doing the wrong problem REVIEWED

Doing the Wrong Problem PhL 11.25.13 Comments 12/5/13. Here is a summary of this doc: I consider the Helmholtz Az problem with an assumed uniform J in the wire, but for arbitrary ω. I realize that this problem is "illogical" since the current cannot be uniform unless dramatic external means are used to maintain it so. This is why the title refers to this as "the wrong problem" if you are trying to get something that matches Chapter 2 where I treat "the right problem" -- the true wire problem. In Section 1 I go ahead and compute Az using the Helmholtz integral, both inside and outside the wire, even though the problem is illogical. I make no mention of homo solutions and boundary conditions on Az because at this time I thought the Helmholtz integral always gave the exact answer, just the way King presents it. This lack of homo solutions is probably OK if you assume μ1 = μ2. I then think maybe the low-ω limit of the above calculation is "logical" even though the general ω result is "illogical". If the low-ω limit is valid, then it ought to agree with the low-ω limit of my Chapter 2 calculation. So I then take the low-ω limit of the Chapter 2 in Section 3 and the low-ω limit of my Section 1 Helmholtz result in Section 4. I do find a certain amount of agreement, though constant terms in Az do not match and probably they are not significant. Some fancy math is used in this doc, and for that reason alone I should keep this doc. 0. Overview comments 2 1. Calculation of Az for round wire with uniform-J using the Helmholtz integral. 2 2. Calculate Az(r) implied by the Chapter 2 result. 5 3. Question: what is the small-β1 limit of this Chapter 2 expression? 6 4. Calculate Az(r) from the Helmholtz integral in the limit of small β1 for r < a 7 0. Overview comments I started out doing the low frequency problem in Appendix M (currently M) (22D + β12) Az(r) = - [Iμ2/(πa2)] θ(r≤a) B = Bθ with Bθ = - ∂rAz . (M.5.1) Here I am assuming a uniform current density! That is why this problem makes no sense for large ω. Now in Chapter 2, I solve the problem I think for any ω, not just small ω, and the skin effect does come out of it. Here I start with a forced uniform Jz and as ω gets large, the problem really makes no sense, because that is not how the current density ends up. You cannot prescribe it that way. [ maybe you could with a set of filament current sources? ] Nevertheless, I did this problem as if the current density could be prescribed that way, and I did get a result. The result does not agree with Chapter 2 because there J does not end up uniform. I want to record this calculation here because it involved doing some integrals that I may have to do again in some other context! So here is that entire calculation which I will just store here: 1. Calculation of Az for round wire with uniform-J using the Helmholtz integral. Note: Here I just compute the Helmholtz integral. I don't consider boundary conditions on Az and I therefore don't talk about homogeneous solutions that might be required. Start with: (22D + β12) Az(r) = - [Iμ2/(πa2)] θ(r≤a) B = Bθ with Bθ = - ∂rAz . (M.5.1) The 2D free-space Helmholtz Green's function propagator is shown in (I.1.7) to be g(x|x') = (j/4) H0(1)(β1R) R = R = |x-x'| where H0(1) is a Hankel function. Thus, we may write the particular solution to (M.5.1) as the following Helmholtz integral, AzH(r) = ∫dA' [(j/4) H0(1)(kR) ] [Iμ2/(πa2)] θ(r≤a) R = R = |x-x'| = [Iμ2/(πa2)] (j/4)!Syntax Error, Ir' dr' !Syntax Error, Idθ' H0(1)(β1) . (M.5.2) = [Iμ2/(πa2)] (j/4)!Syntax Error, Ir' dr' [2I3]. The angle integral may be evaluated as I3 = (1/2) !Syntax Error, Idθ' H0(1)(β1) = !Syntax Error, Idx H0(1)(β1) = !Syntax Error, Idx J0(β1) + j !Syntax Error, Idx Y0(β1) where in the last line we use this NIST fact, Now try this β1) = = α = β1r β = β1r' Then we have I3 = !Syntax Error, Idx J0() + j !Syntax Error, Idx Y0() Then use GR7 page 726 Set ν = 0 to get !Syntax Error, Idx J0(β1) = Γ(1/2) J0(α)J0(β) = π J0(α)J0(β) !Syntax Error, Idx Y0(β1) = Γ(1/2) J0(α)Y0(β) = π J0(α)Y0(β) both for r < r' Then I3 = π J0(α)J0(β) + j π J0(α)Y0(β) = π J0(α) [J0(β) + j Y0(β) ] = π J0(α)H0(1)(β) = π J0(β1r)H0(1)(β1r') for r < r' . But I3 is symmetric so then I3 = π J0(β1r<) H0(1)(β1r>) = π J0(β1r) H0(1)(β1r')θ(r'>r) + π J0(β1r') H0(1)(β1r)θ(r'<r) and then we have AzH(r) = [Iμ2/(πa2)] (j/4) !Syntax Error, Ir' dr' [2I3]. = [Iμ2/(πa2)] (j/4) 2 !Syntax Error, Ir' dr' { πJ0(β1r) H0(1)(β1r') θ(r'>r) + πJ0(β1r') H0(1)(β1r)θ(r'<r)} = [Iμ2/(πa2)](j/4)2π { J0(β1r)!Syntax Error, Ir' dr' H0(1)(β1r') θ(r'>r) + H0(1)(β1r) !Syntax Error, Ir' dr'J0(β1r') θ(r'<r) } The required integrals are !Syntax Error, Ir' dr' H0(1)(β1r') θ(r'>r) = θ(a>r) !Syntax Error, Ir' dr' H0(1)(β1r') = θ(a>r)(1/β1)2!Syntax Error, I dx x H0(1)(x) = θ(a>r)(1/β1)2 [x H1(1)(x) ]β1aβ1r GR7 p 630 = θ(a>r)(1/β1)2 [β1a H1(1)(β1a) - β1r H1(1)(β1r) ] = θ(a>r)(1/β1) [a H1(1)(β1a) - r H1(1)(β1r) ] . !Syntax Error, Ir' dr'J0(β1r') θ(r'<r) = !Syntax Error, Ir' dr'J0(β1r') = (1/β1)2!Syntax Error, Idx x J0(x) = (1/β1)2 [x J1(x) ]β1min(a,r)0 = (1/β1)2 [ β1min(a,r)J1(β1min(a,r))] = (1/β1) Then AzH(r) = [Iμ2/(πa2)](j/4)2π (1/β1) * { J0(β1r) θ(a>r) [a H1(1)(β1a) - r H1(1)(β1r) ] + H0(1)(β1r) } For the r<a region we then have AzH(r) = [Iμ2/(πa2)](j/4)2π (1/β1) { J0(β1r) [a H1(1)(β1a) - r H1(1)(β1r) ] + H0(1)(β1r) rJ1(β1r) } r<a while for the r>a region, AzH(r) = [Iμ2/(πa2)](j/4)2π (1/β1) {H0(1)(β1r) aJ1(β1a) } r > a Do things match at r = a? AzH(a) = [Iμ2/(πa2)](j/4)2π (1/β1) { H0(1)(β1a) aJ1(β1a) } r<a AzH(a) = [Iμ2/(πa2)](j/4)2π (1/β1) {H0(1)(β1a) aJ1(β1a) } r > a Yes, they do match. 2. Calculate Az(r) implied by the Chapter 2 result. Suppose I start with the known Chapter 2 result, B(r) = B(a) J1(β1r) / J1(β1a) = -∂rAz(r) ∂rAz(r) = - B(a) J1(β1r) / J1(β1a) = -k J1(β1r) k = B(a)/ J1(β1a) Then integrate Az(r) = -k !Syntax Error, IJ1(β1r)dr = -k (1/β1) !Syntax Error, I J1(β1r)d(β1r) = -k (1/β1) !Syntax Error, I J1(x) dx = + k (1/β1) J0(β1r) + constant r < a = [B(a)/ J1(β1a)] (1/β1) J0(β1r) + constant where I used this result from p 630 of GR7, 3. Question: what is the small-β1 limit of this Chapter 2 expression? J1(β1a) ≈ (β1a/2) J0(β1r) ≈ 1 So then I get Az(r) = [B(a)/ J1(β1a)] (1/β1) J0(β1r) + constant ≈ [2B(a)/(β1a)] (1/β1) 1 + constant ≈ [2B(a)/(a)] (1/β12) + constant B(a) = Bθ(a) = μ2I/(2πa) so Az(r) ≈ [μ2I/(πa2)] (1/β12) + constant What is my interpretation of this result? It is a very large constant term since β1→ 0 . Now let's add another term and do it again J1(β1a) ≈ (β1a/2) J0(β1r) ≈ 1 - (β1r/2)2 Az(r) = [B(a)/ J1(β1a)] (1/β1) J0(β1r) + constant ≈ [2B(a)/(β1a)] (1/β1)( 1 - (β1r/2)2) + constant ≈ [2B(a)/(a)] (1/β12) ( 1 - (β1r/2)2) + constant B(a) = Bθ(a) = μ2I/(2πa) so Az(r) ≈ [μ2I/(πa2)] (1/β12) ( 1 - (β1r/2)2) + constant ≈ [μ2I/(πa2)] (1/β12) - [μ2I/(2πa2)] (1/β12) (β1r/2)2) + constant ≈ [μ2I/(πa2)] (1/β12) - [μ2I/(2πa2)] (r/2)2 + constant ≈ [μ2I/(πa2)] (1/β12) - [μ2I/(2πa2)] (r2/4) + constant ≈ - [μ2I/(πa2)] (r2/4) + constant' The new term is finite, there is no smallness parameter multiplying it. and it is exactly the term which appears in my Poisson result (M.4.15) for r<a. So I am "happy" with this limit of the Chapter 2 result. 4. Calculate Az(r) from the Helmholtz integral in the limit of small β1 for r < a What does my new calculation say for small β1 ? AzH(r) = [Iμ2/(πa2)](j/4)2π (1/β1) { J0(β1r) [a H1(1)(β1a) - r H1(1)(β1r) ] + H0(1)(β1r) r J1(β1r) } r<a (***) First, let's compute the small β1 limit of [a H1(1)(β1a) - r H1(1)(β1r) ]. I will save time and use Maple to get a series for this thing for small x: So I read this as H1(1)(x) = (-2j/π)(1/x) + (j/π) x ln(x) + Ax + O(x2) A = (1/2π) (π -2j ln(2)+2jγ - j) Then x H1(1)(x) = (-2j/π) + (j/π) x2 ln(x) + Ax2 + O(x3) Then [x H1(1)(x) - y H1(1)(y) ] = [ (-2j/π) + (j/π) x2 ln(x) + Ax2 + O(x3)] - [ (-2j/π) + (j/π) y2 ln(y) + Ay2 + O(y3)] = [(j/π) x2 ln(x) + Ax2 + O(x3)] - [(j/π) y2 ln(y) + Ay2 + O(y3)] = (j/π)[ x2 ln(x) - y2 ln(y)] + A(x2-y2) + O(x3) - O(y3) Now set x = β1a and y = β1r to get [β1a H1(1)(β1a) - β1r H1(1)(β1r) ] = (j/π)[ (β1a)2 ln(β1a) - (β1r)2 ln(β1r)] + A((β1a)2- (β1r)2) + O((β1a)3) - O((β1r)3) = β12 { (j/π)[ (a)2 ln(β1a) - (r)2 ln(β1r)] + A(a2- r2) + O(β1) } or [a H1(1)(β1a) - r H1(1)(β1r) ] = β1 ( (j/π)[ (a)2 ln(β1a) - (r)2 ln(β1r)] + A(a2- r2) + O(β1) ) Now this entire object has smallness order β1, keep that in mind. Look back now at AzH(r) = [Iμ2/(πa2)](j/4)2π (1/β1) { J0(β1r) [a H1(1)(β1a) - r H1(1)(β1r) ] + H0(1)(β1r) r J1(β1r) } r<a (***) Due to the (1/β1) factor, this object is going to contribute to the finite result! You cannot throw it out. Now let's work on the small β1 limit of second grouping H0(1)(β1r) r J1(β1r). I will invoke Maple for the first factor which I interpret to say H0(1)(x) ≈ (2j/π) ln(x) + A' + O(x2) A' = (1/π)(π - 2j ln(2) + 2jγ) so H0(1)(β1r) ≈ (2j/π) ln(β1r) + A' + O((β1r)2) The last term is order β12 so I will likely ignore it. Next, J1(x) ≈ (x/2) - x3/16 J1(β1r) ≈ (β1r /2) - (β1r)3/16 and I will surely drop this last term. Thus we have H0(1)(β1r) r J1(β1r) ≈ [(2j/π) ln(β1r) + A' + O(β12)] r [(β1r /2) ] = β1 r2/2 [(2j/π) ln(β1r) + A' + O(β12)]] so I think I can drop the last term here and end up with H0(1)(β1r) r J1(β1r) ≈ β1 (r2/2) [(2j/π) ln(β1r) + A' ] and again, this is an order β1 term and it will contribute! So now install these both into our Az(r) expression above which is for r < a AzH(r) = [Iμ2/(πa2)](j/4)2π (1/β1) { J0(β1r) [a H1(1)(β1a) - r H1(1)(β1r) ] + H0(1)(β1r) r J1(β1r) } AzH(r) = [Iμ2/(πa2)](j/4)2π (1/β1) { J0(β1r) [β1 ( (j/π)[ (a)2 ln(β1a) - (r)2 ln(β1r)] + A(a2- r2) + O(β1) ) ] + β1 (r2/2) [(2j/π) ln(β1r) + A' ] } AzH(r) = [Iμ2/(πa2)](j/4)2π { J0(β1r) [ (j/π)[ (a)2 ln(β1a) - (r)2 ln(β1r)] + A(a2- r2) + O(β1) ] + (r2/2) [(2j/π) ln(β1r) + A' ] } Now we don't want higher β1 terms, so we now set J0(β1r) = 1 and this then gives AzH(r) = [Iμ2/(πa2)](j/4)2π { [ (j/π)[ (a)2 ln(β1a) - (r)2 ln(β1r)] + A(a2- r2) ] + (r2/2) [(2j/π) ln(β1r) + A' ] } AzH(r) = [Iμ2/(πa2)](j/4)2π { [ (j/π)[ (a)2 ln(β1a) - (r)2 ln(β1r)] + A(a2- r2) ] + (r2/2) [(2j/π) ln(β1r) + A' ] } AzH(r) = [Iμ2/(πa2)](j/4)2π { [ (j/π)[ (a)2 ln(β1a) - (r)2 ln(β1r)] + A(a2- r2) ] + r2 [(j/π) ln(β1r) + A' /2] } AzH(r) = [Iμ2/(πa2)](j/4)2π { (j/π)a2 ln(β1a) - (j/π) r2 ln(β1r) + A(a2- r2) + r2 (j/π) ln(β1r) + A' (r2/2) } AzH(r) = [Iμ2/(πa2)](j/4)2π { (j/π)a2 ln(β1a) + A(a2- r2) + A' (r2/2) } AzH(r) = [Iμ2/(πa2)](j/4)2π { (j/π)a2 ln(β1a) + Aa2- Ar2 + (A'/2) r2 } AzH(r) = [Iμ2/(πa2)](j/4)2π { (j/π)a2 ln(β1a) + Aa2+ r2 (A'/2 - A) } Now go back to A' = (1/π)(π - 2j ln(2) + 2jγ) A = (1/2π) (π -2j ln(2)+2jγ - j) or A'/2 = (1/2π)(π - 2j ln(2) + 2jγ) A = (1/2π) (π -2j ln(2)+2jγ - j) Then (A'/2 - A) = (1/2π)(π - 2j ln(2) + 2jγ) - (1/2π) (π -2j ln(2)+2jγ - j) = (1/2π) [(π - 2j ln(2) + 2jγ)- (π -2j ln(2)+2jγ - j) ] = (1/2π) [π - 2j ln(2) + 2jγ - π +2j ln(2)-2jγ + j ] = (1/2π) [+ j ] = (j/2π) Then we have AzH(r) = [Iμ2/(πa2)](j/4)2π { (j/π)a2 ln(β1a) + Aa2+ r2 (A'/2 - A) } AzH(r) = [Iμ2/(πa2)](j/4)2π { (j/π)a2 ln(β1a) + Aa2+ r2 (j/2π) } AzH(r) = [Iμ2/(πa2)](j/4)2π { (j/π)a2 ln(β1a) + Aa2+ r2 (j/2π) } = [Iμ2/(πa2)](j/4)2π {(j/π)a2 ln(β1a) + Aa2}+ [Iμ2/(πa2)](j/4)2π {r2 (j/2π)} = [Iμ2/(π)](j/2) {(j) ln(β1a) + A}+ [Iμ2/(πa2)](j/4) {r2 (j)} = [Iμ2/(π)](j/2) {(j) ln(β1a) + A} - [Iμ2/(πa2)](r2/4) = - [Iμ2/(πa2)](r2/4) + [Iμ2/(π)](j/2) {(j) ln(β1a) + A} = - [Iμ2/(πa2)](r2/4) + [Iμ2/(π)](j/2) {(j) ln(β1a) + A} As a reminder, here was the Chapter 2 result calculated above Az(r) ≈ - [μ2I/(πa2)] (r2/4) + constant' Absolutely amazingly, these two terms have the exact same form in terms of the r2 non constant piece. This was the result I have spent maybe 3 hours trying to get. Everything is right including the sign. Comment: In this limit, the Helmholtz integral result has a very strange constant term which includes the γ constant! Question: Do I already know the B solution for Chapter 2 outside the wire? Can I just use Ampere's Law, or do I have to worry about jωD ? Carry on here tomorrow! ****************** I will not use this but save****************** The next task is to compute the B field using Bθ = - ∂rAz. We first note that ∂r [ r J1(β1r)] = r ∂r J1(β1r) + J1(β1r) = r β1 J1'(β1r) + J1(β1r) ∂r [ r H1(1)(β1r)] = r ∂r H1(1)(β1r) + H1(1)(β1r) = r β1 H1(1)'(β1r) + H1(1)(β1r) Then for r < a, -Bθ = ∂rAz = C β1J0'(β1r) [a H1(1)(β1a) - r H1(1)(β1r) ] + C J0(β1r) [ -{ r β1 H1(1)'(β1r) + H1(1)(β1r)}] -Bθ = ∂rAz = - C β1J1(β1r) [a H1(1)(β1a) - r H1(1)(β1r) ] + C J0(β1r) [ -{ r β1 H1(1)'(β1r) + H1(1)(β1r)}] r < a and for r > a -Bθ = ∂rAz = C β1 H0(1)'(β1r) [a J1(β1a) - r J1(β1r)] + C H0(1)(β1r) [- { r β1 J1'(β1r) + J1(β1r)}] r > a Now the $64 question: I computed Bθ for r < a in Chapter 2 and got this result Bθ(r) = + B(a) [J1(β1r) / J1(β1a)] r<a Why are these results so different looking? Well, as noted at the start, the problem I solved here is an unphysical problem because it assumes a constant current density, but in reality we know there will be skin effect and a non-uniform Jz therefore! BUT, the results should agree for small β1 and they don't agree! Suppose I start with the known Chapter 2 result, Bθ(r) = k J1(β1r) = -∂rAz(r) Then integrate Az(r) = -k !Syntax Error, IJ1(β1r)dr = -k (1/β1) !Syntax Error, I J1(β1r)d(β1r) = -k (1/β1) !Syntax Error, I J1(x) dx = + k (1/β1) J0(β1r) + constant where I used this result from p 630 of GR7, Now compare this with my r < a result above AzH(r) = C J0(β1r) [a H1(1)(β1a) - r H1(1)(β1r) ] r < a For small β1, this should equal what I just did above, but it does not work due to the bracket factor. For small β1 we in fact know that so H1(1)(z) ≈ -(j/π)(2/z) => z H1(1)(z) ≈ -(j/π)(2) = (-2j/π) Then we get [...] ≈ 0. So why does my full solution have the wrong small-β1 limit?