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Equipartition Theorem Degrees of Freedom

  1. Jan 28, 2014 #1
    Hello everyone,

    Today, in class, I learned about the equipartition, and the degrees of freedom a thing can possess. My professor had said that if you were to consider a gas composed of single atoms, then the atoms couldn't have any degrees of freedom due to rotation, because you wouldn't be able to distinguish the rotation.

    What exactly does that mean, and why does that warrant our not considering it as a degree of freedom.

    Also, would knowing this also explain why, when counting the degrees of freedom, you don't consider clockwise and counterclockwise rotation?
     
  2. jcsd
  3. Jan 28, 2014 #2

    DrClaude

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    It comes from quantum mechanics. Rotation is quantized, which means that you can only have certain discrete amounts of rotation (or, more correctly, angular momentum). You can have 0, 1, 2, ... quantas of rotation . Zero is the ground state, and corresponds to no rotation. For an atom, the next level, 1 quantum of rotation, is so higher up in energy that you will break apart the atom before excitating its rotation. Therefore, atoms are always in the rotational ground state.

    The same happens with linear molecules (for instance, diatomics like N2 and O2). The internuclear axis can rotate around two different axes, but the molecule cannot rotate along the axis that goes through the two atoms.

    For the same reason that you don't count going forwards and going backwards as two distinct translational degrees of freem. There is one degree of freedom, the motion along which can be either positive or negative (counterclockwise or clockwise).
     
  4. Jan 29, 2014 #3
    So, if the discipline of thermal physics is older than quantum mechanics, how did the founders of this field of physics deal with the concept of degrees of freedom, if they didn't know about such quantum mechanical ideas?
     
  5. Jan 29, 2014 #4

    DrClaude

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    Most of this was derived long before the idea itself of atoms existing was widely accepted. If you consider atoms to be point particles, then they can only have translational degrees of freedom.

    At the end of the day, what is important is that the theory should match experimental results. A model where an ideal gas is made up of non-interacting point particles was a good starting point.
     
  6. Jan 29, 2014 #5

    D H

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    The kinetic theory of gases is a semi-classical theory. Atoms exist, but there's no quantum weirdness in this theory. Atoms are point masses in the kinetic theory of gases. The inertia tensor of a point mass is zero, which means atoms have zero angular momentum and zero rotational energy. In a monatomic gas, there is no way to partition energy into rotation because monatomic gases have no rotational energy.

    A diatomic gas has two rotational degrees of freedom. Why not three? The answer again lies in the inertia tensor. The moment of inertia tensor for a pair of point masses separated by some fixed distance is a rank 2 matrix. There is no rotational energy about the line that connects the two atoms. It's only the non-linear polyatomic gases that exhibit three rotational degrees of freedom in this theory.

    The kinetic theory of gases has some very nice successes, but also exhibits some failures. Ideal gases have constant heat capacity, the difference between the constant pressure and constant volume heat capacities is R, and they obey PV=nRT. None of these is strictly true for a real gas. The noble gases come very, very close to exhibiting the behaviors predicted by kinetic theory. Some other gases are also close to ideal, at least over some temperature range.
     
  7. Jan 29, 2014 #6

    sophiecentaur

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    The basics of classical thermal physics (kinetic theory) starts off by assuming the simplest gas particles have point dimensions. That means they can't have rotational energy. Going from that to a diatomic particle, it's reasonable to make an assumption that there is still no rotational energy around the axis of the two (point) atoms. It fits measurements, as well, so the model works. (And you can't ask more than that; "but what's really happening?" is not an allowed question)
     
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