H=joeV×me (5.63) Here Ji can represent either an impressed electric current source or an impressed polar ization current source J= joP. The electric Hertzian potential obeys (5.64) When J≠0andJ=0 we let Ah= joRE Ih jojv×I H=V(V·m)+k2=V×(Vxm) Jm Here J can represent either an impressed magnetic current source or an impressed magnetization current source J i= joAM. The magnetic Hertzian potential obeys (5.67) When both electric and magnetic sources are present we have by superposition E=V(V·真)+k2- join x In 风 H= joe V Ile+V(V·Ih)+kI joe v le+vx(xIn)-ioii 5.2.1 Solution for potentials in an unbounded medium: the retarded Under the Lorentz condition each of the potential functions obeys the wave equation. This equation can be solved using the method of Green's functions to determine the potentials, and the electromagnetic fields can therefore be determined. We now examine the solution for an unbounded medium. Solutions for bounded regions are considered in §5.2.2. e Consider a linear operator L that operates on a function of r and t. If we wish to solve C{(r,t)}=S(r,t), (5.68) C{G(r,|r,t)=8(r-r)(t-t) ②2001
and H˜ = jω˜ c ∇ × Π˜ e. (5.63) Here J˜i can represent either an impressed electric current source or an impressed polarization current source J˜i = jωP˜i . The electric Hertzian potential obeys (∇2 + k2 )Π˜ e = − J˜i jω˜ c . (5.64) When J˜i m = 0 and J˜i = 0 we let A˜ h = jωµ˜ ˜ c Π˜ h and find E˜ = − jωµ˜ ∇ × Π˜ h (5.65) and H˜ = ∇(∇ · Π˜ h) + k2 Π˜ h =∇× (∇ × Π˜ h) − J˜i m jωµ˜ . (5.66) Here J˜i m can represent either an impressed magnetic current source or an impressed magnetization current source J˜i m = jωµ˜ M˜ i . The magnetic Hertzian potential obeys (∇2 + k2 )Π˜ h = − J˜i m jωµ˜ . (5.67) When both electric and magnetic sources are present we have by superposition E˜ = ∇(∇ · Π˜ e) + k2 Π˜ e − jωµ˜ ∇ × Π˜ h =∇× (∇ × Π˜ e) − J˜i jω˜ c − jωµ˜ ∇ × Π˜ h and H˜ = jω˜ c ∇ × Π˜ e + ∇(∇ · Π˜ h) + k2 Π˜ h = jω˜ c ∇ × Π˜ e +∇× (∇ × Π˜ h) − J˜i m jωµ˜ . 5.2.1 Solution for potentials in an unbounded medium: the retarded potentials Under the Lorentz condition each of the potential functions obeys the wave equation. This equation can be solved using the method of Green’s functions to determine the potentials, and the electromagnetic fields can therefore be determined. We nowexamine the solution for an unbounded medium. Solutions for bounded regions are considered in § 5.2.2. Consider a linear operator L that operates on a function of r and t. If we wish to solve the equation L{ψ(r, t)} = S(r, t), (5.68) we first solve L{G(r, t|r , t )} = δ(r − r )δ(t − t )
and determine the green's function G for the operator C. Provided that s resides within S(r, !G(r, tr, t')dr S(r, !CIG(r, tr,, t)dr'dv S(r, td(r-r)s(t-tdt'dv y (r, t) S(r, !G(r, tr, tdr'dv (5.69) by comparison with(5. 68) We can also apply this idea in the frequency domain. The solution to Cly(r, o)=S(r, o) y(r, o) where the green's fu CIG(rr; o))=8(r-r) Equation(5.69)is the basic superposition integral that allows us to find the potentials in an infinite, unbounded medium. We note that if the medium is bounded then we must use Greens theorem to include the effects of sources that reside external to the bound aries. These are manifested in terms of the values of the potentials on the 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 fields on the boundary as the boundary recedes to infinity. In the fre- quency domain we find that an additional "radiation condition" is required to ensure uniqueness. The retarded potentials in the time domain. Consider an unbounded, homoge- neous, lossy, isotropic medium described by parameters u, E, o. In the time domain the vector potential Ae satisfies(5. 30). The scalar components of Ae must obey VAe n(r, t)-Ao 0A(.D-123a(2=-ur,n=xy,x at We may write this in the form a u2 at u2 at y(r, t=-S(r, t) where y= Ae.m, u2=1/uE, S=0/2E, and S=HJi. The solution is y(r, t)= S(r, t)G(r, tr, t)dt'dV ②2001
and determine the Green’s function G for the operator L. Provided that S resides within V w e have L V ∞ −∞ S(r , t )G(r, t|r , t ) dt dV = V ∞ −∞ S(r , t )L{G(r, t|r , t )} dt dV = V ∞ −∞ S(r , t )δ(r − r )δ(t − t ) dt dV = S(r, t), hence ψ(r, t) = V ∞ −∞ S(r , t )G(r, t|r , t ) dt dV (5.69) by comparison with (5.68). We can also apply this idea in the frequency domain. The solution to L{ψ( ˜ r,ω)} = S˜(r,ω) (5.70) is ψ( ˜ r,ω) = V S˜(r ,ω)G(r|r ; ω) dV where the Green’s function G satisfies L{G(r|r ; ω)} = δ(r − r ). Equation (5.69) is the basic superposition integral that allows us to find the potentials in an infinite, unbounded medium. We note that if the medium is bounded then we must use Green’s theorem to include the effects of sources that reside external to the boundaries. 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 fields on the boundary as the boundary recedes to infinity. In the frequency domain we find that an additional “radiation condition” is required to ensure uniqueness. The retarded potentials in the time domain. Consider an unbounded, homogeneous, lossy, isotropic medium described by parameters µ,,σ. In the time domain the vector potential Ae satisfies (5.30). The scalar components of Ae must obey ∇2Ae,n(r, t) − µσ ∂ Ae,n(r, t) ∂t − µ ∂2Ae,n(r, t) ∂t 2 = −µJ i n (r, t), n = x, y,z. We may write this in the form ∇2 − 2 v2 ∂ ∂t − 1 v2 ∂2 ∂t 2 ψ(r, t) = −S(r, t) (5.71) where ψ = Ae,n, v2 = 1/µ, = σ/2, and S = µJ i n . The solution is ψ(r, t) = V ∞ −∞ S(r , t )G(r, t|r , t ) dt dV (5.72)
where g satisfies 29a p)r,r,1= -r)8(t-t) (573) In§A.1 we find that G(r, tr, t) 8(-t'-R/u) h(s2√(t-1)2-(R/u)2 R t aTv s2√(-t)2-(R/u) or lossless media where g =0 this becomes G(r, tr, t) 8(t-t-R/u) and thus 8(t-t-R/u) y(r, t) dt dv R S(r, t-R/u) For lossless media, the scalar potentials and all rectangular components of the vector ials obey the same wave equation. Thus we have, for instance, the solution J(r,t-R/u) I p(r, t-R/u) 中(r,t)= These are called the retarded potentials since their values at time t are determined by the ralues of the sources at an earlier(or retardation) time t-R/u. The retardation time is determined by the propagation velocity v of the potential waves The fields are determined by the potentials E(r, t)=-V p(r, t- R/o)dv'-au(, t-R/v)av J'(r, t-R/u) H(r,t=Vx R The derivatives may be brought inside the integrals, but some care must be taken when the observation point r lies within the source region. In this case the integrals must be performed in a principal value sense by excluding a small volume around the observation point. We discuss this in more detail below for the frequency-domain fields. For details regarding this procedure in the time domain the reader may see Hansen 81 ②2001
where G satisfies ∇2 − 2 v2 ∂ ∂t − 1 v2 ∂2 ∂t 2 G(r, t|r , t ) = −δ(r − r )δ(t − t ). (5.73) In § A.1 we find that G(r, t|r , t ) = e−(t−t ) δ(t − t − R/v) 4π R + + 2 4πv e−(t−t ) I1 (t − t )2 − (R/v)2 (t − t )2 − (R/v)2 , t − t > R v , where R = |r − r |. For lossless media where σ = 0 this becomes G(r, t|r , t ) = δ(t − t − R/v) 4π R and thus ψ(r, t) = V ∞ −∞ S(r , t ) δ(t − t − R/v) 4π R dt dV = V S(r , t − R/v) 4π R dV . (5.74) For lossless media, the scalar potentials and all rectangular components of the vector potentials obey the same wave equation. Thus we have, for instance, the solutions to (5.51): Ae(r, t) = µ 4π V Ji (r , t − R/v) R dV , φe(r, t) = 1 4π V ρi (r , t − R/v) R dV . These are called the retarded potentials since their values at time t are determined by the values of the sources at an earlier (or retardation) time t − R/v. The retardation time is determined by the propagation velocity v of the potential waves. The fields are determined by the potentials: E(r, t) = −∇ 1 4π V ρi (r , t − R/v) R dV − ∂ ∂t µ 4π V Ji (r , t − R/v) R dV , H(r, t) =∇× 1 4π V Ji (r , t − R/v) R dV . The derivatives may be brought inside the integrals, but some care must be taken when the observation point r lies within the source region. In this case the integrals must be performed in a principal value sense by excluding a small volume around the observation point. We discuss this in more detail belowfor the frequency-domain fields. For details regarding this procedure in the time domain the reader may see Hansen [81].
The retarded potentials in the frequency domain. Consider an unbounded, ho- mogeneous, isotropic medium described by A(o)and E(). If y(r, o)represents a scalar potential or any rectangular component of a vector or Hertzian potential then it must (5.75) here k =o(ie)2. This Helmholtz equation has the form of(5. 70)and thus y(r,a)=/S(r, a)G(rlr'; o he (V2+k2)Gr;o)=-8(r-r) (5.76) This is equation(A46 )and its solution, as given by(A49),is 4丌R Here we use u=1/Re and 32=0/2E in(A 47) j2ω2 The solution to(5.75)is therefore e-jkR v(r,o)=/s(r,o)元_nd (5.78) When the medium is lossless, the potential must also satisfy the radiation condition +ik)v(r)=0 79 to guarantee uniqueness of solution. In$ 5.2.2 we shall show how this requirement arises from the solution within a bounded region. For a uniqueness proof for the helmholtz equation, the reader may consult Chew 133 We may use(5.78)to find that Ae(r, o) ](r (5.80) Comparison with(5.74)shows that in the frequency domain, time retardation takes the form of a phase shift. Similarly p(r The electric and magnetic dyadic Greens functions. The frequency-domain elec- tromagnetic fields may be found for electric sources from the electric vector potential using(5.60)and (5.61) E(r, a)=-joit(o)/j(r, o)G(rlr: a)dv'-Jol(vv. /j(r, o)G(rlr: o)dv H=vx(r,o)G(rr;o)dv (5.82) ②2001
The retarded potentials in the frequency domain. Consider an unbounded, homogeneous, isotropic medium described by µ(ω) ˜ and ˜ c(ω). If ψ( ˜ r,ω) represents a scalar potential or any rectangular component of a vector or Hertzian potential then it must satisfy (∇2 + k2 )ψ( ˜ r,ω) = −S˜(r,ω) (5.75) where k = ω(µ˜ ˜ c)1/2. This Helmholtz equation has the form of (5.70) and thus ψ( ˜ r,ω) = V S˜(r ,ω)G(r|r ; ω) dV where (∇2 + k2 )G(r|r ; ω) = −δ(r − r ). (5.76) This is equation (A.46) and its solution, as given by (A.49), is G(r|r ; ω) = e− jkR 4π R . (5.77) Here we use v2 = 1/µ˜ ˜ and = σ/˜ 2 in (A.47): k = 1 v ω2 − j2ω = ω µ˜ ˜ − j σ˜ ω = ω µ˜ ˜ c. The solution to (5.75) is therefore ψ( ˜ r,ω) = V S˜(r ,ω) e− jkR 4π R dV . (5.78) When the medium is lossless, the potential must also satisfy the radiation condition lim r→∞ r ∂ ∂r + jk ψ( ˜ r) = 0 (5.79) to guarantee uniqueness of solution. In § 5.2.2 we shall showhowthis requirement arises from the solution within a bounded region. For a uniqueness proof for the Helmholtz equation, the reader may consult Chew[33]. We may use (5.78) to find that A˜ e(r,ω) = µ˜ 4π V J˜i (r ,ω) e− jkR R dV . (5.80) Comparison with (5.74) shows that in the frequency domain, time retardation takes the form of a phase shift. Similarly, φ(˜ r,ω) = 1 4π˜ c V ρ˜i (r ,ω) e− jkR R dV . (5.81) The electric and magnetic dyadic Green’s functions. The frequency-domain electromagnetic fields may be found for electric sources from the electric vector potential using (5.60) and (5.61): E˜(r,ω) = − jωµ(ω) ˜ V J˜i (r ,ω)G(r|r ; ω) dV − jωµ(ω) ˜ k2 ∇∇ · V J˜i (r ,ω)G(r|r ; ω) dV , H˜ =∇× V J˜i (r ,ω)G(r|r ; ω) dV . (5.82)
As long as the observation point r does not lie within the source region we may take the derivatives inside the integrals. Using v J(r,a)G(rr; o)='(r,o).VG(rlr; o)+G(rlr; o)VJ(r,o vG(rr;ω)·J(r,a) E(r, o)=-jou 于r,o)Gr;o)+2vrvcr;o)J(r,o)]dv This can be written more compactly as E(r,o)=-jop(o)/ Ge(rr;).J'(r,o)dV G-(rr;)=I+ +刘]oro (5.83) is called the electric dyadic Green's function. Using ⅴ×JG]=VG×+GV×J=VGxj ve have for the magnetic field H(r,o)=VG(rr;o)xJ'(r,o)dV Now, using the dyadic identity(B 15) we may show that is called the magnetic dyadic Green's function Proceeding similarly for magnetic sources (or using duality) we have rir;o)·Jn(r,a)dV When both electric and magnetic sources are present we simply use superposition and When the observation point lies within the source region, we must be much more careful about how we formulate the dyadic Greens functions. In(5.82)we encounter the J(r,oG(rr; o)dv
As long as the observation point r does not lie within the source region we may take the derivatives inside the integrals. Using ∇ · J˜i (r ,ω)G(r|r ; ω) = J˜i (r ,ω) · ∇G(r|r ; ω) + G(r|r ; ω)∇ · J˜(r ,ω) = ∇G(r|r ; ω) · J˜i (r ,ω) we have E˜(r,ω) = − jωµ(ω) ˜ V J˜i (r ,ω)G(r|r ; ω) + 1 k2 ∇ ∇G(r|r ; ω) · Ji (r ,ω) dV . This can be written more compactly as E˜(r,ω) = − jωµ(ω) ˜ V G¯ e(r|r ; ω) · J˜i (r ,ω) dV where G¯ e(r|r ; ω) = ¯ I + ∇∇ k2 G(r|r ; ω) (5.83) is called the electric dyadic Green’s function. Using ∇ × [J˜i G] = ∇G × J˜i + G∇ × J˜i = ∇G × J˜i we have for the magnetic field H˜ (r,ω) = V ∇G(r|r ; ω) × J˜i (r ,ω) dV . Now, using the dyadic identity (B.15) we may show that J˜i × ∇G = (J˜i × ∇G) · ¯ I = (∇G × ¯ I) · Ji . So H˜ (r,ω) = − V G¯ m(r|r ; ω) · J˜i (r ,ω) dV where G¯ m(r|r ; ω) = ∇G(r|r ; ω) × ¯ I (5.84) is called the magnetic dyadic Green’s function. Proceeding similarly for magnetic sources (or using duality) we have H˜ (r) = − jω˜ c V G¯ e(r|r ; ω) · J˜i m(r ,ω) dV , E˜(r) = V G¯ m(r|r ; ω) · J˜i m(r ,ω) dV . When both electric and magnetic sources are present we simply use superposition and add the fields. When the observation point lies within the source region, we must be much more careful about how we formulate the dyadic Green’s functions. In (5.82) we encounter the integral V J˜i (r ,ω)G(r|r ; ω) dV .