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Orthogonal Transformations

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Orthogonal Transformations Orthogonal transformations October 12, 2014 1 Defining property The squared length of a vector is given by taking the dot product of a vector with itself, v2 = v · v i j = gijv v An orthogonal transformation is a linear transformation of a vector space that preserves lengths of vectors. This defining property may therefore be written as a linear transformation, v0 = Ov such that v0 · v0 = v · v Write this definition in terms of components using index notation. The transformation O must have mixed indices so that the sum is over one index of each type and the free indices match, 0i i j v = O jv we have 0i 0j i j gijv v = gijv v i k j m i j gij O kv O mv = gijv v i j k m k m gijO kO m v v = gkmv v or, bringing both terms to the same side, i j k m gijO kO m − gkm v v = 0 Because vk is arbitrary, we can easily make six independent choices so that the resulting six matrices vkvm span the space of all symmetric matrices. The only way for all six of these to vanish when contracted on i j gijO kO m − gkm is if i j gijO kO m − gkm = 0 This is the defining property of an orthogonal transformation. In Cartesian coordinates, gij = δij, and this condition may be written as OtO = 1 where Ot is the transpose of O. This is equivalent to Ot = O−1. 1 Notice that nothing in the preceeding arguments depends on the dimension of v being three. We conclude that orthogonal transformations in any dimension n must satisfy OtO = 1, where O is an n × n matrix. Consider the determinant of the defining condition, det 1 = det OtO 1 = det Ot det O and since the determinant of the transpose of a matrix equals the determinant of the original matrix, we have det O = ±1. Any orthogonal transformation with det O = −1 is just the parity operator, P = 0 −1 1 @ −1 A times an orthogonal transformation with det O = +1, so if we treat parity independently −1 we have the special orthogonal group, SO (n), of unit determinant orthogonal transformations. 1.1 More detail If the argument above is not already clear, it is easy to see what is happening in 2-dimensions. We have v1v1 v1v2 vkvm = v2v1 v2v2 i j k m i which is a symmetric matrix. But gijO kO m − gkm v v = 0 must hold for all choices of v . If we make the choice vi = (1; 0) then 1 0 vkvm = 0 0 0 0 and g Oi Oj − g vkvm = 0 implies g Oi Oj = g . Now choose vi = (0; 1) so that vkvm = ij k m km ij 1 1 11 0 1 1 1 and we see we must also have g Oi Oj = g . Finally, let vi = (1; 1) so that vkvm = . This gives ij 2 2 22 1 1 i j i j i j i j gijO 1O 1 − g11 + gijO 1O 2 − g12 + gijO 2O 1 − g21 + gijO 2O 2 − g22 = 0 but by the previous two relations, the first and fourth term already vanish and we have i j i j gijO 1O 2 − g12 + gijO 2O 1 − g21 = 0 Because gij = gji, these two terms are the same, and we conclude i j gijO kO m = gkm for all km. In 3-dimensions, six different choices for vi will be required. 2 Infinitesimal generators We work in Cartesian coordinates, where we can use matrix notation. Then, for the real, 3-dimensional representation of rotations, we require Ot = O−1 Notice that the identity satisfies this condition, so we may consider linear transformations near the identity which also satisfy the condition. Let O = 1 + " 2 where ["]ij = "ij are all small, j"ijj 1 for all i; j. Keeping only terms to first order in "ij, we have: Ot = 1 + "t O−1 = 1 − " where we see that we have the right form for O−1 by computing OO−1 = (1 + ") (1 − ") = 1 − "2 ≈ 1 correct to first order in ". Now we impose our condition, Ot = O−1 1 + "t = 1 − " "t = −" so that the matrix " must be antisymmetric. Next, we write the most general antisymmetric 3×3 matrix as a linear combination of a convenient basis, i " = w Ji 0 0 0 0 1 0 0 0 −1 1 0 0 1 0 1 1 2 3 = w @ 0 0 1 A + w @ 0 0 0 A + w @ −1 0 0 A 0 −1 0 1 0 0 0 0 0 0 0 w2 −w3 1 = @ −w2 0 w1 A w3 −w1 0 i where w 1. Notice that the components of the three matrices Ji are neatly summarized by [Ji]jk = "ijk 0 0 0 0 1 where "ijk is the totally antisymmetric Levi-Civita tensor. For example, [J1]ij = "1ij = @ 0 0 1 A. 0 −1 0 The matrices Ji are called the generators of the transformations. The most general antisymmetric is then a linear combination of the three generators. Knowing the generators is enough to recover an arbitrary rotation. Starting with O = 1 + " we may apply O repeatedly, taking the limit O (θ) = lim On n−!1 = lim (1 + ")n n−!1 i n = lim 1 + w Ji n−!1 Let w be the length of the infinitesmal vector wi, so that wi = wni, where ni is a unit vector. Then the limit is taken in such a way that lim nw = θ n−!1 3 n Pn n! n−k k where θ is finite. Using the binomial expansion, (a + b) = k=0 k!(n−k)! a b we have n i n lim O = lim 1 + w Ji n−!1 n−!1 n X n! n−k i k = lim (1) w Ji n−!1 k!(n − k)! k=0 n X n (n − 1) (n − 2) ··· (n − k + 1) 1 i k = lim (nw) n Ji n−!1 k! nk k=0 n n n−1 n−2 n−k+1 X n n n ··· n i k = lim θn Ji n−!1 k! k=0 1 X 1 k = θniJ k! i k=0 i ≡ exp θn Ji We define the exponential of a matrix by the power series for the exponential, applied using powers of the matrix. This is the matrix for a rotation through an angle θ around an axis in the direction of n. i To find the detailed form of a general rotation, we now need to find powers of n Ji. This turns out to be straightforward: i j i j n Ji k = n "i k m h i 2i i m j k n Ji = n "i k n "j n n i j m k = n n "i k"j n i j m k = −n n "i k"jn i j m m = −n n δijδ n − δinδ j i j m i m j = − δijn n δ n + δinn δ j n m m = −δ n + n nn m m = − (δ n − n nn) m h i 3i m m i k n Ji = − (δ k − n nk) n "i n n m i k m i k = −δ k n "i n + n nkn "i n i m m k i = −n "i n + n n n "ikn i m = − n Ji n m k i m i where n n n "ikn = n (n × n) = 0. The powers come back to n Ji with only a sign change, so we can divide the series into even and odd powers. For all k > 0, m h i 2ki k m m n Ji = (−1) (δ n − n nn) n m h i 2k+1i k i m n Ji = (−1) n Ji n n m h i 0i m For k = 0 we have the identity, n Ji = δ n. n We can now compute the exponential explicitly: m ^ m i [O (θ; n)] n = exp θn Ji n 4 " 1 #m X 1 k = θniJ k! i k=0 n " 1 #m " 1 #m X 1 2l X 1 2l+1 = θniJ + θniJ (2l)! i (2l + 1)! i l=0 n l=0 n " 1 #m " 1 #m X 1 2l X 1 2l+1 = 1 + θniJ + θniJ (2l)! i (2l + 1)! i l=1 n l=0 n 1 l 1 l X (−1) X (−1) m = δm + θ2l (δm − nmn ) + θ2l+1 niJ n (2l)! n n (2l + 1)! i n l=1 l=0 m m m i m = δ n + (cos θ − 1) (δ n − n nn) + sin θn "i n where we get (cos θ − 1) because the l = 0 term is missing from the sum. To see what this means, let O act on an arbitrary vector v, and write the result in normal vector notation, m n m n m m n i m n [O (θ; n^)] n v = δ nv + (cos θ − 1) (δ n − n nn) v + sin θn "i nv m m n m n i n m = v + (cos θ − 1) (δ nv − n nnv ) − sin θn v "in = vm + (cos θ − 1) (vm − (n · v) nm) − [n × v]m sin θ Going fully to vector notation, O (θ; n^) v = v + (cos θ − 1) (v − (n · v) n) − (n × v) sin θ Finally, define the components of v parallel and perpendicular to the unit vector n: vk = (v · n) n v? = v − (v · n) n Therefore, O (θ; n^) v = v − (v − (v · n) n) + (v − (v · n) n) cos θ − sin θ (n × v) = vk + v? cos θ − sin θ (n × v) This expresses the rotated vector in terms of three mutually perpendicular vectors, vk; v?; (n × v).
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