# Tag Info

26

The angular deflection caused by the lensing of a distant background object by the Moon is given by $$\theta \simeq 4 \frac{GM}{Rc^2},$$ where $M$ is the mass of the lensing object and $R$ is the closest projected distance of the ray from the centre of the mass. For the Moon, the maximum deflection would occur when the ray just grazes the limb of the Moon ...

25

The local dark matter density is actually quite tiny, on the order of $\rho\sim10^{-19}\text{ g/cm}^3$ (see e.g. Bovy & Tremaine (2012)). This means that there is roughly $0.001$-$0.01M_{\odot}$ of dark matter per cubic parsec - a staggeringly small amount. 1000 cubic parsecs would contain about one solar mass of dark matter - and that's a cube 10 ...

23

Hot dark matter would be made from very light, fast moving particles. Such particles could not possibly be gravitationally bound to any structure, but rather would be dispersed all across the universe. But dark matter is always "found" (or "inferred") either gravitationally bound to some visible structure (e.g. weak lensing detection of dark matter ...

23

Yes, observations of this kind are within the technical scope of amateur astronomers. Several groups succeeded in replicating the experiment during the 2017 eclipse that crossed the USA. For example Donald Bruns measured deflections of 2.8 arcseconds of multiple stars. Nasa published a "How To" page for anyone wanting to test GR themselves.

16

The quasar gives out light in all directions. The light spreads out in space. Only a very small amount of that light would be pointed exactly in the direction of your telescope. But if a large galaxy or galaxy cluster is between the quasar and us, it bends some of the light towards us, making the quasar brighter (it would also distort the shape, but quasars ...

15

What you do is cross-correlate the observational datasets for the multiple sources and look for the "lag" that maximises the cross-correlation function. Generally speaking, the "events" are not really individual flares or dips, but the summation of all the time variability that is seen. The variability in question usually comes about from the central ...

11

The cosmic horseshoe is beyond amateur instruments. It is a magnitude 20 object. In a large (2.5 m) professional telescope it looks like: This image taken from the SDSS III data. It is small (10'', half the size of Mars at opposition) but that is not insurmountable for amateur equipment. But is it very dim. If an object like this was visible in moderate ...

10

Yes, it would be possible. There are two roads here: Visible light In case of detecting light in the range of visible wavelength, perhaps you would consider that rare occasion when there is a solar eclipse. And it may also be possible at times of 'early' dawn and 'late' dusk. Invisible light (outside visible range, beyond the IR and the UV) Now, other than ...

9

The gravitational focus you are talking about is actually a minimum value, defined by parallel rays of light from a very distant star just skimming past the Sun as they are bent according to General Relativity. The general formula for such lensing is that light is bent through an angle (in radians) of $$\alpha = \frac{4 GM}{c^2 r},$$ where $M$ is the mass ...

8

It is simply not true that gravity can only interact with mass. Rather, any long-range spin-2 force interacts with all energy-momentum equally, and it source is the stress-energy-momentum tensor. That is one way to state the equivalence principle. Note that a massive object in its own rest frame has an associated energy $E = mc^2$, which under ordinary ...

7

A photon is an entity defined in the context of a relativistic field theory, and so it doesn't really make sense to talk about the Newtonian bending of a photon. Necessarily, we need to substitute an analogous question that's sensible in the Newtonian framework. To do so, we can imagine a classical corpuscle of light--appropriately enough, a theory of light ...

7

As mentioned in the comments, a ring will (ideally) only occur if you have perfect alignment of source, lens, and observer. This is true for point sources, for extended sources things get more complicated. Double and quadruple images occur when the source is small and not in perfect alignment - typically a quasar, as in your example. For the sake of ...

7

Gravitational lensing of background galaxies and quasars is used to probe the amount of gravitating matter in foreground objects such as clusters of galaxies. Here is an example, taken with the Hubble Space Telescope. The foreground object is a cluster of galaxies and the lensed background objects, some of which are marked, are galaxies and quasars that ...

7

Gravitational lensing works from anywhere beyond the focus, so in that sense, we could use any star as a gravitational lens. The problem is that the field of view is tiny. We only get any useful information from alpha centauri as a gravitational lens if the target object is almost exactly behind alpha centauri from our point of view. To look in a slightly ...

7

The two most general, publicly available packages for galaxy image fitting (other than GALFIT) are probably Imfit and ProFit. (Note that I am the developer of Imfit.) There is also Lenstronomy, which is specialized for fitting gravitational lenses; this might be more relevant to your particular needs, though I know very little about it.

7

The answer is no, because these effects have been looked for and not found. The microlensing surveys of the 80s and 90s specifically set out to look for the lensing signatures of compact, massive objects, including black holes. Whilst some lensing events were seen, the numbers were low enough that it could be concluded that there was no significant dark ...

7

The redshift of the quasar is 1.975, so it is nowhere near the edge of the observable universe. 17 billion light years is the comoving distance (i.e. where it is now), as you can confirm with this cosmology calculator.

6

Yes it has been observed. There are numerous papers covering this concept. Before I get into weak-lensing (because that is what you reference in your question) I'll note that cosmic shear is simply a specific type of distortion of an object by a lensing event. It is not limited to the weak lensing regime and cases of strong lensing exhibit cosmic shear as ...

6

Firstly, gravitational waves (GWs) are not an echo - we measure the direct signal. The process you describe here is known as gravitational lensing, the deviation of (usually) light rays due to massive objects between the source and the observer. This also applies to GWs. The result will be similar - the direction of the waves can be changed, resulting ...

6

Gravitational lensing is just geometry, as is optical lensing. The amplification factor describes the increase of the area of the image (at constant surface brightness). BTW, the amplification factor does depend on the mass $M$ of the lens, since $u=\theta/\theta_E$ and $\theta_E$ depends on $M$. Here $\theta$ and $\theta_E$ are the relative angular ...

6

You are right that the stars seen on the sky are within the Milky Way. Only with a large telescope is it possible to resolve individual stars in other galaxies, and only for the nearest ones. I don't know which sources you refer to, by I think perhaps you are confusing the different types of gravitational lensing. I cannot explain them better than the ...

6

Actually, one of the first confirmation of GR, is by Sir A. Eddington et al., who measured the deviation of light trays of an unsaid-as-far-as-i-know stars, on May 29, 1919. Here is an original snapshot from their experiment: They took advantage of a solar eclipse to measure the light bending from the expected position of the sources. The experiment has ...

6

That would be possible theoretically. However, in practice, the lensed images are highly non-linearly distorted and due to the limits in resolution, they are often limited to a thickness of a few pixels at best. Due to the non-linear transformations, this would cause the reconstructed original image to be even blurrier. Having said that, I don't see any ...

6

No, there is no effect here. Why? Gravitational lensing magnification works by increasing the observed surface area of the lensed object while preserving the surface brightness. An exoplanet angular diameter is much smaller than that of its host star (even if we can't resolve either). Hence, all the magnification happens across a uniformly lit background, ...

6

Instead of reading half-comprehensible newspaper articles, I found it more helpful to go to the original source. There, the authors explain their method very clearly: There is a background galaxy, the Quasar galaxy. This object, being a Quasar, has an actively accreting SMBH. The SMBH, being an actively accreting one, will emit strongly in X-Rays. Those X-...

6

OK, guessing rather wildly in the absence of solid information.... The press-release page is here; that provides a bit more information than the news article, and includes two images of (different regions of) the cluster and a hint that the cluster is named MACS0940. I think the main image is a color composite from HST imaging. It shows a galaxy cluster (...

6

The answer is no. But there's an interestign alternative. The Eddington's experiment was not about seen a gravitationally lensed star but to see a small displacement of its position (just below 2 arcseconds) due to the solar gravitational field. Lensing involves considerably magnifying the source and that can only be done if the star is directly behind the ...

6

tl;dr: seen at a distance, the lensed object will appear as an annulus or "ring" around the lensing object, and while that will be brighter than if there were only empty space, sadly it won't be death-ray bright! Let's first think about what makes a familliar lens a lens. Near the center the thickness doesn't vary much, but as you move farther ...

5

Using this formula from Wikipedia, $$\theta = \frac{4GM}{rc^2}$$ and plugging in the mass and radius of the Moon gives a deflection angle of $1.2567 \times 10^{-10}$ radians = $2.592 \times 10^{-5}$ arc-seconds. In comparison, the deflection for the Sun is just under 1.75 arc-seconds, using the Sun's equatorial radius. I expect that it would be very ...

5

Solar energy - from our Sun at Jupiter's focal distance? A negligible amount. Gravitational lensing doesn't have a focal point so much as a focal region that begins roughly at a point, and that point (which really shouldn't be called a focal point) can be calculated, ProfRob gives the formula here. Jupiter's gravitational lensing distance is so far (...

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