Section 1 Formalism and Rank-2 Tensors

Section 1 Formalism and Rank-2 Tensors PhL 3.11.17 I want to argue that in the Passive View a rank-2 tensor transforms as follows (T)' = RTRT (1a) which is a shorthand for this more precise statement (T)'ij = RiaTab(RT)bj (1b) In the Passive View, there is no new tensor (T') as there is in the Active View. Here is the Dirac version of (2) (T)'ij = <e'i| T | e'j> = <e'i|en><en| T |em><em| e'j> = Rin(T)nmRjm = [RTRT]ij (2) It is hard to deny this Dirac derivation of (1b) Question #1. What then can be said about a new object (T') which is defined as follows (T') ≡ RTRT (3a) or (T')ij = RiaTab(RT)bj (3b) Comparing the right sides of (1b) and (3b) one would have to conclude that (T')ij = (T)'ij If you are free to define (T') as in (3a) due to no name conflicts, then yes, this is the expected result. Example: Consider T = Jk , one of the rotation generators. Does this transform as a passive rank-2 tensor, yes or no? If yes, we would expect to have (Jk)' = RJkRT (4a) (Jk)'ij = Ria(Jk)ab(RT)bj (4b) If no, we might instead expect to have (Jk)'ij = (Jk)ij . In this case we might vaguely think of Jk as just a matrix of constants which is the same in any Frame. In some sense it is a scalar. How does one answer the question: which is it? Yes or no? More nubbins questions. One method is to assume one answer and find a contraction. The usual issue is having clean definitions of things. Let's try one of the solutions and see where it leads. Hypothesis: Jk does not transform as a passive rank-2 tensor. It is just a matrix of constants. Then (Jk)' = RJkRT not true (Jk)'ij = Ria(Jk)ab(RT)bj not true (Jk)'ij = (Jk)ij true In the context of this hypothesis, we could define a new object as follows (J'k) ≡ RJkRT or (J'k)ij ≡ Ria(Jk)ab(RT)bj // = (R-1)kn(Jn)ij by Theorem 1 This is a matrix whose elements are functions of angles like θ,φ,ψ. Claim: You could say (as in tensor doc) that the pair of objects { Jk, (J'k)} transform as an active rank-2 tensor. But in the passive view, the object (J'k) does not exist. Only (Jk)' exists. You cannot just define (J'k) ≡ (Jk)' because (J'k) is already defined as RJkRT and this is different from (Jk)' = (J)k. Fact: Nothing stops you from defining (J'k) ≡ RJkRT if you want to do that. But you cannot make any statement about { Jk, (J'k)} transforming as a passive tensor. Review of Active and Passive Rank-2 tensors In the Active View, you have the original tensor T in Frame S and you create a new tensor T' also in Frame S such that (T') = RTRT, The pair {T,T'} transforms as an active rank-2 tensor. There is no Frame S', there is no new basis e'n for Frame S' In the Passive View, you have the original tensor T in Frame S and there is no tensor T'. You define Frame S' with rotated basis vectors e'n and the rank-2 transformation rule is then (T)' = RTRT . Application to object A of the ω calculation. In this calculation I have the following matrix A = aiJi as follows: A ≡ - [ ( cosψ + sinθsinψ)(iJ1) + ( sinψ - sinθcosψ)(iJ2) + ( + cosθ)(iJ3) ] The matrix elements are Aij ≡ - [ ( cosψ + sinθsinψ)(iJ1)ij + ( sinψ - sinθcosψ)(iJ2)ij + ( + cosθ)(iJ3)ij ] According to the Hypothesis above, since the Jk do NOT transform as passive rank-2 tensors, we must conclude that the object Aij also does NOT transform as a passive rank-2 tensor. Furthermore, since we know that (Jk)'ij = (Jk)ij we also know that (A)'ij = Aij . We could if we wanted define the object (A') = RTRT . It turns out that when we do the ω calculation we encounter the object B ≡ RTAR so we cannot associate this (A') with B. But suppose we were to give the matrix A a different name like this A' ≡ - [ ( cosψ + sinθsinψ)(iJ1) + ( sinψ - sinθcosψ)(iJ2) + ( + cosθ)(iJ3) ] (A')ij ≡ - [ ( cosψ + sinθsinψ)(iJ1)ij + ( sinψ - sinθcosψ)(iJ2)ij + ( + cosθ)(iJ3)ij ] Then we would end up with B = RTA'R (A') = RBRT and in this case we could identify B with A where {A, (A') } form an active rank-2 tensor. Does this add any clarity to the calculation? I think it does NOT add clarity, because then you have to mention active rank-2 tensor in a passive view world of frames doc. So don't bring this up, and just use A and B!!! Well, in Frames Doc I do write things like T' = RTRT. Is that active or passive? OK, I have now made Sections 1.1,2,3 more consistent by always saying (T)' = RTRT . I am trying to be totally consistent. In doing this in a separate file, I found various spelling errors! Question: What should one make of this development ? (T')mn ≡ <em| T' |en> = <em| RTR-1 |en> = <R-1em|T |R-1en> = <e'm| T |e'n> = (T)'mn Well, here I defined a new tensor T' ≡ RTR-1 or (T') ≡ RTR-1 . If you can do this with no naming conflicts, then yes, you will get (T')mn = (T)'mn . Same as what happens with vectors. You have (V') = RV and if you can define V' ≡ RV with no meaning conflict, you get (V')i = (V)'i.