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It is said that absorption lines originate from regions higher in the photosphere where the gas is cooler.

The gas should be absorbing photons, then re-emitting them... absorb, re-emit, absorb, re-emit...

My question is where are the photons produced by re-emission? Why are there absorption lines?

Although photons could be scattered, photons at other places could be scattered to our line our sight. Surely the total number of photons does not drop the further out you go? So while photons originally headed in the direction of an observer are possibly absorbed and re-emitted in a direction away from the observer, surely this will be made up for by a photon being emitted from another location in the direction of the observer? Otherwise wouldn't there be a location somewhere else where a bunch of extra scattered photons can be observed?

If we integrate all the photons at the lower surface of the photosphere A and integrated all the photons at the upper surface of the chromosphere C, surely there are the same number of photons? Otherwise, where have they gone? Where is the lost energy?

Are there less photons emitted from the surface C because some have been emitted back down into the Sun? But that has to reach an equilibrium, or there would be a cascading increase in the number of photons inside the sun.

Diagram showing photosphere, chromosphere, and photons

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  • $\begingroup$ These can be two different questions. Since you seem to actually understand the answer to your own second question, the short answer to the first one is: they are scattered away from the line of sight. $\endgroup$
    – Py-ser
    Apr 16, 2014 at 3:24
  • $\begingroup$ I've redrafted the question to add to it comments by @questionhang made in the answers already provided. I have created a fresh diagram so we aren't using a drawing we don't have rights to. I have avoided using the terms ionization and recombination as emission and absorption lines are not limited to ionization and recombination, as only the excitation of an electron to a different energy state is required. $\endgroup$
    – Jeremy
    Apr 21, 2014 at 8:23
  • $\begingroup$ Highly relevant astronomy.stackexchange.com/a/38864/2531 $\endgroup$
    – ProfRob
    Mar 13, 2022 at 9:04

4 Answers 4

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All the photons being emitted do not 'fill in the blank' of the absorption line, because when they are emitted, most do not come towards us. You suppose that there will be some emitted from a different hydrogen atom elsewhere, that will just happen to aim a photon at us, but where will it come from? Think about how much of the sky the Sun subtends from our perspective. Now think about how much of the sky the Earth subtends from the perspective of an atom of hydrogen in the Sun's photosphere that is thinking about popping a photon out and wondering if it will head in the direction of Earth. Here is a 2D diagram - bear in mind that the atom actually is in 3D space and instead of being limited to only 360 degrees, can pop that photon off in any direction at all. enter image description here So it doesn't matter where ANY of the hydrogen atoms are in the photosphere, they have next to no chance of actually hitting Earth with a photon that they emit. Even though there are a lot of them, they are ALL firing off photons in almost every direction EXCEPT towards Earth.

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    $\begingroup$ Do I need to say that diagram is not to scale? $\endgroup$
    – Jeremy
    Apr 17, 2014 at 4:52
  • $\begingroup$ ＠LDC3 @Jeremy If we make two integrations at the whole surface of the photosphere. One is at the bottom surface of photosphere. The other is at the bottom surface of the photosphere. The Halpha photon numbers are the same? If they are not the same,where is the lost energy? $\endgroup$
    – questionhang
    Apr 17, 2014 at 13:00
  • $\begingroup$ I assume you mean 'one is at the bottom... The other is at the top...' $\endgroup$
    – Jeremy
    Apr 19, 2014 at 19:56
  • $\begingroup$ yes, you are right. My mistake. $\endgroup$
    – questionhang
    Apr 20, 2014 at 13:43
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Emission and Absorption lines are due to Kirchhoff's Laws that explain what happens when light passes through a gas. This is due to photons being absorbed by an atom or ion causing an electron to jump to a higher energy state, and photons being emitted when an electron returns to a lower energy state.

Photons will only be absorbed by an atom or ion if they are of precisely the energy required to bump an electron up to an excited state. If they are not of the right energy, they pass unimpeded.

Photons of particular wavelengths that are not absorbed by ions in the photosphere come from deep within the photosphere where it is hotter, and travel straight through unimpeded.

Photons of the right energy to be absorbed by a gas present in the photosphere, say hydrogen seeing as you have mentioned h-alpha, have a high probability of being absorbed by an atom on the way through the photosphere, and then being re-emitted in a random direction (including back down!)

Photons that have a high probability of being absorbed and re-emitted are therefore more likely to be coming from the top of the photosphere, or the lower chromosphere, where it happens to be cooler. (During the travel from the lower to upper photosphere, they will have been absorbed and re-emitted).

According to the Stephan-Boltzmann Law for black bodies, the energy radiated per unit of surface area is related to the temperature of the black body. A black body that is cooler will be less bright than a black body that is hotter. So when photons are radiated from a cooler part of the photosphere, they will appear as dips in the spectrum.

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  • $\begingroup$ The main reason is 'including back down!',right? $\endgroup$
    – questionhang
    Apr 20, 2014 at 13:31
  • $\begingroup$ We can argue every layer of photosphere is a black body,right? $\endgroup$
    – questionhang
    Apr 20, 2014 at 13:31
  • $\begingroup$ Not dissimilar to a black body, except for the absorption lines. Did you read what I wrote regarding temperature at different levels of the photosphere? Follow those links I provided. $\endgroup$
    – Jeremy
    Apr 20, 2014 at 20:13
  • $\begingroup$ I add Figure16.9 to my post. We just compare the total amount of Halpha photons at different layers. The amount of Halpha photons at A is larger than B. If the main reason is not your 'including back down!', where are the reduced Halpha photons? $\endgroup$
    – questionhang
    Apr 21, 2014 at 3:01
  • $\begingroup$ @questionhang Where did Figure 16.9 come from? We want it in the correct context and not have to guess what the author is showing us. $\endgroup$
    – LDC3
    Apr 21, 2014 at 3:08
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In the upper photosphere, the electrons join with the ions to neutralize the atom. When a photon strikes this atom (the atom absorbs the photon), it promotes a electron from a low energy shell to an higher energy shell. The electron then falls back to the lower energy shell emitting a photon in a different direction. When you break the sunlight into wavelengths (by a prism), you will see dark bands where the light was absorbed by the atoms and emitted in a different direction. These bands correspond to the energy absorbed and released when an electron changes energy shells. Here is the sun's light spread out to show the absorption bands.

Added: Here is a high resolution of the wavelengths from the sun.

Apr 20: You have the wrong interpretation of Figure 16.9. At level A, there is about the same number of photons at all wavelengths coming at the observer on Earth. As the photons pass through the Chromosphere, the photons with the wavelengths of the atom's electron transition are absorbed and redirected away from the observer. The photons with different wavelengths of the atom's electron transition eventually arrive at the observer. This is why there are dark lines in the spectrum. It is because most of the photons are absorbed and redirected in a different direction.

The absorption lines are formed at a higher region since the photons are removed from the spectrum.

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  • $\begingroup$ Let's take Halpha as an example. The total amount of Halpha photons drops after they leave photosphere? Although photons could be scattered, photons at other places could be scattered to our line our sight. $\endgroup$
    – questionhang
    Apr 16, 2014 at 3:31
  • $\begingroup$ Jeremy means the absorption is not the main reason，but the recombination？ The photons produced by recombination will have a wavelength dispersion? $\endgroup$
    – questionhang
    Apr 16, 2014 at 3:38
  • $\begingroup$ @Jeremy link $\endgroup$
    – questionhang
    Apr 16, 2014 at 3:43
  • $\begingroup$ @questionhang I'm don't understand what you mean by H-alpha. But the photons from the sun (at the absorption by hydrogen) are nearly all absorbed (maybe 90%) and less than 1% are emitted at Earth, so the total light at this one wavelength is about 1/10 (or less) as bright as the wavelengths that have no absorption. $\endgroup$
    – LDC3
    Apr 16, 2014 at 3:49
  • $\begingroup$ @Jeremy The absorption and emission are at the same wavelength. The energy levels don't have the time to change wavelengths as what happens in fluorescence. $\endgroup$
    – LDC3
    Apr 16, 2014 at 3:50
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A photon emmited by the photospere has almost 50% chances to be oriented outwards so as to leave the Sun and 50% to be shot towards the Sun and this means a good chance to be absorbed in the Sun: converted to another frequency or form of energy. Of the photons that leave the Sun, those of absorbable frequencies are going to have another 50% chance to be redirected towards the Sun when being re-emitted by the atoms that took them. But those non-absorbable don't have this second chance to be turned back, thus they can leave the Sun in greater proportions.

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