# Relativistic refrigerator

1. Mar 13, 2014

### jartsa

Let's say we have a long cylinder and two pistons inside the cylinder, and some vapour between the pistons.

Now we make the pistons accelerate with some coordinate acceleration, along the cylinder axis, same coordinate acceleration for both pistons. Cylinder is static.

Going to the frame of the pistons, we can see that vapour will cool and then turn to liquid, as the vapour is doing work, as distance between the pistons increases, in the pistons frame.

Now let's go to inertial frame and observe our vapour condenser. We see entropic vapour turning into less entropic liquid. So the question is: Were does the entropy go, in the inertial frame?

Last edited: Mar 13, 2014
2. Mar 13, 2014

### ghwellsjr

Do the pistons move inside the cylinder? If so, how? If not, would there be any difference between a cylinder with two pistons and a closed cylinder or any other enclosed volume?

Also, what do you mean by "some coordinate acceleration"?

I assume in your second sentence that "cylinders" is a typo, correct?

3. Mar 13, 2014

### jartsa

It's a tank that does not lorentz-contract, full of vapour. There's lot of Bell's spaceship paradox influence in this thing.

And yes I got cylinder and piston mixed two times.

An alternative formulation: Two spaceships accelerate, there's some gas between the spaceships, instead of a thread.

Last edited: Mar 13, 2014
4. Mar 13, 2014

### Staff: Mentor

I am not aware of a generally-accepted framework for relativistic thermodynamics. Specifically, I don't know what thermodynamic properties are invariants or tensors.

5. Mar 13, 2014

### Staff: Mentor

How? By letting the vapor push them? If so, I don't think you can specify how they will accelerate, since that will be determined by the vapor's properties. If you accelerate them by some other means, then the vapor isn't doing work on them, so your description of what happens to the vapor is incorrect.

6. Mar 13, 2014

### jartsa

Can't I just say "they are accelerated". Well ok, it happens the same way as Maglev trains are accelerated. Identical accelerations for both pistons, in an inertial frame.

But this may be a bad thought experiment anyway.

So how about if we have the two spaceships from the Bell's spaceship paradox. And then we have a Bob who is trying to push the two accelerating spaceships apart using a long stick.

Now Bob says "I'm successfully pushing those ships apart, the work I do becomes kinetic energy of the spaceships"

What does an inertial observer say about the work done by Bob?

Last edited: Mar 13, 2014
7. Mar 13, 2014

### Staff: Mentor

You can certainly specify this; what you can't do is specify this and then assume that the vapor does work.

I assume you mean by this that the spaceships are firing their engines, and the thrust produced *by the engines* is exactly the right amount to produce exactly the observed acceleration of the spaceships (since that's how the Bell spaceship paradox in its usual form is specified). That's the crucial assumption.

The same thing any other observer says: Bob is actually doing zero work, because the actual acceleration of the spaceships is completely accounted for by the thrust being produced by their engines.

8. Mar 13, 2014

### jartsa

Well, the spaceships are identical in every way, and then there's the Bob making things just a little bit non-symmetric. So this is not the Bells spaceship scenario. I thought it was, but it isn't

But if we have some understanding about Bells spaceship paradox, we can use that understanding here. In the real Bell's spaceship paradox an observer co-moving with the spaceships sees with his eyes the distance of the spaceships increasing. In this Bell's spaceship paradox resembling thing Bob feels in his muscles the distance of the spaceships increasing. Let's say Bob quits when he feels like he's pushed 100 meters with a force of 1000 Newtons.

Inertial observer disagrees about how much the distance between the spaceships increased. He says the distance increased by 1 mm.

9. Mar 13, 2014

### Staff: Mentor

Then you need to specify what the scenario *is*. How much thrust do the spaceship's engines exert? The same amount as in the standard spaceship paradox? And Bob's push is in addition to that?

But that's without Bob there, so Bob's presence is not required to explain why the distance increases.

No, he feels in his muscles the distance increasing *more* than it would if it were just the spaceships' engines firing. He does *not* feel in his muscles the work that the engines are doing. So the work he does with his muscles can't be translated directly into an increase in the distance between the spaceships, because his muscles are not the only force present.

10. Mar 14, 2014

### jartsa

Yes.

No, I changed my mind, because I looked up John Bell's version of the Bell's spaceship paradox, so I'll go along that version: The thrust profile of the spacesips is programmed so that the space ships have the same coordinate acceleration. Bob's push is known beforehand and is taken into account.

Yes

No

Last edited: Mar 14, 2014
11. Mar 14, 2014

### Staff: Mentor

In bell's spaceship paradox there is not and cannot be an "observer co-moving with the spaceships". Think about it for a moment: there is an observer for whom the distance between the ships does not increase - but he's not co-moving with the ships, they're accelerating identically relative to him. All other observers see the distance between the ships changing; therefore the ships have relative acceleration and it's not possible to be co-moving with both of them.

This is not just a quibble: the resolution of the paradox requires understanding that "co-moving" is coordinate-dependent in non-inertial coordinates. Introducing Bob and his stick is equivalent to rewriting the problem so that the accelerations of the ships are not the same for the original ground guy. That doesn't change the basic physics any.

12. Mar 14, 2014

### JesseM

What do you mean by "accelerating identically relative to him" (relative to him in what frame? his instantaneous comoving inertial frame, or some non-inertial frame?) and "see the distance between the ships changing" (again, what frame is being used to define what he 'sees'?) Whatever the precise meaning, I don't think your statement above will turn out to be correct--if the two ships have the correct proper accelerations and distance to give them fixed position coordinate in a Rindler coordinate system, then any other accelerating observer who is also at a fixed position coordinate in the same Rindler coordinate system will 'see' the two ships as having a constant separation. Of course this is trivial if we are using Rindler coordinates to define what the observer 'sees', but it's still true if we use the observer's instantaneous comoving inertial frame--the ship's separation in the observer's instantaneous comoving frame at time T1 will be the same as the ship's separation in the observer's instantaneous comoving frame at time T2. Rindler coordinates are specifically constructed so that this will be true for a family of 'Rindler observers' who all have a fixed position coordinate in the Rindler system, see the discussion here which says "We can imagine a flotilla of spaceships, each remaining at a fixed value of s by accelerating at 1/s. In principle, these ships could be physically connected together by ladders, allowing passengers to move between them. Although each ship would have a different proper acceleration, the spacing between them would remain constant as far as each of them was concerned. The easiest way to see this is to note that increasing q is equivalent to performing a boost on everything, leaving intervals in spacetime invariant."

edit: Sorry Nugatory, I misremembered the Bell spaceship paradox and so misunderstood you, see below.

Last edited: Mar 14, 2014
13. Mar 14, 2014

### WannabeNewton

Except that the spaceships in the Bell spaceship paradox do not follow orbits of the same time-like Killing field in a Rindler coordinate system comoving with either spaceship. Any instantaneous rest frame of one spaceship will fail to be comoving with the other spaceship because of the non-vanishing divergence of the tangent field the spaceships follow orbits of.

So what Nugatory said is perfectly fine.

14. Mar 14, 2014

### JesseM

Ah, I was misremembering the scenario imagined in the Bell spaceship paradox, I thought it was actually dealing with ships that accelerate in a "Born rigid" way (constant Rindler coordinate) but now I see it's just dealing with a situation where they have the same coordinate acceleration in some inertial frame, so I misunderstood what Nugatory was saying.

Looking at the OP's original question, I'm not sure why there's supposed to be anything paradoxical about the entropy changing when the volume changes in the rest frame of an average molecule--even without considering relativity, if you increase the volume of a thermodynamically isolated box filled with an ideal gas, the entropy should increase according to the Sackur-Tetrode equation. Not sure whether the same is true of the OP's scenario, as DaleSpam said you'd need a relativistic formulation of thermodynamics, but this shows there is nothing strange about the idea that entropy can change when you change an isolated system's volume (and of course the box in the OP scenario isn't really 'isolated' since in any inertial frame the kinetic energy of the walls of the box must be changing as they are accelerated). Also, in this nonrelativistic example, it's not actually true that the "vapour will cool" as you increase the volume--the energy U and particle number N won't change since the system is isolated, and the page I linked to mentions that for an ideal gas temperature is given by T=2U/3Nk (which is just the equipartition theorem for an ideal monoatomic gas)--so I don't know what the OP's reasoning is in thinking the temperature should go down in the relativistic case.

Last edited: Mar 14, 2014
15. Mar 14, 2014

### jartsa

Actually the OP says that the vapour turns into liquid, which is a decrease of entropy. The OP should think to himself "does that kind of thing actually happen?" Then the OP would realise that only part of the vapour will turn ito liguid, and the entropy went into the vapour that was warmed by the condensation process.

Well how about this alternative: The temperature can go down, because the system is not really isolated.

16. Mar 14, 2014

### Staff: Mentor

All of the thermodynamics stuff is just guesswork unless you have a relativistic theory of thermodynamics. How does entropy transform between reference frames? How does temperature transform? I would be surprised if either were invariant.

I am sure that some work has been done on the subject, I just don't know the details.

17. Mar 14, 2014

### pervect

Staff Emeritus
MTW has a decent approach, based on statistical thermodynamics adaquate to handle most problems of interest (it's applied mostly to stellar models in the book).

Entropy can be identified with (the logarithim of) the number of states per particle, then entropy per particle then becomes a frame invariant quantity. (I believe MTW uses entropy per unit baryon). Particle density transforms via the number-flux 4-vector, of course. Temperature and the relationship dS = dQ/T is defined only in the rest frame of the fluid. Trying to generalize temperature as part of a tensor quantity seems to be what causes most of the arguments, it's simpler to avoid the issue.

I think it's also reasonable to regard entropy as a number density in and of itself, that transforms in the same way as a number-flux 4-vector, rather than tie it to particles. The issue that seems to cause all the arguments is temperature.

Trying to do thermo clasically rather than statistically also becomes highly confusing.

18. Mar 15, 2014

### yuiop

You are right that it is not possible to be comoving with both spaceships, but in the reference frame of either spaceship the radar distance to the other spaceship is increasing and perhaps that was what jartsa intended. Now consider tis thought experiment that I hope is the spirit that jartsa intended.

Let's call the comoving reference frame S' and the spaceship that is stationary in this reference frame A. A cylinder that is open at one end is attached to A at the closed end. The cylinder has a thermal jacket in which liquid is circulated to keep the temperature constant as measured by A. Another spaceship (B) has a piston that closes off the open end of the cylinder. Initially the enclosed volume within the cylinder is small and contains a volatile liquid with a small vapour space and the vapour is in equilibrium with the liquid.

The spaceship accelerate in such a way that the enclosed volume within the cylinder slowly increases as measured by A but remains constant as measured in a given inertial frame (S) in which both spaceships have coordinate acceleration. The volume increase is intended to be slow enough to maintain thermodymic equilibrium of the system at all times (quasi-static). According to A the volatile liquid evaporates as the volume of the cylinder increases until none of the volatile substance is left in the liquid phase and the cylinder contains vapour only. Now the interesting part is that the experiment predicts that in the inertial reference frame S, the the volatile liquid completely evaporates without any change in the enclosed volume. The Newtonian prediction is that this would only be possible with a significant increase in temperature and pressure within the constant volume.

19. Mar 15, 2014

### pervect

Staff Emeritus
It's been a while since I've done thermo. But I believe that an adiabaitic compression of an ideal gas is perfectly reversible. Which means no entropy change occurs during such an adiabatic compression. The form of the compression wasn't specified, so I'm assuming the simplest one.

Assuming my recollection is correct, then, applying MTW's prescription, and additionally assuming that the gas being compressed is compressed adiabatically, then there entropy per particle stays constant. The number of particles per unit volume transforms in the usual manner, via the number-flux 4-vector. I'm not sure the OP knows about 4-vectors in general, but I sense a lack of interest on their part, so I don't currently see a point into going into more detail.

Wiki says:
The mathematical equation for an ideal gas undergoing a reversible (i.e., no entropy generation) adiabatic process is

$P V^{\gamma} = \operatorname{constant} \qquad$

which agrees with my recollection that the adiabatic compression is reversible and doesn't generate entropy.

Last edited: Mar 15, 2014
20. Mar 16, 2014

### jartsa

Very good

Observer A will measure the weight of thermal jacket decreasing, because the volatile liquid sucks energy from the thermal jacket.

Does the weight of vapour + volatile liquid increase, according to observer A?

Last edited: Mar 16, 2014
21. Mar 16, 2014

### jartsa

Have you perhaps mixed up compression and expansion?

When ideal gas expands, doing the maximum work, entropy stays constant. (my question about where the entropy goes was answered there)

Gas cools when it does some work. (that's were the "refrigerator" came from)

The gas loses energy, the energy goes somewhere.

Last edited: Mar 16, 2014
22. Mar 16, 2014

### Staff: Mentor

In that case, entropy would be a vector and if it were purely timelike in one frame then in another frame it would have spacelike components. I am not sure what spacelike entropy components would mean thermodynamically.

23. Mar 16, 2014

### yuiop

What you say is true, but not all adiabatic compression/expansion is reversible. If there is a vacuum outside the cylinder, no work is done during the expansion phase, and this results in a drop in pressure and no change in temperature, but there is an increase in entropy making the process spontaneous and irreversible. Unfortunately, because there is no change in temperature in this case, the refrigeration effect jartsa mentions would be absent.

Since the total number of particles in the system is unchanged by a transformation and since the volume transforms the same way as length, the number of particles per unit volume in the transformed frame would be a factor of $y = 1/\sqrt{1-v^2}$ greater than in the instantaneous rest frame of the system. (I am not using the usual relativistic gamma symbol ($\gamma$) to avoid confusion with the thermodynamic gamma factor which is already in use. What does MTW have to say about how entropy transforms? I assume that is what is jartsa is seeking the answer to. Presumably a zero change in entropy in the rest frame would be a zero change of entropy in any other frame. Causality and time's arrow would also suggest an increase in entropy in any reference frame would be an increase in entropy in any other reference frame. To better understand how entropy transforms, we need a thought experiment where there is a change of entropy, such as the external vacuum case I mentioned above. A none zero pressure outside the cylinder piston/s would be a nightmare to analyse anyway, because external gas molecules would be colliding with the external face of the leading piston would increase the effective external pressure and for a trailing piston there would be a reduction in external pressure. .

Last edited: Mar 16, 2014
24. Mar 16, 2014

### WannabeNewton

25. Mar 16, 2014

### yuiop

Tolman states that in the reversible adiabatic case there is no change in entropy (which is correct), but then directly concludes the entropy transforms as $S = S_0$. I think that argument is flawed. For example if the change in length in one reference frame is zero in one reference frame, then the change in length in all other reference frames is also zero, but it does not follow that generally, $L=L_0$ except in the special case that $L=0$. Similarly, $S = S_0$ may not be generally true and may only apply to the special case of an adiabatic reversible process with zero entropy change.

P.S. I am not saying Tolman is wrong, just that the justification he presents is not very good.

Last edited: Mar 16, 2014