Appendix G v2

This has been replaced by Appendix G v.3 doc. do not edit here! Appendix G: Rotation Matrices and Goldstein's body-frame ω 1 G.1 Generators and finite rotation matrices 1 G.2 About the general rotation matrix 3 G.3 The Baker-Campbell-Hausdorff and Sandwich Formulas 6 G.4 Two more theorems for the rotation matrix toolbox 9 G.5 Goldstein's Euler Angles 11 G.6 Euler angles which change in time : computation of ω (Method 1) 22 G.7 Computation of ω (Method 2) 26 G.9 Generalizations of the Rotation Group 37 Appendix G: Rotation Matrices and Goldstein's body-frame ω Here we present still more information on rotation matrices and demonstrate some typical manipulations done with such matrices leading to a derivation of Goldstein body-frame ω vector expressions. G.1 Generators and finite rotation matrices The general active rotation matrix for rotation by angle θ about some n unit vector axis is given by, R(φ) = exp(-i θ n J) , (G.1.1) where the Jk are 3x3 matrices known as the rotation generator matrices, J1 = J2 = J3 = . (G.1.2) The numbers in these three matrices can be summarized in this single statement, (Ja)bc = - i abc // for example, (J1)23 = - i 123 = -i or (iJa)bc = abc (G.1.3) where the εabc permutation tensor is described in (1.5.3). Defining the commutator of two square matrices X,Y as [X, Y] = XY-YX, one can show that the above three matrices satisfy this "commutation relation", [Ji, Jj] = iεijkJk // for example, [J1,J2] = iJ3 (G.1.4) Proof: These easily demonstrated facts will be used in the proof (implied sum on repeated indices) , εabcεABc = δaAδbB - δaBδbA δab ≡ δa,b εabc = εbca = εcab (G.1.5) LHSac = (Ji)ab(Jj)bc - (Jj)ab(Ji)bc = (-i)2[εiabεjbc - εjabεibc ] = - [εiabεcjb - εjabεcib] = - [(δicδaj - δijδac) - (δjcδai - δjiδac)] = δjcδai - δicδaj . RHSac = iεijk(Jk)ac = iεijk[-iεkac] = εijkεkac = εijkεack = δiaδjc - δjaδic . QED The expression shown in (G.1.1) involves the exponentiation of a square matrix to produce a new square matrix of the same dimension. This notion of exponentiating a matrix is straightforward as we demonstrate with a simple example: Rx(θ) = exp(-i θ J) = exp(-i θJ1) ≡ Σn=0∞ (-iθ)n (J1)n /n! = 1 + Σ2,4,6.. (-iθ)n (J1)n /n! + Σ1,3,5.. (-iθ)n (J1)n /n! It is easy to show that (J1)n = J1 for odd n, while (J1)n = for even n > 0. Thus, Rx(φ) = + Σ2,4,6.. (-iθ)n/n! + Σ1,3,5.. (-iθ)n/n! . But Σ2,4,6.. (-iθ)n/n! = - θ2/2! + θ4/4! + ... = (cosθ - 1) Σ1,3,5.. (-iφ)n/n! = (-iθ) + (-iθ)3/3! + ... = (-i) [ θ - θ3/3! + ..] = -i sinθ (G.1.6) and therefore Rx(φ) = + (cosθ - 1) + (-isinθ) = + = . (G.1.7) The three axis-aligned rotations are found in this manner to be Rx(θ) = Ry(θ) = Rz(θ) = . (G.1.8) The general rotation (G.1.1) applied to a vector v produces a forward right-hand-rule rotation of vector v by an angle θ about an arbitrary axis to create v' = Rv. We call this an active rotation. For example, if one applies Rz(θ) shown in (G.1.8) to the unit vector = (1,0,0) one gets (cosθ,sinθ,0) which for small θ is a vector in the first quadrant of the x,y plane. Some authors define rotation matrices which rotate vectors backwards according to the right hand rule, with the connection to our matrices then being θ ↔ -θ. The motivation for doing this is the fact that, whereas v' = Rv in the active view where the vector moves and the axes stay put, one can instead do e'n = R–1en and have the vector stay put and the axes are back-rotated, which is the passive viewpoint. The definition of the rotation matrices is just a convention and (G.1.8) shows our definitions. One says that the matrices Ji "generate" the finite rotations when they are exponentiated. An alternative approach: First, one can show that Ri(dθ) ≈ 1 - idθJi describes a 3x3 matrix which, when applied to a vector, causes that vector to rotate by the small amount dθ about the i axis (an "infinitesimal rotation"). Setting dθ = θ/n, one shows that limn→∞ (1-i[θ/n]Ji)n = exp(-iθJi). This last limit is analogous to limn→∞(1+x/n)n = ex for a scalar value x. G.2 About the general rotation matrix An explicit expression for the general rotation matrix We present this subsection as a Reader Exercise with a set of steps. The result is stated in (G.2.8). We are interested in the following general rotation matrix: Rn(θ) = exp(-i θ n J) ≡ Σn=0∞ (-iθ)n (n J)n/n! . n = = a unit vector (G.2.1) (a) using (G.1.3) and (G.1.5), show that (nJ)2 = T where Tab = (δab - nanb) . (G.2.2) (b) show that (nJ)n = (nJ) T for n = 3,5,7,.... (nJ)n = T for n = 2,4,6... (G.2.3) (c) show that exp(-iθnJ) = cos(θnJ) - i sin(θnJ) where each term is defined by its series (d) show that cos(θnJ) = 1 + T(cosθ - 1) sin(θnJ) = (θnJ) + (nJ)T (sinθ - θ) (G.2.4) (e) show therefore that exp(-iθnJ) = 1 + T (cosθ - 1) -i (nJ){ θ + T (sinθ - θ) } (G.2.5) (f) show that the terms linear in θ all cancel leaving this simpler result exp(-iθnJ) = 1 + (cosθ - 1) T + sinθ[ -i(nJ)] (G.2.6) (g) show that T = -i(nJ) = (G.2.7) giving this final result, Rn(θ) = exp(-iθnJ) = + (cosθ - 1) + sinθ . (G.2.8) Note that the first two terms are symmetric matrices, while the third is antisymmetric. (h) verify the special cases shown in (G.1.8) above. These cases have n = (1,0,0), (0,1,0, (0,0,1). Finding n and θ Problem: One is handed a 9-element rotation matrix R and one wants to find n,θ so R = exp(-iθnJ) . Solution: We are unaware of a single magic formula that solves this problem, so we use brute force. One can break R into the sum of two pieces, one symmetric and one antisymmetric by constructing Sab ≡ (Rab+Rba)/2 Aab ≡ (Rab-Rba)/2 Rab = Sab + Aab R = S + A . (G.2.9) Looking at (G.2.9) one then has, S = = + (cosθ - 1) A = = sinθ (G.2.10) which can be written as these nine equations A12 = -n3sinθ S11 = 1 + (cosθ-1)(1-n12) S12 = (cosθ-1)(-n1n2) A13 = n2sinθ S22 = 1 + (cosθ-1)(1-n22) S13 = (cosθ-1)(-n1n3) A23 = -n1sinθ S33 = 1 + (cosθ-1)(1-n32) S23 = (cosθ-1)(-n2n3) . (G.2.11) If θ = 0, then the original matrix must be R = 1 and then we have no work to do. If θ = π, the entire A matrix vanishes according to (G.2.9) leaving only S. The equations above are then S11 = 1 + (-2)(1-n12) S12 = 2n1n2 S22 = 1 + (-2)(1-n22) S13 = 2n1n3 S33 = 1 + (-2)(1-n32) S23 = 2n2n3 . Since n12 + n22 + n32 = 1, not all three ni can vanish. The left column of equations says ni2 = 1 - (1-Sii)/2 i = 1,2,3 . Inspect the three ni2 and select one nr2 which is non-zero. Then select the plus sign to get, nr = +. To be specific, assume nr = n1. Then from the right column, n2 = S12/(2n1) and n2 = S13/(2n1) and we are done. So at least one of the following solutions must be viable: θ = π n1 = n2 = S12/(2n1) n3 = S13/(2n1) θ = π n2 = n1 = S12/(2n2) n3 = S23/(2n2) θ = π n3 = n1 = S13/(2n3) n2 = S23/(2n3) (G.2.12) If θ ≠ 0 or π, then note from (G.2.10) that A122 + A132 + A232 = sin2θ S11+ S22 + S33 = 1 + 2 cosθ . (G.2.13) Using the sign ambiguity in the direction of n, we can assume that 0 < θ < π so sinθ > 0. Then from (G.2.13) and the left column of (G.2.11) we have this solution, sinθ = cosθ = [ S11+ S22 + S33 - 1]/2 n1 = -A23/sinθ n2 = A13/sinθ n3 = -A12/sinθ (G.2.14) and thus a viable θ and n have been found such that R = exp(-iθnJ) . Example: Let R = so then S = and A = . sinθ = = = sinα cosθ = [ S11+ S22 + S33 - 1]/2 = [cosα + cosα + 1 - 1]/2 = cosα θ = α n1 = -A23/sinθ = 0 n2 = A13/sinθ = 0 n3 = -A12/sinθ = sinα/sinθ = 1 n = (0,0,1) Therefore R = exp(-iαJ3) = Rz(α) , as verified in (G.1.8). G.3 The Baker-Campbell-Hausdorff and Sandwich Formulas Statement and proof of the Baker-Campbell-Hausdorff (BCH) formula We now quote a fascinating fact known as the Baker-Campbell-Hausdorff formula involving two square matrices A and B : e-ABeA = B + [B,A]/1! + [[B,A],A]/2! + [[[B,A],A],A]/3! + ..... (G.3.1) Outline of BCH proof: LHS = RHS (a) Show that LHS = Σn=0∞Σm=0∞ (-A)nBAm / (n!m!) . (b) Set k = n+m to rewrite as LHS = Σk=0∞Σn=0k (-A)nBAk-n / (n![k-n]!) . (c) Rewrite again as LHS = Σk=0∞Tk/k! where Tk ≡ Σn=0k (-A)nBAk-n . (d) Define C0 = B, C1 = [B,A], C2 = [[B,A],A] , etc. so that RHS = Σk=0Ck/k! . Note that [Ck,A] = Ck+1. The proof LHS = RHS is complete if one can show that Tk = Ck. (e) Show Tk= Ck by induction: show T0 = C0 and Tk= Ck Tk+1= Ck+1 . QED Statement and proof of the Sandwich Formula Now using this BCH formula, along with the commutation relation (G.1.4) that [Ji, Jj] = iεijkJk, one can show that exp(- iθnJ) J exp(+ iθnJ) = cosθ J + sinθ J x n + (1 - cosθ) n(nJ) . (G.3.2) This "vector of matrices" notation is just a shorthand for the following equations for k = 1,2,3 : exp(- iθnJ) Jk exp(+ iθnJ) = cosθJk + sinθ[J x n ]k + (1 - cosθ) nk (nJ) = cosθ Jk + sinθ εkmsns Jm + (1 - cosθ ) nk (nJ) . (G.3.3) This is the "sandwich formula" since Jk on the left is sandwiched between two rotations. Outline of Sandwich proof: LHS = RHS (a) We will use the BCH formula with A = iθ nJ and B = Jk. First, define Ck as in (d) above. (b) show that C0 = Jk, C1 = -θniεkijJj, C2 = θ2 (nkniJi - Jk), C3 = -θ2C1, C4 = -θ2C2 (c) deduce (or use induction to show) that in general, Cn = - (-1)n/2 θn ( nk ni Ji - Jk ) n = 2,4,6.... Cn = - (-1)(n-1)/2 θn ni εkij Jj n = 1,3,,5... At this point we have from the BCH formula, exp(- iθnJ) Jk exp(+ iθnJ) = Σn=0∞ Cn/n! with Cn as in (c) above (d) Show that Σn=0∞ Cn/n! = = Jk - ni εkij Jj Σn=1,3,5.. (-1)(n-1)/2 θn/n! - ( nk ni Ji - Jk ) Σn=2,4,6.. (-1)n/2 θn/n! = Jk - ni εkij Jj [θ - θ3/3! + ...] - ( nk ni Ji - Jk ) [ -θ2/2! + θ4/4! + .... ] = Jk - ni εkij Jj sinθ - ( nk ni Ji - Jk ) (cosθ - 1) = Jkcosθ + εkji Jj ni sinθ + nk (nJ)(1-cosθ) . QED Comments A vector v (rank-1 tensor) transforms ("rotates") according to v' = Rv. For a matrix M (rank-2 tensor) the corresponding transformation is M' = RMR-1, and this is what one sees on the left side of (G.3.3) where M = Jk and R = exp(-iθ J). In the expression RJkR-1 the rotation generator Jk is "sandwiched" between the two rotations. The above sandwich formulas play a major role in Magnetic Resonance Imaging. The connection is that protons in your body have magnetic moments (spins) which can be lined up by a strong magnetic field. When the proton spins are slammed with a certain radio frequency pulse, they do conical rotation (precession) about the magnetic field axis at the so-called Larmor frequency. After the pulse this proton precession decays away (T1) and bulk-decoheres (T2) producing a certain return RF signal which can be analyzed. These return signals are sensitive to the local environment of the protons. The location of a particular response is determined by giving the magnetic field a spatial gradient which affects the Larmor frequency. In this manner, an image can be formed. The sandwich formulas are not applied directly to individual spin angular momenta J, but to the average spin (polarization) density in the object being scanned. The analysis is quite complicated since it must take into account thermal and statistical effects which are managed with the use of the density matrix formalism. See the excellent text of Levitt for all the details. Special cases of the sandwich formula We shall have our own purposes for the sandwich formulas in the next Section. For rotations about the i axis we set ns = δs.i n J = Ji and kmsJmns = kmiJm = εikmJm (G.3.4) Then from the sandwich formula (G.3.3), exp(- iθnJ) Jk exp(+ iθnJ) = cosθ Jk + sinθ εkmsnsJm + (1 - cosθ) nk(n J) (G.3.3) we find that (there is no implied sum on i in the rightmost term and εkmi = εikm ) exp(- iθJi) Jk exp(+ iθJi) = cosθ Jk + sinθεikmJm + (1 - cosθ) δk,i Ji so then Ri(θ) Jk Ri(-θ) = cosθ Jk + sinθ εikm Jm + (1 - cosθ) δk,i Ji . (G.3.5) In the case i = k we know that the left side is just Jk since everything then commutes. This is verified on the right since εiim = 0 and the other two terms then add up to Jk. So although obvious, we state: Rk(θ) Jk Rk(-θ) = Jk i = k (G.3.6) In the case i ≠ k the third term in (G.3.5) does not contribute and we have Ri(θ) Jk Ri(-θ) = cosθ Jk + εikmsinθ Jm i ≠ k We shall now construct a table of all the cases for which i ≠ k : (G.3.7) R1(θ) J2R1(-θ) = cosθ J2 + ε12m sinθ Jm = cosθ J2 + sinθ J3 R1(θ) J3R1(-θ) = cosθ J3 + ε13m sinθ Jm = cosθ J3 - sinθ J2 R2(θ) J1R2(-θ) = cosθ J1 + ε21m sinθ Jm = cosθ J1 - sinθ J3 R2(θ) J3R2(-θ) = cosθ J3 + ε23m sinθ Jm = cosθ J3 + sinθ J1 R3(θ) J1R3(-θ) = cosθ J1 + ε31m sinθ Jm = cosθJ1 + sinθ J2 R3(θ) J2R3(-θ) = cosθ J2 + ε32m sinθ Jm = cosθJ2 - sinθ J1 . (G.3.8) 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 ambiguous 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) or R Rn(θ) R-1 = Rn'(θ) where n' = Rn The matrix R is an arbitrary rotation. On the left we have a product of three 3x3 matrices, while the right side is a 3x3 matrix. One proof of this theorem might be to say "what else could it be? ". We shall provide a more substantial proof below. 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 1 (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 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 (1.1.30) . 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) G.5 Goldstein's Euler Angles As the reader no doubt knows, it is possible to specify an arbitrary rotation in terms of three Euler angles, Rz(φ)Rx(θ)Rz(ψ) . (G.5.1) As shown below, the matrix shown in (G.5.1) will be identified with R-1 of our Section 1 formalism. Some authors use other letters for the angles, and some put Ry in the middle in place of Rx. We have in mind that the matrices shown are specifically the active rotation matrices shown in (A.1). Rx(θ) = Ry(θ) = Rz(θ) = . (G.5.2) To get Maple warmed up for activities below, we enter the three matrices of (G.5.2), (G.5.3) These (G.5.2) matrices are "active" because, using the right-hand-rule, they rotate a vector forward by angle θ in the Active View described at the start of Section 1.3. For example, for small θ the rotation Rz(θ) acting on produces a vector in the first quadrant of the x-y plane: (G.5.4) Here now is Goldstein's Euler Angle picture (G p 107, GPS p152), enhanced a bit for readability, (G.5.5) Astronomy Footnote: Imagine that the white disk is a "reference plane" (perhaps the equatorial plane of the Earth) and the grey disk perimeter denotes the orbit of some object about another (perhaps the Moon about the Earth). That orbiting object crosses the reference plane in two places called "nodes" (ascending and descending by some convention). The intersection of the two planes is called the "line of nodes". In Goldstein's drawing this coincides with the ξ' axis. In this application, the Euler angles θ,φ define the plane of the Moon's orbit, and Euler angle ψ shows the progress of the Moon in this orbit. As noted in Section 8.8, the Moon's orbital plane precesses, so the line of nodes rotates in the white disk plane. Fig (G.5.5) shows how one can start with x,y,z axes at the top, and end up with x',y',z' axes on the lower right. There are really three sequential transformations occurring here, and we can just read off the effects on unit vectors by looking at the pictures: ( ξ = xi = "zeye", η = eta = "ate'uh, ζ = zeta = "zee'ta") top (,,) = Rz(φ) (,,) where = left (',',') = Rξ(θ) (,,) where ' = right (',',') = Rζ'(ψ)(',',') where ' = ' (G.5.6) For example, the first of these nine equations says = Rz(φ) which seems clear from the top picture. The rotation Rξ(θ) is an active rotation of θ about the axis, and similarly for Rζ'(ψ). We can combine transformations in various obvious ways : (,,) = Rz(φ) (,,) // each equation is like e'n = R-1en (',',') = Rξ(θ) (,,) = Rξ(θ) Rz(φ) (,,) (',',') = Rζ'(ψ) (',',') = Rζ'(ψ) Rξ(θ) (,,) = Rζ'(ψ) Rξ(θ) Rz(φ) (,,) . (G.5.7) For example, ' = Rζ'(ψ)' = Rζ'(ψ) Rξ(θ) = Rζ'(ψ) Rξ(θ)Rz(φ) . These equations are all of the template form e'n = (R-1)en appearing in our Basis Theorem (1.1.30). The meaning is |e'n> = R-1|en> = |R-1en>. If one takes Frame S components of these equations, then in e'n = (R-1)en one can interpret (R-1) as a matrix. For example, for = Rz(φ) one can write ()i = [Rz(φ)]ij()i in which case [Rz(φ)]ij is the matrix shown in (G.5.2). In going all the way from the Frame S basis en to the Frame S' basis e'n we see from the last line in (G.5.7) that in order to interpret this last line as e'n = (R-1)en, we must make the identification R-1 = Rζ'(ψ) Rξ(θ) Rz(φ) . (G.5.8) The R symbol here is the R that appears in our Section 1 formalism. In particular, recall the Basis Theorem (1.1.29) and (1.1.30), and the alternate notation of (1.1.32), e'n = R-1en e'n = Rnm em or = R . (1.1.29,30) + (1.1.32) Define Q ≡ R-1 and rewrite the Basis Theorem as, e'n = Qen e'n = (Q-1)nm em or = [Q]-1 . (G.5.9) This form of the Basis Theorem then serves as a template with which we can convert the equations of (G.5.7) to the corresponding linear combination equations of basis vectors : = [Rz(φ)]-1 = Rz(-φ) Q = Rz(φ) = [Rξ(θ) Rz(φ)]-1 = Rz(-φ) Rξ(-θ) Q = Rξ(θ) Rz(φ) = [Rζ'(ψ) Rξ(θ)Rz(φ)]-1 = Rz(-φ) Rξ(-θ)Rζ'(-ψ) . (G.5.10) We constantly use facts like [ABC]-1 = C-1B-1A-1 and R-1s(α) = Rs(-α). These equations can be inverted in the obvious manner. The last one would give = Rζ'(ψ) Rξ(θ)Rz(φ) . (G.5.11) If we install this equation on the right of the first two equations in (G.5.10), the results are = Rz(-φ) = Rz(-φ) Rζ'(ψ) Rξ(θ)Rz(φ) = Rz(-φ) Rξ(-θ) = Rz(-φ) Rξ(-θ) Rζ'(ψ) Rξ(θ)Rz(φ) . (G.5.12) We know all about the rotations Rx, Ry and Rz since they are specifically stated in (G.5.2). But what about the strange rotations Rξ and Rζ' which dot the landscape above? In Theorem 3 below we shall show that each of these rotations may be written as a certain product of the Rx, Ry and Rz. Specifically, in Theorem 3 we shall prove the first three results below : (1) Rξ(θ) = Rz(φ) Rx(θ) Rz(-φ) (2) Rζ'(ψ) = Rz(φ) Rx(θ)Rz(ψ)Rx(-θ)Rz(-φ) (3) Rζ'(ψ) Rξ(θ) Rz(φ) = Rz(φ)Rx(θ)Rz(ψ) = R-1 Using these equations one can clear out all the strange rotations from (G.5.10,11,12) as follows : (1) (4) Rz(-φ) Rξ(-θ) = Rz(-φ) [Rz(φ) Rx(-θ) Rz(-φ)] = Rx(-θ) Rz(-φ) (3) (5) Rz(-φ) Rξ(-θ)Rζ'(-ψ) = [Rζ'(ψ) Rξ(θ) Rz(φ)]-1 = [Rz(φ)Rx(θ)Rz(ψ)]-1 = Rz(-ψ)Rx(-θ)Rz(-φ) (2) (1) (6) Rz(-φ) Rζ'(ψ) Rξ(θ)Rz(φ) = Rz(-φ) [Rz(φ) Rx(θ)Rz(ψ)Rx(-θ)Rz(-φ)] [Rz(φ) Rx(θ) Rz(-φ)]Rz(φ) = Rx(θ)Rz(ψ) // use Rz(-φ)Rz(φ) = 1 in three places, then Rx(-θ)Rx(θ) = 1 (7) Rz(-φ) Rξ(-θ) Rζ'(ψ) Rξ(θ)Rz(φ) = Rz(-φ) [Rz(φ) Rx(-θ) Rz(-φ)] [Rz(φ) Rx(θ)Rz(ψ)Rx(-θ)Rz(-φ)][Rz(φ) Rx(θ) Rz(-φ)]Rz(φ) = Rz(ψ) (1) (3) (1) (G.5.13) One can then rewrite (G.5.10,11,12) as (a) = Rz(-φ) (b) = Rx(-θ) Rz(-φ) // using (4) (c) = Rz(-ψ)Rx(-θ)Rz(-φ) // using (3) (d) = Rz(φ)Rx(θ)Rz(ψ) // using (5) (e) = Rx(θ)Rz(ψ) // using (6) (f) = Rz(ψ) // using (7) (G.5.14) Using these equations, one has explicit formulas for writing any of the nine basis vectors either as a linear combination of ,, or as a linear combination of ',',' . By inspection one can rewrite the above six equations in the form shown on the right side of the Basis Theorem (G.5.9), = [Q]-1 e'n = Qen (G.5.9) Therefore, (a) (,,) = Rz(φ) (,,) (b) (',',') = Rz(φ)Rx(θ) (,,) (c) (',',') = Rz(φ)Rx(θ)Rz(ψ) (,,) (d) (,,) = Rz(-ψ)Rx(-θ)Rz(-φ) (',',') (e) (,,) = Rz(-ψ)Rx(-θ) (',',') (f) (',',') = Rz(-ψ) (',',') . (G.5.15) We shall now prove the three facts quoted above as (G.5.13) (1), (2) and (3), and then we shall resume our discussion of the Euler Angles . _____________________________________________________________________________________ Theorem 3 : (G.5.16) (1) Rξ(θ) = Rz(φ) Rx(θ) Rz(-φ) (2) Rζ'(ψ) = Rz(φ) Rx(θ)Rz(ψ)Rx(-θ)Rz(-φ) (3) Rζ'(ψ) Rξ(θ) Rz(φ) = Rz(φ)Rx(θ)Rz(ψ) = the Euler angle rotation R-1 of (G.5.8) Recall from (G.1.1) that a rotation of α about axis may be written R(α) = exp(-i α n J) . Proof of (1) : Recall Theorem 2 of (G.4.7) which says, R exp(-iθnJ) R-1 = exp(-iθn'J) where n' = Rn . (G.4.7) Note from line 1 of (G.5.6) that = Rz(φ) . We take n' = , R = Rz(φ), n = to get Rz(φ) exp(-iθJ) Rz(-φ) = exp(-iθJ) or Rξ(θ) = exp(-iθJ) = Rz(φ) Rx(θ) Rz(-φ) . QED (1) and we have thus proved item (1). One can see intuitively how this works, as in our example of (G.4.14). Instead of rotating θ about the axes, we first back-rotate around by -φ, use the aligned Rx(θ) to create a tilted disk in the top drawing of Fig (G.5.5), then forward rotate that result by Rz(φ) to get the tilted disk in the left picture. The good news is that we don't have to rely on such visualizations to get the result right. Proof of (2) : Recall again Theorem 2 of (G.4.7) which says (now with dummy argument θ → ψ) R exp(-iψnJ) R-1 = exp(-iψn'J) where n' = Rn . (G.4.7) Note from lines 2,1 of (G.5.6) that that ' = Rξ(θ) = Rξ(θ). We take n' = ', R = Rξ(θ), n = to get Rξ(θ)exp(-iψJ)Rξ(-θ) = exp(-iψ' J) = Rζ'(ψ) . Therefore Rζ'(ψ) = Rξ(θ)Rz(ψ)Rξ(-θ) . Then installing result (1) twice we get Rζ'(ψ) = [Rz(φ) Rx(θ) Rz(-φ)] Rz(ψ) [Rz(φ) Rx(-θ) Rz(-φ)] = Rz(φ) Rx(θ) Rz(ψ) Rx(-θ) Rz(-φ) QED (2) and we have thus proved item (2). Reader Exercise: Interpret this result in terms of back-rotations and Fig (G.5.5). Proof of (3) : (2) (1) Rζ'(ψ) Rξ(θ) Rz(φ) = [Rz(φ) Rx(θ)Rz(ψ)Rx(-θ)Rz(-φ)] [Rz(φ) Rx(θ) Rz(-φ)] Rz(φ) = Rz(φ) Rx(θ) Rz(ψ) Rx(-θ) Rz(-φ) Rz(φ) Rx(θ) Rz(-φ) Rz(φ) = Rz(φ) Rx(θ) Rz(ψ) Rx(-θ) Rx(θ) = Rz(φ) Rx(θ) Rz(ψ) . QED (3) ____________________________________________________________________________________ Resuming the Euler angle discussion, from (G.5.8) and Theorem 3 (3) we know that R-1 = Rζ'(ψ) Rξ(θ) Rz(φ) = Rz(φ) Rx(θ) Rz(ψ) so R = Rz(-ψ) Rx(-θ) Rz(-φ) . (G.5.17a) On Goldstein p 109 (GPS p 153) this last equation R = Rz(-ψ)Rx(-θ)Rz(-φ) appears as A = BCD which is written out in detail in (4-46) (GPS 4.46) , and which Maple verifies, (G.5.17b) The transpose R-1 = RT then appears in Goldstein (4-47) (GPS 4.47) . Interpretation of the Goldstein's Triple Concatenation In our Section 1 formalism, we discuss the idea of three concatenated transformations near (1.1.41). We can compare the equations there to those of (G.5.6), e"'n = U-1e''n e"n = S-1e'n e'n = R-1en (G.5.18a) (',',') = Rζ'(ψ) (',',') (',',') = Rξ(θ) (,,) (,,) = Rz(φ) (,,) . It follows that the three "back-rotations" are U-1 = Rζ'(ψ) S-1 = Rξ(θ) R-1 = Rz(φ) . (G.5.18b) In Fig (1.3.4) we show an example where R-1 = Rz(-α) = "back-rotation" and we draw the figure for some small α > 0. In Goldstein's back rotations, the role of α is played by -ψ ,-θ and -φ. Figure (G.5.5) shows that the basis vector "back rotations" are really forward rotations by ψ, θ and φ. Doing the triple concatenation one gets, e"'n = U-1S-1R-1en = Rζ'(ψ)Rξ(θ) Rz(φ)en or (',',') = Rζ'(ψ) Rξ(θ) Rz(φ) (,,) (G.5.19) in agreement with (G.5.7). Transformation of Kinematic Vectors The corresponding Passive View transformations of Kinematic Vectors for the three concatenations are, (V)''' = UV" (V)" = SV' (V)' = RV . (G.5.20) Doing the concatenation and then changing the Section 1 triple-prime to Goldstein's single-prime, we get (V)' = USR V = Rζ'(-ψ)Rξ(-θ)Rz(-φ) V . (G.5.21) If we now redefine R to be our notation for Goldstein's overall transformation, we get (',',') = R-1(,,) R-1 = Rζ'(ψ) Rξ(θ) Rz(φ) = Rζ'(ψ) Rξ(θ) Rz(φ) (,,) (V)' = R V R = [Rζ'(ψ) Rξ(θ) Rz(φ)]-1 = [Rζ'(ψ) Rξ(θ) Rz(φ)]-1 V . (G.5.22) We can rewrite the above lines making use of Theorem 3 item (3) to get (',',') = R-1(,,) R-1 = Rz(φ)Rx(θ)Rz(ψ) = Rz(φ)Rx(θ)Rz(ψ) (,,) (V)' = R V R = [ Rz(φ)Rx(θ)Rz(ψ)]-1 = [ Rz(φ)Rx(θ)Rz(ψ)]-1 V = Rz(-ψ)Rx(-θ)Rz(-φ) V . (G.5.23) This last item is the rule for finding the Frame S' components (V)'i of a vector V in terms of its Frame S components Vi . Figure (G.5.5) shows the Kinematic Vector V = r which is the position of some point in Frame S . Exercise 1: Compute the Frame S' components of the vector r which in Frame S has components (x,y,z) . (r)' = R r = Rz(-ψ)Rx(-θ)Rz(-φ) r (G.5.24) (G.5.25) Therefore (r)' = = R and so x' = (cosψcosφ - sinψcosθsinφ) x + (cosψsinφ + sinψcosθcosφ) y + sinψsinθ z y' = (- sinψcosφ - cosψcosθsinφ) x + (-sinψsinφ + cosψcosθcosφ) y + cosψsinθ z z' = sinθsinφ x - sinθcosφ y + cosθ z . (G.5.26) Application: What are the Frame S' components of ? Apply the previous equation to (x,y,z) = (1,0,0): x' = cosψcosφ - sinψcosθsinφ = ()'1 y' = - sinψcosφ - cosψcosθsinφ = ()'2 z' = sinθsinφ = ()'3 (G.5.27) Inspect the basis vector in the lower right drawing of Fig (G.5.5). For the small Euler angles used in the figure, appears to have positive x' and z' components, but a negative y' component. This is confirmed by looking at (G.5.27). We now reverse Exercise 1 to get Exercise 2. Exercise 2: Compute the Frame S components of the vector r which in Frame S' has components (x',y',z') . r = R-1(r)' = Rz(φ)Rx(θ)Rz(ψ)(r)' . (G.5.28) The matrix R-1 = RT is just the transpose of the matrix shown above; Maple computes it anyway, (G.5.29) Therefore r = = R-1 and so x = (cosφcosψ-cosθsinφsinψ) x' + (-cosφsinψ-cosθsinφcosψ) y' +sinθsinφ z' y = (sinφcosψ + cosθcosφsinψ) x' + (-sinφsinψ+cosθcosφcosψ) y' - sinθcosφ z' z = sinθsinψ x' + sinθcosψ y' + cosθ z' . (G.5.30) Application: What are the Frame S components of ' ? Apply the previous equation to (x',y',z') = (1,0,0): 'x = cosψcosφ - sinψcosθsinφ = (')1 'y = sinφcosψ + cosθcosφsinψ = (')2 'z = sinθsinψ = (')3 (G.5.31) Inspect the basis vector ' in the lower right drawing of Fig (G.5.5). For the small Euler angles used in the figure, ' appears to have positive x,y and z components. This is confirmed in (G.5.31). Exercise 3: Express (,,) in terms of (',',') According to (G.5.14) (d) we know that = Rz(φ)Rx(θ)Rz(ψ) // = R-1 (G.5.14) (d) where Maple computes the matrix, . Therefore = (cosψcosφ - sinψsinφcosθ) ' + (-sinψcosφ - cosψsinφcosθ)' + (sinφsinθ) ' = (cosψsinφ + sinψcosφcosθ) ' + (-sinψsinφ + cosψcosφcosθ)' + (-cosφsinθ) ' = (sinψsinθ)' + (cosψsinθ)' + (cosθ) ' (G.5.32) And now we go the other direction: Exercise 4: Express (',',') in terms of (,,) Inverting (G.5.14) (d) quoted just above we find = Rz(-ψ)Rx(-θ)Rz(-φ) // = R (G.5.14) (d) inverse We can then use the Exercise 3 result with φ,θ,ψ → -ψ,-θ,-φ. But to avoid errors, we just use Maple again to get Therefore, ' = (cosψcosφ - sinψsinφcosθ) + (cosψsinφ + sinψcosφcosθ) + (sinθsinψ) ' = (-sinψcosφ - cosψsinφcosθ) + (-sinψsinφ + cosψcosφcosθ) + (sinθcosψ) ' = (sinφsinθ)' + (-cosφsinθ) + (cosθ) (G.5.33) Exercise 5: Compute the unit vectors , , . Looking at the polar coordinates equation (E.5.3), and then looking at Fig (G,.5.5), we can read off = -sinφ +cosφ = -sinψ '+cosψ ' = -sinθ +cosθ (G.5.34) We could then use results (G.5.32) or (G.5.33) to express these all in terms of (,,) or (',',') G.6 Euler angles which change in time : computation of ω (Method 1) Suppose now that all the Euler angles are changing in time. The combination of all these movements creates an overall ω angular rotation vector relating the relative motion of the two Frames (as in Fig 1). Looking at Fig (G.5.5) we see that ω will have three contributions, one from each Euler angle movement, ωφ = ωθ = ' ωψ = ' . (G.6.1) If we want to know these contributions in Frame S components, we have to replace ' and ' with their appropriate linear combinations of , and . From (G.5.14b) = Rx(-θ) Rz(-φ) (G.5.14b) Therefore ' = cosφ + sinφ . // this particular fact is obvious from Fig (G.5.5) top (G.6.2) From (G.5.33), ' = sinθsinφ - sinθcosφ + cosθ . (G.5.33) Inserting these last two results into (G.6.1) gives, ωφ = ωθ = cosφ + sinφ ωψ = sinθsinφ - sinθcosφ + cosθ . (G.6.3) We add up to get ω = ωφ + ωθ + ωψ = [ sinθsinφ + cosφ] + [- sinθcosφ + sinφ ] + [ cosθ + ] or (ω)x = sinθsinφ + cosφ (ω)y = - sinθcosφ + sinφ (ω)z = cosθ + . // Frame S (G.6.4) These then are the Frame S components of the ω vector. Conversely, suppose we want (as Goldstein does want) the components of ω in Frame S' components, Frame S' being the rotating frame in which a "rigid body" might lie. From (G.5.14e), = Rx(θ)Rz(ψ) (G.5.14e) Therefore ' = = cosψ ' - sinψ ' . (G.6.5) This fact can be verified by staring for a while at the lower right drawing in Fig (G.5.5). There we see that ' = Rz'(-ψ)' which implies the above. The author is prone to making errors staring at drawings and for this reason prefers the bulletproof Maple approach to computing things. From (G.5.32), = sinθsinψ ' + sinθcosψ ' + cosθ ' . (G.5.32) Inserting these last two results into (G.6.1) gives, ωφ = = sinθsinψ ' + sinθcosψ ' + cosθ ' ωθ = ' = cosψ ' - sinψ ' ωψ = ' . (G.6.6) We add up to get ω = ωφ + ωθ + ωψ = [ sinθsinψ + cosψ ] ' + [ sinθcosψ - sinψ] ' + [ cosθ + ] ' or (ω)'x = sinθsinψ + cosψ ≡ ωx' (ω)'y = sinθcosψ - sinψ ≡ ωy' (ω)'z = cosθ + ≡ ωz' // Frame S' (G.6.7) These then are the Frame S' components of the ω vector. This result is in agreement with Goldstein page 134 (GPS page 174), A quick verification of (G.6.4) From (G.5.19) we have R = Rz(-ψ)Rx(-θ)Rz(-φ) . (G.5.19) The rule for transformation of a kinematic vector is given by (1.3.3) as ω' = (ω)' = Rω (G.6.8) which from (1.3.10) we can write as (ω')i = (ω)'i = Rij(ωj) = [ Rz(-ψ)Rx(-θ)Rz(-φ) ]ij(ωj) . (G.6.9) Inverting, (ω)i = [ Rz(φ)Rx(θ)Rz(ψ)]ij (ω)'j . (G.6.10) We enter the (ω)'j components (G.6.7) into Maple and then compute the (ω)i using (G.6.10), (G.6.11) Transcribing the result gives, (ω)x = cosφ + sinθsinφ (ω)y = sinφ - sinθcosφ (ω)z = + cosθ (G.6.12) This agrees with (G.6.4) above, providing some verification for our Frame S components of ω. G.7 Computation of ω (Method 2) In this section our approach to computing ω for the Euler Angle rotation is to find an equation which involves ω and solve it for ω! More or less at random, we choose (1.7.4), (den/dt)S' = – ω x en . (1.7.4) (G.7.1) Unlike our Method 1 computation of ω in the previous section, here we shall have no need for Goldstein's geometric Figure (G.5.5) or intermediate angles like ξ and ζ. We will, however, need various algebraic results developed in Section G.3. This calculation is a bit slippery and requires care and precision in the use of notation -- it is easy to go astray. We shall use both matrix and Section 1.1 Dirac notations as seem convenient. A silver lining is that we shall be able to apply many of the results derived earlier in this document. Recall from (G.5.17a) that the Goldstein Euler angle rotation is given by R = Rz(-ψ)Rx(-θ)Rz(-φ) . (G.5.17a) (G.7.2) In order to avoid a hundred minus signs, we shall temporarily negate all three angles. Then when we are done, we will undo this negation. We therefore temporarily take R to be, R(ψ,θ,φ) = Rz(ψ)Rx(θ)Rz(φ) ≡ R(Φ) = exp(-i Φ J) . // temp (G.7.3) Comment: To find Φ we could write out the matrix Rz(ψ)Rx(θ)Rz(φ), and decompose it into its symmetric and antisymmetric components S and A. In theory we could then compute Φ (=θ) and (=) from (G.2.14) and come up with an explicit expression for Φ = Φ. The reader is just reminded that this is mechanically possible, but luckily we have no need for the result (which is quite complicated). We could also compute dΦ from the following R(ψ+dψ,θ+dθ,φ+dφ) = R(Φ+dΦ) = exp(-i [Φ+dΦ] J) but again there is no need to do this. Note that Φ and dΦ will generally not be in the same direction. We now set about constructing the left side of (G.7.1) starting with our fundamental equation from (1.1.29) which relates the Frame S and Frame S' basis vectors, en(Φ) = R(Φ)e'n or |en(Φ)> = |R(Φ)e'n> = R(Φ) |e'n> . (G.7.4) Since we have in mind that Φ = Φ(t), one sees that the Frame S basis vectors change in time as viewed from Frame S'. In contrast, the basis vectors e'n are static. The basis vectors |en(Φ)> are complete at time t so we can write, in analogy with (1.1.20), 1 = |en(Φ)><en(Φ)| completeness of the en at time t (a) 1 = |en(Φ+dΦ)><en(Φ+dΦ)| completeness of the en at time t+dt (b) 1 = |e'n><e'n| completeness of the e'n at any time (c) (G.7.5) In (b) the rotation vector has changed from Φ to some Φ+dΦ as time moved from t to d+dt. A key point is that the en basis vectors are complete at any point in time. The rotation operator R(Φ) similarly is real orthogonal at any time, analogous to (1.1.37), RT(Φ) = R-1(Φ) and similarly for matrices RT(Φ) = R-1(Φ) . (G.7.6) Now we close the Dirac equation in (G.7.4) on the left with <e'm| to get <e'm| en(Φ) > = <e'm|R(Φ) |e'n> = [R(Φ)]'mn . (G.7.7a) Since this is true for any Φ, one also has <e'm| en(Φ+dΦ) > = <e'm|R(Φ+dΦ) |e'n> = [R(Φ+dΦ)]'mn . (G.7.7.b) Recall from (1.1.35) that (R)'ij = Rij. Here we confirm that fact in the current fancier notation, [R(Φ)]mn = <em(Φ)| R(Φ) |en(Φ)> = <em(Φ) |e'i><e'i| R(Φ) |e'j><e'j|en(Φ)> = [R(Φ)]'im [R(Φ)]'ij[R(Φ)]'jn = [ RT(Φ)R(Φ)R(Φ)]'mn = [ R-1(Φ)R(Φ)R(Φ)]'mn = [R(Φ)]'mn . (G.7.8a) Since this is true for any Φ, one also has [R(Φ+dΦ)]mn = [R(Φ+dΦ)]'mn . (G.7.8b) DETERMINATION OF THE FRAME S' COMPONENTS OF ω First expression for: [(den/dt)S']'i We now examine (from Frame S') a small change in the basis vector en(Φ) , |(den(Φ))S'> ≡ |en(Φ+dΦ)> - |en(Φ)> = |e'i><e'i|en(Φ+dΦ)> - |e'i><e'i|en(Φ)> // completeness twice = |e'i> [R(Φ+dΦ)]'in - |e'i>[R(Φ)]'in // (G.7.7b,a) = ( [R(Φ+dΦ)]'in - [R(Φ)]'in ) |e'i> // reorder = ( [R(Φ+dΦ)]in - [R(Φ)]in ) |e'i> // (G.7.8b,a) to remove primes = ( R(Φ+dΦ) - R(Φ) )in |e'i> = ( RT(Φ+dΦ) - RT(Φ) )ni |e'i> . (G.7.9) The next step is to replace |e'i> as follows |e'i> = |ej(Φ)><ej(Φ) | e'i> = |ej(Φ)>[R(Φ)]'ij = [R(Φ)]ij |ej(Φ)> . (G.7.10) Then |(den(Φ))S'> = ( RT(Φ+dΦ) - RT(Φ) )ni[R(Φ)]ij |ej(Φ)> = ( RT(Φ+dΦ)R(Φ) - RT(Φ)R(Φ) )nj |ej(Φ)> = ( RT(Φ+dΦ)R(Φ) - 1 )nj |ej(Φ)> . (G.7.11) We then add dt/dt to the left side to get dt |(den(Φ)/dt )S'> = ( RT(Φ+dΦ)R(Φ) - 1 )nj |ej(Φ)> . (G.7.12) We wish to evaluate the above vector equation in Frame S' components. To do this, we close both sides with <e'i |, obtaining dt <e'i |(den(Φ)/dt )S'> = ( RT(Φ+dΦ)R(Φ) - 1 )nj <e'i |ej(Φ)> or dt [(den/dt)S']'i = ( RT(Φ+dΦ)R(Φ) - 1 )nj R(Φ)ij = ( RT(Φ+dΦ)R(Φ) - 1 )njRT(Φ)ji = [( RT(Φ+dΦ)R(Φ) - 1 )RT(Φ)]ni = [ RT(Φ+dΦ) - RT(Φ) ]ni = [ R(Φ+dΦ) - R(Φ) ]in ≡ (dR)in where dR ≡ R(Φ+dΦ) - R(Φ) . (G.7.13) Divide both sides by dt to obtain [(den/dt)S']'i = (dR/dt)in dR ≡ R(Φ+dΦ) - R(Φ) . (G.7.14) Second expression for: [(den/dt)S']'i Recall (G.7.1), (den/dt)S' = – ω x en . (1.7.4) (G.7.1) We evaluate the above vector equation in Frame S' components, [(den/dt)S']'i = - εikc(ω)'k(en)'c = - εikc(ω)'k [R(Φ)]'cn = -εikc(ω)'k [R(Φ)]cn . (G.7.15) Equate the two expressions for: [(den/dt)S']'i At this point we have shown that doing component evaluations of (den/dt)S' in Frame S' gives, [(den/dt)S']'i = (dR/dt)in dR ≡ R(Φ+dΦ) - R(Φ) (G.7.14) [(den/dt)S']'i = - εikc(ω)'k [R(Φ)]cn . (G.7.15) Setting the right sides equal, one obtains, - εikc(ω)'k [R(Φ)]cn = (dR/dt)in . Multiply both sides on the right by [R-1(Φ)]nj to get - εikc(ω)'k [R(Φ)]cn [R-1(Φ)]nj = (dR/dt)in [R-1(Φ)]nj or - εikc(ω)'kδcj = [ (dR/dt)R-1(Φ) ]ij or - εikj(ω)'k = [ (dR/dt)R-1(Φ) ]ij or εijk(ω)'k = Aij where A ≡ (dR/dt)R-1(Φ) . (G.7.16) It seems that the object Aij must be antisymmetric in its two indices, so the matrix Aij has only three significant elements. Setting ijk = 231 gives the first line below, then the next two lines follow from cyclic permutation : (ω)'1 = A23 (ω)'2 = A31 (ω)'3 = A12 . (G.7.17) Thus we have succeeded in solving for the components of ω in Frame S' . Recall that Frame S' is rotating at rate ω relative to Frame S as in Fig 1. It remains to compute the three Aij matrix elements so we can learn the specific expressions for the (ω)'i in terms of Euler Angles. Computation of A and Statement of Final Result Our task is to compute A ≡ (dR/dt)R-1(Φ) where our "temporary" R(Φ) is given by R(Φ) = Rz(ψ)Rx(θ)Rz(φ) (G.7.3) and where dR = R(Φ+dΦ) - R(Φ) . (G.7.13) The first step is to compute dR in terms of the Euler angles. We make use of the obvious fact that Ri(α +dα) = Ri(α)Ri(dα) and then the fact (1.5.6) that Ri(dα) ≈ 1 - idα Ji for small dα. Keeping only terms of first order in the differential angles, one finds dR = R(Φ+dΦ) - R(Φ) = Rz(ψ+dψ) Rx(θ+dθ) Rz(φ+dφ) - Rz(ψ) Rx(θ) Rz(φ) = 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)J3Rx(θ)Rz(φ) - idθ Rz(ψ) Rx(θ)J1Rz(φ) - idφ Rz(ψ) Rx(θ)Rz(φ)J3 (G.7.18) where the leading terms exactly cancel. Dividing by dt one then has (dR/dt) = -i Rz(ψ)J3Rx(θ)Rz(φ) - i Rz(ψ)Rx(θ)J1Rz(φ) - i Rz(ψ)Rx(θ)Rz(φ)J3 . (G.7.19) The next step is to compute A ≡ (dR/dt)R-1(Φ) using R-1(Φ) = Rz(-φ)Rx(-θ)Rz(-ψ). In doing so, we shall three times in blue use the fact that Ji commutes with Ri as formally stated in (G.3.6) : A ≡ (dR/dt) R-1(Φ) = [ -i Rz(ψ)J3Rx(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)J3Rz(-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) J1 Rz(-ψ) - i Rz(ψ) Rx(θ)J3Rx(-θ) Rz(-ψ) ] = [ -i J3 - i Rz(ψ) J1 Rz(-ψ) - i Rz(ψ) Rx(θ)J3Rx(-θ) Rz(-ψ) ] = [ -i J3 - i M1 - i M2 ] where (G.7.20) 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) . We now call upon our non-trivial sandwich formulas in (G.3.8) to simplify thing further : M1 = R3(ψ) J1R3(-ψ) = cosψJ1 + sinψ J2 // (G.3.8) line 5 M2 = R3(ψ) [ R1(θ)J3R1(-θ) ] R3(-ψ) = R3(ψ) [ cosθ J3 - sinθ J2 ] R3(-ψ) // (G.3.8) line 2 = cosθ [ R3(ψ) J3 R3(-ψ)] - sinθ [R3(ψ)J2R3(-ψ)] = cosθ J3 - sinθ (cosψJ2 - sinψ J1) // (G.3.6) and (G.3.8) line 6 = sinθsinψJ1 - sinθcosψJ2 + cosθ J3 . (G.7.21) Then, A = [ -i J3 - i M1 - i M2 ] = [ -i J3 - i (cosψJ1 + sinψ J2) - i (sinθsinψJ1 - sinθcosψJ2 + cosθ J3) ] = [ -i J3 - i cosψJ1 - i sinψ J2 - i sinθsinψJ1 + i sinθcosψJ2 - 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) ] . (G.7.22) We now take the ij element of this matrix using the fact (G.1.3) that (iJk)ij = kij , Aij = - [ ( cosψ + sinθsinψ)1ij + ( sinψ - sinθcosψ)2ij + ( + cosθ)3ij ] . (G.7.23) The tensor kij is antisymmetric in i↔j and therefore the entire matrix A is antisymmetric (as conjectured earlier) and thus has only three distinct matrix elements. They are: A23 = - ( cosψ + sinθsinψ)123 = - ( cosψ + sinθsinψ) A31 = - ( sinψ - sinθcosψ)231 = - ( sinψ - sinθcosψ) A12 = - ( + cosθ) 312 = - ( + cosθ) . (G.7.24) From (G.7.17) we then conclude that (ω)'1 = A23 = - ( cosψ + sinθsinψ) (ω)'2 = A31 = - ( sinψ - sinθcosψ) (ω)'3 = A12 = - ( + cosθ) . // angles still negated We now undo the temporary negation of the angles ψ,θ,φ enacted below (G.7.2). The velocities and sines then negate. Our final result for the Frame S' components of ω is then, (ω)'1 = sinθsinψ + cosψ ≡ ωx' (ω)'2 = sinθcosψ - sinψ ≡ ωy' (ω)'3 = cosθ + ≡ ωz' . (G.7.25) This is in agreement with our Method 1 calculation (G.6.7) and with Goldstein's result which we again quote from Goldstein page 134 (GPS page 174), DETERMINATION OF THE FRAME S COMPONENTS OF ω Here we shall review three different Plans for computing the Frame S components of ω. Plan A: Since we just computed the (ω)'i in Frame S', we just use ω = R-1(ω)' to get the Frame S components. This was done at the end of Section G.6 with the result stated in (G.6.12). Plan B: Start with (1.7.1) that (de'n/dt)S = ω x e'n in place of (G.7.1) that (den/dt)S' = – ω x en, which adds an overall minus sign to the result. The new (G.7.4) becomes e'n(Φ) = R-1(Φ)en with en = constant in Frame S. Things go through as presented above, but since R(Φ) → R-1(Φ) and since R(Φ) = Rz(ψ)Rx(θ)Rz(φ), one has R-1(Φ) = Rz(-φ)Rx(-θ) Rz(-ψ). Thus, to convert the result (G.7.25) to the Frame S result, we have to make these changes: (1) φ, ψ, θ → -ψ, -φ , -θ ; (2) add the overall minus sign just noted. We do that right here: φ, ψ, θ → -ψ, -φ , -θ (ω)'1 = sinθsinψ + cosψ ≡ ωx' (ω)'2 = sinθcosψ - sinψ ≡ ωy' (ω)'3 = cosθ + ≡ ωz' . (G.7.25) → -(ω)1 = [-][-sinθ][-sinφ] + [-] cosφ ω1 = sinθsinφ +cosφ -(ω)2 = [-][-sinθ]cosφ - [-][-sinφ] ω2 = - sinθcosφ + sinφ -(ω)3 = [-] cosθ + [-] ω3 = cosθ + (G.7.26) and this agrees with (G.6.12). Plan C: Compute the ωi directly using the machinery developed earlier in this section, but take Frame S components instead of Frame S' components. We regard this as a "stress test" of the machinery. First expression for: [(den/dt)S']i Start with (G.7.11) |(den(Φ))S'> = ( RT(Φ+dΦ)R(Φ) - 1 )nj |ej(Φ)> . (G.7.11) Instead of closing on the left with <e'i | to get Frame S' components, this time close on the left with <ei | to get Frames S components, [(den)S']i = <ei |(den(Φ))S'> = ( RT(Φ+dΦ)R(Φ) - 1 )nj <ei(Φ) |ej(Φ)> = ( RT(Φ+dΦ)R(Φ) - 1 )nj δji // (1.1.14) = ( RT(Φ+dΦ)R(Φ) - 1 )nj (R-1(Φ)R(Φ))ji = [( RT(Φ+dΦ)R(Φ) - 1 )R-1(Φ)R(Φ)]ni = [( RT(Φ+dΦ) - RT(Φ) ) R(Φ)]ni = [RT(Φ)( R(Φ+dΦ) - R(Φ) )]in = [RT(Φ)(dR)]in = [RT(Φ)(dR) R-1(Φ)R(Φ)]in and dividing by dt, [(den/dt)S']i = [RT(Φ)(dR/dt)R-1(Φ) R(Φ)]in = [RT(Φ)A R(Φ)]in (G.7.27) where recall that A ≡ (dR/dt)R-1(Φ) from (G.7.16). Second expression for: [(den/dt)S']i Recall (G.7.1), (den/dt)S' = – ω x en . (1.7.4) (G.7.1) We evaluate the above vector equation in Frame S components, [(den/dt)S']i = - εikcωk(en)c = - εikcωkδnc = - εiknωk . (G.7.28) Equate the two expressions for: [(den/dt)S']i At this point we have shown that doing component evaluations of (den/dt)S' in Frame S gives, [(den/dt)S']i = [RT(Φ)A R(Φ)]in . (G.7.27) [(den/dt)S']i = - εiknωk . (G.7.28) Setting the right sides equal, we obtain - εiknωk = [RT(Φ)A R(Φ)]in ≡ Bin or εinkωk = Bin or εijkωk = Bij B = RT(Φ)A R(Φ) , (G.7.29) Thus we arrive at (G.7.16) with (ω)'i replaced by ωi and with A replaced by B, ω1 = B23 ω2 = B31 ω3 = B12 . (G.7.30) It remains only to compute the Bij. Computation of B and Statement of Final Result Recall from (G.7.22) that A = - [ ( cosψ + sinθsinψ)(iJ1) + ( sinψ - sinθcosψ)(iJ2)+ ( + cosθ)(iJ3) ] . (G.7.22) We may write this as A = aiJi where (G.7.31) a1 = - i( cosψ + sinθsinψ) a2 = - i( sinψ - sinθcosψ) a3 = - i( + cosθ) . Entering into Maple, . (G.7.32) Then the matrix B in (G.7.29) can be written, B = R-1(Φ)A R(Φ) = R-1(Φ)[aiJi] R(Φ) = ai [R-1(Φ)JiR(Φ)[ = aiR(Φ)ijJj // Theorem 1 of (G.4.1) with R→R-1 = aiQi where Qi ≡ R(Φ)ijJj . (G.7.33) Using our "temporary" R(Φ) in (G.7.3) we compute the vector Q as follows (a vector of matrices), (G.7.34) The quantity B = aiQi is then, (G.7.35) We move the factors of i (Maple I) next to the Jk and transcribe the last line above: B = (- cosφ - sinθsinφ)(iJ1) + (sinφ - sinθcosφ)(iJ2) + (- cosθ - )(iJ3) (G.7.36) Using (G.1.3) that (iJk)ij = kij the matrix elements of B are then, Bij = (- cosφ - sinθsinφ)ε1ij + (sinφ - sinθcosφ)ε2ij + (-cosθ - )ε3ij . (G.7.37) Therefore, B23 = - cosφ - sinθsinφ B31 = sinφ - sinθcosφ B12 = -cosθ - . (G.7.38) Then from (G.7.30), ω1 = B23 = - cosφ - sinθsinφ ω2 = B31 = sinφ - sinθcosφ ω3 = B12 = -cosθ - . // angles still negated We now undo the temporary negation of the Euler angles ψ,θ,φ enacted below (G.7.2). The velocities and sines then negate. Our final result for the Frame S components of ω is then, ω1 = B23 = cosφ + sinθsinφ ω2 = B31 = sinφ - sinθcosφ ω3 = B12 = cosθ + (G.7.39) This is in agreement which (G.6.11) which we now quote, (ω)x = cosφ + sinθsinφ (ω)y = sinφ - sinθcosφ (ω)z = + cosθ (G.6.11) Review of this Section Frame S and Frame S' are related by some angular velocity ω as shown in Fig 1. The actual rotation relating the two frames is R = Rz(-ψ)Rx(-θ)Rz(-φ) as in (G.7.2) and (G.5.17a). If the three Euler angles are all static, ω would be 0. It is the fact that one or more of these Euler angles varies in time which causes ω to be non-zero. Our task was to solve the following equation for ω (den)S' = – ω x en dt . (1.7.4) (G.7.1) Acting as an Observer in Frame S', we studied the change in the basis vectors (den)S' caused by the time-varying Euler angles. In (G.7.11) we obtained the following vector equation for this change, (den)S' = [ RT(Φ+dΦ)R(Φ) -1 ]nj ej . (G.7.11) Equating the right sides of the above two equations, we obtained the following vector equation, – ω x en dt = ( RT(Φ+dΦ)R(Φ) -1 )nj ej . (G.7.40) We found that, to lowest order, the quantity (RT(Φ+dΦ)R(Φ) -1 )nj was linear in the differential angles dψ,dθ and dφ and when both sides were divided by dt, these became rates , and . Although we were observing things from Frame S', we were allowed to take components of any vector equation in either Frame S or in Frame S'. By taking the Frame S' components of (G.7.40) we obtained the Frame S' components of ω as in (G.7.25). By taking the Frame S components of (G.7.40) we obtained the Frame S components of ω as in (G.7.39). Along the way we got to exercise some earlier results of this Appendix: the sandwich formulas (G.3.6) and (G.3.8) in the computation of A, Theorem 1 of (G.4.1) saying RJiR-1= R–1ijJj in the computation of B, and the rotation generator matrix representations of (G.1.3). Finally, we were able to exercise the Dirac notation of Section 1.1. The approach of this section made use of the "linear combination" side of the Basis Theorem (1.1.29) rather than the "operator" side, so matrices appeared right from the get-go. G.9 Generalizations of the Rotation Group Here we consider some generalizations of the ideas presented above in Section G.1 N dimensional representations of the rotation group The generators Ji shown in (G.1.2) are a special case of a more general idea which starts with the commutation relation [Ji, Jj] = iεijkJk . // for example, [J1,J2] = iJ3 (G.1.4) (G.9.1) One first thinks of the Ji as abstract "operators" in some abstract "operator space". One can show that it is possible to find a set of three NxN matrices of any integer dimension N which satisfy (G.1.4). The matrices are not unique, so (G.1.2) for the Ji in three dimensions is not unique, but it is a standard form. The three NxN generator matrices Ji are said to form an N-dimensional "irreducible representation" of the abstract generators Ji. One can always create new viable generator matrices by taking a "direct sum" of existing viable generator matrices, such as in this block-diagonal-form picture (G.9.2) This generator matrix is "reducible" into a direct sum of S, T and R. An "irreducible" representation is one that cannot be reduced in this manner. The same comment applies to rotation matrices. The commutation relation (G.9.1) is an example of a Lie Algebra. Our particular Lie Algebra is called so(3), so we can represent the algebra elements Ji of this algebra by three NxN matrices Ji. It is possible to write down a formula analogous to (G.1.3) (iJa)bc = abc which works for any N, but (G.1.3) itself only applies to N = 3. This is so because εabc has no meaning for N ≠ 3. But it always has meaning in (G.9.1) because there are only three generators regardless of the value of N. For general N, the object Ri(θ) = exp(-iθJi) is an NxN matrix which represents the action of a rotation of an N-vector in a Euclidean space EN. The set of such rotation matrices forms an "irreducible representation" of the rotation group SO(3) in N dimensions. The integer N is usually written N = 2j+1 where j = 0, 1/2, 1, 3/2 ... and this j then serves as a label for a given matrix representation. The value of j in N = 2j+1 is associated with "angular momentum" or "spin". In the case N=2 (having j=1/2) the generator matrices are the 2x2 "Pauli matrices". In this case the 2x2 matrices exp(-iθJi) describe the rotations of spin-1/2 particles such as electrons or protons. Here are the details for N=2 : Jx = (1/2) Jy = (1/2) Jz = (1/2) Rx = Ry = Rz = (G.9.3) The "vectors" for spin-1/2 particles have two components . The special case is called "spin up" and is "spin down". For the Lie Algebra so(3) one can show that J2 ≡ J12 + J22 + J32 and any particular Ji commute with each other, so [J2,Ji] = 0. J2 is called a Casimir operator of this Algebra, and fancier Lie Algebras can have several such Casimirs. A Differential operator representation of the rotation group It is also possible to "represent" the three rotation generators Ji by three differential operators in spherical coordinates θ and φ. These operators satisfy [Ji, Jj] = iεijkJk and in this context they are usually called Li but we shall stick with Ji. In this case, one can compute the differential operator J2 and one can ponder differential equations which take the form J2 fjm(θ,φ) = j(j+1) fjm(θ,φ) and Jz fjm(θ,φ) = m fjm(θ,φ). [The facts that the eigenvalue of J2 is j(j+1) and not j2, and that m runs from -j to j, derive from the structure of the Lie Algebra.] The solutions fjm(θ,φ) are able to have well-defined eigenvalues j(j+1) and m because [J2,J3] = 0. If we instead had [J2,J3] ≠ 0, then [J2,J3] fjm(θ,φ) = j(j+1)m - mj(j+1) = 0 is a contradiction and the two eigenfunction equations could not exist. The solutions fjm(θ,φ) of these equations are called the spherical harmonics and are usually written Yjm(θ,φ). Just for the record, here is what the differential operators look like, where C = cos and S = sin (for example, Sφ = sinφ and ∂φ = ∂/∂φ) : J1 = i [ Sφ ∂θ + cotθ Cφ ∂φ] J = eiφ [ ∂θ + i cotθ ∂φ ] J2 = i [ Cφ ∂θ cotθ Sφ ∂φ] J2 = [ ∂θ2 + cotθ ∂θ + (1/Sθ)2 ∂φ2] J3 = -i ∂φ J2 = [ (1/S) ∂θ [ Sθ ∂θ ] + (1/Sθ)2 ∂φ2 ] (G.9.4) Reader Exercise: Verify that the three operators on the left satisfy the Lie Algebra [Ji, Jj] = iεijkJk. For the hydrogen atom with a spinless electron, there are three mutually commuting quantities H, J2 and J3 where H is the Hamiltonian. This means that the solution eigenfunctions can have well defined E, j and m values and these eigenfunctions are those painful "orbitals" appearing in chemistry books. When the Hamiltonian commutes with some other operator like J2, that operator is called a "symmetry". Solution functions then bear a label for each such symmetry, such as j for J2. The Lie Algebra so(3) is isomorphic (one-to-one related) to another Lie Algebra called su(2). The Lie Group SO(3) is isomorphic to another Lie Group called SU(2). In the above we discuss only the Lie Group SO(3) with its three generators Ji. There are many other Lie Groups which have physics applications. Some Other Lie Groups of Interest The group SO(n) is the orthogonal group in n dimensions and it has n(n-1)/2 generators. The group SO(3,1) is the Lorentz Group which has 6 generators Ji and Ki which generate 3 rotations and 3 "boosts" (velocity transformations). The Lie Algebra is this, [ Ji, Jj] = +i εijkJk [ Ji, Kj] = +i εijk Kk [ Ki, Kj] = -i εijkJk (G.9.5) There are two Casimirs : J2 - K2 and JK . In the Lorentz Group "vector representation" known as 1/21/2 the generators are represented as 4x4 matrices. When these are exponentiated, one obtains the finite rotation and boost matrices used in special relativity. For example, with space-time vectors ordered xμ = (ct,x,y,z) one has = exp(-irJ1) where (J1)μν = // rotation Rx(r) (G.9.6) = exp(-ibK1) where (K1)μν = // boost Bx(b) The Poincare Group is basically the Lorentz group SO(3,1) bolted onto the group T(4) of translations in four directions ct,x,y,z. It thus has 10 generators Ji, Ki and Pμ where these last four are momentum and energy. The Poincare algebra has two Casimir operators whose eigenvalues are associated with mass and spin. For example, the mass Casimir is PμPμ . The irreducible representations of the Poincare Group are associated with "elementary particles" which have well defined mass and spin. The group SU(3) has 8 generators and 2 Casimirs. In one key representation the generators are represented by 3x3 matrices which act on 3-vectors. Instead of having up and down states as with spin-1/2 noted above, these vectors have up, down and strange states called u,d.s which are associated with quarks. The full symmetry group for the Standard Model of elementary particles is SU(3) x SU(2) x U(1) in which SU(3) plays its part. Special Groups have Traceless Generators It is desirable that "rotation matrices" have unit determinant because such matrices then do not change the "length" of a vector on which they act. When the representation matrices are restricted to have unit determinant, they are called "special" and the group name is prefixed by the letter S, as in SO(3) for the rotation group. For the 3x3 representation of the rotation group we know that the matrices are real orthogonal which means RRT = 1 which in turn means [det(R)]2 =1 and in SO(3) we select only those R with det(R) = +1. A "rotation" which just negates z (reflection) still satisfies RRT = 1 but has det(R) = -1. An elegant theorem concerning exponentiated square matrices is this (proved below): det(eA) = etr(A) (G.9.7) where det is the determinant and tr(A) ≡ ΣiAii is the "trace" or "spur" of the matrix A -- the sum of the diagonal elements. If we want the matrix exp(-i θ J) to have unit determinant so it is "Special", the exponent must be traceless, and in this case that means that the generators Ji must all be traceless. One can see from examples (G.1.2) and (G.9.3) and (G.9.6) that this is indeed the case. Offhand, the matrix identity det(eA) = etr(A) seems very unlikely and almost too simple. For that reason, we include here a straightforward proof which we feel is "one for the Book". Proof of (G.9.7) There are no implied sums in this proof! Write A = ΣijAij s(ij) where [s(ij)]ab ≡ δiaδjb . To verify, Aab = ΣijAij [s(ij)]ab = ΣijAijδiaδjb = Aab . The matrix s(ij) is all zeros except for a single 1 located in row a and column b. Using ex+y.. = exey ... we can write eA = exp( ΣijAij s(ij)) = Πi,j exp(Aij s(ij)) . Then using det(XY..) = det(X)det(Y).... , det(eA) = det {Πi,j exp(Aij s(ij)) } = Πi,j det [ exp(Aij s(ij)) ] . (G.9.8) [Case i ≠j :] We first note that [s(ij)]2 = 0 : [s(ij)]2ac = Σb[s(ij)]ab [s(ij)]bc = Σbδiaδjb δibδjc = δiaδjiδjc = 0 since i ≠ j Then [s(ij)]n = 0 for n ≥ 2. In this case we have exp(Aij s(ij)) = Σn=0∞ (Aij)n [ s(ij)]n/n! = 1 + Aij [ s(ij)] and so det [exp(Aij s(ij)) ] = det [ 1 + Aij s(ij)] = 1 . This is so because the matrix in question has all 1's on the diagonal and one non-vanishing off-diagonal element at (i,j). In fact, any triangular matrix (one side all zeros) with 1's on the diagonal has det = 1. [Case i=j : ] In this case we have [s(ii)]2 = [s(ii)] : [s(ii)]2ac = Σb[s(ii)]ab [s(ii)]bc = Σbδiaδib δibδic = δiaδiiδic = δiaδic = [s(ii)]ac . Note that [s(ii)]ab = δiaδib = 1 only when a = b = i, so s(ii) is an all-zero matrix with a single 1 at location i on the diagonal. Since [s(ii)]2 = s(ii) it follows that [s(ii)]n = s(ii) for n ≥ 1. Then, exp(Aii s(ii)) = Σn=0∞ (Aii)n [ s(ii)]n / n! = 1 + s(ii) [Σn=1∞(Aii)n / n!] = 1 + s(ii) [ -1 + Σn=0∞(Aii)n/n! ] = 1 + s(ii)[-1 + exp(Aii) ] . This last item is the unit matrix with the ith diagonal 1 replaced by exp(Aii) . Therefore, det [exp(Aii s(ii)) ] = det { 1 + s(ii)[-1 + exp(Aii)] } = exp(Aii) . Now go back to (G.9.8), det(eA) = Πi,j det [exp(Aij s(ij)) ] = (Πi≠j det [exp(Aij s(ij)) ] ) * (Πi=j det [exp(Aij s(ij)) ] ) = (1*1*1*1...... ....*1*1) * ( exp(A11)exp(A22) ..... ) = 1 * exp(A11 + A22 + ...) = exp (tr(A)) = etr(A) . QED