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    $\begingroup$ ahh this makes sense. So it is only guaranteed that the same amount of energy will be released, but it's form may not be a single photon and it may be launched in a direction outside of the scope of the lense $\endgroup$
    – Luke
    Commented Feb 19, 2015 at 23:59
  • $\begingroup$ This is not the correct (or at least not the full) explanation in stars. A temperature gradient is required to produce absorption lines. $\endgroup$
    – ProfRob
    Commented Dec 12, 2020 at 9:44
  • $\begingroup$ @ProfRob "A temperature gradient is required to produce absorption lines.", can you please elaborate on this? $\endgroup$
    – Árpád Szendrei
    Commented Nov 16, 2021 at 21:42
  • $\begingroup$ @ÁrpádSzendrei If there is no temperature gradient then your line of sight ends on material at the same temperature and hence with the same surface brightness. The argument in this question takes the simple case of a cold object illuminated from behind by a hot object (i.e. implicitly assumes a temperature gradient) and where the optical depth is tuned so that you can see the hot object in the continuum but not in the line. $\endgroup$
    – ProfRob
    Commented Nov 16, 2021 at 23:16