gradv expressed v2

Expressing v in curvilinear coordinates PhL 1.24.12 Here I obtained the correct Lai results for polar and spherical coordinates for the first time using Maple. The work here is entirely replaced by Appendix F of tensor doc which gets the result for non-orthogonal as well as orthogonal. So this doc is historical only, don't ever read it again! Part I: Grind down to something we can compare to Lai. 1 Part II: Show that in fact Pij(last) has the desired form shown just above with lai2.mws 4 Part III: Summary and start over using lai3.mws 6 Part IV: For other systems. 10 Part IV: Misc Questions. 11 Part I: Grind down to something we can compare to Lai. In Cartesian coordinates we have (v)ab = ∂bva(x) (v) = Σab (v)ab uaubT = Σab(∂bva) uaubT = Σcd(∂cvd) uducT I have shown in various places now that 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 Therefore we can write (v) = Σcdef(∂cvd) h'e-1 Red h'f-1 Rfc e fT (*) We want to express (∂cvd) entirely in terms of x'-space coordinates and objects. To this end we write ∂cvd = (Ric∂'i)( Rjdv'j) where sums are on the first index since we are going "backwards" from x'-space to x-space. Here is a little in-line proof that this indexing is right: v'e= Recvc => Reb v'e= Reb Recvc = δbcvc = vb QED Notice that we do in fact have ∂dvc expressed as desired, since we can regard Rij as being functions or r and θ which are x'-space coordinates. But we go ahead and expand a little ∂cvd = (Ric∂'i)(Rjdv'j) = Ric ∂'i(Rjdv'j) = Σij Ric [(∂'iRjd) v'j + Rjd (∂'iv'j)] Without further ado, we insert this into (*) above to get (v) = Σcdef(∂cvd) h'e-1 Red h'f-1 Rfc e fT = Σcdef{ Σij Ric [(∂'iRjd) v'j + Rjd (∂'iv'j)]} h'e-1 Red h'f-1 Rfc e fT = Σcdefij Ric [(∂'iRjd) v'j + Rjd (∂'iv'j)] h'e-1 Red h'f-1 Rfc e fT = Σcdefij (Rfc Ric) [(∂'iRjd) v'j + Rjd (∂'iv'j)] h'e-1 Red h'f-1 e fT = Σcdefij (Rfc Ric) [Red (∂'iRjd) v'j + (Red Rjd) (∂'iv'j)] h'e-1 h'f-1 e fT = Σcdefij (δfi) [Red (∂'iRjd) v'j + (δej) (∂'iv'j)] h'e-1 h'f-1 e fT = Σcdefj [Red (∂'fRjd) v'j + (δej) (∂'fv'j)] h'e-1 h'f-1 e fT = [Red (∂'fRjd) v'j + (∂'fv'e)] h'e-1 h'f-1 e fT = [(∂'fv'e) + Red (∂'fRjd) v'j] h'e-1 h'f-1 e fT = [(∂'fv'e) + Red (∂'fRkd) v'k] h'e-1 h'f-1 e fT = [(∂'jv'i) + Rid (∂'jRkd) v'k] h'i-1 h'j-1 i jT and this is the form we seek. I did one change per line above to allow error tracking. Write then as (v) = [(∂'jv'i) + Rid (∂'jRkd) v'k] h'i-1 h'j-1 i jT Our matrix of interest is this one: Pij ≡ [(∂'jv'i) + Rid (∂'jRkd) v'k] h'i-1 h'j-1 where (v) = Σij Pij i jT Now we know that v'i are the covariant x'-space velocity components. To compare with Lai, we really want to see our unit-vector components appear. We know that (App C previous doc) v'n ≡ v'n h'n and v'n = g'nnv'n = h'n-2v'n Therefore v'n = h'n2 v'n = h'n2 (v'n/ h'n ) = h'n v'n Therefore we can write Pij = [(∂'j[h'i v'i]) + Rid (∂'jRkd) [h'k v'k]] h'i-1 h'j-1 = [(∂'jh'i) v'i + h'i (∂'jv'i) + Rid (∂'jRkd) h'k v'k] h'i-1 h'j-1 Let's check the middle term right now: Pij(middle) = h'i (∂'jv'i) h'i-1 h'j-1 = (∂'jv'i) h'j-1 I can write this four terms out: P11(middle) = (∂'1v'1) h'1-1 = (∂'1v'1) = (∂rvr) P12(middle) = (∂'2v'1) h'2-1 = (∂'2v'1) (1/r) = (∂θvr) (1/r) P21(middle) = (∂'1v'2) h'1-1 = (∂'1v'2) = (∂r vθ) P22(middle) = (∂'2v'2) h'2-1 = (∂'2v'2)(1/r) = (∂θvθ)(1/r) Therefore, this part of our matrix has this form: Pij(middle) = For comparison, here is the Lai matrix: I am extremely happy to see that this all matches. In order to verify Lai, here is what we need to show: Pij(other) = Right now this is not very obvious, since we have Pij(other) = [(∂'jh'i) v'i + Rid (∂'jRkd) h'k v'k] h'i-1 h'j-1 = (∂'jh'i) h'i-1 h'j-1 v'i + h'i-1 h'j-1 Σkd Rid (∂'jRkd) h'k v'k Let's just do the first term right here: Pij(first) = (∂'jh'i) h'i-1 h'j-1 v'i P11(first) = (∂'1h'1) h'1-1 h'1-1 v'1 = (∂r1) v'1 = 0 Pij(first) = (∂'jh'i) h'i-1 h'j-1 v'i P12(first) = (∂'2h'1) h'i-1 h'2-1 v'1 = (∂θh'i) h'2-1 v'1 = 0 Pij(first) = (∂'jh'i) h'i-1 h'j-1 v'i P21(first) = (∂'1h'2) h'2-1 h'1-1 v'2 = (∂rr) (1/r) vθ = vθ/r Pij(first) = (∂'jh'i) h'i-1 h'j-1 v'i P22(first) = (∂'2h'2) h'2-1 h'2-1 v'2 = 0 So we find that Pij(first) = We must have then Pij(first) + Pij(last) = Pij(other) or Pij(last) = Pij(other) – Pij(first) = – = So I will have succeeded if I can show that Pij(last) = where Pij(last) = h'i-1 h'j-1 Σkd Rid (∂'jRkd) h'k v'k Part II: Show that in fact Pij(last) has the desired form shown just above with lai2.mws Write Pij(last) = h'i-1 h'j-1 Σkd Uid (∂'jDkd) h'k v'k U11 = cosθ U12 = sinθ U21 = -rsinθ U22 = rcosθ D11 = cosθ D12 = sinθ D21 = -sinθ / r D22= cosθ/r For each ij choice, there are four terms to add up. Just do manually: P11(last) = h'1-1 h'1-1 Σkd U1d (∂'1Dkd) h'k v'k = Σkd U1d (∂'1Dkd) h'k v'k P11(last,kd=11) = U11 (∂'1D11) h'1 v'1 = cosθ (∂r cosθ ) ... = 0 P11(last,kd=12) = U12 (∂'1D12) h'1 v'1 = sinθ (∂r sinθ) ... = 0 P11(last,kd=21) = U11 (∂'1D21) h'2 v'2 = cosθ(∂r[-sinθ / r]) r vθ = -sinθcosθ (∂rr-1)r vθ P11(last,kd=22) = U12 (∂'1D22) h'2 v'2 = sinθ(∂r[cosθ/r])r vθ = sinθ cosθ (∂rr-1)r vθ The sum of these four terms is 0. and that is the desired result! I will get Maple to compute these things, too much room for PL error here. // That effort was successful! First, I will report out the evaluation of Pij(last) by Maple, then in the next section, I will do the complete Pij in a single shot. The setting up Maple entries were these, where the qi are defined and then h, D and U entered: I had to use inert Sum and Diff operators to get things to work, and I put this as an example in my Maple doc matrix section. The first step is to create matrix Pij(last) willed with symbolic expressions: The remaining code evaluated this matrix: and the final result is what we were hoping to find above, namely, Pij(last) = Part III: Summary and start over using lai3.mws So finally I have verified Lai's v matrix for polar coordinates "the hard way". I think this took me Jan 20 through 9 AM Jan 25, today. I remember guessing 6 days when I started. Here is a review of what I did. 1. I started with v expressed in Cartesian coordinates, using the quasi-dyadic notation (v) = Σab (v)ab uaubT = Σab(∂bva) uaubT = Σcd(∂cvd) uducT 2. I then used ud = Σe h'e-1 Red e to rewrite the unit vector part this way uducT = [Σe h'e-1 Red e] [Σf h'f-1 Rfc fT] = Σef h'e-1 Red h'f-1 Rfc e fT so that then (v) = Σcdef(∂cvd) h'e-1 Red h'f-1 Rfc e fT (*) I did this in analogy with my tensor doc method of expression things in terms of e vectors in x-space. In this step 2 we have assumed we have orthogonal curvilinear coordinates in x'-space. 3. I then expressed (∂cvd) in terms of x'-space objects and coordinates to get, ∂cvd = (Ric∂'i)( Rjdv'j) = Σij Ric [(∂'iRjd) v'j + Rjd (∂'iv'j)] which shows the "extra term" which appears in this kind of derivative. It is this extra term recall which causes ∂cvd NOT to be a rank-2 tensor. The above equation is valid for non-orthogonal as well as for orthogonal x' coordinates. Installing this into (v) gave (and we are back now only to orthogonal x' coordinates) (v) = [(∂'jv'i) + Rid (∂'jRkd) v'k] h'i-1 h'j-1 i jT which means the Lai matrix of interest is this Pij ≡ [(∂'jv'i) + Rid (∂'jRkd) v'k] h'i-1 h'j-1 where (v) = Σij Pij i jT Once again, this result is valid for any orthogonal curvilinear coordinates. 4. I then changed from covariant vi' to unit-vector according to v'n = h'n v'n to get Pij = [(∂'jh'i) v'i + h'i (∂'jv'i) + Rid (∂'jRkd) h'k v'k] h'i-1 h'j-1 or Pij = [(∂'jh'i) v'i + h'i (∂'jv'i) + Uid (∂'jDkd) h'k v'k] h'i-1 h'j-1 P1 P2 P3 Once again, this result is valid for any orthogonal curvilinear coordinates. 5. I then wrote Maple program lai3.mws which does the following steps: In Maple I first set things up like this: Then I had to split P into three matrices for technical reasons, so Then I constructed these three matrices Then I added them together to get the final result which can be compared to the Lai result, where νn|Maple = v'n = (vr,vθ)|Lai and these are seen to agree. Part IV: For other systems. The easiest matrix to obtain is Uab = Rab = Sab = (∂xa/∂x'b) because one always has x(x') and one can make Maple compute this directly. For example, for sphericals x = r sinθ sinφ y = r sinθ cosφ z = rcosθ Note that Sab = Sab = Rba = Uba => U = ST Rab = Rab = Dab = (S-1)ab => D = R = S-1 In program lea4.mws I did all this and obtained this result for Pij which compare to Lai and they agree! So I think I can use my "method" to compute v in any orthogonal coordinate system. Part IV: Misc Questions. 1. How do we interpret this result: (v) = Σij Pij i jT I know I can apply n to the right side and Tm to the left to get Tm (v) n = Pmn = < m | v | n > which is a matrix element of operator v in the e-basis. The Cartesian matrix element would be (v) = Σij Cij i jT Cmn = < m | v | n> = (v)mn This does suggest a simple connection. ud = Σe h'e-1 Red e = Σe Qed e Qed = h'e-1 Red = a rotation (v)mn = < m | v | n> = < Σf Qfm f | v | Σe Qen e > = Σf Qfm Σe Qen< f | v | e > = Σfe Qfm Qen Pfe = Σfe Q-1mf Pfe Qen = [Q-1PQ]mn One could invert this for Pfe, but that does not yield Pfe in the x'-space coordinates! One cannot avoid the little "extra term" business. How do we interpret this result : V = ΣiVi i Here I can draw a little picture of V as an arrow with projections on the i axes being Vi. But it is just less obvious to me what this means (v) = Σij Cij i jT = Σij Pij i jT A reason it is useful to have this second thing is this. Suppose you have an equation a = (v) c and suppose you know vectors a and c in curvilinear coordinates. Then c = Σici i a = Σiai i Then you can write Σiai i = Σij Pij i jT (Σncn n) = Σij Pij i (Σncn) δj,n = Σij Pij i cj = Σi[ Σj Pij cj] i Then you know that ai = Σj Pij cj So THIS is the motivation for expressing it in the form Σij Pij i jT : you end up with a simple connection between the curvilinear projections of a and c. Notice that v is NOT a differential operator the way all the other ones are, so I am not sure it makes sense to include it in my tensor doc discussion. 2. What about the other notation: From my tensor expansions doc I know that [(a) v]i = (ai) v = (∂jai)vj But in this notation, with the ai appearing we have committed to Cartesians I think. You really have to do something like ai(x) = Rji a'j(x') so then [(a) v]i = (ai) v = (∂jai)vj = [ (Raj ∂'a)(Rbi a'b(x'))] Rci v'c(x') and then you are off and running on the big mess that we have to do either way. So I guess this does explain to me the utility of the matrix (a) in various coordinate systems. 2. What about the differential notation: We start with (v)ij = ∂jvi dvj(xn) = (∂ivj(xn))dxi // just chain rule on a function dvj = (v)ji dxi dvj= (v)ji dxi = [(v)dr]j dv = (v)dr So this is the derivation of this simple vector equation which then can be studied in any system. Suppose we write dv = Σidv'i ei dr = Σidx'i ei and suppose we have (v) = Σab pab ea ebT Then we get dv = (v)dr v = Σiv'iei dv = Σi(dv'i)ei + Σiv'i(dei) Now there are going to be coefficients f(i)ab such that (dei) = Σab f(i)abdx'a eb and this equation is "the hard part" of this method I think. For sphericals it is not too bad, and I could probably write a general formula. For example, go back to un = Σm amn em Do not assume orthogonal, but assume x-space Cartesian, and then write un ei = Σm amn em ei = Σm amn g'mi un ei = gab(un)a (ei)b = gab δna Rib = gnb Rib = Rin // p 80 and g = δ Then we get Σm amn g'mi = Rin g'im amn = Rin Apply g'ji from the left g'ji g'im amn= g'ji Rin δjm amn= g'ji Rin ajn = g'ji Rin => amn = g'mi Rin so our general result is then un = Σmi g'mi Rin em and this can be inverted to get some en = Σmi Fmi um den = Σmi (dFmi) um = Σmi (dFmi) Σai g'ai Rin ea en ≡ Se'n => Σadv'a ea = [Σab pab ea ebT ][ Σidx'i ei] = [Σab pab ea ][ Σidx'i] ebT ei = [Σab pab ea ][ Σidx'i] eb ei = [Σab pab ea ][ Σidx'i] g'bi = Σa{ Σb pab Σidx'i g'bi} ea so the coefficients must match dv'a = Σb pab Σidx'i g'bi Now suppose g'bi = δbi h'i2 for orthogonal, then we have dv'a = Σb pab Σidx'i δbi h'i2 = Σb pab dx'b h'b2 ∂v'a/∂x'b = pab h'b2 but this seems too simple, you would never got those other linear terms.