31.

If the scattering did not affect the radiation, the total energy reaching $d\sigma$ would necessarily consist of the quantities of energy emitted by the different volume-elements of the pencil, allowance being made, however, for the losses due to absorption

on the way. For ${\beta }_{\nu }=0$ expressions [eqn:(33)] (33) and [eqn:(30)] (30) are identical, as may be seen by comparison with [eqn:(27)] (27). Generally, however, [eqn:(30)] (30) is larger than [eqn:(33)] (33) because the energy reaching $d\sigma$ contains also some rays which were not at all emitted from elements inside of the pencil, but somewhere else, and have entered later on by scattering. In fact, the volume-elements of the pencil do not merely scatter outward the radiation which is being transmitted inside the pencil, but they also collect into the pencil rays coming from without. The radiation $E^{\prime }$ thus collected by the volume-element at $r_0$ is found, by putting in [eqn:(29)] (29), $dt=1,\nu =d{r}_{0}d\Omega r_0^2,d\Omega =\frac{d\sigma }{r_0^2},$ to be $E^{\prime }=d{r}_{0}d\Omega d\sigma {\beta }_{\nu }{𝖪}_{\nu }d\nu .$

This energy is to be added to the energy $E$ emitted by the volume-element, which we have calculated in [eqn:(31)] (31). Thus for the total energy contributed to the pencil in the volume-element at $r_0$ we find: $E+E^{\prime }=d{r}_{0}d\Omega d\sigma ({ϵ}_{\nu }+{\beta }_{\nu }{𝖪}_{\nu })d\nu .$ The part of this reaching $O$ is, similar to [eqn:(32)] (32): $d{r}_{0}d\Omega d\sigma ({ϵ}_{\nu }+{\beta }_{\nu }{𝖪}_{\nu })d\nu e^{-r_0({\alpha }_{\nu }+{\beta }_{\nu })}.$ Making due allowance for emission and collection of scattered rays entering on the way, as well as for losses by absorption and scattering, all volume-elements of the pencil combined give for the energy ultimately reaching $d\sigma$ $d\Omega d\sigma ({ϵ}_{\nu }+{\beta }_{\nu }{𝖪}_{\nu })d\nu {\int }_0^{\infty }d{r}_0{e}^{-r_0({\alpha }_{\nu }+{\beta }_{\nu })}=d\Omega d\sigma \frac{{ϵ}_{\nu }+{\beta }_{\nu }{𝖪}_{\nu }}{{\alpha }_{\nu }+{\beta }_{\nu }}d\nu ,$ and this expression is really exactly equal to that given by [eqn:(30)] (30), as may be seen by comparison with [eqn:(26)] (26).