S-96.510 Advanced Field Theory Course for Graduate Students Lecture Viewgraphs, Fall Term 2004

S-96.510 Advanced Field Theory Course for graduate students Lecture viewgraphs, fall term 2004

I.V.Lindell Helsinki University of Technology Electromagnetics Laboratory Espoo, Finland

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 00.01

1 Contents

[01] Complex Vectors and Dyadics [02] Dyadic Algebra [03] Basic Electromagnetic Equations [04] Conditions for Fields and Media [05] Duality Transformation [06] Aﬃne Transformation [07] Electromagnetic Field Solutions [08] Singularities and Complex Sources [09] Plane Waves [10] Source Equivalence [11] Huygens’ Principle [12] Field Decompositions Vector Formulas, Dyadic Identites as an appendix I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 00.02

2 Foreword

This lecture material contains all viewgraphs associated with the graduate course S-96.510 Advanced Field Theory given at the Department of Electrical and Communications Engineering, fall 2004. The course is based on Chapters 1–6 of the book Methods for Electromagnetic Field Analysis (Oxford University Press 1992, 2nd edition IEEE Press, 1995, 3rd printing Wiley 2002) by this author. The ﬁgures drawn by hand on the blackboard could not be added to the present material. Otaniemi, September 13 2004 I.V. Lindell

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 00.03

3 I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 00.04

4 S-96.510 Advanced Field Theory 1. Complex Vectors and Dyadics

I.V.Lindell

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 01.00

5 Complex Vectors

• Complex vectors a = ar + jai,(ar, ai real vectors)

• Time-harmonic vectors A(t) = A1 cos ωt + A2 sin ωt

• Sense of rotation: A1 → A2 shortest way • Correspondence a ↔ A(t) through two mappings • Mapping a → A(t)

jωt A(t) = <{ae } = ar cos ωt − ai sin ωt

• Inverse mapping A(t) → a

a = ar + jai = A(0) − jA(π/2ω)

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 01.01

6 Special Complex Vectors

• Correspondence

a = ar + jai ↔ A(t) = ar cos ωt − ai sin ωt

• Circularly polarized (CP) vectors a · a = 0

• CP implies ar · ai = 0 and |ar| = |ai| • Linearly polarized (LP) vectors a × a∗ = 0

• LP implies ar × ai = 0 (parallel vectors ar, ai) • Elliptical polarization in general • a and b have same ellipse iﬀ b = ejθa, θ real

B(t) = <{bejωt} = <{aej(ωt+θ)} = A(t + θ/ω)

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 01.02

7 Axial Representation

• To ﬁnd axes of the ellipse of a complex vector a (a · a 6= 0) • Solution through another complex vector b √ | a · a| b = b + jb = √ a = ejθa r i a · a

• a and b have same ellipse, same axes (θ real)

b · b = br · br + 2jbr · bi − bi · bi = |a · a| > 0

• b · b real and positive ⇒ br · bi = 0, |br| > |bi|

• br, bi deﬁne the axes of the ellipse of a

• br on major axis, bi on minor axis

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 01.03

8 Helicity Vector 1

• Helicity vector p(a) (’polarization vector’) of a = ar + jai

a × a∗ 2a × a p(a) = = i r ∗ 2 2 ja · a |ar| + |ai|

• Properties: • p(a) = [p(a)]∗ is a real vector • a → a∗ changes sense of rotation: p(a) → p(a∗) = −p(a) • Linearly polarized vector a × a∗ = 0 ⇒ p(a) = 0 • Circularly polarized vector a · a = 0 ⇒ |p(a)| = 1 • Elliptically polarized vector 0 < |p(a)| < 1

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 01.04

9 Helicity Vector 2

• More properties:

a × a∗ 2a × a p(a) = = i r ∗ 2 2 ja · a |ar| + |ai|

• p(a) orthogonal to plane of a, points RH direction • p(αa) = p(a), α 6= 0, magnitude of a has no eﬀect • |p(a)| = 2e/(e2 + 1), e = ellipticity (axial ratio) • p(a) gives info on ellipticity, plane, and sense of rotation of a • p(a) does not give info on magnitude or orientation of ellipse on its plane

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 01.05

10 Vector Bases

• Three complex vectors a1, a2, a3 form a basis if a1 × a2 · a3 6= 0

• Gibbs’ identity by expanding (a1 × a2) × (a3 × b) in two ways:

(a1 × a2 · a3)b = a1(a2 × a3 · b) + a2(a3 × a1 · b) + a3(a1 × a2 · b)

• Deﬁne reciprocal basis

0 a2 × a3 0 a3 × a1 0 a1 × a2 a1 = , a2 = , a3 = a1 × a2 · a3 a1 × a2 · a3 a1 × a2 · a3

• Expansion for any vector b

3 3 X 0 X 0 b = ai(ai · b) = ai(ai · b) i=1 i=1

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 01.06

11 Dyadic algebra

• Josiah Willard Gibbs 1884: dyadic algebra • Dyadic = linear mapping from vector to vector • Example 1: projection on line parallel to unit vector u

b = u(u · a) = (uu) · a

• uu = projection dyadic • Example 2: projection on plane transverse to unit vector u

b = a − (uu) · a = (I − uu) · a

• It = I − uu projection dyadic (I unit dyadic, mapping to oneself)

• uu axial unit dyadic, It transverse (planar) unit dyadic

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 01.07

12 Dyadic polynomial • Dyad = dyadic product of two vectors ab 6= ba • Dyadic = polynomial of dyads N X A = a1b1 + a2b2 + ··· + aN bN = aibi i=1 • Dyadic as a mapping N N X X A · c = ( aibi) · c = ai(bi · c) = a1(b1 · c) + ··· + aN (bN · c) i=1 i=1 N N X X c · A = c · ( aibi) = (c · ai)bi = (c · a1)b1 + ··· + (c · aN )bN i=1 i=1 N T T X c · A = A · c, transposed dyadic A = biai i=1

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 01.08

13 Dyadic expressions • There is no unique expression for dyadics. Two expressions represent same dyadic if they map all vectors in the same way:

A1 · c = A2 · c, for all c, ⇒ A1 = A2

• any dyadic can be expressed as a sum of three dyads

• Example: take a vector ONB (u1, u2, u3)

N N X X A = aibi = (u1u1 · ai + u2u2 · ai + u3u3 · ai)bi i=1 i=1

N N N X X X = u1 (u1 · ai)bi + u2 (u2 · ai)bi + u3 (u3 · ai)bi i=1 i=1 i=1 = u1c1 + u2c2 + u3c3

• (u1, u2, u3) arbitrary ON base ⇒ inﬁnite number of representations

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 01.09

14 Dyadic classiﬁcation

• Any dyadic can be expressed as a1b1 + a2b2 + a3b3

• Planar dyadic can be expressed as sum of two dyads a1b1 + a2b2

• Linear dyadic can be expressed as a single dyad a1b1 • Dyadic which cannot be expressed as a planar dyadic is complete • Complete dyadic maps volumes to volumes • Planar dyadic maps volumes to plane • Linear dyadic maps volumes to line • (Note: complex vectors ⇒ planes and lines in complex space) • Inverse mapping exists only for complete dyadics

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 01.10

15 Symmetric dyadics

• Symmetric dyadic A = AT = (1/2)(A + AT )

3 3 3 X 1 X 1 X a b = (a b +b a ) = [(a +b )(a +b )−(a −b )(a −b )] i i 2 i i i i 4 i i i i i i i i i=1 i=1 i=1 • Symmetric dyadic can be expressed as sum of symmetric dyads P cici but not necessarily in three terms

• Unit dyadic symmetric I = u1u1 + u2u2 + u3u3 independent of ONB

• Examples: I = uxux + uyuy + uzuz = urur + uθuθ + uϕuϕ

3 3 X X I · a = ( uiui) · a = ui(ui · a) i=1 i=1

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 01.11

16 Antisymmetric dyadics

• Antisymmetric dyadic A = −AT = (1/2)(A−AT ) operates through a vector d(A)

3 3 3 X 1 X 1 X A·c = a (b ·c) = [a (b ·c)−b (a ·c)] = (b ×a )×c = d(A)×c i i 2 i i i i 2 i i i=1 i=1 i=1 3 1 X 1 d(A) = b × a = (b × a + b × a + b × a ) 2 i i 2 1 1 2 2 3 3 i=1 • Denoting a × b = a × (I · b) = (a × I) · b,

3 3 X 1 X A = a b = (b × a ) × I = d(A) × I i i 2 i i i=1 i=1

• Any dyadic of the form a × I = I × a is antisymmetric

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 01.12

17 Products of dyadics and vectors

• Dot products A · c and c · A give vectors X X A · c = ( aibi) · c = ai(bi · c) X X c · A = c · ( aibi) = (c · ai)bi

• Example: antisymmetric dyadic A = a × I = I × a A · c = (a × I) · c = a × c, c · A = c · (a × I) = c × a

• Cross products A × c and c × A give dyadics X X A × c = ( aibi) × c = ai(bi × c) X X c × A = c × ( aibi) = (c × ai)bi

• Note: a · (b × A) = (a × b) · A but 6= a × (b · A)!

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 01.13

18 Dot-product of dyadics

• Dot product A · B gives a dyadic X X X A · B = ( aibi) · ( cjdj) = (bi · cj)aidj i j i,j

• Dot product is associative but not commutative like matrix product

A · (B · C) = (A · B) · C, A · B 6= B · A (in general)

• Powers of dyadics

A2 = A · A, An = A · An−1 = An−1 · A, A0 = I

• Inverse of a dyadic possible for complete dyadics only

A · a = b, ⇒ a = A−1 · b

(A · B)−1 = B−1 · A−1

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 01.14

19 Double-cross product of dyadics

× • Double-cross product A×B gives a dyadic

× X × X X A×B = ( aibi)×( cjdj) = (ai × cj)(bi × dj) i j i,j

• Double-cross product is commutative but not associative

× × × × × × A×B = B×A A×(B×C) 6= (A×B)×C (in general)

• Double-cross square 1 A(2) = A×A, I(2) = I 2 × • Inverse of a dyadic can be expressed as

A(2)T A−1 = detA

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 01.15

20 Double-dot product of dyadics

• Double-dot product A : B gives a scalar X X X A : B = ( aibi):( cjdj) = (ai · cj)(bi · dj) i j i,j

A : B = B : A = AT : BT , A : BT = B : AT

• If A antisymmetric and B symmetric, A : B = 0

ab : A = a · A · b, ab : I = a · b, A : B = (A · BT ): I X X A : I = ( aibi): I = ai · bi = trA trI = 3, trace 1 A(2) : I = (A×A): I = spmA 0sum of principal minors0 2 × 1 1 detA = (A×A): A = A(2) : A determinant 6 × 3

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 01.16

21 Dyadic identities

• Dyadic identities are needed in dyadic analysis. They can be formed (1) through vector expansions or (2) from other identities. × × Example of (1) A×(B×C) =? Procedure:

• Expand the dyadics. For brevity omit indices and sum signs, A → ab, B → cd, C → ef. Apply vector identities.

× × (ab)×[(cd)×(ef)] = [a×(c×e)][b×(d×f)] = [c(a·e)−e(a·c)][d(b·f)−f(b·d)] • Write result in terms of dyadic products ab, cd and ef

= (ab : ef)cd − (ef) · (ba) · (cd) − (cd) · (ba) · (ef) + (ab : cd)ef

• Replace ab → A, cd → B, ef → C, results identity

× × T T A×(B×C) = (A : C)B − C · A · B − B · A · C + (A : B)C

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 01.17

22 More dyadic identities

× • Another dyadic identity (A×B): I =? similarly

× [(ab)×(cd)] : I = [(a × c)(b × d)] : I = (a × c) · (b × d)

= (a · b)(c · d) − (a · d)(c · b) = (ab : I)(cd : I) − (ab):(cd)T

× T • Resulting identity (A×B): I = (A : I)(B : I) − A : B valid for any dyadics A and B. A new identity obtained by writing [note × × that (A×B): C = A :(B×C)]

× T A :[B×I − (B : I)I + B ] = 0

• A : C = 0 for all dyadics A, implies C = 0. × T New identity: B×I = (B : I)I − B

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 01.18

23 The inverse dyadic 1

• The general dyadic involves two vector bases {ai}, {bi}:

3 X A = aibi = a1b1 + a2b2 + a3b3 i=1

• The determinant and double-cross square become 1 detA = A×A : A = a b ×a b : a b = (a ×a ·a )(b ×b ·b ) 6 × 1 1× 2 2 3 3 1 2 3 1 2 3 1 A(2) = A×A = a b ×a b + a b ×a b + a b ×a b 2 × 2 2× 3 3 3 3× 1 1 1 1× 2 2 0 0 0 0 0 0 X 0 0 = (a1 × a2 · a3)(b1 × b2 · b3)[a1b1 + a2b2 + a3b3] = detA aibi i

0 0 • Here {ai} and {bi} are bases reciprocal to {ai} and {bi}

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 01.19

24 The inverse dyadic 2

0 • Orthogonality of reciprocal bases bi · bj = δij and Gibbs’ identity P 0 c = aiai · c give an identity

(2)T X X 0 0 X 0 A · A = detA aibi · bjaj = detA aiai = detA I i j i

A · A(2)T = A(2)T · A = detA I

• For complete dyadic satisfying detA 6= 0 the inverse becomes

A(2)T 1 AT ×AT A−1 = = 2 × 1 × detA 6 A×A : A

• For planar dyadics detA = 0 no inverse exists

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 01.20

25 Example of an inverse dyadic

• Find the inverse of B = αI + a × I 1 B(2) = [(αI)×(αI)+2(αI)×(a×I)+(a×I)×(a×I)] = α2I+α(a×I)+aa 2 × × × • To evaluate we apply the properties

(2) × T × I = I, (a × I)×I = −(a × I) = a × I, (a × I)×(a × I) = 2aa

(a × I):(a × I) = 2a · a, I : I = 3, I :(a × I) = 0 1 detB = B(2) : B = α(α2 + a · a) 3 • The result becomes 1 B−1 = [α2I − α(a × I) + aa] α(α2 + a · a)

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 01.21

26 Problems 1.1 Derive the dyadic identity

(a × A)(2) = aa · A(2)

× by starting from the expression (a × bc)×(a × de).

1.2 Given a symmetric dyadic S and a vector a show that one can ﬁnd a vector b satisfying

S × a + a × S = b × I,

i.e., the dyadic on the left is antisymmetric. Find the vector b.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 01.22

27 S-96.510 Advanced Field Theory 2. Dyadic Algebra

I.V.Lindell

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 02.00

28 Dyadics and matrices 1

• Dyadics can be expanded in any ONB (u1, u2, u3) as X X XX A = I · A · I = uiui · A · ujuj = Aijuiuj,

• Matrix elements Aij = A : uiuj depend on chosen vector basis. Some quantities (invariants) are independent of the basis: X X trA = A : I = A : uiui = Aii i i 1 X spmA = A(2) : I = A A (u × u ) · (u × u ) 2 ij k` i k j ` i,j,k,` 1 X 1 X = A A (δ δ − δ δ ) = (A A − A A ) 2 ij k` ij k` i` jk 2 ii kk ik ki i,j,k,` i,k 1 = A A − A A + ··· = [(A : I)2 − A : AT ] 11 22 12 21 2

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 02.01

29 Dyadics and matrices 2 • Third scalar invariant = determinant 1 X detA = A A A (u × u · u )(u × u · u ) 6 ij k` mn i k m j ` n 1 X = A A A = det(A ) 6 ij k` mn ikm j`n ij • Dot-product of dyadics corresponds to the matrix product: X X X A · B = AijBk`uiuj · uku` = uiu` AijBj` i,` j

• Double-cross product = ’mixed subdeterminant’ in matrix algebra

× X A×B = AijBk`(ui × uk)(uj × u`) i,j,k,`

= u1u1(A22B33 + A33B22 − A23B32 − A32B23) + ···

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 02.02

30 Important identity

• A useful identity valid for all A, a, b: A(2) · (a × b) = (A · a) × (A · b)

or [A(2) × a − (A · a) × A] · b = 0 for all b can be generalized to A(2) × a = (A · a) × A

• Further generalization: substitute A → A + B and cancel terms transforms the quadratic identity to a bilinear one:

× (A×B) × a = (A · a) × B + (B · a) × A

• Proof of the identity quadratic in A through identity linear in A, B × (bc×de)×a = (b×d)[e(c·a)−c(e·a)] = (bc·a)×(de)+(de·a)×(bc)

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 02.03

31 Classiﬁcation of dyadics

• A is a complete dyadic iﬀ detA 6= 0, ⇒ A−1 exists

• A is a planar dyadic iﬀ detA = 0 or exists a 6= 0, A · a = 0

• A is a linear dyadic iﬀ A(2) = 0 or exists a 6= 0, A × a = 0

identity A(2) × a = (A · a) × A

• conclusion: if A planar, A(2) linear q identity (A(2))(2) = AdetA = A detA(2)

• conclusion: if A(2) planar it is also linear

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 2.04

32 Eigenvalue problems

• Eigenvalues λ and (right) eigenvectors a of a dyadic A satisfy A · a = λa, a 6= 0, ⇒ (A − λI) · a = 0

• A − λI planar ⇒ det(A − λI) = 0 3rd degree equation for λ 1 (A − λI)×(A − λI):(A − λI) = detA − λspmA + λ2trA − λ3 = 0 6 × • Three roots satisfy the conditions

trA = λ1 +λ2 +λ3, spmA = λ1λ2 +λ2λ3 +λ3λ1, detA = λ1λ2λ3

• General case: three diﬀerent eigenvalues λ1, λ2, λ3 • Two eigenvalues same if A of the form αI+bc (generalized uniaxial dyadic)

• Three eigenvalues same if A of the form αI (isotropic dyadic)

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 2.05

33 Finding the eigenvectors

• Single eigenvalues λi ⇒ eigenvectors ai from

(2) (A − λiI) · ai = 0, ⇒ (A − λiI) × ai = 0

(2) • (A − λiI) is a linear dyadic of the form biai (bi left eigenvector)

(2) ai = c · (A − λiI) , when 6= 0

• For generalized uniaxial dyadic A = αI + bc double eigenvalue (2) λ2,3 = α, ⇒ A − αI = bc linear dyadic, (A − αI) = 0 any vectors a2,3 satisfying c · a2,3 = 0 are eigenvectors

• For isotropic dyadic A = αI triple eigenvalue λ1 = λ2 = λ3 = α, all vectors are eigenvectors

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 2.06

34 Uniaxial dyadics 1 • Uniaxial dyadics (symmetric) are encountered in medium equations

and interface conditions. Axial real unit vector u, It = I − uu (Notation diﬀerent from that in the book!)

U(α, β) = αIt +βuu, [book :D(α, β) = αI +βuu = U(α, α+β)]  α 0 0  Matrix notation : U(α, β) →  0 α 0  0 0 β • Uniaxial dyadics form a two-dimensional linear space:

U(α1, β1) + U(α2, β2) = U(α1 + α2, β1 + β2), τU(α, β) = U(τα, τβ)

U(1, 1) = I, U(1, 0) = It, U(0, 1) = uu special cases

• Eigensolutions a1 = u, λ1 = β and a2,3⊥u, λ2,3 = α

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 02.07

35 Uniaxial dyadics 2

• Properties of symmetric uniaxial dyadics

U(α1, β1) · U(α2, β2) = U(α1α2, β1β2),

U n(α, β) = U(αn, βn), U −1(α, β) = U(α−1, β−1),

× (2) 2 U(α1, β1)×U(α2, β2) = U(α1β2+α2β1, 2α1α2) ⇒ U (α, β) = U(αβ, α )

2 U(α1, β1): U(α2, β2) = 2α1α2 + β1β2, detU(α, β) = α β • Checking the inverse

U (2)T (α, β) U(αβ, α2) U −1(α, β) = = = U(α−1, β−1) detU(α, β) α2β

• Generalized uniaxial dyadics αI + bc obey more complicated rules

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 02.08

36 Reﬂection dyadic

• Reﬂection dyadic C giving mirror image of vector a in the plane u · r = 0 is uniaxial: C · a = a − 2u(u · a) = (I − 2uu) · a,  1 0 0  C = It − uu = U(1, −1), ⇒  0 1 0  0 0 −1 • Properties of the reﬂection dyadic

trC = 1, spmC = −1, detC = −1, C(2) = −C, C−1 = C

• C2 = I, ⇒ C = I1/2 • Square root of a dyadic is not unique!

• Eigensolutions λ1 = −1, a1 = u, λ2,3 = 1, a2,3⊥u

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 2.09

37 Gyrotropic dyadics 1 • Gyrotropic dyadic = uniaxial dyadic + co-axial antisymmetric dyadic (Notation in the book again obtained by substituting β → α + β)

G(α, β, γ) = αIt + βuu + γJ

2 3 J = u×I, J = (u×I)·(u×I) = u×(u×I) = −It, J = −J, ···  α γ 0  G(α, β, γ) →  −γ α 0  , G(α, β, 0) = U(α, β) 0 0 β • Gyrotropic dyadics are encountered in magnetoplasmas and ferrites

2 ! 2 ! 2 ωgωp ωp jωgωp r = 1 − 2 2 It + 1 − 2 uu + 2 2 J ω(ωg − ω ) ω ω(ωg − ω ) ωoωm ωωm µr = 1 − 2 2 It + uu + j 2 2 J ωo − ω ωo − ω

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 02.10

38 Gyrotropic dyadics 2

• Two-dimensional (planar) part of gyrotropic dyadic can be expressed as Jθ Gt(α, γ) = αIt + γJ = Ge ,

• eJθ deﬁned here as two-dimensional exponential function

θ2 2 θ3 eJθ = I + θJ + J + J 3 + ··· = cos θI + sin θJ t 2! 3! t

• Planar rotation deﬁned through dyadic Rt(θ) as

Jθ Rt(θ) = cos θ It + sin θJ = e

p 2 2 Gt(α, γ) = GRt(θ),G = α + γ , tan θ = γ/α • Note the similarity to operating with complex numbers!

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 02.11

39 Gyrotropic dyadics 3

• Denoting cos θ = α/G, sin θ = γ/G, G = pα2 + γ2

G(α, β, γ) = βuu + GeJθ,

• gives the multiplication rule

J(θ1+θ2) G(α1, β1, γ1) · G(α2, β2, γ2) = β1β2uu + G1G2e =

β1β2uu+G1G2[(cos θ1 cos θ2−sin θ1 sin θ2)It+(sin θ1 cos θ2+cos θ1 sin θ2)J]

= G(α1α2 − γ1γ2, β1β2, α1γ2 + α2γ1) G2(α, β, γ) = β2uu + G2eJ2θ = G(α2 − γ2, β2, 2αγ),

Gn(α, β, γ) = βnuu + GneJnθ, α 1 −γ G−1(α, β, γ) = β−1uu + G−1e−Jθ = G( , , ) α2 + γ2 β α2 + γ2 I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 02.12

40 Eigenvalues of a gyrotropic dyadic

• Basic properties of the two-dimensional dyadics It, J:

× × × It×uu = It, J ×uu = J, It×J = 0, (2) (2) J : J = 2, It : It = 2, It : J = 0, It = J = uu • lead to the following properties of the gyrotropic dyadic

G(2)(α, β, γ) = G(αβ, α2+γ2, βγ), spmG = trG(2) = 2αβ+α2+γ2,

detG(α, β, γ) = β(α2 + γ2), trG(α, β, γ) = 2α + β • Eigenvalues from the characteristic equation det(G−λI) = detG−spmG λ+trG λ2 −λ3 = −(λ−β)((λ−α)2 +γ2) = 0

• solutions λ1 = β, λ2,3 = α ± jγ

• Special case uniaxial medium: γ = 0 ⇒ λ2 = λ3 = α I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 02.13

41 Eigenvectors of a gyrotropic dyadic

• Eigenvector a1 corresponding to λ1 = β obtained from

(2) (2) 2 2 (G−λ1I) = ((α−β)It +γJ) = ((α−β) +γ )uu, ⇒ a1 = u,

• Eigenvectors a2,3 corresponding to λ2,3 = α ± jγ with γ 6= 0 obtained from

(2) (G − λ2,3I) = ∓jγ(β − α ∓ jγ)(It ± jJ)

• Expanding dyadics It ± jJ in an ONB (v, w, u = v × w) as

It ± j(v × w) × I = vv + ww ± j(wv − vw) = (v ± jw)(v ∓ jw)

(2) ⇒ (G − λ2,3I) = ∓jγ(β − α ∓ jγ)(v ± jw)(v ∓ jw)

• Eigenvectors a2,3 = v ∓ jw = (I ∓ jJ) · v are circularly polarized v can be chosen as any vector ⊥u

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 02.14

42 Problems

2.1 Derive the inverse of the dyadic ab + c × I through vector algebra by solving the following linear equation for the vector x:

a(b · x) + c × x = y, x =?

2.2 Derive the two-dimensional inverse of the dyadic A = ab + cd whose vectors a ··· d are orthogonal to a unit vector n by studying T × the expression A · (A ×nn) The dyadic I − nn serves as the two- dimensional unit dyadic.

2.3 Assuming that the inverse of the dyadic A is known, ﬁnd an expression for the inverse of the dyadic B = A + ab and check the result.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 02.15

43 S-96.510 Advanced Field Theory 3. Basic Electromagnetic Equations

I.V.Lindell

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 03.00

44 Electromagnetic quantities

• Sinusoidal time dependence ejωt assumed • Electric field intensity E • Magnetic field intensity H • Electric flux density D • Magnetic flux density B • Electric current density J • Magnetic current density M • Linear, time-invariant media, medium equations ! D ξ E = · B ζ µ H

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 03.01

45 Classiﬁcation of electromagnetic media

• Medium dyadics for diﬀerent types of media:

• Isotropic = I, ξ = 0, ζ = 0, µ = µI

• Bi-isotropic = I, ξ = ξI, ζ = ζI, µ = µI

• Anisotropic , ξ = 0, ζ = 0, µ

• Bi-anisotropic , ξ, ζ, µ • Homogeneous: constant medium dyadics • Inhomogeneous: (r), ··· • Dispersive: (ω), ··· • Lossless, reciprocal, etc...

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 03.02

46 Examples of artiﬁcial media

• Medium dyadics can be realized through microscopic elements pro- ducing electric and magnetic dipole moments

D = oE + Pe, B = µoH + Pm

Pe = χee · E + χem · H, Pm = χme · E + χmm · H

• Metal needles χem = χme = χmm = 0, dielectric medium

• Metal rings χem = χme = 0, dielectric-magnetic medium

• Metal helices, Pe and Pm parallel, chiral medium

• Omega-shaped particles, Pe · Pm = 0, omega medium

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 03.03

47 The Maxwell Equations

• For time-harmonic ﬁelds only curl equations needed. Homogeneous medium assumed: constant parameter dyadics • Isotropic ∇ × H(r) = jωE(r) + J(r) −∇ × E(r) = jωµH(r) + M(r)

• Anisotropic ∇ × H(r) = jω · E(r) + J(r) −∇ × E(r) = jωµ · H(r) + M(r)

• Bi-anisotropic

∇ × H(r) = jω · E(r) + jωξ · H(r) + J(r)

−∇ × E(r) = jωµ · H(r) + jωζ · E(r) + M(r)

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 03.04

48 Operator notation

• Maxwell equations in terms of six-vectors and six-dyadics ! ! 0 ∇ × I E ξ E J · −jω · = −∇ × I 0 H ζ µ H M

• More compact notation through the Maxwell six-dyadic operator

L(∇) · e(r) = j(r), ! −jω (∇ × I − jωξ) L(∇) = −(∇ × I + jωζ) −jωµ

• Electric and magnetic ﬁelds are coupled in the Maxwell equations decoupling through elimination or operator diagonalization

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 03.05

49 Diagonalization of operators

• Decoupling through adjoint Maxwell operators ! jωI (∇ × I − jωξ) · µ−1 La(∇) = −(∇ × I + jωζ) · −1 jωI ! jωI −1 · (∇ × I − jωξ) La(∇) = −µ−1 · (∇ × I + jωζ) jωI

• Diagonalization of the six-dyadic operator ! a He(∇) 0 L (∇) · L(∇) = L(∇) · La(∇) = 0 Hm(∇)

• Dyadic Helmholtz operators He(∇), Hm(∇)

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 03.06

50 Helmholtz operators • Dyadic Helmholtz operators are of second order

−1 2 He(∇) = −(∇ × I − jωξ) · µ · (∇ × I + jωζ) + ω

−1 2 Hm(∇) = −(∇ × I + jωζ) · · (∇ × I − jωξ) + ω µ • Special cases: anisotropic medium

−1× 2 −1× 2 He(∇) = µ ×∇∇ + ω , Hm(∇) = ×∇∇ + ω µ √ • Bi-isotropic medium, k = ω µ

2 H(∇) = µHe(∇) = Hm(∇) = −(∇×I−jωξI)·(∇×I+jωζI)+k I

• Isotropic medium, k2 = ω2µ

× 2 H(∇) = µHe(∇) = Hm(∇) = I×∇∇ + k I

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 03.07

51 Helmholtz equations

• Operating the Maxwell equations by the adjoint operator La(∇)

La(∇) · L(∇) · e(r) = La(∇) · j(r)

• gives the Helmholtz equations with dyadic operators (2nd order)

−1 He(∇) · E(r) = jωJ(r) + (∇ × I − jωξ) · µ · M(r),

−1 Hm(∇) · H(r) = jωM(r) − (∇ × I + jωζ) · · J(r)

(2)T • Operating by He,m (∇)· gives equations with scalar Helmholtz de-

terminant operators detHe,m(∇) (4th order)

(2)T −1 detHe(∇) E(r) = He (∇) · [jωJ(r) + (∇ × I − jωξ) · µ · M(r)]

(2)T −1 detHm(∇) H(r) = Hm (∇) · [jωM(r) − (∇ × I + jωζ) · · J(r)]

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 03.08

52 Potentials

• Expressing the ﬁeld six-dyadic as e(r) = La(∇) · f(r) in terms of two vector potentials f = (FG)

E(r) = jωF(r) + −1 · (∇ × I − jωξ) · G(r),

H(r) = −µ−1 · (∇ × I + jωζ) · F(r) + jωG(r),

• leads to Helmholtz equations with decoupled sources:

L(∇) · La(∇) · f(r) = j(r)

He(∇) · F(r) = J(r), Hm(∇) · G(r) = M(r)

• Helmholtz determinant equations (4th order)

(2)T (2)T detHe(∇) F(r) = jωHe (∇)·J(r), detHm(∇) G(r) = jωHm (∇)·M(r)

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 03.09

53 Helmholtz operators for isotropic medium 1

• Helmholtz dyadic operator

× 2 2 2 H(∇) = I×∇∇ + k I = (∇ + k )I − ∇∇

• Helmholtz adjoint operator 1 H(2)T (∇) = (I×∇∇ + k2I)×(I×∇∇ + k2I) 2 × × ×

2 2 2 × 4 2 2 2 = ∇ ∇∇ + k (2∇ I − I×∇∇) + k I = (∇ + k )(∇∇ + k I)

• Helmholtz determinant operator 1 detH(∇) = H(2)(∇): H(∇) = k2(∇2 + k2)2 3

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 03.10

54 Helmholtz equation for isotropic medium 2

• Helmholtz determinant equation (fourth order)

3 2 2 2 2 µ detHe(∇)E(r) = detH(∇)E(r) = k (∇ + k ) E(r)

2 (2)T = µ He (∇) · [jωµJ(r) + ∇ × M(r)] = (∇2 + k2)(∇∇ + k2I) · [jωµJ(r) + ∇ × M(r)]

• reduces to a second-order equation (uniqueness of solution assumed!) 1 (∇2 + k2)E(r) = (I + ∇∇) · [jωµJ(r) + ∇ × M(r)] k2

• Simpler equation for vector potential E = −jω(I + ∇∇/k2) · A(r) j (∇2 + k2)A(r) = −µJ(r) + ∇ × M(r) ω

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 03.11

55 Helmholtz operators for bi-isotropic medium 1 √ √ • Denote ξ = (χr − jκr) µ, ζ = (χr + jκr) µ κr= relative chirality parameter, χr= relative Tellegen parameter • Dyadic Helmholtz operators factorizable in bi-isotropic medium:

2 2 2 µHe(∇) = Hm(∇) = H(∇) = −(∇ × I − kκrI) + k (1 − χr)I

= −L+(∇) · L−(∇) = −L−(∇) · L+(∇)

• Two auxiliary ﬁrst-order operators L±(∇)

L+(∇) = ∇ × I − k+I, L−(∇) = ∇ × I + k−I

p 2 p 2 k+ = ( 1 − χr + κr)k, k− = ( 1 − χr − κr)k

• Special case: isotropic medium, k± = k, L±(∇) = ∇ × I ∓ kI

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 03.12

56 Helmholtz operators for bi-isotropic medium 2

• Auxiliary operators can be evaluated as

(2) 2 2 2 L± (∇) = ∇∇ ∓ k±∇ × I + k±I, detL±(∇) = ∓k±(∇ + k±)

• Product can be expanded as a sum

2 2 2 2 (2) (2) (2) ∇ + k− (2) ∇ + k+ L+ (∇) · L− (∇) = k− L+ (∇) + k+ L− (∇) k+ + k− k+ + k−

• Using (A · B)(2) = A(2) · B(2) and det(A · B) = detA detB gives

2 2 2 2 (2) ∇ + k− (2) ∇ + k+ (2) [µHe(∇)] = k− L+ (∇) + k+ L− (∇) k+ + k− k+ + k−

2 2 2 2 det(µHe(∇)) = k+k−(∇ + k+)(∇ + k−)

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 03.13

57 Helmholtz equation for bi-isotropic medium

• The Helmholtz equations have the form

µHe(∇) · E(r) = jωµJ(r) + (∇ × I − jωξI) · M(r) = g(r)

(2)T det[µHe(∇)]E(r) = [µHe(∇)] · g(r) • The Helmholtz determinant equation can be expanded to

2 2 2 2 2 2 2 2 ∇ + k− (2)T ∇ + k+ (2)T (∇ +k+)(∇ +k−)E(r) = [ L+ (∇)+ L− (∇)]·g(r) k+(k+ + k−) k−(k+ + k−)

• Field can be solved in two parts as E = E+ + E− from

(2)T 2 2 1 (∇ + k±)E±(r) = L± (∇) · g(r) k±(k+ + k−)

• Two second-order equations for the bi-isotropic medium! General fourth-order equation (bi-anisotropic medium) does not reduce.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 03.14

58 Electromagnetic problems • Basic problem: find field from given source in homogeneous medium • Solution in integral form if field from point source is known

• Field from point source → dyadic Green function G(r − r0) Z E(r) = G(r − r0) · J(r0)dV 0 r ∈/ V

V • Problems involving inhomogeneous media and/or boundaries often handled through integral equations Z G(r − r0) · J(r0)dV 0 = E(r) = known

V • Unknown = (equivalent) source J(r) in certain region V • Problems of uniqueness due to non-radiating sources

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 03.15

59 Discontinuities in ﬁelds 1 • Heaviside unit step U(z) generates the delta function δ(z)

U(z) = 0, z < 0,U(z) = 1, z > 0, ⇒ ∂zU(z) = δ(z)

• Step discontinuity in function F (r) at surface S separating regions V1 and V2 generates the surface delta function δs(r)

F (r) = F1, r ∈ V1,F (r) = F2, r ∈ V2,

∇F (r) = n2(F2−F1)δs(r) = n1(F1−F2)δs(r) = (n2F2+n1F1)δs(r) R R • Integration of surface delta g(r)δs(r)dV = g(r)dS V S • If f(r) discontinuous on surface S:

∇f(r) = {∇f(r)}cont+(∇sf(r))δs(r), ∇sf(r) = n1f(r1)+n2f(r2)

• cont = no delta discontinuity, ∇s = surface nabla operator

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 03.16

60 Discontinuities in ﬁelds 2 • Surface gradient, divergence and curl:

∇sf = n1f(r1) + n2f(r2)

∇s · f = n1 · f(r1) + n2 · f(r2)

∇s × f = n1 × f(r1) + n2 × f(r2)

• δs-discontinuous sources create step-discontinuous ﬁelds:

∇×E = {∇×E}cont +(∇s ×E)δs(r) = −jωB−{M}cont −Msδs(r)

∇ × H = {∇ × H}cont + (∇s × H)δs(r) = jωD + {J}cont + Jsδs(r)

• Equating δs-discontinuous terms gives conditions on interfaces

∇s × E = n1 × E1 + n2 × E2 = −Ms

∇s × H = n1 × H1 + n2 × H2 = Js

∇s · D = n1 · D1 + n2 · D2 = %s

∇s · B = n1 · B1 + n2 · B2 = %ms

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 03.17

61 Boundary conditions 1 • Boundary: ﬁelds vanish on side 2 of surface S

∇s × E = n1 × E1 = −Ms, ∇s × H = n1 × H1 = Js

∇s · D = n1 · D1 = %s, ∇s · B = n1 · B1 = %ms

• Boundary relation between Js, Ms. E.g., linear relation:

Ms = −n1 × Zs · Js, or E1t = Zs · Js

• Impedance boundary condition between tangential ﬁeld compo-

nents, Zs= two-dimensional dyadic

× −1 −1 T Et = Zs·(n×Ht), ⇒ Ht = −(nn×Zs) ·(n×Et) = −(spmZs) Zs ·(n×Et) • Two-dimensional inverse! (spm = two-dimensional determinant)

T × −1 T × At · (At ×nn) = (spmAt)It, At = At ×nn/spmAt

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 03.18

62 Boundary conditions 2 • Local tangential ONB {v, w = n × v} on S

Zs = Zvvvv + Zvwvw + Zwvwv + Zwwww

• Isotropic impedance surface, perfect electric/magnetic conductor:

Zs = ZsIt = Zs(I − nn),Zs = 0, (PEC),Zs = ∞, (PMC)

• Self-dual impedance surface

−1 × −1 Zs = Z(αvv + α vv×nn) = Z(αvv + α ww)

• Limit α → 0 = soft-and-hard surface (SHS) = perfect anisotropic surface, gives symmetric boundary conditions v · E = 0, v · H = 0 • Generalized SHS boundary (conditions a · E = 0, b · H = 0)

−1 × Zs = Z(αba + α ab×nn), a · b = 1, α → 0

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 03.19

63 Interface conditions • Interface S between regions 1,2. Conditions for tangential ﬁelds ! E Z Z n × H 1t = 11 12 · 1 1 E2t Z21 Z22 n2 × H2

• For Z12 = Z21 interface can be represented by a T circuit. • Example: isotropic impedance sheet = thin layer of material on S

E1t = E2t = Et, Js = Et/Zs = n1 × H1 + n2 × H2

• Circuit parameters Zij = ZsIt ⇒ shunt element in the T-circuit. • Example: thin dielectric layer, thickness t → 0, → ∞ assume ﬁnite impedance 0 < |(r − 1)kot| < ∞

1 ηo Js = jω( − o)tEt,Zs = = −j jω( − o)t (r − 1)kot

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 03.20

64 Problems 3.1 Assuming uniaxial anisotropic medium with medium dyadics

= tIt + zuzuz µ = µtIt + µzuzuz, It = I − uzuz, show that the Helmholtz determinant operator can be factorized

as detHe(∇) = H1(∇)H2(∇) where Hi(∇) are two second-order scalar operators. Hint: you may need the expansion (prove it)

(2)× (2) ×µ = µtt(zµ + µz)

3.2 Assuming an anisotropic medium whose medium dyadics satisfy the relation µ = τT where τ is a scalar, show that the inverse of the Helmholtz dyadic operator can be expressed in the simple form

−1 He (∇) = L(∇)/L(∇), where the dyadic operator L(∇) and the scalar operator L(∇) are both of the second order. Hint: use result of Problem 2.3.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 03.21

65 S-96.510 Advanced Field Theory 4. Conditions for ﬁelds and media

I.V.Lindell

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 04.00

66 Uniqueness

• Linear diﬀerential equation and boundary conditions

L(∇)f(r) = g(r),B(∇)f(r) = s(r),

• If two solutions f1(r), f2(r)

L(∇)[f1(r) − f2(r)] = 0,B(∇)[f1(r) − f2(r)] = 0

• Unique f(r) if homogeneous (sourceless) problem has only the null solution

L(∇)fo(r) = 0,B(∇)fo(r) = 0, ⇒ fo(r) = 0

• Homogeneous problem similar to an eigenvalue problem • Uniqueness in integral equations more complicated because they involve sources as unknowns

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 04.01

67 Eigenvalue problem

• Example: sourceless Maxwell equations ! ! 0 ∇ × I E (r) ξ E (r) · o = jω · o −∇ × I 0 Ho(r) ζ µ Ho(r)

• Of the form of an eigenvalue equation, ω=eigenvalue parameter

L(∇) · fo(r) = jωM · fo(r),

• Impedance conditions on surface S:

Eot(r) = Zs · (n × Ho(r)), r ∈ S

• If ω = ωi in the eigenvalue spectrum {ωj}, solution fo(r) 6= 0 multiple of the eigenvector foi(r), otherwise fo(r) = 0

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 04.02

68 Example: Resonance cavity

• Closed PEC surface S with Zs = 0

• Real resonance frequencies ω1, ω2, ···, resonance modes E1(r), E2(r), ··· satisfy the homogeneous equation + boundary conditions

• Mathematics: source problem nonunique at resonance ω = ωi • Physics: without losses ﬁelds become inﬁnite at resonance

• Reality: losses make {ωi} complex ⇒ ω 6= ωi, uniqueness at real frequencies • Practice: if losses small, trouble in numerical computation (almost nonunique due to roundoﬀ errors) • For some problems uniqueness can be proved through uniqueness theorems of the form |α| = β,(β not positive real) ⇒ α = β = 0

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 04.03

69 Uniqueness in electrostatics 1

• f = φ(r) potential, g = −%(r)/ charge, assume isotropic volume V bounded by surface S

2 2 ∇ f = g, ∇ fo = 0, in V Z Z Z ∗ 2 ∗ 2 0 = fo (∇ fo)dV = ∇ · (fo ∇fo)dV − |∇fo| dV V V V Z I 0 2 ∗ Gauss law ⇒ |∇fo| dV = fo (n · ∇fo)dS V S • Different boundary conditions on S making the surface integral vanish will ensure uniqueness for the electrostatic field problem Z 2 |∇fo| dV = 0, ⇒ ∇fo = 0 ⇒ field vanishes V

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 04.04

70 Uniqueness in electrostatics 2

• Boundary conditions for uniqueness

fo = 0 on S, f = s, Dirichlet condition

n · ∇fo = 0 on S, n · ∇f = s, Neumann condition

• Mixed Dirichlet (on S1) and Neumann (on S −S1) gives uniqueness

• What about condition αfo + βn · ∇fo = 0 on S ? I I I Z 2 2 2 2 2 ∗ 2 |αfo+βn·∇fo| dS = |α| |fo| dS+|β| |n·∇fo| dS+2<{α β} |∇fo| dV = 0

S S S V ∗ • If <{α β} > 0, ⇒ ∇fo = 0 in V • ⇒ Impedance boundary condition of the form αf + βn · ∇f = s on S gives uniqueness for <{α∗β} > 0

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 04.05

71 Uniqueness in electrodynamics (Example)

• Uniqueness for lossy isotropic medium with lossless boundary at real frequencies when µ complex (complex resonance frequencies) I Z Z Z ∗ ∗ 2 ∗ 2 n·Eo×Ho dS = ∇·(Eo×Ho )dV = −jωµ |Ho| dV +jω |Eo| dV S V V V H ∗ • Assume S ideal boundary ⇒ n · Eo × Ho dS = 0 S Z Z 2 2 2 ={µ}= 6 0, µ |Ho| dV = || |Eo| dV ⇒ Eo = 0, Ho = 0 V V • Uniqueness for the resonator with lossless medium and boundaries only when frequency not one of eigenfrequencies, ω 6= ωi.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 04.06

72 Power conditions for medium parameters

• Complex Poynting vector for time-harmonic ﬁelds: 1 S = E × H∗ 2

• <{S} = average power flow in the field [Watts/m2] • P = ∇ · <{S} = average power created by medium [Watts/m3] • Active medium P > 0: Medium gives energy to field • Passive medium P < 0: Field gives energy to medium • Lossless medium P = 0: No exchange of energy • Nature of power exchange depends on medium parameters

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 04.07

73 Power exchange

• Power created in the medium 1 P = ∇ · <{S} = <{(∇ × E) · H∗ − (∇ × H)∗ · E} 2 1 ω = <{−jωB · H∗ + jωD∗ · E} = ={E∗ · D + H∗ · B} 2 2 • In six-vector notation ! ω E ξ P = ={e∗ · M · e}, e = , M = 2 H ζ µ

• Lossless medium: P = 0 for all e • Lossy (passive) medium: P < 0 for all e

• Active medium if P > 0 for some ﬁeld e1

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 04.08

74 Lossless bi-anisotropic medium

• Condition for lossless medium: P = 0 for all possible ﬁelds E, H

2j={e∗·M·e} = e∗·M·e−e·M∗·e∗ = e·(MT −M∗)·e∗ = 0 for all e

• To prove: e · A · e∗ = 0 for all e implies A = 0 • Take e = a + b ⇒ a · A · b∗ + b · A · a∗ = 0 • Take e = a + jb ⇒ -ja · A · b∗ + jb · A · a∗ = 0 • Follows a · A · b∗ = 0 for all a, b ⇒ A = 0 • Condition for lossless medium: M is a Hermitian six-dyadic

MT − M∗ = 0 ⇒ M∗T = M

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 04.09

75 Parameters of lossless medium • Medium six-dyadic Hermitian for lossless media

!∗T ! ! ξ ∗T ζ∗T ξ = = ζ µ ξ∗T µ∗T ζ µ

∗T = , µ∗T = µ, ξ∗T = ζ,

• and µ Hermitian dyadics and ξ, ζ a Hermitian pair of dyadics √ √ • Deﬁne ξ = (χ − jκ) µoo, ζ = (χ + jκ) µoo,

⇒ χ∗T = χ, κ∗T = κ

• Lossless bi-anisotropic medium: , µ, χ, κ are Hermitian dyadics

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 04.10

76 Examples of lossless media • Bi-isotropic medium: , µ, χ, κ are all real for a lossless medium ∗ = , µ∗ = µ, χ∗ = χ, κ∗ = κ

• Gyrotropic medium: , µ, χ, κ are of the general gyrotropic form

G = Gzuzuz + GtIt + Gguz × I

• Lossless medium: gyrotropic dyadics are Hermitian G∗T = G ∗ ∗ ∗ Gz = Gz,Gt = Gt,Gg = −Gg

• Hermitian gyrotropic dyadic in a form where Gz,Gt,Gg are real

G = Gzuzuz + GtIt + jGguz × I

• General lossless medium: parameter dyadics D = S + a × I with symmetric part S real and antisymmetric part a × I imaginary

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 04.11

77 Lossy medium

• Condition for lossy media: <{∇ · S} < 0 for any ﬁelds ⇒ 2={e∗ · M · e} = e · [(−jM)T + (−jM)∗] · e∗ < 0 for all e

• Hermitian dyadic H negative definite if e · H · e∗ < 0 for all e • ⇒ H = (−jM)T + (−jM)∗ must be negative definite ∗ • Take H = 0, ⇒ j( − T ) must be negative definite ∗ • Take E = 0, ⇒ j(µ − µT ) must be negative definite

• Relation for ξ, ζ complicated [see MOTL 29(3)175-178 May 2001]

• Isotropic medium: im < 0, µim < 0

2 • Isotropic chiral medium: |κim| < |im| |µim|/µoo (obtained after some algebra). Note: if im = 0 or µim = 0 ⇒ κim = 0

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 04.12

78 Surface impedance conditions

• Impedance condition Et = Zs · (n × H), or n × H = Y s · Et • Power ﬂow from impedance surface to the ﬁeld (direction n) 1 1 P = <{n · E × H∗} = − <{E∗ · n × H} 2 2 1 1 = − <{E∗ · Y · E} = − E∗ · Y · E 2 s 2 sH

1 ∗T • Y sH = 2 (Y s + Y s ), Hermitian part of Y s

• Lossless surface: P = 0 ⇒ Y sH = 0, Y s antihermitian dyadic

• Example: Y s symmetric, Y s = jBs imaginary (reactive surface)

• Lossy medium: P < 0 ⇒ Hermitian part of Y s positive deﬁnite

• E.g.: Y s symmetric, ⇒ Y s = Gs + jBs with Gs pos. deﬁnite

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 04.13

79 Ideal boundary

• Ideal boundary: Poynting vector has no normal component 1 n · S = n · E × H∗ = 0 2 • Ideal boundary is lossless. Examples: PEC, PMC and SHS ∗ • Relation between tangential ﬁelds: Et and Ht parallel • Anisotropic ideal surface: exists (complex) tangential vector a such that a · E = 0, a∗ · H = 0

• Generalized SHS, for a = a∗ = v gives real classical SHS • Isotropic ideal surface: exists a (complex) scalar Z such that ∗ Et = ZHt , not a linear condition!

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 04.14

80 Energy condition • Lossless, nondispersive medium assumed (otherwise complicated) • Condition: energy stored in the medium W positive for all ﬁelds ! ∗ 1 ξ E 1 W = (EH) · · = e · M · e∗ > 0 4 ζ µ H 4

• M Hermitian, condition requires: M pos.deﬁnite ⇒ M−1 PD ! ( − ξ · µ−1 · ζ)−1 −−1 · ξ · (µ − ζ · −1 · ξ)−1 M−1 = −µ−1 · ζ · ( − ξ · µ−1 · ζ)−1 (µ − ζ · −1 · ξ)−1

• ⇒ dyadics , µ, − ξ · µ−1 · ζ, µ − ζ · −1 · ξ must all be PD • Example: bi-isotropic medium (lossless ⇒ , µ, χ, κ real) 2 2 > 0, µ > 0, ξζ < µ ⇒ χ + κ < µ/µoo • Condition limits magnitudes of χ, κ parameters

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 04.15

81 Reciprocity conditions 1 • Reaction of source and ﬁeld deﬁned as [Rumsey 1954] Z J2 < 1, 2 >= (E1 H1) · dV −M2

• Medium reciprocal when reaction is symmetric < 1, 2 >=< 2, 1 > Z Z J2 J1 0 = (E1 H1) · dV − (E2 H2) · dV −M2 −M1

• Replace sources from the Maxwell equations, apply Gauss’ law

I Z T T − ξ + ζ E2 n·(E1×H2−E2×H1)dS = jω (E1 H1)· · dV −ζ − ξT −µ + µT H2 S V • Must be valid for any ﬁelds in any volume V bounded by S. Inte- grals must vanish separately ⇒ conditions for medium and boundary parameters.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 04.16

82 Reciprocity conditions 2

• Volume integral vanishes for any ﬁelds under reciprocity conditions for the medium:

= T , ξ = −ζT , ζ = −ξT , µ = µT

• For example, bi-isotropic medium reciprocal if ξ = −ζ, ⇒ χ = 0. Tellegen parameter χ = nonreciprocity parameter • When is medium with gyrotropic dyadics reciprocal? , µ symmet- T ric ⇒ g = µg = 0, ξ = −ζ ⇒ ξu = −ζu, ξt = −ζt, ξg = ζg • Surface integral vanishes under reciprocity condition for symmetric surface impedance dyadic

T Zs = Zs

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 04.17

83 Reciprocal and lossless media

• Conditions for a medium being both lossless and reciprocal: ! ! ! T ∗ ζT ∗ ξ T −ζT = = ξT ∗ µT ∗ ζ µ −ξT µT

∗ • The permittivity and permeability dyadics satisfy = T = , ∗ µ = µT = µ or and µ are real and symmetric dyadics ∗ • The magnetoelectric dyadics satisfy ξ = −ξ = −ζ and ∗ ζ = −ζ = −ξ, whence ξ and ζ are imaginary dyadics satisfying ξT = −ζ.

• Writing ζ = χ + jκ, ξ = χ − jκ, κ must be real and symmetric and χ imaginary and antisymmetric.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 04.18

84 Problems 4.1 Derive the condition for the imaginary parts of the medium parameters of a lossy bi-isotropic chiral medium:

2 µimim κim < µoo by requiring that the condition <{∇·S} < 0 be valid for all possible ﬁelds in the chiral medium. 4.2 Consider a plane wave in a lossless non-reciprocal bi-isotropic medium √ with real nonzero magnetoelectric parameters ξ = ζ 6= ± µ,

−jkz −jkz E(r) = Eoe , H(z) = Hoe .

Show that this is a TEM wave and solve the factor k. Also show that linearly polarized electric and magnetic ﬁeld vectors are not orthogonal for ξ 6= 0.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 04.19

85 S-96.510 Advanced Field Theory 05. Field Transformations Duality transformation

I.V.Lindell

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 05.00

86 Field transformations • Electromagnetic transformations make it possible to ﬁnd solutions to new problems in terms of old problems with known solutions. The solution process (often tedious) can be avoided:

Problem → Solution process → Result

⇓

Problem → Transf. → Known solution → Inverse transf. → Result

• Duality transformation: linear transformation of fields induces transformation of sources and media • Affine transformation: linear transformation of space induces transformation of fields, sources and media

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 05.01

87 Simple duality (Duality substitution)

• Symmetry in Maxwell equations (e.g., isotropic media):

M1 : ∇ × H = jωE + M

M2 : − ∇ × E = jωµH + J

• Replace

E → H, H → E, → −µ, µ → −, M → −J, J → −M

• Maxwell equations invariant: M1→M2, M2→M1 • Not a transformation, dimensions are changed • Can be used for transforming formulas

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 05.02

88 Example: transformation of formulas • Electric ﬁeld radiated by electric current Z E(r) = −jωµ G(r − r0) · J(r0)dV 0

• Duality substitution: E → H, H → E, → −µ, µ → −, M → −J, J → −M √ • Green dyadic G(r − r0) depends on k = ω µ, which is invariant • Magnetic ﬁeld from magnetic current Z H(r) = −jω G(r − r0) · M(r0)dV 0

• Electric ﬁeld obtained through ∇× ∇ × H(r) Z E(r) = = −∇ × G(r − r0) · M(r0)dV 0 jω

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 05.03

89 Classical duality

• Deﬁne duality transformation E → Ed etc (B scalar constant):

Ed = BH, Bd = −BD, Md = −BJ

• Transforms one Maxwell equation to another:

−∇ × Ed = jωBd + Md ⇔ ∇ × H = jωD + J

• Similarly

Hd = CE, Dd = −CB, Jd = −CM

∇ × Hd = jωDd + Jd ⇔ −∇ × E = jωB + M

• Works for arbitrary scalars B,C 6= 0 (with correct dimensions)

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 05.04

90 Medium transformation • Transformation for medium dyadics ! ! D Cµ · H + Cζ · E µC/B ζ E d = − = − d Bd B · E + Bξ · H ξ B/C Hd • Transformed medium dyadics ! ! ξ −µC/B −ζ d d = ζd µd −ξ −B/C

• Dual ﬁelds and sources exist in dual medium! • Example: dual of isotropic medium = another isotropic medium: C B = − µ, µ = − d B d C

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 05.05

91 Choice of parameters • Duality transformation works for any two parameters B,C 6= 0: E 0 B E d = , Hd C 0 H

• Usually B,C chosen to satisfy two conditions: • (1) Duality transformation in an involution (equals its inverse) E 0 B E E d = = BC ⇒ BC = 1, H C 0 H H d d d

• (2) Free-space medium = oI, µ = µoI and ξ = ζ = 0 is self dual:

d = = −µC/B and µd = µ = −B/C 2 o = −µoC/B, ⇒ B/C = −µo/o = −ηo p • Two possible solutions: B± = 1/C± = ∓jηo = ∓j µo/o

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 05.06

92 Two duality transformations

• Right-hand transformation (+) and left-hand transformation (-) [explained later]

E 0 −η E d± = ±j o Hd± 1/ηo 0 H

• Medium transformation rules same for both cases: d± = d etc ! ! ! ξ µ/η2 −ζ (µ/µ ) −ζ d d = o = o o 2 ζd µd −ξ ηo −ξ µo(/o)

• Relative dual permittivity and permeability: rd = µr, µrd = r √ √ • Isotropic medium: wave number: kd = ω µdd = ω µoroµr = k

p p 2 • Wave impedance ηd = µd/d = µor/oµr = ηo/η

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 05.07

93 Example

• Incident plane-wave Ei = uAe−jk·r, Hi = vBe−jk·r scattering s s from a dielectric object (r = α, µr = 1) ⇒ scattered ﬁelds E , H

• Duality transformation (e.g., right-hand d+) ⇒ dual plane-wave

Ei = −jη Hi = −jη vBe−jk·r, d+ o o

Hi = −Ei/jη = −u(A/jη )e−jk·r d+ o o

• Scattering from a dual object (µrd = α, rd = 1) • Scattered fields Es = −jη Hs and Hs = −Es/jη are dual to d+ o d+ o the original scattered fields • If incident field self dual ⇒ same incident field in dual problem

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 05.08

94 Self-dual quantities • Self-dual ﬁelds invariant in right/left-hand duality transformation • Self-dual ﬁelds in right-hand transformation

E+ = (E+)d+ = −jηoH+, H+ = (H+)d+ = −E+/jηo

• Self-dual ﬁelds in left-hand transformation

E− = (E−)d− = jηoH−, H− = (H−)d− = E−/jηo

• Self-dual decomposition of any ﬁeld: 1 E = E + E , E = (E ∓ jη H), + − ± 2 o 1 1 H = H+ + H−, H± = (H ∓ E) 2 jηo

• Decomposition of sources: J± = (J ± M/jηo)/2

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 05.09

95 Self-dual plane wave

−jku·r 1 −jku·r • Plane wave in air E = Eoe , H = u × Eoe (real u) ηo • Self dual in right-hand transformation:

E = Ed+ ⇒ Eo = −jηoHo = −ju × Eo

• Satisﬁes u · Eo = 0 and Eo · Eo = 0 (circular polarization) E × E∗ −j(u × E ) × E∗ p(E ) = o o = o o = u o ∗ 2 jEo · Eo j|Eo|

• Right-hand circularly polarized plane wave is self dual in right-hand transformation and anti-self-dual in left-hand transformation.

• Decomposition E = E+ + E−,

• Transformations Ed+ = E+ − E−, Ed− = −E+ + E−

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 05.10

96 Surface impedance condition

• Surface impedance condition (Zs = 2D dyadic)

Et = Zs · (n × H)

• Applying 2D inverse of a 2D dyadic A

T × −1 A ×nn −1 −1 A = , A · A = A · A = It spmA

T × T −1 Zs ×nn Zs n × H = Zs · Et = · Et = −n × · (n × E) spmZs spmZs • ⇒ Another form of the same impedance condition:

T Zs Ht = − · (n × E) spmZs

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 05.11

97 Dual surface impedance

• Duality transformations of Et = Zs · (n × H):

Ed±t = Zsd± · (n × Hd± )

1 ZT ⇒ ∓jη H = ∓ Z · (n × E) = ±jη s · (n × E) o t jη sd± o o spmZs

• Identifying the dual of the surface impedance dyadic Zsd± = Zsd (same for both transformations)

T 2 Zs Zsd = ηo spmZs

× • Another form follows from spm(Zs×nn) = spmZs

2 × −1 Zsd = ηo(Zs×nn)

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 05.12

98 Self-dual surface impedance • Self-dual condition: 2 ! 4 ηo T ηo Zs = Zsd = Zs ⇒ spmZs = spmZs spmZs

2 T • Two conditions: spmZs = ±ηo ⇒ Zs = ±Zs

• Antisymmetric (nonreciprocal) solution Zs = ±jηon × I • Symmetric (reciprocal) surface impedance in terms of ONB (v, w, n)

2 Zs = Zvvv + Zwww, ⇒ spmZs = ZvZw = ηo • Self-dual reciprocal impedance surface has the form

−1 Zs = ηo(αvv + α ww)

• Limiting cases α = 0, ∞: soft and hard (tuned corrugated) surface

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 05.13

99 Self-dual media • Free space = self dual medium. Other possibilities? ! ! ! ξ µ/η2 −ζ ξ d d = o = 2 ζd µd −ξ ηo ζ µ

√ √ • Denote ξ = (χ − jκ) µoo, ζ = (χ + jκ) µoo

r = µr = α, ξ = −ζ ⇒ χ = 0,

• Self-dual medium can be deﬁned in terms of two arbitrary dyadics: ! √ ξ α −jκ µ = √ o o o ζ µ jκ µoo αµo

• More general duality transformation ⇒ more general self-dual medium

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 05.14

100 General duality transformation

• General duality transformation T for the ﬁelds (six-vector form) E AB e = T e, e = ,T = , AD − BC 6= 0, d H CD

• Transformed Maxwell equations can be expanded as 0 1 J∇ × e = jωM · e + j ,J = d d d d −1 0

−1 JT ∇ × e = [JTJ ]J∇ × e = jωMdT · e + jd −1 −1 −1 −1 J∇ × e = jω[JT J ]MdT · e + [JT J ]jd

−1 −1 • Identify dual sources from j = [JT J ]jd: J D −C J d = JTJ −1j = Md −BA M

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 05.15

101 Duality transformation of media

• Identify dual medium six-dyadic:

−1 −1 −1 −1 M = [JT J MdT ], ⇒ Md = JTJ MT

   2 2    d D −CD −CDC 1 −BD AD BC −AC  ξ   ξd        =      ζd  AD − BC  −BD BC AD −AC  ζ 2 2   µd B −AB −AB A µ

• Dual medium dyadics linear combinations of original medium dyadics • Self-dual medium with respect to a transformation (A, B, C, D):

d = , ξd = ξ, ζd = ζ, µd = µ

• Four conditions reduce to only two

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 05.16

102 Self-dual media • Self-dual medium conditions for general A, B, C, D reduce to

(A − D) + C(ξ + ζ) = 0, (D − A)µ + B(ξ + ζ) = 0

√ p • Deﬁne ξ, ζ = (χr ∓ jκr) µ, η = µ/ • ⇒ , µ and χ must be multiples of the same dyadic, say, α (Previously A = D = 0, whence χ = 0) √ • ξ − ζ = −2jκr µ is always invariant: κrd = κr • General self-dual medium: medium dyadics have the form

! √ √ ξ χr µ 0 −j µ = √ α + √ κr ζ µ χr µ µ j µ 0

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 05.17

103 Special case: involutory duality √ • Require: T −1 = T, ⇒ T 2 = I ⇒ A = −D = ± 1 − BC

• Assuming self-dual medium with parameters , µ, χr = sin θ,(α, κr do not matter), transformation matrix T can be solved:

p 2 D = −A = ±jχr/ 1 − χr = ±j tan θ

B/η = −ηC = A/χr = ∓j/ cos θ − tan θ −η/ cos θ T = ±j , (η = η , θ = 0 ⇒ classical duality) ± 1/η cos θ tan θ o

• Self-dual ﬁelds: E = E+ + E−, H = H+ + H− E± E± T± = H± H±

∓jθ • Relation E± = Z±H±, wave impedances Z± = ∓jηe

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 05.18

104 Self-dual (Bohren) decomposition 1

• Field decomposition in self-dual parts E = E++E−, H = H++H− E 1 1 E+ E+ = jθ −jθ = K , H −e /jη e /jη E− E−

−jθ E+ 1 e −jη E −1 E = jθ = K E− 2 cos θ e jη H H • Maxwell equations are decoupled for homogeneous self-dual media: E E ∇ × + = K−1∇ × E− H ! −jωζ −jωµ E −M = K−1 K + + K−1 jω jωξ E− J kτ + 0 E+ M+ = − , τ ± = κr ± cos θα 0 kτ − E− M−

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 05.19

105 Self-dual (Bohren) decomposition 2

• Maxwell equations for self-dual ﬁelds

∇ × E± = ±kτ ± · E± − M± 1 M = (e∓jθM ± jηJ) = ±jηe∓jθJ ± 2 cos θ ± • They can also be expressed as four equations (twice the same)

−∇ × E± = jωµ± · H± + M±, ∇ × H± = jω± · E± + J±

• Equivalent medium dyadics for the decomposed ﬁelds

k ±jθ k ∓jθ ± = ± τ ± = e τ ±, µ± = ∓ τ ±Z± = µe τ ± jωZ± jω

• Problem in a self-dual bi-anisotropic medium splits in two parts each associated with an eﬀective anisotropic medium.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 05.20

106 Power propagation in self-dual medium

• Assume homogeneous and lossless self-dual medium: , µ, α, θ, κr real • Propagating power = real part of Poynting vector

1 ∗ 1 E+ E− ∗ <{E × H } = <{(E+ + E−) × + } = 2 2 Z+ Z− 1 <{−jE ×E∗ e−jθ+jE ×E∗ ejθ+j[E ×E∗ ejθ+E∗ ×E e−jθ]} 2η + + − − + − + − |E |2 |E |2 = cos θ[p(E ) + − p(E ) − ],Z = ∓jηe∓jθ + 2η − 2η ± • Decomposed ﬁelds do not couple power to one another!

• p(E+) right-hand polarization, −p(E−) left-hand polarization

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 05.21

107 Problems 5.1 Find conditions for the bi-anisotropic medium which can be trans-

formed to an anisotropic medium, i.e., satisfying ξd = ζd = 0. Deﬁne the parameters A, B, C, D required for the transformation.

−jkoz 5.2 Study the reﬂection of a plane wave Eoe propagating in air from a self-dual surface impedance Zs = ηo(uxuxα + uyuy/α) at z = 0. Find the two-dimensional reﬂection dyadic R giving the

jkoz reflected field as R · Eoe . Show that the handedness of the wave is not changed in the reflection.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 05.22

108 S-96.510 Advanced Field Theory 6. Aﬃne transformation

I.V.Lindell

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109 Affine transformation • Affine transformation = linear deformation of space (stretching, compressing, rotating, mirror-imaging etc.) • Affine transformation makes microscopic distortion to medium. For example: isotropic medium may become anisotropic

• Aﬃne transformation deﬁned by complete dyadic A (detA 6= 0)

space r → ra = A · r aﬃne space

• Property of nabla operator

∇r = ∇xux + ∇yuy + ∇zuz = uxux + uyuy + uzuz = I

• Nabla operator in aﬃne space

−1T T −1T I = ∇ara = ∇a(A · r) = A · ∇r · A , ⇒ ∇a = A · ∇

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 06.01

110 Basic tools • Applying six-vector representation ! E D J ξ 0 1 e = , d = , j = , M = ,J = H B M ζ µ −1 0 • Original Maxwell equations J∇ × e(r) = jωd(r) + j(r), d = M · e

• are aﬃne-transformed to

J∇a × ea(ra) = jωda(ra) + ja(ra), da = Ma · ea

• Analytic tools needed: identity

(2) −1T (D · a) × (D · b) = D · (a × b) = (detD)D · (a × b)

−1 (−2) T • Dyadic relation [D ](2) = D = D /detD

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 06.02

111 Aﬃne transformation of ﬁelds and sources • Transforming Maxwell equations

(2) (2) (D · ∇) × (D · e(r)) = D · (∇ × e(r)) = D · [jωd(r) + j(r)]

−1T −1T • Because ∇a = A · ∇, choose D = A which gives

J∇a × ea(ra) = jωda(ra) + ja(ra)

• Identify term by term ⇒ transformed ﬁelds and sources:

−1T −1 ea(ra) = D · e(r) = A · e(A · ra)

(2) (−2)T −1 −1 da(ra) = D · d(r) = A · d(A · ra) = A · d(A · ra)/detA (2) (−2)T −1 −1 ja(ra) = D · j(r) = A · j(A · ra) = A · j(A · ra)/detA

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 06.03

112 Aﬃne transformation of media • Medium equations

A A A T Ma · ea = da = · d = · M · e = · M · A · ea detA detA detA • Identify the medium six-dyadic: T A · M · A Ma = detA • Transformation rules for all medium dyadics are similar: T T A · · A A · ξ · A a = , ξa = , detA detA T T A · ζ · A A · µ · A ζa = , µa = , detA detA

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 06.04

113 Rotation transformation • Rotation transformation around an axis deﬁned by u

θu×I A = R(θ) = uu + cos θIt + sin θu × I = e , R(0) = I T −1 (2) R (θ) = R (θ) = R(−θ), R (θ) = R(θ), detR(θ) = 1 • Field vectors and source vectors are transformed as

Ea(ra) = R(θ) · E(R(−θ) · r), Ja(ra) = R(θ) · J(R(−θ) · r)

• All medium parameter dyadics are transformed as T X a = A · · A /detA = R(θ) · · R(−θ) = (R(θ) · ai)(R(θ) · bi)

• R(θ) · I · R(−θ) = I ⇒ bi-isotropic media are invariant in rotation • Rotation transformation is not very interesting! General transformation A = S · R. Interest in symmetric transformations A = S.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 06.05

114 Uniaxial transformation

• Uniaxial transformation: stretching in uz and ⊥uz directions

−1 −1 −1 2 S = αIt + βuzuz, S = α It + β uzuz, detS = α β

• Field vectors are transformed as

−1 −1 −1 −1 −1 −1 Ea(ra) = S · E(S · ra) = [α Et + β uzEz](α ρ + β uzz)

• Source vectors are transformed similarly • All medium parameter dyadics are transformed as

−1 −1 −2 a = β It ··It +α [It ··uzuz +uzuz ··It]+α βuzuz( : uzuz)

• (Bi-)isotropic medium is transformed to uniaxially (bi-)anisotropic medium

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 06.06

115 Generalization: triaxial transformation • Triaxial transformation: stretching in three orthogonal directions −1 X X −1 S = αiuiui, S = αi uiui, detS = α1α2α3

• Field and source vectors are transformed as X −1 X −1 Ea(ra) = αi uiui · E( αi uiui · r)

• All medium parameter dyadics are transformed as XX αiαj a = uiuj(uiuj : ) α1α2α3 • Bi-isotropic medium is transformed to a bi-anisotropic medium. 2 All medium dyadics multiples of the same dyadic S . • Conversely: if medium dyadics are multiples of the same symmetric dyadic, bi-anisotropic medium can be transformed to a bi-isotropic medium. I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 06.07

116 Use of affine transformation • Affine transformation changes sources, fields, media and boundaries both geometrically and electrically • Point source remains point source ⇒ field solutions (Green dyadics) can be obtained for transformed media • Boundaries are changed: Real symmetric A: isotropic sphere ⇒ anisotropic ellipsoid −1 • Surface equation f(r) = 0 becomes f(A · ra) = fa(ra) = 0 −1T • Normal n ∼ ∇f(r) and E, H transform similarly: na = A · n

−1T −1T −1T T T Zsa·(na×Ha) = Eat = A ·Et = A ·Zs·(n×H) = A ·Zs·[(A ·na)×(A ·H)] • Surface impedance dyadic transforms as

−1T (2)T −1T −1 Zsa = A · Zs · A = A · Zs · A detA

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 06.08

117 Medium transformations 1 • Isotropic medium → anisotropic medium (aﬃne-isotropic medium)

T 0 0 0 A · A I → a = 0 µ 0 µa 0 µ detA

• Dyadics a and µa symmetric and related by condition µa = µa

• Conversely: anisotropic medium → isotropic medium aI, µaI possible only if = S, µ = µS, S symmetric. Transformation dyadic 1/2 p p A = αS with µa = µa and α = a/( detS) = µa/(µ detS) • Medium is aﬃne-bi-isotropic if medium six-dyadic is the form

! ξ ξ M = = S, ζ µ ζ µ

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 06.09

118 Medium transformations 2 • What kind of medium transformations are possible? • Symmetric dyadics transform to symmetric dyadics, antisymmetric to antisymmetric • Magnetic anisotropy to electric anisotropy when µ symmetric ⇒ a symmetric

0 I 0 → a 0 µ 0 µaI

• Ferrite ↔ magnetoplasma not possible!

• Non-bi-anisotropic (ξ = ζ = 0) ↔ bi-anisotropic not possible! • Combination with duality transformation gives more possibilities

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 06.10

119 Simple use of aﬃne transformation

• Scalar Helmholtz equation with symmetric S

(S : ∇∇ + k2)f(r) = g(r)

−1/2 −1/2 • Deﬁne ∇ = S · ∇a, ⇒ ra = S · r equation simpliﬁed to

1/2 1/2 2 2 (∇a +k )fa(ra) = g(ra), fa(ra) = f(S ·ra), ga(ra) = g(S ·ra)

• If solution function fa(ra) can be found

−1/2 ⇒ f(r) = fa(S · r)

• Square root of symmetric√ dyadic not unique: e.g., same eigenvectors, eigenvalues = ± λi. Boundary conditions (radiation conditions) require uniqueness.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 06.11

120 Duality-aﬃne transformation

• Aﬃne transformation can be connected to duality transformation

AB −1T −1 e(r) → e (r ) = A · e(A · r ) da a CD a

−1T −1 −1T −1 Eda(ra) = AA · E(A · ra) + BA · H(A · ra) −1T −1 −1T −1 Hda(ra) = CA · E(A · ra) + DA · H(A · ra)

• The medium dyadics become

  2 2   da  D −CD −CDC  −BD AD BC −AC ξ T  ξda  1   A     = AD−BC ·   · A  ζda   −BD BC AD −AC  detA  ζ  B2 −AB −AB A2 µda µ

• Medium dyadics da ··· linear combinations of a, ···

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 06.12

121 Reﬂection transformation • Reﬂection in plane normal to n

C = I − 2nn, C · a = It · a − n(n · a)

• Properties

2 T −1 C = I, C = C = C, detC = −1

• Reﬂection-transformed medium dyadics 1 T Ma = C · M · C = −C · M · C detC • Media invariant in reﬂection are strange. For example isotropic medium is not invariant: Ma = −M • Can be combined with duality transformation!

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 06.13

122 Duality-reﬂection transformations • Special duality transformation with B = C = 0 can be combined with reﬂection:       da (D/A) −(D/A) C  ξ  T  −ξ   ξda        = ·   · C = C ·   · C  ζda  detC  ζ   −ζ  µda (A/D)µ −(A/D)µ

• Isotropic medium invariant for ⇒ D/A = −1. Involution required ⇒ two possibilities D = −A = ±1

• Deﬁne two transformations, rc = C · r ±1 0 ±1 0 e (r ) = C·e(C·r ), j (r ) = C·j(C·r ) c c 0 ∓1 c c c 0 ∓1 c

• = Electric and magnetic reﬂection transformations.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 06.14

123 Electric and magnetic reﬂection

• Two reflection transformations with rc denoted by r: • Electric reflection ! ! E (r) C · E(C · r) J (r) C · J(C · r) c = , c = Hc(r) −C · H(C · r) Mc(r) −C · M(C · r) • Magnetic reflection ! ! E (r) −C · E(C · r) J (r) −C · J(C · r) c = , c = Hc(r) C · H(C · r) Mc(r) C · M(C · r)

• Tangential ﬁelds at n · r = 0 satisfying C · r = C · ρ = ρ:

Ect(ρ) = Et(ρ), Hct(ρ) = −Ht(ρ), ER

Ect(ρ) = −Et(ρ), Hct(ρ) = Ht(ρ), MR • Normal components with opposite signs

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 06.15

124 Summed ﬁelds in isotropic media

• Signs for electric/magnetic reﬂection transformation ! E(r) + E (r) E(r) ± C · E(C · r) c = H(r) + Hc(r) H(r) ∓ C · H(C · r) ! J(r) + J (r) J(r) ± C · J(C · r) c = M(r) + Mc(r) M(r) ∓ C · M(C · r)

• Conditions on plane n · r = 0 respectively

PMC E(ρ)+E (ρ) = (I ±C)·E(ρ) = 2 It ·E(ρ), condition c nn PEC

nn PMC H(ρ)+Hc(ρ) = (I∓C)·H(ρ) = 2 ·H(ρ), condition It PEC

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 06.16

125 Mirror image principles

• PEC conditions satisfied for EM problem + magnetic reflection • PEC mirror image of a source is its magnetic reflection ! J(r) −C · J(C · r) ⇒ M(r) C · M(C · r)

• PMC conditions satisfied for EM problem + electric reflection • PMC mirror image of a source is its electric reflection ! J(r) C · J(C · r) ⇒ M(r) −C · M(C · r)

• Image principles transform boundary problems to source problems • Can be extended to problems with two parallel planes

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 06.17

126 Medium condition • PEC/PMC image principle valid if medium invariant in reﬂection transformation (electric or magnetic) ! −ξ c ξc = C · · C = ξ ζc µc −ζ µ ζ µ

• Conditions for medium dyadics

C · · C = , C · ξ · C = −ξ, C · ζ · C = −ζ, C · µ · C = µ

• Denote transverse dyadics t · u = u · t = 0 and vectors at · u = 0. Medium dyadics must be of the form

= t + uuu, µ = µt + µuuu, ξ = atu + ubt, ζ = ctu + udt

• For example, bi-isotropic medium invariant only if ξ = ζ = 0

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 06.18

127 Object above PEC/PMC plane

• An object in air can be replaced by equivalent polarization source o 0 jp(r) = jω[M(r) − Mo] · e(r), Mo = I 0 µo

• PEC/PMC image of the polarization source obtained through the magnetic/electric reﬂection transformation ! ∓C 0 jpc(r) = · jp(C · r) 0 ±C ! ∓C 0 = jω · [M(C · r) − Mo] · e(C · r) = 0 ±C " ! ! # ! ∓C 0 ∓C 0 ∓C 0 jω · M(C · r) · − Mo · ·e(C·r) 0 ±C 0 ±C 0 ±C

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 06.19

128 Image of object

• Identifying the image of the obstacle: ! ! ∓C 0 ∓C 0 Mc(r) = · M(C · r) · 0 ±C 0 ±C ! (C · r) −ξ(C · r) = C · · C −ζ(C · r) µ(C · r)

• For example: image of an isotropic half sphere of parameters , µ on PEC/PMC plane u · r = 0 is the other half sphere on the plane. Problem of complete sphere and original + image sources • Image of a chiral half sphere of parameters , µ, κ is the other half sphere of parameters , µ, −κ with changed handedness. The problem does not reduce to that of a complete homogeneous sphere.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 06.20

129 Problems 11 Media defined by medium dyadics of the form ! √ √ ξ χ µ 0 −jκ µ = √ r α + √ r β ζ µ χr µ µ jκr µ 0 are self dual in some duality transformation. Show that they are also self dual after any affine transformation. Under what condition it is possible to transform α to the unit dyadic I? How to define the transformation dyadic A? 12 Find the expression for the dipole moment p of a small dielectric sphere (radius a, permittivity ) at r = uzh in air above a PEC plane at z = 0, when the incident field is the plane wave Ei(r) = −jk·r Eoe . The scatterer can be replaced by the equivalent current 3 dipole J = pδ(r − uzh), p = αE, α = jω( − o)(4πa /3), where 0 E = 3oE /( + 2o) is the field inside the scatterer when put in the field E0. h can be assumed large in wavelengths.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 06.21

130 S-96.510 Advanced Field Theory 7. Electromagnetic Field Solutions

I.V.Lindell

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 07.00

131 The Green function • Green function = ﬁeld from a point source of unit amplitude

L(∇)G(r) = −δ(r)

• Field from distributed source = integral of Green function Z L(∇)f(r) = g(r), f(r) = − G(r − r0)g(r0)dV 0

• Green function = mapping source → field • Scalar source, scalar field, → scalar Green function • Vector source, vector field, → dyadic Green function • Six-vector source, six-vector field, → six-dyadic Green function

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 07.01

132 Green six-dyadic • Maxwell equations and Maxwell six-dyadic operator L(∇) ! 0 1 E ξ E J ∇ × − jω · = −1 0 H ζ µ H M

L(∇) · e(r) = j(r), L(∇) = J∇ × I − jωM • Green six-dyadic = four dyadic fields from dyadic unit sources L(∇) · G(r) = −Iδ(r), ! ! I 0 G (r) G (r) I = , G(r) = ee em 0 I Gme(r) Gmm(r) • Additional conditions: flow of energy towards infinity (lossless media) or decay of fields towards infinity (lossy media) • Formal solution G(r) = −L−1(∇)δ(r)

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 07.02

133 Use of Green functions • Six-vector ﬁeld from a six-vector source L(∇) · e(r) = j(r), L(∇) · G(r − r0) = −δ(r − r0)I

• can be expressed in terms of the Green six-dyadic Z e(r) = − G(r − r0) · j(r0)dV 0

• which is valid outside the sources. Check: Z Z L(∇)·e(r) = − [L(∇)·G(r−r0)]·j(r0)dV 0 = δ(r−r0)j(r0)dV 0 = j(r)

V V

• When r inside the source region V , order of integration and dif- ferentiation not interchangeable. Singularity of Green function requires more careful consideration.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 07.03

134 Helmholtz operators • Adjoint Maxwell operators ! jωI (∇ × I − jωξ) · µ−1 La(∇) = −(∇ × I + jωζ) · −1 jωI ! jωI −1 · (∇ × I − jωξ) La(∇) = −µ−1 · (∇ × I + jωζ) jωI • diagonalize the Maxwell operator ! a He(∇) 0 L (∇) · L(∇) = L(∇) · La(∇) = , 0 Hm(∇) • Helmholtz second-order dyadic operators

−1 2 He(∇) = −(∇ × I − jωξ) · µ · (∇ × I + jωζ) + ω

−1 2 Hm(∇) = −(∇ × I + jωζ) · · (∇ × I − jωξ) + ω µ

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 07.04

135 Two dyadic Green functions

• Deﬁne Green six-dyadic in terms of another Green six-dyadic g(r):

G(r) = La(∇)g(r)

L(∇) · G(r) = L(∇) · La(∇) · g(r) = −Iδ(r)

• Diagonal six-dyadic operator L(∇) · La(∇) has diagonal solution ! Ge(r) 0 g(r) = , He,m(∇) · Ge,m(r) = −Iδ(r) 0 Gm(r)

• The original Green six-dyadic is obtained as G(r) = La(∇)g(r):

−1 ! jωGe(r) · (∇ × I − jωξ) · Gm(r) G(r) = −1 −µ · (∇ × I + jωζ) · Ge(r) jωGm(r)

• It is suﬃcient to solve only two Green dyadics Ge(r), Gm(r)

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 07.05

136 Two scalar Green functions

• Green dyadics Ge(r), Gm(r) obtained formally as −1 (2)T −1 Ge,m(r) = −He,m(∇) · δ(r) = He,m (∇) δ(r) detHe,m(∇) • Can be solved in terms of two scalar Green functions −1 Ge,m(r) = δ(r), detHe,m(∇) • satisfying scalar fourth-order diﬀerential equations

detHe,m(∇)Ge,m(r) = −δ(r) • Helmholtz Green dyadics can be expressed as (2)T Ge,m(r) = He,m (∇)Ge,m(r) • In some special cases solving 4th order equations can be avoided

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 07.06

137 Summary of the general method • Solve two scalar Green functions satisfying 4th order Helmholtz determinant equations

detHe,m(∇)Ge,m(r) = −δ(r) • Form two dyadic Green functions which are solutions to 2nd order dyadic Helmholtz equations (2)T Ge,m(r) = He,m (∇)Ge,m(r), He,m(∇) · Ge,m(r) = −Iδ(r) • Form the Green six-dyadic which satisﬁes the original 1st order six-dyadic equation ! Ge(r) 0 G(r) = La(∇) · , L(∇) · G(r) = −Iδ(r), 0 Gm(r) • Diﬃculty: solutions to fourth-order equations not available ⇒ works only for some special cases of the bi-anisotropic medium.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 07.07

138 Isotropic medium • Isotropic medium is the simplest special case. Deﬁne

× 2 H(∇) = µHe(∇) = Hm(∇) = I×∇∇ + k I

(2)T detH(∇) = k2(∇2 + k2)2, H (∇) = (∇2 + k2)(∇∇ + k2I) • Helmholtz Green dyadic solved from 2nd order equation: −1 1 G(r) = −H (∇)δ(r) = (I+ ∇∇)G(r) = µ−1G (r) = −1G (r) k2 e m −1 e−jkr G(r) = δ(r) = (outgoing wave) ∇2 + k2 4πr • The Green six-dyadic becomes ! ! G (r) G (r) jωµG(r) ∇G(r) × I G(r) = ee em = Gme(r) Gmm(r) −∇G(r) × I jωG(r)

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 07.08

139 Bi-isotropic medium 1 • For bi-isotropic media dyadic Helmholtz operators factorizable

2 µHe(∇) = Hm(∇) = H(∇) = −[(∇×I−jωξI)·(∇×I+jωζI)−k I]

H(∇) = −L+(∇) · L−(∇) = −L−(∇) · L+(∇)

L±(∇) = ∇ × I ∓ kτ±I, τ± = cos θ ± κr

• Only one Green dyadic needed: Ge(r) = µG(r), Gm(r) = G(r) −1 −1 −1 G(r) = −H (∇)δ(r) = L+ (∇) · L− (∇)δ(r)

• Dyadic partial fraction expansion applicable in this case −1 −1 −1 −1 L+ (∇) · L− (∇) = A+L+ (∇) + A−L− (∇) −1 −1 1 = [A L (∇)+A L (∇)]·L (∇)·L (∇) ⇒ A = ± + − − + + − ± 2k cos θ

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 07.09

140 Bi-isotropic medium 2

• Green dyadic can be expressed

−1 −1 G(r) = [A+L+ (∇) + A−L− (∇)]δ(r) = −[A+G+(r) + A−G−(r)]

• in terms of two auxiliary Green dyadics G±(r)

(2)T −1 (2)T L± G± = −L± (∇)δ(r) = − δ(r) = L± (∇)G±(r), detL±(∇)

(2)T 2 2 L± (∇) = ∇∇ ± kτ±∇ × I + k τ±I

• Scalar Green functions G±(r) satisfy second-order Helmholtz equations

2 2 2 detL±(∇)G±(r) = ∓kτ±(∇ + k τ±)G±(r) = −δ(r)

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 07.10

141 Bi-isotropic medium 3

• Solution to the scalar Helmholtz equation (outgoing waves assume kτ± > 0)

−jkτ±r 2 2 2 1 e (∇ + k τ±)G±(r) = ± δ(r),G±(r) = ∓ kτ± 4πkτ±r

• Solutions to Green dyadics

Ge(r) = µG(r), Gm(r) = G(r),

1 (2)T e−jkτ+r (2)T e−jkτ−r G(r) = L+ (∇) + L− (∇) 2k cos θ 4πkτ+r 4πkτ−r • Two terms correspond to two self-dual ﬁelds • This result can also be also obtained through the general method with some more eﬀort

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 07.11

142 Self-dual medium 1 • Self-dual medium is a generalization of the bi-isotropic medium √ √ sin θ µ 0 −j µ M = √ α + √ κ sin θ µ µ j µ 0 r

• Denote µHe(∇) = Hm(∇) = H(∇) which can be factorized as

H(∇) = −(∇ × I − jωξ) · α−1 · (∇ × I + jωζ) + k2α

−1 −1 = −L+(∇) · α · L−(∇) = −L−(∇) · α · L+(∇)

L±(∇) = ∇ × I ∓ kτ ±, τ ± = cos θα ± κr

• Formal solution for Ge(r) = µG(r), Gm(r) = G(r) from

−1 −1 −1 G(r) = −H (∇)δ(r) = L+ (∇) · α · L− (∇)δ(r)

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 07.12

143 Self-dual medium 2 • Dyadic partial fraction expansion possible also in this case

−1 −1 −1 −1 L+ (∇) · α · L− (∇) = A+L+ (∇) + A−L− (∇)

−1 −1 = L+ (∇) · [A+L−(∇) + A−L+(∇)] · L− (∇) 1 A L (∇) + A L (∇) = α ⇒ A = ± + − − + ± 2k cos θ • Solution in terms of auxiliary Green functions

−1 −1 1 G(r) = L (∇) · α · L (∇)δ(r) = − [G (r) − G (r)] + − 2k cos θ + −

(2)T −1 (2)T L± (∇) G±(r) = −L± (∇)δ(r) = − δ(r) = L± (∇)G±(r) detL±(∇)

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 07.13

144 Self-dual medium 3 • Auxiliary Green dyadics solved from

detL±(∇)G±(r) = det(∇ × I ∓ kτ ±)G±(r) = −δ(r)

• Expand τ ± in symmetric and antisymmetric parts:

τ ± = S± + a± × I

2 detL±(∇) = ∓k[S± :(∇ ∓ ka±)(∇ ∓ ka±) + k detS±]

• Helmholtz equation becomes

2 [S± :(∇ ∓ ka±)(∇ ∓ ka±) + k detS±][∓kG±(r)] = −δ(r)

2 ∓ka±·r [S± : ∇∇ + k detS±][∓ke G±(r)] = −δ(r)

• This can be transformed to a more familiar form

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 07.14

145 Self-dual medium 4 • Apply aﬃne transformation 1/2 −1/2 q 0 0 0 ∇ = S± · ∇, r = S± · r, δ(r ) = δ(r) detS± q 02 2 ∓ka±·r 0 (∇ + k detS±) ∓ ke G±(r) detS± = −δ(r )

√ q −1 0 0 0 • Solution with r = r · r = S± : rr

q (2) p 0 −jk S :rr q e−jk detS±r e ± ∓ka±·r ke G±(r) detS± = = 4πr0 q −1 4π S± : rr • Solutions for the two auxiliary scalar Green functions r e−jkD± (2) ±ka±·r G±(r) = ∓e ,D±(r) = S± : rr 4πkD±

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 07.15

146 Self-dual medium 5

• The dyadic Green functions Ge(r) = µG(r), Gm(r) = G(r) are ﬁnally obtained from −1 G(r) = [G (r) − G (r)] 2k cos θ + − 1 (2)T (2)T = [−L (∇)G (r) + L (∇)G (r)], 2k cos θ + + − − • by inserting (2)T e−jkD± (2) ±ka±·r ∓L± (∇)G±(r) = [−(∇ ∓ ka±) × I ∓ S±] e 4πkD± (2) e−jkD± ±ka±·r 2 = e [∇∇ ± k(S± · ∇) × I + k S± ] 4πkD± • Again the two terms correspond to the self-dual ﬁelds

• a± = 0, S± = τ±I gives the previous result for bi-isotropic space

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 07.16

147 Uniaxial medium 1 • Helmholtz dyadic factorizable only in self-dual media. However, Helmholtz determinant factorizable in uniaxial anisotropic medium

= zuzuz + tIt, µ = µzuzuz + µtIt, ξ = ζ = 0 • Electric Green dyadic −1 −1× 2 Ge(r) = −He (∇)δ(r), He(∇) = µ ×∇∇ + ω (2)T 1 Ge(r) = He (∇)G(r),G(r) = − δ(r) detHe(∇)

• The operator detHe(∇) can be factorized (after some eﬀort): k2 detH (∇) = t H (∇)H (∇), k2 = ω2µ e detµ µ t t t

2 z 2 2 2 µz 2 2 H(∇) = ∇t + (∂z + kt ),Hµ(∇) = ∇t + (∂z + kt ) t µt

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 07.17

148 Uniaxial medium 2 • Scalar Green function detµ 1 detµ G(r) = − 2 δ(r) = 2 [gµ(r) − g(r)], kt Hµ(∇)H(∇) ω (zµt − tµz) 1 gα(r) = − δ(r), α = µ, (∂2 + k2)(∇2 + αz (∂2 + k2) z t t αt z t

• It can be shown that gα(r) can be solved analytically as j −jktz jktz gα(r) = [E1(jkt(Dα − z))e + E1(jkt(Dα + z))e ], 8πktαt

• E1(x) = exponential integral function, Dα(r) = distance function ∞ Z −y r e αz 2 2 E1(x) = dy, Dα = ρ + z y αt x

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 07.18

149 Uniaxial medium 3 • The electric Green dyadic becomes (2)T detµ (2)T Ge(r) = He (∇)G(r) = 2 He (∇)[gµ(r) − g(r)] ω (zµt − tµz) • After diﬀerentiations (quite tedious)

−jktD −1 1 e jτuzq Ge(r) = z + 2 ∇∇ + ∇ × kt 4πD 4πkt u × r −jktD −jktDµ z τ = e − e , q = 2 (uz × r)

• Solution not singular at z axis uz × r = 0 because τq is ﬁnite

• Solution generalizable to ξ = ξzuzuz, ζ = ζzuzuz. • Factorizability of detH(∇) does not warrant analytic solution for Green dyadic!

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 07.19

150 A class of anisotropic media 1 • Helmholtz determinant operator factorizable for the class of anisotropic media deﬁned by µ = τT , ξ = ζ = 0 1 detH (∇) = det(µ−1×∇∇ + ω2) = det( −1T ×∇∇ + ω2) e × τ × ω2 ω4 ω2 = ( : ∇∇)2+ : ∇∇+ω6det = ( : ∇∇+ω2τdet)2 τ 2det τ τ 2det (2) • Leads to a second-order PDE because also He (∇) factorizable: (2) 1 ω2 H (∇) = ( : ∇∇)∇∇ + ((2)×∇∇)× + ω4(2) e τ 2det τdet × × 1 = ( : ∇∇ + ω2τdet)(∇∇ + ω2τ(2)) τ 2det (2)T −1 1 ∇∇ + ω2τ ⇒ H (∇) = e ω2 : ∇∇ + ω2τdet

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 07.20

151 A class of anisotropic media 2

• Solution of the scalar Green function (s = symmetric part of ) 1 e−jkD q G(r) = − δ(r) = ,D = −1 : rr 2 p s s : ∇∇ + k 4π detsD

• Electric Green dyadic obtained in analytic form

(2)T −1 1 ∇∇ + ω2τ G (r) = −H (∇)δ(r) = − δ(r) e e ω2 : ∇∇ + ω2τdet √ p −1 −jω τdet s :rr 2 (2)T e = (∇∇ + ω τ ) q 2p −1 4πω dets s : rr

• Check: isotropic case = s = I, τ = µ/: 2 −jkr √ ⇒ Ge = µ(I + ∇∇/k )(e /4πr), k = ω µ

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 07.21

152 Problems 13 Derive the electric Green dyadic for a bi-anisotropic medium whose medium dyadics are of the form

= S, µ = µS, ξ = x × I, ζ = z × I,

where S is a complete symmetric dyadic, x, z two vectors and , µ two scalars. 14 Show that the Helmholtz determinant operators for the anisotropic medium deﬁned by the medium dyadics of the form

T = A, µ = µA + αab, ξ = ζ = 0

can be factorized. A is a dyadic, a, b two vectors and , µ, α three scalars.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 07.22

153 S-96.510 Advanced Field Theory 8. Green dyadic singularities and complex-space sources

I.V.Lindell

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 08.00

154 Fields outside sources • Isotropic medium Green six-dyadic ! ! G (r) G (r) jωµG(r) ∇G(r) × I ee em = Gme(r) Gmm(r) −∇G(r) × I jωG(r) e−jkr 1 G(r) = , G(r) = (I + ∇∇)G(r) 4πr k2 Fields outside source area V ! E(r) Z G (r − r0) G (r − r0) J(r0) = − ee em · dV 0 0 0 0 H(r) Gme(r − r ) Gmm(r − r ) M(r ) V • For example, electric ﬁeld from electric source Z E(r) = −jωµ G(r − r0) · J(r0)dV 0

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 08.01

155 Singularity of Green dyadic • When D = |r − r0| small, G(D) ≈ 1/4πD large but integrable:

2π π a a Z 1 Z Z Z 1 Z a2 dV = r2drdθdϕ = rdr = 4πr 4πr 2 0 0 0 0

3 • Green dyadic more singular |G(D)| ≈ |∇∇G(∇)| ≈ |urur/D | Not integrable! How to compute ﬁelds inside the source r ∈ V ? Z 1 E(r) = −jωµ (I + ∇∇)G(r − r0) · J(r0)dV 0 k2 V Z H(r) = − ∇G(r − r0) × J(r0)dV 0

• Solution: split V = Vδ + V1. Small volume Vδ around ﬁeld point r containing the singularity is computed separately.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 08.02

156 Principal-value integrals • Outside r = 0 we have (∇2 + k2)G(r) = 0 and 1 1 1 [I + ∇∇]G(r) = (−∇2I + ∇∇)G(r) = − I×∇∇G(r) k2 k2 k2 ×

• The electric ﬁeld integral over V1 = V − Vδ is well deﬁned: 1 Z E (r) = − I× ∇∇G(r − r0) · J(r0)dV 0 1 jω × V1

• The limit Vδ → 0 is denoted as principal-value integral PVδ 1 Z E (r) → − I×PV [∇0∇0G(r − r0)] · J(r0)dV 0 1 jω × δ V

• Total ﬁeld E = E1 + Eδ. The contribution Eδ from Vδ → 0 is obtained separately by assuming that the current is constant 0 J(r ) = J(r) over the small volume Vδ.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 08.03

157 Field from constant current

• If constant current Jc ﬁlls all space V∞, ﬁelds are constant Ec, Hc J ∇ × H = 0 = jωE + J , ⇒ E = − c , H = 0 c c c c jω c

• No singularity for ﬁeld inside the source!

• Field from volume Vδ of constant current Jc around point r: Eδ(r) = Ec − E1(r). Limit Vδ → 0 ⇒ principal-value integral: 1 Z E (r) = E + I×PV ∇0∇0G(r − r0)dV 0 · J δ c jω × δ c V∞ 1 Z = − (I − I×L ) · J , L = PV ∇0∇0G(r − r0)dV 0 jω × δ c δ δ V∞

• Lδ = depolarization dyadic depends on the volume Vδ

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 08.04

158 Depolarization dyadic

• Depolarization dyadic Lδ as surface integral from Gaussian law Z I 0 0 0 0 0 0 0 0 Lδ = PVδ ∇ ∇ G(r − r )dV = − lim n ∇ G(r − r )dS

V∞ Sδ

0 • Sδ = surface of the small volume Vδ, n = outside unit normal Limit δ → 0 static approximation valid G(R) → 1/4πR 0 0 0 02 0 0 0 R = r − r, dS = R dΩ /n · uR I I 0 0 0 0 1 0 1 n uR 0 Lδ = − n ∇ 0 dS = 0 0 dΩ 4π|r − r | 4π n · uR Sδ 4π

• Depolarization dyadic Lδ is symmetric and satisﬁes Lδ : I = 1 Independent of size of Sδ, depends on form of Sδ and location of origin of R0.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 08.05

159 Singularity of Green dyadic

• Field inside a small volume Vδ → 0 of constant current Jc 1 1 E (r) = − (I − L ×I) · J = − L · J δ jω δ× c jω δ c • Applied to ﬁeld integral for point r inside the source region: 1 Z E(r) = E (r)+E (r) = − L ·J(r)−jωµPV G(r−r0)·J(r0)dV 0 δ 1 jω δ δ V Z 1 = −jωµ [PV G(r − r0) − L δ(r − r0)] · J(r0)dV 0 δ k2 δ V • The Green dyadic consists of a regular part (principal value) and a singular delta-function part 1 G(r − r0) = PV G(r − r0) − L δ(r − r0) δ k2 δ

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 08.06

160 Diﬀerent contracting surfaces

0 0 • Spherical Sδ, origin at center: n = uR I 0 0 I 1 n uR 0 1 0 0 0 1 Lδ = 0 0 dΩ = uRuRdΩ = I 4π n · uR 4π 3 4π 4π

• Symmetric surface ⇒ Lδ = αI. From Lδ : I = 1 ⇒ α = 1/3. Valid for cube, tetrahedron, etc when ﬁeld point at center

• Long circular cylinder, axis uz 1 L = (I − u u ) δ 2 z z

• Ellipsoid, Lx,Ly,Lz depolarizing factors, A = 1/(Lx + Ly + Lz)

Lδ = A(Lxuxux + Lyuyuy + Lzuzuz)

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 08.07

161 Example of depolarization dyadic • Spheroid with normal vector n ρ2 z2 ∇f(r) a2ρ + b2zu f(r) = + = 1, n = = z b2 a2 |∇f(r)| pa4ρ2 + b4z2

• Has one special direction uz ⇒ dyadic must be of uniaxial form Lδ = Lzuzuz + LtIt

• Lδ : I = 1 ⇒ Lz = 1 − 2Lt, only one parameter to be found I 0 0 1 n ur 0 Lδ = 0 0 dΩ = Lzuzuz + LtIt, ⇒ Lδ : uzuz = Lz 4π n · ur 4π

2π π π/2 Z Z Z 2 2 1 (∇f · uz)(uz · r) b cos θ sin θdθ Lz = sin θdθdϕ = 4π ∇f · r a2 sin2 θ + b2 cos2 θ 0 0 0 • Can be solved in terms of elementary functions.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 08.08

162 Scattering problem

i • Dielectric scatterer Vs set in incident ﬁeld E (r)

• Scatterer can be replaced by equivalent polarization current Jp(r):

∇ × H(r) = jω(r) · E(r) = jωo · E(r) + Jp(r)

⇒ Jp(r) = jωo[r(r) − 1] · E(r) • Scattered ﬁeld = ﬁeld from polarization current

sc R 0 0 0 2 R 0 0 0 0 E (r) = −jωµo G(r−r )·Jp(r )dV = ko [r(r )−1]G(r−r )·E(r )dV Vs Vs • Total ﬁeld E(r) = Ei(r) + Esc(r) Z i 2 0 0 0 0 E(r) = E (r) + ko [r(r ) − 1]G(r − r ) · E(r )dV

• r ∈ Vs ⇒ integral equation for the total ﬁeld inside the scatterer. I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 08.09

163 Singularity extraction

2 • Insert G = PVδG(r) − Lδδ(r)/ko ⇒ no singularity in integral: Z i 2 0 0 0 0 [I+(r(r)−1)Lδ]·E(r) = E (r)+koPVδ [r(r )−1]G(r−r )·E(r )dV

• For example, constant r and choose symmetric Sδ with Lδ = I/3 Simpliﬁed integral equation + 2 Z r E(r) = Ei(r) + k2( − 1)PV G(r − r0) · E(r0)dV 0 3 o r δ Vs

• Interpretation: Total field − field from current in Vδ = field in the cavity Vδ = incident field + field from polarization current outside the cavity

• In numerical analysis the cell of discretization can be taken as Vδ.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 08.10

164 Numerical analysis

• Simple numerical computation: P – Discretize the scatterer to cells ∆Vi, i = 1 ··· N. – Approximate ﬁeld E(r) in each cell by constant value at the center of the cell E(ri). – Solve linear system of equations for i = 1 ··· N

2 X i (I +(r −1)L)·E(ri)−ko(r −1) G(ri −rj)·E(rj)∆Vj = −E (ri) j6=i – First term: diagonal element of matrix, principal-value integral: off-diagonal terms • More efficient computation through moment methods with field approximated by suitable basis functions in cells

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 08.11

165 Field from source in complex space

0 0 0 • Deﬁne: distance from complex point r = rre + jrim D = p(r − r0) · (r − r0) (no conjugation!)

• Green function G(D) = e−jkD/4πD satisﬁes Helmholtz equation (can be checked) (∇2 + k2)G(D) = 0,D 6= 0

• Singularity of Green function at D = 0. What does it mean? 2 0 2 0 0 02 D = (r − rre) − 2j(r − rre) · rim − rim = 0 0 0 0 0 ⇒ |r − rre| = |rim|, (r − rre)⊥ rim 0 0 0 • Singularity circle: center at r = rre, plane ⊥ rim, radius |rim|. 0 0 Reduces to a point r = rre for rim → 0. • Source of G(D) = −δ(r − r0), delta function at complex point?

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 08.12

166 Point source in complex space • Delta function of complex variable z: limit of the Gaussian function

p −κz2 δκ(z) = κ/π e , κ → ∞ • Denote z = x + jy, expand

p −κ(x2−y2) −2jκxy δκ(x, y) = κ/π e e e−2jκxy = cos(2κxy) − j sin(2κxy) oscillating function • When κ → ∞, vanishes in region |x| > |y|. Oscillates wildly with inﬁnite amplitude in region |x| < |y|. • Corresponds to ordinary delta function in integration if path can be changed to one along the real axis through the origin: b Z f(z)dz = f(z), <{a} < −|={a}|, |={b}| < <{b}

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 08.13

167 Double-valued Green function 1

0 • Complex distance function between points r and r = juzb, b > 0

D(r) = p(r − jr0) · (r − jr0) = pρ2 + (z − jb)2

• Assume r circulates a path in xz plane of radius q b around and through the circle of singularity:

r = uxb + q(ux cos ψ + uz sin ψ), ψ = 0 → 2π

2 2 −jψ D = [b(ux − juz) + q(ux cos ψ + uz sin ψ)] ≈ 2bqe

• The distance becomes two-valued function of geometric angle ψ

D → p2bq e−jψ/2

• For full circulation ψ = 0 → 2π, D → −D nonuniqueness of D(r)!

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 08.14

168 Double-valued Green function 2 • Uniqueness of D(r) and G(D(r)) by deﬁning two Riemann spaces glued together at the circle of singularity

p p 2 2 D = (r + juzb) · (r + juzb) = r − b + 2jbz

• Crossing branch-cut surface bounded by the singularity circle changes branch of distance function • Two obvious choices for branch cut surfaces: (1) disk z = 0, ρ < b ⇒ D imaginary and (2) holey plane z = 0, ρ > b ⇒ D real • Branch-cut surface (1) = disk corresponds to <{D} = 0, separates outgoing (<{D} > 0) and ingoing (<{D} < 0) solutions in the function e−jkoD • Branch-cut surface (2) = holey plane corresponds to ={D} = 0, separates solutions propagating along uz and −uz (Gaussian beam- type of solutions)

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 08.15

169 Gaussian beam (G.A. Deschamps, 1971)

• Field from point source in complex space ≈ Gaussian beam close 0 to z axis. Assume r = −juzb, r = ρ + uzz, ρ z, z > 0

p p 2 2 D = (r + juzb) · (r + juzb) = ρ + (z + jb)

ρ2 ρ2 D ≈ z + jb + = z + jb + (z − jb) 2(z + jb) 2(z2 + b2) ρ2z −ρ2b e−jkoD ≈ ekob e−jkoz exp −jk exp o 2(z2 + b2) 2(z2 + b2) • Interpretation of terms: constant, plane-wave term, deviation from equi-phase plane, Gaussian ﬁeld distribution for z > 0 when b > 0: Gaussian decay outwards from z axis.

0 • Complex point source at r = −juzb, b > 0 creates Gaussian beam propagating in +uz direction.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 08.16

170 Field from dipole in complex space

• Field from dipole J(r) = uxILδ(r + juzb), b > 0

1 e−jkoD E(r) = −jωµoIL(ux + 2 ux · ∇∇) ko 4πD

• Far ﬁeld approximation D = pr2 + 2jzb − b2 ≈ r + jb cos θ

e−jkor E(r) ≈ −jωµ ILu · (I − u u ) ekob cos θ, θ < π/2 o x r r 4πr |E(θ)| ≈ cos θe−kob(1−cos θ) (xz plane) ≈ e−kob(1−cos θ) (yz plane) |E(0)|

• Directivity increases for increasing b. When kob large, in both p planes θ3dB ≈ 2 ln 2/kob. • Simple way to represent directive sources with sources in complex space giving Gaussian beam expansion for radiation ﬁelds.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 08.17

171 Problems 15 Apply the volume integral equation approach to scattering from two small dielectric scatterers at the points r1, r2, with respective constant permittivities r1, r2. Assume that the contracting surfaces Sδ coincide with the boundaries of the scatterers and their distance is much larger than the sizes of the scatterers. Find the expression for the field inside each scatterer E(r1), E(r2) when the incident field is Ei(r). 16 Find the surface of constant amplitude for a scalar field arising 0 from a point source at the complex point r = −juzb, b > 0. What is the surface for the amplitude corresponding to 1/e times the maximum amplitude of the beam?

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 08.18

172 S-96.510 Advanced Field Theory 9. Plane waves

I.V.Lindell

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 09.00

173 Plane-wave ﬁelds • Plane wave = ﬁelds with exponential dependence on r E(r) E = e−jk·r, H(r) H

• Maxwell equations with ∇ → −jk become (sources in inﬁnity) ! 0 1 E ξ E k × + ω · = 0 −1 0 H ζ µ H

• In six-vector representation e(r) = ee−jk·r,Jk × e + ωM · e = 0

• Eigenvalue equation (Jk × I + ωM) · e = 0

• Eigenvalue parameter can be chosen freely: ω, component of k etc.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 09.01

174 Dispersion equation 1 • Dispersion equation = equation for k ! ω ωξ + k × I det(Jk × I + ωM) = det = 0 ωζ − k × I ωµ

• Eigenvalue equations for H and E:

De(k) · E = 0, Dm(k) · H = 0,

2 −1 De(k) = He(−jk) = ω − (ωξ + k × I) · µ · (ωζ − k × I), 2 −1 Dm(k) = Hm(−jk) = ω µ − (ωζ − k × I) · · (ωξ + k × I). • Dispersion equations in 3-dyadic form

detDe(k) = 0, detDm(k) = 0

• Same equation because det(I + A · B) = det(I + B · A)

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 09.02

175 Dispersion equation 2

• Order of the dispersion equation?

6 4 (2) −1 detDe(k) = ω det − ω : [(ωξ + k × I) · µ · (ωζ − k × I)]

+ω2 : [(ωξ + k × I)(2) · µ(−2) · (ωζ − k × I)(2)] 1 − det(ωξ + k × I)det(ωζ − k × I) = 0 detµ • To ﬁnd sixth-order terms:

(2) det(ωξ + k × I) = ω3detξ + ω2ξ :(k × I) + ωξ : kk

(2) det(ωζ − k × I) = ω3detζ − ω2ζ :(k × I) + ωζ : kk

• Sixth-order and ﬁfth-order terms absent ⇒ fourth-order equation! Dispersion equation very complicated for the general medium.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 09.03

176 Solution of eigenwaves

• Assume k = ku, real unit vector u, dispersion equation k = k(u) • Deﬁnes 4th order k-vector surface (dispersion surface) corresponding to eigensolutions (possible plane waves). k may be complex. • In some media splits to two 2nd order surfaces (complex quadrics) • Procedure for solution:

– Roots k = ki(u) from 4th order equation detDe(ku) = 0

– Corresponding eigenﬁelds Ei from De(kiu) · Ei = 0: (2) (2) – If De(kiu) 6= 0 (ki single eigenvalue) Ei = a · De (kiu) (2) – If De(kiu) = 0, (ki double eigenvalue) Ei⊥a · De(kiu) – Vector a chosen so that result 6= 0

P −jkiu·r • General plane-wave in direction u = sum of eigenwaves Eie

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 09.04

177 Isotropic medium, eigenvalues

• Isotropic medium (o, µo). Dispersion dyadics

2 2 µoDe(k) = oDm(k) = D(k) = koI + (k × I)

D(k) = D+(k) · D−(k), D±(k) = k × I ∓ jkoI

• Dispersion equation splits in two 2nd order equations:

2 detD±(k) = ∓ko(k · k − ko) = 0 k± = kou

• Double eigenvalue: k+ = k− = ko. • Dispersion surfaces: two surfaces coincide to a single sphere of radius ko

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 09.05

178 Isotropic medium, eigenpolarizations

• Eigenpolarizations satisfy

D(k±) · E± = ko(u × E± ∓ jE±) = 0 ⇒ E± = ∓ju × E±

• Circularly polarized eigenvectors: E± ·E± = 0, u·E± = 0. Helicity vector gives the handedness: ∗ ∗ E± × E± j(u × E±) × E± p(E±) = ∗ = ∓ ∗ = ±u jE± · E± jE± · E±

• E+ right-hand, E− left-hand circularly polarized • All transverse polarizations possible because eigenwaves have same eigenvalues k± = ko. General plane wave can be expressed as linear combination −jkou·r E(r) = (αE+ + βE−)e

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 09.06

179 Bi-isotropic medium, eigenvalues • Dispersion dyadics

2 µDe(k) = Dm(k) = D(k) = ω µI − (ωξI + k × I) · (ωζI − k × I) √ • Denote K = ω µ, ωξ = K(sin θ − jκr), ωζ = K(sin θ + jκr),

2 D(k) = K I−[K sin θI+(k×I−jκrKI)]·[K sin θI−(k×I−jκrKI)] 2 2 2 = K cos θI + (k × I − jκrKI)

= [k × I + jK(cos θ − κr)I] · [k × I − jK(cos θ + κr)I]

= D−(k) · D+(k) = D+(k) · D−(k), D±(k) = k × I ∓ jk±I. • Dispersion equation splits in two equations:

2 √ detD±(k) = ∓jk±(k·k−k±) = 0 k± = uk± = uω µ(cos θ±κr)

• k-vector surface = two spheres of radii k+, k−. The spheres coincide when no chirality: κr = 0, ⇒ k+ = k−.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 09.07

180 Bi-isotropic medium, eigenvectors

• Eigenvectors satisfy D±(k±) · E± = 0

(2) (2) 2 D± (uk±) = (k±u × I ∓ jk±I) = −k±(It ± ju × I)

2 × 2 = −k±[vv+uu×vv±j(u×vv−vu×v)] = −k±(v±ju×v)(v∓ju×v) • Here v is any transverse unit vector: v · v = 1, v · u = 0.

(2) • Eigenvectors of the form E± = a · D±(k±) = α(v ∓ ju × v) • Helicity vectors of eigenvectors: ∗ E± × E± p(E±) = ∗ = ±v × (u × v) = ±u, jE± · E±

• E+ right-hand, E− left-hand circularly polarized

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 09.08

181 Self-dual medium, factorization • Self-dual medium = generalization of bi-isotropic medium √ again deﬁning K = ω µ ! ξ ω K sin θ 0 −jKκ ω = α + r ζ µ K sin θ ωµ jKκr 0

2 −1 µDe(k) = K α+[k×I+K(sin θα−jκr)]·α ·[k×I−K(sin θα+jκr)]

• De(k) = He(−jk) is factorizable like the Helmholtz operator He(∇) −1 −1 De(k) = D+(k) · µ · D−(k) = D−(k) · µ · D+(k)

D±(k) = k × I ∓ jKτ ±, τ ± = cos θ α ± κr • Eigenvalue equation splits to two equations:

D±(k) · E± = (k × I ∓ jKτ ±) · E± = 0

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 09.09

182 Self-dual medium, eigenvalues

• Write τ ± = S± + a± × I in symmetric and antisymmetric parts

D±(k) · E± = (k ∓ jKa±) × E± ∓ jKS± · E± = 0

−1/2 1/2 q (2) • Aﬃne transformation: multiply by S± = (S± ) / detS±

1/2 q 1/2 1/2 [S± · (k ∓ jKa±)/ detS± ] × (S± · E±) = ±jK(S± · E±)

0 0 0 • This is of the form k± × E± = ±jKE±.

0 0 2 2 k± · k± = K , ⇒ S± :(k± ∓ jKa±)(k± ∓ jKa±) = K detS±

• Second order equations for k± = uk± can be solved for k±(u).

• Dispersion surfaces spheroids if S± positive deﬁnite, decentered by ±jKa± (real vectors for lossless medium).

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 09.10

183 Self-dual medium, eigenvectors

0 0 0 • Equation for the eigenvectors k± × E± = ±jKE± 0 0 0 0 0 0 2 imply k± · E± = 0, E± · E± = 0, k± · k± = K 0 0 • ⇒ E± are circularly polarized orthogonal to k± with general form k0 E0 = v ∓ j ± × v , v · k0 = 0 ± ± K ± ± ± 0 0 • Expressions can be checked by inserting in k± × E± 1/2 0 0 • Substituting k± and E± = S± · E±, the eigenvectors become  1/2  −1/2 S± · (k± ∓ jKa±) E± = S± · v± ∓ j q × v± K detS±

• Eigenvalues k± = uk±(u) must be inserted. Vectors v± can be 0 expressed as k± × w.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 09.11

184 General anisotropic medium

• For an anisotropic medium (ξ = ζ = 0) the eigenvalue equation is

2 −1× De(k) · E = (ω − µ ×kk) · E = 0

• The dispersion equation for k can be expanded as ω2 detD (k) = ω6det − ω4((2)×µ−1): kk + ( : kk)(µ : kk) = 0 e × detµ

• Biquadratic equation for k(u)! If k is a solution, also −k is a solution ⇒ symmetric 4th order dispersion surface. General solution:  s  2 (2)× (2)T (2)× (2)T 2 k±(u) 1 ×µ : uu ( ×µ : uu) 4detµdet =  ± −  ω2 2 ( : uu)(µ : uu) ( : uu)2(µ : uu)2 ( : uu)(µ : uu)

• Eigenvectors through the previous rule.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 09.12

185 Uniaxially anisotropic medium: TE/TM waves

• Uniaxially anisotropic medium: , µ of the form α = αzuzuz +αtIt • Polarization condition directly from Maxwell equations

−jk × E = −jωµ · H, ⇒ E · µ · H = 0

−jk × H = jω · E, ⇒ H · · E = 0

• Combine:

E · (tµ − µt) · H = (tµz − µtz)(E · uz)(uz · H) = 0

• tµz − µtz = 0 ⇒ aﬃne-isotropic medium satisfying zµ = µz.

• tµz − µtz 6= 0 ⇒ Plane wave satisﬁes either uz · E = 0 (TE wave) or uz · H = 0 (TM wave).

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 09.13

186 Uniaxially anisotropic medium: eﬀective media

• For TE and TM fields the original medium can be replaced by effective affine-isotropic media

TE TE t TE TE t · E = tE = µ · E , ⇒ = µ µt µt

TM TM µt TM TM µt µ · H = µtH = · H , ⇒ µ = t t • Special cases of self-dual media! ( and µ multiples of same dyadic) • Dispersion dyadics for eﬀective media can be factorized:

TE TE p −1 p TE De (k)·E = (k×I+jω t/µt µ)·µ ·(k×I−jω t/µt µ)·E = 0 TM TM p −1 p TM Dm (k)·H = (k×I+jω µt/t )· ·(k×I−jω µt/t )·H = 0 • Previous solution procedure for self-dual media applicable.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 09.14

187 Uniaxially anisotropic medium: eigenvalues • Dispersion equations for TE and TM polarized waves

TE 2 ω t 2 2 detDe (k) = 3 (µ : kk − ω µttµz) = 0 µt µz

TM 2 ω µt 2 2 detDm (k) = 3 ( : kk − ω µttz) = 0 t z

• Wave propagating in direction u = uz cos θ + ut sin θ: rµ µ r µ µ kTE = ω t t z = ω t t z 2 2 µ : uu µz cos θ + µt sin θ rµ r µ kTM = ω t t z = ω t t z 2 2 : uu z cos θ + t sin θ • Double eigenvalues for the eﬀective media correspond to two single eigenvalues for the original medium.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 09.15

188 Uniaxially anisotropic medium: eigenvectors

TE TM TE TM • Double eigenvalues ⇒ linear dyadics De (k ), Dm (k ):

TE 2 TE ω t 1 (µ · k)(µ · k) De (k ) = µ− [(µ : kk)µ−(µ·k)(µ·k)] = − µt detµ detµ

TM 2 TM ω µt 1 ( · k)( · k) Dm (k ) = − [( : kk)−(·k)(·k)] = − t det det • TE and TM ﬁeld vectors obtained from

TE TE TE E ⊥uz, E ⊥µ · k ⇒ E ∼ uz × µ · k ∼ uz × u TM TM TM H ⊥uz, H ⊥ · k ⇒ H ∼ uz × · k ∼ uz × u • Also: directly from Maxwell equations TE TE TE k · · E = ktu · E = 0, ⇒ E ∼ uz × u TM TM TM k · µ · H = kµtu · H = 0, ⇒ H ∼ uz × u

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 09.16

189 Uniaxial chiral medium √ • Uniaxial chiral medium with parallel helices: ζz = −ξz = jκz µoo ! ! ξ u u + I ξ u u = z z z t t z z z ζ µ ζzuzuz µzuzuz + µtIt

• Maxwell equations ⇒ orthogonality conditions E · B = H · D = 0

E · µ · H + E · ζ · E = 0, H · · E + H · ξ · H = 0 2 2 µtEt · Ht + µzEzHz + ζzEz = 0, tHt · Et + zHzEz + ξzHz = 0

• Eliminate Et · Ht, solve for parameter A deﬁned by µ H + ζ E E + ξ H A = z z z z = z z z z µtHz tEz s 2 1 µz z 1 µz z ξzζz ⇒ A± = + ± − + 2 µt t 4 µt t µtt

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 09.17

190 Uniaxial chiral medium: eﬀective media

• Solutions A± correspond to two eigenwaves. Each has a special impedance relation between axial ﬁeld components

A±µt − µz A±t − z Ez± = Hz±, ⇔ Hz± = Ez± ζz ξz • Nonchiral eﬃcient media can be deﬁned for the two eigenwaves

D± = · E± + uzξzHz± = (tIt + tA±uzuz) · E± = ± · E±

B± = µ · H± + uzζzEz± = (µtIt + µtA±uzuz) · H± = µ± · H± • ⇒ Aﬃne-isotropic media with uniaxial medium dyadics:

± = tA±, µ± = µtA±, A± = It + A±uzuz • The eﬀective media are also of the self-dual form • Eigenproblems in equivalent media can be solved with previous methods. I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 09.18

191 Uniaxial chiral medium: eigenvalues • Dispersion equations for the two eigenwaves −1 2 1 × detDe±(k) = det(ω tA± − A ×kk) µt 2 ω t 2 2 √ = (A± : kk − kt detA±) = 0, kt = ω µtt 2 µt detA±

• Solutions for k = k± = uk± v u s u detA± A± k = k t = k = ± t t 2 2 A± : uu A± cos θ + sin θ

r 2 2ω tµt(z µz −ξ√z ζz ) 2 2 2 2 cos θ(z µz −ξz ζz )+sin θ(tµz +µtz ∓ (tµz −µtz ) +4tµtξz ζz )

• Uniaxial anisotropic case obtained in the limit ξz → ζz → 0, whence A+ → µz/µt (TE wave) and A− → z/t (TM wave) I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 09.19

192 Problems 17 Show that the plane waves propagating in a bi-anisotropic medium with parameters

= zuzuz +tIt, µ = µzuzuz +It, ξ = ξuz ×I, ζ = ζuz ×I

can be decomposed in TE and TM parts. 18 Find solutions to the wave vectors k for a plane wave propagating √ in the medium of Problem 17. Write ξ = (χ − jκ) µ , ζ = √ √ o o (χ + jκ) µoo and ko = ω µoo for simpler notation.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 09.20

193 S-96.510 Advanced Field Theory 10. Source equivalence

I.V.Lindell

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 10.00

194 Uniqueness of sources • Sources are uniquely deﬁned by ﬁelds inside the source region ! J(r) 0 1 E(r) ξ E(r) = ∇× −jω · M(r) −1 0 H(r) ζ µ H(r)

• Sources are not uniquely deﬁned by ﬁelds outside the source region.

• Two sources g1, g2 are equivalent (g1 ∼ g2) with respect to ﬁeld f in region V outside the sources if their ﬁelds satisfy f1 = f2 in V .

• Difference of equivalent sources g1 −g2 = nonradiating (NR) source which creates zero field in V • Source can be replaced by an equivalent source for fields in V . • Problem of remote sensing: how to determine sources by measuring fields outside the source region? → Additional knowledge of sources needed or to find ’simplest’ of equivalent sources.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 10.01

195 Nonradiating sources • Nonradiating (NR) sources can be expressed in certain form. As- sume linear differential equation (g is the source of the field f) L(∇)f(r) = g(r) (+ uniqueness condition for fields)

• (1) Any NR source, is obviously of the form (by deﬁnition) gNR(r) = L(∇)f(r), f(r) = 0, r ∈ V

• (2) Conversely, any source of the form gNR(r) = L(∇)h(r), h(r) = 0 r ∈ V must be nonradiating because of field uniqueness. The field f(r) radiated by the source satisfies L(∇)f(r) = gNR(r) whence L(∇)[f(r)− h(r)] = 0. Because of the uniqueness condition, the field in brackets is the field with no sources and, hence, it must vanish. Thus, the field f(r) = 0 in V and gNR(r) is nonradiating.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 10.02

196 Nonradiating electromagnetic sources

• Nonradiating electric and magnetic currents have the general form (in isotropic medium)

NR × 2 J (r) = (I×∇∇ + k I) · F(r), F(r) = 0, r ∈ V

NR × 2 M (r) = (I×∇∇ + k I) · G(r), G(r) = 0, r ∈ V

• Nonradiating combination of electric and magnetic currents in bi- anisotropic medium is of the form [F(r) = 0 and G(r) = 0 when r ∈ V ]

NR ! J(r) −jω ∇ × I − jωξ F(r) = . M(r) −∇ × I − jωζ −jωµ G(r)

• Here J(r) and M(r) need not be nonradiating sources separately. Radiation from J cancels the radiation from M in V .

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 10.03

197 Examples of NR sources • One-dimensional example (Θ(z) is the step function): NR J (r) = uzJ[Θ(z + a) − Θ(z − a)]

• From symmetry: E(r) = uzE(z), ⇒ ∇ × E(r) = 0, ⇒ H(r) = 0 everywhere • ∇×H(r) = 0 ⇒ E(r) = −J(r)/jω, ﬁeld vanishes outside current, whence J(r) does not radiate outside its support −a < z < a.

NR • Three-dimensional example: J (r) = urJ(r)Θ(a − r). Radi- ally symmetric current does not radiate ﬁelds outside its support (sphere of radius a). Symmetry ⇒ E(r) = urE(r), ∇ × E(r) = 0. • In both cases ∇ × JNR(r) = 0 and JNR(r) = 0, r ∈ V , whence they are of the general NR form:

NR × 2 NR 2 J (r) = (I×∇∇ + k I) · F(r), F(r) = J (r)/k

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 10.04

198 Equivalence of sources

NR • Equivalent sources g1 ∼ g2 radiate same ﬁeld in V . g1 − g2 ∼ g • Equivalence of electric and magnetic sources J ∼ M if (J, −M) is NR source of the form (F(r) = 0, G(r) = 0 when r ∈ V ) ! J −jω ∇ × I − jωξ F(r) = . −M −∇ × I − jωζ −jωµ G(r)

• Take G = 0 and eliminate F or conversely ⇒ equivalence relations 1 J(r) ∼ M = − (∇ × I + jωζ) · −1 · J(r) jω 1 M(r) ∼ J = (∇ × I − jωξ) · µ−1 · M(r) jω • In isotropic medium J(r) ∼ M(r) = −∇ × J(r)/jω, M(r) ∼ J(r) = ∇ × M(r)/jωµ

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 10.05

199 Properties of equivalence

• Transitivity: J ∼ M, M ∼ J0 ⇒ J ∼ J0

J ∼ M = −∇×J/jω, M ∼ J0 = ∇×M/jωµ = −∇×(∇×J)/(−k2)

• Because J ∼ ∇ × (∇ × J)/k2, we have −∇ × (∇ × J) + k2J ∼ 0, whence this is a NR source • A source of the form J(r) = ∇φ(r), with φ(r) = 0, r ∈ V is a NR source because J(r) ∼ M(r) = −∇ × J(r)/jω = 0 • A source of the form J(r) = (∇2 + k2)F(r), with F(r) = 0 in V , is nonradiating because adding the NR source −∇(∇ · F(r)) we have J(r) ∼ −∇ × (∇ × F) + k2F, which is a NR source

2 2 2 ⇒ J1 = ∇t F(r) ∼ J2(r) = −(∂z + k )F(r)

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 10.06

200 Example of equivalence

2 2 • Cylindrical magnetic current M(r) = uzMΘ(a − ρ)Θ(h − z ) • Equivalent electric current 1 M M(r) ∼ J(r) = ∇ × M(r) = ∇Θ(a − ρ) × u Θ(h2 − z2) jωµ jωµ z M = −u × u δ(a − ρ)Θ(h2 − z2) = u J δ(ρ − a)Θ(h2 − z2) ρ z jωµ ϕ s • Equivalent circumferential surface current on the cylinder ρ = a M J = s jωµ

2 • Small magnetic dipole M = Im/πa length L = 2h moment ImL can be replaced by a current loop of current I = 2hJs = 2hM/jωµ. Relation can be written for the ratio of magnetic dipole moment 2 and current loop moment as ImL/Iπa = jωµ.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 10.07

201 Equivalent sources in electrostatics • Static magnetic current satisfying ∇ · M(r) = 0 is a source in electrostatics ∇ · D(r) = ∇ · E(r) = %(r), ∇ × E(r) = −M(r)

• Nonradiating combination source of the general form NR M(r) = ∇ × I · F(r), F(r) = 0, r ∈ V %(r) ∇

• Equivalence: if M(r) = ∇ × F(r), M(r) ∼ %(r) = −∇ · F(r) • Static charge and magnetic current are sources of potentials ∇2ϕ(r) = −%(r)/, ∇2A(r) = M(r), ∇ · A(r) = 0

• Nonradiating sources are of the general form %NR(r) = ∇2ψ(r), MNR(r) = ∇2F(r), ∇ · F(r) = 0.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 10.08

202 Multipole expansion • Multipole = point source with structure • Expressable as delta function series through a vector operator

J(r) = F(∇)δ(r), F(∇) = P0 + P1 · ∇ + P2 : ∇∇ + ···

• Multipole deﬁned by coeﬃcient polyadics (n + 1-adic Pn)

• Basic term P0 = p0: electric dipole P0δ(r) = p0δ(r)

• Next term P1 = P 1 = P 1s + p1 × I (symmetric + antisymmetric parts): electric quadrupole + electric loop

P1 · ∇δ(r) = P 1 · ∇δ(r) = P 1s · ∇δ(r) + p1 × ∇δ(r)

• Magnetic dipole is equivalent to an electric loop

M(r) ∼ J(r) = ∇ × M(r)/jωµ, ⇒ M = −jωµp1

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 10.09

203 Shifting operator • Shifting operator e−a·∇ produces Taylor expansion

2 3 −a·∇ (−1) (−1) e = 1 − a · ∇ + 2! (aa):(∇∇) + 3! (aaa) 3 (∇∇∇) + ··· • Taylor expansion ⇒ shifting of dipole from origin to r = a e−a·∇f(r) = f(r − a) ⇒ e−a·∇pδ(r) ∼ pδ(r − a)

• Dipole at r = a equivalent to multipole at origin: 1 pδ(r − a) ∼ pδ(r) − pa · ∇δ(r) + paa : ∇∇δ(r) + ··· 2 • Same dipole moment! In addition magnetic dipole + electric quadrupole + higher order multipoles. Second-order terms: 1 1 −pa · ∇δ(r) = (p × a) × ∇δ(r) − (pa + ap) · ∇δ(r) 2 2 • No magnetic dipole (electric current loop) if a parallel to p!

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 10.10

204 Multipole of localized source

• Element of current J(r) at r0= point source ∼ multipole at origin

0 J(r0)dV 0δ(r − r0) ∼ J(r0)dV 0e−r ·∇δ(r)

• Localized source in V = integral of elements ∼ multipole 0 J(r) = R J(r0)δ(r − r0)dV 0 ∼ R J(r0)e−r ·∇dV 0δ(r) = L(∇)δ(r) V V

0 • L(∇) = R J(r0)e−r ·∇dV 0 operator corresponding to source J(r), n = n-dot product, convergence of series assumed Z ∞ 0 −r0·∇ 0 X L(∇) = J(r )e dV = Pn n ∇∇ · · · ∇, V n=0 Z (−1)n P = J(r0)r0r0 ··· r0dV 0 n vectors r0 n n! V

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 10.11

205 Another deﬁnition of multipole • Vector potential from a small source in isotropic space (r 6= V ) Z e−jkr A(r) = µ G(r − r0)J(r0)dV 0,G(r) = 4πr V • Apply shifting operator to G(r − r0)

∞ n 0 X (−1) G(r − r0) = e−r ·∇G(r) = [∇∇ · · · ∇]G(r) n [r0r0 ··· r0] n! n=0 ∞ X (−1)n Z A(r) = µ [∇∇ · · · ∇]G(r) n [r0r0 ··· r0]J(r0)dV 0 n! n=0 V ∞ X (−1)n A(r) = µ [∇∇ · · · ∇]G(r) n P n! n n=0 • Derivatives of the Green function for diﬀerent multipole terms

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 10.12

206 Lowest-order multipole terms • Dipole term Z 0 0 P0 = J(r )dV = jωpe V

• Magnetic dipole term: antisymmetric part of P1 1 Z 1 Z P = (J(r0)r0 −r0J(r0))dV 0 = (r0 ×J(r0))dV 0 ×I = p ×I 1a 2 2 m V V

• Electric quadrupole term: symmetric part of P1 1 Z jω P = (J(r0)r0 + r0J(r0))dV 0 = Q 1s 2 2 e V

• Usually approximations do not proceed further - enough for suﬃ- ciently concentrated sources

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 10.13

207 Location of the multipole

• Field from multipole at origin Z Z A(r) = µG(r) J(r0)dV 0 − µ[∇G(r)] · r0J(r0)dV 0 + ···

V V

• Multipole shifted to r = a: Z Z A(r) = µG(r−a) J(r0)dV 0−µ[∇G(r−a)]· (r0−a)J(r0)dV 0+···

V V

= µG(r − a)p0 − µ[∇G(r − a)] · [P 1 − ap0] + ···

• Basic term (dipole p0) remains the same, other terms are changed • Best dipole approximation: choose a so that the second term is

small ⇒ to minimize norm of dyadic P 1 − ap0 (assuming p0 6= 0).

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 10.14

208 Optimal location of multipole

q ∗ PP 2 • Choose Erhard Schmidt norm kAk = A : A = |Aij| ∗ ∗ ∗ [P 1 − ap0]:[P 1 − a p0 ] = minimum

∗ ∗ ∗ ∗ • Gradient in a space = −[P 1 − a p0 ] · p0 = 0 ⇒ solve for a

P · p∗ R r0J(r0)dV 0 · R J∗(r0)dV 0 a = 1 0 = , (p 6= 0) ∗ R 0 0 R ∗ 0 0 0 p0 · p0 J(r )dV · J (r )dV

• Example: if J(r) = uJ(r), constant u (direction of current) R r0J(r0)dV 0 a = , 0center of gravity0 R J(r0)dV 0

• In general a is a complex vector

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 10.15

209 Example

−jkx • Origocentric cubic current (side = L): J(r) = uzJe , Z sin τ p = J(r0)dV 0 = u JL3 , τ = kL/2 = πL/λ 0 z τ V Z jL3 P = r0J(r0)dV 0 = u u J (cos τ − sin τ/τ) 1 x z k3 V ∗ P 1 · p0 j a = ∗ = ux (τ cos τ − 1) imaginary vector! p0 · p0 k • For a small cube, kL 1 ⇒ τ 1: τ cos τ − sin τ τ 2 τ cot τ − 1 = ≈ − sin τ 3 2 a ≈ −ux(jkL /12), small and imaginary

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 10.16

210 Radiation patterns • Exact radiation pattern in far ﬁeld obtained through integration: sin(τ(1 − sin θ cos ϕ)) sin(τ sin θ sin ϕ) sin(τ cos θ) F (θ, ϕ) = sin θ τ(1 − sin θ cos ϕ) τ sin θ sin ϕ τ cos θ • Compare with approximation by dipole at origin

F0(θ, ϕ) = sin θ, no dependence on ϕ • Dipole at complex point a:

Fa(θ, ϕ) = sin θ exp[(1 − τ cot τ)(1 − sin θ cos ϕ)] • Gives correct two-term approximation for small τ: τ 2 F (θ, ϕ) → F (θ, ϕ) → sin θ[1 − (1 − sin θ cos ϕ)] a 3 • Multipole with optimized location gives one order better approximation than origocentric multipole

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 10.17

211 Extension of multipoles • Multipole expression of a localized source g(r) corresponds to Tay- lor expansion of the operator L(∇):

Z Z 0 g(r) = δ(r − r0)g(r0)dV 0 = e−r ·∇g(r0)dV 0δ(r) = L(∇)δ(r),

Vg Vg Z Z 1 Z L(∇) = g(r0)dV 0 − g(r0)r0dV 0 · ∇ + g(r0)r0r0dV 0 : ∇∇ · · · 2! Vg Vg Vg • Conversely, multipole can be extended to an equivalent volume source if the multipole series can be expressed in terms of an analytic operator L(∇) as ∞ X Ln n (∇ · · · ∇)δ(r) = L(∇)δ(r), n=0 and if this expression can be interpreted as a function of r

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 10.18

212 Example of multipole extension

• To ﬁnd the extended source of the multipole 1 1 sinh(a · ∇) g(r) = δ(r)+ (a·∇)2δ(r)+ (a·∇)4δ(r)+··· = δ(r) 3! 5! a · ∇

• Solution: assume for simplicity that a = uza and integrate 1 1 g(r) = (ea∂z − e−a∂z )δ(r) = (δ(z + a) − δ(z − a))δ(ρ) 2a∂z 2a∂z 1 ∂ g(r) = (δ(z + a) − δ(z − a))δ(ρ) z 2a 1 g(r) = (Θ(z + a) − Θ(z − a))δ(ρ) 2a • Resulting equivalent source = line source of constant amplitude 1/2a between −a < z < a. (Θ(z) = unit step function)

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 10.19

213 Fourier transforms

• Relation between the operator expression L(∇) and the source function g(r) can be expressed as Fourier transformation

Z 0 L(∇) = ejr ·j∇g(r0)dV 0 = h(j∇)

• The inverse transformation is

1 Z 0 g(r0) = e−jr ·ph(p)dV (2π)3 p

• If the function h(p) = L(−jp) corresponding to an operator L(∇) has an inverse function in Fourier transform tables, source function g(r) can be found in analytic form.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 10.20

214 Example • For example, the previous operator gives sinh(a · ∇) sin(a · p) L(∇) = ⇒ h(p) = L(−jp) = a · ∇ a · p ∞ ∞ ∞ 1 Z Z Z sin(ap ) −jxpx −jypy −jzpz z g(r) = 3 e dpx e dpy e dpz (2π) apz −∞ −∞ −∞ • Using known results from transform tables: ∞ ∞ Z Z sin ap −jxpx −jzpz z π/a |z| < a e dpx = 2πδ(x), e dpz = apz 0 |z| > a −∞ −∞ • the previous result is obtained: 1 g(r) = δ(x)δ(y) (Θ(z + a) − Θ(z − a)) 2a

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 10.21

215 Problems

19 Find the multipole at the origin, equivalent to the line charge qi(z) component of the image %i(r) of a point charge Q at z = d, in front of a dielectric sphere of radius a and relative permittivity r:

αQi α−1 %i(r) = [−Qiδ(z − dK ) + qi(z)]δ(ρ), q(z) = (z/dK ) dK

r − 1 a 2 Qi = Q , α = 1/(r + 1), dK = a /d. r + 1 d

The image is zero outside 0 ≤ z ≤ dK . What is the best location of the multipole (no dipole term)? 20 Find the extended source equivalent to the multipole

1 2 sinh2(a · ∇) g(r) = δ(r)+ (a·∇)2δ(r)+ (a·∇)2δ(r)+··· = δ(r) 3 45 (a · ∇)2

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 10.22

216 S-96.510 Advanced Field Theory 11. Huygens’ Principle

I.V.Lindell

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 11.00

217 Truncated sources and ﬁelds • Maxwell equations in six-vector notation " !# 0 1 ξ E(r) J(r) ∇ × I − jω · = −1 0 ζ µ H(r) M(r)

[J∇ × I − jωM] · e(r) = j(r) • Huygens’ source = source on a closed surface equivalent to original sources behind the surface

• Assume closed surface S dividing all space in V1 and V2 with respective unit normal vectors n1 = −n2. Deﬁne complementary pulse functions (P1(r) + P2(r) = 1)

P1(r) = 1,P2(r) = 0, r ∈ V1,P2(r) = 1,P1(r) = 0, r ∈ V2 • Deﬁne complementary truncated sources and ﬁelds

e1,2(r) = P1,2(r)e(r), j1,2(r) = P1,2(r)j(r)

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 11.01

218 Sources of truncated ﬁelds

• Consider Maxwell equations for e1(r) = P1(r)e(r):

[J∇ × I − jωM] · e1(r) =

= P1(r)[J∇ × I − jωM] · e(r) + J∇P1(r) × e(r)

= P1(r)j(r) + J∇P1(r) × e(r) = j1(r) + jH1(r)

• Source of the truncated ﬁeld e1(r) is the truncated source j1(r) + Huygens’ source jH1(r) ∇P1(r) × H(r) jH1(r) = J∇P1(r) × e(r) = −∇P1(r) × E(r)

• Huygens’ electric and magnetic current sources for e1:

JH1(r) = ∇P1(r) × H(r), MH1(r) = −∇P1(r) × E(r)

• Similar expressions for index 2

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 11.02

219 Huygens’ sources • Gradient of the pulse function = surface delta function (assume f(r) = 0 at inﬁnity) Z Z Z f(r)∇P1(r)dV = ∇[f(r)P1(r)]dV − [∇f(r)]P1(r)dV

V1+V2 V1+V2 V1+V2 Z I = − ∇f(r)dV = − n2f(r)dS, ⇒ ∇P1(r) = n1δS(r)

V1 S • Huygens’ sources: surface sources on S

JH1(r) = n1 × H(r)δS(r), MH1(r) = −n1 × E(r)δS(r)

• Huygens’ principle: source j2(r) = j(r) − j1(r) can be replaced by jH1(r) for the ﬁeld in V1. • Huygens’ principle does not depend on the medium! Can be isotropic, anisotropic, bi-anisotropic, even nonlinear.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 11.03

220 Properties of Huygens’ sources

• Because n2 = −n1, Huygens’ sources satisfy jH2(r) = −jH1(r) • Equivalence of sources (with respect to volume in brackets)

jH1(r) ∼ j2(r), (V1), jH2(r) ∼ j1(r), (V2)

• Nonradiating sources (with respect to volume in brackets)

NR NR j2 (r) = jH2(r) + j2(r), (V1), j1 (r) = jH1(r) + j1(r), (V2)

• If all sources are in V1, j2(r) = 0

⇒ jH1(r) = NR source (V1) ⇒ JH1(r) ∼ −MH1(r)(V1)

• Absorbing boundary can be realized by NR Huygens’ source

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 11.04

221 Huygens’ principle for planar S

• If S is plane, image principle simpliﬁes Huygens’ sources

• Since j1(r) + jH1(r) is NR (V2), volume V2 can be ﬁlled with any medium, for example, PEC or PMC • PEC or PMC half space can be replaced by the corresponding image of the source j1(r) + jH1(r)

• If j1(r) = 0 (original source in V2) ⇒ jH1(r) is NR source (V2)

• In this case JH1(r) ∼ −MH1(r) with respect to V2

• From symmetry: JH1(r) ∼ +MH1(r) with respect to V1

• Huygens’ source jH1(r) ∼ 2JH1(r) ∼ 2MH1(r) with respect to V1 • Simpliﬁes problems of wave transmission through apertures in PEC plane ⇒ only electric ﬁeld in apertures needs to be solved.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 11.05

222 How to use Huygens’ principle?

• Fields from complicated sources and structures can be computed by replacing sources and structures through Huygens’ sources on a surface S • Problem: fields not known on S ⇒ Huygens’ sources not known • Principle can be applied approximatively, two main methods: 1. In some problems fields can be approximated on a surface S with good accuracy. For example, if S is a conducting plane with a thin slot, field in a slot can be easily approximated. 2. In more complex cases surface integral equations on S can be formed for unknown Huygens’ sources. Solution through numerical methods. • Forming the surface integral equation will be considered through an example of scattering by an object

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 11.06

223 Scattering problem

• Assume source j(r) = j1(r) in isotropic medium V1. Scatterer V2 is replaced by the Huygens’ source jH1(r) on its boundary S • Incident ﬁeld ei(r) gives rise to scattered ﬁeld esc(r): Z I i 0 0 0 sc 0 0 0 e (r) = − G(r−r )·j(r )dV , e (r) = − G(r−r )·jH1(r )dV

V1 S

• Integral equation for jH1(r) similar to volume integral equation?

i I 0 0 0 E(r) E (r) Gee(r − r ) Gem(r − r ) n1 × H(r ) 0 = − · 0 dS H(r) Hi(r) 0 0 −n × E(r ) Gme(r − r ) Gmm(r − r ) 1 S • When r → S, Green dyadics become too singular for surface sources! Integral equation must be reformulated.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 11.07

224 Singularity problem • Principal-value decomposition is not valid for surface sources be-

cause L · J(r) = L · Js(r)δS(r) is infinite due to the delta function • However: fields at the surface sources are finite! • For example, constant time-harmonic planar surface current J(r) = Jsδ(z) creates the plane wave η 1 E(z) = − J e−jk|z|, H(z) = − sgn(z)u × J e−jk|z| 2 s 2 z s • E(z) is symmetric and continuous, H(z) is antisymmetric and discontinuous at z = 0, both are finite.

• For planar magnetic surface current M(r) = Msδ(z): 1 1 H(z) = − M e−jk|z|, E(z) = sgn(z)u × M e−jk|z| 2η s 2 z s

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 11.08

225 Dissolution of Green dyadic • Extracting the most singular term ∇∇G of Green function. I I 1 I G(r − r0) · J (r0)dS0 = GJ dS0 + (∇0∇0G) · J dS0 s s k2 s S S S • Partial integration on closed surface S: I I I 0 0 0 0 0 0 0 0 0 (∇ ∇ G) · JsdS = ∇ · (Js∇ G)dS − (∇ · Js)∇ GdS S S S

• H ∇0 · F dS = 0 on closed surface when F is continuous I I 1 I G(r − r0) · J (r0)dS0 = GJ dS0 − (∇0 · J )∇0GdS0 s s k2 s S S S • ∇G(r − r0) is not too singular like ∇∇G(r − r0). Drawback: dif- 0 ferentiation of Js(r ) on S required ⇒ more numerical work.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 11.09

226 Fields from surface sources • Define: electric field function, gives electric field due to electric and magnetic surface sources on closed surface S: I 0 0 E[Js|Ms]S = −jωµ G(r − r )JsdS S 1 I I − (∇0 · J (r0))∇0G(r − r0)dS0 + ∇0G(r − r0) × M (r0)dS0 jω s s S S • Similarly: magnetic field function I 0 0 0 H[Js|Ms]S = −jω G(r − r )Ms(r )dS S 1 I I − (∇0 · M (r0))∇0G(r − r0)dS0 − ∇0G(r − r0) × J (r0)dS0 jωµ s s S S

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 11.10

227 Fields from surface sources 2

•E[Js|Ms]S, and H[Js|Ms]S give the radiated electric and magnetic ﬁelds E(r), H(r) when r outside S. • Questions: (1) How to compute E and H when r ∈ S? (2) What is their relation to limit r → S of E(r) and H(r) from both sides of surface?

H 0 • GJsdS can be taken in principal-value sense (extraction of ﬁeld point by a small disk) because integrand is nonsingular: G(r)dS = e−jkrdrdϕ/4π. Limit from both sides gives the same ﬁeld.

H 0 0 0 • (∇ · Js)∇ GdS can be understood in terms of surface charge: 0 ∇ · Js = −jω%s. Field discontinuity in normal component: n · (E+ − E−) = %s/. (For constant %s on plane E± = ±n%s/2.) No discontinuity for tangential components: Et = Et+ = Et− ⇒ principal-value integral when r ∈ S, because integral over small symmetric circular disk R ∇0GdS0 = H m0Gdc0 = R m0dϕ0/4π = 0. c

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 11.11

228 Field from surface sources 3

H 0 0 • ∇ G × MsdS gives discontinuous tangential ﬁeld: Et+ − Et− = n × Ms. (For a constant planar source E± = ±n × Ms/2.) When r ∈ S, small symmetric disk integrates as in previous case to = 0 ⇒ principal-value integral.

• Functions E[Js|Ms]S, H[Js|Ms]S can be applied to find the electromagnetic fields for r outside S (no problem) and for r ∈ S by interpreting integrals in principal-value sense with symmetric disk. For asymmetric disk, extra terms must be added. • Connections between values of E, H for r ∈ S and of the fields E(r), H(r) in the limit r → S from the two sides are n n E(r ) = PVE[J |M ] + 1,2 ∇ · J − 1,2 × M , 1,2 s s S 2jω s 2 s n n H(r ) = PVH[J |M ] + 1,2 ∇ · M + 1,2 × J , 1,2 s s S 2jωµ s 2 s

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 11.12

229 Huygens’ principle and scattering problem

• Huygens’ sources JH1, MH1 give scattered ﬁelds in V1 and cancel i i the incident ﬁelds E , H in V2. Points r1, r2 on each side of S:

i n1 n1 E +PVE[n1×H|−n1×E]S + 2jω ∇·(n1×H)+ 2 ×(n1×E) = E(r1),

i n1 n1 H +PVH[n1 ×H|−n1 ×E]S − 2jωµ ∇·(n1 ×E)+ 2 ×(n1 ×H) = H(r1),

i n2 n2 E +PVE[n1 ×H|−n1 ×E]S + 2jω ∇·(n1 ×H)+ 2 ×(n1 ×E) = 0,

i n2 n2 H +PVH[n1 ×H|−n1 ×E]S − 2jωµ ∇·(n1 ×E)+ 2 ×(n1 ×H) = 0. • For r ∈ S (smooth surface) gives average of ﬁelds: 1 Ei + PVE[n × H| − n × E] = E(r), 1 1 S 2 1 Hi + PVH[n × H| − n × E] = H(r). 1 1 S 2

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 11.13

230 Integral equations for PEC scatterer

• PEC scatterer in V2, boundary condition n×E = 0 on S. Unknown Js = n1 × H 1 Ei(r) + PVE[n × H|0] = E(r), 1 S 2 1 Hi(r) + PVH[n × H|0] = H(r). 1 S 2 • Taking only tangential component ⇒ integral equations on S:

i −n1 × PV E[Js|0]S = n1 × E (r), EFIE, 1 J (r) − n × PV H[J |0] = n × Hi(r), MFIE. 2 s 1 s S 1 • EFIE = electric ﬁeld integral equation = Fredholm of 1st kind • MFIE = magnetic ﬁeld integral equation = Fredholm of 2nd kind

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 11.14

231 Integral equations for PEC scatterer 2

• More explicit forms for EFIE and MFIE (principal-value integrals)

H 0 0 1 0 0 0 0 0 i jωµn1 × G(r − r )Js(r ) + k2 ∇ · Js(r )∇ G(r − r ) dS = n1 ×E (r), S 1 H 0 0 0 0 i 2 Js(r) − n1 × Js(r ) × ∇ G(r − r )dS = n1 × H (r). S

• EFIE (Fredholm of 1st kind) contains diﬀerentiation of Js ⇒ less eﬃcient in general. MFIE (Fredholm of 2nd kind) numerically better behaved. Not suitable for thin objects because Js and ∇G almost parallel, Js × ∇G small quantity ⇒ EFIE better. • Problems of uniqueness: currents of resonator modes of any amplitude can be added if frequency equals resonator frequency. (Res- onator = cavity of V2). Linear combination of EFIE and MFIE may have complex resonances ⇒ uniqueness for real ω.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 11.15

232 Cavity resonator

• Taking V2 resonator cavity, S = PEC boundary, Huygens’ source formulations with no incident ﬁeld are I 1 jωµn ×PV G(r − r0)J (r0) + ∇0 · J (r0)∇0G(r − r0) dS0 = 0, 2 s k2 s S 1 I J (r) − n × PV J (r0) × ∇0G(r − r0)dS0 = 0. 2 s 2 s S

• Solutions Js(r) 6= 0 for certain frequencies (k values, G depends on k) • Integral equation has similar form for interior and exterior problems! • Resonances can be found from the scattering problem by consider- ing singularities of the the moment matrix.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 11.16

233 Scattered ﬁelds

• After solving for the Huygens’ source Js, the scattered electric ﬁeld in V1 can be found through integration: I 0 0 0 E(r) = −jωµ G(r − r ) · Js(r )dS S

• Resonator modes in Js(r) do not aﬀect the ﬁeld E(r), because they are nonradiating sources

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 11.17

234 Dielectric scatterer

• Dielectric scatterer (homogeneous, r constant) V2 ⇒ both n × H and n × E unknown on S

• Huygens’ principle must be applied to V1 and V2 with respective Green functions (diﬀerent media 1,2)

• For example, MFIE formulation for points r1, r2 → S: 1 n × H(r ) − n × PV H [n × H| − n × E] = n × Hi, 2 1 1 1 1 1 1 S 1 1 n × H(r ) − n × PV H [n × H| − n × E] = 0, 2 2 2 2 2 2 2 S • Pair of equations for two unknown functions. • EFIE formulation in a corresponding way.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 11.18

235 Problems 21 Express Huygens’ principle in a transmission line satisfying the equations

U(z) 0 ω` U(z) u(z) ∂ = −j + , z I(z) ωc 0 I(z) i(z)

where u(z), i(z) are the voltage and current source functions. As- sume all sources are in the region z < 0 and replace the line z < 0 by Huygens’ sources at z = 0. 22 Extend the previous Huygens’ principle by assuming a pulse function P (z) as

P (z < 0) = 0,P (z > zo) = 1,P (0 < z < zo) = 1/2

and assuming sources are in the region z < 0.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 11.19

236 S-96.510 Advanced Field Theory 12. Field and Source Decompositions

I.V.Lindell

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 12.00

237 Decomposition of ﬁeld problems

• Decomposition = splitting a problem in two simpler problems • Fields of decomposed problems have restricted polarizations (example: TE and TM decomposition: axial components missing) • Decomposed fields do not ’see’ some medium parameters and they can be replaced (example · uz for a TE field with Ez = 0 can be replaced by any value) • If media of a decomposed problem can be simplified, solution becomes easier • Also some boundaries can be simplified in decomposition • Drawback: simple source may decompose to more complicated sources (example: point source = sum of two line sources)

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 12.01

238 Simple operator factorization

• Factorization of operator leads to decomposition

2 2 (∂z + k )f(z) = (∂z − jk)(∂z + jk)f(z) = g(z) 1 1 1 1 f(z) = g(z) = − g(z) (∂z − jk)(∂z + jk) 2jk ∂z − jk ∂z + jk

• General solution is decomposed: f(z) = f+(z) + f−(z)

−1 −1 (∂z − jk)f−(z) = (2jk) g(z), (∂z + jk)f+(z) = −(2jk) g(z)

• Homogeneous boundary conditions must be added. Outgoing-wave condition at z = ±∞ requires f+(−∞) = f−(∞) = 0

−jkz jkz • For g(z) = δ(z) solutions f+ ∼ e U(z) and f−(z) ∼ U(−z)e

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 12.02

239 TE/TM source decomposition in isotropic medium • What are sources of TE and TM ﬁelds and how to decompose a given source? • Helmholtz equations with scalar operators

(∇2 + k2)E(r) = jωµ(I + (1/k2)∇∇) · J(r) + ∇ × M(r)

(∇2 + k2)H(r) = jω(I + (1/k2)∇∇) · M(r) − ∇ × J(r)

• Take uz· component of equations. Ez = 0, Hz = 0 ⇒ conditions for TE/TM sources 2 TE TE jωµ(uz + (1/k )∂z∇) · J (r) − ∇ · uz × M (r) = 0 2 TM TM jω(uz + (1/k )∂z∇) · M (r) + ∇ · uz × J (r) = 0 • Simplest examples of TE/TM sources: TM TM TE TE J (r) = uzJ (r), M (r) = uzM (r)

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 12.03

240 TE/TM decomposition of dipole 1

• How to decompose a dipole J(r) = uILδ(r)?

• If u = uz ⇒ TM source, no TE ﬁeld

• How to decompose a transverse dipole with u · uz = 0 ?

TM TM • Find J (r) such that J(r) − uzJ (r) = TE source. Equation for J TM (r) from TE condition:

2 TM (k uz + ∂z∇) · [J(r) − uzJ (r)] = 0

2 2 TM 0 ⇒ (∂z + k )J (r) = ∂z∇ · J(r) = ILδ (z)u · ∇δ(ρ) • Solution in terms of one-dimensional Green function e−jk|z| IL J TM (r) = −IL∂ u · ∇δ(ρ) = sgn(z)e−jk|z|u · ∇δ(ρ) z 2jk 2

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 12.04

241 TE/TM decomposition of dipole 2

TE TM • Decomposed transverse dipole: J(r) = J (r) + uzJ (r) IL J TM (r) = sgn(z)e−jk|z|u · ∇δ(ρ) 2

• TM-component = current wave on a two-wire transmission line parallel to z axis; source = −I, wire distance = L • TE-component = original current minus TM-component = transmission-line with current wave reversed

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 12.05

242 Uniaxial Anisotropic Medium 1 • TE/TM ﬁeld decomposition possible in uniaxial anisotropic media Plane-wave relations E · B = 0 and H · D = 0 lead to E H 0 z t z z = . µz µt Et · Ht 0

• zµt − µzt = 0 ⇒ aﬃne-isotropic medium

• zµt−µzt 6= 0 ⇒, every plane wave satisfies EzHz = 0, Et·Ht = 0 • Plane waves either TE or TM to z independent of k vector • Any linear combinations (integrals) of plane waves can be decomposed to TE and TM parts • Any fields outside sources decomposable to TE and TM parts • Problem: how to decompose given sources in uniaxially anisotropic media to radiate TE/TM fields?

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 12.06

243 Uniaxial Anisotropic Medium 2 • Helmholtz determinant equations (4th order) for axial ﬁelds

(2)T −1 [detHe(∇)]Ez(r) = uz · He (∇) · [jωJ(r) + ∇ × µ · M(r)]

(2)T −1 [detHm(∇)]Hz(r) = uz · Hm (∇) · [jωM(r)∇ × · J(r)] 2 2 • Operators on both sides have common operators (kt = ω µtt)

2 detµ detHe(∇) = det detHm(∇) = ω H(∇)Hµ(∇)

(2)T 2 detµ uz · He (∇) = Hµ(∇)(uz · ∇∇ + kt uz) (2)T 2 det uz · Hm (∇) = H(∇)(uz · ∇∇ + kt uz) 2 2 H(∇) = : ∇∇ + zkt ,Hµ(∇) = µ : ∇∇ + µzkt

• Canceling operators H(∇), Hµ(∇), equations reduced to 2nd order

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 12.07

244 Uniaxial Anisotropic Medium 3

• Second-order equations for axial ﬁeld components

2 −1 H(∇)Ez(r) = (uz · ∇∇ + kt uz) · [jωJ(r) + ∇ × µ · M(r)]

2 −1 Hµ(∇)Hz(r) = (uz · ∇∇ + kt uz) · [jωM(r)∇ × · J(r)]

• Vanishing right sides produces TE and TM ﬁelds (no source for Ez,Hz). Equations for TE/TM sources

2 TE 2 TE jωµt(uz · ∇∇ + kt uz) · J (r) + kt uz · ∇ × M (r) = 0

2 TM 2 TM jωt(uz · ∇∇ + kt uz) · M (r) − kt uz · ∇ × J (r) = 0

• Same equations as for the isotropic medium with µ → µt, → t • Similar source decomposition as for the isotropic medium!

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 12.08

245 Uniaxial Anisotropic Medium 4 • Medium can be replaced by eﬀective media for decomposed ﬁelds

TE TE TE TE TE D = · E = tE = (tIt + z uzuz) · E TE TM TM TM TM B = µ · H = µtH = (µtIt + µz uzuz) · H TE TM • Parameters z , µz can be chosen so that the two eﬀective media become aﬃne isotropic: TE TM z = µzt/µt, µz = zµt/t

• Eﬀective medium dyadics (TE, µ) and (, µTM ) satisfy TE TM = tµ/µt, µ = µt/t

• Green dyadics exist is affine-isotropic media in analytic form. • Decomposed problems can be solved in effective affine-isotropic media more easily than in the original medium.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 12.09

246 Uniaxial Bi-anisotropic Medium 1

• Bianisotropic with uniaxial , µ and ξ = ξuzuz, ζ = ζuzuz

• Eliminating Et · Ht from plane-wave orthogonality conditions

E · B = 0, H · D = 0 ⇒ µtHzDz = tEzBz

• For convenience deﬁne an auxiliary parameter A D B E + ξH µ H + ζE A = z = z = z z z = z z z tEz µtHz tEz µtHz

• A can be solved by eliminating Ez/Hz ⇒ two solutions A±

(tA − z)(µtA − µz) = ξζ s 2 1 µz z 1 µz z ξζ A± = + ± + + 2 µt t 4 µt t µtt

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 12.10

247 Uniaxial Bi-anisotropic Medium 2

• The decomposed plane waves E±, H± satisfy conditions

(uz · E − Z+uz · H)(uz · E − Z−uz · H) = 0 • They are deﬁned by constant axial impedances

E± A±µt − µz ξ Z± = = = H± ζ A±t − z

• Eﬀective medium dyadics can be deﬁned with ξ± = ζ± = 0:

−1 D± = [tIt + (zξZ± uzuz] · E± = ± · E±

B± = [µtIt + (µzζZ±uzuz] · H± = µ± · E±

± = tA±, µ± = µtA±, A± = It + A±uzuz • Any field outside its source can be decomposed in ± fields with vanishing axial field functions F±(r) = uz ·[E±(r)−Z±H±(r)] = 0

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 12.11

248 Uniaxial Bi-anisotropic Medium 3

• After some algebra Helmholtz determinant can be factorized as

2 −1 −1 ω H(∇) = det[Hm(∇) · µ ] = det[He(∇) · ] = H+(∇)H−(∇) µzµtzt 2 2 2 H±(∇) = ∇t + A±(∂z + kt )

• 4th order Helmholtz equations for the axial ﬁeld functions F±(r)

−1 −1 H(∇)F±(r) = det[He(∇)· ]Ez±(r)−Z±det[Hm(∇)·µ ]Hz±(r)

u (2)T = z · H (∇) · [jωJ + (∇ × I − jωξu u ) · M] det e z z u (2)T −Z z · H (∇) · [jωM − (∇ × I − jωζu u ) · J] ± detµ m z z

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 12.12

249 Uniaxial Bi-anisotropic Medium 4

• 4th order equations for F±(r) can be reduced to 2nd order equations by eliminating operators from both sides (uniqueness of solutions assuming)

• F+(r) = 0 and F−(r) = 0 lead to conditions for sources J±, M±:

2 2 uz ·[(∇∇+kt I)·(jωµtJ± −jωtZ±M±)+kt ∇×(M± +Z±J±)] = 0

• Example: axial sources J± = uzJ±, M± = uzM± satisfying

µtJ±(r) = EtZ±M±(r)

• Given axial current J(r) = uzJ(r) can be decomposed as

Z± µt/t J±(r) = ±uz J(r), M±(r) = ±uz J(r) Z+ − Z− Z+ − Z−

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 12.13

250 Bi-anisotropic Medium 1 • The previous theory can be generalized to a certain class of bi- anisotropic media called decomposable media • Deﬁning the medium: plane wave condition becomes

E · B = 0, H · D = 0 ⇒ (a1 · E + a2 · H)(b1 · E + b2 · H) = 0

• Decomposition to a-waves and b-waves • Decomposable media must be of the form 1 T 1 = [−B + a b + b a ], ξ = x × I + [a b + b a ], 2τ 2 1 2 1 2τ 2 2 2 2 1 1 ζ = z × I + [a b + b a ], µ = [B + a b + b a ] 2η 1 1 1 1 2η 1 2 1 2 • See IEEE Trans. Ant. Prop, 46(10)1584-1585,1998 for details. • Can be generalized through duality transformation.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 12.14

251 Bi-anisotropic Medium 2 • Decomposable media can be replaced by eﬀective media for the a- and b-ﬁelds: 1 T 1 = [−B + a b − b a ], ξ = [x − a × b ] × I, a 2τ 2 1 2 1 a 2τ 2 2 1 1 ζ = [z − a × b ] × I, µ = [B + a b − b a ] a 2η 1 1 a 2η 1 2 1 2 1 T 1 = [−B − a b + b a ], ξ = [x + a × b ] × I, b 2τ 2 1 2 1 b 2τ 2 2 1 1 ζ = [z + a × b ] × I, µ = [B − a b + b a ] b 2η 1 1 b 2η 1 2 1 2

• The medium dyadics ξa, ζa, ξb, ζb are antisymmetric and T T T ηµ + τ = a1b2 + b1a2, ηµa + τa = ηµb + τb = 0 • Analytic Green dyadics can be constructed for the eﬀective media!

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 12.15

252 Bi-anisotropic Medium 3

• Helmholtz determinants are factorizable in decomposable media!

ω2ηH (∇)H (∇) H(∇) = det[µ−1·H (∇)] = det[H (∇)·−1] = − a b m e τdet detµ

2 Ha(∇) = µa :(∇ + jωza)(∇ − jωxa) − ω (η/τ)detµa 2 Hb(∇) = µb :(∇ + jωzb)(∇ − jωxb) − ω (η/τ)detµb

• Source decomposition can be done in principle as for the uniaxial case, see J. Electro Waves Appl. 13(4)429-444, 1999. • Plane waves in decomposable media have two second-order wave surfaces (instead of one 4th order surface). • Also Hertzian potentials can be generalized to decomposable media, see J. Electro Waves Appl. 15(1)3-18,2001.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 12.16

253 Problems 23 Check that the plane-wave fields in a decomposable medium defined by the medium equations on viewgraph 12.14 satisfy the conditions (a1 · E + a2 · H)(b1 · E + b2 · H) = 0 24 Find the equations for the wave-vector k in the above medium using the effective medium concept. Show that the two wave-vector surfaces for each of the effective media coincide as one surface.

I.V.Lindell: Advanced Field Theory, 2004 Helsinki University of Technology 12.17

254 Vector formulas General formulas ∇(αf(r)) = α∇f(r) ∇[f(r)g(r)] = g(r)∇f(r) + f(r)∇g(r) ∇ · [αf(r)] = α∇ · f(r) ∇ · [f(r)g(r)] = [∇f(r)] · g(r) + f(r)[∇ · g(r)] ∇ × [αf(r)] = α∇ × f(r) ∇ × [f(r)g(r)] = [∇f(r)] × g(r) + f(r)[∇ × g(r)] ∇ · [f(r) × g(r)] = [∇ × f(r)] · g(r) − f(r) · [∇ × g(r)] ∇ × [f × g] = f[∇ · g] − g[∇ · f] + [g · ∇]f − [f · ∇]g ∇ × ∇f(r) = 0 ∇ · [∇ × f(r)] = 0 ∇ × (∇ × f) = ∇(∇ · f) − ∇ · ∇f = ∇(∇ · f) − ∇2f

255 Z I ∇ · fdV = f · dS (Gauss)

V S Z I ∇ × f · dS = f · d` (Stokes)

S C

Cartesian coordinates x, y, z ∂ ∂ ∂ ∇f = u f + u f + u f x ∂x y ∂y z ∂z ∂ ∂ ∂ ∇ · f = f + f + f ∂x x ∂y y ∂z z

ux uy uz ∂ ∂ ∂ ∇ × f = ∂x ∂y ∂z fx fy fz ∂2f ∂2f ∂2f ∇2f = + + ∂x2 ∂y2 ∂z2

256 Cylindrical coordinates ρ, ϕ, z ∂ 1 ∂ ∂ ∇f = u f + u f + u f ρ ∂ρ ϕ ρ ∂ϕ z ∂z 1 ∂ 1 ∂ ∂ ∇ · f = (ρf ) + f + f ρ ∂ρ ρ ρ ∂ϕ ϕ ∂z z

uρ ρuϕ uz 1 ∂ ∂ ∂ ∇ × f = ρ ∂ρ ∂ϕ ∂z fρ ρfϕ fz 1 ∂ ∂f 1 ∂2f ∂2f ∇2f = ρ + + ρ ∂ρ ∂ρ ρ2 ∂ϕ2 ∂z2

∇ρ = ∇(r − uzz) = I − uzuz = It, ∇ · ρ = 2, ∇ × ρ = 0

ρ = |ρ| = |r − uzz|, ∇ρ = uρ 1 1 1 ∇∇ρ = ∇u = (I − u u − u u ) = u u ×u u = u u ρ ρ z z ρ ρ ρ z z× ρ ρ ρ ϕ ϕ

257 1 ∇2ρ = ∇ · u = , ∇ × u = 0 ρ ρ ρ 1 1 1 ∇ϕ = u , ∇u = − u u , ∇ · u = 0, ∇ × u = u ρ ϕ ϕ ρ ϕ ρ ϕ ϕ ρ z 1 u g (ρ) = − ln(k|ρ|), ∇g (ρ) = − ρ 2 2π 2 2πρ 1 1 ∇∇g (ρ) = PV (u u − u u ) − I δ(ρ) 2 2πρ2 ρ ρ ϕ ϕ 2 t 2 ∇ g2(ρ) = −δ(ρ) 1 1 u u ×∇∇g (ρ) = −PV (u u − u u ) − I δ(ρ) z z× 2 2πρ2 ρ ρ ϕ ϕ 2 t 1 k G (ρ) = H(2)(k|ρ|), ∇G (ρ) = −u H(2)(k|ρ|) 2 4j o 2 ρ 4j 1 k 1 ∇∇G (ρ) = −u u k2G (k|ρ|)+PV(u u −u u ) H(2)(k|ρ|)− I δ(ρ) 2 ρ ρ 2 ρ ρ ϕ ϕ 4jρ 1 2 t 2 2 ∇ G2(ρ) = I : ∇∇G2(ρ) = −k G2(ρ) − δ(ρ)

258 Spherical coordinates r, θ, ϕ ∂ 1 ∂ 1 ∂ ∇f = u f + u f + u f r ∂r θ r ∂θ ϕ r sin θ ∂ϕ 1 ∂ 1 ∂ 1 ∂ ∇ · f = (r2f ) + (sin θf ) + f r2 ∂r r r sin θ ∂θ θ r sin θ ∂ϕ ϕ

ur ruθ r sin θuϕ 1 ∂ ∂ ∂ ∇ × f = r2 sin θ ∂r ∂θ ∂ϕ fr rfθ r sin θfϕ 1 ∂ ∂f 1 ∂ ∂f 1 ∂2f ∇2f = r2 + sin θ + r2 ∂r ∂r r2 sin θ ∂θ ∂θ r2 sin2 θ ∂ϕ2 ∇r = I, ∇ · r = 3, ∇ × r = 0 1 2 ∇r = u , ∇u = (I − u u ), ∇ · u = , ∇ × u = 0 r r r r r r r r ∇(a × r) = −I × a = −a × I ∇ · (a × r) = 0, ∇ × (a × r) = 2a

259 e−jkr G(r) = , r = |r| 4πr u ∇G(r) = − r (1 + jkr)G(r) r 1 1 ∇∇G(r) = −u u k2G(r) − PV (1 + jkr)(I − 3u u )G(r) − Iδ(r) r r r2 r r 3 ∇2G(r) = I : ∇∇G(r) = −k2G(r) − δ(r)

260 Dyadic identities

In the following table, A, B, C, D denote arbitrary dyadics while S is a symmetric dyadic. a, b, c, d are arbitrary vectors and u is a unit vector.

Deﬁnitions (ab) · (cd) = (b · c)ad (ab):(cd) = (a · c)(b · d) · (ab)×(cd) = (a · c)(b × d) × (ab) · (cd) = (a × c)(b · d) × (ab)×(cd) = (a × c)(b × d) I · a = a · I = a A2 = A · A, A−2 = (A−1)2 = (A2)−1 1 A(2) = A×A, A(−2) = (A−1)(2) = (A(2))−1 2 ×

261 trA = A : I 1 spmA = trA(2) = A×A : I 2 × 1 detA = A×A : A 6 × detA 6= 0 ↔ A complete detA = 0 ↔ A planar A(2) = 0 ↔ A linear.

Identities Basic identities T T T T A : B = B : A = A : B = (A · B ): I = (A · B): I a × I = I × a antisymmetric dyadic (a × I)T = −a × I

262 T · S = S, S×I = 0, S symmetric dyadic (a × I): S = 0,

(a × I):(b × I) = (b × I):(a × I) = 2a · b

A :(a × B) = −(a × A): B = (a × B): A

A :(B × a) = −(A × a): B = (B × a): A

Double-cross products

× × A×B = B×A = h i = (A : I)(B : I) − A : BT I − (A : I)BT − (B : I)AT + [A · B + B · A]T

× T A×I = (A : I)I − A

× · A×(a × I) = a(A×I) + I × (a · A)

× I×I = 2I

263 × (a × I)×I = a × I

× S×I = −S (S symmetric, trace free)

× (a × I)×(b × I) = ab + ba

× · A×(a × I) = (a · A) × I − a(I×A)

× S×(a × I) = (S · a) × I (S symmetric)

× × (A × a)×(B × a) = (A×B) · aa

× × (a × A)×(a × B) = aa · (A×B)

× (a × I)×(a × I) = 2aa

Multiple double-cross products

× × T T A×(B×C) = (A : C)B + (A : B)C − B · A · C − C · A · B

× × I×(A×B) = (A : I)B + (B : I)A − (A · B + B · A)

264 × × I×(I×A) = A + (A : I)I

× × I×(I×I) = 4I

× × × (2) (2) (A×A)×(A×A) = 8(A ) = 8detA A (A(2))(2) = AdetA

× 2 det(A×A) = 8det A

Mixed products

× × × × (A×B) · (C×D) = (A · C)×(B · D) + (A · D)×(B · C)

× × × (A×A) · (B×B) = 2(A · B)×(A · B)

× 2 2 × 2 × (A×B) = (A )×(B ) + (A · B)×(A · B)

× 2 2 × 2 (A×A) = 2(A )×(A )

× 2 2 × × (A×I) = (A )×I + A×A

265 Inverses 1 (A×A)T · A = A · (A×A)T = (A×A : A)I × × 3 × A(2)T · A = A · A(2)T = detA I A(2)T 3(A×A)T A−1 = = × (A complete) × detA A×A : A (A×A(2)∗)T A−1 = × (planar inverse) A(2) : A(2)∗ A(2)∗ · A(2)T A−1 · A = I − (A planar) A(2) : A(2)∗ AT ×uu A−1 = × (two − dimensional inverse) spmA

−1 A · A = It = I − uu (A two − dimensional)

266 Invariants × × (A×B): C = A :(B×C) (with all permutations) 1 spmA = trA(2) = [(A : I)2 − A : AT ] 2 spm(A · B) = tr(A(2) · B(2)) = A(2) : B(2)T 1 1 1 detA = A3 : I − (A2 : I)(A : I) + (A : I)3 3 2 6 × (2) 2 det(A×A) = 8det(A ) = 8(detA) det(A · B) = detA detB

det(A · B + αI) = det(B · A + αI)

Other identities × a × (A×B) = B × (a · A) + A × (a · B)

× (A×B) × a = (A · a) × B + (B · a) × A

267 1 (A · a) × (A · b) = (A×A) · (a × b) = A(2) · (a × b) 2 × 1 (a · A) × (b · A) = (a × b) · (A×A) = (a × b) · A(2) 2 × −1 1 T a × A = A × (A · a) detA −1 1 T A × a = (a · A) × A detA

268