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The Equipartion principle for diatomic molecules

  1. Apr 12, 2014 #1
    It is taught that the classical treatment of the diatomic atom would give a heat capacity of $7/2$ due to 7 degrees of freedom, (three translational momentum, two rotational momentum, on vibration momentum and on vibration position).

    This is based on the Hamiltonian looking like:
    H = \frac{\mathbf{P}^2}{2m} + \frac{L_1^2}{2I_1} + \frac{L_2^2}{2I_2} + \frac{p^2}{2m} + \frac{m\omega r^2} {2}
    To compute the partition function

    But the original Hamiltonian could be written directly as
    H = \frac{\mathbf{P_1}^2}{2M_1} + \frac{\mathbf{P_2}^2}{2M_2} + k(\mathbf{r_1} - \mathbf{r_2})^2
    Using this there is clearly 9 degrees of freedom that could contribute to the heat capacity (two independent three dimensional momentums, and one three dimensional position).
    Is there something I am missing?
  2. jcsd
  3. Apr 13, 2014 #2
    Yes, you are missing the fact that k(r1−r2)2 has only one degree of freedom because of the symmetries of the Hamiltonian. Only the magnitude of the distance between the particles matter. The Hamiltonian is completely symmetric under rotations of one particle around the other.
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