# relativistic thermodynamics (temperature in inertial frames)

*From*: srwenner@xxxxxxxxxxx*Date*: Sat, 1 Dec 2007 20:23:05 -0800 (PST)

Hi, I'm a layman with an interest in physics. I just read an

interesting report on Physics News Update (focus.aps.org) 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,

SRWenner

.

**Follow-Ups**:**Re: relativistic thermodynamics (temperature in inertial frames)***From:*Tom Roberts

**Re: relativistic thermodynamics (temperature in inertial frames)***From:*Dono

**Re: relativistic thermodynamics (temperature in inertial frames)***From:*Juan R.

**Re: relativistic thermodynamics (temperature in inertial frames)***From:*N:dlzc D:aol T:com \(dlzc\)

**Re: relativistic thermodynamics (temperature in inertial frames)***From:*Androcles

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