App M on the subject of derivatives of Ax and Az REVIEWED

App M on the subject of derivatives of Ax and Az PhL 9.15.14 Here I do some calculations in an attempt to see what Appendix M says about the size of ∂xAx versus ∂zAz in terms of the Helmholtz Integrals. My first results are ∂zAz(x,ω) = - (1+jβd) Σi∫μi Jzi(x',y',ω) I1(x',y') dx' dy' ∂xAx(x,ω) = - (1+jβd) Σi∫μi Jxi(x',y',ω) I2(x',y') (x-x') dx' dy' where I1 = -2 j e-jkz !Syntax Error, Idz" [z" sin(kz")] (1+jβdR) e-jβR/R3 = even in (x-x') I2 = 2 e-jkz !Syntax Error, Idz" [cos(kz")] (1+jβdR) e-jβR/R3 = even in (x-x') I continue on, but these integrals are a bit messy though they can be done. I would associate large ω with large k since k = ω/vd, so large number of waves per meter at large ω. Then large ω chops up the I2 integral, making I2 small, and then ∂xAx(x,ω) becomes small, which is the desired result. In contrast, the I1 integral seems small at small k and at large k, since large k just chops up zero. As ω→ 0 and k→ 0, the I2 integral grows larger, causing ∂xAx(x,ω) to increase in size. So this supports my general conjecture. I just added this conjecture as Observation 3 at the end of Appendix M just now ! Does App M say something about these transverse derivatives by implication? That appendix shows that the current ratio Jr/Jz is small, and then it uses At(x) = - Σi μi∫Jt,i(x',y') ln(s2) dx' dy' to argue that Ar/Az is therefore small. You could perhaps consider ∂xAx = - Σi μi∫Jx,i(x',y') ∂x ln(s2) dx' dy' ∂zAz = - Σi μi∫Jz,i(x',y') ∂z ln(s2) dx' dy' = 0 The second line gives 0 only after we have done the approximate process leading from (M.2) to (M.3), and this suggests the strange result that ∂xAx >> ∂zAz . So maybe I have to back up and write Az(x,ω) = Σi∫μi Jzi(x',y',ω) e-jkz' dV' R = | x - x' | (1.5.9) ∂zAz(x,ω) = Σi∫μi Jzi(x',y',ω) e-jkz' ∂z[ ] dV' // using e-jkz ansats What do I know about this ∂z object in there? I don't think ∂z[] has ever appeared before in lines doc! I would have to compute this out, ∂z[] = { R [∂z e-jβR] - e-jβR∂zR} /R2 But ∂z e-jβR = e-jβR [-jβd ∂zR] so then ∂z[] = { R e-jβR(-jβd) ∂zR - e-jβR∂zR} /R2 = e-jβR (∂zR) { -jβdR - 1 }/R2 // Maple verified Now R2 = (x-x')2 + (y-y')2 + (z-z')2 ∂zR2 = 2(z-z') ∂zR2 = 2R(∂zR) => (∂zR) = (z-z')/R Therefore (∂zR) = 2(z-z')/(2R) = (z-z')/R = very simple and then ∂z[] = - e-jβR (z-z') (jβdR + 1)/R3 So we are then left with ∂zAz(x,ω) = Σi∫μi Jzi(x',y',ω) e-jkz' ∂z[ ]dV' = - Σi∫μi Jzi(x',y',ω) e-jkz'e-jβR (1+jβdR) (z-z') /R3 dV' The same processing would yield, ∂xAx(x,ω) = Σi∫μi Jzi(x',y',ω) e-jkz' ∂x[ ]dV' = - Σi∫μi Jxi(x',y',ω) e-jkz'e-jβR (1+jβd) (x-x') /R3 dV' To summarize ∂zAz(x,ω) = - Σi∫μi Jzi(x',y',ω) e-jkz'e-jβR (1+jβdR) (z-z') /R3 dV' ∂xAx(x,ω) = - Σi∫μi Jxi(x',y',ω) e-jkz'e-jβR (1+jβdR) (x-x') /R3 dV' Now we have two different dz' integrals here which are then integrands in the above, I1 ≡ !Syntax Error, Idz' e-jkz' e-jβR (1+jβdR)/R3 * (z-z') I2 ≡ !Syntax Error, Idz' e-jkz' e-jβR (1+jβdR)/R3 R2 = (x-x')2 + (y-y')2 + (z-z')2 = s2 + (z-z')2 In terms of these integrals, we can write ∂zAz(x,ω) = - Σi∫μi Jzi(x',y',ω) I1(x',y') dx' dy' ∂xAx(x,ω) = - Σi∫μi Jxi(x',y',ω) I2(x',y') (x-x') dx' dy' We now study the two integrals. Rewrite the first using z" = z-z' and dz" = -dz' so z' = z - z" I1 = !Syntax Error, Idz" e-jkz e+jkz" (1+jβdR) e-jβR/R3 * (z") R2= s2 + z"2 or I1 = e-jkz !Syntax Error, Idz" z" ejkz" (1+jβdR) e-jβR/R3 = e-jkz !Syntax Error, Idz" z" [cos(kz") + jsin(kz")] e-jβR/R3 = j e-jkz !Syntax Error, Idz" z" [jsin(kz")] (1+jβdR) e-jβR/R3 = -j e-jkz !Syntax Error, Idz" [z" sin(kz")] (1+jβdR) e-jβR/R3 I expect this integral to be dominated by the region near z" = 0, but there we have z" sin(kz") active. Meanwhile, I2 = !Syntax Error, Idz' e-jkz' (1+jβdR)e-jβR/R3 = = e-jkz !Syntax Error, Idz" e+jkz" (1+jβdR)e-jβR/R3 = e-jkz !Syntax Error, Idz" [cos(kz") + jsin(kz")] (1+jβdR)e-jβR/R3 = e-jkz !Syntax Error, Idz" [cos(kz")] (1+jβdR)e-jβR/R3 and here we have no such zeroing out factor. So I expect I2 = large and I1 = small. To summarize, I1 = -j e-jkz !Syntax Error, Idz" [z" sin(kz")] (1+jβdR)e-jβR/R3 = -2 j e-jkz !Syntax Error, Idz" [z" sin(kz")] (1+jβdR) e-jβR/R3 I2 = e-jkz !Syntax Error, Idz" [cos(kz")] (1+jβdR) e-jβR/R3 = 2 e-jkz !Syntax Error, Idz" [cos(kz")] (1+jβdR) e-jβR/R3 I wonder if these integrals can be done? Skip this, it is all wrong since I failed to have the (1+jβdR) factor when I did them! Write J1 ≡ !Syntax Error, Idx [x sin(kx)] e-jbR/R3 R2 = s2 + x2 J2 ≡ !Syntax Error, Idx [cos(kx)] e-jbR/R3 R2 = s2 + x2 Maple can do both as some hypergeometric series things. Wow. So maybe then it is worth trying to look these up in GR7. Perhaps change from x to R as integration variable R2 = s2 + x2 RdR = xdx x = J1 = !Syntax Error, I RdR sin(k) e-jbR/R3 = !Syntax Error, I dR sin(k) e-jbR/R2 I cannot find anything like this in any of the forms. BUT, maybe I can do a contour integration. Write K2 ≡ !Syntax Error, Idx [x sin(kx)] e-jbR/R3 This thing has zeros on the real axis which are fine. The only singularities are where R vanishes and these unfortunately are branch points at R = ± s. OK I give up. How about method of stationary phase? Consider T1 ≡ !Syntax Error, Idt ejkt [t e-jβR/R3] R = s2 + t2 let x = k T1 ≡ !Syntax Error, Idt ejxt f(t) f(t) = t e-jβR/R3 ψ(t) = t ψ'(t) = 1 Look at the boundaries, and note that f(t)/ψ'(t) = f(t). At t = 0 we have f(t) = 0 so it is zero at one of the boundaries and this invalidates my little set of notes. But let's just follow the notes anyway parts = [ f(t)/(ix) * eixt]t=∞0 = (1/ix) [ f(t) eixt]t=∞0 As t→∞ we have R→ t≠∞ so then f(t) = t e-jbt/t3 ~ 1/t2 = 0. As t→-∞ same thing. So the parts then just vanish and you get no answer. Suppose we assume that most of the integral comes from some central region. Then try R1 ≡ !Syntax Error, Idt ejkt [t e-jβR/R3] R2 = s2 + t2 let x = k R1 ≡ !Syntax Error, Idt ejxt f(t) f(t) = t e-jβR/R3 ψ(t) = t ψ'(t) = 1 Now do the parts and we have parts1 = [ f(t)/(ix) * eixt]t=a-a = (1/ix) [ f(a)eiax - f(-a)e-iax ] = (1/ix) [a e-jβR/R3 eiax - (-a) e-jβR/R3 e-iax ] R2 = s2 + a2 = (1/ix) a e-jβR/R3 [eiax + e-iax ] = (1/ix) a e-jβR/R3 [2 cos(ax)] = (1/ix) cos(ax) 2a e-jβR/R3 R2 = s2 + a2 = (1/ik) cos(ak) 2a e-jβR/R3 R2 = s2 + a2 Well, this just says integral is small as k gets large which I already imagine. The other integral is missing the t factor so it might be parts2 = [ f(t)/(ix) * eixt]t=a-a = (1/ix) [ f(a)eiax - f(-a)e-iax ] = (1/ix) e-jβR/R3 [2j sin(ax)] = (1/ik) 2j sin(ak) e-jβR/R3 Then one might argue that parts1 = (1/ik) cos(ak) 2a e-jβR/R3 parts2 = (1/ik) sin(ak) 2j e-jβR/R3 * D D = transverse dimension Then somehow parts2/parts1 ≈ D/a = small if I take a large and then I2 << I1 which is the desired result. So somehow I have managed to get my desired result. Conclusion: It is not impossible to imagine based on the above that | ∂xAx | >> | ∂zAz| at low ω but just the opposite at large ω. Should I put anything into App M on this issue? J1 = !Syntax Error, I[(R/)dR] sin(k) e-jbR/R3 For the usual reasons (domination by small z" region) I expect I2 >> I1. Now going back ∂zAz(x,ω) = - (1+jβd) Σi∫μi Jzi(x',y',ω) I1(x',y') dx' dy' ∂xAx(x,ω) = - (1+jβd) Σi∫μi Jxi(x',y',ω) (x-x') I2(x',y') dx' dy' Now just based on this idea that I2 >> I1 you might conclude that ∂xAx(x,ω) >> ∂zAz(x,ω) which is a rather amazing conclusion, and this confirms the impression noted above for the 2D analysis. Let's write out the above in more detail ∂zAz(x,ω) = - (1+jβd) Σi∫μi Jzi(x',y',ω) { -j e-jkz !Syntax Error, Idz" [z" sin(kz")] e-jβR/R3} dx' dy' ∂xAx(x,ω) = - (1+jβd) Σi∫μi Jxi(x',y',ω) { e-jkz !Syntax Error, Idz" [cos(kz")] e-jβR/R3} (x-x') dx' dy'