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I have read that the reason why the Sun produces an absorption spectrum is because the temperature drops as you go away from the center, such that as the various layers of the atmosphere of the sun absorb certain wavelengths, the re-emitted light will have a smaller intensity than the absorbed one, causing a dip in the spectrum (i.e., an absorption spectrum). This is consistent with Kirchhoff's laws and Planck's law for blackbody radiation. However, I checked and found that as you go from the photosphere and into the chromosphere and corona, the temperature rises instead. So then, why doesn't the opposite happen where instead of dips in the spectrum, we get peaks in it (i.e., an emission spectrum)?

(I will note that there was a temperature drop inside of the photosphere itself, but to my understanding the photosphere is opaque to all wavelengths, meaning that it can't pick out the specific wavelengths of the absorption spectrum).

Now, I already have some doubts about the logic above applying here. It seems to me that one of the assumptions being made is that the energy that is absorbed by the various layers are first distributed among the various atoms in the layers, such that the re-emitted light that comes out is the ordinary thermal radiation/blackbody radiation. However, considering the fact that the chromosphere and corona is so dilute, it would seem that that wouldn't be the case. Is it more correct to say that the radiation is rather being absorbed and re-emitted (i.e., scattered) so many times on its journey through the Sun's atmosphere that it continually loses small "bits" of energy that once it reaches us, the intensity of the re-emitted light is much smaller than the absorbed one (causing a dip in the spectrum)? Or is the gas perhaps so dilute that even that wouldn't work? If not, then again, why do we see an absorption spectrum?

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The hotter layers above the solar photosphere do have an emission spectrum. The emission spectrum is much fainter than the visible photosphere and so is not easily seen through broadband filters in the optical spectrum, though it can be observed through very narrow filters centred on the emission lines (e.g. H$\alpha$ from the chromosphere) in question.

The hotter gas is much less dense than the photosphere and is essentially transparent to photons from the photosphere.

The situation becomes much easier in the UV and X-ray part of the spectrum, where there is very little radiation from the photosphere, but a more significant amount from the chromosphere and corona, because their hotter temperatures excite higher energy transitions. The Sun has an obvious emission line spectrum in the UV and X-ray range. Here is an example (from del Zanna & Mason 2018).

solar UV spectrum

The (optically) thin chromosphere and corona are not heated by radiation from below. That could not produce the temperature inversion that is seen. Instead the chromosphere and corona are heated by the dissipation of magnetic fields in the form of currents and the acceleration of charged particles.

A shorthand way of thinking about photospheric absorption lines is that we can see to different depths in the Sun at different wavelengths. Where there is a strong atomic transition, light is readily absorbed, we cannot see very far. Thus our sight only penetrates to relatively cool layers (note again that the chromosphere and corona are transparent). In contrast, when we look at a "continuum" wavelength we see to deeper and hotter layers in the Sun. Since the brightness is very temperature dependent this is why we get a continuous spectrum with absorption lines.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Connor Garcia Jun 20 at 18:36
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The photosphere of the sun does produce an emission spectrum (a Planck spectrum according to its temperature of about 6000K). It is only that the atmosphere above the photosphere (the chromosphere) scatters light out of the line of sight at those frequencies where the scattering cross section is very high (i.e. at the atomic resonance frequencies of a given element). This is schematically illustrated by the following diagram

absorption lines

from https://courses.lumenlearning.com/astronomy/chapter/formation-of-spectral-lines/)

We are of course unable to see the pure continuous spectrum (bottom left) in case of the sun, as the scattering 'cloud' is a layer all around its surface, but we can see the 'bright line spectrum' emitted by the chromosphere during a total solar eclipse for instance as then the direct sunlight is blocked out.

The 'missing bits' of energy are thus due to the photons reflected back by the chromosphere into the photosphere where they are absorbed (the photosphere can be considered a black body and a black body absorbs all incoming radiations). They are not truly lost though but are merely thermalized again and re-emitted as continuum photons. The geometry may be different to that shown in the diagram but the result is the same.

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  • $\begingroup$ That diagram makes a lot of sense, however I have some doubts about using this principle for the sun. The thing is that the sun is (approximately) spherically symmetric (around its center), meaning that if you calculate the total power radiated out of a shell with a radius of 1 AU, with the usual absorption lines, and compare that with the total power without absorption lines, then the power with absorption lines is clearly lower. So then where did that lost energy go? It would seem that it was absorbed by certain layers in the sun (which is why my proposed explanation included "losing small $\endgroup$ – Physics2718 Jun 10 at 19:54
  • $\begingroup$ bits of energy". $\endgroup$ – Physics2718 Jun 10 at 19:55
  • $\begingroup$ That is because that diagram is inapplicable to explaining the spectrum of the Sun. @Physics2718 The geometry and temperature structure is quite different. $\endgroup$ – ProfRob Jun 10 at 20:13
  • $\begingroup$ @Physics2718 As the chromosphere is optically thick at frequencies corresponding to the centers of spectral lines, the radiation is reflected back into the photosphere and absorbed there (the photosphere can be considered a black body and a black body absorbs all incoming radiations). So whilst the geometry is obviously different to that shown in the diagram, the physics is the same. The absorption lines arise due to light being scattered out of the beam at those frequencies and being lost (well, not actually lost in this case, but thermalized again into the continuous spectrum).. $\endgroup$ – Thomas Jun 10 at 21:09
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I have read that the reason why the sun produces an absorption spectrum is because the temperature drops as you go away from the center, such that as the various layers of the atmosphere of the sun absorb certain wavelengths, the re-emitted light will have a smaller intensity than the absorbed one, causing a dip in the spectrum (ie an absorption spectrum). This is consistent with Kirchhoff's laws and Planck's law for blackbody radiation.

The further away you go from the solar surface, the photons from the stellar surface/or within have to travel longer through the solar atmosphere. Star already emits a blackbody radiation but Most of the emission and absorption spectra of a star comes from the gases in this solar atmosphere.

However, I checked and found that as you go from the photosphere and into the chromosphere and corona, the temperature rises instead. So then, why doesn't the opposite happen where instead of dips in the spectrum, we get peaks in it (ie an emission spectrum)?

The Stellar atmosphere is made of gases which are atoms in ground state, atoms in excited state and free electrons. Different wavelength/frequency of photons from the star interacts in multiple ways with different types of atoms. These are some major interactions:

  1. Line Scattering: The photon is absorbed by an atom $\rightarrow$ excitation of electron $\rightarrow$ de-excitation of electrons $\rightarrow$ photon is released. The photon could be released in any direction and although it would be released with same energy as initial photon, doppler effect plays an important role here.
  2. Emission by Recombination: The energy of photon is used to de-excite a free electron. Each de-excitation results in emission of a photon.
  3. Emission by collisional or photo-excitation: An atom absorbs the photon and the electron makes a transition from excited state to ground state.
  4. Pure Absorption: Photon is absorbed by an atom to excite an electron. Further de-excitation of that same electron to a lower level will cause emission.
  5. Masering: Atom in excited state emit the same energy photon as the stellar photon. Such a process will result in same photons travelling in same direction.

Now, getting back to why sun or most sun-like stars don't have a strong emission spectrum. We can notice, that for an emission spectrum, we need an excited or free electron to de-excite and emit photons. Usually, this condition needs a very high temperature such as a state of plasma. For sun (and many other stars), the temperature is just not high enough to have atoms in their excited state. Most atoms in solar atmosphere are in their ground state and hence sun shows much stronger absorption spectrum. Moreover, the elements found in sun's atmosphere have more transitions that support absorption rather than emission. Some of the astronomical bodies that show emission spectrum would be massive stars, Black Hole/Neutron Star accretion disk (both have high temperatures and presence of heavy metals like Fe).

Within the stellar surface, there are free electrons, plasma and excited atoms but sun's outer surface being too cold for radiative transport and is convective zone, which is opaque to internal radiation, we are unable to see the emission spectrum from within the solar surface.

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  • $\begingroup$ But the Sun does have an emission spectrum, just not at visible wavelengths. $\endgroup$ – ProfRob Jun 10 at 18:02
  • $\begingroup$ But i never said it doesn't! I said the absorption spectrum is much stronger due to Answered reasons. $\endgroup$ – Aryan Bansal Jun 10 at 19:34
  • $\begingroup$ But the absorption spectrum isn't much stronger (whatever that means) at UV and X-ray wavelengths. $\endgroup$ – ProfRob Jun 10 at 20:09
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    $\begingroup$ Hot, high mass stars, such as O-type main sequence stars and Wolf-Rayet stars, have emission spectrum, due to their strong winds (especially if they are metal rich). $\endgroup$ – Daddy Kropotkin Jun 10 at 20:43
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The sun is the source of emission as well the source of absorption. The hot ball beneath the surface produces a continuous spectrum of black body radiation. Photons in the ball beneath the surface layer are continuously absorbed and reemitted (scattered). This causes a continuous distribution of photon wavelengths, i.e., the radiation conformal to a black body. Upon entering the outer gas layer, part of these photons are absorbed, depending on their wavelengths. The rest of the huge energy escapes and travels to us (in part). We see an almost continuous spectrum (as can be seen in a prism). The radiation coming from the surface gas layer is almost identical to the radiation coming from the inner ball. The difference is that a part of certain wavelength photons is absorbed. So a part of the continuous inner ball radiation is absorbed. So if you subtract the outer layer radiation from the inner ball radiation, you will be left with absorption lines (which can be said to be emission lines at the same time). Where do the absorbed photons go? They are sent back to the inside, while the non-absorbed travel to us.

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