Mystery 1 REVIEWED

Mystery 1 PhL 2.12.17 Notes 2.24.17. Here I ask about meaning of [Bc]'i and this leads to clarity on this question and also leads one to Dirac notation. This is now all understand and encoded into frames doc Section 1.1. Complaint: I try to make my rotation notation as clear, clean, sharp, unambiguous, laser-focused as I can possibly make it. But as soon as I ask the simplest question, I cannot answer the question because I don't understand the notation, or it is incomplete or unclear or ambiguous. Assume that we have Frame S and Frame S' as described in frames doc where en = R e'n for a fixed R. Example 1: In frames doc I write in (1.5.7) : a(t+dt) = Rn(dφ) a(t) where Rn(dφ) = exp(-i dφ n J) where a is an arbitrary vector. Question #1: Presumably the above equation is written in Frame S. What does the equation look like in Frame S' ? Question #1 Answer Attempt A: If the equation is "covariant" under rotations, maybe it has this form in some Frame S' a'(t+dt) = Rn'(dφ) a'(t) If a is an arbitrary vector, then we should have a'(t+dt) = Ra(t+dt) a'(t) = Ra(t) n' = Rn If the above conjectured equation in Frame S' is valid, then it says Ra(t+dt) = Rn'(dφ)Ra(t) Apply R-1 to both sides to get a(t+dt) = [R-1 Rn'(dφ)R] a(t) = Rn(dφ) a(t) If this is all correct, then we must have [R-1 Rn'(dφ)R] = Rn(dφ) and [R Rn(dφ)R-1] = Rn'(dφ) If the above is correct, this is how you would know how to compute Rn'(dφ). I am not sure this is all correct, but it might be correct. One lesson here is that you cannot just think of Rn as a dumb matrix of constants which has no transformation rule. It really is a rank-2 tensor and transforms like one! [ that fact is now quite clear in Section 1.1 where we have matrices like R and R'. It happens they are the same. ] Question #2: How do you take components of the above vector equation in Frame S and Frame S' This has to do with when do matrices have primes on them! Question #2 Answer Attempt A: Start with a(t+dt) = Rn(dφ)a(t) [a(t+dt)]i = [Rn(dφ)a(t)]i = [Rn(dφ)]ij [a(t)]i Frame S' [a(t+dt)]'i = [Rn(dφ)a(t)]'i = ??? [ good question! ] Here we have one of my high-ambiguity issues? What happens on the above line? Suppose we had a = Bc B = rank-2 tensor ai = [Bc]i = Bijcj Frame S a' = B'c' covariant, Frame S' (a')i = [B'c']i = B'ij(c')j Frame S components of vector equation a' = B'c' (a')'i = [B'c']'i = ??? So despite years of tensor doc work, I run aground on the above simple situation. [ This was a good question. When you do [Bc]'i do you get B'ij(c')j ? Answer is yes. ] Let's try Dirac notation Question #3: How do the basis vectors work in Dirac Notation? [ In order to understand primes on matrices, it is helpful to go into Dirac notation. All this stuff and more is now installed into Section 1.1. ] This topic all by itself is so completely confusing that a whole doc is warranted just on it!!! [ is there some other doc? ] Answer: en = R e'n |en> = |R e'n> = R| e'n> <e'm|en> = <e'm|R| e'n> ≡ R'mn = R(p,p)mn Rmn = <e'm|R| ei><ei| e'n> = R(p,n)mi Rni = R(p,n)mi RTin so Rmn = R(p,n)mi RTin RmnRnk = R(p,n)mi RTin Rnk = R(p,n)mk = (R2)mk e'n = R-1en |e'n> = |R-1 en> = R-1| en> <em|e'n> = <em|R-1 en> = <em|R-1| en> = R-1mn = Rnm = R(n,n)nm transpose this last line to get (since real) : <e'n|em> = Rnm compare to earlier line then to get: <e'n|em> = Rnm = R'nm so matrix R' = R or R(p,p) = R(n,n)nm Do I really believe this last result? R(p,p)mn = <e'm|R| e'n> = <e'm|ei><ei|R| ej><ej|e'n> [ p means prime ] = R(p,p)mi R(n,n)ij R(p,p)nj [ n means noprime ] or simply Rmn = Rmi Rij Rnj = Rmi Rij R-1jn = Rmi δin = Rmn so OK Start over with |en> = |R e'n> = R| e'n> <em|en> = <em|R e'n> = <em|R| e'n> δm,n = <em|R| e'n> = R(n,p)mn = R(n,p)nm (n,p) = (noprime, prime) mixed basis |e'n> = |R-1 en> = R-1| en> <e'm|e'n> = <e'm|R-1 en> = <e'm|R-1| en> = <e'm|RT| en> = <en|R |e'm> = R(n,p)nm δm,n = R(n,p)nm = R(n,p)mn Notice the this mixed-basis matrix R(n,p) is symmetric! Also R(n,p)nm = <em|R| e'n> = <e'n|RT |em> = RT But since everything is real, we must have <e'm|en> = R'mn = Rmn Then: a = Bc <ei|a> = <ei| B |ej> <ej|c> // components in Frame S ai = Bij cj Alternately we could say <e'i|a> = <e'i| B |e'j> <e'j|c> (a)'i = B'ij (c)'j // components in Frame S' where B'ij = <e'i| B |e'j> = <e'i|ej><ej| B |ek><ek|e'j> = R'ij Bjk OK, early fiddlings with Dirac, now all encoded into Section 1.1.