Here we explain the transfer of electromagnetic radiation through a medium and the resulting absorption, emission and scattering of the radiation.¹⁾

Specific or spectral intensity or brightness $I_\nu$ is conserved, does not change along the way, so

$$ \frac{dI_\nu}{ds} = 0 $$

where $s$ is the coordinate along the direction of propagation. If there is an intervening medium between the distances $s_{in}$ and $s_{out}$, the radiation will be affected depending on the optical depth or opacity $\tau$ which has the opposite sign compared to $s$. As $s$ increases from left to right, $\tau$ increases from right to left.

Note that $\tau=0$ at $s_{out}$.

Let $d\mathcal{P}$ be the probability of a photon getting absorbed within a thin layer of thickness $ds$ as shown in the diagram above. Then the linear absorption coefficient

$$ \kappa \equiv \frac{d\mathcal{P}}{ds} $$

in units of m$^{-1}$. We consider thin layers because otherwise the relation becomes non-linear. If the layer is thick, more photons will be absorbed in the beginning, so the probability of absorption will decrease wit $s$, meaning $\kappa$ will depend on $s$ leading to non-linearity. So the fraction of absorbed intensity

$$ \frac{dI_\nu}{I_\nu} = -\kappa ds $$

were the negative sign is used to keep the optical depth and distance opposite. Integrating both sides we can show that

$$ I_\nu^{out} = I_\nu^{in} e^{-\tau}. $$

Optically thin absorbing media have $\tau\ll 1$ and vice versa. Note that the optical depth

$$ \tau = \int_{s_{out}}^{s_{in}} \kappa(s) ds. $$

The medium can emit and add more photons. In an infinitesimal volume $ds\ d\sigma$ of cross section $d\sigma$, the probability per unit time of emitting a photon into a solid angle $d\Omega$

$$ \frac{d\mathcal{P}}{dt} \propto ds\ d\sigma\ d\Omega. $$

The emission coefficient is defined as

$$ j_\nu \equiv \frac{dI_\nu}{ds}. $$

The radiative transfer equation combines the absorption and emission coefficients:

\begin{equation} \frac{dI_\nu}{ds} = -\kappa I_\nu + j_\nu. \end{equation}

In thermodynamic equilibrium (TE), $\kappa$ and $j_\nu$ are not independent. Note that the radiation from a cavity (shown below) TE depends neither on the walls nor on the content of the cavity.

Gustav Kirchhoff determined the relation between the absorption and emission coefficients using a thought experiment like the one shown above. In this case $dI_\nu/ds=0$ and $I_\nu=B_\nu(T)$, the blackbody radiation at an equilibrium temperature $T$. So

$$ \frac{j_\nu(T)}{\kappa(T)} = B_\nu(T) $$

meaning the ratio of the two coefficients in a medium is just the blackbody radiation of that medium at a particular equilibrium T.

Turns out this is valid for both FTE (full TE) and LTE (local TE). The atmosphere of Earth is in LTE, but not in FTE because it is located in the atmosphere of the Sun which has a much higher T. But Kirchhoff’s law is still applicable to Earth’s atmosphere when it acts as an absorbing medium.