Kirchhoff's Law of Thermal Radiation - When is a (radiative) process in thermodynamic equilibrium?

Question

I'm very confused/struggling to understand the scope of Kirchhoff's law of radiation, put simply my understanding of the law is \begin{gather} \varepsilon = a \end{gather} where $\varepsilon$ is the emissivity of the material, and $a$ is the absorptivity of the material (fraction of energy emitted at temperature $T$ relative to a black body, and fraction of energy absorbed at $T$ relative to black body, respectively) when an object is at thermodynamic equilibrium.

(I'm taking $\varepsilon, a$ to be averages over wavelength, also just considering gray-bodies, or if I'm not explaining this correctly, I mean $\varepsilon$ is the factor in the stefan-boltzmann equation: emitted radiative flux = $\varepsilon \sigma T^4$)

OK, so I understand why this must be true if we consider an isolated system of two objects in a vacuum, both with temperature $T$. If they are both temperature $T$, then we expect this to not change (it would violate the second law if they were not the same temperature). But if one of the objects had $\varepsilon \neq a$, then it's net energy flux would be nonzero, changing its temperature and thus violating the second law. (Any other explanation would be appreciated too!)

But now for my question:

How can we intuit when Kirchhoff's law of radiation applies and when it doesn't apply?

I recognize the criteria for thermodynamic equilibrium, one of which is that the temperature is constant. That being said, if there are two bodies at different temperatures they are in radiative exchange and thus their temperatures are changing, so they are not at thermodynamic equilibrium, thus $\varepsilon$ does not necessarily have to equal $a$. Is this correct? Would someone be able to explain to me the limits of this law as to when it applies, and when it doesn't/when radiative heat transfer processes occur irreversibly (not in thermodynamic equilibrium) and when they don't?

Thanks!

Answer 1

This post pretty much answers the question that I had.

To reiterate, the emissivity in the Stefan-Boltzmann law is a spectral-averaged emissivity, that is

\begin{gather} \varepsilon = \int_0^\infty \epsilon(\lambda, T) \; d \lambda \end{gather} where $\epsilon(\lambda, T)$ is the (true) emissivity. So when people claim that the absorptivity of a material differs from the emissivity of the material, they mean that the spectral averaged (in the EM spectrum where the material is absorbing) absorptivity differs from the spectral averaged emissivity (again, in the region of the EM spectrum where the material is emitting).

So in practicality (in this averaged sense) the absorptivity can be different from the emissivity. My confusion that led to me posting this question stemmed from the gloss-over of a lot of literature to distinguish the two.

For the actual answer, however, I would assume that in order for two bodies to be in radiative equilibrium, they must be in thermodynamic equilibrium. Since emission in this sense is more of an "averaged" (by this I mean we are not concerned with emission on the quantum scale of atom's electron being promoted to an excited state and then falling back to a relaxed state) phenomena, and emissivity and absorptivity are in general functions of temperature, then in order for two bodies to reach "steady state", their temperatures must not be fluctuating.