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1
Maxwell Equations&
Propagation inAnisotropic Electric Media
SOLO HERMELINUpdated 12.06.06
23.12.12http://www.solohermelin.com
2
Maxwell Equations & Propagation in Anisotropic MediaSOLO
TABLE OF CONTENT
Maxwell’s EquationsHistory of Maxwell’s EquationsSymmetric Maxwell’s EquationsConstitutive RelationBoundary Conditions Energy and MomentumMonochromatic Planar Wave Equations
Planar Waves in an Source-less Anisotropic Electric Media
Lorentz’s Lemma
Planar Wave Group VelocityPhase Velocity, Energy (Ray) Velocity for Planar Waves
Energy Flux and Poynting VectorEnergy Flux and Poynting Vector for a Bianisotropic MediumWave equation Wave-Vector Surface
Double Refraction at a Uniaxial Crystal BoundaryRefractive Index of Some Typical Crystals
Polarization History
3
Maxwell Equations & Propagation in Anisotropic MediaSOLO
TABLE OF CONTENT
Planar Waves in an Source-less Anisotropic Electric MediaOrthogonality Properties of the EigenmodesPhase Velocity, Energy (Ray) Velocity for Planar Waves
Ray-Velocity Surface Index Ellipsoid
Conical Refraction on the Optical Axis Equation of Wave Normal
References
4
POLARIZATION
Erasmus Bartholinus,doctor of medicine and professor of mathematics at the University of Copenhagen, showed in 1669 that crystals of “Iceland spar” (which we now call calcite, CaCO3) produced two refracted rays from a single incident beam. One ray, the “ordinary ray”, followed Snell’s law, while the other, the “extraordinary ray”, was not always even in the plan of incidence.
SOLO
History
Erasmus Bartholinus1625-1698
http://www.polarization.com/history/history.html
5
POLARIZATIONSOLO
History
Étienne Louis Malus1775-1812
Etienne Louis Malus, military engineer and captain in the army ofNapoleon, published in 1809 the Malus Law of irradiance through aLinear polarizer: I(θ)=I(0) cos2θ. In 1810 he won the French AcademyPrize with the discovery that reflected and scattered light also possessed“sidedness” which he called “polarization”.
6
POLARIZATION
Arago and Fresnel investigated the interference of polarized rays of light and found in 1816 that tworays polarized at right angles to each other never interface.
SOLO
History (continue)
Dominique François Jean Arago1786-1853
Augustin Jean Fresnel
1788-1827
Arago relayed to Thomas Young in London the resultsof the experiment he had performed with Fresnel. This stimulate Young to propose in 1817 that the oscillationsin the optical wave where transverse, or perpendicular to the direction of propagation, and not longitudinal as every proponent of wave theory believed. Thomas Young
1773-1829
7
POLARIZATIONSOLO
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
Methods of Achieving Polarization
Polarization is based on one of four fundamental physical mechanisms:
1. Dichroism or selective absorbtion 3. Reflection
2. Scattering 4. Birefrigerence
To obtain Polarization we must have some asymmetry in the optical process.
8
POLARIZATIONSOLO
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
Methods of Achieving Polarization
Polarization by Birefrigerence (continue – 1)
Polarization can be achieved with crystalline materials which have a different index ofrefraction in different planes. Such materials are said to be birefringent or doubly refracting.
Nicol Prism The Nicol Prism (1828) is made up from two prisms of calcite cemented with Canada balsam. The ordinary ray can be made to totally reflect off the prism boundary, leving only the extraordinary ray.
William Nicol(1768 ?– 1851) Scottish physicist
9
POLARIZATIONSOLO
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
Methods of Achieving Polarization
Polarization by Birefrigerence (continue – 2)
Polarization can be achieved with crystalline materials which have a different index ofrefraction in different planes. Such materials are said to be birefringent or doubly refracting.
Wollaston Prism
William HydeWollaston1766-1828
10
POLARIZATIONSOLO
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
Methods of Achieving Polarization
Polarization by Birefrigerence (continue – 3)
Polarization can be achieved with crystalline materials which have a different index ofrefraction in different planes. Such materials are said to be birefringent or doubly refracting.
Glan-Foucault Polarizer
12
POLARIZATIONSOLOPolarizations Prisms Overview
http://www.unitedcrystals.com/POverview.htmlReturn to Table of Content
13
MAXWELL’s EQUATIONSSOLO
Magnetic Field Intensity H
[ ]1−⋅mA
Electric Displacement D [ ]2−⋅⋅ msA
Electric Field Intensity E [ ]1−⋅mV
Magnetic InductionB [ ]2−⋅⋅ msV
Current Density J [ ]2−⋅mA
Free Charge Distributionρ [ ]3−⋅⋅ msA
James Clerk Maxwell(1831-1879)
A Dynamic Theory of Electromagnetic Field 1864 Treatise on Electricity and Magnetism 1874
Return to Table of Content
14
SOLOElectrostatics
Charles-Augustin de Coulomb1736 - 1806
In 1785 Coulomb presented his three reports on Electricity and Magnetism:
- Premier Mémoire sur l’Electricité et le Magnétisme [2]. In this publication Coulomb describes “How to construct and use an electric balance (torsion balance) based on the property of the metal wires of having a reaction torsion force proportional to the torsion angle”. Coulomb also experimentally determined the law that explains how “two bodies electrified of the same kind of
Electricity exert on each other”.- Sécond Mémoire sur l’Electricité et le Magnétisme [3]. In this
publication Coulomb carries out the “determination according to which laws both the Magnetic and the Electric fluids act, either
by repulsion or by attraction”.- Troisième Mémoire sur l’Electricité et le Magnétisme [4]. “On
the quantity of Electricity that an isolated body loses in a certain time period , either by contact with less humid air, or in the
supports more or less idio-electric”.
15
SOLOElectrostatics
Charles-Augustin de Coulomb1736 - 1806
1 212 123
0 12
1
4
q qF r
rπ ε=
Coulomb’s Law
q1 – electric charge located at 1r
q2 – electric charge located at 2r
12 1 2r r r= −
The electric force that the chargeq2 exerts on q1 is given by:
If the two electrical charges have the same sign the force isrepulsive, if they have opposite signs is attractive.
120 8.854187817 10 /Farad mε −= ×Permittivity of
vacuum
Accord to the Third Newton Law of mechanics: 21 12F F= −
12 1 12F q E=
Define 212 123
0 12
1
4
qE r
rπ ε=
where is the Electric Field Intensity [N/C]
16
SOLO
During an evening lecture in April 1820, Ørsted discovered experimental evidence of the relationship between electricity and magnetism. While he was preparing an experiment for one of his classes, he discovered something that surprised him. In Oersted's time, scientists had tried to find some link between electricity and magnets, but had failed. It was believed that electricity and magnetism were not related. As Oersted was setting up his materials, he brought a compass close to a live electrical wire and the needle on the compass jumped and pointed to the wire. Oersted was surprised so he repeated the experiemnt several times. Each time the needle jumped toward the wire. This phenomenon had been first discovered by the Italian jurist Gian Domenico Romagnosi in 1802, but his announcement was ignored.
1820Electromagnetism
Hans Christian Ørsted 1777- 1851
17
SOLO 1820Electromagnetism
André-Marie Ampère 1775 - 1836
Danish physicist Hans Christian Ørsted's discovered in 1820 that a magnetic needle is deflected when the current in a nearby wire varies - a phenomenon establishing a relationship between electricity and magnetism. Ørsted's work was reported the Academy in Paris on 4 September 1820 by Arago and a week later Arago repeated Ørsted's experiment at an Academy meeting. Ampère demonstrated various magnetic / electrical effects to the Academy over the next weeks and he had discovered electrodynamical forces between linear wires before the end of September. He spoke on his law of addition of electrodynamical forces at the Academy on 6 November 1820 and on the symmetry principle in the following month. Ampère wrote up the work he had described to the Academy with remarkable speed and it was published in the Annales de Chimie et de Physique.
Ampère and Arago investigate magnetism
dluu
an infinitesimal element of the contour C
J
curent density A/m2
dSuu
a differential vector area of the surface S enclosed by contour C
Ampère’s Law
Magnetic Field Intensity H [ ]1−⋅mA
18
SOLO
1820
ElectromagnetismBiot-Savart Law
02
1
4rI dL
dBr
µπ
×=uu u
uuMagnetic Field of a current element
Jean-Baptiste Biot1774 - 1862
( )( )1
1 2
21 1 1 21
1 2 1 201 2 3
1 24
c
c c
F I dl B
dl dl r rI I
r r
µπ
= ×
× × −=
−
∫
∫ ∫
uu
uu uu
Ampère was not the only one to react quickly to Arago's report of Orsted's experiment. Biot, with his assistant Savart, also quickly conducted experiments and reported to the Academy in October 1820. This led to the Biot-Savart Law.
Félix Savart1791 - 1841
19
SOLO
1820Electromagnetism
Biot-Savart Law
Jean-Baptiste Biot1774 - 1862
( ) ( )2 300
''
4 '
J rA J A r d r
r r
µµπ
∇ = − ⇒ =−∫
0
0
H J B H
B B A
µ∇× = =
∇× = → = ∇×
( ) ( ) 20B A A A Jµ∇× = ∇× ∇× = ∇ ∇ × − ∇ =
choose 0A∇ × =
( ) ( ) ( ) ( ) ( )3 30 03
' ' '' '
4 ' 4 'r r
J r J r r rB r A r d r d r
r r r r
µ µπ π
× −= ∇ × = ∇ × = ÷ ÷− −
∫ ∫
Where we used ( ) ( )0
' : 1/ 'r r rJ J r J r rφ φ φ φ∇ × = ∇ × + ∇ × = −
14243
Derivation of Biot-Savart Law from Ampère’s Law
Poison’s EquationSolution for an unbounded volume
Ampère Law
Félix Savart1791 - 1841
20
SOLO 1831Electromagnetism
On 29th August 1831, using his "induction ring", Faraday made one of his greatest discoveries - electromagnetic induction: the "induction" or generation of electricity in a wire by means of the electromagnetic effect of a current in another wire. The induction ring was the first electric transformer. In a second series of experiments in September he discovered magneto-electric induction: the production of a steady electric current. To do this, Faraday attached two wires through a sliding contact to a copper disc. By rotating the disc between the poles of a horseshoe magnet he obtained a continuous direct current. This was the first generator.
Michael Faraday 1791- 1867
Magnetic Field Intensity H
[ ]1−⋅mA
Electric Displacement D [ ]2−⋅⋅ msA
Electric Field Intensity E [ ]1−⋅mV
Magnetic InductionB [ ]2−⋅⋅ msV
→→⋅
∂∂−=⋅ ∫∫∫ dS
t
BdlE
SC
t
BE
∂∂−=×∇
→dl
→dS
C
B
E
The voltage induced in a coil moving through a non-uniform magnetic field was demonstrated by this apparatus. As the coil is removed from the field of the bar magnets, the coil circuit is broken and a spark is observed at the gap.
The first transformer: Two coils wound on an iron toroid.
http://www.ece.umd.edu/~taylor/frame1.htm
21
MAXWELL’s EQUATIONS
1. AMPÈRE’s CIRCUIT LW (A) (1820)
SOLO
2. FARADAY’s INDUCTION LAW (F) (1831)
Electric Displacement D [ ]2−⋅⋅ msA
Magnetic Field Intensity H [ ]1−⋅mA
Current Density J [ ]2−⋅mA
André-Marie Ampère1775-1836
→→⋅
∂∂−=⋅ ∫∫∫ dS
t
BdlE
SC
t
BE
∂∂−=×∇
→dl
→dS
C
B
E
Electric Field Intensity E [ ]1−⋅mV
Magnetic InductionB [ ]2−⋅⋅ msV
Michael Faraday1791-1867
The following four equations describe the Electromagnetic Field and wherefirst given by Maxwell in 1864 (in a different notation) and are known as
MAXWELL’s EQUATIONS
22
MAXWELL’s EQUATIONSSOLO
4. GAUSS’ LAW – MAGNETIC (GM)
dV
→dS
V
0=⋅∫∫→
S
dSB
B
0=⋅∇ B
Magnetic InductionB [ ]2−⋅⋅ msV
3. GAUSS’ LAW – ELECTRIC (GE)
dV
→dS
V∫∫∫∫∫ =⋅
→
VS
dVdSD ρ
D
ρ=⋅∇ D
ρ
Electric Displacement D [ ]2−⋅⋅ msA
Free Charge Distributionρ [ ]3−⋅⋅ msA
GAUSS’ ELECTRIC (GE) & MAGNETIC (GM) LAWS developed by Gaussin 1835, but published in 1867.
Karl Friederich Gauss1777-1855
Return to Table of Content
23
James C. Maxwell(1831-1879)
ELECTROMAGNETICSSOLO
Magnetic Field Intensity H [ ]1−⋅mA
Electric Displacement D [ ]2−⋅⋅ msA
Electric Field Intensity E [ ]1−⋅mV
Magnetic InductionB [ ]2−⋅⋅ msV
Electric Current Density eJ
[ ]2−⋅mA
Free Electric Charge Distributioneρ [ ]3−⋅⋅ msA
Fictious Magnetic Current Density mJ [ ]2−⋅mV
Fictious Free Magnetic Charge Distributionmρ
[ ]3−⋅⋅ msV
MAXWELL’S SYMMETRIC EQUATIONS FOR THE ELECTROMAGNETIC FIELD
e
S
e
C
Jt
DHSd
t
DJdlH
+
∂∂=×∇⇒•
∂∂+=• ∫∫∫
→)1( AMPÈRE’S LAW
t
BJESd
t
BJdlE m
S
m
C ∂∂−−=×∇⇒•
∂∂+−=• ∫∫∫
→
)2( FARADAY’S LAW
e
V
e
S
DdvSdD ρρ =•∇⇒=• ∫∫∫∫∫
)3( GAUSS’ ELECTRIC LAW
m
V
m
S
BdvSdB ρρ =•∇⇒=• ∫∫∫∫∫
)4( GAUSS’ MAGNETIC LAW
24
ELECTROMAGNETICSSOLO
SYMMETRIC MAXWELL’s EQUATIONS
Magnetic Field Intensity H [ ]1−⋅mA
Electric Displacement D [ ]2−⋅⋅ msA
Electric Field Intensity E [ ]1−⋅mV
Magnetic InductionB [ ]2−⋅⋅ msV
Electric Current Density eJ
[ ]2−⋅mA
Free Electric Charge Distributioneρ [ ]3−⋅⋅ msA
Fictious Magnetic Current Density mJ [ ]2−⋅mV
Fictious Free Magnetic Charge Distributionmρ
[ ]3−⋅⋅ msV
1. AMPÈRE’S CIRCUIT LW (A) eJ
t
DH
+
∂∂=×∇
2. FARADAY’S INDUCTION LAW (F)mJ
t
BE
−
∂∂−=×∇
3. GAUSS’ LAW – ELECTRIC (GE) eD ρ=⋅∇
4. GAUSS’ LAW – MAGNETIC (GM) mB ρ=⋅∇
Although magnetic sources are not physical they are often introduced as electricalequivalents to facilitate solutions of physical boundary-value problems.
André-Marie Ampère1775-1836
Michael Faraday1791-1867
Karl Friederich Gauss1777-1855
25
ELECTROMAGNETICSSOLO
SYMMETRIC MAXWELL’s EQUATIONS (continue – 1)
eJt
DH
+
∂∂=×∇mJ
t
BE
−
∂∂−=×∇
eD ρ=⋅∇
mB ρ=⋅∇
From the Symmetric Maxwell’s Equations we can see that we obtain the same equationsby performing the following operations.
DUALITY
⇓
⇓
−⇓
⇓
−
⇓
⇓
−
⇓
⇓
⇓
−
⇓µ
ε
ε
µ
ρ
ρ
ρ
ρ
e
m
m
e
e
m
m
e
J
J
J
J
E
H
H
E
B
D
D
B
The Maxwell’s Equations are symmetric and dual.
eJt
DH
+
∂∂=×∇ mJ
t
BE
−
∂∂−=×∇
mB ρ=⋅∇
eD ρ=⋅∇
26
ELECTROMAGNETICSSOLO
SYMMETRIC MAXWELL’s EQUATIONS (continue – 2)
From the Symmetric Maxwell’s Equations
( ) 00 =∂
∂+⋅∇⇒⋅∇+⋅∇
∂∂=×∇⋅∇=⇒
=⋅∇
+∂∂=×∇
tJJD
tH
D
Jt
DH
eee
e
e ρ
ρ
( ) 00 =∂
∂+⋅∇⇒⋅∇−⋅∇
∂∂−=×∇⋅∇=⇒
=⋅∇
−∂∂−=×∇
tJJB
tE
B
Jt
BE
mmm
m
m ρ
ρ
CONSERVATION OF ELECTRICAL AND MAGNETIC CHARGES
0=∂
∂+⋅∇
tJ e
e
ρ
0=∂
∂+⋅∇t
J mm
ρ
Therefore
Return to Table of Content
27
ELECTROMAGNETICSSOLO
CONSTITUTIVE RELATIONS
Homogeneous Medium – Medium properties do not vary from point to point and are the same for all points. ε µIsotropic Medium – Medium properties are the same in all directions and are scalars. ε µLinear Medium – The effects of all different fields can be added linearly
(Superposition of different fields).
For Linear and Isotropic Medium we have:
ED
ε=
HB
µ=where: 0εε eK=
0µµ mK=
- Dielectric Constant (or Relative Permitivity)eK
- Relative Permeability mK
The simpler case:
28
ELECTROMAGNETICSSOLO
CONSTITUTIVE RELATIONS (continue - 1)
The most general form of Linear Constitutive Relations is:
Classification of Media
dyadicsxwhereH
E
B
D33,,, =
=
µζξε
µζ
ξε
Classification according to the functional dependence of the 6x6 matrix
µζ
ξε
1. Inhomogeneous: function of space coordinates
2. Nonstationary: function of time
3. Time-dispersive: function of time derivatives
4. Space-dispersive: function of space derivatives
5. Nonlinear: function of the electromagnetic field
29
ELECTROMAGNETICSSOLO
CONSTITUTIVE RELATIONS (continue - 3)
Classification of Media (continue - 2)
In general 0,0,0,0
≠==≠ µζξεAnisotropic Electromagnetic - medium described by both 0,0
≠≠ µε
Anisotropic Electric - medium described by 0
≠ε
Anisotropic Magnetic - medium described by 0 ≠µ
If is symmetric, it can be diagonalized0
≠ε
=
z
y
x
εε
εε
00
00
00
Uniaxial zyx εεε ≠= (thetragonal, hexagonal, rombohedral crystals)
Biaxial zyx εεε ≠≠ (orthohombic, monoclinic, triclinic crystals)
Isotropic zyx εεε ==
This is called an Anisotropic Medium.
If or is Hermitian; i.e.0
≠ε 0 ≠µ
( )
Transposeconjugate
conjugateTranspose
a
aj
jaT
z
-T,-*
,-H
UUUU *H ==
−=
00
0
0
αα
Is gyroelectric or gyromagnetic depending on whether stands for or . If both tensors are of this form the medium is gyroelectromagnetic.
ε µU
µζ
ξε
30
ELECTROMAGNETICSSOLO
CONSTITUTIVE RELATIONS (continue - 4)
Classification of Media (continue - 3)
In general 0,0,0,0
≠≠≠≠ µζξε This is called an Bianisotropic Medium.
Such properties have been observed in antiferromagnatic chromium oxide ( antiferromagnetic materials are ones in which it is enerically favorable for neighboring dipoles to take an antiparallel orientation; see Ramo, Whinery, and Van Duzer (1965), pg. 145)
−−
−−−=
**
**
4 µµζζ
ξξεεω
j
G
mediumlosslessundefinite
mediumpassivedefinitenegative
mediumactivedefinitepositive
⇒⇒⇒
G
G
G
In this case, we have the following classification (see development later)
Return to Table of Content
31
SOLO ELECTROMAGNETICS BOUNDARY CONDITIONS
Boundary Conditions
( ) ( ) ldtHtHhldtHldtHldHh
C
2211
0
2211ˆˆˆˆ ⋅+⋅=Θ+⋅+⋅=⋅
→→
∫
where are unit vectors along C in region (1) and (2), respectively, and 21ˆ,ˆ tt
2121 ˆˆˆˆ−×=−= nbtt
- a unit vector normal to the boundary between region (1) and (2)21ˆ −n- a unit vector on the boundary and normal to the plane of curve Cb
Using we obtainbaccba ⋅×≡×⋅
( ) ( ) ( )[ ] ldbkldbHHnldnbHHldtHH eˆˆˆˆˆˆ
21212121121 ⋅=⋅−×=×⋅−=⋅− −−
Since this must be true for any vector that lies on the boundary between regions (1) and (2) we must have:
b
( ) ekHHn
=−×− 2121ˆ
∫∫∫ ⋅
∂∂+=⋅
→
S
e
C
Sdt
DJdlH
( ) dlbkbdlht
DJSd
t
DJ e
h
e
S
eˆˆ
0
⋅=⋅
∂∂+=⋅
∂∂+
→
∫∫
AMPÈRE’S LAW
[ ]1
0lim: −
→⋅
∂∂+= mAh
t
DJk e
he
32
SOLO ELECTROMAGNETICS BOUNDARY CONDITIONS
Boundary Conditions (continue – 1)
( ) ( ) ldtEtEhldtEldtEldEh
C
2211
0
2211ˆˆˆˆ ⋅+⋅=Θ+⋅+⋅=⋅
→→
∫
where are unit vectors along C in region (1) and (2), respectively, and 21ˆ,ˆ tt
2121 ˆˆˆˆ−×=−= nbtt
- a unit vector normal to the boundary between region (1) and (2)21ˆ −n- a unit vector on the boundary and normal to the plane of curve Cb
Using we obtainbaccba ⋅×≡×⋅
( ) ( ) ( )[ ] ldbkldbEEnldnbEEldtEE mˆˆˆˆˆˆ
21212121121 ⋅−=⋅−×=×⋅−=⋅− −−
Since this must be true for any vector that lies on the boundary between regions (1) and (2) we must have:
b
( ) mkEEn
−=−×− 2121ˆ
∫∫∫ ⋅
∂∂+−=⋅
→
S
m
C
Sdt
BJdlE
( ) dlbkbdlht
BJSd
t
BJ m
h
m
S
mˆˆ
0
⋅=⋅
∂∂+=⋅
∂∂+
→
∫∫
FARADAY’S LAW
[ ]1
0lim: −
→⋅
∂∂+= mVh
t
BJk m
hm
33
SOLO ELECTROMAGNETICS BOUNDARY CONDITIONS
Boundary Conditions (continue – 2)
( ) ( ) SdnDnDhSdnDSdnDSdDh
S
2211
0
2211 ˆˆˆˆ ⋅+⋅=Θ+⋅+⋅=⋅→
∫∫
where are unit vectors normal to boundary pointing in region (1) and (2), respectively, and
21 ˆ,ˆ nn
2121 ˆˆˆ −=−= nnn
- a unit vector normal to the boundary between region (1) and (2)21ˆ −n
( ) ( ) SdSdnDDSdnDD eσ=⋅−=⋅− −2121121 ˆˆ
Since this must be true for any dS on the boundary between regions (1) and (2) we must have:
( ) eDDn σ=−⋅− 2121ˆ
( ) dSdShdv e
h
e
V
e σρρ0→
==∫∫∫
GAUSS’ LAW - ELECTRIC
[ ]1
0lim: −
→⋅⋅= msAhe
he ρσ
∫∫∫∫∫ =•V
e
S
dvSdD ρ
34
SOLO ELECTROMAGNETICS BOUNDARY CONDITIONS
Boundary Conditions (continue – 3)
( ) ( ) SdnBnBhSdnBSdnBSdBh
S
2211
0
2211 ˆˆˆˆ ⋅+⋅=Θ+⋅+⋅=⋅→
∫∫
where are unit vectors normal to boundary pointing in region (1) and (2), respectively, and
21 ˆ,ˆ nn
2121 ˆˆˆ −=−= nnn
- a unit vector normal to the boundary between region (1) and (2)21ˆ −n
( ) ( ) SdSdnBBSdnBB mσ=⋅−=⋅− −2121121 ˆˆ
Since this must be true for any dS on the boundary between regions (1) and (2) we must have:
( ) mBBn σ=−⋅− 2121ˆ
( ) dSdShdv m
h
m
V
m σρρ0→
==∫∫∫
GAUSS’ LAW – MAGNETIC
[ ]1
0lim: −
→⋅⋅= msVhm
hm ρσ
∫∫∫∫∫ =•V
m
S
dvSdB ρ
35
SOLO ELECTROMAGNETICS BOUNDARY CONDITIONS
Boundary Conditions (summary)
( ) mkEEn
−=−×− 2121ˆ FARADAY’S LAW
( ) ekHHn
=−×− 2121ˆ AMPÈRE’S LAW [ ]1
0lim: −
→⋅
∂∂+= mAh
t
DJk e
he
[ ]1
0lim: −
→⋅
∂∂+= mVh
t
BJk m
hm
( ) eDDn σ=−⋅− 2121ˆ GAUSS’ LAW
ELECTRIC [ ]1
0lim: −
→⋅⋅= msAhe
he ρσ
( ) mBBn σ=−⋅− 2121ˆ GAUSS’ LAW
MAGNETIC [ ]1
0lim: −
→⋅⋅= msVhm
hm ρσ
36
SOLO ELECTROMAGNETICS BOUNDARY CONDITIONS
Boundary Conditions for Perfect Electric Conductor (PEC)
( ) 0ˆ 2121
=−×− EEn FARADAY’S LAW
( ) ekHHn
=−×− 2121ˆ AMPÈRE’S LAW
( ) eDDn σ=−⋅− 2121ˆ GAUSS’ LAW
ELECTRIC
( ) 0ˆ 2121 =−⋅− BBn GAUSS’ LAW
MAGNETIC
ekHn
=×− 121ˆ
0ˆ 121
=×− En
eDn σ=⋅− 121ˆ
0ˆ 121 =⋅− Bn
02
=H
02
=B
02
=E
02
=D
37
SOLO ELECTROMAGNETICS BOUNDARY CONDITIONS
Boundary Conditions for Perfect Magnetic Conductor (PMC)
( ) mkEEn
−=−×− 2121ˆ FARADAY’S LAW
( ) 0ˆ 2121
=−×− HHn AMPÈRE’S LAW
( ) 0ˆ 2121 =−⋅− DDn GAUSS’ LAW
ELECTRIC
( ) mBBn σ=−⋅− 2121ˆ GAUSS’ LAW
MAGNETIC
0ˆ 121
=×− Hn
mkEn
−=×− 121ˆ
0ˆ 121 =⋅− Dn
mBn σ=⋅− 121ˆ
02
=H
02
=B
02
=E
02
=D
Return to Table of Content
38
SOLO ELECTROMAGNETICS BOUNDARY CONDITIONS
Energy and Momentum
Let start from Ampère and Faraday Laws
∂∂−=×∇⋅
+∂∂=×∇⋅−
t
BEH
Jt
DHE e
EJt
DE
t
BHHEEH e
⋅−
∂∂⋅−
∂∂⋅−=×∇⋅−×∇⋅
( )HEHEEH
×⋅∇=×∇⋅−×∇⋅But
Therefore we obtain
( ) EJt
DE
t
BHHE e
⋅−
∂∂⋅−
∂∂⋅−=×⋅∇
First way
This theorem was discovered by Poynting in 1884 and later in the same year by Heaviside.
39
SOLO ELECTROMAGNETICS BOUNDARY CONDITIONS
Energy and Momentum (continue -1)
We identify the following quantities
- Power density of the current density EJ e
⋅
( )HEDEt
BHt
EJ e
×⋅∇−
⋅
∂∂−
⋅
∂∂−=⋅
2
1
2
1
⋅
∂∂=⋅= BHt
pBHw mm
2
1,
2
1
⋅
∂∂=⋅= DEt
pDEw ee
2
1,
2
1
( )HEpR
×⋅∇=
eJ
- Magnetic energy and power densities, respectively
- Electric energy and power densities, respectively
- Radiation power density
For linear, isotropic electro-magnetic materials we can write ( )HBED
00 , µε ==
( )DEtt
DE
ED
⋅∂∂=
∂∂⋅
=
2
10ε
( )BHtt
BH
HB
⋅∂∂=
∂∂⋅
=
2
10µ
40
SOLO ELECTROMAGNETICS BOUNDARY CONDITIONS
Energy and Momentum (continue – 3)
Let start from the Lorentz Force Equation (1892) on the free charge
( )BvEF e
×+= ρ
Free Electric Chargeeρ [ ]3−⋅⋅ msA
Velocity of the chargev [ ]1−⋅sm
Electric Field Intensity E [ ]1−⋅mV
Magnetic InductionB [ ]2−⋅⋅ msV
Hendrik Antoon Lorentz1853-1928
eρ
Force on the free chargeF
[ ]Neρ
Second way
41
SOLO ELECTROMAGNETICS BOUNDARY CONDITIONS
Energy and Momentum (continue – 4)
The power density of the Lorentz Force the charge
( ) ( )EJBvEvp e
Bvv
Jve
ee
⋅=×+⋅==×⋅
=
0
ρρ
or
( ) ( ) ( ) ( ) ( )[ ]
( )HEt
BHE
t
D
Et
DHEEH
Et
DHEJp
t
BE
HEHEEH
Jt
DH
e
e
×⋅∇−
∂∂⋅+⋅
∂∂−=
⋅∂∂−×⋅∇−×∇⋅=
⋅
∂∂−×∇=⋅=
∂∂−=∇×
×⋅∇=∇×⋅−∇×⋅
+∂∂=∇×
eρ
42
SOLO ELECTROMAGNETICS BOUNDARY CONDITIONS
Energy and Momentum (continue – 5)
( )HEDEt
BHt
EJ e
×⋅∇−
⋅
∂∂−
⋅
∂∂−=⋅
2
1
2
1
Let integrate this equation over a constant volume V
∫∫∫∫∫∫∫∫∫∫∫∫ ×∇−
⋅−
⋅−=⋅
VVVV
e dvSdvDEtd
ddvBH
td
ddvEJ
2
1
2
1
If we have sources in V then instead of we must use
E
sourceEE
+
Use Ohm Law (1826)
( )sourceee EEJ
+= γ
=
∂∂
∫∫∫∫∫∫VV td
d
t
Georg Simon Ohm1789-1854
sourcee
e
EJE
−=γ1
For linear, isotropic electro-magnetic materials ( )HBED
00 , µε ==
43
SOLO ELECTROMAGNETICS BOUNDARY CONDITIONS
Energy and Momentum (continue – 6)
∫∫∫∫∫∫∫∫∫∫∫∫∫ ⋅∇+
⋅+
⋅+=⋅
VVVR
n
V
sourcee dvSdvDE
td
ddvBH
td
ddRIdvEJ
2
1
2
12
∫∫∫
⋅=
V
FieldMagnetic dvBHtd
dP
2
1
∫∫∫
⋅=
V
FieldElectric dvDEtd
dP
2
1∫∫∫∫∫ ⋅=⋅∇=SV
Radiation SdSdvSP
∫∫∫ ⋅=V
sourceeSource dvEJP
( ) ( )
∫∫∫∫
∫∫∫∫ ∫∫∫∫∫∫ ∫∫∫∫∫⋅−=
⋅−⋅=⋅−⋅⋅=⋅
V
sourcee
R
n
V
sourcee
L S eee
V
sourcee
L S eee
V
e
dvEJdRI
dvEJdS
dldSJdSJdvEJldSdJJdvEJ
2
11
γγ
∫=R
nJoule dRIP 2
RadiationFieldMagneticFieldElectricJouleSource PPPPP +++=
For linear, isotropic electro-magnetic materials( )HBED
00 , µε ==
R – Electric Resistance
Define the Umov-Poynting vector: [ ]2/mwattHES
×= The Umov-Poynting vector was discovered by Umov in 1873, and rediscovered by
Poynting in 1884 and later in the same year by Heaviside.
44
ELECTROMAGNETICSSOLO
EM People
John Henry Poynting1852-1914
Oliver Heaviside1850-1925
Nikolay Umov1846-1915
1873 “Theory of interaction on final
distances and its exhibit to conclusion of electrostatic and
electrodynamic laws”
1884 1884
Umov-Poynting vector
HES
×=
Return to Table of Content
45
ELECTROMAGNETICSSOLO
Monochromatic Planar Wave Equations
Let assume that can be written as: ( ) ( )trHtrE ,,,
( ) ( ) ( ) ( ) ( ) ( )tjrHtrHtjrEtrE 00 exp,,exp, ωω ==
where are phasor (complex) vectors.
( ) ( ){ } ( ){ } ( ) ( ){ } ( ){ }rHjrHrHrEjrErE
ImRe,ImRe +=+=
We have ( ) ( ) ( ) ( ) ( )tjrEjtjt
rEtrEt 00 expexp, ωωω
=∂∂=
∂∂
Hence
( )
( )
( )
=⋅∇
=⋅∇
−−=×∇
+=×∇
⇒
=⋅∇
=⋅∇
−∂∂−=×∇
+∂∂=×∇
=∂∂
m
e
m
e
jt
m
e
m
e
B
D
JBjE
JDjH
BGM
DGE
Jt
BEF
Jt
DHA
ρ
ρ
ω
ω
ρ
ρ
ω
)(
46
ELECTROMAGNETICSSOLO
NoteThe assumption that can be written as: ( ) ( )trHtrE ,,,
( ) ( ) ( ) ( ) ( ) ( )tjrHtrHtjrEtrE 00 exp,,exp, ωω ==
is equivalent to saying that has a Fourier Transform; i.e.: ( ) ( )trHtrE ,,,
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )∫∫
∫∫∞
∞−
∞
∞−
∞
∞−
∞
∞−
=−=
=−=
ωωωπ
ωω
ωωωπ
ωω
dtjrHtrHdttjtrHrH
dtjrEtrEdttjtrErE
exp,2
1,&exp,,
exp,2
1,&exp,,
A Sufficient Condition for the Existence of the Fourier Transform is:
( ) ( )
( ) ( ) ∞<=
∞<=
∫∫
∫∫∞
∞−
∞
∞−
∞
∞−
∞
∞−
ωωπ
ωωπ
drHdttrH
drEdttrE
22
22
,2
1,
,2
1,
( ) ( ) ( ) ( ) ( )[ ] ( )
( ) ( )[ ] ( ) ( )ωωδωω
ωωωω
−=−=
−=−=
∫
∫∫∞
∞−
∞
∞−
∞
∞−
00
0
exp
expexpexp,,
rEdttjrE
dttjtjrEdttjtrErE
End Note
47
ELECTROMAGNETICSSOLO
Fourier Transform
The Fourier transform of can be written as: ( ) ( )trHtrE ,,,
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )∫∫
∫∫∞
∞−
∞
∞−
∞
∞−
∞
∞−
−==
−==
dttjtrHrHdtjrHtrH
dttjtrErEdtjrEtrE
ωωωωωπ
ωωωωωπ
exp,,&exp,2
1,
exp,,&exp,2
1,
( ) ( )
( ) ( ) ∞<=
∞<=
∫∫
∫∫∞
∞−
∞
∞−
∞
∞−
∞
∞−
ωωπ
ωωπ
drHdttrH
drEdttrE
22
22
,2
1,
,2
1,
JEAN FOURIER
1768-1830
A Sufficient Condition for the Existence of the Fourier Transform is:
48
ELECTROMAGNETICSSOLO
( )( )
×∇−×∇−=×∇×∇×∇+×∇=×∇×∇
−−=×∇+=×∇
⇒
−−=×∇×∇
+=×∇×∇ =
=
m
e
m
e
ED
HBm
e
JHjE
JEjH
JHjE
JEjH
JBjE
JDjH
ωµωε
ωµωε
ω
ω ε
µ
( ) me JJjEkE
×∇−−=−×∇×∇ ωµ2
( ) em JJjHkH
×∇+−=−×∇×∇ ωε2 λππµεω
λ22 f
c
c
fk
=∆
===
Using the vector identity ( ) ( ) ( ) AAA
∇⋅∇−⋅∇∇=×∇×∇
For a Homogeneous, Linear and Isotropic Media:
=⋅∇
=⋅∇⇒
=⋅∇=⋅∇ =
=
µρερ
ρρ ε
µm
e
ED
HBm
e
H
E
B
D
ερωµ e
me JJjEkE∇+×∇+=+∇
22
µρωε m
em JJjHkH∇+×∇−=+∇
22
and
we obtain
Monochromatic Planar Wave Equations (continue - 1)
49
ELECTROMAGNETICSSOLO
Assume no sources:
we have
Monochromatic Planar Wave Equations (continue - 2)
0,0,0,0 ==== meme JJ ρρ
022 =+∇ EkE
022 =+∇ HkH
nkk
n
k
0
00
00
0
=
==
∆
εµ
µεεµωµεω
( ) ( ) ( )
( ) ( ) ( )
==
==⋅−
⋅−
rktjtj
rktjtj
eHerHtrH
eEerEtrE
ωω
ωω
ω
ω
0
0
,,
,,
( ) 022 =+∇⇒
⋅−=∇⋅∇⇒−=∇⋅−
⋅−⋅−⋅−⋅−
rkj
rkjrkjrkjrkj
ek
ekkeekje
Helmholtz Wave Equations
satisfy the Helmholtz wave equations( ) ( )ωω ,,, rHrE
( )( )
=
=⋅−
⋅−
rkj
rkj
eHrH
eErE
0
0
,
,
ω
ω
Assume a progressive wave of phase ( )rkt
⋅−ω (a regressive wave has the phase ) ( )rkt
⋅+ω
For a Homogeneous, Linear and Isotropic Media
k
0E
0H
r t
k
Planes for whichconstrkt =⋅−
ω
50
ELECTROMAGNETICSSOLO
To satisfy the Maxwell equations for a source free media we must have:
Monochromatic Planar Wave Equations (continue - 3)
we haveUsing: 1ˆˆ&ˆˆ =⋅== sssc
nsk ωεµω
=⋅∇=⋅∇
−=×∇=×∇
0
0
H
E
HjE
EjH
ωµωε
=⋅=⋅
=×
−=×
0ˆ
0ˆ
ˆ
ˆ
0
0
00
00
Hs
Es
HEs
EHs
εµ
µε
sPlanar Wave
0E
0Hr
=⋅−
=⋅−
−=×−
=×−
⇒
⋅−
⋅−
⋅−⋅−
⋅−⋅−
−=∇ ⋅−⋅−
0
0
0
0
00
00
rkj
rkj
rkjrkj
rkjrkj
ekje
eHkj
eEkj
eHjeEkj
eEjeHkjrkjrkj
ωµ
ωε
=⋅
=⋅
=×
−=×
0
0
0
0
00
00
Hk
Ek
HEk
EHk
µω
εω
For a Homogeneous, Linear and Isotropic Media: Return to Table of Content
51
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
0
0
0
0
m
e
m
e
B
D
Jt
BE
Jt
DH
ρ
ρ
=⋅∇
=⋅∇
−∂∂−=×∇
+∂∂=×∇
( ) ( ) ( )∫+∞
∞−
⋅−= dkekEtrE rktj
ω0,
ωωωω
jt
ejet
rsc
ntjrs
c
ntj
→∂∂⇒=
∂∂
⋅−
⋅−
ˆˆ
kjsc
njes
c
nje
rsc
ntjrs
c
ntj
−=−→∇⇒−=∇
⋅−
⋅−
ˆˆˆˆ
ωωωω
0ˆ
0ˆ
ˆ
ˆ
=⋅=⋅
=×
−=×
Bs
Ds
BEsc
n
DHsc
n
( )( )DEBH
tt
DE
t
BH
HEEHHESHB
ED
⋅+⋅∂∂−=
∂∂⋅−
∂∂⋅−=
×∇⋅−×∇⋅=×⋅∇=⋅∇⋅=
⋅= 2
1µ
ε
( ) emUDEBHSk
Ssc
n =⋅+⋅=⋅=⋅2
1ˆ
ω
Maxwell’s Symmetric Equations
Planar Wave
0
0
=⋅
=⋅
=×
−=×
Bk
Dk
BEk
DHk
ω
ω
ωjt
→∂∂
sc
nj ˆω−→∇
ωjt
→∂∂
kj
−→∇ vector phasor
vector phasor
sc
nk ˆω=
52
SOLO ELECTROMAGNETICS
Planar Wave Group Velocity
Consider a Planar Wave composed of frequencies centered around ω0.
( ) ( ) ( )( )∫+∞
∞−
⋅−= dkekEtrE rktkj
ω0,
( ) ( ) +−
+= 0
00
kkkd
dk
ωωω
Expand ω(k) , using Taylor series around k0
( ) ( ) ( )( )
∫∞+
∞−
−
−⋅−−
⋅−= dkekEetrEkk
kk
rkkt
kd
dj
rktj0
0
0
0000,
ω
ω
The Phase of the Group of Waves is ( )
00
0
0
kkkk
rkkt
kd
dg
−
−⋅−−
= ωϕ
To find the velocity of phase of the group of waves consider the points for which φg = const
( )0
0
0
0
=−
⋅−−
⇒
kk
rdkktd
kd
dg
d
ω
ϕ
The Group Velocity of the Planar Wave is ( )
0
0
0 kk
kk
kd
dkrd
g td
rdv
−−
=
==
ωδ
Return to Table of Content
53
SOLO ELECTROMAGNETICS
Phase Velocity, Energy (Ray) Velocity for Planar Waves
The phase of the field is given by
⋅−=⋅−= rs
c
ntrkt
ˆωωφ
For a constant phase front we have
⋅−=⋅−=⋅−== rds
c
ndtrdskdtrdkdtd
ˆˆ0 ωωωφ
sn
cs
kdt
rdv
srrconst
pˆˆ:
ˆ
=====
ωφ
Phase VelocityB
D
pv
E
H
HES
×=α
αk
ev
Planar Waves
n
c
ktd
rds ==⋅ ωˆ
n
c
S
Svs e =⋅ :ˆ ( ) α
α
cosˆ
ˆcos
pS
Ss
e
v
Ss
S
n
cv
⋅=
=⋅
=
Define the phase velocity as
Define the energy flow velocity in the Poynting vector direction as S
SvSv ee =
→1
( ) emUDEBHSk
Ssc
n =⋅+⋅=⋅=⋅2
1ˆ
ω
( ) em
ee U
SS
Ssn
c
S
Svv =
⋅==
ˆ1
:
Energy (Ray) Velocity
The electro-magnetic energy propagates along the Poynting vector .For anisotropic media the Poynting vector is not collinear with . k
HES ×=
Return to Table of Content
Electromagnetic Energy Density
54
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
A wave packet can be viewed as a superposition of monochromatic waves, with differentfrequencies ω and wave vector that must satisfy the Maxwell’s equationsk
EHk
HEk
⋅−=×
⋅=×
εω
µω
Suppose that the wave vector changes by an infinitesimal amount , that determineschanges
k
k
δHE δδωδ ,,
HHHEkEk ⋅⋅+⋅=×+× δµωµωδδδ
( )EEEHkHk −⋅⋅−⋅−=×+× δεωεωδδδ
( ) ( ) ( ) ( )( ) ( ) ( ) ( )EEEEEHkHEk
HHHHHEkHEk
δεωεωδδδ
δµωµωδδδ
⋅⋅+⋅⋅=×⋅−×⋅
⋅⋅+⋅⋅=×⋅+×⋅
( ) ( )( ) ( ) 0
2
00
=⋅+×⋅+⋅−×⋅−=
⋅⋅+⋅⋅−×⋅
EHkEHEkH
HHEEHEk
εωδµωδ
µεωδδ
( ) ( ) ( )BACACBCBA
×⋅≡×⋅≡×⋅
55
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
( )( )
eem
vkU
Sk
HHEE
HEk
⋅=⋅=⋅⋅+⋅⋅
×⋅= δδµε
δωδ
21
( ) gk vkk
⋅=∇⋅= δωδωδ
- Energy (Ray) Velocityev
From the definition of Group Velocity : gv
since those relations are true for all : k
δ eg vv =
For a monochromatic wave δω=0, therefore
( ) 0=×⋅=⋅ HEkSk
δδ
( ) ( ) 02 =⋅⋅+⋅⋅−×⋅ HHEEHEk µεωδδ
56
ELECTROMAGNETICSSOLO
Consider an Electric Anisotropic Media
General Symmetric
=⋅∇=⋅∇
−=×∇⋅=×∇
0
0
B
D
HjE
EjH
µωεω
For a Homogeneous, Linear and Anisotropic Media:
=
333231
232221
131211
εεεεεεεεε
ε
=
z
y
x
D
εε
εε
00
00
00
DiagonalizedPrincipal Axis
TTD
εε 1−=
−−=×∇×∇
+⋅=×∇×∇
m
e
JHjE
JEjH
µω
εω ( )( )
×∇−×∇−=×∇×∇×∇+⋅×∇=×∇×∇
m
e
JHjE
JEjH
ωµεω
( ) me JJjEE ×∇−×∇−=×∇×∇ µωεεµεω0
02
Planar Waves in an Source-less Anisotropic Electric Media
Return to Table of Content
57
ELECTROMAGNETICSSOLO
According to the principle of superposition these two fields can exist separately or be superimposed without disturbing each other.
( )1221 HEHE
×−×⋅∇
Consider two distinct fields and . 11,HE
22 ,HE
Assumptions:
1. The two fields are harmonic functions of time and of the same frequency.
2. The medium is linear.
3. The point considered is not inside a source and Ohm Law applies.
∂
∂+⋅+∂
∂⋅+
∂
∂+⋅−∂
∂⋅=t
DJE
t
BH
t
DJE
t
BH ee
112
21
221
12
12212112 HEEHHEEH
×∇⋅+×∇⋅−×∇⋅−×∇⋅=
( ) ( )1122122112 DjJEBHjDjJEBHj ee
jt
ωωωωω
+⋅+⋅++⋅−⋅−==
∂∂
( ) ( ) 00
1122122112
21 ==⋅=
⋅==⋅+⋅+⋅⋅+⋅+⋅−⋅⋅−=
ee JJ
ee
ED
HBEjJEHHjEjJEHHj εωµωεωµω
ε
µ
Lorentz’s Lemma
58
ELECTROMAGNETICSSOLO
Lorentz’s Lemma (continue)
( ) ( )1221 HEHE
×⋅∇=×⋅∇
For the two distinct fields and . 11,HE
22 ,HE
Hendrik Antoon Lorentz1853-1928
For a Monochromatic Planar Wave
( ) ( ) ( )
( ) ( ) ( )
==
==⋅−
⋅−
rktjtj
rktjtj
eHerHtrH
eEerEtrE
ωω
ωω
ω
ω
0
0
,,
,,
( ) ( ) ( )rktjrktjrktj eskjekje
⋅−⋅−⋅− −=−=∇ ωωω ˆ
Lorentz’s Lemma can be written
( ) ( )1221 HEkHEk
×⋅=×⋅
59
ELECTROMAGNETICSSOLO
Lorentz’s Reciprocity Theorem
( )( )222
111
se
se
EEJ
EEJ
+=
+=
σ
σ
This is the more general case that does not exclude the sources.
( ) 12211221 ee JEJEHEHE
⋅−⋅=×−×⋅∇Ohm’s Law:
are the applied electric field intensities within the sources21, ss EE
( )1221 HEHE
×−×⋅∇
∂
∂+⋅+∂
∂⋅+
∂
∂+⋅−∂
∂⋅=t
DJE
t
BH
t
DJE
t
BH ee
112
21
221
12
12212112 HEEHHEEH
×∇⋅+×∇⋅−×∇⋅−×∇⋅=
( ) ( )1122122112 DjJEBHjDjJEBHj ee
jt
ωωωωω
+⋅+⋅++⋅−⋅−==
∂∂
( ) ( )122112211212
21122211122112
esesssssss
ssssee
JEJEEEEEEEEE
EEEEEEEEEEJEJE
⋅−⋅=⋅−⋅−⋅+⋅=
⋅−⋅=+⋅−+⋅=⋅−⋅=
σσσσ
σσσσ
( ) ( )1122122112 EjJEHHjEjJEHHj ee
ED
HB⋅+⋅+⋅⋅+⋅+⋅−⋅⋅−=
⋅=
⋅=εωµωεωµω
ε
µ
60
ELECTROMAGNETICSSOLO
Lorentz’s Reciprocity Theorem (continue - 1)
Integrate over a very large volume:
( ) 12211221 ee JEJEHEHE
⋅−⋅=×−×⋅∇
( ) ( )∫∫ ×−×⋅∇=⋅−⋅VV
eses dvHEHEdvJEJE 12211221
( ) 11
22
21
21
0
0
0
01221 0
VoutsideE
VoutsideE
EE
HHS
s
s
S
S
sdHEHE
=
=
==
==⇐=⋅×−×=
→∞
→∞∫
We obtained
( ) ( )∫∫ ⋅=⋅V
es
V
es dvJEdvJE 1221
( ) 212121 bs
V
es
V
es IVdsJdlEdvJE =⋅=⋅ ∫∫
( ) 121212 bs
V
es
V
es IVdsJdlEdvJE =⋅=⋅ ∫∫
61
ELECTROMAGNETICSSOLO
Lorentz’s Reciprocity Theorem (continue - 2)
The Reciprocity Theorem
1221 bsbs IVIV =
The current induced in 2 when 1 is energized, divided by the voltage applied on 1, is the same as the current induced in 1 when 2 is energized, divided by the applied voltage on 2, as long as the frequency and the impedances remain unchanged.
1221 bbss IIVV =⇔=
Return to Table of Content
62
SOLO
( ) ( ) ( ) ( )
( ) ( )[ ]{ }
( ) ( ) ( ) ( )[ ] ( ) ( ) ( ) ( )[ ]
( ) ( ) ( ) ( ) ( ) ( )[ ]∫
∫
∫
∫∫
−+⋅+=
−+⋅−+=
=
⋅=⋅==
T
T
T
TEDT
e
dttjrErErEtjrET
dttjrEtjrEtjrEtjrET
dttjrEalT
dttrEtrET
dttrDtrET
w
0
2**2
0
**
0
2
00
2exp,,,22exp,4
1
exp,exp,exp,exp,4
1
exp,Re1
,,1
,,1
ωωωωωωε
ωωωωωωωωε
ωωε
εε
But( ) ( )[ ] ( )
( ) ( )[ ] ( )0
2
2exp2exp
2
12exp
1
02
2exp2exp
2
12exp
1
0
0
00
∞→
∞→
→−=−=−
→==
∫
∫
T
TT
T
TT
Tj
Tjtj
Tjdttj
T
Tj
Tjtj
Tjdttj
T
ωωω
ωω
ωωω
ωω
Therefore
( ) ( ) *00
*00
0
*
22
1,,
2EEeEeEdt
TrErEw rkjrkj
T
e
εεωωε === ⋅−⋅∫
Let compute the time averages of the electric and magnetic energy densities
ELECTROMAGNETICS
Energy Flux and Poynting Vector
For a Homogeneous, Linear and Isotropic Media
63
SOLO
In the same way
( ) ( ) ( ) ( ) *00
00 2,,
1,,
1HHdttrHtrH
TdttrBtrH
Tw
TT
m
µµ === ∫∫
Using the relations
( ) 00 ˆ HsEA ×−=εµ
( ) 00 ˆ EsHF ×=µε
since and are real values , where * is the complex conjugate, we obtain
S∇ )**,( SS ∇=∇=
( ) ( )
( ) ( )
( ) ( )[ ] ( ) e
m
e
wHkEHkEHkE
EkHEkHHHw
HkEHkEEEw
=×⋅=×⋅=×⋅=
×⋅=×⋅=⋅=
×⋅=×⋅=⋅=
*
00
*
0*00
**0
*
00
*
00*00
*
00
*
00*00
ˆ2
ˆ2
ˆ2
ˆ2
ˆ22
ˆ2
ˆ22
µεµεµε
µεµεµµ
µεεµεε
ELECTROMAGNETICS
( ) ( ) *00
*00
0
*
22
1,,
2EEeEeEdt
TrErEw rkjrkj
T
e
εεωωε === ⋅−⋅∫
Energy Flux and Poynting Vector (continue – 1)
For a Homogeneous, Linear and Isotropic Media:
Time-averaged Electric Energy Density
Time-averaged Magnetic Energy Density
64
SOLO
Therefore( ) ( )( )*00
ˆ2
rHkrEww me
×⋅==µ ε
Within the accuracy of Geometrical Optics, the Time-averaged Electric and Magnetic Energy Densities are equal.
( ) ( ) ( ) ( ) ( ) ( )( )*0000*00
ˆ22
rHkrErHrHrErEwww me
×⋅=⋅+⋅=+= ∗ µ εµεThe total energy will be:
The Poynting vector is defined as: ( ) ( ) ( )trHtrEtrS ,,:,
×=
( ) ( ) ( ) ( ) ( ) ( )[ ]
( ) ( )[ ] ( ) ( )[ ]∫
∫∫−− +×+=
×=×=×=
Ttjtjtjtj
Ttjtj
T
dterHerHerEerET
dterHerEalT
dttrHtrET
trHtrES
0
**
00
,,2
1,,
2
11
,,Re1
,,1
,,
ωωωω
ωω
ϖϖϖϖ
ϖϖ
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]
( ) ( ) ( ) ( )[ ]ωωωω
ωωωωωωωω ωω
,,,,4
1
,,,,,,,,4
11
**
0
2****2
rHrErHrE
dterHrErHrErHrEerHrET
Ttjtj
×+×=
∫ ×+×+×+×= −
[ ][ ]0
*0
*00
0*
0*00
4
14
1
HEHE
eHeEeHeE rkjrkjrkjrkj
×+×=
×+×= ⋅−⋅⋅⋅−
The time average of the Poynting vector is:
John Henry Poynting1852-1914
ELECTROMAGNETICS
Energy Flux and Poynting Vector (continue – 2)
For a Homogeneous, Linear and Isotropic Media:
Return to Table of Content
65
SOLO
Assume the linear general constitutive relations
ELECTROMAGNETICS
( )( )( )
=⋅∇=⋅∇
−−=×∇+=×∇
m
e
m
e
BGM
DGE
JBjEF
JDjHA
ρρ
ωω
)(
dyadicsxwhereH
E
B
D33,,, =
=
µζξε
µζ
ξε
( ) ( )( ) ( )
−⋅+⋅−=×∇
+⋅+⋅=×∇
m
e
JHEjEF
JHEjHA
µζω
ξεω
Energy Flux and Poynting Vector for a Bianisotropic Medium
66
SOLO ELECTROMAGNETICS
Let compute the following
( )( )[ ] ( )[ ]
( ) meHH
meHH
JHJEHHEHHEEEj
JHEjHEJHEj
EHHEHE
⋅−⋅−⋅⋅−⋅⋅−⋅⋅+⋅⋅=
−⋅+⋅−⋅+⋅+⋅+⋅−−=
×∇⋅+×∇⋅−=×⋅∇
******
****
***
µζξεω
µζωξεω
( )( )[ ] ( ) ( )[ ]
( ) ******
****
***
meHH
mHH
e
JHJEHHHEHEEEj
HJHEjJHEjE
EHHEHE
⋅−⋅−⋅⋅−⋅⋅−⋅⋅+⋅⋅−=
⋅−⋅+⋅−−++⋅+⋅⋅−=
×∇⋅+×∇⋅−=×⋅∇
µζξεω
µζωξεω
( ) ( )( )( ) ******
******
**
meHH
meHH
JHJEHHHEHEEEj
JHJEHHEHEHEEj
HEHE
⋅−⋅−⋅⋅−⋅⋅−⋅⋅+⋅⋅−
⋅−⋅−⋅⋅−⋅⋅−⋅⋅+⋅⋅=
×⋅∇+×⋅∇
µζξεω
µζξεω
( ) ( )( ) ( )
−⋅+⋅−=×∇
+⋅+⋅=×∇
m
e
JHEjEF
JHEjHA
µζω
ξεω
Energy Flux and Poynting Vector for a Bianisotropic Medium (continue – 1)
67
SOLO ELECTROMAGNETICS
( ) ( )( )( )
( ) ( ) ( ) ( )[ ]( ) ( )
[ ] ( ) ( )mmeeHH
HH
mmee
HHHH
meHH
meHH
JHJHJEJEH
EjHE
JHJHJEJE
HHEHHEEEj
JHJEHHHEHEEEj
JHJEHHEHEHEEj
HEHE
⋅+⋅−⋅+⋅−
⋅
−−
−−−⋅=
⋅+⋅−⋅+⋅+
⋅−⋅+⋅−⋅+⋅−⋅+⋅−⋅−=
⋅−⋅−⋅⋅−⋅⋅−⋅⋅+⋅⋅−
⋅−⋅−⋅⋅−⋅⋅−⋅⋅+⋅⋅=
×⋅∇+×⋅∇
******
****
****
******
******
**
µµζζ
ξξεεω
µµζζξξεεω
µζξεω
µζξεω
( ) ( ) ( ) ( )[ ] ( ) ( ) ( ) ( )[ ]ωωωωωω ,,,,4
1,,Re
2
1,, ** rHrErHrErHrEaltrHtrES ×+×=×=×=
We found the time average of the Poynting vector
We see that
( ) ( )[ ] [ ]
( ) ( )mmee
HH
HH
JHJHJEJE
H
EjHEHEHES
⋅+⋅−⋅+⋅−
⋅
−−
−−−⋅=×⋅∇+×⋅∇=⋅∇
****
****
4
1
4
1
44
1
µµζζξξεεω
Energy Flux and Poynting Vector for a Bianisotropic Medium (continue – 2)
68
SOLO ELECTROMAGNETICS
Let integrate the mean value of the Poynting vector over the volume V
[ ]
( ) ( )
( ) ( )[ ]
( ) ∫∫
∫
∫∫
∫∫
→→⋅=⋅×+×=
×⋅∇+×⋅∇=
⋅+⋅−⋅+⋅−
⋅
−−
−−−⋅=⋅∇
SS
Gauss
V
V
mm
V
ee
VHH
HH
V
dSnSdSnHEHE
dVHEHE
dVJHJHdVJEJE
dVH
EjHEdVS
114
1
4
1
4
1
4
1
4
**1
**
****
**
µµζζ
ξξεεω
( ) ( )[ ] [ ]
( ) ( )mmee
HH
HH
JHJHJEJE
H
EjHEHEHES
⋅+⋅−⋅+⋅−
⋅
−−
−−−⋅=×⋅∇+×⋅∇=⋅∇
****
****
4
1
4
1
44
1
µµζζξξεεω
Energy Flux and Poynting Vector for a Bianisotropic Medium (continue – 3)
69
SOLO ELECTROMAGNETICS
We recognize the following
−−
−−−=
HH
HHj
µµζζ
ξξεεω
4G
mediumlosslessundefinite
mediumpassivedefinitenegative
mediumactivedefinitepositive
⇒⇒⇒
G
G
G
[ ]
( ) ( )∫∫
∫∫
⋅+⋅−⋅+⋅−
⋅
−−
−−−⋅=⋅
→
V
mm
V
ee
VHH
HH
S
dVJHJHdVJEJE
dVH
EjHEdSnS
****
**
4
1
4
1
41
µµζζ
ξξεεω
[ ] dVH
EjHE
V∫
⋅
−−
−−−⋅
**
****
4 µµζζ
ξξεεω
( )∫ ⋅+⋅V
ee dVJEJE **
4
1 ( )∫ ⋅+⋅V
mm dVJHJH **
4
1
∫→
⋅S
dSnS 1 Time average of the Radiated
Energy through S (Irradiance)
Time average of Electromagnetic Energy in V
Time average of Joule Energy in V
Time average of Fictious Joule Energy in V
Energy Flux and Poynting Vector for a Bianisotropic Medium (continue – 4)
Return to Table of Content
70
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
0
0
0
0
m
e
m
e
B
D
Jt
BE
Jt
DH
ρ
ρ
=⋅∇
=⋅∇
−∂∂−=×∇
+∂∂=×∇
( )
⋅−
=⋅− ==
rsc
ntj
sc
nk
rktj eEeEE
ˆ
0
ˆ
0
ωω
ω
ωωωω
jt
ejet
rsc
ntjrs
c
ntj
→∂∂⇒=
∂∂
⋅−
⋅−
ˆˆ
sjks
c
njes
c
nje
k
rsc
ntjrs
c
ntj
ˆˆˆˆˆ
−=−→∇⇒−=∇
⋅−
⋅−
ωωωω
0ˆ
0ˆ
ˆ
ˆ
=⋅=⋅
=×
−=×
Bs
Ds
BEsc
n
DHsc
n
( )( )DEBH
tt
DE
t
BH
HEEHHESHB
ED
⋅+⋅∂∂−=
∂∂⋅−
∂∂⋅−=
×∇⋅−×∇⋅=×⋅∇=⋅∇=
⋅= 2
1µ
ε
( ) emUDEBHSk
Ssc
n =⋅+⋅=⋅=⋅2
1ˆ
ω
Maxwell’s Symmetric Equations
PlanarMonochromatic Wave
0
0
=⋅
=⋅
=×
−=×
Bk
Dk
BEk
DHk
ω
ω
ωjt
→∂∂
sc
nj ˆω−→∇
ωjt
→∂∂
kj
−→∇ vector phasor
vector phasor
c
nk ω=:
71
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
0ˆ
0ˆ
ˆ
ˆ
=⋅=⋅
=×
−=×
Bs
Ds
BEsc
n
DHsc
n
dyadicsxwhereH
E
B
D33,,, µζξε
µζξε
=
HHIB
ED
µµε
=⋅=
⋅=
For Anisotropic Electric Media the Linear Constitutive Relations are
Planar Waves in an Source-less Media
The most general form of Linear Constitutive Relations is:
HES ×= Poynting Vector
B
D
pv
E
H
HES
×=α
αk
ev
Planar Waves
( ) DHsc
nBs
c
nEss
c
n µµ −=×=×=××
ˆˆˆˆ
2
( ) ( )[ ]EssEc
nEss
c
nD ⋅−
=××
−= ˆˆ
1ˆˆ
122
µµ SDEs ,,,ˆare coplanar
and normal to H
I µµζξε === ,0,
72
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
B
D
pv
E
H
HES
×=α
αk
ev
Planar Waves( ) ( )[ ]EssEc
nEss
c
nD ⋅−
=××
−= ˆˆ
1ˆˆ
122
µµ
SDEs ,,,ˆare coplanar
and normal to H
( ) DEnEsDsDEc
nDD
c
⋅=
⋅⋅−⋅
=⋅
=2
0
1
0
202
ˆˆ1 εµ
µε
DE
DDn
⋅⋅=
0
2 1
ε
( )
( )( )
( )( )2222
2
2
2
2
2
ˆDEDED
DDEED
D
DEE
D
DDEE
D
D
D
DEE
D
D
D
DEE
s⋅−
⋅−=⋅−
⋅−=
⋅−
⋅−
=
( )
( )( )
( )( )2222
2
2
2
2
2
:DEDEE
EDEDE
E
DED
E
DDED
E
E
E
EDD
E
E
E
EDD
S
St
⋅−
⋅−−=⋅−
⋅−−=
⋅−
⋅−
−==
If and are not collinear:DE
73
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric MediaPhase Velocity, Energy (Ray) Velocity
The phase of the field is given by
⋅−=⋅−= rs
c
ntrkt
ˆωωφ
For a constant phase front we have
⋅−=⋅−=⋅−== rds
c
ndtrdskdtrdkdtd
ˆˆ0 ωωωφ
sn
cs
kdt
rdv
srrconst
p ˆˆˆ
=====
ωφ
Phase VelocityB
D
pv
E
H
HES
×=α
αk
ev
Planar Waves
n
c
ktd
rds ==⋅ ωˆ
n
c
S
Svs e =⋅ˆ ( ) α
α
cosˆ
ˆcos
pS
Ss
e
v
Ss
S
n
cv
⋅=
=⋅
=
Define the phase velocity as
Define the energy flow velocity in the Poynting vector direction as S
SvSv ee =
→1
( ) emUDEBHSk
Ssc
n =⋅+⋅=⋅=⋅2
1ˆ
ω
( ) emee U
SS
Ssn
c
S
Svv =
⋅==
ˆ1
Energy (Ray) Velocity
The electro-magnetic energy propagates along the Poynting vector .For anisotropic media the Poynting vector is not collinear with . k
HES ×=
Return to Table of Content
sc
nk ˆω=
Electromagnetic Energy Density
74
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
BEk
DHk
ω
ω
=×
−=×
HHIB
ED
µµε
=⋅=
⋅=
Wave Equation ( ) EDHkBkEkkEDDHkHBBEk
⋅−=−=×=×=××⋅=−=×==×
εµωµωµωωεωµω
22
2222
0
0
0
ˆ kkkkkk
kk
kk
kk
k
k
k
k
s
s
s
kskk zyx
xy
xz
yz
z
y
x
z
y
x
=++=⋅
−−
−=×⇒
=
==
( )( )
( )( )
+−
+−
+−
=
−−
−
−−
−
=××22
22
22
0
0
0
0
0
0
yxzyzx
zyzxyx
zxyxzy
xy
xz
yz
xy
xz
yz
kkkkkk
kkkkkk
kkkkkk
kk
kk
kk
kk
kk
kk
kk
=
z
y
x
εε
εε
00
00
00
=
=
2
2
2
00
00
00
2
2
00
00
00
00
00
00
c
n
c
n
c
n
z
x
x
z
y
x
ω
ω
ω
εµεµεµ
εµεµεµ
εµεµεµω
εωµ
In principal dielectric axes:0
2
0
2
0
2 ,,εε
εε
εε z
zy
yx
x nnn ===
( ) 02 =⋅+×× EEkk εµω ( ) 021 =+×× − DDkk µωε
or
75
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
Wave Equation (continue – 1)
( ) 02 =⋅+×× EEkk εµω
( )
( )
( )
0
222
22
2
222
=
+−
+−
+−
z
y
x
yxz
zyzx
zyzxy
yx
zxyxzyx
E
E
E
kkc
nkkkk
kkkkc
nkk
kkkkkkc
n
ω
ω
ω
A nonzero solution exists only when
( )
( )
( )
0det
22
2
22
2
22
2
=
+−
+−
+−
yxz
zyzx
zyzx
y
yx
zxyxzyx
kkc
nkkkk
kkkkc
nkk
kkkkkkc
n
ω
ω
ω
76
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
Wave Equation (continue – 2)
( )
( )
( )
+−
+−
+−
=
+−
+−
+−
++=
22
2
22
2
22
2
22
2
22
2
22
2
2222
det
zz
zyzx
zyy
y
yx
zxyxxx
kkkk
yxz
zyzx
zyzx
y
yx
zxyxzyx
kkc
nkkkk
kkkkc
nkk
kkkkkkc
n
kkc
nkkkk
kkkkc
nkk
kkkkkkc
n
zyx
ω
ω
ω
ω
ω
ω
We obtain
−−
−−
−+
−+
−
−
+−= 2
2
22222
2
22222
2
2222
2
2222
2
222
2
2222
2
22
kc
nkkk
c
nkkk
c
nkk
c
nkk
c
nk
c
nkk
c
n y
zxz
yx
y
zz
yzy
xx
ωωωωωωω
02
2
22
2
2
2222
2
222
2
2222
2
222
2
2222
2
222
2
22
2
2
22
=
−
−+
−
−+
−
−+
−
−
−= k
c
nk
c
nkk
c
nk
c
nkk
c
nk
c
nkk
c
nk
c
nk
c
n yxz
zxy
zy
xzyx
ωωωωωωωωω
Divide by (assumed nonzero !!)
−
−
−− 2
2
222
2
22
2
2
22
kc
nk
c
nk
c
n zyx ωωω
1
2
222
2
2
22
2
2
2
222
2
=−
+−
+−
c
nk
k
c
nk
k
c
nk
k
z
z
y
y
x
x
ωωω
This is a quadratic equation of k2 (k6 terms drop – next slide). For each set sx, sy, sz it yields two solutions for k2: k1
2 and k22
=
z
y
x
z
y
x
s
s
s
k
k
k
k
2
2
222
2
2
22
2
2
2
222
21
k
c
nk
s
c
nk
s
c
nk
s
z
z
y
y
x
x =−
+−
+− ωωω
( )skkk ˆ,, 21 ε
=
77
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
02
2
22
2
2
22222
2
222
2
22222
2
22
2
2
22222
2
222
2
22
2
2
22
=
−
−+
−
−+
−
−+
−
−
− k
c
nk
c
nskk
c
nk
c
nskk
c
nk
c
nskk
c
nk
c
nk
c
n yxz
zxy
yzx
zyxωωωωωωωωω
( ) ( )( ) ( ) ( ) 02222
4
44222
2
2622222
4
44222
2
2622222
4
44222
2
262
222
6
62222222
4
44222
2
26
=++−+++−+++−+
+++−+++−
ksnnc
ksnnc
ksksnnc
ksnnc
ksksnnc
ksnnc
ks
nnnc
knnnnnnc
knnnc
k
zyxzyxzyzxyzxyxzyxzyx
zyxzyzxyxzyx
ωωωωωω
ωωω
( ) ( ) ( ) ( )[ ] 0111 222
6
62222222222
4
44222222
2
2
=+−+−+−−++ zyxxzyyzxzyxzzyyxx nnnc
ksnnsnnsnnc
ksnsnsnc
ωωω
This is a quadratic equation of k2 .For each set sx, sy, sz it yields two solutions for k2: k1
2 and k22
( )skkk ˆ,, 21 ε
=
2
2
222
2
2
22
2
2
2
222
2 1
k
c
nk
s
c
nk
s
c
nk
s
z
z
y
y
x
x =−
+−
+− ωωω
02
2
22
2
2
2222
2
222
2
2222
2
222
2
2222
2
222
2
22
2
2
22
=
−
−+
−
−+
−
−+
−
−
− k
c
nk
c
nkk
c
nk
c
nkk
c
nk
c
nkk
c
nk
c
nk
c
n yxz
zxy
zy
xzyx
ωωωωωωωωω
=
z
y
x
z
y
x
s
s
s
k
k
k
k
Wave Equation (continue – 2a)
1222 =++ zyx sss
78
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
Wave Equation (continue – 3)
( )
( )
( )
0
222
22
2
222
=
+−
+−
+−
z
y
x
yxz
zyzx
zyzxy
yx
zxyxzyx
E
E
E
kkc
nkkkk
kkkkc
nkk
kkkkkkc
n
ω
ω
ω
A solution (self-mode) is:
( ) ( )
( ) ( )
( ) ( ) 0
0
0
222
2
222
2
222
2
=+++
++−
=+++
++−
=+++
++−
⋅
⋅
⋅
Ek
zzyyxxzzzyxz
Ek
zzyyxxyyzyx
y
Ek
zzyyxxxxzyxx
EkEkEkkEkkkc
n
EkEkEkkEkkkc
n
EkEkEkkEkkkc
n
ω
ω
ω
−
−
−
2
2
2
2
2
2
kc
n
k
kc
n
k
kc
n
k
z
z
y
y
x
x
ω
ω
ω
( )Ek ⋅−
( )
( )
( )
−
⋅
−
⋅
−
⋅
−=
22
2
2
22
kcn
Ekk
kc
n
Ekk
kcn
Ekk
E
E
E
z
z
y
y
x
x
z
y
x
ω
ω
ω
In principal dielectric axes
Return to Table of Content
79
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
Wave-Vector Surface
The above equation can be represented by a three-dimensional surface in k space.
( )
( )
( )
0det
222
22
2
222
=
+−
+−
+−
yxz
zyzx
zyzxy
yx
zxyxzyx
kkc
nkkkk
kkkkc
nkk
kkkkkkc
n
ω
ω
ω
The understand how this surface look let find the intersection of the surface with kz = 0.
( )
( ) 0
00
0
0
det
222
2
22
222
222
2
2
22
=
−
−
−
+−
=
+−
−
−
yxxy
yx
yxz
yxz
xy
yx
yxyx
kkkc
nk
c
nkk
c
n
kkc
n
kc
nkk
kkkc
n
ωωω
ω
ω
ω
80
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
Wave-Vector Surface (continue – 1)
The intersection of the surface with kz = 0 is given by two equations, derived from:
( ) 0222
222
2
22
=
+−
−
−
−
yxz
yxxy
yx kk
c
nkkk
c
nk
c
n ωωω2
22
=+
c
nkk z
yx
ω
( ) ( ) 1//
22
=+cn
k
cn
k
x
y
y
x
ωω
( )
0
00
0
0
222
2
2
22
=
+−
−
−
z
y
x
yxz
xy
yx
yxyx
E
E
E
kkc
n
kc
nkk
kkkc
n
ω
ω
ω
=
=
zz
y
x
EE
E
E
E 0
0
2
−
−
=
=
0
22
1 yx
yx
z
y
x
kc
n
kk
E
E
E
Eω
021 =⋅ EE
Ellipse
The electric fields corresponding to those two solutions are:
on Ellipse
on Circle Since the coefficients of the electric fields are real the two
solutions represents two linear polarized planar waves.
Circle
81
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
Wave-Vector Surface (continue -2)
Intersection of the surface with kz = 0, are a circle and an ellipse in kx, ky plane.
2
22
=+
c
nkk z
yx
ω( ) ( ) 1
// 2
2
2
2
=+cn
k
cn
k
x
y
y
x
ωω
Intersection of the surface with ky = 0, are a circle and an ellipse in kx, kz plane.
2
22
=+
c
nkk y
zx
ω
( ) ( ) 1// 2
2
2
2
=+cn
k
cn
k
x
z
z
x
ωω
Intersection of the surface with kx = 0, are a circle and an ellipse in ky, kz plane.
2
22
=+
c
nkk x
zy
ω( ) ( ) 1
// 2
2
2
2
=+cn
k
cn
k
y
z
z
y
ωω
Suppose nx < ny < nz.
( ) 0222
2
2
2
22
2
=
−
−
−
+−
yxx
y
yx
yxz kkk
c
nk
c
nkk
c
n ωωω
0
2
0
2
0
2 ,,εε
εε
εε z
z
y
yx
x nnn ===
82
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
Wave-Vector Surface (continue – 3) The complete k surface is double;that is, it consists of an inner shell and an outer shell.
For any given direction of the vectork, there are two possible values for thewave-number k (eigenmodes). Thereforeare also two values for the phase velocity.
From the figure we can see that the inner and outer shells of the k surface intersect at four points in each quadrant of the kx, kz plane (for nx<ny<nz). Those points define two directions for which the twovalues of k are equal. Those two directions are called Optical Axes.
The structure of an anisotropic mediumpermits two monochromatic plane wavesthat are polarized orthogonally with respect to each other.
If nx,ny and nz are different the crystal hastwo optical axes and is said to be biaxial.
0
2
0
2
0
2 ,,εε
εε
εε z
zy
yx
x nnn ===nx < ny < nz.
83
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
Wave-Vector Surface (continue -4)
Let compute the Optical Axis direction. 0
2
0
2
0
2 ,,εε
εε
εε z
z
y
yx
x nnn ===
We can see that the Optical Axis is inx, z plane, therefore
=
=
=
δ
δω
cos
0
sin
cn
s
s
s
k
k
k
k
k OA
OAz
y
x
OA
OAz
y
x
OA
2
22
=+
c
nkk y
zx
ω( ) ( ) 1
// 2
2
2
2
=+cn
k
cn
k
x
z
z
x
ωω
22
=
c
n
c
n yOAωω 1
cossin2
22
2
22
=+x
OA
z
OA
n
n
n
n δδ
must be at the intersection of OAk
and
yOA nn =( )
1coscos1
2
22
2
22
=+−
x
y
z
y
n
n
n
n δδ22
22
2cos −−
−−
−−
=zx
zy
OAnn
nnδ
22
22
2sin −−
−−
−−
=zx
yx
OAnn
nnδ 22
22
2tan −−
−−
−−
=zy
yx
OAnn
nnδ
If δ is less than 45º, the crystal is said to be a positive biaxial crystal.
If δ is greater than 45º, the crystal is said to be a negative biaxial crystal.
nx < ny < nz.
−−
±= −−
−−
−22
22
1
2,1sin
zx
yx
OAnn
nnδ
84
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
Wave-Vector Surface (continue -5)
0
2
0
2
0
2 ,,εε
εε
εε z
z
y
yx
x nnn ===nx < ny < nz.
85
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
Double Refraction at a Birefringent Crystal Boundary
( ) 0ˆ
=−× ti kkn
At the Crystal Boundary defined by the normalthe following condition must be satisfied
n
The refracted wave is, in general, a mixture of two eigenmodes
( ) ( )222111 ,,, snc
ksnc
k
εωεω ==
We can write
( ) ( ) 222111 sin,sin,sin θεθεθ skskk ii
==
This looks like Snell’s Law but k1 and k2 are not constant, and we must add thequadratic equation in k2:
( ) ( ) ( ) ( )[ ] 0111 222
6
62222222222
4
44222222
2
2
=+−+−+−−++ zyxxzyyzxzyxzzyyxx nnnc
ksnnsnnsnnc
ksnsnsnc
ωωω
0
2
0
2
0
2 ,,εε
εε
εε z
z
y
yx
x nnn ===
86
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
Wave-Vector Surface (continue – 6)
If two of the refraction indices are equalthe crystal has one optical axis and is said to be uniaxial.
zyx εεε ≠=nx = ny =no≠ nz=ne
For each set sx, sy, sz it yields two solutions for k2: k1 and k2 ( )skkk ˆ,, 21 ε=
The quadratic equation of k2is:
2
2
222
2
2
222
2 11
k
c
nk
s
c
nk
s
e
z
o
z =−
+−
−ωω
( )[ ] ( ) ( )[ ] 0111 24
6
6222222
4
442222
2
2
=+++−−+− eozezoozezo nnc
ksnsnnc
ksnsnc
ωωω
( ) ( ) ( ) ( )[ ] 0111 222
6
62222222222
4
44222222
2
2
=+−+−+−−++ zyxxzyyzxzyxzzyyxx nnnc
ksnnsnnsnnc
ksnsnsnc
ωωω
nx = ny =no≠ nz=ne
Uniaxial Crystals
One other way is to substitute in:
87
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
Wave-Vector Surface (continue – 7)zyx εεε ≠=nx = ny =no≠ nz=ne
( )[ ] ( ) ( )[ ] 0111 24
6
6222222
4
442222
2
2
=+++−−+− eozezoozezo nnc
ksnsnnc
ksnsnc
ωωω
( ) ( ) 01 2
2
222222
2
2
2
222 =
−−−+
− oeozoe nc
kknnsnc
knωω
( )[ ] 01 22
2
2222222
2
22 =
−+−
− eoezozo nnc
knsnsnc
kωω
The Wave-Vector surface decomposes into two separate shells
02
2
22 =− on
ck
ω
( )[ ] 022
2
2222222 =−++ eoezoyx nn
cknsnss
ω
02
2
2222 =−++ ozyx n
ckkk
ωSpherical surface
012
2
2
2
2
2
2
22
=−++
o
z
e
yx
nc
k
nc
kk
ωω Ellipsoid of Revolutionsurface
These two surfaces are tangent along the z axis.
Uniaxial Crystals (continue – 1)
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88
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
Double Refraction at a Uniaxial Crystal BoundaryFor the Uniaxial Crystal is easy to solve the Boundary condition graphically:
( ) eeooii skkk θεθθ sin,sinsin 2
==
The Wave-Vector surface decomposes into two separate shells:• the sphere of the ordinary wave with• the ellipsoid of extraordinary wave with
.constko =( )2, ske
ε
is found graphically using the equality: Snell’s Law
ooii kk θθ sinsin =ok
The intersection of the normal to the boundary from the end with the ellipse defines the direction of
ek
ok
The intersection of the plane defined by and passing through the common centers of the sphere and the ellipsoid will give a circle and an ellipse, respectively.
nk i
,
iik θsin
ik
ok
iθ
oθ
Intersection ofnormal surface withplane of incidence
in
iik θsin
UniaxialCrystal
Incidentray
n
OpticalAxis
oE
eo nn >
89
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
Double Refraction at a Uniaxial Crystal Boundary
90
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
Wave-Vector Surface (continue – 8)
nx = ny = nz=n
If all three indices are equal, then theWave-vector surface degenerates to a singlesphere, and the crystal is optically isotropic.
εεεε === zyx
2
222
c
nk
ω=
02
2
2
=
z
y
x
zzyzx
zyyyx
zxyxx
E
E
E
kkkkk
kkkkk
kkkkk
2222zyx kkkk ++=
( ) 02 =⋅+×× EEkk εµω
We can see that:
{ }{ }{ }
0
,,
,,
,,
det2
2
2
≡
=
zyxz
zyxy
zyxx
zzyzx
zyyyx
zxyxx
kkkk
kkkk
kkkk
kkkkk
kkkkk
kkkkk
and: 0=⋅=++ EkEkEkEk zzyyxx
0/ εε=n
Isotropic Crystal
Return to Table of Content
91
SOLO ELECTROMAGNETICS
Optically Isotropic (Cubic) Crystals n
Sodium Chloride (NaCl)DiamondFluoriteCdTeGaAs
1.5442.4171.3922.693.40
Uniaxial Positive Crystals nO nE
IceQuartzZirconRutile ( TiO2 )
BeOZnS
1.309 1.3101.544 1.5531.923 1.9682.616 2.9031.717 1.7322.354 2.358
Uniaxial Negative Crystals nO nE
BerylSodium NitrateCalciteTourmaline
1.598 1.5901.587 1.3361.658 1.4861.669 1.638
Refractive Index of Some Typical Crystals
92
SOLO ELECTROMAGNETICS
Biaxial Crystals n1 n2 n3
GypsumFeldsparMicaTopazNaNO2
SbSIYAlO3
1.520 1.523 1.5301.522 1.526 1.5301.552 1.582 1.5881.619 1.620 1.6271.344 1.411 1.6512.7 3.2 3.81.923 1.938 1.947
Refractive Index of Some Typical Biaxial Crystals
94
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
Crystal Types (continue -1)
95
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
Crystal Types (continue – 2)
96
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
Crystal Types (continue – 3)
A.Mermin, “Solid State Physics”,Holt, Reinhart and Winston, 1976,p.122
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97
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
Orthogonality Properties of the EigenmodesAssume the two monochromatic waves corresponding to a given direction
are defined by ands
1111 ,,, nDEH
2222 ,,, nDEH
Start with Lorentz Reciprocity Theorem
Use
HB
BEsc
n
µ=
=×
ˆ Esc
nH ×= ˆ
µ
( )[ ] ( )[ ]122
211 ˆˆˆˆ EsEs
c
nEsEs
c
n ××⋅=××⋅µµ
( ) ( ) ( ) ( )122
211 ˆˆˆˆ EsEs
c
nEsEs
c
n ×⋅×=×⋅×µµ
( ) ( )1221 ˆˆ HEsHEs ×⋅=×⋅
Since we have21 nn ≠
( ) ( ) 0ˆˆ 21 =×⋅× EsEs ( ) ( ) 0ˆˆ 1221 =×⋅=×⋅ HEsHEs
( ) ( )1221 HEHE
×⋅∇=×⋅∇
sc
nj ˆω−→∇
0=⋅=⋅ sHEH
98
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
Orthogonality Properties of the Eigenmodes (continue – 1)
We obtain
0ˆˆˆˆ221121212121212121 =⋅=⋅=⋅=⋅=⋅=⋅=⋅=⋅=⋅=⋅=⋅ HEHEHsHsDsDsEEEDDEDDHH
( ) ( ) 0ˆˆ 1221 =×⋅=×⋅ HEsHEs
( ) ( )
( ) ( ) 0ˆˆ
0ˆˆ
211
221
2
ˆ
2121
211
ˆ
2121
22
2
11
1
11
=⋅=⋅=×⋅−=×⋅
=⋅=⋅×=×⋅
−
−=×
=×
=
DDn
cDE
n
cHsEHEs
HHn
cHEsHEs
Dn
cHs
Bn
cEs
HB
ε
µµ
( ) ( )
( ) ( ) 0ˆˆ
0ˆˆ
121
112
1
ˆ
1212
122
ˆ
1212
11
1
22
2
22
=⋅=⋅=×⋅−=×⋅
=⋅=⋅×=×⋅
−
−=×
=×
=
DDn
cDE
n
cHsEHEs
HHn
cHEsHEs
Dn
cHs
Bn
cEs
HB
ε
µµ
99
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
Orthogonality Properties of the Eigenmodes (continue – 2)
We obtained
The two solutions correspond to two different possible linear polarizations of a wave propagating in direction and these two solutions have mutually
orthogonal polarizations. s
0ˆˆˆˆ221121212121212121 =⋅=⋅=⋅=⋅=⋅=⋅=⋅=⋅=⋅=⋅=⋅ HEHEHsHsDsDsEEEDDEDDHH
Return to Table of Content
100
SOLO ELECTROMAGNETICS
Phase Velocity, Energy (Ray) Velocity for Planar Waves
The phase of the field is given by
⋅−=⋅−= rs
c
ntrkt
ˆωωφ
For a constant phase front we have
⋅−=⋅−=⋅−== rds
c
ndtrdskdtrdkdtd
ˆˆ0 ωωωφ
sn
cs
kdt
rdv
srrconst
pˆˆ:
ˆ
=====
ωφ
Phase VelocityB
D
pv
E
H
HES
×=α
αk
ev
Planar Waves
n
c
ktd
rds ==⋅ ωˆ
n
c
S
Svs e =⋅ :ˆ ( ) α
α
cosˆ
cosˆ
pS
Ss
e
v
Ss
S
n
cv
=⋅
=⋅
=
Define the phase velocity as
Define the energy flow velocity in the Poynting vector direction as S
SvSv ee =
→1
( ) emUDEBHSk
Ssc
n =⋅+⋅=⋅=⋅2
1ˆ
ω
( ) em
ee U
SS
Ssn
c
S
Svv =
⋅==
ˆ1
:
Energy (Ray) Velocity
The electro-magnetic energy propagates along the Poynting vector .For anisotropic media the Poynting vector is not collinear with . k
HES ×=
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101
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
Ray-Velocity Surface
We defined the Energy-Ray Velocity as
Let compute
( ) HDHEDEHU
DHEU
DvemUemem
e =
⋅−
⋅=××=×
0
11
( ) ( ) DEHHEHEHU
HHEU
HvDvvemUemem
eee1
/0
1111 −−=−=
⋅−
⋅=××=×=×× ε
µµµ
In principal dielectric axes:
=
=−
2
2
2
2
2
2
1
00
00
00
100
01
0
001
1
z
y
x
z
y
x
n
c
n
c
n
c
εµ
εµ
εµ
εµ
( ) em
ee U
SS
Ssn
c
S
Svv =
⋅==
ˆ1
:
102
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
Ray-Velocity Surface (continue – 1)
We can write ( ) 01 1 =+×× − DDvv ee εµ
as
( )
( )
( )
0
22
2
22
2
22
2
=
+−
+−
+−
z
y
x
eyexz
ezeyezex
ezeyezexy
eyex
ezexeyexezeyx
D
D
D
vvn
cvvvv
vvvvn
cvv
vvvvvvn
c
We obtain non-zero solutions only when
( )
( )
( )
0det
22
2
22
2
22
2
=
+−
+−
+−
eyexz
ezeyezex
ezeyezexy
eyex
ezexeyexezeyx
vvn
cvvvv
vvvvn
cvv
vvvvvvn
c
103
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
Intersection of the surface with vez = 0, are a circle and an ellipse in vex, vey plane.
2
22
=+
z
eyex n
cvv ( ) ( ) 1
// 2
2
2
2
=+x
ey
y
ex
nc
v
nc
v
Intersection of the surface with vey = 0, are a circle and an ellipse in vex, vez plane.
2
22
=+
y
ezex n
cvv
( ) ( ) 1// 2
2
2
2
=+x
z
z
ex
nc
k
nc
v
Intersection of the surface with vex = 0, are a circle and an ellipse in vey, vez plane.
2
22
=+
x
ezey n
cvv ( ) ( ) 1
// 2
2
2
2
=+y
ez
z
ey
nc
v
nc
v
Suppose nx < ny < nz.
Ray-Velocity Surface (continue – 2)The previous equation has the same structure as the Wave-Vector Surface, therefore
can be represented by a three-dimensional surface in ve space. vex, vey and vez are the
components of ve in the principal axes of the dielectric.
0
2
0
2
0
2 ,,εε
εε
εε z
z
y
yx
x nnn ===
104
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
nx < ny < nz. The complete ve surface is double;
that is, it consists of an inner shell and an outer shell.
For any given direction of the vector
ve, there are two possible values for the
energy-ray velocity ve (eigenmodes):
ve1 and ve2. From the figure we can see that the inner
and outer shells of the ve surface intersect
at four points in each quadrant of the vex, vez plane (for nx<ny<nz). Those points define two directions for which the two
values of ve are equal. Those directions are
called Ray Axes and are distinct from theOptic Axes of the dielectric.
Ray-Velocity Surface (continue – 3)
105
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
Let compute the Ray Axis direction.
We can see that the Ray Axis is inx, z plane, therefore
==
=
→
RA
RA
RA
RARAe
RAez
ey
ex
RAe n
cSv
v
v
v
v
δ
δ
cos
0
sin
1
2
22
=+
y
ezex n
cvv ( ) ( ) 1
// 2
2
2
2
=+x
ez
z
ex
nc
v
nc
v
22
=
yRA n
c
n
c 1cossin
2
22
2
22
=+RA
RAx
RA
RAz
n
n
n
n δδ
must be at the intersection of RAev
and
yRA nn =( )
1coscos1
2
22
2
22
=+−
y
RAx
y
RAz
n
n
n
n δδ22
22
2coszx
zy
RAnn
nn
−−
=δ22
22
2sinzx
yx
RAnn
nn
−−
=δ 22
22
2tanzy
yx
RAnn
nn
−−
=δ
−−
±= −22
22
1
2,1 sinzx
yx
RAnn
nnδ
0
2
0
2
0
2 ,,εε
εε
εε z
z
y
yx
x nnn ===Ray-Velocity Surface (continue – 4)
RA
x
z
yz
xy
x
zOA
n
n
nn
nn
n
n δδ 2
2
2
22
22
2
2
2 tantan =−−
= Ray Axes are distinct fromOptical Axes.
106
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
( )
( )
( )
+−
+−
+−
=
+−
+−
+−
++=
22
2
22
2
22
2
22
2
22
2
22
2
2222
det
eze
z
ezeyezex
ezeyeye
y
eyex
ezexeyexexe
x
vvvv
eyex
z
ezeyezex
ezeyezex
y
eyex
ezexeyexezey
x
vvn
cvvvv
vvvvn
cvv
vvvvvvn
c
vvn
cvvvv
vvvvn
cvv
vvvvvvn
c
ezeyexe
We obtain
−−
−−
−+
−+
−
−
+−= 2
2
2222
2
2222
2
222
2
222
2
22
2
222
2
2
e
y
ezexe
z
eyexe
y
eze
z
eye
z
e
y
exe
x
vn
cvvv
n
cvvv
n
cvv
n
cvv
n
cv
n
cvv
n
c
02
2
22
2
222
2
22
2
222
2
22
2
222
2
22
2
22
2
2
=
−
−+
−
−+
−
−+
−
−
−= e
y
e
x
eze
z
e
x
eye
z
e
y
exe
z
e
y
e
x
vn
cv
n
cvv
n
cv
n
cvv
n
cv
n
cvv
n
cv
n
cv
n
c
Divide by (assumed nonzero !!(
−
−
−− 2
2
22
2
22
2
2
e
z
e
y
e
x
vn
cv
n
cv
n
c
1
2
22
2
2
22
2
2
22
2
=−
+−
+−
z
e
ez
y
e
ey
x
e
ex
n
cv
v
n
cv
v
n
cv
v
Ray-Velocity Surface (continue – 5(
Return to Table of Content
107
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
Index Ellipsoid
The electric energy density is given by: ∗∗−∗ ⋅⋅=⋅⋅=⋅= EEDDDEU e εε
2
1
2
1
2
1 1
Let find the Surfaces of Equal Energy expressed in the principal dielectric axes:
z
z
y
y
x
xe
DDDDDU
εεεε
22212 ++=⋅⋅= ∗−
We have: ( ) ( ) zyxin iii ,,// 002 === εεεµεµ
Define: ezeyex UEzUEyUEx 2/:2/:2/: 000 εεε ===
The surfaces of equal energy expressed in the principal dielectric axes are:
1/1/1/1 2
2
2
2
2
2
=++zyx n
z
n
y
n
xFresnel’s Ellipsoid
0ˆˆˆˆ221121212121212121 =⋅=⋅=⋅=⋅=⋅=⋅=⋅=⋅=⋅=⋅=⋅ HEHEHsHsDsDsEEEDDEDDHH
We found that the two eigenmodes, of each direction , must satisfy:
s
2222 zzyyxxe EEEEEU εεεε ++=⋅⋅= ∗
108
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
Index Ellipsoid (continue – 1(
The electric energy density is given by: ∗∗−∗ ⋅⋅=⋅⋅=⋅= EEDDDEU e εε
2
1
2
1
2
1 1
z
z
y
y
x
xe
DDDDDU
εεεε
22212 ++=⋅⋅= ∗−
We have: ( ) ( ) zyxin iii ,,// 002 === εεεµεµ
Define: 000 2/:2/:2/: εεε ezeyex UDzUDyUDx ===
The surfaces of equal energy expressed in the principal dielectric axes are:
12
2
2
2
2
2
=++zyx n
z
n
y
n
xIndex Ellipsoid,
Optical Indicatrix, Reciprocal Ellipsoid
0ˆˆˆˆ221121212121212121 =⋅=⋅=⋅=⋅=⋅=⋅=⋅=⋅=⋅=⋅=⋅ HEHEHsHsDsDsEEEDDEDDHH
We found that the two eigenmodes, of each direction , must satisfy:
s
2222 zzyyxxe EEEEEU εεεε ++=⋅⋅= ∗
Let find the Surfaces of Equal Energy expressed in the principal dielectric axes:
109
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
Index Ellipsoid (continue – 2(
Since we found that the index ellipsoid is used to find the two indexesof refraction and the two corresponding directions that are in the planenormal to that passes through ellipsoid center O and defines the ellipse.
0ˆˆ21 =⋅=⋅ DsDs
21 DandDs
Sπ
Assume that is one of the expected solutions.
02/ εeUDOP =→
The normal at P to the ellipsoid PV is:
EU
zD
yD
xD
U
zz
yy
xx
ez
z
y
y
x
x
e
zyx
PV
2111
2
111
00
0
εεεε
ε
εεεε
=
++=
++=
∧∧∧
∧∧∧→
Since are coplanar PV is in the plane defined by and sDπ D
sED
,,
The plane tangent to ellipsoid at Pis normal to PV ( (, intersects plane along PT that will be normal to and , therefore it gives direction.
Tπ
E
E
H
Sπ
s 0ˆˆ 221121 =⋅=⋅=⋅=⋅ HEHEHsHs
110
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
Index Ellipsoid (continue – 3(
PT is at the intersection of the tangent plane to the ellipsoid at P to the planenormal to and containing PO. Therefore PT is tangent to the ellipse containing PO.
Tπ Sπs
- Since PT is in , it is normal to Sπ s
E
- Since PT is in the tangent plane to the ellipsoid, it is normal to PV ( (
Tπ
s
H
EH - Since is also normal to and PT defines the direction of
is also normal toH D
The tangent to an ellipse is only normal to the radius vector if it is one of the axes of the ellipse.
We conclude that the directions of are the directions of the axes of the ellipse obtained by the intersection of the ellipsoid with the plane normal to .
21 DandD
s
111
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
Index Ellipsoid (continue – 4(
To find for the propagation direction we find the ellipse obtained by the intersection of the plane normal to , at the origin O, with the Index Ellipsoid.
21 DandD ssSπ
21 DandD are in the direction of ellipsoid axes, and therefore perpendicular to each other
is on the normal to ellipsoidat OP = and in the planeof and . In the same plane and normal to is
E
sE
DπD
Dπ
HE
HES
××=
→
1
112
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
Index Ellipsoid (continue – 5( 12
2
2
2
2
2
=++zyx n
z
n
y
n
xnx < ny < nz.
For the propagation direction of the Optical Axis the intersection of the plane normal to , at the origin O, with the Index Ellipsoid is a Circle.
OAsSπOAs
1D21 DandD are in any direction on this circle, and therefore
for any we can find the perpendicular to it. 2D
Conical Refraction on the Optical Axis
We want to find the direction of vectors. E
We have: [ ] ET
E
E
E
EsDs
z
y
x
zxOAOA ⋅=
⋅=⋅⋅=⋅=
δεδεε cos0sinˆˆ0
where: [ ]TzxT δεδε cos0sin:=
=
δ
δ
cos
0
sin
ˆOAsFor nx < ny < nz we found that in principal dielectric axes
−−
±= −−
−−
−22
22
1
2,1 sinzx
yx
nn
nnδ
SDEs ,,,ˆare coplanar
and normal toHWe also found:
DE
DDn
⋅⋅=
0
2 1
ε constn
DDE
y
==⋅2
0
2
εWe found: ynn =
113
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
Index Ellipsoid (continue – 6( 12
2
2
2
2
2
=++zyx n
z
n
y
n
xnx < ny < nz.
Conical Refraction on the Optical Axis (continue – 1(
OAs Let draw in principal dielectric axes coordinate O,x,y,z,
starting from O, the vectors and . They define theplane Bπ
T
0=⋅ ET
Since we can find in vectors such thatEBπ
constDE =⋅
and , , and are in the same plane. E OAsDHE
HES
××=
→
1
Tπ Therefore the locus of is the projection of the
circle of from plane to an ellipse in plane having the Oy axis in common.
E
D Sπ
Since are in the same plane with and will be on a conical surface with O as a vertex having
and on its surfacesT
OAsDHE
HES
××=
→
1→
S1
OAs
114
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
Index Ellipsoid (continue – 7(
Conical Refraction on the Optical Axis (continue – 2(
Take a biaxial crystal and cut it so that two parallel facesare perpendicular to the Optical Axis. If a monochromaticunpolarized light is normal to one of the crystal faces, theenergy will spread out in the plate in a hollow cone, thecone of internal conical refraction.
When the light exits the crystal the energy and wave directions coincide, and the light will form a hollow cylinder.
This phenomenon was predicted by William Rowan Hamiltonin 1832 and confirmed experimentally by Lloyd, a year later
(Born & Wolf(.
Because it is no easy to obtain an accurate parallel beam ofmonochromatic light on obtained two bright circles
(Born & Wolf(.
William RowanHamilton
(1805-1855(
115
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
Index Ellipsoid (continue – 8(
Conical Refraction on the Optical Axis (continue – 3(
Return to Table of Content
116
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
BEsc
n
DHsc
n
=×
−=×
ˆ
ˆ
HHIB
ED
µµε
=⋅=
⋅=( ) E
n
cD
n
cHs
n
cBs
n
cEss ⋅−=−=×=×=×× εµµµ
2
2
2
2
ˆˆˆˆ
1ˆˆ
0
0
0
ˆˆ 222 =++=⋅
−−
−
=×⇒
= zyx
xy
xz
yz
z
y
x
sssss
ss
ss
ss
s
s
s
s
s
=
z
y
x
εε
εε
00
00
00
==
02
02
02
22
2
00
00
00
1
εε
εε
εε
εµε
µεµ
n
n
n
nn
c
z
y
x
( )( )
( )( )
+−
+−
+−
=
+−
+−
+−
=
−−
−
−−
−=××
2
2
2
22
22
22
1
1
1
0
0
0
0
0
0
ˆˆ
zzyzx
zyyyx
zxyxx
yxzyzx
zyzxyx
zxyxzy
xy
xz
yz
xy
xz
yz
sssss
sssss
sssss
ssssss
ssssss
ssssss
ss
ss
ss
ss
ss
ss
ss
Equation of Wave Normal
( ) 0ˆˆ2
2
=⋅+×× En
cEss εµ
( ) 0ˆˆ2
21 =+×× − D
n
cDss µε
or
117
SOLO ELECTROMAGNETICSPlanar Waves in an Source-less Anisotropic Electric Media
( ) 0ˆˆ2
2
=⋅+×× En
cEss εµ
0
1
1
1
2
02
2
02
2
02
=
+−
+−
+−
z
y
x
zz
zyzx
zyyy
yx
zxyxxx
E
E
E
sn
ssss
sssn
ss
sssssn
εε
εε
εε
The solution is obtained when
−
−+
−
−+
−
−+
−
−
−= 111111111
02
02
2
02
02
2
02
02
2
02
02
02 ε
εε
εε
εε
εε
εε
εε
εε
εε
εnn
snn
snn
snnn
yxz
zxy
yzx
zyx
+−
+−
+−
=
2
02
2
02
2
02
1
1
1
det0
zz
zyzx
zyyy
yx
zxyxxx
sn
ssss
sssn
ss
sssssn
εε
εε
εε
−−
−−
−+
−+
−
−
+−= 1111111
02
22
02
22
02
2
02
2
02
02
2
02 ε
εε
εε
εε
εε
εε
εε
εn
ssn
ssn
sn
snn
sn
yzx
zyx
yz
zy
zyx
x
Equation of Wave Normal (continue – 1(
118
SOLO ELECTROMAGNETICSPlanar Waves in an Source-less Anisotropic Electric Media
0111111111
1
1
1
det
0
2
0
2
2
0
2
0
2
2
0
2
0
2
2
0
2
0
2
0
2
2
0
2
2
0
2
2
0
2
=
−
−+
−
−+
−
−+
−
−
−=
=
+−
+−
+−
εε
εε
εε
εε
εε
εε
εε
εε
εε
εε
εε
εε
nns
nns
nns
nnn
sn
ssss
sssn
ss
sssssn
yxz
zxy
yzx
zyx
zz
zyzx
zyy
y
yx
zxyxxx
Divide by (assumed nonzero !!(
−
−
−− 111
02
02
02
2
εε
εε
εε
nnnn zyx
2
0
2
2
0
2
2
0
2
2 1
nn
s
n
s
n
s
z
z
y
y
x
x =−
+−
+−
εε
εε
εε 222
2
22
2
22
2 1
nnn
s
nn
s
nn
s
z
z
y
y
x
x =−
+−
+−
0
2
0
2
0
2 ,,εε
εε
εε z
zy
yx
x nnn ===
Equation of Wave Normal (continue – 2(
Fresnel equation of wave normal
Fresnel's wave surface, found by Augustin-Jean Fresnel in 1821, is a quartic surface describing the propagation of light in an optically biaxial crystal
Augustin-Jean Fresnel(1788 – 1827(
119
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
Equation of Wave Normal (continue – 3(
Fresnel equation of wave normal
02
0
2
0
222
0
2
0
222
0
2
0
222
0
2
0
2
0
=
−
−+
−
−+
−
−+
−
−
− nnsnnnsnnnsnnnn yx
zzx
y
yzx
zyx
εε
εε
εε
εε
εε
εε
εε
εε
εε
022
00
42
00
6222
00
42
00
6222
00
42
00
62
000
2
000000
4
000
6
=+
+−++
+−++
+−+
+
++−
+++−
nsnsnsnsnsnsnsnsns
nnn
z
yxz
yxzy
zxy
zxyx
zy
xzy
x
zyxzyzxyxzyx
εε
εε
εε
εε
εε
εε
εε
εε
εε
εε
εε
εε
εε
εε
εε
εε
εε
εε
εε
εε
εε
εε
εε
εε
( ) ( ) ( ) 0111000
22
00
2
00
2
00
42
0
2
0
2
0
=+
−+−+−−
++
εε
εε
εε
εε
εε
εε
εε
εε
εε
εε
εε
εε zyx
xzy
yzx
z
yxz
zy
y
xx nsssnsss
01111111110
2
0
2
2
0
2
0
2
2
0
2
0
2
2
0
2
0
2
0
2=
−
−+
−
−+
−
−+
−
−
−
εε
εε
εε
εε
εε
εε
εε
εε
εε
nns
nns
nns
nnnyx
zzx
y
yzx
zyx 6n
This is a quadratic equation of n2 .For each set sx, sy, sz it yields two solutions for n2: n1
2 and n22
( )snnn ˆ,, 21 ε
=
222
2
22
2
22
2 1
nnn
s
nn
s
nn
s
z
z
y
y
x
x =−
+−
+−
1222 =++ zyx sss
Return to Table of Content
120
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric MediaEquation of Wave Normal (continue – 4(
Intersection of the surface with n sz = 0, are a circle and an ellipse in n sx, n sy plane.
( ) ( ) 222zyx nsnsn =+ ( ) ( )
12
2
2
2
=+x
y
y
x
n
sn
n
sn
Intersection of the surface with n sy = 0, are a circle and an ellipse in n sx, n sz plane.
( ) ( ) 222yzx nsnsn =+ ( ) ( )
12
2
2
2
=+x
z
z
x
n
sn
n
sn
Intersection of the surface with n sx = 0, are a circle and an ellipse in n sy, n sz plane.
( ) ( ) 222xzy nsnsn =+
( ) ( )12
2
2
2
=+y
z
z
y
n
sn
n
sn
Suppose nx < ny < nz. 0
2
0
2
0
2 ,,εε
εε
εε z
z
y
yx
x nnn ===
( ) ( ) ( ) ( )[ ] 0111 22222222222224222222 =+−+−+−−++ zyxxzyyzxzyxzzyyxx nnnnsnnsnnsnnnsnsnsn
121
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
( ) ( ) ( ) ( )[ ] 0111 22222222222224222222 =+−+−+−−++ zyxxzyyzxzyxzzyyxx nnnnsnnsnnsnnnsnsnsn
0
2
0
2
0
2 ,,εε
εε
εε z
z
y
yx
x nnn ===
Equation of Wave Normal (continue – 5(
122
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
( ) ( ) ( ) 0111000
22
00
2
00
2
00
42
0
2
0
2
0
=+
−+−+−−
++
εε
εε
εε
εε
εε
εε
εε
εε
εε
εε
εε
εε zyx
xzy
yzx
z
yxz
zy
y
xx nsssnsss
Start with the quadratic equation in n2 .
The condition to have n12 = n2
2 is:
( ) ( ) ( ) 04111000
2
0
2
0
2
0
2
2
00
2
00
2
00
=
++−
−+−+−
εε
εε
εε
εε
εε
εε
εε
εε
εε
εε
εε
εε zyx
zz
y
y
xx
xzy
yzx
z
yx ssssss
together with: 1222 =++ zyx sss
will define the two directions for which:21 , ss ( ) ( )2211 ,, snsn
εε =
i.e., the Optical Axes
xz
xy
y
z
xz
xy
y
z
zx
yx
xnn
nn
n
n
nn
nns
εεεε
εε
−−
=−−
=−−
= −−
−−
22
22
2
2
22
22
2
xz
yz
y
x
xz
yz
y
x
zx
zy
znn
nn
n
n
nn
nns
εεεε
εε
−−
=−−
=−−
= −−
−−
22
22
2
2
22
22
202 =ys
We found previously that the directions of the two Optical Axes are given by
Dead End Equation of Wave Normal (continue – 6(
123
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
Let find a solution of
0
1
1
1
2
02
2
02
2
02
=
+−
+−
+−
z
y
x
zz
zyzx
zyyy
yx
zxyxxx
E
E
E
sn
ssss
sssn
ss
sssssn
εε
εε
εε
( )
( )
( ) 01
01
01
0
2
0
2
ˆ0
2
=+++
−
=+++
−
=+++
−
⋅
zzyyxxzzz
zzyyxxyy
y
Es
zzyyxxxxx
EsEsEssEn
EsEsEssEn
EsEsEssEn
εε
εε
εε
−
−
−
0
2
0
2
0
2
εε
εε
εε
z
z
y
y
x
x
n
s
n
s
n
s
( )Esn ⋅ˆ2
( )
( )
( )
−
⋅
−
⋅
−
⋅
=
0
2
2
0
2
2
0
2
2
ˆ
ˆ
ˆ
εε
εε
εε
z
z
y
y
x
x
z
y
x
n
Essn
n
Essn
n
Essn
E
E
E
Equation of Wave Normal (continue – 7(
124
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
2
0
2
2
0
2
2
0
2
21
/// nn
s
n
s
n
s
z
z
y
y
x
x =−
+−
+− εεεεεε
( )
( )
( )
−
⋅
−
⋅
−
⋅
=
=
0
2
1
2
1
0
2
1
2
1
0
2
1
2
1
1
1
1
1
ˆ
ˆ
ˆ
εε
εε
εε
z
z
y
y
x
x
z
y
x
n
Essn
n
Essn
n
Essn
E
E
E
E
Using n1 and n2 we find
( )
( )
( )
−
⋅
−
⋅
−
⋅
=
=
0
2
2
2
2
0
2
2
2
2
0
2
2
2
2
2
2
2
2
ˆ
ˆ
ˆ
εε
εε
εε
z
z
y
y
x
x
z
y
x
n
Essn
n
Essn
n
Essn
E
E
E
Eand
Fresnel equation of wave normals
( ) ( )c
snskωεωε ˆ,,ˆ, 11
=
( )snnn ˆ,, 21 ε
=
( ) ( )c
snskωεωε ˆ,,ˆ, 22
=
This is a quadratic equation of n2
(n6 terms drop(.For each set sx, sy, sz it yields two solutions for n2: n1
2 and n22.
Equation of Wave Normal (continue – 8(
125
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
This is a quadratic equation of n2
(n6 terms drop(.For each set sx, sy, sz it yields two solutions for n2: n1
2 and n22.
( )
( )
( )
−
⋅
−
⋅
−
⋅
=
=
==
0
2
1
2
1
0
2
1
2
1
0
2
1
2
1
1
1
1
1
1
1
11
ˆ
ˆ
ˆ
εε
ε
εε
ε
εε
ε
ε
ε
ε
ε
z
zz
y
yy
x
xx
z
y
x
zz
yy
xx
n
Essn
n
Essn
n
Essn
D
D
D
E
E
E
ED
Using n1 and n2 we find
and
( )
( )
( )
−
⋅
−
⋅
−
⋅
=
=
==
0
2
2
2
2
0
2
2
2
2
0
2
2
2
2
2
2
2
2
2
2
22
ˆ
ˆ
ˆ
εε
ε
εε
ε
εε
ε
ε
ε
ε
ε
z
zz
y
yy
x
xx
z
y
x
zz
yy
xx
n
Essn
n
Essn
n
Essn
D
D
D
E
E
E
ED
Must satisfy or0ˆˆ21 =⋅=⋅ DsDs 0
0
2
2
0
2
2
0
2
2
=−
+−
+−
εε
ε
εε
ε
εε
εz
zz
y
yy
x
xx
n
s
n
s
n
s
( )snnn ˆ,, 21 ε
=
( ) ( )c
snskωεωε ˆ,,ˆ, 11
=
( ) ( )c
snskωεωε ˆ,,ˆ, 22
=
2
0
2
2
0
2
2
0
2
21
/// nn
s
n
s
n
s
z
z
y
y
x
x =−
+−
+− εεεεεε
ProofNext slide
Equation of Wave Normal (continue – 9(
126
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
222
2
22
2
22
21
nnn
s
nn
s
nn
s
z
z
y
y
x
x =−
+−
+−
122
22
22
22
22
22
1
22
22
22
22
22
22
22
22
22
22
22
22
222
=−
+−
+−
+
−−
−−
−−
−+
−+
−
=++
z
zz
y
yy
x
xx
sss
z
zz
y
yy
x
xx
z
z
y
y
x
x
nn
sn
nn
sn
nn
sn
nn
sn
nn
sn
nn
sn
nn
sn
nn
sn
nn
sn
zyx
022
22
22
22
22
22
=−
+−
+− z
zz
y
yy
x
xx
nn
sn
nn
sn
nn
sn
Subtract and add
022
2
22
2
22
2
=−
+−
+− z
zz
y
yy
x
xx
nn
s
nn
s
nn
s εεε0
2
0
2
0
2 ,,εε
εε
εε z
zy
yx
x nnn ===
0ˆˆ21 =⋅=⋅ DsDs
Start with
Proof that 0ˆˆ21 =⋅=⋅ DsDs
End Proof
0
0
2
2
0
2
2
0
2
2
=−
+−
+−
εε
ε
εε
ε
εε
εz
zz
y
yy
x
xx
n
s
n
s
n
sEquation of Wave Normal (continue – 10(
127
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
If two of the refraction indices are equalthe crystal has one optical axis and is said to be uniaxial.
zyx εεε ≠=nx = ny =no≠ nz=ne
For each set sx, sy, sz it yields two solutions for n2: n1 and n2 ( )snnn ˆ,, 21 ε=
The quadratic equation of n2is:
( )[ ] ( ) ( )[ ] 0111 2422222242222 =+++−−+− eozezoozezo nnnsnsnnnsnsn
Uniaxial Crystals
One other way is to substitute in:
( ) ( ) ( ) 0111000
22
00
2
00
2
00
42
0
2
0
2
0
=+
−+−+−−
++
εε
εε
εε
εε
εε
εε
εε
εε
εε
εε
εε
εε zyx
xzy
yzx
z
yxz
zy
y
xx nsssnsss
ezoyx εεεεε =≠== nx = ny =no≠ nz=ne 0
2
0
2
0
2 ,,εε
εε
εε z
z
y
yx
x nnn === 1222 =++ zyx sss
2
0
2
2
0
2
21
//
1
nn
s
n
s
e
z
o
z =−
+−−
εεεε
Equation of Wave Normal (continue – 11(
128
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
( ) ( ) ( ) ( ) 01 2222222222 =−−−+− oeozoe nnnnnsnnn
( ) ( )[ ]{ } 01 222222222 =−+−− eoezozo nnnnsnsnn
The quadratic equation of n2 decomposes into two separate shells
022 =− onn
( )[ ] 022222222 =−++ eoezoyx nnnnsnss
Spherical surface
( ) ( ) ( )01
2
2
2
22
=−++
o
z
e
yx
n
sn
n
snsn Ellipsoid of Revolutionsurface
These two surfaces are tangent along the z axis.
Uniaxial Crystals (continue – 1(
( )[ ] ( ) ( )[ ] 0111 2422222242222 =+++−−+− eozezoozezo nnnsnsnnnsnsn
( ) ( ) ( ) 02222 =−++ ozyx nnsnsns
zyx εεε ≠=nx = ny =no≠ nz=ne
onn =
( ) 22221 ezoz
eo
nsns
nnn
+−=
Equation of Wave Normal (continue – 12(
129
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
022
22
22
22
22
22
=−
+−
+− z
zz
y
yy
x
xx
nn
sn
nn
sn
nn
sn ( )2
2
2
2
2
2
2
2
2
2
0111
n
n
ns
n
n
s
n
n
s
z
z
y
y
x
x −×=−
+−
+−
0111111
22
2
22
2
22
2
=−
+−
+−
z
z
y
y
x
x
nn
s
nn
s
nn
s
022
22
22
22
22
22
=−
+−
+− z
zz
y
yy
x
xx
nn
kn
nn
kn
nn
kn
zz
yy
xx
skk
skk
skk
=
==
Equation of Wave Normal (continue – 13(
130
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
222
2
22
2
22
21
nnn
s
nn
s
nn
s
z
z
y
y
x
x =−
+−
+−
Define
z
z
y
y
xxx
x n
cv
n
cv
n
cv ===== :,:,
111:
0
0
εµεµεµεµ
n
cv p =
1
1112
2
2
2
2
2
2
2
2
=−
+−
+−
n
n
s
n
n
s
n
n
s
z
z
y
y
x
x
z
pz
y
py
x
px
v
v
n
n
v
v
n
n
v
v
n
n === ,,
1
111 2
2
2
2
2
2
2
2
2
=−
+−
+−
z
p
z
y
p
y
x
p
x
v
v
s
v
v
s
v
v
s
122
22
22
22
22
22
=−
−−
−−
−zp
zy
yp
yy
xp
xx
vv
sv
vv
sv
vv
sv
PhaseVelocity
Equation of Wave Normal (continue – 14(
131
SOLO ELECTROMAGNETICS
Planar Waves in an Source-less Anisotropic Electric Media
222
2
22
2
22
21
nnn
s
nn
s
nn
s
z
z
y
y
x
x =−
+−
+−
122
22
22
22
22
22
=−
−−
−−
−zp
zy
yp
yy
xp
xx
vv
sv
vv
sv
vv
sv
122
22
22
22
22
22
1
22
22
22
22
22
22
22
22
22
22
22
22
222
=−
−−
−−
−
−+
−+
−+
−−
−−
−−
=++
zp
zp
yp
yp
xp
xp
sss
zp
zp
yp
yp
xp
xp
zp
zy
yp
yy
xp
xx
vv
sv
vv
sv
vv
sv
vv
sv
vv
sv
vv
sv
vv
sv
vv
sv
vv
sv
zyx
022
2
22
2
22
2
=−
+−
+− zp
z
yp
y
xp
x
vv
s
vv
s
vv
s
Subtract and add
2
1
pv−
where n
cv p =
z
pz
y
py
x
px
v
v
n
n
v
v
n
n
v
v
n
n === ,,and
Equation of Wave Normal (continue – 15(
137
OPTICSSOLO
References
A. Yariv, P. Yeh, “Optical Waves in Crystals”, John Wiley & Sons, 1984,
Ch. 4 Electromagnetic Propagation in Anisotropic Media
M. Born, E. Wolf, “Principles of Optics”, Pergamon Press,6th Ed., 1980
E. Hecht, A. Zajac, “Optics”, Addison-Wesley, 1979, Ch.8
C.C. Davis, “Lasers and Electro-Optics”, Cambridge University Press, 1996
G.R. Fowles, “Introduction to Modern Optics”,2nd Ed., Dover, 1975, Ch.2
M.V.Klein, T.E. Furtak, “Optics”, 2nd Ed., John Wiley & Sons, 1986
http://en.wikipedia.org/wiki/Polarization
W.C.Elmore, M.A. Heald, “Physics of Waves”, Dover Publications, 1969
E. Collett, “Polarization Light in Fiber Optics”, PolaWave Group, 2003
W. Swindell, Ed., “Polarization Light”, Benchmark Papers in Optics, V.1, Dowden, Hutchinson & Ross, Inc., 1975
http://www.enzim.hu/~szia/cddemo/edemo0.htm (Andras Szilagyi(
138
ELECTROMAGNETICSSOLO
References
J.D. Jackson, “Classical Electrodynamics”, 3rd Ed., John Wiley & Sons, 1999
R. S. Elliott, “Electromagnetics”, McGraw-Hill, 1966
J.A. Stratton, “Electromagnetic Theory”, McGraw-Hill, 1941
W.K.H. Panofsky, M. Phillips, “Classical Electricity and Magnetism”, Addison-Wesley, 1962
F.T. Ulaby, R.K. More, A.K. Fung, “Microwave Remote Sensors Active and Passive”, Addson-Wesley, 1981
A.L. Maffett, “Topics for a Statistical Description of Radar Cross Section”,John Wiley & Sons, 1988
139
ELECTROMAGNETICSSOLO
References
1. W.K.H. Panofsky & M. Phillips, “Classical Electricity and Magnetism”,
2. J.D. Jackson, “Classical Electrodynamics”,
3. R.S. Elliott, “Electromagnetics”,
4. A.L. Maffett, “Topics for a Statistical Description of Radar Cross Section”,