Electromagnetic Waves in Materials

Electromagnetic Waves in Materials Outline Review of the Lorentz Oscillator Model Complex index of refraction what does it mean? TART Microscopic model for plasmas and metals 1

True / False 1. In the Lorentz oscillator model of an atom, the electron is bound to the nucleus by a spring whose spring constant is the same for any atom. 2. The following is the differential equation described by the Lorentz oscillator model: 3. For dielectrics, we can approximate the index of refraction as the square root of the dielectric constant. = μ0 ε 0 ε0 με ε 2

Microscopic Description of Dielectric Constant B x Electron spring Nucleus Electron mass Restoring force (binding electron & nucleus) E field force Damping 3

Solution using complex variables Lets plug-in the expressions for and y into the differential equation from slide 3: Natural resonance d 2 d q y(t)+γ y(t)+ω y oy(t) 2 = E y (t) dt 2 dt m E y (t) =Re{ E y e jωt } y(t) =Re{ ye jωt } + + + + + + p +q q q ω 2 y + jωγy + ωoy 2 = m q 1 y = m (ωo 2 ω 2 )+jωγ E y E y 4

Oscillator Resonance E y (t)=re{ E e jωt y } q 1 y = E y jωt m (ω0 2 ω2 )+jωγ y(t) =Re{ye } Driven harmonic oscillator: Amplitude and Phase depend on frequency Low frequency At resonance High frequency medium large amplitude vanishing amplitude amplitude Displacement, y Displacement, y 90º out of phase with E y Displacement y and in phase with in antiphase E y E y 5

Polarization Since charge displacement, y, is directly related to polarization, P, of our material we can then rewrite the differential equation: D = ɛ o E + P P y = Nqy ŷ For linear polarization in direction d d ( + γ + ω 2 Nq 2 o)p y (t) = E y (t) =ɛ o ω 2 t 2 dt m pe y ( ) dt ω 2 Nq 2 p = ɛo m P y (t) =Re{P y e jωt } ω 2 p P y = ɛ y (ωo 2 oe ω 2 )+jγω 6

Microscopic Lorentz Oscillator Model ω 2 o = k spring m ωo ω p 7

ω 2 o = k spring m Behavior of a driven (and damped) harmonic oscillator can be summarized as follows Low frequency limit Amplitude amplitude Frequency phase 180 90 Phase Lag This type of response of bound charges is typical for many materials (displacement vs. driving field) Hi frequency limit 8

Complex Refractive Index ɛ = ɛ r jɛ i 1 v p = μɛ c n = v p ɛ = 2 n ɛ o =(n jκ) 2 μɛ ɛ μo ɛ o ɛo = n 2 κ 2 2jnκ ɛ real = n 2 κ 2 ɛ o ɛ imag =2nκ ɛ o (real) (imaginary) 9

Absorption Coefficient 1 n = ɛ r + ɛ 2 r + ɛ 2 i 2 1 κ = ɛ r + ɛ 2 r + ɛ 2 i 2 E(t, z) =Re{E e αz/2 e j(ωt k 0nz) 0 } Absorption Refractive index 2π α =2k κ =2 κ [cm -1 o ] λ o 10

Absorption and Reflection E(t, z) =Re{E 0e αz/2 e j(ωt k0nz) } Absorption coefficient Refractive index I(z) =I o e αz Beer-Lambert Law or Beer s Law 11

T A R T ω γ 2 ω + γ 2 ω 0 0 p Different resonant frequencies ω 0 Photograph by Hey Paul on Flickr. Transmissive ω < ω 0 γ 2 Absorptive ω 0 γ 2 <ω < ω 0 Reflective ω 0 + γ 2 <ω < ω p Transmissive ω > ω p Transmissive Absorp. Reflective Transmissive 12

Plasma in Ionosphere Plasma is an ionized gas consisting of positively charged molecules (ions) and negatively charged electrons that are free to move. Plasma exists naturally in what we call ionosphere (80 km ~ 120 km above the surface of the Earth). Here the UV light from the Sun ionizes air molecules. Aurora Australis Image is in the public domain 13

Plasmas (which we will assume to be lossless, γ =0) have no restoring force for electrons, ω o =0 ɛ = ɛ o ( ω 2 1 p ω 2 ) Reflecting Transparent What happens when the dielectric constant is negative? ω < ω p ɛ<0 ω > ω p ɛ>0 ɛ/ɛo 1.0 0.5 0 ɛ If ε < 0 then n is imaginary -0.5 1 1.5 2 ω/ω p 14

Optical Response of Plasmas Reflecting Transparent ɛ = n ɛ o jκ 1 κ n n, κ 0.5 n = 1 ω2 p ω 2 for ω>ω p κ = ω 2 p ω 2 1 for ω<ω p 0-0.2 0.5 1 1.5 2 ω/ω p 15

Plasma Frequency ω p = Ne2 ɛ o m (10 12 M 3 )(1.602x10 19 C) 2 = (8.854 10 12 F/M)(9.109 10 31 Kg) =564. 10 7 rad/sec =2π (8.98MHz) AM radio is in the range 520-1610 khz FM radio in in the range 87.5 to 108 MHz Reflected Transmitted 16

AM radio transmitter The Ionosphere and Radio Wave Propagation The ionosphere is important for radio wave (AM only) propagation... Ionosphere is composed of the D, E, and F layers. The D layer is good at absorbing AM radio waves. D layer disappears at night...the E and F layers bounce the waves back to the Earth. This explains why radio stations adjust their power output at sunset and sunrise. 17

Metals are Lossy Why is there a discontinuity here? ɛ o or γ must change for this to be true Metals have loss γ =0 but have no restoring force for electrons ω o =0 0.5 Drude Model for metals ɛ ( ) 2 ωp = ɛ o 1 ω 2 = ɛ r + iɛ i jωγ ( ) ωp 2 ɛ r = ɛ o 1 ω 2 + γ 2 ( ) γωp 2 ɛ i = ɛ o ω(ω 2 + γ 2 ) ɛr,ɛi 0.4 0.3 0.2 0.1-5 -10-20 -25 ɛ i ɛ r ω p ɛ r ω -30 19

Behavior of Metals γ/2 6 100 5 κ 80 n, κ 4 3 2 1 n R T n Reflection(%) 60 40 20 R T ω p ω ω p ω 20

Image by Kate Hopkins http://www.flickr.com/photos/accidentalhedonist/5200667428/ on flickr T A R T Dielectric ω 0 γ 2 ω 0 ω 0 + γ 2 ω p Metal ω 0 = 0 ω ω 0 + γ 2 p Plasma ω = 0 0 γ = 0 ω p 21

Key Takeaways Lorentz Oscillator Model d 2 y = mω 2 dy m oy + qe y dt 2 mγ dt ω 2 p P (ω) = ɛ o E(ω) (ω o ω 2 )+jωγ Electron spring 2 ñ = n jκ E(t, z) =Re{Ẽ e αz/2 e j(ωt k 0nz) 0 } Absorption coefficient ɛ = ɛ r jɛ i Refractive index Nucleus Decay α =2k o κ Transmissive ω < ω 0 γ 2 Absorptive ω γ 2 <ω < ω 0 0 I(z) =I e Beer s Law o αz Reflective Transmissive ω 0 + γ 2 <ω < ω p ω > ω p 22

MIT OpenCourseWare http://ocw.mit.edu 6.007 Electromagnetic Energy: From Motors to Lasers Spring 2011 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.