Iris Green's by Dirichlet method Attempt #3

Iris Green's by Dirichlet method Attempt #3 PhL 4.25.10 Here I am assembling the "good parts" of Attempts #1 and #2. I need a better drawing. For overview, see meta review doc. Problem 5: Point charge and an Iris 1 1. Attempting a solution by my Dirichlet method 1 (a) Kinematics 2 (b) The Dirichlet Potential on the iris 2 (c) Charge density and Neumann part of the problem: statement of the Mixed BC's 3 (d) Smythian Form and statement of Mixed BC's in terms of same: the Dual Integral Equations 4 (e) Playing with the Dual Integral Equations 6 (f) The Charge Distribution on the Iris 7 (g) Review and Status of our Dirichlet/Neumann Problem 7 2. Discussion and looking for "leads" on how to do the iris problem 8 3. Plan A: Jackson's red book dual integral equations 10 4. Plan B: Let the "form" be for the total solution. 11 (a) Try a 2-region Smyth form with Am(k) and Bm(k), fitting now the total V. 11 (b) The Pillbox Abutment Theorem 12 (c) Point charge back in the hole, try a pillbox condition. 13 (d) Cast our dual integral equations in red Jackson form and solve for Am(k) 14 (e) Try a different approach to our dual integral equations (goes nowhere) 17 5. Plan C: Try expressing the potential of the point charge in terms of Bessel functions. 18 Toroidal Coordinates. 21 Now back to where we were: 23 6. Plan D: Repeat Plan C with other 1/R form. 23 This section is a continuation of what is in "Finding Green's Functions by the Dirichlet Method.doc" which I have stored in the Stakgold folder at the moment. There I deal with four problems which all come out OK, but this 5th one is much harder so we put it into a separate document and we will now have "attempts" in the usual manner. This was my "first cut" attempt to "summit" the iris mountain. Problem 5: Point charge and an Iris Can the Dirichlet method maybe solve this problem I have been long suffering with? 1. Attempting a solution by my Dirichlet method Comment: I think I need to move the Green's charge out of the hole and into one of the side cavities, so the problem can have a real 3D life. This will make it similar to the previous examples I have done. And then we can talk maybe about ∂V/∂z and the charge density on the iris and the lack of it in the hole. Then if we do all this, maybe we can take a limit of the point charge approaching the hole. I am mildly optimistic for some strange reason. (a) Kinematics Start over with Green's off the z = 0 plane: We now allow the points b,r,r' to have arbitrary z components, so all three points are lifted out of the z = 0 plane where they were above. Think of this as a top view with z axis out of paper and x axis to the right. Then we have the following coordinates, where no label means cylindrical: ( if d = 0 and c<a, point charge is in the hole) 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' r' = (x',y',z')Car = (ρ',θ',z') r'2 = ρ'2 + z'2 rr' = xx'+yy'+zz' x = ρcosθ y = ρsinθ rr' = ρ ρ'(cosθ cosθ' + sinθ sinθ') + zz' x' = ρ'cosθ' y' = ρ'sinθ' = ρ ρ' cos(θ-θ') + zz' What is the potential at point r' due to the point charge at b? It is V = +q/|b-r'| |b-r'|2 = b2 + r'2 - 2 br' = b2 + r'2 - 2(cx'+dz') = b2 + ρ'2 + z'2 - 2cρ'cosθ' - 2dz' |b-r|2 = b2 + r2 - 2 br = b2 + r2 - 2(cx+dz) = b2 + ρ2 + z2 - 2cρcosθ - 2dz I am unsure which coordinates to use, Cartesian or Cylindrical, so keeping both alive. In particular, then, for r on the z = 0 plane we have |b-r|2 = b2 + ρ2 - 2cρcosθ // notice c, not b, so not law of cosines (b) The Dirichlet Potential on the iris We know then that the potential at point r on the z = 0 plane due to our Green's charge is this: V(ρ,θ,0) = +q/ This is of course true on the iris and in the hole. For our Dirichlet problem, we will want V(ρ,θ,0) = - q/ // only on the iris, so ρ ≥ a only When we later add back the point charge potential, the total potential will be V = 0 on the iris, as desired. (c) Charge density and Neumann part of the problem: statement of the Mixed BC's Now, the point charge +q does not create any charge density in the hole. So if we assume in our Dirichlet problem that there is no such hole charge, then that fact will be true in the total solution. So let's add a Neumann condition to our "Dirichlet problem" saying no surface charge in the hole. The question now is this: how is no surface charge in the hole related to our Dirichlet problem potential V? Let's put a flat pillbox centered on the z = 0 plane. The sides are tiny, so contribute nothing to the EdA integral of Gauss's law, so only the round top and bottom of the pillbox make contributions. We have ∫EdA = 4πqenclosed in world where V = q/r EdA = -∂zV(z+ε)dA top surface EdA = -∂zV(z-ε)(-dA) bottom surface [ -∂zV(z+ε) + ∂zV(z-ε)] dA = 4πσdA ∂zV(z+ε)- ∂zV(z-ε) = -4πσ Here σ for the iris is the sum of the surface densities on both sides of the original metal iris, and is the total charge of our Dirichlet-problem sticky iris match surface. I have gone through all this detail just to verify what I know is true: the condition that there is σ = 0 in the hole is stated in terms of the potential in this way: ∂zV(z+iε) = ∂zV(z-iε), in other words, ∂zV is continuous as you move through the hole. Now that I realize (see below) that our Dirichlet V is symmetric in z, even when the Green's charge does not lie in the z = 0 plane, we can reach the more useful conclusion that is the same as that reached in Jackson's charged disk problem. Imagine that we were to plot V(z,ρ,θ) as a function of z in some small range above and below the hole, where ρ < a. This little plot would be symmetric in z. Moreover, we know that the plot does not have a kink at z = 0. The reason for that is that we know that ∂zV is continuous in the hole region, as just noted. It then follows that the plot must have zero slope at z = 0. Therefore, we know that ∂zV = 0 in the hole. We may then summarize our Neumann conclusions: ∂zV(z+ε)- ∂zV(z-ε) = -4πσ ρ > a ∂zV(z+ε) = ∂zV(z-ε) = 0 ρ < a Here then are the mixed Dirichlet/Neumann conditions of our Dirichlet potential problem: (1) V(ρ,θ,0) = - q/ ρ > a (2) ∂z V(ρ,θ,0) = 0 0 < ρ <a Remember that this V is "the potential of the induced charge on the iris" and has no singularities in the hole even if the point charge is moved there, because the point charge is not part of this Dirichlet problem. (d) Smythian Form and statement of Mixed BC's in terms of same: the Dual Integral Equations At this point, we need some kind of Smythian form, and that is going to be our next big problem. Consider first this shot in the dark, V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e-|k|zAm(k) Jm(kρ) Several comments. First, this is a valid atomic form. Second, even though point b is off the z=0 plane, we still have ±θ symmetry so cos(mθ) is still correct. Third, the Dirichlet and "Neumann" specifications are at z = 0 and this is associated with the non-oscillatory coordinate z. Fourth, the potential is symmetric in z, and this is what I (now) expect. See argument at the end of Attempt #2. The basic idea is that on the central boundary (the iris), for the Dirichlet potential V above we have a single value of V everywhere on the central plane, and there is only a single math layer of sticky induced charge σ = σ++ σ-, so each cavity sees the exact same BC so V must be symmetric. I first though this was wrong, which is why I started Attempt #2, but I now think it is right. So all these observations support the above form. We expect that V→0 for large ρ as well, but somehow that has to "come out in the wash" since we are oscillatory in ρ. [ we do know that Jm(kρ) → 1/ ] So here is the Dirichlet boundary condition part of our "Dirichlet problem": (1) Σm=0∞ cos(mθ) !Syntax Error, Idk Am(k) Jm(kρ) = - q/ ρ ≥ a Looking into the Neumann part, we see that ∂z e-k|z| = -k ∂z|z| = -k (±1) for z 0 This says then that ∂zV(ρ,θ,z) = ∓ Σm=0∞ cos(mθ) !Syntax Error, Idk k e-k|z|Am(k) Jm(kρ) and in particular ∂zV(ρ,θ,0±) = ∓ Σm=0∞ cos(mθ) !Syntax Error, Idk k Am(k) Jm(kρ) Therefore our Neumann condition is that this vanishes. So here are our boundary conditions: (1) Σm=0∞ cos(mθ) !Syntax Error, Idk Am(k) Jm(kρ) = - q/ ρ > a (2) Σm=0∞ cos(mθ) !Syntax Error, Idk k Am(k) Jm(kρ) = 0 ρ < a We can instantly simplify the second condition doing our usual θ PWA: (2)' !Syntax Error, Idk k Am(k) Jm(kρ) = 0 for all m ρ < a This is similar to Jackson's m=0 only result 3.172, but we have an iris not a disk, so our Neumann condition applies here for ρ < a, which is different from Jackson. Our next problem is to express RHS of (1) as a cosine series and then do our partial wave analysis. This is again the subject of a whole paper I wrote and the result was this: "Consider the Fourier Analysis of the function 1/ on (0,2π) , εn = 2-δn,0 Neumann factor, f(x) = 1/ = a0/2 + Σn=1∞ an cos(nx) = (1/2) Σn=0∞ εnancos(nx) // expansion I = πan = !Syntax Error, Idx cos(nx)/ // projection an = I/π = (2/π ) Σs=0n (-1)s (2n, 2n-2s) B[ s+1/2, n-s+1/2 ] F [s+1/2, 1/2; n+1; -2b/(a-b) ] ao= 4/[(a-b)] K(k)/π a1= [4/ ] K(k)/π k2 = -2b/(a-b) a2 = [4/ ] [(a/b) K(k) + (1 - (a/b) E(k) ] /π So in the RHS of (1) we say that ( change to α and β to avoid confusion on two meanings for b) α = b2+ ρ2 β = 2cρ - q/ = (-q) (1/2) Σn=0∞ εnan(α,β) cos(nθ) = (-q/2) Σn=0∞ εnan(α,β) cos(nθ) am(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) ] So our two Dirichlet/Neumann conditions are now these: (1) Σm=0∞ cos(mθ) !Syntax Error, Idk Am(k) Jm(kρ) = (-q/2) Σm=0∞ εmam(α,β) cos(mθ) ρ ≥ a (2) Σm=0∞ cos(mθ) !Syntax Error, Idk k Am(k) Jm(kρ) = 0 0 < ρ < a We now do our partial wave analysis and write these as (1)' !Syntax Error, Idk Am(k) Jm(kρ) = (-q/2) εmam(α,β) ρ > a α = b2+ ρ2 β = 2cρ (2)' !Syntax Error, Idk k Am(k) Jm(kρ) = 0 0 < ρ < a This is the closest I have ever been to having a dual integral equation for the iris Green's problem! The difference from Jackson p 91 (3.173) is that the RHS of (1) is a function of ρ so is not a constant, as it was in his application where constant was called V. Another difference is that we have the =0 equation applying inside the hole, whereas Jackson had it applying outside the disk. (e) Playing with the Dual Integral Equations Let's now look more closely at am(α,β). α-β = b2 + ρ2 - 2cρ -2β/(α-β) = -4cp/( b2 + ρ2 - 2cρ) So let's make this definition Gm(ρ) ≡ (-q/2) εm am(α,β) = some known but messy function of ρ, b, and c (and q and m) Our dual integral equations are then these: (1) !Syntax Error, Idk Am(k) Jm(kρ) = Gm(ρ) ρ ≥ a (2) !Syntax Error, Idk k Am(k) Jm(kρ) = 0 0 < ρ < a Our goal is now to make these equations somehow fit into one of the molds Bateman shows on page 76. The first step is to define x = ρ/a so make the ranges look right, and to reverse the order of the equations: (2) !Syntax Error, Idk k Am(k) Jm(kax) = 0 0 < x < 1 (1) !Syntax Error, Idk Am(k) Jm(kax) = Gm(ax) x > 1 Next, let ka = y so we have (2) !Syntax Error, Idy/a y/a Am(y/a) Jm(xy) = 0 0 < x < 1 (1) !Syntax Error, Idy/a Am(y/a) Jm(xy) = Gm(ax) x > 1 (2) !Syntax Error, Idy y Am(y/a) Jm(xy) = 0 0 < x < 1 (1) !Syntax Error, Idy Am(y/a) Jm(xy) = a Gm(ax) x > 1 At this point, in Attempt #1 I note that due to the hole/iris reversal from the disk geometry, this does not fit into the relatively simple mold which Bateman gives at the top of page 76. The problem is that our ranges are reversed! In Attempt #1, I therefore go through the more complicated Tranter dual integral equation pair, and show how it is at least a "prescription" for getting a solution. In practice, even the simple dual integral equation form would be hard to use just because our expression am(α,β) is so messy. I won't repeat the Tranter stuff here, see Attempt #1. So it seems that it is pretty messy to come up with the actual Am(k) coefficients. (f) The Charge Distribution on the Iris The charge distribution on our Dirichlet iris sticky surface is given by this: ∂zV(z+ε)- ∂zV(z-ε) = -4πσ and we know from above that ∂zV(ρ,θ,0±) = ∓ Σm=0∞ cos(mθ) !Syntax Error, Idk k Am(k) Jm(kρ) which tells us then that σ(ρ,θ) = (1/2π) Σm=0∞ cos(mθ) !Syntax Error, Idk k Am(k) Jm(kρ) This equation is true for all ρ, but we know that it must give 0 for ρ < a. Smythe claims this result, for when d = 0 : σ(θ,ρ) = - (q/2π) / [ ( ρ2 + b2 - 2bρ cosθ) ] I don't (yet) know how I might get this result. (g) Review and Status of our Dirichlet/Neumann Problem Here is what we know: V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e-|k|zAm(k) Jm(kρ) (1) Σm=0∞ cos(mθ) !Syntax Error, Idk Am(k) Jm(kρ) = - q/ ρ > a (2) Σm=0∞ cos(mθ) !Syntax Error, Idk k Am(k) Jm(kρ) = 0 ρ < a In the hole region, there is some value of the potential we might call f(ρ,θ) . We don't know what it is (yet), but we can give it this name. Then we would say (3) Σm=0∞ cos(mθ) !Syntax Error, Idk Am(k) Jm(kρ) = f(ρ,θ) ρ < a One thing we do know about f(ρ,θ) is this: f(a,θ) = - q/ since it has to match at the ρ = a perimeter. 2. Discussion and looking for "leads" on how to do the iris problem My problem is that I don't yet know enough to be able to solve this problem. I think I am in the right coordinate system (finally) (maybe not!). The naive green Jackson-like dual integral equation approach does not apply to this problem, I end up with a Tranter style dual integral equation with very horrible factors in it and it certainly would be next to impossible to do. I also have the oblate coordinates solution, but it too is quite messy. I am still motivated by Smythe's simple answer. (1) I need to see some problems like this "worked through" by someone who knows what s/he is doing. The one example which comes to mind is the spherical bowl problem which is done in my canonical PDF thing. That problem is related to the iris one by inversion certainly. (2) Another tack is to think about Stakgold's "other methods" for finding Green's functions. In particular, I keep wondering about the partial or full eigenfunction expansion. I have a PDF which claims to state the Green's function for a disk. (3) You would think there would exist an electrostatics handbook somewhere that gives solutions to all the simple geometry problems, but I have never seen such a thing. I wonder what exists in the Marriot non-arc stacks? (4) I see that there is a Journal of Electrostatics. I cannot get to the articles however with Jim. Marriott has some Gupta, K.C books but not the one mentioned in my article. (5) Maybe I should look at Jackson's hole in plate discussion, but I already did this in Smythe. Smythe used oblates, Jackson probably does not. // Aha! Jackson somehow produces the dual integral equations "the other way" which does not appear in Bateman. And he states the solution. And he gives a reference to Sneddon SNEDDON, l. N., Mixed Boundary Value Problems in Potential Theory (North-Holland Publishing Company—Amsterdam, 1966), viii+282 pp., 80s. The topic discussed in this book has flowered in the last few years. Of the 170 references given only 52 are dated earlier than 1950 and most of these are to standard general mathematical works or classical work of the nineteenth century. The author puts the start of the recent work down to Titchmarsh's study of dual integral equations in 1937. This may be so but the continuing work is largely due to the author, and his students and collaborators. There are those who say that with the advent of high speed computers the analytical work of the book is no longer necessary; this is often untrue as anybody who actually tries to solve some of the problems of the book directly by means of a computer will soon find out. The book discusses how a problem is put into a form which can be solved by a computer, usually as a Fredholm integral equation of the second kind. Much recent work has been done in potential theory alone, as this book shows, though there is a good deal which has a wider application than that given. Much remains to be done, however, even in potential theory. For instance I have yet to see a solution to the simple-seeming problem of the field due to a torus of elliptic cross section charged to a constant potential. In this book an operational method is exploited to great effect. The author is largely responsible for the development of this method though he is too modest to say so. By its judicious use much tedious analysis has only to be done once—a great advance over the laborious work so often repeated in slightly disguised forms by the earlier workers. The analysis has to be done once, however, and here it is done in a valuable chapter in the book called Mathematical Preliminaries, which comes in after an introductory descriptive chapter. The author says that most of it will be familiar but indeed there may be much that is not entirely so, and it is most useful to have all this collected together in a space of 37 pages. After a chapter on the classical problem of the electrified disc there is one on dual integral equations, one on dual series equations and one on triple integral equations and series. Next comes a chapter on integral representations and a final chapter on applications to potential problems, where certain numerical results and approximate methods are discussed. This is an eminently practical book, well designed to meet the needs of the people for whom it is written, namely students of applied mathematics, physics and engineering. The reader is given every help and nowhere is a proof " left to the reader ". As the topic is not an easy one this will be greatly appreciated by the workers in many fields who will no doubt use the book. j. c. COOKE I now see that "mixed boundary value problems" is a good web search handle. (6) Here is a little clip from a Google book: So here is some 1986 work on this subject. I have lots of leads now. 3. Plan A: Jackson's red book dual integral equations Let's now back up to where we were above with our iris problem. I had things at this point: (1) !Syntax Error, Idy Am(y/a) Jm(xy) = a Gm(ax) x > 1 (2) !Syntax Error, Idy y Am(y/a) Jm(xy) = 0 0 < x < 1 Gm(ρ) ≡ (-q/2) εm am(α,β) α = b2+ ρ2 β = 2cρ am(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) ] Here is what red Jackson has to say: But this, although different from green Jackson page 91 bottom of page (it differs only in the location of the y) still does not match what I need. I need the y in the other equation. Red Jackson's "form" is for the total of the externally applied influence plus that charge on the iris, so he gets a 0 in the non-k equation. But this is not the "Dirichlet" approach I have been taking. So let's try V = total instead of Dirichlet. 4. Plan B: Let the "form" be for the total solution. (a) Try a 2-region Smyth form with Am(k) and Bm(k), fitting now the total V. Suppose I give up on my "Dirichlet" approach and just try a total solution of this form for the iris including the point charge: (not in the z plane) V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e-|k|zAm(k) Jm(kρ) But this is no good because now we KNOW things won't be symmetric, so we would have to try this: 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 Our V = 0 conditions would then be this: Σm=0∞ cos(mθ) !Syntax Error, Idk Am(k) Jm(kρ) = 0 ρ > a Σm=0∞ cos(mθ) !Syntax Error, Idk Bm(k) Jm(kρ) = 0 ρ > a which does say that !Syntax Error, Idk Am(k) Jm(kρ) = 0 ρ > a !Syntax Error, Idk Bm(k) Jm(kρ) = 0 ρ > a But now the charge situation would be ∂zV(z+ε) – ∂zV(z-ε) = 0 ρ < a where now we have ∂zV(z+ε) = – Σm=0∞ cos(mθ) !Syntax Error, Idk k Am(k) Jm(kρ) ∂zV(z-ε) = + Σm=0∞ cos(mθ) !Syntax Error, Idk k Bm(k) Jm(kρ) ∂zV(z+ε) – ∂zV(z-ε) = – Σm=0∞ cos(mθ) !Syntax Error, Idk k [ Am(k) + Bm(k)] Jm(kρ) = 0 So OK, we can then write: Σm=0∞ cos(mθ) !Syntax Error, Idk [ Am(k) + Bm(k)] Jm(kρ) = 0 V = 0 on iris, both sides ρ > a Σm=0∞ cos(mθ) !Syntax Error, Idk k [ Am(k) + Bm(k)] Jm(kρ) = 0 σ = 0 in the hole ρ < a which certainly seems to say !Syntax Error, Idk [ Am(k) + Bm(k)] Jm(kρ) = 0 V = 0 on iris, both sides ρ > a !Syntax Error, Idk k [ Am(k) + Bm(k)] Jm(kρ) = 0 σ = 0 in the hole ρ < a Here we have RHS = 0 for both our integral equations. This is something I have never seen, and suspect it may imply that [ Am(k) + Bm(k)] = 0. Suppose we interpret this as saying Am(k) + Bm(k) = 0 so that Bm(k) = - Am(k). [ But we know this is wrong because it would claim that V was anti-symmetric in z which is obviously not true, just think of going close the point charge which is only on one side. ] Then we would have σ = 0 in the hole for sure, and we have as our only equation this one: Σm=0∞ cos(mθ) !Syntax Error, Idk Am(k) Jm(kρ) = 0 ρ > a I think this is insufficient to give a solution. We would need to know Am(k) in the hole. So maybe our form is too simple for this Green's function? (I think that is the case, see pillbox comment below). My conclusion here is that my 2-region Smythe fit is very probably no good. (b) The Pillbox Abutment Theorem What would our pillbox analysis have to say around the point charge at z = d and x = c and y = 0? Let's put the charge on the z > 0 side with d > 0. Maybe think of the charge as flat in the z = d plane. But since V has the same "form" on both sides of the point charge in this case, you get ∂zV1 - ∂zV2 ≡ 0. The pillbox idea requires potentials in the two charge-abutting regions to have a different form. (c) Point charge back in the hole, try a pillbox condition. OK, let's put the point charge back in the hole, and go back to our simpler symmetric form (but this is now trying to model the total solution V, not just the Dirichlet V) V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e-|k|zAm(k) Jm(kρ) so we would say something like this for our pillbox condition ( now in the z = 0 plane) q1, q2, q3 = z, ρ, φ h1 = 1 h2 = 1 h3 = ρ ∂ζVi - ∂ζVo = (q/ε) (h2/h3h1) δ(φ-φ0) δ(ξ - ξ0) q1, q2, q3 = ξ, ζ, φ ∂zV2 - ∂zV3 = (q/c) δ(θ) δ(ρ-c) setting electrostatic ε = 1 now Where V2 is z = +ε and V3 is z = - ε (don't worry about sign for now) ∂zV(z+ε) – ∂zV(z-ε) = – 2 Σm=0∞ cos(mθ) !Syntax Error, Idk k Am(k) Jm(kρ) = (q/c) δ(θ) δ(ρ-c) This is only valid in ρ < a, the hole. On the iris we have charges so this would be wrong. So we have: Σm=0∞ cos(mθ) !Syntax Error, Idk k Am(k) Jm(kρ) = -(q/2c) δ(θ) δ(ρ-c) ρ < a // c < a of course Σm=0∞ cos(mθ) !Syntax Error, Idk Am(k) Jm(kρ) = 0 ρ > a // V = 0 on iris Well, OK, let's go with this famous result from our new "transforms" doc, Σm=0∞ εm cos(mθ) = 2π δ(θ) Then our first equation above becomes this: – 2 Σm=0∞ cos(mθ) !Syntax Error, Idk k Am(k) Jm(kρ) = (q/2πc) δ(ρ-c) Σm=0∞ εm cos(mθ) Hopefully we can do our PWA and write this as: !Syntax Error, Idk k Am(k) Jm(kρ) = -(εmq/4πc) δ(ρ-c) ρ < a c < a Now we have this for our "dual integral equation" which somehow rings possibly valid to me: !Syntax Error, Idk k Am(k) Jm(kρ) = -( εmq /4πc) δ(ρ-c) ρ < a c < a !Syntax Error, Idk Am(k) Jm(kρ) = 0 ρ > a // V = 0 on iris So this is a dual integral equation, but the driving non-zero RHS is a distribution, something I have never seen before. [ But I have seen distributions all the time in pillbox conditions which I was able to invert to find coefficients, cases where we had double orthogonality. ] (d) Cast our dual integral equations in red Jackson form and solve for Am(k) Let's compare our dual integral equations to our red Jackson data Can I write δ(ρ-c) in powers or ρ? Well, how about this: !Syntax Error, Idk e-ik(ρ-c) = 2π δ(ρ-c) // completeness = !Syntax Error, Idk eikc e-ikρ = !Syntax Error, Idk eikc [ Σj=0∞ (-ikρ)j/ j! ] = !Syntax Error, Idk eikc [ Σj=0∞ρj (-ik)j/j ! ] = Σj=0∞ρj!Syntax Error, Idk eikc(-ik)j/j ! = Σj=0∞ρj Cj Cj = (-i)j/j! * !Syntax Error, Ids eisc sj = (-i)j/j! Ej which let's pretend is valid in some sense. Then we have !Syntax Error, Idk k Am(k) Jm(kρ) = Σj=0∞ Dj ρj ρ < a c < a !Syntax Error, Idk Am(k) Jm(kρ) = 0 ρ > a // V = 0 on iris where Dj = -(εmq /4πc) (1/2π) Cj = -(εmq /8π2c) Cj Then the idea would be to assume (if it works, assumption must be OK) Am(k) = Σj=0∞ Dj Amj(k) which would then give !Syntax Error, Idk k Amj(k) Jm(kρ) = ρj ρ < a !Syntax Error, Idk Amj(k) Jm(kρ) = 0 ρ > a and now amazingly we have it in red Jackson form. So let's do the scaling by defining x = ρ/a to give !Syntax Error, Idk k Amj(k) Jm(kax) = ajxj x < 1 !Syntax Error, Idk Amj(k) Jm(kax) = 0 x > 1 Now define y = ak so k = y/a and we get !Syntax Error, Idy/a y/a Amj(y/a) Jm(yx) = ajxj x < 1 !Syntax Error, Idy/a Amj(y/a) Jm(yx) = 0 x > 1 !Syntax Error, Idy y a-j-2Amj(y/a) Jm(yx) = xj x < 1 !Syntax Error, Idy a-j-2 Amj(y/a) Jm(yx) = 0 x > 1 Now define g(y) = a-j-2Amj(y/a) => Amj(z) = aj+2 g(az) and we end up with !Syntax Error, Idy y g(y) Jm(yx) = xj x < 1 !Syntax Error, Idy g(y) Jm(yx) = 0 x > 1 Compare to red Jackson So the solution should then be so I then get g(y) = j! / Γ(j+3/2) * (2y)-1/2 Jj+3/2(y) Amj(z) = aj+2 g(az) = aj+2 j! / Γ(j+3/2) * (2az)-1/2 Jj+3/2(az) Amj(k) = aj+2 g(ak) = aj+2 j! / Γ(j+3/2) * (2ak)-1/2 Jj+3/2(ak) Am(k) = Σj=0∞ Dj Amj(k) = Σj=0∞ -(εmq /8π2c) (-i)j/j! * Ej j! / Γ(j+3/2) * (2ak)-1/2 Jj+3/2(ak) = -(εmq /8π2c) (2ak)-1/2 Σj=0∞ (-i)j * Ej / Γ(j+3/2) * Jj+3/2(ak) Even if total trash, at least this has the look of something that is manageable. This is much better than that impossible Tranter form I got earlier. Let this simmer for a while, and recall our form V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e-|k|zAm(k) Jm(kρ) which says that the charge on the iris is this: ∂zV(z+ε)- ∂zV(z-ε) = -4πσ ρ > a ∂zV(z+ε) – ∂zV(z-ε) = – 2 Σm=0∞ cos(mθ) !Syntax Error, Idk k Am(k) Jm(kρ) = (q/c) δ(θ) δ(ρ-c) so we have [ this result has been stated many times in these Attempt docs ] -4πσ = – 2 Σm=0∞ cos(mθ) !Syntax Error, Idk k Am(k) Jm(kρ) Let's bake this overnight and pick it up tomorrow. It seems too simple, so probably I have done something stupid as per usual. ( Next Day 4.28.10. ) Try to do the Ej integral: Ej = !Syntax Error, Ids eisc sj j = 0,1,2,3.... You could think of this as expo Fourier of a power, so we might consult Bateman ET I. Well, suppose f(x) = 1, then g(y) = 2πδ(y). Then EFT(xn) = (i)n 2π δ(n)(y) which comes just from parts integration. so we then have Ej = (i)j 2π δ(j)(c) But c is the location of our point charge and we have c > 0, so we get Ej = 0 and the whole thing collapses. Too bad. Of course I knew it would collapse, I am just shooting in the dark here. That is to say, if I insert this into our Am(k) we get Am(k) = -(εmq /8π2c) (2ak)-1/2 Σj=0∞ (-i)j * (i)j 2π δ(j)(c) / Γ(j+3/2) * Jj+3/2(ak) = -(εmq /4πc) (2ak)-1/2 Σj=0∞ δ(j)(c) / Γ(j+3/2) * Jj+3/2(ak) This says that somehow Am(k) exists only at c = 0, the location of the point charge. (e) Try a different approach to our dual integral equations (goes nowhere) Go back to our "dual integral equation" !Syntax Error, Idk k Am(k) Jm(kρ) = -( εmq /4πc) δ(ρ-c) ρ < a c < a !Syntax Error, Idk Am(k) Jm(kρ) = 0 ρ > a // V = 0 on iris If the first is really valid, we should be able to integrate it from ρ = 0 to ρ = a, and c is in this range, so !Syntax Error, Idk k Am(k) !Syntax Error, Idρ Jm(kρ) = -( εmq /4πc) Can I do this simple looking integral? GR7 p 629 says so we can say x = kρ and !Syntax Error, Idρ Jm(kρ) = !Syntax Error, Idx/k Jm(x) = (1/k) !Syntax Error, Idx Jm(x) = (2/k) Σr=0∞ Jm+2r+1(x)|ka0 We know that Jm(x=0) = δm0 so this says !Syntax Error, Idρ Jm(kρ) = (2/k) Σr=0∞ Jm+2r+1(ka) If m = 0,1,2.... then m+2r+1 = 1,3,5... or 2,4,,6... and so on , but we never have m+2r+1 = 0, so evaluation at x = 0 never produces anything. so we now have: !Syntax Error, Idk k Am(k) !Syntax Error, Idρ Jm(kρ) = -( εmq /4πc) !Syntax Error, Idk k Am(k) (2/k) Σr=0∞ Jm+2r+1(ka) = -( εmq /4πc) Σr=0∞!Syntax Error, Idk Am(k) Jm+2r+1(ka) = -( εmq /8πc) where now a is just a parameter. I don't see any way to use orthogonality to isolate Am(k). I have just recast things into a worse form. So my Plan B has not quite panned out! continue here 5. Plan C: Try expressing the potential of the point charge in terms of Bessel functions. The toroidal Q function mysteriously appears here! I recall how Smythe did this for oblate spheroidals. Jackson green talks about this page 84. He uses the partial eigenfunction method of Stakgold to attack this problem. His result is on page 86 and involves I and K functions. In our application, we could take x' = the point charge at θ = 0 so we would have 1/R = (2/π) Σm !Syntax Error, Idk e-imθ cos[k(z-d)] Im(kρ<) Km(kρ>) ρ> = max (c,ρ) ρ< = min (c,ρ) Here R = |b-r| is the distance from some arbitrary space point r = x = (ρ,θ,z) to Green's point b = x' = (c,0,d). Earlier we had (just for example) |b-r|2 = b2 + r2 - 2 br = b2 + r2 - 2(cx+dz) = b2 + ρ2 + z2 - 2cρcosθ - 2dz |b-r|2 = b2 + ρ2 - 2cρcosθ // if r on the z=0 plane Now if we model our Dirichlet/Neumann part as we have been doing above, we could say Vtot = Σm=0∞ cos(mθ) !Syntax Error, Idk e-|k|zAm(k) Jm(kρ) + (2q/π) Σm !Syntax Error, Idk e-imθ cos[k(z-d)] Im(kρ<) Km(kρ>) where we have an arbitrary location now x' = (c,0,d) for the Green's charge. Let's switch to Jackson's "real form" in 3.149 which I can write like this: 1/R = (2/π) Σm=0∞ εm cos(mθ) !Syntax Error, Idk cos[k(z-d)] Im(kρ<) Km(kρ>) Then we seem to have Vtot = Σm=0∞ cos(mθ) !Syntax Error, Idk e-|k|zAm(k) Jm(kρ) + (2q/π) Σm=0∞ εm cos(mθ) !Syntax Error, Idk cos[k(z-d)] Im(kρ<) Km(kρ>) or ρ> = max (c,ρ) ρ< = min (c,ρ) Vtot = Σm=0∞ cos(mθ) !Syntax Error, Idk { e-|k|zAm(k) Jm(kρ) + (2qεm/π) cos[k(z-d)] Im(kρ<) Km(kρ>) } The benefit of this approach is that the point charge is added in a "reasonable fashion". We can then write the condition that V = 0 on the iris as follows: ( this means ρ > a, and I assume c < a) 0 = Σm=0∞ cos(mθ) !Syntax Error, Idk { Am(k) Jm(kρ) + (2qεm/π) cos[kd] Im(kc) Km(kρ) } and then our PWA tells us that !Syntax Error, Idk { Am(k) Jm(kρ) + (2qεm/π) cos[kd] Im(kc) Km(kρ) } = 0 ρ > a !Syntax Error, Idk Am(k) Jm(kρ) = - (2qεm/π) !Syntax Error, Idk cos[kd] Im(kc) Km(kρ) ρ > a where we might be able to do this integral somehow. Meanwhile, our σ condition will be developed this way, Vtot = Σm=0∞ cos(mθ) !Syntax Error, Idk { e-|k|zAm(k) Jm(kρ) + (2qεm/π) cos[k(z-d)] Im(kρ<) Km(kρ>) } ∂zV(+) = Σm=0∞ cos(mθ) !Syntax Error, Idk k { – Am(k) Jm(kρ) + (2qεm/π) sin[kd)] Im(kρ<) Km(kρ>) } ∂zV(- ) = Σm=0∞ cos(mθ) !Syntax Error, Idk k { + Am(k) Jm(kρ) + (2qεm/π) sin[kd)] Im(kρ<) Km(kρ>) } The difference of these is 0 in the hole region ρ < a. But in the difference, the second term cancels. I guess the reason is that cos[k(z-d)] is the same for z = ±0 so it's derivative is also the same and we get cancelation. We then have our usual Neumann portion claim: Σm=0∞ cos(mθ) !Syntax Error, Idk k Am(k) Jm(kρ) = 0 ρ < a !Syntax Error, Idk k Am(k) Jm(kρ) = 0 ρ < a So here then are our dual integral equations: (1) !Syntax Error, Idk Am(k) Jm(kρ) = - (2qεm/π) !Syntax Error, Idk cos[kd] Im(kc) Km(kρ) ρ > a (2) !Syntax Error, Idk k Am(k) Jm(kρ) = 0 ρ < a So for the moment (a short one) we at least "look viable". We can interpret the first equation as the Dirichlet condition on our "Dirichlet" form for the potential, where the RHS is just the point charge potential evaluated on the iris, but written in cylindrical functions (the new feature). The second is the same as we had before, it is the Neumann part of our problem. We can compare the above to our earlier results, (1)' !Syntax Error, Idk Am(k) Jm(kρ) = (-q/2) εmam(α,β) ρ > a α = b2+ ρ2, β = 2cρ, b2 = c2+d2 (2)' !Syntax Error, Idk k Am(k) Jm(kρ) = 0 0 < ρ < a It certainly seems likely that am(c,d,ρ) ~ !Syntax Error, Idk cos[kd] Im(kc) Km(kρ). So let's look at this integral? Go to 717 of GR7 to see Bessel + trig. On page 719 I find this integral and perhaps this is the tie-in to oblates (no, toroidals!) ? Let's make these settings: x = k a = ρ b = c c = d ν = m Then !Syntax Error, Idk cos[kd] Im(kc) Km(kρ) = (1/2) 1/ Qm-1/2 [(ρ2+ c2+ d2)/(2cρ)] Re(ρ) > |Re(c)| d>0 Re(m) > -1/2 So the only constraint here is ρ > c so we are outside the Green's charge. We now have: So since a > c, we are OK on that condition just mentioned. So this is just an alternative to our earlier statement of the conditions which was this: (1)' !Syntax Error, Idk Am(k) Jm(kρ) = - (q εm /π) (π/2)am(α,β) ρ > a α = b2+ ρ2 β = 2cρ (2)' !Syntax Error, Idk k Am(k) Jm(kρ) = 0 0 < ρ < a So this says we have found a better way to express our complicated am(α,β) am(α,β) = (2/π)(1/) Qm-1/2 [(ρ2+ c2+ d2)/(2cρ)] α = b2+ ρ2 β = 2cρ - q/ = (-q) (1/2) Σn=0∞ εnan(α,β) cos(nθ) = (-q/2) Σn=0∞ εnan(α,β) cos(nθ) am(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) ] This is the first time ever that I have encountered Qm-1/2 type functions! Toroidal Coordinates. As discussed Bateman page 173, these are the "ring or toroidal functions" which arise when you use toroidal coordinates, something I have never done! That is a whole new world I have never visited. The following fact must be true: (1) Σm=0∞ cos(mθ) !Syntax Error, Idk Am(k) Jm(kρ) = - q/ ρ ≥ a Σm=0∞ cos(mθ) (-qεm/π) 1/ Qm-1/2 [(ρ2+ c2+ d2)/(2cρ)] = - q/ ρ>a Σm=0∞ cos(mθ) (εm/π) 1/ Qm-1/2 [(ρ2+ c2+ d2)/(2cρ)] = 1 / ρ>a Smythe mentions toroidals only briefly, but does present a problem involving them. MF discuss toroidals here: I, 689 table entry for toroidals all other hits in this volume are index II, 1301 discussion of toroidals, the P and Q functions of half integer m-1/2 offsets page 1303 gives the atomic form for toroidals Angle φ is the normal azimuth. Coordinates μ and η are the τ and σ of the wiki page. You start with those Steiner Apollonius circle guys, where the z axis is of course up. As you rotate, the blue circles become toroids (size related to τ = μ) , the red circles become spheres (radius related to σ = η). I know this is the right system to treat the circular wire problem, maybe it is also appropriate for disks and irises, I have not seen that mentioned anywhere. Now back to where we were: (1) !Syntax Error, Idk Am(k) Jm(kρ) = - (qεm/π) 1/ Qm-1/2 [(ρ2+ c2+ d2)/(2cρ)] ρ > a (2) !Syntax Error, Idk k Am(k) Jm(kρ) = 0 ρ < a To get into Jackson form, I would have to expand the RHS in powers of ρ. No doubt a big mess. What about the charge density on the iris? I think from above we have this: σ(ρ,θ) = (1/2π) Σm=0∞ cos(mθ) !Syntax Error, Idk k Am(k) Jm(kρ) ρ > a so there is no obvious way to connect this to the Q function. In other words, we still have the dual integral equation "coupling" which somehow determines everything. 6. Plan D: Repeat Plan C with other 1/R form. This section was added 4.30.10. In doc " iris green's by partial eigenfunctions" I derived this fact 1/R = Σm=0∞ εm cos(mφ)!Syntax Error, Idk Jm(kρ) Jm(kρ') e-k|z-z'| Using this in the above Plan C gives: Vtot = Σm=0∞ cos(mθ) !Syntax Error, Idk e-|k|zAm(k) Jm(kρ) + q Σm=0∞ εm cos(mφ)!Syntax Error, Idk Jm(kρ) Jm(kρ') e-k|z-z'| = Σm=0∞ cos(mθ) !Syntax Error, Idk Jm(kρ) { Am(k) e-|k|z + q εm Jm(kρ') e-k|z-z'|} We can then write the condition that V = 0 on the iris as follows where z = 0 and ρ > a 0 = Σm=0∞ cos(mθ) !Syntax Error, Idk Jm(kρ) { Am(k) + q εm Jm(kρ') e-k|z'|} ρ > a and then our PWA tells us that !Syntax Error, Idk Am(k) Jm(kρ) = - (q εm) !Syntax Error, Idk Jm(kρ) Jm(kρ') e-k|z'| ρ > a If I put z' = d > 0 as in Plan C, this becomes !Syntax Error, Idk Am(k) Jm(kρ) = - (q εm) !Syntax Error, Idk Jm(kρ) Jm(kρ') e-kd ρ > a With not much surprise, I locate this on page 696 of GR7 So that !Syntax Error, Idk Jm(kρ) Jm(kρ') e-kd = (1/π) Qm-1/2 [(ρ2+ ρ'2 + d2)/(2ρρ')] so that our condition V = 0 on the iris then becomes (where set ρ' = c) !Syntax Error, Idk Am(k) Jm(kρ) = – (q εm /π) Qm-1/2 [(ρ2+ c2 + d2)/(2ρc)] ρ>a So using this different form for 1/R gives a different integral which evaluates to the same function we got in Plan C (I am happy to see). We can repeat our analysis of σ in the hole, V = Σm=0∞ cos(mθ) !Syntax Error, Idk Jm(kρ) { Am(k) e-|k|z + q εm Jm(kρ') e-k|z-z'|} The derivative of interest will be this: ∂z e-k|z-z'| = -k e-k|z-z'|∂z|z-z'| = -k e-k|z-z'| sign(z-z') so that ∂z e-k|z-z'| at z = 0± = k e-k|z'|sign(z') = k e-kd since we have z' = d > 0 We then get ∂zV(+) = Σm=0∞ cos(mθ) !Syntax Error, Idk k { – Am(k) Jm(kρ) + q εm Jm(kρ') e-kd } ∂zV(-) = Σm=0∞ cos(mθ) !Syntax Error, Idk k { + Am(k) Jm(kρ) + q εm Jm(kρ') e-kd } and exactly as before, the second term cancels in the difference and we get our same other dual equation. This has to happen because we know the field of the point charge does now imply charge anywhere on the z = 0 plan. We end up then with our exact same dual integral equations (1) !Syntax Error, Idk Am(k) Jm(kρ) = - (qεm/π) 1/ Qm-1/2 [(ρ2+ c2+ d2)/(2cρ)] ρ > a (2) !Syntax Error, Idk k Am(k) Jm(kρ) = 0 ρ < a