Why doesn't the light get out?

You're standing on a gedanken planet holding a laser pointer straight up. The light doesn't curve round, or slow down as it ascends, or fall down. It goes straight up. Now I wave my magic gedanken wand and make the planet denser and more massive. The light still doesn't curve round, or slow down as it ascends, or fall down. I make the planet even denser and more massive. The light still doesn't curve round, or slow down as it ascends, or fall down. I make the planet even denser and more massive, and take it to the limit such that it's a black hole. At no point did the light ever curve round, or slow down as it ascends, or fall down. So why doesn't the light get out?

• I thinm this is a duplicate of Speed of light in a gravitational field?. I won't flag it as such because that would immediately close your question, but I thnk you should have a read through my answer to the question I've linked. – John Rennie Mar 25 '15 at 9:40
• You might also want to have look at If you shoot a light beam behind the event horizon of a black hole, what happens to the light? and for a more technical discussion Would the inside of a black hole be like a giant mirror?. – John Rennie Mar 25 '15 at 9:42
• @JohnRennie well if it's really a duplicate then there's no reason for answers to be posted here, so it should be closed (though I like to say "marked as duplicate", because it's not bad in the way some closed questions are). In any case I somewhat agree - it seems at least pretty close to being a duplicate. – David Z Mar 25 '15 at 9:45
• @JohnRennie The answer I was going to give turned out to be pretty near to yours to "Would the inside of a black hole ...." so I think I read the question like you do: the KS co-ordinates with their null lines at $\pm45^\circ$ are a good way to visualise the laser pointer, and the GP river model is a good explanation too. The only thing i'd add is that an outside observer would see the light as more and more red shifted. It would "vanish" as the photon energy becomes nought as the horizon forms. Now, the interesting question is, I'm sure one (someone if not I) could come up with a toy metric.. – WetSavannaAnimal Mar 25 '15 at 9:56
• @JohnRennie ... reproducing the OPs thought experiment, i.e. with the dynamical mass increase. Obviously, this would violate $\nabla \cdot T=0$ where the mass gathers, but I mean study the solution outside this region. – WetSavannaAnimal Mar 25 '15 at 9:58

2 Answers

The problem with you argument is that you are standing on the planet! When you make it denser and denser and eventually turn it into a black hole, you would still be standing on it, ie be outside the event horizon, so as you are saying correctly the light will escape.

If you would like to think about being inside the event horizon then it's wrong to imagine yourself standing on anything, because there can be no static observer inside an event horizon. You would be falling instead.

The correct analogy would then be this: Imagine falling from a very high mountain and also imagine throwing a heavy rock upwards on your way down. To you, it might look like the rock is moving upwards and the distance between you and the rock will be increasing, but to an observer on the ground the rock is also moving downwards, just at a smaller initial velocity (because of your initial push) and even though you are both moving in same direction, the distance between the two of you is increasing because you are moving at different speeds.

In the same way, light that you think you have emitted outwards when you have crossed a black hole horizon, is in reality also falling in right behind you.

• Thanks for responding Heterotic, but see this in John Rennie's answer to a related question: – John Duffield Mar 26 '15 at 14:34
• The obvious example of this is a black hole, where the speed of light falls as it approaches the event horizon and indeed slows to zero at the event horizon. When I ask this question, responses appear to be contradictory, so much so that some people prefer not to answer, and we can't thrash it out. – John Duffield Mar 26 '15 at 14:40

That's just strong equivalence principle. Standing still on a planet is getting accelerated upward. Light is going at the speed $c$ only for inertial observers, that is, for the ones free falling. It is a big misconception to claim that light doesn't "slow down" as it ascends, for you standing still on the planet. Where do you think that red shift comes from ? It comes from this effect, nothing else. So your ray of light will eventually become darker and darker, and even if by ascending it won't "slow down" (the more it climbs into the sky, the less it feels gravitation, so the less it is falling downward, that is the quickest it escapes from you on the ground), it can happen that at some point it just doesn't ascend at all into the sky after it goes out of your laser beam and fall! (obviously, with you before)

edit: for you to understand better what I mean, you have to understand that if you're stationary ahead of a planet (that is, at a constant radius) without rotation, then you're actually getting accelerated a lot upward, otherwise you would simply fall. This is the same on the ground. The ground is accelerating you a lot upward unless it cannot resist to turn into a black hole. Hence, to answer your comment, inside a black hole there is no stationary point, because it would mean that it is possible to get accelerated more than you physically can, so even light has to fall. This in some sense means that there is some maximum acceleration (actually, a maximum gravitational field)