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.'"
Re:article image (Score:4, Informative)
The second link provides a 78MB TIFF (and a more modest but same-resolution 30MB JPEG) image.
Re:article image (Score:5, Informative)
However, it's a dinky low-resolution image one could have captured with a CCTV camera. Come on, you can do better than that.
I'm sorry, but what? The second link in the story has links to 6480 x 4236 JPGs and TIFs, which calculates to 27MP, the file sizes are 31MB and 79MB, respectively.
Normally, I would agree that web stories normally fall short with photos and multimedia, but it's just not true here.
Re:Hmm... (Score:2, Informative)
not all at once, 78.6MB file could cause the slashdot effect
Comment removed (Score:2, Informative)
Coverage by Bad Astronomer here (Score:5, Informative)
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.
Re:How much area does this cover? (Score:5, Informative)
The image is 14.1 x 26.1 arcminutes according to ESO website. For reference, the moon is about 30 arcminutes.
Re:How much area does this cover? (Score:4, Informative)
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:Coverage by Bad Astronomer here (Score:3, Informative)
Actually, for perspective, this image is approximately 1/500000th of the sky.
Re:article image (Score:2, Informative)
If an image is sharp, then you weren't pushing the limits of the instrument in the first place.
Re:Star photo (Score:2, Informative)
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.
Re:How much area does this cover? (Score:2, Informative)
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.
Re:Google Maps (Score:3, Informative)
Re:Star photo (Score:3, Informative)
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.
errors in the calculations, and fixes: (Score:4, Informative)
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.
Re:Only 27 megapixels? (Score:3, Informative)
Re:Star photo (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.