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Suppose the interior of the cavity is made of a non-black body. Then in thermal equilibrium, the spectral radiance of the non-black body, say $R_T^*(\nu)$, will be different from the spectral radiance of a pure black body, $R_T(\nu)$. So the way I see it, inside the cavity there is a distribution of radiation with spectral radiance $R^*_T(\nu)$ but somehow when it passes through the hole it becomes $R_T(\nu)$ because the hole behaves like a black body? I'm finding these two things to be irreconcilable. Can someone expand on this please?

Also I have a more practical question. How is the cavity brought to thermal equilibrium in practice? From what I've read, it's the hole that really absorbs like a black body so to make it be in thermal equilibrium like a black body does one have to send radiation through the hole so that it absorbs all of them and emits at the same rate to be in thermal equilibrium? How would that work? Since you're already sending radiation in through the hole you can't analyze the emission spectra coming out of it too. Or is there actually a way to do it?

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Then in thermal equilibrium, the spectral radiance of the non-black body, say $R_T^*(\nu)$, will be different from the spectral radiance of a pure black body, $R_T(\nu)$.

This is the part that is confusing you. The body (cavity walls) has different-than-blackbody spectrum of thermal emission when exposed to open space (thus not in equilibrium). But when this radiation is released into cavity where it cannot escape, after some time, when equilibrium is reached, total EM radiation inside the cavity in equilibrium won't have the same spectrum.

The emission-suppressed frequencies of radiation also have suppressed absorption, and in time, this leads to buildup of radiation components at these frequencies. When equilibrium state is reached, spectrum of radiation inside will be that of a blackbody, even if the walls are not blackbody. The radiation at the emission-suppressed frequencies exists with the same intensity as in blackbody radiation. The additional intensity isn't emitted/absorbed by the walls but is being reflected by the walls.

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The explanation for why a hole into a cavity behaves like a black body depends on the reversibility of physical processes. Any radiation entering the hole will scatter (or reflect) off the inside of the cavity a number of times, before eventually being absorbed. It is unlikely to escape through the hole (if the cavity is large compared to the hole), so almost all entering radiation is absorbed. Reversing the process, thermal radiation from the inside of the cavity will scatter/reflect, either being reabsorbed or ultimately escaping from the cavity. Since all frequencies entering the hole are equally likely to be absorbed, all frequencies of thermal radiation will be emitted with likelihood determined only by temperature, i.e. according to the black body curve.

Thermal radiation is spontaneous. The hole emits according to the temperature of the cavity, however that is maintained. It may be heated by any source to a required temperature. This does not require any radiation entering through the hole.

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  • $\begingroup$ "Since all frequencies entering the hole are equally likely to be absorbed, all frequencies of thermal radiation will be emitted with likelihood determined only by temperature, i.e. according to the black body curve." <- Regarding that last part, I read that not all objects emit according to the black body curve, only pure/approximate black bodies do. What if the interior of the cavity is not even close to a black body so that even though it will emit thermal radiation with likelihood determined only by temperature, the distribution wouldn't be similar to the one given by the black body curve? $\endgroup$ May 10, 2020 at 10:04
  • $\begingroup$ Yes, it is only an approximation, but if it emits with equal likelihood determined only by temperature, that is enough to determine the black body curve. In practice there are emission lines and ratio of the size of the cavity to the size of the hole is finite, so it cannot be perfect. I believe stars give better black body curves (after removal of spectral lines). The CMB is the best known black body curve. $\endgroup$ May 10, 2020 at 11:14
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A black body, by definition, absorbs all e-m radiation that falls upon it. A hole leading into a cavity (with walls at a fixed temperature) is a good approximation to a black body. The smaller the hole the better the approximation. The reason that the hole is a black body is that any radiation entering the hole and not absorbed at its first encounter with the wall will be reflected and will hit the wall again. Whatever isn't absorbed will hit the wall again, and so on. To all intents and purposes all the radiation incident on the hole is absorbed. [Because the hole is small, the chance of incident radiation bouncing back out of the hole is negligible.]

That a hole in a cavity is a perfectly black body in this sense may seem little more than a curiosity. But we can use the second law of thermodynamics to show that a black body has unique emission behaviour. For example, the power emitted per unit area in any wavelength range depends only on the temperature of the surface, and is greater than for any non-black surface. This applies, then, to the radiation inside a cavity with walls at a given temperature.

"How is the cavity brought to thermal equilibrium in practice?" Keep the walls at a fixed temperature (e.g. by electrical heating with thermostats).

"Since you're already sending radiation in through the hole you can't analyze the emission spectra coming out of it too." I'm not sure why you're sending radiation in, but I don't see why you can't analyse radiation coming out at the same time. Radiation going in and coming out doesn't have to be travelling along the same line.

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  • $\begingroup$ How do you show that a black body has unique emission spectrum using second law of thermo? $\endgroup$ Feb 4, 2021 at 16:54
  • $\begingroup$ @Brain Stroke Patient Take a small hole in a cavity with walls at constant temperature as your black body. Put two such can cavities with walls at the same temperature, hole-to-hole. It would violate the second law for finitely more radiation be coming (per unit area) out of A (and into B) than vice versa as B would become hotter and A cooler, so heat would be travelling from a hotter body to a cooler. To show that the the spectra must be identical (equal energy emitted per unit wavelength interval) just run the same thought experiment with spectral filters between the two holes. $\endgroup$ Feb 4, 2021 at 19:05
  • $\begingroup$ But that just argues that the spectra emitted by a cavity is unique. How does it show that all black bodies have a unique emission spectrum. $\endgroup$ Feb 8, 2021 at 6:55
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    $\begingroup$ If you are thinking of a black body in the form of a solid body with a perfectly black surface, then you run the thought experiment I've just given but with the body inside a cavity with walls at constant temperature rather than with two communicating cavities. $\endgroup$ Feb 8, 2021 at 8:06
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I think its a really good question, and I didn't see it answered as I would have liked if I had asked it. I am not sure I can answer it better, but here is my input. A cavity with a hole will only emit through that hole such radiation as is caused by the thermal energy of material from which the cavity is made. That is, if the hole and the environment are managed properly, nothing will come out of that hole because it entered and is being reflected back out. Or, the amount of such reflected radiation will be small enough to be ignorable by the experiment. Therefore, with a cavity we have satisfied the basic requirement of a black body...namely, that its radation (as measured at the hole exit) is ONLY due to its thermal energy. THEN, physicists USE the fact that they have a cavity as a crutch to calculate how the radiation due to thermal energy might be distributed in the black body. And when they assumed radition was waves the calculcations didn't fit what the cavity did experimentally, and when they assumed quanta where energy is related to frequency the calculations DID match what the cavity experiments showed. Now, it IS CLEARLY POSSIBLE, as you seem to intuit, that the results of the experiments are somehow dependent on this articifical approximation of a blackbody using a cavity. So that if we had a REAL black body, it would behave entirely differently from our cavity. But it turns out that every experiment we do with objects that approximate as black bodies agrees with the results that were obtained using the cavity model. So we assume that the cavity model is valid, and so the calculations based on the cavity model are valid too. This is more than anything else a tribute to the genius that first decided to use the cavity model to study therrmal radiation. It was undoubtedly one of those inspirations that come from nowhere. Now everyone, except you, and hats off to you, just assumes "oh a cavity is a black body" without thinking about it why it should be so. It didn't have to be so, but it was.

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