Maxwell equations and propagation in anisotropic media

1

Maxwell Equations&

Propagation inAnisotropic Electric Media

SOLO HERMELINUpdated 12.06.06

23.12.12http://www.solohermelin.com

http://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



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





Wollaston Prism

William HydeWollaston1766-1828

10

POLARIZATIONSOLO





Glan-Foucault Polarizer

11

POLARIZATIONSOLOPolarizations Prisms Overview

http://www.unitedcrystals.com/POverview.html

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




→→⋅

∂∂−=⋅ ∫∫∫ 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)



Current Density J [ ]2−⋅mA

André-Marie Ampère1775-1836

→→⋅

∂∂−=⋅ ∫∫∫ dS

t

BdlE

SC

t

BE

∂∂−=×∇

→dl

→dS

C

B

E



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


3. GAUSS’ LAW – ELECTRIC (GE)

dV

→dS

V∫∫∫∫∫ =⋅

→

VS

dVdSD ρ

D

ρ=⋅∇ D

ρ


Free Charge Distributionρ [ ]3−⋅⋅ msA

GAUSS’ ELECTRIC (GE) & MAGNETIC (GM) LAWS developed by Gaussin 1835, but published in 1867.

Karl Friederich Gauss1777-1855


23

James C. Maxwell(1831-1879)

ELECTROMAGNETICSSOLO





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


SYMMETRIC MAXWELL’s EQUATIONS





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


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


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


27


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


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



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



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)


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


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



( ) ( ) 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



( ) ( ) 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


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


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


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


38


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


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


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



Hendrik Antoon Lorentz1853-1928

eρ

Force on the free chargeF

[ ]Neρ

Second way

41



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



( )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



∫∫∫∫∫∫∫∫∫∫∫∫∫ ⋅∇+

⋅+

⋅+=⋅

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


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

×=


45


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


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


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


( )( )

×∇−×∇−=×∇×∇×∇+×∇=×∇×∇

−−=×∇+=×∇

⇒

−−=×∇×∇

+=×∇×∇ =

=

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


Assume no sources:

we have


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


To satisfy the Maxwell equations for a source free media we must have:


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


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


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


53


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 ×=


Electromagnetic Energy Density

54



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



( )( )

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


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



57


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


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


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


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


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 =⇔=


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)


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)



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


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


( ) ( )( )( )

( ) ( ) ( ) ( )[ ]( ) ( )

[ ] ( ) ( )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


68


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


a, 09/02/2005

1. A.L. Maffett, "Topics for a Statistical Description of Radar ross Sections", John Wiley, 1988, pp.48, pp. 2892. L.B. Felsen, N. Marcuvitz, "Radiation and Scttering of Waves", Prentice-Hall, 1973,pp.78-83

69


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



70



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



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 ,,,âre coplanar

and normal to H

I µµζξε === ,0,

72



B

D

pv

E

H

HES

×=α

αk

ev

Planar Waves( ) ( )[ ]EssEc

nEss

c

nD ⋅−

=××

−= ˆˆ

1ˆˆ

122

µµ

SDEs ,,,âre 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


Planar Waves in an Source-less Anisotropic Electric MediaPhase Velocity, Energy (Ray) Velocity


⋅−=⋅−= rs

c

ntrkt

ˆωωφ


⋅−=⋅−=⋅−== 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

⋅=

=⋅

=



SvSv ee =

→1

( ) emUDEBHSk

Ssc

n =⋅+⋅=⋅=⋅2

1ˆ

ω

( ) emee U

SS

Ssn

c

S

Svv =

⋅==

ˆ1



HES ×=


sc

nk ˆω=

Electromagnetic Energy Density

74



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



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




( )

( )

( )

+−

+−

+−

=

+−

+−

+−

++=

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



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




( )

( )

( )

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


79



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



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



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



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




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




0

2

0

2

0

2 ,,εε

εε

εε z

z

y

yx

x nnn ===nx < ny < nz.

85



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


ksnnsnnsnnc

ksnsnsnc

ωωω

0

2

0

2

0

2 ,,εε

εε

εε z

z

y

yx

x nnn ===

86




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


ksnnsnnsnnc

ksnsnsnc

ωωω

nx = ny =no≠ nz=ne

Uniaxial Crystals

One other way is to substitute in:

87



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)


88



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



Double Refraction at a Uniaxial Crystal Boundary

90




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


91


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


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

93



Crystal Types

94



Crystal Types (continue -1)

95



Crystal Types (continue – 2)

96



Crystal Types (continue – 3)

A.Mermin, “Solid State Physics”,Holt, Reinhart and Winston, 1976,p.122


97



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

a, 08/31/2005

1. This development follows A. Yariv, P. Yeh, "Optical Waves in Crystals", John Wiley & sons, 1984, pp.79-80

98



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



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



100


Phase Velocity, Energy (Ray) Velocity for Planar Waves


⋅−=⋅−= rs

c

ntrkt

ˆωωφ


⋅−=⋅−=⋅−== 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

=⋅

=⋅

=



SvSv ee =

→1

( ) emUDEBHSk

Ssc

n =⋅+⋅=⋅=⋅2

1ˆ

ω

( ) em

ee U

SS

Ssn

c

S

Svv =

⋅==

ˆ1

:



HES ×=


101



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



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



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



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



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



( )

( )

( )

+−

+−

+−

=

+−

+−

+−

++=

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(


107



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


We found that the two eigenmodes, of each direction , must satisfy:

s

2222 zzyyxxe EEEEEU εεεε ++=⋅⋅= ∗

108



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


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




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




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




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



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

ÔAsFor nx < ny < nz we found that in principal dielectric axes

−−

±= −−

−−

−22

22

1

2,1 sinzx

yx

nn

nnδ

SDEs ,,,âre coplanar

and normal toHWe also found:

DE

DDn

⋅⋅=

0

2 1

ε constn

DDE

y

==⋅2

0

2

εWe found: ynn =

113



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

a, 06/10/2006

See treatment in 1. Born & Wolf, "Principle of Optics", 6th Ed., pp.686-6902. M.V. Klein,"Optics", 1970, pp.6093. Yariv & Yeh,"Optical Waves in Crystals", pp.91-934. Jenkins & White,"Fundamentals of Optics", pp.556-558

114





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






116



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 ===


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




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


120


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



( ) ( ) ( ) ( )[ ] 0111 22222222222224222222 =+−+−+−−++ zyxxzyyzxzyxzzyyxx nnnnsnnsnnsnnnsnsnsn

0

2

0

2

0

2 ,,εε

εε

εε z

z

y

yx

x nnn ===


122



( ) ( ) ( ) 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



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


124



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


=

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.


125



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


=

( ) ( )c


=

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


126



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



If two of the refraction indices are equalthe crystal has one optical axis and is said to be uniaxial.


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


128



( ) ( ) ( ) ( ) 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


onn =

( ) 22221 ezoz

eo

nsns

nnn

+−=


129



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

=

==


130



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


131



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


132



133

SOLO UNIAXIS CRYSTALS

134


135


136


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


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


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”,

January 7, 2015 140

SOLO

TechnionIsraeli Institute of Technology

1964 – 1968 BSc EE1968 – 1971 MSc EE

Israeli Air Force1970 – 1974

RAFAELIsraeli Armament Development Authority

1974 –2013

Stanford University1983 – 1986 PhD AA

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Maxwell equations and propagation in anisotropic media

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