3 More separation

This is the Title PhL 3.26.05 1. How Helmholtz Equation Separation Works First, we have to review a little of our simple separation theory. In order to get IT to work, we had to make certain assumptions concerning the CL coordinate scale factors hi(123). The first one was this: (H/hn2) = gn(≠n)fn(n) n = 1,2,3 where H = h1h2h2 = H(123) (1) Recall that F(123) means F is an arbitrary function of the three orthogonal variables. The assumption shown above is that the scale factor combination on the LHS has the "functional form" shown on the right. This is all a game of "functional form". We are saying that the dependence on ξn "factors out" as shown. This is of course three conditions, one for each value of n, so it is quite a tight restriction on the scale factors. For n = 1 the condition happens to be h2(123)h3(123)/h1(123) = g1(23)f1(1). When we then introduce the Φ Stackel matrix with its cofactors Mn(≠n), we then show (we do this by inverting a little 3x3 matrix equation) that the Φ elements are constrained by these three conditions 1/hn2 = Mn(≠n)/S (2) where remember that the Mn and S are both functions of the Φ elements. This is not a new restriction; it just arises from giving a new name to the linear term vn(n) in our assumed form for the separated ODE LnXn = (1/fn)∂n[fn(∂nXn)] +vnXn = 0 LnXn = (1/fn)∂n[fn(∂nXn)] +[ k12Φn1(n) + k22Φn2(n) + k32Φn3(n)]Xn = 0 In fact Mp(≠p) = cof(Φp1)/S where S = det(Φ). We can combine (1) and (2) to get H Mn(≠n)/S = gn(≠p)fn(n) (3) and this then shows you the connection between Mn and gn. So no new assumptions here either. We can rewrite (3) in this manner H(123)/S(123) = g1(≠1)f1(1)/ M1(≠1) = g2(≠2)f2(2)/ M2(≠2) = g3(≠3)f3(3)/ M3(≠3) To simplify notation for the moment, let's define N1(≠1) ≡ g1(≠1)/ M1(≠1) F(123) ≡ H(123)/S(123) Then our equations above say F(123) = f1(1) N1(23) F(123) = f2(2) N2(31) F(123) = f3(3) N3(12) => F(123) = f1(1) f2(2) f3(3) * [N1(23) N2(31) N3(12)]/3 This is a mistake! The correct equation here should be F(123)3 = f1(1) f2(2) f3(3) * [N1(23) N2(31) N3(12)] so what follows is wrong. Notice that these Nn(≠n) functions are determined in part by our CL system (which determines gi and fi from (1) above), and in part by our Φ matrix of unknown functions. We now make our second assumption which is that [N1(23) N2(31) N3(12)]/3 = 1 That is to say, we put this restriction on the Φ matrix components: 3 M1(≠1) M2(≠2) M3(≠3) = g1(≠1) g2(≠3) g3(≠3) This restriction is the same as this restriction F(123) = f1(1) f2(2) f3(3) which is to say the following, which M&F call "the Robertson condition", H(123)/S(123) = f1(1) f2(2) f3(3) (4) In this last form, we see that the restriction on the elements of Φ is really just a restriction on the value that S(123) = det(Φ) can have. We could have introduced a new constant C in this last step like so [N1(23) N2(31) N3(12)]/3 = C N = g/M 3C M1(≠1) M2(≠2) M3(≠3) = g1(≠1) g2(≠3) g3(≠3) H(123)/S(123) = C f1(1) f2(2) f3(3) Now recall that fngn only appear as a product pair in our "first restriction" (H/hn2) = gnfn. We could for example replace g1f1 by (g1/C)(f1C) = g1f1'. Then our last two equations above would be 3M1(≠1) M2(≠2) M3(≠3) = g1'(≠1) g2(≠3) g3(≠3) H(123)/S(123) = f1'(1) f2(2) f3(3) Our point is that were we to add a constant C to our picture, it does not buy us anything. It does not give us any more flexibility on solving for the Φ, so we just set C = 1 and be done with it. So now let us review things: We first determine the gn and fn from these three equations (H/hn2) = gn(≠n)fn(n) and this is our "first restriction" on things. We assume that we can find gn and fn , otherwise we give up on simple separation. We assume that the linear term vn in our assumed separated ODE is a function of Φ matrix elements in a certain way, so we have vn(n) = k12Φn1(n) + k22Φn2(n) + k32Φn3(n) and this is how Φ first enters the discussion. Notice that we have quietly put a functional-form restriction on the 9 elements of Φ, in that the Φnm are functions only of the variable ξn. By inverting a 3x3 matrix equation, we find this connection between the gn and Φ Mn/S = gnfn/H where the Φ elements appear in both Mn and S. Using our "first restriction" we can write this as Mn/S = 1/hn2 We can think of this equation as a condition on the elements of Φ and it results from our "first restriction". We then make the following "second restriction" M1M2M3 = g1 g2 g3 which can be rewritten as the Robertson condition H/S = f1 f2 f3 and we can interpret this as a second condition on the elements of Φ. We could have put a constant in here, like this M1M2M3 = K g1 g2 g3 H/S = K f1 f2 f3 So here is where we stand. We need to find the 9 elements of Stackel matrix Φ. If we can do so, then we will have obtained our separated equations, which is our ultimate goal in "separation analysis": LnXn = (1/fn)∂n[fn(∂nXn)] +[ k12Φn1(n) + k22Φn2(n) + k32Φn3(n)]Xn = 0 One condition on the Φ elements is the functional dependence form shown here. The other two conditions on Φ are these Mn(Φ)/S(Φ) = 1/hn2 H/S(Φ) = f1 f2 f3 where fn come from (H/hn2) = gnfn Now, even assuming we can meet our "first restriction" and come up with gn and fn, it may still be that our three conditions on Φ are such that there is no solution Φ ! It turns out that only the 11 classical systems can be solved for Φ in this manner. Since 10 derive from 1 (the ellipsoidal), the significant face is that the ellipsoidal system can be solved for Φ. 2. How R separation Works We just noted in the previous section that our search for a solution Stackel matrix Φ is limited by these conditions Mn(Φ)/S(Φ) = 1/hn2 H/S(Φ) = f1 f2 f3 where fn come from (H/hn2) = gnfn Only if the above conditions are satisfied can we find a solution Φ matrix and Helmholtz separates. In R separation theory, we "soften" these restricting conditions by adding two new functions R and Q so these conditions then become (I am here quoting these, not deriving them. They should emerge from our analysis below. ) Mn(Φ)/S(Φ) = Q/hn2 H/S(Φ) = f1 f2 f3 R2Q where fn come from (H/hn2) = gnfnR2 and at the same time, we assume a "softer" form for our separated Helmholtz solution ψ(123) = X1X2X3/R(123) (2.1) where R=R(123) and Q=Q(123). So this is no longer true "simple separation" because of this extra function R, and M&S call it "R separation", whereas M&F refer to R as a "modulation factor". Now let's review what happens in our R separation theory. We first assume ψ(123) = X1X2X3/R(123) and we put this into our Helmholtz equation and start our "hunt" for separation.: (2 +K12)ψ = H-1{ ∂1[(H/h12)(∂1{X1X2X3/R})] + cyclic} + K12X1X2X3/R = 0 = H-1{ X2X3 ∂1[(H/h12)(∂1{X1/R})] + cyclic} + K12X1X2X3/R = 0 = H-1{ X2X3 ∂1[{(H/h12)(1/R2)}{ (∂1X1)R - X1(∂1R) })] + cyclic} + K12X1X2X3/R = 0 Here is where we must make our "first restriction" which is {(H/h12)(1/R2)} = g1f1 or (H/h12) = g1f1R2 (2.2) Now pause here for a moment. Write this out in more functional-form detail (H(123)/[hn(123)]2 = g1(23)f1(1) [ R(123)]2 The LHS is completely determined by the CL system. The only reason we are adding R ≠1 is for the case that the LHS does not match the functional form g1(23)f1(1). In the worst case, we might have to set both f1=1 and g1 = 1 and let R "carry the whole load". However the 1-equation resolves itself, we still of course have the 2- and 3-equations, and this same R(123) might now work in those equations! So (2.2) is not going to work for ANY CL coordinate system, it is indeed a restriction. I think we should think of (2.2) as determining R(123), so this is not a function you can fiddle with in a significant way. So think then of equation (2.2) as determining all these 6 objects: gn, fn and R. We have to assume that our CLS allows this to happen, otherwise we give up. Now we can resume processing our Helmholtz equation from above: = H-1{ X2X3 ∂1[g1f1{ (∂1X1)R - X1(∂1R) })] + cyclic} + K12X1X2X3/R = 0 = H-1{ X2X3 g1∂1[f1{ (∂1X1)R - X1(∂1R) })] + cyclic} + K12X1X2X3/R = 0 = { X2X3 (g1/H) ∂1[f1{ (∂1X1)R - X1(∂1R) })] + cyclic} + K12X1X2X3/R = 0 = { X2X3 (1/[h12f1R2] ∂1[f1{ (∂1X1)R - X1(∂1R) })] + cyclic} + K12X1X2X3/R = 0 => { (1/[h12f1X1R2] ∂1[f1{ (∂1X1)R - X1(∂1R) })] + cyclic} + K12/R = 0 Σn(1/[hn2fnXnR2] ∂n[fn{ (∂nXn)R - Xn(∂nR) })] + K12/R = 0 // mult by R2 Σn(1/[hn2fnXn] ∂n[fn{ (∂nXn)R - Xn(∂nR) })] + K12R = 0 (2.3) where each line is just trivial algebra the reader can check by inspection. Notice in the 4th line that we again made use of (2.2) but now we are outside the ∂1 operator. I don't see any alternative whatsoever to the above processing. In particular, we really must have (H/hn2) = gnfnR2 with no other factors (such as a Q or S), because this is inside the outer ∂1 operator and we don't want to pick up any other messy terms. By computing out the derivatives in the correct threaded manner, ∂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)} we find that in (2.3) we get cancellation of two terms containing (∂nXn)(∂nR) and we are left with Σn (1/[hn2Xnfn]) R∂n{fn(∂nXn)} = Σn (1/[hn2fn]) ∂n{fn(∂nR)} - K12R and we divide by R to get Σn (1/[hn2fnXn]) ∂n{fn(∂nXn)} = Σn (1/[hn2fnR]) ∂n{fn(∂nR)} - K12 (2.4) which agrees with the "waypoint" of M&F on page 519 and with my previous work, except M&F assume from the start that K12 = 0 from the start, but I have chosen to keep this term for now. This (2.4) is our processed Helmholtz equation! Remember that we are looking for a separated ODE equation for Xn, and we have complete freedom in trying to find R and Q functions to make this work. As in the simple separation case, suppose we seek a separated equation of the same form, LnXn = (1/fn)∂n[fn(∂nXn)] +[ K12Φn1(n) + k22Φn2(n) + k32Φn3(n)]Xn = 0 (1/fn)∂n[fn(∂nXn)] = - [ K12Φn1(n) + k22Φn2(n) + k32Φn3(n)]Xn (1/[fn Xn])∂n[fn(∂nXn)] = - [ K12Φn1(n) + k22Φn2(n) + k32Φn3(n)] (1/[hn2fn Xn])∂n[fn(∂nXn)] = - [ K12Φn1(n) + k22Φn2(n) + k32Φn3(n)]/hn2 Σn (1/[hn2fn Xn])∂n[fn(∂nXn)] = - Σn [ K12Φn1(n) + k22Φn2(n) + k32Φn3(n)]/hn2 (2.5) Notice now that LHS(2.5) = LHS(2.6), so we then have Σn [ K12Φn1(n)+k22Φn2(n)+k32Φn3(n)]/hn2+Σn (1/[hn2fnR]) ∂n{fn(∂nR)} -K12 = 0 (2.6) and this has the form K12 (stuff) + k22(stuff') + k32(stuff'') = - Σn (1/[hn2fnR]) ∂n{fn(∂nR)} which is supposed to be valid for arbitrary values of K12, k22 and k32. But this can only be satisfied if RHS = 0, so we would need to find an R such that Σn (1/[hn2fnR]) ∂n{fn(∂nR)} = 0 (2.7) Let's assume we can find an R satisfying this equation. Then we have K12 [Σn (1/hn2) Φn1(n) - 1] + k22 [Σn (1/hn2) Φn2(n)] + k32 [Σn (1/hn2) Φn3(n)] = 0 and setting the coefficients to zero we get Σn (1/hn2) Φn1(n) = 1 Σn (1/hn2) Φn2(n) = 0 Σn (1/hn2) Φn3(n) = 0 (2.8) Now recall our first restriction(2.2) that (H/hn2) = gnfnR2 => (1/hn2) = gnfn R2/H so we can write our three equations as (R2/H) Σn gnfn Φn1(n) = 1 (R2/H) Σn gnfn Φn2(n) = 0 (R2/H) Σn gnfn Φn3(n) = 0 or Σn gnfn Φn1(n) = H/R2 Σn gnfn Φn2(n) = 0 Σn gnfn Φn3(n) = 0 (2.9) or ΦT V = W where V = W= which is the same as in our simple separation work except for the factor R2. If we now invert the equation as we did in the simple separation analysis and we find from the top row of V = [ΦT]-1W that gnfn = Mn H/(SR2) (2.10) We can then combine this with (2.2) (H/hn2) = gnfnR2 => gnfn = (H/R2)(1/hn2) to get this result Mn H/(SR2) = (H/R2)(1/hn2) Mn / S = (1/hn2) (2.11) which agrees with our simple separation analysis. Now go back to (2.10) which says H/(SR2) = gnfn/Mn We can now repeat our simple separation analysis that led us to the Robertson condition there. That is, we could impose this same condition. Here is a quick review of this last thread. We processed the Helmholtz equation into (2.4) which everybody agrees with. We then wrote out the expected form of our ODE separated equation (same as in the simple case) using the same Φ elements, and when combined with the processed Helmholtz equation this led us to (2.6) which I repeat here, Σn [ K12Φn1(n)+k22Φn2(n)+k32Φn3(n)]/hn2+Σn (1/[hn2fnR]) ∂n{fn(∂nR)} -K12 = 0 (2.6) where we are supposed to have three "free" constants, so the coefficients of each of these constants needs to be 0. I observed that if the sum in red does not vanish, we are dead in the water because we cannot possibly solve an equation of this form K12 (stuff) + k22(stuff') + k32(stuff'') = - Σn (1/[hn2fnR]) ∂n{fn(∂nR)} At this point, I split off from the M&F presentation just to pursue my own private path for a while. I asked "what if" the R(123) we were forced to use from (2.2) happens to result in Σn (1/[hn2fnR]) ∂n{fn(∂nR)} = 0 Now remember that everything in this equation is "forced" to be whatever it is by our CL system, so it is extremely unlikely that this sum would happen to come out being 0. But if it just happened to be 0 for some particular curvilinear system, then we end up with two equations which are the same as in the simple separation world, namely, Mn / S = (1/hn2) H/S = f1(1) f2(2) f3(3) where this second condition would be something we impose on our Φ elements as in the simple case. OK, since the sum shown above is NOT going to be 0, we can now forget this little side path. Resume main flow. Since all the objects in our R sum above are determined by the CL system, the R sum is going to just be some result which we cannot control. We might as well remove the R from the LHS denominator. Then let's assume that the resulting sum has this form Σn (1/[hn2fnR]) ∂n{fn(∂nR)} = - k12/Q(123) (2.7) so what we are saying is that this sum determines function Q(123) which we define in a slightly strange way by including the other two factors shown. Once we settle on a scale for Q(123), k12 will have some specific value. It seems we could have just set it to 1, but leave it as it is. Notice that (2.7) is in no way a restriction of anything, it is the definition of our second function Q which we now carry along with R. Here then is what (2.6) becomes Σn [ K12Φn1(n)+k22Φn2(n)+k32Φn3(n)]/hn2- k12/Q -K12 = 0 Σn [ K12Φn1(n)+k22Φn2(n)+k32Φn3(n)]/hn2 -K12 = k12/Q K12 (stuff) + k22(stuff') + k32(stuff'') = k12/Q where again the three LHS constants are supposed to be free. I claim that this equation cannot be solved. For example, if we set all three constants to 0, there is clearly no solution since k12 is determined. This may be the reason that you cannot solve Helmholtz with R separation. It would appear that even if you were to set K12 = 0 that there is still no solution , but I will ignore that fact for the moment. Instead, we return now to (2.4), and just resume from there, so Σn (1/[hn2fnXn]) ∂n{fn(∂nXn)} = Σn (1/[hn2fnR]) ∂n{fn(∂nR)} - K12 (2.4) and we replace the R-sum using (2.7) to get ( I stubbornly retain K12 ≠0 for now) Σn (1/[hn2fnXn]) ∂n{fn(∂nXn)} = - k12/Q - K12 (2.12) Now we introduce the Φ matrix for the first time and write our separated equation this way LnXn = (1/fn)∂n[fn(∂nXn)] +[ κ12Φn1(n) + k22Φn2(n) + k32Φn3(n)]Xn = 0 (2.13) where we use κ12 to avoid confusion. We make our usual assumption that k22 and k32 are separation constants that can take any values, and κ12 is TBD. So now we have some Φ components to worry about and to try to compute. Now at this point suppose out of the blue we impose this condition on Φ Mn(Φ)/S(Φ) = Q/hn2 (2.14) which differs form our simple theory in that Q is present. We are of course allowed to impose whatever condition we like on Φ, the question is "can we solve for Φ" with the conditions we impose? Let's now multiply (2.12) by Q Σn (Q/[hn2fnXn]) ∂n{fn(∂nXn)} + k12 - K12Q and then make use of (2.14) to write Σn (Mn(Φ)/[ S(Φ)fnXn]) ∂n{fn(∂nXn)} + k12 = K12Q (2.15) Now we pause and go back to our simple-separation analysis and we note that we could write the ODE there for Xn in the following manner Σn (Mn(Φ)/[ S(Φ)fnXn]) ∂n{fn(∂nXn)} +k12 = 0 (a) and we also had there that LnXn = (1/fn)∂n[fn(∂nXn)] +[ k12Φn1(n) + k22Φn2(n) + k32Φn3(n)]Xn = 0 (b) These two equations go together. The first (a) is the processed Helmholtz equation with parameter k12 and (b) is the separated ODE that we obtain, in which k12 appears. Now, if (the big if) we set K12 = 0, then 2.15 looks just like (a), and we know the "solution" ODE for (a) is (b). Therefore, in this case, it must be that the κ12 appearing in (2.13) is k12, which remember is the thing appearing in our R sum (2.7). What happened to the Robertson Condition? I have never used it or mentioned it. What role does it play in the simple separation analysis? We wrote it there as (2.4) [ in the Stackel matrix business doc] and here is what we did, I will quote equations from that doc: gn(≠n)fn(n) = H/hn2 n = 1,2,3 SS(2.1) 1/hn2 = Mn(≠n)/S SS(2.3) H/S = f1(1) f2(2) f3(3) SS(2.4) => M1(≠1) = S/h12 = (S/H) g1(≠1)f1(1) = g1(≠1)f1(1)/[ f1(1) f2(2) f3(3) ] = g1(≠1)/ [f2(2) f3(3) ] SS(2.5) Notice that we used all three quoted equations to get this result! Notice also that the result tells us the value of M1 from data that we know! And we also know S from (2.4). So knowing S and the Mn, we are able to solve for the entire matrix Φ. Now we want to do the analogous thing here in our R separation analysis. So far we already know from our (2.2) that (H/h12) = g1f1R2 => gn(≠n)fn(n) = (H/hn2)(1/R2) Also, from (2.14) we have (we had to do this to make things work above) Mn(Φ)/S(Φ) = Q/hn2 => 1/hn2 = (Mn(≠n)/S)(1/Q) Now suppose we impose a new Robertson condition of the form H/S(Φ) = f1 f2 f3 R2/Q We will then have gn(≠n)fn(n) = (H/hn2) (1/R2) n = 1,2,3 SS(2.1)' 1/hn2 = (Mn(≠n)/S )(1/Q) SS(2.3)' H/S = f1(1) f2(2) f3(3)(R2Q) SS(2.4)' and if we do our little calculation of M1 we now get M1 = (S/h12){Q} = (S/H)(H/h12){Q} = (S/H) g1f1 {QR2} = g1f1 {QR2 } Q/[ f1(1) f2(2) f3(3) R2Q] = g1/ [f2(2) f3(3)] SS(2.5)' Thus we see that if we assume the modified Robertson condition shown in (2.4'), then our expressions for the cofactors Mn are exactly the same as they were in the simple separation theory. Conclusion: Here are the steps for doing an R-separation of a CL system (1) Compute the 7 objects gn, fn and R(123) from (2.2): (H/h12) = g1f1R2 (2) Compute Q and a value for k12 from this expression: Σn (1/[hn2fnR]) ∂n{fn(∂nR)} = - k12/Q(123) (3) Compute the M1 from M1 = g1/ [f2(2) f3(3)] and similarly for M2 and M3. (4) Compute S from 1/hn2 = (Mn/S)(1/Q), using any n. (5) You now know the Mi and S from which you can deduce a Stackel matrix Φ. You then have your separated ODE equations as (1/fn)∂n[fn(∂nXn)] +[ k12Φn1(n) + k22Φn2(n) + k32Φn3(n)]Xn = 0 (6) If you maintain K12 ≠ 0 as in (2.15), this method does not work. Thus, we conclude that we do not know how to separate the Helmholtz equation in this R-separation model, although perhaps it is somehow possible. This method works only for the Laplace equation. M&S state on page 96: "No case is known in which the Helmholtz equation is R-separable." After all, 2.15 says Σn (Mn(Φ)/[ S(Φ)fnXn]) ∂n{fn(∂nXn)} + k12 = K12Q(123) (2.15) and in general Q(123) tangles everything together so you can't separate. If for some system you could perhaps write Q(123) = Σn qn(n), then you would have Σn [ (Mn(Φ)/[ S(Φ)fnXn]) ∂n{fn(∂nXn)} - K12 qn(n) ] + k12 = 0 Then you could separate in this mechanical way (M1(Φ)/[ S(Φ)f1X1]) ∂1{f1(∂1X1)} - K12 q1(1) = c12 (M2(Φ)/[ S(Φ)f2X2]) ∂2{f2(∂2X2)} - K12 q2(2) = c22 (M3(Φ)/[ S(Φ)f3X3]) ∂3{f3(∂3X3)} - K12 q3(3) = c32 Σn [ (Mn(Φ)/[ S(Φ)fnXn]) ∂n{fn(∂nXn)} - K12 qn(n) ] + k12 = c12 + c22 + c32 + k12 = 0 So you would have two free separation constants c22 and c32, and the third is c12 = -k12. But I suspect Q(123) = Σn qn(n) never happens for any known system. Question: is there some way to identify something like k22 as a separation constant in the above mechanical way in the simple separation theory? Do we get k22 = c22 ? Problem: try this plan for toroidals and compare to M&S p 97.