# Tag Info

## Hot answers tagged reversibility

6

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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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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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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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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The question is only getting one point wrong, which is that the Raychoudhuri equation requires the reversed black hole whose horizon shrinks by emitting matter to have negative stress energy. The reason this last point fails in the reversed case is becuase the horizon character is reversed--- the white hole horizon is a past horizon, it's another extension ...

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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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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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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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This is not a direct answer to the question but rather a slightly different perspective on this adiabatic expansion. I am not sure how correct it is. So, let us assume that the piston moves(in direction of $x$-axis) infinitely slowly with velocity $\vec{v}_p$. Let a molecule flies to the piston at a velocity $\vec{v}_k$. With respect to the piston, its ...

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Trivially, we could set the initial conditions so that particles all have velocities pointing directly towards one corner with speeds proportional to their distances from the corner. I don't think this is what you were looking for, though, since it doesn't look like the typical configurations of state space you'd expect. One thing to keep in mind (in ...

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