garbage grad v expressed in curvilinear coordinates

Expressing v in curvilinear coordinates PhL 1.20.12 This entire document is obsolete. It was my first shot on this topic after I saw v appearing in Lai's book. The entire correct presentation is now in Appendix F of tensor doc. What we have here is a typical example of my flailing around and screwing up over and over again as is my custom. This thing should be thrown out really, it is just pathetic garbage. 4. Plan C. 1 (4a) Obtain an expression for (v)cd = ∂dvc(x) entirely in terms of x'-space coordinates and objects. 1 (4b) Change the basis vectors from uuT to eeT. 3 Appendix A: Express un as linear combination of the n 6 Appendix B: I will try to make Maple compute T 7 (1) Put things into matrix notation for Maple: 7 (2) What is the R matrix for polar coordinates with 12 = rθ ? 8 (3) The Maple program. 10 Appendix C: What exactly is the meaning of vi and v'i ? 14 Appendix D1: Attempt to do a dimension check on the following equation: 14 Appendix D2: (back up) Attempt to do a dimension check on the following equation: 16 1. Plan 0. 18 2. Plan A. 18 3. Plan B. 20 4. Plan C. (4a) Obtain an expression for (v)cd = ∂dvc(x) entirely in terms of x'-space coordinates and objects. Suppose we do the brute force approach and just start with the end in mind. Note that the objects vi and v'i here are parts of the true covariant tensor v. On the Cartesian side, there is no ambiguity about vi but there is about v'i. This is discussed in question 1 below. We proceed with the true covariant components: (∂'jv'i) = (Rja∂a) ( Ribvb) = Rja ∂a(Ribvb) = Rja [Rib(∂avb) + (∂aRib) vb ] I am trying to "express the object (v) in terms of x'-space coordinates and objects" and I don't know exactly how to interpret the quoted phrase in this situation. Someone hands me this Cartesian object (v)ab = ∂bva(x) Maybe I want to say this, where I use ua as the Cartesian unit vectors, (v) = Σab (v)ab uaubT = Σab( ∂bva) uaubT = Σcd( ∂dvc) ucudT (*) It is this object I want to "express in terms of x'-space stuff". In other examples, this has meant x'-space objects in terms of x'-space coordinates, BUT in terms of x-space tangent base vectors en. Continuing, I can make use of the solution to the equation above and plug it in here. So let's continue with (∂'jv'i) = Rja Rib(∂avb) + Rja(∂aRib) vb First apply Rjc to both sides from the left and sum on j: Rjc(∂'jv'i) = Rjc Rja Rib(∂avb) + Rjc Rja(∂aRib) vb = δca Rib(∂avb) + δca(∂aRib) vb = Rib(∂cvb) + (∂cRib) vb Now apply Rid to both sides from the left and sum on d: Rid Rjc(∂'jv'i) = Rid Rib(∂cvb) + Rid(∂cRib) vb = δdb(∂cvb) + Rid(∂cRib) vb = (∂cvd) + Rid(∂cRib) vb So we can then solve to get (∂cvd) = Rid Rjc(∂'jv'i) – Rid(∂cRib) vb I want to get primed things wherever possible, so replace vb = Rebv'e , which I will now derive. We know that v'e= Recvc => Reb v'e= Reb Recvc = δbcvc = vb QED Inserting this for vb we then have (∂cvd) = Rid Rjc(∂'jv'i) – Rid(∂cRib)Rebv'e And I now show all the sums (∂cvd) = ΣijRid Rjc(∂'jv'i) – ΣibeRid(∂cRib)Rebv'e To avoid conflict with what follows, I will replace the e sum with k sum, so e→k in 2 places: (∂cvd) = ΣijRid Rjc(∂'jv'i) – ΣibkRid(∂cRib)Rkbv'k (**) dim chk OK App D1 Since the R matrix elements are all in terms of r and θ as used below, it would seem that the entire RHS of the above is indeed in terms of x'-space stuff, as desired!! [ WRONG!! what about ∂c ] We can then insert this into (*) above to get (v) = Σcd( ∂dvc) ucudT = Σcd(∂cvd) uducT // dim chk obvious = [ΣijRid Rjc(∂'jv'i) – Σibk Rid(∂cRib)Rkbv'k] uducT (v)cd= ( ∂dvc) Notice that (∂cRib) is going to involve things like sinθ and cosθ since these appear in R matrices. (4b) Change the basis vectors from uuT to eeT. The next step is that we don't want uducT, we want I think cdT in this position. According to Appendix A below, we know that uc = Σi h'i-1Ric i = sum of vectors i with some coefficients. uc = Σi kic (i) Dim Check: In A Simple Theorem.... I show that Qic ≡ h'i-1 Ric = so Q is our usual rotation and the elements of Q are dimensionless. If we take such a (vector = lincom of vectors) equation and transpose it, the coefficients do not change. For example, consider = α + β => a = αc + βe and b = αd + βf (a b) = α (c d) + β (e,f) => same equations Therefore we can write ud = Σe h'e-1 Red e ucT = Σf h'f-1 Rfc fT uducT = [Σe h'e-1 Red e] [Σf h'f-1 Rfc fT] = Σef h'e-1 Red h'f-1 Rfc e fT Dim Check: Factors h'e-1 Red and h'f-1 Rfc are separately dimensionless for any indices, so the above equation passes dimension check. Then plug this into our start (v) = Σcd(∂dvc) ucudT // dim chk obvious to get (v) = Σcd [ΣijRid Rjc(∂'jv'i) – Σibk Rid(∂cRib)Rkbv'k] uducT = Σcd [ΣijRid Rjc(∂'jv'i) – Σibk Rid(∂cRib)Rkbv'k] Σef h'e-1 Red h'f-1 Rfc e fT Dim check: In App D1 we showed that the square bracket above [.....] indeed has the correct dimensions for it to be equal to (∂dvc). Then we showed just above that the extra factors h'e-1 Red h'f-1 Rfc are dimensionless. Therefore, we have in effect verified that the above huge mess is dimensionally correct. Now first write this as two separate terms since the second term has no j sum, = Σcdijef Rid Rjc(∂'jv'i) h'e-1 Red h'f-1 Rfc e fT – Σcdibkef Rid(∂cRib)Rkb v'k h'e-1 Red h'f-1 Rfc e fT Next, shuffle R factors to get orthogonality abutments where possible = Σcdijef (Red Rid) (Rfc Rjc)(∂'jv'i) h'e-1 h'f-1 e fT – Σcdibkef (Red Rid) Rfc (∂cRib)Rkb v'k h'e-1 h'f-1 e fT There are three orthogonalities so install the δ's = Σijef δei δfj(∂'jv'i) h'e-1 h'f-1 e fT – Σcibkef δei Rfc (∂cRib)Rkbv'k h'e-1 h'f-1 e fT = Σij(∂'jv'i) h'i-1 h'j-1 i jT – Σcibkf Rfc (∂cRib)Rkbv'k h'i-1 h'f-1 i fT Now in the second sum, replace f by j = Σij(∂'jv'i) h'i-1 h'j-1 i jT – Σcibkj Rjc (∂cRib)Rkbv'k h'i-1 h'j-1 i jT Now we can combine the two terms again = Σij { (∂'jv'i) – Σcbk Rjc (∂cRib)Rkbv'k } h'i-1 h'j-1 i jT This is at least "promising". Let's now rewrite as (the LHS has been (v) all along) (v) = Σij { (∂'jv'i) – Tji } h'i-1 h'j-1 i jT Tji ≡ Σcbk Rjc (∂cRib)Rkbv'k Now, as a reminder, reconsider A = Σij Aij uiujT = A11 u1u1T + A12 u1u2T + ... u1u2T = (0 1) = so things are correct since this 1 is in the 12 corner Then I think we can say that (v)ij = { (∂'jv'i) – Tji } h'i-1 h'j-1 which goes then with (v) = Σij (v)ij i jT Thus we have taken the object on the left (Cartesian) and expressed it entirely in terms of x'-space coordinates and objects such as v'i. Comment: I have carefully checked every equation above since my last round of corrections, making equations green. The chances of there still existing errors are perhaps < 5% so I have pretty high confidence. One can NEVER rule out errors even after 100 such checks because one makes the same error on each pass. STOP, because there is a problem in Appendix B In Appendix B I have Maple compute T and this is what I find T = cosθ vθ -cosθ r vr - sinθ vθ -sinθ r vθ sinθ r2 vr + cosθ r vθ where I made these replacements to get outlined unit-vector tensor doc quantities vr = vr vθ = r vθ Now the top left element comes out being [ note that h1 = hr = 1 and h2 = hθ= r ] (v)ij = { (∂'jv'i) – Tji } h'i-1 h'j-1 (v)11 = { (∂'1v'1) – T11} h'1-1 h'1-1 = { (∂rv'r) – cosθ vθ } h'r-1 h'r-1 = { (∂r vr) – cosθ vθ } The top right element is (v)ij = { (∂'jv'i) – Tji } h'i-1 h'j-1 (v)12 = { (∂'2v'1) – T21} h'1-1 h'2-1 = { (∂θv'r) + sinθ r vθ } / r = { (∂θ vr) + sinθ r vθ } / r The lower left element is (v)ij = { (∂'jv'i) – Tji } h'i-1 h'j-1 (v)21 = { (∂'1v'2) – T12 } h'2-1 h'1-1 = { (∂rvθ) – [-cosθ r vr - sinθ vθ] } h'2-1 h'1-1 = { (∂r[r vθ]) – [-cosθ r vr - sinθ vθ] }/r = { (∂rvθ + vθ + [cosθ r vr + sinθ vθ] }/r = { (∂rvθ + [cosθ r vr + (sinθ+1) vθ] }/r Finally, the lower right element is (v)ij = { (∂'jv'i) – Tji } h'i-1 h'j-1 (v)22 = { (∂'2v'2) – T22 } h'2-1 h'2-1 = { (∂θvθ) – sinθ r2 vr + cosθ r vθ } /r2 = { (∂θ[r vθ]) – sinθ r2 vr + cosθ r vθ } /r2 = { r(∂θvθ) – sinθ r2 vr + cosθ r vθ } /r2 = { (∂θvθ) – sinθ r vr + cosθ vθ } /r Therefore I have found that (v)ij = (∂r vr) – cosθ vθ { (∂θ vr) + sinθ r vθ } / r { (∂rvθ + [cosθ r vr + (sinθ+1) vθ] }/r { (∂θvθ) – sinθ r vr + cosθ vθ } /r The Lai result is (v) = ∂r vr (∂θ vr - vθ) /r // Lai ∂r vθ (∂θ vθ + vr)/r (v) = // Lai Now just for fun, if I set θ = -π/2, then sinθ = -1 and cosθ = 0 and my result becomes (v)ij = (∂rvr) { (∂θ vr) - r vθ } / r // me adjusted (∂rvθ }/r { (∂θvθ) + r vr } /r It seems strange to me that even in this case, the dimensions disagree as I highlight in red. Appendix A: Express un as linear combination of the n We want to find the coefficients amn. We assume the x'-space coordinates are orthogonal, and we assume that x-space is Cartesian with g = 1. Start then with un = Σm amn em Assume orthogonal so we then have un ei = Σm amn em ei = Σm amn g'mi = ain g'ii // no sum on i un ei = gab(un)a (ei)b = gab δna Rib = gnb Rib = Rin // p 80 and g = δ Thus we find that ain g'ii = Rin => ain = Rin / g'ii = Rin / (h'i2) // p 43 Therefore the correct expansion is this: un = Σi ain ei = Σi h'i-2 Rin ei Now we know that ei = h'i i so therefore un = Σi h'i-1 Rin i This is the result I really want. [ I have verified this is correct on 2.23.12 and this result now appears as one of the theorem conclusions of a whole separate doc. ] P Check: Consider the matrix appearing in the above sum Lemma: Suppose we can write the above as un = ΣiQin i where Qin = h'i-1 Rin (QT)in = Qni = h'n-1 Rni = h'n-1 (RT)in => (QT)nj = h'j-1 (RT)nj If we think of Qin as a matrix, it MUST describe a simple rotation, and therefore it must be orthogonal, so we must have Σn Qin (QT)nj = δij This in turn means that Σn h'i-1 Rin h'j-1 (RT)nj = δij h'i-1 h'j-1Σn Rin (RT)nj = δij Σn Rin (RT)nj = δij h'i2 RRT = g' // agrees with tensor doc Appendix B: I will try to make Maple compute Tji. First, try to get it into matrix form: (1) Put things into matrix notation for Maple: Tji ≡ Σcbk Rjc (∂cRib)Rkbv'k Let's then define three matrices, Ujc ≡ Rjc // the "up" R matrix Dib = Rib // the "down" R matrix Then the above reads Tji ≡ Σcbk Ujc (∂cUib) Dkb v'k = Σcbk Ujc (∂cUib) DTbk v'k Now define this matrix M(i)cb ≡ (∂cUib) Then we have Tji ≡ T(i)j = Σcbk Ujc (∂cUib) Dkb v'k = Σcbk Ujc M(i)cb DTbk v'k = a vector with label i = [UM(i)DTv']j or T(i) = UM(i)DTv' = a vector So I will go compute these two vectors in Maple T(1) = UM(1)DTv' => T1 = U M1 DT v' v' = (vr,vt) T(2) = UM(2)DTv' => T2 = U M2 DT v' Then I can read off from Tji ≡ T(i)j : T11 = T(1)1 T12 = T(2)1 T21 = T(1)2 T22 = T(2)2 (2) What is the R matrix for polar coordinates with 12 = rθ ? On p27 of tensor doc we compute S for the case 12 = θr. What is S if we reverse these? We would write in this case, if xy = 12 and rθ = 12, S12 = (∂x/∂θ) = -rsinθ S11 = (∂x/∂r) = cosθ Sik ≡ ( ∂xi/∂x'k) S22 = (∂y/∂θ) = rcosθ S21 = (∂y/∂r) = sinθ Sij = We can then have Maple compute S = R-1: from which I conclude that Rij = rθ = 12 Rij = Rij = Dij which we can compare to the matrix which appears in tensor doc Rij = θr = 12 // do not use this! Recall that S = [ e1 e2 ] which means that R = S-1= ST = [ e1 e2 ]T = . Therefore, if you swap the labels 1 ↔ 2, you are going to swap the ROWS of the R matrix, and that is exactly what we see above. [ You do NOT do this: R11→ R22, R12→ R21 , R21→ R12 , R22→ R11 !!! ] While we are at it, we can compute the S matrix as well by swapping the columns of the tensor doc one. So Sij = θr = 12 // do not use this! Sij = rθ = 12 How can I verify this R matrix? I know that Qin ≡ h'i-1 Rin should be a rotation matrix. I also know that for our orthogonal case, Rab = Σa'b'g'aa'Ra'b' gb'b = g'aaRab gbb = g'aaRab = h'a2 Rab => Rab = h'a2 Rab => Rin = h'i2 Rin => Qin ≡ h'i-1 Rin = h'i-1 h'i2 Rin = h'i Rin and from above we have Rij = h'1 = 1 h'2 = r 12 = rθ now Therefore: Q11= h'1R11 = cosθ Q12= h'1R12 =sinθ Q21 = h'2R21 = -sinθ Q22 = h'2R22 = cosθ So that Q = which looks good. Therefore we have shown that Rij = = Rij = Dij rθ = 12 So I now have one of my three matrices: Dij = Now consider that Rjc Ric = δji => Ujc Dic = δji => Ujc DTci = δji => U DT = 1 => U = DT,-1 => DT = U-1 (3) The Maple program. In Maple I first unprotect then unassign symbol D so I can use it, so we start with Then we construct two of our three required matrices as follows (that is to say, DT and U) It remains to construct these two matrices M(i)cb ≡ (∂cUib) i = 1,2 M1ij = (∂iU1j) M2ij = (∂iU2j) q[1] := r; q[2] := theta; M1 := (i,j) -> diff(U[1,j],q[i]); In this manner I have managed to obtain matrices M1 and M2 Now let's get back to T1 = U M1 DT v' v' = (vr,vt) T2 = U M2 DT v' v' = (vr,vt) T11 = T(1)1 T12 = T(2)1 T21 = T(1)2 T22 = T(2)2 Here then is what Maple now says STOP. In Appendix C below I show that vr = vr = m/sec and vt = vθ = m2/sec. Something is therefore wrong with the expressions in T2. We have terms m3/s and m2/s being added! So here then is what I read off T11 = cosθ vθ/r T12 = (-cosθ r2 vr + sinθ vθ)/r T21= -sinθ vθ T22 = sinθ r2vr + cosθ vθ So here is the big conclusion in simple notation: T = cosθ vθ/r (-cosθ r2 vr - sinθ vθ)/r -sinθ vθ sinθ r2vr + cosθ vθ Now according to Appendix C we have vr = vr vθ = r vθ so that then T = cosθ vθ (-cosθ r2 vr - sinθ r vθ)/r -sinθ r vθ sinθ r2 vr + cosθ r vθ or T = cosθ vθ -cosθ r vr - sinθ vθ -sinθ r vθ sinθ r2 vr + cosθ r vθ Something is wrong because vr and vθ should both have the same dimensions: m/sec . Appendix C: What exactly is the meaning of vi and v'i ? We know that vx = dx/dt in Cartesians. But what does vr mean? It depends how you define it! If you want your v'i to be a true covariant vector, you need to do this: v = Σi v'i ei In the first section below, where I use v'i = ( Ribvb), I have already assumed that vb is the covariant vector component! So that is what MY v is. What is Lai's v? It is a bit slippery because Lai uses ei to stand for what I call i . Here are some examples from Lai: So here is a little table: Lai tensor doc this doc ei i i vr vr vθ vθ vr = grrvr = vr all = m/sec vθ = gθθvθ = r2vθ vθ = rad/sec vθ = m2/sec Meanwhile, we know that, v'n ≡ v'n h'n => vr = vrh'r = vr = m/sec => vθ = vθh'θ = r vθ = m/sec Combining the above lines says vr = vr = vr // all three are m/sec in dimensions vθ = r2vθ = r2(vθ/r) = r vθ // vθ is m2/sec since vθ is m/sec So I need to make these replacements in my results vr = vr => vr = m/sec vθ = r vθ => vθ = m2/sec Appendix D1: Attempt to do a dimension check on the following equation: (∂cvd) = ΣijRid Rjc(∂'jv'i) – ΣibkRid(∂cRib)Rkbv'k (**) Can I get this into a matrix-like notation? Define U and D as before, so this reads (∂cvd) = ΣijDid Djc(∂'jv'i) – ΣibkDid(∂cUib)Dkbv'k (**) Put things into matrix order where possible (v)dc = (∂cvd) = ΣijDTdi (∂'jv'i) Djc – ΣibkDTdi(∂cUib)DTbkv'k (**) Now define some special matrix Bij ≡ (∂'jv'i) C(c)ib ≡ (∂cUib) Then we have (v)dc = (∂cvd) = ΣijDTdi Bij Djc – ΣibkDTdi C(c)ib DTbkv'k = [ DTBD]dc – [DT C(c)DT v']d What are the dimensions of B? B11 ≡ (∂'1v'1) = ∂rvr = (1/m) (m/s) = 1/s B12 ≡ (∂'2v'1) = ∂θvr = m/s B21 = (∂'1v'2) = ∂rvθ = (1/m)(m2/s) = m/s B22 = (∂'2v'2) = ∂θvθ = (m2/s) Here is an idea. Just represent a matrix by its dimensions, so that here B = We know that Dij = = = D DT = So now we can examine the first term DTBD = = = This matches the dimensions of (∂cvd), so we have a good dimension check on the first term. We now turn to the second term: [ form of U from next appendix ] [DT C(c)DT v']d C(c)ib ≡ (∂cUib) = (1/m) v' = = So let's build up the result: DT v' = = = (m/s) C(c)DT v' = (1/m) (m/s) = (1/s) DT C(c)DT v' = (1/s) = (1/s) so all four elements are 1/s, so this ALSO matches dimensions of (∂cvd) Conclusion: The equation (∂cvd) = ΣijRid Rjc(∂'jv'i) – ΣibkRid(∂cRib)Rkbv'k (**) has good dimensions in both terms. Appendix D2: (back up) Attempt to do a dimension check on the following equation: (∂'jv'i) = (Rja∂a) ( Ribvb) = Rja ∂a(Ribvb) = Rja Rib(∂avb) + Rja (∂aRib) vb = Uja Uib(∂avb) + Uja (∂aUib) vb Let Eba = (∂avb) all elements are 1/s for this E matrix! DT = U = DT,-1 = -1 I will try do find the dimensions of U manually from U DT = 1 = = So conclude that U = Now back to (∂'jv'i) = Uja Uib(∂avb) + Uja (∂aUib) vb Bij = Uja UibEba + Uja (∂aUib) vb = UibEba Uja + Uja (∂aUib) vb = UibEba UTaj + Uja (∂aUib) vb = [UEUT]ij + Uja (∂aUib) vb Consider just the first term UEUT = = (1/s) = (1/s) = In the last appendix we found that B = so our dimensions verify for this first term! Appendix D3: Attempt to do a dimension check on the following equation: (v) = Σij { (∂'jv'i) – Tji } h'i-1 h'j-1 i jT Tji ≡ Σcbk Rjc (∂cRib)Rkbv'k As in App D1, we define Bij ≡ (∂'jv'i) and Bij = Write: h'i-1 h'j-1 (∂'jv'i) = h'i-1 h'j-1 Bij = h'i-1 Bij h'j-1 Now define this diagonal matrix Λab = h'a-1 δab Then the above becomes = h'i-1 Bij h'j-1 = Λii' Bij' Λj'j = = = This then verifies that the first term in our equation being checked has correct dimensions. We turn now to the second term. Tji ≡ Σcbk Rjc (∂cRib)Rkbv'k = Σcbk Ujc (∂cUib)Dkbv'k = = Σcbk Ujc (∂cUib)DTbkv'k Now define this matrix M(i)cb ≡ (∂cUib) Then we have Tji = Σcbk Ujc M(i)cb DTbkv'k = [ U M(i) DT v' ]j Now let's look first at the i=2 case since this has caused a problem. We ________________________________________________________________________________ 1. Plan 0. First, review how one treats the vector object G = f where f is a scalar field: G = G'1e1 + G'2 e2 +... = ΣnG'n en where en G = G 'n [grad f](x) = ∂'nf'(x') en [grad f](x) = h'i (∂'if '(x')) i = h'i g'ij(x') (∂'jf '(x')) i In the first line, we make the statement that "if Gi is a true covariant vector, then the coefficients of the expansion shown will be G'i where these are the components of the vector G in x'-space. " I comment in tensor doc that if the Gi are NOT the components of a covariant vector, we can still do our expansion above, and we still have G'i = RikGk = SkiGk but we cannot interpret the G'i as components of a covariant vector in x'-space. Therefore, the conclusion above is also valid for f(x) NOT being a scalar function. In that case, it happens that ∂nf is not a covariant vector, etc etc. "As long as we don't run into a contradiction." Notice that the second result above gives the object [grad f](x) expressed in terms of the curvilinear unit vectors. If for some reason we wanted to express this in terms of Cartesian unit vectors, we would have to do this en = Σi(en ) (en ) = Σk (en)k ()k = Σk (en)k δik = (en)i = Rni Then we have en = Rni = Rni ui where ui are the axis-aligned unit vectors in x-space. Therefore, we could write [grad f](x) = ∂'nf'(x') en = [ ∂'nf'(x') ] Rni ui which in general is a fairly messy object and probably note a very useful object. 2. Plan A. We know from the newly added tensor doc section that we can expand a matrix (quasi-rank-2 tensor) in this manner: G = G'ij (ei ej) = G'ij (ei ej) = G'ij ei ejT = h'ih'j G'ij i jT Now apply m on the right and nT on the left to get nT G m = h'ih'j G'ij nT i jTm = h'ih'j G'ij n i j m At this point we must assume our curvilinear system is orthogonal. Then we can continue = h'ih'j G'ij δni δjm = h'nh'm G'nm In QM we would express this as < n | G | m> = h'nh'm G'nm = matrix element in curvilinear coordinates. Now in our case we have G = v Gij = (v)ij = ∂jvi G'ij = ('v')ij = ∂'jv'i But we want to use contravariant vectors and covariant derivative, so we care about Gij = (v)ij = ∂jvi G'ij = ('v')ij = ∂'jv'i But we know that (orthogonal) G'nm = g' mm G'nm = G'nm / (h'm)2 // no sum on m Therefore we have < n | G | m> = nT G m = h'nh'm G'nm = h'nh'm G'nm/ (h'm)2 = (hn'/h'm) G'nm = (hn'/h'm) ∂'mv'n Now let's try this for polar coordinates. Set r = 1 and θ = 2 and we know that h1 = hr = 1 h2 = hθ = r < 1 | G | 1> = (h1'/h'1) ∂'1v'1 = ∂'rvr < 1 | G | 2> = (h1'/h'2) ∂'2v'1 = (1/r) ∂θvr < 2 | G | 1> = (h2'/h'1) ∂'1v'2 = r ∂rvθ < 2 | G | 2> = (h2'/h'2) ∂'2v'2 = ∂θvθ and once again, this is the WRONG result according to Lai. The problem is that for a non-linear F, you don't have ∂'jv'i = G'ij = RikGkj = Rik ∂jvk as I very well know. But I am somehow assuming this is true in my logic flow above. I guess there IS a contradiction in my case. The reason is that ∂'jv'i is not some brand new thing in x'-space, it is a will defined thing made out of two vector things. I could apply my rule to vi,j and I would get a valid result. But then I have a lot of detangling to do. 3. Plan B. We know from the newly added tensor doc section that we can expand a matrix (quasi-rank-2 tensor) in this manner: Gab = Gij (ei ej)ab = Gij (ei)a (ej)b (a) What is the meaning of Gab ? I think it means ∂bva in Cartesian coordinates. That was my intention. (b) Do I know that such an expansion is possible for Gab? Maybe it's only possible for a true tensor? But suppose I assume that it is possible. Then I am able to compute the coefficients! Gab = Gij (ei)a (ej)b = Gij Ria Rjb = Ria Rjb Gij But I know I can invert this to get Gij = RiaRjbGab I think the problem now is that I have no way to identify Gij = ∂'jv'i so I cannot make any connection to x'-space.