Angular velocity tensor: Difference between revisions

Revision as of 08:56, 17 February 2008

Revision as of 17:21, 17 February 2008

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:<math> \mathbf{v} = T\mathbf{r} </math>

where we omitted the <math>t</math> parameter, and regard <math> \mathbf{v} </math> and <math> \mathbf{r} </math> as elements of the same 3-dimensional [[vectorspaceEuclidean vector space]] <math>V</math>.

The relation between this linear map and the angular velocity [[pseudovector]] <math>\omega</math> is the following.

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Because of ''T'' is the derivative of an [[orthogonal transformation]], the

:<math>B(\mathbf{r},\mathbf{s}) = (T\mathbf{r},) \cdot \mathbf{s}) </math>

[[bilinear form]] is [[skew-symmetric]]. (Here <math>\cdot</math> stands for the [[scalar product]]). So we can apply the fact of [[exterior algebra]] that there is a unique [[linear form]] <math>L</math> on <math>\Lambda^2 V </math> that

:<math>L(\mathbf{r}\wedge \mathbf{s}) = B(\mathbf{r},\mathbf{s})</math> ,

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Taking the [[dual vector]] ''L''* of ''L'' we get

:<math> (T\mathbf{r},)\cdot \mathbf{s}) = ( L^*, \cdot (\mathbf{r}\wedge \mathbf{s}) </math>

Introducing <math> \omega := *L^* </math>, as the [[Hodge dual]] of ''L''* , and apply further Hodge dual identities we arrive at

:<math> (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}) </math>

where

:<math>\omega \times \mathbf{r} := *(\omega \wedge \mathbf{r} ) </math>

by definition.