# Reference frame of equipartition theorem kinetic energy

Supose I have a molecule with n degrees of freedom, then by the equipartition theorem, the kinetic energy of the molecule will be $$E = \frac{n}{2} k T$$ for a particle moving on the XYZ plane the energy will be

$$E = \frac{3}{2} k T$$ In what reference frame is the energy measured?

• A reference frame in which the mean velocity of the particle is 0, as well as for the angular momentum. Commented Jul 24 at 11:01
• @Syrocco if i have a system composed of two particles that are not interacting with one another with their velocities being always equal in modulus but of oposite signs how can it be possible to measure both of their energies from a refference frame in witch their mean velocities are 0? Commented Jul 24 at 11:13
• If you have multiples particles, then it's the total average velocity that must be 0 (as in the example you gave). But in any case, concerning your example: the equipartition theorem is something that works for sufficiently complex systems, where a description with random number makes sense. Here it would also works because you can define the temperature through the kinetic energy of the particle (making it a tautology) but if you had to define it with an entropy it would not make sense anymore... Commented Jul 24 at 12:35
• In some sense we require that the velocity and angular momentum of the system are on average 0 so that the only contribution to the energy is internal. Commented Jul 24 at 12:36
• The reference frame defined by the first velocity moment. The equipartition theorem is basically the second velocity moment (of the peculiar velocity) within a few constants. Commented 2 days ago

If $$E$$ is the internal kinetic energy of the molecules of the system, then the reference frame is the center of mass (COM) of the system. This distinguishes the internal kinetic energy of the system at the molecular level from the kinetic energy of the system as a whole with the COM moving respect to some external frame of reference.