6

The Kinetic Energy $mv^2/2$ does depend on the FoR, and hence maybe the internal energy?

I've also seen temperature being defined as "a measure of the average kinetic energy of the particles".

Are thermodynamic quantities based on frame of reference? Is there a distinction in this case between state functions like "entropy" and path functions like "work done"?

Agnius Vasiliauskas
  • 16,916

2 Answers2

10

For simplicity and convenience, we often conduct thermodynamics using a frame of reference in which the system of interest is motionless, but this is not essential. I quote at length from Callen's Thermodynamics and an Introduction to Thermostatics:

In accepting the existence of a conserved macroscopic energy function as the first postulate of thermodynamics, we anchor that postulate directly in Noether's theorem and in the time-translation symmetry of physical laws.

An astute reader will perhaps turn the symmetry argument around. There are seven "first integrals of the motion" (as the conserved quantities are known in mechanics). These seven conserved quantities are the energy, the three components of linear momentum, and the three components of the angular momentum; and they follow in parallel fashion from the translation in "space-time" and from rotation. Why, then, does energy appear to play a unique role in thermostatistics? Should not momentum and angular momentum play parallel roles with the energy? In fact, the energy is not unique in thermostatistics. The linear momentum and angular momentum play precisely parallel roles. The asymmetry in our account of thermostatistics is a purely conventional one that obscures the true nature of the subject.

We have followed the standard convention of restricting attention to systems that are macroscopically stationary, in which case the momentum and angular momentum arbitrarily are required to be zero and do not appear in the analysis. But astrophysicists, who apply thermostatistics to rotating galaxies, are quite familiar with a more complete form of thermostatistics. In that formulation the energy, linear momentum, and angular momentum play fully analogous roles. [emph. added]

Callen then gives an example involving a stellar atmosphere in motion.

Chemomechanics
  • 30,163
4

If the principle of local thermodynamic equilibrium holds, you can define thermodynamic variables around the average local velocity $\overline{\mathbf{v}}(\mathbf{r},t)$ of the medium.

As an example, we can write the total kinetic energy of the microscopic particles in a "small" material volume as

$K^{tot} = \sum_i K_i = \sum_i \frac{1}{2} m_i \mathbf{v}_i \cdot \mathbf{v}_i = \sum_i \frac{1}{2} m_i (\overline{\mathbf{v}} + \mathbf{v}'_i ) \cdot (\overline{\mathbf{v}} + \mathbf{v}'_i ) = $
$ \qquad \qquad \qquad= \dfrac{1}{2} \underbrace{\left( \sum_i m_i \right)}_{= m} |\overline{\mathbf{v}}|^2 + \overline{\mathbf{v}} \cdot \sum_i m_i \mathbf{v}_i' + \sum_i \frac{1}{2} m_i |\mathbf{v}'_i|^2$,

and averaging this expression we get

$\overline{K^{tot}} = \overline{ \dfrac{1}{2} m |\overline{\mathbf{v}}|^2 } + \underbrace{\overline{ \overline{\mathbf{v}} \cdot \sum_i m_i \mathbf{v}_i' }}_{\overline{\mathbf{v}'} = \mathbf{0}} + \overline{ \sum_i \frac{1}{2} m_i |\mathbf{v}'_i|^2 } = K^{macro} + E^{int}$,

being the average total energy equal to the sum of:

  • the macroscopic kinetic energy $K^{macro}$ of the closed material volume
  • its internal energy $E^{int}$, related to the fluctuations $\mathbf{v}'_i$, of the microscopic particles around the average velocity, that can be related to thermodynamics variables bridging kinetic theories with classical thermodynamics.

We can provide a more general qualitative definition of the temperature, if compared with the one you gave in your question

temperature is as a measure of the average kinetic energy of the particles, w.r.t. the local average (macroscopic) velocity.

Thus, in this sense, thermodynamic variables are not dependent on the reference frame.

basics
  • 13,368