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I have a unit vector q that is represented in spherical coordinates: \begin{equation} q= \begin{bmatrix} \sin(\theta)\sin(\varphi) \\ \sin(\theta)\cos(\varphi)\\ \cos(\theta) \end{bmatrix} \end{equation} What I need to get is a transformation that goes from $\;\biggl\{\dfrac{\partial\theta}{\partial t};\dfrac{\partial\varphi}{\partial t}\biggr\}$ to $w$. Where $w$ is the rotation vector of q around the origin.

I know that $\dfrac{\partial q}{\partial t}=w\times q$ but as the cross product does not have an inverse i cannot isolate $w$.

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Given $$ {\bf q} = \left[ {\matrix{ {\sin \theta \sin \phi } \cr {\sin \theta \cos \phi } \cr {\cos \theta } \cr } } \right] $$ then $$ d{\bf q} = \left[ {\matrix{ {\cos \theta \sin \phi } \cr {\cos \theta \cos \phi } \cr { - \sin \theta } \cr } } \right]d\theta + \left[ {\matrix{ {\sin \theta \cos \phi } \cr { - \sin \theta \sin \phi } \cr 0 \cr } } \right]d\phi $$ and of course, being $\bf q$ unitary, we will have: $$ {\bf q} \cdot d{\bf q} = 0 $$

So $|d \bf q|$ represents the rotation angle, which is on the plane containing $\bf q$ and $\bf q+d\bf q$ and any linear combination of them, thus also $\bf q$ and $d \bf q$, which are normal to each other. Therefore their cross product is normal to the rotation plane and is equal to $$ \eqalign{ & {\bf w} = {\bf q} \times d{\bf q} = \left( {\left| {\bf q} \right|\left| {d{\bf q}} \right|\sin \left( {\mathop {{\bf q},d{\bf q}}\limits^ \wedge } \right)} \right)\;{\bf n} = \left| {d{\bf q}} \right|\;{\bf n} = {d\alpha} \; {\bf n} \cr & = \left[ {\matrix{ {\sin \theta \sin \phi } \cr {\sin \theta \cos \phi } \cr {\cos \theta } \cr } } \right] \times \left[ {\matrix{ {\cos \theta \sin \phi } \cr {\cos \theta \cos \phi } \cr { - \sin \theta } \cr } } \right]d\theta + \left[ {\matrix{ {\sin \theta \sin \phi } \cr {\sin \theta \cos \phi } \cr {\cos \theta } \cr } } \right]\times \left[ {\matrix{ {\sin \theta \cos \phi } \cr { - \sin \theta \sin \phi } \cr 0 \cr } } \right]d\phi = \cr & = \left[ {\matrix{ { - \cos \phi } \cr {\sin \phi } \cr 0 \cr } } \right]d\theta + {1 \over 2}\left[ {\matrix{ {\sin \phi \sin (2\theta )} \cr {\cos \phi \sin (2\theta )} \cr {\cos (2\theta ) - 1} \cr } } \right]d\phi \cr} $$ where the sign of $\alpha$ is according to the Right-Hand Rule (bringing "starting" v. to "ending" v.).

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  • $\begingroup$ Why is $ {\bf w} = {\bf q} \times d{\bf q} $? Is it because ${\bf q} \cdot d{\bf q} = 0$? $\endgroup$ – Iñigo Moreno May 5 '18 at 14:03
  • $\begingroup$ @IñigoMoreno: not exactly: in any case the rotation vector is normal to "starting vector" and "rotated vector", i.e. $\bf q$ and $\bf q+d\bf q$. It's value then, in this case, is computed taking into account that $\bf q$ and $d\bf q$ are orthogonal. Note there was a typo in my previous answer involving $|d\bf q|^2$: I amended for that. $\endgroup$ – G Cab May 5 '18 at 15:27

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