# 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 ...

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

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

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

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

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

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

One place to look for relatively direct evidence is in the cross sections of time-reversed nuclear and particle physics reactions. For instance comparing $$A + n \to B + \alpha$$ 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

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

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 ...

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

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

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

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 ...

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

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

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

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

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 ...

3

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 ...

3

For an irreversible expansion or compression, let $P_B(t)$ represent the gas pressure on the portion of its boundary where the work is being done, and let $T_B(t)$ represent represent the gas temperature on the portion of the boundary where heat transfer is taking place, with $t$ representing the time during the irreversible change. If the change is ...

3

It is reversible in the first case because it satisfies the reversibility definition. A thermodynamic process is called reversible if an infinitesimal change of the external condition reverses the process. Consider a system at temperature $T$ in thermal contact with a thermal reservoir at same temperature. By an infinitesimal increase $dT$ of the reservoir's ...

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

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

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

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 ...

3

The difference is that one expansion is quasi-static (the reversible one) while the other is spontaneous because of a dramatic change of the external constraints (the irreversible one). In the quasi-static case, you start off indeed in the state where gas pressure equates external pressure. An external operator then slightly decreases the outside pressure ...

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