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65

No, it's not. The radiation field in the interior of the Sun is very close to a blackbody spectrum. If you look in any particular direction the brightness (power per unit area) you see is $\sigma T^4$, where $\sigma$ is Stefan's constant. Even at any particular wavelength it is always the case that a blackbody of higher temperature is brighter than a ...


15

Coming from a different direction as @Rob's, Opacity and Thermal Radiation are orthogonal properties of a material. The photon flux at the interior of the sun is very high, so it is definitely not dark. However, it is opaque to virtually all light outside the sun. To provide an analogy, if you are in a sealed room with no windows, you cannot see anything ...


9

Both free-free and bound-free absorption are strongly dependent on the atomic number $Z$ of nuclei in the gas. For free-free absorption it is simply that the emissivity per unit volume of electrons accelerating in the field of an ion scales as $Z^2$ (there are $Ze$ electrons accelerating near a charge of $+Ze$). Since by Kirchoff's Law, absorptivity is ...


5

Easily ionised ones like sodium and potassium. Not much from lithium because that is rare. You don't need many free electrons. The fraction of H$^{-}$/H is very small, something like $10^{-7}$ in the solar photosphere. And indeed the number densities of sodium and potassium to hydrogen are of that order.


4

Although I will only tackle one part of the question, I find the following part of a picture from NRAO/AUI/NSF, S. Dagnello, cited from space.com worth sharing: You see the radial structure of Antares, a red supergiant of spectral type M1.5Iab-Ib, and more specifically The average temperatures of photosphere, chromosphere, and above are given. One can see ...


4

If you have a look at the top-left panel of Fig.11 in these lecture notes by Rob Rutten, you will see that the continuum opacity at optical wavelengths at the photosphere is about $10^{-6.7}$ cm$^{-1}$. The inverse of this is the optical depth. You can see stuff through about 2-3 optical depths, so your "horizon", looking horizontally in the solar ...


1

The cooling of a gas giants outer envelope (which in turn cools the convective interior) is dominated by radiative transport. The radiative flux at the photospheric boundary is $$F_{rad}\propto- \frac{ T^3}{\rho \kappa_R}\nabla T,$$ where $\kappa_R$ is the Rosseland-mean opacity (for more, see e.g. these open lecture notes, by K. Dullemond, in 5.5.3). This ...


1

No mass blob of stellar mass is transparent at any wavelength of interest. Opacities $\kappa_{\nu}$(inverse transparency) as function of wavelength becomes really high and broad band at pressures above > 0.1 bars, for all wavelengths. This leads to the optical depths $\tau_{\nu}$ being enormous and as transmission is $T=1-\exp(-\tau)$, you won't be able ...


1

It's a bit subtle, the key thing the partial ionization does is to keep the temperature from changing much. What you really want is an increase in density, not temperature. The reason the kappa mechanism is important is that it allows heat to be added to the gas when it is compressed, and removed when the gas is expanded, that's what allows for the ...


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