Lai section 3_29 F notes

Notes on Lai Section 3.29 page 131 PhL 2.15.12 Recall that F = (x) is the "deformation tensor" and dx = F dX = (x) dX Fij = (x)ij = ∂j(X)xi tells how a "short vector" dX moves through a flow (from time t=t0 to time t=t) and ends up as short vector dx. Certain other tensors are derived from F, such as C = FTF and B = FFT (the right and left Cauchy's). B-1 and C-1 are also of interest. Overview (4 pages, 2/15/12) 1 Raw Notes: 4 _________________________________________________________________________________ Overview (4 pages, 2/15/12) Chronology of where F = (f) appears in Lai: In Section 2.28 we first encounter the (v) notation where v is v(r), and we have then dv = (v)dr => dvi = (v)ij drj = (∂jvi) drj = drj(∂jvi) = (dr ) vi => dv = (dr ) v Object (v) makes its next appearance in Section 3.3 where we have for acceleration in the Eulerian view, where a = a(x,t), a = ∂v/∂t + (v)v Still later in Section 3.7 the object F = (u) = (x) and dx = FdX first appears. Other subjects intervene, and then F reappears in Section 3.18. We can reexamine the math here for dx = FdX: dxi = (x)ij dXj ≡ (∂(X)jxi) dXj = dXj(∂(X)jxi) = (dX (X)) xi(X,t) Isochronous dx. Something Lai mentions but does not emphasize is this: Normally one would write dxi(x,t) = [∂(X)jxi(X,t)] dXj + [∂txi(X,t)] dt but in the discussion of dx = FdX, the dt term fails to appear. The reason is that dx is specified as a distance between two points at the same time t, so dt = 0. If you think of the Stakgold x,t cylinder, this dx is perpendicular to the time axis. Similarly, dX is the separation between the same two points at t = t0, so it too is perpendicular to the time axis. Sections 1,2,3,4,5 Overview Lai wants to be able to use two different curvilinear coordinate systems x↔x' and X↔X'. That is, at the start of the flow, we use curvilinear coordinate system X', but at the end we use x' . The systems x and X are both Cartesian. So we have this little picture time t time t0 x' ↔ x ← X ↔ X' In Cartesian coordinates, we start off with F = Σij [F(x,X)]ij (x)n(X)Tj where then [ F(x,X)]ij = (x)Ti F (X)j where (x)n = Cartesian unit vectors for the x system = n that is to say , and (X)n = Cartesian unit vectors for the X system = n that is to say , and From tensor doc App F (b) I know that un = Σi Sni ei en = Σi Rni ui n = (1/h'n) Σi Rni ui = (1/h'n) Σi Sin ui In the detail section below, I write this for the x and X systems as [(x')n] = (1/ h(x')n) Σi [S(x')]in [(X')n] = (1/ h(X')n) Σi [R(X')]in The expansion Lai wants to do is this one: F = Σij [F(x',X')]ij (x')n(X')Tj where then [ F(x',X')]ij = (x')Ti F (X')j I now know things like (X')j = (X')j since X space is Cartesian and so up and down indices on this F object are all the same. I also know that [ F(x',X')]ij is just a set of coefficients and these coefficients are not the components of a tensor. Even if F = (x) were a tensor, [ F(x',X')]ij would not be tensor components due to the hats. See tensor doc end of Section E (b) for more detail on this observation. The next step I do below is to expand the two "short vectors" (isochronous!) in the usual way, dx = Σn dx'n h(x')n (x')n dX = Σn dX'n h(X')n (X')n If I insert these expansions into dx = F dX and then apply (x')mT from the left to both sides (and here is where I use orthonormality of the (x')m ) I get dx'm = [h(x')m]-1 Σn dX'n h(X')n [ (x')mT F (X')n] in which my "object of interest" [ F(x',X')]mn conveniently appears on the RHS. The problem is then to solve the above equation for this object. Inserting the following into the above, dx'm = Σn (∂x'm/∂X'n) dX'n , and then matching dX'n coefficients on both sides, I find that [ F(x',X')]mn = (x')mT F (X')n = [h(x')m/ h(X')n] (∂x'm/∂X'n) This then is the big result at the end of section 5. Then in section 5a I show that F (X')n = Σm [h(x')m/ h(X')n] (∂x'm/∂X'n)] (x')m FT (x')n = Σm [h(x')n/ h(X')m] (∂x'n/∂X'm)] (X')m which are used below in computations of B and C. In Section 6 I consider the case where both systems x' and X' are cylindrical and I obtain a matrix for the object [ F(x',X')]mn using Maple which agrees with Lai's matrix p 132 3.29.12. This is the first part of my Maple program file lai 3_29 cyl cyl.mws . In Section 7 I then move on to compute B = FFT (the left Cauchy) using those section 5a results above, and here is what I get (and of course this is symmetric, so it has only 6 unique elements) [ B(x',x')]mn = Σi[{ [h(x')n/ h(X')i] [h(x')m/ h(X')i] (∂x'n/∂X'i)(∂x'm/∂X'i) I then show another expression which is [ B(x',x')]mn = [F(x',X') F(x',X')T]mn which says you could naively compute B = FFT in this basis! But in the Maple program, I implement the long expression shown above and I get for B the same thing that Lai does on page 131. In Section 8 I want to find B-1. But B-1 = F-1,T F-1, so I first have to compute F-1≡ G and I find rather obviously that F-1(X',x')mn = [ (X')mT F-1(x')n] = [h(X')m / h(x')n] (∂X'm/∂x'n) = G(X',x')mn which is similar to [ F(x',X')]mn stated above a few inches. Then I use the "easy method" and write [B-1](x',x') = GT(X',x') G(X',x') and this produces a Maple result which replicates Lai's results on pages 134-135 . In Section 9 I then go after C = FTF and I use the "easy method" in Maple to get [ C(X',X')]mn = (X')mT FT (x')i (x')Ti F (X')n] = [ F(x',X')T]mi [ F(x',X')]in = [ F(x',X')T F(x',X')]mn and Maple then generates a C matrix which agrees with Lai page 135. Lai goes on to do C-1 but I did not bother to do this in the Maple program. I will now directly quote my last two sections: Section 10 Conclusion. To compute all these matrices in any combination of bases, you need to first compute these two fundamental matrices in the mixed basis as shown: [ F(x',X')]mn = [ (x')mT F (X')n] = [h(x')m/ h(X')n] (∂x'm/∂X'n) F-1(X',x')mn = [ (X')mT F-1(x')n] = [h(X')m / h(x')n] (∂X'm/∂x'n) // = G Once you do that, you compute the other matrices in the obvious manner using simple matrix multiplication and transposing in Maple: B = FFT B-1 = F-1,T F-1 C = FTF C-1 = F-1 F-1,T In Maple the main work is just to give all 6 coordinates appropriate names, and get the six scale factors entered properly. Section 11. The case X = Cartesian and x = cylinder. I created a new Maple file lai 3_29 cart cyl and it gives all the right results. Section 12. The case X = spherical and x = spherical. Done, just another Maple file. So I did all the cases Lai did, and I could no doubt put check marks everywhere if I wanted. _________________________________________________________________________________ Raw Notes: 1. Concerning the systems x and X. I want both these systems to be Cartesian, but they are different Cartesian systems with separate basis vectors. So: (x)n = Cartesian unit vectors for the x system = n that is to say , and (X)n = Cartesian unit vectors for the X system = n that is to say , and These are the basis vectors that appear in the "normal" expansion of the F tensor. F = Σij [F(x,X)]ij (x)n(X)Tj (*) // = Σij [F(x,X)]ij n Tj In my tensor doc, I never used two different coordinate systems in an expansion like this, by the way, I might want to think about that more. I would like to be able to say that [ F(x,X)]ij = (x)Ti F (X)j (**) since X is "on the right" and is the earlier time. If I close (*) from the right with (X)j and from the left with (x)iT, then we get (**), so I have things in the right ordering in (*). I know that in this basic pair of systems we have that F = (x) and Fij = (x)ij = ∂j(X)xi eg F12= ∂x/∂Y. Comment: We are certainly allowed to consider using the same set of Cartesian unit vectors for the x and X systems. That is, we could select (x)n ≡ (X)n for n = 1,2,3. In this case we could conclude that [F(x,x)]ij = [F(X,X)]ij = [F(x,X)]ij = [F(X,x)]ij = ∂j(X)xi = Fij But when x and X have different unit vectors, these things will not be equal. They will be related by some finite rotation matrix which brings the vectors unit vectors into alignment. 2. The system x' . Since we have x' ← x, things are in my standard order for thinking about curvilinear coordinates x'. In my tensor doc, I write the basis vectors for the x' system as n these being the unit tangent base vectors drawn in x-space. But here I need to refer to these as (x')n so I know which of the four coordinate systems they belong to. Perhaps this is a weakness of my tensor doc notation where the prime indicated in which space the vectors exist. In any event, we then have [(x')n]i = [n]i = [en]i / h'n = Sin / h'n = Rni/ h'n But we need to label things more carefully, so write [(x')n]i = [S(x')]in / h(x')n = [R(x')]ni / h(x')n where we now label R and X and scale factors h with the appropriate labels. Perhaps the information presented on the two lines above is not quite what we really need to know. What will be more useful is to know how to expand the [(x')n] in terms of the (x)n ≡ . We are assuming of course that x' and X' are orthogonal coordinate systems, so this is going to be a simple rotation. I addressed this same problem recently in another doc, but let's try it here again from scratch. Let's go back to the simple tangent base vector and write en = Σiani see want to learn the ai ! en = Σiani = Σiani δij = anj But we know that from item 2 above that, in Cartesian x-space, en = (en)j = Sjn = Rnj = anj and therefore we have shown that en = Σi Sin = Σi Rni n = Σi Sin / h'n = Σi Rni / h'n So we can now translate this into our fancier notation to write [(x')n] = (1/ h(x')n) Σi [S(x')]in and we can leave it at that. I can then use for the X unit vectors. 3. The system X'. Since we have X ← X', we have to talk here about the "inverse" transformation. But I know that means swapping R and S, so I think this is the right answer for this case [(X')n]i = [R(X')]in / h(X')n = [S(X')]ni / h(X')n where you have to pay attention to whether the parenthetical letter is upper or lower case! I then expect the expansion corresponding to the above to be [(X')n] = (1/ h(X')n) Σi [R(X')]in 4. Here is the thing we want to compute: [ F(x',X')]ij = (x')Ti F (X')j and this is the most general case, as desired. One way to write the above is then as follows: [ F(x',X')]ij = (x')Ti F (X')j = { (1/ h(x')i) Σk [S(x')]ki T } F {(1/ h(X')j) Σs [R(X')]sj } = { (1/ h(x')i) Σk [S(x')]ki } {(1/ h(X')j) Σs [R(X')]sj T F } = { (1/ h(x')i) Σk [S(x')]ki } {(1/ h(X')j) Σs [R(X')]sj Fks I think this answer is correct, but it is not in the form that we want. The form here is a linear combination of the Cartesian Fks and of course it is a big mess. We want a result directly in terms of x' and X' derivatives! 5. Computing the differential dx . We know from our tensor doc vector expansions that we can write dx = Σn dx'n en = Σn dx'n h'n n // valid since dx is here a "short vector" which we translate into dx = Σn dx'n h(x')n (x')n Similarly it must be that dX = Σn dX'n h(X')n (X')n On the other hand we know that dx = F dX Inserting the above, we get Σn dx'n h(x')n (x')n = F Σn dX'n h(X')n (X')n Now apply (x')mT from the left to both sides to get Σn dx'n h(x')n (x')mT (x')n = (x')mT F Σn dX'n h(X')n (X')n Σn dx'n h(x')n δm,n = Σn dX'n h(X')n (x')mT F (X')n dx'm h(x')m = Σn dX'n h(X')n [ (x')mT F (X')n] dx'm = [h(x')m]-1 Σn dX'n h(X')n [ (x')mT F (X')n] On the other hand, we know that x' = x'(X') so that dx'm = Σn (∂x'm/∂X'n) dX'n Comparing the above we get (∂x'm/∂X'n) = [h(x')m]-1 h(X')n [ (x')mT F (X')n] or [ F(x',X')]mn = (x')mT F (X')n = [h(x')m/ h(X')n] (∂x'm/∂X'n) and (finally) this is the result I have been seeking! 5a. Some expansions of F acting on unit vectors. A. Consider F (X')n = Σi ani (x')i = Σm anm (x')m Apply (x')Tm to both sides to get (x')Tm F (X')n = Σi ani(x')Tm (x')i = anm Therefore F (X')n = Σm anm (x')m = Σm [(x')Tm F (X')n] (x')m = Σm [h(x')m/ h(X')n] (∂x'm/∂X'n)] (x')m B. Consider FT (x')n = Σi bni (X')i = Σm bnm (X')m Apply (X')Tm to both sides to get (X')Tm FT (x')n = Σi bni(X')Tm (X')i = bnm Therefore FT (x')n = Σm bnm (X')m = Σm [(X')Tm FT (x')n] (X')m = Σm [(x')Tn F (X')m] (X')m = Σm [h(x')n/ h(X')m] (∂x'n/∂X'm)] (X')m So we can summarize now our two little expansions: F (X')n = Σm [h(x')m/ h(X')n] (∂x'm/∂X'n)] (x')m FT (x')n = Σm [h(x')n/ h(X')m] (∂x'n/∂X'm)] (X')m Notice that if one set of basis vectors appears on the left, the other appears on the right. Then we avoid the need for any explicit rotation matrices. 6. Application: Suppose both x' and X' systems are cylindricals with r,θ,z and R,Θ,Z. We know that h(x')1 = h(x')r = 1 h(x')2 = h(x')θ = r h(x')3 = h(x')φ = 1 h(X')1 = h(X')R = 1 h(X')2 = h(X')Θ = R h(X')3 = h(X')Z = 1 Then we have F11 = (∂r/∂R) F12 = (1/R) (∂r/∂Θ) F13 = (∂r/∂Z) F21 = r(∂θ/∂R) F22 = (r/R) (∂θ/∂Θ) F23 = r (∂θ/∂Z) F31 = (∂z/∂R) F32 = (1/R) (∂z/∂Θ) F33 = (∂z/∂Z) If we now replace R = r0 and Θ = θ0 and Z = z0, the above 3x3 matrix agrees with Lai p 132 3.29.12. So now I know how to express the F matrix in any orthogonal coordinate system pair I want. The result is really quite simple. // I then did a Maple program to do this and got I just noted the two colors which seems odd. 7. Try matrix B = FFT . We can write [ B(x',x')]mn = [ (x')mT B (x')n] = [ (x')mT F FT (x')n] Now quote expansion derived in 5a (B) above, FT (x')n = Σm [h(x')n/ h(X')m] (∂x'n/∂X'm)] (X')m = Σi [h(x')n/ h(X')i] (∂x'n/∂X'i)] (X')i Install this to get [ B(x',x')]mn = [ (x')mT F FT (x')n] = [ (x')mT F { Σi [h(x')n/ h(X')i] (∂x'n/∂X'i)] (X')i] = [ (x')mT { Σi [h(x')n/ h(X')i] (∂x'n/∂X'i)] F (X')i] Now quote the expansion derived in 5a (A) above F (X')n = Σm [h(x')m/ h(X')n] (∂x'm/∂X'n)] (x')m = Σj [h(x')j/ h(X')n] (∂x'j/∂X'n)] (x')j or F (X')i = Σj [h(x')j/ h(X')i] (∂x'j/∂X'i)] (x')j Then the above becomes [ B(x',x')]mn = [ (x')mT { Σi [h(x')n/ h(X')i] (∂x'n/∂X'i)] F (X')i] = [ (x')mT { Σi [h(x')n/ h(X')i] (∂x'n/∂X'i)]{ Σj [h(x')j/ h(X')i] (∂x'j/∂X'i)] (x')j }] = Σij[{ [h(x')n/ h(X')i] (∂x'n/∂X'i)]{ [h(x')j/ h(X')i] (∂x'j/∂X'i)] (x')mT (x')j }] = Σij[{ [h(x')n/ h(X')i] (∂x'n/∂X'i)]{ [h(x')j/ h(X')i] (∂x'j/∂X'i)] δm,j] = Σi[{ [h(x')n/ h(X')i] (∂x'n/∂X'i)]{ [h(x')m/ h(X')i] (∂x'm/∂X'i)] = Σi[{ [h(x')n/ h(X')i] [h(x')m/ h(X')i] (∂x'n/∂X'i)(∂x'm/∂X'i) So here is our big result: [ B(x',x')]mn = Σi[{ [h(x')n/ h(X')i] [h(x')m/ h(X')i] (∂x'n/∂X'i)(∂x'm/∂X'i) For example [ B(x',x')]mn = Σi[{ [h(x')n/ h(X')i] [h(x')m/ h(X')i] (∂x'n/∂X'i)(∂x'm/∂X'i) [ B(x',x')]11 = Σi[{ 1/ h(X')i] [1/ h(X')i] (∂x'1/∂X'i)(∂x'1/∂X'i) = (∂x'1/∂X'1)2 + (1/R)2 (∂x'1/∂X'2)2 + (∂x'1/∂X'3)2 = (∂r/∂R)2 + (1/R)2 (∂r/∂Θ)2 + (∂r/∂Z)2 // agrees with 3.29.19 [ B(x',x')]mn = Σi[{ [h(x')n/ h(X')i] [h(x')m/ h(X')i] (∂x'n/∂X'i)(∂x'm/∂X'i) [ B(x',x')]12 = Σi[{ [h(x')2/ h(X')i] [h(x')1/ h(X')i] (∂x'2/∂X'i)(∂x'1/∂X'i) = r(∂x'2/∂X'1)(∂x'1/∂X'1) + (r/R2) (∂x'2/∂X'2)(∂x'1/∂X'2) + r (∂x'2/∂X'3)(∂x'1/∂X'3) = r (∂θ/∂R)(∂r/∂R) + (r/R2) (∂θ/∂Θ) (∂r/∂Θ) + r (∂θ/∂Z) (∂r/∂Z) // agrees with 3.29.20 Here is another derivation: [ B(x',x')]mn = [ (x')mT B (x')n] = [ (x')mT F FT (x')n] Now insert completeness, = [ (x')mT F (X')i (X')Ti FT (x')n] = [(x')mT F (X')i][ (X')Ti FT (x')n] = Σi[(x')mT F (X')i][ (x')Tn F (X')i] = ΣiF(x',X')mi F(x',X')ni = ΣiF(x',X')mi F(x',X')Tin = [F(x',X') F(x',X')T]mn So this is " B with respect to x ". If I throw my formula into Maple, here is the result for matrix B(x',x') : B = FFT is symmetric, so one need only state 6 of the above entries, which is what Lai does. The 6 entries above agree with Lai's quoted results on page 133. 8. Try matrix B-1 with respect to x. Remember that B = FFT where dx = F dX . Obviously B-1 = F-1,T F-1. To make a connection with this F-1 matrix, we can consider the reverse blob flow with dX = F-1dx. We found in section 5 above that dx = Σn dx'n h(x')n (x')n dX = Σn dX'n h(X')n (X')n which we can insert to get Σn dX'n h(X')n (X')n = F-1 { Σn dx'n h(x')n (x')n } Apply (X')mT from the left to both sides to get dX'm h(X')m = Σn dx'n h(x')n [ (X')mT F-1(x')n] or dX'm = [h(X')m]-1 Σn dx'n h(x')n [ (X')mT F-1(x')n] This agrees with the earlier section 5 result just doing x ↔ X and F → F-1. On the other hand, we know that X' = X'(x') so that dX'm = Σn (∂X'm/∂x'n) dx'n Comparing the above we get (∂X'm/∂x'n) = [h(X')m]-1h(x')n [ (X')mT F-1(x')n] and again the comment is the same as just made above. Thus F-1(X',x')mn = [ (X')mT F-1(x')n] = [h(X')m / h(x')n] (∂X'm/∂x'n) So this is a matrix I could write out if I wanted. But from above B-1 = F-1,T F-1 = GTG // just define G = F-1 to make things easy G(X',x')mn = [ (X')mT G (x')n] = [h(X')m / h(x')n] (∂X'm/∂x'n) So we then have [B-1](x',x')mn = (x')mT B-1 (x')n = (x')mT GTG (x')n = (x')mT GT (x')i (x')i TG (x')n = GT(X',x')mi G(X',x')in => [B-1](x',x') = GT(X',x') G(X',x') OK, I guess it is Maple Time again. Here is G = F-1(X',x')mn , and then from B-1 = GTG we find that B-1 is given by It too is symmetric and Lai gives the 6 distinct elements on pages 134-135 and they agree with the above. 9. Try matrix C with respect to X. This is going to be just more of the same. [ C(X',X')]mn = [ (X')mT C (X')n] = (X')mT FTF (X')n] = (X')mT FT (x')i (x')Ti F (X')n] = FTF where F = [ F(x,X)]ij = (x)Ti F (X)j So just tell Maple to compute FTF in this mixed basis. We get 10. Conclusion. To compute all these matrices in any combination of bases, you need to first compute these two fundamental matrices in the mixed basis as shown: [ F(x',X')]mn = [ (x')mT F (X')n] = [h(x')m/ h(X')n] (∂x'm/∂X'n) F-1(X',x')mn = [ (X')mT F-1(x')n] = [h(X')m / h(x')n] (∂X'm/∂x'n) Once you do that, you compute the other matrices in the obvious manner using simple matrix multiplication in Maple: B = FFT B-1 = F-1,T F-1 C = FTF C-1 = F-1 F-1,T In Maple the main work is just to give all 6 coordinates appropriate names, and get the six scale factors entered properly. 11. The case X = Cartesian and x = cylinder. I created a new Maple file lai 3_29 cart cyl and it gives all the right results. 12. The case X = spherical and x = spherical. Done, just another Maple file. So I did all the cases Lai did, and I could no doubt put check marks everywhere if I wanted.