meaning of basis vectors in wedge 10_6 REVIEWED

Meaning of Basis Vectors in wedge Ch 10.6 versus other PL sources. PhL 3.7.16 5.17.16: I had an ongoing battle with the meaning of vector component indices (including those on basis vectors) before things stabilized on the axis-aligned meaning. Also, I was less clear at this time about the four basic kinds of basis vectors. Perhaps the u' and e' ones were not yet entered into Chapter 2. I need all four kinds to deal with the kinematics of transformations in Chapter 10. I think it is all clear now. Also, I am writing my first cut of "basis vectors for tangent space" which now appears in Section 10.6 (e). Docs to be open: tensor doc, wedge doc, and Ch 10_6 doc. 1. Review wedge doc on this subject. First, review the bloated discussion in wedge doc on this subject. First I say en = ∂x/∂x'n = ∂'nx tangent base vectors // taken from tensor doc 7.13.5 which I can translate to ten = ∂t/∂xn = ∂(x)nt tangent base vectors I never write this in Ch 10-6, by the way. Next, in both wedge and tensor doc I say for components, (en)i = Rni (en)i = Rni (en)i = Rni (en)i = Rni . (2.3.4) Wedge doc brings up the question: What kind of components are these components? Wedge doc then introduces the un which are the same as those presented in tensor doc. These are the "axis-aligned basis vectors in x-space" with un um = gnm and so on. Along the way I do show things like (en)i and (um)i see 2.4.4 without commenting on the "kind" of components these are. Ah, but this is wrong thinking. I quote from wedge doc: "The meaning of V(e)n is [V(e)]n. The superscript (e) goes with the tensor, not with the component index." I arrive at wedge 2.5.4 which shows lots of statements about e and u vectors where no components are mentioned at all! Everything is vectors. These are various expansions Then I go from Picture A to Picture E at the start of 2.6 Then in (2.6.4) for the first time I started annotating things with (u) and (e) superscripts. As I read this, it seems that I am associating these superscripts with "the component" as if it were a component type. But I though I trashed that whole idea! (yes, see above) So already things are inconsistent in wedge doc. So I will repair equation (2.6.4) right now. // Done. (placement of the labels and indices) Now by (2.6.7) I have lots of objects to wonder about en(e) un(e) en(u) un(u) What exactly are these things? Well en is a vector no matter where you view it from, but it has different components when probed in different spaces. Here is I think the key idea V = Σn V(u)n un where un V = V(u)n V = Σn V(u)n un where un V = V(u)n V = Σn V(e)n en where en V = V(e)n V = Σn V(e)n en where en V = V(e)n . (2.5.1) (2.6.4) The (*) superscript tells you what basis vectors you are expanding on and this is after I go into (e) and (u) space. I used to have V = Vn un where un V = Vn V = Vn un where un V = Vn V = V'n en where en V = V'n V = V'n en where en V = V'n . (7.13.10)' (2.5.1) Then I applied this to all the basis vectors. Now go back and look at (en)i = Rni (en)i = Rni (en)i = Rni (en)i = Rni . (2.3.4) How do I know which kind of components these are? I look at the detail em = (un em) un = Rmn un (em)n = Rmn So this tells me that the above are really (u) vectors. So it should be relatively easy to move this to my t-space and x-space But what exactly is the difference between un and en? I define the en as en = ∂x/∂x'n = ∂'nx tangent base vectors en = ∂x/∂x'n = ∂'nx . reciprocal base vectors (7.13.5)' (2.3.1) so they are defined in terms of the transformation x = F-1(x') . Later I find their components in (2.3.4) which is really (en(u))i = Rni (en(u))i = Rni (en(u))i = Rni (en(u))i = Rni . (2.3.4) So for these components you would never say (en(u))i = δni ! OK, then how do I define the un? They exist in the same x-space where en exists. I define them by their components! (un)i = δni (un)i = δni (7.18.3)' (2.4.1) Eventually I show that um = Rnm en em = Rmn un um = Rnm en em = Rmn un um = Rnm en em = Rmn un um = Rnm en em = Rmn un . (2.5.5) 2. How does this apply to Section 10.6 which I am trying to write? I need to keep this stuff in mind to keep honest, I really need to do this: Now I guess I would define en ≡ ∂t/∂xn = ∂(x)nt (un)i = δni (un)i = δni (7.18.3)' (2.4.1) Then these en and un would be two kinds of vectors which exist in t-space. So my notation was wrong when I wrote this in Ch 10.6: We first define a standard set of n basis vectors in t-space, {tei } i = 1,2...n basis for t-space (tei )j = δij components of these basis vectors in t-space (10.6.3) I think these vectors should be called ui and the t subscript is then not needed since we know they exist in t-space. But maybe use the t anyway to assist keeping things straight. So I should have said We first define a standard set of n basis vectors in t-space, {tui } i = 1,2...n basis for t-space (tui )j = δij components of these basis vectors in t-space (10.6.3) Next I say, Since these are basis vectors in t-space, we map them into x-space using (xV) = R (tVb) of (10.6.2), (xei) = R(tui) or (xei)j = Σa=1n Rja (tui)a = Σa=1n Rja δia = Rji i = 1,2...n j = 1,2..m . (10.6.4) The general mapping is correct, but I am not sure now what these "e vectors" are, since I define them as the maps under R of the u vectors. My tangent base vectors are related to the u's like this em = Σn Rmn un but that is a vector transformation, not a component one. My desired equation then is this (xei) = R(tui) What does this say in original tensor doc notation? (e'i) = R(ui) I don't see anything like this. I think in tensor doc you would say (u'i) = R(ui) which would be the mapping of u from x-space to x'-space. In 7.18.3 I don't see this as stated. In components this would be (u'n)i = ΣjRij(un)j = Rin and that IS CORRECT, it is in 7.18.3. [ correct ] Note added: The equation (u'i) = R(ui) does appear in the to-be-corrected tensor doc in equation (3.5.3). There the corrected equation says un = S u'n which is the same as u'n = Run . Notice that u'n are clearly stated in tensor doc to be the "tangent base vectors for the inverse transformation". This is what I want! That is the fact I have been wanting to see but could not seem to make fly. Now here is the translation (u'i) = R(ui) is (xui) = R(ut) So now here is my next fix to Ch 10.6: Since these are basis vectors in t-space, we map them into x-space using (xV) = R (tVb) of (10.6.2), (xui) = R(tui) or (xui)j = Σa=1n Rja (tui)a = Σa=1n Rja δia = Rji i = 1,2...n j = 1,2..m . (10.6.4) Since there are m basis vectors in x-space, we define the rest of the xui arbitrarily such that the m basis vectors {xei} in Rm are linearly independent, so xui = as needed i = n+1, n+2 .....m . (10.6.5) Note that equation (xui) = R(tui) or (xui)j = Σa=1n Rja(tui)a is a "component transformation". OK, so this is a quite different notation. I will now to create Section 10-6 v1 and change this stuff. OK, my v1 edit is going fine. Now what happens with metric tensors? In original notation I have u'n u'm = gnm  |u'n| = = hn (scale factor) u'n = gni u'i u'n u'm = δnm u'n = gni u'i u'n u'm = gnm  |u'n| = . (6.2.4) This translates in Picture F to say xun xum = (tg)nm  |xun| = = thn (scale factor) xun = tgni xui xun xum = δnm xun = tgni xui xun xum = (tg)nm  |xun| = . (6.2.4) But these are exactly the same according to 7.18.3 for the t guys. So xui xuj = tui tuj = (tg)ij Now I have n vectors in each group if I forget about the higher xui vectors. Then we have an n x n square situation here that just does not involve the higher vectors and g is an n x n matrix. What about xg?? Back to (7.18.3) I do show this (un)i = gni (un)i = gni which would translate into (tun)i = tgni (tun)i = tgni Well that is pretty simple to figure out (tun)i = δni (tun)i = tgij (tun)j = tgin But what happens for xg? We don't have a δni rule in x-space. But ux is u' and we know g' = R g RT  g'ab= Raa'Rbb'ga'b' → g'ab = Raa'Rbb'ga'b' so this last would say (xg)ab = Raa'Rbb'(tg)a'b Question: Can I interpret my xui as "tangent base vectors"? 1. In Picture A, the axis-aligned basis vectors e'i in x'-space are related to the tangent base vectors ei in x-space according to en ≡ Se'n . (3.2.4) (en)i = Sin = ∂xi/∂x'n = Rni = (∂x'n/∂xi) or en = ∂x/∂x'n = ∂'nx // cannot write this another way (3.2.6) S = [e1, e2, e3 .... eN ] matrix = N columns (3.2.7) e'n = R(x) en . = axis-aligned basis vectors in x'-space (3.3.2) I can add to (3.2.6) (shown above in red) using Sab = Rba = (∂xa/∂x'b) = (∂x'b/∂xa) Sab = Rba = (∂xa/∂x'b) = (∂x'b/∂xa) Sab = Rba = (∂xa/∂x'b) = (∂x'b/∂xa) Sab = Rba = (∂xa/∂x'b) = (∂x'b/∂xa) . (7.5.16) so that Sin = Rni = (∂xi/∂x'n) = (∂x'n/∂xi) which I have added in red above. 2. Translating the main part of item 1 to our x-t world we get ten ≡ Sxen . (3.2.4) (ten)i = Sin = ∂ti/∂xn = Rni = (∂xn/∂ti) or ten = ∂t/∂xn = ∂'nt // cannot write this another way (3.2.6) S = [te1, te2, te3 .... teN ] matrix = N columns (3.2.7) xen = R(t) ten . = axis-aligned basis vectors in x-space (3.3.2) I think this is all correct, BUT I don't care about axis-aligned vectors in x-space, I care about them in t-space. So I want to talk about the inverse transformation. In Picture A, the axis-aligned basis vectors e'i in x'-space are related to the tangent base vectors ei in x-space for the forward transformation x' = F(x) and we have e'n = R(x) en like all vectors. In Picture A, the axis-aligned basis vectors ui in x-space are related to the tangent base vectors u'i in x'-space according to u'n = Rs(x') un = S(x') un, so u'n are the tangent base vectors of the inverse transformation x = F-1(x'). Note that R = (DF). Question: For the second bullet, don't we still have V' = RV as the usual vector rule under transformation F? If so then why is it that for the vector V = un we are getting u'n = Sun ? I don't have to go to Picture F to be confused! Well I guess you would say that the usual vector rule is V = SV' for the inverse transformation since it maps x'-space to x-space. Comment: The usual vector rule goes with e'n = R(x) en for the forward transformation. For the inverse transformation we get u'n = Rs(x') un= S(x') un so u'n = Rs(x') un is "the usual vector rule" relative to the inverse transformation. Now I will translate the above two bullet items to Picture F. In Picture F, the axis-aligned basis vectors xei in x-space are related to the tangent base vectors tei in t-space for the forward transformation x= φ(t) and we have xen = R(t) ten like all vectors In Picture F, the axis-aligned basis vectors tui in t-space are related to the tangent base vectors xui in x-space according to xun = Rs(x) tun = S(x) tun, so xun are the tangent base vectors of the inverse transformation t = φ-1(x). Note that R = (DF) But in sec 10.6 v1 I have written instead that (xui) = R(tui) which disagrees with the last bullet above. I quote from tensor doc 3.5 The inverse tangent base vectors u'n and inverse coordinate lines A complete swap x' ↔ x for a mapping x' = F(x) of course produces the "inverse mapping". This has the effect of causing R ↔ S in the above discussion. The tangent base vectors for the inverse mapping would I say R↔S just because x' ↔ x . For example before doing this swap, we have Sab = Rba = (∂xa/∂x'b) = (∂x'b/∂xa) Then after the swap we have (doing x' ↔ x ) sSab = sRba = (∂x'a/∂xb) = (∂xb/∂x'a) where sS means after the swap. So I would say sSab = Rab and sRab = Sab ****************** junk line G = F-1 SG = RF RG = SF