# Vector Associated with Rotation Matrix

I am reading the $$3^{\mathrm{rd}}$$ edition of Goldstein's Classical Mechanics, and on p. 165 in Chapter 4 around eqs. (4.67)-(4.68), the authors derive the transformation matrix corresponding to a rotation by infinitesimal Eulerian angles:

$$\begin{equation*} \mathbf{A} = \left(\begin{array}{ccc} 1 & \mathrm{d}{\phi}+\mathrm{d}{\psi} & 0\\ -\left(\mathrm{d}{\phi}+\mathrm{d}{\psi}\right) & 1 & \mathrm{d}{\theta}\\ 0 & -\mathrm{d}{\theta} & 1\\ \end{array}\right).\tag{p.165} \end{equation*}$$

The derivation makes complete sense, but then, seemingly with no context, the authors mention that in light of this matrix, one has

$$\begin{equation*} \mathrm{d}\boldsymbol{\Omega} = \hat{\boldsymbol{\imath}}\mathrm{d}\theta + \hat{\boldsymbol{k}}\left(\mathrm{d}\phi + \mathrm{d}\psi\right)\tag{p.165} \end{equation*}$$

where $$\hat{\boldsymbol{\imath}}$$ and $$\hat{\boldsymbol{k}}$$ are the unit vectors in the $$x$$- and $$z$$-directions, respectively. Looking back, they have not defined $$\boldsymbol{\Omega}$$, other than that in the previous chapter, it is used in the context of scattering. I'm trying to figure out what this vector represents and how it is derived. Its components are differential angles, and I suppose its direction is that of the rotation axis, but this is guesswork. Does $$\mathrm{d}\boldsymbol{\Omega}$$ represent the action of applying $$\mathbf{A}$$ to some other vector of interest?

I appreciate any hints.

Indeed, as you suspect, the components of $$\bf \Omega$$ are usually defined along the rotation axis (and the length of the vector then is the angle of rotation). And of course applying the rotation to some other vector then is done with the cross-product. We can assume that this definition is also meant here, because it would fit the claimed expression for $$d \bf \Omega$$.

Applying the infinitesimal rotation to some vector $$\bf v$$ would give: $$\bf{A}\, \bf{v} = \bf{v} - d \bf{\Omega} \times \bf{v}$$ where the minus sign is adapted to the signs in the matrix $$\bf A$$ (I think they sometimes are defined opposite).

• Thank you! Indeed, a few pages later, the vector $\mathrm{d}\boldsymbol{\Omega}$ is defined as $\hat{\mathbf{n}}\mathrm{d}\Phi$, as you say. Commented Apr 23 at 8:04

The transformation matrix between body fixed system and inertial system in your case is:

$$A=A_z(\psi)\,A_x(\theta)\,A_z(\phi)\tag 1$$ where $$A_Z= \left[ \begin {array}{ccc} \cos \left( \psi \right) &-\sin \left( \psi \right) &0\\ sin \left( \psi \right) &\cos \left( \psi \right) &0\\ 0&0&1\end {array} \right] \\ A_x= \left[ \begin {array}{ccc} 1&0&0\\ 0&\cos \left( \theta \right) &-\sin \left( \theta \right) \\ 0& \sin \left( \theta \right) &\cos \left( \theta \right) \end {array} \right]$$

to linearized the rotation matrix $$~A~$$ you take

$$\phi=\phi_0+d\phi\quad, \theta=\theta_0+d\theta\quad, \psi=\psi_0+d\psi$$

for $$~\phi_0=0~,\theta_0=0~,\psi_0=0~$$ and $$~d\phi~,d\theta~,d\psi~$$ are all small. you obtain:

$$A_L= \left[ \begin {array}{ccc} 1&-d\phi -d\psi &0\\ d \phi +d\psi &1&-d\theta \\0&d\theta &1\end {array} \right] \quad\Rightarrow d\Omega= \left[ \begin {array}{c} d\theta \\ 0 \\ d\phi+d\psi \end {array} \right]$$

thus

$$A_L\,v=(I_3+d\Omega^\times )\,v=v+d\Omega\times\,v$$

Goldstein's Book

Goldstein use the transformation matrix between inertial system and body system , this means that $$~A_L\mapsto A_L^T~$$ thus $$~d\Omega=-d\Omega~$$ so you obtain

$$A_{LG}\,v=v-d\Omega\,\times\,v$$

• How do you obtain $\mathrm{d}\Omega$ from $A_{L}$? Commented Apr 23 at 22:45
• you can use this equation $~ \dot A~A^{T}=\Omega ^{x}~$ where $~\Omega^\times~$ is skew symmetric matrix
– Eli
Commented Apr 24 at 7:11