relativistic thermodynamics (temperature in inertial frames)

Hi, I'm a layman with an interest in physics. I just read an
interesting report on Physics News Update ( concerning
unresolved questions in reconciling thermodynamics with special
relativity. In particular it seems that there is no agreement on how
temperature would transform between inertial observers. Apparently,
Planck and Einstein were of the opinion that moving observers would
find temperature reduced, but other physicists thought the reverse
would be true. And the question is still causing confusion.

If I recall correctly from my intro physics course, the temperature of
a gas is just (proportional to) the statistical variance of the speeds
of the atoms. Do I have this right? I also recall that the
distribution of speeds (actually, velocities) is normal (Gaussian).
(Actually, contrary to what our teachers told us, the distribution can
not be strictly normal, since the speeds of the atoms must be less
than the speed of light but the normal distribution is unbounded).
So, what would be the distribution of velocities perceived by a moving
observer? I would think we could get an answer by using the formula
for the relativistic addition of velocities. Consider first those
atoms moving most rapidly in the opposite direction to the moving
observer; the formula would compress the range of those velocities to
keep them all less than the speed of light. Now consider those atoms
moving more rapidly than the observer, and in his direction; in this
case the moving observer would measure a wider range of speeds than
would a stationary observer, since the atoms would not be pushing up
against the universal speed limit in his frame. So, we end up with a
skewed distribution as measured by the moving observer that is tighter
on one end and looser on the other, compared to the symmetric
distribution measured by the stationary observer. Is the variance
(temperature) increased or decreased? Hasn't anyone actually done
this calculation? Why is it so hard? Or, do I have it all wrong.

Thanks a bunch,