Electromagnetics

1

ELECTROMAGNETICS

SOLO HERMELIN

Updated 16.12.06

http://www.solohermelin.com

2

ELECTROMAGNETICS

Maxwell’s Equations

SOLOTABLE OF CONTENT

Conservation of Charges

Magnetic Vector Potential and Electric Scalar Potential V A

Constitutive Relations

Electromagnetic Wave EquationAnti-PotentialsElectromagnetic Wave Equation for Polarized Medium

Electromagnetic Wave Equation for Vector and Scalar Potentials in a Polarized Medium

Symmetric Maxwell’s Equations

Stratton-Chu Solution of Non-homogeneous (Helmholtz) Differential Equations

Wave Equation in a Non-Homogeneous Media

Uniqueness of the Solutions of Maxwell Equations Given Boundary Conditions

Kirchhof’s Solution of the Scalar Helmholtz Non-homogeneous Differential Equations

General Solutions of Maxwell Equations

Dyadic Green’s Function Solution of Non-homogeneous (Helmholtz) Differential Equations

Boundary Conditions

3

ELECTROMAGNETICSSOLO

TABLE OF CONTENT (continue)

Monochromatic Planar Wave Equations

Spherical Waveforms

Cylindrical Waveforms

Energy and Momentum

Energy Flux and Poynting Vector

Energy Flux and Poynting Vector for a Bi-anisotropic Medium

References

Reflections and Refractions Laws

Electrical Dipole (Hertzian Dipole) Radiation

4

MAXWELL’s EQUATIONSSOLO

Magnetic Field Intensity H

1mA

Electric Displacement D 2 msA

Electric Field Intensity E 1mV

Magnetic InductionB 2 msV

Current Density J

2mA

Free Charge Distribution 3 msA

James Clerk Maxwell(1831-1879)

A Dynamic Theory of Electromagnetic Field 1864 Treatise on Electricity and Magnetism 1874

5

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.”

6

SOLOElectrostatics

Charles-Augustin de Coulomb1736 - 1806

1q

2q

1 2r

1r

2r

1 212 123

0 12

1

4

q qF r

r

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 m 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

r

where is the Electric Field Intensity [N/C]

1785

7

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

8

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

dl��

an infinitesimal element of the contour C

J

curent density A/m2

dS��

a differential vector area of the surface S enclosed by contour C

Ampère’s Law


1mA

enc

C S

H dl J dS I

H J

dS

C

J

dl

H

9

SOLO

1820

ElectromagnetismBiot-Savart Law

02

1

4rI dL

dBr

��

Magnetic 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

��

��

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

10

SOLO

1820Electromagnetism

Biot-Savart Law

Jean-Baptiste Biot1774 - 1862

2 300

''

4 '

J rA J A r d r

r r

0

0

H J B H

B B A

20B A A A J

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

Where we used 0

' : 1/ 'r r rJ J r J r r

Derivation of Biot-Savart Law from Ampère’s Law

Poison’s EquationSolution for an unbounded volume

Ampère Law

Félix Savart1791 - 1841

11

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


1mA




dSt

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

12

MAXWELL’s EQUATIONS

1. AMPÈRE’s CIRCUIT LW (A) (1820)

SOLO

2. FARADAY’s INDUCTION LAW (F) (1831)

dSt

DJdlH

SC

t

DJH

dS

C

t

DJ

dl

H



1mA

Current Density J

2mA

André-Marie Ampère1775-1836

dSt

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

13


4. GAUSS’ LAW – MAGNETIC (GM)

dV

dS

V

0

S

dSB

B

0 B


3. GAUSS’ LAW – ELECTRIC (GE)

dV

dS

V

VS

dVdSD

D

D


Free Charge Distribution 3 msA

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

Karl Friederich Gauss1777-1855

14

SOLO

The Electromagnetic Spectrum

15

1864 MAXWELL’S EQUATIONS FOR THE

ELECTROMAGNETIC FIELD

(1) Sdt

DJldH

SL

t

DJH

AMPÈRE’S LAW

(2) Sdt

BldE

SL

t

BE

FARADAY’S LAW

(3) VS

dvSdD

D

GAUSS ELECTRIC

(4) 0S

SdB

0 B

GAUSS MAGNETIC

Maxwell (1831-1879)

T H E M A X W E L L E Q U A T I O N S A R E N O T I N V A R I A N T U N D E R G A L I L E A N T R A N S F O R M A T I O N S .

'

'''

t

DJH

t

DJH


16

ELECTROMAGNETICS

From Maxwell’s Equations

SOLO

CONSERVATION OF CHARGES (CC)

0

0

tJ

D

JDt

HJt

DH

00

V

dVt

Jt

J

t

QdV

tdSJdVJ V

Q

VAV

V

dV

dS

V

J

17

ELECTROMAGNETICS

From Maxwell’s Equations

SOLO

MINIMUM NUMBER OF MAXWELL’s EQUTIONS

0

t

J

t

DJH

t

BE

(A)

(F)

(CC)

Dt

JH

0

Bt

E

0

constD

constB

We recovered (beside the constant) the Gauss’ Laws (Electric and Magnetic)

We can see that only three Maxwell’s Equations are independent

18

ELECTROMAGNETICS

From

SOLO

MAGNETIC VECTOR POTENTIAL AND ELECTRIC VECTOR POTENTIAL V A

0B

- Magnetic Vector Potential that generates the field. Not unique since gives the same result (where is any continuous scalar)

A

A

Also

t

AE

t

AE

t

BE

t

BE

AAB

0

From this equation we can derive the Scalar Potential V such that

t

VAt

Vt

AE

The Electromagnetic Field is fully defined by the Vector Potential andthe Scalar Potential V (or by and - called a gauge transformation).

A

tV

A

“gauge symmetry abstract mathematical symmetry of a field related to the freedomto re-gauge, or re-scale, certain quantities in the theory (potentials) without affecting the values of the observable field quantities”. Paul Davies, Ed.,“The New Physics”, Cambridge University Press, 1989, p. 497

AAB

19


CONSTITUTIVE RELATIONS

PED

0MHB

00

Permitivity of free space0

Permeability of free space0

Polarization Vector Intensity P

2 msA

Magnetization Vector Intensity M

1mA

212292122120 10

36

1108542.8 mNsAmNsA

NA 270 104

20


CONSTITUTIVE RELATIONS (continue -1)

Homogeneous Medium – Medium properties do not vary from point to point and are the same for all points. Isotropic Medium – Medium properties are the same in all directions and are scalars. Linear Medium – The effects of all different fields can be added linearly

(Superposition of different fields).

For Linear and Isotropic Medium we have:

ED

HB

where: 0 eK

0 mK

- Dielectric Constant (or Relative Permitivity)eK

- Relative Permeability mK

21


CONSTITUTIVE RELATIONS (continue - 1)

The most general form of Linear Constitutive Relations is:

Classification of Media

dyadicsxwhereH

E

B

D33,,,

��

��

��

Classification according to the functional dependence of the 6x6 matrix

��

��

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

22



Classification of Media (continue - 2)

In general 0,0,0,0��

Anisotropic Electromagnetic - medium described by both 0,0

��

Anisotropic Electric - medium described by 0��

Anisotropic Magnetic - medium described by 0��

If is symmetric, it can be diagonalized0��

z

y

x

00

00

00�

Uniaxial zyx (thetragonal, hexagonal, rombohedral crystals)

Biaxial zyx (orthohombic, monoclinic, triclinic crystals)

Isotropic zyx

This is called an Anisotropic Medium.

If or is Hermitian; i.e.0��

0��

Transposeconjugate

conjugateTranspose

a

aj

jaT

z

-T,-*

,-H

UUUU *H

00

0

0

Is gyroelectric or gyromagnetic depending on whether stands for or . If both tensors are of this form the medium is gyroelectromagnetic.

� �U

23



Classification of Media (continue - 3)

In general 0,0,0,0��

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

��

��j

G

mediumlosslessundefinite

mediumpassivedefinitenegative

mediumactivedefinitepositive

G

G

G

In this case, we have the following classification (see development later)

24

SOLO ELECTROMAGNETICS BOUNDARY CONDITIONS

Boundary Conditions

2t

1t

h

2H

1H

1

2

C

CS1P2P

3P

4P

b

21ˆ n

ek

ldtHtHhldtHldtHldHh

C

2211

0

2211ˆˆˆˆ

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 e

ˆˆˆˆˆˆ21212121121

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:

mAht

DJk e

he

25


Boundary Conditions (continue – 1)

2t

1t

h

2E

1E

1

2

C

CS1P2P

3P

4P

b

21ˆ n

mk

ldtEtEhldtEldtEldEh

C

2211

0

2211ˆˆˆˆ

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 m

ˆˆˆˆˆˆ21212121121

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:

mVht

BJk m

hm

26



h

2D

1D

1

2

21ˆ n

dS

1n

2n

e

SdnDnDhSdnDSdnDSdDh

S

2211

0

2211 ˆˆˆˆ

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 e 2121121 ˆˆ

Since this must be true for any dS on the boundary between regions (1) and (2) we must have:

eDDn 2121ˆ

dSdShdv e

h

e

V

e 0

GAUSS’ LAW - ELECTRIC

1

0lim:

msAhe

he

V

e

S

dvSdD

27



h

2B

1B

1

2

21ˆ n

dS

1n

2n

m

SdnBnBhSdnBSdnBSdBh

S

2211

0

2211 ˆˆˆˆ

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 m 2121121 ˆˆ

Since this must be true for any dS on the boundary between regions (1) and (2) we must have:

mBBn 2121ˆ

dSdShdv m

h

m

V

m 0

GAUSS’ LAW – MAGNETIC

1

0lim:

msVhm

hm

V

m

S

dvSdB

28


Boundary Conditions (summary)

2t

1t

h

22 , HE

11, HE

1

2

C

CS1P2P

3P

4P

b

21ˆ n

me kk

,21ˆ n

dS

11, BD

22 , BD

me ,

mkEEn

2121ˆ FARADAY’S LAW

ekHHn

2121ˆ AMPÈRE’S LAW 1

0lim:

mAht

DJk e

he

1

0lim:

mVht

BJk m

hm

eDDn 2121ˆ GAUSS’ LAW

ELECTRIC 1

0lim:

msAhe

he

mBBn 2121ˆ GAUSS’ LAW

MAGNETIC 1

0lim:

msVhm

hm

29


Boundary Conditions for Perfect Electric Conductor (PEC)

2t

1t

h

22 , HE

11, HE

1

2

C

CS1P2P

3P

4P

b

21ˆ n

ek21ˆ n

dS

11, BD

0,0,0,0 2222 BHDE

e

0ˆ 2121

EEn FARADAY’S LAW

ekHHn

2121ˆ AMPÈRE’S LAW

eDDn 2121ˆ GAUSS’ LAW

ELECTRIC

0ˆ 2121 BBn GAUSS’ LAW

MAGNETIC

ekHn

121ˆ

0ˆ 121

En

eDn 121ˆ

0ˆ 121 Bn

02

H

02

B

02

E

02

D

30


Boundary Conditions for Perfect Magnetic Conductor (PMC)

2t

1t

h

22 , HE

11, HE

1

2

C

CS1P2P

3P

4P

b

21ˆ n

mk21ˆ n

dS

11, BD

0,0,0,0 2222 BHDE

m

mkEEn

2121ˆ FARADAY’S LAW

0ˆ 2121

HHn AMPÈRE’S LAW

0ˆ 2121 DDn GAUSS’ LAW

ELECTRIC

mBBn 2121ˆ GAUSS’ LAW

MAGNETIC

0ˆ 121

Hn

mkEn

121ˆ

0ˆ 121 Dn

mBn 121ˆ

02

H

02

B

02

E

02

D

31

SOLO

Assume the general constitutive relations

ELECTROMAGNETICS

Assume that a set of sources inside a volume V produces two different field solutions: and11, HE

22 , HE

m

e

m

e

BGM

DGE

JBjEF

JDjHA

1

1

11

11

)(

m

e

m

e

BGM

DGE

JBjEF

JDjHA

2

2

22

22

)(

dyadicsxwhereH

E

B

D33,,,

��

��

��

Subtract the two sets of equations

121212

121212

HHEEjEEF

HHEEjHHA

��

��

Define1212 :&: HHHEEE

Uniqueness of the Solutions of Maxwell Equations Given Boundary Conditions

n

iiSS

1

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

32

SOLO

Uniqueness of the Solutions of Maxwell Equations (continue - 1)

ELECTROMAGNETICS

HEjEF

HEjHA

��

��

Let compute the following

HHEHHEEEj

HEjHHEjE

EHHEHE

��

��

******

*****

***

******

*****

***

HHEHHEEEj

HEHjHEEj

EHHEHE

��

��

******

******

**

HHEHHEEEj

HHEHHEEEj

HEHE

��

��

33

SOLO


ELECTROMAGNETICS

H

EjHE

HHEHHEEEj

HHEHHEEEj

HHEHHEEEj

HEHE

**

****

********

******

******

**

��

��

��

��

��

Let integrate this equation over the volume V

S

Gauss

V

V

dSnHEHE

dVHEHE

dVH

EjHE

1**1

**

**

****

��

��

n

iiSS

1

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

34

SOLO


ELECTROMAGNETICS

But

**

**

��

��jG




G

G

G

SV

dSnHEHEdVH

EjHE 1**

**

****

��

��

EHnEHn

HEnHEnnHEHE

**

****

11

111

Suppose or (but not both) are defined at the surface S that bounds the

volume V, then or

En

1 Hn

1

011 *

SS

EnEn 011 *

SS

HnHn

35

SOLO


ELECTROMAGNETICS

**

**

��

��jG




G

G

G

01**

**

****

SV

dSnHEHEdVH

EjHE

��

��

If or (but not both) are defined at the surface S that bounds the

volume V, then or

En

1 Hn

1

011 *

SS

EnEn 011 *

SS

HnHn

If is positive or negative definite, then the volume integral is zero if and only ifG

VinHHHEEE 0&0 1212 If is not positive or negative definite, it can be treated as the limiting case of passive (or active) medium.G

This proofs the uniqueness of the Solutions of Maxwell Equationsin a finite volume V if or (but not both) are defined at the surface S .En

1 Hn

1

36

SOLO Waves

2 2

2 2 2

10

d s d s

d x v d t Wave Equation

Regressive wave Progressive waverun this

-30 -20 -10

0.6

1.0.8

0.40.2

In the same way for a3-D wave

2 2 2 2 2

22 2 2 2 2 2 2

1 1, , , , , , 0

d s d s d s d s ds x y z t s x y z t

d x d y d z v d t v d t

v

xtfs

v

xts

y

y

v

xtf

yd

d

td

sd

v

xtf

yd

d

vxd

sd

2

2

2

2

2

2

22

2

&1

z

z

v

xt

zd

d

td

sd

v

xt

zd

d

vxd

sd

2

2

2

2

2

2

22

2

&1

37


ELECTROMGNETIC WAVE EQUATIONS

For Homogeneous, Linear and Isotropic Medium

ED

HB

where are constant scalars, we have ,

Jt

EJ

t

DH

t

t

H

t

BE

ED

HB

Since we have also tt

t

J

t

EE

DED

EEE

t

J

t

EE

2

222

2

2

&

38


ELECTROMGNETIC WAVE EQUATIONS (continue 1)

Define

meme KK

c

KKv

00

11

where

smc /103

1036

1104

11 8

9700

is the velocity of light in free space.

The absolute index of refraction n is

me KKv

cn

0

The Inhomogeneous Wave (Helmholtz) Differential Equation for the Electric Field Intensity is

t

J

t

E

vE

2

2

22 1

39


ELECTROMGNETIC WAVE EQUATIONS (continue 2)

In the same way

The Inhomogeneous Wave (Helmholtz) Differential Equation for the Magnetic Field Intensity is

Jt

EJ

t

DH

t

H

t

BE

t

ED

HB

Since are constant andtt

,

J

t

HH

HHB

HHH

Jt

HH

2

222

2

2

0&

Jt

H

vH

2

2

2

2 1

40


ELECTROMGNETIC WAVE EQUATIONS FOR THE VECTOR AND SCALAR POTENTIALS

Now

The Inhomogeneous Wave (Helmholtz) Differential Equation for the Vector Potential is

Vt

AE

AHAB

HB

Jt

EH

Jt

DH

ED

1

AAA

JVtt

AA

2

2

21

A

LJt

A

vA

2

2

2

2 1where

t

VAL

JVt

A

tA

1

t

VAJ

t

AA

2

22

41



Also

The Inhomogeneous Wave (Helmholtz) Differential Equation for the Scalar Potential V is

Vt

AE

ED

ED

t

L

t

VA

tt

V

vV

2

2

22 1

Since we have one degree of freedom in the choice of and V, we can choose:

A

0

t

VAL

LORENZ CONDITION (1867)

2

2

2

2

t

V

t

V

Vt

A

t

L

t

VA

tt

VV

2

22

Ludwig Valentin Lorenz

1829-1891

42



We obtained the following Inhomogeneous Wave (Helmholtz) Differential Equation for the Vector and the Scalar Potential V A

2

2

22 1

t

V

vV

Jt

A

vA

2

2

22 1

Let perform a divergent operation on the first equation and a partial time differentiationon the second

t

V

tvt

V

vV

tt

2

2

2

2

2

2

2

2 111

Atvt

A

vAJ

2

2

2

2

2

2

2

2 11

Add the first equation to the second equation multiplied by με.

011

2

2

2

2

2

2

2

2

t

JLtvt

VA

tv

We can see that the Lorenz (gauge) condition L = 0 is compatible with theconservation of charge 0

t

J

43


ANTI-POTENTIALS

The equations Vt

AE

AB

are not the most general since we can add any particular solution of the Homogeneous Differential Equations

0

0

D

t

DH

t

BE

B

44


ANTI-POTENTIALS (continue-1)

From those equations we obtain

Vt

AH

t

AH

t

DH

ADD

0

0

where

- Electric Vector Potential A

- Magnetic Scalar Potential V

Also

AE

ED

AD

1

Vt

AB

HB

Vt

AH

45




Vt

AB

AE

t

BE

1

and

AAA

Vt

A

tA

2

1

In the same way

2

22

2

22

0 t

V

t

V

Vt

AB

Vt

A

tA

1

t

VA

t

AA

2

22

t

VA

tt

VV

2

22

46



The field vectors are given by the superposition of the potentials and anti-potentials by:

t

VA

t

AA

2

22

t

VA

tt

VV

2

22

Since we have one degree of freedom in the choice of and V*, we can choose:*A

0*

**

t

VAL

LORENZ CONDITION

We obtained the following Homogeneous Wave Differential Equation for the Vector and the Scalar Potential V* *A

01

2

2

22

t

A

vA

01

2

2

22

t

V

vV

AVt

AE

1

Vt

AAH

1

47


ELECTROMGNETIC WAVE EQUATIONS FOR POLARIZED MEDIUM

For Homogeneous, Linear and Polarized Medium

we have

PED

0

MHB

0

t

PJ

t

EJ

t

DH

t

t

M

t

H

t

BE

PED

MHB

00

00

0

0

DPED

EEE

Mt

PJ

tt

EE

&0

2

02

2

00


,

Mt

PJ

t

P

t

EE

0

02

2

002

48


ELECTROMGNETIC WAVE EQUATIONS FOR POLARIZED MEDIUM (continue – 1)

Mt

PJ

t

P

t

EE

0

02

2

002

We defined

smc /103

1036

1104

11 8

9700

as the velocity of light in free space.

The Inhomogeneous Wave (Helmholtz) Differential Equation for the Electric Field Intensity in a Polarized Medium is

Mt

PJ

t

P

t

E

cE

0

02

2

22 1

49


ELECTROMGNETIC WAVE EQUATIONS FOR POLARIZED MEDIUM (continue – 2)

In the same way

Since the Inhomogeneous Wave (Helmholtz) Differential Equation for the Magnetic Field Induction in a Polarized Medium is

t

PJ

t

EJ

t

DH

t

M

t

H

t

BE

t

PED

MHB

0

000

0

0


,

MHMHB

HHH

t

M

t

PJ

t

HH

&00

2

2

2

002

2

00

MHB

0

Mt

PJ

t

B

cB

02

2

22 1

2

2

002

2

002

t

MM

t

PJ

t

HH

2

2

002

2

2

002

t

MMM

t

PJ

t

HH

50


ELECTROMGNETIC WAVE EQUATIONS FOR THE VECTOR AND SCALAR POTENTIALSIN A POLARIZED MEDIUM

00

000

0

0

0

1

Vt

AE

MAHAB

MHB

t

PJ

t

EH

Jt

DH

PED

Also

02

00

0020

2

000

1

AAA

Mt

PJV

tt

AA

Mt

PJV

t

A

tA

00

000

1

t

VAM

t

PJ

t

AA 0

000020

2

0002

51


ELECTROMGNETIC WAVE EQUATIONS FOR THE VECTOR AND SCALAR POTENTIALSIN A POLARIZED MEDIUM (continue – 1)

We obtained the following Inhomogeneous Wave (Helmholtz) Differential Equation for the Vector Potential A

where

t

VAM

t

PJ

t

AA 0

000020

2

0002

0

2

020

2

202 1

LMt

PJ

t

A

cA

t

VAL

00000

52



The Inhomogeneous Wave (Helmholtz) Differential Equation for the Scalar Potential is

Also

00

00

Vt

AE

PED

PED

t

LP

t

VA

t

P

t

V

cV

0

0

0000

020

2

202 1

Since we have one degree of freedom in the choice of and , we can choose:0A

0V

0000

t

VAL

LORENZ CONDITION

20

2

0020

2

00

0

t

V

t

VP

t

LP

t

VA

t

P

t

VV

0

0

0000

020

2

0002

00 Vt

A

53



We obtained the following Inhomogeneous Wave Differential Equation for the Vector and the Scalar Potential 0A

0V

Mt

PJ

t

A

cA

02

02

202 1

020

2

202 1

P

t

V

cV

together with

0000

t

VAL

LORENZ CONDITION

54


ANTI-POTENTIALS

The equations 00

0 Vt

AE

0AB

are not the most general since we can add any particular solution of the Homogeneous Differential Equations

0

0

D

t

DH

t

BE

B


55




000

0

0

0

Vt

AH

t

AH

t

DH

ADD

where

- Electric Vector Potential

0A

- Magnetic Scalar Potential

0V

Also


PAEPED

AD

000

0 1

MV

t

AB

MHB

Vt

AH

000

0

0

00

56




and

In the same way


MVt

AB

PAE

t

BE

000

0

00

1

02

00

000

000

1

AAA

t

MPV

t

A

tA

20

2

0202

02

020

000

00 t

V

t

V

MVt

AB

t

MPV

t

A

tA

000

000

1

t

MP

t

VA

t

AA

000

00020

2

0002

Mt

VA

tt

VV

00002

02

0002

57



The field vectors are given by the superposition of the potentials and anti-potentials by:

t

MP

t

VA

t

AA

000

00020

2

0002

Mt

VA

tt

VV

00002

02

0002

Since we have one degree of freedom in the choice of and V0*, we can choose:*

0A

0*

000

*0

*0

t

VAL

LORENZ CONDITION

We obtained the following Inhomogeneous Wave Differential Equations for the Vector and the Scalar Potential V0*

*0A

00

00 1

AVt

AE

00

00

1V

t

AAH


t

MP

t

A

cA

2

0020

2

202 1 M

t

V

cV

20

2

202 1

58


HERTZ’s POTENTIALS


From Lorenz Conditions: we can define theElectric Hertz’s Vector Potential (1888) such that

00000

t

VA

e

Heinrich Rudolf Hertz1857-1894

tA e

000

eV

0

Also from Lorenz Conditions: we can define theMagnetic Hertz’s Vector Potential (or Hertz’s Anti-potential or FitzGerald Potential such that

0*

000

*0

t

VA

m

tA m

000

mV

0

George Francis FitzGerald1851-1901

Ludwig Valentin Lorenz

1829-1891

59


HERTZ’s POTENTIALS (continue – 1)


The field vectors are given by

tt

AVt

AE m

ee

02

2

0000

00 1

mme

ttV

t

AAH

2

2

00000

0

0

1

60




Using the identity we obtain:

2

tt

tt

ttE

me

ee

mee

ee

me

eee

02

2

002

02

2

2

002

02

2

00

22

2

2

002

0

22

2

002

0

2

2

000

22

tt

tt

ttH

mmm

e

mmm

me

mme

mm

61




Let develop the differential equations for Electric Hertz’s Vector Potential e

ttE

PED

PED

me

e

02

2

00

0

0 1

ee

eee

eee

2

2

2

0

Note: We used the identity

End Note

00

22

2

00

1

Pt e

e

ee

m

Pt e

et

2

002

2

00

0 1

Ptt

E me

e

0

02

2

00

1

62




00

22

2

00

1

Pt e

e

Let assume that , then0

where is any differentiable vector, so it can be chosen .w

0w

wPt

ee

0

0

2

2

002 1

Pt

ee

0

0

2

2

002 1

63




Let develop the differential equations for Magnetic Hertz’s Vector Potential m

mme

ttH

MHB

MHB

2

2

000

0

0

022

2

00

Mt m

m

mm

Mt m

m

2

02

2

00

0

2

2

000

t

mme

e

Mtt

H

64




where is any differentiable vector, so it can be chosen .u

0u

uMt

mm

2

2

002

022

2

00

Mt m

m

Therefore

Mt

mm

2

2

002

65




Let return to the field equations

t

P

ttE

me

me

ee

00

0

02

2

002

1

t

PED me

0

0

0

Mt

ttH

me

mmm

e

0

2

2

002

0

me

tMHB

0000

66


ELECTROMGNETIC WAVE EQUATIONS FOR THE VECTOR AND SCALAR POTENTIALSIN A POLARIZED MEDIUM (continue – 16) (SUMMARY

For Homogeneous, Linear and Isotropic Medium we found the following

Mt

PJ

t

P

t

E

cE

0

02

2

22 1

Mt

PJ

tt

D

cD

002

2

22 1

2

2

22

2

2

22 11

t

M

cMM

t

PJ

t

H

cH

Mt

PJ

t

B

cB

02

2

22 1

Mt

PJ

t

A

cA

02

02

202 1

0

20

2

202 1

P

t

V

cV

Ptc

ee

02

2

22 11

Mtc

mm

2

2

22 1

01 0

20

t

V

cA

LORENZ CONDITION

67

SOLO

Wave Equation in a Non-Homogeneous Media

For non-homogeneous media ε,μ are functions of position

HB

ED

BGM

DGE

Jt

DHA

t

BEF

e

e

0

0HGM

EGE

Jt

EHA

t

HEF

e

e

From those equations we have

EEE

EEE

t

J

t

EH

tE e

Jt

EH

t

HE e

22

2

2

11

111

1

ee

e

e

EEE

EEE

t

JEE

t

EE

ln

ln2

22

From which

ee

t

JEE

t

EE

lnln2

22

ELECTROMAGNETICS

68

SOLO

Wave Equation in a Non-Homogeneous Media (continue – 1)

Also

or

et

HE

e

Jt

EH

J

t

HJE

tH

e

2

21

and

HHHHHH

HHH

HHH

ln0

ln11

111

2

eJHH

t

HH

ln2

22

eJHH

t

HH

lnln2

22

ELECTROMAGNETICS

69

SOLO


Far from sources, in the High Frequencies we can write, using the phasor notation

0000 &, 0

kerErE rSjk

The Wave Equation in a Non-Homogeneous Media, without sources is:

0lnln2

22

EEt

EE

or in phasor notation

kEEEkE &0lnln22

ELECTROMAGNETICS

70

SOLO


Let compute

000000, EeeEerErE rSjkrSjkrSjk

SjkSjk eSjke 000

Sjk

SjkSjkSjk

eESjkESjkESjkESjkE

EeeEeErE0

000

00000002

002

0002 ,

Sjk

SjkSjkSjk

eESjkE

EeeEeErE0

000

000

000,

Sjk

SjkSjkSjk

eESjkE

EeEeeErE0

000

lnln

lnlnlnln,

000

000

0lnln22 EEEkE

02

00000002

002 EkESjkESjkESjkESjkE

SjkeESjkEESjkE 0000000 lnlnln

Starting from

ELECTROMAGNETICS

71

SOLO


02

00000002

002 EkESjkESjkESjkESjkE

SjkeESjkEESjkE 0000000 lnlnln

Since ,by dividing the previous equation bywe obtain

000

00

knk

Sjkek 02

0

But

0lnln

1

lnln1

02

0020

0002

00

2

EEEjk

ESESESjk

ESSn

000

0000

lnln2ln

lnlnlnln

ESSEnES

EESESES

Hence we obtain

0lnln1

ln2ln1

0

2

002

0

00

2

0

0

2

EEEjk

SEnESSjk

ESSn

ELECTROMAGNETICS

72

SOLO


or

0lnln

1

ln2ln1

02

0020

002

00

2

EEEjk

SEnESSjk

ESSn

0,,1

,,,1

,, 020

00

0 EMjk

nSELjk

nSEK

where

0

2000

002

0

02

0

lnln,,

ln2ln,,,

,,

EEEEM

SEnESSnSEL

ESSnnSEK

ELECTROMAGNETICS

73

SOLO


In the same way using

0,,1

,,,1

,, 020

00

0 HMjk

nSHLjk

nSHK

we obtain

0

2000

002

0

02

0

lnln,,

ln2ln,,,

,,

HHHHM

SHnHSSnSHL

HSSnnSHK

0000 &, 0

kerHrH rSjk

0,, 02

0 HSSnnSHK

or

For sufficient large (high frequencies) the second and third term may be neglected and the wave equations becomes

000 k

22 nS Eikonal Equation

ELECTROMAGNETICS

74


SYMMETRIC MAXWELL’s EQUATIONS


1mA




Electric Current Density eJ

2mA

Free Electric Charge Distributione 3 msA

Fictious Magnetic Current Density mJ 2mV

Fictious Free Magnetic Charge Distributionm 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

4. GAUSS’ LAW – MAGNETIC (GM) mB

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


James Clerk Maxwell(1831-1879)

75


SYMMETRIC MAXWELL’s EQUATIONS (continue – 1)

eJt

DH

mJt

BE

eD

mB

From the Symmetric Maxwell’s Equations we can see that we obtain the same equationsby performing the following operations.

DUALITY

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

mJt

BE

mB

eD

76



From the Symmetric Maxwell’s Equations

00

tJJD

tH

D

Jt

DH

eee

e

e

00

tJJB

tE

B

Jt

BE

mmm

m

m

CONSERVATION OF ELECTRICAL AND MAGNETIC CHARGES

0

t

J ee

0

t

J mm

Therefore

77



Assume Homogeneous, Linear and Isotropic Medium.

Using Linearity we can decompose the field in two parts:

- A field due to electric sources denoted by A.

- A field due to magnetic sources denoted by F.

mF

AAthatsuchFA

B

ABBBBB

0

eA

FFthatsuchFA

D

FDDDDD

0

t

DHHB

t

DJHHB

HBF

Fthatsuch

FF

AeA

thatsuchAA

t

BEED

t

BJEED

EDA

Athatsuch

AA

FmF

thatsuchFF

78



Assume Homogeneous, Linear and Isotropic Medium (continue – 1)

mFFFDF

F Vt

FH

t

FH

t

DH F

0

eAAABA

A Vt

AE

t

AE

t

BE A

0

t

DJA

t

DJH A

e

BH

AB

AeA

AA

A

11

t

BJF

t

BJE F

m

DE

FD

FmF

FF

F

11

Therefore, we have

eAF Vt

AFEEE

1

mAF Vt

FAHHH

1

79




Also

eAA

Ae

Vt

AED

t

DJA

Since we have one degree of freedom in the choice of and Ve, we can choose A

0

t

VA e

LORENZ CONDITION

Using this condition we obtain:

eJt

AA

2

22

t

VAJ

t

AA e

e

2

22

ee Vt

A

tJAA

2

80




Let compute

or

E

Vt

AFE

e

e

1

ee

e t

VV

2

22

2

22

t

VV e

e t

VA

ee

e

VAt

F

2

0

1

81




Also

Since we have one degree of freedom in the choice of and Vm, we can choose F

Using this condition we obtain:

mFF

Fm

Vt

FHB

t

BJF

GAUGE CONDITION0

t

VF m

mJt

FF

2

22

t

VFJ

t

FF m

m

2

22

mm Vt

F

tJFF

2

82




Let compute

or

H

Vt

FAH

m

m

1

mmm t

VV

2

22

2

22

t

VV m

m t

VF

mm

m

VFt

A

2

0

1

83




where are constant scalars, we have

ED

HB

,

e

ED

m

HB

Jt

EJ

t

DH

t

Jt

H

t

BE


,

m

ee

e

me

Jt

J

t

EE

DED

EEE

Jt

J

t

EE

2

222

2

2

&

84




In the same way

e

ED

e

m

HB

m

Jt

EJ

t

DH

Jt

HJ

t

BE

t


,

e

mm

m

em

Jt

J

t

HH

HHB

HHH

Jt

J

t

HH

2

222

2

2

&

Thereforem

ee Jt

J

t

EE

2

22

emm Jt

J

t

HH

2

22

85


SYMMETRIC MAXWELL’s EQUATIONS (SUMMARY)

(A) Electric Current Density eJ

2mA

Free Electric Charge Distributione 3 msA

Fictious Magnetic Current Density mJ 2mV

Fictious Free Magnetic Charge Distributionm 3 msV

eJt

DH

mJt

BE

eD

mB

(F)

(GE)

(GM)

e

m

m

e

e

m

m

e

J

J

J

J

E

H

H

E

B

D

D

B

DUALITY

86


SYMMETRIC MAXWELL’s EQUATIONS (SUMMARY) (continue – 1)

Field Equations eAF V

t

AFEEE

1

mAF Vt

FAHHH

1

eJt

AA

2

22

eee t

VV

2

22

0

t

VA e

GAUGE CONDITION

mJt

FF

2

22

mmm t

VV

2

22

0

t

VF m

GAUGE CONDITION

mee Jt

J

t

EE

2

22

emm Jt

J

t

HH

2

22

87


SYMMETRIC MAXWELL’s EQUATIONS (SUMMARY) (continue – 2)

Also

For Homogeneous, Linear and Isotropic Medium we found the following Inhomogeneous Differential Equations known also Helmholtz Equations

trJt

trJtr

t

trEtrE m

ee ,,,,

,2

22

trJt

trJtr

t

trHtrH e

mm ,,,,

,2

22

trJt

trAtrA e ,

,,

2

22

trJt

trFtrF m ,

,,

2

22

tr

t

trVtrV ee

e

,,,

2

22

tr

t

trVtrV mm

m

,,,

2

22

0

t

VA e

0

t

VF m GAUGE CONDITIONS

Hermann von Helmholtz1821-1894

88


2. Stratton-Chu Solution of Non-homogeneous (Helmholtz) Differential Equations

1. Kirchhof’s Solution of the Scalar Helmholtz Non-homogeneous Differential Equations

General Solutions of Maxwell Equations

We will present three way of deriving the general solutions of Maxwell Equations:

3. Dyadic Green’s Function Solution of Non-homogeneous (Helmholtz) Differential Equations

89


KIRCHHOFF’s SOLUTION OF THE SCALAR HELMHOLTZ NONHOMOGENEOUS DIFFERENTIAL EQUATION

The Helmholtz Non-homogeneous Differential Equation for the Electric Scalar Potential Ve is:

trtrVtv

trV eee ,1

,1

,2

2

22

We want to find the Electric Scalar Potential Ve at the point F (field) due to all thesources (S) in the volume V, including its boundaries .

n

iiSS

1

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

F inside V F on the boundary of V

Therefore is the vector from S to F.SF rrr

zzyyxxr 111

zzyyxxr SSSS 111

zzyyxxr FFFF 111

Let define the operator

that acts only on the source coordinate .

zz

yy

xx SSS

S 111

Sr

90



To find the solution we need to prove the following:

• GREEN’s IDENTITY

S

ee

V

ee dSGVVGdVGVVG 22

• GREEN’s FUNCTION

FS

FS

FS rr

v

rrtt

trtrG

'

',;,

This Green’s Function is a particular solution of the following Helmholtz Non-homogeneous Differential Equation:

'4',;,1

',;,2

2

22 ttrrtrtrG

tvtrtrG SFFSFSS

(continue – 1)

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr


91



• Scalar Green’s Identities

S

ee

V

ee dSGVVGdVGVVG 22

(continue – 2)

Let start from the Gauss’ Divergence Theorem

SV

dSAdVA


where is any vector field (function of position and time) continuous and differentiable in the volume V. Let define .

A

eVGA

eee VGVGVGA 2

Then

S

e

Gauss

V

ee

V

e dSVGdVVGVGdVVG 2

If we interchange G with Ve we obtain

S

e

Gauss

V

ee

V

e dSGVdVGVVGdVGV 2

Subtracting the second equation from the first we obtain

First Green’s Identity

Second Green’s Identity

We have

GEORGE GREEN1793-1841

92



• GREEN’s FUNCTION

Define the Green’s Function as a particular solution of the following Helmholtz Non-homogeneous Differential Equation:

'4',;,1

',;,2

2

22 ttrrtrtrG

tvtrtrG SFFSFSS

(continue – 3)

where δ (x) is the Dirac function

1

0

00

dxx

x

x

x

Let use the Fourier Transformation to write

33

3

exp2

1

exp2

1

exp2

1exp

2

1exp

2

1

dkrrkj

dkdkdkzzkyykxxkj

dkzzjkdkyyjkdkxxjk

zzyyxxrr

SF

zyxSFzSFySFx

zSFzySFyxSFx

SFSFSFSF

zyx

zyx

dkdkdkdk

zkykxkk

3

111

where

Paul Dirac1902-1984

Joseph Fourier 1768-1830

93



• GREEN’s FUNCTION (continue – 1)

In the same way:

(continue – 4)

dttjtt 'exp2

1'

Therefore

'expexp2

1' 3

4ttjrrkjddkttrr SFSF

Let use the Fourier Transformation to write

'expexp,',;, 3 ttjrrkjkgddktrtrG SFFS

Hence

'expexp2

4'expexp,

1 3

4

3

2

2

2

2 ttjrrkjddkttjrrkjkgddktv SFSFS

or

'expexp2

4

'expexp1

exp'exp,

3

4

2

2

2

23

ttjrrkjddk

ttjt

rrkjv

rrkjttjkgddk

SF

SFSFS

94



• GREEN’s FUNCTION (continue – 2)

Let compute:

(continue – 5)

SFSFSFS

SFzSFySFxSzyx

SFzSFySFxzyxS

SFzSFySFxSSSFS

rrkjkrrkjkjkjrrkjkj

zzkyykxxkjzjkyjkxjk

zzkyykxxkjzjkyjkxjk

zzkyykxxkjrrkj

expexpexp

exp111

exp111

expexp

2

2

'exp'exp 2

2

2

ttjttjt

Therefore:

'expexp2

4

'expexp,

3

4

2

223

ttjrrkjddk

ttjrrkjv

kkgddk

SF

SF

Because this is true for all k and ω, we obtain

2

22

3

1

4

1,

vk

kg

95



• GREEN’s FUNCTION (continue – 3) (continue – 6)

'expexp

4

1'expexp,',;,

2

22

33

3 ttjrrkj

vk

ddkttjrrkjkgddktrtrG SFSFFS

We can see that the integral in k has to singular points forv

k

Let consider only the progressive wave, i.e. G = 0 for t > t’.

To find the integral let change ω by ω + jδ where δ is a small negative number

rkj

v

jk

ddktrtrG FS

exp4

1',;,

2

22

33

where and .SF rrr

'tt

96




In the plane ω we close the integration path by the semi-circle withand the singular points on the upper side, for τ > 0 (for t > t’)

rkj

v

jk

ddktrtrG FS

exp4

1',;,

2

22

33

r

'00exp ttdjfUPC

'00exp ttdjfDOWNC

0exp

0exp

exp

DOWN

UP

C

C

djf

djf

djf

jvk jvk Re

Im

97




CC jvkjvk

rkjvdrkj

vj

k

drkj

vj

k

dI

expexpexp

2

2

22

2

22

Let use the Cauchy Integral for a complex function f (z) continuous on a closed path C, in the complex z plane: 0

0

2lim20

zfjzfjdzzz

zfzz

C

We have:

k

kvrkjv

vk

jkv

vk

jkvrkjvj

jvk

rkjvj

jvk

rkjvjI

vkvk

sinexp2

2

exp

2

expexp2

exp2lim

exp2lim

2

2

0,

2

0,

Therefore, we can write:

k

vkrkjdk

v

vk

rkjddktrtrG FS

sinexp

2

exp

4

1',;, 3

2

2

22

33

98




Let use spherical coordinates relative to vector:

k

vkrkjdk

v

vk

rkjddktrtrG FS

sinexp

2

exp

4

1',;, 3

2

2

22

33

r

rr

r

r

kk

kk

kk

z

y

x

z

y

x

0

0

cos

sinsin

cossin

dk sindk

dk

dddkkdk sin23

r

xy

z

99




kvdvr

jkvvr

jkvvr

jkvvr

jkv

r

4

expexpexpexp1

r

v

rrtt

v

rrtt SFSF

''

rvr

vr

dxv

rjx

v

rjx

r

expexp

2

11

kvdvr

jkvvr

jkv

r

vkvd

vr

jkvvr

jkv

r

4

expexp

4

expexp1

dk

j

jvkjvk

j

jkrjkr

r

v

2

expexp

2

expexp

dkvkvkrr

vdkvkvkr

r

v

sinsinsinsin2

0

00 0

sin2

expexp2cosexpsin dkvk

j

jkrjkr

r

vdk

jkr

jkrvkk

v

0 0

2cossincosexp2

2

dkdvkjkrkv

0 0

2

0

22

sinsincosexp

2

dkddkk

vkjkrv

k

vkrkjdk

vtrtrG FS

sinexp

2',;, 3

2

100




We can see that represents a progressive waveand represents a regressive wave:

v

rrtt

v

rrtt SFSF

''

v

rrtt

v

rrtt SFSF

''

Hence SF

SFSF

FS rr

v

rrtt

v

rrtt

trtrG

''

',;,

We shall consider only the progressive wave and use:

SF

SF

FS rr

v

rrtt

trtrG

'

',;,

Retarded Green Function

The other solution is:

SF

SF

FS rr

v

rrtt

trtrG

'

',;,

Advanced Green Function

101


KIRCHHOFF’s SOLUTION OF THE SCALAR HELMHOLTZ NONHOMOGENEOUS DIFFERENTIAL EQUATION (continue – 12)

Gustav Robert Kirchhoff1824-1887

Let return to the Helmholtz Non-homogeneous Differential Equation for the Electric Scalar Potential Ve is:

trtrVtv

trV eee ,1

,1

,2

2

22

We want to find the Electric Scalar Potential Ve at the point F (field) due to all the sources (S) in the volume V, includingits boundaries

n

iiSS

1

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr


Hermann von Helmholtz1821-1894

102



Since is no defined at r = 0 we define the volume V’ as the volume V minus a smallsphere of radius and surface around the point F, when F is inside V, or asemi-sphere of radius and surface around the point F, when F is on the boundary of V.

rG

00 r

00 r

204 rSF

202 rSF

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

Let define the operator that acts only on the source coordinate .Sr

zz

yy

xx SSS

S 111

Using the Green’s Identity

FSS

SFSSeSeSSF

V

SFSSeSeSSF dStrtrGtrVtrVtrtrGdVtrtrGtrVtrVtrtrG ',;,',',',;,',;,',',',;,'

22

substitute here

',1

','

1',

2

2

22 trtrV

tvtrV SeSeSeS

'4',;,'

1',;,

2

2

22 ttrrtrtrG

tvtrtrG FSFSFSS

we obtain

FSS

SFSSeeSSSF

V

SFSeS

SF

V

SFSeSeSF

dStrtrGtrVtrVtrtrG

dVttrrtrVtr

trtrG

dVtrtrGt

trVtrVt

trtrGv

',;,',',',;,

'4',',

',;,

',;,'

',','

',;,1

'

'2

2

2

2

2

103



Let integrate the equation between to and choose t such that:1' tt 2' tt 21 /' tvrttt

2

2

1

1

2

1

''4',',

',;,

'',;,'

',','

',;,1

'

'2

2

2

2

2

I

t

t V

SFSeS

SF

I

t

t V

SFSeSeSF

dtdVttrrtrVtr

trtrG

dtdVtrtrGt

trVtrVt

trtrGv

4

2

1

3

2

1

'',;,',',',;,

'',;,',',',;,

I

t

t S

SFSSeeSSSF

I

t

t S

SFSSeeSSSF

dtdStrtrGtrVtrVtrtrG

dtdStrtrGtrVtrVtrtrG

F

104



Integral I1

0

'

',;,',;, 21

SF

SF

SFSF rr

v

rrtt

ttrtrGttrtrG

Since 21 /' tvrttt

then

0',;,'

',','

',;,1

'',;,'

',','

',;,'

1

'',;,'

',','

',;,1

'2

'2

'2

2

2

2

21

2

1

2

1

2

1

V

t

t

SFSeSeSF

V

t

t

SFSeSeSF

t

t V

SFSeSeSF

dVtrtrGt

trVtrVt

trtrGv

dVdttrtrGt

trVtrVt

trtrGtv

dVdttrtrGt

trVtrVt

trtrGv

I

0',;,'

',;,' 21

ttrtrGt

ttrtrGt SFSF

105



Integral I2

2

1

''4',',

',;,'

2

t

t V

SFSeS

SF dtdVttrrtrVtr

trtrGI

2

1

2

1 ''

''',4'',/'

t

t

t

t V

SFSe

V

S

FS

dVdtttrrtrVdtdVtr

rr

vrtt

' /'

0

'' /'

',1'',4

',1

V vrttFS

S

V

SFSe

V vrttFS

S dVrr

trdVttrrtrVdV

rr

tr

The integral is zero since in V’. '

/',V

FSSe dVrrvrttrV

FS rr

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

106



Integral I3

2

1

'',;,',',',;,3

t

t S

SFSSeSeSSF dSdttrtrGtrVtrVtrtrGI

S

t

t SF

SF

SSeSeSSF

SF

dSdtrr

v

rrtt

trVtrVrr

v

rrtt

2

1

'

'

',',

'

S

t

t SFS

SFSeSeS

SF

SF

dSdtrrv

rrtttrVtrV

rr

v

rrtt

2

1

'1

'',',

'

S

t

t SF

SFS

Se dSdtrr

v

rrtt

trV2

1

'

'

',

107



Integral I3 (continue – 1)

But

v

rrvrtt

trrvrtt

rv

rrtt SFS

SFS

rrrSF

S

SF

/''

/''

and

S

t

t

SF

SF

SFSSe

S

t

t SF

SFS

Se dSdtv

rrtt

trrv

rrtrVdSdt

rr

v

rrtt

trV2

1

2

1

'''

','

'

',

S

t

t

SF

SF

SFSSe

S

t

t

SF

SF

SFSSe

partsbyegrating

dSdtv

rrtt

rrv

rrtrV

tdS

v

rrtt

rrv

rrtrV

2

1

2

1

''','

'',

0

int

108




Therefore

S

t

t

SF

Se

SF

SFS

SF

SSe

SF

SeS dSdtv

rrtttrV

trrv

rr

rrtrV

rr

trVI

2

1

''','

1',

',3

S

vrtt

Se

SF

SFS

SF

SSe

SF

SeS dStrVtrrv

rr

rrtrV

rr

trV

/'

','

1',

',

S

vrtt

Se

SF

SF

SF

SSe

SF

Se

dStrVtrrv

rrn

rrtrV

rr

n

trV

/'

','

1',

',

The last equality follows from dSn

UdSnn

n

UdSU SS

11

109



Integral I4

In the same way as for integral I4 we have

2

1

'',;,',',',;,4

t

t S

SFSSeSeSSF

F

dSdttrtrGtrVtrVtrtrGI

FS

vrtt

Se

SF

SFS

SF

SSe

SF

SeS dStrVtrrv

rr

rrtrV

rr

trV

/'

','

1',

',

FS

vrtt

Se

SF

SF

SF

SSe

SF

Se

dStrVtrrv

rrn

rrtrV

rr

n

trV

/'

','

1',

',

110




We use since points inside V normal to and points outside V.

nr

r1

0

0

0

0

r

r

nS

n1

On the sphere or the semi-spherearound the field point F with radius and surface or if the point F is inside V or on the boundary, respectively, we have

FS rrr

002

04 rSF 202 rSF

nr

r

rr

rr

rr

rrrr

SF

SF

SF

FS

SSFSF

10

0

n

rr

r

rrr

rr

rrrr

rr

rrrr SF

SF

SFSF

FS

SFSSFS

F

111111

200

02

022

ndrr

rdrdS

FS

120

0

020

Since we can assume mean values for all field quantities in the integral00 r

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

111




FS

vrtt

Se

SF

SFS

SF

SSe

SF

SeS dStrVtrrv

rr

rrtrV

rr

trVI

/'

4 ','

1',

',

drnnrv

trVt

nr

trVr

trV

vrtt

SeSeSeS

r

2

0

/'0

2

000

111

','

11

',',

lim0

0

0

/'0/'0

0

0/'0',

'lim',lim1',lim

000

dv

rtrV

tdtrVdrntrV

vrtt

ServrttServrttSeSr

trVSonF

VinFtrV

SonFd

VinFd

FeFe ,2

4,

2

0

4

0

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

112



SUMMARIZE

The Kirchhoff’s solution to the Helmholtz Non-homogeneous Differential Equation:

trtrVtv

trV eee ,1

,1

,2

2

22

is

S

v

rrtt

Se

SF

SFS

SF

SSe

SF

SeS

V

v

rrttSF

SFe dStrV

trrv

rr

rrtrV

rr

trVTdV

rr

trTtrV

SFSF

''

','

1',

',

4

',

4,

S

v

rrtt

Se

SF

SF

SF

Se

SF

Se

V

v

rrttSF

SdSndS

nn

dStrVtrrv

rrn

rrntrV

rr

n

trV

TdV

rr

trT

SF

SFS

'

'

ˆ

ˆ

','

1',

',

4

',

4

VoutsidenSonF

VinFT

12

1iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

113



Let choose 0/

:,,, 222

FSFSFSSSS zzyyxxrr

vrtwtzyxG

where w (t) is an arbitrary twice-differentiable function of t and (xF,yF,zF) is a fixed point.

011

,,,23

/2

rr

td

wd

rvr

wr

td

wd

rvr

wtzyxG

rrSSSS

r

td

wd

rvr

r

wtzyxG SSSS

23

2 1,,,

tzyxGtvtd

wd

rvtd

wd

rvtd

wd

rvtd

wd

rvr

w

r

w

td

wd

rv SSS ,,,11312

331

2

2

22

2

222

2

22332

3

22

2

233

343

1123 r

td

wd

rvr

v

r

td

wd

rvrr

td

wd

rvr

r

wrr

r

w

r

r

v

r

td

wd

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

)continue – 24(Second Way

114



Substitute

00,,,1

,,,2

2

2

2

rtzyxGtv

tzyxG SSSSSSS

trtrVtv

trV eee ,1

,1

,2

2

2

2

S

SeeSV

SeeS dSGVVGdVGVVG 22

into 2

2

22

2

2

22 ,1

,1

td

wd

rv

VtrtrV

tvr

wGVVG e

eeSeeS

0

/:,,,

r

r

vrtwtzyxG SSS

Using Second Green’s Identity:

we obtain

V

e

Se

e

V

ee dVr

wdS

n

rwV

n

V

r

wdV

td

wd

r

V

t

V

r

w

v /1

2

2

2

2

2

whereS

S

dSn

dS


115



iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

Let denote by rm the maximum distance from the fixed point (xF,yF,zF) to all the points in the volume V, and choose w (t) such that:

0/202

2

td

wd

td

wdvrttw m

V

e

Se

e

V

ee dVr

wdS

n

rwV

n

V

r

wdV

td

wd

r

V

t

V

r

w

v /1

2

2

2

2

2

Define: vrT m /3:0

2

2

v

rT

td

wd

v

rT

td

wd

v

rTw

02

2

2

2

T

T

ee

T

T

ee

v

rt

td

wdV

t

V

v

rtwdt

td

wd

r

V

t

V

r

w

Therefore:

V V

T

T

eeT

T

e

S

T

Te

e dVtd

wd

r

V

t

V

r

w

vdVdt

r

wdSdt

n

rwV

n

V

r

w0

1/2

2

2

2

2


116



iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

But:

V V

T

T

eeT

T

e

S

T

Te

e dVtd

wd

r

V

t

V

r

w

vdVdt

r

wdSdt

n

rwV

n

V

r

w0

1/2

2

2

2

2

td

wd

n

r

rvw

n

r

r

vrtw

n

1/1/

T

T

eT

T

eT

Te

T

Te dt

t

Vwdt

t

VwvrtwVdt

td

wdV

0

/

T

Te

T

Te

T

Te dt

td

wdV

n

r

rvdtw

n

rVdt

r

vrtw

nV

1/1/

T

T

eT

Te

T

Te dt

t

Vw

n

r

rvdtw

n

rVdt

r

vrtw

nV

1/1/

V

T

T

e

S

T

T

ee

e dVdtr

vrtwdSdtt

V

n

r

rvn

rV

n

V

rvrtw 0/

1/11/


117



iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

V

T

T

e

S

T

T

ee

e dVdtr

vrtwdSdtt

V

n

r

rvn

rV

n

V

rvrtw 0/

1/11/

Introduce the transformation: vrtt /'

T

Tvrtt

vrT

T

T

T

T

vrT

v

rT

rr

vrT

vrTvrtt dttwdttwdttwdttwdttw

m

m

'''''''''' /'

/

0/ 0

3/

//'

V

T

T

vrtte

S

T

Tvrtt

evrtte

vrtt

e dVdtr

twdSdtt

V

n

r

rvn

rV

n

V

rtw 0'''

1/11' /'

/'

/'

/'

But w (t’) and T are independent on space coordinates (xs, ys, zs), therefore we caninterchange between time and space integrals to obtain:

0'

1/11' /'

/'

/'

/'

T

T S V

vrtte

vrtt

evrtte

vrtt

e dtdVr

dSt

V

n

r

rvn

rV

n

V

rtw


vrttw m /20 using:

118



iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

Since this is true for an arbitrary w (t’), that satisfies only the condition:

we must have:

0'

1/11' /'

/'

/'

/'

T

T S V

vrtte

vrtt

evrtte

vrtt

e dtdVr

dSt

V

n

r

rvn

rV

n

V

rtw

0/202

2

td

wd

td

wdvrttw m

0

1/11 /'

/'

/'

/'

S V

vrtte

vrtt

evrtte

vrtt

e dVr

dSt

V

n

r

rvn

rV

n

V

r


rVe

00 r

00 r

204 rSF

202 rSF

On the sphere/semi-sphere SF we have: 02

0

&1/1

&1 rrrn

r

n

rF

FF

S

SS

SonFtrV

VinFtrVdS

t

V

rvrV

n

V

r Fe

Fer

Svrtt

evrtte

vrtt

e

F,2

,4111 0

/'0

2

0

/'

/'0

0

0

0

0


119




trtrVtv

trV eeeS ,1

,1

,2

2

2

2

is

S

v

rrtt

Se

SF

SFS

SF

SSe

SF

SeS

Vv

rrttSF

SFe dStrV

trrv

rr

rrtrV

rr

trVTdV

rr

trTtrV

SFSF

''

','

1',

',

4

',

4,

S

v

rrtt

Se

SF

SF

SF

Se

SF

Se

Vv

rrttSF

SdSndS

nn

dStrVtrrv

rrn

rrntrV

rr

n

trV

TdV

rr

trT

SF

SFS

'

'

ˆ

ˆ

','

1',

',

4

',

4

VoutsidenSonF

VinFT

12

1

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr


120



Let write the solution such that it will show explicitly the contribution of the external surface of V. nS

The first two terms represent traveling waves coming from the sources. The thirdterm however, represents the sum of all waves traveling inwards from the surface ,and must, therefore be finite as recedes to infinity.

nS

nS

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

RADIATION CONDITION

nSF

nSF

SF

S

v

rrtt

Se

SF

SFS

SF

SSe

SF

SeS

SS

v

rrtt

Se

SF

SFS

SF

SSe

SF

SeS

V

v

rrttSF

SFe

dStrVtrrv

rr

rrtrV

rr

trVT

dStrVtrrv

rr

rrtrV

rr

trVT

dVrr

trTtrV

'

'

'

','

1',

',

4

','

1',

',

4

',

4,

n SF

SFn

S

v

rrtt

Se

SF

SFS

SF

SSe

SF

SeS

rrS dStrVtrrv

rr

rrtrV

rr

trVI

'

','

1',

',lim

4

1

121



RADIATION CONDITION (continue – 1)

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

On the sphere around the field point F with radius and surface we have

0r2

04 rSF

nr

r

rr

rr

rr

rrrr

nn

n

SSF

SF

SSF

FS

SSFS 10

0


nr

r1

0

0

0

0

r

r

nS

n1

n SF

SFn

S

v

rrtt

Se

SF

SFS

SF

SSe

SF

SeS

rrS dStrVtrrv

rr

rrtrV

rr

trVI

'

','

1',

',lim

4

1

n

rr

r

rrr

rr

rrrr

rr

rrrr SF

SF

SFSF

FS

SFSSFS

n

111111

200

02

022

ndrr

rdrdS

FS

120

0

020

ntrVn

nr

trV

r

r

r

trVr

r

trVtrV

nnn

n

S

See

S

e

S

Se

SSeS 1',1',',',

',0

0

0

0

0

00

0

0

122



RADIATION CONDITION (continue – 2) Since we can assume mean values for all field quantities in the integral 0r

n SF

SF S

v

rrtt

Se

SF

SFS

SF

SSe

SF

SeS

rrdStrV

trrv

rr

rrtrV

rr

trVI

'

','

1',

',lim

4

1

drnnrv

trVt

nr

trVnr

trVn

v

rrtt

SeSeSerSF

2

0

'0

2

00

111

','

11

',11

',lim4

10

dtrVdtrVtv

trVn

rv

rrttSer

v

rrtt

SeSerSF

SF

'

'

0 ',lim4

1',

'

1',lim

4

100

0',lim4

1',

'

1',lim

4

1'

'

000

dtrVdtrV

tvtrV

nr

v

rrttSer

v

rrtt

SeSerSF

SF

The first integral is finite if:

0','

1',lim',

'

1',lim 00

0

000000

trV

tvtrV

rrtrV

tvtrV

nr eereer

Radiation Condition

The second integral is finite if ;i.e. Ve is regular at infinity. 0',lim 00

trVer

123


Also


trJt

trAtrA e ,

,,

2

22

trJt

trFtrF m ,

,,

2

22

tr

t

trVtrV ee

e

,,,

2

22

tr

t

trVtrV mm

m

,,,

2

22

0

t

VA e

0

t

VF m GAUGE CONDITIONS

KIRCHHOFF’s SOLUTION OF THE HELMHOLTZ NONHOMOGENEOUS DIFFERENTIAL EQUATIONS (continue – 34)

124




is

S

v

rrtt

Sm

SF

SFS

SF

SSm

SF

SmS

V

v

rrttSF

SmFm dStrV

trrv

rr

rrtrV

rr

trVTdV

rr

trTtrV

SFSF

''

','

1',

',

4

',

4,

S

v

rrtt

Sm

SF

SF

SF

Sm

SF

Se

V

v

rrttSF

SmdSndS

nn

dStrVtrrv

rrn

rrntrV

rr

n

trV

TdV

rr

trT

SF

SFS

'

'

ˆ

ˆ

','

1',

',

4

',

4

VoutsidenSonF

VinFT

12

1

tr

t

trVtrV mm

m

,,,

2

22

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

125


KIRCHHOFF’s SOLUTION OF THE VECTOR HELMHOLTZ NONHOMOGENEOUS DIFFERENTIAL EQUATION (continue – 36)


is

S

v

rrtt

SF

SFS

S

SF

SS

SF

SS

V

v

rrttSF

SeF dS

rrv

rrtrA

trrtrA

rr

trATdV

rr

trJTtrA

SFSF

''

','

1',

',

4

',

4,

S

v

rrtt

S

SF

SF

SF

S

SF

S

V

v

rrttSF

SedSndS

nn

dStrAtrrv

rrn

rrntrA

rr

n

trA

TdV

rr

trJT

SF

SFS

'

'

ˆ

ˆ

','

1',

',

4

',

4

VoutsidenSonF

VinFT

12

1

trJt

trAtrA e ,

,,

2

22

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

126


KIRCHHOFF’s SOLUTION OF THE VECTOR HELMHOLTZ NONHOMOGENEOUS DIFFERENTIAL EQUATION (continue – 37)


is

S

v

rrtt

SF

SFS

S

SF

SS

SF

SS

V

v

rrttSF

SmF dS

rrv

rrtrF

trrtrF

rr

trFTdV

rr

trJTtrF

SFSF

''

','

1',

',

4

',

4,

S

v

rrtt

S

SF

SF

SF

S

SF

S

V

v

rrttSF

SmdSndS

nn

dStrFtrrv

rrn

rrntrF

rr

n

trF

TdV

rr

trJT

SF

SFS

'

'

ˆ

ˆ

','

1',

',

4

',

4

VoutsidenSonF

VinFT

12

1

trJt

trFtrF m ,

,,

2

22

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

127


STRATTON-CHU SOLUTION OF THE NONHOMOGENEOUS DIFFERENTIAL EQUATIONS


me Jt

J

t

EE

2

2

em Jt

J

t

HH

2

2

Let assume that can be written as: trHtrE ,,,

tjrHtrHtjrEtrE 00 exp,,exp,

where are phasor (complex) vectors.

rHjrHrHrEjrErE

ImRe,ImRe

We have tjrEjtjt

rEtrEt 00 expexp,

Hence0j

t

128


STRATTON-CHU SOLUTION OF THE NONHOMOGENEOUS DIFFERENTIAL EQUATIONS (continue – 1)

NoteThe assumption that can be written as: trHtrE ,,,


Is equivalent to saying that has a Fourier transform; i.e.: trHtrE ,,,

dtjrHtrHtdtjtrHrH

dtjrEtrEtdtjtrErE

exp,2

1,&exp,,

exp,2

1,&exp,,

This is equivalent to:

drHdttrH

drEdttrE

22

22

,2

1,

,2

1,

00

0

exp

expexpexp,,

rEtdtjrE

tdtjtjrEtdtjtrErE

End Note

129



We can write

m

e

m

e

m

e

m

e

B

D

JBjE

JDjH

jt

BGM

DGE

Jt

BEF

Jt

DHA

)(

And the Nonhomogeneous (Helmholtz) Differential Equations that we want to solve are:

where

me JJjEkE

2

em JJjHkH

2

22 f

c

c

fk

130



GREEN’s IDENTITIES

Start from Gauss’ formula that relates the integral of the flux of a union of closed surfaces to it’s divergence.

n

iiSS

1

S

GAUSS

V

dSnFGdVFG 1

n

iiSS

1

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

and must be continuous and twice differentiable in V.G

F

Using the identity we obtain

FGFGFG

S

GAUSS

V

dSnFGdVFGFG 1

First Vector Green Identity

Interchanging and we obtainG

F

S

GAUSS

V

dSnGFdVGFFG 1

By subtracting the second identity from the first we obtain

SV

dSnGFFGdVFGGF 1

Second Vector Green Identity

131



We want to find the field magnitude at the point F (field) due to all the Sources (S) in the volume V including its boundaries .

n

iiSS

1

HE ,

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr


Let use for the Green functionG 0&

exp

rara

r

jkrG

where is the vector from S to F and is an arbitrary constant vector.SF rrr

a

132



Let define the operator that acts only on the source coordinate .S Sr

r

r

rr

rr

rr

rrrrr

SF

SF

SF

FSSFSS

30 SSSFSS rrrr

0

exp1

r

r

r

r

jkr

rjk

r

r

dr

dr

dr

dr SS

rjkr

rr

jkrr SSSS

exp

132

2

rjkrrr

jkrrjkr

rr

jk

rd

dSS

exp

1exp

13232

3exp1

exp32

3232

2

43

jkr

rr

jkr

r

rjkr

r

jk

r

k

rr

jk

jkrrr

jk

r

r

r

jk

r

k

rr

jk

exp

13

3232

2

32

2

43

0exp 22

rrk

r

jkrk

Therefore 0022 rrkrS

0

exp

r

r

jkrr

133



0022 rrkrS 0

exp

r

r

jkrr


r

00 r

00 r

204 rSF

202 rSF

Using the Second Green Identity let compute:

FSS

SS

V

SSSS

dSnarEEar

dVEararE

1

'

We can see that

arkraar

rkr

ararar

SS

consta

SS

S

SSSSS

2

22

2

0

134



and

mSeSS

SSSS

mSeSS

SSSS

JJjarEarkEarkraE

EararE

JJjEkEar

arkraarE

0

22

2

2

|

|

Using the identity with and we get

AAA

ra S

EA

EraraEraE SSSSS

Hence

''

'

V

SSmSe

V

SS

V

SSSS

dVrErJJjadVraE

dVEararE

135



And now

Using the identity with and we get

AAA

r mJA

mSmSmS JrJrJr

'

''

'

V

SSmSe

V

mS

V

SS

V

SSSS

dVrEJrrJja

dVJradVraE

dVEararE

But

FF SS

S

consta

SS

S

GAUSS

V

SS dSrnEadSnEradVraE 11'

and

FSS

m

GAUSS

V

mS dSJnradVJra 1'

Hence

'

'

'

11V

SSmSe

SS

m

SS

S

V

SSSSV

dVrEJrrJjdSJnrdSrnEa

dVEararEI

FF

136



'

'

'

11V

SSmSe

SS

m

SS

S

V

SSSSV

dVrEJrrJjdSJnrdSrnEa

dVEararEI

FF

We succeed to obtain, for the volume integral an expression with a scalar multiplication of with an other expression. a

F

F

SS

SSSS dSnarEEarI 1

Start with

nEranEar S

CBACBA

S 11

and

arEnarEnarEnnarE SS

CBACBA

S

consta

S

1111

Therefore

FF

F

SS

SS

SS

SSSS dSrEnEnradSnarEEarI 111

Let do the same for the surface integral:

137



'

' 11V

SSmSe

SS

m

SS

SV dVrEJrrJjdSJnrdSrnEaIFF

F

F

SS

SSSS dSrEnEnraI 11

Since the equality is true for an arbitrary vector we can drop it to obtain:a

FSS

SSmS

V

SSmSe

dSrnErEnJEnr

dVrEJrrJj

111

'

FS

SSmSF dSrnErEnJEnrI 111

On the sphere or the semi-sphere around the field point F with radius and surface or if the point F is inside V or on the boundary, respectively.

00 r2

04 rSF 202 rSF

Let compute the integral:

We obtained

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

138



FS


We have 0

0exp

r

jkrFS

n

r

jkr

rjk

r

r

r

jkr

rjkrr

dr

d

F

FS

SFSSS 1exp1exp1

0

0

00

0

0

0

0

ndrdS 120

We use since points outside V’ normal to SF and points outside V’.

nr

r1

0

0

0

0

r

r

n1

Since we can assume mean values for all field quantities in the integral00 r

FS


dr

r

jkrnnEnEn

rjkJEn mS

r

20

0

0

00

exp1111

11lim

0

djkrnnEnnEEjkrJEnr mS

r000

0exp111111lim

0

FFFF ESonF

VinFE

SonFd

VinFd

dEI

2

42

0

4

0

139


STRATTON-CHU SOLUTION OF THE NONHOMOGENEOUS DIFFERENTIAL EQUATIONS (continue – 10)We started from

FFFF ESonF

VinFE

SonFd

VinFd

dEI

2

42

0

4

0

S

SSmSF

V

SSmSe

dSrnErEnJEnrI

dVrEJrrJj

111

'

We obtained

Therefore, the final result is

S

SSmS

V

SSmSeF

dSrnErEnrJEnT

dVrEJrrJjT

E

1114

4

whereVoutsiden

SonF

VinFT

12

1

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

140


STRATTON-CHU SOLUTION OF THE NONHOMOGENEOUS DIFFERENTIAL EQUATIONS (continue – 11)We obtained

S

SSmS

V

SSmSeF

dSrnErEnrJEnT

dVrEJrrJjT

E

1114

4

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

We can see that in order to compute the field vector at a point F inside or on theboundary of a volume V we need to know it’s divergence in the volume V and it’s value on the surface S that bounds the volume V.

E

we obtain

From the Maxwell’s Equations

eS

mS

E

JHjE

1

S

SS

V

SemSeF

dSrnErEnrHnjT

dVrJrrJjT

E

1114

1

4

141



S

SSmS

V

SSmSeF

dSrnErEnrJEnT

dVrEJrrJjT

E

1114

4

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

In the same way, if we start withwe can obtain the solution for

eSmSSS JJjHkH 2

FH

A simpler way is to start with the solution for

FE

and to use the Duality property of Maxwell’s Equations

e

m

m

e

e

m

m

e

J

J

J

J

E

H

H

E

to obtain

S

SSeS

V

SSeSmF

dSrnHrHnrJHnT

dVrHJrrJjT

H

1114

4

142



RADIATION CONDITION Let write the Stratton-Chu solution such that it will show explicitly the contribution of

the external surface of V. nS

n

n

S

SSmS

SS

SSmS

V

SSmSeF

dSrnErEnrJEnT

dSrnErEnrJEnT

dVrEJrrJjT

E

1114

1114

4

The first two terms represent traveling waves coming from the sources. The thirdterm however, represents the sum of all waves traveling inwards from the surface ,and must, therefore be finite as recedes to infinity.

nS

nS

n

SFn

S

SSmSrr

S dSrnErEnrJEnI

111lim4

1

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

143




n

SFn

S

SSmSrr

S dSrnErEnrJEnI

111lim4

1

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

On the sphere around the field point F with radius and surface we have

0r2

04 rSF

nr

r

rr

rr

rr

rrrr

nn

n

SSF

SF

SSF

FS

SSFS 10

0

drdSFS

20

nr

jkr

rjk

r

r

r

jkr

rjk

rrr

jkr

rjk

r

jkrr SFS

SSSS

n

n

1exp1exp1

exp1exp

0

0

00

0

0

0

0

0

0

0


nr

r1

0

0

0

0

r

r

nS

n1

144




Since we can assume mean values for all field quantities in the integral 0r

djkrEjkrJEnr

drr

jkrnnEnEn

rjkJEn

dSrnErEnrJEnI

mSr

mSr

S

SSmSrr

S

nSF

n

000

20

0

0

0

exp11lim4

1

exp1111

11lim

4

1

111lim4

1

0

0

Assuming that ; i.e. are regular at infinity, the integral is finite if the following condition is satisfied:

0limlim00

0

EJrr

mr EJ m ,

nSI

0lim 000

EjkrEr Sr

In the same way using the duality relations we obtain:

kkHE ,

0lim 000

HjkrHr Sr

RADIATION CONDITION

RADIATION CONDITION

145



SUMMARY

Duality property of Maxwell’s Equations

0

00

0

)(

mm

ee

e

e

m

m

mm

ee

e

e

m

m

jJ

jJ

D

JDjH

JBjE

B

jt

tJCMG

tJCEG

DGE

Jt

DHA

Jt

BEF

BGM

e

m

m

e

e

m

m

e

J

J

J

J

E

H

H

E

146



SUMMARY

The Stratton-Chu Solution is

me JJjEkE

2

em JJjHkH

2

S

SSmS

V

SSmSeF

dSrnErEnrJEnT

dVrEJrrJjT

E

1114

4

S

SSeS

V

SSeSmF

dSrnHrHnrJHnT

dVrHJrrJjT

H

1114

4

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

where

SonF

VinFT

2

1

0&

exp

rrrr

r

jkrr SF

jt

dSn

1

22 f

c

c

fk

is directed outside the volume V

147



SUMMARY

The other form of Stratton-Chu Solution is

me JJjEkE

2

em JJjHkH

2

S

SS

V

SemSeF

dSrnErEnrHnjT

dVrJrrJjT

E

1114

1

4

S

SS

V

SmeSmF

dSrnHrHnrEnjT

dVrJrrJjT

H

1114

1

4

where

SonF

VinFT

2

1

0&

exp

rrrr

r

jkrr SF

jt

dSn

1

22 f

c

c

fk

is directed outside the volume V

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

EjJH

HjJE

e

m

148



SUMMARY

If F is outside V then we have

Note

End Note

S

SSmS

V

SSmSe

dSrnErEnrJEn

dVrEJrrJj

111

S

SSeS

V

SSeSm

dSrnHrHnrJHn

dVrHJrrJj

111

149

Dyadic Green’s Function Solution of Non-homogeneous (Helmholtz) Differential Equations


The Dyadic (Matrix) Green’s function is the solution of the vector equation SF rrG�

,

SFSS rrIGkG��

42

where is the unit dyadic or the identity matrix.I�

GrrIGkG

GGG

rrIGkG

SSSFS

SSSSSS

SFSS

��

��

��

4

4

22

2

00

42

GG

rrIGkG

SSSSSSSS

SFSSS��

��

and

SFSS

SFSSFSS

rrk

G

rrrrIGk

�

��

2

2

4

44

SFSSSS rrk

G�

2

4

Therefore

SFSSS

SFSSSS

SSSFS

rrk

IGkG

rrk

G

GrrIGkG

��

�

��

222

2

22

14

4

4

150

Dyadic Green’s Function Solution of Non-homogeneous (Helmholtz) Differential Equations (continue – 1)


SFSSS rrk

IGkG��

2

22 14

The form of the above equation suggests that can be written in terms of a Scalar Green’s function as

SF rrG�

, SF rr

,

SFSSSF rrk

IrrG��

,1

,2

To find let perform the following calculations: SF rr

,

SSSSS

SSSSSS

SS

kIk

kIkGkG

kIG

222

222

2

1

1

1

�

��

��

220

222

2222

2222

22222

kIkkI

kkI

kkI

kIkIk

SSSSSS

SSSSSS

SSSSSSSS

SSSSSSSSSSSS

SS

��

�

�

��

We can see that: SFSSS rrIkIGkG��

4222

151



We found that the solution of this equation is:

SFSSS rrIkIGkG��

4222

Therefore satisfies the scalar wave equation: SF rr

,

SFSFSFS rrrrkrr

4,, 22

SFSF rrrwhere

r

rkjrr

exp,

152



Using the Second Vector Green Identity

S

SSV

SSSSV dSnaGEEaGdVEaGaGEI 1��

where is an arbitrary constant vectora

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

We have

and we get

aGaG SS

consta

SS

��

dVJJjGaaE

dVJJjEkGarraaGkE

EaGaGEI

VmSe

VmSeSF

VSSSSV

�

��

��

4

4 22

We used the fact that, since the sources and the observation point are both in the volume V,

Sr

Fr

aEdVrraEV

SF

S

SS

V

mSe dSaGEEaGndVJJjGaEa��

14

Therefore we obtain

Solution of the equation: mSeSS JJjEkE 2

153



Let develop now the expression

S

SS

V

mSe dSaGEEaGndVJJjGaEa��

14

Solution of the equation: mSeSS JJjEkE 2(continue – 1)

aGEEaGn SS

��

1

a

kaa

kaaaG

S

consta

SSSS

consta

SSSS

�

2

0

2

11

and

aEnaEn

aEnnaEnaGE

SS

SSS

�

11

111

154




Since is symmetric andGk

IG SS

��,

12

GaaG

��

Enak

Ena

Enk

IaEnGa

nEGanEaGnEaG

SSSS

SSSS

SSS

11

1

11

1

111

2

2

��

��

addEnak

Enak

subtractEnak

Ena

SSSSSS

SSSS

11

11

11

1

22

2

But since

0&0 aaconsta SSSSSS

SS

SSSSSSSSSS

a

aaaaa

we can develop the following expression

nEak

nEak

Enak

Enak

SSSSSS

SSSSSS

11

11

11

11

22

22

nEak

nEaEak

SSS

SSSSSS

11

11

2

2

155




We get therefore

nEaG S 1�

Enak

Enak

Enak

Ena

SSSSSS

SSSS

11

11

11

1

22

2

Enak

Ena SSSS

11

12

nEak SSS 11

2

aEnnaGE SS

�

11We found that

therefore aGEEaGn SS

��

1

Enak

Ena SSSS

11

12

aEnnEak SSSS

111

2

Enk

EnEna SSSSS 11

112

nEak SSS 11

2

156




Since and we get

aGEEaGn SS

��

1

Enk

EnEna SSSSS 11

112

nEak SSS 11

2

mSeSS JJjEkE 2 22 k

nEak

kJJjnEnEnEna

aGEEaGn

SSS

SmSeSSS

SS

11

1111

1

2

2

��

157




Let compute

S

SS dSnaGEEaG 1��

S

SmSeSSS dS

kJJjnEnEnEna

21111

dSnEak S

SSS

11

2

In our case the integral is performed over a closed surface S and therefore the lastintegral is (using Gauss’ 5 Theorem: ):

0110

5

VSSSS

Gauss

SSSS

SSSS dvEadSnEadSnEa

Compute (using Gauss’ 4 Theorem: ):

V

Se

V

SSmSe

k

V

mSS

j

SS

V

SSmSe

Gauss

S

mSeS

dvadvk

JJja

dvJJjk

advk

JJja

dSJJjnk

a

e

2

0

22

4

2

2

1

VS

dvAdSAn

1

158




Let substitute this result in

VS

e

V

SSmSe

SmSe

S dvadvk

JJjadSJJjnk

a

221

SSS

VmSe dSnaGEEaGdVJJjGaEa 14

��

dVJJjk

IaEaV

mSeSS

24

�

S

SSS dSEnEnEna 111

V

Se

V

SSmSe dvadv

kJJja

2

we obtain

Since this is true for all constant vectors , after simplification and rearranging terms, we obtain

a

SSSS

VS

emSe dSEnEnEndVJJjE

111

4

1

4

1

159




Using

SSSS

VS

emSe dSEnEnEndVJJjE

111

4

1

4

1

we obtain

We recovered Stratton-Chu solution

Using the duality relations

we can write

VmS

Sm

Gauss

VmS

VmS

VmS dVJdSJndVJdVJdVJ 1

5

SSSmS

VS

emSe dSEnEnJEndVJJjE

1114

1

4

1

e

m

m

e

e

m

m

e

J

J

J

J

E

H

H

E

SSSeS

VS

meSm dSHnHnJHndVJJjH

1114

1

4

1

160



Discontinuous Surface Distribution

Stratton-Chu equations are valid only if the vectors are continuous and have continuous derivatives on the S surface. They cannot be applied, therefore, to the problem of diffraction at a slit.

HE

,

1S

2S 0,0 22

HE

0,0

me JJ 0,0 me

11, HE

ld

n1

C

Suppose we have a slit of surface S1 with the curve C serving as his boundary. Let assume any surface S2 closed at infinity that complements the surface S1 and has in common the curve C. Assume no sources

0,0,0,0 meme JJ

Assume also that on S2 we have 0,0 22

HE

ConkHnkEn me

11 1,1

To overcome the discontinuity problem assume that on curve C we have a distribution of charges such that

161



Discontinuous Surface Distribution (continue – 1)Let return to 1S

2S 0,0 22

HE

0,0

me JJ 0,0 me

11, HE

ld

n1

C

1

21

1

1400

S

SS

SS

SS

V

mSe

dSaGEEaGn

dSaGEEaGndVJJjGaEa

��

��

We found

nEak

kJJjnEnEnEna

naGEEaG

SSS

SmSeSSS

SS

11

1111

1

2

2

00

��

Using Stokes’ Theorem: we have CS

rdASdA

C

SS

C

SS

Stokes

S

SSS rdEardEadSnEa

1

1

Therefore

CSS

SSSS

SSS

rdEak

dSEnEnEna

dSaGEEaGnEa

��

2

1111

14

1

1

162



Discontinuous Surface Distribution (continue – 2)

Using the duality relations

1S

2S 0,0 22

HE

0,0

me JJ 0,0 me

11, HE

ld

n1

C

CSS

SSSS rdEa

kdSEnEnEnaEa

2

11114

1

Since this is true for all constant vectors , we obtaina

C

SS

S

SSS rdEk

dSEnEnEnE

24

1111

4

1

1

Using and we get 22 kHjES

C

S

S

SS rdHj

dSEnEnHnjE

4111

4

1

1

we can write

e

m

m

e

e

m

m

e

J

J

J

J

E

H

H

E

C

S

S

SS rdEj

dSHnHnEnjH

4111

4

1

1

163


Monochromatic Planar Wave Equations

Let assume that can be written as: trHtrE ,,,


where are phasor (complex) vectors.

rHjrHrHrEjrErE

ImRe,ImRe

We have tjrEjtjt

rEtrEt 00 expexp,

Hence

m

e

m

e

jt

m

e

m

e

B

D

JBjE

JDjH

BGM

DGE

Jt

BEF

Jt

DHA

)(

164


Fourier Transform

The Fourier transform of can be written as: trHtrE ,,,

dttjtrHrHdtjrHtrH

dttjtrErEdtjrEtrE

exp,,&exp,2

1,

exp,,&exp,2

1,

This is possible if:

drHdttrH

drEdttrE

22

22

,2

1,

,2

1,

JEAN FOURIER

1768-1830

165


NoteThe assumption that can be written as: trHtrE ,,,


is equivalent to saying that has a Fourier transform; i.e.: trHtrE ,,,

dtjrHtrHdttjtrHrH

dtjrEtrEdttjtrErE

exp,2

1,&exp,,

exp,2

1,&exp,,

This is possible if:

drHdttrH

drEdttrE

22

22

,2

1,

,2

1,

00

0

exp

expexpexp,,

rEdttjrE

dttjtjrEdttjtrErE

End Note

166


m

e

m

e

ED

HBm

e

JHjE

JEjH

JHjE

JEjH

JBjE

JDjH

me JJjEkE

2

em JJjHkH

2

22 f

c

c

fk

Using the vector identity AAA

For a Homogeneous, Linear and Isotropic Media:

m

e

ED

HBm

e

H

E

B

D

e

me JJjEkE

22

m

em JJjHkH

22

and

we obtain

Monochromatic Planar Wave Equations (continue)

167


Assume no sources:

we have


0,0,0,0 meme JJ

022 EkE

022 HkH

nkk

n

k

0

00

00

0

rktjtj

rktjtj

eHerHtrH

eEerEtrE

0

0

,,

,,

022

rkj

rkjrkjrkjrkj

ek

ekkeekje

Helmholtz Wave Equations

satisfy the Helmholtz wave equations ,,, rHrE

rkj

rkj

eHrH

eErE

0

0

,

,

Assume a progressive wave of phase rkt ) a regressive wave has the phase ( rkt

For a Homogeneous, Linear and Isotropic Media

k

0E

0H

r t

k

Planes for whichconstrkt

168


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


we haveUsing: 1ˆˆ&ˆˆ sssc

nsk

0

0

H

E

HjE

EjH

0ˆ

0ˆ

ˆ

ˆ

0

0

00

00

Hs

Es

HEs

EHs

sPlanar Wave

0E

0Hr

0

0

0

0

00

00

rkj

rkj

rkjrkj

rkjrkj

ekje

eHkj

eEkj

eHjeEkj

eEjeHkjrkjrkj

0

0

0

0

00

00

Hk

Ek

HEk

EHk


169


To satisfy the Maxwell equations for a source free media we must have: Monochromatic Planar Wave Equations

we haveUsing: 1ˆˆ&ˆˆ0 kkknkkk

0

0

H

E

HjE

EjH

0ˆ

0ˆ

ˆ

ˆ

0

0

00

00

Hk

Ek

HEk

EHk

kˆPlanar Wave

0E

0Hr

0

0

0

0

00

00

rkj

rkj

rkjrkj

rkjrkj

ekje

eHkj

eEkj

eHjeEkj

eEjeHkjrkjrkj

22

22&

2

ˆ

2

ˆHwEwwcn

kwwcn

kS meme

Time Average Poynting Vector of the Planar Wave

Reflections and Refractions Laws Development Using the Electromagnetic Approach

170

SOLO

rM

iE aE

tM

- Incident IrradianceiE

- Absorbed IrradianceaE

- Reflected Radiant ExcitancerM

- Transmitted Radiant ExcitancetM

Law of Conservation of Energy: trai MMEE

T

i

t

R

i

r

A

i

a

E

M

E

M

E

E11 TRA

i

a

E

EA : - fraction of absorbed energy (absorptivity)

i

r

E

MR : - fraction of reflected energy (reflectivity)

i

t

E

MT : - fraction of transmitted energy (transmissivity)

Opaque body (no transmission): 01 TRA

Blackbody (no reflection or transmission): 0&01 TRA

REFLECTION & REFRACTION

Sharp boundary (no absorption): 01 ART

171

SOLO REFLECTION & REFRACTION

iE

iH

rE

rH

ik rk

tH

tE

tk

21n

z

x yi

r

t

Consider an incident monochromatic planar wave

c

nk

eEkH

eEE

iiii

rktjiii

rktjii

ii

ii

1

00

110011

0

0

The monochromatic planar reflected wave from the boundary is

11

1

1

0

0

&n

cv

vc

nk

eEkH

eEE

rrr

rktjrrr

rktjrr

rr

rr

The monochromatic planar refracted wave from the boundary is

22

2

2

0

0

&n

cv

vc

nk

eEkH

eEE

ttt

rktjttt

rktjtt

tt

tt


172


The Boundary Conditions at z=0 must be satisfied at all pointson the plane at all times, impliesthat the spatial and time variations of

This implies that

iE

iH

rE

rH

ik rk

tH

tE

tk

21n

z

x yi

r

t

Phase-Matching Conditions

yxteEeEeEz

rktjt

z

rktjr

z

rktji

ttrrii ,,,,0

00

00

0

yxtrktrktrktz

ttz

rrz

ii ,,000

ttri

yxrkrkrkz

tz

rz

i ,000

must be the same


173


tri nnn sinsinsin 211

iE

iH

rE

rH

ik rk

tH

tE

tk

21n

z

x yi

r

t


zyxc

nk

zyxc

nk

ttttttt

irirrrr

ˆcossinˆsinsinˆcos


2

1

yyxc

nrk

yxc

nrk

yc

nrk

tttz

t

irrz

r

iz

i

ˆsinsincos

sinsincos

sin

2

0

1

0

1

0

yxrkrkrkz

tz

rz

i ,000

2

tr

ttri

x

y

Coplanar

Snell’s Law

zzyyxxr

zyc

nk iiii

ˆˆˆ

ˆcosˆsin1

Given:

Let find:


174


Second way of writing phase-matching equations

ri 11

22

2

1

1

2

sin

sin

v

v

n

n

t

iRefraction Law

Reflection Law


zzyyxxr

zyc

nk iiii

ˆˆˆ

ˆcosˆsin1

zyxc

nk

zyxc

nk

ttttttt

irirrrr



2

1

ynnync

kkz

ynnync

kkz

ittrti

irrrri

ˆsinsinsinˆcosˆ

ˆsinsinsinˆcosˆ

122

111

ttri

We can see that

tri

tiri kkzkkz 0ˆˆ

tri

tri

tr

nnn sinsinsin

2/

211

iE

iH

rE

rH

ik rk

tH

tE

tk

21n

z

x yi

r

t


175


ri 11

22

2

1

1

2

sin

sin

v

v

n

n

t

iRefraction Law

Reflection Law

Phase-Matching Conditions (Summary)

ttri

tri

tiri kkzkkz 0ˆˆ

tri

tri

tr

nnn sinsinsin

2/

211

iE

iH

rE

rH

ik rk

tH

tE

tk

21n

z

x yi

r

t yxrkrkrk

zt

zr

zi ,

000

yxtrktrktrktz

ttz

rrz

ii ,,000

Vector Notation

ScalarNotation


176


0ˆ 2121

EEn

0ˆ 2121

HHn

0ˆ 2121 DDn

0ˆ 2121 BBn

Boundary conditions for asource-less boundary

0ˆ 00021

tri EEEn

0ˆ/ˆ/ˆ/ˆ 02201101121

ttrrii EkEkEkn

0ˆ 02010121 tri EEEn

0ˆˆˆˆ 02201101121 ttrrii EkEkEkn

In our case ttrrii

tri

EkHEkEkH

EEEEE

ˆ&ˆˆ

&

2

22

1

11

21

iE

iH

rE

rH

ik rk

tH

tE

tk

21n

z

x yi

r

t


Fresnel Equations

177


0ˆ 00021

tri EEEn

0111

ˆ 02

01

01

21

ttrrii EkEkEkn


0ˆ 00021 ttrrii EkEkEkn

Using ,ˆ,ˆ,ˆˆ221111

1ttrriii kkkkkk

c

nk

iE

iH

rE

rH

ik rk

tH

tE

tk

21n

z

x yi

r

t

0ˆ 00021

tri EEEn

0ˆ/ˆ/ˆ/ˆ 02201101121

ttrrii EkEkEkn



Boundary Conditions


Fresnel Equations

178


i r

ttH

tE

tk

rH

rk

rE

iH

iE

ik

21n

Boundary

xEknEEknxEknEEkn

xEknEEnkEkn

tttttirrrrri

iiiiiiii

ir

i

ˆcosˆˆˆˆ&ˆcosˆˆˆˆ

ˆcosˆˆˆˆˆˆ

0

cos

2100210

cos

210021

0

cos

210

0

021021

tttttt

rrrrrriiiiii

EEzzEkn

EEzzEknEEzzEkn

sinsinˆˆˆˆ

sinsinˆˆˆˆ&sinsinˆˆˆˆ

00021

0002100021

zn ˆˆ 21

0ˆ 00021

tri EEEn

0ˆ/ˆ/ˆ/ˆ 02201101121

ttrrii EkEkEkn



1

2

3

4

zykzykzyk tttiiriii ˆcosˆsin&ˆcosˆsin&ˆcosˆsin

Assume is normal o plan of incidence(normal polarization)

E

xEExEExEE ttrrii ˆ&ˆ&ˆ 000000

Boundary Conditions


179


0coscoscos 02

200

1

1 ttirii EEE

1

0000

tri EEE2

0ˆ 00021

tri EEEn

0ˆ/ˆ/ˆ/ˆ 02201101121

ttrrii EkEkEkn



1

2

3

4

0

sinsin

sinsin

0sinsinsin

000

2211

0220011

tri

ti

ri

ttrrii

EEE

EEE

4

Identical to 2

3 00


E


Boundary Conditions


i r

ttH

tE

tk

rH

rk

rE

iH

iE

ik

21n

Boundary

180


0cos1

cos1

000

22

200

00

11

1

21

tt

n

iri

n

EEE

1

0000 tri EEE2

From and

ti

ti

i

r

nn

nn

E

Er

coscos

coscos

2

2

1

1

2

2

1

1

0

0

ti

i

i

t

nn

n

E

Et

coscos

cos2

2

2

1

1

1

1

0

0

For most of media μ1= μ2 , and using refraction law:

1

2

sin

sin

n

n

t

i

ti

ti

i

r

E

Er

sin

sin

0

0

ti

it

i

t

E

Et

sin

cossin2

0

0

1 2


E



i r

ttH

tE

tk

rH

rk

rE

iH

iE

ik

21n

Boundary

181


iE

iH

rErH

ik rk

tH

tE

tk

21n

z

x yi

r

t

i r

ttH

tE

tk

rH

rk

rE

iH

iE

ik

21n

Boundary

ti

ti

i

r

nn

nn

E

Er

coscos

coscos

2

2

1

1

2

2

1

1

0

0

ti

i

i

t

nn

n

E

Et

coscos

cos2

2

2

1

1

1

1

0

0

For most of media μ1= μ2 ,

and using refraction law: 1

2

sin

sin

n

n

t

i

ti

ti

i

r

E

Er

sin

sin21

0

0

ti

it

i

t

E

Et

sin

cossin221

0

0


E



Fresnel Equations

182


Assume is parallel to plan of incidence(parallel polarization)

E

zyEE

zyEE

zyEE

tttt

rrrr

iiii

ˆsinˆcos

ˆsinˆcos

ˆsinˆcos

0||0

0||0

0||0

yEEknyEEkn

yEknEEnkEkn

ttirri

iiiiiii

ii

ˆˆˆ&ˆˆˆ

ˆˆˆˆˆˆˆ

00210021

0

cos

210

sin

021021

0ˆˆˆˆˆˆ 021021021 ttrrii EknEknEkn

zn ˆˆ 21

zykzykzyk tttiiriii ˆcosˆsin&ˆcosˆsin&ˆcosˆsin

i r

t

tH

tE

tk

rH

rk

rE

iH

iE

ik

21n

Boundary

xEEnxEEnxEEn tttirriii ˆcosˆ&ˆcosˆ&ˆcosˆ 002100210021

tttirriii EEnEEnEEn sinˆ&sinˆ&sinˆ 002100210021


183



E

zyEE

zyEE

zyEE

tttt

rrrr

iiii

ˆsinˆcos

ˆsinˆcos

ˆsinˆcos

0||0

0||0

0||0

0ˆ 00021

tri EEEn

0ˆ/ˆ/ˆ/ˆ 02201101121

ttrrii EkEkEkn



1

2

3

4

i r

t

tH

tE

tk

rH

rk

rE

iH

iE

ik

21n

Boundary

0sinsin 02001 ttiri EEE 3

0ˆcoscos 000 xEEE ttiri 2

011

0ˆ 000

22

200

00

11

10

2

200

1

1

21

t

n

ri

n

tri EEEoryEEE

1

4 00

Boundary Conditions


184



E

zyEE

zyEE

zyEE

tttt

rrrr

iiii

ˆsinˆcos

ˆsinˆcos

ˆsinˆcos

0||0

0||0

0||0

0sinsin

sin

sin

/

/sinsin

0ˆ

02001

2

1

22

112211

02

200

1

1

ttiri

t

iti

tri

EEE

yEEE

1

Identical to 3

We have two independent equations

0coscos 000 ttiri EEE 2

002

200

1

1 tri En

EEn

1 ti

ti

i

r

nn

nn

E

Er

coscos

coscos

1

1

2

2

1

1

2

2

||0

0||

ti

i

i

t

nn

n

E

Et

coscos

cos2

1

1

2

2

1

1

||0

0||


185


iE

iH

rE

rHik rk

tH

tE

tk

21n

z

x yi

r

t

i r

t

tH

tE

tk

rH

rk

rE

iH

iE

ik

21n

Boundary


E

zyEE

zyEE

zyEE

tttt

rrrr

iiii

ˆsinˆcos

ˆsinˆcos

ˆsinˆcos

0||0

0||0

0||0

ti

ti

i

r

nn

nn

E

Er

coscos

coscos

1

1

2

2

1

1

2

2

||0

0||

ti

i

i

t

nn

n

E

Et

coscos

cos2

1

1

2

2

1

1

||0

0||

For most of media μ1= μ2 ,

and using refraction law: 1

2

sin

sin

n

n

t

i

ti

ti

i

r

E

Er

tan

tan21

||0

0|| titi

it

i

t

E

Et

cossin

cossin221

||0

0||


Fresnel Equations

186


ti

ti

i

r

nn

nn

E

Er

coscos

coscos

1

1

2

2

1

1

2

2

||0

0||

ti

i

i

t

nn

n

E

Et

coscos

cos2

1

1

2

2

1

1

||0

0||

ti

ti

i

r

nn

nn

E

Er

coscos

coscos

2

2

1

1

2

2

1

1

0

0

ti

i

i

t

nn

n

E

Et

coscos

cos2

2

2

1

1

1

1

0

0

The equations of reflection and refraction ratio are called Fresnel Equations, that first developed them in a slightly less general form in 1823, using the elastic theory of light.

Augustin Jean Fresnel1788-1827

The use of electromagnetic approach to prove those relations, as described above, is due to H.A. Lorentz (1875)


Hendrik Antoon Lorentz1853-1928

187


Discussion of Fresnel Equations

it n

ni

2

10

ti

ti

i

r

E

Er

sin

sin21

0

0

ti

it

i

t

E

Et

sin

cossin221

0

0

ti

ti

i

r

E

Er

tan

tan21

||0

0||

titi

it

i

t

E

Et

cossin

cossin221

||0

0||

0i

21

1200|| nn

nnrr

ii

21

100||

2

nn

ntt

ii

1

2

sin

sin

n

n

t

i

Snell’s law


David Brewster1781-1868

David Brewster , “On the laws which regulate the polarization of light by reflection from transparent bodies”, Philos. Trans. Roy. Soc., London 105, 125-130, 158-159 1815).

For n2>n1 we have from Snell’s law θi > θt

therefore r┴ is negative for all values of θi.In contrast r|| start positive and decrease to zero when tan(θi+θt)=0. The incident angle when this occurs is denoted θp and is referred as polarization or Brewster angle(after David Brewster who found it in 1815).

n2 / n1 =1.5 n2 / n1 =1/1.5

188


ti

ti

i

r

nn

nn

E

Er

coscos

coscos

1

1

2

2

1

1

2

2

||0

0||

ti

i

i

t

nn

n

E

Et

coscos

cos2

1

1

2

2

1

1

||0

0||

ti

ti

i

r

nn

nn

E

Er

coscos

coscos

2

2

1

1

2

2

1

1

0

0

ti

i

i

t

nn

n

E

Et

coscos

cos2

2

2

1

1

1

1

0

0

Discussion of Fresnel Equations (continue)

0cos90 ii

19090||

iirr

09090||

iitt


189



n2 / n1 =1.5


Brewster Angle

190



n2 / n1 =1/1.5


1

2

sin

sin

n

n

t

i

Snell’s law

For n2<n1 we have from Snell’s law θi < θt therefore when θi increases,θt increases until it reaches 90°(no refraction and total reflection ).The incident angle when this occurs is denoted θic and is referred as the critical angle.

1

21sinn

nicCritical Angle

Brewster Angle

191


Energy Reflected and Refracted for Normal Polarization

201

12

ˆ

ii

iE

n

ckS

Time Average Poynting Vectors (Irradiances) of the Planar waves are

201

12

ˆ

rr

rE

n

ckS

202

22

ˆ

tt

tE

n

ckS

2

2

2

1

1

2

2

2

1

1

20

20

coscos

coscos

ˆ

ˆ

ti

ti

i

r

i

r

nn

nn

E

E

zS

zSR

2

2

2

1

1

21

12

2

1

1

2021

2012

coscos

coscos

cos2

cos

cos

ˆ

ˆ

ti

i

ti

ii

tt

i

t

nn

nnn

En

En

zS

zST

titi

cn

cn

i

ti

nn

n

nn

n

nn

coscos4coscos1

4cos

coscos2

21

21

2211

122

1

21

21

12

2

1

1

221

222

2

2

2

1

1

2

2

1

1

2021

2012

coscos

coscos4

cos

cos

ˆ

ˆ

ti

ti

ii

tt

i

t

nn

nn

En

En

zS

zST

Reflectance Transmittance

A

ri

t


192


2

2

2

1

1

2

2

2

1

1

20

20

coscos

coscos

ˆ

ˆ

ti

ti

i

r

i

r

nn

nn

E

E

zS

zSR

2

2

2

1

1

2

2

1

1

2021

2012

coscos

coscos4

cos

cos

ˆ

ˆ

ti

ti

ii

tt

i

t

nn

nn

En

En

zS

zST

Reflectance

Transmittance

We can see that

1 TR

Energy Reflected and Refracted for Normal Polarization

A

ri

t


(no absorption)

193


Energy Reflected and Refracted for Parallel Polarization

2||01

1|| 2

ˆi

i

iE

n

ckS

Time Average Poynting vector of the Planar waves are

2||01

1|| 2

ˆr

r

rE

n

ckS

2||02

2|| 2

ˆt

t

tE

n

ckS

2

1

1

2

2

2

1

1

2

2

2||0

2||0

||

||

||

coscos

coscos

ˆ

ˆ

ti

ti

i

r

i

r

nn

nn

E

E

zS

zSR

2

1

1

2

2

21

12

2

1

1

2||021

2||012

||

||

||

coscos

coscos

cos2

cos

cos

ˆ

ˆ

ti

i

ti

ii

tt

i

t

nn

nnn

En

En

zS

zST

titi

cn

cn

i

ti

nn

n

nn

n

nn

coscos4coscos1

4cos

coscos2

21

21

2211

122

1

21

21

12

2

1

1

221

222

2

1

1

2

2

2

2

1

1

2||021

2||012

||

||

||

coscos

coscos4

cos

cos

ˆ

ˆ

ti

ti

ii

tt

i

t

nn

nn

En

En

zS

zST

Reflectance Transmittance


194


2

1

1

2

2

2

1

1

2

2

2||0

2||0

||

||

||

coscos

coscos

ˆ

ˆ

ti

ti

i

r

i

r

nn

nn

E

E

zS

zSR

Reflectance

Transmittance

2

1

1

2

2

2

2

1

1

2||021

2||012

||

||

||

coscos

coscos4

cos

cos

ˆ

ˆ

ti

ti

ii

tt

i

t

nn

nn

En

En

zS

zST

We can see that

1|||| TR

Average Poynting vector of the Planar waves are

A

ri

t


(no absorption)

195


ti

ti

i

r

zS

zSR

2

2

||

||

|| tan

tan

ˆ

ˆ21

titi

ti

i

t

zS

zST

22

||

||

|| cossin

2sin2sin

ˆ

ˆ21

1|||| TR

A

ri

t


ti

ti

i

r

zS

zSR

2

2

sin

sin

ˆ

ˆ21

ti

ti

i

t

zS

zST

2sin

2sin2sin

ˆ

ˆ21

1 TR

Summary

196



197



Spherical Waveforms z

x

y

rcosr

,,rP

sinsinr cossinr

The Inhomogeneous Wave (Helmholtz) Differential Equation for the Electric Field Intensity is

t

J

t

E

vE

2

2

22 1

In spherical coordinates:

cos

sinsin

cossin

rz

ry

rx

2

2

2222

22

sin

1sin

sin

11

rrr

rrr

For a spherical symmetric wave: rErE

,,

Errrr

E

rr

E

r

Er

rrE

2

2

2

22

22 121

198



SourceSourceSource

Spherical Waveforms z

x

y

rcosr

,,rP

sinsinr cossinr

The Inhomogeneous Wave (Helmholtz) Differential Equation for the Electric Field Intensity is assuming no sources

011

2

2

22

2

t

E

vEr

rr

In spherical coordinates:

cos

sinsin

cossin

rz

ry

rx

01

2

2

22

2

Ertv

Err

or:

A general solution is:

waveregressive

waveeprogressiv

tvrFtvrFEr 21

0,0,0,0 meme JJ

r

eEerEtrE

rktjtj

0,,

Assume a progressive monochromatic wave of phase rkt

) a regressive wave has the phase ( rkt

r

eErE

rkj

0,

199




z

x

yr

zrP ,,

sinr

cosr

In cylindrical coordinates:

zz

ry

rx

sin

cos

2

2

2

2

22 11

zrrr

rr

For a cylindrical symmetric wave: rEzrE

,,

r

Er

rrE

12


011

2

2

2

t

E

vr

Er

rr

0,0,0,0 meme JJ

200



Source


z

x

yr

zrP ,,

sinr

cosr

SourceSource

In cylindrical coordinates:

zz

ry

rx

sin

cos


011

2

2

2

t

E

vr

Er

rr

0,0,0,0 meme JJ

Assume a progressive monochromatic wave of phase rkt

) a regressive wave has the phase ( rkt

tjerEtrE ,,

0

12

2

2

Evr

E

rr

E

k

The solutions are Bessel functions which for larger approach asymptotically to: rkje

r

ErE 0,

201

SOLO

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.

ELECTROMAGNETICS

202

SOLO

Energy and Momentum (continue -1)

We identify the following quantities

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

-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 ,

DEtt

DE

ED

2

10

BHtt

BH

HB

2

10

ELECTROMAGNETICS

-Power density of the current density eJ

203

SOLO

Energy and Momentum (continue – 3)

Let start from the Lorentz Force Equation (1892) on the free charge

BvEF e

Free Electric Chargee 3 msA

Velocity of the chargev 1sm



Hendrik Antoon Lorentz1853-1928

e

Force on the free chargeF

Ne

Second way

ELECTROMAGNETICS

204

SOLO


The power density of the Lorentz Force the charge

EJBvEvp e

Bvv

Jve

ee

0

or

HEt

BHE

t

D

Et

DHEEH

Et

DHEJp

t

BE

HEHEEH

Jt

DH

e

e

e

ELECTROMAGNETICS

205

SOLO


HEDEt

BHt

EJ e

2

1

2

1

dve

E

B

eJv

,

V

FdF

Fd

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

Let integrate this equation over a constant volume V

VV td

d

t

Georg Simon Ohm1789-1854

sourcee

e

EJE

1

For linear, isotropic electro-magnetic materials HBED

00 ,

ELECTROMAGNETICS

206

SOLO


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

R

nJoule dRIP 2

RadiationFieldMagneticFieldElectricJouleSource PPPPP

For linear, isotropic electro-magnetic materials HBED

00 ,

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.

ELECTROMAGNETICS

207


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

208

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

But

0

2

2exp2exp

2

12exp

1

02

2exp2exp

2

12exp

1

00

00

T

TT

T

TT

Tj

Tjtj

Tjdttj

T

Tj

Tjtj

Tjdttj

T

Therefore

*00

*00

0

*

22

1,,

2EEeEeEdt

TrErEw rkjrkj

T

e

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

209

SOLO

In the same way

*00

00 2,,

1,,

1HHdttrHtrH

TdttrBtrH

Tw

TT

m

Using the relations 00 ˆ HsEA

00 ˆ EsHF

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

s

0E

0H

ELECTROMAGNETICS

*00

*00

0

*

22

1,,

2EEeEeEdt

TrErEw rkjrkj

T

e

Energy Flux and Poynting Vector (continue – 1)


210

SOLO

Therefore *00

ˆ2

rHkrEww me

Within the accuracy of Geometrical Optics, the time-averaged electric and magnetic energy densities are equal.

*0000*

00ˆ

22rHkrErHrHrErEwww me

The 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

,,

,,,,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)


211

SOLO

Assume the linear general constitutive relations

ELECTROMAGNETICS

m

e

m

e

BGM

DGE

JBjEF

JDjHA

)(

dyadicsxwhereH

E

B

D33,,,

��

��

��

m

e

JHEjEF

JHEjHA

��

��

Energy Flux and Poynting Vector for a Bianisotropic Medium

212

SOLO ELECTROMAGNETICS

Let compute the following

meHH

meHH

JHJEHHEHHEEEj

JHEjHEJHEj

EHHEHE

******

****

***

��

��

******

****

***

meHH

mHH

e

JHJEHHHEHEEEj

HJHEjJHEjE

EHHEHE

��

��

******

******

**

meHH

meHH

JHJEHHHEHEEEj

JHJEHHEHEHEEj

HEHE

��

��

m

e

JHEjEF

JHEjHA

��

��

Energy Flux and Poynting Vector for a Bianisotropic Medium (continue – 1)

213


mmeeHH

HH

mmee

HHHH

meHH

meHH

JHJHJEJEH

EjHE

JHJHJEJE

HHEHHEEEj

JHJEHHHEHEEEj

JHJEHHEHEHEEj

HEHE

******

****

****

******

******

**

��

��

��

��

��

,,,,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

��

��


214


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

**

****

**

��

��

n

iiSS

1

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

mmee

HH

HH

JHJHJEJE

H

EjHEHEHES

****

****

4

1

4

1

44

1

��

��


215


We recognize the following

n

iiSS

1

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

HH

HHj

��

��

4G




G

G

G

V

mm

V

ee

VHH

HH

S

dVJHJHdVJEJE

dVH

EjHEdSnS

****

**

4

1

4

1

41

��

��

dVH

EjHE

V

**

****

4

��

��

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


216

Note:Since there are not magnetic sources the Magnetic Hertz’s Vector Potential is :

0

m

Electrical Dipole (Hertzian Dipole) RadiationSOLO

Given a dipole monochromatic of electric charges defined by the Polarization Vector Intensity

tq

tq

d

r

dqP

dr

tdqdeqaltP tj

e cosRe 00

we want to find the radiation properties.

We start with the Helmholtz Non-homogeneous Differential Equation of the Electric Hertz’s Vector Potential : te

trPtrtc

tr eee ,1

,1

,0

2

2

22

Heinrich Rudolf Hertz1857-1894

- speed of propagation of the EM wave [m/s]00

1

c

- Polarization Vector Intensity eP 2 msA

- Permitivity of space 2122 mNsA

- Electric Hertz’s Vector Potential (1888)e NsA 11

tA e

000 eV

0

Using the Electric Hertz’s Vector Potential we obtain :

The field vectors are given by ee

tcV

t

AE

2

2

200 1

tAH e

000

1

217

SOLO Electric Dipole Radiation

tq

tq

d

r

zSS rrdqP 10

dr

sinr

cosr

zyx

r

r

rr

111

1

cossinsincossin

r1

1

1

x1

y1

z1

Compute (continue-3) ee

tcE

2

2

2

1

We have

32

0

4

0

2

5

0

2

2

2

2 44

3

4

31

rc

rpr

rc

rprpr

r

rprrp

tcE ee

e

230 44 rc

rp

r

rp

tH e

r

ptre

04,

krtjkrtj epedqp 00

Let use spherical coordinates

zyxr rrr 1111 cossinsincossin

111 sincos00 rz krtjkrtj epepp

krtjeprccr

jr

rc

rprrp

rc

rprrp

r

rprrpE

rr

02

0

2

2

0

3

0

32

0

2

4

0

2

5

0

2

4

sin

4

sincos2

4

sincos2

44

3

4

3

11111

r1

1

1

pckpp

pckjpjp222

218


tq

tq

d

r

zSS rrdqP 10

dr

sinr

cosr

zyx

r

r

rr

111

1

cossinsincossin

r1

1

1

x1

y1

z1

Using we can write

11 0

2

0

2

2sin1

4sin

44

krtjkrtj ep

rk

j

r

kcep

rcr

jH

krtjepr

k

r

kj

r

rccrj

rE

r

rr

0

2

23

0

2

0

2

2

0

3

0

111

11111

sinsincos21

4

1

4

sin

4

sincos2

4

sincos2

We can divide the zones around the source, as function of the relation between dipole size d and wavelength λ, in three zones:

Near, Intermediate and Far Fields

22

: c

f

ck

The Magnetic Field Intensity is transverse to the propagation direction at all ranges, but the Electric Field Intensity has components parallel and perpendicular to .r1

r1

E

However and are perpendicular to each other.H

• Near (static) zone: rd

• Intermediate (induction) zone: ~rd

• Far (radiation) zone: rd

219


tq

tq

d

r

zSS rrdqP 10

dr

sinr

cosr

zyx

r

r

rr

111

1

cossinsincossin

r1

1

1

x1

y1

z1

102sin

4

tj

FieldNear epr

kcjH

tj

FieldNear epr

E r 03

0

11 sincos24

1

Near, Intermediate and Far Fields (continue – 1)

• Near (static) zone: rd

In the near zone the fields have the character of the static fields. The near fields are quasi-stationary, oscillating harmonically as , but otherwise static in character.tje

02

r

rk

220


tq

tq

d

r

zSS rrdqP 10

dr

sinr

cosr

zyx

r

r

rr

111

1

cossinsincossin

r1

1

1

x1

y1

z1

102sin

4

krtj

FieldteIntermedia epr

kcjH

krtj

FieldteIntermedia epr

kj

rE r

023

0

11 sincos21

4

1

Near, Intermediate and Far Fields

• Intermediate (induction) zone: ~rd

• Far (radiation) zone: rd

10

2

sin4

krtj

FieldFar epr

kcH

10

0

2

sin4

krtj

FieldFar epr

kE

r1

FieldFarE

FieldFarH

At Far ranges are orthogonal; i.e. we have a transversal wave.

rHE 1,,

In the Radiation Zone the Field Intensities behave like a spherical wave (amplitude falls off as r-1)

12

r

rk

12010

36

11041

:9

7

0

0

1

0

00

c

FieldFar

FieldFar

cH

EZ

221


http://dept.physics.upenn.edu/courses/gladney/phys151/lectures/lecture_apr_07_2003.shtml#tth_sEc12.1 http://www.falstad.com/mathphysics.html

Electric Field Lines of Force

222


tq

tq

d

r

zSS rrdqP 10

dr

sinr

cosr

zyx

r

r

rr

111

1

cossinsincossin

r1

1

1

x1

y1

z1

The phasors of the Magnetic and Electric Field Intensities are:

10

2

sin4

1

krtjep

cr

j

rH

krtjepcrc

jrc

jrrr

E r

02

2

2

0

11 sin11

cos12

4

1

Poynting Vector of the Electric Dipole Field

The Magnetic and Electric Field Intensities are:

1sincossin4

20

krt

ckrt

rr

pHrealH

11 sinsin

1cos

1cossincos

12

4 2

2

2

0

0 krtrc

krtcr

krtc

krtrrr

pErealE r

1

1

cossincossinsincos1

4

2

sincossinsincos1

4

2

0

32

2

0

22

2

2

2

0

22

2

0

krtc

krtr

krtc

krtrr

p

krtc

krtr

krtrc

krtcrr

pHES r

The Poynting Vector of the Electric Dipole Field is given by:

223


tq

tq

d

r

zSS rrdqP 10

dr

sinr

cosr

zyx

r

r

rr

111

1

cossinsincossin

r1

1

1

x1

y1

z1

Let compute the time average < > of the Poynting vector:


Using the fact that:

1

1

cossincossinsincos1

4

2

sincossinsincos1

42

0

32

2

0

22

2

2

2

0

22

2

0

krtc

krtr

krtc

krtrr

p

krtc

krtr

krtrc

krtcrr

pHES r

T

TdttS

TS

0

1lim

2

12cos

1lim

2

11lim

2

1cos

1limcos

0

0

1

00

22

T

T

T

T

T

Tdtrkt

Tdt

Tdtrkt

Trkt

2

12cos

1lim

2

11lim

2

1sin

1limsin

0

0

1

00

22

T

T

T

T

T

Tdtrkt

Tdt

Tdtrkt

Trkt

02sin1

lim2

1cossin

1limcossin

0

00

T

T

T

Tdtrkt

Tdtrktrkt

Trktrkt

rrc

pS 12

23

0

2

42

0 sin42

11 cossin

4sin

1

42

22

0

32

2

02

2

2

2

2

2

2

0

22

2

0

rcrcr

p

rccrcr

pS r

we obtain:

or: Radar Equation

Irradiance

224


tq

tq

d

r

zSS rrdqP 10

dr

sinr

cosr

zyx

r

r

rr

111

1

cossinsincossin

r1

1

1

x1

y1

z1


rrc

pS 12

23

0

2

42

0 sin42

Radar Equation

45 90 135 1800

0

5

10

15

20

25

30

0

45

90

135

180

225

270

315

z

y5.0 0.1

Polar Angle , in degrees

Rel

ativ

e P

ower

, in

db

The Total Average Radiant Power is:

0

22

23

0

2

42

0 sin2sin42

drrc

pdSSP

Arad

22

0

22

120123

0

42

0

3/4

0

3

23

0

42

0 4012

sin16

0

prc

pd

rc

pP

c

c

rad

3

4

3

2

3

2cos

3

1coscoscos1sin

0

30

2

0

3

dd

225


References

H. Lass, “Vector and Tensor Analysis”, McGraw-Hill, 1950

J.D. Jackson, “Classical Electrodynamics”, 3rd Ed., John Wiley & Sons, 1999

R. S. Elliot, “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

L.B. Felsen, N. Markuvitz, “Radiation and Scattering of Waves”, Pentice-Hall, 1973

226


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,”

April 13, 2023 227

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

Science

Electromagnetics