paradox2 v2

This is the Title PhL 1.11.15 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 or |en(Φ)> = |R(Φ)e'n> = R(Φ) |e'n> (G.7.4) We show the Section 1.1 Dirac notation on the right and it is especially helpful in understanding what follows. Since we have in mind that Φ = Φ(t), one sees that the Frame S basis vectors change in time as viewed from Frame S'. In contrast, the basis vectors e'n are static. The basis vectors |en(Φ)> are complete at time t so we can write, in analogy with (1.1.20), 1 = |en(Φ)><en(Φ)| completeness of the en at time t (a) 1 = |en(Φ+dΦ)><en(Φ+dΦ)| completeness of the en at time t +dt (b) 1 = |e'n><e'n| completeness of the e'n at any time (c) (G.7.5) In (b) the rotation vector has changed from Φ to some Φ+dΦ as time moved from t to d+dt. A key point is that the en basis vectors are complete at any point in time. The rotation operator R(Φ) similarly is real orthogonal at any time, analogous to (1.1.37), RT(Φ) = R-1(Φ) and similarly for matrices RT(Φ) = R-1(Φ) (G.7.6) Now we close the Dirac equation in (G.7.4) on the left with <e'm| to get <e'm| en(Φ) > = <e'm|R(Φ) |e'n> = [R(Φ)]'mn (G.7.7) On the other hand [R(Φ)]mn = <em(Φ)|R(Φ) |en(Φ)> = <em(Φ) |e'i><e'i| R(Φ) |e'j><e'j|en(Φ)> = [R(Φ)]'im [R(Φ)]'ij[R(Φ)]'jn = [ RT(Φ)R(Φ)R(Φ)]'mn = [R(Φ)]'mn (G.7.8) We thus find that [R(Φ)]mn = [R(Φ)]'mn as in (1.1.34). This is also true for Φ+dΦ . Question added 3/10/17: What is the implication for a generator Jk ? Consider [R(Φ)]mn = [R(Φ)]'mn [exp(-iΦ J)]mn = [exp(-iΦ J)]'mn [ 1 - iΦ J)]mn = [ 1 - iΦ J)]'mn // if we assume Φ's components all small (J)mn = (J)'mn Question: Does [R(Φ)]mn = [R(Φ)]'mn say that [R(Φ)]mn is a scalar? I think the answer is that it is a rank-2 tensor because it obeys the rank-2 tensor rule A' = RAR-1 and it just happens that the values are the same in the two frames, so it is not a scalar. It is a very special rank-2 tensor being the basis change matrix in an all-Cartesian world. So each generator matrix is the same on Frame S and in Frame S'. On the other hand, one would expect that (J)' = R (J) R-1 since this is the rule for any tensor. In (G.4.1) I claim that RJR-1 = R-1J Question: Is J really a basis object and does it back-rotate like a basis vector? I now seem to have three facts which completely conflict with each other!! (J)mn = (J)'mn // "proved" just above (J)' = [R J R-1] // rule if each Jk is a rank-2 tensor R-1J = RJR-1 // Theorem 1 of (G.4.1) Notice the prime locations in the first two lines. Not the same. In the vector world we have (V)' = RV in passive view for a kinematic vector, but e'n = R-1en for a basis vector. How is this statement extended to the rank-2 tensor world? Look again at vectors V = Vi ei = (V)'i e'i Now install e'i = R-1ei or e'i = Rim em (Basis Theorem) to get V = (V)'i Rim em = Vm em so conclude that Vm = (V)'iRim or Vm R-1mn= (V)'iRimR-1mn or RnmVm = (V)'iδin = (V)'n and we end up with (V)'n = RnmVm or (V)' = RV Now let's try something analogous for rank-2 tensors. Start with T = (T)ij |ei><ej| = (T)'ij |e'i><e'j| T = (T)'ij |e'i><e'j| = (T)'ij |R-1ei><R-1ej| = (T)'ijR-1 |ei><ej| R = (T)'ij |en><en|R-1 |ei><ej| R |em><em| = (T)'ij |en>RTniRjm<em| = RTni(T)'ijRjm |en><em| On the other hand we also have T = (T)nm |en><em| So comparison shows that (T)nm = RTni(T)'ijRjm = [RT (T)' R]nm and thus (T)' = R(T)R-1 so everything seems to work right here. So if the components of J were kinematic tensors, we would claim that (Jk)' = R(Jk)R-1 or (J)' = R(J)R-1 So having done a few confidence-building exercises, we go back to our Triple Paradox which it is safer to write out without vector notation 1 (Ji)mn = (Ji)'mn // proved from [R(Φ)]mn = [R(Φ)]'mn 2 (Ji)' = R Ji R-1 // rule if each Ji were a rank-2 tensor (all matrices) 3 (R-1ij Jj) = RJiR-1 // Theorem 1 of (G.4.1) (all matrices) Lines 1 and 2 conflict for sure because we have 1 (Ji)mn = (Ji)'mn 2 (Ji)' = R Ji R-1 I will verify that line 1 and line 2 conflict with a specific counterexample. Suppose R = Rz(θ) and take J1. Then line 2 claims that (J1)'mn = Rz(θ)mi(J1)ij Rz(-θ)jn = [ Rz(θ)(J1)Rz(-θ)]mn = [ R3(θ)J1R3(-θ)]mn = cosθ(J1)mn + sinθ (J2)mn // from (G.3.8) So if line 2 were true we would have (J1)'mn = cosθ(J1)mn + sinθ (J2)mn But line 1 says (J1)'mn = (J1)mn so lines 1 and 2 conflict for dead sure. Let's look now at line 1 versus line 3 for this same example. But line 3 makes no reference to (J)'mn so I cannot really compare these two lines. But I can see if I think line 3 is valid. Consider line 3: R-1J = RJR-1 This is a bit ambiguous (!) so you have to write out the intended meaning as shown in (G.4.1), Rz(-θ)1jJj = Rz(θ)J1Rz(-θ) The math for the RHS is already done above, and it gives cosθJ1 + sinθ J2 . The left side is Rz(-θ)11J1 + Rz(-θ)12J2 + Rz(-θ)13J3 = cosθJ1 + sinθ J2 + 0 J3 = agrees where Rz(-θ) = So I have just verified Theorem 1 for a specific case, not a bad thing to do. Ouch! I have just found a counterexample to my Theorem 1 of (G.4.1), that is not good! How about lines 2 and 3: They claim that 2 (Ji)' = R Ji R-1 // rule if each Ji were a rank-2 tensor (all matrices) 3 R-1ij Jj = RJiR-1 // Theorem 1 of (G.4.1) (all matrices) I don't know how to check this because I don't know what (J1)'mn is! I don't really have a definition of the numbers (J1)'mn , but I do have a definition of (J1)mn . Proposed resolution of the Triple Paradox. 1. Line 1 is valid, and one should regard the Ji as matrices of constants, the same in all frames. This then makes my Appendix G.7 operator approach work! 2. Line 2 is wrong, and the matrix Ji does not transform as a kinematic tensor. 3. Line 3 is valid and does not conflict with line 1, since line 3 does not mention (J)'mn. Question: How does this relate to Theorem 2 which says R Rn(θ) R-1 = Rn'(θ) where n' = Rn In this theorem R is an arbitrary rotation. If we take R= Rn(θ) it says Rn(θ) = Rn'(θ) but this is true because in this case n' = Rn(θ)n = n . So no conflicts here. Note: We are NOT saying that RJiR-1 = J'i = Ji !!! The object RJiR-1 is non-trivial because Ji does not commute with a general R. What we know is that RJiR-1 = R-1ij Jj . But now here is another paradox. Consider that we find A = - [ ( cosψ + sinθsinψ)(iJ1) + ( sinψ - sinθcosψ)(iJ2)+ ( + cosθ)(iJ3) ] as an operator statement. Is this a tensor or not? According to the resolution above, it must NOT be a tensor because taking matrix elements and then using (Ji)mn = (Ji)'mn, we get A'ij = Aij. This then fouls up the Section G.7 Frame S evaluation of ω ! Possible resolution: 1 (Ji)mn = (Ji)'mn and (A)'ij = (A)ij both valid 2 (J'i) = R Ji R-1 and (A')ij = [R A R-1]ij both valid We now examine (from Frame S') a small change in the basis vector en |(den(Φ))S'> ≡ |en(Φ+dΦ)> - |en(Φ)> = |e'i><e'i|en(Φ+dΦ)> - |e'i><e'i|en(Φ)> = |e'i> [R(Φ+dΦ)]'in - |e'i>[R(Φ)]'in = ( [R(Φ+dΦ)]'in -[R(Φ)]'in ) |e'i> = ( [R(Φ+dΦ)]in - [R(Φ)]in ) |e'i> // lincom ≡ (dR)in |e'i> dR = R(Φ+dΦ) - R(Φ) (matrices) The next step is to replace |e'i> as follows |e'i> = |ej(Φ)><ej(Φ) | e'i> = |ej(Φ)>[R(Φ)]ij // lincom Then |(den(Φ))S'> = (dR)in[R(Φ)]ij |ej(Φ)> or (den(Φ))S' = (dR)in[R(Φ)]ijej(Φ) Divide by dt to get (den(Φ)/dt)S' = (dR/dt)in[R(Φ)]ijej(Φ) = [R-1(Φ)]ji(dR/dt)in ej(Φ) = [R-1(Φ) (dR/dt)]jn ej(Φ) ok to here 3.8.17 Write this last result out please without the dt, (den(Φ))S' = [R-1(Φ)]ji(dR)in ej(Φ) = [R-1(Φ)]ji( [R(Φ+dΦ)]in - [R(Φ)]in ) ej(Φ) = ( [R-1(Φ)R(Φ+dΦ)]jn - δjn ) ej(Φ) = ( [R-1(Φ)R(Φ+dΦ)] - 1 )jn ej(Φ) (*) or |(den(Φ))S'> = ( [R-1(Φ)R(Φ+dΦ)] - 1 )jn |ej(Φ)> |(den(Φ))S'> = ( [RT(Φ)R(Φ+dΦ)] - 1 )jn |ej(Φ)> |(den(Φ))S'> = ( [RT(Φ)R(Φ+dΦ)] - 1 )Tnj |ej(Φ)> |(den(Φ))S'> = ( RT(Φ+dΦ)R(Φ) - 1 )nj |ej(Φ)> This is a vector equation which you should be able to evaluate in Frame S or in Frame S'. [ I think that might be true, you really do have a true HS vector on each side. ] Maybe write this as |(den(Φ)/dt)S'>dt = ( RT(Φ+dΦ)R(Φ) - 1 )nj |ej(Φ)> Now the other equation of interest is this |(den(Φ)/dt)S'> = – ω x |en(Φ)> or |(den(Φ)/dt)S'>dt = – dt ω x |en(Φ)> // no meaning as vector equation So we should be able to equation the vector right hand sides to get ( RT(Φ+dΦ)R(Φ) - 1 )nj |ej(Φ)> = – dt ω x |en(Φ)> Is the object ω x |en(Φ)> well defined? What does it mean? It is a math vector crossed with a Hilbert Space vector. I suspect this is NOT well defined. Go back to (den/dt)S' = – ω x en Where does this come from? Review: Go back to frames doc Section 1.6. We have da = dφ x a. This is a vector equation, yes. There is only one Space at this point in that section. Then (da/dt) = ω x a same comment. Then I jump to (de'n/dt ) = ω x e'n and this is interpreted as (de'n/dt)S = ω x e'n . Although this is a vector equation, I now today think it is valid only in Frame S! This is really a shorthand notation for [(de'n/dt)S]i = εijk ωj (e'n)k Note: This is not the case I treat in App G.7, but I will get to that case below. Then I should be able to construct a vector equation as follows, (de'n/dt)S = [εijk ωj (e'n)k] ei |(de'n/dt)S> = [εijk ωj (e'n)k] |ei> // this is OK Dirac notation Now suppose I close that equation from the left with <e'm| . That would give <e'm|(de'n/dt)S> = [εijk ωj (e'n)k] <e'm|ei> or [(de'n/dt)S]'m = [εijk ωj (e'n)k] (ei)'m Notice that (ω)'j does not appear in this equation! Writing things out we get [(de'n/dt)S]'m = εijk ωj RnkRmi or [(de'c/dt)S]'b = ωJ εjJk RckRbi = ωJεjJkRTjbRTkc = - ωJεJjkRTjbRTkc My App A result is RJaεabc = εJjkRjbRkc = εJjkRjbRkc Using this I then get [(de'c/dt)S]'b = - ωJεJjkRTjbRTkc = - ωJRTJaεabc = - ωJεabcRaJ = - εabcRaJωJ = - εabc[Rω]a = - εabc(ω')a Not sure of this result, but lets go do things the other way now. Start with (den/dt)S' = – ω x en . (1.7.4) (G.7.1) and this is just a shorthand for [(den/dt)S']'i = - εijk(ω)'j(en)'k and this gives the correct ω in Frame S' Now let's turn this into a vector equation by writing [(den/dt)S']'i e'i = - εijk(ω)'j(en)'k e'i or | (den/dt)S'> = - εijk(ω)'j(en)'k |e'i> // seems an OK vector equation or | (den/dt)S'> = - εajk(ω)'j(en)'k |e'a> // seems an OK vector equation Now take components of this equation in Frame S', <e'i | (den/dt)S'> = - εajk(ω)'j(en)'k <e'i |e'a> or [(den/dt)S']'i = - εajk(ω)'j(en)'k δia or [(den/dt)S']'i = - εijk(ω)'j(en)'k or [(den/dt)S']'i = - εijk(ω)'jRkn Now recall from above that |(den(Φ)/dt)S'>dt = ( RT(Φ+dΦ)R(Φ) - 1 )nj |ej(Φ)> Suppose we close this with <e'i | on the left. That gives, <e'i |(den(Φ)/dt)S'>dt = ( RT(Φ+dΦ)R(Φ) - 1 )nj <e'i |ej(Φ)> or [(den/dt)S']'i = ( RT(Φ+dΦ)R(Φ) - 1 )nj Rij / dt Setting the two right hand sides equal we seem then to get - εijk(ω)'jRkn = ( RT(Φ+dΦ)R(Φ) - 1 )nj Rij / dt or - εijk(ω)'jRkn = ( RT(Φ+dΦ)R(Φ) - 1 )nj RTji / dt or - εijk(ω)'jRkn = ( RT(Φ+dΦ) - RT(Φ) )ni / dt or - εijk(ω)'jRkn = ( R(Φ+dΦ) - R(Φ) )in / dt or - εijk(ω)'jRkn = (dR)in / dt Now multiply both sides by R-1nm and sum on n to get - εijk(ω)'jRkn R-1nm = (dR)in R-1nm / dt or - εijk(ω)'jδkm = (dR * R-1)im/ dt = [(dR/dt) * R-1]im ≡ Aim or - εijm(ω)'j = Aim or εjim(ω)'j = Aim or εkim(ω)'k = Aim or εkij(ω)'k = Aij THIS IS THE DESIRED RESULT!!!!!!! I did this all with "lin comb" stuff instead of "operator stuff". Right now this is the ONLY place I am getting the right answer to this problem!!!! Now take components in Frame S instead of Frame S' and see what happens, <em | (den/dt)S'> = - εijk(ω)'j(en)'k <em |e'i> or [ (den/dt)S']m = - εijk(ω)'jRkn Rim = - (ω)'j εijkRkn Rim = - (ω)'j εjkiRkn Rim = + (ω)'j εjikRimRkn = + (ω)'c εcjkRjmRkn = (ω)'c Rcaεamn = εamn Rca (ω)'c where in the last step I have used App A, Rcaεamn = εcjkRjmRkn . So, if I were to insist on doing things in Frame S components, I have just shown that [ (den/dt)S']m = εamn Rca (ω)'c Now the vector equation I end up with in G.7 is this (den/dt)S' = Ain ei (G.7.8b) (den/dt)S' = – ω x en . (G.7.1) If I take Frame S' components of both, using the above, I get [(den/dt)S']m = Ain (ei)m = Ainδim = Amn [ (den/dt)S']m = εamn Rca (ω)'c I would then end up with εamn Rca (ω)'c = Amn Then for example, I get ε123Rc1 (ω)'c = A23 or Rc1 (ω)'c = A23 or RT1c(ω)'c = A23 or [RTω]'1 = A23 or [R-1ω]'1 = A23 But this does not seem to give the right result! There are no Hilbert Space vectors in this equation so no place for the Dirac notation here. Could you say |(de'n/dt)S> = |ω x e'n> ?? The LHS I think is an OK vector in HS, but what is the meaning of the RHS?? Try this instead |(de'n/dt)S> = ω x | e'n> ?? I claim today this has NO MEANING. You are cross producting vectors in two different spaces. 3.9.17: Let's try a little harder on the meaning question. Start with [(de'n/dt)S]i = εijk ωj (e'n)k Conclusions: 1. The equation (de'n/dt)S = ω x e'n appears to be a "vector equation" which you could evaluate in either Frame S or Frame S'. This is a vector equation in a pure math sense, but not in the Hilbert Space sense of our basis vectors. There is no meaning to objects |ω x e'n> or ω x | e'n> for the RHS in the Hilbert Space which is spanned by the basis vectors. These are cross products between a math vector and a HS vector and such an operation is not defined, though maybe we can provide a definition below. 2. The equation (de'n/dt)S = ω x e'n is nothing more than a shorthand for this equation [(de'n/dt)S]i = εijk ωj (e'n)k in which no Hilbert Space vectors appear. You can write this as ei [(de'n/dt)S] = εijk ωj ek e'n If you could express the RHS as ei Q, then you could write |(de'n/dt)S> = |Q> and then you would have a true HS vector equation. So lets try to do this ei Q = εijk ωj ek e'n ?? The k index on the right is tied to the ε so it just does not work. ********************************************* Now that Dirac has helped out a bit, let's try things without Dirac. We go once again to the opening scene of our movie, showing the Basis Theorem linear combination sums on the right en(Φ) = R(Φ)e'n en(Φ) = [R-1(Φ)]nm e'm //e'n = [R(Φ)]nm em(Φ) en(Φ+dΦ) = R(Φ+dΦ)e'n en(Φ+dΦ) = [R-1(Φ+dΦ)]nm e'm Then (den)S' ≡ en(Φ+dΦ) - en(Φ) = ( [R-1(Φ+dΦ)]nm - [R-1(Φ)]nm )e'm = ( [R-1(Φ+dΦ)]nm - [R-1(Φ)]nm ) [R(Φ)]mk ek(Φ) { R-1(Φ+dΦ)R-1(Φ) - R-1(Φ)R(Φ) }nk ek(Φ) { R-1(Φ+dΦ)R-1(Φ) - 1 }nk ek(Φ) // agrees with (*) above Rewrite as dt (den/dt)S' = [R-1(Φ+dΦ)R-1(Φ) - 1]nk ek(Φ) ok to here 3.8.17 Evaluate this first in Frame S: dt [ (den/dt)S']m = dt (den/dt)S' em = [R-1(Φ+dΦ)R-1(Φ) - 1]nm ≡ Dnm Now evaluate instead in Frame S' : dt [ (den/dt)S']'m = dt (den/dt)S' e'm = [R-1(Φ+dΦ)R-1(Φ) - 1]nk ek(Φ) e'm = [R-1(Φ+dΦ)R-1(Φ) - 1]nk R(Φ)mk = [R-1(Φ+dΦ)R-1(Φ) - 1]nk R-1(Φ)km = Dnk R-1(Φ)km Comment: I think Dnk here is just a "matrix of numbers" which has the same value in Frame S and Frame S'. The two results are different, and are related by a transformation matrix. As it should be. ok to here 3.8.17 Now consider dt(den/dt)S' = – dt ω x en(Φ) Simplify notation and compare dt (den/dt)S' = [R-1(Φ+dΦ)R-1(Φ) - 1]nk ek ≡ Ank ek dt(den/dt)S' = – dt ω x en Pause: Can I show that Ank is antisymmetric? A = [R-1(Φ+dΦ)R-1(Φ) - 1] = [RT(Φ+dΦ)RT(Φ) - 1] AT = [R(Φ)R(Φ+dΦ) - 1] One would have to show that 1 - R(Φ)R(Φ+dΦ) = RT(Φ+dΦ)RT(Φ) - 1 Apply R-1(Φ) from the left to get R-1(Φ) - R-1(Φ)R(Φ)R(Φ+dΦ) = R-1(Φ)RT(Φ+dΦ)RT(Φ) - R-1(Φ) or R-1(Φ) - R(Φ+dΦ) = R-1(Φ)RT(Φ+dΦ)R-1(Φ) - R-1(Φ) Now apply R(Φ) from the right to get R-1(Φ) R(Φ) - R(Φ+dΦ) R(Φ) = R-1(Φ)RT(Φ+dΦ)R-1(Φ) R(Φ) - R-1(Φ) R(Φ) or 1 - R(Φ+dΦ) R(Φ) = R-1(Φ)RT(Φ+dΦ) - 1 or 1 - R(Φ+dΦ) R(Φ) = RT(Φ)RT(Φ+dΦ) - 1 I give up on showing antisymmetry for the moment. Comparing we get – dt ω x en = Ank ek ok to here 3.8.17 Evaluate in Frame S: // I now think this is not meaningful to do – dt [ω x en]i = Ank (ek)i – dt εijkωj (en)k = Ank (ek)i – dt εijkωj δnk = Ank δki – dt εijnωj = Ani – dt εnijωj = Ani – dt εnikωk = Ani – dt εijkωk = Aij This does imply that Aij should be antisymmetric, but not obvious – dt ε12kωk = A12 = -dt ω3 – dt ε13kωk = A13 = +dt ω2 – dt ε23kωk = A23 = -dt ω1 So we get our results dt ω1 = -A23 dt ω2 = A13 dt ω3 = -A12 ok to here 3.8.17 Go back – dt ω x en = Ank ek Evaluate in Frame S': – dt [ω x en]'i = Ank (ek)'i // why no prime on A ?? – dt εijkω'j (en)'k = Ank (ek)'i – dt εijkω'j Rkn = AnkRik – dt εijkω'j RknR-1ns = R-1nsAnkRik – dt εijkω'j δks = RsnAnkR-1ki – dt εijsω'j = [RAR-1]si The result is then DIFFERENT! A = [R-1(Φ+dΦ)R-1(Φ) - 1] RAR-1 = R[R-1(Φ+dΦ)R-1(Φ) - 1]R-1 = [ R(Φ) R-1(Φ+dΦ)R-1(Φ)R-1(Φ) - 1 ] This is surely wrong. OK, enough for this shot. At least I have some "ideas". *************************** Question. Define this as Qnk ≡ [R-1(Φ+dΦ)R-1(Φ) - 1]nk Q = R-1(Φ+dΦ)R-1(Φ) - 1 Then QT = [ R-1(Φ+dΦ)R-1(Φ) - 1]T = [ R-1(Φ+dΦ)R-1(Φ)]T - 1] = [ R(Φ) R(Φ+dΦ) - 1 ] // cannot tell if antisymmetric or not Divide by dt to get (den/dt)S' = ek(Φ) Compare to (den/dt)S' = – ω x en(Φ) This seems to suggest the following vector equation, – ω x en(Φ) = ek(Φ) Let's try tacking the Frame S component of each side – [ω x en(Φ)]i = [ek(Φ)]i - εijkωj[en(Φ)]k = [ek(Φ)]i - εijkωjδnk = δki - εijnωj = ≡ Cni Then we find - ε1j2ωj = - ε132ω3 = ω3 = C12 - ε1j3ωj = - ε123ω2 = - ω2 = C13 - ε2j3ωj = - ε213ω1 = ω1 = C23 And then I have an answer to the problem in Frame S: ω1 = C23 ω2 = - C13 ω3 = C12 Now instead evaluate in Frame S' – [ω x en(Φ)]'i = [ek(Φ)]'i - εijk ω'j (en(Φ))'k = [ek(Φ)]'i - εijk ω'j R(Φ)kn = R(Φ)ik Apply R-1(Φ)ns to both sides of this equation - εijk ω'j R(Φ)knR-1(Φ)ns = R-1(Φ)ns R(Φ)ik or - εijk ω'j δks = R(Φ)sn R(Φ)ik - εijs ω'j = R(Φ)sn R-1(Φ)ki - εijs ω'j = *************************************************** I am now reviewing once again the existing Section G.7 I quickly get to this point, (den/dt)S' = [(dR/dt)R-1(Φ) ] en ≡ A en A ≡ [(dR/dt) R-1(Φ) ] (G.7.8) At this point, A is really an operator. In Dirac we have | (den/dt)S'> = |A en> = A |en> = |ei><ei|A |en> = Ain |ei> OK, so now we can say (den/dt)S' = Ain ei and now we have an actual matrix Ain sitting there. I show that A = - [ ( cosψ + sinθsinψ)(iJ1) + ( sinψ - sinθcosψ)(iJ2)+ ( + cosθ)(iJ3) ] and then Aij = - [ ( cosψ + sinθsinψ)1ij + ( sinψ - sinθcosψ)2ij+ ( + cosθ)3ij ] . (G.7.16)