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S-96.510

Advanced Field TheoryCourse for graduate students

Lecture viewgraphs, fall term 2004

I.V.LindellHelsinki University of Technology

Electromagnetics LaboratoryEspoo, Finland

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

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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


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Foreword

This lecture material contains all viewgraphs associated with the gradu-ate course S-96.510 Advanced Field Theory given at the Departmentof Electrical and Communications Engineering, fall 2004. The course is

based on Chapters 16 of the book Methods for Electromagnetic FieldAnalysis(Oxford University Press 1992, 2nd edition IEEE Press, 1995,3rd printing Wiley 2002) by this author. The figures drawn by hand onthe blackboard could not be added to the present material.

Otaniemi, September 13 2004 I.V. Lindell


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S-96.510 Advanced Field Theory1. Complex Vectors and Dyadics

I.V.Lindell


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Complex Vectors

Complex vectors a= ar+ jai, (ar, ai real vectors) Time-harmonic vectors A(t) = A1cos t + A2sin t

Sense of rotation: A1

A2 shortest way

Correspondence a A(t) through two mappings Mapping a A(t)

A(t) = {aejt} =arcos t aisin t

Inverse mapping A(t) a

a= ar+ jai = A(0) jA(/2)


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Special Complex Vectors

Correspondence

a= ar+ jai A(t) = arcos t aisin t

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 iffb = eja, real

B(t) = {bejt} = {aej(t+)} =A(t + /)


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Axial Representation

To find axes of the ellipse of a complex vector a (a a = 0) Solution through another complex vector b

b= br+ jbi =|

a

a

|a a a= ej

a

a and b have same ellipse, same axes ( 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 ofa br on major axis, bi on minor axis


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Helicity Vector 1

Helicity vector p(a) (polarization vector) ofa = ar+ jai

p(a) = a aja a =

2ai ar|ar|2 + |ai|2

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)|

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Helicity Vector 2

More properties:

p(a) = a aja a =

2ai ar|ar|2 + |ai|2

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

its plane


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Vector Bases

Three complex vectors a1, a2, a3 form a basis ifa1 a2 a3= 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

a1= a2 a3a1 a2 a3 , a

2=

a3 a1a1 a2 a3 , a

3=

a1 a2a1 a2 a3

Expansion for any vector b

b=3i=1

ai(ai b) =

3i=1

ai(ai b)


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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 (Iunit dyadic, mapping to oneself)

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


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Dyadic polynomial

Dyad = dyadic product of two vectors ab =ba Dyadic = polynomial of dyads

A= a1b1+ a2b2+

+ aNbN =

N

i=1

aibi

Dyadic as a mapping

A c= (Ni=1

aibi) c=Ni=1

ai(bi c) = a1(b1 c) + + aN(bN c)

c A= c (Ni=1

aibi) =Ni=1

(c ai)bi = (c a1)b1+ + (c aN)bN

c A= AT c, transposed dyadic AT =Ni=1

biai


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Dyadic expressions

There is no unique expression for dyadics. Two expressions repre-sent 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)

A=Ni=1

aibi =Ni=1

(u1u1 ai+ u2u2 ai+ u3u3 ai)bi

=u1

Ni=1

(u1 ai)bi+ u2Ni=1

(u2 ai)bi+ u3Ni=1

(u3 ai)bi

=u1c1+ u2c2+ u3c3

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


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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 vectorsplanes and lines in complex space) Inverse mapping exists only for complete dyadics


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Symmetric dyadics

Symmetric dyadic A = AT = (1/2)(A + AT)3

i=1

aibi = 1

2

3

i=1

(aibi+biai) = 1

4

3

i=1

[(ai+bi)(ai+bi)(aibi)(aibi)]

Symmetric dyadic can be expressed as sum of symmetric dyadscici but not necessarily in three terms

Unit dyadic symmetric I = u1u1+ u2u2+ u3u3 independent ofONB

Examples: I=uxux+ uyuy+ uzuz =urur+ uu+ uu

I a= (3i=1

uiui) a=3i=1

ui(ui a)


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Antisymmetric dyadics

Antisymmetric dyadic A= AT = (1/2)(AAT) operates througha vector d(A)

Ac=3i=1

ai(bic) = 1

2

3i=1

[ai(bic)bi(aic)] = 1

2

3i=1

(biai)c= d(A)c

d(A) = 1

2

3i=1

bi ai = 12

(b1 a1+ b2 a2+ b3 a3)

Denoting a b= a (I b) = (a I) b,

A=3i=1

aibi = 1

2

3i=1

(bi ai) I=d(A) I

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


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Products of dyadics and vectors

Dot products A cand c A give vectorsA c= (

aibi) c=

ai(bi c)

c

A= c

(aibi) = (c ai)bi

Example: antisymmetric dyadic A = a I=I aA c= (a I) c= a c, c A= c (a I) =c a

Cross products A c and c A give dyadicsA c= (

aibi) c=

ai(bi c)

c A= c (

aibi) =

(c ai)bi

Note: a (b A) = (a b) A but=a (b A)!


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Dot-product of dyadics

Dot product A B gives a dyadicA B= (

i

aibi) (j

cjdj) =i,j

(bi cj)aidj

Dot product is associative but not commutative like matrix productA (B C) = (A B) C, A B=B A (in general)

Powers of dyadicsA2 =A A, An =A An1 =An1 A, A0 =I

Inverse of a dyadic possible for complete dyadics onlyA a= b, a= A1 b

(A B)1 =B1 A1


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Double-cross product of dyadics

Double-cross productAB gives a dyadicAB= (

i

aibi)(j

cjdj) =i,j

(ai cj)(bi dj)

Double-cross product is commutative but not associativeAB= B

A A

(B

C) = (AB)C (in general)

Double-cross square

A(2) = 1

2AA, I

(2) =I

Inverse of a dyadic can be expressed as

A1 =A(2)T

detA


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Double-dot product of dyadics

Double-dot product A : B gives a scalarA: B = (

i

aibi) : (j

cjdj) =i,j

(ai cj)(bi dj)

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

IfA antisymmetric and B symmetric, A : B = 0ab: A = a A b, ab: I=a b, A: B = (A BT) : I

A: I= (

aibi) : I=

ai bi = trA trI= 3, trace

A(2) :I=1

2(AA) : I= spmA

sum of principal minors

detA=1

6(AA) : A =

1

3A(2) :A determinant


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Dyadic identities

Dyadic identities are needed in dyadic analysis. They can beformed (1) through vector expansions or (2) from other identities.

Example of (1) A(BC) =? Procedure:

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

(ab)[(cd)(ef)] = [a(ce)][b(df)] = [c(ae)e(ac)][d(bf)f(bd)]

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

A(BC) = (A: C)B C AT B B AT C+ (A: B)C


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More dyadic identities

Another dyadic identity (AB) :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

Resulting identity (AB) : I = (A : I)(B : I) A : BT valid forany dyadics A and B. A new identity obtained by writing [note

that (AB) :C=A : (BC)]

A: [BI (B: I)I+ BT] = 0

A: C= 0 for all dyadics A, implies C= 0.New identity: BI= (B: I)I BT


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The inverse dyadic 1

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

A=3

i=1aibi = a1b1+ a2b2+ a3b3

The determinant and double-cross square become

detA= 1

6AA: A = a1b1

a2b2: a3b3= (a1a2 a3)(b1b2 b3)

A(2) = 1

2AA= a2b2

a3b3+ a3b3

a1b1+ a1b1

a2b2

= (a1 a2 a3)(b1 b2 b3)[a1b1+ a2b2+ a3b3] = detAi

aibi

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


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The inverse dyadic 2

Orthogonality of reciprocal bases bi bj =ij and Gibbs identityc=

aiai c give an identity

A

A(2)T = detA

i

aibi

jbja

j = detA

i

aiai = detA I

A A(2)T =A(2)T A= detA I

For complete dyadic satisfying detA = 0 the inverse becomes

A1 = A(2)T

detA=

12

ATAT

16 A

A: A

For planar dyadics detA= 0 no inverse exists


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Example of an inverse dyadic

Find the inverse ofB = I+ a I

B(2) = 1

2[(I)(I)+2(I)

(aI)+(aI)(aI)] = 2I+(aI)+aa

To evaluate we apply the propertiesI(2) =I , (a I)I= (a I)T =a I, (a I)(a I) = 2aa

(a I) : (a I) = 2a a, I :I= 3, I : (a I) = 0

detB= 1

3B(2) :B = (2 + a a)

The result becomes

B1 = 1

(2 + a a) [2I (a I) + aa]


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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 dyadicSand a vector a show that one can finda vector b satisfying

S a + a S= b I,

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


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S-96.510 Advanced Field Theory2. Dyadic Algebra

I.V.Lindell


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Dyadics and matrices 1

Dyadics can be expanded in any ONB (u1, u2, u3) asA= I A I=

uiui A

ujuj =

Aijuiuj ,

Matrix elements A

ij=A : u

iuj

depend on chosen vector basis.Some quantities (invariants) are independent of the basis:

trA= A : I=i

A: uiui =i

Aii

spmA= A(2) :I= 1

2

i,j,k,

AijAk(ui uk) (uj u)

= 1

2

i,j,k,

AijAk(ijk ijk) = 12

i,k

(AiiAkk AikAki)

=A11A22 A12A21+ = 12

[(A: I)2 A: AT]


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Dyadics and matrices 2

Third scalar invariant = determinant

detA= 1

6

AijAkAmn(ui uk um)(uj u un)

=

1

6

AijAkAmnikmjn = det(Aij)

Dot-product of dyadics corresponds to the matrix product:A B=

AijBkuiuj uku =

i,

uiuj

AijBj

Double-cross product = mixed subdeterminant in matrix algebraAB=

i,j,k,

AijBk(ui uk)(uj u)

=u1u1(A22B33+ A33B22 A23B32 A32B23) +


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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 termstransforms the quadratic identity to a bilinear one:

(AB) a= (A a) B+ (B a) A

Proof of the identity quadratic in A through identity linear inA, B(bcde)a= (bd)[e(ca)c(ea)] = (bca)(de)+(dea)(bc)


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Classification of dyadics

Ais a complete dyadic iff detA = 0, A1 exists

Ais a planar dyadic iff detA= 0 or exists a = 0, A a= 0

Ais a lineardyadic iffA(2)

= 0 or exists a = 0,A a= 0identity A(2) a= (A a) A

conclusion: ifA planar, A(2) linear

identity (A(2))(2) =AdetA= A

detA(2)

conclusion: ifA(2) planar it is also linear


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Eigenvalue problems

Eigenvalues and (right) eigenvectors a of a dyadicAsatisfyA a= a, a = 0, (A I) a= 0

A

Iplanar

det(A

I) = 0 3rd degree equation for

1

6(A I)(A I) : (A I) = detA spmA + 2trA 3 = 0

Three roots satisfy the conditionstrA= 1 +2 +3, spmA= 12 +23 +31, detA= 123

General case: three different eigenvalues 1, 2, 3 Two eigenvalues same ifAof the formI+bc(generalized uniaxial

dyadic)

Three eigenvalues same ifA of the form I(isotropic dyadic)


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Finding the eigenvectors

Single eigenvalues i eigenvectors ai from

(A iI) ai = 0, (A iI)(2) ai = 0

(A iI)(2) is a linear dyadic of the formbiai (bi left eigenvector)

ai = c (A iI)(2), when = 0

For generalized uniaxial dyadic A = I+ bc double eigenvalue2,3 = , A I = bc linear dyadic, (A I)(2) = 0 anyvectors a2,3 satisfying c a2,3= 0 are eigenvectors

For isotropic dyadic A = Itriple eigenvalue 1 = 2 = 3 = ,all vectors are eigenvectors


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Uniaxial dyadics 1

Uniaxial dyadics (symmetric) are encountered in medium equationsand interface conditions. Axial real unit vector u, It = I uu(Notation different from that in the book!)

U(, ) = It + uu, [book :D(, ) = I+ uu= U(, + )]

Matrix notation : U(, ) 0 00 0

0 0

Uniaxial dyadics form a two-dimensional linear space:U(1, 1) + U(2, 2) = U(1+ 2, 1+ 2),

U(, ) =U(,)

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

Eigensolutions a1= u, 1= and a2,3u, 2,3=


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Uniaxial dyadics 2

Properties of symmetric uniaxial dyadics

U(1, 1) U(2, 2) = U(12, 12),

Un

(, ) = U(n

, n

), U1

(, ) = U(1

, 1

),

U(1, 1)U(2, 2) = U(12+21, 212) U(2)(, ) =U(,2)

U(1, 1) : U(2, 2) = 212+ 12, detU(, ) =2

Checking the inverse

U1(, ) =U(2)T(, )

detU(, )=

U(,2)

2 =U(1, 1)

Generalized uniaxial dyadics I+ bc obey more complicated rules


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Reflection dyadic

Reflection dyadic Cgiving mirror image of vector a in the planeu r= 0 is uniaxial:

C a= a 2u(u a) = (I 2uu) a,

C=It uu= U(1, 1), 1 0 00 1 0

0 0 1

Properties of the reflection dyadic

trC= 1, spmC= 1, detC= 1, C(2) = C, C1 =C

C2 =I , C=I1/2

Square root of a dyadic is not unique! Eigensolutions1= 1, a1= u, 2,3= 1, a2,3u


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Gyrotropic dyadics 1

Gyrotropic dyadic = uniaxial dyadic + co-axial antisymmetric dyadic(Notation in the book again obtained by substituting + )

G(,,) = It+ uu + J

J=uI, J2 = (uI)(uI) = u(uI) = It, J3 = J,

G(,,) 0 0

0 0

, G(,, 0) = U(, )

Gyrotropic dyadics are encountered in magnetoplasmas and ferrites

r =

1 g

2p

(2g 2)

It+

1

2p

2

uu +

jg2p(2g 2)

J

r =

1 om

2o 2

It+ uu +j m2o 2

J


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Gyrotropic dyadics 2

Two-dimensional (planar) part of gyrotropic dyadic can be ex-pressed as

Gt(, ) =It+ J=GeJ,

eJ defined here as two-dimensional exponential function

eJ =It+ J+2

2!J

2+

3

3!J3 + = cos It+ sin J

Planar rotation defined through dyadic Rt() as

Rt() = cos It+ sin J=eJ

Gt(, ) = GRt(), G=

2 + 2, tan = /

Note the similarity to operating with complex numbers!


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Eigenvalues of a gyrotropic dyadic

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

uu= J , It

J= 0,

J :J= 2, It: It = 2, It: J= 0, I

(2)

t =J

(2)

=uu

lead to the following properties of the gyrotropic dyadicG(2)(,,) = G(,2+2, ), spmG= trG(2) = 2+2+2,

detG(,,) = (2 + 2), trG(,,) = 2 +

Eigenvalues from the characteristic equationdet(GI) = detGspmG +trG 23 = ()(()2 +2) = 0

solutions 1= , 2,3= j Special case uniaxial medium: = 0 2= 3=


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Eigenvectors of a gyrotropic dyadic

Eigenvector a1 corresponding to1= obtained from(G1I)(2) = (()It+J)(2) = (()2 +2)uu, a1= u,

Eigenvectors a2,3 corresponding to 2,3 =

j with

= 0 ob-

tained from

(G 2,3I)(2) = j( j)(It jJ)

Expanding dyadics It jJin an ONB (v, w, u= v w) as

It j(v w) I=vv + ww j(wv vw) = (v jw)(v jw)

(G 2,3I)(2) = j( j)(v jw)(v jw)

Eigenvectors a2,3= v jw= (IjJ) v are circularly polarizedv can be chosen as any vectoru


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Problems

2.1 Derive the inverse of the dyadicab + c Ithrough vector algebraby 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 + cdwhose vectorsa dare orthogonal to a unit vector n by studyingthe expression A (ATnn) 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 ex-

pression for the inverse of the dyadic B = A+ab and check theresult.


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S-96.510 Advanced Field Theory3. Basic Electromagnetic Equations

I.V.Lindell


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Electromagnetic quantities

Sinusoidal time dependence ejt 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

B

=

E

H


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Classification of electromagnetic media

Medium dyadics for different types of media:

Isotropic = I, = 0, = 0, = I

Bi-isotropic = I, = I, = I, = I Anisotropic , = 0, = 0,

Bi-anisotropic , , , Homogeneous: constant medium dyadics Inhomogeneous: (r), Dispersive: (), Lossless, reciprocal, etc...


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Examples of artificial media

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

D= oE + Pe, B= oH + Pm

Pe = ee E + em H, Pm = me E + mm H

Metal needles em= me= mm= 0, dielectric medium Metal rings em= me= 0, dielectric-magnetic medium Metal helices, Pe and Pm parallel, chiral medium Omega-shaped particles, Pe Pm= 0, omega medium


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The Maxwell Equations

For time-harmonic fields only curl equations needed.Homogeneous medium assumed: constant parameter dyadics

Isotropic

H(r) = jE(r) + J(r)

E(r) = jH(r) + M(r) Anisotropic

H(r) =j E(r) + J(r) E(r) = j H(r) + M(r)

Bi-anisotropic

H(r) = j E(r) +j H(r) + J(r)

E(r) = j H(r) +j E(r) + M(r)


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Operator notation

Maxwell equations in terms of six-vectors and six-dyadics 0 I

I 0

E

H

j

E

H

=

J

M

More compact notation through the Maxwell six-dyadic operator

L() e(r) = j(r),

L() =

j ( Ij )( I+ j) j

Electric and magnetic fields are coupled in the Maxwell equationsdecoupling through elimination or operator diagonalization


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Diagonalization of operators

Decoupling through adjoint Maxwell operators

La() =

jI ( Ij) 1

(

I+ j)

1 jI

La() =

jI 1 ( Ij)1 ( I+ j) jI

Diagonalization of the six-dyadic operator

La() L() =L() La() =

He() 00 Hm()

Dyadic Helmholtz operators He(),Hm()


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Helmholtz operators

Dyadic Helmholtz operators are of second orderHe() = ( Ij) 1 ( I+ j) + 2

Hm(

) =

(

I+ j )

1

(

I

j) + 2

Special cases: anisotropic mediumHe() = 1 + 2, Hm() = 1 + 2

Bi-isotropic medium, k =

H() = He() = Hm() = (IjI)(I+jI)+k2I

Isotropic medium, k 2 =2H() = He() = Hm() = I + k2I


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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)

He() E(r) =j J(r) + ( Ij) 1 M(r),

Hm() H(r) = jM(r) ( I+ j) 1 J(r)

Operating by H(2)Te,m() gives equations with scalar Helmholtz de-terminant operators detHe,m() (4th order)

detHe() E(r) = He(2)T() [jJ(r) + ( Ij) 1 M(r)]

detHm() H(r) =H(2)T

m () [jM(r) ( I+j) 1 J(r)]


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Potentials

Expressing the field six-dyadic as e(r) = La() f(r) in terms oftwo vector potentials f= (F G)

E(r) =j F(r) + 1 ( Ij) G(r),

H(r) = 1 ( I+ j) F(r) +jG(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)

detHe() F(r) =j H(2)Te ()J(r), detHm() G(r) =j H(2)T

m ()M(r)


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Helmholtz operators for isotropic medium 1

Helmholtz dyadic operator

H() = I + k2I= (2 + k2)I

Helmholtz adjoint operator

H(2)T() = 12

(I + k2I)(I + k2I)

= 2 + k2(22I I) + k4I= (2 + k2)( + k2I)

Helmholtz determinant operator

detH() = 13

H(2)() : H() =k2(2 + k2)2


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Helmholtz equation for isotropic medium 2

Helmholtz determinant equation (fourth order)

3detHe()E(r) = detH()E(r) = k2(2 + k2)2E(r)

=

2

H

(2)T

e () [jJ(r) + M(r)]= (2 + k2)( + k2I) [jJ(r) + M(r)]

reduces to a second-order equation (uniqueness of solution assumed!)

(2 + k2)E(r) = (I+ 1k2

) [jJ(r) + M(r)]

Simpler equation for vector potential E = j(I+ /k2) A(r)

(2 + k2)A(r) = J(r) + j

M(r)


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Helmholtz operators for bi-isotropic medium 1

Denote = (r jr), = (r+ jr)r= relative chirality parameter, r= relative Tellegen parameter

Dyadic Helmholtz operators factorizable in bi-isotropic medium:

He() =Hm() =H() = ( I krI)2 + k2(1 2r)I

= L+() L() = L() L+()

Two auxiliary first-order operators L()

L+() = I k+I, L() = I+ kI

k+= (

1 2r+ r)k, k = (

1 2r r)k

Special case: isotropic medium, k = k, L() = I kI


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Helmholtz operators for bi-isotropic medium 2

Auxiliary operators can be evaluated as

L(2)

() = k I+ k2I, detL() = k(2 + k2)

Product can be expanded as a sum

L(2)+ () L(2) () = k

2 + k2k++ k

L(2)+ () + k+

2 + k2+k++ k

L(2)

()

Using (A B)(2) =A(2) B(2) and det(A B) = detAdetB gives

[He()](2) =k2 + k2k++ k

L(2)+ () + k+

2 + k2+k++ k

L(2) ()

det(He()) = k+k(2 + k2+)(2 + k2)


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Helmholtz equation for bi-isotropic medium

The Helmholtz equations have the form

He() E(r) = jJ(r) + ( IjI) M(r) = g(r)

det[He(

)]E(r) = [He(

)](2)T

g(r)

The Helmholtz determinant equation can be expanded to

(2+k2+)(2+k2

)E(r) = [

2 + k2

k+(k++ k)L(2)T+ ()+

2 + k2+k

(k++ k)L(2)T

()]g(r)

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

(2 + k2)E(r) = 1

k(k++ k)L

(2)T

() g(r)

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


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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 r)

E(r) =V

G(r r) J(r)dV r / V

Problems involving inhomogeneous media and/or boundaries oftenhandled through integral equations

V

G(r r) J(r)dV =E(r) = known

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


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Discontinuities in fields 1

Heaviside unit step U(z) generates the delta function (z)U(z) = 0, z 0, zU(z) = (z)

Step discontinuity in function F(r) at surfaceSseparating regions

V1 andV2 generates the surface delta function s(r)

F(r) = F1, r V1, F(r) = F2, r V2,F(r) = n2(F2F1)s(r) = n1(F1F2)s(r) = (n2F2+n1F1)s(r)

Integration of surface delta V

g(r)s(r)dV =S

g(r)dS

Iff(r) discontinuous on surface S:f(r) = {f(r)}cont+(sf(r))s(r), sf(r) = n1f(r1)+n2f(r2)

cont = no delta discontinuity,s = surface nabla operator


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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)

s-discontinuous sources create step-discontinuous fields: E= {E}cont +(sE)s(r) = jB{M}contMss(r) H= { H}cont+ (s H)s(r) = j D + {J}cont+ Jss(r)

Equating s-discontinuous terms gives conditions on interfacess E= n1 E1+ n2 E2= Mss H= n1 H1+ n2 H2= Js

s D= n1 D1+ n2 D2= ss B= n1 B1+ n2 B2= ms


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Boundary conditions 1

Boundary: fields vanish on side 2 of surface Ss 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

Et = Zs(nHt), Ht = (nnZs)1(nEt) = (spmZs)1ZTs (nEt) Two-dimensional inverse! (spm = two-dimensional determinant)

At (ATt nn) = (spmAt)It, A1t =ATt nn/spmAt


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Boundary conditions 2

Local tangential ONB{v, w= n v} on SZs= 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

Zs = Z(vv + 1vvnn) = Z(vv +

1ww)

Limit 0 = soft-and-hard surface (SHS) = perfect anisotropicsurface, gives symmetric boundary conditions v E= 0, v H= 0

Generalized SHS boundary (conditions a E= 0, b H= 0)Zs= Z(ba +

1abnn), a b= 1, 0


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Interface conditions

InterfaceS between regions 1,2. Conditions for tangential fields E1tE2t

=

Z11 Z12

Z21 Z22

n1 H1n2 H2

ForZ12= 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| <

Js= j( o)tEt, Zs= 1j( o)t = j

o(r 1)kot


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Problems

3.1 Assuming uniaxial anisotropic medium with medium dyadics

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

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

(2)(2) =tt(z + z)

3.2 Assuming an anisotropic medium whose medium dyadics satisfythe relation = T where is a scalar, show that the inverse ofthe Helmholtz dyadic operator can be expressed in the simple form

H1e () = L()/L(),where the dyadic operator L() and the scalar operator L() areboth of the second order. Hint: use result of Problem 2.3.


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S-96.510 Advanced Field Theory4. Conditions for fields and media

I.V.Lindell


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Uniqueness

Linear differential equation and boundary conditionsL()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

Uniquef(r) if homogeneous (sourceless) problem has only the nullsolution

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


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Eigenvalue problem

Example: sourceless Maxwell equations 0 I

I 0

Eo(r)Ho(r)

=j

Eo(r)Ho(r)

Of the form of an eigenvalue equation, =eigenvalue parameter

L() fo(r) =j M fo(r),

Impedance conditions on surface S:

Eot(r) =Zs (n Ho(r)), r S

If = i in the eigenvalue spectrum{j}, solution fo(r)= 0multiple of the eigenvector foi(r), otherwise fo(r) = 0


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Example: Resonance cavity

Closed PEC surface Swith Zs= 0 Real resonance frequencies 1, 2, , resonance modes E1(r), E2(r),

satisfy the homogeneous equation + boundary conditions

Mathematics: source problem nonunique at resonance = i Physics: without losses fields become infinite at resonance Reality: losses make{i} complex = i, uniqueness at real

frequencies

Practice: if losses small, trouble in numerical computation (almostnonunique due to roundoff errors)

For some problems uniqueness can be proved through uniquenesstheorems of the form|| =, (not positive real) = = 0


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Uniqueness in electrostatics 1

f = (r) potential, g =(r)/ charge, assume isotropic volumeV bounded by surface S

2f=g, 2fo = 0, in V

0 =V

fo(2fo)dV =V

(fo fo)dVV

|fo|2dV

Gauss law V

|fo|2dV =S

fo(n fo)dS

Different boundary conditions on S making the surface integralvanish will ensure uniqueness for the electrostatic field problem

V

|fo|2dV = 0, fo= 0 field vanishes


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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 (onS1) and Neumann (onSS1) gives uniqueness What about condition fo+ n fo = 0 on S?

S

|fo+nfo|2dS= ||2

S

|fo|2dS+||2

S

|nfo|2dS+2{}

V

|fo|2dV = 0

If{} >0, fo = 0 in V Impedance boundary condition of the form f+n f = s

onSgives uniqueness for{} >0


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Uniqueness in electrodynamics (Example)

Uniqueness for lossy isotropic medium with lossless boundary atreal frequencies when complex (complex resonance frequencies)

S

n

Eo

Ho dS=

V (Eo

Ho )dV =

j

V |Ho

|2dV+j

V |Eo

|2dV

AssumeS ideal boundary S

n Eo Ho dS= 0

{} = 0, V

|Ho|2dV = ||2V

|Eo|2dV Eo = 0, Ho= 0

Uniqueness for the resonator with lossless medium and boundariesonly when frequency not one of eigenfrequencies, =i.


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Power conditions for medium parameters

Complex Poynting vector for time-harmonic fields:

S=1

2E H

{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

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Power exchange

Power created in the medium

P = {S} = 12{( E) H ( H) E}

= 12{jB H +j D E} =

2{E D + H B}

In six-vector notation

P =

2{e M e}, e=

E

H

, M=

Lossless medium: P= 0 for all e Lossy (passive) medium: P 0 for some field e1


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Lossless bi-anisotropic medium

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

2j{eMe} =eMeeMe = e (MTM)e = 0 for alle

To prove: e A e = 0 for all eimplies 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 MT =M


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Parameters of lossless medium

Medium six-dyadic Hermitian for lossless media

T=

T T

T T

=

T =, T =, T = ,

and Hermitian dyadics and , a Hermitian pair of dyadics

Define= ( j)oo, = ( +j)oo,

T =, T =

Lossless bi-anisotropic medium: ,,,are Hermitian dyadics


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Examples of lossless media

Bi-isotropic medium: ,,,are all real for a lossless medium= , = , = , =

Gyrotropic medium: ,,, are of the general gyrotropic form

G= Gzuzuz+ GtIt+ Gguz I Lossless medium: gyrotropic dyadics are Hermitian GT =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 withsymmetric part S real and antisymmetric part a I imaginary


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Lossy medium

Condition for lossy media:{ S}

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Surface impedance conditions

Impedance condition Et= Zs (n H), or n H= Ys Et Power flow from impedance surface to the field (direction n)

P = 1

2{n

E

H

}=

1

2{E

n

H

}= 1

2{E Ys E} = 1

2E YsH E

YsH= 12 (Ys+ YTs ), Hermitian part ofYs Lossless surface: P = 0 YsH= 0, Ys antihermitian dyadic Example: Ys symmetric, Ys= jBs imaginary (reactive surface) Lossy medium: P

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Ideal boundary

Ideal boundary: Poynting vector has no normal component

n S= 12

n E H= 0

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 thata E= 0, a H= 0

Generalized SHS, for a= a= v gives real classical SHS Isotropicideal surface: exists a (complex) scalar Z such that

Et = ZHt, not a linear condition!


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Energy condition

Lossless, nondispersive medium assumed (otherwise complicated) Condition: energy stored in the medium Wpositive for all fields

W = 1

4

(E H)

E

H

=

1

4

e

M

e> 0

MHermitian, condition requires: Mpos.definite M1 PD

M1 =

( 1 )1 1 ( 1 )1

1 ( 1 )1 ( 1 )1

dyadics , , 1 , 1 must all be PD Example: bi-isotropic medium (lossless , , , real)

> 0, > 0, < 2 + 2 < /oo Condition limits magnitudes of, parameters


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Reciprocity conditions 1

Reaction of source and field defined as [Rumsey 1954]

=

(E1 H1)

J2M2

dV

Medium reciprocal when reaction is symmetric < 1, 2>=0 =

(E1 H1)

J2M2

dV

(E2 H2)

J1M1

dV

Replace sources from the Maxwell equations, apply Gauss lawS

n(E1H2E2H1)dS= j

V

(E1 H1)

T + T

T +T

E2

H2

dV

Must be valid for any fields in any volume Vbounded by S. Inte-grals must vanish separately conditions for medium and bound-ary parameters.


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Reciprocity conditions 2

Volume integral vanishes for any fields under reciprocity conditionsfor the medium:

= T, = T, = T, = T

For example, bi-isotropic medium reciprocal if= , = 0.Tellegen parameter = nonreciprocity parameter

When is medium with gyrotropic dyadics reciprocal? , symmet-ricg =g = 0, = T u = u, t = t, g =g

Surface integral vanishes under reciprocity condition for symmetricsurface impedance dyadic

Zs= ZTs


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Reciprocal and lossless media

Conditions for a medium being both lossless and reciprocal: T T

T T

=

=

T T

T T

The permittivity and permeability dyadics satisfy = T =,= T =

or and are real and symmetric dyadics

The magnetoelectric dyadics satisfy =

= and= = , whence andare imaginary dyadicssatisfyingT = .

Writing = +j, = j, must be real and symmetricand imaginary and antisymmetric.


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Problems

4.1 Derive the condition for the imaginary parts of the medium pa-rameters of a lossy bi-isotropic chiral medium:

2im < imim

oo

by requiring that the condition {S}

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S-96.510 Advanced Field Theory05. Field Transformations

Duality transformation

I.V.Lindell


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Field transformations

Electromagnetic transformations make it possible to find solutionsto 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 fieldsinduces transformation of sources and media

Affine transformation: linear transformation of spaceinduces transformation of fields, sources and media


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Simple duality (Duality substitution)

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

M1 : H= jE + M

M2 : E= j H + J Replace

E H, H E, , , M J, J M

Maxwell equations invariant: M1M2, M2M1 Not a transformation, dimensions are changed Can be used for transforming formulas


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Example: transformation of formulas

Electric field radiated by electric current

E(r) = j

G(r r) J(r)dV

Duality substitution:

E H, H E, , , M J, J M

Green dyadic G(r r) depends on k= , which is invariant Magnetic field from magnetic current

H(r) = j

G(r r) M(r)dV

Electric field obtained through

E(r) = H(r)

j =

G(r r) M(r)dV


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Classical duality

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

Ed= BH, Bd= BD, Md= BJ

Transforms one Maxwell equation to another: Ed= jBd+ Md H= jD + J

Similarly

Hd= CE, Dd= CB, Jd = CM

Hd= jDd+ Jd E= j B + M

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


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Medium transformation

Transformation for medium dyadics

DdBd

=

C H + C EB E + B H

=

C/B

B/C

EdHd

Transformed medium dyadics d dd d

=

C/B

B/C

Dual fields and sources exist in dual medium! Example: dual of isotropic medium = another isotropic medium:

d= CB

, d = BC


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Choice of parameters

Duality transformation works for any two parameters B , C= 0: EdHd

=

0 BC 0

E

H

,

Usually B , Cchosen to satisfy two conditions: (1) Duality transformation in an involution (equals its inverse)

EdHd

d

=

0 BC 0

E

H

d

=BC

E

H

BC= 1,

(2) Free-space medium = oI, = oIand= = 0 is self dual:d= = C/B and d= = B/C

o= oC/B, B/C= o/o= 2o Two possible solutions: B = 1/C = jo= j

o/o


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Two duality transformations

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

EdHd =

j 0 o1/o 0

EH Medium transformation rules same for both cases: d =d etc

d dd d

=

/2o

2o

=

o(/o)

o(/o)

Relative dual permittivity and permeability: rd = r, rd = r Isotropic medium: wave number: kd = dd= oror =k Wave impedance d=

d/d=

or/or = 2o/


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Example

Incident plane-wave Ei = uAejkr, Hi = vBejkr scatteringfrom a dielectric object (r =, r = 1)scattered fields Es, Hs

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

Eid+ = joHi = jovBejkr,

Hid+ = Ei/jo = u(A/jo)ejkr

Scattering from a dual object (rd = , rd = 1) Scattered fields Esd+ =joHs and Hsd+ =Es/jo are dual to

the original scattered fields

If incident field self dual same incident field in dual problem


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Self-dual quantities

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

E+= (E+)d+ = joH+, H+= (H+)d+ = E+/jo Self-dual fields in left-hand transformation

E = (E)d =joH, H = (H)d =E/jo

Self-dual decomposition of any field:

E= E++ E, E =1

2(E joH),

H= H++ H, H =1

2(H 1

joE)

Decomposition of sources: J = (J M/jo)/2


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Self-dual plane wave

Plane wave in air E = Eoejkur, H= 1o u Eoejkur (realu) Self dual in right-hand transformation:

E= Ed+

Eo =

joHo=

ju

Eo

Satisfies u Eo= 0 and Eo Eo= 0 (circular polarization)

p(Eo) = Eo EojEo Eo

=j(u Eo) Eo

j|Eo|2 =u

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

Decomposition E= E++ E, Transformations Ed+ =E+ E, Ed = E++ E


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Surface impedance condition

Surface impedance condition (Zs = 2D dyadic)Et = Zs (n H)

Applying 2D inverse of a 2D dyadic AA1 =

ATnn

spmA, A A1 =A1 A= It

n H= Z1s Et = ZTs

nn

spmZs Et = n Z

Ts

spmZs (n E)

Another form of the same impedance condition:

Ht= ZTs

spmZs (n E)


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Dual surface impedance

Duality transformations ofEt= Zs (n H):Edt = Zsd (n Hd)

joHt = 1

jo Zsd (n E) = jo

ZTs

spmZs (n E)

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

Zsd= 2o

ZTs

spmZs

Another form follows from spm(Zsnn) = spmZsZsd=

2o(Zs

nn)

1


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Self-dual surface impedance

Self-dual condition:

Zs= Zsd=

2o

spmZs

ZTs spmZs =

4o

spmZs

Two conditions: spmZs = 2o ZTs = Zs Antisymmetric (nonreciprocal) solution Zs= jon I Symmetric (reciprocal) surface impedance in terms of ONB (v, w, n)

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

Zs= o(vv + 1ww)

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


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Self-dual media

Free space = self dual medium. Other possibilities? d dd d

=

/2o

2o

=

Denote = ( j)oo, = ( +j)oo

r = r =, = = 0,

Self-dual medium can be defined in terms of two arbitrary dyadics:

=

o joo

j

oo o

More general duality transformation more general self-dual medium


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General duality transformation

General duality transformation T for the fields (six-vector form)

ed = Te, e=

E

H

, T =

A BC D

, AD BC= 0,

Transformed Maxwell equations can be expanded as

J ed= jMd ed+jd, J=

0 11 0

JT e= [JT J1]J e= j MdT e+jdJ e= j[JT1J1]MdT e+ [JT1J1]jd

Identify dual sources from j= [JT1J1]jd: JdMd

=J T J1j=

D CB A

J

M


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Duality transformation of media

Identify dual medium six-dyadic:M= [JT1J1MdT], Md = J T J1MT1

d

ddd

= 1AD BC D

2

CD CD C2

BD AD BC ACBD BC AD AC

B2 AB AB A2

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

d= , d= , d= , d=

Four conditions reduce to only two


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Self-dual media

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

(A D) + C(+ ) = 0, (D A) + B(+ ) = 0

Define , = (r jr), = / , and must be multiples of the same dyadic, say,

(PreviouslyA = D = 0, whence = 0)

= 2jr is always invariant: rd = r General self-dual medium: medium dyadics have the form

=

r

r

+

0 j

j

0

r


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Special case: involutory duality

Require: T1 =T , T2 =I A= D= 1 BC Assuming self-dual medium with parameters , , r = sin , (, r

do not matter), transformation matrix Tcan be solved:

D= A= jr/1 2r = j tan B/ = C=A/r = j/ cos

T = j tan / cos

1/ cos tan

, (= o, = 0 classical duality)

Self-dual fields: E= E++ E, H= H++ H

T

EH

=

EH

Relation E = ZH, wave impedances Z = jej


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Self-dual (Bohren) decomposition 1

Field decomposition in self-dual parts E= E++E, H= H++H E

H

=

1 1

ej/j ej/j

E+E

=K

E+E

,

E+E

= 1

2cos

ej

jej j

EH

=K1

EH

Maxwell equations are decoupled for homogeneous self-dual media:

E+E

=K1

E

H

=K1

j j

j j

K

E+E

+ K1

MJ

=

k+ 0

0 k

E+E

M+M

, = r cos


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Self-dual (Bohren) decomposition 2

Maxwell equations for self-dual fields E = k E M

M =

1

2cos (e

jMj

J) = je

jJ

They can also be expressed as four equations (twice the same) E = j H+ M, H = j E+ J

Equivalent medium dyadics for the decomposed fields

= kjZ

= ej, =

k

jZ = e

j

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


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Power propagation in self-dual medium

Assume homogeneous and lossless self-dual medium: ,,,,rreal

Propagating power = real part of Poynting vector1

2{E H} =1

2{(E++ E)

E+Z+

+E

Z

} =1

2{jE+E+ej+jEEej+j[E+Eej+E+Eej ]}

= cos [p(E+)|E+|2

2 p(E) |E|

2

2 ], Z = jej

Decomposed fields do not couple power to one another! p(E+) right-hand polarization,p(E) left-hand polarization


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Problems

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

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

5.2 Study the reflection of a plane wave Eo

ejkoz propagating in air

from a self-dual surface impedance Zs = o(uxux+uyuy/) at

z = 0. Find the two-dimensional reflection dyadic R giving the

reflected field as R Eoejkoz. Show that the handedness of thewave is not changed in the reflection.


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S-96.510 Advanced Field Theory6. Affine transformation

I.V.Lindell


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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 = 0)

space r ra= A r affine space

Property of nabla operatorr= xux+ yuy+ zuz =uxux+ uyuy+ uzuz =I

Nabla operator in affine space

I= ara = a(A r) =A1T r AT, a = A

1T


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Basic tools

Applying six-vector representation

e=

E

H

, d=

D

B

, j=

J

M

, M=

, J=

0 11 0

Original Maxwell equationsJ e(r) = jd(r) +j(r), d= M e

are affine-transformed toJa ea(ra) = jda(ra) +ja(ra), da = Ma ea

Analytic tools needed: identity

(D a) (D b) = D(2) (a b) = (detD)D1T (a b)

Dyadic relation [D1](2) =D(2) =DT/detD


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Affine transformation of fields and sources

Transforming Maxwell equations

(D ) (D e(r)) = D(2) ( e(r)) = D(2) [jd(r) +j(r)]

Becausea = A1T , choose D = A1T which givesJa ea(ra) =j da(ra) +ja(ra)

Identify term by term transformed fields and sources:

ea(ra) = D e(r) =A1T e(A1 ra)

da(ra) =D(2) d(r) =A(2)T d(A1 ra) =A d(A

1 ra)/detA

ja(ra) = D(2) j(r) =A(2)T j(A1 ra) =A j(A

1 ra)/detA


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Affine transformation of media

Medium equations

Ma ea= da= AdetA

d= AdetA

M e= AdetA

M AT ea

Identify the medium six-dyadic:

Ma = A M AT

detA

Transformation rules for all medium dyadics are similar:

a =A AT

detA, a=

A AT

detA,

a =A AT

detA, a=

A AT

detA,


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Rotation transformation

Rotation transformation around an axis defined by u

A= R() =uu + cos It+ sin u I=euI, R(0) =I

RT

() =R1

() =R(

), R

(2)() = R(), detR() = 1

Field vectors and source vectors are transformed asEa(ra) = R() E(R() r), Ja(ra) = R() J(R() r)

All medium parameter dyadics are transformed as

a= A AT

/detA= R() R() =

(R() ai)(R() bi)

R() I R() =I bi-isotropic media are invariant in rotation Rotation transformation is not very interesting! General transfor-

mationA = S R. Interest in symmetric transformations A= S.


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Uniaxial transformation

Uniaxial transformation: stretching in uz anduz directions

S= It+ uzuz, S1

=1It+ 1uzuz, detS=

2

Field vectors are transformed asEa(ra) = S

1 E(S1 ra) = [1Et+ 1uzEz](1 + 1uzz)

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

a = 1It It+1[It uzuz+uzuz It]+2uzuz(: uzuz)

(Bi-)isotropic medium is transformed to uniaxially (bi-)anisotropicmedium


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Generalization: triaxial transformation

Triaxial transformation: stretching in three orthogonal directions

S=

iuiui, S1

=

1i uiui, detS=123

Field and source vectors are transformed as

Ea(ra) =

1i uiui E(

1i uiui r)

All medium parameter dyadics are transformed asa =

ij123

uiuj(uiuj :)

Bi-isotropic medium is transformed to a bi-anisotropic medium.All medium dyadics multiples of the same dyadic S

2.

Conversely: if medium dyadics are multiples of the same symmetricdyadic, bi-anisotropic medium can be transformed to a bi-isotropicmedium.


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Use of affine transformation

Affine transformation changes sources, fields, media and bound-aries 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 sphereanisotropic ellipsoid

Surface equation f(r) = 0 becomes f(A1 ra) = fa(ra) = 0

Normal n f(r) and E, H transform similarly: na = A1T n

Zsa(naHa) = Eat= A1T

Et = A1T

Zs(nH) = A1T

Zs[(AT

na)(AT

H)]

Surface impedance dyadic transforms as

Zsa= A1T Zs A

(2)T=A

1T Zs A1

detA


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Medium transformations 1

Isotropic medium anisotropic medium (affine-isotropic medium)

00

I

a 0

0 a

=

00

A AT

detA

Dyadics a anda symmetric and related by condition a= a Conversely: anisotropic medium isotropic mediumaI, aIpos-

sible only if = S, = S,Ssymmetric. Transformation dyadic

A= S1/2

witha = a and = a/(

detS) = a/(

detS)

Medium is affine-bi-isotropic if medium six-dyadic is the form

M=

=

S,


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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 symmetric a symmetric I 0

0

a 0

0 aI

Ferrite magnetoplasma not possible!

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


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Simple use of affine transformation

Scalar Helmholtz equation with symmetric S

(S: + k2)f(r) =g(r)

Define =S1/2 a, ra= S1/2 r equation simplified to

(2a+k2)fa(ra) = g(ra), fa(ra) = f(S1/2 ra), ga(ra) = g(S

1/2 ra) If solution function fa(ra) can be found

f(r) =fa(S1/2 r)

Square root of symmetric dyadic not unique: e.g., same eigenvec-tors, eigenvalues =i. Boundary conditions (radiation condi-tions) require uniqueness.


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Duality-affine transformation

Affine transformation can be connected to duality transformation

e(r) eda(ra) =

A BC D

A1T e(A1 ra)

Eda(ra) = AA1T E(A1 ra) + BA1T H(A1 ra)

Hda(ra) = C A1T E(A1 ra) + DA

1T H(A1 ra) The medium dyadics become

da

da

dada

= 1ADBC

D2 CD CD C2

BD AD BC AC

BD BC AD AC

B2 AB AB A2

A

detA

AT

Medium dyadics da linear combinations ofa,


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Reflection transformation

Reflection in plane normal to n

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

Properties

C2

=I , CT

=C1

=C , detC= 1

Reflection-transformed medium dyadics

Ma = 1

detCC M CT = C M C

Media invariant in reflection are strange. For example isotropicmedium is not invariant: Ma = M

Can be combined with duality transformation!


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Duality-reflection transformations

Special duality transformation with B = C = 0 can be combinedwith reflection:

da

dada

da

= CdetC (D/A)

(A/D)

CT =C (D/A)

(A/D)

C Isotropic medium invariant for D/A= 1.

Involution required two possibilitiesD= A= 1 Define two transformations, rc = C r

ec(rc) =

1 00 1

Ce(Crc), jc(rc) =

1 00 1

Cj(Crc)

= Electric and magnetic reflection transformations.


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Electric and magnetic reflection

Two reflection transformations with rc denoted by r: Electric reflection

Ec(r)Hc(r)

= C E(C r)C H(C r) , Jc(r)

Mc(r) = C J(C r)C M(C r)

Magnetic reflection Ec(r)Hc(r)

=

C E(C r)

C H(C r)

,

Jc(r)Mc(r)

=

C J(C r)

C M(C r)

Tangential fields at n r= 0 satisfying C r= C = :Ect() = Et(), Hct() = Ht(), EREct() = Et(), Hct() = Ht(), MR

Normal components with opposite signs


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Summed fields in isotropic media

Signs for electric/magnetic reflection transformation

E(r) + Ec(r)H(r) + Hc(r) =

E(r) C E(C r)H(r)

C

H(C

r)

J(r) + Jc(r)M(r) + Mc(r)

=

J(r) C J(C r)M(r) C M(C r)

Conditions on plane n r= 0 respectively

E()+ Ec() = (IC)E() = 2

Itnn

E(),

PMC

PEC

condition

H()+Hc() = (IC)H() = 2

nn

It

H(),

PMC

PEC

condition


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Mirror image principles

PEC conditions satisfied for EM problem + magnetic reflection PEC mirror image of a source is its magnetic reflection

J(r)M(r)

C J(C 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)M(r)

C J(C r)C M(C r)

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


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Medium condition

PEC/PMC image principle valid if medium invariant in reflectiontransformation (electric or magnetic)

c cc c

=C

C=

Conditions for medium dyadics

C C=, C C= , C C= , C C=

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

= t+ uuu, = t+ uuu, = atu + ubt, =ctu + udt

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


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Object above PEC/PMC plane

An object in air can be replaced by equivalent polarization source

jp(r) =j [M(r) Mo] e(r), Mo =

o 00 o

I

PEC/PMC image of the polarization source obtained through themagnetic/electric reflection transformation

jpc(r) =

C 0

0 C

jp(C r)

=j

C 0

0 C

[M(C r) Mo] e(C r) =

j

C 0

0 C

M(C r)

C 0

0 C

Mo

C 0

0 C

e(Cr)


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Image of object

Identifying the image of the obstacle:

Mc(r) =

C 0

0

C

M(C r)

C 0

0

C

=C

(C r) (C r)(C r) (C r)

C

For example: image of an isotropic half sphere of parameters , 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 , , is the other halfsphere of parameters,, with changed handedness. The prob-lem does not reduce to that of a complete homogeneous sphere.


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Problems

11 Media defined by medium dyadics of the form

=

r

r

+

0 jr

jr

0

are self dual in some duality transformation. Show that they arealso self dual after any affine transformation. Under what condition

it is possible to transform to the unit dyadic I? How to define

the transformation dyadic A?

12 Find the expression for the dipole moment p of a small dielectricsphere (radius a, permittivity ) at r = uzh in air above a PECplane at z = 0, when the incident field is the plane wave Ei(r) =Eoe

jkr. The scatterer can be replaced by the equivalent currentdipole J = p(r uzh), p = E, = j( o)(4a3/3), whereE = 3oE/(+ 2o) is the field inside the scatterer when put inthe field E. hcan be assumed large in wavelengths.


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S-96.510 Advanced Field Theory7. Electromagnetic Field Solutions

I.V.Lindell


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The Green function

Green function = field from a point source of unit amplitude

L()G(r) = (r)

Field from distributed source = integral of Green functionL()f(r) =g(r), f(r) =

G(r r)g(r)dV

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


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Green six-dyadic

Maxwell equations and Maxwell six-dyadic operator L() 0 11 0

E

H

j

E

H

=

J

M

L() e(r) =j(r), L() =J IjM Green six-dyadic = four dyadic fields from dyadic unit sources

L() G(r) = I(r),

I=

I 0

0 I

, G(r) =

Gee(r) Gem(r)

Gme(r) Gmm(r)

Additional conditions: flow of energy towards infinity (lossless me-dia) or decay of fields towards infinity (lossy media)

Formal solution G(r) = L1()(r)


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Use of Green functions

Six-vector field from a six-vector sourceL() e(r) = j(r), L() G(r r) = (r r)I

can be expressed in terms of the Green six-dyadic

e(r) = V

G(r r) j(r)dV

which is valid outside the sources. Check:

L()e(r) = V

[L()G(rr)]j(r)dV =V

(rr)j(r)dV =j(r)

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


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Helmholtz operators

Adjoint Maxwell operators

La() =

jI ( Ij) 1( I+ j) 1 jI

La() =

jI 1 ( Ij)1 ( I+ j) jI

diagonalize the Maxwell operator

La() L() = L() La() =

He() 00 Hm()

,

Helmholtz second-order dyadic operatorsHe() = ( Ij) 1 ( I+ j) + 2Hm() = ( I+ j ) 1 ( Ij) + 2


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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) =

I(r)

Diagonal six-dyadic operator L() La() has diagonal solution

g(r) =

Ge(r) 0

0 Gm(r)

, He,m() Ge,m(r) = I(r)

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

G(r) =

jGe(r)

1 ( Ij) Gm(r)1 ( I+ j) Ge(r) jGm(r)

It is sufficient to solve only two Green dyadicsGe(r), Gm(r)


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Two scalar Green functions

Green dyadics Ge(r), Gm(r) obtained formally as

Ge,m(r) = H1

e,m() (r) = H(2)T

e,m() 1

detHe,m()(r)

Can be solved in terms of two scalar Green functionsGe,m(r) =

1detHe,m()

(r),

satisfying scalar fourth-order differential equationsdetHe,m()Ge,m(r) = (r)

Helmholtz Green dyadics can be expressed as

Ge,m(r) = H(2)T

e,m()Ge,m(r) In some special cases solving 4th order equations can be avoided


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Summary of the general method

Solve two scalar Green functions satisfying 4th order Helmholtzdeterminant equations

detHe,m()Ge,m(r) = (r)

Form two dyadic Green functions which are solutions to 2nd orderdyadic Helmholtz equations

Ge,m(r) = H(2)T

e,m()Ge,m(r), He,m() Ge,m(r) = I(r) Form the Green six-dyadic which satisfies the original 1st order

six-dyadic equation

G(r) = La()

Ge(r) 0

0 Gm(r)

, L() G(r) = I(r),

Difficulty: solutions to fourth-order equations not available works only for some special cases of the bi-anisotropic medium.


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Isotropic medium

Isotropic medium is the simplest special case. DefineH() = He() = Hm() = I + k2I

detH(

) =k2(

2 + k2)2, H(2)T

(

) = (

2 + k2)(

+ k2I)

Helmholtz Green dyadic solved from 2nd order equation:

G(r) = H1()(r) = (I+1k2

)G(r) =1Ge(r) = 1Gm(r)

G(r) = 12 + k2 (r) =

ejkr

4r (outgoing wave)

The Green six-dyadic becomes

G(r) =

Gee(r) Gem(r)

Gme(r) Gmm(r)

=

jG(r) G(r) IG(r) I jG(r)


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Bi-isotropic medium 1

For bi-isotropic media dyadic Helmholtz operators factorizableHe() = Hm() = H() = [(IjI)(I+jI)k2I]

H(

) =

L+(

)

L(

) =

L(

)

L+(

)

L() = I kI, = cos r Only one Green dyadic needed: Ge(r) =G(r), Gm(r) = G(r)

G(r) = H1()(r) =L1+ () L1

()(r)

Dyadic partial fraction expansion applicable in this case

L1

+ () L1

() =A+L1

+ () + AL1

()

= [A+L()+ AL+()]L1

+ ()L1

() A = 1

2k cos


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Green dyadic can be expressed

G(r) = [A+L1

+ () + AL1

()](r) = [A+G+(r) + AG(r)]

in terms of two auxiliary Green dyadics G(r)

G = L1

()(r) = L

(2)T

detL()(r) =L

(2)T

()G(r),

L(2)T

() = k I+ k22I Scalar Green functionsG(r) satisfy second-order Helmholtz equa-

tions

detL()G(r) = k(2 + k22)G(r) = (r)


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Solution to the scalar Helmholtz equation(outgoing waves assume k > 0)

(

2 + k22)G(r) =

1

k

(r), G(r) =

ejkr

4kr

Solutions to Green dyadics

Ge(r) = G(r), Gm(r) = G(r),

G(r) = 1

2k cos

L

(2)T

+ ()ejk+r

4k+r + L

(2)T

()ejkr

4kr

Two terms correspond to two self-dual fields This result can also be also obtained through the general method

with some more effort


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Self-dual medium 1

Self-dual medium is a generalization of the bi-isotropic medium

M=

sin

sin

+

0 j

j

0

r

Denote He() = Hm() = H() which can be factorized as

H() = ( Ij) 1 ( I+ j) + k2

= L+() 1 L() = L() 1 L+()L() = I k, = cos r

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

G(r) = H1()(r) = L1+ () L1

()(r)


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Self-dual medium 2

Dyadic partial fraction expansion possible also in this case

L1

+ () L1

() = A+L1

+ () + AL1

()

=L1

+ () [A+L() + AL+()] L1

()A+L() + AL+() = A = 1

2k cos

Solution in terms of auxiliary Green functions

G(r) = L1

+ () L1

()(r) = 1

2k cos [G+(r) G(r)]

G(r) = L1

()(r) = L

(2)T

()detL()

(r) = L(2)T

()G(r)


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Self-dual medium 3

Auxiliary Green dyadics solved from

detL()G(r) = det( I k)G(r) = (r)

Expand

in symmetric and antisymmetric parts:

= S+ a I

detL() = k[S : ( ka)( ka) + k2detS] Helmholtz equation becomes

[S : ( ka)( ka) + k2detS][kG(r)] = (r)

[S : + k2detS][kekarG(r)] = (r) This can be transformed to a more familiar form


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Self-dual medium 4

Apply affine transformation

=S1/2 , r =S1/2

r, (r) = (r)

detS

(

2 + k2detS) ke

karG(r)detS = (r)

Solution with r = r r =

S1

:rr

kekarG(r)

detS =

ejk

detSr

4r =

ejk

S(2)

:rr

4

S1

:rr

Solutions for the two auxiliary scalar Green functions

G(r) = ekar ejkD

4kD, D(r) =

S

(2)

:rr


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Self-dual medium 5

The dyadic Green functions Ge(r) = G(r), Gm(r) = G(r) arefinally obtained from

G(r) = 12k cos

[G+(r) G(r)]

= 12k cos

[L(2)T+ ()G+(r) + L(2)T

()G(r)], by inserting

L(2)T ()G(r) = [( ka) I S](2)ekarejkD

4kD

=ekar[ k(S ) I+ k2S(2)

]ejkD

4kD

Again the two terms correspond to the self-dual fields a = 0,S = I gives the previous result for bi-isotropic space


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Uniaxial medium 1

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

= zuzuz+ tIt, = zuzuz+ tIt, = = 0

Electric Green dyadicGe(r) = H

1

e ()(r), He() =1 + 2

Ge(r) = H(2)T

e ()G(r), G(r) = 1

detHe()(r)

The operator detHe() can be factorized (after some effort):

detHe() = k2t

detH()H(), k2t =2tt

H() = 2t +zt

(2z + k2t ), H() = 2t +

zt

(2z + k2t )


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Uniaxial medium 2

Scalar Green function

G(r) = detk2t

1

H()H() (r) = det

2(zt tz) [g(r) g(r)],

g(r) = 1

(2z + k2t )(2t + zt (2z + k2t ) (r), = ,

It can be shown that g(r) can be solved analytically as

g(r) = j

8ktt[E1(jkt(D z))ejktz + E1(jkt(D+ z))ejktz],

E1(x) = exponential integral function, D(r) = distance function

E1(x) =

x

ey

y dy, D =

zt

2 + z2


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Uniaxial medium 3

The electric Green dyadic becomes

Ge(r) = H(2)T

e ()G(r) = det

2(zt tz) H(2)T

e ()[g(r) g(r)]

After differentiations (quite tedious)Ge(r) =

z

1+

1

k2t

ejktD

4D+ j uzq

4kt

=ejktD ejktD , q= uz r(uz r)2

Solution not singular at z axis uz r= 0 because q is finite Solution generalizable to= zuzuz, =zuzuz. Factorizability of detH() does not warrant analytic solution for

Green dyadic!


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A class of anisotropic media 1

Helmholtz determinant operator factorizable for the class of anisotropicmedia defined by =

T, = = 0

detHe() = det(1 + 2) = det( 1

1T

+ 2)

= 2

2det(: )2+

4

: +6det=

2

2det(: +2det)2

Leads to a second-order PDE because also H(2)e () factorizable:

H(2)

e () = 1

2det(: ) +

2

det(

(2)) + 4(2)

= 1

2det(: + 2det)( + 2 (2))

H1e () = 1

2 + 2 (2)T

: + 2det


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A class of anisotropic media 2

Solution of the scalar Green function (s = symmetric part of)

G(r) = 1s : + k2

(r) = ejkD

4detsD, D=

1s :rr

Electric Green dyadic obtained in analytic form

Ge(r) = H1

e ()(r) = 1

2 + 2 (2)T

: + 2det (r)

= ( + 2 (2)T) ej

det

1

s :rr

42

dets

1s :rr

Check: isotropic case = s= I, =/: Ge= (I+ /k2)(ejkr/4r), k=


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