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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
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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
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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
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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
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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
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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
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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
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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
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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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Bi-isotropic medium 2
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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Bi-isotropic medium 3
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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