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Consider a box full of an ideal gas at rest relative to you with a thermometer reading a temperature T. Now put this box of gas in a space ship and start it moving to relativistic speeds. From your perspective the box contracts in one dimension, reducing the volume, which should increase the temperature. From your perspective you should read a temperature greater than T. However, on the spaceship, the volume has not changed, so the temperature should remain at T. Somehow this doesn't seem right to me.

Also, with the volume reducing due to length contraction, work has been done on the gas. Where did this energy come from?

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  • $\begingroup$ Does it feel not right to you that a moving space ship has kinetic energy from the perspective of an outside observer but not from the perspective of the spaceship? $\endgroup$ – Raskolnikov Nov 28 '17 at 6:42
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    $\begingroup$ "From your perspective you should read a temperature greater than T." - Why? Thermodynamics is usually done in the rest frame of the system, why do you think its relations hold for all observers? $\endgroup$ – ACuriousMind♦ Nov 28 '17 at 8:49
  • $\begingroup$ I have a photograph of a car. When I look at the photograph on a sharp angle, the car looks to me much shorter, as if the volume inside the car is now much smaller. Does this mean the passengers get squeezed just because I look at the car from a different angle? The Lorentz contraction is a projection, a view of how you see things. Nothing actually happens on the spaceship simply because you look at it differently. $\endgroup$ – safesphere Nov 28 '17 at 9:46
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    $\begingroup$ Still applying simply ideal gas law from Earth gives something different than that obtained in the ship. It seems a clever Q to me. $\endgroup$ – Alchimista Nov 28 '17 at 10:28
  • $\begingroup$ @ACuriousMind, that might be the resolution, that only in the rest frame of the system should the laws of thermodynamics hold. However, it feels slightly like a violation of relativity, that the laws of physics need to be adjusted for the observer on Earth. As if, to calculate the pressure for example, the observer on Earth wouldn't be able to apply the ideal gas law. $\endgroup$ – DaYu1729 Nov 28 '17 at 16:11
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Lorentz contraction goes in hand with time dilation. The faster a jar moves the slower molecules of gas move in it. This way moving observer will not be able to measure any change of pressure or temperature, even if he moves at velocity very close to $c$.

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    $\begingroup$ Thanks for the answer. But I'm not sure about this. The component of the velocities of the particles in the direction of motion should change according to the relativistic addition of velocities. This will skew the velocity distribution, but it won't lead to all the particles slowing down. $\endgroup$ – DaYu1729 Nov 29 '17 at 3:47
  • $\begingroup$ All manifestations of what could be construed as a clock in the moving reference frame slow down as observed in a rest frame relativistic with light speed. So everything. Even molecular velocities. $\endgroup$ – docscience Nov 29 '17 at 20:08
  • $\begingroup$ I disagree. It's true that, for example, the transit time of one particle going across the box in the direction of motion of the space ship will increase by $\gamma$ as viewed from Earth, but this would coincide with the velocity of the particle changing according to the relativistic addition of velocities. Particles moving with the motion will be seen to have a slower velocity relative to the median, and particles moving against the motion will be seen to have a faster velocity relative to the median. $\endgroup$ – DaYu1729 Nov 29 '17 at 20:50
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It happens to be so that gas between two pistons that accelerate with the same acceleration, changes. It may become liquid, for example. And a piston observes the distance of pistons changing.

While in the case where those accelerating pistons are connected with a strong rope, the distance of the two pistons stays constant in both pistons' frames. Everything in the frame of a piston stays constant in this case. An inertial observer observes that proper distances stay constant, and proper temperatures, by which I mean temperatures measured by a thermometer co-moving with the gas. To the question "what is the coordinate-temperatuere" my answer is that such thing as coordinate-temperature does not exist.

The pistons are inside a very long cylinder.

The other question is much better, the question about work. The side walls of the box do work on the gas. The kinetic energy of the side walls does the work.

When a mass hanging on a rope is accelerated by pulling the almost massless rope, the pulling force is highest at the mass end of the rope. That means that the rope gains momentum that points bacwards. If the kinetic energy of the rope is its speed times all the momentum that is in the rope, then we can see how as time passes momentum that has a decreasing effect on kinetic energy gets stored in the rope. By the way the forces at the two ends of the rope are different, there is no error there.

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  • $\begingroup$ I don't think this is correct. Acceleration is tricky because different parts of the box (or in your case the two pistons) will experience different accelerations from the perspective of Earth. If you caused one of the pistons to start accelerating, the piston behind it attached by the rope would be seen to accelerate faster, and the rope would be seen to contract, so that once a constant velocity is achieved the two pistons would look to be closer together and the rope would look shorter as viewed from Earth's perspective. $\endgroup$ – DaYu1729 Nov 29 '17 at 21:07
  • $\begingroup$ And I agree that acceleration would cause the gas inside to behave a bit funky. But the question can equally be asked by comparing the measured temperatures from the ship's perspective and Earth's perspective once equilibrium is already achieved. $\endgroup$ – DaYu1729 Nov 29 '17 at 21:09
  • $\begingroup$ Or, even more simply, consider how a box of ideal gas on Earth looks as seen from a space ship whizzing by. $\endgroup$ – DaYu1729 Nov 29 '17 at 21:40
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If you are going to apply special relativity to thermodynamics then you should use four volume instead of just spatial volume. The conservation of momentum even won't appear true if you didn't use the four momentum there. Instead of air let's assume that your spaceship is filled with solid balls. If the length contraction happens it is not like the size of the balls are going to remain the same and only the spaceship is going to contract. Length contraction will affect everything including the atoms. Therefore there is no any actual increase in pressure even if you are seeing the spatial volume decrease. As I said before you should use the space time volume instead of just the spatial volume.

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