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What part of the photons emitted from a star are from black body radiation and what part originate from fusion reactions?
To my understanding these are the two sources of luminosity for a star, so I'm just wondering which phenomena accounts for the majority of photons that come from a star.

lee
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Miika Vuorio
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Fusion reactions produce high energy gamma radiation. None of those photons reach the surface of the star directly. Over timescales of $10^4-10^5$ years, they scatter around as their energy propagates towards the surface. Thus, all photons from a star are from blackbody radiation.

Some of these blackbody photons get absorbed by atoms or ions in the atmospheres of the star, and get re-emitted. This re-emission shows up as absorption lines in the spectrum. However, since re-emission goes in all directions, some of the photons we see are from such atomic recombination processes.

You can get a qualitative understanding of the numbers when you look at any absorption spectrum from a star. The overall smooth distribution is the blackbody radiation. Absorption lines are subdominant to that, and only a tiny fraction of the difference between an absorption line and the thermal spectrum at that wavelength is going into your direction from re-emission. Hence, overall, that becomes pretty negligible and one can say that essentially all photons are blackbody radiation.

rfl
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    How do photons, moving at the speed of light, take tens to hundreds of thousands of years to make it a few hundreds of thousands of miles? That approximately puts the speed of light in solar plasma somewhere on the order of 1-10 miles per year! – Mason Wheeler Nov 09 '21 at 19:28
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    They scatter. This increases their distance traveled by such a large amount. That is because stars are both rather dense and huge. The speed of light is still of the usual order. – rfl Nov 09 '21 at 19:36
  • Can you model the interior of the star as having a very large refractive index n = c/(10 miles per year)? – Jojo Nov 09 '21 at 21:01
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    That would be a poor model. High-energy gammas mostly Compton scatter, and that means a transfer of some of their energy to electrons (this is one way to heat the plasma). In contrast, photons traversing optical media don't change their color. – rfl Nov 09 '21 at 21:25
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    @MasonWheeler Keep in mind that the center of our Sun is an extremely dense plasma, a mixture of fully ionized hydrogen and helium compressed to over 100 times the density of water. A high energy photon created by fusion in the center of a star doesn't last very long. It travels a centimeter or so before before being absorbed -- and then re-emitted as a photon or two or three or more, each in a random direction. The photons created by fusion take a 3D random walk that eventually leads outward from the center of the Sun. – David Hammen Nov 09 '21 at 22:13
  • @DavidHammen does this suggest that in a 1D or 2D sun, we wouldn't in general expect the photons to make it out? Since "a drunk man finds their way home, but a drunk bird doesn't". – Eugene Shvarts Nov 11 '21 at 05:11
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    Blackbody radiation is not an emission process, it is a description of the spectrum. – ProfRob Nov 11 '21 at 07:59
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    @DavidHammen "The photons created by fusion take a 3D random walk" would be more precisely worded as "The succession of photons which started with fusion follows a 3D random walk". – Peter - Reinstate Monica Nov 11 '21 at 14:28
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This is posed as an either-or question. The answer to "what part of the photons emitted from a star are from black body radiation and what part originate from fusion reactions" is yes.

Our Sun is 4.6 billion years old. Prior to becoming a star that fuses hydrogen, the proto-Sun emitted black body radiation due to the energy generated by gravitational collapse. However, since that happened 4.6 billion years ago, the energy from that gravitational collapse has long since been radiated away. The energy emitted by the Sun from its surface in the form of photons originate from fusion reactions in the Sun's core.

The Sun also emits neutrinos, about 3% of the total energy. Almost all of those neutrinos travel from the core to the surface without interference. In contrast to those neutrinos, the photons created by created by fusion are subject to a lot of interference on the way out. The mean free path of the high energy photons created by fusion in the Sun's core is on the order of a centimeter. In the process of randomly bouncing around on their way out of the core, those high energy photons increase in number but decrease in wavelength.

There are multiple feedback mechanisms that keep the electromagnetic energy generated by fusion at a star's core in equilibrium with the electromagnetic energy radiated by a star from its surface. What this means is that the answer to "what part of the photons emitted from a star are from black body radiation and what part originate from fusion reactions" is yes.

David Hammen
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    Less glibly, all that reaches us is black body radiation, but all the energy for it came from fusion reactions releasing high-energy photons that have since been absorbed and their energy reemitted as black body radiation before leaving the sun. – Mar Nov 09 '21 at 19:28
  • @Martin One has to dsitinguish to two things: (a) the photons were "absorbed and reemitted" or otherwise scattered, and (b) black-body radiation - one can define the latter in different ways, but under almost any definition it is safe to say that nothing that reaches us is black-body radiation . – Roger V. Nov 10 '21 at 08:32
  • https://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain says "This mass has been converted into energy, in the form of kinetic energy of produced particles, gamma rays, and neutrinos released during each of the individual reactions. The total energy yield of one whole chain is 26.73 MeV". It has a bit more info on the energy lost to neutrinos, but it's a bit vague on what proportion of the energy goes into the kinetic energy of the other particles (i.e., not the neutrinos or gammas). I think it's almost 50%. Does that sound right? – PM 2Ring Nov 11 '21 at 00:43
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Most of the photons received from the Sun, in the visible and IR parts of the spectrum are due to recombination radiation as free electrons attach themselves to hydrogen atoms to form $H^{-}$ ions. This occurs within a layer of width just $\sim 1000$ km in the solar photosphere. There are of course additional photons due to atomic transitions like H, Na, Ca, Fe etc. that occur at discrete (or at least over a narrow range of) wavelengths.

The combination of these processes produces a spectrum that approximates to that of a blackbody (the Planck function). Note that "blackbody radiation" is a description of the spectrum, not an emission process.

The photons produced directly by fusion and by other processes associated with the hot gas in the core of the Sun do not get anywhere near the surface. Their mean free path is of order 1 mm and they deposit their energy almost in the same place they were produced. We do not see (at any wavelength) photons that originated in the core of the Sun.

ProfRob
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