# degree of freedom in 6 dimensional space

Let us assume a cartesian space, where the directions are given by $\hat{i},\hat{j},\hat{k}$. The degree of freedom of a rigid body is $6$. The first three correspond to the position coordinates $\left(x,y,z\right)$, and the next three correspond to the velocity coordinates $\left(\dot{x},\dot{y},\dot{z}\right)$. Suppose I know the angular momentum of this body along $\hat{i},\hat{j},\hat{k}$. I also know the total energy of this body. Can I say that having knowledge of these four quantities reduces the degree of freedom to $2$?

• I also have the knowledge of total energy. Does that reduce the degree of freedom by one more, i.e. to 5? On the other hand, since for a conservative system, the angular momentum is conserved, then even if we know the angular momentum in three directions, the equation $l_1^2+l_2^2+l_3^2=constant$ does not make the three angular momentum independent. And so the degree of freedom reduces by only $2$ and if we add energy, then that makes it $3$. Is it correct? – user105997 Jul 4 '14 at 20:06
• Knowing total energy does reduce d.o.f. to five assuming you don't know anything about the velocity. Angular momentum: the fact that they add up to a constant does not reduce the d.o.f. You might be given $l_x$, $l_y$, and $|\vec{L}|$, that's still three d.o.f. If you are given all of those plus $|\vec{L}|$ then those four had better form a consistent set, and the d.o.f. is still only three. – garyp Jul 4 '14 at 20:31
• If I consider $l_1$, $l_2$ and $l_3$ to be random variables, does it mean that they are independent of each other, because they are perpendicular to each other? – user105997 Jul 8 '14 at 23:20
• Yes, they are independent. Unless the total angular momentum is also specified. Given $l_x$ and $l_y$ and $\vec{L}$, one can deduce $l_z$, so they are then not independent. – garyp Jul 9 '14 at 1:32