運動座標系變化方程式
向量微積分版
假設三維空間有固定座標系\(\hat{e}\)，以及相對於\(\hat{e}\)旋轉之運動座標系\(\hat{b}\)，兩者關係可由一旋轉張量\([R]\)導出：
$$ \begin{equation} \hat{b} = [R] \hat{e} \end{equation} $$如果將\(\hat{b}\)對時間微分，我們有
$$ \begin{equation} \frac{d}{dt} \hat{b} = [\dot{R}] [R]^{\mathsf{T}} \hat{b} \end{equation} $$由於旋轉張量所對應之矩陣為對稱之單位正交張量（symmetric and orthonormal tensor），因此我們有
$$ \begin{equation} [R][R]^{\mathsf{T}} = [\mathrm{I}] \end{equation} $$因此
$$ \begin{equation} \frac{d}{dt} \left[ [R][R]^{\mathsf{T}} \right] = [\dot{R}] [R]^{\mathsf{T}} + [R] [\dot{R}]^{\mathsf{T}} = [\dot{R}] [R]^{\mathsf{T}} + \left([\dot{R}] [R]^{\mathsf{T}} \right)^{\mathsf{T}} = 0 \end{equation} $$如果我們定義一角速度張量為
$$ \begin{equation} [\Omega] \equiv [\dot{R}] [R]^{\mathsf{T}} = -\left([\dot{R}] [R]^{\mathsf{T}} \right)^{\mathsf{T}} \end{equation} $$可以知道\([\Omega]\)為反對稱張量（anti-symmetric tensor），因此該二階張量可由Levi-Civita張量和一角速度向量\(\vec{\omega}\)表示
$$ \begin{equation} \Omega_{ij} = -\varepsilon_{ijk} \omega_k \end{equation} $$因此我們除了可以把\(\hat{b}\)對時微分寫成
$$ \begin{equation} \dot{b}_{i} = \Omega_{ij} b_{j} \end{equation} $$也可以將其寫成
$$ \begin{equation} \dot{b}_{i} = -\varepsilon_{ijk} b_{j} \omega_k = \varepsilon_{ijk} \omega_j b_k \end{equation} $$$$ \begin{equation} \frac{d}{dt} \hat{b} = \vec{\omega} \times \hat{b} \end{equation} $$現在，我們以\(\hat{b}\)表示空間上一任意向量\(\vec{v} \equiv v^i \hat{b}_i\)，則\(\vec{v}\) 的對時微分為
$$ \begin{equation} \frac{d \vec{v}}{d t} = \frac{d}{dt}\left( v^i \hat{b}_i \right) = \dot{v}^i \hat{b}_i + v^i \dot{\hat{b}}_i = \dot{v}^i \hat{b}_i + \vec{\omega} \times \left( v^i \hat{b}_i \right) = \frac{\delta \vec{v}}{\delta t} + \omega \times \vec{v} \end{equation} $$幾何代數版
我們可以用幾何代數推導出同樣的結果。已知幾何代數中，\(\hat{b}\)和\(\hat{e}\)的關係可以寫成
$$ \begin{equation} \hat{b} = [R_{- \frac{\theta}{2}}] \hat{e} [R_{\frac{\theta}{2}}] \end{equation} $$其中\([R_{\cdot}]\)為轉子（rotor）複向量（multivector），而\(\theta = \cos^{-1} \left( \hat b \cdot \hat e \right)\)。將上式對時微分則有
$$ \begin{equation} \frac{d}{dt} \hat{b} = [\dot R_{- \frac{\theta}{2}}] \hat{e} [R_{\frac{\theta}{2}}] + [R_{- \frac{\theta}{2}}] \hat{e} [\dot R_{\frac{\theta}{2}}] = [\dot R_{- \frac{\theta}{2}}][R_{\frac{\theta}{2}}] \hat{b} + \hat{b} [R_{- \frac{\theta}{2}}] [\dot R_{\frac{\theta}{2}}] \end{equation} $$又由於
$$ \begin{equation} [R_{- \frac{\theta}{2}}] [R_{\frac{\theta}{2}}] = 1 \end{equation} $$因此
$$ \begin{equation} [\dot R_{- \frac{\theta}{2}}][R_{\frac{\theta}{2}}] + [R_{- \frac{\theta}{2}}] [\dot R_{\frac{\theta}{2}}] = 0 \end{equation} $$所以
$$ \begin{equation} \frac{d}{dt} \hat{b} = \hat{b} [R_{- \frac{\theta}{2}}] [\dot R_{\frac{\theta}{2}}] - [R_{- \frac{\theta}{2}}] [\dot R_{\frac{\theta}{2}}] \hat{b} = \hat{b} \:\rfloor \left( 2 [R_{- \frac{\theta}{2}}] [\dot R_{\frac{\theta}{2}}] \right) \end{equation} $$（上式利用底下的性質：一個向量和一個雙向量的乘積結果為各向量和三向量基底的線性組合，而其中向量的部分anti-commute，三向量的部分則commute。）
這等同於向量\(\hat b\)和雙向量\(2[R_{- \frac{\theta}{2}}] [\dot R_{\frac{\theta}{2}}]\)（可稱之為角速度雙向量\([\Omega_{\theta}]\) ）在三維空間所對應的向量（即\(\vec \omega\)）的外積。
那麼，具體來說\([\Omega_{\theta}]\)到底怎麼求出？我們可以將轉子複向量寫成
$$ \begin{equation} [R_{\frac{\theta}{2}}] = \cos \left( \frac{\theta}{2} \right) + \sin \left( \frac{\theta}{2} \right) [\hat B] = \exp \left( \frac{\theta}{2} [\hat B] \right) \end{equation} $$其中\([\hat B]\)是和\([R_{\frac{\theta}{2}}]\)共平面的單位雙向量（bivector），亦可表示成\(-\frac{\hat b \wedge \hat e} {|\hat b \wedge \hat e|}\)（這裡\({|\hat b \wedge \hat e|} = \sqrt{-(\hat b \wedge \hat e)^2}\)）。這麼一來
$$ \begin{equation} \hat{b} = \exp \left( - \frac{\theta}{2} [\hat B] \right) \hat{e} \exp \left( \frac{\theta}{2} [\hat B] \right) \end{equation} $$而
$$ \begin{equation} \frac{d}{dt} \left[ \exp \left( \frac{\theta}{2} [\hat B] \right) \right] = -\frac{\dot \theta}{2}\sin\left( \frac{\theta}{2} \right) + \frac{\dot \theta}{2}\cos \left( \frac{\theta}{2} \right) [\hat B] + \sin \left( \frac{\theta}{2} \right) \frac{d}{dt}[\hat B] \end{equation} $$$$ \begin{equation} \frac{d}{dt} \left[ \exp \left( \frac{\theta}{2} [\hat B] \right) \right] = \frac{\dot \theta}{2} [\hat B] \exp \left( \frac{\theta}{2} [\hat B] \right) + \sin \left( \frac{\theta}{2} \right) \frac{d}{dt}[\hat B] \end{equation} $$故取\(\hat{b}\)的對時微分可得
$$ \begin{equation} \begin{array}{ccc} \frac{d}{dt} \hat{b} &=& \left( -\frac{\dot \theta}{2} [\hat B] \exp \left( -\frac{\theta}{2} [\hat B] \right) - \sin \left( \frac{\theta}{2} \right) \frac{d}{dt}[\hat B] \right) \hat{e} \exp \left( \frac{\theta}{2} [\hat B] \right) \\ \\ &+& \exp \left( -\frac{\theta}{2} [\hat B] \right) \hat{e} \left( \frac{\dot \theta}{2} [\hat B] \exp \left( \frac{\theta}{2} [\hat B] \right) + \sin \left( \frac{\theta}{2} \right) \frac{d}{dt}[\hat B] \right) \end{array} \end{equation} $$$$ \begin{equation} \begin{array}{ccc} \frac{d}{dt} \hat{b} &=& \left( -\frac{\dot \theta}{2} [\hat B] \exp \left( -\frac{\theta}{2} [\hat B] \right) - \sin \left( \frac{\theta}{2} \right) \frac{d}{dt}[\hat B] \right) \exp \left( \frac{\theta}{2} [\hat B] \right) \hat{b} \\ \\ &+& \hat{b} \exp \left(- \frac{\theta}{2} [\hat B] \right) \left( \frac{\dot \theta}{2} [\hat B] \exp \left( \frac{\theta}{2} [\hat B] \right) + \sin \left( \frac{\theta}{2} \right) \frac{d}{dt}[\hat B] \right) \end{array} \end{equation} $$$$ \begin{equation} \begin{array}{rcl} \frac{d}{dt} \hat{b} &=& -\frac{\dot \theta}{2} [\hat B] \hat{b} - \frac{1}{2} \sin(\theta) \frac{d}{dt}[\hat B] \hat{b} - \sin^2 \left( \frac{\theta}{2} \right) \frac{d}{dt}[\hat B] [\hat B] \hat{b} + \frac{\dot \theta}{2}\hat{b} [\hat B] \\ \\ &+& \frac{1}{2} \sin(\theta) \hat{b} \frac{d}{dt}[\hat B] - \sin^2 \left( \frac{\theta}{2} \right) \hat{b} [\hat B] \frac{d}{dt}[\hat B] \end{array} \end{equation} $$上式中我們可以發現
$$ \begin{equation} \frac{\dot \theta}{2}\hat{b} [\hat B] - \frac{\dot \theta}{2} [\hat B] \hat{b} = \dot \theta \hat{b} \:\rfloor\: [\hat B] \end{equation} $$$$ \begin{equation} \frac{1}{2} \sin(\theta) \hat{b} \frac{d}{dt}[\hat B] - \frac{1}{2} \sin(\theta) \frac{d}{dt}[\hat B] \hat{b} = \sin(\theta) \hat{b} \:\rfloor\: \frac{d}{dt}[\hat B] \end{equation} $$又因為
$$ \begin{equation} [\hat B][\hat B] = -1 \end{equation} $$所以
$$ \begin{equation} \frac{d}{dt}[\hat B][\hat B] + [\hat B]\frac{d}{dt}[\hat B] = 0 \end{equation} $$$$ \begin{equation} \begin{array}{lcl} && -\sin^2 \left( \frac{\theta}{2} \right) \frac{d}{dt}[\hat B] [\hat B] \hat{b} - \sin^2 \left( \frac{\theta}{2} \right) \hat{b} [\hat B] \frac{d}{dt}[\hat B] \\ \\ &=& 2\sin^2 \left( \frac{\theta}{2} \right) \hat{b} \:\rfloor\: \frac{d}{dt}[\hat B][\hat B] \\ \\ &=& (1 - \cos(\theta)) \hat{b} \:\rfloor\: \frac{d}{dt}[\hat B][\hat B] \end{array} \end{equation} $$整理各式結果後便得到
$$ \begin{equation} \frac{d}{dt} \hat{b} = \hat{b} \:\rfloor\: \left( \dot \theta[\hat B] + \sin(\theta) \frac{d}{dt}[\hat B] + (1 - \cos(\theta)) \frac{d}{dt}[\hat B][\hat B] \right) \end{equation} $$因此
$$ \begin{equation} [\Omega_{\theta}] = \dot \theta[\hat B] + \sin(\theta) \frac{d}{dt}[\hat B] + (1 - \cos(\theta)) \frac{d}{dt}[\hat B][\hat B] \end{equation} $$以單位向量\(\hat b\)和\(\hat e\)表示的話則為
$$ \begin{equation} \begin{array}{rcl} [\Omega_{\theta}] &=& -\dot \theta \left( \frac{\hat b \wedge \hat e} {|\hat b \wedge \hat e|} \right) - \sin(\theta) \left( \frac{\dot {\hat b} \wedge \hat e} {|\hat b \wedge \hat e|} - \frac{ \left( \hat b \wedge \hat e \right) \left( \dot {\hat b} \wedge \hat e \right)} {|\hat b \wedge \hat e|^2} \right) \\ \\ &+& (1 - \cos(\theta)) \left( \frac{\dot {\hat b} \wedge \hat e} {|\hat b \wedge \hat e|} - \frac{ \left( \hat b \wedge \hat e \right) \left( \dot {\hat b} \wedge \hat e \right)} {|\hat b \wedge \hat e|^2} \right)\left( \frac{\hat b \wedge \hat e} {|\hat b \wedge \hat e|} \right) \end{array} \end{equation} $$而某個以\(\hat{b}\)表示的向量\(\vec{v} \equiv v^i \hat{b}_i\)對時微分為
$$ \begin{equation} \frac{d \vec{v}}{d t} = \frac{d}{dt}\left( v^i \hat{b}_i \right) = \dot{v}^i \hat{b}_i + v^i \dot{\hat{b}}_i = \dot{v}^i \hat{b}_i + \left( v^i \hat{b}_i \right) \:\rfloor\: [\Omega_{\theta}] = \frac{\delta \vec{v}}{\delta t} + \vec{v} \:\rfloor\: [\Omega_{\theta}] \end{equation} $$運動座標系下的角動量、轉動慣量、與力矩
向量微積分版
一物體相對於其質心C的角動量可以寫成
$$ \begin{equation} \vec{H} = \int_{\mathbf M} (\vec{r}_0 - \vec{r}_C) \times (\dot {\vec{r}}_0 - \dot{\vec{r}}_C) dm = \vec{H}_0 - M \vec{r}_C \times \dot{\vec{r}}_C \end{equation} $$其中\(\vec{r}_0\)和\(\vec{H}_0\)為物體相對於固定座標系的位置和角動量。如果物體為剛體，則相對於質心，其僅有轉動。因此剛體對質心的角動量可寫成
$$ \begin{equation} \vec{H} = \int_{\mathbf M} \vec{r} \times (\vec \omega \times \vec{r}) dm = \int_{\mathbf M} (\vec{r} \cdot \vec{r}) \vec \omega - ((\vec{r} \cdot \vec{\omega})) \vec{r} dm \end{equation} $$上述外積公式可利用Levi-Civita張量，由底下方式得出：
$$ \begin{equation} \varepsilon_{ijk}\varepsilon_{mnl}r_j(\omega_nr_l)\delta_{mk} \end{equation} $$由於
$$ \begin{equation} \begin{array}{ccc} \varepsilon_{ijk}\varepsilon_{mnl} &=& \delta_{im}\delta_{jn}\delta_{kl}+\delta_{il}\delta_{jm}\delta_{kn}+\delta_{in}\delta_{jl}\delta_{km} \\ \\ &-& \delta_{il}\delta_{jn}\delta_{km}-\delta_{in}\delta_{jm}\delta_{kl}-\delta_{im}\delta_{jl}\delta_{kn} \end{array} \end{equation} $$因此外積式僅會有兩非零項
$$ \begin{equation} \varepsilon_{ijk}\varepsilon_{mnl}r_j(\omega_nr_l)\delta_{mk} = (\delta_{in}\delta_{jl}-\delta_{il}\delta_{jn})r_j\omega_nr_l = \omega_ir_jr_j-r_ir_j\omega_j \end{equation} $$這便是\((\vec{r} \cdot \vec{r}) \vec \omega - ((\vec{r} \cdot \vec{\omega})) \vec{r}\)。其中\(((\vec{r} \cdot \vec{\omega})) \vec{r}\)也可以表示成\([\vec{r} \otimes \vec{r}] \vec \omega\)；因此
$$ \begin{equation} \vec{H} = \left( \int_{\mathbf M} (\vec{r} \cdot \vec{r})[I] - [\vec{r} \otimes \vec{r}] dm \right) \vec \omega \end{equation} $$定義
$$ \begin{equation} [\mathcal{I}] = \int_{\mathbf M} (\vec{r} \cdot \vec{r})[I] - [\vec{r} \otimes \vec{r}] dm \end{equation} $$則
$$ \begin{equation} \vec{H} = [\mathcal{I}] \vec \omega \end{equation} $$現在考慮對物體施予相對於質心C的力矩\(\vec{T}\)：
$$ \begin{equation} \vec{T} = \int_{\mathbf M} (\vec{r}_0 - \vec{r}_C) \times \vec{a}_0 dm = \vec{T}_0 - M \vec{r}_C \times \vec{a}_C \end{equation} $$同時
$$ \begin{equation} \begin{array}{rcl} \vec{T} &=& \int_{\mathbf M} \vec{r} \times (\vec{a} + \vec{a}_C) dm \\ \\ &=& \int_{\mathbf M} \vec{r} \times \frac{d}{dt}(\vec{\omega} \times \vec{r}) dm \\ \\ &=& \frac{d}{dt} \left( \int_{\mathbf M} \vec{r} \times (\vec{\omega} \times \vec{r}) dm \right) \end{array} \end{equation} $$所以
$$ \begin{equation} \vec{T} = \frac{d \vec H}{dt} = \frac{\delta [\mathcal{ I}]}{\delta t} \vec{\omega} + [\mathcal{I}] \dot{\vec{\omega}} + \vec{\omega} \times \vec{H} \end{equation} $$幾何代數版
在幾何代數中，角動量為一雙向量；一物體對其質心C的角動量可以寫成
$$ \begin{equation} [H] = \int_{\mathbf M} \vec{r} \wedge \dot {\vec{r}} dm = \int_{\mathbf M} (\vec{r}_0 - \vec{r}_C) \wedge (\dot {\vec{r}}_0 - \dot{\vec{r}}_C) dm = [H]_0 - M \vec{r}_C \wedge \dot{\vec{r}}_C \end{equation} $$其中\(\vec{r}_0\)和\([H]_0\)為物體相對於固定座標系的位置和角動量。如果物體為剛體，則相對於原點在質心上的運動座標系而言，其僅有轉動。因此剛體對質心的角動量可寫成
$$ \begin{equation} [H] = \int_{\mathbf M} \vec{r} \wedge (\vec{r} \:\rfloor [\Omega_{\theta}]) dm = \frac{1}{2}\int_{\mathbf M} \vec{r} \wedge (\vec{r} [\Omega_{\theta}] - [\Omega_{\theta}] \vec{r} ) dm \end{equation} $$但由於\(\vec{r}\) 和\(\vec{r} \:\rfloor [\Omega_{\theta}]\)必相互垂直，所以
$$ \begin{equation} [H] = \int_{\mathbf M} \vec{r} (\vec{r} \:\rfloor [\Omega_{\theta}]) dm = \frac{1}{2}\int_{\mathbf M} \vec{r} (\vec{r} [\Omega_{\theta}] - [\Omega_{\theta}] \vec{r} ) dm \end{equation} $$$$ \begin{equation} [H] = \frac{1}{2}\int_{\mathbf M} (\vec{r} \cdot \vec{r}) [\Omega_{\theta}] - \vec{r}[\Omega_{\theta}] \vec{r} dm \end{equation} $$將\(\vec{r}\) 分解成平行／垂直於\([\Omega_{\theta}]\)的\(\vec{r}_{//}\)和\(\vec{r}_{\perp}\)兩項，則
$$ \begin{equation} [H] = \int_{\mathbf M} \vec{r} \vec{r}_{//} [\Omega_{\theta}] dm \end{equation} $$如果定義轉動慣量為一複向量
$$ \begin{equation} [\mathcal{I}] = \int_{\mathbf M} (\vec{r}_{//} \cdot \vec{r}_{//} - \vec{r}_{//} \wedge \vec{r}_{\perp}) dm \end{equation} $$而角動量和其的關係為
$$ \begin{equation} [H] = [\mathcal{I}] [\Omega_{\theta}] \end{equation} $$不過，由於這裡的轉動慣量複向量會隨著角速度雙向量變化而改變，應用上會較困難。
現在考慮相對於質心C的力矩雙向量\([T]\)：
$$ \begin{equation} [T] = \int_{\mathbf M} \vec{r} \wedge \vec{a}_0 dm = \int_{\mathbf M} (\vec{r}_0 - \vec{r}_C) \wedge \vec{a}_0 dm = [T]_0 - M \vec{r}_C \times \vec{a}_C \end{equation} $$$$ \begin{equation} \begin{array}{rcl} [T] &=& \int_{\mathbf M} \vec{r} \wedge (\vec a +\vec{a}_C) dm \\ \\ &=& \int_{\mathbf M} \vec{r} \wedge \frac{d}{dt} \left( \vec{r} \:\rfloor [\Omega_{\theta}] \right) dm \\ \\ &=& \frac{d}{dt} \left( \int_{\mathbf M} \vec{r} \wedge (\vec{r} \:\rfloor [\Omega_{\theta}]) dm \right) \\ \\ &=& \frac{d}{dt} [H] \end{array} \end{equation} $$那麼在幾何代數中，以運動座標系表示的雙向量的對時微分會是甚麼呢？我們可以把角動量雙向量寫成
$$ \begin{equation} [H] = h_{ij} \hat{b}_i \hat{b}_j \end{equation} $$將其對時微分我們會有
$$ \begin{equation} [\dot H] = \dot h_{ij} \hat{b}_i \hat{b}_j + h_{ij} \dot {\hat{b}_i} \hat{b}_j + h_{ij} \hat{b}_i \dot {\hat{b}_j} \end{equation} $$$$ \begin{equation} \begin{array}{rcl} [\dot H] &=& \dot h_{ij} \hat{b}_i \hat{b}_j \\\\ &+& \frac{1}{2} h_{ij} \left(\hat{b}_i [\Omega_{\theta}] - [\Omega_{\theta}] \hat{b}_i \right)\hat{b}_j \\ \\ &+& \frac{1}{2} h_{ij} \hat{b}_i \left(\hat{b}_j [\Omega_{\theta}] - [\Omega_{\theta}] \hat{b}_j \right) \end{array} \end{equation} $$$$ \begin{equation} [\dot H] = \dot h_{ij} \hat{b}_i \hat{b}_j + \frac{1}{2} h_{ij} \left(\hat{b}_i \hat{b}_j[\Omega_{\theta}] - [\Omega_{\theta}] \hat{b}_i \hat{b}_j \right) \end{equation} $$其中
$$ \begin{equation} \hat{b}_i \hat{b}_j[\Omega_{\theta}] - [\Omega_{\theta}] \hat{b}_i \hat{b}_j = 2 \left\langle \hat{b}_i \hat{b}_j [\Omega_{\theta}] \right\rangle_2 \end{equation} $$在這裡\(\left\langle \cdot \right\rangle_2\)表示某幾何物件中的雙向量。
因此
$$ \begin{equation} [\dot H] = \dot h_{ij} \hat{b}_i \hat{b}_j + \left\langle \left( h_{ij} \hat{b}_i \hat{b}_j \right)[\Omega_{\theta}] \right\rangle_2 \end{equation} $$或者也可以寫成
$$ \begin{equation} \frac{d [H]}{dt} = \frac{\delta [H]}{\delta t} + \left\langle [H] [\Omega_{\theta}] \right\rangle_2 \end{equation} $$針對第一項，如果我們的運動座標系取剛體質心為原點，且跟隨剛體做同步轉動，則剛體各部位的位置向量為定質。在這種情況下我們會有
$$ \begin{equation} \frac{\delta [H]}{\delta t} = \int_{\mathbf M} \vec{r} \left( \vec{r} \:\rfloor\:[\dot \Omega_{\theta}] \right) dm \end{equation} $$補充說明：另一種推導力矩與角動量關係的方式
$$ \begin{equation} \begin{array}{rcl} \frac{d}{dt} \left( \vec r [\Omega_{\theta}] \vec r \right) &=& \dot{\vec r} [\Omega_{\theta}] \vec r + \vec r [\dot{\Omega}_{\theta}] \vec r + \vec r [\Omega_{\theta}] \dot{\vec r} \\ \\ &=& \frac{1}{2} \left( \vec r [\Omega_{\theta}] - [\Omega_{\theta}] \vec r \right) [\Omega_{\theta}] \vec r + \vec r [\dot{\Omega}_{\theta}] \vec r + \frac{1}{2} \vec r [\Omega_{\theta}] \left( \vec r [\Omega_{\theta}] - [\Omega_{\theta}] \vec r \right) \\ \\ &=& \frac{1}{2} \left( \vec r [\Omega_{\theta}] \vec r [\Omega_{\theta}] - [\Omega_{\theta}] \vec r [\Omega_{\theta}] \vec r \right) + \vec r [\dot{\Omega}_{\theta}] \vec r \\ \\ &=& \frac{1}{2} \left( \left( \vec r \cdot \vec r [\Omega_{\theta}] + \vec r [\Omega_{\theta}] \vec r \right) [\Omega_{\theta}] - [\Omega_{\theta}] \left( \vec r \cdot \vec r [\Omega_{\theta}] + \vec r [\Omega_{\theta}] \vec r \right) \right) + \vec r [\dot{\Omega}_{\theta}] \vec r \end{array} \end{equation} $$$$ \begin{equation} \begin{array}{rcl} \frac{d}{dt} [H] &=& \frac{d}{dt} \left( \vec r [\Omega_{\theta}] \vec r \right) + \vec r \cdot \vec r [\dot{\Omega}_{\theta}] + \vec r [\dot{\Omega}_{\theta}] \vec r \\ \\ &=& \frac{1}{2} \left( [H][\Omega_{\theta}] - [\Omega_{\theta}][H] \right) + \frac{d}{dt} [H] \\ \\ &=& \frac{d}{dt} \left( \vec r [\Omega_{\theta}] \vec r \right) + \vec r \cdot \vec r [\dot{\Omega}_{\theta}] + \vec r [\dot{\Omega}_{\theta}] \vec r \end{array} \end{equation} $$