Question 1 v2 REVIEWED

Question 2 2.15.17 This doc seems confused. I was at a confused stage on this topic. I had not yet understood that Frame S and Frame S' both exist in x-space and x'-space is superfluous. Question: How do you apply the full machinery of tensor doc to a transformation which is a simple rotation, such as perhaps R = Rz(α) for some fixed α value. [ good question ] Comment: I never added a section to tensor doc addressing this question (I don't think). [ correct ] When I say "machinery", I mean x' = F(x) becomes x' = Rz(α)x which is a linear transformation. The R matrix is therefore R = Rz(α) according to (2.8.7). The underpinning concept is this Picture A Both metric tensors are δij which means up and down indices are the same in both "spaces". Contravariant and covariant vectors are the same, so we talk just about a "vector". A tensorial vector by definition is something that transforms according to V' = RV or V'i = RijVj In this case, a position vector like r' = Rr is in fact a tensorial vector. The basis vectors in un are Cartesian basis vectors for x-space. I think we can treat x-space as Frame S in the language of frames doc. [ correct ] The covariant a b dot product is just the regular geometric dot product. [ all the above are true, granted ] What is the meaning of Picture A? If you apply it to r' = Rr it seems to map a vector r in x-space into a vector r' in x'-space. For polar coordinates the x'-space has axes labeled r and θ since x' = (r,θ). You then map a point (x,y) in x-space into a point (r,θ) in x'-space. In Picture A, x-space is always the domain of a mapping or transformation, and x'-space is the range of that mapping. For our z rotation, let's try to draw a picture, showing that a point in the domain maps into a point in the range: The arrow on the left is rotated by amount α relative to the one on the right. Note all axis labels. I have not discussed the word "Frame" so far. Let us now digress to look at the infamous passive discussion pictures: PASSIVE VIEW Frame S Frame S' For the first time the word Frame appears. In Frame S, the vector r has coordinates (x,y). On the right in Frame S' it has different coordinates (x',y') that I can calculate by doing r' = Rz(α)r . Note that r = x + y = x' ' + y' ' x = r x' = ' r and so on There is only vector r, and it has different coordinates in the two coordinate systems. Confusion: When I say "calculate r' = Rz(α)r ", I must mean (r)'i = Rij(r)j since the only vector anywhere has the name r. I cannot very well say r = Rz(α)r . Observation: One cannot help but notice the strong similarity between the rightmost picture of the set of 3 and the x'-space picture above. The big difference is that the vector is called r' in one case, and r in the other. Question: How do we make sense of this r and r' labeling confusion? What would the active view pictures look like? Rotate the apparatus within Frame S: ACTIVE VIEW Now we have a new vector r' = Rr but it is still in Frame S. So this is a mapping x-xpace → x-space. The apparatus was rotated. Now compare this to Picture A This is the exact same mapping, but instead of R: X→X, it is R: X→X' so the domain in Picture A has a different name X' from the domain X of the active view business. For polar coordinates, a mapping like the active view does not make much sense. That mapping has to go from (x,y) space to (r,θ) space which must have axes of different labels. Question: Is Picture A closer to the Passive View, or to the Active View? I think the Passive View wins here hands down. The only problem there is the r and r' discrepancy. In the Active View we have no space anywhere whose axes are labeled x' and y'. So let's give up on Active view and do side-by-side with the passive view stuff: PASSIVE VIEW PICTURE A In both these ideas, the x'-space coordinates of the destination point are exactly the same. So that is another positive feature of the comparison. The problem is the NAME of the destination point. r = x + y = x' ' + y' ' x = r x' = ' r and so on I think the only resolution is to say that in r' is a shorthand for (r)' which is a notation I was considering in frames doc, and then you have (r)' = Rr as the means to compute the x'-space coordinates of the only vector in town which is r. You then have (r)'i = Rij(r)j which is exactly what you want to say. This gives you the x'-space coordinates of the vector r. Then (r)'1 = x' and (r)'2 = y' and then everything seems to work. You could create a new vector r' in x'-space such that (r')i = (r)'i , that would not violate any laws of the universe. Then for that new vector you would say r' = Rr so that (r')i = Rij(r)j . I think this passive view interpretation of Picture A must be correct! If so, we then have Implications: The basis vectors called e'n in frames doc are also called e'n in this tensor doc interp. (e'1 = ' ) The basis vectors called en in frames doc are called un in this tensor doc interp. (u1 = ) . Frame S is associated with x-space Frame S' is associated with x'-space. Error noted: Somewhere I said (it may be gone) that "both Frame S and Frame S' are in x-space, and the basis vectors in x-space are the un for Frame S and tangent base vectors en for Frame S. " This idea vaguely applies to the center picture of the triple passive set below (as is shown below in the discussion of the en), but I think I have to reject the entire idea with regard to Picture A. I much prefer the new interpretation arrived at above. Question: In this tensor doc interp, what has become of the tangent base vectors en ??? According to tensor doc (3.2.4), en ≡ Se'n = R-1e'n = Rz(-α)e'n (3.2.4) What exactly does the equation en = Rz(-α)e'n mean? In the Picture A world, I have to consider this to be a mapping from x'-space to x-space which agrees with my repeated comment that the en are in x-space. Then for example e1 = Rz(-α)' I will add these vectors to the Picture A picture above Coordinate line interpretation. If I vary y' while holding x' = K = constant, that is a vertical line in x'-space (parallel to the y' axis) . Into what does this map in x-space? Whatever it is, it is called the y' coordinate line. The tangent to this y' coordinate line will be e2 . So consider, Rz(α) = r' = Rz(α)r = = x' = cosα x - sinα y x = cosα x' + sinα y' y' = sinα x + cosα y y = - sinα x' + cosα y' If x' = K we then have K = cosα x - sinα y sinα y = cosα x - K y = (cotα) x - (K/sinα) = mx + b For small α, this indicates a large positive slope and perhaps a large negative intercept. Here I show two coordinate lines passing through the point r . The directions of the en agree with the previous picture. So then, I asked "what has become of the tangent base vectors en" ? They exist in x-space and I have drawn them. They of course depend on α and so are NOT axis aligned in x-space. They do line up with the directions of the primed axes in my middle picture above from the passive view procedure. Conclusion: We do not associate the tangent base vectors en of tensor doc with either the x-space or the x'-space Cartesian axis-aligned basis vectors in Picture A. However, they are associated with the Frame S' axes in the middle picture above where we have back-rotated the Frame S basis vectors to get the Frame S' ones. But this is not a tensor doc Picture A situation . Next step: how does this apply to vectors other than r' = Rz(α)r ? Like V' = Rz(α)V ? Related question: this require Special Case #4 ?