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Escaping Black Holes

  1. Aug 8, 2005 #1
    It seems to me that whenever I've been taught or told anything about black holes, I'm given the impression that once inside a critical radius, it would be impossible to ever escape. But I've been thinking about this, and that doesn't seem like the whole story. Granted I'm not a physicist, so there may be factors I'm not aware of here, but just see if what I'm saying has any validity.

    Let's just say you're in a space ship inside this critical radius of the black hole, and you want to get back home. Now, the general rule for determining escape velocity and the critical radius is a matter of conservation of energy. i.e. [tex]\Delta K=-\Delta U[/tex]. I'm going to say for the sake of simplicity that you start off still and get to your destination still, so [tex]\Delta K[/tex] is 0. So now you're left with this following equation: [tex]0=-(U_f-U_i)[/tex]. Part of this potential energy is due to the gravity of the black hole, so splitting that up would give you
    [tex]U_i - \frac{{GMm}}{{r_i }} = U_f - \frac{{GMm}}{{r_f }}[/tex]
    [tex]U_i - U_f = GMm\left( {\frac{1}{{r_i }} - \frac{1}{{r_f }}} \right)[/tex]
    And so basically what I am to understand from this simple relationship is that even though the product of the masses may be really large, as long as you have enough potential energy (in the form of chemical potential energy or whatever, such as fuel) then you could get out of a black hole, even if you start out beyond the critical radius of no return.

    Maybe when I've heard discussions of black holes only realistic situations are considered where the photons that fall into a black hole don't have some reserve of potential energy lying around, or maybe I'm just not nearly familiar enough with the subject, but at least hypothetically, wouldn't it be possible to get out of a black hole in a space ship?

    And if this is correct, would it then be possible for some set of chemical or nuclear reactions to occur in the matter of a black hole to allow matter to escape on its own (however unlikely)?
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  3. Aug 8, 2005 #2


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    I won't try to follow your argument, but the reason nothing can escape from a black hole is that the escape velocity is greater than the speed of light. Nothing can go faster than the speed of light.
  4. Aug 8, 2005 #3
    I imagine this would be impossible, that by the time you are inside the event horizon you have already gone through such tremendous gravitational acceleration and so you would be going very fast. To get to a point of rest beyond the event horizon you would need to have a force in the oppositte direction at least equal to the hole's gravitational pull, and such a force I doubt exists no matter how much fuel reserves you had.
  5. Aug 8, 2005 #4


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    I didn't look over your argument too carefully, but it seems to be based on Newtonian assumptions about how gravity and escape velocity work, while statements about black holes are based on the theory of general relativity, a more accurate theory of gravitation that gives almost the same predictions as Newtonian physics in certain limits, but very different predictions in other situations.
  6. Aug 8, 2005 #5
    I apologize that the forum has insulted you with "I haven't bothered to read your argument, but...". I think your question is quite valid (although the mathematics is empty and unecessary).

    Escape velocity is one thing, but we don't have to travel at 11km/s to leave the earth, we could leave the earth at 1 m/s (with a large supply of chemical potential, as you say). The reason we can't do that with a black hole, is because once inside the event horizon the required force is infinite. All worldlines inside the event horizon terminate at the singularity (claassically).
  7. Aug 8, 2005 #6
    That's what I was originally thinking, since as long as you provide a force that keeps up with the gravitational force, any velocity should do.

    This is the part I don't quite understand. I'm only familiar with classical mechanics, so that may be where I get lost, but since both masses are finite and the distance to the center of mass of the black hole (treated as pointlike which I think(?) is an ok assumption) is still nonzero even inside the event horizon, why would the force become infinite (and thus require an infinite force in the opposite direction to escape)?
  8. Aug 8, 2005 #7
    This is based on relativity (don't ask ME to explain it) but basically the black hole is making a rip or "hole" in the fabric of space time, so your concepts of distance and time are out of whack, the reason light can't escape is that it's not fast enough to make it back to regular space-time where it was because this outside distance beyond the event horizon is basically infinite because it's off the map. Actually I'd like to hear this explained too! For a good visual you have to see the Simpsons Holloween that was in 3D.
    Last edited: Aug 8, 2005
  9. Aug 8, 2005 #8
    Well I think it's enough to answer my question that you can't ignore relativity. (I have seen that Simpsons episode, I love that show ^.^) But if anyone can explain the relativistic effects, I'd love to learn.
  10. Aug 8, 2005 #9
    Realistic situations are not the only ones being assumed. It is impossible to escape (in the classical sense) from within the event horizon of a black hole to the outside universe. I'm surprised the others haven't given you a good general relativity explanation for why this is incorrect, but I am not fluent in general relativity by any means, so I will defer the job of explaining it to others. If no one else takes up the explanation, I'll do my best, but surely someone will be along who knows a lot more than me soon.

    As the others have pointed out, and as I have implicated above, this is not correct. However, it is possible for the energy (and matter) inside of a black hole to be returned to the universe, by a process through which black holes "evaporate". This "black hole evaporation" (discovered by Stephen Hawking) has been termed "Hawking radiation". But the process has absolutely nothing to do with how much energy objects inside a black hole have access to, but is rather completely quantum mechanical. In quantum mechanics, empty space is known to go through "vacuum fluctuations", which is where small portions of space borrow energy from other portions of space. In such portions with momentary positive energy, particle pairs momentarily come into existance and then almost immediately destroy each other, but near a black hole, Hawking discovered that one particle could fall into the black hole, leaving the other without a partner to destroy. The particle that falls into the black hole will have negative energy, since its potential energy inside a black hole will be so small, which means that the overall energy (remember that Einstein discovered mass is just a condensed form of energy) of the black hole is reduced. But the other particle which escapes from the black hole's event horizon will return the reduced energy in the black hole as positive energy to the outside world, maintaining the law of conservation of energy. I think Hawking has recently begun to think that a better description of this process can be found in another phenomenon of quantum mechanics called quantum tunnelling, which is a different mechanism, but produces the same results: the black hole will evaporate (after billions of billions of billions of years), so eventually things will "escape" the event horizon of a black hole, but they won't necessarily escape in any form resembling what they were inside the black hole, and they won't escape on their own accord (there is no amount of energy you could add to your rocketship that would create an escape route for you; your escape route will only come by quantum randomness).
  11. Aug 8, 2005 #10


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    I didn't mean it as an insult, I did read enough of his question to get the gist of his argument, and as such I didn't think it was necessary to look at the details of his math. The question is valid, but the statement that Newtonian mechanics gives incorrect predictions in this case is at least a partial answer. Besides, another poster before you already mentioned that the escape velocity would have to be faster than light inside the black hole.
  12. Aug 8, 2005 #11


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    Here's a spacetime diagram from an earlier thread which supports mathman's comment.
    Essentially, once inside the event horizon, the light cones are so tipped over that no observer's worldline can ever find its way back outside the event horizon. This is, of course, classical (non-quantum) general relativity.
  13. Aug 8, 2005 #12
    You could think of it like this:
    As you fall into a black hole, you would spacetime at your location to be "normal", spacetime nearer to the black hole as "compressed" and spacetime farther from your position to the black hole as "stretched". Now, as you fall in, spacetime towards the outside becomes more and more stretched, and as you fall past the event horizon, the rate at which spacetime is expanding towards the outside exceeds that of light. If you sent a light beam outwards, it would be like running forwards on a conveyor belt which is running backwards faster than you can run...Also because of this, it would take an infinite amount of time to actually reach the singularity
  14. Aug 9, 2005 #13
    The important thing here is that anything crossing the Blackhole Horizon is Dimensionally collapsed?..any matter falling inwards is re-configured from 3-D to 2-D or less?

    http://arxiv.org/abs/astro-ph/0508122 for instance is a recent paper that gives a nice introduction to Transitions around blackholes.

    P.S there is another factor to consider, an entangled Particle (and I mean that when a particle arrives at a certain distance close to a BH horizon, there are trapping potentials that create default componant entanglement, ie Particle-Antiparticles ) , then it can happen that one of the componants of the entangled 'photon' states, is scattered away at above lightspeed, they have transisted to a Tachyon Hyper Velocity. These Particles are not detected locally, they appear at the Cosmological Horizon..far..far away, which of course is back in 'Time', they are Dimensionally slingshot across local spacetime, so cannot be detected.
  15. Aug 9, 2005 #14


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    What the heck makes you say that?

    There is no reason to believe that objects suddenly become two-dimensional when they cross the event horizon (I assume that's the horizon you're talking about) of a black hole.

    See for instance the "black hole faq"


  16. Aug 9, 2005 #15


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    Actually, it should only take a finite time for an observer who crosses the event horizon to reach the singularity, according to his own clock anyway.
  17. Aug 9, 2005 #16


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    Good post, pervect. The one thing that might be confusing the OP is what is considered to be 'inside' the black hole. To most of us, that assumes having passed the event horizon. There is, however, a radius outside of that called the 'ergosphere'. Within that area, a mass dumped into the hole will transfer a large part of its energy to an associated mass. If for instance, a 100 kg chunk of rock splits in half in that area, half of it can fall into the hole with maybe 20 kg mass/energy equivalence, while the other half buggers off with 80 kg m/e equivalence. (Sorry about the unprofessional terms; I'm into a lot of beer right now.) It was my understanding that pair-production within this area is the basis for Hawking radiation. Part of the hole's energy is transformed into virtual particles, but one gets sucked in before they can annihilate and the other splits the scene. It seems that a spacecraft could toss its garbage into the hole and gain enough energy to leave faster than it arrived.
    Past the event horizon, of course, nothing can get out.
  18. Aug 9, 2005 #17
    Ok, there are a number of ways to answer this, all centered on the fact that in relativity, space and time are relative. So this can be analyzed from the perspective of someone who has fallen into an event horizon and is trying to escape, someone who is watching someone fall into an event horizon, a light beam trying to escape from a black hole, or a number of other reference frames. But no matter what reference frame you use, the key concepts involved are very different from Newtonian gravity and a "Newtonian black hole".

    In the 18th century, John Mitchell used Newtonian gravity to come up with a theory that if stars reached a certain amount of density, light particles would not be able to escape from the surface from the star. The problem was that the escape velocity for the star was more than the speed of light, so Mitchell imagined particles of light flying off the surface of the star but being pulled back down by gravity. In this Newtonian view of a black hole, of course it would be possible, theoretically, to blast pieces of the star off with enough energy that they could escape. For one thing, in Newtonian physics, the speed of light wasn't a maximum speed, so if you just give a particle more velocity than light, it could escape from a Newtonian black hole. If you were in a rocketship exploring the surface of the star, you too could theoretically escape even without moving faster than the escape velocity, as you point out, assuming you have enough potential energy available. But the relativistic view of a black hole is dramatically different from the Newtonian view.

    For starters, let's analyze what relativity says about the limiting velocity of the speed of light and how that will stop you from ever leaving the event horizon of a black hole. There are many facets to the special theory of relativity, but the only relevant part is that nothing can accelerate past (or to) the speed of light. From the point of view of a spaceship that you are driving, if are racing along behind a light beam, trying to catch it, the faster you speed up, the slower time will tick for you and the more lengths will contract. In fact, the time dilation and length contraction will always be just right so that any time you stop to measure the speed the light beam is travelling away from you, you will always measure it going at the same speed. Space and time seem to conspire against us when trying to catch a beam of light.

    In the general theory of relativity, gravity is not treated as a force. Instead, the laws of physics are formulated in Riemannian geometry, which allows for curvatures in space and time. The fifth postulate of Euclidean geometry states, in a round-about way, that if two lines are parallel in a small region of space, they will remain parallel forever. This postulate was fought by a number of mathematicians who didn't believe it obvious enough to be listed as a postulate. Riemann and Gauss invented one of a number of non-Euclidean geometries, and Einstein used theirs for his formulation of general relativity. In Reimannian geometry, two parallel lines don't have to always remain parallel. For instance, think of the surface of the Earth, which is two dimensional. Although we know the Earth exists in three dimensional space, the surface of the Earth itself can be described in only two dimensions. If we make one line perpendicular to the equator and draw it straight North, it will necessarily intersect another line which is perpendicular to the equator at the North Pole. Although these two lines are parallel at the equator, they follow different curvatures along the Earth. A line you draw across the Earth's surface straight from the equator to the North pole which is the shortest possible line between those two points along the surface of the Earth is called a geodesic. A geodesic is the shortest distance between two points in non-Euclidean geometry; it's analogous to a straight line in Euclidean geometry.

    A geodesic in general relativity takes on a slightly different meaning. General relativity includes both space and time as dimensions, so a geodesic in general relativity is the path between two events which takes the longest possible time. To understand this, we refer back to special relativity which states that if you accelerate, your clock slows down (relative to the frame of reference you would be in if you had remained inertial). Because of this, we get a result that is often called the "twin paradox" by which different people can experience different amounts of time between two events. If there are two twins and one leaves Earth in a very fast space ship and returns, he will find that his twin brother has aged more than him. This is because, relative to his brother (who remained inertial) he has been moving at high speeds, so his clock has ticked slower, and he has also not remained in a non-inertial reference frame. So a geodesic in general relativity really applies to someone who has remained inertial. Although in Newtonian physics, the Earth is seen as accelerating around the Sun, in general relativity the Earth is thought of as moving along a curvature in spacetime caused by the Sun. Since Earth moves along a geodesic, it is not experiencing (local) acceleration. This brings us to why light bends in large gravitational fields. Although light must always travel at a set speed (locally) and therefore cannot accelerate, it does follow geodesics in curved spacetime (indeed, if light did not follow geodesics, it would need to accelerate to leave its geodesic, which we have already stated that light cannot do). Since light follows a geodesic in general relativity, and since the speed of light cannot be surpassed in special relativity, and since light cannot escape a black hole, it is easy to understand why nothing that falls into an event horizon can escape a black hole.

    In general relativity, a black hole is created when gravity becomes so strong and spacetime is warped so much that light cannot escape. Unlike the Newtonian view of black holes, light does not shoot out of a black hole and fall back down. Such ideas would require the light to experience an acceleration toward the hole. Instead, light merely follows geodesics, but since its geodesics are so extremely curved, it never reaches the outside of the black hole. Imagine what this means in the context of special relativity. If you are racing along behind a light beam which is trying to escape a black hole, but cannot, how do you expect to escape a black hole? We have already concluded, from the special theory of relativity, that you can never catch a light beam, so if a light beam is trying its hardest to escape from a black hole and still can't, how do you think you can get out of the black hole? You would have to surpass the speed of light in order to do this, but it would take an infinite amount of energy to even reach the speed of light. Remember that any time you're racing along behind this light beam and stop to measure it, you will always measure it to be moving at the same, very fast speed relative to you. The light beam is not being decelerated as it tries to exit the black hole, because a light beam can only travel at one speed. If a light beam cannot escape a black hole, and you cannot accelerate faster than a light beam, you also cannot escape a black hole.

    I meant to use a couple other viewpoints in this explanation, but this post has already turned out to be a lot larger than I thought it would be, so I'll stop there. If you are not already a little familiar with special relativity or general relativity, then this will probably leave you a little confused and with a lot more questions. Any more questions you have, we'll be glad to answer them at the forum. If you'd like to know more about relativity, you can find a great guide of both layman books and textbooks here: http://math.ucr.edu/home/baez/physics/Administrivia/rel_booklist.html
    Last edited: Aug 9, 2005
  19. Aug 9, 2005 #18
    What possessed you to provide a link that is in need of updating?..no offence but the date is:September 1995
  20. Aug 9, 2005 #19


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    Black holes have been reasonably well understood for a long time (long since 1995). There is nothing in the theory of General Relativity that says that objects become two-dimensional when they cross the event horizon of a black hole. In fact, you'll find numerous remarks that nothing special happens when one crosses the event horizon if you look at the literature.

    There is nothing in the paper you cited which suggests that objects suddenly become two dimensional when they cross the event horizon of a black hole. Furthermore the paper is not particularly relevant to the original poster's question, being about a rather exotic situation (what happens to a black hole in the inflationary phase of the universe). It's actually a rather interesting paper, but it's not particularly relevant to the Original Poster's question. And it doesn't say anywhere what you claimed it says.

    The only thing I can think of that might have prompted your remark is that you might be misunderstanding the holographic principle (which is at this point only a conjecture, and a conjecture about quantum gravity rather than about GR)


    This conjecture would be about things being two-dimensional (in some abstract sense) all the time, not when they cross the event horizon of a black hole.
  21. Aug 9, 2005 #20


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    Rotating black holes do indeed have ergospheres - non-rotating black holes don't, they have a single event horizon.

    There's a fairly good diagram at

    http://www.gothosenterprises.com/black_holes/rotating_black_holes.html [Broken]

    Compare to the diagram of a non-rotating black hole at:

    http://www.gothosenterprises.com/black_holes/static_black_holes.html [Broken]

    Non-rotating black holes are much easier to talk about than rotating black holes, though, and easier to understand, too.
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