# Tag Info

4

You're doing something wrong: the units of $h$ are energy*time, not energy/time. With that said, this ratio $k_B~T/h$ is the higher end of the frequency of the characteristic vibrations created by random thermal excitations. These vibrations could be phonons, for example, but also photons, and if you have electronic excitations (in chemical bonds for ...

3

The first thing to realize is that there are no "true" phase transitions (in the sense of non-analytic behaviour of thermodynamic potentials) in finite systems. This is the main difficulty one faces when analysing phase transitions using (most) computer simulation schemes. In particular, such simulations are only reliable as long as the observed correlation ...

3

Your claim that $p(n)$ is the $\lvert c_n \rvert^2$ from the decomposition $\lvert\psi\rangle = \sum_n c_n \lvert \psi_n \rangle$ is incorrect. In statistical quantum mechanics, we must differentiate between a pure state of the system $\lvert \psi \rangle$ and a mixed state, which is given by a collection of states $\lvert \psi_n\rangle$ and the probability ...

2

My personal favorite is "Mathematical Foundations Of Statistical Mechanics" by A. I. Khinchin (a mathematician) and G. Gamow. The content remains mathematically rigorous throughout, but nonetheless very readable. In chapter two, both the Liouville and Birkhoff theorems are derived, followed up by a long discussion on metric decomposability of phase space and ...

2

Only the symmetric stress tensor is physical, since thermodynamics demands a symmetric stress tensor. Since the symmetric stress tensor is unique, your option 2 is the correct one. (Canonical versions may be simpler but need not be physical; cf. the canonical momentum, which is often different from the physical momenetum.) This is completely unrelated to ...

2

The Liouville equation for the $N$ particle system, describes the time evolution of the phase space N-particle probability density, which you can also neatly rewrite with the Liouville operator: $f^{N}(t)= e^{-iLt}f^{N}(0).$ Now almost always we're interested in a smaller subset of only $n$ particles, for which then we have to define a reduced phase space ...

2

The Maximum Entropy principle, principally popularized by Jaynes, is known by most people having studied statistical physics. The way I see it, although Jaynes considered it as crucial in the foundations of equilibrium statistical mechanics and other people in the field still do (like Roger Balian for instance), it is more taught and thought of as a useful ...

2

Fundamentally it is that the $1/N!$ for the classical system only correctly compensates for overcounting of indistinguishable states if the particles are always in different states. For a system of Bosons at low temperature, where it is quite likely that many particles are in the same state, this breaks down. For a very understandable introduction to this ...

2

In the limit that $T\rightarrow\infty$, the partition function and the multiplicity of states are equal. Why? Well, we have that $Q=\sum_{i} e^{-E_i/kT}$, where $i$ indexes all possible microstates. If $T\rightarrow\infty$, these Boltzman factors all approach one, and we have $Q=\sum_i 1=\Omega$. You might think that in the limit $T\rightarrow\infty$ the ...

2

"Thermal energy" is a bit of a misnomer because "thermal" really refers to a method of energy transfer, not energy storage. When energy moves from one system to another, it can do so via a thermal process (e.g. conduction, convection, radiation) or a mechanical process (something pushes on something else). So technically, I wouldn't call $\frac{3}{2}kT$ the ...

2

Let's suppose you don't have this operator, but you only have the self-adjoint Hamiltonian part. This means you have the usual Schrödinger equation (or Liouville equation since it's for the density matrix) $$i \dot{\rho}=[H,\rho]$$ and the solution will be $\rho(t)=e^{iHt}\rho(0)e^{-iHt}$, hence the solution will evolve according to the (strongly ...

2

General form, properties A Lindblad form $$\dot \rho = -i[\eta, \rho] + A \rho A^\dagger - \frac 12 A^\dagger A \rho - \frac 12 \rho A^\dagger A$$ has three important properties: It is still linear dynamics, in terms of $\rho$. It is trace-free regardless of the trace of $\rho$. This means that the total sum of the eigenvalues, which starts out as 1, does ...

2

At the critical point, we have $\Psi(\vec r) = 0$ because that's the basic way in which this whole Landau theory works. First of all, it's important to realize that while making such statements, we consider the case of zero external field, so the term $Bh(\vec r) \Psi(\vec r)$ drops. Without this term, considering the constant configurations (in space) ...

2

A macrostate is characterized by certain definite values of macroscopic variables (ones that you measure with tools of human scale; often called thermodynamic variables). For a simple hydrostatic system you might choose to have the whole system in the liquid state with temperature between $T$ and $T + \mathrm{d}T$ and pressure between $P$ and $P + ... 1 Extending the other answers to see how "noise averages" are used elsewhere in the book as well, we can think of "noise average" of a dynamical variable${\bf A}({\bf x},t)$as$ \langle {\bf A}({\bf x},t) \rangle_{noise} = \int_\Omega {\bf A}({\bf x},t) \rho({\pmb \epsilon}) d{\pmb \epsilon}$where${\pmb \epsilon}$is a random noise distributed as ... 1 My absolute favorite book on the subject is the one that we used in our Gas Dynamics class: Introduction to Physical Gas Dynamics by Vincenti and Kruger. I had never had an introduction to statistical mechanics prior to this book and it does a great job developing the requirements as they are needed and providing motivation for the path it takes. I also ... 1 OK the first thing to notice is that in a gravitational field the pressure of a gas is not constant, but decreases with altitude. This means simply asking "what happens when we change the volume of the gas?" is not a well defined question; the amount of work done in the expansion is going to depend on exactly how we change the shape of the container. We ... 1 Classical electrodynamics and optics are enough to study the macroscopic properties of light. One needs to consider photons for special situations, as in spectra, or very low illumination, where quantum mechanics has to be used. The classical electromagnetic field, reflection and refraction coefficients are enough for what you describe. If you are ... 1 Consider a harmonic oscillator wherein $$a(t) = a e^{-i \omega t} \quad \text{and} \quad a^\dagger (t) = a^\dagger e^{i \omega t} \, .$$ The derivatives are $$\dot{a}(t) = -i \omega a(t) \quad \text{and} \quad \dot{a}^\dagger (t) = i \omega a ^\dagger (t) \, .$$ Consider the observable$X(t)$and$Y(t)$defined by$^{[1]}\begin{align} X(t) &\equiv ... 1 They look like they could be related. What is the relationship? From your two equations, we have $$k\ln \Omega = k \ln Q + kT\frac{\partial}{\partial T} \ln Q = k \ln Q + \frac{kT}{Q}\frac{\partial}{\partial T}Q$$ but $$Q = \sum_ie^{-\frac{E_i}{kT}}$$ and so $$k\ln \Omega = k \ln Q + \frac{kT}{Q}\frac{1}{kT^2}\sum_i E_ie^{-\frac{E_i}{kT}} = ... 1 The Fermi energy, \epsilon_F, is only equal to the chemical potential, \mu, when the Fermi gas is at zero temperature. The Fermi energy basically means, "chemical potential at zero temperature". At any other temperature you could find \mu via one of the standard thermodynamic relations (i.e. as the appropriate derivative of a free energy). 1 Issues with that derivation: You're missing the extra term \frac 52 k N, which may matter if you have to do any work with chemical potentials. Your students will not necessarily know why to parcel the space into volumes of size \lambda^3. Starting from the definition of entropy and deriving that the thermal volume \lambda^3 is important seems ... 1 As in a previous answer \frac{\hbar}{k_B T} is the coefficient in front of frequency in Planck’s Law (where I am using \omega = 2\pi \nu ).$$ I_\omega(\omega, T) = \frac{ 2 \hbar}{c^2} (\frac{\omega}{2\pi})^3(e^\frac{\hbar\omega}{k_B T} - 1)^{-1} $$c(\frac{\hbar}{k_B T})^{-1} is also the acceleration a that gives the Unruh temperature T.$$ ... 1 Hints:v(a)$is identified with a left multiplication operator$L_{v(a)}f:= v(a)f$.$L_{v(a)}^{\dagger}=L_{\overline{v(a)}}$;$ \left(\frac{\partial}{\partial a}\right)^{\dagger}=-\frac{\partial}{\partial a}$; and$(AB)^{\dagger}=B^{\dagger}A^{\dagger}$. 1 Short answer: The major difficulty lies with the definitions themselves, and none of the possibilities given has a real physical meaning which can be univoquely related to stress in non extensive systems in its conventional mechanical original meaning. The long one: What kind of systems does this apply to? This is not answered by referring to systems with ... 1 Liouville's theorem says the accessible volume in phase space does not increase, but it tends to become narrow filaments that "fill up" a much larger volume. If you think of a particle in a reflecting box, you might start it with a known position$\pm 1$mm in all three axes and a known velocity$\pm 1$mm/sec in all three axes. This is a phase space ... 1 Your logic is actually correct. The discordance between the conservation of phase-space volume according to the Liouville theorem and the Second Law is known as the Ergodic Problem. Heuristic explanations as the one provided by Ross Millikan, or course graining the dynamics for another example, do not hold under closer formal examination, since the math ... 1 It works just like every other kind of thermal energy. If a resistor can give out energy to the environment, it can also receive it. For example, if it gives it out by radiating, it can also absorb radiation; if it gives it out by having its fast-moving atoms smash into air molecules, then fast-moving air molecules can also smash into it. When it's in ... 1 I haven't given this enough thought yet, but at a first glance, I would say no, a potential having$\mu, V, E$as natural variables would not be a valid one. One possible attempt to obtain such a thermodynamic potential$Q$that is a natural function of$\mu,V,E$, would be a Legendre transform of the entropy$S(E,V,N).\$ We have: \begin{align} dS = ...

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