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Space Science

Very Large Telescope Captures New 27-Megapixel Deep Field 131

xyz writes "European Southern Observatory's Very Large Telescope has captured the deepest ground based U-band image of the universe yet. The image contains more than 27 million pixels and is the result of 55 hours of observations with the VIMOS instrument. 'Galaxies were detected that are a billion times fainter than the unaided eye can see and over a range of colours not directly observable by the eye. This deep image has been essential to the discovery of a large number of new galaxies that are so far away that they are seen as they were when the Universe was only 2 billion years old.'"
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Very Large Telescope Captures New 27-Megapixel Deep Field

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  • Hmm... (Score:5, Funny)

    by Anonymous Coward on Sunday November 09, 2008 @10:28AM (#25694231)

    Oh my god. It's full of pixels!

  • Shortages? (Score:4, Insightful)

    by jrq ( 119773 ) on Sunday November 09, 2008 @10:30AM (#25694241)
    Hard to believe, looking at this, that there could ever be a shortage of anything.
  • 80 MB TIFF (Score:4, Insightful)

    by owlstead ( 636356 ) on Sunday November 09, 2008 @10:35AM (#25694253)

    Seriously, Slashdot, pointing to an article that contains a link to the 80 MB TIFF image at full resolution. Feeling a bit sadistic today, are we? Oh well, I'm rather early so I clicked it nonetheless. Feeling like a bit of a egocentric sadist myself today.

    It works without a hitch in the AlternaTIFF TIFF Image Viewer. You can clearly see the galaxies, but otherwise it is a large sheet of colored dots (as expected I suppose).

    • Re:80 MB TIFF (Score:4, Interesting)

      by v1 ( 525388 ) on Sunday November 09, 2008 @11:08AM (#25694431) Homepage Journal

      don't feel sorry for them... feel sorry for the poor schmucks that actually CLICK on the image link and expect their browser to render it before the wheels grind to a halt.

    • by Agripa ( 139780 )

      Seriously, Slashdot, pointing to an article that contains a link to the 80 MB TIFF image at full resolution. Feeling a bit sadistic today, are we? Oh well, I'm rather early so I clicked it nonetheless. Feeling like a bit of a egocentric sadist myself today.

      At least it would explain the flash located near Munich, Germany.

  • Fantastic (Score:2, Insightful)

    by Skiron ( 735617 )
    The mind boggles. How anybody can believe we are here all alone, I don't know.
  • by chanrobi ( 944359 ) on Sunday November 09, 2008 @10:58AM (#25694387)
    This new "picture" is taken in UV for which the Hubble ultra deep field [wikipedia.org] is still the deepest image taken in visible wavelengths. Which provides, if you believe the current age estimate of the universe (13.73 ± 0.12 billion years old) means Hubble is still going back further. 0.73 Billion years vs 2 billion years since the beginning of the universe.

    Just to give a sense of perspective in case you read the title and went so what?
    • Maybe so, but the Hubble is an ST and as I understand that's Space based, not ground based.
    • by raynet ( 51803 )

      Also this 27Mpix image is very blurry, they could probably taken it at 10Mpix and still have the same amount of information.

    • Something I've always wondered is that since we supposedly know how far the farthest recorded star is, and supposedly know the age of the universe, how fast are we moving away from the farthest star? And how fast are we moving relative to the source of the big bang? If there's 10 lightyears from us to that star, the most conservative guess (the slowest speed) would put the big bang starting directly between us and the other star, putting the big bang somewhere else would mean we're moving faster. Does anyo
      • Re: (Score:3, Interesting)

        And how fast are we moving relative to the source of the big bang?

        All points equally in the entire universe are "the source of the big bang". The big bang isn't just the origin point for all matter, it's the origin point for the entire universe and all space and time. If you want to "see" the big bang, or as close as we can see, look at the CMB [wikipedia.org].

  • Don't let them trick you into thinking you need MORE megapixels. It's all a feature bloat trick, sales people love to use to make the devices more expensive. No really, I love the deep field/space research. Amazing imagines. I thought Hubble was broken again? I guess they fixed it.
  • Maybe I missed something, but how is this impressive?

    Considering that there are commercial cameras [wired.com] on the market that have resolutions of 50+ megapixels for "just" $40,000 (not much for professional scientists or astronomers). It seems like a fairly simple thing to modify for use in the UV spectrum (maybe that's the part we are supposed to be impressed with?).

    Perhaps they meant gigapixels?
    • Re: (Score:3, Insightful)

      by Mprx ( 82435 )
      The optics are the limiting factor here. Increasing pixel count wouldn't add any more detail.
      • by dwater ( 72834 )

        then one has to wonder why they bothered mentioning it...is there no way to measure the capability of optics, if that is what is important?

    • Re: (Score:3, Informative)

      by E-Lad ( 1262 )
      You're making a bad assumption here, here's why: As another respondent indicated, optics are one of the limiting factors here. The other major factor is the imaging device itself and the size of the individual pixels on the sensor. The amount of photons a individual pixel can collect is governed by the size of that pixel. The larger the pixel, the more photons it can gather. The more photons, the more light-sensitive it is. Now the sensor used on everything from your pocket P&S camera to a imaging de
      • > Given this optical limit, a point is reached where the addition of more pixels on the sensor becomes, essentially, a useless exercise [...]

        Except for marketing, of course. :-)

  • I know a man can see various things in a random set of dots that are not really there, but what about these `filaments' of galaxies?
    What are these?
  • by superid ( 46543 ) on Sunday November 09, 2008 @11:34AM (#25694569) Homepage

    Yeah, I know..... a lot....

    What I mean is, if I look up in the sky, how big of a patch of the sky does this picture cover? The size of the full moon? Bigger? Smaller than a grain of sand at arms length?

    • by Falkkin ( 97268 ) on Sunday November 09, 2008 @12:21PM (#25694801) Homepage

      The image is 14.1 x 26.1 arcminutes according to ESO website. For reference, the moon is about 30 arcminutes.

    • by Tinlad ( 947666 )
      According to Wikipedia [wikipedia.org], the VIMOS has a 14 x 14 arcminute field of view. If my Sunday afternoon trigonometry is correct, that's equivalent to a 3mm square held at arm's length (~70cm). Under half the diameter of the full moon. Hope that helps to put it into context - it's a tiny area of the sky.
      • If my Sunday afternoon trigonometry is correct, that's equivalent to a 3mm square held at arm's length (~70cm). Under half the diameter of the full moon. Hope that helps to put it into context - it's a tiny area of the sky.

        I don't know about you, but if I held 3 mm up at arms length, it isn't going to come close to covering the moon. Perhaps you meant 3 cm.

        • by Tinlad ( 947666 ) on Sunday November 09, 2008 @01:15PM (#25695209) Homepage

          I think you, like a lot of people, have wildly overestimated the angular diameter of the full moon. It's about 30 arcminutes (0.5 degrees). It's a lot smaller than you think. It's one of the first things we were told in my astrophysics lectures, and it's stuck with me.

          An angle of 0.5 degrees at arms length (~70cm) gives you approximately 70cm * tan(0.5 degrees) = 6.1mm (i.e. a circle of paper 6.1mm in diameter held 70cm from your eye would 'cover' the full moon). Try it.

          3cm at arms length equates to an angle of about 2.5 degrees.

          • Re: (Score:3, Interesting)

            I just tried it by holding a ruler at arms length up at the moon. The moon isn't quite full tonight, but it's pretty close, so I can guesstimate where the rest of it is from the curvature of what's there.
            I found the moon to be about 1.2cm with the ruler held at arms length - about twice what you're suggesting. Perhaps I have very short arms?

            I don't dispute your maths, but I would like to know where the discrepancy in my experimental evidence is coming from...

            • Re: (Score:2, Informative)

              by Anonymous Coward

              From wiki: "The average centre-to-centre distance from the Earth to the Moon is 384,403 km, about thirty times the diameter of the Earth. The Moon's diameter is 3,474 km"

              Roughly 1 in 100. So at 70 cm the moon would appear .7 cm.

              I would check the measurement on your arm.

            • by Tinlad ( 947666 )

              Atmospheric lensing? I haven't a clue to be honest. Where I live the moon's just peeked out from behind the clouds so I gave it a go - it was about 6 or 7mm, so my observations seem to fit the theory.

      • Bah, bad click, meant Informative
  • Nice time for new wallpaper on my laptop :)
  • by RevWaldo ( 1186281 ) on Sunday November 09, 2008 @12:14PM (#25694761)
    Excellent name - simple and straightforward. They should have a contest for naming the next model. Put me down for "Amazing Freaking Ginormous Wonderscope"
  • by Falkkin ( 97268 ) on Sunday November 09, 2008 @12:17PM (#25694773) Homepage

    Phil Plait has quite a bit to say about this image:

    http://blogs.discovermagazine.com/badastronomy/2008/11/07/voyaging-deep-into-the-universe/ [discovermagazine.com]

    "Scanning the full-res image is incredible. There's so much to see! Each dot, each smudge, is a full-blown galaxy, a collection of billions of stars. They're very, very far away; some of these galaxies are estimated to be 10 billion light years distant; you're seeing them as they were just a couple of billion years after the Universe itself began, and the faintest are one-billionth as bright as objects you can see with your own eye."

    He also talks quite a bit about his favorite astronomical event - gamma-ray bursts.

    • It's another excellent analysis by Phil, suitable for all audiences - I shall give that to the wife, I think there's a sporting chance of her actually understanding what is going on (normally keen to show an interest in astronomy but can never fathom out even the basic concepts)
    • by blair1q ( 305137 )

      So we're just a motion-capture system away from the Total Perspective Vortex...

      • Re: (Score:3, Informative)

        by Falkkin ( 97268 )

        Actually, for perspective, this image is approximately 1/500000th of the sky.

        • by ashitaka ( 27544 )

          OK, understanding that figure and looking at the full-res image is the earthy equivalent of the Total Perspective Vortex.

          "GGGGHHAAAAAARRRRRRRGGGGHHHHHHH..."

          Hey, I'm still here! I must be the Coolest Guy In The Universe!!
           

  • Galaxies were detected that are a billion times fainter than the unaided eye can see...

    What's a billion times "I can't see shit?"

  • I count galaxies in 1/8 x 1/14 of that image to be 150. In the whole image there are approx. 16800 galaxies. Since this is 14x21 arcminutes and 1 degree is 60minutes, hence this is 0.3 degree of 360 degree sky, I thinkg there are... 6.752*10^9 galaxies in the visible universe!
    • And, with an average of 40 billion stars in a galaxy (it is conjectured that there are some very small galaxies, making the average much smaller than our own Milky Way), that makes 2.7008*10^20 stars... Ummm... woah.
      • by Anonymous Coward on Sunday November 09, 2008 @06:37PM (#25697817)

        Some corrections, because the GP confused linear and solid angles:

        14 linear arcminutes * 21 linear arcminutes = 294 sqare arcminutes

        1 square degree = (60 linear arcminutes)^2 = 3600 sqare arcminutes

        294 square arcminutes / 3600 sqare arcminutes ~= .08167 square degrees

        there are ~41253 square degrees in a sphere, only this fraction of a sphere is subtended by the picture:

        (294 square arcminutes) / (41253 square degrees) ~= 1.980*10^-6

        As someone stated elsewhere, this is about 1/500,000 of the sky (i.e. the celestial sphere).

        So we count the number of galaxies encountered in this secion, then divide by the fraction subtended; using GP's estimate:

        16,800 / (1.980*10^-6) gives ~8.49*10^9 galaxies

        However, about 2 orders of magnitude more galaxies are in the field, though only ~16,800 galaxies are detected in this particular image of the field. The number of galaxies in the *observable* universe is at least on the order of 100 billion (10^11), per other, more sensitive surveys with more rigorous counting methods than a quick subsampling as performed by a human examining an image visually.

        Next:

        ...with an average of 40 billion stars in a galaxy...

        This is lower than I've encountered. The average galactic mass is about 100 billion solar masses, and the average stellar mass is about .5 solar masses*, so the the average number of stars in a galaxy is is on the order of 100~200 billion.

        ...it is conjectured that there are some very small galaxies, making the average much smaller than our own Milky Way...

        Actually, it is fairly well established that there are indeed many such "small" galaxies. But though the number of "extremely large" (trillions to tens of trillions, versus hundred billions for the Milky Way) galaxies is small, the contribution to the mean ("average") number of stars per galaxy is disproportionately large because they themselves are disproportionately large. This is the nature of the arithmetic mean: a few highly weighted outliers skew the mean more than the median, and the median more than the mode. That's precisely why the "average" number of stars per galaxy is actually on the order of the Milky Way.

        (* Note that the "average" stellar mass is skewed upward by the few but extremely massive stars just as galactic mass is. A "typical" star is smaller than the .5-solar mass "average" star; the vast majority of stars are smallish red dwarfs, with the sun being more massive than at least ~90% of stars, if only by a little in the range of stellar masses from ~.04 to ~150.)

        So:
        ~(10^11 galaxies) * ~(10^11 stars/galaxy) = ~10^22 stars
        The highest *reasonable* estimates I've seen yield a little over 5*10^22 stars, so on the order of 10^23 stars is still conceivable.

  • by Pogue Mahone ( 265053 ) on Sunday November 09, 2008 @12:23PM (#25694819) Homepage

    Did anyone notice the name of the press officer?

    Dr. Henri Boffin.

    Nominative determinism [wikipedia.org] in action.

  • 27 megapixels is almost consumer grade today. Why so low?
    • 27 megapixels is almost consumer grade today. Why so low?

      They knew we would Slashdot 270 megapixels into oblivion.

  • Picture-perfect Crystalline Entity in that "Fistful of Data". But, i somehow think we are still quite a way from a lateral sensor array...

  • Why do star photos have crosses over bigger stars?

    • Re: (Score:2, Informative)

      by Tablizer ( 95088 )

      Why do star photos have crosses over bigger stars?

      Refraction flares caused by the crystalline pattern of molecules in the glass of the lenses. It only shows up in brighter objects because the flares are too dim for dimmer objects to make an impression. Bright stars simply overwhelm the local optics when you are gathering enough light to expose the dimmer objects.
           

      • by Kagura ( 843695 )
        Great explanation, thanks a lot!
      • Re: (Score:3, Informative)

        Why do star photos have crosses over bigger stars?

        Refraction flares caused by the crystalline pattern of molecules in the glass of the lenses.

        Um, no. The spikes are caused by the diffraction of light around the struts supporting the secondary mirror in the telescope. The wave nature of light ensures that no matter how large you build your telescope, you cannot focues stars to a perfect point.

        • by Tablizer ( 95088 )

          Well, it's time to hire Mythbusters to settle this. Maybe its both. If the struts were that big of a problem, then couldn't they use a flat lens-plate(s) to hold the secondary mirror instead?

          • Re: (Score:2, Informative)

            Well, it's time to hire Mythbusters to settle this. Maybe its both. If the struts were that big of a problem, then couldn't they use a flat lens-plate(s) to hold the secondary mirror instead?

            Astronomers hate putting lenses into their optical systems--there is always some light lost to reflection off the glass surface. The VLT is an 8 meter diameter telescope, so supporting a giant lens above the telescope would be a major engineering issue. This isn't really a problem you can solve by adding a new lens or tweaking the secondary support structure--it's a fundamental feature caused by the wave nature of light. Anytime light passes through an aperture, it creates a diffraction pattern.

      • by Trogre ( 513942 )

        Glass? Crystalline? I thought glass was an amorphous solid. I had always believed that the cross-shape on stars is caused by light diffracting around the "spider", the struts supporting the secondary mirror.

  • Somehow, I feel like a feeble eyed old man with a cane, putting on me specks to see those faint, distant lights in the fog. Marvelous!

I tell them to turn to the study of mathematics, for it is only there that they might escape the lusts of the flesh. -- Thomas Mann, "The Magic Mountain"

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