# B Light escaping a black hole

1. Jul 31, 2016

### DiracPool

So I've heard that light escaping a massive body such as the Earth becomes red-shifted the further its gets away from Earth's gravity but always travels at exactly the speed of light (is this correct?)

My question is, though, let's say that we're sitting around the event horizon of a black hole. We have light waves right at the border there trying to escape the grips of the black hole's gravity. Some make it and some don't. Say we have two photons very close together right at the horizon. Both are struggling to get out of the black hole's grip. One makes it and one doesn't. What happens there? Were they both traveling at the speed of c and then one just stops moving and becomes part of the black hole? Do neither of these photons separating their trajectories just around the event horizon ever change their speeds relative to one another?

2. Jul 31, 2016

### mathman

All photons move at c in a vacuum. I won't comment about struggling photons.

3. Jul 31, 2016

### jbriggs444

The event horizon is not a place. It is a surface that moves at the speed of light relative to any nearby inertial frame. A pulse of light (please avoid using the term "photon") that is ahead of the horizon can pull ahead due to the local curvature of space-time. A pulse of light that is behind will fall farther behind due to the local curvature of space time. Even though both pulses are moving at the speed of light, their trajectories will separate.

4. Jul 31, 2016

### t_r_theta_phi

Light does not always appear to travel at c in curved space-time. To an observer far from the black hole, a beam of light appears to travel slower near the black hole, and its observed velocity approaches zero at the event horizon. So light escaping from just outside the horizon will appear so start almost at rest and then accelerate out, eventually reaching c far away from the hole. The light that starts just below the horizon cannot be observed.

Edit: Locally, light always has velocity c, so if you were at the same point in spacetime where the beam was emitted, it would appear to travel at c.

5. Jul 31, 2016

### Staff: Mentor

All of the light that is emitted from a point outside the event horizon, not matter how close, escapes to infinity. All of the light that is emitted from a point at or inside the event horizon does not.
Photons are not what you think they are, so the question as you've phrased it is pretty much meaningless. However, we can fix that by saying "flash of light" instead of "photon" and then the question will make sense.... So from here on, we'll assume that you're asking about flashes of light instead of photons.

If the two flashes of light are emitted anywhere in the gravitational field of any massive body (it doesn't have to be a black hole, and even if it is it doesn't matter it's anywhere near the event horizon) they will move apart. They'll both move at speed $c$ along their respective trajectories, but because of the curvature of spacetime those trajectories will diverge. As an analogy, we could consider two automobiles very near to one another and near the north pole. Both start driving at 100 km/hr due south, but because of the curvature of the earth their paths will diverge and by the time they get to the equator they will be separated by many thousands of kilometers... even though they were moving at the exact same speed, 100 km/hr, the whole time. The same thing happens with your two flashes of light - both are moving at $c$ but because of the curvature they can end up in very different places.

Last edited by a moderator: Aug 1, 2016
6. Aug 1, 2016

### dragoo

You are sitting around the event horizont. This means that you are moving outwards with speed of light

7. Aug 1, 2016

### Orodruin

Staff Emeritus
Only light that is emitted in the radial outward direction.
Consider light that is emitted in the direction of the black hole...

You are here talking about the coordinate speed of light. This is an arbitrary and coordinate dependent quantity without physical significance. A far away observer will not measure the speed of the light pulse to be the coordinate speed. In fact it is unclear how you would measure the speed of the pulse in the first place. Depending on which method you select you might get different results due to the geometry of the space-time. (Although parallel transporting the pulse's 4-momentum to the observer will always give c.)

8. Aug 1, 2016

### Staff: Mentor

True.... I must confess that I interpreted "struggling to escape" as implying that it had been emitted in that direction.

9. Aug 1, 2016

### DiracPool

That's interesting. It's kind of like the metaphor or visualization of the black hole as sink hole in the ocean and your outward "swimming" (or escape) velocity must exceed the inward velocity of the water or else you'll get pulled down the sink hole. So, even though just outside the "event horizon" of the sinkhole you appear to be swimming very slowly to, say, an observer watching on the shore (measuring the "coordinate speed?), you are actually swimming very close to the escape velocity of the sink hole, which is much greater. Is this the way it works at the event horizon of a black hole, where the escape velocity is the speed of light?

To put it in a more familiar frame, say we replaced the Earth with a black hole whose event horizon diameter was that of the Earths. Now someone on the International Space Station (ISS) shines a flashlight (or laser) at the moon. Assuming, of course, that the ISS is in-between the black hole and the moon and 200 miles off the surface of the event horizon, how long will the light beam take to reach the moon? Is it the same 1.5 or so seconds it takes right now, only with a much higher redshift, or will the beam of light take a longer amount of time?

10. Aug 1, 2016

### Orodruin

Staff Emeritus
This is not a very well formulated question. It depends on your simultaneity convention and who is doing the measurement.

11. Aug 1, 2016

### DiracPool

I don't know what a simultaneity convention is and let's just say that the person doing the measurement has the perspective of the situation you see in the gif.

12. Aug 1, 2016

### Staff: Mentor

1.5 seconds according to who? You are using that phrase "how long will it take?" without stopping to think about what it means....

Somewhere in the universe there is a guy holding a clock. A light flash is emitted from the space station at the same time that his clock reads $T_1$. That flash of light reaches the moon at the same time that his clock reads $T_2$. You are asking whether $T_2-T_1$ is equal to 1.5 seconds, and that question cannot be answered until you've explained what you mean by "at the same time" - there are a number of intuitively reasonable definitions (called "simultaneity conventions"), but they don't all lead to the same values of $T_1$ and $T_2$ and there is no reason to think that one them is more correct than another.

13. Aug 1, 2016

### DiracPool

As I just quoted above, lets say our observation post is the one identified in the gif I posted above. We see the light pulse moving from the Earth to moon in a 1.5 second interval. What would we see if the Earth instead were a black hole? Can we not compare these two apples to apples? If not, that's fine, but exactly why not? Does it have to do with the simultaneity conventions? Also, I understand that we cannot see the light beam directly as is displayed in the gif, we can only see some reflection of light advanced in our direction, but I'm trying to not overcomplicate the question. If that overcomplication is necessary to answer the question, though, then have at it.

14. Aug 1, 2016

### Orodruin

Staff Emeritus
As we have already noted, it is not that simple. Your image does not reflect the geometry of the space time properly. It is still undefined what you mean by "moving from A to B in a 1.5 second interval". Unless you specify what you mean by this, there is no way to answer your question.

The problem is that, in doing so, you are oversimplifying the question to an extent that it is not well defined what the question is.

It is not an overcomplication, it is a necessary part of the question.

15. Aug 1, 2016

### DiracPool

I mean a beam (or pulse) of light moving from A, a flashlight (or laser) on the ISS, to B, the moon, in 1.5 seconds. Isn't that (roughly) how long it would take if we were to perform the measurement today?

16. Aug 1, 2016

### jbriggs444

How will you perform the measurement today? If you stand on the observation station today and hold a stopwatch in your hand, how will you know when to start it and how will you know when to stop it?

17. Aug 1, 2016

### Staff: Mentor

They cannot be compared apples-to-apples. The gif is drawn assuming a flat spacetime in which gravitational effects can be ignored, and you are asking about a curved spacetime in which they cannot be ignored. Thus, considering the effects of curvature is not overcomplicating the problem, it is exactly the heart of the problem.

There is a more rigorous way of stating your thought experiment, and when we try it the apples-to-apples problem will become more apparent. We set the ISS in orbit 200 miles above the surface of the earth; we set the moon in orbit 250,000 from the earth; we position the observation post where it goes; we do our light travel time measurement using some simultaneity convention. Then we replace the earth with a black hole without disturbing the moon, the ISS, or the observation post; and we repeat our measurement. But there's a catch: There is no spacetime in which the earth is REPLACED with a black hole - there's no such solution to the Einstein field equations. So we've just assumed something impossible, and we're in the same logical swamp that we'd be in if we had assumed that we were moving faster than light, trying to answer the question "What do the laws of physics say about a situation in which the laws of physics don't apply".

So we have to try a different approach, and that won't be apples-to-apples because we no longer have a scenario in which we replace the earth with a black hole while not changing anything else. We can find a convenient earth-sized black hole, put the observation station, the ISS, and the moon in equivalent positions around the black hole, and do our measurements. But what can we possibly mean by "equivalent"? What does it mean to say things are in "the same place" in a curved spacetime as they are in a completely different flat spacetime? This can't be an apples-to-apples comparison because we've changed the underlying geometry so that none of the distances involved have the same meaning.

18. Aug 1, 2016

### DiracPool

Ok, so I'm taking it that the overcomplication is necessary. Is it not possible to address the abovementioned scenario, gedankenexperiment-style, just on the basic setup I offered above?

I will give it a shot though. As far as the 1.5 seconds it takes a light beam to get from the Earth to moon, I'm just going to go with the default measurement of the speed of light in a vacuum, seeing as the gravitational potentials in this scenario are relatively negligible in modifying that velocity or even the redshift. So the question as to how to know when to start and stop the stopwatch is superfluous in that case.

As far as the scenario of where we transform the Earth into a black hole....Let's try this:

Let's say we're standing on the observation station (OS) as depicted in the gif above. Our OS is situated as such that the ISS, the moon, and our OS form an equilateral triangle. On the ISS, we have a laser that emits a bifurcated beam so that one beam heads out toward the moon, and the other heads out toward the OS. As soon as I get the signal on the OS from the ISS, I know that the signal has also reached the moon simultaneously, seeing as the OS and moon are situated radially from the ISS (which is assumed to represent the defacto singularity of the black hole outside the event horizon).

The moon was instructed to send a beam to the OS once it got the signal from the ISS, so it proceeds to do so. At the exact same time, the OS simultaneously sends out a beam to both the ISS and the moon, each of which are to be reflected back instantly to the OS. Which beam hits the OS first on that return? The moon one or the ISS one? That's the question. If the speed of light were constant in this scenario, they would reach the OS at the same time, which, in the case of the real Earth and moon, they would. But what about in the case where the Earth is turned into a black hole?

Now that I look at it, it does seem as if, in also the black hole case, the two beams would reach the OS at the same time. And although intuitively one might think that the light beam would have to be traveling at a slower speed than in the real Earth-moon case, there's not really a way to test/measure this, since no matter where you place your OS, you're going to get the same result. You're going to get the speed of light as a constant. Will it be 3x10^8 m/s also in the black hole scenario? I don't know could be different, but it could be the same too since I guess we are assuming the OS post time clocks are slowed in that scenario.

19. Aug 1, 2016

### Orodruin

Staff Emeritus
No.

The space-time is curved. You will have to define what you mean by an equilateral triangle. Note that the spatial length between two observers is going to be dependent on your simultaneity convention. You are implicitly assuming that you can impose a spatial grid on top of everything.

In general, this topic goes quite far beyond B-level.

20. Aug 1, 2016

### DiracPool

Yeah, that's a good point, I guess you might be able to get away with that in the real Earth scenario but not the black hole scenario.

A similar sentiment to the one above. I knew things were going to go haywire once I replaced the Earth with the black hole, as in the moon and the OS were going to be sucked in immediately to the black hole. I guess what I was trying to model in the scenario was one where we did the measurement on the black hole transformation instantly (riding the gravitational wave) before the curvature situation changed too radically. But that was a big hope. What I think may have been more informative to my initial query, though, is that it is impossible to compare apples to apples because it seems that everything changes in a symmetric way when you go from say the Earth to the black hole scenario, and everywhere in between. So it's almost imposssible to make a distinction, and therefore the speed of light will always remain constant in the scenarios I presented. I mean, in Orodruin's discussion of the equilateral triangle, is the warping of the triangle going to keep at least the ratio's of the lengths of the 3 sides equal if not the actual lengths themselves regardless of the mass of the center of the Earth object? Keeping all other things equal? I guess relative to a flat or Minkowski space.