G_5 archive stuff

archive off olds stuff from Appendix G.5 PhL 1.11.15 The Meaning of the Euler Angles in Fig (G.5.5) In our Section 1 formalism, we discuss the idea of three concatenated transformations near (1.1.41). We can compare the language there to that of our current section's (G.5.6). e"'n = U-1e''n e"n = S-1e'n e'n = R-1en (',',') = Rζ'(ψ) (',',') (',',') = Rξ(θ) (,,) (,,) = Rz(φ) (,,) (',',') = R-1ζ'(-ψ) (',',') (',',') = R-1ξ(-θ) (,,) (,,) = R-1z(-φ) (,,) We speak of U-1 and S-1 and R-1 as "back rotations" and these then accompany Passive View "forward rotations" by U, S and R of a Kinematic Vector , (V)''' = UV" (V)" = SV' (V)' = RV To make this connection, we must make these identifications U = Rζ'(-ψ) S = Rξ(-θ) R = Rz(-φ) and so it happens that we have negative angles in our transformations of Kinematic vectors (V)''' = Rζ'(-ψ) V" (V)" = Rξ(-θ)V' (V)' = Rz(-φ)V When concatenated, and when we then replace the triple-prime of our Section 1.3 with the single prime of Goldstein's picture, we get (V)' = Rζ'(-ψ)Rξ(-θ)Rz(-φ)V for the overall "forward rotation", corresponding to these "back-rotated" basis vectors e'n = [Rζ'(-ψ)Rξ(-θ)Rz(-φ)]-1 en or How to get Frame S' components of a kinematic vector from its Frame S components In our Section 1 formalism, we back-rotate the basis vectors according to e'n = R-1en and then a kinematic vector V has components (V)' = RV in Frame S' as in (1.3.3). Specifically, (V)'i = Rij(V)j . (G.5.17) This gives the Frame S' components of V in terms of the Frame S components of V. In our current context as shown in (G.5.8) we have R-1 = Rζ'(ψ) Rξ(θ)Rz(φ) . (G.5.8) From Theorem 3 (3) this can be rewritten R-1 = Rz(φ)Rx(θ)Rz(ψ) (G.5.18) and so our R is given by R = Rz(-ψ)Rx(-θ) Rz(-φ) . (G.5.19) Finally we can write ** as (V)'i = [Rz(-ψ)Rx(-θ) Rz(-φ)]ij(V)j = Rij(V)j (G.5.20) As Fig (G.5.5) shows, Goldstein has right-hand-rule-back-rotated the basis vectors by negative angles -φ, -θ then -ψ (that is, he has "forward-rotated" them by positive angles φ,θ,ψ) and that is why we have negative angles in ***. Specifically, the matrix R is this: (G.5.21) We shall apply this below to compute (ω)'i in terms of (ω)j . We now do two brief exercises in computing components of basis vectors in "the other" frame. Exercise 2: Compute the Frame S' components of the basis vector . Before doing this calculation we can see the basis vector in the lower right drawing of Fig (G.5.5). For the small Euler angles used in the figure, appears to have positive x' and z' components, but a negative y' component. This will be confirmed in the result below. METHOD A The equation of interest is (G.5.14d) which says = Rz(φ)Rx(θ)Rz(ψ) (G.5.14d) Maple computes the matrix shown (it is R-1 which we shall call Q), (G.5.22) Then = Q11 ' + Q12' + Q13 ', so it is the top row that is involved. Our conclusion is: ()x' = cosψ cosφ - sinψ cosθ sinφ positive ()y' = -sinψ cosφ - cosψ cosθ sinφ negative ()z' = sinθ sinφ positive (G.5.23) and the signs of the components agree with the observations above. METHOD B Set (x,y,z) =(1,0,0) in ***. Exercise 3: Compute the Frame S components of the basis vector '. Before doing this calculation we can see the basis vector ' in the lower right drawing of Fig (G.5.5). For the small Euler angles used in the figure, ' appears to have positive x,y and z components. This can be confirmed in the result below. The equation of interest is (G.5.14c) which says = Rz(-ψ)Rx(-θ)Rz(-φ) (G.5.14c) Maple computes the matrix shown (which is R), (G.5.24) Then ' = R11 + R12 + R13 , so it is the top row that is involved. Our conclusion is: (')x = cosψ cosφ - sinψ cosθ sinφ positive (')y = cosψ sinφ + sinψ cosθ cosφ positive (')z = sinψ sinθ positive (G.5.25) and the signs of the components agree with the observations above.