# How is a blackbody spectrum formed in the Sun?

Sunlight can be treated as BB radiation. Why is it a continuous spectrum while the sun contains only a few elements and the radiation from the jumps between atomic levels are discrete? How does the photon gas achieve thermal equilibrium while they do not interact with themselves?

• Huh? Thermal radiation has nothing to do with atomic transitions. And what "photon gas"? – ACuriousMind Mar 1 '15 at 18:26
• @ACuriousMind BB radiation is a subset of thermal radiation. Very often thermal radiation is dominated by atomic (or ionic) transitions. E.g any optically thin gas in local thermodynamic equilibrium. – Rob Jeffries Mar 1 '15 at 18:36
• – John Rennie Mar 1 '15 at 18:46
• @JohnRennie Maybe so, but since none if the answers to that duplicate mention what is the dominant source of continuum opacity, it's a good job I got my answer in before this is over-enthusiastically closed. – Rob Jeffries Mar 1 '15 at 19:03
• @RobJeffries: well, Lubos' answer in the linked question is a general description of black body radiation, so it wouldn't address what the major source was in the specific case of the Sun. I'll defer to your knowledge of astrophysics, but doesn't the Sun radiate in basically the same way any plasma radiates? – John Rennie Mar 1 '15 at 19:11

Maybe the simplest way to think about this is that the Sun is in approximate thermal equilibrium and would absorb any photon, of any frequency, that is incident upon it. This is essentially the definition of a BB.

There are many radiative processes that can absorb (and hence emit) radiation at all frequencies, not just those corresponding to atomic transitions. For example there is free-free and bound-free opacity associated with negative H ions in the solar photosphere.

Of course not all frequencies are absorbed equally - that is why the solar spectrum is not a BB at a single temperature. At each frequency you see to a different depth (and hence temperature) meaning that the radiation field at any frequency corresponds roughly to that of a blackbody at a temperature where the optical depth reaches $\sim 1$ (or 2/3 in more exact treatments). In a strong absorption line, the photons that finally escape the Sun come from higher up and at cooler temperatures and hence are not as bright as other frequencies, with lower opacities, that arise in deeper, hotter layers.

• so a pure photon gas will not reach equilibrium. it is through various kinds of interaction with matter that photons finally follow BB spectrum, right? – Shadumu Mar 1 '15 at 20:02
• +1, nice answer. $\mathrm H^-$ plays an important role in determining the opacity in the photosphere, which in turn plays an important role in the mix of temperatures we see. – David Hammen Mar 1 '15 at 20:12
• @user3229471 You should look at the duplicate. Yes, the radiation field comes close to equilibrium with the matter by interactions with the matter. – Rob Jeffries Mar 1 '15 at 22:21
• @user3229471 And at some point near the top of the atmosphere, the material above becomes effectively transparent and the radiation field escapes. My point was that this happens at different depths (and hence different temperatures) at different frequencies depending on how effective the photon-matter interactions/opacity is at that frequency. Hence the radiation is in (local) thermodynamic equilibrium with the matter at a frequency-dependent temperature and is thus only an approximation to a blackbody. – Rob Jeffries Mar 1 '15 at 22:36

Black body radiation is given by Planck's formula