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You can't make an interferometer by combining photographs. You need to combine the light sources so that the light waves can "interfere", which means you need to have not only the intensity of the light, but also the phase of the wavelength. A CCD only detects the intensity. You also need to position your telescopes to sub-wavelength accuracy. For ...

8

The LIGO interferometer uses a homodyne detection technique. Basically, the light travelling in each arm of the interferometer is derived from the same laser source and is combined in the output channel and falls onto a photodiode. The interferometer is operated so that when there is no gravitational wave (GW) passing through the instrument, the beams ...

7

This question amounts to: is optical interferometry possible when detecting photons? The answer to this is yes. Many experiments have been done using interferometers where one photon at a time is passed through the apparatus and still reveals an interference pattern when detected on the other side. The problems with optical interferometry are manifold: ...

6

Millimeter and sub-mm observations (110-300 GHz = 2.7-1.0 mm) are sensitive to the thermal emission and provide a brightness temperature of airless bodies like the Moon or asteroids. The radio emission is from just below the surface/skin of the body (down to $\sim10\lambda$ so ~1 cm in this case) and so is also sensitive to the thermal inertia and ...

6

A 130m baseline operating at 2 microns gives a theoretical resolution of $2\times 10^{-6}/130$ radians. At a distance of 400,000 km this translates to 6m. My guess is that Genzel is referring to the accuracy with which the position of a point source of light can be measured. This not the same thing as the smallest thing you can image. The centroid of a ...

6

This is, indeed, a result of how we measure things in radio astronomy. It's not just interferometry, but radio astronomy in general. The thing they're referring to is a concept called "brightness temperature". In the low frequency limit (valid for radio astronomy), we can use the Rayleigh-Jeans approximation, which gives us the expression  T_B = \frac{...

6

Radio interferometry can combine observations over very large baselines. But optical interferometry cannot. According to a list of interferometry instruments on wikipedia, the largest baseline for optical measurements is less than a kilometer. We can't take optical measurements with continent-sized instruments. Then if you drop down to radio where the ...

4

This is a really interesting question! tl;dr: A definite maybe, but you would have to engineer a clever way to focus the transmitted power to a much smaller spot first; possibly several orders of magnitude smaller than what any one dish can do. Phasing several widely-spaced dishes alone would not be enough. Let's see what can be checked easily. ...

4

Actually to convert from Jy/beam to Jy/pixel you need to divide by the beam size. Let's say you have a quantity of 1 Jy/beam, then $\frac{Jy}{beam} \frac{beam}{\Omega}$, then to go from Jy/beam to Jy/pixel you would need to divide by $\Omega$. The values of the beam major and minor axis must be in pixels. Source: NRAO

3

The blind spots are caused by the way the detectors work. They are sensitive to a gravitational wave (GW) changing the relative path length along interferometer arms at right angles to each other. Gravitational waves come in two polarisations (plus and cross). These polarisations cause alternate perpendicular expansions and contractions in space, but are ...

3

The SIMBAD link might be there just because Osterbrock's 2004 AAS presentation about the interferometer mentioned an observation of Betelgeuse. This would be consistent with the policy stated in Wenger et al. 2000: No assessment is made of the relevance of the citation in terms of astronomical contents: the paper can be entirely devoted to the ...

3

A supplemental answer to probably someone's answer: Why this can indeed be called interferometry: Once one thinks in terms of physical optics (e.g. $\text{exp}(j(\omega t - \mathbf{k} \cdot \mathbf{r} ))$ ) instead of ray optics, imaging is always an interference problem, and the math behind correlating signals from an array of radio telescopes to produce ...

3

Interferometry at infrared and shorter wavelengths is more difficult than at microwave/radio wavelengths for a number of reasons. Radio signals can basically be recorded on tape (or rather hard drives these days) at different sites and then recombined (or correlated) "off-line" at another location. This won't work at optical wavelengths because of the higher ...

3

Yes it is possible that a cloud reduce the visibility because the matter in it will absorb some of the light which is going through. Yes, we cannot see all the universe and so we can see interstellar clouds up to a certain distance. Indeed, light takes time to reach us. If there is a star behind a cloud you will have difficulties to determine its true ...

3

So long as you accurately know the beam size, then yes multiplying your Jy/beam measurement (effectively flux density) by the beam size (effectively area of flux) will give you the total Jy (effectively the flux). See this source as an example.

3

They do use two or more detectors to triangulate the origin of a source. Most recently the Advanced Virgo detector in Italy was used in conjunction with the two LIGO detectors in the USA. This has been published in Physical Review Letters (Phys. Rev. Lett. 119, 141101 – Published 6 October 2017). With two detectors the sky area for "pinpointing" the source ...

3

This article basically seems to answer the question. They quote from an earlier study: "Although no CCSNe have currently been detected by gravitational-wave detectors, previous studies indicate that an advanced detector network may be sensitive to these sources out to the Large Magellanic Cloud (LMC). A CCSN would be an ideal multi-messenger source for ...

2

This would not be a giant version of eLISA, but a small (and complicated) one. The distance between the LISA satellites will be 1 million kilometers, while the moon is on average 380.000km from Earth. So that arm would be less than half as long. Geosynchronous orbit is at 36.000km, not even 10% of the distance to the moon. The difference in length between ...

2

The simple answer is that it isn't resolving in both orthogonal directions equally well. The horizontal dimension is the binocular dimension, and from looking at your animation, it looks to have about ~3 times the resolution. The horizontal banding, I'm pretty sure is not ringing, and is in fact, representative of additional information. This article does a ...

2

The reason you would want to cover most of your aperture is so you can point directly at a huge light source (i.e. a star), but ignore most of the light coming from it. This makes it easier to directly detect faint features around the source that would ordinarily be washed out by the light from the source itself (i.e. planets and the like). I believe this is ...

2

As far as I know the only option here is numerical general relativity. There are no known exact solutions for the general two-body problem in GR, since the field equations are very non-linear. Even there there are limits, since when the fields get strong enough the bodies will start to deform in complex ways that also depend on the equation of state of the ...

1

You can think of the object being photographed as a collection of light-emitting points. The light emitted from each point is spherically symmetric, at least over the small angle that we see. If you sample one of those point sources at two points, and the source lies exactly on the perpendicular bisector of the line segment between the sampling points, then ...

1

Simbad is giving a catalogue of astronomical objects (that are identified in the Simbad catalogue) referred to in the AAS article. This is routinely done for most papers in the recognised astronomical literature. In this case, Betelgeuse appears to be the only star specifically mentioned in that AAS paper (actually just the abstract that is seen on the ADS ...

1

The expression $1.22 \frac\lambda D$ (in radians) represents the distance between the center of an Airy disk and the first minimum of an Airy disk that results from a perfect lens with a perfectly circular aperture and a perfectly focused point source. That assumes a bit too much perfection for some, who also see using the center of the first minimum as a ...

1

$2.06 \times 10^5$ is the radian arcsecond conversion, equal to $180 \times 60 \times 60 / \pi$ . The $1.22$ is a geometrical factor applicable to circular apertures, as outlined in Wikipedia.

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The paragraph you quote continues... The images from SQUEEZE and MiRa have very similar characteristics and are shown in Fig. 1 (see also the Image Reconstruction section in Methods) In that section various tests are described including the reconstruction of simulated data, with similar interferometric coverage and signal to noise, from spotted and ...

1

In principle, would this work? I can't say for sure that "no, in principle it could never work" but the combination in the layers loses information than it would almost certainly decrease image quality somewhat. Is it more efficient and/or easier than traditional methods? It depends what "more efficient" means. If the performance of ...

1

According to this site The 10 telescopes will be optically linked together in order to make images of astronomical objects with unprecedented detail. The interferometer will have a resolution 100 times greater than the Hubble Space Telescope and will be able to make accurate images of complex astronomical objects many times faster than other existing ...

1

There are many radiotelescopes that consist of large arrays of antennas. We usually think of an array individual dishes and VLA and ALMA are two of the most commonly recognized. But if you want to keep an eye on a huge patch of sky in order to try to catch a very rare event, you want a wide field of view more than ultrahigh spatial resolution. And if what ...

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