Physlib.ClassicalMechanics.RigidBody.AngularVelocity

For a rigid body motion $M$ in $d$ dimensions and a time $t$, the **angular velocity tensor** $\Omega(t) \in \mathbb{R}^{d \times d}$ is defined as the product of the time derivative of the body's orientation matrix $R(t)$ and its transpose: $$\Omega(t) = \dot{R}(t) R(t)^\top$$ where $R(t)$ is the rotation matrix representing the orientation of the rigid body at time $t$. This skew-symmetric tensor characterizes the instantaneous rate of change of the orientation and appears in the Landau–Lifshitz decomposition of the velocity of a point in the body: $v = V + \Omega r$.

For a rigid body motion $M$ in $d$ dimensions and a time $t$, the angular velocity tensor $\Omega(t)$ is equal to the matrix product of the time derivative of the orientation matrix $R(t)$ and its transpose: $$\Omega(t) = \dot{R}(t) R(t)^\top$$ where $R(t)$ denotes the orientation of the rigid body at time $t$ and $\dot{R}(t)$ is its derivative with respect to time.

Consider a rigid body motion $M$ in $d$ dimensions. Let $R(t)$ be the orientation matrix of the body at time $t$. If $R(t)$ is differentiable at time $t$, then the angular velocity tensor $\Omega(t) = \dot{R}(t)R(t)^\top$ is skew-symmetric, satisfying: $$\Omega(t)^\top = -\Omega(t)$$ Equivalently, the tensor $\Omega(t)$ lies in the Lie algebra $\mathfrak{so}(d)$.

For a rigid body motion $M$ in $d$ dimensions, if the orientation $R(t)$ of the body is constant for all time $t$, such that $R(t) = R$ for some fixed rotation matrix $R \in \text{SO}(d)$, then the angular velocity tensor $\Omega(t)$ is zero for all $t$.

For a rigid body motion $M$ in three-dimensional space and a time $t$, the **angular velocity vector** $\omega(t) \in \mathbb{R}^3$ is defined as the vector dual to the angular velocity tensor $\Omega(t)$ via the vee map: $$\omega(t) = (\Omega(t))^\vee$$ where $\Omega(t) = \dot{R}(t)R(t)^\top$ is the angular velocity tensor and $R(t)$ is the orientation matrix of the body. This vector $\omega(t)$ satisfies the property that for any vector $r \in \mathbb{R}^3$, the product $\Omega(t)r$ is equal to the cross product $\omega(t) \times r$, appearing in the Landau–Lifshitz decomposition of the velocity of a point in the body.

For a rigid body motion $M$ in three-dimensional space and a time $t$, the angular velocity vector $\omega(t)$ is equal to the result of applying the vee map $(\cdot)^\vee$ to the angular velocity tensor $\Omega(t)$: $$\omega(t) = (\Omega(t))^\vee$$ where the angular velocity tensor is defined as $\Omega(t) = \dot{R}(t)R(t)^\top$ and the vee map `crossProductVee` is the left inverse of the hat map (the cross-product matrix map).

For a rigid body motion $M$ in three dimensions, let $\omega(t)$ be the angular velocity vector and $\Omega(t)$ be the angular velocity tensor. If the orientation matrix $R(t)$ of the rigid body is differentiable at time $t$, then the cross product matrix (or hat map) of the angular velocity vector is equal to the angular velocity tensor: $$[\omega(t)]_{\times} = \Omega(t)$$

For a rigid body motion $M$ in three-dimensional space, if the orientation of the body is constant over time, i.e., there exists a fixed rotation matrix $R \in \mathrm{SO}(3)$ such that the orientation $R(t) = R$ for all $t$, then the angular velocity vector $\omega$ is zero ($\omega = 0$).

For a rigid body motion $M$ in $d$ dimensions, let $R(t)$ be the orientation matrix (a rotation matrix) at time $t$. The time derivative of the orientation matrix, denoted $\dot{R}(t)$, is equal to the product of the angular velocity tensor $\Omega(t)$ and the orientation matrix $R(t)$: $$\Omega(t) R(t) = \dot{R}(t)$$ This relationship allows the orientation path to be recovered from the angular velocity tensor, given the orthogonality condition $R(t) R(t)^\top = I$.

For a rigid body motion $M$ in three-dimensional space, let $\mathbf{v}_y(t)$ be the velocity of a material point $y$ at time $t$. Suppose the orientation $R(t)$ and the center-of-mass trajectory $\mathbf{R}_{\text{com}}(t)$ are differentiable. Then the velocity of the point $y$ is given by the sum of the center-of-mass velocity $\mathbf{V}(t)$ and the cross product of the angular velocity vector $\boldsymbol{\omega}(t)$ with the point's position relative to the center of mass: $$\mathbf{v}_y(t) = \mathbf{V}(t) + \boldsymbol{\omega}(t) \times (\mathbf{r}_y(t) - \mathbf{R}_{\text{com}}(t))$$ where $\mathbf{r}_y(t)$ is the displacement of the point $y$ in the inertial frame at time $t$. In component form, for each $i \in \{1, 2, 3\}$: $$(\mathbf{v}_y(t))_i = (\mathbf{V}(t))_i + \left(\boldsymbol{\omega}(t) \times \left(\mathbf{r}_y(t) - \mathbf{R}_{\text{com}}(t)\right)\right)_i$$

For a rigid body motion $M$ in $d$ dimensions and a time $t$, the body-frame (co-rotating) angular velocity tensor $\Omega_{\text{body}}(t)$ is the $d \times d$ real matrix defined by $$\Omega_{\text{body}}(t) = R(t)^T \frac{d}{dt}R(t)$$ where $R(t)$ is the orientation matrix of the rigid body at time $t$. Unlike the spatial angular velocity tensor, this tensor is expressed with respect to the body's own rotating frame of reference.

For a rigid body motion $M$ in $d$ dimensions, the body-frame angular velocity tensor $\Omega_{\text{body}}(t)$ at time $t$ is equal to the product of the transpose of the orientation matrix $R(t)^T$ and the time derivative of the orientation matrix $\dot{R}(t)$: $$\Omega_{\text{body}}(t) = R(t)^T \frac{d}{dt}R(t)$$

For a rigid body motion $M$ in $d$ dimensions, let $R(t)$ be the orientation matrix at time $t$, and let $\Omega_{\text{body}}(t)$ be the body-frame angular velocity tensor. The time derivative of the orientation matrix, $\dot{R}(t)$, is related to the orientation and the body-frame angular velocity tensor by the equation: $$\dot{R}(t) = R(t) \Omega_{\text{body}}(t)$$

For a rigid body motion $M$ in $d$ dimensions at time $t$, let $R(t)$ be the orientation matrix of the body. The spatial angular velocity tensor $\Omega(t)$ is equal to the body-frame angular velocity tensor $\Omega_{\text{body}}(t)$ rotated into the inertial frame via the conjugation: $$\Omega(t) = R(t) \Omega_{\text{body}}(t) R(t)^\top$$

For a rigid body motion in $d$ dimensions, the body-frame angular velocity tensor $\Omega_{\text{body}}(t)$ at time $t$ is the spatial angular velocity tensor $\Omega(t)$ conjugated into the body frame by the orientation matrix $R(t)$: $$\Omega_{\text{body}}(t) = R(t)^\top \Omega(t) R(t)$$ where $R(t)$ is the rotation matrix representing the orientation of the body at time $t$.

Consider a rigid body $M$ in $d$-dimensional space with an orientation matrix $R(t)$ at time $t$. If the orientation $R(t)$ is differentiable with respect to time at $t$, then the body-frame angular velocity tensor $\Omega_{\text{body}}(t) = R(t)^T \dot{R}(t)$ is skew-symmetric, meaning: $$\Omega_{\text{body}}(t)^T = -\Omega_{\text{body}}(t)$$ This property implies that the tensor $\Omega_{\text{body}}(t)$ resides in the Lie algebra $\mathfrak{so}(d)$.

For a rigid body motion $M$ in three dimensions and a time $t$, the body-frame angular velocity vector $\omega_{\text{body}}(t) \in \mathbb{R}^3$ is defined as the vector dual to the body-frame angular velocity tensor $\Omega_{\text{body}}(t)$ under the vee map: $$\omega_{\text{body}}(t) = (\Omega_{\text{body}}(t))^\vee$$ where $\Omega_{\text{body}}(t) = R(t)^T \dot{R}(t)$ is the $3 \times 3$ skew-symmetric tensor representing the angular velocity in the body's rotating frame of reference, and $R(t)$ is the orientation matrix of the body at time $t$.

For a rigid body $M$ in three-dimensional space at time $t$, the body-frame angular velocity vector $\omega_{\text{body}}(t)$ is equal to the result of applying the vee map to the body-frame angular velocity tensor $\Omega_{\text{body}}(t)$: $$\omega_{\text{body}}(t) = (\Omega_{\text{body}}(t))^\vee$$ where $\Omega_{\text{body}}(t) = R(t)^T \dot{R}(t)$ is the $3 \times 3$ skew-symmetric matrix representing the angular velocity in the body's rotating frame, and the vee map $(\cdot)^\vee$ is the operation that extracts a vector from a $3 \times 3$ matrix.

Consider a rigid body $M$ in three-dimensional space. Let $R(t)$ denote its orientation at time $t$. If $R(t)$ is differentiable at time $t$, then the cross product matrix (also known as the hat map) of the body-frame angular velocity vector $\omega_{\text{body}}(t)$ is equal to the body-frame angular velocity tensor $\Omega_{\text{body}}(t)$: $$[\omega_{\text{body}}(t)]_{\times} = \Omega_{\text{body}}(t)$$ where $\Omega_{\text{body}}(t) = R(t)^T \dot{R}(t)$.

For a rigid body motion $M$ in $d$ dimensions, if the orientation $R(t)$ is constant in time (i.e., $R(t) = R$ for some fixed special orthogonal matrix $R$ and all times $t$), then the body-frame angular velocity tensor $\Omega_{\text{body}}$ is the zero matrix.

For a rigid body motion $M$ in three dimensions, if the orientation $R(t)$ is constant in time (i.e., there exists a fixed rotation matrix $R \in SO(3)$ such that $R(t) = R$ for all $t$), then the body-frame angular velocity vector $\omega_{\text{body}}$ is zero.

For a 3D rigid body motion $M$ with orientation $R(t)$ and body-frame angular velocity vector $\boldsymbol{\omega}_{\text{body}}(t)$, let $\mathbf{v} \in \mathbb{R}^3$ be a constant vector. If the orientation $R$ is differentiable at time $t$, then the time derivative of the orientation matrix $\dot{R}(t)$ applied to $\mathbf{v}$ is given by: $$\dot{R}(t) \mathbf{v} = R(t) (\boldsymbol{\omega}_{\text{body}}(t) \times \mathbf{v})$$ where $\dot{R}(t) = \frac{d}{dt} R(t)$ and $\times$ denotes the cross product in $\mathbb{R}^3$.