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About the equipartition theorem

  1. Apr 25, 2008 #1
    1. The problem statement, all variables and given/known data

    (a) The rotational energy of a diatomic molecule such as [tex]H_{2}[/tex] is given by


    [tex]J^2_{\|}[/tex] stands for the angular momentum with respect to the axis of symmetry of the molecule and [tex]J^2_{\bot}[/tex] the angular momentum perpendicular to that axis. If you have a gas with N molecules held at temperature T, what is the mean energy per molecule?

    (b) A long thin needle floats within a gas at constant temperature. What will be the mean orientation of the needle's angular momentum vector? Parallel or perpendicular to the axis of symmetry of the needle?

    2. Relevant equations

    Equipartition theorem

    [tex]\left\langle x_i \frac{\partial \mathcal{H}}{\partial x_j}
    = kT\delta_{ij} \right\rangle [/tex]

    3. The attempt at a solution

    (a) Because of the equipartition theorem, each degree of freedom gives [tex]\frac{1}{2}kT[/tex] to the energy, so I think the mean energy per molecule should be [tex]\frac{3}{2}kT[/tex] (the free particle contribution) plus [tex]\frac{2}{2}kT[/tex] (the rotational energy).

    (b) I think the angular momentum vector will be oriented perpendicular to the axis of symmetry of the needle, but I don't have an argument for it.

    Thanks for your attention!
  2. jcsd
  3. Apr 26, 2008 #2
    (a) Each harmonic term in the hamiltonian should give [tex]\frac{1}{2}kT[/tex] to the mean energy per molecule, and then the rotational contribution to the mean energy per molecule should be 3kT.

    (b) The needle will rotate about the axis of symmetry of the molecule because the moment of inertia is greater than the moment of inertia about the line that passes trough the atoms.
  4. Apr 26, 2008 #3
    Do you know where I can find information about the ideal diatomic molecule? Thanks a million.
  5. Apr 26, 2008 #4
    I think that the rotation perpendicular to the axis of simetry would contribute twice as much (2*kT/2), since it has two possible directions. This gives an average of 3kT total (translation+rotation) energy per molecule (if vibration is neglected).
  6. Apr 26, 2008 #5
    Maybe I'm making a mistake, but one of those 2 directions was already counted when we considered the rotations about the axis of symmetry.

    By the way, I've been reading about this problem and found that books say that the diatomic molecule has (5/2)*kT, since they consider only the rotations about the axis of symmetry. Why do they do that?
  7. Apr 26, 2008 #6
    Let's say that the centers of H atoms of our molecule lie on x axes (this is the symmetry axes). There are 3 possible rotations: around x, y and z axes. x gives angular momentum parallel to symmetry axes, y and z are two degrees of fredom coresponding to angular momentum perpendicular to axes of symmetry: therefore they get 2*kT/2 and rotation around x gets kT/2, so the total energy is 3 kT (together with translation).

    5/2 kT corresponds to translation+2 rotations perpendicular to the axes of symmetry. The rotation around the axes of symmetry is left out.
    I think it is because of the discreet values of angular momentum in quantum mechanics (equipartition theorem is true only in classical aproximation, where all values are allowed). The formula for rotational energy is gama^2/2*J. J around the axes of symmetry is very small, so it is possible that even the smalles allowed value of gama gives an energy much higher than kT, so it the probability for rotation around that axes is small and can be neglected (exceptions are very high temperatures).
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