《电磁学》电子书（英文版）：chap05

Chapter 5 Field decompositions and the EM potentials 5.1 Spatial symmetry decompositions Spatial symmetry can often be exploited to solve electromagnetics problems. For analyticsolutions,symmetrycanbeusedtoreducethenumberofboundaryconditions thatmustbeapplied. Forcomputersolutionsthestoragerequirementscanbereduced. Typical symmetries include rotation about a point or axis, and re?ection through a plane,alonganaxis,orthroughapoint. Weshallconsiderthecommoncaseofre?ection through a plane. Re?ections through the origin and through an axis will be treated in theexercises. Note that spatial symmetry decompositions may be applied even if the sources and ?elds possess no spatial symmetry. As long as the boundaries and material media are symmetric,thesourcesand?eldsmaybedecomposedintoconstituentsthatindividually mimicthesymmetryoftheenvironment. 5.1.1 Planar ?eld symmetry Consideraregionofspaceconsistingoflinear,isotropic,time-invariantmediahaving material parameters epsilon1(r), μ(r), and σ(r). The electromagnetic ?elds (E,H) within this regionarerelatedtotheirimpressedsources(J i ,J i m )andtheirsecondarysources J s = σE throughMaxwell’scurlequations: ?E z ?y ? ?E y ?z =?μ ?H x ?t ? J i mx , (5.1) ?E x ?z ? ?E z ?x =?μ ?H y ?t ? J i my , (5.2) ?E y ?x ? ?E x ?y =?μ ?H z ?t ? J i mz , (5.3) ?H z ?y ? ?H y ?z = epsilon1 ?E x ?t +σ E x + J i x , (5.4) ?H x ?z ? ?H z ?x = epsilon1 ?E y ?t +σ E y + J i y , (5.5) ?H y ?x ? ?H x ?y = epsilon1 ?E z ?t +σ E z + J i z . (5.6) Weassumethematerialconstantsaresymmetricaboutsomeplane,say z = 0. Then epsilon1(x, y,?z) = epsilon1(x, y, z), μ(x, y,?z) = μ(x, y, z), σ(x, y,?z) = σ(x, y, z). Thatis,withrespectto z thematerialconstantsareevenfunctions. Wefurtherassume thattheboundariesandboundaryconditions,whichguaranteeuniquenessofsolution,are alsosymmetricaboutthe z = 0 plane. Thenwede?netwocasesofre?ectionsymmetry. Conditions for even symmetry. Weclaimthatifthesourcesobey J i x (x, y, z) = J i x (x, y,?z), J i mx (x, y, z) =?J i mx (x, y,?z), J i y (x, y, z) = J i y (x, y,?z), J i my (x, y, z) =?J i my (x, y,?z), J i z (x, y, z) =?J i z (x, y,?z), J i mz (x, y, z) = J i mz (x, y,?z), thenthe?eldsobey E x (x, y, z) = E x (x, y,?z), H x (x, y, z) =?H x (x, y,?z), E y (x, y, z) = E y (x, y,?z), H y (x, y, z) =?H y (x, y,?z), E z (x, y, z) =?E z (x, y,?z), H z (x, y, z) = H z (x, y,?z). Theelectric?eldsharesthesymmetryoftheelectricsource: componentsparalleltothe z = 0 planeareevenin z,andthecomponentperpendicularisodd. Themagnetic?eld sharesthesymmetryofthemagneticsource: componentsparalleltothe z = 0 planeare oddin z,andthecomponentperpendiculariseven. We can verify our claim by showing that the symmetric ?elds and sources obey Maxwell’sequations. Atanarbitrarypoint z = a > 0 equation(5.1)requires ?E z ?y vextendsingle vextendsingle vextendsingle vextendsingle z=a ? ?E y ?z vextendsingle vextendsingle vextendsingle vextendsingle z=a =?μ| z=a ?H x ?t vextendsingle vextendsingle vextendsingle vextendsingle z=a ? J i mx | z=a . Bytheassumedsymmetryconditiononsourceandmaterialconstantweget ?E z ?y vextendsingle vextendsingle vextendsingle vextendsingle z=a ? ?E y ?z vextendsingle vextendsingle vextendsingle vextendsingle z=a =?μ| z=?a ?H x ?t vextendsingle vextendsingle vextendsingle vextendsingle z=a + J i mx | z=?a . Ifourclaimholdsregardingthe?eldbehavior,then ?E z ?y vextendsingle vextendsingle vextendsingle vextendsingle z=?a =? ?E z ?y vextendsingle vextendsingle vextendsingle vextendsingle z=a , ?E y ?z vextendsingle vextendsingle vextendsingle vextendsingle z=?a =? ?E y ?z vextendsingle vextendsingle vextendsingle vextendsingle z=a , ?H x ?t vextendsingle vextendsingle vextendsingle vextendsingle z=?a =? ?H x ?t vextendsingle vextendsingle vextendsingle vextendsingle z=a , and we have ? ?E z ?y vextendsingle vextendsingle vextendsingle vextendsingle z=?a + ?E y ?z vextendsingle vextendsingle vextendsingle vextendsingle z=?a = μ| z=?a ?H x ?t vextendsingle vextendsingle vextendsingle vextendsingle z=?a + J i mx | z=?a . SothiscomponentofFaraday’slawissatis?ed. Withsimilarreasoningwecanshowthat thesymmetricsourcesand?eldssatisfy(5.2)–(5.6)aswell. Conditions for odd symmetry. Wecanalsoshowthatifthesourcesobey J i x (x, y, z) =?J i x (x, y,?z), J i mx (x, y, z) = J i mx (x, y,?z), J i y (x, y, z) =?J i y (x, y,?z), J i my (x, y, z) = J i my (x, y,?z), J i z (x, y, z) = J i z (x, y,?z), J i mz (x, y, z) =?J i mz (x, y,?z), thenthe?eldsobey E x (x, y, z) =?E i x (x, y,?z), H x (x, y, z) = H x (x, y,?z), E y (x, y, z) =?E y (x, y,?z), H y (x, y, z) = H y (x, y,?z), E z (x, y, z) = E z (x, y,?z), H z (x, y, z) =?H z (x, y,?z). Again the electric ?eld has the same symmetry as the electric source. However, in this casecomponentsparalleltothe z = 0planeareoddin z andthecomponentperpendicular iseven. Similarly,themagnetic?eldhasthesamesymmetryasthemagneticsource. Here componentsparalleltothe z = 0 planeareevenin z andthecomponentperpendicular isodd. Fieldsymmetriesandtheconceptofsourceimages. Inthecaseofoddsymmetry the electric ?eld parallel to the z = 0 plane is an odd function of z. If we assume that the ?eld is also continuous across this plane, then the electric ?eld tangential to z = 0 mustvanish: theconditionrequiredatthesurfaceofaperfectelectricconductor(PEC). Wemayregardtheproblemofsourcesaboveaperfectconductorinthe z = 0 planeas equivalent totheproblemofsourcesoddaboutthisplane,aslongasthesourcesinboth casesareidenticalfor z > 0. Werefertothesourceintheregion z < 0 asthe image of thesourceintheregion z > 0. Thustheimagesource (J I ,J I m ) obeys J I x (x, y,?z) =?J i x (x, y, z), J I mx (x, y,?z) = J i mx (x, y, z), J I y (x, y,?z) =?J i y (x, y, z), J I my (x, y,?z) = J i my (x, y, z), J I z (x, y,?z) = J i z (x, y, z), J I mz (x, y,?z) =?J i mz (x, y, z). That is, parallel components of electric current image in the opposite direction, and the perpendicular component images in the same direction; parallel components of the magneticcurrentimageinthesamedirection,whiletheperpendicularcomponentimages intheoppositedirection. In the case of even symmetry, the magnetic ?eld parallel to the z = 0 plane is odd, and thus the magnetic ?eld tangential to the z = 0 plane must be zero. We therefore haveanequivalencebetweentheproblemofasourceaboveaplaneofperfectmagnetic conductor (PMC) and the problem of sources even about that plane. In this case we identifyimagesourcesthatobey J I x (x, y,?z) = J i x (x, y, z), J I mx (x, y,?z) =?J i mx (x, y, z), J I y (x, y,?z) = J i y (x, y, z), J I my (x, y,?z) =?J i my (x, y, z), J I z (x, y,?z) =?J i z (x, y, z), J I mz (x, y,?z) = J i mz (x, y, z). Parallelcomponentsofelectriccurrentimageinthesamedirection,andtheperpendicular component images in the opposite direction; parallel components of magnetic current image in the opposite direction, and the perpendicular component images in the same direction. Inthecaseofoddsymmetry,wesometimessaythatan“electricwall”existsat z = 0. The term “magnetic wall” can be used in the case of even symmetry. These terms are particularlycommoninthedescriptionofwaveguide?elds. Symmetric ?eld decomposition. Field symmetries may be applied to arbitrary sourcedistributionsthroughasymmetrydecompositionofthesourcesand?elds. Con- siderthegeneralimpressedsourcedistributions (J i ,J i m ). Thesourceset J ie x (x, y, z) = 1 2 bracketleftbig J i x (x, y, z)+ J i x (x, y,?z) bracketrightbig , J ie y (x, y, z) = 1 2 bracketleftbig J i y (x, y, z)+ J i y (x, y,?z) bracketrightbig , J ie z (x, y, z) = 1 2 bracketleftbig J i z (x, y, z)? J i z (x, y,?z) bracketrightbig , J ie mx (x, y, z) = 1 2 bracketleftbig J i mx (x, y, z)? J i mx (x, y,?z) bracketrightbig , J ie my (x, y, z) = 1 2 bracketleftbig J i my (x, y, z)? J i my (x, y,?z) bracketrightbig , J ie mz (x, y, z) = 1 2 bracketleftbig J i mz (x, y, z)+ J i mz (x, y,?z) bracketrightbig , isclearlyofevensymmetrictypewhilethesourceset J io x (x, y, z) = 1 2 bracketleftbig J i x (x, y, z)? J i x (x, y,?z) bracketrightbig , J io y (x, y, z) = 1 2 bracketleftbig J i y (x, y, z)? J i y (x, y,?z) bracketrightbig , J io z (x, y, z) = 1 2 bracketleftbig J i z (x, y, z)+ J i z (x, y,?z) bracketrightbig , J io mx (x, y, z) = 1 2 bracketleftbig J i mx (x, y, z)+ J i mx (x, y,?z) bracketrightbig , J io my (x, y, z) = 1 2 bracketleftbig J i my (x, y, z)+ J i my (x, y,?z) bracketrightbig , J io mz (x, y, z) = 1 2 bracketleftbig J i mz (x, y, z)? J i mz (x, y,?z) bracketrightbig , isoftheoddsymmetrictype. Since J i = J ie + J io and J i m = J ie m + J io m ,wecandecompose any source into constituents having, respectively, even and odd symmetry with respect toaplane. Thesourcewithevensymmetryproducesaneven?eldset,whilethesource with odd symmetry produces an odd ?eld set. The total ?eld is the sum of the ?elds fromeach?eldset. Planar symmetry for frequency-domain ?elds. The symmetry conditions intro- ducedaboveforthetime-domain?eldsalsoholdforthefrequency-domain?elds. Because boththeconductivityandpermittivitymustbeevenfunctions,wecombinetheire?ects and require the complex permittivity to be even. Otherwise the ?eld symmetries and sourcedecompositionsareidentical. Exampleofsymmetrydecomposition: linesourcebetweenconductingplanes. Considera z-directedelectriclinesource ? I 0 locatedat y = h, x = 0 betweenconducting planes at y =±d, d > h. The material between the plates has permeability ?μ(ω) and complexpermittivity ?epsilon1 c (ω). Wedecomposethesourceintooneofevensymmetrictype with line sources ? I 0 /2 located at y =±h, and one of odd symmetric type with a line source ? I 0 /2 locatedat y = h andalinesource ? ? I 0 /2 locatedat y =?h. Wesolveeach oftheseproblemsbyexploitingtheappropriatesymmetry,andsuperposetheresultsto ?ndthesolutiontotheoriginalproblem. For the even-symmetric case, we begin by using (4.407) to represent the impressed ?eld: ? E i z (x, y,ω)=? ω ?μ ? I 0 (ω) 2 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 e ?jk y |y?h| + e ?jk y |y+h| 2k y e ?jk x x dk x . For y > h thisbecomes ? E i z (x, y,ω)=? ω ?μ ? I 0 (ω) 2 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 2 cos k y h 2k y e ?jk y y e ?jk x x dk x . Thesecondary(scattered)?eldconsistsofwavespropagatinginboththe±y-directions: ? E s z (x, y,ω)= 1 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 bracketleftbig A + (k x ,ω)e ?jk y y + A ? (k x ,ω)e jk y y bracketrightbig e ?jk x x dk x . (5.7) The impressed ?eld is even about y = 0. Since the total ?eld E z = E i z + E s z must be evenin y (E z isparalleltotheplane y = 0),thescattered?eldmustalsobeeven. Thus, A + = A ? andthetotal?eldisfor y > h ? E z (x, y,ω)= 1 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 bracketleftbigg 2A + (k x ,ω)cos k y y ?ω ?μ ? I 0 (ω) 2 2 cos k y h 2k y e ?jk y y bracketrightbigg e ?jk x x dk x . Nowthe electric ?eld must obey the boundary condition ? E z = 0 at y =±d. However, since ? E z is even the satisfaction of this condition at y = d automatically implies its satisfactionat y =?d. Soweset 1 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 bracketleftbigg 2A + (k x ,ω)cos k y d ?ω ?μ ? I 0 (ω) 2 2 cos k y h 2k y e ?jk y d bracketrightbigg e ?jk x x dk x = 0 andinvoketheFourierintegraltheoremtoget A + (k x ,ω)= ω ?μ ? I 0 (ω) 2 cos k y h 2k y e ?jk y d cos k y d . Thetotal?eldforthiscaseis ? E z (x, y,ω)=? ω ?μ ? I 0 (ω) 2 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 bracketleftbigg e ?jk y |y?h| + e ?jk y |y+h| 2k y ? ? 2 cos k y h 2k y e ?jk y d cos k y d cos k y y bracketrightbigg e ?jk x x dk x . Fortheodd-symmetriccasetheimpressed?eldis ? E i z (x, y,ω)=? ω ?μ ? I 0 (ω) 2 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 e ?jk y |y?h| ? e ?jk y |y+h| 2k y e ?jk x x dk x , whichfor y > h is ? E i z (x, y,ω)=? ω ?μ ? I 0 (ω) 2 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 2 j sin k y h 2k y e ?jk y y e ?jk x x dk x . The scattered ?eld has the form of (5.7) but must be odd. Thus A + =?A ? and the total?eldfor y > h is ? E z (x, y,ω)= 1 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 bracketleftbigg 2 jA + (k x ,ω)sin k y y ?ω ?μ ? I 0 (ω) 2 2 j sin k y h 2k y e ?jk y y bracketrightbigg e ?jk x x dk x . Setting ? E z = 0 at z = d andsolvingfor A + we?ndthatthetotal?eldforthiscaseis ? E z (x, y,ω)=? ω ?μ ? I 0 (ω) 2 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 bracketleftbigg e ?jk y |y?h| ? e ?jk y |y+h| 2k y ? ? 2 j sin k y h 2k y e ?jk y d sin k y d sin k y y bracketrightbigg e ?jk x x dk x . Addingthe?eldsforthetwocaseswe?ndthat ? E z (x, y,ω)=? ω ?μ ? I 0 (ω) 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 e ?jk y |y?h| 2k y e ?jk x x dk x + + ω ?μ ? I 0 (ω) 2π ∞+jDelta1 integraldisplay ?∞+jDelta1 bracketleftbigg cos k y h cos k y y cos k y d + j sin k y h sin k y y sin k y d bracketrightbigg e ?jk y d 2k y e ?jk x x dk x , (5.8) whichisasuperpositionofimpressedandscattered?elds. 5.2 Solenoidal–lamellar decomposition Wenowdiscussthedecompositionofageneralvector?eldintoa lamellar component havingzerocurlandasolenoidal componenthavingzerodivergence. Thisisknownasa Helmholtz decomposition.IfV isanyvector?eldthenwewishtowrite V = V s + V l , (5.9) where V s and V l arethesolenoidalandlamellarcomponentsof V. Formulasexpressing thesecomponentsintermsof V areobtainedasfollows. We?rstwrite V s intermsofa “vectorpotential” A as V s =?×A. (5.10) Thisispossiblebyvirtueof(B.49). Similarly,wewriteV l intermsofa“scalarpotential” φ as V l =?φ. (5.11) Toobtainaformulafor V l wetakethedivergenceof(5.9)anduse(5.11)toget ?·V =?·V l =?·?φ =? 2 φ. Theresult, ? 2 φ =?·V, may be regarded as Poisson’s equation for the unknown φ. This equation is solved in Chapter3. By(3.61)wehave φ(r) =? integraldisplay V ? prime · V(r prime ) 4π R dV prime , where R =|r ? r prime |,andwehave V l (r) =?? integraldisplay V ? prime · V(r prime ) 4π R dV prime . (5.12) Similarly,aformulafor V s canbeobtainedbytakingthecurlof(5.9)toget ?×V =?×V s . Substituting(5.10)wehave ?×V =?×(?×A) =?(?·A)?? 2 A. We may choose any value we wish for ?·A, since this does not alter V s =?×A. (Wediscusssuch“gaugetransformations”ingreaterdetaillaterinthischapter.) With ?·A = 0 weobtain ?? × V =? 2 A. ThisisPoisson’sequationforeachrectangularcomponentof A;therefore A(r) = integraldisplay V ? prime × V(r prime ) 4π R dV prime , and we have V s (r) =?× integraldisplay V ? prime × V(r prime ) 4π R dV prime . SummingtheresultsweobtaintheHelmholtzdecomposition V = V l + V s =?? integraldisplay V ? prime · V(r prime ) 4π R dV prime +?× integraldisplay V ? prime × V(r prime ) 4π R dV prime . (5.13) Identi?cationoftheelectromagneticpotentials. Letuswritetheelectromagnetic ?eldsasageneralsuperpositionofsolenoidalandlamellarcomponents: E =?×A E +?φ E , (5.14) B =?×A B +?φ B . (5.15) One possible form of the potentials A E , A B , φ E , and φ B appears in (5.13). However, because E and B arerelatedbyMaxwell’sequations,thepotentialsshouldberelatedto thesources. Wecandeterminetheexplicitrelationshipbysubstituting(5.14)and(5.15) into Ampere’s and Faraday’s laws. It is most convenient to analyze the relationships usingsuperpositionofthecasesforwhich J m = 0 and J = 0. With J m = 0 Faraday’slawis ?×E =? ?B ?t . (5.16) Since ?×E is solenoidal, B must be solenoidal and thus ?φ B = 0. This implies that φ B = 0, which is equivalent to the auxiliary Maxwell equation ?·B = 0.Now, substitutionof(5.14)and(5.15)into(5.16)gives ?×[?×A E +?φ E ] =? ? ?t [?×A B ]. Using ?×(?φ E ) = 0 andcombiningthetermsweget ?× bracketleftbigg ?×A E + ?A B ?t bracketrightbigg = 0, hence ?×A E =? ?A B ?t +?ξ. Substitutioninto(5.14)gives E =? ?A B ?t + [?φ E +?ξ]. Combiningthetwogradientfunctionstogether,weseethatwecanwritebothE and B intermsoftwopotentials: E =? ?A e ?t ??φ e , (5.17) B =?×A e , (5.18) wherethenegativesignonthegradienttermisintroducedbyconvention. GaugetransformationsandtheCoulombgauge. Wepayapriceforthesimplicity ofusingonlytwopotentialstorepresent E and B. While ?×A e isde?nitelysolenoidal, A e itselfmaynotbe: becauseofthis(5.17)maynotbeadecompositionintosolenoidal andlamellarcomponents. However, acorollaryoftheHelmholtztheoremstatesthata vector?eldisuniquelyspeci?edonlywhenbothitscurlanddivergencearespeci?ed. Here there is an ambiguity in the representation of E and B; we may remove this ambiguity andde?ne A e uniquelybyrequiringthat ?·A e = 0. (5.19) Then A e issolenoidalandthedecomposition(5.17)issolenoidal–lamellar. Thisrequire- menton A e iscalledthe Coulomb gauge. The ambiguity implied by the non-uniqueness of ?·A e can also be expressed by the observationthatatransformationofthetype A e → A e +?Gamma1, (5.20) φ e → φ e ? ?Gamma1 ?t , (5.21) leavestheexpressions(5.17)and(5.18)unchanged. Thisiscalledagaugetransformation, andthechoiceofacertain Gamma1 altersthespeci?cationof ?·A e . Thuswemaybeginwith the Coulomb gauge as our baseline, and allowany alteration of A e according to (5.20) aslongasweaugment?·A e by ?·?Gamma1 =? 2 Gamma1. Once ?·A e is speci?ed, the relationship between the potentials and the current J can be found by substitution of (5.17) and (5.18) into Ampere’s law. At this point we assume media that are linear, homogeneous, isotropic, and described by the time- invariantparameters μ, epsilon1,and σ. Writing J = J i +σE we have 1 μ ?×(?×A e ) = J i ?σ ?A e ?t ?σ?φ e ?epsilon1 ? 2 A e ?t 2 ?epsilon1 ? ?t ?φ e . (5.22) Takingthedivergenceofbothsidesof(5.22)weget 0 =?·J i ?σ ? ?t ?·A ?σ?·?φ e ?epsilon1 ? 2 ?t 2 ?·A e ?epsilon1 ? ?t ?·?φ e . (5.23) Then,bysubstitutionfromthecontinuityequationanduseof(5.19)alongwith?·?φ e = ? 2 φ e weobtain ? ?t parenleftbig ρ i +epsilon1? 2 φ e parenrightbig =?σ? 2 φ e . Foralosslessmediumthisreducesto ? 2 φ e =?ρ i /epsilon1 (5.24) and we have φ e (r,t) = integraldisplay V ρ i (r prime ,t) 4πepsilon1R dV prime . (5.25) Wecanobtainanequationfor A e byexpandingtheleft-handsideof(5.22)toget ?(?·A e )?? 2 A e = μJ i ?σμ ?A e ?t ?σμ?φ e ?μepsilon1 ? 2 A e ?t 2 ?μepsilon1 ? ?t ?φ e , (5.26) hence ? 2 A e ?μepsilon1 ? 2 A e ?t 2 =?μJ i +σμ ?A e ?t +σμ?φ e +μepsilon1 ? ?t ?φ e undertheCoulombgauge. Forlosslessmediathisbecomes ? 2 A e ?μepsilon1 ? 2 A e ?t 2 =?μJ i +μepsilon1 ? ?t ?φ e . (5.27) Observethattheleft-handsideof(5.27)issolenoidal(sincetheLaplaciantermcame from the curl-curl, and ?·A e = 0), while the right-hand side contains a general vector ?eld J i and a lamellar term. We might expect the ?φ e term to cancel the lamellar portionof J i , andthisdoeshappen[91]. By(5.12)andthecontinuityequationwecan writethelamellarcomponentofthecurrentas J i l (r,t) =?? integraldisplay V ? prime · J i (r prime ,t) 4π R dV prime = ? ?t ? integraldisplay V ρ i (r prime ,t) 4π R dV prime = epsilon1 ? ?t ?φ e . Thus(5.27)becomes ? 2 A e ?μepsilon1 ? 2 A e ?t 2 =?μJ i s . (5.28) Thereforethevectorpotential A e ,whichdescribesthesolenoidalportionofboth E and B,isfoundfromjustthesolenoidalportionofthecurrent. Ontheotherhand,thescalar potential,whichdescribesthelamellarportionof E,isfoundfrom ρ i whicharisesfrom ?·J i ,thelamellarportionofthecurrent. From the perspective of ?eld computation, we see that the introduction of potential functions has reoriented the solution process from dealing with two coupled ?rst-order partialdi?erentialequations(Maxwell’sequations),totwouncoupledsecond-orderequa- tions(thepotentialequations(5.24)and(5.28)). Thedecouplingoftheequationsisoften worththeaddedcomplexityofdealingwithpotentials,and,infact,isthesolutiontech- nique of choice in such areas as radiation and guided waves. It is worth pausing for a moment to examine the form of these equations. We see that the scalar potential obeys Poisson’s equation with the solution (5.25), while the vector potential obeys the wave equation. As a wave, the vector potential must propagate away from the source with?nitevelocity. However, thesolutionforthescalarpotential(5.25)showsnosuch behavior. In fact, any change to the charge distribution instantaneously permeates all ofspace. ThisapparentviolationofEinstein’spostulateshowsthatwemustbecareful wheninterpretingthephysicalmeaningofthepotentials. Oncethecomputations(5.17) and(5.18)areundertaken,we?ndthatboth E and B behaveaswaves,andthuspropa- gateat?nitevelocity. Mathematically,theconundrumcanberesolvedbyrealizingthat individuallythesolenoidalandlamellarcomponentsofcurrentmustoccupyallofspace, eveniftheirsum,theactualcurrent J i ,islocalized[91]. The Lorentz gauge. Adi?erentchoiceofgaugeconditioncanallowboththevector and scalar potentials to act as waves. In this case E may be written as a sum of two terms: one purely solenoidal, and the other a superposition of lamellar and solenoidal parts. Letusexaminethee?ectofchoosingthe Lorentz gauge ?·A e =?μepsilon1 ?φ e ?t ?μσφ e . (5.29) Substitutingthisexpressioninto(5.26)we?ndthatthegradienttermscancel,giving ? 2 A e ?μσ ?A e ?t ?μepsilon1 ? 2 A e ?t 2 =?μJ i . (5.30) Forlosslessmedia ? 2 A e ?μepsilon1 ? 2 A e ?t 2 =?μJ i , (5.31) and(5.23)becomes ? 2 φ e ?μepsilon1 ? 2 φ e ?t 2 =? ρ i epsilon1 . (5.32) For lossy media we have obtained a second-order di?erential equation for A e , but φ e must be found through the somewhat cumbersome relation (5.29). For lossless media thecoupledMaxwellequationshavebeendecoupledintotwosecond-orderequations,one involving A e and one involving φ e . Both (5.31) and (5.32) are wave equations, with J i asthesourcefor A e and ρ i asthesourcefor φ e . Thustheexpected?nite-velocitywave natureoftheelectromagnetic?eldsisalsomanifestedineachofthepotentialfunctions. Thedrawbackisthat,eventhoughwecanstilluse(5.17)and(5.18),theexpressionfor E isnolongeradecompositionintosolenoidalandlamellarcomponents. Nevertheless,the choiceoftheLorentzgaugeisverypopularinthestudyofradiatedandguidedwaves. The Hertzian potentials. Withalittlemanipulationandtheintroductionofanew notation,wecanmaintainthewavenatureofthepotentialfunctionsandstillprovidea decompositionintopurelylamellarandsolenoidalcomponents. Inthisanalysisweshall assumelosslessmediaonly. WhenwechosetheLorentzgaugetoremovethearbitrarinessofthedivergenceofthe vector potential, we established a relationship between A e and φ e . Thus we should be abletowriteboththeelectricandmagnetic?eldsintermsofasinglepotentialfunction. FromtheLorentzgaugewecanwriteφ e as φ e (r,t) =? 1 μepsilon1 integraldisplay t ?∞ ?·A e (r,t)dt. By(5.17)and(5.18)wecanthuswritetheEM?eldsas E = 1 μepsilon1 ? integraldisplay t ?∞ ?·A e dt ? ?A e ?t , (5.33) B =?×A e . (5.34) The integro-di?erential representation of E in (5.33) is somewhat clumsy in appear- ance. Wecanmakeiteasiertomanipulatebyde?ningthe Hertzian potential Π e = 1 μepsilon1 integraldisplay t ?∞ A e dt. Indi?erentialform A e = μepsilon1 ?Π e dt . (5.35) Withthis,(5.33)and(5.34)become E =?(?·Π e )?μepsilon1 ? 2 ?t 2 Π e , (5.36) B = μepsilon1?× ?Π e ?t . (5.37) AnequationforΠ e intermsofthesourcecurrentcanbefoundbysubstituting(5.35) into(5.31): μepsilon1 ? ?t parenleftbigg ? 2 Π e ?μepsilon1 ? 2 ?t 2 Π e parenrightbigg =?μJ i . Letusde?ne J i = ?P i ?t . (5.38) Forgeneralimpressedcurrentsources(5.38)isjustaconvenientnotation. However,we can conceive of an impressed polarization current that is independent of E and de?ned through the relation D = epsilon1 0 E + P + P i . Then (5.38) has a physical interpretation as describedin(2.119). Wenowhave ? 2 Π e ?μepsilon1 ? 2 ?t 2 Π e =? 1 epsilon1 P i , (5.39) whichisawaveequationforΠ e . ThustheHertzianpotentialhasthesamewavebehavior asthevectorpotentialundertheLorentzgauge. We can use (5.39) to perform one ?nal simpli?cation of the EM ?eld representation. Bythevectoridentity ?(?·Π) =?×(?×Π)+? 2 Π weget ?(?·Π e ) =?×(?×Π e )? 1 epsilon1 P i +μepsilon1 ? 2 ?t 2 Π e . Substitutingthisinto(5.36)weobtain E =?×(?×Π e )? P i epsilon1 , (5.40) B = μepsilon1?× ?Π e ?t . (5.41) Letusexaminetheseclosely. Weknowthat B issolenoidalsinceitiswrittenasthecurl ofanothervector(thisisalsoclearfromtheauxiliaryMaxwellequation?·B = 0). The ?rst term in the expression for E is also solenoidal. So the lamellar part of E must be containedwithinthesourceterm P i . IfwewriteP i intermsofitslamellarandsolenoidal componentsbyusing J i s = ?P i s ?t , J i l = ?P i l ?t , then(5.40)becomes E = bracketleftbigg ?×(?×Π e )? P i s epsilon1 bracketrightbigg ? P i l epsilon1 . (5.42) Sowehaveagainsucceededindividing E intolamellarandsolenoidalcomponents. Potential functions for magnetic current. Wecanproceedasabovetoderivethe ?eld–potential relationships when J i = 0 but J i m negationslash= 0. We assume a homogeneous, loss- less,isotropicmediumwithpermeabilityμandpermittivityepsilon1,andbeginwithFaraday’s andAmpere’slaws ?×E =?J i m ? ?B ?t , (5.43) ?×H = ?D ?t . (5.44) Wewrite H and D intermsoftwopotentialfunctions A h and φ h as H =? ?A h ?t ??φ h , D =??×A h , andthedi?erentialequationforthepotentialsisfoundbysubstitutioninto(5.43): ?×(?×A h ) = epsilon1J i m ?μepsilon1 ? 2 A h ?t 2 ?μepsilon1 ? ?t ?φ h . (5.45) Taking the divergence of this equation and substituting from the magnetic continuity equationweobtain μepsilon1 ? 2 ?t 2 ?·A h +μepsilon1 ? ?t ? 2 φ h =?epsilon1 ?ρ i m ?t . UndertheLorentzgaugecondition ?·A h =?μepsilon1 ?φ h ?t thisreducesto ? 2 φ h ?μepsilon1 ? 2 φ h ?t 2 =? ρ i m μ . Expandingthecurl-curloperationin(5.45)wehave ?(?·A h )?? 2 A h = epsilon1J i m ?μepsilon1 ? 2 A h ?t 2 ?μepsilon1 ? ?t ?φ h , which,uponsubstitutionoftheLorentzgaugeconditiongives ? 2 A h ?μepsilon1 ? 2 A h ?t 2 =?epsilon1J i m . (5.46) WecanalsoderiveaHertzianpotentialforthecaseofmagneticcurrent. Letting A h = μepsilon1 ?Π h ?t (5.47) andemployingtheLorentzconditionwehave D =?μepsilon1?× ?Π h ?t , H =?(?·Π h )?μepsilon1 ? 2 Π h ?t 2 . ThewaveequationforΠ h isfoundbysubstituting(5.47)into(5.46)togive ? ?t bracketleftbigg ? 2 Π h ?μepsilon1 ? 2 Π h ?t 2 bracketrightbigg =? 1 μ J i m . (5.48) De?ning M i through J i m = μ ?M i ?t , wewritethewaveequationas ? 2 Π h ?μepsilon1 ? 2 Π h ?t 2 =?M i . We can think of M i as a convenient way of representing J i m , or we can conceive of an impressed magnetization current that is independent of H and de?ned through B = μ 0 (H + M + M i ). Withthehelpof(5.48)wecanalsowritethe?eldsas H =?×(?×Π h )? M i , D =?μepsilon1?× ?Π h ?t . Summary of potential relations for lossless media. Whenbothelectricandmag- neticsourcesarepresent,wemaysuperposethepotentialrepresentationsderivedabove. Weassumeahomogeneous,losslessmediumwithtime-invariantparametersμandepsilon1.For thescalar/vectorpotentialrepresentationwehave E =? ?A e ?t ??φ e ? 1 epsilon1 ?×A h , (5.49) H = 1 μ ?×A e ? ?A h ?t ??φ h . (5.50) Herethepotentialssatisfythewaveequations parenleftbigg ? 2 ?μepsilon1 ? 2 ?t 2 parenrightbiggbraceleftbigg A e φ e bracerightbigg = braceleftbigg ?μJ i ? ρ i epsilon1 bracerightbigg , (5.51) parenleftbigg ? 2 ?μepsilon1 ? 2 ?t 2 parenrightbiggbraceleftbigg A h φ h bracerightbigg = braceleftBigg ?epsilon1J i m ? ρ i m μ bracerightBigg , andarelinkedbytheLorentzconditions ?·A e =?μepsilon1 ?φ e ?t , ?·A h =?μepsilon1 ?φ h ?t . WealsohavetheHertzpotentialrepresentation E =?(?·Π e )?μepsilon1 ? 2 Π e ?t 2 ?μ?× ?Π h ?t =?×(?×Π e )? P i epsilon1 ?μ?× ?Π h ?t , (5.52) H = epsilon1?× ?Π e ?t +?(?·Π h )?μepsilon1 ? 2 Π h ?t 2 = epsilon1?× ?Π e ?t +?×(?×Π h )? M i . (5.53) TheHertzpotentialssatisfythewaveequations parenleftbigg ? 2 ?μepsilon1 ? 2 ?t 2 parenrightbiggbraceleftbigg Π e Π h bracerightbigg = braceleftbigg ? 1 epsilon1 P i ?M i bracerightbigg . Potential functions for the frequency-domain ?elds. Inthefrequencydomainit ismucheasiertohandlelossymedia. Consideralossy,isotropic,homogeneousmedium describedbythefrequency-dependentparameters ?μ, ?epsilon1,and ?σ. Maxwell’scurlequations are ?× ? E =? ? J i m ? jω ?μ ? H, (5.54) ?× ? H = ? J i + jω?epsilon1 c ? E. (5.55) Here we have separated the primary and secondary currents through ? J = ? J i + ?σ ? E, and used the complex permittivity ?epsilon1 c = ?epsilon1 + ?σ/jω. As with the time-domain equations we introduce the potential functions using superposition. If ? J i m = 0 and ? J i negationslash= 0 then we mayintroducetheelectricpotentialsthroughtherelationships ? E =?? ? φ e ? jω ? A e , (5.56) ? H = 1 ?μ ?× ? A e . (5.57) AssumingtheLorentzcondition ?· ? A e =?jω ?μ?epsilon1 c ? φ e , we ?nd that upon substitution of (5.56)–(5.57) into (5.54)–(5.55) the potentials must obeytheHelmholtzequation parenleftbig ? 2 + k 2 parenrightbig braceleftbigg ? φ e ? A e bracerightbigg = braceleftbigg ??ρ i /?epsilon1 c ??μ ? J i bracerightbigg . If ? J i m negationslash= 0 and ? J i = 0 thenwemayintroducethemagneticpotentialsthrough ? E =? 1 ?epsilon1 c ?× ? A h , (5.58) ? H =?? ? φ h ? jω ? A h . (5.59) Assuming ?· ? A h =?jω ?μ?epsilon1 c ? φ h , we ?nd that upon substitution of (5.58)–(5.59) into (5.54)–(5.55) the potentials must obey parenleftbig ? 2 + k 2 parenrightbig braceleftbigg ? φ h ? A h bracerightbigg = braceleftbigg ??ρ i m / ?μ ??epsilon1 c ? J i m bracerightbigg . Whenbothelectricandmagneticsourcesarepresent,weusesuperposition: ? E =?? ? φ e ? jω ? A e ? 1 ?epsilon1 c ?× ? A h , ? H = 1 ?μ ?× ? A e ?? ? φ h ? jω ? A h . UsingtheLorentzconditionswecanalsowritethe?eldsintermsofthevectorpotentials alone: ? E =? jω k 2 ?(?· ? A e )? jω ? A e ? 1 ?epsilon1 c ?× ? A h , (5.60) ? H = 1 ?μ ?× ? A e ? jω k 2 ?(?· ? A h )? jω ? A h . (5.61) Wecanalsode?neHertzianpotentialsforthefrequency-domain?elds. When ? J i m = 0 and ? J i negationslash= 0 welet ? A e = jω ?μ?epsilon1 c ? Π e and?nd ? E =?(?· ? Π e )+ k 2 ? Π e =?×(?× ? Π e )? ? J i jω?epsilon1 c (5.62) and ? H = jω?epsilon1 c ?× ? Π e . (5.63) Here ? J i canrepresenteitheranimpressedelectriccurrentsourceoranimpressedpolar- izationcurrentsource ? J i = jω ? P i . TheelectricHertzianpotentialobeys (? 2 + k 2 ) ? Π e =? ? J i jω?epsilon1 c . (5.64) When ? J i m negationslash= 0 and ? J i = 0 welet ? A h = jω ?μ?epsilon1 c ? Π h and?nd ? E =?jω ?μ?× ? Π h (5.65) and ? H =?(?· ? Π h )+ k 2 ? Π h =?×(?× ? Π h )? ? J i m jω ?μ . (5.66) Here ? J i m can represent either an impressed magnetic current source or an impressed magnetizationcurrentsource ? J i m = jω ?μ ? M i . ThemagneticHertzianpotentialobeys (? 2 + k 2 ) ? Π h =? ? J i m jω ?μ . (5.67) Whenbothelectricandmagneticsourcesarepresentwehavebysuperposition ? E =?(?· ? Π e )+ k 2 ? Π e ? jω ?μ?× ? Π h =?×(?× ? Π e )? ? J i jω?epsilon1 c ? jω ?μ?× ? Π h and ? H = jω?epsilon1 c ?× ? Π e +?(?· ? Π h )+ k 2 ? Π h = jω?epsilon1 c ?× ? Π e +?×(?× ? Π h )? ? J i m jω ?μ . 5.2.1 Solution for potentials in an unbounded medium: the retarded potentials UndertheLorentzconditioneachofthepotentialfunctionsobeysthewaveequation. This equation can be solved using the method of Green’s functions to determine the potentials,andtheelectromagnetic?eldscanthereforebedetermined. Wenowexamine thesolutionforanunboundedmedium. Solutionsforboundedregionsareconsideredin § 5.2.2. Consideralinearoperator L thatoperatesonafunctionof r and t. Ifwewishtosolve theequation L {ψ(r,t)}=S(r,t), (5.68) we?rstsolve L {G(r,t|r prime ,t prime )}=δ(r ? r prime )δ(t ? t prime ) anddeterminetheGreen’sfunction G fortheoperator L . Providedthat S resideswithin V wehave L braceleftbiggintegraldisplay V integraldisplay ∞ ?∞ S(r prime ,t prime )G(r,t|r prime ,t prime )dt prime dV prime bracerightbigg = integraldisplay V integraldisplay ∞ ?∞ S(r prime ,t prime ) L {G(r,t|r prime ,t prime )}dt prime dV prime = integraldisplay V integraldisplay ∞ ?∞ S(r prime ,t prime )δ(r ? r prime )δ(t ? t prime )dt prime dV prime = S(r,t), hence ψ(r,t) = integraldisplay V integraldisplay ∞ ?∞ S(r prime ,t prime )G(r,t|r prime ,t prime )dt prime dV prime (5.69) bycomparisonwith(5.68). Wecanalsoapplythisideainthefrequencydomain. Thesolutionto L { ? ψ(r,ω)}= ? S(r,ω) (5.70) is ? ψ(r,ω)= integraldisplay V ? S(r prime ,ω)G(r|r prime ;ω)dV prime wheretheGreen’sfunction G satis?es L {G(r|r prime ;ω)}=δ(r ? r prime ). Equation(5.69)isthebasicsuperpositionintegralthatallowsusto?ndthepotentials inanin?nite,unboundedmedium. Wenotethatifthemediumisboundedthenwemust useGreen’stheoremtoincludethee?ectsofsourcesthatresideexternaltothebound- aries. These are manifested in terms of the values of the potentials on the boundaries in the same manner as with the static potentials in Chapter 3. In order to determine whether (5.69) is the unique solution to the wave equation, we must also examine the behavior of the ?elds on the boundary as the boundary recedes to in?nity. In the fre- quency domain we ?nd that an additional “radiation condition” is required to ensure uniqueness. The retarded potentials in the time domain. Consider an unbounded, homoge- neous,lossy,isotropicmediumdescribedbyparameters μ,epsilon1,σ. Inthetimedomainthe vectorpotential A e satis?es(5.30). Thescalarcomponentsof A e mustobey ? 2 A e,n (r,t)?μσ ? A e,n (r,t) ?t ?μepsilon1 ? 2 A e,n (r,t) ?t 2 =?μJ i n (r,t), n = x, y, z. Wemaywritethisintheform parenleftbigg ? 2 ? 2Omega1 v 2 ? ?t ? 1 v 2 ? 2 ?t 2 parenrightbigg ψ(r,t) =?S(r,t) (5.71) where ψ = A e,n , v 2 = 1/μepsilon1, Omega1 = σ/2epsilon1,and S = μJ i n . Thesolutionis ψ(r,t) = integraldisplay V integraldisplay ∞ ?∞ S(r prime ,t prime )G(r,t|r prime ,t prime )dt prime dV prime (5.72) where G satis?es parenleftbigg ? 2 ? 2Omega1 v 2 ? ?t ? 1 v 2 ? 2 ?t 2 parenrightbigg G(r,t|r prime ,t prime ) =?δ(r ? r prime )δ(t ? t prime ). (5.73) In § A.1we?ndthat G(r,t|r prime ,t prime ) = e ?Omega1(t?t prime ) δ(t ? t prime ? R/v) 4π R + + Omega1 2 4πv e ?Omega1(t?t prime ) I 1 parenleftBig Omega1 radicalbig (t ? t prime ) 2 ?(R/v) 2 parenrightBig Omega1 radicalbig (t ? t prime ) 2 ?(R/v) 2 , t ? t prime > R v , where R =|r ? r prime |. Forlosslessmediawhere σ = 0 thisbecomes G(r,t|r prime ,t prime ) = δ(t ? t prime ? R/v) 4π R andthus ψ(r,t) = integraldisplay V integraldisplay ∞ ?∞ S(r prime ,t prime ) δ(t ? t prime ? R/v) 4π R dt prime dV prime = integraldisplay V S(r prime ,t ? R/v) 4π R dV prime . (5.74) Forlosslessmedia,thescalarpotentialsandallrectangularcomponentsofthevector potentials obey the same wave equation. Thus we have, for instance, the solutions to (5.51): A e (r,t) = μ 4π integraldisplay V J i (r prime ,t ? R/v) R dV prime , φ e (r,t) = 1 4πepsilon1 integraldisplay V ρ i (r prime ,t ? R/v) R dV prime . Thesearecalledtheretarded potentials sincetheirvaluesattime t aredeterminedbythe valuesofthesourcesatanearlier(orretardation)time t ? R/v. Theretardationtimeis determinedbythepropagationvelocity v ofthepotentialwaves. The?eldsaredeterminedbythepotentials: E(r,t) =?? 1 4πepsilon1 integraldisplay V ρ i (r prime ,t ? R/v) R dV prime ? ? ?t μ 4π integraldisplay V J i (r prime ,t ? R/v) R dV prime , H(r,t) =?× 1 4π integraldisplay V J i (r prime ,t ? R/v) R dV prime . Thederivativesmaybebroughtinsidetheintegrals,butsomecaremustbetakenwhen theobservationpoint r lieswithinthesourceregion. Inthiscasetheintegralsmustbe performedinaprincipalvaluesensebyexcludingasmallvolumearoundtheobservation point. Wediscussthisinmoredetailbelowforthefrequency-domain?elds. Fordetails regardingthisprocedureinthetimedomainthereadermayseeHansen[81]. The retarded potentials in the frequency domain. Consideranunbounded,ho- mogeneous,isotropicmediumdescribedby ?μ(ω)and ?epsilon1 c (ω).If ? ψ(r,ω)representsascalar potential or any rectangular component of a vector or Hertzian potential then it must satisfy (? 2 + k 2 ) ? ψ(r,ω)=? ? S(r,ω) (5.75) where k = ω(?μ?epsilon1 c ) 1/2 . ThisHelmholtzequationhastheformof(5.70)andthus ? ψ(r,ω)= integraldisplay V ? S(r prime ,ω)G(r|r prime ;ω)dV prime where (? 2 + k 2 )G(r|r prime ;ω) =?δ(r ? r prime ). (5.76) Thisisequation(A.46)anditssolution,asgivenby(A.49),is G(r|r prime ;ω) = e ?jkR 4π R . (5.77) Hereweuse v 2 = 1/ ?μ?epsilon1 and Omega1 = ?σ/2epsilon1 in(A.47): k = 1 v radicalbig ω 2 ? j2ωOmega1 = ω radicalBigg ?μ parenleftbigg ?epsilon1 ? j ?σ ω parenrightbigg = ω radicalbig ?μ?epsilon1 c . Thesolutionto(5.75)istherefore ? ψ(r,ω)= integraldisplay V ? S(r prime ,ω) e ?jkR 4π R dV prime . (5.78) Whenthemediumislossless,thepotentialmustalsosatisfythe radiation condition lim r→∞ r parenleftbigg ? ?r + jk parenrightbigg ? ψ(r) = 0 (5.79) toguaranteeuniquenessofsolution. In § 5.2.2weshallshowhowthisrequirementarises from the solution within a bounded region. For a uniqueness proof for the Helmholtz equation,thereadermayconsultChew[33]. Wemayuse(5.78)to?ndthat ? A e (r,ω)= ?μ 4π integraldisplay V ? J i (r prime ,ω) e ?jkR R dV prime . (5.80) Comparisonwith(5.74)showsthatinthefrequencydomain,timeretardationtakesthe formofaphaseshift. Similarly, ? φ(r,ω)= 1 4π ?epsilon1 c integraldisplay V ?ρ i (r prime ,ω) e ?jkR R dV prime . (5.81) TheelectricandmagneticdyadicGreen’sfunctions. Thefrequency-domainelec- tromagnetic ?elds may be found for electric sources from the electric vector potential using(5.60)and(5.61): ? E(r,ω)=?jω ?μ(ω) integraldisplay V ? J i (r prime ,ω)G(r|r prime ;ω)dV prime ? jω ?μ(ω) k 2 ??· integraldisplay V ? J i (r prime ,ω)G(r|r prime ;ω)dV prime , ? H =?× integraldisplay V ? J i (r prime ,ω)G(r|r prime ;ω)dV prime . (5.82) Aslongastheobservationpoint r doesnotliewithinthesourceregionwemaytakethe derivativesinsidetheintegrals. Using ?· bracketleftbig ? J i (r prime ,ω)G(r|r prime ;ω) bracketrightbig = ? J i (r prime ,ω)·?G(r|r prime ;ω)+ G(r|r prime ;ω)?· ? J(r prime ,ω) =?G(r|r prime ;ω)· ? J i (r prime ,ω) we have ? E(r,ω)=?jω ?μ(ω) integraldisplay V braceleftbigg ? J i (r prime ,ω)G(r|r prime ;ω)+ 1 k 2 ? bracketleftbig ?G(r|r prime ;ω)· J i (r prime ,ω) bracketrightbig bracerightbigg dV prime . Thiscanbewrittenmorecompactlyas ? E(r,ω)=?jω ?μ(ω) integraldisplay V ˉ G e (r|r prime ;ω)· ? J i (r prime ,ω)dV prime where ˉ G e (r|r prime ;ω) = bracketleftbigg ˉ I + ?? k 2 bracketrightbigg G(r|r prime ;ω) (5.83) iscalledthe electric dyadic Green’s function. Using ?×[ ? J i G] =?G × ? J i + G?× ? J i =?G × ? J i wehaveforthemagnetic?eld ? H(r,ω)= integraldisplay V ?G(r|r prime ;ω)× ? J i (r prime ,ω)dV prime . Now,usingthedyadicidentity(B.15)wemayshowthat ? J i ×?G = ( ? J i ×?G)· ˉ I = (?G × ˉ I)· J i . So ? H(r,ω)=? integraldisplay V ˉ G m (r|r prime ;ω)· ? J i (r prime ,ω)dV prime where ˉ G m (r|r prime ;ω) =?G(r|r prime ;ω)× ˉ I (5.84) iscalledthe magnetic dyadic Green’s function. Proceedingsimilarlyformagneticsources(orusingduality)wehave ? H(r) =?jω?epsilon1 c integraldisplay V ˉ G e (r|r prime ;ω)· ? J i m (r prime ,ω)dV prime , ? E(r) = integraldisplay V ˉ G m (r|r prime ;ω)· ? J i m (r prime ,ω)dV prime . When both electric and magnetic sources are present we simply use superposition and addthe?elds. When the observation point lies within the source region, we must be much more carefulabouthowweformulatethedyadicGreen’sfunctions. In(5.82)weencounterthe integral integraldisplay V ? J i (r prime ,ω)G(r|r prime ;ω)dV prime . Figure5.1: Geometryofexcludedregionusedtocomputetheelectric?eldwithinasource region. If r lieswithinthesourceregionthen G issingularsince R → 0 when r → r prime . However, the integral converges and the potentials exist within the source region. While we run into trouble when we pass both derivatives in the operator ??· through the integral and allowthem to operate on G, since di?erentiation of G increases the order of the singularity,wemaysafelytakeonederivativeof G. Even when we allow one derivative on G we must be careful in how we compute the integral. We exclude the point r by surrounding it with a small volume element V δ as showninFigure5.1andwrite ??· integraldisplay V ? J i (r prime ,ω)G(r|r prime ;ω)dV prime = lim V δ →0 integraldisplay V?V δ ? bracketleftbig ?G(r|r prime ;ω)· ? J i (r prime ,ω) bracketrightbig dV prime + lim V δ →0 ? integraldisplay V δ ?G(r|r prime ;ω)· ? J i (r prime ,ω)dV prime . The?rstintegralontheright-handsideiscalledtheprincipalvalueintegral andisusually abbreviated P.V. integraldisplay V ? bracketleftbig ?G(r|r prime ;ω)· ? J i (r prime ,ω) bracketrightbig dV prime . It converges to a value dependent on the shape of the excluded region V δ ,asdoesthe secondintegral. However,thesumofthesetwointegralsproducesauniqueresult. Using ?G =?? prime G, the identity ? prime · ( ? JG) = ? J ·? prime G + G? prime · ? J, and the divergence theorem, wecanwrite ? integraldisplay V δ ? prime G(r|r prime ;ω)· ? J i (r prime ,ω)dV prime = ? contintegraldisplay S δ G(r|r prime ;ω) ? J i (r prime ,ω)· ?n prime dS prime + integraldisplay V δ G(r|r prime ;ω)? prime · ? J i (r prime ,ω)dV prime where S δ is the surface surrounding V δ . By the continuity equation the second integral on the right-hand side is proportional to the scalar potential produced by the charge within V δ ,andthusvanishesas V δ → 0. The?rsttermisproportionaltothe?eldat r producedbysurfacechargeon S δ ,whichresultsinavalueproportionalto J i .Thus lim V δ →0 ? integraldisplay V δ ?G(r|r prime ;ω)· ? J i (r prime ,ω)dV prime =?lim V δ →0 ? contintegraldisplay S δ G(r|r prime ;ω) ? J i (r prime ,ω)· ?n prime dS prime =? ˉ L · ? J i (r,ω), (5.85) so ??· integraldisplay V ? J i (r prime ,ω)G(r|r prime ;ω)dV prime = P.V. integraldisplay V ? bracketleftbig ?G(r|r prime ;ω)· ? J i (r prime ,ω) bracketrightbig dV prime ? ˉ L · ? J i (r,ω). Here ˉ L isusuallycalledthe depolarizing dyadic [113]. Itsvaluedependsontheshapeof V δ ,asconsideredbelow. Wemaynowwrite ? E(r,ω)=?jω ?μ(ω)P.V. integraldisplay V ˉ G e (r|r prime ;ω)· ? J(r prime ,ω)dV prime ? 1 jω?epsilon1 c (ω) ˉ L · ? J i (r,ω). (5.86) We may also incorporate both terms into a single dyadic Green’s function using the notation ˉ G(r|r prime ;ω) = P.V. ˉ G e (r|r prime ;ω)? 1 k 2 ˉ Lδ(r ? r prime ). Hencewhenwecompute ? E(r,ω)=?jω ?μ(ω) integraldisplay V ˉ G(r|r prime ;ω)· ? J i (r prime ,ω)dV prime =?jω ?μ(ω) integraldisplay V bracketleftbigg P.V. ˉ G e (r|r prime ;ω)? 1 k 2 ˉ Lδ(r ? r prime ) bracketrightbigg · ? J i (r prime ,ω)dV prime we reproduce (5.86). That is, the symbol P.V. on G e indicates that a principal value integralmustbeperformed. Our ?nal task is to compute ˉ L from (5.85). When we remove the excluded region from the principal value computation we leave behind a hole in the source region. The contribution to the ?eld at r by the sources in the excluded region is found from the scalarpotentialproducedbythesurfacedistribution ?n · J i . Thevalueofthis correction term dependsontheshapeoftheexcludingvolume. However,thecorrectiontermalways addstotheprincipalvalueintegraltogivethetrue?eldat r,regardlessoftheshapeof thevolume. Sowemustalwaysmatchtheshapeoftheexcludedregionusedtocompute the principal value integral with that used to compute the correction term so that the true ?eld is obtained. Note that as V δ → 0 the phase factor in the Green’s function becomesinsigni?cant,andthevaluesofthecurrentonthesurfaceapproachthevalueat r (assuming J i iscontinuousat r). Thuswemaywrite lim V δ →0 ? contintegraldisplay S δ ? J i (r,ω)· ?n prime 4π|r ? r prime | dS prime = ˉ L · ? J i (r,ω). Thishastheformofastatic?eldintegral. Forasphericalexcludedregionwemaycom- putetheabovequantityquitesimplybyassumingthecurrenttobeuniformthroughout V δ andbyaligningthecurrentwiththe z-axisandplacingthecenterofthesphereatthe origin. Wethencomputetheintegralatapoint r withinthesphere,takethegradient, andallow r → 0. Wethushaveforasphere lim V δ →0 ? contintegraldisplay S ? J i cosθ prime 4π|r ? r prime | dS prime = ˉ L · [?z ? J i (r,ω)]. Thisintegralhasbeencomputedin § 3.2.7withtheresultgivenby(3.103). Usingthis we?nd lim V δ →0 bracketleftbigg ? parenleftbigg 1 3 ? J i z parenrightbiggbracketrightbiggvextendsingle vextendsingle vextendsingle vextendsingle r=0 = ?z ? J i 3 = ˉ L · [?z ? J i (r,ω)] Figure5.2: GeometryofanelectricHertziandipole. andthus ˉ L = 1 3 ˉ I. Weleaveitasanexercisetoshowthatforacubicalexcludingvolumethedepolarizing dyadicisalso ˉ L = ˉ I/3. ValuesforothershapesmaybefoundinYaghjian[215]. The theory of dyadic Green’s functions is well developed and there exist techniques fortheirconstructionunderavarietyofconditions. Foranexcellentoverviewthereader mayseeTai[192]. Example of ?eld calculation using potentials: the Hertzian dipole. Consider a short line current of length l lessmuch λ at position r p , oriented along a direction ?p in a mediumwithconstitutiveparameters ?μ(ω), ?epsilon1 c (ω),asshowninFigure5.2.Weassume that the frequency-domain current ? I(ω) is independent of position, and therefore this Hertzian dipole mustbeterminatedbypointcharges ? Q(ω) =± ? I(ω) jω as required by the continuity equation. The electric vector potential produced by this shortcurrentelementis ? A e = ?μ 4π integraldisplay Gamma1 ? I ?p e ?jkR R dl prime . Atobservationpointsfarfromthedipole(comparedtoitslength)suchthat|r?r p |greatermuchl wemayapproximate e ?jkR R ≈ e ?jk|r?r p | |r ? r p | . Then ? A e = ?p ?μ ? IG(r|r p ;ω) integraldisplay Gamma1 dl prime = ?p ?μ ? IlG(r|r p ;ω). (5.87) Notethatweobtainthesameanswerifweletthecurrentdensityofthedipolebe ? J = jω?pδ(r ? r p ) where ?p isthe dipole moment de?nedby ?p = ? Ql ?p = ? Il jω ?p. Thatis,weconsideraHertziandipoletobea“pointsource”ofelectromagneticradiation. Withthisnotationwehave ? A e = ?μ integraldisplay V bracketleftbig jω?pδ(r prime ? r p ) bracketrightbig G(r|r prime ;ω)dV prime = jω ?μ?pG(r|r p ;ω), whichisidenticalto(5.87). Theelectromagnetic?eldsarethen ? H(r,ω)= jω?×[?pG(r|r p ;ω)], (5.88) ? E(r,ω)= 1 ?epsilon1 c ?×?×[?pG(r|r p ;ω)]. (5.89) Here we have obtained ? E from ? H outside the source region by applying Ampere’s law. Bydualitywemayobtainthe?eldsproducedbyamagneticHertziandipoleofmoment ?p m = ? I m l jω ?p locatedat r = r p as ? E(r,ω)=?jω?×[?p m G(r|r p ;ω)], ? H(r,ω)= 1 ?μ ?×?×[?p m G(r|r p ;ω)]. Wecanlearnmuchaboutthe?eldsproducedbylocalizedsourcesbyconsideringthe simplecaseofaHertziandipolealignedalongthe z-axisandcenteredattheorigin. Using ?p = ?z and r p = 0 in(5.88)we?ndthat ? H(r,ω)= jω?× bracketleftbigg ?z ? I jω l e ?jkr 4πr bracketrightbigg = ? φ 1 4π ? Il bracketleftbigg 1 r 2 + j k r bracketrightbigg sinθe ?jkr . (5.90) ByAmpere’slaw ? E(r,ω)= 1 jω?epsilon1 c ?× ? H(r,ω) = ?r η 4π ? Il bracketleftbigg 2 r 2 ? j 2 kr 3 bracketrightbigg cosθe ?jkr + ? θ η 4π ? Il bracketleftbigg j k r + 1 r 2 ? j 1 kr 3 bracketrightbigg sinθe ?jkr . (5.91) The ?elds involve various inverse powers of r, with the 1/r and 1/r 3 terms 90 ? out-of- phasefromthe 1/r 2 term. Sometermsdominatethe?eldclosetothesource,whileothers dominatefaraway. Thetermsthatdominatenearthesource 1 arecalledthe near-zone or induction-zone ?elds: ? H NZ (r,ω)= ? φ ? Il 4π e ?jkr r 2 sinθ, ? E NZ (r,ω)=?jη ? Il 4π e ?jkr kr 3 bracketleftBig 2?r cosθ + ? θ sinθ bracketrightBig . 1 Notethatwestillrequire r greatermuch l. We note that ? H NZ and ? E NZ are 90 ? out-of-phase. Also, the electric ?eld has the same spatial dependence as the ?eld of a static electric dipole. The terms that dominate far fromthesourcearecalledthe far-zone or radiation ?elds: ? H FZ (r,ω)= ? φ jk ? Il 4π e ?jkr r sinθ, (5.92) ? E FZ (r,ω)= ? θη jk ? Il 4π e ?jkr r sinθ. (5.93) Thefar-zone?eldsarein-phaseandinfactformaTEMsphericalwavewith ? H FZ = ?r × ? E FZ η . (5.94) Wespeakofthetime-averagepowerradiated byatime-harmonicsourceastheintegral of the time-average power density over a very large sphere. Thus radiated power is the powerdeliveredbythesourcestoin?nity. Ifthedipoleissituatedwithinalossymedium, all of the time-average power delivered by the sources is dissipated by the medium. If the medium is lossless then all the time-average power is delivered to in?nity. Let us computethepowerradiatedbyatime-harmonicHertziandipoleimmersedinalossless medium. Writing (5.90) and (5.91) in terms of phasors we have the complex Poynting vector S c (r) = ˇ E(r)× ˇ H ? (r) = ? θη parenleftBigg | ˇ I|l 4π parenrightBigg 2 j 2 kr 5 bracketleftbig k 2 r 2 + 1 bracketrightbig cosθ sinθ + ?rη parenleftBigg | ˇ I|l 4π parenrightBigg 2 k 2 r 2 bracketleftbigg 1 ? j 1 k 3 r 5 bracketrightbigg sin 2 θ. We notice that the θ-component of S c is purely imaginary and gives rise to no time- average power ?ux. This component falls o? as 1/r 3 for large r and produces no net ?uxthroughaspherewithradius r →∞. Additionally,theangularvariation sinθ cosθ integratestozerooverasphere. Incontrast,the r-componenthasarealpartthatvaries as 1/r 2 andas sin 2 θ. Hencewe?ndthatthetotaltime-averagepowerpassingthrough asphereexpandingtoin?nityisnonzero: P av = lim r→∞ integraldisplay 2π 0 integraldisplay π 0 1 2 Re ? ? ? ?rη parenleftBigg | ˇ I|l 4π parenrightBigg 2 k 2 r 2 sin 2 θ ? ? ? · ?rr 2 sinθ dθ dφ = η π 3 | ˇ I| 2 parenleftbigg l λ parenrightbigg 2 (5.95) where λ = 2π/k isthewavelengthinthelosslessmedium. Thisisthepowerradiatedby theHertziandipole. Thepowerisproportionalto | ˇ I| 2 asitisinacircuit, andthuswe mayde?nea radiation resistance R r = 2P av | ˇ I| 2 = η 2π 3 parenleftbigg l λ parenrightbigg 2 thatrepresentstheresistanceofalumpedelementthatwouldabsorbthesamepoweras radiatedbytheHertziandipolewhenpresentedwiththesamecurrent. Wealsonotethat thepowerradiatedbyaHertziandipole(and,infact,byanysourceof?niteextent)may Figure5.3: Geometryforsolutiontothefrequency-domainHelmholtzequation. be calculated directly from its far-zone ?elds. In fact, from (5.94) we have the simple formulaforthetime-averagepowerdensityinlosslessmedia S av = 1 2 Re braceleftbig ˇ E FZ × ˇ H FZ? bracerightbig = ?r 1 2 | ˇ E FZ | 2 η . Thedipole?eldisthe?rstterminageneralexpansionoftheelectromagnetic?eldsin termsofthemultipolemomentsofthesources. EitheraTaylorexpansionoraspherical- harmonicexpansionmaybeused. ThereadermayseePapas[141]fordetails. 5.2.2 Solution for potential functions in a bounded medium In the previous section we solved for the frequency-domain potential functions in an unboundedregionofspace. Hereweshallextendthesolutiontoaboundedregionand identifythephysicalmeaningoftheradiationcondition(5.79). Consider a bounded region of space V containing a linear, homogeneous, isotropic mediumcharacterizedby ?μ(ω)and ?epsilon1 c (ω).AsshowninFigure5.3wedecomposethe multiply-connectedboundaryintoaclosed“excludingsurface” S 0 andaclosed“encom- passing surface” S ∞ that we shall allow to expand outward to in?nity. S 0 may consist of more than one closed surface and is often used to exclude unknown sources from V. WewishtosolvetheHelmholtzequation(5.75)for ? ψ within V intermsofthesources within V and the values of ? ψ on S 0 . The actual sources of ? ψ lie entirely with S ∞ but mayliepartly,orentirely,within S 0 . We solve the Helmholtz equation in much the same way that we solved Poisson’s equation in § 3.2.4. We begin with Green’s second identity, written in terms of the sourcepoint(primed)variablesandappliedtotheregion V: integraldisplay V [ψ(r prime ,ω)? prime 2 G(r|r prime ;ω)? G(r|r prime ;ω)? prime 2 ψ(r prime ,ω)] dV prime = contintegraldisplay S 0 +S ∞ bracketleftbigg ψ(r prime ,ω) ?G(r|r prime ;ω) ?n prime ? G(r|r prime ;ω) ?ψ(r prime ,ω) ?n prime bracketrightbigg dS prime . We note that ?n points outward from V, and G is the Green’s function (5.77). By inspection,thisGreen’sfunctionobeysthereciprocitycondition G(r|r prime ;ω) = G(r prime |r;ω) andsatis?es ? 2 G(r|r prime ;ω) =? prime 2 G(r|r prime ;ω). Substituting ? prime 2 ? ψ =?k 2 ? ψ ? ? S from (5.75) and ? prime 2 G =?k 2 G ? δ(r ? r prime ) from (5.76) weget ? ψ(r,ω)= integraldisplay V ? S(r prime ,ω)G(r|r prime ;ω)dV prime ? ? contintegraldisplay S 0 +S ∞ bracketleftbigg ? ψ(r prime ,ω) ?G(r|r prime ;ω) ?n prime ? G(r|r prime ;ω) ? ? ψ(r prime ,ω) ?n prime bracketrightbigg dS prime . Hence ? ψ within V maybewrittenintermsofthesourceswithin V andthevaluesof ? ψ and its normal derivative over S 0 + S ∞ . The surface contributions account for sources excludedby S 0 . Let us examine the integral over S ∞ more closely. If we let S ∞ recede to in?nity, we expect no contribution to the potential at r from the ?elds on S ∞ . Choosing a sphere centeredattheorigin,wenotethat ?n prime = ?r prime andthatas r prime →∞ G(r|r prime ;ω) = e ?jk|r?r prime | 4π|r ? r prime | ≈ e ?jkr prime 4πr prime , ?G(r|r prime ;ω) ?n prime = ?n prime ·? prime G(r|r prime ;ω) ≈ ? ?r prime e ?jkr prime 4πr prime =?(1 + jkr prime ) e ?jkr prime 4πr prime . Substitutingthese,we?ndthatas r prime →∞ contintegraldisplay S ∞ bracketleftbigg ? ψ ?G ?n prime ? G ? ? ψ ?n prime bracketrightbigg dS prime ≈ integraldisplay 2π 0 integraldisplay π 0 bracketleftbigg ? 1 + jkr prime r prime 2 ? ψ ? 1 r prime ? ? ψ ?r prime bracketrightbigg e ?jkr prime 4π r prime 2 sinθ prime dθ prime dφ prime ≈? integraldisplay 2π 0 integraldisplay π 0 bracketleftbigg ? ψ + r prime parenleftbigg jk ? ψ + ? ? ψ ?r prime parenrightbiggbracketrightbigg e ?jkr 4π sinθ prime dθ prime dφ prime . Sincethisgivesthecontributiontothe?eldin V fromthe?eldsonthesurfacereceding to in?nity, we expect that this term should be zero. If the medium has loss, then the exponentialtermdecaysanddrivesthecontributiontozero. Foralosslessmediumthe contributioniszeroif lim r→∞ ? ψ(r,ω)= 0, (5.96) lim r→∞ r bracketleftbigg jk ? ψ(r,ω)+ ? ? ψ(r,ω) ?r bracketrightbigg = 0. (5.97) This is called the radiation condition for the Helmholtz equation. It is also called the Sommerfeld radiation condition after the German physicist A. Sommerfeld. Note that wehavenotderivedthiscondition: wehavemerelypostulatedit. Aswithallpostulates itissubjecttoexperimentalveri?cation. The radiation condition implies that for points far from the source the potentials behaveassphericalwaves: ? ψ(r,ω)～ e ?jkr r , r →∞. Substitutingthisinto(5.96)and(5.97)we?ndthattheradiationconditionissatis?ed. With S ∞ →∞we have ? ψ(r,ω)= integraldisplay V ? S(r prime ,ω)G(r|r prime ;ω)dV prime ? ? contintegraldisplay S 0 bracketleftbigg ? ψ(r prime ,ω) ?G(r|r prime ;ω) ?n prime ? G(r|r prime ;ω) ? ? ψ(r prime ,ω) ?n prime bracketrightbigg dS prime , which is the expression for the potential within an in?nite medium having source- excludingregions. As S 0 → 0 weobtaintheexpressionforthepotentialinanunbounded medium: ? ψ(r,ω)= integraldisplay V ? S(r prime ,ω)G(r|r prime ;ω)dV prime , asexpected. Thetime-domainequation(5.71)mayalsobesolved(atleastforthelosslesscase)in aboundedregionofspace. TheinterestedreadershouldseePauli[143]fordetails. 5.3 Transverse–longitudinal decomposition We have seen that when only electric sources are present, the electromagnetic ?elds in a homogeneous, isotropic region can be represented by a single vector potential Π e . Similarly, when only magnetic sources are present, the ?elds can be represented by a single vector potential Π h . Hence two vector potentials may be used to represent the ?eldifbothelectricandmagneticsourcesarepresent. Wemayalsorepresenttheelectromagnetic?eldinahomogeneous,isotropicregionus- ingtwoscalarfunctionsandthesources. Thisfollowsnaturallyfromanotherimportant ?elddecomposition: asplittingofeach?eldvectorinto(1)acomponentalongacertain pre-chosenconstantdirection,and(2)acomponenttransversetothisdirection. Depend- ingonthegeometryofthesources,itispossiblethatonlyoneofthesecomponentswill bepresent. Aspecialcaseofthisdecomposition,the TE–TM ?eld decomposition,holds forasource-freeregionandwillbediscussedinthenextsection. 5.3.1 Transverse–longitudinal decomposition in terms of ?elds Consider a direction de?ned by a constant unit vector ?u. We de?ne the longitudinal component of A as ?uA u where A u = ?u · A, andthe transverse component of A as A t = A ? ?uA u . Wemaythusdecomposeanyvectorintoasumoflongitudinalandtransverseparts. An importantconsequenceofMaxwell’sequationsisthatthetransverse?eldsmaybewritten entirelyintermsofthelongitudinal?eldsandthesources. Thisholdsinboththetime andfrequencydomains;wederivethedecompositioninthefrequencydomainandleave the derivation of the time-domain expressions as exercises. We begin by decomposing the operators in Maxwell’s equations into longitudinal and transverse components. We notethat ? ?u ≡ ?u ·? andde?nea transverse del operator as ? t ≡???u ? ?u . Using these basic de?nitions, the identities listed in Appendix B may be derived. We shall ?nd it helpful to express the vector curl and Laplacian operations in terms of theirlongitudinalandtransversecomponents. Using(B.93)and(B.96)we?ndthatthe transversecomponentofthecurlisgivenby (?×A) t =??u × ?u ×(?×A) =??u × ?u ×(? t × A t )? ?u × ?u × parenleftbigg ?u × bracketleftbigg ?A t ?u ?? t A u bracketrightbiggparenrightbigg . (5.98) The?rsttermintherightmemberiszerobyproperty(B.91). Using(B.7)wecanreplace thesecondtermby ??u braceleftbigg ?u · parenleftbigg ?u × bracketleftbigg ?A t ?u ?? t A u bracketrightbiggparenrightbiggbracerightbigg +(?u · ?u) parenleftbigg ?u × bracketleftbigg ?A t ?u ?? t A u bracketrightbiggparenrightbigg . The?rstofthesetermsiszerosince ?u · parenleftbigg ?u × bracketleftbigg ?A t ?u ?? t A u bracketrightbiggparenrightbigg = bracketleftbigg ?A t ?u ?? t A u bracketrightbigg ·(?u × ?u) = 0, hence (?×A) t = ?u × bracketleftbigg ?A t ?u ?? t A u bracketrightbigg . (5.99) Thelongitudinalpartisthen,byproperty(B.80),merelythedi?erencebetweenthecurl anditstransversepart,or ?u(?u ·?×A) =? t × A t . (5.100) AsimilarsetofstepsgivesthetransversecomponentoftheLaplacianas (? 2 A) t = bracketleftbigg ? t (? t · A t )+ ? 2 A t ?u 2 ?? t ×? t × A t bracketrightbigg , (5.101) andthelongitudinalpartas ?u parenleftbig ?u ·? 2 A parenrightbig = ?u? 2 A u . (5.102) Veri?cationisleftasanexercise. Nowwe are ready to give a longitudinal–transverse decomposition of the ?elds in a lossy, homogeneous, isotropic region in terms of the direction ?u. We write Maxwell’s equationsas ?× ? E =?jω ?μ ? H t ? jω ?μ?u ? H u ? ? J i mt ? ?u ? J i mu , (5.103) ?× ? H = jω?epsilon1 c ? E t + jω?epsilon1 c ?u ? E u + ? J i t + ?u ? J i u , (5.104) where we have split the right-hand sides into longitudinal and transverse parts. Then, using (5.99) and (5.100), we can equate the transverse and longitudinal parts of each equationtoobtain ? t × ? E t =?jω ?μ?u ? H u ? ?u ? J i mu , (5.105) ? ?u ×? t ? E u + ?u × ? ? E t ?u =?jω ?μ ? H t ? ? J i mt , (5.106) ? t × ? H t = jω?epsilon1 c ?u ? E u + ?u ? J i u , (5.107) ? ?u ×? t ? H u + ?u × ? ? H t ?u = jω?epsilon1 c ? E t + ? J i t . (5.108) We shall isolate the transverse ?elds in terms of the longitudinal ?elds. Forming the crossproductof ?u andthepartialderivativeof(5.108)withrespectto u, we have ??u × ?u ×? t ? ? H u ?u + ?u × ?u × ? 2 ? H t ?u 2 = jω?epsilon1 c ?u × ? ? E t ?u + ?u × ? ? J i t ?u . Using(B.7)and(B.80)we?ndthat ? t ? ? H u ?u ? ? 2 ? H t ?u 2 = jω?epsilon1 c ?u × ?E t ?u + ?u × ? ? J i t ?u . (5.109) Multiplying(5.106)by jω?epsilon1 c we have ? jω?epsilon1 c ?u ×? t ? E u + jω?epsilon1 c ?u × ? ? E t ?u = ω 2 ?μ?epsilon1 c ? H t ? jω?epsilon1 c ? J i mt . (5.110) Wenowadd(5.109)to(5.110)andeliminate ? E t toget parenleftbigg ? 2 ?u 2 + k 2 parenrightbigg ? H t =? t ? ? H u ?u ? jω?epsilon1 c ?u ×? t ? E u + jω?epsilon1 c ? J i mt ? ?u × ? ? J i t ?u . (5.111) Thisone-dimensionalHelmholtzequationcanbesolvedto?ndthetransversemagnetic ?eld from the longitudinal components of ? E and ? H. Similar steps lead to a formula for thetransversecomponentof ? E: parenleftbigg ? 2 ?u 2 + k 2 parenrightbigg ? E t =? t ? ? E u ?u + jω ?μ?u ×? t ? H u + ?u × ? ? J i mt ?u + jω ?μ ? J i t . (5.112) We?ndthelongitudinalcomponentsfromthewaveequationfor ? E and ? H. Recallthat the?eldssatisfy (? 2 + k 2 ) ? E = 1 ?epsilon1 c ? ?ρ i + jω ?μ ? J i +?× ? J i m , (? 2 + k 2 ) ? H = 1 ?μ ? ?ρ i m + jω?epsilon1 c ? J i m ??× ? J i . Splittingthevectorsintolongitudinalandtransverseparts,andusing(5.100)and(5.102), weequatethelongitudinalcomponentsofthewaveequationstoobtain parenleftbig ? 2 + k 2 parenrightbig ? E u = 1 ?epsilon1 c ? ?ρ i ?u + jω ?μ ? J i u +? t × ? J i mt , (5.113) parenleftbig ? 2 + k 2 parenrightbig ? H u = 1 ?μ ? ?ρ i m ?u + jω?epsilon1 c ? J i mu ?? t × ? J i t . (5.114) Wenotethatif ? J i m = ? J i t = 0,then ? H u = 0andthe?eldsareTMtotheu-direction;these ?elds may be determined completely from ? E u . Similarly, if ? J i = ? J i mt = 0, then ? E u = 0 andthe?eldsareTEtothe u-direction;these?eldsmaybedeterminedcompletelyfrom ? H u . Thesepropertiesareusedin § 4.11.7,wherethe?eldsofelectricandmagneticline sourcesalignedalongthe z-directionareassumedtobepurelyTM z orTE z ,respectively. 5.4 TE–TM decomposition 5.4.1 TE–TM decomposition in terms of ?elds Aparticularlyuseful?elddecompositionresultsifwespecializetoasource-freeregion. With ? J i = ? J i m = 0 in(5.111)–(5.112)weobtain parenleftbigg ? 2 ?u 2 + k 2 parenrightbigg ? H t =? t ? ? H u ?u ? jω?epsilon1 c ?u ×? t ? E u , (5.115) parenleftbigg ? 2 ?u 2 + k 2 parenrightbigg ? E t =? t ? ? E u ?u + jω ?μ?u ×? t ? H u . (5.116) Settingthesourcestozeroin(5.113)and(5.114)weget parenleftbig ? 2 + k 2 parenrightbig ? E u = 0, parenleftbig ? 2 + k 2 parenrightbig ? H u = 0. Hence the longitudinal ?eld components are solutions to the homogeneous Helmholtz equation,andthetransversecomponentsarespeci?edsolelyintermsofthelongitudinal components. Theelectromagnetic?eldiscompletelyspeci?edbythetwoscalar?elds ? E u and ? H u (and,ofcourse,appropriateboundaryvalues). We can use superposition to simplify the task of solving (5.115)–(5.116). Since each equationhastwoforcingtermsontheright-handside,wecansolvetheequationsusing oneforcingtermatatime,andaddtheresults. Thatis,let ? E 1 and ? H 1 bethesolutionsto (5.115)–(5.116)with ? E u = 0,and ? E 2 and ? H 2 bethesolutionswith ? H u = 0. Thisresults inadecomposition ? E = ? E 1 + ? E 2 , (5.117) ? H = ? H 1 + ? H 2 , (5.118) with ? E 1 = ? E 1t , ? H 1 = ? H 1t + ? H 1u ?u, ? H 2 = ? H 2t , ? E 2 = ? E 2t + ? E 2u ?u. Because ? E 1 has no u-component, ? E 1 and ? H 1 are termed transverse electric (or TE)to the u-direction; ? H 2 hasno u-component,and ? E 2 and ? H 2 aretermedtransverse magnetic (or TM) to the u-direction. 2 We see that in a source-free region any electromagnetic ?eld can be decomposed into a set of two ?elds that are TE and TM, respectively, to some ?xed u-direction. This is useful when solving boundary value (e.g., waveguide and scattering) problems where information about external sources is easily speci?ed using the values of the ?elds on the boundary of the source-free region. In that case ? E u and ? H u aredeterminedbysolvingthehomogeneouswaveequationinanappropriate coordinatesystem,andtheother?eldcomponentsarefoundfrom(5.115)–(5.116). Often the boundary conditions can be satis?ed by the TM ?elds or the TE ?elds alone. This simpli?estheanalysisofmanytypesofEMsystems. 5.4.2 TE–TM decomposition in terms of Hertzian potentials We are free to represent ? E and ? H in terms of scalar ?elds other than ? E u and ? H u .In doing so, it is helpful to retain the wave nature of the solution so that a meaningful physicalinterpretationisstillpossible; wethususeHertzianpotentialssincetheyobey thewaveequation. FortheTMcaselet ? Π h = 0 and ? Π e = ?u ? Pi1 e . Setting ? J i = 0 in(5.64)wehave (? 2 + k 2 ) ? Π e = 0. Since ? Π e ispurelylongitudinal,wecanuse(B.99)toobtainthescalarHelmholtzequation for ? Pi1 e : (? 2 + k 2 ) ? Pi1 e = 0. (5.119) Once ? Pi1 e hasbeenfoundbysolvingthiswaveequation,the?eldscanbefoundbyusing (5.62)–(5.63)with ? J i = 0: ? E =?×(?× ? Π e ), (5.120) ? H = jω?epsilon1 c ?× ? Π e . (5.121) Wecanevaluate ? E bynotingthat ? Π e ispurelylongitudinal. Useofproperty(B.98)gives ?×?× ? Π e =? t ? ? Pi1 e ?u ? ?u? 2 t ? Pi1 e . Then,byproperty(B.97), ?×?× ? Π e =? t ? ? Pi1 e ?u ? ?u bracketleftbigg ? 2 ? Pi1 e ? ? 2 ? Pi1 e ?u 2 bracketrightbigg . By(5.119)then, ? E =? t ? ? Pi1 e ?u + ?u parenleftbigg ? 2 ?u 2 + k 2 parenrightbigg ? Pi1 e . (5.122) The ?eld ? H can be found by noting that ? Π e is purely longitudinal. Use of property (B.96)in(5.121)gives ? H =?jω?epsilon1 c ?u ×? t ? Pi1 e . (5.123) 2 Some authors prefer to use the terminology Emodein place of TM, and Hmodein place of TE, indicatingthepresenceofa u-directedelectricormagnetic?eldcomponent. Similarstepscanbeusedto?ndtheTErepresentation. Substitutionof ? Π e = 0 and ? Π h = ?u ? Pi1 h into(5.65)–(5.66)givesthe?elds ? E = jω ?μ?u ×? t ? Pi1 h , (5.124) ? H =? t ? ? Pi1 h ?u + ?u parenleftbigg ? 2 ?u 2 + k 2 parenrightbigg ? Pi1 h , (5.125) while ? Pi1 h mustsatisfy (? 2 + k 2 ) ? Pi1 h = 0. (5.126) Hertzian potential representation of TEM ?elds. Aninterestingsituationoccurs when a ?eld is both TE and TM to a particular direction. Such a ?eld is said to be transverse electromagnetic (or TEM)tothatdirection. Unfortunately,with ? E u = ? H u = 0 wecannotuse(5.115)or(5.116)to?ndthetransverse?eldcomponents. Itturnsout thatasinglescalarpotentialfunctionissu?cienttorepresentthe?eld,andwemayuse either ? Pi1 e or ? Pi1 h . FortheTMcase,equations(5.122)and(5.123)showthatwecanrepresenttheelectro- magnetic?eldscompletelywith ? Pi1 e . Unfortunately(5.122)hasalongitudinalcomponent, and thus cannot describe a TEM ?eld. But if we require that ? Pi1 e obey the additional equation parenleftbigg ? 2 ?u 2 + k 2 parenrightbigg ? Pi1 e = 0, (5.127) then both E and H are transverse to u and thus describe a TEM ?eld. Since ? Pi1 e must alsoobey parenleftbig ? 2 + k 2 parenrightbig ? Pi1 e = 0, using(B.7)wecanwrite(5.127)as ? 2 t ? Pi1 e = 0. Similarly, for the TE case we found that the EM ?elds were completely described in (5.124) and (5.125) by ? Pi1 h . In this case ? H has a longitudinal component. Thus, if we require parenleftbigg ? 2 ?u 2 + k 2 parenrightbigg ? Pi1 h = 0, (5.128) thenboth ? E and ? H arepurelytransverseto u andagaindescribeaTEM?eld. Equation (5.128)isequivalentto ? 2 t ? Pi1 h = 0. We can therefore describe a TEM ?eld using either ? Pi1 e or ? Pi1 h , since a TEM ?eld is bothTEandTMtothelongitudinaldirection. Ifwechoose ? Pi1 e wecanuse(5.122)and (5.123)toobtaintheexpressions ? E =? t ? ? Pi1 e ?u , (5.129) ? H =?jω?epsilon1 c ?u ×? t ? Pi1 e , (5.130) where ? Pi1 e mustobey ? 2 t ? Pi1 e = 0, parenleftbigg ? 2 ?u 2 + k 2 parenrightbigg ? Pi1 e = 0. (5.131) Ifwechoose ? Pi1 h wecanuse(5.124)and(5.125)toobtain ? E = jω ?μ?u ×? t ? Pi1 h , (5.132) ? H =? t ? ? Pi1 h ?u , (5.133) where ? Pi1 h mustobey ? 2 t ? Pi1 h = 0, parenleftbigg ? 2 ?u 2 + k 2 parenrightbigg ? Pi1 h = 0. (5.134) 5.4.3 Application: hollow-pipe waveguides AclassicapplicationoftheTE–TMdecompositionistothecalculationofwaveguide ?elds. ConsiderahollowpipewithPECwalls,alignedalongthe z-axis. Theinsideis?lled with a homogeneous, isotropic material of permeability ?μ(ω) and complex permittivity ?epsilon1 c (ω),andtheguidecross-sectionalshapeisassumedtobeindependentof z. Weassume thatacurrentsourceexistssomewherewithinthewaveguide,creatingwavesthateither propagate or evanesce away from the source. If the source is con?ned to the region ?d < z < d then each of the regions z > d and z < ?d is source-free and we may decomposethe?eldsthereintoTEandTMsets. Suchawaveguideisagoodcandidate forTE–TManalysisbecausetheTEandTM?eldsindependentlysatisfytheboundary conditionsatthewaveguidewalls. Thisisnotgenerallythecaseforcertainotherguided- wavestructuressuchas?beropticcablesandmicrostriplines. We may represent the ?elds either in terms of the longitudinal ?elds ? E z and ? H z ,or in terms of the Hertzian potentials. We choose the Hertzian potentials. For TM ?elds wechoose ? Π e = ?z ? Pi1 e , ? Π h = 0; forTE?eldswechoose ? Π h = ?z ? Pi1 h , ? Π e = 0. Bothofthe potentialsmustobeythesameHelmholtzequation: parenleftbig ? 2 + k 2 parenrightbig ? Pi1 z = 0, (5.135) where ? Pi1 z represents either ? Pi1 e or ? Pi1 h . We seek a solution to this equation using the separationofvariablestechnique,andassumetheproductsolution ? Pi1 z (r,ω)= ? Z(z,ω) ? ψ(ρ,ω), where ρ is the transverse position vector (r = ?zz + ρ). Substituting the trial solution into(5.135)andwriting ? 2 =? 2 t + ? 2 ?z 2 we?ndthat 1 ? ψ(ρ,ω) ? 2 t ? ψ(ρ,ω)+ k 2 =? 1 Z(z,ω) ? 2 ?z 2 Z(z,ω). Becausetheleft-handsideofthisexpressionhaspositionaldependenceonlyon ρ while theright-handsidehasdependenceonlyonz,wemusthavebothsidesequaltoaconstant, say k 2 z . Then ? 2 Z ?z 2 + k 2 z Z = 0, whichisanordinarydi?erentialequationwiththesolutions Z = e ?jk z z . Wealsohave ? 2 t ? ψ(ρ,ω)+ k 2 c ? ψ(ρ,ω)= 0, (5.136) where k c = k 2 ?k 2 z iscalledthecuto? wavenumber. Thesolutiontothisequationdepends onthegeometryofthewaveguidecross-sectionandwhetherthe?eldisTEorTM. The ?elds may be computed from the Hertzian potentials using u = z in (5.122)– (5.123) and (5.124)–(5.125). Because the ?elds all contain the common term e ?jk z z ,we de?nethe?eldquantities ?e and ? h through ? E(r,ω)= ?e(ρ,ω)e ?jk z z , ? H(r,ω)= ? h(ρ,ω)e ?jk z z . Then,substituting ? Pi1 e = ? ψ e e ?jk z z ,wehaveforTM?elds ?e =?jk z ? t ? ψ e + ?zk 2 c ? ψ e , ? h =?jω?epsilon1 c ?z ×? t ? ψ e . Becausewehaveasimplerelationshipbetweenthetransversepartsof ? E and ? H,wemay alsowritethe?eldsas ?e z = k 2 c ? ψ e , (5.137) ?e t =?jk z ? t ? ψ e , (5.138) ? h t =±Y e (?z × ?e t ). (5.139) Here Y e = ω?epsilon1 c k z isthecomplex TM wave admittance. ForTE?eldswehavewith ? Pi1 h = ? ψ h e ?jk z z ?e = jω ?μ?z ×? t ? ψ h , ? h =?jk z ? t ? ψ h + ?zk 2 c ? ψ h , or ? h z = k 2 c ? ψ h , (5.140) ? h t =?jk z ? t ? ψ h , (5.141) ?e t =?Z h (?z × ? h t ). (5.142) Here Z h = ω ?μ k z isthe TM wave impedance. Modal solutions for the transverse ?eld dependence. Equation(5.136)describes the transverse behavior of the waveguide ?elds. When coupled with an appropriate boundarycondition,thishomogeneousequationhasanin?nitespectrumofdiscreteso- lutions called eigenmodes or simply modes. Each mode has associated with it a real eigenvalue k c thatisdependentonthecross-sectionalshapeofthewaveguide,butinde- pendentoffrequencyandhomogeneousmaterialparameters. Wenumberthemodesso that k c = k cn for the nth mode. The amplitude of each modal solution depends on the excitationsourcewithinthewaveguide. The appropriate boundary conditions can be found by employing the condition that forbothTMandTE?eldsthetangentialcomponentof ? E mustbezeroonthewaveguide walls: ?n × ? E = 0, where ?n is the unit inward normal to the waveguide wall. For TM ?eldswehave ? E z = 0 andthus ? ψ e (ρ,ω)= 0, ρ ∈ Gamma1, (5.143) whereGamma1isthecontourdescribingthewaveguideboundary. ForTE?eldswehave ?n× ? E t = 0,or ?n ×(?z ×? t ? ψ h ) = 0. Using ?n ×(?z ×? t ? ψ h ) = ?z(?n ·? t ? ψ h )?(?n · ?z)? t ? ψ h andnotingthat ?n · ?z = 0,wehavetheboundarycondition ?n ·? t ? ψ h (ρ,ω)= ? ? ψ h (ρ,ω) ?n = 0, ρ ∈ Gamma1. (5.144) The wave nature of the waveguide ?elds. We have seen that all waveguide ?eld components, for both TE and TM modes, vary as e ?jk zn z . Here k 2 zn = k 2 ? k 2 cn is the propagation constant ofthe nthmode. Letting k z = β ? jα we thus have ? E, ? H ～ e ?jβz e ?αz . For z > d wechoosetheminussignsothatwehaveawavepropagatingawayfromthe source;for z < ?d wechoosetheplussign. When the guide is ?lled with a good dielectric we may assume ?μ = μ is real and independentoffrequencyanduse(4.254)toshowthat k z = β ? jα = radicalBig bracketleftbig ω 2 μepsilon1 prime ? k 2 c bracketrightbig ? jω 2 μepsilon1 prime tanδ c = radicalbig μepsilon1 prime radicalBig ω 2 ?ω 2 c radicalBigg 1 ? j tanδ c 1 ?(ω c /ω) 2 where δ c isthelosstangent(4.253)andwhere ω c = k c √ μepsilon1 prime iscalledthe cuto? frequency. Underthecondition tanδ c 1 ?(ω c /ω) 2 lessmuch 1 (5.145) wemayapproximatethesquarerootusingthe?rsttwotermsofthebinomialseriesto showthat β ? jα ≈ radicalbig μepsilon1 prime radicalBig ω 2 ?ω 2 c bracketleftbigg 1 ? j 1 2 tanδ c 1 ?(ω c /ω) 2 bracketrightbigg . (5.146) 0.0 0.5 1.0 1.5 2.0 2.5 β/ω v or α/ω v 0.0 0.5 1.0 1.5 2.0 2.5 ω / ω c Light Line β α c c Figure5.4: Dispersion plot for a hollow-pipe waveguide. Light line computed using v = 1/ √ μepsilon1. Condition (5.145) requires that ω be su?ciently removed from ω c , either by having ω>ω c or ω<ω c . When ω>ω c wesaythatthefrequencyisabove cuto? and?ndfrom (5.146)that β ≈ ω radicalbig μepsilon1 prime radicalBig 1 ?ω 2 c /ω 2 ,α≈ ω 2 μepsilon1 prime 2β tanδ c . Here α lessmuch β and the wave propagates down the waveguide with relatively little loss. When ω<ω c wesaythatthewaveguideiscut o? orthatthefrequencyis below cuto? and?ndthat α ≈ ω radicalbig μepsilon1 prime radicalBig ω 2 c /ω 2 ? 1,β≈ ω 2 μepsilon1 prime 2α tanδ c . Inthiscasethewavehasaverysmallphaseconstantandaverylargerateofattenuation. Forfrequenciesnear ω c thereisanabruptbutcontinuoustransitionbetweenthesetwo typesofwavebehavior. When the waveguide is ?lled with a lossless material having permittivity epsilon1 and per- meability μ, thetransitionacrossthecuto?frequencyisdiscontinuous. For ω>ω c we have β = ω √ μepsilon1 radicalBig 1 ?ω 2 c /ω 2 ,α= 0, andthewavepropagateswithoutloss. Forω<ω c we have α = ω √ μepsilon1 radicalBig ω 2 c /ω 2 ? 1,β= 0, andthewaveisevanescent.ThedispersiondiagramshowninFigure5.4clearlyshows the abrupt cuto? phenomenon. We can compute the phase and group velocities of the waveabovecuto?justaswedidforplanewaves: v p = ω β = v radicalbig 1 ?ω 2 c /ω 2 , 0.0 0.5 1.0 1.5 2.0 2.5 ω/ω 0.0 0.5 1.0 1.5 2.0 2.5 v / v or v / v v /v v /v c p g g p Figure5.5: Phaseandgroupvelocityforahollow-pipewaveguide. v g = dω dβ = v radicalBig 1 ?ω 2 c /ω 2 , (5.147) where v = 1/ √ μepsilon1. Notethat v g v p = v 2 . Weshowlaterthat v g isthevelocityofenergy transport within a lossless guide. We also see that as ω →∞we have v p → v and v g →v.Moreinterestingly,asω→ω c we?ndthatv p →∞andv g → 0.Thisisshown graphicallyinFigure5.5. We may also speak of the guided wavelength of a monochromatic wave propagating withfrequency ˇω inawaveguide. Wede?nethiswavelengthas λ g = 2π β = λ radicalbig 1 ?ω 2 c /ˇω 2 = λ radicalbig 1 ?λ 2 /λ 2 c . Here λ = 2π ˇω √ μepsilon1 ,λ c = 2π k c . Orthogonality of waveguide modes. The modal ?elds in a closed-pipe waveguide obey several orthogonality relations. Let ( ˇ E n , ˇ H n ) be the time-harmonic electric and magnetic ?elds of one particular waveguide mode (TE or TM), and let ( ˇ E m , ˇ H m ) be the ?elds of a di?erent mode (TE or TM). One very useful relation states that for a waveguidecontaininglosslessmaterials integraldisplay CS ?z · parenleftbig ˇe n × ˇ h ? m parenrightbig dS = 0, m negationslash= n, (5.148) where CSistheguidecross-section. Thisisusedtoestablishthatthetotalpowercarried byawaveisthesumofthepowerscarriedbyindividualmodes(seebelow). Otherimportantrelationshipsincludetheorthogonalityofthelongitudinal?elds, integraldisplay CS ˇ E zm ˇ E zn dS = 0, m negationslash= n, (5.149) integraldisplay CS ˇ H zm ˇ H zn dS = 0, m negationslash= n, (5.150) andtheorthogonalityoftransverse?elds, integraldisplay CS ˇ E tm · ˇ E tn dS = 0, m negationslash= n, integraldisplay CS ˇ H tm · ˇ H tn dS = 0, m negationslash= n. Thesemayalsobecombinedtogiveanorthogonalityrelationforthecomplete?elds: integraldisplay CS ˇ E m · ˇ E n dS = 0, m negationslash= n, (5.151) integraldisplay CS ˇ H m · ˇ H n dS = 0, m negationslash= n. (5.152) ForproofsoftheserelationsthereadershouldseeCollin[39]. Power carried by time-harmonic waves in lossless waveguides. Thepowercar- ried by a time-harmonic wave propagating down a waveguide is de?ned as the time- averagePoynting?uxpassingthroughtheguidecross-section. Thuswemaywrite P av = 1 2 integraldisplay CS Re braceleftbig ˇ E × ˇ H ? bracerightbig · ?z dS. The ?eld within the guide is assumed to be a superposition of all possible waveguide modes. Forwavestravelinginthe+z-directionthisimplies ˇ E = summationdisplay m (ˇe tm + ?zˇe zm )e ?jk zm z , ˇ H = summationdisplay n parenleftbig ˇ h tn + ?z ˇ h zn parenrightbig e ?jk zn z . Substitutingwehave P av = 1 2 Re braceleftBigg integraldisplay CS bracketleftBigg summationdisplay m (ˇe tm + ?zˇe zm )e ?jk zm z × summationdisplay n parenleftbig ˇ h ? tn + ?z ˇ h ? zn parenrightbig e jk ? zn z bracketrightBigg · ?z dS bracerightBigg = 1 2 Re braceleftBigg summationdisplay m summationdisplay n e ?j(k zm ?k ? zn )z integraldisplay CS ?z · parenleftbig ˇe tm × ˇ h ? tn parenrightbig dS bracerightBigg . By(5.148)wehave P av = 1 2 Re braceleftBigg summationdisplay n e ?j(k zn ?k ? zn )z integraldisplay CS ?z · parenleftbig ˇe tn × ˇ h ? tn parenrightbig dS bracerightBigg . For modes propagating in a lossless guide k zn = β zn . For modes that are cut o? k zn = ?jα zn . However,we?ndbelowthattermsinthisseriesrepresentingmodesthatarecut o?arezero. Thus P av = summationdisplay n 1 2 Re braceleftbiggintegraldisplay CS ?z · parenleftbig ˇe tn × ˇ h ? tn parenrightbig dS bracerightbigg = summationdisplay n P n,av . Henceforwaveguides?lledwithlosslessmediathetotaltime-averagepower?owisgiven bythesuperpositionoftheindividualmodalpowers. Simple formulas for the individual modal powers in a lossless guide may be obtained bysubstitutingtheexpressionsforthe?elds. ForTMmodesweuse(5.138)and(5.139) toget P av = 1 2 Re braceleftbigg |k z | 2 Y ? e e ?j(k z ?k ? z ) integraldisplay CS ?z · parenleftbig ? t ˇ ψ e × [?z ×? t ˇ ψ ? e ] parenrightbig dS bracerightbigg = 1 2 |k z | 2 Re braceleftbig Y ? e bracerightbig e ?j(k z ?k ? z ) integraldisplay CS ? t ˇ ψ e ·? t ˇ ψ ? e dS. Herewehaveused(B.7)and ?z·? t ˇ ψ e = 0. Thisexpressioncanbesimpli?edbyusingthe two-dimensionalversionofGreen’s?rstidentity(B.29): integraldisplay S (? t a ·? t b + a? 2 t b)dS = contintegraldisplay Gamma1 a ?b ?n dl. Using a = ˇ ψ e and b = ˇ ψ ? e andintegratingoverthewaveguidecross-sectionwehave integraldisplay CS (? t ˇ ψ e ·? t ˇ ψ ? e + ˇ ψ e ? 2 ˇ ψ ? e )dS = contintegraldisplay Gamma1 ˇ ψ e ? ˇ ψ ? e ?n dl. Substituting ? 2 t ˇ ψ ? e =?k 2 c ˇ ψ ? e andrememberingthat ˇ ψ e = 0 on Gamma1 wereducethisto integraldisplay CS ? t ˇ ψ e ·? t ˇ ψ ? e dS = k 2 c integraldisplay CS ˇ ψ e ˇ ψ ? e dS. (5.153) Thusthepoweris P av = 1 2 Re braceleftbig Y ? e bracerightbig |k z | 2 k 2 c e ?j(k z ?k ? z )z integraldisplay CS ˇ ψ e ˇ ψ ? e dS. Formodesabovecuto?wehave k z = β and Y e = ωepsilon1/k z = ωepsilon1/β. Thepowercarriedby thesemodesisthus P av = 1 2 ωepsilon1βk 2 c integraldisplay CS ˇ ψ e ˇ ψ ? e dS. (5.154) For modes belowcuto? we have k z =?jα and Y e = jωepsilon1/α.ThusRe{Y ? e }=0 and P av = 0. For frequencies belowcuto? the ?elds are evanescent and do not carry power inthemannerofpropagatingwaves. ForTEmodeswemayproceedsimilarlyandshowthat P av = 1 2 ωμβk 2 c integraldisplay CS ˇ ψ h ˇ ψ ? h dS. (5.155) Thedetailsareleftasanexercise. Stored energy in a waveguide and the velocity of energy transport. Consider asource-freesectionoflosslesswaveguideboundedonitstwoendsbythecross-sectional surfaces CS 1 and CS 2 . Setting ˇ J i = ˇ J c = 0 in(4.156)wehave 1 2 contintegraldisplay S ( ˇ E × ˇ H ? )· dS = 2 jω integraldisplay V [〈w e 〉?〈w m 〉] dV, where V istheregionoftheguidebetween CS 1 and CS 2 . Theright-handsiderepresents thedi?erencebetweenthetotaltime-averagestoredelectricandmagneticenergies. Thus 2 jω [〈W e 〉?〈W m 〉] = 1 2 integraldisplay CS 1 ??z ·( ˇ E × ˇ H ? )dS+ 1 2 integraldisplay CS 2 ?z ·( ˇ E × ˇ H ? )dS? 1 2 integraldisplay S cond ( ˇ E × ˇ H ? )· dS, where S cond indicatestheconductingwallsoftheguideand ?n pointsintotheguide. For apropagatingmodethe?rsttwotermsontheright-handsidecancelsincewithnoloss ˇ E× ˇ H ? isthesameon CS 1 and CS 2 . Thethirdtermiszerosince( ˇ E× ˇ H ? )·?n = (?n× ˇ E)· ˇ H ? , and ?n × ˇ E = 0 onthewaveguidewalls. Thuswehave 〈W e 〉=〈W m 〉 foranysectionofalosslesswaveguide. We may compute the time-average stored magnetic energy in a section of lossless waveguideoflength l as 〈W m 〉= μ 4 integraldisplay l 0 integraldisplay CS ˇ H · ˇ H ? dSdz. ForpropagatingTMmodeswecansubstitute(5.139)to?nd 〈W m 〉/l = μ 4 (βY e ) 2 integraldisplay CS (?z ×? t ˇ ψ e )·(?z ×? t ˇ ψ ? e )dS. Using (?z ×? t ˇ ψ e )·(?z ×? t ˇ ψ ? e ) = ?z · bracketleftbig ? t ˇ ψ ? e ×(?z ×? t ˇ ψ e ) bracketrightbig =? t ˇ ψ e ·? t ˇ ψ ? e we have 〈W m 〉/l = μ 4 (βY e ) 2 integraldisplay CS ? t ˇ ψ e ·? t ˇ ψ ? e dS. Finally, using (5.153) we have the stored energy per unit length for a propagating TM mode: 〈W m 〉/l =〈W e 〉/l = μ 4 (ωepsilon1) 2 k 2 c integraldisplay CS ˇ ψ e ˇ ψ ? e dS. Similarlywemayshow thatforaTEmode 〈W e 〉/l =〈W m 〉/l = epsilon1 4 (ωμ) 2 k 2 c integraldisplay CS ˇ ψ h ˇ ψ ? h dS. Thedetailsareleftasanexercise. Aswithplanewavesin(4.261)wemaydescribethevelocityofenergytransportasthe ratioofthePoynting?uxdensitytothetotalstoredenergydensity: S av =〈w T 〉v e . ForTMmodesthisenergyvelocityis v e = 1 2 ωepsilon1βk 2 c ˇ ψ e ˇ ψ ? e 2 μ 4 (ωepsilon1) 2 k 2 c ˇ ψ e ˇ ψ ? e = β ωμepsilon1 = v radicalBig 1 ?ω 2 c /ω 2 , whichisidenticaltothegroupvelocity(5.147). ThisisalsothecaseforTEmodes,for which v e = 1 2 ωμβk 2 c ˇ ψ h ˇ ψ ? h 2 epsilon1 4 (ωμ) 2 k 2 c ˇ ψ h ˇ ψ ? h = β ωμepsilon1 = v radicalBig 1 ?ω 2 c /ω 2 . Example: ?elds of a rectangular waveguide. Consider a rectangular waveguide withacross-sectionoccupying 0 ≤ x ≤ a and 0 ≤ y ≤ b. Thematerialwithintheguide isassumedtobealosslessdielectricofpermittivity epsilon1 andpermeability μ. Weseekthe modal?eldswithintheguide. BothTEandTM?eldsexistwithintheguide. Ineachcasewemustsolvethedi?er- entialequation ? 2 t ? ψ + k 2 c ? ψ = 0. A product solution in rectangular coordinates may be sought using the separation of variablestechnique(§ A.4). We?ndthat ? ψ(x, y,ω)= [A x sin k x x + B x cos k x x] bracketleftbig A y sin k y y + B y cos k y y bracketrightbig where k 2 x + k 2 y = k 2 c . Thissolutioniseasilyveri?edbysubstitution. ForTMmodesthesolutionissubjecttotheboundarycondition(5.143): ? ψ e (ρ,ω)= 0, ρ ∈ Gamma1. Applyingthisat x = 0 and y = 0 we?nd B x = B y = 0. Applyingtheboundarycondition at x = a wethen?nd sin k x a = 0 andthus k x = nπ a , n = 1,2,.... Notethat n = 0 correspondstothetrivialsolution ? ψ e = 0. Similarly,fromthecondition at y = b we?ndthat k y = mπ b , m = 1,2,.... Thus ? ψ e (x, y,ω)= A nm sin parenleftBig nπx a parenrightBig sin parenleftBig mπy b parenrightBig . From(5.137)–(5.139)we?ndthatthe?eldsare ? E z = k 2 c nm A nm bracketleftBig sin nπx a sin mπy b bracketrightBig e ?jk z z , ? E t =?jk z A nm bracketleftBig ?x nπ a cos nπx a sin mπy b + ?y mπ b sin nπx a cos mπy b bracketrightBig e ?jk z z , ? H t = jk z Y e A nm bracketleftBig ?x mπ b sin nπx a cos mπy b ? ?y nπ a cos nπx a sin mπy b bracketrightBig e ?jk z z . Here Y e = 1 η radicalBig 1 ?ω 2 c nm /ω 2 with η = (μepsilon1) 1/2 . Eachcombinationof m,n describesadi?erent?eldpatternandthusadi?erentmode, designatedTM nm . Thecuto?wavenumberoftheTM nm modeis k c nm = radicalbigg parenleftBig nπ a parenrightBig 2 + parenleftBig mπ b parenrightBig 2 , m,n = 1,2,3,... andthecuto?frequencyis ω c nm = v radicalbigg parenleftBig nπ a parenrightBig 2 + parenleftBig mπ b parenrightBig 2 , m,n = 1,2,3,... where v = 1/(μepsilon1) 1/2 . Thus the TM 11 mode has the lowest cuto? frequency of any TM mode. ThereisarangeoffrequenciesforwhichthisistheonlypropagatingTMmode. ForTEmodesthesolutionissubjectto ?n ·? t ? ψ h (ρ,ω)= ? ? ψ h (ρ,ω) ?n = 0, ρ ∈ Gamma1. At x = 0 we have ? ? ψ h ?x = 0 leadingto A x = 0.Aty = 0 we have ? ? ψ h ?y = 0 leadingto A y = 0.Atx = a werequire sin k x a = 0 andthus k x = nπ a , n = 0,1,2,.... Similarly,fromtheconditionat y = b we?nd k y = mπ b , m = 0,1,2,.... Thecase n = m = 0 isnotallowedsinceitproducesthetrivialsolution. Thus ? ψ h (x, y,ω)= B nm cos parenleftBig nπx a parenrightBig cos parenleftBig mπy b parenrightBig , m,n = 0,1,2,..., m + n > 0. From(5.140)–(5.142)we?ndthatthe?eldsare ? H z = k 2 c nm B nm bracketleftBig cos nπx a cos mπy b bracketrightBig e ?jk z z , ? H t =±jk z B nm bracketleftBig ?x nπ a sin nπx a cos mπy b + ?y mπ b cos nπx a sin mπy b bracketrightBig e ?jk z z , ? E t = jk z Z h B nm bracketleftBig ?x mπ b cos nπx a sin mπy b ? ?y nπ a sin nπx a cos mπy b bracketrightBig e ?jk z z . Here Z h = η radicalBig 1 ?ω 2 c nm /ω 2 . InthiscasethemodesaredesignatedTE nm . Thecuto?wavenumberoftheTE nm mode is k c nm = radicalbigg parenleftBig nπ a parenrightBig 2 + parenleftBig mπ b parenrightBig 2 , m,n = 0,1,2,..., m + n > 0 andthecuto?frequencyis ω c nm = v radicalbigg parenleftBig nπ a parenrightBig 2 + parenleftBig mπ b parenrightBig 2 , m,n = 0,1,2,..., m + n > 0 where v = 1/(μepsilon1) 1/2 . Modeshavingthesamecuto?frequencyaresaidtobedegenerate. ThisisthecasewiththeTEandTMmodes. However,the?elddistributionsdi?erand thus the modes are distinct. Note that we may also have degeneracy among the TE or TM modes. For instance, if a = b then the cuto? frequency of the TE nm mode is identicaltothatoftheTE mn mode. If a ≥ b thentheTE 10 modehasthelowestcuto? frequency and is termed the dominant mode in a rectangular guide. There is a ?nite bandoffrequenciesinwhichthisistheonlymodepropagating(althoughthebandwidth issmallif a ≈ b.) Calculation of the time-average power carried by propagating TE and TM modes is leftasanexercise. 5.4.4 TE–TM decomposition in spherical coordinates It is not necessary for the longitudinal direction to be constant to achieve a TE–TM decomposition. Itispossible,forinstance,torepresenttheelectromagnetic?eldinterms ofcomponentseitherTEorTMtotheradialdirectionofsphericalcoordinates. Thismay beshownusingaprocedureidenticaltothatusedforthelongitudinal–transversedecom- position in rectangular coordinates. We carry out the decomposition in the frequency domainandleavethetime-domaindecompositionasanexercise. TE–TM decomposition in terms of the radial ?elds. Considerasource-freere- gionofspace?lledwithahomogeneous,isotropicmaterialdescribedbyparameters ?μ(ω) and ?epsilon1 c (ω). We substitute the spherical coordinate representation of the curl into Fara- day’sandAmpere’slawswithsourceterms ? J and ? J m setequaltozero. Equatingvector componentswehave,inparticular, 1 r bracketleftbigg 1 sinθ ? ? E r ?φ ? ? ?r (r ? E φ ) bracketrightbigg =?jω ?μ ? H θ (5.156) and 1 r bracketleftbigg ? ?r (r ? H θ )? ? ? H r ?θ bracketrightbigg = jω?epsilon1 c ? E φ . (5.157) Weseektoisolatethetransversecomponentsofthe?eldsintermsoftheradialcompo- nents. Multiplying(5.156)by jω?epsilon1 c r weget jω?epsilon1 c 1 sinθ ? ? E r ?φ ? jω?epsilon1 c ?(r ? E φ ) ?r = k 2 r ? H θ ; next,multiplying(5.157)by r andthendi?erentiatingwithrespectto r weget ? 2 ?r 2 (r ? H θ )? ? 2 ? H r ?θ?r = jω?epsilon1 c ?(r ? E φ ) ?r . Subtractingthesetwoequationsandrearranging,weobtain parenleftbigg ? 2 ?r 2 + k 2 parenrightbigg (r ? H θ ) = jω?epsilon1 c 1 sinθ ? ? E r ?φ + ? 2 ? H r ?r?θ . This is a one-dimensional wave equation for the product of r with the transverse ?eld component ? H θ . Similarly parenleftbigg ? 2 ?r 2 + k 2 parenrightbigg (r ? H φ ) =?jω?epsilon1 c ? ? E r ?θ + 1 sinθ ? 2 ? H r ?r?φ , and parenleftbigg ? 2 ?r 2 + k 2 parenrightbigg (r ? E φ ) = 1 sinθ ? 2 ? E r ?φ?r + jω ?μ ? ? H r ?θ , (5.158) parenleftbigg ? 2 ?r 2 + k 2 parenrightbigg (r ? E θ ) = ? 2 ? E r ?θ?r + jω ?μ 1 sinθ ? ? H r ?φ . (5.159) Hence we can represent the electromagnetic ?eld in a source-free region in terms of thetwoscalarquantities ? E r and ? H r . SuperpositionallowsustosolvetheTEcasewith ? E r = 0 andtheTMcasewith ? H r = 0,andcombinetheresultsforthegeneralexpansion ofthe?eld. TE–TM decomposition in terms of potential functions. If we allow the vector potential(orHertzianpotential)tohaveonlyan r-component,thentheresulting?elds areTEorTMtothe r-direction. Unfortunately,thisscalarcomponentdoesnotsatisfy the Helmholtz equation. If we wish to use a potential component that satis?es the Helmholtz equation then we must discard the Lorentz condition and choose a di?erent relationshipbetweenthevectorandscalarpotentials. 1. TM ?elds. To generate ?elds TM to r we recall that the electromagnetic ?elds maybewrittenintermsofelectricvectorandscalarpotentialsas ? E =?jω ? A e ??φ e , (5.160) ? B =?× ? A e . (5.161) Inasource-freeregionwehavebyAmpere’slaw ? E = 1 jω ?μ?epsilon1 c ?× ? B = 1 jω ?μ?epsilon1 c ?×(?× ? A e ). Here ? φ e and ? A e mustsatisfyadi?erentialequationthatmaybederivedbyexamining ?×(?× ? E) =?jω?× ? B =?jω(jω ?μ?epsilon1 c ? E) = k 2 ? E, where k 2 = ω 2 ?μ?epsilon1 c . Substitutionfrom(5.160)gives ?× parenleftbig ?×[?jω ? A e ?? ? φ e ] parenrightbig = k 2 [?jω ? A e ?? ? φ e ] or ?×(?× ? A e )? k 2 ? A e = k 2 jω ? ? φ e . (5.162) Wearestillfreetospecify ?· ? A e . At this point let us examine the e?ect of choosing a vector potential with only an r-component: ? A e = ?r ? A e . Since ?×(?r ? A e ) = ? θ r sinθ ? ? A e ?φ ? ? φ r ? ? A e ?θ (5.163) weseethat B =?× ? A e hasno r-component. Since ?×(?× ? A e ) =? ?r r sinθ bracketleftbigg 1 r ? ?θ parenleftbigg sinθ ? ? A e ?θ parenrightbigg + 1 r sinθ ? 2 ? A e ?φ 2 bracketrightbigg + ? θ r ? 2 ? A e ?r?θ + ? φ r sinθ ? 2 ? A e ?r?φ we see that ? E ～?×(?× ? A e ) has all three components. This choice of ? A e produces a ?eldTMtothe r-direction. Weneedonlychoose ?· ? A e sothattheresultingdi?erential equationisconvenienttosolve. Substitutingtheaboveexpressionsinto(5.162)we?nd that ? ?r r sinθ bracketleftbigg 1 r ? ?θ parenleftbigg sinθ ? ? A e ?θ parenrightbigg + 1 r sinθ ? 2 ? A e ?φ 2 bracketrightbigg + ? θ r ? 2 ? A e ?r?θ + ? φ r sinθ ? 2 ? A e ?r?φ ? ?rk 2 ? A e = ?r k 2 jω ? ? φ e ?r + ? θ r k 2 jω ? ? φ e ?θ + ? φ r sinθ k 2 jω ? ? φ e ?φ . (5.164) Since ?· ? A e onlyinvolvesthederivativesof ? A e withrespectto r,wemayspecify ?· ? A e indirectlythrough ? φ e = jω k 2 ? ? A e ?r . Withthis(5.164)becomes 1 r sinθ bracketleftbigg 1 r ? ?θ parenleftbigg sinθ ? ? A e ?θ parenrightbigg + 1 r sinθ ? 2 ? A e ?φ 2 bracketrightbigg + k 2 ? A e + ? 2 ? A e ?r 2 = 0. Using 1 r ? ?r bracketleftbigg r 2 ? ?r parenleftbigg ? A e r parenrightbiggbracketrightbigg = ? 2 ? A e ?r 2 wecanwritethedi?erentialequationas 1 r 2 ? ?r bracketleftbigg r 2 ?( ? A e /r) ?r bracketrightbigg + 1 r 2 sinθ ? ?θ bracketleftbigg sinθ ?( ? A e /r) ?θ bracketrightbigg + 1 r 2 sin 2 θ ? 2 ( ? A e /r) ?φ 2 + k 2 ? A e r = 0. The ?rst three terms of this expression are precisely the Laplacian of ? A e /r.Thuswe have (? 2 + k 2 ) parenleftbigg ? A e r parenrightbigg = 0 (5.165) andthequantity ? A e /r satis?esthehomogeneousHelmholtzequation. The TM ?elds generated by the vector potential ? A e = ?r ? A e may be found by using (5.160)and(5.161). From(5.160)wehavetheelectric?eld ? E =?jω ? A e ?? ? φ e =?jω?r ? A e ?? parenleftbigg jω k 2 ? ? A e ?r parenrightbigg . Expandingthegradientwehavethe?eldcomponents ? E r = 1 jω ?μ?epsilon1 c parenleftbigg ? 2 ?r 2 + k 2 parenrightbigg ? A e , (5.166) ? E θ = 1 jω ?μ?epsilon1 c 1 r ? 2 ? A e ?r?θ , (5.167) ? E φ = 1 jω ?μ?epsilon1 c 1 r sinθ ? 2 ? A e ?r?φ . (5.168) Themagnetic?eldcomponentsarefoundusing(5.161)and(5.163): ? H θ = 1 ?μ 1 r sinθ ? ? A e ?φ , (5.169) ? H φ =? 1 ?μ 1 r ? ? A e ?θ . (5.170) 2. TE ?elds. Togenerate?eldsTEto r werecallthattheelectromagnetic?eldsin asource-freeregionmaybewrittenintermsofmagneticvectorandscalarpotentialsas ? H =?jω ? A h ??φ h , (5.171) ? D =??× ? A h . (5.172) Inasource-freeregionwehavefromFaraday’slaw ? H = 1 ?jω ?μ?epsilon1 c ?× ? D = 1 jω ?μ?epsilon1 c ?×(?× ? A h ). Here ? φ h and ? A h mustsatisfyadi?erentialequationthatmaybederivedbyexamining ?×(?× ? H) = jω?× ? D = jω?epsilon1 c (?jω ?μ ? H) = k 2 ? H, where k 2 = ω 2 ?μ?epsilon1 c . Substitutionfrom(5.171)gives ?× parenleftbig ?×[?jω ? A h ?? ? φ h ] parenrightbig = k 2 [?jω ? A h ?? ? φ h ] or ?×(?× ? A h )? k 2 ? A h = k 2 jω ? ? φ h . (5.173) Choosing ? A h = ?r ? A h and ? φ h = jω k 2 ? ? A h ?r we?nd,aswiththeTM?elds, (? 2 + k 2 ) parenleftbigg ? A h r parenrightbigg = 0. (5.174) Thusthequantity ? A h /r obeystheHelmholtzequation. Wecan?ndtheTE?eldsusing(5.171)and(5.172). Substitutingwe?ndthat ? H r = 1 jω ?μ?epsilon1 c parenleftbigg ? 2 ?r 2 + k 2 parenrightbigg ? A h , (5.175) ? H θ = 1 jω ?μ?epsilon1 c 1 r ? 2 ? A h ?r?θ , (5.176) ? H φ = 1 jω ?μ?epsilon1 c 1 r sinθ ? 2 ? A h ?r?φ , (5.177) ? E θ =? 1 ?epsilon1 c 1 r sinθ ? ? A h ?φ , (5.178) ? E φ = 1 ?epsilon1 c 1 r ? ? A h ?θ . (5.179) Example of spherical TE–TM decomposition: a plane wave. Consider a uni- form plane wave propagating in the z-direction in a lossless, homogeneous material of permittivity epsilon1 andpermeability μ,suchthatitselectromagnetic?eldis ? E(r,ω)= ?x ? E 0 (ω)e ?jkz = ?x ? E 0 (ω)e ?jkr cosθ , ? H(r,ω)= ?y ? E 0 (ω) η e ?jkz = ?x ? E 0 (ω) η e ?jkr cosθ . Wewishtorepresentthis?eldintermsofthesuperpositionofa?eldTEto r anda?eld TM to r. We ?rst ?nd the potential functions ? A e = ?r ? A e and ? A h = ?r ? A h that represent the?eld. Thenwemayuse(5.166)–(5.170)and(5.175)–(5.179)to?ndtheTEandTM representations. From(5.166)weseethat ? A e isrelatedto ? E r ,where ? E r isgivenby ? E r = ? E 0 sinθ cosφe ?jkr cosθ = ? E 0 cosφ jkr ? ?θ bracketleftbig e ?jkr cosθ bracketrightbig . Wecanseparatether andθ dependencesoftheexponentialfunctionbyusingtheidentity (E.101). Since j n (?z) = (?1) n j n (z) = j ?2n j n (z) we have e ?jkr cosθ = ∞ summationdisplay n=0 j ?n (2n + 1)j n (kr)P n (cosθ). Using ?P n (cosθ) ?θ = ?P 0 n (cosθ) ?θ = P 1 n (cosθ) we thus have ? E r =? j ? E 0 cosφ kr ∞ summationdisplay n=1 j ?n (2n + 1)j n (kr)P 1 n (cosθ). Herewestartthesumatn = 1 since P 1 0 (x) = 0. Wecannowidentifythevectorpotential as ? A e r = ? E 0 k ω cosφ ∞ summationdisplay n=1 j ?n (2n + 1) n(n + 1) j n (kr)P 1 n (cosθ) (5.180) sincebydirectdi?erentiationwehave ? E r = 1 jω ?μ?epsilon1 c parenleftbigg ? 2 ?r 2 + k 2 parenrightbigg ? A e = ? E 0 k jω 2 ?μ?epsilon1 c cosφ ∞ summationdisplay n=1 j ?n (2n + 1) n(n + 1) P 1 n (cosθ) parenleftbigg ? 2 ?r 2 + k 2 parenrightbigg [rj n (kr)] =? j ? E 0 cosφ kr ∞ summationdisplay n=1 j ?n (2n + 1)j n (kr)P 1 n (cosθ), whichsatis?es(5.166). Herewehaveusedthede?ningequationofthesphericalBessel functions(E.15)toshowthat parenleftbigg ? 2 ?r 2 + k 2 parenrightbigg [rj n (kr)] = r ? 2 ?r 2 j n (kr)+ 2 ? ?r j n (kr)+ k 2 rj n (kr) = k 2 r bracketleftbigg ? 2 ?(kr) 2 + 2 kr ? ?(kr) bracketrightbigg j n (kr)+ k 2 rj n (kr) =?k 2 r bracketleftbigg 1 ? n(n + 1) (kr) 2 bracketrightbigg j n (kr)+ k 2 rj n (kr) = n(n + 1) r j n (kr). Wenoteimmediatelythat ? A e /r satis?estheHelmholtzequation(5.165)sinceithasthe formoftheseparationofvariablessolution(D.113). Wemay?ndthevectorpotential ? A h = ?r ? A h inthesamemanner. Notingthat ? H r = ? E 0 η sinθ sinφe ?jkr cosθ = ? E 0 sinφ ηjkr ? ?θ bracketleftbig e ?jkr cosθ bracketrightbig = 1 jω ?μ?epsilon1 c parenleftbigg ? 2 ?r 2 + k 2 parenrightbigg ? A h , wehavethepotential ? A h r = ? E 0 k ηω sinφ ∞ summationdisplay n=1 j ?n (2n + 1) n(n + 1) j n (kr)P 1 n (cosθ). (5.181) WemaynowcomputethetransversecomponentsoftheTM?eldusing(5.167)–(5.170). Forconvenience,letusde?neanewfunction ? J n by ? J n (x) = xj n (x). Thenwemaywrite ? E r =? j ? E 0 cosφ (kr) 2 ∞ summationdisplay n=1 j ?n (2n + 1) ? J n (kr)P 1 n (cosθ), (5.182) ? E θ = j ? E 0 kr sinθ cosφ ∞ summationdisplay n=1 a n ? J prime n (kr)P 1 n prime (cosθ), (5.183) ? E φ = j ? E 0 kr sinθ sinφ ∞ summationdisplay n=1 a n ? J prime n (kr)P 1 n (cosθ), (5.184) ? H θ =? ? E 0 krη sinθ sinφ ∞ summationdisplay n=1 a n ? J n (kr)P 1 n (cosθ), (5.185) ? H φ = ? E 0 krη sinθ cosφ ∞ summationdisplay n=1 a n ? J n (kr)P 1 n prime (cosθ). (5.186) Here ? J prime n (x) = d dx ? J n (x) = d dx [xj n (x)] = xj prime n (x)+ j n (x) and a n = j ?n (2n + 1) n(n + 1) . (5.187) Similarly,wehavetheTE?eldsfrom(5.176)–(5.179): ? H r =? j ? E 0 sinφ η(kr) 2 ∞ summationdisplay n=1 j ?n (2n + 1) ? J n (kr)P 1 n (cosθ), (5.188) ? H θ = j ? E 0 ηkr sinθ sinφ ∞ summationdisplay n=1 a n ? J prime n (kr)P 1 n prime (cosθ), (5.189) ? H φ =?j ? E 0 ηkr sinθ cosφ ∞ summationdisplay n=1 a n ? J prime n (kr)P 1 n (cosθ), (5.190) ? E θ =? ? E 0 kr sinθ cosφ ∞ summationdisplay n=1 a n ? J n (kr)P 1 n (cosθ), (5.191) ? E φ =? ? E 0 kr sinθ sinφ ∞ summationdisplay n=1 a n ? J n (kr)P 1 n prime (cosθ). (5.192) Thetotal?eldisthenthesumoftheTEandTMcomponents. Example of spherical TE–TM decomposition: scattering by a conducting sphere. ConsideraPECsphereofradius a centeredattheoriginandimbeddedina homogeneous,isotropicmaterialhavingparameters ?μ and ?epsilon1 c . Thesphereisilluminated byaplanewaveincidentalongthe z-axiswiththe?elds ? E(r,ω)= ?x ? E 0 (ω)e ?jkz = ?x ? E 0 (ω)e ?jkr cosθ , ? H(r,ω)= ?y ? E 0 (ω) η e ?jkz = ?x ? E 0 (ω) η e ?jkr cosθ . Wewishto?ndthe?eldscatteredbythesphere. Theboundaryconditionthatdeterminesthescattered?eldisthatthetotal(incident plusscattered)electric?eldtangentialtothespheremustbezero. Wesawintheprevious example that the incident electric ?eld may be written as the sum of a ?eld TE to the r-direction and a ?eld TM to the r-direction. Since the region external to the sphere is source-free, we may also represent the scattered ?eld as a sum of TE and TM ?elds. Thesemaybefoundfromthefunctions ? A s e and ? A s h ,whichobeytheHelmholtzequations (5.165)and(5.174). ThegeneralsolutiontotheHelmholtzequationmaybefoundusing theseparationofvariablestechniqueinsphericalcoordinates,asshownin § A.4,andis givenby braceleftbigg ? A s e /r ? A s h /r bracerightbigg = ∞ summationdisplay n=0 n summationdisplay m=?n C nm Y nm (θ,φ)h (2) n (kr). Here Y nm isthesphericalharmonicandwehavechosenthesphericalHankelfunction h (2) n as the radial dependence since it represents the expected outward-going wave behavior of the scattered ?eld. Since the incident ?eld generated by the potentials (5.180) and (5.181)exactlycancelsthe?eldgeneratedby ? A s e and ? A s h onthesurfaceofthesphere,by orthogonalitythescatteredpotentialmusthave φ and θ dependenciesthatmatchthose oftheincident?eld. Thus ? A s e r = ? E 0 k ω cosφ ∞ summationdisplay n=1 b n h (2) n (kr)P 1 n (cosθ), ? A s h r = ? E 0 k ηω sinφ ∞ summationdisplay n=1 c n h (2) n (kr)P 1 n (cosθ), where b n and c n areconstantstobedeterminedbytheboundaryconditions. Bysuper- positionthetotal?eldmaybecomputedfromthetotalpotentials,whicharethesumof theincidentandscatteredpotentials. Thesearegivenby ? A t e r = ? E 0 k ω cosφ ∞ summationdisplay n=1 bracketleftbig a n j n (kr)+ b n h (2) n (kr) bracketrightbig P 1 n (cosθ), ? A t h r = ? E 0 k ηω sinφ ∞ summationdisplay n=1 bracketleftbig a n j n (kr)+ c n h (2) n (kr) bracketrightbig P 1 n (cosθ), where a n isgivenby(5.187). Thetotaltransverseelectric?eldisfoundbysuperposingtheTEandTMtransverse ?eldsfoundfromthetotalpotentials. Wehavealreadycomputedthetransverseincident ?elds and may easily generalize these results to the total potentials. By (5.183) and (5.191)wehave ? E t θ (a) = j ? E 0 ka sinθ cosφ ∞ summationdisplay n=1 bracketleftbig a n ? J prime n (ka)+ b n ? H (2)prime n (ka) bracketrightbig P 1 n prime (cosθ)? ? ? E 0 ka sinθ cosφ ∞ summationdisplay n=1 bracketleftbig a n ? J n (ka)+ c n ? H (2) n (ka) bracketrightbig P 1 n (cosθ)= 0, where ? H (2) n (x) = xh (2) n (x). By(5.184)and(5.192)wehave ? E t φ (a) = j ? E 0 ka sinθ sinφ ∞ summationdisplay n=1 bracketleftbig a n ? J prime n (ka)+ b n ? H (2)prime n (ka) bracketrightbig P 1 n (cosθ)? ? ? E 0 ka sinθ sinφ ∞ summationdisplay n=1 bracketleftbig a n ? J n (ka)+ c n ? H (2) n (ka) bracketrightbig P 1 n prime (cosθ)= 0. Thesetwosetsofequationsaresatis?edbytheconditions b n =? ? J prime n (ka) ? H (2)prime n (ka) a n , c n =? ? J n (ka) ? H (2) n (ka) a n . Wecannowwritethescatteredelectric?eldsas ? E s r =?j ? E 0 cosφ ∞ summationdisplay n=1 b n bracketleftbig ? H (2)primeprime n (kr)+ ? H (2) n (kr) bracketrightbig P 1 n (cosθ), ? E s θ = ? E 0 kr cosφ ∞ summationdisplay n=1 bracketleftbigg jb n sinθ ? H (2)prime n (kr)P 1 n prime (cosθ)? c n 1 sinθ ? H (2) n (kr)P 1 n (cosθ) bracketrightbigg , ? E s φ = ? E 0 kr sinφ ∞ summationdisplay n=1 bracketleftbigg jb n 1 sinθ ? H (2)prime n (kr)P 1 n (cosθ)? c n sinθ ? H (2) n (kr)P 1 n prime (cosθ) bracketrightbigg . Letusapproximatethescattered?eldforobservationpointsfarfromthesphere. We mayapproximatethesphericalHankelfunctionsusing(E.68)as ? H (2) n (z) = zh (2) n (z) ≈ j n+1 e ?jz , ? H (2)prime n (z) ≈ j n e ?jz , ? H (2)primeprime n (z) ≈?j n+1 e ?jz . Substitutingthesewe?ndthat ? E r → 0 asexpectedforthefar-zone?eld,while ? E s θ ≈ ? E 0 e ?jkr kr cosφ ∞ summationdisplay n=1 j n+1 bracketleftbigg b n sinθ P 1 n prime (cosθ)? c n 1 sinθ P 1 n (cosθ) bracketrightbigg , ? E s φ ≈ ? E 0 e ?jkr kr sinφ ∞ summationdisplay n=1 j n+1 bracketleftbigg b n 1 sinθ P 1 n (cosθ)? c n sinθ P 1 n prime (cosθ) bracketrightbigg . 012345678910 ka 0.01 0.10 1.00 10.00 σ / π a 2 Figure5.6: Monostaticradarcross-sectionofaconductingsphere. From the far-zone ?elds we can compute the radar cross-section (RCS) or echo area ofthesphere,whichisde?nedby σ = lim r→∞ parenleftbigg 4πr 2 | ? E s | 2 | ? E i | 2 parenrightbigg . (5.193) Carryingunitsofm 2 ,thisquantitydescribestherelativeenergydensityofthescattered ?eldnormalizedbythedistancefromthescatteringobject.Figure5.6showstheRCSof aconductingsphereinfreespaceforthemonostatic case: whentheobservationdirection isalignedwiththedirectionoftheincidentwave(i.e., θ = π),alsocalledthebackscatter direction. AtlowfrequenciestheRCSisproportionalto λ ?4 ;thisistherangeofRayleigh scattering,showingthathigher-frequencylightscattersmorestronglyfrommicroscopic particlesintheatmosphere(explainingwhytheskyisblue)[19]. Athighfrequenciesthe resultapproachesthatofgeometricaloptics,andtheRCSbecomestheinterceptionarea of the sphere, πa 2 . This is the region of optical scattering. Between these two regions liestheresonance region,ortheregionofMie scattering,namedforG.Miewhoin1908 published the ?rst rigorous solution for scattering by a sphere (followed soon after by Debyein1909). Severalinterestingphenomenaofspherescatteringarebestexaminedinthetimedo- main. We may compute the temporal scattered ?eld by taking the inverse transform ofthefrequency-domain?eld.Figure5.7shows E θ (t) computed in the backscatter direction (θ = π) when the incident ?eld waveform E 0 (t) is a gaussian pulse and the sphere is in free space. Two distinct features are seen in the scattered ?eld waveform. The?rstisasharppulsealmostduplicatingtheincident?eldwaveform,butofopposite polarity. This is the specular re?ection produced when the incident ?eld ?rst contacts the sphere and begins to induce a current on the sphere surface. The second feature, called the creeping wave, occurs at a time approximately (2 + π)a/c seconds after the -0.5 0.0 0.5 1.0 1.5 2.0 t/(2πa/c) -1.2 -0.8 -0.4 -0.0 0.4 0.8 1.2 Relativ e amplitude incident field waveform A: specular reflection B: creeping wave A B Figure5.7: Time-domain?eldback-scatteredbyaconductingsphere. specular re?ection. This represents the ?eld radiated back along the incident direction by a wave of current excited by the incident ?eld at the tangent point, which travels aroundthesphereatapproximatelythespeedoflightinfreespace. Althoughthiswave continues to traverse the sphere, its amplitude is reduced so signi?cantly by radiation dampingthatonlyasinglefeatureisseen. 5.5 Problems 5.1 Verifythatthe?eldsandsourcesobeyingevenplanarre?ectionsymmetryobeythe component Maxwell’s equations (5.1)–(5.6). Repeat for ?elds and sources obeying odd planarre?ectionsymmetry. 5.2 Wewishtoinvestigatere?ectionsymmetrythroughtheorigininahomogeneous medium. Underwhatconditionsonmagnetic?eld,magneticcurrentdensity,andelectric currentdensityareweguaranteedthat E x (x, y, z) = E x (?x,?y,?z), E y (x, y, z) = E y (?x,?y,?z), E z (x, y, z) = E z (?x,?y,?z)? 5.3 We wish to investigate re?ection symmetry through an axis in a homogeneous medium. Underwhatconditionsonmagnetic?eld,magneticcurrentdensity,andelectric currentdensityareweguaranteedthat E x (x, y, z) =?E x (?x,?y, z), E y (x, y, z) =?E y (?x,?y, z), E z (x, y, z) = E z (?x,?y, z)? 5.4 Consider an electric Hertzian dipole located on the z-axis at z = h. Showthat ifthedipoleisparalleltotheplane z = 0,thenaddinganoppositely-directeddipoleof the same strength at z =?h produces zero electric ?eld tangential to the plane. Also showthat if the dipole is z-directed, then adding another z-directed dipole at z =?h produces zero electric ?eld tangential to the z = 0 plane. Since the ?eld for z > 0 is unalteredineachcaseifweplaceaPECinthe z = 0 plane,weestablishthattangential componentsofelectriccurrentimageintheoppositedirectionwhileverticalcomponents imageinthesamedirection. 5.5 Considera z-directedelectriclinesource ? I 0 locatedat y = h, x = 0 betweencon- ducting planes at y =±d, d > h. The material between the plates has permeability ?μ(ω) andcomplexpermittivity ?epsilon1 c (ω). Writetheimpressedandscattered?eldsinterms of Fourier transforms and apply the boundary conditions at z =±d to determine the electric?eldbetweentheplates. Showthattheresultisidenticaltotheexpression(5.8) obtained using symmetry decomposition, which required the boundary condition to be appliedonlyonthetopplate. 5.6 Consider a z-directed electric line source ? I 0 located at y = h, x = 0 in free space aboveadielectricslaboccupying ?d < y < d, d < h. Theslabhaspermeability μ 0 and permittivity epsilon1. Decomposethesourceintoevenandoddconstituentsandsolveforthe electric?eldeverywhereusingtheFouriertransformapproach. Describehowyouwould usetheevenandoddsolutionstosolvetheproblemofadielectricslablocatedontopof aPECgroundplane. 5.7 Consideranunbounded,homogeneous,isotropicmediumdescribedbypermeabil- ity ?μ(ω) andcomplexpermittivity ?epsilon1 c (ω). Assumingtherearemagneticsourcespresent, butnoelectricsources,showthatthe?eldsmaybewrittenas ? H(r) =?jω?epsilon1 c integraldisplay V ˉ G e (r|r prime ;ω)· ? J i m (r prime ,ω)dV prime , ? E(r) = integraldisplay V ˉ G m (r|r prime ;ω)· ? J i m (r prime ,ω)dV prime , where ˉ G e isgivenby(5.83)and ˉ G m isgivenby(5.84). 5.8 Showthatforacubicalexcludingvolumethedepolarizingdyadicis ˉ L = ˉ I/3. 5.9 Compute the depolarizing dyadic for a cylindrical excluding volume with height anddiameterboth 2a,andwiththelimittakenas a → 0. Showthat ˉ L = 0.293 ˉ I. 5.10 Showthatthesphericalwavefunction ? ψ(r,ω)= e ?jkr 4πr obeystheradiationconditions(5.96)and(5.97). 5.11 VerifythatthetransversecomponentoftheLaplacianof A is (? 2 A) t = bracketleftbigg ? t (? t · A t )+ ? 2 A t ?u 2 ?? t ×? t × A t bracketrightbigg . VerifythatthelongitudinalcomponentoftheLaplacianof A is ?u parenleftbig ?u ·? 2 A parenrightbig = ?u? 2 A u . 5.12 Verifytheidentities(B.82)–(B.93). 5.13 Verifytheidentities(B.94)–(B.98). 5.14 Derivetheformula(5.112)forthetransversecomponentoftheelectric?eld. 5.15 Thelongitudinal/transversedecompositioncanbeperformedbeginningwiththe time-domainMaxwell’sequations. Showthatforahomogeneous,lossless,isotropicregion described by permittivity epsilon1 and permeability μ the longitudinal ?elds obey the wave equations parenleftbigg ? 2 ?u 2 ? 1 v 2 ? 2 ?t 2 parenrightbigg H t =? t ?H u ?u ?epsilon1?u ×? t ?E u ?t +epsilon1 ?J mt ?t ? ?u × ?J t ?u , parenleftbigg ? 2 ?u 2 ? 1 v 2 ? 2 ?t 2 parenrightbigg E t =? t ?E u ?u +μ?u ×? t ?H u ?t + ?u × ?J mt ?u +μ ?J t ?t . Alsoshowthatthetransverse?eldsmaybefoundfromthelongitudinal?eldsbysolving parenleftbigg ? 2 ? 1 v 2 ? ?t 2 parenrightbigg E u = 1 epsilon1 ?ρ ?u +μ ? J u ?t +? t × J mt , parenleftbigg ? 2 ? 1 v 2 ? ?t 2 parenrightbigg H u = 1 μ ?ρ m ?u +epsilon1 ? J mu ?t ?? t × J t . Here v = 1/ √ μepsilon1. 5.16 Considerahomogeneous,lossless,isotropicregionofspacedescribedbypermittiv- ity epsilon1 and permeability μ. Beginning with the source-free time-domain Maxwell equa- tionsinrectangularcoordinates,choose z asthelongitudinaldirectionandshowthatthe TE–TMdecompositionisgivenby parenleftbigg ? 2 ?z 2 ? 1 v 2 ? 2 ?t 2 parenrightbigg E y = ? 2 E z ?z?y +μ ? 2 H z ?x?t , (5.194) parenleftbigg ? 2 ?z 2 ? 1 v 2 ? 2 ?t 2 parenrightbigg E x = ? 2 E z ?x?z ?μ ? 2 H z ?y?t , (5.195) parenleftbigg ? 2 ?z 2 ? 1 v 2 ? 2 ?t 2 parenrightbigg H y =?epsilon1 ? 2 E z ?x?t + ? 2 H z ?y?z , (5.196) parenleftbigg ? 2 ?z 2 ? 1 v 2 ? 2 ?t 2 parenrightbigg H x = epsilon1 ? 2 E z ?y?t + ? 2 H z ?x?z , (5.197) with parenleftbigg ? 2 ? 1 v 2 ? 2 ?t 2 parenrightbigg E z = 0, (5.198) parenleftbigg ? 2 ? 1 v 2 ? 2 ?t 2 parenrightbigg H z = 0. (5.199) Here v = 1/ √ μepsilon1. 5.17 ConsiderthecaseofTM?eldsinthetimedomain. Showthatforahomogeneous, isotropic, lossless medium with permittivity epsilon1 and permeability μ the ?elds may be derived from a single Hertzian potential Π e (r,t) = ?u ? Pi1 e (r,t) that satis?es the wave equation parenleftbigg ? 2 ? 1 v 2 ? 2 ?t 2 parenrightbigg Pi1 e = 0 andthatthe?eldsare E =? t ?Pi1 e ?u + ?u parenleftbigg ? 2 ?u 2 ? 1 v 2 ? 2 ?t 2 parenrightbigg Pi1 e , H =?epsilon1?u ×? t ?Pi1 e ?t . 5.18 ConsiderthecaseofTE?eldsinthetimedomain. Showthatforahomogeneous, isotropic, lossless medium with permittivity epsilon1 and permeability μ the ?elds may be derived from a single Hertzian potential Π h (r,t) = ?u ? Pi1 h (r,t) that satis?es the wave equation parenleftbigg ? 2 ? 1 v 2 ? 2 ?t 2 parenrightbigg Pi1 h = 0 andthatthe?eldsare E = μ?u ×? t ?Pi1 h ?t , H =? t ?Pi1 h ?u + ?u parenleftbigg ? 2 ?u 2 ? 1 v 2 ? 2 ?t 2 parenrightbigg Pi1 h . 5.19 Showthat in the time domain TEM ?elds may be written for a homogeneous, isotropic,losslessmediumwithpermittivity epsilon1 andpermeability μ intermsofaHertzian potentialΠ e = ?uPi1 e thatsatis?es ? 2 t Pi1 e = 0 andthatthe?eldsare E =? t ?Pi1 e ?u , H =?epsilon1?u ×? t ?Pi1 e ?t . 5.20 Showthat in the time domain TEM ?elds may be written for a homogeneous, isotropic,losslessmediumwithpermittivity epsilon1 andpermeability μ intermsofaHertzian potentialΠ h = ?uPi1 h thatsatis?es ? 2 t Pi1 h = 0 andthatthe?eldsare E = μ?u ×? t ?Pi1 h ?t , H =? t ?Pi1 h ?u . 5.21 ConsideraTEMplane-wave?eldoftheform ? E = ?x ? E 0 e ?jkz , ? H = ?y ? E 0 η e ?jkz , where k = ω √ μepsilon1 and η = √ μ/epsilon1. Showthat: (a) ? E maybeobtainedfrom ? H usingtheequationsfora?eldthatisTE y ; (b) ? H maybeobtainedfrom ? E usingtheequationsfora?eldthatisTM x ; (c) ? E and ? H maybeobtainedfromthepotential ? Π h = ?y( ? E 0 /k 2 η)e ?jkz ; (d) ? E and ? H maybeobtainedfromthepotential ? Π e = ?x( ? E 0 /k 2 )e ?jkz ; (e) ? E and ? H maybeobtainedfromthepotential ? Π e = ?z(j ? E 0 x/k)e ?jkz ; (f) ? E and ? H maybeobtainedfromthepotential ? Π h = ?z(j ? E 0 y/kη)e ?jkz . 5.22 Prove the orthogonality relationships (5.149) and (5.150) for the longitudinal ?eldsinalosslesswaveguide. Hint: Substitute a = ˇ ψ e and b = ˇ ψ h intoGreen’ssecond identity(B.30)andapplytheboundaryconditionsforTEandTMmodes. 5.23 Verifythewaveguideorthogonalityconditions(5.151)-(5.152)bysubstitutingthe ?eldexpressionsforarectangularwaveguide. 5.24 Showthatthetime-averagepowercarriedbyapropagatingTEmodeinalossless waveguideisgivenby P av = 1 2 ωμβk 2 c integraldisplay CS ˇ ψ h ˇ ψ ? h dS. 5.25 Showthatthetime-averagestoredenergyperunitlengthforapropagatingTE modeinalosslesswaveguideis 〈W e 〉/l =〈W m 〉/l = epsilon1 4 (ωμ) 2 k 2 c integraldisplay CS ˇ ψ h ˇ ψ ? h dS. 5.26 Considerawaveguideofcircularcross-sectionalignedonthez-axisand?lledwith alosslessmaterialhavingpermittivity epsilon1 andpermeability μ. SolveforboththeTEand TM?eldswithintheguide. Listthe?rsttenmodesinorderbycuto?frequency. 5.27 ConsiderapropagatingTMmodeinalosslessrectangularwaveguide. Showthat thetime-averagepowercarriedbythepropagatingwaveis P av nm = 1 2 ωepsilon1β nm k 2 c nm |A nm | 2 ab 4 . 5.28 ConsiderapropagatingTEmodeinalosslessrectangularwaveguide. Showthat thetime-averagepowercarriedbythepropagatingwaveis P av nm = 1 2 ωμβ nm k 2 c nm |B nm | 2 ab 4 . 5.29 Considerahomogeneous,losslessregionofspacecharacterizedbypermeabilityμ andpermittivity epsilon1. Beginningwiththetime-domainMaxwellequations, showthatthe θ and φ components of the electromagnetic ?elds can be written in terms of the radial components. FromthisgivetheTE r –TM r ?elddecomposition. 5.30 Considertheformulafortheradarcross-sectionofaPECsphere(5.193). Show thatforthemonostaticcasetheRCSbecomes σ = λ 2 4π vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle ∞ summationdisplay n=1 (?1) n (2n + 1) ? H (2)prime n (ka) ? H (2) n (ka) vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 2 . 5.31 BeginningwiththemonostaticformulafortheRCSofaconductingspheregiven inProblem5.30,usethesmall-argumentapproximationtothesphericalHankelfunctions toshowthattheRCSisproportionalto λ ?4 when ka lessmuch 1. 5.32 BeginningwiththemonostaticformulafortheRCSofaconductingspheregiven inProblem5.30,usethelarge-argumentapproximationtothesphericalHankelfunctions toshowthattheRCSapproachestheinterceptionareaofthesphere, πa 2 ,aska →∞. 5.33 Amaterialsphereofradius a haspermittivity epsilon1 andpermeability μ. Thesphere iscenteredattheoriginandilluminatedbyaplanewavetravelinginthe z-directionwith the?elds ? E(r,ω)= ?x ? E 0 (ω)e ?jkz , ? H(r,ω)= ?y ? E 0 (ω) η e ?jkz . Findthe?eldsinternalandexternaltothesphere.