1 the Stackel matrix business

The Stackel matrix business PhL 8.13.11 These are the raw notes, with lots of clumsy stuff and exploratory sections all retained. See meta notes for a better presentation. Explanation of the Stackel Method of Separating Helmholtz 1 Start Off 1 First required assumption 3 Some Side Comments 3 Showing that the rn must vanish 4 Note added: 6 Introduction of the Nine Φ functions 8 Stating the processed Helmholtz equation as a matrix equation 9 Inverting this matrix equation 10 Introduction of Gpb as notation for cof(Φpq), and then Mp as Gp1. 10 Summary of key equations obtained in the previous section 12 Ellipsoidal Coordinates 16 What do we know about extruded conformal 3D systems (cylindricals)? 17 What do we know about extruded other 3D systems (cylindricals)? 19 Symmetry operations on matrix Φ revisited. 19 Comments on the "separation constants". 20 How to find the matrix Φ 21 Finding the Stackel matrix for the ellipsoidal coordinate system 24 The need to use F1,F2, F3 for our Stackel analysis 24 The equations for the Mn cofactors 25 Trial and Error Solution for the last 2 columns of the Stackel matrix. 26 Finding the 1st column of the Stackel matrix. 27 The alternate and simpler solution for the 1st column 30 What are the "symmetry" operations for a Stackel matrix? 31 Plan G: The symmetry transformations 33 Verify that my two alternate 1st column forms have the same determinant 35 What about Laplace separation? 37 History. See http://www.apprendre-math.info/anglais/historyDetail.htm?id=Stackel . The person here is in fact German Paul Stäckel 1862-1919 ( I think it is "shtay'kel") Die Integration der Hamilton-Jacobischen Differentialgleichung mittelst Separation der Variablen, 1891, Habilitation Here is a footnote from somewhere "1 By now, these potentials had come to be known as St¨ackel potentials in the astronomical literature. This seems unwarranted. First, it is poor practice in physics to associate a name with an equation if a perfectly adequate descriptive term exists. On these grounds alone, the term ‘separable potential’ is preferable to ‘St¨ackel potential’. And, second, there is no reason to associate the name of Paul St¨ackel with coordinate systems and potentials that he never wrote down! St¨ackel was a prominent differential geometer, latterly Professor of Mathematics at Heidelberg. In his Habilitationschrift in 1891 at Halle, St¨ackel wrote down the condition for the Hamilton-Jacobi equation to separate in a given coordinate system on a general Riemannian manifold in the form of the vanishing of a determinant (which has reasonably enough come to be called the St¨ackel determinant). St¨ackel did not derive the coordinate systems in Euclidean 3-space for which his determinant vanishes, far less the form of the separable potentials in these coordinates. This work was left to Weinacht (1924) and Eisenhart (1948). In fact, St¨ackel’s result is limited, as it does not even provide a comprehensive test for separability. St¨ackel’s determinant for a separable system only vanishes if it is written down in the separable coordinate system itself. The finding of a general criterion for identifying whether a potential is separable in some coordinate system remains an outstanding research problem. " Explanation of the Stackel Method of Separating Helmholtz Start Off I have been putting this off for a long time, but now the time is right. The subject is specifically how to try to find the separated equations for the Helmholtz equation in arbitrary 3D orthogonal curvilinear coordinates. I will be crudely following the presentation of M&F on page 509 of Vol I, section 5.1. This work is pretty high up in my building of bricks, a lot had to happen before. For example appropriate matrix theorems understanding of curvilinear coordinates and how differential operators look in same. some understanding of PDE's and ODE's understanding the significance of the Helmholtz equation! Consider our known form of 2 given an orthog curvilinear system: (see "tensors and curvilinear doc") 2u = (1/ [h1h2h3]) { ∂1 [ (h1h2h3/h12 ) (∂1u)] + cyclic } (1) Let's define H = h1h2h3 to make things more compact, then we have 2ψ = H-1{ ∂1[(H/h12)(∂1ψ)] + cyclic} = H-1 Σn ∂n[(H/hn2)(∂nψ)] (2) If we now assume that ψ(ξ1,ξ2,ξ3) = X1(ξ1) X2(ξ2) X3(ξ3) At this point, we adopt a certain shorthand notation. First, we show coordinate dependence of a function by replacing ξn with just the integer n and remove the commas. For example, the above reads ψ(123) = X1(1) X2(2) X3(3) Second, if we have a function f1(1) which depends on the variable corresponding to its subscript, then we will shorten it and write just f1, and this implies that it is really f1(ξ1). Then we have ψ(123) = X1X2X3 (3) If we insert (3) into (2) we get 2ψ = H-1{ ∂1[(H/h12)(∂1{X1X2X3})] + cyclic} = H-1 { X2X3{ ∂1[(H/h12)(∂1X1)] + cyclic} (2 +k12)ψ = H-1 { X1X2X3{ ∂1[(H/h12)(∂1X1)] /X1 + cyclic} + k12X1X2X3 or (2 +k12)ψ = H-1 { ∂1[(H/h12)(∂1X1)] /X1 + cyclic}ψ + k12ψ = 0 (4) Now make a new definition w1(123) = (H/h12) // w1 is determined by curvilinear system! (5) Then we have (2 +k12)ψ = H-1 { ∂1[w1(∂1X1)] /X1 + cyclic}ψ + k12ψ = 0 (6) Now, suppose we look for separated equations of this completely general 2nd order form L1X1 = 0 L1X1 = ∂1[p1(∂1X1)] + r1∂1X1 + q1X1 (7) Right here we are going to change a symbol name to make future equations more compatible with the literature on "separation". Stakgold and others like to use p and q for the two functions shown above in (7). We will soon be discarding r, and q will become part of something else, but we now want to change p to f. In this way, our earlier equations below will look more in line with this literature. Thus L1X1 = 0 L1X1 = ∂1[f1(∂1X1)] + r1∂1X1 + q1X1 (7a) When some subject is new to "the student", it helps provide a solid platform for learning the new stuff if different sources use the same notation. Once it is understood, then we can change to any notation we want. It is in the early phases that the symbol name choices seem to help. So this f notation is what you see in the two heavy-weight books on the subject of separation, M&F and M&S. First required assumption We want L1 to appear inside (6). To make this happen, it seems that the only way would be to arrange to have w1(123) = g1(23)f1(1) (8) which them means that g1(23)f1(1) = (H/h12) (8a) Note that this is a restriction on our curvilinear system -- it will rule out some systems. If we find a system that meets this restriction, then both g1(23) and f1 are determined, so we now regard these as given quantities. If condition (8) were met, we would then have (2 +k12)ψ = H-1 { g1(23) ∂1[f1 (∂1X1)] /X1 + cyclic}ψ + k12ψ = 0 (9) Notice that ∂1[f1 (∂1..)] = L1 - r1∂1 + q1 so we then have (2 +k12)ψ = H-1 { g1(23) [(L1 - r1∂1 - q1)X1] /X1 + cyclic}ψ + k12ψ = 0 (10) Some Side Comments We have now achieved our goal of causing the Ln to appear in (6). If for the moment we assume that Xn are some arbitrary functions, and then ψ(123) = X1X2X3 is then not a solution of Helmholtz, and if we think of the L3D = (2 +k12), then we are saying that L3D [X1X2X3] = [ H-1 { g1(23) [(L1 - r1∂1 - q1)X1] /X1 + cyclic} + k12 ] [X1X2X3] If we think of X1X2X3 as spanning a certain subspace of 3D functions, then within this subspace we have found a particular form for the action of the L3D operator in terms of operators L1, L2 and L3. But we are not able to express the operator L3D itself directly in terms of the Ln due to the appearance of the Xn functions inside the RHS operator in front of [X1X2X3]. Now if we further restrict X1X2X3 to the space of functions for which L3D [X1X2X3] = 0 AND for which we have LnXn = 0, AND if we also assume that rn = 0, then the above becomes L3D [X1X2X3] = [ H-1 { g1(23) [(- q1) + cyclic} + k12 ] [X1X2X3] = 0 where now the Xn no longer appear "inside", and acting on this subspace we then have L3D = [ H-1 { g1(23) [(- q1) + cyclic} + k12 ] which is then just a multiple of the unity operator. Our task will then be to arrange to have this multiple be 0, and then we will have L3D = 0 on this solution subspace, and hence Lψ= 0. Showing that the rn must vanish Now suppose we assume that X1 is a solution of L1X1 = 0 and ψ(123) = X1X2X3 is a solution of Helmholtz. Then we have from (10) (2 +k12)ψ = H-1 { g1(23) [(- r1∂1 - q1)X1] /X1 + cyclic}ψ + k12ψ = 0 (11) What we need now is this { g1(23) [(- r1∂1 - q1)X1] /X1 + cyclic} + H(123) k12 = 0 (12) Let's write this out g1(23) [(r1∂1+q1)X1] /X1 + g2(31) [(r2∂2+q2)X2] /X2 + g3(12) [(r3∂3+q3)X3] /X3 = H(123) k12 Now define F1(1; X1) ≡ [(r1∂1+q1)X1] /X1, then we have g1(23) F1(1; X1)) + g2(31) F2(2; X2)) + g3(12) F3(3; X3)) = k12H(123) (13) where now all coordinate dependences are shown. Remember that the gn are already determined by the curvilinear system. Can we now find functions rn, qn, Xn which make this be true for all ξ1,ξ2,ξ3 ? What can we say about the Xn? If we are given rn and qn ( the pn are already determined, recall), then since the Xn must be solutions to LnXn = 0, the Xn are also determined, at least as a linear combination of independent solutions. We require that our equation (13) be true for ANY solutions Xn of LnXn = 0, that is what we mean by our whole separation idea. So let's write Xn = anun+bnvn where an and bn are arbitrary constants and un and vn are any independent solutions of LnXn = 0 . Then our equation (13) becomes g1(23) F1(1; a1u1+b1v1)) + g2(31) F2(2; a2u2+b2v2)) + g3(12) F3(3; a3u3+b3v3)) = k12H(123) where we have six constants on the LHS that are allowed to take any values whatsoever! If you found a solution to the above, if you were to slightly change one of the constants, it would no longer be a solution. So I conclude that there IS no solution of the above equation valid for all values of the 6 constants. Now, suppose we assume that all three rn = 0. Then something magic happens! We then have g1(23) [(r1∂1+q1)X1] /X1 + g2(31) [(r2∂2+q2)X2] /X2 + g3(12) [(r3∂3+q3)X3] /X3 = H(123) k12 g1(23) q1(1) + g2(31) q2(2) + g3(12)q3(3) = H(123) k12 (14) Now we no longer have dependence on the Xn functions, and we ask: Can we find functions qn which make the equation (14) be true? At least we have a chance of the answer being yes we can. Remember: if we can satisfy the equation we keep processing along in this doc, then we have found a separated solution to our PDE. As a reminder, we had to set rn = 0 to even have a chance, and then our assumed form for the separated operators is L1X1 = ∂1[f1(∂1X1)] + q1X1 = 0 We shall now modify this slightly by dividing by p1 L1X1 = (1/f1)∂1[f1(∂1X1)] + (q1/f1)X1 = 0 Let's now change names (q1/f1) → v1 so we then have L1X1 = (1/f1)∂1[f1(∂1X1)] +v1X1 = 0 q1 = f1v1 and (14) becomes g1(23) v1(1) f1(1) + g2(31) v2(2) f2(2) + g3(12)v3(3) f3(3) = H(123) k12 (15) or Σn gn(≠n) fn(n)vn(n) = H(123) k12 _________________________________________________________________________________ Note added: We could if we like combine these last two equations to get Σn gn fnvn(n) = H k12 Σn gn (1/Xn)∂n[fn(∂nXn)] = - H k12 and we could then use (2.2) which will appear below, and which says H/S = gn(≠n)fn(n)/ Mn(≠n) , to get gn = HMn/[S fn] // H/(gnfn) = S/Mn to obtain this processed form of the Helmholtz equation Σn Mn/[S fn Xn] ∂n[fn(∂nXn)] = - k12 This form does lead us to imagine there are two free "separation constants" in the traditional sense: (M1(Φ)/[ S(Φ)f1X1]) ∂1{f1(∂1X1)} = c12 (M2(Φ)/[ S(Φ)f2X2]) ∂2{f2(∂2X2)} = c22 (M3(Φ)/[ S(Φ)f3X3]) ∂3{f3(∂3X3)} = c32 where we have the constraint that c12 + c22 + c32 = -k12 giving then two free separation constants. One has to wonder how these might be related to the two free constants we see in the separated equation we will soon display below (1/fn)∂n[fn(∂nXn)] +[ k12Φn1(n) + k22Φn2(n) + k32Φn3(n)]Xn = 0 We can rewrite a few lines back as (1/H) Σn gn (1/Xn)∂n[fn(∂nXn)] = - k12 So this is another way to write things (1/H) (g1/X1) ∂1{f1(∂1X1)} = c12 (1/H) (g2/X2) ∂2{f2(∂2X2)} = c22 (1/H) (g3/X3) ∂3{f3(∂3X3)} = c32 Question: is this correct? Maybe this conclusion only applies when you have two things in a sum. Like so a(x1) + b(x2) = k As you vary x2 and hold x1 fixed, suppose b(x2) were NOT a constant. Then the LHS would vary, but it cannot vary, so b(x2) must be a constant. a(x1) + b(x2) + c(x3) = k Suppose c(x3) is NOT a constant and hold x1 and x2 fixed. Then the LHS would vary, but again it cannot vary, so c(x3) is a constant. So yes, it does work no matter how many you add. So here we go: (1/f1X1) ∂1{f1(∂1X1)} = (H/g1f1)c12 = - [ k12Φ11(1) + k22Φ12(1) + k32Φ13(1)] (1/f2X2) ∂2{f2(∂2X2)} = (H/g2f2)c22 = - [ k12Φ21(2) + k22Φ22(2) + k32Φ23(2)] (1/f3X3) ∂3{f3(∂3X3)} = (H/g3f3)c32 = - [ k12Φ31(3) + k22Φ32(3) + k32Φ33(3)] We can see that if k2 and k3 are specified, then c12 and c22 and c32 will be determined, and it will turn out that they will be constants, and they will add up to -k12 . We can write as (S/M1)c12 = - [ k12Φ11(1) + k22Φ12(1) + k32Φ13(1)] (S/M2)c22 = - [ k12Φ21(2) + k22Φ22(2) + k32Φ23(2)] (S/M3)c32 = - [ k12Φ31(3) + k22Φ32(3) + k32Φ33(3)] using our usual friend above. Well, let's go ahead with the matrix form - = (Φ/S) where (Φ/S) is a matrix of determinant 1, so I think this means length is preserved so we will have k14 + k24 + k34 = (c12/M1)2 + (c22/M2)2 + (c32/M3)2 This does not seem possible to me because the LHS is a constant and the RHS is not. I guess I am wrong, you would need Φ to have the form of a rotation matrix, ie, Hermitian, and no doubt Φ does not have that property. Forms I have seen such as ellipsoidals certainly are not symmetric hence not Hermitian! If we do invert, we would get - /S = Φ-1 and in our notation below we will have Gpq(≠p) ≡ [ΦT]-1pq => GpqT(≠p) = [Φ]-1pq Gpq(≠p) = cof(Φpq) - /S = GT Mp = Gp1 The ki2 will then be a function of all the cofactors of Φ and will be some mess, stop. Go back c12 = - (M1/S) [ k12Φ11(1) + k22Φ12(1) + k32Φ13(1)] c22 = - (M2/S) [ k12Φ21(2) + k22Φ22(2) + k32Φ23(2)] c32 = - (M3/S) [ k12Φ31(3) + k22Φ32(3) + k32Φ33(3)] All I can say is that in some specific system, once we compute Φ and S and the Mi, we can compute these c12 in terms of the ki2 and they have to come out being constants. I need to try this in some system. It just looks wrong, doesn't it! In the first equation, suppose the three functions are different and none are constants. Well, M1 and S are pretty complicated and could somehow offset this? Well, suppose k2 and k3 are both 0. Then it certainly looks wrong. How can (M1/S) Φ11(1) be a constant? Something must be wrong here, another day. On the other hand, in this case look what happens if we add the three rows. The MΦ sum is just S, so we get c12 + c22 + c32 = -k12 which is correct. ________________________________________________________________________________ Introduction of the Nine Φ functions Now nothing stops us from writing v1 in this rather obscure form: v1 = AΦ11(1) + BΦ12(1) + CΦ13(1) We have just replaced one unknown function with a linear combination of three unknown functions with some unknown coefficients! Furthermore, nothing stops us from selecting A = k12 which is our Helmholtz parameter. So let's do this, and then let's rename B and C to be k22 and k32 where these last two are arbitrary. So we then have v1 = k12Φ11(1) + k22Φ12(1) + k32Φ13(1) Thus, we can write our separated equation as L1X1 = (1/f1)∂1[f1(∂1X1)] +[ k12Φ11(1) + k22Φ12(1) + k32Φ13(1)]X1 = 0 Suppose we then do this same thing for the other two separated equations, so we cause 6 more unknown functions of the Φ type to appear. Since these functions are all arbitrary so far (and have, for example, an arbitrary overall scale), we can select the same three constants k12, k22, k32 to be the linear coefficients. We then have LnXn = (1/fn)∂n[fn(∂nXn)] +vnXn = 0 appearing as LnXn = (1/fn)∂n[fn(∂nXn)] +[ k12Φn1(n) + k22Φn2(n) + k32Φn3(n)]Xn = 0 (16) where again, all 9 of the Φ functions are unknown, and where k22 and k32 are arbitrary constants. Note that each Φ function depends on a single variable corresponding to its first index. We have shown above that Ln must be writable in this manner in order to even have a distant chance of having equation (15) be satisfied. This equation (15) has to be satisfied for us to conclude that X1X2X3 is a solution of L3D[X1X2X3] = 0 so we can say we have found a separated solution! We can now rewrite (15) in our new assumed notation as g1(23) v1(1) f1(1) + g2(31) v2(2) f2(2) + g3(12)v3(3) f3(3) = H(123) k12 g1(23) v1(1) f1(1) + cyclic = H(123) k12 Σn gn(≠n)vn(n)fn(n) = H(123)k12 Σn gn(≠n) [ k12Φn1(n) + k22Φn2(n) + k32Φn3(n)]fn(n) = H(123)k12 (17) where now for example g1(23) is written as g1(≠1). We wonder now if we can find 9 functions Φij such that equation (17) will be true when k22 and k32 are arbitrary constants. Of course k12 being the Helmholtz parameter is also arbitrary. Comparing coefficients of the k2 terms, we would then have to have Σn gn(≠n) Φn1(n) fn(n) = H(123) Σn gn(≠n) Φn2(n) fn(n) = 0 Σn gn(≠n) Φn3(n) fn(n) = 0 (18) At this point we have sort of caught up with M&F. Here is the form they "assume" for the separated equation, and it is identical to my equation (16) above which I repeat here LnXn = (1/fn)∂n[fn(∂nXn)] +[ k12Φn1(n) + k22Φn2(n) + k32Φn3(n)]Xn = 0 (16) Stating the processed Helmholtz equation as a matrix equation Now let's try to cast (18) as a matrix equation. To get it into standard form, we write Σn ΦT1n(n) gn(≠n)fn(n) = H(123) Σn ΦT2n(n) gn(≠n)fn(n) = 0 Σn ΦT3n(n) gn(≠n)fn(n) = 0 (19) or ΦT V = W where V = W= Following tradition, we define to S = det(Φ). Inverting this matrix equation If the above equation is to have a solution for V, we must have S ≠ 0 and we know that V = [ΦT]-1W where [ΦT]-1pq = cof(Φpq)/S = (-1)p+q min(Φpq)/S where we use a standard formula for an inverse matrix. For example, we have S [ΦT]-121 = (-1)2+1 min(Φ21) = – min(Φ21) = - [ Φ12(1)Φ23(3) – Φ13(1)Φ32(3) ] where we need this visual aid to verify this is correct Φ = A minor determinant is formed by crossing out a row and a column. In our example above we crossed out the second row and the first column of Φ. Since the second row is gone, we have nothing that is a function of ξ2 in our matrix element [ΦT]-121 . It will always be the case that [ΦT]-1pq will not be a function of variable ξp so in our previous notation we would use ≠p to indicate this. Introduction of Gpb as notation for cof(Φpq), and then Mp as Gp1. So for another shorthand, lets define Gpq(≠p) ≡ [ΦT]-1pq // = cof(Φpq) = a number, not a matrix Then our solution vector's components (times S) are given by SVp = Gpq(≠p)Wq But Wq = δq,1H(123) so we have SVp = Gpq(≠p) δq,1H(123) = Gp1(≠p) H(123) But we defined V such that Vp = gp(≠p)fp(p). Thus we have gp(≠p)fp(p) = Gp1(≠p) H(123)/S p = 1,2,3 (20) This then is our processed version of equation (19) that we are trying to solve, and if we can solve, then we have found a separated solution! Now let us recall some earlier equations: w1(123) ≡ (H/h12) // w1 is determined by our of curvilinear system! (5) w1(123) = g1(23)f1(1) // a restriction on our curvilinear system (8) Therefore we know that g1(23)p1(1) = g1(23)f1(1) = H(123)/h12 or more generally gp(≠p)fp(p) = H(123)/hp2 p = 1,2,3 (21) So insert this into (20) and we get 1/hp2 = Gp1(≠p)/S = cof(Φp1)/S ≡ Mp(≠p)/S (22) where to save one useless index we define Mp(≠p) as shown. Note: M&F use the word "minor" to mean what we call "cofactor". The terms cofactor or co-factor do not appear in either M&F volume! Was this a standard thing at one point in history? In any event, M&F use this notation Mp which they call a "minor" but which we call a "cofactor". Thus we have 1/hp2 = Mp(≠p)/S (23) and we can then quote (to show we are on track now with M&F) Now let's look back again at (20) above where we had 1/hp2 = gp(≠p)fp(p)/H(123) Combining this with (22) says H(123) Mp(≠p)/S = gp(≠p)fp(p) H(123)/S = gp(≠p)fp(p)/ Mp(≠p) or H(123)/S(123) = g1(≠1)f1(1)/ Mp(≠1) = g2(≠2)f2(2)/ M2(≠2) = g3(≠3)f3(3)/ M3(≠3) (24) Remember that the gifi factors are determined by our curvilinear system, since we said in (21) that gp(≠p)fp(p) = H(123)/hp2 whereas the factors Mp depend on our choice for 6 of the 9 functions of the matrix Φ. Now stare at (24). The first equality says that, as a function of ξ1 , H(123)/S = K(ξ2,ξ3) f1(ξ1). In other words, H(123)/S is proportional to f(1). But it is also proportional to f(2) and f(3) as the next two equalities tell us. We must therefore be able to write H(123)/S(123) = K f1(1) f2(2) f3(3) The constant K cannot depend on ξ1 because then the first equality would not be true! Similarly for ξ2 and ξ3. So K is just a constant. Now look back at our separated equations LnXn = (1/fn)∂n[fn(∂nXn)] +[ k12Φn1(n) + k22Φn2(n) + k32Φn3(n)]Xn = 0 (16) If we multiply fn by 3, this equation is unchanged. Thus, the scale of the fn functions is irrelevant. If we have some scale where H(123)/S = K f1(1) f2(2) f3(3) for some K, then we can just change the scale say of f1 to cancel K. Thus, we loose no generality by setting K = 1 to get H(123)/S(123) = f1(1) f2(2) f3(3) (25) and again we quote M&F Summary of key equations obtained in the previous section We define H = H(123) = h1h2h3. We define S = S(123) = det(Φ). Our first assumption was this: [ from (8a) ] gn(≠n)fn(n) = H/hn2 n = 1,2,3 (2.1) The RHS is forced by our CL system, and we have to assume that it has the functional form shown on the LHS. This may or may not work. If it does not work, our CL system is not separable. This equation determines both fn(n) and gn(≠n) [ ignoring a relative scaling constant] from the CL system. Our next interesting equation was this [ from (20) ] gn(≠n)fn(n) = Mn(≠n) H/S n = 1,2,3 which can be written this way so the LHS does not depend on n, H/S = gn(≠n)fn(n)/ Mn(≠n) n = 1,2,3 (2.2) Combining the above two equations gives this one [ appears as (23) ] 1/hn2 = Mn(≠n)/S (2.3) We note in passing that this says Mnhn2 = S => Mihi2 = Mjhj2 => Mi/Mj = hj2/hi2 (2.3a) which appears in M&S page 7 1.27. The next equation is the Robertson condition [ from (25) ] H/S = f1(1) f2(2) f3(3) (2.4) Since we found the fi from 2.1, this last tells us S. Again in passing, for the last two equations above we can solve each for 1/S and set these equal to get f1(1) f2(2) f3(3)/H = 1/Mnhn2 => Mn f1(1) f2(2) f3(3) = 1/hn2 (2.4a) which is the other equation appearing in M&S page 7 1.27. We can combine (2.4) with the second last to get H/S = f1(1) f2(2) f3(3) = g1(≠1)f1(1)/ M1(≠1) => f2(2) f3(3) = g1(≠1) / M1(≠1) so we then have these results M1(≠1) = g1(≠1)/[ f2(2) f3(3)] M2(≠2) = g2(≠2)/[ f3(3) f1(1)] M3(≠3) = g3(≠3)/[ f1(1) f2(2)] (2.5) We found the gn and the fn from (2.1), so this last result tells us the Mn. So given some CL system, we know from these few equations how to find: fn gn S Mn More work is needed to actually determine the 9 functions of matrix Φ, in theory it can be done with the above information. Review of what has happened to this point: We sought operators Ln of the most general 2nd order form such that LnXn = ∂n[fn(∂nXn)] + rn∂nXn + qnXn = 0 n=1,2,3 In order to have any chance of success, we had to make our first assumption [ (5) with (8) ], gn(≠n)fn(n) = H(123)/hn2 This is a functional-form restriction for the RHS! Some systems won't give a RHS which has the functional form on the LHS. If our system does meet this condition, then, apart from a multiplicative constant, which we shall see does not matter, the fn(n) are determined, and hence also the gn(≠n) are determined. With this assumption, we were able to write the Helmholtz equation in various forms: (2 +k12)ψ = H-1 { g1(23) ∂1[p1 (∂1X1)] /X1 + cyclic}ψ + k12ψ = 0 (9) (2 +k12)ψ = H-1 { g1(23) [(L1 - r1∂1 - q1)X1] /X1 + cyclic}ψ + k12ψ = 0 (10) { g1(23) [(- r1∂1 - q1)X1] /X1 + cyclic} + H(123) k12 = 0 (12) We showed that if Xn are arbitrary solutions of their LnXn = 0, then it was impossible to solve this last equation unless we select all rn = 0. In that case the above became g1(23) q1(1) + g2(31) q2(2) + g3(12)q3(3) = H(123) k12 (14) At this point our candidate Ln were rescaled [p1 → f1 and (q1/p1) → v1 ] and had this self-adjoint form LnXn = (1/fn)∂n[fn(∂nXn)] +vnXn n = 1,2,3 which shows why the scaling of fn does not matter. We do not know the vn, we just have this self-adjoint form. Note that L3D was formally adjoint in its 3D space. We WANT to have self-adjoint Ln operators so we can do SL work with are separated equations so we are not unhappy to have been forced this way. Next we chose to write the functions vn(n) in a very obscure but certainly allowable manner vn = k12Φn1(n) + k22Φn2(n) + k32Φn3(n) where k12 is the Helmholtz parameter, and where we hope to find that k22 and k32 are arbitrary parameters, which we shall then refer to as our "separation constants". We have now suddenly involved a set of 9 as yet unknown functions which form a matrix known as the Stackel Matrix Φ = det(Φ) = S = S(123) This allows us to write our "equations we have to satisfy to get separability" in matrix form (that is to say, the 3D Helmholtz equation with our assumed ODE's for the Xi ) ΦT V = W where V = W= As long as we select our functions so that S ≠ 0, we can invert our matrix equation to get V = [ΦT]-1W where W is a multiple of the unit vector as shown. We then are able to show these two facts: 1/hp2 = Mp(≠p)/S(123) // Fact 1 where Mp(≠p) = Gp1(≠p) = cof(Φp1) = a cofactor of element p in the first column of Φ. The second fact is this H(123)/S(123) = f1(1) f2(2) f3(3) " the Robertson Condition" // Fact 2 I think our goal is now to find a viable Φ matrix which satisfies these two Facts. Recall that the functions fn were determined by our earlier condition gn(≠n)fn(n) = H(123)/hn2 // Fact 0 If we look at M&S page 6, we see Fact 1 and Fact 2 written as above, where H(123) = h1h2h3 = g1/2 where g ≡ det(gij) = H2 and where gii = hi2 as usual. Thus M&S state Fact 1: 1/gii = Mi/S = Gi1/S => gii = S/ Gi1 Fact 2: g1/2/S = f1(1) f2(2) f3(3) p 6 (1.25) Here is how you are supposed to proceed with a given candidate curvilinear system: Determine the three fn(ξn) from Fact 0. Determine S(123) from Fact 2. Fact 1 then tells you the three Mp = cof(Φp1) which are the cofactors of the three elements of the leftmost column of the matrix. We have for example M1 = cof(Φ11) = Φ22(2)Φ33(3) - Φ23(2)Φ32(3) We seem then to have only four conditions on the nine Φ functions, this being the fourth S(123) = Φ11 M1 + Φ21 M2 + Φ31 M3 Thus, there will be very many possible matrices that work. But, there are serious "functional form" restrictions, so we don't really have 4 conditions on 9 functions in a general sense. When all is said and done, it turns out that an important 3D orthogonal curvilinear system that works is the ellipsoidal system involving as it does confocal quadric surfaces (ellipsoids, hyperboloids of one and two sheets). If you don't have confocal quadric surfaces, you don't get separation. Any system that is a degenerate limit of the ellipsoidals will be separable as well. This then is why all 11 classical systems are separable. All along here we are talking about Euclidean coordinate systems, by the way. The reason you need to know a workable Stackel matrix is of course that its 9 member functions then tell you your separation equations! Recall, LnXn = (1/fn)∂n[fn(∂nXn)] +vnXn n = 1,2,3 vn = k12Φn1(n) + k22Φn2(n) + k32Φn3(n) Question: It seems that different choices for a viable Stackel matrix might give you different separated equations! Is that possible? I don't see either M&F or M&S addressing this question. I cannot quickly find comments on this. Let's write out the equations in more detail: M1(23) = Φ22(2)Φ33(3) - Φ23(2)Φ32(3) M2(31) = Φ33(3)Φ11(1) - Φ31(3)Φ13(1) M3(12) = Φ11(1)Φ22(2) - Φ12(1)Φ21(2) S(123) = Φ11(1) M1(23) + Φ21(2) M2(31) + Φ31(3) M3(12 = Φ11(1)[Φ22(2)Φ33(3) - Φ23(2)Φ32(3)] + Φ21(2)[ Φ33(3)Φ11(1) - Φ31(3)Φ13(1)] + Φ31(3)[ Φ11(1)Φ22(2) - Φ12(1)Φ21(2)] Ellipsoidal Coordinates Let's examine the ellipsoidal system as an example. M&F claim that Here is the solution M&F give Notice that Φn1(1) = 1 Φn2(n) = (ξn2-a2)-1 Φn3(n) = (ξn2-b2)-1(a2-b2)-1 so only three "functional forms" appear, not 9. M1 = (ξ22-a2)-1(ξ32-b2)-1(a2-b2)-1 - (ξ32-a2)-1(ξ22-b2)-1(a2-b2)-1 = (a2-b2)-1 { (ξ22-a2)-1(ξ32-b2)-1 - (ξ32-a2)-1(ξ22-b2)-1 } = (a2-b2)-1 {(ξ32-a2)(ξ22-b2) – (ξ22-a2) (ξ32-b2) } (ξ22-a2)-1(ξ32-b2)-1 (ξ32-a2)-1(ξ22-b2)-1 } But {(ξ32-a2)(ξ22-b2) – (ξ22-a2) (ξ32-b2) } = ξ22(b2-a2) - ξ32(b2-a2) = (a2-b2)(ξ32-ξ2)2 so M1 = (ξ32-ξ2)2 (ξ22-a2)-1(ξ32-b2)-1 (ξ32-a2)-1(ξ22-b2)-1 } which confirms the M1 quoted above (note the leading minus sign). Somehow I think you could argue that the three rows have to have the same three functions due to the usual very heavy ellipsoidal cyclic symmetry everywhere. The Stackel matrix must have some family of symmetry transformations that leave the 4 conditions unaltered. M1(23) = Φ22(2)Φ33(3) - Φ23(2)Φ32(3) M2(31) = Φ33(3)Φ11(1) - Φ31(3)Φ13(1) M3(12) = Φ11(1)Φ22(2) - Φ12(1)Φ21(2) S(123) = Φ11(1) M1(23) + Φ21(2) M2(31) + Φ31(3) M3(12) But in this case, I think symmetry forces your hand pretty much. You cannot double every entry. In order to get the right M1,2,3 you cannot play much with the last two columns! You might swap constants a and b or negate the two last columns, but it looks pretty constrained to me. So maybe in simpler systems it is easier to see things. [ I later found rules for S matrix transformations. ] What do we know about extruded conformal 3D systems (cylindricals)? If 3 is the z coordinate, we know that h1 = h2 and h3= 1 (still Cartesian). This is because in all our normal such systems, the 2D pattern is obtained by a conformal map of Cartesians. This applies to regular cylindricals only if you do it with r = eξ. We also know that h1 = h1(12). So Fact 0 tells us g1(≠1)f1(1) = H(123)/h12 = 1 = g2(≠2)f2(2) // Fact 0 so we know then that f1 = f2 = 1 and g1 = g2 = 1. Then g3(≠3)f3(3) = H(123)/h32 = h12/h3 = h12 = h12(12) so we conclude that g3(≠3) = h12(12) and f3 = 1. Fact 2 then says S(123) = H(123)/(f1f2f3) = h12 So we arrive at some pretty simple results: f1 = f2 = f3 = 1 g1 = g2 = 1 g3= h12(12) S(123) = h12(12) does not depend on z. Now don't let's forget Fact 1 1/hp2 = Mp(≠p)/S(123) // Fact 1 which says 1 = M3/ h12(12) M3 = h12(12) 1/h12 = M1/ h12(12) M1 = 1 1/h22 = M2/ h12(12) M2 = 1 S = h12(12) This suggests to me a matrix of this form Then M1 = 1 and M2 = 1 and M3 = - β(2) - α(1) and S = M3 so we end up with hopefully writing - β(2) - α(1) = h12(12). Important fact: this is only going to work in some cylinder systems!! If you cannot write h12(12) as the sum of two functions in this manner, you cannot get separation! An example is M&S p 86 for tangent-cylinder coordinates where h12(12) = 1/(ξ12- ξ22)2. I think only the classical cylinder systems are separable! That is why they are classical! Now look at M&S p 17 on elliptical cylinder η = ξ1. This has h12 = a2(ch2η- cos2ψ) so we get - β(ψ) - α(η) = a2ch2η- a2cos2ψ β(ψ) = a2cos2ψ α(η) = -a2ch2η and I then get and this agrees exactly with M&S page 17. What do we know about extruded other 3D systems (cylindricals)? Suppose we had a cylindrical system in which h1(12) ≠ h2(12) and h3 = 1. Then what happens? g1(≠1)f1(1) = H(123)/h12 = h2(12)/h1(12) // Fact 0 Well the only case I know of is regular cylindrical in which case r,ψ,z we gave h1(12) = hr = 1 h2(12) = hψ = r = h2(1) so we really then have g1(≠1)f1(1) = h2(1)/1 => f1(1) = h2(1) = r and everything gets pretty trivial as on page 12 M&S. This does have the general form of p 7 1.28, and so do my simpler h1 = h2 cases. Symmetry operations on matrix Φ revisited. We know that negating a column OR swapping two columns negates the determinant. Thus, we have a few simple symmetry operations we can perform on S. On is to swap the second two columns and negate one of them. The second is to negate both columns. We cannot swap two rows because that shuffles the Mi. If two of the M's are the same you might do something, but I don't think in the general case. So I guess symmetry is limited to the first thing I said in the general case. Comments on the "separation constants". In this analysis, they don't appear in the pleasing obvious way that they appear, say, in the separation of the H atom in QM. Our SE there becomes something like this f(r) + g(θ,φ) = 0 and then we are allowed to say [ functions here represent things like LrR(r) / R(r) ] f(r) = -k g(θ,φ) = k and we refer to k as a "separation constant". Quantization from the θ,φ system later causes k = n(n+1). Then within that g(θ,φ) equation we find hk(θ) + w(φ) = 0 so we are allowed to say w(φ) = m hk(θ) = -m and that m is our second separation constant, and we end up with a solution of the form ak(r) bk,m(θ) cm(φ) and we get a characteristic "linkage" of the separation constants. This is the "traditional method" of showing that there ARE two separation constants, and of finding out "how they quantize" in the SL problems of the individual coordinates. In contrast, in our method above we just ASSUME there are two separation constants when we write our candidate separated ODE in this manner involving 9 functions of a matrix of functions Φ, LnXn = (1/fn)∂n[fn(∂nXn)] +[ k12Φn1(n) + k22Φn2(n) + k32Φn3(n)]Xn = 0 (16) With this assumption, and few other assumptions made along the path, the Helmholtz equation becomes a matrix equation ΦT V = W where V = W= When we simply invert this equation to get V = [ΦT]-1W we find that gp(≠p)fp(p) = Mp(≠p) H(123)/S p = 1,2,3 (20) This and the earlier assumptions lead us to this set of equations which must be true gp(≠p)fp(p) = H(123)/hp2 Mp(≠p) = S(123)/hp2 S(123) = H(123)/[ f1(1) f2(2) f3(3)] If the system is viable, the first equation tells us the three fp(p) [ and in passing the gp(≠p) ], the second equation gives us the three cofactors Mp, and the third gives us S = det(Φ). We then do trial and error to find a viable matrix Φ that has these cofactors and this determinant. How to find the matrix Φ Let's assume now that Φ has this form (h and g are now overloaded) Φ = 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) Let's focus on the first equation: e(2)i(3)-f(2)h(3) = M1(23) = [S(123)/H(123)] g1(≠1)f1(1) = [ f1(1) f2(2) f3(3)]-1 g1(≠1)f1(1) = g1(23)/ [f2(2) f3(3)] At least we have the RHS in a form that shows it depends only on 2 and 3. If at this point we discovered that g1(23) = , we would conclude that things were non-separable. So just getting this far does not guarantee success. What we need to assume is that g1(23) can be written as a sum of at most two terms each of which factors [ try putting 3 terms and see what happens] . So write g1(23) = j1(2)k1(3) + m1(2)n1(3) // we assume this can be done If this is the case, then we have e(2)i(3)-f(2)h(3) = [j1(2)k1(3) + m1(2)n1(3)]/ [f2(2) f3(3)] = (j1(2)/f2(2)) (k1(3)/f3(3)) + (m1(2)/f2(2)) (n1(3)/f3(3)) Later on where I worked on this more, I realized you can think of this last equation as e2i3 - f2h3 = q2r3 – s2 t3 If you are given the 4 functions of the RHS, it at first seems there are very many choices for the 4 functions of the LHS. In fact, if you think of each side as the det of a 2x2 matrix, det = det then the left matrix could be the right matrix times any unimodular third matrix you choose (right or left multiplied). But this turns out to be wrong due to the functional form requirements. In fact, I think the possibilities are very limited. For example, we can write det = det => e2i3 - f2h3 = (αq2)(r3/α) – (βs2)( t3/β) where we now have to arbitrary scaling constants which obviously do not change the 2x2 determinant. So I think this is the most general solution you can have to our problem e2i3 - f2h3 = q2r3 – s2 t3 Solution #1 is this: e2i3 = (αq2)(r3/α) and f2h3 = (βs2)(t3/β) Solution #2 is this: - f2h3 = (αq2)(r3/α) and -e2i3 = (βs2)(t3/β) So we can write out these solutions this way solution #1: e2 = αq2 i3 = r3/α f2 = βs2 h3 = t3/β solution #2: -f2 = αq2 i3 = t3/β -e2 = βs2 h3 = r3/α In each solution we are free to select α,β as any constants, and these solutions really are different. Let us now summarize what we have just learned: If we assume these labels for the functions of Φ Φ = we have to select the rightmost 6 functions so they satisfy these three equations 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) where, from (2.5) above, we know that M1(≠1) = g1(≠1)/[ f2(2) f3(3)] M2(≠2) = g2(≠2)/[ f3(3) f1(1)] M3(≠3) = g3(≠3)/[ f1(1) f2(2)] (2.5) So our problem then is to solve e(2)i(3)-f(2)h(3) = + g1(≠1)/[ f2(2) f3(3)] b(1)i(3)-c(1)h(3) = – g2(≠2)/[ f3(3) f1(1)] b(1)f(2)-c(1)e(2) = + g3(≠3)/[ f1(1) f2(2)] In order to even stand a chance, we must be able to write the gn in this manner g1(23)/(f2f3) = u12(2)u13(3) – v12(2)v13(3) g2(31)/(f3f1) = u23(3)u21(1) – v23(3)v21(1) g3(12)/(f1f2) = u31(1)u32(2) – v31(1)v32(2) which basically says we can write gn(≠n) as a sum of at most two factored terms. Assuming we can do this (if we cannot, they we cannot separate!), we then have e(2)i(3)-f(2)h(3) = + u12(2)u13(3) – v12(2)v13(3) b(1)i(3)-c(1)h(3) = – u23(3)u21(1) + v23(3)v21(1) b(1)f(2)-c(1)e(2) = + u31(1)u32(2) – v31(1)v32(2) At this point it is probably good to negate the LHS of the middle equation, and also to reverse the order in each of its products. Now we have the three RHS's all looking the same e(2) i(3)-f(2)h(3) = + u12(2)u13(3) – v12(2)v13(3) // = q2r3 – s2 t3 of previous analysis h(3)c(1)- i(3)b(1) = + u23(3)u21(1) – v23(3)v21(1) b(1)f(2)-c(1)e(2) = + u31(1)u32(2) – v31(1)v32(2) We can now state the most general solutions of the first equation: solution #1: e2 = α1 u12(2) i3 = u13(3)/α1 f2 = β1 v12(2) h3 = v13(3)/β1 solution #2: -f2 = α1 u12(2) i3 = v13(3)/β1 -e2 = β1 v12(2) h3 = u13(3)/α1 where we have now subscripted the general constants α and β, and installed our more generalized function names. We can similarly write the possible solutions for second equation. To do this, we make these changes: e2→h3, i3→ c1, f2→ i3, h3→ b1 then cyclic all the RHS's solution #1: h3 = α2 u23(3) c1 = u21(1)/α2 i3 = β1 v23(3) b1 = v31(1)/β2 solution #2: -i3= α2 u23(3) c1 = v21(1)/β2 -h3 = β2 v23(3) b1 = u21(1)/α2 When you make the substitutions with the arrows, it is good to turn each one red after you have done it, then you won't get confused, then make them black again when done. Now, rather than write the general solutions for the third equation, we will just insert the solutions we have already found and "see what happens". Let's use our "solution #1" for both the first rows. Then the third row becomes b(1)f(2)-c(1)e(2) = + u33(1)u32(2) – v31(1)v32(2) [v31(1)/β2][ β1 v12(2)] - [u21(1)/α2][ α1 u12(2)] = + u31(1)u32(2) – v31(1)v32(2) This is one of four choices we could have made for the LHS. This choice requires that [v31(1)/β2][ β1 v12(2)] = – v31(1)v32(2) => (β1/β2) v12(2) = –v32(2) [u21(1)/α2][ α1 u12(2)] = + u31(1)u32(2) I think we have now reached the end of the road. In general, u31(1) could be completely unrelated to all other functions with a 1 argument in its functional form. So no matter which of the four choices of solutions to the first equations we take, we end up with an equation involving u31(1). It might be that this cannot be expressed as constant multiple of the other argument-1 function appearing in the second equation, and then there will be no solution. So I was hoping to show that a general solution was always possible, but I see that is not the case if we think of completely general gn(≠n) functional forms as sums of two factored terms. The upshot is this: we will solve the problem in ellipsoidals using trial and error, and then we will know that a solution must exist in all systems obtained from ellipsoidals, which includes all the classical 11 systems. These are the only systems in which Helmholtz is separable! In the ellipsoidal case, we will find that we don't need the most general form of two terms for the gi as used above, one term will do, which means we can set all the v's to 0 in the above. Finding the Stackel matrix for the ellipsoidal coordinate system The need to use F1,F2, F3 for our Stackel analysis Before continuing, let's quote values from the ellipsoidal system, just to "keep us honest". Here are some facts from that system as presented in MF: f12 = (ξ12-a2)( ξ12-b2) = + + = + G1 = (ξ22- ξ32) > 0 f22 = (ξ22-a2)( ξ22-b2) = - + = - G2 = (ξ32- ξ12) < 0 f32 = (ξ32-a2)( ξ32-b2) = - - = + G3 = (ξ12- ξ22) > 0 Q12 = h12 = G3(-G2)/f12 Q22 = h22 = G3G1/(-f22) Q33 = h32 = (-G2)G1/f32 h1 = h2 = h3 = where all radical arguments are positive. M&F like these forms because they are regular and cyclic. The problem however is that f2 is imaginary and this causes problems with the Stackel matrix analysis which basically thinks of things as real. So let us make these replacements F12 = (ξ12-a2)( ξ12-b2) F22 = - (ξ22-a2)( ξ22-b2) = -f22 F32 = (ξ32-a2)( ξ32-b2) so that our three Fn are now all real. We continue to let G2 be negative so the G's maintain their cyclic form. Looking at the MF data, we can write h1 = = /F1 h12 = - G2G3/F12 h2 = = /F2 h22 = + G3G1/F22 h3 = = /F3 h32 = - G1G2/F32 Things are NOT cyclic now with the hi so be careful. For the hi2 things are cyclic except we have to change the sign of on the 2 equation. For H we get H = /F1 * /F2 * /F3 = G1(-G2)G3/[F1F2F3] which is real and positive, the way we like it. The equations for the Mn cofactors We now turn to (2.1) which says g1(≠1)F1(1) = H/h12 = G1(-G2)G3/[F1F2F3] / [- G2G3/F12] = G1F1/[F2F3] Keeping the 2-equation sign in mind, we get (the gi are all real) g1(≠1) = + G1/[F2F3] g2(≠2) = – G2/[F3F1] g3(≠3) = + G3/[F1F2] Now (2.5) tells us that M1(≠1) = g1(≠1)/[ F2(2) F3(3)] M2(≠2) = g2(≠2)/[ F3(3) F1(1)] M3(≠3) = g3(≠3)/[ F1(1) F2(2)] from which we conclude that ( these are all real and positive ) M1 = + G1/(F22F32) = - (ξ22- ξ32)/[f22f32] M2 = – G2/(F32F12) = - (ξ32- ξ12)/[f32f12] M3 = + G3/(F12F22) = - (ξ12- ξ22)/[f12f22] // cyclic when using f's Trial and Error Solution for the last 2 columns of the Stackel matrix. Now recall from above that (checked) 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) So we end up with these 3 equations on our 6 members of the Φ matrix e(2)i(3)-f(2)h(3) = G1/(F22F32) = (ξ22- ξ32)/(F22F32) b(1)i(3)-c(1)h(3) = G2/(F32F12) = (ξ32- ξ12)/(F32F12) b(1)f(2)-c(1)e(2) = G3/(F12F22) = (ξ12- ξ22)/(F12F22) where we no longer have a minus sign in the middle equation. Now come the Big Questions: (1) To what extent do these three equations nail down our 6 functions? (2) Can we even find 6 functions that solve these 3 equations? As we shall learn later, the equations do not completely nail down the 6 functions, despite the rather tight functional form requirements. So later we shall deal indirectly with question (1). Now try (2). Let's try to find a solution by trial and error. First, write the three equations like this: e(2)i(3)-f(2)h(3) = (ξ22/F22)(1/F32) – (1/F22) (ξ32/F32) b(1)i(3)-c(1)h(3) = (ξ32/F32)(1/F12) – (1/F32) (ξ12/F12) b(1)f(2)-c(1)e(2) = (ξ12/F12)(1/F22) – (1/F12) (ξ22/F22) where the RHS is all cyclic. Now define the obvious factors and compress this to e2i3 - f2h3 = q2r3 – r2 q3 qi ≡ (ξi2/Fi2) ri = (1/Fi2) b1i3 - c1h3 = q3r1 – r3 q1 b1f2 - c1e2 = q1r2 – r1 q2 Now try this solution of the first equation e2 = sq2 => i3 = sr3 where s is a sign, TBD f2 = t r2 => h3 = t q3 t is another sign TBD Now insert this solution into the second equation to get b1 sr3 - c1 t q3 = q3r1 – r3 q1 => sb1 = -q1 and - c1 t = r1 => b1 = -sq1 c1 = -tr1 Now insert all these solutions into the third equation to get LHS = -sq1 t r2 + tr1 sq2 = -st(q1r2 – r1 q2) This matches the RHS's if we select phases so that st=-1. So arbitrarily I will take s=1 and t=-1 so we then have e2 = q2 i3 = r3 f2 = - r2 h3 = - q3 b1 = -q1 c1 = r1 with qi ≡ (ξi2/Fi2) ri = (1/Fi2) These then are our proposed 6 elements of the matrix Φ. Finding the 1st column of the Stackel matrix. But now we have to find three first-column entries such that a(1)M1 + d(2)M2 + g(3)M3 = S From 2.4 we know that S = H/[F1F2F3] = {G1(-G2)G3/[F1F2F3]/ [F1F2F3]} = G1(-G2)G3/[F1F2F3]2 Recall again that M1 = + G1/(F22F32) M2 = – G2/(F32F12) M3 = + G3/(F12F22) So this says a(1)M1 + d(2)M2 + g(3)M3 = S a(1) G1/(F22F32) – d(2) G2/(F32F12) + g(3) G3/(F12F22) = G1(-G2)G3/[F1F2F3]2 a(1) G1F12 – d(2) G2F22 + g(3) G3F32 = G1(-G2)G3 It is not obvious that there even is a solution. Let's expand everything, and I install the fi functions at this point for calculations: a(1) G1f12 + d(2) G2f22 + g(3) G3f32 = - G1G2G3 a(1) (ξ22- ξ32) (ξ12-a2)( ξ12-b2) + d(2) (ξ32- ξ12) (ξ22-a2)( ξ22-b2) + g(3) (ξ12- ξ22) (ξ32-a2)( ξ32-b2) = - (ξ22- ξ32) (ξ32- ξ12) (ξ12- ξ22) Let's now define A(1) = a(1) (ξ12-a2)( ξ12-b2) D(2) = d(2) (ξ22-a2)( ξ22-b2) G(3) = g(3) (ξ32-a2)( ξ32-b2) Then our equation reads A(ξ1) (ξ22- ξ32) + D(ξ2) (ξ32- ξ12) + G(ξ3) (ξ12- ξ22) = - (ξ22- ξ32) (ξ32- ξ12) (ξ12- ξ22) The RHS is this - (ξ22- ξ32) (ξ32- ξ12) (ξ12- ξ22) = - (ξ22- ξ32)[ ξ32 ξ12 - ξ14 - ξ22 ξ32 + ξ12 ξ22] = - [ ξ32 ξ12 ξ22 - ξ14 ξ22 - ξ24 ξ32 + ξ12 ξ24] + [ ξ34 ξ12 - ξ14 ξ32 - ξ22 ξ34 + ξ12 ξ22 ξ32] = -ξ32 ξ12 ξ22 + ξ14 ξ22 + ξ24 ξ32 - ξ12 ξ24 + ξ34 ξ12 - ξ14 ξ32 - ξ22 ξ34 + ξ12 ξ22 ξ32 = + ξ14 ξ22 + ξ24 ξ32 - ξ12 ξ24 + ξ34 ξ12 - ξ14 ξ32 - ξ22 ξ34 = ξ14 (ξ22 - ξ32) + ξ24(ξ32- ξ12) + ξ34(ξ12 - ξ22) // note is cyclic We then have A(ξ1) (ξ22- ξ32) + D(ξ2) (ξ32- ξ12) + G(ξ3) (ξ12- ξ22) = ξ14 (ξ22 - ξ32) + ξ24(ξ32- ξ12) + ξ34(ξ12 - ξ22) and we can luckily read off our solution A(ξ1) = ξ14 D(ξ2) = ξ24 G(ξ3) = ξ34 and this tells us our last three Φ matrix elements a(1) = ξ14/ [(ξ12-a2)( ξ12-b2)] = ξ14/f12 d(2) = ξ24/ [(ξ22-a2)( ξ22-b2)] = ξ24/f22 g(3) = ξ34/ [(ξ32-a2)( ξ32-b2)] = ξ34/f32 Here then is our Stackel matrix Φ = ξ14/f12 - (ξ12/f12) (1/f12) ξ24/f22 - (ξ22/f22) (1/f22) ξ24/f32 - (ξ32/f32) (1/f32) ξ14/(ξ12-a2)( ξ12-b2) - (ξ12/(ξ12-a2)( ξ12-b2)) (1/(ξ12-a2)( ξ12-b2)) ξ24/(ξ22-a2)( ξ22-b2) - (ξ22/(ξ22-a2)( ξ22-b2)) (1/(ξ22-a2)( ξ22-b2)) ξ24/(ξ32-a2)( ξ32-b2) - (ξ32/(ξ32-a2)( ξ32-b2)) (1/(ξ32-a2)( ξ32-b2)) which we can compare to the official M&F matrix How can these be so different? Let's compute the cofactor of MY top left entry M1 = - (ξ22/f22) (1/f32) - (1/f22) (-1)(ξ32/f32) = (ξ32-ξ22)/(f22f32) = (ξ32-ξ22)/ [(ξ22-a2)( ξ22-b2) (ξ32-a2)( ξ32-b2)] which agrees with my earlier computation of this same quantity. Now do the same for THEIR top left entry (ξ22-a2)-1 (ξ32-b2)-1(a2-b2)-1 - (ξ32-a2)-1 (ξ22-b2)-1 (a2-b2)-1 = -(a2-b2)-1{ (ξ22-b2)-1(ξ32-a2)-1 - (ξ22-a2)-1 (ξ32-b2)-1 } = -(a2-b2)-1 {(ξ22-a2) ( ξ32-b2) - ( ξ22-b2) (ξ32-a2)} / [(ξ22-a2)( ξ22-b2) (ξ32-a2)( ξ32-b2)] = -(a2-b2)-1 {(ξ22-a2) ( ξ32-b2) - ( ξ22-b2) (ξ32-a2)} / [(ξ22-a2)( ξ22-b2) (ξ32-a2)( ξ32-b2)] = -(a2-b2)-1 {(ξ22-ξ32) (a2-b2)} / [(ξ22-a2)( ξ22-b2) (ξ32-a2)( ξ32-b2)] = (ξ32-ξ22) / [(ξ22-a2)( ξ22-b2) (ξ32-a2)( ξ32-b2)] We agree! So I think both of our matrices have the same cofactors Mi . We are still alive! The alternate and simpler solution for the 1st column Before continuing, we point out that there is in fact an alternate and in fact much simpler solution for the first column! Go back to where we were a(1) (ξ22- ξ32) (ξ12-a2)( ξ12-b2) + d(2) (ξ32- ξ12) (ξ22-a2)( ξ22-b2) + g(3) (ξ12- ξ22) (ξ32-a2)( ξ32-b2) = - (ξ22- ξ32) (ξ32- ξ12) (ξ12- ξ22) My claim is that the simpler solution is a(1) = d(2) = g(3) = 1. If this is indeed a solution, then this must be true (ξ22- ξ32) (ξ12-a2)( ξ12-b2) + (ξ32- ξ12) (ξ22-a2)( ξ22-b2) + (ξ12- ξ22) (ξ32-a2)( ξ32-b2) = - (ξ22- ξ32) (ξ32- ξ12) (ξ12- ξ22) or (ξ22- ξ32) (ξ12-a2)( ξ12-b2)+ cyclic = - (ξ22- ξ32) (ξ32- ξ12) (ξ12- ξ22) This seems rather preposterous because the LHS appears to be a function of a and b. But let's examine that appearance by looking at the coefficient of a2 of the LHS coeff(a2) = - (ξ22- ξ32) ( ξ12-b2) + cyclic = - [ ξ12(ξ22- ξ32) + cyclic] + b2[(ξ22- ξ32) + cyclic ] But both square brackets vanish due to the following two trivial theorems: f(1)(f(2) - f(3)) +cyclic = 0 // makes first [] = 0 (f(2) - f(3)) +cyclic = 0 // makes second [] = 0 Therefore, since we know the coefficient of a2 (and by similar argument that of b2) on the LHS above is 0, we can select any values we want for a and b to do the evaluation. So take a = ξ12 and b = ξ22 , then the LHS has only the third term which is third term of LHS = (ξ12- ξ22) (ξ32-a2)( ξ32-b2) = (ξ12- ξ22) (ξ32- ξ12)( ξ32- ξ22) = ( ξ32- ξ22) (ξ32- ξ12) (ξ12- ξ22) = - (ξ22- ξ32) (ξ32- ξ12) (ξ12- ξ22) = RHS Let this be a reminder that "functional form" equations can have multiple solution, some of which might be simpler than others! Therefore, we can take 1,1,1 for our first Stackel column in place of what we got above. Our derived Stackel matrix is then Φ = 1 - (ξ12/f12) (1/f12) 1 - (ξ22/f22) (1/f22) 1 - (ξ32/f32) (1/f32) 1 - (ξ12/(ξ12-a2)( ξ12-b2)) (1/(ξ12-a2)( ξ12-b2)) 1 - (ξ22/(ξ22-a2)( ξ22-b2)) (1/(ξ22-a2)( ξ22-b2)) 1 - (ξ32/(ξ32-a2)( ξ32-b2)) (1/(ξ32-a2)( ξ32-b2)) which we again compare to the "official" matrix given in MF: at least now the first column agrees, though the other two do not. What are the "symmetry" operations for a Stackel matrix? What transformation on a matrix preserves the cofactors of the first column of elements, and also preserves the determinant! I flailed away before getting to Plan G !! I will keep the earlier "plans" here, but reader should skip to Plan G. Plan A: Here is an idea. Suppose you take a matrix of this shape Q = where the upper right matrix is unitary. Then we would expect that Q-1SQ would not alter M1 which is the det of the upper right of S. But it would surely alter the two other minors. Plan B: Can you make a 3x3 unitary matrix such that the three right-side 2x2 matrices are unitary? I tried, I think not. Plan C: Make a 3x3 matrix with the three right-side 2x2 matrices all unimodular. Suppose we select the 6 numbers such that the three minors here are 1. that is to say ad-bc = 1 af-be =1 cf-de = 1 This is 3 conditions on 6 numbers, should not be too hard to solve. Then consider Plan D: Start with a matrix with unimodular upper right piece (note detA3 = 1) A3 ≡ ad-bc = 1 Apply this to Stackel matrix S like so R1 = A3SA3-1 The resulting matrix R1 will have the same M1 as S, but M2 and M3 will be damaged. Plan E: Consider e2i3 - f2h3 = q2r3 – r2 q3 You are given the RHS. What is the most general solution set of functions on the left? Think of it this way det = det It seems clear that you can have = U V where U and V are any unimodular matrices, except you have to mind "the functional form" U = is too general = detU=1 so γ3 = 1/α2 So all you really can do is scale up one row and scale down its partner row. But even this doesn't work if α2 is a function of ξ2, so α2 can only be a constant. Plan F. Consider again ξ14/f12 - (ξ12/f12) (1/f12) ξ24/f22 - (ξ22/f22) (1/f22) ξ24/f32 - (ξ32/f32) (1/f32) ξ14/(ξ12-a2)( ξ12-b2) - (ξ12/(ξ12-a2)( ξ12-b2)) (1/(ξ12-a2)( ξ12-b2)) ξ24/(ξ22-a2)( ξ22-b2) - (ξ22/(ξ22-a2)( ξ22-b2)) (1/(ξ22-a2)( ξ22-b2)) ξ24/(ξ32-a2)( ξ32-b2) - (ξ32/(ξ32-a2)( ξ32-b2)) (1/(ξ32-a2)( ξ32-b2)) and we can compare this with the official matrix which is I can make his look a little like mine as follows S = (ξ12-a2)( ξ12-b2)/f12 ( ξ12-b2)/f12 (a2-b2)-1( ξ12-a2)/f12 me = ξ14/f12 - (ξ12/f12) (1/f12) Plan G: The symmetry transformations Theorem: If you add a multiple of one of the last two columns to the other, then (1) none of the minors of the elements of the first column change; (2) the overall determinant does not change. Item (2) is a standard matrix theorem. Item (1) is easy to demonstrate → M1 = (b+kc)f - (e+kf)c = bf - ec QED → M1 = b(f+ke)-e(c+kb) = bf - ec QED Seeing is believing! So this is something you can do to a Stackel matrix to change its form, and you can do it multiple times. You can of course also add a multiple of either of the last two columns to the first without changing either of these things! (1) is obvious since you haven't changed the last two columns, and (2) is the same matrix theorem. So we generalize the theorem Theorem 1: If you add a constant multiple of one of the last two columns to any other column, then (1) none of the minors of the elements of the first column change; (2) the overall determinant does not change. What happens if k instead of being a constant is a function of the variable of its row? Well, then in each row k would be different, since k(1) ≠k(2), and then item (2) fails. Item (1) also fails. Theorem 2: You can change the signs of the last two columns without affecting the determinant or the minors of the first column. (1) → M1 = (-b)(-f) - (-c)(-e) = bf-ce QED (2) We know the negating one column negates the determinant, so doing two column negations leaves it the same. Theorem 3 ? → M1 = (b+ka)(f+kd) - (c+ka)(e+kd) = bf-ce + { bkd+kaf+k2ad - kae-ckd- k2ad } = bf-ce + { kbd+kaf - kae-kcd } = NOGO So let's try to transform my form into their form. We only need to deal with the first row. Here are our starting positions: they = 1 ( ξ12-b2)/f12 (a2-b2)-1( ξ12-a2)/f12 me = 1 - (ξ12/f12) (1/f12) Let's start by negating my last two columns: me1 = 1 (ξ12/f12) -(1/f12) Next, let's add a multiple b2 of the third column to the second column me2 = 1 (ξ12-b2)/f12 -(1/f12) This gets the middle columns to agree. Let's try another column operation on me2 : -(1/f12) → -(1/f12) + k (ξ12-b2)/f12 = (-1 + kξ12-kb2)/f12 =? (a2-b2)-1( ξ12-a2)/f12 I would then need: (-1 + kξ12-kb2) = (a2-b2)-1( ξ12-a2) This requires that k = (a2-b2)-1 to make the ξ12 match, then LHS is (-1 + (a2-b2)-1ξ12-(a2-b2)-1b2) = (-(a2-b2) + ξ12-b2) (a2-b2)-1 = (-a2+b2 + ξ12-b2) (a2-b2)-1 = (ξ12-a2) (a2-b2)-1 and it works! So this is then where we stand: ( I now show both possible first columns) they = 1 ( ξ12-b2)/f12 (a2-b2)-1( ξ12-a2)/f12 me3 = 1 or ξ14/f12 (ξ12-b2)/f12 (a2-b2)-1( ξ12-a2)/f12 So we are done, we have derived the official Stackel matrix for ellipsoidal coordinates. Verify that my two alternate 1st column forms have the same determinant It certainly seems unlikely my two matrices have the same determinant if I use the radically different first column values. . But let's just check : (I am still using M to mean cofactor Sme31 = M1 + M2 + M3 Sme32 = ξ14/f12 M1 + ξ24/f22 M2 + ξ34/f32 M3 If these are the same, then this must be true (ξ14/f12 - 1)M1 + (ξ24/f22 - 1)M2 + (ξ34/f32 - 1)M3 = 0 (ξ14/f12 - 1)(-M1) + (ξ24/f22 - 1)(-M2) + (ξ34/f32 - 1)(-M3) = ? 0 But I know from earlier calculation that M1 = - (ξ22- ξ32)/[f22f32] M2 = - (ξ32- ξ12)/[f32f12] M3 = - (ξ12- ξ22)/[f12f22] // cyclic when using f's So we need to evaluate (ξ14/f12 - 1)(-M1) + (ξ24/f22 - 1)(-M2) + (ξ34/f32 - 1)(-M3) (ξ14/f12 - 1) (ξ22- ξ32)/[f22f32] + (ξ24/f22 - 1) (ξ32- ξ12)/[f32f12] + (ξ34/f32 - 1) (ξ12- ξ22)/[f12f22] Multiply through by [f12f22f32] : (ξ14 - f12) (ξ22- ξ32) + (ξ24 - f22) (ξ32- ξ12) + (ξ34 - f32) (ξ12- ξ22) This is a job for Maple and there it is! I have just shown that (ξ14 - f12) (ξ22- ξ32) + (ξ24 - f22) (ξ32- ξ12) + (ξ34 - f32) (ξ12- ξ22) = 0 which means (ξ14/f12 - 1)M1 + (ξ24/f22 - 1)M2 + (ξ34/f32 - 1)M3 = 0 which means me31 = 1 ( ξ12-b2)/f12 (a2-b2)-1( ξ12-a2)/f12 me32 = ξ14/f12 (ξ12-b2)/f12 (a2-b2)-1( ξ12-a2)/f12 are both valid Stackel matrices! Just a little more. Go back to Q(a,b) = (ξ14 - f12) (ξ22- ξ32) + (ξ24 - f22) (ξ32- ξ12) + (ξ34 - f32) (ξ12- ξ22) where f12 = (ξ12-a2)( ξ12-b2) We can show that Q(a,b) is independent of both a and b, and so can be evaluated at a=b=0 where we clearly get 0. We can examine for example the coefficient of a2 coef (a2, Q(a,b)) = ( ξ12-b2) (ξ22- ξ32) + cyclic and here we get to make use of two famous results (from ancient memory) (ξ22- ξ32) + cyclic = 0 2-3 + 3-1 + 1-2 = 0 ξ12 (ξ22- ξ32) + cyclic = 0 12-13 + 23-21 + 31-32 = 0 (f(2) - f(3)) +cyclic = 0 // a little more general f(1)(f(2) - f(3)) +cyclic = 0 Thus, coef (a2, Q(a,b)) = 0. Since Q(a,b) = Q(b,a), we know coef (b2, Q(a,b)) = 0 as well. QED. _______________________________________________________________________________ just storing this here: What about Laplace separation? First of all, any Helmholtz separable CL system will also be Laplace separable, and this includes the 11 classical systems. So what we really wonder is: are there some CL systems that are NOT Helmholtz separable but which are Laplace separable? The answer really is that there are none. There are some that are R separable for Laplace but not R separable for Helmholtz, but that is not our subject in this document. Method 1. (Limit k12→0) We can take the existing analysis and compute the 6 rightmost functions in the Stackel matrix using the Mi information and be done with it. This is because with k12 = 0, the first column of the matrix does not play a role in anything other than affecting the Stackel determinant number S. We would proceed as follows. First, use (2.1) to determine the gn and the fn gn(≠n)fn(n) = H/hn2 n = 1,2,3 (2.1) Then use these and "trial and error" to find the rightmost 6 Stackel matrix functions, Φ = = e(2)i(3)-f(2)h(3) = + g1(≠1)/[ f2(2) f3(3)] b(1)i(3)-c(1)h(3) = – g2(≠2)/[ f3(3) f1(1)] b(1)f(2)-c(1)e(2) = + g3(≠3)/[ f1(1) f2(2)] Our separated equations are then these: LnXn = (1/fn)∂n[fn(∂nXn)] +[ k22Φn2(n) + k32Φn3(n)]Xn = 0 (16) So here we don't ever mention the number S, although it no doubt still exists. Method 2, start with k12= 0. Now let's take a look at bit back at our Helmholtz development. We worked our way down to this Σn gn(≠n) [ k12Φn1(n) + k22Φn2(n) + k32Φn3(n)]fn(n) = H(123)k12 (17) Now if k1= 0 all along , this would really say Σn gn(≠n) [k22Φn2(n) + k32Φn3(n)]fn(n) = 0 If we want this to be true for arbitrary separation constants, then our set of three equations in (18) would be reduced to just two equations Σn gn(≠n) fn(n)Φn2(n) = 0 Σn gn(≠n) fn(n)Φn3(n) = 0 (18)' We then no longer have a 3x3 matrix equation ΦT V = W. We can still of course talk about 3x3 matrix Φ. Suppose we define these 6 functions An (n) = fn(n)Φn2(n) Bn(n) = fn(n)Φn3(n) Then our equations above are these g1(23) A12(1) + g2(31) A(2) + g3(12) A3(3) = 0 g1(23) B1(1) + g2(31) B2(2) + g3(12) B3(3) = 0 How do we even know solutions exist to these equations? Certainly if the three gi have completely random functional form, there will be no non-trivial solution and we will say that the Laplace equation is not separable in such a system. Let's pause here to look at an example. Look at Bicylindrical (=bipolar = E.4) coordinates M&S page 81 where we have h12 = h22 = a2(chη - cosψ)-2 h3 = 1 Consider then gn(≠n)fn(n) = H/hn2 n = 1,2,3 (2.1) So that g1(≠1)f1(1) = H/h12 = h2h3/h1 = h2/h1 = 1 Then we find that f1(1) = 1 and g1(≠1) = 1. Since 1 and 2 are symmetric, same for 2. Then for 3, g3(≠3)f3(3) = H/h32 = h12 = a2(chη - cosψ)-2 I suppose we can say (η,ψ,z) - (1,2,3) so we then have g3(≠3)f3(3) = a2(coshξ1 - cosξ2)-2 So here are the things we learn from this very first step f1(1) = 1 f2(2) = 1 f3(3) = 1 g1(23) = 1 g2(31) = 1 g3(12)= a2(coshξ1 - cosξ2)-2 Now we can look at our two equations g1(23) A1(1) + g2(31) A2(2) + g3(12) A3(3) = 0 g1(23) B1(1) + g2(31) B2(2) + g3(12) B3(3) = 0 A1(1) + A2(2) + a2(coshξ1 - cosξ2)-2 A3(3) = 0 B1(1) + B2(2) + a2(coshξ1 - cosξ2)-2 B3(3) = 0 Consider the first of this last pair. If we set A3(3) = 0, then we need A1(1) + A2(2) = 0, but due to coordinate dependence, this means A1(1)= c1 and A2(2)= -c1 which gives the kind of solution discussed below. If you assume some non-vanishing A3(3), you then have A1(1) + A2(2) = stuff(12) B3(3) and there is no solution. So the only possibility here is that A3(3) = B3(3) = 0 and that means L3X3 = (1/f3)∂3[f3(∂3X3)] +[ k22Φ32(3) + k32Φ33(3)]X3 = 0 (16) L3X3 = (1/f3)∂3[f3(∂3X3)] +[ 0 + 0 ]X3 = 0 (16) L3X3 = ∂3[(∂3X3)] = 0 (16) X3 = aξ3+ b So in this case, we have 2X3 = 0 and X3 is a linear function at the fanciest. Now, suppose we know that ψ will have the form just X1X2 because we know there will be no dependence at all on ξ3 = z in this case. Then we have a 2D problem and our equations above would become A1(1) + A2(2) = 0 B1(1) + B2(2) = 0 The solutions here would be A1(1) = c1 A2(2) = -c1 B1(1) = d1 B2(2) = -d1 But for our system An (n) = Φn2(n) Bn(n) = Φn3(n) so we have our 4 elements of the 2D case Φ12(n) = c1 Φ22(n) = -c1 Φ13(n)= d1 Φ23(n) = -d1 Now look again at LnXn = (1/fn)∂n[fn(∂nXn)] +[ k22Φn2(n) + k32Φn3(n)]Xn = 0 (16) L1X1 = (1/f1)∂1[f1(∂1X1)] +[ k22Φ12(2) + k32Φ13(3)]X1 = 0 L1X1 = (1/f1)∂1[f1(∂1X1)] +[ k22c1 + k32d1)]X1 = 0 L2X2 = (1/f2)∂2[f2(∂2X2)] +[ k22Φ22(2) + k32Φ23(3)]X2 = 0 L2X2 = (1/f2)∂2[f2(∂2X2)] +[ k22(-c1) + k32(-d1))] X2 = 0 So define α2 = k22c1 + k32d1 and set the f's to one and we have L1X1 = ∂1[ (∂1X1)] +α2 X1 = 0 L2X2 = ∂2[ (∂2X12)] -α2 X1 = 0 and these are our separated equations in this case. But we know these solutions pretty well. α2> 0 : X1 = [sin(αξ1), cos(αξ1)] X2= exp(±αξ2) α2< 0 : X2 = [sin(αξ2), cos(αξ2)] X1= exp(±αξ1) This is what M&S are talking about on page 78 equation (3.07). So the most general solution you can have here is sine and expo and linear in z. In Laplace then the linear in z portion ∂z2 of the Laplacian gives 0 on X3 and contributes nothing to 2ψ. That is ψ = X1X2X3 with X1,2 as the above atoms, and X3 = linear. If you look at M&S, you see in their CYL chapter that none other than the classical ones are separable for either Helmholtz or Laplace. They are all 2D Laplace Separable in the sense outlined above since they are all conformal maps of the Cartesians and therefore only differ from the above example in the g3(12) function. Morse & Feshbach sign error search On page 509 M&F start talking Stackel. Equation 5.1.28 is of the form have always used. (1/fn)∂n(fn∂nXn) + [ k12Φn1(n) + k22Φn2(n) + k32Φn3(n)]Xn = 0 (5.1.28) They then tell us to multiply this equation by factor (Mn/S)ψ/Xn . That gives me ( Mn/S fnXn)∂n(fn∂nXn)ψ + [ k12Φn1(n) + k22Φn2(n) + k32Φn3(n)] (Mn/S)ψ = 0 We can then divide out the ψ and move the Mn to get ( Mn/S fnXn)∂n(fn∂nXn) + [ k12Φn1(n) Mn + k22Φn2(n) Mn + k32Φn3(n) Mn] (1/S)= 0 Now we are instructed to sum over n, which gives me Σn( Mn/S fnXn)∂n(fn∂nXn) + [ k12 Σn Φn1(n) Mn + k22 Σn Φn2(n) Mn + k32 Σn Φn3(n) Mn] (1/S)= 0 Now the red sum goes down the first column, and therefore gives S, and we know the other sums give 0 due to a weird theorem, so this gives Σn( Mn/S fnXn)∂n(fn∂nXn) + k12 = 0 or Σn( Mn/S fn)∂n(fn∂nXn) + k12Xn = 0 (P1) If we then multiply the n=1 equation by X2X3 and so on, we then get Σn( Mn/S fn)∂n(fn∂nψ) + k12ψ = 0 (5.1.29) So there is no sign error at least in this little M&F section. Notice that in (5.1.28) he has a + sign on both LHS terms, as I have always had. So where do I think I have a problem? In my #5 document, I process the Helmholtz equation which starts having the form (2+k12)ψ = 0 and I get it to this form: Σn(1/[hn2Xn]) (1/fn)∂n[fn(∂nXn)] + k12/Q + K12 = 0 (3.10) Signs here agree with my little (P1) intermediate result above. So it seems that (3.10) is compatible with M&F. Now suppose I take M&F 5.1.28 above and write it as (1/fn)∂n(fn∂nXn) = – [ k12Φn1(n) + k22Φn2(n) + k32Φn3(n)]Xn If I then insert this into my (3.10) I get – Σn(1/[hn2Xn]) [ k12Φn1(n) + k22Φn2(n) + k32Φn3(n)]Xn + k12/Q + K12 = 0 and now the first term has a minus sign. But this then disagrees with my #5 paper (4.9). THAT is my sign error problem! OK, I have found and fixed the error in my #5 doc!