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Basically if you take a gray iron bar, and heat it red hot, it would still reflect a dullish white light, even though all you could see is the red light it emits as radiation. In fact if you watch your stove burner, there's even a phase where the burner looks reddish dark gray as it's reflected color mixes with it's emitted color in equal amounts, at least in terms of human perception. Thus things have a color they glow in based on temperature (usually infrared for earthly temperatures), and a color they reflect based on structure and composition.

So what I want to know is, if you could filter out the sun's radiation from it glowing from being really hot, and shine a giant spot light on it, what color would it be?

Edit: removed the term black body for clarity

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    $\begingroup$ A true blackbody does not reflect anything. All frequencies are completely absorbed $\endgroup$ – Andreas H. May 22 '16 at 18:47
  • $\begingroup$ Ignorance faught, thank you. Edited it out of the question for clarity $\endgroup$ – netsplit May 22 '16 at 18:52
  • $\begingroup$ Now I don't understand your question .... Specifically the difference between the "color things glow in" and the "color they reflect". What they reflect does depend on the color of the light source and the color they emit on the temperature... $\endgroup$ – Andreas H. May 22 '16 at 18:59
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    $\begingroup$ @AndreasH. I have an LED light bulb at home. When it glows, it glows with a harsh, bluish-white light. When it is turned off it looks yellow. OP is asking what color the Sun would be if you switched it off. Of course, the question assumes that you'd have a really bright floodlight to shine on the switched-off Sun so that you could see it. $\endgroup$ – Solomon Slow May 23 '16 at 17:42
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If the sun were a true black body, then it would be black. The extent to which it differs from a true black body can be seen from the sun-light spectrum outside the earth's atmosphere. Overall, it is pretty close to a 5778 K black body, so the sun will likely be very dark if you remove its own thermal radiation. Upon closer inspection, you can see that it lacks a bit in the UV and it has excessive radiation around blue-green visible light (500 nm). The explanation for the excess can be found in "How can it be that the sun emits more than a black body?": some short-wavelength light from deeper, hotter layers penetrates through the "cooler" top layer.

So, the outermost layers of the sun are either partially transparent, or black. They do not reflect. Remember that for everyday objects to reflect, they need to involve abrupt changes of the optical properties. Paper and sugar consist of transparent, colorless material; it appears white because of the reflections at the grain boundaries. The sun does not have a well defined surface; rather, there is a very gradual transition from interstellar vacuum to the more dense parts of the sun.

Update after comment The black-body deficit in the UV could be the result of Rayleigh scattering, which would suggest that the sun may be a bit bluish. Only a fraction 4e-4 of the hydrogen in the photosphere is ionized (source: What is the degree of ionization is the solar photosphere?); I would expect that molecular or atomic hydrogen has a very small Rayleigh cross section compared to the earth's atmosphere. I'm not sure what the small plasma fraction in the photosphere would do, exactly. In any case, if the cross sections for inelastic scattering (absorption and thermal emission) are much larger than the cross sections for elastic scattering (Rayleigh scattering), then you won't see much of the elastic scattering, just like adding a drop of milk to black ink will not give the ink a gray appearance.

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  • $\begingroup$ last paragraph: what about scattering ? (at least Rayleigh scattering). Then, (possibly colored) absorption along the multiscattering path could occurs. But I'm concerned about something else: Sun is largely plasma, or very ionised / excited mater. How does this interact with incoming photons ? $\endgroup$ – Fabrice NEYRET May 23 '16 at 16:30
  • $\begingroup$ answer updated. $\endgroup$ – Han-Kwang Nienhuys May 23 '16 at 17:03

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