Active vs Passive in tensor doc

Active vs Passive in tensor doc PhL 11.23.16 This doc shows a tensor-doc model for the active/passive story using ei = R-1ui as the back-rotated vectors. Then Frame S = ui and Frame S' = ei and x'-space is not even involved. For me, this is a rather difficult issue. In tensor doc when I write V' = RV, I always say that V is a vector in x-space and V' is a vector in x'-space. So I don't have the notion of V' as a new-named vector in x-space which is the active rotation in x-space of the vector V. So "active rotation" in my usual sense does not really appear in tensor doc. And "passive rotation" is also quite confusing to me there as well, but I hope to clear it up below. [ all reasonable comments ] 1. Statement of a tensor doc Fact which I failed to realize was true 1 2. A neutral review of active and passive rotations using c,c' vectors 2 3. A connection to tensor doc 4 1. Statement of a tensor doc Fact which I failed to realize was true First, I note that I added Section 7.19 to tensor doc to clarify the meaning of things like e'i by adding superscripts like (u). I first ran into this in wedge doc, and it is written up in the wedge folder in a slightly different superscript notation in "tensor doc and component types", but the Section 7.19 writeup is fine. I think I did this 7.19 after encountering the issue in wedge doc involving various paradoxes. In that writeup and in 7.19 I note that (un)i really means (un)(u)i or (un(u))i which is <ui | un> (en)i really means (en)(u)i or (en(u))i which is <ui | en> Since these two basis vectors are unprimed, they are implicitly expressed in the ui basis. Now it is a fact that un = Ren but I never write this equation anywhere in tensor doc. First here is a proof that this is valid, [ a newly realized and correct fact, it does appear I think in wedge doc ] (un)i = δni // from tensor doc (7.18.3) (Ren)i = Rij(en)j = Rij Rnj // from tensor doc (7.18.1) = δin // orthog rule #3 of (7.6.4) I always say that un and en are vectors in x-space, and in that space un = Ren shows how they are related. I think I avoid writing this because I always want the thing on the left to be in x'-space, and so what appears in tensor doc is e'n = Ren where (e'n)i =(un)i = δni . I like this form for vector fields, V'(x') = RV(x), but for the non-field basis vectors it is just a simple fact that un = Ren . Therefore: Fact: In x-space of tensor doc, we have en = R-1un so the en are back-rotated versions of the un. [right] Of course R in general is not a pure rotation, so "back-transformed" would be better in that case. This is a simple fact that I just never realized was true. In that doc I did not use un until late in my development, perhaps that is why I did not see this. I was always more interested in transformations between x-space and x'-space. 2. A neutral review of active and passive rotations using c,c' vectors BASIS VECTORS We use the following notation: Frame S basis ci Frame S' basis: c'i (0) I have picked a "neutral" basis vector name c to avoid any inadvertent contact with tensor doc. The ci are assumed to be axis-aligned in Frame S so we know that (ci)j = δij (1) The ci' are back-rotated versions of ci so that c'i = R-1ci (2) which one can of course also write as ci = R c'i . (3) In components equation (2) says (c'i)k = (R-1)kn (ci)n = (R-1)kn δin = (R-1)ki = (RT)ki = Rik and this is true even if R is not a rotation (used tensor doc Section 7.9). So we have this simple fact for the components of the back-rotated basis vectors, (c'i)k = Rik (4) Fact: Equation (3) is equivalent to c'i = Rijcj . // right side is sum of vectors (5) Proof: [Rijcj]k = Rij (cj)k // all is in the c-basis here = Rij δjk // from (1) = Rik = (c'i)k // from (4) [ I do the above proof also in frames doc Section 1 (a) ] VECTORS We can expand the vector V in either basis, V = (V(c))ici V ci = (V(c))i Frame S expansion (6) V = (V(c'))ic'i V c'i = (V(c'))i Frame S' expansion (7) Now using just the ci basis, I can create a new vector in Frame S called V' where V' = RV or (V'(c))i = Rij(V(c))j // active rot in Frame S (8) where Rij is the rotation matrix in the ci basis. We can expand this new vector V' in two ways V' = (V'(c))ici V' ci = (V'(c))i Frame S expansion (9) V' = (V'(c'))ic'i V' c'i = (V'(c'))i Frame S' expansion (10) At this point we can say [ok] 1. The components of V' in Frame S are: (V'(c))i 2. The components of V in Frame S' are: (V(c'))i (11) Fact: The active/passive claim is that these two components are exactly the same. Proof: (V'(c))i = Rij(V(c))j // from (8) = Rij(V cj) // from (6) = V (Rijcj) // a rewrite = V c'i // from (5) tilted = (V(c'))i // from (7) 3. A connection to tensor doc I will cut and paste the entire Section 2 above showing what I think is the connection between Section 2 above and tensor doc. [ in this stuff below un and en are exactly the tensor doc basis vectors ] BASIS VECTORS We use the following notation: Frame S basis ui Frame S' basis: ei (0) The ui are axis-aligned in Frame S so we know that (ui)j = δij (1) The ei are back-rotated versions of ui so that ei = R-1ui // as in Section 1 above (2) which one can of course also write as ui = R ei . (3) In components equation (2) says (ei)k = (R-1)kn (ui)n = (R-1)kn δin = (R-1)ki = (RT)ki = Rik and this is true even if R is not a rotation (used tensor doc Section 7.9). So we have this simple fact for the components of the back-rotated basis vectors, (e)k = Rik (4) which agrees with tensor doc (7.18.1) Fact: Equation (3) is equivalent to ei = Rijuj . // right side is sum of vectors (5) Proof: [Rijuj]k = Rij (uj)k // all is in the c-basis here = Rij δjk // from (1) = Rik = (ei)k // from (4) Note that equation (5) appears in wedge doc (2.4.4) second line! VECTORS We can expand the vector V in either basis, V = (V(u))iui V ui = (V(u))i Frame S expansion (6) V = (V(e))iei V ei = (V(e))i Frame S' expansion (7) Now using just the ui basis, I can create a new vector in Frame S called V' where V' = RV or (V'(u))i = Rij(V(u))j // active rot in Frame S (8) where Rij is the rotation matrix in the ui basis. We can expand this new vector V' in two ways V' = (V'(u))iui V' ui = (V'(u))i Frame S expansion (9) V' = (V'(e))iei V' ei = (V'(e))i Frame S' expansion (10) At this point we can say 1. The components of V' in Frame S are: (V'(u))i 2. The components of V in Frame S' are: (V(e))i (11) Fact: The active/passive claim is that these two components are exactly the same. Proof: (V'(u))i = Rij(V(u))j // from (8) = Rij(V uj) // from (6) = V (Rijuj) // a rewrite = V ei // from (5) tilted = (V(e))i // from (7) In tensor doc we write (V'(u))i = (V')i and also (V(e))i = (V')i so we can rewrite the expansions above V = Viui V ui = Vi Frame S expansion (6)' V = V'iei V ei = V'i Frame S' expansion (7)' V' = V'iui V' ui = V'i Frame S expansion (9)' V' = (V'(e))iei V' ei = (V'(e))i Frame S' expansion (10) Lines 1 and 2 above agree with lines 1 and 3 from tensor doc, The other two expansions do not appear in tensor doc but they are valid. I think I could show that the last line above (10') is equivalent to this V' = Viei V' ei = V'i Frame S' expansion (10) but I will leave that for now as just a conjecture. The main fact above then is this association Frame S basis ui Frame S' basis: ei These two frames are associated only with x-space in tensor doc. Remember that these two basis vectors exist in x-space, as I say many times in tensor doc. My Big Confusion I think was that I was making this wrong association Frame S = x-space Frame S' = x'-space // wrong The two spaces are not "frames", they are "spaces". Within each space you could construct whatever "frames" you like, and above I have constructed two "frames" in x-space.; Two comments: (1) I have above realized a version of the active passive story in relation to tensor doc by using ei = R-1ui as the back-rotation starting point, and then I just say Frame S = un and Frame S' = ei and yes both the Frames are in x-space and having nothing to do with x'-space. (2) Thus for the active/passive story, I give up the notion that Frame S' is x'-space and Frame S is x-space.