Better expression for R and Li

Better expression for R and Li PhL 9.6.14 This is the original doc version. It has the same problem as version 2, so it is irrelevant! R in k(ω) is not the bulk R, it is real part of Zs. 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 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