Wave Phenomena Physics 15c. Lecture 17 EM Waves in Matter

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1 Wave Phenomena Physics 15c Lecture 17 EM Waves in Matter

$What We Did Last Time Reviewed reflection and refraction Total internal reflection is more subtle than it looks Imaginary waves extend a few beyond the surface Studied how to create EM waves$

2 What We Did Last Time Reviewed reflection and refraction Total internal reflection is more subtle than it looks Imaginary waves extend a few beyond the surface Studied how to create EM waves Accelerated charge radiates EM waves Power given by Larmor formula Proportional to (acceleration) 2 Instability of atoms QM Polarization parallel to the acceleration Explains Brewster s angle Rayleigh scattering E T = q sinθ a(t r 4πε 0 c 2 c r ) P = q2 a 2 6πε 0 c 3 a

3 Oscillating Charge E T = Now we consider an oscillating charge x = x 0 cosωt Transverse component of E at (r,θ) is Calculate the Poynting vector Total power radiated is Increases with (frequency) 4 a = x = ω 2 x 0 cosωt q sinθ a(t r 4πε 0 c 2 c r )ˆθ = qω 2 x 0 sinθ 4πε 0 c 2 r S = q2 ω 4 2 x 0 sin 2 θ cos 2 ω t r 16π 2 ε 0 c 3 r 2 c 2π π P = dφ S r 2 sinθ dθ = q2 ω 4 x πε 2 c 3 q cos( ω(t r ) ) ˆθ c ( ( )) ˆr S = q2 ω 4 x 0 average 2 π θ 2 E T r 2 32π 2 ε 0 c 3 sin 2 θ sin 3 θ dθ = q2 ω 4 x πε 0 c 3 2 ˆr

4 Rayleigh Scattering Sunlight passing through the air makes air molecules oscillate Incoming light has a broad spectrum All frequencies are more or less equal Molecules radiate power according to P = q2 ω 4 2 x 0 12πε 0 c 3 More power is absorbed and re-emitted at higher frequencies This is why the sky looks blue, and why the sky turns red at sunset E q F = qe

Scattered light (what you see in the blue sky) is polarized

5 Polarization Sunlight contains two polarizations Viewed from right Only one causes radiation that reaches the observer Scattered light (what you see in the blue sky) is polarized Photographers use polarizing filters to deepen the color of the sky

6 Goals for Today We know an accelerated charge radiates EM waves Matter is made of charged particles EM waves passing through matter accelerates them They radiate EM waves in return What happens in the end? It should explain how EM waves behave in matter In particular, why does it travel slower? We put together many charges and accelerate them And try to figure out what happens to the EM field around

7 Sheet of Charges Imagine an infinite array of charges making a sheet All charges are oscillating together as x = x 0 e iωt Acceleration a = ω 2 x 0 e iωt ˆx The charge density is σ E? z y x What kind of EM radiation would they make at a distance z above the sheet?

8 Sheet of Charges Consider a little piece dxdy Charge of this piece is σdxdy E due to this piece is de T = σdxdyω 2 x 0 4πε 0 c 2 sinθ r e iω t r c z r y dx dy x Consider a similar piece at x y-z components cancel We only need the x component of E T de x = σdxdyω 2 x 0 4πε 0 c 2 sin 2 θ r e iω t r c 2E x r We have to integrate this over the sheet

9 Radiation From the Sheet Integrating E x over dxdy is pain E x (z,t) = σω 2 x 0 4πε 0 c 2 e iωt Calculation is difficult, but irrelevant We know that E is parallel to the x axis E does not depend on x or y E oscillates with e ωt + + sin 2 θ e ikr dx dy r Solution must be plane waves in ±z direction r = x 2 + y 2 + z 2 sinθ = y 2 + z 2 r Actual solution is E x (z,t) = iσω x 0 ( k z ωt) σ 2ε 0 c ei = 2ε 0 c v ( t z ) c Proportional to the velocity, in the opposite direction

10 Wall of Charges Imagine the sheet has a thickness dz Density of charge is n 0 per unit volume σ = n 0 qdz per unit area Plane EM waves are arriving from z as Suppose E in moves the charge as x = qe (z = 0) in This is simplistic Velocity is The radiation from the sheet is E rad (z,t) = force v(t) = iωqe 0 σ 2ε 0 c v t z c spring constant e iωt ( ) = E 0 E in iωn 0 q 2 dze i ( k0 z ωt) 2ε 0 c E in = E 0 e i(k 0 z ωt ) ˆx dz z

11 Radiation From the Wall Charge in the wall radiates Add this to the incoming waves E in + E rad = E 0 1+ iωq2 n 0 dz 2ε 0 ck s ei(k 0 z ωt ) for z > 0 For small dz, e αdz = 1+ αdz + E just after the wall (z = dz) is ( ) z=dz = E 0 e iωq 2 n 0 iωn E rad (z,t) = E 0 q 2 0 dze i ( k0 z ωt) 2ε 0 c i k 0 + ωq2 n 0 2ε 0 c dz ωt dz 2ε E in + E 0 c i(k rad e 0 dz ωt ) = E 0 e dz Compared with the incoming waves, the sum seems to have an increased value of k k = k which means decreased velocity! 0 + ωq2 n 0 = ω 2ε 0 c c 1+ q2 n 0 2ε 0 k s z

12 Thin Thick Wall Passing the thin wall changes the E field E z=0 = E 0 e iωt Add more walls E z=0 = E 0 e iωt z = Ndz E z=dz = E 0 e E z=ndz = E 0 e i ω c 1+ q2 n 0 2ε 0 dz e iωt = E 0 e i ω c 1+ q2 n 0 2ε 0 Ndz e iωt ω i c 1+ q 2 n 0 2ε 0 z ωt Solution looks just like plane waves, but with a modified wave number Multiply the same factor N times

13 Phase Velocity Given the modified k, we get Index of refraction is n = 1+ q2 n 0 2ε 0 > 1 c p is constant No dispersion c p = ω k = c 1+ q2 n 0 2ε 0

14 Quick Summary We started from accelerated charged particles Each of them radiates EM waves Matter is densely populated by those Incoming EM waves make them oscillate Collective radiation from a thin sheet makes plane waves Phase of the EM waves is shifted slightly by adding it Accumulating the phase shift over finite thickness makes plane waves again, but with modified phase velocity We end up with a slower speed of light We found origin of refraction!

15 Correction There was a small omission in the above discussion Thin sheet radiates EM waves in both sides I ignored the backward-going waves Backward waves complicate analysis But the conclusion changes little We still find plane waves with modified phase velocity Index of refraction turns out to be n true = 1+ q2 n 0 instead of n = 1+ q ε 0 approx. For material with small n, 2 n 0 2ε 0 n true n approx.

16 Maxwell s Equations Now we go back to Maxwell s equations E = ρ ε 0 B = 0 E = B t B = 1 c 2 E t + µ 0 J J = qn 0 v Movement of the charges in matter Current We assumed x = qe v = q E t Usual trick with BAC-CAB rule gives us the wave equation: 2 E = 1 c 2 2 E t 2 + µ 0 q2 n 0 2 E 1 t 2 c = 1 1+ q2 n 0 2 w c 2 ε 0 = n 2

17 Maxwell s Equation We can do this even faster Take the equation We are assuming J = qn 0 v We can now define ε = ε 0 + q2 n 0 We absorbed the J term into the matter s permittivity ε Now it s trivial to get B = ε 0 µ 0 E t + µ 0 J v = q E t n = E B = εµ 0 t B = ε 0 µ 0 E t + µ 0 q2 n 0 2 n 0 ε = 1+ q ε 0 ε 0 = ε 0 + q2 n 0 µ 0 E t E t

18 Quick Summary Again We took three approaches to get the same answer Microscopic: what does each electron do? Gave us a glimpse of really why light goes slower Integration was tough Current density in vacuum Represent electrons collective movement by J Wave equation is easy enough to solve Permittivity of matter Absorb J by replacing ε 0 with ε Solution is totally trivial They represent different levels of abstraction

19 Realistic Examples In the previous analysis, we used a simplistic model for the movement of the charge Let s try something more realistic We discuss plasma first I know it sounds unfamiliar But it s a lot simpler stuff than most matter Then we talk about more ordinary insulator I ll take air for an example x = qe

20 Plasma Imagine a space filled with free electrons We neutralize the electron gas by adding positive ions The ions are heavy Ignore their movements Such a mixture is called a plasma You can make them by heating stuff up really hot It s rather easy to analyze Each electron s equation of motion is m x = qe = qe 0 e iωt x = q mω 2 E 0 e iωt Current density J = qn 0 v = iq2 n 0 mω E 0 e iωt

21 Plasma Frequency Wave Equation is Dispersion relation is 2 E = 1 c 2 2 E t 2 + µ 0 ω p = q2 n 0 ε 0 m is called the plasma frequency Plasma is a dispersive medium: ω p is the cut-off frequency For ω < ω p, k becomes imaginary Waves disappear J = iq2 n 0 mω E 0 e iωt J k 2 E = ω 2 t c E + µ 0 q2 n 0 2 m E k 2 = ω 2 1 q2 n 0 c 2 ε 0 mω 2 = ω 2 1 ω 2 p c 2 ω 2 c p = ω k = c 1 ω p 2 ω 2

22 Phase and Group Velocities ω(k) = ω p 2 + c 2 k 2 ω = ck c p = c 1 ω p 2 ω 2 c ω p k ω p c g = dω dk = c 1 ω 2 ω 2 p Slope = c p Slope = c g c p is greater than c, but c g remains less than c I hope you remember this from Lecture #11

23 Ionosphere Sunlight ionizes air in the upper atmosphere Ionosphere has free electron density n 0 ~ /m 3 Varies with sunspots, season, day/night, latitude, etc ω p = q2 n 0 ε 0 m = rad/s ν p = ω p 2π = Hz Visible light (n ~ Hz) passes through Shortwave radio (n ~ 5 MHz) is reflected You can hear BBC, Deutsche Welle, Voice of Russia, etc.

24 Insulators Unlike in plasma or in metal, electrons in insulators are bound to the molecules k Binding force is similar to a spring s x qe M m Rest of the molecule M is much heavier Ignore the movement Equation of motion Forced oscillation We know the solution x = x 0 e iωt Current density is m x = qe 0 e iωt x mω 2 x 0 = qe 0 x 0 x 0 = qe 0 mω 2 = qe 0 m(ω 0 2 ω 2 ) J = qn 0 v = iωq2 n 0 m(ω 0 2 ω 2 ) E 0 e iωt ω 0 = m

25 Dispersion Relation Wave Equation is 2 E = 1 c 2 2 E t 2 + µ 0 Dispersion relation is k 2 = ω 2 1+ c 2 It s dispersive again Phase velocity: J t q 2 n 0 ε 0 m(ω 2 0 ω 2 ) = ω 2 c 2 c p = ω k = k 2 E = ω 2 c E µ 0 q2 n 0 2 m ω 2 ω 0 2 ω 2 E 2 ω 1+ ρ 0 ω 2 0 ω 2 ρ q2 n 0 ε 0 c 1+ ρω 0 2 (ω 0 2 ω 2 ) Index of refraction: n = ω ρ ω 2 0 ω 2

26 Example: Air Air is a mixture of N 2, O 2, H 2 O, Ar, etc Many resonances exist in UV and shorter wavelengths Things are simpler in visible light, where ω ω 0 c p = c 1+ ρω 0 2 (ω 0 2 ω 2 ) c 1+ ρ n 1+ ρ ρ We don t really know n 0 and, but we do know n = at STP ρ = ρ q2 n 0 ε 0 ρ is proportional to density n 0 Using the ideal gas formula PV = n mol RT n = K T P 1atm Index of refraction of air for low-freq. EM waves

27 Mirage In a hot day, air temperature is higher near the ground n = K P T 1atm Light travels faster near the ground slow fast cool hot T Refraction makes light bend upward You may see the ground reflect light as if there is a patch of water

28 Summary Studied microscopic origin of refraction in matter Started from EM radiation due to accelerated charges Uniform distribution of such charges make EM waves appear to slow down Three levels of describing EM waves in matter Point charges in vacuum Current density in vacuum Permittivity (and permeability) of matter Analyzed EM waves in plasma and in insulator n plasma = 1 ω p 2 ω 2 n insulator = 1+ ρω 0 2 (ω 0 2 ω 2 ) Plasma is reflective for ω < ω p = q2 n 0 ε 0 m

Problem set 3. Electromagnetic waves

Problem set 3. Electromagnetic waves Second Year Electromagnetism Michaelmas Term 2017 Caroline Terquem Problem set 3 Electromagnetic waves Problem 1: Poynting vector and resistance heating This problem is not about waves but is useful to