Viscosity calculation of a rarefied gas I am studying the rotational-translational relaxation of a diatomic gas (like oxygen) using a GPU in order to accelerate the calculations; during the calculations I get the translational temperature, the rotational temperature and the global temperature; my goal is to calculate the viscosity of the gas; I know viscosity is function of the temperature but I was wandering which one (translational, rotational or global?); what formula should be more suitable for this type of calculation? I know the Sutherland's formula exists but I don't know if it would apply in this case... any idea is welcome
 A: 1) I am not exactly sure what you have in mind. The Sutherland formula is a phenomenological formula, but if you do molecular dynamics, or use a Boltzmann solver, then you can compute the viscosity microscopically, using the Kubo relation
$$
\eta = \frac{V}{T}\int dt\,  \langle \Pi_{xy}(0)\Pi_{xy}(t)\rangle
$$
where $\Pi_{xy}$ is the off-diagonal stress in your simulation. This can be computed using the velocities of particles in molecular dynamics, or using the distribution function in the Boltzmann equation. 
2) If you observe that the translational and rotational temperatures are not the same, then your simulation is not yet equilibrated, and what you measure is not the equilibrium viscosity. You can give a hand-waving argument that translational degrees of freedom are more relevant for shear viscosity, and that it could be that the rotational degrees of freedom show an abnormally long relaxation time (possibly related to bulk viscosity), so that the $T$ in the Kubo formula should be the one extracted from translational motion. However, in the end you should make sure that the code actually does equilibrate eventually.
A: I spent quite a lot of time trying to find bibliography on the subject, but I'm having trouble because the main reference is a very expensive book and most of the literature is less about the ideas and the implementation discussion and more about the aplications. This method seems to be a bit too closed for my taste.
@Frederico Gentile,
If you have any good reference, please send it to me. I'll try to formulate my answer based on particle ansatz to Kinectic Theory methods (Boltzmann Equation and the like). If I find any good reference I can update the answer, thus, please take the following with a grain of salt.
Short Answer:
As Floris said, it should be related to the translational temperature because the stress tensor is usually only composed from kinectic degrees of freedom. If there were translational or vibrational contributions to the stress, those should be taken into account and they would modify the temperature that one would associate to the viscosity of the system.
Long answer:
When one discusses viscosity and the like in the, in the context of extracting averages of microscopic particles, the approach that I'm used to is the following.
A system of N particles $\{\vec x_n(t),\vec p_n(t),m_n(t)\}$ have momentum density/energy current $\vec P = \sum \vec p_n \delta(\vec x-\vec x_n)$ and stress given by $\sigma = \sum \frac{\vec p_n \otimes\ \vec p_n}{m} \delta (\vec x - \vec x_n)$ and (kinectic) energy density $e= \sum_n \frac{\vec p_n^2}{2m}\delta(\vec x-\vec x_n)$
Which need to be smeared in order to have any practical numerical aplicability. The smearing is done via convolution with an smoothing kernel, $W(\vec x,h)$ that implements the idea of coarse graining on the interpretation of your results. So, for example, the smeared version of the kinectic energy is given by:
$\tilde e(t,\vec x)_h = \int d^3\vec x'\ e(t,\vec x') W(\vec x - \vec x',h) = \sum_n  \frac{\vec p_n^2}{2m}W(\vec x-\vec x_n,h)$
Where the h is a parameter that regulates the smearing scale. Two simple choices are to get 
$W(\vec x,h) = \frac{1}{(2\pi h^2)^{3/2}}e^{-|\vec x|^2/2h^2}$ 
or
$W(\vec x,h) = 1/h^3\ {\rm if}\ |x_i| < h/2\ \ {\rm else}\ 0$
These 2 options give a gaussian smearing and a box-like smearing around the $\vec x$ observation point. As far as I could find, people from DSMC use a prescription similar to the 2o one (calculate the average energy/momentum on a given cell), and then skip the 'tilde' notation and simply call this average quantity the energy associated with the cell.
Ok, till now, we have only treated structure-less point particles and dicussed their energy. Since you have translational, rotational and vibrational energy, its evident that you have internal structure, but before discussing that, let's dicuss how to extract information on the viscosity, then dicuss how temperature is related to all this.
So, let's look to the stress tensor $\sigma$, it can be decomposed as:
$\sigma_{ij} = -(p+\Pi) \delta_{ij} + \pi_{ij}$
where usually $p$ is interpreted as the thermodynamic/hydrostatic pressure, $\Pi$ and $\pi_{ij}$ are the bulk and shear/deviatoric viscosity contributions to the stress tensor. On simple/newtonian fluid regimes, one has $\Pi \propto (\nabla \cdot \vec v)$ and $\pi_{ij} \propto (\partial_i v_j + \partial_j v_i) - \frac{2}{3} (\nabla \cdot \vec v) \delta_{ij}$, where $\vec v$ correspond to the fluid velocity, not the microscopic particle velocity. 
With the newtonian fluid hypothesis, one identifies the proportionality constant as the bulk and shear viscosity coeficients, which is probably what your code calculates for you. This is very important, because one can model with kinectic theory non-newtonian fluids, and if this is the case, it's necessary to specify how the viscosity coeficients are calculated from the stress and velocity gradient tensors. 
Now, to discuss temperature, it's necessary to understand a bit of thermodynamics and statistical mechanics. Strictly speaking, for thermodynamics to work it's (conceptually) necessary to have a stationary, (essentially) homogeneous system with infinite volume and infinite number of particles. This is what is called thermodynamical limit. Of course this never happens in real systems, what does happen is that the samples the are analyzed are big, homogeneous enough to the microscopic fluctuations average out, and our coarse instruments are only able to pic the average thermodynamic values. Even if one usually measures only averages, fluctuations are also important (e.g. specific heat), thus it's the system have distribution of momentum/degrees of freedom, and that's what is linked to the temperature.
If one would have a purely thermal distribution of point-like classical particles at temperature $T$ (cannonical ensamble), the momentum distribution would the the boltzmann one:
$f(t,\vec x,\vec p) \propto e^{-|\vec p|^2/2m k_b T}$
The idea behind the connection of thermodynamics and Boltzmann equation is to extend this solution that is independent of both position and time, i.e., it's a global equilibrium solution, to a local solution, thus, having a position and time dependent temperature. The temperature of the system is identified, if the system have a boltzmann like momentum distribution, to the momentum width of this distribution. Note that one can relate the RMS average of this distribution to the width (consequently to the temperature), but it's not always true the other way around. 
So, the use the equipartition theorem, which relates the RMS average of quadratic quantities of the energy to the temperature, is conditioned to the system being on local thermodynamical equilibrium. In this mindset, there makes no sense of defining temperature of very far out of equilibrium (i.e., very non-boltzmann like) systems, because it's possible to get all kinds of non-sense and almost sensible but confusing answers.
As far as I found, they use this RMS average to find each kind of temperature, kinectic, rotational or vibrational. Since their systems can be very far from equilibrium, the RMS average of each one can give completely diferent answers, which for me proves that it makes no sense talking about the temperature of this system at all, at least without looking carefully to the distribution of each of those quantities.
It's possible to calculate RMS averages with any kind of momentum distribution, but it doesn't mean that it relates to anything that can be called temperature. This is why I don't  like this kind of prescription to calculate temperature.
Comming back to the viscosity of the system. One can calculate the stress tensor of the system with local thermal equilibrium, and that appears is a pure diagonal stress, without any relevant dependence on the velocity field, meaning there is only thermodynamical pressure, but no viscosity. This gives a hint that viscosity is linked to deformations of the above distribution.
Given the above argument, it's becomes delicate to discuss the temperature associated with a system with viscosity, because of the momentum distribution is no longer perfectly boltzmann-like. In order to discuss this properly, one need to have the viscous terms small, in a certain sense, and to look the equilibrium distribution closest to the system's distribution, and define the system temperature as equal to this closest equilibrium system. If one has something way too far from equilibrium, this kind of prescription makes no more sense.
Now, again to your answer. The kinectic/translational degrees of freedom are the ones that usually contribute to the stress tensor, so, as to retain any kind of consistency, if you are to calculate any kind of temperature that would apply to the viscosity, it should be the one associated with the ones that contributes to the stress, that usually are the translational ones. 
The fact that rotational and vibrational degrees of freedom may contribute to the collision kernel or to the energy of the system is not an imediate problem, but it leave space for this kind of ambiguity.
