paradox1 with vector w

Paradox with the vector ω PhL 2.12.17 I think this paradox gets resolved within this doc, so no paradox. On the One Hand: Consider this picture where Frame S is inertial. Suppose it happened that e1 is coming out of the plane of paper on the left. Then you would write ω = ωe1 and then ω = (ω,0,0) and these are the components of ω in Frame S. If you treat ω like any other vector, you could ask: "What are the components of ω as measured in Frame S' "? Since the axes of Frame S' are at some weird orientation, the answer will be: "the components of vector ω as seen from Frame S' are complicated". More formally, suppose ei = R(Φ)e'i and suppose ω = ωe1 . Then (ω)'i = ω e'i = ω [R(Φ)-1ei] = [R(Φ)ω] ei = [R(Φ)ω]i = [R(Φ)]ij ωj = [R(Φ)]ij ω δj,1 = ω [R(Φ)]i1 and in general this is some complicated expression for each of the (ω)'i components. On the Other Hand: Start again with the above picture Here you would say: The axes of Frame S' are rotating at ω = (ω,0,0) as seen from Frame S. Now hop onto a camera platform which is rotating CCW at ω about the same axis and this picture becomes Now Frame S' is at rest and Frame S is rotating in effect at -ω . So you would say The axes of Frame S as seen from Frame S' are rotating at -ω = (-ω,0,0) Resolution: The second item I think is valid, and you end up with the statement as claimed, The axes of Frame S as seen from Frame S' are rotating at -ω = (-ω,0,0). However, the components shown of -ω = (-ω,0,0) are still Frame S components, not Frame S' components. The Frame S' components are still something messy. I jump to Goldstein rigid body here because he is going to use ω components in Frame S' !! I work further with this stuff in some other doc. What is the meaning of the ωi in the GPS presentation of rotating objects? So what is the meaning of ω in this equation (5.37) ? The equation says (dL/dt)S' + ω x L = N Since this is a vector equation, we have not yet talked about "components", so we keep going. Now GPS is going to take the components of the above equation!!! This is a tricky operation. If we do this within Frame S' (the rotating frame) then my Commutation Theorem (1.11.1) shows that things commute, so you can say (∂S'L)'i = ∂S'[ (L)'i] = ∂t(L)'i // using below 1.11.1 Then we really should write rigid body doc ∂t(L)'i + [ ω x L]'i = N'i or d(L)'i/dt + εijk(ω)'j(L)'k = (N)'i and indeed it is the Frame S' components that you want!! This last equation appears as and you see how they are using a very compact notation. But I agree, their ωi really are the Frame S' components!!!