Paradox 1 of 2_23_17 v2

Paradox 1 of 2_23_17 v2. PhL 2.23.17 Consider these pictures a Frame S b Frame S passive c Frame S active (1) 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". In particular, ' = Rz(-α) . This is the "passive view Picture". If you rotate year head to the right and look at b, you are in Frame S'. The Frame S components of the equation ' = Rz(-α) are ()'i = [Rz(-α)]ij()j. (2) In picture c, "Frame S active", we write ' = Rz(α) as we would write for any vector V, and vector ' points northeast. This is the "active view Picture". Components are (')i = [Rz(α)]ij()j All three drawings are in Frame S which we associate with our aligned piece of paper. 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. So what do I say next? Here is one option 1. In frames doc, we want to take the passive view because: If we have a vector V in Frame S, we want to know: What are its V's components in Frame S'? In the above example, if we have a vector in Frame S, we get its components in Frame S' using paragraph (1), not paragraph (2). Paragraph (1) says ()'i = [Rz(-α)]ij()j which gives the back-rotated vector components and this points to the southeast. 2. Question: Adopting this frames doc passive view, (a) is there any role for the equation ' = Rz(α) ? (b) is there any role for an equation V' = Rz(α)V ? For item (b), we can certainly define V' = Rz(α)V as some new vector V' in Frame S. But we cannot define ' = Rz(α) because ' is already defined as ' = Rz(-α). We might define V' = Rz(α)V for some general vector V, but we cannot apply this to the position vector,. We cannot say r' = Rz(α)r because for r = we get the wrong answer.