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Theorem M&S page 78: "No cylindrical system supports R separation " Parabolic cylinder does, but R = 1. We want to prove the assertion that you cannot do R separation with a cylindrical system. Therefore we shall follow the four steps of our doc #5 We know that h3 = 1. I don't think they are assuming conformal. So which equation cannot be satisfied? I guess here is what we know h1= h1(12) h2= h2(12) h3= 1 H = h1h2 H/h12= h2/h1 H/h22 = h1/h2 H/h32 = h1h2 (1) Is there a problem with (3.5) ? (H/[R2hn2]) = fn(n)gn(≠n) (h2/h1) = f1(1) g1(23)R2 => g1(23) = G1(2) (h1/h2) = f2(2) g2(13)R2 => g2(13) = G2(1) (h1h2) = f3(3) g3(12)R2 => f3(3) = 1 Since the h's don't depend on ξ3, we can define G1 and G2 as shown. Recall that none of the fn or gn can vanish. Multiply the first two lines to get 1 = f1(1) g1(23)R2 * f2(2) g2(31)R2 = f1g1f2g2R4 Therefore R-4 = f1g1f2g2 = f1(1)G1(2)f2(2)G2(1) = [f1(1)G2(1)][G1(2)f2(2)] ≡ J1(1) J2(2) so we know that R = R(12). Note that R-2 = We can rewrite the third line above as (h1h2) = g3(12)R2 so that g3(12) = (h1h2)/R2 = (h1h2) which has this functional form g3(12) = (h1(12)h2(12) (2) We are supposed to compute Q using (3.9) which says Σn (1/[hn2fnR]) [∂n{fn(∂nR)}] = - k12/Q(123) (3.9) Since ∂3R = 0, there will be only two terms in the sum - k12/Q(123) = (1/[h12f1R]) [∂1{f1(∂1R)}] + (1/[h22f2R]) [∂2{f2(∂2R)}] = (1/[h12f1R]) [∂1{f1(∂1R)}] + (1/[h22f2R]) [∂2{f2(∂2R)}] If I put these two terms into Maple as T1 and T2 and have R-4 = f1g1f2g2, we just get this mess where the first line has lots of ∂1's and the second lots of ∂2's. There is really nothing more I can do without knowing more about the various functions. But one thing I notice. Because R = R(12), you might think that T1 and T2 would each have complicated (12) dependence. But what we seeing is this T1 = ( A1(1)/B1(1)) / h12 T2 = ( A2(2)/B2(2)) / h22 So our computation is telling us that - k12/Q(12) = ( A1(1)/B1(1)) / h12 + ( A2(2)/B2(2)) / h22 so let's leave it at that form the moment. (3) We now look at equation (7.5) which reads 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) and in our current situation this says G1(2) = E(2)I(3)-F(2)H(3) - G2(1) = B(1)I(3)-C(1)H(3) g3(12) = B(1)F(2)-C(1)E(2) Let's try H(3) = 0 and F(2) = 0 since that worked in our cylinder example, then G1(2) = E(2)I(3) - G2(1) = B(1)I(3) g3(12) = -C(1)E(2) We must then have E(2) = k2G1(2) and B(1) = k1G2(1), apart from constants, so we get G1(2) = k2G1(2)I(3) - G2(1) = k1G2(1)I(3) g3(12) = -C(1) k2G1(2) or 1 = k2I(3) - 1 = k1I(3) g3(12) = -C(1) k2G1(2) Adding the first two equations tells us k2 = -k1 so we select k2= 1 and k1= -1 to get 1 = I(3) - 1 = -I(3) g3(12) = -C(1) G1(2) So I(3) = 1 and we are left with the single equation g3(12) = -C(1) G1(2) This equation implies that g3(12) factors as shown. Assuming it does, we get -C(1) = g3(12)/ G1(2) so our only requirement so far is that g3(12) must factor, which it might in some cylindrical system. So in step (3) we are pretty much still alive with some R(12) ≠ 1. Let's look again at our factorization requirement g3(12) = -C(1) G1(2) We learned earlier that g3(12) = (h1(12)h2(12) so we are now requiring that the product of our two hi be a factorizable function. This is certainly the case for cylindrical coordinates, but not for elliptic cylinder coordinates. (4) Now we consider 1/Q = a(1) (1/h12) + d(2) (1/h22) + g(3) (1/h32) or 1/Q(12) = a(1) (1/h12) + d(2) (1/h22) + g(3) or 1/Q(12) = a(1) (1/h12) + d(2) (1/h22) where we remove the last term since there is no 3-dependence in the LHS. So now the question is whether this last equation is compatible with our Q(12) calculation! Above we got - k12/Q(12) = ( A1(1)/B1(1)) / h12 + ( A2(2)/B2(2)) / h22 So we could solve this by saying k12 = -1 a(1) = ( A1(1)/B1(1)) d(2) = ( A2(2)/B2(2)) so in theory we have no problem at all with step (4) !!! Conclusion: I don't see any reason why many cylindrical systems could allow R separation! Any system for which g3(12) factors will be R separable! I am mystified by the M&S comment. But let's go back to my conclusion above we are now requiring that the product of our two hi be a factorizable function. So in order to get R separation, we need to have h1(12) = a1(1)b1(2) h2(12) = a2(1)b2(2) and in M&S that means g11 and g22 have to factor. In his table starting on page 79, we can just look at these gii factors and see if any of his systems meet my requirement. Page 79: no Page 80,81: regular cylindricals do separate and our condition is met. E3C at first looks like a problem until we see ρ = a mess there. So nothing on these pages. Page 82.83: no Page 84,85: no So all the cylindrical systems in the M&S table (their 21 systems have h1 = h2 AND h1 does not factor. Question: Why is it, then, that elliptical cylinder coordinates don't have factoring h's, but still are separable? Go back to this point G1(2) = E(2)I(3)-F(2)H(3) - G2(1) = B(1)I(3)-C(1)H(3) g3(12) = B(1)F(2)-C(1)E(2) What conditions are thus imposed on the scale factors? We can perhaps regard the first two equations here as requiring that H(3) = 0 or I(3) = 0, so lets try the first: H(3) = 0 I think "without loss of generality". We then get G1(2) = E(2)I(3) - G2(1) = B(1)I(3) g3(12) = B(1)F(2)-C(1)E(2) Now pick E(2) = k2 G1(2) and I(3) = 1/k2 so the first line is OK. The next two lines are then - G2(1) = B(1)1/k2 g3(12) = B(1)F(2)-C(1) k2 G1(2) In the top line here we must select B(1) = -k2 G2(1) so our third line is then g3(12) = -k2 G2(1) F(2)-C(1) k2 G1(2) But we can absorb our constants into F(2) and C(1) in general (or set k2 = -1) to get g3(12) = G2(1) F(2)- C(1) G1(2) and THIS then is our requirement on g3(12), and it is pretty specific. Let's examine this relating to elliptic-cylinder coordinates M&S page 17. We have there h12 = h22 = a2[ ch2(ξ1) - cos2(ξ2)] Our equations (3.5) become (h2/h1) = f1(1) g1(23)R2 => g1(23) = G1(2) (h1/h2) = f2(2) g2(13)R2 => g2(13) = G2(1) (h1h2) = f3(3) g3(12)R2 => f3(3) = 1 R-4 = f1g1f2g2 = f1(1)G1(2)f2(2)G2(1) = [f1(1)G2(1)][G1(2)f2(2)] ≡ J1(1) J2(2) 1 = f1(1) g1(23)R2 => g1(23) = G1(2) 1 = f2(2) g2(13)R2 => g2(13) = G2(1) h12 = f3(3) g3(12)R2 => f3(3) = 1 1 = f1(1) G1(2)R2 1 = f2(2) G2(1)R2 h12 = g3(12)R2 R-4 = f1g1f2g2 = f1(1)G1(2)f2(2)G2(1) or R-2 = f1(1) G1(2) = f2(2) G2(1) = h12/ g3(12) One of these equations says f1(1) G1(2) = h12/ g3(12) => g3(12) = h12 /[ f1(1) G1(2)] But we know h12 for this system so we get g3(12) = a2[ ch2(ξ1) - cos2(ξ2)] /[ f1(1) G1(2)] = a2[ ch2(ξ1)/ [ f1(1) G1(2)] - a2 cos2(ξ2)/ [ f1(1) G1(2)] and this fits our mold which was g3(12) = G2(1) F(2)- C(1) G1(2) so that is WHY this system is R separable. Now how can we compute R or show it to be constant? If we try M&S suggestion that f1 = f2 = 1 we get this 1 = G1(2)R2 1 = G2(1)R2 h12 = g3(12)R2 R-2 = G1(2) = G2(1) This last line says R = constant, so let's set it to 1 and we get 1 = G1(2) 1 = G2(1) h12 = g3(12) This in turn says G's are one, so we get this solution: f1 = f2 = f3 = 1 g1 = g2= 1 g3 = h12 = a2[ ch2(ξ1) - a2 cos2(ξ2)] R = 1 Our cofactors are M1 = g1 /f2f3 = 1 = M2 M3 = g3 /f1f2 = g3= a2[ ch2(ξ1) - a2 cos2(ξ2)] // agree M&S p 17 *********************************************** Suppose we now limit ourselves to h1 = h1 which means the 2D is conformal. Then what? h1= h2 = h1(12) h3= 1 H = h12 H/h12= 1 H/h22 =1 H/h32 = h12 (1) Is there a problem with (3.5) ? (H/[R2hn2]) = fn(n)gn(≠n) (h2/h1) = f1(1) g1(23)R2 => g1(23) = G1(2) (h1/h2) = f2(2) g2(13)R2 => g2(13) = G2(1) (h1h2) = f3(3) g3(12)R2 => f3(3) = 1 where we set f3(3) = 1 since below we will find that R = R(12) 1 = f1(1) g1(23)R2 1 = f2(2) g2(13)R2 h12 = f3(3) g3(12)R2 1 = f1(1) G1(2)R2 1 = f2(2) G2(1))R2 h12 = g3(12)R2 R-2 = f1(1) G1(2) = f2(2) G2(1)) = g3(12)/h12 Our solution of (7.5) condition is that g3(12) = -C(1) G1(2) = h12 f1(1) G1(2) => -C(1) = h12 f1(1) which says h1 must be h1(1)! If this were the case, what happens to h1 = h2 ? We conclude that h2 = h2(1) as well. Meanwhile we then have R-2 = g3(12)/h12 = h12 f1(1) G1(2)/h12 = f1(1) G1(2) Our three (3.5) equations then say 1 = f1(1) G1(2)/ f1(1) G1(2) OK 1 = f2(2) G2(1)) / f1(1) G1(2) already knew this h12 = g3(12) / f1(1) G1(2) already know this too So nothing much happening here. (h1h2) = g3(12)R2 = [W(2)G2(1) +V(1)G1(2)] R2 = [W(2)G2(1) +V(1)G1(2)] / = W(2)G2(1)/ + V(1)G1(2)/ = W'(2) G2(1)/ + V'(1) G1(2) / = W'(2) + V'(1) = W'(2) T1(1) + V'(1) T2(2)