$$\newcommand{\u}[1]{\boldsymbol{\mathsf{#1}}} \renewcommand{\b}[1]{\boldsymbol{#1}} \newcommand{\t}[1]{\textsf{#1}} \newcommand{\m}[1]{\mathbb{#1}} \def\RR{\bf R} \def\bold#1{\bf #1} \def\mbf#1{\mathbf #1} \def\uv#1{\hat{\usf {#1}}} \def\dl#1{\underline{\underline{#1}}} \newcommand{\usf}[1]{\boldsymbol{\mathsf{#1}}} \def\bs#1{\usf #1}$$

Angular Velocities: Velocity field

Recall that $\bs{x}(\bs{X},t)$ gives the position vector of the material particle as a function of time $t$. The velocity of the material particle can simply be obtained by differentiating the position vector of the particle $\bs{X}$ w.r.t. time, i.e., by partially differentiating $\bs{x}(\bs{X},t)$ w.r.t. time $t$ and holding $\bs{X}$ fixed. If we denote the velocity of particle $\bs{X}$ at time $t$ as $\bs{V}(\bs{X},t)$, then

$$\begin{align} \bs{V}(\bs{X},t)&=\frac{\partial \bs{x}(\bs{X},t)}{\partial t} \end{align}$$ $$\begin{align} \boldsymbol{V}(\boldsymbol{X},t)&= \dot{\boldsymbol{R}}(t)\boldsymbol{X}+\dot{\bs{t}}(t) \end{align}$$

The above velocity field is called the material velocity field, since it depends on the material position vector (and also time, of course). Recall that $\bs{X}$ and $\bs{x}$ are related to each other as

$$\begin{align} \bs{x}(\bs{X},t)&=\bs{R}(t)\bs{X}+\bs{t}(t)\\ \end{align}$$

We can invert the above equation to write $\begin{align} \bs{X}(\bs{x},t)&=\bs{R}^{-1}(t)\left(\bs{x}-\bs{t}(t)\right)\\ \bs{X}(\bs{x},t)&=\bs{R}^{\textsf{T}}(t)\left(\bs{x}-\bs{t}(t)\right) \end{align}$

Thus we can write the velocity field as a function of the spatial position vector as,

$$\begin{align} \boldsymbol{v}(\boldsymbol{x},t)&= \dot{\boldsymbol{R}}(t) \left( \bs{R}^{\textsf{T}}(t)\left(\bs{x}-\bs{t}(t)\right) \right)+\dot{c}(t)\\ \boldsymbol{v}(\boldsymbol{x},t)&= \dot{\boldsymbol{R}}(t) \left(\bs{R}^{\textsf{T}}(t)\left(\bs{x}-\bs{t}(t)\right)\right)+\dot{q}(t)\\ \end{align}$$ $$\begin{align} \bs{v}(\bs{x},t)&=\dot{\bs{R}}\bs{R}^{\text{T}}\bs{x}-\dot{\bs{R}}\bs{R}^{\text{T}}\bs{t}(t)+\dot{\bs{t}}(t)\\ \bs{v}(\bs{x},t)&=\bs{\Omega}(t)\bs{x}+\bs{c}(t)\\ \bs{\Omega}(t)&=\dot{\bs{R}}\bs{R}^{\text{T}}\\ \bs{c}(t)&=-\dot{\bs{R}}\bs{R}^{\text{T}}\bs{t}(t)+\dot{\bs{t}}(t) \end{align}$$

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