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The Deepest Photo Ever Taken
Astroturtle writes "Astronomers using the Hubble Space Telescope's powerful new Advanced Camera for Surveys (ACS) have taken the deepest visible-light image ever made of the sky. The 3.5-day (84-hour) exposure captures stars as faint as 31st magnitude, according to Tom M. Brown (Space Telescope Science Institute), who headed the eight-person team that took the picture."
Imagine a Beowulf... um. Seriously, how do you cope with reciprocity failure in a 3.5 day exposure. I would have thought that stray heat or electron flow would turn the whole image to static with such a long exposure. HST must consist of unfathomably cool (literally and figuratively) electronics.
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Just how many photons they detected for the faintest star.
"Eve of Destruction", it's not just for old hippies anymore...
3.4 day exposure? Even for a space-based platform, that has to be really stable to produce a good image. Does anyone out there have any info on how they maneuver the telescope to keep it pointing at the same point while minimizing shifts in the field?
Derek
Don't Panic...
It is still pretty incredible...pointing an object the size of a bus and accurately focusing it on something the size of a spec of sand...really, really, really far away. All while moving at a relative 26,000 miles an hour or whatever to keep it up in the sky...Not to mention the orbial speed of the earth itself... Only took 8 guys, several computers, and millions of dollars worth of equipment. Oh yeah, and that one maintenance run made a few years back to keep it pointing straight.
Who is this that even the wind and the waves obey Him? Surely this computer must submit also!
Something I've wondered for a while... what's up with the points coming off the stars? I've always accepted it when I see it with my own eyes because I don't expect my own eyes to be optically perfect, so I always thought it was distortion, but looking at the full image I see that the brightest stars once again have points coming off of them in four directions. Typically they are directly up, down, left, and right, but in that image, they appear to be about five to ten degrees off that.
The biggest example I see is about 3/4s of the way to the right and about 1/5 of the way down on the image, where there is a huge-looking star.
Why four points? Why do we see them even when the star itself is not in the picture (look on the top border for examples, like the one almost directly in the middle)? I guess I would expect that if the light source is too bright the spread would be in a circular formation and simply blur the star, not blur it in just those four directions so much stronger then the rest.
Is it just QM at play? If so, why it is almost always directly up, down, left, and right, instead of random and perhaps even changing over time directions (which probably would get right back to simply looking blurred)? Detector flaws?
I just thought about how detailed the full size pic is, and how to appreciate that without a poster printer. I'd like to see something similar to osx's default screensavers (with the softly zooming pictures of trees/beach, etc) and have it use this picture.
I'd like to see it zoom in to the picture, while also changing x/y of the camera on a spline (etc). And each time choose a different starting point, and make it's speed adjustable.
All of those high-res pics are beautiful! maybe i'm 'a gonna dust off the 'gl..
If you look at the image, there are some odd streaks that go from red to blue (or blue to red).
I'm just curious here, what are they? I thought maybe it could be a bit of space debris that whizzed in front of the camera, but with an exposure of 3.4 days, the streak would go from one side or another.
The streaks are probably something that moved, though some of them seem brighter in the center, which would indicate it was oscillating. I'm not exactly sure. Anything could move any distance in 3.4 days.
And that big bright cluster in the lower bottom, what's that? It looks pretty close galaxy-wise.
The bright cluster is probably a globular cluster, which is a tight grouping of old stars. It's most likely in our galaxy.
It's a neat pic for sure, a little blurry, which makes it less jawdropping than other hubble efforts but makes sense for a 3.4 day exposure.
Scale it down and it's definitely not blurry. At least not the 6116x7014 image.
lower bottom
What the hell?
Does anyone know if there is a BitTorrent file out for the 128mb TIFF? the nasa servers are a bit slow and I feel my hardware cycles and bandwidth could be of use...
Actually, the really unique thing about this image is the stellar populations. The stars you see in the image are almost all in the Andromeda galaxy (aka M 31), seen here.
M 31 is 2.2 million light-years away. This is the galaxy that Hubble originally resolved into stars, thereby settling the Shapley-Curtis debate on the true scale of the Universe. However, the stars Hubble saw were the very brightest supergiants in M 31. In this HST image, we see stars 2 magnitudes fainter than the ancient main-sequence turn-off; i.e., stars which are intrinsically fainter than our Sun! This lets us learn a lot about the ages and chemical composition of M 31's halo stars, which turn out to be quite different from the stars in our halo (our halo is entirely composed of ancient, metal-poor stars; M 31's halo contains stars that are only 6 Gyr old, and much more metal-rich than our halo).
I heard Tom Brown give a talk on this work last week; very cool stuff.
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The most intersting thing about these images now is the fact that they are not blurred:
This Nature article describes how....hmmm I had better quote:
"As a beam of starlight hops towards us through countless Planck times, its speed varies. This would smear the beam out so that different parts arrive at different times and distort our picture of where it came from. The longer the journey, the bigger the smear."
So that means that these deep Hubble photographs should all get more blurry the deeper you look and not razor sharp like we have come to love.
Its a fascinating problem!
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a = angle subtended by capture (radians) = 2*pi*3.1/(360*60) = .000901
.0000008118
.0000000646
b = area of capture on a sphere of radius 1 =approx= a^2 =
c = fraction of entire sphere = b/(4*pi) =
d = number of visible stars in entire sky = 300000/c = 4,643,000,000,000
So that means almost 5 trillion stars are visible by Hubble in the entire sky. That's a lot of stars to catalog. (Assuming I didn't err like they did in the $97 trillion RIAA calculation... someone pls double check and flame me if appropriate.)
Ok, here's the calculation for you curious types, regarding how many photons arrived from the faintest star in the picture:
Let's suppose that the picture was taken in the "V" filter. I just happen to have the number of photons per second per meter squared that arrive from a star of 20th magnitude: 86.157. (taken from here ).
So the faintest stars in this picture are 31st magnitude? That's 11 mags fainter than 20, which by the handy old formula
mag1-mag2 = -2.5 * log(flux1/flux2)
which means that the 30th magnitude star puts out about 4x10^(-5) times as much flux.
Using the reference star's flux from above, this means that 0.0034299 photons per second per meter squared arrived at Hubble. The exposure was 84 hours, and the area of Hubble is (2.5m)^2*pi, so tada:
The total number of photons in the picture from the faintest star is: 20365.83
Still not too shabby. They probably could have found even fainter stuff.
Don't forget, this is only 3.1 arcminutes^2 of the sky. that's the size of a grain of sand held at arms-length against the sky, according to the article. That leaves a ton of sky still unseen.
You like your new Mac more than you like me, don't you, Dave? Dave? I asked...She said Yes.