<p>There is no "up" direction within the event horizon.</p>

<p>Most people get fixated on the speed of light, or energy or whatever. They're like, if light was faster, could it escape the black hole? If my rocket had bigger engines, could I escape? The problem is, all these questions make no sense. You can't get out because there is no way out.</p>

<p>A black hole is formed when gravity is so strong, it ties spacetime into a giant knot. Space is tied into itself. It's not just a little bent; it gets curved until it closes onto itself.</p>

<p>Inside the event horizon, all paths lead to the center. No matter how you're turning, which direction you're looking, you're actually facing the center. It's hard to visualize, but that's how it is. This is not normal spacetime, it's something unlike anything you've thought about.</p>

<p>Starting from point A inside the event horizon, there is no path you could draw that leads to point B outside. All paths lead to the center. Spacetime is really sick and broken.</p>

<p>This is the real reason why nothing gets out of a black hole.</p>

CC BY-SA 4.0
2019-01-14T21:57:36.217Z
So Florin, at what point during my thought experiment does spacetime tie itself into a knot?
John Duffield
CC BY-SA 4.0
2019-01-15T01:05:40.477Z
@JohnDuffield - When it collapses into a black hole.
Florin Andrei
CC BY-SA 4.0
2019-01-15T05:18:56.127Z
I mostly like this answer because it states that all paths inside the EH lead to the centre, i.e., futurewards motion on all timelike and lightlike paths heads towards the centre. However, there is no knotting involved, and although I realise you're just being metaphorical I think that it may mislead some readers.
PM 2Ring
CC BY-SA 4.0
2019-01-15T17:01:21.270Z
@PM 2Ring : I too think this answer may mislead some readers. I'm sorry Florin, but there is _no_ point in the gedankenexperiment at which the upward light beam suddenly stops being vertical and somehow instantly flicks into some backward curve. Space is not curved where a gravitational field is, instead it's [neither inhomogeneous nor isotropic](https://einsteinpapers.press.princeton.edu/vol7-trans/192?highlightText=%22neither%20homogeneous%22).
John Duffield
CC BY-SA 4.0
2019-01-15T18:22:47.000Z
@PM2Ring - You are correct, some metaphors were used, this is an explanation for laypeople, not a scientific paper.
Florin Andrei
CC BY-SA 4.0
2019-01-15T18:23:08.380Z
@JohnDuffield - You may want to start from the basics before you read more advanced literature, there's less chance of confusion this way.
Florin Andrei
CC BY-SA 4.0
2019-01-15T20:58:36.587Z
@Florin Andrei : I'm not confused. I've read the basics thanks. Such as this [Preliminaries](http://math.ucr.edu/home/baez/einstein/node2.html) article on the Baez website. Note this: _"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"_. I've also read the advanced literature.
John Duffield
CC BY-SA 4.0
2019-01-15T21:20:18.583Z
@JohnDuffield - Take an actual physics class. It's all I can say at this point. It will save you a lot of head-scratching of the kind exhibited here. You don't understand fairly important notions of general relativity, and yet you think you can pass judgment on these matters. You need to resolve that big discrepancy first. Quoting random paragraphs from books, and using terms of art such as gedankenexperiment are not proof of knowledge and understanding.
Florin Andrei
CC BY-SA 4.0
2019-01-15T21:28:30.803Z
@Florin Andrei : I've got a physics A level, and I've read physics textbooks, the Einstein digital papers, and a whole lot more. As you can see from articles I've written such as [The Hawking Papers](http://physicsdetective.com/the-hawking-papers/). Nowhere did anything ever say that as a black hole forms, spacetime ties itself into a giant knot. If you have a reference for that, I'd be interested to read it.
John Duffield
CC BY-SA 4.0
2019-01-14T19:59:46.983Z
2019-01-14T19:59:46.983Z
Florin Andrei
<p>Florin Andrei's answer is correct in my opinion, and flows directly from the mathematics of GR. But here is an alternative way of thinking about it.</p>

<p>Light always travels at the speed of light when measured locally. An observer inside the event horizon can emit light moving radially outward (according to them). The problem is that they and everything else is falling inwards. The OP's "gedanken experiment" is simply not possible; within the event horizon there can be no stationary observer that launches a light beam.</p>

<p>An oft-used analogy is drifting in a boat on a river. You release fish into the water that swim at constant speed, upstream or downstream <em>relative to your boat</em>. However, if the river flows fast enough, the fish can never make their way upstream as far as an observer on the bank is concerned, and both boat and fish will end up going over the waterfall.</p>

<p>The event horizon marks the point where the river flows too fast for the fish to escape.</p>

<p>For anyone interested in exploring the so-called "river model" of thinking about dynamics in and around black holes, there is an excellent (though somewhat mathematical) introduction by <a href="https://arxiv.org/abs/gr-qc/0411060" rel="nofollow noreferrer">Hamilton &amp; Lisle (2006)</a>.</p>

CC BY-SA 4.0
2019-01-16T16:50:19.030Z
I'm sorry Rob, but Florin's answer is not correct. Nor is the waterfall analogy. That's based on Gullstrand-Painleve coordinates which Einstein rejected for good reason. He described a gravitational field as a place where space was "neither homogeneous nor isoptropic", not a place where space is falling down. Space is not falling inwards in a gravitational field, and nor is light. By the by, the strength of a gravitational field is related to the local gradient in the "coordinate" speed of light. At the event horizon the coordinate speed of light is zero, and _it can't go lower than that_.
John Duffield
CC BY-SA 4.0
2019-01-16T18:10:21.960Z
@JohnDuffield The waterfall analogy is indeed imprecise in detail, but captures the essential picture. Your views on what is right and wrong about GR and coordinates in GR I can safely ignore.
ProfRob
CC BY-SA 4.0
2019-01-17T14:43:31.063Z
Rob: Ignore what you wish, but the waterfall analogy is a lies-to-children myth. It's totally wrong too, because the ascending light beam speeds up. See what Einstein said in 1920: _“As a simple geometric consideration shows, the curvature of light rays occurs only in spaces where the speed of light is spatially variable”_. Also see Is The Speed of Light Everywhere the Same? by PhysicsFAQ editor Don Koks. Note where he says _“light speeds up as it ascends from floor to ceiling”_. Optical clocks go slower when they’re lower because light goes slower when it’s lower. Not for any other reason.
John Duffield
CC BY-SA 4.0
2019-01-17T20:35:06.303Z
@JohnDuffield you should ask your question on Physics SE and see what others with a clear understanding of GR think, instead of trying to fool a less knowledgeable readership that there is some kind of debate about this. Though it would be closed as a duplicate e.g. https://physics.stackexchange.com/questions/28297/why-is-a-black-hole-black
ProfRob
CC BY-SA 4.0
2019-01-16T08:02:14.023Z
2019-01-17T20:54:46.247Z
2019-01-17T20:54:46.247Z
ProfRob
<p>Using the General Relativity and Schwarzschild metric we can define the redshift of the photon as,</p>
<p><span class="math-container">$$\frac {v_{\infty}} {v_e} = (1-r_s/R_e)^{1/2}$$</span> in this equation <span class="math-container">$v_{\infty}$</span> represents the frequency of the light measured by an observer at infinity, <span class="math-container">$v_e$</span> is the frequency of the emitted wavelength, <span class="math-container">$r_s$</span> is the schwarzschild radius, <span class="math-container">$r_S=2GM/c^2$</span>, and finally <span class="math-container">$R_e$</span> is the radius which photon is emitted.</p>
<p>When we set <span class="math-container">$r_s=R_e$</span> we can see that <span class="math-container">$\frac {v_{\infty}} {v_e}=0$</span> which means that the redshift will be infinitely large and the photon cannot escape from the black hole.</p>
<p>For more information, you can look here, <a href="https://en.wikipedia.org/wiki/Gravitational_redshift#Prediction_by_the_equivalence_principle_and_general_relativity" rel="nofollow noreferrer">Gravitational redshift</a></p>
<blockquote>
<p>For an object compact enough to have an event horizon, the redshift is not defined for photons emitted inside the Schwarzschild radius, both because signals cannot escape from inside the horizon and because an object such as the emitter cannot be stationary inside the horizon, as was assumed above. Therefore, this formula only applies when <span class="math-container">$R_{e}$</span>  is larger than <span class="math-container">$r_{s}$</span> . When the photon is emitted at a distance equal to the Schwarzschild radius, the redshift will be infinitely large, and it will not escape to any finite distance from the Schwarzschild sphere.</p>
</blockquote>
<p>For the energy change in the photon, there has been done many experiments which some of them are explained in the Wikipedia page and I have found another experiment which is, <a href="https://en.wikipedia.org/wiki/Pound%E2%80%93Rebka_experiment" rel="nofollow noreferrer">Pound–Rebka experiment</a>. That perfectly explains the energy change and for the math part, you can look <a href="http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/gratim.html" rel="nofollow noreferrer">here</a>.</p>

CC BY-SA 4.0
2019-01-14T20:03:35.660Z
This is a semi-classical explanation that is basically pre-Einstein and it isn't correct. The answer from General Relativity is what was posted by @Florin Andrei.
Mark Olson
CC BY-SA 4.0
2019-01-14T20:28:12.477Z
What you might do is look up a bit about the classical idea of a black hole (a body whose escape velocity a la Newton is greater than the speed of light) and expand the answer to cover both that and the one based on gravitational red shift -- just indicate that there are still better explanations even if they're not at all intuitive. (Though doubtless other people would recommend other courses of action.)
Mark Olson
CC BY-SA 4.0
2019-01-14T22:01:35.343Z
I'm sorry Reign, but the energy of the photon doesn't decrease. Take a look at [page 149](https://einsteinpapers.press.princeton.edu/vol6-trans/401) of _Relativity, the Special and General Theory_. Einstein said _“an atom absorbs or emits light at a frequency which is dependent on the potential of the gravitational field in which it is situated"_. When the ascending photon ascends, its E=hf energy does not reduce, and nor does its frequency. There is no outflow of energy from the photon. Instead the photon was _emitted_ at a lower frequency at a lower elevation, with less energy.
John Duffield
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2019-01-15T05:07:56.360Z
@JohnDuffield I edited my post, Its better now
seVenVo1d
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2019-01-15T16:50:17.727Z
Noted Reign. But again, the ascending photon doesn't change energy. It doesn't change frequency. So redshift is not the reason why the light doesn't get out. But have an upvote from me for trying anyway.
John Duffield
CC BY-SA 4.0
2019-01-15T16:58:17.377Z
@JohnDuffield Doenst change energy ?? Then how can you explain the result of the experiment ??
seVenVo1d
CC BY-SA 4.0
2019-01-15T17:04:53.937Z
Reign: when I lift you up, I do work on you. I add energy to you. Ditto for all your equipment. You then measure the ascending photon as having less energy. Because you've got more. Like Einstein said, _"an atom absorbs or emits light at a frequency which is dependent on the potential of the gravitational field in which it is situated"_. That light doesn't lose energy as it ascends. It was emitted with less energy at a lower elevation.
John Duffield
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2019-01-15T19:35:06.267Z
If I open another thread about it is that okay ? Cause I think it cannot be that simple
seVenVo1d
CC BY-SA 4.0
2019-01-15T21:00:30.150Z
Go for it Reign. The point to note is that when you send a 511keV photon into a black hole, the black hole mass increases by 511keV/c². That photon didn't actually gain any energy on the way down. Gravitational blueshift doesn't add any energy to the descending photon, and gravitational redhift doesn't remove any energy from the ascending photon.
John Duffield
CC BY-SA 4.0
2019-01-16T06:16:33.507Z
But we can measure the frequency of the photon when is emitted and it will not be the same when its observed. Another point is relativity, you can argue that in some referance frame thats the case but I am also right to say that energy of photon increased or decreased due to the some referance frame that i measured.. I discussed this in another forum and i know that I am right.
seVenVo1d
CC BY-SA 4.0
2019-01-16T06:16:54.377Z
Here is the link https://www.physicsforums.com/threads/gravitational-redshift-and-its-effects-on-photon.964286/#
seVenVo1d
CC BY-SA 4.0
2019-01-16T06:26:31.827Z
But you are right that this explanation is not good to explain why light doeant come out from a black hole cause it seems it not the case. I can delete my post after you read the link
seVenVo1d
CC BY-SA 4.0
2019-01-16T09:32:17.067Z
@JohnDuffield "Gravitational blueshift doesn't add any energy to the descending photon, and gravitational redshift doesn't remove any energy from the ascending photon." -- there is no universal notion of how much energy a photon has. That depends on who is measuring it and how.
Steve Linton
CC BY-SA 4.0
2019-01-16T16:31:03.510Z
@Steve Linton : but you know that when you move towards a stream of photons, you measure them to be blueshifted because _you_ changed, not the photons. It's similar for gravitational blueshift. If you fall to a lower elevation you measure the descending photons to be blueshifted because you changed, not the photons. It's similar but the other way round for gravitational redshift. LIght is emitted at a lower frequency at a lower elevation, Einstein made this clear. But for some reason people think light is emitted at the same frequency and loses energy as it ascends.
John Duffield
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2019-01-14T09:08:15.940Z
2019-01-15T05:14:00.730Z
2020-06-17T09:47:40.653Z
seVenVo1d
<p>I'm asking this question as a follow-up to <a href="https://astronomy.stackexchange.com/questions/29122/if-light-has-no-mass-why-is-it-affected-by-gravity">if light has no mass why is it affected by gravity?</a></p>

<p>Imagine you’re standing on a gedanken planet, shining a laser beam straight up into space. The light goes straight up. It doesn’t curve, and it doesn’t fall back down. Now imagine it’s a denser more massive planet. The light still goes straight up. It still doesn’t curve, and it still doesn’t fall back down. Let’s make it a really massive planet. That light still goes straight up. It <em>still</em> doesn’t curve, and it <em>still</em> doesn’t fall back down:</p>

<p>But when we make our gedanken planet so massive that it’s a black hole, all of a sudden light can’t escape. Why? Why doesn’t the light get out? <em>Why doesn't the vertical light beam get out of a black hole?</em> </p>

CC BY-SA 4.0
2019-01-14T11:36:35.193Z
The planet inside the black hole cannot be stationary. Everything worldline which enters a black hole event horizon hits the singularity (at least in the non-rotating case) within a finite amount of time (as measured by an observer following that worldline).So the photon falls into the "centre" of the black hole, just a little more slowly than the planet does.
Steve Linton
CC BY-SA 4.0
2019-01-15T03:59:39.330Z
I think he is quite logical here, light should bend atleast becuase unlike black hole, large bodies too bend space time a little compared to black hole, so light should bend a little concerning that much magnitude, rule should be followed by everyone, this is the law. It is abnormal, question is quite good here.
Rahul Singh
CC BY-SA 4.0
2019-01-15T17:15:24.333Z
@Steve Linton : that's a stock answer, but I'm afraid it's wrong. The  vertical light beam doesn't fall down in a gravitational field. You will never find an ascending photon going slower and slower then falling backwards like a stone. See the GR section of [this](http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/speed_of_light.html). The ascending photon doesn't slow down. Instead _it speeds up_.
John Duffield
CC BY-SA 4.0
2019-01-15T17:16:51.707Z
@Rahul Singh : the question is intended to make you think about some of the explanations you've heard about black holes, and appreciate that some of them are flawed.
John Duffield
CC BY-SA 4.0
2019-01-16T08:05:20.777Z
A gedanken experiment has to be possible in principle. It is not possible to have a stationary observer within an event horizon.
ProfRob
CC BY-SA 4.0
2019-01-16T08:43:00.340Z
Also see https://physics.stackexchange.com/questions/454499/gravitational-redshift-and-energy-of-a-photon
PM 2Ring
CC BY-SA 4.0
2019-01-16T16:41:10.030Z
@PM 2Ring : I like the way Peter Donis on physics forums flatly contradicts Einstein in post #16. He must have missed what Arman777 said in post #3.
John Duffield
CC BY-SA 4.0
2019-01-16T18:13:47.717Z
Possible duplicate of [Why can't light escape from a black hole?](https://astronomy.stackexchange.com/questions/7/why-cant-light-escape-from-a-black-hole) (though all current answers there are awful).
ProfRob
CC BY-SA 4.0
2019-01-17T00:47:21.390Z
I wonder if there's an analogy with light from beyond the cosmological horizon, in which the photon is eternally heading towards me but can never reach me because the distance between us is expanding FTL. Is it reasonable to approximate that in a black hole, the photon can never reach the event horizon because the gravity well (spacetime curvature) "deepens" faster than light?
Chappo Hasn't Forgotten
CC BY-SA 4.0
2019-01-17T14:49:54.543Z
@Chappo : No, I don't think there's an analogy with the cosmological horizon. And no, I don't think it's reasonable to say the photon can never reach the event horizon because the gravity well deepens faster than light. For a clue see [Propagation of light in non-inertial reference frames](https://en.wikipedia.org/wiki/Propagation_of_light_in_non-inertial_reference_frames) then follow the link to [Shapiro time delay](https://en.wikipedia.org/wiki/Shapiro_time_delay). Note the quote!
John Duffield
<black-hole><gravity><general-relativity>
Why doesn't the vertical light beam get out of a black hole?
John Duffield
# Why doesn't the vertical light beam get out of a black hole?

(Asked by: username_0 on 2019-01-14)

I'm asking this question as a follow-up to [if light has no mass why is it affected by gravity?](https://astronomy.stackexchange.com/questions/29122/if-light-has-no-mass-why-is-it-affected-by-gravity)

Imagine you’re standing on a gedanken planet, shining a laser beam straight up into space. The light goes straight up. It doesn’t curve, and it doesn’t fall back down. Now imagine it’s a denser more massive planet. The light still goes straight up. It still doesn’t curve, and it still doesn’t fall back down. Let’s make it a really massive planet. That light still goes straight up. It *still* doesn’t curve, and it *still* doesn’t fall back down:

But when we make our gedanken planet so massive that it’s a black hole, all of a sudden light can’t escape. Why? Why doesn’t the light get out? *Why doesn't the vertical light beam get out of a black hole?*

**Tags:** black-hole, gravity, general-relativity

> **Comments:**
> **username_1**: The planet inside the black hole cannot be stationary. Everything worldline which enters a black hole event horizon hits the singularity (at least in the non-rotating case) within a finite amount of time (as measured by an observer following that worldline).So the photon falls into the "centre" of the black hole, just a little more slowly than the planet does.  
> **username_2**: I think he is quite logical here, light should bend atleast becuase unlike black hole, large bodies too bend space time a little compared to black hole, so light should bend a little concerning that much magnitude, rule should be followed by everyone, this is the law. It is abnormal, question is quite good here.  
> **username_0**: @username_1 : that's a stock answer, but I'm afraid it's wrong. The  vertical light beam doesn't fall down in a gravitational field. You will never find an ascending photon going slower and slower then falling backwards like a stone. See the GR section of [this](http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/speed_of_light.html). The ascending photon doesn't slow down. Instead _it speeds up_.  
> **username_0**: @username_2 : the question is intended to make you think about some of the explanations you've heard about black holes, and appreciate that some of them are flawed.  
> **username_3**: A gedanken experiment has to be possible in principle. It is not possible to have a stationary observer within an event horizon.  
> **username_4**: Also see https://physics.stackexchange.com/questions/454499/gravitational-redshift-and-energy-of-a-photon  
> **username_0**: @username_4 : I like the way Peter Donis on physics forums flatly contradicts Einstein in post #16. He must have missed what Arman777 said in post #3.  
> **username_3**: Possible duplicate of [Why can't light escape from a black hole?](https://astronomy.stackexchange.com/questions/7/why-cant-light-escape-from-a-black-hole) (though all current answers there are awful).  
> **username_5**: I wonder if there's an analogy with light from beyond the cosmological horizon, in which the photon is eternally heading towards me but can never reach me because the distance between us is expanding FTL. Is it reasonable to approximate that in a black hole, the photon can never reach the event horizon because the gravity well (spacetime curvature) "deepens" faster than light?  
> **username_0**: @Chappo : No, I don't think there's an analogy with the cosmological horizon. And no, I don't think it's reasonable to say the photon can never reach the event horizon because the gravity well deepens faster than light. For a clue see [Propagation of light in non-inertial reference frames](https://en.wikipedia.org/wiki/Propagation_of_light_in_non-inertial_reference_frames) then follow the link to [Shapiro time delay](https://en.wikipedia.org/wiki/Shapiro_time_delay). Note the quote!  

---

## 3 answers

### Answer by username_6 on 2019-01-14. Score: 13

There is no "up" direction within the event horizon.

Most people get fixated on the speed of light, or energy or whatever. They're like, if light was faster, could it escape the black hole? If my rocket had bigger engines, could I escape? The problem is, all these questions make no sense. You can't get out because there is no way out.

A black hole is formed when gravity is so strong, it ties spacetime into a giant knot. Space is tied into itself. It's not just a little bent; it gets curved until it closes onto itself.

Inside the event horizon, all paths lead to the center. No matter how you're turning, which direction you're looking, you're actually facing the center. It's hard to visualize, but that's how it is. This is not normal spacetime, it's something unlike anything you've thought about.

Starting from point A inside the event horizon, there is no path you could draw that leads to point B outside. All paths lead to the center. Spacetime is really sick and broken.

This is the real reason why nothing gets out of a black hole.

> **Comments:**
> **username_0**: So Florin, at what point during my thought experiment does spacetime tie itself into a knot?  
> **username_6**: @JohnDuffield - When it collapses into a black hole.  
> **username_4**: I mostly like this answer because it states that all paths inside the EH lead to the centre, i.e., futurewards motion on all timelike and lightlike paths heads towards the centre. However, there is no knotting involved, and although I realise you're just being metaphorical I think that it may mislead some readers.  
> **username_0**: @username_4 : I too think this answer may mislead some readers. I'm sorry Florin, but there is _no_ point in the gedankenexperiment at which the upward light beam suddenly stops being vertical and somehow instantly flicks into some backward curve. Space is not curved where a gravitational field is, instead it's [neither inhomogeneous nor isotropic](https://einsteinpapers.press.princeton.edu/vol7-trans/192?highlightText=%22neither%20homogeneous%22).  
> **username_6**: @PM2Ring - You are correct, some metaphors were used, this is an explanation for laypeople, not a scientific paper.  
> **username_6**: @JohnDuffield - You may want to start from the basics before you read more advanced literature, there's less chance of confusion this way.  
> **username_0**: @username_6 : I'm not confused. I've read the basics thanks. Such as this [Preliminaries](http://math.ucr.edu/home/baez/einstein/node2.html) article on the Baez website. Note this: _"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"_. I've also read the advanced literature.  
> **username_6**: @JohnDuffield - Take an actual physics class. It's all I can say at this point. It will save you a lot of head-scratching of the kind exhibited here. You don't understand fairly important notions of general relativity, and yet you think you can pass judgment on these matters. You need to resolve that big discrepancy first. Quoting random paragraphs from books, and using terms of art such as gedankenexperiment are not proof of knowledge and understanding.  
> **username_0**: @username_6 : I've got a physics A level, and I've read physics textbooks, the Einstein digital papers, and a whole lot more. As you can see from articles I've written such as [The Hawking Papers](http://physicsdetective.com/the-hawking-papers/). Nowhere did anything ever say that as a black hole forms, spacetime ties itself into a giant knot. If you have a reference for that, I'd be interested to read it.  

---

### Answer by username_3 on 2019-01-16. Score: 7

username_6's answer is correct in my opinion, and flows directly from the mathematics of GR. But here is an alternative way of thinking about it.

Light always travels at the speed of light when measured locally. An observer inside the event horizon can emit light moving radially outward (according to them). The problem is that they and everything else is falling inwards. The OP's "gedanken experiment" is simply not possible; within the event horizon there can be no stationary observer that launches a light beam.

An oft-used analogy is drifting in a boat on a river. You release fish into the water that swim at constant speed, upstream or downstream *relative to your boat*. However, if the river flows fast enough, the fish can never make their way upstream as far as an observer on the bank is concerned, and both boat and fish will end up going over the waterfall.

The event horizon marks the point where the river flows too fast for the fish to escape.

For anyone interested in exploring the so-called "river model" of thinking about dynamics in and around black holes, there is an excellent (though somewhat mathematical) introduction by [Hamilton & Lisle (2006)](https://arxiv.org/abs/gr-qc/0411060).

> **Comments:**
> **username_0**: I'm sorry Rob, but Florin's answer is not correct. Nor is the waterfall analogy. That's based on Gullstrand-Painleve coordinates which Einstein rejected for good reason. He described a gravitational field as a place where space was "neither homogeneous nor isoptropic", not a place where space is falling down. Space is not falling inwards in a gravitational field, and nor is light. By the by, the strength of a gravitational field is related to the local gradient in the "coordinate" speed of light. At the event horizon the coordinate speed of light is zero, and _it can't go lower than that_.  
> **username_3**: @JohnDuffield The waterfall analogy is indeed imprecise in detail, but captures the essential picture. Your views on what is right and wrong about GR and coordinates in GR I can safely ignore.  
> **username_0**: Rob: Ignore what you wish, but the waterfall analogy is a lies-to-children myth. It's totally wrong too, because the ascending light beam speeds up. See what Einstein said in 1920: _“As a simple geometric consideration shows, the curvature of light rays occurs only in spaces where the speed of light is spatially variable”_. Also see Is The Speed of Light Everywhere the Same? by PhysicsFAQ editor Don Koks. Note where he says _“light speeds up as it ascends from floor to ceiling”_. Optical clocks go slower when they’re lower because light goes slower when it’s lower. Not for any other reason.  
> **username_3**: @JohnDuffield you should ask your question on Physics SE and see what others with a clear understanding of GR think, instead of trying to fool a less knowledgeable readership that there is some kind of debate about this. Though it would be closed as a duplicate e.g. https://physics.stackexchange.com/questions/28297/why-is-a-black-hole-black  

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### Answer by username_7 on 2019-01-14. Score: 3

Using the General Relativity and Schwarzschild metric we can define the redshift of the photon as,

$$\frac {v\_{\infty}} {v\_e} = (1-r\_s/R\_e)^{1/2}$$ in this equation $v\_{\infty}$ represents the frequency of the light measured by an observer at infinity, $v\_e$ is the frequency of the emitted wavelength, $r\_s$ is the schwarzschild radius, $r\_S=2GM/c^2$, and finally $R\_e$ is the radius which photon is emitted.

When we set $r\_s=R\_e$ we can see that $\frac {v\_{\infty}} {v\_e}=0$ which means that the redshift will be infinitely large and the photon cannot escape from the black hole.

For more information, you can look here, [Gravitational redshift](https://en.wikipedia.org/wiki/Gravitational_redshift#Prediction_by_the_equivalence_principle_and_general_relativity)

> For an object compact enough to have an event horizon, the redshift is not defined for photons emitted inside the Schwarzschild radius, both because signals cannot escape from inside the horizon and because an object such as the emitter cannot be stationary inside the horizon, as was assumed above. Therefore, this formula only applies when $R\_{e}$ is larger than $r\_{s}$ . When the photon is emitted at a distance equal to the Schwarzschild radius, the redshift will be infinitely large, and it will not escape to any finite distance from the Schwarzschild sphere.

For the energy change in the photon, there has been done many experiments which some of them are explained in the Wikipedia page and I have found another experiment which is, [Pound–Rebka experiment](https://en.wikipedia.org/wiki/Pound%E2%80%93Rebka_experiment). That perfectly explains the energy change and for the math part, you can look [here](http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/gratim.html).

> **Comments:**
> **username_8**: This is a semi-classical explanation that is basically pre-Einstein and it isn't correct. The answer from General Relativity is what was posted by @username_6.  
> **username_8**: What you might do is look up a bit about the classical idea of a black hole (a body whose escape velocity a la Newton is greater than the speed of light) and expand the answer to cover both that and the one based on gravitational red shift -- just indicate that there are still better explanations even if they're not at all intuitive. (Though doubtless other people would recommend other courses of action.)  
> **username_0**: I'm sorry Reign, but the energy of the photon doesn't decrease. Take a look at [page 149](https://einsteinpapers.press.princeton.edu/vol6-trans/401) of _Relativity, the Special and General Theory_. Einstein said _“an atom absorbs or emits light at a frequency which is dependent on the potential of the gravitational field in which it is situated"_. When the ascending photon ascends, its E=hf energy does not reduce, and nor does its frequency. There is no outflow of energy from the photon. Instead the photon was _emitted_ at a lower frequency at a lower elevation, with less energy.  
> **username_7**: @JohnDuffield I edited my post, Its better now  
> **username_0**: Noted Reign. But again, the ascending photon doesn't change energy. It doesn't change frequency. So redshift is not the reason why the light doesn't get out. But have an upvote from me for trying anyway.  
> **username_7**: @JohnDuffield Doenst change energy ?? Then how can you explain the result of the experiment ??  
> **username_0**: Reign: when I lift you up, I do work on you. I add energy to you. Ditto for all your equipment. You then measure the ascending photon as having less energy. Because you've got more. Like Einstein said, _"an atom absorbs or emits light at a frequency which is dependent on the potential of the gravitational field in which it is situated"_. That light doesn't lose energy as it ascends. It was emitted with less energy at a lower elevation.  
> **username_7**: If I open another thread about it is that okay ? Cause I think it cannot be that simple  
> **username_0**: Go for it Reign. The point to note is that when you send a 511keV photon into a black hole, the black hole mass increases by 511keV/c². That photon didn't actually gain any energy on the way down. Gravitational blueshift doesn't add any energy to the descending photon, and gravitational redhift doesn't remove any energy from the ascending photon.  
> **username_7**: But we can measure the frequency of the photon when is emitted and it will not be the same when its observed. Another point is relativity, you can argue that in some referance frame thats the case but I am also right to say that energy of photon increased or decreased due to the some referance frame that i measured.. I discussed this in another forum and i know that I am right.  
> **username_7**: Here is the link https://www.physicsforums.com/threads/gravitational-redshift-and-its-effects-on-photon.964286/#  
> **username_7**: But you are right that this explanation is not good to explain why light doeant come out from a black hole cause it seems it not the case. I can delete my post after you read the link  
> **username_1**: @JohnDuffield "Gravitational blueshift doesn't add any energy to the descending photon, and gravitational redshift doesn't remove any energy from the ascending photon." -- there is no universal notion of how much energy a photon has. That depends on who is measuring it and how.  
> **username_0**: @username_1 : but you know that when you move towards a stream of photons, you measure them to be blueshifted because _you_ changed, not the photons. It's similar for gravitational blueshift. If you fall to a lower elevation you measure the descending photons to be blueshifted because you changed, not the photons. It's similar but the other way round for gravitational redshift. LIght is emitted at a lower frequency at a lower elevation, Einstein made this clear. But for some reason people think light is emitted at the same frequency and loses energy as it ascends.  

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