Derive Hobson p 479 B,C and D

Derive Hobson p 479 B,C and D and p 480 A PhL 11.10.10 I have flailed at this for perhaps 6 hours, so it is time to move it this separate document where I can find out what I am doing wrong. // It is all done right now, there was a lot to do. I summarize the work done here in the main Hobson notes doc. Overview (2 pages) 1 Problem 1: 2 Problem 2: 10 __________________________________________________________________________________ Overview (2 pages) The conical harmonics are given in Hobson p 478 B as: rn En,p(μ) En,p(ν) = { 1 x y z xy xz yz xyz } κ" Πs=1J [ x2/ ρs2 + y2/(ρs2-h2) + z2/(ρs2-k2)] K K L M M L N N // class e o o o e e e o // n = even or odd // Hobson p 478 A 1 2 2 2 3 3 3 4 // species s where J = (n + s - 1)/2 which is always an integer for the allowed n even or odd values shown. For any given n, there will be 2n+1 harmonics and those will be associated with 4 of the 8 forms shown, depending on the parity of n. The number of harmonics associated with each class is of course the number of E functions associated with each class, and those counts are given by (Hobson notation here) K r+1 where r = n/2 if n even, else (n-1)/2 L n-r M n-r N r 2n+1 solutions!! and these are labeled in some manner by the p index on En,p(μ). We know that we can write the E functions in this manner n even: Kn,p = Πs=1J (μ2- ρs2) J = (n-0)/2 = n/2 Ln,p = μ Πs=1J (μ2- ρs2) J = (n-1-1)/2 = (n-1)/2 Mn,p = μ Πs=1J (μ2- ρs2) J = (n-1-1)/2 = (n-1)/2 Nn,p = Πs=1J (μ2- ρs2) J = (n-2)/2 = (n-2)/2 n odd: Kn,p = μ Πs=1J (μ2- ρs2) J = (n-0-1)/2 = (n-1)/2 Ln,p = Πs=1J (μ2- ρs2) J = (n-1)/2 = (n-1)/2 Mn,p = Πs=1J (μ2- ρs2) J = (n-1)/2 = (n-1)/2 Nn,p = μΠs=1J (μ2- ρs2) J = (n-2-1)/2 = (n-3)/2 so we "know all there is to know" about the Lamé functions once we know the roots ρs. We know the exact form for all the classes, and we know the number of harmonics (number of E functions) in each class. The purpose of the calculation here is to determine the θs values which make 2(harmonic) = 0, and then we know the values ρs2 = a2 + θs of the (squared) roots. There is perhaps some issue with how we allocate the roots to the various functions in a given class. [ See META notes about θ(k) for clarification of this issue, there is no ambiguity at all about how to allocate the roots. ] In Problem 1 we compute this quantity 2 [ {1} { Πs=1J [x2/(a2+θs) + y2/ (b2+ θs) + z2/( c2+ θs)] } ] which is simply a product of quadratic factors There are J quadratic factors. J is a function of harmonic degree n as shown above, but that connection is not needed here. The above [ ] shows the first family of "conical harmonics" showing in Hobson p 478B ( K class, n even). We shall then set the result to 0 to get a system of equations which we can then solve to find the root values θs and then ρs2 = a2 + θs which are the actual squared roots of the Lamé functions in the K class for even n. In the main doc, we do cases n = 2 and n = 4 explicitly showing results for all classes. Problem 1 is fairly complicated, so it is presented in 13 numbered subsections. In Problem 2 we show how the result gets modified to apply to any of the other 7 harmonic families. There we need to compute 2 [ { x y z xy xz yz xyz }{ Πs=1J [x2/(a2+θs) + y2/ (b2+ θs) + z2/( c2+ θs)] } ] and set to 0 and deduce the θs in these cases. We treat n=3 explicitly in the main notes, for all four classes K,L,M,N. Problem 2 is much easier than Problem 1 since it uses the hard-earned results of Problem 1. Also, we are able really to do all the cases at once assuming a certain xαxβxγ form for the primitives. ________________________________________________________________________________ Problem 1: What happens if we apply 2 to { Πs=1J [x2/(a2+θs) + y2/ (b2+ θs) + z2/( c2+ θs)] } ? (1) Let's define ai = a,b,c and write this as 2 Πs=1J [ Σi xi2/(ai2 + θs) ] = Σj ∂j2 { Πs=1J [ Σi xi2/(ai2 + θs) ] } (2) now define fs = Σi xi2/(ai2 + θs) Then our problem is to compute 2 Πs=1J [fs] = Σj ∂j2 { Πs=1J [fs] } (3) Compute the first derivative. We want to compute this object, ∂j{ Πs=1J [fs] } It seems to me this is just a multifactor product rule and we get this ∂j (Πs=1J fs) = ∂j ( f1f2....fJ) = f1' f2..... + f1f2'f3.... etc = Σk=1J (∂j fk) (Πs≠k fs) Just to make sure, let's write out the last sum = (∂j f1) f2f3.... + (∂j f2) f1f3f4... + etc I am pretty sure I have step 3 correct. (4) Apply the second derivative: 2 Πs=1J [fs] = Σj ∂j2 { Πs=1J [fs] } = Σj ∂j { ∂j (Πs=1J [fs]) } Now insert into this our result from step 3 = Σj ∂j { Σk=1J (∂j fk) (Πs≠k fs) } = Σk=1J Σj ∂j { (∂j fk) (Πs≠k fs) } For the moment, we just think of {..} containing just two factors, and we use the product rule to get = Σk=1J Σj [(∂j2 fk) (Πs≠k fs) + (∂j fk) ∂j (Πs≠k fs) = Σk=1J [2fk (Πs≠k fs) + Σj (∂j fk) ∂j (Πs≠k fs) ] (5) We now need to very carefully compute this quantity ∂j (Πs≠k fs) = ∂j( f1f2f3 f5f6....) where we assume for the moment that k = 4. We continue: = (∂jf1)f2f3 f5f6... + f1(∂jf2)f3 f5f6... + ... but the (∂jf4) term is missing. = Σn≠4 (∂jfn) (Πs≠n,s≠4 fs) To verify this result, let's go the other way and expand it = (∂jf1) (Πs≠1,s≠4 fs) + (∂jf2) (Πs≠2,s≠4 fs) + ... This certainly seems right. It seems then that instead of using k = 4, we could use general k = k and say ∂j (Πs≠k fs) = Σn≠k (∂jfn) (Πs≠n,s≠k fs) (6) If we install this result back into our computation we get 2 Πs=1J [fs] = Σk=1J [2fk (Πs≠k fs) + Σj (∂j fk) ∂j (Πs≠k fs) ] = Σk=1J [2fk (Πs≠k fs) + Σj (∂j fk) Σn≠k (∂jfn) (Πs≠n,s≠k fs) ] = Σk2fk (Πs≠k fs) + Σk Σj (∂j fk) Σn≠k (∂jfn) (Πs≠n,s≠k fs) That second term is certainly looking pretty opaque. There are three sums and a product. Only the j sum is 1,2,3. It certainly seems that we should rearrange the sums in this way = Σk (2fk) (Πs≠k fs) + Σk Σn≠k Σj (∂j fk)(∂jfn) (Πs≠n,s≠k fs) and we could write (∂j fk)(∂jfn) = fkfn . (7) Let's work only on the first term for the moment. We have 2fk = 2[ Σi xi2/(ai2 + θs)] = Σi (∂j2xi2)/(ai2 + θs) ∂jxi2 = 2xiδij => ∂j2xi2 = Σj ∂j 2xiδij = Σj 2δijδij = 2 δii = 2 In passing, we have just shown that ∂jfs = Σi (∂jxi2)/(ai2 + θs) = Σi 2xiδij/(ai2 + θs) = 2xj / (aj2 + θs) Then we have 2fk = Σi (∂j2xi2)/(ai2 + θk) = Σi 2/(ai2 + θk) = 2 { Σi 1/(ai2 + θk)} Therefore our "first term" is this: 2 Πs=1J [fs]first term = Σk2fk (Πs≠k fs) = = Σk2 { Σi 1/(ai2 + θk)} (Πs≠k fs) = 2 Σk Σi [1/(ai2 + θk)] (Πs≠k fs) Let's write out some terms: = 2 1/(ai2 + θ1) * f2f3.... + 2 1/(ai2 + θ2) f1 f3f4.... + ... There are a total of J terms (J = Hobson's m). So my first term is exactly twice the first line of Hobson p 477 B. (8) If his result and my results so far are both correct, it must be possible to show that 2 Πs=1J [fs]second term = Σk Σn≠k Σj (∂j fk)(∂jfn) (Πs≠n,s≠k fs) is equal to twice Hobson's second line in B. Now we know that from step 7 that ∂jfk = 2xj / (aj2 + θk) ∂jfn = 2xj / (aj2 + θn) so we then have 2 Πs=1J [fs]second term = Σk Σn≠k Σj (∂j fk)(∂jfn) (Πs≠n,s≠k fs) = Σk Σn≠k Σj [2xj / (aj2 + θk)][ 2xj / (aj2 + θn] (Πs≠n,s≠k fs) = 4 Σk Σn≠k Σj [xj2 / [(aj2 + θk) (aj2 + θn) ] (Πs≠n,s≠k fs) At least we are getting denominators that look like Hobson's second line B denominators. Let's now define Fkn ≡ Σj [xj2 / [(aj2 + θk) (aj2 + θn) ] and we note that this thing is symmetric under n↔k. So we have 2 Πs=1J [fs]second term = 4 Σk Σn≠k Fkn (Πs≠n,s≠k fs) The product (Πs≠n,s≠k fs) is also symmetric (Πs≠n,s≠k fs). (9) So what can we say about sum = Σk Σn≠k Snk where S is symmetric? If we draw a 2D sum space picture with axes k and n, we get a sum over the entire square less the diagonal. We can think of this as two triangles. [ Note: in usual matrix pictures, the n axis points down but here it points up. In a matrix picture, symmetric is about a \ directed diagonal, but in our picture symmetric is about the / directed axis. So the numbers in the lower triangle are the same as those in the upper, reflected in the / diagonal. ] Breaking the sum into the two triangular contributions. We then get two ways to write each triangle lower triangle sum = Σk=1J Σn=1k-1Snk = Σn=1J Σk=n+1JSnk upper triangle sum = Σk=1J Σn=k+1JSnk = Σn=1J Σk=1n-1Snk We can then pick two and say sum = lower triangle + upper triangle = Σk=1J Σn=1k-1Snk + Σn=1J Σk=1n-1Snk We can swap the dummy summation indices in the second sum and say sum = Σk=1J Σn=1k-1Snk + Σk=1J Σn=1k-1Skn = Σk=1J Σn=1k-1 [Snk + Skn] But at this point we can use the assumed symmetry of S to get sum = 2 Σk=1J Σn=1k-1 Skn = 2 Σk Σn<k Skn Suppose we swap the summation indices here to get sum = 2 Σn Σk<n Snk If we then use the symmetry of S, we find that sum = 2 Σk Σn<k Skn = 2 Σn Σk<n Skn (*) which just tells us we can swap the two indices with abandon. (10) Now the big question: does this help us in some way? We were doing this sum, which we then rewrite: 4 Σk Σn≠k Fkn (Πs≠n,s≠k fs) = 8 Σk Σn<k Fkn (Πs≠n,s≠k fs) Let's write out some of the terms: Σk Σn<k Fkn (Πs≠n,s≠k fs) = We see that there is no term with k= 1 because you cannot then have n < 1. So start with k = 2 k = 2 n = 1 k = 3 n = 1 k = 3 n = 2 = F21 ( * * f3f4.....fJ) + F31 ( * f2 *f4.....fJ) + F32 ( f1* *f4.....fJ) + ... This does look promising. Recall that Fkn ≡ Σj [xj2 / [(aj2 + θk) (aj2 + θn) ] so on Hobson's second like, that first {...} thing is exactly F12 . His notation really is sufficient to show the full result I think. How many terms are there? How many ways can you pick 2 out of J? So (J,2) is the answer. That would be (1/2) J!/(J-2)! which is J(J-1)/2 terms. So we how have 2 Πs=1J [fs]second term = 4 Σk Σn≠k Fkn (Πs≠n,s≠k fs) = 8 Σk Σn<k Fkn (Πs≠n,s≠k fs) And in total we have 2 Πs=1J [fs] = 2 Σk Σi [1/(ai2 + θk)] (Πs≠k fs) + 8 Σk Σn<k Fkn (Πs≠n,s≠k fs) If we require this be 0, our condition is then Σk Σi [1/(ai2 + θk)] (Πs≠k fs) + 4 Σk Σn<k Fkn (Πs≠n,s≠k fs) = 0 (*) which we write as Σk { Σi [1/(ai2 + θk)] (Πs≠k fs) + 4 Σn<k Fkn (Πs≠n,s≠k fs) } = 0 J terms J(J-1)/2  terms where Fkn ≡ Σj [xj2 / [(aj2 + θk) (aj2 + θn) ] This is Hobson p 479 B, finally ! (11) While we are here, let's also derive Hobson p 479 C. fn - fk = Σi xi2/(ai2 + θn) – Σi xi2/(ai2 + θk) = Σi xi2 [(ai2 + θk) - (ai2 + θn) ] / [(ai2 + θk) (ai2 + θn)] = Σi xi2 [θk-θn] / [(ai2 + θk) (ai2 + θn)] = [θk-θn] { Σi xi2 / [(ai2 + θk) (ai2 + θn)] } = [θk-θn] Fkn So we have that Fnk = { Σi xi2 / [(ai2 + θn) (ai2 + θk)] } = (fn - fk)/( θk-θn) // n↔k symmetric and if we set n = 1 and k = 2, we get Hobson p 479 C. (12) Now we want to get to equation D. Suppose we stuff in this result (11) into (10) (*): Σk Σi [1/(ai2 + θk)] (Πs≠k fs) + 4 Σk Σn<k Fkn (Πs≠n,s≠k fs) = 0 so Σk Σi [1/(ai2 + θk)] (Πs≠k fs) + 4 Σk Σn<k [(fn - fk)/( θk-θn)] (Πs≠n,s≠k fs) = 0 Consider this second term some more: 4Σk Σn<k [(fn - fk)/( θk-θn)] (Πs≠n,s≠k fs) = 4Σk Σn<k [fn /( θk-θn)] (Πs≠n,s≠k fs) - 4Σk Σn<k [fk/( θk-θn)] (Πs≠n,s≠k fs) = 4Σk Σn<k [1/( θk-θn)] (fn Πs≠n,s≠k fs) - 4Σk Σn<k [1/( θk-θn)] (fk Πs≠n,s≠k fs) = 4Σk Σn<k [1/( θk-θn)] ( Πs≠k fs) - 4Σk Σn<k [1/( θk-θn)] (Πs≠n fs) If we look back at the k-n triangle situation again, we see that Σk Σn<k is a sum over the lower right triangle. We can rewrite this same sum in this way Σk Σn<k = Σn Σk>n Lets do this in our second term above to get = 4Σk Σn<k [1/( θk-θn)] ( Πs≠k fs) - 4 Σn Σk>n [1/( θk-θn)] (Πs≠n fs) Now swap the dummy index names in the second term and rewrite = 4Σk Σn<k [1/( θk-θn)] ( Πs≠k fs) - 4 Σk Σn>k [1/( θn-θk)] (Πs≠k fs) = 4Σk Σn<k [1/( θk-θn)] ( Πs≠k fs) + 4 Σk Σn>k [1/( θk-θn)] (Πs≠k fs) = 4Σk (Σn<k + Σn>k ) [1/( θk-θn)] ( Πs≠k fs) = 4Σk Σn≠k [1/( θk-θn)] ( Πs≠k fs) So now lets rejoin our two terms again to say Σk Σi [1/(ai2 + θk)] (Πs≠k fs) + 4 Σk Σn<k [(fn - fk)/( θk-θn)] (Πs≠n,s≠k fs) = 0 // before Σk Σi [1/(ai2 + θk)] (Πs≠k fs) + 4Σk Σn≠k [1/( θk-θn)] ( Πs≠k fs) = 0 // now We can factor out some things and write this now as Σk (Πs≠k fs) { Σi [1/(ai2 + θk)] + 4Σk Σn≠k [1/( θk-θn)] } = 0 And now, once again -- finally --, we can solve this equation if we require that {...} = 0, that is,. Σi [1/(ai2 + θk)] + 4Σk Σn≠k [1/( θk-θn)] = 0 Notice how we have avoided the zero denominator situation in the second term. Rewrite as Σi [1/(ai2 + θk)] + 4Σn≠k [1/( θk-θn)] = 0 k = 1...J where we have slightly changed our meaning of Σn≠k so it now means a double sum excluding the diagonal elements. The first of these equations is Σi [1/(ai2 + θ1)] + 4/( θ1-θ2) + 4/( θ1-θ3) + ... 4/( θ1-θJ) = 0 and this then is the first equation you see in Hobson p 479 D. There are J of these equations, and there are J variables which are θ1....θJ , so it would seem that there is a unique solution for the set {θi}, and that of course is what we have been trying to do. This is not a simple Cramer's rule situation since our variables are all in denominators, so each equation is highly non-linear. Now let's backtrack just a bit. We had 2 Πs=1J [fs] = 2 Σk Σi [1/(ai2 + θk)] (Πs≠k fs) + 8 Σk Σn<k Fkn (Πs≠n,s≠k fs) We then showed that 4 Σk Σn<k Fkn (Πs≠n,s≠k fs) = 4Σk Σn≠k [1/( θk-θn)] ( Πs≠k fs) = 0 Therefore we know that 2 Πs=1J [fs] = 2 Σk Σi [1/(ai2 + θk)] (Πs≠k fs) + 8Σk Σn≠k [1/( θk-θn)] ( Πs≠k fs) = 2 Σk (Πs≠k fs) { Σi [1/(ai2 + θk)] + 4Σk Σn≠k [1/( θk-θn)] } so we can record this as the action of 2 on our Πs=1J [fs], which we will use below. ________________________________________________________________________________- Problem 2: We want here to look at all the other cases { x y z xy xz yz xyz } Πs=1J [fs] . In each case, we can thing or our harmonic as the product of a first term {...} times the second term. We already know certain very useful facts from Problem 1: ∂j Πs=1J[fs] = Σk(∂j fk) (Πs≠k fs) // from (3) ∂j2 Πs=1J[fs] = 2 Σk (Πs≠k fs) { Σi [1/(ai2 + θk)] + 4Σk Σn≠k [1/( θk-θn)] } In the first result we can make use of ∂jfk = 2xj / (aj2 + θk) => ∂j Πs=1J[fs] = 2Σkxj / (aj2 + θk) * (Πs≠k fs) Now let's roll out our vector identity which says (you can here it being trundled out of TK) 2(uv) = u2(v) + v2(u) + 2u v and apply it here to say 2({..}Πs=1J [fs]) = {..}2(Πs=1J [fs]) + Πs=1J [fs]2({..}) + 2{..} Πs=1J [fs] Now the good news is that 2({..}) = 0 for all these primitive function, so that term goes away, 2({..}Πs=1J[fs]) = {..}2(Πs=1J[fs]) + 2{..} Πs=1J[fs] = {..} [ 2 Σk (Πs≠k fs) { Σi [1/(ai2+θk)] + 4Σk Σn≠k [1/( θk-θn)] } ] + 2 (∂j {..}) 2Σkxj / (aj2+θk) * (Πs≠k fs) where we have now installed our "very useful facts" from above. We can move the k sum and Π factor off to the left and write this as 2({..}Πs=1J[fs]) = Σk (Πs≠k fs) {..} * { 2 Σi [1/(ai2+θk)] + 8 Σn≠k [1/( θk-θn)] + 4 [(∂j {..})/{...}] xj / (aj2+θk) } Rewrite as 2({..}Πs=1J[fs]) = 2 Σk (Πs≠k fs) {..} * { Σi [1/(ai2+θk)] + 2 [(∂j {..})/{...}] xj / (aj2+θk) + 4 Σn≠k [1/( θk-θn)] } And if we require this to be 0, we simply have Σi [1/(ai2+θk)] + 2 [(∂j {..})/{...}] xj / (aj2+θk) + 4 Σn≠k [1/( θk-θn)] k=1.2..J We can see right away that the last term is the same as we had before, so it is not affected by our changing from function {1} to function {...}. So now finally we are forced to consider the individual cases. But here is possible idea to treat them all uniformly. Let {...} = xαxβxγ where x0 = 1 and the others are the usuals. Two of the indices can be equal only if they are 0. Two non-zero indices must be different. Then we have ∂j {..} = ∂j { xαxβxγ } = δjαxβxγ + xαδjβxγ + xαxβδjγ ∂j {..}/{...} = [δjαxβxγ + xαδjβxγ + xαxβδjγ]/ (xαxβxγ) = δjα/xα + δjβ/xβ + δjγ/xγ Our middle term is then Σj=132 [(∂j {..})/{...}] xj / (aj2+θk) = 2 Σj=13 [δjα/xα + δjβ/xβ + δjγ/xγ ] xj / (aj2+θk) But now for example (δjα/xα)xj =(δjα/xj)xj = δjα so we can write this as = Σj=13 [2δjα + 2δjβ + 2δjγ ] / (aj2+θk) Our first term is just Σj=13 1 / (aj2+θk) so we can combine these two sums as Σj=13 [ 1 + 2δjα + 2δjβ + 2δjγ] / (aj2+θk) = [ 1 + 2δ1α + 2δ1β + 2δ1γ]/ (a12+θk) + [ 1 + 2δ2α + 2δ2β + 2δ2γ]/ (a22+θk) + [ 1 + 2δ3α + 2δ3β + 2δ3γ]/ (a32+θk) Now we know from our starting {...} = xαxβxγ the no two indices in the 1..3 range can be the same. Therefore, in each [....] above we have at most one 2δ term that will not vanish. Therefore, each [...] can be either a 1 or a 3. We have three terms above written on separate lines. In the first term, if one of the three indices αβγ is a 1, the bracket is 3, otherwise it is 1. Of course only one of the three indices can be a 1 as noted above. This says "if {...} contains x, then first [] = 3, otherwise first [] = 1. The second term has [] = 3 only if one of the indices is a 2, meaning our {...} has a y in it. So here is our conclusion. The sum of the first two terms in our Laplace = 0 result are these k1/(a12+θk) + k2/(a22+θk) + k3/(a32+θk) where ki = 3 if xi appears in {...}, otherwise ki = 1 // Hobson p 480 A For example, if {} = xyz, then all three k's are 3's.