Infinitesimal rotation matrix: Difference between revisions

Revision as of 03:05, 23 July 2024

Revision as of 15:28, 8 September 2024

Line 60:

:<math> dA_\mathbf{x}^\textsf{T} \, dA_\mathbf{x} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 + d\theta^2 & 0 \\ 0 & 0 & 1 + d\theta^2 \end{bmatrix},</math>

differing from an identity matrix by second -order infinitesimals, discarded here. So, to first order, an infinitesimal rotation matrix is an orthogonal matrix.

Next, examine the square of the matrix,

Line 66:

:<math> dA_{\mathbf{x}}^2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 - d\theta^2 & -2d\theta \\ 0 & 2\,d\theta & 1 - d\theta^2 \end{bmatrix}.</math>

Again discarding second -order effects, note that the angle simply doubles. This hints at the most essential difference in behavior, which we can exhibit with the assistance of a second infinitesimal rotation,

:<math>dA_\mathbf{y} = \begin{bmatrix} 1 & 0 & d\phi \\ 0 & 1 & 0 \\ -d\phi & 0 & 1 \end{bmatrix}.</math>

Line 83:

again to first order. In other words, {{em|the order in which infinitesimal rotations are applied is irrelevant}}.

This useful fact makes, for example, derivation of rigid body rotation relatively simple. But one must always be careful to distinguish (the first -order treatment of) these infinitesimal rotation matrices from both finite rotation matrices and from Lie algebra elements. When contrasting the behavior of finite rotation matrices in the [[Baker–Campbell–Hausdorff formula]] above with that of infinitesimal rotation matrices, where all the commutator terms will be second -order infinitesimals one finds a bona fide vector space. Technically, this dismissal of any second -order terms amounts to [[Group contraction]].

== Generators of rotations ==