# Tag Info

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There are two definitions of entropy, which physicists believe to be the same (modulo the dimensional Boltzman scaling constant) and a postulate of their sameness has so far yielded agreement between what is theoretically foretold and what is experimentally observed. There are theoretical grounds, namely most of the subject of statistical mechanics, for our ...

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What is entropy really? I want to answer(!) this question from a different point of view. First off, I focus on your title and the phrase “really”. We don’t know what entropy is really. We don’t know what energy is really also, and any thing or concept else too. Entropy, like all other concepts created by humans, is a convention between some people to ...

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I read in the documentation that came with this simulator that when you change the number of particles "N", the total energy of the system "E" or the number of dimensions, the field will turn yellow (which indicates that the simulator has not incorporated the changes) until you hit enter. I made changes to these values, hit enter with the cursor still in ...

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Why does a critical point even exist? I think this question is equal to this one: "Why the width of the two phase region is bigger at lower temperatures and pressures?" Specific volume of liquids mostly depends on the temperature of them in comparing with their pressure. This means, for a well-defined increment of the pressure, we can neglect its effect ...

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Attempting to answer the "why" question intuitively: In a liquid, the molecules experience significant intermolecular force - so much so, that the average energy of the molecules is insufficient to escape the attractive force of the surrounding materials. The result is that it energetically favorable for them to remain close together, even if that means ...

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I will try to answer these questions from different views. Macroscopic view The "quantitative" rather than qualitative difference in a liquid-gas phase transition is due to the fact that the molecules arrangement does not change so much (there is no qualitative difference) but the value of the compressibility changes a lot (quantitative difference). This ...

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For a pure substance that can exist in the solid, liquid, and vapor states (i.e., wood is not in this category), let's assume that a closed container is half full of liquid and half full of vapor. As the temperature rises, the liquid expands and the liquid density falls. Also, as the temperature rises, the pressure in the container rises due to the vapor ...

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Good question. I don't have my Widom around, but I'll try to answer from memory. I think the consensus is to say a substance is at its gas state if it could be a liquid at the same temperature. This, as opposed to same pressure, same volume, etc. If the temperature is supercritical, there is no transition between liquid and gas, and the generic term "fluid"...

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I'll try to explain why there could be a critical line and not just a critical point, and hopefully that will answer your question. If you think about the Ising model, we have the standard Hamiltonian: $$-\beta H = J_1\sum_{<i,j>}s_i s_j + h\sum_{i}s_i$$ where $\sum_{<i,j>}$ is a sum over nearest neighbors. This model ...

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The dimension issue is solved easily by defining the probability density function(PDF) as $$P(\{q,p\})=\frac{E_0}{h^{3N}} \ \delta (H(\{q,p\})-E)$$ where $E_0$ is an arbitrary constant which will not affect any thermodynamic quantity or equilibrium property. Actually, this definition is incomplete. We have to take into account the indistinguishability of ...

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I think the key here is that you're misunderstanding these integrals. Let's look at the following integral you wrote: $$\Omega(E,V,N)=\frac{1}{h^{3N}}\cdot\int_{H=E} d\Gamma$$ And let's look at a common example when H is a function of p and q. Now this definition becomes: $$\Omega(E,V,N)=\frac{1}{h^{3N}}\cdot\int_{H(p,q)=E} d\Gamma(p,q)$$ Can you see ...

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Particle on a rotating ring For further discussion purpose, let's considere the dynamics of a quantum particle on a $r_0$ radius rotating circle at a constant angular velocity $\mathbf{\Omega}=\Omega\,\hat{e}_z$ . In cylindrical coordinates, we fix $z=0$, and we have the azimuthal angle $\theta$, which is the "good" degree of freedom describing the ...

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In CFT, we are interested in the continuum limit, where we can classify classes of models at their critical points. By means of the Jordan-Wigner transformation one can construct a fermion operator out of the spin operators of the usual 2D Ising model. Then, the continuum critical Ising model is described by a massless real fermion: $$S=\frac 12\int d^2 z\... 0 That calculation has restrictions, but, one in particular should be mentioned, that master equation is supposed to be connected to this entropy, but is not necessarily, the master equation can be connected to general entropic form, and that is a fundamental idea for a more complete proff. 9 Preliminaries: How do we define 'localized?' For a single particle, or for multiple non-entangled particles, it is easy to tell from the expressions for the wavefunctions whether they are localized or delocalized. For example, you might say that if the wavefunction is falling off exponentially or faster for large x, that is with a form like \psi(x)\sim e^... 2 In the context of solid-state physics, a closely related question has been an area of active research in the past few years. Most interacting systems do indeed thermalize (and thus delocalize) over long time scales. However, certain systems whose disorder is much stronger than their interactions experience "Many-Body Localization," in which the individual ... 3 To solve your problem exactly, you would have to solve the Schrödinger equation$$i \frac{\partial}{\partial t} \Psi (\vec r_1 \dots \vec r_N,t)= H \ \Psi(\vec r_1 \dots \vec r_N,t)$$where \Psi (\vec r_1 \dots \vec r_N,t) is the wave function of the N particles and$$H=\sum_i^N \frac{p_i^2}{2 m} + \sum_{i<j}^N u_{ij}+V_{\text{ext}}$$where u_{ij}... 2 I agree that the language is very confusing - I'm a native English speaker, and it also took me a while to understand what they were saying. When they talk about the dimension of "the statistical system itself," they mean the spacetime dimension. So if a system has two spatial dimensions, then it has three dimensions total (including time), and the ... 0 I suppose you mean that the gas is contained in a magic box. Otherwise the walls become part of the system, exchanging momentum/energy with the 'particles'. I have no answer for you; I don't know. What I do know is that none of the particle-particle collisions can be characterized other than by using a probability distribution. Common sense demands that ... 2 Quantum effects appear if the concentration of particles satisfies,$$\frac{N}{V} \ge n_q$$where N is the number of particles, V is the volume, and n_q is the quantum concentration, for which the interparticle distance is equal to the thermal de Broglie wavelength, so that the wavefunctions of the particles are barely overlapping. As the quantum ... 3 It really depends on the boundary conditions. For boundary conditions like a 3D box with reflecting walls, the initial quantum state \Psi will stay a quantum state with the unique wave function depending on variables of each particle:$$\Psi({\bf{r}}_1,...,{\bf{r}}_n, t).$$If the boundary conditions are such that allow exchange with the environment, then ... 0 The behaviour of the molecules in your gedanken experiment can be approached by using decoherence. But I do not believe you can get a definitive answer until somebody makes a full scale simulation (or until some expert's answer can make a formal proof of what really happens, but I am not skilled to do that). The decoherence effects can be argued ... 2 As @valerio92 points out, your mistake is that S = k \ln (\omega\, \delta E), not \delta S. To get \delta S, you differentiate the right-hand expression to get \delta S = k \frac{\delta \omega}{\omega}, and the \delta E drops out and you get an expression with the right dimensions. The notation is a bit misleading, because the \delta in the \... 5$$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 ...

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Your equation (2) is trivially a solution of (1), because $v$ and $T$ are constant. This is a disappointing answer, because it leaves unanswered the question what makes the Boltzmann distribution unique. The answer is that you only wrote down the collision-less Boltzmann equation, but in the real world collisions are always present (and indeed, systems ...

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There is no single point where this becomes true - it is a very gradual change. The buzzwords are microscopic $\to$ mesoscopic $\to$ macroscopic. There is no special kind of mathematics involved; in the mesoscopic domain one uses a mix of quantum mechanics and statistical mechanics. See https://en.wikipedia.org/wiki/Mesoscopic_physics

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Density is the average amount of mass per unit volume $\rho(\vec{r},t) = \frac{M}{V_r}$. Distribution function is defined as a number of particles per unit phase space volume $f(\vec{r}, \vec{c},t) = \frac{N}{V_r V_c}$ (which is space volume times velocity volume). Each particle has mass $M$, so to get the total mass density, we need to sum distribution ...

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The information paradox is 40-45 years old. AdS/CFT is not even 20 years old, modern string theory (after D-branes) in probably 25 years old. This is to say that the information paradox doesn't need any of these, even though of course you can try to solve it in the context of string theory. In a nutshell, the information paradox is a sharp theoretical ...

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This has been open for a while so I will bite. The information paradox has two versions or iterations. The previous one is that information is demolished by black holes by the entropy of its event horizon and that Hawking radiation that is emitted is in a pure blackbody distribution. A blackbody distribution of radiation is maximally random. If you make a ...

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Clausius' statement about heat not being able to flow spontaneously from a cold body to a warm body is sufficient to prove that no engine can have an efficiency greater than that of a perfectly reversible engine. But it's not enough to prove that the Carnot engine is the only reversible engine. For example, there could be a perfectly reversible engine where ...

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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 ...

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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).

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The Maxwell-Boltzmann distribution and the Boltzmann distributions are probability distributions, i.e. functions $\rho(\vec x,\vec v)$ of velocity and position of a particle, that say what is the probability density that the velocity and position belong to the small cube around the given value of them. The Boltzmann distribution is the more general one, \$\...

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Greiner's Thermodynamics and Statistical Mechanics is pretty good from a few short readings I did. Also, it has better reviews from almost all of the other popular textbooks on the subject in goodreads.com

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You said it yourself the molecules have direction rather than randomly moving about. Picture a wide spot in a river, slow water, maybe eddy currents, random water flow. River narrows, water is directed through the slot. I agree with you, speed before pressure differential. What do you think about this, molecules entering Venturi creating vacuum which sucks ...

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