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23

The resolution to Maxwell's demon paradox is mostly understood to be through Landauer's principle, and it is one of the most compelling applications of information science to physics. Landauer's principle asserts that erasing information from a physical system will always require performing work, and particularly will require at least $$k_B T \ln(2)$$ of ...


13

The low-entropy initial state of the universe is an open problem without a satisfactory answer. Your question is the first time I've heard the suggestion that the initial state should have been a crystal; you remind me that the quark-gluon plasma, which was the state of the universe while it was too hot for nucleons to be stable, has been shown to be a ...


8

the two paradigmatic cases that illustrate these two possibilities is a gas, for the first, and a crystal for the second. Paradigms and examples are well and good, but be careful not to assume they are the only possibilities. In particular, black holes have entropy -- a lot of entropy. In fact they saturate the Beckenstein Bound. The entropy of a black ...


6

Your question is an excellent one and I think your IT approach is spot on. The contradiction you rightly point out arises partly from the subtle differences between the different Conditional Entropies (Conditional Informations) at play in this discussion. Leaving aside Universes for the moment, let's instead think about a truly thermodynamically isolated ...


4

As always the answer is a simple thing. You calculated the change in entropy using the definition of entropy \begin{equation} \mathrm{d}S = \frac{\mathrm{d}Q_\mathrm{rev}}{T} \end{equation} Note that this applies to heat transferred reversibly. More generally we must use Clausius theorem \begin{equation} \mathrm{d}S \ge \frac{\mathrm{d}Q}{T}\end{equation} ...


4

What I will state is speculative and based on the statistical mechanics derivation of entropy, and just the way I view it and do not consider that there exists a problem. After all thermodynamic theory emerges from the underlying statistical level of atomic and molecular interactions. where p_i is the probabability of microstate i. Setting aside quantum ...


3

Four additions to other answers and your questions: I agree with your thoughts about the door. In principle, it can be arbitrarily near to lossless. The work cost does not arise in knowing when to open the door, i.e. in measuring the state of gas particles. This was actually what Leo Szilard thought, as he discussed in 1929 in L Szilard, "Über die ...


3

This has everything to do with entropy: when the temperature is higher, the benefit of having more water molecules in the air (giving rise to greater entropy) become energetically more favored. This is why water "dissolves better in air" at higher temperatures. Another way of looking at this (pure statistical thermodynamics): when water is cold, few ...


2

Firstly, the OP is forgetting that the classic microwave polariser experiment is done with EM radiation in a pure state, not a mixture. We simply have polarised light from, say, a Gunn diode and this pure quantum superposition is forced into a polarisation eigenstate by the polariser. So we begin with near to zero entropy light, absorb some of it (adding ...


2

The formula is actually better written $$ \Delta S = \frac{Q}{T}. $$ That is, the change in entropy associated with the flow of heat is inversely proportional to the temperature at which the heat flow occurs. Note that $Q$ is already a change itself: it is not a state variable, but rather something more like $\Delta W$. Physically, this is because adding ...


2

I think the factor you are ignoring is that the polariser will emit thermal radiation. If we continue with the ideal polariser, then it should only emit the polarisation which it absorbs (ideal components are weird). This means that there will still be a component of the absorbed polarisation in the beam after the polariser and so there will always be some ...


2

Boltzmann's version of entropy require a finite number of states, and Planck had asserted that probability has no meaning without a “finite number of equally likely configurations.”{1} That is, in order to be able to use Boltzmann's equation and obtain finite results, he needed to use a discrete number of states, but light was supposed to be continuous. ...


2

You can also think of this in terms of information only, without invoking thermodynamics right from the start. So, you just have a system, and you don't know the exact physical state it is in. If we then consider that system including all the features needed to operate Maxwell's demon as a totally isolated system, such that even quantum decoherence is ...


2

I think it's not purely an issue of degeneracy, from what I understand a system can have a residual entropy even with a unique ground state. I'm not sure, but I think the way it works is that every system would end up in its lowest-energy ground state at exactly absolute zero, but in practice if we are interested in the behavior as you approach absolute ...


1

Macroscopically, entropy $S$ is a function of energy $U$ and maybe some other extensive quantities (volume, particle number, etc.) Temperature can be defined as the derivative of $S$ with respect to energy with the other extensive quantities held fixed, $$\beta\equiv\frac{1}{T} \equiv \frac{\partial S}{\partial U}. $$ So $\beta$ determines how much change ...


1

In the ideal case, at zero kelvin the system must be in a state with the minimum possible energy, and this statement of the third law holds true if the perfect system has only one minimum energy state, called the ground state. Entropy is related to the number of possible microstates. If the ground state of the system is degenerate system containing a ...


1

The problem you are trying to solve is different from the standard textbook situation where $A$ and $B$ are two heat reservoirs. When considering equilibration of temperatures, you have to account for the finite heat capacity of the two objects. This allows you to consider a series of quasi-static reversible heat exchanges with a series of heat reservoirs, ...


1

My two cents to add to the great response from wetsavanna: a short explanation would be the following. Technically, information never increases or decreases at the microscopic level. Entropy is a measure of information about how much do we know about the microscopic levels when making a macroscopic observation. But the relationship between entropy and ...


1

If it is adiabatic then $ \Delta Q_i $ will be always zero. The fact that it is irreversible doesnt matter. Any path that thakes you from A to B will result in the same change of entropy, as both initial and final states are in equilibrium. If you choose what is called a quasistatic path, which is idealized as a tranformation that occurrs slow enough so that ...


1

The first law of thermodynamics, is as the name suggest a law. It states that if you consider some process a thermodynamic system undergoes then $$\Delta U = W + Q$$ The point is that $W$ and $U$ are things that depend on what you are studying. Pick the infinitesimal version just for simplicity $$dU = \delta W + \delta Q$$ Then for a gas it trully makes ...


1

I believe the error is in assuming that the polarized beam is a pure state of zero entropy. If you characterize it in terms of polarization only, then the characterization is not complete. You need a complete set of commuting observables to charactherize a pure state. The macroscopic polarized beam is still compatible with many different quantum microstates ...


1

This phenomenon has nothing to do with the properties of the air, but the properties of the water in it. Hot air means hot water in the air. Cold air means cold water in the air. Cooling water causes it to condense. This is considering a constant volume.


1

In the early Universe, entropy is preserved (dS=0). This comes out of the equations of general relativity, but it can be also understood by thinking in terms of classical dynamics: the Universe is a closed system, no heat is exchanged when expanding, so its entropy must not variate.


1

"High" and "low" are relative terms that usually also carry an anthropocentric connotation. What "high" means depends on what humans think of as a large quantity but to thermodynamics the absolute scale does not matter! What matters is only that there is a change from one entropic state to another. As long as there is such a change, no matter how slow, there ...



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