new Appendix G_4 INSTALLED

this has been installed, to not edit here. G.4 Two more theorems for the rotation matrix toolbox Theorem 1: RJR-1 = R-1J (G.4.1) This is another vector/matrix notation theorem which makes a claim about rotating a vector of matrices. The above notation is a shorthand for the following, RJiR-1 = [R-1J]i = R–1ijJj i = 1,2,3 . (G.4.2) The object on the left is a product of three 3x3 matrices, while the right side is a linear combination of 3x3 matrices, so at least the theorem's claim is dimensionally reasonable. More generally Ji might be an abstract "operator" and R a rotation which acts on that operator, see ** below. Proof: Start with the general rotation form given in (G.2.6), R = exp(-iθnJ) = 1 + (cosθ - 1) T + sinθ[ -i(nJ)] where Tab = (δab - nanb) . (G.4.3) Sandwich this rotation around Ji and then use the sandwich formula (G.3.3), RJiR-1 = exp(- iθn J) Ji exp(+ iθn J) = cosθ Ji + sinθ εijknk Jj + (1 - cosθ ) ni (n J) = cosθ δijJj + sinθ εijknk Jj + (1 - cosθ ) ninjJj = [ cosθ δij + sinθ εijknk + (1 - cosθ ) ninj ] Jj . (G.4.4) Our theorem is proved if, comparing (G.4.2) and (G.4.4), we can show that R–1ij = cosθ δij + sinθ εijknk + (1 - cosθ ) ninj . (G.4.5) Looking back at (G.4.3), one has R-1 = exp(+iθnJ) = 1 + (cosθ - 1) T + sinθ[ +i(nJ)] so R-1ij = δij + (cosθ - 1) Tij + sinθ[+nk (iJk)ij] = δij + (cosθ - 1) (δij - ninj) + sinθ nk εkij // (G.1.3) and def of Tij = δij cosθ + (1-cosθ) ninj + sinθ εijknk (G.4.6) But this is the same as (G.4.5) so the theorem is proved. Theorem 2: R exp(-iθnJ) R-1 = exp(-iθn'J) where n' = Rn (G.4.7) On the left we have a product of three 3x3 matrices, while the right side is a 3x3 matrix. The matrix R is an arbitrary rotation. One proof of this theorem might be to say "what else could it be? ". We shall provide a more substantial proof below. One preliminary test is that we can take the determinant of both sides to make sure we get the same thing. Below in ** we shall prove that det(eA) = etr(A) in any number of dimensions, so our check is then det[R exp(-iθnJ) R-1] = det[ exp(-iθn'J) ] ? det[exp(-iθnJ)] = det[ exp(-iθn'J) ] ? exp[tr(-iθnJ)] = exp[tr(-iθn'J)] ? exp[-iθnk(trJk)] =exp[-iθn'k(trJk)] ? // the Ji generators (G.1.2) are traceless exp[0] = exp[0] ? 1 = 1 ? yes So we are at least encouraged (and we got to apply the fancy trace theorem). Proof: Start by inserting the general rotation form (G.2.6) into the left side of (G.4.7), R exp(-iθnJ) R-1 = R [ 1 + (cosθ - 1) T - i sinθ (nJ) ] R-1 = 1 + (cosθ - 1)RTR-1 - i sinθ nk RJkR-1 . (G.4.8) Now consider [RTR-1]ad = RabTbcR-1cd = Rab(δbc - nbnc)RTcd = δad - Rab nbncRdc = δad - ( Rab nb)(Rdcnc) = δad - n'an'd n' = Rn ≡ T'ad . // in other words, RTR-1 = T' (G.4.9) From Theorem (G.4.2) we know that RJkR-1 = R–1kjJj . (G.4.2) Inserting this last item and (G.4.9) into (G.4.8) gives R exp(-iθnJ) R-1 = 1 + (cosθ - 1)T' - i sinθ nkR–1kjJj = 1 + (cosθ - 1)T' - i sinθ Rjknk Jj = 1 + (cosθ - 1)T' - i sinθ [Rn]j Jj = 1 + (cosθ - 1)T' - i sinθ n'jJj = 1 + (cosθ - 1)T' - i sinθ (n' J) = exp(-iθn'J) // using (G.2.6) with n → n' (G.4.10) Thus the theorem is proved. The fact (G.4.9) that RTR-1 = T' is in fact just the standard rule for the transformation of a rank-2 tensor under rotations, as shown for example in Lucht Tensor (5.7.1). That is to say, if an Observer in Frame S sees T, an Observer in rotated Frame S' will see T' = RTR-1. This same statement applies at the higher level of our theorem R exp(-iθnJ) R-1 = exp(-iθn'J) or R Rn(θ) R-1 = Rn'(θ) n' = Rn // Theorem 2 restated (G.4.11) The rotation Rn(θ) seen by an Observer in Frame S appears as Rn'(θ) in Frame S' where e'n = R-1en as in (1.1.1) . Like T, the rotation Rn(θ) transforms as a rank-2 tensor under rotations. An equivalent active statement of the theorem is the following: R Rn(θ) R-1v = Rn'(θ)v where n' = Rn (G.4.12) Instead of rotating a vector v by amount θ about axis ', one can first back-rotate the vector v, then rotate by amount θ about axis , the forward rotate the result. Example: Rz(π/2)Rx(θ)Rz(-π/2) = ?? Here n = and n' = Rz(π/2) = so we conclude that Rz(π/2)Rx(θ)Rz(-π/2) = Ry(θ) . (G.4.13) Here is a graphical interpretation of (G.4.13) applied to a particular vector v : (G.4.14)