《电磁学》电子书（英文版）：chap04

Chapter 4 Temporal and spatial frequency domain representation 4.1 Interpretation of the temporal transform Whena?eldisrepresentedbyacontinuoussuperpositionofelementalcomponents,the resultingdecompositioncansimplifycomputationandprovidephysicalinsight. Suchrep- resentation is usually accomplished through the use of an integral transform. Although several di?erent transforms are used in electromagnetics, we shall concentrate on the powerful and e?cient Fourier transform. Let us consider the Fourier transform of the electromagnetic ?eld. The ?eld depends on x, y, z,t,andwecantransformwithrespecttoanyorallofthesevariables. However, a consideration of units leads us to consider a transform over t separately. Let ψ(r,t) representanyrectangularcomponentoftheelectricormagnetic?eld. Thenthetemporal transform will be designated by ? ψ(r,ω): ψ(r,t) ? ? ψ(r,ω). Here ω is the transform variable. The transform ?eld ? ψ is calculated using (A.1): ? ψ(r,ω)= integraldisplay ∞ ?∞ ψ(r,t)e ?jωt dt. (4.1) The inverse transform is, by (A.2), ψ(r,t) = 1 2π integraldisplay ∞ ?∞ ? ψ(r,ω)e jωt dω. (4.2) Since ? ψ is complex it may be written in amplitude–phase form: ? ψ(r,ω)=| ? ψ(r,ω)|e jξ ψ (r,ω) , where we take ?π<ξ ψ (r,ω)≤ π. Since ψ(r,t) must be real, (4.1) shows that ? ψ(r,?ω) = ? ψ ? (r,ω). (4.3) Furthermore,thetransformofthederivativeofψ maybefoundbydi?erentiating(4.2). We have ? ?t ψ(r,t) = 1 2π integraldisplay ∞ ?∞ jω ? ψ(r,ω)e jωt dω, hence ? ?t ψ(r,t) ? jω ? ψ(r,ω). (4.4) Byvirtueof(4.2),anyelectromagnetic?eldcomponentcanbedecomposedintoacontin- uous, weighted superposition of elemental temporal terms e jωt . Note that the weighting factor ? ψ(r,ω), often called the frequencyspectrum of ψ(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 ?μepsilon1 ? 2 ?t 2 parenrightbigg 1 2π integraldisplay ∞ ?∞ ? ψ(r,ω)e jωt dω = 0. Di?erentiating under the integral sign we have 1 2π integraldisplay ∞ ?∞ bracketleftbigparenleftbig ? 2 ? jωμσ +ω 2 μepsilon1 parenrightbig ? ψ(r,ω) bracketrightbig e jωt dω = 0, hence by the Fourier integral theorem parenleftbig ? 2 + k 2 parenrightbig ? ψ(r,ω)= 0 (4.5) where k = ω √ μepsilon1 radicalbigg 1 ? j σ ωepsilon1 isthewavenumber. Equation(4.5)iscalledthescalarHelmholtzequation,andrepresents 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)e jωt dω where ? E(r,ω)= 3 summationdisplay i=1 ? i i ? E i (r,ω)= 3 summationdisplay i=1 ? i i | ? E i (r,ω)|e jξ E i (r,ω) . (4.6) Allother?eldquantitieswillbewrittensimilarlywithanappropriatesuperscriptonthe phase. Substitution into Ampere’s law gives ?× 1 2π integraldisplay ∞ ?∞ ? H(r,ω)e jωt dω = ? ?t 1 2π integraldisplay ∞ ?∞ ? D(r,ω)e jωt dω + 1 2π integraldisplay ∞ ?∞ ? J(r,ω)e jωt dω, hence 1 2π integraldisplay ∞ ?∞ [?× ? H(r,ω)? jω ? D(r,ω)? ? J(r,ω)]e jωt dω = 0 after we di?erentiate under the integral signs and combine terms. So ?× ? H = jω ? D + ? J (4.7) bytheFourierintegraltheorem. ThisversionofAmpere’slawinvolvesonlythefrequency- domain ?elds. 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)governthetemporalspectraoftheelectromagnetic?elds. Wemay manipulatethemtoobtainwaveequations,andapplytheboundaryconditionsfromthe following section. After ?nding the frequency-domain ?elds we may ?nd the temporal ?eldsbyFourierinversion. Thefrequency-domainequationsinvolveonefewerderivative (the time derivative has been replaced by multiplication by jω), hence may be easier to solve. However, the inverse transform may be di?cult to compute. 4.3 Boundary conditions on the frequency-domain ?elds Severalboundaryconditionsonthesourceandmediating?eldswerederivedin § 2.8.2. For example, we found that the tangential electric ?eld must obey ?n 12 × E 1 (r,t)? ?n 12 × E 2 (r,t) =?J ms (r,t). The technique of the previous section gives us ?n 12 × [ ? E 1 (r,ω)? ? E 2 (r,ω)] =? ? J ms (r,ω) astheconditionsatis?edbythefrequency-domainelectric?eld. Theremainingboundary conditions are treated similarly. Let us summarize the results, including the e?ects of ?ctitious magnetic sources: ?n 12 ×( ? H 1 ? ? H 2 ) = ? J s , ?n 12 ×( ? E 1 ? ? E 2 ) =? ? J ms , ?n 12 ·( ? D 1 ? ? D 2 ) = ?ρ s , ?n 12 ·( ? B 1 ? ? B 2 ) = ?ρ ms , and ?n 12 ·( ? J 1 ? ? J 2 ) =?? s · ? J s ? jω?ρ s , ?n 12 ·( ? J m1 ? ? J m2 ) =?? s · ? J ms ? jω?ρ ms . Here ?n 12 points 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 ?eld applied to a material undergoes a su?ciently rapid change, there is a time lag in the response of the polarization or magnetization of the atoms. It has been found that such materials have constitutive relations involving products in the frequency domain, and that the frequency-domain constitutive parameters are complex, frequency-dependent quantities. We shall restrict ourselves 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 ? ˉχ e · ? E, (4.11) ? M = ? ˉχ m · ? H, (4.12) ? D = ? ˉepsilon1· ? E = epsilon1 0 [ ˉ I + ? ˉχ e ] · ? E, (4.13) ? B = ? ˉμ· ? H = μ 0 [ ˉ I + ? ˉχ m ] · ? H, (4.14) ? J = ? ˉσ · ? 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) = epsilon1 0 parenleftbigg E(r,t)+ integraldisplay t ?∞ ˉχ e (r,t ? t prime )· E(r,t prime )dt prime parenrightbigg , B(r,t) = μ 0 parenleftbigg H(r,t)+ integraldisplay t ?∞ ˉχ m (r,t ? t prime )· H(r,t prime )dt prime parenrightbigg , J(r,t) = integraldisplay t ?∞ ˉσ(r,t ? t prime )· E(r,t prime )dt prime . These describe the essential behavior of a dispersive material. The susceptances and conductivity,describingtheresponseoftheatomicstructuretoanapplied?eld,depend not only on the present value of the applied ?eld 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 ˉepsilon1(r,t), ˉμ(r,t), and ˉσ(r,t). Thus, applying (4.3) to each entry we must have ? ˉχ e (r,?ω) = ? ˉχ ? e (r,ω), ? ˉχ m (r,?ω) = ? ˉχ ? m (r,ω), ? ˉσ(r,?ω) = ? ˉσ ? (r,ω), (4.16) and hence ? ˉepsilon1(r,?ω) = ? ˉepsilon1 ? (r,ω), ? ˉμ(r,?ω) = ? ˉμ ? (r,ω). (4.17) If we write the constitutive parameters in terms of real and imaginary parts as ?epsilon1 ij = ?epsilon1 prime ij + j ?epsilon1 primeprime ij , ?μ ij = ?μ prime ij + j ?μ primeprime ij , ?σ ij = ?σ prime ij + j ?σ primeprime ij , these conditions become ?epsilon1 prime ij (r,?ω) = ?epsilon1 prime ij (r,ω), ?epsilon1 primeprime ij (r,?ω) =??epsilon1 primeprime 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 eitherdissipative (lossy), transforming some of the electro- magneticenergyinthe?eldsintothermalenergy,oractive,transformingthechemicalor mechanical energy of the material into energy in the ?elds. We investigate this further in § 4.5 and § 4.8.3. We can also write the constitutive equations in amplitude–phase form. Letting ?epsilon1 ij =|?epsilon1 ij |e jξ epsilon1 ij , ?μ ij =|?μ ij |e jξ μ ij , ?σ ij =|?σ ij |e jξ σ ij , and using the ?eld notation (4.6), we can write (4.13)–(4.15) as ? D i =| ? D i |e jξ D i = 3 summationdisplay j=1 |?epsilon1 ij || ? E j |e j[ξ E j +ξ epsilon1 ij ] , (4.18) ? B i =| ? B i |e jξ B i = 3 summationdisplay j=1 |?μ ij || ? H j |e j[ξ H j +ξ μ ij ] , (4.19) ? J i =| ? J i |e jξ J i = 3 summationdisplay j=1 |?σ ij || ? E j |e j[ξ E j +ξ σ ij ] . (4.20) Here we remember that the amplitudes and phases may be functions of both r and ω. For isotropic materials these reduce to ? D i =| ? D i |e jξ D i =|?epsilon1|| ? E i |e j(ξ E i +ξ epsilon1 ) , (4.21) ? B i =| ? B i |e jξ B i =|?μ|| ? H i |e j(ξ H i +ξ μ ) , (4.22) ? J i =| ? J i |e jξ J i =|?σ|| ? E i |e j(ξ E i +ξ σ ) . (4.23) 4.4.1 Thecomplexpermittivity Asmentionedabove,dissipativee?ectsmaybeassociatedwithcomplexentriesinthe permittivitymatrix. Sinceconductione?ectscanalsoleadtodissipation,thepermittivity and conductivity matrices are often combined to form a complexpermittivity. Writing the current as a sum of impressed and secondary conduction terms ( ? J = ? J i + ? J c ) and substituting (4.13) and (4.15) into Ampere’s law, we ?nd ?× ? H = ? J i + ? ˉσ · ? E + jω ? ˉepsilon1· ? E. De?ning the complex permittivity ? ˉepsilon1 c (r,ω)= ? ˉσ(r,ω) jω + ? ˉepsilon1(r,ω), (4.24) we have ?× ? H = ? J i + jω ? ˉepsilon1 c · ? E. Usingthecomplexpermittivitywecanincludethee?ectsofconductioncurrentbymerely replacingthetotalcurrentwiththeimpressedcurrent. SinceFaraday’slawisuna?ected, any equation (such as the wave equation) derived previously using total current retains its form with the same substitution. By (4.16) and (4.17) the complex permittivity obeys ? ˉepsilon1 c (r,?ω) = ? ˉepsilon1 c? (r,ω) (4.25) or ?epsilon1 cprime ij (r,?ω) = ?epsilon1 cprime ij (r,ω), ?epsilon1 cprimeprime ij (r,?ω) =??epsilon1 cprimeprime ij (r,ω). For an isotropic material it takes the particularly simple form ?epsilon1 c = ?σ jω + ?epsilon1 = ?σ jω +epsilon1 0 +epsilon1 0 ?χ e , (4.26) and we have ?epsilon1 cprime (r,?ω) = ?epsilon1 cprime (r,ω), ?epsilon1 cprimeprime (r,?ω) =??epsilon1 cprimeprime (r,ω). (4.27) 4.4.2 Highandlowfrequencybehaviorofconstitutiveparameters At low frequencies the permittivity reduces to the electrostatic permittivity. Since ?epsilon1 prime is even in ω and ?epsilon1 primeprime is odd, we have for small ω ?epsilon1 prime ～ epsilon1 0 epsilon1 r , ?epsilon1 primeprime ～ ω. If the material has some dc conductivity σ 0 , then for low frequencies the complex per- mittivity behaves as ?epsilon1 cprime ～ epsilon1 0 epsilon1 r , ?epsilon1 cprimeprime ～ σ 0 /ω. (4.28) If E or H changesveryrapidly,theremaybenopolarizationormagnetizatione?ectat all. This occurs at frequencies so high that the atomic structure of the material cannot respond to the rapidly oscillating applied ?eld. Above some frequency then, we can assume ? ˉχ e = 0 and ? ˉχ m = 0 so that ? P = 0, ? M = 0, and ? D = epsilon1 0 ? E, ? B = μ 0 ? H. In our simple models of dielectric materials (§ 4.6) we ?nd that as ω becomes large ?epsilon1 prime ?epsilon1 0 ～ 1/ω 2 , ?epsilon1 primeprime ～ 1/ω 3 . (4.29) Ourassumptionofamacroscopicmodelofmatterprovidesafairlystrictupperfrequency limit to the range of validity of the constitutive parameters. We must assume that the wavelengthoftheelectromagnetic?eldislargecomparedtothesizeoftheatomicstruc- ture. This limit suggests that permittivity and permeability might remain meaningful even at optical frequencies, and for dielectrics this is indeed the case since the values of ? P remain signi?cant. However, ? M becomes insigni?cant at much lower frequencies, and at optical frequencies we may use ? B = μ 0 ? H [107]. 4.4.3 TheKronig–Kramersrelations The principle of causality is clearly implicit in (2.29)–(2.31). We shall demonstrate thatcausalityleadstoexplicitrelationshipsbetweentherealandimaginarypartsofthe frequency-domainconstitutiveparameters. Forsimplicityweconcentrateontheisotropic case 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 complex permittivity 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 identi?ed. Writing (2.29) and (2.31) after setting u = t ? t prime and then u = t prime , we have D(r,t) = epsilon1 0 E(r,t)+epsilon1 0 integraldisplay ∞ 0 χ e (r,t prime )E(r,t ? t prime )dt prime , J(r,t) = integraldisplay ∞ 0 σ(r,t prime )E(r,t ? t prime )dt prime . We see that there is no contribution from values of χ e (r,t) or σ(r,t) for times t < 0.So we can write D(r,t) = epsilon1 0 E(r,t)+epsilon1 0 integraldisplay ∞ ?∞ χ e (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 χ e (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 ?epsilon1 c (r,ω)?epsilon1 0 = 1 jω integraldisplay ∞ 0 σ(r,t prime )e ?jωt prime dt prime +epsilon1 0 integraldisplay ∞ 0 χ e (r,t prime )e ?jωt prime dt prime . (4.31) In order to derive the Kronig–Kramers relations we must understand the behavior of ?epsilon1 c (r,ω)? epsilon1 0 in the complex ω-plane. Writing ω = ω r + jω i , we need to establish the following two properties. Property 1: The function ?epsilon1 c (r,ω)? epsilon1 0 is analytic in the lower half-plane (ω i < 0) except at ω = 0 where 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 bracketleftbiggintegraldisplay ∞ 0 σ(r,t prime )e ?jωt prime dt prime bracketrightbigg dω = integraldisplay ∞ 0 σ(r,t prime ) bracketleftbiggcontintegraldisplay Gamma1 e ?jωt prime dω bracketrightbigg dt prime . (4.32) Note that an exchange in the order of integration in the above expression is only valid for ω inthelowerhalf-planewhere lim t prime →∞ e ?jωt prime = 0. Sincethefunction f (ω) = e ?jωt prime is analyticinthelowerhalf-plane,itsclosedcontourintegraliszerobytheCauchy–Goursat theorem. Thus, by (4.32) we have contintegraldisplay Gamma1 ?σ(r,ω)dω = 0. Then, since ?σ may be assumed to be continuous in the lower half-plane for a physical medium,andsinceitsclosedpathintegraliszeroforallpossiblepathsGamma1,itisbyMorera’s theorem [110] analytic in the lower half-plane. By similar reasoning χ e (r,ω)is analytic inthelowerhalf-plane. Sincethefunction 1/ω hasasimplepoleat ω = 0,thecomposite function ?epsilon1 c (r,ω)?epsilon1 0 given by (4.31) is analytic in the lower half-plane excluding ω = 0 where it has a simple pole. Figure 4.1: ComplexintegrationcontourusedtoestablishtheKronig–Kramersrelations. Property2: We have lim ω→±∞ ?epsilon1 c (r,ω)?epsilon1 0 = 0. ToestablishthispropertyweneedtheRiemann–Lebesguelemma[142],whichstatesthat if f (t) is absolutely integrable on the interval (a,b) where a and b are ?nite or in?nite constants, then lim ω→±∞ integraldisplay b a f (t)e ?jωt dt = 0. From this we see that lim ω→±∞ ?σ(r,ω) jω = lim ω→±∞ 1 jω integraldisplay ∞ 0 σ(r,t prime )e ?jωt prime dt prime = 0, lim ω→±∞ epsilon1 0 χ e (r,ω)= lim ω→±∞ epsilon1 0 integraldisplay ∞ 0 χ e (r,t prime )e ?jωt prime dt prime = 0, and thus lim ω→±∞ ?epsilon1 c (r,ω)?epsilon1 0 = 0. To establish the Kronig–Kramers relations we examine the integral contintegraldisplay Gamma1 ?epsilon1 c (r,Omega1)?epsilon1 0 Omega1?ω dOmega1 whereGamma1isthecontourshowninFigure4.l.SincethepointsOmega1= 0,ωareexcluded, theintegrandisanalyticeverywherewithinandonGamma1,hencetheintegralvanishesbythe Cauchy–Goursattheorem.ByProperty2wehave lim R→∞ integraldisplay C ∞ ?epsilon1 c (r,Omega1)?epsilon1 0 Omega1?ω dOmega1 = 0, hence integraldisplay C 0 +C ω ?epsilon1 c (r,Omega1)?epsilon1 0 Omega1?ω dOmega1+ P.V. integraldisplay ∞ ?∞ ?epsilon1 c (r,Omega1)?epsilon1 0 Omega1?ω dOmega1 = 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 C 0 and C ω , consider a function f (Z) analyticinthelowerhalfofthe Z-plane(Z = Z r + jZ i ). If the point z lies on the real axisasshowninFigure4.1,wecancalculatetheintegral F(z) = lim δ→0 integraldisplay Gamma1 f (Z) Z ? z dZ through the parameterization Z ? z = δe jθ . Since dZ = jδe jθ dθ we have F(z) = lim δ→0 integraldisplay 0 ?π f parenleftbig z +δe jθ parenrightbig δe jθ bracketleftbig jδe jθ bracketrightbig dθ = jf(z) integraldisplay 0 ?π dθ = jπ f (z). Replacing Z by Omega1 and z by 0 we can compute lim Delta1→0 integraldisplay C 0 ?epsilon1 c (r,Omega1)?epsilon1 0 Omega1?ω dOmega1 = lim Delta1→0 integraldisplay C 0 bracketleftBig 1 j integraltext ∞ 0 σ(r,t prime )e ?jOmega1t prime dt prime +Omega1epsilon1 0 integraltext ∞ 0 χ e (r,t prime )e ?jOmega1t prime dt prime bracketrightBig 1 Omega1?ω Omega1 dOmega1 =? π integraltext ∞ 0 σ(r,t prime )dt prime ω . We recognize integraldisplay ∞ 0 σ(r,t prime )dt prime = σ 0 (r) as the dc conductivity and write lim Delta1→0 integraldisplay C 0 ?epsilon1 c (r,Omega1)?epsilon1 0 Omega1?ω dOmega1 =? πσ 0 (r) ω . If we replace Z by Omega1 and z by ω we get lim δ→0 integraldisplay C ω ?epsilon1 c (r,Omega1)?epsilon1 0 Omega1?ω dOmega1 = jπ ?epsilon1 c (r,ω)? jπepsilon1 0 . Substituting these into (4.33) we have ?epsilon1 c (r,ω)?epsilon1 0 =? 1 jπ P.V. integraldisplay ∞ ?∞ ?epsilon1 c (r,Omega1)?epsilon1 0 Omega1?ω dOmega1+ σ 0 (r) jω . (4.34) Ifwewrite ?epsilon1 c (r,ω)= ?epsilon1 cprime (r,ω)+ j ?epsilon1 cprimeprime (r,ω)andequaterealandimaginarypartsin(4.34) we ?nd that ?epsilon1 cprime (r,ω)?epsilon1 0 =? 1 π P.V. integraldisplay ∞ ?∞ ?epsilon1 cprimeprime (r,Omega1) Omega1?ω dOmega1, (4.35) ?epsilon1 cprimeprime (r,ω)= 1 π P.V. integraldisplay ∞ ?∞ ?epsilon1 cprime (r,Omega1)?epsilon1 0 Omega1?ω dOmega1? σ 0 (r) ω . (4.36) ThesearetheKronig–Kramersrelations,namedafterR.deL.KronigandH.A.Kramers who derived them independently. The expressions show that causality requires the real and imaginary parts of the permittivity to depend upon each other through the Hilbert transform pair [142]. It is often more convenient to write the Kronig–Kramers relations in a form that employsonlypositivefrequencies. Thiscanbeaccomplishedusingtheeven–oddbehavior of 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 ?epsilon1 cprime (r,ω)?epsilon1 0 =? 2 π P.V. integraldisplay ∞ 0 Omega1?epsilon1 cprimeprime (r,Omega1) Omega1 2 ?ω 2 dOmega1, (4.37) ?epsilon1 cprimeprime (r,ω)= 2ω π P.V. integraldisplay ∞ 0 ?epsilon1 cprime (r,Omega1) Omega1 2 ?ω 2 dOmega1? σ 0 (r) ω . (4.38) The symbol P.V. in this case indicates that values of the integrand around both Omega1 = 0 and Omega1 = ω must 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. Wecannotspecify ?epsilon1 cprime arbitrarily;forapassivemedium ?epsilon1 cprimeprime mustbezeroornegativeat all values of ω, and (4.36) will not necessarily return these required values. However, if wehaveagoodmeasurementorphysicalmodelfor ?epsilon1 cprimeprime ,asmightcomefromstudiesofthe absorbingpropertiesofthematerial,wecanapproximatetherealpartofthepermittivity using (4.35). We shall demonstrate this using simple models for permittivity in § 4.6. The Kronig–Kramers properties hold for μ as well. We must for practical reasons consider the fact that magnetization becomes unimportant at a much lower frequency than does polarization, so that the in?nite integrals in the Kronig–Kramers relations should be truncated at some upper frequency ω max . If we use a model or measured values of ?μ primeprime to determine ?μ prime , the form of the relation (4.37) should be [107] ?μ prime (r,ω)?μ 0 =? 2 π P.V. integraldisplay ω max 0 Omega1 ?μ primeprime (r,Omega1) Omega1 2 ?ω 2 dOmega1, where ω max isthefrequencyatwhichmagnetizationceasestobeimportant,andabove which ?μ = μ 0 . 4.5 Dissipated and stored energy in a dispersive medium LetuswritedownPoynting’spowerbalancetheoremforadispersivemedium. Writing J = J i + J c we have (§ 2.9.5) ? J i · E = J c · 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 di?culty 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 describesboththeenergystoredintheelectromagnetic?eldandtheenergydissipatedby thematerialbecauseoftimelagsbetweentheapplicationof E and H andthepolarization or magnetization of the atoms (and thus the response ?elds D and B). In principle this term can also be used to describe active media that transfer mechanical or chemical energy of the material into ?eld energy. Instead of attempting to interpret (4.40), we concentrate on the physical meaning of ?? · S(r,t) =??·[E(r,t)× H(r,t)]. Weshallpostulatethatthistermdescribesthenet?owofelectromagneticenergyintothe point r attime t. Then(4.39)showsthatintheabsenceofimpressedsourcestheenergy ?ow must act to (1) increase or decrease the stored energy density at r, (2) dissipate energyinohmiclossesthroughtheterminvolving J c ,or(3)dissipate(orprovide)energy through the term (40). Assuming linearity we may write ??·S(r,t) = ? ?t w e (r,t)+ ? ?t w m (r,t)+ ? ?t w Q (r,t), (4.41) wherethetermsontheright-handsiderepresentthetimeratesofchangeof,respectively, stored electric, stored magnetic, and dissipated energies. 4.5.1 Dissipationinadispersivematerial Althoughwemay,ingeneral,beunabletoseparatetheindividualtermsin(4.41),we can examine these terms under certain conditions. For example, consider a ?eld that builds from zero starting from time t =?∞and then decays back to zero at t =∞. Then by direct integration 1 ? integraldisplay ∞ ?∞ ?·S(t)dt = w em (t =∞)?w em (t =?∞)+w Q (t =∞)?w Q (t =?∞) where w em = w e +w m isthevolumedensityofstoredelectromagneticenergy. Thisstored energy is zero at t =±∞since the ?elds are zero at those times. Thus, Delta1w Q =? integraldisplay ∞ ?∞ ?·S(t)dt = w Q (t =∞)?w Q (t =?∞) representsthevolumedensityofthenetenergydissipatedbyalossymedium(orsupplied 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 = ? ˉepsilon1· ? E, ? B = ? ˉμ· ? H, ? J c = ? ˉσ · ? E, 1 Note that in this section we suppress the r-dependence of most quantities for clarity of presentation. we ?nd 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)e jωt dω, D(r,t) = 1 2π integraldisplay ∞ ?∞ ? D(r,ω prime )e jω prime t dω prime , and thus ?nd J c · E + E · ?D ?t = 1 (2π) 2 integraldisplay ∞ ?∞ integraldisplay ∞ ?∞ ? E(ω)· ? ˉepsilon1 c (ω prime )· ? E(ω prime )e j(ω+ω prime )t jω prime dω dω prime where ? ˉepsilon1 c is the complex dyadic permittivity (4.24). Then Delta1w Q = 1 (2π) 2 integraldisplay ∞ ?∞ integraldisplay ∞ ?∞ bracketleftbig ? E(ω)· ? ˉepsilon1 c (ω prime )· ? E(ω prime )+ ? H(ω)· ? ˉμ(ω prime )· ? H(ω prime ) bracketrightbig · · bracketleftbiggintegraldisplay ∞ ?∞ e j(ω+ω prime )t dt bracketrightbigg jω prime dω dω prime . (4.42) Using (A.4) and integrating over ω we obtain Delta1w Q = 1 2π integraldisplay ∞ ?∞ bracketleftbig ? E(?ω prime )· ? ˉepsilon1 c (ω prime )· ? E(ω prime )+ ? H(?ω prime )· ? ˉμ(ω prime )· ? H(ω prime ) bracketrightbig jω prime dω 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 ∞ ?∞ braceleftbigbracketleftbig j ?epsilon1 cprime (ω prime )? ?epsilon1 cprimeprime (ω prime ) bracketrightbig ? E(?ω prime )· ? E(ω prime )+ + bracketleftbig j ?μ prime (ω prime )? ?μ primeprime (ω prime ) bracketrightbig ? H(?ω prime )· ? H(ω prime ) bracerightbig ω prime dω prime . Usingthefrequencysymmetrypropertyforcomplexpermittivity(4.17)(whichalsoholds for permeability), we ?nd that for isotropic materials ?epsilon1 cprime (r,ω)= ?epsilon1 cprime (r,?ω), ?epsilon1 cprimeprime (r,ω)=??epsilon1 cprimeprime (r,?ω), (4.44) ?μ prime (r,ω)= ?μ prime (r,?ω), ?μ primeprime (r,ω)=??μ primeprime (r,?ω). (4.45) Thus, the products of ω prime and the real parts of the constitutive parameters are odd functions,whilefortheimaginarypartstheseproductsareeven. Sincethedotproducts ofthevector?eldsareevenfunctions,we?ndthattheintegralsofthetermscontaining the real parts of the constitutive parameters vanish, leaving Delta1w Q = 2 1 2π integraldisplay ∞ 0 bracketleftbig ??epsilon1 cprimeprime | ? E| 2 ? ?μ primeprime | ? H| 2 bracketrightbig ω dω. (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)leadsustoassociatetheimaginarypartsoftheconstitutiveparameters with dissipation. Moreover, a lossy isotropic material for which Delta1w Q > 0 must have at least one of epsilon1 cprimeprime and μ primeprime less 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: ?epsilon1 cprimeprime < 0, ?μ primeprime < 0,ω>0. (4.48) Alossless medium must have ?epsilon1 primeprime = ?μ primeprime = ?σ = 0 for all ω. Thingsarenotassimpleinthemoregeneralanisotropiccase. Anintegrationof(4.42) over ω prime instead of ω produces Delta1w Q =? 1 2π integraldisplay ∞ ?∞ bracketleftbig ? E(ω)· ? ˉepsilon1 c (?ω)· ? E(?ω)+ ? H(ω)· ? ˉμ(?ω)· ? H(?ω) bracketrightbig jω dω. 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 ? · ? ˉepsilon1 c · ? E ? ? E · ? ˉepsilon1 c? · ? E ? + ? H ? · ? ˉμ· ? H ? ? H · ? ˉμ ? · ? H ? bracketrightbig jω dω. Finally, using the dyadic identity (A.76), we have Delta1w Q = 1 4π integraldisplay ∞ ?∞ bracketleftBig ? E ? · parenleftBig ? ˉepsilon1 c ? ? ˉepsilon1 c? parenrightBig · ? E + ? H ? · parenleftBig ? ˉμ? ? ˉμ ? parenrightBig · ? H bracketrightBig jω dω wherethedagger(?)denotesthehermitian(conjugate-transpose)operation. Thecondi- tion for a lossless anisotropic material is ? ˉepsilon1 c = ? ˉepsilon1 c? , ? ˉμ = ? ˉμ ? , (4.49) or ?epsilon1 ij = ?epsilon1 ? ji , ?μ ij = ?μ ? ji , ?σ ij = ?σ ? ji . (4.50) Theserelationshipsimplythatinthelosslesscasethediagonalentriesoftheconstitutive 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 ?eld has a permittivity of the form [ ? ˉepsilon1] = ? ? epsilon1 ?jδ 0 jδepsilon10 00epsilon1 z ? ? where epsilon1, epsilon1 z , and δ are real functions of space and frequency. Since ? ˉepsilon1 is hermitian it describes a lossless plasma. Similarly, a gyrotropic medium such as a ferrite exposed to a z-directed magnetic ?eld has a permeability dyadic [ ? ˉμ] = ? ? μ ?jκ 0 jκμ 0 00μ 0 ? ? , which also describes a lossless material. 4.5.2 Energystoredinadispersivematerial In the previous section we were able to isolate the dissipative e?ects for a dispersive material under special circumstances. It is not generally possible, however, to isolate a term describing the stored energy. The Kronig–Kramers relations imply that if the constitutiveparametersofamaterialarefrequency-dependent,theymusthavebothreal andimaginaryparts;suchamaterial,ifisotropic,mustbelossy. Sodispersivematerials are generally lossy and must have both dissipative and energy-storage characteristics. However, many materials have frequency ranges calledtransparencyranges over which ?epsilon1 cprimeprime and ?μ primeprime are small compared to ?epsilon1 cprime and ?μ prime . If we restrict our interest to these ranges, wemayapproximatethematerialaslosslessandcomputeastoredenergy. Animportant special case involves a monochromatic ?eld oscillating at a frequency within this range. To study the energy stored by a monochromatic ?eld in a dispersive material we must consider the transient period during which energy accumulates in the ?elds. The assumption of a purely sinusoidal ?eld variation would not include the e?ects described by the temporal constitutive relations (2.29)–(2.31), which show that as the ?eld builds the energy must be added with a time lag. Instead we shall assume ?elds with the temporal variation E(r,t) = f (t) 3 summationdisplay i=1 ? i i |E i (r)|cos[ω 0 t +ξ E i (r)] (4.51) where f (t) is an appropriate function describing the build-up of the sinusoidal ?eld. 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 ?α 2 t 2 ? ? F(ω) = radicalbigg π α 2 e ? ω 2 4α 2 . (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 ?nd that ? E(r,ω)= 1 2 3 summationdisplay i=1 ? i i |E i (r)|e jξ E i (r) ? F(ω ?ω 0 )+ 1 2 3 summationdisplay i=1 ? i i |E i (r)|e ?jξ E i (r) ? F(ω +ω 0 ). We can simplify this by de?ning ˇ E(r) = 3 summationdisplay i=1 ? i i |E i (r)|e jξ E i (r) (4.54) as thephasor vector ?eld to obtain ? E(r,ω)= 1 2 bracketleftbig ˇ E(r) ? F(ω ?ω 0 )+ ˇ E ? (r) ? F(ω +ω 0 ) bracketrightbig . (4.55) We shall discuss the phasor concept in detail in § 4.7. The?eld E(r, t)isshowninFigure4.2asafunctionof t,while ? E(r,ω)isshownin Figure4.2asafunctionofω.Asαbecomessmallthespectrumof E(r, t)concentrates around ω =±ω 0 . We assume the material is transparent for all values α of interest so -40 -20 0 20 40 ω t -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 ω/ω 0 0 Figure 4.2: Temporal (top) and spectral magnitude (bottom) dependences of E used to compute energy stored in a dispersive material. that we may treat epsilon1 as 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 e?ects of ?eld build-up. Hence the interpretation 2 E · ?D ?t = ?w e ?t , H · ?B ?t = ?w m ?t . We shall concentrate on the electric ?eld term and later obtain the magnetic ?eld term by induction. Since for periodic signals it is more convenient to deal with the time-averaged stored energythanwiththeinstantaneousstoredenergy,wecomputethetimeaverageofw e (r,t) over the period of the sinusoid centered at the time origin. That is, we compute 〈w e 〉= 1 T integraldisplay T/2 ?T/2 w e (t)dt (4.56) where T = 2π/ω 0 . With α → 0,thistime-averagevalueisaccurateforallperiodsofthe sinusoidal wave. Because the most expedient approach to the computation of (4.56) is to employ the Fourier spectrum of E,weuse E(r,t) = 1 2π integraldisplay ∞ ?∞ ? E(r,ω)e jωt dω = 1 2π integraldisplay ∞ ?∞ ? E ? (r,ω prime )e ?jω prime t dω prime , ?D(r,t) ?t = 1 2π integraldisplay ∞ ?∞ (jω) ? D(r,ω)e jωt dω = 1 2π integraldisplay ∞ ?∞ (?jω prime ) ? D ? (r,ω prime )e ?jω prime t dω prime . 2 Note 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 ?nd that ?w e ?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 ?epsilon1 and ?μ are associated with power dissipation in the medium, we shall approximate ?epsilon1 and ?μ as purely real. Then (4.57) becomes ?w e ?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 ?w e ?t = 1 8 integraldisplay ∞ ?∞ dω 2π integraldisplay ∞ ?∞ dω prime 2π bracketleftbig jω?epsilon1(ω)? jω prime ?epsilon1(ω prime ) bracketrightbig · · bracketleftbig ˇ E · ˇ E ? ? F(ω ?ω 0 ) ? F(ω prime ?ω 0 )+ ˇ E · ˇ E ? ? F(ω +ω 0 ) ? F(ω prime +ω 0 )+ + ˇ E · ˇ E ? F(ω ?ω 0 ) ? F(ω prime +ω 0 )+ ˇ E ? · ˇ E ? ? F(ω +ω 0 ) ? F(ω prime ?ω 0 ) bracketrightbig e ?j(ω prime ?ω)t . Let ω →?ω wherever the term ? F(ω + ω 0 ) appears, and ω prime →?ω prime wherever the term ? F(ω prime +ω 0 ) appears. Since ? F(?ω) = ? F(ω) and ?epsilon1(?ω) = ?epsilon1(ω), we ?nd that ?w e ?t = 1 8 integraldisplay ∞ ?∞ dω 2π integraldisplay ∞ ?∞ dω prime 2π ? F(ω ?ω 0 ) ? F(ω prime ?ω 0 )· · bracketleftBig ˇ E · ˇ E ? [ jω?epsilon1(ω)? jω prime ?epsilon1(ω prime )]e j(ω?ω prime )t + ˇ E · ˇ E ? [ jω prime ?epsilon1(ω prime )? jω?epsilon1(ω)]e j(ω prime ?ω)t + + ˇ E · ˇ E[ jω?epsilon1(ω)+ jω prime ?epsilon1(ω prime )]e j(ω+ω prime )t + ˇ E ? · ˇ E ? [?jω?epsilon1(ω)? jω prime ?epsilon1(ω prime )]e ?j(ω+ω prime )t bracketrightBig . (4.58) For small α the spectra are concentrated near ω = ω 0 or ω prime = ω 0 . For terms involving thedi?erenceinthepermittivitieswecanexpand g(ω) = ω?epsilon1(ω)inaTaylorseriesabout ω 0 to obtain the approximation ω?epsilon1(ω)≈ ω 0 ?epsilon1(ω 0 )+(ω ?ω 0 )g prime (ω 0 ) where g prime (ω 0 ) = ?[ω?epsilon1(ω)] ?ω vextendsingle vextendsingle vextendsingle vextendsingle ω=ω 0 . Thisisnotrequiredfortermsinvolvingasumofpermittivitiessincethesewillnottend to cancel. For such terms we merely substitute ω = ω 0 or ω prime = ω 0 . With these (4.58) becomes ?w e ?t = 1 8 integraldisplay ∞ ?∞ dω 2π integraldisplay ∞ ?∞ dω prime 2π ? F(ω ?ω 0 ) ? F(ω prime ?ω 0 )· · bracketleftBig ˇ E · ˇ E ? g prime (ω 0 )[ j(ω ?ω prime )]e j(ω?ω prime )t + ˇ E · ˇ E ? g prime (ω 0 )[ j(ω prime ?ω)]e j(ω prime ?ω)t + + ˇ E · ˇ E?epsilon1(ω 0 )[ j(ω +ω prime )]e j(ω+ω prime )t + ˇ E ? · ˇ E ? ?epsilon1(ω 0 )[?j(ω +ω prime )]e ?j(ω+ω prime )t bracketrightBig . By integration w e (t) = 1 8 integraldisplay ∞ ?∞ dω 2π integraldisplay ∞ ?∞ dω prime 2π ? F(ω ?ω 0 ) ? F(ω prime ?ω 0 )· · bracketleftBig ˇ E · ˇ E ? g prime (ω 0 )e j(ω?ω prime )t + ˇ E · ˇ E ? g prime (ω 0 )e j(ω prime ?ω)t + + ˇ E · ˇ E?epsilon1(ω 0 )e j(ω+ω prime )t + ˇ E ? · ˇ E ? ?epsilon1(ω 0 )e ?j(ω+ω prime )t bracketrightBig . Our last step is to compute the time-average value of w e and let α → 0. Applying (4.56) we ?nd 〈w e 〉= 1 8 integraldisplay ∞ ?∞ dω 2π integraldisplay ∞ ?∞ dω prime 2π ? F(ω ?ω 0 ) ? F(ω prime ?ω 0 )· · bracketleftbigg 2 ˇ E · ˇ E ? g prime (ω 0 )sinc parenleftbigg [ω ?ω prime ] π ω 0 parenrightbigg + braceleftbig ˇ E ? · ˇ E ? + ˇ E · ˇ E bracerightbig ?epsilon1(ω 0 )sinc parenleftbigg [ω +ω prime ] π ω 0 parenrightbiggbracketrightbigg where sinc(x) is de?ned in (A.9) and we note that sinc(?x) = sinc(x). Finally we let α → 0 anduse(4.53)toreplace ? F(ω)bya δ-function. Uponintegrationtheseδ-functions set ω = ω 0 and ω prime = ω 0 . Since sinc(0) = 1 and sinc(2π) = 0, the time-average stored electric energy density becomes simply 〈w e 〉= 1 4 | ˇ E| 2 ?[ω?epsilon1] ?ω vextendsingle vextendsingle vextendsingle vextendsingle ω=ω 0 . (4.59) Similarly, 〈w m 〉= 1 4 | ˇ H| 2 ?[ω ?μ] ?ω vextendsingle vextendsingle vextendsingle vextendsingle ω=ω 0 . This approach can also be applied to anisotropic materials to give 〈w e 〉= 1 4 ˇ E ? · ?[ω ? ˉepsilon1] ?ω vextendsingle vextendsingle vextendsingle vextendsingle ω=ω 0 · ˇ E, (4.60) 〈w m 〉= 1 4 ˇ H ? · ?[ω ? ˉμ] ?ω vextendsingle vextendsingle vextendsingle vextendsingle ω=ω 0 · ˇ H. (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) to simplify this and obtain 〈w e 〉= 1 4 ˇ E ? · ˉepsilon1· ˇ E = 1 4 ˇ E · ˇ D ? , (4.62) 〈w m 〉= 1 4 ˇ H ? · ˉμ· ˇ H = 1 4 ˇ H · ˇ B ? , (4.63) in the anisotropic case and 〈w e 〉= 1 4 epsilon1| ˇ E| 2 = 1 4 ˇ E · ˇ D ? , (4.64) 〈w m 〉= 1 4 μ| ˇ H| 2 = 1 4 ˇ H · ˇ B ? , (4.65) in the isotropic case. Here ˇ E, ˇ D, ˇ B, ˇ H are all phasor ?elds as de?ned by (4.54). 4.5.3 Theenergytheorem Aconvenientexpressionforthetime-averagestoredenergies(4.60)and(4.61)isfound by manipulating the frequency-domain Maxwell equations. Beginning with the complex conjugates of the two frequency-domain curl equations for anisotropic media, ?× ? E ? = jω ? ˉμ ? · ? H ? , ?× ? H ? = ? J ? ? jω ? ˉepsilon1 ? · ? E ? , we di?erentiate with respect to frequency: ?× ? ? E ? ?ω = j ?[ω ? ˉμ ? ] ?ω · ? H ? + jω ? ˉμ ? · ? ? H ? ?ω , (4.66) ?× ? ? H ? ?ω = ? ? J ? ?ω ? j ?[ω ? ˉepsilon1 ? ] ?ω · ? E ? ? jω ? ˉepsilon1 ? · ? ? 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 ?× ? E and ?× ? H by Maxwell’s equations we have 1 4 ?· parenleftbigg ? E × ? ? H ? ?ω + ? ? E ? ?ω × ? H parenrightbigg = j 1 4 ω parenleftbigg ? E · ? ˉepsilon1 ? · ? ? E ? ?ω ? ? ? E ? ?ω · ? ˉepsilon1· ? E parenrightbigg + j 1 4 ω parenleftbigg ? H · ? ˉμ ? · ? ? H ? ?ω ? ? ? H ? ?ω · ? ˉμ· ? H parenrightbigg + +j 1 4 parenleftbigg ? E · ?[ω ? ˉepsilon1 ? ] ?ω · ? E ? + ? H · ?[ω ? ˉμ ? ] ?ω · ? H ? parenrightbigg ? 1 4 parenleftbigg ? E · ? ? J ? ?ω + ? J · ? ? E ? ?ω parenrightbigg . Let us assume that the sources and ?elds are narrowband, centered on ω 0 , and that ω 0 lieswithinatransparencyrangesothatwithinthebandthematerialmaybeconsidered lossless. Invokingfrom(4.49)thefactsthat ? ˉepsilon1 = ? ˉepsilon1 ? and ? ˉμ = ? ˉμ ? ,we?ndthatthe?rsttwo termsontherightarezero. Integratingoveravolumeandtakingthecomplexconjugate of both sides we obtain 1 4 contintegraldisplay S parenleftbigg ? E ? × ? ? H ?ω + ? ? E ?ω × ? H ? parenrightbigg · dS = ?j 1 4 integraldisplay V parenleftbigg ? E ? · ?[ω ? ˉepsilon1] ?ω · ? E + ? H ? · ?[ω ? ˉμ] ?ω · ? H parenrightbigg dV ? 1 4 integraldisplay V parenleftbigg ? E ? · ? ? J ?ω + ? J ? · ? ? E ?ω parenrightbigg dV. Evaluating each of the terms at ω = ω 0 and using (4.60)–(4.61) we have 1 4 contintegraldisplay S parenleftbigg ? E ? × ? ? H ?ω + ? ? E ?ω × ? H ? parenrightbiggvextendsingle vextendsingle vextendsingle vextendsingle ω=ω 0 · dS = ?j [〈W e 〉+〈W m 〉] ? 1 4 integraldisplay V parenleftbigg ? E ? · ? ? J ?ω + ? J ? · ? ? E ?ω parenrightbiggvextendsingle vextendsingle vextendsingle vextendsingle ω=ω 0 dV (4.68) where 〈W e 〉+〈W m 〉 isthetotaltime-averageelectromagneticenergystoredinthevolume region V. This is known as theenergytheorem. 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 ?elds has been restricted to macroscopic phenomena. Althoughwerecognizethatmatteriscomposedofmicroscopicconstituents, we have chosen to describe materials using constitutive relationships whose parameters, suchaspermittivity,conductivity,andpermeability,areviewedinthemacroscopicsense. 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, and moleculardipolemoment. Thesemodelsallowustopredictthebehaviorofbroadclasses of materials, such as dielectrics and conductors, over wide ranges of frequency and ?eld strength. Accurate models for the behavior of materials under the in?uence of electromagnetic ?eldsmustaccountformanycomplicatede?ects,includingthosebestdescribedbyquan- tummechanics. However,manysimplemodelscanbeobtainedusingclassicalmechanics and ?eld theory. We shall investigate several of the most useful of these, and in the process try to gain a feeling for the relationship between the ?eld applied to a material and the resulting polarization or magnetization of the underlying atomic structure. Forsimplicityweshallconsideronlyhomogeneousmaterials. Thefundamentalatomic descriptorof“numberdensity,” N,isthustakentobeindependentofpositionandtime. The result may be more generally applicable since we may think of an inhomogeneous material in terms of the spatial variation of constitutive parameters originally deter- mined assuming homogeneity. However, we shall not attempt to study the microscopic conditions that give rise to inhomogeneities. 4.6.1 Complexpermittivityofanon-magnetizedplasma A plasma is an ionized gas in which the charged particles are free to move under the in?uence of an applied ?eld and through particle-particle interactions. A plasma di?ers from other materials in that there is no atomic lattice restricting the motion of theparticles. However,eveninagastheinteractionsbetweentheparticlesandthe?elds give rise to a polarization e?ect, causing the permittivity of the gas to di?er from that of free space. In addition, exposing the gas to an external ?eld will cause a secondary current to ?ow as a result of the Lorentz force on the particles. As the moving particles collide with one another they relinquish their momentum, an e?ect describable in terms of a conductivity. In this section we shall perform a simple analysis to determine the complex permittivity of a non-magnetized plasma. To make our analysis tractable, we shall make several assumptions. 1. Weassumethattheplasmaisneutral: i.e.,thatthefreeelectronsandpositiveions are of equal number and distributed in like manner. If the particles are su?ciently densetobeconsideredinthemacroscopicsense,thenthereisnonet?eldproduced by the gas and thus no electromagnetic interaction between the particles. We also assumethattheplasmaishomogeneousandthatthenumberdensityoftheelectrons N (numberofelectronsper m 3 )isindependentoftimeandposition. Incontrastto this are electronbeams, whose properties di?er signi?cantly from neutral plasmas because of bunching of electrons by the applied ?eld [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 as large as the ratio of a proton to an electron (m p /m e = 1837) and thus the ions accelerate much more slowly. 3. We assume that the applied ?eld 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 ?eld is |H|/|E|= √ epsilon1 0 /μ 0 , so that |B| |E| = μ 0 radicalbigg epsilon1 0 μ 0 = √ μ 0 epsilon1 0 = 1 c . Thus, in the Lorentz force equation we may approximate the force on an electron as F =?q e (E + v × B) ≈?q e E as long as v lessmuch c. Here q e is the unsigned charge on an electron, q e = 1.6021 × 10 ?19 C. Note that when an external static magnetic ?eld accompanies the ?eld of thewave,asisthecaseintheearth’sionosphereforexample,wecannotignorethe magnetic 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 acollisionfrequency 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. Let v(r,t) representthemacroscopicvelocityoftheplasmamedium. Then,byNewton’s secondlaw,theforceactingateachpointonthemediumisbalancedbythetime-rateof changeinmomentumatthatpoint. Becauseofcollisions,thetotalchangeinmomentum density is described by F(r,t) =?Nq e E(r,t) = d?(r,t) dt +ν?(r,t) (4.69) where ?(r,t) = Nm e v(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 ν describes the rate at which the electron velocities move toward a random state, producing a macroscopic plasma velocity v of zero. The time derivative in (4.69) is the total derivative as de?ned 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. Since the electron velocity is induced by the electromagnetic ?eld, we anticipate that for a sinusoidal wave the spatial variation will be on the order of the wavelength of the ?eld: λ = 2πc/ω. Thus, while the ?rst term in (4.70) is proportional to ω, the second term is proportional to ωv/c and 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 = jωm e ?v + m e ν?v (4.71) and thus ?v =? q e m e ? E ν + jω . (4.72) The secondary current associated with the moving electrons is (since q e is unsigned) ? J s =?Nq e ?v = epsilon1 0 ω 2 p ω 2 +ν 2 (ν ? jω) ? E (4.73) where ω 2 p = Nq 2 e epsilon1 0 m e (4.74) is called theplasmafrequency. Thefrequency-domainAmpere’slawforprimaryandsecondarycurrentsinfreespace is merely ?× ? H = ? J i + ? J s + jωepsilon1 0 ? E. Substitution from (4.73) gives ?× ? H = ? J i + epsilon1 0 ω 2 p ν ω 2 +ν 2 ? E + jωepsilon1 0 bracketleftBigg 1 ? ω 2 p ω 2 +ν 2 bracketrightBigg ? E. We can determine the material properties of the plasma by realizing that the above expression can be written as ?× ? H = ? J i + ? J s + jω ? D with the constitutive relations ? J s = ?σ ? E, ? D = ?epsilon1 ? E. Here we identify the conductivity of the plasma as ?σ(ω)= epsilon1 0 ω 2 p ν ω 2 +ν 2 (4.75) and the permittivity as ?epsilon1(ω)= epsilon1 0 bracketleftBigg 1 ? ω 2 p ω 2 +ν 2 bracketrightBigg . We can also write Ampere’s law as ?× ? H = ? J i + jω?epsilon1 c ? E where ?epsilon1 c is the complex permittivity ?epsilon1 c (ω) = ?epsilon1(ω)+ ?σ(ω) jω = epsilon1 0 bracketleftBigg 1 ? ω 2 p ω 2 +ν 2 bracketrightBigg ? j epsilon1 0 ω 2 p ν ω(ω 2 +ν 2 ) . (4.76) If we wish to describe the plasma in terms of a polarization vector, we merely use ? D = epsilon1 0 ? E + ? P = ?epsilon1 ? E to obtain the polarization vector ? P = (?epsilon1 ? epsilon1 0 ) ? E = epsilon1 0 ?χ e ? E, where ?χ e is the electric susceptibility ?χ e (ω) =? ω 2 p ω 2 +ν 2 . We note that ? P is directed opposite the applied ?eld ? E, resulting in ?epsilon1<epsilon1 0 . The plasma is dispersive since both its permittivity and conductivity depend on ω. As ω → 0 we have ?epsilon1 cprime → epsilon1 0 epsilon1 r where epsilon1 r = 1 ? ω 2 p /ν 2 , and also ?epsilon1 cprimeprime ～ 1/ω, as remarked in (4.28). As ω →∞we have ?epsilon1 cprime ? epsilon1 0 ～ 1/ω 2 and ?epsilon1 cprimeprime ～ 1/ω 3 , 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. Weseethattheplasmaconductivity(4.75)isproportionaltothecollisionfrequencyν, and that, since ?epsilon1 cprimeprime < 0 by the arguments of § 4.5, the plasma must be lossy. Loss arises fromthetransferofelectromagneticenergyintoheatthroughelectroncollisions. Ifthere arenocollisions(ν = 0),thereisnomechanismforthetransferofenergyintoheat,and the conductivity of a lossless (or “collisionless”) plasma reduces to zero as expected. Inalowlossplasma(ν → 0)wemaydeterminethetime-averagestoredelectromagnetic energyforsinusoidalexcitationatfrequency ˇω. Wemustbecarefultouse(4.59),which holds for materials with dispersion. If we apply the simpler formula (4.64), we ?nd that for ν → 0 〈w e 〉= 1 4 epsilon1 0 | ˇ E| 2 ? 1 4 epsilon1 0 | ˇ E| 2 ω 2 p ˇω 2 . For those excitation frequencies obeying ˇω<ω p we have 〈w e 〉 < 0, implying that the materialisactive. Sincethereisnomechanismfortheplasmatoproduceenergy,thisis obviously not valid. But an application of (4.59) gives 〈w e 〉= 1 4 | ˇ E| 2 ? ?ω bracketleftBigg epsilon1 0 ω parenleftBigg 1 ? ω 2 p ω 2 parenrightBiggbracketrightBigg vextendsingle vextendsingle vextendsingle vextendsingle ω=ˇω = 1 4 epsilon1 0 | ˇ E| 2 + 1 4 epsilon1 0 | ˇ E| 2 ω 2 p ˇω 2 , (4.77) which is always positive. In this expression the ?rst term represents the time-average energy stored in the vacuum, while the second term represents the energy stored in the kineticenergyoftheelectrons. Forharmonicexcitation,thetime-averageelectronkinetic energy density is 〈w q 〉= 1 4 Nm e ˇv · ˇv ? . Substituting ˇv from (4.72) with ν = 0 we see that 1 4 Nm e ˇv · ˇv ? = 1 4 Nq 2 e m e ˇω 2 | ˇ E| 2 = 1 4 epsilon1 0 | ˇ E| 2 ω 2 p ˇω 2 , which matches the second term of (4.77). Figure 4.3: IntegrationcontourusedinKronig–Kramersrelationsto?nd ?epsilon1 cprime from ?epsilon1 cprimeprime for 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 for a causal material. From (4.76) we see that the imaginary part of the complex plasma permittivity is ?epsilon1 cprimeprime (ω) =? epsilon1 0 ω 2 p ν ω(ω 2 +ν 2 ) . Substituting this into (4.37) we have ?epsilon1 cprime (ω)?epsilon1 0 =? 2 π P.V. integraldisplay ∞ 0 bracketleftBigg ? epsilon1 0 ω 2 p ν Omega1(Omega1 2 +ν 2 ) bracketrightBigg Omega1 Omega1 2 ?ω 2 dOmega1. We can evaluate the principal value integral and thus verify that it produces ?epsilon1 cprime by 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 ?epsilon1 cprime (ω)?epsilon1 0 = 1 π P.V. integraldisplay ∞ ?∞ epsilon1 0 ω 2 p ν (Omega1? jν)(Omega1+ jν) dOmega1 (Omega1?ω)(Omega1+ω) . WeintegratearoundtheclosedcontourshowninFigure4.3.Sincetheintegrandfalls o?as 1/Omega1 4 thecontributionfrom C ∞ iszero. Thecontributionsfromthesemicircles C ω and C ?ω are given by πj times the residues of the integrand at Omega1 = ω and at Omega1 =?ω, respectively, which are identical but of opposite sign. Thus, the semicircle contributions cancel and leave only the contribution from the residue at the upper-half-plane pole Omega1 = jν. Evaluation of the residue gives ?epsilon1 cprime (ω)?epsilon1 0 = 1 π 2πj epsilon1 0 ω 2 p ν jν + jν 1 (jν ?ω)(jν +ω) =? epsilon1 0 ω 2 p ν 2 +ω 2 and thus ?epsilon1 cprime (ω) = epsilon1 0 parenleftBigg 1 ? ω 2 p ν 2 +ω 2 parenrightBigg , which matches (4.76) as expected. 4.6.2 Complexdyadicpermittivityofamagnetizedplasma When an electron plasma is exposed to a magnetostatic ?eld, as occurs in the earth’s ionosphere,thebehavioroftheplasmaisalteredsothatthesecondarycurrentisnolonger aligned with the electric ?eld, requiring the constitutive relationships to be written in terms of a complex dyadic permittivity. If the static ?eld is B 0 , the velocity ?eld of the plasma is determined by adding the magnetic component of the Lorentz force to (4.71), giving ?q e [ ? E + ?v × B 0 ] = ?v(jωm e + m e ν) or equivalently ?v ? j q e m e (ω ? jν) ?v × B 0 = j q e m e (ω ? jν) ? E. (4.78) Writing this expression generically as v + v × C = A, (4.79) we can solve for v as follows. Dotting both sides of the equation with C we quickly establishthat C·v = C·A. Crossingbothsidesoftheequationwith C,using(B.7),and substituting C · A for C · v,wehave v × C = A × C + v(C · C)? C(A · C). Finally, substituting v × C back into (4.79) we obtain v = A ? A × C +(A · C)C 1 + C · C . (4.80) Let us ?rst consider a lossless plasma for which ν = 0. We can solve (4.78) for ?v by setting C =?j ω c ω , A = j epsilon1 0 ω 2 p ωNq e ? E, where ω c = q e m e B 0 . Here ω c = q e B 0 /m e =|ω c | is called theelectroncyclotronfrequency. Substituting these into (4.80) we have parenleftbig ω 2 ?ω 2 c parenrightbig ?v = j epsilon1 0 ωω 2 p Nq e ? E + epsilon1 0 ω 2 p Nq e ω c × ? E ? j ω c ω epsilon1 0 ω 2 p Nq e ω c · ? E. Since the secondary current produced by the moving electrons is just ? J s =?Nq e ?v,we have ? J s = jω bracketleftBigg ? epsilon1 0 ω 2 p ω 2 ?ω 2 c ? E + j epsilon1 0 ω 2 p ω(ω 2 ?ω 2 c ) ω c × ? E + ω c ω 2 epsilon1 0 ω 2 p ω 2 ?ω 2 c ω c · ? E bracketrightBigg . (4.81) Now, by the Ampere–Maxwell law we can write for currents in free space ?× ? H = ? J i + ? J s + jωepsilon1 0 ? E. (4.82) Considering the plasma to be a material implies that we can describe the gas in terms of a complex permittivity dyadic ? ˉepsilon1 c such that the Ampere–Maxwell law is ?× ? H = ? J i + jω ? ˉepsilon1 c · ? E. Substituting (4.81) into (4.82), and de?ning the dyadic ˉω c so that ˉω c · ? E = ω c × ? E,we identify the dyadic permittivity ? ˉepsilon1 c (ω) = bracketleftBigg epsilon1 0 ?epsilon1 0 ω 2 p ω 2 ?ω 2 c bracketrightBigg ˉ I + j epsilon1 0 ω 2 p ω(ω 2 ?ω 2 c ) ˉω c + epsilon1 0 ω 2 p ω 2 (ω 2 ?ω 2 c ) ω c ω c . (4.83) Note that in rectangular coordinates [ ˉω c ] = ? ? 0 ?ω cz ω cy ω cz 0 ?ω cx ?ω cy ω cx 0 ? ? . (4.84) To examine the properties of the dyadic permittivity it is useful to write it in matrix form. Todothiswemustchooseacoordinatesystem. Weshallassumethat B 0 isaligned along the z-axis such that B 0 = ?zB 0 and ω c = ?zω c . Then (4.84) becomes [ ˉω c ] = ? ? 0 ?ω c 0 ω c 00 000 ? ? (4.85) and we can write the permittivity dyadic (4.83) as [ ? ˉepsilon1(ω)] = ? ? epsilon1 ?jδ 0 jδepsilon10 00epsilon1 z ? ? (4.86) where epsilon1 = epsilon1 0 parenleftBigg 1 ? ω 2 p ω 2 ?ω 2 c parenrightBigg ,epsilon1 z = epsilon1 0 parenleftBigg 1 ? ω 2 p ω 2 parenrightBigg ,δ= epsilon1 0 ω c ω 2 p ω(ω 2 ?ω 2 c ) . 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) con?rms this. We also note that since the sign of ω c is determined by the sign of B 0 , the dyadic permittivity obeys the symmetry relation ?epsilon1 c ij (B 0 ) = ?epsilon1 c ji (?B 0 ) (4.87) asdoesthepermittivitymatrixofanymaterialthathasanisotropicpropertiesdependent on an externally applied magnetic ?eld [141]. We will ?nd later in this section that the permeability matrix of a magnetized ferrite also obeys such a symmetry condition. We can let ω → ω ? jν in (4.81) to obtain the secondary current in a plasma with collisions: ? J s (r,ω)= jω bracketleftBigg ? epsilon1 0 ω 2 p (ω ? jν) ω[(ω ? jν) 2 ?ω 2 c ] ? E(r,ω)+ + j epsilon1 0 ω 2 p (ω ? jν) ω(ω? jν)[(ω ? jν) 2 ?ω 2 c )] ω c × ? E(r,ω)+ + ω c (ω ? jν) 2 epsilon1 0 ω 2 p (ω ? jν) ω[(ω ? jν) 2 ?ω 2 c ] ω c · ? E(r,ω) bracketrightBigg . From this we ?nd the dyadic permittivity ? ˉepsilon1 c (ω) = bracketleftBigg epsilon1 0 ? epsilon1 0 ω 2 p (ω ? jν) ω[(ω ? jν) 2 ?ω 2 c ] bracketrightBigg ˉ I + j epsilon1 0 ω 2 p ω[(ω ? jν) 2 ?ω 2 c )] ˉω c + + 1 (ω ? jν) epsilon1 0 ω 2 p ω[(ω ? jν) 2 ?ω 2 c ] ω c ω c . Assumingthat B 0 isalignedwiththe z-axiswecanuse(4.85)to?ndthecomponentsof the dyadic permittivity matrix: ?epsilon1 c xx (ω) = ?epsilon1 c yy (ω) = epsilon1 0 parenleftBigg 1 ? ω 2 p (ω ? jν) ω[(ω ? jν) 2 ?ω 2 c ] parenrightBigg , (4.88) ?epsilon1 c xy (ω) =??epsilon1 c yx (ω) =?jepsilon1 0 ω 2 p ω c ω[(ω ? jν) 2 ?ω 2 c )] , (4.89) ?epsilon1 c zz (ω) = epsilon1 0 parenleftBigg 1 ? ω 2 p ω(ω? jν) parenrightBigg , (4.90) and ?epsilon1 c zx = ?epsilon1 c xz = ?epsilon1 c zy = ?epsilon1 c yz = 0. (4.91) We see that [?epsilon1 c ] is not hermitian when ν negationslash= 0. We expect this since the plasma is lossy when collisions occur. However, we can decompose [ ? ˉepsilon1 c ] as a sum of two matrices: [ ? ˉepsilon1 c ] = [ ? ˉepsilon1] + [ ? ˉσ] jω , where [ ? ˉepsilon1] and [ ? ˉσ] 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 symmetry condition ?epsilon1 c ij (B 0 ) = ?epsilon1 c ji (?B 0 ). 4.6.3 Simplemodelsofdielectrics We de?ne an isotropic dielectric material (also called aninsulator) as one that obeys the macroscopic frequency-domain constitutive relationship ? D(r,ω)= ?epsilon1(r,ω) ? E(r,ω). Since the polarization vector P was de?ned in Chapter 2 as P(r,t) = D(r,t)?epsilon1 0 E(r,t), an isotropic dielectric can also be described through ? P(r,ω)= (?epsilon1(r,ω)?epsilon1 0 ) ? E(r,ω)= ?χ e (r,ω)epsilon1 0 ? E(r,ω) where ?χ e is the dielectric susceptibility. In this section we shall model a homogeneous dielectric consisting of a single, uniform material type. WefoundinChapter3thatforadielectricmaterialimmersedinastaticelectric?eld, the polarization vector P can be viewed as a volume density of dipole moments. We choose to retain this view as the fundamental link between microscopic dipole moments and the macroscopic polarization vector. Within the framework of our model we thus describe the polarization through the expression P(r,t) = 1 Delta1V summationdisplay r?r i (t)∈B p i . (4.92) Here p i isthedipolemomentofthe ithelementarymicroscopicconstituent,andweform the macroscopic density function as in § 1.3.1. We may also write (4.92) as P(r,t) = bracketleftbigg N B Delta1V bracketrightbigg bracketleftBigg 1 N B N B summationdisplay i=1 p i bracketrightBigg = N(r,t)p(r,t) (4.93) where N B is the number of constituent particles within Delta1V. We identify p(r,t) = 1 N B N B summationdisplay i=1 p i (r,t) as the average dipole moment within Delta1V, and N(r,t) = N B Delta1V asthedipolemomentnumberdensity. Inthismodeladielectricmaterialdoesnotrequire higher-order multipole moments to describe its behavior. Since we are only interested in homogeneous materials in this section we shall assume that the number density is constant: N(r,t) = N. Tounderstandhowdipolemomentsarise,wechoosetoadoptthesimpleideathatmat- terconsistsofatomicparticles,eachofwhichhasapositivelychargednucleussurrounded by a negatively charged electron cloud. Isolated, these particles have no net charge and nonetelectricdipolemoment. However,thereareseveralwaysinwhichindividualpar- ticles, or aggregates of particles, may take on a dipole moment. When exposed to an externalelectric?eldtheelectroncloudofanindividualatommaybedisplaced,resulting in an induced dipole moment which gives rise to electronicpolarization. When groups ofatomsformamolecule,theindividualelectroncloudsmaycombinetoformanasym- metricstructurehavingapermanent dipolemoment. Insomematerialsthesemolecules are randomly distributed and no net dipole moment results. However, upon application of an external ?eld the torque acting on the molecules may tend to align them, creating aninduced dipole moment andorientation,ordipole, polarization. In other materials, the asymmetric structure of the molecules may be weak until an external ?eld causes the displacement of atoms within each molecule, resulting in aninduced dipole moment causingatomic,ormolecular, polarization. If a material maintains a permanent polar- ization without the application of an external ?eld, it is called anelectret (and is thus similar in behavior to a permanently magnetized magnet). Todescribetheconstitutiverelations,wemustestablishalinkbetween P (nowdescrib- ableinmicroscopicterms)and E. Wedothisbypostulatingthattheaverageconstituent dipole moment is proportional to thelocalelectric?eldstrength E prime : p = αE prime , (4.94) whereα iscalledthepolarizabilityoftheelementaryconstituent. Eachofthepolarization e?ectslistedabovemayhaveitsownpolarizability: α e forelectronicpolarization, α a for atomicpolarization,and α d fordipolepolarization. Thetotalpolarizabilityismerelythe sum α = α e +α a +α d . In a rare?ed gas the particles are so far apart that their interaction can be neglected. Here the localized ?eld E prime is the same as the applied ?eld E. In liquids and solids where particlesaretightlypacked, E prime dependsonthemannerinwhichthematerialispolarized and may di?er from E. We therefore proceed to determine a relationship between E prime and P. The Clausius–Mosotti equation. We seek the local ?eld at an observation point within a polarized material. Let us ?rst assume that the ?elds are static. We surround theobservationpointwithanarti?cialsphericalsurfaceofradius a andwritethe?eldat theobservationpointasasuperpositionofthe?eld E applied,the?eld E 2 ofthepolarized molecules external to the sphere, and the ?eld E 3 of the polarized molecules within the sphere. Wetake a largeenoughthatwemaydescribethemoleculesoutsidethespherein terms of the macroscopic dipole moment density P, but small enough to assume that P isuniformoverthesurfaceofthesphere. Wealsoassumethatthemajorcontributionto E 2 comesfromthedipolesnearesttheobservationpoint. Wethenapproximate E 2 using theelectrostaticpotentialproducedbytheequivalentpolarizationsurfacechargeonthe sphere ρ Ps = ?n · P (where ?n points 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 = P 0 ?z, we ?nd that ?n · P =?cosθ and thus the electrostatic potential at any point r within the sphere is merely Phi1(r) =? 1 4πepsilon1 0 contintegraldisplay S P 0 cosθ 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) =? P 0 3epsilon1 0 r cosθ =? P 0 3epsilon1 0 z and therefore E 2 = P 3epsilon1 0 . (4.95) Note that this is uniform and independent of a. The assumption that the localized ?eld varies spatially as the electrostatic ?eld, even when P may depend on frequency, is quite good. In Chapter 5 we will ?nd that for a frequency-dependentsource(or,equivalently,atime-varyingsource),the?eldsverynear the source have a spatial dependence nearly identical to that of the electrostatic case. We now have the seemingly more di?cult task of determining the ?eld E 3 produced by the dipoles within the sphere. This would seem di?cult since the ?eld produced by dipoles near the observation point should be highly-dependent on the particular dipole arrangement. As mentioned above, there are various mechanisms for polarization, and the distribution of charge near any particular point depends on the molecular arrange- ment. However, Lorentz showed [115] that for crystalline solids with cubical symmetry, orforarandomly-structuredgas,thecontributionfromdipoleswithinthesphereiszero. Indeed, it is convenient and reasonable to assume that for most dielectrics the e?ects of the dipoles immediately surrounding the observation point cancel so that E 3 = 0. This was ?rst suggested by O.F. Mosotti in 1850 [52]. With E 2 approximated as (4.95) and E 3 assumed to be zero, we have the value of the resulting local ?eld: E prime (r) = E(r)+ P(r) 3epsilon1 0 . (4.96) This is called the Mosotti ?eld. Substituting the Mosotti ?eld into (4.94) and using P = Np, we obtain P(r) = NαE prime (r) = Nα parenleftbigg E(r)+ P(r) 3epsilon1 0 parenrightbigg . Solving for P we obtain P(r) = parenleftbigg 3epsilon1 0 Nα 3epsilon1 0 ? Nα parenrightbigg E(r) = χ e epsilon1 0 E(r). So the electric susceptibility of a dielectric may be expressed as χ e = 3Nα 3epsilon1 0 ? Nα . (4.97) Using χ e = epsilon1 r ? 1 we can rewrite (4.97) as epsilon1 = epsilon1 0 epsilon1 r = epsilon1 0 3 + 2Nα/epsilon1 0 3 ? Nα/epsilon1 0 , (4.98) which we can arrange to obtain α = α e +α a +α d = 3epsilon1 0 N epsilon1 r ? 1 epsilon1 r + 2 . ThishasbeennamedtheClausius–Mosottiformula,afterO.F.Mosottiwhoproposedit in1850andR.Clausiuswhoproposeditindependentlyin1879. Whenwrittenintermsof theindexofrefraction n (where n 2 = epsilon1 r ),itisalsoknownastheLorentz–Lorenzformula, after H. Lorentz and L. Lorenz who proposed it independently for optical materials in 1880. TheClausius–Mosottiformulaallowsustodeterminethedielectricconstantfrom thepolarizabilityandnumberdensityofamaterial. Itisreasonablyaccurateforcertain simplegases(withpressuresupto1000atmospheres)butbecomeslessreliableforliquids and solids, especially for those with large dielectric constants. Theresponseofthemicroscopicstructureofmattertoanapplied?eldisnotinstanta- neous. Whenexposedtoarapidlyoscillatingsinusoidal?eld,theinduceddipolemoments maylagintime. Thisresultsinalossmechanismthatcanbedescribedmacroscopically by a complex permittivity. We can modify the Clausius–Mosotti formula by assuming thatboththerelativepermittivityandpolarizabilityarecomplexnumbers,butthiswill not model the dependence of these parameters on frequency. Instead we shall (in later paragraphs) model the time response of the dipole moments to the applied ?eld. AninterestingapplicationoftheClausius–Mosottiformulaistodeterminethepermit- tivity of a mixture of dielectrics with di?erent permittivities. Consider the simple case in which many small spheres of permittivity epsilon1 2 , radius a, and volume V are embedded within a dielectric matrix of permittivity epsilon1 1 . If we assume that a is much smaller than thewavelengthoftheelectromagnetic?eld,andthatthespheresaresparselydistributed withinthematrix,thenwemayignoreanymutualinteractionbetweenthespheres. Since the expression for the permittivity of a uniform dielectric given by (4.98) describes the e?ect produced by dipoles in free space, we can use the Clausius–Mosotti formula to de?ne an e?ective permittivity epsilon1 e for a material consisting of spheres in a background dielectric by replacing epsilon1 0 with epsilon1 1 to obtain epsilon1 e = epsilon1 1 3 + 2Nα/epsilon1 1 3 ? Nα/epsilon1 1 . (4.99) In this expression α is the polarizability of a single dielectric sphere embedded in the background dielectric, and N is the number density of dielectric spheres. To ?nd α we use the static ?eld solution for a dielectric sphere immersed in a ?eld (§ 3.2.10). Remembering that p = αE and that for a uniform region of volume V we have p = V P, we can make the replacements epsilon1 0 → epsilon1 1 and epsilon1 → epsilon1 2 in (3.117) to get α = 3epsilon1 1 V epsilon1 2 ?epsilon1 1 epsilon1 2 + 2epsilon1 1 . (4.100) De?ning f = NV as thefractionalvolume occupied by the spheres, we can substitute (4.100) into (4.99) to ?nd that epsilon1 e = epsilon1 1 1 + 2 fy 1 ? fy where y = epsilon1 2 ?epsilon1 1 epsilon1 2 + 2epsilon1 1 . This is known as theMaxwell–Garnettmixingformula. Rearranging we obtain epsilon1 e ?epsilon1 1 epsilon1 e + 2epsilon1 1 = f epsilon1 2 ?epsilon1 1 epsilon1 2 + 2epsilon1 1 , which is known as theRayleighmixingformula. As expected, epsilon1 e → epsilon1 1 as f → 0.Even though as f → 1 the formula also reduces to epsilon1 e = epsilon1 2 , our initial assumption that f lessmuch 1 (sparsely distributed spheres) is violated and the result is inaccurate for non-spherical inhomogeneities [90]. For a discussion of more accurate mixing formulas, see Ishimaru [90] or Sihvola [175]. Thedispersionformulaofclassicalphysics. Wemaydeterminethefrequencyde- pendenceofthepermittivitybymodelingthetimeresponseofinduceddipolemoments. ThiswasdonebyH.Lorentzusingthesimpleatomicmodelweintroducedearlier. Con- siderwhathappenswhenamoleculeconsistingofheavyparticles(nuclei)surroundedby cloudsofelectronsisexposedtoatime-harmonicelectromagnetic wave. Usingthesame arguments we made when we studied the interactions of ?elds with a plasma in § 4.6.1, we assume that each electron experiences a Lorentz force F e =?q e E prime . We neglect the magneticcomponentoftheforcefornonrelativisticchargevelocities,andignorethemo- 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 a plasmaandthosewithinasolidorliquidmaterial. Becauseofthesurroundingpolarized matter, any molecule responds to the local ?eld E prime instead of the applied ?eld E. Also, as the electron cloud is displaced by the Lorentz force, the attraction from the positive nuclei provides a restoring force F r . In the absence of loss the restoring force causes the electron cloud (and thus the induced dipole moment) to oscillate in phase with the applied?eld. Inaddition,therewillbelossduetoradiationbytheoscillatingmolecules and collisions between charges that can be modeled using a “frictional force” F s in the same manner as for a mechanical harmonic oscillator. Wecanexpresstherestoringandfrictionalforcesbytheuseofamechanicalanalogue. Therestoringforceactingoneachelectronistakentobeproportionaltothedisplacement from equilibrium l: F r (r,t) =?m e ω 2 r l(r,t), where m e is the mass of an electron and ω r is 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 m e v: F s (r,t) =?2Gamma1m e v(r,t) where Gamma1 isamaterialconstant. WiththesewecanapplyNewton’ssecondlawtoobtain F(r,t) =?q e E prime (r,t)? m e ω 2 r l(r,t)? 2Gamma1m e v(r,t) = m e dv(r,t) dt . Using v = dl/dt we ?nd that the equation of motion for the electron is d 2 l(r,t) dt 2 + 2Gamma1 dl(r,t) dt +ω 2 r l(r,t) =? q e m e E prime (r,t). (4.101) We recognize this di?erential equation as the damped harmonic equation. When E prime = 0 we have the homogeneous solution l(r,t) = l 0 (r)e ?Gamma1t cos parenleftbigg t radicalBig ω 2 r ?Gamma1 2 parenrightbigg . Thus the electron position is a damped oscillation. The resonant frequency radicalbig ω 2 r ?Gamma1 2 is usually only slightly reduced from ω r since radiation damping is generally quite low. Since the dipole moment for an electron displaced from equilibrium by l is p =?q e l, and the polarization density is P = Np from (93), we can write P(r,t) =?Nq e l(r,t). Multiplying(4.101)by?Nq e andsubstitutingtheaboveexpression,wehaveadi?erential equation for the polarization: d 2 P dt 2 + 2Gamma1 dP dt +ω 2 r P = Nq 2 e m e E prime . To obtain a constitutive equation we must relate the polarization to the applied ?eld E. Wecanaccomplishthisbyrelatingthelocal?eld E prime tothepolarizationusingtheMosotti ?eld (4.96). Substitution gives d 2 P dt 2 + 2Gamma1 dP dt +ω 2 0 P = Nq 2 e m e E (4.102) where ω 0 = radicalBigg ω 2 r ? Nq 2 e 3m e epsilon1 0 istheresonancefrequencyofthedipolemoments. Weseethatthisfrequencyisreduced from the resonance frequency of the electron oscillation because of the polarization of the surrounding medium. Wecannowobtainadispersionequationfortheelectricalsusceptibilitybytakingthe Fourier transform of (4.102). We have ?ω 2 ? P + jω2Gamma1 ? P +ω 2 0 ? P = Nq 2 e m e ? E. Thus we obtain the dispersion relation ?χ e (ω) = ? P epsilon1 0 ? E = ω 2 p ω 2 0 ?ω 2 + jω2Gamma1 where ω p is the plasma frequency (4.74). Since ?epsilon1 r (ω) = 1 + ?χ e (ω) we also have ?epsilon1(ω)= epsilon1 0 +epsilon1 0 ω 2 p ω 2 0 ?ω 2 + jω2Gamma1 . (4.103) If more than one type of oscillating moment contributes to the permittivity, we may extend (4.103) to ?epsilon1(ω)= epsilon1 0 + summationdisplay i epsilon1 0 ω 2 pi ω 2 i ?ω 2 + jω2Gamma1 i (4.104) where ω pi = N i q 2 e /epsilon1 0 m i is the plasma frequency of the ith resonance component, and ω i and Gamma1 i are the oscillation frequency and damping coe?cient, respectively, of this component. This expression is the dispersion formula for classical physics, so called because it neglects quantum e?ects. When losses are negligible, (4.104) reduces to the Sellmeierequation ?epsilon1(ω)= epsilon1 0 + summationdisplay i epsilon1 0 ω 2 pi ω 2 i ?ω 2 . (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)?epsilon1 0 = epsilon1 0 summationdisplay i ω 2 pi ω 2 i ?ω 2 [ω 2 i ?ω 2 ] 2 + 4ω 2 Gamma1 2 i , ?epsilon1 primeprime (ω) =?epsilon1 0 summationdisplay i ω 2 pi 2ωGamma1 i [ω 2 i ?ω 2 ] 2 + 4ω 2 Gamma1 2 i . As ω → 0 the permittivity reduces to epsilon1 = epsilon1 0 parenleftBigg 1 + summationdisplay i ω 2 pi ω 2 i parenrightBigg , which is the static permittivity of the material. As ω →∞the permittivity behaves as ?epsilon1 prime (ω) → epsilon1 0 parenleftBigg 1 ? summationtext i ω 2 pi ω 2 parenrightBigg , ?epsilon1 primeprime (ω) →?epsilon1 0 2 summationtext i ω 2 pi Gamma1 i ω 3 . 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 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 ε ε ε W Region of anomalous dispersion ω/ω 0 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 epsilon1 0 (ω p /ω 0 ) 2 . The major characteristic of the dispersion relation (4.104) is the presence of one or moreresonances.Figure4.4showsaplotofasingleresonancecomponent,wherewe have normalized the permittivity as (?epsilon1 prime (ω)?epsilon1 0 )/(epsilon1 0 ˉω 2 p ) = 1 ? ˉω 2 bracketleftbig 1 ? ˉω 2 bracketrightbig 2 + 4 ˉω 2 ˉ Gamma1 2 , ??epsilon1 primeprime (ω)/(epsilon1 0 ˉω 2 p ) = 2 ˉω ˉ Gamma1 bracketleftbig 1 ? ˉω 2 bracketrightbig 2 + 4 ˉω 2 ˉ Gamma1 2 , with ˉω = ω/ω 0 , ˉω p = ω p /ω 0 , and ˉ Gamma1 = Gamma1/ω 0 . We see a distinct resonance centered at ω = ω 0 . Approaching this resonance through frequencies less than ω 0 , we see that ?epsilon1 prime increases slowly until peaking at ω max = ω 0 √ 1 ? 2Gamma1/ω 0 where it attains a value of ?epsilon1 prime max = epsilon1 0 + 1 4 epsilon1 0 ˉω 2 p ˉ Gamma1(1 ? ˉ Gamma1) . After peaking, ?epsilon1 prime undergoes a rapid decrease, passing through ?epsilon1 prime = epsilon1 0 at ω = ω 0 , and then continuing to decrease until reaching a minimum value of ?epsilon1 prime min = epsilon1 0 ? 1 4 epsilon1 0 ˉω 2 p ˉ Gamma1(1 + ˉ Gamma1) at ω min = ω 0 √ 1 + 2Gamma1/ω 0 .Asω continues to increase, ?epsilon1 prime again increases slowly toward a ?nal value of ?epsilon1 prime = epsilon1 0 . The regions of slow variation of ?epsilon1 prime are called regions ofnormal dispersion,whiletheregionwhere ?epsilon1 prime decreasesabruptlyiscalledtheregionofanomalous 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 o? monotonically in each direction away from the peak. The width of the curve is an important parameter that we can most easily determine by approximating the behavior of ?epsilon1 primeprime near ω 0 . Letting Delta1 ˉω = (ω 0 ?ω)/ω 0 and using ω 2 0 ?ω 2 = (ω 0 ?ω)(ω 0 +ω) ≈ 2ω 2 0 Delta1 ˉω, we get ?epsilon1 primeprime (ω) ≈? 1 2 epsilon1 0 ˉω 2 p ˉ Gamma1 (Delta1 ˉω) 2 + ˉ Gamma1 2 . This approximation has a maximum value of ?epsilon1 primeprime max = ?epsilon1 primeprime (ω 0 ) =? 1 2 epsilon1 0 ˉω 2 p 1 ˉ Gamma1 located at ω = ω 0 , and has half-amplitude points located at Delta1 ˉω =± ˉ Gamma1. Thus the width of the resonance curve is W = 2Gamma1. Note that for a material characterized by a low-loss resonance (Gamma1 lessmuch ω 0 ), the location of ?epsilon1 prime max can be approximated as ω max = ω 0 radicalbig 1 ? 2Gamma1/ω 0 ≈ ω 0 ?Gamma1 while ?epsilon1 prime min is located at ω min = ω 0 radicalbig 1 + 2Gamma1/ω 0 ≈ ω 0 +Gamma1. The region of anomalous dispersion thus lies between the half amplitude points of ?epsilon1 primeprime : ω 0 ?Gamma1<ω<ω 0 +Gamma1. As Gamma1 → 0 the resonance curve becomes narrower and taller. Thus, a material charac- terizedbyaverylow-lossresonancemaybemodeledverysimplyusing ?epsilon1 primeprime = Aδ(ω?ω 0 ), where A isaconstanttobedetermined. Wecan?nd A byapplyingtheKronig–Kramers formula (4.37): ?epsilon1 prime (ω)?epsilon1 0 =? 2 π P.V. ∞ integraldisplay 0 Aδ(Omega1?ω 0 ) Omega1dOmega1 Omega1 2 ?ω 2 =? 2 π A ω 0 ω 2 0 ?ω 2 . Sincethematerialapproachesthelosslesscase,thisexpressionshouldmatchtheSellmeier equation (4.105): ? 2 π A ω 0 ω 2 0 ?ω 2 = epsilon1 0 ω 2 p ω 2 0 ?ω 2 , giving A =?πepsilon1 0 ω 2 p /2ω 0 . Hencethepermittivityofamaterialcharacterizedbyalow-loss resonance may be approximated as ?epsilon1 c (ω) = epsilon1 0 parenleftBigg 1 + ω 2 p ω 2 0 ?ω 2 parenrightBigg ? jepsilon1 0 π 2 ω 2 p ω 0 δ(ω?ω 0 ). 7 8 9 10 11 12 log ( f ) 0 20 40 60 ?ε /ε 0 ε /ε 0 10 Figure 4.5: Relaxation spectrum for water at 20 ? C found using Debye equation. DebyerelaxationandtheCole–Coleequation. In solids or liquids consisting of polarmolecules(thoseretainingapermanentdipolemoment,e.g.,water),theresonance e?ect is replaced by relaxation. We can view the molecule as attempting to rotate in response to an applied ?eld within a background medium dominated by the frictional termin(4.101). Therotatingmoleculeexperiencesmanyweakcollisionswhichcontinu- ouslydraino?energy,preventingitfromacceleratingundertheforceoftheapplied?eld. J.W.P.Debyeproposedthatsuchmaterialsaredescribedbyanexponentialdampingof their polarization and a complete absence of oscillations. If we neglect the acceleration term in (4.101) we have the equation of motion 2Gamma1 dl(r,t) dt +ω 2 r l(r,t) =? q e m e E prime (r,t), which has homogeneous solution l(r,t) = l 0 (r)e ? ω 2 r 2Gamma1 t = l 0 (r)e ?t/τ where τ is Debye’srelaxationtime. By neglecting the acceleration term in (4.102) we obtain from (4.103) the dispersion equation, orrelaxationspectrum ?epsilon1(ω)= epsilon1 0 +epsilon1 0 ω 2 p ω 2 0 + jω2Gamma1 . Debyeproposedarelaxationspectrumabitmoregeneralthanthis,nowcalledtheDebye equation: ?epsilon1(ω)= epsilon1 ∞ + epsilon1 s ?epsilon1 ∞ 1 + jωτ . (4.106) Figure4.6:ArcplotsforDebyeandCole–Coledescriptionsofapolarmaterial. Hereepsilon1 s istherealstaticpermittivityobtainedwhenω → 0,whileepsilon1 ∞ isthereal“optical” permittivitydescribingthehighfrequencybehaviorof ?epsilon1. Ifwesplit(4.106)intorealand imaginary parts we ?nd that ?epsilon1 prime (ω)?epsilon1 ∞ = epsilon1 s ?epsilon1 ∞ 1 +ω 2 τ 2 , ?epsilon1 primeprime (ω) =? ωτ(epsilon1 s ?epsilon1 ∞ ) 1 +ω 2 τ 2 . Forapassivematerialwemusthave ?epsilon1 primeprime < 0,whichrequires epsilon1 s >epsilon1 ∞ . Itisstraightforward to show that these expressions obey the Kronig–Kramers relationships. The details are left as an exercise. AplotoftheDebyespectrumofwaterat T = 20 ? CisshowninFigure4.5,wherewe have used epsilon1 s = 78.3epsilon1 0 , epsilon1 ∞ = 5epsilon1 0 , and τ = 9.6 × 10 ?12 s [49]. We see that ?epsilon1 prime decreases overtheentirefrequencyrange. Thefrequencydependenceoftheimaginarypartofthe permittivityissimilartothatfoundintheresonancemodel,formingacurvewhichpeaks at thecriticalfrequency ω max = 1/τ where it obtains a maximum value of ??epsilon1 primeprime max = epsilon1 s ?epsilon1 ∞ 2 . At this point ?epsilon1 prime achieves the average value of epsilon1 s and epsilon1 ∞ : epsilon1 prime (ω max ) = epsilon1 s +epsilon1 ∞ 2 . Since the frequency label is logarithmic, we see that the peak is far broader than that for the resonance model. Interestingly,aplotof??epsilon1 primeprime versus ?epsilon1 prime tracesoutasemicirclecenteredalongtherealaxis at (epsilon1 s +epsilon1 ∞ )/2andwithradius(epsilon1 s ?epsilon1 ∞ )/2.Suchaplot,showninFigure4.6,was?rst describedbyK.S.ColeandR.H.Cole[38]andisthuscalledaCole–Colediagramor“arc 0204060 ε /ε 0 20 40 60 - ε / ε 0 0 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 δ is described by thelosstangent of the material: tanδ =? ?epsilon1 primeprime ?epsilon1 prime = ωτ(epsilon1 s ?epsilon1 ∞ ) epsilon1 s +epsilon1 ∞ ω 2 τ 2 . (4.107) The Cole–Cole plot shows that the maximum value of ??epsilon1 primeprime is (epsilon1 s ? epsilon1 ∞ )/2 and that ?epsilon1 prime = (epsilon1 s +epsilon1 ∞ )/2 at this point. ACole–Coleplotforwater,showninFigure4.7,displaysthetypicalsemicircular nature of the arc plot. However, not all polar materials have a relaxation spectrum that follows the Debye equation as closely as water. Cole and Cole found that for many materialsthearcplottracesacirculararccenteredbelow therealaxis,andthattheline throughitscentermakesanangleofα(π/2)withtherealaxisasshowninFigure4.6. This relaxation spectrum can be described in terms of a modi?ed Debye equation ?epsilon1(ω)= epsilon1 ∞ + epsilon1 s ?epsilon1 ∞ 1 +(jωτ) 1?α , called theCole–Coleequation. A nonzero Cole–Cole parameter α tends 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 su?cient, but for other materials α may be much higher. For instance, consideratransformeroilwithameasuredCole–Coleparameterof α = 0.23,alongwith ameasuredrelaxationtimeofτ= 2.3 × 10 ?9 s,astaticpermittivityofepsilon1 s = 5.9epsilon1 0 , and anopticalpermittivityof epsilon1 ∞ = 2.9epsilon1 0 [4].Figure4.8showstheCole–Coleplotcalculated usingbothα= 0andα= 0.23,demonstratingasigni?cantdivergencefromtheDebye model.Figure4.9showstherelaxationspectrumforthetransformeroilcalculatedwith these same two parameters. 2.0 2.5 3.0 3.5 4.0 4.5 5.0 ε /ε 0 1 2 3 - ε / ε Debye equation Cole-Cole equation 0 0 Figure 4.8: Cole–ColediagramfortransformeroilfoundusingDebyeequationandCole– Cole equation with α = 0.23. 579 log ( f ) 0 1 2 3 4 5 Debye equation Cole-Cole equation ?ε /ε 0 ε /ε 0 10 Figure 4.9: Relaxation spectrum for transformer oil found using Debye equation and Cole–Cole equation with α = 0.23. 4.6.4 Permittivityandconductivityofaconductor The free electrons within a conductor may be considered as an electron gas which is freetomoveunderthein?uenceofanapplied?eld. Sincetheelectronsarenotboundto the atoms of the conductor, there is no restoring force acting on them. However, there is a damping term associated with electron collisions. We therefore model a conductor as a plasma, but with a very high collision frequency; in a good metallic conductor ν is typically in the range 10 13 –10 14 Hz. We therefore have the conductivity of a conductor from (4.75) as ?σ(ω)= epsilon1 0 ω 2 p ν ω 2 +ν 2 and the permittivity as ?epsilon1(ω)= epsilon1 0 bracketleftBigg 1 ? ω 2 p ω 2 +ν 2 bracketrightBigg . Since ν is so large, the conductivity is approximately ?σ(ω)≈ epsilon1 0 ω 2 p ν = Nq 2 e m e ν and the permittivity is ?epsilon1(ω)≈ epsilon1 0 wellpastmicrowavefrequenciesandintotheinfrared. Hencethedcconductivityisoften employed by engineers throughout the communications bands. When approaching the visible spectrum the permittivity and conductivity begin to show a strong frequency dependence. In the violet and ultraviolet frequency ranges the free-charge conductivity becomesproportionalto 1/ω andisdriventowardzero. However,atthesefrequenciesthe resonances of the bound electrons of the metal become important and the permittivity behaves more like that of a dielectric. At these frequencies the permittivity is best described using the resonance formula (4.104). 4.6.5 Permeabilitydyadicofaferrite The magnetic properties of materials are complicated and diverse. The formation of accurate models based on atomic behavior requires an understanding of quantum mechanics, but simple models may be constructed using classical mechanics along with very 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]. Themagneticpropertiesofmatterultimatelyresultfromatomiccurrents. Inoursim- ple microscopic view these currents arise from the spin and orbital motion of negatively chargedelectrons. Theseatomiccurrentspotentiallygiveeachatomamagneticmoment m.Indiamagnetic materials the orbital and spin moments cancel unless the material is exposedtoanexternalmagnetic?eld,inwhichcasetheorbitalelectronvelocitychanges to produce a net moment opposite the applied ?eld. Inparamagneticmaterials the spin momentsaregreaterthantheorbitalmoments,leavingtheatomswithanetpermanent magneticmoment. Whenexposedtoanexternalmagnetic?eld,thesemomentsalignin the same direction as an applied ?eld. In either case, the density of magnetic moments M is zero in the absence of an applied ?eld. In most paramagnetic materials the alignment of the permanent moment of neigh- boring atoms is random. However, in the subsets of paramagnetic materials known as ferromagnetic,anti-ferromagnetic,andferrimagneticmaterials,thereisastrongcoupling betweenthespinmomentsofneighboringatomsresultingineitherparallelorantiparal- lel alignment of moments. The most familiar case is the parallel alignment of moments withinthedomainsofferromagneticpermanentmagnetsmadeofiron,nickel,andcobalt. Anti-ferromagnetic materials, such as chromium and manganese, have strongly coupled moments that alternate in direction between small domains, resulting in zero net mag- netic moment. Ferrimagnetic materials also have alternating moments, but these are unequal and thus do not cancel completely. Ferritesformaparticularlyusefulsubgroupofferrimagneticmaterials. Theywere?rst 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- tivitiesrangingfrom 10 ?4 to 10 0 S/m(comparedto 10 7 foriron),relativepermeabilities inthethousands,anddielectricconstantsintherange10–15. Theirlowlossmakesthem useful for constructing transformer cores and for a variety of microwave applications. TheirchemicalformulaisXO·Fe 2 O 3 ,where X isadivalentmetalormixtureofmetals, such as cadmium, copper, iron, or zinc. When exposed to static magnetic ?elds, ferrites exhibitgyrotropicmagnetic(orgyromagnetic)propertiesandhavepermeabilitymatrices of the form (2.32). The properties of a wide variety of ferrites are given by von Aulock [204]. Todeterminethepermeabilitymatrixofaferritewewillmodelitselectronsassimple spinningtopsandexaminethetorqueexertedonthemagneticmomentbytheapplication of an external ?eld. Each electron has an angular momentum L and a magnetic dipole moment m, with these two vectors anti-parallel: m(r,t) =?γL(r,t) where γ = q e m e = 1.7592 × 10 11 C/kg is called thegyromagneticratio. Letus?rstconsiderasinglespinningelectronimmersedinanappliedstaticmagnetic ?eld B 0 . Anytorqueappliedtotheelectronresultsinachangeofangularmomentumas 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 ?eld experiences a torque m × B. Thus, when ?rst placed into a static magnetic ?eld B 0 an electron’s angular momentum obeys the equation dL(r,t) dt =?γL(r,t)× B 0 (r) = ω 0 (r)× L(r,t) (4.108) where ω 0 = γB 0 . This equation of motion describes theprecession of the electron spin axis about the direction of the applied ?eld, which is analogous to the precession of a gyroscope[129]. ThespinaxisrotatesattheLarmorprecessionalfrequency ω 0 = γ B 0 = γμ 0 H 0 . We can use this to understand what happens when we insert a homogeneous ferrite materialintoauniformstaticmagnetic?eld B 0 = μ 0 H 0 . Theinternal?eld H i experienced byanymagneticdipoleisnotthesameastheexternal?eld H 0 ,andneednotevenbein the same direction. In general we write H 0 (r,t)? H i (r,t) = H d (r,t) where H d is the demagnetizing?eld produced by the magnetic dipole moments of the material. Each electron responds to the internal ?eld by precessing as described above until the precession damps out and the electron moments align with the magnetic ?eld. At this point the ferrite issaturated. Because the demagnetizing ?eld depends strongly on the shape of the material we choose to ignore it as a ?rst approximation, and this allows 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 N isthenumberdensityofelectrons. Sinceweareassumingtheferriteishomoge- neous, we take N to be independent of timeand position. Multiplying(4.108) by ?Nγ, we obtain an equation describing the evolution of M: dM(r,t) dt =?γM(r,t)× B i (r,t). (4.109) To determine the temporal response of the ferrite we must include a time-dependent component of the applied ?eld. We now let H 0 (r,t) = H i (r,t) = H T (r,t)+ H dc where H T is the time-dependent component superimposed with the uniform static com- ponent H dc . Using B = μ 0 (H + M) we have from (4.109) dM(r,t) dt =?γμ 0 M(r,t)× [H T (r,t)+ H dc + M(r,t)]. With M = M T (r,t)+ M dc and M × M = 0 this becomes dM T (r,t) dt + dM dc dt =?γμ 0 [M T (r,t)× H T (r,t)+ M T (r,t)× H dc + + M dc × H T (r,t)+ M dc × H dc . (4.110) Letusassumethattheferriteissaturated. Then M dc isalignedwith H dc andtheircross product vanishes. Let us further assume that the spectrum of H T is small compared to H dc at all frequencies: | ? H T (r,ω)|lessmuchH dc . This small-signal assumption allows us to neglect M T ×H T . Usingtheseandnotingthatthetimederivativeof M dc iszero,wesee that (4.110) reduces to dM T (r,t) dt =?γμ 0 [M T (r,t)× H dc + M dc × H T (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 ?nd that jω ? M T (r,ω)=?γμ 0 [ ? M T (r,ω)× H dc + M dc × ? H T (r,ω)]. De?ning γμ 0 M dc = ω M , where ω M =|ω M | is thesaturationmagnetizationfrequency, we ?nd that ? M T + ? M T × bracketleftbigg ω 0 jω bracketrightbigg = bracketleftbigg ? 1 jω ω M × ? H T bracketrightbigg , (4.112) where ω 0 = γμ 0 H dc with ω 0 now called thegyromagneticresponsefrequency. This has theform v+v×C = A,whichhassolution(4.80). Substitutingintothisexpressionand remembering that ω 0 is parallel to ω M , we ?nd that ? M T = ? 1 jω ω M × ? H T + 1 ω 2 braceleftbig ω M [ω 0 · ? H T ] ?(ω 0 ·ω M ) ? H T bracerightbig 1 ? ω 2 0 ω 2 . If we de?ne the dyadic ˉω M such that ˉω M · ? H T = ω M × ? H T , then we identify the dyadic magnetic susceptibility ? ˉχ m (ω) = jω ˉω M +ω M ω 0 ?ω M ω 0 ˉ I ω 2 ?ω 2 0 (4.113) with which we can write ? M(r,ω)= ˉχ m (ω) · ? H(r,ω). In rectangular coordinates ˉω M is represented by [ ˉω M ] = ? ? 0 ?ω Mz ω My ω Mz 0 ?ω Mx ?ω My ω Mx 0 ? ? . (4.114) Finally, using ? B = μ 0 ( ? H + ? M) = μ 0 ( ˉ I + ? ˉχ m )· ? H = ? ˉμ· ? H we ?nd that ? ˉμ(ω) = μ 0 [ ˉ I + ? ˉχ m (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 H dc is aligned with the z-axis so that H dc = ?zH dc and thus ω M = ?zω M and ω 0 = ?zω 0 . Then (4.114) becomes [ ˉω M ] = ? ? 0 ?ω M 0 ω M 00 000 ? ? and we can write the susceptibility dyadic (4.113) as [ ? ˉχ m (ω)] = ω M ω 2 ?ω 2 0 ? ? ?ω 0 ?jω 0 jω ?ω 0 0 000 ? ? . The permeability dyadic becomes [ ? ˉμ(ω)] = ? ? μ ?jκ 0 jκμ 0 00μ 0 ? ? (4.115) where μ = μ 0 parenleftbigg 1 ? ω 0 ω M ω 2 ?ω 2 0 parenrightbigg , (4.116) κ = μ 0 ωω M ω 2 ?ω 2 0 . (4.117) Because its permeability dyadic is that for a losslessgyrotropic material (2.33), we call the ferritegyromagnetic. Since the ferrite is lossless, the dyadic permeability must be hermitian according to (4.49). The speci?c form of (4.115) shows this explicitly. We also note that since the sign of ω M is determined by that of H dc , the dyadic permittivity obeys the symmetry relation ?μ ij (H dc ) = ?μ ji (?H dc ), 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 [M T (r,t)× H dc + M dc × H T (r,t)] +α M dc M dc × dM T (r,t) dt , where α is the damping parameter [40, 204]. This term tends to reduce the angle of precession. Fourier transformation gives jω ? M T = ω 0 × ? M T ?ω M × ? H T +α ω M ω M × jω ? M T . Remembering that ω 0 and ω M are aligned we can write this as ? M T + ? M T × ? ? ω 0 parenleftBig 1 + jα ω ω 0 parenrightBig jω ? ? = bracketleftbigg ? 1 jω ω M × ? H T bracketrightbigg . This is identical to (4.112) with ω 0 → ω 0 parenleftbigg 1 + jα ω ω 0 parenrightbigg . Thus, we merely substitute this into (4.113) to ?nd the susceptibility dyadic for a lossy ferrite: ? ˉχ m (ω) = jω ˉω M +ω M ω 0 (1 + jαω/ω 0 )?ω M ω 0 (1 + jαω/ω 0 ) ˉ I ω 2 (1 +α 2 )?ω 2 0 ? 2 jαωω 0 . Making the same substitution into (4.115) we can write the dyadic permeability matrix as [ ? ˉμ(ω)] = ? ? ?μ xx ?μ xy 0 ?μ yx ?μ yy 0 00μ 0 ? ? (4.118) where ?μ xx = ?μ yy = μ 0 ?μ 0 ω M ω 0 bracketleftbig ω 2 (1 ?α 2 )?ω 2 0 bracketrightbig + jωα bracketleftbig ω 2 (1 +α 2 )+ω 2 0 bracketrightbig bracketleftbig ω 2 (1 +α 2 )?ω 2 0 bracketrightbig 2 + 4α 2 ω 2 ω 2 0 (4.119) and ?μ xy =??μ yx = 2μ 0 αω 2 ω 0 ω M ? jμ 0 ωω M bracketleftbig ω 2 (1 +α 2 )?ω 2 0 bracketrightbig bracketleftbig ω 2 (1 +α 2 )?ω 2 0 bracketrightbig 2 + 4α 2 ω 2 ω 2 0 . (4.120) In the case of a lossy ferrite, the hermitian nature of the permeability dyadic is lost. 4.7 Monochromatic ?elds and the phasor domain The Fourier transform is very e?cient for representing the nearly sinusoidal signals produced by electronic systems such as oscillators. However, we should realize that the elemental term e jωt by itself cannot represent any physical quantity; only a continuous superpositionofsuchtermscanhavephysicalmeaning,becausenophysicalprocesscan be truly monochromatic. All events must have transient periods during which they are established. Even “monochromatic” light appears in bundles called quanta, interpreted as containing ?nite numbers of oscillations. Arguments about whether “monochromatic” or “sinusoidal steady-state” ?elds can actually exist may sound purely academic. After all, a microwave oscillator can create a wave train of 10 10 oscillations within the ?rst second after being turned on. Such a waveformissurelyasclosetomonochromaticaswewouldcaretomeasure. Butaswith 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 been inthesteadystate. Sinusoidalsteady-statesolutionstoMaxwell’sequationscanleadto troublesomein?nitieslinkedtothein?niteenergycontentofeachelementalcomponent. Forexample,anattempttocomputetheenergystoredwithinalosslessmicrowavecavity understeady-stateconditionsgivesanin?niteresultsincethecavityhasbeenbuildingup energysince t =?∞. Wehandlethisbyconsideringtime-averagedquantities, buteven then must be careful when materials are dispersive (§ 4.5). Nevertheless, the steady- state concept is valuable because of its simplicity and ?nds widespread application in electromagnetics. Since the elemental term is complex, we may use its real part, its imaginary part, or some combination of both to represent a monochromatic (ortime-harmonic) ?eld. We choose the representation ψ(r,t) = ψ 0 (r)cos[ ˇωt +ξ(r)], (4.121) where ξ isthetemporalphaseangleofthesinusoidalfunction. TheFouriertransformis ? ψ(r,ω)= integraldisplay ∞ ?∞ ψ 0 (r)cos[ ˇωt +ξ(r)]e ?jωt dt. (4.122) Here we run into an immediate problem: the transform in (4.122) does not exist in the ordinarysensesince cos( ˇωt +ξ)isnotabsolutelyintegrableon (?∞,∞). Weshouldnot be surprised by this: the cosine function cannot describe an actual physical process (it extends 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 ?eld (4.121) is viewed as having the generalized transform ? ψ(r,ω)= ψ 0 (r)π bracketleftbig e jξ(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 ∞ ?∞ ψ 0 (r)π bracketleftbig e jξ(r) δ(ω? ˇω)+ e ?jξ(r) δ(ω+ ˇω) bracketrightbig e jωt dω. (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) = ψ 0 (r) e j ˇωt e jξ(r) + e ?j ˇωt e ?jξ(r) 2 = ψ 0 (r)cos[ ˇωt +ξ(r)] as expected. 4.7.1 Thetime-harmonicEM?eldsandconstitutiverelations The time-harmonic ?elds are described using the representation (4.121) for each ?eld component. The electric ?eld is E(r,t) = 3 summationdisplay i=1 ? i i |E i (r)|cos[ ˇωt +ξ E i (r)] forexample. Here |E i | isthecomplexmagnitudeofthe ithvectorcomponent,and ξ E i is the phase angle (?π<ξ E i ≤ π). Similar terminology is used for the remaining ?elds. Thefrequency-domainconstitutiverelations(4.11)–(4.15)maybewrittenforthetime- harmonic ?elds 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)|e ξ epsilon1 (r,ω) , ?μ(r,ω)=|?μ(r,ω)|e ξ μ (r,ω) , we can write D(r,t) = 3 summationdisplay i=1 ? i i |D i (r)|cos[ ˇωt +ξ D i (r)] = 1 2π integraldisplay ∞ ?∞ 3 summationdisplay i=1 ? i i ?epsilon1(r,ω)|E i (r)|π bracketleftBig e jξ E i (r) δ(ω? ˇω)+ e ?jξ E i (r) δ(ω+ ˇω) bracketrightBig e jωt dω = 1 2 3 summationdisplay i=1 ? i i |E i (r)| bracketleftBig ?epsilon1(r, ˇω)e j( ˇωt+jξ E i (r)) + ?epsilon1(r,?ˇω)e ?j( ˇωt+jξ E i (r)) bracketrightBig . Since (4.25) shows that ?epsilon1(r,?ˇω) = ?epsilon1 ? (r, ˇω), we have D(r,t) = 1 2 3 summationdisplay i=1 ? i i |E i (r)||?epsilon1(r, ˇω)| bracketleftBig e j( ˇωt+jξ E i (r)+jξ epsilon1 (r, ˇω)) + e ?j( ˇωt+jξ E i (r)+jξ epsilon1 (r, ˇω)) bracketrightBig = 3 summationdisplay i=1 ? i i |?epsilon1(r, ˇω)||E i (r)|cos[ ˇωt +ξ E i (r)+ξ epsilon1 (r, ˇω)]. (4.125) Similarly B(r,t) = 3 summationdisplay i=1 ? i i |B i (r)|cos[ ˇωt +ξ B i (r)] = 3 summationdisplay i=1 ? i i |?μ(r, ˇω)||H i (r)|cos[ ˇωt +ξ H i (r)+ξ μ (r, ˇω)]. 4.7.2 Thephasor?eldsandMaxwell’sequations Sinusoidalsteady-statecomputationsusingtheforwardandinversetransformformulas are unnecessarily cumbersome. A much more e?cient approach is to use the phasor concept. If we de?ne the complex function ˇ ψ(r) = ψ 0 (r)e jξ(r) asthephasorformofthemonochromatic?eld ? ψ(r,ω),thentheinverseFouriertransform is easily computed by multiplying ˇ ψ(r) by e j ˇωt and taking the real part. That is, ψ(r,t) = Re braceleftbig ˇ ψ(r)e j ˇωt bracerightbig = ψ 0 (r)cos[ ˇωt +ξ(r)]. (4.126) Usingthephasorrepresentationofthe?elds,wecanobtainasetofMaxwellequations relating the phasor components. Let ˇ E(r) = 3 summationdisplay i=1 ? i i ˇ E i (r) = 3 summationdisplay i=1 ? i i |E i (r)|e jξ E i (r) represent the phasor monochromatic electric ?eld, with similar formulas for the other ?elds. Then E(r,t) = Re braceleftbig ˇ E(r)e j ˇωt bracerightbig = 3 summationdisplay i=1 ? i i |E i (r)|cos[ ˇωt +ξ E i (r)]. Substituting these expressions into Ampere’s law (2.2), we have ?×Re braceleftbig ˇ H(r)e j ˇωt bracerightbig = ? ?t Re braceleftbig ˇ D(r)e j ˇωt bracerightbig + Re braceleftbig ˇ J(r)e j ˇωt bracerightbig . Since the real part of a sum of complex variables equals the sum of the real parts, we can write Re braceleftbigg ?× ˇ H(r)e j ˇωt ? ˇ D(r) ? ?t e j ˇωt ? ˇ J(r)e j ˇωt bracerightbigg = 0. (4.127) If we examine for an arbitrary complex function F = F r + jF i the quantity Re braceleftbig (F r + jF i )e j ˇωt bracerightbig = Re{(F r cos ˇωt ? F i sin ˇωt)+ j(F r sin ˇωt + F i cos ˇωt)}, we see that both F r and F i must be zero for the expression to vanish for all t.Thus (4.127) requires that ?× ˇ H(r) = j ˇω ˇ D(r)+ ˇ J(r), (4.128) which is the phasor Ampere’s law. Similarly we have ?× ˇ E(r) =?j ˇω ˇ B(r), (4.129) ?· ˇ D(r) = ˇρ(r), (4.130) ?· ˇ B(r) = 0, (4.131) and ?· ˇ J(r) =?j ˇωˇρ(r). (4.132) The constitutive relations may be easily incorporated into the phasor concept. If we use ˇ D i (r) = ?epsilon1(r, ˇω) ˇ E i (r) =|?epsilon1(r, ˇω)|e jξ epsilon1 (r, ˇω) |E i (r)|e jξ E i (r) , then forming D i (r,t) = Re braceleftbig ˇ D i (r)e j ˇωt bracerightbig we reproduce (4.125). Thus we may write ˇ D(r) = ?epsilon1(r, ˇω) ˇ E(r). Notethatweneverwrite ˇepsilon1 orrefertoa“phasorpermittivity”sincethepermittivitydoes not vary sinusoidally in the time domain. An obvious bene?t of the phasor method is that we can manipulate ?eld quantities withoutinvolvingthesinusoidaltimedependence. Whenourmanipulationsarecomplete, 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 = ˇω in the phasor equations. This is sensible, since the phasor ?elds represent a single component of the complete frequency-domain spectrum of the arbitrary time-varying ?elds. Thus, if the phasor ?elds are calculated for some ˇω, we can make the replacements ˇω → ω, ˇ E(r) → ? E(r,ω), ˇ H(r) → ? H(r,ω), ..., andobtainthegeneraltime-domainexpressionsbyperformingtheinversion(4.2). Simi- larly,ifweevaluatethefrequency-domain?eld ? E(r,ω)at ω = ˇω,weproducethephasor ?eld ˇ E(r) = ? E(r, ˇω) for this frequency. That is Re braceleftbig ? E(r, ˇω)e j ˇωt bracerightbig = 3 summationdisplay i=1 ? i i | ? E i (r, ˇω)|cos parenleftbig ˇωt +ξ E (r, ˇω) parenrightbig . 4.7.3 Boundaryconditionsonthephasor?elds The boundary conditions developed in § 4.3 for the frequency-domain ?elds may be adapted for use with the phasor ?elds by selecting ω = ˇω. Let us include the e?ects of ?ctitious magnetic sources and write ?n 12 ×( ˇ H 1 ? ˇ H 2 ) = ˇ J s , (4.133) ?n 12 ×( ˇ E 1 ? ˇ E 2 ) =? ˇ J ms , (4.134) ?n 12 ·( ˇ D 1 ? ˇ D 2 ) = ˇρ s , (4.135) ?n 12 ·( ˇ B 1 ? ˇ B 2 ) = ˇρ ms , (4.136) and ?n 12 ·( ˇ J 1 ? ˇ J 2 ) =?? s · ˇ J s ? j ˇωˇρ s , (4.137) ?n 12 ·( ˇ J m1 ? ˇ J m2 ) =?? s · ˇ J ms ? j ˇωˇρ ms , (4.138) where ?n 12 points into region 1 from region 2. 4.8 Poynting’s theorem for time-harmonic ?elds WecanspecializePoynting’stheoremtotime-harmonicformbysubstitutingthetime- harmonic ?eld 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 of storedenergy. Butiftheoperatingfrequencylieswithintherealmofmaterialdispersion and loss, then we can no longer identify an explicit stored energy term. 4.8.1 GeneralformofPoynting’stheorem Webeginwith(2.301). Substitutingthetime-harmonicrepresentationsweobtainthe term E(r,t)· ?D(r,t) ?t = bracketleftBigg 3 summationdisplay i=1 ? i i |E i |cos[ ˇωt +ξ E i ] bracketrightBigg · ? ?t bracketleftBigg 3 summationdisplay i=1 ? i i |D i |cos[ ˇωt +ξ D i ] bracketrightBigg =?ˇω 3 summationdisplay i=1 |E i ||D i |cos[ ˇωt +ξ E i ] sin[ ˇωt +ξ D i ]. Since 2 sin A cos B ≡ sin(A + B)+ sin(A ? B) we have E(r,t)· ? ?t D(r,t) =? 1 2 3 summationdisplay i=1 ˇω|E i ||D i |S DE ii (t), where S DE ii (t) = sin(2 ˇωt +ξ D i +ξ E i )+ sin(ξ D i ?ξ E i ) describes the temporal dependence of the ?eld product. Separating the current into an impressed term J i and a secondary term J c (assumed to be the conduction current) as J = J i + J c and repeating the above steps with the other terms, we obtain ? 1 2 integraldisplay V 3 summationdisplay i=1 |J i i ||E i |C J i E ii (t)dV = 1 2 contintegraldisplay S 3 summationdisplay i,j=1 |E i ||H j |( ? i i × ? i j )· ?nC EH ij (t)dS+ + 1 2 integraldisplay V 3 summationdisplay i=1 braceleftbig ?ˇω|D i ||E i |S DE ii (t)? ˇω|B i ||H i |S BH ii (t)+|J c i ||E i |C J c E ii (t) bracerightbig dV, (4.139) where S BH ii (t) = sin(2 ˇωt +ξ B i +ξ H i )+ sin(ξ B i ?ξ H i ), C EH ij (t) = cos(2 ˇωt +ξ E i +ξ H j )+ cos(ξ E i ?ξ H j ), and so on. We see that each power term has two temporal components: one oscillating at fre- quency 2 ˇω,andoneconstantwithtime. Theoscillatingcomponentdescribespowerthat cyclesthroughthevariousmechanismsofenergystorage,dissipation,andtransferacross the boundary. Dissipation may be produced through conduction processes or through polarizationandmagnetizationphaselag,asdescribedbythevolumetermontheright- hand side of (4.139). Power may also be delivered to the ?elds either from the sources, as described by the volume term on the left-hand side, or from an active medium, as describedbythevolumetermontheright-handside. Thetime-averagebalanceofpower suppliedtothe?eldsandextractedfromthe?eldsthroughouteachcycle,includingthat transported across the surface S, is given by the constant terms in (4.139): ? 1 2 integraldisplay V 3 summationdisplay i=1 |J i i ||E i |cos(ξ J i i ?ξ E i )dV = 1 2 integraldisplay V 3 summationdisplay i=1 braceleftbig ˇω|E i ||D i |sin(ξ E i ?ξ D i )+ +ˇω|B i ||H i |sin(ξ H i ?ξ B i )+|J c i ||E i |cos(ξ J c i ?ξ E i ) bracerightbig dV + + 1 2 contintegraldisplay S 3 summationdisplay i,j=1 |E i ||H j |( ? i i × ? i j )· ?n cos(ξ E i ?ξ H j )dS. (4.140) We associate one mechanism for time-average power loss with the phase lag between applied ?eld and resulting polarization or magnetization. We can see this more clearly if we use the alternative form of the Poynting theorem (2.302) written in terms of the polarization and magnetization vectors. Writing P(r,t) = 3 summationdisplay i=1 |P i (r)|cos[ ˇωt +ξ P i (r)], M(r,t) = 3 summationdisplay i=1 |M i (r)|cos[ ˇωt +ξ M i (r)], and substituting the time-harmonic ?elds, we see that ? 1 2 integraldisplay V 3 summationdisplay i=1 |J i ||E i |C JE ii (t)dV + ˇω 2 integraldisplay V 3 summationdisplay i=1 bracketleftbig |P i ||E i |S PE ii (t)+μ 0 |M i ||H i |S MH ii (t) bracketrightbig dV =? ˇω 2 integraldisplay V 3 summationdisplay i=1 bracketleftbig epsilon1 0 |E i | 2 S EE ii (t)+μ 0 |H i | 2 S HH ii (t) bracketrightbig dV + + 1 2 contintegraldisplay S 3 summationdisplay i,j=1 |E i ||H j |( ? i i × ? i j )· ?nC EH ij (t)dS. (4.141) Selection of the constant part gives the balance of time-average power: ? 1 2 integraldisplay V 3 summationdisplay i=1 |J i ||E i |cos(ξ J i ?ξ E i )dV = ˇω 2 integraldisplay V 3 summationdisplay i=1 bracketleftbig |E i ||P i |sin(ξ E i ?ξ P i )+μ 0 |H i ||M i |sin(ξ H i ?ξ M i ) bracketrightbig dV + + 1 2 contintegraldisplay S 3 summationdisplay i,j=1 |E i ||H j |( ? i i × ? i j )· ?n cos(ξ E i ?ξ H j )dS. (4.142) Here the power loss associated with the lag in alignment of the electric and magnetic dipolesiseasilyidenti?edasthevolumetermontheright-handside,andisseentoarise throughtheinteractionofthe?eldswiththeequivalentsourcesasdescribedthroughthe phasedi?erencebetween E and P andbetween H and M. Ifthesepairsareinphase,then the time-average power balance reduces to that for a dispersionless material, equation (4.146). 4.8.2 Poynting’stheoremfornondispersivematerials Fornondispersivematerials(2.299)isappropriate. Weshallcarryoutthedetailshere so that we may examine the power-balance implications of nondispersive media. We have, substituting the ?eld expressions, ? 1 2 integraldisplay V 3 summationdisplay i=1 |J i i ||E i |C J i E ii (t)dV = 1 2 integraldisplay V 3 summationdisplay i=1 |J c i ||E i |C J c E ii (t)dV + + ? ?t integraldisplay V 3 summationdisplay i=1 braceleftbigg 1 4 |D i ||E i |C DE ii (t)+ 1 4 |B i ||H i |C BH ii (t) bracerightbigg dV + + 1 2 contintegraldisplay S 3 summationdisplay i,j=1 |E i ||H j |( ? i i × ? i j )· ?nC EH ij (t)dS. (4.143) Here we remember that the conductivity relating E to J c must also be nondispersive. Notethattheelectricandmagneticenergydensities w e (r,t) and w m (r,t) havethetime- average values 〈w e (r,t)〉 and 〈w m (r,t)〉 given by 〈w e (r,t)〉= 1 T integraldisplay T/2 ?T/2 1 2 E(r,t)· D(r,t)dt = 1 4 3 summationdisplay i=1 |E i ||D i |cos(ξ E i ?ξ D i ) = 1 4 Re braceleftbig ˇ E(r)· ˇ D ? (r) bracerightbig (4.144) and 〈w m (r,t)〉= 1 T integraldisplay T/2 ?T/2 1 2 B(r,t)· H(r,t)dt = 1 4 3 summationdisplay i=1 |B i ||H i |cos(ξ H i ?ξ B i ) = 1 4 Re braceleftbig ˇ H(r)· ˇ B ? (r) bracerightbig , (4.145) where T = 2π/ˇω. Wehavealreadyidenti?edtheenergystoredinanondispersivematerial (§ 4.5.2). If(4.144)istomatchwith(4.62),thephasesof ˇ E and ˇ D mustmatch: ξ E i = ξ D i . We must also have ξ H i = ξ B i . Since in a dispersionless material σ must be independent of frequency, from ˇ J c = σ ˇ E we also see that ξ J c i = ξ E i . Upondi?erentiationthetime-averagestoredenergytermsin(4.143)disappear,giving ? 1 2 integraldisplay V 3 summationdisplay i=1 |J i i ||E i |C J i E ii (t)dV = 1 2 integraldisplay V 3 summationdisplay i=1 |J c i ||E i |C EE ii (t)dV ? ?2 ˇω integraldisplay V 3 summationdisplay i=1 braceleftbigg 1 4 |D i ||E i |S EE ii (t)+ 1 4 |B i ||H i |S BB ii (t) bracerightbigg dV + + 1 2 contintegraldisplay S 3 summationdisplay i,j=1 |E i ||H j |( ? i i × ? i j )· ?nC EH ij (t)dS. Equating the constant terms, we ?nd the time-average power balance expression ? 1 2 integraldisplay V 3 summationdisplay i=1 |J i i ||E i |cos(ξ J i i ?ξ E i )dV = 1 2 integraldisplay V 3 summationdisplay i=1 |J c i ||E i |dV + + 1 2 contintegraldisplay S 3 summationdisplay i,j=1 |E i ||H j |( ? i i × ? i j )· ?n cos(ξ E i ?ξ H j )dS. (4.146) This can be written more compactly using phasor notation as integraldisplay V p J (r)dV = integraldisplay V p σ (r)dV + contintegraldisplay S S av (r)· ?n dS (4.147) where p J (r) =? 1 2 Re braceleftbig ˇ E(r)· ˇ J i? (r) bracerightbig is the time-average density of power delivered by the sources to the ?elds in V, p σ (r) = 1 2 ˇ E(r)· ˇ J c? (r) isthetime-averagedensityofpowertransferredtotheconductingmaterialasheat,and S av (r)· ?n = 1 2 Re braceleftbig ˇ E(r)× ˇ H ? (r) bracerightbig · ?n is the density of time-average power transferred across the boundary surface S. Here S c = ˇ E(r)× ˇ H ? (r) iscalledthecomplexPoyntingvector and S av iscalledthetime-averagePoyntingvector. Comparisonof(4.146)with(4.140)showsthatnondispersivematerialscannotmanifest the dissipative (or active) properties determined by the term 1 2 integraldisplay V 3 summationdisplay i=1 braceleftbig ˇω|E i ||D i |sin(ξ E i ?ξ D i )+ ˇω|B i ||H i |sin(ξ H i ?ξ B i )+|J c i ||E i |cos(ξ J c i ?ξ E i ) bracerightbig dV. This term can be used to classify materials as lossless, lossy, or active, as shown next. 4.8.3 Lossless,lossy,andactivemedia In § 4.5.1 we classi?ed materials based on whether they dissipate (or provide) energy over the period of a transient event. We can provide the same classi?cation based on their steady-state behavior. Weclassifyamaterialaslossless ifthetime-average?owofpowerenteringahomoge- 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 dissipate powerthatenters,orthatthereisamechanismthatcreatesenergytoexactlybalancethe dissipation. If the time-average power entering is positive, then the material dissipates power and is termed lossy. If the time-average power entering is negative, then power must originate from within the body and the material is termedactive. (Note that the power associated with an active body is not described as arising from sources, but is rather 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 in?ux of power density is given by ?? · S av (r) = p in (r) = 1 2 3 summationdisplay i=1 braceleftbig ˇω|E i ||D i |sin(ξ E i ?ξ D i )+ ˇω|B i ||H i |sin(ξ H i ?ξ B i )+ +|J c i ||E i |cos(ξ J c i ?ξ E i ) bracerightbig . Materials are then classi?ed as follows: p in (r) = 0, lossless, p in (r)>0, lossy, p in (r) ≥ 0, passive, p in (r)<0, active. We see that if ξ E i = ξ D i , ξ H i = ξ B i , and J c = 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) to obtain p in (r) =? ˇω 2 3 summationdisplay i=1 braceleftbigg |E i | 2 bracketleftbigg |?epsilon1|sin(ξ epsilon1 )? |?σ| ˇω cos(ξ σ ) bracketrightbigg +|?μ||H i | 2 sin(ξ μ ) bracerightbigg . (4.148) The ?rst two terms can be regarded as resulting from a single complex permittivity (4.26). Then (4.148) simpli?es to p in (r) =? ˇω 2 3 summationdisplay i=1 braceleftbig |?epsilon1 c ||E i | 2 sin(ξ epsilon1 c )+|?μ||H i | 2 sin(ξ μ ) bracerightbig . (4.149) Now we can see that a lossless medium, which requires (4.149) to vanish, has ξ epsilon1 c = ξ μ = 0 (or perhaps the unlikely condition that dissipative and active e?ects within the electric and magnetic terms exactly cancel). To have ξ μ = 0 we need B and H to be in phase, hence we need ?μ(r,ω)to be real. To have ξ epsilon1 c = 0 we need ξ epsilon1 = 0 (?epsilon1(r,ω)real) and ?σ(r,ω)= 0 (orperhapstheunlikelyconditionthattheactiveanddissipativee?ects of the permittivity and conductivity exactly cancel). A lossy medium requires (4.149) to be positive. This occurs when ξ μ < 0 or ξ epsilon1 c < 0, meaningthattheimaginarypartofthepermeabilityorcomplexpermittivityisnegative. The complex permittivity has a negative imaginary part if the imaginary part of ?epsilon1 is negative or if the real part of ?σ is positive. Physically, ξ epsilon1 < 0 means that ξ D <ξ E and thus the phase of the response ?eld D lags that of the excitation ?eld E. This results fromadelayinthepolarizationalignmentoftheatoms,andleadstodissipationofpower within the material. Anactivemediumrequires(4.149)tobenegative. Thisoccurswhenξ μ > 0 orξ epsilon1 c > 0, meaningthattheimaginarypartofthepermeabilityorcomplexpermittivityispositive. The complex permittivity has a positive imaginary part if the imaginary part of ?epsilon1 is positive or if the real part of ?σ is 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 lossy whenoneormoreofthefollowingholds: thepermittivityiscomplexwithnegativeimag- inarypart,thepermeabilityiscomplexwithnegativeimaginarypart,ortheconductivity has a positive real part. Finally, a complex permittivity or permeability with positive imaginary part or a conductivity with negative real part indicates anactive medium. For anisotropic materials the interpretation of p in is not as simple. Here we ?nd that the permittivity or permeability dyadic may be complex, and yet the material may still belossless. Todeterminetheconditionforalosslessmedium,letusrecompute p in using the constitutive relations (4.18)–(4.20). With these we have E · bracketleftbigg ?D ?t + J c bracketrightbigg + H · ?B ?t = ˇω 3 summationdisplay i,j=1 |E i ||E j | bracketleftbigg ?|?epsilon1 ij |sin( ˇωt +ξ E j +ξ epsilon1 ij )cos( ˇωt +ξ E i )+ + |?σ ij | ˇω cos( ˇωt +ξ E j +ξ σ ij )cos( ˇωt +ξ E i ) bracketrightbigg + + ˇω 3 summationdisplay i,j=1 |H i ||H j | bracketleftBig ?|?μ ij |sin( ˇωt +ξ H j +ξ μ ij )cos( ˇωt +ξ H i ) bracketrightBig . Usingtheangle-sumformulasanddiscardingthetime-varyingquantities,wemayobtain the time-average input power density: p in (r) =? ˇω 2 3 summationdisplay i,j=1 |E i ||E j | bracketleftbigg |?epsilon1 ij |sin(ξ E j ?ξ E i +ξ epsilon1 ij )? |?σ ij | ˇω cos(ξ E j ?ξ E i +ξ σ ij ) bracketrightbigg ? ? ˇω 2 3 summationdisplay i,j=1 |H i ||H j ||?μ ij |sin(ξ H j ?ξ H i +ξ μ ij ). The reader can easily verify that the conditions that make this quantity vanish, thus describing a lossless material, are |?epsilon1 ij |=|?epsilon1 ji |,ξ epsilon1 ij =?ξ epsilon1 ji , (4.150) |?σ ij |=|?σ ji |,ξ σ ij =?ξ σ ji +π, (4.151) |?μ ij |=|?μ ji |,ξ μ ij =?ξ μ ji . (4.152) Note that this requires ξ epsilon1 ii = ξ μ ii = ξ σ ii = 0. The condition (4.152) is easily written in dyadic form as ? ˉμ(r, ˇω) ? = ? ˉμ(r, ˇω) (4.153) where “?” stands for the conjugate-transpose operation. The dyadic permeability ? ˉμ is hermitian. The set of conditions (4.150)–(4.151) can also be written quite simply using the complex permittivity dyadic (4.24): ? ˉepsilon1 c (r, ˇω) ? = ? ˉepsilon1 c (r, ˇω). (4.154) Thus, an anisotropic material is lossless when the both the dyadic permeability and the complexdyadicpermittivityarehermitian. Since ˇω isarbitrary,theseresultsareexactly 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 ?epsilon1 and ?μ are real and ?σ is 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, although certainlysatis?edbythephasor?elds,doesnotreplacePoynting’stheoremasthepower- balance equation for time-harmonic ?elds. We shall be careful to contrast the interpre- tation of the phasor expression with the actual time-harmonic Poynting theorem. Webeginbydottingbothsidesofthephasor-domainFaraday’slawwith ˇ H ? toobtain ˇ H ? ·(?× ˇ E) =?j ˇω ˇ H ? · ˇ B. Taking the complex conjugate of the phasor-domain Ampere’s law and dotting with ˇ E, we have ˇ E ·(?× ˇ H ? ) = ˇ E · ˇ J ? ? j ˇω ˇ E · ˇ D ? . We subtract these expressions and use (B.44) to write ? ˇ E · ˇ J ? =?·( ˇ E × ˇ H ? )? j ˇω[ ˇ E · ˇ D ? ? ˇ B · ˇ H ? ]. Finally, integrating over the volume region V and dividing by two, we have ? 1 2 integraldisplay V ˇ E · ˇ J ? dV = 1 2 contintegraldisplay S ( ˇ E × ˇ H ? )· dS ? 2 j ˇω integraldisplay V bracketleftbigg 1 4 ˇ E · ˇ D ? ? 1 4 ˇ B · ˇ H ? bracketrightbigg dV. (4.155) ThisisknownasthecomplexPoyntingtheorem,andisanexpressionthatmustbeobeyed by the phasor ?elds. As a power balance theorem, the complex Poynting theorem has meaning only for dispersionlessmaterials. Ifwelet J = J i +J c andassumenodispersion,(4.155)becomes ? 1 2 integraldisplay V ˇ E · ˇ J i? dV = 1 2 integraldisplay V ˇ E · ˇ J c? dV + 1 2 contintegraldisplay S ( ˇ E × ˇ H ? )· dS ? ? 2 jω integraldisplay V [〈w e 〉?〈w m 〉] dV (4.156) where 〈w e 〉 and 〈w m 〉 are 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 V Re braceleftbig ˇ E · ˇ J i? bracerightbig dV = 1 2 integraldisplay V ˇ E · ˇ J c? dV + 1 2 contintegraldisplay S Re braceleftbig ˇ E × ˇ H ? 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, orreactive power: ? 1 2 integraldisplay V Im braceleftbig ˇ E · ˇ J i? bracerightbig dV = 1 2 contintegraldisplay S Im braceleftbig ˇ E × ˇ H ? bracerightbig · dS ? 2 ˇω integraldisplay V [〈w e 〉?〈w m 〉] dV. (4.158) Ingeneral, thereactivepowerbalancedoesnothaveasimplephysicalinterpretation(it isnot the balance of the oscillating terms in (4.139)). However, an interesting concept can be gleaned from it. If the source current and electric ?eld are in phase, and there is no reactive power leaving S, then the time-average stored electric energy is equal to the time-average stored magnetic energy: integraldisplay V 〈w e 〉dV = integraldisplay V 〈w m 〉dV. Thisistheconditionfor“resonance.”AnexampleisaseriesRLCcircuitwiththesource current and voltage in phase. Here the stored energy in the capacitor is equal to the stored energy in the inductor and the input impedance (ratio of voltage to current) is real. Such a resonance occurs at only one value of frequency. In more complicated electromagnetic systems resonance may occur at many discrete eigenfrequencies. 4.9.1 Boundaryconditionforthetime-averagePoyntingvector In § 2.9.5 we developed a boundary condition for the normal component of the time- domain Poynting vector. For time-harmonic ?elds we can derive a similar boundary condition using the time-average Poynting vector. Consider a surface S across which the electromagnetic sources and constitutive parameters are discontinuous, as shown in Figure2.6.Let ?n 12 be 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 two separatesurfacesshowninFigure2.6,weobtain 1 2 integraldisplay V bracketleftbigg ˇ E · ˇ J ? ? 2 j ˇω parenleftbigg 1 4 ˇ E · ˇ D ? ? 1 4 ˇ B · ˇ H ? parenrightbiggbracketrightbigg dV + 1 2 contintegraldisplay S S c · ?n dS = 1 2 integraldisplay S 10 ?n 12 ·(S c 1 ? S c 2 )dS (4.159) where S c = ˇ E × ˇ H ? 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 surface of discontinuity, and include the surface current contribution, we have 1 2 integraldisplay V bracketleftbigg ˇ E · ˇ J ? ? 2 j ˇω integraldisplay V parenleftbigg 1 4 ˇ E · ˇ D ? ? 1 4 ˇ B · ˇ H ? parenrightbiggbracketrightbigg dV + 1 2 contintegraldisplay S S c · ?n dS =? 1 2 integraldisplay S 10 ˇ J ? s · ˇ E dS. (4.160) If we wish to have the integrals over V and S in (4.159) and (4.160) produce identical results, then we must postulate the two conditions ?n 12 ×( ˇ E 1 ? ˇ E 2 ) = 0 and ?n 12 ·(S c 1 ? S c 2 ) =? ˇ J ? s · ˇ E. (4.161) The ?rst condition is merely the continuity of tangential electric ?eld; it allows us to be nonspeci?c as to which value of E we use in the second condition. If we take the real part of the second condition we have ?n 12 ·(S av,1 ? S av,2 ) = p J s, (4.162) where S av = 1 2 Re{ ˇ E × ˇ H ? } is the time-average Poynting power ?ow density and p J s = ? 1 2 Re{ ˇ J ? s · ˇ E} isthetime-averagedensityofpowerdeliveredbythesurfacesources. This is the desired boundary condition on the time-average power ?ow density. 4.10 Fundamental theorems for time-harmonic ?elds 4.10.1 Uniqueness If we think of a sinusoidal electromagnetic ?eld as the steady-state culmination of a transienteventthathasanidenti?ablestartingtime,thentheconditionsforuniqueness established in § 2.2.1 are applicable. However, a true time-harmonic wave, which has existed since t =?∞and thus has in?nite energy, must be interpreted di?erently. Our approach is similar to that of § 2.2.1. Consider a simply-connected region of space V bounded by surface S, where both V and S contain only ordinary points. The phasor-domain?eldswithin V areassociatedwithaphasorcurrentdistribution ˇ J,which may be internal to V (entirely or in part). We seek conditions under which the phasor electromagnetic ?elds are uniquely determined. Let the ?eld set ( ˇ E 1 , ˇ D 1 , ˇ B 1 , ˇ H 1 ) satisfy Maxwell’s equations (4.128) and (4.129) associated with the current ˇ J (along with an appropriate set of constitutive relations), and let ( ˇ E 2 , ˇ D 2 , ˇ B 2 , ˇ H 2 ) be a second solution. To determine the conditions for uniqueness of the ?elds, we look for a situation that results in ˇ E 1 = ˇ E 2 , ˇ H 1 = ˇ H 2 , and so on. The electromagnetic ?elds must obey ?× ˇ H 1 = j ˇω ˇ D 1 + ˇ J, ?× ˇ E 1 =?j ˇω ˇ B 1 , ?× ˇ H 2 = j ˇω ˇ D 2 + ˇ J, ?× ˇ E 2 =?j ˇω ˇ B 2 . Subtracting these and de?ning the di?erence ?elds ˇ E 0 = ˇ E 1 ? ˇ E 2 , ˇ H 0 = ˇ H 1 ? ˇ H 2 , and so on, we ?nd that ?× ˇ H 0 = j ˇω ˇ D 0 , (4.163) ?× ˇ E 0 =?j ˇω ˇ B 0 . (4.164) Establishing the conditions under which the di?erence ?elds vanish throughout V,we shall determine the conditions for uniqueness. Dotting (4.164) by ˇ H ? 0 and dotting the complex conjugate of (4.163) by ˇ E 0 ,wehave ˇ H ? 0 · parenleftbig ?× ˇ E 0 parenrightbig =?j ˇω ˇ B 0 · ˇ H ? 0 , ˇ E 0 · parenleftbig ?× ˇ H ? 0 parenrightbig =?j ˇω ˇ D ? 0 · ˇ E 0 . Subtraction yields ˇ H ? 0 · parenleftbig ?× ˇ E 0 parenrightbig ? ˇ E 0 · parenleftbig ?× ˇ H ? 0 parenrightbig =?j ˇω ˇ B 0 · ˇ H ? 0 + j ˇω ˇ D ? 0 · ˇ E 0 which, by (B.44), can be written as ?· parenleftbig ˇ E 0 × ˇ H ? 0 parenrightbig = j ˇω bracketleftbig ˇ E 0 · ˇ D ? 0 ? ˇ B 0 · ˇ H ? 0 bracketrightbig . Adding this expression to its complex conjugate, integrating over V, and using the di- vergence theorem, we obtain Re contintegraldisplay S bracketleftbig ˇ E 0 × ˇ H ? 0 bracketrightbig · dS =?j ˇω 2 integraldisplay V bracketleftbigparenleftbig ˇ E ? 0 · ˇ D 0 ? ˇ E 0 · ˇ D ? 0 parenrightbig + parenleftbig ˇ H ? 0 · ˇ B 0 ? ˇ H 0 · ˇ B ? 0 parenrightbigbracketrightbig dV. Breaking S into two arbitrary portions and using (??), we obtain Re contintegraldisplay S 1 ˇ H ? 0 ·(?n × ˇ E 0 )dS? Re contintegraldisplay S 2 ˇ E 0 ·(?n × ˇ H ? 0 )dS = ?j ˇω 2 integraldisplay V bracketleftbigparenleftbig ˇ E ? 0 · ˇ D 0 ? ˇ E 0 · ˇ D ? 0 parenrightbig + parenleftbig ˇ H ? 0 · ˇ B 0 ? ˇ H 0 · ˇ B ? 0 parenrightbigbracketrightbig dV. (4.165) Now if ?n × E 0 = 0 or ?n × H 0 = 0 over all of S, or some combination of these conditions holds over all of S, then integraldisplay V bracketleftbigparenleftbig ˇ E ? 0 · ˇ D 0 ? ˇ E 0 · ˇ D ? 0 parenrightbig + parenleftbig ˇ H ? 0 · ˇ B 0 ? ˇ H 0 · ˇ B ? 0 parenrightbigbracketrightbig dV = 0. (4.166) Thisimpliesarelationshipbetween ˇ E 0 , ˇ D 0 , ˇ B 0 ,and ˇ H 0 . Since V isarbitraryweseethat onepossiblerelationshipissimplytohaveoneofeachpair ( ˇ E 0 , ˇ D 0 ) and ( ˇ H 0 , ˇ B 0 ) equalto zero. Then, by (4.163) and (4.164), ˇ E 0 = 0 implies ˇ B 0 = 0, and ˇ D 0 = 0 implies ˇ H 0 = 0. Thus ˇ E 1 = ˇ E 2 , etc., and the solution is unique throughout V. However, we cannot in generalruleoutmorecomplicatedrelationships. Thenumberofpossibilitiesdependson the additional constraints on the relationship between ˇ E 0 , ˇ D 0 , ˇ B 0 , and ˇ H 0 that we must supplytodescribethematerialsupportingthe?eld—i.e.,theconstitutiverelationships. For a simple medium described by ?μ(ω) and ?epsilon1 c (ω), equation (4.166) becomes integraldisplay V parenleftbig | ˇ E 0 | 2 [?epsilon1 c ( ˇω)? ?epsilon1 c? ( ˇω)] +| ˇ H 0 | 2 [ ?μ( ˇω)? ?μ ? ( ˇω)] parenrightbig dV = 0 or integraldisplay V bracketleftbig | ˇ E 0 | 2 ?epsilon1 cprimeprime ( ˇω)+| ˇ H 0 | 2 ?μ primeprime ( ˇω) bracketrightbig dV = 0. For a lossy medium, ?epsilon1 cprimeprime < 0 and ?μ primeprime < 0 as 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 ˇ E 0 = ˇ H 0 = 0, and uniqueness is guaranteed. Whenestablishingmorecomplicatedconstitutiverelationswemustbecarefultoensure thattheyleadtoauniquesolution,andthattheconditionforuniquenessisunderstood. Inthecaseabove,theassumption ?n× ˇ E 0 vextendsingle vextendsingle S = 0 impliesthatthetangentialcomponentsof ˇ E 1 and ˇ E 2 areidenticalover S —thatis,wemustgivespeci?cvaluesofthesequantities on S to ensure uniqueness. A similar statement holds for the condition ?n × ˇ H 0 vextendsingle vextendsingle S = 0. In summary, the conditions for the ?elds within a region V containing lossy isotropic materials to be unique are as follows: 1. the sources within V must be speci?ed; 2. the tangential component of the electric ?eld must be speci?ed over all or part of the bounding surface S; 3. thetangentialcomponentofthemagnetic?eldmustbespeci?edovertheremainder of S. We may question the requirement of alossy medium to demonstrate uniqueness of the phasor?elds. Doesthismeanthatwithinavacuumthespeci?cationoftangential?elds is insu?cient? Experience shows that the ?elds in such a region are indeed properly described by the surface ?elds, and it is just a case of the mathematical model being slightlyoutofsync withthe physics. Aslongaswe recognizethatthesinusoidalsteady state requires an initial transient period, we know that speci?cation of the tangential ?elds is su?cient. We must be careful, however, to understand the restrictions of the mathematical model. Any attempt to describe the ?elds within a lossless cavity, for instance, is fraught with di?culty if true time-harmonic ?elds are used to model the actual physical ?elds. A helpful mathematical strategy is to think of free space as the limit of a lossy medium as the loss recedes to zero. Of course this does not represent the physical state of “empty” space. Although even interstellar space may have a few particles for every cubic meter to interact with the electromagnetic ?eld, the density of these particles invalidates our initial macroscopic assumptions. Another important concern is whether we can extend the uniqueness argument to all of space. If we let S recede to in?nity, must we continue to specify the ?elds over S,or is it su?cient to merely specify the sources within S? Since the boundary ?elds 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 S and thus no need for the boundary ?elds. This is indeed the case. If all sources are localized, the ?elds they producebehaveinjusttherightmannerforthesurfaceintegralin(4.165)tovanish,and thus uniqueness is again guaranteed. Later we will ?nd that the electric and magnetic ?eldsproducedbyalocalizedsourceatgreatdistancehavetheformofasphericalwave: ˇ E ～ ˇ H ～ e ?jkr r . Ifspaceistakentobeslightlylossy,then k iscomplexwithnegativeimaginarypart,and thusthe?eldsdecreaseexponentiallywithdistancefromthesource. Aswearguedabove, itmaynotbephysicallymeaningfultoassumethatspaceislossy. Sommerfeldpostulated that even for lossless space the surface integral in (4.165) vanishes as S →∞. This has been veri?ed experimentally, and provides the following restrictions on the free-space ?elds known as theSommerfeldradiationcondition: lim r→∞ r bracketleftbig η 0 ?r × ˇ H(r)+ ˇ E(r) bracketrightbig = 0, (4.167) lim r→∞ r bracketleftbig ?r × ˇ E(r)?η 0 ˇ H(r) bracketrightbig = 0, (4.168) where η 0 = (μ 0 /epsilon1 0 ) 1/2 . Later we shall see how these expressions arise from the integral solutions to Maxwell’s equations. 4.10.2 Reciprocityrevisited 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 ?elds. In this section we shall apply the reciprocity concept to time-harmonic sources and ?elds, and investigate the properties a material must display to be reciprocal. Thegeneralformofthereciprocitytheorem. Asin § 2.9.3,weconsideraclosed surface S enclosing a volume V. Sources of an electromagnetic ?eld are located either inside or outside S. Material media may lie within S, and their properties are described intermsoftheconstitutiverelations. Toobtainthetime-harmonic(phasor)formofthe reciprocitytheoremweproceedasin § 2.9.3butbeginwiththephasorformsofMaxwell’s equations. We ?nd ?·( ˇ E a × ˇ H b ? ˇ E b × ˇ H a ) = j ˇω[ ˇ H a · ˇ B b ? ˇ H b · ˇ B a ] ? j ˇω[ ˇ E a · ˇ D b ? ˇ E b · ˇ D a ] + + [ ˇ E b · ˇ J a ? ˇ E a · ˇ J b ? ˇ H b · ˇ J ma + ˇ H a · ˇ J mb ], (4.169) where( ˇ E a , ˇ D a , ˇ B a , ˇ H a )arethe?eldsproducedbythephasorsources( ˇ J a , ˇ J ma )and( ˇ E b , ˇ D b , ˇ B b , ˇ H b ) are the ?elds produced by an independent set of sources ( ˇ J b , ˇ J mb ). As in § 2.9.3, we are interested in the case in which the ?rst two terms on the right- hand side of (4.169) are zero. To see the conditions under which this might occur, we substitute the constitutive equations for a bianisotropic medium ˇ D = ? ˉ ξ· ˇ H + ? ˉepsilon1· ˇ E, ˇ B = ? ˉμ· ˇ H + ? ˉ ζ · ˇ E, into (4.169), where each of the constitutive parameters is evaluated at ˇω. Setting the two terms to zero gives j ˇω bracketleftBig ˇ H a · parenleftBig ? ˉμ· ˇ H b + ? ˉ ζ · ˇ E b parenrightBig ? ˇ H b · parenleftBig ? ˉμ· ˇ H a + ? ˉ ζ · ˇ E a parenrightBigbracketrightBig ? ?j ˇω bracketleftBig ˇ E a · parenleftBig ˇ ˉ ξ· ˇ H b + ? ˉepsilon1· ˇ E b parenrightBig ? ˇ E b · parenleftBig ? ˉ ξ· ˇ H a + ? ˉepsilon1· ˇ E a parenrightBigbracketrightBig = 0, which holds if ˇ H a · ? ˉμ· ˇ H b ? ˇ H b · ? ˉμ· ˇ H a = 0, ˇ H a · ? ˉ ζ · ˇ E b + ˇ E b · ? ˉ ξ· ˇ H a = 0, ˇ E a · ? ˉ ξ· ˇ H b + ˇ H b · ? ˉ ζ · ˇ E a = 0, ˇ E a · ? ˉepsilon1· ˇ E b ? ˇ E b · ? ˉepsilon1· ˇ E a = 0. These in turn hold if ? ˉepsilon1 = ? ˉepsilon1 T , ? ˉμ = ? ˉμ T , ? ˉ ξ =? ? ˉ ζ T , ? ˉ ζ =? ? ˉ ξ T . (4.170) These are the conditions for areciprocalmedium. For example, an anisotropic dielectric is a reciprocal medium if its permittivity dyadic is symmetric. An isotropic medium described by scalar quantities μ and epsilon1 is certainly reciprocal. In contrast, lossless Gy- rotropicmediaarenonreciprocalsincetheconstitutiveparametersobey ? ˉepsilon1 = ? ˉepsilon1 ? or ? ˉμ = ? ˉμ ? rather than ? ˉepsilon1 = ? ˉepsilon1 T or ? ˉμ = ? ˉμ T . For a reciprocal medium (4.169) reduces to ?·( ˇ E a × ˇ H b ? ˇ E b × ˇ H a ) = bracketleftbig ˇ E b · ˇ J a ? ˇ E a · ˇ J b ? ˇ H b · ˇ J ma + ˇ H a · ˇ J mb bracketrightbig . (4.171) At points where the sources are zero, or are conduction currents described entirely by Ohm’s law ˇ J = σ ˇ E,wehave ?·( ˇ E a × ˇ H b ? ˇ E b × ˇ H a ) = 0, (4.172) knownasLorentz’slemma. Ifweintegrate(4.171)over V andusethedivergencetheorem we obtain contintegraldisplay S bracketleftbig ˇ E a × ˇ H b ? ˇ E b × ˇ H a bracketrightbig · dS = integraldisplay V bracketleftbig ˇ E b · ˇ J a ? ˇ E a · ˇ J b ? ˇ H b · ˇ J ma + ˇ H a · ˇ J mb bracketrightbig dV. (4.173) ThisisthegeneralformoftheLorentzreciprocitytheorem,andisvalidwhen V contains reciprocal media as de?ned in (4.170). Note that by an identical set of steps we ?nd that the frequency-domain ?elds obey an identical Lorentz lemma and reciprocity theorem. Theconditionforreciprocalsystems. The quantity 〈 ˇ f a , ˇg b 〉= integraldisplay V bracketleftbig ˇ E a · ˇ J b ? ˇ H a · ˇ J mb bracketrightbig dV iscalledthereaction betweenthesource?elds ˇg ofset b andthemediating?elds ˇ f ofan independentset a. Notethat ˇ E a · ˇ J b isnotquiteapowerdensity,sincethecurrentlacks a complex conjugate. Using this reaction concept, ?rst introduced by Rumsey [161], we can write (4.173) as 〈 ˇ f b , ˇg a 〉?〈 ˇ f a , ˇg b 〉= contintegraldisplay S bracketleftbig ˇ E a × ˇ H b ? ˇ E b × ˇ H a bracketrightbig · dS. (4.174) We see that if there are no sources within S then contintegraldisplay S bracketleftbig ˇ E a × ˇ H b ? ˇ E b × ˇ H a bracketrightbig · dS = 0. (4.175) Whenever (4.175) holds we say that the “system” within S is reciprocal. Thus, for instance, a region of empty space is a reciprocal system. Asystemneednotbesource-freeinorderfor(4.175)tohold. Supposetherelationship between ˇ E and ˇ H on S is given by theimpedanceboundarycondition ˇ E t =?Z(?n × ˇ H), (4.176) where ˇ E t isthecomponentof ˇ E tangentialto S sothat ?n×E = ?n×E t ,andthecomplex wallimpedance Z may depend on position. By (4.176) we can write ( ˇ E a × ˇ H b ? ˇ E b × ˇ H a )· ?n = ˇ H b ·(?n × ˇ E a )? ˇ H a ·(?n × ˇ E b ) =?Z ˇ H b · [?n ×(?n × ˇ H a )] + Z ˇ H a · [?n ×(?n × ˇ H b )]. Since ?n ×(?n × ˇ H) = ?n(?n · ˇ H)? ˇ H, theright-handsidevanishes. Hence(4.175)stillholds even though there are sources within S. The reaction theorem. When sources lie within the surface S, and the ?elds on S obey (4.176), we obtain an important corollary of the Lorentz reciprocity theorem. We have from (4.174) the additional result 〈 ˇ f a , ˇg b 〉?〈 ˇ f b , ˇg a 〉=0. Hence a reciprocal system has 〈 ˇ f a , ˇg b 〉=〈 ˇ f b , ˇg a 〉 (4.177) (which holds even if there are no sources within S, since then the reactions would be identicallyzero). Thisconditionforreciprocityissometimescalledthereactiontheorem and has an important physical meaning which we shall explore below in the form of the Rayleigh–Carson reciprocity theorem. Note that in obtaining this relation we must assume 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 involveboth a reciprocal medium and a boundary over which (4.176) holds. Itisimportanttonotethattheimpedanceboundarycondition(4.176)iswidelyappli- cable. If Z → 0,thentheboundaryconditionisthatforaPEC: ?n× ˇ E = 0.IfZ →∞,a PMCisdescribed: ?n× ˇ H = 0. Suppose S representsasphereofin?niteradius. Weknow from(4.168)thatifthesourcesandmaterialmediawithin S arespatially?nite,the?elds far removed from these sources are described by the Sommerfeld radiation condition ?r × ˇ E = η 0 ˇ H where ?r is the radial unit vector of spherical coordinates. This condition is of the type (4.176) since ?r = ?n on S, hence the unbounded region that results from S receding to in?nity is also reciprocal. Summary of reciprocity for reciprocal systems. We can summarize reciprocity asfollows. Unboundedspacecontainingsourcesandmaterialsof?nitesizeisareciprocal systemifthemediaarereciprocal;aboundedregionofspaceisareciprocalsystemonly if the materials within are reciprocal and the boundary ?elds obey (4.176), or if the region is source-free. In each of these cases contintegraldisplay S bracketleftbig ˇ E a × ˇ H b ? ˇ E b × ˇ H a bracketrightbig · dS = 0 (4.178) and 〈 ˇ f a , ˇg b 〉?〈 ˇ f b , ˇg a 〉=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, each oscillating with frequency ˇω and located within an empty box consisting of PEC walls. These dipoles can be described in terms of volume current density as ˇ J a (r) = ˇ I a δ(r ? r prime a ), ˇ J b (r) = ˇ I b δ(r ? r prime b ). Sincethe?eldsonthesurfaceobey(4.176)(speci?cally, ?n× ˇ E = 0),andsincethemedium withintheboxisemptyspace(areciprocalmedium),the?eldsproducedbythesources must obey (4.179). We have integraldisplay V ˇ E b (r)· bracketleftbig ˇ I a δ(r ? r prime a ) bracketrightbig dV = integraldisplay V ˇ E a (r)· bracketleftbig ˇ I b δ(r ? r prime b ) bracketrightbig dV, hence ˇ I a · ˇ E b (r prime a ) = ˇ I b · ˇ E a (r prime b ). (4.180) This is theRayleigh–Carsonreciprocitytheorem. It also holds for two Hertzian dipoles locatedinunboundedfreespace,becauseinthatcasetheSommerfeldradiationcondition satis?es (4.176). As an important application of this principle, consider a closed PEC body located in freespace. Reciprocityholdsintheregionexternaltothebodysincewehave ?n× ˇ E = 0 attheboundaryoftheperfectconductorandtheSommerfeldradiationconditiononthe boundary at in?nity. Now let us place dipole a somewhere external to the body, and dipole b adjacent and tangential to the perfectly conducting body. We regard dipole a asthesourceofanelectromagnetic?eldanddipole b as“sampling”that?eld. Sincethe tangential electric ?eld is zero at the surface of the conductor, the reaction between the two dipoles is zero. Now let us switch the roles of the dipoles so that b is regarded as the source and a is regarded as the sampler. By reciprocity the reaction is again zero and thus there is no ?eld produced by b at the position of a. Now the position and orientation of a are arbitrary, so we conclude that an impressed electric source current placedtangentiallytoaperfectlyconductingbodyproducesno?eldexternaltothebody. ThisresultisusedinChapter6todevelopa?eldequivalenceprincipleusefulinthestudy of antennas and scattering. 4.10.3 Duality A duality principle analogous to that found for time-domain ?elds in § 2.9.2 may be established for frequency-domain and time-harmonic ?elds. 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 ? J m . The ?elds ( ? E 1 , ? D 1 , ? B 1 , ? H 1 ) within the region (which may also contain arbitrary media) are described by ?× ? E 1 =? ? J m ? jω ? B 1 , (4.181) ?× ? H 1 = ? J + jω ? D 1 , (4.182) ?· ? D 1 = ?ρ, (4.183) ?· ? B 1 = ?ρ m . (4.184) Suppose we have been given a mathematical description of the sources ( ? J, ? J m ) and have solved for the ?eld vectors ( ? E 1 , ? D 1 , ? B 1 , ? H 1 ). Of course, we must also have been supplied with a set of boundary values and constitutive relations in order to make the solution unique. We note that if we replace the formula for ? J with the formula for ? J m in (4.182) (and ?ρ with ?ρ m in (4.183)) and also replace ? J m with ? ? J in (4.181) (and ?ρ m with ??ρ in (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 ?× ? E 2 = ? J ? jω ? B 2 , (4.185) ?× ? H 2 = ? J m + jω ? D 2 . (4.186) If we can resolve the question of how the constitutive parameters must be altered to re?ect these replacements, then we can conclude by comparing (4.185) with (4.182) and (4.186) with (4.181) that ? E 2 = ? H 1 , ? B 2 =? ? D 1 , ? D 2 = ? B 1 , ? H 2 =? ? E 1 . The discussion regarding units in § 2.9.2 carries over to the present case. Multiplying Ampere’s law by η 0 = (μ 0 /epsilon1 0 ) 1/2 , we have ?× ? E =? ? J m ? jω ? B, ?×(η 0 ? H) = (η 0 ? J)+ jω(η 0 ? D). Thus if the original problem has solution ( ? E 1 ,η 0 ? D 1 , ? B 1 ,η 0 ? H 1 ), then the dual problem with ? J replaced by ? J m /η 0 and ? J m replaced by ?η 0 ? J has solution ? E 2 = η 0 ? H 1 , (4.187) ? B 2 =?η 0 ? D 1 , (4.188) η 0 ? D 2 = ? B 1 , (4.189) η 0 ? H 2 =? ? E 1 . (4.190) Aswithdualityinthetimedomain,theconstitutiveparametersforthedualproblem mustbealteredfromthoseoftheoriginalproblem. Forlinearanisotropicmediawehave from (4.13) and (4.14) the constitutive relationships ? D 1 = ? ˉepsilon1 1 · ? E 1 , (4.191) ? B 1 = ? ˉμ 1 · ? H 1 , (4.192) for the original problem, and ? D 2 = ? ˉepsilon1 2 · ? E 2 , (4.193) ? B 2 = ? ˉμ 2 · ? H 2 , (4.194) for the dual problem. Substitution of (4.187)–(4.190) into (4.191) and (4.192) gives ? D 2 = parenleftbigg ? ˉμ 1 η 2 0 parenrightbigg · ? E 2 , (4.195) ? B 2 = parenleftbig η 2 0 ? ˉepsilon1 1 parenrightbig · ? H 2 . (4.196) Comparing (4.195) with (4.193) and (4.196) with (4.194), we conclude that ? ˉμ 2 = η 2 0 ? ˉepsilon1 1 , ? ˉepsilon1 2 = ? ˉμ 1 /η 2 0 . (4.197) For a linear, isotropic medium speci?ed by ?epsilon1 and ?μ, the dual problem is obtained by replacing ?epsilon1 r with ?μ r and ?μ r with ?epsilon1 r . The solution to the dual problem is then ? E 2 = η 0 ? H 1 ,η 0 ? H 2 =? ? E 1 , 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 magnetic properties numerically equal to the electric properties of the medium in the original problem. AlternativelywemaydivideAmpere’slawby η = ( ?μ/?epsilon1) 1/2 insteadof η 0 . Then the dual problem has ? J replaced by ? J m /η, and ? J m replaced by ?η ? J, and the solution is ? E 2 = η ? H 1 ,η ? H 2 =? ? E 1 . (4.198) There is no need to swap ?epsilon1 r and ?μ r since 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 = ? J i + ?σ ? E. With this the curl equations become ?×η ? H = η ? J i + jωη?epsilon1 c ? E, ?× ? E =? ? J m ? jω ?μ ? H. Thesolutiontothedualproblemisagaingivenby(4.198),exceptthatnowη = ( ?μ/?epsilon1 c ) 1/2 . As we did near the end of § 2.9.2, we can consider duality in a source-free region. We let S enclose a source-free region of space and, for simplicity, assume that the medium within S is linear, isotropic, and homogeneous. The ?elds within S are described by ?× ? E 1 =?jω ?μ ? H 1 , ?×η ? H 1 = jω?epsilon1η ? E 1 , ?·?epsilon1 ? E 1 = 0, ?·?μ ? H 1 = 0. Thesymmetryoftheequationsissuchthatthemathematicalformofthesolutionfor ? E is the same as that for η ? H. Since the ?elds ? E 2 = η ? H 1 , ? H 2 =? ? E 1 /η, also satisfy Maxwell’s equations, the dual problem merely involves replacing ? E by η ? H and ? H by ? ? E/η. 4.11 The wave nature of the time-harmonic EM ?eld Time-harmonicelectromagneticwaveshavebeenstudiedingreatdetail. Narrowband waves are widely used for signal transmission, heating, power transfer, and radar. They share 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 and group velocity, and polarization. 4.11.1 Thefrequency-domainwaveequation Webeginbyderivingthefrequency-domainwaveequationfordispersivebianisotropic materials. A solution to this equation may be viewed as the transform of a general time-dependent ?eld. If one speci?c frequency is considered the time-harmonic solution is 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. We canovercomethisrequirement,andthusdealwithdispersivematerials,byusingaFourier transform approach. We solve a frequency-domain wave equation that includes the fre- quencydependenceoftheconstitutiveparameters,andthenuseaninversetransformto return to the time domain. Thederivationoftheequationparallelsthatof § 2.10.2. Wesubstitutethefrequency- domain constitutive relationships ? D = ? ˉepsilon1· ? E + ? ˉ ξ· ? H, ? B = ? ˉ ζ · ? E + ? ˉμ· ? H, into Maxwell’s curl equations (4.7) and (4.8) to get the coupled di?erential equations ?× ? E =?jω[ ? ˉ ζ · ? E + ? ˉμ· ? H] ? ? J m , ?× ? H = jω[ ? ˉepsilon1· ? E + ? ˉ ξ· ? H] + ? J, for ? E and ? H. Here we have included magnetic sources ? J m in Faraday’s law. Using the dyadic operator ˉ ? de?ned in (2.308) we can write these equations as parenleftBig ˉ ?+jω ? ˉ ζ parenrightBig · ? E =?jω ? ˉμ· ? H ? ? J m , (4.199) parenleftBig ˉ ??jω ? ˉ ξ parenrightBig · ? H = jω ? ˉepsilon1· ? E + ? J. (4.200) We can obtain separate equations for ? E and ? H by de?ning the inverse dyadics ? ˉepsilon1· ? ˉepsilon1 ?1 = ˉ I, ? ˉμ· ? ˉμ ?1 = ˉ I. Using ? ˉμ ?1 we can write (4.199) as ?jω ? H = ? ˉμ ?1 · parenleftBig ˉ ?+jω ? ˉ ζ parenrightBig · ? E + ? ˉμ ?1 · ? J m . Substituting this into (4.200) we get bracketleftBigparenleftBig ˉ ??jω ? ˉ ξ parenrightBig · ? ˉμ ?1 · parenleftBig ˉ ?+jω ? ˉ ζ parenrightBig ?ω 2 ? ˉepsilon1 bracketrightBig · ? E =? parenleftBig ˉ ??jω ? ˉ ξ parenrightBig · ? ˉμ ?1 · ? J m ? jω ? J. (4.201) Thisisthegeneralfrequency-domainwaveequationfor ? E. Using ? ˉepsilon1 ?1 wecanwrite(4.200) as jω ? E = ? ˉepsilon1 ?1 · parenleftBig ˉ ??jω ? ˉ ξ parenrightBig · ? H ? ? ˉepsilon1 ?1 · ? J. Substituting this into (4.199) we get bracketleftBigparenleftBig ˉ ?+jω ? ˉ ζ parenrightBig · ? ˉepsilon1 ?1 · parenleftBig ˉ ??jω ? ˉ ξ parenrightBig ?ω 2 ? ˉμ bracketrightBig · ? H = parenleftBig ˉ ?+jω ? ˉ ζ parenrightBig · ? ˉepsilon1 ?1 · ? J ? jω ? J m . (4.202) This is the general frequency-domain wave equation for ? H. Waveequationforahomogeneous,lossy,isotropicmedium. Wemayspecialize (4.201) and (4.202) to the case of a homogeneous, lossy, isotropic medium by setting ? ˉ ζ = ? ˉ ξ = 0, ? ˉμ = ?μ ˉ I, ? ˉepsilon1 = ?epsilon1 ˉ I, and ? J = ? J i + ? J c : ?×(?× ? E)?ω 2 ?μ?epsilon1 ? E =??× ? J m ? jω ?μ( ? J i + ? J c ), (4.203) ?×(?× ? H)?ω 2 ?μ?epsilon1 ? H =?×( ? J i + ? J c )? jω?epsilon1 ? J m . (4.204) Using (B.47) and using Ohm’s law ? J c = ?σ ? E to describe the secondary current, we get from (4.203) ?(?· ? E)?? 2 ? E ?ω 2 ?μ?epsilon1 ? E =??× ? J m ? jω ?μ ? J i ? jω ?μ?σ ? E which, using ?· ? E = ?ρ/?epsilon1, can be simpli?ed to (? 2 + k 2 ) ? E =?× ? J m + jω ?μ ? J i + 1 ?epsilon1 ? ?ρ. (4.205) This is thevectorHelmholtzequation for ? E. Here k is thecomplexwavenumber de?ned through k 2 = ω 2 ?μ?epsilon1 ? jω ?μ?σ = ω 2 ?μ bracketleftbigg ?epsilon1 + ?σ jω bracketrightbigg = ω 2 ?μ?epsilon1 c (4.206) where ?epsilon1 c is the complex permittivity (4.26). By (4.204) we have ?(?· ? H)?? 2 ? H ?ω 2 ?μ?epsilon1 ? H =?× ? J i +?× ? J c ? jω?epsilon1 ? J m . Using ?× ? J c =?×(?σ ? E) = ?σ?× ? E = ?σ(?jω ? B ? ? J m ) and ?· ? H = ?ρ m / ?μ we then get (? 2 + k 2 ) ? H =??× ? J i + jω?epsilon1 c ? J m + 1 ?μ ? ?ρ m , (4.207) which is the vector Helmholtz equation for ? H. 4.11.2 Fieldrelationshipsandthewaveequationfortwo-dimensional ?elds Many important canonical problems are two-dimensional in nature, with the sources and ?elds invariant along one direction. Two-dimensional ?elds have a simple structure compared to three-dimensional ?elds, and this structure often allows a decomposition into even simpler ?eld structures. Consider a homogeneous region of space characterized by the permittivity ?epsilon1, perme- ability ?μ,andconductivity ?σ. Weassumethatallsourcesand?eldsare z-invariant,and wish to ?nd the relationship between the various components of the frequency-domain ?eldsinasource-freeregion. Itisusefultode?nethetransversevectorcomponentofan arbitrary vector A as the component of A perpendicular to the axis of invariance: A t = A ? ?z(?z · A). For the position vector r, this component is the transverse position vector r t = ρ.For instance we have ρ = ?xx + ?yy, ρ = ?ρρ, in the rectangular and cylindrical coordinate systems, respectively. Becausetheregionissource-free,the?elds ? E and ? H obeythehomogeneousHelmholtz equations (? 2 + k 2 ) braceleftbigg ? E ? H bracerightbigg = 0. Writing the ?elds in terms of rectangular components, we ?nd that each component must obey a homogeneous scalar Helmholtz equation. In particular, we have for the axialcomponents ? E z and ? H z , (? 2 + k 2 ) braceleftbigg ? E z ? H z bracerightbigg = 0. But since the ?elds are independent of z we may also write (? 2 t + k 2 ) braceleftbigg ? E z ? H z bracerightbigg = 0 (4.208) where ? 2 t is the transverse Laplacian operator ? 2 t =? 2 ? ?z ? 2 ?z 2 . (4.209) In rectangular coordinates we have ? 2 t = ? 2 ?x 2 + ? 2 ?y 2 , while in circular cylindrical coordinates ? 2 t = ? 2 ?ρ 2 + 1 ρ ? ?ρ + 1 ρ 2 ? 2 ?φ 2 . (4.210) Withourconditionon z-independencewecanrelatethetransverse?elds ? E t and ? H t to ? E z and ? H z . By Faraday’s law we have ?× ? E(ρ,ω)=?jω ?μ ? H(ρ,ω) and thus ? H t =? 1 jω ?μ bracketleftbig ?× ? E bracketrightbig t . The transverse portion of the curl is merely bracketleftbig ?× ? E bracketrightbig t = ?x bracketleftbigg ? ? E z ?y ? ? ? E y ?z bracketrightbigg + ?y bracketleftbigg ? ? E x ?z ? ? ? E z ?x bracketrightbigg =??z × bracketleftbigg ?x ? ? E z ?x + ?y ? ? E z ?y bracketrightbigg since the derivatives with respect to z vanish. The term in brackets is the transverse gradient of ? E z , where the transverse gradient operator is ? t =???z ? ?z . In circular cylindrical coordinates this operator becomes ? t = ?ρ ? ?ρ + ? φ 1 ρ ? ?φ . (4.211) Thus we have ? H t (ρ,ω)= 1 jω ?μ ?z ×? t ? E z (ρ,ω). Similarly, the source-free Ampere’s law yields ? E t (ρ,ω)=? 1 jω?epsilon1 c ?z ×? t ? H z (ρ,ω). These results suggest that we can solve a two-dimensional problem by superposition. We ?rst consider the case where ? E z negationslash= 0 and ? H z = 0, calledelectricpolarization. This case is also calledTM ortransversemagnetic polarization because the magnetic ?eld is transverse to the z-direction (TM z ). We have (? 2 t + k 2 ) ? E z = 0, ? H t (ρ,ω)= 1 jω ?μ ?z ×? t ? E z (ρ,ω). (4.212) Once we have solved the Helmholtz equation for ? E z , the remaining ?eld components follow by simple di?erentiation. We next consider the case where ? H z negationslash= 0 and ? E z = 0. Thisisthecaseofmagneticpolarization,alsocalledTEortransverseelectricpolarization (TE z ). In this case (? 2 t + k 2 ) ? H z = 0, ? E t (ρ,ω)=? 1 jω?epsilon1 c ?z ×? t ? H z (ρ,ω). (4.213) A problem involving both ? E z and ? H z is solved by adding the results for the individual TE z and TM z cases. Note that we can obtain the expression for the TE ?elds from the expression for the TM ?elds, and vice versa, using duality. For instance, knowing that the TM ?elds obey (4.212) we may replace ? H t with ? E t /η and ? E z with ?η ? H z to obtain ? E t (ρ,ω) η = 1 jω ?μ ?z ×? t [?η ? H z (ρ,ω)], which reproduces (4.213). 4.11.3 Planewavesinahomogeneous,isotropic,lossymaterial The plane-wave ?eld. In later sections we will solve the frequency-domain wave equation with an arbitrary source distribution. At this point we are more interested in the general behavior of EM waves in the frequency domain, so we seek simple solutions to the homogeneous equation (? 2 + k 2 ) ? E(r,ω)= 0 (4.214) that governs the ?elds in source-free regions of space. Here [k(ω)] 2 = ω 2 ?μ(ω)?epsilon1 c (ω). Many properties of plane waves are best understood by considering the behavior of a monochromatic ?eld oscillating at a single frequency ˇω. In these cases we merely make the replacements ω → ˇω, ? E(r,ω)→ ˇ E(r), and apply the rules developed in § 4.7 for the manipulation of phasor ?elds. Forour?rstsolutionswechoosethosethatdemonstraterectangularsymmetry.Plane waves have planar spatial phase loci. That is, the spatial surfaces over which the phase ofthecomplexfrequency-domain?eldisconstantareplanes. Solutionsofthistypemay be obtained using separation of variables in rectangular coordinates. Writing ? E(r,ω)= ?x ? E x (r,ω)+ ?y ? E y (r,ω)+ ?z ? E z (r,ω) we ?nd that (4.214) reduces to three scalar equations of the form (? 2 + k 2 ) ? ψ(r,ω)= 0 where ? ψ is representative of ? E x , ? E y , and ? E z . 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 Z are chosen from the list (A.102). Since the exponentials describe propagating wave functions, we choose ? ψ(r,ω)= A(ω)e ±jk x (ω)x e ±jk y (ω)y e ±jk z (ω)z where A is theamplitudespectrum of the plane wave and k 2 x + k 2 y + k 2 z = k 2 . Using this solution to represent each component of ? E, we have a propagating-wave solution to the homogeneous vector Helmholtz equation: ? E(r,ω)= ? E 0 (ω)e ±jk x (ω)x e ±jk y (ω)y e ±jk z (ω)z , (4.215) where E 0 (ω) is the vector amplitude spectrum. If we de?ne thewavevector k(ω) = ?xk x (ω)+ ?yk y (ω)+ ?zk z (ω), then we can write (4.215) as ? E(r,ω)= ? E 0 (ω)e ?jk(ω)·r . (4.216) Note that we choose the negative sign in the exponential function and allow the vector components of k to be either positive or negative as required by the physical nature of aspeci?cproblem. Alsonotethatthemagnitudeofthewavevectoristhewavenumber: |k|=k. Wemayalwayswritethewavevectorasasumofrealandimaginaryvectorcomponents k = k prime + jk primeprime (4.217) which must obey k · k = k 2 = k prime2 ? k primeprime2 + 2 jk prime · k primeprime . (4.218) Whentherealandimaginarycomponentsarecollinear,(4.216)describesauniformplane wave with k = ? k(k prime + jk primeprime ). When k prime and k primeprime have di?erent directions, (4.216) describes a nonuniformplanewave. Weshall?ndin §4.13thatanyfrequency-domainelectromagnetic?eldinfreespace may berepresentedasacontinuoussuperpositionofelementalplane-wavecomponentsof the type(4.216),butthatbothuniformandnonuniformtermsarerequired. The TEM nature of a uniform plane wave. Given the plane-wave solution to the wave equation for the electric ?eld, it is straightforward to ?nd the magnetic ?eld. 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; see the 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 = 0 by examining Gauss’ law and employing (B.77): ?· ? E =?jk · ? Ee ?jk·r = ?ρ ?epsilon1 = 0. (4.221) Using this and k · k = k 2 = ω 2 ?μ?epsilon1 c , we obtain from (4.220) ? E =? k × ? H ω?epsilon1 c . (4.222) Now for a uniform plane wave k = ? kk, so we can also write (4.219) as ? H = ? k × ? E η = ? k × ? E 0 η e ?jk·r (4.223) and (4.222) as ? E =?η ? k × ? H. Here η = ω ?μ k = radicalbigg ?μ ?epsilon1 c is the complex intrinsic impedance of the medium. Equations(4.223)and(4.221)showthattheelectricandmagnetic?eldsandthewave vector are mutually orthogonal. The wave is said to be transverseelectromagnetic or TEM to the direction of propagation. Thephaseandattenuationconstantsofauniformplanewave. For a uniform plane wave we may write k = k prime ? k + jk primeprime ? k = k ? k = (β ? jα) ? k where k prime = β and k primeprime =?α. Here α iscalledtheattenuationconstant and β isthephase constant. Since k is de?ned through (4.206), we have k 2 = (β ? jα) 2 = β 2 ? 2 jαβ ?α 2 = ω 2 ?μ?epsilon1 c = ω 2 ( ?μ prime + j ?μ primeprime )(?epsilon1 cprime + j ?epsilon1 cprimeprime ). Equating real and imaginary parts we have β 2 ?α 2 = ω 2 [ ?μ prime ?epsilon1 cprime ? ?μ primeprime ?epsilon1 cprimeprime ], ?2αβ = ω 2 [ ?μ primeprime ?epsilon1 cprime + ?μ prime ?epsilon1 cprimeprime ]. We assume the material is passive so that ?μ primeprime ≤ 0, ?epsilon1 cprimeprime ≤ 0. Letting β 2 ?α 2 = ω 2 [ ?μ prime ?epsilon1 cprime ? ?μ primeprime ?epsilon1 cprimeprime ] = A, 2αβ = ω 2 [|?μ primeprime |?epsilon1 cprime + ?μ prime |?epsilon1 cprimeprime |] = B, we may solve simultaneously to ?nd that β 2 = 1 2 bracketleftBig A + radicalbig A 2 + B 2 bracketrightBig ,α 2 = 1 2 bracketleftBig ?A + radicalbig A 2 + B 2 bracketrightBig . Since A 2 + B 2 = ω 4 (?epsilon1 cprime2 + ?epsilon1 cprimeprime2 )( ?μ prime2 + ?μ primeprime2 ),wehave β = ω radicalbig ?μ prime ?epsilon1 cprime radicaltp radicalvertex radicalvertex radicalbt 1 2 bracketleftBigg radicalBigg parenleftbigg 1 + ?epsilon1 cprimeprime2 ?epsilon1 cprime2 parenrightbiggparenleftbigg 1 + ?μ primeprime2 ?μ prime2 parenrightbigg + parenleftbigg 1 ? ?μ primeprime ?μ prime ?epsilon1 cprimeprime ?epsilon1 cprime parenrightbigg bracketrightBigg , (4.224) α = ω radicalbig ?μ prime ?epsilon1 cprime radicaltp radicalvertex radicalvertex radicalbt 1 2 bracketleftBigg radicalBigg parenleftbigg 1 + ?epsilon1 cprimeprime2 ?epsilon1 cprime2 parenrightbiggparenleftbigg 1 + ?μ primeprime2 ?μ prime2 parenrightbigg ? parenleftbigg 1 ? ?μ primeprime ?μ prime ?epsilon1 cprimeprime ?epsilon1 cprime parenrightbigg bracketrightBigg , (4.225) where ?epsilon1 c and ?μ are functions of ω.If?epsilon1(ω) = epsilon1, ?μ(ω) = μ, and ?σ(ω)= σ are real and frequency independent, then α = ω √ μepsilon1 radicaltp radicalvertex radicalvertex radicalbt 1 2 bracketleftBigg radicalbigg 1 + parenleftBig σ ωepsilon1 parenrightBig 2 ? 1 bracketrightBigg , (4.226) β = ω √ μepsilon1 radicaltp radicalvertex radicalvertex radicalbt 1 2 bracketleftBigg radicalbigg 1 + parenleftBig σ ωepsilon1 parenrightBig 2 + 1 bracketrightBigg . (4.227) These values of α and β are valid for ω>0. For negative frequencies we must be more careful in evaluating the square root in k = ω(?μ?epsilon1 c ) 1/2 . Writing ?μ(ω) = ?μ prime (ω)+ j ?μ primeprime (ω) =|?μ(ω)|e jξ μ (ω) , ?epsilon1 c (ω) = ?epsilon1 cprime (ω)+ j ?epsilon1 cprimeprime (ω) =|?epsilon1 c (ω)|e jξ epsilon1 (ω) , we have k(ω) = β(ω)? jα(ω)= ω radicalbig ?μ(ω)?epsilon1 c (ω) = ω radicalbig |?μ(ω)||?epsilon1 c (ω)|e j 1 2 [ξ μ (ω)+ξ epsilon1 (ω)] . Now for passive materials we must have, by (4.48), ?μ primeprime < 0 and ?epsilon1 cprimeprime < 0 for ω>0. Since we also have ?μ prime > 0 and ?epsilon1 cprime > 0 for ω>0, we ?nd that ?π/2 <ξ μ < 0 and ?π/2 <ξ epsilon1 < 0, and thus ?π/2 <(ξ μ +ξ epsilon1 )/2 < 0. Thus we must have β>0 and α>0 for ω>0.Forω<0 we have by (4.44) and (4.45) that ?μ primeprime > 0, ?epsilon1 cprimeprime > 0, ?μ prime > 0, and ?epsilon1 cprime > 0.Thusπ/2 >(ξ μ + ξ epsilon1 )/2 > 0, and so β<0 and α>0 for ω<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)>0 when ω>0. From this we ?nd a condition on ? E 0 in (4.216). Since by (4.47) we must have ? E(ω) = ? E ? (?ω), we see that the uniform plane-wave ?eld obeys ? E 0 (ω)e [?jβ(ω)?α(ω)] ? k·r = ? E ? 0 (?ω)e [+jβ(?ω)?α(?ω)] ? k·r or ? E 0 (ω) = ? E ? 0 (?ω), since β(?ω) =?β(ω)and α(?ω) = α(ω). Propagation of a uniform plane wave: the group and phase velocities. We havederivedtheplane-wave solutiontothewaveequationinthefrequencydomain,but can discover the wave nature of the solution only by examining its behavior in the time domain. Unfortunately,theexplicitformofthetime-domain?eldishighlydependenton the frequency behavior of the constitutive parameters. Even the simplest case in which epsilon1, μ,and σ arefrequencyindependentisquitecomplicated,aswediscoveredin § 2.10.6. To overcome this di?culty, it is helpful to examine the behavior of a narrowband (but non-monochromatic) signal in a lossy medium with arbitrary constitutive parameters. We will ?nd that the time-domain wave ?eld propagates as a disturbance through the surrounding medium with a velocity determined by the constitutive parameters of the medium. The temporal wave shape does not change as the wave propagates, but the amplitude of the wave attenuates at a rate dependent on the constitutive parameters. Forclarityofpresentationweshallassumealinearlypolarizedplanewave(§ ??)with ? E(r,ω)= ?e ? E 0 (ω)e ?jk(ω)·r . (4.229) Here ? E 0 (ω) is the spectrum of the temporal dependence of the wave. For the temporal dependence we choose the narrowband signal E 0 (t) = E 0 f (t)cos(ω 0 t) where f (t) has a narrowband spectrum centered about ω = 0 (and is therefore called a basebandsignal). Anappropriatechoicefor f (t) istheGaussianfunctionusedin(4.52): f (t) = e ?a 2 t 2 ? ? F(ω) = radicalbigg π a 2 e ? ω 2 4a 2 , producing E 0 (t) = E 0 e ?a 2 t 2 cos(ω 0 t). (4.230) We think of f (t) asmodulating the single-frequency cosinecarrierwave, thus providing theenvelope. Byusingalargevalueof a weobtainanarrowbandsignalwhosespectrum is centered about ±ω 0 . Later we shall let a → 0, thereby driving the width of f (t) to in?nity and producing a monochromatic waveform. By (1) we have ? E 0 (ω) = E 0 1 2 bracketleftbig ? F(ω ?ω 0 )+ ? F(ω+ω 0 ) bracketrightbig where f (t) ? ? F(ω).AplotofthisspectrumisshowninFigure4.2.Weseethat the narrowband signal is centered at ω =±ω 0 . Substituting into (4.229) and using k = (β ? jα) ? k for a uniform plane wave, we have the frequency-domain ?eld ? E(r,ω)= ?eE 0 1 2 bracketleftBig ? F(ω ?ω 0 )e ?j[β(ω)?jα(ω)] ? k·r + ? F(ω +ω 0 )e ?j[β(ω)?jα(ω)] ? k·r bracketrightBig . (4.231) The ?eld at any time t and position r can now be found by inversion: ?eE(r,t) = 1 2π integraldisplay ∞ ?∞ ?eE 0 1 2 bracketleftBig ? F(ω ?ω 0 )e ?j[β(ω)?jα(ω)] ? k·r + + ? F(ω +ω 0 )e ?j[β(ω)?jα(ω)] ? k·r bracketrightBig e jωt dω. (4.232) Weassumethat β(ω)and α(ω) varyslowlywithinthebandoccupiedby ? E 0 (ω). With this assumption we can expand β and α near ω = ω 0 as β(ω)= β(ω 0 )+β prime (ω 0 )(ω ?ω 0 )+ 1 2 β primeprime (ω 0 )(ω ?ω 0 ) 2 +···, α(ω) = α(ω 0 )+α prime (ω 0 )(ω ?ω 0 )+ 1 2 α primeprime (ω 0 )(ω ?ω 0 ) 2 +···, where β prime (ω) = dβ(ω)/dω, β primeprime (ω) = d 2 β(ω)/dω 2 , and so on. In a similar manner we can expand β and α near ω =?ω 0 : β(ω)= β(?ω 0 )+β prime (?ω 0 )(ω +ω 0 )+ 1 2 β primeprime (?ω 0 )(ω +ω 0 ) 2 +···, α(ω) = α(?ω 0 )+α prime (?ω 0 )(ω +ω 0 )+ 1 2 α primeprime (?ω 0 )(ω +ω 0 ) 2 +···. Since we are most interested in the propagation velocity, we need not approximate α with great accuracy, and thus use α(ω) ≈ α(±ω 0 ) within the narrow band. We must consider β to greater accuracy to uncover the propagating nature of the wave, and thus use β(ω)≈ β(ω 0 )+β prime (ω 0 )(ω ?ω 0 ) (4.233) near ω = ω 0 and β(ω)≈ β(?ω 0 )+β prime (?ω 0 )(ω +ω 0 ) (4.234) near ω =?ω 0 . Substituting these approximations into (4.232) we ?nd ?eE(r,t) = 1 2π integraldisplay ∞ ?∞ ?eE 0 1 2 bracketleftBig ? F(ω ?ω 0 )e ?j[β(ω 0 )+β prime (ω 0 )(ω?ω 0 )] ? k·r e ?[α(ω 0 )] ? k·r + + ? F(ω +ω 0 )e ?j[β(?ω 0 )+β prime (?ω 0 )(ω+ω 0 )] ? k·r e ?[α(?ω 0 )] ? k·r bracketrightBig e jωt dω. (4.235) By (4.228) we know that α is even in ω and β is odd in ω. Since the derivative of an odd function is an even function, we also know that β prime is even in ω. We can therefore write (4.235) as ?eE(r,t) = ?eE 0 e ?α(ω 0 ) ? k·r 1 2π integraldisplay ∞ ?∞ 1 2 bracketleftBig ? F(ω ?ω 0 )e ?jβ(ω 0 ) ? k·r e ?jβ prime (ω 0 )(ω?ω 0 ) ? k·r + + ? F(ω +ω 0 )e jβ(ω 0 ) ? k·r e ?jβ prime (ω 0 )(ω+ω 0 ) ? k·r bracketrightBig e jωt dω. Multiplying and dividing by e jω 0 t and rearranging, we have ?eE(r,t) = ?eE 0 e ?α(ω 0 ) ? k·r 1 2π integraldisplay ∞ ?∞ 1 2 bracketleftbig ? F(ω ?ω 0 )e jφ e j(ω?ω 0 )[t?τ] + + ? F(ω +ω 0 )e ?jφ e j(ω+ω 0 )[t?τ] bracketrightbig dω where φ = ω 0 t ?β(ω 0 ) ? k · r,τ= β prime (ω 0 ) ? k · r. Setting u = ω ?ω 0 in the ?rst term and u = ω +ω 0 in the second term we have ?eE(r,t) = ?eE 0 e ?α(ω 0 ) ? k·r cosφ 1 2π integraldisplay ∞ ?∞ ? F(u)e ju(t?τ) du. Finally, the time-shifting theorem (A.3) gives us the time-domain wave ?eld ?eE(r,t) = ?eE 0 e ?α(ω 0 ) ? k·r cos parenleftbig ω 0 bracketleftbig t ? ? k · r/v p (ω 0 ) bracketrightbigparenrightbig f parenleftbig t ? ? k · r/v g (ω 0 ) parenrightbig (4.236) where v g (ω) = dω/dβ = [dβ/dω] ?1 (4.237) is called thegroupvelocity and v p (ω) = ω/β is called thephasevelocity. To interpret (4.236), we note that at any given time t the ?eld is constant over the surface described by ? k · r = C (4.238) where C issomeconstant.Thissurfaceisaplane,asshowninFigure4.10,withits normal along ? k. It is easy to verify that any point r on this plane satis?es (4.238). Let r 0 = r 0 ? k describe the point on the plane with position vector in the direction of ? k, and let d be a displacement vector from this point to any other point on the plane. Then ? k · r = ? k ·(r 0 + d) = r 0 ( ? k · ? k)+ ? k · d. But ? k · d = 0,so ? k · r = r 0 , (4.239) which is a ?xed distance, so (238) holds. Letusidentifytheplaneoverwhichtheenvelope f takesonacertainvalue,andfollow its motion as time progresses. The value of r 0 associated with this plane must increase with increasing time in such a way that the argument of f remains constant: t ? r 0 /v g (ω 0 ) = C. Figure 4.10: Surface of constant ? k · r. Di?erentiation gives dr 0 dt = v g = dω dβ . (4.240) So the envelope propagates along ? k at a rate given by the group velocity v g . Associated with this propagation is anattenuation described by the factor e ?α(ω 0 ) ? k·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 ω 0 bracketleftbig t ? ? k · r/v p (ω 0 ) bracketrightbig = C and di?erentiate to get dr 0 dt = v p = ω β . (4.241) This shows that surfaces of constant carrier phase propagate along ? k with velocity v p . Cautionmustbeexercisedininterpretingthetwovelocitiesv g andv p ;inparticular,we mustbecarefulnottoassociatethepropagationvelocitiesofenergyorinformationwith v p . Since envelope propagation represents the actual progression of the disturbance, v g hastherecognizablephysicalmeaningofenergyvelocity.KrausandFleisch[105]suggest that we think of a strolling caterpillar: the speed (v p ) of the undulations along the caterpillar’sback(representingthecarrierwave)maybemuchfasterthanthespeed(v g ) of the caterpillar’s body (representing the envelope of the disturbance). Infact, v g isthevelocityofenergypropagationevenforamonochromaticwave(§??). However,forpurelymonochromaticwaves v g cannotbeidenti?edfromthetime-domain ?eld, whereas v p can. This leads to some unfortunate misconceptions, especially when v p exceeds the speed of light. Since v p is not the velocity of propagation of a physical quantity,butisrathertherateofchangeofaphasereferencepoint,Einstein’spostulate of c as the limiting velocity is not violated. We can obtain interesting relationships between v p and v g by manipulating (4.237) and (4.241). For instance, if we compute dv p dω = d dω parenleftbigg ω β parenrightbigg = β ?ω dβ dω β 2 Figure 4.11: An ω–β diagram for a ?ctitious material. we ?nd that v p v g = 1 ?β dv p dω . (4.242) Henceinfrequencyrangeswherev p decreaseswithincreasingfrequency,wehavev g <v p . Theseareknownasregionsofnormaldispersion. Infrequencyrangeswhere v p increases with increasing frequency, we have v g >v p . These are known as regions ofanomalous dispersion. Asmentionedin § 4.6.3,theword“anomalous”doesnotimplythatthistype 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 √ μepsilon1, α(ω) = 0. In this case (4.233) becomes β(ω)= ω 0 √ μepsilon1 + √ μepsilon1(ω ?ω 0 ) = ω √ μepsilon1 and (4.236) becomes ?eE(r,t) = ?eE 0 cos parenleftbig ω 0 bracketleftbig t ? ? k · r/v p (ω 0 ) bracketrightbigparenrightbig f parenleftbig t ? ? k · r/v g (ω 0 ) parenrightbig . Since the linear approximation to the phase constant β is in this case exact, the wave packet truly propagates without distortion, with a group velocity identical to the phase velocity: v g = bracketleftbigg d dω ω √ μepsilon1 bracketrightbigg ?1 = 1 √ μepsilon1 = ω β = v p . Examples of wave propagation in various media; the ω–β diagram. A plot of ω versus β(ω) can be useful for displaying the dispersive properties of a material. Figure4.11showssuchanω–βplot,ordispersiondiagram,fora?ctitiousmaterial.The 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000 β (r/m) 0 20 40 60 80 100 120 140 160 180 200 ω /2 ≠ (GHz) Light Line: ε = ε 0 ε s Light Line: ε = ε 0 ε i 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 shows many of the di?erent characteristics of electromagnetic waves (although not necessarily of plane waves). For instance, there may be a minimum frequency ω c called thecuto? frequency atwhich β = 0 andbelowwhichthewavecannotpropagate. Thisbehavioris characteristicofaplanewave propagatinginaplasma(asshownbelow)orofawave in ahollowpipewaveguide(§ 5.4.3). Overmostvaluesof β wehave v g <v p sothematerial demonstrates normal dispersion. However, over a small region we do have anomalous dispersion. 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. Suchbackward waves are encountered in certain guided-wave structures used in microwave oscillators. The ω–β plot also includes thelightline as a reference curve. For all points on this line v g = v p ; 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 e?ects of water. By the Debye formula (4.106) we have ?epsilon1(ω)= epsilon1 ∞ + epsilon1 s ?epsilon1 ∞ 1 + jωτ . Assuming epsilon1 ∞ = 5epsilon1 0 ,epsilon1 s = 78.3epsilon1 0 ,andτ= 9.6 × 10 ?12 s[49],weobtaintherelaxation spectrumshowninFigure4.5.Ifwealsoassumethatμ=μ 0 , we may compute β as a functionofωandconstructtheω–βplot.ThisisshowninFigure4.12.Sinceepsilon1 prime varies with frequency, we show both the light line for zero frequency found using epsilon1 s = 78.3epsilon1 0 , and the light line for in?nite frequency found using epsilon1 i = 5epsilon1 0 . We see that at low values of frequency the dispersion curve follows the low-frequency light line very closely, and thus v p ≈ v g ≈ c/ √ 78.3. As the frequency increases, the dispersion curve rises up and 9101 12 log 10 (f) 0.0 0.2 0.4 0.6 v/c v v g 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 v p and v g showninFigure4.13verifythatthevelocitiesstartoutat c/ √ 78.3 for low frequencies, and approach c/ √ 5 for high frequencies. Because v g >v p at 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 = 0 in (4.76) to ?nd k = ? ? ? ω c radicalBig 1 ? ω 2 p ω 2 ,ω>ω p , ?j ω c radicalBig ω 2 p ω 2 ? 1,ω<ω p . Thus, when ω>ω p we have ? E(r,ω)= ? E 0 (ω)e ?jβ(ω) ? k·r and so β = ω c radicalBigg 1 ? ω 2 p ω 2 ,α= 0. Inthiscaseaplanewave propagatesthroughtheplasmawithoutattenuation. However, when ω<ω p we have ? E(r,ω)= ? E 0 (ω)e ?α(ω) ? k·r with α = ω c radicalBigg ω 2 p ω 2 ? 1,β= 0, and a plane wave does not propagate, but only attenuates. Such a wave is called an evanescentwave. We say that for frequencies below ω p the wave iscuto?, and call ω p thecuto?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) 0 1 2 3 4 5 6 7 8 9 10 ω /2 ≠ (MHz) Light Line β α Figure 4.14: Dispersion plot for the ionosphere computed using N e = 2×10 11 m ?3 , ν = 0. Light line computed using epsilon1 = epsilon1 0 , μ = μ 0 . Consider, for instance, a plane wave propagating in the earth’s ionosphere. Both the electron densityand the collision frequencyare highlydependent on such factors as 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 1 layer of the ionosphere), as measured for a mid-latitude region, we ?nd that during the daythe electron densityis approximately N e = 2×10 11 m ?3 ,whilethecollisionfrequencyisonly ν = 100s ?1 [16]. Theattenuationis so small in this case that the ionosphere maybe considered essentiallylossless above the cuto? frequency(we will develop an approximate formula for the attenuation constant below).Figure4.14showstheω–βdiagramfortheionosphereassumingν= 0,along with the light line v p = c. We see that above the cuto? frequencyof f p = ω p /2π = 4.0 MHz the wave propagates and that v g < c while v p > c. Below the cuto? frequencythe wave does not propagate and the ?eld decays very rapidly because α is large. A formula for the phase velocityof a plane wave in a lossless plasma is easilyderived: v p = ω β = c radicalBig 1 ? ω 2 p ω 2 > c. Thus, our observation from the ω–β plot that v p > c is veri?ed. Similarly, we ?nd that v g = parenleftbigg dβ dω parenrightbigg ?1 = ? ? 1 c radicalBigg 1 ? ω 2 p ω 2 + 1 c ω 2 p /ω 2 radicalBig 1 ? ω 2 p ω 2 ? ? ?1 = c radicalBigg 1 ? ω 2 p ω 2 < c and our observation that v g < c is also veri?ed. Interestingly, we ?nd that in this case of an unmagnetized collisionless plasma v p v g = c 2 . Since v p >v g , this model of a plasma demonstrates normal dispersion at all frequencies above cuto?. For the case of a plasma with collisions we retain ν in (4.76) and ?nd that k = ω c radicalBigg bracketleftbigg 1 ? ω 2 p ω 2 +ν 2 bracketrightbigg ? jν ω 2 p ω(ω 2 +ν 2 ) . When ν negationslash= 0 a true cuto? e?ect is not present and the wave maypropagate at all frequencies. However, when ν lessmuch ω p the attenuation for propagating waves of frequency ω<ω p is quite severe, and for all practical purposes the wave is cut o?. For waves of frequency ω>ω p there is attenuation. Assuming that ν lessmuch ω p and that ν lessmuch ω,wemay approximate the square root with the ?rst two terms of a binomial expansion, and ?nd that to ?rst order β = ω c radicalBigg 1 ? ω 2 p ω 2 ,α= 1 2 ν c ω 2 p /ω 2 radicalBig 1 ? ω 2 p ω 2 . Hence the phase and group velocities above cuto? are essentiallythose of a lossless plasma, while the attenuation constant is directlyproportional to ν. 4.11.4 Monochromaticplanewavesinalossymedium Manyproperties of monochromatic plane waves are particularlysimple. In fact, cer- tain properties, such as wavelength, onlyhave meaning for monochromatic ?elds. And since monochromatic or nearlymonochromatic waves are employed extensivelyin radar, communications, andenergytransport, itisusefultosimplifytheresultsofthepreceding section for the special case in which the spectrum of the plane-wave signal consists of a single frequencycomponent. In addition, plane waves of more general time dependence can be viewed as superpositions of individual single-frequencycomponents (through the inverse Fourier transform), and thus we mayregard monochromatic waves as building blocks for more complicated plane waves. We can view the monochromatic ?eld 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 ?eE(r,t) = ?eE 0 e ?α(ω 0 )[ ? k·r] cos(ω 0 t ? jβ(ω 0 )[ ? k · r]). (4.243) It is convenient to represent monochromatic ?elds with frequency ω = ˇω in phasor form. The phasor form of (4.243) is ˇ E(r) = ?eE 0 e ?jβ( ? k·r) e ?α( ? k·r) (4.244) where β = β(ˇω) and α = α(ˇω). We can identifya surface of constant phase as a locus of points obeying ˇωt ?β( ? k · r) = C P (4.245) for some constant C P .Thissurfaceisaplane,asshowninFigure4.10,withitsnormal in the direction of ? k. It is easyto verifythat anypoint r on this plane satis?es (4.245). Let r 0 = r 0 ? k describe the point on the plane with position vector in the ? k direction, and let d be a displacement vector from this point to anyother point on the plane. Then ? k · r = ? k ·(r 0 + d) = r 0 ( ? k · ? k)+ ? k · d. But ? k · d = 0,so ? k · r = r 0 , (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( ? k · r) = C A where C A is some constant. As with the phase term, this requires that ? k · 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 plane waves have planar surfaces that are not parallel. The cosine term in (4.243) represents atravelingwave.Ast increases, the argument of thecosinefunctionremainsunchangedaslongas ? k·r increasescorrespondingly. Thusthe planarwavefrontspropagatealong ? k. Asthewavefrontprogresses, thewaveisattenuated because of the factor e ?α( ? k·r) . This accounts for energytransferred from the propagating wave to the surrounding medium via Joule heating. Phasevelocityofauniformplanewave. The propagation velocityof the progress- ing wavefront is found bydi?erentiating (4.245) to get ˇω ?β ? k · dr dt = 0. By(4.246) we have v p = dr 0 dt = ˇω β , (4.247) where the phase velocity v p represents 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 √ μepsilon1 , hencethephasevelocityinaconductingmediumcannotexceedthatinalosslessmedium with the same parameters μ and epsilon1. We cannot draw this conclusion for a medium with frequency-dependent ?μ and ?epsilon1 c , since by(4.224) the value of ˇω/β might be greater or less than 1/ √ ?μ prime ?epsilon1 cprime , depending on the ratios ?μ primeprime / ?μ prime and ?epsilon1 cprimeprime /?epsilon1 cprime . 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 ?eld amplitude maynot be the same on these two surfaces because of possible attenuation of the wave. Let r 1 and r 2 be points on adjacent wavefronts. We require β( ? k · r 1 ) = β( ? k · r 2 )? 2π or λ = ? k ·(r 2 ? r 1 ) = r 02 ? r 01 = 2π/β. We call λ the wavelength. Polarizationofauniformplanewave. Plane-wave polarization describes the tem- poral evolution of the vector direction of the electric ?eld, which depends on the manner in which the wave is generated. Completelypolarized waves are produced byantennas or other equipment; these have a deterministic polarization state which maybe described completelybythree parameters as discussed below. Randomlypolarized waves are emit- ted bysome natural sources. Partiallypolarized 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 simpler plane waves that propagate along the same direction but with di?erent phases and spa- tiallyorthogonal electric ?elds. Without loss of generalitywe maystudypropagation along the z-axis and choose the orthogonal ?eld directions to be along ?x and ?y.Sowe are interested in the behavior of a wave with electric ?eld ˇ E(r) = ?xE x0 e jφ x e ?jkz + ?yE y0 e jφ y e ?jkz . (4.248) The time evolution of the direction of E must be examined in the time domain where we have E(r,t) = Re braceleftbig ˇ Ee jωt bracerightbig = ?xE x0 cos(ωt ? kz+φ x )+ ?yE y0 cos(ωt ? kz+φ y ) and thus, bythe identity cos(x + y) ≡ cos x cos y ? sin x sin y, E x = E x0 [cos(ωt ? kz)cos(φ x )? sin(ωt ? kz)sin(φ x )], E y = E y0 bracketleftbig cos(ωt ? kz)cos(φ y )? sin(ωt ? kz)sin(φ y ) bracketrightbig . The tip of the vector E moves cyclically in the xy-plane with temporal period T = ω/2π. Its locus maybe found byeliminating the parameter t to obtain a relationship between E x0 and E y0 . Letting δ = φ y ?φ x we note that E x E x0 sinφ y ? E y E y0 sinφ x = cos(ωt ? kz)sinδ, E x E x0 cosφ y ? E y E y0 cosφ x = sin(ωt ? kz)sinδ; squaring these terms we ?nd that parenleftbigg E x E x0 parenrightbigg 2 + parenleftbigg E y E y0 parenrightbigg 2 ? 2 E x E x0 E y E y0 cosδ = sin 2 δ, whichistheequationfortheellipseshowninFigure4.15.By(4.223 )themagnetic?eld of the plane wave is ˇ H = ?z × ˇ E η , hence its tip also traces an ellipse in the xy-plane. The tip of the electric ?eld vector cycles around the polarization ellipse in the xy- plane once every T seconds. The sense of rotation is determined bythe sign of δ, and is described bythe terms clockwise/counterclockwise or right-hand/left-hand. There is some disagreement about how to do this. We shall adopt the IEEE de?nitions (IEEE Standard145-1983[189])andassociatewithδ<0 rotationintheright-handsense:if Figure 4.15: Polarization ellipse for a monochromatic plane wave. the right thumb points in the direction of wave propagation then the ?ngers curl in the direction of ?eld rotation for increasing time. This isright-handpolarization (RHP). We associate δ>0 with left-handpolarization (LHP). The polarization ellipse is contained within a rectangle of sides 2E x0 and 2E y0 , and has its major axis rotated from the x-axis bythe tilt angle ψ, 0 ≤ ψ ≤ π. The ratio of E y0 to E x0 determines an angle α, 0 ≤ α ≤ π/2: E y0 /E x0 = tanα. The shape of the ellipse is determined bythe three parameters E x0 , E y0 , 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 also inscribe the ellipse within a box measuring 2a by 2b, where a and b are the lengths of the semimajor and semiminor axes. Then b/a determines an angle χ, ?π/4 ≤ χ ≤ π/4, that is analogous to α: ±b/a = tanχ. Here the algebraic sign of χ is used to indicate the sense of polarization: χ>0 for LHP, χ<0 for RHP. The quantities a,b,ψ can 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 , E y0 ,δ), we ?nd that a 2 + b 2 = E 2 x0 + E 2 y0 , tan 2ψ = (tan 2α)cosδ = 2E x0 E y0 E 2 x0 ? E 2 y0 cosδ, sin 2χ = (sin 2α)sinδ = 2E x0 E y0 E 2 x0 + E 2 y0 sinδ. Alternatively, we can describe the polarization ellipse by the anglesψ and χ and one of the amplitudes E x0 or E y0 . Figure 4.16: Polarization states as a function of tilt angle ψ and 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 di?eramongtheparameters. In1852G.Stokesintroducedasystemofthreeindependent quantities with identical dimension that can be used to describe plane-wave polarization. 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 E 2 x0 + E 2 y0 bracketrightbig , (4.249) s 1 = 1 2η bracketleftbig E 2 x0 ? E 2 y0 bracketrightbig = s 0 cos(2χ)cos(2ψ), (4.250) s 2 = 1 η E x0 E y0 cosδ = s 0 cos(2χ)sin(2ψ), (4.251) s 3 = 1 η E x0 E y0 sinδ = s 0 sin(2χ). (4.252) Onlythree of these four parameters are independent since s 2 0 = s 2 1 + s 2 2 + s 2 3 . Often the Stokes parameters are designated (I, Q,U, V) rather than (s 0 ,s 1 ,s 2 ,s 3 ). Figure4.16summarizesvariouspolarizationstatesasafunctionoftheanglesψand χ. Two interesting special cases occur when χ = 0 and χ =±π/4. The case χ = 0 corresponds to b = 0 and thus δ = 0. In this case the electric vector traces out a straight line and we call the polarization linear. Here E = parenleftbig ?xE x0 + ?yE y0 parenrightbig cos(ωt ? kz+φ x ). When ψ = 0 we have E y0 = 0 and refer to this as horizontal linear polarization (HLP); when ψ = π/2 we have E x0 = 0 and verticallinearpolarization (VLP). The case χ =±π/4 corresponds to b = a and δ =±π/2.ThusE x0 = E y0 , and E traces out a circle regardless of the value of ψ.Ifχ =?π/4 we have right-hand rotation of E and thus refer to this case as right-hand circular polarization (RHCP). If χ = π/4 we have left-handcircularpolarization (LHCP). For these cases E = E x0 [?x cos(ωt ? kz)? ?y sin(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 χ result in the general cases of left-hand or right-hand ellipticalpolarization. The French mathematician H. Poincar′e realized that the Stokes parameters (s 1 ,s 2 ,s 3 ) describe a point on a sphere of radius s 0 , and that this Poincar′e sphere is useful for visualizing the various polarization states. Each state corresponds uniquelyto one point on the sphere, and by(4.250)–(4.252) the angles 2χ and 2ψ are the spherical angular coordinatesofthepointasshowninFigure4.17.Wemaytherefor emapthepolarization statesshowninFigure4.16directlyo ntothesphere:left-andright-handpolarizations appear in the upper and lower hemispheres, respectively; circular polarization appears at the poles (2χ =±π/2); linear polarization appears on the equator (2χ = 0), with HLP at 2ψ = 0 and VLP at 2ψ = π. The angles α and δ also have geometrical interpretations on the Poincar′e sphere. The spherical angle of the great-circle route between the point of HLP and a point on the sphere is 2α, while the angle between the great-circle path and the equator is δ. Uniformplanewaves in a good dielectric. We maybase some useful plane-wave approximations on whether the real or imaginarypart of ?epsilon1 c dominates at the frequency of operation. We assume that ?μ(ω) = μ is independent of frequencyand use the notation epsilon1 c = ?epsilon1 c ( ˇω), σ = ?σ(ˇω), etc. Remember that epsilon1 c = parenleftbig epsilon1 prime + jepsilon1 primeprime parenrightbig + σ j ˇω = epsilon1 prime + j parenleftBig epsilon1 primeprime ? σ ˇω parenrightBig = epsilon1 cprime + jepsilon1 cprimeprime . Byde?nition, a “good dielectric” obeys tanδ c =? epsilon1 cprimeprime epsilon1 cprime = σ ˇωepsilon1 prime ? epsilon1 primeprime epsilon1 prime lessmuch 1. (4.253) Here tanδ c is the losstangent of the material, as ?rst described in (4.107) for a material without conductivity. For a good dielectric we have k = β ? jα = ˇω radicalbig μepsilon1 c = ˇω radicalbig μ[epsilon1 prime + jepsilon1 cprimeprime ] = ˇω radicalbig μepsilon1 prime radicalbig 1 ? j tanδ c , hence k ≈ ˇω radicalbig μepsilon1 prime bracketleftbigg 1 ? j 1 2 tanδ c bracketrightbigg (4.254) bythe binomial approximation for the square root. Therefore β ≈ ˇω radicalbig μepsilon1 prime (4.255) and α ≈ β 2 tanδ c = σ 2 radicalbigg μ epsilon1 prime bracketleftbigg 1 ? ˇωepsilon1 primeprime σ bracketrightbigg . (4.256) We conclude that α lessmuch β. Using this and the binomial approximation we establish η = ˇωμ k = ˇωμ β 1 1 ? jα/β ≈ ˇωμ β parenleftbigg 1 + j α β parenrightbigg . Finally, v p = ˇω β ≈ 1 √ μepsilon1 prime and v g = bracketleftbigg dβ dω bracketrightbigg ?1 ≈ 1 √ μepsilon1 prime . To ?rst 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 tanδ c ≈ σ ˇωepsilon1 greatermuch 1. In a good conductor the conduction current σ ˇ E is much greater than the displacement current j ˇωepsilon1 prime ˇ E, and epsilon1 primeprime is usuallyignored. Now we mayapproximate k = β ? jα = ˇω radicalbig μepsilon1 prime radicalbig 1 ? j tanδ c ≈ ˇω radicalbig μepsilon1 prime radicalbig ?j tanδ c . Since √ ?j = (1 ? j)/ √ 2 we ?nd that β = α ≈ radicalbig π f μσ. (4.257) Hence v p = ˇω β ≈ radicalBigg 2 ˇω μσ = 1 √ μepsilon1 prime radicalBigg 2 tanδ c . To ?nd v g we must replace ˇω by ω and di?erentiate, obtaining v g = bracketleftbigg dβ dω bracketrightbigg ?1 vextendsingle vextendsingle vextendsingle vextendsingle ω=ˇω ≈ bracketleftbigg 1 2 radicalbigg μσ 2 ˇω bracketrightbigg ?1 = 2 radicalBigg 2 ˇω μσ = 2v p . 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 dv p dω = d dω radicalBigg 2ω μσ = 1 2 radicalBigg 2 ωμσ and β dv p dω = radicalbigg ωμσ 2 1 2 radicalBigg 2 ωμσ = 1 2 , we see that v p v g = 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 × 10 7 S/m, μ = μ 0 , epsilon1 = epsilon1 0 ) at 1 MHz, we have v p = 415 m/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 √ π f μσ . (4.258) We call δ the skin depth. A good conductor is characterized bya small skin depth. For example, copper at 1 MHz has δ = 0.066 mm. Power carried by a uniform plane wave. Since a plane wavefront is in?nite in extent, we usuallyspeak of the power density carried bythe wave. This is identical to the time-average Poynting ?ux. Substitution from (4.223) and (4.244) gives S av = 1 2 Re{ ˇ E × ˇ H ? }= 1 2 Re braceleftBigg ˇ E × parenleftBigg ? k × ˇ E η parenrightBigg ? bracerightBigg . (4.259) Expanding the cross products and remembering that k · ˇ E = 0, we get S av = 1 2 ? k Re braceleftBigg | ˇ E| 2 η ? bracerightBigg = ? k Re braceleftbigg E 2 0 2η ? bracerightbigg e ?2α ? k·r . Hence a uniform plane wave propagating in an isotropic medium carries power in the direction of wavefront propagation. Velocityofenergytransport. Thegroupvelocity(4.237)hasanadditionalinterpre- tation as the velocityof energytransport. If the time-average volume densityof energy is given by 〈w em 〉=〈w e 〉+〈w m 〉 and the time-average volume densityof energy?ow is given bythe Poynting ?ux density S av = 1 2 Re braceleftbig ˇ E(r)× ˇ H ? (r) bracerightbig = 1 4 bracketleftbig ˇ E(r)× ˇ H ? (r)+ ˇ E ? (r)× ˇ H(r) bracketrightbig , (4.260) then the velocityof energy?ow, v e , is de?ned by S av =〈w em 〉v e . (4.261) Let us calculate v e for a plane wave propagating in a lossless, source-free medium where k = ? kω √ μepsilon1. By(4.216) and (4.223) we have ? E(r,ω)= ? E 0 (ω)e ?jβ ? k·r , (4.262) ? H(r,ω)= parenleftBigg ? k × ? E 0 (ω) η parenrightBigg e ?jβ ? k·r = ? H 0 (ω)e ?jβ ? k·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 ? parenrightbiggvextendsingle vextendsingle vextendsingle vextendsingle ω=ˇω = 4 j〈w em 〉. (4.264) Upon substitution of (4.262) and (4.263) we ?nd that we need to compute the frequency derivatives of ? E and ? H. Using ? ?ω e ?jβ ? k·r = parenleftbigg ? ?β e ?jβ ? k·r parenrightbigg dβ dω =?j ? k · r dβ dω e ?jβ ? k·r and remembering that k = ? kβ,wehave ? ? E(r,ω) ?ω = d ? E 0 (ω) dω e ?jk·r + ? E 0 (ω) parenleftbigg ?jr · dk dω parenrightbigg e ?jk·r , ? ? H(r,ω) ?ω = d ? H 0 (ω) dω e ?jk·r + ? H 0 (ω) parenleftbigg ?jr · dk dω parenrightbigg e ?jk·r . Equation (4.264) becomes ?? · braceleftbigg ? E ? 0 (ω)× d ? H 0 (ω) dω + d ? E 0 (ω) dω × ? H ? 0 (ω)? ?jr · dk dω bracketleftbig ? E ? 0 (ω)× ? H 0 (ω)+ ? E 0 (ω)× ? H ? 0 (ω) bracketrightbig bracerightbiggvextendsingle vextendsingle vextendsingle vextendsingle ω=ˇω = 4 j〈w em 〉. The ?rst 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)× ? H 0 ( ˇω)+ ? E 0 ( ˇω)× ? H ? 0 ( ˇω) bracketrightbig ·? parenleftbigg r · dk dω parenrightbiggvextendsingle vextendsingle vextendsingle vextendsingle ω=ˇω = 4〈w em 〉. The gradient term is merely ? parenleftbigg r · dk dω parenrightbiggvextendsingle vextendsingle vextendsingle vextendsingle ω=ˇω =? parenleftbigg x dk x dω + y dk y dω + z dk z dω parenrightbiggvextendsingle vextendsingle vextendsingle vextendsingle ω=ˇω = dk dω vextendsingle vextendsingle vextendsingle vextendsingle ω=ˇω , hence bracketleftbig ? E ? 0 ( ˇω)× ? H 0 ( ˇω)+ ? E 0 ( ˇω)× ? H ? 0 ( ˇω) bracketrightbig · dk dω vextendsingle vextendsingle vextendsingle vextendsingle ω=ˇω = 4〈w em 〉. (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 S av = 1 2 Re braceleftbig ˇ E × ˇ H ? bracerightbig = 1 4 bracketleftbig ? E 0 ( ˇω)× ? H ? 0 ( ˇω)+ ? E ? 0 ( ˇω)× ? H 0 ( ˇω) bracketrightbig and thus we can write (4.265) as S av · dk dω vextendsingle vextendsingle vextendsingle vextendsingle ω=ˇω =〈w em 〉. Sinceforauniformplanewaveinanisotropicmedium k and S av areinthesamedirection, we have S av = ? k dω dβ vextendsingle vextendsingle vextendsingle vextendsingle ω=ˇω 〈w em 〉 and the velocityof energytransport for a plane wave of frequency ˇω is then v e = ? k dω 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. Nonuniformplanewaves. A nonuniform plane wave has the same form (4.216) as a uniform plane wave, but the vectors k prime and k primeprime described in (4.217) are not aligned. Thus ˇ E(r) = E 0 e ?jk prime ·r e k primeprime ·r . In the time domain this becomes ˇ E(r) = E 0 e k primeprime ·r cos[ ˇωt ? k prime ( ? k prime · r)] where k prime = ? k prime k prime . The surfaces of constant phase are planes perpendicular to k prime and propagating in the direction of ? k prime . The phase velocityis now v p = ˇω/k prime and the wavelength is λ = 2π/k prime . In contrast, surfaces of constant amplitude must obey k primeprime · r = C and thus are planes perpendicular to k primeprime . In a nonuniform plane wave the TEM nature of the ?elds is lost. This is easilyseen bycalculating ˇ H from (4.219): ˇ H(r) = k × ˇ E(r) ˇωμ = k prime × ˇ E(r) ˇωμ + j k primeprime × ˇ E(r) ˇωμ . Thus, ˇ H isnolongerperpendiculartothedirectionofpropagationofthephasefront. The power carried bythe wave also di?ers from that of the uniform case. The time-average Poynting vector S av = 1 2 Re braceleftBigg ˇ E × parenleftBigg k × ˇ E ˇωμ parenrightBigg ? bracerightBigg can be expanded using the identity(B.7): S av = 1 2 Re braceleftbigg 1 ˇωμ ? bracketleftbig k ? ×( ˇ E × ˇ E ? )+ ˇ E ? ×(k ? × ˇ E) bracketrightbig bracerightbigg . (4.266) Since we still have k · E = 0, we mayuse the rest of (B.7) to write ˇ E ? ×(k ? × ˇ E) = k ? ( ˇ E · ˇ E ? )+ ˇ E(k · ˇ E) ? = k ? ( ˇ E · ˇ E ? ). Substituting this into (4.266), and noting that ˇ E × ˇ E ? is purelyimaginary, we ?nd S av = 1 2 Re braceleftbigg 1 ˇωμ ? bracketleftbig jk ? × Im braceleftbig ˇ E × ˇ E ? bracerightbig + k ? | ˇ E| 2 bracketrightbig bracerightbigg . (4.267) Thus the vector direction of S av is 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 k maybe complex when k is real, and the implication is important for the plane-wave expansion of complicated ?elds in free space. By(4.218), real k requires that if k primeprime negationslash= 0 then k prime · k primeprime = 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-average power to the lossless case we note that μ is purelyreal and that E × E ? = (E 0 × E ? 0 )e 2k primeprime ·r . Then (4.267) becomes S av = 1 2 ˇωμ e 2k primeprime ·r Re braceleftbig j(k prime ? jk primeprime )× Im braceleftbig E 0 × E ? 0 bracerightbig +(k prime ? jk primeprime )| ˇ E| 2 bracerightbig or S av = 1 2 ˇωμ e 2k primeprime ·r bracketleftbig k primeprime × Im braceleftbig E 0 × E ? 0 bracerightbig + k prime ˇ E| 2 bracketrightbig . We see that in a lossless medium the direction of energypropagation is perpendicular to the surfaces of constant amplitude (since k primeprime · S av = 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 re?ection and refrac- tion of a plane wave from a planar interface in the next section. We shall also ?nd in § 4.13 that nonuniform plane waves are a necessaryconstituent of the angular spectrum representation of an arbitrarywave ?eld. 4.11.5 Planewavesinlayeredmedia A useful canonical problem in wave propagation involves the re?ection of plane waves byplanar interfaces between di?ering material regions. This has manydirect applica- tions, from the design of optical coatings and microwave absorbers to the probing of underground oil-bearing rock layers. We shall begin by studying the re?ection of a plane wave at a single interface and then extend the results to anynumber of material layers. Re?ection of a uniform plane wave at a planar material interface. Consider twolossymedi aseparatedbythe z = 0planeasshowninFigure4.18.Themediaareas- sumedtobeisotropicandhomogeneouswithpermeability ?μ(ω)andcomplexpermittivity ?epsilon1 c (ω). Both ?μ and ?epsilon1 c maybe complex numbers describing magnetic and dielectric loss, respectively. We assume that a linearly-polarized plane-wave ?eld of the form (4.216) is created within region 1 bya process that we shall not studyhere. We take this ?eld to be the known “incident wave” produced byan impressed source, and wish to compute the total ?eld in regions 1 and 2. Here we shall assume that the incident ?eld is that of a uniform plane wave, and shall extend the analysis to certain types of nonuniform plane waves subsequently. Since the incident ?eld is uniform, we maywrite the wave vector associated with this ?eld as k i = ? k i k i = ? k i (k iprime + jk iprimeprime ) where [k i (ω)] 2 = ω 2 ?μ 1 (ω)?epsilon1 c 1 (ω). We can assume without loss of generalitythat ? k i lies in the xz-plane and makes an angle θ i withtheinterfacenormalasshowninFigure4.18.Werefertoθ i astheincidenceangle of the incident ?eld, and note that it is the angle between the direction of propagation of the planar phase fronts and the normal to the interface. With this we have k i = ?xk 1 sinθ i + ?zk 1 cosθ i = ?xk i x + ?zk i z . Using k 1 = β 1 ? jα 1 we also have k i x = (β 1 ? jα 1 )sinθ i . The term k i z is written in a somewhat di?erent form in order to make the result easily applicable to re?ections from multiple interfaces. We write k i z = (β 1 ? jα 1 )cosθ i = τ i e ?jγ i = τ i cosγ i ? jτ i sinγ i . Thus, τ i = radicalBig β 2 1 +α 2 1 cosθ i ,γ i = tan ?1 (α 1 /β 1 ). We solve for the ?elds in each region of space directlyin the frequencydomain. The incident electric ?eld has the form of (4.216), ? E i (r,ω)= ? E i 0 (ω)e ?jk i (ω)·r , (4.268) while the magnetic ?eld is found from (4.219) to be ? H i = k i × ? E i ω ?μ 1 . (4.269) The incident ?eld maybe decomposed into two orthogonal components, one parallel to the plane of incidence (the plane containing ? k and the interface normal ?z) and one perpendicular to this plane. We seek unique solutions for the ?elds in both regions, ?rst for the case in which the incident electric ?eld has onlya parallel component, and then for the case in which it has onlya perpendicular component. The total ?eld is then determined bysuperposition of the individual solutions. For perpendicular polarization we have from (4.268) and (4.269) ? E i ⊥ = ?y ? E i ⊥ e ?j(k i x x+k i z z) , (4.270) ? H i ⊥ = ??xk i z + ?zk i x k 1 ? E i ⊥ η 1 e ?j(k i x x+k i z z) , (4.271) Figure 4.18: Uniform plane wave incident on planar interface between two lossyregions of space. (a) TM polarization, (b) TE polarization. asshowngraphicallyi nFigure4.18.Hereη 1 = ( ?μ 1 /?epsilon1 c 1 ) 1/2 is the intrinsic impedance of medium 1. For parallel polarization, the direction of ? E is found byremembering that the wave must be TEM. Thus ? E bardbl is perpendicular to k i . Since ? E bardbl must also be perpendicular to ? E ⊥ , we have two possible directions for ? E bardbl . Byconvention we choose the one for which ? H lies in the same direction as did ? E for perpendicular polarization. Thus we have for parallel polarization ? H i bardbl = ?y ? E i bardbl η 1 e ?j(k i x x+k i z z) , (4.272) ? E i bardbl = ?xk i z ? ?zk i x k 1 ? E i bardbl e ?j(k i x x+k i z z) , (4.273) asshowninFigure4.18.Because ? E liestransverse(normal)totheplaneofincidence under perpendicular polarization, the ?eld set is often described astransverseelectric or TE. Because ? H lies transverse to the plane of incidence under parallel polarization, the ?elds in that case are transversemagnetic or TM. Uniqueness requires that the total ?eld obeythe boundaryconditions at the planar interface. We hypothesize that the total ?eld within region 1 consists of the incident ?eld superposed with a “re?ected” plane-wave ?eld having wave vector k r , while the ?eld in region 2 consists of a single “transmitted” plane-wave ?eld having wave vector k t . We cannot at the outset make anyassumption regarding whether either of these ?elds are uniform plane waves. However, we do note that the re?ected and transmitted ?elds cannot have vector components not present in the incident ?eld; extra components would preclude satisfaction of the boundaryconditions. Letting ? E r be the amplitude of the re?ected plane-wave ?eld we maywrite ? E r ⊥ = ?y ? E r ⊥ e ?j(k r x x+k r z z) , ? H r ⊥ = ??xk r z + ?zk r x k 1 ? E r ⊥ η 1 e ?j(k r x x+k r z z) , ? H r bardbl = ?y ? E r bardbl η 1 e ?j(k r x x+k r z z) , ? E r bardbl = ?xk r z ? ?zk r x k 1 ? E r bardbl e ?j(k r x x+k r z z) , where (k r x ) 2 +(k r z ) 2 = k 2 1 . Similarly, letting ? E t be the amplitude of the transmitted ?eld we have ? E t ⊥ = ?y ? E t ⊥ e ?j(k t x x+k t z z) , ? H t ⊥ = ??xk t z + ?zk t x k 2 ? E t ⊥ η 2 e ?j(k t x x+k t z z) , ? H t bardbl = ?y ? E t bardbl η 2 e ?j(k t x x+k t z z) , ? E t bardbl = ?xk t z ? ?zk t x k 2 ? E t bardbl e ?j(k t x x+k t z z) , where (k t x ) 2 +(k t z ) 2 = k 2 2 . Therelationshipsbetweenthe?eldamplitudes ? E i , ? E r , ? E t ,andbetweenthecomponents of the re?ected and transmitted wave vectors k r and k t , can be found byapplying the boundaryconditions. The tangential electric and magnetic ?elds are continuous across the interface at z = 0: ?z ×( ? E i + ? E r )| z=0 = ?z × ? E t | z=0 , ?z ×( ? H i + ? H r )| z=0 = ?z × ? H t | z=0 . Substituting the ?eld expressions, we ?nd that for perpendicular polarization the two boundaryconditions require ? E i ⊥ e ?jk i x x + ? E r ⊥ e ?jk r x x = ? E t ⊥ e ?jk t x x , (4.274) k i z k 1 ? E i ⊥ η 1 e ?jk i x x + k r z k 1 ? E r ⊥ η 1 e ?jk r x x = k t z k 2 ? E t ⊥ η 2 e ?jk t x x , (4.275) while for parallel polarization theyrequire k i z k 1 ? E i bardbl e ?jk i x x + k r z k 1 ? E r bardbl e ?jk r x x = k t z k 2 ? E t bardbl e ?jk t x x , (4.276) ? E i bardbl η 1 e ?jk i x x + ? E r bardbl η 1 e ?jk r x x = ? E t bardbl η 2 e ?jk t x x . (4.277) For the above to hold for all x we must have the exponential terms equal. This requires k i x = k r x = k t x , (4.278) and also establishes a relation between k i z , k r z , and k t z . Since (k i x ) 2 +(k i z ) 2 = (k r x ) 2 +(k r z ) 2 = k 2 1 , we must have k r z =±k i z . In order to make the re?ected wavefronts propagate away from the interface we select k r z =?k i z . Letting k i x = k r x = k t x = k 1x and k i z =?k r z = k 1z , we maywrite the wave vectors in region 1 as k i = ?xk 1x + ?zk 1z , k r = ?xk 1x ? ?zk 1z . Since (k t x ) 2 +(k t z ) 2 = k 2 2 , letting k 2 = β 2 ? jα 2 we have k t z = radicalBig k 2 2 ? k 2 1x = radicalBig (β 2 ? jα 2 ) 2 ?(β 1 ? jα 1 ) 2 sin 2 θ i = τ t e ?jγ t . Squaring out the above relation, we have A ? jB= (τ t ) 2 cos 2γ t ? j(τ t ) 2 sin 2γ t where A = β 2 2 ?α 2 2 ?(β 2 1 ?α 2 1 )sin 2 θ i , B = 2(β 2 α 2 ?β 1 α 1 sin 2 θ i ). (4.279) Thus τ t = parenleftbig A 2 + B 2 parenrightbig 1/4 ,γ t = 1 2 tan ?1 B A . (4.280) Renaming k t z as k 2z , we maywrite the transmitted wave vector as k t = ?xk 1x + ?zk 2z = k prime 2 + jk primeprime 2 where k prime 2 = ?xβ 1 sinθ i + ?zτ t cosγ t , k primeprime 2 =??xα 1 sinθ i ? ?zτ t sinγ t . Since the direction of propagation of the transmitted ?eld phase fronts is perpendicular to k prime 2 , a unit vector in the direction of propagation is ? k prime 2 = ?xβ 1 sinθ i + ?zτ t cosγ t radicalBig β 2 1 sin 2 θ i +(τ t ) 2 cos 2 θ i . (4.281) Similarly, a unit vector perpendicular to planar surfaces of constant amplitude is given by ? k primeprime 2 = ?xα 1 sinθ i + ?zτ t sinγ t radicalBig α 2 1 sin 2 θ i +(τ t ) 2 sin 2 γ t . (4.282) In general ? k prime is not aligned with ? k primeprime and thus the transmitted ?eld is a nonuniform plane wave. With these de?nitions of k 1x ,k 1z ,k 2z , equations (4.274) and (4.275) can be solved si- multaneouslyand we have ? E r ⊥ = ? Gamma1 ⊥ ? E i ⊥ , ? E t ⊥ = ? T ⊥ ? E i ⊥ , where ? Gamma1 ⊥ = Z 2⊥ ? Z 1⊥ Z 2⊥ + Z 1⊥ , ? T ⊥ = 1 + ? Gamma1 ⊥ = 2Z 2⊥ Z 2⊥ + Z 1⊥ , (4.283) with Z 1⊥ = k 1 η 1 k 1z , Z 2⊥ = k 2 η 2 k 2z . Here ? Gamma1 is a frequency-dependentre?ectioncoe?cient that relates the tangential compo- nents of the incident and re?ected electric ?elds, and ? T is a frequency-dependent trans- missioncoe?cient thatrelatesthetangentialcomponentsoftheincidentandtransmitted electric ?elds. These coe?cients are also called the Fresnelcoe?cients. For the case of parallel polarization we solve (4.276) and (4.277) to ?nd ? E r bardbl,x ? E i bardbl,x = k r x k i x ? E r bardbl ? E i bardbl =? ? E r bardbl ? E i bardbl = ? Gamma1 bardbl , ? E t bardbl,x ? E i bardbl,x = (k t z /k 2 ) ? E t bardbl (k i z /k 1 ) ? E i bardbl = ? T bardbl . Here ? Gamma1 bardbl = Z 2bardbl ? Z 1bardbl Z 2bardbl + Z 1bardbl , ? T bardbl = 1 + ? Gamma1 bardbl = 2Z 2bardbl Z 2bardbl + Z 1bardbl , (4.284) with Z 1bardbl = k 1z η 1 k 1 , Z 2bardbl = k 2z η 2 k 2 . Note that we mayalso write ? E r bardbl =? ? Gamma1 bardbl ? E i bardbl , ? E t bardbl = ? T bardbl ? E i bardbl parenleftbigg k i z k 1 k 2 k t z parenrightbigg . Let us summarize the ?elds in each region. For perpendicular polarization we have ? E i ⊥ = ?y ? E i ⊥ e ?jk i ·r , ? E r ⊥ = ?y ? Gamma1 ⊥ ? E i ⊥ e ?jk r ·r , (4.285) ? E t ⊥ = ?y ? T ⊥ ? E i ⊥ e ?jk t ·r , and ? H i ⊥ = k i × ? E i ⊥ k 1 η 1 , ? H r ⊥ = k r × ? E r ⊥ k 1 η 1 , ? H t ⊥ = k t × ? E t ⊥ k 2 η 2 . (4.286) For parallel polarization we have ? E i bardbl =?η 1 k i × ? H i bardbl k 1 e ?jk i ·r , ? E r bardbl =?η 1 k r × ? H r bardbl k 1 e ?jk r ·r , ? E t bardbl =?η 2 k t × ? H t bardbl k 2 e ?jk t ·r , (4.287) and ? H i bardbl = ?y ? E i bardbl η 1 e ?jk i ·r , ? H r bardbl =??y ? Gamma1 bardbl ? E i bardbl η 1 e ?jk r ·r , ? H t bardbl = ?y ? T bardbl ? E i bardbl η 2 parenleftbigg k i z k 1 k 2 k t z parenrightbigg e ?jk t ·r . (4.288) The wave vectors are given by k i = (?xβ 1 sinθ i + ?zτ i cosγ i )? j(?xα 1 sinθ i + ?zτ i sinγ i ), (4.289) k r = (?xβ 1 sinθ i ? ?zτ i cosγ i )? j(?xα 1 sinθ i ? ?zτ i sinγ i ), (4.290) k t = (?xβ 1 sinθ i + ?zτ t cosγ t )? j(?xα 1 sinθ i + ?zτ t sinγ t ). (4.291) We see that the re?ected wave must, like the incident wave, be a uniform plane wave. We de?ne the unsigned re?ection angle θ r as the angle between the surface normal and thedirectionofpropagationofthere?ectedwavefronts(Figure4.18).Since k i · ?z = k 1 cosθ i =?k r · ?z = k 1 cosθ r and k i · ?x = k 1 sinθ i = k r · ?x = k 1 sinθ r we must have θ i = θ r . This is known asSnell’slawofre?ection. We can similarlyde?ne the transmissionangle to be the angle between the direction of propagation of the transmitted wavefronts and the interface normal. Noting that ? k prime 2 · ?z = cosθ t and ? k prime 2 · ?x = sinθ t , we have from (4.281) and (4.282) cosθ t = τ t cosγ t radicalBig β 2 1 sin 2 θ i +(τ t ) 2 cos 2 γ t , (4.292) sinθ t = β 1 sinθ i radicalBig β 2 1 sin 2 θ i +(τ t ) 2 cos 2 γ t , (4.293) and thus θ t = tan ?1 parenleftbigg β 1 τ t sinθ i cosγ t parenrightbigg . (4.294) Depending on the properties of the media, at a certain incidence angle θ c , called the critical angle, the angle of transmission becomes π/2. Under this condition ? k prime 2 has only an x-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-average power is carried bya monochromatic transmitted wave into the second medium. We also see that although the transmitted ?eld maybe a nonuniform plane wave, its mathematical form is that of the incident plane wave. This allows us to easilygeneralize the single-interface re?ection problem to one involving manylayers. Uniformplane-wavere?ectionforlosslessmedia. Wecanspecializethepreceding results to the case for which both regions are lossless with ?μ = μ and ?epsilon1 c = epsilon1 real and frequency-independent. By (4.224) we have β = ω √ μepsilon1, while (4.225) gives α = 0. Wecaneasilyshowthatthetransmitted wave must be uniformunlesstheincidenceangle exceeds the critical angle. By(4.279) we have A = β 2 2 ?β 2 1 sin 2 θ i , B = 0, (4.295) while (4.280) gives τ = bracketleftbig A 2 bracketrightbig 1/4 = radicalBig |β 2 2 ?β 2 1 sin 2 θ i | and γ t = 1 2 tan ?1 (0). We have several possible choices for γ t . To choose properlywe note that γ t represents the negative of the phase of the quantity k t z = √ A.IfA > 0 the phase of the square root is 0.IfA < 0 the phase of the square root is ?π/2 and thus γ t =+π/2. Here we choose the plus sign on γ t to ensure that the transmitted ?eld decays as z increases. We note that if A = 0 then τ t = 0 and from (4.293) we have θ t = π/2. This de?nes the critical angle, which from (4.295) is θ c = sin ?1 parenleftbigg β 2 2 β 2 1 parenrightbigg = sin ?1 parenleftbigg μ 2 epsilon1 2 μ 1 epsilon1 1 parenrightbigg . Therefore γ t = 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): k t = k tprime + jk tprimeprime = braceleftBigg ?xβ 1 sinθ i + ?z √ |A|,θ i <θ c , ?xβ 1 sinθ i ? j ?z √ |A|,θ i >θ c . (4.296) By(4.293) we have sinθ t = β 1 sinθ i radicalBig β 2 1 sin 2 θ i +β 2 2 ?β 2 1 sin 2 θ i = β 1 sinθ i β 2 or β 2 sinθ t = β 1 sinθ i . (4.297) This is known as Snell’slawofrefraction. With this we can write for θ i <θ c A = β 2 2 ?β 2 1 sin 2 θ i = β 2 2 cos 2 θ t . Using this and substituting β 2 sinθ t for β 1 sinθ i , we mayrewrite (4.296) for θ i <θ c as k t = k tprime + jk tprimeprime = ?xβ 2 sinθ t + ?zβ 2 cosθ t . (4.298) Hence the transmitted plane wave is uniform with k tprimeprime = 0. When θ i >θ c we have from (4.296) k tprime = ?xβ 1 sinθ i , k tprimeprime =??z radicalBig β 2 1 sin 2 θ i ?β 2 2 . Since k tprime and k tprimeprime are not collinear, the plane wave is nonuniform. Let us examine the cases θ i <θ c and θ i >θ c in greater detail. Case1: θ i <θ c . By(4.289)–(4.290) and (4.298) the wave vectors are k i = ?xβ 1 sinθ i + ?zβ 1 cosθ i , k r = ?xβ 1 sinθ i ? ?zβ 1 cosθ i , k t = ?xβ 2 sinθ t + ?zβ 2 cosθ t , and the wave impedances are Z 1⊥ = η 1 cosθ i , Z 2⊥ = η 2 cosθ t , Z 1bardbl = η 1 cosθ i , Z 2bardbl = η 2 cosθ t . The re?ection coe?cients are ? Gamma1 ⊥ = η 2 cosθ i ?η 1 cosθ t η 2 cosθ i +η 1 cosθ t , ? Gamma1 bardbl = η 2 cosθ t ?η 1 cosθ i η 2 cosθ t +η 1 cosθ i . (4.299) So the re?ection coe?cients are purelyreal, with signs dependent on the constitutive parameters of the media. We can write ? Gamma1 ⊥ = ρ ⊥ e jφ ⊥ , ? Gamma1 bardbl = ρ bardbl e jφ bardbl , where ρ and φ are real, and where φ = 0 or π. Under certain conditions the re?ection coe?cients vanish. For a given set of constitu- tive parameters we mayachieve ? Gamma1 = 0 at an incidence angle θ B , known as the Brewster or polarizing angle. A wave with an arbitrarycombination of perpendicular and paral- lel polarized components incident at this angle produces a re?ected ?eld with a single component. A wave incident with onlythe appropriate single component produces no re?ected ?eld, regardless of its amplitude. For perpendicular polarization we set ? Gamma1 ⊥ = 0, requiring η 2 cosθ i ?η 1 cosθ t = 0 or equivalently μ 2 epsilon1 2 (1 ? sin 2 θ i ) = μ 1 epsilon1 1 (1 ? sin 2 θ t ). By(4.297) we mayput sin 2 θ t = μ 1 epsilon1 1 μ 2 epsilon1 2 sin 2 θ i , resulting in sin 2 θ i = μ 2 epsilon1 1 epsilon1 2 μ 1 ?epsilon1 1 μ 2 μ 2 1 ?μ 2 2 . The value of θ i that satis?es this equation must be the Brewster angle, and thus θ B⊥ = sin ?1 radicalBigg μ 2 epsilon1 1 epsilon1 2 μ 1 ?epsilon1 1 μ 2 μ 2 1 ?μ 2 2 . When μ 1 = μ 2 thereisnosolutiontothisequation, hencethere?ectioncoe?cientcannot vanish. When epsilon1 1 = epsilon1 2 we have θ B⊥ = sin ?1 radicalbigg μ 2 μ 1 +μ 2 = tan ?1 radicalbigg μ 2 μ 1 . For parallel polarization we set ? Gamma1 bardbl = 0 and have η 2 cosθ t = η 1 cosθ i . Proceeding as above we ?nd that θ Bbardbl = sin ?1 radicalBigg epsilon1 2 μ 1 epsilon1 1 μ 2 ?epsilon1 2 μ 1 epsilon1 2 1 ?epsilon1 2 2 . This expression has no solution when epsilon1 1 = epsilon1 2 , and thus the re?ection coe?cient cannot vanish under this condition. When μ 1 = μ 2 we have θ Bbardbl = sin ?1 radicalbigg epsilon1 2 epsilon1 1 +epsilon1 2 = tan ?1 radicalbigg epsilon1 2 epsilon1 1 . We ?nd that when θ i <θ c the total ?eld in region 1 behaves as a traveling wave along x, but has characteristics of both a standing wave and a traveling wave along z (Problem 4.7).Thetraveling-wavecomponentisassociatedwithaPoyntingpower?ux, while the standing-wave component is not. This ?ux is carried across the boundary into region 2 where the transmitted ?eld consists onlyof a traveling wave. By (4.161) the normal component of time-average Poynting ?ux is continuous across the boundary, demonstrating that the time-average power carried bythe wave into the interface from region1passesoutthroughtheinterfaceintoregion2(Problem 4.8). Case2: θ i <θ c . The wave vectors are, from (4.289)–(4.290) and (4.296), k i = ?xβ 1 sinθ i + ?zβ 1 cosθ i , k r = ?xβ 1 sinθ i ? ?zβ 1 cosθ i , k t = ?xβ 1 sinθ i ? j ?zα c , where α c = radicalBig β 2 1 sin 2 θ i ?β 2 2 is the criticalangleattenuationconstant. The wave impedances are Z 1⊥ = η 1 cosθ i , Z 2⊥ = j β 2 η 2 α c , Z 1bardbl = η 1 cosθ i , Z 2bardbl =?j α c η 2 β 2 . Substituting these into (4.283) and (4.284), we ?nd that the re?ection coe?cients are the complex quantities ? Gamma1 ⊥ = β 2 η 2 cosθ i + jη 1 α c β 2 η 2 cosθ i ? jη 1 α c = e jφ ⊥ , ? Gamma1 bardbl =? β 2 η 1 cosθ i + jη 2 α c β 2 η 1 cosθ i ? jη 2 α c = e jφ bardbl , where φ ⊥ = 2 tan ?1 parenleftbigg η 1 α c β 2 η 2 cosθ i parenrightbigg ,φ bardbl = π + 2 tan ?1 parenleftbigg η 2 α c β 2 η 1 cosθ i parenrightbigg . We note with interest that ρ ⊥ = ρ bardbl = 1. So the amplitudes of the re?ected waves are identical to those of the incident waves, and we call this the case of total internal re?ection. The phase of the re?ected wave at the interface is changed from that of the incident wave byan amount φ ⊥ or φ bardbl . The phase shift incurred bythe re?ected wave upon total internal re?ection is called the Goos–H¨anchenshift. In the case of total internal re?ection the ?eld in region 1 is a pure standing wave while the ?eld in region 2 decays exponentially in the z-direction and is evanescent (Problem 4.9).Sinceastandingwavetransportsnopower,thereisnoPoynting?uxintoregion2. We ?nd that the evanescent wave also carries no power and thus the boundarycondition onpower?uxattheinterfaceissatis?ed(Problem4.10 ).Wenotethatforanyincide nt angle except θ i = 0 (normal incidence) the wave in region 1 does transport power in the x-direction. Re?ectionoftime-domainuniformplanewaves. Solution for the ?elds re?ected and transmitted at an interface shows us the properties of the ?elds for a certain single excitation frequencyand allows us to obtain time-domain ?elds byFourier inversion. Under certain conditions it is possible to do the inversion analytically, providing physical insight into the temporal behavior of the ?elds. Asasimpleexample,consideraperpendicularly-polarized,uniformplanewaveincident fromfreespaceatanangle θ i ontheplanarsurfaceofaconductingmaterial(Figure4.18). Thematerialisassumedtohavefrequency-independentconstitutiveparameters ?μ=μ 0 , ?epsilon1 = epsilon1, and ?σ = σ. By(4.285) we have the re?ected ?eld ? E r ⊥ (r,ω)= ?y ? Gamma1 ⊥ (ω) ? E i ⊥ (ω)e ?jk r (ω)·r = ?y ? E r (ω)e ?jω ? k r ·r c (4.300) where ? E r = ? Gamma1 ⊥ ? E i ⊥ . We can use the time-shifting theorem (A.3)to invert the transform and obtain E r ⊥ (r,t) = F ?1 braceleftbig ? E r ⊥ (r,ω) bracerightbig = ?yE r parenleftBigg t ? ? k r · r c parenrightBigg (4.301) where we have bythe convolution theorem (12) E r (t) = F ?1 braceleftbig ? E r (ω) bracerightbig = Gamma1 ⊥ (t)? E ⊥ (t). Here E ⊥ (t) = F ?1 braceleftbig ? E i ⊥ (ω) bracerightbig is the time waveform of the incident plane wave, while Gamma1 ⊥ (t) = F ?1 braceleftbig ? Gamma1 ⊥ (ω) bracerightbig is the time-domain re?ection coe?cient. By(4.301) the re?ected time-domain ?eld propagates along the direction ? k r at the speed of light. The time waveform of the ?eld is the convolution of the waveform of the incident ?eld with the time-domain re?ection coe?cient Gamma1 ⊥ (t). In the lossless case (σ = 0), Gamma1 ⊥ (t) is a δ-function and thus the waveforms of the re?ected and incident ?elds are identical. With the introduction of loss Gamma1 ⊥ (t) broadens and thus the re?ected ?eld waveform becomes a convolution-broadened version of the incident ?eld waveform. To understand the waveform of the re?ected ?eld we must compute Gamma1 ⊥ (t). Note that by choosing the permittivityof region 2 to exceed that of region 1 we preclude total internal re?ection. We can specialize the frequency-domain re?ection coe?cient (4.283) for our problem bynoting that k 1z = β 1 cosθ i , k 2z = radicalBig k 2 2 ? k 2 1x = ω √ μ 0 radicalbigg epsilon1 + σ jω ?epsilon1 0 sin 2 θ i , and thus Z 1⊥ = η 0 cosθ i , Z 2⊥ = η 0 radicalBig epsilon1 r + σ jωepsilon1 0 ? sin 2 θ i , where epsilon1 r = epsilon1/epsilon1 0 and η 0 = √ μ 0 /epsilon1 0 . We thus obtain ? Gamma1 ⊥ = √ s ? √ Ds + B √ s + √ Ds + B (4.302) where s = jω and D = epsilon1 r ? sin 2 θ i cos 2 θ i , B = σ epsilon1 0 cos 2 θ i . 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 ? Gamma1 0 ⊥ (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 ? Gamma1 0 ⊥ (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 Gamma1 0 ⊥ (t) = F ?1 braceleftbig ? Gamma1 0 ⊥ (ω) bracerightbig = f 1 (t)+ f 2 (t)+ f 3 (t) (4.303) where f 1 (t) =? 2B (1 + √ D)(D ? 1) e ? Bt D?1 U(t), f 2 (t) =? B 2 √ D(D ? 1) 2 U(t) integraldisplay t 0 e ? B(t?x) D?1 I parenleftbigg Bx 2D parenrightbigg dx, f 3 (t) = B √ D(D ? 1) I parenleftbigg Bt 2D parenrightbigg U(t). Here I(x) = e ?x [I 0 (x)+ I 1 (x)] where I 0 (x) and I 1 (x) are modi?ed Bessel functions of the ?rst kind. Setting u = Bx/2D we can also write f 2 (t) =? 2B √ D (D ? 1) 2 U(t) integraldisplay Bt 2D 0 e ? Bt?2Du D?1 I(u)du. Polynomialapproximationsfor I(x) maybefoundinAbramowitzandStegun[?], making the computation of Gamma1 0 ⊥ (t) straightforward. The complete time-domain re?ection coe?cient is Gamma1 ⊥ (t) = 1 ? √ D 1 + √ D δ(t)+Gamma1 0 ⊥ (t). 0.0 2.5 5.0 7.5 10.0 12.5 t (ns) -0.5 -0.4 -0.3 -0.2 -0.1 10 -9 Γ ⊥ (t) =3, σ=0.01 =80, σ=4 r r Figure 4.19: Time-domain re?ection coe?cients. If σ = 0 then Gamma1 0 ⊥ (t) = 0 and the re?ection coe?cient reduces to a single δ-function. Since convolution with this term does not alter wave shape, the re?ected ?eld has the same waveform as the incident ?eld. A plot of Gamma1 0 ⊥ (t)fornormalincidence(θ i = 0 0 )isshowninFigure4.19.Heretwo material cases are displayed: epsilon1 r = 3, σ = 0.01 S/m, which is representative of drywater ice, and epsilon1 r = 80, σ = 4 S/m, which is representative of sea water. We see that a pulse waveform experiences more temporal spreading upon re?ection from ice than from sea water, but that the amplitude of the dispersive component is less than that for sea water. Re?ectionofanonuniformplanewavefromaplanarinterface. Describing the interaction of a general nonuniform plane wave with a planar interface is problematic because of the non-TEM behavior of the incident wave. We cannot decompose the ?elds into two mutuallyorthogonal cases as we did with uniform waves, and thus the analysis is more di?cult. However, we found in the last section that when a uniform wave is incident on a planar interface, the transmitted wave, even if nonuniform in nature, takes on the same mathematical form and maybe decomposed in the same manner as the incident wave. Thus, we may studythe case in which this refracted wave is incident on a successive interface using exactlythe same analysis 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 regionsofspaceseparatedby N planarinterfacesasshowninFigure4.20,and assumethatauniformplanewaveisincidentonthe?rstinterfaceatangleθ i . Eachregion is assumed isotropic and homogeneous with a frequency-dependent complex permittivity and permeability. We can easily generalize the previous analysis regarding re?ection from a single interface byrealizing that in order to satisfythe boundaryconditions each Figure 4.20: Interaction of a uniform plane wave with a multi-layered material. region, except region N, contains an incident-type wave of the form ? E i (r,ω)= ? E i 0 e ?jk i ·r and a re?ected-type wave of the form ? E r (r,ω)= ? E r 0 e ?jk r ·r . In region n we maywrite the wave vectors describing these waves as k i n = ?xk x,n + ?zk z,n , k r n = ?xk x,n ? ?zk z,n , where k 2 x,n + k 2 z,n = k 2 n , k 2 n = ω 2 ?μ n ?epsilon1 c n = (β n ? jα n ) 2 . We note at the outset that, as with the single interface case, the boundaryconditions are onlysatis?ed when Snell’s law of re?ection holds, and thus k x,n = k x,0 = k 0 sinθ i (4.304) where k 0 = ω(?μ 0 ?epsilon1 c 0 ) 1/2 is the wavenumber of the 0th region (not necessarilyfree space). From this condition we have k z,n = radicalBig k 2 n ? k 2 x,0 = τ n e ?jγ n where τ n = (A 2 n + B 2 n ) 1/4 ,γ n = 1 2 tan ?1 parenleftbigg B n A n parenrightbigg , and A n = β 2 n ?α 2 n ?(β 2 0 ?α 2 0 )sin 2 θ i , B n = 2(β n α n ?β 0 α 0 sin 2 θ i ). Provided that the incident wave is uniform, we can decompose the ?elds 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 ?eld in region n, 0 ≤ n ≤ N ? 1,as ? E ⊥n = ? E i ⊥n + ? E r ⊥n where ? E i ⊥n = ?ya n+1 e ?jk x,n x e ?jk z,n (z?z n+1 ) , ? E r ⊥n = ?yb n+1 e ?jk x,n x e +jk z,n (z?z n+1 ) , and the magnetic ?eld as ? H ⊥n = ? H i ⊥n + H r ⊥n where ? H i ⊥n = ??xk z,n + ?zk x,n k n η n a n+1 e ?jk x,n x e ?jk z,n (z?z n+1 ) , ? H r ⊥n = +?xk z,n + ?zk x,n k n η n b n+1 e ?jk x,n x e +jk z,n (z?z n+1 ) . When n = N there is no re?ected wave; in this region we write ? E ⊥N = ?ya N+1 e ?jk x,N x e ?jk z,N (z?z N ) , ? H ⊥N = ??xk z,N + ?zk x,N k N η N a N+1 e ?jk x,N x e ?jk z,N (z?z N ) . Since a 1 is the known amplitude of the incident wave, there are 2N unknown wave am- plitudes. We obtain the necessary 2N simultaneous equations byapplying the boundary conditions at each of the interfaces. At interface n located at z = z n , 1 ≤ n ≤ N ? 1,we have from the continuityof tangential electric ?eld a n + b n = a n+1 e ?jk z,n (z n ?z n+1 ) + b n+1 e +jk z,n (z n ?z n+1 ) while from the continuityof magnetic ?eld ?a n k z,n?1 k n?1 η n?1 + b n k z,n?1 k n?1 η n?1 =?a n+1 k z,n k n η n e ?jk z,n (z n ?z n+1 ) + b n+1 k z,n k n η n e +jk z,n (z n ?z n+1 ) . Noting that the wave impedance of region n is Z ⊥n = k n η n k z,n and de?ning the region n propagation factor as ? P n = e ?jk z,n Delta1 n where Delta1 n = z n+1 ? z n , we can write a n ? P n + b n ? P n = a n+1 + b n+1 ? P 2 n , (4.305) ?a n ? P n + b n ? P n =?a n+1 Z ⊥n?1 Z ⊥n + b n+1 Z ⊥n?1 Z ⊥n ? P 2 n . (4.306) We must still applythe boundaryconditions at z = z N . Proceeding as above, we ?nd that (4.305) and (4.306) hold for n = N if we set b N+1 = 0 and ? P N = 1. The 2N simultaneous equations (4.305)–(4.306) maybe solved using standard matrix methods. However, through a little manipulation we can put the equations into a form easilysolvedbyrecursion,providingaverynicephysicalpictureofthemultiplere?ections thatoccurwithinthelayeredmedium. Webeginby eliminating b n bysubtracting(4.306) from (4.305): 2a n ? P n = a n+1 bracketleftbigg 1 + Z ⊥n?1 Z ⊥n bracketrightbigg + b n+1 ? P 2 n bracketleftbigg 1 ? Z ⊥n?1 Z ⊥n bracketrightbigg . (4.307) Figure 4.21: Wave ?ow diagram showing interaction of incident and re?ected waves for region n. De?ning ? Gamma1 n = Z ⊥n ? Z ⊥n?1 Z ⊥n + Z ⊥n?1 (4.308) as the interfacial re?ection coe?cient for interface n (i.e., the re?ection coe?cient as- suming a single interface as in (4.283)), and ? T n = 2Z ⊥n Z ⊥n + Z ⊥n?1 = 1 + ? Gamma1 n as the interfacialtransmissioncoe?cient for interface n, we can write (4.307) as a n+1 = a n ? T n ? P n + b n+1 ? P n (? ? Gamma1 n ) ? P n . Finally, if we de?ne the global re?ection coe?cient R n for region n as the ratio of the amplitudes of the re?ected and incident waves, ? R n = b n /a n , we can write a n+1 = a n ? T n ? P n + a n+1 ? R n+1 ? P n (? ? Gamma1 n ) ? P n . (4.309) For n = N we merelyset R N+1 = 0 to ?nd a N+1 = a N ? T N ? P N . (4.310) If we choose to eliminate a n+1 from (4.305) and (4.306) we ?nd that b n = a n ? Gamma1 n + ? R n+1 ? P n (1 ? ? Gamma1 n )a n+1 . (4.311) For n = N this reduces to b N = a N ? Gamma1 N . (4.312) Equations(4.309)and(4.311)havenicephysicalinterpretations.ConsiderFigure4.21, which shows the wave amplitudes for region n. We maythink of the wave incident on interface n + 1 with amplitude a n+1 as consisting of two terms. The ?rst term is the wave transmitted through interface n (at z = z n ). This wave must propagate through a distance Delta1 n to reach interface n+1 and thus has an amplitude a n ? T n ? P n . The second term is the re?ection at interface n of the wave traveling in the ?z direction within region n. The amplitude of the wave before re?ection is merely b n+1 ? P n , where the term ? P n results from the propagation of the negatively-traveling wave from interface n + 1 to interface n. Now, since the interfacial re?ection coe?cient at interface n for a wave incident from region n is the negative of that for a wave incident from region n ? 1 (since the wave is traveling in the reverse direction), and since the re?ected wave must travel through a distance Delta1 n from interface n back to interface n +1, the amplitude of the second term is b n+1 ? P n (?Gamma1 n ) ? P n . Finally, remembering that b n+1 = ? R n+1 a n+1 , we can write a n+1 = a n ? T n ? P n + a n+1 ? R n+1 ? P n (? ? Gamma1 n ) ? P n . This equation is exactlythe same as (4.309) which was found using the boundarycon- ditions. Bysimilar reasoning, we maysaythat the wave traveling in the ?z direction in region n ? 1 consists of a term re?ected from the interface and a term transmitted throughtheinterface. Theamplitudeofthere?ectedtermismerely a n ? Gamma1 n . Theamplitude of the transmitted term is found byconsidering b n+1 = ? R n+1 a n+1 propagated through a distance Delta1 n and then transmitted backwards through interface n. Since the transmission coe?cient for a wave going from region n to region n ? 1 is 1 +(? ? Gamma1 n ), the amplitude of the transmitted term is ? R n+1 ? P n (1 ? ? Gamma1 n )a n+1 . Thus we have b n = ? Gamma1 n a n + ? R n+1 ? P n (1 ? ? Gamma1 n )a n+1 , which is identical to (4.311). We are still left with the task of solving for the various ?eld amplitudes. This can be done using a simple recursive technique. Using ? T n = 1 + ? Gamma1 n we ?nd from (4.309) that a n+1 = (1 + ? Gamma1 n ) ? P n 1 + ? Gamma1 n ? R n+1 ? P 2 n a n . (4.313) Substituting this into (4.311) we ?nd b n = ? Gamma1 n + ? R n+1 ? P 2 n 1 + ? Gamma1 n ? R n+1 ? P 2 n a n . (4.314) Using this expression we ?nd a recursive relationship for the global re?ection coe?cient: ? R n = b n a n = ? Gamma1 n + ? R n+1 ? P 2 n 1 + ? Gamma1 n ? R n+1 ? P 2 n . (4.315) The procedure is now as follows. The global re?ection coe?cient for interface N is, from (4.312), ? R N = b N /a N = ? Gamma1 N . (4.316) This is also obtained from (4.315) with ? R N+1 = 0. We next use (4.315) to ?nd ? R N?1 : ? R N?1 = ? Gamma1 N?1 + ? R N ? P 2 N?1 1 + ? Gamma1 N?1 ? R N ? P 2 N?1 . This process is repeated until reaching ? R 1 , whereupon all of the global re?ection coe?- cients are known. We then ?nd the amplitudes beginning with a 1 , which is the known incident ?eld amplitude. From (4.315) we ?nd b 1 = a 1 ? R 1 , and from (4.313) we ?nd a 2 = (1 + ? Gamma1 1 ) ? P 1 1 + ? Gamma1 1 ? R 2 ? P 2 1 a 1 . This process is repeated until all ?eld 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 = k z,n η n k n when computing the interfacial re?ection coe?cients. See Problem??. As a simple example, consider a slab of material of thickness Delta1 sandwiched between two lossless dielectrics. A time-harmonic uniform plane wave of frequency ω = ˇω is normallyincident onto interface 1, and we wish to compute the amplitude of the wave re?ected byinterface 1 and determine the conditions under which the re?ected wave vanishes. In this case we have N = 2, with two interfaces and three regions. By(4.316) we have R 2 = Gamma1 2 , where R 2 = ? R 2 ( ˇω), Gamma1 2 = ? Gamma1 2 ( ˇω), etc. Then by(4.315) we ?nd R 1 = Gamma1 1 + R 2 P 2 1 1 +Gamma1 1 R 2 P 2 1 = Gamma1 1 +Gamma1 2 P 2 1 1 +Gamma1 1 Gamma1 2 P 2 1 . Hence the re?ected wave vanishes when Gamma1 1 +Gamma1 2 P 2 1 = 0. Since the ?eld in region 0 is normallyincident we have k z,n = k n = β n = ˇω √ μ n epsilon1 n . If we choose P 2 1 =?1, then Gamma1 1 = Gamma1 2 results in no re?ected wave. This requires Z 1 ? Z 0 Z 1 + Z 0 = Z 2 ? Z 1 Z 2 + Z 1 . Clearing the denominator we ?nd that 2Z 2 1 = 2Z 0 Z 2 or Z 1 = radicalbig Z 0 Z 2 . This condition makes the re?ected ?eld vanish if we can ensure that P 2 1 =?1.Todo this we need e ?jβ 1 2Delta1 =?1. The minimum thickness that satis?es this condition is β 1 2Delta1 = π. Since β = 2π/λ, this is equivalent to Delta1 = λ/4. A layer of this type is called aquarter-wavetransformer. Since no wave is re?ected from the initial interface, and since all the regions are assumed lossless, all of the power carried bythe incident wave in the ?rst region is transferred into the third region. Thus, two regionsofdi?eringmaterialsmaybe“matched”byinsertinganappropriateslabbetween Figure 4.22: Interaction of a uniform plane wave with a conductor-backed dielectric slab. them. This technique ?nds use in optical coatings for lenses and for reducing the radar re?ectivityof objects. Asasecondexample,consideralosslessdielectricslabwith ?epsilon1 = epsilon1 1 =epsilon1 1r epsilon1 0 ,and ?μ=μ 0 , backedby aperfectconductorandimmersedinfreespaceasshowninFigure4.22.A perpendicularlypolarized uniform plane wave is incident on the slab from free space and we wish to ?nd the temporal response of the re?ected wave by?rst calculating the frequency-domain re?ected ?eld. Since epsilon1 0 <epsilon1 1 , total internal re?ection cannot occur. Thus the wave vectors in region 1 have real components and can be written as k i 1 = k x,1 ?x + k z,1 ?z, k r 1 = k x,1 ?x ? k z,1 ?z. From Snell’s law of refraction we know that k x,1 = k 0 sinθ i = k 1 sinθ t and so k z,1 = radicalBig k 2 1 ? k 2 x,1 = ω c radicalBig epsilon1 1r ? sin 2 θ i = k 1 cosθ t where θ t is the transmission angle in region 1. Since region 2 is a perfect conductor we have ? R 2 =?1. By(4.315) we have ? R 1 (ω) = Gamma1 1 ? ? P 2 1 (ω) 1 ?Gamma1 1 ? P 2 1 (ω) , (4.317) where from (4.308) Gamma1 1 = Z 1 ? Z 0 Z 1 + Z 0 is not a function of frequency. By the approach we used to obtain (4.300) we write ? E r ⊥ (r,ω)= ?y ? R 1 (ω) ? E i ⊥ (ω)e ?jk r 1 (ω)·r . So E r ⊥ (r,t) = ?yE r parenleftBigg t ? ? k r 1 · r c parenrightBigg where bythe convolution theorem E r (t) = R 1 (t)? E i ⊥ (t). (4.318) Here E i ⊥ (t) = F ?1 braceleftbig ? E i ⊥ (ω) bracerightbig is the time waveform of the incident plane wave, while R 1 (t) = F ?1 braceleftbig ? R 1 (ω) bracerightbig is the global time-domain re?ection coe?cient. To invert ? R 1 (ω), we use the binomial expansion (1 ? x) ?1 = 1 + x + x 2 + x 3 +···on the denominator of (4.317), giving ? R 1 (ω) = [Gamma1 1 ? ? P 2 1 (ω)] braceleftbig 1 + [Gamma1 1 ? P 2 1 (ω)] + [Gamma1 1 ? P 2 1 (ω)] 2 + [Gamma1 1 ? P 2 1 (ω)] 3 +... bracerightbig = Gamma1 1 ? [1 ?Gamma1 2 1 ] ? P 2 1 (ω)? [1 ?Gamma1 2 1 ]Gamma1 1 ? P 4 1 (ω)? [1 ?Gamma1 2 1 ]Gamma1 2 1 ? P 6 1 (ω)?···. (4.319) Thus we need the inverse transform of ? P 2n 1 (ω) = e ?j2nk z,1 Delta1 1 = e ?j2nk 1 Delta1 1 cosθ t . Writing k 1 = ω/v 1 , where v 1 = 1/(μ 0 epsilon1 1 ) 1/2 is the phase velocityof the wave in region 1, and using 1 ? δ(t) along with the time-shifting theorem (A.3) we have ? P 2n 1 (ω) = e ?jω2nτ ? δ(t ? 2nτ) where τ = Delta1 1 cosθ t /v 1 . With this the inverse transform of ? R 1 in (4.319) is R 1 (t) = Gamma1 1 δ(t)?(1 +Gamma1 1 )(1 ?Gamma1 1 )δ(t ? 2τ)?(1 +Gamma1 1 )(1 ?Gamma1 1 )Gamma1 1 δ(t ? 4τ)?··· and thus from (4.318) E r (t) = Gamma1 1 E i ⊥ (t)?(1 +Gamma1 1 )(1 ?Gamma1 1 )E i ⊥ (t ? 2τ)?(1 +Gamma1 1 )(1 ?Gamma1 1 )Gamma1 1 E i ⊥ (t ? 4τ)?···. There?ected?eldconsistsoftime-shiftedandamplitude-scaledversionsoftheincident ?eld waveform. These terms can be interpreted as multiple re?ections of the incident wave.ConsiderFigure4.23.The?rsttermisthedirectre?ectionfrominterface1and thus has its amplitude multiplied by Gamma1 1 . The next term represents a wave that pene- trates the interface (and thus has its amplitude multiplied bythe transmission coe?cient 1 +Gamma1 1 ), propagates to and re?ects 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 coe?cient for passage from region 1 to region 0, 1 ? Gamma1 1 ). The time delaybetween this wave and the initially-re?ected wave is given by 2τ, as discussed in detail below. The third term represents a wave that penetrates the interface, re?ects from the conductor, returns to and re?ects from the interface a second time, again re?ects from the conductor, and then passes through the interface in the opposite direction. Its amplitude has an ad- ditional multiplicative factor of ?Gamma1 1 to account for re?ection from the interface and an additional factor of ?1 to account for the second re?ection from the conductor, and is time-delayed by an additional 2τ. Subsequent terms account for additional re?ections; Figure 4.23: Timing diagram for multiple re?ections from a conductor-backed dielectric slab. the nth re?ected wave amplitude is multiplied byan additional (?1) n and (?Gamma1 1 ) n and is time-delayed by an additional 2nτ. It is important to understand that the time delay 2τ is not just the propagation time for the wave to travel through the slab. To properlydescribe the timing between the initially-re?ected wave and the waves that re?ect from the conductor we must consider the?eldoveridenticalobservationplanesasshowninFigure4.23.Forexample,consider the observation plane designated P-P intersecting the ?rst “exit point” on interface 1. To arrive at this plane the initially-re?ected wave takes the path labeled B, arriving at a time D sinθ i v 0 after the time of initial re?ection, where v 0 = c is 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 2Delta1 1 v 1 cosθ t where v 1 is the wave velocityin region 1 and θ t is the transmission angle. Noting that D = 2Delta1 1 tanθ t , the time delaybetween the arrival of the two waves at the plane P-P is T = 2Delta1 1 v 1 cosθ t ? D sinθ i v 0 = 2Delta1 1 v 1 cosθ t bracketleftbigg 1 ? sinθ t sinθ i v 0 /v 1 bracketrightbigg . BySnell’s law of refraction (4.297) we can write v 0 v 1 = sinθ i sinθ t , which, upon substitution, gives T = 2 Delta1 1 cosθ t v 1 . This is exactlythe time delay 2τ. 4.11.6 Plane-wavepropagationinananisotropicferritemedium 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, and note that this behavior is shared bywaves propagating in a magnetized plasma, because of the similarityin the dyadic constitutive parameters of the two media. Consider a uniform ferrite material having scalar permittivity ?epsilon1 = epsilon1 and dyadic per- meability ? ˉμ. We assume that the ferrite is lossless and magnetized along the z-direction. By(4.115)– (4.117) the permeabilityof the medium is [ ? ˉμ(ω)] = ? ? μ 1 jμ 2 0 ?jμ 2 μ 1 0 00μ 0 ? ? where μ 1 = μ 0 bracketleftbigg 1 + ω M ω 0 ω 2 0 ?ω 2 bracketrightbigg ,μ 2 = μ 0 ωω M ω 2 0 ?ω 2 . The source-free frequency-domain wave equation can be found using (4.201) with ? ˉ ζ = ? ˉ ξ = 0 and ? ˉepsilon1 = epsilon1 ˉ I: bracketleftbigg ˉ ?· parenleftbigg ˉ I 1 epsilon1 parenrightbigg · ˉ ??ω 2 ? ˉμ bracketrightbigg · ? H = 0 or, since ˉ ?·A =?×A, 1 epsilon1 ?× parenleftbig ?× ? H parenrightbig ?ω 2 ? ˉμ· ? 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 ?eld. We thus seek solutions of the form ? H(r,ω)= ? H 0 (ω)e ?jk·r (4.321) where k = ?zβ and ?z · ? H 0 = 0. We can ?nd β byenforcing (4.320). From (B.7) we ?nd that ?× ? H =?jβ?z × ? H 0 e ?jβz . ByAmpere’s law we have ? E = ?× ? H jωepsilon1 =?Z TEM ?z × ? H, (4.322) where Z TEM = β/ωepsilon1 is the wave impedance. Note that the wave is indeed TEM. The second curl is found to be ?× parenleftbig ?× ? H parenrightbig =?jβ?× bracketleftbig ?z × ? H 0 e ?jβz bracketrightbig . After an application of (B.43) this becomes ?× parenleftbig ?× ? H parenrightbig =?jβ bracketleftbig e ?jβz ?×(?z × ? H 0 )?(?z × ? H 0 )×?e ?jβz bracketrightbig . The ?rst term on the right-hand side is zero, and thus using (B.76) we have ?× parenleftbig ?× ? H parenrightbig = bracketleftbig ?jβe ?jβz ?z ×(?z × ? H 0 ) bracketrightbig (?jβ) or, using (B.7), ?× parenleftbig ?× ? H parenrightbig = β 2 e ?jβz ? H 0 since ?z · ? H 0 = 0. With this (4.320) becomes β 2 ? H 0 = ω 2 epsilon1 ? ˉμ· ? H 0 . (4.323) We can solve (4.323) for β bywriting the vector equation in component form: β 2 H 0x = ω 2 epsilon1 bracketleftbig μ 1 H 0x + jμ 2 H 0y bracketrightbig , β 2 H 0y = ω 2 epsilon1 bracketleftbig ?jμ 2 H 0x +μ 1 H 0y bracketrightbig . In matrix form these are bracketleftbigg β 2 ?ω 2 epsilon1μ 1 ?jω 2 epsilon1μ 2 jω 2 epsilon1μ 2 β 2 ?ω 2 epsilon1μ 1 bracketrightbiggbracketleftbigg H 0x H 0y bracketrightbigg = bracketleftbigg 0 0 bracketrightbigg , (4.324) and nontrivial solutions occur onlyif vextendsingle vextendsingle vextendsingle vextendsingle β 2 ?ω 2 epsilon1μ 1 ?jω 2 epsilon1μ 2 jω 2 epsilon1μ 2 β 2 ?ω 2 epsilon1μ 1 vextendsingle vextendsingle vextendsingle vextendsingle = 0. Expansion yields the two solutions β ± = ω √ epsilon1μ ± (4.325) where μ ± = μ 1 ±μ 2 = μ 0 bracketleftbigg 1 + ω M ω 0 ?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 H 0x and H 0y that can be found from (4.324). Substituting β + into the ?rst equation we have ω 2 epsilon1μ 2 H 0x ? jω 2 epsilon1μ 2 H 0y = 0 or H 0x = jH 0y . Similarly, substitution of β ? produces H 0x =?jH 0y . Thus, by(4.321) the magnetic ?eld maybe expressed as ? H(r,ω)= H 0y [±j ?x + ?y]e ?jβ ± z . By(4.322) we also have the electric ?eld ? E(r,ω)= Z TEM H 0y [?x + e ?j π 2 ?y]e ?jβ ± z . This ?eld has the form of (4.248). For β + we have φ y ? φ x =?π/2 and thus the wave exhibits RHCP. For β ? we have φ y ?φ x = π/2 and the wave exhibits LHCP. 012345 β/( /v ) 0 1 2 3 4 ω / ω 0 Light Line RHCP stopband RHCP RHCP LHCP 0 c Figure 4.24: Dispersion plot for unmagnetized ferrite with ω M = 2ω 0 . Light line shows ω/β = v c = 1/(μ 0 epsilon1) 1/2 . ThedispersiondiagramforeachpolarizationcaseisshowninFigure4.24,wherewe havearbitrarilychosen ω M = 2ω 0 . Herewehavecombined(4.325)and(4.326)toproduce the normalized expression β ± ω 0 /v c = ω ω 0 radicalBigg 1 + ω M /ω 0 1 ?ω/ω 0 where v c = 1/(μ 0 epsilon1) 1/2 . Except at low frequencies, an LHCP plane wave passes through the ferrite as if the permeabilityis close to that of free space. Over all frequencies we have v p <v c and v g <v c . In contrast, an RHCP wave excites the electrons in the ferrite and a resonance occurs at ω = ω 0 . For all frequencies below ω 0 we have v p <v c and v g <v c and both v p and v g reduce to zero as ω → ω 0 . Because the ferrite is lossless, frequencies between ω = ω 0 and ω = ω 0 + ω M result in β being purelyimaginaryand thus the wave being evanescent. We thus call the frequencyrange ω 0 <ω<ω 0 + ω M a stopband; within this band the plane wave cannot transport energy. For frequencies above ω 0 +ω M the RHCP wave propagates as if it is in a medium with permeabilityless than that of free space. Here we have v p >v c and v g <v c , with v p → v c and v g → v c as ω →∞. Faradayrotation. The solutions to the wave equation found above do not allow the existence of linearlypolarized plane waves. However, bysuperposing LHCP and RHCP waves we can obtain a wave with the appearance of linear polarization. That is, over any z-plane the electric ?eld vector maybe written as ? E = K(E x0 ?x + E y0 ?y) where E x0 and E y0 are real (although K maybe complex). To see this let us examine ? E = ? E + + ? E ? = E 0 2 [?x ? j ?y]e ?jβ + z + E 0 2 [?x + j ?y]e ?jβ ? z = E 0 2 bracketleftbig ?x parenleftbig e ?jβ + z + e ?jβ ? z parenrightbig + j ?y parenleftbig ?e ?jβ + z + e ?jβ ? z parenrightbigbracketrightbig = E 0 e ?j 1 2 (β + +β ? )z bracketleftbigg ?x cos 1 2 (β + ?β ? )z + ?y sin 1 2 (β + ?β ? )z bracketrightbigg or ? E = E 0 e ?j 1 2 (β + +β ? )z [?x cosθ(z)+ ?y sinθ(z)] where θ(z) = (β + ? β ? )z/2. Because β + negationslash= β ? , the velocities of the two circularly polarized waves di?er and the waves superpose to form a linearlypolarized wave with a polarization that depends on the observation plane z-value. We maythink of the wave as undergoing a phase shift of (β + +β ? )z/2 radians as it propagates, while the direction of ? E rotates to an angle θ(z) = (β + ?β ? )z/2 as the wave propagates. Faraday rotation can onlyoccur at frequencies where both the LHCP and RHCP waves propagate, and therefore not within the stopband ω 0 <ω<ω 0 +ω M . Faradayrotation is non-reciprocal. That is, if a wave that has undergone a rotation of θ 0 radians bypropagating through a distance z 0 is made to propagate an equal distance back in the direction from whence it came, the polarization does not return to its initial state 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 2θ 0 . This e?ect is employed in a number of microwave devices including gyrators, isolators, and circulators. The interested reader should see Collin [40], Elliott [67], or Liao [111] for details. We note that for ω greatermuch ω M we can approximate the rotation angle as θ(z) = (β + ?β ? )z/2 = 1 2 ωz √ epsilon1μ 0 bracketleftbiggradicalbigg 1 + ω M ω 0 ?ω ? radicalbigg 1 + ω M ω 0 +ω bracketrightbigg ≈? 1 2 zω M √ epsilon1μ 0 , 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 ?nd 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 ≈ ω 0 )[40].SeeProblem 4.16. 4.11.7 Propagationofcylindricalwaves Bystudying 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 having cylindrical and spherical symmetries. Uniformcylindricalwaves. 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. Considerahomogeneousregionofspacedescribedbythepermittivity ?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 uniformcylindrical waves. Since the ?elds 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 ?nd ? H φ (ρ,ω) = 1 jω ?μ(ω) ? ? E z (ρ,ω) ?ρ . (4.327) For TE polarization we have from (4.213) ? E φ (ρ,ω) =? 1 jω?epsilon1 c (ω) ? ? H z (ρ,ω) ?ρ (4.328) where ?epsilon1 c = ?epsilon1 + ?σ/jω is the complex permittivityintroduced in § 4.4.1. Since ? E = ? φ ? E φ + ?z ? E z and ? H = ? φ ? H φ + ?z ? H z , we can always decompose a cylindrical electromagnetic wave into cases of electric and magnetic polarization. In each case the resulting ?eld is TEM ρ since ? E, ? H, and ?ρ are mutuallyorthogonal. Wave equations for ? E z in the electric polarization case and for ? H z in the magnetic polarization case can be derived bysubstituting (4.210) into (4.208): parenleftbigg ? 2 ?ρ 2 + 1 ρ ? ?ρ + k 2 parenrightbiggbraceleftbigg ? E z ? H z bracerightbigg = 0. Thus the electric ?eld must obeythe ordinarydi?erential equation d 2 ? E z dρ 2 + 1 ρ d ? E z dρ + k 2 ? 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 J 0 (kρ), Y 0 (kρ), H (1) 0 (kρ), H (2) 0 (kρ). We ?nd that J 0 (kρ) and Y 0 (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 behavior of outward propagating waves and so we choose ? E z (ρ,ω) =? j 4 ? E z0 (ω)H (2) 0 (kρ). (4.330) As explained in § 2.10.7, ? E z0 (ω) is the amplitude spectrum of the wave, while the term ?j/4 is included to make the conversion to the time domain more convenient. By(4.327) we have ? H φ = 1 jω ?μ ? ? E z ?ρ = 1 jω ?μ ? ?ρ bracketleftbigg ? j 4 ? E z0 H (2) 0 (kρ) bracketrightbigg . (4.331) Using dH (2) 0 (x)/dx =?H (2) 1 (x) we ?nd that ? H φ = 1 Z TM ? E z0 4 H (2) 1 (kρ) (4.332) where Z TM = ω ?μ k is called the TMwaveimpedance. For the case of magnetic polarization, the ?eld ? H z must satisfyBessel’s equation (4.329). Thus we choose ? H z (ρ,ω) =? j 4 ? H z0 (ω)H (2) 0 (kρ). (4.333) From (4.328) we ?nd the electric ?eld associated with the wave: ? E φ =?Z TE ? H z0 4 H (2) 1 (kρ), (4.334) where Z TE = k ω?epsilon1 c is the TEwaveimpedance. It is not readilyapparent that the terms H (2) 0 (kρ) or H (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 possible directions. Each of these components does have the expected wave behavior, 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/(μepsilon1) 1/2 . For time-harmonic ?elds, the cylindrical wave takes on a familiar behavior when the observation point is su?cientlyremoved from the source. We mayspecialize (4.330) to the time-harmonic case bysetting ω = ˇω and using phasors, giving ˇ E z (ρ) =? j 4 ˇ E z0 H (2) 0 (kρ). If |kρ|greatermuch1 we can use the asymptotic representation (E.62) for the Hankel function H (2) ν (z) ～ radicalbigg 2 πz e ?j(z?π/4?νπ/2) , |z|greatermuch1, ?2π<arg(z)<π, to obtain ˇ E z (ρ) ～ ˇ E z0 e ?jkρ √ 8 jπkρ (4.335) and ˇ H φ (ρ) ～? ˇ E z0 1 Z TM e ?jkρ √ 8 jπkρ (4.336) for |kρ|greatermuch1. Except for the √ ρ term 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 ˇ H z (ρ) ～ ˇ H z0 e ?jkρ √ 8 jπkρ (4.337) and ˇ E φ (ρ) ～ Z TE ˇ H z0 e ?jkρ √ 8 jπkρ (4.338) for |kρ|greatermuch1. To interpret the wave nature of the ?eld (4.335) let us substitute k = β ? jα into the exponential function, where β is the phase constant (4.224) and α is the attenuation constant (4.225). Then ˇ E z (ρ) ～ ˇ E z0 1 √ 8 jπkρ e ?αρ e ?jβρ . Assuming ˇ E z0 =|E z0 |e jξ E , the time-domain representation is found from (4.126): E z (ρ,t) = |E z0 | √ 8πkρ e ?αρ cos[ ˇωt ?βρ ?π/4 +ξ E ]. (4.339) We can identifya surface of constant phase as a locus of points obeying ˇωt ?βρ ?π/4 +ξ E = C P (4.340) where C P is 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 = C A where C A is some constant, are also cylinders. The cosine term in (4.339) represents a traveling wave. As t is increased the argument of the cosine function remains ?xed as long as ρ is increased correspondingly. Hence the cylindrical wavefronts propagate outward as time progresses. As the wavefront travels outward, the ?eld is attenuated because of the factor e ?αρ . The velocityof propagation of the phase fronts maybe computed bya now familiar technique. Di?erentiating (4.340) with respect to t we ?nd that ˇω ?β dρ dt = 0, and thus have the phase velocity v p of the outward expanding phase fronts: v p = 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 ?nd βρ 1 = βρ 2 ? 2π or λ = ρ 2 ?ρ 1 = 2π/β. Computation of the power carried bya cylindrical wave is straightforward. Since a cylindrical wavefront is in?nite in extent, we usually speak of the power per unit length carried bythe wave. This is found byintegrating the time-average Poynting ?ux given in (4.157). For electric polarization we ?nd the time-average power ?ux densityusing (4.330) and (4.331): S av = 1 2 Re{ ˇ E z ?z × ˇ H ? φ ? φ}= 1 2 Re braceleftbigg ?ρ j 16Z ? TM | ˇ E z0 | 2 H (2) 0 (kρ)H (2)? 1 (kρ) bracerightbigg . (4.341) For magnetic polarization we use (4.333) and (4.334): S av = 1 2 Re{ ˇ E φ ? φ× ˇ H ? z ?z}= 1 2 Re braceleftbigg ??ρ jZ TE 16 | ˇ H z0 | 2 H (2)? 0 (kρ)H (2) 1 (kρ) bracerightbigg . For a lossless medium these expressions can be greatlysimpli?ed. By(E.5) we can write jH (2) 0 (kρ)H (2)? 1 (kρ)= j[J 0 (kρ)? jN 0 (kρ)][J 1 (kρ)+ jN 1 (kρ)], hence jH (2) 0 (kρ)H (2)? 1 (kρ)= [N 0 (kρ)J 1 (kρ)? J 0 (kρ)N 1 (kρ)] + j[J 0 (kρ)J 1 (kρ)+ N 0 (kρ)N 1 (kρ)]. Substituting this into (4.341) and remembering that Z TM = η = (μ/epsilon1) 1/2 is real for lossless media, we have S av = ?ρ 1 32η | ˇ E z0 | 2 [N 0 (kρ)J 1 (kρ)? J 0 (kρ)N 1 (kρ)]. Bythe Wronskian relation (E.88) we have S av = ?ρ | ˇ E z0 | 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: P av /l = integraldisplay 2π 0 S av · ?ρρ dφ = | ˇ E z0 | 2 8kη . (4.342) Foralosslessmediumthereisnomechanismtodissipatethepowerandsothewave prop- agates unabated. A similar calculation for the case of magnetic polarization (Problem ??) gives S av = ?ρ η| ˇ H z0 | 2 16πkρ and P av /l = η| ˇ H z0 | 2 8k . Foralossymediumtheexpressionsaremoredi?culttoevaluate. Inthiscaseweexpect the total power passing through a cylinder to depend on the radius of the cylinder, since the ?elds 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 2 Re{ ˇ E z ?z × ˇ H ? φ ? φ}= 1 2 Re braceleftbigg ?ρ e ?2αρ 8πρ|k|Z ? TM | ˇ E z0 | 2 bracerightbigg . Therefore P av /l = integraldisplay 2π 0 S av · ?ρρ dφ = Re braceleftbigg 1 Z ? TM bracerightbigg | ˇ E z0 | 2 e ?2αρ 8|k| . We note that for a lossless material Z TM = η and α = 0, and the expression reduces to (4.342) as expected. Thus for lossymaterials the power depends on the radius of the cylinder. In the case of magnetic polarization we use (4.337) and (4.338) to get S av = 1 2 Re{ ˇ E φ ? φ× ˇ H ? z ?z}= 1 2 Re braceleftbigg ?ρZ ? TE e ?2αρ 8πρ|k| | ˇ H z0 | 2 bracerightbigg and P av /l = Re braceleftbig Z ? TE bracerightbig | ˇ H z0 | 2 e ?2αρ 8|k| . Example of uniform cylindrical waves: ?elds of a line source. The simplest example of a uniform cylindrical wave is that produced byan electric or magnetic line source. Consider ?rst an in?nite 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 ?eld produced bythe line source into TE and TM cases according to § 4.11.2. It turns out that an electric line source onlyexcites TM ?elds, as we shall show in § 5.4, and thus we need only ? E z to completelydescribe the ?elds. Bysymmetrythe ?elds are φ-independent and thus the wave produced bythe line source is a uniform cylindrical wave. Since the wave propagates outward from the line source we have the electric ?eld from (4.330), ? E z (ρ,ω) =? j 4 ? E z0 (ω)H (2) 0 (kρ), (4.343) and the magnetic ?eld from (4.332), ? H φ (ρ,ω) = k ω ?μ ? E z0 (ω) 4 H (2) 1 (kρ). We can ?nd ? E z0 byusing Ampere’s law: contintegraldisplay Gamma1 ? H · dl = integraldisplay S ? J · dS + jω integraldisplay S ? D · dS. Since ? J is the sum of the impressed current ? I and the secondaryconduction current ?σ ? E, we can also write contintegraldisplay Gamma1 ? H · dl = ? I + integraldisplay S (?σ + jω?epsilon1) ? E · dS = ? I + jω?epsilon1 c integraldisplay S ? E · dS. Choosingourpathofintegrationasacircleofradius a inthe z = 0 planeandsubstituting for ? E z and ? H φ , we ?nd that k ω ?μ ? E z0 4 H (2) 1 (ka)2πa = ? I + jω?epsilon1 c 2π ?j ? E z0 4 lim δ→0 integraldisplay a δ H (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 integraldisplay a δ H (2) 0 (kρ)ρdρ = a k H (2) 1 (ka)? 1 k lim δ→0 δH (2) 1 (kδ). The limit maybe found byusing H (2) 1 (x) = J 1 (x) ? jN 1 (x) and the small argument approximations (E.50) and (E.53): lim δ→0 δH (2) 1 (δ) = lim δ→0 δ bracketleftbigg kδ 2 ? j parenleftbigg ? 1 π 2 kδ parenrightbiggbracketrightbigg = j 2 πk . Substituting these expressions into (4.344) we obtain k ω ?μ ? E z0 4 H (2) 1 (ka)2πa = ? I + jω?epsilon1 c 2π ?j ? E z0 4 bracketleftbigg a k H (2) 1 (ka)? j 2 πk 2 bracketrightbigg . Using k 2 = ω 2 ?μ?epsilon1 c we ?nd that the two Hankel function terms cancel. Solving for ? E z0 we have ? E z0 =?jω ?μ ? I and therefore ? E z (ρ,ω) =? ω ?μ 4 ? I(ω)H (2) 0 (kρ)=?jω ?μ ? I(ω) ? G(x, y|0,0;ω). (4.345) Here ? G is called the two-dimensionalGreen’sfunction and is given by ? G(x, y|x prime , y prime ;ω) = 1 4 j H (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 ?eld 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 ? E z vanishes as a → 0 without having to perform the calculation directly[83, 8]. For a magnetic line source ? I m (ω) aligned along the z-axis we proceed as above, but note that the source onlyproduces TE ?elds. By(4.333) and (4.334) we have ? H z (ρ,ω) =? j 4 ? H z0 (ω)H (2) 0 (kρ), ? E φ =? k ω?epsilon1 c ? H 0z 4 H (2) 1 (kρ). We can ?nd ? H z0 by applying Faraday’s law contintegraldisplay C ? E · dl =? integraldisplay S ? J m · dS ? jω integraldisplay S ? B · dS about a circle of radius a in the z = 0 plane. We have ? k ω?epsilon1 c ? H z0 4 H (2) 1 (ka)2πa =? ? I m ? jω ?μ bracketleftbigg ? j 4 bracketrightbigg ? H z0 2π lim δ→0 integraldisplay a δ H (2) 0 (kρ)ρdρ. Proceeding as above we ?nd that ? H z0 = jω?epsilon1 c ? I m hence ? H z (ρ,ω) =? ω?epsilon1 c 4 ? I m (ω)H (2) 0 (kρ)=?jω?epsilon1 c ? I m (ω) ? G(x, y|0,0;ω). (4.347) Note that we could have solved for the magnetic ?eld of a magnetic line current by usingthe?eldofanelectriclinecurrentandtheprincipleofduality. Lettingthemagnetic current be equal to ?η times the electric current and using (4.198), we ?nd that ? H z0 = parenleftbigg ? 1 η ? I m (ω) ? I(ω) parenrightbiggparenleftbigg ? 1 η bracketleftbigg ? ω ?μ 4 ? I(ω)H (2) 0 (kρ) bracketrightbiggparenrightbigg =? ? I m (ω) ω?epsilon1 c 4 H (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), ? H ρ = j Z TM k 1 ρ ? ? E z ?φ , (4.349) ? H φ =? j Z TM k ? ? E z ?ρ , (4.350) where Z TM = ω ?μ/k. For the TE case we have, by(4.213), ? E ρ =? jZ TE k 1 ρ ? ? H z ?φ , (4.351) ? E φ = jZ TE k ? ? H z ?ρ , (4.352) where Z TE = k/ω?epsilon1 c . By(4.208) the wave equations are parenleftbigg ? 2 ?ρ 2 + 1 ρ ? ?ρ + 1 ρ 2 ? 2 ?φ 2 + k 2 parenrightbiggbraceleftbigg ? E z ? H z bracerightbigg = 0. Because this has the form of A.177 with ?/?z → 0,wehave braceleftbigg ? E z (ρ,φ,ω) ? H z (ρ,φ,ω) bracerightbigg = P(ρ,ω)Phi1(φ,ω) (4.353) where Phi1(φ,ω) = A φ (ω)sin k φ φ + B φ (ω)cos k φ φ, (4.354) P(ρ) = A ρ (ω)B (1) k φ (kρ)+ B ρ (ω)B (2) k φ (kρ), (4.355) and where B (1) ν (z) and B (2) ν (z) are anytwo independent Bessel functions chosen from the set J ν (z), N ν (z), H (1) ν (z), H (2) ν (z). Inboundedregionswegenerallyusetheoscillatoryfunctions J ν (z) and N ν (z) torepresent standing waves. In unbounded regions we generallyuse H (2) ν (z) and H (1) ν (z) to represent outward and inward propagating waves, respectively. Boundaryvalueproblemsincylindricalcoordinates: scatteringbyamaterial cylinder. A varietyof problems can be solved using nonuniform cylindrical waves. We shall examine two interesting cases in which an external ?eld is impressed on a two-dimensional object. The impressed ?eld creates secondarysources within or on the object, and these in turn create a secondary?eld. Our goal is to determine the secondary ?eld byapplying appropriate boundaryconditions. As a ?rst example, consider a material cylinder of radius a, complex permittivity ?epsilon1 c , andpermeability ?μ,alignedalongthe z-axisinfreespace(Figure4.25).Anincident plane wave propagating in the x-direction is impressed on the cylinder, inducing sec- ondarypolarization and conduction currents within the cylinder. These in turn produce Figure 4.25: TM plane-wave ?eld incident on a material cylinder. secondaryor scattered ?elds, which are standing waves within the cylinder and outward traveling waves external to the cylinder. Although we have not yet learned how to write the secondary?elds in terms of the impressed sources, we can solve for the ?elds as a boundaryvalue problem. The total ?eld must obeythe boundaryconditions on tangen- tial components at the interface between the cylinder and surrounding free space. We need not worryabout the e?ect of the secondarysources on the source of the primary ?eld, since byde?nition impressed sources cannot be in?uenced bysecondary?elds. Thescattered?eldcanbefoundusingsuperposition. WhenexcitedbyaTMimpressed ?eld, the secondary?eld is also TM. The situation for TE excitation is similar. By decomposing the impressed ?eld into TE and TM components, we maysolve for the scattered ?eld in each case and then superpose the results to determine the complete solution. We ?rst consider the TM case. The impressed electric ?eld maybe written as ? E i (r,ω)= ?z ? E 0 (ω)e ?jk 0 x = ?z ? E 0 (ω)e ?jk 0 ρ cosφ (4.356) while the magnetic ?eld is, by(4.223), ? H i (r,ω)=??y ? E 0 (ω) η 0 e ?jk 0 x =?(?ρsinφ + ? φcosφ) ? E 0 (ω) η 0 e ?jk 0 ρ cosφ . Here k 0 = ω(μ 0 epsilon1 0 ) 1/2 and η 0 = (μ 0 /epsilon1 0 ) 1/2 . The scattered electric ?eld takes the form of a nonuniform cylindrical wave (4.353). Periodicityin φ implies that k φ is an integer, say k φ = n. Within the cylinder we cannot use any of the functions N n (kρ), H (2) n (kρ), or H (1) n (kρ) to represent the radial dependence of the ?eld, since each is singular at the origin. So we choose B (1) n (kρ)= J n (kρ) and B ρ (ω) = 0 in (4.355). Physically, J n (kρ) rep- resents the standing wave created bythe interaction of outward and inward propagating waves. External to the cylinder we use H (2) n (kρ)to represent the radial dependence of the secondary?eld components: we avoid N n (kρ) and J n (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 φ of each individual nonuniform wave cannot match the more complicated dependence of the impressed ?eld (4.356). Since the sinusoids are complete, an in?nite series of the functions (4.353) can be used to represent the scattered ?eld. So we have internal to the cylinder ? E s z (r,ω)= ∞ summationdisplay n=0 [A n (ω)sin nφ + B n (ω)cos nφ] J n (kρ) where k = ω(?μ?epsilon1 c ) 1/2 . External to the cylinder we have free space and thus ? E s z (r,ω)= ∞ summationdisplay n=0 [C n (ω)sin nφ + D n (ω)cos nφ] H (2) n (k 0 ρ). Equations (4.349) and (4.350) yield the magnetic ?eld internal to the cylinder: ? H s ρ = ∞ summationdisplay n=0 jn Z TM k 1 ρ [A n (ω)cos nφ ? B n (ω)sin nφ] J n (kρ), ? H s φ =? ∞ summationdisplay n=0 j Z TM [A n (ω)sin nφ + B n (ω)cos nφ] J prime n (kρ), where Z TM = ω ?μ/k. Outside the cylinder ? H s ρ = ∞ summationdisplay n=0 jn η 0 k 0 1 ρ [C n (ω)cos nφ ? D n (ω)sin nφ] H (2) n (k 0 ρ), ? H s φ =? ∞ summationdisplay n=0 j η 0 [C n (ω)sin nφ + D n (ω)cos nφ] H (2)prime n (k 0 ρ), where J prime n (z) = dJ n (z)/dz and H (2)prime n (z) = dH (2) n (z)/dz. We have two sets of unknown spectral amplitudes (A n , B n ) and (C n , D n ). These can be determined byapplying the boundaryconditions at the interface. Since the total ?eld outside the cylinder is the sum of the impressed and scattered terms, an application of continuityof the tangential electric ?eld at ρ = a gives us ∞ summationdisplay n=0 [A n sin nφ + B n cos nφ] J n (ka) = ∞ summationdisplay n=0 [C n sin nφ + D n cos nφ] H (2) n (k 0 a)+ ? E 0 e ?jk 0 a cosφ , which must hold for all ?π ≤ φ ≤ π. To remove the coe?cients from the sum we apply orthogonality. Multiplying both sides bysin mφ, integrating over [?π,π], and using the orthogonalityconditions (A.129)–(A.131) we obtain π A m J m (ka)?πC m H (2) m (k 0 a) = ? E 0 integraldisplay π ?π sin mφe ?jk 0 a cosφ dφ = 0. (4.357) Multiplying by cos mφ and integrating, we ?nd that 2π B m J m (ka)? 2π D m H (2) m (k 0 a) = ? E 0 epsilon1 m integraldisplay π ?π cos mφe ?jk 0 a cosφ dφ = 2π ? E 0 epsilon1 m j ?m J m (k 0 a) (4.358) where epsilon1 n is 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 ?eld ? H φ at ρ = a.Thus ? ∞ summationdisplay n=0 j Z TM [A n sin nφ + B n cos nφ] J prime n (ka) = ? ∞ summationdisplay n=0 j η 0 [C n sin nφ + D n cos nφ] H (2)prime n (k 0 a)? cosφ ? E 0 η 0 e ?jk 0 a cosφ must hold for all ?π ≤ φ ≤ π. Byorthogonality π j Z TM A m J prime m (ka)?π j η 0 C m H (2)prime m (k 0 a) = ? E 0 η 0 integraldisplay π ?π sin mφ cosφe ?jk 0 a cosφ dφ = 0 (4.359) and 2π j Z TM B m J prime m (ka)? 2π j η 0 D m H (2)prime m (k 0 a) = epsilon1 m ? E 0 η 0 integraldisplay π ?π cos mφ cosφe ?jk 0 a cosφ dφ. The integral maybe computed as integraldisplay π ?π cos mφ cosφe ?jk 0 a cosφ dφ = j d d(k 0 a) integraldisplay π ?π cos mφe ?jk 0 a cosφ dφ = j2πj ?m J prime m (k 0 a) and thus 1 Z TM B m J prime m (ka)? 1 η 0 D m H (2)prime m (k 0 a) = ? E 0 η 0 epsilon1 m j ?m J prime m (k 0 a). (4.360) We now have four equations for the coe?cients A n , B n ,C n , D n . We maywrite (4.357) and (4.359) as bracketleftbigg J m (ka) ?H (2) m (k 0 a) η 0 Z TM J prime m (ka) ?H (2)prime m (k 0 a) bracketrightbiggbracketleftbigg A m C m bracketrightbigg = 0, (4.361) and (4.358) and (4.360) as bracketleftbigg J m (ka) ?H (2) m (k 0 a) η 0 Z TM J prime m (ka) ?H (2)prime m (k 0 a) bracketrightbiggbracketleftbigg B m D m bracketrightbigg = bracketleftbigg ? E 0 epsilon1 m j ?m J m (k 0 a) ? E 0 epsilon1 m j ?m J prime m (k 0 a) bracketrightbigg . (4.362) Matrix equations (4.361) and (4.362) cannot hold simultaneouslyunless A m = C m = 0. Then the solution to (4.362) is B m = ? E 0 epsilon1 m j ?m bracketleftBigg H (2) m (k 0 a)J prime m (k 0 a)? J m (k 0 a)H (2)prime m (k 0 a) η 0 Z TM J prime m (ka)H (2) m (k 0 a)? H (2)prime m (k 0 a)J m (ka) bracketrightBigg , (4.363) D m =? ? E 0 epsilon1 m j ?m bracketleftBigg η 0 Z TM J prime m (ka)J m (k 0 a)? J prime m (k 0 a)J m (ka) η 0 Z TM J prime m (ka)H (2) m (k 0 a)? H (2)prime m (k 0 a)J m (ka) bracketrightBigg . (4.364) With these coe?cients we can calculate the ?eld inside the cylinder (ρ ≤ a) from ? E z (r,ω)= ∞ summationdisplay n=0 B n (ω)J n (kρ)cos nφ, ? H ρ (r,ω)=? ∞ summationdisplay n=0 jn Z TM k 1 ρ B n (ω)J n (kρ)sin nφ, ? H φ (r,ω)=? ∞ summationdisplay n=0 j Z TM B n (ω)J prime n (kρ)cos nφ, and the ?eld outside the cylinder (ρ > a) from ? E z (r,ω)= ? E 0 (ω)e ?jk 0 ρ cosφ + ∞ summationdisplay n=0 D n (ω)H (2) n (k 0 ρ)cos nφ, ? H ρ (r,ω)=?sinφ ? E 0 (ω) η 0 e ?jk 0 ρ cosφ ? ∞ summationdisplay n=0 jn η 0 k 0 1 ρ D n (ω)H (2) n (k 0 ρ)sin nφ, ? H φ (r,ω)=?cosφ ? E 0 (ω) η 0 e ?jk 0 ρ cosφ ? ∞ summationdisplay n=0 j η 0 D n (ω)H (2)prime n (k 0 ρ)cos nφ. We can easilyspecialize these results to the case of a perfectlyconducting cylinder by allowing ?σ →∞. Then η 0 Z TM = radicalBigg μ 0 ?epsilon1 c ?μepsilon1 0 →∞ and B n → 0, D n →? ? E 0 epsilon1 m j ?m J m (k 0 a) H (2) m (k 0 a) . Inthiscaseitisconvenienttocombinetheformulasfortheimpressedandscattered?elds when forming the total ?elds. Since the impressed ?eld is z-independent and obeys the homogeneousHelmholtzequation, wemayrepresentitintermsofnonuniformcylindrical waves: ? E i z = ? E 0 e ?jk 0 ρ cosφ = ∞ summationdisplay n=0 [E n sin nφ + F n cos nφ] J n (k 0 ρ), where we have chosen the Bessel function J n (k 0 ρ) since the ?eld is ?nite at the origin and periodic in φ. Applying orthogonality we see immediately that E n = 0 and that 2π epsilon1 m F m J m (k 0 ρ)= ? E 0 integraldisplay π ?π cos mφe ?jk 0 ρ cosφ dφ = ? E 0 2πj ?m J m (k 0 ρ). Thus, F n = ? E 0 epsilon1 n j ?n and ? E i z = ∞ summationdisplay n=0 ? E 0 epsilon1 n j ?n J n (k 0 ρ)cos nφ. Adding this impressed ?eld to the scattered ?eld we have the total ?eld outside the cylinder, ? E z = ? E 0 ∞ summationdisplay n=0 epsilon1 n j ?n H (2) n (k 0 a) bracketleftbig J n (k 0 ρ)H (2) n (k 0 a)? J n (k 0 a)H (2) n (k 0 ρ) bracketrightbig cos nφ, while the ?eld within the cylinder vanishes. Then, by (4.350), ? H φ =? j η 0 ? E 0 ∞ summationdisplay n=0 epsilon1 n j ?n H (2) n (k 0 a) bracketleftbig J prime n (k 0 ρ)H (2) n (k 0 a)? J n (k 0 a)H (2)prime n (k 0 ρ) bracketrightbig cos nφ. 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 ? J s = ?n × ? H| ρ=a = ?ρ× [ ?ρ ? H ρ + ? φ ? H φ ]| ρ=a = ?z ? H φ | ρ=a we have J s (φ,ω) =? j η 0 ?z ? E 0 ∞ summationdisplay n=0 epsilon1 n j ?n H (2) n (k 0 a) bracketleftbig J prime n (k 0 a)H (2) n (k 0 a)? J n (k 0 a)H (2)prime n (k 0 a) bracketrightbig cos nφ, and an application of (E.93) gives us J s (φ,ω) = ?z 2 ? E 0 η 0 k 0 πa ∞ summationdisplay n=0 epsilon1 n j ?n H (2) n (k 0 a) cos nφ. (4.365) Computation of the scattered ?eld for a magnetically-polarized impressed ?eld pro- ceeds in the same manner. The impressed electric and magnetic ?elds are assumed to be ? E i (r,ω)= ?y ? E 0 (ω)e ?jk 0 x = (?ρsinφ + ? φcosφ) ? E 0 (ω)e ?jk 0 ρ cosφ , ? H i (r,ω)= ?z ? E 0 (ω) η 0 e ?jk 0 x = ?z ? E 0 (ω) η 0 e ?jk 0 ρ cosφ . For a perfectlyconducting cylinder, the total magnetic ?eld is ? H z = ? E 0 η 0 ∞ summationdisplay n=0 epsilon1 n j ?n H (2)prime n (k 0 a) bracketleftbig J n (k 0 ρ)H (2)prime n (k 0 a)? J prime n (k 0 a)H (2) n (k 0 ρ) bracketrightbig cos nφ. (4.366) The details are left as an exercise. Boundaryvalueproblemsincylindricalcoordinates: scatteringbyaperfectly conductingwedge. As a second example, consider a perfectlyconducting wedge im- mersedinfreespaceandilluminatedby alinesource(Figure4.26)carry ingcurrent ? I(ω) and located at (ρ 0 ,φ 0 ). 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) ?eld. This scattered ?eld, also z-invariant, can be found bysolving a bound- aryvalue problem. We do this byseparating space into the two regions ρ<ρ 0 and ρ>ρ 0 , 0 <φ<ψ. Each of these is source-free, so we can represent the total ?eld using nonuniform cylindrical waves of the type (4.353). The line source is brought into the problem byapplying the boundarycondition on the tangential magnetic ?eld across the cylindrical surface ρ = ρ 0 . Since the impressed electric ?eld has onlya z-component, so do the scattered and total electric ?elds. We wish to represent the total ?eld ? E z in terms of nonuniform cylindrical waves of the type (4.353). Since the ?eld is not periodic in φ, the separation constant k φ need not be an integer; instead, its value is determined bythe positions of the wedge boundaries. For the region ρ<ρ 0 we represent the radial dependence of the ?eld using the functions J ν since the ?eld must be ?nite at the origin. For ρ>ρ 0 we use the outward-propagating wave functions H (2) δ .Thus ? E z (ρ,φ,ω) = braceleftBigg summationtext ν [A ν sinνφ+ B ν cosνφ] J ν (k 0 ρ), ρ < ρ 0 , summationtext δ [C δ sinδφ+ D δ cosδφ] H (2) δ (k 0 ρ), ρ > ρ 0 . (4.367) The coe?cients A ν , B ν ,C δ , D δ and separation constants ν,δ maybe found byapplying the boundaryconditions on the ?elds at the surface of the wedge and across the surface ρ = ρ 0 . On the wedge face at φ = 0 we must have ? E z = 0, hence B ν = D δ = 0.On the wedge face at φ = ψ we must also have ? E z = 0, requiring sinνψ = sinδψ = 0 and therefore ν = δ = ν n = nπ/ψ, n = 1,2,.... So ? E z = braceleftBigg summationtext ∞ n=0 A n sinν n φJ ν n (k 0 ρ), ρ < ρ 0 , summationtext ∞ n=0 C n sinν n φH (2) ν n (k 0 ρ), ρ > ρ 0 . (4.368) The magnetic ?eld can be found from (4.349)–(4.350): ? H ρ = braceleftBigg summationtext ∞ n=0 A n j η 0 k 0 ν n ρ cosν n φJ ν n (k 0 ρ), ρ < ρ 0 , summationtext ∞ n=0 C n j η 0 k 0 ν n ρ cosν n φH (2) ν n (k 0 ρ), ρ > ρ 0 , (4.369) ? H φ = braceleftBigg ? summationtext ∞ n=0 A n j η 0 sinν n φJ prime ν n (k 0 ρ), ρ < ρ 0 , ? summationtext ∞ n=0 C n j η 0 sinν n φH (2)prime ν n (k 0 ρ), ρ > ρ 0 . (4.370) The coe?cients A n and C n are found byapplying the boundaryconditions at ρ = ρ 0 . Bycontinuityof the tangential electric ?eld ∞ summationdisplay n=0 A n sinν n φJ ν n (k 0 ρ 0 ) = ∞ summationdisplay n=0 C n sinν n φH (2) ν n (k 0 ρ 0 ). We now applyorthogonalityover the interval [0,ψ]. Multiplying by sinν m φ and inte- grating we have ∞ summationdisplay n=0 A n J ν n (k 0 ρ 0 ) integraldisplay ψ 0 sinν n φ sinν m φ dφ = ∞ summationdisplay n=0 C n H (2) ν n (k 0 ρ 0 ) integraldisplay ψ 0 sinν n φ sinν m φ dφ. Setting u = φπ/ψ we have integraldisplay ψ 0 sinν n φ sinν m φ dφ = ψ π integraldisplay π 0 sin nu sin mu du = ψ 2 δ mn , thus A m J ν m (k 0 ρ 0 ) = C m H (2) ν m (k 0 ρ 0 ). (4.371) The boundarycondition ?n 12 ×( ? H 1 ? ? H 2 ) = ? J s requires the surface current at ρ = ρ 0 .We can write the line current in terms of a surface current densityusing the δ-function: ? J s = ?z ? I δ(φ?φ 0 ) ρ 0 . This is easilyveri?ed as the correct expression since the integral of this densityalong the circular arc at ρ = ρ 0 returns the correct value ? I for the total current. Thus the boundarycondition requires ? H φ (ρ + 0 ,φ,ω)? ? H φ (ρ ? 0 ,φ,ω)= ? I δ(φ?φ 0 ) ρ 0 . By(4.370) we have ? ∞ summationdisplay n=0 C n j η 0 sinν n φH (2)prime ν n (k 0 ρ 0 )+ ∞ summationdisplay n=0 A n j η 0 sinν n φJ prime ν n (k 0 ρ 0 ) = ? I δ(φ?φ 0 ) ρ 0 and orthogonalityyields ? C m ψ 2 j η 0 H (2)prime ν m (k 0 ρ 0 )+ A m ψ 2 j η 0 J prime ν m (k 0 ρ 0 ) = ? I sinν m φ 0 ρ 0 . (4.372) The coe?cients A m and C m thus obeythe matrix equation bracketleftbigg J ν m (k 0 ρ 0 ) ?H (2) ν m (k 0 ρ 0 ) J prime ν m (k 0 ρ 0 ) ?H (2)prime ν m (k 0 ρ 0 ) bracketrightbiggbracketleftbigg A m C m bracketrightbigg = bracketleftbigg 0 ?j2 ? I η 0 ψ sinν m φ 0 ρ 0 bracketrightbigg and are A m = j2 ? I η 0 ψ sinν m φ 0 ρ 0 H (2) ν m (k 0 ρ 0 ) H (2)prime ν m (k 0 ρ 0 )J ν m (k 0 ρ 0 )? J prime ν m (k 0 ρ 0 )H (2) ν m (k 0 ρ 0 ) , C m = j2 ? I η 0 ψ sinν m φ 0 ρ 0 J ν m (k 0 ρ 0 ) H (2)prime ν m (k 0 ρ 0 )J ν m (k 0 ρ 0 )? J prime ν m (k 0 ρ 0 )H (2) ν m (k 0 ρ 0 ) . Using the Wronskian relation (E.93), we replace the denominators in these expressions by 2/(jπk 0 ρ 0 ): A m =? ? I η 0 ψ πk 0 sinν m φ 0 H (2) ν m (k 0 ρ 0 ), C m =? ? I η 0 ψ πk 0 sinν m φ 0 J ν m (k 0 ρ 0 ). Hence (4.368) gives ? E z (ρ,φ,ω)= braceleftBigg ? summationtext ∞ n=0 ? I η 0 2ψ πk 0 epsilon1 n J ν n (k 0 ρ)H (2) ν n (k 0 ρ 0 )sinν n φ sinν n φ 0 ,ρ<ρ 0 , ? summationtext ∞ n=0 ? I η 0 2ψ πk 0 epsilon1 n H (2) ν n (k 0 ρ)J ν n (k 0 ρ 0 )sinν n φ sinν n φ 0 ,ρ>ρ 0 , (4.373) where epsilon1 n is Neumann’s number (A.132). The magnetic ?elds can also be found by substituting the coe?cients into (4.369) and (4.370). The ?elds produced byan impressed plane wave maynow be obtained byletting the line source recede to in?nity. For large ρ 0 we use the asymptotic form (E.62) and ?nd that ? E z (ρ,φ,ω) =? ∞ summationdisplay n=0 ? I η 0 2ψ πk 0 epsilon1 n J ν n (k 0 ρ) bracketleftBiggradicalBigg 2 j πk 0 ρ 0 j ν n e ?jk 0 ρ 0 bracketrightBigg sinν n φ sinν n φ 0 ,ρ<ρ 0 . (4.374) Since the ?eld of a line source falls o? as ρ ?1/2 0 , the amplitude of the impressed ?eld approaches zero as ρ 0 →∞. We must compensate for the reduction in the impressed ?eld byscaling the amplitude of the current source. To obtain the proper scale factor, we note that the electric ?eld produced at a point ρ bya line source located at ρ 0 may be found from (4.345): ? E z =? ? I k 0 η 0 4 H (2) 0 (k 0 |ρ?ρ 0 |) ≈? ? I k 0 η 0 4 radicalBigg 2 j πk 0 ρ 0 e ?jk 0 ρ 0 e jkρ cos(φ?φ 0 ) , k 0 ρ 0 greatermuch 1. But if we write this as ? E z ≈ ? E 0 e jk·ρ then the ?eld looks exactlylike that produced bya plane wave with amplitude ? E 0 trav- eling along the wave vector k =?k 0 ?x cosφ 0 ?k 0 ?y sinφ 0 . Solving for ? I in terms of ? E 0 and substituting it back into (4.374), we get the total electric ?eld scattered from a wedge with an impressed TM plane-wave ?eld: ? E z (ρ,φ,ω) = 2π ψ ? E 0 ∞ summationdisplay n=0 epsilon1 n j ν n J ν n (k 0 ρ)sinν n φ sinν n φ 0 . Here we interpret the angle φ 0 as the incidence angle of the plane wave. To determine the ?eld produced byan impressed TE plane- wave ?eld, we use a mag- netic line source ? I m located at ρ 0 ,φ 0 and proceed as above. Byanalogywith (4.367) we write ? H z (ρ,φ,ω) = braceleftBigg summationtext ν [A ν sinνφ+ B ν cosνφ] J ν (k 0 ρ), ρ < ρ 0 , summationtext δ [C δ sinδφ+ D δ cosδφ] H (2) δ (k 0 ρ), ρ > ρ 0 . By(4.351) the tangential electric ?eld is ? E ρ (ρ,φ,ω) = braceleftBigg ? summationtext ν [A ν cosνφ? B ν sinνφ] j Z TE k 1 ρ νJ ν (k 0 ρ), ρ < ρ 0 , ? summationtext δ [C δ cosδφ? D δ sinδφ] j Z TE k 1 ρ δH (2) δ (k 0 ρ), ρ > ρ 0 . Application of the boundaryconditions on the tangential electric ?eld at φ = 0,ψresults in A ν = C δ = 0 and ν = δ = ν n = nπ/ψ, and thus ? H z becomes ? H z (ρ,φ,ω) = braceleftBigg summationtext ∞ n=0 B n cosν n φJ ν n (k 0 ρ), ρ < ρ 0 , summationtext ∞ n=0 D n cosν n φH (2) ν n (k 0 ρ), ρ > ρ 0 . (4.375) Application of the boundaryconditions on tangential electric and magnetic ?elds across the magnetic line source then leads directlyto ? H z (ρ,φ,ω) = braceleftBigg ? summationtext ∞ n=0 ? I m η 0 2ψ πk 0 epsilon1 n J ν n (k 0 ρ)H (2) ν n (k 0 ρ 0 )cosν n φ cosν n φ 0 ,ρ<ρ 0 ? summationtext ∞ n=0 ? I m η 0 2ψ πk 0 epsilon1 n H (2) ν n (k 0 ρ)J ν n (k 0 ρ 0 )cosν n φ cosν n φ 0 ,ρ>ρ 0 . (4.376) For a plane-wave impressed ?eld this reduces to ? H z (ρ,φ,ω) = 2π ψ ? E 0 η 0 ∞ summationdisplay n=0 epsilon1 n j ν n J ν n (k 0 ρ)cosν n φ cosν n φ 0 . Behaviorofcurrentnearasharpedge. In § 3.2.9 we studied the behavior of static charge near a sharp conducting edge bymodeling the edge as a wedge. We can follow the same procedure for frequency-domain ?elds. Assume that the perfectly conducting wedgeshowninFigure4.26isimmersedina?nite, z-independentimpressed?eldofa sort that will not concern us. A current is induced on the surface of the wedge and we wish to studyits behavior as we approach the edge. Because the ?eld is z-independent, we mayconsider the superposition of TM and TE ?elds as was done above to solve for the ?eld scattered bya wedge. For TM polarization, if the source is not located near the edge we maywrite the total ?eld (impressed plus scattered) in terms of nonuniform cylindrical waves. The form of the ?eld that obeys the boundaryconditions at φ = 0 and φ = ψ is given by(4.368): ? E z = ∞ summationdisplay n=0 A n sinν n φJ ν n (k 0 ρ), where ν n = nπ/ψ. Althoughthe A n dependontheimpressedsource,thegeneralbehavior of the current near the edge is determined bythe properties of the Bessel functions. The current on the wedge face at φ = 0 is given by ? J s (ρ,ω) = ? φ× [ ? φ ? H φ + ?ρ ? H ρ ]| φ=0 =??z ? H ρ (ρ,0,ω). By(4.349) we have the surface current ? J s (ρ,ω) =??z 1 Z TM k 0 ∞ summationdisplay n=0 A n ν n ρ J ν n (k 0 ρ). For ρ → 0 the small-argument approximation (E.51) yields ? J s (ρ,ω) ≈??z 1 Z TM k 0 ∞ summationdisplay n=0 A n ν n 1 Gamma1(ν n + 1) parenleftbigg k 0 2 parenrightbigg ν n ρ ν n ?1 . The sum is dominated bythe smallest power of ρ. Since the n = 0 term vanishes we have ? J s (ρ,ω) ～ ρ π ψ ?1 ,ρ→ 0. For ψ<πthe current density, which runs parallel to the edge, is unbounded as ρ → 0. A right-angle wedge (ψ = 3π/2) carries ? J s (ρ,ω) ～ ρ ?1/3 . Another important case is that of a half-plane (ψ = 2π) where ? J s (ρ,ω) ～ 1 √ ρ . (4.377) This square-root edge singularitydominates the behavior of the current ?owing parallel to any?at edge, either straight or with curvature large compared to a wavelength, and is useful for modeling currents on complicated structures. In the case of TE polarization the magnetic ?eld near the edge is, by(4.375), ? H z (ρ,φ,ω) = ∞ summationdisplay n=0 B n cosν n φJ ν n (k 0 ρ), ρ < ρ 0 . The current at φ = 0 is ? J s (ρ,ω) = ? φ× ?z ? H z | φ=0 = ?ρ ? H z (ρ,0,ω) or ? J s (ρ,ω) = ?ρ ∞ summationdisplay n=0 B n J ν n (k 0 ρ). For ρ → 0 we use (E.51) to write ? J s (ρ,ω) = ?ρ ∞ summationdisplay n=0 B n 1 Gamma1(ν n + 1) parenleftbigg k 0 2 parenrightbigg ν n ρ ν n . The n = 0 term gives a constant contribution, so we keep the ?rst two terms to see how the current behaves near ρ = 0: ? J s ～ b 0 + b 1 ρ π ψ . Here b 0 and b 1 depend on the form of the impressed ?eld. For a thin plate where ψ = 2π this becomes ? J s ～ b 0 + b 1 √ ρ. This is the companion square-root behavior to (4.377). When perpendicular to a sharp edge, the current grows awayfrom the edge as ρ 1/2 . In most cases b 0 = 0 since there is no mechanism to store charge along a sharp edge. 4.11.8 Propagationofsphericalwavesinaconductingmedium We cannot obtain uniform spherical wave solutions to Maxwell’s equations. Any?eld dependent onlyon r produces the null ?eld 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 ?elds dependent on r and θ. These waves displaythe fundamental properties of all spherical waves: theydiverge from a localized source and expand with ?nite velocity. Consider a homogeneous, source-free region characterized by ?epsilon1(ω), ?μ(ω), and ?σ(ω). We seek wave solutions that are TEM r in spherical coordinates ( ? H r = ? E r = 0) and φ-independent. Thus we write ? E(r,ω)= ? θ ? E θ (r,θ,ω)+ ? φ ? E φ (r,θ,ω), ? H(r,ω)= ? θ ? H θ (r,θ,ω)+ ? φ ? H φ (r,θ,ω). To determine the behavior of these ?elds we ?rst examine Faraday’s law ?× ? E(r,θ,ω)= ?r 1 r sinθ ? ?θ [sinθ ? E φ (r,θ,ω)] ? ? θ 1 r ? ?r [r ? E φ (r,θ,ω)] + ? φ 1 r ? ?r [r ? E θ (r,θ,ω)] =?jω ?μ ? H(r,θ,ω). (4.378) Since we require ? H r = 0 we must have ? ?θ [sinθ ? E φ (r,θ,ω)] = 0. This implies that either ? E φ ～ 1/sinθ or ? E φ = 0. We choose ? E φ = 0 and investigate whether the resulting ?elds satisfythe remaining Maxwell equations. Inasource-free, homogeneousregionofspacewehave?· ? D = 0 andthusalso?· ? E = 0. Since we have onlya θ-component of the electric ?eld, this requires 1 r ? ?θ ? E θ (r,θ,ω)+ cotθ r ? E θ (r,θ,ω)= 0. From this we see that when ? E φ = 0, the ?eld ? E θ must obey ? E θ (r,θ,ω)= ? f E (r,ω) sinθ . By(4.378) there is onlya φ-component of magnetic ?eld which obeys ? H φ (r,θ,ω)= ? f H (r,ω) sinθ where ? jω ?μ ? f H (r,ω)= 1 r ? ?r [r ? f E (r,ω)]. (4.379) So the spherical wave is TEM to the r-direction. We can obtain a wave equation for ? f E bytaking the curl of (4.378) and substituting from Ampere’s law: ?×(?× ? E) =? ? θ 1 r ? 2 ?r 2 (r ? E θ ) =?× parenleftbig ?jω ?μ ? H parenrightbig =?jω ?μ parenleftbig ?σ ? E + jω?epsilon1 ? E parenrightbig , hence d 2 dr 2 [r ? f E (r,ω)] + k 2 [r ? f E (r,ω)] = 0. (4.380) Here k = ω(?μ?epsilon1 c ) 1/2 is the complex wavenumber and ?epsilon1 c = ?epsilon1 + ?σ/jω is the complex permittivity. The equation for ? f H is identical. The wave equation (4.380) is merelythe second-order harmonic di?erential equation, with two independent solutions chosen from the list sin kr, cos kr, e ?jkr , e jkr . We ?nd sin kr and cos kr useful for describing standing waves between boundaries, and e jkr and e ?jkr useful for describing waves propagating in the r-direction. Of these, e jkr represents waves traveling inward while e ?jkr represents waves traveling outward. At this point we choose r ? f E = e ?jkr and thus ? E(r,θ,ω)= ? θ ? E 0 (ω) e ?jkr r sinθ . (4.381) By(4.379) we have ? H(r,θ,ω)= ? φ ? E 0 (ω) Z TEM e ?jkr r sinθ (4.382) where Z TEM = ( ?μ/epsilon1 c ) 1/2 is the complex wave impedance. Since we can also write ? H(r,θ,ω)= ?r × ? E(r,θ,ω) Z TEM , the ?eld is TEM to the r-direction, which is the direction of wave propagation as shown below. The wave nature of the ?eld is easilyidenti?ed byconsidering the ?elds in the phasor domain. Letting ω → ˇω and setting k = β ? jα in the exponential function we ?nd that ˇ E(r,θ)= ? θ ˇ E 0 e ?αr e ?jβr r sinθ where ˇ E 0 = E 0 e jξ E . The time-domain representation maybe found using (4.126): E(r,θ,t) = ? θE 0 e ?αr r sinθ cos( ˇωt ?βr +ξ E ). (4.383) We can identifya surface of constant phase as a locus of points obeying ˇωt ?βr +ξ E = C P (4.384) where C P is some constant. These surfaces, which are spheres centered on the origin, are called sphericalwavefronts. Note that surfaces of constant amplitude as determined by e ?αr r = C A , where C A is 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 .By di?erentiation we ?nd that the phase velocityis v p = ˇω/β. The wavelength is given by λ = 2π/β. Our solution is not appropriate for unbounded space since the ?elds have a singularity at θ = 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. Tocomputethepowercarriedbyaspherical wave, we use (4.381)and(4.382)toobtain the time-average Poynting ?ux S av = 1 2 Re{ ˇ E θ ? θ × ˇ H ? φ ? φ}= 1 2 ?r Re braceleftbigg 1 Z ? TEM bracerightbigg E 2 0 r 2 sin 2 θ e ?2αr . The power ?ux is radial and has densityinverselyproportional to r 2 . The time-average power carried bythe wave through a spherical surface at r sandwiched between the cones at θ 1 and θ 2 is P av (r) = 1 2 Re braceleftbigg 1 Z ? TEM bracerightbigg E 2 0 e ?2αr integraldisplay 2π 0 dφ integraldisplay θ 2 θ 1 dθ sinθ = π F Re braceleftbigg 1 Z ? TEM bracerightbigg E 2 0 e ?2αr where F = ln bracketleftbigg tan(θ 2 /2) tan(θ 1 /2) bracketrightbigg . (4.385) This is independent of r when α = 0. For lossymedia the power decays exponentially because of Joule heating. We can write the phasor electric ?eld in terms of the transverse gradient of a scalar potential function ˇ Phi1: ˇ E(r,θ)= ? θ ˇ E 0 e ?jkr r sinθ =?? t ˇ Phi1(θ) where ˇ Phi1(θ) =? ˇ E 0 e ?jkr ln parenleftbigg tan θ 2 parenrightbigg . By ? t we mean the gradient with the r-component excluded. It is easilyveri?ed that ˇ E(r,θ)=?? t ˇ Phi1(θ) =? ? θ ˇ E 0 1 r ? ˇ Phi1(θ) ?θ = ? θ ˇ E 0 e ?jkr r sinθ . Because ˇ E and ˇ Phi1 are related bythe gradient, we can de?ne a unique potential di?erence between the two cones at anyradial position r: ˇ V(r) =? integraldisplay θ 2 θ 1 ˇ E · dl = ˇ Phi1(θ 2 )? ˇ Phi1(θ 1 ) = ˇ E 0 Fe ?jkr , where F is given in (4.385). The existence of a unique voltage di?erence is a propertyof all transmission line structures operated in the TEM mode. We can similarlycompute the current ?owing outward on the cone surfaces. The surface current on the cone at θ = θ 1 is ˇ J s = ?n × ˇ H = ? θ × ? φ ˇ H φ = ?r ˇ H φ , hence ˇ I(r) = integraldisplay 2π 0 ˇ J s · ?rr sinθdφ = 2π ˇ E 0 Z TEM e ?jkr . The ratio of voltage to current at anyradius r is the characteristicimpedance of the bi- conical transmission line (or, equivalently, theinputimpedance of the biconical antenna): Z = ˇ V(r) ˇ I(r) = Z TEM 2π F. If the material between the cones is lossless (and thus ?μ = μ and ?epsilon1 c = epsilon1 are real), this becomes Z = η 2π F 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 thewave maybe found from P av (r) = 1 2 Re braceleftbig ˇ V(r) ˇ I ? (r) bracerightbig = π F Re braceleftbigg 1 Z ? TEM bracerightbigg E 2 0 e ?2αr . The complex power relationship P = VI ? is also a propertyof TEM guided- wave struc- tures. 4.11.9 Nonradiatingsources We showed in § 2.10.9thatnotalltime-varyingsourcesproduceelectromagneticwaves. In fact, a subset of localized sources known as nonradiating sources produce no ?eld external to the source region. Devaneyand Wolf [54] have shown that all nonradiating time-harmonic sources in an unbounded homogeneous medium can be represented in the form ˇ J nr (r) =??× bracketleftbig ?× ˇ f(r) bracketrightbig + k 2 ˇ f(r) (4.386) where ˇ f is anyvector ?eld that is continuous, has partial derivatives up to third order, and vanishes outside some localized region V s . In fact, ˇ E(r) = j ˇωμ ˇ f(r) is preciselythe phasor electric ?eld produced by ˇ J nr (r). The reasoning is straightforward. Consider the Helmholtz equation (4.203): ?×(?× ˇ E)? k 2 ˇ E =?j ˇωμ ˇ J. By(4.386) we have parenleftbig ?×?×?k 2 parenrightbigbracketleftbig ˇ E ? j ˇωμ ˇ f bracketrightbig = 0. Since ˇ f is zero outside the source region it must vanish at in?nity. ˇ E also vanishes at in?nitybythe radiation condition, and thus the quantity ˇ E ? j ˇωμ ˇ f obeys the radiation condition and is a unique solution to the Helmholtz equation throughout all space. Since the Helmholtz equation is homogeneous we have ˇ E ? j ˇωμ ˇ f = 0 everywhere; since ˇ f is zero outside the source region, so is ˇ E (and so is ˇ H). An interesting special case of nonradiating sources is ˇ f = ? ˇ Phi1 k 2 so that ˇ J nr =? parenleftbig ?×?×?k 2 parenrightbig ? ˇ Phi1 k 2 =? 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 z we shall use the notation ψ z (x, y,k z ,t) ? ψ(x, y, z,t). Here the spatial frequencytransform variable k z has units of m ?1 . The forward and inverse transform expressions are ψ z (x, y,k z ,t) = integraldisplay ∞ ?∞ ψ(x, y, z,t)e ?jk z z dz, (4.387) ψ(x, y, z,t) = 1 2π integraldisplay ∞ ?∞ ψ z (x, y,k z ,t)e jk z z dk z , (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 ?eld can be decomposed into a continuous superposition of elemental spatial terms e jk z z with weighting factors ψ z (x, y,k z ,t).In this case ψ z is the spatial frequency spectrum of ψ. The elemental terms are spatial sinusoids along z with rapidityof variation described by k z . As with the temporal transform, ψ z cannot be arbitrarysince ψ must obeya scalar wave equation such as (2.327). For instance, for a source-free region of free space we must have parenleftbigg ? 2 ? 1 c 2 ? ?t 2 parenrightbigg 1 2π integraldisplay ∞ ?∞ ψ z (x, y,k z ,t)e jk z z dk z = 0. Decomposing the Laplacian operator as ? 2 =? 2 t +? 2 /?z 2 and taking the derivatives into the integrand, we have 1 2π integraldisplay ∞ ?∞ bracketleftbiggparenleftbigg ? 2 t ? k 2 z ? 1 c 2 ? 2 ?t 2 parenrightbigg ψ z (x, y,k z ,t) bracketrightbigg e jk z z dk z = 0. Hence parenleftbigg ? 2 t ? k 2 z ? 1 c 2 ? 2 ?t 2 parenrightbigg ψ z (x, y,k z ,t) = 0 (4.389) bythe Fourier integral theorem. Theelementalcomponent e jk z z isspatiallysinusoidalandoccupiesallofspace. Because suchanelementcouldonlybecreatedbyasourcethatspansallofspace, itisnonphysical when taken byitself. Nonetheless it is often used to represent more complicated ?elds. If the elemental spatial term is to be used alone, it is best interpreted physically when combined 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 ∞ ?∞ e jk z z e jωt dω. (4.390) Before attempting to compute this transform, we should note that if the elemental term is to describe an EM ?eld ψ in a source-free region, it must obeythe homogeneous scalar wave equation. Substituting (4.390) into the homogeneous wave equation we have parenleftbigg ? 2 ? 1 c 2 ? 2 ?t 2 parenrightbigg 1 2π integraldisplay ∞ ?∞ e jk z z e jωt dω = 0. Di?erentiation under the integral sign gives 1 2π integraldisplay ∞ ?∞ bracketleftbiggparenleftbigg ?k 2 z + ω 2 c 2 parenrightbigg e jk z z bracketrightbigg e jωt dω = 0 and thus k 2 z = ω 2 c 2 = k 2 . Substitution of k z = k into (4.390) gives the time-domain representation of the elemental component φ(z,t) = 1 2π integraldisplay ∞ ?∞ e jω(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) whichwerecognizeasauniformplanewave propagatinginthe?z-directionwithvelocity 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 t is 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 elemental spatial component is ˇ φ(z) = e jk z z = e jkz . This elemental term is a time-harmonic plane wave propagating in the ?z-direction. Indeed, multiplying by e j ˇωt and 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 de?ne the temporal and spatial transforms using di?ering sign conventions. The temporal transform is de?ned as in (4.1) and (4.2), but the spatial transform is de?ned through ψ z (x, y,k z ,t) = integraldisplay ∞ ?∞ ψ(x, y, z,t)e jk z z dz, (4.392) ψ(x, y, z,t) = 1 2π integraldisplay ∞ ?∞ ψ z (x, y,k z ,t)e ?jk z z dk z . (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. ThedrawbackisthatwemustaltertheformulasfromstandardFouriertransform tables (replacing k by ?k) to re?ect this di?erence. In the following sections we shall show how a spatial Fourier decomposition can be used to solve for the electromagnetic ?elds in a source-free region of space. Byemploying the spatial transform we mayeliminate one or more spatial variables from Maxwell’s equations, making the wave equation easier to solve. In the end we must perform an inversion to return to the space domain. This maybe di?cult or impossible to do analytically, requiring a numerical Fourier inversion. 4.13 Spatial Fourier decomposition of two-dimensional ?elds 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 ?eld maybe decomposed into ?elds TE and TM to the z-direction. In the TM case ? H z = 0, and ? E z obeys the homogeneous scalar Helmholtz equation (4.208). In the TE case ? E z = 0, and ? H z obeys the homogeneous scalar Helmholtz equation. Since each ?eld component obeys the same equation, we let ? ψ(x, y,ω)represent either ? E z (x, y,ω)or ? H z (x, y,ω). Then ? ψ obeys (? 2 t + k 2 ) ? ψ(x, y,ω)= 0 (4.394) where ? 2 t is the transverse Laplacian (4.209) and k = ω(?μ?epsilon1 c ) 1/2 with ?epsilon1 c the 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 boundary conditions are speci?ed at a constant value of a single variable (e.g., over a plane), one transform su?ces. For instance, we mayknow the values of the ?eld in the y = 0 plane (as we will, for example, when we solve the boundaryvalue problems of § ??). Then we maytransform over x and leave the y variable 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 x we have ? ψ x (k x , y,ω)= integraldisplay ∞ ?∞ ? ψ(x, y,ω)e jk x x dx, (4.395) ? ψ(x, y,ω)= 1 2π integraldisplay ∞ ?∞ ψ x (k x , y,ω)e ?jk x x dk x . (4.396) For convenience in computation or interpretation of the inverse transform, we often regard k x as a complex variable and perturb the inversion contour into the complex k x = k xr + jk xi plane. 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 across this pole, the inverse transform does not return the original function. We generally indicate the desire to interpret k x as complex byindicating that the inversion contour is parallel to the real axis but located in the complex plane at k xi = Delta1: ? ψ(x, y,ω)= 1 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 ? ψ x (k x , y,ω)e ?jk x x dk x . (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 = k r + jk i represents 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 k i < 0. The Fourier transform of the function is u x (k x ) = integraldisplay ∞ 0 e ?jkx e jk x x dx = 1 j(k x ? k) bracketleftbig e j(k xr ?k r )x e ?(k xi ?k i )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 k xi > k i , and the transform is u x (k x ) =? 1 j(k x ? k) . Since u(x) is an exponential function, u x (k x ) has a pole at k x = k as anticipated. To compute the inverse transform we use (4.397): u(x) = 1 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 bracketleftbigg ? 1 j(k x ? k) bracketrightbigg e ?jk x x dk x . (4.399) We must be careful to choose Delta1 in such a way that all values of k x along the inversion contour lead to a convergent forward Fourier transform. Since we must have k xi > k i , choosing Delta1>k i ensures proper convergence. This gives the inversion contour shown in Figure4.27,aspecialcaseofwhichistherealaxis.Wecomputetheinversionintegral 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 > 0 we take 0 >Delta1>k i and close the contour in the lower half-plane using a semicircular contour C R of radius R. Then the closed contour integral is equal to ?2πj times the residue at the pole k x = k.AsR →∞we ?nd that k xi →?∞at all points on the contour C R . Thus the integrand, which varies as e k xi x , vanishes on C R and there is no contribution to the integral. The inversion integral (4.399) is found from the residue at the pole: u(x) = (?2πj) 1 2π Res k x =k bracketleftbigg ? 1 j(k x ? k) e ?jk x x bracketrightbigg . Since the residue is merely je ?jkx we have u(x) = e ?jkx . When x < 0 we choose Delta1>0 and close the contour along a semicircle C R of radius R in the upper half-plane. Again we ?nd that on C R the integrand vanishes as R →∞, and thus the inversion integral (4.399) is given by 2πj times 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 > 0 and x < 0. Note that if we had erroneously chosen Delta1<k i we 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 ? ψ in terms of a spatial Fourier integral we will make the equation easier to solve. We have (? 2 t + k 2 ) 1 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 ? ψ x (k x , y,ω)e ?jk x x dk x = 0. Di?erentiation under the integral sign with subsequent application of the Fourier integral theorem implies that ? ψ must obey the second-order harmonic di?erential equation bracketleftbigg d 2 dy 2 + k 2 y bracketrightbigg ? ψ x (k x , y,ω)= 0 where we have de?ned the dependent parameter k y = k yr + jk yi through k 2 x + k 2 y = k 2 . Two independent solutions to the di?erential equation are e ?jk y y and thus ? ψ(k x , y,ω)= A(k x ,ω)e ?jk y y . Substituting this into the inversion integral, we have the solution to the Helmholtz equa- tion: ? ψ(x, y,ω)= 1 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 A(k x ,ω)e ?jk x x e ?jk y y dk x . (4.400) If we de?ne the wave vector k = ?xk x ± ?yk y , we can also write the solution in the form ? ψ(x, y,ω)= 1 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 A(k x ,ω)e ?jk·ρ dk x (4.401) where ρ = ?xx + ?yy is 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 vector lying in the xy-plane. For lossy media the plane wave is nonuniform, and the surfaces of constant phase may not be aligned with the surfaces of constant amplitude (see § 4.11.4). For the special case of a lossless medium we have k i → 0 and can let Delta1 → 0 as long as Delta1>k i . As we perform the inverse transform integral over k x from ?∞ to ∞ we will encounter both the condition k 2 x > k 2 and k 2 x ≤ k 2 .Fork 2 x ≤ k 2 we have e ?jk x x e ?jk y y = e ?jk x x e ?j √ k 2 ?k 2 x y where we choose the upper sign for y > 0 and the lower sign for y < 0 to 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 > 0 along a direction which depends on k x ,makinganangleξwiththe x-axisasshowninFigure4.28.For k x 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 a continuous superposition of plane waves. The amplitude of each plane wave component 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) k 2 x ≤ k 2 ,(b)k 2 x > k 2 . is determined by the values of the ?eld over the boundaries of the solution region. But this is not the whole picture. The inverse transform integral also requires values of k x in the intervals [?∞,k] and [k,∞]. Here we have k 2 x > k 2 and thus e ?jk x x e ?jk y y = e ?jk x x e ? √ k 2 x ?k 2 y , where we choose the upper sign for y > 0 and the lower sign for y < 0 to ensure that the ?eld decays along the y-direction. In these regimes we have an evanescent wave, propagating along x but decaying along y, with surfaces of constant phase and amplitude mutuallyperpendicular(Figure4.28).As k x ranges out to ∞, evanescent waves of all possible decay constants also contribute to the plane-wave superposition. We may summarize the plane-wave contributions by letting k = ?xk x + ?yk y = k r + jk i where k r = braceleftBigg ?xk x ± ?y radicalbig k 2 ? k 2 x , k 2 x < k 2 , ?xk x , k 2 x > k 2 , k i = braceleftBigg 0, k 2 x < k 2 , ??y radicalbig k 2 x ? k 2 , k 2 x > k 2 , where the upper sign is used for y > 0 and 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 k x = k cosξ, k y =±k sinξ, where ξ = ξ r + jξ i is a new complex variable, we have k · ρ = kx cosξ ± ky sinξ and dk x =?k sinξ dξ. With this change of variables (4.401) becomes ? ψ(x, y,ω)= k 2π integraldisplay C A(k cosξ,ω)e ?jkx cosξ e ±jkysinξ sinξ dξ. (4.402) Since A(k x ,ω)is a function to be determined, we may introduce a new function f (ξ,ω) = k 2π A(k x ,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 C f (ξ,ω)e ?jkρ cos(φ±ξ) dξ (4.403) where x = ρ cosφ,y = ρ sinφ, 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 C is 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 k x over [?∞,∞] in the original inversion integral. By the identities cos z = cos(u + jv)= cos u coshv ? j sin u sinhv, sin z = sin(u + jv)= sin u coshv + j cos u sinhv, we?ndthatthecontourshowninFigure4.29providesidenticalvaluesoftheintegrand (Problem 4.24). The portions of the contour [0 + j∞,0] and [?π,?π ? j∞] together correspond to the regime of evanescent waves (k < k x < ∞ and ?∞ < k x < k), while the segment [0,?π] along the real axis corresponds to ?k < k x < k and thus describes contributions from propagating plane waves. In this case ξ represents the propagation angle of the waves. 4.13.1 Boundary value problems using the spatial Fourier represen- tation The ?eld of a line source. As a ?rst example we calculate the Fourier representation of the ?eld of an electric line source. Assume a uniform line current ? I(ω) is aligned along the z-axis in a medium characterized by complex permittivity ?epsilon1 c (ω) and permeability ?μ(ω). We separate space into two source-free portions, y > 0 and y < 0, and write the ?eld in each region in terms of an inverse spatial Fourier transform. Then, by applying the boundary conditions in the y = 0 plane, we solve for the angular spectrum of the line source. Since this is a two-dimensional problem we may decompose the ?elds into TE and TM sets. For an electric line source we need only the TM set, and write E z as a superposition of plane waves using (4.400). For y?0 we represent the ?eld in terms of plane waves traveling in the ±y-direction. Thus ? E z (x, y,ω)= 1 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 A + (k x ,ω)e ?jk x x e ?jk y y dk x , y > 0, ? E z (x, y,ω)= 1 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 A ? (k x ,ω)e ?jk x x e +jk y y dk x , y < 0. The transverse magnetic ?eld may be found from the axial electric ?eld using (4.212). We ?nd ? H x =? 1 jω ?μ ? ? E z ?y (4.404) and thus ? H x (x, y,ω)= 1 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 A + (k x ,ω) bracketleftbigg k y ω ?μ bracketrightbigg e ?jk x x e ?jk y y dk x , y > 0, ? H x (x, y,ω)= 1 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 A ? (k x ,ω) bracketleftbigg ? k y ω ?μ bracketrightbigg e ?jk x x e +jk y y dk x , y < 0. To ?nd the spectra A ± (k x ,ω)we apply the boundary conditions at y = 0. Since tangential ? E is continuous we have, after combining the integrals, 1 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 bracketleftbig A + (k x ,ω)? A ? (k x ,ω) bracketrightbig e ?jk x x dk x = 0, and hence by the Fourier integral theorem A + (k x ,ω)? A ? (k x ,ω)= 0. (4.405) We must also apply ?n 12 ×( ? H 1 ? ? H 2 ) = ? J s . The line current may be written as a surface current density using the δ-function, giving ? bracketleftbig ? H x (x,0 + ,ω)? ? H x (x,0 ? ,ω) bracketrightbig = ? I(ω)δ(x). By (A.4) δ(x) = 1 2π integraldisplay ∞ ?∞ e ?jk x x dk x . Then, substituting for the ?elds and combining the integrands, we have 1 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 bracketleftbigg A + (k x ,ω)+ A ? (k x ,ω)+ ω ?μ k y ? I(ω) bracketrightbigg e ?jk x x = 0, hence A + (k x ,ω)+ A ? (k x ,ω)=? ω ?μ k y ? I(ω). (4.406) Solution of (4.405) and (4.406) gives the angular spectra A + (k x ,ω)= A ? (k x ,ω)=? ω ?μ 2k y ? I(ω). Substituting this into the ?eld expressions and combining the cases for y > 0 and y < 0, we ?nd ? E z (x, y,ω)=? ω ?μ ? I(ω) 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 e ?jk y |y| 2k y e ?jk x x dk x =?jω ?μ ? I(ω) ? G(x, y|0,0;ω). (4.407) Here ? G is the spectral representation of the two-dimensional Green’s function ?rst found in § 4.11.7, and is given by ? G(x, y|x prime , y prime ;ω) = 1 2πj ∞+jDelta1 integraldisplay ?∞+jDelta1 e ?jk y |y?y prime | 2k y e ?jk x (x?x prime ) dk x . (4.408) By duality we have ? H z (x, y,ω)=? ω?epsilon1 c ? I m (ω) 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 e ?jk y |y| 2k y e ?jk x x dk x =?jω?epsilon1 c ? I m (ω)G(x, y|0,0;ω) (4.409) for a magnetic line current ? I m (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 ∞+jDelta1 integraldisplay ?∞+jDelta1 e ?jk y |y| k y e ?jk x x dk x = H (2) 0 (kρ). We have not yet speci?ed the contour appropriate for calculating the inverse transform (4.407). We must be careful because the denominator of (4.407) has branch points at k y = radicalbig k 2 ? k 2 x = 0, or equivalently, k x =±k =±(k r + jk i ). For lossy materials we have k i < 0 and k r > 0,sothebranchpointsappearasinFigure4.30.Wemaytakethebranch cuts outward from these points, and thus choose the inversion contour to lie between the branch points so that the branch cuts are not traversed. This requires k i <Delta1<?k i .It is natural to choose Delta1 = 0 and use the real axis as the inversion contour. We must be careful, though, when extending these arguments to the lossless case. If we consider the lossless case to be the limit of the lossy case as k i → 0, we ?nd that the branch points migrate to the real axis and thus lie on the inversion contour. We can eliminate this problem by realizing that the inversion contour may be perturbed without a?ecting the value of the integral, as long as it is not made to pass through the branch cuts. If we perturbtheinversioncontourasshowninFigure4.30,thenas k i → 0 the branch points do not fall on the contour. Figure 4.30: Inversion contour in complex k x -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’s residue theorem to capture poles of the integrand (which often describe the properties of 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 observation point is far from the source we may use the method of steepest descents to obtain asymptotic forms for the ?elds. 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 = h within a medium having parameters ?μ 1 (ω)and ?epsilon1 c 1 (ω).The y = 0 plane separates this region from a region having parameters ?μ 2 (ω)and ?epsilon1 c 2 (ω).SeeFigure4.31. The impressed line current source creates an electromagnetic ?eld that induces secondary polarization and conduction currents in both regions. This current in turn produces a secondary ?eld that adds to the primary ?eld of the line source to satisfy the boundary conditions at the interface. We would like to solve for the secondary ?eld and give its sources an image interpretation. Since the ?elds are z-independent we may decompose the ?elds into sets TE and TM to z. For a z-directed impressed source there is a z-component of ? E, but no z-component of ? H; hence the ?elds are entirely speci?ed by the TM set. The impressed source is una?ected by the secondary ?eld, and we may represent the impressed electric ?eld using (4.407): ? E i z (x, y,ω)=? ω ?μ 1 ? I(ω) 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 e ?jk y1 |y?h| 2k y1 e ?jk x x dk x , y ≥ 0 (4.410) Figure 4.31: Geometry of a z-directed line source above an interface between two material regions. where k y1 = radicalBig k 2 1 ? k 2 x and k 1 = ω(?μ 1 ?epsilon1 c 1 ) 1/2 . From (4.404) we ?nd that ? H i x =? 1 jω ?μ 1 ? ? E i z ?y = ? I(ω) 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 e jk y1 (y?h) 2 e ?jk x x dk x , 0 ≤ y < h. The scattered ?eld obeys the homogeneous Helmholtz equation for all y > 0, and thus may be written using (4.400) as a superposition of upward-traveling waves: ? E s z1 (x, y,ω)= 1 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 A 1 (k x ,ω)e ?jk y1 y e ?jk x x dk x , ? H s x1 (x, y,ω)= 1 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 k y1 ω ?μ 1 A 1 (k x ,ω)e ?jk y1 y e ?jk x x dk x . Similarly, in region 2 the scattered ?eld may be written as a superposition of downward- traveling waves: ? E s z2 (x, y,ω)= 1 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 A 2 (k x ,ω)e jk y2 y e ?jk x x dk x , ? H s x2 (x, y,ω)=? 1 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 k y2 ω ?μ 2 A 2 (k x ,ω)e jk y2 y e ?jk x x dk x , where k y2 = radicalBig k 2 2 ? k 2 x and k 2 = ω(?μ 2 ?epsilon1 c 2 ) 1/2 . We can solve for the angular spectra A 1 and A 2 by applying the boundary conditions at the interface between the two media. From the continuity of total tangential electric ?eld we ?nd that 1 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 bracketleftbigg ? ω ?μ 1 ? I(ω) 2k y1 e ?jk y1 h + A 1 (k x ,ω)? A 2 (k x ,ω) bracketrightbigg e ?jk x x dk x = 0, hence by the Fourier integral theorem A 1 (k x ,ω)? A 2 (k x ,ω)= ω ?μ 1 ? I(ω) 2k y1 e ?jk y1 h . The boundary condition on the continuity of ? H x yields similarly ? ? I(ω) 2 e ?jk y1 h = k y1 ω ?μ 1 A 1 (k x ,ω)+ k y2 ω ?μ 2 A 2 (k x ,ω). We obtain A 1 (k x ,ω)= ω ?μ 1 ? I(ω) 2k y1 R TM (k x ,ω)e ?jk y1 h , A 2 (k x ,ω)=? ω ?μ 2 ? I(ω) 2k y2 T TM (k x ,ω)e ?jk y1 h . Here R TM and T TM = 1 + R TM are re?ection and transmission coe?cients given by R TM (k x ,ω)= ?μ 1 k y2 ? ?μ 2 k y1 ?μ 1 k y2 + ?μ 2 k y1 , T TM (k x ,ω)= 2 ?μ 1 k y2 ?μ 1 k y2 + ?μ 2 k y1 . These describe the re?ection and transmission of each component of the plane-wave spectrum of the impressed ?eld, and thus depend on the parameter k x . The scattered ?elds are ? E s z1 (x, y,ω)= ω ?μ 1 ? I(ω) 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 e ?jk y1 (y+h) 2k y1 R TM (k x ,ω)e ?jk x x dk x , (4.411) ? E s z2 (x, y,ω)=? ω ?μ 2 ? I(ω) 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 e jk y2 (y?hk y1 /k y2 ) 2k y2 T TM (k x ,ω)e ?jk x x dk x . (4.412) We may now obtain the ?eld produced by an electric line source above a perfect conductor. Letting ?σ 2 →∞we have k y2 = radicalBig k 2 2 ? k 2 x →∞and R TM → 1, T TM → 2. With these, the scattered ?elds (4.411) and (4.412) become ? E s z1 (x, y,ω)= ω ?μ 1 ? I(ω) 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 e ?jk y1 (y+h) 2k y1 e ?jk x x dk x , (4.413) ? E s z2 (x, y,ω)= 0. (4.414) Comparing (4.413) to (4.410) we see that the scattered ?eld 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 > 0 to the problem of the line source above a ground plane. The total ?eld is the sum of the impressed and scattered ?elds: ? E z (x, y,ω)=? ω ?μ 1 ? I(ω) 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 e ?jk y1 |y?h| ? e ?jk y1 (y+h) 2k y1 e ?jk x x dk x , y ≥ 0. We can write this in another form using the Hankel-function representation of the line source (4.345): ? E z (x, y,ω)=? ω ?μ 4 ? I(ω)H (2) 0 (k|ρ? ?yh|)+ ω ?μ 4 ? I(ω)H (2) 0 (k|ρ+ ?yh|) where |ρ± ?yh|=|ρ ?ρ± ?yh|= radicalbig x 2 +(y ± h) 2 . Interpreting the general case in terms of images is more di?cult. Comparing (4.411) and (4.412) with (4.410), we see that each spectral component of the ?eld in region 1 has the form of an image line source located at y =?h in region 2, but that the amplitude of the line source, R TM ? I, depends on k x . Similarly, the ?eld in region 2 is composed of spectral components that seem to originate from line sources with amplitudes ?T TM ? I located at y = hk y1 /k y2 in region 1. In this case the amplitude and position of the image line source producing a spectral component are both dependent on k x . The ?eld 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 ?epsilon1 c (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- directedelectric?eld(Figure4.32).Theprimary?eldinducesasecondarycurrenton the conductor and this in turn produces a secondary ?eld. The total ?eld must obey the boundary conditions at y = 0. Because the z-directed incident ?eld induces a z-directed secondary current, the ?elds may be described entirely in terms of a TM set. The impressed plane wave may be written as ? E i (r,ω)= ?z ? E 0 (ω)e jk(x cosφ 0 +y sinφ 0 ) where φ 0 is the angle between the incident wave vector and the x-axis. By (4.223) we also have ? H i (r,ω)= ? E 0 (ω) η (?y cosφ 0 ? ?x sinφ 0 )e jk(x cosφ 0 +y sinφ 0 ) . The scattered ?elds 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 ?eld we can write ? E s z (x, y,ω)= integraldisplay C f (ξ,ω)e ?jkρ cos(φ±ξ) dξ. (4.415) By (4.404) the x-component of the magnetic ?eld is ? H s x (x, y,ω)=? 1 jω ?μ ? ? E s z ?y =? 1 jω ?μ integraldisplay C f (ξ,ω) ? ?y parenleftbig e ?jkx cosξ e ±jkysinξ parenrightbig =? 1 jω ?μ (±jk) integraldisplay C f (ξ,ω)sinξe ?jkρ cos(φ±ξ) dξ. To ?nd the angular spectrum f (ξ,ω) and ensure uniqueness of solution, we must apply the boundary conditions over the entire y = 0 plane. For x > 0 where the conductor resides, the total tangential electric ?eld must vanish. Setting the sum of the incident and scattered ?elds to zero at φ = 0 we have integraldisplay C f (ξ,ω)e ?jkx cosξ dξ =? ? E 0 e jkx cosφ 0 , x > 0. (4.416) To ?nd the boundary condition for x < 0 we note that by symmetry ? E s z is even about y = 0 while ? H s x , as the y-derivative of ? E s z , is odd. Since no current can be induced in the y = 0 plane for x < 0, the x-directed scattered magnetic ?eld must be continuous and thus equal to zero there. Hence our second condition is integraldisplay C f (ξ,ω)sinξe ?jkx cosξ dξ = 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. Writing ξ = cos ?1 (k x /k) we have d dξ (k cosξ)=?k sinξ = dk x dξ and dξ =? dk x k sinξ =? dk x k radicalbig 1 ? cos 2 ξ =? dk x radicalbig k 2 ? k 2 x . Upon substitution of these relations, the inversion contour returns to the real k x axis (which may then be perturbed by jDelta1). Thus, (4.416) and (4.417) may be written as ∞+jDelta1 integraldisplay ?∞+jDelta1 f parenleftbig cos ?1 kx k parenrightbig radicalbig k 2 ? k 2 x e ?jk x x dk x =? ? E 0 e jk x0 x , x > 0, (4.418) ∞+jDelta1 integraldisplay ?∞+jDelta1 f parenleftbigg cos ?1 k x k parenrightbigg e ?jk x x dk x = 0, x < 0, (4.419) Figure 4.33: Integration contour used to evaluate the function F(x). where k x0 = k cosφ 0 . 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 C R of radius R where R →∞, we ?nd that the contribution from the semicircle is lim R→∞ integraldisplay C R f parenleftbigg cos ?1 k x k parenrightbigg e ?|x|k xi e j|x|k xr dk x = 0 since x < 0. This assumes that f does not grow exponentially with R.Thus contintegraldisplay C f parenleftbigg cos ?1 k x k parenrightbigg e ?jk x x dk x = 0 where C now encloses the portion of the upper half-plane k xi >Delta1. By Morera’s theorem [110],%citeLePage, the above relation holds if f is regular (contains no singularities or branch points) in this portion of the upper half-plane. We shall assume this and investigate the other properties of f that 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 2 jπ ∞+jDelta1 integraldisplay ?∞+jDelta1 h(k x ) h(?k x0 ) 1 k x + k x0 e ?jk x x dk x . Here h(k x ) is some function regular in the region k xi <Delta1, with h(k x ) → 0 as k x →∞. IfwechooseDelta1sothat?k xi >Delta1>?k xi cosθ 0 and close the contour with a semicircle in thelowerhalf-plane(Figure4.33),thenthecontributionfromthesemicirclevanishesfor large radius and thus, by Cauchy’s residue theorem, F(x) =?e jk x0 x . Using this (4.418) can be written as ∞+jDelta1 integraldisplay ?∞+jDelta1 bracketleftBigg f parenleftbig cos ?1 kx k parenrightbig radicalbig k 2 ? k 2 x ? ? E 0 2 jπ h(k x ) h(?k x0 ) 1 k x + k x0 bracketrightBigg e ?jk x x dk x = 0. Setting the integrand to zero and using radicalbig k 2 ? k 2 x = √ k ? k x √ k + k x ,wehave f parenleftbig cos ?1 kx k parenrightbig √ k ? k x (k x + k x0 ) = ? E 0 2 jπ radicalbig k + k x h(k x ) h(?k x0 ) . (4.420) The left member has a branch point at k x = k whiletherightmemberhasabranchpoint at k x =?k.IfwechoosethebranchcutsasinFigure4.30thensince f isregularin the region k xi >Delta1the left side of (4.420) is regular there. Also, since h(k x ) is regular in the region k xi <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 be constant. So f parenleftbig cos ?1 kx k parenrightbig √ k ? k x (k x + k x0 ) = ? E 0 2 jπ radicalbig k + k x h(k x ) h(?k x0 ) = constant. We may evaluate the constant by inserting any value of k x . Using k x =?k x0 on the right we ?nd that f parenleftbig cos ?1 kx k parenrightbig √ k ? k x (k x + k x0 ) = ? E 0 2 jπ radicalbig k ? k x0 . Substituting k x = k cosξ and k x0 = k cosφ 0 we have f (ξ) = ? E 0 2 jπ √ 1 ? cosφ 0 √ 1 ? cosξ cosξ + cosφ 0 . Since sin(x/2) = √ (1 ? cos x)/2, we may also write f (ξ) = ? E 0 jπ sin φ 0 2 sin ξ 2 cosξ + cosφ 0 . Finally, substituting this into (4.415) we have the spectral representation for the ?eld scattered by a half-plane: ? E s z (ρ,φ,ω) = ? E 0 (ω) jπ integraldisplay C sin φ 0 2 sin ξ 2 cosξ + cosφ 0 e ?jkρ cos(φ±ξ) dξ. (4.421) The scattered ?eld inversion integral in (4.421) may be rewritten in such a way as to separate geometrical optics (plane-wave) terms from di?raction terms. The di?raction terms may be written using standard functions (modi?ed Fresnel integrals) and for large values of ρ appear 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 ?elds 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 simpli?es ?eld com- putation, since boundary conditions need only be applied within one period, or cell,of the structure. Examples of situations that lead to periodic ?elds include the guiding of waves in slow-wave structures such as helices and meander lines, the scattering of plane waves from gratings, and the radiation of waves by antenna arrays. In this section we will study the representation of ?elds with in?nite 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 ?eld may be represented in terms of a periodic function ? ψ p that obeys ? ψ p (x, y, z ± mL,ω)= ? ψ p (x, y, z,ω) where m is an integer and L is the spatial period. According to Floquet’s theorem,if ? ψ represents some vector component of the ?eld, then the ?eld obeys ? ψ(x, y, z,ω)= e ?jκz ? ψ p (x, y, z,ω). (4.422) Here κ = β ? jα is a complex wavenumber describing the phase shift and attenuation of the ?eld between the various cells of the environment. The phase shift and attenuation may 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 be produced by the excitation of an antenna array by a distributed terminal voltage. It is also possible to have κ = 0 as when, for example, a periodic antenna array is driven with all elements in phase. Because ? ψ p is periodic we may expand it in a Fourier series ? ψ p (x, y, z,ω)= ∞ summationdisplay n=?∞ ? ψ n (x, y,ω)e ?j2πnz/L where the ? ψ n are found by orthogonality: ? ψ n (x, y,ω)= 1 L integraldisplay L/2 ?L/2 ? ψ p (x, y, z,ω)e j2πnz/L dz. Substituting this into (4.422), we have a representation for the ?eld as a Fourier series: ? ψ(x, y, z,ω)= ∞ summationdisplay n=?∞ ? ψ n (x, y,ω)e ?jκ n z where κ n = β + 2πn/L + jα = β n ? jα. We see that within each cell the ?eld consists of a number of constituents called space harmonics or Hartree harmonics, each with the property of a propagating or evanescent wave. Each has phase velocity v pn = ω β n = 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 v gn = dω dβ = parenleftbigg dβ n dω parenrightbigg ?1 = parenleftbigg dβ dω parenrightbigg ?1 = v g . 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 wave oscillators.” Figure 4.34: Geometry of a periodic strati?ed medium with each cell consisting of two material layers. 4.14.2 Examples of periodic systems Plane-wave propagation within a periodically strati?ed medium. As an exam- ple of wave propagation in a periodic structure, let us consider a plane wave propagating within a layered medium consisting of two material layers repeated periodically as shown inFigure4.34.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 wave amplitudes in any region in terms of the amplitudes in the region immediately preceding it. We may write these in matrix form by eliminating one of the variables a n or b n from each equation: bracketleftbigg T (n) 11 T (n) 12 T (n) 21 T (n) 22 bracketrightbiggbracketleftbigg a n+1 b n+1 bracketrightbigg = bracketleftbigg a n b n bracketrightbigg (4.423) where T (n) 11 = 1 2 Z n + Z n?1 Z n ? P ?1 n , T (n) 12 = 1 2 Z n ? Z n?1 Z n ? P n , T (n) 21 = 1 2 Z n ? Z n?1 Z n ? P ?1 n , T (n) 22 = 1 2 Z n + Z n?1 Z n ? P n . Here Z n represents Z n⊥ for perpendicular polarization and Z nbardbl for 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 the wave amplitudes are often normalized using the wave impedances.We may use these parameters to describe the cascaded system of two layers: bracketleftbigg T (n) 11 T (n) 12 T (n) 21 T (n) 22 bracketrightbiggbracketleftbigg T (n+1) 11 T (n+1) 12 T (n+1) 21 T (n+1) 22 bracketrightbiggbracketleftbigg a n+2 b n+2 bracketrightbigg = bracketleftbigg a n b n bracketrightbigg . Since for a periodic layered medium the wave amplitudes should obey (4.422), we have bracketleftbigg T 11 T 12 T 21 T 22 bracketrightbiggbracketleftbigg a n+2 b n+2 bracketrightbigg = bracketleftbigg a n b n bracketrightbigg = e jκL bracketleftbigg a n+2 b n+2 bracketrightbigg (4.424) where L = Delta1 n +Delta1 n+1 is the period of the structure and bracketleftbigg T 11 T 12 T 21 T 22 bracketrightbigg = bracketleftbigg T (n) 11 T (n) 12 T (n) 21 T (n) 22 bracketrightbiggbracketleftbigg T (n+1) 11 T (n+1) 12 T (n+1) 21 T (n+1) 22 bracketrightbigg . Equation (4.424) is an eigenvalue equation for κ and can be rewritten as bracketleftbigg T 11 ? e jκL T 12 T 21 T 22 ? e jκL bracketrightbiggbracketleftbigg a n+2 b n+2 bracketrightbigg = bracketleftbigg 0 0 bracketrightbigg . This equation only has solutions when the determinant of the matrix vanishes. Expansion of the determinant gives T 11 T 22 ? T 12 T 21 ? e jκL (T 11 + T 22 )+ e j2κL = 0. (4.425) The ?rst two terms are merely T 11 T 22 ? T 12 T 21 = vextendsingle vextendsingle vextendsingle vextendsingle T 11 T 12 T 21 T 22 vextendsingle vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vextendsingle T (n) 11 T (n) 12 T (n) 21 T (n) 22 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle T (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 vextendsingle T (n) 11 T (n) 12 T (n) 21 T (n) 22 vextendsingle vextendsingle vextendsingle vextendsingle = Z n?1 Z n , we have T 11 T 22 ? T 12 T 21 = Z n?1 Z n Z n Z n+1 = 1 where we have used Z n?1 = Z n+1 because of the periodicity of the medium. With this, (4.425) becomes cosκL = T 11 + T 22 2 . Finally, computing the matrix product and simplifying to ?nd T 11 + T 22 , we have cosκL = cos(k z,n Delta1 n )cos(k k,n?1 Delta1 n?1 )? ? 1 2 parenleftbigg Z n?1 Z n + Z n Z n?1 parenrightbigg sin(k z,n Delta1 n )sin(k z,n?1 Delta1 n?1 ) (4.426) or equivalently cosκL = 1 4 (Z n?1 + Z n ) 2 Z n Z n?1 cos(k z,n Delta1 n + k z,n?1 Delta1 n?1 )? ? 1 4 (Z n?1 ? Z n ) 2 Z n Z n?1 cos(k z,n Delta1 n ? k z,n?1 Delta1 n?1 ). (4.427) Note that both ±κ satisfy 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 ω result in cosκL > 1, causing κL to be imaginary and producing evanescent waves. We refer to the frequency ranges over which cosκL > 1 as stopbands, and those over which cosκL < 1 as passbands. This terminology is used in ?lter analysis and, indeed, waves propagating in periodic media experience e?ects similar to those experienced by signals passing through ?lters. Field produced by an in?nite array of line sources. As a second example, consider an in?nite number of z-directed line sources within a homogeneous medium of complex permittivity ?epsilon1 c (ω) and permeability ?μ(ω), aligned along the x-axis with separation L such that ? J(r,ω)= ∞ summationdisplay n=?∞ ?z ? I n δ(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 currents on successive elements, or, if the currents are secondary, from their excitation by an impressed ?eld such as a plane wave.) Thus we write ? I n = ? I 0 e ?jκnL (4.428) where κ is a complex constant. We may represent the ?eld produced by the source array as a superposition of the ?elds of individual line sources found earlier. In particular we may use the Hankel function representation (4.345) or the Fourier transform representation (4.407). Using the latter we have ? E z (x, y,ω)= ∞ summationdisplay n=?∞ e ?jκnL ? ? ? ? ω ?μ ? I 0 (ω) 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 e ?jk y |y| 2k y e ?jk x (x?nL) dk x ? ? ? . Interchanging the order of summation and integration we have ? E z (x, y,ω)=? ω ?μ ? I 0 (ω) 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 e ?jk y |y| 2k y bracketleftBigg ∞ summationdisplay n=?∞ e jn(k x ?κ)L bracketrightBigg e ?jk x x dk x . (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(nk 0 )e jnk 0 x , where k 0 = 2π/D. Letting f (x) = δ(x ? x 0 ) in that expression we have ∞ summationdisplay n=?∞ δ parenleftbigg x ? x 0 ? n 2π L parenrightbigg = L 2π ∞ summationdisplay n=?∞ e jnL(x?x 0 ) . Substituting this into (4.429) we have ? E z (x, y,ω)=? ω ?μ ? I 0 (ω) 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 e ?jk y |y| 2k y bracketleftBigg ∞ summationdisplay n=?∞ 2π L δ parenleftbigg k x ?κ ? n 2π L parenrightbigg bracketrightBigg e ?jk x x dk x . Carrying out the integral we replace k x with κ n = κ + 2nπ/L, giving ? E z (x, y,ω)=?ω ?μ ? I 0 (ω) ∞ summationdisplay n=?∞ e ?jk y,n |y| e ?jκ n x 2Lk y,n =?jω ?μ ? I 0 (ω) ? G ∞ (x, y|0,0,ω) (4.430) where k y,n = radicalbig k 2 ?κ 2 n , and where ? G ∞ (x, y|x prime , y prime ,ω)= ∞ summationdisplay n=?∞ e ?jk y,n |y?y prime | e ?jκ n (x?x prime ) 2 jLk y,n (4.431) is called the periodic Green’s function. We may also ?nd the ?eld produced by an in?nite 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 ? E z (ρ,ω) =? ω ?μ 4 ∞ summationdisplay n=?∞ ? I 0 (ω)e ?jκnL H (2) 0 (k|ρ?ρ n |) =?jω ?μ ? I 0 (ω) ? G ∞ (x, y|0,0,ω) where |ρ?ρ n |=|?yy + ?x(x ? nL)|= radicalbig y 2 +(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 4 j ∞ summationdisplay n=?∞ e ?jκnL H (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 situations in which boundary conditions at constant values of y are to be applied. Both forms are di?cult to compute under certain circumstances, and variants of these forms have been introduced in the literature [203]. 4.15 Problems 4.1 Beginning with the Kronig–Kramers formulas (4.35)–(4.36), use the even–odd be- havior of the real and imaginary parts of ?epsilon1 c to derive the alternative relations (4.37)– (4.38). 4.2 Consider the complex permittivity dyadic of a magnetized plasma given by (4.88)– (4.91). Show that we may decompose [ ? ˉepsilon1 c ] as the sum of two matrices [ ? ˉepsilon1 c ] = [ ? ˉepsilon1] + [ ? ˉσ] jω where [ ? ˉepsilon1] and [ ? ˉσ] are hermitian. 4.3 Show that the Debye permittivity formulas ?epsilon1 prime (ω)?epsilon1 ∞ = epsilon1 s ?epsilon1 ∞ 1 +ω 2 τ 2 , ?epsilon1 primeprime (ω) =? ωτ(epsilon1 s ?epsilon1 ∞ ) 1 +ω 2 τ 2 , obey the Kronig–Kramers relations. 4.4 The frequency-domain duality transformations for the constitutive parameters of an anisotropic medium are given in (4.197). Determine the analogous transformations for the constitutive parameters of a bianisotropic medium. 4.5 Establish the plane-wave identities (B.76)–(B.79) by direct di?erentiation in rect- angular coordinates. 4.6 Assume that sea water has the parameters epsilon1 = 80epsilon1 0 , μ = μ 0 , σ = 4 S/m, and that these parameters are frequency-independent. Plot the ω–β diagram for a plane wave propagatinginthismediumandcomparetoFigure4.12.Describethedispersion:isit normal or anomalous? Also plot the phase and group velocities and compare to Figure 4.13.Howdoestherelaxationphenomenona?ectthevelocityofawaveinthismedium? 4.7 Consider a uniform plane wave incident at angle θ i onto an interface separating twolosslessmedia(Figure4.18).Assumingperpendicularpolarization,writetheexplicit formsofthetotal?eldsineachregionundertheconditionθ i <θ c , where θ c is the critical angle. Show that the total ?eld in region 1 can be decomposed into a portion that is a pure standing wave in the z-direction and a portion that is a pure traveling wave in the z-direction. Also show that the ?eld in region 2 is a pure traveling wave. Repeat for parallel polarization. 4.8 Consider a uniform plane wave incident at angle θ i onto an interface separating twolosslessmedia(Figure4.18).Assumingperpendicularpolarization,usethetotal ?eldsfromProblem4.7toshowthatundertheconditionθ i <θ c the normal component of the time-average Poynting vector is continuous across the interface. Here θ c is the critical angle. Repeat for parallel polarization. 4.9 Consider a uniform plane wave incident at angle θ i onto an interface separating twolosslessmedia(Figure4.18).Assumingperpendicularpolarization,writetheexplicit forms of the total ?elds in each region under the condition θ i >θ c , where θ c is the critical angle. Show that the ?eld in region 1 is a pure standing wave in the z-direction and that the?eldinregion2isanevanescentwave.Repeatforparallelpolarization. 4.10 Consider a uniform plane wave incident at angle θ i onto an interface separating twolosslessmedia(Figure4.18).Assumingperpendicularpolarization,usethe?elds from Problem 4.9 to show that under the condition θ i >θ c the ?eld in region 1 carries no time-average power in the z-direction, while the ?eld in region 2 carries no time-average power. Here θ c is the critical angle. Repeat for parallel polarization. 4.11 Consider a uniform plane wave incident at angle θ i from a lossless material onto agoodconductor(Figure4.18).Theconductorhaspermittivityepsilon1 0 , permeability μ 0 , and conductivity σ. Show that the transmission angle is θ t ≈ 0 and thus the wave in the conductor propagates normal to the interface. Also show that for perpendicular polarization the current per unit width induced by the wave in region 2 is ? K(ω) = ?yσ ? T ⊥ (ω) ? E ⊥ (ω) 1 ? j 2β 2 and that this is identical to the tangential magnetic ?eld at the surface: ? K(ω) =??z × ? H t | z=0 . If we de?ne the surface impedance Z s (ω) of the conductor as the ratio of tangential electric and magnetic ?elds at the interface, show that Z s (ω) = 1 + j σδ = R s (ω)+ jX s (ω). Then show that the time-average power ?ux entering region 2 for a monochromatic wave of frequency ˇω is simply S av,2 = ?z 1 2 ( ˇ K · ˇ K ? )R s . Note that the since the surface impedance is also the ratio of tangential electric ?eld 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-layeredmaterialasshowninFigure4.20.Writingthe?eldsineachregion n, 0 ≤ n ≤ N ? 1,as ? H bardbln = ? H i bardbln + ? H r bardbln where ? H i bardbln = ?ya n+1 e ?jk x,n x e ?jk z,n (z?z n+1 ) , ? H r bardbln =??yb n+1 e ?jk x,n x e +jk z,n (z?z n+1 ) , and the ?eld in region N as ? H bardblN = ?ya N+1 e ?jk x,N x e ?jk z,N (z?z N ) , apply the boundary conditions to solve for the wave amplitudes a n+1 and b n in terms of a global re?ection coe?cient ? R n , an interfacial re?ection coe?cient Gamma1 nbardbl , and the wave amplitude a n . 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 = epsilon1 r epsilon1 0 and permeability μ = μ r μ 0 located in free space between the planes z = z 1 and z = z 2 . A right-hand circularly-polarized plane wave is incident on the slab at angle θ i as shown in Figure 4.22.Determinetheconditions(ifany)underwhichthere?ectedwaveis:(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 = epsilon1 r epsilon1 0 and permeability μ 0 located in free space between the planes z = z 1 and z = z 2 . A transient, perpendicularly- polarizedplanewaveisobliquelyincidentontheslabasshowninFigure4.22.Ifthe temporal waveform of the incident wave is E i ⊥ (t), ?nd the transient re?ected ?eld in region 0 and the transient transmitted ?eld in region 2 in terms of an in?nite superposition of amplitude-scaled, time-shifted versions of the incident wave. Interpret each of the ?rst four terms in the re?ected and transmitted ?elds in terms of multiple re?ection within the slab. 4.15 Consider a free-space gap embedded between the planes z = z 1 and z = z 2 in an in?nite, lossless dielectric medium of permittivity epsilon1 r epsilon1 0 and permeability μ 0 .A perpendicularly-polarized plane wave is incident on the gap at angle θ i >θ c as shown inFigure4.22.Hereθ c is the critical angle for a plane wave incident on the single interface between a lossless dielectric of permittivity epsilon1 r epsilon1 0 and free space. Apply the boundary conditions and ?nd the ?elds in each of the three regions. Find the time- average Poynting vector in region 0 at z = z 1 , in region 1 at z = z 2 , and in region 2 at z = z 2 . Is conservation of energy obeyed? 4.16 A uniform ferrite material has scalar permittivity ?epsilon1 = epsilon1 and dyadic permeability ? ˉμ. 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 wave equation for a TEM plane wave of the form ? H(r,ω)= ? H 0 (ω)e ?jk z z is k 2 z ? H 0 = ω 2 epsilon1 ? ˉμ· ? H 0 where k z = β ? jα. Find explicit formulas for the two solutions k z± = β ± ? jα ± . Show that when the damping parameter α lessmuch 1, near resonance α + greatermuch α ? . 4.17 A time-harmonic, TE-polarized, uniform cylindrical wave propagates in a lossy medium. Assuming |kρ|greatermuch1, show that the power per unit length passing through a cylinder of radius ρ is given by P av /l = Re braceleftbig Z ? TE bracerightbig | ˇ H z0 | 2 e ?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 P av /l = η| ˇ H z0 | 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 the cylinder 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 by the incident wave. Elsewhere, in the “illuminated region,” the induced current is given by ? J s = 2?n × ? H i . Plot the current from (4.365) for various values of k 0 a and compare to the current com- puted from physical optics. How large must k 0 a be for the shadowing e?ect to be signif- icant? 4.19 The radar cross section of a two-dimensional object illuminated by a TM-polarized plane wave is de?ned by σ 2?D (ω,φ) = lim ρ→∞ 2πρ | ? E s z | 2 | ? E i z | 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 σ 2?D (ω,φ) = a π 2 k 0 a 1 ln 2 (0.89k 0 a) 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 ?epsilon1 c (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 ?eld both internal and external to the cylinder. Show that as ?σ →∞the magnetic ?eld external to the cylinder reduces to (4.366). 4.21 A TM-polarized plane wave is incident on a PEC cylinder of radius a aligned 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 free space and determine the total ?eld 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 ,φ 0 ). Expand the ?elds in the regions a <ρ<ρ 0 and ρ>ρ 0 in terms of nonuniform cylindrical waves, and apply the boundary conditions at ρ = a and ρ = ρ 0 to determine the ?elds 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 2π A(k x ,ω)sinξ, use the relations cos z = cos(u + jv)= cos u coshv ? j sin u sinhv, sin z = sin(u + jv)= sin u coshv + j cos u sinhv, toshowthatthecontourinFigure4.29providesidenticalvaluesoftheintegrandin ? ψ(x, y,ω)= integraldisplay C f (ξ,ω)e ?jkρ cos(φ±ξ) dξ as does the contour [?∞+ jDelta1,∞+jDelta1] in ? ψ(x, y,ω)= 1 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 A(k x ,ω)e ?jk x x e ?jk y y dk x . (4.433) 4.25 Verify (4.409) by writing the TE ?elds 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 < 0 contains a perfect electric conductor. Write the ?elds in the regions 0 < y < h and y > h in terms of the Fourier transform solution to the homogeneous Helmholtz equation. Note that in the region 0 < y < h terms representing waves traveling in both the ±y-directions are needed, while in the region y > h only terms traveling in the y-direction are needed. Apply the boundary conditions at y = 0, h to determine the spectral amplitudes. Show that the total ?eld may be decomposed into an impressed term 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 > 0 contains a material with parameters ?epsilon1 c 1 (ω) and ?μ 1 (ω), while the region y < 0 contains a material with parameters ?epsilon1 c 2 (ω) and ?μ 2 (ω). Using the Fourier transform solution to the Helmholtz equation, write the total ?eld for y > 0 as the sum of an impressed ?eld of the magnetic line source and a scattered ?eld, and write the ?eld for y < 0 as a scattered ?eld. Apply the boundary conditions at y = 0 to determine the spectral amplitudes. Can you interpret the scattered ?elds in terms of images of the line source? 4.28 Consider a TE-polarized plane wave incident on a PEC half-plane located at y = 0, x > 0. If the incident magnetic ?eld is given by ? H i (r,ω)= ?z ? H 0 (ω)e jk(x cosφ 0 +y sinφ 0 ) , determine the appropriate boundary conditions on the ?elds at y = 0. Solve for the scattered magnetic ?eld using the Fourier transform approach. 4.29ConsiderthelayeredmediumofFigure4.34withalternatinglayersoffreespace and perfect dielectric. The dielectric layer has permittivity 4epsilon1 0 and thickness Delta1 while the free space layer has thickness 2Delta1. Assuming a normally-incident plane wave, solve for k 0 Delta1 in terms of κDelta1, and plot k 0 versus κ, identifying the stop and pass bands. This type of ω–β plot for a periodic medium is named a Brillouin diagram, after L. Brillouin who investigated energy bands in periodic crystal lattices [23]. 4.30ConsideraperiodiclayeredmediumasinFigure4.34,butwitheachcellcon- sisting of three di?erent layers. Derive an eigenvalue equation similar to (4.427) for the propagation constant.