Why can't you escape a black hole?

Question

I understand that the event horizon of a black hole forms at the radius from the singularity where the escape velocity is $c$. But it's also true that you don't have to go escape velocity to escape an object if you can maintain some kind of thrust. You could escape the earth at 1 km/h if you could maintain the proper amount of thrust for enough time. So if you pass just beneath the event horizon, shouldn't you be able to thrust your way back out, despite the $>c$ escape velocity? Or does this restriction have to do solely with relativistic effects (time stopping for outside observers at the event horizon)?

Answer 1

@Florin is absolutely right, but sometimes a picture is worth a thousand words. This website has multiple pictures and explanations about how the future light cone starts to point only to the inside of the black hole once you pass the event horizon.

Here is one of the images:

Time is vertical, the cylinder represents the event horizon and the cones are the future light cones for the observer as they fall into the black hole.

Note that even if the observer could instantaneously accelerate to almost the speed of light they would still be confined to the future light cone. So once the event horizon is crossed they cannot escape.

UPDATE: As @Florin says in a comment, rotating black holes are even stranger than than the non rotating Schwarzchild black hole described above. In particular there is a region outside of the event horizon, called the ergosphere where space-time itself is dragged around the black hole at faster than the speed of light (relative to distant stars). It is possible to enter the ergosphere and still escape to infinity. In fact some of the black holes rotational energy can be extracted in this region and may be source of energy for gamma ray bursts.

I'd also like to point out that there is no local experiment that can be performed to determine when the observer has crossed the event horizon. It will "feel" perfectly normal - it is only the future destination of light rays that changes when the horizon is crossed and that cannot be determined locally. In fact the "normal Newtonian" concept of the "force" of gravity doesn't have to be particularly strong at the horizon. A more massive black hole has a lower event horizon "gravity force". A hand waving way of seeing this is that the Newtonian force is proportional to $\frac{1}{R^2}$ but the event horizon radius is proportional to the mass, $M$; so the surface force is proportional to $\frac{1}{M}$. See this Wikipedia article for a more rigorous discussion of what it might even mean to talk about the concept of "force" in General Relativity.

Answer 2

I wish those who popularize science would stop talking about the speed of light. It just confuses people.

Forget about speed. Think in terms of energy. You'd need an infinite amount of energy to get out of it. Infinite. Meaning, it's like dealing with the Mafia: no matter how much you're spending, it's still not enough.

Think in terms of topology. From the inside of the event horizon, there are no possible trajectories to the outside. This is a purely geometric issue; spacetime is so mangled that any trajectory you could conceivably draw, starting at your current position, goes only to the inside of the event horizon.

Moreover, any trajectory inside the event horizon is only pointing down. That's right, everywhere you look, you look down towards the center.

The 3 spacelike dimensions and 1 timelike dimension of normal space-time now become 3 timelike and 1 spacelike dimension. All those 3 timelike dimensions terminate at the central singularity. Which means, once inside, you can only go down.

That's how seriously broken spacetime is. It's not a matter of speed, heck even energy doesn't matter, it's that space and time themselves are tied into a big screwball knot that makes no sense and is seemingly designed to suck you in.

Physicists refer to this as the collapse of space-time geometry.

Answer 3

I completely agree with @FrankH's answer. I just want to emphasize the underlying misconception many authors give when talking about escape velocities from black holes. One can have a "black hole" of sorts in Newtonian physics, sans GR, and indeed this has been known since the 18th century (see the Wikipedia page, for instance). All you do is find the radius outside a mass M where the escape velocity is equal to the speed of light.

Unfortunately, this radius corresponds exactly to the Schwarzschild radius, $2GM/c^2$, leading many to think the Newtonian arguments apply somewhat. One key difference is what is meant by "escape." The escape velocity is the speed you'd have to go to head off to infinity with no further thrust. With any less velocity, you could move a certain nonzero distance away from the surface of a Newtonian black hole before being pulled back in. Given a source of thrust, the argument in the original question shows that you could always escape such an object. With GR, things are different, and you are never allowed to go any amount above the event horizon, no matter how much rocket fuel you have on hand.

Answer 4

It is actually possible to get out of the horizon. To get out one only have to reach speed higher than c.

But no object that bears information can reach that speed, this would violate casualty. So only things that bear no information can get out from inside the horizon.

One such thing is the Hawking radiation (not only photons but also particles) which can be viewed as quantum tunneling out of the BH. Note that in all cases of tunelling (say at nucleus decay) the emitted particle reaches higher than c speed. This does not violate casualty because the process is probablistic and as such cannot be used for iunformation transfer.

Answer 5

Suppose we calculate the thrust you need to hover at a fixed distance from the event horizon. The reason for this is that if we can hover at a fixed distance from the horizon then to move away from the black hole we just need to increase the thrust by some amount.

In Newtonian gravity the thrust we need to hover is just equal to the gravitational force. So if our spaceship has a mass $m$ and the black hole mass is $M$ the thrust we need to hover at a distance $r$ is:

$$ F_{\textrm{Newt}} = \frac{GMm}{r^2} \tag{1}$$

If the mass of the black hole $M$ is large and $r$ is small this force can be very large, but for $r \gt 0$ it is never infinite. So as long as our rocket motor is powerful enough we can always hover, and therefore by increasing the thrust slightly we can always move away i.e. we can escape the black hole.

The problem is that the Newtonian equation (1) is an approximation and when we do the calculation using general relativity we find that the thrust required to hover becomes:

$$ F_{\textrm{GR}} = \frac{GMm}{r^2}\frac{1}{\sqrt{1-\frac{r_s}{r}}} \tag{2} $$

where $r_s$ is the event horizon radius. I won't go through the calculation here. If you're interested it is described in detail in the question What is the weight equation through general relativity?

And now we can see the problem. When $r = r_s$, i.e. if you try to hover at the event horizon, equation (2) tells us the hover force becomes infinite. Since no rocket engine is infinitely powerful it is impossible to hover at the horizon, and if you can't even hover you certainly can't move away to escape the black hole.

Answer 6

When scientists say the escape speed of a black hole is greater than the speed of light, they mean that if you leave the surface of black hole at a speed which is near the speed of light, and during the trip you do not get extra boost of speed, then eventually you will fall back. That is absolutely true. But if you get a speed boost during the trip, you can maintain your speed to be near the speed of light, then eventually, you will move to a orbit whose escape speed is lower than the speed of light, then you can escape the black hole. Thus the myth of "nothing can escape black hole" is wrong.