Doesn't the second and third paragraph contradict each other? In the second, you say that a constant temeprature means no absorption lines, and in the third you say that a constant temperature would induce an absorption spectrum.

Your answer is probably correct, as you seem to have a lot of upvotes, but I can't seem to fully understand it. Some analogies would be really great, if there is any to explain this further.

@FelisSuper read it again. Your filling in of absorption lines (to produce a blackbody continuum) would only happen in an isotropic radiation field in a gas of constant temperature. When there is a temperature gradient, you get an anisotropic radiation field and absorption lines.

@RobJeffries Could you provide some reading material so that I can read more about this? I can't say I really understand how a temperature gradient solves any of this, but whenever there's something you don't understand, then it's always better to read about it from several different sources.

@Rob Jeffries - The equation above is not appropriate for a purely scattering atmosphere.The specific intensity J is in this case an implicit function of I and has nothing to do with thermal excitation. See plasmaphysics.org.uk/radiative_transfer.htm for the correct form. The solution of this gives a result similar to (plasmaphysics.org.uk/imgs/scattering2.jpg). Continuum source is to the left here, main curve is local excitation source function (due to the external continuum source), inset curve scattered line intensity looking left (without continuum source).

@Felis Super - If the reference in the above comment is too technical, try this one irina.eas.gatech.edu/EAS8803_Fall2017/petty_13.pdf It is quite technical as well in places but has good explanations and figures, although it does not specifically refer to absorption lines in the sun

@Thomas your comment makes no sense to me. The reference you give does not contain a single example of a radiative transfer equation, only the solutions thereof. The picture you refer to, I have no idea what it means or of what relevance it is to this question.

@RobJeffries Thank you very much the extra information in the answer. I can't say I am familiar with this formula, but assuming it's true, I think I understood the rest. The only thing I don't understand now is why the hot chromosphere and corona isn't considered here. The chromosphere, and especially the corona, tends to be much hotter than the photosphere, in which case there should be emission lines. Shouldn't that cancel out the absorption? Or are these emission lines mostly in other regions of the electromagnetic spectrum, not in the visible range?

@FelisSuper The chromosphere and corona do produce emission lines, from hotter plasma. The process of chromospheric infilling of some lines is a thing that happens (e.g. the Ca II H and K lines), though the Sun's chromosphere is not hot/dense enough to make much of a difference to the disk-averaged spectrum. Most chromospheric/coronal emission is indeed in the UV- X-ray region.

@Rob Jeffries - The first equation in plasmaphysics.org.uk/radiative_transfer.htm is the radiative transfer equation for a scattering atmosphere in integral form. You can call it a formal solution of the differential form if you like, but it is not an actual solution yet (which can only be done numerically). If you have a look at Eq.(13.1) in irina.eas.gatech.edu/EAS8803_Fall2017/petty_13.pdf you can see the same equation in your differential form. The point is that the second term on the right depends again on I, not on some internal thermal source term J.

@Thomas huh? J is the mean specific intensity. As I said.

@RobJeffries Alright, I see. Btw, I found a really great pdf presentation regarding the sun's spectrum, including how a negative temperature gradient yields an absorption spectrum and a positive gradient yields an emission spectrum, and a lot that was previously unclear to me has now been cleared up. Thank you for mentioning this effect, as I was completely unaware of it before reading your answer. Your answer has also been accepted :)

@Rob Jeffries-The sign for the equation is opposite in this reference as they define μ=-1 going into the plane parallel medium. See irina.eas.gatech.edu/EAS8803_Fall2017/petty_11.pdf ; Eqs.(11.9) to (11.11). for the general form. These equations also make clear that the mean specific intensity J (i.e. the local source function in the medium) can consist of two parts, a local thermal emission (related to the Planck function) and an emission due to scattering. You claim your equation treats scattering, then you have to consider the second term only and your temperature argument fails.

@called2voyage - The page I linked to is based on a chapter of my Ph.D. thesis. As this is not generally available, I re-worked it somewhat and put it on my website (the corresponding numerical code linked from that page as well)

@Thomas Ok, just make sure you explicitly disclose here any time you share your own work that is not publicly reviewed.

@RobJeffries - Again, your equation, as used by you, does not describe a scattering problem. It describes continuous (true) absorption (i.e..photo-ionization) and thermal (black body) emission. With this you won't be getting any absorption lines. And if you include scattering in the total absorption coefficient you have to include the scattering term for your source function as well. I am not sure though why you would want to consider continuous absorption in the first place. It is quite evident that above the photosphere the opacity is negligible outside the absorption lines.