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As @valerio92 points out, your mistake is that $S = k \ln (\omega\, \delta E)$, not $\delta S$. To get $\delta S$, you differentiate the right-hand expression to get $\delta S = k \frac{\delta \omega}{\omega}$, and the $\delta E$ drops out and you get an expression with the right dimensions. The notation is a bit misleading, because the $\delta$ in the $\... 4 $$S=k \ln [\Omega(E)] = k \ln [\omega (E) \delta E] = k \ln [\omega(E)] +k \ln (\delta E)$$ Last term is an arbitrary constant, so that we can set $$S = k \ln[\omega (E)]$$ from which $$\delta S = k \frac{\delta \omega}{\omega}$$ If we can ignore the power contribution and set$\omega (E) \simeq e^{\beta E}$, we get $$\delta S = k \frac{\delta(e^{\beta ... 0 It is for the same reason that you need to use a dashed line for irreversible process which is higher than reversible process. This is because we know it is higher but don't know how high. If the dashed line is lower, we will conclude this is not feasible. This is the first thing what we want to know from the diagram. 0 It looks like the reaction \gamma \to 2\gamma is not only dynamically forbidden (Furry's theorem), but also kinematically forbidden. As Dexter Kim points out, the only way to conserve energy and momentum is that the two photons are emitted at 0°, in which case the angular momentum along the direction of motion is given by the coupling of the two photon ... 1 But for an ideal gas, internal energy is only a function of temperature and so internal energy remains constant here,no change in average kinetic energy of gas particles takes place, so where does the chaos come from to increase entropy of the system. 'Chaos' is not a very well defined term in context of statistical physics. It is not necessary to use it ... 0 The chaos comes from by changing of volume or pressure of the system. The average kinetic energy doesn't change, but number of collisions increases (if pressure increase) or length of paths increases (if volume increases). 0 It is certainly thermodynamically possible for a high energy photon to vanish and a multiplicity of lower energy photons to be created. This is observable as a cascade of events (photoelectric absorption of a photon, followed by multiple fluorescence photons) in thermalization of a high energy photon interacting with matter. It isn't a simple photon-in, ... 0 The (physical) concept of entropy is predominantly applied to many-particle systems. We can regard such a system as a many-particle system, whose dynamical variables comprise the positions, momenta, and other variable properties of all particles. It can exhibit, in theory, three types of dynmical behaviours: A low-dimensional regular (i.e., non-chaotic) ... 4 I'll have to disagree that those notions of entropy are disjoint. I'll try to explain my view. In Statistical Mechanics entropy is defined in terms of accessible regions in phase space. It is the logarithm of this volume times a constant. In the process of deriving this formula starting from the number of accessible configurations it is postulated that all ... 7 I would say the connection between chaos and entropy is through ergodic theory, and the fundamental assumption of statistical mechanics that a system with a given energy is equally likely to be found in any 'microstate' with that energy. Although chaos is a very general aspect of dynamical systems, Hamiltonian chaos (encountered in classical mechanics) is ... 0 I have went through a Thermodynamics Course, one of the best book which had used to clear my problem is: Thermal Physics by " Michael Sprackling" it describe the entropy in terms of both "Microstate" and "Macrostate". I have lot of doubts and which was very helpful for me 3 Is this taken to be an additional (and apparently implicit) assumption? You are correct. Take two arbitrary points A,B on the PV (or any other) plane, and draw an arbitrary curve connecting them: you have just defined a reversible transformation connecting A and B. This is because every point in the PV (or any other) plane represents an ... 3 Entropy Demystified (The Second Law Reduced to Plain Common Sense) by Arieh Ben-Naim. Authored discussed not only the thermodynamics origin of entropy but also the same notion in the context of information theory developed by Claude Shannon. 1 So, you want to prove that between any arbitrary two states of a system, it exists at least one reversible path. You can prove this if you accept continuity of properties of substances. I.e. for example, if we have an ideal gas in equilibrium at initial state (P_i,T_i) and final state (P_f,T_f); then certainly there are infinite equilibrium states ... 3 I would like to answer with the words of L.D. Landau, from his book Statistical Physics (first edition 1958): 0 Thermodynamics says that entropy increases in a closed system. The Steady State Theory claimed that the universe was not closed. In 1929 Hubble showed that the universe is expanding. To explain how an expanding universe can be Steady State, they claimed that new matter was created continuously to keep the density constant. This new matter would have very ... 1 If you can take the environment to be one with constant specific heats c_p and c_V, then you can write$$dS_{\textrm{env}}=\frac{\delta Q_{\textrm{env}}}{T_{\textrm{env}}} = \begin{cases} \frac{nc_VdT}{T} & \textrm{if isochoric} \\ \frac{nc_pdT}{T} & \textrm{if isobaric} \end{cases}, $$where the temperatures all refer to the environment. Then, ... 1 Your analysis is totally correct (except for the sign of the change), provided cv refer and cp refer to the heat capacities of the environment, and Tf and Ti are the final and initial temperatures of the environment. In such an analysis, the "environment" becomes your system. The signs are, of course, incorrect. They should be +'s. 3 So far there have been quite a few insightful answers about statistical mechanical entropy, but so far the only mention of thermodynamic entropy has been made by CuriousOne in the comments, so I thought it would be useful to give a short general reminder about the subtle difference between the notion of entropy in thermodynamics and the formulas that come up ... 2 To be honest, I believe this question is not really settled, or at least that there is not yet a consensus in the scientific community about what the answer is. My understanding of the relation is, I think, slightly different than knzhou, rob, or CuriousOne. My understanding is that thermodynamic entropy can be thought of as a particular application of ... 3 So Pratchett's quote seems to be about energy, rather than entropy. I supposed you could claim otherwise if you assume "entropy is knowledge," but I think that's exactly backwards: I think that knowledge is a special case of low entropy. But your question is still interesting. The entropy S in thermodynamics is related to the number of indistinguishable ... 7 Formally, the two entropies are the same thing. The Gibbs entropy, in thermodynamics, is$$S = -k_B \sum p_i \ln p_i$$while the Shannon entropy of information theory is$$H = -\sum p_i \log_2 p_i.$$These are equal up to some numerical factors. Given a statistical ensemble, you can calculate its (thermodynamic) entropy using the Shannon entropy, then ... 2 A Boltzmann brain is not that different from what might be called Boltzmann cheese. Given enough time a set of atoms or particles might arrange themselves by statistical fluctuations into big wheel of cheese. If that happens there is no reason to think the cheese would then rapidly be demolished unless it formed in a star, or falling into a black hole or in ... 0 I just want to add a second option that was proposed by some author whose name I cannot remember (it might be wikipedia, I did not check). Perhaps the laws of physics are such that it is very difficult to form a spontaneous and isolated boltzmann brain. According to this idea the only way to form a brain is to first spontaneously form a universe, like ours, ... 0 I would not say it decays instantaneously, but it will decays quickly, at least the conscious part. There is no body that supports it, so it will lack the oxygen to think in a brief moment. The rest will be bombarded by radiation and other particles until it disintegrates again, the timescale for this to occur will be a function of the temperature. 5 The Boltzmann brain paradox arises due to smaller fluctuations being more probable than larger ones. So, if you contemplate how our universe started out with low entropy initial conditions then it's difficult to explain this in terms of a generic high entropy state. Fluctuation yielding the early universe that in turn would have given rise to you, would be ... 1 Here are some references: Time and chance by david Albert Time's arrow by huw price From eternity to here by sean carroll The direction of time by H. D. Zeh Physical basis of time Asymmetry by paul davies There are many other excellent books or articles about the subject. Especially, in relation to the foundation of statistical mechanics I saw Two ... 0 The problem when you include gravity or other long range forces, is that thermodynamics becomes non extensive. For instance, the energy of the union of two systems is not the sum of the energies of the individual systems. To handle those cases, generalized entropies have been proposed. By generalized it means that these formalisms allow for long range ... 1 I never saw a proof and don't know where to find one. Infinitely stable substances only exist according to a simplified quantum mechanical theory where gravity and nuclear chemistry don't exist. Whether or not there has been a published proof that proves it from that simplified quantum mechanical theory, I don't know. Maybe it has never been proven that the ... 0 As you and others have said, it looks like Boltzmann could plausibly be credited with the idea that the universe had a low-entropy past: The second law will be explained mechanically by means of assumption A (which is of course unprovable) that the universe, considered as a mechanical system—or at least a very large part of it which surrounds us—... 1 Is this true for every reversible cycle? Is the efficiency of all reversible cycle equal to the efficiency of a Carnot Cycle? Yes. They are indeed. The equality in Clausius' Inequality$$\oint \frac{đq_\textrm{sys}}{T_\textrm{source}}=0$$is strictly valid for all reversible cycles. Temperature of a reversible engine is at all times equal to the ... 2 Leaving aside the issue of wavefunction collapse, physics is deterministic. So if you have some system like a gas and you know the exact positions and velocities of all the gas molecules you can predict the evolution of the system forwards and backwards in time. So you can start with a future state and work backwards to desciribe a past state. However ... 1 On top of the other excellent answers I'd like to point out that the accretion rate of dark matter particles is believed to be much smaller. The reason matter in accretion disk is being accreted rapidly is because they lose energy from electromagnetic radiation. For dark matter particles, in practice the only way it can be accreted is if the particle ... 1 So that narrows it down to some time between 1896 and 1979 The second law was known to Clausius, and trivially implies the knowledge that the entropy in the far past was much less than now. (That is, if one is permitted to apply the notion to the universe as a whole; cf. below.) It seems that Clausius stated explicitly (in 1856) only the extrapolation to ... 1 In order to input in your terms 500rpm you are required to input the actual shaft rpm Plus 500 you seek to add. You must charge your input to the systems energy level. So to answer your question yes your input of 500rpm would slow down the shaft rpm. 0 Your question really concerns error analysis. The subject (thermodynamics) is irrelevant. The "usual" custom of 2 significant figures is always subject to the data actually provided. The data here is 8.0kg and 0°C = 273K. The former is 2 sig figs; the latter is 3 sig figs. As a rough guide, if the calculation involves only multiplications and ... 2 An intuitive way of thinking about boiling an egg, is that you start a pool with a bunch of balls of yarn floating in it. Then when you heat the egg, it causes all the balls of yarn to unravel, and rather than a pool with some balls in it now it's a tangled mess of yarn. Before, each ball was free to move around as a unit, now each segment of each strand can ... 0 Entropy is function of state multiplicity. If you have the same gas, you wouldn't change state multiplicity and thus cannot change its entropy. That's different if two gases are different. 2 What is meant by a “change in volume of a system”? "Change in volume of a system" means "change in volume of a system", not anything else. System is a hypothetical concept. There is no specified system before we define it. We ourselves choose and define system. When someone talks about a system defined by himself/herself, he/she talks about that system not ... 1 Tl,dr: Entropy is the right definition, because it's incredibly useful in the description of statistical and thermodynamic systems. Whether or not it quantifies "disorder" in whatever sense of the word is completely irrelevant - it just so happens that it can be interpreted that way. Entropy is not a measure of disorder. At least not really. Then again, ... 3 The hamiltonian of a perfect crystal can be approximated at low temperature as the sum of harmonic oscillator hamiltonians. In 1D we have$$H = \sum_{i=1}^N \frac{p_i^2}{2 m} + \frac 1 2 m \omega^2 \sum_{ij} ( r_i- r_j)^2 $$where the ij sum is over nearest neighbors. It is possible to verify that the eigenvalues of this hamiltonian are$$ E_n = \left( ... 0 This is a very good question. But before I attempt to give you some of the details, the increase in the entropy is not what causes the universe to expand but is rather a consequence of the expanding universe. In fact, to understand why the entropy of the universe was so low before the inflationary epoch is an open question. Based on this question, I am ... 1 The problem that I see with the idea is that it seems contradictory. You used the definition of entropy of the microcanonical ensemble, which is defined by assigning an equal probability to every microstate whose energy falls within a range centered at E. All other microstates are given a probability of zero. That is, the range of energy is reduced in width ... 0 Assume we already know the relation$S\propto\log{W}$. Then we can define a coefficient of proportionality as$k_\text{B}$and write$S=k_\text{B}\log{W}$. So it is defined including all possible multiplication ambiguities from the beginning. Basically, this is a same question asking why we should choose$m$not$2m$or something else in Newton's second law, ... 3 We know that: irreversible+adiabatic =$\Delta S>0$, thus, if$\Delta S=0$the process is either: irreversible+non-adiabatic, reversible+adiabatic, or reversible+non-adiabatic. Then you can conclude that:$\Delta S=0$+adiabatic=reversible. Regarding you example of an irreversible adiabatic cycle: It is impossible and the example is flawed. The ... 4 Gravity can appear to increase order, but it doesn't violate the second law of thermodynamics. For example, take a gas cloud at uniform density and temperature$T_1\$ as our system. If we let it condense under its own gravity we can consider two cases: (1) Assuming it forms a gas giant quickly without radiating any heat (i.e. adiabatically), it will have a ...