Tag Info

New answers tagged

0

Dr. Robitaille says that blackbody radiation is not universal, even inside cavities where the surfaces are all at thermal equilibrium. That is highly controversial since the electromagnetic fields in a cavity are usually considered as an additional substance, called a "photon gas" which is also at thermal equilibrium and hence has a temperature. This ...


0

The thermal conductivity of steel is significantly less than that of aluminum. Heat generated by the friction of asbestos on the surfaces of both discs is carried away much faster by aluminum than by steel. Therefore the steel disc in the viscinity of contact with the pin, as well as the asbestos pin, will retain heat and will measure a higher temperature ...


0

Sometimes, when you have a soft material like aluminum, and a brittle material like asbestos, in a pin/disk configuration, particles of asbestos break off and become embedded in the aluminum. And now you have created a very abrasive disk (a bit like diamond particles in phosphor bronze), and you will wear the pin much more rapidly while apparently protecting ...


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 ...


0

In the example you gave above and in calculating certain thermodynamic variables such as pressure: $P=-\frac{dF}{dV}=\frac{d}{dV}k_bTln(Z)$ the $N!$ cancels: $\frac{d}{dx}ln(Z'/N!)=\frac{d}{dx}(ln(Z')-ln(N!))=\frac{d}{dx}ln(Z')$ you still need to included the N! when you do calculations for chemical potential or entropy because the N! doesn't cancel. What ...


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 = ...


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 ...


0

A very good book along the lines you seem to want is Gallavotti's Statistical Mechanics - A Short Treatise, which can be downloaded from here. He covers many of the classical topics, with a detailed discussion of foundational issues, the role of ergodicity/mixing, etc. From a very different point of view, with a colleague, we are writing a mathematically ...


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 ...


0

resistor is an electric component which is used for the purpose to restrict the electron flow. if you run with 10Km/hour speed now i will stop you is it possible to stop you suddenly. No it is not possible. so that resistor dissipate some energy in the form of heat.


0

In non-eq stat mech the comment above isn't what they mean. We have to establish 3 separate things: Balance This is the condition that all transitions into and out of a state 'balance' ie $$ \sum_x p(x)k(x,y)=p(y) \quad \forall y$$ This ensures a stationary state, but nothing else Detailed balance This is the condition that all transitions ...


0

It "runs out" when it reaches thermal equilibrium with it's environment. In that case, as @Kevin points out, the dissipation equals the absorption from the environment. If you pass a current through the resistor, it will heat up and that will drive extra current/voltage noise in the resistor. FYI: Resistive cooling of ions in a Penning trap uses a cooled ...


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

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

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

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 ...


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 ...


-1

No because gravity is missing. Any macroscopic or microscopic explanation has to include gravity. It sounds trivial but in a world without gravity a hot gas would not rise (if one still can define "rise" in that case). The effect of gravity on each gas molecule is small but its collective effect has to be taken into account.


0

There are 3 directions of vibration: namely the components $x$, $y$ and $z$ and each of these directions has 2 degrees of freedom; one of potential energy and one of kinetic energy. So in total there are $3\times 2 = 6$ degrees of freedom.


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 ...


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}} = ...


0

the fermi energy is such that all states with a energy same or lower are occupied at zero temperature. You have gaps in the spectrum leading to a wider range of "possible" fermi levels. Take the first case (assuming e_3>e_2=e_1): at zero temperature the states e_1 and e_2 are occupied and e_3 is not. Meaning e_f must not be smaller than e_1,e_2 and must be ...


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 ...


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 ...


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 $(AB)^{\dagger}=B^{\dagger}A^{\dagger}$.


0

The quantity $\hbar/k_BT$ comes up in studies of strongly correlated materials. This story is a little complicated but pretty interesting. Empirically, it has been found (1) that many strongly correlated materials have a resistance proportional to temperature, and if one works out the scattering timescale associated with this using something like the Drude ...


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

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 ...


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 ...


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).


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 ...


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 ...


0

I posted the following solution on this board wanting to get opinions of the validity of a solution using only the microcanonical ensemble: Simpler derivation of sarkur-tetrode equation I still haven't received any comments, but if you have any, feel free to chime in. The Sarkur-Tetrode equation is the following without the 5/2 constant term: $$ kn \ln ...


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 ...


0

As already mentioned by others, a macrostate consists of macroscopic observables of a system. The system that I am referring to is normal quite large and may contain a huge number of microscopic constituents, whose exact states we don't really care about when talking about macrostates. For microstates, then we have to specify the states for all of the ...


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 + ...


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) ...


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

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 ...


0

One detail First thing I would like to note is that the operator you are talking about is called the Lindblad superoperator. A superoperator is like an operator that acts on other linear operators (in this case, the density matrix). Lindblad Equation What you have written is known as the Lindblad Equation. The Lindblad Equation is one example of the many ...


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

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 ...


0

I only want to add a few details in the spirit of making the above answer of R. Garcia-Garcia more complete, and also clarifying one detail in the question. Using the questioner's notation, in fact what we've got is the rule $\Psi \rightarrow 3\langle\Psi^2\rangle\Psi - 2\langle\Psi\rangle^3$. By pure and simple algebraic manipulation, this rule can be ...


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 ...


-1

I suggest using the general drag relation: $$F_d=0.5C_d\rho_lv^2D$$ where $F_d$ is the drag force per unit length, $D$ is the diameter of the cylinder and $v$ is the relative velocity norm between fluid and cylinder. The drag coefficient depends on the regime, for $Re<0.1$, $C_d\propto Re^-1$, for $Re>>1$, $C_d\approx1$



Top 50 recent answers are included