# Tag Info

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Is disorder the same as entropy? In statistical physics, the definition of entropy varies and also depends on the particular ensemble chosen, but roughly speaking, is a measure of the number of microstates of a system. A particular definition is given by the expression, $$S=-k_B \log \Omega$$ where $\Omega$ is the number of microstates, and $k_B$ is ...

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There is a universe where the second law doesn't hold. In fact, the opposite holds: entropy always decreases. Things are a bit strange there, though. The people there can easily predict the future many years in advance (although they don't call it "predicting" but rather "gnirebmemer"), but even their best scientists can't remember what the weather was like ...

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The thermodynamic understanding of entropy was arrived at empirically, so a top down deduction of what it means. The statistical mechanics understanding of entropy is a bottom up approach and falls out naturally from an an understanding of the number of different microstates that can result in one macrostate. So I guess they differ in that a ...

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As Jan Lalinsky has already pointed out (and indeed the OP has implied), the concept of macrostate is defined by the macroscopic quantities you have available experimentally. Suppose you know the expectation values of some set of extensive observables $X_i$. Typically these observables will be energy, particle number, average magnetisation, etc. This set ...

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No. Decoherence. No observer necessary.

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What about this derivation? Is this derivation of Black Hole entropy viable? In short, Bekenstein's entropy is integer amount of bits, which cannot be true for variable measured at Planck scale, where the fundamental unit of information is nat, that is $1/\ln 2$ bit

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I believe the author is thinking here of things like electrochemical work that don't involve a change of volume (but rather, in this case, moving a charge through a potential difference). In this case a full thermodynamic description involves the chemical potential $\mu$, as a later equation shows. (I agree that the description of “different free energies ...

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If I understand the question, you are wondering how to justify the statement that a (reverible) adiabatic process is isentropic from the point of view of statistical mechanics (the classical thermodynamics definition makes sense to you). Let us then start with the entropic fundamental relationship, S = S (U, V, N), where U stands for energy, V for volume, N ...

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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 : $dS ... 1 It is common to hear in cosmology that the entropy of the universe is adiabatic (constant) at late times. In the standard model of cosmology,$\Lambda CDM$, the universe is dominated at late times by the cosmological constant that causes exponential expansion of the universe. During this epoch, the formation of new structure is limited by exponential decay, ... 1 Yes and No Yes part If you think the whole Universe as a closed system , then the entropy must be constant. How come ??? According to Quantum mechanics , the evolution of a closed system is unitary. The unitarity forces us to believe that the whole universe can be thought of as a reversible computer and so a big fine quantum computer. Now since the whole ... 2 Entropy is a natural tendency of the Universe to fall apart into disorder. In a reversible process, an increase in the Entropy of the system will be exactly equal to the Entropy decrease of the surroundings. Thus, the net change in the Entropy of the system and its surrounding will be zero. But in an irreversible process in an isolated system (for ... 2 As far as I know, the expansion of the universe contributes into creating more and more microstates. This is almost equivalent to saying that entropy increases (because$S = k_B$ln$(\Omega)$). We cannot be sure that the law of entropy applies to the whole universe (There is debate if the universe is a closed system or not, if its infinite or finite, etc..) ... 2 The paper suggested by Trimok seems to answer your question. The paper gives an entropy for the observable universe of: $$S_{obs U}= 3.1×10^{104} k \approx 10^{104}bits$$ where$k$is the Boltzmann constant and$S_{obs}$is the entropy. However I would like to answer your two questions with a back of the envelope calculation. ... 0 Yes it can be, if the Universe attains equilibrium. 3 The law states that entropy increases in isolated systems. Let us take a cell, as an example of a live organism. If you isolate a cell, it will die, and the entropy of the isolated system will increase , as an isolated cell cannot live. All organisms live by continuously decreasing their own entropy and increasing the entropy of the surroundings, ... 3 You've overlooked gravitational entropy. The entropy of a black hole horizon is given by: $$S = \frac{kA}{4 \ell_p^2}$$ where$A$is the area of the horizon,$k$is Boltzmann's constant and$\ell_p\$ is Planck's constant. This entropy is absolutely huge, so if you take a uniformly distributed gas in thermal equilibrium and concentrate it into a black ...

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