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Just when I thought I was getting an intuition about entropy...

I read in a Q&A here that the only way to lower the entropy of a system is to cool it. I gave that more thought, and I understand it to be true since a hotter substance will have molecules more spread out and randomly dispersed => more possible arrangements of the molecules present => more disorder => higher entropy. So, unusual exceptions aside, higher temperature = higher entropy.

Then I consider the theory that the universe is cooling and its entropy increasing, potentially leading to the 'heat death' of the universe when we reach maximum entropy and absolute zero temperature. Am I misunderstanding this theory?

How can it be true that higher temperature = higher entropy of a system, and yet maximum entropy for the universe would be the maximum dispersal of heat and thus ultimately-low temperature? Is it that for open systems, higher temperature = higher entropy, but for closed systems, gradual cooling and gradual creation of entropy goes together?

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  • $\begingroup$ Hint: gravity. The more or less smooth, hot early universe is, because of gravity, a state of extraordinarily low entropy. Another hint: most of the entropy of the (observable) universe is due, by far, to supermassive black holes. $\endgroup$ – Alfred Centauri Apr 13 '18 at 2:23
  • $\begingroup$ Cooling (e.g., of blackbody radiation) simply implies that the entropy per unit volume is decreasing, but the volume of the universe is scaling up as it expands. $\endgroup$ – Bert Barrois Apr 13 '18 at 11:34
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Entropy $\rm S$ depends on the number of possible ways ($\rm Ω$) a system can be distributed. One way of having different systems organization is by attributing different energies to the particles in it. Increasing temperature $\rm T$ you increase entropy by giving a bigger range of energies from which your particles can share.

Another way to increase $Ω\rm $ is by giving more room from which your particles can move, i.e., by increasing the volume $\rm V$ of your system.

Starting from the definition of entropy $$\rm dS=\frac{δQ}{T}$$ and the first law of thermodynamics $$\rm δQ=dU+pdV$$ Considering that $\rm U$ is the total internal energy (energy inside) the universe, which is constant, it means that $\rm dU=0$ and therefore $$\rm dS=\frac{pdV}{T}=R \frac{dV}{V}$$ here we made use of the relation $\rm pV=RT$.

Integrating $$\rm ΔS=R \ln(\frac{V}{V_0})$$ $V_0$ is current universe’s volume. If the universe continues to expand Indefinitely $V \rightarrow \infty$ and consequently $$\rm ΔS \rightarrow \infty$$

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    $\begingroup$ You have tripped over a paradox about the total energy of the universe. Blackbody radiation cools adiabatically as if it were doing $-PdV$ work against a (nonexistent) moving wall. Newton would say that gravitation potential energy of the universe is increasing, but that concept does not apply in GR. The number of photons and their entropy actually stay constant. $\endgroup$ – Bert Barrois Apr 13 '18 at 11:52
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    $\begingroup$ @BertBarrois Cool. Would you give me some more details about that. Maybe some further reading. About the paradox, I mean. $\endgroup$ – J. Manuel Apr 13 '18 at 12:46
  • $\begingroup$ I’d suggest Weinberg’s oldie-but-goodie Gravitation and Cosmology, which derives $\tfrac{d}{dR}(\rho {{R}^{3}})=-3P{{R}^{2}}$ from Einstein’s field equations on page 472, without reference to imaginary walls. The GR result agrees with the classical thermo result for adiabatic expansion. $\endgroup$ – Bert Barrois Apr 13 '18 at 15:08

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