Let $\left\{\boldsymbol{e}_{i}\right\}$ and $\left\{\widehat{\boldsymbol{e}}_{i}(t)\right\}$ be two coordinate frames for $\mathbb{E}^{3}$, where the first is fixed (time-independent) and the second is moving (time-dependent). Moreover, let $\boldsymbol{Q}(t)$ be the change of basis tensor from $\left\{\boldsymbol{e}_{i}\right\}$ to $\left\{\widehat{\boldsymbol{e}}_{i}(t)\right\}$ so that (omitting arguments $t$ for brevity)
$$
\widehat{\boldsymbol{e}}_{i}=\boldsymbol{Q} \boldsymbol{e}_{i}
$$
(a) Show that the time derivative $d \widehat{\boldsymbol{e}}_{i} / d t$, as measured by an observer in the fixed frame, may be expressed as
$$
\frac{d \widehat{e}_{i}}{d t}=\boldsymbol{\Omega} \widehat{e}_{i}=\boldsymbol{\omega} \times \widehat{\boldsymbol{e}}_{i}
$$
where $\boldsymbol{\Omega}(t)$ is the skew-symmetric tensor defined by
$$
\frac{d \boldsymbol{Q}}{d t}=\boldsymbol{\Omega} Q \quad \text { or equivalently } \quad \boldsymbol{\Omega}=\frac{d \boldsymbol{Q}}{d t} \boldsymbol{Q}^{T}
$$
and $\boldsymbol{\omega}(t)$ is the axial vector of $\boldsymbol{\Omega}(t)$.
(b) For any vector $\boldsymbol{v}=\widehat{v}_{i} \widehat{\boldsymbol{e}}_{i}=v_{i} \boldsymbol{e}_{i}$, show that
$$
\frac{d \widehat{v}_{i}}{d t}=Q_{j i}\left[\frac{d v_{j}}{d t}-\Omega_{j \underline{k}} v_{k}\right],
$$
where $Q_{i j}$ and $\Omega_{j k}$ are the components of $\boldsymbol{Q}$ and $\boldsymbol{\Omega}$ in the fixed frame.
Remark: An observer in the fixed frame $\left\{\boldsymbol{e}_{i}\right\}$ sees the time derivative of $\boldsymbol{v}$ as
$$
\dot{\boldsymbol{v}}=\frac{d v_{i}}{d t} \boldsymbol{e}_{i},
$$
while an observer moving with the frame $\left\{\widehat{e}_{i}(t)\right\}$ sees the time derivative of $\boldsymbol{v}$ as
$$
\stackrel{\circ}{\boldsymbol{v}}=\frac{d \widehat{v}_{j}}{d t} \widehat{\boldsymbol{e}}_{j}
$$
The components of $\dot{\boldsymbol{v}}$ and $\boldsymbol{v}$ in the fixed frame $\left\{\boldsymbol{e}_{i}\right\}$ are
$$
\begin{aligned}
&\dot{v}_{i}=\dot{\boldsymbol{v}} \cdot \boldsymbol{e}_{i}=\frac{d v_{i}}{d t} \\
&\dot{v}_{i}=\stackrel{\boldsymbol{v}} \cdot \boldsymbol{e}_{i}=\frac{d \widehat{v}_{j}}{d t} \widehat{\boldsymbol{e}}_{j} \cdot \boldsymbol{e}_{i}=Q_{i j} \frac{d \widehat{v}_{j}}{d t}
\end{aligned}
$$
In view of $(2.3)$ the vector $\stackrel{\circ}{\boldsymbol{v}}$ can be expressed in the fixed frame as
$$
\stackrel{\circ}{\boldsymbol{v}}=\stackrel{\circ}{v}_{i} \boldsymbol{e}_{i}=\left[\frac{d v_{i}}{d t}-\Omega_{i k} v_{k}\right] \boldsymbol{e}_{i}
$$
The vector $\dot{v}$ is called the co-rotational or Jaumann derivative of $\boldsymbol{v}$ with respect to $\boldsymbol{\Omega}$. It represents the rate of change of $\boldsymbol{v}$ with respect to time as seen by an observer in a moving frame. In tensor notation we have
$$
\stackrel{\circ}{v}=\dot{v}-\Omega v.
$$