Suppose a system has two parts, $A$ & $B$. They were initially at different temperatures and hadn't achieved mechanical equilibrium. After attainting thermodynamic equilibrium, do they continue to exchange heat energy? ie. after attaining thermal equilibrium, ie. Is the heat that $A$ gains from $B$ the same as the heat gained by $B$ from $A$ so that the temperature remains constant ? Meaning to say, is thermodynamic equilibrium dynamic like chemical equilibrium? Or is it static? What is the reason for any of the two?
Thermodynamics – Is Thermodynamic Equilibrium Static or Dynamic?
equilibriumthermodynamics
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First of all I will address your last concern, which translates into: equilibrium doesn't necessarily mean that nothing is moving. As an example, particle in a solution at equilibrium can move from one side to the other as long as almost the same number of particle move the opposite direction (this usually happens, say, because of thermal agitation/Brownian motion...). What doesn't change with time is then the average concentration of particles in the solution. Now using again this analogy, suppose you have an isolated system made of, say, sea water and distilled water. When you first mix them you can still distinguish them, as one has a higher concentration of salts. After a while, internal gradients, that arise because the system is out of equilibrium, will drive the salts to homogenise and the end result is water with a homogeneous concentration of salts, after a suitably long relaxation time. Same happens with temperature: if you replace sea and distilled water with hot and cold water, the same mechanism will homogenize your system. Temperature gradients will mix water around until the temperature is the same everywhere. Then gradients disappears and dynamical equilibrium settles in (water is not completely static, but molecules flow around because of thermal agitation).
I will provide an answer from an astrophysics point of view, in which the term local thermodynamic equilibrium (LTE) is often used. In astrophysics the distinction between 'thermal equilibrium' and 'thermodynamic equilibrium' is not carefully made, because there is rarely if ever a situation in this context in which thermal equilibrium might hold without thermodynamic equilibrium.
The most common situation in which the presence or absence of LTE is considered is for a star. There is a flow of heat from the interior of a star to its atmosphere at large radii, and the temperature varies as a function of radius. So clearly gas near the atmosphere is not in thermodynamic (or thermal) equilibrium with gas in the core.
However, models of stellar interiors can be vastly simplified if one recognizes that there is still a local thermodynamic (thermal) equilibrium in the sense that the kinetic distribution of the free electrons, the plasma ionization state, and the radiation field at each radius can all be very well described by a single number, a local temperature $T$. The electron velocities obey a Maxwell-Boltzmann distribution, the ionization state follows from the Saha equation, and the radiation field is described by a Planck function (blackbody), all evaluated for some common $T$.
As the link you provide explains, this works as long as the mean free path of any particles that might transport heat (e.g. photons, electrons) is very small compared to the length scale over which the temperature is changing. In the atmospheres of stars, where the photon mean free path grows large, LTE in the above sense can break down. However, if the electron mean free path is still small enough, it can helpful to apply an even more limited notion of LTE in which one takes the electron velocities to be in LTE, while acknowledging that the radiation field may depart from a Planck function.
To summarize, LTE is a useful concept to the extent that it simplifies the description of a large system as a collection of local regions each described by a single temperature. The distinction between thermal and thermodynamic equilibrium is rarely needed in the situations in which LTE is a helpful approximation.
Best Answer
It is a dynamic equilibrium. To see this notice that the Boltzmann distribution law is statistical in nature, for instance, the molecules that have energies between $E_1$ and $E_1+\Delta E_1$ at time $t_1$ are not necessarily the same than the ones in between those same energies at time $t_2$. Particles collide (and thus exchange kinetic energy) all the time, and while the distribution stays the same, the particles that contribute to different parts of the distribution keep changing.
Update: It is dynamic, like chemical equilibrium, because the molecules do not all have the same kinetic energy. If, at equilibrium, you randomly pick a molecule, there some probability (from the Boltzmann distribution) that it will have its energy between $E_1+\Delta E_1$. If after some time $\Delta t$ you pick those same molecules again, there is again the same probability to find those molecules with an energy between $E_1$ and $E_1+\Delta E_1$ is the same, but those same molecules will likely have a different energy than the one you measured earlier. this is because they molecules collide all the time and exchange energy with each other. From a macroscopic point of view the two states, at times $t_1$ and $t_1+\Delta t$, are the same; from a microscopic point of view, the two states are different. Thus it is a macroscopic equilibrium not a microscopic one. Same when have ice and water at equilibrium. The molecules that form part of the ice or part of the water keep changing all the time. At equilibrium, you keep the same amounts of water and ice, but the molecules that make up the ice and the water keep being exchanged between the ice and the water.