Betelgeuse captured by ALMA
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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.
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.
[1] http://blogs.discovermagazine.com/badastronomy/2010/01/12/sp...
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.
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.
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.
Edit: Maybe it's just the coffee I drank?
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?
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)
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).
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.
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!
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...>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.
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.
When we talk about time in the sense of things happening "now" or "in the past" or "after", we have to think about reference frame, _not_ location. Reference frames are inherently global, as opposed to an "event", which encodes both location and time relative to all possible inertial reference frames.
The typical metaphor is imagine that you infiltrate space with a three-dimensional grid of clocks that are kept a fixed distance from each other (say by a rigid rod). Those clocks are not moving relative to each other, and it is trivial to synchronize them, because the distance between them is fixed -- fire a light pulse to your neighbor with the current time, and you neighbor will know when you sent the symbol by subtracting off the time it takes light to travel the distance. This grid, that covers all of space, represents a single reference frame.
So if Betelgeuse had in fact exploded six hundred years ago, then the clock grid in Earth's inertial frame would have recorded the event of the Battle of Orewin Bridge on Earth at the same time as the event of the beginning of the Betelgeusian supernova in the Betelgeuse system.
The complexity of this comes in the fact that the inertial frame travelling towards Betelgeuse from Earth at some significant fraction of the speed of light would, with its clock grid, measured Orewin Bridge well before the supernova's start. That's independent of the amount of time it would take for those two clocks to communicate with each other -- we can almost imagine a scientist who finally downloads the logs for all the clocks in a given reference frame collating the data.
What we _can_ say, though, is that once the event of someone on Earth seeing the supernova occurs, then that event is strictly _after_ the supernova -- no inertial reference frame will ever see that event, "observation of supernova", occur before the "initiation of supernova".
No, because it already went nova a long time ago.
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...
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 :)
[1] http://phenomena.nationalgeographic.com/files/2014/05/1RSX_J...
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)
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".
This type of direct detection was one of the first of its kind, so I wouldn't characterize this as an old capability - 2008 is relatively recent. Telescope turnover time is very long; Gemini remains a prominent telescope for science-class observations. Additionally, most new telescope generations don't achieve an order-of-magnitude improvement in resolution, or at least, not anymore. There are a lot of serious, decadal-scale barriers to improving resolution that must be overcome.
In terms of angular resolution, the order-of-magnitude estimates are the minimum improvements, assuming that such a close and large exoplanet exists. (AFAIK, there is no such system.) In practice it is likely that we need even better angular resolution, as there are not many systems within 10 ly away, and extremely large exoplanets are not very common (relatively speaking.)
That sounds spectacular! What's the time frame on that - closer to 10 years or a million years?
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.> 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.
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).
How much visibility are we talking about?
https://en.wikipedia.org/wiki/Sun#After_core_hydrogen_exhaus...
everything is relative
what does that mean?
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