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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?

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    $\begingroup$ Huh? Thermal radiation has nothing to do with atomic transitions. And what "photon gas"? $\endgroup$ – ACuriousMind Mar 1 '15 at 18:26
  • $\begingroup$ @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. $\endgroup$ – Rob Jeffries Mar 1 '15 at 18:36
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    $\begingroup$ possible duplicate of What are the various physical mechanisms for energy transfer to the photon during blackbody emission? $\endgroup$ – John Rennie Mar 1 '15 at 18:46
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    $\begingroup$ @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. $\endgroup$ – Rob Jeffries Mar 1 '15 at 19:03
  • $\begingroup$ @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? $\endgroup$ – John Rennie Mar 1 '15 at 19:11
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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.

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  • $\begingroup$ 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? $\endgroup$ – Shadumu Mar 1 '15 at 20:02
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    $\begingroup$ +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. $\endgroup$ – David Hammen Mar 1 '15 at 20:12
  • $\begingroup$ @user3229471 You should look at the duplicate. Yes, the radiation field comes close to equilibrium with the matter by interactions with the matter. $\endgroup$ – Rob Jeffries Mar 1 '15 at 22:21
  • $\begingroup$ @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. $\endgroup$ – Rob Jeffries Mar 1 '15 at 22:36
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Black body radiation is given by Planck's formula

enter image description here

(see link for variables)

Here is the measured irradiance of the sun and the attempt to fit it with the black body formula:

sun irradiance

The effective temperature, or black body temperature, of the Sun (5,777 K) is the temperature a black body of the same size must have to yield the same total emissive power.

The visible surface of the Sun, the photosphere, is the layer below which the Sun becomes opaque to visible light. Above the photosphere visible sunlight is free to propagate into space, and its energy escapes the Sun entirely. The change in opacity is due to the decreasing amount of H− ions, which absorb visible light easily.Conversely, the visible light we see is produced as electrons react with hydrogen atoms to produce H− ions. The photosphere is tens to hundreds of kilometers thick, being slightly less opaque than air on Earth. Because the upper part of the photosphere is cooler than the lower part, an image of the Sun appears brighter in the center than on the edge or limb of the solar disk, in a phenomenon known as limb darkening.[78] The spectrum of sunlight has approximately the spectrum of a black-body radiating at about 6,000 K, interspersed with atomic absorption lines from the tenuous layers above the photosphere.

The continuum in the spectrum comes because at that high temperature the ions and electrons interact with the fields of each other and the magnetic field of the sun, accelerating/decelerating charges radiate in the continuum.

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  • $\begingroup$ 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? $\endgroup$ – Shadumu Mar 1 '15 at 20:06
  • $\begingroup$ considering the velocity of light and that photon photon interaction is higher order ( very small) only under the conditions of the Big Bang one could consider an equilibrium state in exchanges with all other forces. It is not possible to isolate a "photon gas" so as to talk of equilibrium. $\endgroup$ – anna v Mar 2 '15 at 4:39
  • $\begingroup$ I have been thinking of this: if one had a cavity with totally reflecting walls one could contain photons in space bouncing around. They would not interact with each other so they would not have the attributes of a gas. $\endgroup$ – anna v Mar 2 '15 at 8:02

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