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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 $2$?

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The number of degrees of freedom of a rigid body is nine. The six you mention plus three more, for example the Euler angles of the axis of rotation plus the angular velocity. If you know the angular momentum, that reduces the number to six.

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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
Also, knowing the total angular momentum but none of the individual angular momenta reduces the d.o.f. by one. – garyp Jul 4 '14 at 20:39
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

The degrees of freedom is 9 for rigid body. If we know the three angular momenta, the conversion follows but we don't know the constant value a priori. Energy can be used to solve the problem of knowing values of all the degrees of freedom, I.e., completely defining the system, but the amount of information required to be fed into the system for a complete description of the system is 6. You may acquire this info in the form of knowing value of a particular coordinate or energy of the system. But for a complete description of the system, you need 9 pieces of information, in case of this rigid body. For a particle, we need only 6 coz it is dimensionless and we ignore angular moments.

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Do you mean with a particle a theoretical body with zero dimensions or do you mean electrons, photons, ...? If you mean the last do we really not need the angular moments? – HolgerFiedler Jul 4 '14 at 20:35
I meant a theoretical zero dimension body – Harshfi6 Jul 4 '14 at 20:55

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