text:
Observing the motion of the solar system and galaxy through space?
This is very perplexing.
Conditions:

planets are moving around the sun (solar system)
solar system is moving as part of the galaxy
the galaxy is moving through space on some axis
galaxy is part of a cluster, also moving through space
other stars and constellations are also moving through space, presumably in the same direction and speed as the Milky way.

I appreciate the answer to Why is the Solar Helical (Vortex) model wrong? - but considering the question: how can we observe the solar system's Helical motion through space, I have a new question... please allow me to present both, since they are related.
Question 1:
Again, how can we observe the Helical motion of our solar system (and galaxy) from earth? Is it by observing the passing of planets across/through constellations?
Question 2:
If I can make an analogy: like different ticking clocks stored inside a grandfather clock - is the solar system following its own Helical motion and therefore are all other solar systems within the Milkyway also following their own Helical motion? AND is the galaxy also following its own Helical motion!?
OR!
Is the solar system's motion fixed (Helically speaking), and any such movement is tied to the motion of the parent, i.e. the galaxy? In other words, is the Helical motion of the solar system the same as the galaxy, or are all solar systems moving Helically through space, e.g. many clocks within a grandfather clocks analogy.
Initially, the question was simple: how can be observe the solar system's Helical motion?
But it seemed unintuitive to me that the solar system is moving Helically while the galaxy is also moving as such - both structures on their own trajectory and paths - seems like galactic chaos.
And, all the while planets maintain their fixed orbits.
I would have guessed that the solar system is NOT moving Helically, but the galaxy is, and the solar system is simply a fixed orbital passenger, as so to speak - in the same way 8 grains of sand (planets) are passengers in a bowl of jelly (galaxy), both being carried by another vehicle (cluster).
It seems to me then, the solar system's motion through space is limited by the galaxy's motion, in terms of axis, speed - and therefore the galaxy is also fixed like a passenger on a much larger galaxy cluster.
It seems reasonable to suggest the whole universe is moving Helically(?), and all galaxies merely inherit this, but are more like grains of sand in jelly than independently ticking clocks within a larger grandfather clock.
If our solar system is moving Helically independently from the galaxy, the solar system could move at its own speed, either quicker or slower depending on environmental factors. Could the solar system really skip-out of the galaxy, racing ahead on its own Helical path(?).
So many questions! But Question 1 is the one I wish to request help answering.

Relative to what?

"seems like galactic chaos", yep, that's exactly what it is. There is some order at various levels, but yea. All sorts of things rotating on different axes, in different planes, in different directions. Look up videos of how the positions of stars will change over time in the night sky. Like bees around a hive. Fascinating stuff!

@ProfRob - true. (it seems) everything is moving (solar system), as coblr explains - and then every object is bound together as a single collective object (galaxy). If there is one imaginary absolute point in space to observe the idea that we are not only moving through space, but 'we' form part of a larger structure that is moving (forward), within such an object is the clockwork and chaos of individual galaxies and solar system.

@coblr I think what is interesting about trying to observe the motion of the earth and moon and stars, is the consideration not only of the interconnection of objects (solar system) and considering how the motion of the night sky especially alludes to the fact we are rotating & moving through space - in addition to these, it is the consideration of the smallest objects responsible for structure and matter and their behaviour and motion, and of the the largest objects, such as galaxies and black holes. It is like the 'motion of matter, motion made and caused by matter'. Self-perpetuating

Question 1: "How can we observe the Helical motion of our solar system (and galaxy) from earth?"
Short answer:
We can't observe it directly. Instead, we can observe the position and velocity of other bodies in the Solar System and galaxy to deduce all of our collective relative motion, with help from Newton's law of gravitation.
Long answer:
Within our Solar System, from points on the Earth, we can track the locations of the other planets and moons with telescopes.  We can also determine their distances using relative brightness measures and measures of apparent size in the night sky.  We can also determine their speed and direction relative to the Earth by taking time differences of their positions.  When we put all this together with Newton's law of gravitation we can create a model of the Solar System with the Sun near the center that has incredible descriptive and predictive accuracy.
For other stars in the Milky Way galaxy, we can determine their location with high accuracy telescopes.  We can estimate their composition and mass using techniques from radio astronomy.  We can partially estimate their direction and speed by taking a time series of measurements and by comparing their emitted frequencies to known emission spectra to derive Doppler measurements. Similar to the way we created a model of the Solar System, all the star velocity and position data allows us to create a model of the galaxy that is consistent with our measurements.  For the galaxy, however, the rotation rate consistent with the Newton's law of gravitation (as captured by the virial theorem) requires a much higher mass then we have observed, so our model is only consistent with our measurements if we assume the presence of "dark matter."
Using a coordinate system with the supermassive black hole SGR A* at the center, we create a model of motion consistent with our measurements for both the Solar System and the Milky Way galaxy. Further, this model is consistent with Newton's law of gravitation, dark matter, and just a bit of Einstein's relativity. In this model, the movement of planets in the Solar System around the Galactic center form something similar to a bent helix, since the angular momentum vector of the Solar System is inclined about 60 degrees from the angular momentum vector of the galaxy.
It may not make much sense to think about the motion of the galaxy as a whole, since there isn't an advantage of one spatial frame of reference over another at that scale, according to special relativity.  If you choose something like the Great Attractor as your inertial reference frame origin, the motion of stars in the Milky Way will also trace out bent helixes.
Question 2: "Is the solar system following its own Helical motion and therefore are all other solar systems within the Milkyway also following their own Helical motion?"
Yes, the helical motion of our Solar System is independent of the helical motion of other systems in the galaxy and the galaxy's helical motion is independent of the helical motion of the stellar systems within it.  Your analogy of a grandfather clock with many independent small clocks inside of it is more apt than your analogy of grains of sand in jelly.
Perhaps a better analogy yet is the carnival ride tilt-a-whirl, with the children's toys, pin-wheels affixed to the cars. If you put LED lights on the ends of the pinwheels and took a time lapse photo of the tilt-a-whirl running, you would see independent combinations of helical motion from both the pinwheels and cars!

Yes, I see - reminds me of the illustration often used to explain relativity, with objects moving independently within other moving objects, and the light trace (visual effects) often used to show continuous motion. We can observe the rotation of the earth around the sun, and rotation of moon around earth. We can also observe the spherical property of earth by noticing the shape of sunrise of the horizon (like a half waning moon shape). But the greatest observation when noticing the motion of stars is that earth is spinning and moving forward. Time-lapse is a a great way to notice motion.

Hi Connor, did I manage to answer your question? Otherwise feel free to ask for clarification :)

@pela Dylan was the question asker,  you and I were the answerers.

@ConnorGarcia Ah, I'm sorry, I didn't scroll far enough up

@pela Easy mistake to make, especially since I quoted the questions!

All motion is relative. There is no absolute space which can be said to be the correct reference frame in which we must measure all motion. Hence, for any object you're free to construct any reference frame that does, or does not, give that object a helical motion.
Reference frames
Of course, some reference frames make more sense than others. When you cycle to work, it makes sense to measure your speed relative to the bike lane. The reference frame of the Sun makes sense if you're concerned with planets' orbit. In this frame, Earth has no helical motion, only circular motion. Or you could call it helical motion with $v_z = 0$, if you wish.
On the other hand, the Sun and its neighbors follow a path through the Milky Way; the so-called Local Standard of Rest, measured by observing nearby stars and star-forming regions. In this frame, Earth has a (non-$v_z\!\!=\!0$) helical motion.
Similarly, in the reference frame of Milky Way's path toward, say, Andromeda, the Sun has a helical motion. Another star will have a similar, but different, motion.
Comoving coordinates
That said, there is one reference frame which has a special status, namely the Universe's comoving frame. These are the coordinates that expand along with the Universe, and they are the coordinates in which everything on average is at rest. If you hear the phrase "Milky Way's motion through space", this is probably what is referred to. Since time flows differently in different frames, it's also the frame of choice when you talk about the age of the Universe. Likewise, since lengths are contracted in moving frames, the comoving frame is the frame in which we measure cosmological distances.
It's important to realize, however, that motions, positions, etc. measured in the comoving frame are no more "correct" than in any other frame.
The cosmic microwave background
But how can we measure a velocity in this reference frame, if all motion is relative?
One way would be to measure the motion of a million galaxies with respect to us, and take the average. But a much easier method, which is in fact how we do it, is to measure the dipole of the cosmic microwave background (CMB). This radiation was emitted when the Universe was very young, and has the special property that it was emitted isotropically (i.e. the same in all directions) with a near-perfect Planck spectrum.
Hence, if you standing still in the comoving frame, you will see exactly the same CMB spectrum in all directions, whereas if you're moving you will observe the CMB to be blueshifted in front of you, and redshifted behind you. This difference between the two opposite directions is called the dipole, and its value was measured by the Planck satellite (Planck Collaboration 2020) to imply a total velocity of the Sun in the comoving reference frame of
$$
v_\odot = 369.82\pm0.11\,\mathrm{km}\,\mathrm{s}^{-1},
$$
toward the Galactic coordinates
$$
\{\ell,b\} = \{264.021^\circ\pm0.011^\circ, 48.253^\circ\pm0.005^\circ\}.
$$
Note the amazing precision, with an uncertainty in speed and direction of only 400 km/h and less than an arcminute, respectively!
This velocity is important to know, since it is used to correct all measured cosmological velocities.

Amazing how scientists have been able to rationalise, and develop technology to gain greater insight into the structure of the universe. Before the rational, calculating, mathematical explanation, there is also the visualisation happening in our intelligent mind as we try to comprehend ideas around vast space and structures. The step from visualisation and imagination through to science is a micro-moment of discovery, which is a more accessible way to learn about everything you've outlined. Comoving and considering the motion of objects forming part of moving objects: galaxy & solar system.

@Dylan Yes, for some people (e.g. me) visualization is an important part of understanding. Others are able to understand from pure math, but I often find it helpful to make little drawings of my problems.


text:
Understanding better the factors on $C_l$ in Angular power spectrum and Relation with Matter power spectrum
I am looking for an explanation on the angular power spectrum. I found this extract that is interesting but not fully understood for me (I will cite the step that I didn't understand)
"what is done is we take the map of the CMB sky, and perform a spherical harmonic transform on it. A spherical harmonic transform is basically the same general concept as a Fourier transform, but the spherical harmonics are mutually orthogonal on the surface of a sphere, like so:
$$
f(\theta, \phi)=\sum_{l m} a_{l m} Y(\theta, \phi)_{l}^{m}
$$
Here the "I" index denotes the number of oscillations, and the "m" index is a way of encoding the direction of oscillation on the sphere, and varies from "-l" to "I". For example, $I=0$ is zero oscillation: this is the monopole that sets the overall scale. $I=1$ is a dipole: one full oscillation over the sphere, and there are three possible directions $(x, y$, and $z)$. Go to higher and higher I values, and you get more (and therefore smaller) oscillations and more possible directions for those oscillations. The way they are typically written, the spherical harmonic functions
$$
Y(\theta, \phi)_{l}^{m}
$$
are complex functions, and the coefficients are therefore complex. This is a minor issue, though. In order to build the power spectrum, we average over directions. This is done as follows:
$$
C_{l}=\frac{1}{2 l+1} \sum_{m=-l}^{l} a_{l m} a_{l m}^{*}
$$
Finally, in building that plot, you may notice that the vertical axis is not
$$
C_{l} l
$$
but is instead
$$
C_{l} l(l+1) / 2 \pi
$$
It turns out if we had a power spectrum that was uniform in a logarithmic interval in I, then multiplying said function by
$$
l(l+1) / 2 \pi
$$
would give us a constant. Thus this multiplication allows us to interpret the function more easily, because inflation predicts that the primordial power spectrum, the one initially generated by inflation, would be nearly a constant in this space.
If inflation is true, then, all of the features you see in a power spectrum written as above that deviate from a constant stem from the dynamics of the universe between inflation and the emission of the big bang (plus some very slight modification between us and the CMB). For example, the long damping tail at high I stems from the fact that the surface of emission of the CMB is not instantaneous: the phase transition from a plasma to a gas happened over time, and the resultant blurring of the signal damps the small-scale fluctuations. There's also the ratio between the even and odd peaks of the power spectrum. This comes about because of the differences in the physics between normal matter and dark matter: dark matter just falls into potential wells, while normal matter bounces. The failure of dark matter to bounce causes a reduction of the evennumbered peaks relative to normal matter."
1) First problem :
What I have difficulties to grasp is the signification of "vertical axis" is not :
$$
C_{l} l
$$
but is instead
$$
C_{l} l(l+1) / 2 \pi
$$
What does "vertical axis" mean in this context ?
How to proove that ?
2) Second problem :
And after they say : "It turns out if we had a power spectrum that was uniform in a logarithmic interval in I, then multiplying said function by
$$l(l+1) / 2 \pi$$
would give us a constant. Thus this multiplication allows us to interpret the function more easily, because inflation predicts that the primordial power spectrum, the one initially generated by inflation, would be nearly a constant in this space."
I don't understand which is the trick to make the product by
$$l(l+1) / 2 \pi$$ being constant. It is very disturbing since the angular power spectrum is not constant for example with the angular spectrum of CMB.
3) Just a precision :
in one link of my previous post ( previous link ) Why does the writter say "How to write the 3D power spectrum, $P_{k}$, as an integral of the angular power spectrum, $C_{\ell}$ ?" whereas from another author, it is not the direct relation beween angular power spectrum and matter power spectrum.
By, the way, could anyone write the complete formula that links these 2 quantities ($C_{\ell}$ and $P_k$) ?
I would be grateful since I am a little confused between angualr correlation function and matter power spectrum and spherical Bessel functions.
If someone could give me clarifications on these 2 points, I would be grateful.

I think I can sort of answer your first/second question.
It's a bit hard to guess what your background is, but I hope you have seen or derived somewhere that the $a_{lm}$ coefficients can be written as
$$\oint \Theta(\hat{x}) Y_{lm}^*(\hat{x}) d\hat{x}$$
where $\Theta$ is the temperature fluctuation (as seen in the CMB) and $Y_{lm}^*$ is (the complex conjugate of) a spherical harmonic. Note that $\vec{x} = (\theta, \phi)$, $\hat{x}$ is the unit vector parallel to $\vec{x}$ and $\hat{k}$ is the unit vector parallel to $\vec{k}$. $k$ is the length of $\vec{k}$.
In cosmology there is an effect called the Ordinary Sachs-Wolfe effect, which has to do with redshifting of blueshifting due to matter overdensities. In this theory, you find an expression for $\Theta$ which is
$$\Theta = \frac{1}{3}\Phi = \sum_k \frac{1}{3}\Phi_k e^{i\vec{k}\cdot \vec{x}}$$
where $\Phi$ is the Bardeen potential. You can plug this expression into the equation for the $a_{lm}$ and then you can use the expansion of a plane wave, which goes like this:
$$e^{i\vec{k}\cdot \vec{x}} = 4\pi\sum_{lm} i^l j_l(kx) Y_{lm}(\hat{x})Y_{lm}^*(\hat{k})$$
So then you get
$$a_{lm} = \frac{4\pi}{3}\sum_k \Phi_\vec{k} \sum_{lm} i^l Y_{l'm'}^*(\hat{k}) j_{l'}(kx)\oint Y_{lm}^*(\hat{x}) Y_{l'm'}(\hat{x}) d\Omega$$
Due to orthonormality of the spherical harmonics this becomes
$$a_{lm} = \frac{4\pi}{3}i^l \sum_k \Phi_\vec{k} j_l(kx) Y_{lm}^*(\hat{k})$$
You may have also come across the expression for the $C_l$:
$$ C_l = \frac{1}{2l + 1}\sum_m E[|a_{lm}|^2]$$
which due to the above is proportional to
$$C_l \propto  \frac{1}{2l + 1}\sum_m \sum_{kk'} E[\Phi_\vec{k} \Phi{\vec{k}'}] j_l(k'x) j_l(kx) Y_{lm}^*(\hat{k}) Y_{lm}(\hat{k}')$$
The E[ ] is the covariance matrix.
If $\Phi_\vec{k}$ is a random variable, then $E[\Phi_\vec{k} \Phi_\vec{k'}] \propto k^{n - 4}\delta_{kk'}$
I'm not sure how to show this quickly. I'm afraid you need quite a few derivations to get to this point. It follows in part from the assumption of a power spectrum of the form $P(k) \propto k^n$.
Using this you find
$$C_l \propto \sum_k k^{n - 4}j_l(kx)^2 \rightarrow \int_0^\infty k^{n - 2} j_l(kx)^2 dk$$
if we approximate the sum by an integral. For a Harrison-Zel'dovich spectrum, which is a special kind of powerspectrum, defined as $P(k) \propto k$ (so n=1), this becomes
$$ C_l \propto \int_0^\infty j_l(kx)^2 \frac{dk}{k} = \frac{1}{2l(l + 1)}$$
so in the case of a Harrison-Zel'dovich spectrum, the quantitiy $l(l + 1)C_l/2\pi$ is constant for the Ordinary Zachs-Wolfe effect. They put $l(l + 1)C_l/2\pi$ on the y-axis to make spotting this effect easy, I suppose.
I'd be very interested to see a detailed answer to your last question.

Thanks a lot for your help, it is kind from your part to give a detailled answer. I am waiting also like you for a potential answer of my last question. Best regards

Could you replace please all the $x$ variable by $\theta$ for more visibility and clarity : sorry I don't want to be boring but this is for convenience 2) the $k$ is the wavenumber, we are allright ? Thanks in advance. Best regards

@youpilat13, yes k is the wavenumer. It's a vector, actually, and so is x, so each time you see them multiplied, I'm referring to a dot product where $\vec{x} = (\theta, \phi)$

Just a last precision : when you write $j_l(kx)^2$, how can I understand it ? : is it equal to write $j_l(k\theta)^2$ or $j_l(k\phi)^2$ ? Regards

@youpilat13 sorry, may have confused myself a little with my own notation. I checked my cosmology notes and added all the proper symbols now. I hope this clears up the confusion somewhat.

Sorry, it seems that it is me whos is confused to not understand whith your notations. If you could replace everywhere the variables $\theta$ and $\phi$ instead of $\vec{x} = (\theta, \phi)$, , there are steps where it is ambiguous, especially for the Bessel functions $j_l(kx)^{2}$. I would be grateful if you could do it to me this favor. Best regards

.Isn't possible for you to write all the equations with $\theta$ and $\phi$ variables ? Sorry, this is for my convenience, I don't want to be boring.

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text:
Is threre a relation between relative or absolute error and standard deviation for ratio of power spectra?
I have to compute the variance on this ratio, that is to say on the observable $O$ :
$$O=\left(\frac{C_{\ell, \mathrm{gal}, \mathrm{sp}}^{\prime}}{C_{\ell, \mathrm{gal}, \mathrm{ph}}^{\prime}}\right)=\left(\frac{b_{s p}}{b_{p h}}\right)^{2}\quad(1)$$
where $C'_{\ell}$ are angular power spectra (or matter power spectra).
Can I apply for this the computation of relative or absolute error like in electricity, we have $U=RI$ and the error on the quantity $R$ :
$$\dfrac{\Delta R}{R}= \dfrac{\Delta U}{U}+\dfrac{\Delta I}{I}$$
Is there a relation between absolute or relative error with the standard deviation of theses ratio ?
I would like to apply it to equation($1$) to compute this error or variance : is the right method ?
If not, which solution would be possible ?
I have difficulties to grasp the subtilities between uncertainty and standard deviation : I just need to compute $\sigma_{o}^{2}$.

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text:
Cosmic variance - Fourier Transform - a few precisions
I have posted initially on physics exchange but none answer, so I try my chance here (I am going to delete the physics exchange one) :
I cite an original report of a colleague :

If we are interested in power spectrum, we want to estimate the
variance of the amplitude of the modes $\mathrm{k}$ of our Fourier
decomposition. If one observes the whole observable Universe and we
do the Fourier transformation we get a cube whose center is the mode
$ |\vec{k}| = 0$ which corresponds to the mean value of the observed
field.
This mode has only one pixel. How do we measure the variance of the
process at $ |\vec{k}| = 0 $? We can not.
In fact we can but the value of doesn't mean anything because the
error is infinite. In other words, we have an intrinsic (statistical)
error which depends on the number of achievements to which we have
access.
one will be able to consider spheres of sizes $ dk $ between $ [k: k +
> dk] $ which will contain a number of pixels $ N_ {k} = V_ {k}/(dk)^{3}$ where $ V_{k} = 4 \pi k^{2} dk $ is the volume of the sphere
and $ (dk)^{3} $ is the volume of a pixel in our Fourier transform
cube.
So one can estimate how many values one can use to calculate our
power spectrum for each value of $ k $. The greater the $ k $, the
greater the number of accessible values and therefore the statistical
error decreases. The power spectrum is a variance estimator so the
statistical error is basically a relative error:
$ \dfrac{\sigma (P(k))}{P(k)} = \sqrt{\dfrac{2}{N_{k}
> -1}}_{\text{with}} N_{k} \approx 4\pi \left (\dfrac{k}{dk}\right)^{2} $
So we can see that for the case $|\vec{k}| = 0$ we have an infinite error because $N_{k} = 1$

I can't manage to proove that the statistical error is formulated like :

$\dfrac{\sigma (P (k))}{P(k)} = \sqrt{\dfrac {2}{N_{k} -1}}_{\text{with}} N_{k} \approx 4\pi \left(\dfrac{k}{dk}\right)^{2}$
and why it is considered like a relative error ?

Which are the conditions to assimilate a statistical error (standard deviation) to a relative error ($\dfrac{\Delta x}{x}$) ?

How to prove that $N_{k} = 1$ in the case $|\vec{k}| = 0$ ?

Any help is welcome.

Your question on Physics SE was only posted 14 hours before you cross-posted here.  That's just impatience on your part.  I am particularly confused why you do not simply ask your colleague.


text:
Poisson noise for a squared contrast of density - expression of variance equal to $1/n^2$

I manage to compute the Poisson noise of a density field like this :

If I take $N$ the density of galaxies and compute the Shot noise with a Poisson distribution, I get :
$\langle N^2\rangle - \langle N\rangle^2 = \langle N\rangle$ so :
$\langle N^2\rangle = \langle N\rangle + \langle N\rangle^2$
Let's take the variable : $X=\dfrac{N}{\langle N\rangle}-1$
So, I get $\langle X\rangle=0$.
Then : $\langle X^2\rangle = \left\langle\left (\dfrac{N}{\langle N\rangle}-1\right )^2\right\rangle=\dfrac{\langle N^2\rangle}{\langle N\rangle^2}-2+1 = \dfrac{\langle N^2\rangle}{\langle N\rangle^{2}}-1$
Finally, I get : $\sigma_x^2 = \langle X^2\rangle=\dfrac{1}{\langle N\rangle}+1-1$
$$\Rightarrow\quad \sigma_x^2=\dfrac{1}{\langle N\rangle}$$

Now, I would like to do the same kind of calculation with variable  $Y=\left(\left(\dfrac{N}{\langle N\rangle}\right)-1\right)^2$, and conclude normally that :

$$\sigma_y^2 = \dfrac{1}{\langle N\rangle^2}\quad(1)$$
But following the same reasoning with $N$ following the Poisson distribution, I can't manage to get this expression $(1)$.
UPDATE : i think that I have the proof :
$$\langle X^4\rangle = \Bigg\langle\bigg(\dfrac{N}{\langle N\rangle}-1\bigg)^2\,\bigg(\dfrac{N}{\langle N\rangle}-1\bigg)^2\Bigg\rangle$$
$$=\dfrac{\langle N^2\rangle^2}{\langle N\rangle^4}-2\dfrac{\langle N^2\rangle}{\langle N\rangle^2}+1$$
I) with $\langle N^2\rangle = \langle N\rangle + \langle N\rangle^2$
II) and squared :
$$\langle N^2\rangle^2 = \langle N\rangle^2 + 2\langle N\rangle^3 + \langle N\rangle^4$$
We conclude :
$$\langle X^4\rangle=\dfrac{1}{\langle N\rangle^2}+\dfrac{2}{\langle N\rangle} +1 - \dfrac{2}{\langle N\rangle} -2 + 1 = \dfrac{1}{\langle N\rangle^2}$$

It's not clear to me what you mean by "Shot Noise", particularly as this is usually a synonym for noise that follows Poisson statistics. (e.g., https://en.wikipedia.org/wiki/Shot_noise)

@PeterErwin the OP notes the connection "since Shot Noise is often assimilated to a Poisson noise" but at least the basic Poisson distribution requires a $(k!)^{-1}$ where $k$ is an integer, and I think the analysis discussed here is of density fluctuations over various scales and so there aren't discreet volumes with integer numbers of galaxies in them, except perhaps within some integral over volume sizes.

@uhoh A Poisson distribution has (using the symbols from the Wiki page) an expectation value of $\lambda$ (not $k$) and a variance of $\lambda$. For the "density of galaxies" problem, $\lambda = \bar{n}{\rm gal}$, so I don't understand why the OP says the "Shot Noise" should be $1/\bar{n}{\rm gal}$.

@PeterErwin in backwards order; yes I see, the variance $\sigma^2$ of a Poisson distribution is equal to $\lambda$ just like the normal distribution. About $\lambda$ and $k$, I'm just thinking of how I've used them; I had data of integer values $k$ and looked for the likelihood that it was the result of some value $\lambda$; finding the most likely $\lambda$ to have produced the measured $k$ (maximum likelihood).

@uhoh . Thanks for your answer. Unfortunately, I don't know yet how to justify that the expectation of Shot Noise $\bar{N}{sp}$ should be equal to 0 like a white noise. I am also interested in the variance of Shot Noise : if I take a fixed value for the spectroscopic Shot Noise, then, the variance is equal to : $<(N{sp}-\bar{N}{sp})^{2}>=1/\bar{n}{\rm gal}^{2}$. Could anyone confirm this result ? Regards

@youpilat13 I'm afraid I have no idea what you're trying to get at. Your last comment mentions "spectroscopic Shot Noise" -- whatever that's supposed to mean -- and then tries to relate that to the mean density of galaxies. That makes no sense.

@PeterErwin . For a simple contrast of galaxies, we have for the spectroscopic probe the noise equal to $\dfrac{1}{\bar{n}{gal}}$ with $\bar{n}{gal}$ the density of galaxies detected by spectrocopic detector. What I am looking for is the expression if this noise but with the squared constrat, always following the Poisson law. If you could help me, this would be great ! I know already the result ($\dfrac{1}{\bar{n}_{gal}^{2}}$) but not the demonstration. Regards

@youpilat13 Have you tried asking about this on statistics.stackexchange? That seems like a much more appropriate place, since your question appears to be about estimating variances by calculations and manipulation of expectation values, and not really anything astronomical.

@youpilat13 One additional issue/problem is that you appear to be working within a very specialized subset of cosmology, and aren't attempting to translate the jargon you're using. (E.g., "spectroscopic probe", which means nothing to me.)

@PeterErwin . Sorry, spectroscopic probe means simply measures coming from a spectroscopic detector.

I agree this is better suited for statistics SE.


text:
Can we get a picture of Oumuamua with Hubble Space Telescope?
Can we get a picture of Oumuamua with Hubble Space Telescope?
If not, what are the reasons?
Answers why other telescopes might fail to get a picture of Oumuamua are welcomed as well.

Welcome to astronomy SE! Great question BTW.

No we can't, The Wikipedia article states that:

the size and shape have not been directly observed as ʻOumuamua appears as nothing more than a point source of light even in the most powerful telescopes

In addition, note c states that:

Brightness peaked at 19.7 mag on 18 October 2017, and faded below 27.5 mag (the limit of Hubble Space Telescope for fast-moving objects) around 1 January 2018.

So the HST has a limit of 27.5 magnitude for fast-moving objects, and ʻOumuamua is now fainter than that limit.

Generating an ephemeris for 1I/'Oumuamua for HST in JPL HORIZONS shows that it is now magnitude 36...


text:
Looking for analytical expression of Cosmic Variance - Poisson distribution?
I have an expression of Matter Angular power spectrum which can be computed numerically by a simple rectangular integration method (see below). I make appear in this expression the spectroscopic cosmological bias $b_{s p}^{2}$ and the Cosmic variance $N^{C}$.
$$\mathcal{D}_{\mathrm{gal}, \mathrm{sp}} =\left[\int_{l_{\min }}^{l_{\max }} C_{\ell, \mathrm{gal}, \mathrm{sp}}(\ell) \mathrm{d} \ell\right]+N^{C}=b_{s p}^{2}\left[\int_{l_{m i n}}^{l_{\max }} C_{\ell, \mathrm{DM}} \mathrm{d} \ell+N^{C}\right]=b_{s p}^{2}\left[\mathcal{D}_{\mathrm{DM}}+N^{C}\right] \\
\simeq \Delta \ell \sum_{i=1}^{n} C_{\ell, \mathrm{gal}, \mathrm{sp}}\left(\ell_{i}\right)+b_{s p}^{2} N^{C}$$
I have a code that computes the terms $C_{\ell, \mathrm{gal}, \mathrm{sp}}\left(\ell_{i}\right)$ for each multipole $\ell_{i}$. But to compute $\mathcal{D}_{\mathrm{gal}, \mathrm{sp}}$, I have also to compute the term $b_{s p}^{2} N^{C}$ and especially the Cosmic Variance $N^{C}$ :
The only documentation I have found is the following slide from Nico Hamaus :

But as you can see, I have no explicit expression for Cosmic Variance : Could I consider the relation $\dfrac{\sigma_{p}}{P}=\dfrac{2}{N_{k}^{1/2}}$ as a SNR (Signal Noise Ratio) ?
Which expression of Cosmic Variance could I use to compute the whole expression $\mathcal{D}_{\mathrm{gal}, \mathrm{sp}}$ ?


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I am getting triple image of the moon from a 50mm telescope
I just bought a telescope and saw the moon but with the bright moon at the center there are two faint images of the moon above and below it. I figured out it was due to the periscope mirror. If I see my finger through that periscope it has a faint outline around it. Which is getting magnified when I try to look the moon.then I tried to see my finger from a normal mirror diagonally and I first time noticed that every mirror does the same , there is a faint outline around my finger if I see it diagonally through a mirror. So my question is, how do I fix it now?

If it was built cheaply, the "periscope"(?) might have used back-surface mirrors; normal mirrors with glass on the front and the reflective material on the back side where it's protective.
In this case, you'll get roughly a 5% reflection from the front glass surface at 45° and maybe 80% reflection from the metal on the back (only 95% is transmitted and maybe the metal is 85% reflective)
A periscope has two mirrors in a dogleg configuration, so that will produce one "ghost image" above and one below.
Our eyes are non linear, so even though the reflections are roughly 40 times dimmer than the main image they will still be irritatingly visible.
Good optics would have used front surface = first surface mirrors. Glass lenses used in telescopes have anti-reflection coatings otherwise we'd see lots of lens flare in every image from the curved surfaces.
But since mirror surfaces are flat, they make ghosts instead of flare.
I think there's not much you can do unless you replace the mirrors with front-surface mirrors.
Probably the periscope was meant to be used in daytime so the weak reflections weren't as noticeable. But in astronomy the background is black and even tiny reflections are easy to see.


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How to solve numerical problems based on diurnal parallax method (formulas, examples and calculations)?
In physics, I have to solve some numerical problems based on the parallax method. I learnt the parallax method from Khan Academy and Wikipedia. (I found a really helpful Khan Academy video parallax in observing stars and stellar distance using parallax.) For stellar parallax, I  know that the distance $d=\tan(90^{\circ}-\theta) \ \text{AU} = \cot \theta \ \text{AU} =1/\tan\theta \ \text{AU}$. But I am not sure how to solve problems based on diurnal parallax.
For the parallax method,

Knowing the distance between the observation points L ( basis ) and the displacement angle α, you can determine the distance to the object: $D = {\frac {L} {2 \ \sin \ \alpha / 2}}$.
For small angles (α - in radians): $D = {\frac {L} {\ \alpha}}$.

But how did we derived the formula $D = {\frac {L} {2 \ \sin \ \alpha / 2}}$?
Using parallax method, we can measure:

the distance of sun or moon or any planet (Diurnal parallax method: solar parallax, lunar parallax)
the diameter of sun or moon or any planet
the distance of a nearby star (annual parallax or stellar parallax method)

For diurnal parallax (sun, moon, and planets within the solar system),
What is the formula for measuring the distance and parallax angle (with an example)? In my textbook, it is derived from the arc length formula $s=r\theta$. Now once we know the distance we can solve for the diameter of the sun or moon or any planet using the same parallax formula.
(This parallax angle or parallactic angle is usually given in the question/numerical problem and we have to solve for the distance or the diameter)

EDIT: Since I have started a bounty I want to make it clear what I want in the answer. I want the formulas, derivation of formulas,  related examples and calculations. The following is an example numerical problem based on the parallax method (You don't necessarily need to solve them, just explain how to apply the formula):

(a) The earth-moon distance is about 60 earth radius. What will be the diameter of the earth (approximately in degrees) as seen from the moon?
(b) Moon is seen to be of (½)° diameter from the earth. What must be the relative size compared to the earth?
(c) From parallax measurement, the sun is found to be at a distance of about 400 times the earth-moon distance. Estimate the ratio of sun-earth diameters.

Not necessarily duplicates, but some of the answers to some of these may be helpful: How to actually obtain the parallax angles that are used to determine distances of stars? and How does a telescope measure parallax angle? and Stellar parallax

About "I  am confused between the formulas: $d=\tan(90^{\circ}-\theta) \ \text{AU}$, $d=1/\tan\theta \ \text{AU}$" Just to double check are you confused that $\tan(90^{\circ}-\theta) = 1/\tan\theta$? That's just how $\tan$ works.

@uhoh I am sorry for the confusion I just realized that $\tan(90^{\circ}-\theta)  = \cot \theta = \ 1/\tan\theta $. Thanks for pointing out, I have edited my question accordingly.

The derivation
Carefully observe the following triangle:

We (the observers) are moving between the points $A$ and $B$, our object is in $C$.
By the definition of the sine:
$$\sin\frac{\alpha}{2}=\frac{L}{2D}$$
and with expressing $D$:
$$D=\frac{L}{2\sin(\alpha/2)}$$
but for small angles in radians:
$$\sin x = x$$
so
$$D=\frac{L}{2\alpha/2}=\frac{L}{\alpha}$$

Examples
Note that there are better methods to do these exercises, but we will use the ones we derived above.

(a) The Earth–Moon distance is about 60 Earth radii. What will be the diameter of the Earth (approximately in degrees) as seen from the Moon?

Here the total diameter of Earth is $L=2\rm\,R_E$. The distance is $D=60\rm\,R_E$. We express $\alpha$ in the formula above:
$$\alpha=2\arcsin\frac{L}{2D}=2\arcsin\frac{2\rm\,R_E}{2\cdot 60\rm\,R_E}=1.91°$$

(b) Moon is seen to be of (½)° diameter from the Earth. What must be the relative size compared to the Earth?

Now, the situation is reverse. We have $\alpha=0.5°$ and again $D=60\rm\,R_E$. Now, we need to express $L$:
$$L=2D\sin\frac{a}{2}=2\cdot 60\rm\,R_E\sin\frac{0.5°}{2}=0.52\rm\,R_E$$
We now divide both diameters:
$$\frac{2R_M}{2R_E}=\frac{0.52\rm\,R_E}{2\rm\,R_E}=0.26$$

(c) From parallax measurement, the Sun is found to be at a distance of about 400 times the Earth–Moon distance. Estimate the ratio of Sun-Earth diameters.

Here the required data is the angular size of the Sun: $\alpha=0.5°$ (of course, you need to know that number). The distance is now $D=400\cdot 60\rm\,R_E=24000\rm\,R_E$. As before, we express $L$:
$$L=2D\sin\frac{a}{2}=2\cdot 24000\rm\,R_E\sin\frac{0.5°}{2}=209.4\rm\,R_E$$
The ratio is now
$$\frac{2R_S}{2R_E}=\frac{209.4\rm\,R_E}{2\rm\,R_E}=105\rm\,R_E$$

It's not an isosceles triangle unless the observers happen to be in just the right positions.

@stretch It is. Observe the symmetry with rotating Earth.

Does that mean you'd have a single observer who waits twelve hours between observations? The moon is moving: it's right ascension will have changed by about six degrees between those observations.  Over the course of 24 hours today the Moon's right ascension will have changed by 11.7 degrees, and its declination by 4.6 degrees. The Earth - Moon distance in those same 24 hours will have decreased by 0.8 Earth radii.

@stretch The two observers observe at the same time on the antipodal points on the Earth's sphere.

How do you find the antipodal points? The  Moon's motion is irregular, north or south 25 degrees in declination.

@stretch Draw a line between Moon's and Earth's centre. Select a point on the Earth's sphere that lies on the great circle that is perpendicular to the above-mentioned line. Another point is just antipodal to the one before.

The first point is known as the Moon's geographical position (GP). You won't know where it is unless you look it up before the observation. Then your observers would have to travel great distances to place themselves 180 degrees of arc apart and 90 degrees of arc from the Moon's GP. I'm not saying there's anything wrong with your geometry. It's just it can't be implemented. BTW they don't have to be 180 degrees apart for your method. They just have to have equal zenith distances from the Moon. This is my final word on the question. If you're still not satisfied, try it out with almanac values.

@stretch Such problems are always idealistic and theoretical. In reality, one can obtain Moon's position on any two locations using simple trigonometry (using the angular distance between the observers and the parallax (and possibly the approximate value of Moon's position)).


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How to find the the diameter of the moon during a total solar eclipse?
I am not able to find the diameter of the moon. I have the following question in my physics textbook:

During a total solar eclipse the moon almost entirely covers the sphere of the sun. Write the relation between the distances and sizes of the sun and moon.

Similar question from my textbook:

It is a well known fact that during a total solar eclipse the disk of the moon almost completely covers the disk of the Sun. ,determine the approximate diameter of the moon.

I have been provided the following information:

Earth-Moon distance = $3.84\times10^{8}m$
Sun-Earth distance = $1.496\times10^{11}m$
Sun's diameter = $1.39\times10^{9}m$

The moon does not follow an exactly circular orbit, but a slightly elliptical one. So its apparent size does vary slightly. The same is true, to a slight extent, about the orbit of the earth around the sun - so its apparent size also varies slightly.

I don't think the diameter of the moon changes simply because its position or distance from Earth changes.  Perhaps you meant the 'Apparent diameter'?

A simple web search will find the diameter of the moon.  What have you tried?  You are expected to use similar triangles.  -1

@BillyC. The diameter of the moon does not change, I know that. I wanted to measure the same diameter during a total solar eclipse.

Draw a picture of Sun, moon, observer, as viewed from the side:
SSSSSSSSSSSS
\          /
 \        /
  \      /
   \    /
    MMMM
     \/

Since the triangles formed by the diameter of the sun and the observer, and the diameter of the moon and the observer are similar, you can write down the connection between the given distances using length ratios, or scale factor. (That is a mathematical problem of proportion) Solving the resulting equation will give you the answer.

Come for the physics, stay for the ascii art

If a sphere with a radius $R$ lies at a distance $d$ away from you, it can be covered exactly by an sphere with radius $\frac{1}{2}R$ lying away$\frac{1}{2}d$ away from you. This holds for all $R$ and $d$. Take it from there.

@huzaifaabedeen That's right. The radius of the sun divided by the ratio "distance to the sun" and "distance to the moon" (or multiplication by the inverse ratio).

From @JamesK's answer, I used similar triangles (△EMH ∼ △ESJ) to solve for the diameter of the moon.
$${EM \over ES}={MH \over SJ}$$
$$\Rightarrow {MH}= {{EM \times SJ} \over ES}$$
$$\Rightarrow {MH}= {{(3.84×10^8)×(1.39×10^9)} \over {2×(1.496×10^{11})}}$$
$$\Rightarrow {MH}= 1784 \ km.$$

BTW, the tangent to a circle (or sphere) meets the radius of the circle at a right angle, like this. The triangles are still similar, so it doesn't affect your calculation.


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Special case of rotation period
Rotation period is either prograde, like the Earth, or retrograde, like Venus. How likely is it for a body orbiting its parent object (either a planet around a star or a moon around a planet) to have a rotation period that is neither prograde nor retrograde?
By that I mean, this body has an infinite rotation period, and behaves like the Hubble space telescope when imaging distant stars, it doesn't rotate relative to distant stars.
Seen from the surface of the parent body, this object would look like it spins retrograde. (Seen from the Earth's surface, the Moon looks like it doesn't spin, yet it spins prograde.)
Seen from the object's surface, the night sky would be identical each night, and each day/night cycle is also a year.
Is this kind of infinite rotation period likely to exist and what would be the ideal conditions for it to be stable for a long period of time? What could be the shortest stable year or day duration for such a body?
i.e. avoiding short term tidal locking.
(for instance, Earth won't ever be tidally locked to the Sun because the Sun will become a supernova billions of years before it could happen.)

Statistically, this is what is called a space of measure zero.

This is extremely unlikely. And if it happens, it's a state of limited duration.
To explain "space of measure zero" David Hammen mentions in his comment, broken down for this example: there is exactly ONE state (exactly matching rotation) while there are infinitely many other possible rotation periods: just count all numbers, fractional and irrational included, and compare that to the amount of 0s among that.
Also it wouldn't be able to exist over an extended period of time due to every planet being subject to tides thus subject to slow secular or even longer-term changes in its rotation. And even a tiny bit slower or faster rotation would lead to the night sky changing slowly over the years.
Further, if you look at a planet where each side faces always the same side of the sky, you realize two things:

it is a VERY SLOW rotator, thus the the day-night cycle would match its annular cycle. That's slower than any planet in the solar system.
it would be on the "retrograde side" or rotational motion as it is slower than tidally-locked (tidally-locked means one revolution around its own axis during one year). This makes it even less likely to occur as usually planets and planetesimal get prograde spin during the initial phases of formation (and is changed by collisions in some cases in the later part of formation).

The temporal evolution of the spin would be such, that tides would spin-up its own rotation at the expense of orbital rotation (thus it would move slightly inward in order to start rotating faster - the speed depends on the strength of the tidal interaction with its host star. For comparison: on Earth, 1/3 of the tides is currently caused by the Sun)


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is weight of two identical objects same when they have the same exact distance from earth's surface but one is on a mountain and the other on a plane?
Is the weight of two identical objects the same when they have the same exact distance from the earth's surface but one is on a mountain and the other on a plane?
The distance between the plane and the earth's surface contains only air.
Anyway can anyone test this? I have seen some documentaries where they test the weight of the same person in different heights always on a hill or mountain but I have not seen anyone doing what I ask above. When that is tested, are there any data worth observing?

Do the two objects have the same mass?

They have the same mass. That can be tested with the same object as well.

Similar questions on physics.SE: https://physics.stackexchange.com/questions/458327/does-gravity-get-stronger-when-you-climb-a-mountain https://physics.stackexchange.com/questions/652752/does-gravity-get-stronger-the-higher-up-you-are-on-a-mountain https://physics.stackexchange.com/questions/374132/mountain-gravity

Rant about moderator egos removed.

I’m voting to close this question because there is no astronomical question here.  This is Physics or Earth Science

I know about the ranting part and i am sorry about that. As for the voting part, it is your vote and how you choose to use it. As for the reason for using that vote, i understand. Good luck!

What kind of plane? Do you mean an aeroplane / airplane?

Yes, that is what i meant but i now think that the airplane assuming not moving but just having the same height as the mountain  will act the same as the mountain does. Only if weighing the object being on its own in that specific height might work.

The weight of an object is the force due to gravity an object feels. Consider the force of gravity sourced by Earth's mass acting on two particles of equal mass, $m_1 = m_2$, but that are located at different heights $h$ in Earth's gravitational field, i.e. $h_1 \neq h_2$. Using Newton's second law, the force on each particle due to the gravity of Earth is,
$$ F_1 = m_1 a(h_1)$$
and
$$ F_2 = m_2 a(h_2)$$.
where m is the mass and a is the net acceleration. Since we're only considering accelerations due to gravity here, then $a$ is just the gravitational acceleration.
$a(h)$ is a function of height from Earth's center. So, if two objects of equal mass are at different altitudes, then they will necessarily experience different gravitational acceleration, and thus have different weight. Thus, if $m_1=m_2$ and $h_1 \neq h_2$, then we may conclude that $F_1 \neq F_2$.

And the force or whatever is called that the object is acting on the mountain is cancelled out from the force that the mountain acts back to that object, right?

They do not cancel. Please examine Newton's three laws of motion: the first states that an object will remain in motion unless acted upon; the second is the statement of $F = ma$; the third is the statement that forces between two objects exist in pairs. So, using Newton's third law of motion, there is a gravitational force that is equal in magnitude and opposite in direction acting on the Earth due to the gravity of the object. These are called third law force pairs (remember, for them to be third law force pairs, they have to act on/by the same objects!).

So ask yourself as an analogy: in a car crash between a small car and a big truck, which experiences the larger force? (this might be a trick question! ;) )

I read about Newton's third law on Wikipedia ( i have not understood it 100% yet). Regarding your analogy: small car and big truck have same speed. The force that they act onto the other is proportionate to their mass thus by newton's third law, the big truck experiences larger force. That is hard to understand.

You fell into my trap! ;D Newton's third law states that the force pairs are equal in magnitude and opposite in direction, so the force on the truck by the car has to be equal in magnitude to the force on the car by the truck! The difference is in the $acceleration$ that they feel: the car feels a larger acceleration since its mass is smaller but it feels the same force as the truck does. Does that make sense?

So with your object in the Earth's gravity field, the force on that object by the Earth and the force onthe Earth by that object are equal in magnitude, but the object feels a much larger acceleration than the Earth because the object is (very) less massive than the Earth.

Lastly, do not confuse these third law force pairs with your original question, which is about comparing the force on object 1 by the Earth to the force on object 2 by the Earth.

When you say Earth, you include the mountain right? Also, how can the forces by the vehicles to on another be equal in magnitude? In that sense, if we have one object of steel and an object of cotton with the same mass and height from earth's core and same speed, they should apply a force of equal magnitude to one another, right?

I did not mention a mountain. It depends on what you mean in the problem you're trying to solve. You did not define your problem rigorously enough for me to comment on it directly. In my answer, I am treating the Earth as a sphere and talking about the gravitational acceleration that objects feel at different heights from the surface of that spherical Earth. You could consider one such point to be a mountain. Try reading about third law force pairs here https://www.physicsclassroom.com/class/newtlaws/Lesson-4/Identifying-Action-and-Reaction-Force-Pairs and try to solve their examples.

I understand. Read that article and is a good one. Can you explain to me in simple terms how do you understand the concept of a force? I will be grateful!

It depends on how sophisticated of a description you want. If you're at introductory level, then Newtonian mechanics is sufficient, where a force is represented by a vector defined as the time derivative of the linear momentum, https://en.wikipedia.org/wiki/Newton%27s_laws_of_motion#Newton's_second_law

If you're interested in more sophisticated treatment, you should first read about a 4-force in the theory of Special Relativity https://en.wikipedia.org/wiki/Four-force , and even more advanced is how that is generalized in Einstein's theory of gravitation (called General Relativity) https://iopscience.iop.org/article/10.1088/0143-0807/31/4/024/meta

You have been very helpful and i thank you very much for that. I do not want to waste your time anymore so thank you very much and good luck! " An Atom or Particle can only walk forward never backward" :)

PS. Related to the commend above: Forget "Time" . Time does not exist. Had to add this

This ďoes not answer the question.

It clearly did. Look above at the convo I had with the OP...... @DescheleSchilder you're clearing targeting me b/c I spoke against you yesterday, which is against site policy to seek vengeance like you're doing. Please stop or else I'll be forced to ask for moderator intervention.

Typically, when talking about gravity, we assume the objects to be dimensionless points with all their mass concentrated in their center of mass. This makes working with gravity much easier, and it is generally a "good enough"™ approximation when the two objects are "far away"™ from each other (as is usually the case in Astronomy). If we need a little more precision, we assume that the two objects are spheres. For celestial bodies, specifically, we can add even more precision by assuming they are ellipsoids or oblate spheroids. And for even more precision, we actually have geoid models of Earth, for example.
However, for this particular example, the object and the Earth are not far away from each other, so the shape of the Earth definitely is relevant. In fact, even the composition of the Earth is relevant: a mountain made of aerogel and a mountain made of Uranium will behave very different.
The object on top of the mountain will have more weight, because there is "more mass" below it that is "pulling" on it due to gravity. Even two objects on two mountains that have the same shape but different density will experience a different weight.
While I don't think we have measured this on planes and mountains, we definitely have measured this on satellites. For example, on Earth, the twin GRACE satellites have created "gravity maps" of Earth. Similarly, the GRAIL satellites have produced equivalent results for the Moon. Areas with a high concentration of mass are called mascons.
More narrowly, mascon refers to an area, where topographically you would think that gravity is lower than average (e.g. a depression such as a crater or a valley), but due to the density of the material the ground is made up of, the gravity is actually higher than average.
With regards to your specific question about standing on a mountain, here is a map of gravity anomalies of Earth:

Can you spot the Andes? Looking at the map, imagine you are standing on top of the Aconcagua at an elevation of 6961 m in one of the deep red areas of the anomaly map, and then you are standing in an aeroplane at the same altitude above the North Atlantic north-east of Puerto Rico or the Indian Ocean, above one of the deep blue areas of the map. You will be lighter above the Indian Ocean than above the Andes, even at the same altitude.
As mentioned above, the Earth is not a perfect sphere, it is "fatter" at the equator, so gravity and thus weight also differs with latitude: when you are standing at the equator, there is more mass below your feet than when you are standing on one of the poles, even though at the equator you are actually further away from the center of mass.
The Moon is one of the celestial bodies we have measured that has the most extreme "lumpy" gravity field. Its mascons are so big that there are places on the Moon where a plumb bob will actually deflect sideways by one third of a degree!

Thank you for spending some of your time for that greatly structured answer! I look at the image with the Earth's Gravity Field Anomalies and this question pops up in my head: what is the distance that is needed to be between Earth and another object  in order for Earth's gravity field to act as one ( a single unit or without anomalies) ?  Also, can you describe in simple terms Earth's Elliptical path during its orbit around the sun? If you provide the math for it after the explanation, that would be great and really appreciated.

Sometimes the wrong answers are accepted. Great answer!


text:
Solar Rotation Varies with Latitude. Effects?
From the Wikipedia article of "Solar Rotation":

Solar rotation varies with latitude. The Sun is not a solid body, but is composed of a gaseous plasma. Different latitudes rotate at different periods. The source of this differential rotation is an area of current research in solar astronomy. The rate of surface rotation is observed to be the fastest at the equator (latitude φ = 0°) and to decrease as latitude increases. The solar rotation period is 24.47 days at the equator and almost 38 days at the poles. The average rotation is 28 days.

What are the effects of different rotation period between latitudes on Sun itself?

The sun is thousands of degrees on its surface and millions in the center. Without gravity, it explodes.

It is curious that there seems to be equator to pole shear near the surface. I do think you need to remove the part about removing gravity, nothing meaningful can be said in that case; everything that makes the Sun what it is comes from gravity.

You're asking two questions. Please note for future reference that one of Stack Exchange's reasons for closing a question is asking more than one question.

@JBH  Thank you for telling me that. I will remember not to do that in the future.

I've removed the question about gravity.  It wasn't clear what it was about, how it related to the main question.

The sun is made of plasma, which is a conductive gas. As such, magnetic field lines will get trapped in the plasma and twisted by the differential rotation.
Tangles of magnetic field in the sun have produce very powerful magnetic fields.  Where these tangles meet the surface of the sun there are active regions that have sunspots and prominences. The breaking and reconnection of these fields releases large amounts of energy into the solar atmosphere, causing coronal mass ejections.
The 11 year cycle of the sun's activity seems to be linked to a cycle of "tangling, breaking and relaxation" of magnetic fields, caused by the differential rotation of the sun. (source)
If you turn of gravity, the sun explodes, rotation is not the cause of gravity

Thank you for your answer. I agree with you that rotation is not the cause of gravity. The differential rotation of the sun though i think that can be broken down. Maybe it is not necessary for us to know how a star is formed in order to understand why that differential rotation happens. A constant is that all the particles of the sun are charged, therefore there must be work. Is that right?

If you have a new question about why differential rotation occurs you can ask it.  The question you posed was on the consequences, not the causes.

Thank you James, i will create a new question.


text:
What is the exact reason for the Sun's differential rotation to occur?
The following is an excerpt from the Wikipedia article for "Differential Rotation":

Differential rotation is seen when different parts of a rotating
object move with different angular velocities (rates of rotation) at
different latitudes and/or depths of the body and/or in time. This
indicates that the object is not solid. In fluid objects, such as
accretion disks, this leads to shearing. Galaxies and protostars
usually show differential rotation; examples in the Solar System
include the Sun, Jupiter and Saturn.
The cause of differential rotation
Stars and planets rotate in the first place because conservation of  angular momentum turns random drifting of parts of the molecular cloud that they form from into rotating motion as they coalesce. Given this average
rotation of the whole body, internal differential rotation is caused
by convection in stars which is a movement of mass, due to steep
temperature gradients from the core outwards. This mass carries a
portion of the star's angular momentum, thus redistributing the
angular velocity, possibly even far enough out for the star to lose
angular velocity in stellar winds. Differential rotation thus depends
on temperature differences in adjacent regions.

Can you please explain the cause of differential rotation described above in a very simple way?

Are you referring to the differential rotation between the sun's equator and poles, or the differential rotation between the sun's core and envelope?

@DaddyKropotkin I am referring to the differential rotation between the sun's equator and poles.

Duplicate: https://astronomy.stackexchange.com/questions/40494/why-does-the-outer-layer-of-the-sun-have-differential-rotation

The short answer is that the differential rotation is the result of combining convection, viscosity and a global rotation.
A ball of honey
Imagine you were on the ISS and you had a jar of honey (do not actually try this ;). You take out the honey and make a floating ball of honey. Then you gently star to spin it. You would notice that if at the start you gave the ball some amount of differential rotation, in some time the viscosity of the honey smears the rotation and if you wait a bit you will see that the ball is now rotating uniformly, with no differential rotation.
A rotating ball of viscous fluid tends to rotate uniformly, like a rigid body, because if two layers rotate at a different speed, friction tends to slow the faster layer and drag the slower one until they move at the same speed.
A star
A star is different from a ball of honey, because it generates heat in its core, due to nuclear fusion. This heat propagates from the center outwards. In most stars the heat flow is so intense that some part of the plasma starts to boil, just like water in a pot (the scientific term is convection). Hot bubbles of plasma form deep inside the star and rise towards the surface.
If the star is rotating, as the plasma bubble moves radially outwards it is also pushed sideways, just like when you throw a ball while standing on a merry-go-round. (It is called Coriolis effect). The sideways push will be stronger for plasma bubbles on the equatorial plane, because their velocity is orthogonal to the axis of rotation. The effect will be less intense for bubbles that move towards higher latitudes and negligible for bubbles that move from the centre to the poles.
The equatorial bubbles will develop a greater sideways velocity and will drag with them the plasma in the equatorial region. As a consequence, we see that the plasma at the equator of the Sun rotates faster than at higher latitudes.
This explanation is extremely simplifying, and leaves out lots of details. For a rigorous mathematical model of differential rotation in stars see the review in reference.
Reference
Kitchatinov, L. L., “REVIEWS OF TOPICAL PROBLEMS:  The differential rotation of stars”, Physics Uspekhi, vol. 48, no. 5, pp. 449–467, 2005. doi:10.1070/PU2005v048n05ABEH002099.

Great explanation, thank you! I am studying more and more and i am happy to learn and understand new things. Again, thank you and good luck!

You are welcome, I learned a lot too by answering your question

In a rotating solid body, regions that are adjacent at one point in
time will remain adjacent as the body rotates. This means that points
further from the rotation centre will travel at greater speeds than
those closer in.  If the rotating body is not solid, however,
regions that are adjacent at one point in time do not necessarily
maintain that configuration. This is known as ‘differential rotation’.
Examples of differential rotation are found throughout astronomy. In
stars (including the Sun) and the gas giant planets, the equatorial
regions rotate faster than regions closer to the poles, meaning that
equatorial sunspots and cloud formations will move across the face of
the object faster than their polar ones.
...
In the disks of spiral galaxies, all of the material orbits at roughly
the same speed. However, the outer stars have further to travel in
their orbit around the galactic centre than the inner stars. The
result is that the outer stars lag behind the stars in the inner
reaches of the galaxy.All objects in the disk of a spiral galaxy are
moving at roughly the same orbital speed. Since the outer objects have
further to travel in their orbits than the inner ones, they lag
behind.

Source: https://astronomy.swin.edu.au/cosmos/d/Differential+Rotation

This only repeats the observation that differential rotation can and does exist, but fails to answer the OP's question as to what the physical cause is.


text:
What would happen if the Sun, Moon and Earth were isolated?
I am curious to know what would happen to Earth and life on Earth if all other heavenly bodies outside of our Sun, Moon and our planet Earth ceased to exist in some non-destructive way? (this includes other stars, other planets, asteroids, etc.)
I have tried looking for the answer, but keep getting stuck in different unrelated rabbit holes. I did find a video on the topic (

 ) but the video simply makes assertions (e.g. if Mercury and Venus were gone, Earth would drift closer to the Sun) and lists consequences, but makes zero mention of how Earth's orbit would change, making it really difficult to believe or understand.
So simple question; if the universe were as big as the Sun, Earth and Moon and nothing else changed, how would our lives be impacted?
As caveats; assume this means in the "relatively" short term (within 1 million yrs) and that the Earth, Moon and Sun came to be in their exact current state through some other means.
Thanks!

Yeah, that's a very, very silly video. (As in: pretty much complete nonsense.)

Yeah the video seems to be very much of the "edu"-tainment variety, leaning more towards tainment than edu... lol

@elemein Would the universe you mention above be able to expand?

To start with, I'll just say that I won't really address anything the posted video said (whether right or wrong) and just stick to strictly answering the question:

What would happen to Earth and life on Earth if all other heavenly bodies outside of our Sun, Moon and our planet Earth ceased to exist in some non-destructive way?

That being said, let's be clear about the parameters here, and what is happening in this scenario. I'm assuming that the universe, our solar system, and Earth all formed exactly as they already have, with the single exception that now some god-like being, simulation admin, or whatever controls out universe just removes everything in the universe except the Sun, the Earth, and the Moon. And by remove I mean just blink out of existence (against all logic and physics). Not blow up, not throw into a black hole, just remove. As with any question that takes the form "What happens to physics when I violate physics?", the answers can be a bit hard to specify. But, to the best of my ability, what would happen to the continued existence of Earth and life?
There are two main factors to consider I think. What we'd lose within our solar system and what we'd lose outside our solar system.
Inside Our Solar System
In this scenario, we're losing all the planets except Earth, all the comets and asteroids, and all the gas and dust out there that isn't strictly "the Sun". To be honest, this isn't a hugely impactful change because we're only losing ~0.13% of all the mass of the solar system. What changes internal to our solar system might we expect?

The Sun's orbital motion would suddenly and almost completely cease. This would be almost entirely due to the disappearance of Jupiter. Aside from our Sun, Jupiter is the big heavyweight in our solar system and does actually have a measurable affect on the Sun's motion within our solar system, with respect to the "center" of the solar system. You'd see kind of a discontinuous kick that might be jarring to Earth, potentially adjusting the orbit.
No more meteor showers. At the time this answer was written, Earth was at the height of the Perseid meteor shower, but all that debris would be gone so no more showers. For that matter no more Haley's comet or any other comets that one can sometimes see in the night sky.
As a tongue-in-cheek observation, all Martian meteorites are also gone and visiting Antarctica and the UK National History Museum becomes slightly less fun.
I'm assuming because all spacecraft and rovers derived from Earth they do not disappear, so now we've got a variety of rovers and spacecraft just drifting through space. Perseverance would be unlikely to tell us anything interesting though because I doubt the developers who wrote it handled the case where Mars blinked out of existence. While Perseverance can technically talk directly to Earth or else relay to nearby spacecraft, it's unlikely it could orient itself and point at a target long enough to say anything useful, so sorry Perseverance, looks like you're on your own.

Otherwise, the Moon would continue to orbit the Earth, Earth would continue orbiting the Sun (albeit, potentially in a slightly different orbit) which would continue to shine. Frankly I can't think of much more that would happen inside our Solar System due to losing the planets.
Outside Our Solar System
Here's where things get interesting. We're losing a lot of stuff outside our solar system - in fact most stuff. Your question wasn't very specific though. You simply specified "heavenly bodies". Depending on how that is interpreted, the answer could vary.
Heavenly Bodies == Objects with Mass
If I interpret "heavenly bodies" to just mean objects with mass, but not objects of pure energy, then we're losing all the stars in our galaxy, the rest of our galaxy, all nearby galaxies, and all the other galaxies out there. All black holes are gone.  All interstellar and intergalactic gas and dust is gone. Suddenly the universe doesn't have Dark Matter. What we're keeping are things like all the photons out there, Dark Energy, Vacuum Energy, and the energy in the curvature of space-time.
This kind of has a somewhat boring answer under this scenario. Because it takes time for things like light and gravitational affects to get to Earth and because the Universe is just mind-bogglingly large, for most of the rest of humanity's existence, we don't really see much of a change at all. Even though all the stars are now gone, it'll take a very long time before we stop receiving light from them because most of the light is streaming towards us still and will be for many long years to go. Over time though, some of the nearest stars will blink out of the night sky as their last rays of light finally hit Earth.
A few notable examples include Alpha Centauri (4 years), Sirius (8 years) (the brightest star in the night sky is gone! The next brightest, Canopus, will be around for an additional 302 years before disappearing itself), and Procyon (11 years).
Otherwise, most everything you can see in the night sky will still be observable for most of the rest of human existence, even if the objects don't exist anymore. What's more, the gravitational effects will still be felt since gravitational changes travel at the speed of light. Our solar system will, for the most part, continue to orbit the center of our galaxy, even though the mass it's orbiting doesn't exist anymore! It'll take tens of thousands of years before our orbit will start to change (due to the gravitational disturbances of the missing mass finally catching up to us), but who knows if humanity will be around then.
What we may experience though are sudden jolts in our path through space as these blips in the gravitational fields pass over us - the most prominent coming from the disappearance of the SMBH at the center of our galaxy, although that won't occur for about 26 thousand years.
Outside of our galaxy, it'll be millions or billions of years before light from distance objects stops reaching us, so likely that'll never come to pass.
Finally, it's possible the universe will explode! With basically no mass, but still tons of dark energy (in this scenario) the expansion of the universe might be kicked into overdrive and very possibly the very atoms of our bodies will be ripped apart with that super-expansion! But there are a lot of if's behind this idea so who can really say if that will happen.
One interesting additional change would be that because the interstellar gas and dust has disappeared, our sun's atmosphere will have free reign to expand as much as it wants. Voyager I and II, with what little life they have left, will possibly notice that they've once again passed into the protective sheath of the Sun's magnetic field and solar wind (doubtful if it they the energy anymore to support such a measurement).
Heavenly Bodies == Objects with Mass-Energy
Now we're upping the game. I'm now interpretting "Heavenly Bodies" to include not just the objects described in the previous section, but all the other objects in the universe which have no mass, but have energy. So what does that add to the mix of disappearing objects? All the light pervading the universe is just gone (most of which is light radiated from the recombination era in the form of the Cosmic Microwave Background radiation). The energy built up into the curvature of the universe is gone (essentially meaning the universe stops being curved (or so I'm interpreting it)). And finally, Dark Energy is gone. This last one is a bit dubious as we don't actually know what Dark Energy is so it's hard to say what taking it away means.
This scenario is a bit more scary because the universe will be drastically different. Some changes I can think of off the top of my head.

No more light at night from stars. During a new moon, night time will be very dark, almost unnaturally dark. Although, away from city lights, you'd have a much better chance of seeing the Gegenschein which would be pretty cool.
The static of the universe, the CMB, would be gone. That'd be unfortunate since we can learn a hugely large amount of information about the nature of the universe from this light.
Because I'm assuming dark energy goes away now, the expansion of the universe would likely stop (assuming our ideas about dark energy are correct). This really won't have any noticeable affect on our day-to-day back on Earth. Just something to note.
The universe might still just explode. It's hard to say what may happen when you take all the energy out of the curvature of the universe. It may spring back like a rubber band and destroy everything (which would just be the Earth-Sun-Moon system at that point).

Overall I'm sure there are more effects one could dream up, but this list is getting pretty extensive already.


text:
What happens inside a massive collapsing shell?
I have already asked this on the physics site. I am not sure where it fits better.
Consider a massive spherical shell collapsing under its own gravity. Say the mass is that of one galaxy. The Schwarzschild radius is, say, 300 billion kilometers.
The mass will collapse accelerating. So there are gravitons (or in any case gravitational waves) sent inward and outward. Or will there be not? A rotating spherical mass will not send gravitons, unless accelerated.
In the center of the shell, time will go slower and slower wrt faraway observers outside the shell, if there is no mass present there. This doesn't happen in Newtonian gravity. According to Newtonian gravitation, the inside will always stay without gravity.
What happens if the shell passes the event horizon? Will the whole singularity be formed? For a far away observer it looks as if the shell stops there. But for a free falling observer inside the shell will the singularity be there already at that moment? It falls on towards the center where nothing is yet. It takes little time, while for an outside observer it takes pretty long before the horizon is met (all seems to slow down).
Basic question: What will someone at the center see or feel? His time runs slower and slower but the space around him will get stretched and stretched. Because the inward accelerating mass will send a stationary (standing)  gravitational wave. Will he get torn apart before the mass hits him, even though the spacetime around him stays flat?

Welcome to Astronomy SE. You have asked a lot of questions in this post. Try gathering your thoughts and focus on the problem you are having. Then I suggest you edit and write one clear and concise question that addresses the issue. In this way, it is easier for other users to understand what you are asking

@Prallax I posed a basic question. I have a similar question for an oscillatory rotating solid sphere (how it radiates gravity waves). Shall I put it here too? Thank for the advice!

Let's see how it goes without the oscillatory rotating sphere. You can ask it as a separate question

@Prallax Thanks for the edit! Looks better! I just added an n in collapsig.

You're welcome. I'm really curious about the answer, because it is not trivial. Birkhoff's theorem is tricky to apply inside a fixed spherical shell (see this paper https://arxiv.org/pdf/1203.4428.pdf). I think that (in a region encompassing both internal and external parts) the metric won't be stationary if the shell is collapsing.

I can say one thing for sure: there will not be gravitational waves outside the shell (thanks to Birkhoff)

@Prallax That were my thoughts also!

@Prallax. Outside there will just grow new Schwarzschild metric but inside? Time grows smaller (the pace) everywhere but space? Will it grow larger everywhere inside (will it remain flat?). Will there be some kind of standing gravity wave?


text:
Do some physicists believe that the universe is a three-dimensional space wrapped on a four-dimension doughnut?

The astrophysicists believe if this is true, the universe would be
finite. The entire cosmos may be only three or four times larger than
the limits of the observable universe, which is about 45 billion
light-years away. If true, a doughnut-shaped universe also has the
possibility of allowing a spaceship that goes in one direction to
eventually return to where it started without turning around.
The shape of the universe is something that astronomers have been
debating for decades. Some believe the universe is flat where parallel
lines stay parallel forever. Others believe the universe is closed,
being the parallel lines eventually intersect. Astronomers say the
geometry of the universe dictates its fate.
While open universes continue to expand forever, a closed universe
eventually collapses in on itself. Observations focusing on cosmic
microwave background, which is the flash of light released when the
universe is only 380,000 years old, have established that our universe
is flat and parallel lines will stay parallel forever with an
ever-evolving universe. However, there’s more to shape than geometry,
and topology has to be considered. Topology allows shapes to change
while maintaining the same geometric rules.
An example is a sheet of flat paper that has parallel lines that stay
parallel. If you roll the paper into a cylinder, the parallel lines
are still parallel. If you take that sheet of paper and connect the
opposite ends while it’s rolled like a cylinder, you get the shape of
a doughnut, which is still geometrically flat. The team believes the
warping occurs beyond observational limits and will be very difficult
to detect. The team was looking at perturbations, which describe bumps
and wiggles in the cosmic microwave background radiation temperature.
They believe there could be a maximum size to the perturbations that
could reveal the universe’s topology. Buchert and his team emphasize
their results are preliminary and note that instrument effects could
explain some of their results.

https://www.slashgear.com/astrophysicists-believe-the-universe-may-be-shaped-like-a-giant-3d-doughnut-20683133/
Do some physicists believe that the universe is a three-dimensional space wrapped on a four-dimension doughnut? I was reading this article and was confused about the geometry being discussed, they say the universe is flat, not in the sense that it's two-dimensional, but in the sense that parallel lines remain parallel indefinitely, then they mention that it's wrapped on a doughnut, which is impossible unless they mean a four-dimensional doughnut which would allow parallel lines to join, is this what is being said here? A topological three-dimensional plane wrapped on a fourth-dimensional doughnut would retain its geometrical feature like a two-dimensional plane wrapped on a three-dimensional doughnut would.

Yes, that model has the topology of a 4D doughnut. It's like a 3D version of a game where if you go off the top of the screen you reappear at the bottom. But that doesn't necessarily mean that there's a 4th spatial dimension.


text:
Is the claim that the star Sirius does not precess like other stars true?
There is this popular claim on Internet that the star Sirius does not precess like other stars.
The idea comes from mainly Greeks and Egyptians using Sirius to track seasons, as part of the Sothic cycle.
Wikipedia says that Sirius is used for the Sothic cycle because it's ~40 degrees away from the ecliptic, but there are other stars which are 40 degrees and even further away from ecliptic, why only Sirius? Why is this property only present in Sirius?.
This and this website, of questionable reputation conjectures that our Sun revolves around Sirius with the period  of 24,000 years which is approximately equal to time taken for one cycle of precession of equinoxes, therefore Sirius always staying still in the sky.

Given that the distance between Sun and Sirius is around 8 light years, 24000 years seems to be too quick a time period, but then to calculate the time period of twin star systems is complicated as the distance changes dramatically. Also there is the Centauri system which is around 4 light years away from the Sun. Can the Sun, Sirius and Centauri systems together form a bigger system which has a time period of 24000 years?
Is this model even feasible? If not, why does Sirius not precess like other stars?

Wikipedia doesn't say that.

Why the downvotes? Is it because the question is based on a totally wrong idea? Is it because the OP hasn't done prior research? I've seen other people on the internet confused about this, probably because there are many sites proposing this fantasy as accepted scientific theory. I think that it may be difficult do distinguish truth from someone without a scientific background, also because Wikipedia can be difficult to understand. Is this kind of question to be down voted in general?

@ProfRob Wikipedia does say that, please go to Sothic cycle amd go to Mechanics. It's given there.

The wikipedia page on "Sothic cycle" says that the period between heliacal risings is 365.25 days. I am not sure how you equate that to "does not undergo precession". Could you explain in the question.

@ProfRob yes I have edited it and made it clear

I'm not sure it is clearer. Now the paragraph about Sothic cycle looks unrelated to the rest of the question

Sirius precesses like the rest of the sky.  Precession is caused by changes in the orientation of the Earth. Every star precesses as the Earth wobbles.
Sirius also has a fairly high proper motion. But not so high that it "stands still".
The precession of Sirius was the first observation of precession, as it was noted that the helical rising of Sirius and the tropical year (the year counted from vernal equinox to vernal equinox) were not synchronized, and the Heliacal rising of Sirius was coming slightly later in the season each year. This change in the day of the Heliacal rising, compared to the tropical year (which is well approximated by the Gregorian calendar) is due to the precession of Sirius.
The websites you link to are pure cloud cuckoo land fantasy.

Thank you for the answer.


text:
Cosmology context plotting : Covariance matrix - Script to plot Omega_m vs Omega_Lambda diagram in Python
I have covariance matrices and Markov chains coming from the SCP (Supernova Cosmology Project) and I would like to plot all these data
in the particular diagram Omega_Lambda vs Omega_m or w vs Omega_m.
For example, I would like to get this kind of figures :

and

I would prefer to write a Python script.
Update 1
I managed to get the covariance matrice available on: http://supernova.lbl.gov/Union/figures/SCPUnion2.1_covmat_nosys.txt
from the SCP (SuperNova Cosmology Project): http://supernova.lbl.gov/Union/
Once downloaded, I computed the associated Fisher matrix. But I realize that the size of the matrix is 307x307.
I don't know how to exploit this big and only one covariance/Fisher matrix.
On the figures above, there are 3 probes (CMB, SNIa, and BAO) and in grey, I guess the cross of the 3 previously cited probes.
How can I handle this large matrix? Maybe the authors have made a synthesis of the 3 probes + cross-correlation in a unique matrix. If this is the case, how to break down this only one matrix to plot individually each probe?
Update 2
I recall that Covariance matrix is equal to the inverse of Fisher matrix (matrix representing the information that we have on each cosmological parameter). Then, the covariance matrix gives variances on diagonal elements and covariances on off-diagonal elements.
Update 3
Below the Python script used with getdist tool (pip install getdist) with Fisher matrix coming the inverse of Covariance downloaded above :
# Show plots inline, and load main getdist plot module and samples class
from __future__ import print_function
import sys, os
sys.path.insert(0,os.path.realpath(os.path.join(os.getcwd(),'..')))
import numpy as np
import getdist
from getdist import plots
import matplotlib.pyplot as plt
import matplotlib as mpl
# The plotting scripts also let you plot Gaussian (or Gaussian mixture) contours
from getdist.gaussian_mixtures import GaussianND
from getdist.gaussian_mixtures import Mixture2D
print('GetDist Version: %s, Matplotlib version: %s'%(getdist.__version__, plt.matplotlib.__version__))

## Load Fisher matrix and invert it to get Covariance array
# Load Fisher matrix
File1 = np.loadtxt('Fisher_OmegaL_vs_Omegam.txt')
# Invert to get Covariance matrix
COV_File1 = np.linalg.inv(File1)[2:4,2:4]

#  Example of pre-defined location for legend
# Mean of each cosmo parameters : respect order of generated "Big" Fisher matrix
mean = [0.28, 0.72]
# Names and labels
names = ['Omeega_m', 'Omega_L']
labels = [ r'$\Omega_m$','$\Omega_{\Lambda}$']
matrix1 = GaussianND(mean, COV_File1, labels = labels, names = names)
# Plot triplot
plt.rcParams['text.usetex'] = True
plt.rc_context({'axes.autolimit_mode': 'round_numbers'})
g = plots.get_subplot_plotter()
g.settings.figure_legend_frame = True
g.settings.legend_fontsize = 8
g.settings.axes_labelsize = 15
g.settings.axes_fontsize  = 10
g.settings.alpha_filled_add = 0.85
g.settings.title_limit_fontsize = 18
g.settings.axis_tick_x_rotation = 45
g.settings.axis_tick_y_rotation = 45
g.settings.fig_width_inch = 5
g.settings.fig_width_inch = 5
g.settings.axis_tick_max_labels = 10
g.triangle_plot([matrix1],
                names,
                filled = True,
                legend_labels = [ r'$\Omega_{\Lambda}$'+' vs '+r'$\Omega_{m}$'],
                legend_loc = 'upper right',
                contour_colors = ['blue'],
                )
# Save triplot
g.export('output_OmegaL_Omegam.pdf')

Unfortunately with this current script, I get degenerated 2D joint distribution between Omega_Lambda and Omega_m :

I have to question if this is the appropriate exchange for this question. You're plotting astronomy related concepts, but fundamentally your question is about plotting in python. It might be more appropriate to ask this on another exchange like Stack Overflow.

@zephyr . I know but people on stack Overflow told me that background was astrophysics and advised me to ask on another forum. So I try my chance here


text:
Can stars really be seen after sunrise from Mount Everest?
I came across this image taken on the "Balcony" at over 27,000 ft (8,200 m) elevation, beneath the summit of Mount Everest. So many stars can be seen on it even though the Sun has risen. Do Everest climbers really see stars after sunrise or is the image a long-exposure one or something?

The more I look the more I'm certain that's a composite image.  The shadows don't point away from the sun, and I don't think the fisheye lens is enough to explain that (looking at the most central shadows especially.  And the host page is an advertising company after all. It's linked from a page about an image capture/augmented reality project, which certainly means some processing and implies some combination of images

Related video.  The last scene matches some of this image, but with a tighter crop and a different camera angle, though the preview if you hover over the progress bar looks like a closer match (the preview appears to be of a 3d version that I can't enable here)

@ChrisH Oh it's just snow. blush

It's not implausible that you'd see planets near sunrise (even at sea level, possibly even a few stars at such high altitude) and the shadows are long enough that the sun is low in the sky. Without the distortion we could work out the sun elevation. Wikipedia has examples that might be of interest. Note that astronomical magnitude is one of the strangest scales you'll ever meet: it's logarithmic, with negative being brighter, and an arbitrary zero point (Vega).

@ChrisH One can reportedly see Venus in daytime, if you know where to look and have eagle eyes. But I meant seeing stars easily, with likelihood of spotting one randomly without searching the sky. Just like after sunset or before sunrise.

I've certainly seen Venus with the sun low in the sky, but haven't looked in broad daylight

Despite the fact you have identified the dots on that image being snow, I think the general question is okay and may be on topic on Astronomy so will migrate it.

As I mentioned here, when Venus is near its maximum brightness it's certainly possible to see it with the naked eye at any hour of the day, if you know where to look (assuming the sky is sufficiently clear).

German Wikipedia has a page on daylight star observations, which states that with a 4“ instrument mag 1-2 stars can be found easily (and regardless of observer height): https://de.wikipedia.org/wiki/Tagbeobachtung (use your favourite translator). I‘ve looked for weather ballon footage or high altitude ballon observations (as done in the „early“ days, when cosmic rays were discovered), but didn‘t find anything. Maybe your googlefu is better.

For ISS I believe you can see stars, even if it is in daylight (as long as you‘re not looking into the direction of the sun). This leads to the existence of a height where it is possible (for the atmosphere changing continuously)

The reason that you can't see stars in the daytime is that the atmosphere scatters sunlight and effectively blots them out.  But even where I live, a dark sky area, the brightest stars are visible during twilight.  Not just the planets, Sirius for example becomes visible before the sky is dark.
With the sun on or near the horizon, the effect of scattering is reduced.
At high altitudes the atmosphere is much thinner.  The scattering is therefore similarly reduced and yes, you can sometimes see the brightest stars in daytime.  This would not necessarily be the case every day and at all hours of the day.
Although I haven't been to Everest myself, friends who have report that some stars can be seen on clear days.  I haven't interrogated them as to whether it was near sunrise or sunset.
(But your picture does look like a composite)

My picture actually shows snow as ChrisH found out. Why do you consider Wiltshire a dark sky area? The highest point of the UK is the Ben Nevis at just over 4,400 ft. The daytime sky becomes considerably darker above ~25,000 ft I guess. When in an airliner at cruise altitude the sky above me is much darker up there, but I think I've never seen stars out of a plane during day. Once I have seen a star at twilight out of a plane, when the Sun might still have been above the horizon, but it's a long time so I don't remember exactly. I've seen possible stars on a picture of a view out of Concorde.

From https://nightblight.cpre.org.uk/maps/ there are some pretty dark region of Wiltshire (e.g. much of Salisbury Plain).  But when looking low in the sky that's a bit misleading as I found the other riding round the edge of the Cotswolds Gloucestershire when the Northern Lights came much further S than normal: I was in the 2nd darkest band, maybe even the darkest, but the light pollution from Gloucester, Stroud etc. was scattered by a few low clouds.  Car lights didn't help either.  Even straight up, I couldn't see as many stars as in the Scottish Highlands, but far more than in a city

So you're talking about light pollution, but during day we can't see stars because of the brightness of the closest hydrogen-burning star, the Sun, and therefore the lit blue atmosphere.

@Giovanni that's the only way I can interpret "dark sky area", but this still helps when the sun is low