<p>What you're missing is that the resolution of a prism isn't high enough to resolve the relatively narrow spectral lines. What's more, the light that gets generated by fusion reactions doesn't reach the surface of the sun for a very long time, and it get's scattered and split a large number of times along the way, removing any signature of fusion from their origins. The reason we can be confident that fusion is happening in there is because we can detect <A HREF="https://en.wikipedia.org/wiki/Solar_neutrino" rel="noreferrer">solar neutrinos</A> that exit the sun almost completely unimpeded, and have the expected spectrum from the fusion process.</p>

<p>For more, have a look at <A HREF="http://www.astro.cornell.edu/academics/courses/astro201/kirchhoff.htm" rel="noreferrer">Kirchhoff's laws of spectroscopy</A>.</p>

<p>Back to the sun's spectrum. We have a few reasons to be confident in our ability to measure absorption lines of the sun independent of the atmosphere. First, the amount of atmosphere we look at the sun through depends on the angle of the sun in the sky (a concept called <A HREF="https://en.wikipedia.org/wiki/Air_mass_(astronomy)" rel="noreferrer">airmass</A>). So, if we observe the sun at different angles, and keep track of how the absorption changes with airmass, we can extrapolate the spectrum back to zero airmass. Second, there is at least one element that was first discovered by it's absorption lines in the solar spectrum: helium (named for the Greek god of the sun, Helios), so we can be confident that the measurements we're making correspond to physical reality. Third, we have performed measurements using satellites that are above the atmosphere. </p>

CC BY-SA 3.0
2017-10-19T19:18:50.073Z
2017-10-19T19:27:53.537Z
2017-10-19T19:27:53.537Z
Sean Lake
<p>Two questions are asked here. Sean Lake addresses the first - there is nothing unusual about sunlight; it's spectrum contains many dark absorption lines due to various chemical elements. You just need a reasonably high spectral resolution (high dispersion) in order to see them.</p>

<p>The light from the Sun comes from the photosphere. We do not see what is going on in the centre with light. Light that is emitted at the centre of the Sun is re-absorbed almost immediately. It is when photons have a reasonable chance of escaping that defines where the visible surface is. In other words, the photosphere marks the point where (moving inwards) the opacity to light increases rapidly.</p>

<p>This opacity is not the same at all wavelengths. Where there is a spectral line, due to an electron transition in an atom, it is higher. At these wavelengths, photons can only escape if they are emitted higher in the photosphere, at cooler temperatures. Such light is less intense than at other wavelengths, hence we see an "absorption line".</p>

<p>Outside of absorption lines the photospheric opacity must be lower, but non-zero. The main opacity here is caused by bound-free transitions involving the H$^{-}$ ion. Electrons have a continuum of free states they can occupy outside of ions and atoms, so such processes occur over a continuous range of wavelengths. The reverse of these absorption processes (i.e. free-bound emission) is what provides a continuum emission spectrum.</p>

<p>The Sun's spectrum is therefore due to material at a variety of temperatures (that is why it <em>isn't</em> a blackbody). Photons in absorption lines (they are not black, just faint) are emitted at maybe 4500K, whilst photons in the continuum come from hotter regions at perhaps a bit more than 6000K.</p>

<p>If it were not for the H$^{-}$ ion, the Sun would appear slightly smaller, hotter and with deeper absorption lines. But there would still be other bound-free or free-free processes occurring in the deeper, hotter layers of the Sun that would provide sources of continuous absorption and emission.</p>

CC BY-SA 3.0
2017-10-20T06:10:52.747Z
2017-10-20T06:10:52.747Z
ProfRob
<p>When we look at spectral lines in a star's spectrum, we're actually looking for absorption lines, not emission lines. A star's spectrum usually resembles that of a <a href="https://en.wikipedia.org/wiki/Black_body" rel="nofollow noreferrer">black body</a>, continuous and smooth. However, there are elements in the star's atmosphere that absorb some of the emitted light; these create characteristic absorption lines in the spectrum we observe.</p>

<p>There's a reason that a star's spectrum closely resembles that of a black body. H<sup>-</sup> opacity means that some of the solar photosphere is optically thick for a wide range of frequencies. This in turn means that the emission no longer comes from just a few frequencies corresponding to a few elements. This spectrum appears continuous, and therefore like a black body.</p>

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2017-10-19T20:59:04.203Z
@RobJeffries Which part in particular is incorrect? I don't think I really discussed the source of the opacity.
HDE 226868
CC BY-SA 3.0
2017-10-19T23:30:43.040Z
What are you discussing pressure broadening for? The reason the Sun at all resembles a blackbody (it is not that close) is H minus opacity, which means it is optically thick over a broad range of wavelengths *at a similar temperature*.
ProfRob
CC BY-SA 3.0
2017-10-20T00:11:33.097Z
@RobJeffries I suppose I misinterpreted a sentence in my textbook: "In a dense gas, pressure broadening becomes so significant that the medium becomes optically thick at all frequencies, and photons of all energies can be absorbed and emitted". That leads to a spectrum described by the Planck function. I think I incorrectly worked backwards from that to assume that the optical thickness in the *Sun* was due to pressure broadening.
HDE 226868
CC BY-SA 3.0
2017-10-20T05:45:20.797Z
Photospheric pressure is about 0.1 of Earth's atmospheric pressure...
ProfRob
CC BY-SA 3.0
2017-10-19T18:55:40.357Z
2017-10-20T00:50:41.277Z
2017-10-20T00:50:41.277Z
HDE 226868
<p>So the visible light from stars can be used to identify the elements in that star by looking at the spectral emission lines and comparing those lines to emission lines of various elements (and their isotopes) here on earth.  These well-known line wavelengths can also be used to determine the relative motion of that star to the earth by looking at the red (or blue) shift of wavelengths of those spectral lines.  It would then be reasonable to assume that our own sun emits those same well-defined spectral lines, which should show up at specific wavelengths (based on the elements in the sun).  </p>

<p>However, sunlight appears (mostly) white, and using a prism shows a rather consistent rainbow of color across the visible spectrum, rather than what one would expect to see (based on analysis of emission spectrum from other stars) of spikes at specific wavelengths.  So, what is causing the full-spectrum light from the sun to appear at the surface of the earth if what I'm assuming is specific spectral lines being generated from fusion on the sun itself?   </p>

<p>I'm sure I'm missing something in my understanding, but I'm not sure what it is.  I understand (from this question: <a href="https://astronomy.stackexchange.com/questions/18032/how-deep-is-the-a-fraunhofer-line-in-the-solar-spectrum-is-it-from-the-suns">How deep is the &quot;A&quot; Fraunhofer line in the solar spectrum? Is it from the Sun&#39;s or Earth&#39;s atmosphere?</a>) that the earth's atmosphere absorbs specific wavelengths of light, but why would the sun's light be consistent across its visible spectrum if the typical star light is generated at specific spectrum wavelengths?</p>

<the-sun><spectroscopy><sun-rays>
Solar visible light spectrum
Milwrdfan
# Solar visible light spectrum

(Asked by: username_0 on 2017-10-19)

So the visible light from stars can be used to identify the elements in that star by looking at the spectral emission lines and comparing those lines to emission lines of various elements (and their isotopes) here on earth. These well-known line wavelengths can also be used to determine the relative motion of that star to the earth by looking at the red (or blue) shift of wavelengths of those spectral lines. It would then be reasonable to assume that our own sun emits those same well-defined spectral lines, which should show up at specific wavelengths (based on the elements in the sun).

However, sunlight appears (mostly) white, and using a prism shows a rather consistent rainbow of color across the visible spectrum, rather than what one would expect to see (based on analysis of emission spectrum from other stars) of spikes at specific wavelengths. So, what is causing the full-spectrum light from the sun to appear at the surface of the earth if what I'm assuming is specific spectral lines being generated from fusion on the sun itself?

I'm sure I'm missing something in my understanding, but I'm not sure what it is. I understand (from this question: [How deep is the "A" Fraunhofer line in the solar spectrum? Is it from the Sun's or Earth's atmosphere?](https://astronomy.stackexchange.com/questions/18032/how-deep-is-the-a-fraunhofer-line-in-the-solar-spectrum-is-it-from-the-suns)) that the earth's atmosphere absorbs specific wavelengths of light, but why would the sun's light be consistent across its visible spectrum if the typical star light is generated at specific spectrum wavelengths?

**Tags:** the-sun, spectroscopy, sun-rays

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## 3 answers

### Answer by username_1 on 2017-10-19. Score: 7

What you're missing is that the resolution of a prism isn't high enough to resolve the relatively narrow spectral lines. What's more, the light that gets generated by fusion reactions doesn't reach the surface of the sun for a very long time, and it get's scattered and split a large number of times along the way, removing any signature of fusion from their origins. The reason we can be confident that fusion is happening in there is because we can detect [solar neutrinos](https://en.wikipedia.org/wiki/Solar_neutrino) that exit the sun almost completely unimpeded, and have the expected spectrum from the fusion process.

For more, have a look at [Kirchhoff's laws of spectroscopy](http://www.astro.cornell.edu/academics/courses/astro201/kirchhoff.htm).

Back to the sun's spectrum. We have a few reasons to be confident in our ability to measure absorption lines of the sun independent of the atmosphere. First, the amount of atmosphere we look at the sun through depends on the angle of the sun in the sky (a concept called [airmass](https://en.wikipedia.org/wiki/Air_mass_(astronomy))). So, if we observe the sun at different angles, and keep track of how the absorption changes with airmass, we can extrapolate the spectrum back to zero airmass. Second, there is at least one element that was first discovered by it's absorption lines in the solar spectrum: helium (named for the Greek god of the sun, Helios), so we can be confident that the measurements we're making correspond to physical reality. Third, we have performed measurements using satellites that are above the atmosphere.


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### Answer by username_2 on 2017-10-20. Score: 6

Two questions are asked here. Sean Lake addresses the first - there is nothing unusual about sunlight; it's spectrum contains many dark absorption lines due to various chemical elements. You just need a reasonably high spectral resolution (high dispersion) in order to see them.

The light from the Sun comes from the photosphere. We do not see what is going on in the centre with light. Light that is emitted at the centre of the Sun is re-absorbed almost immediately. It is when photons have a reasonable chance of escaping that defines where the visible surface is. In other words, the photosphere marks the point where (moving inwards) the opacity to light increases rapidly.

This opacity is not the same at all wavelengths. Where there is a spectral line, due to an electron transition in an atom, it is higher. At these wavelengths, photons can only escape if they are emitted higher in the photosphere, at cooler temperatures. Such light is less intense than at other wavelengths, hence we see an "absorption line".

Outside of absorption lines the photospheric opacity must be lower, but non-zero. The main opacity here is caused by bound-free transitions involving the H$^{-}$ ion. Electrons have a continuum of free states they can occupy outside of ions and atoms, so such processes occur over a continuous range of wavelengths. The reverse of these absorption processes (i.e. free-bound emission) is what provides a continuum emission spectrum.

The Sun's spectrum is therefore due to material at a variety of temperatures (that is why it *isn't* a blackbody). Photons in absorption lines (they are not black, just faint) are emitted at maybe 4500K, whilst photons in the continuum come from hotter regions at perhaps a bit more than 6000K.

If it were not for the H$^{-}$ ion, the Sun would appear slightly smaller, hotter and with deeper absorption lines. But there would still be other bound-free or free-free processes occurring in the deeper, hotter layers of the Sun that would provide sources of continuous absorption and emission.


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### Answer by username_3 on 2017-10-19. Score: 6

When we look at spectral lines in a star's spectrum, we're actually looking for absorption lines, not emission lines. A star's spectrum usually resembles that of a [black body](https://en.wikipedia.org/wiki/Black_body), continuous and smooth. However, there are elements in the star's atmosphere that absorb some of the emitted light; these create characteristic absorption lines in the spectrum we observe.

There's a reason that a star's spectrum closely resembles that of a black body. H- opacity means that some of the solar photosphere is optically thick for a wide range of frequencies. This in turn means that the emission no longer comes from just a few frequencies corresponding to a few elements. This spectrum appears continuous, and therefore like a black body.

> **Comments:**
> **username_3**: @RobJeffries Which part in particular is incorrect? I don't think I really discussed the source of the opacity.  
> **username_2**: What are you discussing pressure broadening for? The reason the Sun at all resembles a blackbody (it is not that close) is H minus opacity, which means it is optically thick over a broad range of wavelengths *at a similar temperature*.  
> **username_3**: @RobJeffries I suppose I misinterpreted a sentence in my textbook: "In a dense gas, pressure broadening becomes so significant that the medium becomes optically thick at all frequencies, and photons of all energies can be absorbed and emitted". That leads to a spectrum described by the Planck function. I think I incorrectly worked backwards from that to assume that the optical thickness in the *Sun* was due to pressure broadening.  
> **username_2**: Photospheric pressure is about 0.1 of Earth's atmospheric pressure...  

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