j!t - University of Colorado Boulder | University of ...ecee.colorado.edu/~ecen3410/LECTURES/em-nup.pdf · Relection and Transmission of Plane Waves from Planar Bound aries { Normal

Electromagnetic Waves and TransmissionECEN 3410, Spring 2017

MWF 12:00-12:50PM ECEN 265

Professor Kelvin Wagnerphone: x24661

email: [email protected]: ECEN 232 inside OCS

http://ecee.colorado.edu/~ecen3410/

TAs: Charlie Rackson and Bohan Zhang. Possible office hours T,Th 11AM

Textbook: B. Notaros, Electromagnetics, Pearson, 2011. Chapters 8-14Additional References:Introductory Electromagnetics, Popovic and Popovic, on ECEN3400 web pageRamo, Whinnery and Van Duzer, Fields and Waves in Communication ElectronicsBalanis, Advanced Engineering Electromagnetics

Grades based on Homework (35%), Midterm and Quizes (30%), and Final (35%).

Final Wednesday May 10, 4:30-7:00PM

Kelvin Wagner, University of Colorado EM Waves ECEN 3410 Spring 2017 1

EM Fields and Waves Topics

• Maxwell’s Equations and Waves. Monochromatic Maxwell’s Eqations

– Boundary Conditions. Poyntings Theorem and Energy flow

• Wave Equation. Plane Wave Solutions. Time-Harmonic Plane Waves

– Lossy Media, Conductors, Skin Effect, Plasmas, Dispersion

– Transverse EM Polarization

• Relection and Transmission of Plane Waves from Planar Boundaries

– Normal and Oblique incidents on Conductors and Dielectrics

– Brewster Angle, TIR, Rectangular Waveguides, TE&TM modes

• Field Analysis of Transmission Lines, Nonuniform waves

– Smith Charts. Coaxial Lines, Microstrip and Coplanar Waveguides

• Waveguides and Resonators, Modes and cutoff frequencies

– Dispersion, Couplers. Diectric Waveguides and Optical Fibers

• Antennas and Propagation, Dipoles, Receiving Antennas

– Directivity and Gain, Arrays

• Optical and THz Gaussian Beam Propagation


Maxwell’s equations were a great synthesis ofprevious Optics and E&M

Time-Varying E&MFaraday (1797-1867)Henry (1797-1878)Weber (1804-1891)Helmholtz (1821-1894)Thomson (Kelvin) (1824-1907)Rieman (1826-1866)Lorenz (1829-1891)

ElectrostaticsFranklin (1706-1790)Cavendish (1731-1810)Coulomb (1736-1806)Volta (1745-1827)Laplace (1749-1827)Gauss (1777-1855)Poisson (1781-1840)Green (1793-1841)

OpticsGalileo Galilei (1564-1642)Willebrord Snell (1591-1626)Rene Descartes (1596-1650)Francesco Grimaldi (1618-1663)Pierre de Fermat (1601-1665)Christian Huygens (1629-1695)Robert Hooke (1635-1703)Isaac Newton (1642-1727)Leonhard Euler (1707-1783)Thomas Young (1773-1829)Etienne Malus (1775-1812)Dominique Arago (1786-1853)David Brewster (1781-1868)Joseph von Fraunhofer (1787-1826)Augustin Jean Fresnel (1788-1827)Christian Doppler (1803-1853)William Hamilton (1805-1865)Jean Foucalt (1819-1868)Armand Fizeau (1819-1896)George Stokes (1819-1903)

James Clerk Maxwell (1831-1879)

Lord Rayleigh (1842-1919)Oliver Heavyside (1850-1925)John Poynting (1852-1914)Hendrik Lorentz (1853-1928)Heinrich Hertz (1857-1894)Guglielmo Marconi (1874-1937)

MagneticsGalvani (1737-1798)Biot ((1774-1862)Ampere (1775-1836)Oesrsted (1777-1851)Ohm (1787-1854)Savart (1791-1841)Kirchoff (1824-1887)


Maxwell’s eqns: All fields functions of ~r, tTime domain, t, or time harmonic ejωt

Differential Maxwell’s eqns

∇× ~H(~r, t) = ~J (~r, t) +∂ ~D(~r, t)

∂t

∇× ~E(~r, t) = −∂~B(~r, t)

∂t∇ · ~D(~r, t) = ρ(~r, t)

∇ · ~B(~r, t) = 0

~D(~r, t) =

∫ǫ(~r, T, t− t′) · ~E(~r, t) dt

~B(~r, t) =

∫µ(~r, t− t′) · ~H(~r, t) dt

~J (~r, t) = ~J c(~r, t) + ~J v(~r, t)~J c(~r, t) = σ(~r, t) · ~E(~r, t)~J v(~r, t) = ρ(~r, t)~v(~r, t)

Monochromatic Maxwell’s eqns

∇× ~H(~r, ω) = ~J(~r, ω) + jω~D(~r, ω)

∇× ~E(~r, ω) = −jω~B(~r, ω)

∇ · ~D(~r, ω) = ρ(~r, ω)

∇ · ~B(~r, ω) = 0

~D(~r, ω) = ǫ(~r, ω)~E(~r, ω)

~B(~r, ω) = µ(~r, ω)~H(~r, ω)

~Jc(~r, ω) = σ(~r, ω)~E(~r, ω)

Monochromatic Time-Harmonic form

~E(~r, t) = ℜ~E(~r, ω)ejωt

= ℜ

√2~Erms(~r, ω)e

jωt


Microscopic Maxwell Postulates

Primitive Fields ~e(~r, t) and ~b(~r, t) vary on atomic scale in matter. Fields develop dueto point charges ql at position ~rl(t) with velocity ~vl(t) and microscopic current

Q(~r, t) =∑

l

qlδ(~r−~rl(t)) ~j(~r, t) =∑

l

ql~vlδ(~r−~rl(t))

according to microscopic Maxwell

∇× ~e(~r, t) = −∂~b(~r, t)

∂t

∇× ~b(~r, t) = ǫoµ0∂~e(~r, t)

∂t+ µ0~j(~r, t)

∇ · ~e(~r, t) =1

ǫoQ(~r, t)

∇ · ~b(~r, t) = 0

Enormous numbers of discrete particles in matter require spatial averaging of microscopicfield quantities to produce averaged fields ~E(~r, t), ~B(~r, t), ~J(~r, t), and ρ(~r, t) no longerdisplaying atomic granularity. These macroscopic fields will be what we treat in thisclass.


Induced Macroscopic Fields

Bound source densities produce polarization ~P(~r, t) and magnetization ~M(~r, t) whichcombine with free charges to give effective sources

ρe(~r, t) = ρ(~r, t) +∇ · ~P(~r, t)

~Je(~r, t) = ~J(~r, t)− ∂

∂t~P(~r, t)−∇× ~M(~r, t)

~P(~r, t) → ~P(~r, t) − ∇ × ~A(~r, t) and ~M(~r, t) → ~M(~r, t) + ∂∂t~A(~r, t) not unique.

Concepts of Polarization and Magnetization gives rise to 2 more macroscopic fields

~D(~r, t) = ǫo~E(~r, t) + ~P(~r, t)~H(~r, t) = 1

µ0~B(~r, t)− ~M(~r, t)

Lorentz force on a point charge q(~r, t) traveling at velocity ~v(~r, t)

~F(~r, t) = q(~r, t)[~E(~r, t) + ~v(~r, t)× ~B(~r, t)]

means ~E(~r, t) and ~B(~r, t) are directly measurable primitive fields.Finally, conservation of the charge yields

∇ · ~Je(~r, t) +∂

∂tρe(~r, t) = 0


Macroscopic Maxwell’s eqns

∇× ~H = ~J +∂ ~D

∂t

∇× ~E = −∂~B

∂t

∇ · ~D = ρ

∇ · ~B = 0

~D = ǫ ~E = ǫo~E + ~P = ǫo~E + ǫo

[χ ~E + χ

(2)~E~E + χ

(3)~E~E~E

]

~B = µ0~H

~J = ~Jc + ~Jv

~Jc = σ~E


Maxwell’s Equations:Faraday’s law of induction

~E ≡ Electric Field Vector~B ≡ Magnetic Induction = Magnetic Flux vector

Time varying magnetic flux passing through a closed con-ducting loop generates current around the loop

Magnetic flux φB =sA~B · d~S

Generates emf (electromotive force)

emf =

∮

C

~E · dl = −∂φB∂t

⇒∮

C

~E · dl = − ∂

∂t

∫ ∫

A

~B · d~S

A

dS

B

C

time varyingB

E

MKS

∇× ~E(~r, t) = − ∂

∂t~B(~r, t)

[1

m

] [V

m

]=

[1

s

] [Vs

m2

]

Vs=Weber, VS/m2=Tesla

CGS

∇× ~E = −1

c

∂

∂t~B

[1

cm

] [dyn

esu

]=[ s

cm

] [1s

] [dyn

esu

]

esu = electrostatic unit


Maxwell’s Eqn 2:Ampere’s Law

~H ≡ Magnetic Vector~D ≡Electric Displacement~J ≡ Electric Current Density

Line integral of ~B tangent to closed curve C equals total current ~J passing throughsurface bound by C plus an effective current due to time varying ~D

∮

C

~H · d~l =∫ ∫

A

(~J +

∂ ~D

∂t

)· d~S

MKS

∇× ~H(~r, t) =∂

∂t~D(~r, t) + ~J(~r, t)

[1

m

] [A

m

]=

[1

s

] [As

m2

] [A

m2

]

As=Coulomb

CGS

∇× ~H =1

c

∂

∂t~D +

4π

c~J

[1

cm

][Oe] =

[ s

cm

] [1s

] [Sv

cm

] [ s

cm

] [ esu

scm2

]

[1

cm

] [√g

cm

1

s

]=[ s

cm

] [1s

] [√g

cm

1

s

] [ s

cm

] [√ g

cm

1

s2

]

esu = electrostatic unit


Coulomb’s LawGauss’ Law - Electric

Flux of ~E through a closed surface equals total enclosed charge.Flux

φE =

A

~E · d~S

Enclosed Charge

q =1

ǫ

y

V

ρdV

ǫ

A

~E · d~S =y

V

ρdV

Coulomb’s Law

MKS

div~D = ∇ · ~D(~r, t) = ρ(~r, t)[1

m

] [As

m2

]=

[C

m3

]

CGS∇ · ~D = 4πρ


Gauss’ Law - MagneticNo magnetic monoploes

Any closed surface has an equal number of lines of ~B entering and leaving

φB =

A

~B · d~S = 0

div~B = ∇ · ~B(~r, t) = 0 ∇ · ~B = 0

Vector Identity ∇ · ∇× = 0

∇ · Faraday′s Law ∇ · (∇× ~E) = ∇ · ~B

∂∂tGauss′s Law formagnetic field ∂

∂t(∇ · ~B) = ∇ · ~B = 0

Conservation of fictitious magnetic monopoles density ρm(~r, t) = 0

∇ · ~Jm +∂ρm∂t

= 0


Constitutive Relations

displacement = permitivity · field Polarization susceptibility

~D = ǫ~E = ǫo~E + ~P = ǫo(1 + χL)~E + ǫoχ

(2)~E~E + ǫoχ

(3)~E~E~E = ǫoǫr~E

[Cm2

]=[Fm

] [Vm

]=[C/V·Vm2

]

ǫo = 8.854× 10−12 Fm≈ 1

36π× 10−9 F

mFree space permitivity

Dispersion ǫr(ω) ǫ dielectric permitivity tensor

~B = µ~H = µ0~H + ~M

[Vs

m2

]=

[H

m

] [A

m

]=

[Vs/A ·A

m2

]µ ≡ Permeability tensor

~M ≡ Magnetization

~J = ~Jc + ~Jv Conduction current density + convection current

~Jc = σ~E

[A

m2

]=

[S

m

] [V

m

]=

[A/V · V

m2

]

σ Conductivity tensor [Siemens =1/Ω=A/V]

σ = ne2τm

= neµ0

Metals: σ ↓ T ↑semiconductors: σ ↑ T ↑


Maxwell’s eqns: All vector fields implicitlyfunctions of ~r, t

∇× ~H(~r, t) = ~J(~r, t) +∂ ~D(~r, t)

∂t

∇× ~E(~r, t) = −∂~B(~r, t)

∂t∇ · ~D(~r, t) = ρ(~r, t)

∇ · ~B(~r, t) = 0

~D(~r, t) =

∫ǫEH(~r, T, t− t′) · ~E(~r, t)dt +

∫ξEH(t− t′) · ~H(~r, t)dt

~B(~r, t) =

∫ζEH(t− t′) · ~E(~r, t)dt +

∫µEH(t− t′) · ~H(~r, t)dt

~J(~r, t) = ~Jc(~r, t) + ~Jv(~r, t)~Jc(~r, t) = Jc(~E, ~H) = σ(~r) · ~E(~r, t)

Material Tensors describe medium. Transform as the product of coordinates.

ǫ dielectric permitivity, µ magnetic permeability, ξ, ζ magnetoelectric

ζ = 0, ξ = 0, µ = µ0I σ = 0 ǫ Lossless, Dielectrically anisotropic.

ζ = −ξ Chiral medium (imaginary). ζ = ξ Tellegen medium


Constitutive Relations in Frequency Domain

Temporal Fourier transformation avoids complexities of convolution integral in consti-tutive relation. We can express the full constitutive relation in Tellegen form

~D(~r, ω) = ǫEH(~r, ω) · ~E(~r, t) + ξEH(~r, ω) · ~H(~r, ω)

~B(~r, ω) = ζEH(~r, ω) · ~E(~r, t) + µEH(~r, ω) · ~H(~r, ω)

Or primitive fields can be sees as giving rise to induction fields (Post form)

~D(~r, ω) = ǫEB(~r, ω) · ~E(~r, t) + ξEB(~r, ω) · ~B(~r, ω)

~H(~r, ω) = ζEB(~r, ω) · ~E(~r, t) + µEB(~r, ω) · ~B(~r, ω)

Interrelated as

ǫEB(~r, ω) = ǫEH(~r, ω)− ξEH(~r, ω) · µ−1EH(~r, ω) · ζEH(~r, ω)

ξEB(~r, ω) = ξEH(~r, ω) · µ−1EH(~r, ω)

ζEB(~r, ω) = −µ−1EH(~r, ω) · ζEH(~r, ω)

µEB(~r, ω) = µ−1EH(~r, ω)

and vice versa with EB ↔ EH .For non magnetoelectric media, just have simple susceptibility and permeability.


MKSA

3 ~E Vm = J/C

m = Ws/Asm = m2Kg/s3A

m = NC Electric Field (Amplitude)

3 ~B Vsm2 =

JC

sm2 =

m2Kg/s3As

m2 Magnetic Induction (Flux)

3 ~D Cm2 =

Asm2 Electric Displacement (Flux)

3 ~H Am

Magnetic Field (Amplitude)~M Vs

m2 Magnetization~P C

m2 Polarization vector field

3 ~J Am2 Current Density

1 ρ Cm3 Charge Density

ǫ(ω) Fm = C/V

m = As/J/Asm = A2s2

m3Kg/s2Dielectric permitivity

ǫo = 8.854× 10−12 Fm Free space permitivity

µ VsAm Magnetic Permeabilityµ0 = 4π × 10−7 Vs

Am Free space permeability

σ 1/Ωm conductivity


Current Continuity

Vector Identity ∇ · ∇× = 0

∇ · Ampere′sLaw ∇ · (∇× ~H) = ∇ · ~D +∇ · ~J

∂∂tCoulomb′sLaw

∂

∂t(∇ · ~D) = ∇ · ~D = ρ

Conservation of Charge ∇ · ~J +∂ρ

∂t= 0

y∇ · ~JdV +

∂

∂t

yρdV =

x~J · ndS +

∂

∂tρ = 0

outward flow change ofof current enclosed charge

Force Eqn ~v ≡ velocity of charged particles

~F = q(~E + ~v× ~B)

For a distribution of charge ρ and currents ~J

~F = ρ~E + ~J× ~B


Solving Maxwell’s eqns

Since the divergence of a curl of any vector field is 0

∇ · Faraday′sLaw ∇ · (∇× ~E) = 0 = ∇ ·(− ~B)= − ∂

∂t∇ · ~BSo ∇ · ~B =constant independent of time (which thus must be zero)

⇒ ∇ · ~B = 0 is not an independent eqn

Similarly conservation of charge plus continuity eqn ∇ · ~J + ρ = 0 plus Ampere’s law

(∇× ~H = ~D + ~J) implies Coulomb’s law (∇ · ~D = ρ)

So 4 Maxwell’s eqn in 16 unknowns is only actually 6 independent eqns

~E 3~H 3~B 3~D 3~J 3ρ 1

16 unknowns

Maxwell’s eqns:Faraday’s law +Ampere’s laws

∇× ~E 3

∇× ~H 36 eqns

3 constitutive eqns add 9 more eqns, plusconservation of charge

~D 3~B 3~J 3

cons q 110 eqns

16 unknowns and 16=6+9+1 equations is sufficient to solve


Differential Vector operator definitionsCartesian coordinates (Jackson)

Gradient of a scalar field is a vector field oriented in the direction of maximal increase(which is normal to the equivalue surface) with magnitude given by the derivative inthis direction.

gradΨ = ∇Ψ =3∑

i=1

uidqidli

∂A

∂qi= x

∂

∂xΨ + y

∂

∂yΨ + z

∂

∂zΨ

Divergence of a vector field is the ratio of the net outward flux through an infiniesimalsurface surrounding the point to the volume enclosed. Gives the volume density ofsources of the vector field.

div~A = lim∆V→0

∮~A · ~dS∆V

= ∇ · ~A =[

∂∂x

∂∂y

∂∂z

]A1

A2

A3

=

∂

∂xA1 +

∂

∂yA2 +

∂

∂zA3

Curl of a vector field is a vector pointing in the direction of the normal to an infinitesimalsurface which is oriented so that the limit of the ratio of the line integral of the vectorfield around the perimeter to the area enclosed is maximal with a value equal to thatlimit.

∇× ~A = lim∆a→0

∮~A · ~dl∆a

∣∣∣∣∣max

=

∣∣∣∣∣∣

x y z∂∂x

∂∂y

∂∂z

A1 A2 A3

∣∣∣∣∣∣=

x(∂A3∂y − ∂A2

∂z

)+ y

(∂A1∂z − ∂A3

∂x

)

+z(∂A2∂x − ∂A1

∂y

)


Cascaded Differential Operators

Del2 ∇2Ψ = ∇ · ∇Ψ =[

∂∂x

∂∂y

∂∂z

]·

∂Ψ∂x∂Ψ∂y∂Ψ∂z

=

∂2Ψ

∂x2+∂2Ψ

∂y2+∂2Ψ

∂z2

∇2~E = x∇2Ex + y∇2Ey + z∇2Ez

curl curl Use the mnemonic for cross cross

~A× ~B× ~C = ~B(~A · ~C)− ~C(~A · ~B)

to produce the correct form of curl curl (~C(~A · ~B) = (~A · ~B)~C ⇒ ∇ · ∇~C)

∇× (∇× ~C) = ∇(∇ · ~C)−∇ · ∇~C = ∇(∇ · ~C)−∇2~C = (∇∇ · −∇2)~C

But caution since the rotation/permutation rule for dot cross

~A · (~B× ~C) = ~C · (~A× ~B) = ~B · (~C× ~A)

Does not give the correct generalization for div cross

∇ · (~B× ~C) = ~C · (∇× ~B)− ~B · (∇× ~C)

Differential forms that are always Zero ∇·∇× = 0 ∇× (∇ ) = 0


3 2nd order differential operators of a scalarfield

operating on a scalar and vector field with div, grad, and curl:

produces when produces whenoperating on scalar operating on vector

gradient ∇( ) vector dyad rank raisingdivergence ∇ · ( ) X scalar rank loweringcurl ∇× ( ) X vector keep rank

For a scalar field, can only take gradient. Thus there are only 3 second order differentialoperators of a scalar field, by taking div, grad, and curl of grad of scalar field

∇ · ∇φ = ∇2φ = scalar Laplacian

∇×∇φ = ~0

∇∇φ= dyad


5 2nd order differential operators of vectorfields

For a vector field ~A there are a total of 9 composed second order differential operators,2 are undefined.2 are zero,1 dyad,and 1 triad,leaving 3 vector forms

∇· ∇× ∇∇· ∇ · (∇× ~A) = 0 ∇ · (∇~A) = ∇2~A

vector Laplacian

∇× ∇× (∇× ~A) ∇× (∇~A) =~~0vector dyad

∇ ∇(∇ · ~A) ∇(∇× ~A) ∇(∇~A)vector dyad triadic

B. Maxum, Field Mathematics for EM, Photonics and MS, SPIE TT64, 2005


Circuit-Field RelationsC.A. Balanis, Advanced Engineering Electromagnetics

Magnetic Flux equal to the product of the Inductance and current

~B = µ~H ↔ ψm = LiL

Magnetic displacement current density (analogous to ~Jd =∂ ~D∂t ) proportional to

~B

~Md =∂ ~B

∂t=∂

∂tµ~H = µ

∂ ~H

∂t

Faraday’s law leads to inductor circuit relation∫C~E · d~l = − ∂

∂t

sA~B · d~S

∂

∂t

x

A

~B · d~S =∂ψm∂t

=∂

∂t(LiL) = vL

Giving field circuit relation

~Md = µ∂ ~H

∂t↔ vL = L

∂iL∂t


Ohm’s law as a consequence of conductive constitutive relation

~J = σ~E ↔ iR = 1RvR = GvR

Similarly, charge on a capacitor leads to circuit relation for a capacitor

~D = ǫ~E ↔ Q = Cvc

Current-Voltage relation for a capacitor

~Jd = ǫ∂~E

∂t↔ ic = C

∂vc∂t


Kirchoff’s Voltage Law

Faraday’s law∫C~E · d~l = − ∂

∂t

sA~B · d~S

RHS: Electric field integrated around a loop gives voltage sum∮~E · d~l =

∑

i

vi

LHS: Time rate of change of magnetic flux ψm = LsiL

− ∂

∂t

x

S

~B · d~s = −∂ψm∂t

= − ∂

∂t(LsiL) = −Ls

∂iL∂t

where Ls will be the stray inductance of the wiring around the circuit loop. This givesKirchoff law including stray inductance

∑

i

vi = −Ls∂iL∂t

For lumped circuits the stray inductance ≈ 0 giving∑

i vi = 0Corresponding Field-Circuit relation

∫

C

~E · d~l = − ∂

∂t

x

A

~B · d~S = −∂ψm∂t

↔∑

i

vi = −∂ψm∂t

= −Ls∂iL∂t


Kirchoff’s Current Law

enclose a circuit node with a closed surface S∑

n

in =

S

~Jc · d~s

In terms of stray capacitance Cs from node∑

n

in = −∂Q∂t

= − ∂

∂t(Csv) = −Cs

∂v

∂t

Sum of currents crossing a surface that encloses a circuit node is equal to time rateof change of total enclosed charge, or equal to the current thru stray capacitance Cs.Field-Circuit relation

S

~J · d~s = − ∂

∂t

yρdV = −∂Q

∂t↔

∑

n

in = −∂Q∂t

= −Cs∂v

∂t


Wave Propagation in Isotropicand Anisotropic Media

∇× (Faraday’s law) gives Wave Eqn for ~E

∇× (Ampere’s law) gives Wave Eqn for ~H

∇× (∇× ~E) = ∇×(−∂

~B

∂t

)= − ∂

∂t∇× µ~H = − ∂

∂t∇× (µ0

~H + ~M)

= −µ0

∂2~D

∂t2− µ0

∂~J

∂t−∇× ∂ ~M

∂tuse ∇× ~H = ∂ ~D

∂t +~J

vector identity~A× ~B× ~C = ~B(~A · ~C)− ~C(~A · ~B)

∇× (∇× ~A) = ∇(∇ · ~A)−∇ · ∇~A = ∇(∇ · ~A)−∇2~A

curl curl ~E = grad (div ·~E) - del2~E

∇(∇ · ~E)−∇2~E + µ0

∂2

∂t2(ǫo~E + ~P) = −µ0

∂~J

∂t−∇× ∂ ~M

∂t

∇2~E−∇(∇ · ~E)− 1

c2∂2~E

∂t2= µ0

∂2~P

∂t2+ µ0

∂~J

∂t+∇× ∂ ~M

∂t︸︷︷︸effective current ~J′ = ∂ ~P

∂t +~J +∇× ~M µ0

∂~J′∂t


Optics - Source free, nonmagnetic

µ0ǫ0 =1c2

c = 2.9979246× 108m/s~J = 0ρ = 0~M = 0

∇ · ~D = ρ = 0 = ∇ · ǫ~E = ǫ∇ · ~E + ~E · ∇ǫ(~r) = 0

scalar case

∇ · ~E = −~E · ∇ǫǫ∇ log ǫ∇ǫ = 1

ǫ ⇒ ∇ǫǫ = ∇ log ǫ

tensor case

η ǫ = I ∇ · ~E = −η~E · ∇ǫ 6= 0


Isotropic case

ǫ = ǫoI∇(∇ · ~E) = −∇(~E · ∇ log ǫ)

∇2~E− 1

c2∂2~E

∂t2︸︷︷︸= µ0

∂2~P

∂t2−∇(~E · ∇ log ǫ)

free space material dielectric inhomogeneitywave eqn source term source term

since ~P = ǫoχ~E for linear isotropic media

∇2~E− 1 + χ

c2∂2~E

∂t2︸︷︷︸= −∇(~E · ∇ log ǫ)

isotropic material wave eqn

Index of refraction, relative dielectric constant, susceptibility

1 + χ = εr = n2


homogeneous media

ǫ(~r) = ǫ ⇒ ∇ǫ = 0 ∇ log ǫ = 0

Isotropic, homogensous, source-free, non-magnetic wave eqn in media with index n

∇2~E− n2

c2∂2~E

∂t2= 0

Each component of this vector wave eqn obeys the Scalar wave equation

∇2Ei −n2

c2∂2Ei

∂t2= 0


EM Energy

~H· (Faraday’s Law)

−~E· (Ampere’s Law)

~H · ∇× ~E = −~H · ∂~B

∂t

−~E · ∇× ~H = −~E · ∂~D

∂t− ~E · ~J

add

~H · ∇× ~E− ~E · ∇× ~H = ∇ · (~E× ~H) = −~H · ∂~B

∂t− ~E · ∂

~D

∂t− ~E · ~J

Gauss’ Divergence Theoremy

V

dV∇ · (~E× ~H) =x

S

dS (~E× ~H)︸︷︷︸~S

·n=−y

V

~H·∂~B

∂t+ ~E·∂

~D

∂t︸︷︷︸U

+ ~E · ~J︸︷︷︸Vdz IdA

dV

= −y

V

∂U

∂tdV − Pd

where U is the stored EM energy density [J/m3] and Pd is dissipated power [W].

U =1

2~E · ǫ · ~E +

1

2~H · µ · ~H = Ue + Um

note ~E · ~D = ~E · ∂∂t(ǫ

~E) = 12∂∂t(ǫ

~E2) = 12(~Eǫ~E + ~Eǫ~E) = 1

2∂∂t(~E · ~D)


Poynting vector

and the dissipated power Pd [W] (in absence of sources)

Pd =y

V

~E · ~JdV =y

V

~E · σ · ~EdV

where σ is the conductivity tensorPoynting vector defines the outward power flow

~S = ~E× ~H

[V

m

] [A

m

]=

[W

m2

]

Thus

∇ · ~S = −∂U∂t

− ~E · ~JComplex form of Poynting’s vector

~S =1

2R[~E× ~H∗

]=

1

2R[~S]

U =1

4R[~E · ~D∗ + ~B · ~H∗

]


Intensity

time average of the Poynting vector

I =⟨|~E× ~H|

⟩=

1

T

∫

T

|~E× ~H|dt

η is impedance of media [ohms], η0 =√

µ0ǫo= 377Ω

I =1

2η|E|2 =

n

2η0|E|2

[W

m2

]=

A

V

(V

m

)2

=1

2√µ/ǫ

~E · ~E∗ =1

2

√ǫ

µ~E · ~E∗ =

1

2

ǫ√ǫµ~E · ~E∗ =

1

2

εrǫo√εrǫoµ

~E · ~E∗

=1

2

√εrǫoc|~E|2 =

nǫoc

2|~E|2

n =√εr

c = 1ǫoµ0[

Fm

] [ms

] [V2

m2

]=[AZs/V

ZsV2

m2

]=[AVm2

]=[Wm2

]


Monochromatic Waves

All fields vary harmonically in time at frequency ν = ω2π

A(~r, t) = A + A∗ = A(~r)e−iωt + cc

Time derivative of a single sideband ∂∂tA = −iωA(~r)e−iωt

Monochromatic Maxwell’s eqns∇× ~H = −iω ~D∇× ~E = +iω~B

∇ · ~D = ρ

∇ · ~B = 0

Power flow governed by time average of Poynting vector⟨~S⟩

=⟨~E × ~H

⟩=⟨R[~Ee−iωt

]×R

[~He−iωt

]⟩

=⟨12

(~Ee−iωt + ~E∗eiωt

)× 1

2

(~He−iωt + ~H∗eiωt

)⟩

=⟨14

(~E× ~H∗ + ~E∗ × ~H + ~E× ~He−i2ωt + ~E∗ × ~H∗ei2ωt

)⟩

= 14

(~E× ~H∗ + ~E∗ × ~H

)= 1

2(~S + ~S∗) = 1

2R[~S] = 12R[~E× ~H∗]


Scalar Wave Eqn

A(~r, t) = A(~r)e−iωt + cc = a(r)eiφ(~r)e−iωt + cc

Substitute into scalar wave eqn

∇2A− 1

v2∂2A∂t2

= 0

∇2A +ω2n2

c2∂2A

∂t2= 0

Phase velocity v = cn |k0| = ω

c =2πνλν = 2π

λ |k| = 2πnλ = ωn

c

∇2A + k2A = 0 Helmholtz Eqn

Plane waveφ(~r) = ω(~r · ~k)− δ

a(~r) = a(~r · ~k) k k

Monochromaticplane wave

chirped plane wave variable amplitudeplane wave

Not a plane wave(actually inhomogeneous

plane wave)

Inhomogeneous plane waves: ~k = ~kR + i~kI ⇒ e−~kI ·~r+i~kR·~r


Monochromatic Vectorial Plane Wave solnsin Isotropic Media

∇2~E− µǫ∂2~E

∂t2= 0

∇2 ~H− µǫ∂2 ~H

∂t2= 0

Trial plane wave solns

~E = u1E0e−i(ωt−~k·~r) + cc

~H = u2H0e−i(ωt−~k·~r) + cc

k

Su2

u1

E(z,t )H(z,t )0

0

Homogeneous, source-free media: ρ = 0, ~J = 0Coulomb’s law Transverse Fields

∇ · ~D = ∇ · ~E = 0 = i~k · u1E0e−i(ωt−~k·~r) ⇒ u1 · ~k = 0

∇ · ~H = 0 = i~k · u2H0e−i(ωt−~k·~r) ⇒ u2 · ~k = 0


~B

∇× ~E = +i~k× u1E0e−i(ωt−~k·~r)

︸︷︷︸ = +iω︷︸︸︷µ u2H0e

−i(ωt−~k·~r)︸︷︷︸

~E ~H

~k× ~E = ωµ~H u2 =~k× u1

|~k× u1|=~k× u1

|~k|⇒ u1 ⊥ u2

E0 =ωµ

|~k|H0 =

2πc/λ02π/λ

µH0 =c

nµH0 =

µ√µǫH0 =

√µ

ǫH0 = ηH0 =

η0nH0 =

c

nB

H0 =E0

ηB0 =

n

cE0 η0 =

√µ0

ǫo= 377Ω Impedance η =

η0n

~E(~r, t) = u1E0e−i(ωt−~k·~r) + cc

~H(~r, t) = k× uE0

ηe−i(ωt−

~k·~r) + cc


Plane Waves

direction cosines of a plane wave

E(x, y, z) = E0pei2πλ (αx+βy+γz) = E0pe

i(kxx+kyy+kzz)

α2 + β2 + γ2 = 1 k2x + k2y + k2z = k20 =

(2πn

λ

)2

where k0 = |~k| = 2πn/λ in medium of index n.

k-space

k

2πn/λ

kz

kx

x

zθ

k

λ

kz

kx

In 2-dimensionsE(x, z; t) = A0pe

ik0(x sin θ+z cos θ)e−i2πνt + cc

where α = sin θ/λ and γ = cos θ/λ.ν ≈ 5× 1014 Hz λ ≈ .63× 10−6m (HeNe)


Poynting Vector

Time Average Poynting vector for a monochromatic wave

⟨~S⟩=ω

2π

∫ 2π/ω

0

~Sdt = ω

2π

∫ 2π/ω

0

12R[~E× ~H∗

]dt

~E = u1E0e−i(ωt−~k·~r) + cc

~H =~k× u1

|~k|E0

ηe−i(ωt−

~k·~r) + cc

u1 × k× u1 = k |~E× ~H∗| = E20/η

= kω

2π

∫ 2π/ω

0

E20

2ηdt =

E20

2ηk = ǫE2

0

1

ǫ

√ǫ

µk = (Ue + Um)

k√ǫµ

= Uc

nk

U = EM Energy densitycn = velocity in media with index n

Leads to Hydrodynamical continuity eqn for EM Energy

U +∇ · S = 0


Monochromatic Plane-Waves: Maxwell’s Eqnsin Spatio-Temporal Fourier Space

Consider plane wave with wavevector ~k varing harmonically in time at frequency ν = ω2π

~A(~r, t) = ~A(~k, ω)e−i(ωt+~k·~r) + ~A∗

(~k, ω)ei(ωt+~k·~r)

Example of Time derivative of a monochromatic plane wave

∂

∂t~A(~r, t) = −iω ~A(~k, ω)e−i(ωt−

~k·~r) + iω ~A∗(~k, ω)ei(ωt−

~k·~r)

Differential operators operating on a single side band as vector algebraic operations∂∂t

→ −iω ∇· → i~k·∇× → i~k× ∇ → i~k

Maxwell’s eqns in Spatio-temporal Fourier Space

i~k× ~H(~k, ω) = −iω ~D(~k, ω)

i~k× ~E(~k, ω) = +iω~B(~k, ω) + iω:0~J (~k, ω)

i~k · ~D(~k, ω) =70ρ

i~k · ~B(~k, ω) = 0~D(~k, ω) = ǫ(~k, ω) ∗ ∗ ∗ ~E(~k, ω) = ǫ(ω)~E(~k, ω) ~B(~k, ω) = µ0

~H(~k, ω)


Scalar Waves

Each rectangular component of the field satisfies (homogeneous, source free isotropic)

∇2V − n2

c2∂2V

∂t2= 0

Plane Wavesz′

V (~r, t) = V (︷︸︸︷~r · k, t) represents a plane wave in the direction k = z′

∇2V − n2

c2V = 0 ⇒ ∂2V

∂z′2− n2

c2∂2V

∂t2= 0

z′ − cnt = p z′ + c

nt = q

∂2V

∂p∂q= 0

∂2V

∂(z′ − vt)∂(z′ + vt)=∂2V

∂z′2− 1

v2∂2V

∂t2= 0

soln

V = V1(p) + V2(q) = V1

(~r · k− c

nt)+ V2

(~r · k + c

nt)

=⇒ ⇐=vp =

cn

vp = − cn

V (z’-vt)1 V (z’+vt)2


Scalar Spherical Waves

V = V (r, t)r = |~r| =

√x2 + y2 + z2

Spherical coordinates

∂∂x = ∂r

∂x∂∂r = 2x1

2u−12 ∂∂r =

xr∂∂r

∂2

∂x2= ∂

∂x

(xr∂∂r

)= 1

r∂∂r − x

r2xr∂∂r

∂∂r

similarly for ∂∂y ,

∂2

∂y2, ∂∂z ,

∂2

∂z2giving

∇2V =1

r

∂2

∂r2(rV ) =

1

r

∂

∂r

(V + r

∂V

∂r

)=

1

r

[2∂V

∂r+ r

∂2V

∂r2

]

So wave eqn becomes eqn for U = (rV )

∂2(rV )

∂r2− 1

v2∂2(rV )

∂t2= 0

U = U1(r − vt) + U2(r + vt)

V =V1(r − vt)

r+V2(r + vt)

rKelvin Wagner, University of Colorado EM Waves ECEN 3410 Spring 2017 41

Potentials

Since ∇ · ~B = 0 we can use the identity ∇ ·∇× (any vector field) = 0 to introduce the

vector potential ~A (to which an arbitrary ∇Λ can be added since ∇×∇Λ = 0)

~B = ∇× ~A

Now from Faraday’s law ∇× ~E = − ~B = −∇× ~A ∇×(~E + ~A

)= 0

The identity ∇×∇φ = 0 suggests the scalar potential −∇φ = ~E + ~A

~E = −∇φ− ∂ ~A

∂t

Now plugging ~D = ǫ~E into Coulomb’s law and ~H = ~B/µ0 into Ampere’s law

∇ ·(−ǫ[∇φ +

∂ ~A

∂t

])homogeneous

= −ǫ(∇2φ +

∂∇ · ~A∂t

)= ρ

∇× (∇× ~A) + ǫµ0

∂∇φ + ∂ ~A∂t

∂t= −

(∇2A− µ0ǫ

∂2~A

∂t2−∇

[∇ · ~A + µ0ǫ

∂φ

∂t

])= ~J


Choice of gaugeand retarded solutions

These equations can be simplified by choice of Lorentz gauge

∇ · ~A + µ0ǫ∂φ

∂t= 0

Giving source driven wave eqns

∇2φ− µ0ǫ∂2φ

∂t2= −ρ

ǫ

∇2~A− µ0ǫ∂2~A

∂t2= −µ0

~J

with solutions (v = 1√µǫ and R = |~r−~r′|)

φ(~r, t) =y

V

ρ(t−R/v)

4πǫRdV

~A(~r, t) = µ0

y

V

~J(t−R/v)

4πRdV

Alternativey, can choose ∇ · ~A any way we wish

∇ · ~D = ∇ · ǫ~E = ρ ⇒ ǫ(−∇ · ∇φ− ∂

∂t∇ · ~A) = ρ

eg Coulomb gauge ∇ · ~A = 0

∇ · ∇φ = ∇2φ = −ρǫ


Spherical wave from an EM dipole

Use vector potential as a function of distance ~r from dipole source

~A(~r) = A0U (~r)z

where U (~r) = U (r) = 1re

−ikr is a scalar spherical wave that satisfies scalar Helmholtz

eqn, thus ~A(~r) satisfies vector Helmholtz

∇2~A + k2~A = 0

Magnetic and Electric field defined as~H = 1

µ∇× ~A ~E = 1iωǫ∇× ~H

This ensures ∇ · ~E = 0 and ∇ · ~H = 0 since ∇ · ∇× = 0

because ~A(~r) = 0 satisfies Helmholtz can show ∇× ~E = −iωµ~H as well

For points r ≫ λ using spherical coordinates approximate forms are

~E(~r) = E0 sin θU (~r)θ ~H(~r) = H0 sin θU (~r)φ

where E0 = (jk/µ0)A0, H0 = E0/η and θ = cos−1 xr

Thus wavefronts are spherical, fields are tangential and orthogonal,however unlike scalar spherical wave amplitude varies as sin θ

θz

x

y

r

rφ

θ

^^

^

φ

E

E

E

EH

H

H

H

sphericalwavefront


Hertz Potential ~Π

~A =n2

c2∂ ~Π

∂tφ = −∇ · ~Π

∇ · ~A + n2

c2∂φ∂t = 0 ⇒ ∇ · n2

c2∂ ~Π∂t − n2

c2∂∇·~Π∂t = 0

~B = ∇× ~A = ∇× n2

c2∂ ~Π

∂t= µ0ǫ

∂

∂t(∇× ~Π) ~H = ǫ

∂

∂t(∇× ~Π)

~E = −∇φ− ∂ ~A

∂t= ∇(∇ · ~Π)− µ0ǫ

∂2 ~Π

∂t2

Hertz potential obeys the polarization driven wave eqn

∇2~Π− µǫ∂2~Π

∂t2= −

~P

ǫ

and J = ∂ ~P∂t ρ = −∇ · ~P

Retarded solution

~Π(~r, t) =y

V

~P(~r′, t−R/v)

RdV

R = |~r−~r′| , v = c/n


Hertz Vector

In source free region Hertz vector satisfies the Helmholtz eqn

∇2 ~Π− µǫ∂2 ~Π

∂t2= 0

This represents 3 mathematically independent solutions for Πx,Πy,Πz so it is sufficientto solve for one component

Πx(x, y, z) =1

k2v(x, y, z)e−iωt

where 1/k2 is introduced for convenienceThe x-component of the Hertz potential at any z is represented in terms of its angularspectrum at z = 0

v(x, y, z) =1

(2π)2

xV (kx, ky; z = 0)ei(kxx+kyy+kzz)dkx dky

whereV (kx, ky; z = 0) =

xv(x, y, 0)e−i(kxx+kyy)dx dy


Plane Waves

direction cosines of a plane wave

E(x, y, z) = E0pei2πλ (αx+βy+γz) = E0pe

i(kxx+kyy+kzz)

α2 + β2 + γ2 = 1 k2x + k2y + k2z = k20 =

(2πn

λ

)2

where k0 = |~k| = 2πn/λ in medium of index n.

k-space

k

2πn/λ

kz

kx

x

zθ

k

λ

kz

kx

In 2-dimensionsE(x, z; t) = A0pe

ik0(x sin θ+z cos θ)e−i2πνt + cc

where α = sin θ/λ and γ = cos θ/λ.ν ≈ 5× 1014 Hz λ ≈ .63× 10−6m (HeNe)


Monochromatic Wave Eqn: Helmholtz eqn

Each component of the vector wave eqn satisfies the scalar wave eqn

∇2u(~r, t)− 1

c2u(~r, t) = 0

For monochromatic waves

u(~r, t)=a(~r) cos[ω0t+Φ(x, y, z)]=ae−i[ω0t+Φ]+a∗e+i[ω0t+Φ]=Ae−iω0t+A∗e+iω0t= u+u∗

Since wave eqn is linear, we can just solve for one sideband, add the other later by takingreal part

˙u = (−iω)u = (−iω)Ae−iω0t ¨u = (−iω)2u = −ω2u = −ω2Ae−iω0t

Helmholtz eqn for monocromatic envelope A(~r)

∇2A(~r)e−iω0t +ω2

c2A(~r)e−iω0t = 0 ⇒ (∇2 + k20)A(~r) = 0

Note if wave contains multiple temporal frequencies (say 2 to start, then arbitrary dis-tribution later). We can solve monochromatic isotropic Helmholtz for each temporalfrequency component seperately using an identical solution method and then find thetotal field amplitude by summing monochromatic components.


Postulate and propagate a plane wave soln

u(x, y, z, t) = a0e−i(ωt−~k·~r) = a0e

i~k·~re−iωt = A(~r)e−iωt

where ~k = (kx, ky, kz) =2πλ(α, β, γ) = 2π

λk = 2π(u, v, w) = 2π(fx, fy, fz)

direction cosines

α = k · x = cos θx

β = k · y = cos θy α2 + β2 + γ2 = 1

γ = k · z = cos θz

Plug into Helmholtz eqn, use ∇ · A = i~k · A and ∇2A = (i~k) · (i~k)A = −k2A

−k2A +ω2

c2A = −

(k2 − ω2

c2

)A = 0

⇒ k2 =ω2

c2=

(2πν)2

c2=

(2π

λ

)2

|k| = 2π

λ

So we must choose the magnitude of the wavevector ~k appropriately and with such achoice any monochromatic plane wave is a solution ⇒ sphere of allowed ~k


Propagation of a plane wave

Thus if we have a plane wave at any location (eg on a plane) we know how it propagatesboth forward and backwards eg between 2 planes

zθz

k

Plane wave produces equal 2-D linear phase factors across any parallel plane that simplyphase advance with propagation

At z = 0 linear phase factor due to a plane wave

A(x, y) = ei(kxx+kyy) u(x, y, t) = ei(kxx+kyy)e−iωt

At any other z

A(x, y : z) = ei(kxx+kyy)eikzz u(x, y, z, t) = ei(kxx+kyy)eikzze−iωt

Where kz =√k20 − k2x − k2y


Boundaries: Gauss’ Divergence Theorem

y

V

∇ · ~FdV =x

S

~F · ~d~S

∇ · ~B = 0 n = n2 = −n1 0 as walls → 0x

S2

~B · ndS −x

S1

~B · ndS +︷︸︸︷x

walls

~B · nwdS = 0

δS

ε 2

ε1

S1

nw^

n1

^

n

n2^

S2

D2

D1

n · (~B2 − ~B1) = 0 B1n = B2n Normal ~B conserved across boundary

Coulombs Law ∇ · ~D = ρ 0y

V

ρdV =x

S

σdS =x

S

(~D2 − ~D1) · ndS +︷︸︸︷x

walls

~D · nwdS

where σ = surface charge density (usually 0 in optics)

δh->0

n1

^

n2

^

ε2

ε1

S2

S1

D 1

D2

Dn2

Dt2

Dt1

Dn1

n · (~D2− ~D1)=σ D2n −D1n=σ=0 usually Normal ~D conserved across boundary


Stokes Theorem

x

S

∇× ~F · ~dS =

∮

C

~F · ~dl

∇× ~E = −∂~B

∂t

x∇× ~E · bdS =

∮

C

~E · ~dl =∮

C2−C1+δ

~E · ~dl = ∂

∂t

x

S

~B · bdS

(~E2 · t− ~E1 · t)δS = − limδS→0∂ ~B∂t

· bδS → 0 since ~B is finite

Et2 = Et1 Tangential ~E is conserved across boundary

H1t

H 2

H1

H2t

δl

∆l

n

t

b Surface

Current

Density K

S

1

2

∇× ~H =∂ ~D

∂t+ ~J

x∇× ~H · bdS =

∮

C2−C1

~H · ~dl = ∂

∂t

x

S

~D · bdS +x

S

~J · bdS = 0 +K∆l

where K is impulsive surface current density (usually 0 in optics)

H2t −H1t = K = 0 usually so tangential ~H conserved across boundary


Snell’s law

Boundary

normal

θi

θt

rkθi

θt

θrn=1

n=1.6

n=1

n=1.6 kz tk

ik

kx

Snell’s law -- Conservation of transverse momentum

b

t

n

~ki × n = ~kr × n = ~kt × n

~ki · t = ~kr · t = ~kt · tkix = krx = ktx

θi = θr

n1 sin θi = n2 sin θ2

~Ei(~r, t) = eiEie−i(ωt−kixx−kiyy−kizz) + cc = eiEie

−i(ωt−~ki·~r) + cc

~Er(~r, t) = erEre−i(ωt−krxx−kryy−krzz) + cc = erEre

−i(ωt−~kr·~r) + cc

~Et(~r, t) = etEte−i(ωt−ktxx−ktyy−ktzz) + cc = etEte

−i(ωt−~kt·~r) + cc

|~ki| = 2πn1λ

= ω√µǫ1 = n1

ωc= |~kr| |~kt| = 2πn2

λ= ω

√µǫ2 = n2

ωc

Rotate coordinates so that kiy = kry = kty = 0 and all beams coplanar

θi = tan−1 kixkiz

= sin−1 kix|~ki|

θr = tan−1 krxkrz

= sin−1 krx|~kr|

θt = tan−1 ktxktz

= sin−1 ktx|~kt|


Along interface

~Ei(x, t) = eiEie−i(ωt−~ki·x) t=0⇒= eiEie

ikixx

~Er(x, t) = erEre−i(ωt−~kr·x) t=0⇒= erEre

ikrxx

~Et(x, t) = etEte−i(ωt−~kt·x) t=0⇒= etEte

iktxx

The phase velocity of these waves along the interface must be equal as must the slowness

vpx =ω

kix=ω

krx=ω

ktx

1

vpx=

sin θic/n1

=sin θrc/n1

=sin θtc/nt

To satisfy BC at each x must have same periodicity along interface thus tangentialcomponents of k are equal across interface kix = krx = ktx.

The magnitudes of the ~k vectors are |~ki| = |~kr| = n1ωc and |~kt| = n2

ωc .

Thus the angle of incidence is equal to the angle of reflection θi = θrand the refracted angle is given by the wavevectors projected onto the interface

~ki × n = ~ki · t = |~ki| sin θi = |~kt| sin θt = ~kt · t = ~kt × n

n1 sin θi = n2 sin θ2


Reflected and Refracted amplitudess-polarization

Resolve ~E into components ⊥ and ‖ to the plane of incidence

E⊥ = ~E · b ~E‖ = ~E× b

The tangential components of ~E are equal across boundary

Et+ = Et− ⇒ Eis + Er

s = Ets

Normal component of ~D is conserved

x

Btr

Er

Et t

tB

tiB

Ei

zn1

n2n

t

^

^

b=x z^ ^

Dn− = Dn+ ǫ1En− = ǫ2En+ En− 6= En+

Similarly for ~B = µ~HBn− = Bn+

Ht− = Ht+1

µ1Bt− =

1

µ2Bt+ Bt− 6= Bt+

Notice labeling of ~B’s (handedness of ~k, ~B, ~E preserved)

~Bi · x < 0 ~Bt · x < 0 ~Br · x > 0 −H tt = −H i

t +Hrt

−Bt

µ2cos θt = −B

i

µ1cos θi +

Br

µ1cos θr


Non-magnetic interface µ1 = µ2

Bi cos θi − Br cos θr = Bt cos θt

now θi = θr so cos θi = cos θrand

Bi = Ein1c

Br = Ern1c

Bt = Etn2c

n1c(Ei

0 − Er0) cos θi =

n2cEt

0︸︷︷︸ cos θtEi

0 + Er0 = Et

0

n1c (E

i0 − Er

0) cos θi =n2c (E

i0 + Er

0) cos θtEi0c(n1 cos θi−n2 cos θt) = Er0

c(n1 cos θi+n2 cos θt)

Solving for the ratios of reflected and transmitted fields to the incident gives

rs = r⊥ =Esr

Esi

=n1 cos θi − n2 cos θtn1 cos θi + n2 cos θt

ts = t⊥ =Est

Esi

=2n1 cos θi

n1 cos θi + n2 cos θt= 1 + rs

Using Snell’s law, n1 sin θi = n2 sin θt gives an alternative form

rs = −sin(θi − θt)

sin(θi + θt)ts = −2 sin θt cos θi

sin(θi + θt)Kelvin Wagner, University of Colorado EM Waves ECEN 3410 Spring 2017 56

Fresnel reflection coefficients: s-polarization

x

Btr

Er

Et t

tB

tiB

Ei

zn1

n2n

t

^

^

b=x z^ ^

rs = r⊥ =Esr

Esi

=n1 cos θi − n2 cos θtn1 cos θi + n2 cos θt

=kiz − ktzkiz + ktz

= −sin(θi − θt)

sin(θi + θt)

ts = t⊥ =Est

Esi

=2n1 cos θi

n1 cos θi + n2 cos θt=

2kizkiz + ktz

=2 sin θt cos θisin(θi + θt)

= 1 + rs

Where it can also be expressed in terms of the k-vector components

kiz =√k20n

2i − k2x ktz =

√k20n

2t − k2x


Fresnel reflection coefficients: p-polarization

i

tt

E c

osθ

ki

k rkt

θi

θrθttB

tΕ

i

ii

E c

osθ

ΕBi

rB

E c

osθ

rrrΕ

Since E changes sign for Ei and Er as θ -->0, wont agree with rs

i

tt

E c

osθ

ki

k rkt

θi

θrθttB

tΕ

i

ii

E c

osθ

ΕBi

E c

osθ

rr

rΕ

Alternative Coordinate Systems

rB

Fresnel noted “petit difficulte” with signs

Oevres Completes d’Augistin Fresnel, Paris 1876, V 1 p 787

Verdet changed his manuscript (p 789)

Kelvin labeled change “manifest absurdity”

But was just to today’s coordinate system

Became encumbered with complictaed coordinatedependent rules for finding reflected wave phase

W.T. Doyle, Graphical approach to Fresnel’s equationsfor reflection and refraction of light, Am. J. Phys. 48(8), p.643, 1980

Et− = Et+ Ei cos θi − Er cos θr = Et cos θtHt− = Ht+ niEi + niEr = ntEt

rp = r‖ =Epr

Epi

= ±n2 cos θi − n1 cos θtn2 cos θi + n1 cos θt

= ±n2/ cos θt − n1/ cos θin2/ cos θt + n1/ cos θi

= ±tan(θi − θt)

tan(θi + θt)

θi + θt = 90 denominator → 0 Brewster’s angle

tp = t‖ =Ept

Epi

=2n1 cos θi

n2 cos θi + n1 cos θt=

cos θicos θt

2n1/ cos θin2/ cos θt + n1/ cos θi

=2 sin θt cos θi

sin(θi + θt) cos(θi − θt)= (1− rp)

cos θicos θt


Fresnel Reflection and TransmissionCoefficients: Air to Glass


Conservation of Power

|rs|2 + |ts|2 6= 1 |rp|2 + |tp|2 6= 1

But can show that

ts − rs = 1n2n1tp − rp = 1

Intensity goes as I = nǫoc2|E|2, and energy flux across

boundary scaled by ratio of cross-sectional beam ar-eas cos θt

cos θi, thus Reflectance

R =Ir cos θrIi cos θi

=IrIi

=n1ǫoc2

|Er|2n1ǫoc2 |Ei|2

= |r|2

Transmittance

T =It cos θtIi cos θi

=n2ǫoc2

|Et|2n1ǫoc2 |Ei|2

cos θtcos θi

=n2 cos θtn1 cos θi

|t|2

Conservation of Energy

R + T = |r|2 + n2 cos θtn1 cos θi

|t|2 = 1

θi

θr

A

Acosθi

Acosθt

n1

n2

Energy incident, reflected, andtransmitted per unit area onsurface given by component ofPoynting vectors normal to sur-face Ji=Jr + Jt

Ji = Si cos θi =n12η0

|Ei|2 cos θi

Jr = Sr cos θr =n12η0

|Er|2 cos θr

Jt = St cos θt =n22η0

|Et|2 cos θt


Brewster’s angle

E c

osθ

rr

it

tE

cos

θk

i

k rkt

θi

θrθt

tΕi

iE

cos

θiΕ

rΕ

rp = 0 when θB = tan−1 n2n1

n2n1

= 1.6 ⇒ θB = 57.9946

geometry when transmitted dipoles with polarization ~Et are parallel to the reflected ~krsince a layer of dipoles along the surface is unable to radiate along their axis.θi+ θt+ 90 = 180 θi + θt = 90 ⇒ tan(θi+ θt) → ∞ ⇒ rp → 0


Graphical Treatment of Fresnel ReflectionCoefficients

W.T. Doyle, Graphical Approach to Frenel’s Eqn for refrection and refraction, Am. J. phys. 48(8) p 643 1980

Boundary

b

t

n

i

t tE cosθ

ki

k r

kt

θi θr

θt

tΕ

i iE cosθ E cosθr r

rΕ

rBBi

tB

iΕ

θi

θt

θr

n=1

n=1.6θi

θt

n=1

n=1.6

ik

kx

rk

tkkz

θtθi

θ1 θr

Eip

Erp

E1p

Etp E2

p=

Graphical Construction of Fields

reflected

total1

Field Refraction laws for Electric and magnetic fields come from BC:

~E1 × n = ~E2 × n ~H1 × n = ~H2 × n ~D1 · n = ~D2 · n ~B1 · n = ~B2 · nWhere θ2 = θt and H = 1

µB = 1

µncE

~E1 = ~Ei + ~Er~H1 = ~Hi + ~Hr

~E2 = ~Et~H2 = ~Ht

Since BC give conservation of ~D · n and ~E · t total fields at angles θ1 and θ2 obeyD1 cos θ1 = ǫ1E1 sin θ1 = ǫ2E2 sin θ2 = D2 cos θ2 E1 sin θ1 = E2 sin θ2

p : ǫ1 tan θ1 = ǫ2 tan θ2 s : µ1 tan θ1 = µ2 tan θ2


Graphical Treatment of Fresnel Coefficientsn2 = 1.73, n1 = 1.0 p-polarization



Graphical Treatment of Fresnel Coefficientsn2 = 1.73, n1 = 1.0 s-polarization



Evanescent waves n1 > n2refracting from more dense media to less

n1 sin θi = n2 sin θt has no soln for θi > θc = sin−1 n2n1

Propagation constant in the 2nd rarer medium becomes imaginary

sin θt =sin θin2/n1

cos θt =√1− sin2 θt = ±i

√sin2 θt − 1

Et(x, z, t) = eE1e−iωteikxxe−Γz

kx =ω sin θiv2n2/n1

= 2πc/λc/n2

sin θin2/n1

= n1k0 sin θi

Γ = ωv2

√sin2 θi

(n2/n1)2− 1 = 2π

λ0

√n21 sin

2 θi − n22

characteristic decay distance into media < λo2π

planes of constant

amplitude

planes ofconstant

phase

1/eλ /2π0

kTIR

s-polarization

rs =n1 cos θi − n2 cos θtn1 cos θi + n2 cos θt

=cos θi − i

√sin2 θi − (n2/n1)2

cos θi + i√sin2 θi − (n2/n1)2

=a− ib

a + ib=q

q∗


Phase Difference of δs and δp

⇒ |rs|2 = 1 = qq∗

(qq∗

)∗= qq∗

q∗q = 1 rs = eiδs = qq∗ =

aeiα

ae−iα = ei2α

tanδs2= tanα =

IqRq =

−√

sin2 θi − (n2/n1)2

cos θip-polarization

rp ==n2 cos θi − n1 cos θtn2 cos θi + n1 cos θt

=(n2/n1)

2 cos θi − i√sin2 θi − (n2/n1)2

(n2/n1)2 cos θi + i√sin2 θi − (n2/n1)2

=c− id

c + id=p

p∗

rp = eiδp =p

p∗=

beiβ

be−iβ= ei2β

tanδp2= tan β =

IpRp = −

√sin2 θi − (n2/n1)2

(n2/n1)2 cos θiRelative Phase Difference δ = δs − δp

tanδ

2=

tan δs2 − tan

δp2

1 + tan δs2 tan

δp2

= −

(1− 1(

n2n1

)2

)√sin2 θi −

(n2n1

)2

1 + sin2 θi−(n2/n1)2

(n2/n1)2 cos2 θi

= −cos θi√sin2 θi − n2

sin2 θi


Exponential decrease with penetration depth

~k1 =2πn1λ (sin θ1, 0, cos θ1) k2x1 + k2z1 = n21k

20 ⇒ k2z1 = n21k

20 − k2x1

~k2 =2πn2λ (sin θ2, 0, cos θ2) k2x2 + k2z2 = n22k

20

Snell’s lawn1 sin θ1 = n2 sin θ2 ⇒ kx1 = kx2

Solving for imaginary component of kz when kx1 > n2k0 in TIR regime

kz2 =√n22k

20 − k2x2 =

√n22k

20 − k2x1 = ±iγ = ±i

√k2x1 − n22k

20

k2z2 = −(k2x1 − n22k20) = −[(n21k

20 − k2z1)− n22k

20] = −(n21 − n22)k

20 + k2z1

Exponential decay constant

γ2 = (n21 − n22)k20 − k2z1 ⇒ γ2 + k2z1 = (n21 − n22)k

20

γ

kz

kT

1n 2πλ

2n 2πλ

(n -n )2πλ

1 222


Relation of γ to phase

let n = n2/n1

tanδs2=

√sin2 θi − n2

cos θi

n1k0n1k0

tanδp2=

√sin2 θi − n2

n2 cos θi

n1k0n1k0

Now compare with evanescent wave decay constant

γ = n1k0√sin2 θi − n2

tanδs2=

γ

n1k0 cos θ1=

γ

k1z

tanδp2=

γ

n2n1k0 cos θ1=

γ

n2k1zγ

kz

kT

1n 2πλ

2n 2πλ

(n -n )2πλ

1 222

α

n1co

sθi

n1sinθi

n1sin θi -

n2

2

2

2

2n

α

n1sinθi


Fresnel Reflection Coefficients: Glass to Airn1 > n2: Critical Angle

θc

n2

n1

2πn1

λ2πn2

λθc

n1n2

TIR region

Critical angle θc = sin−1 n2n1

Phase in TIR region

tanφp2

=

√sin2 θi − sin2 θc

cos θi

tanφs2

=

√sin2 θi − sin2 θc

cos θi sin2 θc


Producing a known helicity with a FresnelRhomb

p comes out advanced by 90Left Circularly polarized

like a waveplate with fast propagation along p

slow along s

FS

HVfast

R

L

Right handed rotation around fast axis(when R is on N pole)

takes +45 linear (S2=1) to LHC (S3=-1)

fast axis

LHC

+45

RHC

-45

fast axis


Fields near boundariesand the Goos-Hanchen shift

Incident field reflects off TIR boundary with a phase shift . In region z < 0

E(x, z) = E0

[ei(k1xx+k1zz) + ei(k1xx−k1zz−δ)

]e−iωt = 2E0 cos(k1zz + δ/2)ei(k1xx−ωt−δ/2)

For z > 0 continuity at boundary requires |E2(x, 0)| to be identital to |E1(x, 0)|E2(x, z) = 2E1e

−iδ/2 cos(δ/2)e−γzei(kx1x−ωt)

θ1 ↓ from 90 to θc

Max shifts to 0

γ ↓ so Penetrates more

Leads to evanescentmode coupling, FTIR

Causes Goos-Hanchenshift

Bigger for p than s

θc

δ/kz

e-γz

Z

D

pD

Evanescentfield

penetrationin TIR

metal strip

Glass prism

Light Slit

Also a tiny focal shift, but ray picture and path length gives opposite answer


Goos-Hanchen ShiftM. McGuirk and C.K.Carniglia, Am angular spectrum representation approach to the GH shift, JOSA 67(1), p103, 1977

θ0 θ0

xz

x’

z’θθ0

α

kx

Incident beam

EI(x, z) =

∫A(kx)e

i(kxx+kzz)dkz A(kx, z) = A(kx)eikzz kz =

√k20 − k2x

with well defined center on axis∫kx|A(kx)|2dkx = 0

Reflected beam in TIR regime (r(kx) = eiδ(kx) ). kx′ = kx, kz′ = kz

ER(x′, z′) =

∫r(kx)A(kx, z0)e

i(kxx′+kzz′)dkz

Assume “all” components of angular spectrum arrive at θ > θc and < 90 so

δ(kx) ≈ δ(0) + kxdδ

dkx

∣∣∣∣kx=0

+k2x2!

d2δ

dk2x

∣∣∣∣kx=0

+ · · · = δ0 +Dkx + k2xF/2


Goos-Hanchen Shift

Keeping only first 2 terms shows shift of the reflection predicted by geometrical optics

ER(x′, z′) = eiδ0

∫A(kx, z0)e

i(kx(x′+D)+kzz

′)dkz = eiδ0EI(x′ +D, z)

Angle of incidence around the central angle θ0 is θ = θ0 − sin−1 kxkThus

D = − λ

2π

dδ

dθ

∣∣∣∣θ=θ0

δs(θ) = −2 tan−1

[√n2 sin2 θ − 1

n cos θ

]+ π Ds =

λ

π

n sin θ0√n2 sin2 θ − 1

δp(θ) = −2 tan−1

[n√n2 sin2 θ − 1

cos θ

]+ π

Dp =Ds

(n2 + 1) sin2 θ0 − 1=λ

π

n sin θ0

[(n2 + 1) sin2 θ0 − 1]√n2 sin2 θ − 1

Can also analyze the focal shift along z-axis due to quadratic term F , Fs =Ds cot θ0

n2 sin2 θ0−1,

but answer is opposite to backwards shift predicted by ray model (Fr = −D cot θ0)


Imaginary Wavevectors

~k = ~kR + i~kI = k1kR + ik2kI ~kR wavevector (phase) ~kI attenuation vector

E(~r, t) = ~E0ei[(~kR+i~kI)·~r−ωt] = e−

~kI ·~r(~E0ei(~kR·~r−ωt) + cc)

Plug into monochromatic Maxwell’s eqn (use ~v · (~v× ) = 0

~k× ~E = ωµ~H ⇒ ωµ~k · ~H = ~k · (~k× ~E) = 0 ⇒ ~k · ~H = 0

~k× ~H = −ωǫ~E ⇒ −ωǫ~k · ~E = ~k · (~k× ~H) = 0 ⇒ ~k · ~E = 0

But since these are complex vectors, this doesnt mean spatial orthogonality

Derive usual ω, ~k wave eqn

~k× ~k× ~E = ωµ~k× ~H = −ω2µǫ~E = ~k2~E− ~k(~k · ~E)

(~k2 − k20µrεr)~E = 0 k20 =

ω2

c2= ω2µ0ǫ0


For solution for ~E to exist, complex ~k must satisfy disperion relation

~k2 = |k1kR + ik2kI|2 = k21 − k22 + i~kR · ~kI = k20µrεr

equating real and imaginary parts

k21 − k22 = k20µrεr~kR · ~kI = 0 ⇒ ⊥

New phase velocity ~vp =ωk1kR

k21 = k22 + k20µrεr > k20µrεr

⇒ slower phase velocity for nonuniform wave than planewave bigger momentum vectors accessible in TIR regime

Amplitude of the wave varies as ~E0e−~kI ·~r

planes of constant

amplitude planes ofconstant phase

1/eλ /2π0

Rk

Ik


Absorbing Boundary

complex index N = n + ia = n(1 + iκ)

Complex propagation vector ~K = ~kt + i~α

Incident ei(~ki·~r−ωt)

Reflected ei(~kr·~r−ωt)

Transmitted ei(~K·~r−ωt) = e−~α·~rei(

~kt·~r−ωt)

periodicity along boundary requires κ

γ

rkik

tktα planes of

constant amplitude

planes of

constant

phase

t

n

^

^

~ki ·~r|z=0 = ~kr ·~r|z=0 ⇒ θi = θr

~ki ·~r|z=0 = ~K ·~r|z=0 = (~kt + i~αt) ·~r|z=0

Equating real and imaginary components also a fnc of γ

~kt · t = ~ki · t ki sin θi =︷︸︸︷kt sin γ

~αt · t = 0 ~αt ‖ n


Plug into wave eqn

∇2~E =N 2

c2∂2~E

∂t2

Plane wave ∇· → i~K monochromatic harmonic component ∂∂t → iω

~K · ~K = (~kt + i~α) · (~kt + i~α) =N 2ω2

c2= N 2k20 = (n + ia)2k20

~kt · ~kt − ~α · ~α + 2i~α · ~kt = (n2 − a2 + 2ina)k20

Equate real and imaginary parts

k2t − α2t = (n2 − a)2k20 (1)

~kt · ~α = ktαt cos γ = nak20 (2)

From Snell’s law and the required conservation of transvers momentum we know thetransverse component of the wavevector, but we need to derive the complex wavevectorperpendicular to the surface (2) (1)

( ~K · n)2 = (kt cos γ + iαt)2 = k2t cos

2 γ + 2i︷︸︸︷αtkt cos γ−

︷︸︸︷α2t

= k2t (1− sin2 γ) + 2inak20 + (n2 − a2)k20 − k2t = −k2i sin2 θi + k20(n

2 + 2ina− a2)

= k20(n− ia)2 − n2ik20 sin

2 θi = k20(N 2 − n2i sin2 θi)


kt cos γ + iαt = k0

√N 2 − n2i sin

2 θi (3)

This yields the angle γ between ~kt and ~αt which is parallel to the normalGeneralized Snell’s law

ni sin θi = N sin Φ

where Φ is complex eg sin Φ = ei(p+iq)−e−i(p+iq)2i

n21 sin2 θi = N 2 sin2Φ = N 2(1− cos2Φ)

cos Φ =

√1− n21 sin

2 θiN 2

⇒ k0N cos Φ = k0

√N 2 − n21 sin

2 θi (4)

combining with (3) gives an eqn for Nkt cos γ + iαtk0 cos Φ

= N (5)

k+iαiα

k co

sγt

k sinθ0

k


Solving for the absorbing boundary conditions

Incident~Hi =

1µ0ω~ki × ~Ei

Reflected~Hr =

1µ0ω~kr × ~Er

Transmitted into absorbing media~Ht =

1µ0ω

~K× ~Et =1µ0ω

(~kt × ~Et + i~αt × ~Et)

s-polarized case.Tangential E conserved

xEr

Et t

tH

tiH

Ei

zn1

n2+ia

Htr

αγθr

θi

ik

tkrk

Ei + Er = Et

Tangential H conserved.

Hi cos θi −Hr cos θi = Htangentialt

kiEi cos θi − krEr cos θi = ktEt cos γ + iαtEt = Et(kt cos γ + iαt)

= Et(Nk0 cos Φ)

Where the last step is from (5)


Solve for reflection and transmission coefficients by eliminating Et

n1k0Ei cos θi − n1k0Er cos θi = (Ei + Er)(Nk0 cos Φ)

Ei(n1 cos θi −N cos Φ) = Er(N cos Φ + n1 cos θi)

rs =Esr

Esi

=n1 cos θi −N cos Φ

n1 cos θi +N cos Φ=us1 − us2us1 + us2

Introduce the tangential admittance of an absorbing medium for s-polarization

us2 = N cos Φ

Analogously for p-polarization

up2 = N/ cos Φ

p-polarized reflection from an absorbing medium

rp =Epr

Epi

=up1 − up2up1 + up2

=n1/ cos θi −N/ cos Φ

n1/ cos θi +N/ cos Φ

At normal incidence0 90

0

1

rp

rs

Reflection from absorbing media

rs = rp =n1 −Nn1 +N =

n1 − n− ia

n1 + n + ia

n1 + n− ia

n1 + n− ia=n21 − n2 − ia(n1 − n + n1 + n)

(n1 + n)2 + a2


Loss adds conductivity termH.C. Chen, Theory of EM Waves, Mcgraw Hill, 1983

~D = ǫ~E ~B = µ~H ~Jc = σ~E

Monochromatic wave e−iωt εc

∇× ~E = iωµ~H ∇× ~H =∂ ~D

∂t+ ~J = −iωǫ~E + σ~E = −iωǫo

︷︸︸︷εr(1 + iτ ) ~E

Loss tangent: ratio of conduction to displacement current τ = σωǫoεr

σ = ωǫoεrτ

Trial solution use complex ~K. Then dot with ~K and note ~a · (~a× ~b) = (~a×~a) · ~b = 0

~E = ~E0ei~K·~r ⇒

~H = ~H0ei~K·~r ⇒

~K× ~E0 = ωµ~H0

~K× ~H0 = −ωǫoεc~H0

~K · ~H0 = 0~K · ~E0 = 0

~K× ~K× ~E0 = −ωµ~K× ~H0 = ωµ(−ωǫoεc~E0)

k20~K(*0~K · ~E0)− ~K2~E0 = −

︷︸︸︷ω2µ0ǫo εc~E0 ⇒ (~K2 − k20εc)

~E0 = 0

New dispersion relation for ~K = ~kR + i~kI~K2 = k20εc

~k2R − ~k2

I = k20µrεr~kR · ~kI = k1k2 cos γ = 1

2k20µrεrτ


Uniform Lossy Plane WaveH.C. Chen, Theory of EM Waves, Mcgraw Hill, 1983

γ = 0 ~kR = k1k ~kI = k2k

~K = (k1 + ik2)k = Kk K = k1 + ik2 = k0√µrεc

complex wave impedance ηc =ωµ0µrK

=√

µ0ǫo

µrεr(1+iτ)

= |ηc|eiξ

~H0 =1

ηc(K× ~E0) E0 = ~E1 + i~E2

~E0 = −ηc(K× ~H0) H0 = ~H1 + i~H2

all ~E1, ~E2, ~H1, ~H2 ⊥ k

φ

~E = e−~kI ·~rR

~E0e

i(

︷︸︸︷~kR ·~r− ωt)

= e−

~kI ·~r(~E1 cosφ− ~E2 sinφ)

~H = e−~kI ·~rR

k× ~E0

ηcei(

~kR·~r−ωt)

= e−~kI ·~r[k× ~E1 cos(φ− ξ)− k× ~E2 sin(φ− ξ)

]

Note that just as in lossless caseE andH trace out orthogonal ellipses, but in lossy case

~E · ~H =e−

~kI ·~r

|ηc|~E1 × ~E2 · k [cosφ sin(φ− ξ)− sinφ cos(φ− ξ)]

lossless case→ 0 ⇒⊥

So instantaneous E and H are perpendicular in lossless case, but not in lossy, unless ~E0

is linearly polarized, in which case ~E · ~H = 0Kelvin Wagner, University of Colorado EM Waves ECEN 3410 Spring 2017 82

Determine phase constant k1 and attenuationconstant k2

k21 − k22 = k20µrεr (6)

k1k2 =12k20µrεrτ (7)

square (6) and add to 4× (7) squared

k41 − 2k21k22 + k42 + 4k21k

22 = (k21 + k22)

2 = k40µ2rε

2r(1 + 4

τ 2

4) (8)

Add/subtract (6) and√(8) and note that kj are real ⇒ k2j > 0.Then take

√

k1 = k0√µrεr

√12

(√1 + τ 2 + 1

)≥ k0

√µrεr k2 = k0

√µrεr

√12

(√1 + τ 2 − 1

)

Nonuniform Plane WaveFor nonuniform waves we had (7) replaced by ~kR · ~kI = k1k2 cos γ = 1

2k20µrεrτ

k1 = k0√µrεr

√12

(√1 + τ2

cos2 γ+ 1

)k2 = k0

√µrεr

√12

(√1 + τ2

cos2 γ− 1

)

but we must use boundary conditions to find γ

Can show ~E0 and ~H0 can not simultaneously be linear for nonuniform plane wave, butcan simultaneously be circularly polarized.


Solving for the z-components of thecomplex wavevector

~K2 = k20ur2εr2(1 + iτ ) =

∣∣∣~kT + (qr + iqi)n∣∣∣2

= |~kT |2 + (qr + iqi)2 since n ⊥ ~kT

⇒ qr + iqi =

√K2 − |~kT |2 = k0

√µrεr(1 + iτ )− n1 sin

2 θi =√k20µrεr − k2T + ik20µrεrτ

let a + ib =√A + iB thus a2 − b2 + 2iab = A + iB

a2 = 12

(√B2 + A2 + A

)a =

√12

√√B2 + A2 + A

b2 = 12

(√B2 + A2 − A

)b =

√12

√√B2 + A2 − A

ab = 12

√(B2 + A2)− A2 = 1

2

√B2 = 1

2B

a2 − b2 = 12

(√B2 + A2 − A

)− 1

2

(√B2 + A2 + A

)= A

⇒ a2 − b2 + 2iab = A + iB

qr =√

12

[√k20µ

2rε

2rτ

2 + (k20µrεr − k2T )2 + (k20µrεr − k2T )

2

]12

qi =√

12

[√k20µ

2rε

2rτ

2 + (k20µrεr − k2T )2 − (k20µrεr − k2T )

2

]12


Optics of Conductive Media

∇× ~H =∂ ~D

∂t+ ~J

monochromatic−−−−−−−−→ ∇× ~H = iω ~D + ~J

Constitutive reln ~D = ǫ~E = ǫo(1 + χ)~E Conductivity ~J = σ~E

∇× ~H = iωǫ~E + σ~E = (iωǫ + σ)~E = iω(ǫ +

σ

iω

)~E = iωǫe~E

where ǫe = ǫ+σ

iωk = β−iα

2= ω

√ǫeµ0

η =

√µ0ǫe

≈√iωµ0σ

n− iα

2k0=

√ǫeǫo

≈√

σ

iωǫo=

1 + i√2

√σ

ωǫowhen σ

ω> ǫ conductivity dominates at low freq

n ≈√

σ2ωǫo

α = 2k0√2

√σωǫo

= ω√µ0ǫo

√2σ

x2ωǫo=√2ωµ0σ


Optics of MetalsBorn and Wolf, Principles of Optics, ch 14. G.R. Fowles, Intro to Modern Optics.

Conductive electrons are not bound, move according to

mx +m

τx = −eE

m~v +m

τ~v = −e~E

Current density ~J = −Ne~v⇒ ~J +

~J

τ=Ne2

m~E

τ is relaxation time of homogeneous soln ~J = ~J0e−t/τ

static conductivity σNe2

mτ

Drive with harmonic field e−iωt therefore ~J responds harmonically as e−iωt

(−iω +

1

τ

)~J =

Ne2

m~E =

σ

τ~E

~J = σ1−iωτ

~E when ω = 0, ~J = σ~E


Plug into wave eqn

∇2~E− 1

c2∂2~E

∂t2= µ0

∂~J

∂t=

µ0σ

1− iωτ

∂~E

∂tǫe = ǫ + σ0

iω(1+iωτ)

Plane wave soln ~E = E0ei(Kz−ωt)

K2 =ω2

c2+

iωµ0σ

1− iωτµoǫoµ0

N2 = 1 +

︷︸︸︷µ0c

2 Ne2

m iωτ

ω2 − iωτ= 1− ω2

p

ω2 + iωτ−1

plasma frequency ωp =√

Ne2

mǫo=√

µ0σc2

τ

√σǫoτ

n2 − a2 = 1− ω2p

ω2 + τ−22na =

ω2p

ω2 + τ−2

1

ωτ

typically τ ≈ 10−13s 2πωp

≈ 10−15s

1

na

ωωp

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