Stackel Separation of the Helmholtz Equation

1 Stäckel Separation of the Helmholtz Equation Phil Lucht Rimrock Digital Technology, Salt Lake City, Utah 84102 last update: November 30, 2011 Contents: 1. Introduction and Summary.................................................................................................... ............ 3 2. Notation and Cast of Players................................................................................................ .............. 6 3. Initial Processing of the Helmholtz Equation................................................................................ ...9 Comments on (3.5) and example of toroidal coordinates ................................................................... 10 The Definition of Q............................................................................................................ ................. 12 Example: Compute Q for the toroidal system using (3.9) .................................................................. 13 Finish Processing .............................................................................................................. .................. 14 Comment on separation in Cartesian coordinates............................................................................... 15 4. Starting from the other end: The Stäckel Matrix Φ and a Formulation of the Problem.......... 16 Introduction of the Stäckel Matrix............................................................................................. ......... 16 Formulation of separation Problems A, B and C ................................................................................ 17 5. Recasting Problem B into a new form......................................................................................... ....19 6. The non-uniqueness of Stäckel matrix Φ: Equivalence Operations............................................ 22 7. Solving Problem B........................................................................................................... .................. 24 (a) Conditions required for Separability ....................................................................................... ......25 (b) Steps for finding Φ and related quantities ..................................................................................... 26 (c) Problem B Example (R-separation of Laplace): Toroidal Coordinates........................................ 27 The Φ matrix................................................................................................................................... 27 The separated equations and their solutions: toroidal harmonics .................................................. 29 (d) Problem A Example (simple-separation of Helmholtz): Circular Cylindrical Coordinates......... 30 The Φ matrix................................................................................................................................... 30 The separated equations and their solutions: cylindrical harmonics.............................................. 32 8. Separability in Cylindrical Systems when h 1 ≠ α h2...................................................................... 33 Example: Circular cylindrical coordinat es revisited .......................................................................... 37 9. Separability in Cylindrical Systems when h 2 = α h1...................................................................... 39 The Conditions.................................................................................................................................... 39 The Steps............................................................................................................................................. 40 Special Case h 1 = h2............................................................................................................................ 41 Example 1: elliptic cylinder coordinates...................................................................................... ..42 Example 2: parabolic cylinder coordinates.................................................................................... 42 Example 3: circular cylinder coordinates...................................................................................... .42 The 21 cylindrical system s of Moon and Spencer .............................................................................. 42 10. Separability in Rotational Systems......................................................................................... ....... 44 A look at the Conditions of Section 7 (a): ..................................................................................... .....45 A look at the Steps of Section 7 (b): .......................................................................................... ......... 47 Example 1: toroidal coordinates revisited..................................................................................... .....49 2 The 11 separable rotational sy stem s of Moon and Spencer................................................................ 51 11. Separation of the Schrodinger equation. .................................................................................... ..52 (a) Schrodinger meets Helmholtz ................................................................................................ ....... 52 (b) Separation of the Schrodinger Equation in Curvilinear coordinates ............................................. 53 (c) Central potentials and th e hydrogen at om problem....................................................................... 54 12. Separation in Ellipsoidal Coordinates...................................................................................... .....56 (a) Some details about e llipsoidal coordinates................................................................................. ...56 (b) Notation and comparison to that of Morse and Feshbach ............................................................. 58 (c) Running through the Steps of Section 7 (b) ................................................................................. .60 (d) Summary of Results ......................................................................................................... ............. 66 (e) The separated solutions: Lamé functions ................................................................................... ..67 (f) Comment on a missing minus sign on page 512 of Morse and Feshbach ..................................... 70 13. Stäckel Theory in N dimensions............................................................................................. ........ 72 Altered Development Equations ......................................................................................................... 72 Equivalence Rules.............................................................................................................. ................. 73 Conditions........................................................................................................................................... 74 Steps.................................................................................................................................................... 76 14. Stäckel Theory in N=2 dimensions ................................................................................................ 78 Conditions........................................................................................................................................... 78 Steps.................................................................................................................................................... 79 Example: Polar Coordinates ( ξ1, ξ2) = (r,θ) ...................................................................................... 80 Using Sturm-Liouville to solve a specific Helmholtz problem in polar coordinates...................... 82 The special case h 1 = h2...................................................................................................................... 84 Example : elliptical coordinates.............................................................................................. ........... 86 Appendix A. Review of 1D Sturm-Liouville Th eory; the Kantorovich-Lebedev transform......... 87 References.............................................................................................................................................. 90 3 1. Introduction and Summary Nothing "ne w" is presented in this document. The Stäckel separation theory is treated in Morse & Feshbach (see References) and elsewhere. It is probab ly fair to say, however, that the treatments are far and few between, and the subject is not always presented in a systematic fashion. It is the purpose of this monograph to provide a simple, complete and syst ematic presentation without too much technical baggage. It is assumed only that the reader has some rudimentary knowledge of differential equations. References to the original papers of Stäckel (1891,1893,1897), Robert son (1927) and Eisenhart (1934) and to a later work of Moon & Spencer (1952) are given at the end. The subject at hand is separation of the Helmholtz equation (and its special case the Laplace equation) in various 3D curvilinear coordina te systems whose coordinates we shall call ξ 1,ξ2,ξ3. The term "separation" means that one starts with the 3D Helmholtz equation ( ∇2 + K12) ψ = 0, which is of course a partial differential equation (PDE ), and tries to find solutions of the form ψ(ξ 1,ξ2,ξ3) = X1(ξ1)X2(ξ2)X3(ξ3)/ R(ξ1,ξ2,ξ3) , (1.1) where an ordinary differential equation (ODE n) can be produced for each of the functions X n(ξn). Thus, one can search for solutions "separately" for each X n. If the X n are any solutions of their respective ODEn, then the ψ shown above is a solution of the Helmholtz equation. First, this is useful way generically to solve the Helmholtz problem, and second, because boundary conditions are often specified separatel y for the three coordinates, this separation (usually) turns at least one of the ODE n into its own little 1D boundary value probl em. These ODE's always contain self-adjoint differential operators and are therefore (usually) am enable to normal 1D "Sturm-Liouville theory", a subject reviewed in Appendix A. We say "usually" because in some cases the enta nglement of the separation constants forces one to consider 2D or even 3D Sturm-Liouville situations , where the problem cannot be factored into 1D problems each of which causes quantization of its isolat ed separation constant. Hopefully this statement will become clearer below. A classic 3D case occurs in ellipsoidal coordinates as discussed in Section 12, while a 2D case occurs in conical coordinates. Both these systems involve the Lamé functions. If the separation can be done as outlined above for a certain orthogonal curvilinear coordinate system, then one says that the system is R-separable . If this can be done with R=1, it is simple-separable . Some authors exclude simple-separable cases from those that are R-separable, but we shall not. It turns out that only the 11 "classical" 3D Euclidean orthogonal coordinate systems are simple separable for the Helmholtz equation (and therefor e also for the Laplace equation). These systems are discussed in the first chapter (called Section I) of Moon & Spencer (1961). This separability is why they are the classical systems. It also turns out that with R ≠ constant the Helmholtz PDE is not separable unless K 12 = 0, in which case it is the Laplace PDE. In other words, no Helmholtz equation with K 12 ≠ 0 is R-separable unless R = constant. In this case one might as well assume R=1 since the rest of the constant could be absorbed into the X i functions. The upshot is that there are systems like toroidal coordinates (not one of the classical 11 systems) in which the Laplace equation is R-separable but not simple separable. The Helmholtz equation is extremely significan t because it arises very naturally in problems involving the heat conduction (diffusion) equation and th e wave equation, where the time derivative term in the PDE is replaced by a constant parameter by applying a Laplace or Fourier time transform to the PDE. A huge swath of mathematical physics is dominat ed by these two PDE equation types. One always 4 needs to solve the Helmholtz equatio n that results, and that then involves the notion of "separation". Section 11 provides as an example the Schrodinger Equation PDE of quantum mechanics, which is the heat conduction equation with scaled imaginary time. Methods of generating equivalent Stackel matrices are given in Section 6, and Section 13 generalizes the whole theory to N dimensions. A brief summary of the document is in order. Section 2 sets up our "systematic" notation and conventions, and discusses the notion of a functional- form equation which differs in an ope rational sense from a normal equation. Section 3 starts with the Helmholtz equation and gr inds away on it, making one functional-form assumption (3.5) along the way. We define a support function Q, compute it for toroidal coordinates, and then digress to discuss the notion of separation consta nts as they appear in Cartesian coordinates. Section 4 assumes a most-general self-adjoint form for the desired separated L n which defines ODE n and then inserts this form into the ground-down Helmholtz equation, t hus grinding it down even more to equation (4.4). Finally one is in a position to introduce the Stäckel Matrix Φ and with it one can formulate two potentially solvable problems, called Problem A and Problem B, and one insolvable Problem C. Problem A (simple-separation) is seen to be a special case of Problem B (R-separation). The problem is to "find the Stäckel matrix Φ" and various supporting func tions, and in so doing to "solve" Problem B, and thus also Problem A. Here "solve" means to achie ve separation and thus to have at hand all the ODE n which can then be solved subject to their boundary conditions. Basically Problems A and B are treated at the same time, something not do ne in the referenced books. Section 5 takes Problem B and, using elementary linear algebra, recasts it into a new form which appears as certain conditions on the matrix Φ. The Robertson Condition appears here. Section 6 digresses momentarily to derive a set of "equi valence rules" allowing one to obtain new valid Stäckel matrices from an existing valid Stäckel ma trix. This matrix is thus not unique. We have not seen these simple rules stated anywhere, but they are doubtless out there somewhere. Section 7 discusses the solution of Problem B as recast at the end of Section 5. Section 7 (a) specifies a set of Conditions which must be met for a solution to exist, and Section 7 (b) gives a sequence of Steps one can follow to find the solution. These Steps are then applied to two examples. The first is a Problem B problem: R-separation of toroidal coordinates. The second is a Problem A problem: simple-separation of circular cylindrical coordinates. The examples are just exercises in turning a crank. In each case the solution functions are stated, these being the toroidal and cylindrical harmonics. Section 8 attempts to specialize the solution method to a class of cylindrical systems in which the non-trivial scale factors h 1 and h2 are not multiples of each other. Such systems are uncommon with the exception of the circular cylindrical system considered in Section 7. Section 9 first considers cylindrical systems for which h 2 = αh1, and then sets α = 1 as a further special case. Most practical cylindrical systems have h 1 = h2 since they come from conformal maps of the Cartesian system. The Stäckel matrix is computed for three cylindrical systems of this type and the 21 such systems considered by Moon and Spencer are mentioned. Section 10 specializes the Section 7 Conditions and Steps to rotational systems, and again toroidal coordinates are considered as an example. The 11 such systems listed in Moon and Spencer are mentioned. Section 11 considers how separation is affected when so me function is rudely added to the constant in the Helmholtz equation. If this function has a certa in simple form, the resulting equation can still be separated. A prototype of this situation is the Sc hrodinger Equation of quantum mechanics where that added function is the potential field imposed on a quant um particle. As an example we mention the notion 5 of a central potential in spherical coordinates and take a passing glance at the solution of the hydrogen atom. Section 12 , as a solid exercise of our systematic mach inery, carries out the separation of the Helmholtz equation in ellipsoidal coordinates, the mo st complicated of the classical systems. Two very different-looking Stäckel matrices are obtained, both of which appear in the literature, and it is shown they are equivalent by the rules of Section 6. Our notation matches Morse & Feshbach and a sign error in that book is noted. Section 13 generalizes the Stäckel formalism from 3 to N dimensions, which is quite easy to do, though carrying out the computational Steps is more complicated for N > 3. Section 14 examines the Stäckel theory for N = 2 dimens ions, writes some general results, and then treats polar and elliptical coordinates as examples for h 1≠h2 and h1=h2. A certain polar coordinates Helmholtz Green's Function problem is considered to illustrate how the Sturm-Liouville analysis can be carried out in either coordinate. Appendix A gives a concise review of the 1D Sturm-Li ouville Problem and the transform associated with such a problem. The Kantorovich-Lebedev transform is presented as an example. A short list of References is then provided. 6 2. Notation and Cast of Players The entire analysis is a study of functional forms and this makes it a bit slippery. By functional form is meant the way a function of multiple variables de pends on those variables in terms of possible factorization of the form. For example, some u(x,y,z) might be expressible as v(x)w(y,z) and this would then be the functional form of u. Another sense of func tional form is simply what variables a function is a function of. One might write v(x) simply as v if an expressions get complicated, but one must remember than that its functional form is v(x). The symbols ξ1, ξ2, ξ3 (these are "xi", pronounced "zeye", different from ζ = zeta and χ = chi ) are often used for ellipsoidal coordinates since those coordinates are so closely related to each other. Below ξ 1, ξ2, ξ3 shall represent an arbitrary triplet of orth ogonal curvilinear coordinates. The notation ∂n is used for a partial derivative, ∂f ∂ξn = ∂f/∂ξn = ∂nf n = 1,2,3 and the following other symbols Σ n ≡ Σn=13 LHS = left hand side of some equation RHS = right hand side PDE = partial differential equation ODE = ordinary diff eq There will be many function symbols used below, and each symbol has an implied functional form. At first one needs to show the forms in full detail, bu t eventually one learns to work without doing this. Here is a triplet of function symbols eac h having four different notations, g 1(ξ2, ξ3) = g1(23) = g 1(≠1) = g1 g2(ξ3, ξ1) = g2(31) = g 2(≠2) = g2 g3(ξ1, ξ2) = g3(31) = g 3(≠3) = g3 Each row is the forward cyclic permutation of the previous row. Only the notations of the last two columns are amenable to the generic notation gn(≠n) = gn n = 1,2,3 Notice these facts: ∂ 1g1 = 0 ∂1(g1F) = g1(∂1F) ∂ngn = 0 ∂n(gnF) = gn(∂nF) These follow, for example, since g 1 is a function only of ξ2 and ξ3. "Facts" like these will be crucial in the analysis below. The functions g n are "helper functions" which have no particular significance. Only one other function symbol set will have this same functional form M n(≠n) = Mn n = 1,2,3 7 The letter M stands for "minor" as in the minor of a 3x3 matrix, but in fact M is really a cofactor. This confusion seems to go back to Morse and Feshbach who use the word "minor" to mean what is now usually called a "cofactor". In current terminology cofactor pq = (-1)p+q minorpq where p and q label the rows and columns of a matrix, as in A pq. To be a little more specific, the M n will be the cofactors of the elements of the first column of a certa in 3x3 matrix called the Stäckel matrix Φ discussed soon below. Certain function symbols imply a very simple functional form as follows fn(ξn) = fn(n) = fn n = 1,2,3 Xn(ξn) = Xn(n) = Xn n = 1,2,3 The X n are the factors of the ψ solution shown in the introduction. The f n functions are just more helper functions that will be used in the analysis a nd which appear in the separated ODE equations. There are several function symbols that are assumed to have in general no factored functional form, and here they are R(ξ1,ξ2,ξ3) = R(123) = R // the function appearing in (1) above ("modulation factor") Q(ξ1,ξ2,ξ3) = Q(123) = Q // a helper function (called u by Morse & Feshbach) S(ξ1,ξ2,ξ3) = S(123) = S // the determinant of Φ (coming soon) hn(ξ1,ξ2,ξ3) = hn(123) = h n // the curvilinear system scale factors h n2 = gnn (metric tensor) H(ξ1,ξ2,ξ3) = H(123) = H ≡ h1h2h3 The symbol H is "non standard" but it is convenient to use it as the product of the three curvilinear scale factors. One might say that all the quantities listed he re have a generic (123) functional form. This does not mean they cannot also have some kind of factored form. The symbol ψ, the Helmholtz equation solution function, has the special functional form noted in (1.1). That is, one seeks solutions ψ of this functional form, ψ(ξ 1,ξ2,ξ3) = X1(ξ1)X2(ξ2)X3(ξ3)/ R(ξ1,ξ2,ξ3) ψ(123) = X 1(1)X2(2)X3(3)/ R(123) ψ = X1X2X3/R Historically it seems that R was defined "in the denominator of ψ". Certain constants shall appear below: K 12 = the parameter appearing in the Helmholtz equation ( ∇2 + K12)ψ=0 k22, k32 = two generic constants which will be called "separation constants" Although K 12 and the k n2 are written "squared", they are really meant to be arbitrary real numbers, so if some ki2 < 0 then the corresponding k i is imaginary. The squared notation arises from the way the Helmholtz equation looks when it derives from a tr ansformed wave equation. It is probably an unfortunate notation, but we use it to be compatible with Morse and Feshbach. For equations with three terms where the last two te rms are cyclic permutations of the first term, we sometimes write first term " + cyclic" to save writing obvious extra terms. That is to say, 8 f(123) + f(231) + f(312) = f(123) + cyclic We now come finally to the Stäckel matrix which Morse and Feshbach (and Stäckel) call Φ Φ = ⎝⎜⎜⎛ ⎠⎟⎟⎞ Φ11(1) Φ12(1) Φ13(1) Φ21(2) Φ22(2) Φ23(2) Φ31(3) Φ32(3) Φ33(3) " the Stäckel Matrix" S = d e t ( Φ) " the Stäckel Determinant" A key fact to recognize is that the three functions of row n are functions only of ξ n , as indicated by the notation Φ nm(n) = Φnm(ξn) so the first index of Φ matches the argument. The cofactors of th e elements of the first column of the Φ matrix are these, M n ≡ cof(Φn1) = (-1)n+1 minor(Φn1) For example, M 2 = (-1)2+1 minor(Φn1) = – ⎪⎪ ⎪⎪ Φ12(1) Φ13(1) Φ32(3) Φ33(3) = – [Φ12(ξ1) Φ33(ξ3) – Φ13(ξ1) Φ32(ξ3)] Notice that M 2 has the functional form M 2(≠2) = M2(ξ1,ξ3), and in general M n = Mn(≠n), as it was presented earlier in this section. Some times the following notations are used M n(Φ) S(Φ ) to stress that these quantities are functions of the Stäckel matrix elements. This Φ matrix is named after German mathematician Paul Stäckel (shtay'kel), 1962-1919. Our symbols match exactly those of Morse and Feshbach except for Q which they call u, and except for our added symbol H for h 1h2h3. Our symbols match Moon and Spencer except they use Un(un) in place of our X n(ξn). Morse and Feshbach address this 3D separation subj ect in two places in their first volume: simple separation is treated pp 508-511, while R-separation is treated pp 518-519. They do not use the term "R- separation" and call the R function a "modulation factor". Moon and Spencer discuss simple separation on pp 5- 7, and R-separation on p 96. They provide more detail in some of their other books and papers. One last notational item: when an already-number ed equation is later replicated for the reader's convenience, the equation number in the replicated line is put into italics to show that this is not the first occurrence of the equation. 9 3. Initial Processing of the Helmholtz Equation We are going to treat R-se paration and simple separati on at the same time, and branch off later into the two cases. Remember that simple separation just means R = 1. In orthogonal curvilinear coordinates, using our notation as defined above, the Helmholtz equa tion can be written (see e.g. Morse and Feshbach p 115) L ψ ≡ (∇ 2+K12)ψ = 0 L ≡ (∇2+K12) = the Helmholtz operator L ψ = H-1{ ∂1[(H/h12)(∂1ψ)] + cyclic} + K 12ψ = 0 H ≡ h1h2h3 L ψ = H -1 Σn ∂n[(H/hn2)(∂nψ)] + K12ψ = 0 ( 3 . 1 ) where sometimes the + cyclic notation is more useful for observing functional forms. We seek a solution of this functional form ψ = X1X2X3/ R ( 3 . 2 ) where the four functions X n and R are as yet unknown. Inserting this ψ form into (3.1) gives Lψ = H -1{ ∂1[(H/h12)(∂1{ X1X2X3/R })] + cyclic} + K 12ψ = 0 Lψ = H -1{ X2X3∂1[(H/h12) ∂1{ X1/R }] + cyclic} + K 12ψ = 0 (3.3) Here one sees how the implicit functional forms come into play, as the factors X 2X3 quietly slip to the left through both ∂1 operators. The following is then inserted into (3.3) ∂ 1(X1/R) = ∂1(R-1X1) = R-1∂1X1 - R-2X1∂1R = R-2{ (∂1X1)R - X1(∂1R) } to get Lψ = H-1{ X2X3∂1[ (H/[R2h12]){ (∂1X1)R - X1(∂1R) } ] + cyclic} + K 12ψ = 0 (3.4) In order to move toward a "separ ated form" wherein the terms above are less coordinate-entangled, we shall attempt to select function R so the following equation is satisfied in terms of functional form (comments below): (H/[R 2hn2]) = fn(n)gn(≠n) n = 1,2,3 (3.5) (H/[R 2h12]) = f1(1)g1(23) // for example because , if one inserts this into (3.4), one can pull g 1(23) to the left through ∂1 to get H-1{ X2X3∂1[ f1(1)g1(23) { (∂1X1)R - X1(∂1R) } ] + cyclic} ψ + K12ψ = 0 H-1{ X2X3 g1∂1[ f1 { (∂1X1)R - X1(∂1R) } ] + cyclic}ψ + K12ψ = 0 10 and, recalling that ψ = X1X2X3/R from (3.2), one can then divide by ψ to get (Lψ)/ψ = (R/H){(1/X 1) g1∂1[f1 { (∂1X1)R - X1(∂1R) }] + cyclic} + K 12 = 0 (Lψ)/ψ = (R/H) Σn[(gn/Xn) ∂n[fn { R(∂nXn) - Xn(∂nR) }] + K 12 = 0 Now using (3.5) to replace g n in favor of f n, our processed Helmholtz equation appears as (Lψ)/ψ = (1/R) Σ n[ (1/[hn2Xn]) (1/fn)∂n[fn { R(∂nXn) - Xn(∂nR) }] + K 12 = 0 (3.6) It is not obvious from the discussion above that having (3.5) be true is the only possible pathway to finding a separated solution, though probably that can be proven. It will be shown that this (3.5) pathway does in fact lead to separated solutions. For a coordinate system which satisfies (3.5), where one has found the f n , gn and R, one can write ∇2 in an alternate form. In (3.1) ∇2 is given in terms of the scale factors h n, ∇ 2 = H-1 Σn ∂n[(H/hn2) ∂n ] From (3.5) one can replace (H/h n2) = fn(n)gn(≠n)R2 to get ∇2 = H-1 Σn ∂n[fn(n)gn(≠n)R2∂n ] = H-1 Σn gn(≠n)∂n[fn(n) R2∂n ] = H-1 Σn (H/[hn2R2]) (1/fn) ∂n[fn R2∂n ] = Σn (1/[hn2R2]) (1/fn) ∂n[fn R2∂n ] ( 3 . 7 ) Then in the special case that R = 1, one has ∇ 2 = Σn (1/hn2) (1/fn) ∂n[fn∂n ] ( 3 . 7 a ) showing ∇ 2 in terms of the f n and the scale factors h n. In either case L = ∇2+ K12. Comments on (3.5) and exampl e of toroidal coordinates Conditions (3.5) above, (H/[R2hn2]) = fn(n)gn(≠n) n = 1,2,3 (3.5) require that the LHS factor in a certain specific mann er. One can certainly find an R(123) that works for the first equation with n=1: one could just select two arbitrary functions f 1(1) and g 1(23) and then define R by 11 [R(123) ]-2 ≡ f1(1)g1(23) (H/h 12) But if this is done, it is unlikely that (3.5) w ill be viable for n=2 and 3! So one has to assume one can find a set of 7 functions (three f n, three g n and one R) which make (3.5) valid. Any curvilinear system for which one cannot find a happy set of 7 functions is therefore not separable. So one views (3.5) as a restriction or condition on the curvilinear coordinate sy stem which must be met to obtain any separation. We do not address the question of whether there migh t be some other possible separation solution where (3.5) is not assumed. Example . Since this is perhaps a confusing concept, consider the toroidal coordinate system as an example. The Moon and Spencer page 112 notation is used, where η labels toroids, θ labels bowls, ψ labels azimuthal half planes, a is the ra dius of the limiting toroid, and then ( ξ1,ξ2,ξ3) = (η,θ,ψ). In this system the basic parameters come out being R -2 ≡ [ch(ξ1)-cos(ξ2)] h1 = h2 = a/[ch(ξ 1)-cos(ξ2)] => h 1 = h2 = aR2 h3 = a sh(ξ 1)/[ch(ξ1)-cos(ξ2)] => h 3 = a sh(ξ 1)R2 As a candidate for R one tries R = R and finds then that (H/h 12) = h2h3/h1 = h3 = a sh(ξ 1)/[ch(ξ1)-cos(ξ2)] (H/h 22) = h3h1/h2 = h3 = a sh(ξ 1)/[ch(ξ1)-cos(ξ2)] (H/h 32) = h1h2/h3 = a2/[ch(ξ1)-cos(ξ2)]2 * [ch(ξ1)-cos(ξ2)] /a sh(ξ1) = a/{[ch( ξ1)-cos(ξ2)] sh(ξ1)} Can a set of 6 functions f n and gn be found which satisfy (3.5) ? fn(1)gn(23) = (1/R2) (H/hn2) = [ch(ξ1)-cos(ξ2)] (H/hn2) for n = 1,2,3 The answer is yes: f 1(1)g1(23) = [ch( ξ1)-cos(ξ2)] a sh(ξ 1)/[ch(ξ1)-cos(ξ2)] = a sh( ξ1) = [sh( ξ1) ] [a ] f2(2)g2(31) = [ch( ξ1)-cos(ξ2)] a sh(ξ 1)/[ch(ξ1)-cos(ξ2)] = a sh( ξ1) = [ 1 ] [a sh(ξ 1) ] f3(3)g3(12) = [ch( ξ1)-cos(ξ2)] a/{[ch(ξ1)-cos(ξ2)] sh(ξ1)} = a /sh( ξ1) = [ a ] [1/ sh( ξ1) ] Thus, with this candidate R, a solution set for the 6 functions has been found: f 1(1) = sh(ξ1) g 1(23) = a R = R = [ch(ξ1)-cos(ξ2)]-1/2 f2(2) = 1 g 2(31) = a sh( ξ1) f3(3) = a g 3(12) = 1/ sh( ξ1) Therefore, the toroidal system at least has a chance of being R-separable. It is in fact R-separable (for Laplace), but we don't know that yet since there might be other restrictions that will need checking. 12 Had one tried R = 1 in the above toroidal discussion, the problem would have been to find 6 functions f n and gn which satisfy (3.5). The n=1 condition would read, f1(1)g1(23) = (H/h 12) = a sh(ξ 1)/[ch(ξ1)-cos(ξ2)] The only chance is to select f 1(1) = a sh( ξ1) but then one is stuck with g 1(23) = 1/[ch( ξ1)-cos(ξ2)] which involves coordinate ξ1 which violates the functional form g 1(23)! Therefore the toroidal system is not simple-separable! The Definition of Q Processing of the Hel mholtz equation now resumes where it left off, which was here (Lψ)/ψ = (1/R) Σn[ (1/[hn2Xn]) (1/fn)∂n[fn { R(∂nXn) - Xn(∂nR) }] + K 12 = 0 (3.6) where (3.5) has been assumed. Note th at there is serious variable entanglement in each term of the sum, since R = R(123) and h n= hn(123). To further process (3.6), compute the ∂n derivatives, ∂n[R fn(∂nXn)] = ∂n[R{fn(∂nXn)}] = R∂n{fn(∂nXn)} + (∂nR) {fn(∂nXn)} ∂n[fnXn(∂nR)] = ∂n[Xn{fn(∂nR)}] = X n∂n{fn(∂nR)} + (∂nXn) {fn(∂nR)} Therefore, for the quantity appearing in (3.6), ∂n[fn { R(∂nXn) - Xn(∂nR) })] = R ∂n{fn(∂nXn)} – Xn∂n{fn(∂nR)} since the two second terms cancel. Then (3.6) for ( Lψ)/ψ becomes (1/R) Σn (1/[hn2Xnfn]) [R∂n{fn(∂nXn)} – Xn∂n{fn(∂nR)}] + K 12 = 0 (1/R) Σ n[ (1/[hn2Xnfn]) [R∂n{fn(∂nXn)}] – (1/R) Σn[ (1/[hn2Xnfn]) [Xn∂n{fn(∂nR)}] + K 12 = 0 or (Lψ)/ψ = Σ n[ (1/[hn2Xnfn]) [∂n{fn(∂nXn)}] – Σn[ (1/[hn2Rfn]) [∂n{fn(∂nR)}] + K 12 = 0 (3.8) Notice that the X n part is segregated from the R part. Since a viable set of functions f n and R has already been determined from (3.5), the second term sum shown in (3.8) can be computed. It is going to be some general function of 123 which we now wr ite in a fairly strange manner, Σn (1/[hn2fnR]) ∂n{fn(∂nR)} ≡ - k12/Q(123) (3.9) Here, it is the RHS which is fully determined by the LHS and we choose to partition the RHS into two factors, a constant -k 12 (Moon and Spencer call this constant - α1) and 1/Q(123) (Morse and Feshbach call this 1/u). Remember that the Helmholtz parameter is K 12 and has nothing to do with this new constant 13 k12 just introduced. In theory, k 12 can be any constant, but one usually chooses it to make the resulting Q(123) have some simple form. The reason for calling this constant k 12 will be seen later. For the simple separation case with R = 1, we will choose Q(123) = 1 and k 12 = 0 in (3.9), which of course makes it still valid since ( ∂nR) = 0. Equation (3.8,9) appears as (5.1.46,47) in Morse and Feshbach p 519, with Q → u. Example: Compute Q for the toroidal system using (3.9) Start with the num erators in (3.9). First, (∂1R) = ∂1[ch(ξ1)-cos(ξ2)]-1/2 = (-1/2)sh( ξ1) [ch(ξ1)-cos(ξ2)]-3/2 = (-1/2)sh( ξ1) R3 (∂2R) = ∂2[ch(ξ1)-cos(ξ2)]-1/2 = (-1/2)sin(ξ 2) [ch(ξ1)-cos(ξ2)]-3/2 = (-1/2)sin(ξ 2) R3 (∂3R) = 0 => entire third term in the sum is 0, so ignore it from now on The following tedious algebra must then be done: [∂ 1{f1(∂1R)}] = [∂1{ sh(ξ1) * (-1/2)sh( ξ1) R3}] = (-1/2) ∂ 1[sh2(ξ1)R3] = (-1/2) [ 2sh( ξ1)ch(ξ1) R3 + sh2(ξ1)3R2(∂1R) ] = (-1/2) [ 2sh( ξ1)ch(ξ1) R3 + sh2(ξ1)3R2(-1/2)sh(ξ1) R3 ] = (-1/2) [ 2sh( ξ1)ch(ξ1) R3 – sh3(ξ1)(3/2)R5 ] = (-1/4)sh(ξ 1)R5 [ 4ch(ξ1) R-2 – 3sh2(ξ1) ] = (-1/4)sh(ξ 1)R5 [ 4ch(ξ1) [ch(ξ1)-cos(ξ2)] – 3sh2(ξ1) ] = (-1/4)sh(ξ 1)R5 [ 4ch2(ξ1) - 4 ch(ξ1) cos(ξ2)] – 3sh2(ξ1) ] = (-1/4)sh(ξ 1)R5 [ 3 + ch2(ξ1) - 4 ch(ξ1) cos(ξ2)] [∂ 2{f2(∂2R)}] =[∂2{1 *(∂2R)}] = ∂2[(-1/2) sin( ξ2) R3 = (-1/2)[ cos( ξ2)R3 + sin(ξ2)3R2 (∂2R) ] = (-1/2)[ cos( ξ2)R3 + sin(ξ2)3R2 (-1/2)sin( ξ2) R3 ] = (-1/2)[ cos( ξ2)R3 – sin2(ξ2)(3/2)R5 ] = (-1/2) R5 [ cos(ξ2)R-2 – sin2(ξ2)(3/2)] = (-1/4) R5 [ 2cos(ξ2) [ch(ξ1)-cos(ξ2)] – 3sin2(ξ2)] = (-1/4) R5 [ 2cos(ξ2) ch(ξ1)-2cos2(ξ2) – 3sin2(ξ2)] = (-1/4) R5 [ 2cos(ξ2) ch(ξ1) - 2 - sin2(ξ2)] which can be summarized as [∂ 1{f1(∂1R)}] = (-1/4) R5 [ 3 + ch2(ξ1) - 4 ch(ξ1) cos(ξ2)] sh(ξ1) [∂2{f2(∂2R)}] = (-1/4) R5 [ 2cos(ξ2) ch(ξ1) - 2 - sin2(ξ2)] The denominator factors of the Q expression are these, h12f1R = a2R4 sh(ξ1) R = a2 sh(ξ1) R5 h22f2R = a2R4 1 R = a2 R5 14 Now the LHS of (3.9) can be evaluated: Σn (1/[hn2fnR]) [∂n{fn(∂nR)}] = [ a 2 sh(ξ1) R5]-1 * (-1/4)sh( ξ1)R5 [ 3 + ch2(ξ1) - 4 ch(ξ1) cos(ξ2)] + [a2 R5]-1 * (-1/4) R5 [ 2cos(ξ2) ch(ξ1) - 2 - sin2(ξ2)] = (-1/4a 2) [ 3 + ch2(ξ1) - 4 ch(ξ1) cos(ξ2)] + (-1/4a2) [ 2cos(ξ2) ch(ξ1) - 2 - sin2(ξ2)] = (-1/4a2) [3 + ch2(ξ1) - 4 ch(ξ1) cos(ξ2)+ 2cos(ξ2) ch(ξ1) - 2 - sin2(ξ2) ] = (-1/4a2) [1 + ch2(ξ1) - 4 ch(ξ1) cos(ξ2)+ 2cos(ξ2) ch(ξ1) - sin2(ξ2) ] = (-1/4a2) [cos2(ξ2) + ch2(ξ1) - 2 ch(ξ1) cos(ξ2)] = (-1/4a2) [ch(ξ1)-cos(ξ2)]2 = (-1/4a2) R-4 Therefore (3.9) says -k12/Q(123) = Σn (1/[hn2fnR]) [∂n{fn(∂nR)}] = (-1/4a2) R-4 = (-1/4) / {a2R4} so one can make this simple partitioning k12 = α1 = (1/4) Q = a2R4 = a2[ch(ξ1)-cos(ξ2)]2 which agrees with Moon and Spencer page 97. So after this brute force sum computation, Q come s out being a very simple function. The toroidal information found so far can now be summarized : h 1 = h2 = a/[ch(ξ 1)-cos(ξ2)] => h 1 = h2 = aR2 H = sh( ξ1)a3R6 h3 = a sh(ξ 1)/[ch(ξ1)-cos(ξ2)] => h 3 = a sh(ξ 1)R2 f 1(1) = sh(ξ1) g 1(23) = a R = [ch(ξ 1)-cos(ξ2)]-1/2 f2(2) = 1 g 2(31) = a sh( ξ1) Q = a2[ch(ξ1)-cos(ξ2)]2 = a2R4 f3(3) = a g 3(12) = 1/ sh( ξ1) k 12 = (1/4) Finish Processing So here is where things stood pri or to the above example : (Lψ)/ψ = Σn[ (1/[hn2Xnfn]) [∂n{fn(∂nXn)}] – Σn[ (1/[hn2Rfn]) [∂n{fn(∂nR)}] + K 12 = 0 (3.8) Σ n (1/[hn2fnR]) ∂n{fn(∂nR)} = - k 12/ Q (3.9) Therefore, (Lψ)/ψ = Σ n(1/[hn2Xn]) (1/fn)∂n[fn(∂nXn)] + k12/Q + K12 = 0 (3.10) 15 For later reference, notice in (3.10) the position of the two copies of function f n. Equation (3.10) is not separated because h n= hn(123) and Q = Q(123). At least in the simple separation case the Q term goes away, since as noted above k 12 = 0 and Q = 1. This concludes the "initial processing of the Helmholtz equation". The PDE (3.10) essentially is the Helmholtz equation Lψ = (∇2+K12)ψ = 0 where various assumptions have been made: the form ψ = X1X2X3/R, equation (3.5), and the definition of Q and k 12 from (3.9). At this point, then, for a given curvilinear coordinate system, we know all these items in full detail: K12 hn(123) H(123) R(123) f n(n) gn(≠n) Q(123) k 12 Comment on separation in Cartesian coordinates Cartesian coordinates have h n = 1 and so f n = 1, R=1, Q = 1, k 12 = 0 and (3.10) reads Σ n (1/Xn) (∂n2Xn) + K12 = 0 ( 3 . 1 1 ) or (1/X 1) (∂12X1) + (1/X2) (∂22X2) + (1/X3) (∂32X3) + K12 = 0 (3.11a) If one were to fix ξ 2 and ξ3 and vary just ξ1, one would conclude that (1/X 1) (∂12X1) is a constant, since the other three terms in (3.11a) are then constant. By this argument, one is led to claim that (1/X 1) (∂12X1) = c1 (1/X 1) (∂12X1) = c2 (1/X 1) (∂12X1) = c3 ( 3 . 1 2 ) and then (3.11a) says c 1 + c2 + c3 = -K12 ( 3 . 1 3 ) One can then regard any two of the c i as free parameters, say c 1 and c2, then the third is not free and is constrained to be c 3 = K12 - c1- c2. The two free parameters are called separation constants . This is the simple way separation works in Cartesian coordinates, and it works this way because each term in the sum (3.11) is completely "detangled" from the other two terms in the sense that the first term does not involve variables ξ2 and ξ3 whatsoever. In general curvilinear coordinates, there will still be two separation constants, which will be called k 22 and k32, but the simple arrangement shown in (3.12) does not obtain, and both the free parameters will (in the general case) appear in each of the three separated equations! As shown below, in curv ilinear coordinates the separated equations end up having this form (1/f nXn)∂n[fn(∂nXn)] + [ K12Φn1(ξn) + k22Φn2(ξn) + k32Φn3(ξn)] = 0 n = 1,2,3 where the Φ ni(ξn) are a set of nine functions which can in principle all be different (and they are all different in ellipsoidal coordinates, for example, see Section 12). Here the two separation constants k 22 and k32 appear in each of the three separated equations as advertised, and they cannot be simply extracted as in (3.12). Nevertheless, they are free parameters a nd are still called separation constants. That is to say, all three equations above will be valid for any values of k 22 and k32. 16 4. Starting from the other end: The Stäckel Matrix Φ and a F ormulation of the Problem The goal is to find differential operators L n in terms of which (3.10) can be separated. The most general second order linear L n can be written this way, where p n, rn and qn are real functions, LnXn = (1/pn)∂n[pn(∂nXn)] + rn (∂nXn) + qnXn = 0 ( 4 . 1 ) For several reasons, one wants L n to be formally self-adjoint, L=L*. If (f,g) is the scalar product for the space of functions upon which L n acts, L=L* means that (u,Lv) = (Lu,v), where one ignores the contributions of the "parts" terms as L is swung from one side to the other in integration which this scalar product represents. Since L = r n∂nXn => L* = - r n∂nXn, one must have r n= 0 so then L nXn = (1/pn)∂n[pn(∂nXn)] + qnXn = 0 . ( 4 . 1 A ) If one takes r n ≠ 0, the process given below leads to an intractab le set of equations which it is not hard to show is insolvable. Secondly, L n being self-adjoint is essential to the application of Sturm-Liouville theory to the separated equations. Looking now at (3.10), (Lψ)/ψ = Σ n(1/[hn2Xn]) (1/fn)∂n[fn(∂nXn)] + k12/Q + K12 = 0, (3.10) one is highly motivated to set p n = fn in (4.1A) and then the candidate form for L n is L nXn = (1/fn)∂n[fn(∂nXn)] + qnXn = 0 ( 4 . 2 ) which says (1/X n)(1/fn)∂n[fn(∂nXn)] = - qn ( 4 . 3 ) Inserting (4.3) into (3.10), (Lψ)/ψ = - Σ n(1/hn2)qn + k12/Q + K12 = 0 . (4.4) This then is the final form of the processed Helmholtz equation, assuming that X n satisfies the equation (4.2) . Introduction of the Stäckel Matrix Now co mes a fairly inspired and unexpected step in the development. Suppose one chooses to write the qn(n) function of (4.2) in this manner, wh ich is to say, the unknown function q n(n) is simply written as a linear combination of three other unknown functions, q n(n) = [ κ12Φn1(n) + k22Φn2(n) + k32Φn3(n)] n = 1,2,3 (4.5) 17 where κ12, k22 and k32 are at the moment three arbitrary real constants, and the Φnm(n) are the 9 functions of what is called the Stäckel matrix mentioned in Section 2 above, Φ = ⎝⎜⎜⎛ ⎠⎟⎟⎞ Φ11(1) Φ12(1) Φ13(1) Φ21(2) Φ22(2) Φ23(2) Φ31(3) Φ32(3) Φ33(3) S( Φ) ≡ det(Φ) (4.6) M n(Φ) ≡ cof(Φn1) = (-1)n+1 minor(Φn1) (4.7) With (4.5) used in (4.2 ), the assumed form for L n becomes LnXn = (1/fn)∂n[fn(∂nXn)] + [ κ12Φn1(n) + k22Φn2(n) + k32Φn3(n)]Xn = 0 (4.8) and the processed Helmholtz equation (4.4) becomes Q(Lψ)/ψ = - Σ n(Q/hn2) [ κ12Φn1(n) + k22Φn2(n) + k32Φn3(n)] + k 12 + QK12 = 0 (4.9) This then is the form the processed Helm holtz equation takes if one expands the q n of (4.4) as shown in (4.5). The goal is to find a set of functions Φnm which makes (4.9) be true. It is not obvious that such a set of Φnm exists. The sum is still entangled due to h n(123) and Q(123). Formulation of separation Problems A, B and C Problem A: (simple separation of the Helmholtz equation). In this case R = 1, so Q = 1 and k 12 = 0 in (4.9), and set κ12 = K12 in both (4.8) and (4.9) to get L nXn = (1/fn)∂n[fn(∂nXn)] + [ K12Φn1(n) + k22Φn2(n) + k32Φn3(n)]Xn = 0 (4.8A) ( Lψ)/ψ = - Σn(1/hn2) [ K12Φn1(n) + k22Φn2(n) + k32Φn3(n)] + K 12 = 0 (4.9A) Problem B: (R-separation of the Laplace Equation). In this case K 12 = 0 in (4.9), and set κ12 = k12 in both (4.8) and (4.9) to get L nXn = (1/fn)∂n[fn(∂nXn)] + [ k12Φn1(n) + k22Φn2(n) + k32Φn3(n)]Xn = 0 (4.8B) Q(L 0ψ)/ψ = - Σn(Q/hn2) [k12Φn1(n) + k22Φn2(n) + k32Φn3(n)] + k 12 = 0 (4.9B) where L0 = ∇2 is the Laplace operator. Comparing the A and B equations, one sees that Problem A is a special case of Problem B having Q = 1 and k 12 = K12. Below in the systematic solution of Pr oblem B, we may apply any intermediate equation of that solution to Problem A by making th ese two substitutions (and R= 1). This explains why we (and Morse and Feshbach) gave the constant in (3.9) the strange name k 12. 18 One now treats Problem B as an abstract mathematical problem. That mathematical problem is to find a set of 9 functions Φnm and 3 functions f n that makes (4.8B) and (4.9B) be consistent. The method used for solving this problem is straig htforward, the details appear in the next section. One requires that the coefficient of each k i2 in (4.9B) vanish, since the equation should be true for all real values of the three k i2 parameters. One then writes (4.9B) as a vector equation ΦT V = W where V n = (Q/hn)2 and Wn= δn,1. This equation is inverted to obtain a simple expression for the cofactors M n of the elements of the first column of Φ . From these cofactors one can (hopefully) deduce the 6 components of the rightmost two columns of Φ. We postulate a second condition as an ansatz (the Robertson Condition), and this condition then leads (potentially) to an ev aluation of the elements of the first column of Φ. One then has all of Φ! The functions f n are determined from (3.5). If this program can be completed successfully, the separated L n are obtained, and the Robertson ansat z is justified. The solution will be summarized as a sequence of Steps that one must execute one at a time. A list of Conditions is first stated which, if met, ensure that the Steps can be carried out. Problem C: (R-separation of the Helmholtz Equation with K 12 ≠0). Here one has the full bore (4.9) to contend with - Σ n(Q/hn2) [ κ12Φn1(n) + k22Φn2(n) + k32Φn3(n)] + k 12 + QK12 = 0 (4.9C) Remember that Q = Q(123) is (in general) a function of all three coordinates. It is not clear how to decouple that last two terms in this equation. What should κ12 be set to? The method just outlined above for solving Problem B does not work for Problem C. Moon and Spencer state (p 96, 1961) that no curvilinear system has ever been found in which the Helmholtz equation with K 12 ≠ 0 separates via R- separation (by which they mean with R ≠ constant). So having at least stated it, we shall give up on Problem C and continue now to solve Problem B. 19 5. Recasting Problem B into a new form Equations (4. 9B) and (4.8B) are replicated here with new equations numbers, - Σn(Q/hn2) [k12Φn1(n) + k22Φn2(n) + k32Φn3(n)] + k 12 = 0 (5.1) L nXn = (1/fn)∂n[fn(∂nXn)] + [ k12Φn1(n) + k22Φn2(n) + k32Φn3(n)]Xn = 0 (5.2) If one can find the 3 functions f n appearing in (5.2) and the 9 functions Φnm of the Stäckel matrix that solve (5.1), then (5.2) gives th e separated ODE for functions X n. The plan is to find a solution of (5.1) which is valid for all three k n2 constants having arbitrary (complex in fact) values. Such a solution requires that the coefficients of each k n2 be set to 0 so that Σ n (Q/hn2) Φn1( n ) = 1 Σn (Q/hn2) Φn2(n) = 0 Σn(Q/hn2) Φn3(n) = 0 To put this into a standard vector equation form, use [ Φ T]mn(n) ≡ Φnm (n) so that Σ n ΦT 1n(n) (Q/hn2) = 1 Σn ΦT 2n(n) (Q/hn2) = 0 Σn ΦT 3n(n) (Q/hn2) = 0 and the vector equation is then Φ T V = W w h e r e ( 5 . 3 ) V = ⎝⎜⎜⎛ ⎠⎟⎟⎞ Q/h12 Q/h22 Q/h32 W = ⎝⎜⎛ ⎠⎟⎞1 0 0 Φ = ⎝⎜⎜⎛ ⎠⎟⎟⎞ Φ11(1) Φ12(1) Φ13(1) Φ21(2) Φ22(2) Φ23(2) Φ31(3) Φ32(3) Φ33(3) S ≡ det(Φ) This equation can easily be solved for V using the rule that A-1 = cof(AT)/det(A) : V = { cof(Φ T,T)/det(ΦT)} W = cof( Φ) W /det(Φ) = (1/ S(Φ )) cof(Φ) W In components V n = (1/ S(Φ)) [cof(Φ)]nm Wm = (1/S(Φ)) [cof(Φnm)] δm,1 = (1/S(Φ )) [cof(Φn1)] (5.4) Following tradition, define M n according to M n = cof(Φ n1) = the cofactor of the first element in the nth row of Φ (5.5) and insert this on the right of (5.4) and V n = (Q /hn2) on the left, giving the Cofactor Conditions , 20 Mn/S = (Q /h n2) ( 5 . 6 ) Alternatively, write (5.6) as M n(Φ)/S(Φ) = (Q/h n2) n=1,2,3 " Cofactor conditions" (5.6a) and this emphasizes that there are 3 conditions that the Φ matrix must satisfy for a given coordinate system with scale factors h n. Notice that this conditions arise from the assumed form for L n in (5.2), they are not "imposed". The reason that cofactors appear in (5.6) is because one had to invert the matrix ΦT and matrix inversion always gives rise to cofactors. Since W was the unit vector e1, only the cofactors of the first column of Φ play a role. If the three cofactor conditions are satisfied by some matrix Φ , then equation (5.1) is satisfied. But equation (5.1) is the "ground down" Helmholtz equation, so this provides a solution to the initial problem. Moreover, equation (5.2) then gives the L n operators such that L nXn = 0 from which the X n may be determined. [Repeated Warning : Morse and Feshbach use cofactor M n exactly as described above, but unfortunately they refer to M n as a "minor", and they never use the word "c ofactor". In modern parlance, cofactors are minors with signs added, which make a lot of difference. ] The next step is to impose an extra condition on the Φ matrix known as the Robertson Condition . This is done just as an ansatz to see if it helps find a solution. The requirement is that S = det( Φ) take this form, S(Φ) = H / (f 1f2f3QR2) "Robertson condition" (5.7) It is not obvious that, for a given coordinate system, one can even find a matrix Φ which satisfies the Cofactor and Robertson conditions, but we shall try to construct a solution Φ. Combining (5.6) and (5.7) gives M n = (SQ/h n2) = (H/h n2) /(f1f2f3R2) ( 5 . 8 ) The Robertson condition causes the solution M n to be expressed directly in terms of known quantities. If a full solution matrix Φ can be found, then this Robertson conditi on is retroactively justified. Using (3.5) which says (H/h n2) = fngn R2, (5.8) can be written in an alternate form M n = gnfn/(f1f2f3) ( 5 . 8 a ) or M 1 = g1 /(f2f3) M2 = g2 /(f3f1) M3 = g3 /(f1f2) ( 5 . 8 b ) These are the three cofactors of the elements of the first column of Φ. Notice there are no factors of R or Q in (5.8b). This was the motivation for installing th e Q and R factors in the Robertson condition as done above in (5.7). 21 It remains to find the elements of the first column of Φ. They appear in this standard expansion of the determinant of Φ S = Φ11(1) M1 + Φ21(2) M2 + Φ31(3) M3 ( 5 . 9 ) Install now S from (5.7) and the M n from (5.8) to get [ H / (f1f2f3QR2)] = Φ11(1) (H/h12) /(f1f2f3R2) + Φ21(2) (H/h22) /(f1f2f3R2) + Φ31(3) (H/h32) /(f1f2f3R2) or 1/Q = Φ11(1) (1/h12) + Φ21(2) (1/h22) + Φ31(3) (1/h32) ( 5 . 1 0 ) This is a non-trivial functional-form equation that may or may not have solutions for the Φ n1(n). One is now left with this problem: Given S(Φ) = H / (f 1f2f3QR2) // Robertson Mn(Φ) = S(Φ ) (Q/hn2) / / C o f a c t o r ( 5 . 1 1 ) how exactly does one find the 9 elements of Φ ? This is Problem B restated in a simple form. If one can solve for the elements of matrix Φ , then the Robertson condition is justified because it led to a solution of equation (5.1). Having thus reformulated the problem of finding a separable solution to the Helmholtz equation as a problem of finding the matrix Φ solving (5.11), we put our development in temporary stasis to discuss the non-uniqueness of the Stäckel matrix Φ. The development then resumes in Section 7. 22 6. The non-uniqueness of Stäckel matrix Φ: Equivalence Ope rations Suppose one has successfully found a solution for Φ satisfying (5.11). Write it as Φ = ⎝⎜⎛ ⎠⎟⎞ a(1) b(1) c(1) d(2) e(2) f(2) g(3) h(3) i(3) S = det(Φ ) The cofactors of the elements of the first column are given by M 1(23) = e(2)i(3)-f(2)h(3) -M2(31) = b(1)i(3)-c(1)h(3) M3(12) = b(1)f(2)-c(1)e(2) (1) Look what happens if one scales up the elements of column 2 by constant factor α, and scales down those of column 3 by the same factor. The above three equations become M' 1(23) = αe(2) (1/α)i(3)- (1/α )f(2)α h(3) = e(2)i(3)-f(2)h(3) = M 1(23) -M'2(31) = αb(1) (1/α)i(3)-(1/α)c(1)α h(3) = b(1)i(3)-c(1)h(3) = -M 2(31) M'3(12) = αb(1) (1/α)f(2)- (1/α )c(1)α e(2) = b(1)f(2)-c(1)e(2) = M 3(12) Since S = a(1) M 1 + d(2) M 2 + g(3) M 3 and the M's stayed the same, S also stays the same, and (5.11) is still satisfied. Therefore, this up/down scaling of the last two columns is an "equivalence" operation for Φ -- it creates a new Stäckel matrix th at is just as good as the original Φ . In the special case α = -1, one sees that negating the last two columns of a Stäckel matrix is an equivalence operation. (2) Next, suppose one swaps columns 2 and 3 and then negates either of the swapped columns. For example, swapping columns 2 and 3 and then negating column 2 gives, Φ ' = ⎝⎜⎛ ⎠⎟⎞ a(1) -c(1) b(1) d(2) -f(2) e(2) g(3) -i(3) h(3) S = det(Φ ) // Φ = ⎝⎜⎛ ⎠⎟⎞ a(1) b(1) c(1) d(2) e(2) f(2) g(3) h(3) i(3) A quick inspection shows that M' n = Mn. Also, swapping two columns negates the determinant, and then negating a column restores it. Negating the third column instead of the second is the same as taking Φ' as shown above and negating both colu mns 2 and 3. But from (1) that is an equivalence operation. (3) Adding a multiple of one of the last two columns to the other is also an equivalence operation as is now shown. Linear algebra says that such an opera tion does not change S. Suppose one adds a multiple α of column 2 to column 3: 23 Φ = ⎝⎜⎛ ⎠⎟⎞ a(1) b(1) c(1) d(2) e(2) f(2) g(3) h(3) i(3) S = det(Φ ) c'(1) = c(1) + α b(1) f'(2) = f(2) + α e(2) i'(3) = i(3) + α h(3) Then : M 1'(23) = e(2)i'(3)-f'(2)h(3) = e(2)[ i(3) + α h(3)]-[ f(2) + α e(2)]h(3) = e(2)i(3)-f(2)h(3) + α [e(2) h(3) - e(2) h(3)] = e(2)i(3)-f(2)h(3) = M 1(23) and similarly for the other two M n. Of course one could add a multiple of one of the last two columns to the first column, and that would also be an equivalence operation since the M n and S are both unaltered. Summary of Stäckel Matrix Equivalence Operations (1) multiply one of the last tw o columns by any (nonzero) constant α and the other by 1/ α. (6.1) (2) swap the last two columns of Φ and then negate either of these columns. (6.2) (3) add any multiple of one of the last two columns to a different column. (6.3) These rules are useful if one computes a Stäckel ma trix and it does not agree with a Stäckel matrix appearing in the literature. Some examples will be shown in Section 12. 24 7. Solving Problem B To save writing lots of subscripts, write the to-be-determined Φ matrix as shown above, Φ = ⎝⎜⎛ ⎠⎟⎞ a(1) b(1) c(1) d(2) e(2) f(2) g(3) h(3) i(3) S = d e t ( Φ) (7.1) The cofactors of interest appear in these equations M 1(23) = e(2)i(3)-f(2)h(3) -M2(31) = b(1)i(3)-c(1)h(3) M3(12) = b(1)f(2)-c(1)e(2) (7.2) and now insert the M n expressions from (5.8b) to get g 1/(f2f3) = e(2)i(3)-f(2)h(3) - g2/(f3f1) = b(1)i(3)-c(1)h(3) g3/(f1f2) = b(1)f(2)-c(1)e(2) (7.3) Now define rescaled upper-case functions of the form E(2) = e(2) f 2( 2 ) ( 7 . 4 ) and similarly for the other five functi ons appearing above, so (7.3) becomes, g1(23) = E(2)I(3)-F(2)H(3) - g2(31) = B(1)I(3)-C(1)H(3) g3(12) = B(1)F(2)-C(1)E(2) (7.5) where the g n are those helper functions appearing in (3.5). Recall that it was assumed at the start that the curvilinear coordinate system and its scale factors h n were compatible with the three equations of (3.5) and that one could find a set of 7 functions f n, gn and R, all unique apart from constant allocation. In particular, the analysis of (3.5) yields the three g n which appear on the LHS of equations (7.5). The question is now whether these g n have the functional form shown in (7.5) ! For example, does the g 1(23) obtained from (3.5) have a form which is at most tw o terms each showing simple factorization? One has to regard (7.5) as three functional-form conditions for the chosen coordinate system! If the g n don't have the functional form shown in (7.5), then given the three cofactors M n , one cannot successfully obtain the elements of the rightmost two columns of Φ and the entire plan collapses. Suppose the g n do in fact have the functional form shown in (7.5). Even in this case, it is not clear that (7.5) can be solved for the cap ital letter functions. The reason is that there is correlation in the three equations in that each function shows up in two equa tions. So even if the functional form requirements are met, one still have to actually be able to solve 7.5 for the 6 capital letter functions. If this can be done, then one uses (7.4) to find the lo wer case functions which are the actual Φ matrix elements for the rightmost two columns. 25 In many examples studied below, each of equations (7.5) will have only a single term on the RHS. In this case, at least two of the capital letter functions will vanish. (But this does not happen in Section 12.) We now turn to the S equation which is this from (5.10) 1/Q = a(1) (1/h 12) + d(2) (1/h 22) + g(3) (1/h 32) ( 7 . 6 ) Of course the (1/h n2) and Q(123) objects are completely determin ed by the choice of coordinate system, so once again one has a functional-form question con cerning the solvability of (7.6). To see if it is solvable, we have to examine a specific curvilinear sy stem, or perhaps a family of such systems. When (7.6) has a solution, one often finds that two of the thr ee matrix elements a(1), d(2), g(3) can be set to 0. In the next two subsections, we summarize first th e Conditions needed for a successful separation, then the sequence of Steps needed to actually do the separation. (a) Conditions required for Separability If the following conditi ons are all met, then Problem B (and therefore Problem A) has a solution, and the corresponding curvilinear coordinate system is separable. Condition (1) Equations (3.5) must be solvable for the 7 functions f n, gn and R. If some f n is a constant, that constant is set to 1. ( If one is doing Problem A, R=1.) (H/h 12) = f1(1)g1(23) R2 H = h 1h2h3 (H/h 22) = f2(2)g2(31) R2 (H/h 32) = f3(3)g3(12) R2 (3.5) Assuming one can solve equations (3.5), one computes the M n as follows, M 1 = g1 /(f2f3) / / = g 1f1 / (f2f2f3) M2 = g2 /(f3f1) M3 = g3 /(f1f2) (5.8b) Condition (2) Equations (7.2) must have a solution. This is a triplet of functional-form conditions, where the Mn are as given above, M1(23) = e(2)i(3)-f(2)h(3) -M2(31) = b(1)i(3)-c(1)h(3) M3(12) = b(1)f(2)-c(1)e(2) (7.2) Condition (3) If conditions (1) and (2) are met, one th en does the work of computing Q and k 12 from (3.9), Σn (1/[hn2fnR]) ∂n{fn(∂nR)} = - k 12/Q(123) (3.9) 26 For Problem A, Q = 1 and k 12 = 0 and no work is needed. Then with this Q expression, equation (7.6) mu st have a solution for a, d and g, 1/Q = a(1) (1/h 12) + d(2) (1/h 22) + g(3) (1/h 32) (7.6) and this is another functional-form condition. (b) Steps for finding Φ and related quantities Here it is assumed that the Conditions of 7(a) are all met, so we proceed with our Stäckel solution. Some of the Steps listed here were already carried out to check the Conditions, but they are listed here anyway. If one knows ahead of time that separation is going to work, one can ignore the Conditions and then just carry out the Steps listed here. Step 0: Write down the h n for the curvilinear system of interest and compute H = h 1h2h3. Perhaps write down other useful facts concerning the system of interest. Step (1) As noted above, the first task is to solve (3.5) for the 7 functions f n, gn and R (H/h 12) = f1(1)g1(23) R2 (H/h 22) = f2(2)g2(31) R2 (H/h 32) = f3(3)g3(12) R2 (3.5) This task is pretty much just one of "insp ection" when the LHS's of (3.5) are inserted. Step (2) One can then immediately write down the three first-column cofactors from (5.8b) M1 = g1 /(f2f3) M2 = g2 /(f3f1) M3 = g3 /(f1f2) (5.8b) Step (3) If Problem A, then Q = 1. Otherwise compute Q and k 12 from (3.9), Σn (1/[hn2fnR]) ∂n{fn(∂nR)} = - k 12/Q(123) ( 3 . 9 ) Step (4) Knowing Q, compute S from the Robertson condition, S(Φ) = H / (f 1f2f3QR2) "Robertson condition" (5.7) 27 Step (5) One must next find the rightmost two columns of the Stäckel matrix, Φ = ⎝⎜⎛ ⎠⎟⎞ a(1) b(1) c(1) d(2) e(2) f(2) g(3) h(3) i(3) S = d e t ( Φ) (7.1) This can be done by solving the following equation set, using the M n found in Step (2) M 1 = e(2)i(3)-f(2)h(3) -M2 = b(1)i(3)-c(1)h(3) M3 = b(1)f(2)-c(1)e(2) (7.2) Step (6) One must next find the first column of the Stäckel matrix by solving (7.6) or (5.9) 1/Q = a(1) (1/h 12) + d(2) (1/h 22) + g(3) (1/h 32) (7.6) S = a(1) M 1 + d(2) M 2 + g(3) M 3 (5.9) Step (7) At this point, one may want to apply some of the Section 6 equivalence operations to obtain a Stäckel matrix Φ that is of the simplest possible form, or of a form that matches the literature. (c) Problem B Example (R-separation of Laplace): Toroidal Coordinates The Φ ma trix Section 3 ( above (3.8)) states our accumulated facts about toroidal coordinates and their separation, and those facts are copied here : h 1 = h2 = a/[ch(ξ 1)-cos(ξ2)] => h 1 = h2 = aR2 H = sh( ξ1)a3R6 h3 = a sh(ξ 1)/[ch(ξ1)-cos(ξ2)] => h 3 = a sh(ξ 1)R2 f 1(1) = sh(ξ1) g 1(23) = a R = [ch(ξ 1)-cos(ξ2)]-1/2 f2(2) = 1 g 2(31) = a sh( ξ1) Q = a2[ch(ξ1)-cos(ξ2)]2 = a2R4 f3(3) = a g 3(12) = 1/ sh( ξ1) k 12 = (1/4) Steps (0),(1) and (3) of Section 7 (b) have already been carried out. For Step (2) compute the cofactors, M 1 = g1 /(f2f3) = a/a = 1 M2 = g2 /(f3f1) = a sh(ξ 1)/ a sh(ξ1) = 1 M3 = g3 /(f1f2) = (1/sh( ξ1))/(sh(ξ1) = 1/sh2(ξ1) (5.8b) 28 For Step (4) one has S(Φ) = H / ([f 1f2f3]QR2) = sh(ξ1)a3R6/ ([a sh(ξ 1)] a2R4 R2) = 1 For Step (5) : M1(23) = 1 = e(2)i(3)-f(2)h(3) -M2(31) = -1 = b(1)i(3)-c(1)h(3) M3(12) = 1/sh2(ξ1) = b(1)f(2)-c(1)e(2) (7.2) We try h(3) = 0 and f(2) = 0 (trial and error!) 1 = e(2)i(3) -1 = b(1)i(3) 1/sh 2(ξ1) = -c(1)e(2) The first two lines say e(2) = 1, i(3) = 1 and b(1) = -1. The third line is then 1/sh 2(ξ1) = -c(1) The Stäckel matrix at this point is then Φ = ⎝⎜⎛ ⎠⎟⎞ a(1) b(1) c(1) d(2) e(2) f(2) g(3) h(3) i(3) = Φ = ⎝⎜⎜⎛ ⎠⎟⎟⎞ a(1) -1 -1/sh2(ξ1) d(2) 1 0 g(3) 0 1 For Step (6) write 1/Q = a(1) [1/h 12] + d(2) [1/h 22] + g(3) [1/h 32] (7.6) 1/( a 2R4) = a(1) [1/ (a2R4)] + d(2) [1/ (a2R4)] + g(3) [1/ (a2R4 sh2(ξ1))] 1= a(1) + d(2) + g(3) [1/ (sh 2(ξ1))] A solution here is a(1) = 1 and d(2) = g(3) = 0. So here is the filled-in Φ matrix along with our results found above: Φ = ⎝⎜⎜⎛ ⎠⎟⎟⎞ a(1) -1 -1/sh2(ξ1) d(2) 1 0 g(3) 0 1 = ⎝⎜⎜⎛ ⎠⎟⎟⎞ 1 -1 -1/sh2(ξ1) 0 1 0 0 0 1 ( 7 . 7 ) f1(1) = sh(ξ1) g 1(23) = a R = [ch(ξ 1)-cos(ξ2)]-1/2 f2(2) = 1 g 2(31) = a sh( ξ1) Q = a2[ch(ξ1)-cos(ξ2)]2 = a2R4 f3(3) = a g 3(12) = 1/ sh( ξ1) k 12 = (1/4) 29 M1 = 1 S = 1 M2 = 1 M3 = 1/sh2(ξ1) These results are in agreement with Moon and Sp encer p 112, so Step (7) is not needed. The separated equations and their solutions: toroidal harmonics Here then are the separated equations from (5.2) LnXn = (1/fn)∂n[fn(∂nXn)] + [ k12Φn1(n) + k22Φn2(n) + k32Φn3(n)]Xn = 0 (5.2) We now install the Φ matrix elements from (7.7) above showing enough lines for the reader to quickly verify each step: L1X1 = (1/f1)∂1[f1(∂1X1)] + [ k12Φ11(1) + k22Φ12(1) + k32Φ13(1)]X1 = 0 L1X1 = (1/f1)∂1[f1(∂1X1)] + [ k12 1 + k22(-1) + k32(-1/sh2(ξ1)]X1 = 0 L1X1 = (1/sh(ξ1))∂1[sh(ξ1) (∂1X1)] + [ k12 - k22 + k32(-1/sh2(ξ1)]X1 = 0 L1X1 = (1/sh(ξ1)){ sh(ξ1) ∂12X1 + ch(ξ1) ∂1X1} + [ k12 - k22 - k32/sh2(ξ1)]X1 = 0 L1X1 = ∂12X1 + coth(ξ 1) ∂1X1 + [(1/4) - k 22 - k32/sh2(ξ1)]X1 = 0 L 2X2 = (1/f2)∂2[f2(∂2X2)] + [ k12Φ21(2) + k22Φ22(2) + k32Φ23(2)]X2 = 0 L2X2 = (1/1)∂2[f1(∂2X2)] + [ k120 + k221 + k320]X2 = 0 L2X2 = ∂22X2 + k22X2 = 0 L 3X3 = (1/f3)∂3[f3(∂3X3)] + [ k12Φ31(3) + k22Φ32(3) + k32Φ33(3)]X3 = 0 L3X3 = (1/a)∂3[a(∂3X3)] + [ k120 + k220 + k321]X3 = 0 L3X3 = ∂32X3 + k32X3 = 0 So here are the separated equations : L 1X1 = ∂12X1 + coth(ξ 1) ∂1X1} + [(1/4) - k 22 - k32/sh2(ξ1)]X1 = 0 L2X2 = ∂22X2 + k22X2 = 0 L3X3 = ∂32X3 + k32X3 = 0 in agreement with Moon and Spencer page 114. For this system, the appear ance of the separation constants in the last two separated equations looks just like our c 2 and c3 equations shown in (3.12) for Cartesian coordinates. However, th e first equation involves both separati on constants in a tangled manner. For ellipsoidal coordinates in Section 12, one finds full entanglement in all three separation equations. Moon and Spencer use k 22 = α2 = p2 and k32 = α3 = q2. Whereas the X 2 and X3 are simple trig functions, the X1 (it turns out) are Legendre functions of the type Pq p-1/2 (chξ1). Thus, in a most-general toroidal problem, the "toroidal harmonics" would be ( we have added 1/R on the right to get ψ = X1X2X3/R) 30 [Pq p-1/2 (chξ1), Qq p-1/2 (chξ1) [ sin(pξ2), cos(pξ2) ] [ sin(q ξ3), cos(qξ3) ] * [ch( ξ1)-cos(ξ2)]1/2 This notation means that each bracket [f 1, f2] can be an arbitrary linear combination Af 1+Bf2 and the constants can be different for each bracket. The functions are called "harmonics" because they are functions which solve the Laplace equation. Elsewhere we refer to the above combinations as "atoms" or "atomic forms" with the idea that one can assemble a problem solution by linearly combining the atoms. Since ξ3 is an azimuthal coordinate, in a problem with a full 2 π azimuthal range one would find that k 3 = q was an integer, and in an problem with azimuthal symmetry one finds q = 0. Legendre functions of this kind are called toroidal or ring functions. If k 22 = p2 < 0, one writes p = i τ to get [P q iτ-1/2(chξ1), Qq iτ-1/2(chξ1) [ sh(τξ2), ch(τξ2) ] [ sin(q ξ3), cos(qξ3) ] * [ch( ξ1)-cos(ξ2)]1/2 and Legendre functions of this ty pe are called Mehler functions. (d) Problem A Example (simple-separation of He lmholtz ): Circular Cylindrical Coordinates The Φ ma trix Here are the Steps of Section 7 (b) : Step (0) 1,2,3 = ρ,φ,z h 1 = 1 h 2 = ξ1 h 3 = 1 H = ξ1 R = 1 Step (1) Examine equation (3.5) with R=1 (H/h n2) = fn(n)gn(≠n) n = 1,2,3 ξ 1 = f1(1)g1(23) = [ ξ 1 ] [ 1 ] => f 1 = ξ1 g 1 = 1 1/ξ1 = f2(2)g3(31) = [ 1 ] [ 1/ ξ1] => f 2 = 1 g 2 = 1/ξ1 ξ1 = f3(3)g3(12) = [1] [ ξ1] => f 3 = 1 g 3 = ξ1 Step (2) M 1 = g1 /(f2f3) = 1/1 = 1 M2 = g2 /(f3f1) = (1/ξ1)/ξ1 = 1/ξ12 M3 = g3 /(f1f2) = ξ1/ α1 = 1 (5.8b) Step (3) Q = 1 k 12 = 0 Step (4) S( Φ) = H / ([f 1f2f3]QR2) = ξ1/ (ξ1) = 1 31 Step (5) M1(23) = 1 = e(2)i(3)-f(2)h(3) -M2(31) = -1/ ξ12 = b(1)i(3)-c(1)h(3) M3(12) = 1 = b(1)f(2)-c(1)e(2) (7.2) Try h(3) = 0 and f(2) = 0 1 = e(2)i(3) => e(2) = 1 and i(3) = 1 1/ξ 12 = -b(1)i(3) 1 = -c(1)e(2) => c(1) = -1 The second line then says -1/ ξ12 = b(1). The Stäckel matrix at this point is then Φ = ⎝⎜⎛ ⎠⎟⎞ a(1) b(1) c(1) d(2) e(2) f(2) g(3) h(3) i(3) = ⎝⎜⎜⎛ ⎠⎟⎟⎞ a(1) -1/ ξ12 -1 d(2) 1 0 g(3) 0 1 Step (6) 1/Q = a(1) [1/h 12] + d(2) [1/h 22] + g(3) [1/h 32] (7.6) 1 = a(1) 1 + d(2) [1/ ξ12] + g(3) 1 One can then take g(3) = 1 and a(1) = d(2) = 0. So here is the filled-in Φ matrix along with the results found above: Φ = ⎝⎜⎜⎛ ⎠⎟⎟⎞ a(1) -1/ ξ12 -1 d(2) 1 0 g(3) 0 1 = ⎝⎜⎜⎛ ⎠⎟⎟⎞ 0 -1/ξ12 -1 0 1 0 1 0 1 f1(1) = ξ1 g 1(23) = 1 R = 1 f2(2) = 1 g 2(31) = 1/ξ1 Q = 1 ( 7 . 8 ) f3(3) = 1 g 3(12) =ξ1 k 12 = 0 M 1 = 1 S = 1 M2 = 1/ξ12 M3 = 1 These results are in agreement with Moon and Spencer p 12, so we don't need Step (7). 32 The separated equations and their solutions: cylindrical harmonics Here then are the separated equations from (5.2) LnXn = (1/fn)∂n[fn(∂nXn)] + [ k12Φn1(n) + k22Φn2(n) + k32Φn3(n)]Xn = 0 (5.2) We now install the Φ matrix elements from (7.8), L1X1 = (1/f1)∂1[f1(∂1X1)] + [ K12Φ11(1) + k22Φ12(1) + k32Φ13(1)]X1 = 0 L1X1 = (1/ξ1)∂1[ξ1(∂1X1)] + [ -k22/ξ12 - k32]X1 = 0 L1X1 = ∂12X1 +(1/ξ1)(∂1X1) - [ k22/ξ12 + k32]X1 = 0 L 2X2 = (1/f2)∂2[f2(∂2X2)] + [ K12Φ21(2) + k22Φ22(2) + k32Φ23(2)]X2 = 0 L2X2 = (1/1)∂2[1(∂2X2)] + [ K120 + k221 + k320]X2 = 0 L2X2 = ∂22X2 + k22X2 = 0 L 3X3 = (1/f3)∂3[f3(∂3X3)] + [ K12Φ31(3) + k22Φ32(3) + k32Φ33(3)]X3 = 0 L3X3 = (1/1)∂3[1(∂3X3)] + [ K121 + k220 + k321)]X3 = 0 L3X3 = ∂32X3 + (K12 + k32)]X3 = 0 So here are the separated equations : L 1X1 = ∂12X1 +(1/ξ1)(∂1X1) - [ k22/ξ12 + k32]X1 = 0 L2X2 = ∂22X2 + k22X2 = 0 L3X3 = ∂32X3 + (K12 + k32)]X3 = 0 in agreement with Moon and Spencer page 15 (they use K 12 = κ2, k22 = α2 = p2, and k32= α3 = q2). The solutions of the X 1 equation have the form J p(iqξ1). The X 2 is trig of the form sin(k 2ξ2), and X 3 is also trig of the form sin( K12 + k32 ξ3). The "cylindrical harmonics" can then be written (R=1 so ψ = X1X2X3) [Jp(ik3ξ1), Yp(i k3ξ1)] [sin(k2ξ2), cos(k2ξ2)] [sin( K12 + k32 ξ3), cos( K12 + k32 ξ3)] or (ignoring constants) one can put in the usual modified Bessel functions, [Ip(k3ξ1), Kp(k3ξ1)] [sin(k2ξ2), cos(k2ξ2)] [sin( K12 + k32 ξ3), cos( K12 + k32 ξ3)] all of which agrees with Moon and Spencer page 15. Noti ce that if we had selected a(1) = 1 back in Step (6), the constants get shuffled around in the three factors in a trivial manner. The point here is that k 2 and k3 are arbitrary separation constants and one can "shuffle" them however one wants. If k 32 < 0, then k 3 = -iσ3 to get these harmonics instead. [Jp(σ3ξ1), Yp(σ3ξ1)] [sin(k2ξ2), cos(k2ξ2)] [sin( K12 - σ32 ξ3), cos( K12 - s32 ξ3)] 33 8. Separability in Cylindrical Systems when h 1 ≠ α h2 For the moment, assume some general h 1 and h2 where one h is not just a multiple of the other. They define a 2D orthogonal system of some sort. Then just extrude in the "z direction" ( ξ3) to get a 3D cylindrical coordinate system. This gives h1= h1( 1 2 ) ( 8 . 1 ) h2= h2(12) h3=1 H = h 1h2 H/h 12= h2/h1 H/h22 = h1/h2 H/h32 = h1h2 Here then is an examination of the Conditions stated in Section 7 (a) applied to this situation, with a summary at the end: Condition (1) From (3.5) (H/h 12) = f1(1)g1(23) R2 (H/h 22) = f2(2)g2(31) R2 (H/h 32) = f3(3)g3(12) R2 (3.5) One can define some new functions G 1(2) and G 2(1) that might simplify things in terms of functional form, since the h 1 and h2 scale factors don't depend on coordinate 3: (h2/h1) = f1(1) g1(23)R2 => g 1(23) ≡ G1(2) (h1/h2) = f2(2) g2(13)R2 => g 2(13) ≡ G2(1) (h1h2) = f3(3) g3(12)R2 => f3(3) = 1 (8.2) or (h2/h1) = f1(1) G1(2)R2 (h1/h2) = f2(2) G2(1)R2 (h1h2) = g3(12)R2 ( 8 . 3 ) One is still faced with the problem of solving (8.3) for the 6 functions f 1, f2, G1, G2, g3 and R. It is not clear that a solution set exists for some very strange h 1 and h2. For example, suppose h 1 = 1 and h 2 = ξ12-ξ22 . The first equation of (8.3) forces f 1= G1= 1 and then R2= ξ12-ξ22 . Then the second equation says 1/ ξ12-ξ22 = f2(2) G2(1) ξ12-ξ22 and one ends up with f2(2) G2(1) = 1/(ξ12-ξ22) and there is no solution for f 2 and G2 that works. So the duty to find the solutions to (3.5) remains for cylindrical systems. If there is a solution, one can multiply the first two equations of (8.3) to find that R must have this form 1/R 4 = [f1(1) G2(1)] [G1(2) f2( 2 ) ] ( 8 . 4 ) so R is seen to factorize into R = r 1(1) r2(2). One can also compute the three cofactors 34 M1 = g1 /(f2f3) = G1(2)/f2(2) M2 = g2 /(f3f1) = G2(1)/f1(1) M3 = g3 /(f1f2) = g3(12) /[f1(1)f2(2) ] (5.8b) Condition (2) requires the solution of M1(23) = e(2)i(3)-f(2)h(3) -M2(31) = b(1)i(3)-c(1)h(3) M3(12) = b(1)f(2)-c(1)e(2) (7.2) and in the current situation this says G 1(2)/f2(2) = e(2)i(3)-f(2)h(3) - G2(1)/f1(1) = b(1)i(3)-c(1)h(3) g3(12) /[f1(1)f2(2)] = b(1)f(2)-c(1)e(2) From the first line we must choose either i(3) = 0 or h(3) = 0 and we choose the latter. The reader can show that selecting i(3) = 0 introduces nothing new. Then the above becomes G 1(2)/f2(2) = e(2)i(3) - G2(1)/f1(1) = b(1)i(3) g3(12) /[f1(1)f2(2)] = b(1)f(2)-c(1)e(2) From the first two lines select e(2) = G 1(2)/f2(2) i(3) = 1 b(1) = - G 2(1)/f1(1) i(3) = 1 The third line then says g 3(12) /[f1(1)f2(2)] = [- G 2(1)/f1(1)]f(2) - c(1)[ G 1(2)/f2(2) ] g3(12) = [- G 2(1) f2(2) ]f(2) - c(1)f 1(1)G1(2) or - g 3(12) = c(1)f 1(1)G1(2) + f(2) f 2(2) G2(1) (8.5) This is a very strict requirement on the form g 3(12) must have in order to allow the Stäckel separation process to succeed. One is free to select c(1) and f(2), but G 1(2), G2(1) and f 1,f2 are already determined and none of these factors vanishes. Installation of (8.5 ) and then (8.4) into the last line of (8.3) gives 35 -(h1h2) = -g3(12)R2 = [c(1) f 1(1)G1(2) + f(2) f 2(2) G2(1)] R2 = [c(1) f 1(1)G1(2) + f(2) f 2(2) G2(1)] / { [f 1(1) G2(1)]1/2 [f2(2)G1(2)]1/2} = c(1) f1(1)G1(2) / { [f 1(1) G2(1)]1/2 [f2(2)G1(2)]1/2} + f(2) f 2(2) G2(1)] / { [f 1(1) G2(1)]1/2 [f2(2)G1(2)]1/2} = c(1) [f 1(1) G2(1)]1/2 / [f2(2)G1(2)]1/2 + f(2) [f 2(2)G1(2)]1/2 / [f1(1) G2(1)]1/2 Absorbing [f 1(1) G2(1)]1/2 into c(1) and similarly for f(2), c(1) [f 1(1) G2(1)]1/2 ≡ c'(1) [f 2(2)G1(2)]1/2 f(2) ≡ f'(2) , produces this functional form requirement -(h 1h2) = c'(1) / [f 2(2)G1(2)]1/2 + f'(2) / [f 1(1)G2(1)]1/2 ( 8 . 6 ) so one must find c'(1) and f'(2) to satisfy this equation. So condition (8.5) is replaced with a condition directly on the product of the scale factors, and it still looks very strict for some completely arbitrary pair of functions h 1(12) and h 2(12). Equation (8.6) can be regarded as Condition 2. Assuming Conditions (1) and (2) are met as just outlined, one can construct the Stäckel matrix Φ Φ = ⎝⎜⎛ ⎠⎟⎞ a(1) b(1) c(1) d(2) e(2) f(2) g(3) h(3) i(3) = ⎝⎜⎜⎛ ⎠⎟⎟⎞ a(1) - G 2(1)/f1(1) c(1) d(2) G 1(2)/f2(2) f(2) g(3) 0 1 Condition (3) One is supposed to compute Q using (3.9) which says Σn (1/[hn2fnR]) ∂n{fn(∂nR)} = - k 12/Q(123) (3.9) Since ∂ 3R = 0, there will be only two terms in the sum - k 12/Q(123) = (1/[h 12f1R]) [∂1{f1(∂1R)}] + (1/[h 22f2R]) [∂2{f2(∂2R)}] = ( 1 / [ h 12f1R]) [∂1{f1(∂1R)}] + (1/[h 22f2R]) [∂2{f2(∂2R)}] = T1 + T2 Putting these terms into Maple as T1 and T2 with R -4 = f1g1f2g2 from (8.4) , Maple does the derivatives: 36 The details are less important than the general form of the result, visible from the last lines above, T1 = t 1(1) / h12 T2 = t 2(2) / h22 => - k 12/Q(12) = t 1(1) / h12 + t2(2) / h22 ( 8 . 7 ) Now having to some extent computed Q, one must examine condition (7.6), 1/Q = a(1) (1/h 12) + d(2) (1/h 22) + g(3) (1/h 32) ( 8 . 8 ) 1/Q(12) = a(1) (1/h 1(12)2) + d(2) (1/h 2(12)2) + g(3) 1 Since Q = Q(12), g(3) must be a constant. As an ansatz set g(3)=0 to see if a solution results. Then, 1/Q = a(1) (1/h 12) + d(2) (1/h 22) Comparing this to (8.7) one can make these choices a(1) = - t 1(1)/k12 d(2) = - t 2(2)/k12 and of course g(3) = 0 so the Φ matrix now has this form Φ = ⎝⎜⎜⎛ ⎠⎟⎟⎞ a(1) - G 2(1)/f1(1) c(1) d(2) G 1(2)/f2(2) f(2) g(3) 0 1 = ⎝⎜⎜⎛ ⎠⎟⎟⎞ - t1(1)/k12 - G2(1)/f1(1) c(1) - t2(2)/k12 G1(2)/f2(2) f(2) 0 0 1 37 Summary of the three conditions: (1) One must be able to find f 1, f2, G1, G2, g3 and R which satisfy these functional-form equations, (h 2/h1) = f1(1) G1(2)R2 (h1/h2) = f2(2) G2(1)R2 (h1h2) = g3(12)R2 (8.3) (2) One must be able to find c'(1) and f'(2 ) which satisfy this functional form equation, -(h1h2) = c'(1) / [f 2(2)G1(2)]1/2 + f'(2) / [f 1(1) G2(1)]1/2 (8.6) (3) One can meet condition (3) for sure, and the first column of Φ is as shown above . If Q = 1, choose g(3) = 1 and the two elements above it 0. Example: Circular cylindrical coordinates revisited Cy lindrical systems with h 1 ≠ α h2 are uncommon for the reason discussed at the end of the next section, but there is one notable exception: circular cylinder coordinates, which we shall call ρ,φ,z. If one defines the ξn according to ρ,φ,z = exp(ξ1),ξ2,ξ3, then this system fits into the conformal-mapping-of-Cartesians framework and one finds h 1 = h2 = exp(2ξ1), as in Moon and Spencer p 13. But normally one takes ρ,φ,z = ξ1,ξ2,ξ3 so that h 1 = 1 and h 2 = ξ1, so one can consider this system as an example of a cylindrical system with h 1 ≠ αh2. This system was fully analyzed in Secti on 7 (d), so we just examine the first two conditions summarized above to see how they work out: Condition (1): h 1 = 1 h2 = ξ1 and we know the R = 1 (h 2/h1) = f1(1) G1(2)R2 (h1/h2) = f2(2) G2(1))R2 (h1h2) = g3(12)R2 (8.3) ξ1 = f1(1) G1(2) 1/ξ1 = f2(2) G2(1) ξ1 = g3(12) Inspection shows a simple solution to equation (3.5), f 1(1) = ξ1 G 1(2) = 1 R=1 => Q=1 and k 12 = 0 f2(2) = 1 G2(1)= 1/ξ1 f3(3) = 1 g 3(12) = ξ1 38 Condition (2): -(h1h2) = c'(1) / [f 2(2)G1(2)]1/2 + f'(2) / [f 1(1) G2(1)]1/2 (8.6) - ξ 1 = c'(1) / [1 * 1]1/2 + f'(2) / [ ξ1 * ξ1-1]1/2 - ξ 1 = c'(1) + f'(2) So this condition, noted in the ge neral case to be "strict", has in this system the simple solution, c'(1) = - ξ1 f'(2) = 0 => c(1) = -1 f(2) = 0 Since Q=1, one can take g(3)=1 and a(1) = d(2) = 0 for the first column, so Φ comes out exactly as found in Section (7d), Φ = ⎝⎜⎛ ⎠⎟⎞ a(1) b(1) c(1) d(2) e(2) f(2) g(3) h(3) i(3) = ⎝⎜⎜⎛ ⎠⎟⎟⎞ a(1) - G 2(1)/f1(1) c(1) d(2) G 1(2)/f2(2) f(2) g(3) 0 1 = ⎝⎜⎜⎛ ⎠⎟⎟⎞ 0 -1/ξ12 -1 0 1 0 1 0 1 39 9. Separability in Cylindrical Systems when h 2 = α h1 We review the three Conditions summarized just abov e (end of Section 8), but now instead of general h 1 and h2, it is assumed that h 2 = α h1, α = a fixed constant. The Conditions (1) It m ust be possible to solve equations (8.3) for the 6 functions f 1, f2, G1, G2, g3 and R (h 2/h1) = f1(1) G1(2)R2 (h1/h2) = f2(2) G2(1)R2 (h1h2) = g3(12)R2 (8.3) In this case, α = f1(1) G1(2)R2 α-1 = f2(2) G2(1)R2 αh12 = g3(12)R2 ( 9 . 1 ) Try R = 1 and search for a solution to α = f 1(1) G1(2) 1/α = f2(2) G2(1) αh12 = g3(12) The solution is pretty clear (recall f 3 = 1 for any cylindrical system), f 1=f2=f3=1 G 1(2) = α G2(1) = 1/α g 3 = α h12 R=Q=1 (9.2) Since a solution was found, the R=1 assumption is justified. (2) This condition was -(h 1h2) = c'(1) / [f 2(2)G1(2)]1/2 + f'(2) / [f 1(1) G2(1)]1/2 (8.8) It looked fairly strict in general, but in the current situation it says - αh 12 = c'(1) / α + f'(2) α ( 9 . 3 ) Tracing back we had c(1) [f 1(1) G2(1)]1/2 ≡ c'(1) f(2) [f 2(2)G1(2)]1/2 ≡ f'(2) or c(1) [1/ α] 1/2 ≡ c'(1) f(2) [ α]1/2 ≡ f'(2) 40 so that Condition 2 above becomes - αh 12 = c(1) /α + f(2) α or - h12 = c(1) /α2 + f(2) (9.4) (3) We shall redo this condition fro m scratch. The general condition is 1/Q = a(1) (1/h 12) + d(2) (1/h 22) + g(3) (1/h 32) (8.6) but since Q = 1 and h 3 = 1 this becomes 1 = a(1) (1/h 12) + d(2) (1/ α2h12) + g ( 3 ) ( 9 . 5 ) The obvious solution here is a(1) = 0 , d(2) = 0 and g(3) = 1. The Stäckel matrix from Section 8 now takes this form Φ = ⎝⎜⎜⎛ ⎠⎟⎟⎞ a(1) - G 2(1)/f1(1) c(1) d(2) G 1(2)/f2(2) f(2) g(3) 0 1 = ⎝⎜⎛ ⎠⎟⎞ 0 - 1/α c(1) 0 α f(2) 1 0 1 (9.6) This agrees in its general form with M oon and Spencer on page 7 equation (1.28). Summary of Conditions: The only real condition for separation of this kind of coordinate system is the Condition 2 requirement found above, - h 12 = c(1) /α2 + f ( 2 ) (9.4) and this, assuming it can be satisfied, fill s in the two missing elements in the Φ matrix above. Condition (9.4) of course rules out something like α =1 and h 12 = (ξ12 + ξ2)-2 as occurs in "tangent cylinder coordinates" (Moon and Spencer p 79). That is to say, the general Laplace equation cannot be separated in such coordinates. The Steps From Section 7 (b) we track our 7 steps for the case h 2 = α h1 : Step (1) : Already done above where it was found that f1=f2=f3=1 G 1(2) = α G2(1) = 1/α g 3 = α h12 R=Q=1 (9.2) 41 Step (2): M1 = g1 /(f2f3) = α/1 = α M2 = g2 /(f3f1) = (1/α)/1 = 1/α M3 = g3 /(f1f2) = α h12 (5.8b) Step (3): R = 1 so Q = 1 Step (4): S = [ H / (f 1f2f3QR2)] = H = h 1h2h3 = αh12 Step (5) and Step (6): Already done above where it was found that Φ = ⎝⎜⎛ ⎠⎟⎞ 0 - 1/α c(1) 0 α f(2) 1 0 1 where - h 12 = c(1) /α2 + f(2) Moon and Spencer page 7 e quation (1.28) gives a general form for the Stäckel matrix associated with a cylindrical system Φ = ⎝⎜⎛ ⎠⎟⎞ 0 * * 0 * * 1 0 1 and this is seen to agree with our result above. Special Case h 1 = h2 Practical curvilinear cylindrical coordinate systems have h 1 = h2 so α = 1, and for such systems the results above can be summarized : f1=f2=f3=1 G 1(2) = 1 G 2(1) = 1 g 3 = h12 R=Q=1 (9.2) M 1 = 1 M2 = 1 M3 = h12 S = h 12 (5.8b) Φ = ⎝⎜⎛ ⎠⎟⎞ 0 - 1 c(1) 0 1 f(2) 1 0 1 where - h 12 = c(1) + f(2) This says g 3 = M3 = S = h 12 so we won't repeat those results in the following examples. The critical functional-form condition is that - h 12 = c(1) + f(2). 42 Example 1: elliptic cylinder coordinates For this s ystem, h12 = (ξ12 + ξ22) = - c(1) - f(2) => c(1) = - ξ12 f(2) = - ξ22 Φ = ⎝⎜⎛ ⎠⎟⎞ 0 - 1 c(1) 0 1 f(2) 1 0 1 = ⎝⎜⎜⎛ ⎠⎟⎟⎞ 0 - 1 - a2ch2(ξ1) 0 1 a2cos2(ξ2) 1 0 1 // agrees with Moon and Spencer p 17 η,ψ,z Example 2: parabolic cylinder coordinates For this s ystem, h 12 = a2ch2(ξ1) - a2cos2(ξ2) = - c(1) - f(2) => c(1) = - a2ch2(ξ1) f(2) = a2cos2(ξ2) Φ = ⎝⎜⎛ ⎠⎟⎞ 0 - 1 c(1) 0 1 f(2) 1 0 1 = ⎝⎜⎜⎛ ⎠⎟⎟⎞ 0 - 1 - ξ12 0 1 - ξ22 1 0 1 // agrees with Moon and Spencer p 21 μ,ν,z Example 3: circular cylinder coordinates This case was treated in Section 7 (d) above, where h 1 = 1 and h 2 = ξ1= r. One can do an alternate analysis defining ξ1 by r = exp( ξ1), and in this case it turns out that h 1 = h2 = exp(ξ1) so, h 12 = exp(2ξ1) = - c(1) - f(2) => c(1) = - exp(2 ξ1) f(2) = 0 Φ = ⎝⎜⎛ ⎠⎟⎞ 0 - 1 c(1) 0 1 f(2) 1 0 1 = ⎝⎜⎜⎛ ⎠⎟⎟⎞ 0 - 1 - exp(2ξ 1) 0 1 0 1 0 1 // see Moon and Spencer p 13 ξ ,ψ,z Moon and Spencer have a rare typo in this matrix . They show its upper right element as -exp(-2 ξ) which then disagrees with their M 1 = exp(2ξ) and S = exp(2ξ ). The 21 cylindrical systems of Moon and Spencer In their third chapter (called Section III) Moon and Sp encer consider 21 cylindrical curvilinear coordinate systems which presumably have proven useful in some applications. Each of these systems is obtained by doing some conformal mapping w = f(z) of the Cartesian coordinates z = x+iy to get w = ξ1+iξ2. Such mappings preserve scaling, which is why h 1 = h2, and also preserve angles, so the orthogonality of the Cartesian system maps into the orthogonality of the ( ξ1,ξ2) system. Of these 21 systems, 18 do NOT meet our Condition 2 which says - h 12 = c(1) + f(2), and are therefore not separable! Only 3 of the 21 systems are separable, and they are the th ree cylindrical examples presented above, all simple-separable for Helmholtz. These 3 systems are among the 11 classical systems Moon and Spencer discuss in their first chapter. In addition to Cartesian, the remaining 7 cl assical systems are: ellipsoidal and its two subcases 43 prolate and oblate spheroidal and their special case sphe rical; parabolic-cylinder, parabolic (paraboloidal) and conical. Moon and Spencer give excellent data sets for all 11 systems including good drawings, which seem clearer than the interesting but hard to focus stereoscopic images of Morse and Feshbach. For the 21 cylindrical systems Moon and Spencer also have collected data including detailed 2D drawings of the level curves. On page 78 Moon and Spencer state bluntly that "no cylindrical system allows R-separability". What they mean by this is that no h 1 = h2 cylindrical system is separable with R ≠ constant. Certainly our three cylindrical systems noted above are R-separable with R = 1, and the more general case might be R separable for an obscure contrived system. 44 10. Separability in Rotational Systems First of all, 2D orthogonal curvilinear systems often have one or two symmetry axes. If the 2D system is created by a conformal mapping of the Cartesian system, so w= f(z) or ξ1 + iξ2 = f(x +iy), then the 2D level curve drawing will be symmetric in x if it happens that f(x +iy) = f(-x +iy), and it will be symmetric in y if f(x +iy) = f(x -iy). A 3D rotational system is formed by rotating such a 2D system about one of its symmetry axes. Such a system of course then has an azimuthal variable ξ 3. One could for example rotate any of the 21 2D systems that Moon and Spencer use to generate cylindr ical systems, and some of those can be rotated about two different symmetry axes. All these systems have h 1= h2 since the starting 2D systems are derived from conformal maps of the Cartesian 2D system. An example is the rotation of a 2D bipolar system about one of its symmetry axes to form toroidal coordinates. Rotation about the other symmetry axis gives bispherical coordinates. A simpler example is the rotation of 2D polar coordinates to create 3D spherical coordinates. Since h 3 = 1 is no longer true, one doesn't get much sp ecialization of the general case treated back in Section 7. Since ξ3 is an azimuth angle, the defining e quations must have this general form, x = A(ξ 1,ξ2)cosξ3 y = A(ξ 1,ξ2)sinξ3 z = B( ξ1,ξ2) The scale factors are given by (see derivation a few lines below) h n2 = (∂nx)2 + (∂ny)2 + (∂nz)2 ∂n = ∂/∂ξn Therefore ( see also Moon and Spencer p 50) h 12 = [∂1A(ξ1,ξ2)]2cos2ξ3 + [∂1A(ξ1,ξ2)]2sin2ξ3 + [∂1B(ξ1,ξ2)]2 = [ ∂1A(ξ1,ξ2)]2 + [∂1B(ξ1,ξ2)]2 h 22 = [∂2A(ξ1,ξ2)]2cos2ξ3 + [∂2A(ξ1,ξ2)]2sin2ξ3 + [∂2B(ξ1,ξ2)]2 = [ ∂2A(ξ1,ξ2)]2 + [∂2B(ξ1,ξ2)]2 h 32 = [A(ξ1,ξ2) (-sinξ3)]2 + [A(ξ1,ξ2) (cosξ3)]2 = [A(ξ1,ξ2)]2 If h 1 = h2 one has an obvious condition on the various derivatives, [∂ 1A(ξ1,ξ2)]2 – [∂2A(ξ1,ξ2)]2 = [∂2B(ξ1,ξ2)]2 – [∂1B(ξ1,ξ2)]2 but the main point is that the scale factors are functions only of ξ 1 and ξ2 so one has h n(12). This is the same form arising in all cylindrical systems except there h 3 = 1. Derivation of the above h n2 sum. The curvilinear covariant metric tensor g ij can be related to S in ≡ ∂xi ∂ξn by the matrix equation g = STS. Since h n2 = gnn one finds that, 45 hn2 = gnn = Σi (ST)niSin = Σi (Sin)2 = Σi (∂xi ∂ξn )2 = (∂nx1)2 + (∂nx2)2 + (∂nx3)2 . This is true for any curvilinear coordinates, ort hogonal or not. Morse and Feshbach derive it for a 3D orthogonal system on page 24, equation (1.3.4). We shall now restrict our interest to rotational system s of the type considered in Moon and Spencer which are derived from conformal map 2D systems, so from now on h 1= h2. A look at the Conditions of Section 7 (a): Condition (1) Equations (3.5) must be solvable for the 7 functions f n, gn and R. If some f n is a constant, that constant is set to 1. ( If one is doing Problem A, R=1.) (H/h 12) = f1(1)g1(23) R2 H = h 1h2h3 (H/h 22) = f2(2)g2(31) R2 (H/h 32) = f3(3)g3(12) R2 (3.5) which becomes, assuming h 1 = h2, h 3(12) = f1(1)g1(23) R2 h3(12) = f2(2)g2(31) R2 h12(12)/h3(12) = f3(3)g3(12) R2 As with the cylindrical case, define G 1(2) in the obvious way and set f 3 = 1 to get h 3(12) = f1(1)G1(2) R2 h3(12) = f2(2)G2(1) R2 h12(12)/h3(12) = g3(12) R2 Divide the first two equations by R2 f1(1)G1(2) = f2(2)G2(1) which has this viable solution G 2(1) = f1(1) G1(2) = f2(2) allowing the three equations to be written as 46 h3(12) = f1(1) f2(2) R2 h3(12) = f2(2) f1(1) R2 h12(12)/h3(12) = g3(12) R2 Since the first two are the same, Condition 1 requires the solution of the following pair of equations for the 4 quantities f 1, f2, g3 and R h3(12) = f1(1) f2(2) R(12)2 h12(12)/h3(12) = g3(12) R(12)2 Assuming this problem has a solution, one then has R 2 = h3/f1f2 M 1 = g1 /(f2f3) = G1(2) /f2(2) = f 2(2) /f2(2) = 1 M2 = g2 /(f3f1) = G2(1) /f1(1) = f1(1) /f1(1) = 1 M3 = g3 /(f1f2) = g3(12) /(f1(1)f2(2)) (5.8b) Condition (2) Equations (7.2) must have a solution: M1(23) = 1 = e(2)i(3)-f(2)h(3) -M2(31) = -1 = b(1)i(3)-c(1)h(3) M3(12) = g3(12) /(f1(1)f2(2))= b(1)f(2)-c(1)e(2) (7.2) Select h(3) = 0 based on the first two lines (nothing new if i(3)=0 instead) to get 1 = e(2)i(3) -1 = b(1)i(3) g 3(12) /(f1(1)f2(2))= b(1)f(2)-c(1)e(2) From the first pair select i(3) = 1, e(2) = 1 and b(1) = -1. The last line is then g3(12) /(f1(1)f2(2))= - f(2)-c(1) or - g 3(12) = c(1)[f 1(1)f2(2)] + f(2)[f 1(1)f2(2)] and this then is Condition 2. The Stäckel matrix now has this form Φ = ⎝⎜⎛ ⎠⎟⎞ a(1) b(1) c(1) d(2) e(2) f(2) g(3) h(3) i(3) = ⎝⎜⎛ ⎠⎟⎞ a(1) -1 c(1) d(2) 1 f(2) g(3) 0 1 47 Condition (3) If conditions (1) and (2) are met, we then do the work of computing Q and k 12 from (3.9), Σn (1/[hn2fnR]) ∂n{fn(∂nR)} = - k 12/Q(123) (3.9) If R = 1, we then set Q = 1 and k 12 = 0 since in this case ∂nR = 0. Then with this Q expression, equation (7.6) must have a solution for a, d and g, 1/Q = a(1) (1/h 12) + d(2) (1/h 22) + g(3) (1/h 32) (7.6) 1/Q(12) = (1/h 12) [ a(1) + d(2) ] + g(3) (1/h 32) and this is another functional-fo rm condition. Since nothing other than g(3) is a function of ξ 3, one must set g(3) = α, a constant. Then 1/Q(12) = (1/h 12) [ a(1) + d(2) ] + α (1/h32) so Condition 3 requires soluti on of the above equatino for α, a(1) and d(2). The Stäckel matrix is now Φ = ⎝⎜⎛ ⎠⎟⎞ a(1) -1 c(1) d(2) 1 f(2) α 0 1 Adding -α times the third column to the first column , Equi valence Rule (6.3) gives us this new equivalent matrix Φ ≈ ⎝⎜⎛ ⎠⎟⎞ a(1)-αc(1) -1 c(1) d(2)-αf(2) 1 f(2) 0 0 1 Moon and Spencer page 7 e quation (1.29) gives a general form for the Stäckel matrix associated with a rotational system Φ = ⎝⎜⎛ ⎠⎟⎞ * * * * * * 0 0 1 and this is seen to agree with our last result above. A look at the Steps of Section 7 (b): Step (1) As noted above in Condition 1, the first step is to solve the following 2 equations for the quantities f 1, f2, g3 and R h3(12) = f1(1) f2(2) R(12)2 h12(12)/h3(12) = g3(12) R(12)2 48 Step (2) These items are taken from Condition 2 above R2 = h3/f1f2 M 1 = 1 M2 = 1 M3 = g3(12) /(f1(1)f2(2)) Step (3) If R=1 works in step (1), then Q = 1. Otherwise compute Q from (3.9), Σn (1/[hn2fnR]) ∂n{fn(∂nR)} = - k 12/Q(123) (3.9) Step (4) Having Q, compute S from the Robertson condition S(Φ) = H / (f 1f2f3QR2) "Robertson condition" (5.7) But f3 = 1 and R2f1f2 = h3 so this really says S = [ H / (Qh 3)] = h12/Q Step (5) Find the rightmost two columns of the Stäckel matrix, Φ = ⎝⎜⎛ ⎠⎟⎞ a(1) b(1) c(1) d(2) e(2) f(2) g(3) h(3) i(3) S = d e t ( Φ) (7.1) But in Condition 2 above (with a little Condition 3) we already know that Φ = ⎝⎜⎛ ⎠⎟⎞ a(1) -1 c(1) d(2) 1 f(2) α 0 1 and entries c(1) and f(2) are then determined by - g 3(12) = c(1)[f 1(1)f2(2)] + f(2)[f 1(1)f2(2)] Step (6) In Condition 3 above this was the same as solving 1/Q(12) = (1/h 12) [ a(1) + d(2) ] + α (1/h32) 49 Example 1: toroidal coordinates revisited This case has already been fully solv ed, but we shall do it again using th e systematic steps outlined above for rotational systems. From earlier work the followi ng facts were obtained for the toroidal system, h1 = h2 = a/[ch(ξ 1)-cos(ξ2)] => h 1 = h2 = aR2 H = sh( ξ1)a3R6 h3 = a sh(ξ 1)/[ch(ξ1)-cos(ξ2)] => h 3 = a sh(ξ 1) R2 f1(1) = ash(ξ 1) g 1(23) = 1 R ≡ [ch(ξ1)-cos(ξ2)]-1/2 f2(2) = 1 g 2(31) = a sh( ξ1) Q = a2[ch(ξ1)-cos(ξ2)]2 = a2R4 f3(3) = 1 g 3(12) = a/sh(ξ 1) k 12 = (1/4) We shall here pretend not to know the "blue facts" (6 lower left equations), but the earlier calculation of Q and k12 will be utilized. Notice first that h 1 and h3 are functions only of 12 and not 3, as expected. We use here the symbol R ≡ [ch(ξ1)-cos(ξ2)]-1/2 to distinguish it from the separation R below. And (in the blue) the constant a has been moved from f 3 over to g 3, and also from g 1 over to f 1 -- the allocation of constants is immaterial. Here then are all our Stäckel calculation steps for rotational systems given above: Step (1) As noted above in Condition 1, the first step is to solve the following 2 equations for the quantities f 1, f2, g3 and R h 3 = f1 f2 R2 h12/h3 = g3R2 and these now say a sh(ξ1)R2 = f1 f2 R2 (a R 2)2/ [a sh(ξ1) R 2] = g3R2 or a sh(ξ 1) R2 = f1 f2 R2 a R2/ sh(ξ1) = g3R2 The obvious choice is R = R and then a sh(ξ 1) = f1(1) f2(2) a/sh( ξ1) = g3(12) These imply that f1(1) = a sh( ξ1) f2(2) = 1 g 3(12) = a/sh(ξ 1) R = R and this agrees with the "blue data" above collect ed from earlier work. Done with Step 1. 50 Step (2) These items are taken from Condition 2 above M1 = 1 M2 = 1 M3 = g3(12) /(f1(1)f2(2)) = a/sh(ξ 1) / [a sh(ξ1)] = 1/ sh2(ξ1) Step (3) This slightly painful calculation was done back in Section 3 with results Q = a2R4 k 12 = (1/4) Note that Q = h 12 = (aR2)2. Step (4) S = h 12/Q = 1 Step (5) Φ = ⎝⎜⎛ ⎠⎟⎞ a(1) -1 c(1) d(2) 1 f(2) α 0 1 - g3(12) = c(1)[f 1(1)f2(2)] + f(2)[f 1(1)f2(2)] This last equation becomes - a/sh(ξ 1) = c(1)[ a sh( ξ1)] + f(2)[ a sh( ξ1)] or - 1 = c(1)[ sh 2(ξ1)] + f(2)[ sh2(ξ1)] A solution is c(1) = -1/sh2(ξ1) and f(2) = 0. The Φ matrix at this point is then Φ = ⎝⎜⎛ ⎠⎟⎞ a(1) -1 c(1) d(2) 1 f(2) α 0 1 = ⎝⎜⎜⎛ ⎠⎟⎟⎞ a(1) -1 -1/sh2(ξ1) d(2) 1 0 α 0 1 Step (6) 1/Q(12) = (1/h 12) [ a(1) + d(2) ] + α (1/h32) which says 1 = [ a(1) + d(2) ] + α (Q/h 32) Pick a(1) = 1 and d(2) = 0 and g(3) = α = 0 to get the final matrix 51 Φ = ⎝⎜⎜⎛ ⎠⎟⎟⎞ a(1) -1 -1/sh2(ξ1) d(2) 1 0 α 0 1 = ⎝⎜⎜⎛ ⎠⎟⎟⎞ 1 -1 -1/sh2(ξ1) 0 1 0 0 0 1 All these results agree with those found earlie r and with Moon and Spencer p 112-113. The 11 separable rotational systems of Moon and Spencer Moon and Sp encer in their fourth chapter (called Sec tion IV) review a set of 11 rotational systems which can be R-separated. Evidently, if one tries to rota te the 21 conformal map 2D systems of their second chapter, 10 of them fail to meet one or more of our three Conditions. The hardest Step is computing Q, and perhaps ther e is a more elegant way to do than by brute force from (3.9). (Of course Maple is happy to compute Q if it is given the f n, hn and R.) In particular, it is not obvious why Q comes out being a constant times a power of R. The fact seems to arise since R is always a power of some "atom" as shown in this pa rtial table from Moon and Spencer Section III: q ( a t o m ) h 12 R Q tangent sphere ξ12+ξ22 q-2 q-1/2 q-2 cardioid ξ12+ξ22 q-3 q-1 q-4 bispherical ch ξ1-cosξ2 a2q-2 q-1/2 a2q-2 toroidal ch ξ1-cosξ2 a2q-2 q-1/2 a2q-2 // differ in h 3 inverse prolate ch2ξ1-sin2ξ2 see M&S q-1/2 q-2 The atom appears first in the scale factors and then pr opagates to R through our Step 1. The fact that Q and the scale factors are both powers of the same at om is what allows Condition 3 to be met with α = 0, 1/Q(12) = (1/h 12) [ a(1) + d(2) ] + α (1/h32) Notice how this is satisfied in the cardioid case q 4 = q3 [ a(1) + d(2) ] where then [ a(1) + d(2) ] is taken to be the atom itself, q, to get q4 = q4. In most other cases the two powers match and we just take a(1) = 1 and d(2) = 0. 52 11. Separation of the Schrodinger equation. (a) Schrodinger meets Helmholtz The time- dependent Schro dinger equation of non-relativistic ( no spin) quantum mechan ics has this form for a single point particle existing in 3D space under the influence of a potential V, HΨ = ih∂tΨ Ψ = Ψ(r, t ) ( 1 1 . 1 ) where H = KE + PE = p 2/2m + V = (-i h∇)2/2m + V = - ( h2/em)∇2 + V (11.2) H is the Hamiltonian, the sum of kinetic and potential energy for the particle. Here we have used the quantum mechanical "rabbit in the hat" fact that p = -ih∇, which is to say, the classical particle momentum p is represented by operator -ih ∇ in "coordinate space" r . This results in H being a differential operator as shown. The solution Ψ(r,t) is called a wavefunction, and the probability of the particle being in some small volume d3r at time t is | Ψ(r,t)|2d3r. Assuming a monochromatic time dependence for Ψ, Ψ(r,t) = ψ (r) e-iωt => ∂tΨ (r,t) = (-iω) Ψ (r,t) (11.3) equation (11.1) becomes HΨ = ih ((-iω) Ψ(r,t) = hω Ψ(r,t) Defining E ≡ hω, one finds HΨ = EΨ E = hω Assuming that V( x,t) = V( x), then H does not involve t, and the above equation becomes Hψ E(r) = EψE(r) ( 1 1 . 4 ) which is known as the time- independent Schrodinger equation. The energies E and solutions ψE are thus the eigenvalues and eigenfunctions of the differential operator H subject to some kind of boundary conditions. It is traditional to refer to these eigenfunctions as "eigenstates" or just "states", while the eigenvalues are called "eigenenergies" or "energy levels". The continuous energy E of a classical particle ends up being "quantized" into th ese allowed eigenenergies. Of course the spectrum of E depends on the boundary conditions of the particular problem a nd might be discrete, continuous, or both. 53 Putting (11.2) into (11.4), (- h2∇2/2m + V)ψ = Eψ (∇2 - 2mV/h)ψ = -2mE/hψ (∇2 + 2Em/h - 2mV/h )ψ = 0 One can now define a scaled potential φ and replace constant E with constant K 12 φ ≡ - (2m/h)V K 12 ≡ (2m/h) E ( 1 1 . 5 ) to get (∇ 2 + K12 + φ )ψ = 0 ( 1 1 . 6 ) And so the Schrodinger equation ha s become our Helmholtz equation with the addition of the scaled potential term φ. (b) Separation of the Schrodinger Equation in Curvilinear coordinates The previous section showed that the Schrodinger equation can be written as (∇2 + [K12 + φ(123)] ) ψ(123) = 0 (11.6) Tracing through our "processing of the Helmholtz equati on" in Sections 3 and 4, we find that the K 12 term just "sits there" all the way through, and we end up with this new version of (4.4) in which K 12 is replaced by K 12 + φ, - Σ n(1/hn2)qn + k12/Q + K12 + φ = 0 . (11.7) This suggests, following Morse and Fes hbach, that the most general form for φ which would allow separation is this φ(123) = Σ n sn(n)/hn2 ( 1 1 . 8 ) and putting (11.8) into (11.7) gives - Σ n(1/hn2)(qn-sn) + k12/Q + K12 + φ = 0 (11.9) We shall now modify (4.5) and introduce the St äckel matrix this new way (a new LHS) qn-sn = [ κ12Φn1(n) + k22Φn2(n) + k32Φn3(n)] n = 1,2,3 (11.10) and then (11.9) becomes (multiplying through by Q) 54 - Σn(Q/hn2) [ κ12Φn1(n) + k22Φn2(n) + k32Φn3(n)] + k 12 + QK12 = 0 (11.11) which is identical to equation (4.9) and (5.1). One therefor e solves for the Stäckel matrix exactly as before. Recall that the problem of finding the Stäckel matrix Φ is based only on this equation. What about the separated equations? The most general self-adjoint form is still as in (4.2), LnXn = (1/fn)∂n[fn(∂nXn)] + qnXn = 0 (4.2) and replacing q n from (11.10) gives L nXn = (1/fn)∂n[fn(∂nXn)] + [ κ12Φn1(n) + k22Φn2(n) + k32Φn3(n) + sn(n)]Xn = 0 (11.12) which is the same as before but with the extra potential function s n(ξn) as shown. Summary : In order to achieve separation of the Schr odinger Equation, the potential must have the restricted form φ = Σn sn(n)/hn2shown in (11.8). This leads to the ex act same Stäckel matrix problem we had before, and the separated equations are the same as before but with an added s n term from the potential. Since the Schrodinger E parame ter is normally not 0, and since K 12 ≡ (2m/h)E, the Stäckel matrix problem is that for the pure Helmholtz equation with K 12 ≠ 0. We know from above that we can only achieve simple-separation (R = 1, Problem A) fo r the Helmholtz equation, and that only happens in the 11 classical curvilinear systems. In their review of these 11 classical systems on pages 656-666, for each system Morse and Feshbach state the most genera l form the potential can have ( "General form for V"). (c) Central potentials and the hydrogen atom problem The most famous instance of the Schrodinger Equa tion separation discussed here occurs in spherical coordinates where the potential V(r, θ,φ) is taken as V(r), known as a central potential. This potential meets the requirement of (11.8) where h 1= 1, s1(1) = V(r), and s 2(2)=s3(3)=0 where 123 = rθφ. Thus, only the separated equation for r (the "radial equati on", (11.12) for n=1) has a modification from the pure Helmholtz case. If the spatial region includes all 4π steradians of angle, then the θ and φ separated equations yield the usual spherical harmonics Y lm(θ,φ) in which the two separation constants k 22 and k32 have been shuffled into l and m, both forced to certain integer values: l = 0,1,2... ( to make Y lm finite at θ = π) and m = - l..l (to make Y lm the same at φ = 0 and 2 π) . When dimensionless separation constants get forced to certain discrete values, those valu es are traditionally called "quantum numbers". 55 A particular example is the hydrogen atom problem where V(r) = -|e|2/r and the solution is then the solution to the radial equation R(r) times the spherical harmonics. Only l appears in the radial function along with K 12 ~ En. Specifically, one finds this wavefunction solution to the hydrogen atom problem, ψ(r,θ,φ) = Rnl(r) Y lm(θ,φ) where (see Schiff p 93) R nl(r) = constant * ρl L2l+1 n+l(ρ) e-ρ/2 ρ = (2/n)(r/a 0) Lp q = associated Laguerre a 0= h2/(μe2) = Bohr radius μ = "reduced mass" of electron |e| = charge on electron E n = -e2/(2a0)/n2 = eigenstate energies n = 1,2,3... l = 0,1...(n-1) A third quantum number n has appeared, and it arises from the need for the wave function ψ to be normalizable such that ∫ d3r |ψ(r,θ,φ)|2 = 1, meaning the probability of finding the electron somewhere in all space must be 1. This requirement in turn cr eates a need for the power series expansion for the function L to truncate at some maximum power, and that power is related to n. Then function L becomes a polynomial, known as an associated Laguerre polynomial. The exponential factor e-ρ/2 can then overwhelm the polynomial at large r ~ ρ no matter how large n gets, providing normalization. We can look quickly at a few solutions: ( = states = orbitals) With n=1 we have l=m=0 and the energy E 1 is the most negative of all the E n. This spherically symmetric state is called 1S and is the hydrogen atom "ground state". With n = 2 we get l = 0,1. The l=0 (m=0) case gives a spherically symmetric state (called 2S) while the l=1 cases with m = -1,0,1 are usually linearly co mbined into three perpendicular dumbbell shaped states called 2P x,y,z. All four n=2 states have the same energy E 2 in this non-relativistic model. 56 12. Separation in Ellipsoidal Coordinates The ellipsoidal coordinate sy stem is the most complicated of the classical curvilinear systems and provides a good exercise in applying the apparatus of the previous sections. We shall make use of some of the equivalence operations of Section 6, and shall end up with three separated equations. The good news is that the three separated equations turn out to be identical, a fact tr aceable to cyclic symmetry of this system. The bad news is that all of the Stäckel matrix elements are non-zero, so the two separation constants are fully entangled into each separated e quation, precluding a simple 1D Sturm-Liouville type solution except in very simplest cases, one of which is quoted below. This section is not meant to be a complete monogr aph on ellipsoidal coordinates. Enough facts are given to hopefully make the reader comfortable with this system and to support the calculations below. See Morse and Feshbach for full details. (a) Some details about ellipsoidal coordinates Convenientl y, our generic variable names ξ1, ξ2, ξ3 match the notation of Morse and Feshbach's ellipsoidal coordinate discussion. In this system the level surfaces ( ξn=constant) are ellipsoids and certain hyperboloids. The ellipsoids are the surfaces of constant ξ1 and have this equation in Cartesian space, x2/(ξ12- a2) + y2/(ξ12- b2) + z2/(ξ12) = 1 , (12.1) which is an ellipsoid centered at th e origin. The largest semimajor axis of this ellipsoid is obviously ξ 1, and the other two semimajor axes ξ12- a2 and ξ12- b2 are smaller. The quantities a and b are focal distances associated with the x and y directions. Here ar e crude pictures showing slices of this ellipsoid in the x=0, y=0 and then z=0 plane: Assuming an eye position such that the +x axis is "up", as these pict ures are drawn, then the left and middle pictures show that b is the focal distance of the horizontal ellipse = the intersection of the ellipsoid with the x=0 plane. a is the focal distance of the vertical ellipse = th e intersection of the ellipsoid with the y=0 plane. Notice that both these focal distances are along the z axis . The right picture then shows the other vertical slice ellipse= the intersection of the ellipsoid with the z=0 plane. It will be assumed that a > b. 57 Sweeping ξ1 through some range produces a family of ellipsoids, each labeled by its value of ξ1, the longest semimajor axis length. All these ellipsoids have the same a and b focal distances. For this reason, such a family of ellipsoids is called "confocal". As ξ1 gets very large, the ellipsoid gets long in th e z direction, and has the shape of a cigar whose cross section is elliptical. The lower limit for ξ1 is a, since a semimajor axis cannot be smaller than a focal distance (and a is the larger of the two focal distances). If ξ1 = a +ε , the ellipsoid is crushed vertically, lies close to the x=0 plane, and looks a thin elliptical cookie. The upshot is that one must have a < ξ 1 < ∞. The other two surface families (those of constant ξ2 and constant ξ3) are types of infinite hyperboloids. Here are some graphs showing one surface of constant ξ2 (left) and one of constant ξ3 (right), These hyperboloids are not surfaces of revolution, th ey are generally asymmetric, though these pictures don't make that clear. The surface on the left is ca lled a one-sheeted hyperboloid and the one on the right two-sheeted. The pictures show that the coordinates ξ2 and ξ3 also have certain legal ranges, and these ranges can be summarized as follows: 0 ≤ ξ 3 ≤ b≤ ξ2≤ a≤ ξ1 ( 1 2 . 2 ) These hyperboloid families are also "confocal" in the sense noted above. Replacing ξ1 in (12.1) by ξ2 or ξ3 gives the other two level-surface e quations. Is the fact that one or two of the terms in (12.1) then change sign, accordin g to 12.2, that causes these surfaces to be the one and two sheeted hyperboloids. In fact, the three coordinates ξn2 for n=1,2,3 can be regarded as the three solutions of equation (12.1) written with ξ1→ξn and treated as a cubic equation in ξn2. One could in this manner find ( ξ1,ξ2,ξ3) for a given (x,y,z), though the expressions are quite ugly. It is much easier to go the other direction using (12.3) below. The ellipsoid family intersects these two hyperboloid families such that everythi ng is at right angles -- this is an orthogonal coordinate system! Here is an attempt to show one member of each surface family in the same drawing: ( We call this picture "the Q Bomb", in fond memory of The Mouse that Roared .) 58 Examination of the Q Bomb shows that the surfaces labeled by ξ1, ξ2, ξ3 intersect at 8 places in Cartesian space. These locations are given by (±x, ±y, ±z) where x 2 = (ξ12-a2) (ξ22-a2) (ξ32-a2)/ [ a2(a2-b2)] = + - - / + = + y2 = (ξ12-b2) (ξ22-b2) (ξ32-b2)/ [ b2(b2-a2)] = + + - / - = + // = x2(a↔b) z = ξ1ξ2ξ3/(ab) = + + +/+ = + (12.3) The quantities on the RHS's of these equations are all positive. (b) Notation and comparison to that of Morse and Feshbach Right off the bat we are going to m ake a notational ch ange to our separation machinery. Recall that in Step 1 of our solution method we are supposed to analyze equations (3.5), (H/h 12) = f1(1)g1(23) R2 (H/h 22) = f2(2)g2(31) R2 (H/h 32) = f3(3)g3(12) R2 (3.5) We replace f n(n) by Fn(n), in order to use f n(n) for another purpose, so the new equations (3.5) read, (H/h 12) = F1(1)g1(23) R2 (H/h 22) = F2(2)g2(31) R2 (H/h 32) = F3(3)g3(12) R2 (3.5) 59 and of course we make this same change in all our separation equations where the f n(n) appear. The reason for doing this is to try and maintain comp atibility with the notation of Morse and Feshbach regarding ellipsoidal coordinate separation, so we will use f n for their purpose (see Morse and Feshbach p 512 5.1.35), which purpose is the following set of definitions: f 12 ≡ (ξ12-a2)( ξ12-b2) = + + = + G 1 ≡ (ξ22- ξ32) = + f22 ≡ (ξ22-a2)( ξ22-b2) = - + = - G 2 ≡ (ξ32- ξ12) = - f32 ≡ (ξ32-a2)( ξ32-b2) = - - = + G 3 ≡ (ξ12- ξ22) = + (12.4) These G n functions have nothing to do with G n functions we used in Sections 8, 9 and 10. What we have here are G n(≠n) type functions, such as G 1(23). [ G n is another Morse and Feshbach notation, p 512 5.1.37. ] Running down the lines above, everything is seen to be nice and cyclic. However, the inequality chain in (12.2) gives expression signs as shown to the right above. One implication is that f 2 must be imaginary, and then we have to worry about branches a nd factors of ±i and such things. We don't like having imaginary stuff floating around unnecessarily, so we now introduce some new italicized capital F and G functions as follows (notice the minus signs in front of f 22 and G2 ) F 12 ≡ + (ξ12-a2)( ξ12-b2) = f12 G1 ≡ + (ξ22- ξ32) = G1 F22 ≡ – (ξ22-a2)( ξ22-b2) = - f22 G2 ≡ – (ξ32- ξ12) = - G2 F32 ≡ + (ξ32-a2)( ξ32-b2) = f32 G3 ≡ +(ξ12- ξ22) = G3 (12.5) All the F n2 and Gn quantities are positive and have unambiguous s quare roots. Note by the way that both functions f n and Fn are of the functional form type f n(n) and F n(n). Finally we are ready to write dow n the ellipsoidal scale factors h n. In terms of the f i and Gi functions one finds these unpleasant looking forms (e.g., Morse and Feshbach page 663), h12 = -G2G3/f12 h 1 = -G2G3/f12 ( 1 2 . 6 ) h22 = -G3G1/f22 h 2 = -G3G1/f22 h32 = -G1G2/f32 h 3 = -G1G2/f32 H = h 1h2h3 = -G22G22G32/(f12 f22 f32 ) We hope the reader feels as uncomfortable as the wr iter about scale factors that are not obviously positive real numbers (though of course they are). We prefer to write the above our using our non-Morse and Feshbach symbols which are all positive quantities, h12 = G2G3/F12 h 1 = G2G3 / F1 => h n2 = G1G2G3/( GnFn2) h22 = G3G1/F22 h 2 = G3G1 / F2 h32 = G1G2/F32 h 3 = G1G2 / F3 H = h 1h2h3 = G1G2G3/(F1F2F3)z (H/h 12) = [ G1G2G3/(F1F2F3)] / [ G2G3/F12 ] = G1F1/ (F2F3) (H/h 22) = [ G1G2G3/(F1F2F3)] / [ G3G1/F32 ] = G2F2/ (F3F1) (H/h 32) = [ G1G2G3/(F1F2F3)] / [ G1G2/F32 ] = G3F3/ (F1F2) (12.7) 60 All the equation sets in (12.7) are cyclic. (c) Running through the Steps of Section 7 (b) Step (1) As noted above, the first task is to solve (3.5) for the 7 functions F n, gn and R (H/h 12) = F1(1)g1(23) R2 (H/h 22) = F2(2)g2(31) R2 => (H/h n2) = FngnR2 (H/h 32) = F3(3)g3(12) R2 (3.5) We know the ellipsoidal system is a classical system that allows simple -separation for Helmholtz so R=1. The left sides of the three equations above are provided in (12.7) to give G 1(23)F1/ (F2F3) = F1(1)g1(23) G2(23)F2/ (F3F1) = F2(2)g2(31) G3(23)F3/ (F1F2) = F3(3)g3(12) A visual "functional form inspection" of these (again cyclic) equations tells us what we need to know: F 1 = F1 g1(23) = G1(23)/(F2F3) F2 = F2 g2(31) = G2(31)/(F3F1) => g n = GnFn/(F1F2F3) F3 = F3 g3(12) = G3(12)/(F1F2) (12.8) So the F functions turn out to be exactly our separati on functions F, and we can now dispense with the F. We can now collect our "data" from (12.7) and (12.8) into one place, using (-1)n-1 to handle the minus signs for the case n=2, h n2 = G1G2G3/( GnFn2) H = G1G2G3/(F1F2F3) Gn = (-1)n-1Gn F n2 ≡ (-1)n-1 (ξn2-a2)( ξn2-b2) g n = GnFn/(F1F2F3) (12.9) H/hn2 = GnFn2/(F1F2F3) g n/H = GnFn/ (G1G2G3) 61 Step (2) The three first-column cofactors of Φ are, from (5.8) and (12.9), M n = gnFn/(F1F2F3) = { GnFn/(F1F2F3)} Fn/(F1F2F3) = Gn Fn2/ (F12F22F32) or M 1 = G1/ (F22F32) = - G1/(f22f32) M 2 = G2/ (F32F12) = - G2/(f32f12) M 3 = G3/ (F12F22) = - G3/(f12f22) ( 1 2 . 1 0 ) Step (3) R=1 => Q = 1. Step (4) Knowing Q, one can compute S from the Robertson condition, S(Φ) = H / (F 1F2F3QR2) "Robertson condition" (5.7) = H / (F 1F2F3) = [G1G2G3/(F1F2F3)] / (F1F2F3) = G1G2G3 / (F12F22F32) = G1G2G3/(f12f22f32) (12.11) Step (5) The next step is to find the rightmost two columns of the Stäckel matrix, Φ = ⎝⎜⎛ ⎠⎟⎞ a(1) b(1) c(1) d(2) e(2) f(2) g(3) h(3) i(3) S = d e t ( Φ) (7.1) One can do this by solving the following equation set, using the M n found in Step (2), M1 = e(2)i(3)-f(2)h(3) -M2 = b(1)i(3)-c(1)h(3) M3 = b(1)f(2)-c(1)e(2) (7.2) G 1(23)/F22F32 = e(2)i(3)-f(2)h(3) - G2(31)/F32F12 = b(1)i(3)-c(1)h(3) G3(12)/F12F22 = b(1)f(2)-c(1)e(2) G1(23)/F22F32 = e(2)i(3)-f(2)h(3) G2(31)/F32F12 = b(1)i(3)-c(1)h(3) G3(12)/F12F22 = b(1)f(2)-c(1)e(2) (ξ22- ξ32)/F22F32 = e(2)i(3)-f(2)h(3) (ξ32- ξ12)/F32F12 = b(1)i(3)-c(1)h(3) (ξ12- ξ22)/F12F22 = b(1)f(2)-c(1)e(2) Unlike all our previous examples, both terms on the RHS's of these equations are activated and come into play. Looking at the first equation, one can try the following choices, 62 e(2) = s ξ22/F22 i(3) = s/F 32 f(2) = t/F 22 h(3) = t ξ32/ F32 where s and t are signs ±1 to be determined be low. The second and third equations then become (ξ 32- ξ12)/F32F12 = b(1) * s/F 32-c(1)* t ξ32/ F32 (ξ12- ξ22)/F12F22 = b(1) * t/F 22-c(1) * sξ22/F22 Try b(1) = u ξ12/F12 and c(1) = v/F 12 where u and v are two new signs, (ξ 32- ξ12)/F32F12 = uξ12/F12 * s/F32 - v/F12* t ξ32/ F32 (ξ12- ξ22)/F12F22 = uξ12/F12 * t/F22 - v/F12 * sξ22/F22 (ξ 32- ξ12)/F32F12 = usξ12/F12F32 - vt ξ32/ F32F12 (ξ12- ξ22)/F12F22 = ut ξ12/F12F22 - vs ξ22/F12/F22 us = -1 - vt = 1 ut = 1 vs = 1 Picking one sign s = +1 we find u = -1 - vt = 1 ut = 1 v = 1 then u = -1 - t = 1 -t = 1 v = 1 and everything works. The solution is this b(1) = uξ12/F12 c(1) = v/F 12 e(2) = sξ22/F22 f(2) = t/F 22 h(3) = t ξ32/ F32 i ( 3 ) = s / F 32 b(1) = - ξ12/F12 = - ξ12/f12 c(1) = 1/F 12 = 1/f12 e(2) = ξ22/F22 = - ξ22/f22 f(2) = -1/F 22 = 1/f22 h(3) = - ξ32/ F32 = - ξ32/f32 i(3) = 1/F 32 = 1/f32 and we now have the rightmost two columns of the Stäckel matrix Φ = ⎝⎜⎛ ⎠⎟⎞ a(1) b(1) c(1) d(2) e(2) f(2) g(3) h(3) i(3) = ⎝⎜⎜⎛ ⎠⎟⎟⎞ a(1) - ξ12/f12 1/f12 d(2) -ξ22/f22 1/f22 g(3) - ξ 32/f32 1/f32 ≈ ⎝⎜⎜⎛ ⎠⎟⎟⎞ a(1) 1/f 12 ξ12/f12 d(2) 1/f 22 ξ22/f22 g(3) 1/f 32 ξ32/f32 (12.12) To get to the rightmost form, we have used the Sect ion 6 equivalence rule which allows us to swap the last two columns and then negate one of them. 63 Step (6) Find the first column of the Stäckel matrix by solving (7.6) : 1/Q = a(1) (1/h 12) + d(2) (1/h 22) + g(3) (1/h 32) (7.6) 1 = a(1) F 12/ G2G3 + d(2) F 22/ G3G1 + g(3) F 32/ G1G2 G1G2G3 = a(1) G 1 F12 + d(2) G2F22 + g(3) G3F32 – G 1G2G3 = a(1) G 1 f12 + d(2) G 2f22 + g(3) G 3f32 This formidable-looking equation has at least two solutions. The first is this a(1) = ξ 14/f12 d(2) = ξ24/f22 g(3) = ξ34/f32 ( 1 2 . 1 3 ) which we can verify by showing that – G 1G2G3 = a(1) G 1 f12 + d(2) G 2f22 + g(3) G 3f32 – G 1G2G3 = ξ14G1 + ξ24G2+ ξ34G3 ? (12.14) – (ξ22- ξ32) (ξ32- ξ12) (ξ12- ξ22) = ξ14(ξ22- ξ32) + ξ24(ξ32- ξ12) + ξ34(ξ12- ξ22) ? – (x2- x3) (x3- x1) (x1- x2) = x12(x2- x3) + x22 (x3- x1) + x32 (x1- x2) ? Expansion of both sides into 6 terms shows this l ast equation is true, so the question marks can be removed. The other solution is this, a(1) = 1 d(2) = 1 g(3) = 1 ( 1 2 . 1 5 ) which can be verified by showing that – G 1G2G3 = a(1) G 1 f12 + d(2) G 2f22 + g(3) G 3f32 – G 1G2G3 = G1 f12 + G2f22 + G3f32 ? – (ξ 22- ξ32) (ξ32- ξ12) (ξ12- ξ22) = (ξ22- ξ32) (ξ12-a2)( ξ12-b2) + cyclic ? (12.16) It seems hard to imagine how (12.16) can be true since the RHS depends on a and b while the LHS does not! But let's examine the coefficient of a2 on the RHS 64 coeff(a2) = - (ξ22- ξ32) ( ξ12-b2) + cyclic = - [ ξ 12(ξ22- ξ32) + cyclic] + b2[(ξ22- ξ32) + cyclic ] But both square brackets vanish due to the following two trivial theorems valid for any function f f(1)[f(2) - f(3)] +cyclic = 0 // makes first [] = 0 in the line above [f(2) - f(3)] +cyclic = 0 // makes second [] = 0 in the line above (12.17) Thus coeff(a 2) = 0 and similarly coeff(b2) = 0 . In other words, the RHS of (12.16) depends on neither a2 nor b2, despite appearances. Thus one can select any value of a2 or b2 to evaluate the RHS. If one takes a2 = ξ22 and b2 = ξ32, then both the terms that make up "+ cyclic" in (12.16) vanish because each of those terms then has a zero factor. Th en the RHS of (12.16) becomes (ξ 22- ξ32)(ξ12- ξ22)( ξ12- ξ32) = (ξ22- ξ32)( ξ12- ξ32)(ξ12- ξ22) = - (ξ22- ξ32)( ξ32- ξ12)(ξ12- ξ22) But this matches the LHS of (12.16), so we can erase the question marks above. Installing the first solution (12.13) gives this for the full Stäckel matrix, Φ = ⎝⎜⎛ ⎠⎟⎞ a(1) b(1) c(1) d(2) e(2) f(2) g(3) h(3) i(3) ≈ ⎝⎜⎜⎛ ⎠⎟⎟⎞ a(1) 1/f 12 ξ12/f12 d(2) 1/f 22 ξ22/f22 g(3) 1/f 32 ξ32/f32 ≈ ⎝⎜⎜⎛ ⎠⎟⎟⎞ ξ14/f12 1/f12 ξ12/f12 ξ24/f22 1/f22 ξ22/f22 ξ34/f32 1/f32 ξ32/f32 w h e r e f 12 ≡ (ξ12-a2)( ξ12-b2) (12.18) f 22 ≡ (ξ22-a2)( ξ22-b2) f 32 ≡ (ξ32-a2)( ξ32-b2) Let's calculate the first column cofactors of this result. They are ( cofactor = (-1)n+1 minor ) M1 = (1/f22)(ξ32/f32) - (ξ22/f22)(1/f32) = (ξ32- ξ22) / f22f32 = (- G1/ f22f32) M2 = (1/f32)(ξ12/f12) - (ξ32/f32)(1/f12) = (ξ12- ξ32) / f32f12 = (- G2/ f32f12) M3 = (1/f12)(ξ22/f22) - (ξ12/f12)(1/f22) = (ξ22- ξ12) / f12f22 = (- G3/ f22f32) and these results agree with (12.10) in Step 2 above. The determinant should be S = (ξ 14/f12)M1 + (ξ24/f22)M2 + (ξ34/f32)M3 = (ξ14/f12) (- G1/ f22f32) + (ξ24/f22) (- G2/ f32f12) + (ξ34/f32) (- G3/ f22f32) = - { ξ 14 G1 + ξ24G2 + ξ34G3} /( f12f22f32) 65 But we showed in (12.14) that { ξ14 G1 + ξ24G2 + ξ34G3} = – G1G2G3 so we find that S = G 1G2G3/ ( f12f22f32) which agrees with Step 4 result (12.11) above. So we know this Stäckel matrix is valid. Our result (12.18) agrees with Moon and Spencer page 41, but the agreement is a bit hard to see because Moon and Spencer use the following mapping of our symbols which then hides the cyclic nature of things: a →c, ξ1, ξ2, ξ3 → η, θ, λ . Step (7) Finally we get to do a Step 7! First, we w ill use our second solution (12.15) for the Stäckel first column and write Φ = ⎝⎜⎛ ⎠⎟⎞ a(1) b(1) c(1) d(2) e(2) f(2) g(3) h(3) i(3) ≈ ⎝⎜⎜⎛ ⎠⎟⎟⎞ a(1) 1/f 12 ξ12/f12 d(2) 1/f 22 ξ22/f22 g(3) 1/f 32 ξ32/f32 ≈ ⎝⎜⎜⎛ ⎠⎟⎟⎞ 1 1/f12 ξ12/f12 1 1/f22 ξ22/f22 1 1/f32 ξ32/f32 (12.19) which by itself is not such a bad Stäckel matrix. But there is a simpler form that Morse and Feshbach quote and we will use our Section 6 equivalence rules to "get to it". In the (12.19) Φ matrix above add (-b 2) times the second column to th e third column. The new third column top row entry is ξ12/f12 - b2/f12 = (ξ12-b2)/f12 = (ξ12-b2)/ [(ξ12-a2)( ξ12-b2)] = 1/(ξ 12-a2) and the equivalent Φ matrix is now Φ ≈ ⎝⎜⎜⎛ ⎠⎟⎟⎞ 1 1/f12 1/(ξ12-a2) 1 1/f22 1/(ξ22-a2) 1 1/f32 1/(ξ32-a2) ( 1 2 . 2 0 ) Now add - (a2-b2)-1 times the third column to the second column. The new second column top row entry is 1/f12 - 1/[(ξ12-a2) (a2-b2)] = 1/[(ξ12-a2)( ξ12-b2)] - 1/[(ξ12-a2) (a2-b2)] = 1 / ( ξ 12-a2) * { 1/(ξ12-b2) - 1/(a2-b2) } = = 1 / ( ξ 12-a2) * { (a2-b2) - (ξ12-b2)/ }/ [( ξ12-b2) (a2-b2)] = 1 / ( ξ 12-a2) * { (a2- ξ12)}/ [( ξ12-b2) (a2-b2)] = - 1 / [ ( ξ 12-b2) (a2-b2)] 66 The new equivalent Φ matrix is then Φ ≈ ⎝⎜⎜⎛ ⎠⎟⎟⎞ 1 -1/ [( ξ 12-b2) (a2-b2)] 1/(ξ12-a2) 1 -1/ [( ξ 22-b2) (a2-b2)] 1/(ξ22-a2) 1 -1/ [( ξ 32-b2) (a2-b2)] 1/(ξ32-a2) Now we reuse another one of our rules: swap the last two columns and then negate one of them, Φ ≈ ⎝⎜⎜⎛ ⎠⎟⎟⎞ 1 1/(ξ12-a2) 1/ [( ξ 12-b2) (a2-b2)] 1 1/(ξ22-a2) 1/ [( ξ 22-b2) (a2-b2)] 1 1/(ξ32-a2) 1/ [( ξ 32-b2) (a2-b2)] S = det(Φ ) (12.21) and this form agrees with Morse and Feshbach page 663: (d) Summary of Results Step 1: The functional-form solution to (3.5) is : F1 = F1 g1 = G1/(F2F3) F2 = F2 g2 = G2/(F3F1) F3 = F3 g3 = G3/(F1F2) Step 2: The first column cofactors are M1 = - G1/f22f32 M2 = - G2/f32f12 M3 = - G3/f12f22 Step 3: Q = 1 since R = 1 Step 4: S ≡ det(Φ) = G1G2G3/(f12f22f32) Step (5): Use the cofactor information to obtain the rightmost two columns of a Φ matrix and get Φ = ⎝⎜⎛ ⎠⎟⎞ a(1) b(1) c(1) d(2) e(2) f(2) g(3) h(3) i(3) ≈ ⎝⎜⎜⎛ ⎠⎟⎟⎞ a(1) 1/f 12 ξ12/f12 d(2) 1/f 22 ξ22/f22 g(3) 1/f 32 ξ32/f32 67 Step 6: Obtain two alternative solutions for the first column entries: a(1) = ξ14/f12 a ( 1 ) = 1 d(2) = ξ24/f22 d(2) = 1 g(3) = ξ34/f32 g(3) = 1 Use the first solution to obtain the Φ matrix of Moon and Spencer: Φ ≈ ⎝⎜⎜⎛ ⎠⎟⎟⎞ ξ14/f12 1/f12 ξ12/f12 ξ24/f22 1/f22 ξ22/f22 ξ34/f32 1/f32 ξ32/f32 // Moon and Spencer page 41 (12.18) Step 7: Use the second solution and do some equivalence operations to obtain the Φ matrix of Morse and Feshbach: Φ ≈ ⎝⎜⎜⎛ ⎠⎟⎟⎞ 1 1/(ξ12-a2) 1/ [( ξ 12-b2) (a2-b2)] 1 1/(ξ22-a2) 1/ [( ξ 22-b2) (a2-b2)] 1 1/(ξ32-a2) 1/ [( ξ 32-b2) (a2-b2)] // Morse and Feshbach page 663 (12.19) Notice that in all our Stäckel matrices, the three ro ws are identical except for the coordinate label. (e) The separated solutions: Lamé functions Any of these Stäckel matrices may be used to obtain the separated equations of the Helmholtz equation, which are (4.8) with κ1= K1. Thus, (∇2+K12)ψ = 0 ψ = X1X2X3 LnXn = (1/Fn)∂n[Fn(∂nXn)] + [ K12Φn1(n) + k22Φn2(n) + k32Φn3(n)]Xn = 0 n = 1,2,3 or L iXi = (1/fi)∂i[fi(∂iXi)] + [ K12Φi1(ξi) + k22Φi2(ξi) + k32Φi3(ξi)]Xi = 0 i = 1,2,3 (12.22) In the second line, we can think of f 2 = +i F2 and then the +i's cancel up and down. ( We have changed the summation index from n to i, because n w ill have a completely new meaning below. ) Because the three functions in each row of Φ are different and non-zero, one finds that the separation constants k 22 and k32 are "fully entangled" in each separated equation. Also, and most impressively, we find that each of the separated equations is the same equation, and so has the same solutions, because the three rows of Φ are the same apart from the coordinate. Using the Morse and Feshbach (12.19) form of the Φ matrix, one can install the Φ functions into (12.22) to get L iXi = (1/fi)∂i[fi(∂iXi)] + { K12 + k22/(ξi2-a2) + k32/ [(ξi2-b2)(a2-b2)] } Xi = 0 (12.23) 68 where f i2 ≡ (ξi2-a2)( ξi2-b2) which appears as in Morse and Feshbach (5.1.37) with K 12 → k12. The solutions X i are Lamé functions, and the "ellipsoidal harmonics" are products of three Lamé functions ψ = X1X2X3 ~ [Enp(ξ1),Fnp(ξ1)] [Enp(ξ2),Fnp(ξ2)] [Enp(ξ3),Fnp(ξ3)] (12.24) where E and F are first and second kind Lamé func tions, somewhat analogous to the Legendre P and Q functions. The two separation constants k 22 and k32 have now been shuffled into the function parameters n and p which one sees are fully cross-linked between the three factors of each harmonic. The n values get quantized to be integers n = 0,1, 2,3... and for each value of n, there are certain peculiar quantized values of p for which normali zable solutions exist. Moreover, solutions E np can be partitioned into four classes called K np, Lnp, Mnp, Nnp which have these forms (each series truncates) Kn,p(x) = 1 [ xn + αxn-2 + βxn-4 + .. ] Ln,p(x) = x2 - b2 [ xn-1 + αxn-3 + βxn-5 + ..] Mn,p(x) = x2 - a2 [ xn-1 + αxn-3 + βxn-5 + ..] Nn,p(x) = x2 - b2 x2 - a2 [ xn-2 + αxn-4 + βxn-6 + ..] where α,β are generic constants, different on each line. These E np functions are all quite simple, having the usual Frobenius form (x-x 0)r Σj Ajxj where x 0 is a regular singular point of the ODE. In our case, r = 1/2 and the regular singular points are ±a and ±b. The functions may be simple, but the theory underlying them is quite complicated, though each element of the theory is reasonably straightforward. Here are a few E np functions taken from Byerly for n = 0,1,2 and 3 (a,b are here c,b ): 69 Notice in Byerly's table that there are always (2n+1) solutions for a given n. Hobson considers conical coordinates (r, μ,ν) and shows that the harmonics in that system are rn Enp(μ)Enp(ν) which can be compared to rnYnm(θ,φ) in spherical coordinates. The linearly independent functions E np(μ) Enp(ν) for fixed n are thus linear combinations of the (2n+1) Y nm(θ,φ), and that is why there are (2n+1) E np functions. The label n is of course the quantum number associated with angular momentum. If one writes the angular momentum operator L2 in these two coordinate systems, one finds that E np(μ) Enp(ν) and Ynm(θ,φ) are both eigenfunctions of L2 with eigenvalue n(n+1). 70 A classic problem in ellipsoidal coordinates is determining the electrostatic potential outside a charged metal ellipsoid having label ξ1= c and potential V 0. The solution is this (see Morse and Feshbach p 1308 10.3.91) ψ = constant * [F 00(ξ1)] [E00(ξ2)] [E00(ξ3)] = constant * [F 00(ξ1)] [1] [1] = constant * F 00(ξ1) = constant * sn-1(1/ξ1,b/a) = V 0 sn-1(1/ξ1,b/a)/ sn-1(1/c,b/a) (12.25) where the sn -1 is an inverse Jacobi function which is equal to the elliptic integral of the first kind, sn -1(x,k) = F(sin-1x,k) = F(sin-1x | m) = F(sin-1x \ α) k = sin α m = k2 Taking the limit c →a, ψ becomes the potential of a charged thin metal elliptical plate, and then taking b→a ψ is the potential a charged metal disk. These three problems, especially the first two, are quite difficult to solve in any other coordinate system. The reader interested in learning about Lamé functions luckily has a very excellent source: the 43-page final chapter of a 1931 book by E.W. Hobson. Even in 2011 this chapter is crisp and clear, though there are a few typos. The latter part of the chapter deals with the issue of expressing the ellipsoidal harmonics in Cartesian coordinates. (f) Comment on a missing minus sign on page 512 of Morse and Feshbach Lookin g back at the start of Section 3, we had (take K 12 → k12 for this subsection) ( ∇2+k12)ψ = 0 Σ n(1/H)∂n[(H/hn2)(∂nψ)] + k12ψ = 0 H ≡ h1h2h3 (3.1) For cases with R=1 we know from (3.5) that (H/h n2) = Fngn, so (3.1) may be written Σ n (gn/H) ∂n[Fn(∂nψ)] + k12ψ = 0 . ( 1 2 . 2 6 ) From (12.9), gn/H = GnFn/ (G1G2G3) = - GnFn/ (G1G2G3) and then (12.26) becomes - Σ n GnFn/ (G1G2G3) ∂n[Fn(∂nψ)] + k12ψ = 0 . (12.27) From (12.5) make these replacements in (12.27), G 1F1 .... F1 → G 1f1.... f1 G2F2 .... F2 → (-G2)(±if2).... (±if2) = G2f2.... f2 G3F3 .... F3 → G 3f3.... f3 71 to get – Σ n Gnfn/ (G1G2G3) ∂n[fn(∂nψ)] + k12ψ = 0 (12.28) where notice the minus sign on the LHS. We now compare (12.28) with Morse and Feshbach p 512 (5.1.37), and we see that Morse and Feshbach should ha ve a minus sign to the left of the large Σ in the above equation. Their G n are the same as ours in (12.4), defined cy clically. There is no disagreement about the sign of k 12 as this Morse and Feshbach quote from p 509 shows This rare Morse and Feshbach sign error created mu ch confusion for the author until it was detected. 72 13. Stäckel Theory in N dimensions Altered Development Equations The generalization of the above analysis from N = 3 to N = N is completely straightforward, one can just march down the development and make the necessary alterations. It seemed best not to do this initially to keep things simple. Rather than create new equati on numbers, we show the old ones and the reader can assume that future references to these numbers in this section imply use of the equations as modified here. We start at the beginning, ( ∇ 2+K12)ψ = 0 H-1{ ∂1[(H/h12)(∂1ψ)] + cyclic} + K 12ψ = 0 H ≡ h1h2h3...hN H -1 Σn=1N ∂n[(H/hn2)(∂nψ)] + K12ψ = 0 (3.1) ψ = X 1X2X3...XN/ R (3.2) The "+ cyclic" notation now brings in N-1 other term s obtained by cyclic permutation of the first term. Continuing on, we find (H/[R 2hn2]) = fn(n)gn(≠n) n = 1,2,3...N (3.5) (1/R) Σ n=1N [ (1/[hn2Xn]) (1/fn)∂n[fn { R(∂nXn) - Xn(∂nR) }] + K 12 = 0 (3.6) Σn=1N (1/[hn2fnR]) ∂n{fn(∂nR)} ≡ - k12/Q(123) (3.9) Σn=1N (1/[hn2Xn]) (1/fn)∂n[fn(∂nXn)] + k12/Q + K12 = 0 (3.10) – Σ n=1N (1/hn2)qn + k12/Q + K12 = 0 (4.4) The Stäckel matrix is now an NxN matrix defined in the obvious manner, and one has q n(n) = [ κ12Φn1(n) + k22Φn2(n) + k32Φn3(n) + ... + k N2ΦnN(n)] n = 1,2,3..N (4.5) L nXn = (1/fn)∂n[fn(∂nXn)] + [ κ12Φn1(n) + k22Φn2(n) + k32Φn3(n) + ... + k N2ΦnN(n)]Xn = 0 This represents a set of N separa ted equations, n = 1,2....N (4.8) There are now N-1 separation constants k 22, k32....kN2, each a real number (possibly negative). Recall that κ12 is a stand-in for either the Helmholtz parameter K 12 for Problem B (R-separation), or the constant k 12 associated with Q for Problem A (simple-separation). The matrix equation (5.3) is the same, but the matr ices are NxN. The solution of this equation is still the Cofactor condition as in (5.6a), Mn(Φ)/S(Φ) = (Q/h n2) n=1,2,3..N " Cofactor condition" (5.6a) 73 The Robertson condition is slightly generalized now and says S(Φ) = H / (f 1f2f3...fN QR2) "Robertson condition" (5.7) and then (5.8a) is similarly generalized Mn = gnfn/(f1f2f3...fN) (5.8a) 1/Q = Σ n=1N (1/hn2) Φn1( n ) (5.10) Problem B recast at the end of Section 5 is this. Given S(Φ) = H / (f 1f2f3...fNQR2) // Robertson Mn(Φ) = S(Φ ) (Q/hn2) / / C o f a c t o r (5.11) how exactly do we find the N 2 elements of Φ ? Equivalence Rules The equivalence rules of Section 6 are slightly m odified and their derivation just slightly more complicated. It helps to have a 4x4 example for illustration: Φ = ⎝⎜⎛ ⎠⎟⎞ a b c d e f g h i j k l m n o p Suppose we multiply one of the last N-1 columns by α and another by 1/ α . For example, doing this for the 2nd and 4th columns gives Φ = ⎝⎜⎛ ⎠⎟⎞ a αb c d/α e αf g h/α i αj k l/α m αn o p/α We know that if we take a matrix and multiply any column (or row) by a scalar α, the determinant gets multiplied by α. So here we have done that once with α and once with 1/ α so the determinant S is unchanged. Now consider the cofactor of a first column element. This cofactor is a signed minor with the sign fixed by position of the element, and the minor is the determinant of a 3x3 matrix. By the same argument just stated, this determinant is unaltered by having one column scaled by α and another by 1/ α. Therefore, the cofactors of the elements of th e first column are unaltered by our scaling process. Therefore this process generates an equivalent Stäckel matrix. So we modify Rule (1) to read 74 (1) multiply one of the last N-1 columns by any (nonzero) constant α, and multiply another of the last N-1 columns by 1/ α. (6.1) Now what happens if we swap a pair of columns among the last N-1 columns and then negate one of them? We know that this action does not ch ange the overall determinant S, but by the same argument it does not change any of the 3x3 minors involved in the four cofactors of the first column elements. Thus, this action creates an equivalent Stäckel matrix. So we have Rule (2) (2) swap any pair of the last N-1 columns of Φ and then negate either of these columns. (6.2) Finally, what happens if we add a multiple of one of the last N-1 columns to another of the last N-1 columns. For example, let's add λ times the last column to the second last: Φ = ⎝⎜⎛ ⎠⎟⎞ a b c+λd d e f g+λh h i j k+λl l m n o+λp p We know S = det Φ is unchanged. And once again, this opera tion does not change any of the minors of the first column elements. So again we have an e quivalence operation. Adding a multiple of one of the last N-1 columns to the first is also OK since it changes neither S nor the M n cofactors. Rule (3) becomes: (3) add any multiple of one of the last N-1 columns to a different column. (6.3) Summary of Stäckel Matrix Equivalence Operations for NxN matrices (1) multiply one of the last N-1 columns by any (nonzero) constant α, and multiply another of the last N-1 columns by 1/ α. (6.1) (2) swap any pair of the last N-1 columns a nd then negate either of these columns. (6.2) (3) add any multiple of one of the last N-1 columns to a different column. (6.3) Conditions We generalize Section 7 (a): Condition (1) Equations (3.5) must be solvable for the 2N+1 functions f n, gn a n d R . I f s o m e f n is a constant, set that constant to 1. For Problem A, set R=1. (H/h 12) = f1(1)g1(≠1) R2 H = h 1h2h3...hN (H/h 22) = f2(2)g2(≠2) R2 .... (H/h N2) = fN(N)gN(≠N) R2 (3.5) Assuming equations (3.5) can be solved, compute the M n cofactors as follows from (5.8a) 75 Mn(≠n) = gnfn /(f1f2...fN) n = 1,2...N (5.8a) Condition (2) Equations (7.2) must have a solution. This is a set of N functional-form conditions, where the Mn are as given above, M1(≠1) = ... -M2(≠2) = ... M3(≠3) = ... -M4(≠4) = ... (7.2) where the RHS's of these equations are the appropriate (N-1)x(N-1) minors. So the RHS of each equation has N-1 terms each of which is the product of N-1 Φ elements. We can write these expressions using the totally antisymmetric ε tensor having N indices: det(Φ) = Σnabc.. εnabc.. Φn1 Φa2 Φb3 Φc4 .... = Σ n Φn1 [ Σabc.. εnabc.. Φa2 Φb3 Φc4.... ] = Σ n Φn1Mn so that the above equations are these, where there are indeed N-1 Φ factors in each term, M n(≠n) = Σabc.. εnabc.. Φa2(a) Φb3(b) Φc4(c).... n = 1,2...N As noted earlier, it is not obvious that, given the M n as stated in Condition 1, the non-first-column Φnm elements can be found to satisfy these N conditio ns due to the functional form restriction and due to correlations between the equations since each Φnm appears in N-1 equations. Condition (3) If conditions (1) and (2) are met, we then do the work of computing Q and k 12 from (3.9), Σ n=1N (1/[hn2fnR]) ∂n{fn(∂nR)} = - k 12/ Q ( 1 2 3 . . N ) (3.9) For Problem A, Q = 1 and k 12 = 0 and no work is needed. Then with this Q expression, equation (7.6) must have a solution for the first column Φ elements, 1/Q = Φ 11(1) (1/h12) + Φ21(2) (1/h22) + ..... + Φ2N(N) (1/hN2) (7.6) and this is another functional-form condition. 76 Steps We generalize Section 7 (b ): Step 0: Write down the h n for the curvilinear system of interest and compute H = h 1h2h3 .. hn. Perhaps write down other useful facts concerning the system of interest. Step (1) As noted above, the first task is to solve (3.5) for the 2N+1 functions f n, gn and R (H/h 12) = f1(1)g1(≠1) R2 H = h 1h2h3...hN (H/h 22) = f2(2)g2(≠2) R2 .... (H/h N2) = fN(N)gN( ≠N) R2 (3.5) This task is pretty much just one of "inspecti on" when the LHS's of (3.5) are inserted (assuming Condition 1 that a solution exists! ) Step (2) Write down the N first-column cofactors from (5.8b), Mn(≠n) = gnfn /(f1f2...fN) n = 1,2...N Step (3) If Problem A, Q = 1. Otherwise compute Q and k 12 from (3.9), Σn=1N (1/[hn2fnR]) ∂n{fn(∂nR)} = - k 12/Q(123) (3.9) Step (4) Knowing Q, compute S from the Robertson condition, S(Φ) = H / (f 1f2f3...fN QR2) "Robertson condition" (5.7) Step (5) Find the rightmost N-1 columns of the Stäckel matrix by solving the following N functional-form equations for a viable set of N*(N-1) Φnm elements ( see Condition 2 above) g n(≠n)fn(n) /(f1(1)f2(2)...fN(N)) = Σ abc.. εnabc.. Φa2(a) Φb3(b) Φc4(c).... n = 1,2....N Step (6) Find the first column of the Stäckel matrix by solving (7.6) or (5.9) 1/Q(12...N) = Φ11(1) (1/h12) + Φ21(2) (1/h22) + ..... + Φ2N(N) (1/hN2) (7.6) which is of course another functional-form equation. Step (7) At this point, we may want to apply some of our Section 6 equivalence operations to obtain a Stäckel matrix Φ that is of the simplest possible form, or of a form that matches the literature. Examples? N=2 examples are given in the next section. We shall not work out any examples here for N>3, but two candidates would be "hyper-ellipsoidal coordinates in N dimensions" and "hyper-spherical 77 coordinates in N dimensions". In the first case the confocal hypersurface families would have N-1 focal distances a 1, a2..... aN-1 and things are just a generalization of what appears in Section 12. For example, the inequality chain (12.2) would be 0 ≤ ξN≤ aN-1 ≤ ξN-1..... a2 ≤ ξ2≤ a1 ≤ ξ1 For hyper-spherical coordinates ξ 1 would be the radial coordinate r, then ξ2 through ξN-1 are polar angles θ1 through θN-2, and finally ξN is an azimuth φ . It seems likely to the author that both these N dimensional coordinate systems will be separ able, but that is just a conjecture. Non-Euclidian Stäckel Theory? For an orthogonal non-Euclidian curvilinear coordinate system one has a diagonal metric tensor which cont ains elements of both signs. This means that some of the h n2 (squared scale factors) are negative and H might be negative. These facts should not affect the development as presented above in this section. 78 14. Stäckel Theory in N=2 dimensions As noted ear lier, Moon and Spencer deal with various 3D cylindrical systems each with its own 2D orthogonal system in the ξ1,ξ2 coordinates. Moon and Spencer then state the nature of the Helmholtz and Laplace solution functions in various scenarios, one of which is that the solution has no z dependence due to an "extrusion symmetry" of the problem. Th is scenario yields separated equations for ξ1,ξ2 which agree with those we find below using the 2D Stäckel method, so Moon and Spencer provide a catalog of 2D separated equations, even though they don't explicitly state the 2D Stäckel matrices. It should be realized that the 2x2 Stäckel matrix is not in general the "upper right 2x2 piece" of the 3x3 Stäckel matrix for the corresponding cylindrical system. We cannot just "delete" the 3 coordinate in a naive way, because in the 3x3 case ∂ z2 is an active part of ∇2. This fact will be demonstrated in the second Example below. We assume at first arbitrary h 1 and h2 and then later specialize to h 1= h2. There will be only one separation constant, k 22. We shall now restate the Conditions and Steps as given in Sections 7(a) and 7(b) for our case N = 2. Conditions Condition (1) Equations (3.5) must be solvable for the 5 functions f n, gn and R. If some f n is a constant, set that constant to 1. For Problem A, set R=1. (h 2/h1) = f1(1)g1(2) R2 H = h 1h2 (h1/h2) = f2(2)g2(1) R2 (3.5) Assuming (3.5) can be solved, compute the M n as follows : M 1 = g1 /f2 M2 = g2 /f1 (5.8b) Condition (2) Equations (7.2) must have a solution. For N = 2 each first column cofactor is just a signed multiple of an element in the second column, so then Φ = ⎝⎛ ⎠⎞a(1) b(1) c(2) d(2) S = det(Φ ) M1 = d(2) => d(2) = M 1 = g1 /f2 M2 = -b(1) => b(1) = -M 2= - g2 /f1 Thus Condition 2 is satisfied in any N=2 system so we can ignore it, and we know that Φ has the form Φ = ⎝⎛ ⎠⎞a(1) b(1) c(2) d(2) = ⎝⎛ ⎠⎞a(1) - g 2 /f1 c(2) g 1 /f2 79 Condition (3) If condition (1) is met, compute Q and k 12 from (3.9), Σ n=12 (1/[hn2fnR]) ∂n{fn(∂nR)} = - k 12/ Q ( 1 2 ) (3.9) For Problem A, set Q = 1 and k 12 = 0 and no work is needed. Then with this Q expression, equation (7.6) must have a solution for a and c 1/Q = a(1) (1/h 12) + c(2) (1/h 22) (7.6) Steps Step (0) Write down the h 1 and h2. Step (1) As noted above, the first task is to solve (3.5) for the 5 functions f n, gn and R (h2/h1) = f1(1)g1(2) R2 (h1/h2) = f2(2)g2(1) R2 (3.5) This task is pretty much just one of "insp ection" when the LHS's of (3.5) are inserted. Step (2) Write down the two first-column cofactors from (5.8b) M 1 = g1 /f2 M2 = g2 /f1 (5.8b) Step (3) If Problem A, Q = 1. Otherwise compute Q and k 12 from (3.9), Σ n=12 (1/[hn2fnR]) ∂n{fn(∂nR)} = - k 12/Q(12) (3.9) Step (4) Knowing Q, compute S from the Robertson condition, S(Φ) = (h 1h2) / (f1f2QR2) (5.7) Step (5) We did this in Condition 2 above and found Φ = ⎝⎛ ⎠⎞a(1) - g 2 /f1 c(2) g 1 /f2 . Step (6) Find the first column of the Stäckel matrix by solving (7.6) or (5.9) 1/Q = a(1) (1/h 12) + c(2) (1/h 22) ( 7 . 6 ) or S = a(1) M 1 + c(2) M 2 (5.9) Step (7) The only equivalence rule that survives for N = 2 is to add a multiple of the second column to the first column. 80 Example: Polar Coordinates ( ξ1, ξ2) = (r,θ) Step 0: h1 = 1 h2 = ξ1 Step (1) We know R = 1 so get (h2/h1) = f1(1)g1(2) R2 (h1/h2) = f2(2)g2(1) R2 (3.5) ξ 1 = f1(1)g1(2) => f 1 = ξ1 g1= 1 1/ξ1= f2(2)g2(1) => f 2 = 1 g2= 1/ξ1 Step (2) Write down the two first-column cofactors from (5.8b), M 1 = g1 /f2 = 1/1 = 1 M2 = g2 /f1 = ξ1-1/ξ1 = 1/ξ12 (5.8b) Step (3) R = 1 => Q = 1 Step (4) Knowing Q, compute S from the Robertson condition S(Φ) = [ (h1h2) / (f1f2) ] (5.7) S(Φ) = [ξ1/ ξ1] = 1 Step (5) From Condition 3 above we found Φ = ⎝⎛ ⎠⎞a(1) - g 2 /f1 c(2) g1 /f2 , therefore Φ = ⎝⎛ ⎠⎞a(1) - ξ1-2 c(2) 1 Step (6) We have to solve S = a(1) M 1 + c(2) M 2 (5.9) 1 = a(1) 1 + c(2) 1/ ξ 12 This has a solution a(1) = 1 and c(2) = 0 so that Φ = ⎝⎛ ⎠⎞a(1) - ξ1-2 c(2) 1 = ⎝⎛ ⎠⎞1 - ξ1-2 0 1 // Stäckel matrix for polar coordinates. The separated equations are ( since simple-separation, κ12 = K12 = the Helmholtz parameter ) L nXn = (1/fn)∂n[fn(∂nXn)] + [K12Φn1(n) + k22Φn2(n) ]Xn = 0 (4.8) (1/f1)∂1[f1(∂1X1)] + [K121 + k22(-ξ1-2) ]X1 = 0 (1/f2)∂2[f2(∂2X2)] + [K120 + k221 ]X2 = 0 81 (1/ξ1)∂1[ξ1(∂1X1)] + [K12 – k22/ξ12 ]X1 = 0 ( ∂22X2) + [ k22 ]X2 = 0 (1/r)∂r[r(∂rR)] + [K 12 – k22/r2 ]R = 0 or - (rR')' + k 22 (R/r) - K 12rR = 0 ( ∂θ2Θ) + [ k22 ]Θ = 0 r 2R" + rR' + [ r2 K12 – k22 ]R = 0 R = X 1 (14.1) Θ" + [ k22 ]Θ = 0 Θ = X2 These agree with Moon and Spencer p 16 ("For φ independent of z") with K 12 → κ2 and k22 → α2. To simplify notation a bit, we define a 2 = K12= the Helmholtz parameter // = - α2 b2 = k22 = the separation constant // = - β2 We then rewrite the above pair of separated equations as r 2R" + rR' + [ r2 a2 – b2 ]R = 0 or - (rR')' + b2 (R/r) - a2rR = 0 (14.2) Θ" + b2Θ = 0 This radial equation is a scaled Bessel equation with soluti on of the type J b(ar). Consider now various situations: a2 > 0 a = real a2< 0 a = i α = imaginary, α real b 2 > 0 b = real b2< 0 b = iβ = imaginary, β real With these symbols, one can write the separated solu tion in a variety of ways. Here are the possible "atomic forms" (they cannot be called harmonics b ecause they are Helmholtz solutions, not Laplace solutions ) ψ ~ X 1X2 ~ [Jb(ar), Yb(ar)] [sin(b θ),cos(bθ)] a = real b = real (14.3a) ψ ~ X 1X2 ~ [Ib(αr), Kb(αr)] [sin(b θ),cos(bθ)] a = imaginary b = real (14.3b) ψ ~ X 1X2 ~ [Jiβ(ar), Yiβ(ar)] [sinh( βθ),cosh(βθ )] a = real b = imaginary (14.3c) ψ ~ X1X2 ~ [Iiβ(αr), Kiβ(αr)] [sinh( βθ),cosh(βθ )] a = imaginary b = imaginary (14.3d) 82 Using Sturm-Liouville to solve a specific Hel mholtz problem in polar coordinates A prototype problem in 2D polar coordinates is to fi nd the Helmholtz Green's Function for an infinite 2D wedge-shaped region with vertex at the origin. If this is a normal "Dirichlet" Green's Function, then the solution must vanish on both faces of the wedge, say θ=0 and θ=ψ, and the point source lies somewhere inside the wedge at r', θ'. This then turns the Θ equation above into a simple 1D Sturm Liouville problem and we find that the separation constant called b2 gets quantized to specific values b n2 where b n = nπ/ψ, n = 0,1,2... The complete set of eigenfunctions associated with this SL problem is { sin(b nθ), n = 1.2.3...} and the "transform" associated with this Sturm Liouville problem is the Fourier Sine Series transform. If we start off assuming that the Helmholtz parameter a2 is negative, then we can try to find a solution using atoms in (14.3b) where a = i α, and we try this "Smythian form", g(r,θ|r',θ') = Σn=1∞ Cn(r',θ') sin(bnθ) Ib(αr<) Kb(αr>) r < = min(r,r') r > = max(r,r') where C n is a coefficient to be determined. One can in fact solve for C n to obtain this result ( see Stakgold vol 2 p 273 (7.174)), g(r,θ|r',θ') = Σn=1∞ (2/ψ) sin(nπθ/ψ) sin(nπθ'/ψ) Inπ/ψ(αr<) Knπ/ψ(αr>) (14.4) An interesting special case is also treated by Stakgol d. If one lets the wedge angle increase all the way to 2π, then the two wedge sides meet along the +x axis, and the problem is to find the Helmholtz Green's Function where this half line x>0 is "grounded", me aning g=0. He further places the Green's unit point source at location x = -r 0 on the x axis. The reader can extrude this picture out of the plane of paper to find a 3D problem solvable by 2D methods in which there is an infinite line source lying off th e edge of a grounded half plane. The solution to this special case is of course just the above formula with ψ = 2π and θ' = π, so we get [ Stakgold vol 2 p 275 (7.183) ] g(r, θ|r 0,π) = Σn=1∞ (1/π) sin(nθ/2) sin(nπ/2) In/2(αr<) Kn/2(αr>) (14.5) Remember that α encodes the Helmholtz parameter according to α2 = -K12 and for this problem we can assume that K 12 < 0 so α2 > 0, at least for a start, then we can analytically continue in α as needed. Stakgold then proceeds to do this same sp ecial case problem by instead treating the radial R equation as a 1D Sturm-Liouville problem in r on the interval (0, ∞). In this case, taking b = iβ and a = i α, we can use atomic form (14.3d) to set things up. In this Sturm-Liouville problem, it turns out that the complete set of eigenfunctions is { K iβ(αr), β = real values 0 to ∞ }. Looking at (14.3d) and looking at the symmetry of the problem relative to θ = π, we are led to try the following Smythian form (this form is symmetric under x axis reflection which takes θ → 2π-θ ) 83 g(r,θ|r0,π) = ∫0 ∞ dβ Kiβ(αr) [Fα(β, r0) ch(β |θ-π|) + Gα(β, r0) sh(β |θ-π|)] where now F α and Gα are the unknown coefficients. As in the pr evious case, we are just expanding on the appropriate Sturm-Liouville basis functi ons, which for the r coordinate are K iβ(αr). It is then possible to find these coefficients and express the answer to this problem as follows: g(r,θ|r0,π) = (1/π2) ∫0 ∞ dβ Kiβ(kr) Kiβ(kr0) sh(βπ ) { th(βπ ) ch(β (|θ-π|) - sh(β(|θ-π|) } (14.6) As must be the case for any Green's Function (as we are using them), g must be symmetric under the interchange r ↔r0 and θ ↔ π. In this particular problem, g is symmetric under each of these swaps separately. Notice how all our previous g's have been symmetric as well. An interesting exercise is to show that (14.5) a nd (14.6) are in fact the same. This can be done by rewriting (14.6) this way g(r,θ|r0,π) = - (i/2π) ∫-∞ ∞ dβ I-iβ(kr<) K-iβ(kr>) { th(βπ ) ch(β (|θ-π|) - sh(β(|θ-π|) } The contour can be closed up or down, and then as it is deformed, it picks up the pole residues of th( βπ) which are located where ch( βπ) = 0 = cos(iβπ ), which are points on the imaginary β axis at iβ = ±n/2. This sum of residues becomes the sum shown in (14.5). The point is that in polar coordinates, where there is one separation constant b2= -β2, it is possible to do the 1D Sturm-Liouville analysis in either of the separated functions. The transform associated with the complete set of functions just mentioned is known as the Kantorovich-Lebedev Transform. The functions Kiβ(αr) are the usual modified Bessel functions (McD onald functions) but of imaginary argument, and are in fact real and oscillatory for real argument αr (see Appendix A). We want to emphasize a key point. In our first so lution of the "half plane problem", the spectrum of β2 was discrete, being i βn = bn = nπ/ψ with ψ =2π. In our second solution, the spectrum of this very same separation constant β2 was continuous with β being the continuum of the positive real axis. The reader interested in studying this problem s hould be advised there are 5 typos in this section of Stakgold (1968) Volume 2 which we can just nail down right here : p 273 middle of page should say "by K iγ(kr) and " , not "by r K iγ(kr) and" bottom equation should contain sinh, not sin p 275 the A+B equation should contain sinh, not sin same for the following equation the equation following that is wrong and should read ν~ = – { cosh[ γ(φ-π)] / [γ sinh(2γπ )]} K iγ(kr0) Only the first of these five errors is correct ed in the 2000 SIAM edition of the book. 84 The special case h 1 = h2 Step 0: Let h12 = h22 ≡ R2 Step (1) (h2/h1) = f1(1)g1(2) R2 (h1/h2) = f2(2)g2(1) R2 1 = f 1(1)g1(2) R2 1 = f 2(2)g2(1) R2 An obvious solution is f 1 = 1 g1= 1 R = 1 f2 = 1 g2= 1 Step (2) Write down the two first-column cofactors from (5.8b) M1 = g1 /f2 = 1 M2 = g2 /f1 = 1 Step (3) R = 1 => Q = 1 Step (4) Knowing Q, compute S from the Robertson condition S(Φ) = (h 1h2) / (f1f2) "Robertson condition" (5.7) S(Φ) = (h12) / (1*1) = R2 Step (5) Find the rightmost column of the Stäckel matrix, d(2) = g 1 /f2 = 1 b(1) = - g 2 /f1 = - 1 so that Φ = ⎝⎛ ⎠⎞a(1) b(1) c(2) d(2) = ⎝⎛ ⎠⎞a(1) -1 c(2) 1 Step (6) We have now to solve S = a(1) M 1 + c(2) M 2 or R2 = a(1) + c(2) This will only be possible if our 2D system is such that 85 R2 = R12(1) + R22(2) in which case we have our final Stäckel matrix Φ = ⎝⎛ ⎠⎞a(1) b(1) c(2) d(2) = ⎝⎜⎛ ⎠⎟⎞ R12 -1 R22 1 Step (7) The only equivalence rule that survives for N = 2 is to add a multiple of the second column to the first column. Separated equations: LnXn = (1/fn)∂n[fn(∂nXn)] + [ κ12Φn1(n) + k22Φn2(n) ]Xn = 0 (4.8) (1/f 1)∂1[f1(∂1X1)] + [ κ12Φn1(n) + k22Φn2(n) ]X1 = 0 (1/f2)∂2[f2(∂2X2)] + [ κ12Φn1(n) + k22Φn2(n) ]X2 = 0 (1/1)∂1[1(∂1X1)] + [ κ12R12 - k22]X1 = 0 (1/1)∂1[1(∂1X1)] + [ κ12R22 + k22 ]X2 = 0 ( ∂ 12X1) + [ κ12R12 - k22]X1 = 0 ( ∂22X2) + [ κ12R22 + k22 ]X2 = 0 Summary of 2D systems with h 1 = h2 Define R2 ≡ h12 = h22. Separability is only possible if one can write R2(12) = R12(1) + R22(2). In this case, we get simple-separation for the Helmholtz equation with this data f1 = 1 g1= 1 R = 1 M 1 = 1 S = R2 f2 = 1 g2= 1 Q = 1 M 2 = 1 Φ = ⎝⎜⎛ ⎠⎟⎞ R12 -1 R22 1 (∂12X1) + [ κ12R12 - k22]X1 = 0 (∂22X2) + [ κ12R22 + k22 ]X2 = 0 86 Example : elliptical coordinates In this s ystem one has h12 = h22 =R2 = a2(ch2ξ1 - cos2ξ2) so R 12(1) = a2ch2ξ1 R22(2) = -a2cos2ξ2 Φ = ⎝⎛ ⎠⎞a2ch2ξ1 -1 -a2cos2ξ2 1 ≈ ⎝⎛ ⎠⎞a2sh2ξ1 -1 -a2sin2ξ2 1 (∂12X1) + [ κ12 a2ch2ξ1 - k22]X1 = 0 (∂22X2) + [ - κ 12a2cos2ξ2 + k22 ]X2 = 0 We added a 2 times the last column to the first to get an equivalent form of Φ. The first form agrees with Moon and Spencer p 20 ("For φ independent of z") with κ12 → κ2 and k22 → α2. The solutions to these equations are Mathieu functions. Recall from Section 9 Example 1 the Stäckel matrix for elliptic cylinder coordinates Φ = ⎝⎜⎜⎛ ⎠⎟⎟⎞ 0 - 1 - a2ch2(ξ1) 0 1 a2cos2(ξ2) 1 0 1 ≈ ⎝⎜⎜⎛ ⎠⎟⎟⎞ 0 a2ch2(ξ1) -1 0 -a2cos2(ξ2) 1 1 -1 0 where to get the second form we have swapped the last two columns and then negated the 2nd column. In this special form, it happens that the upper right 2x2 matrix of Φ 3x3 is the same as our first Φ2x2 matrix, but it is pretty clear than one cannot just take any form of Φ3x3 and assume this will be the case! 87 Appendix A. Review of 1D Sturm-Liouville Theory; the Kantorovich-Lebedev transform The requirements of a 1D Sturm -Liouville problem are these: [ Stakgold vol 1 p 268 + p 295 ] (1) L is a self-adjoint differential operator of the form L u = (pu')'+q. (2) L acts on L 2-normalizable (possibly complex) functions on some interval (a,b). L2 is a Hilbert Space of such normalizable functions with scalar product (physics convention) <f,g> = ∫a b dx f*(x)g(x). (3) The eigenvalue problem is Lφλ = s(x)λφ λ where s(x) is some weight function. The functions p,q,s are all real and "reasonable" (C 1), and p and s must be non-negative on (a,b). At each end of the interval we have an unmixed boundary condition such as A φλ(a) + B φλ'(a) = 0, where A and B are real. If either of the endpoints is "singular", such as b= ∞ or p(b) = 0 , the boundary conditio n for that endpoint is replaced by a requirement that φ λ be finite at that endpoint. This can ha ppen in two ways called "limit circle" and "limit point" as discussed by Stakgold. The eigenfunctions of such a problem form a complete set: the φλ span the infinite dimensional Hilbert Space of L2 of functions on (a,b) which meet the boundary conditions. The spectrum of eigenvalues of λ can in general be "mixed", consisting of a point sp ectrum and a continuous spectrum, both on the positive real axis in the λ plane, though in practice one usually has either discrete or continuous. We will assume the general mixed case, so we shall write the eigenvalue problem on two lines Lφλ (x) = λ s(x) φλ (x) // spectrum λn = continuous range of real values (B.3) Lφλi (x) = λi s(x) φλi(x) // spectrum λni = discrete set of real values i = 1,2,3.... The statements of completeness and orthonormality of the eigenfunctions are these Σ i φλi*(x)φλi(ξ) + ∫dλ φλ(x)* φλ(ξ) = δ(x-ξ)/s(x) // completeness (B.4) <φλi,sφλj> = δi,j <φλ,sφλ'> = δ(λ-λ') <φλ,sφλn> = 0 // orthogonality (B.5) where one should take note of the we ight function s(x) in all equations. Such relations immediately imply the existence of a "transform" which we write as F λi ≡ ∫a b dx s(x) φλi*(x) f(x) Fλ ≡ ∫a b dx s(x) φλ*(x) f(x) // projections (B.6) f(x) = Σi Fλi φλi(x) + ∫dλ Fλ φλ(x) // expansion (B.7) 88 where we ignore issues of convergence of sums and in tegrals and assume f(x) is suitable such that its projections exist, and that the expansion converges . The above transform can easily be "verified" by inserting either line into the other and using th e completeness and orthogonality given above. The reader is reminded that every Sturm-Liouv ille problem is associated with its own private transform, though frequently occurring transforms have people's names attached to them such as Fourier Sine Series Transform, Fourier Integral Transform, Mellin Transform, Hankel Transform, and so on. Having been explicit with the mixed spectrum, we now adopt a more compact notation where the reader understands that the spectrum can be discrete, continuous, or mixed: Fλi ≡ ∫a b dx s(x) φ λi*(x) f(x) // projection f(x) = Σλi Fλi φλi(x) // expansion Here the λ i in the projection includes discrete and continuous, and the notation Σλi implies a sum on the discrete part plus an integral on the continuous part. If we knew there was only a continuous spectrum, we might write this as F(λ) ≡ ∫a b dx s(x) φ λ*(x) f(x) // projection f(x) = ∫dλ F(λ) φλ(x) // expansion but we shall retain the F λi notation. The F λi are sometimes called "the coefficients", or "the transform". Basically the variable x of f(x) is traded out for the variable λi of Fλi, just as in a spatial Fourier transform the variable x is traded out for λ = k2. One must be a little careful with the distinction between λ and a convenient variable used to label λ, in this case k. For example, d λ = 2kdk. In the same vein, we can compact down our completeness and orthogonality this way Σ λi φλi*(x)φλi(ξ) = δ(x-ξ)/s(x) // completeness ∫a b dx s(x) φλi*(x) φλi'(x) = δ λi,λi' // orthogonality The notations Σ λi and δλi,λi' are just shorthands for the fuller e quations. We can gather up the above results and summarize the 1D Sturm Liouville situation as follows: Lφλi (x) = λi s(x) φλi(x) // eigenvalue problem (with BC's) on (a,b) Σλi φλi*(x)φλi(ξ) = δ(x-ξ)/s(x) // completeness ∫a b dx s(x) φλi*(x) φλi'(x) = δ λi,λi' // orthogonality Fλi ≡ ∫a b dx s(x) φ λi*(x) f(x) // transform projection f(x) = Σλi Fλi φλi(x) // transform expansion (B.8) 89 We should mention that there is a standard technique used to find the eigenfunctions φλ. One first solves this Green's Function problem, [ Lx - λs(x)] g(x|ξ; λ) = δ(x-ξ) ( B . 9 ) using the usual method of finding a boundary-condition- matching homogeneous solution to the left and to the right of x = ξ , and then matching the jump condition in the first derivative at x = ξ. g(x|ξ; λ) = A(λ ) uleft(x<) uright (x>) Δg'|x=ξ = -1/p Once this g(x| ξ; λ) is found, one can deduce the normalized ei genfunctions by matching the two sides of this equation, where C is a countercl ockwise great circle contour in the λ plane, - (1/2 πi) ∫C dλ g(x|ξ;λ) = Σλn φλn*(x)φλn(ξ) + ∫dλ φλ(x)* φλ(ξ) (B.10) If g(x|ξ;λ) has a branch cut along the real axis, the contour wr aps this cut and one obtains a real integral of the discontinuity of g across the cut, and that beco mes the second term on the RHS. Again, care is needed to distinguish the use of λ on the LHS as a complex contour integral variable, and on the RHS as the variable of a real axis integration. Poles in g(x| ξ;λ) give rise to the first term on the RHS. We would be remiss to omit the following standard expansion for g(x| ξ;λ), g(x|ξ;λ) = Σi φλi (x)* φ λi(ξ)/ (λi-λ) + ∫dλ' φλ(x)* φλ'(ξ)/(λ'-λ)] (B.11) In the second term it is the "continuum of poles" that creates the branch cut whose discontinuity is then picked up as the second term in (B.10). As an example, here is the Sturm Liouville data for equation (14.2) on the interval (0, ∞). This is the radial equation for separated polar coordinat es. We used this data slightly in the discussion leading to (14.6) and it is the basis of the derivation of (14.6), which is somewhat involved. The spectrum in this case is purely continuous and eigenvalue λ= β2 spans the entire positive real axis of the λ plane. -(xu')' + α2xu -β2x-1u = 0 or Lu = β2x-1u where L = -(xu')' + α2xu or L λu = 0 where L λ = L - β2x-1 λ = β2 s(x) = x-1 φβ(x) = (1/π) sh(πβ ) Kiβ(αx) // eigenfunctions (2/π2) ∫0 ∞ dβ β sh(πβ) Kiβ(αx) Kiβ(αx') = x δ (x-x') // completeness ∫0 ∞ dx x-1 Kiβ(αx) Kiβ'(αx) = δ(β-β') π2/[2βsh(πβ )] // orthogonality Fα(β) ≡ ∫0 ∞ dx x-1 f(x) Kiβ(αx) // projection (KL) f(x) = (2/ π2) ∫0 ∞ dβ β sh(πβ) Kiβ(αx) Fα(β) // expansion (KL) 90 As noted earlier, for this particular operator L, the transform is called the Kantorovich-Lebedev transform (see e.g. Stakgold vol 1 p 317 4.30). It is always possible to write the halves of a transform some other way by replacing f(x) = jobob(x) F(x). The entire discussion of this Appe ndix can be generalized such that the transform projections are onto eigenfunctions of multiple operators , which functions are the unitary irreducible representation functions of a continuous group. The spherical harmonics are a well known example associated with the rotation group SO(3). References W.E. By erly, An elementary treatise on Fourier series and spherical, cylindrical and ellipsoidal harmonics, (Dover Publications, New York, 1893). L.P. Eisenhart, "Separable systems of Stäckel", Ann. of Math. vol. 35 (1934) p. 284. E.W. Hobson, The Theory of Spherical and Ellipsoidal Harmonics, (Cambridge University Press, Cambridge, 1931). Chapter XI concerns ellipsoidal ha rmonics, about 43 pages. This presentation seems more readable than Chapter 23 of Whittaker and Watson (see below). P. Moon and D.E. Spencer, "Theorems on Separability in Riemannian n-space." Proc. Amer. Math. Soc. 3, 635-642, 1952. P. Moon and D.E. Spencer, Field Theory Handbook, Including Coordinate Systems, Differential Equations and their Solutions (Springer-Verlag, Berlin, 1961). Th is book is not about quantum field theory or anything like that, it is about curvilinear coordinate systems, how the Laplace and Helmholtz equations appear in each system, and what the solu tions of these equations lo ok 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 2-volume classic behemoth is simply amazing. H.P. Robertson, "Bemerkung über separierbare Syst eme in der Wellenmechanik", Math. Ann. vol. 98 (1927) p. 749. L.I. Schiff, Quantum Mechanics, 3rd Ed ., (McGraw Hill, New York, 1968) P. Stäckel, "Uber die Integration der Hamilton–Jacobischen Diff erential Gleichung mitte lst Separation der Variabel, Habilitationsschrift", University of Halle, 1891 . This 26 page habilitation thesis can be viewed on line at the Hathi Trust repository, http://catalog.ha thitrust.org/Record/010688645. Stäckel arrived at his separation matrix in the context of the Hamilton-Jacobi formulation of classical mechanics. Two subsequent related Stäckel references are: "Sur une classe de problèmes de dynamique", C.R. Acad. Sci. Paris, vol. 116 (1893) p. 485. "Über die Integration der Hamiltonschen Differen tialgleichung mittels Sepa ration der Variabelen", Math. Ann. vol. 49 (1897) p. 145. 91 I. Stakgold, Boundary Value Problems of Mathematical Physics, Volumes 1 and 2 (MacMillan, London, 1967). These are astoundingly good books, but the high level of detail (the subject is intrinsically complex) makes them hard to use in a normal "course" , which is why the author later put out a condensed single-volume version Green's Functions and Boundary Value Problems, now in a third edition. The original two volumes were reprinted with some corrections in 2000 ( SIAM, Philadelphia). The author is, I presume, alive and well and a working emeritus at the University of Delaware. E.T. Whittaker and G.N. Watson, A Course in Modern Analysis, 4th Ed. , (Cambridge University Press, Cambridge, 1927). The 2nd edition is being sold by Merchant Press and does not include Chapter 23 on Ellipsoidal Harmonics and Lamé Functions.