App G_7 on computing w INSTALLED

This has been installed into Appendix G, do not edit here. 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) . (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 = Re'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) where dR' ≡ [ Rz(ψ+dψ) Rx(θ+dθ) Rz(φ+dφ) - Rz(ψ) Rx(θ) Rz(φ) ] . (G.7.6) The matrix dR' has a prime because it is a rank-2 tensor within Frame S'. The same comment applies to the matrix A' below. Replace e'n in (G.7.5) using (G.7.4) to get (den)S' = dR' e'n = dR' [R-1(Φ)en ] = [dR' R-1(Φ) ] en . (G.7.7) Next, divide by dt to get (den/dt)S' = [(dR'/dt)S'R-1(Φ) ] en ≡ A' en A' ≡ [(dR'/dt)S'R-1(Φ) ] . (G.7.8) 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 must 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)S'R-1(Φ) // where R-1(Φ) = Rz(φ)Rx(θ)Rz(ψ) from (G.7.3) = [ -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 (sinθcsinψJ1 - 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) We now take the ij element of this matrix using the fact (G.1.3) that (iJk)ij = kij , A'ij = - [ ( 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: A'23 = - ( cosψ + sinθsinψ)123 = - ( cosψ + sinθsinψ) A'31 = - ( sinψ - sinθcosψ)231 = - ( sinψ - sinθcosψ) A'12 = - ( + cosθ) 312 = - ( + cosθ) . (G.7.17) The work is nearly done. In order to determine ω, we now want to compare, (den/dt)S' = A' en (G.7.8) (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, [A' en]'i = – [ω x en]'i 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)