I'm standing on a black hole, is light approaching red or blue?

In summary, the conversation discusses the effects of extreme gravity on light and time. It is mentioned that an observer cannot physically stand on a black hole, but in a thought experiment, they would see light redshifted due to stretching of the wavelength. The Pound-Rebka experiment contradicts this, indicating that light coming from an outside observer would be blueshifted. The conversation also delves into the concept of the Schwarzschild radius and how it is related to mass and time. It is mentioned that the visible universe has a Schwarzschild radius equivalent to its age, and the conversation ends with a calculation of the Schwarzschild radius for various objects.
  • #1
elevin
19
0
Little thought experiment. Hypothetically let's say I'm standing on the surface of an imaginary black hole and not completely crushed to death, or better yet, I'm in the center of the thing (never mind the singularity) and looking outward into deep space. Is the light coming toward me red or blue shifted? I'm going with red, that the immense gravity surrounding me will elongate the wavelength of light before it reaches my eye.
 
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  • #2
You as a physical observer can't "stand on the black hole" if you mean "staying at the event horizon at rest". The event horizon is a light-like surface, so only photons can stay at rest at the horizon, you as a physical observer with mass will pass the horizon with the speed of light (better: the horizon as a light-like surface will pass with speed of light)

Falling towards a black hole you will see infalling light blue-shifted. When you start a rocket engine slowing down your free-fall towards to horizon, the blue-shift will increase.
 
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  • #3
tom.stoer said:
Falling towards a black hole you will see infalling light blue-shifted.

Actually, if you started falling from a great distance, you will see the light red-shifted.
 
  • #4
Why?

Radiation created near the horizon and observed far away is redshifted. So in the opposite direction I expect blueshift.
 
  • #5
tom.stoer said:
Why?

Radiation created near the horizon and observed far away is redshifted. So in the opposite direction I expect blueshift.

This is for two static observers. The falling observer is not a static observer.

Suppose observer that A hovers at a great distance from a black hole, and that observer B hovers very close to the event horizon. The light that B receives from A is tremendously blueshifted. Now suppose that observer C falls freely from a great distance. C whizzes by B with great speed, and, just past B, light sent from B to C is tremendously Doppler reshifted. What about light from A to C? The gravitation blueshift from A to B is less that the Doppler redshift from B to C. As C crosses the event horizon, C sees light from distant stars redshifted, not blueshifted.

If observer A, who hovers at great distance from the black hole, radially emits light of wavelength [itex]\lambda[/itex], then observer C, who falls from rest freely and radially from A, receives light that has wavelength

[tex]\lambda' = \lambda \left( 1+\sqrt{\frac{2M}{R}}\right).[/tex]
The event horizon is at [itex]R = 2M[/itex], and the formula is valid for all [itex]R[/itex], i.e., for [itex]0 < R < \infty[/itex]. In particular, it is valid outside, at, and inside the event horizon.

See posts 5 and 7 in

https://www.physicsforums.com/showthread.php?p=861282#post861282

I have since done the calculations using Painleve=Gullstrand coordinates.
 
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  • #6
great, thanks!
 
  • #7
tom.stoer said:
You as a physical observer can't "stand on the black hole" if you mean "staying at the event horizon at rest". The event horizon is a light-like surface, so only photons can stay at rest at the horizon, you as a physical observer with mass will pass the horizon with the speed of light (better: the horizon as a light-like surface will pass with speed of light)

Totally get that one can't "stand there" but this is a thought experiment :) Can we say for a minute that hypothetically I'm an observing mass at rest at the center of the black hole looking outward (and I can see through all the layers of dense material and outward through the event horizon). I'm seeing the cosmos through a tremendous gravitational lens. It would seem that the wavelength of light coming toward the black hole would be accelerated or stretched towards me. Time dilation aside, and considering that this light would cross an event horizon on its way toward me, wouldn't that stretch this wavelength a considerable amount?
 
  • #8
elevin said:
Totally get that one can't "stand there" but this is a thought experiment :)
You are asking what do the laws of physics predict in a situation where the laws of physics are being broken. There's no answer to that!

The best we can do is tell you about an observer hovering just outside the event horizon, or one inside the event horizon who is falling inwards.
 
  • #9
Schwarzschild and Pound-Rebka

DrGreg said:
You are asking what do the laws of physics predict in a situation where the laws of physics are being broken. There's no answer to that!

The best we can do is tell you about an observer hovering just outside the event horizon, or one inside the event horizon who is falling inwards.

The Schwarzschild gets exponentially bigger the more mass you pump into the hole. For example, our universe of 10^55kg (give or take an exponent) has an event horizon roughly approximate to its age, and yet we don't consider the universe a black hole. Seems to be a bit of a gray area on our definition of a black hole.

It would appear that the Schwarzschild is a measurement of mass/density on the path of light (and thus time/matter, etc). Time dilation aside, it would seem that in a theoretical pocket of extreme gravity (lets remove ourselves from the term black hole) of which we were at the center, the observer would experience redshift because of a stretching of a wavelength. The Pound-Rebka experiment would indicate otherwise however so perhaps I'm just talking nonsense. Of course, Pound-Rebka was done on earth. Can anyone clear this up?
 
  • #10
elevin said:
... our universe of 10^55kg (give or take an exponent) has an event horizon ...

Say what ?
 
  • #11
HA! If you calculate the Schwarzschild radius of the Mass of the visible universe (roughly 10^80 atoms or 10^55kg) you ironically get its age equivalent. rs=2GM/c^2 (rs=schwarz, G is Newtons G constant, M=mass).

Done my own calc, but below is from Wiki.
"The Schwarzschild radius of an object is proportional to the mass. Accordingly, the Sun has a Schwarzschild radius of approximately 3.0 km (1.9 mi) while the Earth's is only about 9.0 mm, the size of a peanut. The observable universe's mass has a Schwarzschild radius of approximately 10 billion light years.[citation needed]
(m) (g/cm3)
Universe 4.46×10^25[citation needed] (~10B ly)
Milky Way 2.08×10^15 (~0.2 ly)
Sun 2.95×103 1.84×10^16
Earth 8.87×10−3 2.04×10^27
An object whose radius is smaller than its Schwarzschild radius is called a black hole. The surface at the Schwarzschild radius acts as an event horizon in a non-rotating body (a rotating black hole operates slightly differently). Neither light nor particles can escape through this surface from the region inside, hence the name "black hole". The Schwarzschild radius of the (currently hypothesized) supermassive black hole at our Galactic Center would be approximately 13.3 million kilometres.
 
  • #12
elevin said:
our universe of 10^55kg (give or take an exponent) has an event horizon roughly approximate to its age

No, it doesn't. Plugging numbers for the universe into a formula that doesn't even apply to the universe doesn't prove anything.

elevin said:
and yet we don't consider the universe a black hole. Seems to be a bit of a gray area on our definition of a black hole.

No, just in your understanding of what a black hole is. It's a particular static solution to the Einstein Field Equation. Note carefully the "static" part; the universe is not static, so the black hole solution does not apply to it.
 
  • #13
Yes, I'm not arguing whether or not you can compute a Schwarzschild radius for anything, but do you really want to call it an "event horizon" ?
 
  • #14
elevin said:
An object whose radius is smaller than its Schwarzschild radius is called a black hole.

No, this is not correct. A black hole is a region of spacetime from which light cannot escape. Black holes can be formed by the gravitational collapse of massive objects, and once such a collapsing object's radius is smaller than the Schwarzschild radius for its mass, it is indeed inside the event horizon of the black hole that forms from its collapse. But that does not mean the object is identical to the black hole. The object collapses to a singularity and disappears, but the black hole remains.
 
  • #15
phinds said:
I'm not arguing whether or not you can compute a Schwarzschild radius for anything

You should, at least if "compute" means "compute something with physical meaning". Computing the "Schwarzschild radius" of an object that is not described by the Schwarzschild solution to the Einstein Field Equation is computing something that has no physical meaning.
 
  • #16
PeterDonis said:
No, just in your understanding of what a black hole is. It's a particular static solution to the Einstein Field Equation. Note carefully the "static" part; the universe is not static, so the black hole solution does not apply to it.

Obviously Schwarzschild metrics get far more complicated such as that of a spinning black hole, however in the most simple sense it seems to say that the mass of an object will affect the path of light. An object is what? A cluster of gravitating constituents. So just as our bodies are made of constituent cells/atoms/subatomic particles (and we indeed have mass), so are stars/galaxies/clusters and our universe, a body of collective constituents. It's only our perspective that puts them into discrete packets. That gravity has a falloff of 1/r^2, would we not find that another solar system in a very weak way would affect ours, that our galaxy, affect another galaxy (gravitationally)? Gravity is everywhere. With respect to this weakly gravitating field interspersed with pinpoints of mass (stars, galaxies), there seems to be an ironic comparison between (my thoughts on) residual gravity, the simple Schwarzschild metric (of which a center of mass would be located at the position of the observer), and the Hubble Constant.
 
  • #17
elevin said:
Obviously Schwarzschild metrics get far more complicated such as that of a spinning black hole

The correct general term is "Kerr-Newman metrics"; this covers the whole range of possible black holes with angular momentum and electric charge as well as mass. The Schwarzschild solution is a special case of a Kerr-Newman metric with zero angular momentum and zero charge.

However, all of these metrics share a key property: they describe *isolated* systems surrounded by vacuum. (The technical term is "asymptotically flat".) The universe is not an isolated system, so it can't be described by any Kerr-Newman metric.

elevin said:
however in the most simple sense it seems to say that the mass of an object will affect the path of light.

An *isolated* object, yes. As above, the universe is not an isolated object.

elevin said:
An object is what? A cluster of gravitating constituents. So just as our bodies are made of constituent cells/atoms/subatomic particles (and we indeed have mass), so are stars/galaxies/clusters and our universe, a body of collective constituents.

But the universe is not isolated, as all of your other examples are. Subatomic particles, our bodies, stars, galaxies, and clusters are all isolated objects surrounded by empty space. The universe is not.
 
  • #18
PeterDonis said:
You should, at least if "compute" means "compute something with physical meaning". Computing the "Schwarzschild radius" of an object that is not described by the Schwarzschild solution to the Einstein Field Equation is computing something that has no physical meaning.

Fair enough.
 
  • #19
PeterDonis said:
Subatomic particles, our bodies, stars, galaxies, and clusters are all isolated objects surrounded by empty space. The universe is not.

Is it known what the universe IS surrounded by?
If not, any ideas?
 
  • #20
laurub said:
Is it known what the universe IS surrounded by?

It's not surrounded by anything. According to our best-fit current model, the universe is spatially infinite; but even in models with a spatially finite universe, the universe is unbounded (its spatial topology is a 3-sphere), so there is nothing surrounding it.
 
  • #21
laurub said:
Is it known what the universe IS surrounded by?
If not, any ideas?

Yes, the universe is not surrounded by anything. It is all there is. If there were anything "around it" that would be PART of it, because it is all that there is.
 
  • #22
There's a paradox in that the universe appears to be infinite although according to big bang / cosmic inflation it's limited in age. Personally (and I am an extreme minority here) I don't like big bang / inflation. Soooo many paradoxes that require very complex physics to resolve. This is from NASA web site:

- Recent measurements (c. 2001) by a number of ground-based and balloon-based experiments, including MAT/TOCO, Boomerang, Maxima, and DASI, have shown that the brightest spots are about 1 degree across. Thus the universe was known to be flat to within about 15% accuracy prior to the WMAP results. WMAP has confirmed this result with very high accuracy and precision. We now know (as of 2013) that the universe is flat with only a 0.4% margin of error. This suggests that the Universe is infinite in extent; however, since the Universe has a finite age, we can only observe a finite volume of the Universe. All we can truly conclude is that the Universe is much larger than the volume we can directly observe. http://map.gsfc.nasa.gov/universe/uni_shape.html
 
  • #23
elevin said:
There's a paradox in that the universe appears to be infinite although according to big bang / cosmic inflation it's limited in age. Personally (and I am an extreme minority here) I don't like big bang / inflation. Soooo many paradoxes that require very complex physics to resolve. This is from NASA web site:

- Recent measurements (c. 2001) by a number of ground-based and balloon-based experiments, including MAT/TOCO, Boomerang, Maxima, and DASI, have shown that the brightest spots are about 1 degree across. Thus the universe was known to be flat to within about 15% accuracy prior to the WMAP results. WMAP has confirmed this result with very high accuracy and precision. We now know (as of 2013) that the universe is flat with only a 0.4% margin of error. This suggests that the Universe is infinite in extent; however, since the Universe has a finite age, we can only observe a finite volume of the Universe. All we can truly conclude is that the Universe is much larger than the volume we can directly observe. http://map.gsfc.nasa.gov/universe/uni_shape.html

That bolded part is an exaggeration. I don't think it suggests infinite extent so much as allows for the possibility. You could just as well say that it suggests that the universe is finite but unbounded.
 
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  • #24
elevin said:
There's a paradox in that the universe appears to be infinite although according to big bang / cosmic inflation it's limited in age. Personally (and I am an extreme minority here) I don't like big bang / inflation.
First of all, lots of "complex physics" doesn't make something a paradox. Secondly, what exactly is paradoxical about your first sentence? Not understanding != paradoxical

Also, in your above posts, you are conflating particle horizons with event horizons. The former does apply to FLRW solutions to the EFEs.
 
  • #25
phinds said:
Yes, I'm not arguing whether or not you can compute a Schwarzschild radius for anything, but do you really want to call it an "event horizon" ?

I guess it depends on what your modelling, in the case of Schwarzschild it makes little sense. However in the case of Unruh/ Hawking radiation. there have been papers that treat the cosmological horizons as apparent event horizons as opposed to absolute event horizons. Obviously there is no papers beyond the observable.

http://arxiv.org/abs/1007.4044
http://physics.stackexchange.com/questions/10811/do-apparent-event-horizons-have-hawking-radiation

I guess the point is whether or not your referring to an absolute event horizon as opposed to an apparent event horizon.

I haven't seen anything recent on QM vs apparent horizons, the most recent I've seen is the first article.
 
  • #26
laurub said:
Is it known what the universe IS surrounded by?
If not, any ideas?
Of course I have an idea, and it's the "most right one" there is... :devil:

Our universe is surrounded by a parallel universe, which is exactly the same as ours except everyone is wearing a cowboy hat... :approve:

Lol... You know I'm just kidding, but I googled "what the universe IS surrounded by", and that's the first hit... :biggrin:

http://www.google.com/search?q=what...a:en-US:official&client=firefox-a&channel=rcs

Anyway, you have to go with the definition by the boys in the above posts.

And phinds, in an unusual moment of restrained passion, was particularly well-spoken...
If there were anything "around it" that would be PART of it, because it is all that there is.
Yup... I like the way he worded that... :cool:



Here's the definition from Wikipedia.

http://en.wikipedia.org/wiki/The_Universe

And... some fun stuff to read...

http://en.wikipedia.org/wiki/Absolute_time_and_space

http://en.wikipedia.org/wiki/Vacuum_genesis

http://en.wikipedia.org/wiki/Einstein-Cartan_gravity

http://en.wikipedia.org/wiki/Dennis_Sciama



OCR
 
  • #27
Mordred said:
I guess it depends on what your modelling, in the case of Schwarzschild it makes little sense. However in the case of Unruh/ Hawking radiation. there have been papers that treat the cosmological horizons as apparent event horizons as opposed to absolute event horizons. Obviously there is no papers beyond the observable.

http://arxiv.org/abs/1007.4044
http://physics.stackexchange.com/questions/10811/do-apparent-event-horizons-have-hawking-radiation

I guess the point is whether or not your referring to an absolute event horizon as opposed to an apparent event horizon.

I haven't seen anything recent on QM vs apparent horizons, the most recent I've seen is the first article.

There was a long recent thread on apparent event horizons and I both went into the discussion and came out of the discussion with the belief that for all practical purposes, apparent event horizons are mathematical mumbo jumbo which I feel free to equate with the tooth fairy. :smile:
 
  • #28
no bang

WannabeNewton said:
First of all, lots of "complex physics" doesn't make something a paradox. Secondly, what exactly is paradoxical about your first sentence? Not understanding != paradoxical

Also, in your above posts, you are conflating particle horizons with event horizons. The former does apply to FLRW solutions to the EFEs.
Humans have a history of being geocentric (egocentric) when it comes to physics. We've created a model where, by admitted limits in instrumentation, we find ourselves ironically smack in the "center" of a giant expanding balloon of space where 13.77 billion years ago there was some explosion of matter that didn't exist previously. Now we're looking at galaxies that didn't have time to form in that 13.77 billion year time span and have rotational velocities too high for their measured mass density. We're looking at supernovas from 13.2 billion years ago that hardly had enough time to form a star let alone go through an entire solar cycle in the 700k years.

Lets talk about the homogenous distribution of matter not matching what would result from a massive explosion or a faster-than-light inflation. Let's talk about CMBR anisotropies and the cold spot which doesn't match with the sachs-wolf effect or quantum fluctuations in the early stages of the universe. Let's talk about why we're magically at the center of all this.

To quote Steven Weinberg “Why is the dark energy density comparable to the matter energy density at this particular moment in the history of the universe?” Ironic? You bet. Philosophical paradox, absolutely!

The simplest and most elegant solutions are usually right. From Ptolemy to present cosmology has a history correcting complex and or egocentric models. String theory, Dark Energy, MOND, and onward are the flavors of the century because science has no other way to explain what the heck it's looking at (I had a pop punk band when that was popular back in the day too, but that doesn't make it right). Today's cosmologist wants me to believe in 11 or more dimensions, imaginary strings in the order of 10^500 (brian green), an imaginary "inflatron" field, and that 10^80 measurable atoms came from nothing just because we can simulate energies close to what would be required to make a big bang by injecting a couple of photons with 6TeV of energy.

I'm tired of it. There's a more simple and elegant solution.

Mordred said:
I guess it depends on what your modelling, in the case of Schwarzschild it makes little sense. However in the case of Unruh/ Hawking radiation. there have been papers that treat the cosmological horizons as apparent event horizons as opposed to absolute event horizons. Obviously there is no papers beyond the observable.

http://arxiv.org/abs/1007.4044
http://physics.stackexchange.com/questions/10811/do-apparent-event-horizons-have-hawking-radiation

I guess the point is whether or not your referring to an absolute event horizon as opposed to an apparent event horizon.

I haven't seen anything recent on QM vs apparent horizons, the most recent I've seen is the first article.
Thanks! I look forward to reading. Definitely talking about a hypothetical here. Has nothing to do with a real black hole event horizon. Basically talking about a version of "tired light" caused by gravitational redshift instead of some kind of scattering or gray dust.
 
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  • #29
elevin said:
... we find ourselves ironically smack in the "center" of a giant expanding balloon of space ...

This is just plain nonsense and I expect that you know it.
 
  • #30
elevin said:
(and I am an extreme minority here)

Lol... With myself, at least it's a minority of two.



OCR
 
  • #31
phinds said:
There was a long recent thread on apparent event horizons and I both went into the discussion and came out of the discussion with the belief that for all practical purposes, apparent event horizons are mathematical mumbo jumbo which I feel free to equate with the tooth fairy. :smile:


I would be interested in that thread sounds intriguing. Particularly since I have numerous Unruh/Hawking, Parker, Schach-wolf effect etc articles on various forms of particle production.

How one defines a boundary and types of vacuum takes some digging into with regards to virtual particles. QM I'm still weak on but I'm studying it along with particle physics atm.
 
  • #32
phinds said:
That bolded part is an exaggeration. I don't think it suggests infinite extent so much as allows for the possibility. You could just as well say that it suggests that the universe is finite but bounded.

No, you can't; at least, you can't if you are talking about the actual models used in cosmology, which is what the quoted passage was talking about. Those models require that if the universe is flat (more precisely, if it is *spatially* flat), it is infinite in extent (more precisely, in *spatial* extent). If the universe is finite but unbounded [Edit: corrected from "bounded"], it must have positive spatial curvature.

Of course all this assumes that our current cosmological models are basically correct; but AFAIK nobody has proposed an alternative model that allows the universe to be both spatially flat and spatially finite but unbounded [Edit: corrected as above]. (People have speculated about the universe having a flat geometry but a different spatial topology, like a flat 3-torus; but AFAIK these speculations have not led to any actual testable models.)
 
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  • #33
PeterDonis said:
No, you can't; at least, you can't if you are talking about the actual models used in cosmology, which is what the quoted passage was talking about. Those models require that if the universe is flat (more precisely, if it is *spatially* flat), it is infinite in extent (more precisely, in *spatial* extent). If the universe is finite but bounded, it must have positive spatial curvature.

Of course all this assumes that our current cosmological models are basically correct; but AFAIK nobody has proposed an alternative model that allows the universe to be both spatially flat and spatially finite but bounded. (People have speculated about the universe having a flat geometry but a different spatial topology, like a flat 3-torus; but AFAIK these speculations have not led to any actual testable models.)

I don't understand your talking about finite but bounded. It is my understanding that there is wide agreement that if the universe IS finite then it is UNbounded. Finite and bounded would imply an edge and that just doesn't make sense.
 
  • #34
Help! I'm standing on a black hole and I can't get out!
 
  • #35
The terms "finite" and "unbounded" are so ambiguous that it boggles my mind why they are used in some contexts. In the FLRW cosmological model, the simplest 3-manifold corresponding to the ##k = +1## constant sectional curvature case is the 3-sphere ##S^{3}## which is a closed manifold meaning it is compact and has empty manifold boundary. This is what "finite" and "unbounded" commonly seem to refer to in whatever context they show up with regards to the aforementioned cosmological model. However it is of course true that ##S^{3}\subseteq \mathbb{R}^{4}## is bounded in the metric sense (a subset of a metric space is bounded if it can be contained in some open ball) and is also closed in the sense that ##S^{3}\subseteq \mathbb{R}^{4}## is closed in the euclidean topology on ##\mathbb{R}^{4}## (this in fact is one way to verify that ##S^{3}## is compact since every closed and bounded subset of ##\mathbb{R}^{n}## is compact by Heine-Borel). Like I said, the terminology is definitely ambiguous from a mathematical standpoint so it may very well be that Peter was referring to how ##S^{3}## is bounded in the second sense mentioned above.
 
<h2>1. How does a black hole affect the color of light?</h2><p>A black hole has such a strong gravitational pull that it warps the fabric of space-time. This warping effect causes light to be stretched, or redshifted, as it approaches the black hole. This means that the color of light is shifted towards the red end of the spectrum.</p><h2>2. Why does light approaching a black hole turn red?</h2><p>The redshift of light near a black hole is due to the gravitational time dilation effect. As light travels towards the black hole, it has to overcome the strong gravitational pull, causing its wavelength to stretch and its frequency to decrease, resulting in a shift towards the red end of the spectrum.</p><h2>3. Is light approaching a black hole always redshifted?</h2><p>Yes, the gravitational pull of a black hole is so strong that it will always cause light to be redshifted as it approaches. However, the amount of redshift can vary depending on the distance from the black hole and the strength of its gravitational pull.</p><h2>4. Can light approaching a black hole ever turn blueshifted?</h2><p>No, light approaching a black hole can never turn blueshifted. The gravitational pull of a black hole is so strong that it always causes light to be redshifted, regardless of its initial color or direction of travel.</p><h2>5. How does the redshift of light near a black hole affect our perception of it?</h2><p>The redshift of light near a black hole can make it appear dimmer and redder to an outside observer. This can also cause time to appear to slow down for objects near the black hole, as their light is being stretched and their movements appear slower. However, these effects are only noticeable for objects very close to the black hole.</p>

1. How does a black hole affect the color of light?

A black hole has such a strong gravitational pull that it warps the fabric of space-time. This warping effect causes light to be stretched, or redshifted, as it approaches the black hole. This means that the color of light is shifted towards the red end of the spectrum.

2. Why does light approaching a black hole turn red?

The redshift of light near a black hole is due to the gravitational time dilation effect. As light travels towards the black hole, it has to overcome the strong gravitational pull, causing its wavelength to stretch and its frequency to decrease, resulting in a shift towards the red end of the spectrum.

3. Is light approaching a black hole always redshifted?

Yes, the gravitational pull of a black hole is so strong that it will always cause light to be redshifted as it approaches. However, the amount of redshift can vary depending on the distance from the black hole and the strength of its gravitational pull.

4. Can light approaching a black hole ever turn blueshifted?

No, light approaching a black hole can never turn blueshifted. The gravitational pull of a black hole is so strong that it always causes light to be redshifted, regardless of its initial color or direction of travel.

5. How does the redshift of light near a black hole affect our perception of it?

The redshift of light near a black hole can make it appear dimmer and redder to an outside observer. This can also cause time to appear to slow down for objects near the black hole, as their light is being stretched and their movements appear slower. However, these effects are only noticeable for objects very close to the black hole.

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