Lai's grad-v method p 54-66

Lai's (v) method p 54-66 PhL 2.7.12 Here I attempt to generalize Lai's method for computing (Δv) to arbitrary orthogonal systems. The big problem is that in general writing dn = Σj anjj gives a big mess for anj, Since Lai's method hinges critically on the idea that v = vn n and thus dv = dvn n + vn dn, you cannot get anything simple for this second term. For polar, cylinder and spherical, the anj are trivially simple which is why his method works in those cases. This was all just due diligence to make sure I was not missing some "super trick" that makes everything easy. There is not such trick. NOTE ADDED 3.6.12. I learned how to compute den from the affine connection, and then showed in Appendix G that Lai's (v) Method is the same as my covariant tensor method, so all is well! 1. Writing x as a linear combination of the n 1 2. Handling the RHS of dv = T dx, no problem. 4 3. Plan A: Cartesian basis LHS = RHS (makes a big mess). 5 4. Finding the expansion for dn 7 4(a). Finding the expansion for dn , revisited (2.11.12) 8 5. Plan B : The Phil method for LHS of dv = T dx. 8 I think Lai uses the same general method for computing (v) from dv = (v)dr in his three systems of interest. Lai does two things which do not generalize except in a very ugly way to general curvilinear systems. (a) He expands x on the n , the subject of the first section below. As it turns out, his only reason for doing this is to obtain an expansion for dx which I know how to do by other means. Lai does this only because he has avoided talking about the metric tensor, scale factors, and expansions of vectors. So this is not much of a problem for me. (b) He expands dn on the n. He then writes v = vn n and then dv = dvn n + vn dn and this then is the LHS of the equation dv = T dx. So it is here in this dv that he needs his expansion of dn on the n. This is not something I have a simple workaround for. I show in Section 4 below that in general this expansion is very ugly and this is where the "affine connection" complication enters his method. I have to "give up" on generalizing his dn as a useful tool (though I do it to see the mess it is), but I am able to treat dv by a different method and thereby obtain his results, as I have shown in other documents, as shown in Section 5 below. 1. Writing x as a linear combination of the n Lai begins each of his little sections by expanding x. For example, [ my x and n are Lai's r and en ] polar: x = r 1 // 1 = Lai's er = cylindrical: x = r 1 + z 3 spherical: x = r 1 He then uses these little equations to compute dx which he then inserts into dv = (v)dx. In doing this, he is obliged to also compute things like dn. For these simple systems, we all know how to compute d for example, but this starting position / approach does not lend itself at all to fancier curvilinear systems. In what follows, we shall explore this kind of expansion for x in a general system. But the upshot of this beginning for Lai is merely to obtain the following equation to insert into dv = (v)dx : dx = Σn dx'n en = Σn dx'n h'n n . But for me, this equation is obvious (in terms of tensor doc, just a routine vector expansion) since (my) en are the tangent base vectors and since dxi is a true contravariant vector. In effect Lai uses my developmental notation. The Lai book never mentions the metric tensor, nor does it mention scale factors, so Lai has to improvise. His three systems polar, polar x Z and spherical all have non-trivial parts which are really "spherical", and for a spherical in ANY number of dimensions we are always going to have x = r1 . So these are very special systems indeed. Even though we don't need it, let's see what we get expanding x on the en in a general system. I show below my original notes on this subject, leaving in the "failed plans" (which are always instructive). Everything is pretty brief here. Assume x-space is Cartesian, but do not assume x' are orthogonal coordinates. Work in developmental notation. Consider this expansion of the vector x in x-space, using "my" notation: x = an en an = En x => x = (En x) en Since we assume F is non-linear, ai is not x'i. Plan A. Now this dot product can be evaluated in Cartesian x-space and we get that an = (Enx) = (En)i xi = giaRna xi = Rna xa // (En)i = g'naSia = giaRna td p 53 This is not useful to me, because I want to see things in terms of the x'i coordinates, not xi. Plan B. So instead, let's use the td sec 6(a) result that En ≡ g'ni ei, then we have an = En x = g'ni ei x but again, I don't want to see xi stuff. Plan C. Suppose instead we try an = En x = E'n x' but this is wrong because x is not a true vector (!!!). [ As noted many times in tensor doc.] Plan D. The only connection I have to x' is this: x' = F(x) and x = F-1(x') the latter one always being "the simpler" equation direction. So then I have to write an = En x = [g'ni ei] [F-1(x')] = [g'ni ei] [x(x')] = g'ni ei x(x') I am completely stumped as usual. In sphericals we have, for example (td p 29) x(x') = : x1 = r sinθcosφ x2 = r sinθsinφ x3 = rcosθ e1 = e2 = r e3 = rsinθ g'ni = g'nn δni = h'n-2 δni For an orthogonal system, this is my result ( where I specialize now to an orthogonal x' system) an = g'ni ei x(x') = h'n-2 en x(x') = h'n-2 [en]i xi(x') = h'n-2 Sin xi(x') and this just gives a big mess. The mess is very likely correct, but things are very indirect. My main point here is that you cannot avoid involving the full non-linear transformation equations x = x(x'). Example: I entered things for ellipsoidal coordinates, just to pick a complicated system, and I found the three an coefficients to be [ see lai ellipsoidals.mws which has no comments ] But it sure looks to me that an = ± ξn / h'n2 so why does this come out being so "simple"? No doubt there is some simple reason for this, but that is not the point here. In any event, perhaps we then have for our expansion x = Σn an en = Σn (ξn / h'n2) h'n n = Σn (ξn / h'n) n I am not vouching for the signs here, but the coefficients are pretty complicated functions of the coordinates, as you see in blue above. If you used an expansion like this as a Lai-like starting point, in computing dx you would have to deal for example with dn and with d (ξn / h'n)which is a big mess. Conclusion: We know of course that is always possible to write x = Σn αn n where αn is some function of the x' coordinates. This is so because set n is complete. It happens that in polar, cylindrical and spherical coordinates this sum is very simple. But in a general system it is not going to be simple. When you then go try to write dx = Σn dαn n + Σn αn dn (as Lai always does), you are going to find that dαn will be a huge mess and the method will run aground quickly. Now let's look at his spherical case and try to generalize each of his equations, starts p 61. 2. Handling the RHS of dv = T dx, no problem. As noted in "Lai's expansion-of-x starting point..doc" we can just skip to 2.35.11 which has the following generalization, [ my x and n are Lai's r and en ] [ we will both use T ≡ (v) ] dx = Σn dx'n en = Σn dx'n h'n (x')n 2.35.11 We then skip down to his section (ii) where he computes (v). I would write then v = Σn v'n (x')n 2.35.16 He next computes T dx = T Σn dx'n h'n (x)n = Σn dx'n h'n T (x')n 2.35.17 RHS Next, insert the completeness for (x')n twice to get T (x')n = { (x')k (x')Tk }T { (x')m (x')Tm } (x')n = (x')k { (x')Tk T (x')m }(x')Tm (x')n = (x')k T(x')km (x')Tm (x')n T(x')km ≡ (x')Tk T (x')m = (x')k T(x')km δm,n = (x')k T(x')kn = TT(x')nk (x')k 2.35.18 and you do see the transposed elements like Tθr in his first equation. So insert this result into 2.35.17 to get T dx = Σn dx'n h'n T (x')n = Σnk dx'n h'n TT(x')nk (x')k = Σk { Σn dx'n h'n TT(x')nk } (x')k 2.35.19 and this then is our RHS for equation dv = (v)dx = T dx . To this point, I have successfully generalized every equation that Lai has presented. The problem is with the LHS dv ! Note in Passing. If v is a true vector under non-linear F, then dv is not a true vector under same! Proof: v' = R v => dv' = R dv + dR v I am used to saying that dx is a true vector but x is not. So yes, I knew this, but it just never came up anywhere until now. 3. Plan A: Cartesian basis LHS = RHS (makes a big mess). I now run into trouble. He next calls upon 2.35.10 which I want to avoid because it concerns the d(x')k objects and I don't have an expansion for this. As an alternative step for dv, I can write various things v = S v' v is a true vector dv = S dv' + dS v' dvi = Sij dv'j + (dS)ij v'j Comment: In 2.35.19 above we have T dx expanded on the (x')k basis. Lai uses that 2.35.10 stuff to come up with an expansion of dv on this same basis. I don't know how to do that because the only way I know to get the (x')k basis to appear is to expand a true vector like v. [ But see mess below. ] So I will now deviate from his path as follows: I will instead obtain the Cartesian component of each side and set these equal. Thus, my RHS will be this: [T dx ]i = Σk { Σn dx'n h'n TT(x')nk } [ (x')k]i RHS dvi = Sij dv'j + (dS)ij v'j LHS First I need this fact [ (x')k]i = [ (1/h'k) e(x')k]i = (1/h'k) [e(x')k]i = (1/h'k) Sik Then I have this for my final equation Sij dv'j + (dS)ij v'j = dx'n h'n TT(x')nk (1/h'k) Sik and the task is then to solve this for TT(x')nk . What is this (dS)ij object? (dS)ij = (∂Sij/∂x'n) dx'n = some kind of affine connection deal dv'j = (∂v'j/∂x'n) dx'n 2.35.21 Then our equation reads Sij (∂v'j/∂x'n) dx'n + (∂Sij/∂x'n) dx'n v'j = dx'n h'n TT(x')nk (1/h'k) Sik On each side we have a Σn implied, and I think we have to have the coefficient of dx'n match on the two sides, so we can then remove the Σn sum and cancel the dx'n to get ΣjSij (∂v'j/∂x'n) + Σj(∂Sij/∂x'n) v'j = Σk h'n TT(x')nk (1/h'k) Sik n,i are fixed Now untranspose T and move Sik to the left ΣjSij (∂v'j/∂x'n) + Σj(∂Sij/∂x'n) v'j = h'n Σk Sik (1/h'k)T(x')kn n,i are fixed Then define a diagonal matrix HINVkk' = (1/h'k)δk,k' to get ΣjSij (∂v'j/∂x'n) + Σj(∂Sij/∂x'n) v'j = h'n Σkk' Sik HINVkk'T(x')k'n = h'n [ S HINV T(x')]in Now define Xjn ≡ (∂v'j/∂x'n)(1/h'n) Yin = Σj(∂Sij/∂x'n) v'j (1/h'n) then we have (SX)in + Yin = [ S HINV T(x')]in SX + Y = S HINV T Apply R from the left X + RY = HINV T Apply H ≡ HINV-1 from the left HX + HRY = T and then we finally have our solution T = HX + HRY T(x)ij = HikXkj + HikRkaYaj = h'i δikXkj + etc etc But now I have v'j and not v'j so everything has to be processed a lot more, so I give up. I am sure if I followed this through in gory detail, I would get the same result obtained below in Plan B. 4. Finding the expansion for dn For sphericals this is given manually in 2.35.10, and I think I have a way, albeit ugly, of writing this expansion in the general case. I can write from tensor doc en ≡ Se'n n = en/h'n = (1/h'n) Se'n dn = d(1/h'n) Se'n + (1/h'n) dS e'n // since e'n is a constant δ thing = d(1/h'n) en + (1/h'n) dS R en = d(1/h'n))(h'n n ) + (1/h'n) dS R (h'n n ) = - h'n-2 dh'n (h'n n ) + dS Rn = - (dh'n/h'n) n + dS Rn so at least I have something for this mess. Then you have to say dh'n = (∂h'n/ ∂x'k) dx'k dS = (∂S/ ∂x'k) dx'k and then we have dn = - (1/h'n) (∂h'n/ ∂x'k) dx'k n + (∂S/ ∂x'k) dx'k Rn Suppose we want to write this in the form dn = Σi Cni i => Cni = dn i Then Cni = { - (1/h'n) (∂h'n/ ∂x'k) dx'k n + (∂S/ ∂x'k) dx'k Rn } i = Σk dx'k [ - (1/h'n) (∂h'n/ ∂x'k) δn,i + i (∂S/ ∂x'k) Rn ] This then would give the expansion coefficients in dn = Σi Cni i for an arbitrary curvilinear system. As you can see, that "affine connection" type term always appears somewhere, and here it appears in the three index non-tensor (∂Sij/ ∂x'k). This is all quite ugly. This would be a horrible way to find the matrix T for a general system. Conclusion: I did my due diligence on the Lai approach. It is fine for these three systems where we have simple expressions for the dn, but it just does not generalize nicely. 4(a). Finding the expansion for dn , revisited (2.11.12) From Section 3 (b) of tensor doc we have en ≡ Se'n = Σi Sin e'i => n = h'n-1 Σi Sin e'i e'n ≡ Ren = Σi Rin ei => e'n = Σi Rinh'i i Then dn = d(h'n-1) Σi Sin e'i + h'n-1 Σi dSin e'i = [ d(h'n-1) Σi Sin + h'n-1 Σi dSin ] e'i = [ d(h'n-1) Σi Sin + h'n-1 Σi dSin ] Σj Rjih'j j = Σj [ d(h'n-1) Σi Rji Sin + h'n-1 Σi Rji dSin ] h'j j = Σj [ d(h'n-1) (RS)jn + h'n-1 Σi Rji dSin ] h'j j = Σj { [ d(h'n-1) δj,n + h'n-1 Σi Rji dSin ] h'j} j So it is still a mess. You still have to go off and compute dSin = (∂kSin)dxk. Again, I could go off and make Maple do all this and make it all work, but I already have a viable method! 5. Plan B : The Phil method for LHS of dv = T dx. Let's go back now and try again to get dv expressed in the x'-basis. It is HERE that we encounter what I call the affine connection issue. That is, we need to write dvi = (∂avi) dxa = (∂avi) Sab(x')dx'b since dxa is a true vector dv = dvi (x)i = dvi ui = dvi Qki (x')k Qki ≡ h'k-1 Rki from "a simple theorem" Then we end up with the following LHS for our equation dv = T dx dv = (∂avi) Sab(x')dx'b Qki (x')k LHS T dx = Σk { Σn dx'n h'n TT(x')nk } (x')k RHS At this point then we can equate the x'-space components of the two sides to get (∂avi) Sab(x')dx'b Qki = dx'n h'n TT(x')nk k fixed (∂avi) San(x')dx'n Qki = dx'n h'n TT(x')nk change b sum to n (∂avi) San(x')Qki = h'n TT(x')nk coeff's of dx'n must be the same, now k,n fixed And then we get our solution written as T(x')kn = h'n-1(∂avi) San(x')Qki(x') h'n T(x')kn = STna(x') (∂avi) Qki(x') = STna(x') (∂avi) h'k-1 Rki = STnc (∂cvd) h'k-1 Rkd Now comes the affine connection problem. We have to write out (∂avi) in primed stuff! Looking at my tensor doc section, I show there that ∂cvd = Ric [(∂'iRjd) v'j + Rjd (∂'iv'j)] which then gives h'n T(x')kn = STnc Ric [(∂'iRjd) v'j + Rjd (∂'iv'j)] h'k-1 Rkd Then write STnc = Scn → Scn = Rnc and then Rnc Ric = δni and we get h'n T(x')kn = [(∂'nRjd) v'j + Rjd (∂'nv'j)] h'k-1 Rkd = [(∂'nRjd) Rkd v'j + Rkd Rjd (∂'nv'j)] h'k-1 = [(∂'nRjd) Rkd v'j + δkj (∂'nv'j)] h'k-1 = [(∂'nRjd) Rkd v'j + (∂'nv'k)] h'k-1 so in the end we get T(x')kn = h'n-1 h'k-1 [(∂'nv'k) + (∂'nRjd) Rkd v'j ] T(x')kn = h'n-1 h'k-1 [(∂'nv'k) + (∂'nRad) Rkd v'a ] Now replace kn→ij to get T(x')ij = h'j-1 h'i-1 [(∂'jv'i) + (∂'jRac) Ric v'a ] T(x')ij = h'i-1 h'j-1 [(∂'jv'i) + (∂'jRkd) Rid v'k ] and this agrees with my tensor doc result for Pij before converting to v' components.