Since the universe's entropy is going to increase indefinitely and it is going to be a continuous, does this mean that it will be equal to some power of pi or e or any other transcendental number?
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$\begingroup$ I cannot see why the entropy will end up being a transcendental number. Could you clarify why are you expecting that? $\endgroup$– user65081Oct 8, 2020 at 19:01
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$\begingroup$ But in any case, the entropy will be finite when it reaches thermodynamical equilibrium, unless the number of particles in the universe is infinite $\endgroup$– user65081Oct 8, 2020 at 19:03
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$\begingroup$ If you identified the entropy of the universe as a single value at any point of time, would it be exact to an infinite degree like how pi is (not taking into account the ability to measure entropy to infinite degrees)? $\endgroup$– yoloOct 8, 2020 at 19:19
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$\begingroup$ No, it will slightly fluctuate, and the probability to reach a given fluctuations will increase exponentially with the entropy difference between equilibrium and out of equilibrium fluctuation $\endgroup$– user65081Oct 8, 2020 at 19:41
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$\begingroup$ see poincare recursion theorem $\endgroup$– user65081Oct 8, 2020 at 19:42
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1 Answer
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If by ‘the universe’ you mean everything within our event horizon (a comoving volume), which exists because the expansion of the universe is accelerating due to the vacuum energy density, then the universal entropy growth will asymptote to the entropy of the future cosmic event horizon.
The expansion of the universe will cause comoving objects like galaxies, black holes, photons etc to recede beyond our cosmic event horizon (CEH), taking their entropy with them. However, this is compensated by the CEH entropy, which continues to asymptotically grow toward the de Sitter far future, where there is only vacuum energy.
- This is all beautifully outlined in this paper by Lineweaver et al A LARGER ESTIMATE OF THE ENTROPY OF THE UNIVERSE.
However, your intuition isn’t wrong, the CEH entropy at the beginning - hot big bang - was a power of pi, see this answer (also implied in Figure 7 of Lineweavers paper)
