Is there a clean way to examine temperature for solids and liquids in classical mechanics like the kinetic theory for gases?

I'd like to get a good explanation that doesn't involve much in the way of quantum mechanics.


You can think of the atoms of the solid as balls attached by springs to each other. Each ball can vibrate in it's own space and one ball vibrating will also make the balls near it vibrate.

For explaining temperature, the higher the temperature, more is the amplitude of vibration of these balls.

This model can also be used to explain other behavior like elasticity, thermal conduction etc.

  • 2
    $\begingroup$ In a balls and spring model, the spectrum of possible frequencies is independent on temperature. Moreover at equilibrium all modes of vibration are present. Therefore there is no link at all between temperature and frequency of vibration within a classic harmonic model. $\endgroup$ – GiorgioP Jan 8 '19 at 6:39
  • $\begingroup$ @GiorgioP So how would you explain conductivity in solids? I always thought of it as vibration being passed down from end to another. $\endgroup$ – harshit54 Jan 8 '19 at 6:43
  • $\begingroup$ Thermal conductivity is not the same phenomenon as thermal equilibrium. It is a mechanism to get it. You have thermal conductivity if the solid is out of equilibrium and different regions have different temperatures. In that case too, if the model is the harmonic model, in a classical treatment it is the amplitude of vibrations which varies with time, not the frequency. Frequencies depend only on elastic constants of the "springs" and mass of the particles. They definitely do not depend on temperature. $\endgroup$ – GiorgioP Jan 8 '19 at 7:12
  • $\begingroup$ So the amplitude does represent the temperature? $\endgroup$ – harshit54 Jan 8 '19 at 7:14
  • $\begingroup$ Yes, in classical mechanics the higher is the maximum velocity of a harmonic oscillator, the higher are its energy and its amplitude. Frequency remains the same. $\endgroup$ – GiorgioP Jan 8 '19 at 7:21

I imagine that the reference to the kinetic theory of gases means the link between temperature and average kinetic energy in the center of mass frame.

In classical statistical mechanics (i.e. using only classical mechanics and ensemble theory) that link corresponds to, and is a trivial consequence of the equipartition theorem (the average kinetic energy per degree of freedom is $\frac{1}{2}k_B T$).

However, equipartition theorem for the kinetic energy is valid without exception for all classical systems, independently on the presence or on the form of the potential energy term in the Hamiltonian. Therefore, doesn't matter what is the thermodynamic phase. The same relation holds for perfect and imperfect gases, dense liquids, and all kind of crystalline and amorphous solids.

The only limits to the validity of equipartition theorem are

  • lack of thermodynamic equilibrium (in that case the concept of a unique temperature characterizing the system may become meaningless);
  • emergence of quantum effects in the statistical treatment of mechanical systems.

A rough but useful way to estimate the limit of a classical treatment is through the degeneracy parameter $\rho \lambda^3$, where $\rho$ is the number density and $\lambda$ the de Broglie thermal wavelength $ \lambda = h/\sqrt{2 \pi m k_B T}$ for particles of mass $m$. A necessary condition for validity of classical statistical mechanics (and the equipartition theorem) requires $\rho \lambda^3 \ll 1$.

Additional remark

One consequence of classical statistical mechanics is that not only the temperature of a system of molecules is proportional to the average kinetic energy of the particles independently on the thermodynamic phase, but also that the Maxwell-Boltzmann (MB) distribution of molecular speeds remains the same in all the phases. This trivial consequence of classical statistical mechanics is often overlooked when the MB distribution is discussed in the context of the kinetic theory of gases, leaving the feeling that its validity is confined to the gas phase.


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