Paradox 1 of 2_23_17 v3

Paradox 1 of 2_23_17 v3. PhL 2.23.17 Statement of Paradox and Resolution are Below 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. Inverting, ()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. Realization A: When I write equations like V' = RV, I have in mind that V is some vector associated with an Apparatus. It could be the position of a point in the Apparatus, so I would then write r' = Rr . It could be a velocity or acceleration. so v' = Rv or a' = Ra for b= 0 I think. In Experiment A (or whatever), Apparatus stays fixed and the basis vectors back-rotate. Here is the key idea: Fact: The vector is a vector, yes, but it is not a point in the Apparatus, it is related to the coordinate system of Frame S. As you back rotate such that ' = Rz(-α), a vector r describing a point in your Apparatus does not change. It stays put in Frame S. It neither forward nor back rotates. Suppose in Frame S you have r = describing a particular point in your Apparatus. In Frame S' this apparatus point appears at r' = Rz(α) r and you then have r' = cosα ' + sinα ' . This goes with the red arrow in b above. ******************************************************************** Statement of Paradox Let's try to state and then resolve our Paradox which is this: 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 it should actively rotate ' = R. In Frame S components, (')i = Rij()j . 3. How can these both be true? ***************************************************************** Resolution Attempt 1: When you talk about "normal vectors" and their rule V' = RV , you are talking about a vector which is "part of" your Apparatus under study. In Experiment A where the Apparatus stays fixed and the basis vectors are back-rotated, the vector V stays put in Frame S and its components in Frame S' is what you mean by the equation V' = RV. If r is a point on the Apparatus, then r would stay put in this experiment. In Frame S' you would have r' = Rr as the new coordinates of that point in Frame S. So the rule V' = RV applies for such a point, and you have r' = Rr . Suppose a point in the Apparatus is located at r = . In Frame S' this same apparatus point is located at r' = R . Yes, ' = R-1. So you conclude that r' ≠ ' . Now examine the sentences in the Paradox. Item 1 is correct, nothing wrong, ' = R-1 . Item 2 says "the vector seems like a normal vector so it should actively rotate ' = R ". This is where mud is thrown onto the windshield. The phrase "normal vector" is not defined anywhere, first of all. Fuzzy thinking. What I mean by "normal vector" is a vector what is somehow "part of" the Apparatus. In a sense a normal vector is one which is "glued to" the Apparatus. This statement works for points of the Apparatus. The velocity of a point in the apparatus is "with the Apparatus". So my first problem is how to describe what I mean by an "apparatus vector". It is any vector that is "associated with" the apparatus. The coordinate system is a second entity, it is not associated with the apparatus. So ************************************************************** Here is the Paradox Resolution 1. Apparatus Vectors are associated with an Apparatus being observed. Examples include points r in the Apparatus, or velocities v of such points, or their accelerations a or angular momenta L or electric field E at some point. If the Apparatus is forward rotated in space by R, all these apparatus vectors actively rotate according to the rule V' = RV. 2. Basis Vectors are associated with a coordinate system or a Frame of reference of an Observer. These might be written ei or for Frame S, and other names for other Frames. If the Frame is backward rotated in space by R-1, all these apparatus vectors actively rotate according to the rule e'i = R-1ei. 3. In the Passive View Experiment, an Apparatus stays fixed in Frame S, and the Basis Vectors are back-rotated e'i = R-1ei. The new coordinates of an apparatus vector are given by V' = RV. 4. In the Active View Experiment, an Apparatus is rotated by R, and the Basis Vectors ei stay put. The new coordinates of an apparatus vector are given by V' = RV. Fact: No vector can be both a Basis Vector and an Apparatus Vector. These are mutually exclusive classes of vectors. It might be that an apparatus point in Frame S is located at r = . In the Passive View Experiment, that apparatus vector r stays put. while the basis vector is back-rotated into ' = R-1. The coordinates of the apparatus vector r in Frame S' are given by r' = Rr =R. There is no equation anywhere which says ' = R because is not an Apparatus vector. We end up with r' = R ≠ ' . In the Active View Experiment, that apparatus vector r rotates into r' = Rr which is some new location in Frame S. The Basis vectors like stay put. There are no vectors named ' and consequently there are no equations like ' = R-1 or ' = R. We now return to that statement 2: Item 2 says "the vector seems like a normal vector so it should actively rotate ' = R ". We now see that "normal vector" means "Apparatus Vector" and in fact is NOT in this class of vectors so the conclusion ' = R is incorrect. ***********************************************************************8 OK, I have written a section in Section 1.2 on all this stuff. The next issue is this: What to do about Fig 1 equations like r = b + r' where you don't have r' = Rr . What kind of "vectors" are these guys? Are these a third class of vectors? Maybe I have to generalize my Active/Passive discussion so it applies more to Fig 1, and not just to Special Case #4. This is very confusing. I will try, while looking at the Fig. In Fig 1 "where is the Apparatus? " It is not shown. We have Observers in the two Frames and each of them sees things. Suppose I glue an Apparatus into Frame S. It has Apparatus Vectors, one of which is r to some point in the Apparatus. I really only care about the Passive View. The Apparatus stays put. WE have two things going on here, rotation and translation. I never made that very clear in frames doc. Suppose V is an apparatus vector in Frame S. If Frame S' had the same origin, we could say that for any apparatus vector, we have V' = RV (passive) as seen in Frame S'. But this does not apply for V = r,v or a in Fig 1