text:
Does the image of the cosmic microwave background radiation show us what the universe used to look like?
Is it possible to interpolate using the cosmic microwave background radiation and currently what we can see in the sky at night to predict what the night sky will look like in the future? I’m guessing the night sky will be increasingly bright regardless of light pollution on earths because more light from galaxies will reach us in more time?  If we can interpolate forwards, would we be also to interpolate backwards?
From my understanding, the cosmic microwave background radiation is what the universe used to look like.

With the naked eye, we can only see our near neighbour galaxies: the Magellanic Clouds, Andromeda, and the Triangulum galaxy. And that's only in areas of low light pollution. We won't see light from more galaxies in the future. The Milky Way will (probably) collide with Andromeda in a few billion years, but Earth will be uninhabitable by then. See https://en.wikipedia.org/wiki/Timeline_of_the_far_future

Additionally expansion causes dilution even if new objects enter our observable universe (which by the way does not mean seen).

Can you relate more clearly to something particular. CMBR means a lot of things all together.


text:
Is it only possible to look at solar systems with stars at least as old as ours to be able to find intelligent life?
Assuming the speed it takes to development intelligent life is the same in all solar systems, would we need to look at exoplanets with stars as old or older than ours to find intelligent life? A young star would be assumed to have been surrounded by planets that are relatively new and haven’t had time to develop intelligent life. Visiting those planets with young stars relative to the sun would most likely only give us planets with prehistoric life.

"Assuming the speed it takes to development intelligent life is the same in all solar systems, would we need to look at exoplanets with stars as old or older than ours " If you assume this, then obviously the answer is yes.  ?

@RobJeffries I suppose a possible counterargument would be "What if intelligent life spread to a younger star system"...

Assuming the speed it takes to development intelligent life is the same in all solar systems,

We don't know if this is true.
We don't know if:

Life has started on more than one planet in the Universe.
How long, on average, it takes for life to start.
Whether life normally evolves past the "bacterial slime" stage to form complex multicellular life, and how long this takes.
Whether multicellular life normally develops intelligence, and how long this takes.
How long intelligent life survives, on average.

So we don't know what time it takes to develop intelligent life. However, planets are the same age as their stars, so it is reasonable to suppose that older stars are more likely to harbour intelligent life.
Moreover, visiting any extrasolar planets is too hard with current technology. Stars are just too far away.

Suggest adding that we have no scientific definition for "intelligent" either.  Do we count Dogs ?  Slugs ?  Insects ?  Mice ?  Dolphins ?  OK, the Dolphins and Mice are a given, but the Hairless Apes ? :-)

source: Stack Exchange

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text:
What planets or exoplanets orbit the Sun’s elder twin HIP 102152?
Has anyone done any research on the planets that orbit the HIP102152? Since that star is similar to ours and older, I postulate that it’s likely those planets are most worthwhile searching for advanced civilizations.

No planets have been detected in orbit around HIP 102152.
That does not mean that no planets exist, but that our current techniques are no able to detect them.  Most planet are detected by the transit method. This observes the very small dip in light when a planet goes in front of the star.  However if the planet's orbit doesn't line up exactly with Earth, then the planet will not be detected.
Other methods can detect large planets that orbit close to the star, or very large planets in orbit very var from the star. Solar systems like ours are harder to detect.
Given what we know about the abundance of planets, it is likely that HIP 102152 has a planetary system. But actual detection might no be possible with current technology.
Moreover there is no obvious reason to think that life or intelligent life is particularly likely just because the star is similar to the sun. The nature of the star may rule out intelligent life. But as far as we know, life just needs a reasonally stable energy source and a lot of luck. It doesn't need a solar twin.

I think that intelligent life requires that their solar system has a sun as old or older than ours assuming the speed of intelligent life evolution was the same on their planet. Is a star’s age correlated with the orbiting planet’s age? It’d be really useless to go to an exoplanet and find only non intelligent as humans life. It’d be only practical for increasing our food supply if we found another planet with only livestock and then there’s the problem of galactic food and livestock farming and transportations

Planets form at the same time as their star, so planets are the same age.  For the rest of the comment, I don't think you've really understood how big space is.  Space is big.  Really big. You just won't believe how vastly, hugely, mindbogglingly big it is.

@Numbers Note that posts and comments about life on other planets are just wild speculation from the point of view of Astronomy SE and we try not to encourage wild speculation on this real science based SE.  SE is a Q&A site, not really for discussing or speculating.  We have essentially no idea what is required to develop intelligence on Earth (or even a widely agreed definition of intelligence), let alone elsewhere in completely unknown biochemical environments.

If you look up the star on SIMBAD and view the references, you will come across some papers about a project called the Solar Twin Planet Search, which as the name implies is a project searching for planets around solar twins. The first paper in the series, Ramírez et al. (2014) "The Solar Twin Planet Search. I. Fundamental parameters of the stellar sample" lists HIP 102152 as one of the target stars for the project, which measures radial velocities using the HARPS spectrograph.
So far, nothing's shown up. If you search the exoplanet catalogues, e.g. the NASA Exoplanet Archive or the Extrasolar Planets Encyclopaedia, there are no known planets around HIP 102152. Data collected during the Solar Twin Planet Search has been used to put upper limits on the masses of planets around the star, see Figure 4 in Monroe et al. (2013) "High Precision Abundances of the Old Solar Twin HIP 102152: Insights on Li Depletion from the Oldest Sun": the limit ranges from a few Earth masses for 1-day orbits to roughly Saturn-mass planets at 1000 days. The abundances of various elements in the star do suggest that there may well be terrestrial planets orbiting the star, but they would be well below the detection limits for all but the shortest orbits.

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text:
If the distance between galaxies is increasing and the speed of light is constant will the most of the light from the universe ever reach us?
Will the light from other galaxies ever reach us if the universe is expanding?

Thanks for the edit, there's nothing wrong with clearly written question being short!

You may enjoy the Scientific American article by Davis & Lineweaver about expansion: https://people.smp.uq.edu.au/TamaraDavis/papers/SciAm_BigBang.pdf For more details, see their classic paper, Expanding Confusion: common misconceptions of cosmological horizons and the superluminal expansion of the Universe.

Related question: https://physics.stackexchange.com/questions/129980/expansion-of-the-universe-will-light-from-some-galaxies-never-reach-us

Related: https://astronomy.stackexchange.com/a/1615/893


text:
Is looking through a telescope to do astronomy a thing of the past?
Do astronomers actually look through a telescope with the naked eye or do they mostly look the the data generated by telescopes on computers?
Are they mostly working behind the computer and pencil and paper or is it still stargazing?
Do they actually point a red laser to observe constellations or do they do it all on the computer, and lasers that can point us at constellations are more of a gimmick for amateurs?
I’m sure astronomers would need to look under the lens to confirm the computers aren’t processing data that’s wrong. Do astronomers actually trust the algorithms that automatically do data entry from telescope observed data to their notebooks enough that they usually don’t bother to under the telescope until they’ve confirmed something?

Scientifically there's little to gain when you look through a telescope with your own eyes. Attach a camera to the telescope and you immediately document what you observe and take out the subjective factor from it at the same time and allow careful analysis and more detailed. Additionally you gain improved sensitivity enhanced observation duration and broaden (or narrow) the range of possible wavelengths which allows to gain additional insights.
That said, many professional astronomers still enjoy looking at the sky with their own eyes, be that with or without telescope.

Amateur comet searchers still look through the lens, and still find some comets that Pan-STARRS has missed. https://en.wikipedia.org/wiki/Pan-STARRS

Many amateurs have put their traditional eyepiece away in favor of the electronic eyepiece.
The camera looks through the telescope the viewer sees the result on a screen. The screen can be anywhere. Beside the telescope or on another continent.
Those caught waving LASERS around the constellations may spend a bit of their time viewing the sky through bars. :)

I think that astronomers, amateur or professional, who use lasers to point out constellations or other astronomical objects are NOT put in prison. Persons who point green lasers at aircraft can be in trouble with the law, but they are not necessarily astronomers.

So, don't go star watching with laser equipment near an airport.    Good advice.   :-)   It does seem that there's a chance for an arrest, though I think most of the laser pointer users who get in trouble point them at planes deliberately.   The risk (I think) is pretty low for astronomy use, since dark skies are the most important thing and airports are well lit.

Having looked into this, lasers can be distracting to pilots as far as 3 km distance.   Most planes fly higher than that, so if you're away from airports, you should be fine.   I think it's a good point, if not strictly an answer, it's a good addendum.     https://skyandtelescope.org/observing/some-pointers-on-the-use-of-laser-pointers/


text:
Is science data from astronomical telescopes analyzed with something analogous to "optical character recognition"?
Do people manually input the science data from telescopes into databases or does some software automatically collect the data from telescopes and put it in servers and databases?
I'd imagine there's too much data from present-day telescopes that they can collect to handwrite or hand by hand input the data from them into spreadsheets therefore computers do it via machine learning?

What do you mean by "data" ? Telemetry from the telescope itself or science data from the instruments ?

@astrosnapper science data from instruments

I've made an edit to your question based on your comment above and to adjust some of the wording. Please feel free to edit further if you like.

https://en.wikipedia.org/wiki/FITS and http://nova.astrometry.net/upload don't answer your question but may help. https://www.cosmos.esa.int/web/gaia/data-processing notes that the GAIA project uses software to process raw images.

google.com/search?q=ai+galaxies&oq=ai+galaxies


text:
When did it take place, and what was the cause of the first blackhole in the universe?
When did it take place, and what was the cause of the first back hole in the universe? Does this time precede stars collapsing as a result of old age?

First we need to find it...

Is it possible to somehow go back and time to find it or do the equations of physics shed insight?

Nobody knows. See https://en.wikipedia.org/wiki/Primordial_black_hole BTW, stellar black holes form from large stars, and large stars have short life spans, so they're much younger than the Sun's present age when they collapse. A 20 solar mass star (about the smallest that can produce a black hole) lives for about 10 million years. It's possible that a slightly less massive star can end up as a BH via a failed supernova, but the lifespan of such stars isn't thst much longer.

Black hole or back hole? Important difference.

We don't know for sure.
It is possible that random fluctuations in the density of matter of the early universe created black holes that could be much smaller than regular black holes formed by stellar collapse. These would have formed in the moments after the big bang.
Otherwise the first stars formed about 400 million years after the big bang, and some of these could form black holes after their (short) lives. Stars over 250 solar masses (which could form in the early universe) would produce large black holes when they collapsed.

https://en.wikipedia.org/wiki/Primordial_black_hole might be an interesting reading about those early black holes.

I think the terminology describing the black holes as tending to "fall towards the centre of galaxies" is problematic.  Do you mean migrate to the center of galaxies? Do you have any sources to cite for this statement?

dynamical friction: gravitational interactions will tend to transfer energy and momenum from massive to smaller bodies.  But this is hardly a key part of the answer and I'll delete it.


text:
What is the last day a waning moon is visible in daytime (the afternoon)?
From the fourth quarter onward, can we see the moon in the afternoon sky?

I think you mean the third quarter. The fourth quarter would be the New Moon.

https://web.archive.org/web/20190611151431/https://aa.usno.navy.mil/faq/docs/crescent.php may or may not be helpful as it discusses the inverse problem: first seeing a new moon after sunset. This is an archive.org copy of a navy.mil page that no longer exists.

At third quarter the moon rises at about midnight and sets at about midday The exact rising times depends on the position of the moon relative to the ecliptic and the season.
So in spring, at third quarter, the moon is low and sets quite early (around 10:00 in my location)  In Autumn the moon is high at third quarter, and sets at about 15:00 (though daylight saving time affects this)
As the month progresses the moon rises later and sets later. The moon is above the horizon in the afternoon, but visibility depends on many factors, and a very thin cresent, close to the sun, may not be visible with the naked eye. There is some analysis of this with respect to the appearance of the new moon, (which has calendrical significance to some religions) There are too many uncertainties in determining visibility to give a simple answer, much depends on the observer.


text:
Can you provide a visual description of the sun rise?
The Sun rises in the east only two times a year. The other days, the Sun rises in the east, but a little bit to the north, or to the south. How can you know the degree of how much north or south it is? Does it vary in different hemispheres?

Elaborating on the answer from James K, the azimuth of the sunrise depends on the declination of the Sun and the observer's latitude. The declination ranges between approximately +23.4 to -23.4 and depends on the date.
$$\cos(\theta_R)=-\frac{\sin(declination)}{\cos(latitude)}$$
where $\theta_R$ is the angle measured around the horizon from due south to the location where the object rises or sets. From this formula, the azimuth of rising is due east when the declination is 0 which occurs on the spring and autumnal equinoxes.
The calculation of the Sun's declination is not too involved, but it is longer than I want to type at the moment. Since the declination does not change very much from year to year, it would be relatively easy to pick a few dates throughout the year, get the declination for those dates, and calculate the azimuth.
Also see the following question: Solar azimuth for celtic fire festivals

It varies from place to place and from day to day, with a more extreme range of angles if you go further away from the equator.
Calculating the angle (called the azimuth of the sun at sunrise) is possible, but not trivial.  Instead you can get a computer to calculate it for you.  The pyephem package or astropy can do this, if you use python. On the other hand Suncalc.com has a web interface on which you can calculate the azimuth of the sun at any time and on any day.

The sun always rises from the south if you're standing on the north pole, and from the north if you're standing on the south pole.


text:
Is the equinox absolutely true east everywhere?
I've read that only only in the equinoxes do you get to see the sun rise in the exact east regardless where you might happen to be. However I think this is not absolutely true everywhere, but only in the equator, and in the south the Sun will rise just a tiny bit to the north, but this tiny bit to the north is negligible due to how far the Sun is from Earth. If this reasoning is correct, when would the 'true' equinox be for any country in the south? Can you provide an example and how you did it, please?

Do you mean "when" or "where"?

You're talking about solar parallax, which is a minor issue, but, within minutes (seconds?) of the equinoxes, the Sun has already moved, so other people would now see the Sun rising due east. It would be a vaguely interesting project to see how long it takes "east rise" to cross from pole to pole.

The sun's apparent latitude is well known as a function of time-of-year; further it doesn't get past roughly the Tropics lines.

Sunrise or sunset is usually defined as the moment when the upper limb of the Sun appears on the horizon.
This is conventionally computed as the time when the center of the Sun is 50' below the horizon, using a 16' angular radius and assuming 34' atmospheric refraction.
Let δ be the Sun's declination and φ be the observer's latitude north of the equator.
Near an equinox, the Sun's azimuth at sunrise or sunset is approximately
δ / cos φ + 50' tan φ
north of due east or west.
Solar parallax changes δ by about -8" sin φ,
shifting the sunrise/sunset azimuths about 8" tan φ southward.
This is about 1/400 as large as 50' tan φ, so we can neglect it.
For example, an observer self-isolating on a yacht at 50°S, 145°W, has φ = -50°.
Sunset on 2020-03-19 occurs within a few minutes of the equinox;
δ = 0°, and the Sun disappears at azimuth 269.0°.
At sunrise on 2020-03-21, δ = +0.59°, and the Sun first appears at azimuth 90.1°.
The NOAA Solar Calculator may be useful for other cases.

The celestial equator's plane intersects the horizon due east and due west, no matter where you are on Earth (except at the poles where the horizon is the celestial equator).
Those that have the sunrise at the exact time of the equinox will see the sun rise in the east (90º). For the rest, it will be too late, the sun will have made the crossing, and it could be something like 90.00001º or 89.99999º, although it will still be perceived as due east.
The last equinox was March 20th, 03:50 (UTC). At that moment the sun was rising very close to Maputo, Mozambique (table), so everybody on that meridian saw the sun rise in the east, exactly at 90º.

This is a great answer, and I notice the linked page even compensates for refraction.


text:
Why is rotational motion absolute, is the universe spinning?
When considering any object, one can say its translational movement is relative, depending on the point of view or reference frame adopted. If it moves at 1/4c relative to some observer, one might say it is the observer that moves at 1/4c and the object is stationary relative to it. It has no internal acceleration.
Experimentation at small scale will tell that a spinning object displays increasing internal acceleration the further away from its center of mass it is measured. -centrifugal force-
This forbids rotational motion to be understood as something relative, in the same way as uniform rectilinear motion is, so that when an object spins around its center of mass, noone could say "No, this object is rotationally stable, it's the entire universe that spins around it."
Fundamentally, why and how is rotational motion absolute in this way, and how could we know if the universe is or is not spinning at larger scale, at some spinning rate we cannot measure since we're trapped in this spinning reference frame, unable to detect it experimentally since it's too huge?

A related question on our sister site: https://physics.stackexchange.com/questions/1048/what-if-the-universe-is-rotating-as-a-whole

The proposed duplicate does not address the question about why rotation is absolute, so I'm voting to leave this open for now.

Fair enough, @antispinwards. Here's a Physics.SE question that focuses on that, with links to other related questions: https://physics.stackexchange.com/q/455418/123208

Thanks @PM2Ring for the links, the last one being the closest to what I was wondering about.


text:
Interpretation and understanding of the relation for the photometric redshift in a given bin
In the context of photometric probe of surveys (like LSST), I need to understand the relation I have to use for photometric bins.
Considering $p_{ph}(z_p|z)$ the  probability to  measure a photometric redshift equal to $z_p$ knowing the real redshift is $z$ and given $n(z)$ the density distribution of objects, I have the following formula which gives the density $n_i$ of galaxies into $i$-th bin and that I would like to understand :
$$
n_{i}(z)=\frac{\int_{z_{i}^{-}}^{z_{i}^{+}} \mathrm{d} z_{\mathrm{p}}\,n(z)\,p_{\mathrm{ph}}(z_{\mathrm{p}} | z)}{\int_{z_{min}}^{z_{max}} \int_{z_{i}^{-}}^{z_{i}^{+}}\,\mathrm{d} z \,\mathrm{d} z_{\mathrm{p}} \,n(z) \,p_{\mathrm{ph}}(z_{\mathrm{p}} | z)}
\quad(1)$$
with $\left(z_{i}^{-}, z_{i}^{+}\right)$ which are the values of both sides (+ and -) of redshift $i$ bin.
Here my attempt to simplify : with Bayes theorem, one has :
$$p_{\mathrm{ph}}(z_{\mathrm{p}} | z) = p_{\mathrm{ph}}(z|z_{\mathrm{p}})\,\dfrac{p_{\mathrm{ph}}(z_{\mathrm{p}})}{p_{\mathrm{ph}}(z)}\quad(2)$$
and Moreover, the bottom term of $(1)$ gives :
$$\int_{z_{min}}^{z_{max}} \int_{z_{i}^{-}}^{z_{i}^{+}}\,\mathrm{d} z \,\mathrm{d} z_{\mathrm{p}} \,n(z) \,p_{\mathrm{ph}}(z_{\mathrm{p}} | z)$$
$$=\int_{z_{i}^{-}}^{z_{i}^{+}}\,n(z)\,p_{\mathrm{ph}}(z_{\mathrm{p}})\,\text{d}z_{\mathrm{p}}\quad(3)$$
So, by combining $(2)$ and $(3)$, we get :
$$
n_{i}(z)=\dfrac{n(z)\,p_{\mathrm{ph}}(z_{\mathrm{p}})}{\int_{z_{i}^{-}}^{z_{i}^{+}}\,n(z)\,p_{\mathrm{ph}}(z_{\mathrm{p}})\,\text{d}z_{\mathrm{p}}}\quad(4)$$
BUT from this $(4)$ I don't know how conclude about the interpretation of this objects density into the redshift of bin $i$.
Surely there is an error in my attempt to simplify the expression $(1$) but I don't where.
If someone could help me to grasp better the subitilities of the relation $(1)$, from a mathematical or physics point of views, this would be nice to explain it.
UPDATE 1:

In particular, I don't agree with Equations (3) and (4). Indeed, using
Bayes theorem in the bottom term of Equation (1) leads to
$$ \int_{z_{min}}^{z_{max}} \int_{z_i^-}^{z_i^+} dz dz_p n(z) p_{ph}(z|z_p)\frac{p_{ph}(z_p)}{p_{ph}(z)} $$
But $p_{ph}(z | z_p) dz$ is not equal to $p_{ph}(z)$, so you can't
simplify the fraction out, like you do in Equation (3)
In addition, $dz_p n(z) p_{ph} (z_p | z)$ has no reason to be
constant, so you can't easily remove the integration sign, like you do
in Equation (4).

For equation $(3)$, I wanted to simplify by :
$$\int_{z_{min}}^{z_{max}} \int_{z_{i}^{-}}^{z_{i}^{+}}\,\mathrm{d} z \,\mathrm{d} z_{\mathrm{p}} \,n(z) \,p_{\mathrm{ph}}(z_{\mathrm{p}} | z)= \int_{z_{i}^{-}}^{z_{i}^{+}}\,n(z)\,p_{\mathrm{ph}}(z_{\mathrm{p}})\,\text{d}z_{\mathrm{p}}\quad(5)$$
since we should have : $$p_{ph}(z) = \int_{z_{min}}^{z_{max}} p_{ph}(z | z_p)\,dz_p\quad(6)$$, shouldn't we ?
In comment above, you talk about the relation : $dz_p\,p_{ph} (z_p | z)$, not the equation $(6)$ above, do you agree ?
From an intuitive point of view, I agree that ratio between density $n_i(z)$ and total density $n(z)$ is equal to the ratio between the probability of getting objects of redshift $z_p$ knowing the true redshift
$z$ by integrating over $z_p$ and the  total density integrated over $z_p$ and $z$.
But as soon as we have to formulate it with mathematics and conditional probabilities, it's more difficult.
Maybe I shoud consider for example the following relation which is the equivalent of Bayes theorem but with density functions (called conditional density) :
$$g\left(x | y_{0}\right)=\frac{f\left(x, y_{0}\right)}{\int f\left(t, y_{0}\right) \mathrm{d} t}\quad(7)$$
But I don't know how to connect this equation $(7)$ with equation $(1)$.
Sorry if I misunderstood : your comment will be precious, that's why I want to clarify this step.
UPDATE 2 :
1)

The numerator of Equation (1) $$ \int_{z_i^-}^{z_i^+} dz_p n(z) p_{ph}(z_p | z) $$ is simply the number of of samples in the $i^{th}$
photometric bin. Indeed, at a given redshift $[z, z+dz]$, there are
$n(z).dz$ samples. Each of these samples has a probability $p_{ph}(z_p|z)$ of ending up in the $i^{th}$ bin. So by integrating over
$[z_i^-, z_i^+]$, you get the number of samples with true redshift $z$
and a photometric redshift in that bin.

The problem is that, in the numerator of equation $(1)$ :
$$\int_{z_i^-}^{z_i^+} dz_p n(z) p_{ph}\,(z_p | z)$$
we integrate over $z_p$ and not $z$, so we can't consider the number of samples $n(z)\,\text{d}z$ and after say that we compute the probability to be in $i-th$ bin to have $n_i(z)$ samples/density value. Indeed, $\text{dz}$ doesn't appear in the numerator of equation $(1)$.
Do you agree with this issue for me ?
2) I provide you a figure that coul help someone to grasp the signification of $eq(1)$ :

You can see each color corresponding to the i-th redshift considered and computed from the $eq(1)$. I hope this will help.
I would like to make the connection between what I have put into my UPDATE 1, i.e the conditional density or maybe I should rather express the condditional probability like this :
$$g\left(x | y_{0}\right)=\dfrac{f\left(x, y_{0}\right)}{\int f\left(t, y_{0}\right) \mathrm{d} t}\quad(8)$$
This relation is qualified of "Bayes theorem" for continuous case :
But If I inegrate this expression $(8)$, I get :
$$\int\,g\left(x | y_{0}\right)\,\text{d}y_0=\int\,\dfrac{f\left(x, y_{0}\right)\,\text{d}y_0}{\int f\left(t, y_{0}\right) \mathrm{d} t}\quad\neq\quad P(X|y_0)\quad(9)$$
How to make appear the term $P(X|y_0)$ from equation $(8)$ ?
UPDATE 3: Really no one to explain problematics cited above in my UPDATE 2 and my attempt to simplify the equation $(1)$ at the beginning of my post (with eventual connection with conditional probability in equations $(7$ or $(9)$ ?
Any help/remark/suggestion is welcome

I don't understand how you get (3)

Nor how you get the numerator in (4)

@usernumber I got them with Bayes theorem

I don't really understand all the equations you wrote out and I'm not sure you can perform the simplification the way you suggest. In particular, I don't agree with Equations (3) and (4). Indeed, using Bayes theorem in the bottom term of Equation (1) leads to
$$
\int_{z_{min}}^{z_{max}} \int_{z_i^-}^{z_i^+} dz dz_p n(z)
p_{ph} (z | z_p)\frac{p_{ph}(z_p)}{p_{ph}(z)}
$$
But $p_{ph}(z | z_p) dz$ is not equal to $p_{ph}(z)$, so you can't simplify the fraction out, like you do in Equation (3)
In addition, $dz_p n(z) p_{ph} (z_p | z)$ has no reason to be constant, so you can't easily remove the integration sign, like you do in Equation (4).

But I can try to explain the Equation (1) you are puzzled about.
The numerator of Equation (1)
$$
\int_{z_i^-}^{z_i^+} dz_p n(z) p_{ph} (z_p | z)
$$
is simply the number of of samples in the $i^{th}$ photometric bin. Indeed, at a given redshift $[z, z+dz]$, there are $n(z).dz$ samples. Each of these samples has a probability $p_{ph}(z_p | z)$ of ending up in the $i^{th}$ bin. So by integrating over $[z_i^-, z_i^+]$, you get the number of samples with true redshift $z$ and a photometric redshift in that bin.
The denominator
$$
\int_{z_{min}}^{z_{max}} \int_{z_i^-}^{z_i^+} dz dz_p n(z) p_{ph} (z_p | z)
$$
provides a normalization term, by integrating the numerator over all redshifts.

Thanks for your quick answer. I put an UPDATE 1  to try grasping subtilities of this $n_i(z)$ definition.

could you please take a look at my UPDATE2 since I have not graspped all the subtilities of the definition of $n_i(z)$ ?

@usenumber . I have provided the results of the analytic formula (1) on a figure in section 2). Could you still think that I can't  make the connection between what I have put into my UPDATE 1, i.e the conditional density or maybe I should rather express the condditional probability like this. Best regards


text:
Why are telescopes positioned in Lagrange points?
In this Wikipedia article about the list of space telescopes to be launched (which I assume is exhaustive), of the 11 telescopes yet to be launched, 6 will be positioned at the Sun-Earth L2 Lagrange point. Why is that so?
I read that it has something to do with lower fuel consumption, but I would like a more detailed answer.

This question might receive better answers on Space Exploration.SE.

The supposed duplicate is not a duplicate of this question. Risk of debris is not the driving factor; it's not even close. Voting to reopen.

This is absolutely not a duplicate of the other question.  This question asks why they are positioned in Lagrange points while the other one asks about debris collecting.

tldr;

L2 is a very stable thermal environment as well as good instantaneous sky visibility and high observing efficiency.

The main reason space telescopes are placed in an L2 orbit is because L2 is a stable thermal environment. Telescopes in Earth orbit can receive sunlight and earthlight in different directions, meaning that the telescope would have to shield in most directions if it wanted to keep itself cool. At L2, the Sunlight and Earthlight come from the same direction, meaning the telescope has to only shield itself from that direction. This is very important for infrared telescopes, as heat radiation acts as interference to measurements. Of the 6 space telescopes going to L2 listed, 3 are infrared telescopes.
For example JWST, one of the infrared telescopes, has to cools some of it's detectors to 7 kelvin to operate properly.

JWST L2 Halo Orbit
The other main benefit from being in L2 is that being so far away from Earth means that there is a much greater available field of view as the Earth blocks a very minor portion of the sky. This increases sky visibility and observation efficiency.
Other Benefits:
- Ease of communication
- Delta-V requirements low (2-4 m/s per year)

Does it also help that the dark side of Earth is the one facing L2, or is the difference minimal in the IR spectrum?

@notovny The difference in brightness in the light and dark side of the Earth is negligible  when compared to the Sun. The reason Earthlight is an important consideration is that it can come in different directions to the Sunlight. L2 orbit removes this consideration. The Athena xray telescope could be placed in a L1 Halo orbit, indicating the thermal difference between the sides isn't that big. (I think they're considering this to avoid having to plan around eclipses, which can occur in L2 orbits).
Constant amounts of light are much easier to manage.

How come ease of communication? Communication is more difficult surely.

@RobJeffries I didn't mention it there, but that's mostly compared to the HCO that Spitzer is in. NASA; "Webb's position out at L2 also makes it easy for us to talk to it. Since it will always be at the same location relative to Earth-in the midnight sky about 1.5 million km away - we can have continuous communications with it as the Earthwith rotates through the Deep Space Network"

For this answer, I'll consider space telescopes to be telescopes that operate in space and that are intended to look at objects at the extremes of the solar system and beyond. This excludes Earth observation satellites, satellites that monitor the Sun, and satellites sent to another planet to observe that other planet as "space telescopes".
Every space telescope has two primary objectives. One is to view objects in space, as commanded from the Earth. The other is communicating with the Earth (receiving commands, relaying observations back to Earth). There are many alternatives regarding where to place such devices. These include

Low Earth orbit

Description:

The satellite that contains the space telescope is injected into a low Earth orbit shortly after launch.

Examples:

Hubble Space Telescope

Advantages:

This is the cheapest of all alternatives in terms of launch ΔV.
Repair / replenishment is possible at this altitude. The Hubble is proof of that.

Disadvantages:

The very high orbital velocity complicates viewing, communications.
Being so close to the Earth means that it obscures a large portion of the sky. The closeness also means the Earth is a significant heat source for infrared and microwave astronomy.
Significant external torques result from high Earth gravity gradient and atmospheric drag and must be countered. LEO is probably the most expensive of all alternatives in terms of attitude maintenance.
This, and to a much less extent, geosynchronous orbits, are the only alternatives where debris poses a threat.

Geosynchronous / geostationary orbit

Description:

The satellite is injected into a geostationary transfer orbit shortly after launch. Later, when the satellite reaches geostationary altitude, another burn is performed to raise perigee to geostationary altitude.

Examples:

International Ultraviolet Explorer
Hipparcos. That was the intent; Hipparcos's apogee boost motor failed to fire, so it remains in a geostationary transfer orbit.

Advantages:

The Earth isn't as big (compared to LEO), but it's still large compared to other alternatives.
Communication is easy; this is the cheapest of all alternatives regarding communications -- assuming the vehicle does make it to the intended geosynchronous orbit.

Disadvantages:

Surprisingly, this alternative is rather expensive in terms of launch ΔV. The ΔV cost to go from the Earth's surface to GEO is higher than is the ΔV cost to achieve escape velocity from the Earth's surface.
Fuel is needed for orbit maintenance as well as for attitude control.
Fuel should be reserved for end-of-life activity.

Highly eccentric Earth orbit

Description:

The satellite is injected into a highly eccentric orbit shortly after launch. Later, when the satellite reaches apogee, another burn might be performed to raise perigee a bit (but not a whole lot).

Examples:

Chandra X-ray Observatory

Advantages:

Can be cheap in terms of launch ΔV, not as cheap as LEO, but cheaper than GEO.

Disadvantages:

Low perigee means orbit might cross Van Allen radiation belts and might even descend into altitudes where atmospheric drag is present.
Dives toward perigee typically interrupt operations.

Very high Earth orbit

Description:

The satellite is injected into a highly eccentric orbit shortly after launch. Later, when the satellite reaches apogee, another burn is be performed that raises perigee by a large amount.

Examples:

I know this option has been discussed, but I can't find any.

Advantages:
Disadvantages:

Very expensive in terms of launch ΔV.
Can be expensive in terms of orbit maintenance ΔV due to perturbations from the Moon and the Sun.

Earth-trailing heliocentric orbit

Description:

The satellite is injected into an Earth escape orbit shortly after launch. The resulting orbit has a period that is a bit longer than one year. Later, when the satellite does escape the Earth's gravity field, another burn might be performed that ensures the vehicle won't come back to Earth years / decades later.

Examples:

Spitzer Space Telescope
Kepler Space Telescope

Advantages:

This alternative is surprisingly cheap in terms of launch ΔV. The only alternatives with a lower launch ΔV are LEO and highly eccentric Earth orbits.
Zero orbit maintenance ΔV.
Low attitude maintenance costs.
Thermal radiation from the Earth more or less is a non-issue.

Disadvantages:

Communication systems are not cheap. The Deep Space Network (or equivalent) is needed on the Earth, and a simple fixed antenna is not sufficient on the satellite.

Finally, the Sun-Earth L2 point

Description:

The satellite is injected into a complex transfer orbit shortly after launch that eventually carries the vehicle close to the Sun-Earth L2 point. When it gets close, the vehicle injects itself into a pseudo orbit (either a halo orbit or a Lissajous orbit) about the Sun-Earth L2 point.

Examples:

Lots and lots of them.

Advantages:

The telescope can always be pointed so that all three of the Sun, Earth, and Moon are always behind the telescope. This alone is a huge advantage.
Communication is simpler than communicating with a satellite that is receding from the Earth. The Deep Space Network is not needed to receive data from a satellite orbiting Sun-Earth L2, and the antenna on the spacecraft can be fairly simple.

Disadvantages:

Somewhat expensive in terms of launch ΔV.
Fuel is needed for stationkeeping.

Sorry for the briefing chart style answer. My brain apparently has become accustomed to writing in bullets.

I know Stack Exchange does not encourage comments that just say 'thanks' but this is an incredibly comprehensive answer! Thanks a lot David!

Satellites positioned at L2 has the sun, earth, and moon all behind it so it gets a continuous view of deep space.  The Webb Space Telescope will be positioned there. A telescope at L1 would have a continuous view of the Sun and the SOHO satellite is currently there. This link explains some of the benefits in general terms.

Is "continuous view of deep space" your answer to "Why are telescopes positioned in Lagrange points?" Is it possible to at least briefly mention "some of the benefits in general terms" from your space.com link to minimize link-onliness? Thanks!

This is incomplete.

I added a link when i exhausted my knowledge! lol.


text:
Would magnetic reversals on Ganymede affect radiation levels at the surface?
Currently the intrinsic magnetic field of Ganymede is anti-aligned with respect to the Jovian magnetic field. If this situation changed to become aligned, either due to magnetic reversals of the Ganymedean dynamo or due to the reversal of Jupiter's magnetic field (which has been predicted to undergo solar-like reversals on timescales of the order of centuries), would there be any changes to the radiation levels at the surface?


text:
Correct units to use while plotting mollweide projection of sky chart with RA and Dec values, matplotlib
I am trying to plot a mollweide projection of a sky chart in Python using the matplotlib package. I am using the RA and Dec values from a database containing all stars in Hipparcos, Yale Bright Star, and Gliese catalogs (almost 120,000 stars).
This is the database: https://drive.google.com/file/d/1sJGUxQpnsw93q48I1v8qo6rbDKCeGYOH/view?usp=sharing
It contains the degree values of RA and Dec (column names are ra and dec respectively) as well as the radian values of RA and Dec (column names are rarad and decrad respectively).
My code assuming I need to use the radian values of Dec and RA is:
figure=plt.figure()
figure.patch.set_facecolor('black') #setting plot background to dark colour
ax = figure.add_subplot(111, projection='mollweide')
plt.scatter(df['decrad'], df['rarad'], s=1, color='red')
plt.show()
figure.savefig("weird.png")

And the output plot I get is:

My question is: Why are all the datapoints plotted within that circle only and not throughout the entire elliptical surface of the celestial sphere? Is it because I have chosen the wrong units (radians instead of degrees) or is the nature of the plot supposed to be this way based on the given star data?

What is your astronomy question?

With stars extending to ±90° on the horizontal axis and rather sparse near those limits, I suspect that RA and Dec are swapped.
Try this instead:
plt.scatter(df['rarad'], df['decrad'], s=1, color='red')
Also if you want the plot to resemble the night sky, filter the star list by apparent magnitude, and reverse the horizontal axis so it increases from right to left.


text:
Units confusion while calculating flux of a star using distance and luminosity, Python
I am trying to calculate the flux of various stars in Python. I am using the information from a database containing all stars in Hipparcos, Yale Bright Star, and Gliese catalogs (almost 120,000 stars).
This is the database. The original source is here and it is also hosted on GitHub.
The formula I am using to calculate the flux is
$F=L/(4\pi d^2)$ where L is the luminosity and D is the distance.
In the database, the distance is provided in the 'dist' column and the luminosity is provided in the 'lum' column.
This is the code I used to calculate the flux data:
df["Flux"]=df['lum']/(4*np.pi*df['dist']**2)
df['Flux'].head(10)

And this is the output I received:
0     1.587863e-05
1     1.357072e-05
2     1.572559e-04
3     4.137163e-05
4     2.635442e-05
5     8.254667e-07
6     9.653330e-06
7     1.662834e-05
8     2.539475e-05
9     2.540078e-05

My question is:

What is the unit of luminosity provided in the database?
Are the calculated flux values correct in terms of units, or does it require a unit conversion?

Where did you get that database from? Otherwise you're just asking people to download a random file off Google Drive.

What are the units for each of these? Add the link requested above and check it for units.

@uhoh sorry for my late response. I got the database from this website: http://www.astronexus.com/hyg. It is the HYG 3.0 database from the website.

@DrishikaNadella Okay I've reversed my down vote to "up" and retracted my close vote, and added those links back to your original question. Thanks for the reply!

Have you read the catalogue information that tells you what each column is (and the units)? https://github.com/astronexus/HYG-Database

Looking at the Readme , I wonder how the table can be unclear... And how any answer here could give more info

As can be seen in the page given by Dr. Jeffries, luminosity units are in solar luminosities. Just for fun though I looked at the Gaia Data Release 2 info in the VizieR catalogue (accessed thru SIMBAD) for the first two stars HIP1 and HIP2. HIP2 is close, but HIP1 has a luminosity in HYG of 9.63 L(sun) and 6.263 L(sun) in VizieR.


text:
Could the unseen companion of HR 6819 be a close binary system?
Recently there was an announcement of a black hole candidate in the HR 6819 system (ESO Instrument Finds Closest Black Hole to Earth). In the proposed system, the black hole forms a close binary with the B3III primary of the system, and a Be star is located on a wide outer orbit forming a hierarchical triple:
 ┌ B3III star
┌┤
│└ black hole
┤
│
└ Be star

Reading through the research paper Rivinius et al., 2020, it is mentioned that there is no X-ray emission detected from the system indicating that it is either not accreting significant quantities of material, or that it may not be a black hole. The paper notes that the companion cannot be a single main-sequence star (which they estimate would be of spectral type B7 or earlier) because this would appear in the spectral lines, which given the mass leads to the inference that the companion is a black hole as it is too massive to be a neutron star or white dwarf.
As far as I can see, I don't see any mention of the possibility that the companion could itself be a close stellar binary, i.e. the following architecture:
 ┌ B3III star
 │
┌┤
││┌ close binary, star 1
│└┤
│ └ close binary, star 2
┤
│
│
│
└ Be star

A binary of near-equal-mass stars would have lower luminosity than a single star of the same total mass, which might help it avoid detection.
I'm wondering if the configuration would be possible given the short period (40 days) of the companion's orbit, but maybe the data is sufficient to rule this out.

Just so I'm visualizing this correctly - HR 6819 is thought to be a triple system, with a Be star in a wide orbit and a B-type giant and this unknown object in an inner orbit. You're talking about whether the unseen object itself could be a binary, making HR 6819 a quadruple-star system?

@HDE226868 - yes, I've added a mobile diagram for clarity.

The absolute minimum mass of the unseen companion is $4.2 M_{\odot}$ (and my reading is that this really is an unlikely minimum, requiring an abnormally faint primary for its spectral type and an orbital inclination of 90 degrees, but let's go with it) and the argument is that a main sequence star of this mass would make its presence known in the spectrum. But let's suppose that instead of a $4.2M_{\odot}$ main sequence star, we had  a pair of A stars with $2.1M_{\odot}$ each, in a tight binary. The luminosity of ths combined binary would be less than 25% that of a $4.2M_{\odot}$ single star, so might have gone undetected.
The orbital period of the B3 giant primary with the putative black hole is 40 days. Assuming the minimum mass of the system as a whole is $9.2M_{\odot}$ (for an inclination of 90 degrees), then Kepler's third law yields a binary separation of 0.48 AU.
The radius of a $2.1M_{\odot}$ A star (like Sirius A) is about $1.7R_{\odot}$, so the minimum separation of a pair of such objects in a binary system is about $3.4R_{\odot}$, or 0.016 AU.
As a rule of thumb, as long as the separation of the closely orbiting stars is less than about ten times that of the wider binary then that ought to be stable to tidal disruption (though it also depends a bit on the masses too).
There are examples of systems where there is a wide binary, where one of the components is itself a triple featuring another close binary, in the catalog of binary stars by Tokovinin (1997). These would be multiples with a level 112 in his notation and there are a dozen examples in this catalogue. Perhaps the most famous is Tau CMa, which consists of an O supergiant in a wide, just resolved, 250 year orbit with another bright B-star. However, the O-type star is in a 154 d binary system with a pair of eclipsing B-stars that orbit each other every 1.28 d.
So I think you have a legitimate case that this may not be what the authors claim.
Edit:
I was interested enough by this to email Rivinius and put this scenario to him. The reply raised one or two interesting objections.
Firstly, he suggests that the reduction in luminosity of the secondary (compared with a main sequence single star of double the mass) might not render it invisible in the spectrum. The reason is that the A-type stars have narrower (metallic) lines that could be observed than the B-type giant. He cites the star 66 Oph as an example where this has been done. I am not totally convinced by this, or at least not convinced that it has been demonstrated.
The second point is better. If the inclination is anywhere near edge-on (as required for the unseen companion to be at its minimum mass), then in a close binary you would expect eclipses or at least significant ellipsoidal modulation due to distortion from spherical symmetry. Using Spica as an example (a 4d binary) Rivinius argues that even with dilution that this modulation would have been picked up in the SMEI and TESS light curves discussed in the paper.
If the binary were more face-on to reduce this ellipsoidal modulation, then of course the companion mass would increase dramatically from the 4.2 solar mass minimum, making even a binary companion impossible to hide.
So my conclusion changes. I'm reasonably convinced by the second argument - that a close binary companion would be revealed through ellipsoidal modulations.

Thanks for that! I actually find the metallic lines reason to be the more convincing. From what I can see, the second reason might still permit a misaligned system where the close binary is nearer to face-on (suppressing ellipsoidal variations) while its orbit around the B-giant is closer to edge-on (keeping the total mass down), although I suspect this is going to end up being quite a low-probability scenario.

@antispinwards That arrangement would I think be unstable to the Kozai mechanism. The other problem with detecting the metallic lines is that the components of a close A-star binary might be rapidly rotating.

The binary hypothesis came up in today's arXiv batch. Although by this point several groups have come to the conclusion that the system is a binary with a low-mass stripped helium star primary with the Be star being the 40-day companion.


text:
How many times can iron be detected for the first time in an exoplanetary atmosphere?
I see today that there's a story about the first detection of iron in an exoplanetary atmosphere, in this case in the atmosphere of the ultra-hot Jupiter KELT-9b.
I remember there being a story a couple of months ago about the detection of iron rain in an exoplanetary atmosphere, the example being another ultra-hot Jupiter, WASP-76b.
It doesn't seem plausible for both of these to be the first detection of iron in an exoplanetary atmosphere. So which one is it?

Neither of these, interestingly enough, is the first time iron has been detected in an exoplanetary atmosphere. Other groups (Hoeijmakers et al. 2018, cited by both papers) have detected absorption lines of iron and other metals in the day-night transition zone of exoplanets, during transit. Ehrenreich et al. 2020, the new ESO paper, did something similar.
Now, Pino et al. 2020 are referred to in the spaceref.com article as having "directly demonstrated" the presence of iron in an exoplanet atmosphere for the first time. What I believe "directly" in this case refers to is the detection of iron emission lines, rather than iron absorption lines. This is what makes Pino et al.'s observations different: they observed KELT-9b shortly before it passed behind the star, allowing them to view the hot, day side of the planet and thus iron emission.
In other words, I think the spaceref.com article is playing fast and loose with its wording - Pino et al.'s results don't constitute the first time iron has been detected in an exoplanet atmosphere, just the first time iron emission lines have been detected. They got around that by characterizing emission lines as "direct"$^{\dagger}$, implying absorption lines are "indirect". Whether you agree with that is up to you, I think.
If you had to write an article labeling one study as the first detection of iron in an exoplanetary atmosphere, you could make a good argument in favor of Hoeijmakers et al., but I think the better option might be to simply not try to describe any of these in that manner. Some nuance here wouldn't be a bad thing.

$^{\dagger}$Hoeijmakers et al. refer to their observations as the first "direct detection" of iron, adding a layer of complexity and irony.


text:
Is Beta Lyrae still thought to be losing mass in a spiral outflow?
In the famous 1960 painting by Chesley Bonestell, Beta Lyrae is depicted with a glowing spiral trail of matter emanating from the binary stars. I'm wondering if that's still thought to be a possibility given the additional study of the system between the 1960s and the present day.
Deschamps et al. (2005) suggest that non-conservative evolution of binary stars might be via a hotspot mechanism, where the mass is lost from the region where the accretion stream impacts the disc. Most studies of Beta Lyrae itself seem to discuss mass loss via bipolar jets, for which there is observational evidence (e.g. Ak et al. 2007). Is there any evidence for the kind of spiral outflow depicted by Bonestell (either via the Deschamps et al. mechanism or via direct loss through the L2 point), or is this no longer thought to be occurring at Beta Lyrae?


text:
Can a star turn into a planet?
I have read in a magazine that a star can actually form back to a planet... is it true?!
Does it mean all the planets in our galaxy can be revived back to what they were before?

"I have read somewhere" is not very clear.  Can you find the exact citation.  Maybe you can link to where you read it.

i have done the edits. you may check.

I have read in a magazine is not very clear.  Can you find the exact name and issue of the magazine.  Maybe you can link to the magazine's website.

@JamesK I am sorry but i can't remember the magazine name. I had this question for a long time. I found about this site now. So, i asked. Really sorry for the trouble caused.

https://wtamu.edu/~cbaird/sq/2019/02/24/can-a-star-turn-into-a-planet/

No. This is not true.
A star is much larger than a planet.  A star has nuclear reactions in its core, that prevent it from collapsing under gravity.
A planet is much smaller. It is small enough for "normal" pressure to be strong enough to stop gravity from making it any smaller, and there are no thermonuclear reactions at its core.
A star cannot form back to a planet. A planet cannot be "revived" back to a star.
(There is one class of bodies that may be star-like when young and planet-like when older: "Brown dwarfs".  These are objects with a mass of 15-75 times the mass of Jupiter. These can support deuterium fusion when young, but not hydrogen or even lithium fusion, when the deuterium is used up they don't have enough mass to collapse to a dense stellar remnant, and so become "planet-like".  Really they are objects sui generis, neither stars nor planets. Non of the solar system planets are brown dwarfs, and you can't revive a brown dwarf.)

To add to James K's answer, there are known examples of stars or stellar remnants that have been reduced to planetary masses due to mass transfer in close binary systems. An example is the so-called "diamond planet" orbiting PSR J1719-1438, which is likely the remnant core of a carbon-oxygen white dwarf which lost its outer layers to the neutron star, leaving behind an object roughly the mass of Jupiter while spinning up the neutron star to become a millisecond pulsar. The title of the discovery paper by Bailes et al. (2011) is "Transformation of a Star into a Planet in a Millisecond Pulsar Binary".
Whether these should be referred to as planets could be seen as debatable: while PSR J1719-1438 is listed in the NASA Exoplanet Catalogue, these kind of stellar remnants are a completely different type of object to the ones usually considered planets. On the other hand, you could make similar arguments for other objects in the exoplanet catalogues: e.g. the terrestrial-mass objects orbiting the pulsar PSR B1257+12 (which might themselves be the result of a white dwarf being disrupted by a neutron star, see Margalit & Metzger 2017), or the population of superjovians at very wide separations from their host stars which might represent either planets scattered onto wide orbits or very low mass brown dwarfs.


text:
If time is not absolute and depends on the gravitational field, then why we keep using our own (particular) time in cosmology?
Sorry if my question is not really clear, but indeed it is not clear in my mind neither.
But maybe some of you can help.
Time is not absolute and can be affected by the gravitational potential.
Now the question is: do concepts like the age of the Universe, time between different cosmic epochs etc, depend on the position of the observer with respect to a gravitational field?
If yes, how much things would change if we were observers, e.g., close to a black hole or in absence of gravity?

The difference in rate of time caused by Earth's gravity is smaller than the error bars on most (if not all) cosmological time estimates. It does affect however for example measurements of pulsars, so in such cases the gravitational time dilation must be and is taken into account.

@tuomas I would write an answer.

@tuomas I see, thanks.. what about an observer far from any (local) gravitational field, e.g., a void?

From https://en.wikipedia.org/wiki/Gravitational_time_dilation a clock on Earth's surface loses about 2.19 seconds per century due to the Earth's gravity.

@PM2Ring From this answer it seems that on Earth time is -0.7 ppb (parts per billion) slower due to Earth's own gravity well and rotational speed, (agrees with your value) but if you take into account that the Earth sits in the Sun's gravitational well and orbits it at about 29 km/sec, the effect is -15.5 ppb! Only slightly related: Parker Solar Probe passing extremely close to the Sun; what relativistic effects will it experience and how large will they be?


text:
Does time flow in a Minkowski spacetime?
In a spacetime where the stress-energy tensor is null (no energy, no matter, no entropy), I wonder whether any motion of the variable time in the phase-space is well defined.
The arrow of time, meant as parametrisation of succession of changing events, might lose validity, since no events can occur without actors (particles or fields) that can generate them, right?
Could this be a paradox, when using the Minkowski approximation to propagate information?

This question seems mote suitable for the Physics site, but please see https://physics.stackexchange.com/q/235511/123208

@PM2Ring thanks for the link.


text:
Why was the diaeresis removed from Pasiphae?
The Gazetteer of Planetary Nomenclature notes that the spelling of Jupiter's moon Pasiphae was changed from Pasiphaë in July 2009, i.e. they removed the diaeresis. I've been unable to track down the rationale for this decision. Is anyone aware of why the spelling was changed?

The name wasn’t changed, just the orthography. The use of diacritics has been diminishing for many decades: you rarely see “coöperation” any more, and even “naïve” is often written with a plain “i”. Is Boötes still the formal spelling?

Note that in the 1975 minutes no dieresis is used, but this may just be an artifact of the technological limitations of the time https://www.iau.org/static/administration/ecminutes/ec40minutes.pdf  (page7

However note that in the report https://books.google.co.uk/books?id=nv7nCAAAQBAJ&lpg=PA337&dq=planetary%20system%20nomenclature%20pasiphae&pg=PA337#v=onepage&q&f=false The dieresis is absent,  but the umlaut on Dr Jürgen Blunk's name is written.  This is almost an answer  "The diaresis was never part of the name, and was only used hypercorrectly"  Probably a report in 2009 clarified that.


text:
Intuition about Hubble's Law
I have recently learnt about Hubble's Law and this is my understanding of it:
Galaxies move away from us at a rate that is proportional to their distance from us. So, considering an arbitrary unit of distance d, if a galaxy at a distance d from us is moving away from us with a velocity v, then a galaxy at a distance 2d is moving away from us with a velocity 2v. This proves that the universe is expanding.
But the part that I can't quite visualize is the fact that it doesn't matter where you are in the universe. Regardless at what point in space we are standing, it will look like the galaxies are moving away from us. How is this possible? Are the galaxies moving outward? Should I visualize this as the galaxies being on the surface of a sphere whose radius is increasing?
How do I gain intuition about this and better visualize this?

See also pela's answer here: "This excellent question continues to confuse laymen as well as professional astronomers. In the standard, general relativistic (GR), cosmological model, the Universe is described as expanding, with galaxies lying approximately still in space and being 'dragged along' ".

Think about it as if you were baking bread (or cake, whatever you prefer). When you bake a bread with raisins, it rises in all directions (due to the yeast). Every raisin in a rising loaf of raisin bread will see every other raisin expanding away from it. Now suppose the raisins as galaxies and the yeast as the Hubble constant.
An illustration from Hyperphysics (Georgia State University) to illustrate this further:

When you now suppose you are the raisin at the edge, from your perspective, you are standing still and the other raisins are receding away. It is all about the perspective, the body which you take as a reference (like with relativity).

Here are some galaxies (in the first line), and the same galaxies a bit later (in the second line)

If you are in the red galaxy (the one in the middle), galaxies that used to be one space away are now two spaces away, so they are moving away with a speed of one space per unit time. Galaxies that used to be two spaces away are now four spaces away, so they are moving away with a speed of two spaces per unit time. All of the galaxies seem to be moving away, and the further away they are, the faster they go.
If you are in the purple galaxy (second from the left), the red galaxy (in the middle) is moving away with a speed of one space per unit time. The turquoise galaxy (fourth from the left) is moving away with a speed of two spaces per unit time. So you also see all the galaxies moving away from you, and the further they are, the faster they go.
Whichever galaxy you are in, you see all the other galaxies getting further and further.

I don't know how colorblind friendly this drawing is. If you have a hard time distinguishing two colors, say so in the comments, I'll try to do better.

another way to think about this is to imagine what some magic observer expanding with the universe would see
they see galaxies as (mostly) fixed in place - with some small individual random motion - but those galaxies, and everything within them, appears to be getting smaller
this frame of reference is what we call the comoving frame and we use it a lot in the mathematics of an expanding universe - like a rotating frame it is arguably aphysical (it introduces extra forces to account for the transformation, and no being following the laws of physics could inhabit it) and thus we rarely talk about observers in it
but, if we did, we'd have an observer who sees a mostly stationary universe in which the physical constants are changing such as to shrink down every bound system - effectively changing the units of measurement
if our magic comoving observer peered into a galaxy and saw a rapidly shrinking physical observer they would agree that the measured distance between galaxies is increasing, but not because they're moving apart - because the meter-long stick used to measure the constant distance keeps shrinking


text:
Expresion of comoving distance
I have a simple question :
How to prove the following relation :

The comoving distance to an object at redshift $z$ can be computed as

$$r(z)=\dfrac{c}{H_{0}} \int_{0}^{z} \dfrac{\mathrm{d} z}{E(z)}$$
from the relation :
$$r(t)=\int_{0}^{t} \dfrac{c\mathrm{d} t}{R(t)}$$
I tried to use with the definition : $1+z= \dfrac{R_{0}}{R(t)}$ but I can't conclude.
Any help is welcome.
UPDATE 1 : @Tosic's demonstration seems to be correct. But the factor $R_{0}$ is not disappearing. Indeed, If I do :
$$\dfrac{\text{d}(1+z)}{\text{d}t} = \dfrac{\text{d}z}{\text{d}t} = -\dfrac{H(t)}{R(t)}\,R_{0}$$
which implies :
$$\int_{0}^{z} \dfrac{c\text{d}z}{H(z)} = \int_{0}^{t}c\text{d}t\dfrac{R_{0}}{R(t)} = R_{0} \int_{0}^{t}\dfrac{c\text{d}t}{R(t)}$$
How to get rid of the factor $R_{0}$ ? Since if I multiply the comoving coordinate $r(t)$ by $R_{0}$, I get the cosmological horizon (the limit of observable universe if I integrate up to $z=1100$), don't I ?

According to this, the Hubble constant for redshift z is $H_0E(z)$. Meaning we need to prove that $$\int_{0}^{z_0}\frac{cdz}{H(z)} = \int_{0}^{t_0}\frac{cdt}{R(t)}$$
Take the first derivative of both sides of your second equation to obtain, by the chain rule, and the equation (this is the definition of the Hubble constant) $H = \frac{R(t)'}{R(t)}$ the following:
$$\frac{dz}{dt} = -\frac{1}{R(t)^2}*R'(t) = -\frac{H(t)}{R(t)}$$
That the redshift is zero for the current time, and some value greater than zero for some time before the current time should explain the minus sign. After multiplying with $c$ and placing the small changes on both sides this becomes $$\frac{cdz}{H(z)}=-\frac{cdt}{R(t)}$$, which looks like it can be integrated to obtain what we need to prove (the integral from z to 0 is the one from 0 to t).This proof is not very formal, but it's the best I can do, so I hope it is somewhat correct and that someone will give a more detailed answer if this is not good enough.

Thanks for your quick answer. Your demonstration seems to be correct. Unfortunately, the factor $R_{0}$ puts the mess. I point this in my UPDATE 1.

AFAIK R0 is 1, that is an agreement regarding the scale factor which has to be relative to sth

from your point of view, what is the dimension of $R(t)$ and $R_{0}$ ? since when we want to compute the cosmological horizon, i.e a physical distance, we multiply the comoving coordinate $r(t)$ by $R_{0}$ with $R_{0}$ the scale factor of today. Or is it the comoving coordinate $r(t)$ that has the demension of a length and $R(t)$ just a scale factor as it is called ?

Yes, the comoving distance has the dimension of length https://en.m.wikipedia.org/wiki/Scale_factor_(cosmology)


text:
Retrogradation movement of planet Mars relatively from Earth by Copernic
I  have a simple question : on the following figure:

I don't understand on the right figure why there is a progressive shift on the right when we start from step 1) to step 9). I guess there is an angle between the rotation plane of Mars and the rotation plane of Earth. Otherwise, we could'nt see clearly the recessing movement if the 2 planes were identical, could we ? We would just draw in the sky a simple line which would be the projection of the curve (on right figure) on a single Oy axis.
Anyone could explain me if this difference in two rotation planes is the cause of the shape of this curve (which is also due to the relative position of Mars from Earth) ?
Any help is welcome.

Is it due to the inclination of the Earth's axis? So the ecliptic appears to oscillate North to South over a year as seen from any point on Earth.

According to https://en.wikipedia.org/wiki/Mars the angle between the orbital planes of Earth & Mars (the inclination to the ecliptic) is 1.850°. That's small, but certainly noticeable: in comparison, the angular diameter of the Moon is about half a degree.

Just another question : why doesn't figure on the left is not rotated by an angle of -90 degrees to be more representative ?. the motions are almost horizontal and not vertical when we look at at planets in the the sky Any precision is welcome. Regards

You are right.  If Mars orbited in exactly the same plane as the Earth, instead of an S or a loop, we would see Mars moving prograde relative for the stars along the ecliptic, then slowing and stopping, moving retrograde for a few months, as Earth overtakes it, still on the ecliptic, then moving prograde again.
But Mars doesn't orbit in the same plane, so it has some motion perpendicular to the ecliptic. when these motions are combined, the usual effect is a "loop" or sometimes an "S".
(note that the axial tilt of the Earth is not relevant here)

Thanks for your quick answer. Nevertheless, the remark of @Steve Linton is interesting : why don't we take into account of the changing of inclination of the Earth's axis ? Indeed, this one is about 23 degree but planet Mars has alo an inclination roughly equal to 25 degree : so both are similar (difference of 2 degree) ? Is there a compensation effect or anything else since difference between the orbital planes between Earth and Mars is only 1.850° as pointed by @PM 2Ring ?  this is the same order of values.

Because the position of Mars on the celestial sphere is independent of your location on Earth, or the time of day, or the inclination of the Earth. Steve's comment is incorrect.  The position in the sky of the ecliptic changes over the year.  But it is the position of Mars on the ecliptic that creates the "S" and not the position of the ecliptic on the sky.


text:
If I know apparent planetary positions how do I calculate the date?
I have planetary position in retrograde  how do I calculate the date?
For eg, I know Mars was in retrograde in Scorpion (Sidereal 210 to 240 deg), how do I calculate the date and year of this?
As of now the only other info I have is that just days before this was a lunar eclipse where Sun was in Libra.
Just a year or two before this
Mars was in retrograde in first half of Leo and at this time Jupiter was placed at Capricorn. These are all sidereal observations.
P.S: This is regarding an ancient scripture citation where it was said Mars was in retrograde in Scorpion, current proposed dates range from 500 BC to 6000 BC. This is all the information I have, that Mars was is in retrograde in sidereal scorpion and I want to find possible date of this happening. The iterative approach would take too long, because the span is about thousands of years

Please be a bit more elaborate on the details and background of your problem. Otherwise a similarily fleshed-out answer would be use the known ephemeris and do an iterative search. Good guide: put as much detail in your question, showing the research you did yourself as you expect in an answer

@planetmaker thank you for your comment. This is all the information I have, that Mars was is in retrograde in sidereal scorpion and I want to find possible date of this happening. I tried the iterative approach and couldn't find it, because the span could be about thousands of years.

You have more information :) What is the accuracy you expect and the one of your input data? What's their source? What is the permissible time span, thus what do you try to (dis)prove? Something archeological? Futuristic? Have the dinosaurs seen it? How did you try the iterative approach? And why is it a problem to iterate over thousands of years?

@planetmaker oh ok, I will update those things

There are several related questions : https://astronomy.stackexchange.com/questions/4815/can-i-look-at-the-sky-and-find-the-day-of-the-week https://astronomy.stackexchange.com/questions/26410/position-of-specific-stars-in-the-sky-as-a-function-of-time https://astronomy.stackexchange.com/questions/34860/encode-place-and-date-in-star-pattern https://astronomy.stackexchange.com/questions/33150/is-it-possible-to-determine-an-observers-position-on-earth-from-a-photo-of-the?noredirect=1&lq=1

You seem to be dripping additional information.  How come you didn't know about the Lunar Eclipse yesterday when you asked the question?  Can you clarify what you mean by "siderial" since scorpio is now between 16h and 18h ra,  or about 240-270 degrees

@JamesK the thing is I'm a noob,  and I didnt know that Moon and Sun position could be related with Mars's retro in finding out dates. But when Mike in the answer said it Mars retro happens like every 2 yrs and asked for other positions,  then I added it. As for Mars I only have so much info.

@JamesK yes scorpio is 240 to 270 deg. Well some people use Tropical measurements. So wanted to explicitly differentiate this. Sidereal meaning reported as seen, if you went out and looked at that tine you would actually see Mars in Scorpion.

Mars's ~72 day retrograde loop occurs every ~2.1 years and intersects a given constellation once or twice every 15 or 17 years.
The more information you have about the positions of other planets, e.g. Jupiter or Saturn, the more you can narrow the set of possible dates.

So this is very frequent happening. As of now the only other info I have is that just days before this was a lunar eclipse where Sun was in Libra.Just a year or two before this Mars was in retrograde in first half of Leo and at this time Jupiter was placed at  Capricorn. These are all sidereal observations.


text:
Get the expression of probed volume between 2 redshifts
1) I can't manage to find/justify the relation (1) below, from the common relation (2) of a volume.
2) It seems the variable r is actually the comoving distance and not comoving coordinates (with scale factor R(t) between both).
The comoving volume of a region covering a solid angle $\Omega$ between two redshifts $z_{\mathrm{i}}$ and $z_{\mathrm{f}},$is given by
$$
V\left(z_{\mathrm{i}}, z_{\mathrm{f}}\right)=\Omega \int_{z_{\mathrm{i}}}^{z_{\mathrm{f}}} \frac{r^{2}(z)}{\sqrt{1-\kappa r^{2}(z)}} \frac{c \mathrm{d} z}{H(z)}\quad\quad(1)
$$
for a spatially flat universe $\kappa=0)$ the latter becomes
$$V\left(z_{\mathrm{i}}, z_{\mathrm{f}}\right)=\Omega \int_{r\left(z_{\mathrm{i}}\right)}^{r\left(z_{\mathrm{f}}\right)} r^{2} \mathrm{d} r=\frac{\Omega}{3}\left[r^{3}\left(z_{\mathrm{f}}\right)-r^{3}\left(z_{\mathrm{i}}\right)\right]
\quad\quad(2)$$
I would like to demonstrate it from the comoving distance with :
$$D_{\mathrm{A}}(z)=\left\{\begin{array}{ll}
{(1+z)^{-1} \frac{c}{H_{0}} \frac{1}{\sqrt{\left|\Omega_{\mathrm{K}, 0}\right|}} \sin \left[\sqrt{\left|\Omega_{\mathrm{K}, 0}\right|} \frac{H_{0}}{c} r(z)\right],} & {\text { if } \Omega_{\mathrm{K}, 0}<0} \\
{(1+z)^{-1} r(z),} & {\text { if } \Omega_{\mathrm{K}, 0}=0} \\
{(1+z)^{-1} \frac{c}{H_{0}} \frac{1}{\sqrt{\Omega_{\mathrm{K}, 0}}} \sinh \left[\sqrt{\Omega_{\mathrm{K}, 0}} \frac{H_{0}}{c} r(z)\right]} & {\text { if } \Omega_{\mathrm{K}, 0}>0}
\end{array}\right.
$$
UPDATE 1: It's been a long time that I posted this question.
I recently took over this isssue and I have done little progress, at least I think.
The FLRW metric can be expressed under following (0,2) tensor form :
$\left[\begin{array}{cccc}1 & 0 & 0 & 0 \\ 0 & -\frac{R^{2}(t)}{1-k r^{2}} & 0 & 0 \\ 0 & 0 & -R^{2}(t) r^{2}  & 0 \\ 0 & 0 & 0 & -R^{2}(t) r^{2} \sin ^{2} \theta\end{array}\right]$
If I consider only slice times constant, my goal is to compute the volume probeb by a satellite between 2 redshifts.

We can easily find that :

$$\int_{0}^{z_0}\frac{cdz}{H(z)} = \int_{0}^{t_0}\frac{cdt}{R(t)}$$

Then, If I consider a volume with $\text{d}r$, $\text{d}\theta$ and $\text{d}\phi$ coordinates, I have the following expression for determinant :

$$g=\text{det}[g_{ij}] = -\dfrac{R(t)^6}{1-kr^2}\,r^4\,\text{sin}^2\theta$$
Which means that I have :
$$\text{d}V=\sqrt{-g}\text{d}^3x = \dfrac{R(t)^3}{\sqrt{1-kr^2}}\,r^2\,\text{d}r\text{sin}\theta\,\text{d}\theta\, \text{d}\phi$$
$$V=\int\text{d}V= \int \dfrac{R(t)^3}{\sqrt{1-kr^2}}\,r^2\,\text{d}r\text{sin}\theta\,\text{d}\theta\, \text{d}\phi$$
$$V = \int \text{d}\Omega \int \dfrac{R(t)^3}{\sqrt{1-kr^2}}\,r^2\,\text{d}r$$
$$\rightarrow\quad V = \Omega \int \dfrac{R(t)^3}{\sqrt{1-kr^2}}\,r^2\,\text{d}r$$
with $\Omega$ the solid angle considered.
But as you can see, I am far away from the expression $(1)$ that I would like to find, i.e :
$$V\left(z_{\mathrm{i}}, z_{\mathrm{f}}\right)=\Omega \int_{z_{\mathrm{i}}}^{z_{\mathrm{f}}} \frac{r^{2}(z)}{\sqrt{1-\kappa r^{2}(z)}} \frac{c \mathrm{d} z}{H(z)}\quad\quad(1)$$
EDIT : Maybe I have found a partial explanation to my issue to determine the expresion of this volume between 2 redshifts. Here below a formula :

The main expression to keep in mind is :
$$d V_{C}=D_{H} \frac{(1+z)^{2} D_{A}^{2}}{E(z)} d \Omega d z\quad(3)$$

Could anyone explain me please the different justifications to introduce all the factors implied in this expression ?

I have not yet with this expression the same expression (1) at the beginning of my post, so could anyone manage to find (1) from (3) ?

Any help would be fine, I am stucked for the moment.

Don't cross-post. Pick one SE for your question.

@RobJeffries . Okay but I don't know which one to use, actually, the issue is about astrophysics/cosmology but given the fact that I can't get answer on physics exchange, I have put it here in the hope to get help. What do you advise me to do ?

uhm. Do I get that right, that you have an expression for the volume of up to a certain redshift, but you want to get the volume between two redshifts? What's wrong with just calculating the difference?


text:
Ecliptic coordinate ranges in radians
In ecliptic coordinates, what are the ranges of ecliptic longitude and latitude in radians? Do they both go from $-\pi$ to $\pi$?

It is purely a matter of convention.  Longitude and latitude are normally expressed in degrees, longitude from 0 to 360⁰ and Latitude from -90 to 90⁰.  If you express these in radians, longitude would be from 0 to $2\pi$ and latitude from $-\frac\pi2$ to $\frac\pi2$.
If you are writing software, your functions should be robust enough to accept any value out of this range, for example when calculating the position of a planet, it may be convenient to have a continuous mapping from time to longitude, this would necessitate longitudes that are greater than $2\pi$.

I agree with @JamesK's answer that it doesn't really matter.
The direction of longitude = 0 is the same in both cases, and it's really a matter of taste if you display in a plot or table 0 to $2 \pi$ or $-\pi$ to $+ \pi$ (0 to 360 degrees or -180 to plus 180 degrees).
For example in Python when I type np.arctan2(-5, 5) I get $-\pi/2$ (-90 degrees) but if I type =atan2(-5, 5) in Excel I get $+ 3\pi/2$ (270 degrees) so while they are mathematically equivalent, different calculators have different conventions.
If you want to convert between on and the other, to move zero between the center and the left side, use the modulo operation as follows:
$$\mod(-1, \ 2 \pi) = 2 \pi - 1 \approx 5.2831853$$
$$\mod(5.2831853 + \pi , \ 2 \pi) - \pi \approx 1$$


text:
How many solar masses is the central black hole in the Whirlpool Galaxy?
I am having trouble finding the mass of the central black hole in the Whirlpool Galaxy M51a/NGC5194, it is stated that its companion galaxy NGC5195 has a supermassive black hole at its centre that is 19 million solar masses but I can't find details on the Whirlpool Galaxy's central black hole, is its mass known?

Maybe it's hard to make a good mass estimate with all the dust obscuring the centre of NGC 5194.

Using the HyperLEDA mean central stellar velocity dispersion for M51a ($\sigma = 88$ km/s) and the $M_{\rm BH}$-$\sigma$ relation from Saglia et al. (2016), one would predict $M_{\rm BH} \sim 3 \times 10^{6} M_{\odot}$. That is not, of course, an actual measurement....

It is not well known, but a paper by M. Brightman et al gives a value of $10^{6.3\pm0.4}$ or between 8 hundred thousand and 5 million solar masses, while noting that this estimate is lower than previous estimates which had $10^{6.95}$ or about 9 million masses.
It seems that, although the galaxy is face-on to us, its black hole is viewed from the edge.  This means that the black hole itself is hidden behind a torus of relatively cool dust and gas. [ESA] which is visible as a dark streak in the image of the central part of the galaxy
There is much that we don't understand about the M51a black hole.  It is active, but not as active as would be expected, given the interaction between the two galaxies in M51. There is a second dark streak making an "X" the cause of which is unknown, and, as suggested by the wide error margins above, the mass of the black hole is not well constrained.


text:
Demonstration of scales for the two components (transverse and perpendicular) of Baryonic Acoustic Oscillations (BAO)
I transfer here a post that has not had any answer pn physics exchange, so I am going to delete this latter. Maybe I will have more luckiness here.
In an astrophysics context about BAO (Baryonic Acoustic Oscillation), it is written on the following paper BAO paper :
"BAO are much smaller in amplitude than the CMB acoustic peaks, and are washed out on small scales. BAO in the observed galaxy power spectrum have the characteristic scale determined by the comoving sound horizon at recombination, which is precisely measured by the CMB anisotropy data $[12] .$ In principle, the BAO scale can be extracted from data both in the transverse direction $\left(s_{\perp}\right),$ and along the line-of-sight $\left(s_{\|}\right)$, i.e radial. Comparing the observed BAO scales with the expected values gives the angular diameter distance $D_{A}(z)=r(z) /(1+z)(\text { where } r(z)$ is the comoving distance $)$ in the transverse direction, and the Hubble parameter $H(z)$ in the radial direction:
\begin{aligned}
s_{\perp} & \propto \frac{s}{D_{A}(z)}\quad(1) \\
s_{\|} & \propto s H(z)\quad(2)
\end{aligned}
]
Seo & Eisenstein (2007) (11] (henceforth SE07) provided simple fitting formulae for forecasting the accuracies of $s / D_{A}(z)$ and $s H(z)$ from future galaxy redshift surveys. In this paper, we will clarify the assumptions made in BAO forecasts, and investigate their implications. We will discuss the method in Sec.2, present results on BAO forecasts in Sec.3, and summarize in Sec.4."
Concerning my question, I don't understand the justification/origin of the 2 equations in $(1)$.
i.e :
$$
\begin{aligned}
s_{\perp} & \propto \frac{s}{D_{A}(z)}\quad(1) \\
s_{\|} & \propto s H(z)\quad(2)
\end{aligned}
$$
Indeed, assuming we have just a pure transverse component $s_{\perp}$ :
then, the angular under which it is observed is :
$\theta \simeq \tan(\theta) = \dfrac{s_{\perp}}{D_{A}(z)}\quad(3)$
with $D_{A}(z)$ the angular diameter distance, i.e the physical distance of "objects" at given redshift ( I think from the paper that it is equal $D_{A}(z)=r(z) /(1+z)(\text { where } r(z)$ is the comoving distance).
By the way, with this notation, $r(z)$ comoving distance is actually the cosmological horizon, i.e $r(z=1100) \approx 44\,\text{Gyr}$, isn't it ?
But back to equations $(1)$ and $(2)$.
So for a pure transverse component, I have a first element of answer for $(1)$ but not totally since I don't know how to make the link between $\theta$ and $s$ BAO scale.
And Moreover, what about the equation $(2)$ :
how to get it ? It makes think to the Hubble-Lemaître law of expansion but $s_{\|}$ is a distance and not a velocity. ok, there is a $\propto$ relation but I would like to  know where this relation comes from.
If someone could help me to proove these 2 relations $(1)$ and $(2)$. Naively, I would have thought that $s_{\|} = s_{\perp}$ if Universe is homogenous but this is not the case actually and I don't understand why ?
UPDATE 1: From this paper BAO scales , I am expecting inverted formulas for $eq(1)$ and $eq(2)$. Indeed, here an image from this given paper illustrating the 2 quantities :

So, compared to what is shown in $eq(1)$ and $eq(2)$, I would rather say that :
$$
\begin{aligned}
d_{A}(z)& =\frac{s_{\perp}}{\Delta\theta}\quad(3) \\
s_{\|} & = \Delta z \frac{c}{H(z)}\quad(4)
\end{aligned}
$$
From $eq(3)$, we could infer for $s_{\perp}$ :
$$s_{\perp} = d_{A}(z)\,\Delta\theta\quad(5)$$
So, as conclusion, how to make the link between (1),(2) definitions and ((4),(5)) : I don't grasp very well the parameter s :
Can we write that $$s^2=s_{\perp}^2+s_{\|}^2$$ ?
If this is the case, how to conclude to be consistent between the 2 papers ?
Any help is welcome.


text:
how long a light year on earth will seem
First of all, sorry if this is not the right StackExchange to ask this question, but i don't know a better place (open to suggestion in case).
So, the question:
Suppose that for some reason, we can travel at the speed of light, and lets suppose that there is no acceleration involved, so it's something like uniform rectilinear motion.
Let's say that today someone leave with this spaceship at the speed of light, the after one year (for him), he will have travelled 9460730777119.56 km, but how much will be spent on the Earth since he has leaved?
So, in other words, I read about a special planet 31 light years from Earth, and I was wondering, if we can travel at the speed of light and someone travels for 31 years to that location, how much older would I be from when he has leaved the Earth?

Sorry, your question cannot be answered in its current form. Nothing with non-zero mass can travel at the speed of light. But if the spacecraft is travelling very close to light speed, then for the passenger the journey would take much less than 31 years, (and slightly more than 31 years for the people on Earth). Please see https://astronomy.stackexchange.com/q/21022/16685

I’m voting to close this question because objects with mass can't travel at light speed.

@PM2Ring so in other words, traveling near the speed of light for 30years (for the traveler) won't be felt a lot longer on the earth?

@JamesK ok so suppose he is traveling with a velocity very close to the speed of light

Okay Berto99,  please [edit] your question.  I'm fairly sure there are similar questions here or on physics stack exchange, so perhaps have a little search around too.

Massive things can not go with the speed of light, but they can go with nearly the speed of the light. So I suggest to edit your question to ask from that. Furthermore, you are asking many questions here, ask only a single one. Third, ask this on the https://physics.stackexchange.com .

With enough speed, in 30 years ship time you could easily cross the galaxy & back. But when you got back to Earth, many thousands of years would have passed. You might like to browse http://math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html There's a fair bit of mathematics on that page, but you can skip over the hard stuff. ;)

You can never reach the speed of light but can get very close to it. Thus, when you travel with, say, 99.99% speed of light it will take you a little more than 31 years to reach the star 31 light years away. This is, however, the time it takes seen from earth. In the spaceship, the time taken will be dilated by
$$\gamma ={1\over \sqrt {1-{v^2\over c^2}}}$$
Which is roughly a factor of 70.7. This means in the space ship it is only 0.44 years or 160 days.


text:
Where did the idea that tidally-locked planets have a big hurricane come from?
I've been noticing a bit of a trend in the depiction of tidally-locked habitable planets, where they are shown having a huge hurricane-like storm over the daylight hemisphere. Here's an example, and another one in the background of this Cool Worlds video.
I'm wondering where the idea that the cloud formation over the daylight hemisphere would take the form of a single rotating storm came from. It doesn't seem particularly realistic: the Coriolis effect should tend to disrupt such a storm as it would lead to opposite rotation directions in the northern and southern hemispheres. The study by Yan & Yang (2020) "Hurricanes on Tidally Locked Terrestrial Planets: Fixed SST Experiments" predicts storms similar in scale to the hurricanes on our own planet, and the overall circulation (figure 3) appears to have separate vortices north and south of the equator rather than the single giant circulation in the artworks.

Isn't this an illustrator misunderstanding of the classic idea of a vast Hadley cell with cold air perpetually flowing from the nightside into the dayside and rising?


text:
Do retrograde spin-orbit resonances exist?
The end state of rotation of an initially fast-spinning prograde terrestrial planet (in the absence of additional forcings such as "thermal tides" in an atmosphere, e.g. Venus) is a spin-orbit resonance of the form $n + 1 : 2,\ n \in \{1,2,3,...\}$. Examples being the common 1:1 resonance ($n=1$), and the 3:2 resonance of Mercury ($n=2$). See also this paper (Correia & Laskar, 2001) discussing the probability of Mercury's capture into higher-order resonances, e.g. 2:1 or 5:2.
What about the case for a planet with retrograde rotation? Do there exist retrograde spin-orbit resonances, e.g. extending the sequence to negative $n$, giving resonances such as $-1:1$ or $-3:2$, and would there be any probability of capture into such a state?

Some papers: 2016 and 2020.


text:
Are there any predictions of what galaxies exist in the Norma cluster/ Abell 3267?
I am working on a map for a science fiction story and using what information is available I have a good idea of locations of nearby galaxy groups and clusters and where the known supermassive black holes are.
As the Norma cluster is a big area of debate and due to that and other areas being obscured by our own galaxies, observing what is in those areas is difficult. Going off Wikipedia this cluster has a binding mass of 1E15 solar masses, which is the same as the Virgo cluster and only one galaxy is mentioned which is ESO 137-001.
Is this mass based of our galaxies motion in that direction so mass behind the cluster (some believe the Shapley supercluster) is giving the value we have or is that a known mass of galaxies in that cluster? If so can we predict if it has a similar number of large galaxies and supermassive black holes as the Virgo cluster or is there other predictions for the Norma cluster?

@planetmaker the OP has now addressed it

VizieR lists two catalogues of Norma cluster members:

J/MNRAS/383/445 Woudt, P.A. et al. (2008) "Radial velocities in the Norma cluster (A3627)"
J/MNRAS/396/2367 Skelton, R.E. et al. (2009) "NIR Ks photometry of Norma cluster (A3627)" (this one contains angular diameter information)

The paper associated with the first of these, "The Norma Cluster (ACO 3627): I. A Dynamical Analysis of the Most Massive Cluster in the Great Attractor" describes the methodology used for estimating the cluster mass from their measurements of the radial velocities (in particular their dispersion) of the galaxies. From the paper:

For the determination of the dynamical mass of the Norma cluster, we have used both the virial theorem ($M_{\rm VT}$) and the projected
mass estimator ($M_{\rm PME}$), see equations (21) and (22) of Pinkney et al. (1996). The use of the biweight velocity centroid and scale (Beers et al. 1990) in the virial theorem (instead of the velocity mean and
standard deviation) leads to a more robust mass estimate ($M_{\rm RVT}$).
The latter is more robust against the effects of contamination by the
inclusion of possible non-members in the analysis. The projected
mass estimator (Bird 1995), on the other hand, is sensitive to the
presence of (spatially separated) subclusters due to its proportionality to the projected distance between galaxy $i$ and the cluster centroid
($R_{\rm \perp,i}$) (see equation 22 in Pinkney et al. 1996).

They also note that the masses they derive are consistent with masses estimated from the X-ray emission by Böhringer et al. (1996) and Tamura et al. (1998).
As regards what types of galaxy these are, unfortunately neither of the catalogues provide morphological types directly. On the other hand, they do include for many of the objects a reference to the WKK98 catalogue (Woudt, P.A. & Kraan-Korteweg, R.C. 1998).
WKK98 is also available on VizieR as J/A+A/380/441 and does include the morphological type.
You can use the VizieR query interface to join the tables (or you can download the tables in their entirety via the FTP page on VizieR and do the joining on your local machine, e.g. via the various lookup functions in your spreadsheet software of choice), which will give you a list of the morphological types of the objects.

Thanks, I found this paper too which goes into further details https://academic.oup.com/mnras/article/383/2/445/992596


text:
Origins of meteorites?
I am aware that the vast majority of meteorites come from the asteroid belt (including 4 Vesta), with some from the Moon and Mars. Are there any other locations from which meteorites have originated (the Kuiper belt? Ceres?). Here it mentions that carbonaceous chondrites formed farthest from the Sun. Is that still within the asteroid belt?

Well there's the ones from Japan, but I'm not sure they count ;-) Why would a box full of 1cm balls released into LEO be so scary to an engineer supporting the ISS?

Kuiper belt (and beyond) objects do sometimes come into the inner solar system: we call them "comets".  However comets have a large proportion of volatiles (water etc) and the non-volatile matter tends to be in the form of small dust particles, neither of which form useful meteorites  Ceres is also covered mostly in ice, so even if some was ejected into space and reach the Earth, it wouldn't survive the descent.
An older paper wonders if very old dead comets might be a source of meteorites, but notes that none have been found. A much more recent paper describes a small (100 µm) particle of cometary dust found embedded in another meteorite.  This dust particle would have come from the outer solar system.


text:
What is the difference between aurorae and electroglow?
One of the discoveries of Voyager 2 at Uranus was a phenomenon called "electroglow", which as I understand it is related to charged particles interacting with the atmosphere that cause the emission of ultraviolet radiation. What I am not clear on is how this process is different to the production of aurorae. Is there some clear difference between the two processes that allows them to be distinguished?

Aurorae are caused by charged particles (usually electrons) being drawn into the polar atmosphere by the planet's magnetic field.
Airglow is caused by just photons interacting with the atmosphere.
Electroglow requires both electrons and photons.
Source: The Free Library  (article is under express copyright, see the paragraph that starts "Voyager 2")

I understand that the copyright dissuade one from even a brief quote, that still makes this a link-only answer. Without going offsite, I still do not know what electroglow is nor why it requires both electrons and photons. You can certainly paraphrase what your source says without violating any copyright.


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Why both sunrise and sunset use the position of the sun's upper limb?
It is my understanding, that astronomical moments of both rise and set use the relative position of background object's upper limb to the foreground object's horizon.
When were these definitions formulated?
Is this done purely because sunrise and sunset are the main examples of such events and it is easier for the naked eye to tell when sun is completely below horizon?
Wouldn't this offset the symmetry of rotations (since the "day" is always longer than the "night" by the duration it takes the background object to move fully below the foreground horizon)?

I would guess that then the upper limb disappears the light simply disappears even if visibility is poor. If you want to determine when the lower limb touches, it requires much better imaging perhaps. e.g. 01:30 here: https://youtu.be/lwus2nqU0SY Compare to the funny business at the bottom of the moon making it difficult to determine when it last touches the horizon here https://youtu.be/LEH0WmCZTAo

Ya, that might in fact be right and my guess is wrong!

When were these definitions formulated?

Multiple religions have rather strict requirements based on sunrise / sunset, for example, when one must start or can stop fasting, or when exactly to sacrifice a lamb (or a human in some religions). The definition has very deep roots.

Is this done purely because sunrise and sunset are the main examples of such events and it is easier for the naked eye to tell when sun is completely below horizon?

Yes.

Wouldn't this offset the symmetry of rotations (since the "day" is always longer than the "night" by the duration it takes the background object to move fully below the foreground horizon)?

Daytime is not longer than nighttime in the middle of winter, in places that do have winter. The definition of sunrise and sunset does make daytime be a bit longer than nighttime on the equinoxes.
Suppose sunrise was instead defined as the time at which the entire Sun first appears above the horizon and sunset was instead defined as the time at which the first bit of the Sun drops below the horizon. Even with these extreme definitions, daytime would still last longer than nighttime on the equinoxes thanks to atmospheric refraction.

The skew towards longer day than night at equinoxes is also - and even more - by the effect of refraction, that is the sun is physically already below the horizon but the atmosphere still bends the sun rays towards the observer. This effect is around arc minutes, thus bigger than the sun's diameter.

@planetmaker - That is exactly what the last paragraph in the answer addresses.

https://www.timeanddate.com/astronomy/refraction.html has a nice picture showing the apparent sun and astronomical sun, and states that the difference may be "several minutes". It would be nice to have some kind of typical value or range here.