Archive old Section G_7

Archive old Section G_7 PhL 3.11.17 G.7 Computation of ω (Method 2) In this section our approach to computing ω for the Euler Angle rotation is to find an equation which involves ω and solve it for ω! More or less at random, we choose (1.7.4), (den/dt)S' = – ω x en . (1.7.4) (G.7.1) Unlike our Method 1 computation of ω in the previous section, here we shall have no need for Goldstein's geometric Figure (G.5.5) or the various variables like ξ and ζ . We will, however, need various algebraic results developed in Section G.3. Recall from (G.5.19) that the Goldstein Euler angle rotation is given by R = Rz(-ψ)Rx(-θ)Rz(-φ) . (G.5.19) (G.7.2) In order to avoid a hundred minus signs, we shall temporarily negate all three angles, then when we are done, we will undo this negation. We therefore temporarily take R to be, R(ψ,θ,φ) = Rz(ψ)Rx(θ)Rz(φ) ≡ R(Φ) = exp(-i Φ J) . // temp (G.7.3) Comment: To find Φ we could write out the matrix Rz(ψ)Rx(θ)Rz(φ), and decompose it into its symmetric and antisymmetric components S and A. In theory we could then compute Φ (=θ) and (=) from (G.2.14) and come up with an explicit expression for Φ = Φ. The reader is just reminded that this is mechanically possible, but luckily we have no need for the result (which "ain't purdy"). We could also compute dΦ from the following R(ψ+dψ,θ+dθ,φ+dφ) = R(Φ+dΦ) = exp(-i [Φ+dΦ] J) but again there is no need to do this. Note that Φ and dΦ will generally not be in the same direction. We now set about constructing the left side of (G.7.1) starting with our fundamental equation from (1.1.29) which relates the Frame S and Frame S' basis vectors, en(Φ) = R(Φ)e'n . (G.7.4) We are a Worker/Observer sitting in Frame S' and we want to compute the change we see in the Frame S basis vector en due to Euler angle changes dψ, dθ and dφ. Then, (den)S' = R(Φ+dΦ)e'n - R(Φ)e'n // e'n is a constant in Frame S' = Rz(ψ+dψ) Rx(θ+dθ) Rz(φ+dφ) e'n - Rz(ψ) Rx(θ) Rz(φ) e'n ≡ dR e'n (G.7.5) // (den)S' = (dR)Tni e'i = (dR)in e'i // apply sep to each rot in dR // this agrees with paradox2 v3 doc where dR ≡ [ Rz(ψ+dψ) Rx(θ+dθ) Rz(φ+dφ) - Rz(ψ) Rx(θ) Rz(φ) ] . (G.7.6) Comment: Equation (G.7.5) is a vector equation. Because we assume that de'n = 0 in writing (G.7.5), that equation is valid only in Frame S' (and not in Frame S). A vector equation is only a shorthand for its component equations, so we are not allowed to take the Frame S components of (G.7.5). When we take the Frame S' components, the rotation matrices are at first of the form R(Φ)'ij , but as shown in (1.1.35) this is the same as R(Φ)ij with no prime. The upshot is that the matrices like Rx(θ) appearing in (G.7.6) are our standard rotation matrices of (E.2.1). We then have dR' = dR, but we shall continue to write dR' (and A' below) as a reminder that everything is only valid when evaluated in Frame S'. Replace e'n in (G.7.5) using (G.7.4) to get e'n = R-1(Φ) en ↔ |e'n> = [R(Φ)]nj |ej(Φ)> ok to here (den)S' = dR e'n = dR [R-1(Φ)en ] = [dR R-1(Φ) ] en . (G.7.7) = |ej(Φ)><ej(Φ) | dR R-1(Φ) |en> = [dR R-1(Φ)]jn |ej(Φ)> = [R dRT ]nj |ej(Φ)> this disagrees with paradox2 v3 which shows [(dRT)R]nj |ej(Φ)> at this point!! More detail: (den)S' = |ej(Φ)><ej(Φ) | [ R(Φ+dΦ) - R(Φ)] R-1(Φ) |en> = |ej(Φ)><ej(Φ) | [ R(Φ+dΦ)RT(Φ) - 1] |en(Φ)> In paradox 2 v2 I use these facts [R(Φ+dΦ)]'in = <e'i|en(Φ+dΦ)> = <e'i| R(Φ+dΦ)|e'n> [R(Φ)]'in = <e'i|en(Φ)> = <e'i| R(Φ)|e'n> So now redo the above (den)S' = dR e'n = dR [R-1(Φ)en ] = [(dR) R-1(Φ) ] en = [(R(Φ+dΦ) - R(Φ)) R-1(Φ) ] en = [ R(Φ+dΦ) RT(Φ) - 1] en = [ |e'r><e'r|R(Φ+dΦ) |e's><e's| RT(Φ) - 1] |e'm><e'm| en = [ |e'r>R(Φ+dΦ)rs<e's| RT(Φ) |e'm> - |e'm>] <e'm| en> = [ |e'r>R(Φ+dΦ)rs RT(Φ)sm - |e'm>] Rmn(Φ) = [ R(Φ+dΦ)rs RT(Φ)smRmn(Φ)|e'r> - Rmn(Φ) |e'm>] = [ R(Φ+dΦ)rs RT(Φ)smRmn(Φ)|e'r> - Rrn(Φ) |e'r>] = [ R(Φ+dΦ)rs RT(Φ)smRmn(Φ) - Rrn(Φ)] |e'r> = [ [R(Φ+dΦ)RT(Φ)R(Φ)]rn - Rrn(Φ)] |e'r> = [ [R(Φ+dΦ)]rn - Rrn(Φ)] |e'r> = [dR]rn |e'r> = [dR]in |e'i> ok to here Now go back to the operator form (den)S' = dR e'n = [(dR) R-1(Φ) ] en I don't know how to matricize this other than what I just did above. You might argue that (dR) R-1(Φ) is just a rotation so you should be able to use the basis theorem. But dR contains rotations at different times!!! Thus you cannot just treat it as an operator at time t and use |er><er| completeness to get the right matrix elements. This is a very subtle issue and the paradox2 method avoids this whole problem. Next, divide by dt to get (den/dt)S' = [(dR/dt)R-1(Φ) ] en ≡ A en A ≡ [(dR/dt) R-1(Φ) ] (G.7.8a) = (1/dt) [dR R-1(Φ) ] en = (1/dt) [(R(Φ+dΦ) - R(Φ)) R-1(Φ) ] en = (1/dt) [(R(Φ+dΦ)RT(Φ) - 1 ] en Insertion 3/10/17: Take components in Frame S' right here: [(den/dt)S']'i = [A en]'i = A'ij (en)'j = A'ij Rjn // note prime on A where A is defined as shown. Recall that we are trying to solve the following equation for ω, (den/dt)S' = – ω x en (G.7.1) so we seem to be on the right track. We will set A en = – ω x en and solve for ω. So far objects like the objects above like R(Φ), dR, Rz(ψ) and A are really operators, not matrices. However, they become matrices when we rewrite (G.7.8a) using completeness of the basis states : | (den/dt)S'> = | A en> = A | en> = |ei><ei| A | en> = |ei> Ain = Ain |ei> . Then (G.7.8) becomes, in non-Dirac notation. (den/dt)S' = Ajn ej , (G.7.8b) Insertion 3/10/17: Take components in Frame S' right here: [(den/dt)S']'i = Ajn (ej)'i = AjnRij // note no prime on A Compare with last insertion: A'ij Rjn = AjnRij or A'ij RjnRTnm = AjnRijRTnm or A'ijδjm = RijAjnRTnm or A'im = [RART]im At this point then everything is consistent. We must now compute this matrix A, so we start with dR which we compute to first order in the differential angles, dR = [ Rz(ψ+dψ) Rx(θ+dθ) Rz(φ+dφ) - Rz(ψ) Rx(θ) Rz(φ) ] = [ Rz(ψ)Rz(dψ) Rx(θ) Rx(dθ)Rz(φ)Rz(dφ) - Rz(ψ) Rx(θ) Rz(φ)] = [ Rz(ψ){1-idψJ3} Rx(θ){1-idθJ1}Rz(φ){1-idφJ3} - Rz(ψ) Rx(θ) Rz(φ)] = [-idψ Rz(ψ)J3Rx(θ)Rz(φ) - idθ Rz(ψ) Rx(θ)J1Rz(φ) - idφ Rz(ψ) Rx(θ)Rz(φ)J3 ] . (G.7.9) We have used the first order expansion (1.5.6) for the small angle rotations. Dividing by dt gives, (dR/dt)S' = [ -i Rz(ψ)J3Rx(θ)Rz(φ) - i Rz(ψ)Rx(θ)J1Rz(φ) - i Rz(ψ)Rx(θ)Rz(φ)J3 ] . (G.7.10) We now assemble the matrix A and process it with vigor, making use three times of the trivial sandwich rule (G.3.6) as shown in blue, A ≡ (dR/dt) R-1(Φ) // where R-1(Φ) = Rz(-φ)Rx(-θ)Rz(-ψ) = [ -i Rz(ψ)J3 Rx(θ)Rz(φ) - i Rz(ψ) Rx(θ)J1Rz(φ) - i Rz(ψ) Rx(θ)Rz(φ)J3 ] Rz(-φ) Rx(-θ) Rz(-ψ) = [ -i Rz(ψ)J3 Rx(θ) - i Rz(ψ) Rx(θ)J1 - i Rz(ψ) Rx(θ)Rz(φ)J3Rz(-φ) ] Rx(-θ) Rz(-ψ) = [ -i Rz(ψ)J3 Rx(θ) - i Rz(ψ) Rx(θ)J1 - i Rz(ψ) Rx(θ)J3 ] Rx(-θ) Rz(-ψ) = [ -i Rz(ψ)J3 - i Rz(ψ) Rx(θ)J1Rx(-θ) - i Rz(ψ) Rx(θ)J3Rx(-θ) ] Rz(-ψ) = [ -i Rz(ψ)J3 - i Rz(ψ) J1 - i Rz(ψ) Rx(θ)J3Rx(-θ) ] Rz(-ψ) = [ -i Rz(ψ)J3 Rz(-ψ) - i Rz(ψ) J1 Rz(-ψ) - i Rz(ψ) Rx(θ)J3Rx(-θ) Rz(-ψ) ] = [ -i J3 - i Rz(ψ) J1 Rz(-ψ) - i Rz(ψ) Rx(θ)J3Rx(-θ) Rz(-ψ) ] = [ -i J3 - i M1 - i M2 ] (G.7.11) where M1≡ Rz(ψ) J1Rz(-ψ) = [ R3(ψ) J1R3(-ψ)] M2 ≡ Rz(ψ) Rx(θ)J3Rx(-θ)Rz(-ψ) = R3(ψ) [ R1(θ)J3R1(-θ) ] R3(-ψ) . (G.7.12) We now call upon our non-trivial sandwich formulas in (G.3.8) to simplify thing further : M1 = R3(ψ) J1R3(-ψ) = cosψJ1 + sinψ J2 // (G.3.8) line 5 (G.7.13) M2 = R3(ψ) [ R1(θ)J3R1(-θ) ] R3(-ψ) = R3(ψ) [ cosθ J3 - sinθ J2 ] R3(-ψ) // (G.3.8) line 2 = cosθ [ R3(ψ) J3 R3(-ψ)] - sinθ [R3(ψ)J2R3(-ψ)] = cosθ J3 - sinθ (cosψJ2 - sinψ J1) // (G.3.6) and (G.3.8) line 6 = sinθsinψJ1 - sinθcosψJ2 + cosθ J3 . (G.7.14) Then, A = [ -i J3 - i M1 - i M2 ] = [ -i J3 - i (cosψJ1 + sinψ J2) - i (sinθsinψJ1 - sinθcosψJ2 + cosθ J3) ] = [ -i J3 - i cosψJ1 - i sinψ J2 - i sinθsinψJ1 + isinθcosψJ2 - i cosθ J3 ] = (-i)[ J3 + cosψJ1 + sinψ J2 + sinθsinψJ1 - sinθcosψJ2 + cosθ J3 ] = (-i)[ ( cosψ + sinθsinψ)J1 + ( sinψ - sinθcosψ)J2+ ( + cosθ)J3 ] = - [ ( cosψ + sinθsinψ)(iJ1) + ( sinψ - sinθcosψ)(iJ2)+ ( + cosθ)(iJ3) ] . (G.7.15) Insertion 3/10/17. The above at this point is only an operator equation! A and the Ji are operators. Operators do not get primes or no primes. I want to evaluate this object [(den/dt)S']'i = [A en]'i = (A)'ij (en)'j = (A)'ij Rjn = [(dR/dt) R-1]'ijRjn = [(dR/dt)]in where (A)'ij = <e'i| A |e'j> so we will want to know <e'i| iJk |e'j> = (iJk)'ij. I need (iJk)'ij = εkij in order for things to work. This means that the Frame S' observer uses *** as his representation of the J's. Doing this, and looking at the other expression, [(den/dt)S']'i = – εijk(ω)'j(en)'k = –εijk(ω)'jRkn = –εikj(ω)'kRjn = εijk(ω)'kRjn we then get the desired result that A'ij = εijk(ω)'k So in this approach, everything hinges on the Frame S' observer using (iJk)'ij = εkij . I have now studied this issue in Paradox2 v2.doc and I conclude there that in fact (Ji)mn = (Ji)'mn . Here that implies that A'ij = Aij so A is then not really a tensor. We then get the correct result above. How then does the Frame S Plane work out ? 3/10/17. (den/dt)S' = Aen (den/dt)S' = – ω x en Take components in Frame S to get [(den/dt)S']i = [Aen]i = Aij(en)j [(den/dt)S']i = -εikj(ω)k(en)j This then gives the WRONG answer for the Frame S result! We now take the ij element of this matrix using the fact (G.1.3) that (iJk)ij = kij , Aij = - [ ( cosψ + sinθsinψ)1ij + ( sinψ - sinθcosψ)2ij+ ( + cosθ)3ij ] . (G.7.16) The tensor kij is antisymmetric in i↔j and therefore the entire matrix A is antisymmetric and thus has only three distinct matrix elements. They are: A23 = - ( cosψ + sinθsinψ)123 = - ( cosψ + sinθsinψ) A31 = - ( sinψ - sinθcosψ)231 = - ( sinψ - sinθcosψ) A12 = - ( + cosθ) 312 = - ( + cosθ) . (G.7.17) The work is nearly done. In order to determine ω, we now want to compare, (den/dt)S' = Ain ei (G.7.8b) (den/dt)S' = – ω x en . (G.7.1) Since the left sides are the same, we take Frame S' components of the right sides and set them equal, Comment: The first equation in the above pair is a true vector equation in the basis state Hilbert Space. It can be written |(den/dt)S'> = Ain |ei> in Dirac notation. One could evaluate the two sides of this vector equation by closing either with <ek| for Frame S, or <e'k| for Frame S' . However, the second equation of the above pair is not a Hilbert Space vector equation, although it is of course mathematically a vector equation. One cannot write |(den/dt)S'> = – |ω x en> because the object on the right has no meaning! Nor does ω x |en> have a meaning, a cross product between a math vector and a Hilbert Space vector. The second equation above is merely a shorthand notation for the following equation (components in Frame S') [(den/dt)S']'i = - εijk(ω)'j(en)'k (den/dt)S' = – ω x en Therefore, the seeming vector equation Ainei = – ω x en has no meaning if one tries to evaluate both sides in Frame S. It is only valid when both sides are evaluated in Frame S'. We now evaluate the equation Ain ei = – ω x en in Frame S' : Ain (ei)'k = –εijk(ω)'j(en)'k RkiAin = –εijk(ω)'jRkn Apply RTnm to both sides to get RkiAin RTnm = –εijk(ω)'jRkn RTnm or RkiAin RTnm = –εijk(ω)'jδkm or [RAR-1]km = –εijm(ω)'j A'km = –εijm(ω)'j This only gives the right answer if I assume that my matrix A computed above has a prime on it! Then I need the crummy arm-waving argument that we have an Observer who is sitting in Frame S' and for whom we have (iJk)'ij = kij . [A' en]'i = – [ω x en]'i // where does LHS come from???? or A'ij (en)'j = - εikj (ω)'k(en)'j or A'ij = - εikj (ω)'k = εijk (ω)'k so A'12 = ε123(ω)'3 = (ω)'3 A'23 = ε231(ω)'1 = (ω)'1 A'31 = ε312(ω)'2 = (ω)'2 . (G.7.18) Then, (ω)'1 = A'23 = - ( cosψ + sinθsinψ) = - sinθsinψ - cosψ (ω)'2 = A'31 = - ( sinψ - sinθcosψ) = sinθcosψ - sinψ (ω)'3 = A'12 = - ( + cosθ) = - cosθ - . // angles still negated We now undo the temporary negation of the angles ψ,θ,φ enacted below (G.7.2). The velocities and all sines then negate. Our result for the components of ω in rotating Frame S' is then (ω)'1 = sinθsinψ + cosψ ≡ ωx' (ω)'2 = sinθcosψ - sinψ ≡ ωy' (ω)'3 = cosθ + ≡ ωz' . (G.7.19) This is in agreement with our Method 1 calculation (G.6.11) and with Goldstein's result which we again quote from Goldstein page 134 (GPS page 174), These are the Frame S' components of ω. We could then obtain the Frame S components of ω from (G.6.12) that ω = R-1ω', which can be written as in (G.6.14), (ω)i = [ Rz(φ)Rx(θ)Rz(ψ)]ij (ω)'j . // ω = R-1ω', R from (G.7.2) (G.6.14) This was done at the end of Section G.6 and the result was (G.6.16), (ω)x = cosφ + sinθsinφ (ω)y = sinφ - sinθcosφ (ω)z = + cosθ . (G.6.16) (G.7.20) Outline of the direct Method 2 derivation of the Frame S result (G.7.20) For equation (G.7.1) we start with (1.7.1) that (de'n/dt)S = ω x e'n in place of (den/dt)S' = – ω x en, which adds an overall minus sign to the result. The new (G.7.4) becomes e'n(Φ) = R-1(Φ)en with en = constant in Frame S. Things go through as presented above, but since R(Φ) → R-1(Φ) and since R(Φ) = Rz(ψ)Rx(θ)Rz(φ), one has R-1(Φ) = Rz(-φ)Rx(-θ) Rz(-ψ). Thus, to convert the result (G.7.19) to the Frame S result, we have to make these changes: (1) φ → -ψ, ψ→ -φ and θ→ -θ; (2) add the overall minus sign just noted. We do that right here: (ω)'1 = sinθsinψ + cosψ ≡ ωx' (ω)'2 = sinθcosψ - sinψ ≡ ωy' (ω)'3 = cosθ + ≡ ωz' . (G.7.19) → -(ω)1 = [-][-sinθ][-sinφ] + [-] cosφ ω1 = sinθsinφ +cosφ -(ω)2 = [-][-sinθ]cosφ - [-][-sinφ] ω2 = - sinθcosφ + sinφ -(ω)3 = [-] cosθ + [-] ω3 = cosθ + (G.7.21) and this does agree with (G.7.20).