APOD: The Sun and Its Missing Colors (2023 Jun 11)

[APOD Robot]

The Sun and Its Missing Colors

Explanation: Here are all the visible colors of the Sun, produced by passing the Sun's light through a prism-like device. The spectrum was created at the McMath-Pierce Solar Observatory and shows, first off, that although our white-appearing Sun emits light of nearly every color, it appears brightest in yellow-green light. The dark patches in the featured spectrum arise from gas at or above the Sun's surface absorbing sunlight emitted below. Since different types of gas absorb different colors of light, it is possible to determine what gasses compose the Sun. Helium, for example, was first discovered in 1868 on a solar spectrum and only later found here on Earth. Today, the majority of spectral absorption lines have been identified - but not all.

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[Ann]

The Sun and Its Missing Colors. Image Credit: Nigel Sharp (NSF), FTS, NSO, KPNO, AURA, NSF

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Color grid.

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Well, the Sun may be missing some of its colors. And yet, we can create all possible colors with the light from the Sun (and by using various tips and tricks, but primarily, with the light from the Sun).

There is a huge difference between additive colors, which light itself is painting with, and subtractive colors, which come from dyes.

Subtractive primary colors. Credit: WIKIMEDIA COMMONS (CC BY-3.0)

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Additive primary colors. Credit: WIKIMEDIA COMMONS (CC BY-3.0)

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Old Superman comic book. The colors would be made from pixels of cyan, magenta and yellow.

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Excessive and incorrect pixellating.

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I remember my favorite childhood comic book, Superman, how you could see the pixels of cyan, magenta and yellow creating the vivid colors.

All right, but where do the dyes come from? Nowadays they are easy to create (I think), but in the past, making some of those dyes was a laborious process indeed.

A polished block of lapis lazuli.

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Wikipedia wrote:

Lapis lazuli (UK: /ˌlæpɪs ˈlæz(j)ʊli, ˈlæʒʊ-, -ˌlaɪ/; US: /ˈlæz(j)əli, ˈlæʒə-, -ˌlaɪ/), or lapis for short, is a deep-blue metamorphic rock used as a semi-precious stone that has been prized since antiquity for its intense color. As early as the 7th millennium BC, lapis lazuli was mined in the Sar-i Sang mines,[1] in Shortugai, and in other mines in Badakhshan province in modern northeast Afghanistan.[2] Lapis lazuli artifacts, dated to 7570 BC, have been found at Bhirrana, which is the oldest site of Indus Valley civilisation.
...

By the end of the Middle Ages, lapis lazuli began to be exported to Europe, where it was ground into powder and made into ultramarine, the finest and most expensive of all blue pigments. Ultramarine was used by some of the most important artists of the Renaissance and Baroque, including Masaccio, Perugino, Titian and Vermeer, and was often reserved for the clothing of the central figures of their paintings, especially the Virgin Mary.

The Virgin Mary wearing ultramarine. Painting by Sandro Botticelli.

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Another popular dye was Tyrian purple, extracted from a mollusc:

A snail whose colorful secretion could be made into a dye.

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Wikipedia wrote:

Tyrian purple (Ancient Greek: πορφύρα porphúra; Latin: purpura), also known as, royal purple, imperial purple, or imperial dye, is a reddish-purple natural dye. The name Tyrian refers to Tyre, Lebanon. It is secreted by several species of predatory sea snails in the family Muricidae, rock snails originally known by the name 'Murex'. In ancient times, extracting this dye involved tens of thousands of snails and substantial labor, and as a result, the dye was highly valued. The colored compound is 6,6′-dibromoindigo.

Important people of the past wearing Tyrian purple.

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Some objects in nature are blue not because of any dye, but because of how their structure reflects light.

Wikipedia wrote:

The brilliant blue color in the butterfly's wings is caused by the diffraction of the light from millions of tiny scales on its wings.

Color is amazing. But for us on the Earth, indeed in the Solar system, it all starts the Sun.

Which isn't yellow. But never mind.

Ann

[revloren]

HAPPY PRIDE!!!

[JohnD]

Ann,
I understand about absorbtion of colours, but am puzzled. For instance, sodium vapour street lights give off the same wavelength of light that identifies sodium in the Sun. But the sodium in the Sun must be at much higher temperature than in the street light, so why does the solar sodium not glow and fill the dark patch in the spectrum?
Thanks for an explanation!
John

[Ann]

Ann,
I understand about absorbtion of colours, but am puzzled. For instance, sodium vapour street lights give off the same wavelength of light that identifies sodium in the Sun. But the sodium in the Sun must be at much higher temperature than in the street light, so why does the solar sodium not glow and fill the dark patch in the spectrum?
Thanks for an explanation!
John

It's easier for me to talk about hydrogen lines and hydrogen emission than to talk about sodium lines, so I'll start with hydrogen. But you must understand that a lot of this has to do with math, and there I'm on very shaky ground!


Take a look at the Trifid Nebula:

The Trifid Nebula: Credit: Jay Ballauer.

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The Trifid Nebula, the pink as well as the blue part of it, was born from a cloud of cold, neutral gas, mostly made of hydrogen. This cloud was mostly invisible, although trace amounts of heavier elements like carbon, silicates and ices may have made it look dark. The cloud contracted and gave birth to a cluster of stars, most notably HD 164492, a star of spectral class O7.5V. This hot star ionized the hydrogen in the part of the Trifid Nebula that is pink.

This is what happens after a hydrogen atom has become ionized:

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You can see the core of the hydrogen atom, the proton. There is a single electron, which will normally be located in the electron shell designated N = 2. (Don't ask me about the electron shell designated N = 1, and please don't ask me what it means, in quantum mechanics terms, what it means that an electron is "located" somewhere.)

Anyway. When an energetic photon from an O-type star hits a hydrogen atom, the electron will gain so much energy from the photon that hit it that it can "jump up" to the next level, that is, to the shell designated N = 3. Now the hydrogen atom has become ionized.

But the electron will not stay in the higher electron shell. It will fall down again, and as it does so, it will release a photon whose exact wavelength is 656.279 nm. This is hydrogen alpha. The color of hydrogen alpha is deep red, ███. But in an emission nebula, the hydrogen alpha will almost always be mixed with some hydrogen beta light. Hydrogen beta is created when an electron is hit by such an energetic photon that the electron "jumps up two levels", and when it falls down, it releases a photon of 486.135 nm. This is a bluish cyan color, ███, and when hydrogen alpha is mixed with some hydrogen beta, the result is the pink color typical of emission nebulas. Maybe this color, ███, but I'm really just guessing.

So much for the emission of hydrogen alpha. The way I understand it, the ionization of sodium should work in a similar manner. A neutral sodium atom should become ionized so that one (or more?) electrons are made to "jump up" to a higher electron shell. When the electron falls down again, it emits yellow sodium light.

Usp.br wrote:

The sodium spectrum is dominated by the bright doublet known as the Sodium D-lines at 588.9950 and 589.5924 nanometers. From the energy level
diagram it can be seen that these lines are emitted in a transition from the 3p to the 3s levels. The line at 589.0 has twice the intensity of the line at 589.6 nm.

I'm sure that made you so much wiser! In any case, sodium light would be this color, ███.

So much for hydrogen or sodium emission of light: It would happen when an electron in a neutral hydrogen or sodium atom is hit by an energetic photon and is "kicked upstairs" to another electron shell. The specific hydrogen alpha or singly ionized sodium light would be emitted when the electron "falls back" again.

I understand the emission mechanism, more or less. I'm so much more vague when it comes to the absorption of wavelengths inside the Sun. But I do know that the same elements and the same wavelengths are involved:

Hydrogen absorption lines and hydrogen emission lines. Source: Khan Academy.

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Khan Academy wrote:

Emission lines refer to the fact that glowing hot gas emits lines of light, whereas absorption lines refer to the tendency of cool atmospheric gas to absorb the same lines of light.

In emission nebulas, cool neutral gas is heated up and ionized by energetic photons. But inside the Sun, almost all the gas is already ionized, except near the surface where the temperature is too low. After all, the temperature at the Sun's core is some 15 million K, but at the photosphere it is less than 6,000 K. If you ask my poor brain to guess, I'd guess that elements that are ionized near the Sun's core become neutral near the Sun's outermost layers. I guess that photons inside the Sun of specific wavelengths, such as 656 nm or 589 nm, get absorbed by relatively (relatively!) cool hydrogen or relatively cool sodium near the Sun's "edge" or photosphere.

The eclipsed Sun, showing hydrogen alpha along its limb. Photo: Randy Attwood

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During a solar eclipse, you can sometimes see hydrogen alpha along the solar limb.

All right, Johnny, that's the best I can do for you!

Ann

[Mel]

JohnD,

The only way I can imagine it is as an issue of direction. The light coming to us from the star is coming directly from the star (duh:). The light emitted by molecules in the dust cloud is subtracted from the starlight, then it's re-radiated in all directions, so less comes to us compared with light at frequencies that don't get absorbed. So light at absorbed+reradiated wavelengths appears dimmer.
( am not a scientist.)

[RJN]

An email alerts me that Helium was first discovered on the Sun in 1868, not in 1870 as originally mentioned. I have now corrected the date on the main NASA APOD and apologize for the oversight. - RJN

[johnnydeep]

I've read Ann's posts above, but have two stupid questions:

1. Ok, the absorption lines are caused by the presence of various elements in the Sun's atmosphere that absorb the corresponding frequencies of a continuous spectrum light, resulting in dark lines, meaning the photons absorbed never get to us. But what happens to those photons? They can't STAY absorbed, ore else the absorption effect would not last. Do they just get immediately reemitted in a different direction and so we don't see them?

2. Why does the nuclear fusion of hydrogen to helium produce a continuous spectrum of photons to be begin with. Why aren't all the photons being emitted restricted to certain specific wavelengths that I would have expected to reflect this particular process of fusion?

[Holger Nielsen]

Ann writes:

This is what happens after a hydrogen atom has become ionized:

... When an energetic photon from an O-type star hits a hydrogen atom, the electron will gain so much energy from the photon that hit it that it can "jump up" to the next level, that is, to the shell designated N = 3. Now the hydrogen atom has become ionized.

Not "ionized", but "excited". An ionized hydrogen atom has lost its single electron and is just a proton. In an exited atom the electron occupies an energy level higher than that of the stable ground state (n = 1). It can then jump back to the ground state, or to a state with lower energy, if such one exists. The energy lost by the atom is emitted as a photon in a random direction. This is indicated in the illustration:
E_initial - E_final = h·ν
where E_initial and E_final are the energies of the upper resp. lower exited state, h is Planck's constant and ν is the frequency of the produced photon (ν is the Greek letter "nu"; in this font it seems to be indistinguishable from Latin "v").

The Balmer lines correspond to transitions to the first exited state (n = 2) from some higher state (n = 3, 4, ...). Balmer lines are weak in stars of a "late" spectral class (M, K) because the temperature in the star's atmosphere is so low, that the radiation field does only contain few photons with energy sufficient to exite hydrogen atoms to the required higher states (n = 3, 4, ...). Balmer lines are also weak in the hottest stars of spectral type O, simply because their radiation field is so energetic, that most hydrogen atoms are ionized: No exites states, no transitions. The Balmer lines happen to be most prominent in stars of spectral type A0.

Similarly, transitions to the ground state (n = 1) from higher up (n = 2, 3...) produce the ultraviolet lines of the Lyman series.

[Holger Nielsen]

Technical question:
Is it not possible to write Greek characters on Asterisk? Or Unicode? I tried to use the "char" button, but I cannot get it to work:

Code: Select all

[char]beta[/char] [char]deg[/char]

produces this output: β °, even though the last example is suggested by the "char" button.

[johnnydeep]

Technical question:
Is it not possible to write Greek characters on Asterisk? Or Unicode? I tried to use the "char" button, but I cannot get it to work:

Code: Select all

[char]beta[/char] [char]deg[/char]

produces this output: β °, even though the last example is suggested by the "char" button.

There's probably some special character sequences you can enter to allow typing them from a keyboard, but just copy/paste from this or other similar page: https://en.wikipedia.org/wiki/Greek_script_in_Unicode

E.g., α, β, φ, ω, etc.

[Holger Nielsen]

johhnydeep writes:

1. Ok, the absorption lines are caused by the presence of various elements in the Sun's atmosphere that absorb the corresponding frequencies of a continuous spectrum light, resulting in dark lines, meaning the photons absorbed never get to us. But what happens to those photons?

I would say they cease to exist.

They can't STAY absorbed, ore else the absorption effect would not last. Do they just get immediately

more or less

reemitted in a different direction and so we don't see them?

Yes!

2. Why does the nuclear fusion of hydrogen to helium produce a continuous spectrum of photons to be begin with. Why aren't all the photons being emitted restricted to certain specific wavelengths that I would have expected to reflect this particular process of fusion?

The photons produced by nuclear fusion are in an environment, where they get scattered zillions of times on atoms, so their energy is statistically changed to produce a Planck spectrum given by the temperature of the gas. The British astrophysicist Arthur Eddington has given a lively description of this ("quantum of aether waves": photon):

The Inside of a Star.

17. The inside of a star is a hurly-burly of atoms, electrons and aether waves. We have to call to aid the most recent discoveries of atomic physics to follow the intricacies of the dance. We started to explore the inside of a star; we soon find ourselves exploring the inside of an atom. Try picture the tumult! Dishevelled atoms tear along at 50 miles a second with only a few tatters left of their elaborate cloaks of electrons torn from them in the scrimmage. The lost electrons are speeding a hundred times faster to find new resting-places. Look out! there is nearly a collision as an electron approaches an atomic nucleus; but putting on speed it sweeps around it in a sharp curve. A thousand narrow shaves happen to the electron in 10^-10 of a second; sometimes there is a side-slip at the curve, but the electron still goes on with increased or decreased energy. Then comes a worse slip than usual; the electron is fairly caught and attached to the atom, and its career of freedom is at an end. But only for an instant. Barely has the atom arranged the new scalp on its girdle when a quantum of aether waves runs into it. With a great explosion the electron is off again for further adventures. Elsewhere two of the atoms are meeting full tilt and rebounding, with further disaster to their scanty remains of vesture.

As we watch the scene we ask ourselves, Can this be the stately drama of stellar evolution? It is more like the jolly crockery-smashing turn of a music-hall. The knockabout comedy of atomic physics is not very considerate towards our aesthetic ideals; but it is all a question of time-scale. The motions of the electrons are as harmonious as those of the stars but in a different scale of space and time, and the music of the spheres is being played on a keyboard 50 octaves higher. To recover this elegance we must slow down the action, or alternatively accelerate our own wits; just as the slow-motion film resolves the lusty blows of a prize-fighter into movements of extreme grace - and insipidity.

And what is the result of all this bustle? Very little. Unless we have in mind an extremely long stretch of time the general state of the star remains steady. Just as many atoms are repaired as are smashed; just as many bundles of radiation are sent out as are absorbed; just as many electrons are captured as are exploded away. The atoms and the electrons for all their hurry never get anywhere; they only change places. The aether waves are the only part of the population which do actually accomplish something; although apparently darting about in all directions without purpose they do in spite of themselves make a slow general progress outwards.

From A. S. Eddington, The Internal Constitution of the Stars, Cambridge University Press, 1930.

[zendae]

[


Color is amazing. But for us on the Earth, indeed in the Solar system, it all starts the Sun.

Which isn't yellow. But never mind.

Ann

I am so fascinated by color. It means so much for all of us. No wonder plants are green; they have to reflect away all that poisonous green light from the Sun! And it is compelling that color mixing via dyes and color mixing via light can be so different. I'm in the A/V field, and mixing red and green gives yellow. Which is why traditional emitters on TV screens are red, blue and green. It's even that way with the modern digital LED concert lamps we use. You can get a nice yellow by programming red and green just right. (Although pure yellow emitters are now being employed to free up the other color emitters). But mixing red and green via dyes gives brown. There is no such thing as a brown light gel, because brown is a very low amplitude yellow. Put a bright enough source through a brown gel (which is not even made) and you'd get yellow. A fascinating brain koan in Zen is visualizing a color we have never seen... There are lots of colors we have never seen. So enticing, compelling, and infinitely elusive!

[Holger Nielsen]

johnnydeep wrote:

just copy/paste

Thank you, that works! I tried that with a Greek "nu", but was unhappy, because it came out resembling a Latin "v". I should have known better!
This leaves the question why "char" does not deliver for instance a degreen symbol, º, as advertized.

[Cousin Ricky]

You can get a nice yellow by programming red and green just right. (Although pure yellow emitters are now being employed to free up the other color emitters). But mixing red and green via dyes gives brown. There is no such thing as a brown light gel, because brown is a very low amplitude yellow. Put a bright enough source through a brown gel (which is not even made) and you'd get yellow.

Brown is no different than any other dark color, such as dark green or dark blue. The only reason brown seems weird or special in the color scheme of things is that the English language just happens to have a common word for it.

[Ann]

You can get a nice yellow by programming red and green just right. (Although pure yellow emitters are now being employed to free up the other color emitters). But mixing red and green via dyes gives brown. There is no such thing as a brown light gel, because brown is a very low amplitude yellow. Put a bright enough source through a brown gel (which is not even made) and you'd get yellow.

Brown is no different than any other dark color, such as dark green or dark blue. The only reason brown seems weird or special in the color scheme of things is that the English language just happens to have a common word for it.

Yes, brown is really just dark orange.

Ann

[orin stepanek]

APOD wrote Today, the majority of spectral absorption lines have
been identified - but not all! Does that mean we still don't know all
the elements in Old Sol?
Ann; I liked all the pictures in your post!

[Igwasborn]

Doesn't some dark patches in the featured spectrum arise from earth's atmosphere gases absorbing sunlight?

[Chris Peterson]

sunspectrum_mpso_960.jpg
APOD wrote Today, the majority of spectral absorption lines have
been identified - but not all! Does that mean we still don't know all
the elements in Old Sol? :shock:

No. It means we don't know all of the possible absorption lines of all the elements. In some cases there are ambiguities, too.

[Ann]

Ann,
I understand about absorbtion of colours, but am puzzled. For instance, sodium vapour street lights give off the same wavelength of light that identifies sodium in the Sun. But the sodium in the Sun must be at much higher temperature than in the street light, so why does the solar sodium not glow and fill the dark patch in the spectrum?
Thanks for an explanation!
John

Okay, John, let's repeat! Emission happens when an ultraviolet photon hits a hydrogen atom and knocks the electron into a higher-energy electron shell. The electron falls back again and releases a photon of 656 nm, hydrogen alpha. The ultraviolet photon that originally hit the hydrogen atom ceases to exist.

Inside the Sun, near the photosphere of the Sun, there are photons whose energy level correspond exactly to hydrogen alpha, or 656 nm. Near the perimeter of the Sun, there are neutral hydrogen atoms. When a photon of 656 nm hits a neutral hydrogen atom, the photon is absorbed. I guess it happens because the hydrogen atom "recognizes" the 656 nm photon as something that belongs to the hydrogen atom and corresponds to one of its own energy states.

Something very slightly similar to the cat who "absorbed" ducklings that she thought belonged to her!

Cat nurses ducklings.

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The ducklings survived and thrived. But the 656nm photons that encounter neutral hydrogen atoms inside the Sun are absorbed forever, leaving only darkness behind!

Ann

[johnnydeep]

Doesn't some dark patches in the featured spectrum arise from earth's atmosphere gases absorbing sunlight?

Hmm. This image uses a spectrum taken by a Solar observatory on the Earth's surface, so it would see to me to have to also show absorption lines created by elements in Earth's atmosphere. Paging Chris to tell us why that's not so...

[johnnydeep]

johnnydeep wrote:

just copy/paste

Thank you, that works! I tried that with a Greek "nu", but was unhappy, because it came out resembling a Latin "v". I should have known better!
This leaves the question why "char" does not deliver for instance a degreen symbol, º, as advertized.

Yeah, I have no idea why the [char] tags don't seem to work, nor all the strings - "deg", "frac12", etc - that are/were supported.

°
½

[Chris Peterson]

Doesn't some dark patches in the featured spectrum arise from earth's atmosphere gases absorbing sunlight?

Hmm. This image uses a spectrum taken by a Solar observatory on the Earth's surface, so it would see to me to have to also show absorption lines created by elements in Earth's atmosphere. Paging Chris to tell us why that's not so...

The path length of light through the atmosphere just isn't long enough. The amount of attenuation from atomic absorption from a few kilometers of air is extremely tiny for the visible range of EM radiation.

[johnnydeep]

Doesn't some dark patches in the featured spectrum arise from earth's atmosphere gases absorbing sunlight?

Hmm. This image uses a spectrum taken by a Solar observatory on the Earth's surface, so it would see to me to have to also show absorption lines created by elements in Earth's atmosphere. Paging Chris to tell us why that's not so...

The path length of light through the atmosphere just isn't long enough. The amount of attenuation from atomic absorption from a few kilometers of air is extremely tiny for the visible range of EM radiation.

Aha! Thanks!