<p>It depends on the size of the star's subphotospheric convection zone and the speed of other (non-convective) mixing processes.</p>
<p>Absorption lines are formed in the photosphere, a thin (few hundred km) layer from which almost all the visible and infrared light emerge (at least for stars lower than about 10 solar masses).</p>
<p>Stars below 2 solar masses have an outer convection zone. This is well-mixed with the photosphere on a very short timescale. i.e. Anything you drop onto the star is diluted into the whole convection zone. The fraction of the star's mass in the convection zone ranges from 100% for stars below 0.35 solar masses to a few percent in solar type stars to a tiny percentage in F- and late A-type stars. Above about 2 solar masses the outer part of the star is radiative and not particularly well-mixed.</p>
<p>There are however non-convective mixing mechanisms - diffusion, thermohaline mixing, gravity waves, rotational turbulence that act on slower (many millions of years) timescales.</p>
<p>So the answer is- it's complicated. Dropping a load of heavy elements (even a few Earth masses) on an early F-star or hotter could have an immediate big effect, but would barely register in a cooler star like the Sun or a giant, because it is mixed into the entire convection zone on a timescale of hours to days.</p>

CC BY-SA 4.0
2024-03-17T19:44:24.637Z
2024-03-18T06:15:19.487Z
2024-03-18T06:15:19.487Z
ProfRob
<p>In principle, assuming that there are not already substantial amounts of an element present, there is actually only a relatively small amount of mass required to produce an absorption line. Absorption lines develop when the optical depth in the line center along the whole path &gt;1, so</p>
<p><span class="math-container">$$n\cdot q \cdot h &gt;1$$</span></p>
<p>with <span class="math-container">$n$</span> the number density of the element, <span class="math-container">$q$</span> the resonant scattering cross section in the center of the line (the line is actually caused by scattering not absorption as such), and h the height of the layer. So this yields the condition</p>
<p><span class="math-container">$$n &gt; \frac{1}{q\cdot h}$$</span></p>
<p>Now the resonance cross section for spectral lines in the optical region is of the order of <span class="math-container">$10^{-12} cm^2$</span>. and with the height <span class="math-container">$h$</span> of the solar photosphere about <span class="math-container">$100  km = 10^7 cm$</span>, this means the number density</p>
<p><span class="math-container">$$n&gt;10^5 cm^{-3} = 10^{11} m^{-3}$$</span></p>
<p>If the density is uniform throughout  the solar photosphere, this results in a total required mass of</p>
<p><span class="math-container">$$M &gt; 4\pi \cdot R^2 \cdot h \cdot n \cdot m$$</span></p>
<p>where <span class="math-container">$R$</span> is the radius of the Sun and <span class="math-container">$m$</span> the mass of the atom producing the line, which, if you insert the values, gives</p>
<p><span class="math-container">$$M &gt; 10^8 kg \cdot N$$</span></p>
<p>where N is the atomic mass number of the element. So for instance for sodium (N=23) we would have <span class="math-container">$M&gt;2.3\cdot 10^9 kg$</span>. The Great Pyramid of Giza would roughly have about this mass (although of course it is not made from pure sodium). The mass of a comet is even about 1000 times higher, so should produce noticeable lines for all elements present. This is of course assuming that there are not much higher amounts of these elements already present in the first place. Most Fraunhofer lines have merely an optical depth of the order of 1000 or so though (see <a href="https://babel.aob.rs/151/pdf/013-016.pdf" rel="nofollow noreferrer">https://babel.aob.rs/151/pdf/013-016.pdf</a> ), so on the basis of the estimate above, comets or other interplanetary material falling into the Sun could well make a significant contribution to the observed lines (although I am saying this here just as an interesting thought rather than a substantiated claim).</p>

CC BY-SA 4.0
2024-03-17T23:30:02.020Z
2024-03-18T22:27:42.727Z
2024-03-18T22:27:42.727Z
Thomas
<p>This question was inspired by a recent Cool Worlds video <a href="https://www.youtube.com/watch?v=maMDGZOD3mI" rel="noreferrer">&quot;The Star That Shouldn't Exist&quot;</a>, in which one of the speculative solutions to Przybylski’s Star's strange spectrum is aliens &quot;salting&quot; the star with heavy elements. This has raised much discussion in the comments section about how much one would need to alter the composition of the star in order to alter is spectral absorption lines, with the discussions making the implicit assumption that one would need to alter the composition of the entire star to make these changes. For the star in question; that would apparently require planet-scale masses of these elements, which to my mind removes any plausibility that the &quot;salting&quot; hypothesis might have had.</p>
<p>But is that true? Could it not be possible that a relatively small amount of an element &quot;salted&quot; in the surface of the star would have the same result as changing the composition of the whole star? Naively, I would expect the spectral lines to be &quot;imprinted&quot; on the spectrum in the photosphere, making the composition of the rest of the star largely irrelevant, at least in terms of absorption lines. So what determines the spectral lines: the composition of the whole star, the photosphere, or some other layer(s)?</p>

<star><spectroscopy>
Where in a star are the spectral absorption lines formed?
asgallant
# Where in a star are the spectral absorption lines formed?

(Asked by: username_0 on 2024-03-17)

This question was inspired by a recent Cool Worlds video ["The Star That Shouldn't Exist"](https://www.youtube.com/watch?v=maMDGZOD3mI), in which one of the speculative solutions to Przybylski’s Star's strange spectrum is aliens "salting" the star with heavy elements. This has raised much discussion in the comments section about how much one would need to alter the composition of the star in order to alter is spectral absorption lines, with the discussions making the implicit assumption that one would need to alter the composition of the entire star to make these changes. For the star in question; that would apparently require planet-scale masses of these elements, which to my mind removes any plausibility that the "salting" hypothesis might have had.

But is that true? Could it not be possible that a relatively small amount of an element "salted" in the surface of the star would have the same result as changing the composition of the whole star? Naively, I would expect the spectral lines to be "imprinted" on the spectrum in the photosphere, making the composition of the rest of the star largely irrelevant, at least in terms of absorption lines. So what determines the spectral lines: the composition of the whole star, the photosphere, or some other layer(s)?

**Tags:** star, spectroscopy


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## 2 answers

### Answer by username_1 on 2024-03-17. Score: 10

It depends on the size of the star's subphotospheric convection zone and the speed of other (non-convective) mixing processes.

Absorption lines are formed in the photosphere, a thin (few hundred km) layer from which almost all the visible and infrared light emerge (at least for stars lower than about 10 solar masses).

Stars below 2 solar masses have an outer convection zone. This is well-mixed with the photosphere on a very short timescale. i.e. Anything you drop onto the star is diluted into the whole convection zone. The fraction of the star's mass in the convection zone ranges from 100% for stars below 0.35 solar masses to a few percent in solar type stars to a tiny percentage in F- and late A-type stars. Above about 2 solar masses the outer part of the star is radiative and not particularly well-mixed.

There are however non-convective mixing mechanisms - diffusion, thermohaline mixing, gravity waves, rotational turbulence that act on slower (many millions of years) timescales.

So the answer is- it's complicated. Dropping a load of heavy elements (even a few Earth masses) on an early F-star or hotter could have an immediate big effect, but would barely register in a cooler star like the Sun or a giant, because it is mixed into the entire convection zone on a timescale of hours to days.


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### Answer by username_2 on 2024-03-17. Score: 8

In principle, assuming that there are not already substantial amounts of an element present, there is actually only a relatively small amount of mass required to produce an absorption line. Absorption lines develop when the optical depth in the line center along the whole path >1, so

$$n\cdot q \cdot h >1$$

with $n$ the number density of the element, $q$ the resonant scattering cross section in the center of the line (the line is actually caused by scattering not absorption as such), and h the height of the layer. So this yields the condition

$$n > \frac{1}{q\cdot h}$$

Now the resonance cross section for spectral lines in the optical region is of the order of $10^{-12} cm^2$. and with the height $h$ of the solar photosphere about $100 km = 10^7 cm$, this means the number density

$$n>10^5 cm^{-3} = 10^{11} m^{-3}$$

If the density is uniform throughout the solar photosphere, this results in a total required mass of

$$M > 4\pi \cdot R^2 \cdot h \cdot n \cdot m$$

where $R$ is the radius of the Sun and $m$ the mass of the atom producing the line, which, if you insert the values, gives

$$M > 10^8 kg \cdot N$$

where N is the atomic mass number of the element. So for instance for sodium (N=23) we would have $M>2.3\cdot 10^9 kg$. The Great Pyramid of Giza would roughly have about this mass (although of course it is not made from pure sodium). The mass of a comet is even about 1000 times higher, so should produce noticeable lines for all elements present. This is of course assuming that there are not much higher amounts of these elements already present in the first place. Most Fraunhofer lines have merely an optical depth of the order of 1000 or so though (see <https://babel.aob.rs/151/pdf/013-016.pdf> ), so on the basis of the estimate above, comets or other interplanetary material falling into the Sun could well make a significant contribution to the observed lines (although I am saying this here just as an interesting thought rather than a substantiated claim).


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