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If I understand correctly, spontaneous decreases in entropy are possible and don't actually violate the 2nd law of thermodynamics -- it's just that, for a system of more than a few particles, the probability of that happening in any reasonable amount of time is sufficiently small as to be negligible. Does that mean such a spontaneous decrease in entropy could theoretically result in a system temporarily reaching absolute zero, even if the probability is so small that we should never expect it to happen?

Basically, assuming our current understanding of thermodynamics is correct, is the probability of a system reaching absolute zero precisely zero, or just negligibly small?

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    $\begingroup$ fluctuations do not change thermodynamic macroscopic equilibrium entropy, constraints do; see physics.stackexchange.com/questions/761468/… $\endgroup$
    – hyportnex
    Jul 1, 2023 at 23:30
  • $\begingroup$ Note that it is impossible in the thermodynamic limit. If you have a tiny system, then it is easy to get it to ground state, which would then likely have zero entropy. Note again, that some ground states have entropy. $\endgroup$ Jul 2, 2023 at 6:38
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    $\begingroup$ Does this answer your question? How exactly does it follow mathematically from the 3rd law of thermodynamics that it's impossible to reach absolute zero? $\endgroup$
    – Farcher
    Jul 2, 2023 at 8:26
  • $\begingroup$ @Farcher, no. I'm the one who posted that other question in the first place, then I realized it wasn't really what I wanted to ask, but it had already received an answer at that point and so it wouldn't have been fair to edit it to change the meaning. $\endgroup$ Jul 2, 2023 at 11:05
  • $\begingroup$ @MikaylaEckelCifrese Your reason is well explained and I thank you for it. $\endgroup$
    – Farcher
    Jul 2, 2023 at 13:35

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A widespread misconception in Statistical Mechanics is using microstates (microscopic states) instead of equilibrium macrostates (states of thermodynamic equilibrium).

A system of a few particles may show fluctuations corresponding to microstates typical of low-entropy macrostates. For example, a finite temperature cluster of 32 atoms may arrive, during its dynamical evolution, at a configuration close to a perfect fcc structure. However, this would be a temporary fluctuation between many other configurations. Thermodynamic entropy is an equilibrium property, i.e., a property of the ensemble of microstates; it is not a property of a single microstate. Therefore, it is improper saying that the cluster at the instantaneous fcc configuration has lowered its entropy. Of course, at very low temperatures, the same system will remain close forever to the fcc structure, provided it is a local minimum of the potential energy.

Similar considerations apply to the question if there is a finite probability a thermodynamic system of a few particles could get the absolute zero of temperature as a temporary fluctuation. Temperature is an equilibrium property, thus depending on the whole set of microstates associated with the equilibrium macrostate. We can follow the evolution of the kinetic energy of a small system which shows fluctuations. However, a local fluctuation cannot be considered representative of the equilibrium average. Therefore, a fluctuation reducing the kinetic energy to its minimum value cannot be considered a fluctuation of the thermodynamic temperature of the system toward zero.

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  • $\begingroup$ I assume you're talking about this FCC structure $\endgroup$
    – PM 2Ring
    Jul 2, 2023 at 7:52
  • $\begingroup$ @PM2Ring almost that. The image you linked contains 14 atoms. One needs to add more atoms in the three directions up to a symmetric structure of 32 atoms. $\endgroup$ Jul 2, 2023 at 11:52
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For what it’s worth, this is a non-statistical mechanics response to the title of your post.

Consider using a Carnot (most efficient) refrigeration cycle in an attempt to remove heat $Q_C$ from a cold reservoir that approaches absolute zero temperature ($T_{C}\rightarrow 0$). The coefficient of performance of the cycle is

$$COP=\frac {T_{C}}{T_{H}-T_{C}}$$

Note that as $T_C$ approaches absolute zero, the COP approaches zero. Now consider that the general definition of the COP, for any cycle, the desired heat transfer divided by the work necessary to cause the transfer, or

$$COP=\frac{Q_C}{W}$$

Therefore as the COP approaches zero it takes an infinite amount of work to remove heat from the cold reservoir. In other words, the work required to remove heat from a substance increases substantially the colder you try and make the substance. An infinite amount of work would be required to reach absolute zero. See https://blogs.scientificamerican.com/observations/racing-toward-absolute-zero/

Hope this helps

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