Question 1 v1 REVIEWED

Question regarding Frames doc and Tensor doc 2.15.17 This doc has been reviewed on 3.18.17 and I am happy with it. Question: How do en and e'n in frames doc relate to basis vectors in tensor doc? [ a good question ] In tensor doc there are 4 basis vector types. un axis aligned in x-space en tangent base vectors in x-space e'n axis aligned in x'-space u'n I guess inverse tangent base vectors in x'-space Linear differentials are S and R. tensor doc frames doc e'n = R(x) en en = Re'n e'n = Sen [ all data above I think is correct, no conclusions yet ] I have to start somewhere. Suppose I make this connection tensor doc: un = axis-aligned in x-space, expansion V = Vi ui , gij [ ok ] frames doc: en = axis-aligned in Frame S, expansion V = Vi ei , δij [ok ] Let us now probe into what this comparison implies. If we restrict to rotations, then x is a vector, and we have the transformation x' = Rx in tensor doc, similar to v' = Rv in frames doc. In tensor doc, I say that x is in x-space while x' is in x'-space. What exactly does that mean? [ good question ]I use it in the sense that x' = curvilinear coordinates, x = Cartesian. The axes in x'-space might be θ and r for polar coordinates. You would think that for a simple rotation as transformation, x' = Rx, the axes of x'-space would be something like x' and y' [ok]. So you would think that x-space was Frame S with axes x and y [ok], while x'-space was Frame S' with axes x' and y' [wrong!]. So in this view, x-space is Frame S with y going up, and x'-space is Frame S' with y' going up. [wrong!] Here is what at least seems to be a reasonable picture x' = x' '+ y' ' x = x+ y [ wrong ] This above picture is then only about coordinates x and x'. There is no V and V'. I guess I would say x'μ = Rμνxν so show components of the mapping. Question: In tensor doc, does V' = RV have anything to do with the above pictures? How about this picture as a candidate: As in the previous picture, here each drawing shows the appropriate components of the vector . In Frame S the components of the velocity vector V are vx and vy. [ok] In Frame S' the components of the velocity vector V' are v'x and v'y. [ok] This seems to be an Active View situation since we have two vectors V' and V. We write V'μ = RμνVν Actually, I think the next picture is better V' = v'x'+ vy'' V = vx+ vy This is compatible with my tensor doc expansions V' = V'1 e'1 + V'2 e'2 + V = V1 u1 + V2 u2 +... But this is not really Active View because in Active View, V and V' are drawn in the same space/Frame which has the same basis vectors. You have V = Viei = V' [ the above picture is reasonable if you say that V is a vector in x-space and V' is a different vector in x'-space with components there of V'i , but we really need Frame S and Frame S' in the same space. ] **************************saved from bottom of Question 2 doc ******************* Now of the following I am not so sure: Conjecture: Maybe the e'n could serve as the Cartesian basis vectors of x'-space and maybe we can associate x'-space with Frame S'. I think I would like things to work out that way, but right now I have no idea. [ again, for Active View need both Frames in the same space, with Frame S' axes back rotated .] What are the en ? According to 3.2.4, en ≡ Se'n . (3.2.4) Therefore en = (R-1)e'n e'n = Ren Now I will use Rz(α) = My frames doc basis theorem (1.1.29) says, for a rotation Q (in this application. Q = R-1 local vars) e'n = Qen en = Qnm e'm = Q-1 = Q Applying that idea here I would use Q = R = Rz(α) and conclude that e'n = Ren en = Rnm e'm = R-1 = R Then explicitly we have = Then e1 = cosα e'1 - sinα e'2 e2 = sinα e'1 + cosα e'2 Now I have always said that the en are vectors in x-space. [ that is correct] So this equation seems a little strange since it says that e1 is a sum of two x'-space vectors. So I am at once confused. The un were supposed to be the Cartesian basis vectors in x-space = Frame S. You see why this is a "hard question". Let's go back to the original equation en ≡ Se'n quoted above. What is the meaning of this equation? It appears early in tensor doc. I start there by defining the e'n to be axis aligned vectors in x'-space. For the polar coordinates case, these are e'r and e'θ in x'-space which is the space whose axes are labeled by r and θ. [ I think all these matters are resolved in Appendix J ! There I have no need really for the e'n and u'n vectors in x'-space because we can completely ignore x'-space. ]