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I want a different proof of 6 degrees of freedom of a solid object made of $\ N$ particles. I am thinking along these lines:

Definition of rigid body is

$\ modulus[\vec{r_i}-\vec{r_j}]=constant \ \forall\ i,j$

This gives me $\ ^NC_2$ constraints. There exist in total $\ 3N$ equations. So the number of free variables should be $\ n= 3N- \ \ ^NC_2=\frac{N(5-N)}{2}$ Which is clearly not the answer as $\ n$ is $\ N$ dependent, but it should be $\ 6$.

What I want to do is show that :

$$\ number\ of\ constraints \ actually\ required= 3N-6$$

which is the correct answer since I know $\ n=6$

I am aware of the proof given in Goldstein, Rana Joag etc. What I am asking is how to do it following this approach.

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Possible duplicates: and links therein. –  Qmechanic Oct 2 '13 at 7:20

1 Answer 1

You're imposing too many constraints. Suppose you have $N=4$ particles. These have $3N=12$ positions, and $N(N-1)/2=6$ constraints, forming a tetrahedron. Thus you have $12-6=6$ degrees of freedom, as expected.

Now add a fifth particle. This adds three more positions, but it is sufficient to put only three constraints on them, e.g. $|\vec{r}_5 - \vec{r}_1|$, $|\vec{r}_5 - \vec{r}_2|$, and $|\vec{r}_5 - \vec{r}_3|$. This will determine the position of particle 5 with respect to 1, 2, and 3, forming another tetrahedron. But it will also automatically determine the position of particle 5 with respect to particle 4.

In other words, for every new particle you need to add three new constraints, so that the number of degrees of freedom remains 6.

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Yes. As I said, I am aware of this proof. Can you show how the minimal number of constraints is (3N-6)? –  Man May 28 '13 at 14:31
I don't know how to formulate it differently. For 4 particles, you need 6 constraints, so they form a tetrahedron, which is a rigid body. For an extra particle, you need to add three constraints so that it forms a rigid tetrahedron with 3 other particles. So the total amount of constraints for $N>4$ is $6 + 3(N-4)= 3N-6$. –  Pulsar May 28 '13 at 14:45

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