Black body radiation and thermal equilibration

Imagine two $$ideal$$ black bodies, one at temperature $$T_1$$ and the other at $$T_2$$, $$T_1 \ne T_2$$, both are in thermal equilibrium with their respective heat baths and now we separate the cavities from them. Both have a small opening through which we connect them with a tube (a waveguide) that is ideal reflecting internally, absorbs nothing and is both a thermal and an electrical insulator having constant wave impedance at all frequencies, etc., let us assume that such thing exists.

After a while I expect that the two cavities (black bodies) thus connected will thermally equilibrate, they will assume the same temperature so that the total flux from one will equal the total flux from the other.

Now let us place a reciprocal $$band-stop$$ (or $$band-pass$$) filter in the tube, and assume that the filter is lossless, absorbs nothing, and either reflects the incoming waves or passes them without loss to other side.

Staying in the domain of classical physics my question is:

Can an ideal band-stop (or band-pass) filter prevent thermal equilibration? I believe excluding a finite band will only slow down equilibration but how? What is the mechanism by which a black-body cavity not in equilibrium converts, if that is the right word, energy at one frequency to another? It is clear that at one end an all-frequency ideal band-stop filter, i.e., an ideal reflector, will prevent equilibrium, for the cavities then do not communicate at all, but how is the transition "from nothing to everything".

• I don't get it, why would it stop thermal equilibrium? Commented Aug 8, 2020 at 17:49
• @OfekGillon I do not know and this is why I am asking. Clearly, if the bandstop is wide enough it will surely slow equilibration down but how? What happens to the reflected waves back to the respective cavity? Take two resistors connected with a transmission line (line and resistors have the same impedance), break the line in the middle and place an ideal 1:1 transformer with an LC resonator. Is it obvious that thermal equilibrium is reached if a particular frequency band is excluded from the exchange, if so why? Commented Aug 8, 2020 at 17:58
• Are the black bodies evacuated and totally insulated from the outside environment? Commented Aug 8, 2020 at 19:41
• @DavidWhite yes, that is what I assume: the $T_1, T_2$ are only initial temperatures of the cavities and once are connected with each other they are disconnected from the thermal baths Commented Aug 8, 2020 at 19:50

After a while I expect that the two cavities (black bodies) thus connected will thermally equilibrate, they will assume the same temperature so that the total flux from one will equal the total flux from the other.

I don't think this is consistent with the description of the setup. If all there is are two perfectly reflecting cavities whose insides are connected by a perfectly reflecting tube, each initially with different equilibrium radiation at $$T_1, T_2$$, there is no reason why, in time, radiation inside would turn into equilibrium radiation. Reflections from nonmoving walls can't change frequency or intensity of radiation, so they do not change its spectrum overall, and radiation does not interact with itself in classical physics. So radiation inside would keep having characteristics of both $$T_1$$ and $$T_2$$, only with smaller intensity than that of equilibrium radiation.

If we introduce some filter into the tube, this will be made of non-perfectly reflecting matter and then we have radiation interacting with real matter. That is what will, in time, turn radiation into equilibrium radiation.

You can find this understanding of equilibration in old works on blackbody radiation by Planck and experimenters - there is always either a piece of carbon, or soot on the walls, or some other matter that speeds up the equilibration. If walls are perfectly reflecting, a piece of matter is theoretically necessary for the equilibration. Perfectly reflecting walls are considered only to contain the radiation energy, but for equilibration some real matter is needed.

• thank you, I think I agree with you but here is my real question that I really wanted to ask but instead made it "unrecognizably global" physics.stackexchange.com/questions/594803/… Commented Nov 18, 2020 at 20:58

Examine the power spectrum from both objects. For any fraction of the spectrum you want to look at, the hotter body has greater power than the cooler body. There is no portion of the spectrum where this fails to hold. At any frequency, the power difference exists and is always in the same direction.

Therefore if any portion is in communication, it will deliver net thermal energy from the warmer to the cooler object. The greater the portion of the spectrum, the faster this transfer can occur.

• what happens to the spectral content that is reflected back (blocked) to the cavity by the filter, how does that energy get converted (if that is the right word) to different frequencies where exchange can take place? Commented Sep 13, 2020 at 0:41
• if you take the simplest electric model example of a transmission line terminated with matched loads but differing initial temperatures can you provide the equations that describe the equilibration in time with and without a lossless filter placed on the line? Commented Sep 13, 2020 at 0:44
• The concept of a temperature is one of thermal equilibrium. That is happening constantly in any body that we consider to have reached thermal equilibrium. So any energy input at one wavelength thermalizes to others. Commented Sep 13, 2020 at 3:11

1. The meaning of having black body at temperature $$T$$ is that it is in continuous energy exchange with a thermal bath at temperature $$T$$.
2. No real significant difference between using black body radiation and thermal gas. The real difference is in using canonical vs grand canonical ensemble, i.e. conservation of particle number $$N$$, which doesn't seem to have any relevance to your question.

Without uncoupling each system from its bath - you cannot reach equilibrium, and you will get thermodiffusion - energy flow from the hot to the cold system.

Now let's suppose you separately equilibrate each system at their proposed temperature, then uncouple them from their respective baths (impose elasatic scattering walls, rather then thermal). Here the initial energy/micro-state occupation/distribution will be thermal, yet the ongoing micro-dynamics will influence what state you will get. The results differ for interacting gas (and black body) vs non interacting, and quantum vs classical treatment.

For quantum treatment - each isolated system evolve unitarily, which preserves the structure of the density matrix, i.e. $$U\rho U^{\dagger}=\rho$$. Classical treatment of non-interacting gas will have the same effect. Classical treatment of interacting gas will evolve the system to the microcanonical ensemble (H theorem and such).

Now what happens when you let energy exchange between the two systems? (energy exchange is the analogous condition for massive particles, as particle number is not conserved for black body and thus useless criterion at this investigation) For classical interacting gas the system will converge to new appropriate microcanonical ensemble. (even with the band pass, which is at energy, assuming the system is ergodic/mixing). For classical non-interacting gas nothing will happen. As the particles don't interact, thus cannot transfer energy to one another thus the filter is simply a wall.

The quantum case is interesting, but I'm not sure I can elaborate much more, as it seems to be very dependent on the specific system state and specific evolution dynamics. Eigenstate thermalization hypothesis may seem as a valid option, but not the only one.

Two suggestions to make the question even more interesting, for you to think about -

1. Single black body, yet the filter is placed between the system and the bath. (This maybe was the real intent of your original question). The bath effectively becomes non-thermal. How the system evolves and what is it's steady state.
2. Filter between the systems that breaks time reversal symmetry (which is a necessary condition for thermal equilibrium)
• When the cavities are connected with each other I assume that they are disconnected from their respective thermal baths. I think this question ought to be solvable within classical physics (thermodynamics and EM) without resorting to QM. Commented Aug 8, 2020 at 20:00
• @hyportnex Black body is essentially a quantum problem (thanks to Plank's solution), and more close to quantum nature due to the wave nature of the constituents. Commented Aug 8, 2020 at 20:13