# Tag Info

Your simple transition model doesn't have any energies in it; $j$ (or $\epsilon j$) is acting as a spatial coordinate and not an energy coordinate. The net effect of this rule is to have the particles diffuse around the sites without any bias towards one or another. A common way to achieve detailed balance is through the Metropolis-Hastings algorithm, ...
(Full disclosure, I didn't RTFA, and I don't have time to.) Just to review, $f\left(x,v,t\right)dxdv$ is the number of particles with positions between $x$ and $x+dx$ and velocities between $v$ and $v+dv$. First, why is there no integral over position? In principle, there should be. However, assuming that $d$ is small, $f\left(x,v,t\right)\approx ... 2 I would say they are not entirely the same, but it depends on the context. First the definitions: the Wigner transform of an operator$\hat{A}$is defined as $$\tilde{W}\left[\hat{A}\right]=\int dz\left[e^{\mathbf{i}pz/\hbar}\left\langle x-z/2\right|\hat{A}\left|x+z/2\right\rangle \right]$$ and this is a strange function. You see that on the left, the ... 1 I have found that for older books, you want to check out Alibris.com, rather than Amazon. So a quick search there turned up a few copies of the 1972 edition. 2 Under a Wick rotation, which is what you do in order to go from Minkowski to Euclidean space, both the partial derivative$\partial_0$with respect to time and the zero-component of the gauge field transform as $$\partial_0\rightarrow i\partial_\tau$$ $$A_0\rightarrow iA_0.$$ This defines the covariant derivative for statistical field theory. 1 So lets start with the low energy configuration. To have the lowest energy all molecules must be lying in the x-y plane. Each one has 2 directions it could be lying in (x or y), and there are$N$particles, hence$2^N$possible configurations. (I always find picturing this with small number of molecules, say 2 or 3, for which counting the states is easy, ... 3 So my question is, "Why should the change in entropy be zero, even if the particles are distinguishable?" In statistical physics, entropy can be defined in many different ways. One possibility is to define it as log of the accessible phase space, given the macroscopic constraints (volume). Such entropy is not a homogeneous function of energy, volume ... 1 Addressing parts of the question out of order: ...the heat capacity would smoothly approach zero around the transition. I have never seen anyone refer to these types of transition... The heat capacity of all substances smoothly approaches zero at absolute zero.$S(E)$has a segment of zero first derivative no, the first derivative is a constant, ... 1 I am going to address the question as to why energy and information have time symmetric conservation properties whereas entropy does not. According to the Wikipedia entry on entropy - "The entropy of an isolated system never decreases, because isolated systems spontaneously evolve towards thermodynamic equilibrium, which is the state of maximum entropy." ... 0 This came to my mind after reading some introduction to maximum entropy probability distributions. Independence can be derived from the following four assumptions: (1) average momentum of particles inside the box is fixed at 0 (2) average kinetic energy of particles inside the box is fixed (3) the gas velocity distribution must be maximum entropy ... 1 I don't know if this answers your question. Have you seen how a cube of ice melts? Focus on one corner and you will see the melting happening on the edges. This is precisely the limit shape that you get from the domino tiling of a hexagon (which can be mapped to the dimer problem). This is called the Wulff shape of a crystal. See also the (theoretical) ... 1 I) If we expect$\Omega(E)$to depend analytically on the variable$\hbar\omega>0$extended to (parts of) the complex plane, then we may regularize by introducing an$i\epsilon\$ prescription, and substitute $$\tag{1} \hbar\omega ~\longrightarrow ~ \hbar\omega (1-i\epsilon).$$ The variable $$\tag{2} q~:=~ e^{-i\hbar\omega k}~\longrightarrow ~ ... 1 The derivation of the Fermi-Dirac distribution using the density matrix formalism proceeds as follows: The setup. We assume that the single-particle hamiltonian has a discrete spectrum, so the single-particle energy eigenstates are labeled by an index i which runs over some finite or countably infinite index set I. A basis for the Hilbert space of the ... 1 Here's an semi-formal explanation. Define$$f(N,V,T)=\frac{p(N,V,T)}{kT}.$$While f(N,V,T) is a function of N,T and V, the variables N and V are partially redundant, and only the ratio \rho=\frac{N}{V} is needed, since pressure is an intensive quantity. Thus we can write$$f(\rho,T)=\frac{p(\rho,T)}{kT}.$$Every smooth multivariate function ... 1 A definitive volume, one that I learned from during graduate school, is Kerson Huang's (of MIT, emeritus of the Physics Dept.) Statistical Mechanics. The book covers both classical and quantum computations of the partition function and observables from it, as well as thermodynamics, kinetic theory, transport, superfluids, critical phenomena, and the Ising ... 0 I suggest you: Statistical Mechanics: Theory and Molecular Simulation Mark E. Tuckerman There are all the necessary prerequisites and the discussion about Ising model and critical points. I don't know if there's online. 1 In Landau's theory, the order parameter M should make G(M) minimal.$$\frac{\partial G}{\partial M} = 2 B(T) M + 4 C(T) M^3=0\frac{\partial^2 G}{\partial M^2} = 2 B(T) + 12 C(T) M^2 > 0$$Hence, M=0 or M = \pm M_0 = \pm \sqrt{ - \frac{B(T)}{2 C(T)}} When T is higher than the critical point T_c, the groud state of system satisfies M=0. ... 0 The following calculation gives the correct answer:$$Z\int_0^{\pi/2}\int_0^\infty 2\pi v \sin\theta\; v\; \mathrm{d}\theta\mathrm{d}v\; e^{-mv^2/2kT}\; v \cos\theta\; \frac{1}{2}mv^2,$$where Z is such that$$Z\int_0^{\pi}\int_0^\infty 2\pi v \sin\theta\; v\; \mathrm{d}\theta\mathrm{d}v\; e^{-mv^2/2kT} = n,where n is the particle number density. The ... 1 Law of equipartition predicts the heat capacity of gases correctly. It assumes that inter-molecular attraction in gases is negligible (which is true). But for solids, inter-molecular attraction is not negligible, the, how come it still predicts the correct value for molar heat capacity? You have to know what "negligible" means in the context of the ... 0 Perfect vacuum can't be created. Even if you somehow get rid of all material particles, there still will be blackbody photons from the container, not to mention virtual gravitons. Generally, you can't be 100% sure that some part of space is perfect vacuum - to know that you should measure precisely the energy of that region, but it's forbidden by ... 1 Given an ideal piston/cylinder, starting with the piston completely inserted and zero volume, the work to make a perfect vacuum is simply: (distance the piston moves) X (force) = (distance) X (area of the piston) X (exterior pressure). So the work to make a vacuum of volume V, is V X P, where P is the exterior pressure, such as atmospheric pressure. ... 0 In quantum scale, particles are appearing and disappearing out of random everywhere all the time, meaning that if you are actually able to create a perfect vacuum at a macro scale, it would be instantly denied by the quantum scale. Also, in the quantum world, there's a small chance of random "teleportation" of any particle or atom from your container to any ... 2 The answer is no, or at least it is in the classical vacuum sense. I also don't see a rationale for why creating a vacuum would require infinite energy. An explicit construction is to use a solid-phase reactive chemical "getter" to eliminate (nearly) all gas molecules present; in experimental practice, virtually all man-made materials still outgas ... 1 Mustafa's answer gives one important reason for the logarithmic dependence: microstates multiply, whereas we'd like an extrinsic property of a system to be additive. So we simply need an isomorphism that turns multiplication into addition. The only continuous one is the "slide rule isomorphism" aka the logarithm. The base e is arbitrary as you can see from ... 2 Let me expand on the previous answer. We expand each term separately. We have, \begin{align} \log \left( 2 \cosh x \right) & = \log \left( e ^x + e ^{ - x } \right) \\ & = \log \left[ e ^{ x } ( 1 + e ^{ - 2 x } ) \right] \\ & = x + \log \left( 1 + e ^{ - 2 x } \right) \\ & \approx x + e^{-2x} \end{align} For the second term we have, ... 0 They probably mean to say "for large x \gg 1". The limit of low temperature is vague if they don't say how quantities are being fixed in the limit. Anyway, your approximations are wrong. For large x, \ln(2 \cosh x) goes like x, and \tanh x goes like 1 - 2 e^{-2x}. Try graphing it if you don't believe me. :) 2 Two obvious desirable features of this definition are: When you put two systems next to each other, considering them as one system, the total number of possible microstates \Omega_t is equal to the product of \Omegas of the two systems, \Omega_t=\Omega_1\times \Omega_2. But for this system the entropy is the sum of the entropies, indicating the ... 2 Entropy was first met in classical thermodynamics and was defined as , where Q comes from the first law of thermodynamics and T is the temperature, W work done by the system. Once it was established experimentally that matter at the micro level is discrete, i.e. is composed of molecules the statistical behavior of matter became the underlying ... 4 You will almost never encounter a calculation that is intended to account for every detail of a phenomenon with perfect accuracy. That isn't possible, and in fact many times adding more detail to a calculation only takes away from the insight it grants. Why make a complicated calculation when a simple one tells you everything you want to know? Gamow is ... 0 Yes, as you say, there is a built-in time symmetry in the mechanical laws that underlie our universe. At the moment the most accurate statement seems to be CPT symmetry. Under a CPT reversal (particles -> antiparticles, flip space, flip time), mechanics works identically. On a practical level though, even time symmetry alone holds to a good degree. It is of ... 0 Full reversibility at the elementary level does not imply what you suggest: new qualitative features appear as the scale of the system and its ability to interact with the rest of the world increase, so that dissipation (irreversibility, loss of "useful" energy) at the level of our everyday experience does not contraddict microscopic reversibility. If a ... 0 For the melting process You should use Q=mc\Delta T_1+mL_f + mc\Delta T_2, assuming that after the ice cube melts, there is still heat exchange between the warm water and the cold water (former ice cube) and \Delta T_2 is the temperature difference between the final temperature of the mixture and the melting temperature. For the total enthropy change I ... 3 The factorial factor 1/N! is exact but does not apply to all statistics. Consider a two level system and let us call \xi_i the grand-canonical partition function for the energy level i (i=0 or 1) and z=\mathrm e^{\beta\mu} the fugacity. Keep in mind that \xi_i is the partition function for a given energy level. Classical particles are ... 3 It is indeed an approximation, in particular an approximation that is usually good at high enough temperatures. For N distinguishable, non-interacting particles the partition function is Z(\mathrm{dist.}) = {Z_0}^N, where Z_0 is the single-particle partition function. If the N particles are all in different quantum states then there are N! ... 1 It is indeed an approximation. As suresh says, the approximation holds only when each particle is in a different single-particle state, or more precisely, when the number of states in which each particle is in a different single-particle state vastly outweighs the number of states in which more than one particle is in the same SP state. The single particle ... 3 In the figure above, consider the different configurations that are possible with 3 particles and 5 energy levels. Dividing by 3! gets the symmetry factor correct only for configurations of type 1 but is wrong for configurations of type 2 and 3. You can see this by explicitly writing out Z and comparing with z^3/3!. That is why the OP's statement that ... 0 Consider the fact that\rho_\beta :=Z^{-1}_\beta e^{-\beta H}$$is trace class by definition in a given (separable) Hilbert space \cal H, and thus it can be expanded as:$$\rho_\beta = \sum_{j\in \mathbb N} p_j(\beta) |\psi_j\rangle \langle \psi_j|\quad \mbox{where } 0 \leq p_j(\beta)\leq 1 \mbox{ and} \sum_j p_j(\beta)=1\quad (1)$$above I use the ... 3 Actually, it's exact. The flaw is "regarding the whole system consisting of these two particles, we can also write" Z = 1 + e^{- \beta E} + e^{-2 \beta E}. Assuming the two particles are distinguishable, we have$$Z=\sum_ig_ie^{-\beta E_i}=1+2e^{-\beta E}+e^{-2\beta E}=Z_0^2,$$with the 2e^{-\beta E} since the state of energy E is doubly-degenerate. ... 2 The expression$$ k_B \frac{\Omega}{\bar{\Omega}} $$equals$$ k_B\frac{1}{\bar{\Omega}}\frac{d\bar{\Omega}}{dE} $$which equals$$ \frac{dS}{dE}. $$In thermodynamics, where S is the Clausius entropy, this is equal to 1/T where T is the Kelvin temperature. In statistical physics, this expression can be taken as a definition of 1/T of a system from ... 0 If you are still around, Nathaniel, you may be interesting in this followup paper, and the summary in the same issue. In essence, the authors argue that to have a maximum energy, which is needed for negative-temperature population inversion, one must necessarily be in the microcanonical ensemble. But in this ensemble, the only definition of entropy that is ... 0 You have$$S=k_b\ln\left(\begin{array}{c} N\\ L_{y} \end{array}\right)$$which for N=400 looks like DiscretePlot[Log[Binomial[400, L]], {L, 0, 400}] 0 You know the starting position [(0,0) I'm guessing?] and the ending position (L_x, L_y). With just this, you could draw same possible arrangements: _ _f _ _ _f | | s_ _ _| s_ _| In the above example, (L_x,L_y)=(5,2). Both of the drawn possibilities have a total of five x's and two y's. There are many others. Are you ... 2 You should use U=q\epsilon_1. With the total number of particles N being constant, we have:$$\frac{\partial S}{\partial q}=\epsilon_1 \frac{\partial S}{\partial E}=\frac{1}{T}\epsilon_1\tag{1}$$As you said:$$\frac{\partial S}{\partial q}=k_B\ln(N/q - 1)=k_B\ln(N\epsilon_1/q\epsilon_1 - 1)=k_B\ln(N\epsilon_1/U - 1)\tag{2}\to ...