# Thread: Black Holes and Light Question

1. In the case of light trying to escape a black hole, is it possible that the light never fully stops, just gets red shifted to the point where it has a negligibly small frequency (or extremely long wavelength) ?

If we tried to detect extremely long wavelength light coming from a black hole.... like maybe on oscillation per year..... do you think we'd have any success?

2.

3. In the case of light trying to escape a black hole, is it possible that the light never fully stops, just gets red shifted to the point where it has a negligibly small frequency (or extremely long wavelength) ?
Firstly, light cannot escape from a black hole. Hawking radiation allows some objects to escape, but only if they are particles with their own antiparticle. The antiparticle of a photon is a photon, so you can see why that will not work.

Secondly, there would have to be a mechanism for light to actually try to get out of the black hole. All light seems to be falling in the black hole; none has ever attempted to go the other way. However, assuming that something managed to do so, such as a highly energetic electromagnetic interaction between charged particles, then I would say that such a light particle will undergo little or no shift.

The reason is that if the light were to do so, and was at the outskirts, it eventually settle into an orbit around the black hole. It would possess a redshift if the black hole was in motion away from it, and a blueshift if it was moving towards a black hole. However, because it doesn't do either, there will be no shift. I

f the graviational attraction is strong enough to send it careening into the hole, however, then it will probably be extremely blueshifted. Either of the two cases above will happen, because light will only be redshifted if it manages to escape the event horizon, which is impossible.

So, if the light settles into an orbit, there will be no shift. If it is attracted back, there will be enormous blueshift. It cannot escape the black hole.

If we tried to detect extremely long wavelength light coming from a black hole.... like maybe on oscillation per year..... do you think we'd have any success?
No. That is because light cannot escape from a black hole.

However, many major physicists (and this is opinion, not fact) believe the light must eventually escape, because of the law of conservation of information. This may not be so, however. If they are proven correct, though, then I think we'd be noticing large bursts of energy, as typically seen by quasars.

4. I guess my confusion is at the notion that light always travels at C. So, in attempting to escape a black hole, I guess it still travels at C through the whole process, right?

It just turns to the side at some point, and then its path curves around until its headed back toward the singularity?

This creates a question, though: What would happen if a photon of light were to depart from the singularity at an exactly, perfectly, perpendicular angle to the surface? Gravity can't make it turn to the right or the left unless there's slightly more force felt from one side than from the other.

Would the photon just disappear? Would it disappear temporarily, and then re-appear headed in the direction of the singularity? Or... would it just red shift down to the point where it had a perfect zero frequency? (Which I guess is the same thing as disappearing).

5. Would the photon just disappear? Would it disappear temporarily, and then re-appear headed in the direction of the singularity? Or... would it just red shift down to the point where it had a perfect zero frequency? (Which I guess is the same thing as disappearing).
You forget that in a black hole, space-time is curved. Even if the photon did escape at a perpendicular angle, the curvature would force the photon to travel back to the hole. So, obviously, either of the two cases I mentioned above will happen.

Further, I think you mean wavelength, not frequency. Redhsift cannot change freqeuncy, thanks to the law of conservation of energy. Besides, the photon will not disappear, thanks to the law of conservation of information, although that is still suspect in a black hole.

6. kojax. Think of gravity as an incline. The greater the gravity, the steeper and longer the incline. Steam can easily escape from a kettle because the spout is small but imagine an ever longer spout. Eventually the steam would not have sufficient energy to get out so it would fall back.

The trouble with trying to detect things coming from a black hole is that infalling matter is whipped up by magnetic fields and sent away at near light speed. Photons I would think could have grazing orbits where they don't come quite close enough to fall in so escape.

7. This leads to exactly my problem.

Suppose you have a beam of light headed directly up the incline, perfectly parallel to its slope. It can't slow down (has to always travel at C). Its path can't curve (would need an unballanced force pulling it either to the right or the left).

All it can do is increase its wave length/decrease its frequency. But how far? Down to zero?

What mechanism allows it to ever turn around and head back down the slope?

8. Instead of a slope, think of it as a 3D version of the 2D surface of the inside of a very large ball. You will never leave it no matter how fast you run.

For the photon, an outside of the event horizon no longer exists. It is completely causely isolated from the outside. From its own perspective no path leads to the outside.

9. So it's moving at C relative to a time perspective where time is standing perfectly still?

It's moving at C, but to us, it's not moving at all? So, it's not that it turns around and goes back to the singularity, it's that it just isn't moving. Is that what you mean?

It clearly can't turn around and go back. (If the BH emitted it, I mean)

10. Originally Posted by kojax
So it's moving at C relative to a time perspective where time is standing perfectly still?

It's moving at C, but to us, it's not moving at all? So, it's not that it turns around and goes back to the singularity; it's that it just isn't moving. Is that what you mean?

It clearly can't turn around and go back. (If the BH emitted it, I mean)
It is perhaps a bit confusing thinking about the photon's movement inside the event horizon from our outside perspective. From the photons perspective, I think, the story should be a different one. I always think of the inside of a black hole as a similar envoronment as our big bang limited universe. From our perspective space-time is flat, as in we don't experience gravity as moving through bent space in terms of the shape of the universe. From our perspective there is a sphere of observable universe around us, but as we move, the edge of the sphere moves too. As you go further away from earth you would eventually start to recognize the things you can see on the edge of the universe in your direction of travel as the same things you saw in the opposite direction back at earth. This is due to the 3D curvature of the larger universe. You could go on traveling in as straight a path as you can, but eventually you will see the earth coming up from the front (if expansion had stopped that is).

The way I think of it then (maybe wrongly) is that a photon's path inside an event horizon would be very similar to this. Since the photon is causally isolated from the outside, it would not make much sense for it to experience curvature which is measurable only relative to the outside. In our universe there is no "outside". Similarly, for the photon there is no "outside" either for all intents and purposes. If the photon is emitted at the event horizon, it would only be severely red shifted by the time you detect it, but it would still have been traveling at C.

AFAIK and IMHO

11. The singularity infinitely stretches space. So my understanding is that a photon just inside the event horizon can't escape because to do so it would have to travel an infinite distance. The photon never changes path or curves or anything, it's always heading out of the black hole.

I don't know if it would Doppler shift.

12. So, we *would* observe it to be moving much slower than C. Except.... it's inside a singularity, and therefore totally unobservable.

Does this fit with our understanding of ordinary gravity situations?

13. Hmm, I hadn't considered the effect of gravitational lensing, which would make the light path apparently curve. A strong enough mass and the light path might orbit that mass...

I don't know if this is the same phenomena as an infinitely stretched space, just from a different perspective, or if it's a separate thing altogether.

Either way, the photon would seem to take longer (infinitely longer) to reach us than it should, so in that naive way it is traveling slower than c. Of course it's actually traveling at c, it just has to cover more ground.

Or mathematically: d = v * t, so if we pretend d is fixed, and time increases, velocity decreases. But what really is happening is that d increases, and v stays the same, so t has to increase.

14. infinitely stretched space
I have never heard of this bit.

15. This talks about it. The infinite curvature isn't necessarily a physical reality (space doesn't magically stretch to infinity at the event horizon), but more of a mathematical construct. But it's still a valid way of understanding why light can't escape.

16. The complicated thing about relativity is the ambiguity between which is being stretched: time, or space? (or space-time?)

Is it fair to say that we observe the light emitted within a black hole to be a future event? It'll take an infinitely long time to reach us (and if it did reach us it would be very very red shifted) , but we will observe it to have been emitted distance/C long ago?

Alright, let's discuss a near singularity, then. Suppose we take a concentrated mass that almost counts as a singularity, but not quite. Suppose light takes 1000 years to escape it. How do we square that with relativity?

17. I don't think it would be red shifted. My understanding of Doppler shift is that it requires the source to be moving. Whether the source is 1000 LY away or 100 meters away, its frequency will be the same, just that the apparent magnitude will drop.

Of course in the real universe other objects are accelerating away from us, so light from far away is red shifted.

For a super dense but not singularity object (Neutron star is a good example), space is "stretched" such that what we would consider something like 1 cubic light year is actually 10 cubic light years, or however the math works out. This stretching affects both space and time, since they're just two sides of the same coin really.

18. if we think of a contracting space near a black hole the problem is solved.
it means the photon will be continuously moving away, untill the black hole gets stronger. then the rate of contraction will be so fast that the photon will be sucked back to the black hole with the space even while the photon will be moving away with a speed of C.

19. Originally Posted by Numsgil
This talks about it. The infinite curvature isn't necessarily a physical reality (space doesn't magically stretch to infinity at the event horizon), but more of a mathematical construct. But it's still a valid way of understanding why light can't escape.
I don't know, I am not very comfortable with it. From your link:

"An example is the Schwarzschild solution which describes a non-rotating, uncharged black hole. In coordinate systems convenient for working in regions far away from the black hole, a part of the metric becomes infinite at the event horizon. However, spacetime at the event horizon is regular."

Then you say:

For a super dense but not singularity object (Neutron star is a good example), space is "stretched" such that what we would consider something like 1 cubic light year is actually 10 cubic light years, or however the math works out. This stretching affects both space and time, since they're just two sides of the same coin really.
Which does not follow IMO. There is a red shift of light emitted from the surface of a neutron star, the same as there is a red shift of light emitted from the surface of the earth. This red shift, though, is a direct consequence of time dilation. It is the same reason that, from an outside observer, an object falling into a black hole will never quite reach the event horizon, since time would stop and the red shift would approach infinity.

Also, the red shift of distant light sources is due to the expansion of space and not because of a Doppler shift.

We need some expert advice. Janus! Dishmaster! *whistles*

20. Originally Posted by KALSTER
Originally Posted by Numsgil
This talks about it. The infinite curvature isn't necessarily a physical reality (space doesn't magically stretch to infinity at the event horizon), but more of a mathematical construct. But it's still a valid way of understanding why light can't escape.
I don't know, I am not very comfortable with it.
You don't have to use it. It's just a mathematical construct that allows for another way of thinking of things.

There is a red shift of light emitted from the surface of a neutron star, the same as there is a red shift of light emitted from the surface of the earth. This red shift, though, is a direct consequence of time dilation. It is the same reason that, from an outside observer, an object falling into a black hole will never quite reach the event horizon, since time would stop and the red shift would approach infinity.
Ah, I was unaware of Gravitational Redshift. So yes, the light would be redshifted towards an infinite wavelength size. That's another way of thinking about it.

Also, the red shift of distant light sources is due to the expansion of space and not because of a Doppler shift.
Okay, so I guess the issue here is whether the light is redshifted because its source is moving away from us, or if the light is redshifted just by traveling through space which is expanding, right? That is, if we were to take some starlight reaching Earth and examine it, and take that same star another 1000 LY further along that light's path, would the frequency be the same?

I don't know the answer. Redshifting isn't my forte, but I imagine the answer would be illustrative.

I don't see any mechanism for the two frequencies to differ.

21. Well, the BBT redshift can't be due to a standard Doppler effect, because Hubble observed that (distance * Hubble constant) = amount of redshift.

For that to happen from a normal Doppler effect would require that a much more complicated and unlikely mathematical relationship between distance and redshift were present, because you have factor in that we're seeing a star 100 light years away, where it was 100 lightyears ago, not where it is now.

Originally Posted by basim
if we think of a contracting space near a black hole the problem is solved.
it means the photon will be continuously moving away, untill the black hole gets stronger. then the rate of contraction will be so fast that the photon will be sucked back to the black hole with the space even while the photon will be moving away with a speed of C.

As far as gravity pulling in space at a speed faster than C, I wonder if Relativity tells us to consider that as still meaning the light is traveling at C through the space. So really, an observer only observes light near a source of gravity to travel at C, once the gravity has been accounted for?

So, by this theory, light actually would not be perceived to travel at C from the perspective of an object outside any gravitational field, when it is traveling against the pull of gravity.

So, for example, if we fire a laser beam from the surface of Earth to the Moon, it should take slightly longer to arrive than the distance/c. And if it's fired from the Moon to the Earth, it should take a slightly shorter time than distance/c.

22. Originally Posted by kojax
So, for example, if we fire a laser beam from the surface of Earth to the Moon, it should take slightly longer to arrive than the distance/c. And if it's fired from the Moon to the Earth, it should take a slightly shorter time than distance/c.
Isnt that so?
is there any experiment done to disprove that?

23. So, by this theory, light actually would not be perceived to travel at C from the perspective of an object outside any gravitational field, when it is traveling against the pull of gravity.

So, for example, if we fire a laser beam from the surface of Earth to the Moon, it should take slightly longer to arrive than the distance/c. And if it's fired from the Moon to the Earth, it should take a slightly shorter time than distance/c.
I don't know if that would be true, but as a more instructive example, in cases of gravitational lensing, the distortion of the gravity source causes the light to take longer to reach us. So in the naive sense, light wouldn't be traveling at c (again, this is actually a case of distance increasing instead of speed decreasing).

24. Originally Posted by Numsgil
I don't know if that would be true, but as a more instructive example, in cases of gravitational lensing, the distortion of the gravity source causes the light to take longer to reach us. So in the naive sense, light wouldn't be traveling at c (again, this is actually a case of distance increasing instead of speed decreasing).
So it should be like that the space is contracting near a black hole.
and that may be the reaction for the big bang expansion.

25. I don't follow

26. Originally Posted by Numsgil
So, by this theory, light actually would not be perceived to travel at C from the perspective of an object outside any gravitational field, when it is traveling against the pull of gravity.

So, for example, if we fire a laser beam from the surface of Earth to the Moon, it should take slightly longer to arrive than the distance/c. And if it's fired from the Moon to the Earth, it should take a slightly shorter time than distance/c.
I don't know if that would be true, but as a more instructive example, in cases of gravitational lensing, the distortion of the gravity source causes the light to take longer to reach us. So in the naive sense, light wouldn't be traveling at c (again, this is actually a case of distance increasing instead of speed decreasing).
Well, one has to be careful with such ambiguities. Otherwise the theory of a C constant becomes "un-disprovable" by evidence, because whatever anomalous behavior we see can be described in either of two ways:

1) - The time it took to travel a certain distance was off.

2) - The distance itself was off.

Since velocity is D/T, a problem with either D or T would have exactly the same result on the apparent velocity.

This opens an interesting discussion, though, about what we can hold as reliable data in our perception of space and time. It seems like what you're saying is that, while light still truly travels at C, that doesn't mean it will always be observed to travel at C.

27. C being a constant was demonstrated by the Michelso-Morley experiment. Of course, if there was some anomaly with c not being constant, we should expect the light beam to take longer to reach us and not bend its path. One of the artifacts of gravitational lensing is that the light path apparently bends. So present theory can demonstrate falsifiable experiments to be done so we're in no danger of being unscientific.

With a light beam, time and distance are strictly related because of light's constant velocity, so if time is off, it's automatically implies distance was off, with distance being the primary cause and time being a secondary effect.

So yes, in a strictly naive sense you won't always measure light traveling at c. However, since we understand the nature of spacetime curvature, it should be easy to match observed data to theory and still arrive at a constant c.

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