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Betelgeuse captured by ALMA

222 points by aaron695 9y ago

jngreenlee 9y ago
Something about the contrast in the image makes it seem to "pulse" as I look at it...eerie! Edit: Maybe it's just the coffee I drank?
- TeMPOraL 9y ago
  No, it's totally the image; I see it too and I didn't drink any coffee recently. It's one of those weird image effects that sometimes happen. Related, a combination of red and dark blue text on a black background tends to jump out from the screen for me, seemingly gaining a third dimension. I wonder how sensitive are those effects to things like ambient light levels and your display's color calibration. I'm also curious if anyone tried to explain them with reproducible steps that could be used for crafting such images on purpose?
  - rangibaby 9y ago
    > I'm also curious if anyone tried to explain them with reproducible steps that could be used for crafting such images on purpose? http://www.ritsumei.ac.jp/~akitaoka/index-e.html http://www.ritsumei.ac.jp/~akitaoka/index-e.html Mr Kitaoka is quite a nice guy and will answer questions if you ask him on Facebook (his English is quite good)
  - cr0sh 9y ago
    > Related, a combination of red and dark blue text on a black background tends to jump out from the screen for me, seemingly gaining a third dimension. This effect is called "chromostereoscopy": https://www2.warwick.ac.uk/fac/sci/physics/research/cfsa/people/erwin/research/3d/principles/ https://www2.warwick.ac.uk/fac/sci/physics/research/cfsa/peo... This has been commercialized into something called ChromaDepth (using special holographic lens glasses to enhance the effect): https://en.wikipedia.org/wiki/ChromaDepth https://en.wikipedia.org/wiki/ChromaDepth Strangely, the above patented system notwithstanding, I recall from being a kid this (or something similar) being marketed (with glasses too) as part of a really cheap and cheesy comic book (I most likely still have copies of that comic book at home), sometime in the 1980s. Unless I am mis-remembering the timeframe (possible), it was long before ChromaDepth (I also recall it being used for firework displays, too).
- morsch 9y ago
  Made me double-check to verify it's not an animation...
- ajnin 9y ago
  I sort of see it too. I think it's caused by a mix of 2 things, first when you stare at something for a while the details tend to fade away, and this image is very susceptible to it since it has very soft colors on the border, and the gradients are from the outside towards the inside so that causes the perceived object to shrink. Second, the eye constantly has small involuntary saccadic movements, and whenever that happens the first effect gets "reset", and the perceived image grows to its real size again.
- woliveirajr 9y ago
  Look just at some point in the middle of it, without moving your eyes, head. Then, after some 30 seconds, slowly move your head towards the screen, and you'll see that the edges get stronger/fainter as you move.
mrspeaker 9y ago
"When [it supernovas], the resulting explosion will be visible from Earth, even in broad daylight." That sounds spectacular! What's the time frame on that - closer to 10 years or a million years?
- maxxxxx 9y ago
  Somewhere in between. In astronomical terms "really soon"
- enimodas 9y ago
  Supernovas visible in daylight have been recorded in history a couple times, for example https://en.wikipedia.org/wiki/SN_1006 https://en.wikipedia.org/wiki/SN_1006
zucchini_head 9y ago
That's amazing resolving power, even if it's been done (maybe not in the exact same way) for quite a while now. Hopefully newer telescopes like the James Webb Telescope will be able to resolve even the _planets_ around other stars, which we are already able to do with the biggest of exoplanets today (good example -> [1]). [1] http://phenomena.nationalgeographic.com/files/2014/05/1RSX_J160929.1-210524.jpg http://phenomena.nationalgeographic.com/files/2014/05/1RSX_J...
- itcrowd 9y ago
  Keep in mind that JWT is a space-based telescope and ALMA is ground based (and an interferometer, not a single dish).
  - SamUK96 9y ago
    Used JWT more as an example of increasingly advanced telescopes, not as a specific type of telescope, but maybe it wasn't the best example...
- semaphoreP 9y ago
  JWST, like ground-based telescopes now, will be able to separate the light from massive Jupiter-like exoplanets from the light of their host stars. But it won't be able to see things at a resolution like this image of Betelguse (so no surfaces of planets... that's quite a long time away).
- contact_fusion 9y ago
  That image is quite stunning. As the 2008 press release [1] states, this image was one of the first successes at direct imaging an exoplanet. It raised some interesting questions, such as why such a massive planet could be found so far out (330 AU!) The scientific paper for this observation can be found in [2] for those interested more astrophysical detail. I feel compelled to offer an astronomer's clarification though. The planet in this image is not "resolved" in the technical sense. A resolved image usually means that fine details about the object are discernible spatially. For example, unresolved images of Betelgeuse provide a point source image, without details; a resolved image of Betelgeuse allows you to find spatial features such as that enormous bubble. Another example is, say, Jupiter: by eye or with a very modest telescope, Jupiter is a (bright) point of light. But with a moderate increase in resolving power, you can see all sorts of interesting features, such as the Great Red Spot, and the various cloud layers that vary with latitude. Individual exoplanets are simply too small to resolve, even with JWST. Even being generous - assuming that the planet is bright enough to detect and that the host star doesn't overwhelm the signal - the angular sizes of exoplanets are miniscule. Lets assume some very generous numbers: a hypothetical exoplanet ten times the diameter of Jupiter (very large), and very, very close to Earth - let's say, 10 lightyears for simplicity and generosity. In arcseconds, the angular diameter of such an object on the sky is about 0.003". Smaller planets at more reasonable distances are even smaller. (The angular size of an object is just small angle trigonometry: in radians, about the width of the object divided by its distance.) Currently, science-class telescopes usually require about 1" resolution. JWST has about 0.1" resolution [3]; an interferometer like ALMA can, at its very best, achieve maybe 0.02" [4], though interferometers (as mentioned in other answers) sacrifice some things in exchange for spatial resolution. This isn't to say you can't just detect exoplanets - you can, even with a ground based telescope like Gemini - but you probably won't resolve them, at least in this generation of telescopes, including JWST. But you can do a lot without spatial resolution - for example, you don't need to resolve the object to measure its spectrum, and spectral analysis can tell you a great deal. [1] http://www.gemini.edu/sunstarplanet http://www.gemini.edu/sunstarplanet [2] https://arxiv.org/abs/0809.1424 https://arxiv.org/abs/0809.1424 [3] https://jwst.nasa.gov/faq.html#webbbetter https://jwst.nasa.gov/faq.html#webbbetter (question 25) [4] https://almascience.eso.org/about-alma/alma-basics https://almascience.eso.org/about-alma/alma-basics (section: spatial resolution)
kijin 9y ago
What's with the weird shape in the bottom left? I don't suppose stars can get squished like that, so it must be some sort of optical artifact?
- jefurii 9y ago
  I also thought maybe that was some gravitational lensing effect but TFA says there's a "giant bubble" on the surface so maybe it really is shaped like that.
- denis1 9y ago
  I guess it is the "vast plume of gas" that the article mentions: The star has been observed in many other wavelengths, particularly in the visible, infrared, and ultraviolet. Using ESO’s Very Large Telescope astronomers discovered a vast plume of gas almost as large as our Solar System. Astronomers have also found a gigantic bubble that boils away on Betelgeuse’s surface.
- ajmurmann 9y ago
  The article mentions some kind of bubble. Maybe that's it?
- chc 9y ago
  I could be interpreting the image wrong, but the star appears to be expelling some material on the left. It's not flattened on the bottom so much as bulging in the middle.
- pyre 9y ago
  From the article: > In this picture, ALMA observes the hot gas of the lower chromosphere of Betelgeuse at sub-millimeter wavelengths — where localised increased temperatures explain why it is not symmetric. I'm guessing the "explain why it is not symmetric" part is related to this.
- privong 9y ago
  > What's with the weird shape in the bottom left? I don't suppose stars can get squished like that, so it must be some sort of optical artifact? It's possible that it could be something physical, as some of the other commenters have mentioned. But it also could be a result of the response of the telescope. Interferometers like ALMA do not directly measure the distribution of emission on the sky. Instead they sample the fourier transform of the sky brightness. Because there are discrete pairs of antennas, the full fourier transform cannot be measured. When images such as this are created, the sampling of the fourier plane is deconvolved. But that process does not create a perfect recovery of the sky emission distribution. One would need to look at the raw data, but another explanation for that extension is that it is an artifact of the fourier plane sampling (e.g., analgous to Gibbs Ringing; https://en.wikipedia.org/wiki/Gibbs_phenomenon https://en.wikipedia.org/wiki/Gibbs_phenomenon).
mabbo 9y ago
Since I had to look it up out of curiosity/paranoia, I figure someone else might also be interested/relieved to know: Betelgeuse has frequently been the subject of scare stories and rumors suggesting that it will explode within a year, leading to exaggerated claims about the consequences of such an event. The timing and prevalence of these rumors have been linked to broader misconceptions of astronomy, particularly to doomsday predictions relating to the Mayan calendar. Betelgeuse is not likely to produce a gamma-ray burst and is not close enough for its x-rays, ultraviolet radiation, or ejected material to cause significant effects on Earth. https://en.wikipedia.org/wiki/Betelgeuse#Approaching_supernova https://en.wikipedia.org/wiki/Betelgeuse#Approaching_superno...
- seren 9y ago
  However according to Wikipedia this is: >the star where the Elder Gods came from to battle the Great Old Ones [...]. Betelgeuse is also mentioned as the homeworld of the 'Ithria, a star-faring fungoid race.
  - throwanem 9y ago
    Sure - if you take Derleth's mankind-involving Manichaeism as canon. That's something I don't recommend doing, because it rather misses the point, which is that we are as ants under the careless tread of powers far older, far greater, and far stranger than anything we can possibly imagine.
  - sqeaky 9y ago
    That is all in the Pop culture and mentions lovecraft right there, this isn't some prank or defacement. This is just real references to literature.
  - _jal 9y ago
    And if you say the star's name three times, you can judge the age of respondents on message boards by their response to the non-sequitir.
    - 9y ago
quotha 9y ago
"causing it to have only a short life expectancy. The star is just about eight million years old" everything is relative
- jbmorgado 9y ago
  Well, when you think that the Sun is now about 5 thousand million years old and happily living the quiet times of its middle age, you can see you "young" is Betelgeuse.
jcims 9y ago
A 1.64 billion km diameter sphere 600 light years away will have an apparent width of 2.9 x 10^-7 radians. That's roughly equivalent to looking at an object 250nm wide at arm's length. A red blood cell is approximately 8000nm wide. Crazy resolving power.
- iamgopal 9y ago
  How many photons ?
  - lisper 9y ago
    Well, let's do the math. ALMA consists of 66 antennas, most of which are 12 meters in diameter. That's about 7000 square meters of receiving area. Betelgeuse is 642 light years away, which is 6x10^18 meters. The area of a sphere with that diameter is about 10^38 square meters. So 10^-34 of the power emitted from Betelgeuse ends up falling on the ALMA array. According to Wikipedia, the luminosity of Betelgeuse is 90-150 thousand solar luminosity units, which is about 4x10^26 watts. Let's call it 10^31 watts. So the total power received from Betelgeuse by ALMA is about a milliwatt. But that's the total power, and the ALMA array only receives at 0.32 to 3.6 mm. To figure out what proportion of Betelgeuse's power falls in this range we need to integrate over Betelgeuse's spectrum, both the total spectrum and then this range in order to find the ratio. That part of the calculation is not so easy. But let's see what we can do. Let's assume that Betelgeuse has a blackbody spectrum. Its temperature is 3500K. We can use this handy dandy blackbody spectrum calculator: http://www.spectralcalc.com/blackbody_calculator/blackbody.php http://www.spectralcalc.com/blackbody_calculator/blackbody.p... When you crunch the numbers it turns out that about 10^-7 of the total power falls in the range 0.32 to 3.6mm. So the total power received by ALMA is about 10^-10 watts. 0.32-3.6mm is in the far infrared. A photon at this wavelength has an energy of about one meV, or about 10^-22 Joules. So 10^-10 watts is about 10^12 photons per second. I don't know how long the exposure times are, but my guess is that they are measured in hours.
    - dnautics 9y ago
      That's a pretty good Fermi calculation!
kmm 9y ago
I wonder what it looks like up close. Betelgeuse's mean density is a few milligrams per cubic meter, probably way less so in the photosphere, comparable to the density of the atmosphere at the edge of space. And yet it still gives light. What would it look like up close? Wispy and ghostlike? Would you even notice you're inside the star?
- lawless123 9y ago
  it's probably more dense as you go in.
- nkrisc 9y ago
  That's only the mean density. I doubt its mass is evenly distributed. You'd probably see through the wispy outer layers and mostly be able to see the denser interior. Much like looking at Earth from space, you don't see the atmosphere (mostly so, anyway) and see the point where the atmosphere transitions to solid planet.
- aruggirello 9y ago
  > And yet it still gives light. ... Would you even notice you're inside the star? I bet at between 90K and 150K times the Sun's luminosity, you'd be too busy boiling away (OK, perhaps exploding in half a millisecond?) to notice :)
0xFFC 9y ago
> When that happens, the resulting explosion will be visible from Earth, even in broad daylight. How much visibility are we talking about?
- aruggirello 9y ago
  IIRC we're talking about it reaching apparent magnitude ~ -12, given or taken a couple magnitudes (or about as bright as the full Moon - Just imagine that much light coming from an infinitely tiny dot instead). BTW with a declination of ~ +7° Betelgeuse is very close to the celestial equator, so the supernova will be visible from anywhere on Earth, save a very small region within 7° from the South Pole.
  - dahart 9y ago
    It's fun to ponder how it will be that bright at every point that is the same distance. Imagine a sphere centered on Betelgeuse with a 600 light year radius with Earth on the surface of this sphere. The fact that enough photons reach my eyeballs to be able to see it at night is difficult to comprehend - the amount of energy needed to sprinkle every square millimeter of a sphere that size is just unimaginable. Now make that 100,000 times brighter during a supernova event -- crazy town. I feel like imagining the energy from a star spread out on a galactic scale helps me understand and appreciate the magnitudes better than any large number of luminosity or watts or photos can.
    - theandrewbailey 9y ago
      Rigel, the blueish star on the other side of Orion, is a bit brighter (magnitude 0.13 vs. 0.5), and is a bit further away (860 ly vs. 640).
placebo 9y ago
At the risk of uncovering my ancientness, I remember reading astronomy books as a kid which specified that stars are so far away that they can't appear as anything more than dots of light even when viewed through the largest telescopes. Always makes me wonder what can be achieved in the future, especially since we're probably somewhere on an exponential progress curve. Of course, assuming a lot of optimism about not cutting off the branch we're sitting on before that happens...
- sdenton4 9y ago
  I think it was in Cixin Liu's 'The Dark Forest' where there's a giant telescope at the edge of the solar system with twelve independently floating lenses focused and corrected by tiny thrusters...
  - Florin_Andrei 9y ago
    What a terrible book. Naively written, leaves the impression that the author is more showing off that he can write about every sci-fi meme in existence ("I'm a big boy now, I know about that topic too!"). Unfocused, meandering plot. Nonsensical end. I regret wasting my time with it.
- pavement 9y ago
  Regarding visible spectrum observations, I've been waiting to see if anyone can come up with a way to develop a consumer-accessible instrument that can sample a high enough resolution, to image all of the moon landing sites. For as long as I can remember, the same thing has been said about the surface of the moon, which is the primary fuel for hoax narratives. With all the buzz about high-resolution arrays being cobbled together from current-generation megapixel digital cameras, I'd love to see someone pull this off. It'd be pretty cool to know that for a budget of maybe tens of thousands of dollars, and some software skills, it'd be within the reach of hobbyists to snap some legit photos of the original moon landing artifacts as they exist.
  - jermaustin1 9y ago
sizzzzlerz 9y ago
If the radius of Betelgeuse is 1400 times that of our sun, it would accommodate 1400^3, or 2,744,000,000, spheres of that size. Humongous doesn't begin to describe just how big it is.
jyriand 9y ago
How come there is a white spot on this orange blob?
- semaphoreP 9y ago
  The colors are only there to represent brightness, so the white spot is a brighter patch on the star (brighter because it's hotter).
- theandrewbailey 9y ago
  I was wondering also. That image looks more like a shiny smooth blob.
tim333 9y ago
Some pictures of and info on the telescope https://directory.eoportal.org/web/eoportal/satellite-missions/a/alma https://directory.eoportal.org/web/eoportal/satellite-missio...
joshvm 9y ago
My girlfriend wrote the copy text for this piece while she was working at ESO - good to see some astronomy on HN :) Also of interest, Cambridge's COAST instrument did this almost 2 decades ago using optical interferometry. http://www.mrao.cam.ac.uk/outreach/radio-telescopes/coast/coast-astronomical-results/surface-imaging-of-betelgeuse/ http://www.mrao.cam.ac.uk/outreach/radio-telescopes/coast/co... http://www.mrao.cam.ac.uk/outreach/radio-telescopes/coast/ http://www.mrao.cam.ac.uk/outreach/radio-telescopes/coast/
grovegames 9y ago
So if our sun were replaced by Betelgeuse, Betelgeuse would engulf all the inner planets. That's huge. It's hard to fathom a star that big.
- finnh 9y ago
  Our star is going to do that in 5 billion years - transform into a red giant & engulf Mercury, Venus, and probably Earth. https://en.wikipedia.org/wiki/Sun#After_core_hydrogen_exhaustion https://en.wikipedia.org/wiki/Sun#After_core_hydrogen_exhaus...
btilly 9y ago
https://en.wikipedia.org/wiki/Angular_resolution https://en.wikipedia.org/wiki/Angular_resolution explains the relevant formula for being able to take pictures like this. Space telescopes like the James Webb are not actually as good as the ground-based arrays that were used here, which put together multiple receivers over a distance to create a much wider "eye". I'm hoping that some day we'll have space-based arrays for this. Imagine if the virtual "eye" on the array was as wide as the orbit of the Moon!
pier25 9y ago
> in the millimeter continuum what does that mean?
- contact_fusion 9y ago
  "Millimeter" refers to the wavelength of light. "Continuum" is a shorthand that in this context refers to thermal emission. All matter emits thermal radiation. The spectral energy distribution of this radiation is determined by the Planck's law [1]. If you measure the spectrum of an object, some part of it will be from this thermal emission, which is a continuous function of wavelength/frequency. In many cases, the conditions are right for spectral lines [2] to be produced, either in emission or absorption. Because these features are centered at specific wavelengths, they are not usually thought of as "continuous" features in the spectrum. (This isn't strictly accurate, as all spectral lines suffer some broadening into extremely narrow, but still continuous, features. Additionally, there are sometimes finite width continuous features called "bands" that arise due to so many lines being present that they blend together.) Generally the continuous part of the spectrum is called "continuum" while the other parts are "lines." [1] https://en.wikipedia.org/wiki/Planck%27s_law https://en.wikipedia.org/wiki/Planck%27s_law [2] https://en.wikipedia.org/wiki/Spectral_line https://en.wikipedia.org/wiki/Spectral_line

222 points by aaron695 9y ago

Something about the contrast in the image makes it seem to "pulse" as I look at it...eerie! Edit: Maybe it's just the coffee I drank?

TeMPOraL 9y ago
No, it's totally the image; I see it too and I didn't drink any coffee recently. It's one of those weird image effects that sometimes happen. Related, a combination of red and dark blue text on a black background tends to jump out from the screen for me, seemingly gaining a third dimension. I wonder how sensitive are those effects to things like ambient light levels and your display's color calibration. I'm also curious if anyone tried to explain them with reproducible steps that could be used for crafting such images on purpose?
- rangibaby 9y ago
  > I'm also curious if anyone tried to explain them with reproducible steps that could be used for crafting such images on purpose? http://www.ritsumei.ac.jp/~akitaoka/index-e.html http://www.ritsumei.ac.jp/~akitaoka/index-e.html Mr Kitaoka is quite a nice guy and will answer questions if you ask him on Facebook (his English is quite good)
- cr0sh 9y ago
  > Related, a combination of red and dark blue text on a black background tends to jump out from the screen for me, seemingly gaining a third dimension. This effect is called "chromostereoscopy": https://www2.warwick.ac.uk/fac/sci/physics/research/cfsa/people/erwin/research/3d/principles/ https://www2.warwick.ac.uk/fac/sci/physics/research/cfsa/peo... This has been commercialized into something called ChromaDepth (using special holographic lens glasses to enhance the effect): https://en.wikipedia.org/wiki/ChromaDepth https://en.wikipedia.org/wiki/ChromaDepth Strangely, the above patented system notwithstanding, I recall from being a kid this (or something similar) being marketed (with glasses too) as part of a really cheap and cheesy comic book (I most likely still have copies of that comic book at home), sometime in the 1980s. Unless I am mis-remembering the timeframe (possible), it was long before ChromaDepth (I also recall it being used for firework displays, too).
morsch 9y ago
Made me double-check to verify it's not an animation...
ajnin 9y ago
I sort of see it too. I think it's caused by a mix of 2 things, first when you stare at something for a while the details tend to fade away, and this image is very susceptible to it since it has very soft colors on the border, and the gradients are from the outside towards the inside so that causes the perceived object to shrink. Second, the eye constantly has small involuntary saccadic movements, and whenever that happens the first effect gets "reset", and the perceived image grows to its real size again.
woliveirajr 9y ago
Look just at some point in the middle of it, without moving your eyes, head. Then, after some 30 seconds, slowly move your head towards the screen, and you'll see that the edges get stronger/fainter as you move.

mrspeaker 9y ago

"When [it supernovas], the resulting explosion will be visible from Earth, even in broad daylight." That sounds spectacular! What's the time frame on that - closer to 10 years or a million years?

maxxxxx 9y ago
Somewhere in between. In astronomical terms "really soon"
enimodas 9y ago
Supernovas visible in daylight have been recorded in history a couple times, for example https://en.wikipedia.org/wiki/SN_1006 https://en.wikipedia.org/wiki/SN_1006

zucchini_head 9y ago

That's amazing resolving power, even if it's been done (maybe not in the exact same way) for quite a while now. Hopefully newer telescopes like the James Webb Telescope will be able to resolve even the _planets_ around other stars, which we are already able to do with the biggest of exoplanets today (good example -> [1]). [1] http://phenomena.nationalgeographic.com/files/2014/05/1RSX_J160929.1-210524.jpg http://phenomena.nationalgeographic.com/files/2014/05/1RSX_J...

itcrowd 9y ago
Keep in mind that JWT is a space-based telescope and ALMA is ground based (and an interferometer, not a single dish).
- SamUK96 9y ago
  Used JWT more as an example of increasingly advanced telescopes, not as a specific type of telescope, but maybe it wasn't the best example...
semaphoreP 9y ago
JWST, like ground-based telescopes now, will be able to separate the light from massive Jupiter-like exoplanets from the light of their host stars. But it won't be able to see things at a resolution like this image of Betelguse (so no surfaces of planets... that's quite a long time away).
contact_fusion 9y ago
That image is quite stunning. As the 2008 press release [1] states, this image was one of the first successes at direct imaging an exoplanet. It raised some interesting questions, such as why such a massive planet could be found so far out (330 AU!) The scientific paper for this observation can be found in [2] for those interested more astrophysical detail. I feel compelled to offer an astronomer's clarification though. The planet in this image is not "resolved" in the technical sense. A resolved image usually means that fine details about the object are discernible spatially. For example, unresolved images of Betelgeuse provide a point source image, without details; a resolved image of Betelgeuse allows you to find spatial features such as that enormous bubble. Another example is, say, Jupiter: by eye or with a very modest telescope, Jupiter is a (bright) point of light. But with a moderate increase in resolving power, you can see all sorts of interesting features, such as the Great Red Spot, and the various cloud layers that vary with latitude. Individual exoplanets are simply too small to resolve, even with JWST. Even being generous - assuming that the planet is bright enough to detect and that the host star doesn't overwhelm the signal - the angular sizes of exoplanets are miniscule. Lets assume some very generous numbers: a hypothetical exoplanet ten times the diameter of Jupiter (very large), and very, very close to Earth - let's say, 10 lightyears for simplicity and generosity. In arcseconds, the angular diameter of such an object on the sky is about 0.003". Smaller planets at more reasonable distances are even smaller. (The angular size of an object is just small angle trigonometry: in radians, about the width of the object divided by its distance.) Currently, science-class telescopes usually require about 1" resolution. JWST has about 0.1" resolution [3]; an interferometer like ALMA can, at its very best, achieve maybe 0.02" [4], though interferometers (as mentioned in other answers) sacrifice some things in exchange for spatial resolution. This isn't to say you can't just detect exoplanets - you can, even with a ground based telescope like Gemini - but you probably won't resolve them, at least in this generation of telescopes, including JWST. But you can do a lot without spatial resolution - for example, you don't need to resolve the object to measure its spectrum, and spectral analysis can tell you a great deal. [1] http://www.gemini.edu/sunstarplanet http://www.gemini.edu/sunstarplanet [2] https://arxiv.org/abs/0809.1424 https://arxiv.org/abs/0809.1424 [3] https://jwst.nasa.gov/faq.html#webbbetter https://jwst.nasa.gov/faq.html#webbbetter (question 25) [4] https://almascience.eso.org/about-alma/alma-basics https://almascience.eso.org/about-alma/alma-basics (section: spatial resolution)

kijin 9y ago

What's with the weird shape in the bottom left? I don't suppose stars can get squished like that, so it must be some sort of optical artifact?

jefurii 9y ago
I also thought maybe that was some gravitational lensing effect but TFA says there's a "giant bubble" on the surface so maybe it really is shaped like that.
denis1 9y ago
I guess it is the "vast plume of gas" that the article mentions: The star has been observed in many other wavelengths, particularly in the visible, infrared, and ultraviolet. Using ESO’s Very Large Telescope astronomers discovered a vast plume of gas almost as large as our Solar System. Astronomers have also found a gigantic bubble that boils away on Betelgeuse’s surface.
ajmurmann 9y ago
The article mentions some kind of bubble. Maybe that's it?
chc 9y ago
I could be interpreting the image wrong, but the star appears to be expelling some material on the left. It's not flattened on the bottom so much as bulging in the middle.
pyre 9y ago
From the article: > In this picture, ALMA observes the hot gas of the lower chromosphere of Betelgeuse at sub-millimeter wavelengths — where localised increased temperatures explain why it is not symmetric. I'm guessing the "explain why it is not symmetric" part is related to this.
privong 9y ago
> What's with the weird shape in the bottom left? I don't suppose stars can get squished like that, so it must be some sort of optical artifact? It's possible that it could be something physical, as some of the other commenters have mentioned. But it also could be a result of the response of the telescope. Interferometers like ALMA do not directly measure the distribution of emission on the sky. Instead they sample the fourier transform of the sky brightness. Because there are discrete pairs of antennas, the full fourier transform cannot be measured. When images such as this are created, the sampling of the fourier plane is deconvolved. But that process does not create a perfect recovery of the sky emission distribution. One would need to look at the raw data, but another explanation for that extension is that it is an artifact of the fourier plane sampling (e.g., analgous to Gibbs Ringing; https://en.wikipedia.org/wiki/Gibbs_phenomenon https://en.wikipedia.org/wiki/Gibbs_phenomenon).

mabbo 9y ago

Since I had to look it up out of curiosity/paranoia, I figure someone else might also be interested/relieved to know: Betelgeuse has frequently been the subject of scare stories and rumors suggesting that it will explode within a year, leading to exaggerated claims about the consequences of such an event. The timing and prevalence of these rumors have been linked to broader misconceptions of astronomy, particularly to doomsday predictions relating to the Mayan calendar. Betelgeuse is not likely to produce a gamma-ray burst and is not close enough for its x-rays, ultraviolet radiation, or ejected material to cause significant effects on Earth. https://en.wikipedia.org/wiki/Betelgeuse#Approaching_supernova https://en.wikipedia.org/wiki/Betelgeuse#Approaching_superno...

seren 9y ago
However according to Wikipedia this is: >the star where the Elder Gods came from to battle the Great Old Ones [...]. Betelgeuse is also mentioned as the homeworld of the 'Ithria, a star-faring fungoid race.
- throwanem 9y ago
  Sure - if you take Derleth's mankind-involving Manichaeism as canon. That's something I don't recommend doing, because it rather misses the point, which is that we are as ants under the careless tread of powers far older, far greater, and far stranger than anything we can possibly imagine.
- sqeaky 9y ago
  That is all in the Pop culture and mentions lovecraft right there, this isn't some prank or defacement. This is just real references to literature.
- _jal 9y ago
  And if you say the star's name three times, you can judge the age of respondents on message boards by their response to the non-sequitir.
  - 9y ago

quotha 9y ago

"causing it to have only a short life expectancy. The star is just about eight million years old" everything is relative

jbmorgado 9y ago
Well, when you think that the Sun is now about 5 thousand million years old and happily living the quiet times of its middle age, you can see you "young" is Betelgeuse.

jcims 9y ago

A 1.64 billion km diameter sphere 600 light years away will have an apparent width of 2.9 x 10^-7 radians. That's roughly equivalent to looking at an object 250nm wide at arm's length. A red blood cell is approximately 8000nm wide. Crazy resolving power.

iamgopal 9y ago
How many photons ?
- lisper 9y ago
  Well, let's do the math. ALMA consists of 66 antennas, most of which are 12 meters in diameter. That's about 7000 square meters of receiving area. Betelgeuse is 642 light years away, which is 6x10^18 meters. The area of a sphere with that diameter is about 10^38 square meters. So 10^-34 of the power emitted from Betelgeuse ends up falling on the ALMA array. According to Wikipedia, the luminosity of Betelgeuse is 90-150 thousand solar luminosity units, which is about 4x10^26 watts. Let's call it 10^31 watts. So the total power received from Betelgeuse by ALMA is about a milliwatt. But that's the total power, and the ALMA array only receives at 0.32 to 3.6 mm. To figure out what proportion of Betelgeuse's power falls in this range we need to integrate over Betelgeuse's spectrum, both the total spectrum and then this range in order to find the ratio. That part of the calculation is not so easy. But let's see what we can do. Let's assume that Betelgeuse has a blackbody spectrum. Its temperature is 3500K. We can use this handy dandy blackbody spectrum calculator: http://www.spectralcalc.com/blackbody_calculator/blackbody.php http://www.spectralcalc.com/blackbody_calculator/blackbody.p... When you crunch the numbers it turns out that about 10^-7 of the total power falls in the range 0.32 to 3.6mm. So the total power received by ALMA is about 10^-10 watts. 0.32-3.6mm is in the far infrared. A photon at this wavelength has an energy of about one meV, or about 10^-22 Joules. So 10^-10 watts is about 10^12 photons per second. I don't know how long the exposure times are, but my guess is that they are measured in hours.
  - dnautics 9y ago
    That's a pretty good Fermi calculation!

kmm 9y ago

I wonder what it looks like up close. Betelgeuse's mean density is a few milligrams per cubic meter, probably way less so in the photosphere, comparable to the density of the atmosphere at the edge of space. And yet it still gives light. What would it look like up close? Wispy and ghostlike? Would you even notice you're inside the star?

lawless123 9y ago
it's probably more dense as you go in.
nkrisc 9y ago
That's only the mean density. I doubt its mass is evenly distributed. You'd probably see through the wispy outer layers and mostly be able to see the denser interior. Much like looking at Earth from space, you don't see the atmosphere (mostly so, anyway) and see the point where the atmosphere transitions to solid planet.
aruggirello 9y ago
> And yet it still gives light. ... Would you even notice you're inside the star? I bet at between 90K and 150K times the Sun's luminosity, you'd be too busy boiling away (OK, perhaps exploding in half a millisecond?) to notice :)

0xFFC 9y ago

> When that happens, the resulting explosion will be visible from Earth, even in broad daylight. How much visibility are we talking about?

aruggirello 9y ago
IIRC we're talking about it reaching apparent magnitude ~ -12, given or taken a couple magnitudes (or about as bright as the full Moon - Just imagine that much light coming from an infinitely tiny dot instead). BTW with a declination of ~ +7° Betelgeuse is very close to the celestial equator, so the supernova will be visible from anywhere on Earth, save a very small region within 7° from the South Pole.
- dahart 9y ago
  It's fun to ponder how it will be that bright at every point that is the same distance. Imagine a sphere centered on Betelgeuse with a 600 light year radius with Earth on the surface of this sphere. The fact that enough photons reach my eyeballs to be able to see it at night is difficult to comprehend - the amount of energy needed to sprinkle every square millimeter of a sphere that size is just unimaginable. Now make that 100,000 times brighter during a supernova event -- crazy town. I feel like imagining the energy from a star spread out on a galactic scale helps me understand and appreciate the magnitudes better than any large number of luminosity or watts or photos can.
  - theandrewbailey 9y ago
    Rigel, the blueish star on the other side of Orion, is a bit brighter (magnitude 0.13 vs. 0.5), and is a bit further away (860 ly vs. 640).

placebo 9y ago

At the risk of uncovering my ancientness, I remember reading astronomy books as a kid which specified that stars are so far away that they can't appear as anything more than dots of light even when viewed through the largest telescopes. Always makes me wonder what can be achieved in the future, especially since we're probably somewhere on an exponential progress curve. Of course, assuming a lot of optimism about not cutting off the branch we're sitting on before that happens...

sdenton4 9y ago
I think it was in Cixin Liu's 'The Dark Forest' where there's a giant telescope at the edge of the solar system with twelve independently floating lenses focused and corrected by tiny thrusters...
- Florin_Andrei 9y ago
  What a terrible book. Naively written, leaves the impression that the author is more showing off that he can write about every sci-fi meme in existence ("I'm a big boy now, I know about that topic too!"). Unfocused, meandering plot. Nonsensical end. I regret wasting my time with it.
pavement 9y ago
Regarding visible spectrum observations, I've been waiting to see if anyone can come up with a way to develop a consumer-accessible instrument that can sample a high enough resolution, to image all of the moon landing sites. For as long as I can remember, the same thing has been said about the surface of the moon, which is the primary fuel for hoax narratives. With all the buzz about high-resolution arrays being cobbled together from current-generation megapixel digital cameras, I'd love to see someone pull this off. It'd be pretty cool to know that for a budget of maybe tens of thousands of dollars, and some software skills, it'd be within the reach of hobbyists to snap some legit photos of the original moon landing artifacts as they exist.
- jermaustin1 9y ago

sizzzzlerz 9y ago

If the radius of Betelgeuse is 1400 times that of our sun, it would accommodate 1400^3, or 2,744,000,000, spheres of that size. Humongous doesn't begin to describe just how big it is.

jyriand 9y ago

How come there is a white spot on this orange blob?

semaphoreP 9y ago
The colors are only there to represent brightness, so the white spot is a brighter patch on the star (brighter because it's hotter).
theandrewbailey 9y ago
I was wondering also. That image looks more like a shiny smooth blob.

tim333 9y ago

Some pictures of and info on the telescope https://directory.eoportal.org/web/eoportal/satellite-missions/a/alma https://directory.eoportal.org/web/eoportal/satellite-missio...

joshvm 9y ago

My girlfriend wrote the copy text for this piece while she was working at ESO - good to see some astronomy on HN :) Also of interest, Cambridge's COAST instrument did this almost 2 decades ago using optical interferometry. http://www.mrao.cam.ac.uk/outreach/radio-telescopes/coast/coast-astronomical-results/surface-imaging-of-betelgeuse/ http://www.mrao.cam.ac.uk/outreach/radio-telescopes/coast/co... http://www.mrao.cam.ac.uk/outreach/radio-telescopes/coast/ http://www.mrao.cam.ac.uk/outreach/radio-telescopes/coast/

grovegames 9y ago

So if our sun were replaced by Betelgeuse, Betelgeuse would engulf all the inner planets. That's huge. It's hard to fathom a star that big.

finnh 9y ago
Our star is going to do that in 5 billion years - transform into a red giant & engulf Mercury, Venus, and probably Earth. https://en.wikipedia.org/wiki/Sun#After_core_hydrogen_exhaustion https://en.wikipedia.org/wiki/Sun#After_core_hydrogen_exhaus...

btilly 9y ago

https://en.wikipedia.org/wiki/Angular_resolution https://en.wikipedia.org/wiki/Angular_resolution explains the relevant formula for being able to take pictures like this. Space telescopes like the James Webb are not actually as good as the ground-based arrays that were used here, which put together multiple receivers over a distance to create a much wider "eye". I'm hoping that some day we'll have space-based arrays for this. Imagine if the virtual "eye" on the array was as wide as the orbit of the Moon!

pier25 9y ago

> in the millimeter continuum what does that mean?

contact_fusion 9y ago
"Millimeter" refers to the wavelength of light. "Continuum" is a shorthand that in this context refers to thermal emission. All matter emits thermal radiation. The spectral energy distribution of this radiation is determined by the Planck's law [1]. If you measure the spectrum of an object, some part of it will be from this thermal emission, which is a continuous function of wavelength/frequency. In many cases, the conditions are right for spectral lines [2] to be produced, either in emission or absorption. Because these features are centered at specific wavelengths, they are not usually thought of as "continuous" features in the spectrum. (This isn't strictly accurate, as all spectral lines suffer some broadening into extremely narrow, but still continuous, features. Additionally, there are sometimes finite width continuous features called "bands" that arise due to so many lines being present that they blend together.) Generally the continuous part of the spectrum is called "continuum" while the other parts are "lines." [1] https://en.wikipedia.org/wiki/Planck%27s_law https://en.wikipedia.org/wiki/Planck%27s_law [2] https://en.wikipedia.org/wiki/Spectral_line https://en.wikipedia.org/wiki/Spectral_line