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I started to think about Maxwell's demon again after learning about it a few months ago, and I've realized that there are a few things I don't understand. Here are my questions:

1: At first, I thought the resolution was that storing information in the demon's memory corresponded to an increase in the demon's entropy. If you store N bits of information, then there are 2^N possible configurations, and thus an entropy of S=Nkln(2). But everyone talks about how deleting information corresponds to an entropy increase in the environment, and since the demon's memory is finite, it must eventually delete info to have space for more. But why is this necessary to save the 2nd law, seeing how storing information is enough? Also, why can't you just imagine having a demon with a memory big enough to store all the N bits of info, making it unneccesary to delete? And even if you insist that info must eventually be deleted, say as the demon eventually forgets it, there is the potential of there being an arbitrary amount of time between the demon's mischief and the deletion, and thus an arbitrary amount of time where the 2nd law would be violated.

2: I have also been starting to think that there is a subtle difference between the statistical mechanical entropy and the Shannon information entropy. In the case of statistical entropy, we talk about the number of microstates in a given macrostate, and an important postulate for this formulation is that each microstate is equally likely, and that due to the internal dynamics of the system, the microstates change with time, such that the system spends an equal amount of time in each state. But in the case of storing memories/information, the configuration of the N bits don't change with time, i.e. there is no internal dynamics to ensure that the configuration of ones and zeros change. So for the same reason that the entropy at absolute zero is zero because the particles are "frozen" into a single microstate, shouldn't the statistical entropy corresponding to either storing or deleting info be zero since the ones and zeros are "frozen" into a single configuration?

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The memory must be thawed before writing, and frozen afterwards. Instead of doing that it's easier to dump information into the environment. This is also easier for the brains of physicists to handle.

Thawing a memory cell means giving it kinetic energy when making it move towards a state. Freezing means turning the kinetic energy into heat when applying a brake that stops the motion towards a state.

How much heat is generated depends on the technical details of the memory, and that amount can be arbitrarily small. So it's not important. So here we have an example of the difficulty I mentioned earlier, I almost got fooled into thinking that it's important.

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  • $\begingroup$ Thank you for an answer, I think I figured it out myself too. But there is one thing I still don't quite understand. I think we can all agree that the entropy of the gas goes down after the demon has finished, although the demon's (or the environment's) entropy has gone up. But doesn't a system have to release heat in order to lose some entropy, such that dS>=dQ/T? Where in this process does the gas release heat? $\endgroup$ May 4, 2021 at 20:00
  • $\begingroup$ @FelisSuper Sorting gas molecules by some property decreases the entropy of the gas. Sorting by energy produces some hot gas and some cold gas. The end result has no definite temperature. By the way, I think your question was vague, and my answer is not very good at all. $\endgroup$
    – stuffu
    May 7, 2021 at 7:13

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