Converting quaternions to spherical angles

Question

Consider a situation where a beam is shot at a cube C from an arbitrary position P. The cube detects the angle of incidence relative to its $ x $ axis. The cube can be rotated and moved, and the orientation (rotation) of the cube is being tracked.

Below is an example in two dimensions. In the reference orientation, $ C_\text{ref} $, the angle of incidence is $ 0° $. The cube is then moved to an arbitrary position and rotated with an arbitrary quaternion rotation $ \mathbf{q} $. The measured angle of incidence is now $ \gamma $. By subtracting the rotation angle $ \beta $, the angle of incidence relative to the reference orientation is found to be $ \alpha $. ($ \alpha = \gamma - \beta $) In three dimensions, the same problem would also have an elevation angle.

My question is this: how do I convert the (three-dimensional) rotation from quaternions $ \mathbf{q}=(q_0,q_1,q_2,q_3) $ to spherical angles $ (\theta,\phi) $?

Answer 1

Any rotation in three-dimensions can be represented as a rotation by an angle $\theta$ about an axis determined by a unit vector $\hat{n}=(n_1,n_2,n_3)$ -- that is three parameters and not two as the OP indicates. Such a rotation can be represented by $$ \exp(\theta\ \sum_{j=1}^3 n_j \cdot T^j)\ , $$ where $(T^1,T^2,T^3)$ are a basis in the fundamental (three) representation of $so(3)$ and are given by the matrices $(T_{i})_{jk}=\epsilon_{ijk}$. The connection with quaternions that I know makes use of the isomorphism of the (complex) Lie algebras $so(3)$ and $su(2)$. We just replace $\mathbf{T}$ with the corresponding generators in the fundamental (two) dimensional representation of $su(2)$. A basis for this Lie algebra are given in terms of the Pauli matrices. Using the identification $$T^j \longleftrightarrow i\sigma^j/2\ ,$$ one obtains $$ \exp(\theta\ \sum_{j=1}^3n_j \tfrac{i\ \sigma^j}2)= \cos \tfrac{\theta}2 \mathbf{1} + i \sin \tfrac\theta2 \sum_{j=1}^3 n_j\ \sigma^j $$ where I have used a standard identity involving Pauli matrices. The right hand side is a quaternion, if you wish. Note that this is not a 1-1 map as group manifold of $SU(2)$ is a double cover of $SO(3)$.

Answer 2

$ \newcommand\Cref{C_{\text{ref}}} $The quaternion $q = q_0 + q_1i + q_2j + q_3k$ takes the $x$-axis of $\Cref$ to that of $C_q$ via $$ qi\bar q = (q_0^2 + q_1^2 - q_2^2 - q_3^2)i + 2(q_1q_2 + q_0q_3)j + 2(q_1q_3 - q_0q_2)k $$ where we represent vectors $(x,y,z)$ as imaginary quaternions $xi + yi + zi$ (and assume $q$ is normalized). Now we just convert from Cartesian to spherical coordinates: $$ \tan\phi = 2\frac{q_1q_2 + q_0q_3}{q_0^2+q_1^2-q_2^2-q_3^2},\quad \tan\theta = \frac12\frac{\sqrt{(q_0^2+q_1^2-q_2^2-q_3^2)^2 + 4(q_1q_2+q_0q_3)^2}}{q_1q_3-q_0q_2}. $$ (Something like this, I'm not going to promise I got this conversion exactly correct.) Here I've used $\phi$ for the azimuthal angle and $\theta$ for the polar angle.