My Math Forum Rotation matrices and exponential map for angular momentum

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 March 1st, 2013, 01:35 PM #1 Newbie   Joined: Mar 2013 Posts: 1 Thanks: 0 Rotation matrices and exponential map for angular momentum Hi guys, I'm having some trouble with rotation matrices. Basically, this is the problem: Let R(na) be a matrix in SO(3), where na specifies the rotation the matrix represents. Namlely, n is the unit vector about which the rotation is performed, and a is the angle of the rotation. Let {J1,J2,J3} be the standard basis of the Lie algebra so(3) (which is the set of 3x3 skew-symmetric matrices). Then prove R(na) = exp[a * ? n_i J_i]. I know that exp[a_1 J_1] is the rotation of a_1 degrees about the x-axis, and similarly for exp[a_2 J_2] and exp[a_3 J_3]. I also know that the exponent map from so(3) to SO(3) is surjective. This gives that any R(na) can be written as R(na)= exp[X(na)] for some X(na) in so(3). I know I'm also supposed to use the fact that the Ji form a basis for so(3), but I can't exactly prove the problem above. Can anyone help?
 March 1st, 2013, 08:41 PM #2 Senior Member   Joined: Feb 2013 Posts: 281 Thanks: 0 Re: Rotation matrices and exponential map for angular moment The infinitesimal generator of a Lie group is defined by $G= (dR(\varphi)/d\phi)_{\varphi=0}$ Otherweise saying we consider the element infinitesimally near to the identity element, i.e. $R(d\varphi)= I + Gd\varphi + O(d\varphi^2)$ The rotation matrix about the z axis is well-known, derivating the $R_3(\varphi)$ you can show $(G_3)_{ij}= -\epsilon_{ij3}$ You can manually show that it's true on the rest axises $(G_k)_{ij}= -\epsilon_{ijk}$ Now consider the general rotation, i.e. an infinitesimal $d\varphi$ rotation about n axis. Heuristically we can say $\mathbf{r}' = \mathbf{r} + d\varphi(\mathbf{n} \times \mathbf{r})$ or $r'_i = r_i + d\varphi\epsilon_{ikj}n_kr_j = \delta_{ij}r_j - d\varphi\epsilon_{ijk}n_kr_j$ Comparing this with the second definition of infinitesimal generator shows that $(G_{\mathbf{n}})_{ij}= -\epsilon_{ijk}n_k$ or reexpressing with the axis generators $G_{\mathbf{n}}= n_kG_k = \mathbf{n}\dot\mathbf{G}$ Thus the general rotation is $R_{\mathbf{n}}(\varphi)= \exp(\varphi\mathbf{n}\dot\mathbf{G})$

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