For a precise solution, one needs to solve this equation separately for each 0 and then integrate over angles to get the net upward and downward fluxes. The angular distribution of radiation changes with distance from the source, since radiation propagating near the direction 0 = 0 or 0 = n decays more gradually than radiation with 0 nearer to n/2. Hence, radiation that starts out isotropic at the source (as is the blackbody emission) tends to become more forward-peaked as it propagates. For some specialized problems, it is indeed necessary to solve for the angular distribution explicitly in this fashion, which is rather computationally demanding. Fortunately, the isotropy of the blackbody source term tends to keep longwave radiation isotropic enough to allow one to make do with a much more economical approximate set of equations.
We can derive an equation for the net upward flux per unit frequency, I+, by multiplying Eq. 4.7 by cos 0 and integrating over all solid angles in the upward-pointing hemisphere. Integrating over the downward hemisphere yields the net downward flux I—. However, because of the factor 1/ cos 0 on the right hand side of Eq. 4.7, the hemispherically averaged intensity appearing on the right hand side is not I+. Instead, it is f I(t*,n, or equivalently /Qn/2 2nI(t*,0, v) sin 0d0.
One cannot proceed further without some assumption about the angular distribution. If we assume that the distribution remains approximately isotropic, by virtue of the isotropic source B, then
I(t*, 0, v) is independent of 0, and hence the problematic integral becomes 2nl f0/2 sin 010 which is equal to 2I+ under the assumption of isotropy. This result yields a closed equation for I+. It states that, if the radiation field remains approximately isotropic, the decay rate is the same as for unidirectional radiation propagating with an angle 0 such that cos 0 = 1, i.e. 0 = 60o. From now on we will deal only with this approximate angle-averaged form of the equations, and use tv = t*/ cos 0 as our vertical coordinate. The choice cos 0 = 2> is by no means a unique consequence of the assumption of isotropy. The fact is that an isotropic distribution is not an exact solution of Eq. 4.7 except in a few very special limits, so that the choice we make is between different errors of roughly the same magnitude. If we had calculated cos 0 by multiplying Eq. 4.7 by (cos0)2 instead of cos 0 before averaging over angles, we would have concluded cos 0 = | and this would be an equally valid choice within the limitations of the isotropic approximation. Sometimes, a judicious choice of cos 0 is used to maximize the fit to an angle-resolved calculation in some regime of particular interest. For the most part we will simply use cos 0 = 1 in our calculations unless there is a compelling reason to adopt a different value.
In terms of tv , the equations for the upward and downward flux are
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