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I was wondering about the hypothetical - and apparently improbable - heat death of the Universe when I stumbled upon this seeming contradiction. A certain volume of space with a uniform distribution of particles has maximum entropy. However, the action of gravity would condense these particles, decreasing the entropy of the system, which would violate the second law of thermodynamics.

My question is simply: what am I missing here? What is the solution to this contradiction?

Dave Coffman
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1 Answers1

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A certain volume of space with a uniform distribution of particles has maximum entropy.

That is correct for non-interacting particles, but wrong for particles with the gravitational interaction. When gravity condenses these particles, it increases the entropy of the system, not decreases it, at least when the Jeans instability condition is satisfied.

To calculate entropy properly, you should consider the phase space volume, and the phase space is built with taking into account all interactions in the system.

firtree
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  • But as Hawking radiation and black hole evaporation prove, gravity wells are, by far, not the highest entropic states, still. – CuriousOne Aug 26 '14 at 01:22
  • @CuriousOne This is a speculation by Penrose, not a rigorous proof. You can take my answer as restricted to classical mechanics and Newtonian gravitation. – firtree Aug 26 '14 at 01:23
  • I must have missed the latest science, where it was proven experimentally, that the universe cares about classical mechanics. – CuriousOne Aug 26 '14 at 01:23
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    @CuriousOne It was proven at least in 17-19th centuries, not the very latest science. – firtree Aug 26 '14 at 01:25
  • Of course, in the 17-19th century, they didn't even know how large and old the universe was... so they were kind of allowed to get it completely wrong. – CuriousOne Aug 26 '14 at 01:29
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    Firtree is correct that gravitational collapse of massive particles to a black hole leads to an increase in entropy. That doesn't contradict the statement that Hawking evaporation into (mostly) zero-mass particles also leads to an increase in entropy. Here is a paper on the latter issue: http://arxiv.org/abs/gr-qc/0609022 –  Aug 26 '14 at 01:39
  • @BenCrowell I improved my answer a bit, please check. I hope it covers also hot gas and photonic gas cases now. But I'm going to sleep so I have to break from further discussion :-) – firtree Aug 26 '14 at 01:45
  • @BenCrowell: I am looking at the article now, but I am not sure about relevance. BH evaporation is only a non-equilibrium process, because BHs are heating up while the universe around them is cooling down. It doesn't have to be that way. There is no physical law against surrounding a BH with a large thermal shield/mirror, that would allow us to control its temperature in a Gedankenexperiment, in which case we can make the evaporation as close to an equilibrium process, as we wish. As long as the universe is colder than the BH, it will evaporate, which increases total entropy, does it not? – CuriousOne Aug 26 '14 at 02:29
  • @BenCrowell: never mind, the author has answered my question, already. He basically does exactly, what I was trying to suggest. – CuriousOne Aug 26 '14 at 02:35
  • @CuriousOne That only means that BHs have negative heat capacity, exactly like ordinary stars. That does not violate anything, that just makes energy transfer speeding up, not slowing down. – firtree Aug 26 '14 at 14:42
  • @BenCrowell: The negative heat capacity doesn't upset me. I was merely wondering about the claim that one had to analyze a BH with non-equilibrium thermodynamics. I don't think that's the case, and I don't think that's what this paper really does. It just takes the right Gedankenexperiment to model the system near equilibrium and the author and I have sort of the same model in mind, after all. – CuriousOne Aug 26 '14 at 17:43
  • @firtree you have plane with gas atmoshere on it, planet gravitation sort gas particals, so T is not uniform, so H is not maximum. Interactions of gas particals, are you serious? – Marat Zakirov Oct 11 '20 at 10:28
  • @MaratZakirov In the case of such planet, the most important would be the interaction between the gas particles and the planet. Also, if the atmosphere is an ideal gas (and there is no sun around to heat it), T in equilibrium would be uniform because of particle collisions. Such atmosphere would follow the wiki:barometric formula. The planet does not sort gas particles by energy, because they exchange energy too much. Though planet does sort particles by mass, in some sense. – firtree Nov 26 '20 at 20:21
  • @firtree $T$ will never be uniform and that is why barometric formula is a mistake. See Lapse Rate in wkipedia or my another question https://physics.stackexchange.com/questions/595758/what-is-the-reason-of-dt-dh-0-in-the-gas-column. I also made a simulation (see link) which shows exactly that $dT/dh < 0$ – Marat Zakirov Nov 27 '20 at 08:33
  • @MaratZakirov This comment thread is not the correct place to discuss your simulations. Your mistake is that your model does not include particle collisions which are crucial for settling the Maxwell & Boltzmann distribution. Thus, the thing you simulate is not the standard ideal gas. It is the collisionless gas and it does not have a thermodynamic equilibrium at all. Instead, each particle make its own isolated subsystem and does not exchange energy (and other quantities) with other subsystems. Please read textbooks before jumping to conclusions. Also... – firtree Nov 30 '20 at 16:14
  • @MaratZakirov Also concerning real-world lapse rate in Earth's atmosphere, its cause is the Sun which heats the Earth's surface and starts a convection. The real atmosphere has mass and energy flows, and it is far from equilibrium. In the equilibrium atmosphere there would be uniform T because the very definition of T is that it is uniform in euqilibrium. Refer to Kittel C., Kroemer H. Thermal Physics for example. – firtree Nov 30 '20 at 16:29
  • @firtree You are completely mistaken. Dry adiabatic lapse rate has no relation with sun because it is adiabatic... See article in wikipedia https://en.wikipedia.org/wiki/Lapse_rate#Dry_adiabatic_lapse_rate. You may also check my simulation of this process in link I have already provided to you. – Marat Zakirov Dec 01 '20 at 10:41
  • @MaratZakirov Ok, this is Q&A site, not a discussion site, so I quit the discussion. Sorry. – firtree Dec 02 '20 at 20:17