App D the Helm phi equation REVIEWED

What about the Helmholtz Eφ equation? This stuff is now incorporated as Appendix section D.3. Here are the Er and Eφ equations in m space: [2E]r + β2 Er = 0 : [r2∂r2 + r∂r - (m2+1) + r2(β2-βd2)] Er(r,m) - 2jm Eφ(r,m) = 0 (D.1.17) [2E]φ + β2 Eφ = 0 : [r2∂r2 + r∂r - (m2+1) + r2(β2-βd2)] Eφ(r,m) + 2jmEr(r,m) = 0 (D.1.18) How do I know that my solutions Eφ and Er satisfy this last equation? Maybe they don't! Why should they? Rewrite the above in this manner [x2∂x2 + x∂x - (m2+1) +x2] Er(r,m) - 2m jEφ(r,m) = 0 (D.1.17) [x2∂x2 + x∂x - (m2+1) + x2] jEφ(r,m) - 2mEr(r,m) = 0 (D.1.18) I solved the first equation for Eφ in my work, and I solved the Hz equation for Er. So what happens when I insert these two solutions into the last equation above? Rewrite the last as [x2∂x2 + x∂x - (m2+1) + x2] jEφ(r,m) = 2m Er(r,m). It would certainly be nice if this were true. I have Er(r,m) = am x-1 Jm(x) + Jm+1(x) . (D.2.11) jEφ(r,m) = - am x-1 Jm(x) + ( + ) Jm+1(x) x = β'r . (D.2.15) Then we get LHS = [x2∂x2 + x∂x - (m2+1) + x2] {- am x-1 Jm(x) + ( + ) Jm+1(x) } RHS = 2m{ am x-1 Jm(x) + Jm+1(x)} You would think this has to balance separately as if Km and am were independent constants. So then LHSK = [x2∂x2 + x∂x - (m2+1) + x2] ( Jm+1(x) ) RHSK = 2m Jm+1(x) Then I would have to show that [x2∂x2 + x∂x - (m2+1) + x2] Jm+1(x) = 2m Jm+1(x) or [x2∂x2 + x∂x - (m2+1) + x2 - 2m] Jm+1(x) = 0 or [x2∂x2 + x∂x - (m2+1+2m) + x2] Jm+1(x) = 0 or [x2∂x2 + x∂x - (m+1)2 + x2] Jm+1(x) = 0 or [x2∂x2 + x∂x + x2- (m+1)2] Jm+1(x) = 0 Bessel's equation does say that [x2∂x2 + x∂x + x2 - (m+1)2 ] Jm+1(x) = 0 so it works!!!!! I am totally amazed. At least the Km terms balance. Now let's try the am balance. LHS = [x2∂x2 + x∂x - (m2+1) + x2] {- am x-1 Jm(x) + Jm+1(x) } RHS = 2m{ am x-1 Jm(x) } LHS = [x2∂x2 + x∂x - (m2+1) + x2] {- x-1 Jm(x) + m-1 Jm+1(x) } RHS = 2m{ x-1 Jm(x) } This is a much messier situation. I could try Maple. Here is Maple evaluating the LHS: But this is exactly the RHS !! I am again astounded to see this working right. Now I can consider the notion of trying to find the m = 0 solution for Eφ using the φ equation, [r2∂r2 + r∂r - (m2+1) + r2β'2] Eφ(r,m) + 2jmEr(r,m) = 0 which for m = 0 becomes [r2∂r2 + r∂r -1 + r2β'2] Eφ(r,m=0) = 0 or [x2∂x2 + x∂x -1 + x2] Eφ(r,m=0) = 0 This seems to say that Eφ(r,0) = Cφ0 J1(x) where Cφ0 is some constant. How does this then figure into my boundary conditions stuff? If I want this to be zero on the surface using my equipotential idea, then for m =0 I must conclude that Cφ0 = 0. But that is not required, and I suppose you could allow an m = 0 rotating current on the surface? Question: Can the φ Helm be derived from the others and div E = 0? [2E]z + β2 Ez = 0 : [r2∂r2 + r ∂r - m2 + r2 β'2] Ez(r,m) = 0 ok (D.1.16) [2E]r + β2 Er = 0 : [r2∂r2 + r∂r - (m2+1) + r2 β'2] Er(r,m) - 2jm Eφ(r,m) = 0 (D.1.17) [2E]φ + β2 Eφ = 0 : [r2∂r2 + r∂r - (m2+1) + r2β'2] Eφ(r,m) + 2jmEr(r,m) = 0 (D.1.18) div E = 0 : ∂r [r Er(r,m)] + jmEφ(r,m) -jβd r Ez(r,m) = 0 (D.1.19) Maybe ponder in Cartesians first. (∂j2 + β2) Ei= 0 ∂jEj = 0 (∂12+∂22+ ∂32 + β2) E1= 0 (∂12+∂22+ ∂32 + β2) E2= 0 (∂12+∂22+ ∂32 + β2) E3= ???? Start with ∂1E1 + ∂2E2+ ∂3E3 = 0 then (∂12+∂22+ ∂32 + β2)∂1E1 + (∂12+∂22+ ∂32 + β2)∂2E2+ (∂12+∂22+ ∂32 + β2)∂3E3 = 0 or ∂1 (∂12+∂22+ ∂32 + β2) E1 + ∂2 (∂12+∂22+ ∂32 + β2) E2+ ∂3 (∂12+∂22+ ∂32 + β2) E3 = 0 or ∂1 0 + ∂2 0+ ∂3 (∂12+∂22+ ∂32 + β2) E3 = 0 or ∂3 [(∂12+∂22+ ∂32 + β2) E3] = 0 Well it certainly works but not a hard proof. Go back to (2+ β2)E = 0 => grad(div E) – curl (curl E) + β2 E = 0 When div E = 0 this says – curl (curl E) + β2 E = 0