2
$\begingroup$

Kirchhoff's law of thermal radiation states that for thermal equilibrium for a particular surface the monochromatic emissivity $\epsilon_{\lambda}$ equals the monochromatic absorptivity $\alpha_{\lambda}$: $$\alpha_{\lambda} = \epsilon_{\lambda}$$ I wonder how this can be only valid for thermal equilibrium, because $\epsilon_{\lambda}$ and $\alpha_{\lambda}$ seem to depend only on the temperature of the surface under consideration, and it seems to me that they don't depend on the other bodies around the surface. So according to this, it should follow that they are equal always, when they are equal in thermal equilibrium, which isn't correct. So I wonder how the lack of thermal equilibrium can change the values of $\alpha$ and $\epsilon$?

$\endgroup$
6
$\begingroup$

Think of those not as a capacity to absorb/emit but as simply absorption/emission. Imagine you put a cold metal cube next to a hot identic cube. The hot one will emit a lot of heat but receive a very little amount of heat from the cold one. Therefore $α_λ<ϵ_λ$, and vice-versa for the other cube. After a while, they will reach the same temperature, and at that moment, the radiation from cube A will be equal as cube B's. At that moment $α_λ$ will equal $ϵ_λ$ since the "output" radiation and "input" radiation are equal.

You are right when saying it doesn't take into account the other object, but this equation becomes true only when both objects reach thermal equilibrium so it's not necessary to explicitly refer to the other one.

It's like if the surface was a mirror, and you were asking with this equation when does the reflect equals the original? You only need the original to answer this question.

$\endgroup$
  • $\begingroup$ You changed question and answered that. You didn't answer the original one. Emissivity is not power or energy of radiation: it is a ratio of intensity of radiation of the body at question and black body of the same temperature. $\endgroup$ – Ján Lalinský Jun 2 '18 at 21:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.