component meaning confusion REVIEWED

Massive Confusion about the Meaning of Components in Wedge Doc PhL 4.3.16 5.16.16. This really was a swirling mess. There were many issues at once: 1. What do you mean by a component index, should you write (V(e))i or (V)(e)i or something else? 2. I had ei in the entire document and had to change them all to ui so could have TI with no primes. 3. Notion of a "default basis" which gives meaning to things like gij (being axis aligned basis). 4. I had a Picture with u-space and e-space, all now gone, with its little paradox thing, also gone. I feel that all this complicated confusion has been rendered clearly in wedge doc, eg, above 4.1.10. Chapter 2 of wedge doc In tensor doc and early Chapter 2 of wedge doc I have lots of expressions with component indices: V'a = RabVb dx'a = Rabdxb M'ab = Raa' Rbb' Ma'b' Rab = g'aa'Ra'b' gb'b Va = gabVb a b = gijaibj = gijaibj = aibi = aibi en = g'ni ei (en)i = Rni (en)i = Rni (en)i = Rni (en)i = Rni . (2.3.4) (en)a(en)b = δab (2.3.5) (un)i = gni = δni (en)i = Rni (u'n)i = Rin (e'n)i = g'in = δin (2.6.6) V = Σn Vn un where un V = Vn = [V]n V = Σn Vn un where un V = Vn = [V]n V = Σn V'n en where en V = V'n V = Σn V'n en where en V = V'n V' = Σn V'n e'n where e'n V' = V'n = [V']n V' = Σn V'n e'n where e'n V' = V'n = [V']n V' = Σn Vn u'n where u'n V' = Vn V' = Σn Vn u'n where u'n V' = Vn (2.7.3) Questions: (1) is the "meaning of an index" the same for all these equations? (2) if so, then what is "the meaning of an index" ? I think the meaning is everywhere consistent and is this: Fact: The meaning of a component index on a vector is that it indicates the projection of that vector onto the axis-aligned basis vector of the space in which the vector lives. (2.7.6) For example: Vn = <un| V> = [V(u)]n V'n = <e'n| V> = [V'(e')]n // correct Let's rewrite the above equations in this "fuller" notation: [V'(e')]a = Rab [V(u)]n [dx'(e')]a = Rab[dx(u)]b [M'(e)]ab = Raa' Rbb' [M(u)]a'b' [V(u)]a = gab[V(u)]b for Va = gabVb [V(e')]a = g'ab[V'(e')]b for V'a = g'abV'b a b = gijaibj = gijaibj = aibi = aibi a b = gij[a(u)]i[b(u)]j = gij[a(u)]i[b(u)]j = [a(u)]i[b(u)]i = [a(u)]i[b(u)]i a' b' = g'ij[a(e')]i[b(e')]j = g'ij[a(e')]i[b(e')]j = [a(e')]i[b(e')]i = [a(e')]i[b(e')]i en = g'ni ei does not notation enhancement, no vector components appear (en)i = Rni (en)i = Rni (en)i = Rni (en)i = Rni . (2.3.4) (en(u))i = Rni (en(u))i = Rni (en(u))i = Rni (en(u))i = Rni . (2.3.4) The above are all x-space vectors so all therefore have (u) indices. (u'n)i = Rin (e'n)i = g'in = δin (2.6.6) are (u'(e')n)i = Rin (e'(e')n)i = g'in = δin Really g is in the same business since it is a tensor, so [g'(e')]ab = Raa' Rbb' [g(u)]a'b' [M'(e')]ab = Raa' Rbb' [M(u)]a'b' Rab ≡ (∂x'a/∂xb) = [∂[x'(e')]a / [∂[x(u)]a x ua = [x(u)]a Try V' = Σn V'n e'n or V' = Σn [V'(e')]n e'n Conclusion: In tensor doc and early chapter 2 of wedge, we have a massive world involving indices which have default meanings according to the rule: Fact: The meaning of a component index on a vector is that it indicates the projection of that vector onto the axis-aligned basis vector of the space in which the vector lives. (2.7.6) For example: Vn = <un| V> = [V(u)]n V'n = <e'n| V> = [V'(e')]n  // correct Now things carry on in wedge doc and we get then to Section 2.8. Right off the bat I am back to the old convention in discussing outer products. (a'i)(b'k) = (ΣjRijaj)(ΣmRkmbm) = Σjm RijRkm(ajbm) But then I start wandering a bit. [(ab)(e)]ij = a(e)ib(e)j [(ab)(e,u)]ij = a(e)ib(u)j . (2.8.14) Here I would use markers to mean something non-default. But this is the default in other parts of the same chapter. I guess I am arguing that the outer product concept applies to any basis. In Section 2.9 I come back to a b = gijaibj = gijaibj = aibi = aibi This starts off in the default tensor doc basis, a b = <ui|uj><ui|a><uj|b> = <a|ui> <ui|uj><uj|b> = <a | b> gij ai bj You can put in any other completenesses you want for x-space, so could say = <a|ei> <ei|ej><ej|b> = [a(e)]i g'ij [b(e)]j = g'ij [a(e)]i[b(e)]j so a b = [g'(e')]ij [a(e)]i[b(e)]j and that looks very strange. Consider though a b = [g'(e')]ij [a'(e')]i[b'(e')]j Is it true that [a'(e')]i = [a(e)]i <e'i| a'> = <ei| a> yes it is true Dirac always tells you the right answer. Then down to Section 2.10 (b) where I am now using the second default convention again. Finally we are at Section 2.11. I say these things λi = <ei| basis functional // λi = (ei)T λi(v) = <ei|v> basis function // λi(v) = (ei)Tv (2.11.c.2) v = Σjvjej or |v> = Σjvj|ej> (2.11.c.3) Last line is convention #2 for components. λi(v) = <ei|v> = vi // = [v(e)]i (2.11.c.5) λi(ej) = <ei|ej> = δij . (2.11.c.6) I just back and forth between conventions!! Now consider T(v1,v2) = <T | v1, v2> = ΣabTab < ea, eb| v1, v2> = ΣabTab <ea| v1> <eb| v2> = ΣabTab (v1)a (v2)b (2.11.e.8) This last line really says T(v1,v2) = ΣabT(e)ab (v1(e))a (v2(e))b Is this really a tensor contraction so you conclude it is a scalar? Ask Dirac: T(v1,v2) = <ea,eb| T> <ea|v1><eb|v2> = <ea,eb|ui,uj><ui,uj| R><ea|v1><eb|v2> = <ea |ui><eb |uj><ui,uj| T><ea|v1><eb|v2> = <ui,uj| T><ea |ui><ea|v1><eb |uj><eb|v2> = <ui,uj| T><ui |ea><ea|v1><uj |eb><eb|v2> = <ui,uj| T><ui |v1><uj |v2> = Tij(v1)i(v2)j = scalar! So why did this work out? I know that um = Rnm en so this is a linear transformation from x-space to another copy of x-space, so I guess that would "preserve a scalar object". Tilt reversed sum would have no transformation effect just as in the u basis,e tc. = <eb |uj><ui,ui| T><ea |ui><v1|ea> <eb |uj><eb|v2> Chapter 4: Changing the default meaning One could extend this default notation into the tensor expansion world like so V = ΣnVn un M = ΣmnMmn um un which is the default of M = Σmn[M(u)] mn um un However , if you want your tensor expansions to be in terms of arbitrary basis vectors bn , to be consistent you must include a marker, or you will have created an inconsistency: V = Σn [V(b)]n bn M = Σmn [M(b)]mn bm bn I have treated en as a generic basis, so I have created a major inconsistency by letting it default to having no marker! I do at least say that I am doing that. Question: What should I do about this inconsistency? It sounds very dangerous when you have any kind of Rab transformations nearby! For example, v w = ( Σiviei)( Σjwjej') = Σijviwj (eie'j) (4.1.6) Here I violate tensor doc by saying v = Σiviei which is really v = Σi[v(e)]iei Similarly |v> |w> = Σijviwj |ei> |ej> . (4.1.6a) At least no Rab are around at this point. But then suddenly Meanings of tensor. The word "tensor" has a weak and a strong meaning. In the weak meaning, a rank-2 tensor is something that has components with two indices like Tij. In the strong meaning, a rank-2 tensor is a set of components Tij which transform in a certain manner with respect to some underlying transformation, T'ab = Raa' Rbb' Ta'b' Picture A T ϵ VV (2.1.6) and here I have the different convention for Ta'b', for example. Then I say Default Notation. In the rest of this document, unless otherwise specified, expansions in Vk shall always be on the ei basis, and the label (e) appearing for example on [T(e)]ab in (2.6.8) or [ei(e)]b in (2.6.4) will be omitted. For example, Tab = [T(e)]ab for T ϵ VV // Tab = [T(e,e')]ab for T ϵ VW (ei)b = [ei(e)]b = δib for V // (e'i)b = [e'i(e')]b = δib for W (eiej)ab = [(eiej)(e)]ab = [ei(e)]a [eb(e)]b = (ei)a (ej)b = δia δjb . (4.1.10) Holy cow. Now suddenly (ei)b = δib since I am defaulting to (ei(e))b = δib I am scanning now through Chapter 4. I still have λi = <ei| . Cut to the Chase and Chapter 10. There I want to be using the Rij stuff, and for that reason there I select the two λi to be the axis-aligned ones! Then λi(v) = vi produces convention #1 components and those are the ones for which Rij applies. I think that is the real endgame. Idea: How about forgetting about picture F and (e) and (u) and just doing this: V = ΣnV'n en M = Σmn M'mn emen T = Σii....i T'ii....i (ei ei ..... ei) . (2.10.14) Just add a tiny little prime that then there is no "second convention" when you do expansions. T = ΣI T'I eI (2.10.14) (2.10.21) Since no new convention is made, I don't have to worry about what basis vectors look like in that new convention. I can forget all the (e) and (u) stuff unless we have mixed basis expansions, then I would show it. In x'-space we have this V' = Σn V'n e'n where e'n V' = V'n = [V']n V' = Σn Vn u'n where u'n V' = Vn So if you want to think of u'n as some "arbitrary basis vectors, then there you are/ V' = ΣnVn u'n M' = Σmn Mmn u'mu'n T' = Σii....i Tii....i (u'i u'i ..... u'i) . (2.10.14) Now I think all is well again.