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Here is a following problem I encountered when chatting about physics with my friend:

Let us imagine a classical example of ordered state of matter in thermodynamic sense: let's take a cylinder filled with air in normal pressure. Now let's make order in the chaotic movement of air particles and arrange their speed so that all the particles are coherently revolving around the axis of the cylinder in such a way, as to change only direction of the movement of each particle, not its velocity; so that the energy (and energy distribution) remains unchanged.

It is clear, that immediately after such magical modification, there is a lot of energy that can be extracted from such system, for instance if we have had installed a wind turbine inside the cylinder.

So the question is:

Will the system of cylinder with air inside change its weight if we change its entropy only?

The question is related to Is a hard drive heavier when it is full?, but substantially different, because there is no potential energy involved here.

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Adam Ryczkowski
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  • You're being challenged about your example. But I think you're on to a good question. Maybe you should change your example into something wherein only the entropy changes? Such as, if the air was swirling inside a cylinder (more order) as opposed to if it was moving around in typical Brownian motion (more chaos)? Keeping all other energies the same, of course. – markovchain Jan 31 '13 at 23:29
  • @markovchain Good example, I'll do that. – Adam Ryczkowski Feb 01 '13 at 06:40
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    This is a surprisingly interesting question. If your example had been mixing two similar gases, or short-circuiting a battery inside an insulated container (so that all the heat is kept inside) then the answer would be clear: the relevant energy is the total energy (or "first law energy"), not the free energy, so the weight would not change. However, your example involves changing the angular momentum, and momentum is deeply connected to energy in relativity. This makes me hesitant to say that the same thing will apply here. – N. Virgo Feb 02 '13 at 04:48
  • @Nathaniel. Yes, I'm aware of that. I'm a physicist myself, and this difference in interpretation of seemingly simple idea of information from the perspective of physicists and from the perspective of statisticians was pondering my mind for over 15 years... it's the first time I've managed to articulate this out loud :-) I personally believe, that in future there will be a "law of equivalence between information and energy" which will shed a completely new light on our perception of the universe... ;-) – Adam Ryczkowski Feb 02 '13 at 09:23
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    @AdamRyczkowski the tricky issue here is not information but momentum. The information has little to do with it. If we were talking about a chemical process or mixing gases, I would say the information appears in the $dS$ term, but to work out the change in weight we need only the $dU$ term, which is zero by construction, hence there is no change in weight, no matter what happens to $S$. But in your example there are momentum changes as well, which means we need to consider more than just $dU$. – N. Virgo Feb 02 '13 at 09:48
  • Yes, you are right. – Adam Ryczkowski Feb 02 '13 at 20:53

2 Answers2

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Now let's make order in the chaotic movement of air particles and arrange their speed so that all the particles are coherently revolving around the axis of the cylinder in such a way, as to change only direction of the movement of each particle, not its velocity; so that the energy (and energy distribution) remains unchanged.

1) By making order you are adding rotational motion ( angular momentum), which was 0 before. This means you have somehow added rotational energy to the system.

2)This physical system is unsustainable after you stop inputting the rotational impulse. The molecules will be going off on a tangent the instant the rotationally constraining field goes off. It cannot be a closed system.

Will the system of cylinder with air inside change its weight if we change its entropy only?

Ignoring your badly formulated example, the answer on whether a change in entropy affects the weight of a mass ensemble is that if the ordered state is reached by the improbable throw of the dice as a possible microstate in the statistical formulation of entropy, it is one of the states that are counted in the sum that defines entropy, the weight will not change. If order happens because of extra binding forces: a crystal will have less mass than the sum of its constituents, yes the weight will change.

anna v
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  • I think you should give him a little more leeway. It isn't impossible, as you said, to get something like that if you're really lucky. I also believe you should elaborate more on the "no, changing the entropy doesnt change the mass." – markovchain Feb 02 '13 at 01:26
  • @markovchain well, i tried elucidating a bit. – anna v Feb 02 '13 at 04:36
  • @anna So, you are implying, that there is no fluctuation of mass... but you realize, that one can store arbitrary large amount of energy from this fashion. – Adam Ryczkowski Mar 28 '13 at 10:28

  • 2)This physical system is unsustainable after you stop inputting the rotational impulse. The molecules will be going off on a tangent the instant the rotationally constraining field goes off. It cannot be a closed system.

    Gas/liquid will still rotate on its own for some time. So let's make it not a gas, but a rotating ring. It doesn't change the issue.

     – Adam Ryczkowski Mar 28 '13 at 10:30
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    A rotating ring in vacuum just needs the original impulse of rotation, lets suppose two jets in opposing sides of the ring. It will turn the energy of the jets into rotational kinetic energy of the ring which will be conserved and it will rotate continually without change of mass or energy. The difference with your gas example is that there are cohesive forces on the molecules and there is no need to keep feeding rotational energy to keep the circular path. – anna v Mar 28 '13 at 12:03
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    Any energy taken out from such a rotating system will be taken out of its initial rotational energy. – anna v Mar 28 '13 at 12:04
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Your example has a tricky issue involving angular momentum (see below), but I can address the spirit of the question using a much simpler example. Let us imagine we have a chamber containing two gases, $A$ and $B$, such that $A$-$A$ interaction terms are equal to both the $B$-$B$ interaction terms and the $A$-$B$ interaction terms. (It doesn't hurt to imagine the molecules as red and blue spheres of the same size and weight.)

Now we imagine swapping some of the $A$ molecules with $B$ molecules in such a way that the $A$ molecules all end up on one side of the chamber and all the $B$ molecules on the other. Due to the assumption above, swapping two particles doesn't change the total energy, but by doing this we have created an ordered system from which work can be extracted. (To extract work from this system you need a piston that's permeable to $A$ molecules but not $B$ molecules, and another piston that's only permeable to $B$ but not $A$. See here for details on how, as well as some other relevant stuff about the relationship between work and entropy.)

Now, since we can extract work from this system, will its weight change due to the $E=mc^2$ relation? Perhaps surprisingly, the answer is no. This is because the relevant $E$ is the internal energy of the system (usually written $U$ in thermodynamics), and that hasn't changed.

The work that can be extracted from a system at a constant temperature is given by $U-TS$. By reordering the atoms we've reduced $S$ but kept $U$ constant. When we extract work from this ordered system, its internal energy $U$ also stays constant, but the energy of the environment reduces. Effectively, we take heat out of the environment and convert it to work. Normally this isn't allowed by the second law, because converting heat into work would cause a reduction in entropy - but through a clever use of semi-permeable pistons we can offset that reduction in entropy by an increase in the entropy of the gas mixture.

The point is that the $S$ term represents the disorder (or, more correctly, it represents the opposite of information - see the paper linked above) and the $U$ term represents the energy. The mass of an object depends only on the $U$ term and not on the $S$ term, so order and mass/energy are actually quite independent things.

The tricky thing with your example is that although you keep the energy constant, changing the velocities in the way you describe adds angular momentum to the system. When momentum is involved, $E=mc^2$ is no longer strictly valid, and you have to use the full energy-momentum relation $$ E^2 = (pc)^2 + (mc^2)^2, $$ where $p$ is the relativistic momentum. I do not know how to do this for your example. It may or may not be the case that the effective mass would change. However, if it does change it's because the momentum has changed and not because the system has become more ordered.

N. Virgo
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