a rotation problem 2012

A Rotation Problem PhL 7.3.12 1. Statement of the problem: 1 2. Plan A: Too Ugly 1 3. Plan B: Works but is a mess. 2 4. Digression On Irrelevant Fact. 4 5. The problem with order reversed 6 1. Statement of the problem: Given vectors θ and dθ , find vector dΩ such that R(dΩ) = R(θ)-1R(θ+dθ) We know that some R(dΩ) exists because, since rotations form a group, the product on the right is some rotation. We anticipate that it is a small rotation, since the two items on the right are so close to being inverses of each other. In terms of generators, the above reads exp(-i dΩ J) = exp(+i θ J) exp(-i [θ+dθ] J) I have been using the following simple notation for the vector parameters θ = θn dθ = dφ m dΩ = dΩ r [θ+dθ] = Ss = θn + dφ m where n,m,r,s are all unit vectors. It seems to me that there should be some easy way to solve this problem, but so far I have been unable to find it. I will review the various methods I have tried but which have failed. In any solution, one must somehow deal with the complicated object exp(-i [θ+dθ] J) . Comment: If we use Euler angles, I know that we can solve the general problem R(ψ3,θ3,φ3) = R(ψ1,θ1,φ1) R(ψ2,θ2,φ2) and this is probably written down somewhere in my notes or in Tinkham or what have you. So we know our little problem is solvable. It's just that I wanted an easy vector way to do it. 2. Plan A: Too Ugly In this plan, I do a brute force attempt to expand exp(-i [θ+dθ] J) exp(-i [θ+dθ] J) = Σn=0∞ (-i[θ+dθ] J)n/n! The quantity of interest then is (-i[θ+dθ] J)n = (-i[θ+dθ]k Jk)n Define the vector q ≡ -i[θ+dθ] Then the quantity of interest is (qkJk)n Here then are the first few powers: (qaJa)1 = qaJa (qaJa)2 = qaJa qbJb = qaqb Ja Jb (qaJa)3 = qaqbqc Ja Jb Jc (qaJa)4 = qaqbqcqd Ja Jb Jc Jd = (-i)4 [θ+dθ]a [θ+dθ]b [θ+dθ]c [θ+dθ]d Ja Jb Jc Jd = (-i)4 θaθbθcθd Ja Jb Jc Jd + (-i)4 dθaθbθcθd Ja Jb Jc Jd + etc This seems a very messy path to follow so I did not pursue it further. The messy part is really taken care of elsewhere in Plan B to follow. 3. Plan B: Works but is a mess. Now in document "exp nJ calculation.doc" I show this fact in 3D, exp(-iθnJ)ab = δab + (1 - Cθ) (δab - nanb) - niεiac { Sθ (δcb - ncnb)} which I translate to say exp(-iSsJ)ab = δab + (1 - CS) (δab - sasb) - sjεjae { SS (δeb - sesb)} So this seems to at least give some reasonable expansion for the ugly exp(-i [θ+dθ] J) object. Then we get [exp(-i dΩ J)]ab = [exp(+i θ J)]ad [exp(-i [θ+dθ] J)]db [δad + (1 - Cθ) (δad - nand) - niεiac { Sθ (δcd - ncnd)}] * [ δdb + (1 - CS) (δdb - sdsb) - sjεjde { SS (δeb - sesb)}] = δab + [ other terms ]ab Expanding the LHS assuming small angle gives [ since i (Jk)ab = kab] δab + (-i dΩ J)ab = δab + (-i dΩk Jk)ab = δab + (-i dΩk (Jk)ab = δab - dΩk kab So the implication is that this must be true - dΩk kab = [ other terms ]ab // = -dΩ rk kab Now define these two symmetric matrices Nab = (δab - nanb) Sab = (δab - sasb) Then we can write [δad + (1 - Cθ) Nad - niεiac SθNcd] * [ δdb + (1 - CS) Sdb - sjεjdeSS Seb] = δab + [ other terms ]ab There are 8 terms in "other terms" and I will just list them δad(1 - CS) Sdb δad (- sjεjdeSS Seb) (1 - Cθ) Nad δdb (1 - Cθ) Nad(1 - CS) Sdb (1 - Cθ)( - sjεjdeSS Seb) (- niεiac SθNcd) δdb (- niεiac SθNcd) (1 - CS) Sdb (- niεiac SθNcd) (- sjεjdeSS Seb) Since the sum of these terms is supposed to be equal to - dΩk kab, the implication is that the above some is antisymmetric under a↔b, but this fact is totally non-obvious. But let's assume it is in fact true. Then we can compute for example - dΩ1 123 = - dΩ1 = δ2d(1 - CS) Sd3 = (1 - CS) S23 δ2d (- sjεjdeSS Se3) = (- sjεj2eSS Se3) (1 - Cθ) N2d δd3 = (1 - Cθ) N23 (1 - Cθ) N2d(1 - CS) Sd3 (1 - Cθ)( - sjεjdeSS Se3) (- niεi2c SθNcd) δd3 = (- niεi2c SθNc3) (- niεi2c SθNcd) (1 - CS) Sd3 (- niεi2c SθNcd) (- sjεjdeSS Se3) It is an ugly mess, but at least it is something for -dΩ1. Similarly we can compute the others and then we have computed dΩ which was the goal. Somehow we think this will be proportional to dφ but that too is not very obvious. To pursue this we would have to compute more things, such as S2 = θ2 + 2 θ dφ = θ2 ( 1 + 2 dφ/θ ) S = θ ( 1 + 2 dφ/θ )1/2 ≈ θ (1 + dφ/θ ) ≈ θ + dφ Ss = θn + dφ m s = θn/S + dφ m/S s ≈ θn/( θ + dφ ) + dφ m/θ s ≈ n/( 1 + dφ/θ ) + dφ m/θ s ≈ n( 1 - dφ/θ ) + dφ m/θ s ≈ n - dφ/θ n + dφ m/θ s ≈ n - (dφ/θ)[ n + m] I could go on for a long time, but since I have an answer, let it be. 4. Digression On Irrelevant Fact. When I started on this, I thought there was some reason to compute R(θ) R(dθ) R(-θ) . But now I don't see that this has any bearing at all on the problem at hand. But since I did this calculation, I will record it here. I did it like this R(θ) R(dθ) R(-θ) = exp(- i J) exp(- i dφ J) exp(+ i J) Now expand exp(- i dφ J) ≈ 1 - i dφ J Then have R(θ) R(dθ) R(-θ) = exp(- i J) [1 - i dφ J] exp(+ i J) = 1 -idφ { exp(- i J) J exp(+ i J) } Now use the sandwich rule for the thing in "Evaluate n dot J rotation of J.doc" exp(- i J) J exp(+ i J) = J cos + J x sin + ( J) (1 - cos) and so we then have R(θ) R(dθ) R(-θ) = 1 -idφ { J cos + J x sin + ( J) (1 - cos) } ≈ exp(-idφ { J cos + J x sin + ( J) (1 - cos) }) Now we know that J x = J x so the above becomes = exp(-idφ { J cos + J x sin + ( J) (1 - cos) }) = exp(-idφ { cos + x sin + () (1 - cos) } J ) I then defined u as u ≡ cos + x sin + () (1 - cos) I showed that u was a unit vector like so: u u = [ cos + x sin + () (1 - cos)]2 = cos2θ + (x )2 sin2θ + ( )2 (1-cosθ)2 + 2 cos sin x + 2 cos (1 - cos)( )2 + 2 sin (1 - cos) x ( ) = cos2θ + (x )2 sin2θ + ( )2 (1-cosθ)2 + 2 cos (1 - cos)( )2 But (x )2= sin2ψ where ψ is the angle between these two vectors, and ( )2 = cos2ψ so = cos2θ + sin2ψ sin2θ + cos2ψ (1-cosθ)2 + 2 cos (1 - cos) cos2ψ = cos2θ + sin2ψ sin2θ + cos2ψ ( 1 + cos2θ - 2cosθ) + 2cosθcos2ψ - 2 cos2θ cos2ψ = cos2θ + sin2ψ sin2θ + cos2ψ ( 1 - cos2θ - 2cosθ) + 2cosθcos2ψ = cos2θ + sin2ψ sin2θ + cos2ψ ( 1 - cos2θ) = cos2θ + sin2ψ sin2θ + cos2ψ sin2θ = cos2θ + sin2θ = 1 And the conclusion then was R(θ) R(dφ ) R(-θ) ≈ exp(-idφ J ) = R(dφ ) where ≡ cosθ + x sinθ + () (1 - cosθ) So basically the sandwich here just changes the axis of the small rotation from to . 5. The problem with order reversed Given vectors θ and dθ , find vector dΩ' such that R(dΩ') = R(θ+dθ) R(θ)-1 ↔ R(θ+dθ) = R(θ) R(dΩ') This problem can be solved in a similar fashion, but we expect that dΩ' ≠ dΩ.