how to compute w

A rotation matrix problem PhL 2.11.17 Statement of the Problem: Suppose you have the following e'i = R(Φ)ei where R(Φ) = Rz(ψ) Rx(θ) Rz(φ) = rotation by Goldstein Euler angles Here ei are the usual Cartesian unit vectors of Frame S and e'i are some unit vectors of Frame S'. Following the path of frames doc, we know that de'n = dΦ x en or (∂Se'n) = ω x e'n where ω = dΦ/dt This is all shown in frames doc for vector a, and we are here applying it to the basis vectors themselves. The Problem is to find an expression for ω which involves and and . Hopefully this expression will look something like Goldstein page 134 (4-103). Note that Φ ≠ (ψ,θ,φ) or anything simple like that. A separate Problem would be to compute the function Φ(ψ,θ,φ) which I am not doing right now. I am not sure how to start, so I will try several Plans. Plan A. Write de'n = R(Φ+dΦ)en - R(Φ)en = Rz(ψ+dψ) Rx(θ+dθ) Rz(φ+dφ) en - Rz(ψ) Rx(θ) Rz(φ) en Not sure if this makes sense, but pursue for a while. I can rewrite the Euler stuff as = [ Rz(ψ)Rz(dψ) Rx(θ) Rx(dθ)Rz(φ)Rz(dφ) - Rz(ψ) Rx(θ) Rz(φ)] en = [ Rz(ψ){1-idψJ3} Rx(θ){1-idθJ1}Rz(φ){1-idφJ3} - Rz(ψ) Rx(θ) Rz(φ)] en ≈ [ -idψ Rz(ψ)J3 Rx(θ)Rz(φ) - idθ Rz(ψ) Rx(θ)J1Rz(φ) - idφ Rz(ψ) Rx(θ)Rz(φ)J3 ] en Then I could say (de'n /dt) = [ -i Rz(ψ)J3 Rx(θ)Rz(φ) - i Rz(ψ) Rx(θ)J1Rz(φ) - i Rz(ψ) Rx(θ)Rz(φ)J3 ] en The [...] thing is a matrix I could have Maple compute. But we are not yet in the proper form because I want to have e'n on the right in the final formula. I can write en = R(-Φ)e'n where R(-Φ) = Rz(-φ) Rx(-θ) Rz(-ψ) = the inverse rotation Then I have (de'n /dt) = [ -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)e'n and now at least it is in the right general form. Let's now examine the three terms' matrices term1 Rz(ψ)J3 Rx(θ)Rz(φ) Rz(-φ) Rx(-θ) Rz(-ψ) = Rz(ψ)J3 Rz(-ψ) = J3 term2 Rz(ψ) Rx(θ)J1Rz(φ) Rz(-φ) Rx(-θ) Rz(-ψ) = Rz(ψ) Rx(θ)J1Rx(-θ) Rz(-ψ) = Rz(ψ)J1Rz(-ψ) term 3 Rz(ψ) Rx(θ)Rz(φ)J3Rz(-φ) Rx(-θ) Rz(-ψ) = Rz(ψ) Rx(θ)J3 Rx(-θ) Rz(-ψ) So things have at least simplified a bit and we have (de'n /dt) = [ -i J3 - iRz(ψ)J1Rz(-ψ) - i Rz(ψ) Rx(θ)J3 Rx(-θ) Rz(-ψ)] e'n (de'n /dt) = ω x e'n or (de'n /dt) = Ae'n A = [ -i J3 - iRz(ψ)J1Rz(-ψ) - i Rz(ψ) Rx(θ)J3 Rx(-θ) Rz(-ψ)] (de'n /dt) = ω x e'n where I show the target on the second line. It sure seems far-fetched but I think is correct. I have basically this statement A e'n = ω x e'n Suppose I just look at the three components of the equations [A e'n]i = [ω x e'n]i Aik(e'n)k = εijk ωj(e'n)k ok It must then be true that Aik = εijk ωj = εkijωj = - εikjωj ok or Aij = - εijkωk Specifically this says A12 = - ε123ω3 = - ω3 A23 = - ε231ω1 = - ω1 A31 = - ε312ω2 = - ω2 So we conclude that - ω1 = A23 - ω2 = A31 - ω3 = A12 . But for this to have a chance of being valid, we must show that A is antisymmetric! Let's look at the three terms' matrices term 1 J3 this matrix IS antisymmetric as are all the generator matrices term 2 [Rz(ψ)J1Rz(-ψ)] [Rz(ψ)J1Rz(-ψ)]T = Rz(-ψ)T J1T Rz(ψ)T = - Rz(ψ) J1Rz(-ψ) = - term 2 OK term 3 [ Rz(ψ) Rx(θ)J3 Rx(-θ) Rz(-ψ) ]T = - Rz(ψ)Rx(θ)J3 Rx(-θ)Rz(-ψ) = - term 3 This is promising. Each matrix term really is antisymmetric. So the task now is to compute A = [ -i J3 - iRz(ψ)J1Rz(-ψ) - i Rz(ψ) Rx(θ)J3 Rx(-θ) Rz(-ψ)] I could have Maple create this matrix and then just read off the results. Now let's try to get a full result here. I think I have formulas for things Rz(ψ)J2Rz(-ψ) = ... Yes I have all these sandwich rules as in exp(- i J) Jk exp(+ i J) = Jk cos + kmsJmns sin + nk ( J) (1 - cos) (1) And I also have this interesting result Rn(θ) = exp(-iθnJ) = + (Cθ - 1) + Sθ Let's apply the above rules to my necessary special cases. Go back to exp(- i J) Jk exp(+ i J) = Jk cos + kmsJmns sin + nk ( J) (1 - cos) (1) as a vector statement exp(- i J) J exp(+ i J) = J cos + J x sin + ( J) (1 - cos) (1) Suppose ns = δs,r pointing in the r direction. Then the above reads exp(- i Jr) Jk exp(+ i Jr) = Jk cos + kmrJm sin + δk,rJr(1 - cos) or Rr(θ) JkRr(-θ) = cosθ Jk - εkrm sinθ Jm + δk,r(1 - cos)Jr When k = r we know that Rr(θ) JrRr(-θ) = Jr since everything commutes. Rr(θ) JrRr(-θ) = Jr . When k ≠r we know δk,r = 0 so Rr(θ) JkRr(-θ) = cosθ Jk - εkrm sinθ Jm k ≠r We can quickly build a complete table of these cases R1(θ) J2R1(-θ) = cosθ J2 - ε21m sinθ Jm = cosθ J2 + sinθ J3 R1(θ) J3R1(-θ) = cosθ J3 - ε31m sinθ Jm = cosθ J3 - sinθ J2 R2(θ) J1R2(-θ) = cosθ J1 - ε12m sinθ Jm = cosθ J1 - sinθ J3 R2(θ) J3R2(-θ) = cosθ J3 - ε32m sinθ Jm = cosθ J3 + sinθ J1 R3(θ) J1R3(-θ) = cosθ J1 - ε13m sinθ Jm = cosθJ1 + sinθ J2 R3(θ) J2R3(-θ) = cosθ J2 - ε23m sinθ Jm = cosθJ2 - sinθ J1 Now consider the matrices which appear in A above. In the second term we have, Rz(ψ)J1Rz(-ψ) = R3(ψ)J1R3(-ψ) = cosψJ1 + sinψ J2 In the third term, Rz(ψ) Rx(θ)J3 Rx(-θ) Rz(-ψ) = R3(ψ) R1(θ)J3 R1(-θ) R3(-ψ) = R3(ψ)[ cosθ J3 - sinθ J2 ] R3(-ψ) = cosθ R3(ψ)J3R3(-ψ) - sinθ R3(ψ)J2 R3(-ψ) = cosθ J3 - sinθ [ cosψJ2 - sinψ J1 ] = sinθsinψ J1 - sinθcosψ J2 + cosθ J3 Then we may write A = [ -i J3 - iRz(ψ)J2Rz(-ψ) - i Rz(ψ) Rx(θ)J3 Rx(-θ) Rz(-ψ)] = -i J3 - i ( cosψJ1 + sinψ J2) - i (sinθsinψ J1 - sinθcosψ J2 + cosθ J3) = - i [ J3 + ( cosψJ1 + sinψ J2) + (sinθsinψ J1 - sinθcosψ J2 + cosθ J3) = - i [ ( cosψ + sinθsinψ ) J1 + (sinψ -sinθcosψ )J2 + ( + cosθ )J3 ] Therefore -A = ( cosψ + sinθsinψ ) (iJ1) + (sinψ -sinθcosψ )(iJ2) + ( + cosθ )(iJ3) Taking the ij matrix element -Aij = ( cosψ + sinθsinψ ) (iJ1)ij + (sinψ -sinθcosψ )(iJ2)ij + ( + cosθ )(iJ3)ij = ( cosψ + sinθsinψ )ε1ij + (sinψ -sinθcosψ )ε2ij + ( + cosθ )ε3ij Then from relations stated earlier, -ω1 = A23 = - ( cosψ + sinθsinψ ) -ω2 = A31 = - (sinψ -sinθcosψ ) -ω3 = A12 = - ( + cosθ ) And we have then found our ω vector: ω = (cosψ + sinθsinψ , sinψ -sinθcosψ , + cosθ ) = (cosψ + sinθsinψ ) e1 + (sinψ -sinθcosψ )e2 + ( + cosθ)e3 or ωx = cosψ + sinθsinψ ωy = sinψ -sinθcosψ ωz = + cosθ This can be compared with GPS 3rd edition page 174 which I have (don't know my printing) We disagree on the sign of ωy which does seem very odd. Do I recall an errata on this? I checked, and I don't see any errata on this equation apart from adding commas. Here is what I have done in the above. I start with this Euler rotation e'i = R(Φ)ei where R(Φ) = Rz(ψ) Rx(θ) Rz(φ) = rotation by Goldstein Euler angles I then write (de'n/dt)S = ω x e'n and I solve for ω. This then gives the ω which appears in my Fig 1. Does Maple Confirms My result ? Since I have this GPS discrepancy, I will try a Maple version and see what it days. Maple gives A: This says, reading off A12 = - - cosθ A13 = sinψ - cosψsinθ A23 = - cosψ - sinψsinθ all three ok -ω1 = A23 -ω2 = A31 -ω3 = A12 all three ok and therefore Maple is saying -ω1 = - cosψ - sinψsinθ -ω2 = - sinψ + cosψsinθ -ω3 = - - cosθ all three ok Now here is what I got above using the non-Maple method: ωx = cosψ + sinθsinψ ωy = sinψ -sinθcosψ ωz = + cosθ So the Maple method and my Hand method give exactly the same results. Question: I have shown by both MY methods that ωx = cosψ + sinθsinψ ωy = sinψ -sinθcosψ ωz = + cosθ where ω = ωxe1+ωye2+ωze3 and so I am showing the Frame S components of ω where Frame S is my inertial frame. What are the Frame S' components? These are the ones Goldstein seems to be talking about. ω = (ω)'xe'1+ (ω)'ye'2+ (ω)'ze'3 ? Well, consider (ω)'x = ω e'1 = ωxe1 e'1+ωye2 e'1+ωze3 e'1 = ωxe1 Re1 + .... = ωx(e1)i (Re1)i + .... = ωx(e1)i Rij (e1)j + .... = ωxδ1i Rij δ1j + .... = ωxR11 + ... But I don't know what R(Φ) = Rz(ψ) Rx(θ) Rz(φ) unless I multiply it out. This method seems painful, there must be a more direct route. Go back to this point: (de'n /dt) = [ -i J3 - iRz(ψ)J1Rz(-ψ) - i Rz(ψ) Rx(θ)J3 Rx(-θ) Rz(-ψ)] e'n = Ae'n (de'n /dt) = ω x e'n The matrix A is a 3x3 matrix of elements which are functions of the generator matrices and the rates. Can I think of this as a matrix of numbers like π ? I think yes. Then: Suppose I take the Frame S' components of both sides at this point. Then [ω x e'n]'i = εijk (ω')j (e'n)'k = εijk (ω')j δn,k = εijn (ω')j [Ae'n]'i = Aij(e'n)'j = Ain A = matrix of constants??? Then I get the same thing I got before Ain = εijn (ω')j = - εinj (ω')j or Aij = εikj (ω')k = - εijk (ω')k Compare this to the earlier result which was Aij = - εijkωk This seems to suggest that the components are the same in both frames ??? I think this is the wrong step (been here before) [Ae'n]'i = Aij(e'n)'j = Ain In general suppose you have A = MB in Frame S How do you write this in Frame S' ? Let A' = RA and B' = RB so then A = MB RA = RMB = RMR-1RB A' = (RMR-1)B' = M' B' So my real equation then is [Ae'n]'i = A'ij(e'n)'j = A'in = [RAR-1]in R = Rz(ψ) Rx(θ) Rz(φ) and that removed the paradox at least. How about ω' = Rω I have argued in (1.3.5) that when this is true for a vector, we have (ω)'n = (ω')n. Now ω' = Rω (ω')i = (Rω)i = Rij (ω)j and then I get (ω)'i = (ω')i = Rij (ω)j and this is an explicit method for computing the desired components, but it just seems very messy. Plan B. Find a more direct path. Suppose I start out differently: ei = R(Φ)e'i where R(Φ) = Rz(ψ) Rx(θ) Rz(φ) = rotation by Goldstein Euler angles where ei are still the Frame S axis-aligned basis vectors, and Frame S is still the inertial frame. Then. den = dΦ x en (den/dt)S' = - ω x en (1.7.4) I think I can track the previous path in this framework. I will now have, as seen from Frame S', den = R(Φ+dΦ)e'n - R(Φ)e'n = [ -idψ Rz(ψ)J3 Rx(θ)Rz(φ) - idθ Rz(ψ) Rx(θ)J1Rz(φ) - idφ Rz(ψ) Rx(θ)Rz(φ)J3 ] e'n Then I install e'i = R-1(Φ)ei and I will get the same result as before (with this new R(Φ) relation) (den /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 ] Rz(-φ) Rx(-θ) Rz(-ψ)en = [ -i J3 - iRz(ψ)J1Rz(-ψ) - i Rz(ψ) Rx(θ)J3 Rx(-θ) Rz(-ψ)] en = [Aen] exactly the same A as before Then we have (den /dt)S' = [Aen] (den/dt)S' = -[ω x en] So I have this equation of interest [Aen] = -[ω x en] or A R(Φ)e'i = - ω x [R(Φ)e'i] R(Φ)-1A R(Φ)e'i = R(Φ)-1 [ - ω x [R(Φ)e'i] ] How about a simple theorem: Theorem: Imagine Frame S and Frame S' with the same origins. Suppose in Frame S we see some rotation vector ω with certain Frame S components ωi. The Frame S' basis vectors are ALL rotating about some axis with rate ω. If we jump onto Frame S', we will see the Frame S basis vectors rotating with the same axis but with rate -ω. They are just rotating backwards. That is why we have (de'n/dt)S = ω x e'n (den/dt)S' = – ω x en Well, this does not tell how to compare the components in the two frames, but OK. Go back again and recall that A = [ -i Rz(ψ)J3 Rx(θ)Rz(φ) - i Rz(ψ) Rx(θ)J1Rz(φ) - i Rz(ψ) Rx(θ)Rz(φ)J3 ] R-1(Φ) Then perhaps we can say A' = R A R-1 ??? Plan C. Think in terms of ω' = Rω and so (ω)'i = (ω')i = Rij(ω)j R = R(Φ) = Rz(ψ) Rx(θ) Rz(φ) Have Maple directly compute the components (ω)'i and just see what they look like before doing any more threading attempts. Use ωx = cosψ + sinθsinψ ωy = sinψ -sinθcosψ ωz = + cosθ And what about the bugaboo : " if the basis vectors rotate one way, then general vectors rotate the other way". Is this a myth or is there some truth to the statement which I have encountered many times? And how would that apply here? Questions for manana to get things rolling again.