E dot B equals 0 REVIEWED

The E dot B equals 0 problem PhL 1.27.14 See red comment in the v1 doc. I have to say something about my little proof around (3.7.1). It is not valid below a certain ω and I would like to make an estimate. I get to this point C E B = C E B cosθ = ( ∂xBy - ∂yBx)Bz + ( ∂yBz - ∂zBy)Bx + ( ∂zBx - ∂xBz)By C E B = curl B B = [ r-1∂θBz - ∂zBθ] + [∂zBr - ∂rBz] + [ r-1∂r(rBθ) - r-1∂θBr ] [ Br + Bθ + Bz ] = Br[ r-1∂θBz - ∂zBθ] + Bθ[∂zBr - ∂rBz] + Bz [ r-1∂r(rBθ) - r-1∂θBr ] Let's just assume Br = 0 for simplicity. Then C E B = curl B B = + Bθ[∂rBz] + Bz[ r-1∂r(rBθ) ] = Bθ[∂rBz] + Bz[ r-1Bθ + ∂rBθ ] where C = j(β2/ω) = j [ωμ( jωε + σ) /j] /ω = μ( jωε + σ) where μ and σ are for the dielectric. So then C E B = C E B cosθ = μ( jωε + σ) E B cosθ = ( jωε + σ) E H cosθ So then, C E B cosθ = Bθ[∂rBz] + Bz[ r-1Bθ + ∂rBθ ] But B ≈ Bθ so then roughly C E cosθ = [∂rBz] + Bz[ r-1 + (∂rBθ )/Bθ] μ( jωε + σ) E cosθ = [∂rBz] + Bz[ r-1 + (∂rBθ )/Bθ] ( jωε + σ) E cosθ = [∂rHz] + Hz[ r-1 + (∂rHθ )/Hθ] I need estimates for E and B. E D = V(z) 2πa H = i(z) at surface D = transverse dimension H = field outside round wire H = i(z)/2πa 1/E = D/V(z) H/E = i(z)/V(z) * D/2πa = (D/2πa) 1/Z0 so then roughly we have H/E = 1/Z0 amps/m // volts/m = mhos so that 1/E = 1/(Z0Hθ) Then above becomes ( jωε + σ) E cosθ = [∂rHz] + Hz[ r-1 + (∂rHθ )/Hθ] ( jωε + σ) cosθ = { [∂rHz] + Hz[ r-1 + (∂rHθ )/Hθ] }/ (Z0Hθ) Now yes, if you set Hz≡ 0, you get cosθ = 0 so θ = 90o . But let's keep only a few terms so that ( jωε + σ) cosθ = { Hz[ D-1] }/ (Z0Hθ) = (1/DZ0) (Hz/Hθ) Dimensions? σ = mho/m DZ0 = m ohm 1/DZ0 = mho/m checks Let's now assume that σ = 0 for dielectric, then we have jωε cosθ = (1/DZ0) (Hz/Hθ) cosθ ≈ (1/DZ0) (Hz/Hθ)(1/ωε0) A condition for E B = 0 is then (1/DZ0) (Hz/Hθ)(1/ωε0) << 1 ω >> (1/DZ0ε0) (Hz/Hθ) Now assume ω > 60 Hz so at least we have Z0 = K 30Ω ω >> (1/DK30ε0) (Hz/Hθ) ω2 >> (1/DK30ε0)2 (Hz/Hθ)2 Then from below ω2 >> (1/DK30ε0)2 = (1/DK30)2 (1/ε0) ω >> (1/DK30)2 (1/σε0) Now (1/σε0) ≈ 1/[ 108 10-11] = 103 So ω >> (1/DK)2 This is a very low frequency! Suppose D = 1 cm = 10-2 m and suppose K = 2. then ω >> [ 2 10-2 ]-1 = 100/2 = 50 Hz. Then D Z0 ε0 = D K 30Ω 10-11 dim = m ohm farad/m = ohm-farad = sec OK Then ω >> 1011 / (30DK) * (Hz/Hθ) Now for my round wire I think I had Bz(r,0) = (β'/ω) J0(x) Bφ(r,0) = (β'/ω) ( + ) J1(x) Then Hz/Hθ= (βd/β) J0(x)/J1(x) ≈ (βd/β) = ( .)-1/2 from below (D.2.2) call this Hz/Hθ ≈ ( )-1/2 Then I get ω >> 1011 / (30DK) * (Hz/Hθ) ω2 >> 1022/ (30DK)2 ω >> 1022/ (30DK)2 = 1022 / (30DK)2 * 10-11 / 108 ω >> 1011 / (30DK)2 / 108 ω >> 103 / (30DK)2 2πf >> 103 / (30DK)2 f >> (1/6)(1000 Hz) / (30* .01m * 5)2 (1/6)(105 Hz) / (30* 1 * 5)2 = 105 / 135000 = 100000/ 135000 = 1 Hz ??? curl B = [ r-1∂θBz - ∂zBθ] + [∂zBr - ∂rBz] + [ r-1∂r(rBθ) - r-1∂θBr ] Now keep only Bθ