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 ?

  • $\begingroup$ Duplicate question, see math.stackexchange.com/questions/1970754/… $\endgroup$
    – SV-97
    Commented Oct 16, 2021 at 8:44
  • $\begingroup$ its not a duplicate question. I am asking about line segment not a line. $\endgroup$ Commented Oct 16, 2021 at 8:45
  • $\begingroup$ 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. $\endgroup$ Commented Oct 16, 2021 at 10:48

1 Answer 1


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).


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