Extreme temperatures, relativity and kinetic theory According to kinetic theory, average kinetic energy is proportional to temperature. Supposing $k_BT/2$ per particle, can we use relativity and kinetic theory to calculate, e.g., the temperature and velocity of quarks in a quark-gluon plasma and hotter/denser states of matter? 
Note: I did some calculations already myself to check this idea
Normal kinetic theory (in 3D)-> $$E_c(av)=\dfrac{1}{2}mv^2=3k_BT/2$$ must be substituted (???) by 
$$E_c(rel)=Mc^2-mc^2$$
with $$M=m\gamma$$
For a quark-gluon plasma, taking the proton mass as m and the critical temperature as 200MeV I get
$$200MeV=3/2k_BT$$ 
so T is about $$2\cdot 10^{12}K$$ or about 4x10¹²K if I drop the factor 3 above. That is OK with the known temperature of the quark-gluon plasma. I am also concerned about the issue of determining the gamma factor for protons(quarks?) at those energies/temperatures...
I get, for protons at this
$$\beta=v/c=\sqrt{1-(mc^2/E)^2}=0.977$$
Am I right?
 A: Perhaps a better way is to redefine your energy equation to:
$$
m \ c^{2} \left( \gamma_{rms} - 1 \right) = \frac{3}{2} k_{B} \ T
$$
where $\gamma_{rms}$ is the Lorentz factor of the rms speed of the relativistic gas.  This can be shown to be:
$$
\left( V_{rms} \right)^{2} = c^2 \left[ 1 - \left( \frac{3 \ k_{B} \ T}{2 \ m \ c^{2}} + 1 \right)^{-2} \right]
$$
This arises because your assumption that energy is proportional to $\left( 3 k_{B} T \right)/2$ comes from the second velocity moment of a Maxwell-Boltzmann velocity distribution function.  If you use the relativistic version of this distribution function, you need to use the relativistic energy form of the equation, not the velocity form.  Here the exponential term goes to:
$$
EXP\left( -\frac{m \ v^{2}}{2 \ k_{B} \ T} \right) \rightarrow EXP\left( -\frac{m \ c^{2} \left( \gamma_{v} - 1 \right)}{k_{B} \ T} \right)
$$
where $\gamma_{v}$ is the Lorentz factor of the particle speed, $v$.  I think this will change your results slightly.  There might be other (or better?) ways to do this but I think this is correct because it prevents any given particle in the velocity distribution from having a speed that exceeds the speed of light.
