Physlib.ClassicalMechanics.RigidBody.Motion

For a rigid body motion $M$ in $d$-dimensional space, let $R(t)$ denote the orientation matrix of the body at time $t$. Then $R(t) R(t)^\top = I$, where $I$ is the $d \times d$ identity matrix.

For a rigid body motion $M$ in $d$-dimensional space, the velocity of its center of mass is the function mapping each time $t \in \text{Time}$ to the time derivative of the center-of-mass trajectory $\mathbf{R}_{\text{com}}(t) \in \text{Space } d$. Mathematically, this is expressed as: $$\mathbf{V}(t) = \frac{d}{dt} \mathbf{R}_{\text{com}}(t)$$ This velocity corresponds to the translational component $\mathbf{V}$ in the Landau–Lifshitz decomposition of the velocity of a point on the body, $\mathbf{v} = \mathbf{V} + \boldsymbol{\Omega} \times \mathbf{r}$, where $\boldsymbol{\Omega}$ is the angular velocity and $\mathbf{r}$ is the position relative to the center of mass.

For a rigid body motion $M$ in $d$-dimensional space, the velocity of the center of mass is equal to the time derivative of the trajectory of the center of mass. This is expressed as: $$\mathbf{V}(t) = \frac{d}{dt} \mathbf{R}_{\text{com}}(t)$$ where $\mathbf{V}$ is the velocity of the center of mass and $\mathbf{R}_{\text{com}}$ is the trajectory of the center of mass.

For a rigid body motion $M$ in $d$-dimensional space, if the trajectory of the center of mass is constant (i.e., $\mathbf{R}_{\text{com}}(t) = \mathbf{c}$ for all times $t$ and some constant vector $\mathbf{c}$), then the velocity of the center of mass is zero ($\mathbf{V}(t) = \mathbf{0}$ for all $t$).

For a rigid body motion $M$ in $d$-dimensional space, the linear momentum is the function that maps each time $t \in \text{Time}$ to the vector obtained by multiplying the body's total mass $m$ by the velocity of its center of mass $\mathbf{V}(t)$. Mathematically, the linear momentum $\mathbf{p}(t)$ is defined as: $$\mathbf{p}(t) = m \mathbf{V}(t)$$ where $m$ is the total mass of the rigid body and $\mathbf{V}(t) \in \text{Space } d$ is the velocity of the center of mass at time $t$.

For a rigid body motion $M$ in $d$-dimensional space, the linear momentum $\mathbf{p}(t)$ is equal to the product of the total mass $m$ of the rigid body and the velocity of its center of mass $\mathbf{V}(t)$. Mathematically, this is expressed as: $$\mathbf{p}(t) = m \mathbf{V}(t)$$ where $m$ is the mass of the rigid body and $\mathbf{V}(t)$ is the center-of-mass velocity.

For a rigid body motion $M$ in $d$-dimensional space $\text{Space } d$, the displacement at time $t$ is the map that carries a point $y$ from the body-fixed frame to its position in the inertial frame. It is defined by the formula: $$\text{displacement}(t, y) = \mathbf{R}(t)(y - \mathbf{r}_{\text{cm}}) + \mathbf{r}_{\text{traj}}(t)$$ where: - $\mathbf{R}(t)$ is the rotation matrix (the `orientation` of the body at time $t$), - $\mathbf{r}_{\text{cm}}$ is the fixed `centerOfMass` of the rigid body in the body frame, - $\mathbf{r}_{\text{traj}}(t)$ is the `comTrajectory` representing the position of the center of mass in the inertial frame at time $t$.

For a rigid body motion $M$ in $d$-dimensional space, the $k$-th coordinate of the displacement of a point $y$ (from the body-fixed frame to the inertial frame) at time $t$ is given by: $$(\text{displacement}(t, y))_k = \sum_{j=1}^{d} R_{kj}(t) (y_j - (\mathbf{r}_{\text{cm}})_j) + (\mathbf{r}_{\text{traj}}(t))_k$$ where: - $R_{kj}(t)$ is the $(k, j)$-th entry of the rotation matrix $\mathbf{R}(t)$ (representing the orientation of the body at time $t$), - $y_j$ is the $j$-th coordinate of the point $y$ in the body frame, - $(\mathbf{r}_{\text{cm}})_j$ is the $j$-th coordinate of the fixed center of mass in the body frame, - $(\mathbf{r}_{\text{traj}}(t))_k$ is the $k$-th coordinate of the center of mass trajectory in the inertial frame at time $t$.

For a rigid body motion $M$ in $d$-dimensional space $\text{Space } d$ and a fixed time $t$, the displacement map $y \mapsto \text{displacement}(t, y)$, which maps a point $y$ from the body-fixed frame to its position in the inertial frame, is infinitely differentiable (of class $C^\infty$).

For a rigid body motion $M$ in $d$-dimensional space $\text{Space } d$, the mass distribution at time $t$ is defined as the pushforward of the body-fixed mass distribution $\rho$ along the rigid displacement map $\Phi_t : \text{Space } d \to \text{Space } d$. Specifically, the distribution $\rho_t$ acts on a test function $f$ according to the formula: $$\rho_t(f) = \rho(f \circ \Phi_t)$$ where $\Phi_t$ is the displacement at time $t$ that maps coordinates from the body-fixed frame to the inertial frame.

For a rigid body motion $M$ in $d$-dimensional space, the total mass of the mass distribution at any time $t$ is equal to the total mass of the reference (body-fixed) rigid body. Mathematically, this is expressed as $\text{mass}(\rho_t) = \text{mass}(\rho)$, where $\rho_t$ is the mass distribution at time $t$ and $\rho$ is the original body-fixed mass distribution.

For a rigid body motion $M$ in $d$-dimensional space with non-zero total mass, the center of mass of the body's mass distribution at any time $t$ is equal to the position specified by the prescribed center of mass trajectory $\text{comTrajectory}(t)$ at that time.

Given a rigid body motion $M$ in $d$-dimensional space $\text{Space } d$ and a material point $y$ in the body-fixed frame, the velocity $\mathbf{v}_y: \text{Time} \to \text{Space } d$ is the function that assigns to each time $t$ the time derivative of the trajectory of $y$ in the inertial frame. It is defined by: $$\mathbf{v}_y(t) = \frac{d}{dt} \text{displacement}(t, y)$$ where $\text{displacement}(t, y)$ is the position of the material point $y$ in the inertial frame at time $t$.

For a rigid body motion $M$ in $d$-dimensional space $\text{Space } d$, let $y$ be a material point in the body-fixed frame and $t$ be a time. The velocity $\mathbf{v}_y(t)$ is equal to the time derivative of the displacement (position in the inertial frame) of $y$ evaluated at time $t$: $$\mathbf{v}_y(t) = \left. \frac{d}{ds} \text{displacement}(s, y) \right|_{s=t}$$

Let $M$ be a rigid body motion in $d$-dimensional space. For a material point $y$ in the body-fixed frame, let $\mathbf{r}(t, y)$ denote its displacement (position in the inertial frame) at time $t$, and let $\mathbf{v}(t, y)$ be its velocity. If the trajectory $t \mapsto \mathbf{r}(t, y)$ is differentiable at time $t$, then the $i$-th component of the velocity is the time derivative of the $i$-th coordinate of the trajectory: $$v_i(t, y) = \frac{d}{dt} r_i(t, y)$$ where $i \in \{0, \dots, d-1\}$.

For a rigid body motion $M$ in $d$-dimensional space, the velocity $\mathbf{v}_{\mathbf{r}_{\text{cm}}}$ of the material point located at the body's center of mass $\mathbf{r}_{\text{cm}}$ is equal to the center-of-mass velocity $\mathbf{V}$ of the motion. Mathematically, this is expressed as: $$\mathbf{v}_{\mathbf{r}_{\text{cm}}}(t) = \mathbf{V}(t)$$ for all times $t$, where $\mathbf{V}(t)$ is the time derivative of the center-of-mass trajectory in the inertial frame.

For a rigid body motion $M$ in $d$-dimensional space, if the orientation is constant over time—meaning there exists a rotation matrix $R$ in the special orthogonal group $SO(d)$ such that $M.\text{orientation}(t) = R$ for all $t$—then for any material point $y$, its velocity $\mathbf{v}_y$ is equal to the velocity of the center of mass $\mathbf{V}$. Mathematically, if $M.\text{orientation}(t) = R$, then $\mathbf{v}_y(t) = \mathbf{V}(t)$.

For a rigid body motion $M$ in $d$-dimensional space and a material point $y$ in the body-fixed frame, if the orientation $\mathbf{R}(t)$ and the center-of-mass trajectory are differentiable, then the $i$-th component of the velocity of point $y$ at time $t$ is given by the decomposition: $$(\mathbf{v}_y(t))_i = \left( \frac{d}{dt} \mathbf{R}(t) \cdot (\mathbf{y} - \mathbf{c}) \right)_i + (\mathbf{V}(t))_i$$ where $\mathbf{R}(t)$ is the rotation matrix (the first component of the orientation) at time $t$, $\mathbf{c}$ is the center of mass of the rigid body in the body-fixed frame, and $\mathbf{V}(t)$ is the velocity of the center of mass in the inertial frame. In this expression, $\mathbf{y} - \mathbf{c}$ represents the position of the point relative to the center of mass in the body frame, and the term $\frac{d}{dt} \mathbf{R}(t) \cdot (\mathbf{y} - \mathbf{c})$ represents the velocity due to the body's rotation.

For a rigid body motion $M$ in $d$-dimensional space, the closed-form velocity at time $t$ for a material point $y$ (given in the body-fixed frame) is defined as: $$\mathbf{v}_{\text{closed}}(t, y) = \left( \frac{d}{dt} \mathbf{R}(t) \right) (\mathbf{y} - \mathbf{c}) + \mathbf{V}(t)$$ where: - $\mathbf{R}(t)$ is the rotation matrix representing the body's orientation at time $t$. - $\mathbf{c}$ is the center of mass of the rigid body in the body-fixed frame. - $\mathbf{V}(t)$ is the velocity of the center of mass in the inertial frame at time $t$. - $\frac{d}{dt} \mathbf{R}(t) (\mathbf{y} - \mathbf{c})$ is the matrix-vector product of the time derivative of the orientation and the position of the point relative to the center of mass.

For a rigid body motion $M$ in $d$-dimensional space, let $\mathbf{R}(t)$ be the rotation matrix representing the body's orientation and $\mathbf{x}_{\text{com}}(t)$ be the trajectory of the center of mass in the inertial frame. If both $\mathbf{R}$ and $\mathbf{x}_{\text{com}}$ are differentiable at time $t$, then for any material point $y$ in the body-fixed frame, the closed-form velocity $\mathbf{v}_{\text{closed}}(t, y)$ (defined by the rotation and center-of-mass velocity) is equal to the honest point velocity $\mathbf{v}_y(t)$ (defined as the time derivative of the point's position in the inertial frame).

Let $M$ be a rigid body motion in $d$-dimensional space and $t$ be a fixed time. The function that maps a point $y \in \mathbb{R}^d$ to its squared speed $\langle v(t, y), v(t, y) \rangle$ (where $v(t, y)$ is the velocity of the point at time $t$ in closed form) is infinitely differentiable (smooth).