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One of the assumptions of the thermo textbook I'm reading is that the probability of any given microstate of a system is equally likely to occur. This does not mean that the macrostates of the system are all equally probable, since there will usually be several microstates corresponding to a particular macrostate which is expressed as the multiplicity of the macrostate.

  1. Under the assumption that every microstate is equally probable, the formula for entropy is $S=k_B\ln(\Omega)$. Hence, $\Omega=e^{S/k_B}$.
  2. The second law states that every thermodynamic system tends toward higher entropy. Since the exponential function is increasing, higher entropy implies higher mutliplicity.
  3. As the multiplicity of a system increases, the number of macrostates with higher multiplicity decreases.

But doesn't this imply that there should be a steadily decreasing probability of entropy increasing as there are fewer and fewer states with higher entropy? And once the system achieves maximum equilibrium, isn't the system almost certainly going to revert to a state with lower entropy? There must be something I don't understand.

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  • $\begingroup$ Entropy is a measure of the number of acceptable states of a system, not the likelihood to be in each state. Some of your later statements are difficult to interpret, e.g. "But doesn't an increase in overall entropy leave fewer possible configurations with higher entropy". A state doesn't have entropy. All of the states define entropy. I'm a bit confused. I recommend you list your statements in order and identify how they imply a contradiction in a more organized manner. $\endgroup$
    – user196418
    Commented Dec 29, 2018 at 13:08
  • $\begingroup$ Or, were you referring to macrostates and I interpreted it as a microstate-ment. $\endgroup$
    – user196418
    Commented Dec 29, 2018 at 13:09
  • $\begingroup$ Don't different macrostates generally have different entropy? I thought entropy was related to the multiplicity of a particular macrostate, not the macrostates as a whole. $\endgroup$
    – Alex S
    Commented Dec 29, 2018 at 13:16
  • $\begingroup$ A macro state is defined by the macro variables, S is one. Yes, I think I misinterpreted your collection of statements. Entropy is related to the number of accessible states with truly independent configurations (i.e. modulo particle exchange). What is multiplicity of a macrostate? $\endgroup$
    – user196418
    Commented Dec 29, 2018 at 13:20
  • $\begingroup$ The multiplicity of a macrostate is the number of microstates that correspond to a particular macrostate and in all the literature I've read it's denoted $\Omega$ as in the formula $S=k_B\ln(\Omega)$. $\endgroup$
    – Alex S
    Commented Dec 29, 2018 at 13:23

2 Answers 2

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One of the assumptions of the thermo textbook I'm reading is that the probability of any given microstate of a system is equally likely to occur.

Without further qualification, it is impossible to check the truth of the above sentence. From what you write after and from the comments, I guess that you are referring to the microstates of an isolated system at equilibrium.

In such a case, the macrostate is not something one can play with unless the equilibrium conditions are changed. For example, for an isolated fluid system, the equilibrium macrostate is characterized by given energy values, volume, and number of molecules ($U,V,N$). Therefore, once the macrostate has been fixed, nothing else may happen to the system at equilibrium. Indeed, the multiplicity of the macrostate cannot vary spontaneously.

Indeed, the logarithm of the multiplicity of a macrostate $\Omega$ is connected to the system's entropy through the celebrated Planck-Boltzmann formula $S=k_B \log \Omega$. But to see an increase in entropy, one has to do something to the system. For example, if the fluid is in an isolated, rigid, and impenetrable container, and at some time the container increases its volume, the multiplicity of the microstates and the entropy will also increase, similarly, for increases of energy or number of particles. However, this example shows no problem with "exhausting" the number of microstates because, with the rise of the state function, the macrostate depends on the increase in number. From the formal point of view, this is granted by the positivity of temperature, pressure, and the negative sign of the chemical potential.

Of course, this is not the only way to increase a macrostate's multiplicity. Even removing some internal constraints can get the same effect (there are plenty of examples of containers divided into two subsystems in statistical mechanics textbooks). The common feature is that more microstates become available for the total system after removing the constraint.

So, in almost all the conditions where one would expect an increase of entropy, there is an accompanying increase of the microstate multiplicity compatible with the new macroscopic conditions.

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  • $\begingroup$ So once a system is in a particular macrostate, it remains fixed, or is this only true if it's isolated and in equilibrium? $\endgroup$
    – Alex S
    Commented Dec 29, 2018 at 14:14
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    $\begingroup$ If there is no change in the macroscopic conditions which define the state it remains fixed. If the system is not isolated, entropy (and then the multiplicity of the macrostate) does not play a central role, it shouldn't be maximum, and other thermodynamic potentials/probability distributions in the phase space would be relevant. If the system is not at equilibrium, in general a thermodynamic description is not enough. But in that case, one has to clarify which kind of non-equilibrium is characterizing the system. $\endgroup$ Commented Dec 29, 2018 at 15:59
  • $\begingroup$ I thought that if a system was not a maximum entropy, then it wasn't at equilibrium, hence why I concluded that the macrostate would be changing to achieve equilibrium. $\endgroup$
    – Alex S
    Commented Dec 30, 2018 at 1:39
  • $\begingroup$ It is quite a common misunderstanding. Maxent principle is ok. But the key point is to have clear in mind maximum with respect to what. It cannot be maximum wrt its macrostate variables because they have to be fixed to specify the macrostate. The maximum should be intended as wrt any additional variable specifying an additional constraint on the system. It turns out that constraints reduce the multiplicity at fixed macrostate variables. As soon as they are removed, the number of available microstates compatible with the macrostate increases until the system is completely unconstrained. $\endgroup$ Commented Dec 30, 2018 at 5:57
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2) The second law states that every thermodynamic system tends toward higher entropy. Since the exponential function is increasing, higher entropy implies higher mutliplicity.

3) As the multiplicity of a system increases, the number of macrostates with higher multiplicity decreases.

I don't think I fully understand your complete argument, but I think your flaw might be in one of these two points.

It is true that entropy and multiplicity are related, but it seems to be like you are considering the change in one of these values as a cause and the change in another one of these values as an effect, when in fact they are exactly the same thing.

For the purposes of this question, you can essentially reword the second law as "the most likely thing will definitely happen". So the system will just go to whatever macrostate has the most microstates because that is what is most likely. It doesn't really matter how many macrostates exist for some multiplicity.

Of course you can get into finer details. If there is a single macrostate with the largest multiplicity, and if there are fluctuations that causes our system to be in a different macrostate, then the entropy would technically decrease. But then the system will just move back to the higher entropy state right after. It will not keep decreasing in entropy. And even then these fluctuations are probably not large enough to be concerned with. Imagine the highest entropy state as a stable equilibrium, where any small perturbations won't kick us out of equilibrium, and on average we can choose to ignore the perturbations.

Like I said, I don't know if I fully understand your question, so I'm sorry if this answer seems like I'm just throwing information and thoughts out to see what sticks.

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  • $\begingroup$ I think I understand now, the equilibrium state is essentially a sharp spike on the PDF of available states, so that it becomes fantastically unlikely for any deviations to occur. $\endgroup$
    – Alex S
    Commented Dec 29, 2018 at 14:37
  • $\begingroup$ @AlexS yes that is right. $\endgroup$ Commented Dec 29, 2018 at 14:59
  • $\begingroup$ @AlexS If by PDF you mean the probability distribution function, with your summary you are not dealing with Boltzmann-Gibbs microcanonical ensemble. There is no fluctuation of the macroscopica state variables. $\endgroup$ Commented Dec 30, 2018 at 6:09
  • $\begingroup$ @GiorgioP But I thought that kinetic theory implies the system is constantly changing, i.e. the molecules are constantly exchanging kinetic and potential energy? $\endgroup$
    – Alex S
    Commented Dec 30, 2018 at 6:19
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    $\begingroup$ @AlexS That's correct. But the total energy of an isolated system must be conserved. No fluctuation at all for total energy. The microscopic dynamics is just making the system visiting all the available phase space compatible with the macrostate. $\endgroup$ Commented Dec 30, 2018 at 6:24

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