text:
Why aren't there eclipses every month when the moon's orbit is aligned with the ecliptic as a result of lunar precession?
I can understand why we don't have eclipses every month when the inclinations of the ecliptic plane and the moon's orbital plane are different. But the inclination of the moon's orbital plane changes from 18.5 degrees to 28.5 degrees and back to 18.5 degrees every 18.6 years (as a result of lunar precession). In other words, once every 9.3 years, the inclination should be 23.5 degrees, aligning the moon's orbital plane with the ecliptic plane. And yet, we don't see eclipses happening at every new moon and full moon when the two planes are supposedly aligned. Why is that?

Moon's plane of rotation around the earth is at an inclination of some 5 degrees from the plane of earth's rotation around the sun. This doesn't change over years, I think

See https://xkcd.com/1878/

The precession of the moon's orbital plane does not align it with the ecliptic plane. The angle between those two planes is approximately 5.14° (it varies by ~±0.15°, mostly due to perturbation by the Sun). Precession causes the orientation of the lunar plane to vary, but the angle between the two planes stays (almost) the same. It's very similar to the precession of this gyroscope (courtesy of Wikipedia), except that the fixed point of the lunar plane is at the centre of the plane (i.e., the Earth-Moon barycentre).

As the gyroscope precesses, the plane of the gold disk maintains a constant angle to the horizontal plane.
So although it's certainly true that the angle between the lunar plane & Earth's equatorial plane varies between ~18° and ~28°, the angle between the lunar & ecliptic planes stays (relatively) constant.
It's difficult to draw a good diagram (or animation) to illustrate this. I've created an interactive 3D animation which may be helpful. ;) By default, it exaggerates the tilt of the lunar plane to 10° because it's a bit hard to see what's going on if you use the correct value of 5.14°.
As mentioned above, the lunar plane actually passes through the Earth-Moon barycentre (as does the ecliptic plane), but in this diagram those two planes pass through the centre of the Earth to keep things simple. That is, all three planes (ecliptic, equatorial, & lunar) pass through the centre of the Earth. The ecliptic plane is grey, the equatorial plane is blue, and the lunar plane is orange. There are matching coloured lines associated to the lunar & equatorial planes which show their axes and where they cross the ecliptic.
The large black circle in the centre of the diagram is in the ecliptic plane. The axis of the lunar plane touches the two small black circles at the top & bottom of the diagram.
You can rotate, pan & zoom on the diagram using the mouse, left mouse button, and scroll wheel. On a touchscreen, use 1 finger to rotate, 2 fingers to pan and zoom.
I recommend first toggling the equatorial plane off so you can more clearly see the basic precession of the lunar plane.
The steps parameter controls the number of animation frames. Increase it for a slower, smoother animation.

The equatorial plane also precesses, but with a period of approximately 26,000 years. The lunar plane precesses with a period of ~18.6 years. Wikipedia has a good article on the various lunar precessions.

Here's a plot (courtesy of JPL Horizons) showing the small variations in the inclination of the Moon's orbit over a couple of years.

"As mentioned above, the lunar plane actually passes through the Earth-Moon barycentre" – it's also the earth-lunar barycentre which is orbiting the sun in the ecliptic plane, so I guess that cancels out.

Good point, @Paŭlo. I'll make that explicit in my answer.

Thank you so much! The 3D interactive animation is very useful!

My pleasure, @Priya! I just added a graph to my answer.

you have several excellent answers about the Moon's inclination; in Space SE this 3D puzzler was just posted: Why does the moon’s transit appear so inclined in this STEREO-B video?

There are few things here I think might be worth to state:

The tilt of Earth is of no importance here. As the comment says what is of importance how much the Moon orbit is inclined to the ecliptic. Now, one can say that if both Sun-orbit and Moon-orbit have the same inclination to Earth equator (namely 23.5 deg) it means the Moon and Sun orbits have no inclination with respect to each-other - and they are actually on the same plane; But this is wrong conclusion. For not only the inclination itself is important but its direction vector also: i.e., where we reach the top/button. and where are the nodes (the points of intersection between the planes; on the Sun-orbit those also called equinoxes). So even if Moon's orbit were always inclined 23.5 to Earth equator, still it would not mean we would have eclipse every month -- we would only if the nodes are in the same position as the nodes of the Sun-orbit;

And this is exactly the catch here - when the Moon-orbit is inclined 23.5 deg to Earth tilt - this always happens in the same two specific directions of the inclination vector (against the equator). Those directions are not as the direction of the ecliptic inclination vector against the equator.

This is because the Moon-orbit is always tilt 5 deg. against the Sun-orbit (aka ecliptic). So in the period of this node precession of 18.6 years: when the limits collide (so also the node - so the direction vector is the same) the inclination to Earth equator is 28.5 deg or 18.5 deg; When they don't - the direction vector is not the same.

There is somewhat subtle issue here that cause me to edit this answer is that the node precession of Moon-orbit with respect to the equator is not complete. in the sense the nodes (again with respect to the equator) are not running over 360 deg in 18 years. But they run over much smaller span of maybe around 25 deg, ~12.5 deg from each side of the solar equinox. The precession works as regular (360 deg in 18 years) when we consider the nodes of the intersection not with Earth equator but with the ecliptic.

Thank you! This was very easy to understand.


text:
Are gravitational waves affected by gravity?
Are gravitational waves affected by gravity? When a gravitational wave passes by objects, do their gravitational fields warp or change it?

A closely related question, perhaps a duplicate.  https://astronomy.stackexchange.com/questions/44297/do-gravitational-waves-produce-a-surrounding-gravity-field?rq=1

No , it even makes a case.

Another closely related question: Redshift for gravitational waves?

Gravitational waves can even be gravitationally lensed https://www.birmingham.ac.uk/news/latest/2021/05/scientists-hunt-for-evidence-of-%27lensed%27-gravitational-waves.aspx

Sorry ,but that didn't help .I think it was just basic info .

Gravitational waves are just the way a change in a gravitational field is communicated. If the change oscillates, you get oscillating wave forms like we see from binary systems that make up the waveforms we all know and love.
Since these waves are (according the the GR way looking at gravity) oscillating propagating parts of spacetime, they propogate through other parts of spacetime such as other curved spacetime. The geometry of the curvature of the change in medium does (contrary to what I previously thought) affect the gravitational waves; one example of this the redshirting of the gravitational waves when they pass through a gravity well. For similar reasons of light, the medium itself is stretched and curved causing gravitational redshift in the gravitational waves themselves. Furthermore, if space is curved, naturally the gravitational waves’ direction will be affected, but in a more complicated way than just curving like a photon since they can interfere with each other and the differences in curvature from one spot would lead to some complicated behavior and nonlinear interference patterns.
In short, yes, gravity does affect gravitational waves, and probably in more ways than listed here.

The answer most certainly is "yes", not "no". That gravitation itself causes gravitation is one of the vexing qualities that make it so hard to make general relativity consistent with quantum physics.

@JustinTacket gave the explanation also

@DavidHammen I personally would appreciate if you could expound upon this in another answer, maybe there’s something I’m overlooking here

Ok did a bit more digging and I see what you mean; editing to fix

@DavidHammen Reminds me of an ancient rock/radio play "Flash Fearless," with the lines "How come we fell UP??"  -- "You see, in this part of the galaxy, gravity is not so much a law as a local ordinance."

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
Great Attractor's gravity vs Universe Expansion
I would like to know if the trajectory of our galaxy has been calculated because it is usually said that the cosmos is emptying due to the expansion of the universe but at the same time there is a region of space that we are traveling to along with many other galaxies in the Laniakea supercluster.
Space is expanding in all directions and getting faster and faster but due to the proximity to the Great Attractor, will the Milky Way end up joining together in a sea of galaxies or will it be left alone with Andromeda and satellite galaxies?

There is a recent paper (Shaya, E., Tully, R. B., Pomarède, D., & Peel, A. (2022). Galaxy flows within 8,000 km/s from Numerical Action methods. arXiv preprint arXiv:2201.12315, more material ) that addresses this, and has some nice animations.  In particular, this video shows an extrapolation of galactic cluster trajectories 10 billion years into the future.
Note that most parts show locations in co-moving coordinates that expand with the universe (much more convenient for examination), but about a minute in there is a demonstration how the distances actually scale when the expansion is plotted.
In particular, it turns out that while the Local Cluster is moving towards the Virgo cluster and the core of Laniakea with the Great Attractor, the speed is not enough to catch up with them. The end result is that we will be isolated.

In the long run, the expansion of space will win out over any gravitational attraction our galaxy feels towards the Great Attractor.
First off, we are not gravitationally bound to the Great Attractor. Even if the universe was not expanding, the fact that we are not gravitationally bound implies that we would eventually leave the gravitational influence of the Great Attractor (though that would likely take billions upon billions of years).
Secondly, the expansion of the universe will one day become so dominant that even the closest clusters (what's left of them anyway) will succumb to the expansion of the universe. Nothing that isn't gravitationally bound will be able to fight that expansion. The expansion will win out in the end. At least, that's what current observations and theory tell us we can expect...

How does your last few sentences work? This is only true if $w<-1$?

@ProfRob Unless I'm much mistaken, it is the general consensus that that is how things will turn out under (debatable) reasonable assumptions. E.g. this source or this one.

Neither source says that the expansion would have an influence on our Galaxy, or any other gravitationally bound system. That would require phantom energy with $w < -1$.

I was under the impression that the coined "great attractor" was more like a force that is attracting the universe within on itself leading to the theorized big crunch of the universe. And that it is the force opposing the expansion of the universe? Or is it the void that seems to surround the universe?

You may be confused by what is being asked in the question.  The question is referring to https://en.wikipedia.org/wiki/Great_Attractor

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


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Time in 0 gravity points
If being close to a supermassive body like a black hole makes time pass more slowly for us than for an observer from a point of view with a weaker gravitational field, if we get to be at a point in space where there is no gravitational influence, would that make X time on earth a long time at that point?
I'm not talking about a Lagrangian point, where as I've studied, the field strengths cancel each other out. (I am a high school student from Spain).
If we could spend our lives at a point with this characteristic, would we be longer lived or would we still live ~85 earth years?
Has it been calculated what would be the time equivalence in such a place?

remember that perspective of time is the same for all the observers irrespective of the time dilation. If a person experiences 1 second on earth, the person next to a SMBH will also experience the same 1 second, in simple terms. if you are familiar with the term spacetime, your time co-ordination is rotated to certain angle respect to another frame in time dilation if you live for 85 years on earth, you will live the same 85 years near a super massive black hole

Your edit, removing the part about life-span, makes the existing answers look odd. Please can you revert the edit (or I can do it for you).

BTW, gravitational time dilation isn't exactly a function of gravitational field strength. It's correlated with the difference in gravitational potential energy. See https://physics.stackexchange.com/a/564031/123208 FWIW, the time dilation due to Earth's gravity (compared to the time where the gravitational potential due to Earth's gravity is zero) is approximately 2.19 seconds per century. You can calculate it using the equation here: https://physics.stackexchange.com/a/528660/123208

@JamesK I reverted the edit. Thank you for the recommendation, I'm new here.

I saw your question on Phil SE. I should warn you that you are unlikely to get good advice there, as many cranks are running wild there and hardly any of the users of Phil SE even notice. To search for truth, analytic philosophy is indeed necessary, together with empirical science, but what is trendy in philosophy today is mostly irrelevant to truth.

makes time pass more slowly for us

This is a fundamental misunderstanding of time dilation, which only says anything about the relative rates that clocks run compared to a clock that is in your own frame of reference.
All observers "experience time" in the same way.
The second problem with your post is that "gravitational time dilation" is not directly connected to the gravitational field strength. What matters is the gravitational potential. The difference in the relative rates at which clocks run (according to observers with those clocks) depends on the difference in gravitational potential.
A region with "zero gravitational field" is just a place where the gradient of the potential is zero. It says nothing about the gravitational potential itself. For example, the gravitational field at the centre of the Earth is zero but clocks here would appear to run slow compared with clocks on the Earth's surface.
An observer on (the surface of the)  Earth will judge that a clock in orbit is running faster (ignoring any time dilation due to relative motion); a clock a long way from the Sun will run a bit faster still and one beyond our Galaxy a little bit faster again.
Edit: For example, the size of the effect can be approximated by saying a clock far from the Galaxy would run at $\sim 1 + GM/Rc^2$ times the rate of a clock orbiting the Galaxy at the distance of the Sun (there is also time dilation due to relative motion, which I'm ignoring), where $M$ is the mass interior to the Sun's Galactic orbit and $r$ is the radius of that orbit. $M \sim 10^{11}$ solar masses and $r \sim 8000$ pc gives an increased rate of $1 + 6\times 10^{-7}$ (or about 20 seconds per year).

Can we quantify this "a little bit faster"?

@PaŭloEbermann about 20 seconds/year - see edit

Question: If we could spend our lives at a point with this characteristic, would we be longer lived or would we still live ~85 earth years?
Simple Answer: No
Abstract

Perspective of time is the same for all the observers irrespective of the time dilation. If a person experiences 1 second on earth, the person next to a SMBH will also experience the same 1 second, in simple terms. if you are familiar with the term spacetime, your time co-ordination is rotated to certain angle respect to another frame

Have look at this for more clarity.
To understand about time dilation, I hope you are familiar with Euclidian Metric, if not, it just about three dimensional graph where you will find some co-ordinates using Pythogoras theorem, along with that you will have a separate one dimensional graph which is time, if you record an event in this four co-ordinates (i.e t,x,y,z), you are recording events in spacetime !
Each person will have their own graphs of spacetime, where each one of them will have their own "cloaks" to say so, their each person's graph is rotated with respect to one another.
Lets visualize it even simpler:
Suppose there is no Time dilation our time graphs would be parallel like this:

if you record an event at Time B, I will also record the event to be happened at B' where these two readings are the same.
But in time dilation my time graph is rotated with respect to your time graph like this:

if you record an even at B, I will record the event slightly ahead of B', So we can say that your clock and my clock are running differently.
Its important to note that the distance between time A and B is same as A' and B', only thing is it is rotated, So the perspective of time is same for all the observers
This will answer your question, if you live 85 years at Earth, you will live the same 85 years near a supermassive blackhole.
Question 2: Has it been calculated what would be the time equivalence in such a place?
Einstein general relativity will show you how time dilates in different circumstances. You cannot calculate time dilation without a proper reference frame.
Again even if you are under the influence of "zero Gravity", you can still experience time dilation if you are moving relatively. Have a look at this


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Accurate formula for calculating the mass of an exoplanet using the transit method
A friend and I did some work on exoplanets with the help of a research institute (IEEC in Spain) for a major high school project. What we did was to "redetect" the exoplanet XO-6b through the transit method. This is the scientific paper of the official discovery https://arxiv.org/pdf/1612.02776.pdf . With the data and information provided we could only extract the flux-time curve and with it calculate the radius of the planet. Once the radius was calculated, we wanted to go further and know more characteristics of the planet such as mass, period, distance to its star... What happened is that we needed the mass and we discovered that the most natural way to calculate it was through the method of radial velocities to which we did not have access. What we decided to do was to look in books and we found the following formula that approximated the mass of the planet:
$$M_P = \left(\frac{R_P}{R_\oplus}\right)^{2.06}M_\oplus$$
$$M_P = \left(\frac{162549\cdot10^3}{6370\cdot10^3}\right)^{2.06} \cdot 5,972\cdot10^{24}$$
$$M_P = 4,230\cdot10^{27} kg$$
$$M_P(M_J) = \frac{4,230\cdot10^{27}}{1,89\cdot10^{27}} = 2,48M_J$$
The problem with this formula is that it is imprecise and the result is not satisfactory as it is not within the range of the official finding which is M XO-6b = 1.9 ±0.5 MJ. Our result was 2.48 MJ (Masses of jupiter). Considering that our radius calculation already deviated a bit from the official one it is not a bad approximation but I would like to know if there is a more correct way to calculate the mass just knowing the radius and having the flux-time curve. With the mass we wanted to calculate other variables like the orbital period or the semi-major axis but the results deviated a lot.

You are accepting answers to hastily. Give others a chance to add their own answers or indeed find the mistakes in mine. There's no rush.

@ProfRob Okay, thank you for the recommendation, I'm new here.

I don't know how you would want to derive the mass of the planet without the radial velocities of the star. And the paper you linked to clearly states that the radial velocities can not be reliably measured as the star rotates too fast for this (the spectral lines are too broad). They only can give an upper limit of $4.4 M_J$. The value of $1.9 M_J$ they mention is not much more than a guess. I also don't know why you need the mass of the planet for the orbital period or the semi-major axis. Both should only depend on the star's mass.

@WiseMode I have converted the inline image to MathJax.  Please check that I have done so correctly and consider using MathJax in the future.

No, there is not. The transit method can only give you the radius of the planet (unless there are multiple planets in the system, in which case there is a method that uses "transit timing variations" that cannot be applied in your case I think).
What your approximate formula does is to assume some relationship between the mass of a planet and its density (or equivalently, between the radius of a planet and its density). It is probable that this formula is empirically derived from measurements of exoplanets that have both a mass and a radius. But that relationship will have considerable scatter, because it seems clear that exoplanets have a diversity of internal structure and composition, even at the same mass.
Even worse, once you get up to Jupiter-mass exoplanets, they all have a very similar radius (to within a factor two) for a broad range of masses. So trying to determine an exact mass from the radius is impossible.
The plot below of mass vs radius for exoplanets (from Santerne et al. 2013) where both parameters have been measured, should illustrate the problem.

If I'm reading your question correctly, the transit gave a radius of 25.5 Earth radii. Looking at the plot above you can see this is a complete outlier. Only a handful of giant exoplanets are as large as this and it is probably due to them being very close to their host stars and being heated by it. If this was the only information  you had, you could probably only argue that your "planet" was more massive than about 0.3 M_J and in fact has a radius that would be more appropriate for a brown dwarf or a small star.
Note that orbital period is just the time gap between adjacent transits in the light curve and that the semi-major axis can then be derived from Kepler's third law under the reasonable assumption that the mass of the planet is much less than the mass of the star.


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What are the applications of cross matching astronomical catalogues?
Why do we cross match astronomical catalogues at different wavelengths? What information can we determine from combining the information from two catalogues?
One of the applications I read was creating a catalogue of Active Galactic Nuclei (AGNs) by searching for optical counterparts to objects appearing in radio catalogues. However I'm not sure if this information is correct. Are there any other such similar applications of cross matching?

I did work in the 90s on cross-matching optical catalogues with catalogues of X-ray emitting sources. The main reason for doing this was to figure out which object was responsible for the X-ray emission, since there could be one, many or no optical counterparts within the "error circle" of the X-ray source position.
Armed with a list of counterparts one could then go to a telescope and see whether the counterparts betrayed other signs that they might be responsible for the X-ray emission. In my case, I was looking for examples of rapidly rotating young stars, with high levels of coronal X-ray activity. Thus the cross-matching was used to locate these stars in the first place. The paired data could then be used to explore the dependence of coronal activity with other stellar parameters like temperature, gravity and rotation.
This is an example. Your question is very (too) broad. If I were to summarise, it is that astrophysical objects emit at multiple wavelengths and a full understanding of what is going on requires the full picture.

If I’m understanding what you’re asking, then it boils down to two things: resource management and different physics.
Let’s say you’re looking at a given variable star. Maybe you just started researching it and you want some observations done on it to study it’s internal structure or something. Getting access to a telescope capable of doing that is not trivial; maybe you’re lucky and you have one at your institution you can use, but maybe ground observations are not very good for the data you want to collect, and so you want to get access to some space telescope time. Imagine the feelings of embarrassment if you submitted a proposal to observe this target, just to find out it’s already been observed indirectly a dozen times, and that the data you needed was already out there, a product of collateral benefit from other surveys/ missions.
Its so important for individuals to know what kind of data exists from different observations, so that we don’t end up looking at the same object a dozen times because we didn’t know we had already looked at it. That’s why websites like MAST collect data from a bunch of different catalogues to give you reliable information of what has already been done for a given object.
The next part has to do with physics; you mentioned AGN specifically. Different kinds of electromagnetic radiation are created in different processes, and these processes can give different insights as to what kind of processes are happening. Maybe high energy photons would be useful for studying the jets while you want to look at lower energy photons to study the accretion disk.
tl;dr: We got to know what we got so we don’t accidentally waste time getting it again, and we want different wave lengths to study different kinds of features
Let me know if this isn’t what you meant and I can adjust.


text:
Can you get motion data using astropy.coordinates.get_body()?
I am trying to use get_body() to get position and motion data for solar system objects.
I’ve tried the following:
from astropy.time import Time
from astropy.coordinates import solar_system_ephemeris, get_body

t = Time('2019.5',format='jyear')

mars = get_body('Mars',t,ephemeris='de432s')
print(mars.ra)
print(mars.dec)
print(mars.pm_dec)

The first two print statements work fine. The last line throws this error:
TypeError: Frame data has no associated differentials (i.e. the frame has no velocity data) - represent_as() only accepts a new representation.

The same thing happens if I try any other solar system body as well.
Can I get motion data (such as the rate of change of the declination) using get_body()?

"Proper motion" is the name used for the motion in the sky of stars. It is not a velocity. It is not measured in metres per second, but in arcseconds per year. If a star was moving directly towards Earth, its proper motion would be zero. Proper motion is "proper" because it is not the apparent motion caused by parallax as the Earth moves around the sun.
So you can't talk about the proper motion of the Earth. The Earth is not a "heavenly body", when viewed from Earth.  We don't talk about "proper motion" of planets either. For a planet, the parallax motion is much larger, and of course the motion of the planet is an orbit. Whereas the proper motion of stars is linear. The proper motion can be found in a database, the planetary motion has to be calculated.
If you want the velocity of a planet relative to it's orbital barycentre you have get_body_barycentric_posvel it returns a tuple of (position, velocity) as cartesian coordinates (x,y,z)
>>> m = get_body_barycentric_posvel('mars',t)
>>> m[1]
<CartesianRepresentation (x, y, z) in AU / d
    (-0.00981358, -0.00759165, -0.00321724)>

I don't know if you posted your response before I made an edit but I realized the contradiction and I changed Earth to Mars.  The fact remains that the error occurs for ALL solar system bodies.

Based on what you said about proper motion of planets, does that mean I can't look up the rate of change of RA and dec, for a given moment in time, of a planet in a database?  I am able to do so using JPL Horizons

The rate of change of declination (or of RA) is a type of velocity.  It's an angular velocity in spherical coordinates

I hadn't seen your edit,  Horizons does have rates of change of ra and dec. but it doesn't look them up in a database, they are calculated. Of course the values of ra and dec are also calculated for planets, so it could be done. Of course a cheap and cheerful approximation can be found using a linear or quadratic approximation to the curve.


text:
Is it possible to block the surrounding light in a solar eclipse if we made the moon bigger or closer to the earth?

Is it possible to block the surrounding light in a solar eclipse if we made the moon bigger or closer to the earth? In an eclipse, you always see a ring of light surrounding the moon, but I am wondering if it's possible to completely block any light so you only see a dark circle in the sky. How much bigger does the moon have to be in order to achieve that?

Just take a photo during totality, not seconds before or after.

Any picture? When I google total solar eclipse, I still see a ring on all photos.

The fact that you see a "dark circle in the sky" is because the moon is silhouetted against the sun's outer atmosphere.

It's not possible to make the moon  bigger or closer to Earth...

@James K.  Possibly Sayaman was asking about the scientific possilbility of an Earthlike planet with a larger or closer moon.  It is possible for a habitable planet to have a moon larger and/or closer than the Moon is.  Since the Moon is slowly moving away from Earth due to tidal effects, the Moon was closer to Earth in the past.  So Earth is an example of a habitable planet with a Moon sized moon that was once closer than the Moon and will in the future be farther than the Moon.

Possibly...I care not to speculate

@James K  Furthermore, there is no law of science making it it impossible to enlarge the Moon, it would merely be a very, very, very, very, big megaproject as I explain in my answer.  For example, if the Moon was terraformed to have a breathable atmosphere a roof would ahave to be built all over the Moon to hold the air in, thus increasing the diameter of the Moon at least slightly.

@Sayaman  You question seems to have two possible meanings to me, so I answer both in my answer.

There seems to be a lot of ultra-concrete responses here.   Let's just answer with a simple example:  if you're in the ISS, the Earth provides a complete eclipse, including the sun's corona.

Totally unrelated to the question itself - that picture keeps making me squint at it, even though I know it's just on my monitor.

@Sayaman sorry, your question is confusing to me. If you "completely block any light" around the moon, then you can't see the moon's silhouette. The background around the moon is complete darkness, and the moon doesn't have any light source so you wouldn't see the moon itself either, it would also be completely dark. "so you only see a dark circle in the sky" - how can you see a dark circle against a dark background? I'm not sure about what is it that you're trying to see.

Let's clarify what "ring of light" you want to cover. In the original photo, the bright spot is the photosphere ("P" in the annotated copy or the original image). The photosphere is the bright disk you see when the Sun is not in an eclipse. This photo was taken seconds before or after totality, so the a small fraction of the photosphere was still visible. The complete photosphere would have been covered a few seconds earlier or later, and the "Sun" would not be visible.

Slightly visible in the original photo is the inner corona. The corona is the outer atmosphere of the Sun that is only visible during an eclipse. During totality, the corona extends out much further than "a ring". As seen in this photo (2017 Aug 21), the corona extends out at least 5 times the diameter of the Moon.

Whether you want to make the apparent diameter of Moon 5% larger (to cover the inner corona) or 400% larger, the technology do either one is not realistic.
A cheaper alternative using existing technology is to ride in a spacecraft to be closer to the Moon. You can choose how much of the Sun you want to be blocked.
Incidentally, millions of people see the Sun completely covered every day. It's called night.

Because you didn't discuss the different types of solar eclipses you give the missleading impression that all solar eclipses are total.

+1 for Incidentally, millions of people see the Sun completely covered every day. It's called night.

"Based on this decrease in the star's brightness, I believe it is orbited by at least one planet."

@Bernat I do not know. After all, they are not actually seeing the Sun :-p.

Part One:
Eclipses of the Sun by the Moon come in four types:

A total eclipse occurs when the dark silhouette of the Moon completely obscures the intensely bright light of the Sun, allowing the much fainter solar corona to be visible. During any one eclipse, totality occurs at best only in a narrow track on the surface of Earth.[6] This narrow track is called the path of totality.[7]

An annular eclipse occurs when the Sun and Moon are exactly in line with the Earth, but the apparent size of the Moon is smaller than that of the Sun. Hence the Sun appears as a very bright ring, or annulus, surrounding the dark disk of the Moon.[8]
A hybrid eclipse (also called annular/total eclipse) shifts between a total and annular eclipse. At certain points on the surface of Earth, it appears as a total eclipse, whereas at other points it appears as annular. Hybrid eclipses are comparatively rare.[8]
A partial eclipse occurs when the Sun and Moon are not exactly in line with the Earth and the Moon only partially obscures the Sun. This phenomenon can usually be seen from a large part of the Earth outside of the track of an annular or total eclipse. However, some eclipses can be seen only as a partial eclipse, because the umbra passes above the Earth's polar regions and never intersects the Earth's surface.[8] Partial eclipses are virtually unnoticeable in terms of the Sun's brightness, as it takes well over 90% coverage to notice any darkening at all. Even at 99%, it would be no darker than civil twilight.[9] Of course, partial eclipses (and partial stages of other eclipses) can be observed if one is viewing the Sun through a darkening filter (which should always be used for safety).

https://en.wikipedia.org/wiki/Solar_eclipse#Types
The orbit of the Earth around the Sun is elliptical, making Earth sometimes closer to and sometimes farther from the Sun. Since the physical diameter of the Sun, doesn't change, that means that the apparant or angular diameter of the Sun as seen from Earth changes.
The orbit of the Moon around the Earth is elliptical, making the Moon sometimes closer to and sometimes farther from the Earth. Since the physical diameter of the Moon, doesn't change, that means that the apparant or angular diameter of the Moon as seen from Earth changes.
So when the center of the Moon passes betweent the Earth and the center of the Sun, the angular diameter of the Moon can be larger than that of the Sun, and there can be a total eclipse, or the angular diameter of the Moon can be less than that of the Sun, and there can be an annular eclipse.
During annular eclipses, the photosphere of the Sun is not totally covered and appears as a thin bright ring around the Moon.  During total eclipses, the photosphere is completely covered and hidden by the Moon.  But the darkening of the Sun's light during total eclipses allows the dim corona, or inner atmosphere of the Sun, to be visible, usually extending a few times the angular diameter of the moon and looking rather irregular.
Part Two;
There are two interpretations of the question.
One) Can humans enlarge the moon and/or bring it closer to the Earth?
Two) Could a possible habitable world similar to Earth have a moon Larger than Earth's Moon or closer to the planet than the Moon is to Earth.
Part Three: Making the Moon bigger and/or closer.
Both would be very, very, very, very, very, very big projects which might take many thousands or millions of years.  But enlarging the Moon would seem to be many thousands of times easier, though still very, very, very  difficult.
For example, it would obviously be possible to take inflatable buildings to the Moon's surface and inflate them with some sort of gases.  And if enough huge inflatable buildings are taken to the Moon, it is possible to cover the entire surface of the Moon with inflated buildings and so increase it's diamater somewhat.  And if a very vast giant project is made, perhaps the Moon's diamter might be increased by thousands of miles.
Such a project would be a form of a shellworld, and there are other theoretical forms of shellworld mega projects:

A shellworld1[3] is any of several types of hypothetical megastructures:
A planet or a planetoid turned into series of concentric matryoshka doll-like layers supported by massive pillars. A shellworld of this type features prominently in Ian M. Banks' novel Matter.
A megastructure consisting of multiple layers of shells suspended above each other by orbital rings supported by hypothetical mass stream technology. This type of shellworld can be theoretically suspended above any type of stellar body, including planets, gas giants, stars and black holes. The most massive type of shellworld could be built around supermassive black holes at the center of galaxies.
An inflated canopy holding high pressure air around an otherwise airless world to create a breathable atmosphere.[4] The pressure of the contained air supports the weight of the shell.
Completely hollow shell worlds can also be created on a planetary or larger scale by contained gas alone, also called bubbleworlds or gravitational balloons, as long as the outward pressure from the contained gas balances the gravitational contraction of the entire structure, resulting in no net force on the shell. The scale is limited only by the mass of gas enclosed; the shell can be made of any mundane material. The shell can have an additional atmosphere on the outside.[5][6]

https://en.wikipedia.org/wiki/Shellworld
So it would be possible to enlarge the Moon using less effort than merely dumping counless gazillions of tons of rock on it.
But it would still be a very vast mega project.  And even enlarging the Moon by dumping gazillions of tons of space rock on it would be a much smaller mega project than trying to move the Moon closer to Earth.
Part Four:  A Hypothetical Earth like planet with a moon that was larger and/or closer than Earth's Moon.
It is certainly possible for an Earth like planet to have no moon, like Mercury or Venus, or to have one moon like Earth, or to have two moons like Mars, or to have more than two moons.  Pluto, a dwarf planet, has five known moons.  And possibly some Earth like planets can have more than five moons.
There are some theories that having a large moon is important to make an Earth like planet habitable.  If those theories are correct - and it is not know if they are correct - then each habitable planet is likely to have one large moon in addition to possible smaller moons.
The Moon is more massive, relative to the mass of the Earth, than any known moon of any other planet in our solar system.  However, Charon, the largest moon of the dwarf planet Pluto, is much larger and more massive compared to Pluto than the Moon is compared to the Earth.  And it is speculated that the moons of Pluto, including the massive Charon and the four small moons, were formed by a gigantic collison between astronomical objects, similar to how the Earth's Moon is believed to have been formed.
So as far as we know now, it is certainly possible for an Earth-like habitable planet to no moon, or a moon much smaller than the Moon, or a moon the same size as the Moon, or a moon much larger than the Moon.
Can the hypothetical large moon of a hypothetical habitable Earth like planet orbit the planet much closer than the Moon orbits Earth?
Yes.  Astronomers studying the orbit of the moon calculated that tidal effects would slow down the rotations of the Earth and the Moon and cause the Moon to slowly spiral outward from the Earth.
Some of the Apollo missions left reflectors on the Moon pointed at the Earth. Astronomers point intense lasers at the Moon, and detect very faint reflections from the reflectors on the Moon.  By detecting the time it takes for the reflections to be detected, they can measure the distance to the Moon and calculate how slowly the Moon is receeding from the Earth.
The Moon is receeding at a rate of about 38.30 plus or minus 0.08 millimeters per year.  So in 500 million years the Moon should have receeded 1.915 times 10 to the 10th power millimiters or 19,150,000 meters, or 19,150 kilometers.
And photosynthic lifeforms created an oxygen rich atmosphere on Earth, making Earth habitable for humans and other lifeforms which need to breathe oxygen, over 500 million years ago.  So it is possible that when Earth first became habitible for beings like humans, the Moon might have been close enough to always cover the entire photosphere of the Sun during central eclipses, though it would not have been close enough to cover the entire solar corona.
Of course the farther the Moon gets from Earth, the weaker the tidal interactions are, and so the slower the Moon moves away from Earth. Thus the Moon must have actually been much more than 19,500 kilometers closer to Earth 500 million years ago.
The tidal interactions between the Moon and the Earth slow down their rotation rates.  The rotation period of the Moon has slowed down until it equals the period of the Moon's orbit around the Earth, making the Moon tidally locked to the Earth.
Because the tidal effect of the Moon on the Earth is weaker than the Earth's effect on the Moon, the rotation of the Earth has not been slowed as much.  But it has been slowed enough to make the Earth's day several times as long as it originally was, and the Earth's rotation is still slowing.  It has been calculated that if the Earth's rotation slows enough for it to be tidally locked to the Moon, the Moon would cease to move away from the Earth and the distance between would remain unchanged.
And if a much larger moon formed close to an Earthlike planet, its much stronger tidal forces would slow the rotation of the Earthlike planet much faster than the Moon slowed the rotation of the Earth.  Thus the two worlds might both become tidally locked while the moon was still close enough to the planet to totally cover the star's photosphere, and maybe also its corona, during total stellar eclipses.
I note that not only is Charon tidally locked to Pluto, but Pluto is also tidally locked to Charon, which means that tidal forces no longer drive Charon outward from pluto.
So it an Earth like planet formed a much larger moon close to it, that moon might cease spiralling outward from the planet, while it still had an angular diameter wide enough to cover their star's photosphere and even the star's corona.
Conclusion.
I don't know if you were asking about the possibility of a super gigantic future project to inclrease the angular diameter of the Moon as seen from Earth, or about the possbiility of a hypothetical habitable planet with a moon having an anglar diameter large enought to cover their star's photospere and maybe even its corona.
So I answered both.
And if you intended to ask the second question for a story you are thinking about writing you should ask a question about it on the Worldbuilding stack Exchange.


text:
Is there a word or short phrase that denotes the apparently moving part of the lunar limb?
Between new moon and full moon, the moon's disc as viewed from the earth is bounded by a semicircular arc on its right and an arc on its left that moves to the right until at full moon it forms the other part of the circle.
Between full moon and new moon, the disc is bounded by a semicircular arc on its left and an arc on its right that moves until it coincides with the leftside arc, making the visible moon disappear in the sky.
Is there a word or short phrase that denotes the apparently mobile part of the lunar limb?

It's called "the terminator line" or simply "the terminator"; where the illumination from the sun terminates.
There's a fuller discussion in EarthSky.org's What is the moon’s terminator line?, with ends:

Bottom line: The terminator line – on Earth’s moon and other planets and moons in space – is caused by sunlight falling on the surfaces of these worlds. It’s the line dividing night and day on the moon and other solar system objects.


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Create an artificial solar eclipse
I have a meade ETX-125 telescope and I have a solar filter, my question is, can I place a small disk in the centre to create an artificial solar eclipse in the hope of seeing corona ejections, if this is possible how big does the disk need to be?

Would [astronomy.se] be a better home for this question?

@Qmechanic  maybe, or even better photography.se .  There's a lot of overlap, seeing as the physics of optical systems is something one should understand when mucking with fancy photographic techniques.

The term for what you want is "coronagraph". They're a bit tricky.

Hi @JimiJo. Is it corona ejections you want to see, or the corona? (The corona is what you see during a total solar eclipse.) If you want to see corona ejections, covering 100% of the sun would block a large percentage of them. Only the ones projecting above the limb would be visible. You should research if a hydrogen alpha telescope is what you want so that you can view the entire disk to see flares.

You can't do this unless your telescope is of a type which has an internal focal plane prior to the last set of lenses.  Think of it this way: if you are looking at the sun thru your car's windshield, any obscuration on the windshield simply blocks some of the incoming wavefront, which is (roughly) the Fourier Transform of the object's spatial intensity pattern.  To block the sun's disk, you need to block light at the object or image plane, not in phase space.
This can be, and is, done in some of the specialized solar observatories such as one of the 'scopes in Hawaii.
For home use, if you are able to open up your imager (the CCD or similar array inside the camera) and physically block the pixels when the solar disk is imaged, you might get enough contrast to see the corona in the unblocked pixels.

What you're describing is a type of Coronagraph. The OP doesn't mention an imager, if they try to do this with an eyepiece they will burn a hole through their eyeball and into their brain. It would be best to add a "Never point your telescope at the Sun and try to look through it!" warning along with a procedure to avoid destroying their CCD as soon as the telescope is slightly out of alignment. Perhaps the simplest thing would be to remind them that this does not replace the function of the solar filter, the filter must be in place at all times.

The once time that  "Never point your telescope at the Sun and try to look through it!" warning doesn't apply is when a proper solar filter is securely fastened to the aperture of the telescope.


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Why do the Andromeda Galaxy images from NASA have some sort of color shifting in it
I was checking the images from this website of the Andromeda Galaxy.
However, I noticed that low res images look good, however, the 1.7Gb images has some sort of color shifting in it:
Is there any simple explanation for this? Or is it just normal and the compression to lower res removes it?
Update:
Here is the zoomed version on one of the red region (rotated)

Without annotations:

I think this is an interesting question and the answer(s) will prove to be interesting as well, but so far I am not sure you've clearly demonstrated the color difference you're asking about. I assume the image in your post is the "low resolution" version which "looks good"? If so, can you some how find a way to explain why the 1.7 Gb image does not "look good"? Without showing the difference you're perceiving it may be difficult to give a good answer, though it's probably easy to speculate that it involves simple averaging...

Human eyes are great at detecting tiny color differences in close proximity, but they are really really lousy at absolute measurements. Eyes will never tell their owners that a mixture of red green and blue dots  on a black background averages to white, but sure enough if you "zoom out" far enough that's what they will become; if you don't believe that go get a magnifying glass or jeweler's loupe and look closely at your screen! :-)

@uhoh hi there, I've posted the one with the problem... I'll post a zoom one one of the "correction/shift" to see if it helps

@uhoh posted, you should now be able to see a sort of horizontal line, which over that line the image is sort of "blue-ish" and below is "yellow-ish"

Okay great, thank you for your speedy update! If you are able, If you still have it, could you also post that same crop without the arrows/annotation? I'd like to try some amateur statistical analysis of the pixels.

@uhoh sure, posted

I can't find the answer, but the image is collected from such a complex patchwork of cameras and filters, I think it can be processing. The weird thing is that they are almost parallel, and there are just two of them and they are irregular sharp edges.

@LifeInTheTrees where does "such a complex patchwork of cameras and filters" come from? Perhaps just one or two extremely well characterized cameras (e.g. the Hubble Wide Field Camera 3 and perhaps 2?) and three very carefully chosen standard filters?

To be honest, I think that you should mail some researchers that have worked on the PHAT image processing and query them about it.

@LifeInTheTrees seems a bit too much for a mere curiosity ahaha

I think they are probably processing artefacts, because there are other artefacts there, and because the lines are parallel.
I sent it through RGB saturation and found other artefacts of linear star zones at the seams of some of the images.
The visible color version is only a small element of the PHAT study and perhaps it is not scientifically accurate.

oh was, I wasn't even thinking that the image was a reconstruction of different images taken at different times, this might explain everything

@AlbertoSinigaglia yes, sure, it's a mosaic :) https://www.popphoto.com/news/2015/01/check-out-nasas-15-billion-pixel-image-hubble-telescope/ claims it's just shy of 8000 individual images.

I'm curious why parallel boundaries can't be the result of shear in a spinning disk of inhomogeneous makeup. https://www.eso.org/public/norway/images/potw1938a/?lang The milky way seen from our location is full of reddish filaments of dust that remove blue light. On what basis have you decided that what's seen in the OP's section of the image can not be dust and must be stitching artifacts, something I have a really hard time believing astronomers are doing?

@AlbertoSinigaglia I suspect this is not the correct answer; the author offers no evidence that these aren't dust other than "I think they are probably processing artifacts, because there are other artifacts there, and because the lines are parallel" which does not actually discount dust. Astronomers go through great pains to make these mosaics as scientifically correct as possible, so while possible "I think ESA simply screwed up" is not the most likely possibility. One would have to exclude the other possibilities, including that it's real, first.

@AlbertoSinigaglia Astronomers don't use photoshop, aligning overlapping images can be done to incredible precision as Hubble space telescope's cameras are extremely well characterized and while the motion of individual stars can sometimes be detected, these lanes of dust are not going to be moving or changing brightness substantially over a few years. A few million years yes. But for these purposes everything is effectively standing still with rock-steady brightness. I think you've accepted a wrong or at least a low-quality, unsupported guess of an answer.

@AlbertoSinigaglia Perhaps consider un-accepting and waiting for something more authoritative in a few days or a week or so?

That's fair enough, although it can't be dust because dust is diffuse and doesn't create 100k ly very sharp line color effects. Here's a list of artefacts, and neither your nor my suggestions is figured precisely on the list. https://illuminateduniverse.org/2021/04/01/image-artifacts/

@LifeInTheTrees 1) a Stack Exchange answer should do that already, especially when challenging such an effort by ESA astronomers 2) dust is not an artifact, dust is real! 3) "and perhaps it is not scientifically accurate" is pure guesswork and while not impossible, is unlikely considering the source.

The depth is crowding limited in the optical and NIR, and thus is a strong function of radius. As a result, photometry in the inner bulge fields is far shallower.

I concur that this is likely the right answer - these look completely like small differences in color balancing from stitching together smaller images.  I suspect that the larger images are balanced on average, leading to small discontinuities at the edges. The straightness of the edge is the giveaway that it’s not real.  Regarding “astronomers don’t use Photoshop,” it’s used all the time for public press release images like this one. See here for example. The science is done on the originals.

Oh! Okay I'm finally getting your question, sorry for being so slow.
We'll wait for a card-carrying astronomer to answer but my guess is that you're seeing either the edge of a thin dust cloud which preferentially scatters & removes blue light, or the edge of a gas cloud that's emitting red light.
With the Python script below I made this image where I "turned off" the stars and then added some additional blurring to visualize the edge better.

I then selected the range from horizontal pixel numbers 528 to 558 and made a 1D projection of the red, green and blue color channels by averaging across the 30 pixel width chosen to avoid bright stars.
It looks like the edge is defined more by a drop in blue than a rise in red, so I think that the most likely guess is the edge of a dust cloud that's reddening the lower half by scattering blue light from stars behind it away from the line of sight.
One of the (many) challenging things about astronomical observation is that the light you are seeing is not only from the objects you think you're observing, but includes extra light from in front of and behind them, and has light "subtracted" by wavelength-selective absorption by "stuff" between them and you, some of which you only know exists by its absorption.

WAW, this is  amazing, thank you so much for the time you took to review my question... do you have any though on why this looks like its "decaying" from left to right? there is also another "shadow" on the left, which makes the image go from (left) yellow, to yellow-blueish, to blue on the right

the "shift on the right" is more visible on the top part, where you see a kind of switch from red to blue

The python graph is fantasic and I think you could be right about some near field interference, although it probably isn't a cloud?

This is information about local dust in the vicinty of Andromeda. It doesn't interact with visible light, is diffuse, smaller, not aligned with the center, and wasn't even documented until about 2018. http://www.deepskycolors.com/archive/2017/01/01/Clouds-Of-Andromeda.html

What about dust that's inside and part of Andromeda?

Because it's resisting the gravity of andromeda and keeping two straight barriers, it's not swirling like dust does.

@LifeInTheTrees "swirling" moves at astronomical timescales, it's not like we can see it moving or that a few Hubble WFC images of Andromeda are going to look different a few years apart.

Theres no evidence of swirling from the last 100 million years either, thats strange. You suggest its a uniform thickness straight edge double cloud. Other than that i agree.


text:
How has the rate of photon arrival from stars been estimated and then confirmed?
My understanding is that we can estimate the total energy emitted over a star’s entire spectrum and from that and the star’s measured spectrum estimate how many photons of a given wavelength we should receive at a given instrument per second.
My question is: How has this been estimated and then measured to validate the estimation method? Are there specific papers or texts that lay out a widely agreed-upon method?

What are you suggesting? That the energy of a photon isn't $h\nu$? All "exposure time calculators" do these calculations; and they work.

"My understanding is that we can estimate the total energy emitted over a star’s entire spectrum" This usually happens after the star's output has been measured at many wavelengths, not before. Until recently photomultiplier tubes were widely used in Astronomy and certainly at low light levels they were used in photon-counting mode, and in some niche applications avalanche photodiodes are still used to count photons. But experiments that call into question the relationship between the energy of a photon and it's wavelength or frequency are rare now as none have shown significant deviations..

from $E = h \nu$ or $hc/\lambda$

I didn’t mean to imply any doubts about the Planck equation. I’m literally wondering how astronomers have estimated then validated the photon flux arriving at Earth, in a particular band, from a specific star. Thanks for the pointer to “exposure time calculators”, a term I wasn’t familiar with. It looks like many of these are developed for specific instruments in order to assist PIs.  I want to learn about the general method/process astronomers follow to create and then validate and refine these ETCs— is there a good published guide to that?

I think it's the "...how astronomers have estimated then validated the photon flux arriving at Earth" that's throwing some of us off. What does "validate" mean here? Are you drawing a distinction between photon flux (number of photons per second per unit area on Earth in a given wavelength range) with spectral intensity (energy per second per unit area on Earth in a given wavelength range)? Some measurement techniques use intensity, others use photon counting, and except for some rare experiments nobody validates one with the other. Instead they're used interchangeably.


text:
What is the difference between the Synodic & the Sidereal Periods?
While studying Spherical Astronomy, I came to know the two terms: 'Synodic Period' & 'Sidereal Period'. Both were referring to the returning of a particular celestial body on the sky or the celestial sphere but I couldn't spot the difference between the two definitions. What is the actual difference between the two periods and what is their importance?

https://en.wikipedia.org/wiki/Orbital_period has a good explanation. Is there something in that article that isn't clear for you?

A synodic period is the time required for a given body to return to a given position relative to Earth. A sidereal period is the time required for a given body to return to the same position relative to the stars.

A planet is in opposition when it is on the opposite side of the sky from the sun, so it will be due south at midnight.
The synodic period is the (average) time between oppositions. Or, more generally, it is the time for a planet to reach a given position relative to the Earth. This latter definition can also be applied to the synodic period of Venus and Mercury.
The sidereal period is the time it takes (on average) for the planet to return to the same position relative to the stars.  Because the stars are fixed, this is the same as the actual period of time for the planet to orbit the sun.
The periods are related by the formula
$$1/P = 1/E \pm 1/S$$
Where S= synodic period, P= sidereal period and E=356.25 days=orbital period of the Earth, and you add for Venus and Mercury, but subtract for Mars and the outer planets.

So does this mean that the Earth doesn't have a synodic period?

No, the synodic period is always relative to observations made on Earth.  So the Earth can't have a synodic period.  You could ask for the synodic period of Earth as viewed from Mars, The synodic period of Earth from Mars is the same as the synodic period of Mars from Earth.

"Synodic" does not, by definition, mean "as observed from Earth".  An observer on the moon could easily define a synodic period for the Earth, likewise an observer on Mars could define synodic periods for its moons.

Yes @greg you're right but that would make the answer needlessly less clear.  In particular, you couldn't talk about the synodic period of Earth (or mars or any other ssb), you would always need to mention the observer.  Michael asks about the synodic period of Earth.  That isn't defined. So "Earth doesn't have a synodic period".  But "Earth has a synodic period when observed from Mars, and a synodic period from Pluto and from Mercury.  but there is no one number.  Similarly when we say "the synodic period of mars" we imply "from earth"

If you think of the Sun, Earth, Moon system.  The Earth orbits the Sun, and the Moon orbits the Earth.  At a full moon, the bodies are essentially lined up, the synodic period is the average time to the next full moon, when they're lined up again (in the same order).
Now, if you ignore the Sun, and just think of the moon's movement relative to the Earth and stars.  E.g. imagine every time the Earth, Moon, and Gemini are nearly lined up.  This is the sidereal period.
Put another way, the sidereal period is measuring a period relative to the distant stars.  And the synodic period is measuring the period relative to some other object.


text:
Is the Vernal Point a fixed point on the sky?
The Vernal Point is a reference point in the Equatorial Coordinate System, from which R.A. is measured. My book says that the zero point of Right Ascension, i.e. the Vernal Point keeps moving in the sky. How can this be? If it moves then how can it be used as a reference point? Also why doesn't the dates of Vernal Equinox keep changing over the years?

Related: https://astronomy.stackexchange.com/q/13870/16685 & https://astronomy.stackexchange.com/q/41377/16685

The date of the vernal equinox keeps sticking around March 21st because the Gregorian Calendar was designed as a refinement to the Julian Calendar to keep it there.

It's best to think of the vernal point as the intersection of the Earth's equator with the Earth's plane of orbit.  Since there are two possible choices, arbitrarily the point where the Sun crosses the equator moving North was chosen as the 0 point.
Since the Earth's poles move, obviously the vernal point will be different depending on where the Earth's pole happens to be pointing.
Now, why would we choose a coordinate system that constantly changes?  It's because it makes celestial navigation easier.  The vernal point doesn't move that fast, so  charts are valid for long periods of time.  The "geographic position" of a star can be determined rather simply.  The lattitude is the declination.  And the longitude is the right ascention minus Greenwhich sidereal time.
Since celestial navigation isn't used much anymore, the IAU has abandoned the "equinox method" and created the International Celestial Reference Frame.  It is based on radio sources so distant, they will show no apparent movement at human timescales.

A nice explanation is given by Bob King in Sky&Telescope in the article "Right Ascension & Declination: Celestial Coordinates for Beginners". You can find it at:
https://skyandtelescope.org/astronomy-resources/right-ascension-declination-celestial-coordinates/
Kind Regards
Klaus

This isn't an explanation, it is more a link to an explanation. You can use links to support an answer, but the answer should be valid even if the linked resource stops working.

Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center.


text:
Are the Vernal Point and the Autumnal Point two antipodal points on the celestial sphere?
Intersections of the elliptic and the equatorial planes on the celestial sphere are the Vernal Point and the Autumnal Point. Are these two points two antipodes on the celestial sphere of the Earth?

Yes… and no. If the Sun–Earth system was alone in the Universe (or at least in the Solar System), it could indeed be the case. However, the Earth’s orbit around the Sun slowly rotates with respect to the rest of the Universe, so that technically, the points where the Sun seems to cross the celestial equator (a projection of the Earth’s equator on the celestial sphere) are not exactly opposite each other.
However, for all practical purposes, yes, these two points are opposite, being defined by the instant orbit of the Earth around the Sun.

When you say the Earth’s orbit around the Sun slowly rotates with respect to the rest of the Universe, it might be easier to just say precession and nutation?

@BarryCarter: Yes indeed, but not everyone knows what these are.

True, but it's nice to provide jargon/terms of art so the OP can do further research.

I’ll keep that in mind for next time.

Yes. The points (in fact directions) are defined by the intersection of the two planes (at a given time), not by the position of the Sun at the real equinoxes (six months apart).


text:
Why is Earth 1/3 the age of the universe?
The Earth is 4.6 billion years old
The universe is 13.8 billion years old.
3 times the age of the Earth
Are there other planets with ages that are close to say 1/4 1/5 etc the age of the universe? Why would this happen?

Would you expect the Earth to be older or younger than it actually is? And for what reasons would you expect that?

The (approximate) 3:1 ratio is nearly entirely co-incidence.  The other planets formed at the same time as the Earth, and are therefore of the same age.
The Earth is quite accurately dated to 4.54 billion years and multiplying by three gives 13.62 billion,  quite a bit less than the age of the universe.
There are other planets around other stars, some formed early in the universe, some formed recently. There is no reason to expect simple ratios of ages between planets and the universe.

The (approximate) 3:1 ratio is nearly entirely co-incidence. Playing devil's advocate, is it? Perhaps older stars with older planets could not have developed intelligent life (which I'll define as life capable of escaping the planet's gravitational field and potentially communicate with life on other planets), and perhaps younger stars with younger planets have not yet had enough time to have enough time to have developed intelligent life. This is admittedly extrapolating from a sample size of one (the Earth), but for now, that sample size one one is all we have to go upon.

But nonetheless, plus one. Extrapolating from a sample size of one is a very bad idea.

What about 2 x 4.6 billion light years  for the anomalous galactic arc 3 billion light years long here at :https://www.sciencenews.org/article/galaxy-giant-arc-3-billion-light-years-long-cosmology-space

The formation of rocky planets would have been tricky for the first stars, since there was not much beyond hydrogen, helium, and a bit of lithium in the universe. Not quite sure when the formation of rocky planets would have become more likely than not...

@jon  yes, a close to 1:1 ratio is not possible, but there is no special reason to suppose that life on planets develops intelligence on a timescale that gives a fixed small ratio.  There is no resonance effect (for example) here.

@DavidHammen I see what you are saying, but I feel that this kind of anthropomorphic speculation is not particularly relevant.  It may be that intelligent life tends to develop at about a 3:1 ratio, but there is certainly no effect to lock the time that life develops intelligence at a small ratio, there is no resonance that causes this.

@JamesK - oh, for sure, playing random numbers game has no place here.

@DaveTheWave What about it?  "There is an object which has a light travel time that is about twice the age of the solar system". . .  But there are lots of things that are not a simple multiple of 4.6 billion, (and as noted, 4.6 is a rounded value)

The temperature of space was around the boiling point of helium 4.5 billion years ago when Earth formed.     https://physics.stackexchange.com/questions/76241/cmbr-temperature-over-time                9 billion years ago (2 x 4.5) there was 3 times more hydrogen present as atomic gas as there was in stars ( hence the period of maximum star formation) .    https://phys.org/news/2021-06-gmrt-atomic-hydrogen-gas-mass.html    Could all be coincidence , I agree. But no harm in seeing what people think.

@DaveTheWave There is no such thing as "the" temperature of space. (What you link to is describing the temperature of the cosmic background radiation, not the temperature of gas in space.) The temperature of gas in space varies from a few degrees Kelvin to tens of millions of degrees, depending on the location.

@DaveTheWave The temperature of the protoplanetary disk from which the Earth formed varied with distance from the protostar that eventually would become the Sun. At about 1 AU (i.e., about where the Earth formed), the temperature probably was about 200 °C (for example, see https://www.sciencedirect.com/science/article/abs/pii/S0012821X18305697).

The closeness to a 1/3 ratio is coincidence but the observation that the Earth is a lot younger than the universe may be quite profound.
Firstly, the Earth cannot be as old as the universe. The first galaxies and stars likely formed about 0.5-1 billion years after the big bang, once primordial hydrogen and helium had time to cool and clump.
Earth-like planets could not form around these stars, since there were no heavy elements like carbon, oxygen, silicon and iron present.
These elements are made inside stars and then dispersed into the interstellar medium for new generations of stars to incorporate that material. This process takes some time.
When the Earth formed, it was extremely hot, it was also being bombarded by asteroids and other sub-planetary bodies. There was no chance of forming life for 0.5 billion years or so.
Once life formed, it didn't have the capacity to estimate the age of the Earth and of the universe. That is a recent development and has apparently taken about 4 billion years of evolution.
If we assume that our galaxy took 1 billion years to form, that adequate chemical enrichment took another 2-3 billion years and that the time taken for a planet to cool and have evolved intelligent life is 4 billion years, then we arrive at the conclusion that the universe as we see it would have to be at least 7-8 billion years old and that our planet would then be less than about half the age of the universe.
A more interesting question is why is the 1/3 fraction quite close to, but still significantly smaller than, the apparent maximum of 1/2? We do know that small, rocky planets have formed around stars with a wide variety of ages - anything from a few million years to estimated ages of around 10 billion years (e.g. Weiss et al. 2021), though how common they are around very old stars is still to be determined. Was the early galaxy inhospitable to life for some reason? Are we just part of a distribution, with many other older planets out there hosting more ancient life-forms?

Why the Earth isn't 1/2 age of universe is a good point I think it relates to the temperature of the cmbr being close to heliums boiling point at 1 atmosphere 4.5 billion years ago  There are exoplanets surrounded by helium .I wonder how old they are . Under oressure helium can be liquid  hotter than it is at 1 atmosphere Perhaps liquid  helium was trapped inside rocks before the planets and sun formed.

@DaveTheWave I'll be blunt: That is nonsense. You are thinking of helium as a superfluid, which only happens at extremely low temperatures. Those extremely low temperatures are not the temperatures at which planets form, whether they are terestrial planets, gas giants, ice giants, or even plutino dwarf planets.

At least 40% of the field stars (thus outside globular clusters) seem to have planets, if not more.
Planets form around the protostar, thus at the same time as the stars. Assuming that the laws of physics are universal, then the distribution of planetary ages will be the same as the stellar ages: there will be planets of all ages, very young ones, very old ones. The very old ones will be smaller in mass in average as the remaining old stars are lower in mass (the higher-mass stars already reached their end of life and doing so likely lost their planet system in the end phase).


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What mass does an evaporating black hole have when it's schwarzschild radius equals the Planck length?
I am referring to Hawking radiation and the decrease in mass of the black hole with time

1/2 Planck mass. See https://www.vttoth.com/CMS/physics-notes/311-hawking-radiation-calculator

The physicallity of such an object is doubtful, it would exist in the realm of quantum gravity, and we don't have a working model of quantum gravity yet.

A black hole's Schwarzschild radius is $$R=\frac{2GM}{c^2}$$
In Planck units, $G=1$ and $c=1$, so the above reduces to $R=2M$ with mass and radius expressed in Planck units. Thus a black hole with a mass of one half Planck masses will have a Schwarzschild radius of one Planck length.

Thanks  .Useful equation.


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Does the core of the sun bounce around because of turbulent hot gas around it?
How much does the core of the sun wobble and does this affect light reaching the Earth?

The core is way larger than one gas scale height, which is thought to set the maximum buoyant eddy size. So the core doesn't wobble around.

Does this mean that gas moving towards the core thins out exponentially the closer it gets to the core?

No, quite the opposite. Have a look at solar structure models (density and temperature as function of radius). As gravity dictates, the mean gas density can only go up as you approach the center of the star. Buyoant instability (which does not happen inside the sun, but at its surface. Higher mass stars have convective cores) will be only a small density perturbation on top of that mean density. The descending convective cells are denser than the mean.

The core of the Sun is stable to convection - the energy transport is dominated by radiative diffusion. The core of the Sun is therefore not turbulent and not wobbling around.
However, the Sun itself does "wobble" because it has characteristic oscillation frequencies that are excited by convective motions in its outer parts. The amplitude of the oscillations are about 0.2 m/s and have a period of about 5 minutes (Kjeldsen et al. 2008). This equates to amplitudes of radial motion of around 60 m, or about 1 part in 10 million compared with the solar radius.


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Is there enough matter orbiting the sun to make a second sun?
I read somewhere a long time ago that there is enough matter in our solar system in the form of rocks and dust to create another sun.Is this correct? Was our solar system trying to create a 2 star system?

If you have an advanced civilization able to "make a sun", you can do starlifting to extract material from the original Sun for that purpose. As it becomes smaller, it will become less bright, and longer-living. (Though then the question is better for worldbuilding.SE.)

What does "orbiting the sun" mean to you? Is that out to the orbit of Pluto, or only of Neptune?
Either way, what do you see as the mass of the sun, and of the rest of the system?

Although the solar system doesn't currently have enough extra mass to form an additional star, star formation is a complex process which typically generates multiple stars together ( http://abyss.uoregon.edu/~js/ast122/lectures/lec13.html ) and some of the initial matter can be slung out of the individual star systems, although it's unlikely that our particular solar system ever had enough for two stars: https://phys.org/news/2015-10-minimum-mass-proto-solar-disk.html

Can you say where "Was our solar system trying to create a 2 star system?" comes from? That doesn't seem related to the OQ.

@RobbieGoodwin, "orbiting the Sun" generally means something that is gravitationally bound, rather than referring to a specific distance.

@Mark Thanks and while that is wholly logical, I feel a need to ask how it is helpful here? Don't you accept that gravitational bounds decrease with distance?

The vast majority of the stuff in the solar system other than the Sun itself  is contained in one body, Jupiter. The total mass of the solar system is estimated to be about 1.0014 solar masses, or about one solar mass plus 1.4 Jupiter masses. (Jupiter's mass is a bit less than 0.001 solar masses.) Using the highest estimates on the mass of the Oort cloud, the total mass of the solar system, excluding the Sun itself, is about 30 Jupiter masses. These early estimates have been shown to be wrong. Current estimates are that the mass of the Oort cloud is one or two Earth masses. Even if the hypothesized planet IX does exist and is as large as some hypothesize (about five Earth masses), that will only budge the estimated 1.0014 solar masses by a tiny, tiny bit.
The smallest possible star, defined as something capable of fusing hydrogen, is about 65 to 80 Jupiter masses. The answer is no.

And a good part of that matter wouldn't be useful for forming a star as it's not useful as fusion fuel (in fact would poison the fusion process). Which is one reason stars die, their fusion process gets poisoned by heavy isotopes.

@Panzercrisis https://www.wolframalpha.com/input?i2d=true&i=Divide%5B%5C%2840%29jupiter+mass%5C%2841%29%2C%5C%2840%29sun+mass%5C%2841%29%5D

@Panzercrisis The mass of the smallest possible star is typically stated in terms of Jupiter's mass rather than the Sun's mass. There are many good reasons for this.

@Panzercrisis 0.001% --> 0.00001, I think the % sign is an error there, or it should be 0.1%.

@NickT Okay, yeah, that must've been a typo.

"Using the highest estimates on the mass of the Oort cloud, the total mass of the solar system, excluding the Sun itself, is about 30 Jupiter masses." That phrasing is confusing. It fails to clearly distinguish between reality and beliefs about reality. Clearer would be "Previously, the highest estimates of the mass of the Oort cloud resulted in the total mass of the solar system, excluding the Sun itself, to be calculated as 30 Jupiter masses."

No.  The total mass of the planets, asteroids, dust etc in the solar system is only about 0.1% of the mass of the sun.  There is not nearly enough to make even a small star.

0.1 % is about 330 Earth masses  .

@DaveTheWave That's not nearly enough. A better number is 0.0014 Sun masses, or about 1.4 Jupiter masses. The smallest possible star is 65 to 80 Jupiter masses, depending on who one reads.

There's probably at least another 0.1% in the "vacuum", by which I mean the 3 to 10 atoms per cubic centimeter in the empty regions of space.  By my calculations, 1 cubic light year would contain about 1,000 Earth masses (assuming those were all protons) per atom per cubic centimeter.

I'm not sure that counts as "rocks and dust". On the other hand, I'm not sure if planets count as "rocks and dust".  However, no matter which way you cut it, there isn't enough "stuff" for a second star.

@Michael 1000 Earth masses is a bit more than 3 Jupiter masses, so not enough stuff to cut it. Moreover, the Sun and the planets are very good at either expelling small material from the solar system or making such material drop into the Sun. (The zodiacal light falls into the latter category.) Even if what you wrote is true (which it isn't), an additional three Jupiter masses will not cut it.

@DavidHammen Yes, I'm aware that 1000 Earth masses isn't enough to cut it... it would need to be 1000 times larger still to equal the mass of the sun.  But why do you say that what I wrote isn't true?  Are my calculations off on the total mass present in the "empty" space of the solar system?

@Michael I suspect your number is too large by a factor of $10^8$. The vast majority of protons in interplanetary space come from the Sun (aka the solar wind, which arises from things such as sunspots and coronal mass ejections). The density of those solar emissions drops with the square of distance from the Sun. Moreover, it is not spherically symmetrical. The majority of those particles are near the plane of the Sun's equator. (continued)

Assuming spherical symmetry and a density of 13 protons / cubic cm at 1 AU I get $10^{-5}$ earth masses for a sphere with a radius of one light year. That is certainly overkill due to the assumption of spherical symmetry and is smaller than your calculation by a factor of $10^{-8}$.

@DavidHammen Thanks! That makes sense.  In the back of my head I was setting 1 proton / cm^3 as a lower bound, forgetting that intergalactic space is 1 proton / m^3, not cm^3!


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Is frame dragging expected to be quantized in a theory of quantum gravity?
Protons and electrons have angular momentum which is quantized. Would the angular momentum of black holes be quantized in quantum gravity? Would this then affect Hawking radiation?

A paper that seems relevant https://arxiv.org/abs/0906.1949 'Quantization of the black hole area as quantization of the angular momentum component'

It was interesting to read that rotating and non rotating black holes have the same quantization for area on the event horizon.


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Could we detect the light of cities at night in other solar systems?
Would it be possible to see electric lights on distant planets using current telescopes?

While detecting and resolving an artificial light itself may not be an option soon, an unusually skewed spectrum of a planet may be easier to detect. On the other hand, in the last years we almost abandoned sodium and mercury lamps that would create a distinct signature for our fellow alien astronomers. Modern LEDs become more and more "natural".

No.  There are very few exoplanets that can even be imaged as "dots".  There are none that can be resolved as discs and it would be well beyond current technology to see lights on the surface of an Earth-like planet.
Most exoplanets are only known from periodic dimming of a star as the planet passes in front of its star, or wobbles in a star's motion, as the planet's gravity causes the star to move forward and back.
We could see lights on Mars, if they existed as they do on Earth,  but they don't.

This answer is focused on direct imaging and ignores the possibility of spectroscopic information. We’re not able to directly image the atmosphere of any exoplanet either, yet we can detect their absorptive spectroscopic influence. Theoretically characteristic emissive spectroscopic information from very large-scale illumination could similarly be detected.

@MichaelMacAskill My answer stands, If you feel you can add something please write your own answer.

You don't need to resolve a planet as a disk to observe light emitted from it.  After all, you can't resolve most stars into disks, but you can see their light without even using a telescope.

I did a few rough calculations.
Let's assume a neighbor system (4 light years), and a population of 1 billion people each transmitting 1000W of light into the night sky.
Then on Earth, a 100 square meters telescope would receive about 2 photons per minute from the electric lights. This is far beyond current technology, especially as there are much, much brighter sources in very close vicinity.

To add to this: Currently planned telescopes (such as Roman) might be able to achieve a contrast of a few parts per billion at Jupiter-like distances from a star. 1e9 people putting out 1e3 watts per person is still a contrast of 1:1e15 against our sun.

No, at least not with currently deployed telescopes. However, it is theoretically possible with current (or near future) technology. We could potentially image the lights of an exoplanet using a Solar Gravitational Lens (SGL), as is currently under consideration by the NASA Institute for Advanced Concepts as a "Phase II" program. "Phase II" is described as signifying that NASA has:

determined that much of the foundational technology exists or is in intermediate levels of readiness due to the proliferation of government and commercial smallsat programs.

The technique works the same as we normally observe gravitational lensing. That is, the light of a distant object is concentrated in a narrow region of space called a focal line by a massive object between the observer and the lensed object. This lensing dramatically increases the resolution and apparent brightness of the observed object, enabling us to see things that are too far away or too dim for us to directly observe with conventional telescopes.
A SGL mission would send probes so that they place the Sun between the probe and the exoplanet. By blocking out the light of the sun, the probe could observe the Einstein Ring of the exoplanet surrounding the Sun. This ring could be reconstructed into a traditional image to view the surface of the exoplanet, such as this simulated image.
At this resolution (estimated by NASA as being up to ~25 km-scale surface resolution for an Earth sized exoplanet at 30 pc), it would be possible to directly observe lights of an advanced alien civilization.

Of course, it takes a long time with current technology to get to the SGL focal distance of ~547.6 AU, especially if you want to remain there (and not just do a flyby). In comparison, Voyager 1 is currently ~158.5 AU from the Sun. https://voyager.jpl.nasa.gov/mission/status/

Per the wikipedia page to which you linked, the SGL was granted a NIAC Phase III award, confirmed here. That's still a smallish amount of money ($2 million), maybe 3 or 4 people at JPL working full time for two years.

Another problem is that this telescope would basically point in a single direction for all its operative lifetime, and then need to work with that. Changing the view direction would take as long time as setting up a new telescope. So this would only be able to investigate one single solar system... you better select one where you have good reason to believe it has life on it!

Can't they  intercept photons closer to Earth by by using a circular array of detectors to intercept photons .

@DaveTheWave Not really. The converging light rays form a cone 547.6 AU long, with the Sun's disk as the base of the cone, and the apex of the cone at the SGL point. If you want to collect photons at half that distance, your circular collector has a diameter of half the Sun's diameter. That's rather large, and the photons are very diffuse, making them hard to detect.

TL;DR:
We could detect but it really depends on the distance you're talking about
Long answer:
Not using current technology (if the distance is huge). The Solar Gravitational lensing mission could do detect city lights from exoplanets but a mass (like a galaxy, neutron star and blackhole) would be required (which also most probably would be available).
If the distance is too far away according to the inverse square law the light's intensity would get smaller to a limit where we can't really detect it.
Because of the Doppler shift it would be tougher because the redshift created from the effect  would be huge if the distance is huge as per Hubble law  but that is kind of manageable because we even have radio telescopes. But one of the most powerful telescopes we have ever made the JWST has the mid-infrared ranges ( 5-28.5 μm) but doppler shift could extend even farther as per the distance so it's a catch 22 situation if we want to see large distances then we would have encounter more redshift too.
We wouldn't see the lights if the planet/asteroid is in the zone of avoidance
Also since different elements (in the Periodic table) in Bulbs emit different Wavelengths of light we're not sure which wavelength to focus on like Argon, Xenon, Incandescent (Tungsten)!
The amount of light emmited is based on the energy usage given by the Karshadev scale
Hope it helps, Thanks

Here there was only sludge, microbial mats and goop for 1.5 billion years, then it was filaments, small specs and sludge for 1.5 billion years, and there has been light for 50 years, and even those look like a volcano or a reflection of moons and stars onto clouds or ice.
i.e. Venus because of it's clouds reflects more than 4 times as much incoming light than the moon, and more than twice of the Earth's.
So astronomers want to detect chemicals rather than lights, we don't know what alien light bulbs use for chemistry, however we can measure high oxygen content on another planet, which means photosynthesis. Planet Rarth's oxygen started rising about 1000 million years ago at the same time as life-form size was rising.
https://astronomy.swin.edu.au/cosmos/a/Albedo

We need hypersensitive instruments for that.
Freeman Dyson postulated an industrial society could be detected on Infrared. Like Earth looking for heat islands in urban areas.
Another would periodic spikes in infrared waves from stars because society was tapping its power.


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How does the Sachs-Wolfe effect confirm the existence of dark energy?
How is the Sachs–Wolfe effect and the existence of voids significant in providing physical evidence for dark energy?

"Void (astronomy) - Wikipedia" https://en.m.wikipedia.org/wiki/Void_(astronomy)

I've rephrased your question because the source you cite does not use "proves" or anything like it. Have you tried reading https://arxiv.org/abs/0805.3695 and  https://arxiv.org/abs/1511.04075 (the two citations on Wikipedia) ...

@DaveTheWave your question would indeed greatly gain attraction if you tried to understand the sources James linked and point out in what way you don't follow or where you have doubts. Show us that and what you tried yourself to find an answer.

I looked at the arxiv papers  . I understand what is going on now. Photons entering voids gain energy and blueshift and on exiting the void don't lose the energy gained and  redshift an equal amount because dark energy has moved surrounding galaxies further away weakening gravity. Although I suppose electrons from hot gas clouds in voids could be passing energy to photons if the electrons move in a preferred direction.

The effect being discussed is the late-time integrated Sachs-Wolfe (ISW) effect. The idea is that, whilst travelling towards us, cosmic microwave background photons gain energy as they fall into the gravitational potential well of a supercluster and then lose energy as they exit that potential. Conversely, photons travelling into a "cosmic void", experience a higher gravitational potential and lose energy. They then gain energy when they leave the void.
If the CMB photons gain energy, the CMB would appear a little hotter and if they lose energy, the CMB appears colder. If there was an imbalance between the energy gain and energy loss processes as photons propagate, then this might be detectable by looking for correlations between fluctuations in the CMB temperature across the sky and large scale structure (LSS) (i.e. concentrations/voids of matter) in the local universe.
The key point is, that in an expanding universe that is "flat", but which is explained by the matter density being just sufficient to halt the expansion at late times (i.e. an Einstein de Sitter universe with $\Omega_M = 1$), then the amounts of energy that are being gained/lost as photons enter a supercluster/void are exactly balanced by the amount of energy lost/gained as they exit the supercluster/void and so no ISW effect would be expected. This is because the density perturbations would grow at exactly the same rate as the scale factor of the universe and there would be no evolution of the gravitational potential.
However, we know that while the universe is almost exactly flat, this is not because $\Omega_M=1$, it is that $\Omega_M + \Omega_\Lambda = 1$. But in this case the accelerating expansion of the universe, due to the dark energy term, means that the potential wells and hills attributable to superclusters and voids do evolve and are somewhat smoothed out during the time it takes a photon to traverse them and the CMB photons will therefore pick up a small amount of energy as they traverse a supercluster and lose a little energy as they traverse a void, resulting in hot and cold spots in the CMB respectively. Note that this is in addition to the usual expansion-associated redshift effects expected.

Then within the framework of general relativity, any detection of the ISW effect [in a flat universe] would serve as a smoking gun of dark energy.
(Dong et al. (2021).

Therefore my original sceptical response to the wording of your question was a bit misplaced. The detection of a significant ISW effect would be really strong evidence for the existence of dark energy (e.g., Raccanelli 2011 uses the word "proof"), though other explanations are possible - a problem with General Relativity for example or more exotic physics (Multamaki & Elgaroy 2004).

Fair enough I deleted it!