disk green attempt 2

Disk Green Attempt #2 PhL 4.12.10 This is a cos(kz) attempt. 1. Setting up the disk problem. 1 2. The potential matching condition between region 1 and region 2. 2 3. The potential matching condition between region 2 and region 3. 2 4. The Gaussian Pillbox condition between region 2 and region 3 3 (a) Digression on the Fourier Cosine Transform 4 Resume Main Flow 7 5. Continuity of electric field Eρ between regions 1 and 2. 7 6. Gather up all four conditions and solve for the coefficients 8 7. Clean things up with better presentation 9 Redo Section 4 for pillbox. 10 Redo Section 5 for non-pillbox. 12 Aside on Neumann's Number or Factor: 15 ... Resume Main Flow 16 1. Setting up the disk problem. Our problem specifically is the Green's function where the point charge is in the plane of the disk or iris. Let's start with the disk and draw this picture showing where things are: I have identified three cylindrically shaped regions that we need to worry about, as shown in the cross section on the right. For all three regions, we want e-k|z| and not e-ik|z| for the z dependence, since we know in the ± z direction the potential must go to 0, so we do not want any I or K functions. However, when I march down this path, I cannot solve for coefficients because I don't have a "complete set" of functions in the z direction. For "cylindrical harmonics" we really need cos(kz) cos(mφ) where we have oscillatory in both directions on our cylinder surface where the pillbox will be located. So even though this seems wrong in terms of z decay, let's try it out and see what happens. This means we DO have I and K functions for the ρ dimension. So, for our Bessel functions I shall choose region 1 Im(kρ) since good at ρ = 0 region 2 Im(kρ) ,Km(kρ) mixture general mix region 3 Km(kρ) since good at ρ = ∞ So we might then write a form for our potential in each region V1(z,ρ,φ) = Σm=0∞ cos(mφ) !Syntax Error, Idk cos(kz)Am(k) Im(kρ) V2(z,ρ,φ) = Σm=0∞ cos(mφ) !Syntax Error, Idk cos(kz) [ Bm(k) Im(kρ) + Cm(k) Km(kρ)] V3(z,ρ,φ) = Σm=0∞ cos(mφ) !Syntax Error, Idk cos(kz) Dm(k) Km(kρ) 2. The potential matching condition between region 1 and region 2. The potential must be continuous at every point on this math cylinder, so we write V1(a,φ,z) = V2(a,φ,z) Σm=0∞ cos(mφ) !Syntax Error, Idk cos(kz) Am(k) Im(ka) = Σm=0∞ cos(mφ) !Syntax Error, Idk cos(kz) [ Bm(k) Im(ka) + Cm(k) Km(ka)] Since the φ and z functions are both complete, this tells us that Am(k) Im(ka) = Bm(k) Im(ka) + Cm(k) Km(ka) which is one condition on our four sets of coefficients. 3. The potential matching condition between region 2 and region 3. The potential must be continuous at every point on this math cylinder, so we write V3(a,φ,z) = V2(a,φ,z) // away from the single point z=0 and φ = 0 and ρ = b Σm=0∞ cos(mφ) !Syntax Error, Idk cos(kz) Dm(k) Km(kb) = Σm=0∞ cos(mφ) !Syntax Error, Idk cos(kz) [ Bm(k) Im(kb) + Cm(k) Km(kb)] Since the φ and z functions are both complete, this tells us that Dm(k) Km(kb) = Bm(k) Im(kb) + Cm(k) Km(kb) which is a second condition on our four coefficients. 4. The Gaussian Pillbox condition between region 2 and region 3 Using the M&M ordering for curvilinear coordinates we have q1, q2, q3 = z, ρ, φ h1 = 1 h2 = 1 h3 = ρ and from our oblate spheroid doc ∂ζVi - ∂ζVo = (q/ε) (h2/h3h1) δ(φ-φ0) δ(ξ - ξ0) q1, q2, q3 = ξ, ζ, φ => ∂ρVi - ∂ρVo = (q/ε) (h2/h3h1) δ(φ-0) δ(z - 0) = (q/ε)(1/b) δ(φ)δ(z) = σ/ε and for our designated inner and outer regions this says ∂ρV2 - ∂ρV3 = (q/ε)(1/b) δ(φ)δ(z) I think I did this right, can come back later and fix it. Dimensions are correct Q/L2 for σ. Now we can compute our radial derivatives in the z = 0 plane. Start with the potentials at z=0: V2(z,ρ,φ) = Σm=0∞ cos(mφ) !Syntax Error, Idk cos(kz) [ Bm(k) Im(kρ) + Cm(k) Km(kρ)] V3(z,ρ,φ) = Σm=0∞ cos(mφ) !Syntax Error, Idk Dm(k) cos(kz) Km(kρ) Then we have, setting ρ = b in the last line, ∂ρV2(z,ρ,φ) = Σm=0∞ cos(mφ) !Syntax Error, Idk k cos(kz) [ Bm(k) Im'(kρ) + Cm(k) Km'(kρ)] ∂ρV3(z,ρ,φ) = Σm=0∞ cos(mφ) !Syntax Error, Idk k cos(kz)Dm(k) K'm(kρ) ∂ρV2 - ∂ρV3 = Σm=0∞ cos(mφ) !Syntax Error, Idk k cos(kz) * { Bm(k) Im'(kb) + Cm(k) Km'(kb) – Dm(k) K'm(kb) } Then, using our pillbox condition above, we get (q/ε)(1/b) δ(φ)δ(z) = Σm=0∞ cos(mφ) !Syntax Error, Idk k cos(kz){ Bm(k) Im'(kb) + Cm(k) Km'(kb) – Dm(k) K'm(kb) } Now we first quote our result !Syntax Error, Idφ cos(m'φ) cos(mφ) = δm,m'2π/(2-δm,0) So apply !Syntax Error, Idφ cos(m'φ) to both sides of the above to get (q/ε)(1/b) δ(z) = Σm=0∞ δm,m'2π/(2-δm,0) !Syntax Error, Idk k cos(kz){ Bm(k) Im'(kb) + Cm(k) Km'(kb) – Dm(k) K'm(kb) } = 2π/(2-δm',0) !Syntax Error, Idk k cos(kz){ Bm'(k) Im''(kb) + Cm'(k) Km''(kb) – Dm'(k)K'm'(kb) } This must be true for every value of m'. Rewrite with m'→ m as usual (q/ε)(1/b) δ(z) = 2π/(2-δm,0) !Syntax Error, Idk k cos(kz){ Bm(k) Im'(kb) + Cm(k) Km'(kb) – Dm(k) K'm(kb) } (a) Digression on the Fourier Cosine Transform The cos(kz) functions are complete. We can start from this point: orthogonality on z in (-∞,∞) completeness on same interval !Syntax Error, Idz eikz e-ik'z = 2π δ(k-k') !Syntax Error, Idk e-ikz e+ik'z = 2π δ(z-z') The usual Fourier Integral transform is then f(z) = (1/2π)!Syntax Error, Idk e+ikz F(k) F(k) = !Syntax Error, Idz e-ikz f(z) Here I am selecting signs as on page 1 of my Spectral Theory chapter 1. Now if f(z) is an even function of z, we can write instead (where we drop sine terms for obvious reasons in both) f(z) = (1/2π)!Syntax Error, Idk cos(kz) F(k) F(k) = !Syntax Error, Idz cos(kz) f(z) The right equation says F(k) is even in k. Therefore, both integrands in the above equations are even with respect to the integration variables, so we can fold both to get f(z) = (1/π)!Syntax Error, Idk cos(kz) F(k) F(k) = 2!Syntax Error, Idz cos(kz) f(z) Note added 1.10.11. The transform appearing on the line above involves only k > 0 and z > 0 ! We don't even talk about f(z) if z < 0. However, if we were to use the expansion shown as the definition of f(z) for negative z, then we would say that the extended function f(z) was even in z, where we mean f(-z) = f(z). Suppose we don't do this extension. Still, if f(z) came out to be, say, a power series in z, the expansion shown for z > 0 tells us that only even powers can occur in f(z). So we can then say the following: in the expansion and projection shown above, we have only z ≥ 0 and k ≥ 0 so f(z) is undefined for z < 0. If the expansion shown is valid for some f(z), then we know that if we were to somehow write f(z) as a power series in z, only even powers would appear, so in that sense, this transform is only valid for "even" functions. So, if our domain of functions of interest f(z) consists only of even functions, then we can write orthogonality and completeness in this way: f(z) = (1/π)!Syntax Error, Idk cos(kz) F(k) = f(z) = (1/π)!Syntax Error, Idk cos(kz) 2!Syntax Error, Idz' cos(kz') f(z') =!Syntax Error, Idz' f(z') !Syntax Error, Idk cos(kz) cos(kz') (2/π) = !Syntax Error, Idz' f(z')δ(z'-z) [ Note added 1.10.11: if z > 0, the last integral equals f(z) unambiguously. If z = 0, we shall regard that as a limiting case z = ε > 0 where we take ε → 0. So there is no issue of only picking up "half a delta function" or anything like that. ] so that !Syntax Error, Idk cos(kz) cos(kz') = (π/2)δ(z - z') // completeness z,z' ≥ 0 If we did this the other way around, we have: F(k) = 2!Syntax Error, Idz cos(kz) f(z) = 2!Syntax Error, Idz cos(kz) (1/π)!Syntax Error, Idk' cos(k'z) F(k') = !Syntax Error, Idk' F(k') !Syntax Error, Idz cos(kz) cos(k'z)(2/π) = !Syntax Error, Idk' F(k') δ(k-k') so that !Syntax Error, Idz cos(kz) cos(k'z)= (π/2) δ(k-k') // orthogonality k,k' ≥ 0 This stuff is just the Fourier Cosine transform as discussed Stak I page 293. As for our transform f(z) = (1/π)!Syntax Error, Idk cos(kz) F(k) F(k) = 2!Syntax Error, Idz cos(kz) f(z) [ Note added 1.10.11. Suppose we consider f(z) = δ(z) as the limit of a narrow Gaussian. Since we are only supposed to consider f(z) for z ≥ 0, when we write δ(z) in this context, we are referring to the right half of the Gaussian, albeit in the limit. It would seem then that we should say F(k) = 2!Syntax Error, Idz cos(kz) δ(z) = 2 * (1/2) = 1 so here we think of our integral as "picking up only the right half" of the full delta function. If we then put this into the expansion, we get δ(z) = (1/π) !Syntax Error, Idk cos(kz) 1 = (1/π) !Syntax Error, Idk cos(kz) z ≥ 0 We can compare this to the Fourier Transform idea that δ(z) = (1/2π) !Syntax Error, Idk e-ikz = (1/2π) !Syntax Error, Idk cos(kz) = (1/π) !Syntax Error, Idk cos(kz) z ≥ 0 and we get the same result! ] We can make things symmetrical by defining G(k) ≡ F(k) so that G(k) = 2!Syntax Error, Idz cos(kz) f(z) = !Syntax Error, Idz cos(kz) f(z) f(z) = (1/π)!Syntax Error, Idk cos(kz) F(k) = (1/π)!Syntax Error, Idk cos(kz) G(k) = !Syntax Error, Idk cos(kz) G(k) So here then are all our results: For functions f(z) which are even and decay fast enough to converge the projection integral: f(z) = !Syntax Error, Idk cos(kz) G(k) // expansion G(k) = !Syntax Error, Idz cos(kz) f(z) // projection !Syntax Error, Idz cos(kz) cos(k'z)= (π/2) δ(k-k') // orthogonality k in (0,∞) !Syntax Error, Idk cos(kz) cos(kz') = (π/2)δ(z-z') // completeness z in (0,∞) The transform now appears as in Stak p 293 (4.71) and completeness as in (4.70). [ See " confusion about completeness relations.doc" for more general comments about these cos cos integrals and their "second terms". Here we get to ignore the second terms since both k and z are on the half line. ] Resume Main Flow We had this pillbox condition: (q/ε)(1/b) δ(z) = 2π/(2-δm,0) !Syntax Error, Idk k cos(kz){ Bm(k) Im'(kb) + Cm(k) Km'(kb) – Dm(k) K'm(kb) } We now apply !Syntax Error, Idz cos(k'z) to both sides. LHS integral is just cos(k'0) = 1, so get (q/ε)(1/b) = 2π/(2-δm,0) !Syntax Error, Idk k (π/2) δ(k-k'){ Bm(k) Im'(kb) + Cm(k) Km'(kb) – Dm(k) K'm(kb) } = 2π/(2-δm,0) (π/2) k' { Bm(k') Im'(k'b) + Cm(k') Km'(k'b) – Dm(k') K'm(k'b) } This must be valid for all k'. Replace k' → k to get (q/ε)(1/b) = 2π/(2-δm,0) (π/2) k { Bm(k) Im'(kb) + Cm(k) Km'(kb) – Dm(k) K'm(kb) } 5. Continuity of electric field Eρ between regions 1 and 2. This analysis is the same as the previous section, with these modifications: q → 0 b → a Dm(k) Km(kρ) → Am(k) Im(ka) // in various forms Thus we end up with 0 = 2π/(2-δm,0) (π/2) k { Bm(k) Im'(kb) + Cm(k) Km'(kb) – Am(k) Im'(ka) } which really says Bm(k) Im'(kb) + Cm(k) Km'(kb) = Am(k) Im'(ka) 6. Gather up all four conditions and solve for the coefficients Here they are from the above sections: Am(k) Im(ka) = Bm(k) Im(ka) + Cm(k) Km(ka) // match V between region 1 and 2 Dm(k) Km(kb) = Bm(k) Im(kb) + Cm(k) Km(kb) // match V between region 2 and 3 Bm(k) Im'(kb) + Cm(k) Km'(kb) = Am(k) Im'(ka) // match E field between region 1 and 2 (q/ε)(1/b) = 2π/(2-δm,0) (π/2) k { Bm(k) Im'(kb) + Cm(k) Km'(kb) – Dm(k) K'm(kb) } // match E field between region 2 and 3 Rewrite them all (change order too) [Am(k) - Bm(k)] Im(ka) = Km(ka) [Am(k) - B(k)] Im'(ka) = Cm(k) Km'(kb) [ Dm(k) - Cm(k) ] Km(kb) = Bm(k) Im(kb) [ Dm(k) - Cm(k) ] Km'(kb) = Bm(k) Im'(kb) - (q/ε)(1/b){ 2π/(2-δm,0) (π/2) k}-1 Rewrite again [Am(k) - Bm(k)] = Km(ka)/ Im(ka) [Am(k) - B(k)] = Cm(k) Km'(kb)/ Im'(ka) [ Dm(k) - Cm(k) ] = Bm(k) Im(kb)/ Km(kb) [ Dm(k) - Cm(k) ] = Bm(k) Im'(kb)/ Km'(kb) - (q/ε)(1/b){ 2π/(2-δm,0) (π/2) k Km'(kb)}-1 Equate the first pair and equate the second pair to get Km(ka)/ Im(ka) = Cm(k) Km'(kb)/ Im'(ka) Bm(k) Im(kb)/ Km(kb) = Bm(k) Im'(kb)/ Km'(kb) - (q/ε)(1/b){ 2π/(2-δm,0) (π/2) k Km'(kb)}-1 We can solve these as follows: Cm(k) = [Km(ka)/ Im(ka)]/[ Km'(kb)/ Im'(ka)] Bm(k) = [Im(kb)/ Km(kb)– Im'(kb)/ Km'(kb)] - (q/ε)(1/b){ 2π/(2-δm,0) (π/2) k Km'(kb)}-1 Then we get D and A from 3 and 1 above Dm(k) = Cm(k) + Bm(k) Im(kb)/ Km(kb) Am(k) = Bm(k) + Km(ka)/ Im(ka) 7. Clean things up with better presentation Now that we think we have solved for all the coefficients, we can start over and think more in terms of Smythian Forms. If we first think just about the potential continuity, we have these facts: V1(z,ρ,φ) = Σm=0∞ cos(mφ) !Syntax Error, Idk cos(kz)Am(k) Im(kρ) V2(z,ρ,φ) = Σm=0∞ cos(mφ) !Syntax Error, Idk cos(kz) [ Bm(k) Im(kρ) + Cm(k) Km(kρ)] V3(z,ρ,φ) = Σm=0∞ cos(mφ) !Syntax Error, Idk cos(kz) Dm(k) Km(kρ) Am(k) Im(ka) = Bm(k) Im(ka) + Cm(k) Km(ka) Dm(k) Km(kb) = Bm(k) Im(kb) + Cm(k) Km(kb) Use the first line to get rid of A, and the second to get rid of D: V1(z,ρ,φ) = Σm=0∞ cos(mφ) !Syntax Error, Idk cos(kz){ Bm(k) Im(ka) + Cm(k) Km(ka)} Im(kρ)/ Im(ka) V2(z,ρ,φ) = Σm=0∞ cos(mφ) !Syntax Error, Idk cos(kz) [ Bm(k) Im(kρ) + Cm(k) Km(kρ)] V3(z,ρ,φ) = Σm=0∞ cos(mφ) !Syntax Error, Idk cos(kz){ Bm(k) Im(kb) + Cm(k) Km(kb)} Km(kρ)/ Km(kb) So we have only B and C coefficients to worry about. Now define: bm(k) ≡ Bm(k) / [Im(ka) Km(kb)] cm(k) ≡ Cm(k) / [Im(ka) Km(kb)] Then our potentials are these ( save space with cos(x) = Cx) V1(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { bm(k)Km(kb)Im(ka) Im(kρ) + cm(k) Km(ka)Km(kb) Im(kρ)} V2(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { bm(k) Im(ka) Km(kb) Im(kρ) + cm(k) Im(ka) Km(kb) Km(kρ)} V3(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { bm(k) Im(kb) Im(ka) Km(kρ) + cm(k) Km(kb) Im(ka) Km(kρ)} Now try to put factors in some reasonable order V1(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { bm(k) Im(ka)Km(kb) Im(kρ) + cm(k) Km(ka)Km(kb) Im(kρ)} V2(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { bm(k) Im(ka) Km(kb) Im(kρ) + cm(k) Km(kρ) Km(kb) Im(ka) } V3(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { bm(k) Im(ka) Km(kρ) Im(kb) + cm(k) Km(kρ) Km(kb) Im(ka) } The red terms are equal, and so are the blue terms. At the 1-2 boundary, we see that the second terms of the first two expressions agree at ρ = a, and at the 2-3 boundary we see that the first terms of the last two expressions agree at ρ = b. Let's write this again, then, as our Smythian Form for this problem: V1(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { bm(k) Im(ka) Km(kb) Im(kρ) + cm(k) Km(ka) Km(kb) Im(kρ)} V2(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { bm(k) Im(ka) Km(kb) Im(kρ) + cm(k) Km(kρ) Km(kb) Im(ka) } V3(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { bm(k) Im(ka) Km(kρ) Im(kb) + cm(k) Km(kρ) Km(kb) Im(ka) } This is fancier that earlier forms we had because we have three regions instead of 2 to worry about. But the idea is the same: cause the boundary matches to be manifestly obvious, and don't have any factors in the denominator. Redo Section 4 for pillbox. So let's take the above three expressions for the potential as our starting point, and re-do the electric field continuity calculations. This is probably faster that plugging things in, and I hope it will expose the Wronskian simplification which we have not seen yet. So here we go for the 2/3 boundary fields: V2(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { bm(k) Im(ka) Km(kb) Im(kρ) + cm(k) Km(kρ) Km(kb) Im(ka) } V3(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { bm(k) Im(ka) Km(kρ) Im(kb) + cm(k) Km(kρ) Km(kb) Im(ka) } ∂ρV2(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk k Ckz { bm(k) Im(ka) Km(kb) Im'(kρ) + cm(k) Km'(kρ) Km(kb) Im(ka) } ∂ρV3(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk k Ckz { bm(k) Im(ka) Km'(kρ) Im(kb) + cm(k) Km'(kρ) Km(kb) Im(ka) } ∂ρV2 - ∂ρV3 = Σm=0∞Cmφ!Syntax Error, Idk k Ckz * { bm(k) Im(ka) Km(kb) Im'(kρ) + cm(k) Km'(kρ) Km(kb) Im(ka) } - { bm(k) Im(ka) Km'(kρ) Im(kb) + cm(k) Km'(kρ) Km(kb) Im(ka) } Evaluate this thing at ρ = b and we have ∂ρV2 - ∂ρV3 = Σm=0∞Cmφ!Syntax Error, Idk k Ckz * { bm(k) Im(ka) Km(kb) Im'(kb) + cm(k) Km'(kb) Km(kb) Im(ka) } - { bm(k) Im(ka) Km'(kb) Im(kb) + cm(k) Km'(kb) Km(kb) Im(ka) } The cm(k) terms cancel each other (as usual in this kind of process) and we have ∂ρV2 - ∂ρV3 = Σm=0∞Cmφ!Syntax Error, Idk k Ckz * bm(k) Im(ka) Km(kb) Im'(kb) - bm(k) Im(ka) Km'(kb) Im(kb) = Σm=0∞Cmφ!Syntax Error, Idk k Ckz bm(k) Im(ka) { Km(kb) Im'(kb) - Km'(kb) Im(kb) } = Σm=0∞Cmφ!Syntax Error, Idk k Ckz bm(k) Im(ka) W[ Km(kb), Im(kb)] Bateman p 80 says this Wronskian is +1/(kb), also Jackson p 86, first item not primed. So ∂ρV2 - ∂ρV3 = Σm=0∞Cmφ!Syntax Error, Idk k Ckz bm(k) Im(ka)/(kb) = (1/b) Σm=0∞Cmφ!Syntax Error, Idk Ckz bm(k) Im(ka) ∂ρV2 - ∂ρV3 = (q/ε)(1/b) δ(φ)δ(z) Equating we get (q/ε)δ(φ)δ(z) = Σm=0∞Cmφ!Syntax Error, Idk Ckz bm(k) Im(ka) Apply !Syntax Error, Idφ Cm'φ to both sides to get (q/ε)δ(z) = Σm=0∞[!Syntax Error, Idφ Cm'φ Cmφ ]!Syntax Error, Idk Ckz bm(k) Im(ka) = Σm=0∞[ δm,m'2π/(2-δm,0) ]!Syntax Error, Idk Ckz bm(k) Im(ka) = 2π/(2-δm,0) !Syntax Error, Idk Ckz bm(k) Im(ka) where in the last line I replaced m' with m. Now apply !Syntax Error, Idz Ck'z to both sides. I think the LHS somehow sees the full δ, not just half, come back later if wrong. Then we have (q/ε) = 2π/(2-δm,0) !Syntax Error, Idk [!Syntax Error, Idz Ck'z Ckz ] bm(k) Im(ka) (*) = 2π/(2-δm,0) !Syntax Error, Idk [(π/2) δ(k-k')] bm(k) Im(ka) = π2/(2-δm,0) bm(k) Im(ka) where in the last line I replaced k' by k. So we have found our result for one of the two coefficients: bm(k) Im(ka) = (2-δm,0) (q/π2ε) IF we are only supposed to pick up half the delta, then we get 1/2 on the LHS of (*) so just replace q with q/2. Redo Section 5 for non-pillbox. V1(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { bm(k) Im(ka) Km(kb) Im(kρ) + cm(k) Km(ka) Km(kb) Im(kρ)} V2(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { bm(k) Im(ka) Km(kb) Im(kρ) + cm(k) Km(kρ) Km(kb) Im(ka) } ∂ρ V1(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk k Ckz { bm(k) Im(ka) Km(kb) Im'(kρ) + cm(k) Km(ka) Km(kb) Im'(kρ)} ∂ρ V2(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk k Ckz { bm(k) Im(ka) Km(kb) Im'(kρ) + cm(k) Km'(kρ) Km(kb) Im(ka) } ∂ρV2 - ∂ρV3 = Σm=0∞Cmφ!Syntax Error, Idk k Ckz * { bm(k) Im(ka) Km(kb) Im'(kρ) + cm(k) Km(ka) Km(kb) Im'(kρ)} - { bm(k) Im(ka) Km(kb) Im'(kρ) + cm(k) Km'(kρ) Km(kb) Im(ka) } Evaluate this thing at ρ = a and we have ∂ρV2 - ∂ρV3 = Σm=0∞Cmφ!Syntax Error, Idk k Ckz * { bm(k) Im(ka) Km(kb) Im'(ka) + cm(k) Km(ka) Km(kb) Im'(ka)} - { bm(k) Im(ka) Km(kb) Im'(ka) + cm(k) Km'(ka) Km(kb) Im(ka) } This time the bm(k) terms cancel each other and we have ∂ρV2 - ∂ρV3 = Σm=0∞Cmφ!Syntax Error, Idk k Ckz * cm(k) Km(ka) Km(kb) Im'(ka) - cm(k) Km'(ka) Km(kb) Im(ka) = Σm=0∞Cmφ!Syntax Error, Idk k Ckz cm(k) Km(kb) [Km(ka) Im'(ka) - Km'(ka) Im(ka)] = Σm=0∞Cmφ!Syntax Error, Idk k Ckz cm(k) Km(kb) W[ Km(ka), Im(ka)] This Wronskian is +1/(ka) so we get ∂ρV2 - ∂ρV3 = Σm=0∞Cmφ!Syntax Error, Idk k Ckz cm(k) Km(kb) 1/(ka) = (1/a) Σm=0∞Cmφ!Syntax Error, Idk Ckz cm(k) Km(kb) ∂ρV2 - ∂ρV3 = 0 // no Green's charge on this boundary Equating we get 0 = (1/a) Σm=0∞Cmφ!Syntax Error, Idk Ckz cm(k) Km(kb) Completeness in both φ and z gives the surprising (to me) result that cm(k) = 0. Put the pieces together . Here was our Smythian form V1(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { bm(k) Im(ka) Km(kb) Im(kρ) + cm(k) Km(ka) Km(kb) Im(kρ)} V2(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { bm(k) Im(ka) Km(kb) Im(kρ) + cm(k) Km(kρ) Km(kb) Im(ka) } V3(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { bm(k) Im(ka) Km(kρ) Im(kb) + cm(k) Km(kρ) Km(kb) Im(ka) } and we found from our field continuity conditions that bm(k) Im(ka) = (2-δm,0) (q/π2ε) cm(k) = 0 Stuff these in to get: V1(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { (2-δm,0) (q/π2ε) Km(kb) Im(kρ) } V2(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { (2-δm,0) (q/π2ε) Km(kb) Im(kρ) } V3(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { (2-δm,0) (q/π2ε) Km(kρ) Im(kb) } which we rewrite as: V1(z,ρ,φ) = (q/π2ε)Σm=0∞ (2-δm,0)Cmφ!Syntax Error, Idk Ckz { Km(kb) Im(kρ) } V2(z,ρ,φ) = (q/π2ε)Σm=0∞ (2-δm,0)Cmφ!Syntax Error, Idk Ckz { Km(kb) Im(kρ) } V3(z,ρ,φ) = (q/π2ε)Σm=0∞ (2-δm,0)Cmφ!Syntax Error, Idk Ckz { Km(kρ) Im(kb) } This says the potential in regions 1 and 2 is the same. This was not obvious to me at the start. I guess the reason is that on this math cylinder, continuity of potential and field forces the potential to be the same in both regions. We have not said much about what happens on the disk itself! I ignored that in the Smythian method, and probably now I am going to get into trouble with it. Indeed, if I set z = 0 in region 1, we should be getting V = constant on the disk. But we get the same result in regions 1 and 2 for z = 0 which we know is wrong. So I have "gone astray" by ignoring this fact. The error is in the "field continuity between region 1 and 2". I think the other three conditions are OK. When we did the pillbox for the Green's point charge, we had to think about the fact that there was charge located somewhere on our math boundary between the two regions. For the 1/2 boundary, there is going to be some charge on the rim of the disk, and I have ignored it! Let's try to maintain our Smythian form (which may be wrong, but maybe not) which is V1(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { bm(k) Im(ka) Km(kb) Im(kρ) + cm(k) Km(ka) Km(kb) Im(kρ)} V2(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { bm(k) Im(ka) Km(kb) Im(kρ) + cm(k) Km(kρ) Km(kb) Im(ka) } V3(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { bm(k) Im(ka) Km(kρ) Im(kb) + cm(k) Km(kρ) Km(kb) Im(ka) } We can install our conclusion from the 2/3 boundary, bm(k) Im(ka) = (2-δm,0) (q/π2ε) = εm (q/π2ε) // see below on Neumann's factor to get V1(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { (2-δm,0) (q/π2ε) Km(kb) Im(kρ) + cm(k) Km(ka) Km(kb) Im(kρ)} V2(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { (2-δm,0) (q/π2ε) Km(kb) Im(kρ) + cm(k) Km(kρ) Km(kb) Im(ka) } V3(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { (2-δm,0) (q/π2ε) Km(kρ) Im(kb) + cm(k) Km(kρ) Km(kb) Im(ka) } But now we have the dual integral equation situation which is, in our current context, ( all at z = 0) ( I wrongly had Vo as in Jackson's disk, but here we have a Green's problem so V = 0 on the disk ) V1(z,ρ,φ) = 0 ρ ≤ a Dirichlet ∂zV2(z,ρ,φ) = 0 a < ρ ≤ b Neumann ∂zV3(z,ρ,φ) = 0 b < ρ < ∞ Neumann, except for the Green's charge The last lines say there is no charge lying in the z plane outside the disk except for the Green's charge. Comment: I have been wondering how the dual integral equation situation was going to appear here. I now am stating the mixed BC at the same time I am providing a guess at the Smythian atomic form. This is pretty close to what Jackson does for the charged disk. We know that ∂z Ckz = -kSkz = 0 at z = 0. This makes it appear that the two Neumann region conditions are already met (away from the Green's charge). The problem is then done if we can find cm(k) such that V1(z=0,ρ,φ) = 0, which is to say, Σm=0∞Cmφ!Syntax Error, Idk { (2-δm,0) (q/π2ε) Km(kb) Im(kρ) + cm(k) Km(ka) Km(kb) Im(kρ)} = 0 Aside on Neumann's Number or Factor: Various papers refer to εm ≡ 2-δm,0 as "Neumann's number". I had to dig very hard to find the origin of this name. If you go to Watson's Bessel function treatise, you find So Watson calls it a Neumann's factor and gives an actual reference, last line above, which I look up: So yes, it is our Bessel guy Carl involved here! On page 32 of Bateman II we see the εn symbol also in use, it appears in the Jn expansion of 1/(x-y). Does not appear in W&W. I see now that the web has I would say that Watson invented this symbol εn just as a convenience, and did so in his 1922 first edition, though I have the second edition. ... Resume Main Flow We can write this out as two conditions, the first line for m = 0, the second for m ≠ 0 : !Syntax Error, Idk I0(kρ) K0(kb){ (q/π2ε) + c0(k) K0(ka) } = 0 !Syntax Error, Idk Im(kρ) Km(kb){ (2) (q/π2ε) + cm(k) Km(ka) } = 0 These equations seem to have simple solutions: cm(k) = - (2) (q/π2ε)/ Km(ka) m ≠ 0 cm(k) = - (1) (q/π2ε)/ Km(ka) m = 0 which we can combine using Neumann's factor εn cm(k) = - εm(q/π2ε)/ Km(ka) Now let's back up and see where we are. I have this Smythian Form V1(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { bm(k) Im(ka) Km(kb) Im(kρ) + cm(k) Km(ka) Km(kb) Im(kρ)} V2(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { bm(k) Im(ka) Km(kb) Im(kρ) + cm(k) Km(kρ) Km(kb) Im(ka) } V3(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { bm(k) Im(ka) Km(kρ) Im(kb) + cm(k) Km(kρ) Km(kb) Im(ka) } and I think I have been able to compute the coefficients: bm(k) Im(ka) = (2-δm,0) (q/π2ε) = εm (q/π2ε) // maybe add factor of 1/2 see above cm(k) Km(ka) = – εm (q/π2ε) Of special interest to me is the potential in the cylinder containing the disk. We then have V1(z,ρ,φ) = εm (q/π2ε) Σm=0∞Cmφ!Syntax Error, Idk Ckz { Km(kb) Im(kρ) - Km(kb) Im(kρ)} but oops, the {...} = 0 so the potential V1 vanishes everywhere in this cylinder, not just on the disk. If the factor of 1/2 from δ(z) is supposed to be used, we get instead: V1(z,ρ,φ) = - (1/2)εm (q/π2ε) Σm=0∞Cmφ!Syntax Error, Idk Ckz Km(kb) Im(kρ) But then if we evaluate at z = 0 we must get Σm=0∞Cmφ!Syntax Error, Idk Km(kb) Im(kρ) => !Syntax Error, Idk Km(kb) Im(kρ) = 0 which seems pretty unlikely. I am not sure what my conclusion is at this point, but I feel the current document is a reasonable attempt on the disk and merits more study. [ 4.21.10]