Physics 504, Spring 2010 Electricity and Magnetism

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1 Resonant cavities do not need to be cylindrical, of course. The surface of the Earth R E 6400 m the ionosphere R = R E + h, h 100 m form concentric spheres which are sufficiently good conductors to form a resonant cavity. Tae E, H e iωt, cavity essentially vacuum. Z 0 = µ 0 /ɛ 0, c = 1/ µ 0 ɛ 0. Set = ω/c. Maxwell: E = iz 0 H, So H H = i Z 0 E, E = 0, H = 0, = i Z0 E = H = H H } {{ } 0

2 So + H = 0, H = 0, E Z 0 = i H. Similiarly we can derive + E = 0, E = 0, H 1 = i E. Z 0 Each cartesian component obeys Helmholtz, but the radial component r A for A either E or H is more suitable to loo at. r A = r r j A j = r j ij i r A j + A j δ ij ij i r i so = r A + A }{{} + r E = 0, =0 for E, H. + r H = 0. Magnetic multipole field: r E 0, Electric multipole field: r H 0. Whichever isn t identically zero satisfies Helmholtz.

3 In spherical coordinates, = 1 r r r r + 1 r [ 1 sin θ sin θ + 1 θ θ sin θ ] φ. Thus solutions of Helmholtz s are found by separation of, F ry θ, φ, where the angular part satisfies [ 1 sin θ θ sin θ θ + 1 sin θ ] θ Y lm = ll + 1Y lm. This you should recognize from Quantum Mechanics as the for the spherical harmonics. Single-valuedness for corresponding values of θ φ require l Z.

4 Thus the solutions are TE: TM: r r H M lm ll + 1 = g l ry lm θ, φ, r E M = 0 ll + 1 f l ry lm θ, φ, r H E = 0. E E lm = Z 0 In fact, let s steal more from quantum mechanics. Define the operators L = i r. L ± = L x ± il y = e ±iφ ± θ + i cot θ, L z = i φ φ, we recall L ± Y lm = l ml±m+1y l,m±1, L z Y lm = my lm, L Y lm = ll+1y lm.

5 Dotting r into the first Maxwell, iz 0 r H = r E = r E = il E, so for the magnetic multipole TE field L E M lm = Z 0 r H = Z 0 g l rl Y lm, which at least hints at E M lm = Z 0g l r LY lm. 1 Also, this is consistent with r r L = i r r = 0. E M lm = 0 as

6 The rest of the fields in a magnetic multipole are H M lm = i Z 0 E M lm. This magnetic multipole field configuration is also called transverse electric TE, as E is transverse to the radial direction. The same holds for the electric multipole TM field: H E lm = f lr LY lm θ, φ, E E lm = iz 0 H E lm = Z 0 But r = r 1 + r r r f l ry lm θ, φ., so E E lm = Z [ 0 r 1 + r ] f l ry lm θ, φ. r

7 the transverse part of the electric field is determined by r 1 + r r r E E lm = Z 0 r = Z 0 = i Z 0 f l ry lm θ, φ r 1 + r f l ry lm θ, φ r [ 1 + r [ L ] f l r] Ylm θ, φ. r Now we need r E E lm = 0 at r = R E r = R E + h. Note for l = 0 we have spherical symmetry, vece H are purely radial angle-independent, so then E = 0 = E c/r, we have a solution only for = 0 this is a static coulomb field. For l 0, vanishing requires 1 + r r fl r = 0 at r = R E r = R E + h. If, instead, we loo for a magnetic multipole solution, we need g l r = 0 at r = R E r = R E + h.

8 Solution of Equation As = r + r mode satisfies r + r r 1 r L, the radial part of an l, m ll + 1 r r + g l r = 0. The same holds for f l r. Solutions are spherical Bessel Hanel functions, similar to sinr cosr. Easy to mae combinations which vanish at two points h apart, with of order π/h. For h 100 m, frequency 10 Hz. Radio waves are higher frequency, we could use geometrical optics to describe what happens. j n j 4 x n 4 x 4 4

9 We could have resonant magnetic multipole TE fields in the ilohertz range. But there are observed resonances at 8, 14, 0 Hertz! Why? We need to loo at solutions more closely. Our is d dr + r d dr + ll + 1 r f l r = 0, This can be transformed in several useful ways. Fiddle the scale of r the multiply by a power of r, f l r = u l,α,ββr βr α d = dx + d x dx + 1 ll + 1 ul,α,β x β x x α = 0 d dx + 1 α x d dx + 1 β αα 1 ll+1 + x u l,α,β x = 0.

10 Two useful choices for α β: Choice 1: α = 1/, β = 1 d dx + 1 d x dx l + 1/ + 1 x u l, 1,1x = 0, This is Bessel s with ν = l + 1, solutions u = aj l+ 1 r + bn l+ 1 r, f l r = a j l r + b n l r, where j n are spherical Bessel spherical Neumann functions: j l x = π x J l+1/x, n l x = π x N l+1/x h 1, π l x = Jl+1/ x ± i N x l+1/ x.

11 Choice : α = 1, β = 1/ ll + 1, d + ll x dx u l = 0, As f u/r, the boundary conditions for an electric multipole TM field at x = βr E x = βr E + h are 1 + r d uβr = 0 = du/dx, with x = βr. dr βr To get du/dx to vanish at nearby x s is now easy. Of course the average value of d u/dx has to be zero between the two zeroes of du/dx, but that is assured by for x = 1 roughly in the center of the interval, so 1 βr E, or ll + 1, f = c ll + 1 = 7.46 ll + 1 R E π R E Hz = 10.5 Hz, 18.3 Hz, 5.8 Hz,.... The observed resonant frequecies are about 0% lower, said to be due to imperfect conductivity of the ground ionosphere.