meaning of x'-space

What is the meaning of x'-space PhL 2.23.12 Several topics in this doc: ( today is 2.28.12) (1) you only need to have g' = RgRT in x'-space if you need to preserve scalars (Lai is different) (2) a comment now added early in Section 5 to this effect was developed here. (3) I pulled section 5 (o), but then later reinstated it with small changes. (4) did a pass through Appendix C with edits, since it addresses the subject of x'-space (5) did a little Section 8 work, but that got completely replaced later. 1. One can define a transformation x' = F(x) where x is a point in x-space and x' is a corresponding point in x'-space. F can be any transformation you want which is 1 to 1 and continuous. This transformation can be linearized around any point to get dx' = R dx. The matrix R could end up being ANY matrix, and the 1 to 1 property of F ensures that we will have detR ≠0, so this is the only restriction on R. If we write this out in the form [dx']i = Rij[dx]j (developmental notation) then the [dx]j are the contravariant components of a tensorial vector, and any object which transforms in this manner is a tensorial vector. Notice that nothing has been said about "metric tensors". Both x-space and x'-space are "vector spaces" at this point, they are not yet "metric" spaces. 2. One can by fiat associate an arbitrary metric tensor g with x-space and another arbitrary metric tensor g' with x'-space. These metric tensors then determine the notion of distance and length in each space. 3. Example 1: Suppose x' = F(x) is some highly non-linear transformation with a very arbitrary R(x). Suppose also that all the xi components of x have dimensions of length L and so do all the x'i components of x'i. [ Note that x does not transform like a vector since we don't have x' = Rx ]. In this case, we could if we liked associate a Cartesian metric tensor with each space so g = 1 and g' = 1. This is just what is done by Lai in his treatment of a flow. Then x-space is the "earlier space" and x'-space is the "later space". 4. Suppose V satisfies the rule V' = RV so that V is a tensorial vector. V is a vector in x-space, and V' is a corresponding vector in x'-space. We have two interesting expansions for each vector: V = Viui = V'iei ei = forward tangent base vectors existing in x-space V' = V'ie'i = Viu'i u'i = inverse tangent base vectors existing in x'-space 5. In each space, the "length" of a vector is determined by the metric tensor in that space. Thus, we have |V|2 = gijViVj |V'|2 = g'ijV'iV'j. 6. If for some reason we want to have |V'| = |V|, then we are forced to have a certain relationship between the two metric tensors g and g'. That relation is this g' = RgRT 7. Example 2: In the "curvilinear coordinates application" we choose g = 1 and g' = RRT and F is the transformation from Cartesian coordinates to curvilinear coordinates. Thus, in this application, the length of a vector is the same in the two spaces. Why do we want this length to be the same? Well, we have some velocity in the physical Cartesian world V and it has some magnitude or speed |V|. No matter what coordinate system we choose to use to describe this velocity, it is the same velocity, and it must have the same magnitude or speed. 8. Example of Example 2. For spherical coordinates, there are several ways we can describe the same velocity vector: V = Viui = V'iei = V'ii For the first and last expansion we usually write for spherical coordinates V = Viui = Vx + Vy + Vzz^ V = V'ii = Vr + Vθ + Vφ The speed is the same no matter how we describe the vector V in terms of some components. 9. Misguided example. In my short added section 5 (o), I talk about having g' = FFT = C and I am then able to "interpret" the Cauchy thing C as a metric tensor g'. This is what you would do if you wanted to have the length of dx' be the same as the length of dx. But that is exactly NOT what we want to do for flow analysis! So this interpretation is completely wrong and meaningless in the flow context!!! [ But I did rescue Section 5 (o) using g' = 1 and would-be g' = FFT = C ] 10. Consider x'-space and some g' it might have. There is nothing wrong with "drawing" three right-angled lines to represent the axes of x'-space ( such as axes r,θ and φ for sphericals, or such as θ and φ for my polar elliptical coordinates). The unit vectors are e'n = (0,0,....1...0) regardless of what g' is set to!!! In general, we will NOT have |e'n| = 1 and we will NOT have e'n e'm = 0. The drawing is only meant as a way to place a point like (r,θ,φ). Each point in x-space maps to some point in x'-space. Only if we choose g' = 1 do we get |e'n| = 1 and e'n e'm = 0 . When we have some non-diagonal g', those 90 degree angles have nothing at all to do with the quantity e'1 e'2 . The right angle grid just gives us a unique place to locate each of our points x'. It could have just as easily been a grid of skewed straight lines. 11. Review of Tensor Doc. I am scanning through it now looking for places where the ideas above might have some implications. I find tiny errors, I highlight in red things that might need attention soon. Section 4 is timely, and it is true that in Sections 1,2,3 distance and length is really not mentioned except in a few places. (a) Now in Section 5, I added a very long comment about the invariance of (ds)2 and got the main ideas above stated, and I have done this pretty early in the metric tensor discussion. [ I just reread this comment, did a few edits, it seems fine.] (b) I think I will delete the following comment about symmetry of g: _______________________________________________________________________________ Another view is to again consider (ds)2 = km dxk dxm. An arbitrary matrix km can always be decomposed into the sum of a symmetric matrix and an antisymmetric matrix. The antisymmetric part makes no contribution to (ds)2 since Aijaiaj = 0. Therefore, any such antisymmetric part might as well be thrown out, leaving a symmetric km . __________________________________________________________________________________ The reason is that this argument does not work for the general dot product A B ! (c) I also have to remove my entire Section 5 (o) because of my misunderstanding of scalarity of ds2 not being part of that theory. So I will just park that here and remove it from the contents. [ But then I fixed it up here, and then cut and pasted the fixed-up section (o) back into tensor doc! ] ________________________________________________________________________ (it was here) __________________________________________________________________ I now need to look at Appendix C and Section 8 at the start of a new day because of my many temp documents floating around. I think today I broke the conceptual logjam regarding "the Lai connection". I see now what my "Cartesian view x'-space" is the one relevant for Lai, and I need to ponder how to phrase this in all my sections. Appendix C work. Section (e) is the main item of interest, at least the first main item. // It took a long time, but I have now rehashed this section, getting into it new a few new ideas. The usefulness of the Cartesian view is less obscure now due to the Lai application as well as integration . Section (f) was easy, no changes made. Section (g) OK, except for the last claim. I will try to verify that here: ds2right, Cart = dx2 + dy2 ds2left, Cart = dθ2 + dρ2 = dθ2 for all these vars Therefore dsright = dsleft = dθ So why do I claim that the ratio is h'θ ? What is my general rule concerning the meaning of hn' ? dx(n)= hn' dx'n dx(θ) = hθ' dθ dx(n) = en dx'n = eθ dθ mag = h'θdθ OK, I added a good derivation and the claim is accurate! [ reread and is good. ] Section (h) is next, and this then is the last section of App C. Done, but I had to highlight in red some sections that depend on Section 8 and are really more general than App C is addressing. I could just delete them, but I will wait until Section 8 has been rewritten. Review of my section 8 rewrite which is now sitting in a separate document. [ this has all been replaced later ] Here is a summary of section (d) 1. The following object (dAn)i = σ (-1)n-1 εiabc..x (e1)a(e2)b.... (eN)x Πi≠ndx'i // en missing transforms as a vector density of weight -1, and therefore dA'n = |J|-(-1) R dAn = |J| R dAn = g'1/2RdAn 2. Since one can write dV = | dAn dx(n) |, it follows that dV transforms as a scalar density of weight -1. dV' = |J| dV 3. I would add that dx is a regular vector and therefore transforms in this manner: dx' = R dx and I could then summarize this section on how my three items of interest transform: dx' = R dx | dx'| = | dx | dA'n = |J| R dAn | dA'n | = |J| | dAn | dV' = |J| dV dV' = |J| dV If someone asks how, in the transformation from Cartesian x-space to Curvilinear-view x'-space, the above three lines would be my answer.