# 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

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