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## Hot answers tagged reversibility

10

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 back....

10

The bottom line is that hot water loses heat at high temperature, giving a small negative entropy change while the cold water gain heat at low temperature resulting in a high entropy change. The net entropy change is positive. We can explicitly see this: At any instant, the infinitesimal change in the entropy of the system is $$dS=\frac{dQ_H}{T_H}+\frac{... 8 As you said, the case of black holes is conceptually totally analogous to the burning books. In principle, the process is reversible, but the probability of the CPT-conjugated process (more accurate a symmetry than just time reversal) is different from the original one because$$ \frac{Prob(A\to B)}{Prob(B^{CPT}\to A^{CPT})} \approx \exp(S_B-S_A ).This is ... 8 Squeezing the wavefunction means confining it to a smaller space. It takes more energy to confine something within a small space than within a big one. 2, 3: These are consequences of the quantum adiabatic theorem: if you take a system in state n of some system, and act on the system sufficiently slowly, it ends up still in state n of the new system. ... 7 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 ... 6 Yes. For a reversible process, we have the relation \begin{align} dS = \frac{\delta Q}{T} \end{align} and for an adiabatic process, we have (by definition) \begin{align} \delta Q = 0, \end{align} which implies that \begin{align} dS=0. \end{align} 6 A reversible process leaves entropy unchanged. Entropy never decreases and so an irreversible process involves an increase in entropy. Increases in entropy absorb work that would otherwise be spent on the environment. 6 Pretty much the entire field of Classical Mechanics comes down to one thing: prediction. Given the initial conditions of a system, and a set of mathematical laws that model reality, we want to be able to tell what state the system will be in after a given time. The principal of reversibility that Susskind mentions is essential to Classical Mechanics ... 6 Although this isn't obvious, the system doesn't return to its initial state. If you were to very slowly remove the weight from the piston, then the gas would do work on the piston as you removed it, which means that its internal energy would be reduced. If you remove the weight very quickly then the gas still does work on it, but it will do less work than it ... 6 One place to look for relatively direct evidence is in the cross sections of time-reversed nuclear and particle physics reactions. For instance comparingA + n \to B + \alpha $$with$$B + \alpha \to A + n$$consistently shows the same (energy dependent) cross-sections for both directions where these reaction can be done between ground states of the ... 6 From a strictly by-the-formulas point of view, entropy change is heat transfer divided by the temperature over which the heat transfer occurs. The heat transfer is clearly the same for both volumes but positive for the cold volume and negative for the hot volume (heat flowed out of the hot volume and into the cold volume) but the average temperature over ... 6 To get the entropy change for a system experiencing an irreversible process, the first step is to forget entirely about the actual irreversible process and, instead, devise a reversible process that takes the system between the same initial and final equilibrium states. That is what is meant by dq_{rev}/T. The reversible process that you devise does not ... 6 The temperature appearing the the Clausius inequality is definitely the temperature of the "boundary interface (with the surroundings)", or simply the temperature of the sources. One of the best places I have seen this discussion is in Fermi's book, chapter 5, section 11. He is explicit about it. To see this you have to recapitulate the steps in obtaining ... 5 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 ... 5 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 ... 5 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 (... 5 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 ... 5 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... 5 Many processes cannot be drawn on a p-V diagram because the pressure is not always defined. Those processes that can be drawn are called "quasi-static". However, you cannot look at a certain path and say whether it represents a reversible or irreversible process for sure. For example, imagine a vertical line on the p-V plot, corresponding to adding heat to ... 4 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 ... 4 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 ... 4 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 ... 4 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 \;,$$... 4 Yes. From Clausius theorem the following inequality can be deduced:$$\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 would like to share my thoughts and questions on the issue. The Boltzmann H theorem based on classical mechanics is well discussed in various literatures, the irreversibility comes from his assumption of molecular chaos, which cannot be justified from the underlying dynamical equation. Here I will try to say something on quantum H theorem, the point I want ...

4

I once read of a science fiction scenario where on a planet the distance from a specific center was time. Life in that format progressed in height from the center, grew contours of certain height and became flat at death. The consciousness of those entities had time defined by their changes but humans just saw a completed contoured landscape unchanging in ...

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Reversible processes are important because they are related to the efficiency of a process. Take for examples a pair hotplates, one at 100C and one at 0C. In a theoretically ideal setting you could extract some work $W_0$ from this system until the two hotplates reached equilibrium. Then we would say the process is reversible, because in the same ideal world ...

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If you want to filter out the grains then certainly you could using normal filter papers in a filter funnel and repeat until the solution is clear of bits. You could also use a sintered glass filter. However, there will still be compounds from the coffee dissolved in the water and so a molecular sieve could be used and/or a chromatography column to separate ...

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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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