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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?

The answer ProfRob gave an excellent answer, as always, but in case someone wants a plain language summary we talked about this:

If the optical depth of the solar plasma is about a few micrometers this means that, when I'm inside the Sun, I'm observing the plasma only in an extremely small spherical volume (bacterium sized in fact) around my eye and that the rest of the Sun is hidden from me, as if it didn't existed. No photons outside that extremely tiny volume around my eyes will reach them. This is because there's high opacity.

But because of opacity we also know that the plasma in that tiny volume must absorb a lot of light, and that energy rises the temperature of the plasma. You might not be seeing a lot of material but this material absorbs so much light from the surroundings that it is itself re-emitting light in the form of thermal radiation. The plasma is in fact a black-body radiator.

In thermal equilibrium the absorbed energy (by the opaque plasma) is the same as the emitted one (as black-body radiation). Thus, if we were outside of thermal equilibrium, like during a supernova event, we might get a different result.

In any case what you would see if your eyes where inside the stellar plasma is the same as what you would see if you pressed your eyeballs to the surface of a star (assuming that the temperature inside is the same as that of the surface, which is only true for a small depth (as you go deeper the temperature increases and you would see it even brighter).

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    $\begingroup$ 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? $\endgroup$
    – lvella
    Mar 6, 2020 at 11:25
  • $\begingroup$ @ivella dunking in glowing luminol might be a better (and less painful) parallel than hammering into an iron bar. $\endgroup$
    – Mindwin Remember Monica
    Mar 6, 2020 at 14:06
  • $\begingroup$ @Mindwin but the process that makes the iron glow red is the same that makes the sun shine: black body radiation. $\endgroup$
    – lvella
    Mar 6, 2020 at 18:21
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    $\begingroup$ Maybe a simpler way: "dark" implies no photons. There are lots and lots of photons, so it cannot be dark. $\endgroup$
    – Carl Witthoft
    Mar 6, 2020 at 18:28
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    $\begingroup$ At night it is. $\endgroup$
    – void_ptr
    Mar 6, 2020 at 20:37

2 Answers 2

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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.

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    $\begingroup$ 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. $\endgroup$
    – uhoh
    Mar 5, 2020 at 22:18
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    $\begingroup$ @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. $\endgroup$
    – ProfRob
    Mar 5, 2020 at 23:51
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    $\begingroup$ @uhoh Or perhaps you could ask your questions as questions (on Physics SE). $\endgroup$
    – ProfRob
    Mar 6, 2020 at 7:59
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    $\begingroup$ 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? $\endgroup$
    – Swike
    Mar 6, 2020 at 11:46
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    $\begingroup$ 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.) $\endgroup$
    – Ilmari Karonen
    Mar 6, 2020 at 12:06
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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 outside the room. If you turn on a flashlight in the room, it is no longer dark but it is still opaque to the outside world.

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