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I read in a popular science book by a respected German science journalist (some 25 years ago) that the core of the Sun is absolutely and utterly pitch black.
The reason is that at the temperature the emission of the Sun's core in the visible wavelengths is nearly nil, as the emission maximum lies far, far away in the gamma range.
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Non sunt multiplicanda entia praeter necessitatem. |
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The photon production in the core is primarily gamma rays as a result of the proton-proton reaction. I think there are several different energies of these produced in the chain. There are other high energy photons produced by the CNO process which constitutes only about 1% of the energy production. I don't know what might take place within the core that would produce much lower energy photons that would allow the production of photons that would fall in the optical range. So, my color view is starting to turn black. But, I want to look at the core's "surface" and consider that the random walk of the photons is indeed random. Thus, photons of all energies can be found here since they are meandering in all directions, though more outward than inward. It is not until the temperature drops enough, when hydrogen can accept a second electron, that photons get trapped (ie the convective zone). Thus the inner sections allow radiative transfer, including the core. There is also Doppler shifts due to the kinetic energy levels of the hydrogen in the core, though I won't bother calculating them. So, thanks mainly to this sunny, photonic Volksmarsch, I will argue against a black core, for now. [It would still be pretty colorful if you are right, admittedly. ] For the sake of this question, and given the colorful spirit it was asked, let us assume a more blackbody result for the energy production of the core, at least along the outer boundary. You may further assume a 15 million degree temperature here.
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Lighten up! This is a stellar board! Author: duh. "The Sun, with all the planets revolving around it, and depending on it, can still ripen a bunch of grapes as though it had nothing else in the universe to do..." Author: Galileo supposedly. |
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There had to be something wrong with the pitch black Sun, as I found out here - a hot black body radiator emits more energy in absolute terms at any wavelength than a colder one. So although the maximum of a black body radiating at 15 million K (if we accept that temperature as that of the surface of a "naked" Solar core) is in the x-ray and gamma, there is still copious energy at the visible range. Enough to blind you, I should say.
Were he still alive, I'd chalk the jourmalist one down for bad astronomy. And me, too, for accepting it. ![]() As for the colour: In the visible spectrum, the intensity curve of a 15 MK black body rises monotonously from red to blue. I suspect we'll see a pure white object becaue of the ferocious intensity of the radiation. Filtered down with a strong neutral grey filter, the colour might have a bluish tinge, as the shorter wavelengths predominate. So I'll suspect we'll have a blue-white object - a lot like in "white dwarf", which according to The Once and Future Sun starts out at about 120000 K. I doubt the difference between 120000 and 15000000 million K makes much of a difference as far as the appearance in the optical spectrum is concerned.
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Non sunt multiplicanda entia praeter necessitatem. |
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Now the why. You are so close, too, but it may involve some good guessing (which is why I through Tyndall in there) or some math.
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Lighten up! This is a stellar board! Author: duh. "The Sun, with all the planets revolving around it, and depending on it, can still ripen a bunch of grapes as though it had nothing else in the universe to do..." Author: Galileo supposedly. |
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George - If you want math, I am afraid you have to get it elsewhere. My ineptitude there was my prime reason not to study physics.
As for guessing - hmmm, Tyndall effect. I don't know. Scattering? In the densely packed core of the Sun? As for black body - I thought it is the point of the core of the Sun that emits all its energy via a perfect thermal spectrum. I don't see what should stop the Sun's core from emitting one. All the distubances (emission and absorption lines) happen in the convective zone and the photosphere at much lower temperatures. Anyway, I'll leave it here. Surely, someone will provide the solution as to the why.
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Non sunt multiplicanda entia praeter necessitatem. |
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![]() And thanks for the sportsmanship. I doubt you would have been caught.
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Non sunt multiplicanda entia praeter necessitatem. |
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Tyndall also missed the math that stood-out in an energy-temperature experiment he did, IIRC. Quote:
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Lighten up! This is a stellar board! Author: duh. "The Sun, with all the planets revolving around it, and depending on it, can still ripen a bunch of grapes as though it had nothing else in the universe to do..." Author: Galileo supposedly. |
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Hint: Tyndall did experiments that allowed another prominent physicist, Stefan, to realize the relationship of radiant energy loss varies to the fourth power of the absolute temperature.
Arneb has already corectly guessed Tyndall scattering as part of the answer. [I think it was Tyndall who was first to claim the sky's color was due to scattering; hint, hint.]
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Lighten up! This is a stellar board! Author: duh. "The Sun, with all the planets revolving around it, and depending on it, can still ripen a bunch of grapes as though it had nothing else in the universe to do..." Author: Galileo supposedly. |
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As a WAG, is the Stefan-Boltzman Law involved in the answer?
E = sigma X T4 But it yields the total amount of radiation being emitted. Weins Law gives the wavelength at which most of the radiation would be emitted. lambda max = 2897 mm / T(K) Dave Mitsky
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So, you have 15+ million degrees, Rayleigh Scattering and Stefan-Boltzman (or Planck). What is the connection for the color determination?
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Lighten up! This is a stellar board! Author: duh. "The Sun, with all the planets revolving around it, and depending on it, can still ripen a bunch of grapes as though it had nothing else in the universe to do..." Author: Galileo supposedly. |
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At 15 million degrees, the black body equation is dominated by the nu^3 term in the visible range. Blue light has a frequency of about 7x10^14 hz, and red light has a frequency of around 4x10^14hz. So there would be about (7/4)^3 = 5.4 times as much blue light as red light.
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]Thus, the rich blue sky we see from the top of a mountain would also be the optical color of anything close to a bb profile if it is at such a high temperature (assuming proper attenuation). However, this does not mean hotter stars would appear bluer because of the saturation issues which would cause blue stars to look white. Consider the color of white dwarfs; are they blue or white? [rhetorical as my question time is up.] Your go Gilese 581c . [Added: it was interesting to learn that it was Tyndall that led us into the fourth power relationship of scattering light and had the data that led us to the fourth power temperature relationship to energy (as realized by Stefan).]
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Lighten up! This is a stellar board! Author: duh. "The Sun, with all the planets revolving around it, and depending on it, can still ripen a bunch of grapes as though it had nothing else in the universe to do..." Author: Galileo supposedly. |
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I think it was 1 %
and why couldn't a space craft go much faster ? I guess because all other forms of propulsion like anti-matter and wormholes were based on theoretical or science fiction concepts, rather than being sceintifically achievable in the near future |