If you have a gas, you can insert a bit of energy $E$ and measure the resulting increase $K$ in the average kinetic energy in your favourite direction. For monatonic gases, $K=E/3$, as the energy is evenly distributed in each translational direction (the "equipartition theorem"). We can plot the ratio $$C\ =\ \frac{1}{2}E/K$$ called the specfic heat capacity, and for diatomic gases it looks like this:

So for a billion $H_2$ in a box $B$ at five hundred Kelvin,

  1. The ratio $E/K=5$,
  2. The state space of this system is $$\mathcal{M}\ =\ T^*(B\times \text{SO}(3)/S^1)^{\text{one billion}}$$ where the thing in brackets is the five-dimensional degrees of freedom of an individual $H_2$.
  3. The flow under Hamilton's equations on $\mathcal{M}$ is probably ergodic, i.e. mixes the atoms sufficiently so inputted energy goes equally into each of the five degrees of freedom (the equiparitition theorem applies).


Now increase the temperature of the box to $1000$K, my question is: what happens?

  1. The ratio is now $E/K\approx 5.4$.
  2. The state space is now presumably $$\mathcal{M}_{1000}\ =\ T^*(X)^{\text{a billion}}$$ where $X$ is some seven dimensional space which includes the two vibrational modes which are now relevant.
  3. But what happens to the equipatition? We have for $i=1,...,5$ and $i=6,7$ that $$K_i/E\ =\ 1/5.4 ,\ \ \ \text{ }\ \ \ K_i/E\ =\ 0.2/5.4,$$ or something similar, so Q1) is there a general form of the equipartition theorem that includes cases like this where inputted energy is not equally mixed among the degrees of freedom?

My following question is: Q2) what does the flow of Hamilton's equations on $\mathcal{M}_{1000}$ look like?

More precisely, it can't be ergodic, because otherwise we'd have the equipartition theorem and $K_i/E=1/7$, but in the limit of high and low temperatures it is ergodic. So, what analogue of ergodicity does this system at $1000$K satisfy? For instance, is there a Hausdorff dimension $5.4$ subspace on which the flow is ergodic?

  • $\begingroup$ n.b. the subspace of $\mathcal{M}_{1000}$ should have dimension $2\cdot 5.4\cdot$ one billion, so that it has "5.4 degrees of freedom", but the question was cluttered enough already so I left it out. $\endgroup$
    – Pulcinella
    Feb 22 at 14:47
  • 1
    $\begingroup$ Your second question assumes that the system is well described by Hamiltonian flow on a manifold, i.e. that it is obeying classical mechanics. This phenomenon, however, is fundamentally the result of quantum mechanics, and indeed was historically one of the first steps towards Planck's work on black body radiation $\endgroup$ Feb 22 at 15:07

1 Answer 1


First, the atoms involved are quantum objects, so their energy states are quantized. There is a minimum non-zero energy associated with a rotational or vibrational mode. This is where the cartoon version of the specific heat plot comes from (your left panel). There needs to be enough thermal energy, $kT$, to excite the first non-zero vibrational state. If $kT$ is less than the $\frac{1}{2}\hbar\omega$ of the quantum harmonic oscillator representing the atomic bond, there is no vibration.

The reason why it doesn't work perfectly is fundamentally how statistical mechanics works. At a given temperature there is a probability that a given energy state is occupied. You can have the same total energy per particle by giving all of it to one particle or evenly distributing it among all of them. The perfectly even distribution of energy corresponds to the most probable outcome, but the probability of exactly that is basically zero. There is a cluster of outcomes with pretty close to (but not exactly) even energy distribution. In practice you are very likely to observe near even energy distribution, but the distribution is never exactly even. Some particles will have a little more energy than others, overcoming the quantized minimum energy barrier and activating a vibrational mode.

The effective degrees of freedom is a statement about the average activated degrees of freedom per particle. But even so, a choice of $n_\mathrm{eff} = 5.4$ doesn't fully determine the number of particles with active vibrations. There will be several possible ratios of particles with $n=3, 5, 7$ degrees of freedom that can result in the observed macroscopic state.

  • $\begingroup$ Thanks for this. To relate to the last part of my question, is there a "quantum ergodic/equipartition" property of thermodynamic systems that makes precise the idea you mentioned that, subject to the constraints of rotational and vibrational energy being quantised and at each moment the energy not being distributed evenly among atoms, the time-averaged energy is spread out "as evenly as possible" in each degree of freedom? $\endgroup$
    – Pulcinella
    Feb 22 at 15:28

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