2 R separation theory

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) /R(123) 2ψ = H-1{ ∂1[(H/h12)(∂1{X1X2X3/R(123)})] + cyclic} Now maybe work backwards with hints from M&S 96,such as (4.01) which adds Q like so Q(123)/hp2 = Gp1(≠p)/S = cof(Φp1)/S ≡ Mp(≠p)/S (22) and a new Roberson's condition which looks like this, which is M&S 4.02 H(123)/S(123) = f1(1) f2(2) f3(3) R2(123)/Q(123) Now the quote p 96: there is NO R-separation for any Helmholtz! Only Laplace has a chance. So maybe I will adjust my tune a little. They are looking for this result -(Q/R) Σn (1/fnhn2)∂n (fn∂nR) = constant So we better go back to the beginning: 2ψ = H-1{ ∂1[(H/h12)(∂1{X1X2X3/R})] + cyclic} 2ψ = H-1{ X2X3 ∂1[(H/h12)(∂1{X1/R})] + cyclic} 2ψ = H-1{ X2X3 ∂1[(H/h12)(X1(-R-2∂1R) + R-1∂1X1)] + cyclic} 2ψ = - H-1{ X2X3 ∂1[(H/h12) R-1X1R-1∂1R) ] + cyclic} + H-1{ X2X3 ∂1[(H/h12) R-1∂1X1)] + cyclic} Looking at the second line, we might propose that (H/h12) R-1 = g1(23)f1(1) like we did last time, but extra factor of R. Then we could write 2ψ = - H-1{ X2X3 ∂1[g1(23)f1(1)X1R-1∂1R) ] + cyclic} + H-1{ X2X3 ∂1[g1(23)f1(1)∂1X1)] + cyclic} 2ψ = - H-1{ X2X3 g1(23) ∂1[f1X1R-1∂1R) ] + cyclic} + H-1{ X2X3 g1(23) ∂1[ f1∂1X1)] + cyclic} Now this is all 2ψ = 0 Laplace so we could divide both sides by whatever. Maybe 2ψ = - { g1(23) ∂1[f1X1R-1∂1R) ]/X1 + cyclic} + { g1(23) ∂1[ f1∂1X1)]/X1 + cyclic} = 0 Maybe do the derivative in the first term a la ∂1[(X1R-1)(f1∂1R) ] = (X1R-1) ∂1(f1∂1R) + ∂1(X1/R) (f1∂1R) Maybe it is too time wasteful to reverse engineer this. I just need to get a start from somebody here! M&F Have it. Their buzzword is the R is the "modulation factor", see page 518 vol I. Start with these "assumptions" within our Stackel matrix formalism: // u = Q of M&S 1/h12 ≡ M1(≠p)/(Su) (22) H/S = f1(1) f2(2) f3(3) R2u h12 ≡ Su/M1 h12/S ≡ u/M1 1/S = f1(1) f2(2) f3(3) R2u/H 1/S = u/(M1 h12) f1 f2 f3 R2u/H = u/(M1h12) f1 f2 f3 R2/H = 1/(M1h12) f1 f2 f3 M1 R2 = H/(h12) So using their two suggestions, I get the last item here which could be maybe useful. Derive the M&F Waypoint stated after 5.1.46 on page 519. (2 +K12)ψ = H-1{ ∂1[(H/h12)(∂1{X1X2X3/R})] + cyclic} + K12X1X2X3/R = H-1{ ∂1[f1 f2 f3 M1 R2 ∂1{X1X2X3/R}] + cyclic}+ K12X1X2X3/R = H-1{ f2 f3 X2X3 M1(23)∂1[f1R2(∂1{X1/R}] + cyclic}+ K12X1X2X3/R = 0 This looks at least a little promising. Multiply by H f2 f3 X2X3 M1(23)∂1[f1R2∂1(X1/R)] + cyclic + K12H X1X2X3/R = 0 Now ready to do the derivative ∂1(X1/R) = ∂1(R-1X1) = R-1∂1X1 - R-2X1∂1R = R-2{ (∂1X1)R - X1(∂1R) } Then we get f2 f3 X2X3 M1(23)∂1[f1{ (∂1X1)R - X1(∂1R) }] + cyclic = - K12H X1X2X3/R Multiply and divide by f1 to get f1f2 f3 X2X3 (1/f1)M1(23)∂1[f1{ (∂1X1)R - X1(∂1R) }] + cyclic = - K12H X1X2X3/R Now divide all terms by the three fi X2X3 (1/f1)M1(23)∂1[f1{ (∂1X1)R - X1(∂1R) }] + cyclic =- K12H X1X2X3/(f1f2f3R) Now pause to use our result above H/S = f1(1) f2(2) f3(3) R2u => H/(f1f2f3R) = SRu and we then have X2X3 (1/f1)M1(23)∂1[f1{ (∂1X1)R - X1(∂1R) }] + cyclic = - K12SRu X1X2X3 Do the same thing with X1 to get (1/[X1f1])M1(23)∂1[f1{ (∂1X1)R - X1(∂1R) }] + cyclic = - K12SRu  Now I will try to get rid of M1 using their hint h12 ≡ Su/M1 => M1 = Su/h12 so the above then becomes Su (1/[h12X1f1]))∂1[f1{ (∂1X1)R - X1(∂1R) }] + cyclic = - K12SRu Question: when we say S, we mean S(123) always. So in the cyclic sum, it does not "cycle". I guess the same for u(123), so maybe we can divide them out to get Σn (1/[hn2Xnfn]) ∂n[fn{ (∂nXn)R - Xn(∂nR) }] = - K12R Σn (1/[hn2Xnfn]) ∂n[fn(∂nXn)R] = Σn (1/[hn2Xnfn]) ∂n[fnXn(∂nR)] - K12R Now we are trying to get to our next waypoint on page 519 and we seem not too bad. Now let's "do out" the two derivatives: ∂n[fn(∂nXn)R] = ∂n[R{fn(∂nXn)}] = R∂n{fn(∂nXn)} + (∂nR) {fn(∂nXn)} ∂n[fnXn(∂nR)] = ∂n[Xn{fn(∂nR)}] = Xn∂n{fn(∂nR)} + (∂nXn) {fn(∂nR)} Now insert this into our equation above to get LHS = Σn (1/[hn2Xnfn])[ R∂n{fn(∂nXn)} +(∂nR) {fn(∂nXn)} ] = Σn (1/[hn2Xnfn])R ∂n{fn(∂nXn)} + Σn (1/[hn2Xnfn]) (∂nR) {fn(∂nXn)} = Σn (1/[hn2Xnfn]) R∂n{fn(∂nXn)} + Σn (1/[hn2Xn]) (∂nR) (∂nXn) RHS = Σn (1/[hn2Xnfn]) [ Xn∂n{fn(∂nR)} + (∂nXn) {fn(∂nR)} ] - K12R = Σn (1/[hn2Xnfn]) Xn∂n{fn(∂nR)} + Σn (1/[hn2Xnfn]) (∂nXn) {fn(∂nR)} - K12R = Σn (1/[hn2fn]) ∂n{fn(∂nR)} + Σn (1/[hn2Xn]) (∂nXn)(∂nR) - K12R The two red sums cancel and we then get Σn (1/[hn2Xnfn]) R∂n{fn(∂nXn)} = Σn (1/[hn2fn]) ∂n{fn(∂nR)} - K12R Divide both sides by R to get Σn (1/[hn2fnXn]) ∂n{fn(∂nXn)} = Σn (1/[hn2fnR]) ∂n{fn(∂nR)} - K12 and we have finally arrived at the page 519 waypoint (if we set K12 = 0 which I think they did all along) Σn (1/[hn2fnXn]) ∂n{fn(∂nXn)} = Σn (1/[hn2fnR]) ∂n{fn(∂nR)} // the waypoint Now suppose we find R that solves this Σn (1/[hn2fnR]) ∂n{fn(∂nR)} = - k12/u or Σn (1/[hn2fn]) ∂n{fn(∂nR)} + k12R/u = 0 // agrees with 5.1.47 WARNING: this k12 is not the Helmholtz parameter K12, it is just "some constant" (I think). Then we have from the waypoint that Σn (1/[hn2fnXn]) ∂n{fn(∂nXn)} = - k12/u or Σn (u/[hn2fnXn]) ∂n{fn(∂nXn)} +k12 = 0 But now we use hn2/S ≡ u/Mn => u/hn2= Mn/S and this becomes Σn (Mn /[SfnXn]) ∂n{fn(∂nXn)} +k12 = 0 // agrees with 5.1.48 Now, beam back to the PL Stackel notes where equation (9) said (2 +k12)ψ = H-1 { g1(23)∂1[f1(∂1X1)] /X1 + cyclic}ψ + k12ψ = 0 (9) or H-1 { g1(23)∂1[f1(∂1X1)] /X1 + cyclic} + k12 = 0 or H-1 { g1(23)/ [X1] ∂1[f1(∂1X1)] + cyclic} + k12 = 0 and where we eventually in those notes had in (2.2) H/S = g1(23)f1/ M1 => g1(23) = M1H/[Sf1] so we can write our last result as H-1 { M1H/[Sf1]/ [X1] ∂1[f1(∂1X1)] + cyclic} + k12 = 0 or { M1/[Sf1]/ [X1] ∂1[f1(∂1X1)] + cyclic} + k12 = 0 or { M1/[Sf1X1] ∂1[f1(∂1X1)] + cyclic} + k12 = 0 and this last result matches 5.1.48. So here is the point: if we find R that solves 5.1.47 with some arbitrary constant we call k12, then we find that we obtain Σn (Mn /[SfnXn]) ∂n{fn(∂nXn)} +k12 = 0 and we see that this is the Helmholtz equation where the parameter is that constant k12 and where the separated equations for the Xn are given in the usual Stackel theory form (1/fn)∂n[fn(∂nXn)] +[ k12Φn1(n) + k22Φn2(n) + k32Φn3(n)]Xn = 0 where now we seem to have three arbitrary constants. Now one more time write the assumed R equation Σn (1/[hn2fnR]) ∂n{fn(∂nR)} = - k12/u Σn (u/[hn2fnR]) ∂n{fn(∂nR)} = - k12 k12 = - (u/R) Σn (1/[hn2fn]) ∂n{fn(∂nR)} And I am happy to see this agree with M&S 4.03 where they call k12 = α2 and u = Q. Now let's go back go our M&F assumptions at the start: H/S = f1(1) f2(2) f3(3) R2u h12 ≡ Su/M1 and these are seen to agree with M&S page 96 equations (4.01) and (4.02), so we have made contact! Now M&S reference their own "Field theory for Engineers" 1961 for this stuff, but I use M&F which is just fine.