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I guess the simplest answer is just to carefully read you own words again. A reversible process is the one that can be made flow backwards. It is intuitive to think that it can be made flow backwards at any time we wish. But if the system were in a non-equilibrium state, one would need to wait a bit until it goes to equilibrium before trying to drive it ...

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The short answer is that it is not technically irreversible. If you wait some huge amount of time, the gas will de-expand, as per the Poincaré recurrence theorem. The problem with this is that the amount of time for a system to evolve from a low-entropy configuration to a high entropy configuration is very small, while you would be waiting several ages of ...

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1/ OK, let's start from an initial condition where all the particles are made to fit a tiny little corner of the room and their initial velocities are chosen randomly, according to a Maxwell-Boltzmann distribution for instance. As we let the system evolve, the gas will expand, that is true because it corresponds to the behaviour the Maxwell-Boltzmann ...

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I am a student so please point out in gory detail anything I did wrong. For a process to be quasistatic, the time scales of evolving the system should be larger than the relaxation time. Relaxation time is the time needed for the system to return to equilibrium. We have an adiabatic process, so equilibrium must be preserved at each point, that is to say ...

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I'm not sure if there's a definitive answer because I've seen it discussed recently at high level. I do think there's some broad agreement that entropy is important because it has an irreversible property: closed systems progress from low entropy states to higher entropy states. So we can define the passage of time more precisely by talking about increasing ...

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Time seems to "pass" because it is not symmetric -- it is T symmetric. This is often called the "arrow of time." The arrow of time points in the direction of increasing entropy. More: http://en.wikipedia.org/wiki/Arrow_of_time The real question you are asking is why our minds perceive this direction...

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The fact that evolutions of quantum mechanics are unitary after finite periods of time can be proven from the Schrödinger equation, and hinges on the characterization of unitary operators as those linear operators which are norm-preserving. Recall the Schrödinger equation: $$\frac{\mathrm d}{\mathrm d t} |\psi\rangle \;=\; -i H |\psi\rangle \;,$$ ...

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As you said, the case of black holes is conceptually totally analogous to the burning books. In principle, the process is reversible, but the probability of the CPT-conjugated process (more accurate a symmetry than just time reversal) is different from the original one because $$\frac{Prob(A\to B)}{Prob(B^{CPT}\to A^{CPT})} \approx \exp(S_B-S_A ).$$ This is ...

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Let's look at your first statement: A thermodynamic transformation that has a path (in its state space) that lies on the surface of its equation of state (e.g., $PV=NkT$) is always reversible I don't think this is right, but there may be some implicit qualifiers in your statement that I'm missing. Here's an example to illustrate why. Consider a ...

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Yes. From Clausius theorem the following inequality can be deduced: $$\delta Q \le TdS$$ where the equality holds in the reversible case. So, a reversible adiabatic process is necessarily isentropic, but irreversible adiabatic processes are not so. To put it in another way, in an irreversible process, according to the above inequality, either entropy ...

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I suppose that what you are thinking about is the principle of causality. We have two events: a cause and an effect, where the second event is a consequence of the first. That is how we perceive all events around us and what we intuitively accept as true. In physics, however, we sometimes obtain two different solutions: first with the cause before the ...

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An open system in the sense you describe is not a system coupled to the environment, but the idealized version of this system where the environment has been eliminated using a Markov approximation, so that there is a closed dynamics on the system itself, without reference to the environment (and hence to measurement), and without memory (which is enforced by ...

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I'll concentrate on cellular automata in this answer, because it's a good example, and should help to give a good intuition about algorithms in general. The answer is: most cellular automata do have an intrinsic time direction, but some don't. The most famous example of a cellular automaton is John Conway's Game of Life. This is an irreversible cellular ...

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There is no contradiction. The Stirling cycle you drew above is reversible but does not operate between two reservoirs at fixed temperatures $T_1$ and $T_2$. The isovolumetric parts of the cycle operate at continuously changing temperatures (think ideal gas law). Addendum. Note that in thermodynamics, a heat engine is said to operate (or work) between (two ...

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Now that your question is a little more clear I believe that the answer is no. What you say cannot be done spontaneously (entropy variation should be >0). You can reverse some things in some systems, but not everything. I don't know if this what you mean but maybe this can help: http://en.wikipedia.org/wiki/Spontaneous_process ...

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Yes, this is always possible. To do this you need to implement, given a function $f:\mathbb{Z}^k\rightarrow\mathbb{Z}^l$, a unitary evolution $U_f$ which will take the register bits to themselves and the ancilla bits to the function: $$U_f|x\rangle|0\rangle=|x\rangle|f(x)\rangle.$$ This is part of a more general problem: is it always possible to execute an ...

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There are two equivalent descriptions for the same process in terms of the time-forward version, and the time-reversed version. Externally, both look the same; some matter in a pure state collapses together into a dense state — a gravitational hole — and slowly, over time, it evaporates Hawking radiation until nothing is left of it. The totality ...

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Things become irreversible when you start ignoring certain degrees of freedom. What we call heat and friction is just our wilful ignorance of the trajectories of countless atoms. But the fact that the underlying equations of motion are time symmetric deals with microscopic phenomena. Sure, the time-reversed process is equally probable, which leads into the ...

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However, the simplest operations in computation, reset as well as the binary AND and OR operators, are irreversible. So? Their implementation in terms of CMOS logic is not irreversible, one can trackback the voltage levels. Sure, we can simulate irreversible systems with computers, but these aren't physically valid. However, because of the ...

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I'm not entirely sure I understand what you're asking but here is how I've interpreted your question: It seems like all of the laws of physics are reversible in time. That is, given the state of a physical system, it's possible to both go forward in time or backwards in time from that state. Assuming this is the case for physical laws and the equations ...

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Think about this is physical rather than algebraic terms for a moment. Notice that the term $\frac{T_L}{T_H}$ is the ratio between the highest and lowest temperature. This ratio tells you how well heat can flow from $T_H$ to $T_L$. The absolute temperatures don't actually matter, only their ratio. So what is the physical meaning of $T_L = 0$? It means ...

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naI think what you are essentially asking is that, "Can we violate or circumvent the second law of thermodyanmics?" The answer is no, based on all the physics we know so far and observations we have made so far. When you freeze the melted ice back to ice you are creating irreversible changes elsewhere in the universe (e.g., outside your refrigerator). ...

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There's some confusion here. Thermodynamics involves changes in going from an initial state to a final state. The path taken between those two states can be either reversible or irreversible. The process that moves the system along the path between the two states is reversible if and only if two conditions are fulfilled: (1) that a microscopic change in ...

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Like all 20th century physics, the formalism is invariant with respect to time reversal. This was true in classical mechanics and it remains true in QM because canonical quantization does not alter the meaning of energy - it just becomes an evolution operator. Unitary operators satisfying $A A^{\dagger} = I$ are associated logarithmically to Hermitian ones ...

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The two-state formalism of Aharanov can answer your question. True, in ordinary QM, there's a limit as to how low the Casimir negative null energy can go. In What is the physical meaning of weak expectation values?, it was pointed out that the weak expectation value $\langle \chi |A|\psi \rangle/\langle \chi |\psi\rangle$ can be much larger in magnitude ...

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The entropy of the surroundings does change infinitesimally. But the surroundings are large and such a change does not change the total entropy of the surroundings in any sensible way. Indeed, one already uses that fact in putting the system through a series of reversible steps. As you point out, if the temperature of system and surrounding were in fact ...

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Think of the Reeh-Schlieder theorem for a vacuum. It states that the vacuum is an entangled state, even between spatially distinct regions. By acting upon a local region here on Earth by a local operator which is appropriately fine-tuned, you can create any arbitrary configuration of matter behind the moon. For a vacuum that is, but we're not living in a ...

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The good answer to your question was indeed a condition on the velocity of the piston much lower than the average molecule velocity. To understand why, you need to study kinetics and fluid theories. From Boltzmann's equation one can deduce the fluid equations that give rise to classical thermodynamics. The passage from the kinetics scale to the fluid scale ...

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