surface current scraps save REVIEWED

scratch save PhL 11.13.13 Reviewed 11.16.13, keep for first part only. Verify GR7: Let's make sure the GR7 integral is correct. Use Stak and (6.22) which claims that ln(1 + a2 - 2a cosx ) = -2 Σn=1∞ (an/n)cos(nx) // where |A| < 1 Then my integral is !Syntax Error, Idx ln[α2 + 1 - 2α cos(x)] cosx = !Syntax Error, Idx {- 2 Σn=1∞ (αn/n)cos(nx)} cosx = - 2 Σn=1∞ (αn/n) !Syntax Error, Idx {cos(nx)} cosx Now according to Spiegel p 96 15.27 this integral vanishes unless n =1 in which case it is π/2, so = -2 (α )π/2 = -πα which agrees with GR7. **************************************************************** This is a big chunk from my surface currents doc where I think I had the wrong propagator: _________________________________________________ Now write curl J = (∂xJy- ∂yJx) + (∂yJz- ∂zJy) + (∂zJx- ∂xJz) Coordinates are given by x = x + y + z x' = x' + y' + z' J(x) = Jx(x) + Jy(x) + Jz(x) J(x') = Jx(x') + Jy(x') + Jz(x') If J has only a z component, then curl J = (∂yJz) + (- ∂xJz) curl' J(x') = (∂y'Jz(x')) + (- ∂x'Jz(x')) Then ** becomes these two equations, where J has no z dependence, Hx(x,y) = (1/4π) ∫d3x' ∂y'Jz(x',y')/R R = |x - x'| wrong propagator. Hy(x,y) = - (1/4π) ∫d3x' ∂x'Jz(x',y')/R Doing parts integration (the parts vanish since Jz has finite x,y extent) we find Hx(x,y) = (1/4π) ∫dx'dz'{ [Jz(x',y')R-1]∞-∞ - ∫dy' Jz(x',y') ∂y'R-1 } Assuming a finite Jz transverse distribution, the "parts" contribution vanishes. Meanwhile, R2 = (x-x')2 + (y-y')2 +(z-z')2 2RdR = 2(y'-y)dy' => ∂y'R-1 = -R-2∂y'R = -R-2 (y'-y)/R = -R-3(y'-y) Thus, Hx(x,y) = (1/4π) ∫dx'dz'{ 0 - ∫dy' Jz(x',y') [-R-3(y'-y)] } = (1/4π) ∫dx'dy'dz'{ Jz(x',y') [R-3(y'-y)] } = (1/4π) ∫dx'dy' (y'-y)dz' Jz(x',y') ∫dz' [R-3] The dz integration over the conductor gives ∫dz' [R-3] = !Syntax Error, Idz' (s2+z'2)-3/2 = 1/s2 where s ≡ Therefore Hx(x,y) = (1/4π) ∫dx'dy' Jz(x',y') Doing similar processing for Hy gives this pair of results: Hx(x,y) = (1/4π) ∫dx'dy' Jz(x',y') Hy(x,y) = (1/4π) ∫dx'dy' Jz(x',y') Let's see how this does for a round wire with origin at center of cross section and radius a. Then x = acosθ y = asinθ x' = r'cosθ' x-x' = acosθ-r'cosθ' y' = r'sinθ' y-y' = asinθ-r'sinθ' R2 = (acosθ-r'cosθ')2 + (asinθ-r'sinθ')2 = a2 + r'2 - 2ar'cos(θ-θ') dx'dy' = r'dr'dθ' Hθ = Hy cosθ -Hx sinθ // scratch drawing So start with Hx(x,y) = (1/4π) ∫dx'dy' Jz(x',y') = - (1/4π) ∫ r'dr'dθ' Jz(x',y') So we then have these two equations Hx(x,y) = - (1/4π) ∫ r'dr'dθ' Jz(x',y') Hy(x,y) = - (1/4π) ∫ r'dr'dθ' Jz(x',y') Now for the round wire example, assume Jz = I/(πa2) so then Hx(x,y) = - (1/4π) I/(πa2) !Syntax Error, I r'dr'!Syntax Error, Idθ' Hy(x,y) = - (1/4π) I/(πa2) !Syntax Error, I r'dr'!Syntax Error, Idθ' There are three angle integrals to evaluate. I1 ≡ !Syntax Error, Idθ' I2 = !Syntax Error, Idθ' I3 = !Syntax Error, Idθ' in terms of which we then have Hx(x,y) = - (1/4π) I/(πa2) !Syntax Error, I r'dr' { asinθ I1 - r' I2 } Hy(x,y) = - (1/4π) I/(πa2) !Syntax Error, I r'dr'{ acosθ I1 - r' I3 } Let x = θ-θ' so sinθ' = sin(θ-x) = sinθcosx - cosθsinx cosθ' = cos(θ-x) = cosθcosx + sinθsinx Then I1 = ∫dx I2 = ∫dx = sinθ ∫dx - cosθ∫dx = sinθ I4 - cosθ I5 I3 = ∫dx = cosθ ∫dx + sinθ ∫dx = cosθ I4 + sinθ I5 We now have to evaluate I1 = ∫dx I4 = ∫dx I5 = ∫dx But taking x from -π to π, we see that I5 has an odd integrand so we write I5 = 0. We then have two to worry about I1 = ∫dx = (1/a2) ∫dx = (1/a2) ∫dx α = r'/a then we have two to worry about I1 =(1/a2) ∫dx I4 = (1/a2) ∫dx Now these appear on page 391 circa of GR7, and I quote Therefore I1 = (1/a2) 2 π α0/(1-α2) α = r'/a < 1 = (2π/a2) /(1-(r'/a)2) = 2π/(a2-r'2) I4 = (1/a2) 2 π α1/(1-α2) = (2π/a2)(r'/a)/(1-(r'/a)2) = = 2π (r'/a) / (a2-r'2) Now we go backwards up the stack: I2 = sinθ I4 - cosθ I5 = sinθ I4 = sinθ 2π (r'/a) / (a2-r'2) I3 = cosθ I4 + sinθ I5 = cosθ I4 = cosθ 2π (r'/a) / (a2-r'2) which we summarize as I1 = 2π/(a2-r'2) I2 = sinθ 2π (r'/a) / (a2-r'2) I3 = cosθ 2π (r'/a) / (a2-r'2) Then go back to Hx(x,y) = - (1/4π) I/(πa2) !Syntax Error, I r'dr' { asinθ I1 - r' I2 } Hy(x,y) = - (1/4π) I/(πa2) !Syntax Error, I r'dr'{ acosθ I1 - r' I3 } Hx(x,y) = - (1/4π) I/(πa2) !Syntax Error, I r'dr' { asinθ 2π/(a2-r'2) - r' sinθ 2π (r'/a) / (a2-r'2) } Hy(x,y) = - (1/4π) I/(πa2) !Syntax Error, I r'dr'{ acosθ 2π/(a2-r'2) - r' cosθ 2π (r'/a) / (a2-r'2) } Hx(x,y) = - (1/2) I/(πa2) sinθ !Syntax Error, I r'dr' { a 1/(a2-r'2) - r' (r'/a) / (a2-r'2) } Hy(x,y) = - (1/2) I/(πa2) cosθ!Syntax Error, I r'dr'{ a 1/(a2-r'2) - r' (r'/a) / (a2-r'2) } Now we have various r' integrals to worry about. J1 = !Syntax Error, I r'dr'/ (a2-r'2) J2 = !Syntax Error, I r'3dr'/(a2-r'2) in terms of which Hx(x,y) = - (1/2) I/(πa2) sinθ { a J1 - (1/a) J2 } Hy(x,y) = - (1/2) I/(πa2)cosθ { a J1 - (1/a) J2 } But oops, both J1 and J2 diverge at the r' ≈ a endpoint. That was completely unexpected by me. ************************* this was at the start of the surface currents deal ********** I am hopeful of generalizing the Chapter 4 development of transmission line parameters to allow for the possibility of magnetic conductors, if it is not too hard! I already know the surface currents on a round conductor with uniform DC current density, but I would like to find an expression for this current in a more general case. For any boundary we know that Ht2 = Ht1 (1/μ1)Bt1 = (1/μ2)Bt2 at the boundary If we have μ2 > μ1 then Bt2 > Bt1.