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Due to nuclear fusion that produces gamma rays which goes through compton scattering, in the end, on the surface, visible, infrared light ends up as before then, gamma rays lost energy and became x-rays, then x-rays to UV and so on.

If we imagine radiative zone for example, we know compton scattering happens and each time, x-rays's wavelengths become bigger. Since compton scattering cause the free electrons to speed up, in this zone, there must be bremsstrahlung happening as well. So it's a little bit interesting. Compton causes the photon to decrease in energy, while increaseing the speed of free electron which in turn each time it gains more speed, would cause higher energy photon emission through bremsstrahlung. You might say these 2 events kind of balance each other out.

Question 1: It's true that from bigger wavelength, we won't be able to get less wavelength, so compton always reduces photos, but since it increases the speed of electrons each time, don't these free electrons produce full light spectrum ? and if so, photon escaped from radiative zone for example or convection zone, can be produced due to compton or bremsstrahlung.

Question 2: It's said bremsstrahlung produces full spectrum radiation, but would this be the case in the sun ? maybe not as electrons wouldn't have this much energy in the first place ? but I could argue they would as gamma ray hits it and definatey would give quite a good speed and each time other gammas hit it as well, its speed would even increase more.

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  • $\begingroup$ The word is 'Bremsstrahlung' - german for 'braking radiation'. "breumsterlung" is not a word. $\endgroup$ Commented May 1, 2023 at 10:41

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The interior of the Sun is almost at perfect thermodynamic equilibrium. The mean free path of a photon before scattering or absorption is on a scale of mm. This is true even in the convective region. Each microscopic process is balanced by its inverse process. Thus there are just as many inverse Compton scatterings as Compton scatterings; just as much bremsstrahlung (note spelling) as inverse bremsstrahlung.

There is a tiny temperature gradient, which means the radiation field is slightly anisotropic, with slightly more photons travelling outwards than inwards.

The overall radiation spectrum at any point is a blackbody at the local temperature.

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  • $\begingroup$ Thanks Rob. I dont get what is inverse compton. In my mind, i call the inverse comptoning as bremsstrahlung. Why not like this ? $\endgroup$ Commented May 1, 2023 at 12:20
  • $\begingroup$ (inverse) Compton scattering is the interaction of a photon with an electron. (Inverse) bremsstralung (absorption) emission is the interaction of photons with an electron and an ion. @GiorgiLagidze $\endgroup$
    – ProfRob
    Commented May 1, 2023 at 15:38
  • $\begingroup$ well, Compton scattering (not inverse) is the same thing as interaction with photon and electron. So not sure what inverse suggests $\endgroup$ Commented May 1, 2023 at 16:43
  • $\begingroup$ Why don't you find out? @GiorgiLagidze A photon gains energy in inverse Compton scattering. $\endgroup$
    – ProfRob
    Commented May 1, 2023 at 17:29

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