Boltzmann Brain Update

My new model universe for Arun et. al. Consider a rectangular box, containing ten molecules. Here the ticks of the state change clock occur whenever a molecule passes from the left side to the right or vice versa. At some point we drop a partition down the middle and find that there are eight molecules on the left side and two on the right. We denote this macro state by (8;2). What is the most likely predecessor state (PS)?

Once again, there are just two possibilities: (9;1) and (7;3). There are 9 ways for the (9;1) state to go to the (8;2), but only three for the (7;3), so can we conclude that the (9;1) is the most likely PS? No, because the two states are not equally likely. There are only 10 ways to make a (9;1) state but 120 = (10x9x8)/(3x2) ways to make the (7;3) state.

No bit of probabilistic hocus pocus is complete without Bayes theorem, so, let A be the event that the initial state is (9;1), B that the initial state is (7;3), and C that the final state is (8;2). P(A)=10/1024, P(B)=120/1024, and P(C)=(10x9/2)/1024 = 45/1024. We have already seen that P(CA)=9/10 and that P(CB)=3/10.
Invoking Bayes Theorem the probability that the PS was A is P(AC)=P(CA)P(A)/P(C) = (9/10)(10/1024)(1024/45) = 1/5. Similarly the probability that the PS was B is
(3/10)(120/1024)(1024/45)= 4/5.

Consequently, it is 4 times as likely that the PS was the higher entropy state.

UPDATE:

I think that the argument is quite general. What does it mean that a macroscopic system is far from equilibrium? It means that the ensemble of states corresponding to the system has a small (coarse grained) volume in phase space compared to the available volume. If we evolve the system forward it tends to spread out in phase space increasing its (coarse grained) volume because most trajectories from the volume spread out – that is the second law of thermodynamics.

Similarly, if we look at trajectories leading into the original volume, those coming from a smaller volume have higher probability of forming our original volume, but the overwhelmingly larger number of trajectories coming from larger volumes means that if you treat all possible predecessor states as equally likely then it is more likely that the PS had higher entropy rather than lower.

From the standpoint of Hamiltonian dynamics the situation is completely symmetric in time. There are a very small number of future trajectories for the ensemble which steadily decrease entropy. There are an identical number of past trajectories that do the same thing. We are used to ignoring this fact because low entropy situations of this type are usually not “found” but created by deliberate action, e.g., placing a cold reservoir in contact with a hot one. (If we put two warm reservoirs in contact and waited for one to get hot and the other to get small, we would have a very long wait, unless we made them very small – like my toy universes.) Even more important, the evolution of the Universe seems to have presented us with a situation a convenient distance from equilibrium – unless, as is a priori more probable, it’s all just a dream in a Boltzmann brain.

This problem was recognized by Boltzmann. The only known ways to overcome it are to either introduce an explicitly time asymmetric principle (like Boltzmann’s “molecular chaos”) or to assume that the whole shebang started out in a very low entropy state. My money is still on the latter, and if cosmological theories find that inconvenient, so much the worse for them. Subject to revision without notice.;)