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Solar radiation spectrum

The Sun emits more at $450-600 \,\rm{nm}$ than a black body of the same effective temperature would, it also emits far less UV. I’ve heard this is due, in part, to the fact that the Sun doesn’t have a surface from which radiation is emitted, instead having a photosphere that is hundreds of kilometres thick.

However, to me that seems like it should make the radiation more spread out with a shorter peak, the exact opposite of what is observed.

Could someone please explain this to me?

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    $\begingroup$ I found in this a good survey of radiation from stars ( and the sun is a star) alevelphysicsnotes.com/astrophysics/black_body_rad.php . At the very end it has data identifying spectral lines in the radiation from a star, that gives information of its content. $\endgroup$
    – anna v
    Jul 19, 2023 at 19:33
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    $\begingroup$ Note that the effective temperature of the Sun is about 5500 C. It's also unclear then how the curve has been normalised. The answer BTW is just that the Sun isn't a blackbody at a single temperature. $\endgroup$
    – ProfRob
    Jul 19, 2023 at 20:39

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The spectrum of the Sun is essentially the Planck function at the temperature where the optical depth reaches around $2/3$ at any particular wavelength. i.e. The radiation we see arises from the layer where the emitted radiation can escape.

The spectrum will be depressed where the opacity is high because the photons come from a higher, cooler layer. Conversely, the spectrum will be raised where the opacity is low and the photons arise from deeper, hotter layers. The effective temperature, defined as $(L_\odot/4\pi \sigma R_\odot^2)^{0.25}$, represents some sort of average.

The high UV opacity is provided by the higher Balmer series bound-free transitions, photoelectric absoorption by metal atoms and ions, and Rayleigh scattering. In the red part of the spectrum it is $H^-$ ions that do the bulk of the absorption.

Here's a handy table and plot published in Wheeler et al. (2023).

Opacity sources in the photosphere

The height of the solar spectrum will be anti-correlated with the opacity at that wavelength. You can see that the opacity is much higher towards the UV and the near-infrared.

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The Sun is not a perfect black body as it has temperature variations over its surface as well as radiative absorption and emission processes by the various elements. This means thermodynamic equilibrium (which requires that collisional processes dominate radiative processes) does not strictly hold anymore.

You can find an extended discussion of the solar spectrum at http://www.ccpo.odu.edu/SEES/ozone/class/Chap_4/4_3.htm (the figures referenced there are not shown on that page; you can get them from http://www.ccpo.odu.edu/SEES/ozone/class/Chap_4/4_thumbs.htm )

Anyway, the deviations from the Planck curve in your graphic seem to be exaggerated by the linear scale used. In the logarithmic plot below (taken from this reference) the difference at 500nm for instance is barely visible anymore (solar spectrum = dark blue line) It also shows that the spectrum in the extreme UV (wavelength less than the Lyman ionization edge (91nm)) is not related to the black body curve at all anymore but stays at a more or less constant (wavelength-independent) level (with some fluctuations)) (don't be confused by the elements written in red on top of the graph at certain wavelengths; these indicate absorbing elements in the Earth's atmosphere, not the Sun)

solar spectrum

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    $\begingroup$ "exaggerated by the linear scale used" — What?! I'd rather say they are hidden by the log scale in your chart. The linear scale shows pretty well that spectral power density is ~10% higher than that of the black body in this wavelength range. 10% is quite a serious departure from the idealized description, worthy of study. $\endgroup$
    – Ruslan
    Jul 20, 2023 at 21:07

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