<p>Coming from a different direction as @Rob's, <em>Opacity</em> and <em>Thermal Radiation</em> are orthogonal properties of a material. The photon flux at the interior of the sun is very high, so it is definitely not <em>dark</em>. However, it is <em>opaque</em> to virtually all light outside the sun.</p>

<p>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.</p>

CC BY-SA 4.0
2020-03-06T22:15:04.097Z
2020-03-06T22:15:04.097Z
MooseBoys
<p>No, it's not. The radiation field in the interior of the Sun is very close to a blackbody spectrum.</p>
<p>If you look in any particular direction the brightness (power per unit area) you see is <span class="math-container">$\sigma T^4$</span>, where <span class="math-container">$\sigma$</span> 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.</p>
<p>Given that the interior temperature might be <span class="math-container">$10^7\ \mathrm K$</span>, then the surface brightness is <span class="math-container">$5.7 \times 10^{20}\ \mathrm{W/m^2}$</span>, compared with the <span class="math-container">$1400\ \mathrm{W/m^2}$</span> you would get by looking directly at the Sun (<strong>please don't do this</strong>). 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).</p>
<p>A possible source of confusion is this term &quot;opacity&quot;. 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.</p>
<p><strong>Details for interest:</strong></p>
<p>The opacity, <span class="math-container">$\kappa$</span> in the solar interior ranges from 1 cm<span class="math-container">$^2$</span> g at the centre to about <span class="math-container">$10^5$</span> cm<span class="math-container">$^2$</span> g just below the photosphere. To estimate the mean free path of photons we need to multiply this by the density <span class="math-container">$\rho$</span> and take the reciprocal:
<span class="math-container">$$ \bar{l} = \frac{1}{\kappa \rho}\ .$$</span>
The density varies from 160 g/cm<span class="math-container">$^3$</span> at the centre to about 0.001 g/cm<span class="math-container">$^3$</span> 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).</p>
<p>Thus your &quot;view&quot; of the stellar interior is of a foggy sphere with radius of no more than a few times <span class="math-container">$\bar{l}$</span>. The fog however is tremendously bright - as outlined above.</p>
<p>The brightness at particular wavelengths is proportional to the Planck function
<span class="math-container">$$B_\lambda = \frac{2hc^2}{\lambda^5} \left(\frac{1}{\exp(hc/\lambda k_B T) -1}\right).$$</span></p>
<p>Thus at <span class="math-container">$\lambda=500$</span> nm (visible light), the ratio of brightness for blackbodies at <span class="math-container">$10^7$</span> K (solar interior) to 6000 K (solar photosphere) is <span class="math-container">$4.2\times 10^{4}$</span>. i.e. Even just considered at visible wavelengths, the interior of the Sun is about 40,000 times brighter than the photosphere.</p>

CC BY-SA 4.0
2020-03-05T22:18:21.863Z
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](https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law) have to be adjusted somehow to account for this? btw [Is there a blackbody spectrum of photons inside a solid?](https://physics.stackexchange.com/q/479644/83380) needs a better answer and [Is the visible light spectrum from “red-hot glass” at least close to Blackbody Radiation?](https://physics.stackexchange.com/q/254099/83380) could use some review as well.
uhoh
CC BY-SA 4.0
2020-03-05T22:42:12.850Z
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
CC BY-SA 4.0
2020-03-05T23:47:17.567Z
@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.
ProfRob
CC BY-SA 4.0
2020-03-05T23:51:53.987Z
@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.
ProfRob
CC BY-SA 4.0
2020-03-06T00:20:54.527Z
An Etalon is not a good analogy, but it may help to just clarify what I mean by "build up". Perhaps an [integrating sphere](https://en.wikipedia.org/wiki/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.
uhoh
CC BY-SA 4.0
2020-03-06T00:28:09.897Z
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
CC BY-SA 4.0
2020-03-06T01:17:45.060Z
@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.
ProfRob
CC BY-SA 4.0
2020-03-06T03:30:47.450Z
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
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2020-03-06T07:59:51.390Z
@uhoh Or perhaps you could ask your questions as questions (on Physics SE).
ProfRob
CC BY-SA 4.0
2020-03-06T11:46:43.143Z
@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.
Swike
CC BY-SA 4.0
2020-03-06T11:46:48.923Z
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?
Swike
CC BY-SA 4.0
2020-03-06T12:06:10.110Z
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.)
Ilmari Karonen
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2020-03-06T13:33:52.523Z
@swike Correct.
ProfRob
CC BY-SA 4.0
2020-03-06T13:57:43.883Z
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.
Peter - Reinstate Monica
CC BY-SA 4.0
2020-03-07T01:43:27.230Z
"compared with the 1400 W/m2 you get by looking directly at the Sun." **Don't look directly at the Sun!**
CJ Dennis
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2020-03-09T12:31:12.987Z
@CJDennis Indeed. Warning added.
ProfRob
CC BY-SA 4.0
2020-03-05T20:27:23.243Z
2020-08-20T16:00:16.733Z
2020-08-20T16:00:16.733Z
ProfRob
<p>This might sound like a strange question, but something got me thinking about it recently.</p>
<p>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.</p>
<p>My questions are:</p>
<ul>
<li>Is this line of reasoning correct?</li>
<li>Are these conditions actually possible inside a star?</li>
<li>Where exactly inside a star are these conditions possible?</li>
</ul>
<hr />
<p><strong>The answer</strong>
ProfRob gave an excellent answer, as always, but in case someone wants a plain language summary we talked about this:</p>
<p>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.</p>
<p>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.</p>
<p>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.</p>
<p>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).</p>

CC BY-SA 4.0
2020-03-06T11:25:28.747Z
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?
lvella
CC BY-SA 4.0
2020-03-06T14:06:56.863Z
@ivella dunking in glowing luminol might be a better (and less painful) parallel than hammering into an iron bar.
Mindwin Remember Monica
CC BY-SA 4.0
2020-03-06T18:21:50.823Z
@Mindwin but the process that makes the iron glow red is the same that makes the sun shine: black body radiation.
lvella
CC BY-SA 4.0
2020-03-06T18:28:16.653Z
Maybe a simpler way:  "dark" implies no photons.  There are lots and **lots** of photons, so it cannot be  dark.
Carl Witthoft
CC BY-SA 4.0
2020-03-06T20:37:35.907Z
At night it is.
void_ptr
CC BY-SA 4.0
2020-03-06T23:09:54.127Z
Blackbody radiation is not a "process".
ProfRob
CC BY-SA 4.0
2020-03-09T20:47:42.487Z
"dark" implies no photons ...I would have thought that "dark" implies no photons in the visible spectrum
Mike H
<the-sun><stellar-astrophysics><plasma-physics><opacity>
Is it dark inside the Sun?
Swike
# Is it dark inside the Sun?

(Asked by: username_0 on 2020-03-05)

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**
username_5 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).

**Tags:** the-sun, stellar-astrophysics, plasma-physics, opacity

> **Comments:**
> **username_1**: 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?  
> **username_2**: @ivella dunking in glowing luminol might be a better (and less painful) parallel than hammering into an iron bar.  
> **username_1**: @Mindwin but the process that makes the iron glow red is the same that makes the sun shine: black body radiation.  
> **username_3**: Maybe a simpler way:  "dark" implies no photons.  There are lots and **lots** of photons, so it cannot be  dark.  
> **username_4**: At night it is.  
> **username_5**: Blackbody radiation is not a "process".  
> **username_6**: "dark" implies no photons ...I would have thought that "dark" implies no photons in the visible spectrum  

---

## 2 answers

### Answer by username_5 on 2020-03-05. Score: 76 | Accepted answer

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.

> **Comments:**
> **username_8**: 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](https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law) have to be adjusted somehow to account for this? btw [Is there a blackbody spectrum of photons inside a solid?](https://physics.stackexchange.com/q/479644/83380) needs a better answer and [Is the visible light spectrum from “red-hot glass” at least close to Blackbody Radiation?](https://physics.stackexchange.com/q/254099/83380) could use some review as well.  
> **username_8**: 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...  
> **username_5**: @username_8 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.  
> **username_5**: @username_8 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.  
> **username_8**: An Etalon is not a good analogy, but it may help to just clarify what I mean by "build up". Perhaps an [integrating sphere](https://en.wikipedia.org/wiki/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.  
> **username_8**: 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!  
> **username_5**: @username_8 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.  
> **username_8**: 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.  
> **username_5**: @username_8 Or perhaps you could ask your questions as questions (on Physics SE).  
> **username_0**: @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.  
> **username_0**: 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?  
> **username_9**: 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.)  
> **username_5**: @swike Correct.  
> **username_10**: 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.  
> **username_11**: "compared with the 1400 W/m2 you get by looking directly at the Sun." **Don't look directly at the Sun!**  
> **username_5**: @CJDennis Indeed. Warning added.  

---

### Answer by username_7 on 2020-03-06. Score: 18

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.


---

