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What if Earth crossed the event horizon of a supermassive black hole

  1. Jul 27, 2013 #1
    From previous threads I have understood that crossing the event horizon of a supermassive black hole is nothing very unusual for the falling observer locally. Usually in these considerations the falling observer has been thought as a "point" without much dimension.

    How about if Earth (and rest of the Solar System) crossed the event horizon of a very supermassive black hole? Moreover, let's assume that Earth approaches the event horizon very slowly so that it would take several days (in Earth citizens clocks) before Earth would be completely below the horizon.

    Would we notice anything unusual (except that the Sun is missing for some time, and so on)? Or would the crossing of EH be dramatic and catastrophic event on Earth, interfering Earth's rotation and blocking signals and travels from parts below EH to parts above EH?
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  3. Jul 27, 2013 #2


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    You have it backwards here - a slow approach is much harder and more disruptive because you need some very powerful rocket motor to slow you down, fight the gravitational field. The way to see nothing unusual is to free-fall towards and through the horizon. Then you will experience only tidal forces (forces that depend not on the strength o fthe gravitational field, but on the difference between its strength at two nearby points) and these can be made arbitrarily small by making the black hole arbitrarily large.

    Free-fall in and the the sun won't even disappear. If the sun goes in first, we'll still be receiving the light that left the sun before it goes through the horizon, and we'll keep receiving it until we go through the horizon ourselves. And if we go in first, there's nothing stopping the light from the sun coming through after us.
  4. Jul 27, 2013 #3


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    Earth is big enough some tidal effects would be noticeable even for biggest known supermassive BH (magically cleared of all surrounding debris). However, nothing dramatic.

    The bigger issue is your proposal of slow descent. This is impossible. In free fall, even starting from relatively near the horizon, the crossing time would be very short. If you imagine holding the earth to slow it down, the parts outside the horizon would need to be held with a force that crush all forms of matter, while the part inside could not be slowed at all, by any means.

    Just go with the flow, quickly through the horizon, then all will be 'fine' for a few hours after crossing, for an ideal supermassive BH. Then tidal forces can no longer be small, and all will be far from fine - earth will soon rupture.

    [I chose to ignore the sun, and just focus on the earth.]
  5. Jul 27, 2013 #4
    Thanks for the answers, it sounds like the slow descend is a problem. However, I thought that the acceleration caused by gravity at EH depends on the black hole radius: larger radius, smaller acceleration. If this is the case, we could (in a thought experiment) make the black hole awfully large e.g. with several lightyears radius and place the earth at rest just above the event horizon, then free fall. Still nothing dramatic observed on earth?
  6. Jul 27, 2013 #5


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    Regardless of the size of the Schwarzschild black hole, the acceleration of a static observer approaches infinity the closer the observer is to the event horizon.
  7. Jul 27, 2013 #6


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    No. You're thinking of the tidal forces, which are indeed smaller as the event horizon radius grows. The gravitational acceleration grows without bound as you approach the horizon regardless of the size; it has to, or light would be able to escape.
  8. Jul 28, 2013 #7


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    If you're very close to the horizon, say a distance d away, that that distance is much less the the black hole's radius r, and you were held there in place by some long cable or something holding you from above, the gravitational acceleration, g, you would feel by being kept in place is g = c2/d.

    As you can see, there's a problem with a slow approach; something's going to be ripped apart as d approaches 0.

    But oddly enough, if were to drop something into the black hole it wouldn't shoot right through. It would speed up as it left your hand, but then gradually slow down, and never actually cross the horizon, as measured in your frame of reference. The position of the object away from you would be:

    [tex] x' = d \left[ 1 - \mathrm{sech} \left( \frac{c}{d} \tau \right) \right] [/tex]
    where [itex] \tau [/itex] is the time as measured in your frame of reference, and c is the speed of light, and sech() is 1 divided by the hyperbolic cosine function. x' is the distance between you and the object. (It's assumed that you release the object at time [itex] \tau = 0 [/itex]. It's also assumed that d [itex] \ll [/itex] r.)

    On the other hand, if you were to reach back and cut the cable (causing you to free fall into the black hole), The horizon effects would vanish (as measured from your frame of reference), and you wouldn't notice anything special about crossing the horizon as it overtakes you at the speed of light; it would be similar to just floating in empty space (at least for the moment -- eventually you'll get spaghettified, but that comes later when you approach the center of the black hole).

    [Edit: corrected a mistake. Originally I had cosh(), but I really meant 1/cosh() = sech(). Doh!]
    Last edited: Jul 28, 2013
  9. Jul 28, 2013 #8


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    This is not possible. The EH is a null surface, meaning that it moves at the speed of light. No matter what, the EH would pass the earth at the speed of light.
  10. Jul 29, 2013 #9
    By the way, is the falling object really there hovering just above the horizon infinitely (in outside observer's frame), or do we see just an image of the falling object, but the object itself has actually fallen below event horizon?

    I think this is physically relevant and related to the question "will the falling observer see universe end". This is a well known question with different answers on the internet (not much to do with the original subject though), any opinions in this forum?

    If the object is really there, it seems to me that a photon (with proper direction and constant speed c) should eventually reach the falling object, no matter how far the photon was send from. So the falling observer would see the universe end in this case, but if it's just an image, then not.
  11. Jul 29, 2013 #10


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    Whatever the 'internet' says, it is readily computable that for any light source ouside the horizon, there is a specific moment when its signal will reach an infalling object at the horizon. There is also a specific, somewhat later moment when its signal will reach the infaller the moment before the infaller reaches the singularity. These statements are true for the ideal, non-rotating, spherical, classical, black hole. The statements about the horizon are true for any purely classical BH - thus an infaller certain sees a specific finite history external before crossing the horizon. However, for rotating black holes, there are possibilities of not hitting the singularity at all after crossing the horizon. When quantum effects are considered (without a complete theory - none exists), there are many other speculations: firewalls at the horizon (this is for some years the hot -pun intended- debate); that BH evaporation will occur before the object crosses horizon (this is a minority view with some notable adherents); that there is no horizon at all due to graviton self interaction deviating from GR predictions (an even smaller minority view).
    Last edited: Jul 29, 2013
  12. Jul 30, 2013 #11
    Free Fall

    Once you cross the event horizon, you lose all contact with the universe, and this is cannot be reversed. The universe become totally unaware of you. This is like going to Heaven. A person who dies and goes to Heaven loses all contact with the universe permanently.

    This is confusing. I thought that science consists only of things that can be objectively measured. The inside of a black hole cannot be objectively measured for the reason mentioned above. Does the inside of a BH involve a radical change in our understanding of physical principles? What is the justification for this radical explanation? Can someone clarify this?
  13. Jul 30, 2013 #12


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    No, it consists of things that can be objectively measured, plus things which are predicted to exist by our best confirmed theories, even if they can't be objectively measured. There are at least two kinds of things that can fall into the latter category.

    First, there are things we can't measure directly, but for which we have indirect evidence. Quarks, for example, can't be measured directly; we can't isolate a single quark and measure its properties, because the laws of the strong interaction don't work that way. But we believe that quarks exist because we have extremely strong indirect evidence for their existence.

    Second, science allows us to extrapolate from what we currently know to what we predict in regimes we haven't yet explored, but the extrapolation has to assume that our best confirmed theories will continue to hold in regimes we haven't yet explored. (Obviously that's an assumption that could be falsified in specific cases when we explore new regimes, but unless and until it is, it's really the only assumption we can make.)

    Both of the above considerations apply to black holes. We have plenty of indirect evidence that there are extremely massive objects in the universe that (a) emit no radiation, so we can't directly see them (but we can tell that they're there indirectly because of their gravitational effects), and (b) are packed into a very small volume for their mass, much smaller than a star or galaxy of comparable mass. The only kind of object that fits within our current theories and is consistent with all this indirect evidence is a black hole.

    Also, even though no signals can escape from inside a black hole's horizon, we can still calculate that objects falling in will reach the horizon in a finite time, and that there is nothing at the horizon that could stop them from falling in still further. So either there is indeed a region of spacetime behind the horizon (even though we can't directly observe it), or the laws of physics suddenly stop working at the horizon, for no apparent reason. In other words, we believe there are regions of spacetime behind black hole horizons because that's what our best current theories, extrapolated to the case of a black hole, predict.

  14. Jul 30, 2013 #13


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    Actually, no. The outside loses all contact with you, but not vice versa. You still continue to receive signals, information, etc. from the outside until you are destroyed by conditions near the singularity.
  15. Jul 30, 2013 #14


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    It's more than just an image. It actually is the object (as measured from your frame of reference).

    But it gets complicated. And my last post made several assumptions and approximations. Allow me to elaborate.

    My [itex] x' = d \left[1 - \mathrm{sech}\left(\frac{c}{d} \tau \right) \right] [/itex] equation (the distance, x', from "stationary" observer to an object dropped at close range into the event horizon of a super-massive black hole [event horizon a distance d away; d [itex] \ll [/itex] r]) doesn't describe a few details, and made some assumptions. Perhaps it would be fair too describe a few more details and the impacts of the assumptions.

    Mass of the object:

    One of the assumptions in my equation is that the object has negligible mass. It ignores the any spacetime curvature caused by the object itself. What impact does this assumption have? Well, if we don't ignore the mass of the object then as the object approaches the event horizon, it can affect the local spacetime around the horizon nearby it, causing a little "dimple." So does the dimple overtake the object (such that you would observe the object cross the horizon)? That's not a simple answer: the event horizon is not "well behaved" in such situations. I don't have the mathematical skills to analyses it from that aspect. But what I can say, is that you wouldn't observe it "shooting" right through. It would slow down to a snail's pace, at the very least. And what you are observing is not an "image" but rather the actual object.

    Time dilatation, Doppler, and resolution:

    A detail that I have yet to describe is the Doppler red-shift you would observe on the object quickly becomes enormous.

    Imagine that the object you drop in the super massive black hole is a clock. (I'm using the same scenario where you are suspended above the event horizon of the black hole by a cable or something, a constant distance d above the horizon, such that d [itex] \ll [/itex] the black hole's radius r. Suppose you release the clock from your hand at [itex] \tau [/itex] = t' = 0. The time that you observe on the clock, from your frame of reference, is

    [tex] t' = \frac{d}{c} \tanh \left( \frac{c}{d} \tau \right) [/tex]
    If you plot that out, the time you observe on the clock approaches d/c, but never reaches it.

    Now imagine the clock is transmitting its time by using some sort of radio source. Every second it transmits a pulse. The time interval between pulses become farther and farther apart. Better than pulses though, let's assume the clock is transmitting its time by continuous gamma rays, and you can observe the clock's time by counting the number of wavelengths received. As the clock approaches the horizon, the gamma rays emitted by the clock come increasingly red-shifted: the frequency becomes smaller and the wavelength becomes larger. Suppose the original frequency and wavelength of the gamma rays, when you are holding the clock are [itex] f_0 [/itex] and [itex] \lambda_0 [/itex] respectively (btw, [itex] c = \lambda f [/itex]). As the clock approaches the horizon, these become,

    [tex] f = f_0 \left[1 - \mathrm{tanh}^2 \left(\frac{c}{d} \tau \right) \right] [/tex]
    [tex] \lambda = \lambda_0 \mathrm{cosh}^2 \left(\frac{c}{d} \tau \right) [/tex]
    The ability to determine the position of an object, purely by measurement/observation, is inversely proportional to the wavelength used for the measurements.

    The point of all of this is as the object approaches the horizon, not only is it red shifted to point of "disappearing" from sight, it's location becomes more difficult to determine by observation alone. (Eventually the wavelength will exceed the diameter of the black hole itself.)

    Other things in the vicinity.

    Being held by a cable from above, that allows you to be close the event horizon of a black hole would not be a pleasant thing at all. The acceleration from the cable holding you there would be so great that you'd be smashed to bits. Also, any light coming from above would be so incredibly blue-shifted that it would fry you instantly.

    The scenario I described assumes that everything around the black hole is very calm (which it wouldn't be in reality), including lack of other stars in the sky creating light, etc.

    There's also an assumption that you are somehow able to survive the extreme forces of being held in place above the horizon (which in reality, you wouldn't).

    Quantum effects:

    Thus far in my simple description, quantum effects are completely ignored. They're outside the scope of this thread anyway, so I'm not going to discuss them here. (Not that I would be able to discuss much anyway.)

    Now back to Ookke's post.

    If you were to reach back and cut the cable, so that you fell into the supermassive black hole, things would be quite different.

    You would cross the event horizon after about d/c (units of time) from when you cut the cable (again this assumes that d [itex] \ll [/itex] r, so we can ignore tidal effects), as measured by your (new) frame of reference. When you did cross the horizon, it would pass you at the speed of light, but wouldn't appear to be anything special.

    Remember that clock/object that you dropped into the black hole at some time before? It would now be moving away from you at a constant velocity. Until you got closer to the center of the black hole(when tidal effects come into play), the rules of special relativity of simple, non-accelerating objects apply.

    I don't know what the "end of the universe" means, but however you define it, if you didn't see it before you cut the cable, you wouldn't likely see it when you crossed the horizon.

    As you've figured out, something falling through a black hole brings up seeming paradoxes. According to the math, observing something from a "stationary" viewpoint is quite different than what is observed by an observer free-falling though the horizon.

    I'm not going to discuss the attempted resolutions to the paradoxes here [doing so might even violate the Forum guidelines]. But if you want to explore further, Leonard Susskind and Gerard t'Hooft might be good resources.

    Here is a link to a paper by Susskind:
    Last edited: Jul 30, 2013
  16. Jul 30, 2013 #15


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    Though this is in your quantum section (except one phrase I grabbed from earlier), the bolded parts appear to be a purely classical observation. But then you refer authors and a paper discussing only paradoxes of the quantum treatment.

    There are no paradoxes in the classical treatment (there may be issues of believability esp. given limits on verifiability, but that is not a paradox).

    I specifically object to the use of 'frame of reference' to describe a particular global coordinate system, or to describe distant objects. There are no global frames in GR. There are only local frames and global coordinate systems. Local frames have a physical basis. Global coordinates are largely arbitrary. It is unambiguously true that that an external stationary observer never detects information from matter crossing a horizon (classically). It is not correct to say that they have some unambiguous point of view encompassing the infalling object, in which you give objective meaning to the claim the object 'never crosses the horizon'.

    IMO, it is correct to say a distant observer is detecting frozen light in the exactly the same sense as a Rindler observer's seeing objects redshift and effectively disappear can reasonably conclude that this is an apparent effect of light never being able to catch up to them, not that planets freeze and pile up behind them. Just as the Rindler observer can translate to an inertial frame to get a more complete map of the universe, a distant observer can pick a dozen different coordinates (more complete maps) to understand parts of the universe 'missing' from their initial map.
    Last edited: Jul 30, 2013
  17. Jul 30, 2013 #16
    Objectively measured

    Science consists of mathematical systems, namely, a collection of arbitrary consistent meaningful statements, along with empirical verification. If we find one inconsistent statement, the entire theory is not valid. However, although we cannot get perfect verification, it must be possible in principle to perform the verification.

    It is not possible in principle to observe what happens as our observer crosses the event horizon. The comment about quarks puzzles me, for we can indeed observe in principle the behavior of quarks, and so why do you say it cannot be objectively measured? Secondly, division by zero is not defined as it leads to inconsistencies. A singularity is basically a division by zero, or so I thought. I am not aware of any other meaning to the word. Einstein's theory based upon the Schwartzchild metric that yields black holes and a definite singularity at the center must necessarily be wrong, if I understand the meaning of math and physics. Why then do people talk about it so much?

    You said something great. Physics cannot stop at the event horizon. Since physics is not valid inside the black hole, the physics outside the black hole cannot possibly be valid. Don't you agree?
    Last edited: Jul 30, 2013
  18. Jul 30, 2013 #17


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    Who said physics is not valid anywhere inside the black hole? GR breaks down at the singularity itself, which is a completely different statement.
  19. Jul 30, 2013 #18
    "GR breaks down at the singularity itself, which is a completely different statement." No, it is not. If there is one point inside the BH for which physics is not valid, then the entire inside is invalid physics. At least, this is what I think. I do not follow the logic of one point being invalid, and the other points valid. Can you explain with an example?
  20. Jul 30, 2013 #19


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    An inconsistency in a formal theory is that it predicts both A and not A. Observability has nothing to do with consistency. Physical theories always make some untestable predictions. For example, Newtonian physics predicts that a spinning bucket of water with nothing else at all in the universe will have a water surface with a concave shape. Ernst Mach did not believe this. It is, in principle, untestable because we can't create an empty universe.
    It is perfectly possible to observe an object crossing the horizon - you just have to fall in yourself to do it. That you may not choose to do so, or that you can't send me your findings is irrelevant.

    You can observe consequences of quark behavior, but QCD says you can never, in principle, observe an isolated quark - yet it continues to pass every experimental test of indirect predictions. GR has also passed all tests so far, including ever more significant indirect tests of BH behavior - detailed observations not consistent with any material surface, but exactly with characteristics of a horizon.

    Wrong and inconsistent are two different things. A singularity is not quite the same as division by zero. The modern mathematical approach is that the singularity is not part of the manifold at all. No part of the manifold has a problem. There is just a feature that limits (of e.g. curvature) along all timelike world lines inside the EH don't converge.

    You should also realize that GR itself says the Schwartzchild metric is not believable in these limits because it has perfect symmetry and no rotation. Any real BH has huge rotation and definite asymmetry. It is, at present, beyond known mathematical techniques to extract what GR says about plausible rotating black hole interior. You may hear of Kerr solutions. However, they are solutions with perfect axial symmetry. It is known that the tiniest perturbations radically change the interior to be very different from Kerr, but the nature what results is simply unknown (even classically).
  21. Jul 30, 2013 #20


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    Your logic is flawed I'm afraid. You're using informal logic to make formal statements. Just because the theory breaks down at a surface of extreme conditions (the singularity) doesn't mean it is invalid elsewhere. If that were true then all of Newtonian theory would be wrong (Newton's law of gravitation or Coloumb's law of electrostatics have true point singularities) which is clearly false.
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