thermodynamics - Is mean kinetic energy related to temperature of a system of interacting classical particles?

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?

Answer

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:

$$H=A\zeta^2+H'$$

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 $d\zeta$.