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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.

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  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/20954/2451 and links therein. $\endgroup$
    – Qmechanic ♦
    Commented Oct 2, 2013 at 7:20
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    $\begingroup$ What does $^N C_2$ mean? I've never seen that symbol before. $\endgroup$
    – DanielSank
    Commented Nov 10, 2015 at 18:19
  • $\begingroup$ The following reference provides a proof of the problem posed above. arxiv.org/abs/1002.2002 $\endgroup$
    – user102980
    Commented Jan 5, 2016 at 12:57

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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.

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  • $\begingroup$ Yes. As I said, I am aware of this proof. Can you show how the minimal number of constraints is (3N-6)? $\endgroup$
    – Man
    Commented May 28, 2013 at 14:31
  • $\begingroup$ 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 $6 + 3(N-4)= 3N-6$. $\endgroup$
    – Pulsar
    Commented May 28, 2013 at 14:45
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    $\begingroup$ @Man I think this answer actually does answer your question. The problem is that you are naively counting $|r_i - r_j| = {\rm const}$ as $^N C_2$ constraints. The point, as Pulsar nicely makes, is that those equations are not all independent. So in other words among the $^N C_2$ equations in $|r_i - r_j| = {\rm const}$ some are degenerate, and so the true number of constraints is $^N C_2 -{\rm (number\ of\ degeneracies)}$. Pulsar has nicely shown explicitly how this works for $N=4,5$. If you want a more complete proof, the challenge is to show the degeneracy is $6 - 3N + ^N C_2$... $\endgroup$
    – Andrew
    Commented Nov 10, 2015 at 15:32
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    $\begingroup$ ... So you could do that for example by constructing a matrix representing the equations $|r_i - r_j| = {\rm const}$ and determining the rank. But, the argument Pulsar gives is actually an extremely clever and efficient way of determining the degeneracy of the equations, basically an inductive proof. So I think it's more a matter of reinterpreting what you are asking, than it is finding a different answer. $\endgroup$
    – Andrew
    Commented Nov 10, 2015 at 15:34

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