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In physics and chemistry we learn that entropy is given by $$S=k\ln\Omega$$

where $S$ is entropy, $k$ is Boltzmann's constant, and $\Omega$ is the number of microstates in the system. We also learn by the 2nd Law of Thermodynamics, that the entropy of the universe always increases. So if we apply both ideas, can we say that the number of microstates in the universe is always increasing? Or is this a naïve approach to the subject?

Chris
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Stoby
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    Formally, this expression of entropy is only valid for an equilibrium macrostate. – Cham Feb 08 '19 at 02:16
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    For one, $\Omega$ should be something more like "the number of microstates consistent with the observed macrostate." This can increase without the total number of microstates itself changing; the usual classroom problem of removing a partition between two gases and allowing them to mix provides an example of this. – Rococo Feb 08 '19 at 02:28

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In statistical mechanics, the word entropy is used for $k\ln\Omega$ where $\Omega$ is the number of microstates that are compatible with specified conditions, such as a given total energy and/or a given total volume. If no conditions are specified, then it's the total number of possible microstates. Therefore, whenever we talk about (or read about) entropy in statistical mechanics, it is important to specify what, if any, conditions are being imposed.

Here's an example. Suppose we start with an airtight box full of gas sitting inside a bigger airtight box with no gas. If the smaller box breaks open, so that the gas molecules are free to move into the larger space, then the number of possible microstates of the gas has suddenly increased. In other words, if we define $S$ in terms of the number of microstates compatible with the given constraints, then $S$ has suddenly increased, at least if we pretend that the change in the available volume is the only information we have.

(By the way, in quantum theory, we count mutually orthogonal microstates.)

Now: is the entropy of the universe always increasing?

Although there are many possible microstates, the universe is in only one of them. We have only partial information about which one we're in, so many (maybe infinitely many) microstates are consistent with the partial information we have. We can define $S=k\ln\Omega$ where $\Omega$ is the number of microstates that are consistent with the information we have, and this $S$ may change with time, if it is finite at all. I don't think our understanding of the laws of nature are complete enough yet to quantify this, at least not for the whole universe, as noted in this post.

Some related thoughts can be found here:

Explain the second principle of thermodynamics without the notion of entropy

Chiral Anomaly
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    Are you assuming the volume of the gas increases as soon as the small box breaks, or are you assuming the new system comes to equilibrium first? – BioPhysicist Feb 08 '19 at 03:07
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    @AaronStevens That's an excellent question/point. I assumed for simplicity that we don't know anything else about what happens to the gas after the inner box breaks, as though it could instantly be in any one of the possible microstates compatible with that larger $V$. In reality, of course, we do have more info. We know that real gas molecules move at finite speeds, etc, and accounting for that knowledge would give an $S$ that increases smoothly. Or, as you said, we could get the same answer by waiting until equilibrium is reached: "equilibrium" = we've lost track of everything except $V$. – Chiral Anomaly Feb 08 '19 at 03:20
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    In what sense can the number of microstates increase if the volume of the "universe" is infinite, or is the increase of the statistical entropy an indication that the eventual volume any gas expands must be finite? – hyportnex Feb 08 '19 at 15:57
  • @hyportnex I don't know how to apply the statistical-mechanics concept of "entropy" to the whole universe. That's why I wrote "I don't think our understanding of the laws of nature are complete enough yet to quantify this, at least not for the whole universe, as noted in this post." However, according to http://arxiv.org/abs/hep-th/0007146, an asymptotically de Sitter spacetime (like ours) has a finite entropy; but this idea involves quantum gravity, so I didn't go there. Practical applic'ns of the 2nd law involve finite systems. – Chiral Anomaly Feb 08 '19 at 17:33
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    Thanks. My question was actually hinting at this statement by Clausius because I have always been a bit surprised by its implication from a historic point of view: "The energy of the universe is constant. The entropy of the universe tends to a maximum." I cannot make any sense out of this unless I assume that energy/entropy/size, etc., are all finite for the whole universe, whatever that may mean in classical physics but in classical physics with "Eucledian 3D space" must be infinite... what am I missing? – hyportnex Feb 08 '19 at 18:02
  • @hyportnex I don't think you're missing anything. I think Clausius was compromising rigor for the sake of being concise. The way Clausius said it is more memorable than this more-careful version: "The energy of any finite closed system, no matter how large, is constant by the definition of 'closed', and its entropy tends to a maximum because in practice we eventually lose track of the motions of all those little molecules." This might have been a more correct way to say it, but then nobody would remember it -- a common theme in the pedagogy of physics. :) – Chiral Anomaly Feb 08 '19 at 18:12