9 Electromagnetic Waves 9.1 Waves in One Dimension 9.1.1 The Wave Equation. 9.1.2 Sinusoidal Waves 9.1.3 Boundary Conditions: Reflection and Transmission 9.1.4 Polarization 9.2 Electromagnetic Waves in Vacuum 9.2.1 The Wave Equation for E and B 9.2.2 Monochromatic Plane Waves 9.2.3 Energy and Momentum in Electromagnetic Waves 9.3 Electromagnetic Waves in Matter 9.3.1 Propagation in Linear Media 9.3.2 Reflection and Transmission at Normal Incidence 9.3.3 Reflection and Transmission at Oblique Incidence 9.4 Absorption and Dispersion 9.4.1 Electromagnetic Waves in Conductors, 9.4.2 Reflection at a Conducting Surface 9.4.3 The Frequency Dependence of Permittivity 9.5 Guided Waves 9.5.1 Wave Guides 9.5.2 TE Waves in a Rectangular Wave Guide 9.5.3 The Coaxial Transmission Line 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.1 Waves in One Dimension Traveling wave 9.1.1 The Wave Equation How to describe a wave that propagates with a fixed shape at constant speed mathematically? Let 𝑓 𝑧,𝑡 represent the displacement of the string at the point z, at time t. 𝑓 𝑧, 0 ≡𝑔 𝑧 𝑓 𝑧,𝑡 =𝑓 𝑧−𝑣𝑡,0 =𝑔(𝑧−𝑣𝑡) The displacement at point 𝑧, at the later time 𝑡, is the same as the displacement a distance 𝑣𝑡 to the left (at 𝑧 − 𝑣 𝑡), back at time t = 0: Any function 𝑓 𝑧,𝑡 that depends only on 𝑧−𝑣𝑡 represents a wave of fixed shape traveling in the 𝑧 direction at constant speed 𝑣. 𝑓 1 𝑧,𝑡 =𝐴 𝑒 −𝑏 𝑧−𝑣𝑡 2 𝑓 2 𝑧,𝑡 =𝐴 sin[𝑏 𝑧−𝑣𝑡 ] 𝑓 3 𝑧,𝑡 = 𝐴 𝑏 𝑧−𝑣𝑡 2 +1 𝑓 4 𝑧,𝑡 =𝐴 𝑒 −𝑏 𝑏 𝑧 2 −𝑣𝑡 2 𝑓 5 𝑧,𝑡 =𝐴 sin 𝑏𝑧 cos 𝑏𝑣𝑡 2 represent travelling wave at constant speed v. do not represent travelling wave at constant speed v. A and b are constants. 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.1 Waves in One Dimension The wave equation 9.1.1 The Wave Equation Why does a stretched string support wave motion? The net transverse force on the segment A string under tension 𝑇 Δ𝐹=𝑇 sin 𝜃′ −𝑇 sin 𝜃 Assuming these angles are small, the sine can be replaced by the tangent: Δ𝐹≅𝑇 tan 𝜃 ′ − tan 𝜃 =𝑇 𝜕𝑓 𝜕𝑧 ​ 𝑧+Δ𝑧 − 𝜕𝑓 𝜕𝑧 ​ 𝑧 ≅𝑇 𝜕 2 𝑓 𝜕 𝑧 2 Δ𝑧 Newton's second law says Δ𝐹=𝑚𝑎=𝜇(Δ𝑧) 𝜕 2 𝑓 𝜕 𝑡 2 𝜇 is the mass per unit length 𝑇 𝜕 2 𝑓 𝜕 𝑧 2 Δ𝑧=𝜇(Δ𝑧) 𝜕 2 𝑓 𝜕 𝑡 2 𝑣= 𝑇 𝜇 𝜕 2 𝑓 𝜕 𝑧 2 = 𝜇 𝑇 𝜕 2 𝑓 𝜕 𝑡 2 𝜕 2 𝑓 𝜕 𝑧 2 = 1 𝑣 2 𝜕 2 𝑓 𝜕 𝑡 2 The wave equation 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.1 Waves in One Dimension The general solution 9.1.1 The Wave Equation The wave equation 𝜕 2 𝑓 𝜕 𝑧 2 = 1 𝑣 2 𝜕 2 𝑓 𝜕 𝑡 2 admits as solutions all functions of the form 𝑓 𝑧,𝑡 =𝑔(𝑧−𝑣𝑡) 𝑢≡𝑧−𝑣𝑡 𝜕𝑓 𝜕𝑧 = 𝑑𝑔 𝑑𝑢 𝜕𝑢 𝜕𝑧 = 𝑑𝑔 𝑑𝑢 𝜕𝑓 𝜕𝑡 = 𝑑𝑔 𝑑𝑢 𝜕𝑢 𝜕𝑡 =−𝑣 𝑑𝑔 𝑑𝑢 𝜕 2 𝑓 𝜕 𝑧 2 = 𝜕 𝜕𝑧 𝑑𝑔 𝑑𝑢 = 𝑑 2 𝑔 𝑑 𝑢 2 𝜕𝑢 𝜕𝑧 = 𝑑 2 𝑔 𝑑 𝑢 2 𝜕 2 𝑓 𝜕 𝑡 2 =−𝑣 𝜕 𝜕𝑡 𝑑𝑔 𝑑𝑢 =−𝑣 𝑑 2 𝑔 𝑑 𝑢 2 𝜕𝑢 𝜕𝑡 = 𝑣 2 𝑑 2 𝑔 𝑑 𝑢 2 𝑑 2 𝑔 𝑑 𝑢 2 = 𝜕 2 𝑓 𝜕 𝑧 2 = 1 𝑣 2 𝜕 2 𝑓 𝜕 𝑡 2 The wave equation involves the square of 𝑣, so 𝑓 𝑧,𝑡 =ℎ(𝑧+𝑣𝑡) is another class of solutions. The most general solution to the wave equation is the sum of a wave to the right and a wave to the left: 𝑓 𝑧,𝑡 =𝑔 𝑧−𝑣𝑡 +ℎ(𝑧+𝑣𝑡) The wave equation is linear: The sum of any two solutions is itself a solution. 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.1 Waves in One Dimension Formula 9.1.2 Sinusoidal Waves 𝑓 𝑧,𝑡 =𝐴 cos [𝑘 𝑧−𝑣𝑡 +𝛿] Phase constant Amplitude Phase Angular wave number At 𝑧=𝑣𝑡− 𝛿 𝑘 , the phase is zero; If 𝛿=0, the central maximum passes the origin at time 𝑡=0; 𝛿 𝑘 is the distance by which the central maximum (and therefore the entire wave) is "delayed." The wavelength 𝜆= 2𝜋 𝑘 At any fixed point z, the string vibrates up and down, undergoing one full cycle in a period 𝑇= 2𝜋 𝑘𝑣 . The frequency 𝜈 (number of oscillations per unit time) is 𝜈= 1 𝑇 = 𝑘𝑣 2𝜋 = 𝑣 𝜆 The angular frequency 𝜔=2𝜋𝜈=𝑘𝑣 𝑓 𝑧,𝑡 =𝐴 cos (𝑘𝑧−𝜔𝑡+𝛿) traveling to the right 𝑓 𝑧,𝑡 =𝐴 cos (𝑘𝑧+𝜔𝑡−𝛿) traveling to the left 𝑓 𝑧,𝑡 =𝐴 cos (−𝑘𝑧−𝜔𝑡+𝛿) traveling to the left we could simply switch the sign of k to produce a wave with the same amplitude, phase constant, frequency, and wavelength, traveling in the opposite direction. 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.1 Waves in One Dimension Euler's formula 9.1.2 Sinusoidal Waves 𝑒 𝑖𝜃 = cos 𝜃+𝑖 sin 𝜃 Euler's formula 𝑖≡ −1 𝑓 𝑧,𝑡 =𝐴 cos [𝑘 𝑧−𝑣𝑡 +𝛿] 𝑓 𝑧,𝑡 =𝑅𝑒[𝐴 𝑒 i 𝑘𝑧−𝜔𝑡+𝛿 ] 𝑅𝑒 𝜉 denotes the real part of the complex number 𝜉. The complex wave function 𝑓 𝑧,𝑡 = 𝐴 𝑒 i 𝑘𝑧−𝜔𝑡 𝐴 ≡𝐴 𝑒 i𝛿 The complex amplitude The actual wave function is the real part of 𝑓 𝑓 𝑧,𝑡 =𝑅𝑒[ 𝑓 𝑧,𝑡 ] The advantage of the complex notation is that exponentials are much easier to manipulate than sines and cosines. 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.1 Waves in One Dimension Example 9.1.2 Sinusoidal Waves Find the amplitude and phase constant of the wave resulted from combing two sinusoidal waves of the same frequency and wave number. 𝑓 3 = 𝑓 1 + 𝑓 2 =Re f 1 +Re f 2 =Re f 1 + f 2 =Re f 3 f 3 = 𝐴 1 𝑒 i 𝑘𝑧−𝜔𝑡 + 𝐴 2 𝑒 i 𝑘𝑧−𝜔𝑡 = 𝐴 3 𝑒 i 𝑘𝑧−𝜔𝑡 𝐴 3 = 𝐴 1 + 𝐴 2 𝐴 3 𝑒 i 𝛿 3 = 𝐴 1 𝑒 i 𝛿 1 + 𝐴 2 𝑒 i 𝛿 2 𝐴 3 𝑒 i 𝛿 3 = 𝐴 1 cos 𝛿 1 +𝑖 𝐴 1 sin 𝛿 1 + 𝐴 2 cos 𝛿 2 +𝑖 𝐴 2 sin 𝛿 2 𝐴 3 𝑒 i 𝛿 3 = 𝐴 1 cos 𝛿 1 + 𝐴 2 cos 𝛿 2 +𝑖( 𝐴 1 sin 𝛿 1 + 𝐴 2 sin 𝛿 2 ) = 𝐴 1 2 + 𝐴 2 2 +2 𝐴 1 𝐴 2 ( cos 𝛿 1 cos 𝛿 2 + sin 𝛿 1 sin 𝛿 2 ) 𝐴 3 = 𝐴 1 cos 𝛿 1 + 𝐴 2 cos 𝛿 2 2 + 𝐴 1 sin 𝛿 1 + 𝐴 2 sin 𝛿 2 2 𝐴 3 = 𝐴 1 2 + 𝐴 2 2 +2 𝐴 1 𝐴 2 cos ( 𝛿 1 − 𝛿 2 ) tan 𝛿 3 = 𝐴 3 sin 𝛿 3 𝐴 3 cos 𝛿 3 𝛿 3 = tan −1 𝐴 1 sin 𝛿 1 + 𝐴 2 sin 𝛿 2 𝐴 1 cos 𝛿 1 + 𝐴 2 cos 𝛿 2 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.1 Waves in One Dimension Linear combinations of sinusoidal waves 9.1.2 Sinusoidal Waves Any wave can be expressed as a linear combination of sinusoidal ones: 𝜔 is a function of 𝑘. 𝑘 runs through negative values in order to include waves going in both directions. 𝑓 (𝑧,𝑡)= −∞ ∞ 𝐴 (𝑘) 𝑒 i 𝑘𝑧−𝜔𝑡 𝑑𝑘 The formula for 𝐴 (𝑘), in terms of the initial conditions 𝑓(𝑧, 0) and 𝑓 (𝑧, 0), can be obtained from the theory of Fourier transforms. Therefore if you know how sinusoidal waves behave, you know in principle how any wave behaves. So from now on we shall confine our attention to sinusoidal waves. 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.1 Waves in One Dimension Two tied strings 9.1.3 Boundary Conditions: Reflection and Transmission Suppose that a string is tied onto a second string. The tension 𝑇 is the same for both stings. The mass per unit length 𝜇 is not the same. The wave velocities 𝑣 1 = 𝑇 𝜇 1 and 𝑣 2 = 𝑇 𝜇 2 are different. reflected pulse transmitted pulse Incident pulse Knot 𝑧=0 knot Assume the incident wave f 𝐼 (𝑧, 𝑡) is a sinusoidal oscillation that extends all the way back to 𝑧=−∞, and has been doing so for all time. Assume the same goes for f 𝑅 and f 𝑇 (except f 𝑇 , extends to 𝑧=+∞). The incident wave 𝑓 𝐼 (𝑧,𝑡)= 𝐴 𝐼 𝑒 i 𝑘 1 𝑧−𝜔𝑡 (𝑧<0) All parts of the string are oscillating at the same frequency 𝜔 (a frequency determined by the person at 𝑧=−∞, who is shaking the string in the first place). The reflected wave 𝑓 𝑅 (𝑧,𝑡)= 𝐴 𝑅 𝑒 i −𝑘 1 𝑧−𝜔𝑡 (𝑧<0) The transmitted wave 𝑓 𝑇 (𝑧,𝑡)= 𝐴 𝑇 𝑒 i 𝑘 2 𝑧−𝜔𝑡 (𝑧>0) 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.1 Waves in One Dimension Boundary conditions 9.1.3 Boundary Conditions: Reflection and Transmission With incident and reflected waves of infinite extent traveling on the same piece of string, it's going to be hard to tell them apart. 𝑓 (𝑧,𝑡)= 𝐴 𝐼 𝑒 i 𝑘 1 𝑧−𝜔𝑡 + 𝐴 𝑅 𝑒 i −𝑘 1 𝑧−𝜔𝑡 , 𝑧<0 & 𝐴 𝑇 𝑒 i 𝑘 2 𝑧−𝜔𝑡 , 𝑧>0 The wavelengths and wave numbers are different. 𝜆 1 𝜆 2 = 𝑘 2 𝑘 1 = 𝑣 1 𝑣 2 𝑓 (𝑧,𝑡)= is continuous at 𝑧 = 0: 𝑓 0 + ,𝑡 =𝑓 0 − ,𝑡 Else there would be a break between the two strings. If the knot itself is of negligible mass, then the derivative of 𝑓 must also be continuous: 𝜕𝑓 𝜕𝑧 ​ 0 − = 𝜕𝑓 𝜕𝑧 ​ 0 + Otherwise there would be a net force on the knot, and therefore an infinite acceleration. Since the imaginary part of 𝑓 differs from the real part only in the replacement of cosine 𝑓 0 + ,𝑡 = 𝑓 0 − ,𝑡 → 𝐴 𝐼 + 𝐴 𝑅 = 𝐴 𝑇 𝜕 𝑓 𝜕𝑧 ​ 0 − = 𝜕 𝑓 𝜕𝑧 ​ 0 + → 𝑘 1 𝐴 𝐼 − 𝐴 𝑅 = k 2 𝐴 𝑇 𝐴 𝑅 =( 𝑘 1 −𝑘 2 𝑘 1 + 𝑘 2 ) 𝐴 𝐼 𝐴 𝑅 =( 𝑣 2 −𝑣 1 𝑣 2 + 𝑣 1 ) 𝐴 𝐼 𝐴 𝑅 𝑒 𝑖 𝛿 𝑅 =( 𝑣 2 −𝑣 1 𝑣 2 + 𝑣 1 ) 𝐴 𝐼 𝑒 𝑖 𝛿 𝐼 𝐴 𝑇 =( 2𝑘 1 𝑘 1 + 𝑘 2 ) 𝐴 𝐼 𝐴 𝑇 =( 2𝑣 2 𝑣 2 + 𝑣 1 ) 𝐴 𝐼 𝐴 𝑇 𝑒 𝑖 𝛿 𝑇 =( 2𝑣 2 𝑣 2 + 𝑣 1 ) 𝐴 𝐼 𝑒 𝑖 𝛿 𝐼 14 Feb 2017 Chapter 9: Electromagnetic Waves

Phase of reflected waves 9.1 Waves in One Dimension Phase of reflected waves 9.1.3 Boundary Conditions: Reflection and Transmission 𝑣 2 <𝑣 1 𝑣 2 >𝑣 1 𝐴 𝑅 𝑒 𝑖 𝛿 𝑅 =( 𝑣 2 −𝑣 1 𝑣 2 + 𝑣 1 ) 𝐴 𝐼 𝑒 𝑖 𝛿 𝐼 𝑣= 𝑇 𝜇 The second string is heavier than the first The second string is lighter than the first 𝐴 𝑅 𝑒 𝑖 𝛿 𝑅 =( 𝑣 2 −𝑣 1 𝑣 2 + 𝑣 1 ) 𝐴 𝐼 𝑒 𝑖 𝛿 𝐼 𝐴 𝑅 𝑒 𝑖 𝛿 𝑅 =( 𝑣 2 −𝑣 1 𝑣 2 + 𝑣 1 ) 𝐴 𝐼 𝑒 𝑖 𝛿 𝐼 <0 Amplitudes always positives 𝐴 𝑅 𝑒 𝑖 𝛿 𝑅 =−( 𝑣 1 −𝑣 2 𝑣 2 + 𝑣 1 ) 𝐴 𝐼 𝑒 𝑖 𝛿 𝐼 𝐴 𝑅 = ( 𝑣 1 −𝑣 2 𝑣 2 + 𝑣 1 ) 𝐴 𝐼 𝐴 𝑅 =( 𝑣 2 −𝑣 1 𝑣 2 + 𝑣 1 ) 𝐴 𝐼 𝑒 𝑖 𝛿 𝑅 =− 𝑒 𝑖 𝛿 𝐼 𝑒 𝑖 𝛿 𝑅 = 𝑒 𝑖 𝛿 𝐼 𝑒 𝑖 𝛿 𝑅 = 𝑒 𝑖𝜋 𝑒 𝑖 𝛿 𝐼 𝛿 𝑅 = 𝛿 𝐼 𝛿 𝑅 = 𝛿 𝐼 +𝜋 The reflected wave is out of phase by 180°. cos − 𝑘 1 𝑧−𝜔𝑡+ 𝛿 𝐼 +𝜋 =− cos − 𝑘 1 𝑧−𝜔𝑡+ 𝛿 𝐼 The reflected wave is inverted. 𝑣 2 <𝑣 1 or 𝑣 2 >𝑣 1 𝐴 𝑇 𝑒 𝑖 𝛿 𝑇 =( 2𝑣 2 𝑣 2 + 𝑣 1 ) 𝐴 𝐼 𝑒 𝑖 𝛿 𝐼 𝐴 𝑇 =( 2𝑣 2 𝑣 2 + 𝑣 1 ) 𝐴 𝐼 𝑒 𝑖 𝛿 𝑇 = 𝑒 𝑖 𝛿 𝐼 𝛿 𝑇 = 𝛿 𝐼 If the second string is infinitely massive or, if the first string is simply fixed at the end 𝐴 𝑅 = 𝐴 𝐼 There is no transmitted wave-all of it reflects back. 𝐴 𝑇 =0 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.1 Waves in One Dimension Transverse waves 9.1.4 Polarization Transverse waves Wave on a string The displacement is perpendicular to the direction of propagation Longitudinal waves The displacement is along the direction of propagation Slinky compression waves Electromagnetic waves are transverse. For transverse waves there are two dimensions perpendicular to any given line of propagation. Transverse waves occur in two independent states of polarization: The polarization vector 𝐧 defines the plane of vibration. Because the waves are transverse, 𝐧 is perpendicular to the direction of propagation: 𝐧 ∙ 𝐳 =0 Vertical polarization In terms of the polarization angle 𝜃, 𝐧 = cos 𝜃 𝐱 + sin 𝜃 𝐲 Horizontal polarization 𝑓 (𝑧,𝑡)= 𝐴 cos 𝜃 𝑒 i 𝑘𝑧−𝜔𝑡 𝐱 + 𝐴 sin 𝜃 𝑒 i −𝑘 1 𝑧−𝜔𝑡 𝐲 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.2 Electromagnetic Waves in Vacuum The Wave Equation 9.2.1 The Wave Equation for E and B 𝛁×𝐄=− 𝜕 𝜕𝑡 𝐁 In regions of space where there is no charge or current, Maxwell's equations read 𝛁∙𝐄=0 𝛁×𝐁= 𝜇 0 𝜖 0 𝜕 𝜕𝑡 𝐄 𝛁∙𝐁=0 𝛁×(𝛁×𝐄)=𝛁×− 𝜕 𝜕𝑡 𝐁 𝛁×(𝛁×𝐁)=𝛁× 𝜇 0 𝜖 0 𝜕 𝜕𝑡 𝐄 𝛁 𝛁∙𝐄 − 𝛻 2 𝐄=− 𝜕 𝜕𝑡 𝛁×𝐁 𝛁 𝛁∙𝐁 − 𝛻 2 𝐁 = 𝜇 0 𝜖 0 𝜕 𝜕𝑡 𝛁×𝐄 𝛻 2 𝐄= 𝜇 0 𝜖 0 𝜕 2 𝜕 𝑡 2 𝐄 𝛻 2 𝐁 = 𝜇 0 𝜖 0 𝜕 2 𝜕 𝑡 2 𝐁 𝛻 2 𝑓= 1 𝑣 2 𝜕 2 𝑓 𝜕 𝑡 2 In vacuum, each Cartesian component of 𝐄 and 𝐁 satisfies the three-dimensional wave equation, 𝛻 2 𝐸 𝑥 = 𝜇 0 𝜖 0 𝜕 2 𝜕 𝑡 2 𝐸 𝑥 𝑣= 1 𝜇 0 𝜖 0 =3.00× 10 8 𝑚/𝑠 𝛻 2 𝐸 𝑦 = 𝜇 0 𝜖 0 𝜕 2 𝜕 𝑡 2 𝐸 𝑦 The velocity of light, c. 𝛻 2 𝐸 𝑧 = 𝜇 0 𝜖 0 𝜕 2 𝜕 𝑡 2 𝐸 𝑧 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.2 Electromagnetic Waves in Vacuum The electromagnetic spectrum 9.2.2 Monochromatic Plane Waves We confine our attention to sinusoidal waves of frequency 𝜔. Since different frequencies in the visible range correspond to different colors, such waves are called monochromatic. 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.2 Electromagnetic Waves in Vacuum Transverse waves 9.2.2 Monochromatic Plane Waves Suppose that the sinusoidal waves are traveling in the z direction and have no 𝑥 or 𝑦 dependence. 𝐄 𝑧,𝑡 = 𝐄 𝟎 𝑒 i 𝑘𝑧−𝜔𝑡 𝐁 𝑧,𝑡 = 𝐁 𝟎 𝑒 i 𝑘𝑧−𝜔𝑡 These are called plane waves because the fields are uniform over every plane perpendicular to the direction of propagation. 𝐸 𝑥 𝑧,𝑡 = 𝐸 0𝑥 𝑒 i 𝑘𝑧−𝜔𝑡 𝐸 𝑦 𝑧,𝑡 = 𝐸 0𝑦 𝑒 i 𝑘𝑧−𝜔𝑡 𝐸 𝑧 𝑧,𝑡 = 𝐸 0𝑧 𝑒 i 𝑘𝑧−𝜔𝑡 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.2 Electromagnetic Waves in Vacuum Transverse waves 9.2.2 Monochromatic Plane Waves 𝐵 𝑥 𝑧,𝑡 = 𝐵 0𝑥 𝑒 i 𝑘𝑧−𝜔𝑡 𝐸 𝑥 𝑧,𝑡 = 𝐸 0𝑥 𝑒 i 𝑘𝑧−𝜔𝑡 𝐸 𝑦 𝑧,𝑡 = 𝐸 0𝑦 𝑒 i 𝑘𝑧−𝜔𝑡 𝐵 𝑦 𝑧,𝑡 = 𝐵 0𝑦 𝑒 i 𝑘𝑧−𝜔𝑡 𝐵 𝑧 𝑧,𝑡 = 𝐵 0𝑧 𝑒 i 𝑘𝑧−𝜔𝑡 𝐸 𝑧 𝑧,𝑡 = 𝐸 0𝑧 𝑒 i 𝑘𝑧−𝜔𝑡 Maxwell's equations impose extra constraints on 𝐄 𝟎 and 𝐁 𝟎 𝛁∙𝐄=0 → ( 𝐸 0 ) 𝑧 =0 Electromagnetic waves are transverse: the electric and magnetic fields are perpendicular to the direction of propagation. 𝛁∙𝐁=0 → ( 𝐵 0 ) 𝑧 =0 𝐱 𝐲 𝐳 𝜕 𝜕𝑥 𝜕 𝜕𝑦 𝜕 𝜕𝑧 𝐸 𝑥 𝐸 𝑦 0 −𝑘 ( 𝐸 0 ) 𝑦 = 𝜔 ( 𝐵 0 ) 𝑥 𝐱 𝐲 𝐳 0 0 1 𝐸 𝑥 𝐸 𝑦 0 𝛁×𝐄=− 𝜕 𝜕𝑡 𝐁 → 𝑘 ( 𝐸 0 ) 𝑥 = 𝜔 ( 𝐵 0 ) 𝑦 𝐁 𝟎 = 𝑘 𝜔 ( 𝐳 × 𝐄 𝟎 ) E and B are in phase and mutually perpendicular; their (real) amplitudes are related by 𝐵 0 = 𝑘 𝜔 𝐸 0 = 1 𝑐 𝐸 0 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.2 Electromagnetic Waves in Vacuum Example 9.2.2 Monochromatic Plane Waves If 𝐄 points in the x direction, then 𝐁 points in the y direction. 𝐁 𝟎 = 𝑘 𝜔 ( 𝐳 × 𝐄 𝟎 ) 𝐄 𝑧,𝑡 = 𝐸 0 𝑒 i 𝑘𝑧−𝜔𝑡 𝐱 𝐁 𝑧,𝑡 = 𝐵 0 𝑒 i 𝑘𝑧−𝜔𝑡 𝐲 𝐄 𝐫,𝑡 = 𝐸 0 cos 𝑘𝑧−𝜔𝑡+𝛿 𝐱 𝐁 𝐫,𝑡 = 1 𝑐 𝐸 0 cos 𝑘𝑧−𝜔𝑡+𝛿 𝐲 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.2 Electromagnetic Waves in Vacuum Propagation vector 9.2.2 Monochromatic Plane Waves The propagation (or wave) vector, 𝐤, points in the direction of propagation and its magnitude is the wave number 𝑘. The scalar product 𝐤∙𝐫 is the generalization of 𝑘𝑧 𝐧 is the polarization vector. Because 𝐄 is transverse. 𝐧 ∙ 𝐤 =0 𝐄 𝐫,𝑡 = 𝐸 0 𝑒 i 𝐤∙𝐫−𝜔𝑡 𝐧 𝐁 𝐫,𝑡 = 1 𝑐 𝐸 0 𝑒 i 𝐤∙𝐫−𝜔𝑡 ( 𝐤 × 𝐧 ) = 1 𝑐 ( 𝐤 × 𝐄 ) The actual (real) electric and magnetic fields in a monochromatic plane wave with propagation vector 𝐤 and polarization 𝐧 are 𝐄 𝐫,𝑡 = 𝐸 0 cos 𝐤∙𝐫−𝜔𝑡+𝛿 𝐧 𝐁 𝐫,𝑡 = 1 𝑐 𝐸 0 cos 𝐤∙𝐫−𝜔𝑡+𝛿 ( 𝐤 × 𝐧 ) 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.2 Electromagnetic Waves in Vacuum Poynting vector 9.2.3 Energy and Momentum in Electromagnetic Waves The energy per unit volume stored in electromagnetic fields is 𝑢= 1 2 ( 𝜖 0 𝐸 2 + 1 𝜇 0 𝐵 2 ) So the electric and magnetic contributions are equal. 1 𝜇 0 𝐵 2 = 1 𝜇 0 1 𝑐 2 𝐸 2 = 1 𝜇 0 𝜖 0 𝜇 0 𝐸 2 = 𝜖 0 𝐸 2 For a monochromatic plane wave As the wave travels, it carries this energy along with it. 𝑢= 𝜖 0 𝐸 2 = 𝜖 0 𝐸 0 2 cos 2 𝑘𝑧−𝜔𝑡+𝛿 𝐒= 1 𝜇 0 (𝐄×𝐁) The energy flux density (energy per unit area, per unit time) transported by the fields is given by the Poynting vector For monochromatic plane waves propagating in the 𝑧 direction 𝐒= 1 𝜇 0 1 𝑐 𝐸 0 2 cos 2 𝑘𝑧−𝜔𝑡+𝛿 𝐳 𝑐 2 = 1 𝜇 0 𝜖 0 𝐒 = 𝜖 0 𝑐 𝐸 0 2 cos 2 𝑘𝑧−𝜔𝑡+𝛿 𝐳 𝐒=𝑐𝑢 𝐳 = Energy density × velocity of the wave In a time Δ𝑡, a length 𝑐Δ𝑡 passes through area 𝐴, carrying with it an energy 𝑢𝐴𝑐Δ𝑡. The energy per unit time, per unit area, transported by the wave is therefore 𝑢𝑐. 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.2 Electromagnetic Waves in Vacuum Momentum 9.2.3 Energy and Momentum in Electromagnetic Waves Electromagnetic fields carry momentum. 𝓹= 𝜇 0 𝜖 0 𝐒= 1 𝑐 2 𝐒 The momentum density stored in the fields is For monochromatic plane waves 𝐒 = 𝜖 0 𝑐 𝐸 0 2 cos 2 𝑘𝑧−𝜔𝑡+𝛿 𝐳 𝓹= 𝜖 0 𝑐 𝐸 0 2 cos 2 𝑘𝑧−𝜔𝑡+𝛿 𝐳 𝑢= 𝜖 0 𝐸 0 2 cos 2 𝑘𝑧−𝜔𝑡+𝛿 𝓹= 1 𝑐 𝑢 𝐳 In the case of light, the wavelength is so short (~ 5× 10 −7 𝑚), and the period so brief (~ 10 −15 s), that any macroscopic measurement will encompass many cycles. 𝑥 denotes the time average of 𝑥 over a complete cycle. cos 2 (𝑘𝑧− 2𝜋𝑡 𝑇 +𝛿) = 1 𝑇 0 𝑇 cos 2 (𝑘𝑧− 2𝜋𝑡 𝑇 +𝛿) 𝑑𝑡= 1 2 We are interested only in the average value. 𝑢 = 1 2 𝜖 0 𝐸 0 2 Another way of finding cos 2 (𝜃) 𝐒 = 1 2 𝜖 0 𝑐 𝐸 0 2 𝐳 cos 2 𝜃 + sin 2 (𝜃) =1 cos 2 (𝜃) = sin 2 (𝜃) = 1 2 𝓹 = 1 2𝑐 𝜖 0 𝐸 0 2 𝐳 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.2 Electromagnetic Waves in Vacuum Intensity 9.2.3 Energy and Momentum in Electromagnetic Waves The intensity is the average power per unit area transported by an electromagnetic wave 𝐼≡ S = 1 2 𝜖 0 𝑐 𝐸 0 2 When light falls on a perfect absorber it delivers its momentum to the surface. In a time Δ𝑡 the momentum transfer is Δ𝐩= 𝓹 𝐴𝑐Δ𝑡, so the radiation pressure (average force per unit area) is 𝓹 = 1 2𝑐 𝜖 0 𝐸 0 2 𝐳 𝑃= 1 𝐴 Δp Δt = 1 2 𝜖 0 𝐸 0 2 = 𝐼 𝑐 On a perfect reflector the pressure is twice as great, because the momentum switches direction, instead of simply being absorbed. We can account for this pressure qualitatively, as follows: The electric field drives charges in the 𝑥 direction, and the magnetic field then exerts on them a force (𝑞 𝐯× 𝐁) in the 𝑧 direction. The net force on all the charges in the surface produces the pressure. 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.3 Electromagnetic Waves in Matter Maxwell’s equations 9.3.1 Propagation in Linear Media 𝛁×𝐄=− 𝜕 𝜕𝑡 𝐁 Inside matter and in regions where there is no free charge or free current, Maxwell's equations 𝛁∙𝐃=0 𝛁×𝐇= 𝜕 𝜕𝑡 𝐃 𝛁∙𝐁=0 𝐇= 1 𝜇 𝐁 If the medium is linear , 𝐃=𝜖𝐄 In homogeneous medium 𝜖 and 𝜇 do not vary from point to point 𝛁×𝐄=− 𝜕 𝜕𝑡 𝐁 𝛁∙𝐄=0 differ from the vacuum analogs only in the replacement of 𝜇 0 𝜖 0 by 𝜇𝜖 𝛁×𝐁=𝜇𝜖 𝜕 𝜕𝑡 𝐄 𝛁∙𝐁=0 Electromagnetic waves propagate through a linear homogeneous medium at a speed 𝑣= 1 𝜖𝜇 = 𝑐 𝑛 𝑛≡ 𝜖𝜇 𝜖 0 𝜇 0 𝑛 is the index of refraction of the material For most materials, 𝜇 is very close to 𝜇 0 , so 𝑛≅ 𝜖 𝑟 where 𝜖 𝑟 is the dielectric constant Since 𝜖 𝑟 is almost always greater than 1, light travels more slowly through matter. 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.3 Electromagnetic Waves in Matter Boundary conditions 9.3.1 Propagation in Linear Media 𝜖 0 →𝜖 𝜇 0 →𝜇 𝑐→𝑣= 𝑐 𝑛 = 𝑐 𝜖𝜇 All of our previous results carry over, with the simple transcription 𝑢= 1 2 (𝜖 𝐸 2 + 1 𝜇 𝐵 2 ) The energy density 𝐒= 1 𝜇 (𝐄×𝐁) The Poynting vector Boundary conditions When a wave passes from one transparent medium into another 𝜔=𝑘𝑣 𝐵= 1 𝑣 𝐸 𝐼= 1 2 𝜖𝑣 𝐸 0 2 For monochromatic plane waves 𝜖 1 𝐸 1 ⊥ = 𝜖 2 𝐸 2 ⊥ 𝐄 1 ∥ = 𝐄 2 ∥ 1 𝜇 1 𝐁 1 ∥ = 1 𝜇 2 𝐁 2 ∥ 𝐵 1 ⊥ = 𝐵 2 ⊥ 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.3 Electromagnetic Waves in Matter Boundary conditions 9.3.2 Reflection and Transmission at Normal Incidence Suppose the 𝑥𝑦 plane forms the boundary between two linear media. A plane wave, traveling in the z direction and polarized in the x direction, approaches the interface from the left 𝐄 𝐼 𝑧,𝑡 = 𝐸 0𝐼 𝑒 i 𝑘 1 𝑧−𝜔𝑡 𝐱 𝐁 𝐼 𝑧,𝑡 = 1 𝑣 1 𝐸 0𝐼 𝑒 i 𝑘 1 𝑧−𝜔𝑡 𝐲 It gives rise to a reflected wave travels back to the left in medium 𝐄 𝑅 𝑧,𝑡 = 𝐸 0𝑅 𝑒 i − 𝑘 1 𝑧−𝜔𝑡 𝐱 𝐄 1 ∥ = 𝐄 2 ∥ 𝐸 0𝐼 + 𝐸 0𝑅 = 𝐸 0𝑇 𝐁 𝑅 𝑧,𝑡 = − 1 𝑣 1 𝐸 0𝑅 𝑒 i − 𝑘 1 𝑧−𝜔𝑡 𝐲 1 𝜇 1 ( 1 𝑣 1 𝐸 0𝐼 − 1 𝑣 1 𝐸 0𝑅 )= 1 𝜇 2 1 𝑣 2 𝐸 0𝑇 and a transmitted wave 1 𝜇 1 𝐁 1 ∥ = 1 𝜇 2 𝐁 2 ∥ 𝐄 𝑇 𝑧,𝑡 = 𝐸 0𝑇 𝑒 i 𝑘 2 𝑧−𝜔𝑡 𝐱 𝐸 0𝐼 − 𝐸 0𝑅 = 𝜇 1 𝜇 2 𝑣 1 𝑣 2 𝐸 0𝑇 𝐁 𝑇 𝑧,𝑡 = 1 𝑣 2 𝐸 0𝑇 𝑒 i 𝑘 2 𝑧−𝜔𝑡 𝐲 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.3 Electromagnetic Waves in Matter Reflection and transmission amplitudes 9.3.2 Reflection and Transmission at Normal Incidence 𝐸 0𝐼 + 𝐸 0𝑅 = 𝐸 0𝑇 𝐸 0𝐼 − 𝐸 0𝑅 = 𝜇 1 𝜇 2 𝑣 1 𝑣 2 𝐸 0𝑇 𝛽≡ 𝜇 1 𝜇 2 𝑣 1 𝑣 2 =𝛽 𝐸 0𝑇 𝐸 0𝑅 = 1−𝛽 1+𝛽 𝐸 0𝐼 𝐸 0𝑇 = 2 1+𝛽 𝐸 0𝐼 If the permittivities 𝜇 are close to their values in vacuum (for most media), 𝛽= 𝑣 1 𝑣 2 𝑣= 1 𝜖𝜇 = 𝑐 𝑛 𝐸 0𝑅 = 𝑣 2 − 𝑣 1 𝑣 2 + 𝑣 1 𝐸 0𝐼 𝐸 0𝑅 =| 𝑣 2 − 𝑣 1 𝑣 2 + 𝑣 1 | 𝐸 0𝐼 𝐸 0𝑅 =| 𝑛 1 − 𝑛 2 𝑛 1 + 𝑛 2 | 𝐸 0𝐼 Identical to the ones for waves on a string 𝐸 0𝑇 = 2 𝑣 2 𝑣 2 + 𝑣 1 𝐸 0𝐼 𝐸 0𝑇 = 2 𝑣 2 𝑣 2 + 𝑣 1 𝐸 0𝐼 𝐸 0𝑇 = 2 𝑛 1 𝑛 1 + 𝑛 2 𝐸 0𝐼 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.3 Electromagnetic Waves in Matter Reflection and transmission coefficients 9.3.2 Reflection and Transmission at Normal Incidence 𝜇 1 ≈ 𝜇 2 ≈ 𝜇 0 𝑛≡ 𝜖𝜇 𝜖 0 𝜇 0 𝑣= 1 𝜖𝜇 = 𝑐 𝑛 𝐸 0𝑅 =| 𝑛 1 − 𝑛 2 𝑛 1 + 𝑛 2 | 𝐸 0𝐼 𝐸 0𝑇 = 2 𝑛 1 𝑛 1 + 𝑛 2 𝐸 0𝐼 What fraction of the incident energy is reflected, and what fraction is transmitted? 𝐼 = 1 2 𝜖𝑣 𝐸 0 2 The intensity (average power per unit area) is If 𝜇 1 = 𝜇 2 = 𝜇 0 , then the ratio of the reflected intensity to the incident intensity is 𝑅≡ 𝐼 𝑅 𝐼 𝐼 = 𝐸 0𝑅 𝐸 0𝐼 2 = 𝑛 1 − 𝑛 2 𝑛 1 + 𝑛 2 2 𝑅 is the reflection coefficient the ratio of the transmitted intensity to the incident intensity is 𝑛 𝑎𝑖𝑟 =1 𝑇≡ 𝐼 𝑇 𝐼 𝐼 = 𝜖 2 𝑣 2 𝜖 1 𝑣 1 𝐸 0𝑇 𝐸 0𝐼 2 = 4 𝑛 1 𝑛 2 𝑛 1 + 𝑛 2 2 𝑛 𝑔𝑙𝑎𝑠𝑠 =1.5 𝑇 is the transmission coefficient When light passes from air into glass, R = 0.04 T = 0.96 Conservation of energy, of course, requires 𝑅+𝑇 =1 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.3 Electromagnetic Waves in Matter Wave numbers 9.3.3 Reflection and Transmission at Oblique Incidence Suppose a monochromatic plane wave approaches from the left 𝐄 𝐼 𝐫,𝑡 = 𝐄 0𝐼 𝑒 i 𝐤 𝐼 ∙𝐫−𝜔𝑡 𝐁 𝐼 𝐫,𝑡 = 1 𝑣 1 ( 𝐤 𝐼 × 𝐄 𝐼 ) giving rise to a reflected wave 𝐄 𝑅 𝐫,𝑡 = 𝐄 0𝑅 𝑒 i 𝐤 𝑅 ∙𝐫−𝜔𝑡 𝐁 𝑅 𝐫,𝑡 = 1 𝑣 1 𝐤 𝑅 × 𝐄 𝑅 All three waves have the same frequency 𝜔 that is determined once and for all at the source and a transmitted wave 𝐄 𝑇 𝐫,𝑡 = 𝐄 0𝑇 𝑒 i 𝐤 𝑇 ∙𝐫−𝜔𝑡 𝐁 𝑇 𝐫,𝑡 = 1 𝑣 2 𝐤 𝑇 × 𝐄 𝑇 The three wave numbers are related 𝑘 𝐼 𝑣 1 = 𝑘 𝑅 𝑣 1 = 𝑘 𝑇 𝑣 2 =𝜔 𝑘 𝐼 = 𝑘 𝑅 = 𝑣 2 𝑣 1 𝑘 𝑇 = 𝑛 1 𝑛 2 𝑘 𝑇 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.3 Electromagnetic Waves in Matter At the boundary, phases are the same 9.3.3 Reflection and Transmission at Oblique Incidence The combined fields in medium (1), 𝐄 𝐼 + 𝐄 𝑅 and 𝐁 𝐼 + 𝐁 𝑅 , must be joined to the fields 𝐄 𝑇 and 𝐁 𝑇 in medium (2), using the boundary conditions. These all share the generic structure 𝑒 i 𝐤 𝐼 ∙𝐫−𝜔𝑡 + 𝑒 i 𝐤 𝑅 ∙𝐫−𝜔𝑡 = 𝑒 i 𝐤 𝑇 ∙𝐫−𝜔𝑡 notice is that the x, y, and t dependence is confined to the exponents. Because the boundary conditions must hold at all points on the plane, and for all times, these exponential factors must be equal. This is another way to show that the transmitted and reflected frequencies must match the incident one 𝐤 𝐼 ∙𝐫−𝜔𝑡= 𝐤 𝑅 ∙𝐫−𝜔𝑡= 𝐤 𝑇 ∙𝐫−𝜔𝑡 When 𝑧=0 𝐤 𝐼 ∙𝐫= 𝐤 𝑅 ∙𝐫= 𝐤 𝑇 ∙𝐫 When 𝑧=0 𝑥 𝑘 𝐼 𝑥 +𝑦 𝑘 𝐼 𝑦 =𝑥 𝑘 𝑅 𝑥 +𝑦 𝑘 𝑅 𝑦 =𝑥 𝑘 𝑇 𝑥 +𝑦 𝑘 𝑇 𝑦 for all 𝑥 and all 𝑦. This can only hold if the components are separately equal, for if 𝑥=0, 𝑘 𝐼 𝑦 = 𝑘 𝑅 𝑦 = 𝑘 𝑇 𝑦 for if 𝑦=0, 𝑘 𝐼 𝑥 = 𝑘 𝑅 𝑥 = 𝑘 𝑇 𝑥 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.3 Electromagnetic Waves in Matter Fundamental laws of geometrical optics 9.3.3 Reflection and Transmission at Oblique Incidence 𝑘 𝐼 𝑦 = 𝑘 𝑅 𝑦 = 𝑘 𝑇 𝑦 We may orient our axes so that 𝐤 𝐼 lies in the 𝑥𝑧 plane. Thus 𝑘 𝐼 𝑦 =0= 𝑘 𝑅 𝑦 = 𝑘 𝑇 𝑦 Hence 𝐤 𝑅 and 𝐤 𝑇 must lie in the 𝑥𝑧 plane. First Law: The incident, reflected, and transmitted wave vectors form a plane (called the plane of incidence), which also includes the normal to the surface(here, the z axis). 𝑘 𝐼 𝑥 = 𝑘 𝑅 𝑥 = 𝑘 𝑇 𝑥 𝜃 𝐼 the angle of incidence 𝜃 𝑅 the angle of reflection 𝜃 𝑇 the angle of transmission, or the angle of refraction 𝑘 𝐼 sin 𝜃 𝐼 = 𝑘 𝑅 sin 𝜃 𝑅 = 𝑘 𝑇 sin 𝜃 𝑇 Second Law: The angle of incidence is equal to the angle of reflection, The law of reflection. 𝜃 𝐼 = 𝜃 𝑅 All of them measured with respect to the normal sin 𝜃 𝑇 sin 𝜃 𝐼 = 𝑛 1 𝑛 2 The law of refraction, or Snell's law. Third Law: All we used was “the phases are same at the boundary”. Therefore, any other waves (water waves or sound waves) can be expected to obey the same "optical“ laws when they pass from one medium into another. 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.3 Electromagnetic Waves in Matter polarization parallel to the plane 9.3.3 Reflection and Transmission at Oblique Incidence 𝑥 𝐄 𝑅 sin 𝜃 𝐼 = sin 𝜃 𝑅 𝜃 𝑅 𝜖 1 − 𝐸 0𝐼 + 𝐸 0𝑅 = 𝜖 2 − 𝐸 0𝑇 sin 𝜃 𝑇 sin 𝜃 𝐼 𝐤 𝑅 𝜃 𝑇 sin 𝜃 𝑇 sin 𝜃 𝐼 = 𝑣 2 𝑣 1 𝐁 𝑅 𝐄 𝑇 𝐤 𝑇 𝐸 0𝐼 − 𝐸 0𝑅 = 𝜖 2 𝜖 1 𝑣 2 𝑣 1 𝐸 0𝑇 𝜃 𝑅 𝐁 𝑇 𝜃 𝑇 𝑧 𝜖𝜇= 1 𝑣 2 𝜃 𝐼 𝐄 𝐼 𝐸 0𝐼 − 𝐸 0𝑅 = 𝜇 1 𝑣 1 𝜇 2 𝑣 2 𝐸 0𝑇 𝜃 𝐼 𝐤 𝐼 𝐁 𝐼 1 2 Suppose that the polarization of the incident wave is parallel to the plane of incidence (𝑥𝑧 plane) Boundary conditions 𝐸 0𝐼 − 𝐸 0𝑅 = 𝜇 1 𝑣 1 𝜇 2 𝑣 2 𝐸 0𝑇 𝜖 1 𝐸 1 ⊥ = 𝜖 2 𝐸 2 ⊥ 𝜖 1 𝐄 0𝐼 + 𝐄 0𝑅 𝑧 = 𝜖 2 𝐄 0𝑇 𝑧 𝜖 1 − 𝐸 0𝐼 sin 𝜃 𝐼 + 𝐸 0𝑅 sin 𝜃 𝑅 = 𝜖 2 − 𝐸 0𝑇 sin 𝜃 𝑇 𝐵 1 ⊥ = 𝐵 2 ⊥ 𝐁 0𝐼 + 𝐁 0𝑅 𝑧 = 𝐁 0𝑇 z 0+0=0 0+0=0 𝐸 0𝐼 + 𝐸 0𝑅 = 𝐸 0𝑇 cos 𝜃 𝑇 cos 𝜃 𝐼 𝐄 1 ∥ = 𝐄 2 ∥ 𝐄 0𝐼 + 𝐄 0𝑅 𝑥,𝑦 = 𝐄 0𝑇 𝑥,𝑦 𝐸 0𝐼 cos 𝜃 𝐼 + 𝐸 0𝑅 cos 𝜃 𝑅 = 𝐸 0𝑇 cos 𝜃 𝑇 1 𝜇 1 𝐁 1 ∥ = 1 𝜇 2 𝐁 2 ∥ 1 𝜇 1 𝐁 0𝐼 + 𝐁 0𝑅 𝑥,𝑦 = 1 𝜇 2 𝐁 0𝑇 𝑥,𝑦 1 𝜇 1 𝑣 1 𝐸 0𝐼 − 𝐸 0𝑅 = 1 𝜇 2 𝑣 2 𝐸 0𝑇 𝐸 0𝐼 − 𝐸 0𝑅 = 𝜇 1 𝑣 1 𝜇 2 𝑣 2 𝐸 0𝑇 𝑥,𝑦 represent pairs of equations, one for the x-component and one for the y-component. 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.3 Electromagnetic Waves in Matter Fresnel's equations 9.3.3 Reflection and Transmission at Oblique Incidence 𝑥 𝐄 𝑅 𝐤 𝑅 𝐸 0𝐼 − 𝐸 0𝑅 = 𝜇 1 𝑣 1 𝜇 2 𝑣 2 𝐸 0𝑇 𝛽≡ 𝜇 1 𝜇 2 𝑣 1 𝑣 2 𝐸 0𝐼 − 𝐸 0𝑅 =𝛽 𝐸 0𝑇 𝐁 𝑅 𝐄 𝑇 𝐤 𝑇 𝜃 𝑅 𝐁 𝑇 𝜃 𝑇 𝑧 𝛼 ≡ cos 𝜃 𝑇 cos 𝜃 𝐼 𝐸 0𝐼 + 𝐸 0𝑅 = 𝐸 0𝑇 cos 𝜃 𝑇 cos 𝜃 𝐼 𝐸 0𝐼 + 𝐸 0𝑅 =𝛼 𝐸 0𝑇 𝜃 𝐼 𝐄 𝐼 𝐤 𝐼 𝐁 𝐼 1 2 Fresnel's equations 𝐸 0𝑅 = 𝛼−𝛽 𝛼+𝛽 𝐸 0𝐼 The reflected and the incident waves are either in phase, if 𝛼>𝛽, or 180° out of phase, if 𝛼<𝛽 The polarization of the incident wave is parallel to the plane of incidence 𝐸 0𝑇 = 2 𝛼+𝛽 𝐸 0𝐼 The transmitted wave is always in phase with the incident one. 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.3 Electromagnetic Waves in Matter Brewster's angle 9.3.3 Reflection and Transmission at Oblique Incidence 𝑥 𝐄 𝑅 𝐸 0𝑅 = 𝛼−𝛽 𝛼+𝛽 𝐸 0𝐼 Fresnel's equations 𝐤 𝑅 𝐁 𝑅 𝐄 𝑇 𝐤 𝑇 𝐸 0𝑇 = 2 𝛼+𝛽 𝐸 0𝐼 𝜃 𝑅 𝐁 𝑇 𝜃 𝑇 𝑧 𝜃 𝐼 𝐄 𝐼 = 1− sin 2 𝜃 𝑇 cos 𝜃 𝐼 = 1− 𝑛 1 / 𝑛 2 sin 𝜃 𝐼 2 cos 𝜃 𝐼 𝛽≡ 𝜇 1 𝜇 2 𝑣 1 𝑣 2 𝛼 ≡ cos 𝜃 𝑇 cos 𝜃 𝐼 𝐤 𝐼 𝐁 𝐼 1 2 At Brewster's angle 𝜃 𝐵 , the reflected wave is completely extinguished. 𝛼=𝛽 Air 𝑛 1 =1.0 Glass 𝑛 2 =1.5 𝜇 1 ≅ 𝜇 2 ≅ 𝜇 0 1− 𝑛 1 / 𝑛 2 sin 𝜃 𝐵 2 cos 𝜃 𝐵 =𝛽 𝛽≅ 𝑣 1 𝑣 2 = 𝑛 2 𝑛 1 1− 𝑛 1 𝑛 2 sin 𝜃 𝐵 2 = cos 2 𝜃 𝐵 𝛽 2 sin 2 𝜃 𝐵 ≅ 𝛽 2 1+𝛽 2 1− 𝑛 1 𝑛 2 sin 𝜃 𝐵 2 = 1− sin 2 𝜃 𝐵 𝛽 2 1− 𝛽 2 = 𝑛 1 𝑛 2 2 −𝛽 2 sin 2 𝜃 𝐵 𝑡𝑎𝑛 𝜃 𝐵 ≅ 𝑛 2 𝑛 1 sin 2 𝜃 𝐵 = 1− 𝛽 2 𝑛 1 𝑛 2 2 −𝛽 2 180° out of phase 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.3 Electromagnetic Waves in Matter Reflection and transmission coefficients 9.3.3 Reflection and Transmission at Oblique Incidence 𝑥 𝐄 𝑅 The power per unit area striking the interface is 𝐒∙ 𝐳 𝐤 𝑅 𝐁 𝑅 𝐄 𝑇 𝐤 𝑇 The average power per unit area of interface, and the interface is at an angle to the wave front. 𝜃 𝑅 𝐁 𝑇 𝜃 𝑇 𝑧 𝜃 𝐼 𝐄 𝐼 𝐼 𝐼 = 1 2 𝜖 1 𝑣 1 𝐸 0𝐼 2 cos 𝜃 𝐼 The incident intensity is 𝐤 𝐼 𝐼 𝑅 = 1 2 𝜖 1 𝑣 1 𝐸 0𝑅 2 cos 𝜃 𝑅 𝐁 𝐼 The reflected intensity is 1 2 𝐼 𝑇 = 1 2 𝜖 2 𝑣 2 𝐸 0𝑇 2 cos 𝜃 𝑇 The transmitted intensity is The reflection coefficient Air 𝑛 1 =1.0 Glass 𝑛 2 =1.5 𝑅≡ 𝐼 𝑅 𝐼 𝐼 = 𝐸 0𝑅 𝐸 0𝐼 2 The polarization of the incident wave is parallel to the plane of incidence The transmission coefficient 𝑇≡ 𝐼 𝑇 𝐼 𝐼 = 𝜖 2 𝑣 2 𝜖 1 𝑣 1 𝐸 0𝑇 𝐸 0𝐼 2 cos 𝜃 𝑇 cos 𝜃 𝐼 =𝛼𝛽 2 𝛼+𝛽 2 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.4 Absorption and Dispersion Transient behavior of free charge 9.4.1 Electromagnetic Waves in Conductors In the case of conductors 𝐉 𝑓 is not zero. 𝜕 𝜌 𝑓 𝜕𝑡 =−𝛁∙ 𝐉 𝑓 The continuity equation for free charge Ohm's law, the free current density in a conductor is proportional to the electric field. 𝜕 𝜌 𝑓 𝜕𝑡 =−𝛁∙𝜎𝐄 Using Ohm's law 𝐉 𝑓 =𝜎𝐄 𝜕 𝜌 𝑓 𝜕𝑡 =−𝜎(𝛁∙𝐄) For a homogeneous linear medium 𝛁∙𝐄= 1 𝜖 𝜌 𝑓 𝜕 𝜌 𝑓 𝜕𝑡 =− 𝜎 𝜖 𝜌 𝑓 Using Gauss's law 𝜌 𝑓 (𝑡)= 𝜌 𝑓 (0) 𝑒 −( 𝜎 𝜖 )𝑡 Any initial free charge density 𝜌 𝑓 (0) dissipates in a characteristic time 𝜏=𝜖⁄𝜎. If you put some free charge on a conductor, it will flow out to the edges. For a "perfect" conductor 𝜎=∞ and 𝜏=0. For a "good" conductor, 𝜏 is much less than the other relevant times in the problem (in oscillatory systems, 𝜏≪1/𝜔) For a "poor" conductor, 𝜏 is greater than the characteristic times in the problem (𝜏≫1/𝜔). We're not interested in this transient behavior. We'll wait for any accumulated free charge to disappear. From then on 𝜌 𝑓 =0. 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.4 Absorption and Dispersion Complex wave number 9.4.1 Electromagnetic Waves in Conductors In the case of conductors 𝐉 𝑓 is not zero. Ohm's law 𝐉 𝑓 =𝜎𝐄 𝛁×𝐄=− 𝜕 𝜕𝑡 𝐁 𝛁×𝐁=𝜇𝜖 𝜕 𝜕𝑡 𝐄+𝜇𝜎𝐄 𝛁∙𝐄=0 𝛁∙𝐁=0 𝛁×(𝛁×𝐄)=𝛁×− 𝜕 𝜕𝑡 𝐁 𝛁× 𝛁×𝐁 =𝛁×𝜇𝜖 𝜕 𝜕𝑡 𝐄+𝛁×𝜇𝜎𝐄 𝛁 𝛁∙𝐄 − 𝛻 2 𝐄=− 𝜕 𝜕𝑡 𝛁×𝐁 𝛁 𝛁∙𝐁 − 𝛻 2 𝐁 =𝜇𝜖 𝜕 𝜕𝑡 𝛁×𝐄+𝜇𝜎 𝛁×𝐄 𝛻 2 𝐄=𝜇𝜖 𝜕 2 𝜕 𝑡 2 𝐄+𝜇𝜎 𝜕 𝜕𝑡 𝐄 𝛻 2 𝐁 =𝜇𝜖 𝜕 2 𝜕 𝑡 2 𝐁+𝜇𝜎 𝜕 𝜕𝑡 𝐁 These equations still admit plane-wave solutions 𝐁 𝑧,𝑡 = 𝐁 0 𝑒 i 𝑘 𝑧−𝜔𝑡 𝐄 𝑧,𝑡 = 𝐄 0 𝑒 i 𝑘 𝑧−𝜔𝑡 "wave number" 𝑘 is complex − 𝑘 2 𝐄 0 𝑒 i 𝑘 𝑧−𝜔𝑡 =𝜇𝜖 − ω 2 𝐄 0 𝑒 i 𝑘 𝑧−𝜔𝑡 +𝜇𝜎(−𝑖𝜔 𝐄 0 𝑒 i 𝑘 𝑧−𝜔𝑡 ) 𝑘 2 =𝜇𝜖 ω 2 +𝑖𝜇𝜎𝜔 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.4 Absorption and Dispersion Real and imaginary parts of 𝑘 9.4.1 Electromagnetic Waves in Conductors "wave number" 𝑘 is complex 𝑘 2 =𝜇𝜖 ω 2 +𝑖𝜇𝜎𝜔 𝑘 =𝑘+𝑖𝜅 𝑘 2 =𝜇𝜖 ω 2 (1+𝑖 𝜇𝜎𝜔 𝜇𝜖 ω 2 ) 𝑘 2 =𝜇𝜖 ω 2 (1+𝑖 𝜎 𝜖𝜔 ) 𝑏≡ 𝜎 𝜖𝜔 𝑎 2 ≡𝜇𝜖 ω 2 𝑘=𝑎𝑥 𝑘 =𝑎 1+𝑖𝑏 1⁄2 =𝑎(𝑥+𝑖𝑦) 𝜅=𝑎𝑦 1+𝑖𝑏= 𝑥 2 − 𝑦 2 +2𝑖𝑥𝑦 𝑥 2 − 𝑦 2 =1 2𝑥𝑦=b 4 𝑥 2 𝑦 2 = 𝑏 2 𝑘=ω 𝜇𝜖 2 1+ 𝜎 𝜖𝜔 2 +1 1/2 4(1+ 𝑦 2 ) 𝑦 2 = 𝑏 2 𝑦 2 2 + 𝑦 2 = 𝑏 2 4 𝑦 2 = −1± 1+ 𝑏 2 2 𝑦 = 1 2 1+ 𝑏 2 −1 1⁄2 𝜅=ω 𝜇𝜖 2 1+ 𝜎 𝜖𝜔 2 −1 1/2 𝑥 2 = +1± 1+ 𝑏 2 2 𝑥 = 1 2 1+ 𝑏 2 +1 1⁄2 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.4 Absorption and Dispersion Skin depth 9.4.1 Electromagnetic Waves in Conductors 𝑘=ω 𝜇𝜖 2 1+ 𝜎 𝜖𝜔 2 +1 1/2 𝐄 𝑧,𝑡 = 𝐄 0 𝑒 i 𝑘 𝑧−𝜔𝑡 plane-wave solutions 𝐁 𝑧,𝑡 = 𝐁 0 𝑒 i 𝑘 𝑧−𝜔𝑡 𝜅=ω 𝜇𝜖 2 1+ 𝜎 𝜖𝜔 2 −1 1/2 𝑘 =𝑘+𝑖𝜅 "wave number" 𝑘 is complex 𝐄 𝑧,𝑡 = 𝐄 0 𝑒 −𝜅𝑧 𝑒 i 𝑘𝑧−𝜔𝑡 𝐁 𝑧,𝑡 = 𝐁 0 𝑒 −𝜅𝑧 𝑒 i 𝑘𝑧−𝜔𝑡 𝑑≡ 1 𝜅 The skin depth is the distance it takes to reduce the amplitude by a factor of 1/𝑒. It is a measure of how far the wave penetrates into the conductor. 𝑘, the real part of 𝑘 , determines 𝜆= 2𝜋 𝑘 the wavelength, 𝑣= 𝜔 𝑘 the propagation speed, 𝑛= 𝑐 𝑣 and the index of refraction. 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.4 Absorption and Dispersion Transverse waves 9.4.1 Electromagnetic Waves in Conductors 𝐄 𝑧,𝑡 = 𝐄 0 𝑒 −𝜅𝑧 𝑒 i 𝑘𝑧−𝜔𝑡 No new constraints from 𝛁×𝐁=𝜇𝜖 𝜕 𝜕𝑡 𝐄+𝜇𝜎𝐄 𝐁 𝑧,𝑡 = 𝐁 0 𝑒 −𝜅𝑧 𝑒 i 𝑘𝑧−𝜔𝑡 𝐱 𝐲 𝐳 𝜕 𝜕𝑥 𝜕 𝜕𝑦 𝜕 𝜕𝑧 0 𝐵 𝑦 0 Maxwell's equations impose extra constraints on 𝐄 𝟎 and 𝐁 𝟎 𝛁∙𝐄=0 → ( 𝐸 0 ) 𝑧 =0 No 𝑧 components: the fields are transverse 𝛁∙𝐁=0 → ( 𝐵 0 ) 𝑧 =0 −𝑖 𝑘 𝐵 0 =(𝜇𝜖(−𝑖𝜔) +𝜇𝜎) 𝐸 0 We may orient our axes so that 𝐄 is polarized along the 𝑥 direction: 𝐵 0 = 𝜇𝜖 −𝑖𝜔 +𝜇𝜎 −𝑖 𝑘 𝐸 0 𝐄 𝑧,𝑡 = 𝐸 0 𝑒 −𝜅𝑧 𝑒 i 𝑘𝑧−𝜔𝑡 𝒙 𝐵 0 = 𝜇𝜖 −𝑖𝜔 +𝜇𝜎 −𝑖 𝑘 𝑘 𝑘 𝜔 𝜔 𝐸 0 𝛁×𝐄=− 𝜕 𝜕𝑡 𝐁 𝑖 𝑘 𝐸 0 =− −𝑖𝜔 𝐁 0 𝑦 𝐵 0 = 𝑘 𝜔 𝐸 0 𝐵 0 = 𝑖(𝜇𝜖 −𝑖 𝜔 2 +𝜇𝜎𝜔) −𝑖 𝑘 2 𝑘 𝜔 𝐸 0 0=−(−𝑖𝜔) 𝐁 0 𝑥 𝐱 𝐲 𝐳 𝜕 𝜕𝑥 𝜕 𝜕𝑦 𝜕 𝜕𝑧 𝐸 𝑥 0 0 𝐁 𝑧,𝑡 = 𝑘 𝜔 𝐸 0 𝑒 −𝜅𝑧 𝑒 i 𝑘𝑧−𝜔𝑡 𝒚 𝐵 0 = 𝜇𝜖 𝜔 2 +𝑖𝜇𝜎𝜔 𝑘 2 𝑘 𝜔 𝐸 0 𝐵 0 = 𝑘 2 𝑘 2 𝑘 𝜔 𝐸 0 E and B are mutually perpendicular 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.4 Absorption and Dispersion Magnetic and electric fields are not in phase 9.4.1 Electromagnetic Waves in Conductors 𝐄 𝑧,𝑡 = 𝐸 0 𝑒 −𝜅𝑧 𝑒 i 𝑘𝑧−𝜔𝑡 𝒙 𝑘=ω 𝜇𝜖 2 1+ 𝜎 𝜖𝜔 2 +1 1/2 𝜅=ω 𝜇𝜖 2 1+ 𝜎 𝜖𝜔 2 −1 1/2 𝐁 𝑧,𝑡 = 𝑘 𝜔 𝐸 0 𝑒 −𝜅𝑧 𝑒 i 𝑘𝑧−𝜔𝑡 𝒚 𝐵 0 = 𝑘 𝜔 𝐸 0 𝐾= ω 𝜇𝜖 2 2 1+ 𝜎 𝜖𝜔 2 +1 + 1+ 𝜎 𝜖𝜔 2 −1 𝐾= 𝑘 = 𝑘 2 + 𝜅 2 𝑘 =𝑘+𝑖𝜅 =𝐾 𝑒 𝑖𝜙 𝜙= tan −1 (𝜅⁄𝑘) 𝐸 0 = 𝐸 0 𝑒 𝑖 𝛿 𝐸 𝐾= ω 𝜇𝜖 2 2 2 1+ 𝜎 𝜖𝜔 2 =ω 𝜇𝜖 1+ 𝜎 𝜖𝜔 2 𝐵 0 = 𝐵 0 𝑒 𝑖 𝛿 𝐵 𝐵 0 𝑒 𝑖 𝛿 𝐵 = 𝐾 𝑒 𝑖𝜙 𝜔 𝐸 0 𝑒 𝑖 𝛿 𝐸 The electric and magnetic fields are no longer in phase. 𝛿 𝐵 − 𝛿 𝐸 =𝜙 The magnetic field lags behind the electric field. = 𝜇𝜖 1+ 𝜎 𝜖𝜔 2 𝐄 𝑧,𝑡 = 𝐸 0 𝑒 −𝜅𝑧 cos 𝑘𝑧−𝜔𝑡+ 𝛿 𝐸 𝒙 𝐵 0 𝐸 0 = 𝐾 𝜔 𝐁 𝑧,𝑡 = 𝐵 0 𝑒 −𝜅𝑧 cos 𝑘𝑧−𝜔𝑡+ 𝛿 𝐸 +𝜙 𝒙 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.4 Absorption and Dispersion Excellent reflectors 9.4.2 Reflection at a Conducting Surface 𝐄 𝑅 𝑧,𝑡 = 𝐸 0𝑅 𝑒 i − 𝑘 1 𝑧−𝜔𝑡 𝐱 𝑥 𝐸 0𝑅 = 1− 𝛽 1+ 𝛽 𝐸 0𝐼 𝐤 𝑅 𝐄 𝑇 𝑧,𝑡 = 𝐸 0𝑇 𝑒 i 𝑘 𝟐 𝑧−𝜔𝑡 𝐱 𝛽 ≡ 𝜇 1 𝑣 1 𝑘 𝟐 𝜇 2 𝜔 a complex number 𝐁 𝑅 𝑧,𝑡 = − 1 𝑣 1 𝐸 0𝑅 𝑒 i − 𝑘 1 𝑧−𝜔𝑡 𝐲 𝐤 𝑇 𝐁 𝑇 𝑧,𝑡 = 𝑘 𝟐 𝜔 𝐸 0𝑇 𝑒 i 𝑘 2 𝑧−𝜔𝑡 𝐲 𝐸 0𝑇 = 2 1+ 𝛽 𝐸 0𝐼 𝐄 𝐼 𝑧,𝑡 = 𝐸 0𝐼 𝑒 i 𝑘 1 𝑧−𝜔𝑡 𝐱 𝑧 𝐤 𝐼 For a perfect conductor, 𝜎=∞, 𝑘 2 =∞, 𝛽 =∞. 𝐁 𝐼 𝑧,𝑡 = 1 𝑣 1 𝐸 0𝐼 𝑒 i 𝑘 1 𝑧−𝜔𝑡 𝐲 𝐸 0𝑅 =− 𝐸 0𝐼 𝐸 0𝑇 =0 nonconducting linear medium 1 2 conductor Boundary conditions 𝜖 1 𝐸 1 ⊥ − 𝜖 2 𝐸 2 ⊥ = 𝜎 𝑓 Since 𝐸 ⊥ =0 on both sides → 𝜎 𝑓 =0 Excellent conductors make good mirrors. 𝐵 1 ⊥ = 𝐵 2 ⊥ Since 𝐵 ⊥ =0 on both sides 𝐄 1 ∥ = 𝐄 2 ∥ 𝐸 0𝐼 + 𝐸 0𝑅 = 𝐸 0𝑇 𝐸 0𝐼 + 𝐸 0𝑅 = 𝐸 0𝑇 The skin depth in silver at optical frequencies is on the order of 100 Å. 1 𝜇 1 𝐁 1 ∥ − 1 𝜇 2 𝐁 2 ∥ = 𝐊 𝑓 × 𝐧 1 𝜇 1 𝑣 1 𝐸 0𝐼 − 𝐸 0𝑅 − 𝑘 𝟐 𝜇 2 𝜔 𝐸 0𝑇 =0 𝐸 0𝐼 − 𝐸 0𝑅 = 𝜇 1 𝑣 1 𝑘 𝟐 𝜇 2 𝜔 𝐸 0𝑇 𝜎 𝑓 is the free surface charge For ohmic conductors ( 𝐉 𝑓 =𝜎𝐄) there can be no free surface current, since this would require an infinite electric field at the boundary. 𝐉 𝑓 = 𝐊 f 𝛿 𝑧 =𝜎𝐄 𝐊 𝐟 is the free surface current density 𝐧 is a unit vector perpendicular to the surface, pointing from medium (2) into medium (1) 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.4 Absorption and Dispersion Dispersion 9.4.3 The Frequency Dependence of Permittivity The permittivity 𝜖, the permeability 𝜇, and the conductivity 𝜎 depends on the frequency of the waves. a typical glass The index of refraction in a transparent medium n is a function of wavelength A prism bends blue light more sharply than red. This phenomenon is called dispersion. The medium is called dispersive if the speed of a wave depends on its frequency. In a dispersive medium, a wave form that incorporates a range of frequencies will change shape as it propagates. A sharply peaked wave typically flattens out. 𝑣= 𝜔 𝑘 Each sinusoidal component travels at the ordinary wave (or phase) velocity, The packet as a whole (the "envelope") travels at the group velocity 𝑣 𝑔 = 𝑑𝜔 𝑑𝑘 The energy carried by a wave packet in a dispersive medium travels at the group velocity, not the phase velocity. It is ok that at some circumstances the phase velocity becomes greater than c. We will stick to monochromatic waves. 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.4 Absorption and Dispersion Model 9.4.3 The Frequency Dependence of Permittivity We want to account for the frequency dependence of 𝜖 in nonconductors, using a simplified model for the behavior of electrons in dielectrics. The electrons in a nonconductor are bound to specific molecules. Picture each electron as attached to the end of an imaginary spring, with force constant 𝑘 𝑠𝑝𝑟𝑖𝑛𝑔 Any binding force can be approximated by a spring force for sufficiently small displacements from equilibrium. 𝐹 𝑏𝑖𝑛𝑑𝑖𝑛𝑔 =− 𝑘 𝑠𝑝𝑟𝑖𝑛𝑔 𝑥=−𝑚 𝜔 0 2 𝑥 𝑥 is displacement from equilibrium. 𝑚 is the electron's mass. 𝜔 0 is the natural oscillation frequency = 𝑘 𝑠𝑝𝑟𝑖𝑛𝑔 ⁄𝑚 . Newton's second law gives There will be some damping force on the electron 𝑚 𝑑 2 𝑥 𝑑 𝑡 2 = 𝐹 𝑏𝑖𝑛𝑑𝑖𝑛𝑔 + 𝐹 𝑑𝑎𝑚𝑝𝑖𝑛𝑔 + 𝐹 𝑑𝑟𝑖𝑣𝑖𝑛𝑔 𝐹 𝑑𝑎𝑚𝑝𝑖𝑛𝑔 =−𝑚𝛾 𝑑𝑥 𝑑𝑡 𝑚 𝑑 2 𝑥 𝑑 𝑡 2 +𝑚 𝜔 0 2 𝑥+𝑚𝛾 𝑑𝑥 𝑑𝑡 =𝑞 𝐸 0 cos 𝜔𝑡 The driving force due to an electromagnetic wave of frequency 𝜔, polarized in the 𝑥 direction Our model describes the electron as a damped harmonic oscillator, driven at frequency 𝜔. 𝐹 𝑑𝑟𝑖𝑣𝑖𝑛𝑔 =𝑞𝐸=𝑞 𝐸 0 cos 𝜔𝑡 𝑞 is the charge of the electron. 𝐸 0 is the amplitude of the wave. We assume that the much more massive nuclei remain at rest. 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.4 Absorption and Dispersion Dipole moment 9.4.3 The Frequency Dependence of Permittivity 𝑚 𝑑 2 𝑥 𝑑 𝑡 2 +𝑚 𝜔 0 2 𝑥+𝑚𝛾 𝑑𝑥 𝑑𝑡 =𝑞 𝐸 0 cos 𝜔𝑡 A damped harmonic oscillator, driven at frequency 𝜔. It is easier to handle if we regard it as the real part of a complex equation: 𝑚 𝑑 2 𝑥 𝑑 𝑡 2 +𝑚 𝜔 0 2 𝑥 +𝑚𝛾 𝑑 𝑥 𝑑𝑡 = 𝑞 𝑚 𝐸 0 e −i𝜔𝑡 In the steady state, the system oscillates at the driving frequency 𝜔 𝑥 = 𝑥 0 𝑒 −𝑖𝜔𝑡 −𝑚 𝑥 0 𝜔 2 e −i𝜔𝑡 +𝑚 𝜔 0 2 𝑥 0 e −i𝜔𝑡 +𝑚𝛾(−𝑖𝜔 𝑥 0 ) e −i𝜔𝑡 = 𝑞 𝑚 𝐸 0 e −i𝜔𝑡 𝑥 0 = 𝑞 𝑚 𝜔 0 2 − 𝜔 2 −𝑖𝜔𝛾 𝐸 0 The dipole moment is the real part of The imaginary term in the denominator means that 𝑝 is out of phase with 𝐸-lagging behind by an angle tan −1 [ 𝛾𝜔 𝜔 0 2 − 𝜔 2 ] that is very small when 𝜔≪ 𝜔 0 and rises to 𝜋 when 𝜔≫ 𝜔 0 𝑝 𝑡 =𝑞 𝑥 (𝑡)= 𝑞 2 𝑚 𝜔 0 2 − 𝜔 2 −𝑖𝜔𝛾 𝐸 0 e −i𝜔𝑡 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.4 Absorption and Dispersion Complex dielectric constant 9.4.3 The Frequency Dependence of Permittivity 𝑝 𝑡 =𝑞 𝑥 (𝑡)= 𝑞 2 𝑚 𝜔 0 2 − 𝜔 2 −𝑖𝜔𝛾 𝐸 0 e −i𝜔𝑡 𝐃≡ ϵ 0 𝐄+𝐏 Linear medium 𝐏≡ ϵ 0 𝜒 𝑒 𝐄 In general, differently situated electrons within a given molecule experience different natural frequencies and damping coefficients. Let's say there are 𝑓 𝑗 electrons with frequency 𝜔 𝑗 and damping 𝛾 𝑗 in each molecule. If there are N molecules per unit volume, the polarization 𝐏 is given by the real part of 𝜒 𝑒 is the electric susceptibility 𝐃= ϵ 0 (1+ 𝜒 𝑒 )𝐄 𝐃= ϵ 0 𝜖 𝑟 𝐄 𝐃=ϵ𝐄 𝐏 𝑡 = 𝑁 𝑞 2 𝑚 𝑗 𝑓 𝑗 𝜔 𝑗 2 − 𝜔 2 −𝑖 𝛾 𝑗 𝜔 𝐄 This is not a linear medium since 𝐏 is not proportional to 𝐄 because of the difference in phase. The complex polarization 𝐏 is proportional to the complex field 𝐄 𝐏 𝑡 = 𝜖 0 𝜒 𝑒 𝐄 𝜒 𝑒 is complex susceptibility 𝐃 = 𝝐 𝐄 𝜖 = 𝜖 0 (1+ 𝜒 0 ) is the complex permittivity The complex dielectric constant 𝜖 𝑟 =1+ 𝑁 𝑞 2 𝑚 𝜖 0 𝑗 𝑓 𝑗 𝜔 𝑗 2 − 𝜔 2 −𝑖 𝛾 𝑗 𝜔 Ordinarily, the imaginary term is negligible; however, when 𝜔 is very close to one of the resonant frequencies ( 𝜔 j ) it plays an important role. 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.4 Absorption and Dispersion Absorption coefficient 9.4.3 The Frequency Dependence of Permittivity Linear homogenous In a dispersive medium the wave equation for a given frequency is 𝛁∙𝐃=0 𝛁∙𝐄=0 𝛁∙𝐁=0 𝛻 2 𝐄 = 𝜖 𝜇 0 𝜕 2 𝜕 𝑡 2 𝐄 𝛁×𝐄=− 𝜕 𝜕𝑡 𝐁 𝛁×𝐇= 𝜕 𝜕𝑡 𝐃 Non-magnetic 𝛁×𝐁= 𝜇 0 𝜕 𝜕𝑡 𝐃 It admits plane wave solutions 𝐄 𝑧,𝑡 = 𝐄 0 𝑒 i 𝑘 𝑧−𝜔𝑡 𝛁×(𝛁×𝐄)=𝛁×− 𝜕 𝜕𝑡 𝐁 − 𝑘 2 = 𝜖 𝜇 0 (− 𝜔 2 ) 𝛁 𝛁∙𝐄 − 𝛻 2 𝐄=− 𝜕 𝜕𝑡 𝛁×𝐁 𝛻 2 𝐄= 𝜇 0 𝜕 2 𝜕 𝑡 2 𝐃 The complex wave number 𝑘 = 𝜖 𝜇 0 𝜔 𝛻 2 𝐄 = 𝜇 0 𝜕 2 𝜕 𝑡 2 𝐃 𝐃 = 𝝐 𝐄 𝛻 2 𝐄 = 𝜖 𝜇 0 𝜕 2 𝜕 𝑡 2 𝐄 𝑘 =𝑘+𝑖𝜅 𝐄 𝑧,𝑡 = 𝐄 0 𝑒 −𝜅𝑧 𝑒 i 𝑘𝑧−𝜔𝑡 The wave is attenuated , since the damping absorbs energy. Because the intensity is proportional to 𝐸 2 and hence to 𝑒 −2𝜅𝑧 , 𝛼≡2𝜅 is called the absorption coefficient. Here 𝑘 and 𝜅 have nothing to do with conductivity. They are determined by the parameters of the damped harmonic oscillator. The wave velocity is 𝑣= 𝜔 𝑘 The index of refraction is 𝑛= 𝑐𝑘 𝜔 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.4 Absorption and Dispersion Absorption coefficient for gases 9.4.3 The Frequency Dependence of Permittivity For gases, this is small << 1 𝜖 𝑟 =1+ 𝑁 𝑞 2 𝑚 𝜖 0 𝑗 𝑓 𝑗 𝜔 𝑗 2 − 𝜔 2 −𝑖 𝛾 𝑗 𝜔 𝑘 = 𝜖 𝜇 0 𝜔= 𝜖 𝑟 𝜖 0 𝜇 0 𝜔= 𝜔 𝑐 𝜖 𝑟 𝜖 𝑟 = 1+𝛿 ≈1+ 1 2 𝛿 𝑘 = 𝜔 𝑐 𝜖 𝑟 ≅ 𝜔 𝑐 1+ 1 2 𝑁 𝑞 2 𝑚 𝜖 0 𝑗 𝑓 𝑗 𝜔 𝑗 2 − 𝜔 2 −𝑖 𝛾 𝑗 𝜔 Most of the time the index of refraction rises gradually with increasing frequency, consistent with our experience from optics. The index of refraction 𝑛= 𝑐𝑘 𝜔 ≅1+ 1 2 𝑁 𝑞 2 𝑚 𝜖 0 𝑗 𝑓 𝑗 𝜔 𝑗 2 − 𝜔 2 𝜔 𝑗 2 − 𝜔 2 2 + 𝛾 𝑗 2 𝜔 2 In the neighborhood of a resonance the index of refraction drops sharply. It is called anomalous dispersion. The region of anomalous dispersion ( 𝜔 1 <𝜔< 𝜔 2 ) coincides with the region of maximum absorption. The absorption coefficient The amplitude of electrons oscillation is relatively large, and hence a large amount of energy is dissipated by the damping mechanism. 𝛼=2𝜅≅ 𝑁 𝑞 2 𝜔 2 𝑚 𝜖 0 𝑐 𝑗 𝑓 𝑗 𝛾 𝑗 𝜔 𝑗 2 − 𝜔 2 + 𝛾 𝑗 2 𝜔 2 n runs below 1 above the resonance, suggesting that the wave speed exceeds c. This is ok, since energy does not travel at the wave velocity but rather at the group velocity. 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.4 Absorption and Dispersion Absorption coefficient for gases 9.4.3 The Frequency Dependence of Permittivity The index of refraction 𝑛= 𝑐𝑘 𝜔 ≅1+ 1 2 𝑁 𝑞 2 𝑚 𝜖 0 𝑗 𝑓 𝑗 𝜔 𝑗 2 − 𝜔 2 𝜔 𝑗 2 − 𝜔 2 2 + 𝛾 𝑗 2 𝜔 2 𝑛≅1+ 1 2 𝑁 𝑞 2 𝑚 𝜖 0 𝑗 𝑓 𝑗 𝜔 𝑗 2 − 𝜔 2 Away from the resonances, the damping can be ignored, For most substances the natural frequencies 𝜔 𝑗 are scattered all over the spectrum. For transparent materials, the nearest significant resonances typically lie in the ultraviolet, 𝜔<𝜔 𝑗 𝑛≅1+ 1 2 𝑁 𝑞 2 𝑚 𝜖 0 𝑗 𝑓 𝑗 𝜔 𝑗 2 + 𝜔 2 1 2 𝑁 𝑞 2 𝑚 𝜖 0 𝑗 𝑓 𝑗 𝜔 𝑗 4 1 𝜔 𝑗 2 − 𝜔 2 = 1 𝜔 𝑗 2 1− 𝜔 2 𝜔 𝑗 2 = 1− 𝜔 2 𝜔 𝑗 2 −1 𝜔 𝑗 2 ≅ 1 𝜔 𝑗 2 1+ 𝜔 2 𝜔 𝑗 2 In terms of the wavelength in vacuum 𝜆=2𝜋𝑐/𝜔 𝑛=1+𝐴 1+ B 𝜆 2 A is the coefficient of refraction Cauchy's formula B is the coefficient of dispersion It applies reasonably well to most gases, in the optical region. 14 Feb 2017 Chapter 9: Electromagnetic Waves

Boundary conditions 9.5 Guided Waves 𝐄 𝑥,𝑦,𝑧,𝑡 = 𝐄 0 (𝑥,𝑦) 𝑒 i 𝑘𝑧−𝜔𝑡 9.5.1 Wave Guides We consider electromagnetic waves confined to the interior of a hollow pipe, or wave guide. We'll assume the wave guide is a perfect conductor. Boundary conditions inside the material Boundary conditions at the inner wall 𝐄 1 ∥ = 𝐄 2 ∥ 𝐄=0 𝐄 ∥ =0 Free charges and currents will be induced on the surface in such a way as to enforce these constraints. 𝐵 1 ⊥ = 𝐵 2 ⊥ 𝐁=0 𝐵 ⊥ =0 𝐱 𝐲 𝐳 𝜕 𝜕𝑥 𝜕 𝜕𝑦 𝜕 𝜕𝑧 𝐸 𝑥 𝐸 𝑦 𝐸 𝑧 For monochromatic waves that propagate down the tube: 𝐄 𝑥,𝑦,𝑧,𝑡 = 𝐄 0 (𝑥,𝑦) 𝑒 i 𝑘𝑧−𝜔𝑡 𝛁×𝐄=− 𝜕 𝜕𝑡 𝐁 𝛁×𝐁= 1 𝑐 2 𝜕 𝜕𝑡 𝐄 𝐁 𝑥,𝑦,𝑧,𝑡 = 𝐁 0 (𝑥,𝑦) 𝑒 i 𝑘𝑧−𝜔𝑡 Confined waves are not in general transverse. We need to include longitudinal components 𝐸 𝑧 and 𝐵 𝑧 . 𝜕 𝐸 𝑧 𝜕𝑦 −𝑖𝑘 𝐸 𝑦 =𝑖𝜔 𝐵 𝑥 𝜕 𝐵 𝑧 𝜕𝑦 −𝑖𝑘 𝐵 𝑦 =− 𝑖𝜔 𝑐 2 𝐸 𝑥 𝐄 0 = 𝐸 𝑥 𝒙 + 𝐸 𝑦 𝒚 + 𝐸 𝑧 𝒛 − 𝜕 𝐸 𝑧 𝜕𝑥 +𝑖𝑘 𝐸 𝑥 =𝑖𝜔 𝐵 𝑦 − 𝜕 𝐵 𝑧 𝜕𝑥 +𝑖𝑘 𝐵 𝑥 =− 𝑖𝜔 𝑐 2 𝐸 𝑦 𝐁 0 = 𝐵 𝑥 𝒙 + 𝐵 𝑦 𝒚 + 𝐵 𝑧 𝒛 𝜕 𝐸 𝑦 𝜕𝑥 − 𝜕 𝐸 𝑥 𝜕𝑦 =𝑖𝜔 𝐵 𝑧 𝜕 𝐵 𝑦 𝜕𝑥 − 𝜕 𝐵 𝑥 𝜕𝑦 =− 𝑖𝜔 𝑐 2 𝐸 𝑧 To avoid cumbersome notation, the subscript 0 and the tilde are left off the individual components. 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.5 Guided Waves Wave equations 9.5.1 Wave Guides 𝜕 𝐸 𝑧 𝜕𝑦 −𝑖𝑘 𝐸 𝑦 =𝑖𝜔 𝐵 𝑥 v 𝜕 𝐵 𝑧 𝜕𝑦 −𝑖𝑘 𝐵 𝑦 =− 𝑖𝜔 𝑐 2 𝐸 𝑥 ×𝑖𝜔 𝑖𝜔 𝜕 𝐵 𝑧 𝜕𝑦 +𝜔𝑘 𝐵 𝑦 = 𝜔 𝑐 2 𝐸 𝑥 ①−② 𝑖𝜔 𝜕 𝐵 𝑧 𝜕𝑦 +𝑖𝑘 𝜕 𝐸 𝑧 𝜕𝑥 + 𝑘 2 𝐸 𝑥 = 𝜔 𝑐 2 𝐸 𝑥 − 𝜕 𝐸 𝑧 𝜕𝑥 +𝑖𝑘 𝐸 𝑥 =𝑖𝜔 𝐵 𝑦 − 𝜕 𝐵 𝑧 𝜕𝑥 +𝑖𝑘 𝐵 𝑥 =− 𝑖𝜔 𝑐 2 𝐸 𝑦 ×𝑖𝑘 −𝑖𝑘 𝜕 𝐸 𝑧 𝜕𝑥 − 𝑘 2 𝐸 𝑥 =−𝜔𝑘 𝐵 𝑦 𝑖 𝜔 𝜕 𝐵 𝑧 𝜕𝑦 +𝑘 𝜕 𝐸 𝑧 𝜕𝑥 = 𝜔 𝑐 2 − 𝑘 2 𝐸 𝑥 𝜕 𝐸 𝑦 𝜕𝑥 − 𝜕 𝐸 𝑥 𝜕𝑦 =𝑖𝜔 𝐵 𝑧 𝜕 𝐵 𝑦 𝜕𝑥 − 𝜕 𝐵 𝑥 𝜕𝑦 =− 𝑖𝜔 𝑐 2 𝐸 𝑧 𝐸 𝑥 = 𝑖 𝜔 𝑐 2 − 𝑘 2 𝜔 𝜕 𝐵 𝑧 𝜕𝑦 +𝑘 𝜕 𝐸 𝑧 𝜕𝑥 𝐸 𝑥 = 𝑖 𝜔 𝑐 2 − 𝑘 2 𝑘 𝜕 𝐸 𝑧 𝜕𝑥 +𝜔 𝜕 𝐵 𝑧 𝜕𝑦 𝛁∙𝐄=0 𝑖 𝜔 𝑐 2 − 𝑘 2 𝑘 𝜕 2 𝐸 𝑧 𝜕 𝑥 2 +𝜔 𝜕 2 𝐵 𝑧 𝜕𝑥𝜕𝑦 + 𝑘 𝜕 2 𝐸 𝑧 𝜕 𝑦 2 −𝜔 𝜕 2 𝐵 𝑧 𝜕𝑦𝜕𝑥 +𝑖𝑘 𝐸 𝑧 =0 𝐸 𝑦 = 𝑖 𝜔 𝑐 2 − 𝑘 2 𝑘 𝜕 𝐸 𝑧 𝜕𝑦 −𝜔 𝜕 𝐵 𝑧 𝜕𝑥 𝜕 2 𝜕 𝑥 2 + 𝜕 2 𝜕 𝑦 2 + 𝜔 𝑐 2 − 𝑘 2 𝐸 𝑧 =0 𝐵 𝑥 = 𝑖 𝜔 𝑐 2 − 𝑘 2 𝑘 𝜕 𝐵 𝑧 𝜕𝑥 − 𝜔 𝑐 2 𝜕 𝐸 𝑧 𝜕𝑦 𝛁∙𝐁=0 𝑖 𝜔 𝑐 2 − 𝑘 2 [ 𝑘 𝜕 2 𝐵 𝑧 𝜕 𝑥 2 − 𝜔 𝑐 2 𝜕 2 𝐸 𝑧 𝜕𝑥𝜕𝑦 + 𝑘 𝜕 2 𝐵 𝑧 𝜕 𝑦 2 + 𝜔 𝑐 2 𝜕 2 𝐸 𝑧 𝜕𝑦𝜕𝑥 +𝑖𝑘 𝐵 𝑧 =0 𝐵 𝑦 = 𝑖 𝜔 𝑐 2 − 𝑘 2 𝑘 𝜕 𝐵 𝑧 𝜕𝑦 + 𝜔 𝑐 2 𝜕 𝐸 𝑧 𝜕𝑥 𝜕 2 𝜕 𝑥 2 + 𝜕 2 𝜕 𝑦 2 + 𝜔 𝑐 2 − 𝑘 2 𝐵 𝑧 =0 It suffices to determine the longitudinal components 𝐸 𝑧 and 𝐵 𝑧 ; if we knew those, we could calculate all the others. 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.5 Guided Waves No TEM waves in a hollow wave guide 9.5.1 Wave Guides 𝜕 2 𝜕 𝑥 2 + 𝜕 2 𝜕 𝑦 2 + 𝜔 𝑐 2 − 𝑘 2 𝐸 𝑧 =0 𝐸 𝑧 =0 TE ("transverse electric") waves. 𝐵 𝑧 =0 TM ("transverse magnetic") waves. 𝐸 𝑧 =0 and 𝐵 𝑧 =0 TEM waves. 𝜕 2 𝜕 𝑥 2 + 𝜕 2 𝜕 𝑦 2 + 𝜔 𝑐 2 − 𝑘 2 𝐵 𝑧 =0 TEM waves cannot occur in a hollow wave guide. Proof: Gauss's law × 𝜕 𝜕𝑥 𝐸 𝑧 =0 𝜕 𝐸 𝑥 𝜕𝑥 + 𝜕 𝐸 𝑦 𝜕𝑦 =0 𝜕 2 𝐸 𝑥 𝜕 𝑥 2 + 𝜕 2 𝐸 𝑦 𝜕𝑥𝜕𝑦 =0 𝛁∙𝐄=0 ①−② 𝜕 2 𝐸 𝑥 𝜕 𝑥 2 + 𝜕 2 𝐸 𝑥 𝜕 2 𝑦 =0 Faraday's law 𝛻 2 𝐸 𝑥 =0 × 𝜕 𝜕𝑦 𝛁×𝐄=− 𝜕 𝜕𝑡 𝐁 𝐵 𝑧 =0 𝜕 𝐸 𝑦 𝜕𝑥 − 𝜕 𝐸 𝑥 𝜕𝑦 =0 𝜕 2 𝐸 𝑦 𝜕𝑦𝜕𝑥 − 𝜕 2 𝐸 𝑥 𝜕 2 𝑦 =0 Since Laplace's equation admits no local maxima or minima, and since, on the boundary, 𝐸 𝑥 =0, this means that 𝐸 𝑥 =0 throughout. The same argument holds for 𝐸 𝑦 . 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.5 Guided Waves No TEM waves in a hollow wave guide 9.5.1 Wave Guides TEM waves cannot occur in a hollow wave guide. 𝛁×𝐄= 𝐱 𝐲 𝐳 𝜕 𝜕𝑥 𝜕 𝜕𝑦 𝜕 𝜕𝑧 𝐸 𝑦 𝐸 𝑦 0 = 𝜕 𝐸 𝑦 𝜕𝑥 − 𝜕 𝐸 𝑥 𝜕𝑦 𝐳 Another Proof: =0 Gauss's law 𝛁∙𝐄=0 𝛁×𝐄=− 𝜕 𝜕𝑡 𝐁 𝜕 𝐸 𝑦 𝜕𝑥 − 𝜕 𝐸 𝑥 𝜕𝑦 =− 𝜕 𝜕𝑡 𝐵 𝑧 Faraday's law =0 The vector 𝐄 0 has zero divergence and zero curl. It can be written as the gradient of scalar potential that satisfies Laplace's equation. 𝛁×𝛁𝑉=0 𝛁×𝐄=0 𝐄≡−𝛁𝑉 𝛁∙𝐄=0 𝛁∙𝛁𝑉=0 𝛻 2 𝑉=0 The boundary condition 𝐄 ∥ =0 requires that the surface be an equipotential. Since Laplace's equation admits no local maxima or minima, this means that the potential is constant throughout, and hence the electric field is zero. This argument applies only to a completely empty pipe. It does not apply if a pipe has a separate conductor in the middle. 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.5 Guided Waves 𝐵 𝑧 (𝑥,𝑦) 9.5.2 TE Waves in a Rectangular Wave Guide 𝜕 2 𝜕 𝑥 2 + 𝜕 2 𝜕 𝑦 2 + 𝜔 𝑐 2 − 𝑘 2 𝐵 𝑧 =0 𝐸 𝑧 =0 Separation of variables: 𝐵 𝑧 𝑥,𝑦 =𝑋 𝑥 𝑌(𝑦) A wave guide of rectangular shape TE waves 𝑌 𝑑 2 𝑋 𝑑 𝑥 2 +𝑋 𝑑 2 𝑌 𝑑 𝑦 2 + 𝜔 𝑐 2 − 𝑘 2 𝑋𝑌=0 1 𝑋 𝑑 2 𝑋 𝑑 𝑥 2 + 1 𝑌 𝑑 2 𝑌 𝑑 𝑦 2 + 𝜔 𝑐 2 − 𝑘 2 =0 Divide by 𝑋𝑌: The x- and y-dependent terms must be constant: 1 𝑋 𝑑 2 𝑋 𝑑 𝑥 2 =− 𝑘 𝑥 2 1 𝑌 𝑑 2 𝑌 𝑑 𝑦 2 =− 𝑘 𝑦 2 − 𝑘 𝑥 2 − 𝑘 𝑦 2 + 𝜔 𝑐 2 − 𝑘 2 =0 𝑋 𝑥 =𝐴 sin 𝑘 𝑥 𝑥 +𝐵 cos 𝑘 𝑥 𝑥 Boundary condition: 𝐵 ⊥ =0 𝐵 𝑥 =0 𝑑𝑋 𝑑𝑥 =𝐴 𝑘 𝑥 cos 𝑘 𝑥 𝑥 − 𝑘 𝑥 𝐵 sin 𝑘 𝑥 𝑥 𝐵 𝑥 = 𝑖 𝜔 𝑐 2 − 𝑘 2 𝑘 𝜕 𝐵 𝑧 𝜕𝑥 − 𝜔 𝑐 2 𝜕 𝐸 𝑧 𝜕𝑦 𝑑𝑋 𝑑𝑥 ​ 𝑥=0 =𝐴 𝑘 𝑥 =0 𝑑𝑋 𝑑𝑥 ​ 𝑥=𝑎 =− 𝑘 𝑥 𝐵 sin 𝑘 𝑥 𝑎 =0 0= 𝑖 𝜔 𝑐 2 − 𝑘 2 𝑘 𝜕 𝐵 𝑧 𝜕𝑥 − 𝜔 𝑐 2 𝜕0 𝜕𝑦 𝐴=0 𝑘 𝑥 = 𝑚𝜋 𝑎 𝑚=0, 1, 2, … 𝜕 𝐵 𝑧 𝜕𝑥 =0 𝑑𝑋 𝑑𝑥 =0 The same goes for 𝑌(𝑦) 𝑘 𝑦 = 𝑛𝜋 𝑏 𝑛=0, 1, 2, … 𝐵 𝑧 𝑥,𝑦 = 𝐵 0 cos 𝑚𝜋𝑥 𝑎 cos ( 𝑛𝜋𝑦 𝑏 ) 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.5 Guided Waves Cutoff frequency 9.5.2 TE Waves in a Rectangular Wave Guide 𝐵 𝑧 𝑥,𝑦 = 𝐵 0 cos 𝑚𝜋𝑥 𝑎 cos ( 𝑛𝜋𝑦 𝑏 ) 𝑚=0, 1, 2, … At least one of the indices must be nonzero 𝑛=0, 1, 2, … This solution is called the 𝑇 𝐸 𝑚𝑛 mode. The first index is conventionally associated with the larger dimension, so we assume a > b. A wave guide of rectangular shape TE waves − 𝑘 𝑥 2 − 𝑘 𝑦 2 + 𝜔 𝑐 2 − 𝑘 2 =0 𝑘 𝑥 = 𝑚𝜋 𝑎 𝑘 𝑦 = 𝑛𝜋 𝑏 𝑘= 𝜔 𝑐 2 − 𝜋 2 [ 𝑚 𝑎 2 + 𝑛 𝑏 2 ] 𝑐 2 𝜋 2 𝑚 𝑎 2 + 𝑛 𝑏 2 ≡ 𝜔 𝑚𝑛 2 𝑘= 1 𝑐 𝜔 2 − 𝜔 𝑚𝑛 2 If 𝜔< 𝜔 𝑚𝑛 , the wave number is imaginary, and instead of a traveling wave we have exponentially attenuated fields. 𝜔 𝑚𝑛 is called the cutoff frequency for the 𝑛𝑚 mode. 𝐁 𝑥,𝑦,𝑧,𝑡 = 𝐁 0 (𝑥,𝑦) 𝑒 i 𝑘𝑧−𝜔𝑡 𝑣= 𝜔 𝑘 = 𝑐 1− 𝜔 𝑚𝑛 𝜔 2 The wave velocity is >𝑐 The lowest cutoff frequency for a given wave guide occurs for the mode 𝑇 𝐸 10 The energy carried by the wave travels at the group velocity 𝜔 10 = 𝑐𝜋 𝑎 𝑣 𝑔 = 𝑑𝜔 𝑑𝑘 =𝑐 1− 𝜔 𝑚𝑛 𝜔 2 Frequencies less than this will not propagate at all. <𝑐 14 Feb 2017 Chapter 9: Electromagnetic Waves

Chapter 9: Electromagnetic Waves 9.5 Guided Waves Another way 9.5.2 TE Waves in a Rectangular Wave Guide Consider an ordinary plane wave, traveling at an angle 𝜃 to the 𝑧 axis, and reflecting perfectly off each conducting surface. In the 𝑥 and 𝑦 directions the (multiply reflected) waves interfere to form standing wave patterns, 𝜆 𝑥 = 2𝑎 𝑚 𝑘 𝑥 = 𝜋𝑚 𝑎 𝜆 𝑦 = 2𝑏 𝑛 𝑘 𝑦 = 𝜋𝑛 𝑏 In the 𝑧 direction there remains a traveling wave, with wave number 𝑘 𝑧 =𝑘 The plane wave travels at speed e, but because it is going at an angle 𝜃 to the 𝑧 axis, its net velocity down the wave guide is 𝑣 𝑔 =𝑐 cos 𝜃 =𝑐 1− 𝜔 𝑚𝑛 ∕𝜔 2 The propagation vector for the "original" plane wave is 𝐤 ′ = 𝜋𝑚 𝑎 𝐱 + 𝜋𝑛 𝑏 𝐲 +𝑘 𝐳 The frequency is 𝜔=𝑐 𝐤 ′ =𝑐 𝑘 2 − 𝜋 2 [ 𝑚 𝑎 2 + 𝑛 𝑏 2 ] = 𝑐𝑘 2 + 𝜔 𝑚𝑛 2 The wave velocity is the speed of the wave fronts down the pipe. Similar to point A. 𝑣= 𝑐 cos 𝜃 = 𝑐 1− 𝜔 𝑚𝑛 𝜔 2 Only certain angles will lead to one of the allowed standing wave patterns cos 𝜃 = 𝑘 𝐤 ′ = 1− 𝜔 𝑚𝑛 ∕𝜔 2 Like the intersection of a line of breakers with the beach 14 Feb 2017 Chapter 9: Electromagnetic Waves