We all know that if we consider a mono-atomic molecule, it has $3$ translational degrees of freedom only, along the $3$ principal coordinates of the Cartesian coordinate system.

In case of a diatomic molecule, it has $6$ total degrees of freedom, $3$ translational, $2$ rotational and finally $1$ vibrational degree of freedom.

I'm thinking that aren't there infinite directions in which a molecule may move (other than the $3$ principal coordinates) like in the direction of $$(x~\hat i+y~\hat j +z~\hat k)$$ in general?

These directions can also become the directions of translation or the axes of rotation for the molecules. They can also vibrate. So why aren't these taken into account? Why aren't rotations about an arbitrary axis or translation along arbitrary directions taken into account while finding energy?

Or is it that in such cases of translation, the direction, and hence the kinetic energy is already taken component-wise?

Is the axis of rotation also found in component form?

If so, how is the energy of a molecule while rotation about an arbitrary axis determined? I'm not sure if it can be added component-wise.

Any help is appreciated.

  • $\begingroup$ I count you gave the diatomic molecule 6 degrees of freedom $\endgroup$ – Bob D Nov 4 '19 at 9:37
  • $\begingroup$ This is general. You can choose a system of reference but you don't need more of them. Like to describe the motion of a projectile $\endgroup$ – Alchimista Nov 4 '19 at 10:26

A degree of freedom in this context is a quadratic term in energy. The translational kinetic energy for a free particle in 3D is $p^2/2m = (p_x^2 + p_y^2 + p_z^2)/2m$, three quadratic terms.

A diatomic has two vibrational degrees of freedom (at sufficiently high temperatures): one for kinetic energy and one for potential energy.

  • $\begingroup$ +1 Thank you very much for the answer. I have a few questions...how can the rotation along an arbitrary axis be broken down into components where the axes lie along the principal axes like we broke the translation into components along the three axes? $\endgroup$ – user8718165 Nov 4 '19 at 11:12
  • $\begingroup$ @user8718165 Just write down the expression for the energy. I am not in the mood to do the algebra and TeX formatting for that, but it is pretty standard. $\endgroup$ – user137289 Nov 4 '19 at 11:39

Motion in any direction would be a combination of motions in three perpendicular directions $x,y,z$. Similarly, rotation around any axis can be regarded as a combination of three rotations around $x,y,z$. That is why you don't need to count other directions.

  • $\begingroup$ $+1$ Thank you very much! I was a bit sure about the translation case. But I don't know how to add rotations component wise. Could you please elaborate on how is the addition done for rotations. $\endgroup$ – user8718165 Nov 4 '19 at 10:53
  • $\begingroup$ See this question physics.stackexchange.com/questions/286/… which says that why angular velocities can be added like vectors. $\endgroup$ – richard Nov 4 '19 at 10:58
  • $\begingroup$ Did you mean this? $\endgroup$ – user8718165 Nov 4 '19 at 11:04
  • $\begingroup$ Yes. Large rotations are not like vectors but the differential rotations are. $\endgroup$ – richard Nov 4 '19 at 11:05
  • 1
    $\begingroup$ Large rotation can be broken into small rotations then you can treat them like vectors at each step. $\endgroup$ – richard Nov 4 '19 at 11:53

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