problems in Appendix B

Problems in Appendix B PhL 10.14.14 I have found some new forms of Maxwell's implied equations. 1. Allocation of E and J in the curl H equation Consider the Maxwell equation normally written as ξ ≡ ε - jσ/ω = ε + σ/jω curl H = ∂tD + J = jωD + J = jωεE + J Using Ohm's Law, we can write the RHS three different ways 1 jωεE + J 2 jωε(J/σ) + J = [ jω(ε/σ) + 1] J = ( jω/σ) [ ε + σ/jω ] = jω(ξ/σ) J 3 jω(ξ/σ)σE = jωξ E So here are our three ways to write the curl H equation: 1 curl H = jωεE + J 2 curl H = jω(ξ/σ) J = 3 curl H = jωξ E I have checked with Maple to make sure the three RHS's are the same. 2. Forms of the Helmholtz equation Consider this identity, curl curl H = grad divH -2H = -2H which then says 2H = - curl curl H This equation can then be written three ways using the three forms shown above. I will do them one at a time: 1 2H = - curl [jωεE + J] = -jωε curl E - curl J = -jωε [-jωB ] - curl J = -ω2εμH - curl J so that we end up with (2 + ω2εμ)H = -curl J or (2 + k2 )H = -curl J with k2 = ω2εμ This form appears in lines doc here: (2+k32)E = μ3jωJ3 (2+k32)B = - μ3 curl J3 // region 3 ki2 = ω2μiεi (1.5.26) 2 2H = - curl [jω(ξ/σ) J] = - jω(ξ/σ) curl J ξ ≡ ε - jσ/ω = ε + σ/jω This form does not appear anywhere in lines doc as far as I know. The ratio ξ/σ never appears. 3 2H = - curl[jωξ E] = -jωξ curl E = -jωξ [-jωB] = -ω2ξ B = -ω2ξμ H so that we end up with (2 + ω2ξμ )H = 0 (2 + β2 )H = 0 This form appears in lines doc here. (2+β32)E = 0 (2+β32)B = 0 // region 3 βi2 = ω2μiξi (1.5.27) Here then are the three different "Helmholtz equations" which are the same: 1 (2 + k2)H = -curl J with k2 = ω2εμ 2 2H = - jω(ξ/σ) curl J ξ ≡ ε + σ/jω jω(ξ/σ) = jω([ε + σ/jω]/σ) = jω(ε/σ) + 1 = - [1 + jω(ε/σ) ] curl J 3 (2 + β2 )H = 0 β2 = ω2ξμ 3. Copper conductor case Question: For copper, what is the size of jω(ε/σ) compared to 1 ? That is, when is ωε << σ ? I answer this question below (2.2.2) and conclude that ωε << σ as long as f << 1018 Hz. Therefore, in Form 2 above, we really can neglect the jω(ε/σ) relative to the 1. Now go back to curl H = ∂tD + J = jωD + J = jωεE + J = [ jω(ε/σ) + 1] J The fact that ωε << σ in this range is the same as saying you can neglect the displacement current in copper relative to the conduction current. This says ξ ≡ ε + σ/jω ≈ σ/jω all the way up to 1018 Hz Then jω(ξ/σ) = jω(1/σ) * σ/jω = 1 β2 = ω2ξμ = ω2(σ/jω)μ = -jμσω as usual Then the three forms shown above become, in the limit ωε << σ, 1 ( 2 + k2 )H = -curl J with k2 = ω2εμ 2 2H = - curl J 3 ( 2 + β2 )H = 0 β2 = -jμσω How do I explain how these can all be true at the same time? Are they if ωε << σ ? First consider β2 = -jμσω k2 = ω2εμ | k2/β2| = ωε/σ << 1 so we do know that k << β in magnitude. Comparing 1 and 2, I would like to show that | k2 H | << | 2 H| Using 3 this is the same as | k2 H | << | β2 H| and this is at once true since k << β . So that explains how 1 and 2 can both be true. 4. NOW take a look at Appendix B.2 (a) I have added the above data to the start of Section B.2. I can then take two different paths. If I assume the e-jkz ansatz, then ∂z2 = -k2 and 2H = -curl J // copper for f << 1018 Hz becomes (22D - k2) H = -curl J but I don't know how to invert this equation very well. But I do know how to invert -2H = curl J from Appendix H.1.8 -2 f(x) = s(x) => f(x) = ∫d3x' [1/4πR] s(x') + homogeneous solutions The Poisson Equation (H.1.8) so then H(x) = ∫d3x' [1/4πR] curl' J(x') I think this result is good for all ω, whereas the result I wrote H(x,y) = ∫d2x' [ ln(1/R) ] curl' J(x') R = |x-x'| (B.2.5) makes the assumption that J = Jz(x,y) so e-jkz ≈ 1 which means low ω. So maybe I will make the replacement above. Let's continue and see what comes next in Appendix B before doing this. 5. Take a look at Appendix B.2 (b) Here the starting equation is (2 + β2)A(x) = - μ J(x) . all of region R (1.5.4) STOP! Who says the above equation is true inside the conductor?? I think the correct equation to be using here is this (2 - με ∂t2)A = - μJ [ = Jackson (6.16) ] (1.3.5) which uses the Lorenz gauge. I don't think the King gauge makes sense for a single conductor? Well, I show that with the large σ assumption, I do get this result (2 - μ1ε1 ∂t2 - μ1σ1∂t ) A = - μ3J3 . region 3 (1.3.21) and I guess I can consider region 3 as my conductor. But in the King gauge, this yields, (2 + βd2)A(x) = - μ J(x) !! Even though we are inside the conductor, βd is what shows! Then it all works! I could do a full 3D solution to this using - (2+k2) f(x) = s(x) => f(x) = ∫d3x' [e-jβR/4πR] s(x') + homogeneous solutions The Helmholtz Equation (H.1.9) so the solution would be A(x) = μ ∫d3x' [e-jβR/4πR] J(x') Now consider B = curl A = μ ∫d3x' curl { [e-jβR/4πR] J(x') } But we know that for Q a constant vector, x (φQ) = φ x Q + φ ( x Q) = φ x Q = grad φ x Q Thus B = curl A = μ ∫d3x' grad [e-jβR/4πR] x J(x') so H = ∫d3x' grad [e-jβR/4πR] x J(x') Hi = ∫d3x' [ grad [e-jβR/4πR] x J(x') ]i = - εirs ∫d3x' [∂'r [e-jβR/4πR] Js(x') = + εirs ∫d3x' [e-jβR/4πR] ∂'r Js(x') = + ∫d3x' [e-jβR/4πR] εirs ∂'r Js(x') = + ∫d3x' [e-jβR/4πR] [curl' J'(x') ]i which then says H = + ∫d3x' [e-jβR/4πR] curl' J'(x') Compare this to the other method which said H(x) = ∫d3x' [1/4πR] curl' J(x') How do I then argue that both these results are the same for a good conductor?