We know that a monatomic compound can only have 3 degrees of freedom as we can consider it to be a point mass. However now that we consider a diatomic molecule, there are 3 degrees of freedom in translational movement, 1 degree in vibration and the last is in rotation. However why is it that rotation can only have 1 degree of freedom when it can rotate in an infinite number of directions?


marked as duplicate by John Rennie, Sebastian Riese, Kyle Kanos, ACuriousMind, Qmechanic Oct 6 '15 at 18:29

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there are 3 degrees of freedom in translational movement, 1 degree in vibration and the last is in rotation

Actually there are 3 translational, 2 rotational, and 1 vibrational degree of freedom for a two-atomic molecule.

Picture source: http://astarmathsandphysics.com/a-level-physics-notes/182-thermal-physics-and-gases/3036-degrees-of-freedom-2.html

The vibrational one is not shown in this picture, although it's easy to see what it is (atoms oscillating along the molecular bond with a phase difference of $\pi$, i.e. in antiphase, for more information see this).

Now for the question: It's true that the molecule can rotate in an infinite number of directions, but (think about it) it can also move translationally in an infinite number of directions as well. The key point is that all the translation (rotation) can be written as a superposition of the basic three (two for rotation) movement directions. So, in a way, any rotation of the molecule you may perceive is actually a combined, simultaneous rotation around both possible axes.

  • $\begingroup$ The vibrational degrees of freedom come into picture only at a higher temperature. At lower temperature, the diatomic molecule has only 5 degrees of freedom. $\endgroup$ – SchrodingersCat Oct 6 '15 at 14:05
  • $\begingroup$ Why are there two vibrational degrees of freedom? $\endgroup$ – Gert Oct 6 '15 at 14:28
  • $\begingroup$ @Aniket True, but irrelevant, since we're not discussing the availability of the degrees of freedom but their geometry. $\endgroup$ – Soba noodles Oct 6 '15 at 14:44
  • $\begingroup$ @Gert Whoops, what I meant is that there are two $\frac{1}{2}k_BT$ terms contributing to the energy of a molecule from its vibrations, one coming from the potential energy, one from kinetic. There is, of couse, only one vibrational degree of freedom, since the other one actually degenerates to translational movement. I'll correct that detail in an hour or two when I get home. $\endgroup$ – Soba noodles Oct 6 '15 at 14:44
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    $\begingroup$ @Sobanoodles: phew. I would have hated to be wrong on that. ;-) Thanks! $\endgroup$ – Gert Oct 6 '15 at 14:57

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