26.

For the calculation of the energy absorbed we shall employ the same reasoning as was illustrated by [fig:1]Fig. 1 ([sect:22.] Sec. 22) and shall retain the notation there used. The radiant energy absorbed by the volume-element $v$ in the time $dt$ is found by considering the intensities of all the rays passing through the element $v$ and taking that fraction of each of these rays which is absorbed in $v$. Now, according to [eqn:(19)] (19), the conical element that starts from $d\sigma$ and cuts out of the volume $v$ a part equal to $fs$ has the intensity (energy radiated per unit time) $d\sigma ·\frac{f}{r^2}·K$ or, according to [eqn:(12)] (12), by considering the different parts of the spectrum separately: $2d\sigma \frac{f}{r^2}{\int }_0^{\infty }{𝖪}_{\nu }d\nu .$ Hence the intensity of a monochromatic ray is: $2d\sigma \frac{f}{r^2}{𝖪}_{\nu }d\nu .$ The amount of energy of this ray absorbed in the distance $s$ in the time $dt$ is, according to [eqn:(4)] (4), $dt{\alpha }_{\nu }s2d\sigma \frac{f}{r^2}{𝖪}_{\nu }d\nu .$ Hence the absorbed part of the energy of this small cone of rays, as found by integrating over all frequencies, is: $dt2d\sigma \frac{fs}{r^2}{\int }_0^{\infty }{\alpha }_{\nu }{𝖪}_{\nu }d\nu .$

When this expression is summed up over all the different cross-sections $f$ of the conical elements starting at $d\sigma$ and passing through $v$, it is evident that $\sum fs=v$, and when we sum up over all elements $d\sigma$ of the spherical surface of radius $r$ we have $\int \frac{d\sigma }{r^2}=4\pi .$ Thus for the total radiant energy absorbed in the time $d$t by the volume-element $v$ the following expression is found: $dtv8\pi {\int }_0^{\infty }{\alpha }_{\nu }{𝖪}_{\nu }d\nu .\mathrm{\backslash Label}[eqn](25)\text{upshape (25)}$ By equating the emitted and absorbed energy we obtain: ${\int }_0^{\infty }{ϵ}_{\nu }d\nu ={\int }_0^{\infty }{\alpha }_{\nu }{𝖪}_{\nu }d\nu .$

A similar relation may be obtained for the separate parts of the spectrum. For the energy emitted and the energy absorbed in the state of thermodynamic equilibrium are equal, not only for the entire radiation of the whole spectrum, but also for each monochromatic radiation.

This is readily seen from the following. The magnitudes of ${ϵ}_{\nu }$, ${\alpha }_{\nu }$, and ${𝖪}_{\nu }$ are independent of position. Hence, if for any single color the absorbed were not equal to the emitted energy, there would be everywhere in the whole medium a continuous increase or decrease of the energy radiation of that particular color at the expense of the other colors.

This would be contradictory to the condition that ${𝖪}_{\nu }$ for each separate frequency does not change with the time. We have therefore for each frequency the relation:

$\epsilon_{\nu} &= \alpha_{\nu} \mathsf{K}_{\nu}, \textrm{ or} \Label[eqn]{(26)}\tag{\upshape (26)}\ \mathsf{K}{\nu} &= \frac{\epsilon\tag}}{\alpha_{\nu}}, \Label[eqn]{(27)$

i.e.: in the interior of a medium in a state of thermodynamic equilibrium the specific intensity of radiation of a certain frequency is equal to the coefficient of emission divided by the coefficient of absorption of the medium for this frequency.