Editing Infinitesimal rotation matrix (section)

==Order of rotations==

These matrices do not satisfy all the same properties as ordinary finite rotation matrices under the usual treatment of infinitesimals.<ref>{{Harv|Goldstein|Poole|Safko|2002|loc=§4.8}}</ref> To understand what this means, consider

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

First, test the orthogonality condition, {{math|1=''Q''<sup>T</sup>''Q'' = ''I''}}. The product is

:<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,

:<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>

Compare the products {{math|''dA''<sub>'''x'''</sub>&nbsp;''dA''<sub>'''y'''</sub>}} to {{math|''dA''<sub>'''y'''</sub>&nbsp;''dA''<sub>'''x'''</sub>}},

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

Since <math>d\theta \, d\phi</math> is second-order, we discard it: thus, to first order, multiplication of infinitesimal rotation matrices is ''commutative''. In fact,

:<math> dA_{\mathbf{x}}\,dA_{\mathbf{y}} = dA_{\mathbf{y}}\,dA_{\mathbf{x}},</math>

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]].