# Continuous Transition of Degrees of Freedom in Thermodynamics

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(Original on SE https://physics.stackexchange.com/questions/459086/continuous-transition-of-degrees-of-freedom-in-thermodynamics-with-simple-exampl, though I usually get better answers here.)

In thermodynamics books I have read, I have often come across statements about how certain degrees of freedom are relevant only at certain temperatures (such as the vibration degrees of freedom of some molecules only being relevant in certain ranges), but I can't recall a convincing quantitative analysis of this. I tried to set up a simple example to explore this issue, but I'm unsure what goes wrong (though I haven't thought TOO much about it).

We know that for a free particle in one dimension at finite temperature, the partition function is given by:

$$Z(\beta)=\frac{L}{h}\int \text{d}p e^{-\beta p^2/2m}=\frac{L}{h}\sqrt{\frac{2 \pi m}{\beta}}$$

And then our expected energy is just:

$$\langle E \rangle =\frac{1}{2}k_B T$$

Which we'd expect for a particle with one degree of freedom.

On the other hand, the partition function for a particle in a 1-D harmonic potential is:

$$Z(\beta)=\frac{1}{h}\int \text{d}x \int \text{d}p e^{-\beta(kx^2+p^2/m)/2}=\frac{2\pi}{\beta h\omega}$$

Which gives the expected energy:

$$\langle E \rangle = k_BT$$

Here's my problem. If we take a limit of the spring constant to zero ($k \rightarrow 0$), doesn't that just correspond to a free particle? The average energies depend ONLY on temperature, so where exactly does this limit come in?

Even though this is a relatively simple example, I encounter similar problems when I try to work out when and how degree of freedom fails to be "relevant" in the examples alluded to in textbooks.

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The answer on SE is quite consistent, but it speaks of classical partition functions of particles "attached" to the coordinate system origin with a quadratic potential.

If your question is about seeing certain degrees of freedom getting "irrelevant", you must consider quantum mechanics. There the energy levels are discrete rather than continuous, so the occupation number for a given energy level $E_n$ is proportional to $\text{e}^{-E_n/k_B T}$ and the partition function is a sum over them. In particular, for a quantum harmonic oscillator at "low" temperatures the contribution of the first excited state $\text{e}^{-E_1/k_B T}$ with respect to the ground state $\text{e}^{-E_0/k_B T}$ is exponentially small: $\propto(1+\text{e}^{-(E_1-E_0)/k_B T})$ due to inequality $(E_1-E_0)/k_B T\gg 1$. They say, the higher energy levels (or the corresponding degrees of freedom) are "frozen" in this limit. If the "oscillator" is a vibrational motion of a two-atomic molecule, then its ground state corresponds to a free point-like particle, for which $\langle E\rangle=0.5k_B T$, as if there were no vibrational degree of freedom at all.

As the enrgy levels $E_n$ contain all the necessary constants, including $\hbar$ and $k$, you can analyse where a quantum oscillator becomes similar to the classical one.

answered Feb 6 by (122 points)
edited Feb 7

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