Lai coordinate systems

Lai Coordinate Systems PhL 2.5.12 This is an early document concerning dv = (v)dr ≡ G dr . It discusses perhaps for my first time how you might expand G as a tensor-like object on a generic basis bibjT, which subject is now codified in tensor doc App E. Later I realized a certain ambiguous dual meaning for dv and cleared that up below on 2.15, which is really the only item of value in this doc. I then continued consideration of dv = G dr and got the RHS of this little equation understood. But I was confused about the LHS because of the "dn issue" and just stopped. I later treated this issue in another doc "Lai's grad-v method p 54-66.doc" and concluded there that it is such a big issue that the Lai method of doing things only works for the three CL systems he deals with: polar, cylinder, spherical. My method written up in App F of tensor doc works for all CL systems. My Lai Section 3.29 notes were originally at the end of this doc, but I moved them to their own separate doc. A. 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 where v is just some "vector function". When we write (v)ij drj = (∂jvi) drj above, we are quietly making the assumption that coordinate r is Cartesian! Suppose we introduce G = (v) and change the coordinate name from r to x, and then we have dv = G dx dvi = [G(x)]ij dxj [G(x)]ij = ∂jvi where I put an (x) superscript on G to indicate that when we think of it in the Cartesian basis, it is equal to the quantity shown. The quantities dvi and dxj are implicitly Cartesian because they have no primes. We could just write Gij and say that this is also Cartesian implicitly because G has no prime. Also implicit in the above is that we have only one Cartesian system with unit vectors I have written in many different notations in my various documents, to wit, = n = n = (x)n and both dv and dr are expanded on these guys. Comment added: Now that I have written App E and F, some slight adjustment of the above is required. Even if x-space were non-Cartesian, one could still define (v)ij = (∂jvi) in x-space. The more important point is that, although vi and ∂jφ are true vectors, ∂jvi is not a true rank-2 tensor, and for that reason, if we go to x'-space we don't get (v(x'))ij = (∂'jv'i) for general F. Lai however always assumes that x-space is Cartesian, which is fine. And we can move indices up and down there at will. OK to here B. Digression to look at tensor doc. I show in my tensor doc how one can think of "expanding" a tensor G in different "bases", in complete analogy with expanding a vector V. One thinks of V or G as being a vector and an operator (respectively) in the usual EN Hilbert space. One can select any basis one wants, and in each basis V or G will have different components. I don't think I made this idea very clear in tensor doc and I probably will improve it. [ I have now done so.] As an example, I write in Section 7 that V = V1 u1 + V2 u2 +... = ΣnVnun where Un V = Vn Un = gni ui where the un are "axis aligned basis vectors" in x-space, meaning (un)i = δn,i. Since we might have g ≠1 in x-space, I don't call these "unit vectors". In Hilbert space bra-ket notation the above appears as | V > = 1 | V > = Σn |un><un| V > = Vn |un> Vn = <un| V > where 1= Σn |un><un| states the fact that this basis un is complete, and <un|um> = δn,m states that the basis vectors are orthonormal. For a rank-2 tensor like G, I write in tensor doc G = Σij Gij (uiuj) = Σij GijuiujT Now, suppose bn are some arbitrary basis for En. Then we must properly write (standard notation) V = ΣnVnbn bn V = Vn where bn is the unique basis which is dual to bn where we know bn bm = δn,m. We can similarly expand the tensor G this way G = Σij Gij (bibj) = Σij GijbibjT Now consider bnTG bm = bnTΣij GijbibjT bm = Gnm Thus, we have G = Σij GijbibjT Gij = biTG bj and to make this a little clearer, we might write G = Σij [G(b)]ijbibjT [G(b)]ij = (bi)TG bj so that the matrix G(b) is "with respect to the basis bi" Now suppose we select a basis bi which is orthonormal. That means bnbm = δnm , and that in turn means that we have the dual basis equal to the basis, so bn = bn . In this case, our results above can be written this way V = ΣnVn bn bn V = Vn G = Σij [G(b)]ijbibjT [G(b)]ij = (bi)TG bj Finally, suppose in this special situation we decide that we shall represent the components of G as in the tensor doc developmental notation, so lower indices (we could use overbars for covariant G). Then V = ΣnVn bn bn V = Vn G = Σij [G(b)]ijbibjT [G(b)]ij = (bi)TG bj Note added. As noted in App F, and as obvious here, (1) [G(b)]ij =[G(b)]ij = [G(b)]ij= [G(b)]ij (2) [G(b)]ij are not the components of a tensor under F. Notice that this does NOT imply that bn is a Cartesian system basis, it is just an orthonormal one, like spherical coordinates with hatted basis vectors. Now, suppose we denote by (x')n the orthonormal basis vectors of some orthogonal curvilinear coordinate system. Then the above become V = ΣnV(x')n (x')n (x')n V = V(x')n G = Σij [G(x')]ij (x')i (x')jT [G(x')]ij = (x')iT G (x')j Admittedly the notation is a bit cluttered, but hopefully some precision is gained at the cost of the clutter! Comment: At the end of tensor doc App E (b) I comment that (x')j = (x')j (vector = vector) when x'-space g' is orthogonal, and I prove that this is so. OK to here C. Resume discussion. So with B above in mind, we can write dv = G dx // below is an expansion of the vector dv. dv = Σn dv(x')n (x')n = Σn dv'n (x')n = Σn dv'n h'n (x')n (*) wrong, see below dx = Σn dx(x')n (x')n = Σn dx'n h'n (x')n G = Σij [G(x')]ij (x')i (x')jT _____________________________________________________________________ Two meanings of dv. How do we reconcile the above dv expansion with this Lai-like differential v = Σn v(x')n (x')n => line 1 dv = Σn dv(x')n (x')n + Σn v(x')n d(x')n line 2 as just derived here dv = Σn dv(x')n (x')n line 3 as quoted from above This is a very good and interesting and revealing question, as many are. To answer it, consider the following simpler "contradiction" (simpler notation) v = Σn v'n en line 1 v is a vector dv = Σn dv'n en line 2 dv is also a vector dv = Σn dv'n en + Σn v'n den line 3 differential of line 1 above The two objects on the LHS of line 2 and line 3 look exactly the same, but they are not the same! On line 3, we assume there is some underlying variation in the coordinate x', call it dx'n. This variation causes a change dv'(x') in the quantity v'(x'), and it causes a change den in en as we well know). So starting with line 1, we have to add both these variations to get the total change dv. On line 2 no variation in x' is involved at all. Rather, dv(x) is just some "short vector" at location x and dv'(x') is a short vector which is the mapping of dv(x) in x'-space. In Lai we use such "short vectors" as probes in a flow. In tensor doc we have the prototyte dv' = R dv contravariant vector. Since there is no variation in x', there is no change in en(x') in line 2. Yes, R is a certain deriviative. The main point: when you write dv, you need to know which dv you mean!!! In the above equation dv = G dx, we ARE varying x, and this dv should be the differential dv. Therefore my line (*) above is wrong because it is missing the second term! _____________________________________________________________________ In this section I will highlight wrong expressions in red. Now, let's insert the last three expansions into the first line to get dv = G dx Σn dv'n h'n (x')n = Σij [G(x')]ij (x')i (x')jT Σn dx'n h'n (x')n LHS wrong Now the RHS contains this combination (x')jT(x')n = δn,j so we then have Σn dv'n h'n (x')n = Σij [G(x')]ij (x')i h'j dx'j Now on the LHS replace index n by index i Σi dv'i h'i (x')i = Σij [G(x')]ij dx'j h'j (x')i (*) Then since (x')i is a complete basis, we obtain h'i dv'i = Σj [G(x')]ij h'j dx'j where [G(x')]ij = (x')iT G (x')j h'i dv'i = Σj [(v)(x')]ij h'j dx'j where [(v)(x')]ij = (x')iT (v) (x')j Thus we arrive at "the matrix" for tensor G in this orthogonal curvilinear x' coordinate system. If we were to now insert 1 = Σn (x)n (x)nT twice as our Cartesian unit expansion, we have [G(x')]ij = (x')iT G (x')j = (x')iT Σn (x)n (x)nT G Σm (x)m (x)mT (x')j = Σn Σm {(x')iT (x)n } {(x)nT G (x)m }{(x)mT (x')j } Now {(x')iT (x)n } = {(x')iT } = [(x')iT]n = [(x')i]n {(x)mT (x')j } = [(x')j]m {(x)nT G (x)m } = [G(x)]nm so we can write [G(x')]ij = [(x')i]n [G(x)]nm[(x')j]m which relates the curvilinear components of G to the Cartesian components. Meanwhile, we know that (x')i = ei / h'i where ei are the tangent base vectors so that (these are all contravariant components) [(x')i]n = (ei)n/ h'i = Sin/ h'i = Rni/h'i = Qni // see p 88 of tensor doc, Q = a rotation mat [(x')j]m = Sjm/ h'j = Rmj/h'j = Qmj Then we have [G(x')]ij = Qni [G(x)]nm Qmj = (QT)in [G(x)]nm Qmj or G(x') = QT G(x)Q This says that the elements of our two matrices are related by a similarity with a rotation matrix Q. Now back up to equation (*) and write it in vector notation as dv = Σi{ Σj [G(x')]ij dx'j } (x')i // see 2.33.15 for an example, T = G Now expand v, since like dv it is also a vector, v = Σi v'i (x')i // see 2.33.11 for example Now one can say [ this time I got the LHS correct! ] dv = Σi (dv'i) (x')i + Σi v'i [d(x')i] // see 2.33.16 for example I am tracing Lai page 56 which I skipped at the time since I did things a different way, but now I want to see what "his way" was. What do I have to say about objects d(x')i in general? This is the part I missed. I know that e'n = R(x) en => en = S e'n but this does not seem helpful. I can also write d(x')i = Σj [∂'j(x')i] dx'j At this point I "ran out of steam" and ground to a halt. OK to here In this section C, my work was mainly on the RHS of the equation dv = Gdr. I think the RHS conclusions above are all correct, namely, the idea of [G(x')]ij = (x')iT G (x')j . But I got nowhere with the LHS. After I finally got the LHS stated correctly, I was unable to "do anything" with it and stopped. This then is where my "Overview and Summary" came to premature end. Now, in all the discussion to this point, dead ending?