Computational Electromagnetics; Chapter 1 1 Chapter 1 Mathematical Foundations 1.1 Maxwell s Equations Electromagnetic phenomena can be described by the electric field E, the electric induction D, the current density J as well as the magnetic field H and the magnetic induction B according to Maxwell s equations given by Faraday s law B where according to the Gauss law and Ampère s law D where, again observing the Gauss law + curl E = 0 in lr 3, (1.1) div B = 0 in lr 3, (1.2) curl H + J = 0 in lr 3, (1.3) div D = ρ in lr 3. (1.4) In particular, Faraday s law describes how an electric field can be induced by a changing magnetic flux. It states that the induced electric field is proportional to the time rate of change of the magnetic flux through the circuit. B E n For D lr 3 the integral form of Faraday s law states: D B dx = D E n dσ. where E n is the tangential trace of E with n denoting the exterior unit normal. Observe the orientation of the induced electric field in Fig. 1. The Stokes theorem implies E n dσ = curl E dx, D D and we thus obtain [ B + curl E] dx = 0. D Figure 1: Faraday s law
2 Ronald H.W. Hoppe n B Figure 2: Gauss law of the magnetic field For D lr 3 the integral form of the Gauss law of the magnetic field states: D n B dσ = 0, where n B is the normal component of B. The Gauss integral theorem implies n B dσ = div B dx, D D and we thus obtain divb = 0 as the differential form of the Gauss law. In other words, the Gauss law of the magnetic field says that the magnetic induction is a solenoidal vector field (source-free). On the other hand, Ampère s law shows that an electric current can induce a magnetic field. It says that the path integral of the magnetic flux around a closed path is proportional to the electric current enclosed by the path. D + J For D lr 3 the integral form of Ampére s law states: ( D + J) dx = H n dσ. D D H n where H n is the tangential trace of H. The Stokes theorem implies H n dσ = curl H dx, D D and we thus obtain [ D curl H + J] dx = 0, D Figure 3: Ampére s law whence D curl H + J = 0 as the differential form of the Ampére law. Finally, the Gauss law of the electric field expresses the fact that the charges represent the source of the electric induction D.
Computational Electromagnetics; Chapter 1 3 n D ρ Figure 2: Gauss law of the electric field For D lr 3 the integral form of the Gauss law of the electric field states: n D dσ = ρ dx, D D where n D is the normal component of D. The Gauss integral theorem implies n D dσ = div D dx, D D from which we deduce divd = ρ as the differential form of the Gauss law. The fields D, E, J, and B, H are related by the material laws D = ε E + P, (1.5) J = σ E + J e, (1.6) B = µ H + µ 0 M, (1.7) where J e, M, and P are the impressed current density, magnetization, and electric polarization, respectively. Here, ε = ε r ε 0 and µ = µ r µ 0 are the electric permittivity and magnetic permeability of the medium with ε 0 and µ 0 denoting the permittivity and permeability of the vacuum ( ε r and µ r are referred to as the relative permittivity and the relative permeability). At interfaces Γ := 1 2, 1 2 =, separating different media 1, 2 lr 3, transmission conditions have to be satisfied. According to (1.4), the sources of the electric field are given by the electric charges. Denoting by n the unit normal on Γ pointing into the direction of 2, the normal component n D of the electric induction experiences a jump [n D] Γ := n (D Γ 2 D Γ 1 ) = η (1.8) with η denoting the surface charge. On the other hand, the tangential trace E n of the electric field behaves continuously [E n] Γ := (E Γ 2 E Γ 1 ) n = 0, (1.9) which is in accordance with the physical laws, since otherwise a nonzero jump would indicate the existence of a magnetic surface current. Since the magnetic induction B is solenoidal, the normal component n B must behave continuously, i.e., [n B] Γ = 0, (1.10)
4 Ronald H.W. Hoppe whereas the tangential trace H n of the magnetic field undergoes a jump according to [H n] Γ = j Γ, (1.11) where j Γ is the surface current. Finally, the continuity of the current at interfaces requires that the normal component n J of the current density satisfies [n J] Γ = η. (1.12) 1.2 Electromagnetic Potentials The special form of Maxwell s equations (1.1)-(1.7) allows to introduce electromagnetic potentials which facilitate the computation of electromagnetic field problems by reducing the number of unknowns. 1.2.1 Electric Scalar Potential In case of an electrostatic field in a medium occupying a bounded simplyconnected domain lr 3, Faraday s law (1.1) reduces to curl E = 0 in. (1.13) Consequently, the electric field E can be represented as the gradient of an electric scalar potential ϕ according to E = grad ϕ. (1.14) 1.2.2 Magnetic Vector Potential The solenoidal character of the magnetic induction B according to (1.2) implies the existence of a magnetic vector potential A such that Moreover, the material law (1.7) gives and hence, B = curl A. (1.15) H = µ 1 B µ 1 r M, H = µ 1 curl A µ 1 r M. (1.16) Replacing H in Ampère s law (1.3) by (1.16), we obtain curl (µ 1 curl A µ 1 r M) = J + D. (1.17)
Computational Electromagnetics; Chapter 1 5 On the other hand, replacing B in Faraday s law (1.1) by (1.15), we get curl (E + A ) = 0. (1.18) From (1.18) we deduce the existence of an electric scalar potential ϕ such that E = A grad ϕ. (1.19) Using (1.19) in (1.17) and observing the material laws (1.5) (with P = 0) and (1.6), we arrive at the following wave-type equation for the magnetic vector potential A: ε 2 A 2 + σ A + curl (µ 1 curl A) = (1.20) = J e + curl µ 1 r M σ grad ϕ ε (grad ϕ) Since the curl-operator has a nontrivial kernel, the magnetic vector potential A is not uniquely determined by (1.15). This can be taken care of by a proper gauging which specifies the divergence of the potential. We distinguish between the Coulomb gauge given by and the Lorentz gauge div A = 0 (1.21) ϕ = diva, (1.22) which is widely used in electromagnetic wave propagation problems. 1.2.3 Magnetic Scalar Potential In case of a magnetostatic problem without currents, i.e., J = 0, D = 0, and vanishing magnetization M = 0, equation (1.3) reduces to curl H = 0. (1.23) As for electrostatic problems, (1.39) implies the existence of a magnetic scalar potential ψ such that H = grad ψ. (1.24) The solenoidal character of the magnetic induction (1.2) and the material law (1.7) imply, that ψ satisfies the elliptic differential equation div (µ grad ψ) = 0. (1.25)
6 Ronald H.W. Hoppe We further note that even problems with nonzero current density J can be cast in terms of the magnetic scalar potential. For this purpose, we decompose the magnetic field H according to H = H 1 + H 2 (1.26) into an irrotational part H 1, i.e., curl H 1 = 0, and a second part H 2 that can be computed by means of the Bio-Savart law H 2 = 1 J x dx. (1.27) 4π x 3 In this case, we have H 1 = grad ψ R, (1.28) with ψ R being referred to as the reduced magnetic scalar potential. It follows readily that ψ R satisfies the elliptic differential equation div (µ grad ψ R ) = div µ H 2. (1.29) 1.3 Electrostatic Problems In case of electrostatic problems, according to (1.14) the electric field E is given by the gradient of an electric scalar potential ϕ. Using (1.4) as well as the material law (1.5), we arrive at the following linear second order elliptic differential equation div ε grad ϕ = ρ div P in. (1.30) On the boundary Γ 1, where the normal component n D of the electric induction D is given by means of a prescribed surface current η, we obtain the Neumann boundary condition n ε grad ϕ = η + n P on Γ 1. (1.31) On the other hand, if the boundary Γ 2, Γ 2 Γ 1 =, only contains metallic contacts, the electric field is perpendicular to Γ 2. In other words, the tangential trace n E vanishes, and we get E n = grad ϕ n = 0 on Γ 2, from which we deduce the Dirichlet boundary condition ϕ = g on Γ 2, (1.32) where the constant g is given by prescribed voltages. The variational formulation of (1.30),(1.31),(1.32) involves the Hilbert space H 1 g,γ 2 () := { v H 1 () v Γ2 = g } (1.33)
Computational Electromagnetics; Chapter 1 7 and can be derived as follows: Multiplying (1.30) by v H0,Γ 1 2 () and integrating over, Green s formula implies div ε grad ϕ v dx = (1.34) = = ε grad ϕ grad v dx ε grad ϕ grad v dx Γ n ε grad ϕ v dσ = Γ 1 (η + n P) v dσ. Moreover, applying Green s formula once more, we have div P v dx = (1.35) = P grad v dx Γ n P v dσ = P grad v dx Using (1.34) and (1.35), the variational problem reads: Find ϕ H 1 g,γ 2 () such that Γ 1 n P v dσ. a(ϕ, v) = l(v), v H 1 0,Γ 2 (), (1.36) where a(, ) is the bilinear form a(ϕ, v) = ε grad ϕ grad v dx (1.37) and the functional l( ) is given by l(v) = ρ v dx + P grad v dx + Γ 1 η v dσ. (1.38) 1.4 Magnetostatic Problems For magnetostatic problems we use the potentials A and ϕ according to (1.15) and (1.19). Equation (1.20) reduces to curl (µ 1 curl A µ 1 r M) = J e σ grad ϕ =: f. (1.39) As far as boundary conditions are concerned, we assume = Γ 1 Γ 2, Γ 1 Γ 2 =. On Γ 1 we assume vanishing tangential trace of A whereas on Γ 2 we suppose that A n = 0 on Γ 1, (1.40) H n = j Γ2 on Γ 2,
8 Ronald H.W. Hoppe where j Γ2 get is the surface current density on Γ 2. Taking (1.16) into account, we (µ 1 curl A µ 1 r M) n = j Γ2 on Γ 2. (1.41) 1.5 The Eddy Currents Equations The eddy current equations represent the quasi-stationary limit of Maxwell s equations and describe the low frequency regime characterized by slowly time varying processes in conductive media. In this case, we have σ E εe, (1.42) which means that the dielectric displacement can be neglected. Hence, (1.20) reduces to the parabolic type equation σ A + curl (µ 1 curl A µ 1 r M) = J e σ grad ϕ. (1.43) 1.6 The Time-Harmonic Maxwell Equations We consider a homogeneous, nonconducting medium (i.e., σ = 0 and J e = M = P = 0) with electric permittivity ε and magnetic permeability µ. In this case, Maxwell s equations (1.1),(1.3) reduce to ε E µ H curl H = 0, (1.44) + curl E = 0. (1.45) Applying the divergence to both equations, we see that which implies div E(x, t) = div H(x, t) = 0, (1.46) div E(x, t) = div H(x, t) = 0, (1.47) provided div E(x, t 0 ) = div H(x, t 0 ) = 0 at initial time t 0. Differentiating (1.44),(1.45) with respect to time, we get 2 E 2 2 H 2 1 ε + 1 µ H curl E curl = 0, (1.48) = 0. (1.49)
Computational Electromagnetics; Chapter 1 9 Replacing H in (1.48) by (1.45) and E in (1.49) by (1.44), we obtain 2 E + 1 curl curl H 2 εµ = 0, (1.50) 2 H + 1 curl curl E 2 εµ = 0. (1.51) Taking (1.47) into account and observing the vectorial identity E = grad div E curl curl E, we finally see that E and H are solutions of the wave equations 2 E 2 c 2 E = 0, (1.52) 2 H 2 c 2 H = 0, (1.53) where the speed of light in the medium is given by c = 1 εµ. (1.54) The time-harmonic solutions of Maxwell equations, also called plane waves, are complex-valued fields E(x, t) = Re ( E(x) exp( iωt) ), (1.55) H(x, t) = Re ( H(x) exp( iωt) ) that satisfy the system of time-harmonic Maxwell equations curl H + iωεe = 0, (1.56) curl E iωµh = 0, where ω stands for the frequency of the electromagnetic waves. Similar computations as done before reveal that E and H satisfy (1.47) and thus the vectorial Helmholtz equations where k = ω εµ is the wave number. E + k 2 E = 0, (1.57) E + k 2 E = 0, (1.58) 1.6.1 Fundamental Solution and Radiation Conditions We will be concerned with the fundamental harmonic solution of Maxwell s equations in a homogeneous medium with given electric permittivity ε and magnetic permeability µ. The time-harmonic system is given by iωεe + curl H = 0, (1.59) iωµh + curl E = 0. (1.60)
10 Ronald H.W. Hoppe The fundamental solution consists of two parts. Both parts are given by 3 3 matrices E (ν) F, H (ν) F C 3 3, 1 ν 2, satisfying iωεe (1) F + curl H (1) F = δ I, (1.61) iωµh (1) F + curl E 1) F = 0, iωεe (2) F + curl H (2) F = 0, (1.62) iωµh (2) F + curl E 2) F = δ I, where I is the 3 3 identity matrix and δ stands for the Dirac mass at the origin. We introduce potentials V F C 3 and A F C 3 3 by means of and the Lorentz gauge where k is the wave number Further, we refer to E F = V F + A F (1.63) div A F k 2 V F = 0, (1.64) k = ω εµ. (1.65) G(x) = 1 4π exp(ik x ) x as the outgoing fundamental solution of the scalar Helmholtz equation. Then, there holds (1.66) Theorem 1.1 Outgoing fundamental solution The outgoing fundamental solution E (1) F, H (1) F associated with (1.61) is given by E (1) F (x) = iωµ G(x) I + i ωε D2 G(x), (1.67) H (1) F (x) = curl(g(x)i), where D 2 G denotes the Hessian of G. Likewise, for the fundamental solution E (2) F, H (2) F associated with (1.62) we get Proof: E (2) E (x) = curl(g(x)i), (1.68) H (2) F (x) = iωε G(x) I i ωµ D2 G(x). From (1.63) we deduce that div E (1) F = V (1) F + div A (1) F. (1.69)
Computational Electromagnetics; Chapter 1 11 On the other hand, the first equation in (1.61) gives div E (1) F = 1 iωε div(δi) = i ωε δ. (1.70) The gauge condition (1.64) allows to eliminate A 1 F (1) from (1.69). Together with (1.70) we thus obtain an equation for V (1) F : V (1) F + k 2 V (1) F = i ωε Consequently, the fundamental solution of (1.71) is given by V (1) F (x) = i ωε Now, applying the curl-operator to (1.63), we get δ. (1.71) G(x). (1.72) curl E (1) F = curl A (1) F. (1.73) On the other hand, applying the curl-operator to the second equation in (1.61) yields Using (1.63),(1.64), we obtain Observing (1.73),(1.74), gives curl curl E (1) F k 2 E (1) F = iωµ δi. (1.74) grad div A (1) F k 2 (E (1) F A (1) F ) = 0. (1.75) grad div A (1) F curl curl A (1) F + k 2 A (1) F = iωµ δi. (1.76) which is equivalent to A (1) F + k 2 A (1) F = iωµ δi. (1.77) It follows that the fundamental solution of (1.77) is given by A (1) F = iωµ G(x)I. (1.78) Using (1.72) and (1.78) in (1.63) and taking advantage of (1.61) gives (1.67). The assertion (1.68) can be shown along the same lines of proof by exchanging E and H, and ε and µ, as well as by replacing i by i. Remark 1.2 Conductive Media Theorem 1.1 also holds true in case of a conductive medium with the only modification that we have to replace ε by ε = ε + i σ ω (1.79)
12 Ronald H.W. Hoppe and the wave number k by k = ω µ ε. (1.80) As far as the asymptotic behavior of the fundamental solution is concerned, we have the following radiation conditions. Theorem 1.3 Radiation Conditions For r := x, the fundamental solution E F, H F as given by either (1.67) or (1.68) satisfies the Sommerfeld radiation conditions E F r ik E F O(r 2 ), (1.81) H F ik H F O(r 2 ) (1.82) r as well as the Silver-Müller radiation conditions ε E F µ H F n O(r 2 ), (1.83) ε E F n + µ H F O(r 2 ), (1.84) where n = x/ x. Any of the four above radiation conditions uniquely determines a fundamental solution E F, H F which satisfies E F (x) = O(r 1 ), H F (x) = O(r 1 ), (1.85) E F (x) x = O(r 1, H F (x) x = O(r 1 ), (1.86) H F (x) E F (x) = O(r 3 ). (1.87) Proof: Since the fundamental solution G of the Helmholtz equation and its partial derivatives satisfy the Sommerfeld radiation condition, (1.81) and (1.82) readily follow from (1.67) and (1.68). Moreover, for any x lr 3 \ {0} the fields E (ν) F and H (ν) F, 1 ν 2, satisfy E (ν) F + k 2 E (ν) F = 0, H (ν) F + k 2 H (ν) F = 0. (1.88) The Sommerfeld radiation condition (1.81) resp. (1.82) then implies the uniqueness of E (ν) F and of H (ν) F. As far as the Silver-Müller radiation condition is concerned, we will prove (1.83) for the fundamental solution E (1) F, H (1) F : We observe G = O( ik exp(ikr) x i x i 4π r r ), (1.89) 2 x i x j G = O( k2 4π exp(ikr) r Using (1.72),(1.78) and (1.89) in (1.63), we find E ij = O( iωµ 4π exp(ikr) r x i x j r 2. (δ ij x ix j )). (1.90) r2
Computational Electromagnetics; Chapter 1 13 On the other hand, using the second equation in (1.67) whence by (1.89) It follows readily that H (1) F (x) = G(x) I, (1.91) H (1) F (x) = O( ik 4π (H (1) F (x) n) ij = O( ik 4π exp(ikr) r exp(ikr) r x r I). (1.92) (δ ij x ix j )), (1.93) r2 which together with (1.90) implies (1.83). The proof of (1.84) follows along the same lines. Furthermore, (1.85)-(1.87) follow directly from (1.83) and (1.84). For the uniqueness we refer to Theorem 5.2.2 in J.-C. Nédélec; Acoustic and Electromagnetic Equations. Integral Representations for Harmonic Problems. Springer, Berlin-Heidelberg-New York, 2001.