Angular velocity tensor

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

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

It allows us to express the cross product

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

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:

${\displaystyle T(t)={\begin{pmatrix}0&-\omega _{z}(t)&\omega _{y}(t)\\\omega _{z}(t)&0&-\omega _{x}(t)\\-\omega _{y}(t)&\omega _{x}(t)&0\\\end{pmatrix}}}$

Coordinate-free description

At any instant, ${\displaystyle t}$ , the angular velocity tensor is a linear map between the position vectors ${\displaystyle \mathbf {r} (t)}$  and their velocity vectors ${\displaystyle \mathbf {v} (t)}$  of a rigid body rotating around the origin:

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

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

The relation between this linear map and the angular velocity pseudovector ${\displaystyle \omega }$  is the following.

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

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

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

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

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

Taking the dual vector L* of L we get

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

Introducing ${\displaystyle \omega :=*L^{*}}$ , as the Hodge dual of L* , and apply further Hodge dual identities we arrive at

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

where

${\displaystyle \omega \times \mathbf {r} :=*(\omega \wedge \mathbf {r} )}$ 

by definition.

Because ${\displaystyle \mathbf {s} }$  is an arbitrary vector, from nondegeneracy of scalar product follows

${\displaystyle T\mathbf {r} =\omega \times \mathbf {r} }$ 

See also

• Angular velocity
• Rigid body dynamics