One of the ideas involved in the concept of entropy is that nature tends from order to disorder in isolated systems. But we even know that Poincare recurrence time also is a particular time after which a system of particles get back to their original position,and entropy is how can a system of particles be arranged. So are these two related?

  • $\begingroup$ Isn't Poincare recurrence time essentially the time it takes for the entropy decrease function of a system to occur? $\endgroup$ – V.H. Belvadi Nov 15 '14 at 8:31
  • $\begingroup$ Also, the order-to-disorder analogy is not accurate; in other words, entropy decrease is not prohibited in nature. Statistically, the likelihood is so small that we nonchalantly claim it never occurs. However, it can and will occur when the Poincare recurrence time is infinitely long, so to speak. The familiar box with gas molecules is the simplest example of this. After time, there is a small chance that the molecules will bunch up, decreasing entropy, the way they started. $\endgroup$ – V.H. Belvadi Nov 15 '14 at 8:43
  • $\begingroup$ Is there any mathematical way to represent this?? $\endgroup$ – sayan chattopadhyay Nov 15 '14 at 10:05

I'll give you a complete mathematical answer at the simpler level of symbolic dynamics. Proved by Ornstein and Weiss:

  1. if $\sigma^+|\Sigma^+$ is a one-sided topological Markov chain and $\mu^+$ is an ergodic $\sigma^+$-invariant probability measure on $\Sigma^+$, then $$ \lim_{k\to\infty} \frac{\log\inf\{n\in\mathbb N:(i_{n+1}\cdots i_{n+k})=(i_1\cdots i_k)\}}{k}=h_{\mu^+}(\sigma^+) $$ for $\mu^+$-almost every $(i_1i_2\cdots)\in\Sigma^+$;

  2. if $\sigma|\Sigma$ is a two-sided topological Markov chain, and $\mu$ is an ergodic $\sigma$-invariant probability measure in~$\Sigma$, then $$ \lim_{k\to\infty} \frac{\log\inf\{n\in\mathbb N:(i_{n-k}\cdots i_{n+k})=(i_{-k}\cdots i_k)\}}{2k+1}=h_\mu(\sigma) $$ for $\mu$-almost every $(\cdots i_{-1}i_0i_1\cdots)\in\Sigma$.

For the general case, I recommend the book "Dimension and Recurrence in Hyperbolic Dynamics", by Barreira.

  • 1
    $\begingroup$ Could you define some of the terms used here? This is almost impossible to parse unless one is already familiar with the subject. $\endgroup$ – Rococo Feb 15 '18 at 18:08
  • $\begingroup$ Could you please be more specific? The terms (not the results) in my answer are really at the rudiments of dynamics (which is the canonical area of Poincaré recurrence). Because of this, my view (apparently of 2016 already) is that one either knows the basic terms or one really needs to study them with time. Still, I will try my best if you tell me exactly what is missing. $\endgroup$ – John B Feb 15 '18 at 18:37
  • 1
    $\begingroup$ It is exactly the opposite: in case you don't know these basic things you should first learn them, and only then you should go to the advanced things that I discuss in my answer. Simply it is too advanced for you since you don't know the extremely basic things that you are asking. Sorry, but there is no way around it. $\endgroup$ – John B Feb 15 '18 at 23:15
  • 2
    $\begingroup$ Yes, that's why I come here- to learn things, including regarding some subject that might be considered to be 'basic topics'. Unfortunately, I did not learn anything at all from your answer, or from your belittling comments when I asked for clarification. $\endgroup$ – Rococo Feb 24 '18 at 20:15
  • 1
    $\begingroup$ That said, as always my downvote only indicates that I did not find a question helpful, and should not be taken too personally. $\endgroup$ – Rococo Feb 24 '18 at 20:22

Yes, entropy and recurrence times are related, indeed proportionally so. The logarithm of the entropy of a well-defined (eg at a single temperature) system contains the density-of-states, the number of quantum levels in the vicinity of the average energy. Entropy increase accompanies an increase in that density- i.e. more states for the system to get lost in. Meanwhile, the recurrence time is also proportional to that same density of states.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.