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From here, how do we say that probability distribution function is constant as we flow in the phase-space? More accurately, value of probability distribution function $f_t(p,q)$ at representative point $p^*(t),q^*(t)$ moving along any Hamiltonian trajectory in phase space is constant in time. The function itself generally changes in time. This value is ...

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The uses of this two theories are completely different. Statistical Mechanics is used to see how by modelling the behavior of microscopic constituents you can predict the macroscopic phenomenas that you observe. On the other hand Many Body Theory uses first principle techniques to see what happens microscopically when you have large no of particles in your ...

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After a long discussion with "curiousone" I would like to like to share the relevant points of our discussion (hopefully I will do them justice) and some extra bits I added after thinking it over First Law of thermodynamics While the equation $$TdS = k_b T \ln N dN + dU -PdV$$ is quite general to any system where particle number is not conserved. We ...

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A CFT is still a QFT, and the way to put it at finite temperature is standard for any quantum system - you take your Hamiltonian $H$ and compute $Z=\mathrm{tr}\,e^{-\beta H}$, where the trace is over the Hilbert space of states living on $\mathbb{R}^{d-1}$ if your CFT is in $d$ dimensions. The thermal correlators are computed in a similar way, ...

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While a funny-looking coincidence, this is not a valid alternative expression for entropy in general, since the entropy of a probability distribution (which are what rigorously hides behind the strange word "macrostate") is more generally given by $$S = - k_B \sum_i p_i\ln(p_i) \tag{1}$$ and becomes only $$S = k_B \ln(\Omega) \tag{2}$$ in the case of a ...

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Let us start with an example, called Langevin paramagnetism, where the magnetic momentum is described classically, as a vector in three dimensions. Calling $\vec\mu$ this momentum, $\vec B$ the magnetic induction and $\theta$ the angle between $\vec\mu$ and $\vec B$. The probability density of the angle $\theta$ is $\rho(\theta)=\frac{1}{\mathcal ... 0 The main reason sum of countable set instead of integral over continuous volume is introduced in statistical physics in the context of classical physics is simplification of mathematics needed to make use of the probability theory to explain basics of information theory and derive the very notion of information entropy, given by the$p\ln p$formula. Or use ... 1 Most things along those lines are just simply the central limit theorem / law of large numbers. For example, when you compress a gas in an insulated container using a piston, its temperature goes up. Why? Because the moving piston accelerates gas molecules that bounce off it. And why does the temperature always go up by the same amount? Because there are ... 2 The whole point of statistical physics is that we don't really care about what the microstate could be, we just know that there's a great collection of similar ones that give rise to the same macrostate. I believe what you're questionning about is closely related to the ergodic hypothesis 1 From a mathematical perspective this means that it is not differentiable. The problem is that you need the discreteness to be able to count states. If you replace the discreteness by something smooth you get something differentiable, but your definition of entropy no longer makes sense. This is just one of the points where mathematicians cringe, but it works ... 0 Your question does not make a lot of sense, which is due to the words "microstate" and "macrostate" floating around without being given a precise meaning. A microstate is a point in the phase space of the system - it refers to a single, unambiguous configuration of the system. For a system of$N$freely moving particles (an ideal gas) the phase space is ... 1 In ideal gas model, temperature is the measure of average kinetic energy of the gas molecules. In the kinetic theory of gases random motion is assumed before deriving anything. If by some means the gas particles are accelerated to a very high speed in one direction, KE certainly increased, can we say the gas becomes hotter? Do we need to ... 3 I would highly recommend the two seminal papers by E. T. Jaynes, http://journals.aps.org/pr/abstract/10.1103/PhysRev.106.620 and, http://journals.aps.org/pr/abstract/10.1103/PhysRev.108.171 Also check out the book by E. T. Jaynes, which has a focus on the foundations in probability but is rather light on applications in physics: ... 2 It says in the notes [...]the factor of 8 arises because we are only counting the states in the quadrangle with positive$n_x$,$n_y$and$n_z$. The eigenfunctions of a particle in a box are $$\Psi(x) \propto\prod_{j\in{x,y,z}} \sin\left(\frac{n_j}{L_j}x_j\right)$$ with$n_j\in\mathbb{N}$. Choosing$n_j$a negative integer does not yield a different ... 1 Yes, the Z partition function acts as a normalizing factor to compute the probability that the system of the N particles has an energy$\epsilon_i$. It is obviously more complex in the case of many particles system since a lot of configurations may lead to that energy level. For example, consider a 2 particles system with 0 and$\epsilon$as the allowable ... 1 A macrostate is a set of microstates. Some microstates are thermal, others are not. Without the assumption of being in thermal equilibrium you can't assume anything since any possible microstate is possible. And lots of possibilities macrostates could be picked. Usually you want to group your macrostates according to a state variables such as pressure, ... 0 In statistical mechanics, a microstate is a specific microscopic configuration of a thermodynamic system that the system may occupy with a certain probability in the course of its thermal fluctuations. In contrast, the macrostate of a system refers to its macroscopic properties, such as its temperature, pressure, volume and density. Treatments on ... 0 In mechanics,$E = \int F.dr$. That is, energy is given by a force integrated over a displacement. You can find analogous quantities in thermodynamics. You find a generalized "force" that with a generalized "displacement" can be used to find an energy. The analogy runs deeper in that if the "force" becomes unbalanced, a "displacement" will occur and the ... 2 If operators$a_j$,$a^\dagger_j$correspond to the system's orthonormal natural orbitals$\phi_j({\bf x})$, such that $$\int{d{\bf x}\; \phi^*_j({\bf x})\phi_k({\bf x})} = \delta_{jk}, \;\;\;\sum_j{\phi^*_j({\bf x})\phi_j({\bf x'})} = \delta({\bf x} - {\bf x'})\\ \hat\psi({\bf x}) = \sum_j{\phi_j({\bf x})\;a_j},\;\;\; a_j = \int{d{\bf x} \;\phi^*_j({\bf ... 2 For both hydrogen and chlorine E = 3/2 kT is only true at very low temperatures since they are diatomic gases. In general you get a contribution of 1/2 kT to the energy for every quadratic degree of freedom. For a monatomic gas that is 3 translational degrees of freedom, hence 3*1/2 kT. For a diatomic gas there are in addition 2 rotational, 1 vibrational ... 2 The weight for a specific state is w(\phi)=e^{-E/T+F/T}\equiv e^{-I(\phi)+F/T}. Due to the definition of the entropy S, we have S=-\int D\phi w\ln w=\frac{1}{T}(\int D\phi wE-F\int D\phi w)=\frac{1}{T}(U-F). Then we have F=U-TS. 0 The Hamiltonians H_S and H_R both implicitly depend on their respective volumes (or confining potential strength). To allow volume exchange between the two systems, you simply impose the constraint V_R = V_{tot}-V_S. The joint Hamiltonian is always given by H_S+H_R. You can check that in mechanical equilibrium, \partial_{V_S} ... 0 Disclaimer The following answer is largely taken from the arXiv paper with e-print number 1512.04015 by Treumann and Baumjohann (from here on I abbreviate references to this paper as TB15). Background It is well known that the Maxwell-Jüttner distribution works well for a momentum/energy distribution with an isotropic, scalar temperature, T. This ... 3 First, let me explain what the notation means. The sum over \{n_\ell\} is a sum over all possible values of n_\ell, for each possible values of \ell (in other words, \{n_\ell\} specifies the values of n_1,n_2,n_3,\ldots). Then, once these values are fixed, you take the product of all the functions f_\ell(n_\ell) = \frac{1}{n_\ell!}\bigl( ... 1 To start, it helps to first visualize the original counting problem and then come up with a physical system that corresponds to the picture you have in mind. There are a few nifty ways to visualize partitions of n. Recall that a partition of n is a sequence of integers \lambda_1\geq\lambda_2\geq\dots\geq\lambda_k such that \sum_j\lambda_j=n. I assume ... 0 As many comments say, there is not a single and best answer, each one uses a different method. The solution that you found is a good one, but how do you define when the equilibrium has been reached? In order to do that you need check the last values of the simulation (Energy, pressure, etc.), so you choose a set of previous configurations that you'll check: ... 2 In statistical mechanics and field theory, the second type is referred to as a "connected" correlation function. You sometimes see the notation$$ g_{\text{c}}(\mathbf{x}-\mathbf{x'},t-t') = \langle s_1(\mathbf{x},t) s_2(\mathbf{x}',t') \rangle_{\text{c}}\,\text{,} $$where \langle\ldots\rangle_{\text{c}} indicates that the product of the averages should ... 2 The second version is the first one but with the observables shifted by their respective averages:$$\langle \tilde A\tilde B\rangle = \langle (A-\langle A\rangle)(B-\langle B\rangle)\rangle = \langle AB\rangle - \langle A\rangle\langle B\rangle$$Oftentimes in e.g. numerical studies, it is easier to just sample the average product of the original ... 2 The general Pearson correlation between two variables is defined as$$ \textrm{cor}(X,Y) = E[XY] - E[X]E[Y] $$up to a denominator containing the standard deviations of the distributions of the two variables. In some field theories the expectation values of the variable itself (one-point function) vanishes, therefore oftentimes the above definition reduces ... 1 A simple way to see this is by considering the fact that the probability to transition from a particular initial state to a particular final state is the same as for the inverse process where one considers the transition from the final to the initial state. This is because the square of the modulus of the matrix element is the same for both cases. This ... 1 I think you have the wrong idea when you ask how specific heat is "defined". In computational physics, the starting point is an experimental measurement that one could measure, or at least, a physical quantity that one might care about ... and then the question is, "how do I compute it?" The wrong approach is to have in mind a certain formula. You should ... 2 You can think of the chemical potential as the amount of free energy needed to add one additional particle to the system. Because the ground state of a BEC is degenerate and can hold an infinite number of particles, there's no energy cost to add another particle to that state. So, \mu = 0. 2 Let us use the definition \langle\eta(s)\eta(t)\rangle=\Gamma\gamma^2\delta(t-s). First of all, C(s,t) depends on t because$$C(s,t)=\Gamma\min(s,t).$$It is clear, from causality, that \frac{\delta x(t)}{\delta \eta(s)}=0 if t<s. If t>s, compute the difference \delta x(t) caused by two realisations of noise that differ only at time s ... 4 To determine the upper limit on chemical potential for a gas of \mathcal N bosons, look at the form of the Bose distribution in the grand canonical ensemble with \langle N \rangle = \mathcal N. When using the GCE, it's easiest to work at chemical potential \mu and to then choose \mu(\mathcal N) so that \langle N\rangle(\mu)=\mathcal N. Each state ... 2 Very interesting question. As you wrote yourself in your Edit it is hard to describe water via the ideal gas model. You have to introduce at least two important improvements of your ideal gas: Dipole - Dipole - Interaction instead of no interaction. Let's call this pair potential V_d and note that for two given molecules V_d is not only distance but ... -1 Warmer water doesn't sink at any temperatur unless it is an iceberg and the coldest water floats. Vibrating atoms tend to move to slower moving atoms evening out. Vortexes could from the temperature change in the pot or in the ocean only then may you see warmer water lower than cold water in a continuous flow. 0 I think I've actually got this after some work. According to (I.4)$$F_{AC}(A_1, A_2, ..., C_1, C_2, ...)-F_{BC}(B_1, B_2, ..., C_1, C_2, ...)=0.$$For this to be equal to (I.5) the above needs to be true no matter what the C coordinates are. Therefore we ought to be able to decompose the above functions as$$F_{AC}(A_1, A_2, ..., C_1, C_2, ... 0 These responses do not really answer the question. Planck's distribution was explicitly provided by Planck to represent the distribution of energy for what are now referred to as a boson. The proposition of Bose, 24 years later, was a quite terse reiteration of the same geometric expansion, and to recommend it as a more general principle for thermal ... 1 Regardless of the system, Cv will be proportional to the variance of energy. If you have peaks at higher energies, that will increase its value. But at high enough energies the occupation of those states will be so low they won't significantly affect the variance. In this case the variance of the distribution isn't just the width² of one of the peaks, you ... 1 The reason why heat cannot flow from cold to warm is that the change in entropy will become negative, and that doesn't happen in a closed system. Negative entropy is by definition not possible. Here's why: Alternative example: gas in a box I think entropy gets a little more intuitive if we think of it in terms of statistical mechanics. If we imagine a box ... 2 The 2nd Law of Thermodynamics is based on an overwhelmingly extensive body of empirical evidence on how thermodynamic systems behave. There are many different statements of the 2nd Law, and all of them are equivalent to one another. Once one of these has been specified, all the other follow. One such statement says that heat cannot flow spontaneously from ... 1 When two balls hit each other elastically, energy and momentum are conserved - but they don't usually leave the collision with the same velocity. Each air molecule is like a ball, and at every collision there is a possible distribution of velocities. There is a very small probability that a molecule will be either much faster than all the others, or much ... 2 Umm...OK well lets see what happens. Lets let$s = \beta\varepsilon$, where$\varepsilon$is some fixed energy and $$Z\left(\frac{s}{\varepsilon}\right) = \zeta(s)$$. To get some kind of idea for what kind of system$Z$describes we need to find the energy levels of the system and to do that we need to express$Z$in the form$\sum_{i} e^{-\beta E_i}$. ... 0 I am sure there are more complete or simpler books on the topic, but I found very enlightening to read the first chapter of Puri's book on Kinetics of Phase Transitions. In particular, I think this is a book written by a researcher which works with phase transitions of mixtures, which can be particularly relevant for those studying non-equilibrium physics in ... 0 The general principle is that macroscopic variables and macrostates are not "real" from the microscopic, Hamiltonian perspective. They're things that we, human beings on the scale of$10^{23}$atoms, make up based on what we can observe. For example, let's take pressure. Given a microstate$\Gamma\$, you can't calculate the pressure, because such a thing ...

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In principle you could, but people don't do this very often for MD simulations of inhomogeneous systems. The transport coefficients (diffusion coefficient, heat conductivity and etc.) are tensors actually. When these tensor acts on the corresponding gradients (temperature, concentration gradients and etc.), we obtained the corresponding flux. In homogeneous ...

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Four of Boltzmann's prominent papers are translated into English on pages Ludwig Boltzmann: Further Studies on the Thermal Equilibrium of Gas Molecules (from Sitzungsberichte der kaiserlichen Akademie der Wissenschaften, Vienna, 1872) & Ludwig Boltzmann: On the Relation of a General Mechanical Theorem to the Second Law of Thermodynamics (from ...

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The thermal conductivity contains the cross-section for particle collisions. The cross-section itself contains the mass of the particles that are carrying the thermal energy (as well as the mass of the "thing" the particle is colliding with). Sometimes one of these masses can be ignored. The exact dependence of the cross-section on the mass of the ...

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