APOD: M57: The Ring Nebula from Hubble (2021 Aug 17)

[APOD Robot]

M57: The Ring Nebula from Hubble

Explanation: Except for the rings of Saturn, the Ring Nebula (M57) is probably the most famous celestial circle. Its classic appearance is understood to be due to our own perspective, though. The recent mapping of the expanding nebula's 3-D structure, based in part on this clear Hubble image,indicates that the nebula is a relatively dense, donut-like ring wrapped around the middle of a (American) football-shaped cloud of glowing gas. The view from planet Earth looks down the long axis of the football, face-on to the ring. Of course, in this well-studied example of a planetary nebula, the glowing material does not come from planets. Instead, the gaseous shroud represents outer layers expelled from the dying, once sun-like star, now a tiny pinprick of light seen at the nebula's center. Intense ultraviolet light from the hot central star ionizes atoms in the gas. The Ring Nebula is about one light-year across and 2,500 light-years away.

<< Previous APOD

This Day in APOD

Next APOD >>

[Ann]

M57: The Ring Nebula from Hubble. Image Credit: NASA, ESA, Hubble Legacy Archive; Processing: Judy Schmidt

---

That's a lovely Ring Nebula from Hubble and Geck. Of course, the Ring Nebula almost always looks good!

Ring Nebula from Hubble Heritage, published on January 6, 1999. Red: F658N ([N II]), Green: F501N ([O III]), Blue: F469N (He II)

---

In this composite image, visible-light observations by NASA's Hubble Space Telescope are combined with infrared data from the ground-based Large Binocular Telescope in Arizona to assemble a dramatic view of the well-known Ring Nebula. Image Credit: Credit: NASA, ESA

---

The picture at left shows the classic Hubble Heritage image of the Ring Nebula. I loved the color, because the Ring Nebula looked surprisingly blue in this image. The filters for the image were chosen before NASA got enamored with its Hubble palette of OIII, Hα and SII. The filters used for the Hubble Heritage image were F658N ([N II], red), F501N ([O III], green), and F469N (He II, blue). Looking good!

The picture at right shows the classic optical Hubble image being combined with infrared data from an Earth-based telescope. We can see an intricate structure of large, faint, infrared shells surrounding the bright nebula.

The outer shells of the Ring Nebula have been highlighted by other photographers.

HaRGB image of the Ring Nebula (M57) showing the faint outer shells. Data from the Liverpool Telescope on La Palma, Islas Canarias (Canary Islands), Spain.

---

Rings Around the Ring Nebula. Image Credit: Hubble, Large Binocular Telescope, Subaru Telescope; Composition & Copyright: Robert Gendler

---

Note that the Liverpool Telescope image of the Ring Nebula did not rely on any infrared data at all. Apparently the outer structures are faintly visible in Hydrogen Alpha. But the Hubble, Large Binocular Telescope and Subaru Telescope image combined visible light, narrowband hydrogen and infrared data and got the best portrait of the outer structures.

Okay, but I would not be me if I didn't show you the Blue Ring Nebula too, would I?

The Blue Ring Nebula consists of two expanding cones of gas ejected into space by a stellar merger. As the gas cools, it forms hydrogen molecules that collide with particles in interstellar space, causing them to radiate far-ultraviolet light. Invisible to the human eye, it is shown here as blue. Credits: NASA/JPL-Caltech/M. Seibert (Carnegie Institution for Science)/K. Hoadley (Caltech)/GALEX Team

---

Click to play embedded YouTube video.

NASA wrote:

The Blue Ring Nebula, which perplexed scientists for over a decade, appears to be the youngest known example of two stars merged into one...

While merged star systems are thought to be fairly common, they are nearly impossible to study immediately after they form because they're obscured by debris the collision kicks up. Once the debris has cleared – at least hundreds of thousands of years later – they're challenging to identify because they resemble non-merged stars. The Blue Ring Nebula appears to be the missing link: Astronomers are seeing the star system only a few thousand years after the merger, when evidence of the union is still plentiful. It appears to be the first known example of a merged star system at this stage.
...
The team concluded that the nebula was the product of a relatively fresh stellar merger that likely occurred between a star similar to our Sun and another star only about one-tenth that size (or about 100 times the mass of Jupiter). Nearing the end of its life, the Sun-like star began to swell, creeping closer to its companion. Eventually, the smaller star fell into a downward spiral toward its larger companion. Along the way, the larger star tore the smaller star apart, wrapping itself in a ring of debris before swallowing the smaller star entirely.

This was the violent event that led to the formation of the Blue Ring Nebula. The merger launched a cloud of hot debris into space that was sliced in two by the gas disk. This created two cone-shaped debris clouds, their bases moving away from the star in opposite directions and getting wider as they travel outward. The base of one cone is coming almost directly toward Earth and the other almost directly away. They are too faint to see alone, but the area where the cones overlap (as seen from Earth) forms the central blue ring GALEX observed.

Millennia passed. The expanding debris cloud cooled and formed molecules and dust, including hydrogen molecules that collided with the interstellar medium, the sparse collection of atoms and energetic particles that fill the space between stars. The collisions excited the hydrogen molecules, causing them to radiate in a specific wavelength of far-UV light. Over time, the glow became just bright enough for GALEX to see.

Ohh... yeah... The Blue Ring Nebula isn't really blue, but far ultraviolet. Spoilsports!

Ann

[GonzoTheGreat]

Except for the rings of Saturn, the Ring Nebula (M57) is probably the most famous celestial circle.

It may be that the Milky Way is an even better known celestial circle. Yes, I know; technically that's a disc. But we're on the inside, and from the inside it looks like a circle. But apart from Saturn's rings and the Milky Way ..
Oh, possibly, just maybe, the number of people familiar with the Zodiac could also be somewhat larger than the numbers who have heard of the Ring Nebula. So, apart from Saturn's rings, the Milky Way, and the Zodiac, ..

[JohnD]

The intense spectrum of colours across the Ring Nebula represent the different elements being ionised and then fluorescing at a distinct frequnency and colour. So the spectrum represents a gradient of elements in the nebula that depends on their distance from the star. How does this gradient form?
If the colours are produced by the elements suggested by Ann (F658N ([N II], red), F501N ([O III], green), and F469N (He II, blue), then it cannot be the atoms' masses, as blue - Helium, by far the lightest - is in the middle, nearest the star and apparently least affected by the solar winds and expansion of its outer layers. What other atomic property has so effectively differentiated the population of different atoms?

JOhn

[Chris Peterson]

The intense spectrum of colours across the Ring Nebula represent the different elements being ionised and then fluorescing at a distinct frequnency and colour. So the spectrum represents a gradient of elements in the nebula that depends on their distance from the star. How does this gradient form?
If the colours are produced by the elements suggested by Ann (F658N ([N II], red), F501N ([O III], green), and F469N (He II, blue), then it cannot be the atoms' masses, as blue - Helium, by far the lightest - is in the middle, nearest the star and apparently least affected by the solar winds and expansion of its outer layers. What other atomic property has so effectively differentiated the population of different atoms?

It's more complex than this, because the different elements also have different ionization energies. And different energies produce different degrees of ionization and therefore result in different emissions bands (which will not be seen in narrowband images).

[neufer]

If the colours are produced by the elements suggested by Ann (F658N ([N II], red), F501N ([O III], green), and F469N (He II, blue), then it cannot be the atoms' masses, as blue - Helium, by far the lightest - is in the middle, nearest the star and apparently least affected by the solar winds and expansion of its outer layers.

M57, The Ring Nebula (Hubble Space Telescope)

---

<<M57 is 2,570 light-years from Earth. Photographs taken over a period of 50 years show the rate of nebula expansion is roughly 1 arcsecond per century, which corresponds to spectroscopic observations as 20–30 km s⁻¹. M57 is illuminated by a central white dwarf or planetary nebula nucleus (PNN) of 15.75v visual magnitude.

All the interior parts of this nebula have a blue-green tinge that is caused by the doubly ionized oxygen emission lines at 495.7 and 500.7 nm. These observed so-called "forbidden lines" occur only in conditions of very low density containing a few atoms per cubic centimeter.

In the outer region of the ring, part of the reddish hue is caused by hydrogen emission at 656.3 nm, forming part of the Balmer series of lines. Forbidden lines of ionized nitrogen or N II contributes to the reddishness at 654.8 and 658.3 nm.>>

[orin stepanek]

Ah! Looks like a Flower!
So this is how the sun might go?

[JohnD]

Chris,
OK, I imagined a simple explanation, and was wrong, but I can do complexity! You will be as short of time as any busy academic - please refer me to some source that will explain the mechanism?

Neufer,
Thank you! So the blue is due to Oxygen, not Helium? But why is that fluorescing near the star, while Nitrogen is, further out?
John

[Ann]

The intense spectrum of colours across the Ring Nebula represent the different elements being ionised and then fluorescing at a distinct frequnency and colour. So the spectrum represents a gradient of elements in the nebula that depends on their distance from the star. How does this gradient form?
If the colours are produced by the elements suggested by Ann (F658N ([N II], red), F501N ([O III], green), and F469N (He II, blue), then it cannot be the atoms' masses, as blue - Helium, by far the lightest - is in the middle, nearest the star and apparently least affected by the solar winds and expansion of its outer layers. What other atomic property has so effectively differentiated the population of different atoms?

JOhn

Well, 658 nm is the wavelength of ionized sulfur. It's seems relatively straightforward that sulfur should be ionized and made to emit phtotons in the relative presence of an extremely hot white dwarf. Similarly, 501 nm is the wavelength of doubly ionized oxygen. Doubly ionized oxygen represents a higher degree of ionization that ionized sulfur.

I do find the 469 nm ionized helium mystifying. I checked the emission lines of helium, and there appears to be no helium line at 469 nm. There is one at 471.3 nm, however.

---

There is as far as I can understand nothing strange about helium being ionized close to an extremely hot star and and being made to emit photons of light as a result. Note that the gases and elements closer to the white dwarfs will receive more far ultraviolet light than gases and elements farther away from the star, and are likely to become more highly ionized and emit more energetic photons as a result.

This is, as well as I can understand, how ionization works:

Emission lines of hydrogen. http://www.physast.uga.edu/~rls/1010/ch5/ovhd.html

---

In the picture above you can see various emission lines of hydrogen. Amateurs looking at pretty pictures are most interested in the Balmer series, since its emission lines are in the visible part of the electromagnetic spectrum.

Here is what happens in the Balmer series. An ultraviolet photon hits the electron of a hydrogen atom. The electron receives an extra helping of energy from the ultraviolet photon and is "kicked upstairs" to a higher "electron shell". The electron quickly "falls down" to its original electron shell again, and as it does so, it rids itself of the extra energy it received from the ultraviolet photon that hit it by emitting a photon of its own. Normally that will be a photon with a wavelength of 656 nm, corresponding to hydrogen alpha. You know, the red light that gives all those red emission nebulas their color?

es, but if the ultraviolet photon that hit the electron of a hydrogen atom was particularly energetic, it could kick the electron "two floors up". And then the electron would fall down again while emitting a photon of hydrogen beta at 486.1 nm. And there are even higher "floors" and progressively more purple photons that could be emitted by the electron as it falls down from its lofty heights.

The way I understand it, a helium atom works more or less like a hydrogen atom when it comes to ionization. The way I understand it, 471.3 nm represents a pretty high level of ionization for a helium electron. It has probably been kicked several floors up.

So, is a photon of 471.3 nm blue? Oh, sure it is!

Is 471.3 nm the dominant color of ionized helium? No, it doesn't look that way at all to me. Look at the first picture of this post, and take a look at that big fat line of yellow at 587.5 nm! Doesn't that look like the dominant color of helium to you? It sure does to me.

So let's put it like this: The blue color of 471.3 nm is not likely to dominate the light output of ionized helium, any more than the 486.1 nm blue light of hydrogen beta dominates any of our well-known ionized hydrogen emission nebulas. Because they are red from the 656.3 nm light of hydrogen alpha!

However, the inner parts of planetary nebulas are very often cyan-colored.

An old but relatively true-color image of the Cat's Eye Nebula. Photo: Photo: Adam Block/NOAO/AURA/NSF.

---

The bright cyan or green color of the inner parts of the Cat's Eye Nebula is real, and it results from doubly ionized oxygen at 501 nm. The red color in the outer parts likely results from hydrogen alpha. The reason why doubly ionized oxygen is so bright in the inner parts of many planetaries has to do with the fact that this oxygen emission requires very rarefied conditions and an onslaught of extremely energetic photons, which is exactly what you get near the central star of planetary nebulas.

So, in short: Is there blue light from helium emission at 471.3 nm in the central part of the Ring Nebula? Yes, undoubtedly. Is this blue helium emission brighter than the green emission from oxygen? No, absolutely not! Quite the opposite!

So the colors in the Hubble picture of the Ring Nebula are not exactly "true", but what we get is a reasonable approximation.

Ann

[Chris Peterson]

Chris,
OK, I imagined a simple explanation, and was wrong, but I can do complexity! You will be as short of time as any busy academic - please refer me to some source that will explain the mechanism?

Neufer,
Thank you! So the blue is due to Oxygen, not Helium? But why is that fluorescing near the star, while Nitrogen is, further out?
John

Sorry, I don't really have any reference to a good source. Of course, you could read about the details of how ionization relaxation and photon emission works on Wikipedia. That concept is pretty straightforward. But the ways that astronomers analyze data to figure out the spatial distribution of different elements in nebulas is more complicated, and would require tracking down a lot of pretty technical papers. Suffice to say that we're seeing a complex combination of gradients and different concentrations, presumably caused by layering in the star's outer layers, by successive ejections, and by interactions after ejection, we are seeing differences because different regions have different densities (with the consequences Art pointed out), and we are seeing differences because different areas are receiving different amounts of ionizing radiation, and different energies of ionizing radiation.

[JohnD]

Thank you, Ann and Chris!

My next assignment is clear! Meanwhile, and if I may, isn't a photon a photon? For photons to have different energies, they will have different frequencies? Sure, near the star, an atom is more likely to receive a photon more frequently, perhaps leading to the doubly ioniosed oxygen?

JOhn

[Ann]

Thank you, Ann and Chris!

My next assignment is clear! Meanwhile, and if I may, isn't a photon a photon? For photons to have different energies, they will have different frequencies? Sure, near the star, an atom is more likely to receive a photon more frequently, perhaps leading to the doubly ioniosed oxygen?

JOhn

Photons do indeed have different energies and frequencies.

A diagram of the electromagnetic spectrum, showing various properties across the range of frequencies and wavelengths. Illustration: Inductiveload, NASA - self-made, information by NASA Based off of File:EM Spectrum3-new.jpg by NASA The butterfly icon is from the P icon set, File:P biology.svg The humans are from the Pioneer plaque, File:Human.svg The buildings are the Petronas towers and the Empire State Buildings, both from File:Skyscrapercompare.svg A diagram of the Milton spectrum, showing the type, wavelength (with examples), frequency, the black body emission temperature. Temporary file for gauging response to an improved version of this file. Adapted from File:EM Spectrum3-new.jpg, which is a NASA image.

---

You wouldn't want to be hit by a gamma ray frequency photon, trust me! Okay, maybe getting hit by ONE such photon wouldn't hurt you. I don't know.

Let me point out, regarding the illustration, that I personally don't believe at all that the the most intense wavelength emitted by an object whose temperature is 10,000 K would be in the yellow-green part of the visual spectrum!

Ann

[neufer]

Ah! Looks like a Flower!
So this is how the sun might go?

• Where have all the graveyards gone?
  Long time passing.
  Where have all the graveyards gone?
  Long time ago.
  Where have all the graveyards gone?
  They're covered with flowers every one.
  Oh, When will they ever learn?
  Oh, When will they ever learn?

[Chris Peterson]

Thank you, Ann and Chris!

My next assignment is clear! Meanwhile, and if I may, isn't a photon a photon? For photons to have different energies, they will have different frequencies? Sure, near the star, an atom is more likely to receive a photon more frequently, perhaps leading to the doubly ioniosed oxygen?

JOhn

A photon may be a photon, but not all photons are the same. They have different energies (which yes, correspond to different frequencies). And the specific energy of a photon is critical to whether it can ionize some particular atom, and which ionization state that atom will enter. And the ionization state of the atom will determine the energy (frequency) of the photon that is re-emitted. The photon flux does not have any impact on the ionization levels or emitted wavelengths, only on how many atoms are ionized, and therefore how bright the region may appear. Doubly ionized oxygen results from a collision with single photon that is energetic enough to knock two electrons away, which is why we see [OIII] emission only around very hot sources. It is rare for an atom to absorb a photon. Absorbing two is vanishingly rare in that environment. The only place that multiphoton ionization can be studied is in the lab using very intense laser beams.

[JohnD]

Thank you all!
John

[NGC3314]

Well, 658 nm is the wavelength of ionized sulfur. It's seems relatively straightforward that sulfur should be ionized and made to emit photons in the relative presence of an extremely hot white dwarf. Similarly, 501 nm is the wavelength of doubly ionized oxygen. Doubly ionized oxygen represents a higher degree of ionization that ionized sulfur.

I do find the 469 nm ionized helium mystifying. I checked the emission lines of helium, and there appears to be no helium line at 469 nm. There is one at 471.3 nm, however.

The 469nm (4686 A) emission line is from He II (with a single electron), most often seen when that single electron hasjust recombined with a bare helium nucleus. It is analogous to Paschen-alpha of hydrogen at four times the wavelength (hydrogen ions with Z protons have their spectral lines at Z² times the energy of hydrogen). It takes 54.6 electron volts (four times the value 13.6 for hydrogen) to fully ionize helium, and only the very hottest stars can manage that over significant volumes. (In contrast, sulfur is practically always ionized at least once, since its first ionization energy is less than that of hydrogen so space is full of radiation that can ionize S.) A nonintuitive feature in the structures of nebulae is that the photons just above an ionization threshold are absorbed more effectively than photons at higher energies, so the average photon energy (say between the edges of H and He) increases going outward because the lower-energy ones were absorbed first. He II and the hydrogen lines come directly from recombination , while such "forbidden" lines as [O III], [N II], [ S II] etc. are collisionally excited as the loose electrons rattle around before actually binding again to a nucleus (so each electron can excite many forbidden-line photons during its free escapades. (Disclosure - I have become a big fan of He II emission as a way to trace photoionization by active galactic nuclei even when the gas has such low oxygen and nitrogen fraction that normal tests might suggest that it's ionized by hot stars. Cough, Hanny's Voorwerp and its kin, cough.)

(Hmm - I just learned that you have to put an extra space in [ S II] to keep BBS from interpreting that as strikethrough markup).

[Ann]

Well, 658 nm is the wavelength of ionized sulfur. It's seems relatively straightforward that sulfur should be ionized and made to emit photons in the relative presence of an extremely hot white dwarf. Similarly, 501 nm is the wavelength of doubly ionized oxygen. Doubly ionized oxygen represents a higher degree of ionization that ionized sulfur.

I do find the 469 nm ionized helium mystifying. I checked the emission lines of helium, and there appears to be no helium line at 469 nm. There is one at 471.3 nm, however.

The 469nm (4686 A) emission line is from He II (with a single electron), most often seen when that single electron hasjust recombined with a bare helium nucleus. It is analogous to Paschen-alpha of hydrogen at four times the wavelength (hydrogen ions with Z protons have their spectral lines at Z² times the energy of hydrogen). It takes 54.6 electron volts (four times the value 13.6 for hydrogen) to fully ionize helium, and only the very hottest stars can manage that over significant volumes. (In contrast, sulfur is practically always ionized at least once, since its first ionization energy is less than that of hydrogen so space is full of radiation that can ionize S.) A nonintuitive feature in the structures of nebulae is that the photons just above an ionization threshold are absorbed more effectively than photons at higher energies, so the average photon energy (say between the edges of H and He) increases going outward because the lower-energy ones were absorbed first. He II and the hydrogen lines come directly from recombination , while such "forbidden" lines as [O III], [N II], [ S II] etc. are collisionally excited as the loose electrons rattle around before actually binding again to a nucleus (so each electron can excite many forbidden-line photons during its free escapades. (Disclosure - I have become a big fan of He II emission as a way to trace photoionization by active galactic nuclei even when the gas has such low oxygen and nitrogen fraction that normal tests might suggest that it's ionized by hot stars. Cough, Hanny's Voorwerp and its kin, cough.)

(Hmm - I just learned that you have to put an extra space in [ S II] to keep BBS from interpreting that as strikethrough markup).

Thank you so much for your great explanation!

Ann