
Betelgeuse captured by ALMA - aaron695
http://www.eso.org/public/unitedkingdom/images/potw1726a/
======
jcims
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
How many photons ?

~~~
lisper
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...](http://www.spectralcalc.com/blackbody_calculator/blackbody.php)

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
That's a pretty good Fermi calculation!

~~~
lisper
Thanks! Ain't math great? :-)

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placebo
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
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.

~~~
abruzzi
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.

~~~
mjb
> we won't have to worry about optics anymore

Right, if you have phase information you have a lot more options for (cheaply)
making a synthetic aperture that's way bigger than your possible physical
aperture.

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jngreenlee
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
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?

~~~
cr0sh
> 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...](https://www2.warwick.ac.uk/fac/sci/physics/research/cfsa/people/erwin/research/3d/principles/)

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).

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btilly
[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!

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mabbo
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...](https://en.wikipedia.org/wiki/Betelgeuse#Approaching_supernova)

~~~
seren
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.

~~~
_jal
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.

~~~
anonymous_iam
And watch out for those sand worms!

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sizzzzlerz
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.

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joshvm
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/coast-astronomical-results/surface-imaging-of-betelgeuse/)

[http://www.mrao.cam.ac.uk/outreach/radio-
telescopes/coast/](http://www.mrao.cam.ac.uk/outreach/radio-telescopes/coast/)

------
kmm
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
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.

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zucchini_head
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...](http://phenomena.nationalgeographic.com/files/2014/05/1RSX_J160929.1-210524.jpg)

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

~~~
SamUK96
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...

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mrspeaker
"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
Somewhere in between. In astronomical terms "really soon"

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kijin
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
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.

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0xFFC
> When that happens, the resulting explosion will be visible from Earth, even
> in broad daylight.

How much visibility are we talking about?

~~~
aruggirello
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
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
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).

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tim333
Some pictures of and info on the telescope
[https://directory.eoportal.org/web/eoportal/satellite-
missio...](https://directory.eoportal.org/web/eoportal/satellite-
missions/a/alma)

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grovegames
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
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...](https://en.wikipedia.org/wiki/Sun#After_core_hydrogen_exhaustion)

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

~~~
semaphoreP
The colors are only there to represent brightness, so the white spot is a
brighter patch on the star (brighter because it's hotter).

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

everything is relative

~~~
jbmorgado
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.

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pier25
> in the millimeter continuum

what does that mean?

~~~
contact_fusion
"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)

