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Jul 14, 2020 at 4:08 history bounty ended CommunityBot
Jul 11, 2020 at 15:36 vote accept Jackson Walters
Jul 7, 2020 at 18:14 comment added Jeppe Rømer Juul Even if it does not cost energy to flip spin around (that is, the total energy of the system is the same before and after the flip) then there will still be an energy barrier associated with the flip. Otherwise, thermal noise would disrupt the state immediately. The rate with which the system approaches the thermal equilibrium could perhaps be approximated by the Arrhenius equation. In that case, the rate will decrease exponentially with inverse temperature so for temperatures that are low compared to the activation energy the times might very well be the lifetime of the universe.
Jul 7, 2020 at 18:02 comment added Jeppe Rømer Juul The Kolmogorov complexity is very hard to quantify, since it depends on the programming language chosen etc. But since the thermal equilibrium is an even distribution over all possible microstates there will be no consistent smart way of programming the microstate a random timepoint regardless of language. The complexity K will, as you say, be something like the capacity C.
Jul 7, 2020 at 15:55 comment added Jackson Walters For this system, I am more interested in $K$, the complexity, not $S$, the entropy which as you say is $S=C\log(2)$. Note that, assuming the answer to this question (physics.stackexchange.com/questions/563838/…) is accurate, it doesn't require any energy to permute bits around, which will change $K$. If we leave a hard drive at room temperature, we would be very disappointed to find its content changed. However, in your final statement you claim that it does. Roughly how long would this take? Longer than the lifetime of the universe?
Jul 7, 2020 at 10:01 review First posts
Jul 7, 2020 at 10:28
Jul 7, 2020 at 10:00 history answered Jeppe Rømer Juul CC BY-SA 4.0