Betelgeuse captured by ALMA

http://www.eso.org/public/unitedkingdom/images/potw1726a/

by aaron695 • 9 years ago

222 77 9 years ago

eso.org

jngreenlee• 9 years 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• 9 years 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• 9 years 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

Mr Kitaoka is quite a nice guy and will answer questions if you ask him on Facebook (his English is quite good)

cr0sh• 9 years 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/peo...

This has been commercialized into something called ChromaDepth (using special holographic lens glasses to enhance the effect):

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• 9 years ago

Made me double-check to verify it's not an animation...

ajnin• 9 years 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• 9 years 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• 9 years 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• 9 years ago

Somewhere in between. In astronomical terms "really soon"

enimodas• 9 years ago

Supernovas visible in daylight have been recorded in history a couple times, for example https://en.wikipedia.org/wiki/SN_1006

zucchini_head2• 9 years 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_J...

itcrowd• 9 years ago

Keep in mind that JWT is a space-based telescope and ALMA is ground based (and an interferometer, not a single dish).

SamUK96• 9 years 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...

semaphoreP3• 9 years 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_fusion3• 9 years 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 [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)

zucchini_head2• 9 years ago

I think maybe your definition of "resolved" is a little skewed. It is not about the features of the object, but more by the Rayleigh Criterion [1][2]

So we can already (and have been able to for a long time) to "resolve" things as apparently-small as exoplanets, but for resolving _surface details_ we are one order of magnitude away for interferometers and two orders of magnitude away for standard single-mirror telescopes. Right?

[1] https://en.wikipedia.org/wiki/Angular_resolution#Explanation [2] Lord Rayleigh, F.R.S. (1879). "Investigations in optics, with special reference to the spectroscope".

kijin• 9 years 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• 9 years 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• 9 years 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• 9 years ago

The article mentions some kind of bubble. Maybe that's it?

chc• 9 years 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• 9 years 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• 9 years 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).

westbywest• 9 years ago

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.

mabbo• 9 years 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_superno...

seren• 9 years 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• 9 years 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• 9 years 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• 9 years 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.

slivanes• 9 years ago

When we notice it exploding it would have actually exploded 600 years prior.

SAI_Peregrinus• 9 years ago

That depends on which frame of reference you pick, of course.

didgeoridoo2• 9 years ago

What is the right way to think about this? If we observe a supernova 600LY away, do we say that event is happening "now" from our frame of reference? Or should we think of it as happening "600 years ago", and the light from the event is only now reaching us?

If you think of causality itself moving at the speed of light (which of course it does), and think in light cones rather than referring to a nonexistent universal frame of reference with an authoritative clock, the "now" language seems more appropriate. Intuitively, it also feels quite wrong — but I attribute this more to a failure of intuition than of language.

quotha• 9 years ago

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

everything is relative

jbmorgado• 9 years 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• 9 years 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.

iamgopal2• 9 years ago

How many photons ?

lisper2• 9 years 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.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.

semaphoreP3• 9 years ago

Optical interferometers have resolved Betelgeuse down to 9 milliarcseconds (4 x 10^-8 radians), so this isn't even the highest resolution image yet of this star[1].

[1] http://blogs.discovermagazine.com/badastronomy/2010/01/12/sp...

kmm• 9 years 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• 9 years ago

it's probably more dense as you go in.

nkrisc• 9 years 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.

aruggirello2• 9 years 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• 9 years ago

When that happens, the resulting explosion will be visible from Earth, even in broad daylight.

How much visibility are we talking about?

aruggirello2• 9 years 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• 9 years 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.

placebo2• 9 years 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• 9 years 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• 9 years 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• 9 years 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• 9 years ago

Interesting read on the subject of capturing the lander sites: http://curious.astro.cornell.edu/about-us/45-our-solar-syste...

abruzzi• 9 years ago

I'm not an expert on this stuff, but my understanding is the real issue isn't the sensor, it's the lens you put in front of it. You essentially have two ways to see smaller things from a fixed vantage point--put a longer focal length in front of the existing sensor, or put a higher resolution sensor behind the existing focal length lens. The problem is longer focal length gets big and expensive very quick, and existing lenses would limit the ability of a high resolution sensor. In many cases, high end professional camera with 36mp or higher sensors are hampered by the lenses that can't resolve that much detail.

Now, maybe in a few decades the CalTech lensless sensor will be commercially available and will work well enough that we won't have to worry about optics anymore, and it will all be silicon, but CalTech's sensor currently has something like 16 pixels total, so it has a long way to go.

jankotek• 9 years ago

Hubble resolved this start to a disc 20 years ago.

sizzzzlerz• 9 years 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• 9 years ago

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

semaphoreP3• 9 years 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).

theandrewbailey2• 9 years ago

I was wondering also. That image looks more like a shiny smooth blob.

tim333• 9 years ago

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

joshvm• 9 years 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/co...

http://www.mrao.cam.ac.uk/outreach/radio-telescopes/coast/

grovegames• 9 years 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• 9 years 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_exhaus...

btilly• 9 years ago

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• 9 years ago

in the millimeter continuum

what does that mean?

contact_fusion3• 9 years 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 [2] https://en.wikipedia.org/wiki/Spectral_line

Betelgeuse captured by ALMA

Related Stories