APOD: Messier 99 (2021 Jun 24)

[APOD Robot]

Messier 99

Explanation: Grand design spiral galaxy Messier 99 looks majestic on a truly cosmic scale. This recently processed full galaxy portrait stretches over 70,000 light-years across M99. The sharp view is a combination of ultraviolet, visible, and infrared image data from the Hubble Space Telescope. About 50 million light-years distant toward the well-groomed constellation Coma Bernices, the face-on spiral is a member of the nearby Virgo Galaxy Cluster. Also cataloged as NGC 4254, a close encounter with another Virgo cluster member has likely influenced the shape of its well-defined, blue spiral arms.

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[Ann]

M99. Image: Hubble/Leo Schatz.

M99. Credit and Copyright: Adam Block/Mount Lemmon SkyCenter/University of Arizona.

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Obviously the Hubble/Leo Schatz image does a much better job than the Adam Block image of giving us great resolution, but Adam Block gives us a clearer idea of the interestingly distorted shape of this galaxy and its very high level of star formation, as evidenced by the very high number of pink emission nebulas.

Can't resist showing you M99 in its galactic neighborhood:

Elegant spiral galaxy M100 (top left), squashed spiral M99 (bottom right), galactic
fraternal twins NGC 4302 (left) and NGC 4298 (right), bottom. Photo: Galactic Hunter
(Antoine & Dalia Grelin)

M100, an elegant grand design galaxy. The designation means that M100 has two long spiral arms. Photo: Isaac Newton Group of Telescopes.

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NGC 4302 (left), a slim edge-on galaxy that doesn't bulge in the middle, and NGC 4298, a flocculent, many-armed galaxy. Photo: NASA / ESA / M. Mutchler, STScI.

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M99 is not a good example of a grand design galaxy, in my opinion. It has sometimes been described as a three-armed spiral galaxy.

Ann

[spinlock]

It could be interesting to combine data of M99 from Adam Block with this image, but I wanted to point out that APOD chose to publish a version of my work which was slightly cropped. The full image is shown here (and also appears on my AstroBin page). I was surprised to see the distant arm of the galaxy carved along almost a straight line - could it be result of interaction with its close galactic neighbor?

Leo Shatz

[XgeoX]

M99_LeoShatz_cropped1024[1].jpg

M99. Image: Hubble/Leo Schatz.

M99. Credit and Copyright: Adam Block/Mount Lemmon SkyCenter/University of Arizona.

---

Obviously the Hubble/Leo Schatz image does a much better job than the Adam Block image of giving us great resolution, but Adam Block gives us a clearer idea of the interestingly distorted shape of this galaxy and its very high level of star formation, as evidenced by the very high number of pink emission nebulas.

Ann

I notice that in the Block image one can make out what looks like a small bar in the galactic center as well.

Eric

[neufer]

Click to play embedded YouTube video.

[johnnydeep]

It could be interesting to combine data of M99 from Adam Block with this image, but I wanted to point out that APOD chose to publish a version of my work which was slightly cropped. The full image is shown here (and also appears on my AstroBin page). I was surprised to see the distant arm of the galaxy carved along almost a straight line - could it be result of interaction with its close galactic neighbor?

Leo Shatz

Interestingly, the pic of M100 that Ann posted above also seems to show a (less well defined to be sure) straight line termination distortion of the spiral arms on the right side. I wonder if the mechanism is the same.

[orin stepanek]

I love the photo; it is beautiful; as are most galaxies! I love the
fact that you can pick out individual stars in a galaxy 50 million light
years away; let alone a (maybe intermediate nova)! Here's to you M99!

[johnnydeep]

M99_LeoShatz_cropped.jpg

I love the photo; it is beautiful; as are most galaxies! I love the
fact that you can pick out individual stars in a galaxy 50 million light
years away; let alone a (maybe intermediate nova)! Here's to you M99!

Is that really true? This galaxy is 20 times more distant than Andromeda. And for that matter, can we even see individual stars - that are not novas - in Andromeda?

EDIT: I mean seeing individual stars with telescopes of course, not with the naked eye.

[Chris Peterson]

M99_LeoShatz_cropped.jpg

I love the photo; it is beautiful; as are most galaxies! 8-) I love the
fact that you can pick out individual stars in a galaxy 50 million light
years away; let alone a (maybe intermediate nova)! Here's to you M99!

Is that really true? This galaxy is 20 times more distant than Andromeda. And for that matter, can we even see individual stars - that are not novas - in Andromeda?

EDIT: I mean seeing individual stars with telescopes of course, not with the naked eye.

Yes. We can see individual stars in galaxies at least tens of millions of light years away. (The most distant Cepheid is in M100, at 56 Mly.) It just requires a bright star in a low density region, which describes the edges of arms in many galaxies.

[orin stepanek]

I can't wait for the JWST to be launched and operational! Then I hope they take a look at these places! Damn; I wish I wasn't so old!

[VictorBorun]

3D turned into 2D at plane perpendicular to our line of view

The galaxy is inclined by 42° to the line-of-sight with a major axis position angle of 68°

I struggle with position angle at an image. I have to rotate the posted pic by -34.2°(the minus sign means clockwise)

Click to view full size image

to fit to old pink&blue image.
I think it means position angle of 68°-34.2° = 33,8° for the posted pic.
Now, to compensate 42° to the line-of-sight we have to

Click to view full size image

compress the longer axis of visual ellipsis with factor of cos42°=0.743.

[johnnydeep]

M99_LeoShatz_cropped.jpg

I love the photo; it is beautiful; as are most galaxies! I love the
fact that you can pick out individual stars in a galaxy 50 million light
years away; let alone a (maybe intermediate nova)! Here's to you M99!

Is that really true? This galaxy is 20 times more distant than Andromeda. And for that matter, can we even see individual stars - that are not novas - in Andromeda?

EDIT: I mean seeing individual stars with telescopes of course, not with the naked eye.

Yes. We can see individual stars in galaxies at least tens of millions of light years away. (The most distant Cepheid is in M100, at 56 Mly.) It just requires a bright star in a low density region, which describes the edges of arms in many galaxies.

I'm amazed! So how far away would a star with the absolute brightness of our Sun have to move to escape detection with our best telescopes?

I know that an apparent magnitude increase of 1 corresponds to a brightness decrease of 2.512 (=100^1/5), and an apparent magnitude increase of 5 is a brightness decrease of 100 (=100^5*1/5). This also means that since if the Sun was moved 10 times farther away, it would appear 100 times dimmer, its apparent magnitude would increase by 5. Now, 1 light year = 63240 AU, and if the Sun was moved 1 ly away it would appear 4e9 times dimmer (=63240²), which is close to an apparent magnitude increase of 24 (2.512²⁴=4e9)...etc.

The apparent magnitude of the Sun is -26.74 so if was moved 1 ly away it would have an apparent magnitude of -26.74 + 24 = -2.74.

And in general, a difference in apparent magnitude m2 − m1 = −2.5 * log₁₀(B), where B is the change in brightness.

So, if the Sun was moved 10 Mly away, that would be 1e7 * 63240 = 6.324e11 times as far, which would make it 4e23 times dimmer. Plugging that in as B in the formula gives an apparent magnitude increase of about 59.3, so it's new apparent magnitude at 10 Mly would be -26.74 + 59.3 = 32.5. Could we detect it?

[ EDIT: my math might be suspect... ]

[Chris Peterson]

Is that really true? This galaxy is 20 times more distant than Andromeda. And for that matter, can we even see individual stars - that are not novas - in Andromeda?

EDIT: I mean seeing individual stars with telescopes of course, not with the naked eye.

Yes. We can see individual stars in galaxies at least tens of millions of light years away. (The most distant Cepheid is in M100, at 56 Mly.) It just requires a bright star in a low density region, which describes the edges of arms in many galaxies.

I'm amazed! So how far away would a star with the absolute brightness of our Sun have to move to escape detection with our best telescopes?

I know that an apparent magnitude increase of 1 corresponds to a brightness decrease of 2.512 (=100^1/5), and an apparent magnitude increase of 5 is a brightness decrease of 100 (=100^5*1/5). This also means that since if the Sun was moved 10 times farther away, it would appear 100 times dimmer, its apparent magnitude would increase by 5. Now, 1 light year = 63240 AU, and if the Sun was moved 1 ly away it would appear 4e9 times dimmer (=63240²), which is close to an apparent magnitude increase of 24 (2.512²⁴=4e9)...etc.

The apparent magnitude of the Sun is -26.74 so if was moved 1 ly away it would have an apparent magnitude of -26.74 + 24 = -2.74.

And in general, a difference in apparent magnitude m2 − m1 = −2.5 * log₁₀(B), where B is the change in brightness.

So, if the Sun was moved 10 Mly away, that would be 1e7 * 63240 = 6.324e11 times as far, which would make it 4e23 times dimmer. Plugging that in as B in the formula gives an apparent magnitude increase of about 59.3, so it's new apparent magnitude at 10 Mly would be -26.74 + 59.3 = 32.5. Could we detect it?

[ EDIT: my math might be suspect... ]

Your math is fine. The magnitude limit for the HST is 31.5, so the Sun at 10 Mly would be right around the sensitivity limit. The Webb telescope's limiting magnitude will be 34.

[johnnydeep]

Yes. We can see individual stars in galaxies at least tens of millions of light years away. (The most distant Cepheid is in M100, at 56 Mly.) It just requires a bright star in a low density region, which describes the edges of arms in many galaxies.

I'm amazed! So how far away would a star with the absolute brightness of our Sun have to move to escape detection with our best telescopes?

I know that an apparent magnitude increase of 1 corresponds to a brightness decrease of 2.512 (=100^1/5), and an apparent magnitude increase of 5 is a brightness decrease of 100 (=100^5*1/5). This also means that since if the Sun was moved 10 times farther away, it would appear 100 times dimmer, its apparent magnitude would increase by 5. Now, 1 light year = 63240 AU, and if the Sun was moved 1 ly away it would appear 4e9 times dimmer (=63240²), which is close to an apparent magnitude increase of 24 (2.512²⁴=4e9)...etc.

The apparent magnitude of the Sun is -26.74 so if was moved 1 ly away it would have an apparent magnitude of -26.74 + 24 = -2.74.

And in general, a difference in apparent magnitude m2 − m1 = −2.5 * log₁₀(B), where B is the change in brightness.

So, if the Sun was moved 10 Mly away, that would be 1e7 * 63240 = 6.324e11 times as far, which would make it 4e23 times dimmer. Plugging that in as B in the formula gives an apparent magnitude increase of about 59.3, so it's new apparent magnitude at 10 Mly would be -26.74 + 59.3 = 32.5. Could we detect it?

[ EDIT: my math might be suspect... ]

Your math is fine. The magnitude limit for the HST is 31.5, so the Sun at 10 Mly would be right around the sensitivity limit. The Webb telescope's limiting magnitude will be 34.

Cool. I would assume that the detection limit is also affected by the duration of the exposure, no? Or do other factors (like, say, the quality of the mirrors and lenses) eventually trump that possible gain?

Finally, in this APOD, can we reasonably say that we CAN actually point out some individual stars, or is that still unlikely for "run of the mill" Hubble images? And how would we know that a suspect single star wasn't really a small cluster?

[Chris Peterson]

Your math is fine. The magnitude limit for the HST is 31.5, so the Sun at 10 Mly would be right around the sensitivity limit. The Webb telescope's limiting magnitude will be 34.

Cool. I would assume that the detection limit is also affected by the duration of the exposure, no? Or do other factors (like, say, the quality of the mirrors and lenses) eventually trump that possible gain?

Finally, in this APOD, can we reasonably say that we CAN actually point out some individual stars, or is that still unlikely for "run of the mill" Hubble images? And how would we know that a suspect single star wasn't really a small cluster?

The major wall that we run up against is background light. On the ground, that's skyglow, which limits us to around mag 27, no matter the size of the telescope or length of exposure. The HST limit is set by zodiacal light and other scatter off of dust and in some cases gas. Webb does better because this background is reduced in the IR where the scope operates.

There are thousands of individual stars captured in this image. Indeed, that's something that backyard imagers can do. It doesn't require a space telescope. (Clusters have different intensity profiles than stars, and different spectroscopic characteristics.)

[neufer]

The magnitude limit for the HST is 31.5, so the Sun at 10 Mly would be right around the sensitivity limit. The Webb telescope's limiting magnitude will be 34.

<<Telescopes at large observatories are typically located at sites selected for dark skies. They also increase the limiting magnitude by using long integration times on the detector, and by using image-processing techniques to increase the signal to noise ratio. Most 8 to 10 meter class telescopes can detect sources with a visual magnitude of about 27 using a one-hour integration time.

Zodiacal light schmutz Mars the limiting magnitude

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Even higher limiting magnitudes can be achieved for telescopes above the Earth's atmosphere, such as the Hubble Space Telescope, where the sky brightness due to the atmosphere is not relevant. For orbital telescopes, the background sky brightness is set by the zodiacal light. The Hubble telescope can detect objects as faint as a magnitude of +31.5, and the James Webb Space Telescope (operating in the infrared spectrum) is expected to have a magnitude limit of 34th magnitude.

Automated astronomical surveys are often limited to around magnitude 20 because of the short exposure time that allows covering a large part of the sky in a night. In a 30 second exposure the 0.7-meter telescope at the Catalina Sky Survey has a limiting magnitude of 19.5. The Zwicky Transient Facility has a limiting magnitude of 20.5, and Pan-STARRS has a limiting magnitude of 24.>>

[johnnydeep]

Your math is fine. The magnitude limit for the HST is 31.5, so the Sun at 10 Mly would be right around the sensitivity limit. The Webb telescope's limiting magnitude will be 34.

Cool. I would assume that the detection limit is also affected by the duration of the exposure, no? Or do other factors (like, say, the quality of the mirrors and lenses) eventually trump that possible gain?

Finally, in this APOD, can we reasonably say that we CAN actually point out some individual stars, or is that still unlikely for "run of the mill" Hubble images? And how would we know that a suspect single star wasn't really a small cluster?

The major wall that we run up against is background light. On the ground, that's skyglow, which limits us to around mag 27, no matter the size of the telescope or length of exposure. The HST limit is set by zodiacal light and other scatter off of dust and in some cases gas. Webb does better because this background is reduced in the IR where the scope operates.

There are thousands of individual stars captured in this image. Indeed, that's something that backyard imagers can do. It doesn't require a space telescope. (Clusters have different intensity profiles than stars, and different spectroscopic characteristics.)

Ok, thanks Chris, and again, Cool! (And thanks neufer for the link to the limiting magnitude article.)

EDIT: actually, I have one more question. Wouldn't it be the case that every pixel of the CCD detector used to make the image would have the combined light of many more than one star? If so, how can we way that we can see individual stars?

[johnnydeep]

[ Note: I had edited my last post with a question, but I'm not sure that causes a notice to be sent to those subscribed to the topic, so I'm posting it separately here. ]

I have one more question for Chris (or anyone!). Wouldn't it be the case that every pixel of the CCD detector used to make an image of a galaxy such as this APOD would have the combined light of many more than one star in the galaxy? If so, how can we way that we can see individual stars?

[neufer]

Wouldn't it be the case that every pixel of the CCD detector used to make an image of a galaxy such as this APOD would have the combined light of many more than one star in the galaxy? If so, how can we way [sic] that we can see individual stars?

I imagine the important thing is that the light from a pixel is dominated by the light from the brightest star in that pixel.

[Chris Peterson]

Wouldn't it be the case that every pixel of the CCD detector used to make an image of a galaxy such as this APOD would have the combined light of many more than one star in the galaxy? If so, how can we way [sic] that we can see individual stars?

I imagine the important thing is that the light from a pixel is dominated by the light from the brightest star in that pixel.

Agreed. This is complicated slightly (in practice, not conceptually) by the fact that no star bright enough to be detected against its background will have all of its light falling on a single pixel.

[johnnydeep]

Wouldn't it be the case that every pixel of the CCD detector used to make an image of a galaxy such as this APOD would have the combined light of many more than one star in the galaxy? If so, how can we way [sic] that we can see individual stars?

I imagine the important thing is that the light from a pixel is dominated by the light from the brightest star in that pixel.

Agreed. This is complicated slightly (in practice, not conceptually) by the fact that no star bright enough to be detected against its background will have all of its light falling on a single pixel.

I don't understand Chris's point here. Certainly in any image of anything, not all the light from an object falls on a single pixel?

As for the light falling on a pixel being dominated by the brightest star, why would that be true: there could be one brightest star, plus 10 not so bright stars all of whose light mixes together on a pixel, thereby overwhelming the one bright star.

[Chris Peterson]

I imagine the important thing is that the light from a pixel is dominated by the light from the brightest star in that pixel.

Agreed. This is complicated slightly (in practice, not conceptually) by the fact that no star bright enough to be detected against its background will have all of its light falling on a single pixel.

I don't understand Chris's point here. Certainly in any image of anything, not all the light from an object falls on a single pixel?

If a point source (like a star) is sufficiently dim, all of its light above the instrumental noise floor might fall on a single pixel.

As for the light falling on a pixel being dominated by the brightest star, why would that be true: there could be one brightest star, plus 10 not so bright stars all of whose light mixes together on a pixel, thereby overwhelming the one bright star.

In which case you would not detect the star, because it would be lost in the noise of the background. The detected star's light would fall on multiple pixels, which would provide an intensity profile- a profile that would be quite different for a point source as opposed to some kind of cluster.

While it's possible we could construct some interesting borderline cases, the point here is that we're seeing stars that very distinctly stand out against their backgrounds. As Art puts it, they dominate. We're not talking about borderline cases of possible stars that are just barely above the noise floor.

[johnnydeep]

Agreed. This is complicated slightly (in practice, not conceptually) by the fact that no star bright enough to be detected against its background will have all of its light falling on a single pixel.

I don't understand Chris's point here. Certainly in any image of anything, not all the light from an object falls on a single pixel?

If a point source (like a star) is sufficiently dim, all of its light above the instrumental noise floor might fall on a single pixel.

As for the light falling on a pixel being dominated by the brightest star, why would that be true: there could be one brightest star, plus 10 not so bright stars all of whose light mixes together on a pixel, thereby overwhelming the one bright star.

In which case you would not detect the star, because it would be lost in the noise of the background. The detected star's light would fall on multiple pixels, which would provide an intensity profile- a profile that would be quite different for a point source as opposed to some kind of cluster.

While it's possible we could construct some interesting borderline cases, the point here is that we're seeing stars that very distinctly stand out against their backgrounds. As Art puts it, they dominate. We're not talking about borderline cases of possible stars that are just barely above the noise floor.

Ok, though I still don't really understand how we can know for sure that any given single pixel represents, even predominantly, a single star. Just to further complicate this, I'm sure most of those "single pixel" stars are really multiple stars (since most stars are).

And it's at this point that I think I would benefit from a short course on astrophotography and CCD imaging theory. Can you point me to a good resource?

[Deathfleer]

3D turned into 2D at plane perpendicular to our line of view

The galaxy is inclined by 42° to the line-of-sight with a major axis position angle of 68°

I struggle with position angle at an image. I have to rotate the posted pic by -34.2°(the minus sign means clockwise)

Click to view full size image

to fit to old pink&blue image.
I think it means position angle of 68°-34.2° = 33,8° for the posted pic.
Now, to compensate 42° to the line-of-sight we have to

Click to view full size image

compress the longer axis of visual ellipsis with factor of cos42°=0.743.

Thanks..
.Formerly, I thought that there were elliptical galaxies.. Now, no more. Every galaxy is spiral and circular

[Ann]

.Formerly, I thought that there were elliptical galaxies.. Now, no more. Every galaxy is spiral and circular

Elliptical galaxy M60 (center), and spiral galaxy NGC 4647 (upper right). NASA / ESA / the Hubble Heritage / STScI / AURA / Hubble Collaboration

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Galaxy cluster Abell 1689. Most of the 'yellow blobs' here are elliptical galaxies. ESA/Hubble.

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There are certainly elliptical galaxies in the Universe. They are very common, especially in large galaxy clusters. Most of the "yellow blobs" in the picture of galaxy cluster Abell 1689 are ellipticals, although some of them have enough structure to be lenticulars or barred ring galaxies.

Elliptical galaxies are typically very pale yellow in color and lacking in structure. The color of the galaxies in the ESA/Hubble picture of Abell 1689 has to do not only with the kind of stars that make up the galaxies - and in ellipticals the stars are yellow and red, whereas in spirals the brightest stars are typically blue - but also the amount of redshift-reddening that the galaxies are affected by.

Ann