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It seems I have figured out an answer for 2 terms in the original state. Suppose that the state is $$\rho = a |\alpha \rangle \langle \alpha | + (1-a) |\beta\rangle \langle \beta|$$ We need to write it in a basis, which is $$|+\rangle = \frac{|\alpha\rangle + |\beta\rangle}{\sqrt{2}}; \quad |-\rangle = \frac{|\alpha\rangle - |\beta\rangle}{\sqrt{2}}. $$ ...


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As stated in the comment by Peter Diehr, the question is in principle no different whether you ask it for electromagnetic, gravitational or any other kind of wave. The wave's entropy is simply the conditional Shannon entropy of the specification needed to define the wave's full state given knowledge of its macroscopically measured variables. A theoretical ...


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The answer is a definite yes and no. Gravitational waves have entropy in that we can think of them travelling from their source to our detector as a channel that sends units of information in the sense of the Shannon formula. The ringing of our detector is then the reception of that information. The Shannon formula $S=-k\sum_n p_n log(p_n)$ would give a ...


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It is being done reversibly, where the gas is only slightly lower in temperature than the reservoir. It can therefore absorb heat from the reservoir very slowly, over a long period of time. Since the gas is passing through virtually a continuous sequence of thermodynamic equilibrium states with virtually no temperature driving force, the process is ...


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This can be concluded by reviewing Gibbs equation for upstream and downstream stagnation conditions. $$T_0ds_0=dh_0-\frac 1{\rho_0}dP_0$$ Because across the shock wave is an adiabatic process, $dh_0=0$ Then Gibbs equation becomes $$ds_0=-\frac 1{\rho_0T_0}dP_0=-\frac {R}{P_0}dP_0$$ We know entropy increases. This leads to conclusion that the stagnant ...


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My main concern is: does this blurring/loss of knowledge come from any well-known physical law/principle? For instance, should we link the quantization of the phase space to ΔxΔp≥ℏ2ΔxΔp≥ℏ2, or to some kind of observer effect? The description in the Wikipedia article is misinformed and misguiding. The blurring, or coarse-graining, is merely one possible ...


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The physical principle being invoked is the finite resolution of any experiment, independent of the value of $\hbar$, together with coupling between observable and microscopic degrees of freedom, i.e. it applies to both classical and quantum systems. Technically energy conservation and Liouville's theorem, or unitarity in QM, are also needed to prevent ...


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Just wanted to point out that the last comment may mislead people. "Entropy, as defined in information theory, is a measure of how random the message is..." Perhaps the writer means a measure of how unpredictable it is, and uses random as a synonym. It's not a good one. If a signal is predictable it is redundant, and redundancy, under scarce resources, is ...


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Entropy has various uses in Physics. It was first used as a state variable for thermodynamics, connected to energy and heat. Later, it was revealed that entropy is a measure of the disorder of a system. By extension, it is a measure of the observer's ignorance or lack of expectation about the precise, microscopic state of a system. This in turn is related to ...


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The information paradox has no particular connection to electromagnetism. Hawking radiation is not just photons, it's any sort of particle. And Hawking radiation in itself doesn't solve the information paradox - the problem is that Hawking radiation is supposed to be thermal, so quantum information of infalling objects has been irreversibly lost, but that ...


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Yes. Black Holes (BH) can grow from accreting anything with energy --- including dark matter (DM). I'm not entirely clear on the second part of your question, but probably the most important thing to keep in mind is that the black hole information paradox is still unresolved. Answering how information is not lost for any type of particle, including DM, ...


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If you can define a perfectly closed system which eventually arrives back at its starting state, that system is reversible. A reversible system is not irreversible, and can never "become" irreversable without changing the system. However, in the real world, there are limits which come into play. The first is that the systems are never actually isolated ...


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Every deterministic system with chaotic trajectories is losing information on account of infinitely small deviations leading to completely different results. In your case, if the ball eventually settles into a stable orbit after some time, no matter where it started, then computing backwards would amplify any uncertainty so that after a few cycles, you know ...


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Let's take the 1D equivalent of your problem for simplicity: a particle bouncing back and forth along a segment, reversing its velocity every time it hits the boundaries of the segment. If we know perfectly the initial state of the particle, i.e. its position and velocity at time $0$, $(x(0),v(0))$, we will know exactly what the motion of the particle will ...


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If you assume that the ball is in the state you describe, and will continue to run through the same path over and over again, then you assume, that your system is deterministic, that means, from one position and momentum of the ball, you will be able to calculate its path for all the times that will come after that. In your special case: If the time goes ...


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The quantity $\left(\frac{\partial S}{\partial T} \right)_{\{\alpha\}} = TC_v$ is essentially proportional to the heat capacity of the thermodynamic system under study. As far as I know, there is no principle of thermodynamics that forbids such a quantity to be negative. Considerations such as "yes otherwise matter would not be stable" lie outside the ...


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No. Under negative thermal temperatures the entropy of a system increases as T is lowered. For example, if the magnetic field is reversed quickly enough around a ferromagnetic material, then the system is in negative thermal temperature.


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No, it doesn't. For example, consider a two-state system with ground state $| 0 \rangle$ and excited state $| 1 \rangle$. At zero tempreature, the entropy is zero, since only state $| 0 \rangle$ is occupied. As the temperature goes to infinity, the occupancies of $|0 \rangle$ and $|1 \rangle$ become closer and closer to equal. When they are equal, we have ...


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Before talking about entropy, we need to discuss what possible states an atom can be in. I will start by the most general case that consists in considering a single-atom gas in a 3D box. In that case, the microstate of the atom is described by: The definite linear momentum states $| \textbf{k} \rangle$ of the atom (that are eigenvectors of the hamiltonian) ...


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You need to consider the surroundings as well. If you go from state A to state B via a reversible process, the change in system's entropy exactly cancels out the opposite change in entropy for the surroundings; so overall there is no change in entropy. On the other hand, if it were an irreversible process, entropy change of the system (though same as the ...


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The entropy of a single atom does not make sense per se, unless you specify the preparation. The entropy of a single isolated atom, fixed at a point, is indeed not defined – the entropy is, after all, a property of an ensemble not of a system. The entropy of an ensemble of isolated atoms prepared at a specific energy, on the other hand, is well defined (this ...


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You state that: there is literally no way to squeeze more information (entropy) into a given volume than that in a black hole occupying that volume But you must keep in mind that the volume occupied by the radiation+BH system is larger than the volume of the black hole by itself. When the black hole initially forms the horizon has a radius $r$ which ...


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It's a confusion of terms. The universe is a closed thermodynamic system whether or not it is 'open' or 'closed' in a cosmological sense. In cosmology, open and closed universes refer to the curvature of the universe, whether positive (closed, finite universe), zero (flat, open, infinite universe) or negative (curved, open, infinite universe). In ...


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If the universe is open, there's obviously more universe that you haven't included in your system. The universe, by definition, contains all energy and matter. An open system, by definition, has an outside system to exchange energy and matter with. If that outside system isn't part of the universe, then where is it?


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There are many ways to think about the entropy of a vacuum (assuming there is no radiation and thus T=0), but all give the same result, the entropy is zero. One easy way is to notice that the walls are made of something (it doesn't matter what) that cannot change its state, so the number of microstates, $\Omega$, is equal to 1. Then $S=k\ln\Omega=0$.


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There is something they forgot to mention in your notes (either from ignorance, or out of omission). The temperature within the system is spatially non-uniform during an irreversible process. So what value of the temperature are you supposed to use in the integral of dq/T? The Clausius inequality calls for the use of the temperature at the boundary ...


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I think you have a typo in your question. A reversible process will have a smaller entropy change than an irreversible process. Your interpretation that the equality refers to a reversible process, while the inequality refers to in irreversible process is correct. Looking at the specific equations in that notes document, the integral in 8.31 applies to an ...


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This is Clausius inequality. The term ds>dQ/T. Holds an impossible reaction which does not obey Clausius statement in 2nd law of thermodynamics. The things in your class thought is about this ds<0. Where you can add c that is entropy generation term. Hope it will help you


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if we assume $E_1$ (zero state) and $E_2$ (state 1), from this question, the system has n units of energy and N particles. The multiplicity is then $$\Omega (N,n) = \frac {(n+N-1)!}{n!(N-1)!}$$ and entropy can be calculated, $$S=k \ln \Omega(N,n)$$


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I would like to add to what Procyon said in his(her) answer, which is right on target. The first equation in the OP should be an equality, not an inequality. For an irreversible process, the temperature of the system is typically non-uniform, and when we write $\int{\frac{dQ}{T}}$ for the Clausius inequality, what we really mean is ...


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Entropy is a property of the system. The change in entropy of a system as it traverses from an initial state(1) to a final state(2) is independent of the path by which the system is taken from state 1 to state 2. The path can be a reversible one, or even irreversible, the change in entropy is always the same as long as the initial and final states are the ...


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From a classical standpoint, it seems pretty clear that information can be easily lost. Nope, classical mechanics is reversible. It's so reversible that you can't even squeeze the state space: states that start out "very different" stay "very different". Those vague terms are formalized by Liouville's theorem. Because classical mechanics is reversible, ...



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