Better expression for R and Li v2

Better expression for R and Li PhL 9.6.14 This entire effort is ill conceived. I was thinking that "R" was the effective bulk R of the round conductor, so here I integrate Jz over the cross section, but really R is the real part of the surface impedance, and I already have an expression for that for the symmetric round wire in Chapter 2! And the inductance is the imaginary part Li. Oy. It is the SURFACE of the conductors that is involved with transmission lines. My current Appendix Q model with simple Heaviside functions seems too crude. In that model, C = C and G = expression valid for all ω. The problem is with Li and R. I think I can come up with a better and smoother model for these two parameters based on physics. At high ω, the current flows in a sheaf and I have expressions for Li and R for such a sheaf. In lines doc this is treated in App C and App K and App R for Belden. See App R section (f). I will now " try a few things" in hopes of coming up with something better. Let's start at relatively high ω so we have this picture where we see the annulus and a crude plot of Jz versus radius. Our model here from (D.10.11) for Jz in a given partial wave m is, Jz(r,m) = -(j/2) σ ηm I Rdc (aβ) e(1+j)(r-a)/δ Also from (D.10.15), Ez(r,θ) = - (jω/σ) e(1+j)(r-a)/δ n(θ) (β/k) Jz(r,θ) = - jωe(1+j)(r-a)/δ n(θ) (β/k) In a separate file (App D filed) I just verified that Jz integrates to I as long as a/δ >> 1. If this last is not true, the erf appears in the integral. Of course I claim to have this more general answer, at least in partial waves, Jz(r,m) = σ (1/4) ηm I Rdc (aβ') fm(r) Note that Jz(r,-m) = Jz(r,m) for Ch 6 case Jz(r,θ) = Σm=-∞∞ Jz(r,m) ejmθ = Jz(r,0) + 2 σ (1/4) I Rdc (aβ') Σm=1∞ ηm fm(r) cos(mθ) But maybe I can just assume a coaxial cable so we have a symmetric in θ situation. Then Jz is what I computed in Ch 2. Here that is just Jz(r,θ) = Jz(r,0) = σ (1/4) I Rdc (aβ') f0(r) = σ (1/4) I Rdc (aβ') f0(r) = σ (1/4) I Rdc (aβ') 2 J0(x)/J1(xa) = σ (1/2) I Rdc xa J0(x)/J1(xa) = (1/2) I (1/πa) (β) J0(x)/J1(xa) β ≠ β' = (I/2πa) (β) J0(x)/J1(xa) which agrees with (2.2.30). So let's perhaps simplify the problem and just work on coaxial cable center conductor. We then have Jz(r,θ) = (I/2πa) (β) J0(β'r)/J1(β'a) β'2 = β2- k2 and perhaps I could add a factor n(θ) / <n(θ)> = n(θ)/ N0 to simulate the θ dependence. So here then is the question: given the above Jz function, what can I say about R and Li for the center conductor? Since k2 appears here, we have a somewhat circular situation since I am trying to create a model from which I can compute k2!! I think I will have to make the ansatz that β' ≈ β. I recently studied this question in " Appendix M rewrite done Aug 19, 2014 original.doc" and it seems reasonable there based on my simple 2-phase model for k. So let's just assume that Jz(r) = (I/2πa) (β) J0(βr)/J1(βa) ansatz From this Jz(r) function, I should be able to compute both R and Li at all ω ! Li = μi ∫in dS (Hi/I)2 (C.3.7) The magnetic field Hi at radius r is determined by 2πr H(r) = Ienc = !Syntax Error, Ir dr!Syntax Error, Idθ Jz(r) = 2π !Syntax Error, Ir dr Jz(r) = 2π !Syntax Error, Ir dr J0(βr) = !Syntax Error, Ir dr J0(βr) Now let x = βr dx = βdr xdx = β2rdr so that 2πr H(r) = !Syntax Error, Ixdx/β2 J0(x) = !Syntax Error, IxJ0(x) dx This I know is a doable integral. GR7 p 630 says so the integral is just [xJ1(x)]|βr0 = (βr)J1(βr), so we find that 2πr H(r) = (βr)J1(βr) = I (r/a) J1(βr)/ J1(βa) which is very simple. When r = a you get I as expected. So: H(r) = I (1/2πa) J1(βr)/ J1(βa) amp/m Then I can compute the internal inductance of the center conductor this way 2Li = μ ∫in dS |H/I|2 // correction made for complex fields, see below = μ 2π !Syntax Error, Ir dr |(1/2πa) J1(βr)/ J1(βa)|2 Do a check right here: For small ω, J1(z) = z/2 so 2Li = μ 2π (1/2πa)2 !Syntax Error, Ir dr (r/a)2 = μ 2π (1/4π2a2) a-2 (1/4)a4 = μ/8π But this seems wrong by factor of 2. So get this cleaned up before continuing. = μ 2π (1/2πa)2 |1/ J1(βa)|2 !Syntax Error, Ir dr |J1(βr)|2 = μ 2π (1/4π2a2) |1/ J1(βa)|2 !Syntax Error, Ir dr |J1(βr)|2 = μ (1/2πa2) |1/ J1(xa)|2 !Syntax Error, Ir dr |J1(βr)|2 = μ (1/2πa2) |1/ J1(xa)|2 K The integral here is K = !Syntax Error, Ir dr |J1(βr)|2 = !Syntax Error, I xdx/β2 * |J1(x)|2 = (1/β2) !Syntax Error, I xdx |J1(x)|2 Check: For small ω, have J1(βr) ≈ (βr/2) and then K = !Syntax Error, Ir dr (|β|r/2)2 = |β2|/4 * !Syntax Error, Ir3 dr = |β2|/4 * (1/4) a4 = |β2| a4 . Then we get 2Li = μ (1/2πa2) (2/|β|a)2 |β2| a4 = μ (1/2π) (2)2 = 2μ/π => Li = μ/π = wrong Something is wrong, but I don't pursue this since topic not relevant. STOP. With first = you can see that K is a real number, but this fact seems lost in the next form. Let's just leave it as is and write K = !Syntax Error, Ir dr |J1(βr)|2 = !Syntax Error, Ir dr J1(βr) J1*(βr) Side Note: If β is very small, we get K ≈ a2/2 and then 2Li = μ (1/2πa2)(2/|xa|) a2/2 = μ (1/2πa2)(2/a|β|) a2/2 = μ (1/4π)(2/a|β|) = wrong!! Now I think we can claim from the series form of J that Jν*(z) = Jν(z*). I thought this is "real analytic" but I see from the web that is a not the right meaning of real analytic, there seems to be no name for this thing. If the power series has real coefficients, then you conclude that Jν*(z) = Jν(z*), so fine. We then have K = !Syntax Error, Ir dr J1(βr) J1(β*r) Luckily I can evaluate this integral using where I will take the first form. Then K = !Syntax Error, Ix dx J1(β*x) J1(βx) = [ β* x J1(β*x)J0(βx) - βx J0(β*x)J1(βx) ] |a0 The zero end evaluation gives nothing since J1 = 0 there, so we then have K = [ β* a J1(β*a)J0(βa) - βa J0(β*a)J1(βa) ] = [ xa*J1(xa*)J0(xa) - xa J0(xa*)J1(xa) ] Now we know that (z2)* = (z*)2 . Reverse order in both top and bottom to get K = [xa J0(xa*)J1(xa) - c.c. ] = 2j Im[xa J0(xa*)J1(xa)] = Im[xa J0(xa*)J1(xa)] = real! Just verifying that K is real. Side Note: If β is very small, we get K = Im[(xa2)/2] = Im[β2a2/2] = a2/2 so OK. But I want to write this a different way. Note that β2 = -jωμσ => Imβ2 = -ωμσ Go back so that K = [xa J0(xa*)J1(xa) - c.c. ] dim K = m2 = [xa J0(xa*)J1(xa) - c.c. ] = [xa J0(xa*)J1(xa) - xa* J0(xa)J1(xa*) ] = [xa J0*(xa)J1(xa) - xa* J0(xa)J1*(xa)] So we then have 2Li = μ (1/2πa2) |1/ J1(xa)|2 K = (μ/2πa2) [xa J0*(xa)J1(xa) - xa* J0(xa) J1*(xa)] = (μ/2πa2) [xa J0*(xa)/J1*(xa) - xa* J0(xa)/J1(xa)] = (1/4πa2) [ - ] = (j/4πa2σω) [ - ] and finally here is my result: Li = [ - ] But OK, now combine the two terms to get 2jIm(..) to get Li = 2j Im [ ] = - Im [ ] // real at least Now evaluate this at small β meaning small xa to get J0 = 1 J1 = xa/2 Im [ ] = Im [ ] = 2 Im [ ] = 2 Im [] Now write β ≡ ej3π/4 from (2.2.12) to get = 2 Im [] = 2 Im[e+j3π/2] = 2 Im[-j] = -2 So our low ω limit is then Li = - [ -2] = wrong!!! *********************** So we then have Li = μ 2π (1/2πa)2 |1/ J1(βa)|2 * (1/β2) !Syntax Error, I xdx |J1(x)|2 = !Syntax Error, I xdx |J1(x)|2 = !Syntax Error, I xdx |J1(x)|2 Maple cannot do this integral. But GR7 can using which says !Syntax Error, I xdx [J1(x)]2 = [ (x2/2) { [J1(x)]2 - J0(x)J2(x) }|βa0 = (β2a2/2) { [J1(βa)]2 - J0(βa)J2(βa) } = (xa2/2) { [J1(xa)]2 - J0(xa)J2(xa) } and so Li = (xa2/2) { [J1(xa)]2 - J0(xa)J2(xa) } = { [J1(xa)]2 - J0(xa)J2(xa) }} = ( 1 - ) xa = βa Check some limits here. At low ω, xa is small and we use Jn(x) = (x/2)n / n! so J0 = 1 J1 = xa/2 J2 = (xa/2)2/2 so then Li = ( 1 - ) = ( 1 - ) = μ/8π hurray!! Fine, but Li is complex, so I think I need to fix something in Appendix C interpretation. There I write u(r) = (1/2) μ H(r)2. (C.3.1) So what does it mean that the energy density has an r-dependent phase? I suspect this should say u(r) = (1/2) μ |H(r)|2. (C.3.1) My appendix is only for DC applications. Jackson says uem = (ED + BH)/2 . But if we have monochrome fields, this usual thing happens H = H1cos(ωt) 2<u>/μ = < HH> = ∫ dt [H1cos(ωt)]2 = (1/2) H1 H1 And if H = H1ejωt then H H* = H1 H1 = 4 <u>/μ so the rule is this <u> = (1/4) μ H H* so you add a * to one of the fields, and you add an extra overall 1/2. To quote from Jackson so that is an example of this extra factor of 1/2. So I will now REDO the above calculation