# Entropy of an isolated system

It seems that entropy of a system is related to the fluctuation in total energy due to exchange with the environment. But in an isolated system, there can't be any fluctuation, which would imply zero entropy. Is this correct?

In detail, in Susskind's lectures on Statistical Mechanics, he derives the Boltzmann distribution with the following setup:

1. A "system" is defined as $$N$$ copies of a "subsystem", each weakly connected to each other such that they can exchange energy. Initially, each copy is given $$E$$ total energy, so that the total energy of the system is $$NE$$. This peculiar setup is meant to simulate the idea that the subsystem could be in an infinite heat bath.
2. Each copy can take on some set (possibly infinite) of discrete states $$i$$, each having a distinct energy level $$E_i$$. The energy levels are simply given to us as a result of some physical laws due to the characteristics of the subsystem.
3. A "configuration of the system" is an assignment of N states, one to each subsystem, e.g. $$(c_1, c_2, ..., c_N)$$. Because each configuration specifies a state, and therefore an energy level to each subsystem, the configuration also determines the total energy of the system, therefore only certain configurations fulfill the energy constraint. We assume that each configuration that satisfies the constraint, is equally probable.
4. An "occupancy state" is a description of a configuration. It is the set $$(n_1, n_2, ..., n_k)$$, ($$k$$ may be infinite) for each energy level, how many subsystems occupy that energy level. The total number of configurations for a given occupancy state is $$\dfrac{N!}{\prod_i^k{n_i!}}$$.
5. By Stirling's formula, $$\lim_{n \rightarrow \infty}{N!} = N^Ne^{-N}$$, and $$\lim_{n \rightarrow \infty}{\log \dfrac{N!}{\prod_i^k{(Np_i)!}}} = -N \sum_i{p_i \log p_i}$$. Therefore, the occupancy state that has the maximal number of configurations is most probable, and this happens to be the one whose occupancy state distribution has maximal entropy.

So far so good. So, zooming out a bit, we can look at the subsystem in its heat bath, and see that the energy of the subsystem fluctuates around an average energy level $$E$$, and that the system in fact spends $$p_i$$ fraction of time with energy level $$E_i$$, with $$\sum_i{p_i E_i} = E$$. The "entropy of the subsystem" is given as $$S = - \sum_i{p_i \log p_i}$$.

But, if the subsystem were not in a heat bath, its energy would not fluctuate at all. By this logic, its entropy would be zero. What???

• "But in an isolated system, there can't be any fluctuation, which would imply zero entropy. Is this correct?". I don't think so. The free expansion of a gas (the system) against a vacuum in a rigid adiabatic vessel is irreversible and generates entropy, yet the system is considered isolated. Sep 21, 2020 at 19:55
• Oh, I should have mentioned that the system I describe above is assumed to be in thermal equilibrium. I'm more concerned with the idea that non-zero entropy relies on non-zero total energy fluctuation. And, such energy fluctuation is still consistent with equilibrium. Sep 21, 2020 at 20:07

The question has set out the ideas in a clear, helpful way, and thus formulated a clear, useful question. I think the essence of what you are missing throughout your argument is the notion of degeneracy. That is, the fact that many different (that is, orthogonal) quantum states of a given system can have the same energy.

For an isolated system of given energy $$E$$, if there were only one quantum state having that energy then if the system were in that state then indeed its entropy would be zero. But this never happens for a macroscopic system (at temperatures above absolute zero temperature). If you found a large system which, by some unusual character, had a non-degenerate energy eigenstate well above the ground state then although the system might be deemed "macroscopic" in other respects, in this respect it would be sufficiently unusual as to be a special case. One could then debate whether or not it should be included under the label "macroscopic". But an ordinary macroscopic isolated system has a huge number of microstates on its energy surface (if the temperature is not near absolute zero).

More generally the relationship between entropy and fluctuation is as follows. We ordinarily treat entropy as the more fundamental idea and define it without regard to fluctuation. The definition is $$S = -\sum_i p_i \ln p_i$$ where $$i$$ runs over microstates which are consistent with the macroscopic parameters defining the macrostate. We then look into the fact that, depending on the constraints, certain system properties may undergo thermal fluctuations. If the energy is constrained (isolated system) then it does not fluctuate, but if the temperature is constrained (system in equilibrium with heat bath) then the energy does fluctuate and the standard deviation of the fluctuations is related to the heat capacity. The heat capacity is in turn related to the entropy. Similar statements can be made about other macroscopic parameters such as volume and total number of particles.

But, if the subsystem were not in a heat bath, its energy would not fluctuate at all. By this logic, its entropy would be zero. What???

Absolute entropy is not set by the degree of energy fluctuations in a system. Absolute entropy is a measure of the number of possible configurations that exist at the same total absolute energy of the system.

When we have but one configuration to a given energy state, the absolute entropy is zero at that energy state. When we have more than one configuration, the absolute entropy is not zero.

The third law of thermodynamics defines the absolute zero at $$T = 0$$ K where a substance is in one configuration (a perfect crystal order). The postulates form of thermodynamics will result in $$S \rightarrow 0$$ as $$T \rightarrow 0$$. Finally, the statistical mechanics form gives $$S = k \ln \Omega = 0$$ as $$\Omega = 1$$.

Whether the system is or is not connected to a heat bath has no concern to the above findings. Alternatively said, we do not require a heat bath to have a system that is at a defined $$T$$ above zero fluctuate among its many possible states at the same energy.

By example, bring a closed system to a temperature $$T$$ above absolute zero using a heat bath. Insulate the system perfectly. It will remain at $$T$$ theoretically forever. The system will not have zero absolute entropy.

• I'd never heard the phrases "absolute entropy" or "absolute energy". How are those different from just entropy and energy? But what you said makes more sense. What I'm hoping to find is a general formula for entropy that applies to both classical and quantum systems. Would it just be the entropy of the probability distribution over all accessible microstates of the system? (Whether they are spatially and/or energetically defined?) Sep 22, 2020 at 19:09
• @HenryBigelow The opposing term is entropy change. I use "absolute" as the definitive qualifier. In truth, entropy and energy are "absolute", and we should use the phrase "change in" as the qualifying prefix. You will not find a common form. Laws: $dS = \delta q/T$; Postulates $S(U,V,n)$, or Statistical Mechanics: $S = k \ln \Omega$. Sep 22, 2020 at 19:54
• Got it, thanks! Sep 23, 2020 at 3:34

Well if the subsystem has fixed energy that we know, say $$E = -5eV$$, then we know it is in one of the states with that energy. For macroscopic systems, there is zillion of such states. For microscopic system, if the energy level is degenerate, there may still be several microstates with that energy, so this still allows for some small non-zero entropy.

But if there is only one such state, then indeed energy $$E$$ implies that system is in that state. Multiplicity $$W$$ is 1, probabilities are all zero except for that single state where it is one, and so both information entropy $$\sum_s - P_s\log P_s$$ and the Boltzmann entropy $$\log W$$ is zero.

This does not happen for macroscopic systems, where knowledge of $$E$$ implies astronomic number of possible states.