<p>The atmospheric layer that produces the absorption lines acts somewhat like a mirror at these frequencies and scatters the light back into the sun (although this is diffuse reflection not specular reflection like an actual mirror). In principle, light is scattered also outwards (with a probability of 1/2 for each scattering event), but since the layer is very dense at the line frequencies it takes many scattering events to get trough. After two scattering events it would only be a fraction 1/2 *1/2 =1/4,  after three 1/2 *1/2 *1/2 =1/8 and so on (this is just to demonstrate the principle, in reality it is bit more complicated due to multiple scattering back and forth in the layer). There are so many scattering events required that very little is getting through. It is being all scattered back into the lower layers of the atmosphere where it is eventually converted to photons of different frequencies.</p>
<p>It is a bit similar to why you have little light from the sun here on our earth under a dense cloud layer compared to a clear sky. If you have ever been in an airplane 5 miles high above the clouds, you realize that this light missing under the clouds is in fact reflected back from the top into space, making the clouds appear blindingly white. It is just the reverse situation in the solar atmosphere (if you could take a spectrum from below the layer responsible for the Fraunhofer lines looking upwards, you would see those lines all in emission)</p>
<p>Edit:
The following diagram (taken from <a href="https://courses.lumenlearning.com/astronomy/chapter/formation-of-spectral-lines/" rel="nofollow noreferrer">https://courses.lumenlearning.com/astronomy/chapter/formation-of-spectral-lines/</a> ) illustrates what happens here
<a href="https://i.sstatic.net/K4B8U.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/K4B8U.jpg" alt="enter image description here" /></a></p>
<p>The specific difference is here only that the geometry of the scattering layer is different, being more like an infinitely extended vertical plane layer than kind of cylindrical. So in this case you can see the emission line (bright line) spectrum only from underneath the solar layer producing absorption lines when looking upwards (this is the emission the OP was missing in the absorption spectrum). In all other directions, you see (for obvious geometrical reasons) always the continuum source behind (which you have to assume as an extended plane layer as well) and thus the absorption spectrum.</p>
<p>Edit 2:
Note that the accepted answer above is incorrect. It claims to describe the scattering of radiation, but the quoted equation effectively neglects the scattering source term when associating the source term later on with the thermal black-body term in order to bring in the temperature argument here. The correct equation is (see <a href="http://irina.eas.gatech.edu/EAS8803_Fall2017/petty_11.pdf" rel="nofollow noreferrer">http://irina.eas.gatech.edu/EAS8803_Fall2017/petty_11.pdf</a> )
<a href="https://i.sstatic.net/snNiZ.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/snNiZ.jpg" alt="enter image description here" /></a>
Note that <span class="math-container">$\beta_e$</span> is here the combined absorption/scattering coefficient going into the loss term (with the minus sign), and <span class="math-container">$\tilde\omega=\beta_s/\beta_e=\beta_s/(\beta_a+\beta_s)$</span> is the relative contribution of scattering to the absorption coefficient. This means for pure scattering we have <span class="math-container">$\tilde\omega=1$</span> and the thermal black-body radiation term vanishes. The temperature argument given in the accepted answer above is therefore not applicable in this case. It is clear from this that the thermal emission is only related to the continuum absorption, which however a) is negligible in the visible region above the photosphere and b) can not produce absorption lines anyway, whether there is temperature gradient or not.</p>
<p>So absorption lines can only be produced by resonance scattering, as already qualitatively explained by the colour illustration above. I have made in this respect some explicit numerical calculation with my own radiative transfer program reproduced at <a href="https://www.plasmaphysics.org.uk/programs/plantrans.htm" rel="nofollow noreferrer">https://www.plasmaphysics.org.uk/programs/plantrans.htm</a> , modified somewhat to show the actual line profile rather than frequency integrated intensities.</p>
<p>This is what you get from a mono-directional continuum source falling from one side onto an isothermal purely scattering plane-parallel layer with a line center optical depth <span class="math-container">$\tau$</span>=10 (assuming a Doppler (Gausian) scattering emissivity) for the transmitted line at the other end (looking vertically into the layer and including the continuum source)</p>
<p>Transmitted <span class="math-container">$\tau$</span>=10
<a href="https://i.sstatic.net/gydzt.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/gydzt.jpg" alt="" /></a></p>
<p>and this is what is being vertically reflected back to the continuum source</p>
<p>Reflected <span class="math-container">$\tau$</span>=10
<a href="https://i.sstatic.net/OynV4.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/OynV4.jpg" alt="enter image description here" /></a></p>
<hr />
<p>Here is the same for an optical depth <span class="math-container">$\tau$</span>=100 instead</p>
<p>Transmitted <span class="math-container">$\tau$</span>=100
<a href="https://i.sstatic.net/1Bhrq.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/1Bhrq.jpg" alt="enter image description here" /></a></p>
<p>Reflected <span class="math-container">$\tau$</span>=100
<a href="https://i.sstatic.net/lzdhw.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/lzdhw.jpg" alt="enter image description here" /></a></p>
<p>If one looks at the actual numerical scale of the graphs, it is obvious that the amount reflected back does not fully explain the amount missing from the continuum on the other side. This is simply due to the fact these plots hold for a fixed (vertical) viewing direction only and are furthermore normalized to a solid angle of 1 steradian (which is only 1/2/<span class="math-container">$\pi$</span> of the full half-space the radiation is scattered back into). If one would add up the back-scattered radiation over the complete half-space, taking also into account that the line shape and intensity varies with the viewing direction, it would exactly account for the radiation that is missing in the transmitted spectrum. The question the OP had can only be answered in this way.</p>

CC BY-SA 4.0
2020-09-13T20:31:52.280Z
This isn't correct.  Outward scattered light will escape from layers where the photospheric absorption lines are formed. That is what the photosphere is. The last part is correct, but happens because of the temperature gradient, not because anything is acting like a mirror.
ProfRob
CC BY-SA 4.0
2020-09-14T02:00:34.037Z
In the end half of the photons will end up going out, half will end up going deeper into hotter layers and the energy likely comes back in a different form.
Loren Pechtel
CC BY-SA 4.0
2020-09-14T19:00:23.017Z
@Rob Jefferies If you have an atmospheric layer that is sufficiently dense, its albedo (reflectivity) will be close to 1. Any light falling on it from one side will effectively be reflected like from a mirror. Very little will come through to the other side. In this case the reflectivity is frequency dependent, being very small in the continuous part of the spectrum but very high at the line frequencies (because the atoms have resonances there). Therefore the latter won't get through but are reflected back. Temperature gradients have nothing to do with it.  // See also my edited post above
Thomas
CC BY-SA 4.0
2020-09-14T19:04:11.453Z
@Loren Pechtel There is much more than just half of the photons missing in the absorption lines
Thomas
CC BY-SA 4.0
2020-09-14T20:18:21.317Z
Perhaps you should think about how heat is transported by radiation. It certainly doesn't happen by reflecting photons back into the Sun. It happens by more radiation travelling outwards than inwards.  The Sun is totally opaque at all wavelengths just below the photosphere, so by your reasoning light would never escape. The analogy to clouds is quite l false.
ProfRob
CC BY-SA 4.0
2020-09-15T21:52:00.730Z
@Rob Jeffries Below the photosphere you won't have any absorption lines anymore as you won't have any neutral atoms anymore. Anyway, the lines constitute only a small fraction of the solar spectrum, and the light that does not manage to get out within the lines will eventually be re-emitted at frequencies where they can get out easier (there is actually nothing as 'totally opaque', all the radiation that is internally produced will get out; the internal radiation density will build up until this is the case.).
Thomas
CC BY-SA 4.0
2020-09-16T18:44:13.810Z
The layer where the line is formed is *less* dense than what lies beneath it  An outward scattered photon has a longer mean free path than an inward scattered photon and is more likely, in fact quite likely, to escape. The analogy with clouds is not appropriate. The air above the cloud is not (I) hotter than the cloud, (II) denser than the cloud, which is the situation for light travelling outwards in the solar photosphere.
ProfRob
CC BY-SA 4.0
2020-09-16T18:45:45.930Z
The -1 is for the comment that temperature gradients have nothing to do with the formation of absorption lines. The only spectrum that can emerge from a uniform temperature body in thermal equilibrium is a blackbody spectrum.
ProfRob
CC BY-SA 4.0
2020-10-03T16:05:43.287Z
$\sigma J$ *is* the scattering source term and is *entirely* equivalent to the integral in your equation for the case of isotropic scattering. $J =\int I d\Omega /4\pi$ and $p=1$. Perhaps you should read the definitions of the terms used.
ProfRob
CC BY-SA 4.0
2020-10-04T08:25:55.620Z
@RobJeffries - Have a look again at the equation I quoted: for pure scattering $\omega$=1 and the thermal black-body emission term vanishes, so your temperature argument can not be applied anymore. The source term is only due to multiple scattered external radiation in this case.
Thomas
CC BY-SA 4.0
2020-10-04T16:10:02.820Z
Omega is solid angle. What are you talking about!? Your integral is over solid angle. Ignoring that typo, if you remove the (undefined) term involving "B", then what is left is identical to what I've written for the case of isotropic scattering.
ProfRob
CC BY-SA 4.0
2020-09-13T19:57:52.230Z
2020-10-05T17:17:03.093Z
2020-10-05T17:17:03.093Z
Thomas
<p>Possibly you are under the misapprehension that the number of photons is a conserved quantity? That isn't true, there are more photons at any given wavelength when you are deeper into the star because there is a <em>temperature gradient</em>. Cooler material further out is less emissive because fewer atoms are in excited states.</p>
<p><strong>The temperature gradient is responsible for the formation of absorption lines</strong>. If the photosphere of the Sun were at a single temperature then we would see a perfect blackbody spectrum, for the reasons you outline.</p>
<p>The filling in of absorption by scattering would only take place if the radiation field that the atoms were in were isotropic. But it isn't isotropic because of the temperature gradient.</p>
<p>A much better way to think about the spectrum of a star is to imagine that you can see to a wavelength-dependent depth into the star. Where there is a strong atomic absorption feature, you cannot see very far into the star at that wavelength.</p>
<p>Since the star gets hotter the deeper you go into it, and the emissivity scales as <span class="math-container">$T^4$</span>, then the deeper we can see into the star, the brighter it will appear at that wavelength (and vice-versa).</p>
<p>EDIT:</p>
<p>More formally. The radiative transfer equation, if you want to consider the absorption and remission as some sort of scattering process, would be
(e.g, equation 1.86 of the canonical textbook <a href="http://minerva.ufsc.br/%7Eandre/George%20B.%20Rybicki,%20Alan%20P.%20Lightman-Radiative%20Processes%20in%20Astrophysics-Wiley-VCH%20(1985).pd" rel="nofollow noreferrer">&quot;Radiative Processes in Astrophysics&quot;</a>, 1979, Rybicki &amp; Lightman or any book detailing radiative transfer)
<span class="math-container">$$\frac{dI_{\nu}}{ds} = -\sigma_\nu I_{\nu} + \sigma_\nu J_\nu\ ,$$</span>
where <span class="math-container">$I$</span> is the specific intensity in the solar photosphere (in this case, directed towards the Earth), <span class="math-container">$J$</span> is the mean specific intensity at a point in the solar photosphere averaged over all directions (i.e. <span class="math-container">$J = \int I d\Omega/4\pi$</span>, where <span class="math-container">$\Omega$</span> is solid angle), <span class="math-container">$\sigma$</span> is the scattering coefficient (assumed to be isotropic) and <span class="math-container">$ds$</span> is a piece of pathlength towards the observer. The <span class="math-container">$\nu$</span> subscript just indicates everything is wavelength/frequency dependent.</p>
<p>To avoid creating an absorption or emission line then <span class="math-container">$dI_\nu/ds$</span> must equal zero (i.e. nothing is added or subtracted from the beam of light).</p>
<p>This will only happen if <span class="math-container">$I_\nu = J_\nu$</span>, which would require that the  specific intensity averaged over all directions is equal to the specific intensity emerging from the Sun and heading towards the observer. This will only be true if the radiation field is isotropic and equal to <span class="math-container">$I_\nu$</span> in all directions.</p>
<p>Whilst this would be true for a blackbody radiation field at a set temperature,
it isn't true in the solar photosphere. The specific intensity heading towards us (generally outwards) is always larger than the specific intensity heading away (generally inwards and true regardless of which portion of the visible solar disc is considered) because of the temperature gradient in the photosphere, which means it is hotter further into the interior. That means that <span class="math-container">$I_\nu$</span> is always greater than <span class="math-container">$J_\nu$</span> and hence <span class="math-container">$dI_\nu/ds &lt; 0$</span> and we have net absorption.</p>

CC BY-SA 4.0
2020-09-21T12:09:17.477Z
Comments are not for extended discussion; this conversation has been [moved to chat](https://chat.stackexchange.com/rooms/113247/discussion-on-answer-by-rob-jeffries-problem-regarding-the-absorption-lines-of-t).
called2voyage
CC BY-SA 4.0
2023-02-18T14:25:04.953Z
@ProfRob So the existence of absorption lines due to intergalactic dust and cold gas clouds also has to be related to a necessarily existing temperature gradient in those?
trynerror
CC BY-SA 4.0
2023-02-18T15:24:43.117Z
@trynerror they exist because they are illuminated by hotter things. The temperature gradient is between the illuminating objects and the absorbing objects. In a star, the illuminating object is a hotter layer beneath a colder layer.
ProfRob
CC BY-SA 4.0
2020-09-13T13:35:16.273Z
2023-02-18T15:38:05.050Z
2023-02-18T15:38:05.050Z
ProfRob
<p>Some of the wavelengths of light that are emitted from the Sun will be absorbed by atoms in the outer layer of the Sun and also the atmosphere of the Sun, and we see this as absorption lines in the spectrum. Now, this absorbed radiation will indeed be re-emitted again, so one might think that thIs emission should &quot;cancel out&quot; the absorption lines. The usual explanation for why this doesn't happen is that the re-emitted light is radiated in all directions, not just towards us, meaning that to us these wavelengths will be much fainter than the other wavelengths.</p>
<p>But the problem I have is that this happens all around the Sun (since the atmosphere is completely surrounding it), and intuitively it seems then that all of this re-emitted light should combine such that far away it would appear that the Sun is radiating at these wavelengths just as it is radiating at all the other wavelengths. And if that is true, then we shouldn't see absorption lines in the spectrum. So what is it that I am missing?</p>

CC BY-SA 4.0
2023-04-23T20:39:48.207Z
@ProfRob I don't see it happen often, but because of your "Star" badge your single vote to close resulted in a closing as duplicate of [this question](https://astronomy.stackexchange.com/q/53513/7982). See [screenshot](https://i.sstatic.net/vGTLh.png). That's fine, I just though you might not know you had this "superpower" :-)
uhoh
<the-sun><astrophysics><light><spectroscopy><radiation>
Problem regarding the absorption lines of the Sun
User3141
# Problem regarding the absorption lines of the Sun

(Asked by: username_0 on 2020-09-13)

Some of the wavelengths of light that are emitted from the Sun will be absorbed by atoms in the outer layer of the Sun and also the atmosphere of the Sun, and we see this as absorption lines in the spectrum. Now, this absorbed radiation will indeed be re-emitted again, so one might think that thIs emission should "cancel out" the absorption lines. The usual explanation for why this doesn't happen is that the re-emitted light is radiated in all directions, not just towards us, meaning that to us these wavelengths will be much fainter than the other wavelengths.

But the problem I have is that this happens all around the Sun (since the atmosphere is completely surrounding it), and intuitively it seems then that all of this re-emitted light should combine such that far away it would appear that the Sun is radiating at these wavelengths just as it is radiating at all the other wavelengths. And if that is true, then we shouldn't see absorption lines in the spectrum. So what is it that I am missing?

**Tags:** the-sun, astrophysics, light, spectroscopy, radiation
> **Comments:**
> **username_1**: @ProfRob I don't see it happen often, but because of your "Star" badge your single vote to close resulted in a closing as duplicate of [this question](https://astronomy.stackexchange.com/q/53513/7982). See [screenshot](https://i.sstatic.net/vGTLh.png). That's fine, I just though you might not know you had this "superpower" :-)  

---

## 2 answers

### Answer by username_3 on 2020-09-13. Score: 18 | Accepted answer

Possibly you are under the misapprehension that the number of photons is a conserved quantity? That isn't true, there are more photons at any given wavelength when you are deeper into the star because there is a *temperature gradient*. Cooler material further out is less emissive because fewer atoms are in excited states.

**The temperature gradient is responsible for the formation of absorption lines**. If the photosphere of the Sun were at a single temperature then we would see a perfect blackbody spectrum, for the reasons you outline.

The filling in of absorption by scattering would only take place if the radiation field that the atoms were in were isotropic. But it isn't isotropic because of the temperature gradient.

A much better way to think about the spectrum of a star is to imagine that you can see to a wavelength-dependent depth into the star. Where there is a strong atomic absorption feature, you cannot see very far into the star at that wavelength.

Since the star gets hotter the deeper you go into it, and the emissivity scales as $T^4$, then the deeper we can see into the star, the brighter it will appear at that wavelength (and vice-versa).

EDIT:

More formally. The radiative transfer equation, if you want to consider the absorption and remission as some sort of scattering process, would be
(e.g, equation 1.86 of the canonical textbook ["Radiative Processes in Astrophysics"](http://minerva.ufsc.br/%7Eandre/George%20B.%20Rybicki,%20Alan%20P.%20Lightman-Radiative%20Processes%20in%20Astrophysics-Wiley-VCH%20(1985).pd), 1979, Rybicki & Lightman or any book detailing radiative transfer)
$$\frac{dI\_{\nu}}{ds} = -\sigma\_\nu I\_{\nu} + \sigma\_\nu J\_\nu\ ,$$
where $I$ is the specific intensity in the solar photosphere (in this case, directed towards the Earth), $J$ is the mean specific intensity at a point in the solar photosphere averaged over all directions (i.e. $J = \int I d\Omega/4\pi$, where $\Omega$ is solid angle), $\sigma$ is the scattering coefficient (assumed to be isotropic) and $ds$ is a piece of pathlength towards the observer. The $\nu$ subscript just indicates everything is wavelength/frequency dependent.

To avoid creating an absorption or emission line then $dI\_\nu/ds$ must equal zero (i.e. nothing is added or subtracted from the beam of light).

This will only happen if $I\_\nu = J\_\nu$, which would require that the specific intensity averaged over all directions is equal to the specific intensity emerging from the Sun and heading towards the observer. This will only be true if the radiation field is isotropic and equal to $I\_\nu$ in all directions.

Whilst this would be true for a blackbody radiation field at a set temperature,
it isn't true in the solar photosphere. The specific intensity heading towards us (generally outwards) is always larger than the specific intensity heading away (generally inwards and true regardless of which portion of the visible solar disc is considered) because of the temperature gradient in the photosphere, which means it is hotter further into the interior. That means that $I\_\nu$ is always greater than $J\_\nu$ and hence $dI\_\nu/ds < 0$ and we have net absorption.

> **Comments:**
> **username_5**: Comments are not for extended discussion; this conversation has been [moved to chat](https://chat.stackexchange.com/rooms/113247/discussion-on-answer-by-rob-jeffries-problem-regarding-the-absorption-lines-of-t).  
> **username_6**: @ProfRob So the existence of absorption lines due to intergalactic dust and cold gas clouds also has to be related to a necessarily existing temperature gradient in those?  
> **username_3**: @trynerror they exist because they are illuminated by hotter things. The temperature gradient is between the illuminating objects and the absorbing objects. In a star, the illuminating object is a hotter layer beneath a colder layer.  

---

### Answer by username_2 on 2020-09-13. Score: 1

The atmospheric layer that produces the absorption lines acts somewhat like a mirror at these frequencies and scatters the light back into the sun (although this is diffuse reflection not specular reflection like an actual mirror). In principle, light is scattered also outwards (with a probability of 1/2 for each scattering event), but since the layer is very dense at the line frequencies it takes many scattering events to get trough. After two scattering events it would only be a fraction 1/2 \*1/2 =1/4, after three 1/2 \*1/2 \*1/2 =1/8 and so on (this is just to demonstrate the principle, in reality it is bit more complicated due to multiple scattering back and forth in the layer). There are so many scattering events required that very little is getting through. It is being all scattered back into the lower layers of the atmosphere where it is eventually converted to photons of different frequencies.

It is a bit similar to why you have little light from the sun here on our earth under a dense cloud layer compared to a clear sky. If you have ever been in an airplane 5 miles high above the clouds, you realize that this light missing under the clouds is in fact reflected back from the top into space, making the clouds appear blindingly white. It is just the reverse situation in the solar atmosphere (if you could take a spectrum from below the layer responsible for the Fraunhofer lines looking upwards, you would see those lines all in emission)

Edit:
The following diagram (taken from <https://courses.lumenlearning.com/astronomy/chapter/formation-of-spectral-lines/> ) illustrates what happens here
[![enter image description here](https://i.sstatic.net/K4B8U.jpg)](https://i.sstatic.net/K4B8U.jpg)

The specific difference is here only that the geometry of the scattering layer is different, being more like an infinitely extended vertical plane layer than kind of cylindrical. So in this case you can see the emission line (bright line) spectrum only from underneath the solar layer producing absorption lines when looking upwards (this is the emission the OP was missing in the absorption spectrum). In all other directions, you see (for obvious geometrical reasons) always the continuum source behind (which you have to assume as an extended plane layer as well) and thus the absorption spectrum.

Edit 2:
Note that the accepted answer above is incorrect. It claims to describe the scattering of radiation, but the quoted equation effectively neglects the scattering source term when associating the source term later on with the thermal black-body term in order to bring in the temperature argument here. The correct equation is (see <http://irina.eas.gatech.edu/EAS8803_Fall2017/petty_11.pdf> )
[![enter image description here](https://i.sstatic.net/snNiZ.jpg)](https://i.sstatic.net/snNiZ.jpg)
Note that $\beta\_e$ is here the combined absorption/scattering coefficient going into the loss term (with the minus sign), and $\tilde\omega=\beta\_s/\beta\_e=\beta\_s/(\beta\_a+\beta\_s)$ is the relative contribution of scattering to the absorption coefficient. This means for pure scattering we have $\tilde\omega=1$ and the thermal black-body radiation term vanishes. The temperature argument given in the accepted answer above is therefore not applicable in this case. It is clear from this that the thermal emission is only related to the continuum absorption, which however a) is negligible in the visible region above the photosphere and b) can not produce absorption lines anyway, whether there is temperature gradient or not.

So absorption lines can only be produced by resonance scattering, as already qualitatively explained by the colour illustration above. I have made in this respect some explicit numerical calculation with my own radiative transfer program reproduced at <https://www.plasmaphysics.org.uk/programs/plantrans.htm> , modified somewhat to show the actual line profile rather than frequency integrated intensities.

This is what you get from a mono-directional continuum source falling from one side onto an isothermal purely scattering plane-parallel layer with a line center optical depth $\tau$=10 (assuming a Doppler (Gausian) scattering emissivity) for the transmitted line at the other end (looking vertically into the layer and including the continuum source)

Transmitted $\tau$=10
[![](https://i.sstatic.net/gydzt.jpg)](https://i.sstatic.net/gydzt.jpg)

and this is what is being vertically reflected back to the continuum source

Reflected $\tau$=10
[![enter image description here](https://i.sstatic.net/OynV4.jpg)](https://i.sstatic.net/OynV4.jpg)

---

Here is the same for an optical depth $\tau$=100 instead

Transmitted $\tau$=100
[![enter image description here](https://i.sstatic.net/1Bhrq.jpg)](https://i.sstatic.net/1Bhrq.jpg)

Reflected $\tau$=100
[![enter image description here](https://i.sstatic.net/lzdhw.jpg)](https://i.sstatic.net/lzdhw.jpg)

If one looks at the actual numerical scale of the graphs, it is obvious that the amount reflected back does not fully explain the amount missing from the continuum on the other side. This is simply due to the fact these plots hold for a fixed (vertical) viewing direction only and are furthermore normalized to a solid angle of 1 steradian (which is only 1/2/$\pi$ of the full half-space the radiation is scattered back into). If one would add up the back-scattered radiation over the complete half-space, taking also into account that the line shape and intensity varies with the viewing direction, it would exactly account for the radiation that is missing in the transmitted spectrum. The question the OP had can only be answered in this way.

> **Comments:**
> **username_3**: This isn't correct.  Outward scattered light will escape from layers where the photospheric absorption lines are formed. That is what the photosphere is. The last part is correct, but happens because of the temperature gradient, not because anything is acting like a mirror.  
> **username_4**: In the end half of the photons will end up going out, half will end up going deeper into hotter layers and the energy likely comes back in a different form.  
> **username_2**: @Rob Jefferies If you have an atmospheric layer that is sufficiently dense, its albedo (reflectivity) will be close to 1. Any light falling on it from one side will effectively be reflected like from a mirror. Very little will come through to the other side. In this case the reflectivity is frequency dependent, being very small in the continuous part of the spectrum but very high at the line frequencies (because the atoms have resonances there). Therefore the latter won't get through but are reflected back. Temperature gradients have nothing to do with it.  // See also my edited post above  
> **username_2**: @Loren Pechtel There is much more than just half of the photons missing in the absorption lines  
> **username_3**: Perhaps you should think about how heat is transported by radiation. It certainly doesn't happen by reflecting photons back into the Sun. It happens by more radiation travelling outwards than inwards.  The Sun is totally opaque at all wavelengths just below the photosphere, so by your reasoning light would never escape. The analogy to clouds is quite l false.  
> **username_2**: @Rob Jeffries Below the photosphere you won't have any absorption lines anymore as you won't have any neutral atoms anymore. Anyway, the lines constitute only a small fraction of the solar spectrum, and the light that does not manage to get out within the lines will eventually be re-emitted at frequencies where they can get out easier (there is actually nothing as 'totally opaque', all the radiation that is internally produced will get out; the internal radiation density will build up until this is the case.).  
> **username_3**: The layer where the line is formed is *less* dense than what lies beneath it  An outward scattered photon has a longer mean free path than an inward scattered photon and is more likely, in fact quite likely, to escape. The analogy with clouds is not appropriate. The air above the cloud is not (I) hotter than the cloud, (II) denser than the cloud, which is the situation for light travelling outwards in the solar photosphere.  
> **username_3**: The -1 is for the comment that temperature gradients have nothing to do with the formation of absorption lines. The only spectrum that can emerge from a uniform temperature body in thermal equilibrium is a blackbody spectrum.  
> **username_3**: $\sigma J$ *is* the scattering source term and is *entirely* equivalent to the integral in your equation for the case of isotropic scattering. $J =\int I d\Omega /4\pi$ and $p=1$. Perhaps you should read the definitions of the terms used.  
> **username_2**: @RobJeffries - Have a look again at the equation I quoted: for pure scattering $\omega$=1 and the thermal black-body emission term vanishes, so your temperature argument can not be applied anymore. The source term is only due to multiple scattered external radiation in this case.  
> **username_3**: Omega is solid angle. What are you talking about!? Your integral is over solid angle. Ignoring that typo, if you remove the (undefined) term involving "B", then what is left is identical to what I've written for the case of isotropic scattering.  

---

