Joel A. Shapiro January 21, 2010

Transcription

1 Joel A. January 21, 20

2 rmation Instructor: Joel Serin X 3886, shapiro@physics Book: Jackson: Classical Electrodynamics (3rd Ed.) Web home page: contains general info, syllabus, lecture other notes, homework assignments, etc. Classes: ARC 207, Monday Thursday, 10:20 (sharp!) - 11:40 Homework: there will be one or two projects, homework assignments every week or so. Due dates to be discussed. Exams: a midterm a final. Office Hour: Tuesdays, 3:30 4:30, in Serin 325.

3 Course Content Last term you covered Jackson, Chapters 1 7 We will cover most of chapters 8 16 Everything comes from Maxwell s Equations the Lorentz Force. We will discuss: EM fields confined: waveguides, cavities, optical fibers Sources of fields: antennas their radiation, scattering diffraction Relativity, relativistic formalism for E&M Relativistic particles other gauge theories

4 Maxwell s Equations B 1 E c 2 t E + B t E = 1 ɛ 0 ρ all Gauss for E B = 0 Gauss for B = µ 0 Jall Ampère (+Max) = 0 Faraday plus the Lorentz force: F = q( E + v B)

5 In Ponderable Media In the last slide, ρ all J all represent all charges, both free induced. Separate free from induced : For the electric field E called electric field P called electric polarization is induced field D called electric displacement is field of free charges D = ɛ0e + P For the magnetic field B called magnetic induction (unfortunately) M called magnetization is the induced field H called magnetic field H = 1 µ 0 B M Then the two Maxwell equations with sources, Gauss for E Ampère, get replaced by H D t D = ρ = J

6 between conductor non-conductor Conductor c: if perfect, no E. Surface charge Σ, eddy currents so no H inside conductor. Just outside the conductor: Faraday on loop Γ E 0 Gauss on pillbox S B 0 n Γ S ξ conductor c

7 For good (not perfect) conductor, take J = σ E with large conductivity σ. Assume time-dependence e iωt Skin depth ξ, H varies rapidly. H c = J + D t σ E, E c = B t = iωµ ch c Rapid variation with depth ξ dominates, = ˆn ξ, E c = 1 J σ = 1 σ ˆn H c ξ, Hc = i ˆn E c ωµ c ξ

8 E c = 1 J σ = 1 σ ˆn H c ξ, Hc = i ˆn E c ωµ c ξ so ˆn H c = 0 ˆn H c = ( i ˆn ˆn E c ωµ c ξ ( [ = i σωµ c ˆn = ˆn ) i 2 ( σωµ c ξ 2 ˆn H ) c. ]) ˆn 2 Hc ξ 2 Simple DEQ, exponential solution, with δ = H c = H e ξ/δ e iξ/δ, 2 µ c ωσ, H is tangential field outside surface of conductor.

9 From H c = H e ξ/δ e iξ/δ, E c = 1 σ ˆn H c ξ = µ c ω 2σ (1 i)ˆn H e ξ/δ e iξ/δ, which means, by continuity, that just outside the conductor µ c ω E = 2σ (1 i)ˆn H.

10 How much power is dissipated (per unit area?). 2 ways: 1) Flow of energy into conductor: Energy flow given by S = E H, for real fields E H. so 1 S ( ) = 1 2 Re E H, dp loss = ˆn S da = 1 µc ω [(1 2 2σ ˆn Re i)(ˆn H ) H ] = µ cωδ 4 H 2 = 1 2σδ H 2 Method 2, Ohmic heating, power lost per unit volume 1 2J E = J 2 /2σ, J = σe c = 2 δ H e ξ/δ, the power loss per unit area is dp loss da = = 1 δ 2 σ H 2 Agrees with method 1. 0 dξ e 2ξ/δ = 1 2δσ H 2. 1 The 1, Re, * will be discussed in lecture 3. 2

11 In terms of surface current K eff = Thus 0 = ˆn H. dξ J(ξ) = 1 δ ˆn H dξ (1 i)e ξ(1 i)/δ dp loss da = 1 2σδ K eff 2. 1 is surface resistance (per unit area) σδ is the surface impediance Z. 0 E K eff = 1 i σδ

12 For electromagnetic fields with a fixed geometry of linear materials, fourier transform decouples, we can work with frequency modes, E( x, t) = E(x, y, z) e iωt B( x, t) = B(x, y, z) e iωt Actually the fields are the real parts of these complex expressions. If ρ = 0, J = 0, Maxwell gives Then E = B t = iω B, E = 0, B = 0, B = µ H = µ D t = µɛ E t = iωµɛ E. 2 E = ( E)+ ( E ) = (iω B) = ω 2 µɛ E.

13 similarly for B, so we get Helmholtz equations ( 2 + ω 2 µɛ ) E = 0, ( 2 + ω 2 µɛ ) B = 0. Consider a waveguide, a cylinder of arbitrary cross section but uniform in z. Fourier transform in z E(x, y, z, t) = E(x, y)e ikz iωt B(x, y, z, t) = B(x, y)e ikz iωt k can take either sign ( a sting wave is a superposition of k = ± k ). The Helmholtz equations give [ 2 t + (µɛω 2 k 2 ) ] ( E(x, ) y) = 0, B(x, 2t := 2 y) x y 2.

14 Decompose longitudinal transverse Let E = E z ẑ + E t B = B z ẑ + B t with E t ẑ B t ẑ ( E) z = ( t E t ) z = iωb z, ( E) = ẑ E t z ẑ te z = iω B t. For any vector V, ẑ (ẑ V ) = V + ẑ(ẑ V ), so for a transverse vector ẑ (ẑ V t ) = V t. Taking ẑ last equation, E t z t E z = iωẑ B t. (1) Similarly decomposition of B = iωµɛe gives ( t B ) t = iωµɛe z z B t z t B z = iωµɛẑ E t. (2)

15 Divergencelessness: t E t + E z z = 0, t B t + B z z = 0. Equations (1) (2), with the fourier transform in z, give ik E t + iωẑ B t = t E z (3) ik B t iωµɛẑ E t = t B z (4) Solving 4 for B t plugging into 3, then the reverse for E t, give E t = i k t E z ωẑ t B z ω 2 µɛ k 2 (5) B t = i k t B z + ωµɛẑ t E z ω 2 µɛ k 2 (6) Unless k 2 = k 2 0 := µɛω2, E z B z determine the rest.

16 We have seen that E z B z largely determine the fields, these satisfy the two-dimensional Helmholtz equation ( 2 t + γ 2) ψ = 0 with γ 2 = µɛω 2 k 2 (7) If the walls of the waveguide are very good conductors, we may impose the perfect conductor conditions E 0 B 0 on the boundary S of the two-dimensional cross section. E z is parallel to the boundary so E z S = 0. Also the component of E t parallel to the boundary vanishes at the wall, so E t is in the ±ˆn direction. Then from the ˆn component of (2) (normal to the boundary) ˆn B t z ˆn t B z = iωµɛˆn (ẑ E ) t = 0 B z n = 0, where / n is the derivative normal to the surface. So we have Dirichlet conditions on E z Neumann conditions for B z.

17 In general, nonzero solutions exist only for discrete values of γ, those values are generally different for Dirichlet for Neumann. So we need to consider, with E z (x, y) = B z (x, y) 0. That is, there are no longitudinal fields, both electric (E) magnetic (M) fields are purely transverse to the direction z of propagation. TE modes, E z (x, y) 0, the transverse fields are determined by the gradiant of B z = ψ, a solution of (7) with Neumann conditions. TM modes, B z (x, y) 0, the transverse fields are determined by E z = ψ, a solution of (7) with zero boundary conditions.

18 With E z (x, y) = B z (x, y) 0, (5) (6) = everything vanishes or the denominator vanishes, k = ±k 0 with k 0 = µɛ ω Wave travels z with speed 1/ µɛ, same as for infinite medium. No dispersion. t E t = 0 t E t = iωb z = 0, so Φ E t = t Φ (though Φ might not be single valued) 2 Φ = 0. As E S = 0, Φ = constant on each boundary. If cross section simply connected, Φ = constant, E = 0 No on simply connected cylinder Yes on coaxial cable, or two parallel wires.

19 TE TM modes Equations (5) (6) simplify for TM modes, B z = 0, γ 2 Et = ik t E z, γ 2 Bt = iµɛωẑ t E z, so H t = ɛωk 1 ẑ E t. TE modes, E z = 0, γ 2 Bt = ik t B z, γ 2 Et = iωẑ t B z, so E t = ωẑ B t /k = ẑ H t = kẑ E t /µω. In either case, H t = 1 Z ẑ E t, with { k/ɛω = (k/k0 ) µ/ɛ TM Z = µω/k = (k 0 /k) µ/ɛ TE

20 To Summarize Solutions given by ψ(x, y), with ( 2 t + γ 2) ψ = 0, γ 2 = µɛω 2 k 2, by TM: E z = ψe ikz iωt, Et = ikγ 2 t ψe ikz iωt with ψ Γ = 0 TE: H z = ψe ikz iωt, Ht = ikγ 2 t ψe ikz iωt with ˆn t ψ Γ = 0 By looking at 0 = A ψ ( 2 t + γ 2) ψ we can show γ 2 0. There are solutions for discrete values γ λ, so only certain wave numbers k λ for a given frequency can propagate: k 2 λ = µɛω2 γ 2 λ, only frequencies ω > ω λ := γ λ / µɛ can propagate, k λ < µɛ ω, the infinite medium wavenumber. Phase velocity v p = ω/k λ is greater than in the infinite medium.

21 : Circular Wave Guide Jackson does rectangle. You should too. Needed to do homework. We will consider a circular pipe of (inner) radius r. Of course we should use polar coordinates ρ, φ, with 2 t = 1 ρ ρ ρ ρ ρ 2, try ψ(ρ, φ) = R(ρ)Φ(φ), φ2 ( 2 t + γ 2) ψ = ( ) 1 ρ ρ ρ R ρ + γ2 R(ρ) Φ(φ) + 1 Φ(φ) ρ 2 R(ρ) 2 φ 2 = 0. Divide by R(ρ)Φ(φ) multiply by ρ 2 : ( 1 ρ ) R(ρ) ρ ρ R ρ + γ2 ρ 2 R(ρ) Φ(φ) Φ(φ) φ 2 = 0.

22 : Circular Wave Guide Jackson does rectangle. You should too. Needed to do homework. We will consider a circular pipe of (inner) radius r. Of course we should use polar coordinates ρ, φ, with 2 t = 1 ρ ρ ρ ρ ρ 2, try ψ(ρ, φ) = R(ρ)Φ(φ), φ2 ( 2 t + γ 2) ψ = ( 1 ρ ρ ρ R ) ρ + γ2 R(ρ) Φ(φ) + 1 ρ 2 R(ρ) 2 Φ(φ) φ 2 = 0. Divide by R(ρ)Φ(φ) multiply by ρ 2 : ( 1 ρ ) R(ρ) ρ ρ R ρ + γ2 ρ 2 R(ρ) = C 1 2 Φ(φ) Φ(φ) φ 2 = C.

23 Solving it Φ first: 2 Φ(φ) φ 2 + CΦ(φ) = 0 Φ(φ) = e ±i Cφ. Periodicity = C = m Z. Now R(ρ): (ρ ρ ρ ρ + γ2 ρ 2 m 2 ) R(ρ) = 0 Bessel equation, solutions regular at origin are R(ρ) J m (γρ), so ψ(ρ, φ) = m,n A m,n J m (γ mn ρ)e imφ. γ mn is determined by boundary conditions...

24 Boundary conditions: For TM, ψ(r, φ) = 0 = J m (γr) = 0, so γ TM mn = x mn /r where x mn is the n th value of x > 0 for which J m (x) = 0, given on page 114. For TE, ˆn t ψ(r, φ) = 0 = djm dr (γr) = 0, so γmn TM = x mn/r where x mn is the n th value of x > 0 for which dj m (x)/dx = 0, given on page 370. Thus the lowest cutoff frequency is the m = 1 TE mode, with x 11 = while the lowest TM mode or circularly symmetric mode has x = For a waveguide 5 cm in diameter, with air or vacuum inside, the cutoff frequencies are f = ω 2π = 3.5 GHz for the lowest TE 4.6 GHz for the lowest TM modes.