# Tag Info

9

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 ...

6

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 ...

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 ...

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

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

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.

5

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 ...

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

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

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

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}

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 ...

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

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...

4

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 ...

4

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 ...

4

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 ...

3

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 ...

3

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 ...

3

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 ...

3

In the process you describe the system won't necessarily return to its original state. Suppose you instantly remove the weight and lift up the piston so the gas expands irreversibly to it's new equilibrium volume. The gas does no work in expanding so its temperature doesn't change - all that happens is that the pressure falls. Now compress the gas back to ...

3

This relation is not true for general processes. For a closed system, the general relation is $\delta Q \leq TdS$, as is illustrated by the Clausius Theorem (http://en.wikipedia.org/wiki/Clausius_theorem). Another way of writing it is $dS=\delta Q/T + dS_{irr}$ where $dS_{irr}$ is the entropy change due to irreversibilitiy in the closed system ...

3

To add to Whelp's Answer. Even though the $$\bar{d}Q=T\,dS$$ does not hold in an irreversible process, it still gives us something general. Consider a thermodynamic system linked to a system of reservoirs and which can only exchange heat and nothing else with the reservoirs. Let's call the thermodynamic system an engine for our convenience. Then the engine ...

3

It is true that classical thermodynamic equations emerge from statistical mechanics. And that the increase in entropy depends on the increase in the number of microstates. Decays also increase the number of microstates. They are irreversible because decay releases energy and the thermodynamic system cannot deliver enough energy and combination of ...

2

This is a very important question, thermodynamics strictly speaking only applies to quasistatic processes. E.g. if heat is transferred from one system to another one, then that process is necessarily an out of equilibrium process during which you cannot rigorously define thermodynamic variables like temperature (not just one temperature for the entire ...

2

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 ...

2

By definition a reversible adiabatic system has $dQ = 0$. We also know the following from the Clausius Theorem : $dS = \frac{dQ}{T}$ Then it is easy to see that there can be no change in entropy. Note that irreversible adiabatic systems CAN see a change in entropy because in that case the above equation is no longer an equality but an inequality : \$dS ...

2

(1) Adiabatic accessibility means that by some purely mechanical, electrical, magnetic, etc. but not thermal method an equilibrium state can be reached from another one. At the heart of Caratheodory's idea is the observation that given an equilibrium state A all other states fall into 3 categories: (a) states that are mutually accessible Ba, (b) states that ...

2

No it doesn't. The expansion of the gas is irreversible, so when then piston finally comes to a rest the entropy will be higher than it was before. There will be no way to restore the initial state without doing work. For example, if you try to push the piston back down again the gas will exert a non-infinitesimal force upon it. If you want an example of a ...

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