Starship Asterisk*

The Great Globular Cluster in Hercules

Explanation: In 1716, English astronomer Edmond Halley noted, "This is but a little Patch, but it shows itself to the naked Eye, when the Sky is serene and the Moon absent." Of course, M13 is now less modestly recognized as the Great Globular Cluster in Hercules, one of the brightest globular star clusters in the northern sky. Sharp telescopic views like this one reveal the spectacular cluster's hundreds of thousands of stars. At a distance of 25,000 light-years, the cluster stars crowd into a region 150 light-years in diameter. Approaching the cluster core, upwards of 100 stars could be contained in a cube just 3 light-years on a side. For comparison, the closest star to the Sun is over 4 light-years away. The deep, wide-field image also reveals distant background galaxies including NGC 6207 at the upper left, and faint, foreground Milky Way dust clouds known to some as integrated flux nebulae.

<< Previous APOD

This Day in APOD

Next APOD >>

I love seeing the Globular star clusters. I've seen them through telescope, it looks amazing. 100 stars within 3 ly cube.. must be so amazing the view of night sky from any planet inside them but i also can't imagine a nightmare (and how devastating it could be unless planet gets flung out..) if some planet gets caught in 2 stars slamming each other being so close to each other (as the description in one of the links says - "These stars are so crowded that they can, at times, slam into each other and even form a new star, called a "blue straggler.")

It's a nice APOD:


What I admire most about it is actually the faint gray Milky Way cirrus clouds of dusty gas, called the IFN. Note that we can see two separate IFN clouds, one to the right of M13 and one to the left of it.


One can almost imagine that these IFN clouds are the remnants of the humongously large and massive nebula that originally gave birth to Messier 13. But of course, the idea that we are seeing some remnants of that nebula is impossible. The estimated age of M13, according to Wikipedia, is 11.65 billion years. Any remnants of the nebula that created M13 must have dissolved and mixed with the interstellar medium of the Milky Way very long ago.


As for M13 itself, my favorite portrait of this great globular is the one by ESA/Hubble that really emphasizes the blue horizontal branch stars of this huge star city. Admittedly though, ESA/Hubble has created another portrait of M13 that is not at all so "blue-forward":


The choice of filters, and the choices made during the processing of the image, determine the colors and the color balance of the final image.

Anyway, M13 really is rich in blue horizontal branch stars. These are evolved low-mass low-metallicity stars that have shrunk mightily and turned blue after they started fuse helium to carbon and oxygen in their cores. First these stars fused hydrogen to helium in their cores, then they ran out of core hydrogen so that their cores became inert and started shrinking, releasing potential gravitational energy that made their outer atmospheres expand mightily, so that the stars became red giants. But then the cores of these stars became hot enough to start fusing helium to carbon and oxygen, and then (due to some process that I don't understand) the outer atmospheres of these stars shrunk quite dramatically, and the light that these "smallish hot giants" emitted was now blue. (And if you are wondering, when these stars have used up their core helium they will become asymtotic branch bright red stars, and when they are done with their asymtotic giant branch phase they will cast off their outer atmospheres and become white dwarfs.


M13 is a very handsome cluster, which looks extra nice in the sky because it is relatively close to us at "only" some 22,200–25,000 light-years away from us. M13 is also big, and according to Wikipedia, it is home to some 300,000-500,000 stars. M13 is metal-poor, which means that its stars contain much lower levels of elements heavier than hydrogen and helium than our own Sun does. This low metallicity is the reason why M13 can host blue horizontal branch stars in the first place, because in metal-rich stars like the Sun the horizontal branch stage just makes the stars become a bit less reddish and a bit more yellow-orange.

Let me say just a little something about the background galaxy that is seen along the line of sight near M13, NGC 6207.


Maybe you can tell from the appearance of this galaxy that it is likely to be smallish but not tiny. It is smallish because its spiral structure is not very well-developed, and it has two long dust lanes full of star formation that seem to have no connection with the spiral arms. On the other hand, the galaxy has a substantial yellow population and an obvious core region, so it is not is not a dwarf galaxy. According to PGC, Principal Galaxy Catalog, NGC 6207 is located some 46 million light-years away, and its luminosity is about half that of the Milky Way. While I can't guarantee that either figure is correct, they seem reasonable to me.

Ann

Many thanks for the feedback and interesting explanations. We were actually planning to create a Hertzsprung-Russell diagram from the image. We will add this to our homepage soon. When we revise the image again, we will take the blue horizontal branch into account. We did use “photometric color calibration”, but as you say, it's always a bit of an interpretation. One interesting thing I would like to add at this point: as far as I know, there are no other images of M13 with IFN, but there are a number of optical observers who have made drawings of the nebulae with large optics. This motivated us to invest a long exposure time in the object. Here is an example: https://www.cloudynights.com/topic/9350 ... ds-of-m13/

Best
Julian

Ann wrote: ↑Thu Sep 26, 2024 7:53 am
But then the cores of these stars became hot enough to start fusing helium to carbon and oxygen, and then (due to some process that I don't understand) the outer atmospheres of these stars shrunk quite dramatically,

Ann

It does seem odd indeed that a red giant grows in size when its core is inert, but shrinks when it's fusing! And both times around at that: during the first ascent the star grows with an inert helium core, then it shrinks when the helium core ignites and begins fusion, and when this core is done fusing and leaves an inert carbon core behind, the star grows bigger than ever! I've read it has to do with the fact that the increasing energy which the inert and contracting core is accumulating finally radiates away when it ignites, and this stops the expansion of the star. But I'm not sure I get this explanation! It seems it could equally explain the opposite result! Of course, I'm missing something.

p.s. BEAUTIFUL APOD!

Christian G. wrote: ↑Thu Sep 26, 2024 3:02 pm

Ann wrote: ↑Thu Sep 26, 2024 7:53 am
But then the cores of these stars became hot enough to start fusing helium to carbon and oxygen, and then (due to some process that I don't understand) the outer atmospheres of these stars shrunk quite dramatically,

Ann

It does seem odd indeed that a red giant grows in size when its core is inert, not when it's fusing! And both times around at that: during the first ascent the star grows with an inert helium core, then it shrinks when the helium core ignites and begins fusion, and when this core is done fusing and leaves an inert carbon core behind, the star no longer shrinks but gets bigger than ever! I've read it has to do with the fact that the increasing energy which the inert and contracting core is accumulating finally radiates away when it ignites, and this stops the expansion of the star. But I'm not sure I get this explanation! It seems it could equally explain the opposite result! Of course, I'm missing something.

p.s. BEAUTIFUL APOD!

My best answer to you is this:

What happens when a star's core becomes inert and starts shrinking, is that a lot of gravitational energy is released. Or, in other words: More energy is released when the core is shrinking than when the core produces energy through fusion.

Take a look at this diagram of a one solar mass pre-main sequence and then main sequence evolution:


Before a star reaches the main sequence and starts fusing hydrogen to helium in its core, it shines by releasing gravitational energy. It does so by shrinking. The process of shrinking releases more energy than the process of fusion, and the star is brighter than it will be when it reaches the main sequence.

As the star shrinks, it gets hotter and hotter. Eventually its core gets hot enough to start fusing hydrogen to helium. When this happens, the star stops shrinking, because the energy generated in the core of the star counteracts the gravity that wants to make the star shrink even further.

Since the star's core also shrinks when the star leaves the main sequence and becomes a red giant, that in itself should be enough to make the star grow larger and brighter. But bear in mind that even during its red giant phase, the star keeps generating energy through fusion, but not in its core. Fusion takes place in a shell outside the core, which is to say that the star is "out of balance". Because when the star was on the main sequence, the outward-pushing energy that the star generated in its core and the gravitational forces that wanted to compress the star balanced each other perfectly. But when the core becomes inert, and fusion moves out away from the core, the forces that want to enlarge the parts of the star that are outside the core are greater than the gravitational energy that wants to compress the star. Therefore, the star grows larger.


Ann

Ann wrote: ↑Thu Sep 26, 2024 3:53 pm

Christian G. wrote: ↑Thu Sep 26, 2024 3:02 pm

Ann wrote: ↑Thu Sep 26, 2024 7:53 am
But then the cores of these stars became hot enough to start fusing helium to carbon and oxygen, and then (due to some process that I don't understand) the outer atmospheres of these stars shrunk quite dramatically,

Ann

It does seem odd indeed that a red giant grows in size when its core is inert, not when it's fusing! And both times around at that: during the first ascent the star grows with an inert helium core, then it shrinks when the helium core ignites and begins fusion, and when this core is done fusing and leaves an inert carbon core behind, the star no longer shrinks but gets bigger than ever! I've read it has to do with the fact that the increasing energy which the inert and contracting core is accumulating finally radiates away when it ignites, and this stops the expansion of the star. But I'm not sure I get this explanation! It seems it could equally explain the opposite result! Of course, I'm missing something.

p.s. BEAUTIFUL APOD!

My best answer to you is this:

What happens when a star's core becomes inert and starts shrinking, is that a lot of gravitational energy is released. Or, in other words: More energy is released when the core is shrinking than when the core produces energy through fusion.

Take a look at this diagram of a one solar mass pre-main sequence and then main sequence evolution:

Pre main sequence stellar evolution physics stackexchange.png

A pre-main sequence solar mass star is brighter than it will be when it reaches
the main sequence. Credit: Possibly physics.stackexchange.com

Before a star reaches the main sequence and starts fusing hydrogen to helium in its core, it shines by releasing gravitational energy. It does so by shrinking. The process of shrinking releases more energy than the process of fusion, and the star is brighter than it will be when it reaches the main sequence.

As the star shrinks, it gets hotter and hotter. Eventually its core gets hot enough to start fusing hydrogen to helium. When this happens, the star stops shrinking, because the energy generated in the core of the star counteracts the gravity that wants to make the star shrink even further.

Since the star's core also shrinks when the star leaves the main sequence and becomes a red giant, that in itself should be enough to make the star grow larger and brighter. But bear in mind that even during its red giant phase, the star keeps generating energy through fusion, but not in its core. Fusion takes place in a shell outside the core, which is to say that the star is "out of balance". Because when the star was on the main sequence, the outward-pushing energy that the star generated in its core and the gravitational forces that wanted to compress the star balanced each other perfectly. But when the core becomes inert, and fusion moves out away from the core, the forces that want to enlarge the parts of the star that are outside the core are greater than the gravitational energy that wants to compress the star. Therefore, the star grows larger.

This red giant star has an inert core made of carbon and oxygen. Outside the core, the star has a helium shell where fusion may be going on, where, if so, energy is generated as helium fuses into carbon and oxygen. Actually, a star like this probably also has a hydrogen shell outside its helium shell where hydrogen is fusing into helium. Credit: Sorry, I don't know.

---

Ann

Thank you very much for this answer, Ann, all the more since it touches on another odd fact for me! Namely that some pre-main sequence stars outshine the main sequence stars they will become.
When you say that in these cases the young star shines by "gravitational energy" instead of nuclear fusion, does that simply refer to the process of gravity increasing pressure and heat at the center, and that it's this heat which makes the star shine? (not sure if I'm stating the obvious or completely misunderstanding)

Wow - 100 stars per 3 cu light years! Just curious - what is the comparable estimated star density of our galactic center?

You can have a look! The Arches cluster is very close to the center of our galaxy and often described as the densest known cluster. With the same density there would be 100 000 stars between us and the nearest star to the Sun! Supposedly. And you could remove the central cluster from the image, only see the vast number of surrounding stars, and still believe you are seeing a cluster! A dense region indeed.
But is this cluster really denser than a rich globular cluster such as M13? There are only some 200 though very massive stars in there, plus a few thousand lesser massive ones, but still this is far from the hundreds of thousands found in globulars. All within an area only a light-year across instead of the 100 ly or more as in globulars, however. Hubble

Christian G. wrote: ↑Thu Sep 26, 2024 5:17 pm

Ann wrote: ↑Thu Sep 26, 2024 3:53 pm

Christian G. wrote: ↑Thu Sep 26, 2024 3:02 pm

It does seem odd indeed that a red giant grows in size when its core is inert, not when it's fusing! And both times around at that: during the first ascent the star grows with an inert helium core, then it shrinks when the helium core ignites and begins fusion, and when this core is done fusing and leaves an inert carbon core behind, the star no longer shrinks but gets bigger than ever! I've read it has to do with the fact that the increasing energy which the inert and contracting core is accumulating finally radiates away when it ignites, and this stops the expansion of the star. But I'm not sure I get this explanation! It seems it could equally explain the opposite result! Of course, I'm missing something.

p.s. BEAUTIFUL APOD!

My best answer to you is this:

What happens when a star's core becomes inert and starts shrinking, is that a lot of gravitational energy is released. Or, in other words: More energy is released when the core is shrinking than when the core produces energy through fusion.

Take a look at this diagram of a one solar mass pre-main sequence and then main sequence evolution:

Pre main sequence stellar evolution physics stackexchange.png

A pre-main sequence solar mass star is brighter than it will be when it reaches
the main sequence. Credit: Possibly physics.stackexchange.com

Before a star reaches the main sequence and starts fusing hydrogen to helium in its core, it shines by releasing gravitational energy. It does so by shrinking. The process of shrinking releases more energy than the process of fusion, and the star is brighter than it will be when it reaches the main sequence.

As the star shrinks, it gets hotter and hotter. Eventually its core gets hot enough to start fusing hydrogen to helium. When this happens, the star stops shrinking, because the energy generated in the core of the star counteracts the gravity that wants to make the star shrink even further.

Since the star's core also shrinks when the star leaves the main sequence and becomes a red giant, that in itself should be enough to make the star grow larger and brighter. But bear in mind that even during its red giant phase, the star keeps generating energy through fusion, but not in its core. Fusion takes place in a shell outside the core, which is to say that the star is "out of balance". Because when the star was on the main sequence, the outward-pushing energy that the star generated in its core and the gravitational forces that wanted to compress the star balanced each other perfectly. But when the core becomes inert, and fusion moves out away from the core, the forces that want to enlarge the parts of the star that are outside the core are greater than the gravitational energy that wants to compress the star. Therefore, the star grows larger.

This red giant star has an inert core made of carbon and oxygen. Outside the core, the star has a helium shell where fusion may be going on, where, if so, energy is generated as helium fuses into carbon and oxygen. Actually, a star like this probably also has a hydrogen shell outside its helium shell where hydrogen is fusing into helium. Credit: Sorry, I don't know.

---

Ann

Thank you very much for this answer, Ann, all the more since it touches on another odd fact for me! Namely that some pre-main sequence stars outshine the main sequence stars they will become.
When you say that in these cases the young star shines by "gravitational energy" instead of nuclear fusion, does that simply refer to the process of gravity increasing pressure and heat at the center, and that it's this heat which makes the star shine? (not sure if I'm stating the obvious or completely misunderstanding)

I think it's called "potential energy".

This rock formation in Norway is called the "Troll's tongue":


The person sitting at the edge of the Troll's tongue is being held up by the strength of the rock formation. The way I understand it, this is the potential energy holding the person up. If the rock formation was to crack open, the person sitting on it would lose his or her potential energy and fall 700 meters (~2000 feet) down.

I think that stars are losing potential energy when they shrink. That's a lot of energy lost, and this energy is turned into heat. I think!!

You'd probably better ask Chris!

Ann

markc wrote: ↑Thu Sep 26, 2024 5:27 pm Wow - 100 stars per 3 cu light years!

Actually it is 100 stars per 27 cu light years.

APOD Robot wrote: ↑Thu Sep 26, 2024 4:06 am
...100 stars could be contained in a cube just 3 light-years on a side.

So the cube is 3 by 3 by 3.

GoodFoodStove wrote: ↑Thu Sep 26, 2024 6:27 pm

markc wrote: ↑Thu Sep 26, 2024 5:27 pm Wow - 100 stars per 3 cu light years!

Actually it is 100 stars per 27 cu light years.

APOD Robot wrote: ↑Thu Sep 26, 2024 4:06 am
...100 stars could be contained in a cube just 3 light-years on a side.

So the cube is 3 by 3 by 3.

Yes. And so, the average distance between stars would be about 0.646 light years (3/0.646)³= 100)

I would like to see a post on UV bright star Barnard 29 (not to be confused with a dark nebula of the same name) with some discussion on the unique place it has on the Hertzsprung-Russell diagram. Thanks. How was this star found in the first place? Did someone have UV film and the star was ultra-bright?

Randall Rathbun wrote: ↑Thu Sep 26, 2024 7:29 pm I would like to see a post on UV bright star Barnard 29 (not to be confused with a dark nebula of the same name) with some discussion on the unique place it has on the Hertzsprung-Russell diagram. Thanks. How was this star found in the first place? Did someone have UV film and the star was ultra-bright?

I doubt the UV is detectable from the ground. The star appears very blue, and by measuring its intensity at two or more wavelengths the shape of the blackbody can be inferred, and therefore the temperature. At something like 21,000 K, the peak output will be around 140 nm, in the UV... even if that can't be measured directly through the atmosphere.

Ann wrote: ↑Thu Sep 26, 2024 6:10 pm
This rock formation in Norway is called the "Troll's tongue":

---

Ann

Great: https://www.youtube.com/watch?v=tUNbhYcY9Ik

Chris Peterson wrote: ↑Thu Sep 26, 2024 7:55 pm

Randall Rathbun wrote: ↑Thu Sep 26, 2024 7:29 pm I would like to see a post on UV bright star Barnard 29 (not to be confused with a dark nebula of the same name) with some discussion on the unique place it has on the Hertzsprung-Russell diagram. Thanks. How was this star found in the first place? Did someone have UV film and the star was ultra-bright?

I doubt the UV is detectable from the ground. The star appears very blue, and by measuring its intensity at two or more wavelengths the shape of the blackbody can be inferred, and therefore the temperature. At something like 21,000 K, the peak output will be around 140 nm, in the UV... even if that can't be measured directly through the atmosphere.

How did the two of you, Randall and Chris, find out about this star? I asked Simbad Astronomical database about Barnard 29, and all it would give me was the dark cloud. The only source I could find that would talk about this star at all was messier.seds.org., and it gave me the designation NGC 6205 222, which was accepted by Simbad. (Simbad itself called the star 2MASS J16413368+3626077, which is a mouthful.)

Unfortunately, Simbad has no information on the U magnitude of this star. It does tell us that the B magnitude of the star is 12.98, the V magnitude is 13.14, and the J, K and H magnitudes are progressively (but slowly) fainter. According to Simbad, the spectral class of this star is B2p D ~. I don't know what D ~ means, but B2p should mean that the star is spectral class B2, but peculiar. Of course, since it is a post-AGB star, it will indeed be peculiar compared to stars of spectral class B2 that are supporting themselves through fusion.

My point, however, is that this star is spectral class B2 and not O. A main sequence star of spectral class B2 will have a temperature of ~21,000 K, according to University of Northern Iowa. Some central stars of planetary nebulas are known to be much, much hotter. And yes, I know that you know that. And all that it really means is that the star is in the process of removing its outer layers, but it has most certainly not finished that process or cast off all of its outer atmosphere.

The Gaia parallax of this star is 0.0775 ± 0.0298 milliarcseconds. A parallax of 0.0775 mas corresponds to a distance of 12,900 parsecs or 42,000 light-years. That's too far away for M13, whose accepted distance is 22,000-25.000 light-years. If we use the uncertainty of the Gaia measurement to make the parallax as large as possible, we get a parallax of 0.1073, corresponding to a distance of about 30,000 light-years, which still seems a bit far away. Using that distance, however, and the star's B and V magnitudes, gives an absolute B magnitude of some 440 solar luminosities and a V magnitude of some 380 solar luminosities. That sounds reasonable for a star of spectral class B2.

I tried to find a reasonably good color photo of a star of spectral class B2, to show it here as a proxy for Barnard 29. Well, you wouldn't believe how hard it was to find a star of spectral class B2 that is well-known enough to have had its portrait taken. Finally I found Alpha Musca, which gets photographed only because it is located next to the Dark Doodad.

Edit: Oh no!!! I showed you a picture not of B2-type alpha Musca, but of B5-type gamma Musca!!!

All right, but here you can really see (at left) both alpha and beta Musca, which are both spectral class B2:


Full size of the image is here: https://apod.nasa.gov/apod/image/2303/D ... 24crop.jpg


Still, I have to ask, how did you, Randall, hear of Barnard 29?

Ann

Ann wrote: ↑Fri Sep 27, 2024 4:16 am

Chris Peterson wrote: ↑Thu Sep 26, 2024 7:55 pm

Randall Rathbun wrote: ↑Thu Sep 26, 2024 7:29 pm I would like to see a post on UV bright star Barnard 29 (not to be confused with a dark nebula of the same name) with some discussion on the unique place it has on the Hertzsprung-Russell diagram. Thanks. How was this star found in the first place? Did someone have UV film and the star was ultra-bright?

I doubt the UV is detectable from the ground. The star appears very blue, and by measuring its intensity at two or more wavelengths the shape of the blackbody can be inferred, and therefore the temperature. At something like 21,000 K, the peak output will be around 140 nm, in the UV... even if that can't be measured directly through the atmosphere.

How did the two of you, Randall and Chris, find out about this star? I asked Simbad Astronomical database about Barnard 29, and all it would give me was the dark cloud. The only source I could find that would talk about this star at all was messier.seds.org., and it gave me the designation NGC 6205 222, which was accepted by Simbad. (Simbad itself called the star 2MASS J16413368+3626077, which is a mouthful.)

Unfortunately, Simbad has no information on the U magnitude of this star. It does tell us that the B magnitude of the star is 12.98, the V magnitude is 13.14, and the J, K and H magnitudes are progressively (but slowly) fainter. According to Simbad, the spectral class of this star is B2p D ~. I don't know what D ~ means, but B2p should mean that the star is spectral class B2, but peculiar. Of course, since it is a post-AGB star, it will indeed be peculiar compared to stars of spectral class B2 that are supporting themselves through fusion.

My point, however, is that this star is spectral class B2 and not O. A main sequence star of spectral class B2 will have a temperature of ~21,000 K, according to University of Northern Iowa. Some central stars of planetary nebulas are known to be much, much hotter. And yes, I know that you know that. And all that it really means is that the star is in the process of removing its outer layers, but it has most certainly not finished that process or cast off all of its outer atmosphere.

The Gaia parallax of this star is 0.0775 ± 0.0298 milliarcseconds. A parallax of 0.0775 mas corresponds to a distance of 12,900 parsecs or 42,000 light-years. That's too far away for M13, whose accepted distance is 22,000-25.000 light-years. If we use the uncertainty of the Gaia measurement to make the parallax as large as possible, we get a parallax of 0.1073, corresponding to a distance of about 30,000 light-years, which still seems a bit far away. Using that distance, however, and the star's B and V magnitudes, gives an absolute B magnitude of some 440 solar luminosities and a V magnitude of some 380 solar luminosities. That sounds reasonable for a star of spectral class B2.

Still, I have to ask, how did you, Randall, hear of this star?

Ann

I just got the most recent temperature estimate from this paper: https://arxiv.org/pdf/1903.00350 . A photometric U filter has its center at around 360 nm, which is still quite a bit longer than the 140 nm peak of a 21,000 K source. Of course, there will be something there. But the star temperature can be calculated from B and V filters, too.

Ann wrote: ↑Fri Sep 27, 2024 4:16 am

Chris Peterson wrote: ↑Thu Sep 26, 2024 7:55 pm

Randall Rathbun wrote: ↑Thu Sep 26, 2024 7:29 pm I would like to see a post on UV bright star Barnard 29 (not to be confused with a dark nebula of the same name) with some discussion on the unique place it has on the Hertzsprung-Russell diagram. Thanks. How was this star found in the first place? Did someone have UV film and the star was ultra-bright?

I doubt the UV is detectable from the ground. The star appears very blue, and by measuring its intensity at two or more wavelengths the shape of the blackbody can be inferred, and therefore the temperature. At something like 21,000 K, the peak output will be around 140 nm, in the UV... even if that can't be measured directly through the atmosphere.

How did the two of you, Randall and Chris, find out about this star? I asked Simbad Astronomical database about Barnard 29, and all it would give me was the dark cloud. The only source I could find that would talk about this star at all was messier.seds.org., and it gave me the designation NGC 6205 222, which was accepted by Simbad. (Simbad itself called the star 2MASS J16413368+3626077, which is a mouthful.)

Unfortunately, Simbad has no information on the U magnitude of this star. It does tell us that the B magnitude of the star is 12.98, the V magnitude is 13.14, and the J, K and H magnitudes are progressively (but slowly) fainter. According to Simbad, the spectral class of this star is B2p D ~. I don't know what D ~ means, but B2p should mean that the star is spectral class B2, but peculiar. Of course, since it is a post-AGB star, it will indeed be peculiar compared to stars of spectral class B2 that are supporting themselves through fusion.

My point, however, is that this star is spectral class B2 and not O. A main sequence star of spectral class B2 will have a temperature of ~21,000 K, according to University of Northern Iowa. Some central stars of planetary nebulas are known to be much, much hotter. And yes, I know that you know that. And all that it really means is that the star is in the process of removing its outer layers, but it has most certainly not finished that process or cast off all of its outer atmosphere.

The Gaia parallax of this star is 0.0775 ± 0.0298 milliarcseconds. A parallax of 0.0775 mas corresponds to a distance of 12,900 parsecs or 42,000 light-years. That's too far away for M13, whose accepted distance is 22,000-25.000 light-years. If we use the uncertainty of the Gaia measurement to make the parallax as large as possible, we get a parallax of 0.1073, corresponding to a distance of about 30,000 light-years, which still seems a bit far away. Using that distance, however, and the star's B and V magnitudes, gives an absolute B magnitude of some 440 solar luminosities and a V magnitude of some 380 solar luminosities. That sounds reasonable for a star of spectral class B2.

I tried to find a reasonably good color photo of a star of spectral class B2, to show it here as a proxy for Barnard 29. Well, you wouldn't believe how hard it was to find a star of spectral class B2 that is well-known enough to have had its portrait taken. Finally I found Alpha Musca, which gets photographed only because it is located next to the Dark Doodad.

B2V-type blue star Alpha Musca is seen next to the Dark Doodad nebula. Globular cluster NGC 4372 is also in the picture. Post-AGB star Barnard 29 is similar in color to Alpha Musca. Credit: Andrey Oreshko

---

Still, I have to ask, how did you, Randall, hear of Barnard 29?

Ann

In the UV, Barnard 29 actually shows a slight halo around the star.
But the star still poses several mysteries, such as why it is practically invisible in the IR.

"A post-AGB star (pAGB, abbreviation of post-asymptotic giant branch) like Barnard 29* (or L2 Puppis, RV Tauri, R Scuti od U Monocerotis) is a type of luminous supergiant star of intermediate mass in a very late phase of stellar evolution. The post-AGB stage occurs after the asymptotic giant branch (AGB or second-ascent red giant) has ended. The stage sees the dying star, initially very cool and large, shrink and heat up.The duration of the post-AGB stage varies based on the star's initial mass, and can range from 100,000 years for a solar-mass star to just over 1,000 years for more massive stars. The timescale gets slightly shorter with lower metallicity.

Towards the end of this stage, post-AGB stars also tend to produce protoplanetary nebulae as they shed their outer layers, and this creates a large infrared excess and obscures the stars in visible light. After reaching an effective temperature of about 30,000 K, the star is able to ionise its surrounding nebula, producing a true planetary nebula.
Properties

Post-AGB stars span a large range of temperatures, as they are in the process of heating up from very cool temperatures (3,000 K or less) up to about 30,000 K. Technically, the post-AGB stage only ends when the star reaches its maximum temperature of 100-200,000 K,[2] but beyond 30,000 K, the star ionises the surrounding gas and would be considered a central star of a planetary nebula more often than a post-AGB star.

On the other hand, the luminosity of post-AGB stars is usually constant throughout the post-AGB stage, and slightly dependent on the star’s core mass, and getting slightly brighter with lower metallicity." https://en.wikipedia.org/wiki/Post-AGB_star and "Iron abundances of B-type post-Asymptotic Giant Branch stars in globular clusters: Barnard 29 in M 13 and ROA 5701 in omega Cen" H. M. A. Thompson, F. P. Keenan, P. L. Dufton, R. S. I. Ryans, J. V. Smoker, D. L. Lambert, A. A. Zijlstra

XMM-Newton-UV and SDSS9-opt jac berne (flickr)
ALLWISE-MidIR and XMM-Newton-UV jac berne (flickr)

AVAO wrote: ↑Fri Sep 27, 2024 5:13 am
In the UV, Barnard 29 actually shows a slight halo around the star.
But the star still poses several mysteries, such as why it is practically invisible in the IR.

"A post-AGB star (pAGB, abbreviation of post-asymptotic giant branch) like Barnard 29 (or L2 Puppis, RV Tauri, R Scuti od U Monocerotis) is a type of luminous supergiant star of intermediate mass in a very late phase of stellar evolution. The post-AGB stage occurs after the asymptotic giant branch (AGB or second-ascent red giant) has ended. The stage sees the dying star, initially very cool and large, shrink and heat up.The duration of the post-AGB stage varies based on the star's initial mass, and can range from 100,000 years for a solar-mass star to just over 1,000 years for more massive stars. The timescale gets slightly shorter with lower metallicity.

Towards the end of this stage, post-AGB stars also tend to produce protoplanetary nebulae as they shed their outer layers, and this creates a large infrared excess and obscures the stars in visible light. After reaching an effective temperature of about 30,000 K, the star is able to ionise its surrounding nebula, producing a true planetary nebula.
Properties

Post-AGB stars span a large range of temperatures, as they are in the process of heating up from very cool temperatures (3,000 K or less) up to about 30,000 K. Technically, the post-AGB stage only ends when the star reaches its maximum temperature of 100-200,000 K,[2] but beyond 30,000 K, the star ionises the surrounding gas and would be considered a central star of a planetary nebula more often than a post-AGB star.

On the other hand, the luminosity of post-AGB stars is usually constant throughout the post-AGB stage, and slightly dependent on the star’s core mass, and getting slightly brighter with lower metallicity." https://en.wikipedia.org/wiki/Post-AGB_star

Click to view full size image 1 or image 2

XMM-Newton-UV and SDSS9-opt jac berne (flickr)

Click to view full size image 1 or image 2

ALLWISE-MidIR and XMM-Newton-UV jac berne (flickr)

Thanks again, Jac, your images contribute so much to Starship Asterisk!

Wikipedia wrote:

The post-AGB stage occurs after the asymptotic giant branch (AGB or second-ascent red giant) has ended. The stage sees the dying star, initially very cool and large, shrink and heat up.

So the post-AGB stars is another class of stars that heat up because they shrink.

Ann

Chris Peterson wrote: ↑Fri Sep 27, 2024 4:35 am

Ann wrote: ↑Fri Sep 27, 2024 4:16 am

Chris Peterson wrote: ↑Thu Sep 26, 2024 7:55 pm

I doubt the UV is detectable from the ground. The star appears very blue, and by measuring its intensity at two or more wavelengths the shape of the blackbody can be inferred, and therefore the temperature. At something like 21,000 K, the peak output will be around 140 nm, in the UV... even if that can't be measured directly through the atmosphere.

How did the two of you, Randall and Chris, find out about this star? I asked Simbad Astronomical database about Barnard 29, and all it would give me was the dark cloud. The only source I could find that would talk about this star at all was messier.seds.org., and it gave me the designation NGC 6205 222, which was accepted by Simbad. (Simbad itself called the star 2MASS J16413368+3626077, which is a mouthful.)

Unfortunately, Simbad has no information on the U magnitude of this star. It does tell us that the B magnitude of the star is 12.98, the V magnitude is 13.14, and the J, K and H magnitudes are progressively (but slowly) fainter. According to Simbad, the spectral class of this star is B2p D ~. I don't know what D ~ means, but B2p should mean that the star is spectral class B2, but peculiar. Of course, since it is a post-AGB star, it will indeed be peculiar compared to stars of spectral class B2 that are supporting themselves through fusion.

My point, however, is that this star is spectral class B2 and not O. A main sequence star of spectral class B2 will have a temperature of ~21,000 K, according to University of Northern Iowa. Some central stars of planetary nebulas are known to be much, much hotter. And yes, I know that you know that. And all that it really means is that the star is in the process of removing its outer layers, but it has most certainly not finished that process or cast off all of its outer atmosphere.

The Gaia parallax of this star is 0.0775 ± 0.0298 milliarcseconds. A parallax of 0.0775 mas corresponds to a distance of 12,900 parsecs or 42,000 light-years. That's too far away for M13, whose accepted distance is 22,000-25.000 light-years. If we use the uncertainty of the Gaia measurement to make the parallax as large as possible, we get a parallax of 0.1073, corresponding to a distance of about 30,000 light-years, which still seems a bit far away. Using that distance, however, and the star's B and V magnitudes, gives an absolute B magnitude of some 440 solar luminosities and a V magnitude of some 380 solar luminosities. That sounds reasonable for a star of spectral class B2.

Still, I have to ask, how did you, Randall, hear of this star?

Ann

I just got the most recent temperature estimate from this paper: https://arxiv.org/pdf/1903.00350 . A photometric U filter has its center at around 360 nm, which is still quite a bit longer than the 140 nm peak of a 21,000 K source. Of course, there will be something there. But the star temperature can be calculated from B and V filters, too.

Wow, Chris, that is so fascinating! I goggled at this:

The post-HB evolution of M13 stars is illustrated in Fig. 13. The coolest (Teff . 8,000 K), most massive (MZAHB & 0.70M) stars populate the red horizontal branch (RHB, red points). As they depart the HB, they evolve to the red and ascend the AGB. Stars on the blue horizontal branch (BHB, blue points) are hotter (8,000 K . Teff . 20,000 K) and have lower masses (0.52 . MZAHB . 0.70M). Most post-BHB stars also climb the AGB and reach the thermal-pulsing AGB phase, but only for a short time. The hottest (Teff & 20,000 K), least-massive (MZAHB . 0.52M) stars populate the extreme horizontal branch (EHB, cyan points). Most of these stars do not climb the AGB after the He-core burning phase, but evolve to high luminosities with little change in temperature. Post-EHB stars are also known as AGB-manqúe stars. Stars near the boundary between the EHB and the BHB (Teff ∼ 20,000 K, MZAHB ∼ 0.52M) follow an intermediate path: they climb the AGB, but depart before reaching the thermal-pulsing phase, becoming post-early AGB (post-EAGB) stars.

That's so fascinating! The red horizontal branch stars become pulsating bright red AGB stars. The blue horizontal branch stars reach the AGB pulsating stage, but only for a short time. And the extreme horizontal branch stars never reach the asymtotic giant branch at all! But they do become much brighter when they have used up their core helium, retaining much the same temperature, before burning out and turning into white dwarfs! Wow!

As for Barnard 29, it never pulsated, so it never underwent the chemical mixing that is expected from AGB and post-AGB stars. Fantastic.

Thanks, Chris!

Ann

Randall Rathbun wrote: ↑Thu Sep 26, 2024 7:29 pm I would like to see a post on UV bright star Barnard 29 (not to be confused with a dark nebula of the same name) with some discussion on the unique place it has on the Hertzsprung-Russell diagram. Thanks. How was this star found in the first place? Did someone have UV film and the star was ultra-bright?

The star was found during the infancy of astrophotography for a very simple reason: A hundred years ago and more, only blue- and ultraviolet-sensitive photographic film was available. Take at look at an early picture of the Southern Cross (Crux) and compare with a recent one:


And because only blue- and ultraviolet-sensitive film was available in the early days of astrophotography, you can imagine that a hot blue star like Barnard 29 would have been really obvious in the old photos of M13.

William V. Dixon et al. wrote:

The bright, blue star Barnard 29 in Messier 13 (NGC 6205) has intrigued astronomers for more than a century. In a series of papers, Barnard (1900, 1909, 1914) pointed out that some of the cluster’s stars “shine with a much bluer light than the great majority” and cited Barnard 29 as the “most striking example” of this group.

Clearly Barnard 29 was much more obvious and striking in the early 1900s than it is today!

Ann

Ann wrote: ↑Fri Sep 27, 2024 2:42 pm
Clearly Barnard 29 was much more obvious and striking in the early 1900s than it is today!

Not really. Only when imaged across a broad spectrum (which wasn't possible 100 years ago). But we don't do that much anymore. We image across multiple spectral ranges. Which means the star is every bit as obvious and striking today in you compare apples to apples.

Chris Peterson wrote: ↑Fri Sep 27, 2024 3:16 pm

Ann wrote: ↑Fri Sep 27, 2024 2:42 pm
Clearly Barnard 29 was much more obvious and striking in the early 1900s than it is today!

Not really. Only when imaged across a broad spectrum (which wasn't possible 100 years ago). But we don't do that much anymore. We image across multiple spectral ranges. Which means the star is every bit as obvious and striking today in you compare apples to apples.

No? Why was this star "discovered" and found to be very remarkably blue compared to the other members of M13 as early as in the year 1900?

Ann

Ann wrote: ↑Fri Sep 27, 2024 3:47 pm

Chris Peterson wrote: ↑Fri Sep 27, 2024 3:16 pm

Ann wrote: ↑Fri Sep 27, 2024 2:42 pm
Clearly Barnard 29 was much more obvious and striking in the early 1900s than it is today!

Not really. Only when imaged across a broad spectrum (which wasn't possible 100 years ago). But we don't do that much anymore. We image across multiple spectral ranges. Which means the star is every bit as obvious and striking today in you compare apples to apples.

No? Why was this star "discovered" and found to be very remarkably blue compared to the other members of M13 as early as in the year 1900?

Ann

Because they didn't have panchromatic film. It was no more striking photographically then than it is today. It would be discovered no less readily today.

Chris Peterson wrote: ↑Fri Sep 27, 2024 4:12 pm

Ann wrote: ↑Fri Sep 27, 2024 3:47 pm

Chris Peterson wrote: ↑Fri Sep 27, 2024 3:16 pm

Not really. Only when imaged across a broad spectrum (which wasn't possible 100 years ago). But we don't do that much anymore. We image across multiple spectral ranges. Which means the star is every bit as obvious and striking today in you compare apples to apples.

No? Why was this star "discovered" and found to be very remarkably blue compared to the other members of M13 as early as in the year 1900?

Ann

Because they didn't have panchromatic film. It was no more striking photographically then than it is today. It would be discovered no less readily today.

Chris, you know what has been said about the bumblebee? It can't fly, and yet it does.


You seem to be saying that the astronomers of the very early 1900s couldn't have detected the UV peculiarity of Barnard 29. And yet they did.

Or perhaps you are just saying that it is a miracle that they did detect it more than a hundred years ago, because it did not stand out in any way back then?

Ann