# Tag Info

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Brown dwarfs have historically been difficult to detect (directly) simply because of how faint they are. Typical luminosities may range from $10^{-3}L_{\odot}$ to $10^{-5}L_{\odot}$ depending on spectral type. Any object that dim will be tough to find, regardless of spectral type or what sort of telescope you're using - you can have the largest, highest-...

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There's a pretty good discussion at this page. There are several factors at work: The smaller isoplanatic angle, as you note. This limits how much of the sky you can observe with AO, since your target needs to be within the isoplanatic angle of a bright enough references star. (Even with laser guide stars, there is still a need for a reference star for "...

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At the risk of self-plagiarism: (https://physics.stackexchange.com/a/532568/43351 with a bit added). Molecular chemistry in the early universe requires species with bound electrons. Helium hydride is the first molecule to form because neutral helium atoms, formed about 120,000 years after the big bang, could combine with plentiful protons; but it was another ...

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This link gives you all the papers that have cited the instrument description paper. The list of papers shows that it has been used for studying: the centres of AGN, close binary systems, discs around young stars, the atmospheres of AGB stars and interferometric imaging of exoplanets at least. Here is a paragraph from the instrument description paper ...

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The Nature paper by Bertoldi et al. (2006) says: Our millimetre observations were performed with the Max-Planck Millimeter Bolometer (MAMBO-2) array detector at the IRAM 30 m telescope on Pico Veleta, Spain. This is a ground-based telescope in the Spanish Sierra Nevada mountains at 2850m to try and get above as much of the precipitable water vapor, which ...

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The IRAS Point Source Catalog, Version 2.0, is a catalog of some 250,000 well-confirmed infrared point sources observed by the Infrared Astronomical Satellite (IRAS), i.e., sources with angular extents less than approximately 0.5, 0.5, 1.0, and 2.0 arcminutes in the in-scan direction at 12, 25, 60, and 100 microns (um), respectively. This includes some ...

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Consider a planet with effective temperature $T$ radiating as a blackbody, emitting a total luminosity $L=4 \pi \sigma R^2 T^4$. At distance $a$ the power per square meter will be $$P = \frac{L}{4\pi a^2} = \sigma \left(\frac{R}{a}\right)^2 T^4.$$ If we demand a sun-like energy flow of $P_{required}$, we get a necessary temperature $$T=\left[\frac{P_{... 5 Mainzer et al. 2014 characterize the performance of the reactivated NEOWISE. Having run out of cryogenic coolant for the original WISE's 12 and 22 μm bands, it only detects in the 3.4 and 4.6 μm bands. The comet looks red in the false-color infrared image because, relative to the 4.6 μm "W2" band, it emits less in the 3.4 μm "W1" band than stars ... 4 Black holes are black. They are only observed directly by telescopes if they are accreting matter. Any radiation observed comes from the matter surrounding the black hole. Generally speaking, the smaller the black hole, the hotter the accreted material becomes. For something of planetary mass, one might expect X-rays and Gamma rays from accreting material. ... 4 Has the amount of energy being emitted from R Hydrae's bow shock been studied? Yes, here are a few papers: "Detection of a Far-Infrared Bow-Shock Nebula Around R Hya: the First MIRIAD Results" (July 13 2006), by T. Ueta, A. K. Speck, R. E. Stencel, F. Herwig, R. D. Gehrz, R. Szczerba, H. Izumiura, A. A. Zijlstra, W. B. Latter, M. Matsuura, M. Meixner, M. ... 4 Partial problem with Anders' answer: 730K is a "hot Jupiter", which I think are usually kept hot by stellar insolation, which would also heat the proposed moon. Removing that heat-source: Shortly after Jupiter formed, assuming it orbited at its present distance from the Sun, at some point it might have had a surface radiation temperature of 730K, ... 3 Generally speaking as many as possible because the number of lenslets determines the lateral wavefront resolution. But in reality there are a few factors to be considered. Wavefront sensor Say your lenslet array is of focal length f and subaperture size D, given a maximum desired detection angle \alpha (in your case is 2'' \approx 1\times10^{-5}\, \... 3 There shouldn't be any correlation. The CMB light that we see is from a spherical region in the early universe. Its homogeneity strongly suggests that the interior of the sphere was just as homogeneous, but we can't actually see CMB light from the interior. The galaxies that we can see formed from matter inside the sphere, and quite far from the edge. ... 3 CMB fluctuations The CMB fluctuations are often analyzed through their power spectrum P(k), which is a measure of the extent to which it is "clumpy" on a given scale \ell, with corresponding wavenumber k = 2\pi/\ell. The origin of this power spectrum is laid in the very early early Universe, just after the Big Bang, and it is of utmost ... 3 The first identified gravitationally lensed object is known as the Twin Quasar. The Twin Quasar (also known as Twin QSO, Double Quasar, SBS 0957+561, TXS 0957+561, Q0957+561 or QSO 0957+561 A/B), was discovered in 1979 and was the first identified gravitationally lensed object. It is a quasar that appears as two images, a result from gravitational lensing ... 3 They're all mirrors, they just operate in different wavelength ranges. A wire fence could be a mirror if the wavelength is big enough. The surface is an easy problem to solve. Pick any material and do vacuum deposition with it on the substrate. It's the substrate, the base material that matters a lot - the thick part that keeps the shape. A metal mirror is ... 3 I would not attribute too much importance to the quote "colder than expected", because that always depends which expectations one had initially. Here the quote you are discussing: Temperatures on the moon's surface plummeted to minus 190 degrees celsius (-310 degrees Fahrenheit) during the probe's first lunar night, which "was colder than ... 3 The simple answer for the wavelength part is that performance of AO systems degrades the shorter in wavelength you look. The basics of what happens is as you go to shorter the wavelengths of light, you need a finer plate scale to detect variations in seeing which requires very expensive (and in some cases non-existant) hardware. You also need a higher AO ... 3 Assuming your question is a reference request: I now about one resource that has a very broad range of spectral data, the NIST Atomic Spectra Database Lines Form This has worked great for me. This page lists a couple of different alternatives. For the near IR case you are interested in, several collections are listed here. 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 Your source is mistaken. Almost all the radiation we see from stars is "thermal". That is, the radiation arises from material where the occupation of energy states in atoms, molecules and particle speeds is characterised by a temperature. This includes emission from the photosphere, chromospheres and coronae of stars. In the case of the 1-5 micron ... 2 This question already discusses the challenges of discovering planet 9 The main problem of the hypothetical planet 9 as compared to Sedna and 2012 VP113 is: Sedna and 2012 VP113 have orbits that reach far out but currently they are near their perihelion so it was possible to observe them in the visible light spectrum. Planet 9's orbit is far out or at ... 2 For all of these direct imaging results, the critical parameter is the contrast as a function of separation. This lets you know how much fainter an object you can see around the much brighter primary object whose light has been suppressed by the coronograph (the black circle in the center of the star). From the change in eclipse timings (Figure 1 in their ... 2 According to this article, the TRAPPIST-1 system is supposed to be studied by JWST with the potential to detect atmospheres' composition in infrared band. So at last 40 ly (approx. 12 pc) should work. Somewhere else I read there is an assumption to study earth-like exoplanetary atmospheres at distance of 10-20 pc. But it might be better to see it from ... 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 Let's assume a uniform laminar emitter, oriented so that it is at right angles to the line to the detector. The specific intensity emitted by a blackbody, at the surface of the blackbody is$$B_{\nu} = \frac{2h\nu^3}{c^2} \frac{1}{\exp[h\nu/kT] -1}\ {\rm Wm}^{-2}{\rm Hz}^{-1},$$where B_{\nu} is the "Planck function". Using the definition of specific ... 2 The GREAT receiver attached to SOFIA to detect far-infrared radiation incorporates elements reminiscent of both radio and optical detectors, but since it works by mixing a signal from a tunable oscillator with the oscillating field from the incoming radiation rather than by the photoelectric effect as in most optical detectors, you might consider it more ... 2 On the basis of this review article by Delbo et al. (2015) about modeling asteroid properties, I think it might work something like this: From optical measurements (where you're seeing reflected sunlight) and knowledge of the asteroid's orbit and its current orientation with respect to the Sun and Earth, you can work out its approximate size, shape, and ... 2 Say that you have an image with N \times N pixels that covers a region of sky of \theta \times \theta steradians. For each pixel i you have a value of flux density J_i, expressed in Jy/sr. The image contains a galaxy and you want to calculate its the total flux. If this were a continuum problem, you would need to do an integral$$F = \int_{\Omega_{...

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A few things: The conversion between magnitudes is dependent on the filters being used, you can see it on the tables from the website you linked. That being said, the conversions available in there will not be useful for you if you are trying to replicate Stern et al 2005, as they use redder filters (in the website with the conversions, the longest ...

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