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Thermodynamics & Degrees of Freedom

  1. Nov 1, 2008 #1

    I am a bit unclear on degrees of freedom in thermodynamics/stat mech, can someone critique my rational? Here I go:

    Essentially, the equipartition theorem states that per each degree of freedom it has an energy of 1/2kT associated with it. So, for a monatomic gas, there are three degrees of freedom associated with the translational components of each atom (along X, Y, or Z). There are no rotational components since nearly all the mass of the atom is located at the center.

    For a diatomic molecule, we can visualize it as simply being two atoms connected with a tiny spring. The molecule still has three translational degrees of freedom (X, Y and Z), but now we must consider the rotational and vibrational components. It won't rotate about its axis for the same reasons a single atom cannot. It can rotate about an axis perpendicular to the spring connecting the molecules though: if both our atoms are on the X-axis, then it can rotate about an axis pointing along either the Y-axis or the Z-axis. So, there are two rotational degrees of freedom. For vibrational, this is only along one axis (the X-axis), so there is only one vibrational degree of freedom. So, there are 3+2+1=6 degrees of freedom in a diatomic molecule.

    For a triatomic molecule, we still have three translational degrees of freedom, but this time we have three rotational degrees of freedom and also three vibrational degrees of freedom.

    Is my thinking correct? Thanks yall


    EDIT: Also, unless we aren't talking about high temperatures, can the vibrational and rotational components be neglected?
    Last edited: Nov 1, 2008
  2. jcsd
  3. Nov 1, 2008 #2


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    To be precise, the equipartition theorem states that each quadratic degree of freedom has an average energy of 1/kT. So, for example, if you were dealing with a nonlinear perturbation to the harmonic oscillator (in 1d) , V(x) = ax^2 + bx^3, this would not have an average energy of (1/2)kT, due to the x^3 term.

    Your counting of the states for a diatomic molecules looks good for including all (classical) translational, rotational and vibrational degrees of freedom. Typically we take diatomic molecules to have 5 (quadratic) degrees of freedom, as we picture they are joined by a rigid rod instead of a spring. As the temperature gets high, we're pumping more energy into the system, which activates the vibrational mode between the atoms. So, at high temperatures all are active. At lower temperatures the first five are active, but you have to be careful as you decrease the temperature, as things begin break down because you start entering the quantum regime and the equipartition theorem no longer applies.
  4. Nov 1, 2008 #3


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    Everything sounds good to me :approve:
    For small and light molecules, one can ignore the vibrational (and electronic) degrees of freedom at room temperature, but one must take into account the rotational degrees of freedom.

    EDIT: Seems that I was a little slow
  5. Nov 1, 2008 #4
    If a diatomic molecule is vibrating along one axis, would the degree of freedom be 1 or 2 (in the limit of high temperatures, of course)? I just read that you should count it as 2, accounting for both the potential and kinetic components.
  6. Nov 1, 2008 #5


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    Right. Good point. I guess we forgot that one. We should be more careful here:

    When considering the translational degrees of freedom, you consider only the translational motion of the centre of mass, not each atom in the molecule, so that contributes three degress of freedom, one for each direction.

    The rotational motion depends on the shape of an object, but for a diatomic molecule there are two ways to rotate it, so you get 2 more degrees of freedom for the two Euler angles you're using.

    Then, considering vibrations about the centre of mass, you have the potential energy piece, but also motion relative to the centre of mass, so you have another potential energy piece that goes as p^2/2\mu[/itex], where mu is the reduced mass.
  7. Nov 1, 2008 #6
    Thanks for your help!
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