For rigid body we need to know position of three points and their velocities to determine everything. So that would make 12 DOF. Why do text books say it has six DOFs?
$\begingroup$
$\endgroup$
1
asked Oct 2, 2013 at 3:30
-
$\begingroup$ Possible duplicates of first title question (v2): physics.stackexchange.com/q/20954/2451 and links therein. Possible duplicate of second title question (v2): physics.stackexchange.com/q/8860/2451 $\endgroup$– Qmechanic ♦Oct 2, 2013 at 7:20
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
A rigid body has 6 configuration degrees of freedom because its most general configuration can be obtained by translating (3 degrees of freedom) and rotating (3 degrees of freedom) its initial configuration. A mathy way of saying this is that its configuration manifold is $\mathbb R^3\times \mathrm{SO}(3)$.
However, you are right that the phase space of a rigid body is 12-dimensional because each independent configuration degree of freedom corresponds to an independent momentum degree of freedom.
answered Oct 2, 2013 at 3:54
-
$\begingroup$ DOF is a vague term at best and should just not be used. $\endgroup$– AHusainAug 8, 2016 at 15:24
-
$\begingroup$ @AHusain A few years ago, I might have agreed with you, but I've come to realize that sometimes vague terms are fine (and even useful) to use in the appropriate context. In particular, in the context of classical mechanics, it's usually clear from the context that the degrees of freedom are the dimensions of a certain manifold (configuration or phase), and with this understanding, the vagueness is eliminated. Once this happens, and you know what you're talking about, the terminology is actually quite intuitively appealing in my opinion. $\endgroup$ Aug 8, 2016 at 15:33