Chapter 6 Integral solutions of Maxwells equations 6. 1 Vector Kirchoff solution: method of stratton and chu One of the most powerful tools for the analysis of electromagnetics problems is the integral solution to Maxwells equations formulated by Stratton and Chu [187, 188 These authors used the vector Green's theorem to solve for e and h in much the same way as is done in static fields with the scalar Green's theorem. An alternative approach is to use the Lorentz reciprocity theorem of$ 4. 10. 2, as done by Fradin[ 74. The reciprocity approach allows the identification of terms arising from surface discontinuities, which must be added to the result obtained from the other approach [187 6.1.1 The Stratton-Chu formula Consider an isotropic, homogeneous medium occupying a bounded region V in space The medium is described by permeability i(o), permittivity E(@), and conductivity o() The region V is bounded by a surface S, which can be multiply-connected so that S is the union of several surfaces S1,..., SN as shown in Figure 6.1: these are used to exclude unknown sources and to formulate the vector Huygens principle. Impressed electric and magnetic sources may thus reside both inside and outside V We wish to solve for the electric and magnetic fields at a point r within V. To do this we employ the Lorentz reciprocity theorem(4. 173), written here using the frequency-domain fields as an integral over primed coordinates ea(r,o)x Hb(r,a)-Eb(r,o)x Ha(r, o) [Eb(r,o)·Ja(r,a)-Ea(r,a)·J6(r,o)- (6.1) Hr,o)·Jma(r,o)+H2(r,o)·Jmb(r,o)]dv (6.2) Note that the negative sign on the left arises from the definition of f as the inward normal to V as shown in Figure 6. 1. We place an electric Hertzian dipole at the point r= rp where we wish to compute the field, and set E,= En and H,= H, in the reciprocity theorem,where Ep and H, are the fields produced by the dipole(5. 88)-(5.89) IpG(rrp: o), Ep(r, o)==Vx(V x IpG(rrp; o))) 4 ②2001 by CRC Press LLC
Chapter 6 Integral solutions of Maxwell’s equations 6.1 Vector Kirchoff solution: method of Stratton and Chu One of the most powerful tools for the analysis of electromagnetics problems is the integral solution to Maxwell’s equations formulated by Stratton and Chu [187, 188]. These authors used the vector Green’s theorem to solve for E˜ and H˜ in much the same way as is done in static fields with the scalar Green’s theorem. An alternative approach is to use the Lorentz reciprocity theorem of § 4.10.2, as done by Fradin [74]. The reciprocity approach allows the identification of terms arising from surface discontinuities, which must be added to the result obtained from the other approach [187]. 6.1.1 The Stratton–Chu formula Consider an isotropic, homogeneous medium occupying a bounded region V in space. The medium is described by permeability µ(ω) ˜ , permittivity (ω) ˜ , and conductivity σ(ω) ˜ . The region V is bounded by a surface S, which can be multiply-connected so that S is the union of several surfaces S1,... , SN as shown in Figure 6.1; these are used to exclude unknown sources and to formulate the vector Huygens principle. Impressed electric and magnetic sources may thus reside both inside and outside V. We wish to solve for the electric and magnetic fields at a point r within V. To do this we employ the Lorentz reciprocity theorem (4.173), written here using the frequency-domain fields as an integral over primed coordinates: − S E˜ a(r ,ω) × H˜ b(r ,ω) − E˜ b(r ,ω) × H˜ a(r ,ω) · nˆ d S = V E˜ b(r ,ω) · J˜a(r ,ω) − E˜ a(r ,ω) · J˜b(r ,ω)− (6.1) H˜ b(r ,ω) · J˜ma(r ,ω) + H˜ a(r ,ω) · J˜mb(r ,ω) dV . (6.2) Note that the negative sign on the left arises from the definition of nˆ as the inward normal to V as shown in Figure 6.1. We place an electric Hertzian dipole at the point r = rp where we wish to compute the field, and set E˜ b = E˜ p and H˜ b = H˜ p in the reciprocity theorem, where E˜ p and H˜ p are the fields produced by the dipole (5.88)–(5.89): H˜ p(r,ω) = jω∇ × [p˜G(r|rp; ω)], (6.3) E˜ p(r,ω) = 1 ˜ c ∇ × ∇ × [p˜G(r|rp; ω)] . (6.4)
J/JA S Figure 6. 1: Geometry used to derive the Stratton-Chu formula We also let Ea=E and Ha=h, where E and h are the fields produced by the impressed sources Ja=J and Jma=Jm within V that we wish to find at r=rp. Since the dipole fields are singular at r=rp, we must exclude the point rp with a small spherical surface Ss surrounding the volume Va as shown in Figure 6. 1. Substituting these fields into(6.2) we obtain ,、匡B一Eד=⊥,国了-B,:(6 A useful identity involves the spatially-constant vector p and the Greens function G(rrp) V××(cp=vv·(Cp]-v2(Gp) =v·(Gp)}-pv2G =V师p·VG)+pk2G (6.6) where we have used V G=-k-G for r*rp We begin by computing the terms on the left side of(6.5). We suppress the rde- pendence of the fields and also the dependencies of G(rrp). Substituting from(6.3)we E x Hp]. nds xVx(p]·的 Using f.xVx(Gp=mx(VG×p)=('xE)·(VG×p) we can write E X Hp]- nds= ja ×E]×vGdS'. ②2001 by CRC Press LLC
Figure 6.1: Geometry used to derive the Stratton–Chu formula. We also let E˜ a = E˜ and H˜ a = H˜ , where E˜ and H˜ are the fields produced by the impressed sources J˜a = J˜i and J˜ma = J˜i m within V that we wish to find at r = rp. Since the dipole fields are singular at r = rp, we must exclude the point rp with a small spherical surface Sδ surrounding the volume Vδ as shown in Figure 6.1. Substituting these fields into (6.2) we obtain − S+Sδ E˜ × H˜ p − E˜ p × H˜ · nˆ d S = V−Vδ E˜ p · J˜i − H˜ p · J˜i m dV . (6.5) A useful identity involves the spatially-constant vector p˜ and the Green’s function G(r |rp): ∇ × ∇ × (Gp˜) = ∇ [∇ · (Gp˜)] − ∇2 (Gp˜) = ∇ [∇ · (Gp˜)] − p˜∇2G = ∇ (p˜ · ∇ G) + p˜ k2G, (6.6) where we have used ∇2G = −k2G for r = rp. We begin by computing the terms on the left side of (6.5). We suppress the r dependence of the fields and also the dependencies of G(r |rp). Substituting from (6.3) we have S+Sδ [E˜ × H˜ p] · nˆ d S = jω S+Sδ E˜ × ∇ × (Gp˜) · nˆ d S . Using nˆ · [E˜ × ∇ × (Gp˜)] = nˆ · [E˜ × (∇ G × p˜)] = (nˆ × E˜) · (∇ G × p˜) we can write S+Sδ [E˜ × H˜ p] · nˆ d S = jωp˜ · S+Sδ [nˆ × E˜ ] × ∇ GdS .
Figure 6.2: Decomposition of surface S, to isolate surface field discontinuit examine Ep×田ndS H× V'xVX(Gp)]·fdS Use of (6.6) along with the identity(B 43) gives 中叵×前d I(H X p)kG S+Sa EC [(·vGm+(pVG 0} We would like to use Stokes's theorem on the second term of the right-hand side. Since the theorem is not valid for surfaces on which h has discontinuities, we break the closed rfaces in Figure 6.1 into open surfaces whose boundary contours isolate the disconti- nuities as shown in Figure 6.2. Then we may write nVx[(p VG)H]ds'= dr.A(pVG) For surfaces not containing discontinuities of h the two contour provide equ and opposite contributions and this term vanishes. Thus the left-hand side of (6.5)is 只、国xB,一ExS= [joe(ixE)xvG+k(×mG+·(J+ joE E)VG]dS where we have substituted J+ joe E for V’ x H and used(×p)·=p·(’×H Now consider the right-hand side of (6.5). Substituting from(6.4 )we have E,.J Using(6.6) and(B 42), we have ②2001 by CRC Press LLC
Figure 6.2: Decomposition of surface Sn to isolate surface field discontinuity. Next we examine S+Sδ [E˜ p × H˜ ] · nˆ d S = − 1 ˜ c S+Sδ H˜ × ∇ × ∇ × (Gp˜) · nˆ d S . Use of (6.6) along with the identity (B.43) gives S+Sδ [E˜ p × H˜ ] · nˆ d S = − 1 ˜ c S+Sδ (H˜ × p˜)k2G − − ∇ × (p˜ · ∇ G)H˜ + (p˜ · ∇ G)(∇ × H˜ ) · nˆ d S . We would like to use Stokes’s theorem on the second term of the right-hand side. Since the theorem is not valid for surfaces on which H˜ has discontinuities, we break the closed surfaces in Figure 6.1 into open surfaces whose boundary contours isolate the discontinuities as shown in Figure 6.2. Then we may write Sn=Sna+Snb nˆ · ∇ × (p˜ · ∇ G)H˜ d S = na+ nb dl · H˜ (p˜ · ∇ G). For surfaces not containing discontinuities of H˜ the two contour integrals provide equal and opposite contributions and this term vanishes. Thus the left-hand side of (6.5) is − S+Sδ E˜ × H˜ p − E˜ p × H˜ · nˆ d S = − 1 ˜ c p˜ · S+Sδ jω˜ c (nˆ × E˜) × ∇ G + k2 (nˆ × H˜ )G + nˆ · (J˜i + jω˜ c E˜)∇ G d S where we have substituted J˜i + jω˜ cE˜ for ∇ × H˜ and used (H˜ × p˜) · nˆ = p˜ · (nˆ × H˜ ). Now consider the right-hand side of (6.5). Substituting from (6.4) we have V−Vδ E˜ p · J˜i dV = 1 ˜ c V−Vδ J˜i · ∇ × ∇ × (p˜G) dV . Using (6.6) and (B.42), we have V−Vδ E˜ p · J˜i dV = 1 ˜ c V−Vδ k2 (p˜ · J˜i )G + ∇ · [J˜i (p˜ · ∇ G)] − (p˜ · ∇ G)∇ · J˜i dV .
Replacing V.J with -jop from the continuity equation and using the divergence theorem on the second term on the right-hand side. we then have 1 E. dV==p (2J'G+jop'v'g)dv (的·Jv"GdS Lastly we examine Jn·Vx(Gp) Use of J.V×(p)=Jn(VG×p)=p(Jn×VG)gi JmV=jp·/Jmxv We now substitute all terms into(6.5)and note that each term involves a dot product with p Since p is arbitrary we have axE×VG+(,EG-(x(dS+ dr.HV'G Jm x V'G+=V'G-jojJ' The electric field may be extracted from the above expression by letting the radius of the excluding volume Vs recede to zero. We first consider the surface integral over Ss Examining Figure 6.3 we see that R=rp-r=8,n=-R R VG(rrp) d VRER Assuming E is continuous at r=rp we can write [(×E)×VG+('EVG-joj(xmG]dS R R 40xE)×+(R.E-xm|82aB 1切~(,E+食RE+(RER]d2=E(r ②2001 by CRC Press LLC
Figure 6.3: Geometry of surface integral used to extract E at rp. Replacing ∇ · J˜i with − jωρ˜i from the continuity equation and using the divergence theorem on the second term on the right-hand side, we then have V−Vδ E˜ p · J˜i dV = 1 ˜ c p˜ · V−Vδ (k2 J˜i G + jωρ˜i ∇ G) dV − S+Sδ (nˆ · J˜i )∇ GdS . Lastly we examine V−Vδ H˜ p · J˜i m dV = jω V−Vδ J˜i m · ∇ × (Gp˜) dV . Use of J˜i m · ∇ × (Gp˜) = J˜i m · (∇ G × p˜) = p˜ · (J˜i m × ∇ G) gives V−Vδ H˜ p · J˜i m dV = jωp˜ · V−Vδ J˜i m × ∇ G dV . We now substitute all terms into (6.5) and note that each term involves a dot product with p˜. Since p˜ is arbitrary we have − S+Sδ (nˆ × E˜) × ∇ G + (nˆ · E˜)∇ G − jωµ(˜ nˆ × H˜ )G d S + + 1 jω˜ c a+ b (dl · H˜ )∇ G = V−Vδ −J˜i m × ∇ G + ρ˜i ˜ c ∇ G − jωµ˜ J˜i G dV . The electric field may be extracted from the above expression by letting the radius of the excluding volume Vδ recede to zero. We first consider the surface integral over Sδ . Examining Figure 6.3 we see that R = |rp − r | = δ, nˆ = −Rˆ , and ∇ G(r |rp) = d d R e− jkR 4π R ∇ R = Rˆ 1 + jkδ 4πδ2 e− jkδ ≈ Rˆ δ2 as δ → 0. Assuming E˜ is continuous at r = rp we can write − lim δ→0 Sδ (nˆ × E˜) × ∇ G + (nˆ · E˜)∇ G − jωµ(˜ nˆ × H˜ )G d S = lim δ→0 1 4π (Rˆ × E˜) × Rˆ δ2 + (Rˆ · E˜) Rˆ δ2 − jωµ(˜ Rˆ × H˜ ) 1 δ δ2 d = lim δ→0 1 4π −(Rˆ · E˜)Rˆ + (Rˆ · Rˆ )E˜ + (Rˆ · E˜)Rˆ d = E˜(rp).
Here we have used a ds2= 4T for the total solid angle subtending the sphere Ss. Finally, assuming that the volume sources are continuous, the volume integral over Vs vanishes E(r,)= JmxV'G+=V'G-joij'Gdv'+ G-jo(×mG]ds (dI·mVG (6.7) A similar formula for H can be derived by placing a magnetic dipole of moment pm at r=rp and proceeding as above. This leads to H(r,)=(J'xVG+mV'G-joe Ji) dv'+ 点x+(B下O+“+ (dI·E)VG (6.8) =i Jop We can also obtain this expression by substituting(6.7)into Faraday's law 6.1.2 The Sommerfeld radiation condition In 8 5.2.2 we found that if the potentials are not to be influenced by effects that are infinitely removed, then they must obey a radiation condition. We can make the same argument about the fields from(6.7) and(6.8). Let us allow one of the excluding surfaces, say Sw, to recede to infinity(enclosing all of the sources as it expands). AS Sw-00 any contributions from the fields on this surface to the fields at r should vanish Letting Sw be a sphere centered at the origin, we note that ft= -f and that as 4r VG(rr; w)=R jki Substituting these expressions into(6.7)we find that xVG+(·EvG [(×E)×P+(f·E)门] jop(fx [rakE E ②2001 by CRC Press LLC
Here we have used d = 4π for the total solid angle subtending the sphere Sδ . Finally, assuming that the volume sources are continuous, the volume integral over Vδ vanishes and we have E˜(r,ω) = V −J˜i m × ∇ G + ρ˜i ˜ c ∇ G − jωµ˜ J˜i G dV + + N n=1 Sn (nˆ × E˜) × ∇ G + (nˆ · E˜)∇ G − jωµ(˜ nˆ × H˜ )G d S − − N n=1 1 jω˜ c na+ nb (dl · H˜ )∇ G. (6.7) A similar formula for H˜ can be derived by placing a magnetic dipole of moment p˜ m at r = rp and proceeding as above. This leads to H˜ (r,ω) = V J˜i × ∇ G + ρ˜i m µ˜ ∇ G − jω˜ c J˜i m G dV + + N n=1 Sn (nˆ × H˜ ) × ∇ G + (nˆ · H˜ )∇ G + jω˜ c (nˆ × E˜)G d S + + N n=1 1 jωµ˜ na+ nb (dl · E˜)∇ G. (6.8) We can also obtain this expression by substituting (6.7) into Faraday’s law. 6.1.2 The Sommerfeld radiation condition In § 5.2.2 we found that if the potentials are not to be influenced by effects that are infinitely removed, then they must obey a radiation condition. We can make the same argument about the fields from (6.7) and (6.8). Let us allow one of the excluding surfaces, say SN , to recede to infinity (enclosing all of the sources as it expands). As SN → ∞ any contributions from the fields on this surface to the fields at r should vanish. Letting SN be a sphere centered at the origin, we note that nˆ = −rˆ and that as r → ∞ G(r|r ; ω) = e− jk|r−r | 4π|r − r | ≈ e− jkr 4πr , ∇ G(r|r ; ω) = Rˆ 1 + jkR 4π R2 e− jkR ≈ −rˆ 1 + jkr r e− jkr 4πr . Substituting these expressions into (6.7) we find that lim SN→S∞ SN (nˆ × E˜) × ∇ G + (nˆ · E˜)∇ G − jωµ(˜ nˆ × H˜ )G d S ≈ lim r →∞ 2π 0 π 0 (rˆ × E˜) × rˆ + (rˆ · E˜)rˆ 1 + jkr r + jωµ(˜ rˆ × H˜ ) e− jkr 4πr r2 sin θ dθ dφ ≈ lim r →∞ 2π 0 π 0 r jkE˜ + jωµ(˜ rˆ × H˜ ) + E˜ e− jkr 4π sin θ dθ dφ .