Is it dark inside the Sun?

  • This might sound like a strange question, but something got me thinking about it recently.



    The opacity of plasma in stellar interiors can get quite high, making for shorter free-paths for photons. In these conditions I guess that the light you could theoretically gather, supposing you have a pair of indestructuble eyes submerged in the solar interior, would be the one emitted by the plasma in your immediate surroundings, right? So if the opacity is high enough I can imagine places inside a star like the Sun where there is the same ambient illumination as a typical moonless night here on Earth.



    My questions are:




    • Is this line of reasoning correct?

    • Are these conditions actually possible inside a star?

    • Where exactly inside a star are these conditions possible?


    It helps to think: If your indestructible eyeballs was hammered into a (constant temperature) red hot iron bar, would it stll be able to see the red light from the iron?

    @ivella dunking in glowing luminol might be a better (and less painful) parallel than hammering into an iron bar.

    @Mindwin but the process that makes the iron glow red is the same that makes the sun shine: black body radiation.

    Maybe a simpler way: "dark" implies no photons. There are lots and **lots** of photons, so it cannot be dark.

    At night it is.

    Blackbody radiation is not a "process".

    "dark" implies no photons ...I would have thought that "dark" implies no photons in the visible spectrum

  • ProfRob

    ProfRob Correct answer

    2 years ago

    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 blackbody at lower temperature.


    Given that the interior temperature might be $10^7\ \mathrm K$, then the surface brightness is $5.7 \times 10^{20}\ \mathrm{W/m^2}$, compared with the $1400\ \mathrm{W/m^2}$ you would get by looking directly at the Sun (please don't do this). Note that most of this power comes out at X-ray wavelengths, but because of the properties of a blackbody, the brightness at visible wavelengths will still be plenty brighter than that of the solar photosphere (see below).


    A possible source of confusion is this term "opacity". When things are in thermal equilibrium, which the interior of the Sun is, then they emit the same amount of radiation as they absorb. So high opacity also means high emissivity.


    Details for interest:


    The opacity, $\kappa$ in the solar interior ranges from 1 cm$^2$ g at the centre to about $10^5$ cm$^2$ g just below the photosphere. To estimate the mean free path of photons we need to multiply this by the density $\rho$ and take the reciprocal:
    $$ \bar{l} = \frac{1}{\kappa \rho}\ .$$
    The density varies from 160 g/cm$^3$ at the centre to about 0.001 g/cm$^3$ just below the photosphere. Thus the mean free path is about 6 micrometres at the center and is actually quite similar just below the photosphere (it peaks at around 2 mm about three quarters of the way out towards the surface).


    Thus your "view" of the stellar interior is of a foggy sphere with radius of no more than a few times $\bar{l}$. The fog however is tremendously bright - as outlined above.


    The brightness at particular wavelengths is proportional to the Planck function
    $$B_\lambda = \frac{2hc^2}{\lambda^5} \left(\frac{1}{\exp(hc/\lambda k_B T) -1}\right).$$


    Thus at $\lambda=500$ nm (visible light), the ratio of brightness for blackbodies at $10^7$ K (solar interior) to 6000 K (solar photosphere) is $4.2\times 10^{4}$. i.e. Even just considered at visible wavelengths, the interior of the Sun is about 40,000 times brighter than the photosphere.


    Since the mean free path is so short in the high density areas photons "build up" inside. When used to estimate the photon flux or radiation field, does the Stefan–Boltzmann law have to be adjusted somehow to account for this? btw Is there a blackbody spectrum of photons inside a solid? needs a better answer and Is the visible light spectrum from “red-hot glass” at least close to Blackbody Radiation? could use some review as well.

    the glass question still bothers me; how can the visible part of the spectrum be black body if glass is not black in visible light (you can still see through it when it's hot). I'm guessing this happens all the time in astronomy...

    @uhoh and I answered that there. If you can see through it, then it isn't emitting blackbody radiation. If the absorption is wavelength-i dependent then it's spectrum may have the shape of a blackbody.

    @uhoh I don't know what you mean by "build up". The energy density of the radiation field is also proportional to $T^4$. It is totally independent of the mean free path or any other property of the gas producing it.

    An Etalon is not a good analogy, but it may help to just clarify what I mean by "build up". Perhaps an integrating sphere a different but similarly imperfect example. If I shine 1 Watt of light into one end of an etalon I'll collect (nearly) 1 Watt out the other. But if I sampled the photon flux inside a long Etalon using something with a small sensor that measures from one side only and used that to estimate the photon flux, I'd say that the flux is consistent with 100 or 1000 Watts, not 1 Watt. This is because of the "build up" of photons.

    I'm not saying that 5.7E+20 is necessarily way off, and I'm not saying that the number would not be proportional to $T^4$, I'm wondering about a multiplicative constant for the photon flux due to the strong scattering/very short mean free path. Suppose I do the following; put an photodiode in the middle of a red hot glass sphere and make a measurement. Then I put the whole thing at the center of a larger sphere with a highly reflective matte surface so that the light "builds up" due to dozens or hundreds of scatterings. My meter reading at the center of my glass sphere will go way up!

    @uhoh a blackbody absorbs all photons that are incident upon it. It doesn't reflect any of them. Etalons are generally monochromatic. Nothing like blackbody radiation. Your second example doesn't achieve any kind of equilibrium; also not a blackbody.

    okay I have said that these are "analogies" (and not particularly good ones at that), not *examples* so of course you can say they are not the same thing. You've asked me to explain "what you mean by 'build up'" and I've simply explained what I meant by 'build up'. You're arguing against a something I didn't say. *I'll look elsewhere* for the photon flux vs temperature, perhaps a book or paper, and compare it to Stefan–Boltzmann to see if there is a substantial proportionality constant required to match your simple $\sigma T^4$ argument.

    @uhoh Or perhaps you could ask your questions as questions (on Physics SE).

    @RobJeffries Excellent! Let me see if I grasped the idea. If the optical depth of the material is about a few micrometers this means that, when I'm inside the Sun, I'm observing the plasma in an extremely small spherical volume (bacterium sized) around my eye and that the rest of the Sun is hidden from me, as if it didn't existed.

    But since the optical depth is related to opacity and we are in thermal equilibrium, this large opacity means that the crazy small sphere of plasma around my eye is also absorbing a crazy large amount of energy and thus heating so hard that it shines as bright or even brighter than the surface of the Sun as viewed from outside it. Is that correct?

    Yes. Basically, it would be somewhat like being inside a brightly lit cloud of dense fog: all you can see is the glowing fog right in front of your eyes. (The microscale mechanisms are slightly different, in that the fog droplets are reflecting and scattering light while the plasma particles are absorbing and thermally re-emitting it, but the end result is the same — a very short mean free path for photons.)

    @swike Correct.

    Interestingly, while the maximum of the wavelength distribution in the black body radiation (if I have ever seen an oxymoron...) of the Sun's core would be X rays there would be plenty of radiation all the way down to radio; in particular, the sun's core is brighter than the surface in the visible spectrum as well. See https://ase.tufts.edu/cosmos/view_picture.asp?id=1332.

    "compared with the 1400 W/m2 you get by looking directly at the Sun." **Don't look directly at the Sun!**

    @CJDennis Indeed. Warning added.

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Content dated before 7/24/2021 11:53 AM