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What is the degrees of freedom of a three dimensional polyatomic molecule when only one vibrational mode is excited?

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For a rigid body, there are three translational degrees of freedom and three rotational degrees of freedom. That is six purely kinetic degrees of freedom. There is also one vibrational mode, which has both potential and kinetic energies associate with it; for equipartition of energy in an ideal gas, this would count as two degrees of freedom. That makes eight total, so heat capacity of a gas of $N$ such molecules would be $4Nk_{B}$, each quadratic term in the energy contributing $\frac{1}{2}Nk_{B}$.

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  • $\begingroup$ If it is a rigid body there is no possibility of vibration. If it is not a rigid body and one is counting the number of degrees of freedom in the Hamiltonian (which is not the same as the number of configurational degrees of freedom) vibrations do not add two degrees of freedom to the configurational ones but only one. $\endgroup$ – GiorgioP Jan 12 at 7:47
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Vibrational modes correspond to a change of variables to describe the possible motions of a molecule, provided interatomic interactions could be modeled via harmonic forces. As an approximation, such a harmonic modeling is very well justified for small deviations from the equilibrium molecular configuration.

Since the passage from the original coordinates to normal modes is nothing but a change of variables, the transformation has to keep the number of independent quantities. A polyatomic molecule made by $N$ atoms requires $3N$ independent cartesian coordinates for the complete description of its spatial configuration. Correspondingly, one has exactly the same number of normal modes (few of them, five or six, depending on the molecule being linear or non-linear, actually correspond to global translations and global rotations of the molecule, but are considered normal modes at zero frequency).

If one counts modes with the same frequency (degenerate modes) as separate (*), each mode depends only on one quantity. Therefore, each mode corresponds to exactly one degree of freedom.

After this clarification on how counting of normal modes goes, I would add that I would never say that a one dimensional polyatomic molecule with only one excited mode is a system with one degree of freedom. It is the normal mode which corresponds to one degree of freedom, not the molecule. The molecule as a whole, still requires $3N$ quantities to specify its configuration, independently on how many of those quantities are in their minimum energy state or not. Describing a molecule with a single excited mode as a system with only one degree of freedom, sounds equivalent to say that a system of two independent atoms in the space becomes a system of $3$ degrees of freedom as soon as one of the atoms is at rest. And even in the cases described by a freezing of degrees of freedom, I would have better to say that there is a reduction of the number of effective (or active) degrees of freedom.

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