Iris Green's by Dirichlet method Attempt #2

Iris Green's by Dirichlet method Attempt #2 PhL 4.24.10 Overview: When I started this doc, I thought wrongly that VDirichlet had to be non-symmetric in z if the Green's point charge was not in the z = 0 plane. So in Section 2 I tried a non-symmetric 2-region Smythe form, but it led to a contradiction: assume Am(k) ≠ Bm(k), but then the z=0 match forces Am(k) = Bm(k) since each form valid for all ρ and you can use Hankel orthogonality. In Section 3 I then tried a 4-region Smythe form to get around this contradiction. But then when I tried to match both V and ∂ρV at ρ = a, I got another contradiction, all coefficients are 0, where it says OUCH in the contents below. In actual fact, I did not discover this until section 4 (e), but I copied that section up to be 3 (b). This allows Section 4 to be completely ignored. In Section 4 I was just trying to write down all potential and electric field matching conditions I could think of. I wrongly was thinking that the tangent E field was 0 on the iris, but of course this is not true for the Dirichlet problem. So basically all of Section 4 is a complete wasted effort, but it is there for the record. Finally, in Section 5 I understood why in fact VDirichlet is symmetric in z even if the Green's is not in the z=0 plane, and I wrote this up in detail. 1. Introduction. 1 2. Trying a two-region Smythian form, above and below z = 0. 2 3. Trying a four-region Smythian form, above and below z = 0. 3 (a) The matching condition on V at ρ = a between our two cylindrical regions: 4 (b) The matching condition on ∂ρV at ρ = a between our two cylindrical regions: OUCH! 6 4. The following sections are just idle spinning of wheels: 7 (a) The matching condition on V at z = 0 regions for both cylinder regions: 7 (b) Recap to this point. 7 (c) Idea #1: 9 (d) Idea #2: 10 (e) Now rewrite just the ones that are independent conditions: 12 5. Realization that V really is symmetric in z and why this is so. 18 1. Introduction. The situation is this: we have our "two cavities" of the Great Sphere separated by a plane with a hole in it, these are the atria and the hole is the hole in the heart. The Green's charge is in one of the cavities, let's say the one on the right. We want to write the Green's function as the sum of two potentials. One is that of the point charge. The other is that of "the induced charge" on the iris. It is this second potential that we want to calculate by our "Dirichlet method". We know that the σ surface charge distribution will be different on the two sides of the iris, it will be larger in general on the side where we have the point charge. And we know that the potential itself of the induced charge will be asymmetrical as well (wrong!). In the extreme limit, as we close down the hole, the potential and charge goes to 0 in the "other" cavity from the point charge. In our Dirichlet method, our starting point is to compute the potential that the point charge makes on our boundary (the central iris). We did that in Attempt #1 and found this result: b = (c, 0, d)Car = (c, 0,d) b2 = c2 + d2 br = cx+dz r = (x,y,z)Car = (ρ,θ,z) r2 = ρ2 + z2 br' = cx'+dz' Vpoint(ρ,θ,0) = +q/ where I have of necessity also shown the definitions of various symbols. Therefore, in our "Dirichlet problem" we want to assume this prescribed potential on the iris: V(ρ,θ,0) = - q/ ρ ≥ a Now an interesting situation arises. This potential, it were the whole game, would imply a symmetric solution for the potential in the ±z cavities. After all, this is the potential on a math plane z = 0 and it is the same potential BC as seen on both sides. But this Dirichlet component is NOT the whole game. The other part of the game, which we could crudely call the Neumann part, has to do with the hole. As outlined in Attempt #1, what we know about the hole region is this: ∂zV(ρ,θ,0+) = ∂zV(ρ,θ,0-) 0 < ρ <a We know, in other words, that ∂zV is continuous through the hole as you move from z = +ε to z = -ε. This does not tell us that ∂zV = 0 in the hole! Recall that in the Jackson charged disk problem, in the region outside the disk we really did have ∂zV = 0 in the outer "iris" region due to symmetry, but here we don't have that situation. Therefore, we don't have a simple "mixed" Dirichlet/Neumann problem in either of our cavity regions. In order to have such a straightforward mixed problem, we have to have ∂zV specified in the hole! So taking a hint from the "canonical paper", let's just assume some function in the hole and say - ∂zV(ρ,θ,0+) = - ∂zV(ρ,θ,0-) = Ez(ρ,θ) ρ < a I was going to call this function N for Neumann, but I realize it is just the electric field z component in the hole, so might as well use that name. There exists some function Ez(ρ,θ) for our "induced charge only" potential problem, but we don't know what that function is! We know that this function is the same on both sides of the hole, ie, at z = ±ε, because there is no charge density in the hole. It is the fact that this function Ez(ρ,θ) will NOT be zero in the hole which causes our Dirichlet potentials in the two cavities to be different! This despite that both cavities see the same prescribed potential on the iris part. ______________________________________________________________________________ Everywhere below this line uses the wrong Smythian form! I only found that out at the end when I tried to get continuity on the Eρ field. 2. Trying a two-region Smythian form, above and below z = 0. Now let's try out the following Smythian Form for the two cavity regions: V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e-kzAm(k) Jm(kρ) z ≥ 0 V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e+kzBm(k) Jm(kρ) z ≤ 0 This cannot be correct for the following reason. We know that the Dirichlet solution potential, like any potential, is continuous through the metal of the iris. The iris is not just a math iris anyway with a prescribed potential on it. If you approach this iris from either cavity, you get the same potential. But that is also true in the iris region. Therefore, using the above Smythe form, if we take the limit of each as z→ 0 we get the following facts, each valid for all ρ : V(0+,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk Am(k) Jm(kρ) z ≥ 0 V(0-z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk Bm(k) Jm(kρ) z ≤ 0 These would have to be equal. If we do the azimuthal PWA we get !Syntax Error, Idk Am(k) Jm(kρ) = !Syntax Error, Idk Bm(k) Jm(kρ) for all ρ We could then apply the Hankel Transform to conclude that Am(k) = Bm(k). This is then a contradiction because if we insert this into our Smythe form, the potential in the two cavities is symmetric, but we know that is not the case. This is telling us that something is wrong with our Smythe form, it cannot be valid for this problem. At this point, it is perhaps useful to think about a simpler problem so we can study what is going on here. Suppose we set c = 0 so that our point charge is on the z axis. In this case, we know that our solution is m=0 only, so we could then repeat the above analysis in this context. We then have for our prescribed iris potential V(ρ,θ,0) = - q/ ρ ≥ a and we would try to use this Smythe form V(z,ρ,θ) = 0 !Syntax Error, Idk e-kzA0(k) J0(kρ) z ≥ 0 V(z,ρ,θ) = 0 !Syntax Error, Idk e+kz B0(k) J0(kρ) z ≤ 0 Then our requirement that the potential be unique on the z = 0 plane tells us that !Syntax Error, Idk A0(k) J0(kρ) = !Syntax Error, Idk B0(k) J0(kρ) // for all ρ and then the Hankel forces (seems to force) A0(k) = B0(k). So we now have our "form problem" stated in a simpler situation, which is closer to Jackson's math. Comment: In this two-region form, things are "uniform" in ρ, that is, the forms are valid on all (0,∞) for ρ. This means you are allowed to use the orthogonality of the Jm(kρ) to conclude that Am(k) = Bm(k). 3. Trying a four-region Smythian form, above and below z = 0. So the big question now is this: what is it that is wrong with this "form"? It worked for Jackson in his symmetric case. Well, the form itself has this contradiction built in! The two cavities have a common boundary, and the potential is the same on this boundary for both cavities, and then the form forces both cavities to have the same potential everywhere. The form is just "insufficient" for the problem. One idea is to break the problem into two cylindrical regions as I have done elsewhere in my disk problem attempts. We might then have V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e-kzAm(k) Jm(kρ) z ≥ 0 ρ < a 1 V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e+kzBm(k) Jm(kρ) z ≤ 0 ρ < a 2 V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e-kz A'm(k) H(1)m(kρ) z ≥ 0 ρ > a 3 V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e+kz B'm(k) H(1)m(kρ) z ≤ 0 ρ > a 4 Then setting z = 0 we get these facts: !Syntax Error, Idk Am(k) Jm(kρ) = !Syntax Error, Idk Bm(k) Jm(kρ) ρ < a !Syntax Error, Idk A'm(k) H(1)m(kρ) = !Syntax Error, Idk B'm(k) H(1)m(kρ) ρ > a Now we have "broken up" the Hankel transform so it no longer forces Am(k) = Bm(k), and we have at least removed out inherent contradiction. This form may be no good, but at least it no longer has this particular problem. It also has the benefit (though this benefit was not needed in Jackson) of large-ρ decay. (a) The matching condition on V at ρ = a between our two cylindrical regions: Σm=0∞ cos(mθ) !Syntax Error, Idk e-kzAm(k) Jm(ka) = Σm=0∞ cos(mθ) !Syntax Error, Idk e-kz A'm(k) H(1)m(ka) and this must be valid for all z including z = 0 since V is continuous even at the iris hole edge. The θ PWA then tells us that !Syntax Error, Idk e-kzAm(k) Jm(ka) = !Syntax Error, Idk e-kz A'm(k) H(1)m(ka) !Syntax Error, Idk e-kz [Am(k) Jm(ka) - A'm(k) H(1)m(ka)] = 0 I have faced this kind of condition several times in the past, and I am always nervous because e-kz don't form an orthogonal set of functions of z. Still, the above has this form: !Syntax Error, Idk e-kz f(k) = 0 This says the Laplace transform of f(k) is 0. We can look at the transform stated on page 161 Schaum. In our case we are saying that f(s) = 0 in Schaum language. The inversion formula says F(t) = 0, there is just no other way! So despite orthogonality lacking, I feel pretty confident in saying this forces f(k) = 0, that is not just "an option". Therefore, our cylinder boundary match says that [Am(k) Jm(ka) - A'm(k) H(1)m(ka)] = 0 We could find a similar result for z ≤ 0 and our two results would then be Am(k) Jm(ka) = A'm(k) H(1)m(ka) Bm(k) Jm(ka) = B'm(k) H(1)m(ka) or Am(k)/ H(1)m(ka) = A'm(k) / Jm(ka) ≡ am(k) Bm(k)/ H(1)m(ka) = B'm(k) / Jm(ka) ≡ bm(k) We have a standard way of dealing with this, but I always have to write it all out every time, so here we go. First, restate the above, V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e-kzAm(k) Jm(kρ) z ≥ 0 ρ < a V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e+kzBm(k) Jm(kρ) z ≤ 0 ρ < a V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e-kz A'm(k) H(1)m(kρ) z ≥ 0 ρ > a V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e+kz B'm(k) H(1)m(kρ) z ≤ 0 ρ > a Then "process" each line by "doing nothing" V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e-kzAm(k) / H(1)m(ka) Jm(kρ) H(1)m(ka) z ≥ 0 ρ < a V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e+kzBm(k) / H(1)m(ka) Jm(kρ) H(1)m(ka) z ≤ 0 ρ < a V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e-kz A'm(k) / Jm(ka) H(1)m(kρ) Jm(ka) z ≥ 0 ρ > a V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e+kz B'm(k) / Jm(ka) H(1)m(kρ) Jm(ka) z ≤ 0 ρ > a So we then have this adjusted Smythian form which satisfies the ρ = a potential match condition. V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e-kz am(k) Jm(kρ) H(1)m(ka) z ≥ 0 ρ < a V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e+kz bm(k) Jm(kρ) H(1)m(ka) z ≤ 0 ρ < a V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e-kz am(k) H(1)m(kρ) Jm(ka) z ≥ 0 ρ > a V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e+kz bm(k) H(1)m(kρ) Jm(ka) z ≤ 0 ρ > a So now we have a candidate Smythian form with two unknown coefficients am(k) and bm(k). ****************** (b) The matching condition on ∂ρV at ρ = a between our two cylindrical regions: OUCH! Look now at the z ≥ 0 pair from above V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e-kz am(k) Jm(kρ) H(1)m(ka) z ≥ 0 ρ < a V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e-kz am(k) H(1)m(kρ) Jm(ka) z ≥ 0 ρ > a If we match the ∂θV azimuthal E field components, the match will be automatic. If we match the ∂zV azimuthal E field components, the match will be automatic. It is the ∂ρV radial E field component that must be considered here. ∂ρV(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk k e-kz am(k) Jm(kρ)' H(1)m(ka) z ≥ 0 ρ < a ∂ρ V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk k e-kz am(k) H(1)m(kρ)' Jm(ka) z ≥ 0 ρ > a ∂ρV(z,a,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk k e-kz am(k) Jm(ka)' H(1)m(ka) z ≥ 0 ρ < a ∂ρV(z,a,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk k e-kz am(k) H(1)m(ka)' Jm(ka) z ≥ 0 ρ > a These things need to be equal, so if we subtract them, we need to get 0. Thus we require that Σm=0∞ cos(mθ) !Syntax Error, Idk k e-kz am(k) W[Jm(ka), H(1)m(ka)] = 0 W[Jm(ka), H(1)m(ka)] = +2i (πka)-1 Bateman p 80 Σm=0∞ cos(mθ) !Syntax Error, Idk k e-kz am(k) 2i (πka)-1= 0 Σm=0∞ cos(mθ) !Syntax Error, Idk e-kz am(k) = 0 !Syntax Error, Idk e-kz am(k) = 0 am(k) = 0 since inverse Laplace of 0 is 0 Oops, we have another problem Houston. Our trial Smythian form cannot support continuity of the radial electric field on the matching cylinder!!! The only possible escape hatch is what happens at z = 0 which means we don't quite have a Laplace transform here. But I don't believe that escape hatch. I think this is a deficiency of the Smythian Form that I just failed to notice until just now. Therefore, all the rest of this doc has no relevance. ****************** 4. The following sections are just idle spinning of wheels: (a) The matching condition on V at z = 0 regions for both cylinder regions: NOW let's try again looking at z = 0 and see what happens. We do this for the first and then the second pair above and we get, after θ PWA, !Syntax Error, Idk am(k) Jm(kρ) H(1)m(ka) = !Syntax Error, Idk bm(k) Jm(kρ) H(1)m(ka) ρ < a !Syntax Error, Idk am(k) H(1)m(kρ) Jm(ka) = !Syntax Error, Idk bm(k) H(1)m(kρ) Jm(ka) ρ > a This is an interesting "crossed" set of relations that the solution coefficients must satisfy. This replaces the A and B form I had above. Write them as !Syntax Error, Idk [am(k) - bm(k)] Jm(kρ) H(1)m(ka) = 0 ρ < a !Syntax Error, Idk [am(k) - bm(k)] H(1)m(kρ) Jm(ka) = 0 ρ > a If we were to allow am(k) = bm(k), we get back to our symmetry result, so these have to be different! (b) Recap to this point. I am proposing the following Smythian form: V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e-kz am(k) Jm(kρ) H(1)m(ka) z ≥ 0 ρ < a V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e+kz bm(k) Jm(kρ) H(1)m(ka) z ≤ 0 ρ < a V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e-kz am(k) H(1)m(kρ) Jm(ka) z ≥ 0 ρ > a V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e+kz bm(k) H(1)m(kρ) Jm(ka) z ≤ 0 ρ > a This form is "manifestly" continuous at the ρ = a boundary. At the z = 0 plane, the forms have to agree with each other, and that gives these two conditions: !Syntax Error, Idk [am(k) - bm(k)] Jm(kρ) H(1)m(ka) = 0 ρ < a !Syntax Error, Idk [am(k) - bm(k)] H(1)m(kρ) Jm(ka) = 0 ρ > a Now we try to move forward some more. In our "Dirichlet" approach, we want the potential on the iris to be this: V(ρ,θ,0) = - q/ ρ > a But the Neumann component of our Dirichlet approach concerns what happens in the hole, and what happens there is that ∂zV is continuous. We wrote above that - ∂zV(ρ,θ,0+) = - ∂zV(ρ,θ,0-) = Ez(ρ,θ) ρ < a Our forms in the central cylinder are these V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e-kz am(k) Jm(kρ) H(1)m(ka) z ≥ 0 ρ < a V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e+kz bm(k) Jm(kρ) H(1)m(ka) z ≤ 0 ρ < a The derivatives are then ∂z V(z,ρ,θ) = - Σm=0∞ cos(mθ) !Syntax Error, Idk k e-kz am(k) Jm(kρ) H(1)m(ka) z ≥ 0 ρ < a ∂z V(z,ρ,θ) = + Σm=0∞ cos(mθ) !Syntax Error, Idk k e+kz bm(k) Jm(kρ) H(1)m(ka) z ≤ 0 ρ < a and at the z = 0 plane of the hole we have ∂z V(z,ρ,θ) = - Σm=0∞ cos(mθ) !Syntax Error, Idk k am(k) Jm(kρ) H(1)m(ka) ρ < a ∂z V(z,ρ,θ) = + Σm=0∞ cos(mθ) !Syntax Error, Idk k bm(k) Jm(kρ) H(1)m(ka) ρ < a These quantities must be equal in the hole, so we get - Σm=0∞ cos(mθ) !Syntax Error, Idk k am(k) Jm(kρ) H(1)m(ka) = Σm=0∞ cos(mθ) !Syntax Error, Idk k bm(k) Jm(kρ) H(1)m(ka) and we do θ PWA to get - !Syntax Error, Idk k am(k) Jm(kρ) H(1)m(ka) = !Syntax Error, Idk k bm(k) Jm(kρ) H(1)m(ka) ρ < a which we can write as !Syntax Error, Idk k [am(k) + bm(k)] Jm(kρ) H(1)m(ka) = 0 ρ < a which is another relation connecting the a and b. So we now have three relations: matching potentials at z = 0: !Syntax Error, Idk [am(k) - bm(k)] Jm(kρ) H(1)m(ka) = 0 ρ < a !Syntax Error, Idk [am(k) - bm(k)] H(1)m(kρ) Jm(ka) = 0 ρ > a matching Ez fields in the hole: !Syntax Error, Idk k [am(k) + bm(k)] Jm(kρ) H(1)m(ka) = 0 ρ < a If we add and subtract the first and third of these, we get some simpler results for ρ < a : !Syntax Error, Idk am(k) Jm(kρ) H(1)m(ka) = 0 ρ < a !Syntax Error, Idk bm(k) Jm(kρ) H(1)m(ka) = 0 ρ < a I am now "thirsty" for more conditions, nothing is seeming to drive the coefficients, we have just general conditions. (c) Idea #1: the radial and azimuthal E fields must be 0 on the iris. [ This is not true for our Dirichlet potentials! ] The relevant potential forms are these: V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e-kz am(k) H(1)m(kρ) Jm(ka) z ≥ 0 ρ > a V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e+kz bm(k) H(1)m(kρ) Jm(ka) z ≤ 0 ρ > a The azimuthal field will be proportional to Eθ = -1/ρ ∂θV. So we get -∂θ V(z,ρ,θ) = Σm=0∞m sin (mθ) !Syntax Error, Idk e-kz am(k) H(1)m(kρ) Jm(ka) = 0 z ≥ 0 ρ > a -∂θ V(z,ρ,θ) = Σm=0∞m sin (mθ) !Syntax Error, Idk e+kz bm(k) H(1)m(kρ) Jm(ka) = 0 z ≤ 0 ρ > a -∂θ V(0+,ρ,θ) = Σm=0∞m sin (mθ) !Syntax Error, Idk am(k) H(1)m(kρ) Jm(ka) = 0 ρ > a -∂θ V(0-,ρ,θ) = Σm=0∞m sin (mθ) !Syntax Error, Idk bm(k) H(1)m(kρ) Jm(ka)= 0 ρ > a I guess we do PWA to get what appear to be some "new facts" which we can add to our growing list. !Syntax Error, Idk am(k) H(1)m(kρ) Jm(ka) = 0 ρ > a !Syntax Error, Idk bm(k) H(1)m(kρ) Jm(ka) = 0 ρ > a As for the radial E field, we have Eρ = -∂ρV so we get ∂ρ V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk k e-kz am(k) H(1)m'(kρ) Jm(ka) z ≥ 0 ρ > a ∂ρ V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk k e+kz bm(k) H(1)m'(kρ) Jm(ka) z ≤ 0 ρ > a ∂ρ V(0+,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk k am(k) H(1)m'(kρ) Jm(ka) = 0 ρ > a ∂ρ V(0-,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk k bm(k) H(1)m'(kρ) Jm(ka) = 0 ρ > a and after the θ PWA we get yet two more "new facts" [ both are wrong ] !Syntax Error, Idk k am(k) H(1)m'(kρ) Jm(ka) = 0 ρ > a !Syntax Error, Idk k bm(k) H(1)m'(kρ) Jm(ka) = 0 ρ > a So we now have all tangential E fields set to 0 on the iris. (d) Idea #2: Something else we could do is match the other components of the E fields in the hole. So start with V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e-kz am(k) Jm(kρ) H(1)m(ka) z ≥ 0 ρ < a V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e+kz bm(k) Jm(kρ) H(1)m(ka) z ≤ 0 ρ < a -∂θ V(z,ρ,θ) = Σm=0∞ m sin(mθ) !Syntax Error, Idk e-kz am(k) Jm(kρ) H(1)m(ka) z ≥ 0 ρ < a -∂θ V(z,ρ,θ) = Σm=0∞ m sin(mθ) !Syntax Error, Idk e+kz bm(k) Jm(kρ) H(1)m(ka) z ≤ 0 ρ < a -∂θ V(0+,ρ,θ) = Σm=0∞ m sin(mθ) !Syntax Error, Idk am(k) Jm(kρ) H(1)m(ka) ρ < a -∂θ V(0-,ρ,θ) = Σm=0∞ m sin(mθ) !Syntax Error, Idk bm(k) Jm(kρ) H(1)m(ka) ρ < a Setting these equal and doing PWA gives !Syntax Error, Idk [am(k)- bm(k)] Jm(kρ) H(1)m(ka) = 0 ρ < a but we already have this condition so it is not new. As for radial, we have V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e-kz am(k) Jm(kρ) H(1)m(ka) z ≥ 0 ρ < a V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e+kz bm(k) Jm(kρ) H(1)m(ka) z ≤ 0 ρ < a ∂ρV(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk k e-kz am(k) Jm'(kρ) H(1)m(ka) z ≥ 0 ρ < a ∂ρV(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk k e+kz bm(k) Jm'(kρ) H(1)m(ka) z ≤ 0 ρ < a ∂ρV(0+,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk k am(k) Jm'(kρ) H(1)m(ka) ρ < a ∂ρV(0-,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk k bm(k) Jm'(kρ) H(1)m(ka) ρ < a and the PWA in θ then gives !Syntax Error, Idk k [am(k)- bm(k)] Jm'(kρ) H(1)m(ka) = 0 ρ < a So let's gather up all our conditions on the coefficients: matching potentials at z = 0: !Syntax Error, Idk [am(k) - bm(k)] Jm(kρ) H(1)m(ka) = 0 ρ < a !Syntax Error, Idk [am(k) - bm(k)] H(1)m(kρ) Jm(ka) = 0 ρ > a matching Ez fields in the hole: !Syntax Error, Idk k [am(k) + bm(k)] Jm(kρ) H(1)m(ka) = 0 ρ < a If we add and subtract the first and third of these, we get some simpler results for ρ < a : !Syntax Error, Idk am(k) Jm(kρ) H(1)m(ka) = 0 ρ < a !Syntax Error, Idk bm(k) Jm(kρ) H(1)m(ka) = 0 ρ < a azimuthal E fields vanish on both sides of the iris: !Syntax Error, Idk am(k) H(1)m(kρ) Jm(ka) = 0 ρ > a !Syntax Error, Idk bm(k) H(1)m(kρ) Jm(ka) = 0 ρ > a radial E fields vanish on both sides of the iris: !Syntax Error, Idk k am(k) H(1)m'(kρ) Jm(ka) = 0 ρ > a !Syntax Error, Idk k bm(k) H(1)m'(kρ) Jm(ka) = 0 ρ > a match radial E fields in the hole: !Syntax Error, Idk k [am(k)- bm(k)] Jm'(kρ) H(1)m(ka) = 0 ρ < a (e) Now rewrite just the ones that are independent conditions: First, the ρ > a conditions: matching potentials at z = 0 on the iris: !Syntax Error, Idk [am(k) - bm(k)] H(1)m(kρ) Jm(ka) = 0 ρ > a azimuthal E fields vanish on both sides of the iris: !Syntax Error, Idk am(k) H(1)m(kρ) Jm(ka) = 0 ρ > a !Syntax Error, Idk bm(k) H(1)m(kρ) Jm(ka) = 0 ρ > a radial E fields vanish on both sides of the iris: !Syntax Error, Idk k am(k) H(1)m'(kρ) Jm(ka) = 0 ρ > a !Syntax Error, Idk k bm(k) H(1)m'(kρ) Jm(ka) = 0 ρ > a Second, the ρ < a conditions: If we add and subtract the first and third of these, we get some simpler results for ρ < a : !Syntax Error, Idk am(k) Jm(kρ) H(1)m(ka) = 0 ρ < a !Syntax Error, Idk bm(k) Jm(kρ) H(1)m(ka) = 0 ρ < a match radial E fields in the hole: !Syntax Error, Idk k [am(k)- bm(k)] Jm'(kρ) H(1)m(ka) = 0 ρ < a If we now scan this list for a pair that would make a dual integral equation, we are in trouble because the ρ < a ones involve Jm(kρ) while the ρ > a ones involve H(1)m(kρ). Now I have left out perhaps a driving force which is this: we have that Dirichlet driving potential on the iris which works for both sides V(ρ,θ,0) = - q/ ρ > a So if we go back to our iris forms, V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e-kz am(k) H(1)m(kρ) Jm(ka) z ≥ 0 ρ > a V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e+kz bm(k) H(1)m(kρ) Jm(ka) z ≤ 0 ρ > a we then have Σm=0∞ cos(mθ) !Syntax Error, Idk am(k) H(1)m(kρ) Jm(ka) = - q/ ρ > a Σm=0∞ cos(mθ) !Syntax Error, Idk e+kz bm(k) H(1)m(kρ) Jm(ka) = q/ ρ > a Now we steal some results from Attempt #1, which were α = b2+ ρ2 β = 2cρ - q/ = (-q) (1/2) Σn=0∞ εnan(α,β) cos(nθ) = (-q/2) Σm=0∞ εmcm(α,β) cos(mθ) cm(α,β) = (2/π ) Σs=0m (-1)s (2m, 2m-2s) B[ s+1/2, m-s+1/2 ] F [s+1/2, 1/2; m+1; -2β/(α-β) ] Then we have for our two equations above Σm=0∞ cos(mθ) !Syntax Error, Idk am(k) H(1)m(kρ) Jm(ka) = (-q/2) Σm=0∞ εmcm(α,β) cos(mθ) ρ > a Σm=0∞ cos(mθ) !Syntax Error, Idk bm(k) H(1)m(kρ) Jm(ka) = (-q/2) Σm=0∞ εmcm(α,β) cos(mθ) ρ > a Doing the PWA then gives us these two conditions (neither of which is solvable on its own for coeff) !Syntax Error, Idk am(k) H(1)m(kρ) Jm(ka) = (-q/2) εmcm(α,β) ρ > a !Syntax Error, Idk bm(k) H(1)m(kρ) Jm(ka) = (-q/2) εmcm(α,β) ρ > a Again this is trying to make us think am(k) = bm(k) but we know that is wrong. We can subtract these to get !Syntax Error, Idk [am(k) - bm(k)] H(1)m(kρ) Jm(ka) = 0 ρ > a which is the first item already on our list, so not a new condition. So let's now restate our complete list of conditions adding the two above: (e) Again rewrite just the ones that are independent conditions: First, the ρ > a conditions: matching potentials at z = 0 on the iris: !Syntax Error, Idk [am(k) - bm(k)] H(1)m(kρ) Jm(ka) = 0 ρ > a azimuthal E fields vanish on both sides of the iris: !Syntax Error, Idk am(k) H(1)m(kρ) Jm(ka) = 0 ρ > a !Syntax Error, Idk bm(k) H(1)m(kρ) Jm(ka) = 0 ρ > a radial E fields vanish on both sides of the iris: !Syntax Error, Idk k am(k) H(1)m'(kρ) Jm(ka) = 0 ρ > a !Syntax Error, Idk k bm(k) H(1)m'(kρ) Jm(ka) = 0 ρ > a Dirichlet potential match at z = 0 due to negative of point charge potential: !Syntax Error, Idk am(k) H(1)m(kρ) Jm(ka) = (-q/2) εmcm(α,β) ρ > a !Syntax Error, Idk bm(k) H(1)m(kρ) Jm(ka) = (-q/2) εmcm(α,β) ρ > a Second, the ρ < a conditions: If we add and subtract the first and third of these, we get some simpler results for ρ < a : !Syntax Error, Idk am(k) Jm(kρ) H(1)m(ka) = 0 ρ < a !Syntax Error, Idk bm(k) Jm(kρ) H(1)m(ka) = 0 ρ < a match radial E fields in the hole: !Syntax Error, Idk k [am(k)- bm(k)] Jm'(kρ) H(1)m(ka) = 0 ρ < a Again, I we have the same problem regarding the formation of dual integral equations! Somehow the solution of this problem is a pair of functions am(k) and bm(k) which satisfy ALL of the above equations. The solution must exist, but I don't have any way to find it. (f) Are there other conditions at the cylinder math wall? (Contradiction originally found here) All I did so far was match the potential, but the three field components also should match away from z = 0. I have not incorporated that fact anywhere that I know of. Once again, the full Smythian form: V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e-kz am(k) Jm(kρ) H(1)m(ka) z ≥ 0 ρ < a V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e+kz bm(k) Jm(kρ) H(1)m(ka) z ≤ 0 ρ < a V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e-kz am(k) H(1)m(kρ) Jm(ka) z ≥ 0 ρ > a V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e+kz bm(k) H(1)m(kρ) Jm(ka) z ≤ 0 ρ > a Look now at the z ≥ 0 pair V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e-kz am(k) Jm(kρ) H(1)m(ka) z ≥ 0 ρ < a V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e-kz am(k) H(1)m(kρ) Jm(ka) z ≥ 0 ρ > a If we match the ∂θV azimuthal E field components, the match will be automatic. If we match the ∂zV azimuthal E field components, the match will be automatic. It is the ∂ρV radial E field component that must be considered here. ∂ρV(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk k e-kz am(k) Jm(kρ)' H(1)m(ka) z ≥ 0 ρ < a ∂ρ V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk k e-kz am(k) H(1)m(kρ)' Jm(ka) z ≥ 0 ρ > a ∂ρV(z,a,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk k e-kz am(k) Jm(ka)' H(1)m(ka) z ≥ 0 ρ < a ∂ρV(z,a,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk k e-kz am(k) H(1)m(ka)' Jm(ka) z ≥ 0 ρ > a These things need to be equal, so if we subtract them, we need to get 0. Thus we require that Σm=0∞ cos(mθ) !Syntax Error, Idk k e-kz am(k) W[Jm(ka), H(1)m(ka)] = 0 W[Jm(ka), H(1)m(ka)] = +2i (πka)-1 Bateman p 80 Σm=0∞ cos(mθ) !Syntax Error, Idk k e-kz am(k) 2i (πka)-1= 0 Σm=0∞ cos(mθ) !Syntax Error, Idk e-kz am(k) = 0 !Syntax Error, Idk e-kz am(k) = 0 am(k) = 0 since inverse Laplace of 0 is 0 Oops, we have another problem Houston. Our trial Smythian form cannot support continuity of the radial electric field on the matching cylinder!!! The only possible escape hatch is what happens at z = 0 which means we don't quite have a Laplace transform here. But I don't believe that escape hatch. I think this is a deficiency of the Smythian Form that I just failed to notice until just now. Let's do it one more time in the original coefficients: V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e-kzAm(k) Jm(kρ) z ≥ 0 ρ < a V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e-kz A'm(k) H(1)m(kρ) z ≥ 0 ρ > a The V match condition says that !Syntax Error, Idk e-kz[Am(k) Jm(ka) - A'm(k) H(1)m(ka)] = 0 The ∂ρV match condition says !Syntax Error, Idk e-kz[Am(k) k Jm'(ka) - A'm(k) k H(1)m'(ka)] = 0 Our Laplace transform argument then says Am(k) Jm(ka) = A'm(k) H(1)m(ka) Am(k) Jm'(ka) = A'm(k) H(1)m'(ka) k ≠ 0 If we divide these two equations, we get Jm(ka)/ Jm'(ka) = H(1)m(ka)/ H(1)m'(ka) which in turn says Jm(ka) H(1)m'(ka) = H(1)m(ka) Jm'(ka) which in turn says W[Jm(ka), H(1)m'(ka)] = 0 which is a contradiction. Conclusion: the Smythian form on which this entire Attempt #2 was based is invalid. I am not sad to see it go, I never liked it in the first place. Also, the Jackson disk did not use it. This then takes us back to the problem if trying to imagine what the valid Smythian form might be! I keep going back here V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e-kzAm(k) Jm(kρ) z ≥ 0 V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e+kzBm(k) Jm(kρ) z ≤ 0 Again, matching at z = 0 gives this logic sequence: V(0+,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk Am(k) Jm(kρ) V(0-z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk Bm(k) Jm(kρ) !Syntax Error, Idk Am(k) Jm(kρ) = !Syntax Error, Idk Bm(k) Jm(kρ) for all ρ Am(k) = Bm(k) => V is symmetric in z 5. Realization that V really is symmetric in z and why this is so. If we were to take the actual Green's problem, take note of the induced charge on the iris, then we replace the metal with charge sticky surface and freeze the charge on the iris, then we delete the point charge. We somehow know that the charge is not the same on the two sides of the iris and is surely greater on the side where the point charge was. On the other hand, for this induced-charge-only problem, there exists some potential everywhere on the z = 0 plane. So from a Dirichlet point of view, why should not this charge as Dirichlet BC result in the same potential on both sides of the iris? Each side is just a Laplace cavity, and the solutions should be mirror reflections. So this argues that the Dirichlet potential IS symmetric. But if it were symmetric, then the induced charge would be symmetric. I am missing some obvious factoid here. I think the Dirichlet potential IS symmetric. Here is the sequence that explains why: start with the full Green's situation, V = 0 on the iris, point charge present, unequal induced charges on the two sides of the iris. the potential everywhere can be thought of as the sum of the potentials of the point charge and these induced charges. now replace the charged metal with a non-conducting sticky charge surface which holds the charge in place without changing it. After doing this, there is no longer a distinction between the two sides of the sticky surface. The sticky charge σ is the sum of the two side-charges so σ = σ+ + σ-. It is a single layer of charge having no thickness. When we make this change, the potential everywhere stays the same because it comes from the exact same charge distribution. For example, we still have V = 0 on the iris. This is the step that requires thinking about. I use the word sticky only anticipating the next step. When we dissolve the metal and replace it with a math surface, we still have V = 0 on this math surface. I guess the term "sticky" does have an implication for this step. The surface does not let the charge diffuse laterally away from the surface. This role is played somehow by the metal's surface potential and we just imagine that our magic sticky surface (it is a non-conducting surface remember, made of vacuum) sticks the charges right where they were. We then delete the point charge. At this point, we no longer have V = 0 on the math sticky iris surface (we in fact have V = -q/R) , so the charge there would like to move tangentially on the surface, but the stickiness stops that motion as well. So this is the situation we have for our "induced charge only" problem. Our cavity is symmetric. The sticky charge is symmetric (left-right we are meaning here). The Dirichlet BC potential on the entire z=0 plane is "symmetric" in the sense that it looks the same from both sides. Therefore, the potential on both sides really is symmetric! So it is only when you add the point charge back in that the potential becomes asymmetric! The upshot of this argument is that we are now back to our very simple Smythian form for the induced-charge potential: V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e-|z|Am(k) Jm(kρ) This think is value everywhere in either cavity, and is symmetric ±z. So now I will go back and reexamine Attempt #1. Maybe it is OK!