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We know that, in non-relativistic case, for Maxwell-Statistics, the local equilibrium distribution is like the following form $$ C\exp\left(-\lambda_{1}|\vec{u}-\vec{U}|^2-\lambda_{2}\vec{\xi}^2\right), $$ where $\vec{u}$ denotes the microscopic gas particle velocity, and $\vec{U}$ denotes the macroscopic velocity, and $\vec{\xi}$ denotes the internal variable.

In three dimension, $\vec{\xi}$ is zero-dimensional for monatomic molecule, and two-dimensional for diatomic molecule.

And in relativistic case, for monatomic molecule, the local equilibrium distribution is like the following form $$ C\exp(-\lambda_{1}U_{\alpha}p^{\alpha}), $$ where $p^{\alpha}$ denotes the momentum 4-vector of particle, and $\vec{U}$ denotes the macroscopic velocity 4-vector.

But what the internal variable and the local equilibrium distribution is like for diatomic molecule or polyatomic molecule?

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  • $\begingroup$ Related: http://physics.stackexchange.com/q/216819/59023 and see doi:10.5194/angeo-34-737-2016 $\endgroup$ – honeste_vivere Oct 20 '16 at 13:18
  • $\begingroup$ As the aboove comment said, the two links are local equilibrium distribution for monatomic molecule, and how about the one for diatomic molecule? $\endgroup$ – Y.Y.Kuang Oct 20 '16 at 13:58
  • $\begingroup$ You would need to include spin (and maybe vibrational modes?) into the equation somehow, which is more symbol gymnastics than I have time with which to wrestle... Technically, Treumann and Baumjohann's work only applies to protons because to deal with a relativistic electron distribution, one would need to include the particle spins which is not trivial. $\endgroup$ – honeste_vivere Oct 20 '16 at 14:25

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