# Is it possible for the entropy in an isolated system to decrease?

As far as I can tell, the concept of entropy is a purely statistical one. In my engineering thermodynamics course we were told that the second law of Thermodynamics states that "the entropy of an isolated system never decreases". However, this doesn't make much sense to me.

By counter-example: Consider a gas-filled isolated system where the gas has maximum entropy (it is at equilibrium). Since the molecular motion is considered to be random, at some point in the future there will be a pressure gradient formed by pure chance. At this point in time, entropy has decreased.

According to Wikipedia, the second law purely states that systems tend toward thermodynamic equilibrium which makes sense. I then ask a) is the second law as we were taught it wrong (in general), and b) what is the use of entropy (as a mathematical value) if it's effectively an arbitrary definition (i.e. what implications can we draw from knowing the change in entropy of a system)?

• – Hannesh Oct 6 '13 at 18:08
• – user4552 Oct 6 '13 at 18:09
• see e.g. physics.stackexchange.com/a/547/671 for a discussion of how small this 'pure chance' typically is for macroscopic systems... (and compare this to the number of atoms in the observable universe which is estimated to be $10^{80}$) – Andre Holzner Oct 6 '13 at 18:56

By counter-example: Consider a gas-filled isolated system where the gas has maximum entropy (it is at equilibrium). Since the molecular motion is considered to be random, at some point in the future there will be a pressure gradient formed by pure chance. At this point in time, entropy has decreased.

Violations of the second law are possible. The law is probabilistic, not absolute or fundamental. In your example, small pressure differences $\Delta p$ will always exist. These will fluctuate randomly about a mean of zero. Because the number of particles is something like Avogadro's number, the probability is extremely high that $\Delta p/p$ will be extremely small -- much too small to be measured with a macroscopic device such as a pressure gauge.

I then ask a) is the second law as we were taught it wrong (in general) [...]

It's right in the sense that you could spend the rest of your life watching for a detectable $\Delta p/p$, and the rest of the human race could also devote their own lives to similar observations, and there would be no meaningful probability that any of you would ever see what you were looking for.

b) what is the use of entropy (as a mathematical value) if it's effectively an arbitrary definition [...]

What do you mean by arbitrary? It doesn't seem arbitrary to me at all.

Historically, the entropy concept was invented precisely because it was useful. It was useful for understanding limits on the efficiency of steam engines.

• I think he means arbitrary in the sense that 'increase' or 'decrease' are arbitrary definitions. You could say: (1) entropy is always decreasing; and (2) entropy never increases. In essence it only means one state goes towards another and dont ever go back to its previous state (in isolation). – eJunior Oct 6 '13 at 18:31
• @eJunior: Sorry, I don't understand your comment. Did you write "decreasing" and "increases" when you meant "increasing" and "never decreases?" Or do you mean that we could arbitrarily change the definition of entropy from $\ln\Omega$ to $-\ln\Omega$, in which case it would always decrease? In essence it only means one state goes towards another and dont ever go back to its previous state (in isolation). This is sort of true, see Lieb and Yngvason, arxiv.org/abs/math-ph/0003028 . But that doesn't mean that the definition of entropy is arbitrary. In fact, L&Y prove that it's unique. – user4552 Oct 6 '13 at 19:40
• Or do you mean that we could arbitrarily change the definition of entropy from lnΩ to −lnΩ, in which case it would always decrease? I meant exactly that. If it has a 'sign' and it cannot go the opposite way what is it's 'mathemathical value'. – eJunior Oct 6 '13 at 21:22
• @eJunior: Sure, in that trivial sense it's arbitrary. It's arbitrary up to a multiplicative constant, and the constant can be negative if you switch the words around in the 2nd law. – user4552 Oct 6 '13 at 21:42

If you consider the case of different configurations of the gas in question will get a number of microstates. With them you can define an equivalence relation that form a macrostate. All this microstates share the general characteristics of the system (as the gas pressure exerted on the walls of the container).

These macrostates can be assigned a probability of occurrence based on the microstates that comprise it. We determine that there is some physical quantity that allows us to define nature's preference for certain states over others and called entropy.

In this way we are not talking about whether the system may or may not be in a particular state, but what is the probability of that happening. Importantly entropy allows us to know what will be the state of the system may not need to develop mechanical equations for each of the particles. Additionally does not tell us how long it will happen, only that a state is more likely to occur will have a higher entropy value.

• You define micro and macro states but then use "states" which kind of states are you referring to? – Rick Sep 22 '15 at 12:05

By the Poincaré recurrence theorem, an isolated system is guaranteed to return arbitrarily close to its initial state after a sufficiently long time.

As you said, entropy is a statistical phenomenon. As with everything statistical, the parameters of your sampling effect the quality of your conclusions.

In your given example - a random molecular gas spontaneously forming a pressure gradient - you examine a brief timescale during which the entropy of the system is less than a previous value. Given that we are considering a process that evolves over time (and assuming that the particle number is "large enough" from a statistical standpoint) "brief period" indicates a small sample. Perhaps a fitting analogy would be demographics: the population of a college town changes dramatically in a week's time several times a year, but a longer study may reveal a much more consistent trend.

Likewise, in the given example, the pressure gradient described can only last a brief time. A statistically sound examination of the system will show that the system is indeed in thermal equilibrium, and at a higher entropy than similar large-sample studies performed at earlier times in the evolution of the system from an initial non-equilibrium state.

Statistical thermodynamics doesn't have the rigor or simplicity or beauty of classical thermodynamics.

Classical thermodynamics distinguishes processes into two types: (1) possible (change in the energy of the universe is zero and change in the entropy of the universe is greater than or equal to zero) and (2) impossible (change in the energy of the universe is zero and change in the entropy of the universe is negative). However, statistical thermodynamics says that the probability of an impossible process occurring is extremely low. It never ever categorically state a process to be impossible!

For example, as you rightly pointed out, in an isolated system of an ideal gas (gas filled system) at equilibrium, it is possible, according to statistical thermodynamics for pressure differences to develop spontaneously, as you are taught in your statistical thermodynamic courses! The defendants of statistical thermodynamics argue that the probability of such an occurrence is extremely low! Classical thermodynamics prohibits spontaneous development of pressure differences in an isolated system at equilibrium.

If you are prepared to dig deeper, you will find many more such instances which don't make sense. Then resolution is sought in quantum mechanics!

• Statistical thermodynamics makes fewer assumptions and as a result is valid in more situations than classical thermodynamics. Thus I would consider statistical thermodynamics more rigorous. Quantum mechanics is not necessary to understanding statistical thermodynamics. – Rick Sep 22 '15 at 12:02
• Dear Rick, Would you please mention the assumptions made by Statistical thermodynamics and classical thermodynamics, so that it would be useful for further clarification/discussion. Also, give one example of a situation where Statistical thermodynamics is valid but classical thermodynamics is inavalid. – Radhakrishnamurty P Sep 24 '15 at 4:16
• Classical thermodynamics assumes a closed system's macro-state will never move away from thermal equilibrium, only towards it. Statistical thermodynamics assumes that the macro-state will shift randomly, but with a distribution based on the number of micro-states associated with each macro-state. When the number of particles is large, the deviation from classical thermodynamics is negligible. However, the rules break down when the particle count is low, for example when there are only a dozen particles in a system. – Rick Sep 24 '15 at 11:52
• I don't find fewer assumptions in statistical thermodynamics, as you stated in your earlier comment. Again, when the number of particle count is low the rules of statistical thermodynamics break down, you say, how then is statistical thermodynamics is valid in more situations than classical thermodynamics, as you stated in your earlier comment? – Radhakrishnamurty P Sep 25 '15 at 15:30