# Why doesn't the vertical light beam get out of a black hole?

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1. Oct 14, 2015

### John Duffield

For this we need a thought experiment: imagine you're on a gedanken planet manning a gedanken laser cannon, and it's pointing straight up. The light doesn't curve round, or slow down as it ascends, or fall down. It goes straight up. Now let's keep you safe in a bubble of artistic licence, and make a few little changes.

Let's make the planet denser and more massive. The laser cannon is still pointing straight up. The light still doesn't curve round, or slow down as it ascends, or fall down. No problem.

Then let's make the planet even denser and more massive. The light still doesn't curve round, or slow down as it ascends, or fall down. Still no problem.

Then let's make the planet denser still and more massive, so much so that we take it to the limit. Let's make that planet so dense and so massive that it's a black hole. At no point did the light ever curve round, or slow down as it ascends, or fall down. But Houston, we have a problem. Because that black hole is black, because the light can't get out. So the \$64,000 dollar question is this:

Why can't the light get out?

PS: If you're tempted to invoke redshift, remember that when you send a 511keV photon into a black hole, the black hole mass increases by 511keV/c² only. The photon E=hf energy didn't actually increase, conservation of energy applies. And if you're tempted to invoke the waterfall analogy, please note that Einstein never ever modelled a gravitational field as a place where space is falling down.

2. Oct 14, 2015

### phinds

For the same reason that you can throw a rock off of an asteroid hard enough to let it escape the asteroid's gravity field, but you can't throw one off of the Earth and get it to escape the Earth's gravity field because the field is too strong. As soon as your photon is created, it heads backwards towards the singularity.

To expand on that just slightly, ALL world-lines inside a black hole point to the singularity.

3. Oct 14, 2015

### rootone

Because it would have to travel faster than light

4. Oct 14, 2015

### SlowThinker

The waterfall analogy seems to explain why the light can't get out. Einstein was a genius, but only had one life. If he didn't do something, it doesn't mean it's wrong.
What bugs me, perhaps the light can't escape to infinity, but it can show above the horizon? As in, you can throw a stone to *some* height, even if it doesn't escape Earth. The waterfall analogy would explain why the light can't get out even in this case - it's falling from the start.

5. Oct 14, 2015

### Grinkle

I think the OP is having difficulty picturing a world line that at once originates pointing away and perpendicular to the surface of a sphere (the singularity) and also curves back to the sphere.

I have the same difficulty. A photon emitted at some non-perpendicular angle to what one (at least me) pictures as the sphere of the singularity (I know its not really a sphere, its not actually describable, its a singularity) is easy to picture falling back. Its harder to picture a photon going straight up, never moving slower than the speed of light, and still somehow falling straight back down.

I know that I am relying way too much on the throwing-a-rock analogy, and where that analogy breaks down is that a rock is allowed to travel at any speed <c, including zero, relative to any observer. Light is not allowed to do that, so its hard to use that analogy to picture this situation. At least for me.

Edit -

Perhaps what I need to picture is that all perpendicular world lines point inward. As a world lines get closer and closer to perpendicular, they start to look like smaller and smaller radius loops until finally at the limit of perpendicular emission a photon is never actually emitted at all. So a photon can never be emitted absolutely perpendicular to a singularity.

That also can't be right because a singularity has no radius, so defining perpendicular seems problematic, but thinking of it that way at least gives me a picture.

6. Oct 14, 2015

### DaveC426913

As you make the planet denser and denser, time dilation due to gravity creeps up. The universe ages faster outside than inside. At some point, you are in a gravity well so strong that, while you see the light rise in a split second, the universe outside ages and dies before the light reaches the event horizon.

7. Oct 14, 2015

### Grinkle

Dave -

For me, that is not a very satisfying response. The question invokes an image of an observer who can see what happens to the photon, from emission through absorption by the singularity. Having the answer be that there is no such observer is disappointing. I suppose an observer inside the SR of the singularity would also never see the photon fall back to the singularity in finite time from their perspective, so perhaps that is the sadly boring answer?

8. Oct 14, 2015

### DaveC426913

I'm not suggesting the photon falls back to the singularity, I'm suggesting it continues to the event horizon, but that, to an outside observer, it takes an infinite time to get there.

9. Oct 14, 2015

### Grinkle

Its the same for an inside observer, right? I think an observer inside an SR can never see a photon reach the SR, they just see it getting more and more red-shifted.

10. Oct 14, 2015

### DaveC426913

Well, in fact, they won't see anything at all. You cannot see a photon unless you intercept it.

11. Oct 14, 2015

### Grinkle

Sigh. Can you make it MORE boring? ;-) You are right, of course.

12. Oct 14, 2015

### SlowThinker

I feel there is something wrong with this idea of infinite time dilation.
For example, the black hole should evaporate before the end of the universe, so perhaps the singularity doesn't even have enough time to form? This idea seems to be posted quite often, but the knowledgeable people around always say that the singularity does form, and you fall into it very quickly.
I suggest we wait for a mentor before going too wild. Sorry for occupying Peter for so long

13. Oct 14, 2015

### DaveC426913

All of that is true, so how is it a problem?

14. Oct 14, 2015

### SlowThinker

It means that saying that the beam would get out after infinite time, is wrong.
Something else is preventing the escape.

15. Oct 14, 2015

### rootone

Yes a lot of people are very uncomfortable about that, and personally I am in the camp with those who see the 'singularity' as something not physically real, just an indication that our best theories while impressive enough, have limits to their accuracy.

16. Oct 14, 2015

### DaveC426913

Wait. Why?

You have to do is ask what an external observer would see, and separately, what an internal observer would see.

In the split second that the internal observer sees his laser fire, the outside universe has aged and died. Neither observer ever saw the light escape.

17. Oct 14, 2015

### SlowThinker

For starters, the inside of a black hole would be a really busy place. Imagine everything that ever fell to that BH, compressed into, say, 20 minutes at most. I'm pretty sure that's not what happens.

Again, the BH evaporates before the universe ends. So how can the ray not hit the horizon, if that eventually shrinks to 0?

18. Oct 14, 2015

### DaveC426913

Yeah, it is. And highly blue-shifted, hard radiation.

Yeah, but the black hole evaporating is certainly a condition in which the thought experiment breaks down. We're talking about the universe ending. The 'light can't escape' claim was not really meant to apply to that edge condition.
Remember, it is a physically impossible scenario, since no observer with any mass could hover inside the event horizon.

19. Oct 14, 2015

### Staff: Mentor

You can't do this. A black hole is not a static object like a planet, that just happens to have an escape velocity of $c$ from its surface. And a black hole's horizon is not a static surface where light just happens to stay put.

If you work out the math using GR, it turns out that it is impossible to have a static object (i.e., an object whose radius does not change with time), like a planet, with a radius smaller than 9/8 of the Schwarzschild radius corresponding to its mass (which is $r_s = GM / c^2$ in conventional units). (This is known as Buchdahl's Theorem.) So at some point in your process of making the planet smaller and denser, it will become unstable and collapse. A black hole will be the object that the collapse process leaves behind.

It is possible in principle, if you have a really technologically advanced spaceship (one far beyond our current level of technology), to "hover" as close to a black hole's horizon as you like; but even such a spaceship can't "hover" exactly at the horizon. Any ordinary object (i.e., anything with nonzero rest mass--note that light has zero rest mass) that is at the horizon can't stay there; it must be falling inward. So asking what things would look like to someone "hovering" at the hole's horizon, or why light emitted by such a person can't get out, is meaningless, because there can't be any such person; it's physically impossible.

The correct way of thinking about the hole's horizon is this: it is an outgoing null surface, i.e., a surface made up of radially outgoing light rays. Anyone falling through the horizon will see these light rays come at them and pass them as they fall. So in fact, light at the horizon is moving outward--relative to the only possible observers that can exist at the horizon, who must be falling inward. Locally, the light at the horizon is not staying in the same place; it is moving radially outward, just as one would expect.

But because of the curvature of spacetime, this outward moving light stays at the same radial coordinate $r = r_s$ forever. The horizon is a global phenomenon, not a local one; it's not something that you can tell is there just by local observations of the light there. Those local observations will show the light moving radially outward at $c$. But globally, when you look at the entire spacetime, you will see that it is curved in such a way as to keep that light at $r = r_s$ forever.

20. Oct 14, 2015

### Staff: Mentor

Light emitted from inside the horizon will do this, yes. But even so, this light will still be moving radially outward at $c$, relative to local observers (who will be falling inward, from a global perspective, faster than the light is).

21. Oct 14, 2015

### Staff: Mentor

No. Light emitted at the horizon stays at the horizon forever. It doesn't go up, stop, and then fall back.

The key thing to remember is that spacetime at and inside the horizon is not static. There are no observers who can "hover" at a constant radial coordinate. So it's simply not correct to think of events there the way you would think of things on Earth, where things that are thrown upward at less than escape velocity move upward, stop, and then fall back. Spacetime at and inside the horizon simply doesn't work that way.

22. Oct 14, 2015

### Staff: Mentor

No, this is not correct. The spacetime at and inside the horizon is not static; the concept of a "gravity well", which requires a static spacetime, does not apply there. (It does apply outside the horizon, but outside the horizon light can escape, though it can take a long time to do so.)

This isn't correct, either. As phinds correctly pointed out, a photon emitted inside the horizon falls into the singularity.

Also, since spacetime at and inside the horizon is not static, it can't be described using the Schwarzschild coordinates that are used in the region outside the horizon to assign an infinite time coordinate to events on the horizon. So there is no meaning to the question of how long it takes light to travel inside the horizon "to an outside observer".

23. Oct 14, 2015

### SlowThinker

Would not that mean that as a star gets sucked into the BH, it is turned into a superdense neutron gas? Otherwise all those stars simply would not fit in. I've never heard of such a scenario.

I think it does apply. If you could send any messages out of a black hole, the very idea of an event horizon would be kind of non-sensical. If it was possible in quantum theory, then the current debate of firewalls and information paradoxes would look very differently.

The waterfall model seems to offer a very intuitive, paradox-less explanation of what's going on. Maybe it's even correct

24. Oct 14, 2015

### Staff: Mentor

Yes, it does. This has been discussed in a number of previous threads; see, for example, these:

https://www.physicsforums.com/threa...gularity-evaporate-before-it-can-form.683755/

In Hawking's original model of an evaporating black hole, the ray hits the singularity before the horizon shrinks to zero and the hole vanishes.

There is a lot of contention still in this area; most physicsts seem to think that Hawking's original model is not correct, but there is no good consensus on what should replace it. In some candidate models, there isn't ever a true event horizon at all; there are only "apparent horizons", surfaces where light, locally, is trapped for some period of time. But these apparent horizons eventually disappear and the light gets out again, and there are never any singularities or any regions from which light can never escape at all. (This doesn't mean it's not risky to fall into these regions, though; strange quantum gravity stuff is going on there that probably will destroy your body if you ever fall in, and it won't be much consolation to you that the light from the process of your destruction will eventually escape again.)

25. Oct 14, 2015

### Staff: Mentor

Not inside the black hole, if you're free-falling towards the singularity. Such observers see radiation falling in from the outside universe redshifted, and the redshift factor increase as they fall towards the singularity.

It's possible to see radiation from the outside universe blueshifted inside the horizon, if you accelerate very hard in an outward radial direction. But that will also greatly decrease the proper time it takes by your clock for you to hit the singularity.