Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

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.

share|cite|improve this question
Possible duplicates: and links therein. – Qmechanic Oct 2 '13 at 7:20
What does $^N C_2$ mean? I've never seen that symbol before. – DanielSank Nov 10 at 18:19

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.

share|cite|improve this answer
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
@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 $6 - 3N + ^N C_2$... – Andrew Nov 10 at 15:32
... 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. – Andrew Nov 10 at 15:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.