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From the book Spacecraft Attitude Dynamics by Peter C. Hughes

A certain scalar $s$, is defined as the inner product if the two vectors:$$s(t) \overset{\Delta}{=} \underset{\vec{}}{u}(t)\cdot\underset{\vec{}}{v}(t)$$ The derivative $\dot s(t)$ is measured in two reference frames that are rotating with respect to each other. Using vector identities, show that both frames record the same $\dot s$.

I solved first considering $\dot s(t)$ in first frame results into $$\dot s_1(t) = \underset{{\vec{}}}{\dot u}(t)\cdot\underset{{\vec{}}}{v}(t) + \underset{{\vec{}}}{u}(t)\cdot\underset{{\vec{}}}{\dot v}(t)$$ and in second Reference frame $$\dot s_2(t) = (R\cdot\underset{{\vec{}}}{\dot u}(t)) \cdot(R\cdot\underset{\vec{}}{v}(t))+(R\cdot\underset{\vec{}}{u}(t)) \cdot (R \cdot \underset{{\vec{}}}{\dot v}(t))$$ where $R$ is the rotation matrix. But how can we say these two are same?

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  • $\begingroup$ Note that $\dot s$ can be coded in MathJax as \dot s. You don't need \overset{.}{s} See my edits. $\endgroup$ – Michael Hardy Aug 28 '17 at 22:34
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\begin{align} & \sum_a\left(\sum_b R_{ab}\dot{u}_b\right)\left(\sum_c R_{ac}v_c\right)= \sum_a\left(\sum_b R_{ab}\dot{u}_b\right)\left(\sum_c R^T_{ca}v_c\right) \\[10pt] = {} & \sum_a\left(\sum_b R_{ab}\dot{u}_b\right)\left(\sum_c R^{-1}_{ca} v_c\right) \end{align} because the transpose of a rotation is its inverse. So \begin{align} & \sum_a\left(\sum_b R_{ab}\dot{u}_b\right)\left(\sum_c R_{ac}v_c\right)=\sum_a\sum_b\sum_c\left( R^{-1}_{ca}R_{ab}\right)\dot{u}_b v_c \\ = {} & \sum_b\sum_c\left( \sum_aR^{-1}_{ca}R_{ab}\right)\dot{u}_b v_c = \sum_b\sum_c\delta_{bc}\dot{u}_b v_c=\sum_b\dot{u}_b v_c= \dot{u}\cdot v \end{align} Similarly for the $\dot{v}\cdot u$ term.

By the way, note that $R$ can even be a function of $t$ and this would still hold.

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