Angular velocity tensor

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In physics, the angular velocity tensor is defined as a matrix T such that:

$\boldsymbol\omega(t) \times \mathbf{r}(t) = T(t) \mathbf{r}(t)$

It allows us to express the cross product

$\boldsymbol\omega(t) \times \mathbf{r}(t)$

as a matrix multiplication. It is, by definition, a skew-symmetric matrix with zeros on the main diagonal and plus and minus the components of the angular velocity as the other elements:

$T(t) = \begin{pmatrix} 0 & -\omega_z(t) & \omega_y(t) \\ \omega_z(t) & 0 & -\omega_x(t) \\ -\omega_y(t) & \omega_x(t) & 0 \\ \end{pmatrix}$

[edit] Coordinate-free description

At a given time instance $t$ , the angular velocity tensor is a linear map between the position vectors $\mathbf{r}(t)$ and their velocity vectors $\mathbf{v}(t)$ of a rigid body rotating around the origo:

$\mathbf{v} = T\mathbf{r}$

where we omitted the $t$ parameter, and regard $\mathbf{v}$ and $\mathbf{r}$ as elements of the same 3-dimensional Euclidean vector space $V$ .

The relation between this linear map and the angular velocity pseudovector $ω$ is the following.

Because of T is the derivative of an orthogonal transformation, the

$B(\mathbf{r},\mathbf{s}) = (T\mathbf{r}) \cdot \mathbf{s}$

bilinear form is skew-symmetric. (Here $\cdot$ stands for the scalar product). So we can apply the fact of exterior algebra that there is a unique linear form $L$ on $Λ 2 V$ that

$L(\mathbf{r}\wedge \mathbf{s}) = B(\mathbf{r},\mathbf{s})$ ,

where $\mathbf{r}\wedge \mathbf{s} \in \Lambda^2 V$ is the wedge product of $\mathbf{r}$ and $\mathbf{s}$ .

Taking the dual vector L* of L we get

$(T\mathbf{r})\cdot \mathbf{s} = L^* \cdot (\mathbf{r}\wedge \mathbf{s})$

Introducing $ω: = * L *$ , as the Hodge dual of L* , and apply further Hodge dual identities we arrive at

$(T\mathbf{r}) \cdot \mathbf{s} = * ( *L^* \wedge \mathbf{r} \wedge \mathbf{s}) = * (\omega \wedge \mathbf{r} \wedge \mathbf{s}) = *(\omega \wedge \mathbf{r}) \cdot \mathbf{s} = (\omega \times \mathbf{r}) \cdot \mathbf{s}$