First attempt calculation of w REVIEWED

A rotation matrix problem v2 PhL 2.12.17 This is where I do my first attempt at computing Goldstein's ω in Frame S' . I got the right answer after some fiddling around. Here I am doing a bit of a rewrite on the previous version. First, note that : Backwards Rule. Consider: for some vector v, (v)'n = v e'n = v [Ren] = [R-1v] en = [R-1v]n = (R-1)ni (v)i This is my "famous fact" that if you rotate the basis vectors one way, you have to rotate the components of a [ kinematic] vector the other way. In frames doc I start with en = R e'n which says e'n = R-1en so there I rotate the basis vectors backwards, and I then end up with (v)'n = v e'n = v [R-1en] = [Rv] en = [Rv]n = (R)ni (v)i and so v' = Rv gives the components of v in Frame S'. An example would be ω' = Rω. [ok] Question: Recall in frames doc that I show that a(t+dt) = R(dφ) a(t) (1.5.7) da = dφ x a (1.5.8) Is this valid for both regular vectors and for basis vectors? [ a reasonable question ] In the first line above a(t) I think is ANY vector you want, so yes it could be a basis vector like en or e'n or a regular vector like v . [ yes ] Statement of the Problem: Suppose you have the following, consistent as noted above with frames doc, e'i = R-1(Φ)ei = R(-Φ)ei where R(Φ) = Rz(ψ) Rx(θ) Rz(φ) R-1(Φ) = Rz(-φ)Rx(-θ)Rz(-ψ) ei = R(Φ)e'i At this point, Ri(α) could refer to either my active convention or Goldstein's passive convention. Only when you write out specific matrices do you have to choose! Here ei are the usual Cartesian unit vectors of Frame S and e'i are some unit vectors of Frame S'. In frames doc I wrote R(φ) = exp(-i φ J) but here I am writing R(Φ) = exp(-i Φ J) to not overload the symbol φ. [ here Φ is meant to be a completely arbitrary rotation ] Following the path of frames doc, we know that for any vector a da = dΦ x a (1.5.8) (da/dt)S = ω x a where ω ≡ dΦ/dt . (1.6.1) (de'n/dt)S = ω x e'n . // selecting a = e'n (1.7.1) (den/dt)S' = – ω x en . // "the other one" (1.7.4) The Problem is to find an expression for ω which involves , and . Hopefully this expression will look something like Goldstein page 134 (4-103). Note that Φ ≠ (ψ,θ,φ) or anything simple like that. [ie, these are not the Cartesian comps of Φ ] A separate Problem would be to compute the function Φ(ψ,θ,φ) which I am not doing right now. [ Here I must have been thinking of Φ as indicating an Euler angle rotation with those params.] Solution: I am not sure how to start, so I will try several Plans. [ I think this is the very first time I went down this path ] en = R(Φ) e'n [This is my standard Section 1.1 formalism R ] (den)S' = R(Φ+dΦ)e'n - R(Φ)e'n = Rz(ψ+dψ) Rx(θ+dθ) Rz(φ+dφ) e'n - Rz(ψ) Rx(θ) Rz(φ) e'n [ above I just made the assumption that R(Φ) = Rz(ψ) Rx(θ) Rz(φ). If I instead use (G.5.19) for R, then all three angles would be negated and that would make the result come out exactly right! ] = [ Rz(ψ+dψ) Rx(θ+dθ) Rz(φ+dφ) - Rz(ψ) Rx(θ) Rz(φ) ]e'n = dR e'n , dR ≡ [ Rz(ψ+dψ) Rx(θ+dθ) Rz(φ+dφ) - Rz(ψ) Rx(θ) Rz(φ) ] Now process dR' as follows dR = [ 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)J3 Rx(θ)Rz(φ) - idθ Rz(ψ) Rx(θ)J1Rz(φ) - idφ Rz(ψ) Rx(θ)Rz(φ)J3 ] Then to first order, (dR/dt)S' = [ -i Rz(ψ)J3 Rx(θ)Rz(φ) - i Rz(ψ) Rx(θ)J1Rz(φ) - i Rz(ψ) Rx(θ)Rz(φ)J3 ] . Claim: I think the matrix elements of R have time derivatives which do not need S or S' Frame label, recall that any component of any tensor has this property of not needing a specifier. But any symbol representing a tensor without components does need a specifier if you try to take a time derivative den = (dR') e'n (den/dt)S' = (dR'/dt)S'e'n This last equation is true because in Frame S' the vector e'n is a constant. We need the S' specifier on the both left and right since en is a rank-1 tensor and dR' is a rank-2 tensor. Now replace e'n = R-1(Φ)en in *** to get (den/dt)S' = (dR/dt)S'e'n = (dR/dt)S'[R-1(Φ)en] = [ (dR/dt)S'[R-1(Φ) ] en ≡ A' en A' = (A)S' = [ (dR/dt)S'[R-1(Φ) ] Since A' is a rank-2 tensor which involves time derivatives within Frame S', it really should have an S' specifier. We abbreviate this by writing it as A'. Now we want to compute the matrix A'. Commutator: Define this operation for square matrices X and Y, [X,Y] ≡ XY-YX. Then consider , [ Ri(α) , Ji ] = [ exp(-iαJi) , Ji ] = 0 since [Ji,Ji] = 0 Therefore, Ri(α) Ji Ri(-α) = Ri(α) Ri(-α)Ji = 1 * Ji = Ji Calculate the matrix A A' = (dR/dt)S'R-1(Φ) = [ -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)J3 Rz(-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) J1Rz(-ψ) - i Rz(ψ) Rx(θ)J3Rx(-θ)Rz(-ψ)] = [ -i J3 - i {Rz(ψ) J1Rz(-ψ) } - i {Rz(ψ) Rx(θ)J3Rx(-θ)Rz(-ψ) }] = [ -i J3 - i M1 - i M2 ] 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) Now we evaluate M1 and M2 using the following results which are derived in a Footnote at the end of this section, R1(θ) J2R1(-θ) = cosθ J2 + sinθ J3 R1(θ) J3R1(-θ) = cosθ J3 - sinθ J2 R2(θ) J1R2(-θ) = cosθ J1 - sinθ J3 R2(θ) J3R2(-θ) = cosθ J3 + sinθ J1 R3(θ) J1R3(-θ) = cosθJ1 + sinθ J2 R3(θ) J2R3(-θ) = cosθJ2 - sinθ J1 Therefore M1 = R3(ψ) J1R3(-ψ) = cosψJ1 + sinψ J2 M2 = R3(ψ) [ R1(θ)J3R1(-θ) ] R3(-ψ) = R3(ψ) [ cosθ J3 - sinθ J2 ] R3(-ψ) = cosθ [ R3(ψ) J3 R3(-ψ)] - sinθ [R3(ψ)J2R3(-ψ)] = cosθ J3 - sinθ (cosψJ2 - sinψ J1) = sinθsinψJ1 - sinθcosψJ2 + cosθ J3 Our calculation of matrix A may now be concluded, 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) ] We now take the ij element of this matrix using the fact (1.5.2) that (iJk)ij = kij : A'ij = - [ ( cosψ + sinθsinψ)1ij + ( sinψ - sinθcosψ)2ij+ ( + cosθ)3ij ] Notice from the properties of ε shown in (1.5.3) that kij is antisymmetric in i↔j and therefore the entire matrix A' is antisymmetric and thus has only three distinct matrix elements. Then are: A'23 = - ( cosψ + sinθsinψ)123 = - ( cosψ + sinθsinψ) A'31 = - ( sinψ - sinθcosψ)231 = - ( sinψ - sinθcosψ) A'12 = - ( + cosθ) 312 = - ( + cosθ) The work is now nearly done. Recall from ** and ** that (den/dt)S' = A' en // LHS = ∂S'(en) (den/dt)S' = – ω x en Therefore A' en = – ω x en Since the tensor elements A'ij as shown above are associated with Frame S', we take the Frame S' component of each side of this equation to get [A' en]'i = – [ω x en]'i or A'ij (en)'j = - εikj (ω)'k(en)'j or A'ij = - εikj (ω)'k or A'ij = εijk (ω)'k or A'12 = ε123(ω)'3 = (ω)'3 A'23 = ε231(ω)'1 = (ω)'1 A'31 = ε312(ω)'2 = (ω)'2 Therefore, (ω)'1 = A'23 = - ( cosψ + sinθsinψ) = - sinθsinψ - cosψ (ω)'2 = A'31 = - ( sinψ - sinθcosψ) = sinθcosψ - sinψ (ω)'3 = A'12 = - ( + cosθ) = - cosθ - [ this result agrees with G if you negate all three angles, I could make connection to his ABC ] In our specification at the start of the Euler angle rotation we wrote R(Φ) = Rz(ψ) Rx(θ) Rz(φ) = Rzact(ψ) Rxact(θ) Rzact(φ) where the three rotations on the right are represented by their active rotation matrices. Racti(α) = exp(-iαJi) with Ji as in (1.5.2) eg, Ractz(θ) = Such a matrix rotates a vector v "actively" in the usual counterclockwise right-hand-rule direction to get v' = Ractz(θ) v. As a quick example Ractz(θ) = = = vector in quadrant I of the z=0 plane. Goldstein ( G p ** and GPS p ***) instead use passive rotation matrices, so for them Rpasi(α) = exp(+iαJi) with Ji as in (1.5.2) eg, Rpasz(θ) = Since Goldstein is then using R(Φ)G = Rzpas(ψ) Rxpas(θ) Rzpas(φ) = Rzact(-ψ) Rxact(-θ) Rzact(-φ) in order to obtain the Goldstein result we have to negate all the angles ψ,θ and φ (and therefore also their time derivatives) . Doing so in ** above gives (ω)'1 = sinθsinψ + cosψ (ω)'2 = sinθcosψ - sinψ (ω)'3 = cosθ + // Goldstein which we compare to GPS page 174, They agree! This was the first time *****************************************************************8 Goldstein addresses this same problem but uses passive rotation matrices instead of our active rotation matrices (G p 109, GPS p 153), so we may convert our results to their his by negating all three Euler angles θ,φ,ψ with this result ω1 = A23 = sinθsinψ + cosψ ω2 = A31 = sinθcosψ - sinψ ω3 = A12 = cosθ + // G and GPS (*) We compare this result with that shown on G page 134 and GPS page 174, Paradox: when I compute the components of ω in Frame S (using Goldstein's rotation matrices) I end up with what he says are components of ω in Frame S' which is his body frame (non-inertial). [ yes, this is a crucial issue. You have to say you are working in Frame S' with the same R and you put a prime on A' as I have done now above and that made things work. ] Here I sort of flail away at this question thru the end of this doc. Can I find some simple examples to see who is right and who is wrong here? θ = 0 and = 0: MY result in this case is ω1 = - sinθsinψ - cosψ = 0 ω2 = sinθcosψ - sinψ = 0 ω3 = - cosθ - = - - // us In this case R = Rz(ψ) Rx(0) Rz(φ) = Rz(φ+ψ), pretty simple. I manually compute that A = Then suppose to have (den/dt)S' = A en so that (de1/dt)S' = A e1 = + (de2/dt)S' = A e2 = - - (de3/dt)S' = A e3 = 0 ************************************ Question: Consider: a(t+dt) = Rz(dφ) a(t) Which of the following is true: 1. This equation is valid only in Frame S 2. This equation is valid only in Frame S' 3. This equation is valid in both Frame S and Frame S' Answer: I start off in Frame S which has a clear z axis and I take Frame S components to get [a(t+dt)]i = [Rz(dφ) a(t)]i = [Rz(dφ)]ij [a(t)]j This seems to be completely well defined and I know the matrix [Rz(dφ)]ij . Here is another way to approach this claim: a(t+dt) ei = [Rz(dφ) a(t)] ei [a(t+dt)]i = [Rz(dφ) a(t)]i = [Rz(dφ)(S)]ij [a(t)]j where I now indicate that the tensor object [Rz(dφ)(S)]ij is "in Frame S". Remember that components of vectors have different values in different frames, and the same is true of a rank-2 tensor. The question this brings up is the following: Is [Rz(dφ)(S)]ij a true tensor, or is it just a matrix of constants like π and 2 ? I think it really is a tensor, and you might say R [Rz(dφ)(S)] R-1 = [Rz(dφ)(S')] or something like that. You want a(t+dt) = Rz(dφ) a(t) to be a true vector equation, so Rz has to then be a true rank-2 tensor. Now go back to a(t+dt) = Rz(dφ) a(t) In Frame S' components we have [a(t+dt)]'i = [Rz(dφ) a(t)]'i = [Rz(dφ)]'ij [a(t)]'j = [Rz(dφ)(S')]ij [a(t)]'j So I think this would be a different matrix! Do it another way a(t+dt) = Rz(dφ) a(t) Ra(t+dt) = RRz(dφ)R-1R a(t) a' = Ra for general vector a'(t+dt) = [RRz(dφ)R-1]a'(t) So my answer I think is this. The equation a(t+dt) = Rz(dφ) a(t) with the usual matrix associated with Rz(dφ) is valid only in Frame S. Similarly, consider da = dφ x a = (dφ) x a For a specific specified in Frame S, this equation is valid only in Frame S. I then get to (da/dt) = ω x a where ω ≡ dφ/dt. dφ = dφ (1.6.1) I write this as (da/dt)S = ω x a ω ≡ dφ/dt. dφ = dφ Is this valid only in Frame S? I certainly can compute components of ω x a in any frame of reference since it is just a vector. But the equation itself may only be true in Frame S ! But (da/dt)S is also a vector, so why can't we get its components in Frame S' ? (da/dt)S = () x a Back up: a(t+dt) = Rn(dφ) a(t) n = a vector in Frame S I guess this equation in Frame S' might be a'(t+dt) = Rn'(dφ) a'(t) n' = Rn and this is definitely a different matrix. So, q = (da/dt)S = () x a q' = [(da/dt)S]' = (') x a'