Euler unit vector exercise INSTALLED

Euler unit vector exercise PhL 3.14.17 After looking at the Euler Angle unit vectors, I show (Question 3) that (L)z = pφ (L)'z = pψ which is something Goldstein claims. This is installed now in Appendix H. Based on (E.5.3), I think the following equations are valid = -sinφ +cosφ = -sinψ '+cosψ ' = -sinθ +cosθ These unit vectors and equations do not appear anywhere in my Section G.5 (for some reason). Probably yes, but they are never really used as a basis by me. Question 1: do these vectors form a basis for Frame S or for Frame S' ? Probably yes. I could write them all out in terms of (,,) or (',','). Question 2: Do , and form a basis for E3 ? Well, basically these are the tangent base vectors which I write there as en apart from scaling factors, and the en do form basis for x-space. Yes. Question 3: In the top analysis, can I make these claims ? Lφ = constant because φ is cyclic Lψ = constant because ψ is cyclic Using the above dot products I could write for example, Lψ = L = -sinψ (L)'x + cosψ (L)'y using body frame Frame S' evaluation. Is this related to Goldstein's pψ = ∂L/∂ = ( cosθ + )I'3 ≡ aI'1 = (ω)'3I'3 ? Recall that L = Iω so for sure, (L)'i = (I)'ij(ω)'j = I'i(ω)'i Then (L)'x = (L)'1 = I'1(ω)'1 (L)'y = (L)'2 = I'2(ω)'2 Now recall that (ω)'1 = sinθsinψ + cosψ (ω)'2 = sinθcosψ - sinψ (ω)'3 = cosθ + // Frame S' (G.6.11) So I then know thaqt (L)'x = (L)'1 = I'1(ω)'1 = I'1 [ sinθsinψ + cosψ] (L)'y = (L)'2 = I'2(ω)'2 = I'2 [ sinθcosψ - sinψ ] Then Lψ = -sinψ (L)'x + cosψ (L)'y = -sinψ [ sinθsinψ + cosψ] I'1 + cosψ [ sinθcosψ - sinψ ] I'2 But for the top we have I'2 = I'1 so write this as Lψ = -sinψ [ sinθsinψ + cosψ] I'1 + cosψ [ sinθcosψ - sinψ ] I'1 = I'1 [ -sinθsin2ψ + sinθcos2ψ ] + I'1 [ -sinψ cosψ - cosψ sinψ ] Compare this to pψ = ∂L/∂ = ( cosθ + )I'3 I don't see any connection!!! So it looks like my conjecture that pψ = Lψ is no good. Go back and review what has happened here. I wrote Lψ = L = -sinψ (L)'x + cosψ (L)'y = I'1 [ -sinθsin2ψ + sinθcos2ψ ] + I'1 [ -sinψ cosψ - cosψ sinψ ] It seems more that, associated with angle ψ, we should have (L)'z being a constant of the motion, and not the quantity Lψ which really is orthogonal to the ' axis. You look for the axis about which your angle is a rotation. So now I think: pψ = (L)'z ??? OK, write this out (L)'z = (L)'3 = I'3 (ω')3 and yes this = pψ exactly. Next conjecture that (L)z is associated with φ, so should have (L)z = constant. Then (L)z = L Now how can I relate to the body frame basis vectors? I have failed to write this down anywhere in G.5. // But I have now added this and the result is = (sinψsinθ)' + (cosψsinθ)' + (cosθ) ' Therefore (L)z = L { (sinψsinθ)' + (cosψsinθ)' + (cosθ) ' } = sinψsinθ (L)'1 + cosψsinθ (L)'2 + cosθ (L)'3 = sinψsinθ I'1(ω)'1 + cosψsinθ I'2(ω)'2 + cosθ I'3(ω)'3 = sinψsinθ I'1[ sinθsinψ + cosψ] + cosψsinθ I'2[ sinθcosψ - sinψ] + cosθ I'3[ cosθ + ] = sinψsinθ I'1[ sinθsinψ + cosψ] + cosψsinθ I'1[ sinθcosψ - sinψ] + cosθ I'3[ cosθ + ] = sinθ I'1 { sinθsin2ψ + cosψsinψ } + sinθ I'1 { sinθcos2ψ - sinψcosψ) + cosθ I'3[ cosθ + ] = sinθ I'1 { sinθ } + cosθ I'3[ cosθ + ] = sin2θI'1 + cos2θI'3 + cosθ I'3 = [sin2θI'1 + cos2θI'3 ] + cosθ I'3 Compare this with pφ = (sin2θ I'1 + cos2θ I'3) + cosθ I'3 So I am now sure that (L)z = pφ (L)'z = pψ Clarity keeps improving !