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2

The formula is actually better written $$ \Delta S = \frac{Q}{T}. $$ That is, the change in entropy associated with the flow of heat is inversely proportional to the temperature at which the heat flow occurs. Note that $Q$ is already a change itself: it is not a state variable, but rather something more like $\Delta W$. Physically, this is because adding ...


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A good, semi-technical discussion of the general problem (how the post-Big-Bang evolution of the universe, including the formation of galaxies, stars, etc., can be reconciled with the 2nd Law) can be found here: http://arxiv.org/abs/0907.0659 It's important to realize that while the ensemble of atoms in the gas cloud does indeed, as your intuition suggest, ...


2

Firstly, the OP is forgetting that the classic microwave polariser experiment is done with EM radiation in a pure state, not a mixture. We simply have polarised light from, say, a Gunn diode and this pure quantum superposition is forced into a polarisation eigenstate by the polariser. So we begin with near to zero entropy light, absorb some of it (adding ...


4

As always the answer is a simple thing. You calculated the change in entropy using the definition of entropy \begin{equation} \mathrm{d}S = \frac{\mathrm{d}Q_\mathrm{rev}}{T} \end{equation} Note that this applies to heat transferred reversibly. More generally we must use Clausius theorem \begin{equation} \mathrm{d}S \ge \frac{\mathrm{d}Q}{T}\end{equation} ...


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I think the factor you are ignoring is that the polariser will emit thermal radiation. If we continue with the ideal polariser, then it should only emit the polarisation which it absorbs (ideal components are weird). This means that there will still be a component of the absorbed polarisation in the beam after the polariser and so there will always be some ...


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I believe the error is in assuming that the polarized beam is a pure state of zero entropy. If you characterize it in terms of polarization only, then the characterization is not complete. You need a complete set of commuting observables to charactherize a pure state. The macroscopic polarized beam is still compatible with many different quantum microstates ...


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No relation. The second law of thermodynamics can be proven on general statistical grounds. It is independent of what particles are in our universe or what equations describe their motion. That is why physicists are very confident about the second law of thermodynamics, even though nobody knows for sure what particles are in our universe or what equations ...


0

A really helpful discussion of this question is given by Daniel F. Styler, "Insight into entropy," Am. J. Phys. 68 (2000), pp. 1090 - 1096. A (macro)state with high entropy is one that corresponds to many microstates, i.e. the system has many ways that it could configure itself on the microstate level to achieve the given macrostate. Styler gives some nice ...


1

The first law of thermodynamics, is as the name suggest a law. It states that if you consider some process a thermodynamic system undergoes then $$\Delta U = W + Q$$ The point is that $W$ and $U$ are things that depend on what you are studying. Pick the infinitesimal version just for simplicity $$dU = \delta W + \delta Q$$ Then for a gas it trully makes ...


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As you wrote $W = \int_{\gamma} p(s) ds$ you are implying that the pressure, I assume you denoted pressure by $p$, is function of only the volume $V$ and the dummy integration variable is in fact the volume. That is not the case, the thermal equation of state even for the simplest system involves the (absolute or empirical) temperature, as well. That is ...


0

we know that every spontaneous process will in the direction of increasing entropy . At poles water spontaneously converted into ice . so total entropy (entropy (ice+water+surrounding)) will increase


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This has everything to do with entropy: when the temperature is higher, the benefit of having more water molecules in the air (giving rise to greater entropy) become energetically more favored. This is why water "dissolves better in air" at higher temperatures. Another way of looking at this (pure statistical thermodynamics): when water is cold, few ...


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This phenomenon has nothing to do with the properties of the air, but the properties of the water in it. Hot air means hot water in the air. Cold air means cold water in the air. Cooling water causes it to condense. This is considering a constant volume.


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Four additions to other answers and your questions: I agree with your thoughts about the door. In principle, it can be arbitrarily near to lossless. The work cost does not arise in knowing when to open the door, i.e. in measuring the state of gas particles. This was actually what Leo Szilard thought, as he discussed in 1929 in L Szilard, "Über die ...


1

Macroscopically, entropy $S$ is a function of energy $U$ and maybe some other extensive quantities (volume, particle number, etc.) Temperature can be defined as the derivative of $S$ with respect to energy with the other extensive quantities held fixed, $$\beta\equiv\frac{1}{T} \equiv \frac{\partial S}{\partial U}. $$ So $\beta$ determines how much change ...


0

Entropy $S$ is a measure of microscopic freedom, temperature $T$ the average energy per degree of freedom and heat $Q$ the amount of energy transferred microscopically. They are related through the equation* $$ Q=T\cdot\Delta S $$ or verbally $$ \text{energy transferred microscopically} = \text{energy per degree of freedom} \times \text{change in ...


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I think it's not purely an issue of degeneracy, from what I understand a system can have a residual entropy even with a unique ground state. I'm not sure, but I think the way it works is that every system would end up in its lowest-energy ground state at exactly absolute zero, but in practice if we are interested in the behavior as you approach absolute ...


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In the ideal case, at zero kelvin the system must be in a state with the minimum possible energy, and this statement of the third law holds true if the perfect system has only one minimum energy state, called the ground state. Entropy is related to the number of possible microstates. If the ground state of the system is degenerate system containing a ...


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The problems of the Demon are coming from the randomness. The second law of thermodynamics is just saying, that randomness is spreading through any hole and any obstacles. Just because of it randomness. If you put very complex lock on the way of randomness, it will found the combination sooner or later. It is just the matter of time. So, if Demon ...


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You can also think of this in terms of information only, without invoking thermodynamics right from the start. So, you just have a system, and you don't know the exact physical state it is in. If we then consider that system including all the features needed to operate Maxwell's demon as a totally isolated system, such that even quantum decoherence is ...


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The resolution to Maxwell's demon paradox is mostly understood to be through Landauer's principle, and it is one of the most compelling applications of information science to physics. Landauer's principle asserts that erasing information from a physical system will always require performing work, and particularly will require at least $$k_B T \ln(2)$$ of ...


0

What you are saying is correct: reversible and adiabatic process between two states $A$ and $B$ does not change the entropy either of the system or of its environment (surroundings). Irreversible and adiabatic process between two states $A'$ and $B'$ increases the system's entropy. The two statements are reconciled by noting that if $A=A'$ then $B\ne B'$ and ...


0

Let me explain it in simple words without equations. When you have a closed system at state A, you have to use energy on it to convert it into state B. If the process of shifting from state A to B required no energy, the system would already be in state B. Now when you have the system at state B and want to return it to state A, you again have to use ...


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The problem you are trying to solve is different from the standard textbook situation where $A$ and $B$ are two heat reservoirs. When considering equilibration of temperatures, you have to account for the finite heat capacity of the two objects. This allows you to consider a series of quasi-static reversible heat exchanges with a series of heat reservoirs, ...


2

Boltzmann's version of entropy require a finite number of states, and Planck had asserted that probability has no meaning without a “finite number of equally likely configurations.”{1} That is, in order to be able to use Boltzmann's equation and obtain finite results, he needed to use a discrete number of states, but light was supposed to be continuous. ...


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My two cents to add to the great response from wetsavanna: a short explanation would be the following. Technically, information never increases or decreases at the microscopic level. Entropy is a measure of information about how much do we know about the microscopic levels when making a macroscopic observation. But the relationship between entropy and ...


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Your question is an excellent one and I think your IT approach is spot on. The contradiction you rightly point out arises partly from the subtle differences between the different Conditional Entropies (Conditional Informations) at play in this discussion. Leaving aside Universes for the moment, let's instead think about a truly thermodynamically isolated ...


1

If it is adiabatic then $ \Delta Q_i $ will be always zero. The fact that it is irreversible doesnt matter. Any path that thakes you from A to B will result in the same change of entropy, as both initial and final states are in equilibrium. If you choose what is called a quasistatic path, which is idealized as a tranformation that occurrs slow enough so that ...


1

I would not write $dS=0$ for a cyclic process but rather its closed contour integral $\oint dS = 0$, then you also have $\oint \frac {\delta Q}{T} \le 0$ and from $T>0$ you will get that in a cyclic process you must have somewhere, sometime, at some stage of the cycle $\delta Q < 0$, that is some amount of heat must be rejected to complete the cycle. ...


0

For a reversible path between two states (1 and 2), entropy change of a system is NOT zero. It is $$\Delta S = \int_a^b \frac{dQ}{T}$$ For reversible path between two states, entropy of the universe (Or any isolated system) is zero. $$\Delta S + \Delta S_\text{surroundings} = 0$$ So You cannot just take any system and say that entropy change between two ...


0

What is important here is that the entropy is also very low today. So, since the Big Bang the entropy has been increasing but we're still quite far removed from the heat death state. So, the reason why at the macroscopic level time reversal inavariance appears to be broken is simply because the entropy today is low compared to the maximum entropy, and the ...


1

Can this scenario be considered as erasure of information as referred to in Landauer's principle? No. The reason is that the whole process is reversible and Landauer's principle specifically needs an irreversible step - otherwise no "erasure" can have taken place. Where's the problem? It's here: The state $$\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$$ ...


1

The second law implies that entropy of an isolated system should be expected to go to a maximum given a sufficiently large length of time, but in the short term, for certain isolated systems it would be possible to have the probability that entropy will decrease over a certain time interval be larger than the probability it will increase over the same ...


2

After reaching thermodinamic equilibrium, if you wait enough time, any system will reaach, by random chance, any state of lower entropy. Of couse, the time needed to for such a fluctuation to occur increase exponentially with the amount of decrease in entropy (search poincare recurrence theorem). My guess is that this is not the kind of answer you were ...


0

I understand that there was zero entropy or at least low entropy that became self agitated to bring about the big bang. The closest example to zero or very low entropy is a solution of Dropleons with quasiparicles structures in a liquid state that are self propagating. This is based on the initial conditions of the solution containing super conducting and ...


1

In the standard homogeneous cosmological models the total energy in an expanding volume is zero. This is true for positive, negative or zero curvature and it must take into account the gravitational energy (which is negative), dark energy, matter and heat. Since the gravitational energy is negative the heat can be positive and increasing as you go back ...


1

In Zero-Energy Model, negative energy associated with Gravity counterbalances positive energy associated with matter, photons, etc. So, No, Big Bang wasn't cold. You are just looking at partial picture (you just ignored Gravity). This is what Zero-Energy Model says: With traditional Big Bang model (which doesn't contain Inflation), the universe started out ...



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