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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:

The definition of rigid body is

$$\left\lvert \vec{r_i}-\vec{r_j} \right\rvert = \text{constant} \ \forall\ i,j \, .$$

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 $6$.

I want to show that

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

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

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

DanielSank
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Man
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1 Answers1

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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.

Pulsar
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