I am trying to visualize and calculate how a unit sphere would appear to an observer if it was spinning so fast that the tangential speed at the equator approached the speed of light.
Here is the setup: A unit sphere is centered at the origin $(0,0,0)$, and an observer is situated at $D = (0,d,0)$ where $d$ is the distance from the center of the sphere along the $y$-axis. The sphere rotates around the $z$-axis such that the motion takes place in the $x,y$ plane.
A moving point on the sphere can be described in spherical coordinates by: $$ r(\theta, t) = \begin{pmatrix} \sin(\theta) \cos(\omega t) \\ \sin(\theta) \sin(\omega t) \\ \cos(\theta) \end{pmatrix} $$ and the corresponding tangential velocity by: $$ v(\theta, t) = \begin{pmatrix} - \omega \sin(\theta) \sin(\omega t) \\ \omega \sin(\theta) \cos(\omega t) \\ 0 \end{pmatrix} $$ from within the sphere. From outside, it would just be $ r(\theta,t) + D $.
Considering relativistic length contraction, we should observe it in the $x,y$ plane, but not when the tangential velocity vector aligns with the observer's line of sight, which is given by $ r(\theta,t) - D $.
My difficulty lies in calculating the apparent contracted lengths, taking into account the directionality of length contraction. I understand that the Lorentz factor $\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ should be applied, but since the tangential speed has $x,y$ components, and these components shouldn't contribute to contraction when aligned with the observer's view, I'm confused about the correct application of $\gamma$.
I'm considering that there must be a dot product or cross product between the tangential velocity vector and the observer's line of sight involved in this calculation, but I'm unsure of how to proceed.
I'd be very grateful for any insights or guidance on how to calculate this.
I have resolved my question with the following approach:
Let us define $$ \mathbf{L} = \frac{\mathbf{r} - \mathbf{D}}{\|\mathbf{r} - \mathbf{D}\|} $$ as the unit vector from the observer to a point on the spinning sphere. Then if we set $$ \mathbf{V}_{\text{perpendicular}} = \mathbf{v} - (\mathbf{v} \cdot \mathbf{L}) \mathbf{L} $$ we have that $ \mathbf{V}_{\text{perpendicular}}$ zero iff it aligns with $\mathbf{L}$ giving the desired property.
The x and y components of $ \mathbf{V}_{\text{perpendicular}}$ can then be easily used to determine the corresponding relativistic length contractions. This effectively solve the problem I had.