I have a collection of systems made by 2 identical line segments in 3D space. Could you help me understand which are the simplest way to describe each system using the least amount of variables possible?
The distance between the geometrical mean of each line and their relative angle seem clear candidates to me, but these do not capture not all the degrees of freedom and I am struggling to find the best descriptors.
I hope the description of the problem is clear enough. Thanks a lot for your help!
Best, Raimon
Two quantities will always appear no matter how we do this: the length of the line segments and the distance between the midpoints. Call these $l$ and $d$ respectively. $l$ and $d$ can take any value in $(0,\infty)$, and if two pairs of segments are isometric, they must have $l$ and $d$ the same. This is clear.
Let's call the two segments $A$ and $B$, and denote their midpoints by $m(A),m(B)$. Translate so that $m(A)$ is at the origin, and rotate so that $A$ lies on the $x$-axis. We can rotate about the $x$-axis, thus preserving $A$, to ensure that $m(B)$ lies in the upper $(x,y)$-half-plane, and then rotate by $\pi$ about the $y$-axis to ensure that it's in the $(+,+)$-quadrant. Then $m(B)$ is just determined by the angle $\theta$ it subtends with the positive $x$-axis, which can take any value in $[0,\pi/2]$ (remember, we already determined its distance $d$ from the origin).
Now we just need to parametrise the position of $B$ centred on $m(B)$ (we already determined its length $l$). If $\theta\in(0,\pi/2)$ then the only isometry we can apply which preserves $A$ and $m(B)$ is reflection in the $(x,y)$-plane. By this we can ensure that one of the endpoints of $B$ lies in the upper right-hand quarter of the sphere of radius $l/2$ around $m(B)$. This endpoint is then determined by 2 angles: $\psi\in[-\pi/2,\pi/2)$, the angle it subtends with the plane normal to $m(B)$ viewed as a vector, and $\varphi\in[0,\pi/2]$, the angle it subtends with the $(x,y)$-plane. Note that if $\psi=-\pi/2$ then the freedom in $\varphi$ disappears, so by convention take $\varphi=0$ if $\psi=-\pi/2$.
If $\theta=0$ then we can also rotate by any angle about the $x$-axis, and so force $\varphi=0$, and that's all. Finally, if $\theta=\pi/2$, we can rotate by $\pi$ about the $y$-axis, or reflect in the $(y,z)$-plane; but neither of these actions removes any freedom.
So in summary, the space of pairs of identical line segments in $\mathbb{R}^3$ up so Euclidean isometry is 5-dimensional, with parameters $l,d,\theta,\psi$ and $\varphi$ as described above. The tuple $(l,d,\theta,\psi,\varphi)$ can take any value in (the most horrible set ever written down): $$(0,\infty)^2\times\Bigg[\bigg[ (0,\pi/2]\times\Big[\big[(-\pi/2,\pi/2)\times[0,\pi/2]\big]\cup\big[\{-\pi/2\}\times\{0\}\big]\Big] \bigg]\cup\bigg[\{0\}\times[-\pi/2,\pi/2)\times\{0\}\bigg]\Bigg].$$
A more reader-friendly version of this is:
I think there is no redundancy in there, but it's always hard to quotient out by everything. But there's certainly nothing missing!
Edit: posting my comment
I've just realised that this doesn't force the line segments to be disjoint, and more than that, if $\theta=\psi=0$ and $d\leq l/2$ you have that $A\cup B$ is a line segment. I don't know if you want to allow that... Also, you should allow $d=0$ if you don't mind the segments having the same midpoint.