# Tag Info

6

In some sense yes. The temperature is defined as an imaginary time in Matsubara Green's functions or some path integrals. Thus, a negative inverse imaginary temperature can be considered as a time. Here is a quotation from Alexander Altland, Ben Simons "Condensed Matter Field Theory": "Thus, real time dynamics and quantum statistical mechanics can be ...

4

It's true. Special equipment and a long time is required to mix helium and nitrogen. According to one study, a mixture of 2.7% He, 93.3% N at 800 p.s.i.g. required a special cradle to repeatedly upend the cylinder, and 20.5 hours to reach equilibrated gas, which then remained mixed: http://pubs.acs.org/doi/abs/10.1021/je60005a002. The helium repeatedly ...

4

wsc's answer (i.e., Onsager's computation of the free energy) provides one alternative road to a proof of a phase transition in the Ising model. It implies the existence of a phase transition in dimension 2 (for the nearest-neighbor model). Combined with correlation inequalities, this implies existence of a phase transition in any dimension d≥2, and ...

4

You can get a nice expression for the leading-order correction to the flat-manifold result via the use of Riemann normal coordinates. Basically, imagine expanding the metric in a power series at the point $x_0$: $$g_{\mu \nu}(x) = g_{\mu \nu} (x_0) + \partial_\rho g_{\mu \nu} (x^\rho - x_0^\rho) + \frac{1}{2} \partial_\rho \partial_\sigma g_{\mu \nu} ... 3 A microstate is just a particular microscopic configuration of the system, where the state of each particle is fixed. For example, take a three-level system with four particles. Treating the particles as indistinguishable, one particular microstate corresponds to two particles in the lowest state, which has energy -\epsilon, one particle in the second ... 3 The Minkowski vacuum is defined on the whole Minkowski space. The Rindler wedge covers only half of a given spacelike surface, the other half being covered by a different Rindler wegde. So that the \tilde{\phi} field defined gives only half od the degrees of freedom corresponding to \phi%. Ehen you trace over the the DOF from the other Rindler wedge the ... 2 There's no issue with the energy having an angular dependency. This is similar to the case of a spin in a magnetic field, in which the energy is$$E = -\mathbf{\mu \bullet B}$$or$$E (\theta) = -\mu B cos(\theta)$$This poses no problem. As you say, the Boltzmann factor is e^{\mu B cos(\theta)/kT}, and the partition function is found by integrating ... 2 Question: The above analysis seems to treat the oscillators as distinguishable. But aren't they indistinguishable? In Einstein's model, the oscillators are supposed to sit at (oscillate around) definite place in space. So you could say they are distinguishable. For example, by their cartesian coordinates with respect to lab frame. 1)That the Gibbs ... 2 A few issues with this argument. First of all, you're clearly right that the RHS can yield an irrational/transcendental number. It would require insane mathematical coincidence for this not to be so. (It's really hard to have e^x be algebraic, for instance.) And there's nothing in the formula about how many particles there are; we could imagine N =2, ... 2 To illustrate the difference take two blocks in contact with each other, with block A at a higher temperature than block B initially. As soon as we bring A in contact with B, energy will flow from A to B, the temperature of A will decrease, and the temperature of B will increase. Given enough time, the two blocks will reach the same temperature, and the ... 2 In general, the density of states can be computed as follows: Find eigenstates of the Hamiltonian, \psi_{s}, so that H\psi_s=E_s\psi_s (except in special cases, this is usually the hardest part). Compute N(\epsilon), defined as the number of states \psi_s with energy E_s<\epsilon. D(\epsilon)=\frac{dN}{d\epsilon} is the density of states. ... 2 This is simply about words. A process can cause a change. For example: A (reversible) adiabatic process can cause a (reversible) temperature change. 1 Based on a quick read of the Wikipedia article on the kinetic theory of gases, it looks like you would need statistical mechanics to derive any of the results in the kinetic theory of gases. For example, the Maxwell distribution of velocities is typically derived using the canonical ensemble. However, the equilibrium velocity distribution is not a dynamical ... 1 Statistical Equilibrium: That state of a closed statistical system in which the average values of all the physical quantities characterizing the state are independent of time. In other words when systems do not evolve in time i.e. when they are in steady state. In thermodynamic equilibrium there are no net macroscopic flows of matter or of energy, either ... 1 You may have seen the reasoning to follow in most textbooks already but apparently it is not emphasized enough so I will say it again here. The crucial starting point is the second law of thermodynamics that claims that the entropy change of the universe \Delta S_{univ} is either zero or strictly positive for any physical change that occurs in it. I ... 1 In principle one can always write down the differential equations for a system of N particles and attempt to solve them: as you pointed out, there is no general solution if usually N>2. As such, nevertheless the need to describe features of general systems remains. The key point here is understanding that, as a matter of fact, whenever dealing with ... 1 In Hamiltonian or Lagrangian Mechanics, we can exactly predict the trajectory of the system. The general conclusion is that for Hamiltonian systems we don't have a sufficient number of analytical constants of motion in order to write the solutions down exactly. The systems are generally non-integrable. We use the tools of statistical mechanics to pay ... 1 Yes, you can find a relation between E_F and other functions. However remember that this value is a parameter, not a function. The average number of particles in state k in Fermi-Dirac statistic is$$<n_k> = \frac{1}{e^{\beta(E-\mu)}+1}$$With \beta = 1/kT. At a very low temperatures, you find that <n_k> can only have two values: 0, ... 1 The d=2 (square lattice) Ising model has a special "duality" property (the high-temperature and low-temperature partition functions can be mapped on to one another) discovered by Kramers & Wannier in 1941. This doesn't rigorously prove that a phase transition exists, but it remarkably predicts the critical point where a phase transition, if it did ... 1 We can use statistics by being willing to ask different questions. No individual particle has a pressure or a temperature. But we can ask about subsystems with particular pressures or temperatures. So we group collections into subsystems that are large enough and regular enough to have collective properties like pressure and temperature. Then we can use ... 1 I am not too familiar with KT transitions yet, but I would like to learn about them myself. I have read in the notes of Prof. Jensen (available online http://www.mit.edu/~levitov/8.334/notes/XYnotes1.pdf) in the end of chapter 4.2 that the divergence in the specific heat is so fast that it is experimentally not observable. Analytically (according to his ... 1 \mu =\frac{\partial G}{\partial N}  i.e. it is free energy per particle in an ensemble, or energy needed for a particle to add it to the system so that the system will stay in equilibrium. (Some call it Gibbs free energy/enthalpy. Free energy is usually the name for helmoltz free energy). Formal definition -$$ G(T,P,N)=E(T,P,N)+P\cdot V(T,P,N)-T\cdot ...

1

Irreversible free expansion of a gas in adiabatic condition is not isentropic. There is an increase of entropy. The equation of dS=dQ/T is not an accurate equation, the actual equation should be dS=dQ(reversible)/T. For dQ of irreversible the equation should be changed into the clausius inequality form which is dS> dQ (irreversible)/T. Under this clausius ...

1

The quantum treatment of this system yields a non zero magnetic moment (although it vanishes at infinite temperatures) in the limit where $k_B T \gg \hbar \omega_c$ while the classical treatment gives strictly zero. I do not understand how does the left-right symmetry argument used in the classical partition function disappear in the quantum ...

1

It sounds like you're looking for an analog to the density of states. The density of states $D\left(E\right)$ tells you about the number of states in a given energy range $\left[E,E+dE\right]$. So the probability of ending up in a state with energy $E$ is $P\left(E\right) \propto e^{E/k_bT} D\left(E\right)$. Here, you would have an angular density of states ...

1

Here is a sketch of a derivation of the Kubo formula: Write the full Hamiltonian $H=H_0+H'e^{-\eta t}$ where $H'$ is an external (time-independent) perturbation and $\eta\rightarrow 0$. We would like to evaluate the expectation value of an operator $A(t)$: $\langle A(t)\rangle=\mathrm{Tr}[\rho(t)A(t)]$. So we need to first determine $\rho(t)$ using ...

1

I am not particularly familiar with the primon gas you are linking to, but similar ideas have been tossed around for a long time; see, for example this page for many references (including the topic you mention). The first two topics (quantum mechanics and statistical mechanics) are particularly relevant to your question; I'll concentrate on the second one, ...

1

Depending on how "real" you want the system to be. In a Casimir plate setup in a way the vacuum gets a negative pressure. But for any system of real particles a negative pressure means, that the system will be unstable and collapse, so you cannot have negative pressure in equilibrium. For example, negative pressure can formally occur in the van der Waals ...

1

The $kT$ comes from the Bose-Einstein statistics. The photons are governed by this statistic, nothing suspicious is here. The power three appears when we go form the variable $E$ to the variable $E/kT$ in the integral.

1

In many situations in statistical mechanics, the configuration space that you sum over in your partition function is coarse-grained in a way that certain microscopic degrees of freedom are ignored. The weight of each coarse-grained configuration $c'$ in the partition function includes the sum of all weights of microscopic configurations $c$ that get mapped ...

Only top voted, non community-wiki answers of a minimum length are eligible