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Suppose we consider a rigid body, which has $N$ particles. Then the number of degrees of freedom is $3N - (\mbox{# of constraints})$.
As the distance between any two points in a rigid body is fixed, we have $N\choose{2}$ constraints giving $$\mbox{d.o.f} = 3N - \frac{N(N-1)}{2}.$$ But as $N$ becomes large the second term being quadratic would dominate giving a negative number. How do we explain this negative degrees of freedom paradox?
You've duplicated constraints because if any one particle is constrainined in all three dimensions with all the other particles this constrains all the particles. The number of constraints is 3(N - 1).
To give an example, take three particles a, b and c. If a is fixed relative to b and is also fixed relative to c, then b and c are fixed relative to each other without having to introduce new constraints.
Each particle that makes up a mechanical system, can be located by three independent variables labelling a point in space.
You can choose any particle in the rigid body to start with and move it any where you want, giving three independent variables needed to specify its location.
Choosing a second particle, you choose another set of three independent variables to specify its location, the obvious being spherical coordinates with the origin at the first particle. The first constraint is that the radius is a constant, leaving two remaining independent variables.
Choosing a third particle, you have complete freedom to rotate it by any angle about the axis through the first and second particles giving just one degree of freedom, the other two variables constrained.
For the remaining (N-3) particles, all three coordinates are constrained.
Therefore, the total number of degrees of freedom for a rigid body is 3+2+1 = 6, with 0+1+2+3(N-3) = (3N-6) constraints.
So that the degrees of freedom becomes 3N - (3N-6) = 6
The problem is that you are double counting a lot of your constraints. If the (vector) displacements between particles A and B, and between B and C is fixed, then the displacement between A and C is fixed. Therefore the constraint on distance between A and C is redundant, and you can't count it separately.
One could do this by mathematical induction. Begin with four particles that have distances between each other that do not change. Simple enumeration will show that there are only six degrees of freedom. Now add another particle that has its distances relative to the others that are fixed. There are no unconstrained degrees of freedom that this particle bring to the system. We can do the same for a system of N particles. This is not rigorously stated in mathematical parlance, but contains the principle of the proof.
As others have already pointed out, you are overcounting the constraints. I'll try to explain that with this illustration.
In the case of $N=2$, the positions of the two points already determine the distance between them, and therefore there are no constraints.
In the case of $N=3$, there are $3$ constraints: the fixed distance between the one point and the two others (in black), and the remaining fixed distance (in blue). For $N=4$, logic is essentially the same.
The things get different with $N=5$. What you were doing is counting the connection between the two points that are unconnected in the picture as a constraint. The reason why you can't do that is that the position of those points is already determined by the three lines that are connected to them. The same conclusion can be reached for $N>5$.
These constraints are not independent.
You're double counting here. Lets take three particles. You're counting $\binom{3}{2}=3$ DOFs, right? But fixing the vector distance between particle 1 and two, and then fixing it between 2 and 3 includes fixing it between 1 and 3. Mathematically, $\vec{d}_{1,3}=\vec{d}_{1,2}+\vec{d}_{2,3}$
The easier way to count DOFs is like this. For a molecule with N particles, number of DOFs is $3N$. Out of these, 3 will be translational. For a point molecule (i.e, a single atom), subtract 3 as it has 0 rotational DOFs. For a perfectly linear molecule, subtract 1, as it has 2 rotational DOFs (Rotation along its axis is irrelevant). Now, we usually neglect vibrational DOFs (at normal temperatures). Vibrational DOFs are whatever DOFs are remaining. Thus, we always have a total of 3N DOFs, out of which we may count only the translational (3) and rotational (2 or 3) DOFs. See the table here.
