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## Hot answers tagged statistical-mechanics

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To find the global minima of a function in a configuration space using Monte Carlo methods there are two main approaches simulated annealing and parallel tempering. Simulated annealing Simulated annealing is single Markov chain starting at high temperature for global exploration. The system is then evolved via Monte Carlo update whose criteria for ...

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It depends if we look at particle as classical ones or quantum ones. In the first case, particles are usually following a Boltzmann statistics. However, things become more interesting when entering the quantum world. Here, the spin of the particles become crucial. We have that particles with integer spin follow a Bose-Einstein statistics. Whereas particles ...

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The assumption of a full eigensystem is usually made for convenience. But it is not always satisfied. If it is not satisfies one gets additional logarithmic contributions to the scaling laws. This is discussed, e.g., in the paper by Wegner and Riedel.

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The first thing to point out is that there are two equivalent ways to describe a quantum statistical distribution: the density matrix, and a probability distribution on the results of measurements of a "complete" set of observables. (It is remarkable that one should think of givin such a probability distribution on the classical phase space of the quantum ...

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The von Neumann entropy, written in terms of the quantum mechanical density operator, is a constant of the motion if you keep track of everything (including entanglement with the environment) and don't have any collapse events (which, depending on your favorite interpretation of quantum mechanics, might not exist anyway). The thing is that this fact already ...

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I think you make it sound much more mysterious than it is. The relativistic distribution function is $$f_p = \frac{1}{(2\pi)^3}\exp(-(\mu-u\cdot p)/T)\,$$ where $u_\alpha$ is the 4-velocity of the fluid, $p_\alpha$ is the 4-momentum of the particle, $T$ is temperature, and $\mu$ is the chemical potential. This is sometimes called the Juttner ...

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Here's how I would come to some intuition for it. I would think about the rate of "probability flow" into a region by integrating the equation over a region in space. For now, let's suppose that no diffusion occurs at all, since that is more complicated (although directly doable and understandable). Then $$\int_a^b dx\frac{\partial p(x,t|x_0)}{\partial ... 2 Therefore, when we say, for example, that the energy of the ideal gas at temperature T is E=32NkBT, we should really be saying "the energy of the ideal gas immersed in a heat bath at temperature T"? Is this reasoning valid? This is true. What is also true is that you can also say that the temperature of the (completely isolated) gas of energy E is T=2/3 ... 2 The minimal counterexample seems to me to be the following: Take two materials, placed next to each other: ____________________ | | | | Material|Material | | 1 | 2 | ____________________ E1 _ _ _ E0 _ _ They have energy levels as indicated above- both have states at E0 and E1, but one has two excited states. ... 2 If the object is at rest it would imply that gravity transfer heat without an increase in potential energy, and there are no other forces that produce work. This would violate the conservation of energy. Regarding your comment "any contact forces imposed on an object will increase that object's heat energy": this is incorrect, friction only results in heat ... 2 A infinite barrier of potential reflects the fact the particle cannot enter a certain region of space. Solving the Langevin equation with such a barrier means that you have to find a way to state that the particle cannot enter the domain, but you also have to describe what happens at the boundary, because several scenarios are possible : the particle stops ... 1 I'll take a stab at this, although I am not an expert in the laser fields. Negative temperatures likely do not work: One concept that may seem closely related to the possible "thermal excitation of a laser" is that of "negative temperature". In a state of negative temperature higher energy levels have higher occupation probabilities than lower ... 1 Here may not be a complete answer, but it may give you (and me) some hints and perhaps some alternative solution. Euler's scheme should not work, even for the deterministic equation where noise is set to zero. This is because in Euler's scheme, one always requires \Delta t small so that the \Delta x is small. When \Delta x is large, one runs into the ... 1 The acceleration of gravity is the weight of the object on the surface. If the structure, either of the object or the surface, is deformed by the weight, then part of the gravitational potential energy of the object will turn into kinetic energy of surface moleucles and therefore heat, and this will be shared between the object and the surface. In this ... 1 From Reif's derivation of the canonical ensemble, the idea in describing the smaller system S is to use the fact that the total system including both S and R is isolated and in thermal equilibrium. That is, the total energy, using the notation you have written, would be E_T=E_1+E_2 and it must be constant. Then we use the assumption that, for this total ... 1 It is just what is needed to describe a pure chemical substance. For other systems you may have fewer or more than 3 degrees of freedom - yes, many dof may occur in practice, think of a complex mixture of chemicals such as in crude oil. The general case is usually discussed in the context of Gibbs' phase rule. 1 The simplest way to resolve this paradox is to require$$\rho_B(t=0)\simeq \tilde{\rho}_B=\frac{1}{Z}e^{-\beta H_B}.$$That is to say, you do not need any time evolution to reach thermal equilibrium, and this is the statement of ETH. This is very different from a classical chaotic system, where you need some time to explore the phase space and reach ... 1 A typical renormalization group flow can be thought of as a smooth vector field \vec V(\mu) defined on parameter space. Starting with parameters \vec\mu(\ell) at scale \ell, you obtain parameters at scale \ell' by solving the differential equation \frac{d\vec\mu}{d\ell}=\vec V(\vec\mu(\ell)). The function R referred to above can be thought of as ... 1 What you have written are all correct. But we should note that P(E) alone is not the partition function. The canonical partition function is$$ Z \propto \Omega_\mathrm{tot}(E_\mathrm{tot}),\qquad (1)  which counts the microstates of the univserse, i.e., system and bath. But $P(E)$ only counts the microstates of the bath! Let us see how to use the ...

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The electric fan increases the velocity and hence the kinetic energy of the molecules in the air. this would mean that the temperature has increased. I think that there is a bit of a problem here. Kinetic energy is not quite the same as thermal energy and temperature. Thermal energy and temperature is a measure of random, thermal motion of atoms or ...

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When normalized, $A$ is just equal to $1,$ so that $f(E)$ varies between $0<f(E)<1.$ Addendum for the edited question: The prefactor $\frac{2}{(2\pi\hbar)^3}$ crops up in the volume integration of density of states performed in k-space for the computation of number of states $N$ (i.e. all available energy states up to a certain maximum (fermi level) ...

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What books usually call "thermo-dynamics" is really "thermo-statics", the time variable is missing. There are all kinds of steady states and ever since Kirchhoff (~160 years) people have been trying to find some extremum principle to describe these. Kirchhoff showed that stationary electric current is distributed so that the dissipation is minimized and ...

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