MadSci Network: Physics

Re: How is PV=NrT affected by relativistic speeds?

Date: Sat Jun 12 15:17:29 1999
Posted By: Ken Wharton, Post-doc/Fellow, Laser/Plasma Physics, I'd rather not say.
Area of science: Physics
ID: 928249035.Ph

Interesting question! I thought solving for P and T would be trivial, but it turned out to be rather involved. There even turned out to be some really fascinating implications...

If you have a bunch of gas in a box, and then you fly past that box at near-lightspeed, you are correct that the length of the box will appear to change, shrinking the volume (V) of the box by some factor 1/gamma.

So what happens to the pressure? The same number of particles are now in a smaller box, so the density goes up by the same factor gamma. But there's also an apparent slowing down of the gas relative to the walls, by a factor gamma-2, so the number of particles hitting the walls (per unit of observed time) actually goes DOWN by a factor of 1/gamma. However, each collision supplies more momentum to the walls (by a factor of gamma -- although this calculation is far from trivial) so all these effects exactly cancel: the pressure (P) stays the same. (All this analysis is just for the walls that are at the "front" and "back" of the moving box; the walls to the side have a different analysis I won't get into, but it gives the same answer...)

Now, if PV=NRT still holds, the above analysis tells us that T must drop by a factor of 1/gamma, to balance the drop in V. Would that mean that water vapor at 101o C could appear to be only 99o C in a different frame and condense to water? Einstein tells us this CAN'T happen -- the same phenomena must have consistent explanations in all reference frames, and the difference between steam and water is too great. But does that mean that water has a different condensing temperature in different frames? Or does PV=NRT not work in some reference frames? There appears to be a paradox here: it looks like SOME physics has to be different... This, I presume, is the heart of your question.

The answer to this paradox lies in the definition of "temperature" itself. Keep in mind that T is a single number that attempts to describe the entire energy distribution of a very large number of gas molecules -- obviously some information is being lost, by reducing all those parameters to a single number. However, in a system at thermal equilibrium (a steady-state energy distribution), the concept of temperature still works remarkably well.

But what happens when one shifts to a moving frame? Instead of a distribution of particle energies that are centered around zero, the energies are now centered around some non-zero energy. And even if the transformation is purely classical -- for example, you only supply a small change in velocity of the box -- the energy distribution still changes shape! This is because (classically) kinetic energy goes as velocity squared; adding the same velocity to every particle changes their energies in rather different ways, resulting in a new energy distribution. So does the Temperature itself change under such small transformations? No -- but one has to be very careful about how one defines the "temperature" of an energy distribution that is not centered around zero.

Doing this relativistically, one finds that the energy distribution changes in TWO ways; one that is analogous to the classical shift described in the last paragraph, and also by another factor that decreases the energy difference between particles by exactly a factor of 1/gamma. So, if you toss out the classical shift in energy distributions, you could argue that the temperature T really has dropped by a factor of 1/gamma, allowing PV= NRT to hold.

But there's another way to define temperature as well: FIRST shift to the center of mass frame and THEN measure the temperature. If one does this, then T (obviously) doesn't change at all. In this case, the constant T would seem to explain the fact that steam doesn't condense when viewed at relativistic speeds.

So which definition of T is correct? How DOES one find the temperature of a moving distribution of gas?

At this point I'm mostly guessing, but I would make the hypothesis that in relativistic thermodynamics (a far from completely understood field of physics) one needs to define temperature differently depending on how one is using the parameter. For thermodynamic analysis, one might need to define T the first way (lowering it by 1/gamma); but for phase-space diagrams one might need to define it the second way (keeping it constant). When the box is at rest, the two definitions of temperatures just happen to be the same, so this distinction is not normally necessary; only from relativistic reference frames does this distinction between different types of "temperature" emerge.

That's my best guess! I assume this has been looked at by experts in the field of relativistic thermodynamics.

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