Chapter 9: Electromagnetic Waves The Wave Equation 9.1 Waves in One Dimension 9.1.1 The Wave Equation How to represent such a “wave” mathematically? What is a “wave”? Hint: The wave at different times, once at t=0, and again at A start: A wave is disturbance of a continuous medium that some later time t --- each point on the wave form simply shifts propagates with a fixed shape at constant velocity. to the right by an amount vt, where v is the velocity. In the presence of absorption, the wave will diminish in initial shapefz ( ,0)= gz ( ) size as it move; If the medium is dispersive different frequencies travel at subsequent formfzt ( , ) = ? different speeds; (capture (mathematically) the Standing waves do not propagate; fzt(,)=− fzvt ( ,0) =− gzvt ( ) essence of wave motion.) Light wave can propagate in vacuum;… The function f(z,t) depends on them only in the very special combination z-vt; When that is true, the function f(z,t) represents a wave of fixed shape traveling in the z direction at speed v. 1 2

The Wave Equation (II) The Wave Equation of a String

2 fzt(,)= Ae−−bz() vt From Newton’sθ second law we have 1 θ fzt(,)=− A sin[( bzvt )] θ ∂2 y Examples: 2 F[sin( +∆ )−sin( )]= (µ∆x) A 2 fzt(,)= ∂t 3 bz()1−+ vt2 Small angle approximation: How about these functions? θ 2 ∂y fzt(,)= Ae−+bz() vt sin ≈ tanθ = 4 ∂x f(,) z t= A sin()cos( bz bvt ) ∂2 y ∂2 y 5 = (µ / F) A ∂x2 ∂t2 =++−[sin(bz ( vt )) sin( bz ( vt ))] a standing wave 2

3 4 The Wave Equation 9.1.2 Sinusoidal Waves

Derive the wave equation that a disturbance propagates wave speed without changing it shape. (i) Terminology fztgzvt( , )=− ( ); Let u ≡− zvt fzt(,)=−+ A cos[( kz vt )δ ]

22 ∂∂f df udgf ∂ ∂dg 2 d g ==−vvv ⇒ =-()= amplitude wave number phase constant ∂∂t du t du ∂∂ t22 t du du ∂∂f df u dg ∂∂22 f dg d g fzt( , )=−+=−+ A cos[ kz ( vt )δ ] A cos( kzωδ t ) ==⇒ 22 = ( )= ∂∂z du z du ∂∂ z z du du 2π k = , λ : wave length dg22211∂∂ f f ∂ 2 f ∂ 2 f λ ==⇒−= 0 qed v duvt222∂∂ z 2 ∂ z 2 vt 22 ∂ ωππ==kv2=2 f

+−vv or  : angular frequency ω λ fzt( , )=−++ gz ( vthzvt ) ( ) the wave equation is linear.  5  f : frequency 6

Sinusoidal Waves Example 9.1 (ii) Complex notation The advantage of the complex notation. Suppose we want to combine two sinusoidal waves: Euler’s formulaδωeiiθ =+ ωcosθ sinθ ffffffff312=+=Re[ 1 ] + Re[ 2 ] = Re[ 12 + ] = Re[ 3 ] fzt(,)=−+= A cos[( kz vt )δ ] Re[ Aeikz()−+ωδ t ] iikzt()−− ikzt () Simply add the corresponding complex wave functions, and ==Re[Ae ] Re[ Ae ] take the real part. In particular, when they have the same frequency and wave ikz()−ω t fAe ≡ complex wave function number AAe ≡ iδ complex amplitude ikzt()−−−ω ikzt ()ωω ikzt () fAe31=+= Aeδ 2 Ae 3 i ii 3 12δδ fzt(,)= Re[(,)] fzt where AAeAeAe33==+ 1 2

The advantage of the complex notation is that exponentials Try doing this without using the complex notation. are much easier to manipulate than sines and cosines. 7 8 9.1.3 Boundary Conditions: Sinusoidal Waves (III) Reflection and Transmission

(iii) Linear combinations of sinusoidal waves ikz()1 −ω t Incident wave: fztII(,)= Ae

∞ ikzt()−−1 ω ikz()−ω t Reflected wave: fztAe (,)= fzt( , )== Ake ( ) dk , where ωω ( k ) RR ∫−∞ ikz()2 −ω t Transmitted wave: fztAeTT(,)= Ak( ) can be obtained in terms of the initial conditions ∗ All parts of the system are oscillating at the same frequency ω. fz( ,0) and fz ( ,0) from the theory of Fourier transforms. The wave velocities are different in two vkλ regimes, which means the wave lengths 121== vkλ Any wave can be written as a linear combination of and wave numbers are also different. 212 sinusoidal waves. The waves in the two regions:

ikzt()11−−−ωω ikzt ( ) So from now on we shall confine our attention to sinusoidal AeIR+ Ae for z < 0 fzt(,)=  waves. ikz()2 − t  AeT for z > 0 9  ω 10

Boundary Conditions Determine the Complex Boundary Conditions Amplitudes Mathematically, f(z,t) is continuous at z=0. ftft(0−+ , )=⇒+= (0 , ) AAA ftft(0−+ , )= (0 , ) IRT df df The derivative of f(z,t) must also be continuous at z=0. =⇒−= kA ( A ) kA dz dz 12IR T 00−+ df df Why?  kk−− vv = A==()()12AA 21 dz00−+ dz  RII AA+= A   kk12++ vv 21 IRT ⇒ kA()−= A kA 22kv 12IR T A==()()12AA  TII  kk12++ vv 21

When v2>v1, all three waves have the same phase angle. The complex wave function obeys the same rules: Why? When v2

Transverse waves occur in two independent states of polarization: ikz()−ω t ikz()−ω t ˆ fxv (,)zt= Ae ˆ fyh (,)zt= Ae

General form: fnnxy(zt , )==+ Aeikz()−ω t ˆˆ , where cosθ ˆˆ sinθ

13 14

9.2 Electromagnetic Waves in Vacuum Right and Left Hand Circular Polarizations 9.2.1 The Wave Equation for E and B In regions of space where there is no charge or current, Maxwell’s equations read ∂B (i) ∇⋅EE =0 (iii) ∇× =− ∂t ∂E (ii) ∇⋅BB = 0 (iV) ∇× =µε 00∂t ∂∇×()BE ∂2 ∇×( ∇×EEE ) =− ⇒ ∇∇⋅ ( ) −∇2 =−µε ∂∂tt00 2 2 ∂∇×()EB2 ∂ ∇×( ∇× µεBBB ) =µε 00 ⇒ ∇ ( ∇⋅ ) −∇ =− 00 2 ∂∂tt2  2 ∂ E ∇=E µε00 2 ∇⋅E =0  ∂t since   2 ∇⋅B =0 ∂ B  ∇=2B 15  00µε∂t 2 16 The Wave Equation for E and B Hertz’s Experiment In vacuum, each Cartesian component of E and B satisfies When Maxwell’s work was published in 1867 it did not the three-dimensional wave equation receive immediate acceptance. It is Hertz who conclusively 2  2 ∂ E demonstrated the existence of electromagnetic wave. ∇=E µε00 2 2  ∂t 2 1 ∂ f  ⇒∇ f = ∂2B vt22∂ ∇=2B  00µε∂t 2 Maxwell’s equations imply that empty space supports the propagation of electromagnetic waves, traveling at a speed 1 v ==×3108 m/s the speed of light µε00

17 18

9.2.2 Monochromatic Plane Waves The Electromagnetic Spectrum Since different frequencies in the visible range correspond to Electromagnetic waves span an immerse range of different colors, such waves are called monochromatic. frequencies, from very long wavelength to extremely high This definition can be applied to the whole spectrum. A wave energy with frequency 1023 Hz. There is no theoretical limit of single frequency is called a monochromatic wave. to the high end.

19 20 Mainly Heating Effect in Micro/mm-Wave Spectrum Windows for Research and Application Opportunities

21 22

Spectrum to Be Exploited Monochromatic Plane Waves --- Significance of the Electron Cyclotron Maser Consider a monochromatic wave of frequency ω and the one photon multiple-photon multiple photon wave is traveling in the z direction and has no x or y per excitation, per electron, per electron, dependence, called plane waves. Plane waves: the fields are uniform over every plane large interaction large interaction interaction space space space ~ wavelength perpendicular to the direction of propagation. ↓ ↓↓ Are these waves common? Yes, very common.  ikz()−ω t EE(,)zt= 0 e  where EB00 and are the complex amplitudes. ikz()− tω BB(,)zt= 0 e THz gap

23 24 Transverse Electromagnetic Waves Transverse Electromagnetic Waves (II) ∂E Q: What is the relation between E and B? Ampere’s law with Maxwell’s correction: ∇×B = µε 00∂t ∂E ∇⋅E =0z = (Eike )ikz()−ω t = 0 ⇒ ( E ) = 0 ∂∂()BE∂()B () ∂z 00zz xˆ : 00zx−=0 µεωy µε ⇒ kB ( ) = ( E ) ∂∂yz00 ∂ t 0yx 00 0 ∂B ikz()− t z ω ∇⋅B =0 = (Bike00 )zz = 0 ⇒ ( B ) = 0 ∂∂()BB () ∂()E ∂z ˆ 00xz 0 y y : −= µεω00µε ⇒ kB ( 0 )x =− 00 ( E 0 ) y That is, electromagnetic waves are transverse: the electric ∂∂zx ∂ t ∂()B ∂∂()BE () and magnetic fields are perpendicular to the direction of zˆ : 0 y −=00xz ⇒= 0 0 ∂B 00µε propagation. Moreover, Faraday’s law ∇×E =− ∂∂xy ∂ t ∂t ω 1 ∂∂()EB∂()E () In free space, the speed of light is c == xˆ : 00zx−=−⇒0 y kE ( ) =−ω ( B ) k µ ε ∂∂yz ∂ t 00yx 00 k 1 ∂()B More compactly, BzEzEEB=×=×⇒⊥()()ˆˆ ∂∂()EE00xz () 0 y 00 0 yˆ : −=−⇒ kE (00 )x = (ω B ) y ω c ∂∂zx ∂ t 1 amplitude relation: BE= ∂()E0 y ∂∂()EB () 00 zˆ : −=−⇒=00xz 0 0 25 c 26 ∂∂xy ∂ t

Example 9.2 Plane Waves Traveling in an Arbitrary Direction Prove: If E points in the x direction, There is nothing special about the z then B points in the y direction. direction---we can generalize to monochromatic plane waves traveling in ikz()−ω t Sol: Ex(,)zt= Ee0 ˆ an arbitrary direction.

k 11ikz()−−ωω t ikz () t BzE000=×=()ˆˆˆˆEe () zx ×= Ee 0 y The propagation (or wave) vector, k: pointing in the ω cc direction of propagation. Take the real part: Generalization of kz: using the scalar product k·r.

it()kr⋅−ω Ex(,)zt=−+ E0 cos( kzωδ t )ˆ Er(,)tEe= 0 nˆ ← the polarization vector

1 11it()kr⋅− B(,)zt=−+ E cos( kzω t δ )yˆ Br(,)tEe=×=×ω ( kˆˆ nˆ ) k E c 0 cc0 Q: Why not use sin function? Q: Can you write down the real electric and magnetic fields? 27 28 9.2.3 Energy and Momentum in Energy Transport and the Poynting Vector Electromagnetic Waves The energy per unit volume stored in the electromagnetic Consider two planes, each of area A, a distance dx apart, field is and normal to the direction of propagation of the wave. The 1122 total energy in the volume between the planes is dU=uAdx. uEB=+()ε0 2 0 The rate at which this energy through a unit area normal to µ 1 Monochromatic plane wave: BE22==µε E 2 the direction of propagation is c2 00 1 dU 1 dx 11122 222 S = = uA = uc uEB=+=+=()()εεεε0000 EEE A dt A dt 220 µ EB Their contributions are equal. S = uc = µ0 222 uE==ε000εωδ Ecos ( kzt −+ ) E×B S = (the vector form) As the wave travels, it carries this energy along with it. µ0

1122 1 Q: How about the momentum? See next slide. gS==22εωδ00E cos (kz −+= t ) zzˆˆ u 29 cc c 30

Average Effect Example In the case of light, the period is so brief, that any A radio station transmits a 10-kW signal at a frequency of 100 macroscopy measurement will encompass many cycles. MHz. For simplicity, assume that it radiates as a point source. At a distance of 1 km from the antenna, find: (a) the amplitude All we want is the average value. of the electric and magnetic field strengths, and (b) the energy 1 2 incident normally on a square plate of side 10 cm in 5 min. uE= ε00 2 Solution: 1 2 2 Sz==uc cε00 E ˆ Average power E0 2 ()aS av ==2 11 42πµrc0 gS==ε E 2 zˆ 2 00 10000π cc2 ⇒×××××=24 10310−782E 410002 π 0 The average power per unit area transported by an E0 = 0.775 V/m electromagnetic wave is called the intensity:  −9 B0 =×2.58 10 T 1 2 −3 IcE≡=S ε ()bUSAt ∆=av ∆=× 2.410 J 2 00 31 32 Momentum and Radiation Pressure Momentum and Radiation Pressure (II) An electromagnetic wave transports linear momentum. The radiation pressure at normal incident is F S The linear momentum carried by an electromagnetic wave is = = u related the energy it transport according to A c U p = Examples: (a) the tail of comet, (b) A “solar sail” c If surface is perfectly reflecting, the momentum change of the wave is double, consequently, the momentum imparted to the surface is also doubled. The force exerted by an electromagnetic wave on a surface may be related to the Poynting vector F ∆p ∆U SA S = = = = = u A A∆t Ac∆t Ac c 33 34

Homework of Chap.9 (I) 9.3 Electromagnetic Waves in Matter 9.3.1 Propagation in Linear Media In regions where there is no free charge and free current, Maxwell’s equations become ∂B ∇⋅DE =0 ∇× =− Prob. 2, 6, 8, 10, 12 ∂t ∂D ∇⋅BH =0 ∇× = ∂t 1 If the medium is linear, DE==ε and H B µ ∂B If the medium is linear and ∇⋅EE =0 ∇× =− homogeneous (ε and µ do not vary ∂t from point to point), ∂E ∇⋅BB =0 ∇× =µε 35 36∂t The Index of Refraction Energy Density, Poynting Vector, and Intensity in Linear Media Electromagnetic waves propagate through a linear All of our previous results carry over, with the simple homogeneous medium at a speed 1 c transcription 1  ∂2E v == 1122 gz= uˆ ∇=2E µε uEB=+()ε  2 2 µε n ε0 → ε 2 v  ∂t 2 1 ∂ f ⇒∇ f =  2 22 µ → µ µ ∂ B vt∂ µε 0 EB× 1 2  2 n ≡ S = IvE≡=S ε ∇=B µε 2 0  ∂t 00µε cv→ µ 2

The index of refraction of the material Q: What happens when a wave passes from one ε transparent medium into another? Boundary conditions. For most material, µ is very close to µ0, so n ≅=ε r ε0 Since ε is almost always greater than 1, light travels ⊥⊥ // // r DD12f−=σ EE 12 −=0 more slowly through matter. ⊥⊥ // // BB12−=0()HH 1 −= 2 Knf ׈

Q: What happens when εr is less than 1 or negative? Very interesting. 37 38

9.3.2 Reflection and Transmission at Normal Incidence The Boundary Conditions A plane wave of frequency ω, traveling in the z direction and Normal incident: no components perpendicular to the surface. polarized in the x direction, approaches the interface from the left. EE00IRT+= E 0 11 Incident wave: ()E00IR−=EE 0 T µµ11vv 2 2 Ex(,)zt= E eikz()1 −ω t ˆ µ v II0 ⇒−= (EE )ββ E , where =11 1 00IR 0 T v ikz()1 − t 22 ByII(,)zt= E0 e ωˆ v1 In terms of the incident amplitude: µ Transmitted wave: Reflected wave: 1− β vv21− ikz()−ω t E = ()E E = ()E ikzt()−−ω 2 00R I 00R I 1 ExTT(,)zt= E0 e ˆ 1+ β vv+ ExRR(,)zt= E0 e ˆ if µ µ ⇒ 12 1 0 1 ikz()2 − t 2 2v ikzt()−−1 ˆ 2 ˆ ByTT(,)zt= E0 e ω E00TI= ()E E = ()E ByRR(,)zt=− E0 e ω 00TI v2 1+ v1 39 β vv12+ 40 Determine the Complex Amplitudes of a String Reflection and Transmission Coefficients

ikz()1 −ω t The reflected wave is in phase if v2 >v1 Incident wave: fztII(,)= Ae and is out of phase if v2

22vn21 Boundary −+df df ftft(0 , )= (0 , ) = E000TI==()()EE I conditions: dz dz vv12++ nn 12 00−+ 1 The intensity (average power per unit area) is: IvE≡=S ε 2  kk−− vv 0 A==()()12AA 21 2  RII Inn− AA+= A   kk12++ vv 21 R 122 IRT ⇒ Reflection coefficient R ≡= ( ) 22kv InnI 12+ kA12()IR−= A kA T  12 ATII==()()AA  IT ε 22vn24 12 nn 12  kk12++ vv 21 Transmission coefficient T ≡= ( ) = I vn++ n() n n2 When v2>v1, all three waves have the same phase angle. I 11ε 1 2 1 2 41 42 When v2

9.3.2 Reflection and Transmission at Oblique Incidence Boundary Conditions Suppose that a monochromatic plane wave of frequency ω, All three waves have the same frequency ω. traveling in the kI direction vn21 ω ===kvI 112 kvRT kv or k IRT === k k k T Incident wave: vn12

it()krI ⋅−ω Using the boundary conditions ErII(,)te= E0 1 ⊥⊥ // // ˆ εεEE−=00EE −= BrIII(,)t =× ( k E ) 11 2 2 1 2 v1 ⊥⊥ 11// // BB12−=00BB 1 − 2 = 12 Reflected wave: Transmitted wave:

it()krR ⋅−ω it()krT ⋅−ω ErRR(,)te= E0 ErTT(,)te= E0 A generic structure for the four boundaryµµ conditions. 1 1 ˆ ˆ it()kr⋅−ω i ( k ⋅− rωω t ) it () kr ⋅− BrR (,)t =× ( kRR E ) BrTTT(,)t =× ( k E ) ( )eeIRT+= ( ) ( ) e v1 v2 43 44 Laws of Reflection and Refraction Boundary Conditions (ii)

θθ, θ , and are angles ( )eeit()krIRT⋅−ω += ( ) i ( k ⋅− rωω t ) ( ) e it () kr ⋅− krkrkr⋅= ⋅= ⋅ at z=0 IR T IRT of incidence, reflection, We have taken care of the exponential factors—they cancel. kksinθ == sinθθ k sin IIRRTTand refraction, respectively. The boundary conditions become: (i) εε10 (EEIRzTz+= 0 ) 20 (E ) Normal D The law of reflection: θIR= θ (ii) (BB00IRzTz+= ) ( B 0 ) Normal B (iii) (EE00,0,IRxyTxy+= ) (E ) Tangential E sinθ kvn The law of refraction: TI11 11 === (iv) (BB00,IRxyTxy+= ) ( B 0, ) Tangential H (Snell’s law) sinθITkvn22 12 µµ 1 Common properties of waves: These equations are where Br ( ,t )=× ( kEˆ ) obtained from their generic form. 00v 45 46

Parallel to the Plane of Incidence Parallel to the Plane of Incidence (ii) Q: If the polarization of the incident wave is parallel to the cosθT plane of incidence, are the reflected and transmitted (iii) (EE00IR+= )αα ( E 0 T ) ≡ waves also polarized in this plane? Yes. cos I v (iv) (EE−= ) ( E ) ≡11 00IRββ 0 T v θ αβ 22 − 2µ ⇒=E( )EE = ( ) EFresnel’s equations 0000RITI++αβ αβ µ

How about the first boundary condition? Normal D (i) ε (−+EE sinθθεθ sin ) =− ( E sin ) 10IIRR 0 20 TT Does this condition contribute anything new? Tangential E (iii) (EE00IIRR cosθθ+= cos ) ( E 0 TT cos θ ) Normal B (ii) 0=0 ε 2 sinθT ε 211sinθµT v (i) (EE00IR−= )εθ (E 0 T ) εθµ= ? 11 1 sin I 1 sin I 22v Tangential H (iv) (µµEE00IR−= ) ( E 0 T ) 47 48 11vv 2 2 Brewster’s Angle Brewster’s Angle (II) αβ αβ  αβ− If µ ≅≅µβ θ , β βnn / and sin222 ≅+ /(1 ) EE= () 12 21 B  00RI θµ n  αβ+ cosθT µ11v 2 where ≡≡ and ⇒≅ tan B  n  2 cos I 22v θ 1 EE00TI= ()  + When α =β, there is no reflected wave. E0R = 0 cosθT µ11v µθθµ== when IB (called Brewster's angle) cos I 22v θθ v sinθθθµθ cosθµθ sin From Snell's lawβθ 21=⇒TTB θθ = v12sinBBT cos sin 22 2 12sin B 22v 222 1− 22= and sinTBB=⇒= ( ) sin sin 22 µ2112sinθβT vnn ( / ) −49 50 β

Transmission and Reflection Perpendicular to the Plane of Incidence 1 ˆ 2 Q: If the polarization of the incident wave is perpendicular IvEIII≡⋅=Sz 11εθ 0 cos I αβ− 2 R ≡=R ()2 to the plane of incidence, are the reflected and transmitted εθ I + 1 22− I αβ waves also polarized in this plane? Yes. IvERRRI==11 0 cos ( ) I 2 αβ+ IT 2 2 T ≡=αβ() 12I α + β IvE==22εθαβcos (αβ ) I I TTTI2 22 0 +

αβ

51 See Problem 9.16 52 9.4 Absorption and Dispersion Electromagnetic Waves in Conductors (II) 9.4.1 Electromagnetic Waves in Conductors ∂ρ When wave propagates through vacuum or insulating The continuity equation for free charge: f =−∇⋅J ∂t f materials such as glass or teflon, assuming no free charge conductivity and no free current is reasonable. JEf = σ ∂ρ ρ ff=−σ () ∇⋅E =−σρ =− ∂t σ f But in conductive media such as metal or plasma, the free ρ f =∇⋅ε E εε charge and free current are generally not zero. t − τ The free current is proportional to the electric field: Ohm’s law For a homogeneous linear medium: ρρff()te= (0) JE= σ where = f σ Classification of conductors: τ ε Maxwell’s equation for linear media assume the form superconductor σ =∞, τ =0 What’s the difference? ρ ∂B στ ∇⋅EE =f ∇× + =0 perfect conductor=∞ , = 0 See Prob. 7.42 ∂t ε τω −19 ∂E good conductor << τ ≈10 s for copper ∇⋅BB =0 ∇× −µεµσ = E τω −14 ∂t 53 poor conductor >> τc ∼ 10 s collision time 54

Electromagnetic Waves in Conductors (III) Electromagnetic Waves in Conductors (IV) Omitting Transient Effect Complex Wave Number Omit the transient behavior. These equations still admit plane-wave solutions, ρ = 0 Assume no charges accumulation: f ∂∂2EE ∇=2E µσµε + ikz() −ω t ∂B 2 EE(,)zt= e ∇⋅EE =00 ∇× + = ∂∂tt  0 ∂t 2  ikz() −ω t 2 ∂∂BB BB(,)zt= 0 e ∂E ∇=B µσµε 2 + ∇⋅BB =0 ∇× =µεµσ + E ∂∂tt ∂t ∂∂∇×BB() Note this time the " wave number" k is complex: ∇×()() ∇×EE + =∇× ∇× + ∂∂tt ki22=+µεω µσω  22 κ ω σεµ 2 ∇×()() ∇×EEEE =∇ ∇⋅ −∇ =−∇ k ≡++1( ) 1 =0  2  ∂∇×()BEE ∂2 ∂ kki =+, where =+ µσµε  εω 2  κω ∂∂∂ttt ≡+− σεµ 1( )2 1 22   22∂∂EE ∂∂ BB 2 ∇=EB µσµε µσ µε +, ∇= + (likewise)55  56 ∂∂tt22 ∂∂ tt εω The Real Parts of The Fields 9.4.2 Reflection at a Conducting Surface k ⊥⊥ // // −−κωzikzt() κω −− zikzt () εε11EE−= 2 2 σ fEE 1 −= 2 0 EE(,)zt=00 e e Faraday's law→= B (,) zt E e e ω 11 BB⊥⊥−=0 BBKn// − // =׈ kki =+κ = Keiφ 12 1 2 f 12 Where σ is the free surface charge, K is the free surface K ≡+=kk221( + ) 2 and ≡ tan(/)− 1 f f κωεµ φ κ current, and n ˆ is a unit vectorµµ perpendicular to the surface, δδ pointing from medium (2) into medium (1). kKe iφ BE(,)zt=⇒ BeiiBE =σ Ee 00 Normal ωωεω δδφ εµ incident B0 K 2 BE−= and == 1 + ( ) E0 ω εω (1) nonconducting (2) conductor −κ z linear medium Ex(,)zt=−+ Ee0 cos( kzωδ t E )ˆ

K − z σ B(,)zt=−++ Eeκ cos( kzω t δφ )yˆ ω 0 E 57 58

Reflection at a Conducting Surface (II) Reflection at a Conducting Surface (III) Incident wave: Tangential components of the fields at z=0: ikz()−−ωω t1 ikz () t 11 // // ExByII(,)zt== Ee00ˆˆ , I (,) zt Ee I EE12−=0 E+=EE v1 00IRT 0 11// // Reflected wave: BBKn−=׈ 1 k2 1 12f ()E00IR−−EEK 0 Tf = ikzt()−−11ωω ikzt () −− µµ12 µµωv ExByRR(,)zt==− Ee00ˆˆ , R (,) zt Ee R 11 2 v1 // with K f = 0, why? KEf ∝=0 1 Transmitted wave: − β κκ E00R = ()E I k E+=EE 1+ β ExBy(,)zt== E e−−zz eikz()22−−ωω tˆˆ , (,) zt2 E e e ikz () t 00IRT 0 00TT µ v 2 ω (E−=EE )ββ , where ≡11 k E= ()E 00IR 0 T 2 00TI1+ µ2ω β Normal components of the fields ⊥⊥ For a perfect conductor (σ=∞), k2=∞ Î EE00RI=− and E 0 T = 0 εε11EE−= 2 2 σf ⊥ ⊥⊥ ⇒=f 0 (0E1,2 = ) BB12−=0 59 That’s why excellent conductors make good mirrors. 60

σ 9.4.3 The Frequency Dependence of Permittivity The Group Velocity and Phase Velocity When the speed of a wave depends on its frequency, the When two waves of slightly different frequencies are supporting medium is called dispersive. superposed, the resulting disturbance varies periodically in amplitude.

Akkzsin((00+∆ ) − (ω +∆ωωω ) tAkkz ) + sin(( 00 −∆ ) − ( −∆ ) t )

=Asin((kz00 −ω t ) +ωω ( ∆ kz −∆ωωω t )) + A sin(( kz 00 − t ) − ( ∆ kz −∆ t ))

=∆−∆−2Akztkzt cos[( )]sin[(00 )]

ω0 Phase velocity vp = k0 ∆ω dω Group velocity v == g ∆kdk

61 62

Interference in Time: Beats Simplified Model for the Frequency Dependence of Permittivity in Nonconductors When two waves of slightly different frequencies are superposed, the resulting disturbance varies periodically in The electrons in a nonconductor are amplitude. bound to specific molecules. y = y1 + y2 =πAsin(2πf1t) + Asin(2πf2t) f − f f + f = 2Acos[2 ( 1 2 )t]sin[2π( 1 2 )t] The simplified binding force: Fkxmx=− =− ω 2 2 2 binding spring 0 Beat frequency (|f1-f2|): frequency of the amplitude Is this model oversimplified to you? dx envelope The damping force on the electron: Fm=− γ damping dt

The driving force on the electron: FqEqEtdriving== 0 cos(ω )

dx2 Newton's law: mFFF2 ==tot binding + damping + F driving 63 dt 64 Permittivity in Nonconductors Permittivity in Nonconductors (II)

The equation of motion N molecules per unit volume; each molecule contains fj electrons with frequency ω and damping γ . dx2 dx j j mmmxqEt++=γ ωω2 cos( ) The polarization P is given by the real part of: dt2 dt 00 ωω γω χNq2 f dx2 dx PEE==j εχ Re mmmxqEe++=γω2 −itω miεωωγω∑ 22−− 0 e 2 00 j jj dt dt Nq2 f Let the system oscillates at the driving frequency ω =←j the complex susceptibility e mi∑ 22−− qm/ 0 j jj −itω xxe==00, where x22 E 0 ωω0 −− γωi theεχ complex permittivity ε =εχ0 (1+ e ) The dipole moment is the real part of pqxt= ( ) the complex dielectric constant 2 2 Nq f j q 1 −itω re=(1+=+ ) 1 22 pEe= 22 0 ∑ miωω−− γω ε0mij ωωjj−− γω 0 65 66

Waves in a Dispersive medium Anomalous Dispersion The wave equation for a given frequency reads The index of refraction:

2 2 2 22 ∂ E Nq f j ck Nq f ()ωω− ∇=2E µε εε=(1) χ ε+=+  n =≅+1 jj ∂t 2 00e mi∑ ω 22−−ωγω ω 2()mεωωγω∑ 22222−+ j jj ακ 0 j jj −−κωzikzt() 222 kki≡=+εµω κ EE(,)zt= e e Nq ω f jj 0 0 =≅2  ∑ 22222 mc0 εωωγωj ()jj−+ −2κ z γ IIe≡=S 0 , ακ ≡ 2 (absorption coefficient) In the immediate neighborhood of a resonance, the index of For gases, the second term of ε is small refraction drops sharply. Í called anomalous dispersion. ωω1 ωNq2 f k ≡≅+=+(1 ) 1 j εεrr ∑ 22 Faster Than Light (FTL): cc22 cmε 0 j ωωjj−− γω i Can we find cases where the waves propagate at a speed The binomial expansion faster than light? Superluminal effect. 67 68 9.5 Guided Waves General Properties of Wave Guides 9.4.1 Wave Guides In the interior of the wave guide, the waves satisfy Maxwell’s Can the electromagnetic waves propagate in a hollow metal equations: ∂B pipe? Yes, wave guide. ∇⋅EE =00 ∇× + = ∂t Why ρ ff==0 and J 0? Waveguides generally made of good 1 ∂E 1 conductor, so that E=0 and B=0 inside ∇⋅BB =0 ∇× = where v = vt2 ∂ the material. εµ We obtain The boundary conditions at the inner wall are: E// ==0 and B⊥ 0 …

The generic form of the monochromatic waves:  ikzt()−−ω ikzt ()ω EE(,xyzt ,,)==++0 (, xye ) ( Exyz xyzˆˆˆ E E ) e  ikzt()−−ω ˆˆˆ ikzt ()ω BB(,xyzt ,,)==++0 (, xye ) ( Bxyz xyz B B ) e The confined waves are not (in general) transverse. 69 70

TE, TM, and TEM Waves No TEM Waves in a Hollow Wave Guide

Determining the longitudinal components Ez and Bz, we could Proof: quickly calculate all the others. ∂Ex ∂Ey If Ez=0, Gauss’s law says +=0 ∂∂xy ∂2 E ∂2 E ⇒+=x y 0 ∂E ∂E ∂∂xy22 If B =0, Faraday’s law says y −=x 0 z ∂∂xy

The boundary condition on E requires that the surface be an equal-potential. Laplace’s equation admits no local maxima or minima.

We obtain Î the potential is constant throughout. Ez=0 — no wave at all. ∂∂222ω ++−kE2 =0 222 z If Ez =⇒ 0 TE (transverse electric) waves; ∂∂xyv A hollow wave guide cannot support the TEM wave. 222 If Bz =⇒ 0 TM (transverse magnetic) waves; ∂∂ ω 2 Can a metal wire support the TEM wave? Yes. 222++−kBz =0 If EBzz==⇒ 0 and 0 TEM waves. ∂∂xyv 71 72 A Diagram of the Optical Setup 9.5.2 TE Waves in a Rectangular Wave Guide

EBxyXxYyzz==←0, and ( , ) ( ) ( ) separation of variables

11∂∂222XYω ++−=()0k 2 Xx∂∂222 Yy v

11∂∂22XY =−kk22 and =− Xx∂∂22x Yy y ω 2 with =++kkk222 v2 xy

X ()xAkxBkx=+ sinxx cos

Y()y =+Ck sinyyy Dk cos y K. Wang and D. M. Mittleman, “Metal wires for terahertz wave guiding”, Nature, vol.432, No. 18, p.376, 2004. 73 74

TE Waves in a Rectangular Wave Guide (II) TE Waves in a Rectangular Wave Guide (III) εε µµ ∂B ECkyDky∝∝z cos − sin Bz (,x y )= Bmxan0 cos(π / )cos(π y /b ) xyy∂y In vacuum, ===00 and , vc . the cutoff frequency Ey(@==⇒= 0) 0 C 0 x 1 nπ kcmanb=−ωω22 ω, where π 2 = 22 [( / ) 2 + ( / ) 2 ] Eyb(@ ==⇒) 0 sin kbk = 0, = ( n = 0,1, 2,...) c mn mn xyyb ωω If < mn , the wave number is imaginary. ∂Bz EAkxBkxyxx∝∝cos − sin ∂x The lowest cutoff frequency of TE10 mode is: ω10 = caπ / Exy (@==⇒= 0) 0 A 0 mπ The wave velocities are: Exayxx(@ ==⇒) 0 sin kak = 0, = ( m = 0,1,2,...) a ω c vcp == > phase velocity k 1−ωω22 Bxyz ( , )=← B0mn cos( mxaπ / )cos( nybπ / ) the TE mode ω mn d vc==−122 < c group velocity kvmanb=−(/)ωπ22 [(/) 2 + (/)] 2 gmndk ωω 75 76 Why the Phase Velocity Greater Than The Field Profiles: Examples The Speed of Light ω c vc== > phase velocity p k 22 1− mn d ωω vc==−ω 122 < c group velocity gmndk 2 vvpg= c ωω

77 78

9.5.3 The Coaxial Transmission Line The Coaxial Transmission Line (II) The problem is reduced to two dimensions. A hollow wave guide cannot support the TEM wave, but a coaxial transmission line can. Electrostatic: the infinite line charge; Magnetostatic: an infinite straight current. A A EsB(,ssφφ )==ˆˆ , (, )φ 00scs

Takingφ the real part: Akztcos(−ω ) Es(,szt ,φ ,)= ˆ s Akztcos(− ) B(,szt , ,)= ωφˆ cs ∇⋅EE =0 and ∇× = 0 E =−∇φ electrostatic ⇒ E ∇⋅BBB =0 and ∇× = 0  =−∇ φB magnetostatic 79 80 Homework of Chap.9 (II)

Prob. 16, 18, 19, 29, 30, 35, 38

81