# Getting rotation vector from spherical coordinates angles derivatives

I have a unit vector q that is represented in spherical coordinates: $$q= \begin{bmatrix} \sin(\theta)\sin(\varphi) \\ \sin(\theta)\cos(\varphi)\\ \cos(\theta) \end{bmatrix}$$ 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 $w$.

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.).
• Why is ${\bf w} = {\bf q} \times d{\bf q}$? Is it because ${\bf q} \cdot d{\bf q} = 0$? – Iñigo Moreno May 5 '18 at 14:03
• @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. – G Cab May 5 '18 at 15:27