TheCharged Bowl in Toroidal Coordinates 11_30_11

1 The Charged Bowl in Toroidal Coordinates Phil Lucht Rimrock Digital Technology, Salt Lake City, Utah 84102 last update: November 30, 2011 Contents: 1. Introduction. ..................................... ................................................... ..........................................2 2. Solution of the Charged Bowl Problem in Toroidal Coordin ates............................................... ..3 (a) Level curves, the charged ellipsoid problem and a dashe d hope.............................................. .....3 (b) Cartesian, spherical, toroidal and ellipsoidal atomic forms .............................................. ............4 (c) Smythian forms and one for the bowl.................. ................................................... .....................5 (d) Solution to the charged bowl potential problem: the Mehl er-Fock Transform.............................6 (e) Derivation of an integral....................... ................................................... ...................................8 3. Solution of the Charged Double-Bowl Problem in Toroidal Coordinates....................................9 (a) The general double bowl solution ......................... ................................................... ...................9 (b) the special case of a bowl with a flat lid. ......... ................................................... ....................... 11 (c) The sessile drop ( lens shaped double-bowl) ......... ................................................... ................. 11 4. The Single Bowl Surface Charge Densities and Capacitan ce ................................................. .... 12 (a) Bowl Charge Densities. .......................... ................................................... ............................... 12 (b) Bowl Capacitance.................................. ................................................... ................................ 14 5. Toroidal coordinates................................. ................................................... ................................ 15 (a) Verbal description ................................. ................................................... ................................ 15 (b) Equations ......................................... ................................................... ..................................... 16 (c) More pictures .................................... ................................................... .................................... 16 (d) plot of the bowl coordinate u versus coordinates ρ and z.............................................. ............. 17 (e) plot of u versus coordinates ρ and z after Lebedev's adjustment.................... ............................. 18 (f) plot of the toroid coordinate ξ versus coordinates ρ and z .............................................. ............ 19 6. A selection of charged bowl potential plots .......... ................................................... ................... 19 7. Doing Mehler integrals. .............................. ................................................... .............................. 21 (a) List of useful Mehler integrals .................. ................................................... ............................. 21 (b) The integral X appearing in Section 4(a)............ ................................................... .................... 22 (c) Where to find Mehler integrals (some errata noted).. ................................................... .............. 23 8. Evaluating the toroidal bowl potential integral into eleme ntary functions................................ 24 9. Using Maple to plot the bowl potential............... ................................................... ...................... 25 10. Potential and capacitance of a charged toroid. ........... ................................................... ........... 28 (a) The potential of a charged toroid................. ................................................... ........................... 28 (b) Capacitance of a toroid......................... ................................................... ................................. 30 (c) Charge density on a toroid ........................ ................................................... ............................. 33 Appendix A. Maple code text for plotting the Bowl Potent ial ................................................. ...... 34 Appendix B: Comments on computing the capacitance of a toroid and its degenerate limit. ...... 35 Appendix C: Kelvin's approach to the Charged Bowl Probl em................................................. ... 39 2 Appendix D: Smythe's approach to the Charged Bowl Problem . ................................................. 41 Appendix E: Dual Equations and their connection to th e Charged Bowl Problem...................... 46 Appendix F: The Abel Transform in various forms ....... ................................................... ............ 49 Appendix G: Solving the Charged Bowl Problem using a dou ble Abel transform....................... 52 (a) Find g 1................................................... ................................................... ............................... 52 (b) Run a check on g 1................................................... ................................................... .............. 55 (c) Find a n................................................... ................................................... ................................ 56 (d) Find Ψ and f 2................................................... ................................................... ...................... 58 References. ........................................ ................................................... ............................................ 58 1. Introduction. This document was motivated by a "stub" section appearing i n an informal 2002 paper by Kirk McDonald (see References) which reviews various solutions of the ch arged bowl problem. Here is that section in its entirety: where [10] is our Lebedev reference. The present document is meant to fill in this stub section! The phrase "charged bowl" refers to an isolated conducting spherical bowl, sometimes called a spherical cap, shell or segment. An arbitrary plane slicing through a f ull spherical shell divides that shell into two spherical bowls, each with the same circular lip. The "ch arged bowl problem" is this: put some charge Q on an isolated conducting bowl and determine the resulting el ectrostatic potential V everywhere as well as the charge densities σ on the inner and outer bowl surfaces. The reader versed in related problems (e.g., the charged disk) would not be surprised to find the charge densities to be divergent (but integrable) at the bowl edge and non-zero everywhere on the bowl. In practice, one nor mally assumes that the bowl is at some potential V 0 relative to infinity, and then the charge on the bowl is Q = C bowl V0. Of course finding Cbowl is part of the problem. In 1869 Kelvin used the now-famous method of inversion (last seen lurking in green Jackson) to solve the charged bowl problem for the inner and outer charg e densities. Kelvin's "bowl paper" requires great patience to read, and is outlined in McDonald's pa per. Modern readers might have "forgotten" things like "the chord theorem" and other geometric properties of c ircles. We comment more on Kelvin's bowl paper in Appendix C and discuss how Kelvin showed the now-well-known fact that the inne r and outer bowl charge densities differ by the constant V 0/(4 πR) which is independent of the bowl's lip angle and of the location on the bowl. Certainly this result is valid fo r a fully closed bowl since σ = 0 inside and on the outside σ = Q/A = C sphere V0/A = R V 0/(4 πR2) = V 0/(4πR). In Appendix D we mention a fascinating variation of the Kelvin inversion approach indicated by Smythe in a set of problems. Besides inversion, various other methods exist to obtain an exact solution of this problem. Perhaps the most well known is the use of "dual" Legendre series equa tions in friendly spherical coordinates as 3 outlined by Sneddon and others. One series (Dirichlet) sets the potential to a constant on the bowl, while a radial derivative of that series (Neumann) sets the char ge density to zero on the cap. The problem is then to find series coefficients which satisfy both equations , neither of which is invertible since it only covers a partial range of polar angle. In matrix language, these dua l equations are A Ψ = f and B Ψ = g where Ψ is an infinite vector of unknown coefficients. Appendix E shows formally how such dual equations may be solved, and Appendix G solves the charged bowl problem using this method. A tool used in the dual equations approach is the Abel transform, a summary of wh ich is given in Appendix F . Our main concern, despite all these spherical coordinate appendices, is to obtain an exact solution of the charged bowl problem using the much less friendly toroida l coordinates. Surely this has been done many hundred times since the mid 19th century, but the author has not found much on the web -- certainly not much that is freely downloadable. After doing the bowl, we turn our attention briefly to the pr oblem of the charged toroid . The general plan is shown in the Contents above. Appendix A provides simple Maple code to allow the reader to make plots of the bowl potential. Appendix B concerns a limit which seems to indicate that the capacitance of a degenerate toroid of unit tube radius i s given by the strange number C = 1.7413 which can be compared to the capacitance C = 2 of its emb edding sphere. In all that follows, we shall refer to the spherical c onducting bowl as "the bowl", and shall refer to the unoccupied remainder of the bowl's sphere as "the cap". The sym bol εn stands for 2- δn,0 , sometimes called Neumann's factor or number. 2. Solution of the Charged Bowl Problem in Toroidal Coordin ates. (a) Level curves, the charged ellipsoid problem and a d ashed hope As one sees scanning through the beautiful pictures in Moon & Spencer's strangely but correctly named Field Theory Handbook , each orthogonal coordinate system has its characteristi c level surfaces. In spherical coordinates these are spheres, polar cones and azimut hal planes, whereas in toroidal coordinates they are "bowls and toroids" and azimuthal planes. In spheri cal coordinates a bowl has a hole in it since the polar angle runs only part of its range, but in toroidal coordinates there is no such "hole". That is to say, a bowl has a label (a value of one of the toroidal c oordinates) and as the other two coordinates sweep their full ranges , a bowl is swept out. The fact that the bowl is a level surface in toroidals makes one at least interested in solving the charged bowl problem in thi s system. In ellipsoidal coordinates the level surfaces are ellipsoid s and asymmetric hyperboloids of one and two sheets, which certainly sounds foreboding. One can solve the "charged ellipsoid problem" and the result is shockingly simple (as Kelvin also showed) and σ on the ellipsoid is simple even in Cartesian coordinates (more magic geometry). A family of confocal elli psoids can be described by the equation x2/( ξ12- a 2) + y 2/( ξ12- b 2) + z 2/( ξ12) = 1 where ξ1 in (0, ∞) is the "label" of an ellipsoid. It happens then that the label ξ1 is also the largest semi-major axis of the ellipsoid, w hile a > b are focal distances associated with the other two axes. Ellipsoidal coordi nates are fully separable, and a Laplace-satisfying potential function which is constant on each of these el lipsoids and which vanishes at infinity is given by V( ξ1)/const = F 00(ξ1) E00(ξ2) E00(ξ3) = [(1/a) sn -1(a/ ξ1,b/a)] * 1 * 1 = (1/a) sn -1(a/ ξ1,b/a) = (1/a) F(sin -1(a/ ξ1),b/a) 4 Here ( ξ1,ξ2,ξ3) are the three ellipsoidal coordinates in Morse & Fesh bach notation, E and F are first and second kind Lamé functions, and a different F is the fi rst kind elliptic integral. By setting V = V 0 on a particular ellipsoid ξ1 = c, one then obtains the potential anywhere outside this c harged ellipsoid, V( ξ1) = V 0 F(sin -1(a/ ξ1),k=b/a) / F(sin -1(a/c),k=b/a) (2.1) Although not immediately obvious, this expression is exactly the same if one swaps a ↔b and such a swap is necessary to show that the above form agrees with that of Kelvin. [ There is some confusion in elliptic F notation to beware: F( φ,k) = F( φ | k 2) = F( φ \ sin -1k ), the first notation being that of GR7 8.111.2.] One might hope that the bowl potential in toroidals could be as simp ly stated as the ellipsoid potential in ellipsoidals, but alas it is not so, but only in the foll owing sense. In toroidals a certain weight factor cross-links the bowl and toroid coordinates (labels), which makes the Laplace equation separable only in the form of three separated functions times the weight fac tor ch ξ-cosu , where ξ is a toroid label and u a bowl label. This weaker kind of separability is called R-sepa rability by Moon and Spencer. A solution ch ξ-cosu [A( ξ)=1]B(u)[C( φ)=1] evaluated on the surface of bowl u 0 gives V = ch ξ-cosu 0 B(u 0) which varies with ξ, not allowing V = V 0 on the bowl. Nevertheless , it will turn out that the charged bowl potential can be expressed in simple inverse trig function s, and is thus even simpler that the charged ellipsoid potential. (b) Cartesian, spherical, toroidal and ellipsoidal atomic f orms "Atomic forms" or just "atoms" are the author's private phra ses for "harmonics", which word means simple solutions of the Laplace equation which can be superpo sed to construct non-simple solutions. The Cartesian atoms illustrate the idea that if you curve toward axis in 2 dimensions, you must curve away from axis in the 3rd: osc osc expo (1) [sin(k xx), cos(k xx)], [sin(k yy), cos(k yy)], [exp( κzz), exp(-κzz) ] k z = imaginary = i κz toward toward away κz = kx2 + k y2 For solving practical problems, 2 of the 3 coordinates have to be oscillatory to allow for functional completeness, here on a surface of x and y, so that expans ions can be inverted and problems solved. For spherical atoms, two interesting atomic forms can be writ ten in which azimuthal φ is oscillatory: r in (0, ∞) z in (-1,1) φ in (0,2 π) expo osc os c (1) [ r n, r -n-1] [ P nm(z), Q nm(z)] [ sin(m φ),cos(m φ)] z = cos θ osc expo osc (2) (1/ r )[ r iτ, r -iτ] [ P iτ-1/2 m(z), Q iτ-1/2 m(z)] [ sin(m φ),cos(m φ)] n = i τ-1/2 osc expo osc ~ (1/ r )[sin( τ lnr), cos( τ lnr)] [ P iτ-1/2 m(z), Q iτ-1/2 m(z)] [ sin(m φ),cos(m φ)] 5 The first is doubtless more familiar to the reader, but the second is appropriate for, say, a Dirichlet problem involving a cone, since the atomic form is oscillatory both ways across the surface of a cone, allowing a prescribed potential there to be inverted. The underlying fact is that each oscillatory coordinate becomes a 1D Sturm-Liouville problem with a complete set of eigenfunctions, and then these two sets provide a complete set of eigenfunctions for a 2D surface spa nned by those coordinates. In passing, we note that the Legendre functions in form (2) above are call ed conical functions and have |z|<1. Without further ado, we can write two similar atomic fo rms for toroidal coordinates, ξ in (0, ∞) u in (0,2 π) φ in (0,2 π) expo osc osc (1) ch ξ - cosu [P n-1/2 m(ch ξ), Q n-1/2 m(ch ξ) ] [ sin(nu),cos(nu)] [ sin(m φ),cos(m φ)] osc expo osc (2) ch ξ - cosu [P iτ-1/2 m(ch ξ), Q iτ-1/2 m(ch ξ) ] [exp( τu), exp(-τu) ] [ sin(m φ),cos(m φ)] To solve bowl problems, we need the two coordinates other th an the bowl label u to be oscillatory, and that means we must use form (2). In the first form, if a problem has a full azimuth, the parameter m gets quantized to integers, and this indirectly causes parameter n on P to be quantized to integers, so the spectra of both Sturm-Liouville problems are discrete. In form (2) even if m is quantized, parameter τ remains unquantized and, since P ν= P -ν-1, the τ spectrum can be restricted to the positive real axis. Again in passing, the Legendre functions in system (1) are sometimes called toroidal functions or ring functions, argument > 1, whereas those appearing in (2) are called Mehler functions, again with argument > 1. At least that is the author's terminology. In reference to our earlier charged ellipsoid comments, we just mention the atomic form for ellipsoidal coordinates, [E mp(ξ1), F mp(ξ1)] [E mp(ξ2), F mp(ξ2)] [E mp(ξ3),F mp(ξ3)] where 0 < ξ3< b < ξ2< a < ξ1 with a>b the confocal ellipsoid focal distances. As noted earlier, E and F are the first and second kind Lamé functions. The ξi are the three roots of the cubic equation which is the equation of the ellipsoid given above. In this system there is no azimuthal coordinate, and the separated solutions are triply cross-linked in that all three funct ions bear the same quantum numbers m and p (separation constants arising when the Laplace equation is se parated). This system, though quite complicated, is well explained in full detail in the last chapter of Hobson's book. (c) Smythian forms and one for the bowl Please forgive the author's predilection for strange phrases . A Smythian form refers to a linear combination (sum and/or integral) of atomic forms, havin g some to-be-determined "coefficients", which provides a candidate solution to some problem. Such a form usual ly "builds in" certain boundary conditions, such as continuity between two regions of space on w hose boundary a Green's point charge lies in a Green's Function problem. Smythe's book makes excell ent use of such forms, and one might even 6 refer to "the method of Smythian forms" in a list of eff ective methods of solving boundary value problems. So, we propose the following Smythian form for the potentia l of our charged bowl (since things are azimuthally symmetric, we have only m = 0 atoms) V( ξ,u) = ch ξ - cosu ∫0 ∞ dτ P iτ-1/2 (ch ξ)[ A( τ)ch(u τ) + B( τ)sh(u τ) ] (2.2) where A and B are to-be-determined coefficient functions . Although it is not immediately obvious, this form builds in the boundary condition that the potential vani sh at infinity, and for that reason the Q function has been rejected. Since the Mehler P functions form a complete orthogonal set, it is possible to set in some prescribed potential V = f( ξ,u) on the surface of a bowl, and invert to find the coefficient functions. At first it seems odd that there are two fun ctions A and B to be found, and only one boundary condition, but the dilemma is quickly resolved by realizing th at we apply the boundary condition separately on each surface of the bowl, so really there ar e two boundary conditions and two unknown coefficient functions. For our charged bowl problem, the prescr ibed Dirichlet potential is f( ξ,u) = V 0. (d) Solution to the charged bowl potential problem: the Mehler-Fock Transform The generalized Mehler-Fock (Mehler-Fok) transform (Oberhet tinger and Higgins page 2) is this, g(y) = ∫0 ∞ dτ P m -1/2+i τ(y) f( τ) // expansion f( τ) = ( τ/π) sh( πτ ) Γ(1/2-m+i τ) Γ(1/2-m-iτ) ∫1 ∞ dy Pm -1/2+i τ(y) g(y) // projection and for m = 0, using Γ(1/2+i τ)Γ(1/2-iτ) = π/cosh( πτ ), one obtains the regular Mehler-Fock transform, g(y) = ∫0 ∞ dτ P -1/2+i τ(y) f( τ) // expansion f( τ) = τ th( πτ ) ∫1 ∞ dy P-1/2+i τ(y) g(y) // projection which can also be written as G( ξ) = ∫0 ∞ dτ P -1/2+i τ(ch ξ) f( τ) // expansion f( τ) = τ th( πτ ) ∫0 ∞ dξ shξ P-1/2+i τ(ch ξ) G( ξ) // projection (2.3) For azimuthally symmetric bowl problems, this transform i s the one associated with the Sturm Liouville problem in the ξ coordinate. Every reasonable Sturm-Liouville problem define s a complete set of functions and therefore defines a transform (there are th ousands of them), and this happens to be the transform for our oscillatory ξ coordinate. Admittedly, this transform is more sparsely found in the literature that, say, the Fourier Integral Cosine Tran sform. 7 Before using this transform to invert the two boundary condi tion equations, we have to first know a little more about the range of the bowl label coordinate u. To that end, here is a picture: This shows in cross section one up-side-down bowl having label u 0. Following Lebedev et. al. 's excellent suggestion, instead of running the bowl label in the usual (0,2 π) range, we shall have it lie in (u 0, u 0+2 π) so it can be continuous away from the two sides of the bowl. As we start off with u 0 as the angle shown, we can imagine a set of mathematical bowl surfaces attac hed to the two "focal points" on the x axis, whose labels run from u 0 up to u 0 + 2 π. We will explain all this below, but it is perhaps useful t o solve the problem first, then see the coordinate details later. The inside surface of the bowl lies at u = u 0 while the outside surface is at u = u 0+2 π. It is convenient to define u 0' ≡ u 0 + 2 π and to note then that cos(u 0') = cos(u 0). Here then are the two boundary conditions: V0 = ch ξ - cosu 0 ∫0 ∞ d τ P iτ-1/2 (ch ξ) [ A( τ)ch(u 0τ) + B( τ)sh(u 0τ) ] (2.4) V0 = ch ξ - cosu0' ∫0 ∞ d τ P iτ-1/2 (ch ξ) [ A( τ)ch(u 0'τ) + B( τ)sh(u 0'τ) ] u 0' ≡ u 0 + 2 π Application of the Mehler-Fock transform (2.3) gives [ A( τ)ch(u 0τ) + B( τ)sh(u 0τ) ] = V 0 τ th( πτ ) ∫0 ∞ dξ sh( ξ) Piτ-1/2 (ch ξ)/ ch ξ - cosu 0 (2.5) [ A( τ)ch(u 0'τ) + B( τ)sh(u 0'τ) ] = V 0 τ th( πτ ) ∫0 ∞ dξ sh( ξ) Piτ-1/2 (ch ξ)/ ch ξ - cosu0 Then using this integral (derived in section (e) below) ∫1 ∞ dx P iτ-1/2 (x) / x - cosu 0 = 2 ch[ τ(u 0-π)]/ ( τ sh( πτ)) 0 ≤ u 0 ≤ 2 π (2.6) one can do a simple Cramer's Rule solution to find the coef ficient functions A( τ) = V 0 2 ch[( π–u0)τ] ch[( π+u 0)τ] / ch 2(πτ ) B( τ) = – V 0 2 ch[( π–u0)τ] sh[( π+u 0)τ] / ch 2(πτ ) (2.7) 8 Inserting these into our Smythian form (2.2) then gives the po tential of the charged bowl of label u 0 V( ξ,u) = V 0 2 ch ξ - cosu ∫0 ∞ dτ P iτ-1/2 (ch ξ) ch[( π–u0)τ] ch[( π+u 0-u) τ]/ch 2(πτ ) (2.8) We are all done! It is shown below that this integral can in fact be evaluated into the following set of elementary functions V( ξ,u)= (V 0/π ) ch ξ - cosu * { (1/ ch ξ -cosu ) cot -1[(-1) η1| 1+cosu | / ch ξ -cosu ] + (1/ ch ξ-cos(2u 0-u) ) cot -1[(-1) η2| 1+cos(2u 0-u) |/ ch ξ -cos(2u 0-u) ] } (2.9) where η1 = floor[(3 π-u)/2 π] u 0 in (0, π) η2 = floor[(2u 0-u+ π)/2 π] u in (u 0, u 0+2 π) where the strange η exponents are related to analytic continuation as describ ed later. We can throw this equation into Maple and obtain very nice plots of the charge d bowl potential across a symmetric slice of the bowl. Here is the plot for a bowl with u 0 = π/4. Along the bowl edge the potential is constant at the value V 0= 1, and outside the region shown it drops off to 0 at infinit y. Kelvin would have liked Maple. (e) Derivation of an integral We wish to verify integral (2.6), which does not seem to appear as such in the usual sources, 9 ∫1 ∞ dx P iτ-1/2 (x) / x - cosu 0 = 2 ch[ τ(u 0-π)]/ ( τ sh( πτ)) 0 ≤ u 0 ≤ 2 π (2.6) Start with the left hand side (LHS) and expand part of the integrand using (10.3) below 1/ x - cosu 0 = (1/ π) 2 Σn=0 ∞ εn Qn-1/2 (x) cos(nu 0) so that LHS = ( 2 /π) Σn=0 ∞ εn cos(nu 0) ∫1 ∞ dx P iτ-1/2 (x) Q n-1/2 (x) But according to GR7 page 770 7.114.1 the PQ integral above is jus t 1/(n 2+τ2) so that LHS = ( 2 /π) Σn=0 ∞ εn cos(nu 0) /(n 2+τ2) Then use GR7 page 47 1.445.2, which can be written as Σn=0 ∞ εn cos(nu0)/(n 2+τ2) = ( π/τ) ch[ τ(π- u0)]/sh( τπ ) 0 ≤ u 0 ≤ 2 π to conclude that LHS = 2 ch[ τ(π- u0)]/ [ τ sh( τπ )] and therefore (2.6) is validated 3. Solution of the Charged Double-Bowl Problem in Toroidal Coordinates. (a) The general double bowl solution This seems a good place to address this related problem. S neddon deals with the problem of two separated bowls having a common symmetry axis. That problem seems to have no closed form solution and Sneddon reduces its solution to solving one or more second-kind Fre dholm integral equations. Our double-bowl problem is much simpler in that both bowls have a common lip and the same potential V 0. The 3D shapes formed in this way can vary from a sort of 3D lune, to a bowl with a flat lid on it, to a "lens" composed of two spherical surfaces (two facing attache d bowls). If the two bowls of the lens have the same size, the solution to this problem describes the ev aporation from the surface of a liquid drop sitting on a flat surface, which drop then is half the le ns shape. See Hu and Larson regarding such "sessile droplets". Quantity π-u is the "contact angle" of such a drop. It is nice to s ee that solutions to 19th century electrostatic and heat flow potential theory problems ha ve 21st century applications. The solution to the double bowl problem follows exactly that of the single-bowl problem. For obscure reasons, we defined the two bowl labels to be u 0 and u 2 rather than the more reasonable u 0 and u 1, forgiveness is requested. Here is the picture showing a lune shaped cross section 10 We use the exact same Smythian form as for the single bowl, V( ξ,u) = ch ξ - cosu ∫0 ∞ dτ P iτ-1/2 (ch ξ)[ A( τ)ch(u τ) + B( τ)sh(u τ) ] (3.1) As before we define u 0' ≡ u 0 + 2 π and note that cos(u 0') = cos(u 0). Here then are the boundary conditions for the double-bowl problem, where for the moment we use V 0 and V 2 : V2 = ch ξ - cosu 2 ∫0 ∞ d τ P iτ-1/2 (ch ξ) [ A( τ)ch(u 2τ) + B( τ)sh(u 2τ) ] (3.2) V0 = ch ξ - cosu0' ∫0 ∞ d τ P iτ-1/2 (ch ξ) [ A( τ)ch(u 0'τ) + B( τ)sh(u 0'τ) ] u 0' ≡ u 0 + 2 π These conditions are for the "exterior problem", we want t he potential outside the lune. Comparison with (2.4) shows that we set 0 →2 in the first condition and the second condition is the same. T he coefficients this time are a little more complicated: ch( πτ ) A( τ) = V 0 2 Qch[ τ(π-u0)] / [C ch(u 0'τ) + D sh(u 0'τ) ] ch( πτ ) B( τ) = V 0 2 Pch[ τ(π-u0)] / [C ch(u 0'τ) + D sh(u 0'τ) ] where (3.3) C = { V 0 sh(u 2τ) ch[ τ(π-u0)] – V 2 sh(u 0'τ) ch[ τ(π-u2)] } D = { V 2 ch(u 0'τ) ch[ τ(π-u2)] – V 0 ch(u 2τ) ch[ τ(π-u0)] } Inserting A( τ) and B( τ) into the Smythian form (3.1), and setting set u 0' ≡ u 0 + 2 π and V 2 = V 0, we find after some algebra this relatively simple result for the potential of the charged double-bowl, V( ξ,u) = V 0 2 ch ξ - cosu ∫0 ∞ d τ P iτ-1/2 (ch ξ) * (3.4) { ch[( π-u0)τ] sh[(u-u 2)τ] + ch[( π-u2)τ] sh[(2 π+u 0-u) τ] } / {ch( πτ ) sh[(2 π+u 0-u2)τ] } 11 This result appears in the Hu paper mentioned above in this form , where β,β1,β2 = u,u 0,u 2 and α = ξ. If one takes the limit u 2→u0, the solution to the single-bowl problem is recovered. We have not made any attempt to evaluate this integral but it may yield an expr ession involving elementary functions. (b) the special case of a bowl with a flat lid. Taking u 2 = π to get the flat lid, the solution (3.4) reduces to the fol lowing, where ξ=α , β=u, and u 0=β1, V( α, β) = V 0 2 ch α - cos β ∫0 ∞ d τ P iτ-1/2 (ch ξ) * { sh[(2 π+β1-β)τ] – ch[( π-β1)τ] sh[( π- β)τ] } / { sh[( π+β1)τ] ch( πτ ) } (3.5) which may be compared with the result quoted in Lebedev e t. al. Problem 503 (page 241) with β0= β1 (c) The sessile drop ( lens shaped double-bowl) This time we take u0 = β1 = [ π-θ] u2 = β2 = [ π+θ] where θ is the small "contact angle" for the drop. Potential (3. 4) then becomes V( α,β) = V 02 ch α - cos β ∫0 ∞ d τ P iτ-1/2 (ch ξ) ch( θτ) ch[(2 π-β)τ] /{ch( πτ ) ch[(π-θ])τ]} (3.6) which appears in the Hu paper as 12 4. The Single Bowl Surface Charge Densities and Capacitanc e (a) Bowl Charge Densities. We start with the potential as stated above in (2.8) and set up to compute the charge densities, V( ξ,u) = V 0 2 ch ξ - cosu ∫0 ∞ dτ P iτ-1/2 (ch ξ) ch[( π–u0)τ] ch[( π+u 0-u) τ]/ ch 2(πτ) (2.8) σ+ = – (1/4 π) (1/h u) ∂uV( ξ,u)| u=u0 1/h u = (ch ξ - cosu)/a σ– = +(1/4 π) (1/h u) ∂uV( ξ,u)| u=u0+2 π (4.1) where the sign difference arises from the direction of approa ch to the surface relative to increasing coordinate u. Quantity h u is the curvilinear scale factor guu in the u direction, and a is the radius of the circular lip of the bowl. Defining the above integral to be f(u) , we find that ∂uf| u = ∓ X at the surfaces, where X = ∫0 ∞ dτ P iτ-1/2 (ch ξ) ch[( π–u0)τ] τ sh( πτ ) / ch 2(πτ ) (4.2) and this time the two signs arise from evaluating sh[( π+u 0-u) τ] = ± sh( πτ ) on the two surfaces. Evaluation of this integral (see section 7 (b) below) gives X = (1/ π) [sin(u 0/2)/B 2] [1 + (A/B) tan -1 (A/B) ] (4.3) where A = 2 cos(u 0/2) and B = ch ξ-cosu 0 Meanwhile, from (2.8) above we have ∂uV = V0 2 B ( ∓ X) + (1/2) sinu { V 02 (1/ B) f(u)} (4.4) and we finally end up with σ+ = σin = ( V0/4π2R) [ (B/A) – tan -1(B/A)] σ– = σout = ( V0/4π2R) [ π + (B/A) – tan -1(B/A)] (4.5) Therefore 13 σout - σin = ( V0/4π2R) [ π ] = (V0/4πR) // the famous fact, see Appendix C (4.6) σin = ( V0/4π2R) { ch ξ-cosu 0 / ( 2 cos(u 0/2)) – tan -1[ ch ξ-cosu 0 / ( 2 cos(u 0/2))] } (4.7) which we can compare with Lebedev's Problem 501 claim ( a = R and α = ξ and u 0 = β0) Finally we compare our "inner" σ result above with that of Kelvin, where now we write u 0=α (which is different from α=ξ in the screen clip above) σL =( V0/4π2R) { ch ξ-cos α / ( 2 cos( α/2)) – tan -1[ ch ξ-cos α / ( 2 cos( α/2))] } // Lebedev (4.8) σK = (V0/4 π2R) { 1-cos α / cos α - cos θ – tan -1 [ 1-cos α /cos α - cos θ ] } // Kelvin (4.9) where angles α = u 0 and θ are explained by this picture One can show from the toroidal coordinate equations that cos α - cos θ = sin α / ch ξ-cos α (4.10) 14 and this makes the Lebedev and Kelvin results agree. In the Lebedev form, the bowl edge lies at ξ = ∞, while in the Kelvin form it lies at θ=α, and in either case σ (the first term) diverges at the edge, as noted in the Introduction. Here is Kelvin's actual statement of his result (4.11) where f = 2R, α(Kelvin) = π-α, and η = π-θ, causing all cosines to negate. (b) Bowl Capacitance We start yet again with our toroidal charged bowl potentia l V( ξ,u)= V 0 2 ch ξ - cosu ∫0 ∞ dτ P iτ-1/2 (ch ξ) ch[( π–u0)τ] ch[( π+u 0-u) τ] / ch 2(πτ ) (2.8) The large-r limit is given in toroidal coordinates by ξ→0 and u →2π, so we set P iτ-1/2 (ch ξ) = 1. Also, we can show from our equations below that, in this limit, ch ξ - cosu ≈ 2 a / r . (4.12) The residual integral is then (after some work) I ≡ ∫0 ∞ dτ ch[( π–u0)τ] ch[(u 0-π)τ] / ch 2(πτ ) = (1/2 π)[ πb/sin( πb) + 1 ] b = 1-u 0/π (4.13) So V( ξ,u) ≈ V 0 2 ch ξ - cosu (1/2 π)[ πb/sin( πb) + 1 ] ≈ ( aV 0 /r) [ b + sin πb/ π ]/sin( πb) (4.14) Thinking of this as V = Q/r and using Q = CV 0 one finds Q = ( aV 0 ) [ b/sin( πb) + 1/ π ] C = ( a ) { b/sin( πb) + 1/ π } b = 1-u 0/π = Rsin(u 0) { (1-u0/π) /[sin(u 0)] + 1/ π } a = Rsin(u 0) = (R/ π) { π - u 0 + sin(u 0) } (4.15) which is indeed the known capacitance of a spherical bowl. In t he limit u 0 → 0 the formula gives C = R which is the capacitance of a sphere of radius R, while the limit u 0→ π (b →0) gives C = 2a/ π which is the capacitance of a disk of radius a. 15 5. Toroidal coordinates (a) Verbal description Here is the basic toroidal picture as a slice in any azim uthal plane, where ρ and z are the usual cylindrical coordinates, In this picture, the level surfaces of u appear to be vertica l set of circles, but each circle is really truncated at the plane z=0 [ since sign(z) = sign(sinu) according to 5.1 below] so we have a set of upper bowls and a set of lower bowls. In this traditional picture, a huge upper bowl has as its limit an iris just above the z=0 plane which is labeled u=0. As these bowls shrink, we pass through the u = 2 π/8 bowl shown, and eventually reach very shallow bowls with the limit of a di sk of radius a at u = π. Continuing u we cover all the lower bowls and end up with an iris just below the z=0 pla ne at u = 2 π. There is then a discontinuity in u at this iris value, as shown on the ri ght. All these u bowls have a common circular lip in the z=0 plane of radius a, so the distance between the two f ocal points in the picture above is 2a. One can interpret u as the external bowl lip angle relative to the z=0 plane, but that fact is not very obvious from the above pictures. Meanwhile the horizontal circles show slices of the toroids labeled by ξ . In the limit ξ → 0 one gets an exceedingly fat toroid with no hole whose inner surface s weeps up against the z axis. Conversely, as ξ→∞ , the toroids shrink to a thin circular wire matching the bowl lips. It might be noted that these horizontal circles are the famous ones of Apollonius where | r - a|/| r + a| = constant, and in this case that constant is e -ξ. Toroidal coordinates start life as 2D bipolar coordinate s with the same picture as above, but the left side circles have ξ < 0. For 3D toroidal coordinates, one takes just the right side of the above picture with ξ > 0 and rotates this around the z axis to get the 3D bowls a nd toroids. Lest it be overlooked, the two sets of circles are orthogo nal at all intersections, and are both orthogonal to the azimuthal unit vector at any point. The t oroidal system is an orthogonal one with a diagonal metric tensor. 16 (b) Equations The defining equations are these for toroidal coordinates (toroid,bowl,plane) = ( ξ,u, φ) : x = a cos φ sh ξ/(ch ξ - cosu) ρ = a sh ξ/(ch ξ - cosu) ρ2 = x 2+ y 2 y = a sin φ sh ξ/(ch ξ - cosu) z = a sinu/(ch ξ - cosu) (5.1) Here then is a summary of all the basic facts of intere st ρ = ash ξ/(ch ξ–cosu) ρ>0 h ξ = h u = a/(ch ξ–cosu) h φ = sh ξ h ξ z = asinu/( ch ξ–cosu) ρ2 = x 2+ y 2 z/ ρ = sinu/sh ξ (5.2) ξ = tanh -1[2a ρ/(a 2+ ρ2+ z 2)] metric tensor = diag(h ξ2, hu2, hφ2) tanu = [ -2az/(a 2-ρ2-z2)] // use algorithm to find u (see later) (5.3) r2 = a 2 (sh 2ξ + sin 2u) / (ch ξ-cosu) 2 = ρ2+z 2 tan φ ≡ z/ ρ = sinu/sh ξ (5.4) ρ2 + (z- acotu) 2 = a 2/sin 2u z c = a cot(u) radius = a/|sinu| // vertical circles (ρ - acoth ξ)2 + z 2 = a 2/sh 2ξ ρc = acoth ξ radius = a/|sh ξ| // horizontal circles (5.5) ranges: ξ in (0, ∞) u in (0,2 π) φ in (0,2 π) Notice that the radius of the sphere on which a bowl u 0 lies is R = a/|sinu 0|. (c) More pictures The following detailed Maple plots of u contours have proved hel pful to the author: 17 (d) plot of the bowl coordinate u versus coordinates ρ and z Here is a plot of u oriented roughly to match the picture s in Section (a) above, At the upper right (yellow, near "4") we start on the lower plane at u = 0. As we rotate CCW about the right focal point we go up the hill and arrive at the top right on the u = 2 π upper plane. Unfortunately this orientation does not show the hill very well, so we rotate t hings to get a better view 18 Here the vertical headwall is the discontinuity noted above. (e) plot of u versus coordinates ρ and z after Lebedev's adjustment If we adjust things so the u coordinate ranges from u 0 to u 0 + 2 π, the above plot changes. Here is the plot for u 0 = π/4 = .78 : 19 Now the spiral voyage begins inside the bowl at the foot of a vertical curved Niagara Falls which marks the inner boundary of our bowl. We climb up the hill and arrive at the upper plane and walk over to the falls and gaze down from the outer surface of the bowl. Her e again is our bowl picture from above We show all these pictures to make the point that toroidal c oordinates are a bit "unfriendly" compared to say spherical coordinates. They have to do violent things to be able to have bowls as level surfaces. (f) plot of the toroid coordinate ξ versus coordinates ρ and z This coordinate is much better behaved, though still a bit unusual. The two mountain peaks go up to infinity at the focal points for ξ = ∞, and there is a grounded crease along the ρ axis where ξ = 0. The contour plot on the right is perhaps better -- a top view of it would show our horizontal circle set. 6. A selection of charged bowl potential plots ____________________________________________________________________________ u0 = .01 π/8: (basically, u 0 = 0) 20 ____________________________________________________________________________ u0 = π/8: a bowl that is almost a complete sphere u 0 = 2 π/8: ____________________________________________________________________________ u0 = 3 π/8: u 0 = 4 π/8, the hemispherical bowl ____________________________________________________________________________ u0= 5 π/8 u 0= 6 π/8 21 ____________________________________________________________________________ u0 = 7 π/8, a very shallow bowl u 0 = π, potential of a disk ____________________________________________________________________________ When u 0 = π, the bowl slice becomes the line between the two foci, and the bowl becomes a perfect disk of radius a. So built into this problem solution is an exact plot of the disk potential. You can see, for example, that if you move 1" off the edge of the disk, V drops faster than if you move 1" out from the disk center. 7. Doing Mehler integrals. (a) List of useful Mehler integrals Mehler integration is the seamy underside of using toroidal coor dinates in the Mehler function atomic form stated earlier, and is worth some comment. The ge neral form of a "Mehler integral" is this: G(y) = ∫0 ∞ dτ P iτ-1/2 (y) f( τ) normally with y ≥1 About the best we can do for P is a form like P-1/2+i τ (ch α) = e -(1/2) αeiατ F(1/2-iτ, 1/2; 1; 1 - e -2α) 22 which at least isolates the variable τ into just one of the hypergeometric function's parameter argum ents. But even so, there is very little literature available o n such integrals, so changing from P to F really buys little. There are some integrals of P iτ-1/2 (y) f( τ) appearing in the literature (see below), but often has to do one's own integrals, especially if tables have typos a s noted below. The starting point for doing a Mehler integral is to eleva te P into one of these integral representations ( see p 962 GR7 8.715.1 and p 961 8.713.3 setting µ=0 ) P-1/2+i τ(y) = ( 2 /π) ∫0 α dx cos( τx) / y - chx y=ch α (7.1) P-1/2+i τ(y) = ( 2 /π ) ch( πτ ) ∫0 ∞ dx cos( τx) / y+chx (7.2) which exposes a simple τ dependence. The τ integration can then usually be done, and one is faced with the final x integration which can hopefully be looked up somewhe re. Here are some integrals the author has calculated in this manner: ∫0 ∞ dτ P iτ-1/2 (y) = (1/ 2 ) (1/ y - 1 ) (7.3) ∫0 ∞ dτ P iτ-1/2 (y) cos(a τ) = (1/ 2 ) (1/ y - cha ) Heaviside(y-cha) (7.4) ∫0 ∞ dτ P iτ-1/2 (y) cos(a τ) sech( πτ ) = (1/ 2 ) 1/ y + cha (7.5) ∫0 ∞ dτ P iτ-1/2 (y) cos(a τ) sech 2(πτ ) = ( 2 /π ) (1/ y-cha ) tan -1[ y-cha / 1+cha ] (7.6) ∫0 ∞ dτ P iτ-1/2 (y) ch(bτ) sech 2(πτ ) = ( 2 /π ) (1/ y-cosb ) cot -1[1+cosb / y-cosb ] (7.7) ∫0 ∞ dτ P iτ-1/2 (y) sh(bτ) th( πτ ) sech( πτ ) = ( 2 /π ) (1/ y+cosb ) tan -1[1-cosb / y+cosb ] (7.8) and this set includes all those needed to obtain results gi ven earlier in this paper. Along the way, the following facts are useful (the first is a corrected GR7 3.197.2 ) ∫u ∞ dx x-λ (x+ β)ν (x-u) µ-1 = u -λ ( β+u) µ+ν B( λ-µ-ν,µ) F( λ,µ; λ-ν; -β/u) (7.9) F(1,1/2; 3/2; z) = [ tanh -1z ] / z (7.10) (b) The integral X appearing in Section 4(a) 23 As a Mehler integral example, the integral (4.2) above has this form, X = ∫0 ∞ dτ P iτ-1/2 (y) ch(b τ) τ th( πτ ) sech(πτ ) // y = ch ξ and b = π-u0 which has an unpleasant τ factor in the integrand. This integral can be done as X = ∂bY where Y = ∫0 ∞ dτ P iτ-1/2 (y) sh(b τ) th( πτ ) sech(πτ ) X = ∂bY = ∂b { ( 2 /π ) (1/ y+cosb ) tan -1[1-cosb / y+cosb ] } // from (7.8) above Setting y = ch ξ and b = π-u0 we then find (after doing the above ∂b and some algebra) that X ≡ ∫0 ∞ dτ P iτ-1/2 (ch ξ) ch([ π-u0]τ) τ th( πτ ) sech(πτ ) = (2/ π) [sin(u 0/2) / (2ch ξ-2cosu 0) ] * [1 + ( 2 cos(u 0/2)/ ch ξ-cosu 0 ) tan -1 ( 2 cos(u 0/2) / ch ξ-cosu 0 ) ] which can be compared with the mysterious integral Lebede v et. al. give in Problem 501 (page 239) which uses α = ξ and β0= u 0 : (c) Where to find Mehler integrals (some errata noted) The largest source of such integrals known to the author is a n extremely obscure 1961 Boeing Report #246 written by no less than Professor Fritz Oberhettinger a nd coworker Higgins. Integral (7.8) appears for example as page 20 #3. Integral (7.6) above as page 20 #5 for y > cha, and the corresponding log form for y<cha has a typo in that the leading factor should b e 1/2 instead of 1/ 2 . Similarly, result (7.7) appears as page 20 #6 expressed as tan -1 and in this form a certain exponent has a wrong sign. Th at erroneous exponent also appears in PBM mentioned next // wrong 24 Using cot -1 = π/2 - tan -1 for the principle branch of the arc trig functions, (7.7) above becomes ∫0 ∞ dτ P iτ-1/2 (y) ch(bτ) sech 2(πτ ) = (1/ 2 ) (1/ y-cosb ) – ( 2 /π ) (1/ y-cosb ) tan -1[1+cosb / y-cosb ] (7.7A) which shows the correct placement of the radicals. It can be shown that 7.7A is only valid for |b| < π, whereas 7.7 is valid for |b| < 2 π. The second largest source is volume 3 of the special functi ons series of PBM which has about 20 integrals of Piτ-1/2 (y) against elementary functions and many more against s pecial functions. Bateman ET 2 has a grand total of 6 Mehler integrals, and GR7 has somewhat more. Perhaps there is some recent collection of Mehler integrals unknown to the aut hor. 8. Evaluating the toroidal bowl potential integral into eleme ntary functions Start with the form derived above V( ξ,u)= V 0 2 ch ξ - cosu ∫0 ∞ dτ P iτ-1/2 (ch ξ) ch[( π–u0)τ] ch[( π+u 0-u) τ] / ch 2(πτ ) (2.8) Expand the ch ch product to get V( ξ,u)= (V 0 / 2 )ch ξ - cosu ∫0 ∞ dτ P iτ-1/2 (ch ξ){cosh[(2 π-u) τ] + cosh[(2u 0-u) τ]} / ch 2(πτ ) (8.1) Use integral (7.7) above to find V( ξ,u)= (V 0 / 2 ) ( 2 /π ) ch ξ - cosu * { (1/ ch ξ -cos(2π-u) ) cot -1[1+cos(2π-u) / ch ξ -cos(2π-u) ] + (1/ ch ξ-cos(2u 0-u) )cot -1[1+cos(2u 0-u) / ch ξ -cos(2u 0-u) ] } (8.2) u0 in (0, π) u in (u 0, 2 π+u 0) ch ξ in (1, ∞) and we can if we like replace cos(2 π-u) by cosu in three places. Before one can do any Maple plots, one needs to understand the meaning of the numerator radicals appearing in the above formula. Consider again the underlying inte gral (7.7) ∫0 ∞ dτ P iτ-1/2 (y) ch(bτ) sech 2(πτ ) = ( 2 /π ) (1/ y-cosb ) cot -1[1+cosb / y-cosb ] (7.7) 25 Written in this form, the LHS converges for |b| < 2 π and so does the RHS and we can regard the equation as matching two functions analytic in parameter b near b = 0. If the RHS is written in tan -1 form, the RHS is only analytic for |b| < π, which is why we use the cot -1 form. As parameter b makes its excursion from b = 0 to b = 2 π, it has to pass by the value b = π, a place where 1+cosb = 0. The correct analytic continuation is to swap branches of 1+cosb when passing through an odd multiple of π, but this fact is not very obvious when one simply writes " 1+cosb ". Maple, for example, always assumes the positive branch. This continuation issue arises in both cot -1 numerators 1+cosu and 1+cos(2u 0-u) appearing above. To clarify the meaning of our integral evaluation, w e formalize this sign change as it occurs in both cot -1 numerators, by writing V( ξ,u)= (V 0/π ) ch ξ - cosu * { (1/ ch ξ -cosu ) cot -1[(-1) η1| 1+cosu | / ch ξ -cosu ] + (1/ ch ξ-cos(2u 0-u) ) cot -1[(-1) η2| 1+cos(2u 0-u) |/ ch ξ -cos(2u 0-u) ] } (8.3) where η1 = floor[(3 π-u)/2 π] u 0 in (0, π) η2 = floor[(2u 0-u+ π)/2 π] u in (u 0, u 0+2π) This form provides the correct analytic continuation of our RHS to any real value of u, even beyond the point where the LHS integral actually converges, though we only use it in the convergent region. In order to rigorously verify this analytic continuation, one h as to study the Riemann sheet structure of the function f(z) = 1 + cos(z) . One first looks at u(z) = cos(z) which is plenty complicated by itself, and then one has to add a square root branch cut to get to f(z). Here is the kind of picture one ends up having to ponder The black arrow in the z plane crosses Re(z) = an odd m ultiple of π, and we can trace this in black to an elliptical "ant trail" in the u-plane. But the green cut is that from the radical in 1 + cos(z) and to analytically continue one must dive through the green cut to a rrive at a "the other sheet" of this function, where it has the opposite sign. See Ahlfors's Chapter 3 on analytic mapping. 9. Using Maple to plot the bowl potential We use Maple because we have it; any computer algebra system will do, though the syntax changes slightly. The code text is given in Appendix A below, here we use screen clips. The first order of business is just to enter the evaluated bowl potential and set a f ew parameter values, 26 Next, we need a special routine which takes a point (x,y) on a circle and returns a tan -1 angle which lies in the range (0,2 π), where the angle is measured CCW away from the x axis. Maple has an internal function that does something like this, but we want to see wh at is happening, thus Lastly, but very importantly, we have to figure out how to c ompute toroidal parameter u from Cartesian coordinates ρ and z ( really cylindrical coordinates, but just think ρ = x). In the toroidal equations section above we wrote in (5.3) tanu = [ -2az/(a 2-ρ2-z2)] => u = tan -1[ -2az/(a 2-ρ2-z2)]] 27 but the letters "tan -1" do not indicate which branch of tan -1() we need to be on for a given value of ρ and z. This information is not present in the equation above. For that reason, we compute cosu and sinu separately from these equations ξ = tanh -1[2a ρ/(a 2+ ρ2+ z 2)] (5.3) ρ = ash ξ/(ch ξ–cosu) z = asinu/( ch ξ–cosu) (5.2) and then use the (0,2 π) arctan2Pi routine shown above to get the proper u. Finally, we have to adjust this value of u so it fits into the Lebedev-adjusted range (u 0, u 0+2 π). And we have to worry about all the special cases that tend to blow things up when ignored. So here then is a routine to compute u: where the type functions relate to an obscure evaluation q uirk of Maple and should be ignored. We used this getu function to plot the surfaces of u shown earlier: (you may have to stare at this picture for awhile, it is a camera shot from below the plane) 28 Finally, here is the code to plot the potential surface: 10. Potential and capacitance of a charged toroid. The solution to the charged toroid problem provides an inter esting comparison to the bowl problem. (a) The potential of a charged toroid 29 Looking back at the two toroidal atomic forms, we can see that for problems involving a toroid as boundary ( ξ = ξ0) we must be oscillatory in the other two coordinates u an d φ. Thus we select expo osc osc (1) ch ξ - cosu [P n-1/2 m(ch ξ), Q n-1/2 m(ch ξ) ] [ sin(nu),cos(nu)] [ sin(m φ),cos(m φ)] It is convenient for this problem to put bowl label u in the ran ge (-π,π) as suggested by this picture Unlike the situation with the bowl problem, the red path of u val ues around the grey toroid core is unobstructed so the potential must be periodic in u with per iod 2 π, and this fact causes quantization to integers of the parameter n appearing in the atomic form. Furthermore, we can see that with this range of u, the potential must be symmetric V( ξ,u) = V( ξ,-u), so only the cos(nu) term survives. Thus we quickly arrive at the following Smythian form for the potentia l of a charged toroid, V( ξ,u) = ch ξ - cosu Σn=0 ∞ P n-1/2 (ch ξ) A ncos(nu) (10.1) where A n are coefficients to be determined. As in the bowl case, the Q function is ruled out since it diverges at ξ = 0, whereas ξ → 0 is part of the condition for being "far away" where the pote ntial must vanish. (The other part of that condition is u → 0. ) The boundary condition of constant potential on the toroid of label ξ0 is this V0/ch ξ0 - cosu = Σn=0 ∞ P n-1/2 (ch ξ0) A n cos(nu) (10.2) Armed with knowledge of the Fourier series expansion and proj ection of 1/ a - b cos(x) 1/ a - b cos(x) = (1/ π) 2/b Σn=0 ∞ εn Qn-1/2 (a/b) cos(nx) // expansion ∫0 π dx cos(nx)/ a - b cos(x) = 2/b Q n-1/2 (a/b) // projection (10.3) (see GR7 p961 8.713.1 with µ = 0 and ν = n-1/2) and using ∫0 π du cos(nu)cos(n'u) = ( π/εn) δnn' where εn = 2-δn,0 = "Neumann's Factor" (Watson) 30 one can easily invert (10.2) to find coefficient A n An = V 0 ( 2 εn/π) [ Q n-1/2 (ch ξ0)/ Pn-1/2 (ch ξ0)] (10.4) and thus one arrives at the potential for a conducting charge d toroid, V( ξ,u) = V 0 ( 2 /π)ch ξ - cosu Σn=0 ∞ εn P n-1/2 (ch ξ) [Q n-1/2 (ch ξ0)/ Pn-1/2 (ch ξ0)]cos(nu) (10.5) which can be compared with Morse and Feshbach p 1304 ( ξ = µ and u = η ) // wrong where the εn factor has been erroneously omitted. I asked Mark Fesh bach if he knew of an errata collection for the massive 2000 page masterwork coauthored by his father, but he did not. (b) Capacitance of a toroid Since large r means ξ→0 and u →0, we can set P n-1/2 (ch ξ) = 1 and cos(nu) = 1 in (10.5) to get V( ξ,u) ≈ V 0 ( 2 /π)ch ξ - cosu Σn=0 ∞ εn [Q n-1/2 (ch ξ0)/ Pn-1/2 (ch ξ0)] (10.6) As in the bowl problem, we know from (4.11) that ch ξ - cosu ≈ 2 a / r in the "far limit", so V( ξ,u) ≈ V 0 ( 2 /π) 2 a / r * Σn=0 ∞ εn [Q n-1/2 (ch ξ0)/ Pn-1/2 (ch ξ0)] (10.7) Thinking of this as V = Q/r and using Q = CV 0 we find Q = ( 2V 0a/ π ) Σn=0 ∞ εn [Q n-1/2 (ch ξ0)/ Pn-1/2 (ch ξ0)] C = ( 2a/ π ) Σn=0 ∞ εn [Q n-1/2 (ch ξ0)/ Pn-1/2 (ch ξ0)] (10.8) Sometimes this result is expressed in terms of R and ρc shown here (R is the tube radius, ρc is the toroid radius to the center line of the tube) 31 Toroidal equations (5.5) show that ρc = a coth ξ0 R = a /sh ξ0 => ρc2 - R 2 = a , ρc/R = ch ξ0 (10.9) so that the C formula can be written C = ( 2/ π) ρc2 - R 2 Σn=0 ∞ εn [Q n-1/2 (ρc/R)/ Pn-1/2 (ρc/R)] (10.10) Setting ρc = 1, we can have Maple plot C as a function of R: R/ ρc = 1 describes a degenerate toroid where the hole just disappea rs (upper right of plot), and R →0 gives a limiting thin wire ring (lower left of plot). Both these limits are quite fascinating as discussed in some detail in Appendix B below. The degenerate toroid limit R=1 gives the mysterious valu e C = 1.7413. The author would love to know if this number is a simple function of π, small integers and simple roots (see Appendix B). A comparison of this degenerate toroid to a sphere which just encloses it (r=2R) 32 reveals these facts Ctoroid = 1.7413 // mystery number Csphere = 2.0000 ratio sphere/toroid = 2.0000/1.741 = 1.149 // C = (2R ) AREA toroid = 4 π2R2 // area = 4 π2Rρc AREA sphere = 16 πR2 ratio sphere/toroid = 4/ π = 1.273 // area = 4 π(2R) 2 So the sphere has 27% more area and 15% more capacitance th an the enclosed degenerate toroid. We expect the sphere to have more capacitance since it is the optimal shape for keeping the charges apart. As for the thin-wire-ring limit (left edge of the above plot), here is more detail (notice that the axes are now switched) If we go from R/ ρc = 10 -3 to R/ ρc = 10 -40 , capacitance drops only by a factor of 10, so one might say that a wire ring "holds its capacitance quite well" as t he wire gets thinner and thinner. Ultimately the capacitance goes to 0 as the wire vanishes out of existence. This can be compared to the capacitance of a metal sphere which is shrunk to a point, in which case C = R → 0 in a more reasonable fashion. 33 If a metal sphere carrying fixed charge Q is gradually s hrunk to a point, work must be done to get the charges closer together which raises the sphere's potenti al V relative to infinity ( E = CV 2/2 = QV/2). The charges are all piled on top of each other in the limit, so C = Q/V goes to 0 quickly. In the toroid case, the charges can stay away from each other to some extent by b eing spread out around the wire ring, so V rises more slowly as the ring is made thinner. [ The slowness of the thin ring limit approach is due to t he fact that the first term in the series for C dominates, and Q -1/2 (z)/ P-1/2 (z) → ( π2/2) (1/ ln[8z] ) as z →∞ , so C(R; ρc=1) ≈ π / ln(8/R). Inverting one gets R = 8 e -π/C so that lnR = ln8 - π/C which is the plot shown above. ] (c) Charge density on a toroid Using this formula ( plus sign because ξ decreases moving outward from the toroid surface ) σ = + (1/4 π) (1/h ξ) ∂ξV( ξ,u)| ξ=ξ0 1/h ξ = (ch ξ - cosu)/a (10.11) and the potential stated above V( ξ,u) = V 0 ( 2 /π)ch ξ - cosu Σn=0 ∞ εn P n-1/2 (ch ξ) [Q n-1/2 (ch ξ0)/ Pn-1/2 (ch ξ0)]cos(nu) (10.5) we get the following expression for the charge density σ on a charged toroid with label ξ0, (R = a/sh ξ0) σ(u) = (V 0 /4 π R) * (10.12) { ( 2 /π)( ch ξ0 - cosu )3 Σn=0 ∞ εn P n-1/2 '(ch ξ0) [Q n-1/2 (ch ξ0)/ Pn-1/2 (ch ξ0)]cos(nu) + (1/2) } where P n-1/2 '(z) means ∂zPn-1/2 (z) = (z 2-1) -1(n+1/2)[ P n+1/2 (z) - z P n-1/2 (z) ]. In the large z limit, it can be shown that σ = (V 0/4 π R)z 0 / ln(8z 0), so for a thin wire the linear charge density is λ = 2 πRσ and then the total charge is Q = 2 πρ c λ = π V 0 / R ln(8/R), consistent with the capacitance limit fou nd above when ρc=1. Here is a plot of σ(u) for ξ0 = π/4, a toroid marked in our bipolar coordinates picture above , where the horizontal axis shows u in units of π/8 and the height is relative units (left picture) One can convert from angle u to angle θ as shown using the following equations 34 sin θ = sh ξ0sinu/(ch ξ0-cosu) cos θ = (ch ξ0cosu-1) /(ch ξ0-cosu) (10.13) The main idea in the σ(u) sum is that near u = 0 the terms are additive since c os(nu) ~ 1, whereas in the "backward direction" especially near u = π there is term interference from cos(nu) ~ (-1) n, causing the charge to concentrate on the outer toroidal surface just as one would expect. This situation is akin to the forward peak in a scattering amplitude in partial wave an alysis. The plot was made from data computed in Maple using 20 terms and confirming term stability as discuss ed in Appendix B. Appendix A. Maple code text for plotting the Bowl Potent ial The following code can be cut from this document and pasted in to the Maple V R5 white worksheet window. It then runs when you hit the enter key and plots should a ppear. Hopefully this also works in more current versions of Maple. See author's Maple User's Guide for how to break code into separate execution groups. restart; # Construct the charged bowl potential in toroidal coordinates A := sqrt(cosh(xi)-cos(u)): B := sqrt(1 + cos(u)): C := sqrt(cosh(xi)-cos(2*u0-u)): E := sqrt(1+cos(2*u0-u)): eta1 := floor((3*Pi-u)/(2*Pi)): eta2 := floor((2*u0-u+Pi)/(2*Pi)): T1 := (1/A)*arccot(((-1)^eta1)*(B/A)): T2 := (1/C)*arccot(((-1)^eta2)*(E/C)): V := (V0/Pi)*A*(T1+T2); V0 := 1: a := 1: u0 := evalf(2*Pi/8); # Routine arctan2Pi # Given (x,y) somewhere on a circle, return the ang le in (0,2Pi) measured CCW from the # axis. # Warning: returned result may include unevaluated multiples of Pi arctan2Pi := proc(x,y) local q; if type(x,numeric) and type(y,numeric) then if x = 0 and y = 0 then print("arctan2Pi(0,0) error." ); RETURN(0) fi; if x = 0 and y > 0 then RETURN(Pi/2) fi; if x = 0 and y < 0 then RETURN(3*Pi/2) fi; if x > 0 and y = 0 then RETURN(0) fi; if x < 0 and y = 0 then RETURN(Pi) fi; if x > 0 and y > 0 then q := 0 fi; if x < 0 and y > 0 then q := Pi fi; if x < 0 and y < 0 then q := Pi fi; if x > 0 and y < 0 then q := 2*Pi fi; RETURN(arctan(y/x)+q); else 'arctan2Pi(x,y)'; fi; end: # Routine getu # Given a,u0, rho,z, compute toroidal bowl-label pa rameter u in range (u0,u0+2Pi) 35 getu := proc(rho,z) global a,u0; local xi,cosu,sinu,t1; if type(rho,numeric) and type(z,numeric) then xi := arctanh(2*a*rho/(a^2+rho^2+z^2)); if rho = 0 then cosu := (z^2- a^2)/ (z^2+a^2); sinu := 2*a*z/(z^2+a^2); else cosu := cosh(xi)-(a/rho)*sinh(xi); sinu := (z/rho)*sinh(xi); fi; t1 := evalf(arctan2Pi(cosu,sinu)); if t1 < u0 then t1 := t1 + evalf(2*Pi) fi; # get into proper range RETURN(t1); else 'getu(rho,z)'; fi; end: # Plot the potential of the charged bowl (an azimut hal slice) xi := arctanh(2*a*abs(rho)/(a^2+rho^2+z^2)): u := getu(rho,z): plot3d(V, rho = -2..2, z = -2..3,numpoints=2000,axe s=BOXED, view=0..1); u := getu(rho,z): plot3d(u, rho = -3..3, z = -4..4,axes=BOXED,numpoin ts=2000); Appendix B: Comments on computing the capacitance of a toroid and its degenerate limit. The capacitance formula (10.10) given above is this, C = (2R/ π) z2 - 1 Σn=0 ∞ εn [Q n-1/2 (z)/ Pn-1/2 (z)] z = ch ξ = ρc/R (B.1) It is convenient to remove the leading factor and talk about this function T(z) ≡ z2 - 1 Σn=0 ∞ εn [Q n-1/2 (z) / Pn-1/2 (z)] (B.2) Basically this is a "special function" that has no nam e. We can call it the "toroidal capacitance function". It is defined for our purposes on the range z in (1, ∞). One can see from this fact Qν(z)/ P ν(z) ≈ π { Γ(1+ ν)2/ [ Γ(ν+3/2) Γ(ν+1/2)] } (2z) -2ν-1 (B.3) that the series converges for z > 1. It is a simple matte r to have Maple compute T(z) for a variable number of terms to make sure enough terms are included to give a stabilized result. As z →1, the number of required terms increases (Q ν(z) has a log singularity at z=1). The magic question is t o find the limit T(z=1), it if exists. This is the limit which determine s the capacitance of a toroid for which the hole has just gone out of existence. This physical interpretation sugge sts that the limit must exist. We use appropriate definitions of the P and Q functions as fo llows, and do some cursory checks: 36 In order to study the behavior of sums involving P and Q func tions, we first considered this "sum rule" (which is the 10.3 expansion above with x = 0, b = 1 and a = z) R(z) ≡ z-1 Σn=0 εn Q n-1/2 (z) = ( π/2 ) = 2.221441469 (B.4) Our Maple program was able to obtain the correct sum to thr ee decimal places down to z = 1.00001 where it had to add 2000 terms. Because this series is in a sense only half as convergent as our intended Q/P series, the partial sums (left) and terms (right) ex hibited characteristics of an asymptotic series, for example [ the horizontal axis shows the number of terms a dded ] 37 The series R(z)/ z-1 really is truly convergent, and this Maple behavior aris es (we think) from the way Maple computes the hypergeometric function for very large para meters and z near 1, an area known to be problematic. The Q/P series partial sums (below) did not e xhibit this asymptotic series behavior. After this warmup exercise, we turned to the T(z) series (B.2) and computed it the same way, always with a stable number of terms. Our results were as follows: z T(z) terms used -log 10 (z-1) 1.1 2.8465 15 1 1.01 2.7 987 140 2 1.001 2.7 414 300 3 1.0001 2.73 59 600 4 1.00001 2.735 3 1200 5 1.000001 2.7353 4000 6 (B.5) suggesting that we are approaching a limit close to 2.7353 for the value of T(1). We can plot the above points as follows: where the x axis is -log 10 (z-1). The point is that we are coming into the limit T( 1) with zero slope. 38 Using the Wronskian of P and Q, one can show that this slope is given by T'(z) = { T(z) – z2 - 1 Σn=0 ∞ εn /[P n-1/2 (z)] 2 } / (z 2-1) (B.6) If we conjecture that T'(1) is not infinite, which the pict ure certainly suggests, we conclude that T(1) = lim z→1{ z2 - 1 Σn=0 ∞ εn /[P n-1/2 (z)] 2 } (B.7) and we have then a second way to compute T(1). This series is more stable because it avoids the ln(z-1) singularity of the Q functions. Using this series Maple yielde d the following results z = 1.1 above = 2.898517346 z = 1.01 above = 2.7 51570937 z = 1.001 above = 2.73 6973437 z = 1.0001 above = 2.735 058209 z = 1.00001 above = 2.7353 62078 z = 1.000001 above = 2.73535 4499 10,000 terms z = 1.0000001 above = 2.735353 885 40,000 terms (B.8) We verified most of these results by jamming strings like the following into the Wolfram Alpha website computation window ( the argument 3 indicates off-the-cut Legendre functions in Mathematica) sqrt(1.0000001^2-1)*(2*sum(1/LegendreP(n-1/2,0,3,1.0000001)^2,n=1 to 39999) + 1/Le gendreP(- 1/2,0,3,1.0000001) ^2) Though it timed out on this case, it did all the other c ases listed, confirming Maple's results. Our conclusion is the following strange fact T(1) ≈ 2.735353 (B.9) and therefore we find that the capacitance of a degenerate toroid is given by C(ρc= R = 1) = (2/ π)T(1) ≈ 1.741380 (B.10) It is interesting that T(1) is slightly larger than e ≈ 2.718282, but for sure T(1) ≠ e. We do not know how to obtain this limit in any other way. We know of no integr al representations of inverse Legendre functions, nor of any published series involving them, though PBM ha s very many series involving Legendre functions and products of same combined with other fu nctions. Perhaps a variation for z ≥1 of Bateman HT I p 149 (11) could be used. We sometimes read about the interchange of limits giving di fferent results in situations where uniform convergence is lacking in both limits, or where one of the li mits does not exist. We have here exactly a case of this second situation: lim z→1 lim N→∞ { z2 - 1 Σn=0 N εn /[P n-1/2 (z)] 2 } = T(1) = 2.735353 39 whereas lim N→∞ lim z→1 { z2 - 1 Σn=0 N εn /[P n-1/2 (z)] 2 } = Σn=0 ∞ εn limz→1 { z2 - 1 } since lim z→1Pν(z) = 1 = ∞ * 0 The situation is even worse with the P/Q series since li m z→1Qν(z)/P ν(z) = ∞/1 = ∞. Appendix C: Kelvin's approach to the Charged Bowl Probl em. Lord Kelvin (William Thomson) published his own Collected Works i n 1872 and again in 1884 and the latter is the form available on the web as a Microsoft d igitized document. The Contents gives section numbers, not page numbers. There are 673 sections filling s ome 600 pages, and the section numbers just increment through all the many Collected papers. In the first collected paper (page 1, the ellipsoid paper, " On the Uniform Motion of Heat...", 14p, 1842) Kelvin derives the potential of and charge distribution on a charged conducting ellipsoid and gets the correct result (2.1) above. He bases things on a certai n geometric distance p which turns out to be proportional to σ on the ellipsoid: p = 1/ x2/a 4 + y2/b 4+ z2/c 4 and σ = Qp/(4 πabc) where a,b,c are the semimajor axes and Q is the total charge on the ellipsoid (see McDonald's paper). His method is very unusual, there is no formal ellipsoidal coordinate system he re, but at age 18 he did the whole thing by brute force. He wanders around between the heat, gravitationa l and electrostatic manifestations of potential theory, but is mostly concerned with heat. This brings us then to his bowl paper (paper XV, page 178, section s 231-248, "Determination of the distribution...", 14p, 1869). He opens by quoting the above σ formula (he refers to σ by the symbol ρ) and then specializes it first to an elliptical plate then t o a circular disk, noting the disk capacitance C = 2a/ π. He finds that σdisk (ρ) = [Q/(4 πa)] (1/ a2-ρ2 ) on each side, where ρ is our modern-day cylindrical coordinate and a the disk radius. This is the key result upon which his bowl theory is built. Kelvin notes retrospectively that George Green obtained this result in 1832. (See green Jackson 3.179 for σ both sides.) Deviating slightly from Kelvin's order of presentation, we tack sideways and ponder the inversion of his charged disk into a grounded spherical bowl with a point charge located on the bowl's cap. The point charge appears at the inversion origin in the bowl space whe n one adds a constant potential in the disk space to bring the disk potential down to 0 so it will map into a grounded bowl in the bowl space. By shifting the charged disk off the symmetry axis, the point charge can be made to appear at any desired point on the bowl's cap: 40 Since Kelvin knows the disk's charge density σdisk (ρ) just stated above, he knows by inversion σ on the bowl. He does not mention the disk's potential, but that maps i nto the bowl potential by inversion as well. So in the bowl space, we have the grounded bowl with a point charge on the cap, and we know σ on the bowl. Since σdisk is the same on both sides of the disk, σ is the same on both sides of the bowl. Basically, the bowl potential is the Green's Function for the bowl wi th a restricted placement of the Green's point charge. Kelvin then does two superpositions. He first superposes an infinite number these Green's Function situations to obtain a target uniform charge density - σ0 on the bowl's cap. In doing this the point charge has to be pro perly scaled for each point in the integration as it is made to wander over the ca p region. The corresponding bowl σ's are also being superposed in this process. One then ends up with the fin al σ on the bowl (let's call it σ2, same on both surfaces) as a superposition integral which can be ev aluated into trig and inverse trig functions. One can interpret this σ2 as the charge induced on each surface of a grounded bowl due to the presence of uniform - σ0 on the cap. This induced charge on the bowl is brought in from i nfinity on the traditional "infinitesimally thin wire" grounding the bowl to The Great Metal Sphere at Infinity. This cap of charge density - σ0 is what we call "sticky charge". It is an infinitely t hin layer of charge that is magically glued in place so it cannot move, just as the point charge in a Green's function problem is glued to its location and cannot move. In a second superposition , Kelvin superposes Problem A and Problem B to get Problem C: Problem A is the grounded conducting bowl of radius R in the pr esence of a cap of uniform sticky charge -σ0 as just described above. The bowl is at V=0 and has σ2 on both surfaces. Problem B is the same conducting bowl carefully prepared to have a uniform free charge density + σ0 on its outer surface along with a cap of sticky charge also of + σ0. Since we have then a full sphere of uniform σ0, the potential on and inside the entire bowl sphere is V 0 = Q/R = ( σ04πR2)/R = 4 πRσ0 = constant. By the same argument used for a full conducting sphere, this bowl must have σ = 0 on its inner surface. One might prepare this Problem B by first shrink-wr apping a neutral conducting bowl's sphere with a full shell of sticky σ0. One then unglues the sticky charge covering just the bowl part of the sphere. The released charges don't move because there is no tangent ial E field to make them move. 41 Each of these two problems represents a valid solution to t he Laplace equation in the presence of the same conductor. Kelvin forms a third solution by superposing these two problems as Problem C. In this superposed Problem C one has: (1) V = 0 + V 0 = 4 πRσ0 on the bowl ; (2) σin ≡ σ2 + 0 = σ2 on the bowl's inner surface; (3) σout ≡ σ2 + σ0 on the bowl's outer surface; (4) exact cancellation of the st icky charge on the cap. Notice that σout – σin = σ0 = V 0/(4 πR) = a constant. But Problem 3 is recognized as exactly our "charged bowl problem " for a bowl with potential V 0, and we have just shown that σout – σin = σ0 = V/(4 πR) = constant, which is the famous result. And of course we know σin = σ2 which came with Problem A. The underlying fact is that a Laplace solution is unique, so if your find something that works, that is the answer. Kelvin's result for σin = σ2 is stated in (4.9) or (4.11) above. Kelvin uses his hard-won formulas to compute σin and σout at several locations on spherical bowls of various shapes. He does this to 5 decimal places and shows the results in a full page graphic. Had he stopped here, he would have had a great paper, but he had more to say. Kelvin now knows all about "the charged bowl" problem. Just as h e started with a "charged disk" and obtained by inversion the Green's function for a bowl, he now sta rts with the "charged bowl" and inverts it into either another bowl or a disk, and this time the "inversion-generated point charge" can be made to appear anywhere relative to that bowl or disk. Here are his drawings for these two cases The inversion sphere is shown dashed, and the point charge is at the inversion origin Q. Thus Kelvin has obtained (in theory) the fully general Green's Function for a bowl or a disk. His paper only discusses the charge densities, but the method applies as well to the poten tial. One can start with the known disk potential (green Jackson p 92 (3.178) with typo corrected) Vdisk (ρ,z) = (q/a) sin -1 [ 2a/ [ (ρ-a)2+z 2 + (ρ+a)2+z 2 ] (C.1) and process it through all of Kelvin's steps above. Appendix D: Smythe's approach to the Charged Bowl Problem . In the Kelvin discussion above we saw an inversion of the on-c ap bowl Green's Function problem to an off-axis charged disk problem. If the bowl and its on-cap poi nt charge are rotated together so the bowl surface touches the inversion origin, the same bowl Green's F unction problem (with the same inversion 42 origin and inversion sphere) can be inverted into an iris Green' s Function problem with the point charge in the hole of the iris ("iris" being an infinite grounded conduc ting plane containing a circular hole). With the bowl in this position, the bowl's cap maps into the hole in the iris and a point charge on the cap maps into a point charge in the hole of the iris: If one could somehow solve this iris Green's Function problem , one could thereby gain full knowledge of the on-cap bowl Green's Function problem, and one could then car ry out Kelvin's two superpositions described in the previous Appendix and thereby obtain the potentia l and charge densities for the charged bowl problem. This is the approach taken by Smythe in an intriguing set of Problems (38 through 42 starting on page 203 of his 2nd Edition book). The starting point is to figure out the Green's Function for the iris with point charge in the hole, and that can be done by inverting the charge d disk using the following planar inversion arrangement (the point charge appears in the iris space at the inversion origin = red crosshair, a distance S from the center of the iris hole) If the iris space has primed coordinates, the solution to thi s inversion problem, starting with the Jackson charged-disk potential quoted above, is as follows: V'( r') = 2q 1/( πr1) cos -1 [ 2 B r1 / { m + (B2-S2) [ (ρ'-B)2+z' 2] [ (ρ'+B)2+z' 2] } ] (D.1) where m = ( ρ'2+z' 2)(S 2-B2) + B 2(B 2- S 2 + 2r 12) 43 r 12 = ρ'2 + S 2 - 2S ρ'cos θ' + z' 2 θ' = 0 in direction of the point charge σ'( r') = - [ q 1 / (2 π2r12) ] B2-S2 / ρ'2-B2 // either side of the iris, z'=0 in r 1 (D.2) where q 1 is the size of the point charge in the hole and B is the ra dius of the hole. The expression for σ' here is remarkably simple and only appears after considerable br ute-force Maple algebra which secretly implements all those Kelvin geometry theorems. In Problem 38 Smythe asks his reader to come up with σ' as shown above. In Problem 39 he asks the reader to integrate the point charge problem aro und a circle, to obtain the charge density on the iris for a circular ring of charge cen tered in the hole. That result is still amazingly simple: σ2(ρ) = - (q/2 π2) ( B2-S2 / ρ2 - B 2 ) / ( ρ2-S2) // either side (D.3) where now we remove the primes, and q is the total charge on the ring in the hole. In Problem 40 we are instructed to invert this grounded iris + ring cha rge in hole into a grounded bowl + ring charge on the cap (see first picture above). Then in Problem 42 we do a weighted integral of this bowl-cum-ring situati on to obtain a uniform charge density on the cap set to the target amount - σ0. This corresponds to Kelvin's first superposition described above. We then do Kelvin's second superposition and out pop all the charged bowl results. Going back now, in Problem 40 Smythe tells his reader to use "Green's Reciprocation Theor em" to find the potential anywhere on the cap of a charged bowl of potenti al V 0. This theorem concerns the charges and potentials on a set of conductors in two "situations" , one primed and one unprimed, Σi V i Q i' = Σi Vi' Q i. In our application there are only two metal objects, o ne is the bowl, the other is an imagined fine wire in the location of the ring of charge on the bowl's cap. One situation is a charged bowl, the other is the bowl + ring-on-cap Green's function. One quickly finds th at V( θ) = V 0(2/ π)sin -1[cos( α/2)/cos( θ/2)], (D.4) where θ and α are shown in our drawing above equation (4.10). This result has great significance which is brought out in Problem 41 . One now knows that V = V 0 on the bowl, and V = V( θ) on the cap, so we have a fully prescribed Dirichlet problem and we can then obtain the potential everywhere by assuming an appropriate Smy thian form and inverting to find the coefficients. For the exterior problem that atomic superpos ition is V(r, θ) = Σn=0 ∞ A n r -n-1Pn(cos θ) and the usual Legendre inversion gives An = R n+1 (2n+1)(1/2) ∫0π Pn(cos θ) sin θ V(R, θ,φ) d θ = V 0 Rn+1 (2n+1)(1/2) {(2/ π) ∫0α dθ sin θ Pn(cos θ) sin -1 [ cos( α/2) /cos( θ/2) ]) + ∫απ dθ sin θ Pn(cos θ) } (D.5) where R is the radius of the bowl's sphere. We find that A 0 = V0 (R/π) { π - α + sin α }. For large r, we have V → A 0/r so A 0 is in fact the total charge on the bowl as seen from far away, and we find that capacitance is C = R (1/ π) {π - α + sin α }, in agreement with (4.15) above. 44 We close this appendix with a remark on a related problem which is the Green's Function problem for a disk with the point charge in the plane of (and outside) the disk. One can solve this disk Green's function problem using an inversion picture like this where R' space contains a charged disk, and R space holds our desired Green's function problem for the disk with point charge at the marked origin. Note that the point charge is distance c from the center of the grounded disk. When this problem is solved, one finds a striking r esemblance between its solution and that of the grounded iris with point charge in the hole. We now compare these two problem solutions the disk problem the iris problem Vdisk(ρ,θ,z) = q1 (2/ πr1) cos -1 { 2 R r1 / m + (c2-R2) [ (R-ρ)2 + z 2] [ (R+ ρ)2 + z 2] } m = ( ρ2+z 2)(c 2-R2) + R 2(R 2- c 2 + 2r 12) r 12 = ( ρ2 + c2 + z2 - 2cρ cos θ ) (D.6) σdisk (r) = - q 1(1/(2 π2r12) c2-R2 /R2-ρ2 // each side r 12 = ( ρ2 + c2 - 2cρ cos θ ) (D.7) 45 total charge induced on the disk = - q 1 (2/ π) tan -1(R/ c2-R2 ) Viris (ρ,θ,z) = q1(2/ πr1) cos -1 [ ( 2 Rr1 / m + (R2- c2) [ (ρ-R)2+z2] [ (ρ+R)2+z2] ] m = ( ρ2+z2)(c 2-R2) + R2(R2- c2 + 2r 12) r 12 = ( ρ2 + c2 + z2 - 2cρ cos θ ) (D.8) σiris (ρ,θ) = -q1[ 1/(2π2r12)] R2-c2 / ρ2 - R2 // each side r 12 = ( ρ2 + c2 - 2cρ cos θ ) (D.9) total charge induced on the iris = -q 1 It turns out that the two problems are related by analy tic continuation of the variable ρ from ρ < R for the disk problem to ρ > R for the iris problem. The path going directly through ρ = R is blocked by a branch cut of f( ρ) = (ρ-R)2 + z 2 = (ρ - a +)( ρ - a -) = ρ - a + ρ - a – joining the points a ± = R ± iz, so one must continue around either of the branch points with the r esult that (ρ-R)2 + z 2 → – (R-ρ)2 + z 2 , and this is the only difference in the disk and iris sol ution sets shown above. Using the iris Green's Function potential (D.8), one can compute an electric field line by computing n^ = ∇∇ ∇∇V and tracking it in space. In this Maple plot, starting points on the iris were selected by a random number generator, and all roads lead to the Green's po int charge in the hole: As expected, each field line launches itself at right a ngles to the iris surface. 46 Appendix E: Dual Equations and their connection to th e Charged Bowl Problem Sneddon uses the term "dual relations" to cover both dual inte gral equations and dual series equations, but we shall just call them "dual equations". We can represen t a pair of dual equations this way: AΨ = f f = (f 1, f 2) presented as [A Ψ]1 = f 1 on I 1 BΨ = g g = (g 1, g 2) presented as [B Ψ]2 = g 2 on I 2 (E.1) Here A and B are invertible linear operators (perhaps Han kel Transform integral operators) and the other symbols stand for functions of a real variable x. The ran ge I of the variable x is partitioned into two regions I 1 and I 2 so the full interval of interest is I = I 1∪I2. The notation f = (f 1, f 2) means that f(x) = f1(x) on I 1 and f(x) = f 2(x) on I 2. The dual equation problem is easily stated: given the pres cribed functions f 1 and g 2 , find the partner functions f 2 and g 1 (so that you then know f and g on all of I), and also find the solution function Ψ(x) on all of I. Although A and B are invertible operators, neither dual eq uation can be inverted because information is missing. For example, one could write Ψ = A -1f, but we don't know f, we only know f 1. The same is true for the second equation: we can write Ψ = B -1g, but we only know g 2. Only by considering both equations of the dual pair can a solution be found. In elect rostatics problems, one usually has a Dirichlet boundary condition on interval I 1 and a Neumann boundary condition on I 2, so a dual equation really represents a mixed boundary value problem. Here is the formal trick used to solve the problem. Althou gh A and B are operators, it is useful to think of them as matrices. And it is useful further to thi nk of these matrices as consisting of submatrices so that the row and column spaces are partitioned in the sen se of I = I 1∪I2 . Then for example we could say A Ψ = f ↔   A 11 A 12 A 21 A 22   Ψ1 Ψ2 =   f1 f2 (E.2) The game is to find invertible lower and upper triangular mat rices (really operators) L and U such that LA = UB. It might seem at first that the existence of s uch L and U would be dubious, but an analysis of matrix decomposition theorems shows that in general such L and U do in fact exist. If we apply L to our first equation in (E.1) and U to the second equation, and i f we define S ≡ LA = UB (E.3) (which S will also be invertible) then our dual equations bec ome SΨ = Lf SΨ = Ug (E.4) We can then examine the right hand sides of these two equati ons in our matrix language Lf =   L 11 0 L 21 L 22   f1 f2 =   L11 f1 L21 f1+ L 22 f2 47 Ug =   U 11 U 12 0 U 22   g1 g2 =   U11 g1+ U 12 g2 U22 g2 Therefore, since both left hand sides are S Ψ, we find that Lf = Ug =   L11 f1 U22 g2 (E.5) where recall that we know L and U, and we know f 1 and g 2 since they are the prescribed functions (the driving terms on the RHS of the dual equations, often one of the se is 0 ). The solution to the problem is then obtained as follows. First take either S Ψ equation in (E.4) (say the first one) and invert to get Ψ = S -1 (Lf) ↔   Ψ1 Ψ2 =   S -1 11 S -1 12 S -1 21 S -1 22   L11 f1 U22 g2 (E.6) Since we know L,U,S,f 1,g 2, we have solved the problem for Ψ. To obtain the unknown partner functions we use the original equations: AΨ = f ↔   A 11 A 12 A 21 A 22   Ψ1 Ψ2 =   f1 f2 so f 2 = A 21 Ψ1 + A 22 Ψ2 BΨ = g ↔   B 11 B 12 B 21 B 22   Ψ1 Ψ2 =   g1 g2 so g 1 = B 11 Ψ1 + B 12 Ψ2 (E.7) Hopefully we have clarified the "basic idea" with this li ttle matrix viewpoint summary. In practice there is of course a lot of fine detail. An illustration of the nature of this det ail appears in Appendix G below. It might be noted that Sneddon also considers Triple Equati ons and the above analysis then involves 3x3 matrices since the variable range is partitioned into I = I 1∪I2∪I3 . An example of such a problem is the "charged barrel problem" where barrel means a spherica l shell with two polar caps removed. As examples of dual equations, we can take a quick look at t wo famous problems. The Beltrami unit disk problem . (Beltrami's 1881 work is reviewed by Sneddon.) An appropriate Smythian form for the potential can be constructed from cylindrical atoms: V( ρ,z) = ∫0 ∞ dk e -k|z| J 0(k ρ) [k -1 a(k) ] (E.8) The prescribed potential on the disk is f 1(ρ), and the prescribed charge density is σ ~ ∂zV = 0 outside the disk in the z=0 plane. Again, this is a "mixed boundary va lue problem". The dual equations are then ∫0 ∞ dk k -1J0(k ρ)a(k) = f 1(ρ) ρ<1 A Ψ = f ∫0 ∞ dk J 0(k ρ) a(k) = 0 ρ>1 B Ψ = g // g 2 = 0 (E.9) 48 Here Ψ(k) = a(k), the function to be solved for. A is an integral operator with kernel K( ρ,k) = k -1J0(k ρ) while B has kernel J 0(k ρ). The matrix sense of A is A ρ,k = K( ρ,k) where both indices of A are continuous real numbers, and similarly for B. Both A and B can be regarded as modified Hankel transforms. The formal matrix method above gives us a(k) which we insert into the Smythian form to find solution V( ρ,z). The partner functions are f 2 which is the potential outside the disk in the z=0 plane, and g 1 which is the charge density on the disk (sum of both sides). The appropriate L and R operators for this problem turn o ut to be gussied-up Abel transform operators (called I and K by Sneddon) known as Erdélyi-Kober operators. These operators can be interpreted as fractional integral operators (fractional meaning α is continued off the integers; the disk problem uses α = 1/2) , Rα{ f(t); x } = (1/ Γ(α)) ∫0 x dt' f(t') (x-t')α-1 Riemann-Liouville 1850 Wα{ f(t); x } = (1/ Γ(α) ∫x ∞ dt' f(t') (t'-x) α-1 Weyl 1917 (E.10) but the fact is that they are really just generalized Abe l transforms with fancy names. Like any respectable transform, the Abel transform is invertible (see Appendix F). The disk problem is then solved as outlined above using the Abel transforms L and U and the Hankel trans forms A and B. Setting f 1(ρ) = V 0= 1 of course gives the "charged conducting disk problem". The Corresponding Bowl Problem. An appropriate Smythian form for the unit-radius bowl poten tial, inside and outside the bowl, can be constructed from spherica l atoms, Vi(r, θ) = Σn=0 ∞ an r n Pn(cos θ) Vo(r, θ) = Σn=0 ∞ an r -n-1Pn(cos θ) (E.11) The prescribed potential is f 1(θ) on the bowl, and the prescribed charge density is σ ~ ∂rV = 0 on the cap. The dual equations are these: ( θ0 is the polar angle which defines the circular lip of the bowl) Σn=0 ∞ Pn(cos θ) an = f 1(θ) θ in (0, θ0) AΨ = f Σn=0 ∞ (2n+1) P n(cos θ) an = 0 θ in ( θ0,π) BΨ = g // g 2 = 0 (E.12) Now we have Ψn = a n and the matrix sense of A is A θ,n = P n(cos θ) where θ is a continuous angle index and n is a discrete index. As in the previous problem, A is an infinite x infinite matrix. Vinogradov et. al show (Section 1.4.3, "Noble's method") that for our bowl probl em the appropriate L,U operators are L h( θ) = (1/ 2 ) ∂θ ∫0 θ dφ sin φ h( φ) /cos φ-cos θ // = K 1 h( θ) R h( θ) = (1/ 2 ) ∫θ π dφ sin φ h( φ) /cos φ-cos θ // = K 2 h( θ) (E.13) where we note that the Volterra θ endpoints cause these operators to have triangular matr ices as kernels. If we apply L to the first equation above and R to the second, we get (after some work) 49 Σn=0 ∞ cos[(n+1/2) θ] a n = L f 1(θ) θ < θ0 or S Ψ = Lf on I 1 Σn=0 ∞ cos[(n+1/2) θ] an = 0 θ > θ0 or SΨ = Ug = 0 on I 2 (E.14) where both left hand sides are now of the form S Ψ as discussed above in (E.4). In the case of the charged bowl, f 1(θ) = V 0 = 1, we find that L f 1(θ) = L 1 = cos( θ/2). Then using (G.4), ∫0 π dθ cos([n+1/2] θ) cos([m+1/2] θ) = δn,m ( π/2) (E.15) one can trivially invert the equations (E.14) treated as a si ngle equation on (0, π) to get Ψ = S -1 Lf , or, an = (2/ π) ∫0 θ0 dθ cos([n+1/2] θ) cos( θ/2) = (1/ π) { sin(n θ0)/n + sin[(n+1) θ0] /(n+1) } (E.16) which is the famous result. In Sneddon's treatment, the P n(cos θ) are replaced by Jacobi polynomials P n(α,β)(cos θ) so the bowl problem is seen to be a special case with α = β = 0. Appendix F: The Abel Transform in various forms Here we just state the generalized Abel Transform in th ree different ways. The historical Abel transform problem (tautochrone) involves α = 1/2, while "generalized" allows 0 < α < 1. The Abel transform can be expressed for α outside this range by doing one parts integration for each in teger step of shift required (Sneddon). A derivation of the Abel Transform is given in Sneddon Section 2.3 and Vinogradov Section 1.5 in terms of a generic monotonic speed function h(u) where de nominators are [h(x) - h(t)] α . In the S1 forms, the upper endpoint is the variable, while in the S2 forms it is the lower endpoint. For the linear form we have explicitly done the ∂t derivatives on the third line of each transform. The starting point is to consider g(x) = ∫a x dt k(x-t)f(t) and assume that the kernel can be written as k(x-t) = ∂x s(x-t) = – ∂t s(x-t). This allows a parts integration which moves the deri vative off the kernel and onto the function f(t), giving these general results, g(x) = ∫a x dt k(x-t)f(t) => ∂x g(x) = f(a) k(x-a) + ∫a x dt k(x-t) f '(t) g(x) = ∫x b dt k(t-x)f(t) => ∂x g(x) = – f(b) k(b-x) + ∫x b dt k(t-x) f '(t) Linear Form of the Generalized Abel Transform: S1: ∫a x dt f(t) / [x - t] α = g(x) 50 => f(t) = (1/ π)sin( πα ) ∂t { ∫a t du g(u) / [t - u]1-α } (F.1) = (1/ π)sin( πα ) { g(a) / [t -a] 1-α + ∫a t du g'(u) / [t - u] 1-α } S2: ∫x b dt f(t)/ [t - x] α = g(x) => f(t) = – (1/ π) sin( πα ) ∂t { ∫t b du g(u) / [u - t]1-α } (F.2) = – (1/ π) sin( πα ) { - g(b) / [b - t] 1-α + ∫t b du g'(u) / [u - t] 1-α } Quadratic Form of the Generalized Abel Transform: S1: ∫a x dt f(t) / [x 2 - t 2]α = g(x) => f(t) = (2/ π)sin( πα ) ∂t { ∫a t du u g(u) / [t 2- u 2]1-α } (F.3) S2: ∫x b dt f(t) / [t 2 - x 2]α = g(x) => f(t) = – (2/ π) sin( πα ) ∂t { ∫t b du u g(u) / [u 2- t 2]1-α } (F.4) Trig Form of the Generalized Abel Transform: S1: ∫a x dt f(t) / [cos(t) - cos(x)]α = g(x) => f(t) = π-1sin( πα ) ∂t { ∫a t du sin(u)g(u) / [cos(u) - cos(t)] 1-α } (F.5) S2: ∫x b dt f(t) / [cos(x) - cos(t)] α = g(x) => f(t) = – π-1sin( πα ) ∂t { ∫t b du sin(u)g(u) / [cos(t) - cos(u)] 1-α } (F.6) and this last case we write again with α = 1/2 S1: ∫a x dt f(t) / cost - cosx = g(x) => f(t) = π-1 ∂t { ∫a t du sinu g(u) / cosu - cost } // Sneddon 2.3.5 (F.7) S2: ∫x b dt f(t) / cosx - cost = g(x) 51 => f(t) = – π-1 ∂t { ∫t b du sinu g(u) / cost - cosu } // Sneddon 2.3.6 (F.8) When α = 1/2, the trig form always involves factors of the form 1/ cosa-cosb . There is doubtless a deeper (perhaps group theoretic) explanation, but due to the f ollowing integral representations of the P function P n(cos θ) = ( 2 /π) ∫0 θ dφ cos[(n+1/2) φ]/ cos φ-cos θ // GR7 8.823 (F.9) P n(cos θ) = ( 2 /π) ∫θ π dφ sin[(n+1/2) φ] / cos θ-cos φ // θ→π -θ, φ→π -φ there is an close connection between Legendre functions and Ab el transforms in the solution of dual series equations of the type discussed in Appendix E. The int egrals (F.9) are called the Mehler-Dirichlet integrals for P n and they are derived in Section 15.231 of Whittaker and Watso n. Corresponding Bessel J 0 representations also expose this Abel transform connection (though in the Abel quadratic form) J 0(k ρ) = (1/ π) ∫0 ρ dx cos(kx) / ρ2-x2 // Vinogradov 1.156 J 0(k ρ) = (1/ π) ∫ρ ∞ dx cos(kx) / x2-ρ2 // Vinogradov 1.157 (F.10) Here are two interesting expansions which in certain ci rcumstances allow for the solution of dual equation problems by the use of two sequential Abel transforms. The first operates in the Legendre wo rld, the second in the Bessel world: Σn=0 ∞ Pn(cos θ) Pn(cos θ') = (1/ π) ∫0 min(θ,θ') dφ { 1/ cos φ-cos θ 1/ cos φ-cos θ' } (F.11) ∫0 ∞ Jn(rx) J n(r'x) dx = (2/ π) (rr') -n ∫0 min(r,r') ds s 2n / [ r2-s2 r' 2-s2 ] (F.12) The main idea is that on the RHS the variables of inter est appear in factorized Abel-transform-ready form. The first appears as Vinogradov 1.107 and the second as 2.169. Vinogr adov et. al have much to say about Abel transforms in the context of dual equations. We shall derive (F.11) in Appendix G, and use it. Since this is supposed to be a document about the toroidal cha rged bowl problem, we note this alternate evaluation of the above double Bessel integral ∫0 ∞ Jn(rx) J n(r'x) dx = (1/ π) (rr') -1/2 Q n-1/2 [ (r 2+r' 2)/(2rr')] ) (F.13) So in the midst of a cylindrical and spherical coordinates discussion, we are somewhat surprised to find ourselves staring at a toroidal Q function (see Section 2 (b) above). 52 Appendix G: Solving the Charged Bowl Problem using a dou ble Abel transform The steps given in this lengthy and tedious Appendix are typical of what one encounters in a dual equations problem. In particular, one often encounters pain ful integrals. Sometimes there are "easy ways" to do integrals, but when the easy ways are not known, one must resort to brute force, as done below. This Appendix closely follows Section 1.4.1 of Vinogradov et. al . They in turn are presenting 1964 work of W.E. Williams. Unlike Vinogradov, we try to show all of the waypoints of the calculation. In Appendix E we outlined a matrix method for solving dual equa tions in this framework AΨ = f f = (f 1, f 2) presented as [A Ψ]1 = f 1 on I 1 BΨ = g g = (g 1, g 2) presented as [B Ψ]2 = g 2 on I 2 (E.1) where the problem is to solve for the "potential" Ψ and the partner functions f 2 and g 1 if one is given the driving functions f 1 and g 2. We placed the V 0=1 R=1 charged bowl problem into this framework according to Σn=0 ∞ Pn(cos θ) an = 1 θ in (0, θ0) AΨ = f // f 1(θ) = 1 Σn=0 ∞ (2n+1) P n(cos θ) an = 0 θ in ( θ0,π) BΨ = g // g 2(θ) = 0 (E.12) In the matrix method, one formally finds Ψ first, and then from that gets the partner functions f 2 and g 1. As one quickly learns from reading Sneddon's reviews, there a re many variations of this general method. Sometimes the batting order is to first find the partner funct ion g 1 and then from it obtain Ψ and f 2. That is exactly what we are going to do here. (a) Find g 1 The starting point is B Ψ = g = (g 1,g 2) where g 2 = 0 since there is no charge density on the cap of a charged bowl. The unknown partner function g 1 is proportional to the total charge density on the bowl. We have g1(θ) = Σn=0 ∞ (2n+1) P n(cos θ) an = ∂rVi - ∂rVo = 4 π[σi(θ)+σo(θ)] θ in (0, θ0) from Gauss's Law where σ is expressed in cgs units. For a bowl of radius R, g1 = 4 πR( σi+σo). (G.0) The second equation of (E.12) above may be written as ( z = cos θ in all that follows ) Σn=0 ∞ (2n+1) P n(z) an =  g1(θ) θ in (0,θ0) 0 θ in (θ0,π) (G.1) Applying ∫-1 1 dz Pn' (z) to both sides and using ∫-1 1 dz P n(z)P n' (z) = δn,n' 2/(2n+1) one finds that 53 an = (1/2) ∫-1 1 dz Pn(z) g( θ) = (1/2) ∫0 θ0 dθ sin θ Pn(z) g 1(θ) (G.2) If we install this a n into our first dual equation of (E.12), Σn=0 ∞Pn(z') an = 1, we get ∫-1 1 dz g( θ) [ (1/2) Σn=0 ∞ Pn(z) Pn(z') ] = 1 or ∫-1 1 dz g( θ) K( θ,θ') = 1 where K( θ,θ') ≡ (1/2) Σn=0 ∞ Pn(z) Pn(z') (G.3) Thus, K( θ,θ') is the kernel of an integral equation we want to sol ve for g. Digressing momentarily, if one considers the "string problem with the left end offset " u"( θ) = -n2u( θ) u(0) = 1 u( π) = 0 one finds that the eigenfunctions cos([n+1/2] θ) form a complete set with completeness Σn=0 ∞ cos([n+1/2]x) cos([n+1/2]x') = ( π/2) δ(x-x') (G.4) and if the right end instead is offset, one gets the same r esult with cos →sin. Now, using the integral representation (F.9) for each P fun ction in (G.3) along with (G,4), we get K( θ,θ') = (1/2 π) ∫0 min(θ,θ') dφ { 1/ cos φ-cos θ 1/ cos φ-cos θ' } (G.5) We have thus derived (F.11) quoted in Appendix F. Equation (G.3) then says 1 = ∫0 θ0 dθ' [sin θ' g 1(θ')] (1/2 π) ∫0 min(θ,θ') dφ { 1/ cos φ-cos θ 1/ cos φ-cos θ' } (G.6) where we have used that fact that g( θ)= 0 on ( θ0,π). Pondering the region of the double integration 54 we can swap the integration order to obtain 2π = ∫0 θ dφ 1/ cos φ-cos θ ∫φ θ0 dθ' [sin θ' g1(θ')] 1/ cos φ-cos θ' θ ≤ θ0 (G.7) The game is now to solve this equation for g 1 using a double Abel transform. We can break (G.7) into two pieces, each an Abel transform, in this way 2π = ∫0 θ dφ h( φ)/ cos φ-cos θ φ ≤ θ ≤ θ0 (G.8) h( φ) ≡ ∫φ θ0 dθ' [sin θ' g1(θ')] 1/ cos φ-cos θ' (G.9) Using Abel transform (F.7) applied to (G.8) we find that h( φ) = 2 sin φ/1-cos φ φ in (0, θ0) (G.10) Using this for the left side of (G.9) we apply Abel tra nsform (F.8) to get [sin θ g1(θ)] = - (2/ π)∂θ I (G.11) where I( θ) = ∫θ θ0 du sin 2u [1/ 1-cosu ] [1 / cos θ - cosu ] (G.12) This integral can be transformed into a simpler form which can be evaluated, I( θ) = ∫θ θ0 du sin 2u [1/ 1-cosu ] [1 / cos θ - cosu ] = ∫a b [ dx/ 1-x2 ] (1-x2) / [ 1-x cos θ - x ] // x = cosu, a=cos θ0, b= cos θ = ∫a b dx 1+x / b - x 55 = ∫0 c dy d-y / y // y = b-x, c = b-a, d = b+1 , d-c = a+1 = 2 ∫0 c dx d-x2 // y = x 2 = c d-c + d sin -1(c /d ) (G.13) Then using ∂θ = -sin θ ( ∂c+ ∂d) we can rewrite (G.11) as [sin θ g1(θ)] = - (2/ π)∂θ I = (2/ π) sin θ(∂c+ ∂d) I so that g1(θ) = (2/ π) (∂c+ ∂d) I = (2/ π) { d-c / c + sin -1(c /d ) } = (2/ π) { a+1 / b-a + sin -1(b-a / b+1 ) } Using the fact that sin -1 ( b - a / b+1 ) = π/2 – sin -1[1+a / 1+b ] we find g1(θ) = (2/ π) {1+a / b-a + π/2 – sin -1[1+a / 1+b ] } = (2/ π) { 2 cos( θ0/2)/ cos θ - cos θ0 + π/2 – sin -1[ cos( θ0/2) / cos( θ/2)] } which is our corrected version of Vinogradov equation 1.111 (whi ch has three typos, something very rare in their book). So the final result can be stated g1(θ) = (2/ π) {2 cos( θ0/2)/ cos θ - cos θ0 + cos -1[ cos( θ0/2) / cos( θ/2)] } θ ≤ θ0 (G.14) To summarize: using a double Abel transform, we have solve d our dual equation problem for the partner function g 1. (b) Run a check on g 1 As a check on (G.14), since g 1 = 4 πR( σi+σo) from (G.0), we can examine the Kelvin bowl charge densities from (4.6) and (4.9) σo+σi = (V0/2 π2R) { 1-cos α / cos α - cos θ – tan -1 [ 1-cos α /cos α - cos θ ] } + V 0/(4 πR) = V 0/(2 π2R) { 1-cos α / cos α - cos θ – tan -1 [ 1-cos α /cos α - cos θ ] + π/2 } But for comparison we have to flip the z axis which takes all cos → -cos, σo+σi = V 0/(2 π2R) { 1+cos θ0 / cos θ - cos θ0 – tan -1 [ 1+cos θ0 /cos θ - cos θ0 ] + π/2 } 56 and finally we use the fact that ("draw a small trian gle") tan -1 [ 1+cos θ0 /cos θ - cos θ0 ] = sin -1[1+cos θ0 /1+cos θ ] = sin -1[ cos( θ0/2) / cos( θ/2)] to get σo+σi = V 0/(2 π2R) { 2 cos( θ0/2)/ cos θ - cos θ0 + π/2 – sin -1[ cos( θ0/2) / cos( θ/2)] } Setting V 0 = 1 and R = 1 we then have g1 = 4 π(σi+σo) = (2/ π) { 2 cos( θ0/2)/ cos θ - cos θ0 + π/2 – sin -1[ cos( θ0/2) / cos( θ/2)] } in agreement with (G.14). (c) Find a n The next step is to determine coefficients a n by inserting (G.14) for g 1(θ) into (G.2). This gives an = (1/ π) ∫0 θ0 dθ sin θ Pn(cos θ) { 1+cos θ0 /cos θ-cos θ0 + cos -1[1+cos θ0 /1+cos θ ] } , (G.15) another unruly integral. The first term can be evaluate d using GR7 p 790 7.225.2 (ignore the P -1/2 typo) ∫cos θ0 1 dz Pn(z) / z-cos θ0 = (n+1/2) -1 (1-cos θ0)-1/2 [ T n(cos θ0) – T n+1 (cos θ0)] = (n+1/2) -1 (1-cos θ0)-1/2 [ cos(n θ0) – cos[(n+1) θ0 ] (G.16) so that an(1st term) = (2/ π) cot( θ0/2) [ cos(n θ0) – cos[(n+1) θ0] / (2n+1) = (2/ π) cot( θ0/2) [2 sin[(n+1/2) θ0] sin( θ0/2)] / (2n+1) = (4/ π) cos( θ0/2) sin[(n+1/2) θ0] / (2n+1) = (2/ π) {sin(n θ0) + sin[(n+1) θ0] } / (2n+1) (G.17) The second term integral an(2nd term) = (1/ π) ∫0 θ0 dθ sin θ P n(cos θ) cos -1[1+cos θ0 /1+cos θ ] (G.18) is an incomplete range integral of P n times an inverse trig function of an algebraic argument , making this integral a bit difficult to look up in a table. This is w here brute force comes in. We put P n into its integral representation (F.9), and then reverse the integration order us ing ∫0 θ0 dθ ∫0 θ dφ = ∫0 θ0 dφ ∫φ θ0 dθ to get 57 a n(2nd term) = ( 2 /π2) ∫0 a dφ cos[(n+1/2) φ] R where R ≡ ∫φ θ0 dθ sin θ/cos φ-cos θ * cos -1[1+cos θ0 /1+cos θ ] (G.19) R may be evaluated by setting x = ch( θ/2), a = cos( φ/2) and b = cos( θ0/2) to get R = (2 2 ) ∫b a dx x / a2-x2 * sec -1(x/b) = –(2 2 ) ∫b a dx ∂x (a 2-x2)1/2 sec -1(x/b) // set up for parts = (2 2 ) ∫b a dx (a 2-x2)1/2 ∂x sec -1(x/b) // the "parts" vanish = (2 2 b) ∫b a dx x -1 (a 2-x2)1/2 (x 2-b2)-1/2 // arc trig function is now gone = ( 2 b) ∫β α dy/(y) ( α-y) 1/2 (y-β)-1/2 // x 2 = y ,a2 = α , b 2 = β = ( 2 b) (c-1) ∫0 ∞ dz z / [(z+1)(z+c)] // z = ( α-y)/(y-β) c = α/β = a 2/b 2 = ( 2 b) π(c -1) // regulated partial fractions, ∞→Λ→∞ = ( 2 b) π ( (a/b) - 1) = ( 2 π) (a - b) = 2 π (cos( φ/2) - cos( θ0/2)) (G.20) After this saga, we are left with an(2nd term) = (2/ π) ∫0 θ0 dφ cos[(n+1/2) φ] [cos( φ/2) - cos( θ0/2) ] ] = (1/ π) [ (n+1-n cos θ0)sin(n θ0) - nsin θ0cos(n θ0) ] / [ n(n+1)(2n+1)] = (1/ π){ [ sin(n θ0)/n - sin[(n+1) θ0]/(n+1) } / (2n+1) (G.21) At this point then we have shown that an(1st term) = (1/ π) { 2sin(n θ0) + 2sin[(n+1) θ0] } / (2n+1) an(2nd term) = (1/ π) { sin(n θ0)/n - sin[(n+1) θ0]/(n+1) } / (2n+1) (G.22) Adding we find an = (1/ π) { [2+1/n] sin(n θ0) + [2-1/(n+1)] sin[(n+1) θ0] } / (2n+1) = (1/ π) { (2n+1)/n * sin(n θ0) + (2n+1)/(n+1)* sin[(n+1) θ0] } / (2n+1) = (1/ π) { sin(n θ0)/n + sin[(n+1) θ0] /(n+1) } (G.23) which agrees with the result found in (E.16). 58 (d) Find Ψ and f 2 For the unit-radius bowl the potential is given by (E.11) wit h (G.23) Vi(r, θ) = (1/ π) Σn=0 ∞ { sin(n θ0)/n + sin[(n+1) θ0] /(n+1) } r n Pn(cos θ) Vo(r, θ) = (1/ π) Σn=0 ∞ { sin(n θ0)/n + sin[(n+1) θ0] /(n+1) } r -n-1Pn(cos θ) (G.24) Finally, the other partner function f 2 is the potential on the cap, θ0 < θ < π, f2(θ) = (1/ π) Σn=0 ∞{ sin(n θ0)/n + sin[(n+1) θ0] /(n+1) } P n(cos θ) (G.25) References. L. A. Ahlfors, Complex Analysis, 2nd Ed. , ( McGraw-Hill, New York, 1966) A. Erdelyi et. al , Higher Transcendental Functions , Volume I, (McGraw-Hill, New York, 1953). This is the 5-volume Bateman Manuscript Project: Higher Transcendent al Functions EH 1,2,3 and Tables of Integral Transforms ET 1,2. (GR7) I.S. Gradshteyn and I.M. Ryzhik, Table of Integrals, Series, and Products, 7th Ed ( Academic Press, New York, 2007). Editor Dan Zwillinger is collecti ng errata. H. Hu and R.G. Larson, " Evaporation of a Sessile Drop let on a Substrate", J. Phys. Chem. B 2002, 106, 1334-1344. Points out that 19th century toroidal potential theory has a role in DNA processing. E.W. Hobson, The Theory of Spherical and Ellipsoidal Harmonics, (Cambridge University Press, Cambridge, 1931). Chapter XI concerns ellipsoidal harmonics, abo ut 50 pages. This presentation seems more readable than Chapter 23 of Whittaker and Watson (se e below). J.D. Jackson, Classical Electrodynamics , (Wiley, New York, 1962). This is the first edition with a green cover. Later editions (red 1975 and blue 1998 ) omit discussion of the inversion method to make room for more modern subjects. Kelvin (W. Thomson), Reprint of Papers on Electrostatics and Magnetism, 2nd. Ed., (Macmillan, London, 1884). As are several other items in this Reference l ist, this out-of-print book is available on the web (this one "Digitized by Microsoft"). Converting PDF to D JVU makes things smaller and faster and maintains searchability. N.N. Lebedev, I.P. Skalskaya and Y.S. Uflyand, Worked Problems in Applied Mathematics , (Dover, New York, 1965). There are not many books like this one. In the first 2/3 of the book, 566 problems requiring various mathematical techniques from various fields of phys ics and engineering are presented with hints as to their solution, and in last 1/3 a subset of these problem s is solved (but not those discussed in the current paper). 59 Maple (Maplesoft), a symbolic computer algebra system, today of ficially Maple 14. See wiki for comments about Maple's history versus Mathematica and other systems. See author's Maple User Guide. K.T. McDonald, "Conducting Spherical Shell with a Circular O rifice", PDF file downloadable from his Princeton website. This paper discusses the bowl problem in several ways, and reviews some of Kelvin's papers noted above. The current paper tries to fill in Sect ion 2.5 of this document. P. Moon and D.E. Spencer, Field Theory Handbook, Including Coordinate Systems, Differential Equations and their Solutions (Springer-Verlag, Berlin, 1961). This book is not about quantu m field theory or anything like that, it is about curvilinear coordin ate systems, how the Laplace and Helmholtz equations appear in each system, and what the solutions of the se equations look like. This husband and wife team wrote several excellent books. Long ago they were strangely involved in an accident involving a test of general relativity. P.M. Morse and H. Feshbach, Methods of Theoretical Physics, ( McGraw-Hill, New York, 1953). This 2000 page behemoth is simply amazing. F. Oberhettinger and T.P. Higgins, Tables of Lebedev, Mehler and Generalized Mehler Transforms , (Boeing Scientific Research Laboratories, Mathematical N o. 246, 1961). A 48 page monograph with about 16 transforms per page. The first author is a member of A. Erdelyi et. al mentioned above. We obtained this book by interlibrary loan from the Coe library a t the University of Wyoming. (PBM) A.P. Prudnikov, Y.A. Brychkov and O.I. Marichev, Integrals and Series: More Special Functions, Vol 3 (Gordon and Breach, New York, 1996). A huge 5 volume set similar to the Bateman / Erdelyi set plus Gradshteyn & Ryzhik. The Russian editions are perhaps more available and can easily be used by non-Russian readers. Here are a few useful words: оглавление = table of contents ряды = series сумма = sum глава = chapter преобразовать = transform интеграл = integral функция = function http://www.babylon.com/define/118/russian-english-dictionary.html W.R. Smythe, Static and Dynamic Electricity, 2nd Ed . ( McGraw-Hill, New York, 1950). I. Sneddon, Mixed Boundary Value Problems in Potential Theory , (Wiley, New York, 1966). This book treats dual and triple integral and series equations wit h examples from electrostatics and from crack theory. A lot of output is obtained from basic Abel and Han kel type transforms. S.S Vinogradov, P.D. Smith, and E.D Vinogradova, Canonical Problems in Scattering and Potential Theory, Part I: Canonical Structures in Potential Theory , (Chapman & Hall/CRC, New York, 2001). The bowl problem is treated in Sections 1.4.1-3 of this excellen t but very advanced monograph. (See (G.14) for a correction to their equation 1.111 for g( θ). ) Chapter 2 incorporates in compact form nearly the entire huge panorama of Sneddon's book. The book is advanced in that it treats impressive variations of already-difficult problems in electrostatics, surfaces with holes and slots and missing pieces. For example, 60 E.T. Whittaker and G.N. Watson, A Course in Modern Analysis, 4th Ed. , (Cambridge University Press, Cambridge, 1927). The Mehler-Dirichlet integral is derived in S ection 15.231 of this and at least two earlier editions. Beware that the 2nd edition is being sol d by Merchant Press and does not include Chapter 23 on Ellipsoidal Harmonics and Lamé Functions. See Kelvin reference above. _____________________________________________________________________________________