Appendix M (now G) REVIEWED

Appendix G (used to be M before 1.17.14) 11.24.13 Comments 12/5/13. This Appendix M was never installed. The opening section gives a good summary of what is done here. Part 1 still has my "illogical" problem which I have to consider valid only in the low-ω limit, but I suppose this is OK to do. In this doc, I am aware of the Az boundary conditions and of the possible need for homo adder solutions. I was pretty excited about this Appendix M when I wrote it, but I feel cooler now and it is just on hold. I am using the King gauge global region R method, and it requires me to put in something for the driving current! This just makes it seem that the Helmholtz Az equation is not a good way to solve the round wire problem for Az in a precise manner. This King method only works if you have a prescribed current configuration. Appendix G: The DC vector potential of a round wire carrying a uniform current 1 G.1 Setup and Assumptions 1 G.2 Direct solution for Az(r) from the differential equation 3 G.3 Instant solution for A using Ampere's Law and computation of Jm 5 G.4 Solution using the 2D Helmholtz Integral 6 G.5 Comments on the low frequency solution 11 Appendix G: The DC vector potential of a round wire carrying a uniform current In this problem, an isolated, infinitely-long and z-aligned round wire ( μ2,ε2,σ2, radius a ) carries a current I. The wire is immersed in an infinite dielectric medium (μ1,ε1,σ1). We begin for general ω, but quickly go to the DC limit ω = 0. We wish to calculate the vector potential A of this wire both inside and outside. Section G.1 sets up the problem, makes some ansatz assumptions, and then ends up with a 2D Poisson equation for the potential which is 22D Az(r) = - [Iμ2/(πa2)] θ(r≤a). Section G.2 directly solves this Poisson equation for the potential Az(r). The solution is required to meet two boundary conditions at r = a. Section G.3 very quickly computes this same Az(r) using Ampere's Law with the same boundary conditions and obtains the same result found in Section G.2. Section G.4 laboriously obtains the same Az(r) result using the 2D Helmholtz integral (which in this case is really just a Poisson integral). This serves as a prototype case for dealing with such integrals, so much detail is provided. It is found that for μ1 ≠ μ2 homogenous terms must be added to the Helmholtz integral in order to meet the boundary conditions. Section G.5 comments on the solution for Az at low frequency ω > 0. G.1 Setup and Assumptions The ω-domain Helmholtz wave equation for A using the King gauge is given by (1.5.4), (2 + β12)A = - μ2J2 β12 = ω2μ1 ξ1 ξ1 ≡ ε1 - jσ1/ω (G.1.1) div A = jωμ1ξ1φ . // King gauge (G.1.2) We take a uniform prescribed current inside the wire ( assume low frequency), J2 = [I/(πa2)] (G.1.3) so the Helmholtz wave equation reads (2 + β12)A(x) = - [Iμ2/(πa2)] θ(< a) where 2 is the vector Laplacian and where θ(B) = 1 if B is true, else 0. Using Cartesian components, this says (2 + β12)Ax(x) = 0 (2 + β12)Ay(x) = 0 (2 + β12)Az(x) = - [Iμ2/(πa2)] θ(< a) where in these three equations 2 is the scalar Laplacian. We shall seek a solution in which both Ax and Ay vanish. In this case we have A(x) = Az(x) . We assume a very low frequency ω for which we know any longitudinal wave that might be going down the wire has a very long wavelength. We then ignore z variations in Az to write A(x) = Az(x,y) . (G.1.4) In this case, one finds that 2Az = 22D Az and then our only equation of interest is this: (22D + β12) Az(x,y) = - [Iμ2/(πa2)] θ(< a) . (G.1.5) We now take ω→0 to get 22D Az(x,y) = - [Iμ2/(πa2)] θ(x2+y2 < a2) which is just a 2D Poisson equation with a constant source limited to a region of space. At this point we are free to replace x,y with polar coordinates r,θ, so we have for r in the range (0,∞), 22D Az(r,θ) = - [Iμ2/(πa2)] θ(r<a) . // θ(r<a) = Heaviside θ(a-r). (G.1.6) From B = curl A in cylindrical coordinates we find that, since only Az is non-vanishing, B = curl A = [ r-1∂θAz - ∂zAθ] + [∂zAr - ∂rAz] + [ r-1∂r(rAθ) - r-1∂θAr ] = [ r-1∂θAz] + [- ∂rAz] . Since we expect the magnetic field lines to be entirely in the direction, we are led to make the assumption that ∂θAz = 0 and then the problem is this 22D Az(r) = - [Iμ2/(πa2)] θ(r≤a) B = Bθ with Bθ = - ∂rAz . (G.1.7) If we can find a solution, then our ansatz assumptions that Ax = Ay = 0 and B = Bθ are justified. G.2 Direct solution for Az(r) from the differential equation Using 22D in polar coordinates our ODE (G.1.7) reads (1/r)∂r(r∂rAz(r)) = - [Iμ2/(πa2)] θ(r≤a) or ∂r(r∂rAz(r)) = - [Iμ2/(πa2)] r θ(r≤a) or r Az"(r) + Az'(r) = - [Iμ2/(πa2)] r θ(r≤a) . (G.2.1) The two regional differential equations are then r Az"(r) + Az'(r) = 0 r>a region 1 r Az"(r) + Az'(r) = - [Iμ2/(πa2)] r r<a region 2 . (G.2.2) The general-form solutions to these ODE's are, Az(r) = C ln(r) + D r>a region 1 Az(r) = - [Iμ2/(4πa2)] r2 + E ln(r) + F r<a region 2 (G.2.3) where there are 4 constants to be determined. For r>a the functions 1 and lnr are the well-known atomic forms (harmonic elements) for the 2D Laplace equation for situations of azimuthal symmetry. The first term in region 2 (G.2.3) is the particular solution of region 2 (G.2.2) to which we have added a possible homogeneous solution E ln(r) + F. In order that Az(r) be finite at r = 0, we must set E = 0. For very large r the round wire looks like a line source and we know the solution of that problem. Using Ampere's Law that 2πrHθ = I we find ( recall that Bθ = - ∂rAz) Hθ = [I/2π](1/r) => Bθ = μ1 [I/2π](1/r) => Az = - [I μ1/2π] ln(r) + D' . r >> a Comparing this solution to our r>a round wire solution we conclude that D' = D and C = -[μ1I/2π] . There are still two unknown constants D and F : Az(r) = -[Iμ1/2π] ln(r) + D r > a region 1 Az(r) = - [Iμ2/(4πa2)] r2 + F r < a region 2 . (G.2.4) The potential Az(r) always has an additive constant which we are free to specify and which affects nothing. We shall choose the zero point of Az(r) by setting D = 0 arbitrarily. This means that Az(r) has the simple form K ln(r) for r>a, and that choice implies that Az(a) = - [Iμ1/2π] ln(a), so we have in effect specified Az(r) on the wire surface to be this value. Notice for future reference that ∂rAz(r) = - [Iμ1/2π] (1/r) r > a region 1 ∂rAz(r) = - [Iμ2/(2πa2)] r r < a region 2 . (G.2.5) Now, since our prescribed current J2 does not specify a free surface current Kz on the round wire surface, which would have the form Js,surface = Kzfree δ(a-r), we conclude that there is no free surface current on the round wire surface; there is only the bulk volume current Jz = I/(πa2)θ(r<a). Therefore the boundary condition (1.1.46) applies (though now in polar coordinates) and we conclude that (1/μ1) [(∂rAz)(a)]1 = (1/μ2) [(∂rAz)(a)]2 . In addition, we shall require that Az itself be continuous at the boundary, so here are our two boundary conditions of interest (superscript 1 means region 1 which is r>a), [Az(a)]1 = [Az(a)]2 . (1/μ1) [(∂rAz)(a)]1 = (1/μ2) [(∂rAz)(a)]2 . (G.2.6) We now require that both these boundary conditions be met by the Az expressions of (G.2.4), [Az(a)]1 = [Az(a)]2 // (G.2.6) repeated (1/μ1) [(∂rAz)(a)]1 = (1/μ2) [(∂rAz)(a)]2 or -[Iμ1/2π] ln(a) = - [Iμ2/(4πa2)] a2 + F // insert expressions, set r = a (1/μ1){- [Iμ1/2π] (1/a)} = (1/μ2){ - [Iμ2/(2πa2)]a } or -[Iμ1/2π] ln(a) = - [Iμ2/(4π)] + F // simplify 1 = 1 The second boundary condition is thus met automatically by our solution. The first says F = [Iμ2/(4π)] - [Iμ1/2π] ln(a) and so the solution is then Az(r) = - [Iμ1/2π] ln(r) r > a region 1 Az(r) = - [Iμ2/(4πa2)] r2 + [Iμ2/(4π)] - [Iμ1/2π] ln(a) r < a region 2 or Az(r) = - [Iμ1/2π] ln(r) r > a region 1 Az(r) = - [Iμ2/(4πa2)] (r2-a2) - [Iμ1/2π] ln(a) r < a region 2 . (G.2.7) This then is the complete solution to the problem for the round wire, 22D Az(r) = - [Iμ2/(πa2)] θ(r≤a) B = Bθ with Bθ = - ∂rAz . (G.1.7) where the potential is "pinned" by the requirement that Az = K ln(r) for r > a. We may now compute the B field from our potential solution, Bθ = - ∂rAz = [Iμ1/2π]∂rln(r) = [Iμ1/2π](1/r) r > a region 1 Bθ = - ∂rAz = [Iμ2/(2πa2)] r = [Iμ2/2π](r/a2) r < a region 2 (G.2.8) Comment: Although only μ2 appears in the differential equation (G.1.7), once the equation is properly solved with attention to boundary conditions, we find that μ2 appears in Bθ in region 1, while μ1 appears in Bθ in region 2 ! This is the main point of this Section G.2. G.3 Instant solution for A using Ampere's Law and computation of Jm For r > a Ampere's law (1.1.37) (converted to ω space with ω = 0) says 2πrHθ = I so Hθ = I/(2πr) => Bθ = μ1I/(2πr) r>a region 1 (G.3.1) => - ∂rAz = μ1I/(2πr) => Az = - [μ1I/(2π)] ln(r) + D . For r < a Ampere's law says 2πrHθ = I(πr2/πa2) [ the "current enclosed" ] Hθ = I(r2/a2)1/(2πr) = I r/(2πa2) => Bθ = [Iμ2/2π](r/a2) r<a region 2 (G.3.2) => - ∂rAz = [Iμ2/2π](r/a2) => Az = - [Iμ2/4π](r2/a2) + E We then set D = 0 to get Az = K ln(r) for r > 0, as done previously, and then we must match at r = a : - [μ1I/(2π)] ln(a) = - [Iμ2/4π] + E => E = [Iμ2/4π] - [μ1I/(2π)] ln(a) so our potential solution is then Az(r) = - [μ1I/(2π)] ln(r) r>a Az(r) = - [Iμ2/4π](r2/a2) + { [Iμ2/4π] - [μ1I/(2π)] ln(a) } r<a or Az(r) = - [μ1I/(2π)] ln(r) r>a Az(r) = - [Iμ2/4πa2]r2 + [Iμ2/4π] - [μ1I/(2π)] ln(a) r<a or Az(r) = - [μ1I/(2π)] ln(r) r>a Az(r) = - [Iμ2/4πa2](r2-a2) - [μ1I/(2π)] ln(a) r<a (G.3.3) This result agrees with result (G.2.7) of the previous section. Magnetization Current: It was mentioned in Section G.2 that there is no free surface current Kzfree at the wire surface. There is in fact a bound magnetization current on this surface and in both bulk regions as well. Luckily, our Helmholtz equation is unable to see these currents so we don't have to worry about them. The magnetization current density is given by Jm = curl M where M = [μ/μ0- 1] H as shown in (1.1.20-23). Here then is the calculation: Mθ1 = [ μ1/μ0 - 1] Hθ1 = [ μ1/μ0 - 1] (I/2πr) r > a Mθ2 = [ μ2/μ0 - 1] Hθ2 = [ μ2/μ0 - 1] (Ir/2πa2) r < a or Mθ = [ μ1/μ0 - 1] (I/2πr) θ(r>a) + [ μ2/μ0 - 1] (Ir/2πa2)θ(r<a) ∂rMθ = [ μ1/μ0 - 1] (I/2πr) δ(r-a) + [ μ2/μ0 - 1] (Ir/2πa2)[ -δ(r-a)] = [ μ1/μ0 - μ2/μ0] (I/2πr) δ(r-a) Jm = curl M = [ r-1∂r{rMθ} ] // see below (G.1.6) above Jmz = r-1∂r{rMθ} = r-1[ Mθ + r∂rMθ] = r-1Mθ + ∂rMθ (G.3.4) = [ μ1/μ0 - 1] (I/2πr2) θ(r>a) + [ μ2/μ0 - 1] (I/2πa2)θ(r<a) + [ μ1 - μ2] (I/2πaμ0) δ(r-a) ~1/r2 outside constant inside on the surface It is the slope discontinuity of the bulk magnetization at the surface which creates the surface current term in Jmz. G.4 Solution using the 2D Helmholtz Integral This method is technically more difficult than the first two methods, so we show a lot of detail. The method is important because our entire Chapter 4 is based on using Helmholtz integrals to compute transmission line parameters, and this is one Helmholtz integral that we can actually compute without too much effort. An important result we find is that, when μ1≠μ2, the Helmholtz integral by itself does not supply the complete solution, and one must add in some amount of homogeneous solution of 22D Az(r) = 0 in order to meet the required boundary conditions at r = a. This subject is then pursued more in Section G.x. We really have a Poisson integral since β12 = 0, but the full Helmholtz integral works the same way so we keep referring to it as a Helmholtz integral. Recall from above: 22D Az(r) = - [Iμ2/(πa2)] θ(r≤a) B = Bθ with Bθ = - ∂rAz . (G.1.7) Using the 2D free-space Green's function (propagator) as reviewed in Appendix I equation (I.1.6), g(x|x') = (1/2π) ln(1/R) = - (1/4π) ln(R2) R = R = |x-x'| (G.4.1) we may write the particular solution to (G.1.7) as the following "Helmholtz" integral [see (I.1.8)] AzH(r) = - ∫dA' [(1/4π) ln(R2) ] [Iμ2/(πa2)] θ(r≤a) = -!Syntax Error, Ir' dr'!Syntax Error, Idθ' [(1/4π) ln(r2+ r'2- 2rr'cos(θ-θ'))] [Iμ2/(πa2)] = - (1/4π) [Iμ2/(πa2)] !Syntax Error, Ir' dr' [ !Syntax Error, Idθ' ln(r2+ r'2- 2rr'cos(θ-θ')) ] = - (1/4π) [Iμ2/(πa2)] !Syntax Error, Ir' dr' [ 2 I1(r,r') ] (G.4.2) where I1(r,r') ≡ (1/2) !Syntax Error, Idθ' ln(r2+ r'2- 2rr'cos(θ-θ')) = (1/2) !Syntax Error, Idθ" ln(r2+ r'2- 2rr'cos(θ")) = !Syntax Error, Idx ln(r2+ r'2- 2rr'cosx) . (G.4.3) Since this integral I1 is quite important, we evaluate it two independent ways: _______________________________________________________________________________ Method 1: I1 = !Syntax Error, Idx ln[r2(1+ r'2/r2- 2r'/rcosx) ] = !Syntax Error, Idx ln[r2(1+ α2- 2αcosx) ] α ≡ r'/r = !Syntax Error, Idx ln[r2] + !Syntax Error, Idx ln (1+ α2- 2α cosx) = π ln(r2) + I2 . (G.4.4) We then use this easily derived expansion Stakgold Vol II p 104 (6.22), ln(1 + α2 - 2α cosx ) = -2 Σn=1∞ (αn/n)cos(nx) |α| < 1 (G.4.5) which is just a Fourier Cosine Series expansion of ln(1 + α2 - 2α cosx ). Then, I2 = !Syntax Error, Idx ln (1+ α2- 2αcosx) = -2 Σn=1∞(αn/n)!Syntax Error, Idx cos(nx) = -2 Σn=1∞(αn/n) [ (1/n)sin(nx)]|π0 = -2 Σn=1∞(αn/n2) [sin(nπ) - sin(0)] = 0 and then I1 = π ln (r2) + I2 = π ln (r2) + 0 = 2π ln(r) α < 1 r'<r Since I1(r,r') is symmetric under r↔r' we must have I1 = 2π = 2π . (G.4.6) Method 2: I1 = !Syntax Error, Idx ln(r2+ r'2- 2rr'cosx) = !Syntax Error, Idx ln (a + b cosx) . This integral may be evaluated using GR7 p 531 4.224, with a = r'2 +r2 and b = -2rr' and a2-b2 = (r'2-r2)2 so that = | r'2-r2 |. The condition a > |b| > 0 is met since (r±r')2 > 0 => r2+r'2 > ±2rr' which says a > ±b so a > |b|. Thus, I1 = !Syntax Error, Idx ln [r'2 +r2-2rr' cosx)] = π ln [ ] = = 2π (G.4.7) which agrees with the result of method 1. ___________________________________________________________________________________ Using the notation θ(x >a) ≡ θ(x-a) [ Heaviside step ] we write (G.4.7) as I1 = 2π { ln(r)θ(r'<r) + ln(r') θ(r'>r) } . (G.4.8) Note: We use notation θ(x >a) in place of the more correct θ(x-a) because it requires "less thinking" to interpret. Our particular "Helmholtz" integral (G.4.2) is then AzH(r) = - (1/4π) Iμ2/(πa2) !Syntax Error, Ir' dr' [ 2 I1(r,r') ] = - (1/4π) Iμ2/(πa2) !Syntax Error, Ir' dr' 4π { ln(r)θ(r'<r) + ln(r') θ(r'>r) } = - Iμ2/(πa2) !Syntax Error, Ir' dr' { ln(r)θ(r'<r) + ln(r') θ(r'>r) } = - Iμ2/(πa2) { ln(r)!Syntax Error, Ir' dr' θ(r'<r) + !Syntax Error, Ir' dr' ln(r') θ(r'>r) } . (G.4.9) The two integrals are !Syntax Error, Ir' dr' θ(r'<r) = !Syntax Error, Ir' dr' = (1/2) [min(a,r)]2 (G.4.10) !Syntax Error, Ir' dr' ln(r') θ(r'>r) = θ(r<a) !Syntax Error, Ir' dr' ln(r') = θ(r<a) !Syntax Error, Ix lnx dx = θ(r<a) [(1/4) (r2-a2) + (1/2) a2lna - (1/2) r2lnr ] . (G.4.11) Then AzH(r) = - Iμ2/(πa2) { ln(r)!Syntax Error, Ir' dr' θ(r'<r) + !Syntax Error, Ir' dr' ln(r') θ(r'>r) } = - Iμ2/(πa2) { ln(r) (1/2) [min(a,r)]2 + θ(r<a) [(1/4) (r2-a2) + (1/2) a2lna - (1/2) r2lnr ] } . For r > a we get AzH(r) = - Iμ2/(πa2) { ln(r) (1/2)a2 } = - Iμ2/(2π) ln(r) (G.4.12) ∂rAzH(r) = - Iμ2/(2π) (1/r) . For r < a the result is AzH(r) = - Iμ2/(πa2) { ln(r) (1/2)r2 + (1/4) (r2-a2) + (1/2) a2lna - (1/2) r2lnr } = - Iμ2/(πa2) { (1/4) (r2-a2) + (1/2) a2lna } (G.4.13) ∂rAzH(r) = - Iμ2/(πa2) (r/2) . Recall the two boundary conditions [Az(a)]1 = [Az(a)]2 . (1/μ1) [(∂rAz)(a)]1 = (1/μ2) [(∂rAz)(a)]2 . (G.2.6) For our particular Helmholtz integral AzH(r) we evaluate these boundary conditions to find - Iμ2/(2π) ln(a) = - Iμ2/(πa2) (1/2) a2lna (1/μ1) [- Iμ2/(2π) (1/a)] = (1/μ2) [- Iμ2/(πa2) (a/2)] or 1 = 1 (μ2/μ1) = 1 Thus, only in the case μ1 = μ2 does the Helmholtz particular solution meet both boundary conditions. If μ1≠ μ2, we must add to the particular solution some amount of homogeneous solution of 22D Az(r,θ) = 0. So we then write generally, Az(r) = AzH(r) + Azhomo(r) (G.4.14) where we know that Azhomo(r) can only have terms α + β ln r. We then write for the two regions Az(r) = - Iμ2/(2π) ln(r) + α + β lnr r>a Az(r) = - Iμ2/(πa2) { (1/4) (r2-a2) + (1/2) a2lna } + α ' + β' lnr r<a As earlier, we choose the zero point for Az(r) by requiring that the large r behavior be K ln(r) without a constant added, which then means α = 0. And for r<a we must have β' = 0 to be finite at r = 0. So Az(r) = [- Iμ2/(2π) + β] lnr r>a Az(r) = - Iμ2/(πa2) { (1/4) (r2-a2) + (1/2) a2lna } + α ' r<a (G.4.15) -∂rAz(r) = [Iμ2/(2π) - β](1/r) r>a -∂rAz(r) = Iμ2/(πa2) (2r) r<a The boundary conditions are then [Az(a)]1 = [Az(a)]2 (1/μ1) [(-∂rAz)(a)]1 = (1/μ2) [(-∂rAz)(a)]2 // (G.2.6) repeated or [- Iμ2/(2π) + β] lna = - Iμ2/(πa2) { (1/4) (a2-a2) + (1/2) a2lna } + α ' (1/μ1)[ Iμ2/(2π) - β](1/a) = (1/μ2) Iμ2/(πa2) (2a) // insert expressions, set r = a or [- Iμ2/(2π) + β] lna = - Iμ2/(2π) lna + α ' (1/μ1)[ Iμ2/(2π) - β](1/a) = I/(2πa) // simplify or β lna = α ' Iμ2/(2π) - β = μ1I/(2π) // simplify some more so we find that β = I/(2π) (μ2-μ1) α' = I/(2π) (μ2-μ1) lna (G.4.16) The full solution is then Az(r) = [- Iμ2/(2π) + β] lnr r>a Az(r) = - Iμ2/(πa2) { (1/4) (r2-a2) + (1/2) a2lna } + α ' r<a or Az(r) = [- Iμ2/(2π) + {I/(2π) ( μ2-μ1)}] lnr r>a Az(r) = - Iμ2/(πa2) { (1/4) (r2-a2) + (1/2) a2lna } + { I/(2π) ( μ2-μ1) lna } r<a or Az(r) = I/(2π) [- μ2 + ( μ2-μ1)] lnr r>a Az(r) = - Iμ2/(πa2) (1/4) (r2-a2) - Iμ2/(πa2) (1/2) a2lna + I/(2π) ( μ2-μ1) lna r<a or Az(r) = - (I/(2π) [μ1] lnr r>a Az(r) = - Iμ2/(πa2) (1/4) (r2-a2) - Iμ1/(π) (1/2) lna r<a or Az(r) = - [Iμ1/2π] lnr r>a Az(r) = - [Iμ2/(4πa2)] (r2-a2) - [Iμ1/2π] lna r<a (G.4.17) This result matches the results (G.2.7) and (G.3.3) of the previous two methods and then results in the B field solution Bθ = - ∂rAz = [Iμ1/2π]∂rln(r) = [Iμ1/2π](1/r) r > a region 1 Bθ = - ∂rAz = [Iμ2/(2πa2)] r = [Iμ2/2π](r/a2) r < a region 2 (G.2.8) G.5 Comments on the low frequency solution In Chapter 2 we compute the E and B fields inside a round wire operating at frequency ω. The results are rather complicated and involve special Bessel functions called Kelvin functions. The vector potential was not used in that Chapter. We shall not carry through the calculation here, but shall merely make some comments. 1. For sufficiently low frequencies (the transmission line limit) we imagine that the ansatz assumptions we made in Section G.1 are still pretty good. The current distribution will be nearly uniform. There will likely be some small Ax and Ay fields which we can ignore, and we still assume roughly that B = Bθ with Bθ = - ∂rAz and that we can ignore the z-dependence of Az, though we know it must vary some small amount in order to have a long-λ wave passing down the wire. Therefore, our problem is basically (G.1.5) for small β12, (22D + β12) Az(r) = - [Iμ2/(πa2)] θ(r≤a) B = Bθ with Bθ = - ∂rAz . (G.5.1) 2. Since 22D = (1/r)∂r(r∂r), one could write out the above differential equation and repeat the work of section G.2 above. The resulting B field obtained from Bθ = - ∂rAz should then agree with the low frequency limit of (2.2.25) which applies inside the round wire, Bθ(r) = Bθ(a) . (2.2.25) That low-frequency limit is Bθ(r) ≈ Bθ(a) = Bθ(a) β12 = ω2μ ξ1 (G.5.2) Certainly as ω → 0 so β1→ 0 the result Bθ(r)= Bθ(a)(r/a) agrees with (G.2.8). 3. 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 (G.5.1) as the following Helmholtz integral [ see (I.1.9) ] , 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) . (G.5.3) The dθ' integral is actually doable with this result (making use of GR7 p 726 6.684 1 and 2) !Syntax Error, Idθ' H0(1)(β1) = (1/2) { π J0(β1r) H0(1)(β1r')θ(r'>r) + π J0(β1r') H0(1)(β1r)θ(r'<r) } . (G.5.4) The two dr' integrals can then be done (using GR7 p 629-630 Section 5.5) with the final result 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) } . (G.5.5) We leave it to the reader to determine the small β1 limit of this result and see if the resulting Bθ = - ∂rAz agrees with (G.5.2) after homogeneous solutions are added to match boundary conditions. Remember that we only expect this result to be meaningful for low ω since we have assumed the uniform current distribution of (G.1.3). If one makes the small-argument approximation H0(1)(x) ≈ (2j/π)ln(x) directly in (G.5.3), the integral replicates the Poisson result (G.4.2), so more expansion terms would be needed for this approach.