I'm trying to intuitively understand what temperature is for a system of classical particles. The usual definitions via Gibbs measure or entropy appear very unintuitive me. But, as for ideal gas temperature is proportional to mean kinetic energy of its molecules (in the rest frame of gas' center of mass), I thought it's a good place to start.

So, suppose we have a system $\Lambda$ of interacting classical particles, such that there's always 3 degrees of freedom per particle (so that molecules would be modelled as multiple bound particles). Assume that their center of mass is at rest. In general they'll not behave as an ideal gas — they could condensate etc.. But when brought to contact with an ideal gas $\Gamma$ (with heat capacity much smaller than that of $\Lambda$) initially at $T_\Gamma=0\,\mathrm K$, the gas $\Gamma$ would ultimately assume the temperature of the system: $T_\Gamma\to T_\Lambda$ as $t\to\infty$. This actually means that the mean kinetic energy of the gas molecules will be proportional to the temperature of the system $\Lambda$.

So, I'd assume that the increase in mean kinetic energy of $\Gamma$ was due to the kinetic energy present in $\Lambda$. Then, isn't mean kinetic energy of the particles in $\Lambda$ actually proportional to the temperature, even despite $\Lambda$ being not an ideal gas, and not even a gas in general?


2 Answers 2


Yes, according to the equipartition theorem, the average kinetic energy of a particle of a classical system is $\frac{3}{2}k_B T$, where $k_B$ is the Boltzmann constant and $T$ is temperature (https://en.wikipedia.org/wiki/Temperature#Kinetic_theory_approach_to_temperature). In general, this is not true for a quantum system.

EDIT (12/31/2016): The relationship between kinetic energy and temperature for classical systems in thermal equilibrium holds no matter what interaction potential. See, e.g., math.nyu.edu/~cai/Courses/MathPhys/Lecture3.pdf : "For example, the Hamiltonian for interacting gas particles is...where $U(\vec{r_i}−\vec{r_j})$ is the potential energy between any two particles. The equipartition theorem tells us that the average kinetic energy is $\frac{3}{2}k_B T$ for every particle (since there are three translational degrees of freedom for each particle)." See also the derivation there (at the beginning):

Equipartition Theorem:

    If the dynamics of the system is described by the Hamiltonian:


where $\zeta$ is one of the general coordinates $q_1,p_1,\cdots,q_{3N},p_{3N}$, and $H'$ and $A$ are independent of $\zeta$, then

$$\langle A\zeta^2\rangle=\frac12k_BT$$

where $\langle\rangle$ is the thermal average, i.e., the average over the Gibbs measure $e^{-\beta H}$.

    This result can be easily seen by the following calculation:

$$\begin{align}\langle A\zeta\rangle&=\frac{\int A\zeta^2e^{-\beta H}d^{3N}qd^{3N}p}{\int e^{-\beta H}d^{3N}qd^{3N}p} \\ &=\frac{\int A\zeta^2 e^{-\beta A\zeta^2}d\zeta e^{-\beta H'}[dpdq]}{\int e^{-\beta A\zeta^2}d\zeta e^{-\beta H'}[dpdq]} \\ &=\frac{\int A\zeta^2 e^{-\beta A\zeta^2}d\zeta}{\int e^{-\beta A\zeta^2}d\zeta} \\ &=-\frac{\partial}{\partial\beta}\ln\int e^{-\beta A\zeta^2}d\zeta \\ &=-\frac{\partial}{\partial\beta}\ln\left[\beta^{-\frac12}\int e^{-Ax^2}dx\right] \\ &=\frac1{2\beta} \\ &=\frac12k_BT \end{align}$$

where $d\zeta[dpdq]=d^{3N}qd^{3N}p$, i.e., $[dpdq]$ stands for the phase-space volume element without $d\zeta$.


Although I originally said "no" because of quantum mechanics and because of classical potential energy, it was pointed out that the answer to the question as posed is actually "yes." Interestingly, systems of classical particles in 3D come to the same average kinetic energy, regardless of their potential energy, as they come to the same temperature. As mentioned above, that's the equipartition theorem, and then the virial theorem implies that even if the magnitude of the potential energy significantly exceeds that of the kinetic energy, it will also end up proportional to kT, so it seems that the kinetic energy is ultimately controlling the path to reaching a given T in all classical systems, just as suggested in the question.

  • $\begingroup$ I was specifically talking about classical particles to avoid the zero point energy, especially for fermions. $\endgroup$
    – Ruslan
    Commented Dec 31, 2016 at 10:47
  • $\begingroup$ OK, but as I added above, if there is not a general rule for nonclassical particles, why would there need to be one for classical particles? The point is, all you need is anything that inhibits energy transfer, which essentially means any interaction that is not between ideal gases where the particles are completely free to share energy as they collide. If you want something that is not ideal, say molecules, then you are already talking about nonclassical particles-- molecules are ruled by quantum mechanics. $\endgroup$
    – Ken G
    Commented Dec 31, 2016 at 10:52
  • $\begingroup$ @Ken G : Actually, there is such a general rule for classical systems (please see my answer). $\endgroup$
    – akhmeteli
    Commented Dec 31, 2016 at 10:55
  • $\begingroup$ I would add that the crucial point to get is that even for an ideal gas, when you put such a gas into thermal contact with a reservoir at high T, the reason the kinetic energy per particle ends up proportional to that T is because the reservoir gives up energy to the ideal gas until they both "equally covet" energy. So the amount of energy that comes across is not just a function of the reservoir, it is also a function of how readily ideal gases accept and hold onto energy. $\endgroup$
    – Ken G
    Commented Dec 31, 2016 at 10:59
  • $\begingroup$ @akhmeteli: that theorem is restricted to systems with only quadratic terms in the Hamiltonian (basically, free particles and springs), and 1/2 kT of energy goes into every mode that corresponds to each of those quadratic terms. But classical systems don't have to have Hamiltonians with only quadratic terms. To inhibit energy transfer to make T not proportional to the kinetic energy, all you need is other kinds of interactions. $\endgroup$
    – Ken G
    Commented Dec 31, 2016 at 11:03

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