# Tag Info

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Short quick answer: From Wikipedia Entropy In thermodynamics, entropy (usual symbol S) is a measure of the number of specific ways in which athermodynamic system may be arranged, commonly understood as a measure of disorder. According to the second law of thermodynamics the entropy of an isolated system never decreases; such a system will spontaneously ...

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It can be rather involved. A lot of technical progress as been on this subject leading up to the modern conformal bootstrap work. Something you can exploit is that these functions should behave like correlation functions and thus are eigenfunctions of the conformal Casimir. That gives you differential equations which in some cases, especially in $D=2$ and ...

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The variables involved here are classical and you resolve them classically. They enter into the operator because they are parameters of the wavefunction. So let's do this a little more broadly. For continuous systems, we want a family of solutions based on some parameters which I'll collectively identify as $\alpha \in A$; the solutions are then labeled ...

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I just want to add to Lumo's answer: The paper by vafa and Strominger instigated a lot of work in determining the statistical formulation of entropy in black holes. Although it must be pointed out that most of these are for cases with supersymmtry and (near) extremal conditions at small couplings. There has also been work in trying to address the microscopic ...

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This is a very profound question in physics. Given that a black hole has an entropy which scales as $$S_{BH} \sim \frac{A}{4},$$ the question is how does this relate to $S_{Boltzmann} = K_B \ln W$. As in, what are the microstates of the theory which hold the information in the black hole. This was answered in part by a series of papers by Vafa, Strominger, ...

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Yes, neutrinos should obey Fermi-Dirac statistics and yes, the Pauli Exclusion Principle should operate for neutrinos. But let's examine how dense the neutrino population has to be for this to be important. The Fermi momentum is given by $$p_F = \left( \frac{3}{8\pi}\right) h n_{\nu}^{1/3}$$ where $n_{\nu}$ is the neutrino number density. In order to be ...

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The most important reason to limit the scope to quasi-static transformations is, because you are doing equilibrium thermodynamics and therefore always (maybe implicitly) assume that the system is in an equilibrium state. The question itself appears to me like you are thinking of the difference between adiabatic (no exchange of heat) and isentropic (no ...

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Not sure the calculation was done correctly (on first glance). However, that is not important here, just think about what you are doing: You have N particles all of which are mutually interacting via a super-long-ranged potential (Coulomb-interaction $\sim r^{-1}$ would be considered long-ranged, you are using a parabolic $\sim r^2$ potential). So, what you ...

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In macroscopic units it should be $$S=-R\alpha \log(\alpha e^{-S_1/R})-R(1-\alpha)\log\Big(1-\alpha)e^{-S_2/R}\Big) \\=\alpha \Big(S_1-R\log\alpha\Big)+(1-\alpha)\Big(S_2-R\log(1-\alpha)\Big),$$ where $R$ is the universal gas constant. In the pure case, this reduces to the textbook formula. But such a formula cannot be true in general. The general formula ...

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I don't know if it will help you, if it doesn't, sorry for wasting your time. But this is the best explanation I have at this moment. The gas must be flowing faster because of the lower cross sectional area. Assuming no density change, the only way to maintain the same volumetric flow rate is to increase the flow speed. Now consider the perspective of a gas ...

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Wave function collapse (or a measurement on the QM system) is not unitary since it is a projection of the state vector.

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To give you the idea of the complexity of this problem, consider that the dynamics of a single fluid is governed by the Navier-Stokes equation, which is already impractical to solve. $$\frac {\partial}{\partial t} (\rho u) + \nabla \cdot (\rho u \otimes u + p I) = \nabla \cdot \boldsymbol \tau + \rho g$$ Then this equation is coupled to probably three ...

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As people have pointed out, the total heat transferred to raise or lower the temperature of an object by $\Delta T$ does not depend on how the object is heated. The rate of temperature change depends on the mode of heat transfer. For example, cooling by convection gives a rate proportional to $T$, while cooling by radiation gives a rate proportional to ...

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You're missing a minus in the entropy definition - $S=-Tr(\rho\ln\rho)$ Entropy of a unitarily evolving system (doesn't matter in which picture) is conserved (The entropy is a trace of a function of the density matrix "operator" thus it depend solely on the eigenvalues of it's input operator, but the eigenvalues of the density matrix don't change under ...

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Mechanism of heat transfer is different in different ways of heating up the system. For example, in Conduction, heat is transferred by collision of neighbouring particles whereas in Radiation heat is transferred by means of electromagnetic radiation. Equations governing these two mechanisms are different. Therefore, heat transfer rate won't be same in both ...

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The equipartition theorem is a mathematical consequence of very specific kind of Hamiltonians. It states that any 'squared' term of deegree of freedom in the Hamiltonian gets $\frac{1}{2}k_bT$ of energy (it is a statement about the energy distribution for this kind of Hamiltonians). For example - classical ideal gas Hamiltonian - ...

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I am not sure if this answer will give you an intuitive understanding of the result, but I think it may be useful as it shows the assumptions behind it. What your result means is that in an idealized situation when the volume gas or solute occupies is shrunk slightly by $\delta << V$ while its energy remains the same (let's say, isothermal compression ...

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This phenomenon has been studied by Vella and Mahadevan and written up in the American Journal of Physics (http://scitation.aip.org/content/aapt/journal/ajp/73/9/10.1119/1.1898523). It's called the Cheerios effect. If the cereal pieces clump together away from the edges of the bowl, they gravitate toward a slight concavity in the surface caused by water ...

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Something analogous to the Fermi-Dirac distribution function will probably work pretty well for the electrons in the Sun; see for example this PDF. You may need to put in some "fudge factors" for their interaction energy with the rest of the proton soup, but you're in some luck, because the interaction energy in the Sun is actually mostly lower than the ...

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Noether's theorem states that if a system has a continuous symmetry, there is a quantity related to this symmetry, called the Noether charge, which is conserved. It does not state anything on the fact that adding a constant term to a measurable quantity may or may not change the physical description of the system. Only some physical quantities in fact are ...

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The route of finding $\frac{\partial S}{\partial x}$ and $\frac{\partial S}{\partial T}$ is the more practical route. $\frac{\partial S}{\partial x}$ can be found using a simple Maxwell relation for the Helmholtz free energy: $$dA=dU-d(TS)=Jdx-SdT$$ $$\left. \frac{\partial A}{\partial x}\right|_T = J \, \, \mathrm{and} \, \, \left. \frac{\partial A}{\partial ... 1 Ideal string cannot be described in terms of canonical ensemble (with Boltzmann probability distribution). The fact it is ideal means it has infinity of degrees of freedom and since each would have, according to the Boltzmann probability distribution, average energy$$\frac{1}{2}k_B T,$$total energy of the string would be, on average, infinite. This is ... 0 If you assume linearity and the string is finite, then the partition function is a sum of discrete energies: it is the same case as a string in 2D ( y=f(x,t) ), but with two uncoupled dynamics: (y,z)=(f(x,t),g(x,t)). This can be shown by considering an oscillator in 2 dimensions with dynamics \ddot{\vec{x}}=-k\vec{x}: you will see you can separate it ... 2 Maybe the important step is to realise that the allowed range of momenta is$$ R=[-p_\text{max}:-p_\text{min}]\cup[p_\text{min}:p_\text{max}].  Then the first two $\Theta$'s give one if $p$ is positive and in the allowed range, whereas the last two $\Theta$'s give a contribution that's only 1 if the $p$ is in the negative allowed range. The term out front ...

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You mentioned that The Boltzmann distribution is given by $p_i =\frac{1}{Z} e^{−\beta E_i}$, where $p_i$ is the probability of the system being in state $i$ , $E_i$ is the energy associated the $i$th state, $\beta=1/k_B T$ is the inverse temperature, and the normalising factor $Z$ is known as the partition function. It would be a good ...

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In principle one can always write down the differential equations for a system of $N$ particles and attempt to solve them: as you pointed out, there is no general solution if usually $N>2$. As such, nevertheless the need to describe features of general systems remains. The key point here is understanding that, as a matter of fact, whenever dealing with ...

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You may have seen the reasoning to follow in most textbooks already but apparently it is not emphasized enough so I will say it again here. The crucial starting point is the second law of thermodynamics that claims that the entropy change of the universe $\Delta S_{univ}$ is either zero or strictly positive for any physical change that occurs in it. I ...

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Ah, but who says that negative absolute temperatures exist at all? This is not without its controversies. There's a nature paper here which challenges the very existence of negative absolute temperatures, arguing that negative temperatures come about due to a poor method of defining the entropy, which in turn is used to calculate the temperature. Other ...

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@Couchyam gave a very nice, detailed explanation. I'll just add that the density of states is just the continuum limit of the degeneracy of energy states. For a discrete system $\Omega\left(E \right)$ gives the number of microstates with energy $E$. When calculating average values of bulk thermodynamic properties, this acts as a weighting function. To find ...

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Statistical Equilibrium: That state of a closed statistical system in which the average values of all the physical quantities characterizing the state are independent of time. In other words when systems do not evolve in time i.e. when they are in steady state. In thermodynamic equilibrium there are no net macroscopic flows of matter or of energy, either ...

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In general, the density of states can be computed as follows: Find eigenstates of the Hamiltonian, $\psi_{s}$, so that $H\psi_s=E_s\psi_s$ (except in special cases, this is usually the hardest part). Compute $N(\epsilon)$, defined as the number of states $\psi_s$ with energy $E_s<\epsilon$. $D(\epsilon)=\frac{dN}{d\epsilon}$ is the density of states. ...

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