Could a contracting Universe create the redshift effect observed by Hubble? What if we just can’t see far enough and the redshift actually shows the galaxies speeding up towards a gigantic black hole to where everything is converging, repeating a cycle that took place countless times “before” the big bang?

Complete text in this link: http://www.therealebook.com/Astronomy/Redshift_ContractingUniverse.html

Even in a contracting Universe, there is a region around the observer where all the galaxies are moving away; the galaxies that are moving towards the observer could be too far to be seen. If we can demonstrate this, we could challenge the conclusion that the Universe is expanding, based on the redshift effect observed in the “nearby” galaxies.

Let’s start with a simpler representation: 1 dimension only (1D). Assume that all the galaxies move along a single line. We could represent an expanding and a contracting Universe as per the images below:

In Situation 1, we have the currently accepted model, where galaxies move away from each other, at increasing speed, which creates the redshift observed by Hubble.

In Situation 2, we assume that in the contracting Universe the galaxies have a speed that is inversely proportional to the distance from the center (as in a sink flow where the particles accelerate towards the drain), galaxies 1 and 3 will display a red shift effect, because they are moving away from each other, while galaxies 4, 5 and 6 will appear with blueshift effect. However, in order to see the blue shifted galaxies it is necessary to see beyond the center of the Universe, which would be not possible either because of the distance or because of the gigantic black hole in the center traps light from galaxies 4, 5 and 6 preventing it from reaching us at galaxy 2.

In fact, assuming that from galaxy 2 we cannot see beyond galaxy 1, it is impossible to distinguish between situation 1 and 2.
With this we have demonstrated that for a 1D Universe it is impossible to determine if it is expanding or contracting, simple because we cannot see far enough. What about a 2D space?

Consider two galaxies (1 and 2); consider that they form a angle beta, relative to the center of the universe (assuming that there is one); Consider that galaxy 1 is at the distance r from the center; consider that galaxy 2 is at a fraction of r (say xr) from the center.

We can calculate the distance between galaxy 1 and 2 as follows:

From the equation above we can calculate alfa and from  we can calculate d.

We also suppose that the speed along the radius can be calculated with the equation v = c / r, where c is constant.

Therefore we can calculate the position of the two galaxies after a period of time t, and measure the distance d between them to verify if d increases (redshift) of decreases (blueshift).

We can see that no matter what the fraction x is, there is always a minimum value for the angle beta, such that any galaxy within this angle will be moving away from the reference galaxy. Now assume that the universe is so big that we cannot see galaxies beyond beta. In this way, all that we can see is redshifted galaxies!

This means that even for a contracting universe, depending on how far we can see, it is possible to detect only redshifted galaxies. Does not this disprove the theory of expanding universe based on redshift?

Aside from being, I suspect, totally incompatible with GR, the model fails to match observations.

The universe appears to be homogeneous and isotropic. In the model you suggest, Hubble's law would be different depending on which direction you were looking.

In particular, looking away from the black hole, galaxy redshifts would increase to some asymptotic value, whereas looking towards the black hole they would increase to infinity.

If the argument is that the central black hole is so far away that we cannot measure this asymmetry (i.e. well beyond the observable universe, given the homogeneity of the cosmic microwave background - but then does the model have an explanation for the CMB at all?), then the model also needs to explain why galaxies would move in this particular way, since they would not "feel" the influence of the black hole.

A contracting universe is an attractive proposition as it seems to do away with the need for dark energy.
Equally a primordial black hole is an attractive proposition as its expansion would seem to indicate that time/events can never be  entirely squeezed out of existence.(ie:if time/events ceased to be, there could never be an initial expansion). Both these propositions are currently unproven.
A dirty initial expansion seems feasible (ie: one where super massive black holes split from the primordial black hole and seeded the universe, allowing the creation of galaxies which could not themselves have produced  black holes of this size in the time available).
Obviously we can only observe/photograph what to the observer are present space/ time events (as their light reaches us) which events have taken place previously in the space/time continuum. Presumably a universal gravitational field has always existed within the space time continuum (even inside an event horizon which may have evolved and expanded as the primordial black hole decayed), which will have affected all events and structures in the universe.
Cosmic background radiation  can perhaps be thought of as the remnant of the original event horizon/expansion now massively red shifted.
It is difficult to see how  observed red shifts can definitively prove whether the universe is expanding or contracting.


Can we (theoretically) spin the black hole so strong that it will be broken apart by centrifugal force? I can't imagine the forces involved in black holes' lives. So please, help me to find out, if it is possible or not to destroy black hole in this specific way.

Can we (theoretically) spin the black hole so strong so it will be broken apart by centrifugal force?

For a Kerr-Newman (rotating, charged, isolated) black hole of mass $M$, angular momentum $J$, and charge $Q$, the surface area of the event horizon is given by
$$A = 8M\left[M^2 + (M^2-a^2-Q^2)^{1/2} - Q^2/2\right]\text{,}$$
where $a = J/M$. An extremal black hole occurs when $M^2 = a^2 + Q^2$. Beyond that, if the black hole is even more overspun or overcharged, is an "overextremal" Kerr-Newman spacetime, which wouldn't really be a black hole at all, but rather a naked singularity.

Thus, I interpret your question as asking whether or not a black hole can be be spun up to the extremal limit and beyond, so as to destroy the event horizon. It's very probable that it can't be done.

Wald proved in 1974 that as one flings matter into a black hole to try to increase its angular momentum, the nearer to an extremal black hole it is, the harder it is to continue this process: a fast-spinning black hole will repel matter that would take it beyond the extremal limit. There are other schemes, and though I'm not aware of any completely general proof within classical general relativity, the continual failure of schemes like this is well-motivated by the connection between black hole dynamics and thermodynamics.

For example, the Hawking temperature of the black hole is $T_\text{H} = \kappa/2\pi$, where
$$\kappa = \frac{\sqrt{M^2-a^2-Q^2}}{2M\left(M+\sqrt{M^2-a^2-Q^2}\right)-Q^2}$$
is the black hole's surface gravity. Thus, even reaching the extremal limit is thermodynamically equivalent to cooling a system to absolute zero.

Let's try the following:

Equate the forces:

$$\begin{eqnarray*}
F_c &=& \frac{mv^2}R\\
F_g &=& \frac{GMm}{R^2}\\
\newline
mv^2/R &=& \frac{GMm}{R^2}\\
v^2 &=& \frac{GM}R
\end{eqnarray*}$$

Since the radius is $R_s = 2GM/c^2$

$$\begin{eqnarray*}
v^2 &=& \frac{GM}{2GM/c^2}\\
v &=& \sqrt{c^2/2}\\
v &=& \frac{c}{\sqrt2}\\
v &=& 0,707c
\end{eqnarray*}$$

I don't know if I made all the considerations, but as far as my knowledge goes, if the blackhole is spinning faster than $0,707c$, the event horizon won't be able to maintain itself on the current radius.

However, when the radius expands, the rotation will slow down by conservation of angular momentum... so I don't think it will rip apart... perhaps become a "grey hole"?

Please forgive me if any fundamental mistake was made, I'm new to all this... :P


How does the amount of dark matter in measured galaxies vary? I have recently viewed two news articles announcing a galaxy composed almost entirely of dark matter and one with hardly any. This made me wonder if there is a "continuum" of percentages of dark matter in galaxies, where it has been measured, or if there is a "preferred" range.

tl;dr: In general, smaller galaxies have relatively more dark matter than larger galaxies. The atypical galaxies that you refer to are discussed in the last paragraph.

Mass fractions in the Universe

Matter in the Universe is dominated by two components; baryons (i.e. atoms in various forms) and dark matter (DM). The baryons predominantly exist in two forms: gas (including plasma) and stars (some gas condenses to form dust, planets, etc., but that's a minor part).

The fraction of the Universe's total energy in the form of matter is $\Omega_\mathrm{M}\simeq0.31$ (most of the rest is dark energy). The baryons and DM have fractions $\Omega_\mathrm{b}=0.05$ and $\Omega_\mathrm{DM}=0.26$, respectively (numbers from Planck Collaboration et al. 2016).

Hence, the baryons and DM comprise fractions
$f_\mathrm{b}  = \Omega_\mathrm{b}  / \Omega_\mathrm{M} = 16$% and
$f_\mathrm{DM} = \Omega_\mathrm{DM} / \Omega_\mathrm{M} = 84$% of the total mass, respectively.

Galaxies have more dark matter than the average Universe

Galaxies formed when DM+gas overdensities decoupled from the Hubble flow and started collapsing, and so were "born" with the "cosmic" fractions. However, observationally it is found that the baryon fraction in galaxies is much smaller; for instance, a "typical" Milky Way-sized galaxy roughly has $f_\mathrm{b}$ only half of the cosmic average, and the DM fraction is correspondingly larger, >$90$% (e.g. Werk et al. 2014).

Gas and dark matter dynamics

The reason is the different mechanisms that affect the dynamics of the particles.
Whereas the DM is collisionless, the gas particles collide and dissipate their energy, making it easier to cool and contract. Thus, the part of the galaxies that we can observe is much smaller than "whole" galaxy; the  typical "half-light radius" is but a few percent of the emcompassing DM halo (Kravtsov 2013, Somerville et al. 2017). The half-light radius is the radius within which half of the light is emitted; when you see an image of a galaxy, it typically looks perhaps 4-5 times larger than this. So, something like this:

A significant fraction of the baryons reside in the halo as well. This is the hot gas of millions of Kelvin, and the high temperature and low density makes it difficult to detect, as it glows only in faint X-rays.

Feedback blows out gas, but dark matter stays

But various physical processes act so as to try to blow the baryons — but not the DM — out of the galaxies. These processes are known as feedback, and are due to energy being "injected" in the baryons.
For massive galaxies with halo masses $M_\mathrm{h}\gtrsim10^{12}\,M_\odot$ this feedback is dominated by their central black hole accreting gas, resulting in extreme luminosities manifesting itself as a quasar, or active galactic nucleus (Silk & Rees 1998; Croton et al. 2006). For smaller galaxies with $M_\mathrm{h}\lesssim10^{12}\,M_\odot$, the feedback is mostly due to stellar winds and exploding stars depositing thermal and kinetic energy in the surrounding gas (Dekel & Silk 1986; Hopkins et al. 2012). At even lower masses, the galaxies are so small that a significant fraction of their gas may be ejected altogether during early epochs of star formation (e.g. Bullock et al. 2000), resulting in galaxies that consist virtually only of DM and a few stars.

The smaller a galaxy is, the shallower its gravitational potential, and thus the easier it is for the gas to escape the galaxy.

Hence, in general the smaller a galaxy is, the larger its amount of dark matter is.

Conversely, the larger a halo is, the more its mass fractions converge toward the cosmic fractions. These are only reached for halos of masses $M_\mathrm{h} \gtrsim 10^{15}\,M_\odot$, which are no longer individual galaxies, but rather groups and clusters. This is seen in the figure below (from a paper from last week by Henden et al. 2018), which shows the stellar mass fraction (left) and gas mass fraction (right) as a function of halo mass:

The baryon fraction is $f_\mathrm{b} = f_\mathrm{stars}+f_\mathrm{gas}$, and the DM fraction — which is what you're asking for — is then given by $f_\mathrm{DM} = 1 - f_\mathrm{stars} - f_\mathrm{gas}$. The stellar fraction is seen to decrease with halo mass, due to the AGN feedback discussed above, but the total star+gas fraction increases.

Note though that although the DM — being collisionless — is not exactly blown out of the galaxy like the baryons, the gravitational attraction between the two components still affects the DM and alters the density profile of the halo (Duffy et al. 2010).

Galaxies with more gas than dark matter (?)

Thus, a galaxy without baryons is not a huge mystery, as long as it's small. Even larger galaxies are apparently able to rid themselves of most the baryons; van Dokkum et al. (2016) reported a MW-sized halo with 98% DM.

A galaxy with baryons, but without DM is more spectacular, but was reported recently by van Dokkum et al. (2018) (yes, the same guy).
In another answer about DF2 I discuss various processes that might lead to such a galaxy, including misinterpretation of the data.

The basic method used to measure dark matter in a galaxy is to observe the extra gravitational pull coming from matter which is not visible. The effect on orbits of stars in a galaxy (determined by observing galaxy rotational curves) and the bending of light by gravitational lensing are the some of the techniques used for measuring the gravitational pull of a galaxy.


Are there other proofs of the expanding universe apart from the redshift? The theory of the expanding universe is so widely accepted, that the redshift is sometimes used as a measure of distance to far away galaxies.

But is it still possible that the redshift is caused by some unknown phenomena and not by galaxies moving away from each other?

Is there any other proof (apart from the redshift) that the universe is indeed expanding and far away galaxies are moving away from us?

Yes, there is direct, non-red-shift evidence of expansion.

The past temperature of the Cosmic Microwave Background Radiation (CMBR) has been directly measured and found to be substantially higher than it is today.  Its reduction in temperature over time is direct evidence of expansion.  Here are the details:

According to this paper, the CMBR was measurably hotter in the past (less-technical synopsis here).  The researchers observed absorption lines in a gas cloud located in a distant galaxy and found that the pattern of lines seen could only be explained if the CMBR temperature at time of absorption was between 6 K  and 14 K (it's 3 K now).  This temperature is consistent with the expected temperature for that galaxy's redshift (9 K).  Note that the temperature was measured from the specific pattern of lines seen and not from how much the lines had been red-shifted; this measurement would yield the same temperature even if there were no red-shifting.  Since a hotter temperature implies a higher density, this cooling of the CMBR over time is direct evidence for expansion of the universe.

Additional comments

What is the relationship between red-shift and absorption lines?

Inspired by a conversation with uhoh in the comments:

In my answer I refer to a "pattern" of "absorption lines".  For those not versed in the topic allow me to explain.

When a light shines through a cloud of gas specific frequencies of light get absorbed.  When this light is then shone through a prism the blocked frequencies will appear as black lines in the spectrum (see illustration below).  The exact lines that appear and their positions in the spectrum (the "pattern" of "absorption lines") depends on the elements present in the gas and the gas's environment.  The effect is most clearly seen with a light that emits photons at all frequencies; this kind of light is known as black-body radiation.  Although emitting light at all frequencies, a black-body radiator will emit the most light at a particular wavelength; the location of this peak is referred to as the black-body's temperature.  The CMBR discussed in the question is an example of black-body radiation.

Source: Doppler Shift, Edward L. Wright
(Excellent site BTW, the FAQ is worth a look for more info on red shifts and cosmology in general)

As light is travels through (expanding) space it's wavelength and the wavelengths of the absorption lines stretches at a fixed rate for all frequencies.  Let's say that at time of emission/absorption a spectrum shows lines at wavelengths of 1, 3, and 5 nm1.  After the photons have traveled for a certain amount of time, all of the spectrum's wavelengths will appear to have doubled2.  The line formerly at 1 nm is now seen at 2 nm, the one formerly at 3 nm is now seen at 6 nm, and the one originally at 5 nm is now seen at 10 nm.  Though their absolute frequencies change over time, the ratio of the lines' wavelengths (and frequencies) relative to one-another remains constant.

The precise amount that a given object's spectrum is shifted directly correlates with its distance.  As seen in the diagram above, close objects (like the Sun) show no red-shifting.  As one looks at objects further and further away one sees increasing amounts of red shift3.

In the discussion in the answer above, it's this pattern of relative positions in the lines that is affected by the CMBR temperature at time of absorption and not the degree to which the lines have been shifted.

1 To put it technically, this point is at $z=0$ where $z$ indicates the magnitude of the shift, positive for red shifts (moving away) and negative for blue shifts (approaching).  A more in-depth discussion of this topic (including the precise definition of $z$) can be found here.
2 The wavelength-doubling (frequency-halving) point is at $z=1$
3 It should be noted that since there is some uncertainty in the rate that the Universe is expanding at, red-shifts do not refer to precisely-known distances.  Thus astronomers and cosmologists rarely refer to the distances to distant objects in absolute terms of, say, light-years or parsecs, preferring, rather, to use the amount of red-shift observed (the $z$ mentioned above).

The mechanism behind red shifting isn't that the photons themselves are changing rather it is that the very space that the electromagnetic waves are moving through is expanding.  (Photons are both particles and waves; no, it's not exactly intuitive.)  This constant stretching of space stretches the light's wavelength giving rise both to the effect of red-shifting and the increase in a given photon's red shift over time.

Douglas Hofstadter, CC A-SA 3.0

How does red-shift relate to the CMBR?

In the comments Alchimista asked "Isn't CMBR actually the quintessence of red-shift anyway?"
(I'm assuming you're using the common, and not cosmological, meaning of "quintessence")

Yes, the current CMBR temperature (3 K) is generally agreed to be the result of relatively high-energy photons (3000 K) emitted about 380,000 years after the Big Bang that have had their wavelengths stretched over time by the expansion of the Universe towards the red (i.e. cooler or lower-energy) end of the spectrum.  This expansion was inferred by Hubble et al. from the observation that smaller and dimmer galaxies (as seen from Earth) have a greater shift in their spectra.  The farther the apparent distance, the greater the observed shift.  Using this apparent distance-correlated red shift we can infer that the Universe was smaller in the past and thus denser with a higher temperature for the CMBR.  Based on observed red shifts of distant galaxies, we can then deduce, but not directly measure, what the CMBR temperature was at each distance.

What the authors of the above paper did was make a direct measurement of the CMBR's temperature at a specific time in the past.  The measured temperature is higher than it is today which implies a denser and thus smaller Universe.  The researchers further found that the directly-measured temperature fit neatly with that inferred from the observed red shift of the galaxy being studied.

In a nutshell, the chain of inference is swapped:

For reasoning based on red-shifting:
Increasing red shifts with apparent distance (directly measured) ⇒ Expansion ⇒ Denser Universe in the past ⇒ Higher CMBR temperature in the past.
For a direct measurement of past temperature (as with this paper):
Higher CMBR temperature in the past (directly measured) ⇒ Denser Universe in the past ⇒ Expansion ⇒ Observed red shift.

These two inference chains based on different sets of evidence neatly complement and support one-another.

One thing to note is that the CMBR wasn't created by expansion (at least not directly) rather it's expansion that explains its current temperature and uniformity.  Per the Big Bang theory, the early universe was very dense; so dense and hot that all matter was a plasma of subatomic particles, opaque to photons. At about 380,000 years after the Big Bang the Universe had cooled (through expansion) enough that protons and electrons could combine to form neutral Hydrogen gas (which is transparent). The CMBR is the light that was set free at this time and has been cooling ever since.

OK, this answer involves red-shifts but hear me out.

Under General Relativity, multiple mechanisms can create red-shifts: expansion of space, objects moving relative to an observer (i.e. us), and light moving out of gravity well.  The latter option is outside the scope of this question and the former is excluded from consideration at the questioner's request.  That leaves only the second option (relative motion, a.k.a. the relativistic Doppler effect) under consideration; this shifting can be (and has been) tested here on Earth and has been shown to exist.

Red-shift is observed in all apparently distant objects (dim, low metallicity, etc.).  From the red-shift of the spectra seen in any given object we can determine how quickly it is moving away from us.  For example, an object with a measured red-shift of $z=0.5$ is moving away from us at about half the speed of light.  So far so good.  The trouble arises when we observe objects with $z>1$.  Many such objects have been found; the current record-holder is GN-z11 with a red-shift of $z=11.09$.  Put another way, if only relativistic shifting were at play this object would be travelling away from us at over 11 times the speed of light.

Given that no object having mass can reach light-speed it is clear that the observed red-shifts cannot be caused by relativistic motion.  Since there are no known mechanisms beyond the three listed above that can cause red-shifts in the spectra (compare extinction), the only explanation matching these observations is the expansion of space.  Put succinctly, the fact that superluminal red-shifts are observed at all is evidence that space is expanding.


With all the stars in the universe, why isn't the sky saturated with light? Space is dark. But if we direct telescopes to seemingly dark portions of sky, we see that it is filled with galaxies. If the universe is theoretically infinite, wouldn't an infinite number of photons be reaching the earth and illuminating the night sky brighter than it appears?

Logic1: If the universe is infinite AND static, the sky will be saturated with light.

Note "AND."

Logic2: If the sky is not saturated with light, the universe is not infinite or not static.

Note "OR."

Logic1 = Logic2

Observation: the sky is not saturated with light.
Therefore: the universe is not infinite or not static.

We know that the universe is dynamic, at least.


If all stars rotate, why was there a theory developed that requires non-rotating stars? According to Penrose's research, a non-rotating star would end up, after gravitational collapse, as a perfectly spherical black hole. However, every star in the universe has some kind of angular momentum.

Why even bother doing that research if that won't ever happen in the universe and does it have any implications for the future of astrophysics?

In a similar way, we could ask...

  No beams can be exactly 1 meter long. No beams can be exactly
  straight. The material making up a beam cannot be truly isotropic.
  So why should we bother calculating the stress in a 1 meter straight
  beam having isotropic material?

Because knowing how to perform this calculation is a building block for doing more complex calculations.

The non-rotating black hole calculation also provides a limiting solution.  The solution for a spinning star's collapse will approach this solution as the spin approaches zero.

Similarly, Newton told us that as external forces approach zero, the path of a moving object will approach a straight line.  This is useful to know even though there is no place in our universe that doesn't have gravitational influence.

You're right: all stars rotate. The only reason I can think of why astrophysicists make calculations for a non-rotating star or black hole is that it makes their calculations a bit easier. Although all stars rotate, some rotate much faster than others, and their masses vary too, so there is a wide degree of uncertainty which is reduced by calculating for a star that does not rotate.


Is the bright red body a star? I was helping classify galaxies at Galaxy Zoo when I came across this picture of a galaxy. The dull red body looks huge compared to the galaxy. Is it because it is closer to the earth? It would be interesting to know what is going on here.

It is a bright star. You can see the "diffraction spikes" that are the signature of a point-like object.

The star will be in our Galaxy, so is certainly closer than the galaxy in the image. However, this is not why it appears to be big.

The image appears big because the starlight will be spread by the telescope objects over a finite area - the point spread function. In addition, when a star is very bright, the central CCD pixels saturate and charge can spill into adjacent pixels. Even if that were not the case, there is enough light in the "wings" of the point spread function to produce a large image.

Is this a photo from the Digitized Sky Survey (DSS)? If so, I have also seen these, but they are not true stars most of the time. They  are mostly "artifacts" left over from stacking or imaging.


Terminology for empty space at the edge of the observable universe? Is there a name for a point at the edge of the observable universe at which there is no visible matter, that appears completely dark?

I'm thinking something like "empty space", but that could equally apply to an empty void between galaxies, which isn't quite what I'm looking for.

So it would refer to a theoretical line in space from the observer to the edge of the universe which contains no visible matter. Is there a term for this?

There is no name for this.

As we look further in space, we also look further back in time. The "edge of the observable universe" is in the early universe. At an early time in the evolution of the universe, the whole of space was so hot and dense that it was opaque.  In whichever direction we look, we can't look back further than this. The is called the cosmic microwave background. There is no direction in which we can look in which we don't see the cosmic microwave background. There is no direction that contains no visible matter.

There are directions that contain no galaxies or other more nearby objects, but there is no name for these directions.

The observable universe is merely how far we can see with current telescope technology.

Space itself is endless. Oh, there's no end of people here who will aggressively argue this. But empty space, by definition, cannot have an end. It can only be infinite. Literally infinite.

There could very well be an edge in the sense that there's no more matter, where stars, planets and galaxies etc simply tapers off. After that, it would be endless black space.


Solar series 142 and 147 started same year - series 147 and 152 will be ending same year. Why? Coincidence? This question was left over from the last question I had, so I am asking it separately here:

I have been looking into two solar eclipses (Saros series 147 and 152) that will both be occurring in 2021. Each of these eclipses will be happening near the opposite poles of the earth.  Both are part of young saros series. Saros 152 is comprised of 70 eclipses, while saros 147 contains a total of 80 eclipses. These series are 10 eclipses difference in length and began 180 years apart. So, I happened to check out their ending dates and found that both series are projected to end during the exact same year - 177 day apart – together in the year 3049. I thought that both of these series ending together one semester apart in the very same year was a pretty amazing coincidence.

So, checking further – it seems new saros series start from as few as 7 years apart to as many as 100 years apart (this is extremely variable spacing, but it averages to about 31 years apart over the usual 1260+ year active series lifespan). I wanted to know why series 147 and 152 would both be ENDING at the same time.  I went back to the NASA website, and checked the 5000 year listing of all solar saros series numbering 0 through 180. I only found one other such instance of two solar saros series ENDING in the same year – series 49 and 54 both ENDED together in 32 CE – so it seems like a pretty rare event.

But then, I checked to see if there were any occasions of two saros series BEGINNING together in the same year. Going back to the NASA website again with this question, I checked the 5000 year listing of all solar saros series numbering 0 through 180. I only found one single example of two saros series both BEGINNING in the same year. This time series 142 and 147 both BEGAN together in the year 1624 CE.

Not only is solar saros 147 involved in BOTH of these "same year" cases, but all three series eclipses will be presenting themselves during a single upcoming semester series in 2021 – solar series 142 (Dec 14, 2020), 147 (June 10, 2021),
and 152 (Dec 4, 2021).

  Can anyone explain to me the actual mechanics behind what causes saros series to begin and end? I am familiar with Saros/Inex
  Panorama – but unfortunately they do not answer this question, and I
  can’t even read them online anyway.

  Does anyone know why solar saros 147 may have STARTED in the same year as saros 142 or why saros 147 will be ENDING together in the same
  year as saros 152? Are these two occasions both very rare
  coincidences, while at the same time, both involve the same series 147?

The reason Saros Cycles begin and end is down to the positions of the orbits of the Sun, the Moon and the Earth. From the main Wikipedia page on Saros:
For an eclipse you need the 3 bodies to line up at a full or new moon,

and repeat occurrences of these lunar phases result from solar and lunar orbits producing the Moon's synodic period of 29.53059 days...however for the majority of these, the shadow of the Earth or Moon fall to the north or south of the other body.

It can only happen when a full or new moon is around the ecliptic.

The period of time for two successive lunar passes through the ecliptic plane (returning to the same node) is termed the draconic month, a 27.21222 day period. The three-dimensional geometry of an eclipse, when the new or full moon is near one of the nodes, occurs every 5 or 6 months when the Sun is in conjunction or opposition to the Moon and coincidentally also near a node of the Moon's orbit at that time, or twice per eclipse year.

Also:

Two Saros eclipses have the same appearance and duration due to the distance between the Earth and Moon being nearly the same for each event because the Saros eclipse cycle is also an integer of the anomalistic month, the period of the eccentricity of lunar orbit, 27.5545 days.
After one saros, the Moon will have completed roughly an integer number of synodic, draconic, and anomalistic periods (223, 242, and 239) and the Earth-Sun-Moon geometry will be nearly identical

The reason we get varying start and ends is because of this. Those roughly integer periods vary.
This diagram from that Wikipedia page illustrates perfectly why they can be different lengths, stop and start at changing intervals etc.


Why do scientists assume they can measure the shape of the universe if it is also widely believed to be infinite? The shape of the universe  is the question of whether the universe is flat, has a positive curvature or a negative one. More recently astronomers have compared radiation coming from extremely distant points of the observable universe and have concluded that it is flat with a 0.4% margin of error, Based on articles I've read elsewhere this seems to have created the consensus that the universe is indeed flat. But isn't the universe also believed to be infinite in size? If it is really infinite then shouldn't such measurements be inconsequential as an infinite curved universe would still appear perfectly flat to a local observer?

Yes, the universe is believed to be infinite in size. That's what you get if the curvature is zero or negative, assuming a simple topology. The curvature has to be positive for a finite universe, once again, assuming a simple topology, and no weird stuff like edges.

Now it's possible that the universe has a very small positive curvature, so that it's finite, but it's so large that it looks flat to us.

However, it's reasonable to assume that the observable universe is representative of the whole thing, and not just coincidentally a region of anomalous curvature. Of course, that's impossible to verify, but if the curvature were significantly greater outside our observable patch we'd expect the curvature to be a bit higher near the edges than it is in the middle of the patch, and we don't see that in the data.

But if the global curvature equals the maximum positive curvature consistent with the 0.4% margin of error of the WMAP, BOOMERanG, and Planck data, then the radius of curvature of the whole universe is (currently) around 150 times larger than the radius of the observable universe. And of course, in the future it will continue to expand.

That figure comes from How Big is the Entire Universe? by astrophysicist Ethan Siegel. That article has a great explanation of curvature, with lots of helpful diagrams.

Why do scientists assume they can measure the shape of the universe if it is also widely believed to be infinite?

Because not everybody thinks it's infinite. Take a look at the timeline of the Big Bang written by Luke Mastin in 2009. He said “The linear dimensions of the early universe increases during this period of a tiny fraction of a second by a factor of at least 10$^{26}$ to around 10 centimetres (about the size of a grapefruit)”. You can find John Gribbin talking about a universe the size of a grapefruit in his 2008 book The Universe: A Biography. You can find Marcus Chown talking about a universe the size of a grapefruit in his review of Alan Guth’s 1997 book The Inflationary Universe. You can also find Jeremiah Ostriker and Paul Steinhardt talking about a universe the size of a grapefruit in The Quintessential Universe in SciAm in 2002. They weren't talking about the observable universe. They were talking about the whole universe. And if it was the size of a grapefruit 13.8 billion years ago, it can't be infinite now.

  The shape of the universe is the question of whether the universe is flat, has a positive curvature or a negative one. More recently astronomers have compared radiation coming from extremely distant points of the observable universe and have concluded that it is flat with a 0.4% margin of error.

That's right. Two out of three options were always going to be wrong. IMHO if you've read the Einstein digital papers it's obvious that the universe was going to be flat.  Yes, Einstein came up the idea of a closed curved universe, see his 2nd February postcard to Willem de Sitter. But by 1932 he and de Sitter had dropped the idea in favour of flat space in their Einstein-de Sitter universe. See John Baez and Emory Bunn’s preliminaries article dating from 2006: “Similarly, in general relativity gravity is not really a ‘force’, but just a manifestation of the curvature of spacetime. Note: not the curvature of space, but of spacetime. The distinction is crucial”. Curved space is not the same thing as curved spacetime. Unfortunately some people confuse the two.

  Based on articles I've read elsewhere this seems to have created the consensus that the universe is indeed flat.

I think that consensus is correct.

  But isn't the universe also believed to be infinite in size?

A lot of people claim that a flat universe must be an infinite universe. But it's a non-sequitur. Going back to Einstein again, he thought of space as a thing. That’s why in his 1929 essay on the history of field theory, he described a field as a state of space. In his Nottingham lecture in 1930 he said space “remains the sole medium of reality”. So I think it's reasonable to envisage the universe as a sphere of space, with no space beyond it. Maybe light rays would undergo something akin to total internal reflection.

  If it is really infinite then shouldn't such measurements be inconsequential as an infinite curved universe would still appear perfectly flat to a local observer?

That's what some people say. But they also say things like "the universe was infinite at the time of the Big Bang and it's even more infinite now". I'm not fond of that because it doesn't seem to fit in with Big Bang cosmology, and I just don't see how an infinite universe can expand.


Could an ultra-hot Jupiter flare? Some of the ultra-hot Jupiters (WASP-33b, KELT-9b) have dayside surface temperatures similar to low-mass stars. Would they be capable of producing flares? To be specific, I am asking about the possibility of flaring on the planet itself, rather than the planet inducing flares on the star.

Flares in space are known for coming from stars. Sunlike stars, red dwarfs, etc, almost any kind of star has flares.

Hot Jupiters can be the size of Jupiter,(hence the name). They can be bigger; however, if they get too big, they become brown dwarfs. This means that hot Jupiters are too small to have flares.

In addition, many hot Jupiters have unstable atmospheres. They are so close to their parent stars, that they lose their atmospheres. No atmosphere, no flares.


What are the RAs of the boundaries between traditional zodiacal signs? I'm trying to figure out at what right ascension (RA) — not celestial longitude — the sun enters the various traditional (Western) astrological signs of the Zodiac; but I can't figure out how to calculate this.

What are the right ascensions of the boundaries between traditional zodiacal signs along the ecliptic?

TL;DR: Calculating the right ascension (i.e. converting from ecliptic to equatorial coordinates) requires some spherical trigonometry, but fortunately there are internet tools that will do this, such as here. The resulting RA coordinates are:
Aries 0h 0m 0s; Taurus 1h 51m 39s; Gemini 3h 51m 16s; Cancer 6h 0m 0s; Leo 8h 8m 44s; Virgo 10h 8m 21s; Libra 12h 0m 0s; Scorpio 13h 51m 39s; Sagittarius 15h 51m 16s; Capricorn 18h 0m 0s; Aquarius 20h 8m 44s; Pisces 22h 8m 21s.

Explanation and background

Since traditional (Western) astrology uses the same tropical year as astronomical celestial coordinates, the tropical zodiac commences with Aries at 0˚ longitude on the ecliptic, and each subsequent zodiac sign (NB not the constellation, which is different) commences exactly 30˚ further along the ecliptic. So, the zodiac division of Taurus is from 30.00˚ to 59.99˚, Gemini from 60.00˚ to 89.99˚, and so on. It's then just a matter of converting ecliptic longitude to celestial coordinates (right ascension).

Further useful information on the relationship between the ecliptic and right ascension can be found in Mike G's answer to "Effect of the obliquity of the ecliptic / tilt of the Earth on the equation of time".

Both the traditional tropical zodiac and modern astronomical coordinates are based on the northern hemisphere vernal equinox. This starting point, also known as the First Point of Aries, is one of the two points at which the celestial equator crosses the ecliptic.

When Hipparchus defined the First Point of Aries in 130 BCE, it actually aligned with the border between the constellations of Pisces and Aries. However, due to the precession of the equinoxes, the tropical zodiac's starting point has a retrograde movement along the ecliptic at the rate of one degree every 71.5 years. As a result, the First Point of Aries is currently deep in Pisces! Using the current official IAU constellation boundaries, the point of 0h RA will cross into Aquarius in the year 2597, so we have quite some time to wait for the "Age of Aquarius".

Note that not all astrological traditions are based on the tropical year. Hindu astrology, for instance, is based on the sidereal zodiac - i.e. it aligns permanently with the visible stars, regardless of the Earth's axial precession. Some modern Western astrologers also use a sidereal zodiac, in which sidereal Aries currently begins on 15 April.

It is going to depend on where you draw the lines, but they're not equally spaced... and it also depends on whether or not you include Ophiuchus.  Since they're not evenly spaced, it's not like you can start with a point on the ecliptic and calculate an even spacing for them.


Invisible Mass Affecting the Expansion of the Universe Are there enough neutrinos & other invisible particles in a cubic light year of space to significantly affect the expansion of the universe,& at a very rough estimate without taking into account WIMPS & other exotic particles for which there is no firm evidence,what would be the total mass of this quantity of matter? Only a tiny proportion of all the neutrinos emitted since the Big Bang have been absorbed,the rest are still flying around in search of a home.

Yes, there is enough of these things, especially dark matter which is accelerating the expansion of the universe.


Could non-supernova carbon, oxygen, or silicon flashes be observed? I was reading about the helium flash, the short but sudden onset of helium fusion in certain red giant stars. As I understand, the upper (nondegenerate) layers of the star absorb the energy as they quickly expand, which leads to the stabilization of the core as thermal pressure takes over, leading to the period of helium burning.

I was curious as to whether this could be generalized to the onset of fusion of heavier elements later in a more massive star's life, and I came across a reference to a hypothetical silicon flash in Woosley & Heger (2015). More digging led me to a passage in Fundamental Astronomy (page 250), which states

  In stars with masses around 10 $M_\odot$ either carbon or oxygen may be ignited explosively just like helium in relatively low-mass stars; there is a carbon or oxygen flash. This is much more powerful than the helium flash, and may make the star explode as a supernova.

Let's say that a star undergoes a relatively low-power carbon, oxygen, or silicon flash, and enough energy is absorbed by the outer layers such that the star does not undergo a supernova. Would enough energy escape the outer layers of the star so that we be able to detect the flash using optical or neutrino telescopes? Obviously, this is distance-dependent, so I'll say that the hypothetical star is within five parsecs of Earth.

As another way to put the question, how luminous would the flash be?

There is also something called a "giant eruption", like eta Carinae, which is a type of "failed supernova" because it didn't destroy the star.  It is not known what causes giant eruptions, but perhaps one thing to think about is a flash of the type you mention.  Giant eruptions certainly are visible, eta Car got quite bright when it blew off much of its envelope.


Are there any known planets whose axis is oriented such that one pole always faces its star? Earth rotates around the sun and around its axis (A North, South axis) giving us days, nights and seasons.

Are there any known planets that rotate with an axis pointing toward its sun so that one half of the planet is always facing it (the other half is always facing away from it)? Like Uranus, but not switching which side is facing the sun every 6 months.

(Would this even be theoretically possible?)

Would this even be theoretically possible?

No. The law of conservation of angular momentum prevents this for a planet with reasonable inner structure in a gravitational field like that around the sun.

The angular momentum of the planet points in a certain direction. Changing this direction requires a torque perpendicular to the rotation axis.

If a torque existed that could force the rotation axis of the planet to always point towards the sun, the corresponding force had to have a component perpendicular to the line from the sun to the planet. The force also had to change periodically with the orbit of the planet. Obviously such a force does not exist in the planet-sun-system.

In theory, there's no reason why a planet couldn't orbit close to the way you suggest.   It's probably extremely rare, but theoretically possible.    This gets a little tricky to explain, but an axis of rotation can be divided into perpendicular vectors.   There's no real benefit to doing it that way but it can be done, just like direction and velocity can be separated into 3 vectors, rotation can too, and the combination of the perpendicular vectors gives you the rotation of the object and it's axis of rotation.   That's discussed in a bit more detail here.

If we use the Moon as an example.   The Moon is oriented so it's heavier side permanently points towards the Earth.  (the side of the Moon facing the Earth has a thinner crust and because crust is lighter than the mantle, the "Earth side" of the Moon has slightly greater density than the Far side.  That greater density is more attracted to the Earth than the less dense far side and over time, the dense side of the Moon was oriented to permanently face the earth, similar to how the dense side of a floating objects orients itself downwards when floating in water.

All planets, not being perfect spheres with perfectly layered density are unbalanced but the Moon is more lopsided than most.  Mars more than the rest of the planets.  Earth's uneven gravity has been measured in great detail, but in terms of effects on Earth's orbit and rotation, it's uneven gravity is pretty insignificant.

Now some objects experience significant tidal deformation, like Io and Enceladus which are measurably squeezed by their elliptical orbits as they move closer to and further from their planets, but that's something else and you probably wouldn't want that on a planet as the volcanic effects would be too big.   But what I'm talking about isn't a tidal bulge but a permanent imbalance of mass where one side of the Moon is denser than the other.

So, if we figure the dense side of the Moon must always point to Earth due to tidal locking, that still leaves one way that the Moon can rotate without affecting the locking and that's along it's 90 degree longitude lines, and that, in theory could happen in addition to it's existing synchronous rotation that keeps the heavy side of the Moon pointed towards earth.

This wouldn't be two axis of rotation, it would be one axis of rotation where the East-West vector of the rotation would still be tidally locked but the rotation around the 90 degree longitude line wouldn't interfere with the tidal locking, so the effect, the familiar face would spin around in a circle but always face the earth.

We could in theory do that artificially if we put a big train on the moon and ran it 24/7 in the same direction around the 90 degree longitude line.   Do that long enough and the moon would begin to spin.

But for such a scenario to actually exist, you'd need a lot of luck, because the forces that tend to tidally lock a planet or moon tend to also reduce any other rotations other than the synchronous 1:1 tidally locked rotation.

Another problem is the Equatorial bulge which is a consequence of rotation, tends to have more mass and that would want to orient itself towards the planet, so for this to work you'd need a slow rotation and a small equatorial bulge where the added mass around the equatorial bulge was small enough to not alter the direction of the heavier side of the planet.

Now it's not hard to imagine a planet that's not tidally locked having just the right orbital velocity and position to do what you want, but in such a scenario with no tidal locking, it would be temporary.   Planets rotations tends to slow down over time, so perfectly matching without tidal locking would be coincidental and temporary.   More likely you'd have a very slow rotation, not a permanent side facing the planet, but a very gradual movement.

There are no planets (or moons) like you describe that we know of.  It's probably a hugely unlikely scenario that would probably only be approximated (a bit like Uranus).


Was the Geocentric Model correct at all? It's easy to find resources stating that the heliocentric model is right and geocentric is wrong.
But how wrong was it? Was it correct in any way?
It was built on incorrect assumptions, but despite that - was it of any use to describe the apparent motion of celestial bodies? Was it more accurate for some things, but less accurate for others? Or was it altogether a flop and astronomers couldn't get anything out of it either way?

I can only find multiple articles proving the heliocentric model, explaining the geocentric one or claiming that it was simply wrong - but I can't find anything about its accuracy and usefulness.
(1) (2) (3)

Edit 1:
I incorrectly used geocentric model when it seems I wanted to say Ptolemaic model - the one with deferents and epicycles, with Earth as its origin.

Ptolemaic model (click for full size)

Thank you for clarifying, and sorry for the confusion. The answers provided regarding any other geocentric models are still valid and useful, so this is just a minor errata.

Ptolemy's epicyclic, geocentric model, in use until the Renaissance, was very accurate in terms of predicting the positions of planets and the times of eclipses. What it couldn't account for were things like the correlations between apparent size and phase of Venus, or to properly account for the variation in brightness of the planets.
Thus the reason for discarding the geocentric model was not really because it lacked precision, but that it failed to explain various other observational facts, especially after the development of telescopes.
No doubt you could tune the Ptolemaic system even further (more epicycles?) to iron out some of the small errors that were revealed by Tycho's positional measurements at the turn of the 16th century, which had a precision unavailable to Ptolemy. However, the advent of Kepler's laws and subsequent explanation by Newton, rendered the geocentric model obsolete.
As you can judge from (well written) articles like this one, geocentrism is actually quite hard to kill-off observationally, if you are prepared to accept that the universe is arranged "just so".

To add to Jerry Coffin's very good answer, there've been experiments that disproved geokinetism and proved geocentricity. They're called the Michelson Morley experiment and Airy's failure and made clear whether absolute gocentrism is correct or not. Since then, it was made effort to disprove the results of these experiments because absolute geocentrism would be proof of God and Scripture, the evil wanted to avoid that.
This proven geocentricity is called the Neo-Tychonic model, after Tycho Brahe who lived in about the same time as Copernicus. The difference to the original Tychonic model is that in the neo-Tychonic model the stars are focused on the Sun, not on the Earth. This explains parallaxes, and this geocentricity has the very same effects like an Earth spinning on its axis, such as the Foucault pendulum, geostationary satellites staying up, and a lower gravity at the equator.
If you don't trust these experiments nor Scripture, the only way to find out which model is true (Copernican or neo-Tychonic) would be to stand outside the universe and look onto it.
For more information see The Principle documentary by Robert Sungenis with various scientists.


How to test direct sun viewing glasses I purchased a 10 pack of glasses for viewing the upcoming eclipse. They show to be ISO 12312-2:2015 certified. How can this be tested and verified?

The American Astronomical Society (AAS) answers this question on their How to Tell If Your Eclipse Glasses or Handheld Solar Viewers Are Safe page.

Basically they say that since anyone can claim ISO 12312-2 compliance even if they haven't been certified, you should make sure you obtained your eclipse glasses from a vendor whose ISO 12312-2 compliance has been verified by the AAS Solar Eclipse Task Force:

  We used to say that you should look for evidence that they comply with the ISO 12312-2 international safety standard for filters for direct viewing of the Sun. But now the marketplace is being flooded by counterfeit eclipse glasses that are labeled as if they're ISO-compliant when in fact they are not. So now we suggest that you make sure you get (or got) your eclipse viewers from one of the suppliers listed on our Reputable Vendors of Solar Filters & Viewers page.

They say that checking eclipse glasses to see if they meet the ISO 12312-2 standard would require a spectrophotometer, which most people do not have access to:

  Unfortunately, you can't check whether a filter meets the ISO standard yourself — doing so requires a specialized and expensive piece of laboratory equipment called a spectrophotometer that shines intense UV, visible, and IR light through the filter and measures how much gets through at each wavelength.

So they have assembled a task force to put together a list of manufacturers and vendors of eclipse glasses that they have verified to be ISO 12312-2 certified (their list is longer than NASA's list of 5 manufacturers):

  The AAS Solar Eclipse Task Force has been working diligently to compile a list of such vendors, now posted on our Reputable Vendors of Solar Filters & Viewers page. We've checked manufacturers' ISO paperwork to make sure it's complete and that it comes from a recognized, accredited testing facility, and we've personally examined manufacturers' products. We've asked manufacturers to identify their authorized resellers, and we've asked dealers to identify the source of the products they're selling. Only when everything checks out do we add a vendor to our listing. If we don't list a supplier, that doesn't mean their products are unsafe — only that we have no knowledge of them or that we haven't convinced ourselves they are safe.

They also give advice on how to tell if your eclipse glasses are not safe to use:

  How can you tell if your solar viewer is not safe? The only thing you can see through a safe solar filter from a reputable vendor is the Sun itself. If you can see ordinary household lights through your eclipse glasses or handheld viewer, it’s no good. Safe solar filters produce a view of the Sun that is comfortably bright (like the full Moon), in focus, and surrounded by black sky. If you glance at the Sun through your solar filter and find it uncomfortably bright, out of focus, and surrounded by a murky haze, it’s no good.

I purchased 100 pair of glasses from Walmart online from AllStar Gadgets to use at school with students. It says they are certified. How can you be sure?


Create Position-Velocity diagram from a velocity field Can someone explain to me how I can generate a Position-Velocity (PV) diagram from a galaxy's radial velocity field? I understand that usually PV diagrams are generated across the major axis of the galaxy, but I am not quite sure why they appear as a distribution. For example:

Such diagrams are usually called rotation curves, and show the velocity of stars/gas in a disk galaxy viewed edge-on along the line of sight (LOS) as a function of distance from the center.

That is, the $x$ axis gives the distance of the stars from the center ($x=0$) of the galaxy, and the $y$ axis gives the measured velocities at a given distance. Along a given LOS you'll find not a single velocity, but a (Gaussian-like) distribution of velocities, both because there is a certain scatter in velocities, but also because you only measure the component of the velocities projected onto your LOS, while both speed and direction change along the LOS, as seen here:

If they all had the same velocity along the LOS, the plot you show would be a thin line; instead it is smeared out.

Due to mass being concentrated in the center, there's a steep increase in velocity at low $x$. If there were only the mass you can see, the velocity would decrease at larger $x$ due to the larger distance from the central mass. The reason it stay more or less constant is thought to be the extra, invisible mass known as dark matter.

The reason the left side of the plot is similar to the right, but mirrored not only across the $y$ axis but also the $x$ axis, is that on this side, the stars move toward you, while on the right side they move away from you. The whole galaxy evidently moves away from you at $500\,\mathrm{km}\,\mathrm{s}^{-1}$, since this is the velocity at $x=0$.

So, to recap: How do you generate such a plot? You measure the distribution of velocities along many lines of sight in the galactic plane, from one side ($-R_\mathrm{gal}$) to the other ($+R_\mathrm{gal}$). For instance, at $x=14\,\mathrm{kpc}$, you'd measure $V=675\pm12\,\mathrm{km}\,\mathrm{s}^{-1}$:

In practice, images like this is made by using a grism which disperses the light according to its wavelength, together with a slit placed along the galactic plane, such that you block the light from the rest of the field. This is called slit spectroscopy, in contrast to "normal" spectroscopy where you block all light except for a little hole placed on some object (e.g. a star), or integral field spectroscopy, where you get the whole spectrum behind every pixel in an image.

If you have some spectral image in the form of a FITS-format file, like you have in the original image to the right of the P-V diagram: I think you will need to look up some reference on Fourier transform. This Fourier transform will take you from the RA-Dec image to the P-V diagram you want to generate. Also, you might need to read about the cross correlation method, as detailed in Tonry and Davis's 1979 paper: http://adsabs.harvard.edu/abs/1979AJ.....84.1511T

Another method is called fitting which involves emission lines.

The IRAF package you will need to use is called "rvsao". There are some references on some webpages you can follow:

http://tdc-www.harvard.edu/iraf/rvsao/papers/rvsao.adass97.html
https://arxiv.org/abs/astro-ph/9803252

To generate a P-V digram you will need to map your radial velocity to a 1D-grid, say one depending on the radial position, the mapping should be one-to-one. Otherwise you can also generate a cube, depending what kind of original FITS data files you have from an observation.

I am only providing the details regarding the technical aspect, and am on purpose on explaining anything regarding "distribution"... Every relationship in physics which depends on space (and maybe time also) is some sort of a distribution.


Why would discovering life on another planet be important/matter to us? I had a funny thought, and would like to pose it to you:

When I was a kid, reading any kind of astronomy or similar books would all say that there was no other life in the universe, and made it pretty clear that anyone thinking there could be was probably off his rocker.

Nowadays, it seems this has become a very important question for astronomers and other scientists, a new deep-seated belief that must now be proven. Some seem to think it would be the most important discovery of all time.

I'm wondering why?

My guess is, if we did 'discover' life on another planet, it'll make the news for a bit, and then the average person will go back to their every day life without much more thought about it.

For me, I would think this is great, only insofar as it means a possible new source of food (i.e., some delicious exotic cuisine assuming it's edible, and possibly even some place to go conquer and colonize if there was any profitability/money in doing so).

For some scientists, though, I get the sense it's more of an axe to grind with the long-dreary he-said she-said argument of science vs. religion.

Why do you think it would matter?

My question assumes any life discovered is probably not going to be very 'interesting' life, at least nothing we're going to have a conversation with.

It will be very important in several different ways. Not as in Messiah having descended to the World, and maybe not during one single human lifetime. But it will profoundly change the world in our minds, and thereby change our collective long term actions. The next generation would grow up taking alien life in space for granted, their minds will be more different than we old guys can change our stiff brain flesh. Like Newton's physics dominate our everyday life now, even for those of us who don't understand how all of physics has been deduced.

If life survives in other places than Earth, then maybe we (as in life from Earth) can do so too in the future.
If there's life in other places than Earth, then maybe we came from it? All life on Earth is family, all living things are closely related with each other. The question of our origin has always had high priority in human minds, see all religions.
If alien life is biologically similar to that on Earth (as in using DNA and proteins) then we could use it as a huge laboratory for medicine and environment control and agriculture and what not in life science. Astronomy today contributes to physics in our everyday life by studying a laboratory (the universe) with scale far beyond what humans could ever construct. If biology could do the same, we would soon notice the results in the shape of useful products.
If alien life does not use DNA and proteins but is very different from us, then I can't even guess what discoveries would come out of that. Zweistein.
If alien life is not found, even when we have searched long and hard, then we can gradually make the greatest discovery of all times: That we are unique and special. Searching for ET without finding it would be the most mind boggling thing imaginable. I don't think this hypothesis has been well addressed yet, I suppose no philosopher knows even where to start.

Since any form of contact, would be next to impossible, because of the distance involved, I think the only importance of such a discovery would be for our curiosity .


Planets visible at night Stars give off light and are visible at nighttime but Planets do not give off light . So why are Planets visible at nighttime ?

Planets and moons reflect sunlight, as mentioned in this comment by AtmosphericPrisonEscape.

Now I'll add a bit more info to make this post more informative :)

That also means that just like the moon, the planets have phases, but they're just too subtle to see with the naked eye, and for any planet further from the Sun than the Earth is, we don't see all the phases.  They're full when the planet is furthest from us (on the other side of the Sun), so it'd be hard to tell the phase just from how bright the planet appears.

But through a telescope, you can easily see the phases of Venus, and you can tell it's smaller when it's full (because it's farther away):

Source: Wikimedia Commons

This is incorrect.  All bodies give off light according to Planck's Law.  The amount of emitted light may be small compared with reflected light, or small compared to the sensitivity of your detector.

This may seem like nitpicking, but failure to incorporate emittance when analyzing spectra will lead to sadness.


Expansion of the universe and visibility of distant galaxies The universe expands at an accelerating rate, which makes me wonder if at a certain point the speed of expansion between two points will exceed the speed of light. In that case, it might eventually become impossible for two gravitationally bound clusters of the universe to exchange light. I recall reading that future civilizations born after a certain point would be unable to observe the distant galaxies that we can now, and would have no way of knowing galaxies besides their own exist.

My question: Is this possible, and if so could it have already happened; the entire observable universe is one of multiple clusters that have since sped away from each other? Or can this be ruled out by other known facts about the history of the universe?

(My answer here isn't correct, as pointed out by the first comment on this post, please refer to it.)

Yes, it certainly is possible - and it already has happened. The age of the universe is 13.8 billion years, yet it is something like 90 to 100 billion light-years in diameter. This means that we definitely can't see all of it, not because of our own observational limitations, but because the light from those parts of the universe just hasn't reached us yet - and perhaps it never will.

As for the future, yes, if cosmological expansion overpowers gravity at close range (i.e. the Milky Way and Andromeda, and the world "close" is relative here) then eventually many galaxies and clusters will disappear out of view due to light not being able to "keep up" with the expansion.


Does the Sun belong to a constellation? Each new star we find is generally considered to be part of the constellation it is nearest to.

Our Sun is obviously a star, just much closer. Is our Sun part of any constellation? If so, which constellation is it a part of?

Constellations are human constructs to make sense of the night sky.  When you are trying to find your way around, it helps to "chunk" stars into patterns and assign those groupings names.  When I want to point out a particular object in the sky (say Polaris, the North Star), I start by pointing out a familiar constellation (say Ursa Major, the Big Dipper).  From there, I can tell my friend to follow this or that line to get them to look where I'm looking:

With the advent of computerized telescopes and large data sets, constellations are less important for professional astronomers.  However, many stellar databases use Flamsteed or Bayer designations, which assign stars to constellations.  In order to include all stars, the sky is divided into irregular regions that encompass the familiar constellations.

So, which constellations is the Sun assigned to?  Well, from the perspective of someone on the Earth, the Sun moves through the constellations throughout the course of the year.  Or rather, Sol moves through the region of the sky where some of the constellations would be seen if its light did not drown out distant stars.  Our moon and the rest of the planets move through those same constellations.  (The Greek phrase which gives us the word "planet" means "wandering star".)

The current position of the sun against the background of distant stars changes over the course of the year.  (This is important for astrology.)  It's a little easier to make sense of with a diagram:

So perhaps a better question is:

  What constellation does the Sun belong to today?

Presumably an observer on an exoplanet would assign Sol to some constellation that is convenient from her perspective.  But from our perspective within the Solar system our sun, moon, and planets are not part of any constellation.

Yes, but it does not stay in the same constellation.

The Sun, Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus and Neptune and all the other solar system objects move across the constellations of the Zodiac: Aries, Taurus, Gemini, Cancer, Leo, Virgo, Libra, Scorpius, Sagittarius, Capricorn, Aquarius, Pices and Ophiuchus.

As of March 28, 2014, the Solar System is most likely traveling through Cancer, Virgo or Libra and will move on to Scorpius.


How deep and shaped is the depth of a black hole? A common representation of a black hole is a black circle which if I'm correct, indicates its breadth.

This representation depicts it as a 'plate', so a really small depth compared to its breadth.

On this other representation, its depth has a funnel shape.

Question:

How deep and what shape has the depth of a black hole ?

The first representation isn't depicting the black hole; it's depicting its accretion disk, which is indeed plate-shaped. The accretion disk is around the black hole, and - if it's a rotating (Kerr) black hole - it's perpendicular to the black hole's axis of rotation. Accretion disks are disk-shaped, and so the representation is disk-shaped.

The other is a common representation of a black hole, but it's misleading. It's an extension of the "rubber sheet" analogy, which describes the bending of spacetime. A heavy object, placed on a semi-taut sheet, will curve it a little bit. This represents ordinary matter curving spacetime. A black hole is represented in that image as bending spacetime by an infinite amount. The issue is that this representation embeds the two-dimensional representation of spacetime in three dimensional space, making it appear as if the black hole has this extra depth. Black holes are not embedded in an extra spatial dimension.

Note also that the small object moving near the edge of the black hole is also represented by a tiny funnel. The creator seems to have tried to take the analogy to the extreme, portraying the object's funnel as finite, and the black hole's as either extremely large or infinite, in an attempt to explain that light cannot escape from a black hole. It's not too accurate, but it works to get the point across.

By the way, black holes do have size (the central point singularities of non-rotating black holes are infinitely tiny, while the ring singularities of rotating black holes are in the shapes of infinitely thin rings). You might want to read up on event horizons. The radius of the event horizon of a non-rotating black hole is a finite number, the Schwarzschild radius. It is calculated as
$$r_s=\frac{2GM}{c^2}$$

In this case, the event horizon is spherical. In rotating black holes, it may be in the shape of an oblate spheroid.

Look at this picture.

!black-hole shape][1]

www.astro.umd.edu

As the star gets bigger, the bigger the black hole will go, it's event horizon goes lower, and its inside grows thinner.

Also, if the star is small, you will fell the gravitational pull from much further away. But with a much larger one like a super-massive black-hole, you will get closer without noticing anything much.


How to Distinguish between Temperature and Doppler effect using Black-body radiation? I understand that the radiation of a body can be described using the curve for black-body radiation. In the sense that a hotter body will be blue shifted and a cooler body will be red shifted.

The doppler effect gives similar results in the sense that a body is approaching us is blue shifted. While an object that is travelling away from us is red shifted.

How is it possible to distinguish between temperature and doppler effect?

Without any other information, you cannot distinguish between the two effects.

$$ T = T_0 (1 + z) $$

A blackbody spectrum of temperature $T$ is identical to a blackbody spectrum of temperature $T_0$ with redshift $z$.

For stellar/galactic radiation, we can use the fact that the radiation is not a perfect blackbody.  For the CMB, we can use the fact that atomic combination happens only below a particular temperature.

How is it possible to distinguish between temperature and doppler
  effect?

The blackbody curve varies in both peak and shape:

As you can see, even if you slide the curves left or right, the 5000k curve is still different than the 3000k one. The peak is at a different point relative to the curve as a whole, and especially the extinction point in the ultraviolet (around 0.2 um in this case).

At a minimum, you could compare the position of the peak to the extinction point and produce a temperature. In this example, blue has a span of about 0.3, while red is about 0.5.

So using the numbers in this example, if you were given a spectrum and measured a span of 0.3, you know it's blue. So if the peak of the curve is at 0.8 instead of 0.6 you would ascribe that to 0.2 um of doppler shifting (or other shifting effects).


9th planet location? I've seen a number of news reports indicating there is likely a 9th planet in our Solar System, something with an orbital period of between 10k-20k years, that is 10 times Earth's mass. I haven't seen any real indication of where this object might be. If I had access to a sufficient telescope, would I be able to find this planet, and what way would I point a telescope to find it? How far is it likely to be, or is that not well known?

It's too dim to be seen during a normal survey during the majority of its
orbit.

Update:  Scientists at the University of Bern have modeled a hypothetical 10 Earth mass planet in the proposed orbit to estimate its detectability with more precision than my attempt below.

The takeaway is that NASAs WISE mission would have probably spotted a planet of at least 50 Earth masses in the proposed orbit and that none of our current surveys would have had a chance to find one below 20 earth masses in most of its orbit.  They put the planets temperature at 47K due to residual heat from formation; which would make is 1000x brighter in infrared than it is in visible light reflected from the sun.

It should however be within reach of the LSST once it is completed (first light 2019, normal operations beginning 2022); so the question should be resolved within a few more years even if its far enough from Batygin and Brown's proposed orbit that their search with the Subaru telescope comes out empty.

My original attempt to handwave an estimate of detectability is below.
The paper gives potential orbital parameters of $400-1500~\textrm{AU}$ for the semi major axis, and $200-300~\textrm{AU}$ for perihelion. Since the paper doesn't give a most-likely case for orbital parameters, I'm going to go with the extreme case that makes it most difficult to find. Taking the most eccentric possible values from that gives an orbit with a $1500~\textrm{AU}$ semi-major axis and a $200~\textrm{AU}$ perihelion has a $2800~\textrm{AU}$ aphelion.

To calculate the brightness of an object shining with reflected light, the proper scaling factor is not a $1/r^2$ falloff as could be naively assumed.  That is correct for an object radiating its own light; but not for one shining by reflected light; for that case the same $1/r^4$ scaling as in a radar return is appropriate. That this is the correct scaling factor to use can be sanity checked based on the fact that despite being similar in size, Neptune is $\sim 6x$ dimmer than Uranus despite being only $50\%$ farther away: $1/r^4$ scaling gives a $5x$ dimmer factor vs $2.25$ for $1/r^2$.

Using that gives a dimming of 2400x at $210~\textrm{AU}\;.$ That puts us down $8.5$ magnitudes down from Neptune at perihelion or $16.5$ magnitude.  $500~\textrm{AU}$ gets us to $20$th magnitude, while a $2800~\textrm{AU}$ aphelion dims reflected light down by nearly $20$ magnitudes to $28$ magnitude. That's equivalent to the faintest stars visible from an 8 meter telescope; making its non-discovery much less surprising.

This is something of a fuzzy boundary in both directions. Residual energy from formation/radioactive material in its core will be giving it some innate luminosity; at extreme distances this might be brighter than reflected light. I don't know how to estimate this. It's also possible that the extreme cold of the Oort Cloud may have frozen its atmosphere out. If that happened, its diameter would be much smaller and the reduction in reflecting surface could dim it another order of magnitude or two.

Not knowing what sort of adjustment to make here, I'm going to assume the two factors cancel out completely and leave the original assumptions that it reflects as much light as Neptune and reflective light is the dominant source of illumination for the remainder of my calculations.

For reference, data from NASA's WISE experiment has ruled out a Saturn-sized body within $10,000~\textrm{AU}$ of the sun.

It's also likely too faint to have been detected via proper motion; although if we can pin its orbit down tightly Hubble could confirm its motion.

Orbital eccentricity can be calculated as:

$$e = \frac{r_\textrm{max} - r_\textrm{min}}{2a}$$

Plugging in the numbers gives:

$$e = \frac{2800~\textrm{AU} - 200~\textrm{AU}}{2\cdot 1500~\textrm{AU}} = 0.867$$

Plugging $200~\textrm{AU}$ and $e = 0.867$ into a cometary orbit calculator gives a $58,000$ year orbit.

While that gives an average proper motion of $ 22~\textrm{arc-seconds/year}\;,$ because the orbit is highly eccentric its actual proper motion varies greatly, but it spends a majority of its time far from the sun where its values are at a minimum.

Kepler's laws tell us that the velocity at aphelion is given by:

$$v_a^2 = \frac{ 8.871 \times 10^8 }{ a }  \frac{ 1 - e }{ 1 + e }$$

where $v_a$ is the aphelion velocity in $\mathrm{m/s}\;,$ $a$ is the semi-major axis in $\mathrm{AU},$ and $e$ is orbital eccentricity.

$$v_a = \sqrt{\frac{ 8.871 \times 10^8 }{ 1500 } \cdot \frac{ 1 - 0.867 }{ 1 + 0.867 }} = 205~\mathrm{m/s}\;.$$

To calculate the proper motion we first need to convert the velocity into units of $\textrm{AU/year}:$

$$205 \mathrm{\frac{m}{s}}\;  \mathrm{\frac{3600 s}{1 h}} \cdot \mathrm{\frac{24 h}{1 d}} \cdot \mathrm{\frac{365 d}{1 y}} \cdot \mathrm{\frac{1\; AU}{1.5 \times 10^{11}m}} = 0.043~\mathrm{\frac{AU}{year}}$$

To get proper motion from this, create a triangle with a hypotenuse of $2800~\textrm{AU}$ and a short side of $0.043~\textrm{AU}$ and then use trigonometry to get the narrow angle.

$$\sin \theta = \frac{0.044}{2800}\\ \implies \theta = {8.799×10^{-4}}^\circ   = 3.17~\textrm{arc seconds}\;.$$

This is well within Hubble's angular resolution of $0.05~\textrm{arc seconds};$ so if we knew exactly where to look we could confirm its orbit even if its near its maximum distance from the sun. However its extreme faintness in most of its orbit means that its unlikely to have been found in any survey. If we're lucky and it's within $\sim 500~\textrm{AU},$ it would be bright enough to be seen by the ESA's GAIA spacecraft in which case we'll located it within the next few years. Unfortunately, it's more likely that all the GAIA data will do is to constrain its minimum distance slightly.

Its parallax movement would be much larger; however the challenge of actually seeing it in the first place would remain.

If you had access to a sufficient telescope, you could theoretically see it, if you looked in the right place (although no one knows where the right place might be). But if it's anywhere near aphelion there are only a handful of sufficient telescopes in the world (let's say an 8m mirror or larger), so I think it highly unlikely that you have access to one of them.


Would a Phobos full impact make Mars temporary habitable? Due to tidal deceleration, Phobos is predicted to be destroyed by Mars in the next 30-50 million years either by collision or by breaking up in a planetary ring. If Phobos falls on Mars in one piece could the impact could be comparable with Chicxulub meteorite? Since Mars is smaller that Earth, would the effects of the impact cause a global restart of volcanic activity and temporary warm the planet? Would this be sufficient to make Mars more friendly to life for a few thousand years?

The impact of Phobos, even in one piece (less likely), would be different from Chicxulub. Phobos is probably a little larger than the Chicxulub impactor, but much slower, and comes in almost perfectly tangential. The impact energy would be less than a 10th of the Chicxulub impact, and the energy would be distributed over a large region around the Martian equator. It wouldn't start volcanism, and it wouldn't warm Mars in a relevant way. It wouldn't make Mars more friendly to life by warming. The only noteworthy contribution to habitability would be additional organic matter which Phobos is presumed to contain.

Phobos will (probably) desintegrate due to tidal forces, before its fragments will collide with Mars. Any loose rocks, dust, regolith on the surface will separate from Phobos below the Roche limit of
$$d=1.26~R_M\cdot\left(\frac{\rho_M}{\rho_m}\right)^\frac{1}{3},$$
with $R_M=3389.5\mbox{ km}$ the radius of Mars, $\rho_M=3.9335 \mbox{ g/cm}^3$ the mean density of Mars, and $\rho_m=1.887\mbox{ g/cm}^3$ the mean density of Phobos. This results in a Roche limit of
$$d=1.26\cdot 3389.5\mbox{ km}\cdot\left(\frac{3.9335 \mbox{ g/cm}^3}{1.887\mbox{ g/cm}^3}\right)^\frac{1}{3}=5455.6\mbox{ km}.$$
That's for Phobos' center about 2059 km above the equatorial surface of Mars (taking 3396.2 km as equatorial radius.)
Therefore desintegration will start long before the impact. Loose rocks will collide with Phobos and unlock more material. Whether there will be a considerable solid remnant to finally cause a big (tangential) impact is difficult to predict.

The orbital speed of the fragments at the time of collision will be about
$$v_0=\frac{v_e}{\sqrt 2},$$
with $v_e=5.03\mbox{ km/s}$ the escape velocity of Mars.
Phobos' inclination of 1.093° (relative to Mars's equator) is almost neglectable for the calculation of the impact velocity.
Mars' equatorial rotation velocity is about 241.17 m/s.

The impact velocity will hence be about
$$\frac{5030\mbox{ m/s}}{\sqrt 2}-241.17 \mbox{ m/s}=3316\mbox{ m/s}.$$
Phobos' mass is about $1.0659\cdot 10^{16}\mbox{ kg}.$
This results in a kinetic impact energy of
$$\frac{1}{2}mv^2=\frac{1}{2}\cdot 1.0659\cdot 10^{16}\mbox{ kg}\cdot (3316\mbox{ m/s})^2=5.86\cdot 10^{22}\mbox{ J}.$$

The heat of fusion of water is about 334 kJ/kg, the heat of vaporization of water at 0°C is about 2504 kJ/kg, together 2838 kJ/kg are needed to evaporate water ice of 0°C.
Hence the impact energy of Phobos would be sufficient to evaporate
$$\frac{5.86\cdot 10^{22}\mbox{ J}}{2.838 \cdot 10^{6}\mbox{ J/kg}}=2.06\cdot 10^{16}\mbox{ kg}$$ of water ice of 0°C, corresponding to $2.06\cdot 10^{13}\mbox{ m}^3$ of liquid water.

Mars' surface area is about 144,798,500 km², or $1.447985\cdot 10^{14}\mbox{ m}^2$.
Equally distributed, we could get a layer of water ice corresponding to
$$\frac{2.06\cdot 10^{13}\mbox{ m}^3}{1.447985\cdot 10^{14}\mbox{ m}^2}=0.142\mbox{ m},$$
(correspoding to 14.2 kg/m² of liquid water) evaporated by the impact energy.

With a surface gravity of 3.711 m/s², this would correspond to
a pressure of $14.2 \mbox{ kg/m}^2\cdot 3.711 \mbox{ m/s}^2= 52.8 \mbox{ N/m}^2=0.528\mbox{ hPa}$.

The current atmospheric pressure on Mars is about 6.3 hPa, with diurnal, seasonal and regional changes.
Thus we could get a maximum of 10% short-term pressure increase in the Martian atmosphere by water vapor, in an ideal case.

At the end, an impact of Phobos, or its fragments, would result in a regional reworking of the Martian surface in the equatorial region, but not in a long-term global change.

Just melting water ice would lead to a global ocean with a depth of about 1.2 m. But this ocean would boil due to the low atmospheric pressure, and freeze out again. Same with local lakes due to the uneven topography.

Model calculatios with a release of carbon dioxide would cause some temporary greenhouse effect, but in the equatorial region a release of carbon dioxide is even more unlikely than a release of water vapor.

I noticed an important error in Gerald's otherwise excellent calculation above. A layer of water 0.142 m thick per meter squared has a mass of 142 kg, not 14.2 kg. This means Gerald's results should be multiplied by 10:

14.2 kg/m2 x 3.711 m/s2 = 528 N/m2 = 5.28 hPa

This is comparable to the current average Martian surface pressure of ~6.3 hPa.

A side note: The orbit at impact could also be an ellipse with current orbit as apoapsis and the Martian surface as periapsis. This hypothetical change in orbit would increase the impact speed to ~4.3 km/s (first equation at https://en.wikipedia.org/wiki/Orbital_speed#Precise_orbital_speed ). However, this should probably be achieved artificially and has nothing to do with the natural decay of the moon.


Does any asteroid we have ever seen have a hot core like earth.? I was wondering if any of the asteroids we have ever seen has a hot core like Earth's.

I'm thinking that Earth has a strong gravitational field and so formation would have involved a lot of gravitational energy being converted to heat, but not so much for the asteroids.

See Could a planet form without a molten state?, where Anders Sandberg calculates that a body with radius ~ 1000 km would melt from the heat of formation.

The largest asteroid, Ceres, has a radius of 473 km, so it's a bit too small to have totally melted just from gravitational accretion, but it may have picked up a little more energy from impacts that were faster than freefall. It could also generate some internal heat from radioactive decay, but it's not expected to contain a significant percentage of radioactive material.

Presumably, Ceres was plastic to some degree during formation, since it's (almost) spherical (and it's the only asteroid large enough for gravity to make it spherical). But that wouldn't have required a very high temperature, since Ceres contains a lot of water ice.

But even if it did totally melt when it formed, it's had over 4 billion years to cool down, so it's extremely unlikely to still have a hot core. However, it does emit more water vapour than expected, and it may be warm enough to contain some liquid water below the surface.

'll explain, the asteroids are small rocky objects that orbit the sun. Although asteroids orbit the sun like planets, they are much smaller than planets. Remember that most of them live in the asteroid belt that is that region between the orbits of Mars and Jupiter. Additionally some asteroids are in the orbital trajectory of the planets. This means that the asteroid and the planet follow the same path around the sun. You should also consider that these asteroids were usually all formed at the same time and that most of them are located at the furthest distances from the Kuipier belt that is closer to the 100UA. The fact is that due to their size and their physical characteristics, these objects can not have active nuclei.


Gravitation - Pulling or Pushing force? According to Newton, gravity is the pulling - or in fact the attracting - force of any heavenly body towards any object to its center. But, to the contrary, Einstein once said that the four dimensions of space and time push the object downwards.

So can which one of them is correct?

I'm guessing that this misunderstanding is a result of the oft-used rubber sheet analogy. The rubber sheet analogy says that, according to general relativity, mass curves space-time like a heavy bowling ball on a near-taut blanket (or rubber sheet) curves the blanket/sheet. This resulting curve makes other bits of matter/energy move in different ways. I'm guessing that this is your confusion.

The rubber sheet analogy fails massively in one area, any demonstration of it involves gravity on Earth. If I use a bowling ball to deform a sheet, and then role a golf ball along the sheet nearby, the golf ball will move a bit towards the bowling ball because of the force of gravity around me - not "gravity" in the simulation. It thus makes it seem like gravity pulls the golf ball "down" because the bowling ball pulls the rubber sheet "down". This is the result of using a two-dimensional analogy of a three-dimensional universe.

The point is, there is no "pushing" going on in the general relativistic model of gravity. Gravity is attractive (so long as the strong energy conditions holds for the object in question), just like Newton postulated.

How about thinking of it as pressure. If you imagine space as a big tank of water in zero gravity. Normally bubbles float up but in a zero gravity environment the idea of floating up gose out the window. The bubble would instead flow to the lower density areas rather then higher density areas.
So if you can think of matter displacing space and creating a low density area, then matter will 'float' up towards it.
Dose that make sense?


Universe isn't infinite theory Forgive me if any of my terminology is off I am no physicist but I have wondered about this question for a while.

I have read in various books that if the universe was infinite, and uniformly created the whole sky would be white. This is because in every possible location in the sky there would be a star emitting light. Sort of like pixels, filling a screen.

I thought about this and wondered, doesn't this therefore imply that light is never absorbed by another medium?

What if a distant stars light is absorbed, therefore never reaching earth, giving the illusion that there are less stars?

Olber's paradox describes an infinite, static, eternal, and homogenous universe. If there were a cloud of opaque gas situated in some region of the sky, light coming from all conceivable directions via the paradox would bombard the cloud until either it begins emitting light itself via the radiation it absorbs or its particles scatter to allow light to pass through.

Those books have it wrong apparently, and are not up-to-date on astronomy.

We can't even see the stars from the other side of our galaxy. Why? There's just not enough photons coming in from them, they're too faint.

If one would want to place so many stars in every line of sight to fill the sky 'white', either the universe would collapse because those stars would represent so much mass that it couldn't be stable, or time-of-flight effects come into play, if the stars were farther apart.