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!