F(r,t)]·dS /Delta1t−lim /Delta1t→0/Delta1t/contintegraltext /Gamma1[v×F(r,t)]·dl /Delta1t. Taking the limit and using Stokes’s theorem on the last integral we have ﬁnally d dt/integraldisplay S(t)F·dS=/integraldisplay S(t)/bracketleftbigg∂F ∂t+v∇·F−∇× (v×F)/bracketrightbigg ·dS, (A.63) which is the Helmholtz transport theorem [190, 43]. In case the surface corresponds to a moving physical material, we may wish to write the Helmholtz transport theorem in terms of the material derivative. We can set v=u and use ∇×(u×F)=u(∇·F)−F(∇·u)+(F·∇)u−(u·∇)F and (A.60) to obtain d dt/integraldisplay S(t)F·dS=/integraldisplay S(t)/bracketleftbiggDF Dt+F(∇·u)−(F·∇)u/bracketrightbigg ·dS. IfS(t)in (A.63) is closed, enclosing a volume region V(t), then /contintegraldisplay S(t)[∇×(v×F)]·dS=/integraldisplay V(t)∇·[∇×(v×F)]dV=0 by the divergence theorem and (B.49). In this case the Helmholtz transport theorem becomes d dt/contintegraldisplay S(t)F·dS=/contintegraldisplay S(t)/bracketleftbigg∂F ∂t+v∇·F/bracketrightbigg ·dS. (A.64) We now come to an essential tool that we employ throughout the book. Using the divergence theorem we can rewrite (A.64) as d dt/integraldisplay V(t)∇·FdV=/integraldisplay V(t)∇·∂F ∂tdV+/contintegraldisplay S(t)(∇·F)v·dS. Replacing ∇·Fby the scalar ﬁeld ρwe have d dt/integraldisplay V(t)ρdV=/integraldisplay V(t)∂ρ ∂tdV+/contintegraldisplay S(t)ρv·dS. (A.65) In this general form of the transport theorem vis an arbitrary velocity. In most appli- cations v=udescribes the motion of a material substance; then D Dt/integraldisplay V(t)ρdV=/integraldisplay V(t)∂ρ ∂tdV+/contintegraldisplay S(t)ρu·dS, (A.66) which is the Reynolds transport theorem [214]. The D/Dtnotation implies that V(t) retains exactly the same material elements as it moves and deforms to follow the materialsubstance. We may rewrite the Reynolds transport theorem in various forms. By the divergence theorem we have d dt/integraldisplay V(t)ρdV=/integraldisplay V(t)/bracketleftbigg∂ρ ∂t+∇·(ρv)/bracketrightbigg dV. Setting v=u, using (B.42), and using (A.59) for the material derivative of ρ, we obtain D Dt/integraldisplay V(t)ρdV=/integraldisplay V(t)/bracketleftbiggDρ Dt+ρ∇·u/bracketrightbigg dV. (A.67) We may also generate a vector form of the general transport theorem by taking ρin (A.65) to be a component of a vector. Assembling all of the components we have d dt/integraldisplay V(t)AdV=/integraldisplay V(t)∂A ∂tdV+/contintegraldisplay S(t)A(v·ˆn)dS. (A.68) A.3 Dyadic analysis Dyadic analysis was introduced in the late nineteenth century by Gibbs to generalize vector analysis to problems in which the components of vectors are related in a linearmanner. It has now been widely supplanted by tensor theory, but maintains a foothold inengineering where the transformation properties of tensors are not paramount (except,of course, in considerations such as those involving special relativity). Terms such as“tensor permittivity” and “dyadic permittivity” are often used interchangeably. Component form representation. We wish to write one vector ﬁeld A(r,t)as a linear function of another vector ﬁeld B(r,t): A=f(B). By this we mean that each component of Ais a linear combination of the components of B: A1(r,t)=a11/primeB1/prime(r,t)+a12/primeB2/prime(r,t)+a13/primeB3/prime(r,t), A2(r,t)=a21/primeB1/prime(r,t)+a22/primeB2/prime(r,t)+a23/primeB3/prime(r,t), A3(r,t)=a31/primeB1/prime(r,t)+a32/primeB2/prime(r,t)+a33/primeB3/prime(r,t). Here the aij/primemay depend on space and time (or frequency). The prime on the second index indicates that Aand Bmay be expressed in distinct coordinate frames (ˆi1,ˆi2,ˆi3) and(ˆi1/prime,ˆi2/prime,ˆi3/prime), respectively. We have A1=/parenleftbig a11/primeˆi1/prime+a12/primeˆi2/prime+a13/primeˆi3/prime/parenrightbig ·/parenleftbigˆi1/primeB1/prime+ˆi2/primeB2/prime+ˆi3/primeB3/prime/parenrightbig , A2=/parenleftbig a21/primeˆi1/prime+a22/primeˆi2/prime+a23/primeˆi3/prime/parenrightbig ·/parenleftbigˆi1/primeB1/prime+ˆi2/primeB2/prime+ˆi3/primeB3/prime/parenrightbig , A3=/parenleftbig a31/primeˆi1/prime+a32/primeˆi2/prime+a33/primeˆi3/prime/parenrightbig ·/parenleftbigˆi1/primeB1/prime+ˆi2/primeB2/prime+ˆi3/primeB3/prime/parenrightbig , and since B=ˆi1/primeB1/prime+ˆi2/primeB2/prime+ˆi3/primeB3/primewe can write A=ˆi1(a/prime 1·B)+ˆi2(a/prime 2·B)+ˆi3(a/prime 3·B) where a/prime 1=a11/primeˆi1/prime+a12/primeˆi2/prime+a13/primeˆi3/prime, a/prime 2=a21/primeˆi1/prime+a22/primeˆi2/prime+a23/primeˆi3/prime, a/prime 3=a31/primeˆi1/prime+a32/primeˆi2/prime+a33/primeˆi3/prime. In shorthand notation A=¯a·B (A.69) where ¯a=ˆi1a/prime 1+ˆi2a/prime 2+ˆi3a/prime 3. (A.70) Written out, the quantity ¯alooks like ¯a=a11/prime(ˆi1ˆi1/prime)+a12/prime(ˆi1ˆi2/prime)+a13/prime(ˆi1ˆi3/prime)+ +a21/prime(ˆi2ˆi1/prime)+a22/prime(ˆi2ˆi2/prime)+a23/prime(ˆi2ˆi3/prime)+ +a31/prime(ˆi3ˆi1/prime)+a32/prime(ˆi3ˆi2/prime)+a33/prime(ˆi3ˆi3/prime). Terms such as ˆi1ˆi1/primeare called dyads, while sums of dyads such as ¯aare called dyadics . The components aij/primeof¯amay be conveniently placed into an array: [¯a]= a11/primea12/primea13/prime a21/primea22/primea23/prime a31/primea32/primea33/prime . Writing [A]= A1 A2 A3 , [B]= B1/prime B2/prime B3/prime , we see that A=¯a·Bcan be written as [A]=[¯a][B]= a11/primea12/primea13/prime a21/primea22/primea23/prime a31/primea32/primea33/prime  B1/prime B2/prime B3/prime . Note carefully that in (A.69) ¯aoperates on Bfrom the left. A reorganization of the components of ¯aallows us to write ¯a=a1ˆi1/prime+a2ˆi2/prime+a3ˆi3/prime (A.71) where a1=a11/primeˆi1+a21/primeˆi2+a31/primeˆi3, a2=a12/primeˆi1+a22/primeˆi2+a32/primeˆi3, a3=a13/primeˆi1+a23/primeˆi2+a33/primeˆi3. We may now consider using ¯ato operate on a vector C=ˆi1C1+ˆi2C2+ˆi3C3from the right: C·¯a=(C·a1)ˆi1/prime+(C·a2)ˆi2/prime+(C·a3)ˆi3/prime. In matrix form C·¯ais [¯a]T[C]= a11/primea21/primea31/prime a12/primea22/primea32/prime a13/primea23/primea33/prime  C1 C2 C3  where the superscript “ T” denotes the matrix transpose operation. That is, C·¯a=¯aT·C where ¯aTis the transpose of ¯a. If the primed and unprimed frames coincide, then ¯a=a11(ˆi1ˆi1)+a12(ˆi1ˆi2)+a13(ˆi1ˆi3)+ +a21(ˆi2ˆi1)+a22(ˆi2ˆi2)+a23(ˆi2ˆi3)+ +a31(ˆi3ˆi1)+a32(ˆi3ˆi2)+a33(ˆi3ˆi3). In this case we may compare the results of ¯a·Band B·¯afor a given vector B= ˆi1B1+ˆi2B2+ˆi3B3. We leave it to the reader to verify that in general B·¯a/negationslash=¯a·B. Vector form representation. We can express dyadics in coordinate-free fashion if we expand the concept of a dyad to permit entities such as AB. Here Aand Bare called theantecedent andconsequent , respectively. The operation rules (AB)·C=A(B·C), C·(AB)=(C·A)B, deﬁne the anterior and posterior products of ABwith a vector C, and give results consistent with our prior component notation. Sums of dyads such as AB+CDare called dyadic polynomials , or dyadics. The simple dyadic AB=(A1ˆi1+A2ˆi2+A3ˆi3)(B1/primeˆi1/prime+B2/primeˆi2/prime+B3/primeˆi3/prime) can be represented in component form using AB=ˆi1a/prime 1+ˆi2a/prime 2+ˆi3a/prime 3 where a/prime 1=A1B1/primeˆi1/prime+A1B2/primeˆi2/prime+A1B3/primeˆi3/prime, a/prime 2=A2B1/primeˆi1/prime+A2B2/primeˆi2/prime+A2B3/primeˆi3/prime, a/prime 3=A3B1/primeˆi1/prime+A3B2/primeˆi2/prime+A3B3/primeˆi3/prime, or using AB=a1ˆi1/prime+a2ˆi2/prime+a3ˆi3/prime where a1=ˆi1A1B1/prime+ˆi2A2B1/prime+ˆi3A3B1/prime, a2=ˆi1A1B2/prime+ˆi2A2B2/prime+ˆi3A3B2/prime, a3=ˆi1A1B3/prime+ˆi2A2B3/prime+ˆi3A3B3/prime. Note that if we write ¯a=ABthen aij=AiBj/prime. A simple dyad ABby itself cannot represent a general dyadic ¯a; only six independent quantities are available in AB(the three components of Aand the three components ofB), while an arbitrary dyadic has nine independent components. However, it can be shown that any dyadic can be written as a sum of three dyads: ¯a=AB+CD+EF. This is called a vector representation of¯a.I fVis a vector, the distributive laws ¯a·V=(AB+CD+EF)·V=A(B·V)+C(D·V)+E(F·V), V·¯a=V·(AB+CD+EF)=(V·A)B+(V·C)D+(V·E)F, apply. Dyadic algebra and calculus. The cross product of a vector with a dyadic produces another dyadic. If ¯a=AB+CD+EFthen by deﬁnition ¯a×V=A(B×V)+C(D×V)+E(F×V), V×¯a=(V×A)B+(V×C)D+(V×E)F. The corresponding component forms are ¯a×V=ˆi1(a/prime 1×V)+ˆi2(a/prime 2×V)+ˆi3(a/prime 3×V), V×¯a=(V×a1)ˆi1/prime+(V×a2)ˆi2/prime+(V×a3)ˆi3/prime, where we have used (A.70) and (A.71), respectively. Interactions between dyads or dyadics may also be deﬁned. The dot product of two dyads ABand CDis a dyad given by (AB)·(CD)=A(B·C)D=(B·C)(AD). The dot product of two dyadics can be found by applying the distributive property. Ifαis a scalar, then the product α¯ais a dyadic with components equal to αtimes the components of ¯a. Dyadic addition may be accomplished by adding individual dyadic components as long as the dyadics are expressed in the same coordinate system. Sub-traction is accomplished by adding the negative of a dyadic, which is deﬁned throughscalar multiplication by −1. Some useful dyadic identities appear in Appendix B. Many more can be found in Van Bladel [202]. The various vector derivatives may also be extended to dyadics. Computations are easiest in rectangular coordinates, since ˆi 1=ˆx,ˆi2=ˆy, and ˆi3=ˆzare constant with position. The dyadic ¯a=axˆx+ayˆy+azˆz has divergence ∇·¯a=(∇·ax)ˆx+(∇·ay)ˆy+(∇·az)ˆz, and curl ∇× ¯a=(∇× ax)ˆx+(∇× ay)ˆy+(∇× az)ˆz. Note that the divergence of a dyadic is a vector while the curl of a dyadic is a dyadic. The gradient of a vector a=axˆx+ayˆy+azˆzis ∇a=(∇ax)ˆx+(∇ay)ˆy+(∇az)ˆz, a dyadic quantity. The dyadic derivatives may be expressed in coordinate-free notation by using the vector representation. The dyadic ABhas divergence ∇·(AB)=(∇·A)B+A·(∇B) and curl ∇×(AB)=(∇× A)B−A×(∇B). The Laplacian of a dyadic is a dyadic given by ∇2¯a=∇(∇·¯a)−∇× (∇× ¯a). The divergence theorem for dyadics is /integraldisplay V∇·¯adV=/contintegraldisplay Sˆn·¯adS. Some of the other common diﬀerential and integral identities for dyadics can be found in Van Bladel [202] and Tai [192]. Special dyadics. We say that ¯aissymmetric if B·¯a=¯a·B for any vector B. This requires ¯aT=¯a, i.e., aij/prime=aji/prime. We say that ¯aisantisymmetric if B·¯a=− ¯a·B for any B. In this case ¯aT=− ¯a. That is, aij/prime=−aji/primeand aii/prime=0. A symmetric dyadic has only six independent components while an antisymmetric dyadic has only three. The reader can verify that any dyadic can be decomposed into symmetric and antisymmetric parts as ¯a=1 2/parenleftbig¯a+¯aT/parenrightbig +1 2/parenleftbig¯a−¯aT/parenrightbig . A simple example of a symmetric dyadic is the unit dyadic ¯Ideﬁned by ¯I=ˆi1ˆi1+ˆi2ˆi2+ˆi3ˆi3. This quantity often arises in the manipulation of dyadic equations, and satisﬁes A·¯I=¯I·A=A for any vector A. In matrix form ¯Iis the identity matrix: [¯I]= 100 010 001 . The components of a dyadic may be complex. We say that ¯aishermitian if B·¯a=¯a∗·B (A.72) holds for any B. This requires that ¯a∗=¯aT. Taking the transpose we can write ¯a=(¯a∗)T=¯a† where “ †” stands for the conjugate-transpose operation. We say that ¯aisanti-hermitian if B·¯a=− ¯a∗·B (A.73) for arbitrary B. In this case ¯a∗=− ¯aT. Any complex dyadic can be decomposed into hermitian and anti-hermitian parts: ¯a=1 2/parenleftbig¯aH+¯aA/parenrightbig (A.74) where ¯aH=¯a+¯a†, ¯aA=¯a−¯a†. (A.75) A dyadic identity important in the study of material parameters is B·¯a∗·B∗=B∗·¯a†·B. (A.76) We show this by decomposing ¯aaccording to (A.74), giving B·¯a∗·B∗=1 2/parenleftBig/bracketleftbig B∗·¯aH/bracketrightbig∗+/bracketleftbig B∗·¯aA/bracketrightbig∗/parenrightBig ·B∗ where we have used (B·¯a)∗=(B∗·¯a∗). Applying (A.72) and (A.73) we obtain B·¯a∗·B∗=1 2/parenleftBig/bracketleftbig¯aH∗·B∗/bracketrightbig∗−/bracketleftbig¯aA∗·B∗/bracketrightbig∗/parenrightBig ·B∗ =B∗·1 2/parenleftbig/bracketleftbig¯aH·B/bracketrightbig −/bracketleftbig¯aA·B/bracketrightbig/parenrightbig =B∗·/parenleftbigg1 2/bracketleftbig¯aH−¯aA/bracketrightbig ·B/parenrightbigg . Since the term in brackets is ¯aH−¯aA=2¯a†by (A.75), the identity is proved. A.4 Boundary value problems Many physical phenomena may be described mathematically as the solutions to bound- ary value problems . The desired physical quantity (usually called a “ﬁeld”) in a certain region of space is found by solving one or more partial diﬀerential equations subject tocertain conditions over the boundary surface. The boundary conditions may specify thevalues of the ﬁeld, some manipulated version of the ﬁeld (such as the normal derivative),or a relationship between ﬁelds in adjoining regions. If the ﬁeld varies with time as wellas space, initial or ﬁnal values of the ﬁeld must also be speciﬁed. Particularly importantis whether a boundary value problem is well-posed and therefore has a unique solution which depends continuously on the data supplied. This depends on the forms of the dif-ferential equation and boundary conditions. The well-posedness of Maxwell’s equationsis discussed in §2.2. The importance of boundary value problems has led to an array of techniques, both analytical and numerical, for solving them. Many problems (such as boundary valueproblems involving Laplace’s equation) may be solved in several diﬀerent ways. Unique-ness permits an engineer to focus attention on which technique will yield the most eﬃcientsolution. In this section we concentrate on the separation of variables technique, which iswidely applied in the solution of Maxwell’s equations. We ﬁrst discuss eigenvalue prob-lems and then give an overview of separation of variables. Finally we consider a numberof example problems in each of the three common coordinate systems. Sturm–Liouville problems and eigenvalues The partial diﬀerential equations of electromagnetics can often be reduced to ordinary diﬀerential equations. In some cases symmetry permits us to reduce the number ofdimensions by inspection; in other cases, we may employ an integral transform (e.g.,the Fourier transform) or separation of variables. The resulting ordinary diﬀerentialequations may be viewed as particular cases of the Sturm–Liouville diﬀerential equation d dx/bracketleftbigg p(x)dψ(x) dx/bracketrightbigg +q(x)ψ(x)+λσ(x)ψ(x)=0, x∈[a,b]. (A.77) In linear operator notation L[ψ(x)]=−λσ(x)ψ(x), (A.78) where Lis the linear Sturm–Liouville operator L=/parenleftbiggd dx/bracketleftbigg p(x)d dx/bracketrightbigg +q(x)/parenrightbigg . Obviously ψ(x)=0satisﬁes (A.78). However, for certain values of λdependent on p, q,σ, and the boundary conditions we impose, (A.78) has non-trivial solutions. Each λ that satisﬁes (A.78) is an eigenvalue ofL, and any non-trivial solution associated with that eigenvalue is an eigenfunction . Taken together, the eigenvalues of an operator form itseigenvalue spectrum . We shall restrict ourselves to the case in which Lisself-adjoint . Assume p,q, and σ are real and continuous on [a,b]. It is straightforward to show that for any two functions u(x)andv(x)Lagrange’s identity uL[v]−vL[u]=d dx/bracketleftbigg p/parenleftbigg udv dx−vdu dx/parenrightbigg/bracketrightbigg (A.79) holds. Integration gives Green’s formula /integraldisplayb a(uL[v]−vL[u])dx=p/parenleftbigg udv dx−vdu dx/parenrightbigg/vextendsingle/vextendsingle/vextendsingleb a. The operator Lis self-adjoint if its associated boundary conditions are such that p/parenleftbigg udv dx−vdu dx/parenrightbigg/vextendsingle/vextendsingle/vextendsingleb a=0. (A.80) Possible sets of conditions include the homogeneous boundary conditions α1ψ(a)+β1ψ/prime(a)=0,α 2ψ(b)+β2ψ/prime(b)=0, (A.81) and the periodic boundary conditions ψ(a)=ψ(b), p(a)ψ/prime(a)=p(b)ψ/prime(b). (A.82) By imposing one of these sets on (A.78) we obtain a Sturm–Liouville problem . The self-adjoint Sturm–Liouville operator has some nice properties. Each eigenvalue is real, and the eigenvalues form a denumerable set with no cluster point. Moreover, eigen-functions corresponding to distinct eigenvalues are orthogonal, and the eigenfunctionsform a complete set. Hence we can expand any suﬃciently smooth function in terms ofthe eigenfunctions of a problem. We discuss this further below. Aregular Sturm–Liouville problem involves a self-adjoint operator Lwith p(x)> 0 andσ(x)>0everywhere, and the homogeneous boundary conditions (A.81). If porσ vanishes at an endpoint of [a,b], or an endpoint is at inﬁnity, the problem is singular . The harmonic diﬀerential equation can form the basis of regular problems, while prob-lems involving Bessel’s and Legendre’s equations are singular. Regular Sturm–Liouvilleproblems have additional properties. There are inﬁnitely many eigenvalues. There isa smallest eigenvalue but no largest eigenvalue, and the eigenvalues can be ordered asλ 0<λ 1<···<λ n···. Associated with each λnis a unique (to an arbitrary multiplicative constant) eigenfunction ψnthat has exactly nzeros in (a,b). If a problem is singular because p=0at an endpoint, we can also satisfy (A.80) by demanding that ψbe bounded at that endpoint (a singularity condition ) and that any regular Sturm–Liouville boundary condition hold at the other endpoint. This is the casefor Bessel’s and Legendre’s equations discussed below. Orthogonality of the eigenfunctions. LetLbe self-adjoint, and let ψ mandψnbe eigenfunctions associated with λmandλn, respectively. Then by (A.80) we have /integraldisplayb a(ψm(x)L[ψn(x)]−ψn(x)L[ψm(x)])dx=0. ButL[ψn(x)]=−λnσ(x)ψn(x)andL[ψm(x)]=−λmσ(x)ψm(x). Hence (λm−λn)/integraldisplayb aψm(x)ψn(x)σ(x)dx=0, andλm/negationslash=λnimplies that /integraldisplayb aψm(x)ψn(x)σ(x)dx=0. (A.83) We say that ψmandψnare orthogonal with respect to the weight function σ(x). Eigenfunction expansion of an arbitrary function. IfLis self-adjoint, then its eigenfunctions form a complete set . This means that any piecewise smooth function may be represented as a weighted series of eigenfunctions. Speciﬁcally, if fand f/primeare piece- wise continuous on [a,b], then fmay be represented as the generalized Fourier series f(x)=∞/summationdisplay n=0cnψn(x). (A.84) Convergence of the series is uniform and gives, at any point of (a,b), the average value [f(x+)+f(x−)]/2of the one-sided limits f(x+)and f(x−)off(x). The cncan be found using orthogonality condition (A.83): multiply (A.84) by ψmσand integrate to obtain /integraldisplayb af(x)ψm(x)σ(x)dx=∞/summationdisplay n=0cn/integraldisplayb aψn(x)ψm(x)σ(x)dx, hence cn=/integraltextb af(x)ψn(x)σ(x)dx /integraltextb aψ2n(x)σ(x)dx. (A.85) These coeﬃcients ensure that the series converges in mean tof; i.e., the mean-square error /integraldisplayb a/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsinglef(x)−∞/summationdisplay n=0cnψn(x)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle2 σ(x)dx is minimized. Truncation to ﬁnitely-many terms generally results in oscillations ( Gibb’s phenomena ) near points of discontinuity of f. The cnare easier to compute if the ψn are orthonormal with /integraldisplayb aψ2 n(x)σ(x)dx=1 for each n. Uniqueness of the eigenfunctions. If both ψ1andψ2are associated with the same eigenvalue λ, then L[ψ1(x)]+λσ(x)ψ1(x)=0,L[ψ2(x)]+λσ(x)ψ2(x)=0, hence ψ1(x)L[ψ2(x)]−ψ2(x)L[ψ1(x)]=0. By (A.79) we have d dx/bracketleftbigg p(x)/parenleftbigg ψ1(x)dψ2(x) dx−ψ2(x)dψ1(x) dx/parenrightbigg/bracketrightbigg =0 or p(x)/parenleftbigg ψ1(x)dψ2(x) dx−ψ2(x)dψ1(x) dx/parenrightbigg =C where Cis constant. Either of (A.81) implies C=0, hence d dx/parenleftbiggψ2(x) ψ1(x)/parenrightbigg =0 so that ψ1(x)=Kψ2(x)for some constant K. So under homogeneous boundary condi- tions, every eigenvalue is associated with a unique eigenfunction. This is false for the periodic boundary conditions (A.82). Eigenfunction expansion then becomes diﬃcult, as we can no longer assume eigenfunction orthogonality. However, theGram–Schmidt algorithm may be used to construct orthogonal eigenfunctions. We referthe interested reader to Haberman [79]. The harmonic diﬀerential equation. The ordinary diﬀerential equation d 2ψ(x) dx2=−k2ψ(x) (A.86) is Sturm–Liouville with p≡1,q≡0,σ≡1, andλ=k2. Suppose we take [a,b]=[0,L] and adopt the homogeneous boundary conditions ψ(0)=0and ψ(L)=0. (A.87) Since p(x)>0andσ(x)>0on[0,L], equations (A.86) and (A.87) form a regular Sturm– Liouville problem. Thus we should have an inﬁnite number of discrete eigenvalues. Apower series technique yields the two independent solutions ψ a(x)=Aasinkx,ψ b(x)=Abcoskx, to (A.86); hence by linearity the most general solution is ψ(x)=Aasinkx+Abcoskx. (A.88) The condition at x=0gives Aasin 0+Abcos 0=0,hence Ab=0. The other condition then requires AasinkL=0. (A.89) Since Aa=0would give ψ≡0, we satisfy (A.89) by choosing k=kn=nπ/Lfor n=1,2,....Because λ=k2, the eigenvalues are λn=(nπ/L)2 with corresponding eigenfunctions ψn(x)=sinknx. Note that λ=0is not an eigenvalue; eigenfunctions are nontrivial by deﬁnition, and sin(0πx/L)≡0. Likewise, the diﬀerential equation associated with λ=0can be solved easily, but only its trivial solution can ﬁt homogeneous boundary conditions: with k=0, (A.86) becomes d2ψ(x)/dx2=0,giving ψ(x)=ax+b;this can satisfy (A.87) only with a=b=0. These “eigensolutions” obey the properties outlined earlier. In particular the ψnare orthogonal, /integraldisplayL 0sin/parenleftBignπx L/parenrightBig sin/parenleftBigmπx L/parenrightBig dx=L 2δmn, and the eigenfunction expansion of a piecewise continuous function fis given by f(x)=∞/summationdisplay n=1cnsin/parenleftBignπx L/parenrightBig where, with σ(x)=1in (A.85), we have cn=/integraltextL 0f(x)sin/parenleftbignπx L/parenrightbig dx /integraltextL 0sin2/parenleftbignπx L/parenrightbig dx=2 L/integraldisplayL 0f(x)sin/parenleftBignπx L/parenrightBig dx. Hence we recover the standard Fourier sine series for f(x). With little extra eﬀort we can examine the eigenfunctions resulting from enforcement of the periodic boundary conditions ψ(0)=ψ(L)and ψ/prime(0)=ψ/prime(L). The general solution (A.88) still holds, so we have the choices ψ(x)=sinkxandψ(x)= coskx. Evidently both ψ(x)=sin/parenleftbigg2nπx L/parenrightbigg and ψ(x)=cos/parenleftbigg2nπx L/parenrightbigg satisfy the boundary conditions for n=1,2,.... Thus each eigenvalue (2nπ/L)2is associated with two eigenfunctions. Bessel’s diﬀerential equation. Bessel’s equation d dx/parenleftbigg xdψ(x) dx/parenrightbigg +/parenleftbigg k2x−ν2 x/parenrightbigg ψ(x)=0 (A.90) occurs when problems are solved in circular-cylindrical coordinates. Comparison with (A.77) shows that λ=k2,p(x)=x,q(x)=−ν2/x, andσ(x)=x. We take [a,b]=[0,L] along with the boundary conditions ψ(L)=0and |ψ(0)|<∞. (A.91) Although the resulting Sturm–Liouville problem is singular, the speciﬁed conditions (A.91) maintain satisfaction of (A.80). The eigenfunctions are orthogonal because (A.80)is satisﬁed by having ψ(L)=0and p(x)dψ(x)/dx→0asx→0. As a second-order ordinary diﬀerential equation, (A.90) has two solutions denoted by J ν(kx)and Nν(kx), and termed Bessel functions . Their properties are summarized in Appendix E.1. The function Jν(x), the Bessel function of the ﬁrst kind and order ν, is well-behaved in [0,L]. The function Nν(x), the Bessel function of the second kind and order ν, is unbounded atx=0; hence it is excluded as an eigenfunction of the Sturm–Liouville problem. The condition at x=Lshows that the eigenvalues are deﬁned by Jν(kL)=0. We denote the mth root of Jν(x)=0bypνm. Then kνm=/radicalbig λνm=pνm/L. The inﬁnitely many eigenvalues are ordered as λν1<λν2<. . .. Associated with eigen- value λνmis a single eigenfunction Jν(√λνmx). The orthogonality relation is /integraldisplayL 0Jν/parenleftBigpνm Lx/parenrightBig Jν/parenleftBigpνn Lx/parenrightBig xd x=0,m/negationslash=n. Since the eigenfunctions are also complete, we can expand any piecewise continuous function fin aFourier–Bessel series f(x)=∞/summationdisplay m=1cmJν/parenleftBig pνmx L/parenrightBig ,0≤x≤L,ν> −1. By (A.85) and (E.22) we have cm=2 L2J2 ν+1(pνm)/integraldisplayL 0f(x)Jν/parenleftBig pνmx L/parenrightBig xd x. The associated Legendre equation. Legendre’s equation occurs when problems are solved in spherical coordinates. It is often written in one of two forms. Letting θbe the polar angle of spherical coordinates ( 0≤θ≤π), the equation is d dθ/parenleftbigg sinθdψ(θ) dθ/parenrightbigg +/parenleftbigg λsinθ−m2 sinθ/parenrightbigg ψ(θ)=0. This is Sturm–Liouville with p(θ)=sinθ,σ(θ)=sinθ, and q(θ)=− m2/sinθ. The boundary conditions |ψ(0)|<∞and |ψ(π)|<∞ deﬁne a singular problem: the conditions are not homogeneous, p(θ)=0at both end- points, and q(θ) < 0. Despite this, the Legendre problem does share properties of a regular Sturm–Liouville problem — including eigenfunction orthogonality and complete-ness. Using x=cosθ, we can put Legendre’s equation into its other common form d dx/parenleftbigg [1−x2]dψ(x) dx/parenrightbigg +/parenleftbigg λ−m2 1−x2/parenrightbigg ψ(x)=0, (A.92) where −1≤x≤1. It is found that ψis bounded at x=±1only if λ=n(n+1) where n≥mis an integer. These λare the eigenvalues of the Sturm–Liouville problem, and the corresponding ψn(x)are the eigenfunctions. As a second-order partial diﬀerential equation, (A.92) has two solutions known as associated Legendre functions . The solution bounded at both x=± 1is the associated Legendre function of the ﬁrst kind, denoted Pm n(x). The second solution, unbounded at x=± 1, is the associated Legendre function of the second kind Qm n(x). Appendix E.2 tabulates some properties of these functions. For ﬁxed m, each λmnis associated with a single eigenfunction Pm n(x). Since Pm n(x)is bounded at x=± 1, and since p(±1)=0, the eigenfunctions obey Lagrange’s identity (A.79), hence are orthogonal on [−1,1]with respect to the weight function σ(x)=1. Evaluation of the orthogonality integral leads to /integraldisplay1 −1Pm l(x)Pm n(x)dx=δln2 2n+1(n+m)! (n−m)!(A.93) or equivalently /integraldisplayπ 0Pm l(cosθ)Pm n(cosθ)sinθdθ=δln2 2n+1(n+m)! (n−m)!. Form=0,Pm n(x)is a polynomial of degree n. Each such Legendre polynomial , denoted Pn(x), is given by Pn(x)=1 2nn!dn(x2−1)n dxn. It turns out that Pm n(x)=(−1)m(1−x2)m/2dmPn(x) dxm, giving Pm n(x)=0form>n. Because the Legendre polynomials form a complete set in the interval [−1,1],w em a y expand any suﬃciently smooth function in a Fourier–Legendre series f(x)=∞/summationdisplay n=0cnPn(x). Convergence in mean is guaranteed if cn=2n+1 2/integraldisplay1 −1f(x)Pn(x)dx, found using (A.85) along with (A.93). In practice, the associated Legendre functions appear along with exponential functions in the solutions to spherical boundary value problems. The combined functions are knownasspherical harmonics , and form solutions to two-dimensional Sturm–Liouville problems. We consider these next. Higher-dimensional SL problems: Helmholtz’s equation. Replacing d/dxby∇, we generalize the Sturm–Liouville equation to higher dimensions: ∇·[p(r)∇ψ(r)]+q(r)ψ(r)+λσ(r)ψ(r)=0, where q,p,σ,ψare real functions. Of particular interest is the case q(r)=0,p(r)= σ(r)=1, giving the Helmholtz equation ∇ 2ψ(r)+λψ(r)=0. (A.94) In most boundary value problems, ψor its normal derivative is speciﬁed on the surface of a bounded region. We obtain a three-dimensional analogue to the regular Sturm–Liouville problem by assuming the homogeneous boundary conditions αψ(r)+βˆn·∇ψ(r)=0 (A.95) on the closed surface, where ˆnis the outward unit normal. The problem consisting of (A.94) and (A.95) has properties analogous to those of the regular one-dimensional Sturm–Liouville problem. All eigenvalues are real. There areinﬁnitely many eigenvalues. There is a smallest eigenvalue but no largest eigenvalue.However, associated with an eigenvalue there may be many eigenfunctions ψ λ(r). The eigenfunctions are orthogonal with /integraldisplay Vψλ1(r)ψλ2(r)dV=0,λ 1/negationslash=λ2. They are also complete and can be used to represent any piecewise smooth function f(r) according to f(r)=/summationdisplay λaλψλ(r), which converges in mean when aλm=/integraltext Vf(r)ψλm(r)dV/integraltext Vψ2 λm(r)dV. These properties are shared by the two-dimensional eigenvalue problem involving an open surface Swith boundary contour /Gamma1. Spherical harmonics. We now inspect solutions to the two-dimensional eigenvalue problem ∇2Y(θ, φ) +λ a2Y(θ, φ) =0 over the surface of a sphere of radius a. Since the sphere has no boundary contour, we demand that Y(θ, φ) be bounded in θand periodic in φ. In the next section we shall apply separation of variables and show that Ynm(θ, φ) =/radicalBigg 2n+1 4π(n−m)! (n+m)!Pm n(cosθ)ejmφ where λ=n(n+1). Note that Qm ndoes not appear as it is not bounded at θ=0,π. The functions Ynmare called spherical harmonics (sometimes zonal ortesseral harmonics, depending on the values of nand m). As expressed above they are in orthonormal form, because the orthogonality relationships for the exponential and associated Legendrefunctions yield /integraldisplay π −π/integraldisplayπ 0Y∗ n/primem/prime(θ, φ) Ynm(θ, φ) sinθdθdφ=δn/primenδm/primem. (A.96) As solutions to the Sturm–Liouville problem, these functions form a complete set on the surface of a sphere. Hence they can be used to represent any piecewise smooth function f(θ, φ) as f(θ, φ) =∞/summationdisplay n=0n/summationdisplay m=−nanmYnm(θ, φ), where anm=/integraldisplayπ −π/integraldisplayπ 0f(θ, φ) Y∗ nm(θ, φ) sinθdθdφ by (A.96). The summation index mranges from −ntonbecause Pm n=0form>n.F o r negative index we can use Yn,−m(θ, φ) =(−1)mY∗ nm(θ, φ). Some properties of the spherical harmonics are tabulated in Appendix E.3. Separation of variables We now consider a technique that ﬁnds widespread application in solving boundary value problems, applying as it does to many important partial diﬀerential equations suchas Laplace’s equation, the diﬀusion equation, and the scalar and vector waveequations. These equations are related to the scalar Helmholtz equation ∇ 2ψ(r)+k2ψ(r)=0 (A.97) where kis a complex constant. If kis real and we supply the appropriate boundary conditions, we have the higher-dimensional Sturm–Liouville problem with λ=k2.W e shall not pursue the extension of Sturm–Liouville theory to complex values of k. Laplace’s equation is Helmholtz’s equation with k=0. With λ=k2=0it might appear that Laplace’s equation does not involve eigenvalues; however, separation of vari-ables does lead us to lower-dimensional eigenvalue problems to which our previous meth-ods apply. Solutions to the scalar or vector wave equations usually begin with Fourier transformation on the time variable, or with an initial separation of the time variable toreach a Helmholtz form. The separation of variables idea is simple. We seek a solution to (A.97) in the form of a product of functions each of a single variable. If ψdepends on all three spatial dimensions, then we seek a solution of the type ψ(u,v,w) =U(u)V(v)W(w), where u,v, and ware the coordinate variables used to describe the problem. If ψ depends on only two coordinates, we may seek a product solution involving two functionseach dependent on a single coordinate; alternatively, may use the three-variable solutionand choose constants so that the result shows no variation with one coordinate. TheHelmholtz equation is considered separable if it can be reduced to a set of independent ordinary diﬀerential equations, each involving a single coordinate variable. The ordinarydiﬀerential equations, generally of second order, can be solved by conventional techniquesresulting in solutions of the form U(u)=A uUA(u,ku,kv,kw)+BuUB(u,ku,kv,kw), V(v)=AvVA(v,ku,kv,kw)+BvVB(v,ku,kv,kw), W(w)=AwWA(w,ku,kv,kw)+BwWB(w,ku,kv,kw). The constants ku,kv,kware called separation constants and are found, along with the am- plitude constants A,B,by applying boundary conditions appropriate for a given problem. At least one separation constant depends on (or equals) k, so only two are independent. In many cases ku,kv, and kwbecome the discrete eigenvalues of the respective diﬀer- ential equations, and correspond to eigenfunctions U(u,ku,kv,kw),V(v,ku,kv,kw), and W(w,ku,kv,kw). In other cases the separation constants form a continuous spectrum of values, often when a Fourier transform solution is employed. The Helmholtz equation can be separated in eleven diﬀerent orthogonal coordinate systems [134]. Undoubtedly the most important of these are the rectangular, circular-cylindrical, and spherical systems, and we shall consider each in detail. We do note,however, that separability in a certain coordinate system does not imply that all prob-lems expressed in that coordinate system can be easily handled using the resulting solu-tions. Only when the geometry and boundary conditions are simple do the solutions lendthemselves to easy application; often other solution techniques are more appropriate. Although rigorous conditions can be set forth to guarantee solvability by separation of variables [119], we prefer the following, more heuristic list: 1. Use a coordinate system that allows the given partial diﬀerential equation to sep- arate into ordinary diﬀerential equations. 2. The problem’s boundaries must be such that those boundaries not at inﬁnity co- incide with a single level surface of the coordinate system. 3. Use superposition to reduce the problem to one involving a single nonhomogeneous boundary condition. Then:(a) Solve the resulting Sturm–Liouville problem in one or two dimensions, with homogeneous boundary conditions on all boundaries. Then use a discreteeigenvalue expansion (Fourier series) and eigenfunction orthogonality to sat-isfy the remaining nonhomogeneous condition. (b) If a Sturm–Liouville problem cannot be formulated with the homogeneous boundary conditions (because, for instance, one boundary is at inﬁnity), usea Fourier integral (continuous expansion) to satisfy the remaining nonhomo-geneous condition. If a Sturm–Liouville problem cannot be formulated, discovering the form of the integral transform to use can be diﬃcult. In these cases other approaches, such as conformalmapping, may prove easier. Solutions in rectangular coordinates. In rectangular coordinates the Helmholtz equation is ∂ 2ψ(x,y,z) ∂x2+∂2ψ(x,y,z) ∂y2+∂2ψ(x,y,z) ∂z2+k2ψ(x,y,z)=0. (A.98) We seek a solution of the form ψ(x,y,z)=X(x)Y(y)Z(z);substitution into (A.98) followed by division through by X(x)Y(y)Z(z)gives 1 X(x)d2X(x) dx2+1 Y(y)d2Y(y) dy2+1 Z(z)d2Z(z) dz2=−k2. (A.99) At this point we require the separation argument . The left-hand side of (A.99) is a sum of three functions, each involving a single independent variable, whereas the right-handside is constant. But the only functions of independent variables that always sum toa constant are themselves constants. Thus we may equate each term on the left to a diﬀerent constant: 1 X(x)d2X(x) dx2=−k2 x, 1 Y(y)d2Y(y) dy2=−k2 y, (A.100) 1 Z(z)d2Z(z) dz2=−k2 z, provided that k2 x+k2 y+k2 z=k2. The negative signs in (A.100) have been introduced for convenience. Let us discuss the general solutions of equations (A.100). If kx=0, the two indepen- dent solutions for X(x)are X(x)=axxand X(x)=bx where axand bxare constants. If kx/negationslash=0, solutions may be chosen from the list of functions e−jkxx,ejkxx,sinkxx,coskxx, any two of which are independent. Because sinx=(ejx−e−jx)/2jand cosx=(ejx+e−jx)/2, (A.101) the six possible solutions for kx/negationslash=0are X(x)=  A xejkxx+Bxe−jkxx, Axsinkxx+Bxcoskxx, Axsinkxx+Bxe−jkxx, Axejkxx+Bxsinkxx, Axejkxx+Bxcoskxx, Axe−jkxx+Bxcoskxx.(A.102) We may base our choice on convenience (e.g., the boundary conditions may be amenable to one particular form) or on the desired behavior of the solution (e.g., standing waves vs. traveling waves). If kis complex, then so may be kx,ky,o r kz; observe that with imaginary arguments the complex exponentials are actually real exponentials, and thetrigonometric functions are actually hyperbolic functions. The solutions for Y(y)and Z(z)are identical to those for X(x). We can write, for instance, X(x)=/braceleftBigg A xejkxx+Bxe−jkxx,kx/negationslash=0, axx+bx, kx=0,(A.103) Y(y)=/braceleftBigg Ayejkyy+Bye−jkyy,ky/negationslash=0, ayy+by, ky=0,(A.104) Z(z)=/braceleftBigg Azejkzz+Bze−jkzz,kz/negationslash=0, azz+bz, kz=0.(A.105) Examples. Let us begin by solving the simple equation ∇2V(x)=0. Since Vdepends only on xwe can use (A.103)–(A.105) with ky=kz=0anday=az=0. Moreover kx=0because k2 x+k2 y+k2 z=k2=0for Laplace’s equation. The general solution is therefore V(x)=axx+bx. Boundary conditions must be speciﬁed to determine axand bx; for instance, the condi- tions V(0)=0and V(L)=V0yield V(x)=V0x/L. Next let us solve ∇2ψ(x,y)=0. We produce a lack of z-dependence in ψby letting kz=0and choosing az=0. Moreover, k2 x=− k2 ysince Laplace’s equation requires k=0. This leads to three possibilities. If kx=ky=0, we have the product solution ψ(x,y)=(axx+bx)(ayy+by). (A.106) Ifkyis real and nonzero, then ψ(x,y)=(Axe−kyx+Bxekyx)(Ayejkyy+Bye−jkyy). (A.107) Using the relations sinhu=(eu−e−u)/2and cosh u=(eu+e−u)/2 (A.108) along with (A.101), we can rewrite (A.107) as ψ(x,y)=(Axsinhkyx+Bxcosh kyx)(Aysinkyy+Bycoskyy). (A.109) (We can reuse the constant names Ax,Bx,Ay,By, since the constants are unknown at this point.) If kxis real and nonzero we have ψ(x,y)=(Axsinkxx+Bxcoskxx)(Aysinhkxy+Bycosh kxy). (A.110) Consider the problem consisting of Laplace’s equation ∇2V(x,y)=0 (A.111) holding in the region 0<x<L1,0<y<L2,−∞<z<∞, together with the boundary conditions V(0,y)=V1,V(L1,y)=V2,V(x,0)=V3,V(x,L2)=V4. The solution V(x,y)represents the potential within a conducting tube with each wall held at a diﬀerent potential. Superposition applies: since Laplace’s equation is linearwe can write the solution as the sum of solutions to four diﬀerent sub-problems. Eachsub-problem has homogeneous boundary conditions on one independent variable andinhomogeneous conditions on the other, giving a Sturm–Liouville problem in one of thevariables. For instance, let us examine the solutions found above in relation to the sub-problem consisting of Laplace’s equation (A.111) in the region 0<x<L 1,0<y<L2, −∞<z<∞, subject to the conditions V(0,y)=V(L1,y)=V(x,0)=0,V(x,L2)=V4/negationslash=0. First we try (A.106). The boundary condition at x=0gives V(0,y)=(ax(0)+bx)(ayy+by)=0, which holds for all y∈(0,L2)only if bx=0. The condition at x=L1, V(L1,y)=axL1(ayy+by)=0, then requires ax=0. But ax=bx=0gives V(x,y)=0, and the condition at y=L2 cannot be satisﬁed; clearly (A.106) was inappropriate. Next we examine (A.109). The condition at x=0gives V(0,y)=(Axsinh 0 +Bxcosh 0 )(Aysinkyy+Bycoskyy)=0, hence Bx=0. The condition at x=L1implies V(L1,y)=[Axsinh(kyL1)](Aysinkyy+Bycoskyy)=0. This can hold if either Ax=0orky=0, but the case ky=0(=kx)was already considered. Thus Ax=0and the trivial solution reappears. Our last candidate is (A.110). The condition at x=0requires V(0,y)=(Axsin 0+Bxcos 0)(Aysinhkxy+Bycosh kxy)=0, which implies Bx=0. Next we have V(L1,y)=[Axsin(kxL1)](Aysinhkyy+Bycosh kyy)=0. We avoid Ax=0by setting sin(kxL1)=0so that kxn=nπ/L1forn=1,2,....(Here n=0is omitted because it would produce a trivial solution.) These are eigenvalues corresponding to the eigenfunctions Xn(x)=sin(kxnx), and were found in §A.4 for the harmonic equation. At this point we have a family of solutions Vn(x,y)=sin(kxnx)[Aynsinh(kxny)+Byncosh(kxny)],n=1,2,.... The subscript non the left identiﬁes Vnas the eigensolution associated with eigenvalue kxn. It remains to satisfy boundary conditions at y=0,L2.A t y=0we have Vn(x,0)=sin(kxnx)[Aynsinh 0 +Byncosh 0] =0, hence Byn=0and Vn(x,y)=Aynsin(kxnx)sinh(kxny),n=1,2,.... (A.112) It is clear that no single eigensolution (A.112) can satisfy the one remaining boundary condition. However, we are guaranteed that a series of solutions can represent the con-stant potential on y=L 2; recall that as a solution to a regular Sturm–Liouville problem, the trigonometric functions are complete (hence they could represent any well-behavedfunction on the interval 0≤x≤L 1). In fact, the resulting series is a Fourier sine series for the constant potential at y=L2. So let V(x,y)=∞/summationdisplay n=1Vn(x,y)=∞/summationdisplay n=1Aynsin(kxnx)sinh(kxny). The remaining boundary condition requires V(x,L2)=∞/summationdisplay n=1Aynsin(kxnx)sinh(kxnL2)=V4. The constants Ayncan be found using orthogonality; multiplying through by sin(kxmx) and integrating, we have ∞/summationdisplay n=1Aynsinh(kxnL2)/integraldisplayL1 0sin/parenleftbiggmπx L1/parenrightbigg sin/parenleftbiggnπx L1/parenrightbigg dx=V4/integraldisplayL1 0sin/parenleftbiggmπx L1/parenrightbigg dx. The integral on the left equals δmnL1/2where δmnis the Kronecker delta given by δmn=/braceleftBigg 1,m=n, 0,n/negationslash=m. After evaluating the integral on the right we obtain ∞/summationdisplay n=1Aynδmnsinh(kxnL2)=2V4(1−cosmπ) mπ, hence Aym=2V4(1−cosmπ) mπsinh(kxmL2). The ﬁnal solution for this sub-problem is therefore V(x,y)=∞/summationdisplay n=12V4(1−cosnπ) nπsinh/parenleftBig nπL2 L1/parenrightBigsin/parenleftbiggnπx L1/parenrightbigg sinh/parenleftbiggnπy L1/parenrightbigg . The remaining three sub-problems are left for the reader. Let us again consider (A.111), this time for 0≤x≤L1,0≤y<∞,−∞<z<∞, and subject to V(0,y)=V(L1,y)=0,V(x,0)=V0. Let us try the solution form that worked in the previous example: V(x,y)=[Axsin(kxx)+Bxcos(kxx)][Aysinh(kxy)+Bycosh(kxy)]. The boundary conditions at x=0,L1are the same as before so we have Vn(x,y)=sin(kxnx)[Aynsinh(kxny)+Byncosh(kxny)],n=1,2,.... To ﬁnd Aynand Bynwe note that Vcannot grow without bound as y→∞. Individually the hyperbolic functions grow exponentially. However, using (A.108) we see that Byn= −Ayngives Vn(x,y)=Aynsin(kxnx)e−kxny where Aynis a new unknown constant. (Of course, we could have chosen this exponential dependence at the beginning.) Lastly, we can impose the boundary condition at y=0 on the inﬁnite series of eigenfunctions V(x,y)=∞/summationdisplay n=1Aynsin(kxnx)e−kxny to ﬁnd Ayn. The result is V(x,y)=∞/summationdisplay n=12V0 πn(1−cosnπ)sin(kxnx)e−kxny. As in the previous example, the solution is a discrete superposition of eigenfunctions. The problem consisting of (A.111) holding for 0≤x≤L1,0≤y<∞,−∞<z<∞, along with V(0,y)=0,V(L1,y)=V0e−ay,V(x,0)=0, requires a continuous superposition of eigenfunctions to satisfy the boundary conditions. Let us try V(x,y)=[Axsinhkyx+Bxcosh kyx][Aysinkyy+Bycoskyy]. The conditions at x=0and y=0require that Bx=By=0.T h u s Vky(x,y)=Asinhkyxsinkyy. A single function of this form cannot satisfy the remaining condition at x=L1.S ow e form a continuous superposition V(x,y)=/integraldisplay∞ 0A(ky)sinhkyxsinkyyd k y. (A.113) By the condition at x=L1 /integraldisplay∞ 0A(ky)sinh(kyL1)sinkyyd k y=V0e−ay. (A.114) We can ﬁnd the amplitude function A(ky)by using the orthogonality property δ(y−y/prime)=2 π/integraldisplay∞ 0sinxysinxy/primedx. (A.115) Multiplying both sides of (A.114) by sink/prime yyand integrating, we have /integraldisplay∞ 0A(ky)sinh(kyL1)/bracketleftbigg/integraldisplay∞ 0sinkyysink/prime yydy/bracketrightbigg dky=/integraldisplay∞ 0V0e−aysink/prime yydy. We can evaluate the term in brackets using (A.115) to obtain /integraldisplay∞ 0A(ky)sinh(kyL1)π 2δ(ky−k/prime y)dky=/integraldisplay∞ 0V0e−aysink/prime yydy, hence π 2A(k/prime y)sinh(k/prime yL1)=V0/integraldisplay∞ 0e−aysink/prime yydy. We then evaluate the integral on the right, solve for A(ky), and substitute into (A.113) to obtain V(x,y)=2V0 π/integraldisplay∞ 0ky a2+k2ysinh(kyx) sinh(kyL1)sinkyyd k y. Note that our application of the orthogonality property is merely a calculation of the inverse Fourier sine transform. Thus we could have found the amplitude coeﬃcient byreference to a table of transforms. We can use the Fourier transform solution even when the domain is inﬁnite in more than one dimension. Suppose we solve (A.111) in the region 0≤x<∞,0≤y<∞,−∞<z<∞, subject to V(0,y)=V 0e−ay,V(x,0)=0. Because of the condition at y=0let us use V(x,y)=(Axe−kyx+Bxekyx)(Aysinkyy+Bycoskyy). The solution form Vky(x,y)=B(ky)e−kyxsinkyy satisﬁes the ﬁniteness condition and the homogeneous condition at y=0. The remaining condition can be satisﬁed by a continuous superposition of solutions: V(x,y)=/integraldisplay∞ 0B(ky)e−kyxsinkyyd k y. We must have V0e−ay=/integraldisplay∞ 0B(ky)sinkyyd k y. Use of the orthogonality relationship (A.115) yields the amplitude spectrum B(ky), and we ﬁnd that V(x,y)=2 π/integraldisplay∞ 0e−kyxky a2+k2ysinkyyd k y. (A.116) As a ﬁnal example in rectangular coordinates let us consider a problem in which ψ depends on all three variables: ∇2ψ(x,y,z)+k2ψ(x,y,z)=0 for 0≤x≤L1,0≤y≤L2,0≤z≤L3, subject to ψ(0,y,z)=ψ(L1,y,z)=0, ψ(x,0,z)=ψ(x,L2,z)=0, ψ(x,y,0)=ψ(x,y,L3)=0. Here k/negationslash=0is a constant. This is a three-dimensional eigenvalue problem as described in§A.4, where λ=k2are the eigenvalues and the closed surface is a rectangular box. Physically, the wave function ψrepresents the so-called eigenvalue ornormal mode solutions for the “TM modes” of a rectangular cavity. Since k2 x+k2 y+k2 z=k2,w e might have one or two separation constants equal to zero, but not all three. We ﬁnd,however, that the only solution with a zero separation constant that can ﬁt the boundaryconditions is the trivial solution. In light of the boundary conditions and because weexpect standing waves in the box, we take ψ(x,y,z)=[A xsin(kxx)+Bxcos(kxx)]· ·[Aysin(kyy)+Bycos(kyy)]· ·[Azsin(kzz)+Bzcos(kzz)]. The conditions ψ(0,y,z)=ψ(x,0,z)=ψ(x,y,0)=0give Bx=By=Bz=0. The conditions at x=L1,y=L2, and z=L3require the separation constants to assume the discrete values kx=kxm=mπ/L1,ky=kyn=nπ/L2, and kz=kzp=pπ/L3, where k2 xm+k2 yn+k2 zp=k2 mnpand m,n,p=1,2,.... Associated with each of these eigenvalues is an eigenfunction of a one-dimensional Sturm–Liouville problem. For the three-dimensional problem, an eigenfunction ψmnp(x,y,z)=Amnpsin(kxmx)sin(kyny)sin(kzpz) is associated with each three-dimensional eigenvalue kmnp. Each choice of m,n,ppro- duces a discrete cavity resonance frequency at which the boundary conditions can besatisﬁed. Depending on the values of L 1,2,3, we may have more than one eigenfunction associated with an eigenvalue. For example, if L1=L2=L3=Lthen k121=k211= k112=√ 6π/L. However, the eigenfunctions associated with this single eigenvalue are all diﬀerent: ψ121=sin(kx1x)sin(ky2y)sin(kz1z), ψ211=sin(kx2x)sin(ky1y)sin(kz1z), ψ112=sin(kx1x)sin(ky1y)sin(kz2z). When more than one cavity mode corresponds to a given resonant frequency, we call the modes degenerate . By completeness, we can represent any well-behaved function as f(x,y,z)=/summationdisplay m,n,pAmnpsin(kxmx)sin(kyny)sin(kzpz). The Amnpare found using orthogonality. When such expansions are used to solve prob- lems involving objects (such as excitation probes) inside the cavity, they are termednormal mode expansions of the cavity ﬁeld. Solutions in cylindrical coordinates. In cylindrical coordinates the Helmholtz equa- tion is 1 ρ∂ ∂ρ/parenleftbigg ρ∂ψ(ρ, φ, z) ∂ρ/parenrightbigg +1 ρ2∂2ψ(ρ,φ, z) ∂φ2+∂2ψ(ρ,φ, z) ∂z2+k2ψ(ρ,φ, z)=0.(A.117) With ψ(ρ,φ, z)=P(ρ)/Phi1(φ) Z(z)we obtain 1 ρ∂ ∂ρ/parenleftbigg ρ∂(P/Phi1Z) ∂ρ/parenrightbigg +1 ρ2∂2(P/Phi1Z) ∂φ2+∂2(P/Phi1Z) ∂z2+k2(P/Phi1Z)=0; carrying out the ρderivatives and dividing through by P/Phi1Zwe have −1 Zd2Z dz2=k2+1 ρ2/Phi1d2/Phi1 dφ2+1 ρPdP dρ+1 Pd2P dρ2. The left side depends on zwhile the right side depends on ρandφ, hence both must equal the same constant k2 z: −1 Zd2Z dz2=k2 z, (A.118) k2+1 ρ2/Phi1d2/Phi1 dφ2+1 ρPdP dρ+1 Pd2P dρ2=k2 z. (A.119) We have separated the z-dependence from the dependence on the other variables. For the harmonic equation (A.118), Z(z)=/braceleftBigg Azsinkzz+Bzcoskzz,kz/negationslash=0, azz+bz, kz=0.(A.120) Of course we could use exponentials or a combination of exponentials and trigonometric functions instead. Rearranging (A.119) and multiplying through by ρ2, we obtain −1 /Phi1d2/Phi1 dφ2=/parenleftbig k2−k2 z/parenrightbig ρ2+ρ PdP dρ+ρ2 Pd2P dρ2. The left and right sides depend only on φandρ, respectively; both must equal some constant k2 φ: −1 /Phi1d2/Phi1 dφ2=k2 φ, (A.121) /parenleftbig k2−k2 z/parenrightbig ρ2+ρ PdP dρ+ρ2 Pd2P dρ2=k2 φ. (A.122) The variables ρandφare thus separated, and harmonic equation (A.121) has solutions /Phi1(φ)=/braceleftBigg Aφsinkφφ+Bφcoskφφ, kφ/negationslash=0, aφφ+bφ, kφ=0.(A.123) Equation (A.122) is a bit more involved. In rearranged form it is d2P dρ2+1 ρdP dρ+/parenleftBigg k2 c−k2 φ ρ2/parenrightBigg P=0 (A.124) where k2 c=k2−k2 z. The solution depends on whether any of kz,kφ,o rkcare zero. If kc=kφ=0, then d2P dρ2+1 ρdP dρ=0 so that P(ρ)=aρlnρ+bρ. Ifkc=0butkφ/negationslash=0, we have d2P dρ2+1 ρdP dρ−k2 φ ρ2P=0 so that P(ρ)=aρρ−kφ+bρρkφ. (A.125) This includes the case k=kz=0(Laplace’s equation). If kc/negationslash=0then (A.124) is Bessel’s diﬀerential equation. For noninteger kφthe two independent solutions are denoted Jkφ(z) and J−kφ(z), where Jν(z)is the ordinary Bessel function of the ﬁrst kind of order ν.F o r kφan integer n,Jn(z)and J−n(z)are not independent and a second independent solution denoted Nn(z)must be introduced. This is the ordinary Bessel function of the second kind, order n. As it is also independent when the order is noninteger, Jν(z)and Nν(z) are often chosen as solutions whether νis integer or not. Linear combinations of these independent solutions may be used to produce new independent solutions. The functions H(1) ν(z)and H(2) ν(z)are the Hankel functions of the ﬁrst and second kind of order ν, and are related to the Bessel functions by H(1) ν(z)=Jν(z)+jNν(z), H(2) ν(z)=Jν(z)−jNν(z). The argument zcan be complex (as can ν, but this shall not concern us). When zis imaginary we introduce two new functions Iν(z)and Kν(z), deﬁned for integer order by In(z)=j−nJn(jz), Kn(z)=π 2jn+1H(1) n(jz). Expressions for noninteger order are given in Appendix E.1. Bessel functions cannot be expressed in terms of simple, standard functions. However, a series solution to (A.124) produces many useful relationships between Bessel functionsof diﬀering order and argument. The recursion relations for Bessel functions serve to connect functions of various orders and their derivatives. See Appendix E.1. Of the six possible solutions to (A.124), R(ρ)=  A ρJν(kcρ)+BρNν(kcρ), AρJν(kcρ)+BρH(1) ν(kcρ), AρJν(kcρ)+BρH(2) ν(kcρ), AρNν(kcρ)+BρH(1) ν(kcρ), AρNν(kcρ)+BρH(2) ν(kcρ), AρH(1) ν(kcρ)+BρH(2) ν(kcρ), which do we choose? Again, we are motivated by convenience and the physical nature of the problem. If the argument is real or imaginary, we often consider large or smallargument behavior. For xreal and large, J ν(x)→/radicalbigg 2 πxcos/parenleftBig x−π 4−νπ 2/parenrightBig , Nν(x)→/radicalbigg 2 πxsin/parenleftBig x−π 4−νπ 2/parenrightBig , H(1) ν(x)→/radicalbigg 2 πxej(x−π 4−νπ 2), H(2) ν(x)→/radicalbigg 2 πxe−j(x−π 4−νπ 2), Iν(x)→/radicalbigg 1 2πxex, Kν(x)→/radicalbiggπ 2xe−x, while for xreal and small, J0(x)→1, N0(x)→2 π(lnx+0.5772157 −ln 2), Jν(x)→1 ν!/parenleftBigx 2/parenrightBigν , Nν(x)→−(ν−1)! π/parenleftbigg2 x/parenrightbiggν . Because Jν(x)and Nν(x)oscillate for large argument, they can represent standing waves along the radial direction. However, Nν(x)is unbounded for small xand is inappropriate for regions containing the origin. The Hankel functions become complex exponentials forlarge argument, hence represent traveling waves. Finally, K ν(x)is unbounded for small x and cannot be used for regions containing the origin, while Iν(x)increases exponentially for large xand cannot be used for unbounded regions. Examples. Consider the boundary value problem for Laplace’s equation ∇2V(ρ, φ) =0 (A.126) in the region 0≤ρ≤∞,0≤φ≤φ0,−∞<z<∞, where the boundary conditions are V(ρ,0)=0,V(ρ, φ 0)=V0. Since there is no z-dependence we let kz=0in (A.120) and choose az=0. Then k2 c=k2−k2 z=0since k=0. There are two possible solutions, depending on whether kφis zero. First let us try kφ/negationslash=0. Using (A.123) and (A.125) we have V(ρ, φ) =[Aφsin(kφφ)+Bφcos(kφφ)][aρρ−kφ+bρρkφ]. (A.127) Assuming kφ>0we must have bρ=0to keep the solution ﬁnite. The condition V(ρ,0)=0requires Bφ=0.T h u s V(ρ, φ) =Aφsin(kφφ)ρ−kφ. Our ﬁnal boundary condition requires V(ρ, φ 0)=V0=Aφsin(kφφ0)ρ−kφ. Because this cannot hold for all ρ, we must resort to kφ=0and V(ρ, φ) =(aφφ+bφ)(aρlnρ+bρ). (A.128) Proper behavior as ρ→∞ dictates that aρ=0.V(ρ,0)=0requires bφ=0.T h u s V(ρ, φ) =V(φ)=bφφ.The constant bφis found from the remaining boundary condition: V(φ0)=V0=bφφ0so that bφ=V0/φ0. The ﬁnal solution is V(φ)=V0φ/φ 0. It is worthwhile to specialize this to φ0=π/2and compare with the solution to the same problem found earlier using rectangular coordinates. With a=0in (A.116) we have V(x,y)=2 π/integraldisplay∞ 0e−kyxsinkyy kydky. Despite its much more complicated form, this must be the same solution by uniqueness. Next let us solve (A.126) subject to the “split cylinder” conditions V(a,φ)=/braceleftBigg V0,0<φ<π , 0,−π<φ< 0. Because there is no z-dependence we choose kz=az=0and have k2 c=k2−k2 z=0. Since kφ=0would violate the boundary conditions at ρ=a, we use V(ρ, φ) =(aρρ−kφ+bρρkφ)(Aφsinkφφ+Bφcoskφφ). The potential must be single-valued in φ:V(ρ, φ+2nπ)=V(ρ, φ) . This is only possible ifkφis an integer, say kφ=m. Then Vm(ρ, φ) =/braceleftBigg (Amsinmφ+Bmcosmφ)ρm,ρ < a, (Cmsinmφ+Dmcosmφ)ρ−m,ρ > a. On physical grounds we have discarded ρ−mforρ< aandρmforρ> a. To satisfy the boundary conditions at ρ=awe must use an inﬁnite series of the complete set of eigensolutions. For ρ< athe boundary condition requires B0+∞/summationdisplay m=1(Amsinmφ+Bmcosmφ)am=/braceleftBigg V0,0<φ<π , 0,−π<φ< 0. Application of the orthogonality relations /integraldisplayπ −πcosmφcosnφdφ=2π /epsilon1nδmn,m,n=0,1,2,..., (A.129) /integraldisplayπ −πsinmφsinnφdφ=πδmn,m,n=1,2,..., (A.130) /integraldisplayπ −πcosmφsinnφdφ=0,m,n=0,1,2,..., (A.131) where /epsilon1n=/braceleftBigg 1,n=0, 2,n>0,(A.132) is Neumann’s number, produces appropriate values for the constants Amand Bm. The full solution is V(ρ, φ) =  V 0 2+V0 π∞/summationdisplay n=1[1−(−1)n] n/parenleftBigρ a/parenrightBign sinnφ, ρ < a, V0 2+V0 π∞/summationdisplay n=1[1−(−1)n] n/parenleftbigga ρ/parenrightbiggn sinnφ, ρ > a. The boundary value problem ∇2V(ρ, φ, z)=0,0≤ρ≤a,−π≤φ≤π,0≤z≤h, V(ρ, φ, 0)=0,0≤ρ≤a,−π≤φ≤π, V(a,φ,z)=0,−π≤φ≤π,0≤z≤h, V(ρ, φ, h)=V0,0≤ρ≤a,−π≤φ≤π, describes the potential within a grounded “canister” with top at potential V0. Symmetry precludes φ-dependence, hence kφ=aφ=0. Since k=0(Laplace’s equation) we also have k2 c=k2−k2 z=− k2 z. Thus we have either kzreal and kc=jkz,o r kcreal and kz=jkc. With kzreal we have V(ρ,z)=[Azsinkzz+Bzcoskzz][AρK0(kzρ)+BρI0(kzρ)]; (A.133) with kcreal we have V(ρ,z)=[Azsinhkcz+Bzcosh kcz][AρJ0(kcρ)+BρN0(kcρ)]. (A.134) The functions K0and I0are inappropriate for use in this problem, and we proceed to (A.134). Since N0is unbounded for small argument, we need Bρ=0. The condition V(ρ,0)=0gives Bz=0,t h u s V(ρ,z)=Azsinh(kcz)J0(kcρ). The oscillatory nature of J0means that we can satisfy the condition at ρ=a: V(a,z)=Azsinh(kcz)J0(kca)=0for 0≤z<h ifJ0(kca)=0.Letting p0mdenote the mth root of J0(x)=0form=1,2,..., we have kcm=p0m/a. Because we cannot satisfy the boundary condition at z=hwith a single eigensolution, we use the superposition V(ρ,z)=∞/summationdisplay m=1Amsinh/parenleftBigp0mz a/parenrightBig J0/parenleftBigp0mρ a/parenrightBig . We require V(ρ,h)=∞/summationdisplay m=1Amsinh/parenleftbiggp0mh a/parenrightbigg J0/parenleftBigp0mρ a/parenrightBig =V0, (A.135) where the Amcan be evaluated by orthogonality of the functions J0(p0mρ/a).I f pνmis themth root of Jν(x)=0, then /integraldisplaya 0Jν/parenleftBigpνmρ a/parenrightBig Jν/parenleftBigpνnρ a/parenrightBig ρdρ=δmna2 2J/prime2 ν(pνn)=δmna2 2J2 ν+1(pνn) (A.136) where J/prime ν(x)=dJν(x)/dx. Multiplying (A.135) by ρJ0(p0nρ/a)and integrating, we have Ansinh/parenleftbiggp0nh a/parenrightbigga2 2J/prime2 0(p0na)=/integraldisplaya 0V0J0/parenleftBigp0nρ a/parenrightBig ρdρ. Use of (E.105), /integraldisplay xn+1Jn(x)dx=xn+1Jn+1(x)+C, allows us to evaluate /integraldisplaya 0J0/parenleftBigp0nρ a/parenrightBig ρdρ=a2 p0nJ1(p0n). With this we ﬁnish calculating Amand have V(ρ,z)=2V0∞/summationdisplay m=1sinh(p0m az)J0(p0m aρ) p0msinh(p0m ah)J1(p0m) as the desired solution. Finally, let us assume that k>0and solve ∇2ψ(ρ,φ, z)+k2ψ(ρ,φ, z)=0 where 0≤ρ≤a,−π≤φ≤π, and −∞<z<∞, subject to the condition ˆn·∇ψ(ρ,φ, z)/vextendsingle/vextendsingle/vextendsingle ρ=a=∂ψ(ρ, φ, z) ∂ρ/vextendsingle/vextendsingle/vextendsingle ρ=a=0 for−π≤φ≤πand−∞<z<∞.The solution to this problem leads to the transverse- electric (TE z) ﬁelds in a lossless circular waveguide, where ψrepresents the z-component of the magnetic ﬁeld. Although there is symmetry with respect to φ, we seek φ-dependent solutions; the resulting complete eigenmode solution will permit us to expand any well-behaved function within the waveguide in terms of a normal mode (eigenfunction) series.In this problem none of the constants k,k z,o r kφequal zero, except as a special case. However, the ﬁeld must be single-valued in φand thus kφmust be an integer m.W e consider our possible choices for P(ρ),Z(z), and /Phi1(φ). Since k2 c=k2−k2 zand k2>0is arbitrary, we must consider various possibilities for the signs of k2 cand k2 z. We can rule outk2 c<0based on consideration of the behavior of the functions Imand Km. We also need not consider kc<0, since this gives the same solution as kc>0. We are then left with two possible cases. Writing k2 z=k2−k2 c, we see that either k>kcand k2 z>0,o r k<kcand k2 z<0.F o r k2 z>0we write ψ(ρ,φ, z)=[Aze−jkzz+Bzejkzz][Aφsinmφ+Bφcosmφ]Jm(kcρ). Here the terms involving e∓jkzzrepresent waves propagating in the ±zdirections. The boundary condition at ρ=arequires J/prime m(kca)=0 where J/prime m(x)=dJm(x)/dx. Denoting the nth zero of J/prime m(x)byp/prime mnwe have kc=kcm= p/prime mn/a.This gives the eigensolutions ψm=[Azme−jkzz+Bzmejkzz][Aφmsinmφ+Bφmcosmφ]kcJm/parenleftbiggp/prime mnρ a/parenrightbigg . The undetermined constants Azm,Bzm,Aρm,Bρmcould be evaluated when the individual eigensolutions are used to represent a function in terms of a modal expansion. Forthe case k 2 z<0we again choose complex exponentials in z; however, kz=− jαgives e∓jkzz=e∓αzand attenuation along z. The reader can verify that the eigensolutions are ψm=[Azme−αz+Bzmeαz][Aφmsinmφ+Bφmcosmφ]kcJm/parenleftbiggp/prime mnρ a/parenrightbigg where now k2 c=k2+α2. We have used Bessel function completeness in the examples above. This property is a consequence of the Sturm–Liouville problem ﬁrst studied in §A.4. We often use Fourier– Bessel series to express functions over ﬁnite intervals. Over inﬁnite intervals we use theFourier–Bessel transform. The Fourier–Bessel series can be generalized to Bessel functions of noninteger order, and to the derivatives of Bessel functions. Let f(ρ)be well-behaved over the interval [0,a]. Then the series f(ρ)=∞/summationdisplay m=1CmJν/parenleftBig pνmρ a/parenrightBig ,0≤ρ≤a,ν> −1 converges, and the constants are Cm=2 a2J2 ν+1(pνm)/integraldisplaya 0f(ρ)Jν/parenleftBig pνmρ a/parenrightBig ρdρ by (A.136). Here pνmis the mth root of Jν(x). An alternative form of the series uses p/prime νm, the roots of J/prime ν(x), and is given by f(ρ)=∞/summationdisplay m=1DmJν/parenleftBig p/prime νmρ a/parenrightBig ,0≤ρ≤a,ν> −1. In this case the expansion coeﬃcients are found using the orthogonality relationship /integraldisplaya 0Jν/parenleftbiggp/prime νm aρ/parenrightbigg Jν/parenleftbiggp/prime νn aρ/parenrightbigg ρdρ=δmna2 2/parenleftbigg 1−ν2 p/prime2νm/parenrightbigg J2 ν(p/prime νm), and are Dm=2 a2/parenleftBig 1−ν2 p/prime2νmJ2ν(p/primeνm)/parenrightBig/integraldisplaya 0f(ρ)Jν/parenleftbiggp/prime νm aρ/parenrightbigg ρdρ. Solutions in spherical coordinates. If into Helmholtz’s equation 1 r2∂ ∂r/parenleftbigg r2∂ψ(r,θ,φ) ∂r/parenrightbigg +1 r2sinθ∂ ∂θ/parenleftbigg sinθ∂ψ(r,θ,φ) ∂θ/parenrightbigg + +1 r2sin2θ∂2ψ(r,θ,φ) ∂φ2+k2ψ(r,θ,φ) =0 we put ψ(r,θ,φ) =R(r)/Theta1(θ)/Phi1(φ) and multiply through by r2sin2θ/ψ( r,θ,φ) , we obtain sin2θ R(r)d dr/parenleftbigg r2dR(r) dr/parenrightbigg +sinθ /Theta1(θ)d dθ/parenleftbigg sinθd/Theta1(θ) dθ/parenrightbigg +k2r2sin2θ=−1 /Phi1(φ)d2/Phi1(φ) dφ2. Since the right side depends only on φwhile the left side depends only on randθ, both sides must equal some constant µ2: sin2θ R(r)d dr/parenleftbigg r2dR(r) dr/parenrightbigg +sinθ /Theta1(θ)d dθ/parenleftbigg sinθd/Theta1(θ) dθ/parenrightbigg +k2r2sin2θ=µ2, (A.137) d2/Phi1(φ) dφ2+µ2/Phi1(φ)=0. (A.138) We have thus separated oﬀ the φ-dependence. Harmonic ordinary diﬀerential equation (A.138) has solutions /Phi1(φ)=/braceleftBigg Aφsinµφ+Bφcosµφ, µ /negationslash=0, aφφ+bφ,µ =0. (We could have used complex exponentials to describe /Phi1(φ), or some combination of exponentials and trigonometric functions, but it is conventional to use only trigonometricfunctions.) Rearranging (A.137) and dividing through by sin 2θwe have 1 R(r)d dr/parenleftbigg r2dR(r) dr/parenrightbigg +k2r2=−1 sinθ/Theta1(θ)d dθ/parenleftbigg sinθd/Theta1(θ) dθ/parenrightbigg +µ2 sin2θ. We introduce a new constant k2 θto separate rfromθ: 1 R(r)d dr/parenleftbigg r2dR(r) dr/parenrightbigg +k2r2=k2 θ, (A.139) −1 sinθ/Theta1(θ)d dθ/parenleftbigg sinθd/Theta1(θ) dθ/parenrightbigg +µ2 sin2θ=k2 θ. (A.140) Equation (A.140), 1 sinθd dθ/parenleftbigg sinθd/Theta1(θ) dθ/parenrightbigg +/parenleftbigg k2 θ−µ2 sin2θ/parenrightbigg /Theta1(θ)=0, can be put into a standard form by letting η=cosθ (A.141) and k2 θ=ν(ν+1)where νis a parameter: (1−η2)d2/Theta1(η) dη2−2ηd/Theta1(η) dη+/bracketleftbigg ν(ν+1)−µ2 1−η2/bracketrightbigg /Theta1(η)=0,−1≤η≤1. This is the associated Legendre equation . It has two independent solutions called as- sociated Legendre functions of the ﬁrst andsecond kinds , denoted Pµ ν(η)and Qµ ν(η), respectively. In these functions, all three quantities µ,ν,η may be arbitrary complex constants as long as ν+µ/negationslash= −1,−2,.... But (A.141) shows that ηis real in our discus- sion;µwill generally be real also, and will be an integer whenever /Phi1(φ) is single-valued. The choice of νis somewhat more complicated. The function Pµ ν(η)diverges at η=±1 unless νis an integer, while Qµ ν(η)diverges at η=±1regardless of whether νis an inte- ger. In §A.4 we required that Pµ ν(η)be bounded on [−1,1]to have a Sturm–Liouville problem with suitable orthogonality properties. By (A.141) we must exclude Qµ ν(η)for problems containing the z-axis, and restrict νto be an integer ninPµ ν(η)for such prob- lems. In case the z-axis is excluded, we choose ν=nwhenever possible, because the ﬁnite sums Pm n(η)and Qm n(η)are much easier to manipulate than Pµ ν(η)and Qµ ν(η). In many problems we must count on completeness of the Legendre polynomials Pn(η)=P0 n(η)or spherical harmonics Ymn(θ, φ) in order to satisfy the boundary conditions. In this book we shall consider only those boundary value problems that can be solved using integervalues of νandµ, hence choose /Theta1(θ)=A θPm n(cosθ)+BθQm n(cosθ). (A.142) Single-valuedness in /Phi1(φ)is a consequence of having µ=m, andφ=constant boundary surfaces are thereby disallowed. The associated Legendre functions have many important properties. For instance, Pm n(η)=  0, m>n, (−1)m(1−η2)m/2 2nn!dn+m(η2−1)n dηn+m,m≤n.(A.143) The case m=0receives particular attention because it corresponds to azimuthal invari- ance ( φ-independence). We deﬁne P0 n(η)=Pn(η)where Pn(η)is the Legendre polynomial of order n. From (A.143), we see that4 Pn(η)=1 2nn!dn(η2−1)n dηn is a polynomial of degree n, and that Pm n(η)=(−1)m(1−η2)m/2dm dηmPn(η). Both the associated Legendre functions and the Legendre polynomials obey orthogonality relations and many recursion formulas. In problems where the z-axis is included, the product /Theta1(θ)/Phi1(φ) is sometimes deﬁned as the spherical harmonic Ynm(θ, φ) =/radicalBigg 2n+1 4π(n−m)! (n+m)!Pm n(cosθ)ejmθ. These functions, which are complete over the surface of a sphere, were treated earlier in this section. Remembering that k2 r=ν(ν+1), the r-dependent equation (A.139) becomes 1 r2d dr/parenleftbigg r2dR(r) dr/parenrightbigg +/parenleftbigg k2+n(n+1) r2/parenrightbigg R(r)=0. (A.144) When k=0we have d2R(r) dr2+2 rdR(r) dr−n(n+1) r2R(r)=0 so that R(r)=Arrn+Brr−(n+1). When k/negationslash=0, the substitution ¯R(r)=√ kr R(r)puts (A.144) into the form r2d2¯R(r) dr2+rd¯R(r) dr+/bracketleftBigg k2r2−/parenleftbigg n+1 2/parenrightbigg2/bracketrightBigg ¯R(r)=0, which we recognize as Bessel’s equation of half-integer order. Thus R(r)=¯R(r)√ kr=Zn+1 2(kr) √ kr. For convenience we deﬁne the spherical Bessel functions jn(z)=/radicalbiggπ 2zJn+1 2(z), nn(z)=/radicalbiggπ 2zNn+1 2(z)=(−1)n+1/radicalbiggπ 2zJ−(n+1 2)(z), h(1) n(z)=/radicalbiggπ 2zH(1) n+1 2(z)=jn(z)+jnn(z), h(2) n(z)=/radicalbiggπ 2zH(2) n+1 2(z)=jn(z)−jnn(z). 4Care must be taken when consulting tables of Legendre functions and their properties. In particular, one must be on the lookout for possible disparities regarding the factor (−1)m(cf., [76, 1, 109, 8] vs. [5, 187]). Similar care is needed with Qm n(x). These can be written as ﬁnite sums involving trigonometric functions and inverse powers ofz. We have, for instance, j0(z)=sinz z, n0(z)=−cosz z, j1(z)=sinz z2−cosz z, n1(z)=−cosz z2−sinz z. We can now write R(r)as a linear combination of any two of the spherical Bessel functions jn,nn,h(1) n,h(2) n: R(r)=  A rjn(kr)+Brnn(kr), Arjn(kr)+Brh(1) n(kr), Arjn(kr)+Brh(2) n(kr), Arnn(kr)+Brh(1) n(kr), Arnn(kr)+Brh(2) n(kr), Arh(1) n(kr)+Brh(2) n(kr).(A.145) Imaginary arguments produce modiﬁed spherical Bessel functions ; the interested reader is referred to Gradsteyn [76] or Abramowitz [1]. Examples. The problem ∇2V(r,θ,φ) =0,θ 0≤θ≤π/2,0≤r<∞,−π≤φ≤π, V(r,θ0,φ)=V0,−π≤φ≤π,0≤r<∞, V(r,π/2,φ)=0,−π≤φ≤π,0≤r<∞, gives the potential ﬁeld between a cone and the z=0plane. Azimuthal symmetry prompts us to choose µ=aφ=0. Since k=0we have R(r)=Arrn+Brr−(n+1). (A.146) Noting that positive and negative powers of rare unbounded for large and small r, respectively, we take n=Br=0. Hence the solution depends only on θ: V(r,θ,φ) =V(θ)=AθP0 0(cosθ)+BθQ0 0(cosθ). We must retain Q0 0since the solution region does not contain the z-axis. Using P0 0(cosθ)=1and Q0 0(cosθ)=ln cot(θ/2) (cf., Appendix E.2), we have V(θ)=Aθ+Bθln cot(θ/2). A straightforward application of the boundary conditions gives Aθ=0and Bθ= V0/ln cot(θ0/2), hence V(θ)=V0ln cot(θ/2) ln cot(θ0/2). Next we solve the boundary value problem ∇2V(r,θ,φ) =0, V(a,θ,φ) =− V0,π / 2≤θ<π , −π≤φ≤π, V(a,θ,φ) =+ V0,0<θ≤π/2,−π≤φ≤π, for both r>aand r<a. This yields the potential ﬁeld of a conducting sphere split into top and bottom hemispheres and held at a potential diﬀerence of 2V0. Azimuthal symmetry gives µ=0. The two possible solutions for /Theta1(θ)are /Theta1(θ)=/braceleftBigg Aθ+Bθln cot(θ/2),n=0, AθPn(cosθ), n/negationslash=0, where we have discarded Q0 0(cosθ)because the region of interest contains the z-axis. The n=0solution cannot match the boundary conditions; neither can a single term of the type AθPn(cosθ), but a series of these latter terms can. We use V(r,θ)=∞/summationdisplay n=0Vn(r,θ)=∞/summationdisplay n=0[Arrn+Brr−(n+1)]Pn(cosθ). (A.147) The terms r−(n+1)and rnare not allowed, respectively, for r<aand r>a.F o r r<a then, V(r,θ)=∞/summationdisplay n=0AnrnPn(cosθ). Letting V0(θ)be the potential on the surface of the split sphere, we impose the boundary condition: V(a,θ)=V0(θ)=∞/summationdisplay n=0AnanPn(cosθ), 0≤θ≤π. This is a Fourier–Legendre expansion of V0(θ). The Anare evaluated by orthogonality. Multiplying by Pm(cosθ)sinθand integrating from θ=0toπ, we obtain ∞/summationdisplay n=0Anan/integraldisplayπ 0Pn(cosθ)Pm(cosθ)sinθdθ=/integraldisplayπ 0V0(θ)Pm(cosθ)sinθdθ. Using orthogonality relationship (A.93) and the given V0(θ)we have Amam 2 2m+1=V0/integraldisplayπ/2 0Pm(cosθ)sinθdθ−V0/integraldisplayπ π/2Pm(cosθ)sinθdθ. The substitution η=cosθgives Amam 2 2m+1=V0/integraldisplay1 0Pm(η)dη−V0/integraldisplay0 −1Pm(η)dη =V0/integraldisplay1 0Pm(η)dη−V0/integraldisplay1 0Pm(−η)dη; then Pm(−η)=(−1)mPm(η)gives Am=a−m2m+1 2V0[1−(−1)m]/integraldisplay1 0Pm(η)dη. Because Am=0formeven, we can put m=2n+1(n=0,1,2,...) and have A2n+1=(4n+3)V0 a2n+1/integraldisplay1 0P2n+1(η)dη=V0(−1)n a2n+14n+3 2n+2(2n!) (2nn!)2 by (E.176). Hence V(r,θ)=∞/summationdisplay n=0V0(−1)n4n+3 2n+2(2n!) (2nn!)2/parenleftBigr a/parenrightBig2n+1 P2n+1(cosθ) forr<a. The case r>ais left to the reader. Finally, consider ∇2ψ(x,y,z)+k2ψ(x,y,z)=0,0≤r≤a,0≤θ≤π,−π≤φ≤π, ψ(a,θ,φ) =0,0≤θ≤π,−π≤φ≤π, where k/negationslash=0is constant. This is a three-dimensional eigenvalue problem. Wave function ψrepresents the solutions for the electromagnetic ﬁeld within a spherical cavity for modes TE to r. Despite the prevailing symmetry, we choose solutions that vary with both θ andφ. We are motivated by a desire to solve problems involving cavity excitation, and eigenmode completeness will enable us to represent any piecewise continuous functionwithin the cavity. We employ spherical harmonics because the boundary surface is asphere. These exclude Q n m(cosθ), which is appropriate since our problem contains the z-axis. Since k/negationslash=0we must choose a radial dependence from (A.145). Small-argument behavior rules out nn,h(1) n, and h(2) n, leaving us with ψ(r,θ,φ) =Amnjn(kr)Ynm(θ, φ) or, equivalently, ψ(r,θ,φ) =Amnjn(kr)Pm n(cosθ)ejmφ. The eigenvalues λ=k2are found by applying the condition at r=a: ψ(a,θ,φ) =Amnjn(ka)Ynm(θ, φ) =0, requiring jn(ka)=0.Denoting the qth root of jn(x)=0byαnq, we have knq=αnq/a and corresponding eigenfunctions ψmnq(r,θ,φ) =Amnqjn(knqr)Ynm(θ, φ). The eigenvalues are proportional to the resonant frequencies of the cavity and the eigen- functions can be used to ﬁnd the modal ﬁeld distributions. Since the eigenvalues areindependent of m, we may have several eigenfunctions ψ mnqassociated with each kmnq. The only limitation is that we must keep m≤nto have Pn m(cosθ)nonzero. This is another instance of mode degeneracy. There are 2ndegenerate modes associated with each resonant frequency (one for each of e±jnφ). By completeness we can expand any piecewise continuous function within or on the sphere as a series f(r,θ,φ) =/summationdisplay m,n,qAmnqjn(knqr)Ynm(θ, φ). Appendix B Useful identities Algebraic identities for vectors and dyadics A+B=B+A (B.1) A·B=B·A (B.2) A×B=−B×A (B.3) A·(B+C)=A·B+A·C (B.4) A×(B+C)=A×B+A×C (B.5) A·(B×C)=B·(C×A)=C·(A×B) (B.6) A×(B×C)=B(A·C)−C(A·B)=B×(A×C)+C×(B×A) (B.7) (A×B)·(C×D)=A·[B×(C×D)]=(B·D)(A·C)−(B·C)(A·D) (B.8) (A×B)×(C×D)=C[A·(B×D)]−D[A·(B×C)] (B.9) A×[B×(C×D)]=(B·D)(A×C)−(B·C)(A×D) (B.10) A·(¯c·B)=(A·¯c)·B (B.11) A×(¯c×B)=(A×¯c)×B (B.12) C·(¯a·¯b)=(C·¯a)·¯b (B.13) (¯a·¯b)·C=¯a·(¯b·C) (B.14) A·(B×¯c)=−B·(A×¯c)=(A×B)·¯c (B.15) A×(B×¯c)=B·(A×¯c)−¯c(A·B) (B.16) A·¯I=¯I·A=A (B.17) Integral theorems Note: Sbounds V,/Gamma1bounds S,ˆnis normal to Satr,ˆland ˆmare tangential to Sat r,ˆlis tangential to the contour /Gamma1,ˆm×ˆl=ˆn,dl=ˆldl, and dS=ˆndS. Divergence theorem/integraldisplay V∇·AdV=/contintegraldisplay SA·dS (B.18) /integraldisplay V∇·¯adV=/contintegraldisplay Sˆn·¯adS (B.19) /integraldisplay S∇s·AdS=/contintegraldisplay /Gamma1ˆm·Adl (B.20) Gradient theorem /integraldisplay V∇adV=/contintegraldisplay SadS (B.21) /integraldisplay V∇AdV=/contintegraldisplay SˆnAdS (B.22) /integraldisplay V∇sadS=/contintegraldisplay /Gamma1ˆmad l (B.23) Curl theorem /integraldisplay V(∇× A)dV=−/contintegraldisplay SA×dS (B.24) /integraldisplay V(∇× ¯a)dV=/contintegraldisplay Sˆn×¯adS (B.25) /integraldisplay S∇s×AdS=/contintegraldisplay /Gamma1ˆm×Adl (B.26) Stokes’s theorem /integraldisplay S(∇× A)·dS=/contintegraldisplay /Gamma1A·dl (B.27) /integraldisplay Sˆn·(∇× ¯a)dS=/contintegraldisplay /Gamma1dl·¯a (B.28) Green’s ﬁrst identity for scalar ﬁelds /integraldisplay V(∇a·∇b+a∇2b)dV=/contintegraldisplay Sa∂b ∂ndS (B.29) Green’s second identity for scalar ﬁelds (Green’s theorem) /integraldisplay V(a∇2b−b∇2a)dV=/contintegraldisplay S/parenleftbigg a∂b ∂n−b∂a ∂n/parenrightbigg dS (B.30) Green’s ﬁrst identity for vector ﬁelds /integraldisplay V{(∇× A)·(∇× B)−A·[∇×(∇× B)]}dV= /integraldisplay V∇·[A×(∇× B)]dV=/contintegraldisplay S[A×(∇× B)]·dS (B.31) Green’s second identity for vector ﬁelds /integraldisplay V{B·[∇×(∇× A)]−A·[∇×(∇× B)]}dV= /contintegraldisplay S[A×(∇× B)−B×(∇× A)]·dS (B.32) Helmholtztheorem A(r)=− ∇/bracketleftbigg/integraldisplay V∇/prime·A(r/prime) 4π|r−r/prime|dV/prime−/contintegraldisplay SA(r/prime)·ˆn/prime 4π|r−r/prime|dS/prime/bracketrightbigg + +∇×/bracketleftbigg/integraldisplay V∇/prime×A(r/prime) 4π|r−r/prime|dV/prime+/contintegraldisplay SA(r/prime)×ˆn/prime 4π|r−r/prime|dS/prime/bracketrightbigg (B.33) Miscellaneous identities /contintegraldisplay SdS=0 (B.34) /integraldisplay Sˆn×(∇a)dS=/contintegraldisplay /Gamma1adl (B.35) /integraldisplay S(∇a×∇b)·dS=/integraldisplay /Gamma1a∇b·dl=−/integraldisplay /Gamma1b∇a·dl (B.36) /contintegraldisplay dl A=/integraldisplay Sˆn×(∇A)dS (B.37) Derivative identities ∇(a+b)=∇a+∇b (B.38) ∇·(A+B)=∇· A+∇· B (B.39) ∇×(A+B)=∇× A+∇× B (B.40) ∇(ab)=a∇b+b∇a (B.41) ∇·(aB)=a∇·B+B·∇a (B.42) ∇×(aB)=a∇× B−B×∇a (B.43) ∇·(A×B)=B·∇× A−A·∇× B (B.44) ∇×(A×B)=A(∇·B)−B(∇·A)+(B·∇)A−(A·∇)B (B.45) ∇(A·B)=A×(∇× B)+B×(∇× A)+(A·∇)B+(B·∇)A (B.46) ∇×(∇× A)=∇(∇·A)−∇2A (B.47) ∇·(∇a)=∇2a (B.48) ∇·(∇× A)=0 (B.49) ∇×(∇a)=0 (B.50) ∇×(a∇b)=∇a×∇b (B.51) ∇2(ab)=a∇2b+2(∇a)·(∇b)+b∇2a (B.52) ∇2(aB)=a∇2B+B∇2a+2(∇a·∇)B (B.53) ∇2¯a=∇(∇·¯a)−∇× (∇× ¯a) (B.54) ∇·(AB)=(∇·A)B+A·(∇B)=(∇·A)B+(A·∇)B (B.55) ∇×(AB)=(∇× A)B−A×(∇B) (B.56) ∇·(∇× ¯a)=0 (B.57) ∇×(∇A)=0 (B.58) ∇(A×B)=(∇A)×B−(∇B)×A (B.59) ∇(aB)=(∇a)B+a(∇B) (B.60) ∇·(a¯b)=(∇a)·¯b+a(∇·¯b) (B.61) ∇×(a¯b)=(∇a)×¯b+a(∇× ¯b) (B.62) ∇·(a¯I)=∇a (B.63) ∇×(a¯I)=∇a×¯I (B.64) Identities involving the displacement vector Note: R=r−r/prime,R=|R|,ˆR=R/R,f/prime(x)=df(x)/dx. ∇f(R)=− ∇/primef(R)=ˆRf/prime(R) (B.65) ∇R=ˆR (B.66) ∇/parenleftbigg1 R/parenrightbigg =−ˆR R2(B.67) ∇/parenleftbigge−jkR R/parenrightbigg =− ˆR/parenleftbigg1 R+jk/parenrightbigge−jkR R(B.68) ∇·/bracketleftbig f(R)ˆR/bracketrightbig =− ∇/prime·/bracketleftbig f(R)ˆR/bracketrightbig =2f(R) R+f/prime(R) (B.69) ∇·R=3 (B.70) ∇·ˆR=2 R(B.71) ∇·/parenleftbigg ˆRe−jkR R/parenrightbigg =/parenleftbigg1 R−jk/parenrightbigge−jkR R(B.72) ∇×/bracketleftbig f(R)ˆR/bracketrightbig =0 (B.73) ∇2/parenleftbigg1 R/parenrightbigg =−4πδ(R) (B.74) (∇2+k2)e−jkR R=−4πδ(R) (B.75) Identities involving the plane- wavefunction Note: Eis a constant vector, k=|k|. ∇/parenleftbig e−jk·r/parenrightbig =− jke−jk·r(B.76) ∇·/parenleftbig Ee−jk·r/parenrightbig =− jk·Ee−jk·r(B.77) ∇×/parenleftbig Ee−jk·r/parenrightbig =− jk×Ee−jk·r(B.78) ∇2/parenleftbig Ee−jk·r/parenrightbig =−k2Ee−jk·r(B.79) Identities involving the transverse/longitudinal decomposition Note: ˆuis a constant unit vector, Au≡ˆu·A,∂/∂u≡ˆu·∇,At≡A−ˆuAu,∇t≡ ∇− ˆu∂/∂u. A=At+ˆuAu (B.80) ∇=∇ t+ˆu∂ ∂u(B.81) ˆu·At=0 (B.82) (ˆu·∇t)φ=0 (B.83) ∇tφ=∇φ−ˆu∂φ ∂u(B.84) ˆu·(∇φ)=(ˆu·∇)φ=∂φ ∂u(B.85) ˆu·(∇tφ)=0 (B.86) ∇t·(ˆuφ)=0 (B.87) ∇t×(ˆuφ)=− ˆu×∇ tφ (B.88) ∇t×(ˆu×A)=ˆu∇t·At (B.89) ˆu×(∇t×A)=∇ tAu (B.90) ˆu×(∇t×At)=0 (B.91) ˆu·(ˆu×A)=0 (B.92) ˆu×(ˆu×A)=−At (B.93) ∇φ=∇ tφ+ˆu∂φ ∂u(B.94) ∇·A=∇ t·At+∂Au ∂u(B.95) ∇× A=∇ t×At+ˆu×/bracketleftbigg∂At ∂u−∇ tAu/bracketrightbigg (B.96) ∇2φ=∇2 tφ+∂2φ ∂u2(B.97) ∇×∇× A=/bracketleftbigg ∇t×∇ t×At−∂2At ∂u2+∇ t∂Au ∂u/bracketrightbigg +ˆu/bracketleftbigg∂ ∂u(∇t·At)−∇2 tAu/bracketrightbigg (B.98) ∇2A=/bracketleftbigg ∇t(∇t·At)+∂2At ∂u2−∇ t×∇ t×At/bracketrightbigg +ˆu∇2Au (B.99) Appendix C Some Fourier transform pairs Note: G(k)=/integraldisplay∞ −∞g(x)e−jkxdx,g(x)=1 2π/integraldisplay∞ −∞G(k)ejkxdk,g(x)↔G(k). rect(x)↔2 sinc k (C.1) /Lambda1(x)↔sinc2k 2(C.2) sgn(x)↔2 jk(C.3) ejk0x↔2πδ(k−k0) (C.4) δ(x)↔1 (C.5) 1↔2πδ(k) (C.6) dnδ(x) dxn↔(jk)n(C.7) xn↔2πjndnδ(k) dkn(C.8) U(x)↔πδ(k)+1 jk(C.9) ∞/summationdisplay n=−∞δ/parenleftbigg t−n2π k0/parenrightbigg ↔k0∞/summationdisplay n=−∞δ(k−nk0) (C.10) e−ax2↔/radicalbiggπ ae−k2 4a (C.11) e−axU(x)↔1 a+jk(C.12) e−a|x|↔2a a2+k2(C.13) e−axcosbx U(x)↔a+jk (a+jk)2+b2(C.14) e−axsinbx U(x)↔b (a+jk)2+b2(C.15) cosk0x↔π[δ(k+k0)+δ(k−k0)] (C.16) sink0x↔jπ[δ(k+k0)−δ(k−k0)] (C.17) 1 2be−1 2bx/bracketleftbigg I0/parenleftbigg1 2bx/parenrightbigg +I1/parenleftbigg1 2bx/parenrightbigg/bracketrightbigg U(x)↔/radicalBigg jk+b jk−1 (C.18) g(x)−ae−ax/integraldisplayx −∞eaug(u)du↔jk jk+aG(k) (C.19) Appendix D Coordinate systems Rectangular coordinate system Coordinate variables u=x,−∞<x<∞ (D.1) v=y,−∞<y<∞ (D.2) w=z,−∞<z<∞ (D.3) Vector algebra A=ˆxAx+ˆyAy+ˆzAz (D.4) A·B=AxBx+AyBy+AzBz (D.5) A×B=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆxˆyˆz A xAyAz BxByBz/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle(D.6) Dyadic representation ¯a=ˆxa xxˆx+ˆxaxyˆy+ˆxaxzˆz+ +ˆyayxˆx+ˆyayyˆy+ˆyayzˆz+ +ˆzazxˆx+ˆzazyˆy+ˆzazzˆz (D.7) ¯a=ˆxa/prime x+ˆya/prime y+ˆza/prime z=axˆx+ayˆy+azˆz (D.8) a/prime x=axxˆx+axyˆy+axzˆz (D.9) a/prime y=ayxˆx+ayyˆy+ayzˆz (D.10) a/prime z=azxˆx+azyˆy+azzˆz (D.11) ax=axxˆx+ayxˆy+azxˆz (D.12) ay=axyˆx+ayyˆy+azyˆz (D.13) az=axzˆx+ayzˆy+azzˆz (D.14) Diﬀerential operations dl=ˆxdx+ˆydy+ˆzdz (D.15) dV=dx dydz (D.16) dSx=dydz (D.17) dSy=dx dz (D.18) dSz=dx dy (D.19) ∇f=ˆx∂f ∂x+ˆy∂f ∂y+ˆz∂f ∂z(D.20) ∇·F=∂Fx ∂x+∂Fy ∂y+∂Fz ∂z(D.21) ∇× F=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆxˆyˆz ∂ ∂x∂ ∂y∂ ∂z FxFyFz/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle(D.22) ∇ 2f=∂2f ∂x2+∂2f ∂y2+∂2f ∂z2(D.23) ∇2F=ˆx∇2Fx+ˆy∇2Fy+ˆz∇2Fz (D.24) Separation of the Helmholtz equation ∂2ψ(x,y,z) ∂x2+∂2ψ(x,y,z) ∂y2+∂2ψ(x,y,z) ∂z2+k2ψ(x,y,z)=0 (D.25) ψ(x,y,z)=X(x)Y(y)Z(z) (D.26) k2 x+k2 y+k2 z=k2(D.27) d2X(x) dx2+k2 xX(x)=0 (D.28) d2Y(y) dy2+k2 yY(y)=0 (D.29) d2Z(z) dz2+k2 zZ(z)=0 (D.30) X(x)=/braceleftBigg AxF1(kxx)+BxF2(kxx),kx/negationslash=0, axx+bx, kx=0.(D.31) Y(y)=/braceleftBigg AyF1(kyy)+ByF2(kyy),ky/negationslash=0, ayy+by, ky=0.(D.32) Z(z)=/braceleftBigg AzF1(kzz)+BzF2(kzz),kz/negationslash=0, azz+bz, kz=0.(D.33) F1(ξ),F2(ξ)=  e jξ e−jξ sin(ξ) cos(ξ)(D.34) Cylindrical coordinate system Coordinate variables u=ρ, 0≤ρ<∞ (D.35) v=φ,−π≤φ≤π (D.36) w=z,−∞<z<∞ (D.37) x=ρcosφ (D.38) y=ρsinφ (D.39) z=z (D.40) ρ=/radicalbig x2+y2 (D.41) φ=tan−1y x(D.42) z=z (D.43) Vector algebra ˆρ=ˆxcosφ+ˆysinφ (D.44) ˆφ=− ˆxsinφ+ˆycosφ (D.45) ˆz=ˆz (D.46) A=ˆρAρ+ˆφAφ+ˆzAz (D.47) A·B=AρBρ+AφBφ+AzBz (D.48) A×B=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆρˆφˆz A ρAφAz BρBφBz/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle(D.49) Dyadic representation ¯a=ˆρaρρˆρ+ˆρaρφˆφ+ˆρaρzˆz+ +ˆφaφρˆρ+ˆφaφφˆφ+ˆφaφzˆz+ +ˆzazρˆρ+ˆzazφˆφ+ˆzazzˆz (D.50) ¯a=ˆρa/prime ρ+ˆφa/prime φ+ˆza/prime z=aρˆρ+aφˆφ+azˆz (D.51) a/prime ρ=aρρˆρ+aρφˆφ+aρzˆz (D.52) a/prime φ=aφρˆρ+aφφˆφ+aφzˆz (D.53) a/prime z=azρˆρ+azφˆφ+azzˆz (D.54) aρ=aρρˆρ+aφρˆφ+azρˆz (D.55) aφ=aρφˆρ+aφφˆφ+azφˆz (D.56) az=aρzˆρ+aφzˆφ+azzˆz (D.57) Diﬀerential operations dl=ˆρdρ+ˆφρdφ+ˆzdz (D.58) dV=ρdρdφdz (D.59) dSρ=ρdφdz, (D.60) dSφ=dρdz, (D.61) dSz=ρdρdφ (D.62) ∇f=ˆρ∂f ∂ρ+ˆφ1 ρ∂f ∂φ+ˆz∂f ∂z(D.63) ∇·F=1 ρ∂ ∂ρ/parenleftbig ρFρ/parenrightbig +1 ρ∂Fφ ∂φ+∂Fz ∂z(D.64) ∇× F=1 ρ/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆρρˆφˆz ∂ ∂ρ∂ ∂φ∂ ∂z FρρFφFz/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle(D.65) ∇ 2f=1 ρ∂ ∂ρ/parenleftbigg ρ∂f ∂ρ/parenrightbigg +1 ρ2∂2f ∂φ2+∂2f ∂z2(D.66) ∇2F=ˆρ/parenleftbigg ∇2Fρ−2 ρ2∂Fφ ∂φ−Fρ ρ2/parenrightbigg +ˆφ/parenleftbigg ∇2Fφ+2 ρ2∂Fρ ∂φ−Fφ ρ2/parenrightbigg +ˆz∇2Fz (D.67) Separation of the Helmholtz equation 1 ρ∂ ∂ρ/parenleftbigg ρ∂ψ(ρ, φ, z) ∂ρ/parenrightbigg +1 ρ2∂2ψ(ρ,φ, z) ∂φ2+∂2ψ(ρ,φ, z) ∂z2+k2ψ(ρ,φ, z)=0(D.68) ψ(ρ,φ, z)=P(ρ)/Phi1(φ) Z(z) (D.69) k2 c=k2−k2 z (D.70) d2P(ρ) dρ2+1 ρdP(ρ) dρ+/parenleftBigg k2 c−k2 φ ρ2/parenrightBigg P(ρ)=0 (D.71) ∂2/Phi1(φ) ∂φ2+k2 φ/Phi1(φ)=0 (D.72) d2Z(z) dz2+k2 zZ(z)=0 (D.73) Z(z)=/braceleftBigg AzF1(kzz)+BzF2(kzz),kz/negationslash=0, azz+bz, kz=0.(D.74) /Phi1(φ)=/braceleftBigg AφF1(kφφ)+BφF2(kφφ), kφ/negationslash=0, aφφ+bφ, kφ=0.(D.75) P(ρ)=  aρlnρ+bρ, kc=kφ=0, aρρ−kφ+bρρkφ, kc=0and kφ/negationslash=0, AρG1(kcρ)+BρG2(kcρ),otherwise .(D.76) F1(ξ),F2(ξ)=  e jξ e−jξ sin(ξ) cos(ξ)(D.77) G1(ξ),G2(ξ)=  J kφ(ξ) Nkφ(ξ) H(1) kφ(ξ) H(2) kφ(ξ)(D.78) Spherical coordinate system Coordinate variables u=r,0≤r<∞ (D.79) v=θ, 0≤θ≤π (D.80) w=φ,−π≤φ≤π (D.81) x=rsinθcosφ (D.82) y=rsinθsinφ (D.83) z=rcosθ (D.84) r=/radicalbig x2+y2+z2 (D.85) θ=tan−1/radicalbig x2+y2 z(D.86) φ=tan−1y x(D.87) Vector algebra ˆr=ˆxsinθcosφ+ˆysinθsinφ+ˆzcosθ (D.88) ˆθ=ˆxcosθcosφ+ˆycosθsinφ−ˆzsinθ (D.89) ˆφ=− ˆxsinφ+ˆycosφ (D.90) A=ˆrAr+ˆθAθ+ˆφAφ (D.91) A·B=ArBr+AθBθ+AφBφ (D.92) A×B=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆrˆθˆφ A rAθAφ BrBθBφ/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle(D.93) Dyadic representation ¯a=ˆra rrˆr+ˆrarθˆθ+ˆrarφˆφ+ +ˆθaθrˆr+ˆθaθθˆθ+ˆθaθφˆφ+ +ˆφaφrˆr+ˆφaφθˆθ+ˆφaφφˆφ (D.94) ¯a=ˆra/prime r+ˆθa/prime θ+ˆφa/prime φ=arˆr+aθˆθ+aφˆφ (D.95) a/prime r=arrˆr+arθˆθ+arφˆφ (D.96) a/prime θ=aθrˆr+aθθˆθ+aθφˆφ (D.97) a/prime φ=aφrˆr+aφθˆθ+aφφˆφ (D.98) ar=arrˆr+aθrˆθ+aφrˆφ (D.99) aθ=arθˆr+aθθˆθ+aφθˆφ (D.100) aφ=arφˆr+aθφˆθ+aφφˆφ (D.101) Diﬀerential operations dl=ˆrdr+ˆθrdθ+ˆφrsinθdφ (D.102) dV=r2sinθdr dθdφ (D.103) dSr=r2sinθdθdφ (D.104) dSθ=rsinθdr dφ (D.105) dSφ=rd rdθ (D.106) ∇f=ˆr∂f ∂r+ˆθ1 r∂f ∂θ+ˆφ1 rsinθ∂f ∂φ(D.107) ∇·F=1 r2∂ ∂r/parenleftbig r2Fr/parenrightbig +1 rsinθ∂ ∂θ(sinθFθ)+1 rsinθ∂Fφ ∂φ(D.108) ∇× F=1 r2sinθ/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆrrˆθrsinθˆφ ∂ ∂r∂ ∂θ∂ ∂φ FrrFθrsinθFφ/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle(D.109) ∇ 2f=1 r2∂ ∂r/parenleftbigg r2∂f ∂r/parenrightbigg +1 r2sinθ∂ ∂θ/parenleftbigg sinθ∂f ∂θ/parenrightbigg +1 r2sin2θ∂2f ∂φ2(D.110) ∇2F=ˆr/bracketleftbigg ∇2Fr−2 r2/parenleftbigg Fr+cosθ sinθFθ+1 sinθ∂Fφ ∂φ+∂Fθ ∂θ/parenrightbigg/bracketrightbigg + +ˆθ/bracketleftbigg ∇2Fθ−1 r2/parenleftbigg1 sin2θFθ−2∂Fr ∂θ+2cosθ sin2θ∂Fφ ∂φ/parenrightbigg/bracketrightbigg + +ˆφ/bracketleftbigg ∇2Fφ−1 r2/parenleftbigg1 sin2θFφ−21 sinθ∂Fr ∂φ−2cosθ sin2θ∂Fθ ∂φ/parenrightbigg/bracketrightbigg (D.111) Separation of the Helmholtz equation 1 r2∂ ∂r/parenleftbigg r2∂ψ(r,θ,φ) ∂r/parenrightbigg +1 r2sinθ∂ ∂θ/parenleftbigg sinθ∂ψ(r,θ,φ) ∂θ/parenrightbigg + +1 r2sin2θ∂2ψ(r,θ,φ) ∂φ2+k2ψ(r,θ,φ) =0 (D.112) ψ(r,θ,φ) =R(r)/Theta1(θ)/Phi1(φ) (D.113) η=cosθ (D.114) 1 R(r)d dr/parenleftbigg r2dR(r) dr/parenrightbigg +k2r2=n(n+1) (D.115) (1−η2)d2/Theta1(η) dη2−2ηd/Theta1(η) dη+/bracketleftbigg n(n+1)−µ2 1−η2/bracketrightbigg /Theta1(η)=0,−1≤η≤1(D.116) d2/Phi1(φ) dφ2+µ2/Phi1(φ)=0 (D.117) /Phi1(φ)=/braceleftBigg Aφsin(µφ)+Bφcos(µφ), µ /negationslash=0, aφφ+bφ,µ =0.(D.118) /Theta1(θ)=AθPµ n(cosθ)+BθQµ n(cosθ) (D.119) R(r)=/braceleftBigg R(r)=Arrn+Brr−(n+1),k=0, ArF1(kr)+BrF2(kr), otherwise .(D.120) F1(ξ),F2(ξ)=  j n(ξ) nn(ξ) h(1) n(ξ) h(2) n(ξ)(D.121) Appendix E Properties of special functions E.1 Bessel functions Notation z= complex number; ν,x= real numbers; n= integer Jν(z)= ordinary Bessel function of the ﬁrst kind Nν(z)= ordinary Bessel function of the second kind Iν(z)= modiﬁed Bessel function of the ﬁrst kind Kν(z)= modiﬁed Bessel function of the second kind H(1) ν= Hankel function of the ﬁrst kind H(2) ν= Hankel function of the second kind jn(z)= ordinary spherical Bessel function of the ﬁrst kind nn(z)= ordinary spherical Bessel function of the second kind h(1) n(z)= spherical Hankel function of the ﬁrst kind h(2) n(z)= spherical Hankel function of the second kind f/prime(z)=df(z)/dz= derivative with respect to argument Diﬀerential equations d2Zν(z) dz2+1 zdZν(z) dz+/parenleftbigg 1−ν2 z2/parenrightbigg Zν(z)=0 (E.1) Zν(z)=  J ν(z) Nν(z) H(1) ν(z) H(2) ν(z)(E.2) Nν(z)=cos(νπ) Jν(z)−J−ν(z) sin(νπ),ν/negationslash=n,|arg(z)|<π (E.3) H(1) ν(z)=Jν(z)+jNν(z) (E.4) H(2) ν(z)=Jν(z)−jNν(z) (E.5) d2¯Zν(x) dz2+1 zd¯Zν(z) dz−/parenleftbigg 1+ν2 z2/parenrightbigg ¯Zν=0 (E.6) ¯Zν(z)=/braceleftbiggIν(z) Kν(z)(E.7) L(z)=/braceleftbiggIν(z) ejνπKν(z)(E.8) Iν(z)=e−jνπ/2Jν(zejπ/2),−π< arg(z)≤π 2(E.9) Iν(z)=ej3νπ/2Jν(ze−j3π/2),π 2<arg(z)≤π (E.10) Kν(z)=jπ 2ejνπ/2H(1) ν(zejπ/2),−π< arg(z)≤π 2(E.11) Kν(z)=−jπ 2e−jνπ/2H(2) ν(ze−jπ/2),−π 2<arg(z)≤π (E.12) In(x)=j−nJn(jx) (E.13) Kn(x)=π 2jn+1H(1) n(jx) (E.14) d2zn(z) dz2+2 zdzn(z) dz+/bracketleftbigg 1−n(n+1) z2/bracketrightbigg zn(z)=0,n=0,±1,±2,... (E.15) zn(z)=  j n(z) nn(z) h(1) n(z) h(2) n(z)(E.16) jn(z)=/radicalbiggπ 2zJn+1 2(z) (E.17) nn(z)=/radicalbiggπ 2zNn+1 2(z) (E.18) h(1) n(z)=/radicalbiggπ 2zH(1) n+1 2(z)=jn(z)+jnn(z) (E.19) h(2) n(z)=/radicalbiggπ 2zH(2) n+1 2(z)=jn(z)−jnn(z) (E.20) nn(z)=(−1)n+1j−(n+1)(z) (E.21) Orthogonality relationships /integraldisplaya 0Jν/parenleftBigpνm aρ/parenrightBig Jν/parenleftBigpνn aρ/parenrightBig ρdρ=δmna2 2J2 ν+1(pνn)=δmna2 2/bracketleftbig J/prime ν(pνn)/bracketrightbig2,ν > −1 (E.22) /integraldisplaya 0Jν/parenleftbiggp/prime νm aρ/parenrightbigg Jν/parenleftbiggp/prime νn aρ/parenrightbigg ρdρ=δmna2 2/parenleftbigg 1−ν2 p/prime2νm/parenrightbigg J2 ν(p/prime νm), ν > −1(E.23) /integraldisplay∞ 0Jν(αx)Jν(βx)xd x=1 αδ(α−β) (E.24) /integraldisplaya 0jl/parenleftBigαlm ar/parenrightBig jl/parenleftBigαln ar/parenrightBig r2dr=δmna3 2j2 n+1(αlna) (E.25) /integraldisplay∞ −∞jm(x)jn(x)dx=δmnπ 2n+1,m,n≥0 (E.26) Jm(pmn)=0 (E.27) J/prime m(p/prime mn)=0 (E.28) jm(αmn)=0 (E.29) j/prime m(α/prime mn)=0 (E.30) Speciﬁc examples j0(z)=sinz z(E.31) n0(z)=−cosz z(E.32) h(1) 0(z)=−j zejz(E.33) h(2) 0(z)=j ze−jz(E.34) j1(z)=sinz z2−cosz z(E.35) n1(z)=−cosz z2−sinz z(E.36) j2(z)=/parenleftbigg3 z3−1 z/parenrightbigg sinz−3 z2cosz (E.37) n2(z)=/parenleftbigg −3 z3+1 z/parenrightbigg cosz−3 z2sinz (E.38) Functional relationships Jn(−z)=(−1)nJn(z) (E.39) In(−z)=(−1)nIn(z) (E.40) jn(−z)=(−1)njn(z) (E.41) nn(−z)=(−1)n+1nn(z) (E.42) J−n(z)=(−1)nJn(z) (E.43) N−n(z)=(−1)nNn(z) (E.44) I−n(z)=In(z) (E.45) K−n(z)=Kn(z) (E.46) j−n(z)=(−1)nnn−1(z),n>0 (E.47) Power series Jn(z)=∞/summationdisplay k=0(−1)k(z/2)n+2k k!(n+k)!(E.48) In(z)=∞/summationdisplay k=0(z/2)n+2k k!(n+k)!(E.49) Small argument approximations |z|/lessmuch1. Jn(z)≈1 n!/parenleftBigz 2/parenrightBign (E.50) Jν(z)≈1 /Gamma1(ν+1)/parenleftBigz 2/parenrightBigν (E.51) N0(z)≈2 π(lnz+0.5772157 −ln 2) (E.52) Nn(z)≈−(n−1)! π/parenleftbigg2 z/parenrightbiggn ,n>0 (E.53) Nν(z)≈−/Gamma1(ν) π/parenleftbigg2 z/parenrightbiggν ,ν > 0 (E.54) In(z)≈1 n!/parenleftBigz 2/parenrightBign (E.55) Iν(z)≈1 /Gamma1(ν+1)/parenleftBigz 2/parenrightBigν (E.56) jn(z)≈2nn! (2n+1)!zn(E.57) nn(z)≈−(2n)! 2nn!z−(n+1)(E.58) Large argument approximations |z|/greatermuch1. Jν(z)≈/radicalbigg 2 πzcos/parenleftBig z−π 4−νπ 2/parenrightBig ,|arg(z)|<π (E.59) Nν(z)≈/radicalbigg 2 πzsin/parenleftBig z−π 4−νπ 2/parenrightBig ,|arg(z)|<π (E.60) H(1) ν(z)≈/radicalbigg 2 πzej(z−π 4−νπ 2),−π< arg(z)<2π (E.61) H(2) ν(z)≈/radicalbigg 2 πze−j(z−π 4−νπ 2),−2π< arg(z)<π (E.62) Iν(z)≈/radicalbigg 1 2πzez,|arg(z)|<π 2(E.63) Kν(z)≈/radicalbiggπ 2ze−z,|arg(z)|<3π 2(E.64) jn(z)≈1 zsin/parenleftBig z−nπ 2/parenrightBig ,|arg(z)|<π (E.65) nn(z)≈−1 zcos/parenleftBig z−nπ 2/parenrightBig ,|arg(z)|<π (E.66) h(1) n(z)≈(−j)n+1ejz z,−π< arg(z)<2π (E.67) h(2) n(z)≈jn+1e−jz z,−2π< arg(z)<π (E.68) Recursion relationships zZν−1(z)+zZν+1(z)=2νZν(z) (E.69) Zν−1(z)−Zν+1(z)=2Z/prime ν(z) (E.70) zZ/prime ν(z)+νZν(z)=zZν−1(z) (E.71) zZ/prime ν(z)−νZν(z)=− zZν+1(z) (E.72) zLν−1(z)−zLν+1(z)=2νLν(z) (E.73) Lν−1(z)+Lν+1(z)=2L/prime ν(z) (E.74) zL/prime ν(z)+νLν(z)=zLν−1(z) (E.75) zL/prime ν(z)−νLν(z)=zLν+1(z) (E.76) zzn−1(z)+zzn+1(z)=(2n+1)zn(z) (E.77) nzn−1(z)−(n+1)zn+1(z)=(2n+1)z/prime n(z) (E.78) zz/prime n(z)+(n+1)zn(z)=zzn−1(z) (E.79) −zz/prime n(z)+nzn(z)=zzn+1(z) (E.80) Integral representations Jn(z)=1 2π/integraldisplayπ −πe−jnθ+jzsinθdθ (E.81) Jn(z)=1 π/integraldisplayπ 0cos(nθ−zsinθ)dθ (E.82) Jn(z)=1 2πj−n/integraldisplayπ −πejzcosθcos(nθ)dθ (E.83) In(z)=1 π/integraldisplayπ 0ezcosθcos(nθ)dθ (E.84) Kn(z)=/integraldisplay∞ 0e−zcosh(t)cosh(nt)dt,|arg(z)|<π 2(E.85) jn(z)=zn 2n+1n!/integraldisplayπ 0cos(zcosθ)sin2n+1θdθ (E.86) jn(z)=(−j)n 2/integraldisplayπ 0ejzcosθPn(cosθ)sinθdθ (E.87) Wronskians and cross products Jν(z)Nν+1(z)−Jν+1(z)Nν(z)=−2 πz(E.88) H(2) ν(z)H(1) ν+1(z)−H(1) ν(z)H(2) ν+1(z)=4 jπz(E.89) Iν(z)Kν+1(z)+Iν+1(z)Kν(z)=1 z(E.90) Iν(z)K/prime ν(z)−I/prime ν(z)Kν(z)=−1 z(E.91) Jν(z)H(1) ν/prime(z)−J/prime ν(z)H(1) ν(z)=2j πz(E.92) Jν(z)H(2) ν/prime(z)−J/prime ν(z)H(2) ν(z)=−2j πz(E.93) H(1) ν(z)H(2) ν/prime(z)−H(1) ν/prime(z)H(2) ν(z)=−4j πz(E.94) jn(z)nn−1(z)−jn−1(z)nn(z)=1 z2(E.95) jn+1(z)nn−1(z)−jn−1(z)nn+1(z)=2n+1 z3(E.96) jn(z)n/prime n(z)−j/prime n(z)nn(z)=1 z2(E.97) h(1) n(z)h(2) n/prime(z)−h(1) n/prime(z)h(2) n(z)=−2j z2(E.98) Summation formulas ✁✁✁✁ ✦✦✦✦✦✦✦✦✦✦ φψ r R ρR, r,ρ,φ,ψas shown. R=/radicalbig r2+ρ2−2rρcosφ. ejνψZν(zR)=∞/summationdisplay k=−∞Jk(zρ)Zν+k(zr)ejkφ,ρ < r,0<ψ<π 2(E.99) ejnψJn(zR)=∞/summationdisplay k=−∞Jk(zρ)Jn+k(zr)ejkφ(E.100) ejzρcosφ=∞/summationdisplay k=0jk(2k+1)jk(zρ)Pk(cosφ) (E.101) Forρ< rand 0<ψ<π / 2, ejzR R=jπ 2√rρ∞/summationdisplay k=0(2k+1)Jk+1 2(zρ)H(1) k+1 2(zr)Pk(cosφ) (E.102) e−jzR R=−jπ 2√rρ∞/summationdisplay k=0(2k+1)Jk+1 2(zρ)H(2) k+1 2(zr)Pk(cosφ) (E.103) Integrals /integraldisplay xν+1Jν(x)dx=xν+1Jν+1(x)+C (E.104) /integraldisplay Zν(ax)Zν(bx)xd x=x[bZν(ax)Zν−1(bx)−aZν−1(ax)Zν(bx)] a2−b2+C,a/negationslash=b(E.105) /integraldisplay xZ2 ν(ax)dx=x2 2/bracketleftbig Z2 ν(ax)−Zν−1(ax)Zν+1(ax)/bracketrightbig +C (E.106) /integraldisplay∞ 0Jν(ax)dx=1 a,ν > −1,a>0 (E.107) Fourier–Bessel expansion of a function f(ρ)=∞/summationdisplay m=1amJν/parenleftBig pνmρ a/parenrightBig ,0≤ρ≤a,ν > −1 (E.108) am=2 a2J2 ν+1(pνm)/integraldisplaya 0f(ρ)Jν/parenleftBig pνmρ a/parenrightBig ρdρ (E.109) f(ρ)=∞/summationdisplay m=1bmJν/parenleftBig p/prime νmρ a/parenrightBig ,0≤ρ≤a,ν > −1 (E.110) bm=2 a2/parenleftBig 1−ν2 p/prime2νmJ2ν(p/primeνm)/parenrightBig/integraldisplaya 0f(ρ)Jν/parenleftbiggp/prime νm aρ/parenrightbigg ρdρ (E.111) Series of Bessel functions ejzcosφ=∞/summationdisplay k=−∞jkJk(z)ejkφ(E.112) ejzcosφ=J0(z)+2∞/summationdisplay k=1jkJk(z)cosφ (E.113) sinz=2∞/summationdisplay k=0(−1)kJ2k+1(z) (E.114) cosz=J0(z)+2∞/summationdisplay k=1(−1)kJ2k(z) (E.115) E.2 Legendre functions Notation x,y,θ= real numbers; l,m,n= integers; Pm n(cosθ)= associated Legendre function of the ﬁrst kind Qm n(cosθ)= associated Legendre function of the second kind Pn(cosθ)=P0 n(cosθ)=Legendre polynomial Qn(cosθ)=Q0 n(cosθ)=Legendre function of the second kind Diﬀerential equation x=cosθ. (1−x2)d2Rm n(x) dx2−2xdRm n(x) dx+/bracketleftbigg n(n+1)−m2 1−x2/bracketrightbigg Rm n(x)=0,−1≤x≤1(E.116) Rm n(x)=/braceleftbiggPm n(x) Qm n(x)(E.117) Orthogonality relationships /integraldisplay1 −1Pm l(x)Pm n(x)dx=δln2 2n+1(n+m)! (n−m)!(E.118) /integraldisplayπ 0Pm l(cosθ)Pm n(cosθ)sinθdθ=δln2 2n+1(n+m)! (n−m)!(E.119) /integraldisplay1 −1Pm n(x)Pk n(x) 1−x2dx=δmk1 m(n+m)! (n−m)!(E.120) /integraldisplayπ 0Pm n(cosθ)Pk n(cosθ) sinθdθ=δmk1 m(n+m)! (n−m)!(E.121) /integraldisplay1 −1Pl(x)Pn(x)dx=δln2 2n+1(E.122) /integraldisplayπ 0Pl(cosθ)Pn(cosθ)sinθdθ=δln2 2n+1(E.123) Speciﬁc examples P0(x)=1 (E.124) P1(x)=x=cos(θ) (E.125) P2(x)=1 2(3x2−1)=1 4(3 cos 2 θ+1) (E.126) P3(x)=1 2(5x3−3x)=1 8(5 cos 3 θ+3 cosθ) (E.127) P4(x)=1 8(35x4−30x2+3)=1 64(35 cos 4 θ+20 cos 2 θ+9) (E.128) P5(x)=1 8(63x5−70x3+15x)=1 128(63 cos 5 θ+35 cos 3 θ+30 cos θ)(E.129) Q0(x)=1 2ln/parenleftbigg1+x 1−x/parenrightbigg =ln/parenleftbigg cotθ 2/parenrightbigg (E.130) Q1(x)=x 2ln/parenleftbigg1+x 1−x/parenrightbigg −1=cosθln/parenleftbigg cotθ 2/parenrightbigg −1 (E.131) Q2(x)=1 4(3x2−1)ln/parenleftbigg1+x 1−x/parenrightbigg −3 2x (E.132) Q3(x)=1 4(5x3−3x)ln/parenleftbigg1+x 1−x/parenrightbigg −5 2x2+2 3(E.133) Q4(x)=1 16(35x4−30x2+3)ln/parenleftbigg1+x 1−x/parenrightbigg −35 8x3+55 24x (E.134) P1 1(x)=−(1−x2)1/2=− sinθ (E.135) P1 2(x)=−3x(1−x2)1/2=−3 cosθsinθ (E.136) P2 2(x)=3(1−x2)=3 sin2θ (E.137) P1 3(x)=−3 2(5x2−1)(1−x2)1/2=−3 2(5 cos2θ−1)sinθ (E.138) P2 3(x)=15x(1−x2)=15 cos θsin2θ (E.139) P3 3(x)=−15(1−x2)3/2=−15 sin3θ (E.140) P1 4(x)=−5 2(7x3−3x)(1−x2)1/2=−5 2(7 cos3θ−3 cosθ)sinθ (E.141) P2 4(x)=15 2(7x2−1)(1−x2)=15 2(7 cos2θ−1)sin2θ (E.142) P3 4(x)=−105x(1−x2)3/2=−105 cos θsin3θ (E.143) P4 4(x)=105(1−x2)2=105 sin4θ (E.144) Functional relationships Pm n(x)=/braceleftBigg 0, m>n, (−1)m(1−x2)m/2 2nn!dn+m(x2−1)n dxn+m,m≤n.(E.145) Pn(x)=1 2nn!dn(x2−1)n dxn(E.146) Rm n(x)=(−1)m(1−x2)m/2dmRn(x) dxm(E.147) P−m n(x)=(−1)m(n−m)! (n+m)!Pm n(x) (E.148) Pn(−x)=(−1)nPn(x) (E.149) Qn(−x)=(−1)n+1Qn(x) (E.150) Pm n(−x)=(−1)n+mPm n(x) (E.151) Qm n(−x)=(−1)n+m+1Qm n(x) (E.152) Pm n(1)=/braceleftBigg 1,m=0, 0,m>0.(E.153) |Pn(x)|≤Pn(1)=1 (E.154) Pn(0)=/Gamma1/parenleftbign 2+1 2/parenrightbig √π/Gamma1/parenleftbign 2+1/parenrightbigcosnπ 2(E.155) P−m n(x)=(−1)m(n−m)! (n+m)!Pm n(x) (E.156) Power series Pn(x)=n/summationdisplay k=0(−1)k(n+k)! (n−k)!(k!)22k+1/bracketleftbig (1−x)k+(−1)n(1+x)k/bracketrightbig (E.157) Recursion relationships (n+1−m)Rm n+1(x)+(n+m)Rm n−1(x)=(2n+1)xRm n(x) (E.158) (1−x2)Rm n/prime(x)=(n+1)xRm n(x)−(n−m+1)Rm n+1(x) (E.159) (2n+1)xRn(x)=(n+1)Rn+1(x)+nRn−1(x) (E.160) (x2−1)R/prime n(x)=(n+1)[Rn+1(x)−xRn(x)] (E.161) R/prime n+1(x)−R/prime n−1(x)=(2n+1)Rn(x) (E.162) Integral representations Pn(cosθ)=√ 2 π/integraldisplayπ 0sin/parenleftbig n+1 2/parenrightbig u√cosθ−cosudu (E.163) Pn(x)=1 π/integraldisplayπ 0/bracketleftbig x+(x2−1)1/2cosθ/bracketrightbigndθ (E.164) Addition formula Pn(cosγ)=Pn(cosθ)Pn(cosθ/prime)+ +2n/summationdisplay m=1(n−m)! (n+m)!Pm n(cosθ)Pm n(cosθ/prime)cosm(φ−φ/prime), (E.165) cosγ=cosθcosθ/prime+sinθsinθ/primecos(φ−φ/prime) (E.166) Summations 1 |r−r/prime|=1/radicalbig r2+r/prime2−2rr/primecosγ=∞/summationdisplay n=0rn < rn+1 >Pn(cosγ) (E.167) cosγ=cosθcosθ/prime+sinθsinθ/primecos(φ−φ/prime) (E.168) r<=min/braceleftbig |r|,|r/prime|/bracerightbig ,r>=max/braceleftbig |r|,|r/prime|/bracerightbig (E.169) Integrals /integraldisplay Pn(x)dx=Pn+1(x)−Pn−1(x) 2n+1+C (E.170) /integraldisplay1 −1xmPn(x)dx=0,m<n (E.171) /integraldisplay1 −1xnPn(x)dx=2n+1(n!)2 (2n+1)!(E.172) /integraldisplay1 −1x2kP2n(x)dx=22n+1(2k)!(k+n)! (2k+2n+1)!(k−n)!(E.173) /integraldisplay1 −1Pn(x)√ 1−xdx=2√ 2 2n+1(E.174) /integraldisplay1 −1P2n(x)√ 1−x2dx=/bracketleftBigg /Gamma1/parenleftbig n+1 2/parenrightbig n!/bracketrightBigg2 (E.175) /integraldisplay1 0P2n+1(x)dx=(−1)n(2n)! 2n+21 (2nn!)2(E.176) Fourier–Legendre series expansion of a function f(x)=∞/summationdisplay n=0anPn(x),−1≤x≤1 (E.177) an=2n+1 2/integraldisplay1 −1f(x)Pn(x)dx (E.178) E.3 Spherical harmonics Notation θ,φ= real numbers; m,n= integers Ynm(θ, φ) = spherical harmonic function Diﬀerential equation 1 sinθ∂ ∂θ/parenleftbigg sinθ∂Y(θ, φ) ∂θ/parenrightbigg +1 sin2θ∂2Y(θ, φ) ∂φ2+1 a2λY(θ, φ) =0 (E.179) λ=a2n(n+1) (E.180) Ynm(θ, φ) =/radicalBigg 2n+1 4π(n−m)! (n+m)!Pm n(cosθ)ejmθ(E.181) Orthogonality relationships /integraldisplayπ −π/integraldisplayπ 0Y∗ n/primem/prime(θ, φ) Ynm(θ, φ) sinθdθdφ=δn/primenδm/primem (E.182) ∞/summationdisplay n=0n/summationdisplay m=−nY∗ nm(θ/prime,φ/prime)Ynm(θ, φ) =δ(φ−φ/prime)δ(cosθ−cosθ/prime) (E.183) Speciﬁc examples Y00(θ, φ) =/radicalbigg 1 4π(E.184) Y10(θ, φ) =/radicalbigg 3 4πcosθ (E.185) Y11(θ, φ) =−/radicalbigg 3 8πsinθejφ(E.186) Y20(θ, φ) =/radicalbigg 5 4π/parenleftbigg3 2cos2θ−1 2/parenrightbigg (E.187) Y21(θ, φ) =−/radicalbigg 15 8πsinθcosθejφ(E.188) Y22(θ, φ) =/radicalbigg 15 32πsin2θe2jφ(E.189) Y30(θ, φ) =/radicalbigg 7 4π/parenleftbigg5 2cos3θ−3 2cosθ/parenrightbigg (E.190) Y31(θ, φ) =−/radicalbigg 21 64πsinθ/parenleftbig 5 cos2θ−1/parenrightbig ejφ(E.191) Y32(θ, φ) =/radicalbigg 105 32πsin2θcosθe2jφ(E.192) Y33(θ, φ) =−/radicalbigg 35 64πsin3θe3jφ(E.193) Functional relationships Yn0(θ, φ) =/radicalbigg 2n+1 4πPn(cosθ) (E.194) Yn,−m(θ, φ) =(−1)mY∗ nm(θ, φ) (E.195) Addition formulas Pn(cosγ)=4π 2n+1n/summationdisplay m=−nYnm(θ, φ) Y∗ nm(θ/prime,φ/prime) (E.196) Pn(cosγ)=Pn(cosθ)Pn(cosθ/prime)+ +n/summationdisplay m=−n(n−m)! (n+m)!Pm n(cosθ)Pm n(cosθ/prime)cos/bracketleftbig m(φ−φ/prime)/bracketrightbig (E.197) cosγ=cosθcosθ/prime+sinθsinθ/primecos(φ−φ/prime) (E.198) Series n/summationdisplay m=−n|Ynm(θ, φ)|2=2n+1 4π(E.199) 1 |r−r/prime|=1/radicalbig r2+r/prime2−2rr/primecosγ =4π∞/summationdisplay n=0n/summationdisplay m=−n1 2n+1rn < rn+1 >Y∗ nm(θ/prime,φ/prime)Ynm(θ, φ), (E.200) r<=min/braceleftbig |r|,|r/prime|/bracerightbig ,r>=max/braceleftbig |r|,|r/prime|/bracerightbig (E.201) Series expansion of a function f(θ, φ) =∞/summationdisplay n=0n/summationdisplay m=−nanmYnm(θ, φ) (E.202) anm=/integraldisplayπ −π/integraldisplayπ 0f(θ, φ) Y∗ nm(θ, φ) sinθdθdφ (E.203) References [1] Abramowitz, M., and Stegun, I., Handbook of Mathematical Functions , Dover Pub- lications, New York, 1965. [2] Adair, R., Concepts in Physics , Academic Press, New York, 1969. [3] Anderson, J., and Ryon, J., Electromagnetic Radiation in Accelerated Systems , Physical Review, vol. 181, no. 5, pp. 1765–1775, May 1969. 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