Determining sunrise and sunset times based on azimuth and elevation If I observe the sun and measure the local azimuth and elevation, and I also have my latitude, can I calculate the local sunrise and local sunset times? If so, how would I calculate it?

Further, can the same calculations be used for the moonrise and moonset?

$
\cos ^{-1}\left(-\frac{\tan (\lambda ) (\cos (\lambda ) \cos (\phi ) \cos
(Z)+\sin (\lambda ) \sin (Z))}{\sqrt{(\cos (\lambda ) \sin (Z)-\sin (\lambda)
\cos (\phi ) \cos (Z))^2+\sin ^2(\phi ) \cos ^2(Z)}}\right)-\tan^{-1}
(\cos (\lambda ) \sin (Z)-\sin (\lambda ) \cos (\phi ) \cos (Z),\sin
(\phi ) (-\cos (Z)))
$

is the amount of time from now (see important notes below) a celestial object will set, where:

$\phi$ is the azimuth of the object
$Z$ is the altitude of the object above the horizon
$\lambda$ is the latitude of the observer

Note the two-argument form of arctangent is required for accuracy: https://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Two-argument_variant_of_arctangent

Important notes (MUST READ!):

The result is in radians. To convert to sidereal hours, multiple by $\frac{12}{\pi }$
If the object has a fixed right ascension and declination, divide sidereal hours by 1.002737909350795 to get clock hours.
For the Sun (which doesn't have fixed right ascension/declination), do NOT divide as above. By not dividing, you compensate for the Sun's change in right ascension.
For the Moon, see "The Moon" section below.
The time computed above is for geometric midpoint setting. In reality, refraction near the horizon means the object will set later. Additionally, for objects that have an angular diameter (eg, the Sun), setting usually means when the top edge disappears over the horizon, which will make the set time even later. Accounting for both effects should be possible, but might require numerical methods instead of a closed form formula as above.
If the quantity inside the arccosine is greater than 1, the object is always in the sky and never sets or rises.
If the quantity inside the arccosine is less than -1, the object is never in the sky and thus also never sets or rises.
I did only minimal testing. As always, do not rely on my answers for anything important.

The Moon:

The Moon's right ascension and declination change rapidly, so this calculation does not work well for the moon.
You could compensate for the change in right ascension (and thus hour angle) by approximating the increase as 24 hours every 27.32158 days (its sidereal period) and do an iterative calculation.
An even better compensation for the change in right ascension would be to approximate the moon's movement in ecliptic longitude (which is more constant than its movement its right ascension) as 360 degrees per 27.32158 days and then project the ecliptic longitude back to right ascension, and then iterate.
Compensating for the moon's change in declination is more difficult. The moon's ecliptic latitude (which can be converted to declination) varies sinusoidally, but the equation $\sin (x)=a$ normally has two solutions. Unless you know whether the moon's ecliptic latitude is increasing or decreasing (ie, whether it's between ascending and descending nodes or vice versa), you won't know the direction of declination change.

Less important notes (optional):

See Need Simple equation for Rise, Transit, and Set time for more general equations on when an object rises/sets/etc.
Calculations for this answer at: https://github.com/barrycarter/bcapps/blob/master/STACK/bc-object-riset-from-az-elt.m
Related calculations: https://github.com/barrycarter/bcapps/blob/master/STACK/bc-rst.m
Mathematica was unable to find a simpler form for the "time to set" above, though I sense a simpler form does exist (I could be wrong).
If you know the Sun's declination (which you can get from its azimuth and elevation as above), you can almost determine the date. However, the sun reaches a given declination twice a year (example: it reaches 0 degrees declination on both equinoxes, by definition), so you can only know it's one of two days.

Solution notes:

I learned quite a bit answering this question, and thought it was only solvable numerically until I figured out the shortcut:

To convert from the azimuth/elevation sphere to the hour angle/declination sphere, you just rotate around the y axis by $\frac{\pi }{2}-\lambda$ (90 degrees minus the latitude) and then rotate $pi$ (180 degrees) around the z axis.
Once you have the declination, computing the hour angle when an object sets is easy.
You then subtract the setting time from the current hour angle to get the answer.

TODO: I was hoping to add some graphs/visualizations to this answer but couldn't quite get things working.


What is frequency range? Using the NED, you can search via frequency range corresponding to the selected pass band filter. I'm a beginner here and don't really understand. Here's the quote:

NOTES: Objects with observations in the frequency range corresponding to the selected passband filter will be returned. For
example, selection of Spitzer MIPS 24 µm will also return objects with
photometry from IRAS 25 µm because these filters overlap. This feature
permits the location of objects with photometry in the selected region
of the spectrum, while cutting across multiple surveys, detectors and
filters that have been cross-correlated by NED. A future upgrade will
include an option to restrict the search to only the selected passband
filter.

There are then a list of 'passbands', all the way up to gamma. What is this actually mean? The amount of radiation given off by a celestial body?
I'm looking at the categories I'm more familiar with:
Helio. Radial Velocity
Redshift
Major Diameter (arcmin)
Minor Diameter (arcmin)
Magnitude and Filter
Classifications
And I'm not sure where this radiation fits in. Is it
Spectral Energy Distribution (SED)

? Could someone explain the following terms to me if this is the case:

Observed Passband
Uncertainty Units
Freq (Hz)
Mode Aperture/Qualifiers
Reference Code   No.

Wnad what about published and homogenised variables? Are homogenised variables the avarage of all observations?

Take for instance radio, when listening to radio you gather static on the frequency. A passband filters the static which in turn provides a clearer signal. Objects radiate in multiple frequencies/wavelengths, for instance heat/thermal (infrared) and visible. Not all object radiate in a range of frequencies, humans for example can be seen well in infrared and visible, but our skin and circulatory system are opaque in x-ray and radio. Spitzer MIPS 24µm is the wavelength corresponding to mid-infrared (between 3µm-50µm).

What the statement is saying, is multiple imaging sources have been used for the same frequency. When choosing the Spitzer MIPS 24µm result, it additionally includes results from other imaging sources such as the IRAS 25µm result. It does this because the filters of the passband overlap and return the same result, as a result the process is changing where it only displays the result of the filter you choose.

All radiation is apart of the electromagnetic spectrum, spectral energy distribution is a way to present the data results as a plot, in turn that can be used to identify the source of the radiation.

The observed passband, is just stating which frequencies are being filtered, such as perhaps within a small range. For example, a Quasar is a highly active radio source typically found in the center of young galaxies. In correlation, you can determine what kind of stellar objects you would expect given a certain range of frequencies or the observed passband.

Frequency is the simple measure of how frequent something happens, for example if you flash a torch once a second, the time between each flash is 1 second long, the frequency 1Hz. If you flash a second torch twice a second, the time between each flash is a 1/2 second long, the frequency is 2Hz. More relevant in physics, is the correlation between wavelength, frequency and photon energy. Waves oscillate and light acts as waves, and the frequency between each wave correlates to how fast the wave is moving and what wavelength the radiation is.

Homogenized results are just a collection of results which are made to fit a pattern, sometimes to better represent the data (this may be filtering results in a database to exclude results outside a range of study). Published results on the other hand are raw results which include a full spectrum of the data. I am unsure about the mode aperture, this may refer to the sensitivity and correlate to how much interference/static is captured with the wave. Best if someone else can explain this.


Is there Dark Matter inside the Earth? Since Dark Matter is assumed to interact gravitationally, it follows that the gravitation of massive regular matter bodies should have an effect on it. Even if it is otherwise weakly interacting with regular matter, one might naively expect concentrations of it enter orbit around massive objects, like neutron stars, regular stars, and even (why not) our own planet.

So, a two-fold question:
1) Could there be Dark Matter in orbit around (or within) the Earth?
2) If that were possible, would there be any way for humans to tell if such is actually the case?

Yes, there should be dark matter within the Earth, but at very low densities - a few $10^{-22}$ kg/m$^3$ (Bovy & Tremaine 2012), something like one hundredth of the density of the interplanetary medium. The whole of the Earth would contain few hundred grammes!

Of course, this density averaged over a sphere a lot larger than the Galaxy adds up to a lot of mass, but unlike luminous matter, dark matter should not be particularly concentrated in the plane of the Galaxy (and observations of the dynamics of local stars confirm this).

Neither do we expect dark matter to be especially concentrated within the Earth. Although it does interact gravitationally, the kinetic energy it receives from falling in to the Earth's gravitational potential will be exactly sufficient for it to escape again, unless it suffers some inelastic interaction inside the Earth. Some researchers do think that it may be possible that weak interactions may trap dark matter in the core of the Sun (e.g. Vincent et al. 2015) or even more likely, in neutron stars (Guver et al. 2012). But the Earth is not large or dense enough for this to be a likely scenario.


Why do pair production and photodisintegration have such different effects in collapsing supermassive stars? Stars of at least 100 solar masses or so can reach core temperatures (and, thus, core photon energies) great enough that pair production (where a very-high-energy photon strikes another particle, transforming the photon into a matched particle-antiparticle pair - usually an electron-positron pair, though other types are possible - and causing the other particle to recoil slightly) starts to occur in earnest.  As this is an endothermic (energy-absorbing) process, it reduces the temperature and pressure within the star’s core, causing the star to start to collapse under its own weight.

One of three things can now happen:

In high-metallicity stars, and in low-metallicity stars up to ~130 MS, the resultant increase in core temperature and pressure halts the collapse before it can do anything irrevocable, and the star heats up and expands back out again, blowing off significant mass in the process.  These pulsations will continue until either the star becomes too small and cold for much pair production or it explodes for some other reason.
In low-metallicity stars from ~130 to ~250 MS, the collapse compresses and heats the interior of the star quickly and vigorously enough that the resulting increase in core reaction rate and energy release is sufficient to unbind the entire star.
In low-metallicity stars exceeding ~250 MS, the collapse is so rapid, and the rise in core temperature so great, that an increasingly-large fraction of the photons produced have energies high enough to cause photodisintegration (where an extremely-high-energy photon is absorbed by an atomic nucleus, causing the nucleus to break apart into two or more smaller pieces), robbing the star’s core of energy and causing the entire star to collapse directly into a black hole.

Both pair production and photodisintegration are endothermic (energy-absorbing) processes, and, thus, tend to cause the star to collapse.  However, their eventual effects on the collapsing star are different; pair production causes only a partial collapse, followed by runaway thermonuclear fusion and a supernova explosion, while photodisintegration causes a complete collapse into a black hole, with nothing escaping.

Why does core energy loss resulting from pair production result in a partial collapse, runaway fusion, and a supernova, while core energy loss resulting from photodisintegration results in a complete collapse to a black hole?  Why don’t these two endothermic processes either both result in a partial collapse with resulting explosive fusion, or both result in total collapse to a black hole?

A bit of wild speculation that I'll need to check.

Photodisintegration produces simpler particles. These simpler particles are able to "neutronise" (i.e. undergo inverse beta decay/electron-capture) with electrons, because the energy threshold for neutronisation is much lower for free protons and alpha particles than it is for say oxygen or iron nuclei. This process removes free electrons from a core that is supported by electron degeneracy pressure and it collapses.

Pair production converts photon energy into electrons and positrons. The net effect will be to take kinetic energy out of the gas and turn it into rest mass. This will lower the pressure, but an increase in density/temperature can restore it.

I think the process of neutronisation/electron-capture is the key difference. The threshold electron energy for neutronising a free proton is 1.3 MeV, but is about 11 MeV for a proton inside an oxygen nucleus, so photodisintegration is required to trigger collapse (or the Fermi energy of the electrons reaching the neutronisation threshold).


What is the actual black hole merger speed? I have seen models of 2 black hole mergers.

https://www.theverge.com/2017/11/16/16665132/gravitational-waves-ligo-virgo-black-holes-merger-multi-messenger-astronomy

When we see a space ship appear to be frozen in time as it approaches the event horizon then would a back hole merger be faster then perceived in our time?

John put it well in the comments:  People say merging black holes end up moving at half the speed of light, but don't say anything about the gravitational time dilation wherein the "coordinate" speed of light goes to zero at the event horizon. Which would mean black holes would have to be moving faster than the local speed of light. How can this be right?

Can things move faster than light inside the event horizon of a black hole?

What is the actual black hole merger speed?

It varies per object, and throughout the merger (and doesn't neatly increase linearly).
An object with a lot mass can not come close to the speed of light, that would require almost infinite energy. Extremely tiny objects, particles, can come close to the speed of light but not actually reach it. Light can go the speed of light, only in a vacuum. Gravity always travels at the speed of light, unlike everything else which goes slower.
The first animation shows what happens when two large objects approach each other, miss, and then don't slingshot away from each other - they become trapped in an extreme elliptical orbit.
The second animation shows what happens when the orbits are relatively close, > ~0.01 parsecs - instead of an elliptical orbit they have a somewhat circular-like orbit, around the barycenter.
There are many different trajectories and resulting orbits, the binaries are also affected by: inflow, outflow, and other objects entering and leaving the area of their orbits. No one answer, no two are alike.
The paper "Observation of Gravitational Waves from a Binary Black Hole Merger" (Feb 11 2016), by B. P. Abbott et al. offers one study of one merger, they provide this illustration:

Top: Estimated gravitational-wave strain amplitude from GW150914 projected onto H1. This shows the full bandwidth of the waveforms, without the filtering used for Fig. 1. The inset images show numerical relativity models of the black hole horizons as the black holes coalesce. Bottom: The Keplerian effective black hole separation in units of Schwarzschild radii ($R_S=2GM/c^2$) and the effective relative velocity given by the post-Newtonian parameter $v/c=(GMπf/c^3)^{1/3}$, where $f$ is the gravitational-wave frequency calculated with numerical relativity and $M$ is the total mass (value from Table 1).

In-between those distances occurs a situation referred to as the final parsec problem.

Final parsec problem
When two galaxies collide, the supermassive black holes at their centers do not hit head-on, but would shoot past each other if some mechanism did not bring them together. The most important mechanism is dynamical friction, which brings the black holes to within a few parsecs of each other. At this distance, they form a bound, binary system. The binary system must lose orbital energy somehow, for the black holes to orbit more closely or merge.
Initially, the explanation is easy. The black holes transfer energy to gas and stars between them, ejecting matter at high speed via a gravitational slingshot and thereby losing energy. However, the volume of space subject to this effect shrinks as the orbits do, and when the black holes reach a separation of about one parsec, there is so little matter left between them that it would take billions of years to orbit closely enough to merge - more than the age of the universe. Gravitational waves can be a significant contributor, but not until the separation shrinks to a much smaller value, roughly 0.01–0.001 parsec.
Nonetheless, supermassive black holes appear to have merged, and what appears to be a pair in this intermediate range has been observed, in PKS 1302-102. The question of how this happens is the "final parsec problem".
A number of solutions to the final parsec problem have been proposed. Most involve the interaction of the massive binary with surrounding matter, either stars or gas, which can extract energy from the binary and cause it to shrink. For instance, if enough stars pass close by to the orbiting pair, their gravitational ejection can bring the two black holes together much quicker than would otherwise be the case.

See the "final parsec problem" link for the whole webpage.

"... the gravitational time dilation ..."

Means: In the black hole's timeframe time moves forward normally for both the black hole and the observer in their own frames. In each other's frames the outside observer see the black hole's time seemingly come to a standstill while the black hole observer sees the outside observer's time pass very rapidly.


Did the Babylonians believe in the Heliocentric version? It's mentioned that the Babylonians decreed that there were exactly 360 days in a year. This statement was based upon the utilization of 360 as a standard (based on the sexagesimal system) i.e., they divided the circle into 360 equal parts.
Although, there is a mention about the Sun making one full circle across the sky in one year (as estimated by them), but, how were they able to measure it? And did they believe in the Heliocentric version of the solar system?

In 150 BC a babylonian heliocentric model is documented. Heliocentric models were not uncommon historically, but Ptolemaios geocentric model was actually better at predicting the movements of the planets until Kepler came up with the idea of elliptic rather than circular orbits in the early 1600s. In the 1500s Copernicus hesitated to publish his heliocentric ideas because they did not improve predictions.


Eclipses of Jupiter's Moon during Retrograde motion When Jupiter is undergoing retrograde motion as seen from the Earth, would you expect the eclipses of Jupiter’s moons to occur several minutes early, several minutes late, or neither?
Can someone please explain?

I am a novice in astronomy, and I read a couple of articles on the retrograde motion. I understand it is an illusion. But how would I expect the eclipses of Jupiter's moon?

The fact that eclipses are either "early", "on time" or "late" only depend on the distance between Jupiter and the Earth. Since the speed of light is not infinite, a longer distance means a longer delay between the actual eclipse and the time it is observed from Earth.

For superior planets, retrograde motion occurs around opposition, at a time when the distance between the planet and the Earth is minimal. So you should expect eclipses to appear early during retrograde motion. But note that the early eclipses are not a consequence of retrograde motion: they are both consequences of keplerian motion.


How astronomers distinguishes between natural and artificial signals coming from outer space? How astronomers really distinguishes between the natural and artificial signals? From time to time astronomers detects many signals from outer space and SETI is always trying to find signals that carries the sign of intelligent beings. The Wow signals was such a much hyped signal that we received. So what properties a signal should carry so that we can conclude it is from some alien civilizations?

We could begin to conclude if a signal is from some Extra Terrestrial Intelligence (ETI) if the signals are very difficult or impossible to explain by non-life origins or terrestrial origins. One commonly cited example is with a very narrowband radio signal. In natural radio emission, the frequencies are spread out due to processes like Doppler shift which raise and lower the frequency or radio emission. The Wow signal was so narrowband that it behaved just like one would expect for ETI, but having only one detection that disappeared makes it hard to be sure it wasn't interference or a problem with that particular antenna.

I would suggest that ETI detection should have some of these properties, but feel free to add more:

No natural abiotic terrestrial phenomenon that can produce it
Verifiability - multiple detection methods, instruments and teams that can confirm the signal is not due to errors in processing, analysis or local interference
Evidence for non-random patterns such as prime numbers

KIC 8462852 currently has a feasible solution that dust escaping from a comet or multiple comets can obscure light from its host star. This natural explanation has ways of being confirmed (such as wavelength dependence to the stellar blockage), which will help with the first criterion. Hopefully, the flux dips observed in Kepler's days ~800 and ~1500 will repeat in the future but if they don't, it fails the second criterion. As with definition life, it is hard to come up with a rule book. However, if we see a lot of the properties we associate with ETI, we can start the classification process. As more evidence comes in, like polarization, multi-wavelength and Doppler shift data for KIC 8462852, we can evaluate which criteria are satisfied.


Is Blue Shift just as provable as Red Shift? I've heard a lot about red shift and it sounds pretty legit. But it made it sound as if blue shift was a lot less likely. Can anyone help me out?

The expansion of the Universe can be expressed using Hubble's law. This is that the recession velocity that a galaxy has with respect to our own is proportional to its distance from us. i.e.
$$ v = H_0 D, $$
where $v$ is the recession velocity, $D$ is the distance and H_0 is the Hubble "constant", which has a value of about 70 km/s per Megaparcsec.

Hubble's law is attributed tot he overall expansion of the universe and means that once we get further away than about 10 Mpc (or about 30 million light years), that other galaxies recede from us with velocities of 700 km/s or more. This recession produces a redshift.

However, superimposed on Hubble's law at a more local scale, individual galaxies are gravitationally interacting with each other and hence have their own peculiar motions with respect to the general "Hubble flow". In addition, don't forget our Sun rotates around the centre of our Galaxy with a velocity of some 230 km/s, so is heading in the direction of some galaxies and away from others.
These local peculiar motions means that some nearby galaxies are moving towards us (at relatively modest speeds) and do exhibit a blue shift. An example would be the Andromeda galaxy at 2 million light years, which is approaching our Galaxy at about 110 km/s (or at a relative closing speed of 300 km/s with respect to the Sun).

But there are only about 50 galaxies in our local group and a further few tens of galaxies in some other structures out to about 30 million light years, and it is only from this small sample that any can exhibit blueshifts. The vast majority of known galaxies are further way and are swept along in the general universal expansion and hence away from us.


Binary Star question Is it POSSIBLE that we are a binary system? I have been doing research into this topic and it seems we may infact be a binary system. If we are, what effects would a massive body comming into our system be? And are these effects happening now?

Since every object in the known universe which has mass has an infinite gravitational field every object in the universe affects one another.
Every star in our surrounding affects Our solar system but the relative proximity of the sun masks these effects to us.
However if the sun would be a part of a binary system of stars

Our planet should have been affected by it because a star is a much larger mass when compared to the planets.
Since the star and our sun interact with each other they would produce a noticeable wobble.
Any such star would definitely be very close to the sun so as to be able to influence it and hence easily visible yet no evidence has been found for this.

So for now one can say that we and the sun are not part of any binary star system which has been (if it exists) undiscovered till recently


How does the Event Horizon Telescope implement the interferometry? The BBC article Event Horizon Telescope ready to image black hole describes the Event Horizon Telescope, a coordinated observing technique with several radio telescope arrays across the globe forming a synthetic aperture with an Earth-sized baseline.

$$\frac{\lambda}{r_{Earth}} \sim \frac{r_{Sag A*}}{D_{Sag A*}} \sim  10^{-11}$$

...when plugging in 1 millimeter for $\lambda$, and with $r_{Sag A*}$ and $D_{Sag A*}$ the radius of, and distance from Earth to Sagittarius A*, the black hole in the center of the Milky Way Galaxy of 20 million km and 26,000 light years, respectively. (values taken from the article).

The equation I've written shows that millimeter wavelength interferometry with an Earth-sized baseline has the possibility to resolve the existence of some structure with scale of the black hole's event horizon.

My question is How does the Event Horizon Telescope implement the interferometry? It would certainly be impossible to bring all signals together to a central site and perform the interference there in real time as down-converted analog signals, and quite difficult/expensive to do it with dedicated, synchronized digital optical fiber lines. Is the massive amounts of data sent as IP packets over the internet to a central correlator (numerical interferometer)?

The article mentions atomic clocks and lots of hard drives, and I have a hunch these have something to do with it.

above: "The eventual EHT array will have 12 widely spaced participating radio facilities". From BBC.

Assuming I understand your question:

As in actually mentioned in the article, all data collected are stored on hard drives, timestamped with an atomic clock, and then flown to a central location where the interferometry actually takes place.

Further reading:

R2DBE: A Wideband Digital Backend for the Event Horizon Telescop
eventhorizontelescope.org/publications


Why do these Hubble images of Neptune look like Cheela? The open access paper Formation of a New Great Dark Spot on Neptune in 2018 (cited in Gizmodo's Formation of Dark Vortex on Neptune Captured For the Very First Time and Phys.org's Hubble captures birth of giant storm on Neptune) shows images of Neptune taken by the Hubble Space Telescope.

The projections are called "global cylindrical maps" and those usually result in weirdly stretched features near the poles but do not result in the spiky hairdo or Cheela-like appearance in the Hubble  images below.

Source

What causes this effect seen in these projections?

  Individual maps from Hubble images in 2017 through 2018, in the same filters and processing as Figure 1. These maps span ±90° of latitude and longitude, and tick marks indicate 30° latitude and longitude intervals.

  False‐color global cylindrical maps from Hubble, spanning 180° of latitude and 360° of longitude, from images at 845 (red), 547 (green), and 467 nm (blue), and all use the same contrast scaling with a light unsharp mask applied. The maps are centered on 180°W longitude, except 2015, which was centered on 0° longitude. Dotted lines are shown at ~10° and 35°N to indicate the active region prior to the discovery of NDS‐2018.

Reference Cheela, from here:

Neptune has an axial tilt of 28.3 degrees and an orbital period of 164.8 Earth years; currently, the north pole is tilted away from Earth.
The maps in your question were each constructed from twelve exposures of Neptune: four each in 845 (red), 547 (green), and 467 nm (blue) wavelengths.  Each of the exposures was then converted from an effectively-orthographic view of a hemisphere of Neptune to the equirectangular projection used in the maps.  The precise technique is cited to https://iopscience.iop.org/article/10.3847/1538-3881/aaa6d6 and https://iopscience.iop.org/article/10.1088/0004-637X/812/1/55.
The bumpy appearance comes from using four discrete observations rather than a single continuous one; the black area is portions of Neptune that were never visible.  Each of those spikes is the result of projecting a single color of a single pixel on the edge of Neptune's visible disk onto an equirectangular map.


Effect of black coating on the inside of a hobby telescope's tube? I was watching this (rather old) episode of The Making, where they show how a telescope is built in a factory. At 7:34, the cylindrical body of the telescope is shown to have a dark coating on the inside and the video explains that this prevents internal reflection of the light received from whatever you're observing.
I saw an interesting comment from Izumikawa Fukumi who asked if the telescope would become more accurate if they coated the inside with the blackest black in the world. How would such a dark coating affect the telescope's performance?

Telescopes have the inner part of the tube blackened to minimize scattered light reaching the eyepiece. Such scattered light reduces contrast, and hence reduces the ability to see very faint objects, or low contrast detail.  A complementary approach for refractor telescopes is to use baffles in the tube. The baffles form a physical barrier to scattered light.  However baffles are costly to make accurately and add to the weight of the telescope.
The more that scattered light is reduced, the better the performance.
Alternatives to super absorbing paint are to attach to the tube, either matt black velvet, or sand paper sprayed with matt black paint.


What constellation is closest to the Zenith with RA=12 and Dec =47.6? I have several options, but I do not know how to calculate which option is closest to the zenith in title. Could you please help me with this? Thank you.

Canes Venatici,CVN,13.14763.+40.04187,M
Cygnus,CYG,20.61006,+42.53633,A
Lynx,LYN,8.01440,+46.23233,M
Perseus,PER,3.49832,+43.88451,A
Ursa Major,UMA,10.98253,+52.52900,A

With Stellarium, it appears that the point with coordinates RA=12, DEC=47.6 is in Ursa Major.

In the screenshot from Stellarium below, the constellation boundaries are in red. Ursa Major is at the bottom, Canis is on top. The blue grid is the equatorial coordinates.


Is this the Cancer constellation? In this photo taken by SpaceX during one of their Falcon 9 launches, I noticed a familiar group of stars but I'm not sure of what it really is. I encircled it in red, is this group the Cancer constellation?

Original Image


Could pulsar navigation replace GPS as a follow-up to my previous question I found out that the Voyager Golden record uses a star map of pulsars to precisely show where our solar system is...

Is it possible/feasible to base a navigation system on pulsars in order to replace GPS?

Pulsar, or X-ray navigation is being actively tested right now. Currently an X-ray telescope mounted on the International Space Station completed a test where a small navigation computer was able to solve for it's own position and orbit in space relying only timing information from X-rays produced by a group of pulsars.
I've written more about it in the question Is NICER/SEXTANT the first civilian “spacecraft” to determine it's own position in space without GPS or uplinked data? Spoiler alert, hold your cursor over the box:

 Yes, it is the first

You can also read about an investigation for X-ray communication in How will NavCube (actually) be important for the XCOM testing and demonstration? I believe this will be a significantly longer term technology development path.
NICER, NASA’s Neutron star Interior Composition Explorer, will of course also be used for pure Astronomy research on neutron stars.
Will this replace GPS? Not really, No.
GPS serves the Earth's surface, and near Earth orbit (NEO) very nicely and efficiently. However, once you get a little farther, to MEO and GEO, it gets much more difficult to receive several simultaneous GPS signals because the satellites concentrate most of their power directly on to the Earth's disk, with only some side-lobes "leaking" past the terminator and into space.
See the questions and of course their answers:

Will Glonass, Galileo, or BeiDou-2 satellites provide better cis-lunar navigation than GPS?
Has GPS been used beyond GEO? (duplicate)
How far up have satellites used a GNSS for positioning, and how does the precision degrade with altitude?

So for spacecraft in farther Earth's orbits it may be possible to augment or replace GPS or GNSS in general, and for deep space, there is currently no way to autonomously navigate today, so it will be the first.

above: NICER X-ray timing telescope/concentrator array, from About NICER.


Does near light-speed travel doppler-shift the light from the target into gamma rays? If a space ship travels at near the speed of light towards a star, wouldn't the light from the star doppler shift into gamma rays (or worse?) and become lethal to life and destructive to space ship structure?

The equation for Doppler shifting towards the blue spectrum (i.e. when you are traveling towards it) is:

$\lambda_b = \lambda_c/(c + v_b)$

where $\lambda_b$ is the shifted wavelength, $\lambda_c$ is the original wavelength, $c$ is the speed of light and $v_b$ is your velocity towards the source.

BUT! When you get closer to the speed of light you need to go relativistic!  When moving closer to the speed of light you also need to take other effects into account like time dilation.  So we need to move to an equation that looks more like this:

$v_{ob} = v_{source}\times\sqrt{(1+B)/(1-B)}$

Where $v_{ob}$ is the observed wavelength, $v_{source}$ is the emitted source radiation and $B$ is your velocity/speed of light.

So lets say that someone is shining some UV light on us at about $4\times10^{-7}~\rm{m}$ and we want to see at what velocity this light is shifted into gamma i.e. about $1\times10^{-11}~\rm{m}$.  We can re-arange that formula to:

$B = [(\frac{v_o}{v_s})^2-1]/[(\frac{v_o}{v_s})^2+1]$

where $B$ will be a decimal of the speed of light.
The term $(v_o/v_s)^2$ is ridiculously small, we are talking magnitudes of $10^{-10}$.

Then when you add or subtract 1, you really dont move that far from 1. Just under 1/ just over 1 you get the answer that You have to be going ridiculously close to the speed of light for this to happen.  Like so ridiculously close MATLAB wont calculate the difference and just gives me the value of 1, i.e. the speed of light.

IN SUMMARY:

Yes but you have to be going so close to the speed of light it would be physically impossible.


What is this "time travel" these popular blogs refer to? I am getting confused by some "popular science" about time travel, such as

or "Time Travel Is Real. Here Are the People and Spacecraft Who Have Done It":

It is indeed an interesting property of physics as we know it, and a great gift from the theory of relativity. It leads to useful perspectives on the nature of time as exemplified in the Twin Paradox example: When you travel for instance at a Lorentz factor of γ = 100 (99.995% the speed of light) for 10 years, time "dilates" to thousand years. For instance it has been proven experimentally that muons would "tick" slower, and therefore for a measure of time in seconds. Similarly, I would assume that extending this property at a macroscopic scale, your whole body would slow down...

My question is to know if will you "live for a thousand years" (internet abounds of such statements as in this example forum and may be interpreted as such in the headlines above), or rather that during the 10 years of your travel (in your spaceship) and which would be measured as 1000 years in another frame of reference (the  earth), you would have aged by "only" 10 years in your spaceship - time "froze" during the travel.

My question is to know if will you "live for a thousand years" (...) or "only" 10 years in your spaceship - time "froze" during the travel.

If you were to travel to close the speed of light (not doing the math, plainly speaking), you would appear to live for a thousand years to the people that are outside of your frame of reference. Time would actually pass the same to you and you would observe the time outside your frame of reference to pass significantly faster.

  Is it a scientific fact?

Human time travel is a very real scientific fact, but the implications are too complex to be presented to the public sphere since it has no practical applications yet. If we were to sent people to another star system (Alpha Centaury A/B - at 4.3 light-years from Earth), accelerating their ship close to the speed of light would make the time in their frame of reference pass slower relative to the outside of their frame of reference, thus allowing them to reach the destination in a shorter time from their perspective. It would be useful when managing supplies on this vast journey.


Is there a star simulation software that can handle mass ejections and supernovae? I use MESA right now to play around with stellar dynamics, but it can't do mass ejections and therefore, I can't think of a way to let the simulated stars explode as supernovae.

What simulation software would I use for these?

Regarding Terminology:
'Mass ejections' (at least semantically) are very different from supernovae.  A "mass ejection" would generally refer to something like what the sun does as part of its normal activity --- ejecting very small amounts of plasma, or on the other end of the spectrum, the ejection of massive shells of material by very massive (e.g. $M \gtrsim 40 M_\odot$) stars$^{[1]}$.  Supernovae, on the other hand, are the explosion and destruction of an entire star (or at least everything outside of the core).

Simulations
The sad fact of the matter is that there are no simulations which can fully self-consistently explode a star.  In other words, there is no code that can start with a star and have it naturally explode$^{[2]}$.  There are many specialized codes, however, which deal with particular aspects of the explosions.  The initial collapse and explosion is usually done with very specialized, complex hydrodynamic codes by people like Christian Ott, Adam Burrows, Chris Fryer, etc.  These codes are not public.

More general hydrodynamic codes are usually used to study the effects of supernovae, i.e. how the blast waves evolve under different circumstances.  In these simulations, the user artificially deposits a huge amount of energy in the center of a stellar model (mimicking the energy produced by an explosion), and then sees how the system evolves.  One of the most popular and advanced codes for this is called FLASH, produced and managed at the U of Chicago.

[1] For a technical article see, e.g.: http://arxiv.org/abs/1010.3718
[2] There are numerous reasons for this, first and foremost that it is an incredibly difficult computational task.  Stars are generally about $10^6$ km in size, while their cores are only a few km --- to understand the explosion, the code needs to model this entire dynamic range which is incredibly challening.  Additionally, the densities and temperatures involved in the cores of supernovae are way outside of the range of whats ever been explored in a laboratory --- so information about that materials behavior is still largely uncertain.  There are lots of other big challenges (e.g. incorporating general relativity, advanced nuclear reactions, etc) but these are some of the key issues people are exploring.


Constructing a periscope/telescope - trouble with lenses Not sure if this is the best place for this, maybe physics would be a better fit, but I'm trying to build a periscope and am running into an issue.

Traditional mirror periscopes work well, but when I look into the one end of the periscope, I see the reflected image at the end of a long tube. I figure that using some combination of lenses, you should be able to make it look as if the position of your eye/head was actually at the top of the tube looking out at the world rather than looking through a tube.

My thinking here is that when you look through binoculars or a telescope, you don't see the inside of the tube, just the enlarged image. So shouldn't you be able to do the same thing but without the magnification?

I purchased some lenses from Amazon with focal lengths of 200mm and -100mm and based on some optics simulators assumed(incorrectly apparently) that I could arrange them 10cm apart and then another set reversed on the other side like so:

`````()````````````````)(```````````)(``````````````()
200mm````````-100mm``-100mm```````200mm

() = convex lens
)( = concave lens

The thinking here was that one set would condense the image down so that the walls of the tube were no longer visible within that viewing angle, and then the other reversed set would reverse that and the image would be back at normal size to my eye, it would just look as though my eye were actually at the opposite end of the tube.

This didn't work. The image is close to life size, but the tube looks even longer: a foot long tube has turned into a 3 foot long tube, and the circle I can see the world through is very small.

Is this a pipe dream? I feel like it must be possible if they used periscope in military applications for years before electronics and fiber-optic cables were invented.

Where did I go wrong? Any guidance would be hugely appreciated.

Edit: some follow up questions

Am I stuck having a small field of view out of the end? Or would using a lower focal point(wide angle) lens allow me to see a wider angle?

Also, is there any way to have a variable length tube? For instance if I'm looking around a wall, is there a way to install the lenses so that the tube could be "telescoped" out to look around a thicker wall?

You're close!  What you built is a back-to-back set of Galilean telescopes (with an overall magnification of 1:1).  The reason you're getting a smaller circle, but with a life-size image, is that the light bundles in the middle are diverging faster, and so you lose the edges if the Galilean assemblies are too far apart.

Military perisopes use optical relays to get the image down the tube without blocking out any of the view.  In simple terms, they have one relatively short-focusing  group of achromatic/apochromatic lenses (called an Objective) to focus the light to a crosshair, like in a rifle scope.  From there, the image is picked up by another long-focusing lens group called a relay.  The relay re-collimates the light (but at low angles).  The light is focused and collimated several times through the tube, until it reaches a final image plane, which is then looked at with an eyepiece (like you might have in a telescope).

Overall, you can think of a periscope as a telescope/monocular/riflescope with relatively low magnification with several pairs of relays in between.  A point of note, you need an even number of image planes (0, 2, etc) to avoid having an
upside-down image.  I can go into a lot more detail or provide some resources, but I think this most directly answers your question.  Please let me know if there's any other questions on this.

Source: I design optics for periscopes for naval applications.

Edit for additional questions:
First, the most comprehensive resource I found in my lazy search is this page: http://taylortechassoc.com/?page_id=97, which is actually written by a previous engineer at my company.  There's a ton of information there, mostly intended for people who already know about optical design, but there's probably a good amount you can get from it as well.

To the question in your comment: in your preliminary design, the positive and negative lenses are taking collimated light input, and outputting collimated light.  However, since they are acting as a telescope with magnification power, the angle of the rays coming out is wider than the angle going in (by a factor equal to your magnification).  The trick to a relay lens design (used in periscopes and some kinds of endoscopes, as well as other applications) is to get the collimated spaces short enough that you don't lose that light before it spreads out.  Or more accurately, that you get the angles of the collimated light to be small enough that you can spread out your lenses quite a bit while still maintaining this condition.

Regarding field of view: you're absolutely not stuck with a small FOV.  That depends entirely on the focal length of your objective.  A short focal length lens as your objective will give you a wider field of view (so long as your light isn't clipped by your top mirror).  You could also have an objective that changes focal length (a zoom lens) to change your field of view as desired.  For your application, an SLR camera lens could work quite well for this.

Regarding tube length: You could vary the spacing in the sections where the light is collimated.  This will impact how much light is being lost from it spreading out too much (typically limited by whichever section is longest).  The result will be a dropoff in the relative illumination (how bright the edge of the FOV is compared to the center).  This is an effect known as vignetting, and can be mitigated, but not without tradeoffs.


Statistically, what would the average distance of the closest black hole be? The closest confirmed black hole is several thousand light years away from earth. Our galaxy has about 100 billion stars. I didn't find any reliable information on the black hole count of ratio versus stars for our galaxy. Some sources say about one in a thousand.

I would like to estimate how close a black hole would be given the data we have. If I did the calculation myself, I would use the volume of the galaxy and the number of stars in it. This would give the average number of stars per some volume, and out of that, the average number of black holes per volume. The distribution of stars would also be have to taken into account, since our part of the galaxy isn't as densely populated. Out of this the statistically approximate distance of the closest black hole could be obtained.

Is anyone equipped with enough information to give me an estimate?

Basically what is needed is the ratio of black holes to stars for our galaxy and then list of x nearest stars and their distances, where x is the ratio.

(I'm asking this because black holes are object of interest, and will have to be visited in the future. Given that our space capabilities might be limited to a distance of a couple of dozens of light years in the near future, this number is interesting.)

Let us assume that $N$ stars have ever been born in the Milky Way galaxy, and given them masses between 0.1 and 100$M_{\odot}$. Next, assume that stars have been born with a mass distribution that approximates to the Salpeter mass function - $n(m) \propto m^{-2.3}$. Then assume that all stars with mass $m>25M_{\odot}$ end their lives as black holes.

So, if $n(m) = Am^{-2.3}$, then
$$N = \int^{100}_{0.1} A m^{-2.3}\ dm$$
and thus $A=0.065N$.

The number of black holes created will be
$$N_{BH} = \int^{100}_{25} Am^{-2.3}\ dm = 6.4\times10^{-4} N$$
i.e 0.06% of stars in the Galaxy become black holes. NB: The finite lifetime of the galaxy is irrelevant here because it is much longer than the lifetime of black hole progenitors.

Now, I follow the other answers by scaling to the number of stars in the solar neighbourhood, which is approximately 1000 in a sphere of 15 pc radius $\simeq 0.07$ pc$^{-3}$. I assume that as stellar lifetime scales as $M^{-2.5}$ and the Sun's lifetimes is about the age of the Galaxy, that almost all the stars ever born are still alive. Thus, the black hole density is $\simeq 4.5 \times 10^{-5}$ pc$^{-3}$ and so there is one black hole within 18 pc.

OK, so why might this number be wrong? Although the number is very insensitive to the assumed upper mass limit of stars, it is very sensitive to the assumed lower mass limit. This could be higher or lower depending on the very uncertain details of the late stellar evolution and mass-loss from massive stars. This could drive our answer up or down.

Some fraction $f$ of these black holes will merge with other black holes or will escape the Galaxy due to "kicks" from a supernova explosion or interactions with other stars in their dense, clustered birth environments (though not all black holes require a supernova explosion for their creation). We don't know what this fraction is, but it increases our answer by a factor $(1-f)^{-1/3}$.

Even if they don't escape, it is highly likely that black holes will have a much higher velocity dispersion and hence spatial dispersion above and below the Galactic plane compared with "normal" stars. This is especially true considering most black holes will be very old, since most star formation (including massive star formation) occurred early in the life of the Galaxy, and black hole progenitors die very quickly. Old stars (and black holes) have their kinematics "heated" so that their velocity and spatial dispersions increase.

I conclude that black holes will therefore be under-represented in the solar neighbourhood compared with the crude calculations above and so you should treat the 18pc as a lower limit to the expectation value, although of course it is possible (though not probable) that a closer one could exist.


Finding the suns position in the sky anywhere at any time What am I looking for?

from any point Latitude, Longitude and date D

I want to know where the sun is in the sky at any given time anywhere on our planet. I even don't know what the outcome will be for this calculation. I Imagine an angle and a height.

Can someone send me on the right path for doing this?

I am not looking for a quick formula that let's me calculate this bur rather understand the components :)

There are many resources that will let you compute the position of the Sun in a coordinate system with its center at the center of the Earth. The answer will be an angle above or below the equator, an angle eastward from the vernal equinox (first point of Aries, and a distance. Books such as Astronomical Algorithms 2nd ed. by Jean Meeus have information on how to transform such a position to a horizon system centered at an observer on the surface of the Earth. The answer will be an angle from the north, an angle above the horizon, and the distance. Wikipedia has articles about software libraries that will do this kind of calculation, such as "SOFA (astronomy)" and "Naval Observatory Vector Astrometry Subroutines". These subroutines are quite complicated because they are meant to be suitable for the most critical work, with angles accurate to a few milliarcseconds. The Meeus formulas are much simpler.


So where are these measurements of galaxies moving faster than light? https://physics.stackexchange.com/questions/400457/what-does-general-relativity-say-about-the-relative-velocities-of-objects-that-a

"we can actually observe galaxies that are moving away from us at >c"

Um, I think I missed the groundbreaking headline that said scientists have directly measured a galaxy moving away from us at speeds faster than light.

So the first answer seems to contradict this other answer

https://physics.stackexchange.com/questions/107748/how-are-galaxies-receding-faster-than-light-visible-to-observers

"If recession velocity at the location of a traveling photon were greater than the speed of light the entire time the photon from a distance galaxy were traveling, we would never observe the photon."

It seems like common sense that at a certain distance, space is expanding faster than light, thus preventing light beyond a certain distance from reaching Earth which means we can't directly measure galaxies traveling faster than light, but we can imply that they do by projecting their velocity beyond the observable universe. But who's right?

If a galaxy used to be within the observable universe, we can measure photons from that galaxy reaching Earth despite that that galaxy may be currently outside the observable radius of the universe. Galaxies can be projected as travelling faster than light due to the expansion of space, but I haven't seen that we can directly measure them as doing such since the redshift of photons is proportional do the amount of space that is expanding between us and a certain distance.

So where are these measurements of galaxies moving faster than light?

They're redshift measurements. Check out the Wikipedia redshift article. It's good stuff.

  "we can actually observe galaxies that are moving away from us at >c"

It's true. You might think it cannot be, but it can.

  Um, I think I missed the groundbreaking headline that said scientists have measured a galaxy moving away from us at speeds faster than light.

It's not groundbreaking at all. It's simpler than you think.

  So the first answer seems to contradict this other answer https://physics.stackexchange.com/questions/107748/how-are-galaxies-receding-faster-than-light-visible-to-observers "If recession velocity at the location of a traveling photon were greater than the speed of light the entire time the photon from a distance galaxy were traveling, we would never observe the photon".

Yes, that's a bit unfortunate, because he also says this: "galaxies with red shifts greater than ~3 were and are receding from us faster than light".

  It seems like common sense that at a certain distance, space is expanding faster than light, thus preventing light beyond a certain distance from reaching Earth which means we can't directly measure galaxies traveling faster than light. But who's right?

The guy who says we can see galaxies that were and are receding from us faster than light. As per Rob's comment, see Expanding Confusion: common misconceptions of cosmological horizons and the superluminal expansion of the Universe by Tamara Davis and Charles Lineweaver. Note this: "We show that we can observe galaxies that have, and always have had, recession velocities greater than the speed of light." See page 8: "Amongst those who acknowledge that recession velocities can exceed the speed of light, the claim is sometimes made that objects with recession velocities faster than the speed of light are not observable [App. B: 9–13]."

Also have a read of the Wikipedia ant on a rubber rope article. The rubber rope is being stretched as the ant walks along it: "At first consideration it seems that the ant will never reach the end of the rope, but in fact it does (although in the form stated above the time taken is colossal). Whatever the length of the rope and the relative speeds of the ant and the stretching, providing the ant's speed and the stretching remain steady the ant will always be able to reach the end given sufficient time. Once the ant has begun moving, the rubber rope is stretching both in front of and behind the ant, conserving the proportion of the rope already walked by the ant and enabling the ant to make continual progress".

There's a section on the metric expansion of space. It says this: "By thinking of photons of light as ants crawling along the rubber rope of space between the galaxy and us, we can see that just as the ant can eventually reach the end of the rope, so light from distant galaxies, even some that appear to be receding at a speed greater than the speed of light, can eventually reach Earth, given sufficient time. However, the metric expansion of space is accelerating. An ant on a rubber rope whose expansion increases with time is not guaranteed to reach the endpoint.[3] The light from sufficiently distant galaxies may still therefore never reach Earth".


Can the Earth be used as a gravitational lens? Could the Earth be used as a gravitational lens to image very far distances with an effective aperture that would be far greater than any telescope? How far away from Earth would an observing spacecraft have to get to enter the focal point, and see very far at a resolution? Would that enable the direct imaging of small planets around other stars?

Yes, it's possible in theory, but beyond current technology to achieve. The focal point of the Earth is 15300 AU away.  By contrast, Neptune is about 30AU.
As a gravitational lens is not like a glass lens, you don't get an image formed and at any distance greater than 15300 AU an Einstein ring would be formed around the Earth. But at that distance, the Earth would be very close to the sun in the sky, and the sun has its own gravity. Moreover the Einstein ring would be have the same radius as the Earth at that distance.  It would be very small. No Einstein ring is formed at less than 15300; the ring that would be formed by a body with Earth's mass is smaller than the diameter of the Earth at distances less than 15300
Larger bodies have a nearer focus, the Sun has a gravitational focus starting at 550 AU, which is perhaps just about within current solar sail technology. It might be possible to make direct observations using the sun as a gravitational lens.  But you don't get a steerable telescope. It would still be a very challenging mission.
See a related question about the mission on space exploration stack and this forum discussion


How is information preserved on the surface of a Black Hole? Is there an compression algorithm for storing the information?

What if there is a uniformaly complex object dropped to the black hole? Then there will be no compression of information.

How is there enough space on the surface (2D) to encode something which is in 3D. The planck distance is the same in the black hole as in the universe. So how can the black hole encode information about an object in 3D that is both uniformly complex and as dense as the black hole? Or is there a law that says 3D objects cannot be more dense then some level - which is sufficiently sparse so that when collapsed to a black hole all its information can be encoded on the surface of a sphere? Has this limit been found - given that a sphere has the lowest surface to volume of any object: https://en.wikipedia.org/wiki/Surface-area-to-volume_ratio ?

Edit: added '3D' to first sentence of 2. question.

How is information preserved on the surface of a black hole?

We have no evidence that it is. We have some claims of this nature, associated with the holographic principle. But I'd say the physics concerned is hypothetical and speculative. It isn't in the same league as general relativity, which is one of the best tested theories we've got. See Clifford M Will's paper on the confrontation between general relativity and experiment for details of that.

  Is there an compression algorithm for storing the information?

Not that I know of.

  What if there is a uniformly complex object dropped to the black hole? Then there will be no compression of information.

Instead there will be a gamma ray burst. Friedwardt Winterberg wrote about this. See the 2013 AMPS paper an apologia for firewalls. Tucked away in the conclusion is footnote 31, containing a reference 87 to Winterberg’s 2001 paper gamma ray bursters and Lorentzian relativity. I'm confident that Winterberg is essentially correct.

  How is there enough space on the surface (2D) to encode something which is in. The planck distance is the same in the black hole as in the universe. So how can the black hole encode information about an object in 3D that is both uniformly complex and as dense as the black hole?

I don't think it can.

  Or is there a law that says 3D objects cannot be more dense then some level - which is sufficiently sparse so that when collapsed to a black hole all its information can be encoded on the surface of a sphere? Has this limit been found - given that a sphere has the lowest surface to volume of any object: https://en.wikipedia.org/wiki/Surface-area-to-volume_ratio?

As far as I know there is no law or limit. And as far as I know there is no evidence to support the holographic principle. I for one do not take it seriously. But please advise if you receive any useful answers about this subject on any other website.


Can we see Earth by looking into space? It is known that spacetime is curved,so I am wandering to know if its possible to look into space and we see earth (light from earth long long ago). Did I understood spacetime correctly and if its possible then can some one after million years can see dinosaurs by looking into space?

The answer, for the universe we live in, is almost certainly No. The best guess is that the universe, on the largest scale, is very nearly flat and probably extends infinitely in all directions.

The physics of general relativity does permit a hypothetical universe that curves back on itself in the way you describe, and if it were small enough, we could indeed see ourselves. But I don't know whether stars and planets could evolve in such a universe.

Incidentally, it doesn't have to even have curvature to wrap around as you describe. The universe could be like the game Asteroids, where you wrap off one edge and come back on the other. That has the topology of a torus, but is geometrically flat -- for instance, the angles inside a triangle still add up to 180 degrees.

But it doesn't look like our universe is like that.


Does Hawking Radiation really exist and Why "negative energy" in virtual particles? I'm doing some research on black holes for a science video contest - more specifically, on Hawking Radiation. As far as I know, there are these constant quantum fluctuations of "virtual" particle pairs (basically particle-antiparticle) that appear, then almost instantaneously mutually annihilate. Near the event horizon of a BH, if one virtual particle gets trapped inside, then it acquires negative energy, while the other particle escapes away and is emitted as radiation. My question is, why does the virtual particle that's sucked in get negative energy? Is it due to some rotating vector stuff or is it inherent in the particle?
And, as per my logic, if NASA detected this image of a BH, it's clear that it's emanating some radiation, so Hawking Radiation has to exist, right?
Any help in clarifying or correcting my understanding would be appreciated.Thank you.

Unless everything we know about general relativity and quantum field theory is wrong, Hawking radiation almost certainly exists.
It is also entirely undetectable. The thermal output of a 3 solar mass black hole by way of Hawking radiation amounts to $\sim 10^{-29}$ W. This, of course, is not only completely undetectable but the corresponding Hawking temperature, $\sim 2\times 10^{-8}$ K, is far, far below the temperature of the CMB; so any astrophysical black hole today would absorb many orders of magnitude more radiation from the CMB than it emits by way of Hawking radiation.
The explanation with virtual particle-antiparticle pairs and negative energy particles absorbed by the black hole originates, I believe, from Hawking's popular science book A Brief History of Time, but it really is at odds with Hawking's own groundbreaking paper on the subject (Nature 248, 1974). This is explained quite nicely in an informative article by Ethan Siegel, published last year online: https://www.forbes.com/sites/startswithabang/2020/07/09/yes-stephen-hawking-lied-to-us-all-about-how-black-holes-decay/?sh=52d60ad44e63 .
Adding to Ethan's explanation (which is well worth the read) I ought to mention two things. First, there is the closely related concept Unruh radiation: thermal radiation seen by an accelerating observer from the direction of his apparent Rindler-horizon, in spacetime that is seen as empty by an inertial observer. This alone stresses the important point that in quantum field theory, the concept of a particle becomes observer-dependent once we permit accelerating observers and/or gravitational fields. Second, the wavelength of Hawking radiation alone should tell us that it is not originating from the event horizon. For the aforementioned three solar mass black hole, the peak wavelength is almost 180 km. Comparing this to the less than 10 km Schwarzschild radius of the black hole, I think it is quite evident that this is not about particle-antiparticle pairs at the horizon itself.
Finally, regarding the images of the M87* supermassive black hole, first, these are not NASA images but images produced by an international collaboration called the Event Horizon Telescope project. Second, though they are often described as images of the black hole, that is not strictly correct. What you actually see is radio frequency emissions of infalling matter surrounding the black hole in the form of an accretion disk. The central, black shadow corresponds to the black hole's photon sphere (radius at which unstable closed photon orbits become possible). And it is important to remember that this is not exactly a photograph, but rather, a synthesized image that is fitted to a large number of radio frequency observations using an optimization method.


Will the shape of the constellations change if we observe from another planet, say Mars? Constellations are dominant features of night sky. But these are basically group of stars that forms particular shapes when viewed from Earth.

How much the shape of a constellation will change if we observe them from another planet, say, Mars?

If no significant change be visible, then how much distance we need to travel to observe a change?

There will be no change observable by naked eye from any planet in the Solar System.

It's easy to estimate the distances for which change is visible.  There are two key points. First, changes in the constellations are due to parallax.  Second, by definition, a baseline of 2 AU (astronomical units) produces a parallax of 1 second of arc at a parsec's distance (wihc turns out to be about 3 light years) and half an arc second at two parsecs, and so forth.  Double the displacement to four AU and you double the parallax.

The nearest visible star to the Sun (Alpha Centauri) is just over a parsec away.  For it to move a barely noticeable distance (say a tenth of a degree or 1/5 the diameter of the Moon) in the night sky, the baseline would have to be 360 times as long, since 0.1 degree is 360 arc seconds.  So the closest star in the night sky would have moved a just-perceptible amount if observed from 500 AU away from Earth.

The brighter stars which make up the constellations range from the distance of Alpha Centauri out to thousands of light years, but many of them are within 100 light years (30 parsecs).  While a movement of 0.1 degree is perceptible, for the constellations to be obviously distorted, you'd probably want many stars to move by more than a degree (two full Moons).

So: 30 parsecs and 1 degree requires a movement of 1AU * 30 * 3600 (arc second/degree) or about 100,000 AU.  That's about 1.5 light years.

If you voyaged away from the Sun, a keen observer would notice an slight effect somewhere out past the Kuiper Belt, and a few stars would be massively displaced by the time you were out in the Oort cloud, but for wholesale disruption of the constellations, you'd have to go about a light year.

Two points: First, at a light year's distance from the Sun, the constellations would still be somewhat recognizable, though there'd be no doubt at all that things were scrambled.

Secondly, the shifts would be largest in the direction perpendicular to your line of travel (the baseline for parallax) and much smaller fore and aft.


Giant variation of the proposed eLISA mission using reflectors on Earth and the Moon possible? I was thinking about the proposed ESA mission eLISA, which is essentially a space version of the LIGO experiment, and it occurred to me that trailing three satellites behind the Earth's orbit seemed excessively complicated in comparison to using a reflector on the ground and another on the moon (if I am not mistaken, there is even already a mirror on the moon, from the Lunar Laser Ranging experiment) and a single satellite. Is there not a geosynchronous orbit with a line of sight to the moon at all times? It seems to me the added range would provide better sensitivity. Provided you know the exact distance to the reflector, you can calculate individual frequencies for the Earth and Moon lasers, downconvert the faster frequency through a PLL of sorts, and still use the resulting data in an interferometry analysis. Is there something major I am missing here that would make this arrangement more difficult than the proposed eLISA mission? It seems to shift a lot of the physical risk and cost of the mission to computational costs, which by comparison are much, much cheaper.

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 your proposed detector arms is much shorter than the coherence length of any laser, so your interferometer would not work.

Further problems:

Positioning of mirror on Earth: Where do you want to put the mirror such that the moon is always visible in the sky?
Positioning of satellite: I doubt (without checking) that you can see the moon at all times from geosynchronous orbit. Maybe you could work with one of the Earh-Moon Lagrange points.
Constant mirror adjustment: Earth is rotating, and the Moon revolves around the Earth. So your interferometer arms would change angles constantly. (And you'd need to dig lots of tunnels straight through Earth, see above.)
Vibration: The mirrors have to be as free from vibration as possible. That is much easier to achieve with a free floating satellite than on earth or the moon. (Did you know there are moon quakes?)
Ellipticity: The distance to the moon varies between 360.000km and just over 400.000km during its orbit. So you'd be constantly adjusting the interferometer. A constant drift is not necessarily problematic, but the huge difference in distance might be.

Lastly, it is not clear to me how you want to replace precision measurement by calculation. What is PLL? How do you get meaningful interference patterns with lasers of different wavelengths? Also note that the "exact distance", if you want to count waves, must be better than half the wavelength of your laser, so a few hundred nanometers. At the distance of the moon, that is better than $1 : 10^{15}$ (hope that calculation is correct).


What exactly is a "conclusive association" in Astronomy? Phys.org's Astronomers use slime mold model to reveal dark threads of the cosmic web says

A computational approach inspired by the growth patterns of a bright yellow slime mold has enabled a team of astronomers and computer scientists at UC Santa Cruz to trace the filaments of the cosmic web that connects galaxies throughout the universe.
Their results, published March 10 in Astrophysical Journal Letters, provide the first conclusive association between the diffuse gas in the space between galaxies and the large-scale structure of the cosmic web predicted by cosmological theory.

This links to Joseph N. Burchett et al, Revealing the Dark Threads of the Cosmic Web, The Astrophysical Journal (2020)
Kudos for the use of the slime mold model, this organism is a gift that keeps on giving to not only biologists but a wide range of other fields https://www.google.com/search?q=slime+mold+growth+modeling which includes computing, network analysis, transportation and even cosmology.
Astronomy differs from some other fields of science in that most hypotheses must be tested by observation of naturally occurring phenomena, it's hard to test them through carefully designed experiments. While particle physicists can "throw large hadrons" at each other under controlled conditions, astronomers can't similarly "collide white dwarfs" in the laboratory.
Question: In the results described in the ApJ Letter, was there such a thing as a "conclusive association"? Is that even a thing? Or was there simply a similarity as defined by some metric which the authors believe to be reasonably objective?

A "conclusive association" is an oxymoron; for example, "brawling love", "loving hate" (these first two are from Shakespeare), "act naturally", or "jumbo shrimp". An association is a vague term while conclusive is very specific and means proven without a doubt.
The underlying journal article does not claim the authors have provided a "conclusive association between the diffuse gas in the space between galaxies and the large-scale structure of the cosmic web predicted by cosmological theory." That is what the press release claims. Press releases are not subject to peer review. Press releases can, and oftentimes do, use over flowery speech.
The underlying journal article does claim that "statistics of absorption by the intergalactic medium (IGM) via spectroscopy of distant quasars support the model yet have not conclusively tied the diffuse IGM to the web." This is true. The implication is that the authors' model has conclusively tied the diffuse IGM to the web. The author do not explicitly make this claim in the journal paper. Suppose the authors had make this claim in early versions of the paper. If that was the case, peer reviewers would inevitably have said to remove this claim on the basis that "conclusive" is an extremely strong word in science.
Even if peer review had allowed that claim, it would still be dubious. Many scientific press releases claim that the research being reported is ground shaking and conclusive. Don't take press releases as fact. Only a tiny few scientific papers show something that is ground shaking and conclusive. Those that do do so (e.g., Einstein's annus mirabilis papers) are very few and very far between.


How often do stars form? There are many more stars in the universe than the years that the universe has been in existence, so wouldn't this mean that stars were/are forming at an astounding rate?

By an estimated order of magnitude there are about $10^{22}$ in our observable universe (hundred billion galaxies with hundred billion stars, each, as a rough estimate), and the universe exits since (order of magnitude) $10^{10}$ years. Hence the average formation rate of stars has been about $10^{12}$ per year, meaning more than 10,000 per second, give or take a factor of 100. This calculation ignores stars which don't exist any more, but for an order-of-magnitude estimate numbers should be ok.

Although the star formation rate has been much higher in the younger universe than today. (One of the press releases about star formation in the early universe)


regarding spectral type of a star consider two stars of the spectral type and subtype, but not necessarily both on the main sequence. Describe whether this information is sufficient to tell you how the stars differ in surface temperature, size, and distance?

so if they have they same spectral type and subtype i think they're supposed to have the same surface temperature. however this is not enough information to tell us about their sizes and distances. is this correct? any insight would be great, thanks!

The key for stellar classification is the Hertzsprung-Russell diagram (or HRD for brevity), also know as the color-magnitude diagram.

This shows the relation between the surface temperature (on the x-axis) of a star and its luminosity (on the y-axis). The temperature is an index of the Spectral type. For distant stars also spectroscopy is taken into account, to classify spectral types according to their absorption lines here some examples. The usual classification is called MKK (from the astronomers that had developed it - Morgan, Keenan and Kellman in 1943), and divides the stars in 7 stellar types O, B, A, F, G, K, M. Please note that from the 1943 other stellar types have added, but these are the most used. This sequence goes from the hottest (O) to coolest (M) stars. Each letter class is then subdivided using a numeric digit with 0 being hottest and 9 being coolest. Our Sun is a G2 type star. This classification also uses an additional notation to avoid the "degeneracy" you mentioned: a roman numeral between "I" (one) and "V" (five). The higher the number, the wider the lines in the observed spectrum. However, despite of the quantity these roman numbers indicate, they are an index of luminosity class. This is based on the width of certain absorption lines in the star's spectrum which vary with the density of the atmosphere and so distinguish giant stars from dwarfs. Luminosity class I stars are supergiants, class III regular giants, and class V dwarfs or main-sequence stars, with II for bright giants, IV for sub-giants, and VI for sub-dwarfs (from here). Then, our Sun is "completely" identified by the type G2V. You can take a look at the figure below, and check out by yourself where it belongs:
.
Also note that other indexes are used to characterize the stars, as for the width of the lines in the luminosity class, or for some peculiarities.

Finally, if you are interested in the study of stars, you should always keep in mind the Stefan-Boltzmann law:

$L=4\pi\sigma R^2 T^4$

By this you can identify isoradii lines in the HR diagram:

Also, there is a proportionality between luminosity and mass:

$L\sim M^{3.5}$

Then you can identify "isomasses" lines in the HR diagram, but I can't find such a figure to show.


What is the distance from Alpha Centauri to Barnard's Star? Alpha Centauri AB is the closest star system to Earth (4.366 ly), followed closely by Barnard's star (5.988 ly). The closest star system to Alpha Centauri is Luhman 16 (3.8 ly from α Cen). So I am wondering, what is the distance from α Centauri AB to Barnard's star, and more generally, from one star to a different one (all below 10 pc)?

To find the distance from one star to another, we need three things for both of the stars: their right ascensions, declinations, and the distance from Earth to those stars.
So, let's get those things:
From the Wikipedia page on Alpha Centauri:
$RA = 14^h\:39^m\:36.49400^s$
$DEC = -60^{\circ}\:50'\:0.23737''$
$R = 4.37\:\rm{ly}$ (you gave 4.366, some other sources give 4.367... I'm going to stick with 4.37)
and for Barnard's Star:
$RA = 17^h\: 57^m\: 48.49303^s$
$DEC = +04^{\circ}\: 41'\: 36.2072''$
$R = 5.958 \: \rm{ly}$ (again, you gave a slightly different value, I'm sticking with Wikipedia for now)
where RA is right ascension, DEC is declination, and R is radial distance from Earth to the target star.
Now, by themselves, it is relatively difficult for us to obtain an actual distance. What I would do is convert these to rectangular coordinates, and then it's a matter of using the 3-d distance formula.
First, however, we need to convert RA and DEC into units like radians or degrees.
For right ascension, we can use the general formula:
$degrees = 15  (h + \dfrac{m}{60} + \dfrac{s}{3600})$
and for declination:
$degrees = deg + \dfrac{m}{60} + \dfrac{s}{3600}$
(when the declination is negative, however, multiply all terms in the formula by -1)
So, for Alpha Centauri AB, we have:
$RA = 15  (14 + \dfrac{39}{60} + \dfrac{36.49400}{3600}) \approx 219.902^{\circ}$
$DEC = -1  (60 + \dfrac{50}{60} + \dfrac{0.23737}{3600}) \approx -60.833^{\circ}$
and for Barnard's Star, we have:
$RA = 15  (17 + \dfrac{57}{60} + \dfrac{48.49303}{3600}) \approx 269.452^{\circ}$
$DEC = 04 + \dfrac{41}{60} + \dfrac{36.2072}{3600} \approx 4.693^{\circ}$
Now, to convert from spherical to rectangular coordinates, we have to define which of RA, DEC, and R can be assigned to $r$, $\theta$, and $\phi$. R should be $r$ - that's pretty straightforward. Since RA can be thought of as "celestial longitude", we'll assign it to $\theta$, and thus declination will be $\phi$.
To clarify, I'm defining $\phi$ as the angle from the xy-plane - so a $\phi$ of $\dfrac{\pi}{2}$ would mean pointing straight upwards. I know some sources define $\phi$ as the angle complementary to that angle (so, $\dfrac{\pi}{2}$ -
the angle from the xy-plane), but for astronomical purposes, I think the definition I'm using is more intuitive and easier to work with.
We can then use the conversions:
$x = r\cos{\theta}\cos{\phi}$
$y = r\sin{\theta}\cos{\phi}$
$z = r\sin{\phi}$
So, for Alpha Centauri AB:
$x = 4.37  \cos{219.902^{\circ}}  \cos{−60.833^{\circ}} \approx -1.634\: \rm{ly}$
$y = 4.37  \sin{219.902^{\circ}}  \cos{−60.833^{\circ}} \approx -1.366\: \rm{ly}$
$z = 4.37  \sin{−60.833^{\circ}} \approx 3.816\: \rm{ly}$
and for Barnard's Star:
$x = 5.958  \cos{269.452^{\circ}}  \cos{4.693^{\circ}} \approx -0.057\: \rm{ly}$
$y = 5.958  \sin{269.452^{\circ}}  \cos{4.693^{\circ}} \approx -5.938\: \rm{ly}$
$z = 5.958  \sin{4.693^{\circ}} \approx 0.487\: \rm{ly}$
And now, finally, we can use the distance formula for 3-d:
$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2}$
So, the distance between Alpha Centauri AB and Barnard's Star is:
$d = \sqrt{(-1.643 + 0.057)^2 + (-1.366 + 5.938)^2 + (3.816 - 0.487)^2} \approx\mathbf{5.873\,ly}$
Well, that was certainly tedious - but it's a process that you can standardize to pretty much any star, or really, any two astronomical objects:
First, convert RA and DEC to degrees.
Second, assign R, RA, and DEC to the spherical coordinates $r$, $\theta$, and $\phi$.
Third, convert spherical coordinates to rectangular coordinates.
Lastly, use the distance formula with the two sets of $x$, $y$, and $z$ coordinates.
Hope this helps. :)


is there any theory or observational evidence that our universe is electrically neutral or not? It seems our universe is neutral in large scale.

There is CP violationwiki problem about matter and dark matter.
Similarly is there any theory about whether our universe is electrically neutral?

Cosmic rays are not neutral. However the atomic nucleus and electrons are just separated. There should be no charge excess in total.

What is the largest scale that is severely non-neutral?

By observation, gravity dominates the universe on large scales.  If there were a significant disparity in positive and negative charges then we would expect electromagnetism—which is approximately $10^{39}$ times more powerful than gravity—to dominate.  So by this observation we can conclude that the universe is approximiately electrically neutral.  Exactly why this should be so is not well-understood, and ultimately ties into the matter-antimatter problem, aka the baryon asymmetry.

Protons and electrons are produced through distinct processes.  That we have many protons is not particularly remarkable, as this would come about by conservation of baryon numbers as other particles decayed.  Why we have any electrons at all is a fair bit more mysterious.

In the early universe, we understand the energy was well above the electron-positron pair production threshold, and so the early universe had large amounts of both electrons and positrons.  When the universe cooled below this threshold, the production would have stopped and we would expect the electrons and positrons to then collide and annihilate each other, leaving essentially none of either.  But we obviously see large amounts of electrons: approximately as many electrons as protons by the aforementioned observational evidence.  And we also do not see large amounts of antimatter—positrons in particular.  If there were large amounts of positrons out there, we would expect to see tell-tale signatures in space as they annihilate with regular matter (or in any weak force interactions, such as would be evident in supernovae), and we have no such observations.

So somehow the early universe must have produced significantly more electrons than positrons.  In other words, we have strong observational evidence that there must be an asymmetry in physics between matter and antimatter production.  There are ways to work this into the theory, but the Standard Model by default does not support it, and experts have not really settled on any one particular modification.


Structure of the torus of the Active Galactic Nucleus How much do we know about the structure of the torus surrounding the Active Galactic Nucleus (AGN)? Is the shape defined or still under speculations? And are there different structures for this torus possible i.e., does it vary from galaxy to galaxy?

This question was asked more roughly 8 years ago, and for 2 years there is a review article around which I found worth mentioning and summarizing. The manuscript is called Redefining the torus:  A unifying view of AGN in the infrared and sub mm and work of Sebastian Florian Hönig. Here an excerpt of the abstract:

The advent of high-angular resolution IR and sub-mm interferometry allows for spatially-resolvedobservations  of  the  parsec-scale  environment  of  active  galactic  nuclei  (AGN),  commonly  referred  toas the “torus.”  While molecular lines show the presence of large, massive disks, the IR observationsappear to be dominated by a strong polar component that has been interpreted.

The main result seems to be the following schematic picture (Fig. 4 of the manuscript) of the AGN which is an overlay of various observations:

The purple ring are observations in the near-IR.
The yellow torus and the gray dashed lines are from ${\rm H}_2$ emission.
The main brownish cone is from mid-IR.
The redish wobbly plane with the red (solid) arrows and the purple (dashed) arrows are sub-mm obseravtions (${\rm CO/HCN/HCO^+}$)
Bright green dots on the $x$-axis are ${\rm H_2 O}$ maser emissions.


What about the magnetic field strength inside of an old accreting neutron star? As the magnetic field strength outside an old accreting neutron star is thought to be small (about $10^8-10^{10}$ Gauss), what about it inside the star? Because of the superconductivity and degeneration, can the strength reach $10^{17}$ Gauss or even more? Is there any article that describe such problem?

This turns out to be tricky to answer definitively and is a topic of contemporary research.
The interior fields may be quite high ($\sim 10^{12}$ G) or approaching zero, depending on details of the interior microphysics.
The argument that fields might be close to zero is based on the probable presence of a superconducting, superfluid of protons. The interior of a neutron star cannot be just neutrons, because of the weak decay mechanism into protons, electons and (anti)neutrinos. Neutron star interiors are therefore expected to have an equilibrium concentration of $\sim 1$% protons (and electrons). If temperatures are low enough, and by that we mean $< \sim 10^9$ K, then a long-range interaction between the protons can create bosonic pairs in much the same way that electrons pair up in a low-temperature superconductor. A similar pairing mechanism can create a superfluid of neutrons in the interior.
In neutron stars, despite neutrino cooling being strongly inhibited by the degeneracy of neutrons, protons and electrons, the cooling rate is very rapid because the same degeneracy means neutron stars have an extremely low heat capacities. This means the core cools on timescales of tens of seconds after the supernova, slowing to decades after a year or two. The result is that the interior temperature falls below $10^9$ K and superfluidity (with accompanying superconductivity in the case of protons) will probably set in on timescales of hundreds of years.
When materials make a phase tranition to a superconducting state it is possible for the magnetic flux that threads them to be expelled. This is known as the Meissner effect.
On that basis, one might expect the magnetic field to be expelled from the interior of the neutron star once the protons become superconducting.
However, the superconductive state in a neutron star may be akin to a type II superconductor. This does allow the penetration of magnetic flux tubes (or "fluxoids") into the interior (Baym, Pethick & Pines 1969 - sorry, can't find a free version) if the magnetic field is greater than some critical value.
There seems to be healthy debate about exactly what state the interior of a neutron star is in as a function of time and whether the timescale for magnetic flux expulsion is short enough for the Meissner effect to take place or whether there is effectively a "frozen in" field in older neutron stars (e.g. Wynn et al. 2017).