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I'm asking this question as a follow-up to 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?

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    $\begingroup$ 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. $\endgroup$
    – Steve Linton
    Jan 14, 2019 at 11:36
  • $\begingroup$ 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. $\endgroup$
    – Rahul Singh
    Jan 15, 2019 at 3:59
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    $\begingroup$ @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. The ascending photon doesn't slow down. Instead it speeds up. $\endgroup$
    – John Duffield
    Jan 15, 2019 at 17:15
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    $\begingroup$ A gedanken experiment has to be possible in principle. It is not possible to have a stationary observer within an event horizon. $\endgroup$
    – ProfRob
    Jan 16, 2019 at 8:05
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    $\begingroup$ Possible duplicate of Why can't light escape from a black hole? (though all current answers there are awful). $\endgroup$
    – ProfRob
    Jan 16, 2019 at 18:13

3 Answers 3

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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.

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  • $\begingroup$ So Florin, at what point during my thought experiment does spacetime tie itself into a knot? $\endgroup$
    – John Duffield
    Jan 14, 2019 at 21:57
  • $\begingroup$ @JohnDuffield - When it collapses into a black hole. $\endgroup$
    – Florin Andrei
    Jan 15, 2019 at 1:05
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    $\begingroup$ @JohnDuffield - You may want to start from the basics before you read more advanced literature, there's less chance of confusion this way. $\endgroup$
    – Florin Andrei
    Jan 15, 2019 at 18:23
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    $\begingroup$ @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. $\endgroup$
    – Florin Andrei
    Jan 15, 2019 at 21:20
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    $\begingroup$ @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. 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. $\endgroup$
    – John Duffield
    Jan 15, 2019 at 21:28
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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.

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).

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  • $\begingroup$ 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. $\endgroup$
    – John Duffield
    Jan 16, 2019 at 16:50
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    $\begingroup$ @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. $\endgroup$
    – ProfRob
    Jan 16, 2019 at 18:10
  • $\begingroup$ 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. $\endgroup$
    – John Duffield
    Jan 17, 2019 at 14:43
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    $\begingroup$ @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. physics.stackexchange.com/questions/28297/… $\endgroup$
    – ProfRob
    Jan 17, 2019 at 20:35
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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

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. That perfectly explains the energy change and for the math part, you can look here.

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    $\begingroup$ 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. $\endgroup$
    – Mark Olson
    Jan 14, 2019 at 20:03
  • $\begingroup$ 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.) $\endgroup$
    – Mark Olson
    Jan 14, 2019 at 20:28
  • $\begingroup$ I'm sorry Reign, but the energy of the photon doesn't decrease. Take a look at page 149 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. $\endgroup$
    – John Duffield
    Jan 14, 2019 at 22:01
  • $\begingroup$ @JohnDuffield I edited my post, Its better now $\endgroup$
    – seVenVo1d
    Jan 15, 2019 at 5:07
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    $\begingroup$ @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. $\endgroup$
    – Steve Linton
    Jan 16, 2019 at 9:32

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