How large (that is, radius) could a planet be? I know that normal planets don't get any larger than Jupiter (or 2 Jupiter radii if hot), as adding more gas just increases density, not radius, until you reach the point of being a brown dwarf star.

That compressability, I suppose, is a property of the overall average material that forms star systems, as beyond a critical size it accretes the ambiant gas as well as dust in the cloud.  That is, it's mostly hydrogen and a quarter helium with a bit of other stuff.

But what about a body made of other stuff, such as rock or metal? Without worrying about planetary formation processes, just that atoms are piled together and feel self gravity. Perhaps it needs to be grown slowly so it has time to cool before adding more. Otherwise, no special techniques: just what would happen if material of suitable composition was heaped together?

I suspect that normal ideas of minerals would not exist under that kind of pressure, even if carefully allowed to cool. But can elements other than H and He still compress their volume in the same way, or would a rocky world be able to reach sizes of millions of miles?

What about more exotic cases, like the "puffy" planet I heard of that has the density of styrofoam?

Puffy planets tend to be Jupiter or Saturn like, probably lower mass than Jupiter, perhaps lower metalicity but the most important factor is heat.  Either close to the sun or recently formed.   Heat expands gas planets.  You're correct that as you add more mass the planet of Jupiter mass tends not to grow larger, but if there's enough internal heat, gas giant planets can get a bit larger than Jupiter.   Planets as much as 2 Jupiter Radii have been observed (though there's some inaccuracy in those estimates), but growing larger than Jupiter is largely a factor of high temperature.

Rocky worlds will never grow as large as Jupiter.  The greater mass will prevent it, and past a certain mass, a rocky world is unlikely to remain what we consider rocky.  Above a certain mass it holds onto hydrogen which is the most abundant gas in the universe, and that would give the massive rocky world an appearance more like a gas giant.

But to answer your question in theory, if you had a rocky world with Jupiter mass and a negligible atmosphere, it could never get close to Jupiter diameter.  The mass crushes the inside of the planet.  Mercury for example is made of denser material than the earth.  Higher Iron content, but it's less dense than the Earth because the Earth's mass crushes its rocky mantle and metallic core.  When you start to get a Jupiter-mass rocky world, the crushing becomes significant and you could never build a rocky world close to as big as Jupiter.  As with hydrogen, it would reach a certain maximum size, then it would begin to get smaller under the crush of gravity.   Even so called "non compactible" material, does compact at the pressure inside large planets.   Iron has a density of 7.874 g/cm3 and slightly less than that at high temperature, but the Earth's metallic core has a density of 12.6–13.0, and that's primarily due to crushing.  When you've reached Jupiter mass, the crushing and density is significantly greater.

At about the mass of the sun the planet would be smaller than the Earth, and it would basically resemble a white dwarf star, and no longer be distinguishable as a rocky world.

I can add a few links to back this up if needed and if someone wants to run the math or give an answer with more detail, feel free.


Why the habitable zone of a binary star is a disc and not a sphere? I understand that the orbit of a planet will always be flat, so there is no point on making a sphere out of the HZ when we consider one star, but what about binary stars? For example, this binary system (taken from here):

The habitable zone of this binary surrounds both stars, but what if the orbit of the planet isn't contained in the binary's orbit plane? For example, orbiting the star on the left with an inclination of, say, 20 degrees? Wouldn't it have to be two spheres and not two disks? Or is the HZ calculated assuming coplanar orbits?

If you go to the associated paper Müller & Haghighipour (2014), it shows that what they're plotting are the points which satisfy:
$$\frac{L_\odot}{l_{\rm In-Sun}^2} \le \sum_{i=1}^{N} W_i (T_{\rm star}) \frac{L_i / L_\odot}{d_i^2} \le \frac{L_\odot}{l_{\rm Out-Sun}^2}$$
where:

$L_\odot$ is the luminosity of the Sun
$L_i$ is the luminosity of the $i$th star (of a total $N$ stars)
$l_{\rm In-Sun}$ and $l_{\rm Out-Sun}$ are the inner and outer boundaries of the Sun's habitable zone (two sets of distances are used for the conservative and extended habitable zones, plotted as dark and light green respectively)
$d_i$ is the distance from the point to the $i$th star
$W_i(T_{\rm star})$ is a weight factor that accounts for the different spectral energy distribution at different temperatures $T_{\rm star}$. Details of the calculation of this weight factor are given in the paper.

This formula does not rely on the objects being co-planar. If the binary were inclined relative to the plane of the plot, the $d_i$ values would be altered at each plotted point to account for the component of the distance out of the plane, but essentially this would just be a different slice of a 3-dimensional volume.


Why is Mars considered the outer edge of the "goldilocks zone"? In both books and documentaries I often see the Goldilocks Zone as described between Venus and Mars with the Earth "just right." This seems sort of ill thought out to me, because it assumes that Venus' atmosphere comes from its distance from the sun and the same with Mars. I'm not too familiar with current ideas about how Venus' atmosphere formed, but with Mars I'm fairly sure it's widely understood the reason the atmosphere is so small is because the gravity is so weak. Even some discussions about terra forming Mars mention that within a few million years, perhaps, the atmosphere would drift off into space.

So how can we be so certain that's the limits of the zones, especially with Mars? If Mars was as large as Earth/Venus and could hold on to a thicker atmosphere and thus retain more heat, then wouldn't that automatically push out the Goldilocks Zone? It seems like the edges have been places arbitrarily without consideration as to why they're the edges instead of "they just are." This comes up especially because even with its weak atmosphere, at the equator, it can reach up to 20 degrees which is well within the realm of liquid water.

The circumstellar habitalbe zone can be defined as the distance range around a star, where the mean temperature of a rotating planet would be between 0 and 100 centigrades, if radiation (heat) received from the star and thermal radiation emitted by the planet form an equilibrium.

But that's only a rule of thumb. It has been redefined, and is still disputed.
Hence your criticism is justified to some degree.
There may exist exceptions of habitable planets outside the presumed habitable zone; and there exist certainy many planets within the habitable zone which aren't habitable.
But the probability for a planet similar to Earth to be habitable on its surface is thought to be highest in the habitable zone. When looking for habitable exoplanets, it's therefore more efficient to focus on the habitable zone.


What will it finally take to accurately measure the distance to Betelgeuse? Comments below What is the maximum distance measurable with parallax? discuss challenges associated with parallax measurements of Betelgeuse and link to Wikipedia’s Betelgeuse; Distance measurements which contains the intriguing paragraph:

In 2008, using the Very Large Array (VLA), produced a radio solution of 5.07±1.10 mas, equaling a distance of 197±45 pc or 643±146 ly.83 As the researcher, Harper, points out: "The revised Hipparcos parallax leads to a larger distance (152±20 pc) than the original; however, the astrometric solution still requires a significant cosmic noise of 2.4 mas. Given these results it is clear that the Hipparcos data still contain systematic errors of unknown origin." Although the radio data also have systematic errors, the Harper solution combines the datasets in the hope of mitigating such errors.83 An updated result from further observations with ALMA and e-Merlin gives a parallax of 4.51±0.8 mas and a distance of 222 (+34/−48) pc.[10] Further observations have resulted in a slightly revised parallax of 4.51±0.80.10
83Harper et al. (2008) A New VLA-Hoppocaros Distance to Betelgeuse and its Implications
10Harper et al. (2017)A Updated 2017 Astrometric Solution for Betelgeuse

Looking at section 3.1 of Harper et al. (2008):

As discussed in Section 1, Betelgeuse required a significant additional error source (cosmic noise) to obtained the expected χ2. The actual nature of this cosmic noise is not known but it may be related to photocenter movement which is unlikely to be random in position on short timescales and would likely provide a systematic error in the astrometric solution. The position angle of the star's rotation axis has been measured from spatially resolved ultraviolet Hubble Space Telescope spectra. Uitenbroek et al. (1998) found ~55° (measured East of North) from absorption features in Goddard High Resolution Spectrograph data, and Harper & Brown (2006) found ~65° from emission features in multi-epoch Space Telescope Imaging Spectrograph data. The distribution of photocenters, perhaps driven by convective and Coriolis terms, may have a special relation to the rotation axis. The stellar proper motion vector also has a position angle of 68° and brightness fluctuations that occurred preferentially near the stellar equator might induce scatter in the parallax displacements, although fluctuations due to stellar rotation itself, would be unlikely because of the ~17 yr rotational period (Uitenbroek et al. 1998).

If I understand correctly; the problem with measuring Betelgeuse's position accurately is that it is so large and non-uniform that its "photocenter" (radio, optical, or otherwise) can be significantly different than its center of mass.
And the problem with obtaining accurate parallax determinations is that its non-uniformity changes on timescales comparable to a year. If it were much faster or much slower, it would not be so much of a challenge.
Interferometric optical imaging of Betelgeuse confirm dramatic non-uniformity across its disk.
Question: What will it finally take to accurately measure the distance to Betelgeuse?

To see how nonuniform Betelgeuse appears optically: (feel free to edit and add better links)

this answer to What else could the Event Horizon Telescope Observe?
this answer to How far have individual stars been seen by radio telescopes?
Why can't the surfaces of stars be observed? links to Wikipedia's List of stars with resolved images
This answer to Betelgeuse in false colors, is this common?

The problem is that the apparent diameter of Betelguese is about 50 mas (milli arc seconds --- 1 mas  is about 5 nano-radians) while its parallax is about 5 mas and its shape and surface brightness are both irregular and variable.
Given that, the current measurement are amazingly accurate. So I can identify about three approaches to doing this measurement, and I don't know enough to know which one(s) might really work, or which might be expected to work first.

Measure the distance without using parallax. This most likely means getting an accurate estimate of the absolute brightness or all, or part of, the disk of the star and comparing that with the apparent brightness. Spectroscopy might give an accurate idea of the temperature, which should in turn allow the brightness to be calculated and interferometry might allow the measurement to be restricted to a small area  near the centre of the disk which was not obviously a star-spot or anything.

Simultaneously image the disk and measure its position relative to background stars, allowing the position of the centre of mass to be estimated accurately.

Increase the baseline of the parallax measurement. A pair of telescopes on opposite sides of Saturn's orbit, for instance would have 10 times the baseline, and more or less ten times the accuracy.

Update: It seems the Gaia consortium now expects to work around the brightness constraints and get data for all stars. From the wikipedia page

Although it was originally planned to limit Gaia's observations to stars fainter than magnitude 5.7, tests carried out during the commissioning phase indicated that Gaia could autonomously identify stars as bright as magnitude 3. When Gaia entered regular scientific operations in July 2014, it was configured to routinely process stars in the magnitude range 3 – 20.[57] Beyond that limit, special procedures are used to download raw scanning data for the remaining 230 stars brighter than magnitude 3; methods to reduce and analyse these data are being developed; and it is expected that there will be "complete sky coverage at the bright end" with standard errors of "a few dozen µas".[58]

I don't know how much that will help with Betelgeuse though, due to its large apparent diameter.


Declination and Ascension - the Sun and Andromeda If Andromeda is overhead at midnight, what is the right ascension of the sun? (andromeda is ar RA =1hr, Dec = +40).

I thought I understood what Dec and RA were, but I can't think of how to start this question?

Thanks.

If you mean solar time midnight than the answer is simple (as commented by @rgettman). The Sun should then be at the opposite side of the sky at $\alpha = 13^{\textrm{h}}$, where $\alpha$ is the right ascension. There is a problem, however, when you use local time (using a timezone). If, for instance, you mean midnight GMT in Liverpool, which is at 3° west then solar midnight will happen at 12 minutes past midnight GMT and the Sun would have $\alpha \approx 13^{\textrm{h}}12^\prime$.


Will I be able to see the "Beads of Baily" at 99%+ totality? Where I am (Vancouver, WA), the Sun will be 99%+ covered. Will I be able to see the "Beads of Baily" when the sun is 99%+ covered in my area ?

Downtown Portland, OR is expected to be 99.4% (according to local news). I expect Vancouver, WA, a suburb just north of downtown Portland, to be over 99% as well (99.3-99.2%??)

No, at 99% a narrow sliver of sun will be visible, the width of that sliver will be about 1.5% of the diameter of the sun at its thickest point.

Baily's beads are caused by the sun shining through valleys on the edge of the moon. Such a valley would have to be about 50km deep (or surrounded by hills 50km high).  You do get some beads at the thin edge of the sliver but these would be too bright to see against the sliver of sun (which would still be blinding bright)

To get Bailey's beads you need totality, with the sun shining through valleys. Baily's which is why they are normally only visible for a few seconds before true totality, usually a couple of seconds after the predicted start of totality.

You can see the worksheet I used to make these estimations at https://ggbm.at/sZ3EYp7z however it is undocumented.


Infinite universe? How? So it seems the overall thought is that the universe is infinite. I don't get how.

Most depictions I read mention something to the effect of "at 1^-20 seconds the universe was the size of _". (Essentially mentioning a finite size at a very small length of time). So it came from a singularity, and had a finite size at some point. No matter how fast inflation went, any multiple of that finite size is NOT infinity though. If it was finite once, it is still finite.

If it's infinite then it was always infinite which means it was infinite density, infinitely large, etc. That's too crazy to swallow.

Isn't it more likely that the universe is really really big but not infinite? And we are at the center of the observable universe (to us), however we are off center in some grander universe that's maybe 1 trillion light years across? And nothing is past that because space and time are the same thing. Therefore if there is no light/matter/events, there is no time and hence no space.

Depending on the mass of the universe, the universe has three different possible shapes.
http://en.wikipedia.org/wiki/Shape_of_the_universe

It can be finite or infinite depending on the shape it has, and there are some ways we can discard possible shapes (if it started to shrink instead of expand).

Hope this helps


Origins and most frequently used; perinigricon vs peribothron? Gizmodo.com's Astronomers Spot Unprecedented Flashes From Our Galaxy's Black Hole mentions Sgr A*'s companion gas cloud G2 and that Wikipedia article uses the term perinigricon, but that mentions peribothron as a synonym.

So which term for an object's orbit's closest approach to a black hole is most frequently used; perinigricon or peribothron? And what are the origins of these competing terms?

I don't work with black holes, but I've been in the field of astrophysics for several years now, and until I read this question, I had never heard of either the two terms.

But a query on NASA/ADS, the primary database for "professional" astronomical papers (both refereed and non-refereed) yields 54 hits for "peribothron" vs. only 10 hits for "perinigricon". Not a lot.

As your links reveal, the etymological origin of peribothron/-nigricon is "near the black hole" in Greek and Latin, respectively. The reason has been to conform with terms such as perihelion, meaning "near the Sun". The first to use the Greek version in the literature was Frank & Rees (1976) who write in a footnote:

  $^\star$We are grateful to W. R. Stoeger for suggesting this word, derived from the Greek bothros, a pit.

William Stoeger was the second author Martin Rees' student (and Stephen Hawking's classmate).

The first to use the Latin version seems to have been Schödel et al. (2002), but they don't offer any explanation for the term.


Linear extent of an accretion disk I have a homework assignment question on accretions discs (essentially an estimation of the number of electron scatterings, but this is just for background).

There are a few parameters, one of them being $L$, which is the linear size of the medium (the medium in this case being an accretion disc around a blackhole)

Now, I have been given the mass of the black hole. Other than that, nothing else which could give me the linear size of the accretion disc.

Could I assume that the linear extent of the accretion disc is perhaps of the order of a few Schwarzschild radii? Which could be calculated from the mass, which is given.

If anyone could shed some light on this I would be very appreciative. I need a nudge in the right direction on this.

I think the outer edge of an accretion disk is not well-defined, and observationally the radius will depend on which wavelength you consider, since the farther you get from the BH, the softer the radiation will be. But if you look in the UV, then Morgan et al. (2010) find the following relation between $R_{2500}$ (the radius when observed at $\lambda = 2500$ Å) and the mass $M_\mathrm{BH}$ of the black hole:
$$
\log\left( \frac{R_{2500}}{\mathrm{cm}}\right) = 15.78 + 0.80
\log\left( \frac{M_\mathrm{BH}}{10^9M_\odot}\right),
$$
(modulo some uncertainties that you can look up in the paper).

That is, if your BH has a mass of $10^8 M_\odot$, its radius will be $R_{2500}\sim64\,\mathrm{AU}$, or roughly 1/3 lightdays.

For comparison, its Schwarzschild radius is $\sim2\,\mathrm{AU}$, so your estimate was actually pretty good.

This result is consistent with Edelson et al. (2015), who find 0.35 lightdays, also in the UV. However, in you look in longer wavelengths, the disk is much, much larger. If you're interested beyond your homework assignment, take a look at the accretion disk theory review by Armijo (2013), who shows that in the radio regime, the disk is thousands of AU, and even up to ~100 pc.


The # of Estimated Galaxies increased 10 fold. Where does that leave our # of stars estimate? I have found oddly different answers through my multiple google searches.

I found one article that mentioned this did not alter the estimated number of stars in the universe. Link: http://www.universetoday.com/30305/how-many-galaxies-in-the-universe/

Yet other article gave the estimate of stars an increase. Link: http://gizmodo.com/we-were-very-wrong-about-the-number-of-galaxies-in-the-1787750693

There were other examples but they also had similar contradictory answers.
Thank you.

The sensationalist headlines mask a rather more prosaic result that has absolutely no effect on the number of stars we might estimate are present in the observable universe today.

The study in question has discovered (actually it was always supposed and suggested that this was the case), that the distribution of galaxy masses was much more skewed to lower mass galaxies in the past. That is, there were many more small galaxies in the past. But this happens at the expense of the large galaxies - there were fewer large galaxies in the past. To first order, the overall mass contained within galaxies stays the same and so long as the ratio of stellar mass to luminosity (since this is actually what is measured) remains roughly constant, than any estimate of stellar numbers stays the same.

Actually, a better estimate for the number of stars in the observable universe uses the present day galactic mass function and then extrapolates that to the size of the observable universe today. This is a far more honest estimate, since we have a much better understanding of how many stars there are in our local part of the universe and can then invoke the cosmological principle to say that other places (but not times) in the universe are similar. This cannot possibly be affected by the new study since that is dealing with the numbers of galaxies in the past.


Is remote measurement of isotope ratios possible, or must a sample be aquired Is it possible to measure to measure isotopic abundance of remote astronomical objects - ie measurement without having a sample to feed into a mass spectrometer? Do different isotopes show any differences in absorption or emission spectra which can be used for this purpose?

Yes, isotopic ratios can be readily established by spectroscopy.

Molecules where the constituent atoms differ by one mass unit have different vibrational and rotational energy levels and the gaps between them depend on the reduced mass of the molecule.

For example, in a diatomic molecule
$$ \mu = \frac{m_1 m_2}{m_1 + m_2}$$
and the vibrational frequencies depend on $\mu^{-1/2}$.

This technique is used for example to look at carbon monoxide feature in the infrared and establish the ratio of $^{13}$C/$^{12}$C.

The isotopic shift for atoms is much smaller in general, but is still detectable. The different mass of the nucleus changes the reduced mass of the electron (a little) and this changes the wavelength of the characteristic transition.

An interesting example of this has been the search for $^6$Li in the spectra of population II stars (e.g. Asplund et al. 2006). This isotope of Li should be produced in the big bang along with the more common $^7$Li but is more readily burned in stellar interiors.

The main optical resonance line of lithium occurs at 670.8 nm. The difference between the  wavelength of this line (actually, it is an unresolved doublet) is 0.015nm. This separation cannot be resolved in stellar spectra and instead the shape of the absorption line has to be modelled to estimate the $^6$Li/$^7$Li ratio.

The isotopic shifts become smaller for heavier elements.


Refraction by Saturn's atmosphere - how dense is it here? This tweet from CassiniSaturn shows an image of Saturn's rings being optically distorted by the dense atmosphere. It is explained more fully on this page and is available here.

The refraction is visible because at at least some depth, the density and path length is sufficient to bend the light significantly while the density and composition is still sufficiently transparent.

Roughly speaking, what is the density where this is taking place? Is it rarefied compared to earth but simply a very very long path length, or is this happening at a density far larger than Earth's surface atmosphere? How large is this refraction -  like 0.1 degrees or 10 degrees? I can't picture the actual geometry very well.

This one may be a bit of a judgement call, unless this image happens to be further documented somewhere.

above: cropped portion of the NASA's Cassini image from here

This is a hard question to answer without hard numbers. When you look at this image, you can clearly see that the refractive power of the atmosphere changes as you look closer to the limb of Saturn. The rings start off being refracted slightly, then get more refracted as you get closer to the limb. That essentially implies the portion of the ring closer to the limb is traveling through lower and denser regions of Saturn's atmosphere and thus being refracted more. In that sense, there isn't a single answer here since there is a range of refraction going on.

My first approach in considering this problem is to consider optical depth. In case you don't know what this is, optical depth, usually denoted by $\tau$, is a dimensionless number which is a measure of how much material light must pass through and interact with when traveling through a medium. The way optical depth is defined is such that a medium with $\tau\lt1$ means light can get through, while $\tau\gt1$ means light cannot get through. A more conceptual way of thinking about optical depth is to say that it represents, on average, how many atoms a particle of light will hit when trying to travel through a medium. That is, if $\tau=0.5$, then about half the light particles will hit something and be lost, while the other half will get through unscathed. If $\tau = 2$, then on average the light particles will hit two atoms and most light will be scattered/absorbed out of your line of sight, making you unable to see through that medium.

A simplified definition of optical depth is given by

$$\tau=\sigma N \ell$$

where $\sigma$ is the collision cross section of the atoms in the medium, in $m^2$, $N$ is the number density of atoms in the medium, in $m^{-3}$, and $\ell$ is the path length of light through that medium, in $m$. So let's finally take a look at your questions.

  Roughly speaking, what is the density where this is taking place? Is it rarefied compared to earth but simply a very very long path length, or is this happening at a density far larger than Earth's surface atmosphere?

Ultimately, you're asking what $N$ and $\ell$ are such that $\tau=1$, since this would be the point where the refraction stops and you don't see any more of the rings. Unfortunately, this is difficult to answer based on that image alone. At best you can look up a radial density profile for Saturn such as in Nagy et al. 2009 (ref. fig. 8.1). Note in that paper, they refer to height as the height above the 1 bar pressure level. This paper also gives you an idea for common atoms within the upper atmosphere and you can look up $\sigma$ for these atoms. Predominantly though, I imagine hydrogen will dominate the interactions so using the cross section of hydrogen will be a good approximation. Then assume some sort of approximate path length $\ell$ based on the geometry (maybe use the formula for chord length). Putting actual numbers into this problem could be a research project all to its own, but you can play around with what you consider to be good values for $N$ and $\ell$ such that you're constrained by the fact that $\tau=\sigma N\ell=1$.

  How large is this refraction - like 0.1 degrees or 10 degrees?

This is even harder to answer. The only way I know to determine something like this is to run a full ray tracing model of light through Saturn's atmosphere. I can hazard a guess that the most refraction you're seeing there is on the order of 10 degrees. This is supported by Dalba et al. 2015 who actually did a ray tracing analysis of light through Saturn (in regards to planetary transits) and found refraction angles on that order (ref. fig. 2).


Asteroid visible during the day? In a story I'm working on, a huge, 20 by 14 kilometer, Ni-Fe asteroid attains a 1760 km. orbit around earth (why/how? Long story). I figure it has an H magnitude of ~12. Would this body be visible naked-eye from the surface during the day?

The key to answering your question is to know the albedo of the asteroid compared to the Moon. The Moon is rather dark (reflecting 7% to 12% of the light if I remember correctly, about as black as worn asphalt). I think, but do not know with certainty, that the asteroid would not be much darker.

For the dimensions that you give, the asteroid would be approximately the same apparent size as the moon: 0.45 x 0.65° compared to the Moon's 0.5°.

So both have approximately the same albedo and approximately the same size. Therefore I say that the asteroid would be visible in the daylight.

Also this is a sci-fi story, correct? If you can get the asteroid into that orbit, then anything is possible. :-)


Problem about finding distance with magnitude given A star in Pleiades cluster has apparent magnitude of +12 ($m_{star} = +12$) determine it's distance from the sun? (We know the Sun properties such apparent and absolute magnitudes, distance from earth, luminosity, brightness (energy flux) and others but nothing is given about the other star expect that apparent magnitude).

My problem is that we don't have luminosity of the star. We can say:

$m_{sun} - m_{star} = -2.5 * \log_{10}{\frac{b_{sun}} {b_{star}}} \Rightarrow 10^{(-26.8 - 12)/(-2.5)} \approx 3.31131 * 10^{15} = \frac{b_{sun}} {b_{star}} \approx \frac{1370 Wm^-2}{L_{star} / (4  \pi  d^2)} $

As we can see, we don't have luminosity of the star. Can this question be solved with this information?

Short answer: no, you can't solve the question if all you know is its apparent magnitude. It could be a faint star that's nearby, or a very luminous star that's far away. (The Sun's parameters are irrelevant.)

(I'm ignoring the bit about it being "a star in the Pleiades", which implies that it's a physical member of the Pleiades cluster, in which you already know its distance: it's the same as the distance of the Pleiades cluster. Maybe you just meant "it's in the same part of the sky".)


If light has no mass, why is it affected by gravity? Light doesn’t accelerate in a gravitational field, which things with mass would do, because light has a universally constant velocity. Why is that exception?

Another way to answer this question is to apply the Equivalence Principle, which Einstein called his "happiest thought" (so you know it has to be good).  The equivalence principle says that if you are in an enclosed box in the presence of what Newton would call a gravitational field, then everything that happens in that box must be the same as if the box was not in a gravitational field, but accelerating upward instead.  So when you release a ball, you can imagine the ball is accelerated downward by gravity, or you can imagine everything but the ball is accelerated upward, and the ball is simply being left behind (which, ironically, checks better with the stresses you can easily detect on every object around you that are not present on the ball, including the feeling you are receiving from your bottom right now).

Given that rule, it is easy to see how light would be affected by gravity-- simply imagine shining a laser horizontally.  In the "left behind" reference frame, we see what would happen-- the beam would start from a sequentially higher and higher point, and that raising effect is accelerating.  So given the finite speed of light, the shape of the beam would appear to curve downward, and the beam would not strike the point on the wall of the box directly opposite the laser.  Therefore, this must also be what is perceived from inside the box-- the beam does not strike the point directly across from the laser (as that point is getting higher then the point across from it where the light was emitted), and its path appears to curve downward.  Ergo, light "falls."

Indeed, this is the crucial simplification of the Equivalence Principle-- you never need to know what the substance is, all substances "fall the same" because it's nothing happening to the substance, it is just the consequences of being "left behind" by whatever actually does have forces on it and is actually accelerating.

Incidentally, it is interesting to note that even in Newtonian gravity, massless objects would "fall the same" as those with mass, but to see it requires taking a limit.  Simply drop a ball in a vacuum, then a lower mass ball, then a lower still mass.  All objects fall the same under Newtonian gravity.   So simply proceed to the limit of zero mass, you will not see any difference along the path of that limit.  Nevertheless, Newtonian gravity doesn't get the answer quite right for the trajectory of light in gravity, because Newtonian physics doesn't treat the speed of light correctly.


Night-Day Cycles in S-Type Planetary Systems I am a writer and space enthusiast. For a story I am currently writing I would like to have a planet which orbits a star which resembles the sun, which in turn orbits another star. In this situation there would be multiple types of "days" including "bright day," when both suns are visible, "dim day," when only the further sun is visible, and "day," where only the nearer sun is visible. Is there any way to determine how this cycle would work? Any formulas or papers done? If so, what information would I need to have to figure it out? Thank you.

Although this question would probably be more suited for the worldbuilding.se, you're asking for scientific references, so I guess this is ok to ask here.

What you describe is a hierarchical system, akin to Sun-Jupiter-Anymoon. I assume the planet orbits star $S_1$ directly, and $S_2$ is the larger star, and the planet will be abbreviated with $P$.

I don't know any references, but just from geometric considerations I'll state a formula here that should work in a quasi-2D setting (what I mean by this, later).
The surface illumination function $I(t)$ should be a simple sum of the two-star contributions $$I(t)=I_1\,\sin(\omega_0 t) + I_2(t)\, \sin(\omega_1\,t), $$ where $\omega_0$ is the spin rotation frequency of P around itself. The amplitudes of the sinii are $I_1$, the constant amplitude related to the constant distance of $S_1$ to $P$, and $I_2(t)$ being the irradiation amplitude coming $S_2$. As example, $I_1$ could be a number like the familiar $1360 \, kW/m^2$ for the sun.
I wrote $I_2(t)$ as a function of time, because distance from the planet to $S_2$ varies over time, as it orbits $S_1$. Therefore, as simplest approximation, I would assume that this is also a sinus, hence $$I_2(t)=I_{2,flct}\; \sin(\omega_1\,t)+I_{2,avg},$$ with $I_{2,avg}$ being the average irradiation at $S_1$'s orbit around $S_2$, and $I_{2,flct}$ is given by the minimum and maximum distance of the planet towards $S_2$, and finally $\omega_1$ being the orbital frequency of P around $S_1$.

Once you have some numbers for the constants, you can use python, or any other routine, in order to investigate the behaviour of your day/night cycles. Here it is important that your routine can replace the negative contributions in $I(t)$ with zero values, otherwise the negative values of $I_1$ could 'eat' the positive values of $I_2(t)$ and instead of perpetual day you'd get a perpetual night from the formula, when P would be directly between $S_1$ and $S_2$. In python you would do this via some command like "I_updated = [0 if i < 0 else i for i in I_raw]", where "I_raw" would be the values from the individual $S_1$ or $S_2$ contributions, before summing them up to $I(t)$.

If you want to know 'which sun is up' at any given moment, you can plot the sinii separately.

This quickly scribbled formula ignores possible 3D-effects on the illumination, such as a possible Kozai resonance in your system, but also misrepresents 'shadowing' of $S_2$ by $S_1$. Therefore the formula represents a very slightly inclined orbital plane w.r.t $S_1$.

Please observe that your system, as you want to build it, might not be stable for too long:
For the hierarchical structure of your orbits to exist, the stars must have a mass ratio of at least $m_2/m_1\approx 23$. If $m_1$ is a sun-like G0 star, then $m_2$ will be an O9, which has a lifetime of only 4 million years on the main sequence.
However, if you assign $m_2$ to be the G0 star, then $m_1$ could be a M2 dwarf, and that should work out fine. Also remember, that in your chosen configuration, there's always one part of the planets orbit around $S_1$, namely when it's exactly between $S_1$ and $S_2$, where the whole planet experiences perpetual day, and depending on the choice of stellar type, might get quite cooked.

For more on three-body motion, I'd recommend diving into Murray & Dermott: Solar system dynamics.


Calculate eclipse days and duration of satellite If a satellite is revolving in a GEO orbit, then what will be the eclipse duration for it? How to calculate when the satellite will enter the dark side of the Earth?

Any formula or online tool would be appreciated.

I found a convenient formula to determine Eclipse (satellite in shadow). From Ismail, et al., Equation 10 here.

Image from same article:

So time in shadow = ((Theta-E)/360)*Orbital Period.

But how do you calculate Theta-E? Well, for my part, I used the law of similar triangles. You will have two similar isosceles triangles - one from the Earth to the point of the umbra cone and one from the Sun to the point of the umbra cone. Cut the isosceles triangles in half and use the right triangles. The sides of similar triangles are proportional and using the distance of 1 AU and the radii of the Sun and Earth, you can find the distance to the point of the umbra cone with a little algebra. Then use the arc-tangent to find the half-angle to the distance at GEO, then doubled it to find Theta-E.

You should be aware, however, that you are dealing with (1) circular orbits with (2) 0 inclination. The time of year also can change things.

My result was very close to that of James K. with only a few seconds difference.


If a black hole has a mass of a universe what will be the volume of it? Will it suck the entire universe in? What will the black hole look like to us, assuming we do not immediately get sucked in?

According to measurements of the cosmic microwave background the universe is geometrically flat - which means that the mass/energy density of the universe is close to the "critical value" of $\sim 10^{-26}$ kg/m$^{3}$.

The radius of the observable universe is 46.6 billion light years, so the mass/energy contained within it is equivalent to $3.6\times 10^{54}$ kg.

The Schwarzschild radius of a black hole is $2GM/c^2$. If the mass/energy of the universe is spherically symmetric then its Schwarzschild radius is 560 billion light years and thus larger than the observable universe.

Note though that the Schwarzschild solution in General Relativity is static. The universe is definitely not static.


Split telescope into two eyes I am looking for an appendage to attach to the end of a telescope which splits the image into two eyes instead of one.

Does such an appendage exist?

Absolutely.  You are looking for a binocular eyepiece, or a binocular viewer.  Most of them require you to insert two matching ocular eyepieces, so it can be fairly expensive.

Most major telescope eyepiece manufacturers sell them.  I'm not sure what the rules are for mentioning brand names,  but a quick google for 'telescope binocular viewer' will turn up many examples.

You will require a decent telescope that can take standard 1.25" or 2" eyepieces.  Also, the telescope needs a lot of 'back focus', which many Newtonian telescopes do not have. You may also need a barlow lens for focus, which will increase magnification but cut down the field of view and light gathering.

It would be best to check the specifications of the specific bino viewer and your telescope to see if they are compatible.  If you have a Schmidt-Cassegrain scope, you should be fine and don't need the barlow.


How to query a region on the sky when given only ecliptic coordinates? I am new at using the Python packages astropy/astroquery and what I want to achieve is to use it to find stars that lay on the TESS continuous view zone (CVZ).

I know the CVZ is centered at the south pole of the ecliptic an I would like to use the function astroquery.mast.Observations.query_region() to query a region centered at the coordinates of the pole. Such function uses coordinates in the international celestial reference system (ICRS) so first I have to convert from the ecliptic frame the pair $(\text{lon}=0^\circ , \text{lat}=-90^\circ)$.

I tried the conversion using the function transform_to(ICRS),however, it returns the following error:

  UnitsError: The input HeliocentricTrueEcliptic coordinates do not have length units. This probably means you created coordinates with lat/lon but no distance.  Heliocentric<->ICRS transforms cannot function in this case because there is an origin shift.

Could someone elaborate a bit more about the "distance" and "origin shift" mentioned in the error message and how to correct it?

This is because the ICRS has an origin at the barycenter and this is not the same as the center of the Sun (the heliocenter). The heliocenter is offset from the barycenter (although it's still within the radius of the Sun) and varies with time, primarily due to the perturbations of Jupiter and Saturn.

If you use BarycentricTrueEcliptic it will work fine:

from astropy.coordinates import BarycentricTrueEcliptic
from astropy import units as u

cvz = BarycentricTrueEcliptic(lon=0*u.deg, lat=-90*u.deg)
cvz.transform_to(ICRS)
Out[9]:
  <ICRS Coordinate: (ra, dec) in deg
  (89.9999853, -66.56071866)>

Which agrees with the Coordinate conversions using the LAMBDA website at NASA's HEASARC


IAU 2018 Exoplanet definition In 2018 the IAU produced a definition of an exoplanet:

The current official wokring definition of an exoplanet as amended by IAU Comission F2: Exoplanets and the Solar System in August 2018, reads as follows:

Objects with true masses below the limiting mass for thermonuclear fusion of deuterium (currently calculated to be 13 Jupiter masses for objects of solare metallicity) that orbit stars, brown dwarfs or stellar remnants and that have a mass ratio with the central object below the L4/L5 instability ($M/M_{\rm central} < 2/(25+ \sqrt{621})\approx 1/25 $) are "planets" (no matter how they formed).
The minimal mass/ size required for an extrasolar object to be considered should be the same as that used in our Solar System.
We can expect this definition to evolve as our knowledge improves.

What is the L4/L5 instability and why was it chosen as part of the exoplanet definition?

"What" is straightforward: L4 and L5 are two points on an orbit that are 60 degrees ahead and 60 degrees behind the planet, and so move as the planet orbits.  If the planet is less than about 1/25 of the mass of the central object, then a combination of centrifugal and Coriolis forces will cause the L4 and L5 points to be places where a third body can have a stable orbit.
Asteroids that have orbits at (or near) the L4 and L5 points are known as "Trojans" Jupiter has hundreds of Trojans that share its orbit, even Earth has one 2010 TK7 (and perhaps more).
"Why" is harder.  It seems that this is chosen to ensure that the "planet" is significantly smaller in mass than the central "star". So that we can clearly claim that "the planet orbits the star" and not "there are two roughly equal-sized bodies in orbit around each other".  Such configurations would be a binary system and not a planet in orbit.
The significance would be in the nomenclature.  In a binary (or multiple) system, the objects are named "on the same level" (for example Alpha-Cen A, Alpha Cen B and Alpha Cen C)  Planets, however, are named at a lower level, with lower case letters (Alpha Cen Cb) There isn't really another reason for specifying an upper limit to the size of second body.
It also seems to be a proxy for the mode of formation (this is hinted at by the parenthetical comment). Two roughly equal bodies will have formed by gravitational collapse of a cloud of gas that split in two as it collapsed to form two objects mutually orbiting each other.  Whereas classical "planets" form in the disc surrounding a proto-star as a secondary process. But rather than use the unobservable "mode of formation" as a definition, the IAU (in its infinte wisdom) chooses the astronomically meaningful L4/L5 instability


safeguards against comets? I read one of the related questions here, but what i'm curious to know is how do we deflect imminent collision courses of comets, if we detect them at a later stage? I read somewhere else that there are systems which use powerful lasers (radiation pressure) to deflect them, but honestly, i don't understand why something like radiation pressure will work on something travelling at about 40km/s through space, and still manage to cause a large enough deflection. Are there other methods ?

Fortunately it is not something that we have yet had to face.

Diverting the impact of a comet would not be easy, especially if it was a newly discovered comet on a nearly parabolic orbit. We would probably have less than 2 years to prepare.

Remember that the Earth is orbiting the sun at 30 km/s. So if we can delay the comet by three or four minutes, that turns a direct hit into a miss. But most of the plans to divert objects apply to asteroids whose orbit is well known and for which we could have tens of years in which to act. For a new comet there may not be time.


Fusion of elements inside heavy stars I am confused with nucleosynthesis inside supernovae. I have read that the heavier elements are made through fusion of lighter element's namely hydrogen and helium.

Does the star "store" all the elements they fuse inside themselves or are they radiated out?

Example, A star fused hydrogen into helium and 2 helium atoms into carbon. Would it keep fusing until it hits the limit and store it inside only to be released as an explosion.

You are correct to say that all the heavier elements in the universe were formed in stars.

Stars like the sun fuse hydrogen into helium. When they get older they can fuse the helium into carbon (it actually takes 3 helium to make one carbon). Larger stars can fuse carbon into oxygen, and neon and elements in the first half of the periodic table.

When the star runs out of fuel, the outer layers are gently ejected in what is called a planetary nebula (though it has directly to do with actual planets). The ejected gas is enriched with the heavier elements that the star has fused. The heavier elements in the planetary nebula mix with the hydrogen and helium gas in outer space and can later go on to be found in new stars. Most of the carbon and oxygen and nitrogen on earth was formed by this process.

Very large stars will fuse all the way up to iron, and then collapse in a supernova. This releases a very large amount of energy, some of which is used to form elements heavier than iron. All the heavier elements (copper gold, uranium for example) are formed in supernovae.

The supernovae explosion distributes these elements back into space, mixed in with the debris from the star, as time goes on, and stars are born and die, the gas in space gets enriched with more and more heavier elements

So the heavier elements are formed in the cores of stars, and are stored there until the star dies, when some of the star's atoms are released into space. The heavier elements are not "radiated out" of the star, until the star dies. Some stars die in a supernova explosian, but most have a more peaceful death in a planetary nebula.


How to convert from FELO-HEL to Km/s? I am using DS9 to calculate the rotational velocity of a galaxy. I have 21cm cube and  I used DS9 to plot this graph

I do not understand what is FELO-HEL! and I want to convert the units from FELO-HEL to Km/s?

The axis type ‘FELO’ is regularly gridded in frequency but expressed in velocity units in the optical convention. The unit here is   $m/s$.

This means that the wavelength/frequency has already been expressed as velocity corresponding to a Doppler shift around a reference wavelength. This velocity is given by
$$
v = c \ \frac{\lambda - \lambda_0}{\lambda_0  } ,
$$
where $\lambda$ is the measured wavelength and $\lambda_0$ the reference wavelength in a rest frame.

The HEL term on the axis means that velocities have been transferred to the solar systems  barycenter (HELiocentric). This means, the velocity of earth around the sun and its rotation has already been removed.

Your distribution, therefore, has a width of about 200 km/s.
And the projected rotational velocity of the galaxy $v_\mathrm{rot} \sin(i) \approx 100\,$km/s, where $i$ is the galaxies inclination.

The offset of 550 km/s probably comes from the relative speed between our sun and the galaxy, but that is hard to tell without knowing which galaxy it is.

Some more information about frequency and velocity formats can be found in this paper


Could the mineral composition of planets explain the absence of megastructures? The Fermi paradox concerns why we don't observe evidence of alien life, such as megastructures. If we were to build a Dyson swarm, the best source material may be Mercury (Armstrong and Sandberg 2013): not only is it close, with plenty of solar power for exponential disassembly, but its mineral composition is suitable. But could planets made of suitable minerals, with plenty of solar power, be rare enough to prevent most advanced civilizations building Dyson spheres, or other detectable megastructures?

It's a mundane solution to the Fermi paradox, but it's conceivable on my limited knowledge of exoplanets, if only because our planetary system is atypical. For example, where we have Mercury, many systems would have hot Jupiters instead (or possibly as well, if that's consistent with how most systems form). Can we begin to answer this with our current understanding of what's typical for exoplanets?

In our paper we just used Mercury because it was convenient (close, airless, low binding energy). But we did not try to optimize the Dyson sphere design at all. It turns out that one can likely construct a far more efficient sphere out of very thin aluminium foil reflectors balancing light pressure and solar gravity to stay in place. That only takes the mass of the asteroid Vesta or so. Plus, were the reflectors manufatured out in the asteroid belt one could light-sail them into close orbits with modest effort.

Aluminium is just one possible material. Essentially any metal works for the reflectors (the trick is to keep the mass low enough), while energy gathering subsystems could be based on semiconductors or even heat turbines - silicates and water are not uncommon in the solar system, and presumably similarly available in others. Even if the inner region has been swept clean by a hot jupiter parts can be manufactured and spiraled in from the outer system.

So unless we expect even asteroids to be rare (a pretty implausible idea) there is likely plenty of building material. As an answer to the Fermi paradox lack of resources is not reliable enough.


Why does time get slow near a black hole? Why exactly does time get slow around a black hole? And does it provide a way to go into the future?

The simplified answer is: Due to gravitational time dilation.

That's a consequence of general relativity. One fundamental idea of general relativity is the equivalence of inertial and gravitational mass, meaning forces induced by acceleration aren't distinguishable from gravity.
Persuing this principle together with the constant speed of light lead to Einstein's field equations. One of the solutions of these equations is the Schwarzschild solution for non-rotating, electrically neutral black holes.
This solution can be used to derive gravitational time dilation, as part of spacetime warp near mass/energy.
(Introductory video to general relativity)

A black hole provides a way to go faster into the future of a region of higher gravitational potential away from the black hole.
But the body near the black hole needs to withstand the high gravity. Circular orbits around a black hole are only possible down to 3/2 times, stable only down to 3 times, the Schwarzschild radius.


What are criteria to compare and evaluate different 12 inch truss Dobsonian telescopes? What are criteria to compare and evaluate different 12 inch truss Dobsonian telescopes? I'm primarily interested in comparing mirror quality and weight (price I can do myself of course). Two examples of the category I'm asking about are the Sky-watcher 300P Flextube and Meade Lightbridge.

One of them appears to have a fan to cool the mirror, how can I evaluate that in terms of a positive or negative towards performance?

The cooldown fan is purely to speed getting the primary mirror to the local ambient temperature. It is not used - and should not be left on during observation, as it'll undoubtedly be a vibration source.

That shouldn't be a significant factor in your final decision.  I would go with total weight, and perhaps adaptability to an equatorial mount for star-tracking.


Has any moon achieved "retrograde equatorial orbit"? There are many moons which have low (almost negligible) inclination and can be considered to rotate at the planet's equatorial plane. For instance, Galilean moons have almost negligible inclination (< 1$^\circ$) and so they are considered to have a "prograde equatorial orbit".
OTOH, "Retrograde equatorial orbit" is that orbit which has an inclination of exactly 180° (or near 180°). Has any moon achieved such kind of orbit? I have found some Saturnian moons (or moonlets) which have inclination around 177° and didn't found anything beyond that. Why is it hard to achieve a "Retrograde equatorial orbit"?

Tangentially related questions: Can a moon rotate perpendicular to planet's plane? [90° inclination] (Theoretically yes, practically hard to achieve; See Kozai Mechanism)

Physics.SE
WorldBuilding.SE

Also related:  Are all satellites of all planets in the same plane?

This is directly related to another question: Why are asteroids with zero orbital inclination rare?
If captured, irregular moons are randomly oriented in space then there is very little chance of them having either inclination angles near zero or near $180^\circ$. This is because, if they are uniformly distributed in space, the fraction of orbits within a given inclination range $i_1 < i < i_2$ is given by
$$ f = \frac{\int^{i_2}_{i_1} \sin i\ di}{\int_{0}^{180^\circ} \sin i\ di}$$
$$f = \frac{1}{2}\left(\cos i_1 - \cos i_2\right)$$
Thus for $i_1 =177^\circ$ and $i_2=180^\circ$, $f= 0.00068$ (i.e. about 0.07%).
Given that there are only $\sim 100$ irregular satellites, it is not that surprising that none are found with $i>177^\circ$.
Note there are physical reasons why the regular satellites have low inclinations - they are not uniformly distributed and therefore they can have very low inclinations.


Image sets for testing stacking algorithms? I am looking for sets of astronomical images for testing different kinds of stacking algorithms.

The idea is simple: if one has $N$ images of the same object, the signal-to-noise ratio of the averaged image increases with $\sqrt{N}$ under certain conditions.

Where could I obtain data for testing?

PS. I would propose adding the tags "image", "ccd", "research", "signal processing", and "machine learning".

It's pretty easy to create a dataset for yourself using remotely controlled telescopes. I've used iTelescope in the past and had a good experience. Their "starter" account costs $20 and should be good to get you going.


What was the nature of the signal received from the direction of HD164595? There's been a story in the news recently of a potential SETI candidate signal received from the vicinity of star HD164595.  What's the nature of this signal?  Is it a single pulse of energy, or does it change over time?  Can someone point me to a link or paper than shows it in more detail?

The page you link to contains what is, to my knowledge, the most descriptive graph of the signal from the RATAN-600 data that is open to the public. Bursov et al. have not yet presented their results, which will be explained at the International Academy of Astronautics SETI Permanent Committee meeting late this September. Chances seem to be good that we won't see much more information until then.
Here is the graph at a frequency of 11.1 GHz, as will be presented by Bursov et al. and as used in Croft et al. (2016), who very quickly put together a paper with an analysis of some basic follow-up observations using the Green Bank Telescope:

The main blip lasted only for two to three seconds - two seconds seems to be the popularly cited figure. Since then, no other group looking at HD 164595 has detected a repeat of the signal, and Croft et al. were unable to find anything similar in records in the NVSS catalog. Continuum imaging revealed nothing, and the authors concluded that the signal may have been due to observational error - although a one-time event is not out of the question.
Again, though, there is no more information forthcoming from the RATAN-600 group, and I expect this will be the case until the results are formally presented in full this September. Until then, this is all the data we have, but it appears to be the most important bit of data.

Update, 14 September 2016
The Russian Academy of Sciences recently released a statement saying

Subsequent processing and analysis of the signal revealed its most probable terrestrial origin.

I'm not aware if the "processing and analysis" was done by the RATAN team or an outside group, but it seems more and more like the signal is not in fact from an extraterrestrial civilization. We'll probably find out more details two weeks from now.

Update, 25 October 2016
I have yet to see any published results from the RATAN group after the IAA meeting about a month ago, and I haven't seen any articles on the findings. The IAA has some information on the meeting, but it has yet to add details to its Minutes page. Perhaps that will be released at a later date - and I assume it will be - but that could be a while. In the meantime, I'm taking the lack of information to mean that the RATAN group agrees with the analysis by Croft et al., and that the signal was a false positive.


Can the centripetal force be inverted? Using the centripetal force to create artificial gravity has been proposed for space exploration. From our perspective, the force is an outward force, pushing objects away. Could it be inverted to attract them instead?

Could we, instead of having a spacecraft, or a "dish," rotate to create the force that would push the travelers to the inside of the hull - could we instead have a cluster of dishes rotating in such a manner that the travelers are pressed to the outside of a sphere?

Or, could we construct an artificial mini-planet with a gravity exceeding by far the gravity of a natural celestial body with the same mass?

Close to a black hole, but still outside the event horizon, orbiting arbitrarily fast, wouldn't result in a circular orbit, but in a trajectory towards the black hole.
In this extreme scenario, centrifugal force could be interpreted as attractive.

In more usual settings, centrifugal forces point outward.
Since the centrifugal force (in usual settings) $m \omega^2r$ is proportional to square of the angular velocity $\omega$, you would need an imgaginary angular velocity, or alternatively a negative mass to get an inward pointing centrifugal force.

I'm not aware of a physical realization of imaginary angular velocities, at least not for circular orbits,
although this paper introduces imaginary velocities for hyperbolic orbits. This may be derived from imaginary circular angles.
Negative masses aren't entirely impossible, but those potentially feasible are too tiny to be of any practical value in your sense.


What does a black dwarf look like? A black dwarf is a hypothetical object that is the end result of the cooling of a white dwarf. None yet exist, because there hasn't been enough time in the age of the universe for them to cool down sufficiently. The oldest known white dwarfs are at ~3800 K and would glow orange. It will take many billions of years before any are cool enough (~750 K I believe) to not emit light visible to the naked eye.

So, what would a black dwarf look like? I'm assuming it is illuminated by some other source of light such as a star.

The question could be rephrased: What is the visible absorption (and emission) spectrum of electron degenerate matter?

This is my attempt at an answer: Probably not black. A lot would depend on whether electron-degenerate matter behaves as metallic or dielectric, based on its band-structure. My guess would be metallic, considering the electrons are dissociated from their parent nuclei in electron-degenerate matter.

Because the surface gravity of a black dwarf would be so strong, I'd expect the surface to be very smooth and relatively free of Lambertian scattering.

From these two features, I'd make a tentative prediction that a black dwarf would appear as a mirror-like sphere.

The material at the surface of a white dwarf is not degenerate.
The "visible" surface is defined as where the optical depth exceeds some threshold and this will occur at a low enough density that even at a few hundred kelvin, the ratio of the Fermi energy to the thermal energy is too low for significant degeneracy.

In addition, at these temperatures, the electrons attached to atomic hydrogen and helium would not be ionised and the material would mostly be in atomic and molecular states.

In terms of appearance a reasonable comparison would be with that of the very cool T- and Y-dwarfs; although some of the more complex chemistry that occurs in those atmospheres would not occur in very cool white dwarfs because the trace quantities of heavier elements would probably have sunk out of the atmosphere.

Edit: I note that there probably are white dwarfs cooler than 3000K. This is the upper limit to the temperature of an invisible companion of to the pulsar PSR 2222-0137. Kaplan et al. (2014) say that this most likely a massive ($\sim 1 M_{\odot}$) old white dwarf that has cooled so much that they cannot detect it at visible wavelengths - so probably you would call this a "black dwarf", though the term is not used in the scientific literature.


What is the farthest object we've been able to bounce signals off of to date? I know people bounced lasers off of Venus and Mercury in 1960s in order to test the Shapiro delay. Since then, we've developed higher power lasers and radio transmitters (Bolshakovo transmitter).

Are there any farther objects where we've been able to detect the reflected wave?

1977, Arecibo: Galilean Satellites of Jupiter: 12.6-Centimeter Radar Observations Radar echoes.

1973, 12.6 cm: Radar Observations of the rings of Saturn Content is hidden behind Elsevier's paywall, so this might not be radar echoes.

NASA, 2006 implies the limit for radar echoes is at the moons of Saturn:

  Radar astronomy employs the world's most massive dish-shaped antennas, which beam directed microwave signals at their targets, which can be as close as our moon and as far away as the moons of Saturn.

Looked for studies w radar echoes from moons of Uranus and Neptune, as well as Pluto/Charon, none found. The data may be out there.


Which Chandrasekhar Limit do I use? 1.39 or 1.44? Different sources online say that the Chandrasekhar Limit is either 1.40 or 1.39, or 1.44 solar masses. Why the discrepancy? I heard it might have to do with the composition of the white dwarf, but, correct me if I'm wrong, don't pretty much all white dwarf stars have the same composition?

The "classic" Chandrasekhar mass is given by
$$ M_{\rm Ch} = 1.445 \left(\frac{\mu_e}{2}\right)^{-2}\ M_{\odot},$$
where $\mu_e$ is the number of mass units per electron ($\mu_e =2$ for ionised carbon, oxygen or helium; $\mu_e = 1$ for hydrogen, $\mu_e= 56/26$ for iron (56)).

This assumes a white dwarf star of uniform, ionised composition and an equation of state given by that for ideal fermions (electrons in this case). It also ignores rotation and it uses Newtonian gravity. When defined in this way, the Chandrasekhar mass is obviously composition dependent - but most white dwarfs are expected to be formed from something with $\mu_e =2$.

In practice, a white dwarf can never become this massive. There are (at least) 4 effects that can mean the de facto Chandrasekhar mass (if it is taken to mean the maximum mass of a stable white dwarf) and at least one effect that can increase it.

The effects that reduce it:

Electrostatic interactions The electrons and ions do not form an ideal Fermi gas because of Coulomb interactions. The net result is to make a white dwarf slightly more compressible and the maximum mass about 2% lower. The correction is composition dependent; it is stronger for white dwarfs made of material with a larger atomic number.

General Relativity Massive white dwarfs are strongly affected by General Relativity. White dwarfs will become unstable at a finite density when the hydrostatic equilibrium is treated with GR. This finite density is reached at about $1.38-1.39 M_{\odot}$.

Inverse beta decay At high densities the electron Fermi energy becomes high enough to initiate inverse beta decay reactions. Electrons are captured by protons in the nuclei and this renders the star unstable. For a Carbon white dwarfs this will also happen at about $1.39 M_{\odot}$, for oxygen the threshold is lower at $\sim 1.37M_{\odot}$.

Pyconuclear reactions At high densities, even at low temperatures, fusion reactions can be initiated by quantum tunnelling. This changes the composition of the white dwarf and can change $\mu_e$ or lower the density threshold for inverse beta decay making the star unstable.

The effect that can increase the maximum mass for stability is rotation. Some authors have claimed that the limit can be increased to as high as $2.6M_{\odot}$ in certain circumstances, though these are usually referred to as Super-Chandrasekhar mass (e.g. Das & Mukhopadhyay 2013) and are not stable. A stability analysis of rotating white dwarfs in GR found that the maximum stable mass was only increased to around $1.47 M_{\odot}$ for a Carbon white dwarf (Boshkayev et al. 2012), but these would be rotating faster than any observed white dwarfs.


Was Milky Way formed after Sagittarius A Explosion? Is it possible that most galaxies (and our galaxy) were formed from leftovers of supernova explosions?

In order to have a black hole 4,000,000 x solar mass the initial mass of the star should have been even greater.

So I was thinking, is it possible that after the big bang, gas formed much bigger stars 1,000,000,000,000,000,000,000x bigger than the sun, which later exploded forming:

black holes in the middle of future galaxies
Gas, which later created smaller stars, planets, etc.

Am I trying to reinvent a bicycle here?

This is ruled out because that's not what we see when we look back in the past.

If this was the case at the beginning, we would see these huge stars, with probably different characteristics from the ones formed only recently. But we don't see anything like that.

From what we see when we look very far away - that's few hundreds of millions of years after the beginning - we see signatures of first generation stars forming, very big stars indeed, but nothing close to what you're talking about.


How can artificial gravity hold the water in swimming pool? I was watching Passengers (2016). In this movie, a swimming pool exists in the space station. In middle of the movie, Artificial Gravity is turned off and the water rises, with the swimmer suffocating as they try to get out of the water.

Suddenly Artificial Gravity turned itself on again and the swimmer survives.

A clip of this scene is found on YouTube.

However, water is an anti-ferromagnetic material - it can't attract to magnetic bodies.

How then, can artificial gravity hold the water in swimming pool?

My question is totally different from this question.

"Artificial gravity" should act like gravity. So just as gravity has a direction on earth and holds water in a pool, so should  "artificial gravity" hold water in a pool on a space station with  "artificial gravity." Notice that bubbles rise in the space station pool just as they would on earth.

If gravity were loss on a space station with a pool, then the surface tension of the water would want to pull the water into a sphere.

I'll also point out that water is an anti-ferromagnetic material in earth swimming pools, and gravity works just fine. So using magnetic boots on a steel deck is very different that having gravity which would work with an aluminum deck.


Solar Azimuth angle Derivation I tried posting this on math stack exchange to no avail so I thought maybe it's more relevant here :
I know a similar question has been posted here before but no links to any mathematical derivations given on the following post. I was not able to access the citations on the wiki page either.
How do I calculate the sun's azimuth based on zenith, hour angle, declination, and latitude?
I was recently trying to derive the position of the Sun in the sky given the latitude, time of year and time of day.
I used the following definitions:
$\phi :$ I used this to denote the angle of latitude, as per convention
$t:$ I used this to denote the angle of the year completed since Summer solstice
$\theta:$ I used this to denote the angle of the day completed since solar noon
$\Omega:$ I used this to denote the solar zenith angle.
$\Gamma:$ I used this to denote the azimuth of the Sun clockwise from due North
$k:$ I used this to denote the angle of 66.56 degrees between the Earth's axis and plane of revolution
$\lambda:$  I used this to denote the angle between the Earth's axis and the vector in the direction of the sun, this is complementary to the more commonly used declination angle $\delta$
My goal was to get a function $f(\phi, t, k,\theta)=\Omega,\Gamma$
I was able (with some help) to derive the following formulae:
$\cos(\lambda)= \cos(t)*\cos(k)$
$\cos(\Omega)=\cos(\lambda)*\sin(\phi)+\cos(\phi)*\sin(\lambda)*\cos(\theta)$
However, I was not able to then derive the formula given on the following Wikipedia page for $\Gamma=f(\theta,\lambda,\Omega)$ which I have transcribed into my variables below:
https://en.wikipedia.org/wiki/Solar_azimuth_angle
$\sin(\Gamma)=-\dfrac{(\sin(\theta)*\sin(\lambda))}{\sin(\Omega)}$
I think in order to derive this I need to convert to a local co-ordinate system with due North along the y-axis, due East along the x-axis and zenith along the z-axis.
Can someone help me to proceed?

What I think you're trying to do is workable, but approximate. You first find the Sun's geographical position (GP), the point on the Earth where the Sun is in zenith at the time of the observation. You start by finding the Sun's GP latitude (its declination). It's approximately 23 degrees times the sine of ((days since the vernal equinox)times 360/365). Then you find its longitude as the ((hours since local apparent noon)times(15 degrees) + observer's west longitude). You will have three points, vertices of a spherical triangle: observer's location, Sun's GP, and the north pole. The angle at the pole is the difference between the longitudes of the Sun's GP and the observer. The sides between the pole and the other two points are 90 degrees minus their latitudes. All that's needed to solve a spherical triangle is the arc length of two sides and their included angle. Using the law of cosines for spherical triangles, find the arc distance between the observer and the Sun's GP. The altitude of the sun will be 90 degrees minus that number. Find the angle at the observer using the law of sines for spherical triangles. That will be the Sun's azimuth.


Can a black hole turn back into a star? I understand that a black hole happens after a star exhaustions elements capable of undergoing fusion.  Without the energy produced by fusion the remaining mass of the star collapses on itself and forms a black hole.  I conclude black holes must be deficient in hydrogen, helium and elements that undergo fusion in stars.  I understand that hydrogen and the like undergoes fusion in stars in part (or completely?) because of the gravitational pressure which occurs in stars.

Suppose a young star replete with hydrogen merged with a black hole.  My question: would the result be a larger black hole, or a radiant star with a very massive heart?  Could a massive infusion of (fusible?) elements reboot the black hole back into a star?

No.

A black hole is characterised by the formation of an event horizon. Matter which travels past an event horizon cannot return to the space outside the event horizon, to do so would require the matter to travel faster than light, and so would require infinite energy.

There is nothing like a regular star "inside" a black hole. The general relativity model of a black hole contains a singularity. Matter which enters a black hole is certain to reach this singularity in a (usually short) amount of time. Adding hydrogen to a black hole would just make the black hole more massive. There is a theorem in mathematical physics that the only properties that a black hole can have are mass, angular momentum and charge. In other words, a black hole made of collapsed hydrogen would be identical to one made of collapsed iron. There is no matter in a black hole, only mass.

The formation of a black hole normally occurs when a core in a massive star has a certain amount of iron, which saps the core of energy causing collapse. If the resulting object has more than about 3 solar masses, then nothing in the universe can stop it from collapsing all the way to a singularity.

As with much about general relativity, your intuition is a very bad guide to how black holes work. However General relativity is a very good model for how black holes should form and behave


Timekeeping on other celestial bodies than Earth I have looked for months for any site capable of calculating solar altitude w.r.t an observer located on the surface of a generic celestial bodies, but with no luck as of now. Even NASA does not provide such information in its Horizons page/software, although technically Horizons contains all the needed data to calculate "local solar time"; indeed, Horizons allows asking for such an output... but currently output is jut "n/a" for bodies other than Earth:
*******************************************************************************
 Date__(UT)__HR:MN, , , L_Ap_Sid_Time,  L_Ap_SOL_Time,  L_Ap_Hour_Ang,
**********************************************************************
$$SOE
 2020-Aug-09 10:51,*,x,          n.a.,           n.a.,           n.a.,
 2020-Aug-09 11:51,*,x,          n.a.,           n.a.,           n.a.,

Url to ask for quantities 7, 34 and 42 (note: only 42 is declared as "Earth surface only").
https://ssd.jpl.nasa.gov/horizons_batch.cgi?batch=1&COMMAND=%27-85%27&CENTER=%27coord@301%27&OBJ_DATA=%27yes%27&MAKE_EPHEM=%27yes%27&TABLE_TYPE=%27OBSERVER%27&REF_PLANE=%27ECLIPTIC%27&COORD_TYPE=%27GEODETIC%27&SITE_COORD=%2723.4729700,0.67417634,0%27&START_TIME=%272020/8/9%2010:51%27&STOP_TIME=%272020/8/10%2010:51%27&STEP_SIZE=%271h%27&QUANTITIES=%277,34,42%27&FIXED_QUANTITIES=%27Custom%27&REF_SYSTEM=%27J2000%27&OUT_UNITS=%27KM-S%27&VECT_TABLE=%273%27&VECT_CORR=%27NONE%27&CAL_FORMAT=%27CAL%27&ANG_FORMAT=%27HMS%27&APPARENT=%27AIRLESS%27&TIME_TYPE=%27UTC%27&TIME_DIGITS=%27MINUTES%27&RANGE_UNITS=%27AU%27&SUPPRESS_RANGE_RATE=%27no%27&SKIP_DAYLT=%27no%27&EXTRA_PREC=%27yes%27&CSV_FORMAT=%27yes%27&VEC_LABELS=%27yes%27&ELM_LABELS=%27yes%27&TP_TYPE=%27ABSOLUTE%27&R_T_S_ONLY=%27NO%27&CA_TABLE_TYPE=%27STANDARD%27
So I tried to setup my own page ( http://win98.altervista.org/space/exploration/moon/moontime.html ), which makes its calculations starting from "colongitude" and moon terminator longitude, but it completely disregards observer latitude and altitude.
Any suggestion about pages which perform these calculations?

My GUI for NASA Horizons:
http://win98.altervista.org/space/exploration/NHUGUI.html
To specify a location on another body, write "coord@body" in "CENTER" field, without changing "COMMAND" field, which will be ignored, and write the triplet "lon, lat, alt" in field SITE_COORD.
To find predefined sites on a body use *@body , for example *@301 will list Moon landing aites.

The HORIZONS web interface allows for you to change the observer location to other bodies in the solar system.  Select "Observer location [change] and read the instructions carefully.  I was able to generate an ephemeris showing the Sun's position for the Viking lander location by entering in "Viking 1@499".  Mars is designated by "@499" and it allows you to enter in any Mars surface coords (Mars lat/long) before the @499 to generate the ephemeris.  There is a search function to find the code for nearly anything in the solar system; it will always be @ something.
To determine when the Sun rises and sets on Mars, you'll also need to change the columns returned to include "Apparent AZ/EL" (azimuth and elevation over the horizon).  Also, you'll want to have it run for the accuracy you're looking for, such as every 5 mins or every 1 min, etc. Here is a screenshot using 10 mins for the step time.


How do scientists name space objects? I am curious that there are lots of celestial objects (planets, comets, stars, galaxies, etc.) in space. How do we name them?

Millennia ago, when only a few thousand stars, a handful of planets and nebulae, and some transient objects like supernovae and comets, were known, people usually named these objects after gods and heros, but also after everyday objects. Stellar constellations that vaguely resembled something known, was named after this. Mars is red, so it was named after the god of war; Mercury, being the fastest to complete one cycle around the Sun, was named the god of messaging and traveling.
With the advent of telescopes, so many objects are now known that we had to come up with a more pragmatic convention, so usually an object is named according to its place in some catalogue, its type, and/or it coordianates on the sky. For instance, a quasar detected at the "J2000" coordinates $\{\mathrm{dec.},\mathrm{RA}\} = \{22^\circ\, 22'\, 56'', -09^\mathrm{h}\, 46^\mathrm{m}\, 36^\mathrm{s}\}$ may be called J2222–0946. An x-ray source observed in the constellation Cygnus may be called Cyg X-1, while its binary counterpart is number 226,868 in the Henry Draper Extension Catalogue and is thus named "HDE 226868" (not to be confused with one of our site moderators).
Oftentimes, an object appears in multiple catalogues, and will thus have multiple names. For instance, a Lyman $\alpha$-emitting galaxy appears in a catalogue of Ly$\alpha$ emitters, but due to its infrared properties, it will also appear in some catalogue over dusty galaxies (though Ly$\alpha$ emitters tend to contain little dust).


What is the significance of minus sign in the expression of apparent magnitude? The apparent magnitude of a distant object in the sky is measured as
m=(-)2.5 log (I/I0)
Where I is the intensity of the object whose apparent magnitude is to be measured and I0 is the reference magnitude. My question is why we put the minus sign in the expression.

So if I read your question correctly, you're asking why as stars get dimmer, they are given higher magnitudes?

The reason is purely historical. The ancient greeks assigned stars with 6 brightness levels or magnitudes. The brightest stars were of first magnitude and the least bright stars (to the naked eye) were of 6th magnitude.

In the 1800's this system was formalised by an astronomer named Norman Robert Pogson. By then they were (presumably) able to more accurately measure the brightness of a star, and he decided to use a system that fitted well with the ancient measurements, but could be used to interpolate and extrapolate to brighter and dimmer objects. In order to do this, the formula has to have a negative in front of the log.

More information here


What's that limit, where we start to see infrared galaxies? All the nearby galaxies we see come in visible range, many are visible from ground based telescopes. And the farthest galaxy that hubble discovered recently, 13.4 billion light years away is in infra red.
Was this Galaxy ever in visible part of spectrum?
How long will it take galaxies like andromeda or whirlpool to go infrared so that they won't be visible to us if we are not using in infraRed lens?
What's the specific limit where the light makes transition from visible to infrared part of spectrum?

Galaxies give off a range of frequencies of light: Visible, but also infrared, microwave, and radiowaves. at longer wavelengths, ultraviolet, Xrays and gamma rays at short wavelengths. But stars give off quite a lot of their radiation in visible light, so nearby galaxies are quite bright in visible light and rather less bright in Ultraviolet.

Very distant galaxies are moving away from us due to the general expansion of the galaxy. This means that the light is stretched. Visible light gets stretched to infra-red, ultraviolet gets stretched to visible light. The distant galaxy is very very far away, and so very dim. It is a little bit brighter in infrared, although there is also a little visible light (and other wavelengths too)

Galaxies never go completely invisible in visible light, even if they are red-shifted, because there is always ultraviolet light to red-shift to visible.

The specific limit is often taken to be 700nm as the boundary between red and infrared. Although this is fuzzy because some people can see a little further than others.


What is the difference between our time and space time? I am trying to understand the phenomenon of space-time. But, everything on internet seems to be too complicated for me since I do not have a background in physics. Can anyone give me simple explanations for the following:

What is the difference between time and space-time?
How does gravity affect the passage of time?
What is the speed of light and how does it relate to time?
How do scientists deal with timescales on the order of billions of years if time is not constant for all observers in the universe?
How is time, or for example the age of the universe, actually measured experimentally?

What is the difference between time and space-time?

Space-time is time plus space.

  How does gravity affect the passage of time?

The higher the gravity of a planet or star and the closer to that body the slower the time.

  What is the speed of light and how does it relate to time?

The speed of light is 299,792.4580 km/s in vacuum, the speed at which light propagates, roughly 1.3 seconds from Earth to Moon. Velocity is distance divided by time; this applies also to the speed of light.

  How do scientists deal with timescales on the order of billions of years if time is not constant for all observers in the universe?

They treat time dependent of the observer. For different observers on Earth variations are tiny, in many cases neglectable in comparison to measurement errors, although not for precision measurements.

  How is time, or for example the age of the universe, actually measured experimentally?

There are many ways to measure or estimate ages. The age of the earth can be estimated by ratios of certain radionuclides in the oldest rocks. Estimates of the age of the universe are obtained by simulations based on observations (redshift and distance) of distant galaxies, and on observations of the cosmic microwave background. The redshift is used for a velocity estimate; together with a distance estimate (obtained by stars of known brightness) one can calculate back, when all objects of the universe should have been close together.


what is the percentage of stars with planetary systems? I'm having hard times understanding the equation of Drake, i can´t find the number of the variable that refers to the stars with planetary systems (fp)

For the purposes of the Drake equation you may as well assume that every star has a solar system.

At present, the exact fraction is unknown since the search techniques are limited to finding planets with certain characteristics. For example, transit searches tend to find close-in, giant planets; doppler shift surveys are also most sensitive to massive planets with short orbital periods, and so-on.

One can try and account for this incompleteness, but although some examples have been found, we are in the dark about the fraction of stars that have an Earth-sized planet at distances further than the Earth is from the Sun, or the fraction of stars that have any kind of planet orbiting beyond where Saturn is in our solar system.

For the Drake equation, you need to know the fraction of stars with planetary systems times the number of "habitable" planets per system is. There have been attempts to estimate this from Kepler data.

Possibly the best estimate at present is from Petigura et al. (2013), who estimate that 22% of sun-like stars have an earth-sized (1-2 Earth radii) planet that receives between 0.25-4 times the radiative flux of the Earth.

Obviously this is a lower limit because it doesn't include planets smaller than the Earth. It also doesn't include the moons of giant planets, which may also be habitable.

We still know very little about this fraction when it comes to M-dwarf stars, which are the most common type of star in the Galaxy...


Where can I find a catalog for stars within 450 light-years of Earth? Where can I find a catalog for stars within 450 light-years of Earth? I'd prefer one that has approximated masses as well.

The Hipparcos catalogue (use the revision by van Leeuwen 2007) contains trigonometric parallaxes and is complete down to about 9th magnitude.

What this means is that you will have a complete set of measurements for F-type main sequence stars and hotter, and all giant stars within 450 light years. However, the sample will not be complete for G dwarfs or cooler, since these are fainter than 9th magnitude at 450 light years.

The catalogue does not contain mass, since mass is not a measurable quantity for single stars. If you want masses you will have to use relationships between absolute magnitude and mass for main sequence stars (e.g. Kroupa 2007), but note that these depend on metallicity and a little bit on stellar age. Masses are extremely difficult to estimate for giants.

If you can wait a couple of years, the Gaia mission will report parallax distances for all stars to magnitudes of 19 or 20. This will be almost complete at 450 light years, even for M dwarfs.


Finding luminosity of a star with magnitude I'm working on my assessment for Physics in the IB, and I'm planning to investigate stars in spiral galaxies and determine their ages. For this I'm going to use data from the SDSS (Sloan digital sky survey http://skyserver.sdss.org/dr13/en/tools/chart/navi.aspx).

I need to get the luminosity of a star from the absolute bolometric magnitude. However the SDSS only gives magnitude in 5 values, u g r i and z. I've tried to understand what to do with these and what they mean but I can't figure it out.

How would I get one number the absolute bolometric magnitude from 5 different values, and what do they mean?

The ugriz values are just measures of the stellar brightness taken in 5 bands from the near ultraviolet u through to the near infrared z.

You need the distance to the star to convert the apparent SDSS magnitudes to absolute magnitudes.

Then you need to find a relationship (or relationships) between bolometric correction in a particular band and the colour of the star (by colour I mean something like g-i). Possibly instead you might find relationships between bolometric correction and temperature and will need to first estimate the temperatures of the stars from their colours.

Once you have absolute magnitudes and a bolometric correction, then the absolute bolometric magnitude follows.

In principle you could try this with a number of different colours and see if the results agree. Alternatively, you would fit a library of stellar models to the spectral energy distribution defined by the 5 magnitude values.

A table of bolometric corrections for the SDSS filters was calculated by Girardi et al. (2004) and can be found here. (Other tables/formulations are available, there is no universally accepted version).

Some bolometric correction tables can be found in Schmidt et al. (2014) for very cool objects, which are more problematic.


How to calculate Longitude from Right Ascension? Considering that a star of certain declination is crossing the local meridian at the observer's zenith at an unknown location on the earth. Here, the declination of the star is equal to the observer's latitude. Knowing the Right Ascension of the star, how could the observer calculate his longitude?

Example: The star Miaplacidus (Dec: -69° 42' and RA: 9h 13m) crossing at the observer's zenith!

Latitude = 69° 42' S

[ Longitude= RA(Decimal Conversion) X 15° ... I'm not sure how to proceed from here ... ]

In other "words", the connection between the time of transit $t_\mathrm{tr}$ of an object, its right ascension $\alpha$ and the geographical longitude of an observer $l$ is the (apparent) siderial time at Greenwich $\theta_0$ (if you know your local siderial time $\theta$, you don't need $\theta_0$ or $l$):
$$t_\mathrm{tr} = \alpha - \theta_0 - l = \alpha - \theta.$$


Could different views of the universe simply be different points in time? I'm not sure if this is the correct stackexchange site for this question.

I've been reading this article on Vice Motherboard There’s Growing Evidence That the Universe Is Connected by Giant Structures which while admittedly simplified for simpler folk like me put me wondering:

Is it at all possible that when we look out in to the distant universe in one direction, we are seeing the same galaxies and stars at a different point in time than if we look out in another direction? Could this occur if the distant body is moving laterally at a fast enough velocity relative to our position, or if the shape of the universe itself loops around?

Or is this not a proper interpretation of current theories of cosmology?

Could this occur if the distant body is moving laterally at a fast enough velocity relative to our position

Nope. I mean if the universe is flat it doesnt matter how fast the object goes you cannot see it twice in the sky.

  or if the shape of the universe itself loops around?

Yes. If the universe has a positive curvature. When we look at the sky, we can see the same object in two opposite directions.

In our universe this is not possible since if the universe is positively curved then the radius of curvature is much larger then the size of the observable universe. So its highly unlikely that we will see the same object in two different positions in the sky.


I saw TWO moving stars going at the same speed! What could this be? This weekend me and my family took the opportunity to go camping. My dad and I went star watching because where we lived we didn’t get many stars. I was looking at the stars when I saw a bright star moving across the sky. It not as fast as a falling star or a meteor but it faster and brighter than airplane or a satellite. My dad and I kept watching the star when another star like thing moving as fast, in the same direction, but not as bright as the other star, started trailing the brighter star. We watched in disbelief as the two stars went across and soon behind a tree, out of our view. I can’t explain that. Can you?

The International Space Station (ISS) and several other satellites can be seen early in the morning or late in the evening, even from urban locations.

Go to this site.

https://www.heavens-above.com/main.aspx

Enter your location then click on ISS. The site will give you a list of the times when the ISS will be visible and the area of the sky that it will move through.

The periods when the ISS or other satellites are visible come in cycles. If nothing is listed as visible in the ten day period wait a week then try again. It might take several weeks. If you live somewhere between 50 deg North latitude and 50 deg South latitude the ISS will be visible sooner or later.

Once you have seen the ISS passing overhead you can compare it to what you saw on your camping trip.

Some people are using electronic eyepieces to view the sky from light polluted areas. They often broadcast their observing sessions on the NightSkysNetwork

https://www.nightskiesnetwork.ca/

If you live in a light polluted area and are interested in astronomy you might find the site interesting.


Why are wavelengths shorter than visible light neglected by new telescopes? The diagram below, which I stole from this post by @HDE226868, shows that angular resolution as a function of wavelength suddenly drops by three orders of magnitudes from visible to UV-light. The resolution of wavelengths shorter than what the Very Large Telescope Interferometer or the European Extremely Large Telescope detect, in the near UV, suddenly cuts off to a factor of a thousand.

This is obviously because of the properties of Earth's atmosphere. But major space telescopes like the JWST and WFIRST will fill in the far infrared gap. Why aren't there any as ambitious space telescopes planned for UV and shorter wavelengths? (Or is the sudden cut off in that diagram misleading?)

Is it because it is more difficult, even from in-space observatories, or is it because the angular resolution of UV and shorter wavelengths are of lesser scientific value?

There are some technological issues to solve with putting any large telescope into space - and a space telescope is required at UV wavelengths. It is not possible to optimise such an instrument to work at both UV and IR wavelengths because of issues like cooling, mirror coatings and such-like. The simple angular resolution limit of a telescope goes $\lambda/D$, so on the face of it, to get equivalent resolution to an optical telescope, a UV telescope can be smaller. However, you also have to have optics that are good to a small fraction of a wavelength, so much better than the visible/IR. At even shorter wavelengths then conventional "optics" doesn't work because photons get absorbed and you move to the grazing incidence technologies of X-ray telescopes, which is a whole different game and much harder to achieve a given angular resolution.

Given all that, back in the 80s/90s I would guess that a decision was taken about the wavelength range to be covered by the successor to HST (ie JWST at a cost of approximately 10 billion USD) The real reason that no major UV successor to HST or IUE is ready to go now is simply that it is considered that the most important science priorities are achievable at near and mid-IR wavelengths. These are: observing the high redshift universe (essentially no UV light is detected from galaxies beyond a redshift of 3), observing star and planet formation (mostly in dusty environments where UV light cannot emerge and protoplanetary discs emit mostly at IR wavelengths) and doing exoplanetary science (planets are cooler than stars and emit mostly in the IR).

Thus, I don't think there are any technological showstoppers to a big UV telescope (at least the equivalent of JWST), it just comes down to science prorities.


How to correct observed flux densities for redshift Say I have a spectrum of a galaxy at a redshift $z$, in flux density units of erg/s/cm^2/Angstrom. I'd like to recover the spectrum (in the same flux density units) at z=0, i.e. at its rest wavelengths. How do I do this?

I saw this paper (Distance measures in cosmology, Hogg (2000)) but am still confused, because of their use of luminosity and luminosity densities, and in my class I only have fluxes.

You are familiar with the relationship between flux and luminosity, right? This is usually taken as the definition of luminosity distance in extragalactic astronomy,
$$ F = \frac{L}{4\pi D_L^2}.$$
The quickest way to convert this into a relationship with spectral variables is to use differential 1-forms. So, $L \rightarrow L_\lambda \,\mathrm{d}\lambda_e$ and $F \rightarrow F_\lambda \,\mathrm{d}\lambda_o$, with $\lambda_e$ the photon wavelength at emission (also called the rest frame wavelength) and $\lambda_o$ the observed wavelength at $z=0$ (also called the comoving wavelength). Thus
\begin{align}
  F_\lambda \,\mathrm{d}\lambda_o & = \frac{L_\lambda \,\mathrm{d}\lambda_e}{4\pi D_L^2} \Rightarrow\\
  F_\lambda & = \frac{L_\lambda}{4\pi D_L^2} \times \frac{\mathrm{d}\lambda_e}{\mathrm{d}\lambda_o}.
\end{align}
Now you just need to take your derivative using the definition of redshift, $\lambda_o = (1+z)\lambda_e$, and use some way of calculating the luminosity distance in your cosmology.

Formally, a lot of people like to use $\nu F_\nu = \lambda F_\lambda$ as though it were a total flux, but it's not. The derivation will work, but only because redshifting scales all frequencies/wavelengths by a single constant. $\nu F_\nu$ isn't a measure of the flux under any part of the spectrum, it's a spectral flux using a different coordinate system. Consider the following
\begin{align}
   F & = F_\nu \,\mathrm{d}\nu \\
    &= \nu F_\nu \frac{\mathrm{d}\nu}{\nu} \\
    &= F_{\ln \nu} \,\mathrm{d}\ln \nu.
\end{align}
In other words, $\nu F_\nu$ is still a spectral flux. If $F_\nu$ is in $\mathrm{ergs}\,\mathrm{cm}^{-2}\,\mathrm{Hz}^{-1}$ then $\nu F_\nu$ will be in $\mathrm{ergs}\,\mathrm{cm}^{-2}\,(e\text{-fold})^{-1}$. The $e\text{-fold}$ is a pseudo-unit, like radians, dex, octave, decibel, or magnitudes. In principle it can be omitted, just like how angular frequencies can be in radians per second, or just $\mathrm{s}^{-1}$.

Fun math exercise: show that $\ln(10) \nu F_\nu$ has units of $\mathrm{ergs}\,\mathrm{cm}^{-2}\,\mathrm{dex}^{-1}$.


Is our Universe inside a black hole? Could the known universe be inside a black hole and it's curvature really be the event horizon of the black hole?

No. The metrics look very differently.

The metrics of the the Universe is determined by the FLRW metrics. Its spacelike sections are planar.
The spacelike sections of the spacetime around a Schwarzschild (non-rotating) BH is spherical.

More elementarily formulated: the black hole is like a static sphere. The Universe is like an expanding sheet. However, this is what we currently see, matched by our current models. These are not so strong facts, like the density of the iron, what anybody could measure on his desk. I think the correct explanation is roughly this: "it doesn't really looks so, but no one knows".