# Absorption vs emission not in thermodynamic equilibrium

I have read that good absorbers are good emitters. The argument goes that at thermodynamic equilibrium, the amount of radiation a body emits must be the same as it absorbs, otherwise the body will get hotter and the surroundings will get colder which violates 2nd low of thermodynamics. The question is why the same holds in non-equilibrium? Suppose we heat a body to a high temperature (via conduction) and the surroundings are cold, then it will emit a lot more radiation than it absorbs. If the most energy used for emission comes via conduction and not absorption of radiation, why the ability to absorb would influence emissivity at all in this case?

E.g. it is said that a cavity can approximate black body because the light will be trapped in it and almost all of it will be absorbed. If we heat the same cavity to high temperature (and have cold surroundings), how does it follow that it will be a good emitter as well? Or will it have the same emissivity as a flat surface of the same material in this case?

## 2 Answers

The question is why the same holds in non-equilibrium?

In general, it doesn't hold. It only holds for thermal radiation emission (when the radiation is due to microscopic thermal motions characterized by thermodynamic temperature, not due to other non-equilibrium processes that are not characterized by thermodynamic temperature).

Consider a fluorescent mercury gas discharge lamp. When supplied with electric energy, its phosphor layer (a chemical compound, not the element) produces much more visible radiation energy than it absorbs and ratio of these energies isn't determined by temperature of the radiating surface. So this radiating surface doesn't obey Kirchhoff's radiation law. A different way to describe this is to say that light of a discharge lamp is not due to thermal emission. Similarly for the laser light.

In some non-equilibrium cases Kirchhoff's radiation law is nevertheless obeyed. For example, if the body surface radiates just thermal radiation (such as in your example of energy supplied via heat conduction), then there is no reason why the law should not be obeyed. The surface elements do not "know" that their temperature is due to heat conduction from a hotter body and not from incoming radiation.

• Thanks. Yeah it makes sense that for thermal radiation it will hold whether there is equilibrium or not, when emissivity depends on the intrinsic properties of the material. However, i m still confused as to how changing the shape (e.g. going from smooth to non-smooth or making a cavity as above) can improve emissivity in this kind of non-equilibrium situation when the absorption from the surroundings is almost not a factor. Is there any sort of layman-terms explanation for this? Feb 23, 2022 at 12:58
• "Changing shape" (arranging reflecting walls into a cavity) slows down escape of radiation energy inside, and increases "the number of interactions" between radiation and the material inside the cavity (walls and other bodies inside). This leads to radiation inside changing its spectrum away from that defined by the wall emissivity (metals are bad thermal emitters at IR) towards the spectrum of equilibrium radiation at some temperature (broad spectrum). This makes radiation coming off an opening in the cavity stronger at most wavelengths, as long as there is enough energy inside the cavity. Feb 23, 2022 at 19:44

It does not really matter how the body absorbs energy. It can be by radiation, conduction or other processes. The crucial point is that in a state of equilibrium, the total energy absorbed is, by definition, equal to the total energy emitted (the latter may not necessarily be just by radiation either). In a state of non-equilibrium, the temperature, and possibly other physical parameters, would be time-dependent (for instance if you turn off the explicit heating, the temperature of the body will drop until a new equilibrium with the ambient background temperature/energy is reached).