2.6: Net Flux and Exitance

Formerly known as emittance, the exitance $M$ refers to a point on a reflecting or emitting surface and is defined as the total power emitted in all directions per unit physical area, so that

$$M = \int_0^{2 \pi} \int_{0}^{ \pi/2} L ( \vartheta, \varphi) \sin \vartheta \cos \vartheta d \vartheta d \varphi$$

where it may be seen from the limits of integration that “in all directions” means over a hemisphere. The factor $\sin \vartheta d \vartheta d \varphi$ is an element of solid angle, dω, and the factor $\cos \vartheta$ is needed to convert the projected area of radiance back into physical area. Using the notation of Chapter 1., i.e. let $\mu = \cos \vartheta,~ d \mu = - \sin \vartheta d \vartheta$, we have

$$M = \int_0^{2 \pi} \int_0^1 L ( \mu,~ \varphi ) \mu d \mu d \varphi .$$

If we compare M to Chandrasekhar’s quantity the net flux πF, which, in particular, he uses for a plane parallel beam of radiation

$$\begin{array}{l} \pi F & = \int_0^{2 \pi} \int_0^{2 \pi} L ( \vartheta,~ \varphi ) \sin \vartheta \cos \vartheta d \vartheta d \varphi \\ ~ & = \int_0^{2 \pi} \int_{-1}^1 L ( ( \mu,~ \varphi ) \mu d \mu d \varphi \end{array}$$

we see that the net flux is indeed the result of integration over all directions, i.e. over a sphere. It follows that net flux and exitance are not the same thing (although there may be situations in which they amount to the same), and nor does $πF$ always mean the strength of a plane parallel beam of radiant flux density F. Indeed, we can calculate the net flux of a plane parallel beam incident on a surface in the direction $\left( \mu_0,~ \varphi_0 \right)$, using the radiance of a plane parallel beam given by Chapter1, equation (7), as

$$\pi F = \int_0^{2 \pi} \int_{-1}^1 \textbf{F} \delta \left( \mu - \mu_0 \right) \delta \left( \varphi - \varphi_0 \right) \mu d \mu d \varphi,$$

which results in

$$\pi F = \bf{F} \mu_0,$$

this result being the irradiance E of the surface, as we knew it should be!