Escaping a black hole

Factual Questions
1

I’m having trouble with the idea of nothing being able to escape from a black hole (and to avoid unnecessary complications, let’s just talk about classical black holes, and sufficiently large ones so tidal forces wouldn’t rip you apart just inside the Schwarzchild radius).

At the Shwarzchild radius, the escape velocity reaches the speed of light, and no light can escape. An explanation I’ve seen many places is that since the escape velocity inside a black hole is greater than the speed of light nothing else can escape either. To me, that doesn’t necessarily follow.

As has been mentioned in at least another thread you don’t need to reach escape velocity to get as far away from an object as you wish as long as you are constantly supplying energy to avoid falling back (think of a space elevator extending out from earth). Since the gravitational force only becomes infinite at the centre of the black hole, why can’t I go just inside the Schwarzchild radius and then “push” myself out?

There is obviously a flaw in my reasoning since if I understand the time dilation correctly, from the point of view of an outside observer I’d come out sometime after eternity, but where is the flaw from the point of view of someone actually going in and trying to come out?

2

A seat-of-the-pants way of explaining it is this: when you take an object and apply a force to it, the acceleration that you get isn’t just given by F = ma, but rather by the more complicated formula

F = m a / (1 - v[sup]2[/sup]/c[sup]2[/sup])[sup]3/2[/sup],

where c is the speed of light, v is the velocity of the object, and a is its acceleration.

The important thing about this formula is that as v gets closer and closer to c, the amount of force needed to accelerate it further gets bigger and bigger, and eventually diverges. So you can’t apply a force to an object that would make it go faster than light. That means that if you’re standing just inside the Schwarzchild radius of a black hole, you would need an infinite “push” to be able to push yourself out of the black hole as you describe, since it would require travelling faster than light to do so.

3

Yes, but I’m not asking to go faster than the speed of light, just to apply continual force (i.e. not bothering to try for escape velocity) to “creep” out. I still can’t see the fallacy in my argument (even though my comment about an outside observer makes it pretty obvious that there is one).

Am I not correct that gravity does not actually affect the speed of light (which is c in a vacuum regardless of anything else going on) but that it’s more a case that light doesn’t have the energy to escape (i.e. it gets redshifted to zero)? In that case I don’t see how your formula applies.

4

Hmm… let me take a crack at this.

F= [(ma)/x], where x is (1 - v2/c2)3/2.

Let’s rewrite this to make a the subject, therefore a=Fx/m

as v increases, x becomes closer to 0. This is because the speed limit is c, and v tends towards c. The most v you can get is c, and at that point v/c will be one, and 1-1=0.

Therefore, as V increases, x, and therefore Fx/m, tends toward zero. In order for there to be any a, you need to get a F that returns you a positive number when multiplied by zero. Aint gonna happen. What that means in english is that you need infinite F to get any acceleration at all. In other words, you can’t accelerate anything beyong the speed of light.

I think you get that part.

But the problem with your reasoning is that even outside the swartzchild radius, there is still a gravitational field. When people say that you could be “just inside” the radius and still feel nothing, what they are talking about is the gravitational differential, that is, gravity is different at your feet and at your head. This would result in a very stretchy you.

However, outside the radius, you are still being affected by the gravity field. Still escapable, but just barely. Therefore, your acceleration is already negative. It’s much like jumping when your falling elevator is about to hit the ground - you’re still accelerating downwards, just not so quickly. Since we’re talking about a g (think of it as a negative a) that’s so big that you need to be moving at the speed of light just to balance it to zero, you need to move faster than light to move out of it.

Think of satelites. They stay up there seemingly without any F, right? So if we just add a bit more a, we can move them a bit further out? Can’t we apply that to the spaceship that’s orbiting the black hole as just inside the radius?

The part about satelites sorta right, but it’s not correct to treat v as zero. v is actually pretty big, because they’re orbiting the earth. a doesn’t really care which direction you’re going, as long as you’re accelerating. So what happens with the satelites is that a is not 0, but some positive number. They’re just conveniently missing the earth by falling “to the side”. In order to move the satelite to some higher orbit, you need to increase their v by increasing their a for a while. The bigger their v to start with, the more F you need to increase their a.
Does that sorta clear things up?

5

It does not, in general, pay off to think of the geometry inside (or even near) a black hole as the normal, flat Euclidean geometry (with time tacked on) that you’re used to. Spacetime is distorted enough inside the Schwarzschild radius that, speaking somewhat loosely here, it is impossible to stay in the same place, let alone to creep out slowly. Even applying a very large force will not stop your approach to the singularity.

The word I had to use loosely in that explanation is “place,” which might give you an idea of just how strange things are inside a black hole. The way a relativist might describe it would be to say that once you pass the event horizon, the direction “toward the outside” is now into your past, not a spatial direction. Once you pass the horizon, you have to travel backwards in time, or at least faster than light, to get back out. This is straightforward to show using the Schwarzschild metric, but rather confusing to visualize, although it becomes easier once you learn to love Penrose diagrams like this one, which represents an astrophysical black hole of the type you might expect to encounter. Note that once you pass within the Schwarzschild radius (the shaded area) you’d have to travel at an angle more than 45° from vertical, faster than light, to get out again before hitting the singularity.

6

The easiest way to escape from a black hole? Turn around and tell it that it doesn’t exist. (Of course, the dark star wouldn’t be much fun either!)

7

The tipping of the Lightcone where you can no more avoid the singularity then you can avoid tomorrow is shown very clearly at this site on page 7. Unfortunately it’s a pdf but it explains very well why you can’t get out a hole no matter what you do.

http://ls.poly.edu/~jbain/philrel/philrellectures/13.BlackHoles1.pdf

Note: for anyone who might care I believe the Lightcones in the drawing on page 7 tip gradually as they approach the horizon rather than abruptly as they pass it is because the drawing uses Finkelstein coordinates, but Omphaloskeptic or Chronos would have to explain this.

8

His question is more simple than that. He’s asking why he couldn’t travel at 3 MPH to eventualy leave the black hole.

You’re operating under a misconception. You’re thinking of escape velocity as initial velocity. For example, fire an object upwards at 25,000 MPH. It will just barely leave earths atmosphere. This is the initial escape velocity needed to leave earth without applying power during the trip upwards. Of course you could have a rocket with an unlimited supply of fuel that went away from earth at a rate of 5 MPH. It never reaches this velocity, but it still escapes. So why couldn’t the same principle apply to the black hole? I never reach the speed of light, but I still make it out slowly.

In the black hole, you are falling inward always. When you apply force to try and leave this black hole, you would slow your decent into it down, but are still travelling inward. You’d need to reach the speed of light just to overcome the gravity pull inwards, so you could never leave the black hole at 3 MPH.

In other words, you cannot travel in a vector that takes you away from the singularity of the black hole. Doing so would require more velocity than light and would therefore be impossible.

9

I’m going to have to “meditate on” everybody’s answers for a while to completely wrap my head around them, but I basically get the picture. I have a few books that talk about light cones and similar topics; I just had not been able to see the relationship between them and “creeping” out of a black hole. That’s clearly the flaw in my reasoning that I knew had to be there.

Thanks! I’ll muse on the topic some more the next time I want to exercise my brain. :slight_smile:

10

Maybe this will help.

In other words inside the EH firing your rocket engine trying to get out of the hole is really an attempt to get back to yesterday. No matter what you do you’ve got a date with the singularity, just as in normal space you’ve inevitably got a date with tomorrow. This is all, of course, just a way of visualizing the bizarre space time curvature of a black hole.

http://www.faculty.iu-bremen.de/course/fall02/GeneralGeoAstro1/students/BlackHoles/Black%20holes%20and%20Schwartzschild%20geometry.htm

11

Even worse than that. The best possible thing you can do inside of a black hole is to turn off your thrusters entirely. Anything at all that you do, no matter which direction you point your thrusters, will hasten your demise, not postpone it. At least, according to your comoving reference frame, which is probably the one most significant to you.

Back to the OP, the “escape speed equals c” argument does give the correct value for the Schwartzschild radius, but it’s really only a coincidence that it does so. Basically, you’re making a couple of errors of a factor of 2 which conveniently cancel out (but there are some similar relativity calculations where they don’t cancel out). The actual calculation bears no particular resemblence to an escape speed calculation, aside from the final answer.

On the question of coordinates, there are a few coordinate systems which I can’t spell which do not go screwey at the event horizon. These are very useful for discussing an observer falling into the hole, since said infalling observer would not observe anything screwey at the crossing. For a small black hole (one of stellar mass), you would be spaghettified (which is most assuredly noticeable), but that’ll happen well outside of the horizon. For a very large hole, however, the point of spaghettification could be well inside the horizon. There’s no relation between the horizon and the spaghettification. In fact, it’s even possible to cross the event horizon of a black hole while your local spacetime is completely, 100% flat, and therefore have no detectable indications at all of your plight.

12

Is that a technical term? :stuck_out_tongue:

13

Actually, I recall seeing that term (spaghettification) in an article in Physical Review a decade or so back. It was referring to gravity gradients so large they would shear molecules asunder. So, I guess it is. So are “shakes”, as in “two shakes of a lambs tail”.

Physicists have some cockeyed notions about humor. They’re fun at cocktail parties, but only ones hosted by Larry Niven. Otherwise, people just back away while looking for heavy objects with which to defend themselves against the obviously raving lunatic who is going on about the non-local nature of entanglement and so forth.

Stranger

14

Well, I’ve had two astronomy professors who’ve used it, and it was in my textbook, so it may well be.

This may be a simple way to think about it: Doing your “move up at 5mph” solution, you’re not just pushing forward at 5mph, you’re also overcoming the pull of gravity. On earth, that means you’re pushing with enough force to accelerate at 10m/s[sup]2[/sup] just to stay where you are and keep from falling back to the ground. In the area of a black hole, that force just to keep from falling back gets much larger. Using the equations that others have posted above, when you cross the Schwarzschild radius, the force needed just to keep from falling becomes infinite.

15

I believe the black hole hypothesized (or confirmed?) to exist at the center of the Milky Way is large enough to produce this effect — or lack of effect, as the case may be.

The mass of this black hole is a whopping 3.6 million solar masses, giving it a Schwarzschild radius of about 11 million km. The tidal force on a human-sized body at that distance would be about 0.0015 m/s[sup]2[/sup], which is a tiny fraction of the “1 g” force you feel on the surface of the earth.

So, no spaghettification when passing through that monster’s event horizon. In fact, you’d hardly feel a thing.

(Someone feel free to check my math, or reasoning.)

16

I think I have a handle on the not escaping part, but I really don’t get this.

17

I have a question - if you are at a great distance from the EH, and observe someone falling into it, you will see time slow down for him as he approaches it (as is my understanding).

What if you are both falling in, but he is 100 meters closer to it than you are? Will you see strange effects at the same location in space that a faraway observer would, or would the perceived event horizon be relative to your position, so that you two would have to be much closer to the singularity before you see him approach it? I base this perception on the notes here and in other threads that you can cross the EH without noticing it - i.e., you don’t suddenly see your feet disappear. So is the EH at an absolute position, or is it relative to your distance from it?

18

No, that’s just a case of space being really really close to flat at the horizon. What I was referring to was space which is completely, absolutely identically zero curvature. Not just really close to flat, but completely flat. It’s a bit contrived, but it’s possible.

As to why you don’t want to fire your engines, if you’re inside the horizon, the singularity (and certain death) lies a certain proper time in your future (that is to say, that’s the time it’ll take you to reach it if you just sit tight and wait: Like this time tomorrow is 24 hours away, and if you wait that long, you’ll be there). You don’t care about the proper time, though: You care about the time in your reference frame, which is what determines how many more breaths you can take. If you accelerate in any spatial direction, inside or outside a black hole, you’ll decrease the amount of your time that it takes to reach “proper tomorrow”. And inside a black hole, that “proper tomorrow” is The End.

If you have two infalling observers, they’ll be able to interact more or less normally, as long as the tidal forces remain small. When the tidal forces get large enough, they’ll start seeing Doppler shifts and the like, but that point need not correspond to the horizon. In fact, the event horizon of a black hole cannot be locally defined: You need to consider the entire Universe to define it. A person outside the horizon can (eventually) reach any point in the Universe; a person inside can not reach any point outside.

19

A futher question: I love this stuff! Is there another message board out there where people sit around explaining/talking about stuff like this (Black Holes, Twin Paradox, etc.) ?

These topics come up from time to time on the SDMB, but they can be few and far between.

20

I still didn’t get it, so I googled “black hole geometry” and came up with this site that has diagrams and explanations. I still don’t fully understand this weirdness, but I think I have the general idea.