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## Hot answers tagged statistical-mechanics

10

Temperature is related to kinetic energy in the rest frame of the fluid/gas. In non-relatvistic kinetic theory the distribution function is $$f(p) \sim \exp\left(-\frac{(\vec{p}-m\vec{u})^2}{2mT}\right)$$ where $\vec{u}$ is the local fluid velocity. The velocity can be found by demanding that the mean momentum in the local rest frame is zero. Then $\vec{u}... 10 A system is in a heat bath of temperature T so we work with the canonical ensemble. We consider$N$degrees of freedom$x_1, x_2, ..., x_N$and$x$is the vector$(x_1~ x_2 ~ ... ~ x_N)^T$. The potential energy is quadratic so it can be expressed as a function of its second derivatives:$ U=\sum_{i,j} x_i ~H_{i,j} ~x_j = x^T H x ~~ $with$H_{i,j}=\frac{\...

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[June 19,2016: thoroughly revised, giving a more detailed, comparative presentation and better references] General case. In relativistic thermodynamics, inverse temperature $\beta^\mu$ is a vector field, namely the multipliers of the 4-momentum density in the exponent of the density operator specifying the system in terms of statistical mechanics, using the ...

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While the trace is invariant under a transform to another basis, you need to take into account here that the coherent state basis is not an orthogonal basis and it is overcomplete. We can evaluate the trace of an operator $A$ by inserting identity operators in front of and after the operator and then using resolution of identity in terms of the coherent ...

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This is essentially a result of the equipartition theorem where each degree of freedom contributes $k_B T/2$ to the energy. Given that the specific heat in this context is just ${\partial E}/{\partial T}$ then each degree of freedom contributes $k_B/2$ to the specific heat. For the classical model of lattice vibrations in solids this leads to the Dulong-...

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The Boltzmann brain paradox arises due to smaller fluctuations being more probable than larger ones. So, if you contemplate how our universe started out with low entropy initial conditions then it's difficult to explain this in terms of a generic high entropy state. Fluctuation yielding the early universe that in turn would have given rise to you, would be ...

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If you want to filter out the grains then certainly you could using normal filter papers in a filter funnel and repeat until the solution is clear of bits. You could also use a sintered glass filter. However, there will still be compounds from the coffee dissolved in the water and so a molecular sieve could be used and/or a chromatography column to separate ...

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$$S=k \ln [\Omega(E)] = k \ln [\omega (E) \delta E] = k \ln [\omega(E)] +k \ln (\delta E)$$ 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 $$\delta S = k \frac{\delta(e^{\beta ... 4 Your equation (2) is trivially a solution of (1), because v and T are constant. This is a disappointing answer, because it leaves unanswered the question what makes the Boltzmann distribution unique. The answer is that you only wrote down the collision-less Boltzmann equation, but in the real world collisions are always present (and indeed, systems ... 3 This combination represents the following. f in your reference is the number of particles per cubic space volume per cubic velocity volume. When you multiply it by some volume in velocity space, you obtain the regular density — number of particles per volume. But, as f itself depends on velocity, to obtain the total density of all particles with all ... 3 In the kinetic theory of gases, you only really define the temperature for molecules that are in constant, random, and rapid motion. So if you have a container with a gas at temperature T you don't change the internal energy of the gas by uniformly moving the container. Uniformly moving the container gives all the molecules a non-zero average motion, but ... 3 I am not sure if you're getting your concepts right !! In statistical mechanics, you don't study how the system evolves to a equilibrium state, you just take a system in equilibrium. In the Grand Canonical ensemble, the system is still in equilibrium although there is exchange of energy and particles. Lastly, the chemical potential and temperature are ... 3 The relations between thermodynamics and quantum field theories is treated in books and papers about non-equilibrium statistical mechanics. Here macroscopic many-particle systems are considered, and the focus is either on equilibrium, or on a dynamical description at finite times. For a readable introduction see, e.g., J. Berges, Introduction to ... 3 The hamiltonian of a perfect crystal can be approximated at low temperature as the sum of harmonic oscillator hamiltonians. In 1D we have$$H = \sum_{i=1}^N \frac{p_i^2}{2 m} + \frac 1 2 m \omega^2 \sum_{ij} ( r_i- r_j)^2 $$where the ij sum is over nearest neighbors. It is possible to verify that the eigenvalues of this hamiltonian are$$ E_n = \left( ...

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Entropy Demystified (The Second Law Reduced to Plain Common Sense) by Arieh Ben-Naim. Authored discussed not only the thermodynamics origin of entropy but also the same notion in the context of information theory developed by Claude Shannon.

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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, $\... 2 Yes, this can be done as an application of the transfer matrix'' method. You can, for example, find an explicit computation for the XY model (N=2) in the book The theory of magnetism II: Thermodynamics and Statistical Mechanics by Daniel Mattis (p. 75 ff.). Another reference, in which the general case is treated, is Section 9.2 in the book Elements of ... 2 I think it is wrong to define the temperature by the average energy of the molecule in all frames of reference. The reason for that is clear: take all of your particles and send them at$100 m/s$to the north. This won't make the gas hotter, just like the fan does not cool/heat the air (another great mystery!). The organized movement does not participate in ... 2 No, you don't need to work in the basis where the Hamiltonian is diagonal. It's a fact of linear algebra that the sum of the diagonal elements of a matrix is the same no matter what basis you're in, so you can easily evaluate the trace in whichever basis is convenient. 2 Leaving aside the issue of wavefunction collapse, physics is deterministic. So if you have some system like a gas and you know the exact positions and velocities of all the gas molecules you can predict the evolution of the system forwards and backwards in time. So you can start with a future state and work backwards to desciribe a past state. However ... 2 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 2 Consider a cube with side length$a$. Then its perimeter is proportional to$a$, its surface area$A$is proportional to$a^2$, and its volume$V$is proportional to$a^3$. It's clear that any reasonable 3D shape satisfies the same kind of scaling, which implies$A \propto V^{2/3}$. However, it's possible for fractals to have perimeter/area/volume that ... 2 Let's think of the problem this way...it's a bit rough but should give you an idea: We know that the energy levels of a particle in a 1D box of length$L$are given by the formula $$E_n = \frac{\pi^2 \hbar^2}{2 m L^2} n^2$$ where$n=1,2,3,\dots$. We also know that the average kinetic energy of a particle$\langle E\rangle$is related to the temperature by ... 2 I'd think of effusion effect, centrifugal techniques or other methods used for uranium enrichment. You could possibly select particles with the right mass (mass of the water) out of this mixture in a lot of iterations. It would be long and very expensive but probably possible. 2 A Boltzmann brain is not that different from what might be called Boltzmann cheese. Given enough time a set of atoms or particles might arrange themselves by statistical fluctuations into big wheel of cheese. If that happens there is no reason to think the cheese would then rapidly be demolished unless it formed in a star, or falling into a black hole or in ... 2 Coffee is a homogeneous solution hence it can not be separated by usual methods. I have few suggestions may not be very accurate but good for brainstorming. Distillation : Although you have mentioned not to mention it but I would like to add that coffee has several aromatic organic compounds that make the smell of coffee hence you can not get rid of ... 2 I will address the stat mech part. Over-counting refers only to indistinguishable particles, i.e. those with all the same properties. It also only comes into play when you are performing calculations by "labeling" or "tracking" indistinguishable particles. This is not the same as using the number of particles in each energy state. In other words, if you ... 2 The derivation by Pols is correct. Ryden makes the strange decision to plug the relativistic rest energy$\varepsilon = \rho c^2$into the classical ideal gas law. Surely it makes more sense to define a classical kinetic energy $$u = \frac{1}{2}\rho\langle v^2\rangle$$ so that $$P = \frac{2kTu}{\mu\langle v^2\rangle} = \frac{2}{3}u.$$ 1 Writing the partition function in terms of the density of states boils down to simply ordering our summations to count up all the states with the same energy first, and then summing over the different possible energies. This procedure can be done regardless of the coordinates being used to describe the states; there is no need to introduce an effective ... 1 I think you can avoid all these troubles if you define the temperature as proportional to the variance of velocity, i.e. $$E[(v-\overline{v})\cdot(v-\overline{v})]=E(v\cdot v)-\overline{v}\cdot\overline{v}$$ Here$E$means expected value,$v$ranges over the velocities of the individual particles, and$\overline{v}=E(v)\$. Clearly this is frame-...

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