## Some special solutions of the Two Stream equations

Beer's law

Suppose that the atmosphere is too cold to radiate significantly at the frequency under consideration. In that case, B(v, T) « 0 and the internal source vanishes. This would be the case, for example, if v is in the visible light range and the temperature of the atmosphere is Earthlike. In this case, the solutions are simply I+ = I+ (0)exp(-tv) and I_ = I_(TTO)exp(Tv - tto). The exponential attenuation of radiation is known as Beer's Law. Here we've neglected scattering, but in Chapter 5 we'll see that a form of Beer's law still applies even if scattering is taken into account.

### Infinite isothermal medium

Consider next an unbounded isothermal medium. In this case, it is readily verified that 1+ = 1— = nB(v,T) is an exact solution to 4.9. The right hand sides of the equations vanish, but the derivatives on the left hand sides vanish also, because T is independent of tv . Hence, in an unbounded isothermal medium, the radiation field reduces to uniform blackbody radiation.

Since the fluxes are independent of tv, the radiative heating rate vanishes, from which we recover the fact that blackbody radiation is in equilibrium with an extended body of isothermal matter.

Exercise 4.2.1 Derive this result from Eq. 4.9; from Eq. 4.13. Finite-thickness isothermal slab

Now let's consider an isothermal layer of finite thickness, embedded in an atmosphere which is completely transparent to radiation at frequency v. We suppose further that there is no radiation at this frequency incident on the layer from either above or below. We are free to define tv = 0 at the center of the layer, so that tv = 2at the top of the layer and tv = — 2at the bottom. The boundary conditions corresponding to no incident flux are 1— = 0 at the top of the layer and 1+ = 0 at the bottom of the layer. The solution 1+ = 1— = nB is still a particular solution within the layer, since the layer is isothermal, but it does not satisfy the boundary conditions. A homogeneous solution must be added to each flux in order to satisfy the boundary conditions. The homogeneous solutions are just exponentials, and so we easily find that the full solution within the layer is

### Exercise 4.2.2 Derive this result from Eq. 4.13.

The radiation emitted out the top of the layer is 1+(tto, v), or (1 — exp(—TTO))nB, which reduces to the blackbody value nB when the layer is optically thick for the frequency in question, i.e. tto(v) ^ 1. The same applies for the emission out of the bottom of the layer, -mutatis mutandum. Note that in the optically thick limit, 1+ = 1— = nB through most of the layer, and the inward-directed intensities only fall to zero in the two relatively thin skin layers near the top and bottom of the slab.

In the opposite extreme, when the slab is optically thin, both t and are small. Using the first order Taylor expansion of the exponentials, we find that the emission out the top of the layer is t^-kB, and similarly for the bottom of the layer. Hence, in this case is just the bulk emissivity of the layer. This is consistent with the way we constructed the Schwarzschild equations, which can be viewed as a matter of stacking a great number of individually optically thin slabs upon each other.

Substituting into Eq. 4.14, the heating rate for this solution is