Electromagnetic Theory (Hecht Ch. 3)

Transcription

1 Phys 531 Lecture 2 30 August 2005 Electromagnetic Theory (Hecht Ch. 3) Last time, talked about waves in general wave equation: 2 ψ(r, t) = 1 v 2 2 ψ t 2 ψ = amplitude of disturbance of medium For light, medium = EM field 1 This time: derive wave equation from Maxwell s Equations study properties of EM waves in vacuum Next time: consider EM waves in matter 2

2 Maxwell s Equations: Basic postulates of electromagnetism Physical quantities: Electric field E(r, t) (Volts/m) Magnetic field B(r, t) (Tesla) Charge density ρ(r, t) (Coulombs/m 3 ) Current density J(r, t) (Amperes/m 2 ) Gauss s Laws (charge produces a field): E ds = 1 ρ dv B ds = 0 ɛ 0 ɛ 0 = permittivity of free space = 8.8 pf/m (from capacitor measurements) 3 Faraday s Law (changing B produces E): B E dl = t ds Ampere s Law (current, changing E produce B): B dl = µ 0 ( J + ɛ 0 E t ) ds µ 0 = permeability of free space = 1.3 µh/m 4π 10 7 H/m These are integral form of Maxwell s equations 4

3 Will be more useful in differential form Use Gauss s Theorem: for any F(r) F ds = F dv Here ˆx x + ŷ y + ẑ z Also Stoke s Theorem: F dl = ( F) ds Analogous to ordinary calculus: f(x 2 ) f(x 1 ) = function on boundary x2 x 1 df dx dx integral of derivative 5 Apply to Maxwell: E ds = E dv = 1 ρ dv ɛ 0 True for any volume V, so at each point r E(r) = 1 ɛ 0 ρ(r) Similarly, get B = 0 6

4 Also, for any surface S, so E dl = E = B t and similarly E ds = B = µ 0 J + ɛ 0 µ 0 E t B t ds These are differential form of Maxwell s equations 7 Light Waves (Hecht 3.2) Light propagates in vacuum: ρ = J = 0 Maxwell equations become: E = 0 (1) B = 0 (2) E = B (3) t E B = ɛ 0 µ 0 (4) t Where s the wave equation? 8

5 Derivation of wave equation Take / t of (4): or t ( B) = ɛ 2 E 0µ 0 t 2 B t = ɛ 2 E 0µ 0 t 2 Then using (3): ( E) = ɛ 0 µ 0 2 E t 2 Need to simplify ( E) 9 Cross product rule: so a (b c) = b(a c) (a b)c ( E) = ( E) ( )E But E = 0, and = 2, so 2 E = ɛ 0 µ 0 2 E t 2 wave equation, v = (ɛ 0 µ 0 ) 1/2 10

6 Similarly, take / t of (3), end up with 2 B = ɛ 0 µ 0 2 B t 2 So expect EM waves to exist. Do they correspond to light? Compare speeds: (ɛ 0 µ 0 ) 1/2 = m/s Measured light speed c = m/s Conclude light is EM wave 11 EM waves are vector waves Previously considered scalar waves Actually six coupled wave equations: (E x, E y, E z, B x, B y, B z ) Components must still obey Maxwell s equations example: B x x + B y y + B z z = 0 and E x y E y x = B z t 12

7 Can simplify for plane wave solutions: E(r, t) = E 0 e i(k r ωt) B(r, t) = B 0 e i(k r ωt) with E 0, B 0 = complex vector amplitudes Sometimes confusing, so write out: E 0 = E 0x e iφxˆx + E 0y e iφyŷ + E 0z e iφzẑ and actual wave is E = E 0x ˆx cos(k r ωt + φ x ) + E 0y ŷ cos(k r ωt + φ y ) + E 0z ẑ cos(k r ωt + φ z ) 13 Plane Waves Have E x = x E 0e i(k xx+k y y+k z z ωt) Also E y = ik ye, and B x = ik xb, = ik x E 0 e i(k xx+k y y+k z z ωt) = ik x E(r, t) E z = ik z E B y = ik y B, B z = ik z B Thus, for plane waves, can replace = ˆx x + ŷ y + ẑ z i(ˆxk x + ŷk y + ẑk z ) = ik 14

8 Also, t iω So wave equation becomes 2 E = 1 2 E c 2 t 2 k2 E = ω2 c 2 E Solution if k = ω/c, as before. Maxwell equations become k E = 0 k B = 0 k E = ωb k B = ω c 2 E 15 Exponentials factors e i(k r ωt) drop out, so ˆk E 0 = 0 ˆk B 0 = 0 ˆk E 0 = cb 0 ˆk B 0 = 1 c E 0 with ˆk = propagation direction Dot products indicate E 0, B 0 k Say EM waves are transverse Cross products indicate B 0 = 1 c ˆk E 0 So B E as well B 0 = E 0 /c 16

9 So (E, B, k) form orthogonal basis E k B Picture: (snapshot at fixed t; ˆk = ˆx) E x B 17 Question: If a wave has E along ˆx and B along ẑ, what is the direction of propagation? Question: How would my snapshot picture evolve in time? 18

10 Frequencies (Hecht 3.6) EM waves observed over large range of ω: ν (Hz) λ Name < 10 5 > 3000 m ELF wave m 30 cm radio wave cm 3 mm microwave mm 30 µm terahertz µm 750 nm infrared nm visible light nm ultraviolet nm 30 pm x-rays > < 30 pm gamma rays 19 Optics particularly deals with light: infrared to ultraviolet Lower ν: approximations not valid ray optics fails wave approximations fail Higher ν: quantum effects important wave effects hard to see ray optics OK, but no optical materials 20

11 Properties of plane waves Sometimes write E 0 = E 0 ĵ with ĵ = 1 E 0 = complex amplitude (V/m) ĵ = polarization vector = Jones vector - More on polarization later - Amplitude related to energy in wave: discuss now 21 General EM energy density (J/m 3 ): u(r) = ɛ 0 2 E µ 0 B 2 Energy in volume V = V u(r) dv For plane wave B = E /c. So u = ɛ 0 2 E µ 0 c 2 E 2 = ɛ 0 2 E 2 + ɛ 0 2 E 2 = ɛ 0 E 2 22

12 But here E refers to real electric field If E = Re [E 0 ĵe i(k r ωt) ], then really and E = E 0x ˆx cos(k r ωt + φ x ) + E 0y ŷ cos(k r ωt + φ y ) + E 0z ẑ cos(k r ωt + φ z ) u =ɛ 0 E 0x 2 cos 2 (k r ωt + φ x ) + ɛ 0 E 0y 2 cos 2 (k r ωt + φ y ) + ɛ 0 E 0z 2 cos 2 (k r ωt + φ z ) Energy oscillates in time. 23 For light, oscillation is rapid: usually average over many periods Average of cos 2 () over many periods = 1 2 So time-average u = ɛ 0 2 ( E 0x 2 + E 0y 2 + E 0z 2 ) In terms of complex fields, u = ɛ 0 2 E 2 where E 2 = E E = E 0x 2 + E 0y 2 + E 0z 2 Question: What is the total energy in a plane wave with amplitude E 0? 24

13 Also interested in energy flow: Use Poynting vector S = 1 µ 0 E B Can show that W = Σ S dσ is energy per unit time crossing surface Σ Units of S are W/m 2 25 Plane waves: B = 1 c ˆk E, so S = 1 µ 0 cˆk ( E 0x 2 cos 2 (k r ωt + φ x ) + E 0y 2 cos 2 (k r ωt + φ y ) + E 0z 2 cos 2 (k r ωt + φ z ) Direction of propagation ˆk = direction of energy flow Ŝ Question: What do you think S(r) looks like for a spherical wave? 26

14 Define irradiance (aka intensity) = time-average of magnitude of S I = S = 1 2µ 0 c E 0 2 = 1 2η 0 E 0 2 where η 0 (µ 0 /ɛ 0 ) 1/2 = 377 Ω impedance of free space Units check: (W/m 2 ) = (V/m) 2 /Ω (Recall P = V 2 /R from electronics) I is most common measure of optical field strength 27 In terms of complex fields: S = 1 2µ 0 E B (implicit time average!) Poynting vector gives other properties too: Energy density u = S c Linear momentum p = 1 c 2 S dv Angular momentum L = 1 c 2 r S dv 28

15 Summary: Light is EM wave Coupled E, B fields Plane waves, complex vector amplitudes Irradiance I = 1 2η 0 E 0 2 (W/m 2 ) 29