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Betelgeuse captured by ALMA

222 points| aaron695 | 8 years ago |eso.org | reply

77 comments

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[+] jcims|8 years ago|reply
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|8 years ago|reply
How many photons ?
[+] placebo|8 years ago|reply
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...
[+] pavement|8 years ago|reply
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.

[+] sdenton4|8 years ago|reply
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...
[+] jankotek|8 years ago|reply
Hubble resolved this start to a disc 20 years ago.
[+] jngreenlee|8 years ago|reply
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|8 years ago|reply
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?

[+] ajnin|8 years ago|reply
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|8 years ago|reply
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.
[+] morsch|8 years ago|reply
Made me double-check to verify it's not an animation...
[+] btilly|8 years ago|reply
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!

[+] mabbo|8 years ago|reply
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_superno...
[+] seren|8 years ago|reply
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.

[+] slivanes|8 years ago|reply
When we notice it exploding it would have actually exploded 600 years prior.
[+] sizzzzlerz|8 years ago|reply
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.
[+] kmm|8 years ago|reply
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?
[+] nkrisc|8 years ago|reply
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|8 years ago|reply
> 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 :)

[+] lawless123|8 years ago|reply
it's probably more dense as you go in.
[+] zucchini_head|8 years ago|reply
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_J...

[+] semaphoreP|8 years ago|reply
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).
[+] itcrowd|8 years ago|reply
Keep in mind that JWT is a space-based telescope and ALMA is ground based (and an interferometer, not a single dish).
[+] contact_fusion|8 years ago|reply
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 [2] https://arxiv.org/abs/0809.1424 [3] https://jwst.nasa.gov/faq.html#webbbetter (question 25) [4] https://almascience.eso.org/about-alma/alma-basics (section: spatial resolution)

[+] mrspeaker|8 years ago|reply
"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|8 years ago|reply
Somewhere in between. In astronomical terms "really soon"
[+] kijin|8 years ago|reply
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?
[+] westbywest|8 years ago|reply
Despite its immense size, Betelgeuse can be described as a "red hot vacuum." The density of its outer atmosphere / corona is very thin compared to our sun, so it's readily distorted into irregular shapes by the star shedding gas/material.
[+] pyre|8 years ago|reply
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|8 years ago|reply
> 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).

[+] chc|8 years ago|reply
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.
[+] denis1|8 years ago|reply
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.
[+] jefurii|8 years ago|reply
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.
[+] ajmurmann|8 years ago|reply
The article mentions some kind of bubble. Maybe that's it?
[+] 0xFFC|8 years ago|reply
> When that happens, the resulting explosion will be visible from Earth, even in broad daylight.

How much visibility are we talking about?

[+] aruggirello|8 years ago|reply
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.
[+] jyriand|8 years ago|reply
How come there is a white spot on this orange blob?
[+] quotha|8 years ago|reply
"causing it to have only a short life expectancy. The star is just about eight million years old"

everything is relative

[+] jbmorgado|8 years ago|reply
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.
[+] pier25|8 years ago|reply
> in the millimeter continuum

what does that mean?

[+] contact_fusion|8 years ago|reply
"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 [2] https://en.wikipedia.org/wiki/Spectral_line