Dear Eigenstate,

Whew, that's a pretty technical question, but I think I can explain. As you point out, the line source function is just the ratio of the level populations. Statistical mechanics tells us how to calculate those. I don't know whether you were trained as a physical scientist or engineer, but I'll provide a little bit of math here. Disclaimer: I'm not a theorist and provide no warranty for this information, just a little physical insight (maybe).

Often in supernova work, you use an approximation called "local thermodynamic equilibrium" (LTE), which means you can approximate the atomic levels in any chunk of gas as though the atoms have a well-defined temperature. This makes things easier since you can approximate the transfer of energy in the SN atmosphere in terms of heat diffusion. It breaks down in cases where the SN ejecta can't transfer heat efficiently -- for example, when the density is very low (so that collisions between atoms don't happen often enough to exchange energy between them) -- or when you have a lot of non-thermal radiation e.g. from accelerated particles in the ejecta. You call those cases "NLTE" and it means you have to solve the full radiation transport equation, which is very complicated and takes lots of time to solve numerically even on very powerful computers. But for a lot of SN work, LTE is good enough.

So let's try an example, say, in a SN Ia atmosphere around maximum light. Let's estimate the source function for a particular line, Si II 6355, which would have a wavelength of lambda = 6355 angstroms or 635.5 nm; this is the line most commonly used to identify type Ia supernovae. If you pull out your thermal physics book (e.g. Kittel & Kroemer), you'll read that the level populations in thermal equilibrium are given by the Boltzmann factor exp(-(E_upper - E_lower)/kT). The SN photosphere, which is hotter and denser than the region where the lines are emitted, is probably about 10000 K, and the energy level difference is just hc/lambda, where h = Planck's constant and c = speed of light. If you copy and paste the expression

exp(-(h*c)/(6355 angstroms * k * 10000 K))

into Google (Calculator), you'll get about 0.1. This is a plausible

*upper* bound to the source function, and if you watch Dan's animation again, you'll see that there's still a healthy absorption trough at 0.1 or so. At 6000 K, the value is more like 0.02.

So I think the answer (in our example) is just that the line-forming region is cool enough that absorption dominates over emission, if you have a photosphere backlighting it. Of course, in the limbs there is no photosphere behind for us to view, but light from the photosphere at the wavelength of the line will be scattered into our line of sight, creating the emission part of the P Cygni profile. You've got a point though, you wouldn't expect this always to be true in general; it depends on the temperature, the energy levels, the density, and possibly other things.

-- Richard