D.28 Kirchhoff’s law

Suppose you have a material in thermodynamic equilibrium at a given temperature that has an emissivity at a given frequency that exceeds the corresponding absorptivity. Place it in a closed box. Since it emits more radiation at the given frequency than it absorbs from the surrounding blackbody radiation, the amount of radiation at that frequency will go up. That violates Plank’s blackbody spectrum, because it remains a closed box. The case that the emissivity is less than the absorptivity goes similarly.

Note some of the implicit assumptions made in the argument. First, it assumes linearity, in the sense that emission or absorption at one frequency does not affect that at another, that absorption does not affect emission, and that the absorptivity is independent of the amount absorbed. It assumes that the surface is separable from the object you are interested in. Transparent materials require special consideration, but the argument that a layer of such material must emit the same fraction of blackbody radiation as it absorbs remains valid.

The argument also assumes the validity of Plank’s blackbody spectrum. However you can make do without. Kirchhoff did. He (at first) assumed that there are gage materials that absorb and emit only in a narrow range of frequencies, and that have constant absorptivity $a_{\rm {g}}$ and emissivity $e_{\rm {g}}$ in that range. Place a plate of that gage material just above a plate of whatever material is to be examined. Insulate the plates from the surrounding. Wait for thermal equilibrium.

Outside the narrow frequency range, the material being examined will have to absorb the same radiation energy that it emits, since the gage material does not absorb nor emit outside the range. In the narrow frequency range, the radiation energy $\dot{E}$ going up to the gage plate must equal the energy coming down from it again, otherwise the gage plate would continue to heat up. If $B$ is the blackbody value for the radiation in the narrow frequency range, then the energy going down from the gage plate consists of the radiation that the gage plate emits plus the fraction of the incoming radiation that it reflects instead of absorbs:

\begin{displaymath}
\dot E = e_{\rm {g}} B + (1-a_{\rm {g}}) \dot E
\quad \Longrightarrow \quad \dot E/B = e_{\rm {g}} / a_{\rm {g}}
\end{displaymath}

Similarly for the radiation going up from the material being examined:

\begin{displaymath}
\dot E = e B + (1-a) \dot E
\quad \Longrightarrow \quad \dot E/B = e/a
\end{displaymath}

By comparing the two results, $e$$\raisebox{.5pt}{$/$}$$a$ $\vphantom0\raisebox{1.5pt}{$=$}$ $e_{\rm {g}}$$\raisebox{.5pt}{$/$}$$a_{\rm {g}}$. Since you can examine any material in this way, all materials must have the same ratio of emissivity to absorptivity in the narrow range. Assuming that gage materials exist for every frequency range, at any frequency $e$$\raisebox{.5pt}{$/$}$$a$ must be the same for all materials. So it must be the blackbody value 1.

No, this book does not know where to order these gage materials, [37]. And the same argument cannot be used to show that the absorptivity must equal emissivity in each individual direction of radiation, since direction is not preserved in reflections.