Recurrence time - A Nietzschean problem

What if some day or night a demon were to steal after you into your loneliest loneliness and say to you: 'This life as you now live it and have lived it, you will have to live once more and innumerable times more' (Friedrich Nietzsche, Die fröhliche Wissenschaft)

At any instant of time, the system (gas of particles in the box) can be in any configuration with $N_{L}$ particles in the left chamber and $N_{R}$ particles in the right chamber, such that $N_{L}+N_{R}=100$ . We will call these configurations the microstates of the system. The system starts in the microstate $N_{L}=100$ and $N_{R}=0$ . After each second, the system evolves to a new microstate, characterised by a different value of $N_L$ and $N_{R}$ .

As an order of magnitude estimate, we can imagine that if the system starts in a microstate, it will return to that precise microstate after it has visited all the others. Therefore the Poincaré recurrence time $t_r$ is of the order $t_r \sim \Omega$ (seconds), where $\Omega$ is the total number of microstates for our system. Since we have 100 particles and each particle can either be in the left or right chamber, $\Omega=2^{100}$ . We conclude that $t_r \sim 2^{100} \, s\sim 1.27 \times 10^{30} \, s$ , therefore the correct answer is $x=30$ .

Notice that this timescale is a trillion times larger than the age of the observable universe ... and only for 100 particles! Macroscopic gases contain a whopping $10^{23}$ particles (the famous Avogadro's number), and the recurrence time (in appropriate units) in this case is the insanely large number $10^{10^{23}}$ , which means that observing a recurrence in macroscopic systems is practically impossible.

We are looking at the Markov chain with states $0,1,2,...,N$ and transition probabilities $p_{n,n+1} \; = \; \tfrac{n}{N} \hspace{2cm} p_{n,n-1} \; = \; \tfrac{N-n}{N} \hspace{3cm} 1 \le n \le N$ and $p_{0,1} = p_{N,N-1} = 1$ . The state $n$ is where $n$ particles are on the left of the partition. This Markov chain has the invariant distribution $\pi_n$ for $0 \le n \le N$ , where $\pi_n \; = \; \tfrac{n-1}{N}\pi_{n-1} + \tfrac{N-n-1}{N}\pi_{n+1} \hspace{2cm} 1 \le n \le N-1$ with, in addition $\pi_0 \; = \; \tfrac{1}{N}\pi_1 \hspace{2cm} \pi_N \; = \; \tfrac{1}{N}\pi_{N-1}$ These equations can be solved to obtain $\pi_n \; = \; \binom{N}{n}\pi_0 \hspace{2cm} 0 \le n \le N$ Since the probabilities add up to $1$ , we have $1 \; = \; \sum_{n=0}^N \pi_n \; = \; 2^N\pi_0$ so that $\pi_n \; = \; 2^{-N}\binom{N}{n} \hspace{2cm} 0 \le n \le N$ By standard Markov chain theory, the mean recurrence time of the state $0$ is $\pi_0^{-1} = 2^N$ .

For this problem, the mean recurrence time is $2^{100} \approx 1.2676506 \times 10^{30}$ seconds, making the answer $\boxed{30}$ .