# Tag Info

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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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Getting cooler is the key phrase in you question. If you vigorously stir cold water you might get it to warm up if it was well insulated - duplicating Joule's clasic Mechanical equivalent of heat experiment. But if you stir a cup of hot coffee or a pot of hot water the far more significant effect effect it that the stirring accelerates cooling. Stirring ...

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Moving water convects heat better than static water. Take the reference frame of the water being static and the air moving. In this case, new cooler air is always sweeping in to take heat away. If the air is static, hot air remains at the interface and will not accept heat as well as cooler air. The same argument can be made for the moving water. Edit: The ...

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Yes. I believe it is done from Lagrange multipliers. This whole maximum entropy is in most textbooks on statistical mechanics. The configurational entropy of a system will tend to increase... Or have i misunderstood your question?

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To do it reversibly, you can heat the body from $T_1$ to $T_2$ (i.e., over a finite temperature change) using an infinite sequence of constant temperature reservoirs, in which each reservoir in turn is only dT higher in temperature than the body at any time (and also only dT higher in temperature than the reservoir before it in the sequence). Each increment ...

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 ...

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What justifies identifying this entropy with the one found in thermodynamics? Suppose parameter $X_1$ is contrained to some value $x$. The corresponding number of available microstates is $Ω(X_1=x,X_2,\ldots,X_n)$. This part of phase space is a subspace of the whole one, consisting of all microstates without constraint. So ...

3

We do observe spontaneous symmetry restorations in nature. This is called an emergent symmetry. See e.g. this post. A system posses an emergent symmetry if it appears symmetric at large (coarse-grained) scales although the apparent symmetry is explicitly broken by the microscopic description (typically the Hamiltonian or Lagrangian). I can give two examples ...

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It's maximizing the angular entropy in the sense that far more "available" angle vectors are inhabited. I suspect you need to be careful with the word "entropy" here. For example, the photons are not down-converted into a larger number of photons of longer wavelength (aka 'heat death of the universe').

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The thermodynamic definition of entropy of a system is where T is the absolute temperature of the system, dividing an incremental reversible transfer of heat into that system (δQrev). (If heat is transferred out the sign would be reversed giving a decrease in entropy of the system.) You may consider a vacuum as a system under study, but in this ...

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Your definition of Holevo information is wrong. It corresponds to $C_{ea}$, the entanglement assisted capacity of the channel. See equation (5) of the paper. The Holevo information is defined for a probabilistic mixture of density matrices, or for a cq-state (cq = classical quantum state).

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To understand the origin of black body radiation, which is what every material body with a temperature T has been measured to emit, one has to go to quantum statistical mechanics. Atoms and molecules, to start with, in any ensemble, interact with the electromagnetic radiation. At that level, the processes are quantum mechanical. In quantum mechanics the ...

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So why do we say entropy is extensive? It is a convention that is possible and useful for weakly interacting systems. Multiplicity of state of the system made of two systems of the same kind and size and energy $E$ is $$\Omega'(2E) = \Omega(E)\Omega(E)$$ with very good accuracy, the other terms in the sum over $x$ you indicated can be neglected. ...

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The entropy $S$ is extensive as long as you're consistent about what you mean by entropy. In your case you've mixed up two different definitions. One definition of the entropy has the system at fixed energy $E$ -- the other, a fixed temperature $T$. Fixed-E entropy For a system with fixed energy $E$ the entropy is defined to be $$S = \log\Omega(E) ... 0 Simple-Every physical or chemical process always aims at decreasing the energy and increasing the entropy(randomness of things).Lets take some examples When an object falls on the ground then it would break and result in a large number of pieces completely scattered(randomness) thereby increasing entropy.The pieces would never come together to make the ... 0 The entropy defined by Boltzmann can be applied in systems with fluctuating energy. The only requirement is the sharp definition of \Delta\Gamma, the number of microstate inside the macrostate. When the time is very long, the distribution of probability in phase space tend to be constant at every allowed point. This justifie the microcanonical ensemble. ... 9 I think there is a misunderstanding. You are perfectly right when you write that the total micro canonical entropy of a combined system will be $$S_\textrm{combined}(2E) = k_B\ln \sum_x \Omega(x)\Omega(2E-x)$$ The micro canonical entropy ought to be a function of only the total energy, total amount of matter and total volume of ... 6 Major edit: In @gatsu's answer, it is pointed out that only the amount of energy should matter, which is correct, as there's no such thing as distinguishable microstates with only rearranged energy (think stars-and-bars-type entropy calculations). So, I've edited out that part of the first paragraph and equations (in the first draft, I dropped that part of ... 1 All comments miss a point, or maybe I didn't get it right. I see it this way, lets start with other forms of energy (Kinetic and potential for example) of an harmonic oscillator. We know for a non damped HO the relation between position "s" and velocity "v" is given by total energy Eo = 1/2kx^2 + 1/2mv^2, k spring constant, m mass. this relationship (since ... 0 Asking a question about the foundations of statistical mechanics is a good way to start a fight, so don't expect a clear consensus on this. Stripping away the hype, what these papers try to do is establish that an isolated quantum system will, under certain (but general) conditions, evolve to a state that looks like thermodynamic equilibrium locally, even ... 0 Your derivation is almost correct, and as I'll show below, one does sometimes make use of it. However, it relies on an unwritten assumption that makes it applicable only in a limited range of situations. That assumption is that the heat capacity c doesn't depend on the temperature. It is this assumption that allowed you to take the factor of mc outside ... 0 Your equation is correct only if:$$\mathrm dQ = mc\,\mathrm dT$$which is not generally true, indeed, common sense tells you that a change in temperature leads to conclusion that an object being heated up. But we do not encounter gases much in our life, which could be regarded as a general case. In reality your assumption is generally false, a good example ... -1 Here is a link that might be useful http://hyperphysics.phy-astr.gsu.edu/Hbase/thermo/temper2.html#c2 1 Entropy is something related to thermodynamics, which is essentially a study about energy interactions. So you can use the concepts where ever you see an energy interaction. Entropy is a measure of randomness resulting from the increase in heat energy supplied to a system. When you consider an electric circuit, you have an energy source which is the emf of ... 3 At the power station electricity is generated as work from a heat engine. Work is entropy free, so we have an entropy free electron-gas at the point of generation. However, a thermodynamic gas will always equilibrate to the available degrees of freedom. In this case it is the electronic states of the conductor in the transmission wire. There will be a ... 5 The energy of an electrical wave certainly undergoes energy loss through heat, so in that way is entropic. This is a result of the material's resistivity through which the electricity is conducted. The mechanism is primarily scattering of the energy through electron-phonon interactions, but electron-electron interactions do also occur. In metals, the main ... 0 The book in question is self-published, and a cursory glance through it and some of the author's other writings convinces me that he is almost certainly a crackpot (although perhaps self-educated enough to sound convincing to one who might not know better). If you have a true interest in topics about space and particle physics (and, if so, good for you!), I ... 2 Let \lambda_i be the eigenvalues of \rho_A. Then$$ \log \text{tr} \rho_A^n = \log \left( \sum_i \lambda_i^n \right) $$Now, let is differentiate w.r.t. n. We find$$ - \frac{\partial }{ \partial n} \log \text{tr} \rho_A^n \bigg|_{n=1} = - \frac{\sum_i \lambda_i^n \log \lambda_i }{\sum_i \lambda_i^n } \bigg|_{n=1} = - \frac{ \sum_i \lambda_i \log ...

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The wikipedia article about the holographic principle does not suggest, that this is well-accepted and founded theory :) The ideas seem to be somehow based on the concept of the Bekenstein-Hawking entropy for a black hole. This is supposed to be the entropy of a black hole from outside (according to the no-hair-theorem the black hole has only three ...

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