Degrees of freedom in a molecule of $N$ atoms

Question

I am trying to understand why the number of degrees of freedom in a molecule, $3N$, is the same as the number of degrees of freedom of $N$ independent particles. Why do the molecular constraints like bond length and bond angle not take away any degrees of freedom?

My question stems from explaining why a molecule has a certain number of vibrational modes/vibrational degrees of freedom: $3N-5$ for linear molecules and $3N-6$ for nonlinear molecules, and the easiest way to show that is by postulating that the total number of degrees of freedom is $3N$ and going from there. But why are there $3N$ degrees of freedom? A proof would be nice, but an intuitive explanation is sufficient.

I understand that the degrees of freedom in a molecule are coupled, that makes sense, but what am I missing? Are these "constraints" not actual constraints since there is no definitive bond lengths and bond angles?

Answer 1

Your final questions go in the right direction to answer your question. Indeed, even though we speak about bond angles or bond distance, such quantities generally do not have a fixed value. They oscillate around some average value. Therefore, we do not have the typical situation of rigid constraints that allow us to reduce the degrees of freedom. Instead, each atom in a molecule keeps its freedom of moving in three independent directions, and the total number of independent displacements remains $3N$.

Answer 2

If an atom is free to move in any direction within a molecule, even thought there may be constraints on how much it can move, then it contributes $3$ degrees of freedom to the molecule's total number of degrees of freedom.

Degrees of freedom are only reduced if a bond length or bond angle has to take a fixed value. For example, a molecule with $3$ atoms normally has $3 \times 3 = 9$ degrees of freedom; $3$ translational, $3$ rotational and $9-6=3$ vibrational. However, if two of the bond lengths are fixed then we only need $7$ numbers to define the configuration of the molecule; $3$ to locate its centre of mass, $3$ to specify its orientation in space and $1$ to specify the distance between the two atoms that are not a fixed distance apart. Fixing two bond lengths has removed two vibrational degrees of freedom.

Answer 3

The fact that the atoms interact with one another does not restrict the number of ways there are for those atoms to move about a particular “point” (or, more aptly, molecular geometry): each atom can still move in 3-D. But the fact that we have a well-defined molecule means that often the most appropriate set of coordinates is a body-fixed set of coordinates containing three translational coordinates for the whole molecule, three rotational coordinates for the whole molecule (two in linear cases due to symmetry), and 3$N-6$ ($3N-5$ in the linear case), where $N$ is the number of atoms, coordinates for the internal vibrational motions of the molecule. I say most appropriate because this choice essentially “de-couples” the motions of the degrees of freedom (approximately in the case of vibrations and rotations; see Wilson’s Molecular Vibrations) so that the dimensionality of the differential equations needed to describe these motions is reduced. This saves exponential computational cost when applying this in the real world. The choice of coordinates that we use is certainly not unique. Different authors use sometimes rather exotic choices like Jacobi-Radau coordinates or even just the primitive Cartesians to describe systems depending on their needs. The point is that the translational, rotational, and vibrational coordinates align nicely with chemical intuition about how a molecule “should” move.

You point out that there is a sense of restriction in the sense that bonds restrict the degree of internal deformation of the molecule and you are correct; the chemically intuitive set of coordinates introduces many problems when the motions move far away from the equilibrium geometry of the molecule. Nevertheless, for well-defined small-amplitude motions these coordinates work quite well, which allows their use in most contexts of chemical interest.

Answer 4

As you know, the number 3N comes from the fact that each of the N atoms may have translational, rotational and vibrational degrees of freedom.

Bonding couples the atoms together, however each are still free to move in all of the above manners, albeit in a somewhat restricted manner, i.e. an atom may not translate itself too far away, etc. However, limitations on the motion are not the same as removing the atom's ability to move (a true constraint), thus bonding does not reduce the number of variables in each atom's configuration space.

In practice, however, not each of the degrees of freedom are equally relevant in all physical situations, for example at low energies, there is not as much energy partitioned to the vibrational aspect as there might be at higher energies, but in each case the atoms can actually vibrate; as they can also engage in the other forms of motion, thus providing for a total of 3N degrees of freedom.

Answer 5

One free particle has 3 degrees of freedom. Now attach that particle to a spring, say in a 3D harmonic oscillator. It still has 3 degrees of freedom, even though it bound.

Extrapolate from there.