0
$\begingroup$

I'm solving the following problem from the book "Analytical Mechanics for Relativity and Quantum Mechanics" by Oliver Davis Johns:

Let $R(t)=R_a(t)R_b(t)$ be a rotation matrix. Show that the angular velocity associated to the product is \begin{gather*} \vec{w}=\vec{w_a}+R_a(t)\vec{w_b}(t) \end{gather*}

So, my attempt, which seems correct, is: \begin{gather*} \vec{x}(t)=R(t)\vec{x}(0)\tag{1} \\ R(t)=R_a(t)R_b(t)\tag{2} \\ \dot{R}(t)=\dot{R_a}(t)R_b(t)+R_a(t)\dot{R_b}(t)\tag{3} \\ \dot{\vec{x}}(t)=\dot{R}(t)R^{t}(t)\vec{x}(t)\tag{4} \end{gather*} Using some matrices properties, we have
\begin{gather*} \dot{R}(t)R^{t}(t)=(\dot{R_a}(t)R_b(t)+R_a(t)\dot{R}_b(t))(R_b^{t}(t)R_a^{t}(t)) \tag{5} \\ =\dot{R_a}(t)IR_a(t)+R_a(t)\dot{R}_b(t)R_b(t)R_a^{t}(t)\tag{6} \\ =\dot{R_a}(t)R_a(t)+T(t) \\ where \\ T(t)=R_a(t)\dot{R}_b(t)R_b(t)R_a^{t}(t) \end{gather*}

On equation 6, how do I manipulate the term $T(t)$ so when mulplitying it by $x(t)$ I get $\vec{w_b}(t)$?

$\endgroup$

1 Answer 1

0
$\begingroup$

You are correct that $$\dot{R}(t)R(t)^T = \dot{R}_a(t)R_a(t)^T + R_a(t)\Big(\dot{R}_b(t)R_b(t)\Big)R_a(t)^T$$ Then, define the skew-symmetric matrices $\omega(t) = \dot{R}(t)R(t)^T,\,\,\, \omega_a(t) = \dot{R}_a(t)R_a(t)^T$ and $\omega_b(t) = \dot{R}_b(t)R_b(t)^T$. Then your expression can be written as $$\omega(t) = \omega_a(t) + R_a(t) \,\omega_b(t) \,R_a(t)^T$$ Then, by a theorem from the Lie group theory of the orthogonal group of three space, there are unique vectors $\vec{w}(t), \,\,\, \vec{w}_a(t)$ and $\vec{w}_b(t)$ such that for any three dimensional vector $\vec{x}$ \begin{align} &\omega(t)\, \vec{x} = \vec{w}(t) \times \vec{x}\\ &\omega_a(t)\, \vec{x} = \vec{w}_a(t) \times \vec{x}\\ &\omega_b(t)\, \vec{x} = \vec{w}_b(t) \times \vec{x} \end{align} Moreover, \begin{align} \Big(R_a(t)\, \omega_b(t) \, R_a(t)^T\Big) \, \vec{x} &= R_a(t)\Big( \,\omega_b(t)\big(\, R_a(t)^T \, \vec{x}\,\big)\,\Big) = R_a(t)\Big( \,\vec{w}_b(t)\times \big(\, R_a(t)^T \, \vec{x}\,\big)\,\Big)\\ &= \Big( R_a(t) \,\vec{w}_b(t)\times \big(\,R_a(t) R_a(t)^T \, \vec{x}\,\big)\,\Big) \\ &=\Big( R_a(t) \,\vec{w}_b(t)\Big) \times \vec{x} \end{align} That is why for any vector $\vec{x}$ $$\omega(t)\, \vec{x} = \omega_a(t)\, \vec{x} + \Big(R_a(t) \,\omega_b(t) \,R_a(t)^T\Big) \vec{x}$$ can be written as $$\vec{w}(t)\times \vec{x} = \vec{w}_a(t) \times \vec{x} + \Big( R_a(t) \,\vec{w}_b(t)\Big) \times \vec{x}$$ which implies that $$\vec{w}(t) = \vec{w}_a(t) + R_a(t) \,\vec{w}_b(t)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.