Paradox 1 of 2_23_17

Paradox 1 of 2_23_17. PhL 2.23.17 1. On the one hand, I say that basis vectors are back-rotated so e'n = R-1en. ( ' = R-1) In Frame S components this says (')i = (R-1)ij()j . 2. On the other hand, the vector seems like a normal vector so you should actively rotate ' = R. In Frame S components, (')i =Rij()j . ' 3. How can these both be true? My notation must not be very good if such a simple paradox can exist within its confines. Where are my most recent active/passive pix? OK it is "Interpreting Goldstein's Euler Angle Picture" doc and here is what I say there: ___________________________________________ My usual picture is this a Frame S b Frame S c Frame S active The axes in b are right-hand-rule back-rotated by some angle positive α around 15 degrees. The coordinates (components) of vector V are different in Frame S versus Frame S' as you can see. In the passive view, I can get the Frame S' coordinates from the Frame S ones by this rule: V' = RV. I would write in my frames doc notation, (V)'i = Rij(α) (V)i (1) V' = Rz(α)V and e'i = Rz(-α)ei in Frame S In the active view all within Frame S I would again write V' = RV but I would express this as (V')i = Rij(α) (V)i (2) R = Rz(α) In frames doc, it is item (1) that you normally want to do: express Frame S' components in terms of Frame S components, and maybe vice versa. You think "passive". ________________________________________ Reading it now, it all seems very reasonable for the generic vector V. But what happens if I set V = ??? Let's draw the exact pictures corresponding to those above: a Frame S b Frame S passive c Frame S active Now look at item (1) above: Frame S active Frame S passive ()'i = Rij(α) ()i (1) ' = Rz(α) and e'i = Rz(-α)ei in Frame S ok ' = Rz(-α) in Frame S ok In picture b, "Frame S passive", the vector ' points southeast and we think e'i = Rz(-α)ei and we say that the basis vectors are "back-rotated". We have ()'i = Rij(α) ()i as the Frame S' components of the vector . This is the "passive view Picture". If you rotate year head and look at b, you are in Frame S'. In picture c, "Frame S active", the vector ' points northeast. This is the "active view Picture". Both these drawings are "in Frame S". You obviously cannot have "both situations at once". In the first b situation ' in Frame S points southeast, while in the second c picture ' in Frame S points northeast. It seems to be an either/or situation, but both situations are in Frame S. The paper is Frame S. Look again at item (1) above, The passive problem solved Frame S active Frame S passive ()'i = Rij(α) ()i (1) ' = Rz(α) or e'i = Rz(-α)ei in Frame S The equation ()'i = Rij(α) ()i is the solution to the canonical problem Frames doc is addressing: "What are the coordinates of as observed from Frame S' ?" The statement ' = Rz(α) is simply not true in this passive picture. So how can I write ω' = Rω or in general how can I be writing a' = Ra in Section 1? These are inherently active Picture equations. In the passive picture the vectors ω' and a' don't even exist. So what happens to all of tensor doc where I always write V' = RV ? I say that V is a vector in x-space and then V' is the corresponding vector in x'-space. I am getting worried now. V'i(x') = RijVi(x) In tensor doc I expand vector V on two different bases: V = V1 u1 + V2 u2 +... = ΣnVn un where Un V = Vn Un = gni ui V = V'1 e1 + V'2 e2 +... = Σn V'n en where En V = V'n En = g'ni ei This seems to be a Passive Picture statement. If I say the ui define Frame S and the ei define Frame S', then the V'i are the components of vector V in Frame S' and that is the canonical question solved. Note that both Frame S and Frame S' then exist within x-space. Am I confusing "coordinates" (like x and x') with "Frames" (like S and S') ? How does one reconcile the notion of Active and Passive Pictures with tensor doc? I better stay in frames doc for now, don't want to metastasize my paradox into a huge unmanageable mess which is spinning out of control . I will come back later to tensor doc. Let's get stable first in the frames doc world. And right now things are not stable there. I suspect there are big problems in Section 1.2 or 1.3 where I write a' = Ra which are Active Picture statements. I have not even discussed Pictures there. I am not ready to even think about Euler angles because basic Section 1 stuff is very unstable. So once again, I have to go back to Section 1 and try to get it stabilized. All else drops by the wayside. ************************ = ( ()x, ()y, ()z ) ' = ( (')x, (')y, (')z )