# Degrees of freedom of a line segment in 3D

Suppose there is a line segment of fixed length $$l$$ in 3$$D$$. THen How many degrees of freedom does it have ?

To specify the line segment first we need to specify one endpoint which requires 3 parameters. Now the other endpoint lies in the circumference of the sphere of radius $$l$$ whose centre is the first end point. Now to choose the second endpoint we still need all 3 parameters ($$r,\theta,\phi$$) because for a specific $$\theta,\phi$$ there can be two solutions in which specifying $$r$$ makes it clear. Hence a line segment should have 6 degree of freedoms.

Am I right ?

• Duplicate question, see math.stackexchange.com/questions/1970754/… Commented Oct 16, 2021 at 8:44
• its not a duplicate question. I am asking about line segment not a line. Commented Oct 16, 2021 at 8:45
• You don't have two solutions for a specific θ, ϕ. But more importantly, even if you did, that would not constitute a degree of freedom. Commented Oct 16, 2021 at 10:48

Okay so think about it like this: You specify one point $$p$$ as the start for your line. As you noted you can define the other point as laying on the sphere or radius $$l$$ around $$p$$ - so naturally in this spherical coordinate system you'll only need two parameters to describe the other point and thus end up with $$5$$ degrees of freedom.
Another way (it's very similar though) to arrive at this answer is the following: Lets first disregard that you want to fix the length. Note that the general parametrisation for a line is $$\gamma : \Bbb R \to \Bbb R^3, t \mapsto t v + a$$ for some $$a, v \in \Bbb R^3$$. You can easily pick out a (directed) segment by restricting the domain (requiring two parameters for a total of 6), or by fixing the length of $$v$$ to some value and fixing the domain of $$\gamma$$ to $$[0,1]$$. So a directed line segment is uniquely identified by the parametrisation $$\gamma : [0, 1] \to \Bbb R^3, t \mapsto t v + a, \quad \text{for }a,v\in \Bbb R^3.$$ So we have three degrees of freedom for $$a$$ and three for $$v$$ for a total of $$6$$. By specifying the length we drop one degree of freedom and end up with $$5$$ - we can achieve the same by restricting $$v$$ to $$lS^2 \subseteq \Bbb R^3$$. Note that for any such $$\gamma$$ the function $$\widetilde{\gamma}(t) = t (-v) + \gamma(1)$$ will be the same line segment, just traversed backwards. But this doesn't change the number of degrees of freedom (see linked post).